GRADUATE STUDIES I N M AT H E M AT I C S
181
A First Course in Sobolev Spaces Second Edition Giovanni Leoni
American Mathematical Society
A First Course in Sobolev Spaces Second Edition
GRADUATE STUDIES I N M AT H E M AT I C S
181
A First Course in Sobolev Spaces Second Edition
Giovanni Leoni
American Mathematical Society Providence, Rhode Island
EDITORIAL COMMITTEE Dan Abramovich Daniel S. Freed (Chair) Gigliola Staffilani Jeff A. Viaclovsky 2010 Mathematics Subject Classification. Primary 46E35; Secondary 26A27, 26A30, 26A42, 26A45, 26A46, 26A48, 26B30, 30H25.
For additional information and updates on this book, visit www.ams.org/bookpages/gsm-181
Library of Congress Cataloging-in-Publication Data Names: Leoni, Giovanni, 1967– Title: A first course in Sobolev spaces / Giovanni Leoni. Description: Second edition. | Providence, Rhode Island : American Mathematical Society, [2017] | Series: Graduate studies in mathematics; volume 181 | Includes bibliographical references and index. Identifiers: LCCN 2017009991 | ISBN 9781470429218 (alk. paper) Subjects: LCSH: Sobolev spaces. Classification: LCC QA323 .L46 2017 | DDC 515/.782–dc23 LC record available at https://lccn.loc.gov/2017009991
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established to ensure permanence and durability. Visit the AMS home page at http://www.ams.org/ 10 9 8 7 6 5 4 3 2 1
22 21 20 19 18 17
In memory of my Ph.D. advisor, James Serrin
Contents
Preface
xiii
Preface to the Second Edition
xiii
Preface to the First Edition
xv
Acknowledgments
xxi
Second Edition
xxi
First Edition
xxii
Part 1. Functions of One Variable Chapter 1. Monotone Functions
3
§1.1. Continuity
3
§1.2. Differentiability
9
Chapter 2. Functions of Bounded Pointwise Variation
29
§2.1. Pointwise Variation
29
§2.2. Continuity
34
§2.3. Differentiability
40
§2.4. Monotone Versus BPV
44
§2.5. The Space BP V (I; Y )
47
§2.6. Composition in BP V (I; Y )
55
§2.7. Banach Indicatrix
59
Chapter 3. Absolutely Continuous Functions
67
§3.1. AC(I; Y ) Versus BP V (I; Y )
67
§3.2. The Fundamental Theorem of Calculus
71 vii
viii
§3.3. §3.4. §3.5. §3.6. §3.7.
Contents
Lusin (N ) Property Superposition in AC(I; Y ) Chain Rule Change of Variables Singular Functions
84 91 95 100 103
Chapter 4. Decreasing Rearrangement §4.1. Definition and First Properties §4.2. Function Spaces and Decreasing Rearrangement
111 111 126
Chapter §5.1. §5.2. §5.3. §5.4. §5.5.
5. Curves Rectifiable Curves Arclength Length Distance Curves and Hausdorff Measure Jordan’s Curve Theorem
133 133 143 146 149 152
Chapter §6.1. §6.2. §6.3.
6. Lebesgue–Stieltjes Measures Measures Versus Increasing Functions Vector-valued Measures Versus BP V (I; Y ) Decomposition of Measures
157 157 168 177
Chapter §7.1. §7.2. §7.3.
7. Functions of Bounded Variation and Sobolev Functions BV (Ω) Versus BP V (Ω) Sobolev Functions Versus Absolutely Continuous Functions Interpolation Inequalities
183 183 188 196
Chapter §8.1. §8.2. §8.3. §8.4. §8.5.
8. The Infinite-Dimensional Case The Bochner Integral Lp Spaces on Banach Spaces Functions of Bounded Pointwise Variation Absolute Continuous Functions Sobolev Functions
205 205 212 220 224 229
Part 2. Functions of Several Variables Chapter §9.1. §9.2. §9.3.
9. Change of Variables and the Divergence Theorem Directional Derivatives and Differentiability Lipschitz Continuous Functions The Area Formula: The C 1 Case
239 239 242 249
Contents
ix
§9.4. The Area Formula: The Differentiable Case
262
§9.5. The Divergence Theorem
273
Chapter 10. Distributions
281
§10.1. The Spaces DK (Ω), D(Ω), and
D (Ω)
§10.2. Order of a Distribution
281 288
§10.3. Derivatives of Distributions and Distributions as Derivatives 290 §10.4. Rapidly Decreasing Functions and Tempered Distributions
298
§10.5. Convolutions
302
§10.6. Convolution of Distributions
305
§10.7. Fourier Transforms
309
§10.8. Littlewood–Paley Decomposition
316
Chapter 11. Sobolev Spaces
319
§11.1. Definition and Main Properties
319
§11.2. Density of Smooth Functions
325
§11.3. Absolute Continuity on Lines
336
§11.4. Duals and Weak Convergence
344
§11.5. A Characterization of
349
W 1,p (Ω)
Chapter 12. Sobolev Spaces: Embeddings
355
§12.1. Embeddings: mp < N
356
§12.2. Embeddings: mp = N
372
§12.3. Embeddings: mp > N
378
§12.4. Superposition
387
§12.5. Interpolation Inequalities in RN
399
Chapter 13. Sobolev Spaces: Further Properties
411
§13.1. Extension Domains
411
§13.2. Poincar´e Inequalities
430
§13.3. Interpolation Inequalities in Domains
449
Chapter 14. Functions of Bounded Variation
459
§14.1. Definition and Main Properties
459
§14.2. Approximation by Smooth Functions
462
§14.3. Bounded Pointwise Variation on Lines
468
§14.4. Coarea Formula for BV Functions
478
§14.5. Embeddings and Isoperimetric Inequalities
482
x
Contents
§14.6. Density of Smooth Sets
489
§14.7. A Characterization of BV (Ω)
493
Chapter 15. Sobolev Spaces: Symmetrization §15.1. Symmetrization in
Lp
497
Spaces
497
§15.2. Lorentz Spaces §15.3. Symmetrization of
502 W 1,p
and BV Functions
§15.4. Sobolev Embeddings Revisited
504 510
Chapter 16. Interpolation of Banach Spaces
517
§16.1. Interpolation: K-Method
517
§16.2. Interpolation: J-Method
526
§16.3. Duality
530
§16.4. Lorentz Spaces as Interpolation Spaces
535
Chapter 17. Besov Spaces §17.1. Besov Spaces
539
Bqs,p
539
§17.2. Some Equivalent Seminorms
545
§17.3. Besov Spaces as Interpolation Spaces
551
§17.4. Sobolev Embeddings §17.5. The Limit of
Bqs,p
561
as s →
0+
and s →
m−
565
§17.6. Besov Spaces and Derivatives
571
§17.7. Yet Another Equivalent Norm
578
§17.8. And More Embeddings
585
Chapter 18. Sobolev Spaces: Traces
591
§18.1. The Trace Operator §18.2. Traces of Functions in
592 W 1,1 (Ω)
598
§18.3. Traces of Functions in BV (Ω) §18.4. Traces of Functions in
W 1,p (Ω),
§18.5. Traces of Functions in
W m,1 (Ω)
605 p>1
606 621
§18.6. Traces of Functions in W m,p (Ω), p > 1
626
§18.7. Besov Spaces and Weighted Sobolev Spaces
626
Appendix A.
Functional Analysis
635
§A.1. Topological Spaces
635
§A.2. Metric Spaces
638
§A.3. Topological Vector Spaces
639
Contents
xi
§A.4. Normed Spaces
643
§A.5. Weak Topologies
645
§A.6. Hilbert Spaces
648
Appendix B. Measures
651
§B.1. Outer Measures and Measures
651
§B.2. Measurable and Integrable Functions
655
§B.3. Integrals Depending on a Parameter
662
§B.4. Product Spaces
663
§B.5. Radon–Nikodym’s and Lebesgue’s Decomposition Theorems 665 §B.6. Signed Measures
666
§B.7. Lp Spaces
668
§B.8. Modes of Convergence
673
§B.9. Radon Measures §B.10. Covering Theorems in Appendix C.
676 RN
The Lebesgue and Hausdorff Measures
678 681
§C.1. The Lebesgue Measure
681
§C.2. The Brunn–Minkowski Inequality
683
§C.3. Mollifiers
687
§C.4. Maximal Functions
694
§C.5. BMO Spaces
695
§C.6. Hardy’s Inequality
698
§C.7. Hausdorff Measures
699
Appendix D. Notes
703
Appendix E. Notation and List of Symbols
711
Bibliography
717
Index
729
Preface The Author List, I: giving credit where credit is due. The first author: Senior grad student in the project. Made the figures. — Jorge Cham, www.phdcomics.com
Preface to the Second Edition There are a lot of changes in the second edition. In the first part of the book, starting from Chapter 2, instead of considering real-valued functions, I treat functions u : I → Y , where I ⊆ R is an interval and Y is a metric space. This change is motivated by the addition of a new chapter, Chapter 8, where I introduce the Bochner integral and study functions mapping time into Banach spaces. This type of functions plays a crucial role in evolution equations. Another important addition in the first part of the book is Section 7 in Chapter 7, which begins the study of interpolation inequalities for Sobolev functions of one variable. One wants to estimate some appropriate norm of an intermediate derivative in terms of the norms of the function and the highest-order derivative. Except for Chapter 8, the first part of the book is meant as a textbook for an advanced undergraduate course or beginning graduate course on real analysis or functions of one variable. One should simply take Y to be the real line R so that H1 reduces to the Lebesgue measure L1 . All the results needed from measure theory are listed in the appendices at the end of the book. The second part of the book starts with Chapter 9, which went through drastic changes. In the revised version I give an overview of classical results for functions of several variables, which are somehow scattered in the literature. These include Rademacher’s and Stepanoff’s differentiability theorems,
xiii
xiv
Preface
Whitney’s extension theorem, and Brouwer’s fixed point theorem. The focus of the chapter is now the divergence theorem for Lipschitz domains. While this fundamental result is quoted and used in every book on partial differential equations, it’s hard to find a thorough proof in the literature. To introduce the surface integral on the boundary I start by proving the area formula, first in the C 1 case, and then, using Whitney’s extension theorem in the Lipschitz case. In the chapter on distributions, Chapter 10, I added rapidly decreasing functions, tempered distributions, and Fourier transforms. This was long overdue. The book is structured in such a way that an instructor of a course on Sobolev spaces could actually skip Chapters 9 and 10, which serve mainly as reference chapters and jump to Chapter 11. Chapters 11, 12, and 13, the first part of Chapter 18 could be used as a textbook on a course on Sobolev spaces. They are mostly self-contained. One of the main changes in these chapters is that I caved in and decided to include higher order derivatives. The reason why I did not do it in the first edition was because standard operations like the product rule and the chain formula become incredibly messy for higher order derivatives and there are so many multi-indices to take into account that even elementary proofs become unnecessarily complicated. My compromise is to present proofs first in W 1,p (Ω) (first-order derivatives) or in W 2,p (Ω) (second-order derivatives) and only after, when the idea of the proof is clear, to do the general case W m,p (Ω). I did not always follow this rule, since sometimes there was no significant change in difficulty in treating W m,p (Ω) rather than W 1,p (Ω). The advantage in having higher order derivatives is that I can now prove the classical interpolation inequalities of Gagliardo and Nirenberg. These are done in Section 12.5 in Chapter 12 for RN and in Section 13.3 in Chapter 13 for uniformly Lipschitz domains. The main novelty with respect to other textbooks is that in the case of uniformly Lipschitz domains I do not assume that functions are in W m,p (Ω) but only that u is in Lq (Ω) and the weak derivatives of order m are in Lp (Ω). Another new section is Section 12.4 in Chapter 12, where I study superposition in Sobolev spaces. The last major departure from the first edition is the chapter on Besov spaces, Chapter 17. This chapter was completely rewritten in collaboration with Ian Tice. The main motivation behind the changes was the proof that the trace space of functions in W 2,1 (RN ) is given by the Besov space B 1,1 (RN −1 ) (see Chapter 18). I am only aware of one complete and simple
Preface to the First Edition
xv
proof, which is in a recent paper of Mironescu and Russ [173]. It makes use of two equivalent norms for B 1,1 (RN −1 ), one using second order difference quotients and the other the Littlewood-Paley norm. To introduce the second norm, I went through several different versions of Chapter 17. Eventually, to study Besov spaces I used heavily the K method of real interpolation introduced by Peetre. The interpolation theory needed was added in a new chapter, Chapter 16. Webpage for mistakes, comments, and exercises: The AMS is hosting a webpage for this book at http://www.ams.org/bookpages/gsm-181/ where updates, corrections, and other material may be found.
Preface to the First Edition There are two ways to introduce Sobolev spaces: The first is through the elegant (and abstract) theory of distributions developed by Laurent Schwartz in the late 1940s; the second is to look at them as the natural development and unfolding of monotone, absolutely continuous, and BV functions1 of one variable. To my knowledge, this is one of the first books to follow the second approach. I was more or less forced into it: This book is based on a series of lecture notes that I wrote for the graduate course “Sobolev Spaces”, which I taught in the fall of 2006 and then again in the fall of 2008 at Carnegie Mellon University. In 2006, during the first lecture, I found out that half of the students were beginning graduate students with no background in functional analysis (which was offered only in the spring) and very little in measure theory (which, luckily, was offered in the fall). At that point I had two choices: continue with a classical course on Sobolev spaces and thus lose half the class after the second lecture or toss my notes and rethink the entire operation, which is what I ended up doing. I decided to begin with monotone functions and with the Lebesgue differentiation theorem. To my surprise, none of the students taking the class had actually seen its proof. I then continued with functions of bounded pointwise variation and absolutely continuous functions. While these are included in most books on real analysis/measure theory, here the perspective and focus are rather different, in view of their applications to Sobolev functions. Just to give an example, most books study these functions when the domain is either the closed interval [a, b] or R. I needed, of course, open intervals (possibly unbounded). 1 BV
functions are functions of bounded variation.
xvi
Preface
This changed things quite a bit. A lot of the simple characterizations that hold in [a, b] fall apart when working with arbitrary unbounded intervals. After the first three chapters, in the course I actually jumped to Chapter 7, which relates absolutely continuous functions with Sobolev functions of one variable, and then started with Sobolev functions of several variables. In the book I included three more chapters: Chapter 5 studies curves and arclength. I think it is useful for students to see the relation between rectifiable curves and functions with bounded pointwise variation. Some classical results on curves that most students in analysis have heard of, but whose proof they have not seen, are included, among them Peano’s filling curve and the Jordan curve theorem. Section 5.4 is more advanced. It relates rectifiable curves with the H1 Hausdorff measure. Besides Hausdorff measures, it also makes use of the Vitali–Besicovitch covering theorem. All these results are listed in Appendices B and C. Chapter 6 introduces Lebesgue–Stieltjes measures. The reading of this chapter requires some notions and results from abstract measure theory. Again it departs slightly from modern books on measure theory, which introduce Lebesgue–Stieltjes measures only for right continuous (or left) functions. I needed them for an arbitrary function, increasing or with bounded pointwise variation. Here, I used the monograph of Saks [201]. I am not completely satisfied with this chapter: I have the impression that some of the proofs could have been simplified more using the results in the previous chapters. Readers’ comments will be welcome Chapter 4 introduces the notion of decreasing rearrangement. I used some of these results in the second part of the book (for Sobolev and Besov functions). But I also thought that this chapter would be appropriate for the first part. The basic question is how to modify a function that is not monotone into one that is, keeping most of the good properties of the original function. While the first part of the chapter is standard, the results in the last two sections are not covered in detail in classical books on the subject. As a final comment, the first part of the book could be used for an advanced undergraduate course or beginning graduate course on real analysis or functions of one variable. The second part of the book starts with one chapter on absolutely continuous transformations from domains of RN into RN . I did not cover this chapter in class, but I do think it is important in the book in view of its ties with the previous chapters and their applications to the change of variables formula for multiple integrals and of the renewed interest in the subject in recent years. I only scratched the surface here.
Preface to the First Edition
xvii
Chapter 10 introduces briefly the theory of distributions. The book is structured in such a way that an instructor could actually skip it in case the students do not have the necessary background in functional analysis (as was true in my case). However, if the students do have the proper background, then I would definitely recommend including the chapter in a course. It is really important. Chapter 11 starts (at long last) with Sobolev functions of several variables. Here, I would like to warn the reader about two quite common misconceptions. Believe it or not, if you ask a student what a Sobolev function is, often the answer is “A Sobolev function is a function in Lp whose derivative is in Lp .” This makes the Cantor function a Sobolev function :( I hope that the first part of the book will help students to avoid this danger. The other common misconception is, in a sense, quite the opposite, namely to think of weak derivatives in a very abstract way not related to the classical derivatives. One of the main points of this book is that weak derivatives of a Sobolev function (but not of a function in BV!) are simply (classical) derivatives of a good representative. Again, I hope that the first part of the volume will help here. Chapters 11, 12, and 13 cover most of the classical theorems (density, absolute continuity on lines, embeddings, chain rule, change of variables, extensions, duals). This part of the book is more classical, although it contains a few results published in recent years. Chapter 14 deals with functions of bounded variation of several variables. I covered here only those parts that did not require too much background in measure theory and geometric measure theory. This means that the fundamental results of De Giorgi, Federer, and many others are not included here. Again, I only scratched the surface of functions of bounded variation. My hope is that this volume will help students to build a solid background, which will allow them to read more advanced texts on the subject. Chapter 17 is dedicated to the theory of Besov spaces. There are essentially three ways to look at these spaces. One of the most successful is to see them as an example/by-product of interpolation theory (see [7], [232], and [233]). Interpolation is very elegant, and its abstract framework can be used to treat quite general situations well beyond Sobolev and Besov spaces. There are two reasons for why I decided not to use it: First, it would depart from the spirit of the book, which leans more towards measure theory and real analysis and less towards functional analysis. The second reason is that in recent years in calculus of variations there has been an increased
xviii
Preface
interest in nonlocal functionals. I thought it could be useful to present some techniques and tricks for fractional integrals. The second approach is to use tempered distributions and Fourier theory to introduce Besov spaces. This approach has been particularly successful for its applications to harmonic analysis. Again it is not consistent with the remainder of the book. This left me with the approach of the Russian school, which relies mostly on the inequalities of Hardy, H¨older, and Young, together with some integral identities. The main references for this chapter are the books of Besov, Il in, and Nikol ski˘ı [26], [27]. I spent an entire summer working on this chapter, but I am still not happy with it. In particular, I kept thinking that there should be easier and more elegant proofs of some of the results, but I could not find one. In Chapter 18 I discuss traces of Sobolev and BV functions. Although in this book I only treat first-order Sobolev spaces, the reason I decided to use Besov spaces over fractional Sobolev spaces (note that in the range of exponents treated in this book these spaces coincide, since their norms are equivalent) is that the traces of functions in W k,1 (Ω) live in the Besov space B k−1,1 (∂Ω), and thus a unified theory of traces for Sobolev spaces can only be done in the framework of Besov spaces. Finally, Chapter 15 is devoted to the theory of symmetrization in Sobolev and BV spaces. This part of the theory of Sobolev spaces, which is often missing in classical textbooks, has important applications in sharp embedding constants, in the embedding N = p, as well as in partial differential equations. In Appendices A, B, and C I included essentially all the results from functional analysis and measure theory that I used in the text. I only proved those results that cannot be found in classical textbooks. What is missing in this book: For didactic purposes, when I started to write this book, I decided to focus on first-order Sobolev spaces. In my original plan I actually meant to write a few chapters on higher-order Sobolev and Besov spaces to be put at the end of the book. Eventually I gave up: It would have taken too much time to do a good job, and the book was already too long. As a consequence, interpolation inequalities between intermediate derivatives are missing. They are treated extensively in [7]. Another important theorem that I considered adding and then abandoned for lack of time was Jones’s extension theorem [122]. Chapter 14, the chapter on BV functions of several variables, is quite minimal. As I wrote there, I only touched the tip of the iceberg. Good
Preface to the First Edition
xix
reference books of all the fundamental results that are not included here are [10], [72], and [251]. References: The rule of thumb here is simple: I only quoted papers and books that I actually read at some point (well, there are a few papers in German, and although I do have a copy of them, I only “read” them in a weak sense, since I do not know the language). I believe that misquoting a paper is somewhat worse than not quoting it. Hence, if an important and relevant paper is not listed in the references, very likely it is because I either forgot to add it or was not aware of it. While most authors write books because they are experts in a particular field, I write them because I want to learn a particular topic. I claim no expertise on Sobolev spaces. Webpage for mistakes, comments, and exercises: In a book of this length and with an author a bit absent-minded, typos and errors are almost inevitable. I will be very grateful to those readers who write to
[email protected] indicating those errors that they have found. The AMS is hosting a webpage for this book at http://www.ams.org/bookpages/gsm-105/ where updates, corrections, and other material may be found. The book contains more than 200 exercises, but they are not equally distributed. There are several on the parts of the book that I actually taught, while other chapters do not have as many. If you have any interesting exercises, I will be happy to post them on the web page. Giovanni Leoni
Acknowledgments The Author List, II. The second author: Grad student in the lab that has nothing to do with this project, but was included because he/she hung around the group meetings (usually for the food). The third author: First year student who actually did the experiments, performed the analysis and wrote the whole paper. Thinks being third author is “fair”. — Jorge Cham, www.phdcomics.com
Second Edition I am profoundly indebted to Ian Tice for months spent discussing several parts of the book, especially Section 12.5 and Chapters 16 and 17. In particular, Chapter 17 was really a collaborative effort. Any mistake is purely due to me. Thanks, Ian, I owe you a big one. I would like to thank all the readers who sent corrections and suggestions to improve the first edition over the years. I would also like to thank the Friday afternoon reading club (Riccardo Cristoferi, Giovanni Gravina, Matteo Rinaldi) for reading parts of the book. I am really grateful to to Sergei Gelfand, AMS publisher, and to the all the AMS staff I interacted with, especially to Christine Thivierge, for her constant help and technical support during the preparation of this book, and to Luann Cole and Mike Saitas for editing the manuscript. I would like to acknowledge the Center for Nonlinear Analysis (NSF PIRE Grant No. OISE-0967140) for its support during the preparation of this book. This research was partially supported by the National Science Foundation under Grants No. DMS-1412095 and DMS-1714098. Also for this edition, many thanks must go to all the people who work at the interlibrary loan of Carnegie Mellon University for always finding in a timely fashion all the articles I needed.
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Acknowledgments
The picture on the back cover of the book was taken by Adella Guo, a student from Carnegie Mellon School of Design, whom I would like to thank. Finally, I would like to thank Jorge Cham for giving me permission to continue to use quotes from www.phdcomics.com for the second edition. They are of course the best part of the book :)
First Edition I am profoundly indebted to Pietro Siorpaes for his careful and critical reading of the manuscript and for catching 2ℵ0 mistakes in previous drafts. All remaining errors are, of course, mine. Several iterations of the manuscript benefited from the input, suggestions, and encouragement of many colleagues and students, in particular, Filippo Cagnetti, Irene Fonseca, Nicola Fusco, Bill Hrusa, Bernd Kawohl, Francesco Maggi, Jan Mal´ y, Massimiliano Morini, Roy Nicolaides, Ernest Schimmerling, and all the students who took the Ph.D. courses “Sobolev spaces” (fall 2006 and fall 2008) and “Measure and Integration” (fall 2007 and fall 2008) taught at Carnegie Mellon University. A special thanks to Eva Eggeling who translated an entire paper from German for me (and only after I realized I did not need it; sorry, Eva!). The picture on the back cover of the book was taken by Monica Montagnani with the assistance of Alessandrini Alessandra (always trust your high school friends for a good laugh. . . at your expense). I am really grateful to Edward Dunne and Cristin Zannella for their constant help and technical support during the preparation of this book. I would also like to thank Arlene O’Sean for editing the manuscript, Lori Nero for drawing the pictures, and all the other staff at the AMS I interacted with. I would like to thank three anonymous referees for useful suggestions that led me to change and add several parts of the manuscript. Many thanks must go to all the people who work at the interlibrary loan of Carnegie Mellon University for always finding in a timely fashion all the articles I needed. I would like to acknowledge the Center for Nonlinear Analysis (NSF Grant Nos. DMS-9803791 and DMS-0405343) for its support during the preparation of this book. This research was partially supported by the National Science Foundation under Grant No. DMS-0708039. Finally, I would like to thank Jorge Cham for giving me permission to use some of the quotes from www.phdcomics.com. They are really funny.
Part 1
Functions of One Variable
Chapter 1
Monotone Functions Undergradese, I: “Is it going to be an open book exam?” Translation: “I don’t have to actually memorize anything, do I?” — Jorge Cham, www.phdcomics.com
In this chapter we study continuity and differentiability properties of monotone functions. The central result of this chapter is the Lebesgue differentiation theorem.
1.1. Continuity In this section we study regularity properties of monotone functions. Definition 1.1. Let E ⊆ R. A function u : E → R is called (i) increasing if u(x) ≤ u(y) for all x, y ∈ E with x < y, (ii) strictly increasing if u(x) < u(y) for all x, y ∈ E with x < y, (iii) decreasing if u(x) ≥ u(y) for all x, y ∈ E with x < y, (iv) strictly decreasing if u(x) > u(y) for all x, y ∈ E with x < y, (v) monotone if any of the above holds. A monotone function is not continuous in general, so a natural question is how discontinuous it can be. The answer is given by the following theorem. In what follows an interval I ⊆ R is a set of R such that if x, y ∈ I and x < y, then [x, y] ⊆ I. Theorem 1.2. Let I ⊆ R be an interval and let u : I → R be a monotone function. Then u has at most countably many discontinuity points. Conversely, given a countable set E ⊂ R, there exists a monotone function u : R → R whose set of discontinuity points is exactly E. 3
4
1. Monotone Functions
We begin with a preliminary result. Given a set X and a function v : X → [0, ∞], the infinite sum of v over X is defined as v(x) := sup v(x) : E ⊆ X, E finite . x∈X
x∈E
The following lemma shows that every infinite sum which is finite can be written as a series. Lemma 1.3. Given a set X and a function v : X → [0, ∞], if v(x) < ∞, x∈X
then the set {x ∈ X : v(x) > 0} is countable. Moreover, v does not take the value ∞. Proof. Define L :=
v(x) < ∞.
x∈X
For n ∈ N set Xn := {x ∈ X : v(x) > 1/n} and let E be a finite subset of Xn . Then 1 v(x) ≤ L, n number of elements of E ≤ x∈E
which shows that E cannot have more than nL elements, where · is the integer part. In turn, Xn has a finite number of elements, and so Xn {x ∈ X : v(x) > 0} = n
is countable. We turn to the proof of Theorem 1.2.
Proof. Step 1: Assume that I = [a, b] and, without loss of generality, that u is increasing. For every x ∈ (a, b) there exist lim u(y) =: u+ (x),
y→x+
lim u(y) =: u− (x).
y→x−
Let S(x) := u+ (x) − u− (x) ≥ 0 be the jump of u at x. Then u is continuous at x if and only if S(x) = 0. Moreover, since u is increasing, for every a ≤ x1 < · · · < xn ≤ b, u(a) ≤ u− (x1 ) ≤ u+ (x1 ) ≤ u− (x2 ) ≤ u+ (x2 ) ≤ · · · ≤ u− (xn ) ≤ u+ (xn ) ≤ u(b), and so,
x∈[a,b]
S(x) ≤ u(b) − u(a).
1.1. Continuity
5
It now follows by the previous lemma that u has at most countably many discontinuity points. Step 2: If I is an arbitrary interval, construct an increasing sequence of intervals [an , bn ] such that I = n [an , bn ]. Since the union of countable sets is countable and on each interval [an , bn ], the set of discontinuity points of u is at most countable; by the previous step it follows that the set of discontinuity points of u in I is at most countable. Step 3: Conversely, let E ⊂ I be a countable set. If E is finite, then an increasing function with discontinuity set E may be constructed by hand. Consider now the more interesting case in which E is denumerable, so that E = {xn : n ∈ N}. For each n ∈ N define the increasing function un : R → R as −1/n2 if x < xn , un (x) := 1/n2 if x ≥ xn . Note that un is discontinuous only at the point xn . Set u(x) :=
∞
un (x),
x ∈ R.
n=1
Since |un (x)| ≤ 1/n2 for all x ∈ R, the series of functions is uniformly convergent, and so it is continuous at every point at which all the functions un are continuous. In particular, u is continuous in R \ E. We now prove that u is discontinuous at every point of E. Indeed, for every m ∈ N write un . u = um + n=m
Then n=m un is continuous at xm while um is not. Hence, u is discontinuous at each x ∈ E. To conclude, observe that u is increasing, since it is the pointwise limit of a sequence of increasing functions. By taking E := Q, we obtain the following result. Corollary 1.4. There exists an increasing function u : R → R that is continuous at all irrational points and discontinuous at all rational points. Definition 1.5. Given an interval I ⊆ R, an increasing function u : I → R is called an increasing saltus or jump function if for every [a, b] ⊆ I it can be written in the form un (x), x ∈ [a, b], u(x) = c + n∈E
6
1. Monotone Functions
where E ⊆ N, c ∈ R, and
⎧ ⎨ 0 if x < an , bn if x = an , un (x) = ⎩ cn if x > an ,
with an ∈ [a, b] and 0 ≤ bn ≤ cn , with at least one of the two inequalities strict, for all n ∈ E. Exercise 1.6 (The jump function). Let I ⊆ R be an interval and let u : I → R be increasing. For each x ∈ I define (u+ (y) − u− (y)) + u(x) − u− (x), uJ (x) := y∈I, y inf J. If v(y) = v(y0 ) for all y ∈ J with y < y0 , then there is nothing to prove. Thus assume that there is y1 ∈ J with y1 < y0 and v(y1 ) < v(y0 ). Let 0 < ε < v(y0 ) − v(y1 ) and fix x0 ∈ [v(y0 ) − ε, v(y0 )). Then u(x0 ) < y0 by the definition of v(y0 ). Hence, if y ∈ (u(x0 ), y0 ), we have that u(x) ≤ u(x0 ) < y for all x ∈ I with x < x0 , since u is increasing, and so v(y) = inf{x ∈ I : u(x) ≥ y} ≥ x0 ≥ v(y0 ) − ε, which shows that v is left continuous at y0 .
8
1. Monotone Functions
(ii) Assume that v(y0 ) < v+ (y0 ) for some y0 ∈ J, with y0 < sup J = supI u. By the definition of v(y0 ) and the fact that u is increasing, we have that u(x) ≥ y0 for all x > v(y0 ). On the other hand, if v(y0 ) < x < v+ (y0 ), then, since v is increasing, v(y) > x for all y > y0 , and so u(x) < y for all y > y0 , which implies that u(x) ≤ y0 . Thus, u(x) ≡ y0 for all x ∈ (v(y0 ), v+ (y0 )). Conversely, assume that u(x) ≡ y0 for all x in some interval (x1 , x2 ) ⊆ I, with x1 < x2 and y0 < sup J = supI u. Then v(y0 ) ≤ x1 , by the definition of v(y0 ). On the other hand, if y ∈ (y0 , sup J), then for every x ∈ (x1 , x2 ), we have that u(x) = y0 < y, and so v(y) ≥ x by the definition of v(y). Letting x → x+ 2 , we get that v(y) ≥ x2 for all y ∈ (y0 , sup J). In particular, v+ (y0 ) ≥ x2 . Thus, we have shown that v(y0 ) ≤ x1 < x2 ≤ v+ (y0 ). (iii) Taking y = u(x) in the definition of v(y), yields v(u(x)) ≤ x. If v(u(x0 )) < x0 for some x0 ∈ I, then there exists x ∈ I, with x < x0 , such that u(x) ≥ u(x0 ), but since u is increasing, it follows that u ≡ u(x0 ) in [x, x0 ]. Conversely, if u ≡const. in some interval [x, x0 ] ⊆ I, then v(u(x0 )) ≤ x < x0 . (iv) Assume that v ≡ x0 ∈ I ◦ in some interval (y1 , y2 ) ⊆ J, with y1 < y2 . If x ∈ I with x > x0 , then u(x) ≥ y for all y ∈ (y1 , y2 ) by the definition of v(y). Letting y → y2− , we get that u(x) ≥ y2 , and so u+ (x0 ) ≥ y2 . On the other hand, if x ∈ I with x < x0 , then u(x) < y for all y ∈ (y1 , y2 ) by the definition of v(y). Letting y → y1+ , we get that u(x) ≤ y1 , and so u− (x0 ) ≤ y1 . Conversely, let y ∈ (u− (x0 ), u+ (x0 )). If x < x0 , then u(x) < y, and so v(y) ≥ x0 . On the other hand, since u and v are increasing and by part (iii), for every x ∈ I with x > x0 , x0 ≤ v(y) ≤ v(u+ (x0 )) ≤ v(u(x)) ≤ x. Letting x → x+ 0 , we get that v(y) = x0 for all y ∈ (u− (x0 ), u+ (x0 )). Finally, assume that u is strictly increasing. Then by parts (i) and (ii) the function v is continuous. By part (iii), v(u(x)) = x for every x ∈ I, which proves that v is a left inverse of u. This concludes the proof. Remark 1.9. The fact that I is bounded from below is used to guarantee that the function v does not take the value −∞. Thus, (a modified version of) the previous theorem continues to hold if I is unbounded from below, provided the function v is allowed to take the value −∞. We leave the details as an exercise. Exercise 1.10. Let I = [a, b], let u : [a, b] → R be increasing and left continuous, and let v : [u(a), u(b)] → R be defined as in the previous theorem.
1.2. Differentiability
9
Prove that u(x) = inf{y ∈ [u(a), u(b)] : v(y) ≥ x},
x ∈ [a, b].
Remark 1.11. The previous theorem continues to hold if u is decreasing, with the only changes being that I is assumed to be bounded from above, the function v : J → R is now defined by v(y) := inf{x ∈ I : u(x) ≤ y}, that in part (i) one should replace increasing and left continuous with decreasing and right continuous, that in part (ii) one should take J \ {inf I u} in place of J \ {supI u}, and that in part (iv) the interval (u− (x0 ), u+ (x0 )) should be replaced by (u+ (x0 ), u− (x0 )).
1.2. Differentiability We now study the differentiability of monotone functions. Definition 1.12. Let E ⊆ R and let x0 ∈ E be an accumulation point of E. Given a function u : E → R, if there exists in [−∞, ∞] the limit lim
x→x0
u(x) − u(x0 ) , x − x0
then the limit is called the derivative of u at x0 and is denoted u (x0 ) or du (x0 ). The function u is differentiable at x0 if u (x0 ) exists in R. dx The central theorem of this section is Lebesgue’s theorem, which shows that a monotone function is differentiable everywhere except possibly on a set of Lebesgue measure zero. Note that this result relies strongly on monotonicity. Indeed, most (continuous) functions are nowhere differentiable. A classical example is due to Weierstrass and is given by (1.2)
u(x) :=
∞
an cos(bn πx),
x ∈ R,
n=0
where 0 < a < 1 and b ∈ N is an odd integer such that ab > 1 + 32 π (see [108]). In Theorem 1.15 below we present a simpler example. Definition 1.13. Given two metric spaces (X, dX ) and (Y, dY ) and E ⊆ X, a function u : E → Y is (i) Lipschitz continuous if Lip(u; E) =
sup x, y∈E, x=y
dY (u(x), u(y)) < ∞, dX (x, y)
10
1. Monotone Functions
(ii) H¨ older continuous of exponent α > 0 if |u|C 0,α :=
dY (u(x), u(y)) < ∞. α x, y∈E, x=y (dX (x, y)) sup
The number Lip(u; E) is called the Lipschitz constant of u. We usually write Lip u in place of Lip(u; E), whenever the underlying set is clear. Exercise 1.14. Let I ⊆ R be an interval and let u : I → R be H¨older continuous of exponent α > 1. Prove that u is constant. Theorem 1.15. Let v(x) = |x| for x ∈ [−1, 1] and extend v to R as a periodic function of period 2. Then the function (1.3)
u(x) =
∞
( 34 )n v(4n x),
x ∈ R,
n=1
is real-valued, continuous, and nowhere differentiable. Proof. Let vn (x) = ( 34 )n v(4n x), x ∈ R. Since v is periodic, supy∈R |v(y)| = supy∈[−1,1] |v(y)| = 1, and so |vn (x)| ≤ ( 34 )n . It follows that the series of functions converges uniformly to a continuous bounded function u. Next we prove that u is nowhere differentiable. Fix x ∈ R. We are going to construct a sequence hm → 0 such that u(x+hhmm)−u(x) → ∞ as m → ∞. We take hm = ± 12 41m , where the sign is chosen in such a way that in the open interval of endpoints 4m x and 4m (x + hm ) there is no integer. Let’s prove that we can always do this. We have 4m (x + 12 41m ) − 4m (x − 12 41m ) = 1. If both 4m (x + 12 41m ) and 4m (x − 12 41m ) are integers, then in the interval (4m (x − 12 41m ), 4m (x + 12 41m )) there is no integer and so we can take the sign of hm as we like. If 4m (x + 12 41m ) and 4m (x − 12 41m ) are not both integers, then in the interval (4m (x − 12 41m ), 4m (x + 12 41m )) there is exactly one integer. If this integer is 4m x, then we take the sign of hm as we like. If the integer is in (4m (x − 12 41m ), 4m x), then we take hm = 12 41m , while if the integer is in (4m x, 4m (x + 12 41m )), then we take hm = − 12 41m . We now study (1.4)
( 3 )n v(4n (x + hm )) − ( 34 )n v(4n x) vn (x + hm ) − vn (x) = 4 hm hm 1 n−m n v(4 x ± 2 4 ) − v(4n x) . = ( 34 )n ± 12 41m
If n > m, then 12 4n−m is an even integer and so by the periodicity of v the difference quotient is zero. If n = m, then since by construction in the open interval of endpoints 4m x and 4m (x + hm ) there is no integer, we have that
1.2. Differentiability
11
the points (x + hm , v(4m (x + hm ))) and (x, v(4m x)) lie in the same line of the graph of v with slope either 1 or −1. Hence by (1.4), (1.5)
|vm (x + hm ) − vm (x)| 4m |hm | = ( 34 )m = 3m . |hm | |hm |
Finally, if n < m, then using the fact that v is Lipschitz continuous with Lipschitz constant 1, we get (1.6)
4n |hm | |vn (x + hm ) − vn (x)| ≤ ( 34 )n = 3n . |hm | |hm |
Hence, u(x + hm ) − u(x) vn (x + hm ) − vn (x) = hm hm m
n=1
and using the inequality |a + b| ≥ |b| − |a|, (1.5), and (1.6), we get u(x + h ) − u(x) v (x + h ) − v (x) m−1 vn (x + hm ) − vn (x) m m m m + = hm hm hm n=1
v (x + h ) − v (x) m−1 m vm (x + hm ) − vm (x) m m ≥ − hm hm n=1
≥ 3m −
m−1
3n = 3m − 12 3m +
3 2
= 12 3m +
3 2
→∞
n=1
as m → ∞. This concludes the proof.
Exercise 1.16. Study the H¨older continuity of the function u in (1.3). Remark 1.17. It is actually possible to construct a continuous function u : (−1, 1) → R such that u(x + h) − u(x) u(x + h) − u(x) < lim sup lim inf =∞ + h h h→0 h→0+ for all x ∈ (−1, 1) (see [177]). In contrast to Theorem 1.15, a monotone function is differentiable everywhere except possibly on a set of Lebesgue measure zero. Note that this result implies, in particular, that the Weierstrass function and the function u in (1.3) are monotone on no interval. Theorem 1.18 (Lebesgue). Let I ⊆ R be an interval and let u : I → R be a monotone function. Then there exists a set E ⊂ R, with L1 (E) = 0, such that u is everywhere differentiable on I \ E.
12
1. Monotone Functions
We recall that the Lebesgue outer measure of a set E ⊆ R is defined as ∞ ∞ 1 Lo (E) := inf (bn − an ) : E ⊆ (an , bn ) . n=1
n=1
The proof will use the following properties of the outer measure L1o (see Proposition C.4): (L1 ) If {En }n∈N is a sequence of subsets of R, then
∞ ∞ 1 Lo En ≤ L1o (En ). n=1
n=1
(L2 ) If {En }n∈N is a sequence of subsets of R, with En ⊆ En+1 for every n, then
∞ 1 1 En . lim Lo (En ) = Lo n→∞
n=1
(L3 ) If E ⊆ R, then L1o (E) = inf{L1o (U ) : U open , E ⊆ U }. (L4 ) If I1 , . . . , Im are disjoint intervals, then m 1 length In . Lo (I1 ∪ · · · ∪ Im ) = n=1
We begin with some auxiliary results. Lemma 1.19. Let E ⊂ R be a bounded set and let F be a family of open intervals with the property that each x ∈ E is the left endpoint of an interval (x, x + hx ) in F . Then for every ε > 0 there exist disjoint intervals I1 , . . . , In ∈ F and a set n Ik F ⊆E∩ k=1
such that
L1o (F )
≥
L1o (E)
− ε.
Proof. Define En := {x ∈ E : hx > 1/n}. Then En ⊆ En+1 and ∞ 1 1 n=1 En = E. By property (L2 ), Lo (En ) → Lo (E) as n → ∞. Hence we can find m ∈ N so large that L1o (Em ) > L1o (E) − 2ε . Let a1 := inf Em , b1 := sup Em , and let := b1 − a1 > 0. Given ε , η= 2(m + 1) by the definition of infimum we can find x1 ∈ Em with a1 < x1 < a1 + η. By the definition of Em there exists an interval I1 = (x1 , x1 + h1 ) in F , with h1 > 1/m. If x1 + h1 ≥ b1 , then we can take F = Em ∩ (x1 , x1 + h1 ).
1.2. Differentiability
13
If x1 + h1 < b1 , let a2 := inf{x ∈ Em : x ≥ x1 + h1 }. Then by the definition of infimum we can find x2 ∈ Em with a2 < x2 < a2 +η. By the definition of Em there exists an interval I2 = (x2 , x2 + h2 ) in F , with h2 > 1/m. If x2 + h2 ≥ b1 , we stop, while if x2 + h2 < b1 we define a3 := inf{x ∈ Em : x ≥ x2 + h2 }. We continue in this way constructing intervals Ik until xk + hk < b1 . Since 1 , we have that we will find at most each interval Ik has length larger than m n intervals with n < m + 1. Let n n Ik , T := (xk − η, xk ]. S := k=1
k=1
Then x1 − η ≤ a1 = inf Em and so Em ⊆ S ∪ T . Moreover, the intervals Ik are disjoint by construction. Now by property (L1 ), L1o (E) −
ε 2
< L1o (Em ) ≤ L1o (Em ∩ S) + L1o (Em ∩ T ) ≤ L1o (Em ∩ S) + L1o (T ) ≤ L1o (Em ∩ S) +
n
L1o ((xk − η, xk ])
k=1
=
L1o (Em
∩ S) + nη ≤
L1o (Em
∩ S) +
ε 2,
and so Em ∩ S has the desired property.
Lemma 1.20. Let E ⊂ R be a bounded set and let F be a family of open intervals with the property that each x ∈ E is the left endpoint of an interval (x, x + hx ) in F with hx arbitrarily small (that is, for every η > 0 there is at least one interval with hx < η). Then for every ε > 0 there exist disjoint intervals I1 , . . . , In ∈ F and a set F ⊆E∩
n
Ik
k=1
such that L1o (F )
≥
L1o (E)
− ε,
n
length Ik ≤ L1o (E) + ε.
k=1
Proof. By property (L3 ) there exists an open set U ⊇ E such that L1o (U ) ≤ L1o (E) + ε. Let F be the subfamily of intervals (x, x + hx ) in F contained in U . Note that for each x ∈ E ⊆ U there must exist at least one such interval, since U contains a ball centered at x and there are intervals of arbitrarily small length.
14
1. Monotone Functions
Apply Lemma 1.19 to the family F to find disjoint intervals I1 , . . . , In ∈ F and a set n F ⊆E∩ Ik k=1
such that L1o (F ) ≥ L1o (E) − ε. Since the intervals are disjoint and contained in U , it follows from properties (L1 ) and (L4 ) that n
length Ik ≤ L1o (U ) ≤ L1o (E) + ε.
k=1
This concludes the proof.
Given a set E ⊆ R and function u : E → R, for every x0 ∈ E such that x0 is an accumulation point for the sets E ∩ (−∞, x0 ) and E ∩ (x0 , ∞), the four Dini’s derivatives of u are given by u(x) − u(x0 ) , x − x0
D + u(x0 ) := lim sup
u(x) − u(x0 ) , x − x0
D+ u(x0 ) := lim inf
D − u(x0 ) := lim sup x→x− 0
D − u(x0 ) := lim inf x→x− 0
x→x+ 0
x→x+ 0
u(x) − u(x0 ) , x − x0
u(x) − u(x0 ) . x − x0
We now turn to the proof of Lebesgue’s differentiation theorem. Proof. Step 1: Assume that u is increasing and that I is bounded and let E := {x ∈ I ◦ : D + u(x) < D+ u(x)}. We claim that E has Lebesgue measure zero. To see this we write E as a countable union of sets Er,s , Er,s := {x ∈ E : D + u(x) < r < s < D+ u(x)}. E= r,s∈Q, 0 0 there exist disjoint intervals I1 , . . . , In ∈ F and a set (1.7)
Fr,s ⊆ Er,s ∩
n k=1
Ik
1.2. Differentiability
15
such that (1.8)
L1o (Fr,s ) ≥ L1o (Er,s ) − ε,
n
length Ik ≤ L1o (Er,s ) + ε.
k=1
Write Ik = (xk , xk + hk ). Then by (1.7) and (1.8), n n u(xk + hk ) − u(xk ) < r hk ≤ rL1o (Er,s ) + rε. (1.9) k=1
k=1
n
Let V := k=1 Ik . Note that V is open. Since Fr,s ⊆ Er,s ∩ V , we have that D+ u(x) > s for each x ∈ Fr,s , and so we can find an open interval (x, x + t) ⊆ V , where t > 0 is arbitrarily small and such that u(x + t) − u(x) > s. t Let G be the family of all such intervals. By Lemma 1.20 for every ε > 0 there exist disjoint intervals J1 , . . . , Jm ∈ G and a set m Ji (1.11) Gr,s ⊆ Fr,s ∩
(1.10)
i=1
such that L1o (Gr,s ) ≥ L1o (Fr,s ) − ε.
(1.12)
Write Ji = (yi , yi + ti ). Then by (1.8), (1.10), (1.11), and (1.12), m m (1.13) u(yi + ti ) − u(yi ) > s ti ≥ sL1o (Gr,s ) i=1
i=1 ≥ sL1o (Fr,s ) − sε ≥ sL1o (Er,s ) − 2sε. But since each Ji is contained in V = nk=1 Ik , and since the intervals Ik are disjoint, it follows that each interval Ji is contained in some interval Ik .
Since u is increasing it follows that m n u(yi + ti ) − u(yi ) ≤ u(xk + hk ) − u(xk ). i=1
k=1
Combining this inequality with (1.9) and (1.13) gives m u(yi + ti ) − u(yi ) sL1o (Er,s ) − 2sε < i=1
≤
n
u(xk + hk ) − u(xk ) < rL1o (Er,s ) + rε,
k=1
that is, (s − r)L1o (Er,s ) < 2sε + rε. Since s − r > 0, letting ε → 0+ we conclude that L1o (Er,s ) = 0.
16
1. Monotone Functions
Hence, we have shown that L1o (E) = 0. It follows that for all x ∈ I ◦ \ E there exists the right derivative u+ (x) (possibly infinite). With a similar proof we can show that the left derivative exists (possibly infinite) for all x ∈ I ◦ except for a set of Lebesgue measure zero. Step 2: Let E∞ := {x ∈ I ◦ : u+ (x) = ∞}. We leave as an exercise to prove that L1o (E∞ ) = 0. Step 3: If I is an arbitrary interval, construct an increasing sequence of intervals [an , bn ] such that I = n [an , bn ]. Since the union of sets of Lebesgue outer measure zero still has Lebesgue outer measure zero and since on each interval [an , bn ] the set of points in which u is not differentiable has Lebesgue outer measure zero by the previous step, it follows that the set of points of I in which u is not differentiable has Lebesgue outer measure zero. The proof is concluded by the following exercise. Exercise 1.21. Let I ⊆ R be an interval and let u, v : I → R. (i) Prove that the set of x ∈ I ◦ such that there is ε > 0 such that v(y) > v(x) for all y ∈ I ∩ (x − ε, x + ε), y = x, is countable. (ii) Prove that the sets {x ∈ I ◦ : D+ u(x) < D− u(x)} and {x ∈ I ◦ : D− u(x) < D + u(x)} are countable. Hint: Apply part (i) to the function v(x) = u(x)−rx to prove that the set {x ∈ I ◦ : D − u(x) < r < D + u(x)} is countable. (iii) Prove that the set {x ∈ I ◦ : u+ (x) = u− (x)} is countable. Exercise 1.22. Let I ⊆ R be an interval and let u : I → R be continuous in I and differentiable for all x ∈ I except at most a countable set E. (i) Prove that if u (x) > 0 for all x ∈ I \E, then u is strictly increasing. Hint: If u(b) < u(a), let u(b) < t < u(a) with t ∈ / u(E) and consider the set {x ∈ [a, b] : u(x) = t}. (ii) Prove that if u (x) ≥ 0 for all x ∈ I \ E, then u is increasing. The next theorem shows that Lebesgue’s differentiation theorem is sharp, to be precise, that there exist monotone functions that are not differentiable on sets of Lebesgue measure zero. Theorem 1.23. For every set E ⊂ R of Lebesgue (outer) measure zero there exists a continuous increasing function that is not differentiable at every point of E.
1.2. Differentiability
17
Proof. Since L1o (E) = 0, for every l ∈ N we may cover E with a countable family of open intervals {(al,m , bl,m )}m such that ∞
(1.14)
(bl,m − al,m ) ≤
m=1
1 . 2l
Let I := {(al,m , bl,m )}l,m = {In }n , with In = (an , bn ). Note that by (1.14), ∞
(1.15)
(bn − an ) ≤ 1.
n=1
Moreover, for every x ∈ E we may find infinitely many In such that x ∈ In . For each n ∈ N define the continuous increasing function un : R → [0, ∞) as follows: ⎧ if x < an , ⎨ 0 x − an if an ≤ x ≤ bn , un (x) := ⎩ bn − an if x > bn . Note that 0 ≤ un (x) ≤ bn − an for all x ∈ R. Set (1.16)
u(x) :=
∞
un (x),
x ∈ R.
n=1
Then for all x ∈ R and l ∈ N, 0 ≤ u(x) −
l n=1
un (x) ≤
∞
(bn − an ),
n=l+1
and so the series of functions is uniformly convergent by (1.15). In particular, this implies that u is continuous. Since each un is nonnegative and increasing, it follows that u has the same properties. It remains to show that u is not differentiable in E. Thus, fix x0 ∈ E and l ∈ N. By the way we constructed the sequence I, we may find l positive integers n1 < n2 < · · · < nl such that x0 ∈ Ini for all i = 1, . . . , l. Let x ∈ In1 ∩ · · · ∩ Inl , x = x0 . Since each un is increasing, the difference quotient is nonnegative, and so u(x) − u(x0 ) uni (x) − uni (x0 ) x − ani − (x0 − ani ) ≥ = = l. x − x0 x − x0 x − x0 l
l
i=1
i=1
It follows that lim sup x→x0
u(x) − u(x0 ) ≥l x − x0
for every l ∈ N. Hence, lim sup x→x0
u(x) − u(x0 ) = ∞, x − x0
which implies that u is not differentiable at x0 .
18
1. Monotone Functions
Remark 1.24. Note that the set of points at which the increasing function u (defined in (1.16)) is not differentiable contains the set E, but in general it may be larger. To the author’s knowledge an exact characterization of the class of sets that are sets of nondifferentiability for increasing functions is still an open problem (see Exercise 1.29 below for some properties of this set). We refer to [37] and [237] for more information on this subject. As a simple consequence of Lebesgue’s theorem we obtain that the derivative of a monotone function is always locally Lebesgue integrable. Corollary 1.25. Let I ⊆ R be an interval and let u : I → R be a monotone function. Then u is a Lebesgue measurable function and for every [a, b] ⊆ I, b (1.17) |u (x)| dx ≤ |u(b) − u(a)|. a
Moreover, if u is bounded, then u is Lebesgue integrable and |u (x)| dx ≤ sup u − inf u. (1.18) I
I
I
Proof. Assume that u is increasing, fix [a, b] ⊆ I, and let u(x) if x ≤ b, (1.19) v(x) := u(b) if x ≥ b. Let E := {x ∈ [a, b] : v is not differentiable at x}. By Lebesgue’s theorem, L1 (E) = 0. Moreover, (1.20)
v (x) = u (x) for all x ∈ (a, b) \ E.
Since v is increasing, we have that v (x) ≥ 0 for all x ∈ [a, b] \ E and v (x) = lim
n→∞
v(x + 1/n) − v(x) . 1/n
By Theorem 1.2 the nonnegative functions, wn (x) :=
v(x + 1/n) − v(x) , 1/n
x ∈ [a, b],
are continuous except for a countable set, and so they are Borel measurable functions. Hence, v : [a, b] \ E → [0, ∞) is a Lebesgue measurable function. Since v is increasing, by (1.19) for every h > 0 we have b+h a+h 1 1 b (1.21) [v(x + h) − v(x)] dx = v(x) dx − v(x) dx h a h b a 1 ≤ {(u(b) − u(a))h} = u(b) − u(a). h
1.2. Differentiability
19
Taking h := 1/n, it follows from Fatou’s lemma, (1.20), and (1.21) that b b u (x) dx = v (x) dx 0≤ a
a
b
≤ lim inf n n→∞
[v(x + 1/n) − v(x)] dx ≤ u(b) − u(a).
a
This concludes the proof of the first part. Assume next that u is bounded and construct an increasing sequence of intervals [an , bn ] such that I = n [an , bn ]. By the previous part, bn u (x) dx ≤ u(bn ) − u(an ) ≤ sup u − inf u, 0≤ an
I
I
and so, letting n → ∞ and using the Lebesgue monotone convergence theorem, we obtain (1.18). Remark 1.26. In Exercise 1.29 below we will show that u is actually a Borel function. Exercise 1.27. Prove that if equality holds in (1.17) (respectively, in (1.18)), then u is continuous in [a, b], (respectively, in I). Using the previous corollary, we obtain the following result. We will see its usefulness for functions of bounded pointwise variation in Corollary 2.51. Corollary 1.28. Let I ⊆ R be an interval, let u : I → R be a monotone function, and let h > 0. Then for every [a, b] ⊆ I, with b − a ≥ h, 1 b−h |u(x + h) − u(x)| dx ≤ |u(b) − u(a)|. (1.22) h a Moreover, if u is bounded, then 1 |u(x + h) − u(x)| dx ≤ sup u − inf u, (1.23) I h Ih I where Ih := {x ∈ I : x + h ∈ I}. Proof. Assume that u is increasing, fix [a, b] ⊆ I, with b − a ≥ h, and let v be defined as in (1.19). Then by (1.21), 1 b 1 b−h [u(x + h) − u(x)] dx ≤ [v(x + h) − v(x)] dx h a h a ≤ u(b) − u(a). Assume next that u is bounded and construct an increasing sequence of intervals [an , bn ] such that I = n [an , bn ]. Then 1 bn −h [u(x + h) − u(x)] dx ≤ u(bn ) − u(an ) ≤ sup u − inf u, 0≤ I h an I
20
1. Monotone Functions
and so, letting n → ∞ and using the Lebesgue monotone convergence theorem, we obtain (1.23). In Corollary 1.25 we have shown that the derivative of a monotone function is Lebesgue measurable. The next exercise shows that the derivative of an arbitrary function is actually Borel measurable. We recall that a set E ⊆ R is called an Fσ set if it is a countable union of closed sets, a Gδ set if it is a countable intersection of open sets. Exercise 1.29. Let I ⊆ R be an open interval, let u : I → R be an arbitrary function, and let E := {x ∈ I : u is continuous at x}, F := {x ∈ I : u is differentiable at x}. Prove that (i) for each n ∈ N the set Un := {x ∈ I : there is δ > 0 such that (x − δ, x + δ) ⊆ I and if y, z ∈ (x − δ, x + δ), then |u(y) − u(z)| < 1/n} is open, (ii) the set E is a Gδ set, (iii) for each ε > 0, δ > 0 the set Uε,δ := {x ∈ E : there is y ∈ I such that 0 < |x − y| < δ and
u(y)−u(x) y−x
> ε}
is a relatively open subset of E, that is, the intersection of E with an open set, Uε,δ , where (iv) {x ∈ E : Du(x) > 0} = ε>0, ε∈Q δ>0, δ∈Q
Du(x) := lim sup y→x
u(y) − u(x) , y−x
(v) the sets {x ∈ E : Du(x) > α}, {x ∈ E : Du(x) < α} are Borel sets for every α ∈ R, where u(y) − u(x) , Du(x) := lim inf y→x y−x (vi) the set F is a Borel set, (vii) the function v : I → R, defined by u (x) if x ∈ F, v(x) := 0 otherwise, is Borel measurable.
1.2. Differentiability
21
Exercise 1.30 (Expansion in base b). Let b ∈ N be such that b ≥ 2 and let x ∈ [0, 1]. (i) Prove that there exists a sequence {an }n of nonnegative integers such that 0 ≤ an < b for all n ∈ N and ∞ an . x= bn n=1
Prove that the sequence {an }n is uniquely determined by x, unless x is of the form x = bkm for some k, m ∈ N, in which case there are exactly two such sequences. (ii) Conversely, given a sequence {an }n of nonnegative integers ∞ such an that 0 ≤ an < b for all n ∈ N, prove that the series n=1 bn converges to a number x ∈ [0, 1]. The next example shows that the inequality in (1.17) may be strict, and so for continuous monotone functions the fundamental theorem of calculus for Lebesgue integration fails. Example 1.31 (The Cantor function). Divide [0, 1] into three equal subintervals, and remove the middle interval I1,1 := ( 13 , 23 ). Divide each of the two remaining closed intervals [0, 13 ] and [ 23 , 1] into three equal subintervals, and remove the middle intervals I1,2 := ( 312 , 322 ) and I2,2 := ( 372 , 382 ). Continuing in this fashion, at each step n remove 2n−1 middle intervals I1,n , . . . , I2n−1 ,n , each of length 31n . The Cantor set D is defined as ∞ 2
n−1
D := [0, 1] \
Ik,n .
n=1 k=1
The set D is closed (since its complement is given by a family of open intervals), and ∞ 2
n−1
L (D) = 1 − 1
n=1 k=1
∞ 2 ∞ 1 2n−1 =1− = 1 − = 0. 3n 3n n−1
diam Ik,n
n=1 k=1
n=1
Thus, D has Lebesgue measure zero. To prove that D is uncountable, observe that x ∈ D if and only if ∞ cn , (1.24) x= 3n n=1
where each cn ∈ {0, 2} (see the previous exercise). For every x ∈ D of the form (1.24) define ∞ dn , u(x) := 2n n=1
22
1. Monotone Functions
y 1 3 4 1 2 1 4 1 9
2 9
1 3
2 3
7 9
8 9
1
x
Figure 2. The Cantor function.
where
1 if cn = 2, 0 if cn = 0. The function u : D →[0, 1] is well-defined (why?), increasing, continuous, and has the same values at the endpoints of each removed interval Ik,n and thus u extends to a continuous function on [0, 1]. This function is called the Cantor function (see Figure 2). Since u(D) = [0, 1], it follows that D is uncountable. Note also that u (x) = 0 for all x ∈ Ik,n , k, n ∈ N, so that u = 0 except on a set of Lebesgue measure zero. Since u(0) = 0 and u(1) = 1, it follows that the inequality in (1.17) is strict, and so the fundamental theorem of calculus for Lebesgue integration fails. dn :=
Exercise 1.32. Consider the complete metric space X := {u : [0, 1] → R : u is continuous, u(0) = 0, and u(1) = 1}, where we take the metric induced by supremum norm u∞ := max |u(x)|. x∈[0,1]
Consider the operator T : X → X, defined by ⎧ 1 if 0 ≤ x ≤ 13 , ⎪ ⎨ 2 u(3x) 1 if 13 < x < 23 , T (u)(x) := 2 ⎪ ⎩ 1 1 2 2 + 2 u(3x − 2) if 3 ≤ x ≤ 1. (i) Prove that T is a contraction, to be precise, that T (u1 ) − T (u2 )∞ ≤ 12 u1 − u2 ∞ for all u1 , u2 ∈ X. (ii) Prove that the Cantor function is the unique fixed point of T .
1.2. Differentiability
23
(iii) Define recursively the sequence of functions u0 (x) := x, un+1 (x) := T (un )(x), x ∈ [0, 1]. Prove that for all n ∈ N, |un (x1 ) − un (x2 )| ≤ |x1 − x2 |α 2 for all x1 , x2 ∈ [0, 1], where α := log log 3 . Deduce that the Cantor function u is H¨older continuous with exponent α.
(iv) Prove that the exponent α and the constant one are the best possible for the Cantor function u, in the sense that the inequality |u(x1 ) − u(x2 )| ≤ c|x1 − x2 |β does not hold for all x1 , x2 ∈ [0, 1] if either β > α or β = α but c < 1. Using the Cantor set, we can construct Lebesgue measurable sets that are not Borel sets. Exercise 1.33 (A non-Borel, Lebesgue measurable set). Consider the function v : [0, 1] → R defined by v(x) := x + u(x),
x ∈ [0, 1].
(i) Prove that L1 (v(D)) = 1. (ii) Let E ⊆ v(D) be a set that is not Lebesgue measurable (see Exercise C.2 in Appendix C) and consider the set F := v −1 (E) ⊆ D. Prove that F is Lebesgue measurable but not a Borel set. We have seen that the Cantor function is increasing with u = 0, L1 -a.e. in [0, 1]. It is actually possible to construct strictly increasing continuous functions whose derivative is zero except on a set of Lebesgue measure zero. To prove this result, we will need the following result. Theorem 1.34 (Fubini). Let I ⊆ R be an interval and let {un }n be a that the series of sequence of increasing functions, un : I → R. Assume functions n=1 un converges pointwise in I. Then n=1 un converges uniformly on compact sets of I, the sum of the series u(x) :=
∞
un (x),
x ∈ I,
n=1
is differentiable L1 -a.e. in I, and u (x) =
∞
un (x)
for L1 -a.e. x ∈ I.
n=1
Proof. Since the result is local, without loss of generality, we may assume that I = [a, b].
24
1. Monotone Functions
Step 1: Assume that un (a) = 0 for all n ∈ N so that un ≥ 0. For every l ∈ N set l sl := un . n=1
Then for x ∈ [a, b], ∞
0 ≤ u(x) − sl (x) =
n=l+1
and so 0 ≤ sup |u(x) − sl (x)| ≤ [a,b]
∞
un (x) ≤
un (b),
n=l+1 ∞
un (b) → 0
n=l+1
as l → ∞. Hence, the series is uniformly convergent. Since u and sl are increasing, by Lebesgue’s differentiation theorem (Theorem 1.18) they are differentiable L1 -a.e. in [a, b], and since the countable union of sets of Lebesgue zero measure still has Lebesgue measure zero, we may find a Lebesgue measurable set E ⊂ [a, b], with L1 (E) = 0, such that u and all the functions sl are everywhere differentiable in [a, b] \ E. Since the functions sl+1 − sl and u − sl are increasing for all l ∈ N, we have that (sl+1 − sl ) ≥ 0 and (u − sl ) ≥ 0 in [a, b] \ E. Hence, 0 ≤ s1 (x) ≤ s2 (x) ≤ · · · ≤ u (x) for all x ∈ [a, b] \ E, and so there exists lim s (x) n→∞ n
≤ u (x) < ∞.
In turn, (1.25)
lim un (x) = lim [sn (x) − sn−1 (x)] = 0
n→∞
n→∞
for all x ∈ [a, b] \ E. It remains to show that limn→∞ sn (x) = u (x) for L1 -a.e. x ∈ [a, b]. Since limn→∞ sn (x) exists for all x ∈ [a, b] \ E, it is enough to show this for a subsequence {snk }k of {sn }n . So let nk ∞ be such that 1 2k 1 and define vk (x) := u(x) − snk (x) ≤ 2k . Then vk is increasing and 0 ≤ u(b) − snk (b) ≤
0≤
∞ k=1
∞ 1 vk (x) ≤ . 2k k=1
Hence, we can apply the first part of the proof to the sequence {vk }k to conclude that lim vk (x) = 0 k→∞
1.2. Differentiability
25
for L1 -a.e. x ∈ [a, b] by (1.25). This completes the proof. Step 2: Write un = un − un (a) + un (a) =: wn + un (a) and apply Step 1 to wn . Exercise 1.35. Let C ⊆ R be a closed set and let u(x) := dist(x, C) = inf{|x − y| : y ∈ C,
x ∈ R}.
Prove that (i) u is 1-Lipschitz continuous, that is, |u(x) − u(y)| ≤ |x − y| for all x, y ∈ R, (ii) there exists an increasing function v : R → R such that v is continuous and C = {x ∈ R : v (x) = 0}, (iii) if C contains no intervals, then every such v is strictly increasing. Exercise 1.36. Let I ⊆ R be an interval, let u : I → R be an increasing function, and let uJ be the jump function of u. Prove that uJ (x) = 0 for L1 -a.e. x ∈ I. Theorem 1.37. There exists a strictly increasing continuous function u : R → R whose derivative is zero except on a set of Lebesgue measure zero. Proof. Let v : R → R be the Cantor function extended to be 1 for x ≥ 1 and 0 for x ≤ 0. Consider a countable dense set {an }n in R (e.g., the rational numbers) and define ∞ 1 v(2n (x − an )), u(x) := 2n
x ∈ R.
n=1
Since 0 ≤ v ≤ 1, the series of functions converges uniformly to a continuous bounded function. To prove that u is strictly increasing, let x < y. By density there exists x < am < y. Then y − am > 0 > x − am and so v(2m (y − am )) > 0 = v(2m (x − am )), since the Cantor function is positive for x > 0, while v(2n (y−an )) ≥ v(2n (x− an )), since v is increasing. It now follows by Fubini’s theorem that v (x) = 0 for L1 -a.e. x ∈ R.
Exercise 1.38. Let u : [a, b] → R be differentiable at x0 ∈ (a, b) and let xn , yn ∈ (a, b) \ {x0 } be such that xn = yn and xn , yn → x0 .
26
1. Monotone Functions
(i) Prove that if xn < x0 < yn for all n ∈ N, then u(yn ) − u(xn ) = u (x0 ). n→∞ yn − xn (ii) Prove that if the condition xn < x0 < yn is violated, then the limit n) limn→∞ u(yynn)−u(x may not exist or may be different from u (x0 ). −xn lim
Exercise 1.39. For each x ∈ (0, 1] find positive integers a0 < · · · < an < · · · such that ∞ 1 . x= 2an n=0
(see Exercise 1.30). Fix r > 0, r = 1. For x = 0 define u(0) := 0, while for each x ∈ (0, 1] define ∞ 2n u(x) := . 3an n=0
(i) Prove that u is strictly increasing. (ii) Given x ∈ (0, 1], for every m ∈ N0 let km be the number of subscripts n = 0, . . . , m for which an ≤ m. Define xm
km 1 := , 2an
zm := xm +
n=0
1 . 2m
Prove that xm < x < zm and that u(zm ) − u(xm ) → 0 as m → ∞. (iii) Prove that u is continuous. (iv) Prove that if u is differentiable at some x ∈ (0, 1], then necessarily u (x) = 0. Hint: Compute u(zm ) − u(xm ) . z m − xm Exercise 1.40. Let {rn }n be an enumeration of the rationals in [a, b] and define ∞ 1 (x − rn )1/3 , x ∈ [a, b]. u(x) := 2n n=1
(i) Prove that u is continuous and strictly increasing. (ii) Prove that if the series ∞
1 1 1 n 3 2 (x − rn )2/3 n=1
converges, then there exists ∞ 1 1 1 . u (x) = n 3 2 (x − rn )2/3 n=1
1.2. Differentiability
27
(iii) Prove that at all other points there exists u (x) = ∞. (iv) Let u([a, b]) = [α, β]. Prove that the inverse of u−1 : [α, β] → R is differentiable at every point and the set {x ∈ [α, β] : (u−1 ) (x) = 0} is dense in [α, β].
Chapter 2
Functions of Bounded Pointwise Variation Undergradese, II: “Is grading going to be curved?” Translation: “Can I do a mediocre job and still get an A?” — Jorge Cham, www.phdcomics.com
Let I ⊆ R be an interval. The set of monotone functions u : I → R is not a vector space, since in general the difference of monotone functions is not monotone. In this chapter we characterize the smallest vector space of functions u : I → R that contains all monotone functions. In many applications such as abstract evolution equations, gradient flows (see, e.g., [33], [71]), optimal transport (see, e.g., [11]), length spaces, (see, e.g., [38]), it is important to consider functions u : I → Y , where Y is a metric space, possibly infinite-dimensional. In this book we will focus mainly on the cases in which Y is RM or an infinite-dimensional normed space. For readers without a solid background in functional analysis, simply take Y = RM in the entire chapter.
2.1. Pointwise Variation In what follows, given an interval I ⊆ R, a partition of I is a finite set P := {x0 , . . . , xn } ⊂ I, where x0 < · · · < xn . Definition 2.1. Let I ⊆ R be an interval, let (Y, d) be a metric space, and let u : I → Y be a function. The pointwise variation of u on the interval I is n d(u(xi ), u(xi−1 )) , (2.1) Var u := sup i=1
29
30
2. Functions of Bounded Pointwise Variation
where the supremum is taken over all partitions P := {x0 , . . . , xn } of I, n ∈ N. A function u : I → Y has finite or bounded pointwise variation if Var u < ∞. The space of all functions u : I → Y of bounded pointwise variation is denoted by BP V (I; Y ). When Y = R, for simplicity we write BP V (I) for BP V (I; R)1 . Remark 2.2. Note that if I reduces to a singleton, then it has no partitions. In this case we set Var u := 0. In the remaining of the chapter it is tacitly understood that inf I < sup I, with some obvious exceptions. Remark 2.3. Note that if one of the endpoints, say, b := sup I, belongs to I, then in the definition of Var u it suffices to restrict our attention to partitions P := {x0 , . . . , xn } such that x0 < · · · < xn = b. Indeed, if P := {x0 , . . . , xn } is a partition with xn < b, then Q := {x0 , . . . , xn , b} is also a partition of I and so n n d(u(xi ), u(xi−1 )) ≤ d(u(xi ), u(xi−1 )) + d(u(b), u(xn )) ≤ Var u. i=1
i=1
In the sequel we will often use this property without further mention. To highlight the dependence on the interval I, we will sometimes write VarI u. Given an interval I ⊆ R and a metric space (Y, d), a function u : I → Y has locally finite or locally bounded pointwise variation if Var[a,b] u < ∞ for all intervals [a, b] ⊆ I. The space of all functions u : I → Y of locally bounded pointwise variation is denoted by BP Vloc (I; Y ). When Y = R, as before we write BP Vloc (I) for BP Vloc (I; R). It almost goes without saying that if I = [a, b], then BP Vloc ([a, b]; Y ) = BP V ([a, b]; Y ). If Ω ⊆ R is an open set, then we can write Ω as a countable union of pairwise disjoint open intervals In . Ω= n 1 This is not the standard notation for this space, which is usually denoted BV (I) in books on real analysis and measure theory. Although we do not like changing standard notation, unfortunately, in the literature the notation BV (I) is also used for a quite different (although strictly related) function space. This book studies both spaces, so we really had to change the notation for one of them.
2.1. Pointwise Variation
31
Given a function u : Ω → Y , where (Y, d) is a metric space, we define the pointwise variation of u on Ω as VarIn u, Var u := n
and we say that u has bounded pointwise variation in Ω if Var u < ∞. The space of functions u : Ω → Y of bounded pointwise variation is denoted by BP V (Ω; Y ). Exercise 2.4. Given u : [a, b] → R, the extended real numbers n (u(xi ) − u(xi−1 ))+ PVar u := sup i=1
and
n − NVar u := sup (u(xi ) − u(xi−1 )) , i=1
where the suprema are taken over all partitions P := {x0 , . . . , xn } of [a, b], n ∈ N, are called the positive and negative pointwise variation of u in [a, b], respectively. Prove that if u ∈ BP V ([a, b]), then Var u = PVar u + NVar u and that u(b) − u(a) = PVar u − NVar u. Remark 2.5. Note that a function u : I → RM belongs to BP V (I; RM ) (respectively, BP Vloc (I; RM )) if and only if each component ui : I → R, i = 1, . . . , M , belongs to BP V (I) (respectively, BP Vloc (I)). Indeed, for all i = 1, . . . , M and for every interval J ⊆ I we have (2.2)
VarJ ui ≤ VarJ u ≤ VarJ u1 + · · · + VarJ uM .
We now discuss the relation between functions of bounded pointwise variation with other important classes of functions. We begin by showing that Lipschitz continuous functions have locally bounded pointwise variation. Proposition 2.6. Let I ⊆ R be an interval, let (Y, d) be a metric space, and let u : I → Y be a Lipschitz continuous function. Then u belongs to BP Vloc (I; Y ). Moreover, if I is bounded, then u belongs to BP V (I; Y ). Proof. Since u is Lipschitz continuous, there exists L ≥ 0 such that d(u(x), u(z)) ≤ L|x − z|
32
2. Functions of Bounded Pointwise Variation
for all x, z ∈ I. Given [a, b] ⊆ I and a partition x0 < · · · < xn of [a, b], we have that n
d(u(xi ), u(xi−1 )) ≤ L
i=1
n
|xi − xi−1 | = L(xn − x0 ) ≤ L(b − a).
i=1
Taking the supremum over all partitions of [a, b] gives Var[a,b] u ≤ L(b − a) < ∞. Similarly, if I is bounded, then VarI u ≤ L length I < ∞.
The next exercise shows that continuous functions of bounded pointwise variation need not be Lipschitz continuous. Exercise 2.7. Let u : [0, 1] → R be defined by a x sin x1b if 0 < x ≤ 1, u(x) := 0 if x = 0, where a, b ∈ R. Study for which a, b the function u has bounded pointwise variation. In contrast to the previous proposition, there is no relation between a H¨ older continuous function and a function of bounded pointwise variation. This fact is illustrated by the following two exercises. Exercise 2.8. Let
u(x) :=
1 log x2
if 0 < x ≤ 1,
0
if x = 0.
Prove that u has finite pointwise variation in [0, 1] but it is not H¨ older continuous of any exponent 0 < α ≤ 1. Exercise 2.9. Let 1 < α < 2 and ∞ 1 < ∞. L := nα n=1
Subdivide the interval [0, L] into countably many intervals with x0 = 0 and xn − xn−1 = n1α . In each interval [xn−1 , xn ] define √ n−1 x − xn−1 if xn−1 ≤ x ≤ xn +x , 2 u(x) := √ n−1 xn − x if xn +x ≤ x ≤ x . n−1 2 Prove that the function u is H¨older continuous of exponent 1/2 but Var[0,L] u = ∞.
2.1. Pointwise Variation
33
Exercise 2.10. Let {rn }n be an enumeration of [0, 1] ∩ Q and define 1 2n if x = rn for some n, u(x) := 0 otherwise in [0, 1]. Compute Var[0,1] u. The next proposition shows that monotone functions have locally finite pointwise variation. Proposition 2.11. Let I ⊆ R be an interval and let u : I → R be a monotone function. Then for every interval J ⊆ I, VarJ u = sup u − inf u. J
J
In particular, u belongs to BP Vloc (I). Moreover, u belongs to BP V (I) if and only if it is bounded. Proof. Assume that u is increasing and let J ⊆ I be an interval. Then for every partition P := {x0 , . . . , xn } of J, n
|u(xi ) − u(xi−1 )| =
i=1
n
(u(xi ) − u(xi−1 ))
i=1
= u(xn ) − u(x0 ) ≤ sup u − inf u. J
J
Taking the supremum over all partitions of J yields VarJ u ≤ sup u − inf u. J
J
To prove the opposite inequality, it suffices to consider the case in which J is nondegenerate. Consider the partition P := {a, b} ⊂ J, where inf J ≤ a < b ≤ sup J. Then u(b) − u(a) = |u(b) − u(a)| ≤ VarJ u. / J, If sup J ∈ J, then taking b = sup J gives u(b) = supJ u, while if sup J ∈ letting b sup J gives u(b) → supJ u. Reasoning in a similar way for the left endpoint, we obtain that sup u − inf u ≤ VarJ u. J
This concludes the proof.
J
The previous proposition will be complemented by Theorem 2.36 below. Finally we discuss the relation between bounded functions and functions of bounded pointwise variation.
34
2. Functions of Bounded Pointwise Variation
Proposition 2.12. Let I ⊆ R be an interval, let (Y, d) be a metric space, and u : I → Y . For every x0 ∈ I and every interval J ⊆ I containing x0 , d(u(x), u(x0 )) ≤ VarJ u for all x ∈ J. In particular, if u belongs to BP Vloc (I; Y ) (respectively, BP V (I; Y )), then u is locally bounded (respectively, bounded). Proof. Fix an interval J ⊆ I containing x0 . For every x ∈ J with x = x0 , consider the partition P := {x0 , x}. Then d(u(x), u(x0 )) ≤ VarJ u,
which concludes the proof.
Remark 2.13. If Y is a normed space, then it follows from the previous proposition that sup u(x) ≤ u(x0 ) + VarJ u. x∈J
Example 2.14. The function u(x) := sin x, x ∈ R, is bounded and belongs to BP Vloc (R), but it does not belong to BP V (R) (why?).
2.2. Continuity In this section we study continuity properties of functions of bounded pointwise variation. Proposition 2.15. Let I ⊆ R be an interval, let (Y, d) be a metric space and let u : I → Y be a function. If c ∈ I, then VarI∩(−∞,c] u + VarI∩[c,∞) u = Var u. Proof. Let I1 := I ∩ (−∞, c] and I2 := I ∩ [c, −∞). Let P := {x0 , . . . , xn } and Q := {y0 , . . . , ym } be partitions of I1 and I2 , respectively. Then P ∪ Q is a partition of I, and so n i=1
d(u(xi ), u(xi−1 )) +
m
d(u(yi ), u(yi−1 )) ≤ Var u.
i=1
Taking first the supremum over all partitions P of I1 and then over all partitions Q of I2 yields VarI1 u + VarI2 u ≤ Var u. Conversely, let P := {x0 , . . . , xn } be a partition of I. Let m ∈ {1, . . . , n} be such that xm−1 ≤ c ≤ xm . Then P1 := {x0 , . . . , xm−1 , c} and P2 :=
2.2. Continuity
35
{c, xm , . . . , xn } are partitions of I1 and I2 , respectively, and so n
d(u(xi ), u(xi−1 )) ≤
m−1
i=1
d(u(xi ), u(xi−1 )) + d(u(c), u(xm−1 ))
i=1
+ d(u(xm ), u(c)) +
n
d(u(xi ), u(xi−1 ))
i=m+1
≤ VarI1 u + VarI2 u. Taking the supremum over all partitions P of I gives Var u ≤ VarI1 u + VarI2 u. Exercise 2.16. Let I ⊆ R be an interval, let (Y, d) be a metric space and let u : I → Y be a function. Prove that if I does not contain its right endpoint, then VarI∩(−∞,x] u = Var u, lim x→(sup I)−
while if I does not contain its left endpoint, then lim
x→(inf I)+
VarI∩[x,∞) u = Var u.
Proposition 2.15 implies that the function x ∈ I → VarI∩(−∞,x] u is increasing. However, for functions in BP Vloc (I) it may be infinite. Thus, we modify it as follows. Theorem 2.17 (Indefinite pointwise variation). Let I ⊆ R be an interval, let (Y, d) be a metric space, let t0 ∈ I, and let u ∈ BP Vloc (I; Y ). For every x ∈ I define Var[t0 ,x] u if x ≥ t0 , (2.3) V (x) := − Var[x,t0 ] u if x < t0 . Then for all x, y ∈ I, with x < y, (2.4)
d(u(y), u(x)) ≤ V (y) − V (x) = Var[x,y] u.
In particular, V is increasing and u is continuous at all but countably many points of I. Moreover, if Y is a complete metric space, then there exist u− (x) = lim u(y), y→x−
u+ (t) = lim u(y) y→x+
for all x ∈ I.2 Finally, if Y = R, the functions V ± u are increasing. Proof. Fix x, t ∈ I, with x < t. By Proposition 2.15, (2.5) ⎧ if t0 ≤ x < t, ⎨ Var[t0 ,t] u − Var[t0 ,x] u = V (t) − V (x) Var[x,t0 ] u − Var[t,t0 ] u = −V (x) + V (t) if x < t ≤ t0 , Var[x,t] u = ⎩ Var[x,t0 ] u + Var[t0 ,t] u = −V (x) + V (t) if x ≤ t0 ≤ t. 2 With
the obvious changes if x is an endpoint of I.
36
2. Functions of Bounded Pointwise Variation
Since d(u(t), u(x)) ≤ Var[x,t] u, the inequality (2.4) follows. Note that (2.4) implies that V (x) ≤ V (t) for all x, t ∈ I, with x < t. Hence, the function V is increasing. It follows by Theorem 1.2 that V is continuous at all but countably many points of I. Let x0 ∈ I be such that V is continuous at x0 . Then for every ε > 0 there exists δ = δ(x0 , ε) > 0 such that |V (x) − V (x0 )| ≤ ε for all x ∈ I with |x0 − x| ≤ δ. By (2.4), u is continuous at x0 . Thus, u is continuous at all but countably many x ∈ I. Next assume that Y is complete. Since V is increasing, it has a left and a right limit at every interior point of I. Hence, given x0 ∈ I ◦ for every ε > 0 there exists δ = δ(x0 , ε) > 0 such that V (x) − V + (x0 ) ≤ ε for all x ∈ I with x0 < x < x0 + δ. Taking x0 < x < t < x0 + δ, we have that V (t) − V (x) ≤ V (t) − V + (x0 ) ≤ ε and so by (2.4), d(u(t), u(x)) ≤ ε for all x0 < x < t < x0 + δ. Since Y is complete, this implies that there exists u+ (x0 ). Similarly, we can show that there exists u− (x0 ). Finally, if Y = R, (2.6)
±(u(t) − u(x)) ≤ |u(t) − u(x)| ≤ V (t) − V (x).
Hence, the functions V ± u are increasing.
The function V is called the indefinite pointwise variation of the function u. Remark 2.18. To highlight the dependence on t0 or on u, when necessary, we will write V t0 or V t0 ,u . Remark 2.19. Note that when inf I ∈ I, the choice t0 = inf I gives the simpler function (2.7)
V ∞ (x) := VarI∩(−∞,x] u,
x ∈ I.
However, if inf I does not belong to I or is −∞, then the function V ∞ may not be finite, unless u ∈ BP V (I). Using the function V we can establish the analog of Corollary 1.28. Theorem 2.20. Let I ⊆ R be an interval, let (Y, d) be a metric space, and let u ∈ BP Vloc (I; Y ). Then for h > 0, 1 d(u(x + h), u(x)) dx ≤ Var u, (2.8) h Ih where Ih := {x ∈ I : x + h ∈ I}.
2.2. Continuity
37
Proof. If Var u = ∞, then there is nothing to prove. Thus, assume that Var u < ∞. Fix [a, b] ⊆ I with 0 < h ≤ b − a. By (2.4) and Corollary 1.28 applied to V we have 1 b−h 1 b−h d(u(x + h), u(x)) dx ≤ (V (x + h) − V (x)) dx h a h a ≤ V (b) − V (a) = Var[a,b] u. Construct an increasing sequence of intervals [an , bn ] such that I = n [an , bn ]. If diam Ih > 0, then for all n sufficiently large we have that 0 < h < bn − an , and so by the previous inequality we have that 1 bn −h d(u(x + h), u(x)) dx ≤ Var[an ,bn ] u ≤ Var u. h an Letting n → ∞ and using the Lebesgue monotone convergence theorem gives 1 d(u(x + h), u(x)) dx ≤ Var u. h Ih The previous inequality continues to hold if diam Ih = 0, since in this case the integral on the left-hand side is zero. This completes the proof. In Corollary 2.51 we will prove a converse of Theorem 2.20. Inequality (2.4) shows that if the indefinite pointwise variation V (see (2.3)) is right continuous at some x ∈ I (respectively, left continuous), then so is u. Next we show that the opposite is also true. Proposition 2.21. Let I ⊆ R be an interval, let (Y, d) be a metric space, and let u ∈ BP Vloc (I; Y ). If u is right continuous at some x ∈ I (respectively, left continuous), then so is V . Proof. Assume that u is right continuous at x0 ∈ I. Fix r > 0 such that x0 + r ∈ I. Then given ε > 0 there exists 0 < δ ≤ r such that (2.9)
d(u(x), u(x0 )) ≤ ε
for all x ∈ I with x0 ≤ x ≤ x0 + δ. Let x0 < x1 < · · · < xn = x0 + r be a partition of [x0 , x0 + r] such that Var[x0 ,x0 +r] u ≤
n i=1
d(u(xi ), u(xi−1 )) + ε.
38
2. Functions of Bounded Pointwise Variation
By adding the point x0 + δ to the partition, without loss of generality, we may assume that x0 < x1 ≤ x0 + δ. Then by (2.9), Var[x0 ,x0 +r] u ≤ d(u(x1 ), u(x0 )) +
n
d(u(xi ), u(xi−1 )) + ε
i=2
≤ Var[x1 ,x0 +r] u + 2ε. By (2.4), it follows that V (x1 ) − V (x0 ) = Var[x0 ,x1 ] u ≤ 2ε. Note that x1 depends on δ and so on ε. Let δ1 := x1 − x0 > 0. Since V is increasing, for all x0 < x < x0 + δ1 , V (x0 ) ≤ V (x) ≤ V (x0 + δ1 ) = V (x1 ) ≤ V (x0 ) + 2ε, which shows that V is right continuous at x0 .
Exercise 2.22. Let I ⊆ R be an interval and let u ∈ BP V (I). Prove that if I contains its right endpoint b := sup I and u is left continuous at b, then lim VarI∩(−∞,x] u = Var u.
x→b−
Exercise 2.23. Let I ⊆ R be an interval and let u ∈ BP V (I). Let V be the function defined in (2.3). Prove that supI V − inf I V = Var u. The following result will be useful in the parametrization of curves in Chapter 5. Corollary 2.24. Let I ⊆ R be an interval, let (Y, d) be a metric space, and let u ∈ BP Vloc (I; Y ) be a continuous function. Then we can write u=w◦V, where w : J → Y is Lipschitz continuous with Lipschitz constant at most 1 and VarI u = VarJ w, where J := V (I). Proof. By Theorem 2.17 and Proposition 2.21 the function V : I → V (I) is continuous, increasing, and onto. Hence, J = V (I) is an interval. Note that the function V is constant on an interval if and only if u is constant there. Hence, for every s ∈ J there exists either a single point xs such that V (xs ) = s or there exists a maximal interval of endpoints αs and βs such that V (x) = s for all x ∈ [αs , βs ] ∩ I. In the latter case, we let xs be any element in [αs , βs ] ∩ I. For s ∈ J define w(s) := u(xs ). Note that by construction (2.10)
w(V (x)) = u(x)
2.2. Continuity
39
for all x ∈ I. We claim that w is Lipschitz continuous with Lipschitz constant less than or equal to 1. To see this, note that for all s, t ∈ J, with s < t, we have d(w(t), w(s)) = d(u(xt ), u(xs )) ≤ V (xt ) − V (xs ) = t − s, by (2.4) and the facts that V (xt ) = t and V (xs ) = s. Thus, w is Lipschitz continuous with Lipschitz constant less than or equal to 1. In particular, if s0 < · · · < sn is a partition of J, then n
d(w(si ), w(si−1 )) ≤
n (si − si−1 ) ≤ length J = VarI u.
i=1
i=1
On the other hand, since V is increasing, for every partition of I, x0 < · · · < xn , we have that V (x0 ) ≤ · · · ≤ V (xn ) and so if at least one of these inequalities is strict, then by (2.10), n i=1
d(u(xi ), u(xi−1 )) =
n
d(w(V (xi )), w(V (xi−1 ))) ≤ VarJ w.
i=1
On the other hand, if V (x0 ) = V (xn ), then u is constant in [x0 , xn ] and so the left-hand side of the previous inequality is zero. Thus the previous inequality continues to hold. In turn, VarI u ≤ VarJ w, which completes the proof. Exercise 2.25. Let p ≥ 1. Given a function u : [a, b] → R, we define the p-variation of u as n 1/p |u(xi ) − u(xi−1 )|p , Varp u := sup i=1
where the supremum is taken over all partitions P = {x0 , . . . , xn } of I, n ∈ N. (i) Prove that if Varp u < ∞, then u is bounded and for every x ∈ (a, b) there exist in R the left and right limits u+ (x) and u− (x). (ii) Prove that if Varp u < ∞, then u has at most a countable number of discontinuity points {tn }n and that 1/p |u+ (tn ) − u− (tn )|p . Varp u ≥ n
(iii) Prove that if Varp u < ∞ and q > p, then Varq u ≤ (2u∞ )
q−p q
p
(Varp u) q .
(iv) Prove that for all u, v : [a, b] → R, Varp (u + v) ≤ Varp u + Varp v.
40
2. Functions of Bounded Pointwise Variation
(v) Prove that if u : [a, b] → R is H¨older continuous of exponent then Varp u < ∞.
1 p,
Exercise 2.26. Let p ≥ 1. Given a function u : [a, b] → R for every ε > 0 let n 1/p p Var(ε) u := sup |u(x ) − u(x )| , i i−1 p i=1
where the supremum is taken over all partitions P = {x0 , . . . , xn } of I such that 0 < xi − xi−1 ≤ ε, i = 1, . . . , n, n ∈ N. Define (ε) Var(0) p u := lim Varp u. ε→0+
(0)
(i) Prove that Varp u < ∞ if and only if Varp u < ∞. (0)
(ii) Prove that there exists a function u for which Varp u < Varp u < ∞. (0)
(0)
(iii) Prove that if u is continuous and Varp u < ∞, then Varq u = 0 for q > p. Hint: Use uniform continuity. (iv) Find a continuous function u : [0, 1] → R such that Var u = ∞ and Var2 u < ∞. 2k Exercise 2.27. Let u ∈ BP V ([0, 2π]) and let ak := 0 u(x)e−ikx dx, k ∈ Z, be its Fourier coefficients. Prove that there exists a constant c > 0 such that |ak | ≤ c/|k| for all k ∈ Z \ {0}. Hint: Consider a special partition of [0, 2π] and construct a suitable step function.
2.3. Differentiability In this section we discuss the differentiability of functions of bounded pointwise variation in the case in which Y is a finite dimensional vector space. We begin by studying the relation between u and the derivative of the indefinite pointwise variation V of u (see (2.3)). Theorem 2.28. Let I ⊆ R be an interval, let (Y, · ) be a normed space, let u ∈ BP Vloc (I; Y ), and let E := {x ∈ I : u is differentiable at x}. Then (2.11)
u (x) = V (x)
for L1 -a.e. x ∈ E.
Proof. Since V is increasing by Theorem 2.17, it is differentiable at L1 -a.e. x ∈ I, in view of the Lebesgue differentiation theorem (Theorem 1.18). Let x ∈ E be such that V has a derivative (possibly infinite) at x. Consider the case in which x ∈ I ◦ . By (2.4), u(x + h) − u(x) ≤ V (x + h) − V (x)
2.3. Differentiability
41
for all h > 0 such that x + h ∈ I. By dividing both sides by h > 0 and letting h → 0+ we obtain u (x) ≤ V (x).
(2.12)
To prove the converse inequality, assume first that I = [a, b], so that Var u < ∞, and let E0 := {x ∈ I : u and V are differentiable at x and V (x) > u (x)}. For every k ∈ N, let Ek be the set of all points x ∈ E0 such that u(x2 ) − u(x1 ) 1 V (x2 ) − V (x1 ) ≥ +k (2.13) x2 − x1 x2 − x1 for all intervals [x1 , x2 ] such that x ∈ [x1 , x2 ] and 0 < x2 − x1 ≤ 1/k. Since ∞ Ek , E0 = k=1
to prove (2.11), it suffices to show that L1 (Ek ) = 0. Let ε > 0 and consider a partition P := {x0 , . . . , xn } of I such that n (2.14) u(xi ) − u(xi−1 ) > Var u − ε. i=1
By adding more points to the partition, without loss of generality, we may assume that 0 < xi − xi−1 ≤ k1 for all i = 1, . . . , n. Fix i ∈ {1, . . . , n}. We distinguish two cases. If the interval [xi−1 , xi ] contains points of Ek , then multiplying (2.13) by xi − xi−1 yields xi − xi−1 , V (xi ) − V (xi−1 ) ≥ u(xi ) − u(xi−1 ) + k while if [xi−1 , xi ] does not intersect Ek , then V (xi ) − V (xi−1 ) ≥ u(xi ) − u(xi−1 ) by (2.4). Summing these two inequalities and using (2.14) gives n n Var u = (V (xi ) − V (xi−1 )) ≥ u(xi ) − u(xi−1 ) + k1 L1o (Ek ) i=1
≥ Var u − ε +
i=1 1 1 k Lo (Ek ),
which implies that L1o (Ek ) ≤ kε. Given the arbitrariness of ε > 0, we conclude that L1 (Ek ) = 0. This proves (2.11) in the case I = [a, b]. In the general case it suffices to write I as an increasing countable union of closed intervals and apply the previous case in each closed interval. Since the countable union of sets of Lebesgue measure zero has Lebesgue measure zero, we conclude the proof. In the case Y = R an alternative proof is given in the following exercise.
42
2. Functions of Bounded Pointwise Variation
Exercise 2.29. Let u ∈ BP V ([a, b]) and for every a ≤ x ≤ b define V (x) := Var[a,x] u. Let P be a partition of [a, b], with a = x0 < x1 < · · · < xn = x, and define v : [a, b] → R inductively as follows. If u(x0 ) ≤ u(x1 ), for every x ∈ [x0 , x1 ] set v(x) := u(x) − u(x0 ), while if u(x0 ) > u(x1 ), for every x ∈ [x0 , x1 ] set v(x) := u(x0 ) − u(x). Let 1 ≤ k < n and suppose that v has been defined on [x0 , xk ]. If u(xk ) ≤ u(xk+1 ), for every x ∈ (xk , xk+1 ] set v(x) := u(x) + v(xk ) − u(xk ), while if u(xk ) > u(xk+1 ), for every x ∈ (xk , xk+1 ] set v(x) := −u(x) + v(xk ) + u(xk ). Prove that (i) on each interval [xk−1 , xk ], 1 ≤ k ≤ n, either v − u or v + u is constant, (ii) V (x) = Var[a,x] v for every a ≤ x ≤ b and v(b) =
n
|u(xk ) − u(xk−1 )|,
k=1
(iii) for every n ∈ N there is a function vn : [a, b] → R such that V − vn is increasing, V (a) = vn (a), V (b) − vn (b) ≤
1 2n ,
and |vn (x)| = |u (x)| for L1 -a.e. x ∈ [a, b], (iv) V (x) = |u (x)| for L1 -a.e. x ∈ [a, b]. The next exercise shows that in general the inequality (2.12) can be strict. Exercise 2.30. Let u : [0, 1] → R be defined by a x cos x1 if 0 < x ≤ 1, u(x) := 0 if x = 0, where a > 0. (i) Prove that if a = 2, then u (0) = 0 = V (0) < ∞. (ii) Prove that if 1 < a < 2, then u (0) = 0, while V (0) = ∞. Corollary 2.31. Let I ⊆ R be an interval and let u ∈ BP Vloc (I; RM ). Then u is differentiable L1 -a.e. in I, and for every interval J ⊆ I, u (x) dx ≤ VarJ u. (2.15) J
In particular, if u ∈ BP V (I; RM ), then u is Lebesgue integrable.
2.3. Differentiability
43
Proof. Step 1: By (2.2), if u ∈ BP Vloc (I; RM ), then ui ∈ BP Vloc (I) for all i = 1, . . . , M . Let V ui be the indefinite pointwise variation of ui (see (2.3)). By Theorem 2.17 the functions V ui ± ui are increasing, and so ui = 12 (V ui + ui ) − 12 (V ui − ui ) is the difference of two increasing functions. It follows from Lebesgue’s differentiation theorem (Theorem 1.18) that ui is differentiable L1 -a.e. in I for all i = 1, . . . , M . In turn, u is differentiable L1 -a.e. in I. To prove (2.15), let [a, b] ⊆ I and let V u be the indefinite pointwise variation of u (see (2.3)). Since V u is increasing, by Corollary 1.25 applied to V u , Theorem 2.17, and (2.11), we obtain b b u dx = V u dx ≤ V u (b) − V u (a) = Var[a,b] u. (2.16) a
a
Now, if J ⊆ I is any nonempty interval, construct an increasing sequence of intervals [an , bn ] such that J = n [an , bn ]. Then from (2.15), bn u dx ≤ Var[an ,bn ] u ≤ VarJ u. an
Letting n → ∞ and using the Lebesgue monotone convergence theorem concludes the proof. Remark 2.32. Note that if u ∈ BP V (I; RM ), one could have used the function V ∞ defined in (2.7) instead of V u . Exercise 2.33. Let un (x) =
1 −n2 x2 e , n2
x ∈ R.
(i) Calculate Var un . (ii) Let u(x) := ∞ n=1 un (x), x ∈ R. Prove that u ∈ BP V (R). ∞ (iii) Prove that n=1 un (x) does not converge uniformly in [−1, 1]. (iv) Find a formula for u . (v) What is the relevance of this exercise? Exercise 2.34. Let u : [a, b] → RM . (i) Let g : [α, β] → RM be a continuous function. Prove that if x0 ∈ [α, β], then β β β g(x) dx ≥ g(x) dx − 2 g(x) − g(x0 ) dx. α
α
α
(ii) Using part (i), prove that if u is of class C 1 ([a, b]; RM ), then b u (x) dt. Var u = a
44
2. Functions of Bounded Pointwise Variation
(iii) Prove that if u : [a, b] → RM is (everywhere) differentiable and u is Riemann integrable, then b Var u = u (x) dx. a
Compare this with Theorem 2.40 below. Exercise 2.35. Let I = [a, b]. (i) Prove that if v : I → R has bounded pointwise variation and the intermediate value property (that is, if it takes two values, it takes also all the values in between), then v is continuous. (ii) Prove that if u : I → R is differentiable and u has bounded pointwise variation, then u is continuous.
2.4. Monotone Versus BPV Another consequence of Theorem 2.17 is the result mentioned at the beginning of the chapter, namely, the characterization of the smallest vector space of functions u : I → R that contains all monotone functions. Theorem 2.36. Let I ⊆ R be an interval. The smallest vector space of functions u : I → R that contains all monotone functions (respectively, bounded monotone functions) is given by the space BP Vloc (I) (respectively, BP V (I)). Moreover, every function in BP Vloc (I) (respectively, BP V (I)) may be written as a difference of two increasing functions (respectively, two bounded increasing functions). Proof. Let u, v : I → R and let t ∈ R. By (2.1) for every interval J ⊆ I we have (2.17)
VarJ (tu) = |t| VarJ u,
VarJ (u + v) ≤ VarJ u + VarJ v.
Hence, the functions in BP Vloc (I) (respectively, BP V (I)) form a vector space. By Propositions 2.11 and 2.12, the space BP Vloc (I) (respectively, BP V (I)) contains all monotone functions (respectively, bounded monotone functions). To prove that it is the smallest, it suffices to show that every function u in BP Vloc (I) (respectively, in BP V (I)) may be written as the difference of two increasing functions (respectively, bounded increasing). In view of Theorem 2.17, it suffices to write u = V − (V − u) This concludes the proof.
or
u = 12 (V + u) − 12 (V − u).
2.4. Monotone Versus BPV
45
Exercise 2.37. Let I ⊆ R be an interval and let u ∈ BP Vloc (I) be right continuous (respectively, left continuous). Prove that for every x ∈ I ◦ the functions 12 (V + u) and 12 (V − u) cannot both be discontinuous at x. Exercise 2.38. Let u, v ∈ BP V ([a, b]). Prove the following. (i) u ± v ∈ BP V ([a, b]). (ii) uv ∈ BP V ([a, b]). (iii) If v(x) ≥ c > 0 for all x ∈ [a, b] and for some c > 0, then u/v ∈ BP V ([a, b]). (iv) max{u, v}, min{u, v} ∈ BP V ([a, b]). (v) What happens if we replace [a, b] with an arbitrary interval I ⊆ R (possibly unbounded)? Theorem 2.36 shows that every function in BP Vloc (I) may be written as the difference of two increasing functions. The next result shows that the difference of two increasing functions can lose every kind of monotonicity and be rather pathological. Exercise 2.39. Let (X, d) be a metric space and let E be the intersection of countably many Gδ sets dense in X. Prove that E is dense in X. Theorem 2.40 (Weil). There exists a Lipschitz continuous function u : [0, 1] → R, which is everywhere differentiable, monotone on no interval of [0, 1], and such that u is bounded but not Riemann integrable over any interval of [0, 1]. Proof. Step 1: Let X be the vector space of all bounded functions v : [0, 1] → R with the property that there exists a differentiable function u : [0, 1] → R such that u = v. Then X is a Banach space when endowed with the sup norm. To see this, it suffices to show that it is a closed subspace of the space of bounded functions. Let {vn }n be a sequence in X such that vn − v∞ → 0, for some bounded function v : [0, 1] → R . Then un − v∞ → 0. By replacing un with un − un (0), we may assume that un (0) = 0 for all n. It follows (why?) that there exists a differentiable function u such that u = v. This shows that v ∈ X. Let Z be the subspace of all v ∈ X such that the set v −1 ({0}) is dense in [0, 1]. Note that the set v −1 ({0}) is a Gδ set. Indeed, let u be such that u = v. Then v −1 ({0}) = x ∈ [0, 1] : lim n(u(x + n1 ) − u(x)) = 0 n→∞
=
∞ ∞ k=1 n=k
x ∈ [0, 1] : n|u(x + n1 ) − u(x)| <
1 k
=:
∞ ∞ k=1 n=k
Un,k .
46
2. Functions of Bounded Pointwise Variation
Since u is continuous, the sets Un,k are relatively open, which shows that v −1 ({0}) is a countable intersection of relatively open sets. We claim that Z is a Banach space. Indeed, let {vn }n be a sequence in Z such that vn − v∞ → 0 for some v ∈ X. Then vn−1 ({0}) is a Gδ set dense in [0, 1] for every n. It follows by the previous exercise that E0 := ∞ −1 −1 n=1 vn ({0}) is dense in [0, 1]. But since E0 ⊆ v ({0}}, we have that −1 v ({0}} is dense in [0, 1], which shows that v belongs to Z. Step 2: Let W be the set of all v ∈ Z such that there exists a proper interval I ⊆ [0, 1] such that either v ≥ 0 in I or v ≤ 0 in I. Let {In }n be the family of all closed proper intervals with rational endpoints and set Wn+ := {v ∈ W : v ≥ 0 in In },
Wn− := {v ∈ W : v ≤ 0 in In }.
Then the sets Wn± are closed and W :=
∞
(Wn+ ∪ Wn− ).
n=1
Fix n. We claim that Wn± is nowhere dense. We only prove it for Wn+ . Let v ∈ Wn+ and let ε > 0. Since v −1 ({0}} is dense in [0, 1] and inf In < sup In , there exists x0 ∈ In◦ such that v(x0 ) = 0. By Exercise 1.40 the space Z contains a nonzero function w ∈ Z with w(x0 ) = 0. By multiplying w by ε/(2w∞ ), without loss of generality, we may assume that w∞ < ε. Moreover, by possibly changing sign to w, we can also assume that w(x0 ) < 0. Consider the function v + w. Since w∞ < ε, we have that / Wn+ . This shows v + w ∈ B(v, ε). Moreover, (v + w)(x0 ) < 0. Thus, v + w ∈ that Wn+ has empty interior. In turn, by the Baire category theorem (see Corollary A.13) we have that W is a proper subset of Z. Let v ∈ Z \ W . Given any proper interval I ⊆ [0, 1], the function v must change sign in I. Thus, if u is such that u = v, necessarily u cannot be monotone in I. By the mean value theorem, u is Lipschitz continuous, and so by Proposition 2.6, u is of bounded pointwise variation on bounded sets. The proof of the fact that u is not Riemann integrable on any closed interval [a, b] will make use of absolutely continuous functions, and so it will be postponed until after Theorem 3.12. Exercise 2.41. Let In := (an , bn ), n ∈ N, be a sequence of pairwise disjoint intervals in [0, 1] and choose an < αn < βn < bn such that αn −an = bn −βn . (i) For each n ∈ N construct a differentiable function un : In → R such that (a) un = 0 in In \ (αn , βn ), (b) 0 ≤ un ≤ (αn − an )2 ,
2.5. The Space BP V (I; Y )
47
(c) max |un | = 1. In
(ii) Define
un (x) if x ∈ In for some n ∈ N, 0 elsewhere. Prove that if x0 ∈ [0, 1] \ In , then for all x ∈ [0, 1], x = x0 , u(x) :=
n
|u(x) − u(x0 )| ≤ |x − x0 |. |x − x0 | (iii) Prove that u is differentiable and that |u | ≤ 1.
(iv) Prove that the intervals In can be chosen so that [0, 1] \ n In is closed, has positive Lebesgue measure, and is nowhere dense (that is, its closure has empty interior). u is discon(v) Prove that with such a choice of {In }n the function tinuous at every accumulation point of [0, 1] \ n In and u is not Riemann integrable.
2.5. The Space BP V (I; Y ) In this section we prove that BP V (I; Y ) is a Banach space, provided Y is a Banach space. Let I ⊆ R be an interval and let (Y, · ) be a normed space. By the properties of the supremum of a set and of the norm · we have that Var(u + v) ≤ Var u + Var v and Var(su) = |s| Var u for all functions u : I → Y and v : I → Y , and for all s ∈ R. However, the pointwise variation Var is not a norm since a constant function has zero variation. To overcome this problem, it suffices to consider u(x0 ) + Var u for some x0 ∈ I. Indeed, the following result holds. Theorem 2.42. Let I ⊆ R be an interval, let (Y, · ) be a Banach space, and let x0 ∈ I. Then the space BP V (I; Y ) endowed with the norm uBP V := u(x0 ) + Var u
(2.18) is a Banach space.
Proof. Let {un }n be a Cauchy sequence in BP V (I; Y ). Given ε > 0, there exists nε ∈ N such that un (x0 ) − um (x0 ) + Var(un − um ) ≤ ε
48
2. Functions of Bounded Pointwise Variation
for all n, m ≥ nε . Hence, by Remark 2.13, un −um ∞ ≤ ε for all n, m ≥ nε . This implies that {un }n is a Cauchy sequence in the space B(I; Y ) = {v : I → Y, v bounded} with the norm · ∞ . Using the completeness of Y , we have that B(I; Y ) is a Banach space (exercise). Hence, the sequence {un }n converges uniformly to some bounded function u : I → Y . Given a partition x0 < x1 < · · · < x of I we have that
(un − um )(xi ) − (un − um )(xi−1 ) ≤ Var(un − um ) ≤ ε
i=1
for all n, m ≥ nε . Letting m → ∞ and using the fact that um (xi ) → u(xi ) for all i = 0, . . . , (since uniform convergence implies pointwise convergence) gives (un − u)(xi ) − (un − u)(xi−1 ) ≤ ε i=1
for all n ≥ nε . Taking the supremum over all partitions of I gives Var(un − u) ≤ ε for all n ≥ nε . Since un (x0 ) − um (x0 ) → un (x0 ) − u(x0 ) as m → ∞, we have shown that un − uBP V ≤ 2ε for all n ≥ nε . Moreover, Var u ≤ Var(un − u) + Var un < ∞. Thus, u ∈ BP V (I; Y ) and un → u in BP V (I; Y ). This completes the proof. The next exercise shows that the space BP V (I) endowed with the norm defined in (2.18) is not separable. Exercise 2.43. Consider the space BP V ([0, 1]) endowed with the norm (2.18). For every u ∈ BP V ([0, 1]) and r > 0 let BBP V (u, r) := {v ∈ BP V ([0, 1]) : u − v < r}. For every t ∈ [0, 1] define ut := χ{t} . (i) Prove that BBP V (ut , 1) ∩ BBP V (us , 1) = ∅ for every t, s ∈ [0, 1], with t = s. (ii) Prove that for every t ∈ [0, 1], BBP V (ut , 1) ⊆ BBP V (0, 3). (iii) Prove that the Banach space (BP V ([0, 1]), · ) is not separable. We conclude this section with an important compactness result.
2.5. The Space BP V (I; Y )
49
Theorem 2.44 (Helly’s selection theorem). Let I ⊆ R be an interval and let F be an infinite family of functions in BP V (I). Assume that there exist x0 ∈ I and a constant c > 0 such that |u(x0 )| + Var u ≤ c
(2.19)
for all u ∈ F . Then there exist a sequence {un }n in F and a function v ∈ BP V (I) such that un (x) → v(x) for all x ∈ I. We divide the proof into a few lemmas, which are of independent interest. Lemma 2.45. Let I ⊆ R be an interval and let F be an infinite family of functions u : I → R. Assume that there exists a constant c > 0 such that |u(x)| ≤ c for all x ∈ I and all u ∈ F . Then for every countable set E ⊂ I, there exists a sequence {un }n in F such that the limit limn→∞ un (x) exists in R for all x ∈ E. Proof. The proof makes use of the Cantor diagonal argument. Without loss of generality, we can assume that E is denumerable, so that E = {xk : k ∈ N}. Since the set {u(x1 ) : u ∈ F } is bounded, by the Bolzano–Weierstrass (1) theorem we can find a sequence {un }n in F for which there exists the limit lim u(1) n (x1 ) = 1 ∈ R.
n→∞ (1)
Since the set {un (x2 ) : n ∈ N} is bounded, again by the Bolzano–Weierstrass (2) (1) theorem we can find a subsequence {un }n of {un }n for which there exists the limit lim u(2) n (x2 ) = 2 ∈ R. n→∞
(k)
By induction for every k ∈ N, k > 1, we can find a subsequence {un }n of (k−1) }n for which there exists the limit {un lim u(k) n (xk ) = k ∈ R.
n→∞
We now consider the diagonal elements of the infinite matrix, that is, the (n) sequence {un }n . For every fixed xk ∈ E we have that the sequence (n) (k) {un (xk )}∞ n=k is a subsequence of {un (xk )}n , and thus it converges to k as n → ∞. This completes the proof. Lemma 2.46. Let I ⊆ R be an interval and let F be an infinite family of increasing functions u : I → R. Assume that there exists a constant c > 0 such that |u(x)| ≤ c
50
2. Functions of Bounded Pointwise Variation
for all x ∈ I and all u ∈ F . Then there exist a sequence {un }n in F and an increasing function v : I → R such that limn→∞ un (x) = v(x) for all x ∈ I. Proof. Apply the previous lemma with E given by the union of all the rationals in I and the endpoints of I that belong to I (if any) to find a sequence {un }n in F such that the limit limn→∞ un (x) exists in R for all x ∈ E. Define w : E → R by (2.20)
x ∈ E.
w(x) := lim un (x), n→∞
Since each function un is increasing, if x, y ∈ E and x < y, then un (x) ≤ un (y) for all n ∈ N, and so letting n → ∞, we conclude that w(x) ≤ w(y). Thus, w is increasing. To extend w to I \ E, for x ∈ I \ E we define w(x) :=
w(y).
sup y∈E∩(−∞,x)
Note that w is increasing by construction. We claim that lim un (x) = w(x)
(2.21)
n→∞
for all those x ∈ I such that w is continuous at x. To see this, fix ε > 0 and any such x. In view of (2.20) it is enough to consider the case in which x ∈ I \ E. Since x is not an endpoint of I, using the continuity of w at x, together with the density of the rational numbers, we may find x1 , x2 ∈ E such that x1 < x < x2 and w(x2 ) − w(x1 ) < 2ε .
(2.22)
By (2.20) there exists an integer n0 ∈ N such that (2.23)
|un (x1 ) − w(x1 )| < 2ε ,
|un (x2 ) − w(x2 )| <
ε 2
for all n ≥ n0 . Since w and un are increasing, w(x) − ε ≤ w(x2 ) + w(x1 ) − w(x1 ) − ε ≤ w(x1 ) −
ε 2
≤ un (x1 ) ≤ un (x) ≤ un (x2 ) ≤ w(x2 ) +
= w(x2 ) + w(x1 ) − w(x1 ) + ≤ w(x) + w(x2 ) − w(x1 ) +
ε 2
ε 2
ε 2
≤ w(x) + ε
for all n ≥ n0 , where we have used (2.22) and (2.23). This implies that |un (x) − w(x)| ≤ ε for all n ≥ n0 . Thus the claim is proved. Now let E1 be the set of points of discontinuity of w, which is at most countable. Applying the previous lemma once more, with E1 and {un }n in
2.5. The Space BP V (I; Y )
51
place of E and F , we may find a subsequence {unk }k of {un }n that converges at every point of E1 . The function w(x) if x ∈ I \ E1 , v(x) := limk→∞ unk (x) if x ∈ E1 has all the desired properties. Indeed, in view of (2.20) and (2.21), we have that w(x) = lim un (x) = lim unk (x) n→∞
k→∞
for all x ∈ I \ E1 . We now turn to the proof of the Helly theorem. Proof of Helly’s selection theorem. For every u ∈ F write u = V u − (V u − u),
where V u is the function defined in (2.3). By Theorems 2.17 and 2.36, V u and V u − u are increasing functions, and by (2.19), |V u (x)| ≤ c,
|V u (x) − u(x)| ≤ |V u (x)| + |u(x) − u(x0 )| + |u(x0 )| ≤ 2c
for all x ∈ I, and thus we may apply the previous lemma to the family {V u : u ∈ F } to find a sequence {un }n in F and an increasing function v1 : I → R such that lim V un (x) = v1 (x) n→∞
for all x ∈ I. Applying the previous lemma to the family {V un −un }, we may find a subsequence {unk }k of {un }n and an increasing function v2 : I → R such that lim (V unk − unk )(x) = v2 (x) k→∞
for all x ∈ I. By Theorem 2.36, the function v := v1 −v2 belongs to BP V (I) and lim unk (x) = lim [V unk (x) − (V unk − unk )(x)]
k→∞
k→∞
= v1 (x) − v2 (x) = v(x) for all x ∈ I.
The fact that the function v belongs to BP V (I) can also be derived by the lower semicontinuity of the variations with respect to pointwise convergence. To be precise, we have the following result. Proposition 2.47. Let I ⊆ R be an interval, let (Y, d) be a metric space and let {un }n be a sequence of functions un : I → Y converging pointwise to a function u : I → Y . Then (2.24)
Var u ≤ lim inf Var un . n→∞
52
2. Functions of Bounded Pointwise Variation
In particular, if the right-hand side of the previous inequality is finite, then u belongs to BP V (I; Y ). Proof. Let P = {x0 , . . . , xm } be a a partition of I. Since un (xi ) → u(xi ) as n → ∞ for all i = 0, . . . , m, for every ε > 0 we may find n ¯ ∈ N such that ε d(un (xi ), u(xi )) ≤ 2m for all n ≥ n ¯ and i = 0, . . . , m. Hence, for n ≥ n ¯, m
d(u(xi ), u(xi−1 )) ≤
i=1
=
m ε d(un (xi ), un (xi−1 )) + m i=1 m
d(un (xi ), un (xi−1 )) + ε ≤ Var un + ε.
i=1
Taking the limit inferior on both sides, we obtain m
d(u(xi ), u(xi−1 )) ≤ lim inf Var un + ε. n→∞
i=1
Letting ε → 0+ and taking the supremum over all partitions yields the desired result. Remark 2.48. If u is continuous, then it is enough to assume that un (x) → u(x) for all x in a dense set E of I. To see this, let P = {x0 , . . . , xm } be as in the previous proof. By the continuity of u and the density of E there exists a partition P = {y0 , . . . , ym } ⊆ E such that ε d(u(yi ), u(xi )) ≤ m for all i = 0, . . . , m. We may now continue as in the previous proof (working with P in place of P ) to conclude that m
d(u(xi ), u(xi−1 )) ≤ lim inf Var un + 2ε. n→∞
i=1
The next exercise shows that the equality sign does not necessarily hold in (2.24). Exercise 2.49. For x ∈ [0, 1], let u(x) := (x, 0) and un (x) := (x, n1 sin 2πnα x), n ∈ N, α > 0. (i) Estimate Var un for all n ∈ N and α > 0. (ii) Prove that if α = 12 , then Var u = lim Var γ2i . i→∞
(iii) Prove that for α = 1 or α = 2 we have strict inequality in (2.24).
2.5. The Space BP V (I; Y )
53
Exercise 2.50. Consider the space BP V ([0, 1]) endowed with the metric
1
d(u, v) :=
|u(x) − v(x)| dx + | Var u − Var v|.
0
(i) Prove that if {un }n is in BP V ([0, 1]) and supn d(un , 0) < ∞, then there exist a subsequence {unk }k of {un }n and a function u ∈ BP V ([0, 1]) such that d(un , u) → 0. Hint: Use Helly’s theorem. (ii) Prove that (BP V ([0, 1]), d) is a complete metric space. (iii) Prove that (BP V ([0, 1]), d) is separable. Hint: Any function can be approximated by an inscribed polygonal function. (iv) Is there any relation between convergence in the norm (2.18) and in the metric d? Is there any difference in the answer if the limit function u is continuous? An important consequence of the Helly selection theorem is the following result. Corollary 2.51. Let I ⊆ R be an interval and let u : I → R be an integrable function such that 1 c := lim sup h→0+ h
|u(x + h) − u(x)| dx < ∞, Ih
where Ih := {x ∈ I : x + h ∈ I}. Then there exists a function v ∈ BP V (I) such that v(x) = u(x) for L1 -a.e. x ∈ I and Var v = c. Proof. For every n ∈ N we partition the interval I into subintervals of equal length δn > 0, with δn → 0 as n → ∞. Let {xj,n }j be the family of endpoints of the intervals. For every x ∈ (xj−1,n , xj,n ) define the piecewise constant function xj,n 1 u(t) dt. un (x) := δn xj−1,n Extend un at the endpoints in such a way that it is right continuous and if b ∈ I, extend un at b in such a way that it is left continuous at b. Since un
54
2. Functions of Bounded Pointwise Variation
is piecewise constant, xj,n 1 xj+1,n u(t) dt − u(t) dt Var un = δn xj,n xj−1,n j xj,n xj,n 1 = u(x + δn ) dx − u(t) dt δn xj−1,n xj−1,n j x j,n 1 |u(x + δn ) − u(x)| dx ≤ δn xj−1,n j 1 |u(x + δn ) − u(x)| dx, ≤ δn Iδn and so, by hypothesis, (2.25)
lim sup Var un ≤ c. n→∞
Moreover, since un → u in L1loc (I) (why?), we may find x0 ∈ I and a subsequence, not relabeled, such that supn |un (x0 )| < ∞, sup Var un ≤ c + ε, n
and un (x) → u(x) for L1 -a.e. x ∈ I. Thus, we are in a position to apply Helly’s selection theorem to find a further subsequence, not relabeled, and a function v ∈ BP V (I) such that un (x) → v(x) for all x ∈ I. Hence v(x) = u(x) for L1 -a.e. x ∈ I, and so 1 1 |u(x + h) − u(x)| dx = |v(x + h) − v(x)| dx ≤ Var v h Ih h Ih by Theorem 2.20. Letting h → 0+ , we conclude that c ≤ Var v. On the other hand, by Proposition 2.47 and (2.25), Var v ≤ lim inf Var un ≤ c, n→∞
and so Var v = c.
Results of this type are often useful in the regularity theory of solutions of differential equations (see, e.g., [33] and [71]). Exercise 2.52. Prove that the sequence {un }n constructed in the previous corollary converges to u in L1loc (I).
2.6. Composition in BP V (I; Y )
55
2.6. Composition in BP V (I; Y ) In this section we study the composition of functions with bounded pointwise variation. Exercise 2.38 shows that BP V ([a, b]) is closed under the operations of addition and multiplication. The next exercise shows that it is not closed under composition (see, however, Exercise 2.54). Exercise 2.53. Consider the functions f : [0, 1] → R and u : [0, 1] → [0, 1], defined by 2 21 √ x sin x if 0 < x ≤ 1, f (z) := z, u(x) := 0 if x = 0. Prove that f and u have bounded pointwise variation, but f ◦ u does not. Exercise 2.54. Let I, J ⊆ R be two intervals, let f ∈ BP Vloc (J), and let u : I → J be monotone. Prove that f ◦ u belongs to BP Vloc (I). Next we discuss necessary and sufficient conditions for a given function f to be such that f ◦ u is a function of bounded pointwise variation for every function u of bounded pointwise variation. Theorem 2.55 (Josephy). Let I ⊆ R be an interval and let f : RM → R. Then f ◦ u belongs to BP Vloc (I) (respectively, BP V (I)) for all functions u ∈ BP Vloc (I; RM ) (respectively, BP V (I; RM )) if and only if f is locally Lipschitz continuous. Proof. We only prove the result in the case of BP Vloc (I; RM ), the case of BP V (I; RM ) being completely similar. Step 1: Assume that u ∈ BP Vloc (I; RM ) and that f is locally Lipschitz continuous. Fix an interval [a, b] ⊆ I. Since u is bounded in [a, b] (see Proposition 2.12), we have that m := supx∈[a,b] u(x) < ∞, and so we may find L > 0 such that |f (z1 ) − f (z2 )| ≤ Lz1 − z2 for all z1 , z2 ∈ B(0, m). Let P := {x0 , . . . , xn } be a partition of [a, b]. Then n i=1
|(f ◦ u)(xi ) − (f ◦ u)(xi−1 )| =
n
|f (u(xi )) − f (u(xi−1 ))|
i=1 n
≤L
u(xi ) − u(xi−1 ) ≤ L Var[a,b] u,
i=1
which implies that Var[a,b] (f ◦ u) ≤ L Var[a,b] u < ∞. Step 2: Conversely, assume that f ◦ u ∈ BP Vloc (I) for every u ∈ BP Vloc (I; RM ). Fix [a, b] ⊆ I. We begin by showing that f is locally bounded. Fix r > 0. We claim that f is bounded in B(0, r). Indeed, assume by
56
2. Functions of Bounded Pointwise Variation
contradiction that there exists {zn }n in B(0, r) such that |f (zn )| → ∞ as n → ∞. By compactness we can assume that zn → z0 , and by taking a further subsequence, also that zn − z0 ≤ 1/n2 b−a for every n. For every n ∈ N, set In := a + n+1 ,a + function zn if x ∈ In◦ , n ∈ N, (2.26) u(x) := z0 otherwise in I.
b−a n
. Consider the
Then VarI n u = 2zn − z0 and hence, by Exercise 2.22 (which holds for the left endpoint with the obvious changes) and Proposition 2.15 we have that Var[a,b] u = lim Var
u = lim
b−a a+ n+1 ,b
n→∞
n→∞
n
VarI k u ≤
k=1
∞
2 k2
< ∞.
k=1
Since u is constant outside [a, b], it follows that u belongs to BP Vloc (I; RM ), and so by hypothesis the function f ◦ u belongs to BP V ([a, b]). However, Var[a,b] (f ◦ u) ≥ VarI n (f ◦ u) = 2|f (zn ) − f (z0 )| → ∞ as n → ∞, which is a contradiction. Step 3: Next we claim that f is Lipschitz continuous in B(0, r). Indeed, assume by contradiction that this is not the case. Then we may find two sequences {yn }n , {zn }n in B(0, r) such that yn = zn and (2.27)
|f (yn ) − f (zn )| > 2(n2 + n) yn − zn
for all n ∈ N. Since {yn }n is bounded, we may extract a subsequence (not relabeled) such that yn → y∞ . Take a further subsequence (not relabeled) such that (2.27) continues to hold and yn − y∞ <
(2.28)
1 . (n+1)2
Since f is bounded in B(0, r) by some constant cr > 0 by Step 2, by (2.27) for all n ∈ N, we have (2.29)
2cr ≥ |f (yn ) − f (zn )| > 2(n2 + n)yn − zn .
Hence, (2.30)
0 < δn :=
yn − zn (b − a) b−a → 0. < 2cr 2(n2 + n)
2.6. Composition in BP V (I; Y )
57
We now define the piecewise constant function u : I → RM in the following way: (2.31) ⎧ y∞ if x ≤ a, ⎪ ⎪ ⎪ ⎪ b−a ⎨ z if x ∈ In , x − a + n+1 is not a multiple of δn , n ∈ N, n u(x) := b−a ⎪ yn if x ∈ In , x − a + n+1 is a multiple of δn , n ∈ N, ⎪ ⎪ ⎪ ⎩ y1 if x ≥ b. Here In is as in Step 2. For every n ∈ N, let (2.32)
n :=
2cr diam In b−a = >2 = 2 δn δn (n + n) yn − zn (n2 + n)
by (2.29) and set mn := max{j ∈ N0 : j < n }. Note that mn is the number of times that the function u takes the value yn in the interval In . Since n > 2, we have ≤ mn < n . b−a ,a + Consider the partition Pn of I n = a + n+1 n 2
(2.33)
b−a n
given by
1 b−a Pn : = a + + jδn : j = 0, . . . , 2mn n+1 2 =: {x0,n , . . . , x2mn ,n }. Then by (2.29)–(2.33), VarI n (f ◦ u) ≥
2m n
|f (u(xi,n )) − f (u(xi−1,n ))|
i=1
≥ 2mn |f (yn ) − f (zn )| ≥ 2n (n2 + n)yn − zn = 4cr , and so for every n ∈ N by Proposition 2.15 we obtain that Var[a,b] (f ◦ u) ≥
n
VarI k (f ◦ u) ≥ 4cr n → ∞
k=1
as n → ∞. On the other hand, since u is a step function on I n , by (2.28)–(2.33) we have that (why?) VarI n u ≤ 2mn yn − zn + yn+1 − yn + yn+1 − zn < 2n yn − zn + 2yn+1 − y∞ + 2y∞ − yn + yn − zn 4 5cr + 4 cr 4cr + ≤ + 2 . < 2 2 n + n (n + 1) n +n n2 Since yn → y∞ and zn → y∞ , it follows by (2.31) that u is right continuous at a. Hence, by Exercise 2.22 (which holds for the left endpoint with the
58
2. Functions of Bounded Pointwise Variation
obvious changes) and Proposition 2.15 we have that Var[a,b] u = lim Var n→∞
u = lim
b−a a+ n+1 ,b
n→∞
n k=1
VarI k u ≤
∞ 5cr + 4 k=1
k2
< ∞,
which shows that u ∈ BP Vloc (I). Hence, we have obtained a contradiction. Remark 2.56. The previous proof continues to hold if f : RM → Z, where (Z, dZ ) is a metric space. On the other hand, reasoning as in Step 1 we have that if (Y, dY ) and (Z, dZ ) are metric spaces and f : Y → Z is locally Lipschitz continuous, then f ◦ u belongs to BP Vloc (I; Z) (respectively, BP V (I; Z)) for all functions u ∈ BP Vloc (I; Y ) (respectively, BP V (I; Y )). The next exercise gives necessary and sufficient conditions for a function h to have the property that u ◦ h belongs to BP V (I) for every function u ∈ BP V (I). Exercise 2.57. For each ∈ N let J := {E ⊆ [0, 1] : E can be written as a union of intervals}, where we allow singletons as degenerate closed intervals. A function g : [0, 1] → R is said to be of -bounded pointwise variation if g −1 ([a, b]) ∈ J for all [a, b] ⊂ R. (i) Prove that if g : [0, 1] → R is of -bounded pointwise variation and if |g(x)| ≤ c for all x ∈ [0, 1], then Var g ≤ 4c( + 1). (ii) Prove that if u : [0, 1] → R is increasing and if g : [0, 1] → [0, 1] is of -bounded pointwise variation for some ∈ N, then u ◦ g is bounded and of -bounded pointwise variation. (iii) Prove that if u : [0, 1] → R is of bounded pointwise variation and if g : [0, 1] → [0, 1] is of -bounded pointwise variation for some ∈ N, then u ◦ g is of bounded pointwise variation. ∞ (iv) For each n ∈ N let gn : [0, 1] → 0, 31n and let g := n=1 gn . Var g . Prove that Var g ≥ 16 ∞ n n=1 (v) Prove that if g : [0, 1] → [0, 1] is such that u ◦ g is of bounded pointwise variation for all u : [0, 1] → R of bounded pointwise variation, then g : [0, 1] → R is of -bounded pointwise variation for some ∈ N. Hint: If not, then for each n ∈ N find an interval / J3n and take u := ∞ In ⊆ [0, 1] such that g −1 (In ) ∈ n=1 un for an appropriate choice of un .
2.7. Banach Indicatrix
59
2.7. Banach Indicatrix We conclude this chapter with a characterization of continuous functions of bounded pointwise variation due to Banach [18]. It expresses the fact that the graph of a function in BP V (I) cannot intersect a horizontal line too many times. This result paved the way to the area formula (see [10], [72], [76], and [251]). Definition 2.58. Given two nonempty sets X, Y , a function u : X → Y , and a subset E ⊆ X, for every y ∈ Y let Nu (y; E) be the number of elements (if any) of the set {x ∈ E : u(x) = y}. The function Nu (·; E) : Y → N0 ∪{∞} is called the counting function or Banach indicatrix of u in the set E. Note that Nu (y; E) = H0 (E ∩ u−1 ({y})). When E = X, we sometimes omit the dependence on E; namely, we write Nu := Nu (·; X). We begin with a preliminary result due to Federer [75]. Theorem 2.59 (Federer). Let I ⊆ R be an interval, let (Y, d) be a metric space, let u : I → Y be such that u(J) is a Borel set for every interval J ⊆ I, and let μ : B(Y ) → [0, ∞] be a measure. For every n ∈ N let Fn be a countable family of pairwise disjoint intervals with the properties that Fn is a partition of I and sup diam J → 0
J∈Fn
as n → ∞. Then Nu (·; I) is a Borel function and Nu (y; I) dμ(y) = lim μ(u(J)). n→∞
Y
J∈Fn
Proof. We divide the proof into two parts. Step 1: We claim that for every y ∈ Y , χu(J) (y). (2.34) Nu (y; I) = lim n→∞
J∈Fn
To see this, if Nu (y; I) > 0, let m ∈ N be such that m ≤ Nu (y; I). Then there exist m distinct points x1 , . . . , xm ∈ I such that u(xi ) = y. Assume that x1 < · · · < xm and let δm := min{xi − xi−1 : i = 2, . . . , m} > 0. Since supJ∈Fn diam J → 0 as n → ∞, there exists nδ ∈ N such that sup diam J < δm J∈Fn
for all n ≥ nδ . Using the fact that each Fn is a partition of I, it follows that for n ≥ nδ there must exist m distinct intervals J1,n , . . . , Jm,n ∈ Fn such
60
2. Functions of Bounded Pointwise Variation
that xi ∈ Ji,n for all i = 1, . . . , m and n ≥ nδ . Thus, for all n ≥ nδ , m
m=
χu(Ji,n ) (y) ≤
i=1
χu(J) (y),
J∈Fn
and consequently, m ≤ lim inf n→∞
χu(J) (y).
J∈Fn
If Nu (y; I) < ∞, take m := Nu (y; I), while if Nu (y; I) is infinite, let m → Nu (y; I) in the previous inequality to conclude that χu(J) (y). Nu (y; I) ≤ lim inf n→∞
J∈Fn
This inequality continues to hold if Nu (y; I) = 0. On the other hand, by the definition of Nu (y; I), for every integer n ∈ N, we have that χu(J) (y) ≤ Nu (y; I), (2.35) J∈Fn
where we have used again the fact that Fn is a partition of I. Letting n → ∞ in the previous inequality gives χu(J) (y) ≤ Nu (y; I). lim sup n→∞
J∈Fn
Hence (2.34) holds. Note that since by hypothesis χu(J) is a Borel function for every J ∈ Fn , (2.34) implies that Nu (·; I) is a Borel function. Step 2: We claim that lim inf n→∞
μ(u(J)) ≥
Nu (y; I) dμ(y). Y
J∈Fn
By Corollary B.42 in Appendix B, for every n ∈ N, μ(u(J)) = χu(J) dμ(y) = χu(J) dμ(y). (2.36) J∈Fn
J∈Fn
Y
Y J∈F n
It now follows from (2.35) that μ(u(J)) ≤ Nu (y; I) dμ(y). lim sup n→∞
Y
J∈Fn
On the other hand, from (2.36), Fatou’s lemma, and Step 1, in this order, we get μ(u(J)) = lim inf χu(J) dμ(y) lim inf n→∞
n→∞
J∈Fn
≥
lim inf Y
This completes the proof.
Y J∈F n
n→∞
J∈Fn
χu(J) dμ(y) =
Nu (y; I) dμ(y). Y
2.7. Banach Indicatrix
61
The following theorem was first proved by Banach [18] in the case Y = R and extended by Federer [75] to metric spaces. Theorem 2.60 (Banach). Let I ⊆ R be an interval, let (Y, d) be a metric space, and let u : I → Y be a continuous function. Then Nu (·; I) is a Borel function and Nu (y; I) dH1 (y) = Var u. (2.37) Y
In particular, u belongs to BP V (I) if and only if Nu (·; I) is Lebesgue integrable with respect to H1 . Compare the following lemma with Exercise 2.26. Lemma 2.61. Let (Y, d) be a metric space and let u : [a, b] → Y be a continuous function. For every δ > 0 let Pδ := {x0,δ , . . . , xnδ ,δ } be a partition of [a, b], with x0,δ := a < x1,δ < · · · < xnδ ,δ =: b, such that xi,δ − xi−1,δ ≤ δ for all i = 1, . . . , nδ . Then Var u = lim
δ→0+
nδ
d(u(xi,δ ), u(xi−1,δ )).
i=1
Proof. Without loss of generality, we may assume that Var u > 0, since otherwise u is constant and there is nothing to prove. Fix any 0 < t < Var u and find a partition P = {x0 , . . . , xm } of [a, b], with x0 = a and xm = b (see Remark 2.3), such that (2.38)
S :=
m
d(u(xj ), u(xj−1 )) > t.
j=1
Since u is uniformly continuous, there exists η > 0 such that S−t (2.39) d(u(x), u(x )) ≤ 2m for all x, x ∈ [a, b] with |x − x | ≤ η. Let 0 < δ0 := 12 min η, min (xj − xj−1 ) . j=1,...,m
If 0 < δ ≤ δ0 , then for each j = 1, . . . , m − 1 there exists a unique index ij ∈ {1, . . . , nδ } such that xj ∈ (xij −1,δ , xij ,δ ]. Moreover, if j = k, then ij = ik , and i1 < · · · < im−1 . Set i0 := 0 and im := nδ . Since |xij ,δ − xj | ≤ xij ,δ − xij−1 ,δ ≤ δ ≤ η,
62
2. Functions of Bounded Pointwise Variation
by (2.39) we have d(u(xj ), u(xj−1 )) ≤ d(u(xj ), u(xij ,δ )) + d(u(xij ,δ ), u(xij −1,δ )) + d(u(xij −1,δ ), u(xj−1 )) S−t + d(u(xij ,δ ), u(xij−1 ,δ )). ≤ m Summing over all j = 1, . . . , m and using (2.38), we get S=
m
d(u(xj ), u(xj−1 )) ≤ S − t +
j=1
m
d(u(xij ,δ ), u(xij−1 ,δ ))
j=1
≤S−t+
nδ
d(u(xi,δ ), u(xi−1,δ )),
i=1
and so t≤
nδ
d(u(xi,δ ), u(xi−1,δ )) ≤ Var u.
i=1
In turn, t ≤ lim inf δ→0+
nδ
d(u(xi,δ ), u(xi−1,δ ))
i=1 nδ
≤ lim sup δ→0+
d(u(xi,δ ), u(xi−1,δ )) ≤ Var u.
i=1
Letting t Var u gives the desired result.
Note that in the previous lemma we are not assuming that Var u < ∞. Exercise 2.62. Prove that the previous lemma continues to hold if u : [a, b] → Y is only assumed to be left continuous (respectively, right continuous). Hint: Use left continuity (respectively, right continuity) at each point xj to find δ0 . Lemma 2.63. Let (Y, d) be a metric space and let u : [a, b] → Y be a continuous function. Then d(u(a), u(b)) ≤ H1 (u([a, b])) ≤ Var u. Proof. By the triangle inequality the function f (y) := d(y, u(a)), y ∈ Y , is Lipschitz continuous with Lipschitz constant 1. Hence, by Proposition C.44, H1 (f (u([a, b]))) ≤ H1 (u([a, b])).
2.7. Banach Indicatrix
63
The function f ◦ u : [a, b] → R is continuous and so f (u([a, b])) contains the interval of endpoints (f ◦ u)(a) and (f ◦ u)(b). Since H1 = L1 in R, we have d(u(a), u(b)) = (f ◦ u)(b) − (f ◦ u)(a) ≤ L1 (f (u([a, b]))) = H1 (f (u([a, b]))) ≤ H1 (u([a, b])). This proves the first inequality. To prove the second, it suffices to assume that Var u < ∞, since otherwise there is nothing to prove. By Corollary 2.24 we can write u=w◦V, where w : V ([a, b]) → Y is Lipschitz continuous with Lipschitz constant at most 1. Hence, by Proposition C.44, H1 (u([a, b])) = H1 (w(V ([a, b]))) ≤ H1 (V ([a, b])) = L1 ([0, Var u]) = Var u, where we used the fact that H1 = L1 in R.
Proof of Banach’s theorem. Step 1: Assume that I = [a, b]. For every n ∈ N let Fn be the partition of [a, b] given by I1,n := a, a + b−a 2n , b−a for k = 2, . . . , 2n . Ik,n := a + (k − 1) b−a 2n , a + k 2n By Lemma 2.63 and Proposition 2.15, we get d(u(max J), u(min J)) ≤ H1 (u(J)) ≤ VarJ u = Var[a,b] u. J∈Fn
J∈Fn
J∈Fn
Letting n → ∞ and using Lemma 2.61, we obtain Var[a,b] u = lim d(u(max J), u(min J)) n→∞
≤ lim
n→∞
J∈Fn
H1 (u(J)) ≤ Var[a,b] u,
J∈Fn
and so, by Theorem 2.59 we have that Nu (y; [a, b]) dH1 (y) = lim H1 (u(J)) = Var[a,b] u. Y
n→∞
J∈Fn
Step 2: If I is an arbitrary interval, construct an increasing sequence of intervals [ak , bk ] such that I = k [ak , bk ]. By the definition of Nu (y; ·) for every y ∈ R we have that Nu (y; [ak , bk ]) ≤ Nu (y; [ak+1 , bk+1 ])
64
2. Functions of Bounded Pointwise Variation
and Nu (y; [ak , bk ]) → Nu (y; I). Hence, by Proposition 2.15, Exercise 2.22, the previous step, and the Lebesgue monotone convergence theorem, we get VarI u = lim Var[ak ,bk ] u = lim Nu (y; [ak , bk ]) dH1 (y) k→∞ k→∞ Y Nu (y; I) dH1 (y). = Y
This concludes the proof.
Corollary 2.64. Let I ⊆ R be an interval, let (Y, d) be a metric space, and let u ∈ BP V (I; Y ) be continuous. Then the set of values y ∈ Y for which u−1 ({y}) is infinite has H1 measure zero. Proof. Since by the previous theorem the function Nu is integrable, it fol lows that it must be finite for H1 -a.e. y ∈ Y . The previous corollary expresses the fact that a function in BP V (I; Y ) cannot oscillate too much. Exercise 2.65. Consider the Cantor set ∞ 2
n−1
D := [0, 1] \
Ik,n .
n=1 k=1
For every interval Ik,n = (ak,n , bk,n ), n ∈ N, k = 1, . . . , 2n−1 , define gk,n (x) :=
2π(x − ak,n ) 1 sin , n 2 bk,n − ak,n
x ∈ [ak,n , bk,n ].
(i) Prove that the function g : [0, 1] → R, defined by gk,n (x) if x ∈ Ik,n , n ∈ N, k = 1, . . . , 2n−1 , g(x) := 0 if x ∈ D, is continuous. (ii) Let f := g +u, where u is the Cantor function. Prove that for every n ∈ N, Ik,n . [0, 1] = f k≤2n−1
(iii) Prove that the counting function Nf : R → N∪{0, ∞} is identically ∞ in [0, 1]. Exercise 2.66. Let u : [a, b] → R be continuous. Using Banach’s theorem, give an alternative proof of the fact that if c ∈ (a, b), then Var[a,c] u + Var[c,b] u = Var u.
2.7. Banach Indicatrix
65
Exercise 2.67. Let (Y, d) be a metric space and let u : [a, b] → Y . Prove that Var u = Nu (y; I) dH1 (y) + v+ (x) + v− (x), Y
a≤x 0 such that (3.1)
n
d(u(bi ), u(ai )) ≤ ε
i=1
for every finite number of nonoverlapping intervals (ai , bi ), i = 1, . . . , n, with [ai , bi ] ⊆ I and n (bi − ai ) ≤ δ. i=1
67
68
3. Absolutely Continuous Functions
The space of all absolutely continuous functions u : I → Y is denoted by AC(I; Y ). When Y = R we write AC(I) for AC(I; R). Remark 3.2. Note that since n is arbitrary, we can also take n = ∞, namely, replace finite sums by series. Let I ⊆ R be an interval and let (Y, d) be a metric space. A function u : I → Y is locally absolutely continuous if it is absolutely continuous in [a, b] for every interval [a, b] ⊆ I. The space of all locally absolutely continuous functions u : I → Y is denoted by ACloc (I; Y ). As before, when Y = R we write ACloc (I) for ACloc (I; R). Note that ACloc ([a, b]; Y ) = AC([a, b]; Y ). If Ω ⊆ R is an open set and (Y, d) is a metric space, then we define the notion of absolute continuity for a function u : Ω → Y as in Definition 3.1, with the only change that we now require the intervals [ai , bi ] to be contained in Ω in place of I. The space of all absolutely continuous functions u : Ω → Y is denoted by AC(Ω; Y ). Exercise 3.3. Let I ⊆ R be an interval and let u : I → R. (i) Prove that u belongs to AC(I) if and only if for every ε > 0 there exists δ > 0 such that n (u(bi ) − u(ai )) ≤ ε i=1
for every finite number of nonoverlapping intervals (ai , bi ), i = 1, . . . , n, with [ai , bi ] ⊆ I and ni=1 (bi − ai ) ≤ δ. (ii) Assume that for every ε > 0 there exists δ > 0 such that n (u(bi ) − u(ai )) ≤ ε i=1
for every finite number of intervals (ai , bi ), i = 1, . . . , n, with [ai , bi ] ⊆ I and ni=1 (bi − ai ) ≤ δ. Prove that u is locally Lipschitz continuous. Part (ii) of the previous exercise shows that in Definition 3.1 we cannot remove the condition that the intervals (ai , bi ) are pairwise disjoint. Remark 3.4. By taking n = 1 in Definition 3.1, it follows that an absolutely continuous function u : I → Y is uniformly continuous. The next exercise shows that the converse is false.
3.1. AC(I; Y ) Versus BP V (I; Y )
69
Exercise 3.5. Let u : (0, 1] → R be defined by 1 , xb where a, b ∈ R. Determine for which a, b the function u is absolutely continuous. Prove that there exist a, b for which u is uniformly continuous but not absolutely continuous. u(x) := xa sin
Exercise 3.6. Let I ⊆ R be an interval and let u : I → R be differentiable with bounded derivative. Prove that u belongs to AC(I). Exercise 3.7. Let u, v ∈ AC([a, b]). Prove the following. (i) u ± v ∈ AC([a, b]). (ii) uv ∈ AC([a, b]). (iii) If v(x) > 0 for all x ∈ [a, b], then u/v ∈ AC([a, b]). (iv) max{u, v}, min{u, v} ∈ AC([a, b]). (v) What happens if the interval [a, b] is replaced by an arbitrary interval I ⊆ R (possibly unbounded)? We now turn to the relation between absolutely continuous functions and functions of bounded pointwise variation. In Corollary 2.31 we have proved that if u : I → R has bounded pointwise variation, then u is bounded and u is Lebesgue integrable. However, the function u(x) := x, x ∈ R, is absolutely continuous, but it is unbounded and u (x) = 1, which is not Lebesgue integrable. Also the function u(x) := sin x, x ∈ R, is absolutely continuous, bounded, but u is not Lebesgue integrable. These simple examples show that an absolutely continuous function may not have bounded pointwise variation. Proposition 3.9 below will show that this can happen only on unbounded intervals. Exercise 3.8. Let I ⊆ R be an interval, let (Y, d) be a complete metric space, and let u : I → Y be uniformly continuous. (i) Prove that u may be extended uniquely to I in such a way that the extended function is still uniformly continuous. (ii) Prove that if u belongs to AC(I; Y ), then its extension belongs to AC(I; Y ). (iii) Let Y be a normed space. Prove that there exist A, B > 0 such that for all x ∈ I, u(x) ≤ A + B|x|. The previous exercise shows that although an absolutely continuous function may be unbounded, it cannot grow faster than linear when |x| → ∞.
70
3. Absolutely Continuous Functions
Hence, the functions u(x) = |x|p , p > 1, and v(x) = |x| log(1 + |x|) cannot be absolutely continuous in an unbounded interval I. Next we show that ACloc (I; Y ) ⊆ BP Vloc (I; Y ). Proposition 3.9. Let I ⊆ R be an interval, let (Y, d) be a metric space, and let u ∈ ACloc (I; Y ). Then u belongs to BP Vloc (I; Y ). In particular, if Y = RM , then u is differentiable L1 -a.e. in I and u is locally Lebesgue integrable. Proof. Let [a, b] ⊆ I. Take ε = 1, and let δ > 0 be as in Definition 3.1. Let n be the integer part of 2(b − a)/δ and partition [a, b] into n intervals [xi−1 , xi ] of equal length (b − a)/n. Since (b − a)/n ≤ δ, in view of (3.1), on each interval [xi−1 , xi ] we have that Var[xi−1 ,xi ] u ≤ 1, and so by Proposition 2.15, n Var[xi−1 ,xi ] u ≤ n ≤ 2(b − a)/δ < ∞, Var[a,b] u = i=1
where we have used the fact that (b − a)/n ≥ δ/2. The last part of the statement follows by Corollary 2.31. Corollary 3.10. Let I ⊆ R be a bounded interval and let (Y, d) be a metric space. Then AC(I; Y ) ⊆ BP V (I; Y ). In particular, if u ∈ AC([a, b]; RM ), then u is differentiable L1 -a.e. in [a, b] and u is Lebesgue integrable. The converse of Corollary 3.10 is false, since absolutely continuous functions are always continuous while monotone functions may not be. Even more, there exist continuous monotone functions that are not absolutely continuous. The Cantor function and the function constructed in Exercise 1.39 are such examples. What is missing for a continuous function in BP V (I; Y ) to belong to AC(I; Y ) is the fundamental theorem of calculus. In view of Proposition 3.9, if u ∈ ACloc (I; Y ), then u belongs to BP Vloc (I; Y ) and thus its indefinite pointwise variation V (see (2.3)) is continuous in view of Proposition 2.21. We now prove that V is actually in ACloc (I). Theorem 3.11. Let I ⊆ R be an interval, let (Y, d) be a metric space, and let u : I → Y . Then u ∈ ACloc (I; Y ) (respectively, AC(I; Y )) if and only if V ∈ ACloc (I) (respectively, AC(I)). Proof. Assume that u ∈ ACloc (I; Y ) and let J ⊆ I be an interval such that u ∈ AC(J; Y ) (in particular any closed interval in I will have this property). Given ε > 0, let δ > 0 be as in Definition 3.1 with J in place of I. Consider a finite number of nonoverlapping intervals (ai , bi ), i = 1, . . . , n, with [ai , bi ] ⊆ n J and i=1 (bi − ai ) ≤ δ. Given a partition Pi = {x0,i , . . . , xmi ,i } of [ai , bi ],
3.2. The Fundamental Theorem of Calculus
since
mi n
|xk,i − xk−1,i | =
i=1 k=1
n
71
(bi − ai ) < δ,
i=1
it follows from the fact that u ∈ AC([a, b]; Y ) that mi n
d(u(xk,i ), u(xk−1,i )) ≤ ε.
i=1 k=1
Taking the supremum over every partition Pi of [ai , bi ] for each i, we have that n Var[ai ,bi ] u ≤ ε, i=1
that is, by (2.4),
n (V (bi ) − V (ai )) ≤ ε, i=1
which shows that V belongs to AC(J). By the arbitrariness of J it follows that V belongs to ACloc (I). Conversely, assume that V belongs to ACloc (I) and let J ⊆ I be an interval such that V belongs to AC(J). Given ε > 0, let δ > 0 be as in Definition 3.1, but now for V . Consider a finite number of nonoverlapping intervals (ai , bi ), i = 1, . . . , n, with [ai , bi ] ⊆ J and ni=1 (bi − ai ) ≤ δ. Then by (2.4), n n d(u(bi ), u(ai )) ≤ (V (bi ) − V (ai )) ≤ ε, i=1
i=1
which concludes the proof.
3.2. The Fundamental Theorem of Calculus In this section we will show that absolutely continuous functions u : I → RM can be characterized as the family of all functions for which the fundamental theorem of calculus holds (for the Lebesgue integral). We have seen that the Cantor function u : [0, 1] → R has zero derivative L1 -a.e. in I, but u(1) = 1, u(0) = 0. The following theorem shows that this cannot happen for absolutely continuous functions. Theorem 3.12. Let I ⊆ R be an open interval, let (Y, · ) be a normed space, and let u ∈ ACloc (I; Y ) be such that there exists u (x) = 0 for L1 -a.e. x ∈ I. Then u is constant. Proof. Let a, b ∈ I with a < b. We claim that u(a) = u(b). Given ε > 0, let δ > 0 be the number given in the definition of absolute continuity in [a, b]. Let E := {x ∈ (a, b) : u (x) = 0}. Then L1o (E) = b − a.
72
3. Absolutely Continuous Functions
For every x ∈ E, we have that u(x + h) − u(x) = 0, h→0 h lim
and so there exists hx > 0 such that [x − hx , x + hx ] ⊂ (a, b) and u(x + h) − u(x) ≤ ε|h|
(3.2)
for all x + h ∈ I with |h| ≤ hx . Consider the family F of intervals (x, x + h), where x ∈ E and 0 < h ≤ hx . By Lemma 1.20 there exist disjoint intervals (x1 , x1 + h1 ), . . . , (xn , xn + hn ) ∈ F and a set F ⊆E∩
n
(xi , xi + hi )
i=1
such that L1o (F ) ≥ b − a − δ,
n
hi ≤ b − a + δ.
i=1
Without loss of generality, assume that x1 < x2 < · · · < xn . Since n
hi ≥ L1o (F ) ≥ b − a − δ,
i=1
the sum of the lengths of the intervals [a, x1 ], [x1 + h1 , x2 ], . . . , [xn + hn , b] is less than or equal to δ. Since u is absolutely continuous, we have that u(a) − u(x1 ) +
n−1
u(xi+1 ) − u(xi + hi ) + u(b) − u(x + hn ) ≤ ε.
i=1
On the other hand, by (3.2), u(xi + hi ) − u(xi ) ≤ εhi , and so u(a) − u(b) ≤ u(a) − u(x1 ) +
n−1
u(xi+1 ) − u(xi + hi )
i=1
+
n
u(xi ) − u(xi + hi ) + u(b) − u(xn + hn )
i=1
≤ε+ε
n
hi ≤ ε + ε(b − a).
i=1
Letting ε →
0+
gives u(a) = u(b). Hence, u is constant.
Using the previous theorem, we are in a position to complete the proof of Theorem 2.40. We begin with some well-known results on Riemann integration.
3.2. The Fundamental Theorem of Calculus
73
Exercise 3.13 (Riemann integration, I). Let u : [a, b] → R be a bounded function and for x ∈ [a, b] define ω(x) := lim sup{|u(x1 ) − u(x2 )| : x1 , x2 ∈ [a, b], δ→0+
|x1 − x| ≤ δ, |x2 − x| ≤ δ}. (i) Prove that u is continuous at x ∈ [a, b] if and only if ω(x) = 0. (ii) Prove that if the set E := {x ∈ [a, b] : u is discontinuous at x} has positive Lebesgue outer measure, then there exists a constant α > 0 such that the set E1 := {x ∈ E : ω(x) ≥ α} has positive Lebesgue outer measure. (iii) Let t := L1o (E1 ) > 0 and prove that for every partition P the intervals containing points of E1 in their interior have total length greater than or equal to t. (iv) Deduce that if u is Riemann integrable, then E has Lebesgue measure zero. Exercise 3.14 (Riemann integration, II). Let u : [a, b] → R be a bounded function. (i) Let
x
g(x) := a
x
u(t) dt −
u(t) dt,
a < x ≤ b,
g(a) := 0,
a
x x where a and a are the upper and lower Riemann integrals. Prove that g is Lipschitz continuous. (ii) Prove that if u is continuous at x, then g is differentiable at x and g (x) = 0. (iii) Deduce that if u is continuous L1 -a.e. in [a, b], then u is Riemann integrable. Proof of Theorem 2.40, continued. We claim that u cannot be Riemann integrable on any closed interval [a, b]. Indeed, by the previous exercises, this would imply that u is continuous except for a set of Lebesgue outer measure zero. But, since u is nowhere monotone, if u is continuous at x, then necessarily u (x) = 0. Hence, u = 0 L1 -a.e. on [a, b]. By the previous theorem, we would get that u is a constant in [a, b], which is a contradiction.
74
3. Absolutely Continuous Functions
Remark 3.15. Note that the function constructed in Theorem 2.40 is Lipschitz continuous. Hence, it provides an example of a differentiable function with bounded derivative for which the fundamental theorem of calculus holds for Lebesgue integration (see Theorem 3.20 below) but not for Riemann integration. We now show that the primitive of an integrable function is absolutely continuous. Theorem 3.16. Let I ⊆ R be an interval and let v : I → RM be a Lebesgue integrable function. Fix x0 ∈ I and let x v(t) dt, x ∈ I. u(x) := x0
Then the function u is absolutely continuous in I and u (x) = v(x) for L1 -a.e. x ∈ I. We begin with an auxiliary result. Lemma 3.17. Let v : [a, b] → R be a Lebesgue integrable function such that x v(t) dt = 0 a
for all x ∈ [a, b]. Then v(x) = 0 for L1 -a.e. x ∈ [a, b]. Proof. Observe that for every a ≤ α < β ≤ b we have β β α v(t) dt = v(t) dt − v(t) dt = 0. α
a
a
Let U ⊂ [a, b] be open. Then U can be written as a countable union of disjoint open intervals (αn , βn ). Since χ
n=1 (αn ,βn )
(t)|v(t)| ≤ |v(t)|
for every t ∈ [a, b] and lim→∞ χ
n=1 (αn ,βn )
(t)v(t) = χU (t)v(t) we can apply
the Lebesgue dominated convergence theorem to conclude that b b v(t) dt = χU (t)v(t) dt = lim χ (t)v(t) dt U
a
= lim
b
→∞ a n=1
→∞ a
n=1 (αn ,βn ) βn
χ(αn ,βn ) (t)v(t) dt = lim
→∞
In turn, for every compact set K ⊆ [a, b], b v(t) dt = v(t) dt − K
a
n=1 αn
v(t) dt = 0. [a,b]\K
v(t) dt = 0.
3.2. The Fundamental Theorem of Calculus
75
Now let E + := {x ∈ [a, b] : v(x) > 0}. Since L1 (E + ) = sup{L1 (K) : K ⊆ E + , K compact}, + 1 if L1 (E + ) > 0, then there exists a compact set K ⊆ E with L (K) > 0. Define Kn := {x ∈ K : v(x) ≥ 1/n}. Then K = n Kn . Since v > 0 in K, v(t) dt ≥ v(t) dt ≥ n1 L1 (Kn ), 0= K
Kn
= 0 for all n, which implies that L1 (K) = 0. Thus, we have and so a contradiction, and so L1 (E + ) = 0. Similarly, we can prove that the set E − := {x ∈ [a, b] : v(x) < 0} has Lebesgue measure zero. Hence, we have shown that v(x) = 0 for all x ∈ [a, b] \ (E + ∪ E − ). L1 (Kn )
Remark 3.18. The proof is simpler if v is Riemann integrable. In this case v is continuous for L1 -a.e. x ∈ [a, b] (see Exercise 3.13). Take a point x0 ∈ (a, b) such that v is continuous at x0 . If v(x0 ) > 0, then taking ε = v(x0 )/2 we can find δ > 0 such that v(x) ≥ v(x0 ) − ε = ε for all x ∈ (x0 − δ, x0 + δ) ⊆ (a, b). Then x0 +δ v(t) dt ≥ 2δε > 0, 0= x0 −δ
which is a contradiction. Similarly, if v(x0 ) < 0, then we get a contradiction. Hence, v = 0 at every continuity point of v. We now turn to the proof of Theorem 3.16. Proof. It is enough to give the proof for M = 1. Step 1: Assume that v is bounded, with |v(x)| ≤ c for L1 -a.e. x ∈ [a, b]. Then u is absolutely continuous, since n n bi n bi ≤ |u(bi ) − u(ai )| = v(t) dt |v(t)| dt i=1
ai i=1 n
i=1
ai
(bi − ai ) ≤ ε
≤c
i=1
for every finite number of nonoverlapping intervals (ai, bi ), i = 1, . . . , n, with [ai , bi ] ⊆ I and ni=1 (bi − ai ) ≤ δ := ε/(c + 1). Extend v to be zero outside [a, b] and define x+1/n u(x + 1/n) − u(x) =n v(t) dt. wn (x) := 1/n x Then |wn (x)| ≤ nc(x + 1/n − x) = c for all x ∈ [a, b]. Since u is absolutely continuous, it is differentiable for all x ∈ [a, b] except a set of measure zero
76
3. Absolutely Continuous Functions
and so wn (x) → u (x) as n → ∞ for all x ∈ [a, b] except a set of measure zero. Hence, by the Lebesgue dominated convergence theorem, for all x0 ∈ [a, b], x0 x0 u (x) dx = lim wn (x) dx. n→∞ a
a
On the other hand, x0 wn (x) dx = n a
x0
(u(x + 1/n) − u(x)) dx
a
=n
x0 +1/n
a+1/n
u(x) dx −
x0
u(x) dx a
→ u(x0 ) − u(a) = u(x0 ) − 0, where we used the fact that u is continuous. Hence, x0 x0 u (x) dx = u(x0 ) = v(x) dx, a
a
that is,
x0
(u (x) − v(x)) dx = 0
a
for all x0 ∈ [a, b]. By the previous lemma, it follows that u (x) = v(x) for L1 -a.e. x ∈ [a, b]. Step 2: Assume that v ≥ 0 and define v(x) if v(x) ≤ n, vn (x) := 0 if v(x) > n. Then
x
v(t) dt =
u(x) = a
x
x
vn (t) dt + a
(v(t) − vn (t)) dt =: un (x) + sn (x).
a
By Step 1 we have that un (x) = vn (x) for all x ∈ [a, b] except a set of Lebesgue measure zero. On the other hand, since v ≥ vn we have that sn is increasing and so sn (x) ≥ 0 for L1 -a.e. x ∈ [a, b]. Hence, since u = un + sn , by differentiating, we obtain that u (x) = un (x) + sn (x) = vn (x) + sn (x) ≥ vn (x) + 0 for L1 -a.e. x ∈ [a, b], say, for all x ∈ [a, b] \ En , where L1 (En ) = 0. Since countable union of sets of Lebesgue measure zero have Lebesgue measure zero, we have that E := n En has Lebesgue measure zero. If x ∈ [a, b] \ E, then u (x) ≥ vn (x) for all n and so, letting n → ∞ we obtain that u (x) ≥ v(x). In turn, b b u (x) dx ≥ v(x) dx = u(b) − u(a). a
a
3.2. The Fundamental Theorem of Calculus
77
On the other hand, by Theorem 1.25, b u (x) dx ≤ u(b) − u(a), a
which shows that
b
b
u (x) dx = a
Hence,
v(x) dx. a
b
(u (x) − v(x)) dx = 0,
a
but since u ≥ v, it follows that u (x) = v(x) for L1 -a.e. x ∈ [a, b]. Step 3: In the general case it is enough to write v = v + − v − , where s+ and s− are the positive and negative parts of s ∈ R, and observe that since x x + v (t) dt − v − (t) dt =: u1 (x) + u2 (x), u(x) = x0
x0
the result follows by applying Step 2 to u1 and u2 .
By applying the previous theorem to each [a, b] ⊆ I we obtain Corollary 3.19. Let I ⊆ R be an interval and let v : I → RM be a locally Lebesgue integrable function. Fix x0 ∈ I and let x u(x) := v(t) dt, x ∈ I. x0
Then the function u is locally absolutely continuous in I and u (x) = v(x) for L1 -a.e. x ∈ I. We are now ready to prove the main theorem of this section. Theorem 3.20 (Fundamental theorem of calculus). Let I ⊆ R be an interval. A function u : I → RM belongs to ACloc (I; RM ) if and only if (i) u is continuous in I, (ii) u is differentiable L1 -a.e. in I, and u belongs to L1loc (I; RM ), (iii) the fundamental theorem of calculus is valid; that is, for all x, x0 ∈ I, x u (t) dt. u(x) = u(x0 ) + x0
Proof. Assume that u ∈ ACloc (I; RM ). In view of Proposition 3.9, it remains to prove (iii). By Corollary 3.19 the function x u (t) dt, x ∈ I, (3.3) w(x) := x0
78
3. Absolutely Continuous Functions
belongs to ACloc (I; RM ) with w (x) = u (x) for L1 -a.e. x ∈ I. Since u − w ∈ ACloc (I) and (u − w) (x) = u (x) − w (x) = 0 for L1 -a.e. x ∈ I, by Theorem 3.12, we have that u − w is constant in I. Thus, there exists c ∈ R such that x u(x) = c + u (t) dt x0
for all x ∈ I. Taking x = x0 gives c = u(x0 ). Conversely, assume that (i)–(iii) are satisfied. Then by Theorem 3.16, the function w defined in (3.3) belongs to ACloc (I; RM ), and hence, by (iii), so does u. Exercise 3.21. Let I ⊆ R be an interval and let u : I → R be a convex function. (i) Prove that the left and right derivatives u− (x) and u+ (x) exist in R for all x ∈ I ◦ . (ii) Prove that the functions u− and u+ are increasing. (iii) Prove that (3.4)
u(x) − u(x0 ) =
x x0
u− (t) dt
x
= x0
u+ (t) dt
for all x, x0 ∈ I ◦ with x0 < x. Exercise 3.22. Prove that there exists an increasing absolutely continuous function u : [0, 1] → R which is not constant and with the property that for every x ∈ [0, 1] with u (x) > 0, u is not strictly increasing in a neighborhood of x. As a corollary of Theorem 3.20 we recover the formula for integration by parts. Corollary 3.23 (Integration by parts). Let I ⊆ R be an interval and let u, v ∈ ACloc (I). Then for all x, x0 ∈ I, x x uv dt = u(x)v(x) − u(x0 )v(x0 ) − u v dt. x0
x0
Proof. Since u, v ∈ ACloc (I), by Exercise 3.7, uv ∈ ACloc (I), and thus, by part (iii) of Theorem 3.20, x (uv) dt. u(x)v(x) − u(x0 )v(x0 ) = x0
Since the functions u, v, and uv are differentiable L1 -a.e. in I, the standard calculus rule for a product now gives the desired result.
3.2. The Fundamental Theorem of Calculus
79
Another consequence of Theorem 3.20 is the following result, which says that for monotone functions, to prove absolute continuity in a fixed interval [a, b], it is sufficient to test the fundamental theorem of calculus only at the endpoints a and b. Corollary 3.24 (Tonelli). Let I ⊆ R be an interval, let u : I → R be a monotone function, and let [a, b] ⊆ I. Then u belongs to AC([a, b]) if and only if b |u (t)| dt = |u(b) − u(a)|. a
Moreover, if u is bounded, then u belongs to AC(I) if and only if |u (t)| dt = sup u − inf u. I
I
I
Proof. Without loss of generality, we may assume that u is increasing. Step 1: If u ∈ AC([a, b]), then by Theorem 3.20 we have b u (t) dt = u(b) − u(a). (3.5) a
Conversely, assume that (3.5) holds and let x w(x) := u (t) dt, x ∈ [a, b]. a
If a ≤ x0 < x ≤ b, then by Corollary 1.25, x (3.6) w(x) − w(x0 ) = u (t) dt ≤ u(x) − u(x0 ), x0
and so the function u − w is increasing in [a, b]. But by (3.5), b u (t) dt = w(b) − w(a), u(b) − u(a) = a
and so (u − w)(b) = (u − w)(a). This implies that the increasing function u − w must be constant in [a, b]. Since w ∈ AC([a, b]) by Theorem 3.16, it follows that u ∈ AC([a, b]). Step 2: If u ∈ AC(I), then by the previous step for every inf I < x0 < x < sup I, we have x u (t) dt = u(x) − u(x0 ). x0
and x → (sup I)− and using the Lebesgue monotone Letting x0 → (inf convergence theorem and the fact that u is continuous and increasing gives u (t) dt = sup u − inf u. (3.7) I)+
I
I
I
80
3. Absolutely Continuous Functions
Conversely, assume that (3.7) holds. Then u is Lebesgue integrable. Moreover, by Exercise 1.27, u is continuous in I. Fix c ∈ R and define x w(x) := c + u (t) dt, x ∈ I. inf I
As in the previous step we have that u − w is increasing (and bounded) and that (3.6) holds for all inf I < x0 < x < sup I. Letting x0 → (inf I)+ and x → (sup I)− in (3.6) and using the Lebesgue monotone convergence theorem, the monotonicity and the continuity of u and w, we obtain that sup I u(x) − lim u(x) = u (t) dt lim x→(sup I)−
x→(inf I)+
inf I
=
lim
x→(sup I)−
w(x) −
lim
x→(inf I)+
w(x),
and so, since u − w is increasing and continuous, sup(u(x) − w(x)) = x∈I
=
lim
(u(x) − w(x))
lim
(u(x) − w(x))
x→(sup I)− x→(inf I)+
= inf (u(x) − w(x)), x∈I
which implies as before that u − w must be constant. By choosing c appropriately, we have that u = w, and so u belongs to AC(I), since w does (see Theorem 3.16). Since functions with bounded pointwise variation are differences of bounded monotone functions, the previous corollary implies the following result. Theorem 3.25 (Tonelli). Let I ⊆ R be an interval, let u ∈ BP Vloc (I; RM ), and let [a, b] ⊆ I. Then u belongs to AC([a, b]; RM ) if and only if b u (t) dt = Var[a,b] u. (3.8) a
In addition, if u belongs to BP V (I; RM ), then u belongs to AC(I; RM ) if and only if u (t) dt = Var u. (3.9) I
Proof. Step 1: Let V be the indefinite pointwise variation of u (see (2.3)). By Corollary 1.25 and Theorem 2.28, b b u (t) dt = V (t) dt ≤ V (b) − V (a) = Var[a,b] u. (3.10) a
a
3.2. The Fundamental Theorem of Calculus
Hence, if
b
81
u (t) dt = Var[a,b] u,
a
then all the previous inequalities are equalities, and so b V (t) dt = V (b) − V (a). a
Since V is increasing, it follows by Corollary 3.24 that the function V belongs to AC([a, b]). In turn, by Theorem 3.11, u ∈ AC([a, b]). Conversely, if u ∈ AC([a, b]; RM ), then by Theorem 3.20 for every partition {x0 , . . . , xn } of [a, b], b n n xi n xi ≤ u(xi ) − u(xi−1 ) = u dt u dt = u dt. i=1
i=1
xi−1
i=1
xi−1
a
Taking the supremum over all partitions, we get b u dt, Var[a,b] u ≤ a
which, together with (3.10) yields the desired equality. Step 2: Assume that u ∈ BP V (I; RM ). Then by Corollary 1.25 and Theorem 2.28, u (t) dt = V (t) dt ≤ sup V − inf V = Var u, I
and so if (3.9) holds, then
I
I
I
V (t) dt = sup V − inf V .
I
I
I
In turn, by the previous corollary, V ∈ AC(I). As in the previous step we conclude that u ∈ AC(I; RM ). Conversely, if u ∈ AC(I; RM ), then by the previous step, (3.8) holds for every [a, b] ⊆ I. If inf I ∈ I, define an := inf I, and otherwise construct a sequence an inf I. Similarly, if sup I ∈ I, define bn := sup I, and otherwise construct a sequence bn sup I. It suffices to apply the previous step in [an , bn ] and then to let n → ∞ using Proposition 2.15 and the Lebesgue monotone convergence theorem. Theorem 3.20 characterizes functions in ACloc (I; RM ) in terms of the fundamental theorem of calculus. We now prove a similar result for functions in AC(I; RM ). We recall the following definition.
82
3. Absolutely Continuous Functions
Definition 3.26. If E ⊆ R is a Lebesgue measurable set and v : E → RM is a Lebesgue measurable function, then v is equi-integrable if for every ε > 0 there exists δ > 0 such that v(x) dx ≤ ε F
for every Lebesgue measurable set F ⊆ E, with L1 (F ) ≤ δ. Exercise 3.27. Let E ⊆ R be a Lebesgue measurable set, let 1 ≤ p ≤ ∞, and let v ∈ Lp (E; RM ). Prove that v is equi-integrable. Prove that if we only assume that v ∈ L1loc (E; RM ), then the result may no longer be true. Exercise 3.28. Let E ⊆ R be a Lebesgue measurable set with finite measure and let v : E → RM be equi-integrable. Prove that v ∈ L1 (E; RM ). Theorem 3.29 (Fundamental theorem of calculus, II). Let I ⊆ R be an interval. A function u : I → RM belongs to AC(I; RM ) if and only if (i) u is continuous in I, (ii) u is differentiable L1 -a.e. in I, and u belongs to L1loc (I; RM ) and is equi-integrable, (iii) the fundamental theorem of calculus is valid; that is, for all x, x0 ∈ I, x u (t) dt. u(x) = u(x0 ) + x0
Proof. Assume that u ∈ AC(I; RM ). In view of Theorem 3.20 it remains to show that u is equi-integrable. By Exercise 3.8 we may assume, without loss of generality, that I is closed. Fix ε > 0 and let δ > 0 be as in Definition 3.1. Consider a Lebesgue measurable set E ⊆ I, with L1 (E) ≤ 2δ . By the outer regularity of the Lebesgue measure we may find an open set U ⊇ E such that L1 (U ) < δ. Decompose U into a countable family {Ji }i of pairwise disjoint that Ji ⊆ I. Let intervals. By replacing each Ji with Ji ∩ I, we may assume Ji = [ai , bi ]. Using the fact that L1 (U ) < δ, we have that i |bi − ai | < δ. Reasoning as in the first part of the proof of Theorem 3.11 we have that Var[ai ,bi ] u ≤ ε. i
Hence, by Tonelli’s theorem applied to each interval [ai , bi ], u dt ≤ u dt = u dt = Var[ai ,bi ] u ≤ ε. E
This proves that
U
u
i
[ai ,bi ]
i
is equi-integrable.
Conversely, assume that u satisfies (i)–(iii). Since u is equi-integrable, given ε > 0 let δ > 0 be the number given in Definition 3.26. Given a finite
3.2. The Fundamental Theorem of Calculus
83
number n of nonoverlapping intervals (ai , bi ), i = 1, . . . , n, with [ai , bi ] ⊆ I and i=1 (bi − ai ) ≤ δ, by Theorem 3.20 in [ai , bi ] we have n n bi u(bi ) − u(ai ) = u dt ai i=1 n bi
i=1
≤
u dt =
i=1
ai
n
u dt ≤ ε.
i=1 (ai ,bi )
This shows that u is absolutely continuous.
Remark 3.30. If I is bounded, then in view of Exercises 3.27 and 3.28 in part (iii) we can replace u equi-integrable with u integrable. The next corollary will be useful in the study of Sobolev spaces in Chapter 7. The proof follows from Exercise 3.27 and Theorem 3.29. Corollary 3.31. Let I ⊆ R be an interval and let u : I → RM be such that (i) u is continuous on I, (ii) u is differentiable L1 -a.e. in I, and u ∈ Lp (I; RM ) for some 1 ≤ p ≤ ∞, (iii) the fundamental theorem of calculus is valid; that is, for all x, x0 ∈ I, x u (t) dt. u(x) = u(x0 ) + x0
Then u belongs to AC(I; RM ). Exercise 3.32. Let p > 1 and let ACp ([a, b]) be the class of all functions u : [a, b] → R such that for every ε > 0 there exists δ > 0 such that
1/p n |u(bi ) − u(ai )|p ≤ε i=1
for every finite number of nonoverlapping intervals (ai , bi ), i = 1, . . . , n, with [ai , bi ] ⊆ I and
1/p n p (bi − ai ) ≤ δ. i=1
(i) Prove that if u ∈ ACp ([a, b]), then Varp u < ∞ (see Exercise 2.25). (ii) Prove that the function u(x) =
∞ 1 cos 2n πx, n/p 2 n=0
x ∈ [0, 1],
is such that Varp u < ∞, but it does not belong to ACp ([0, 1]).
84
3. Absolutely Continuous Functions
In the discussion before Exercise 3.8 we have shown that an absolutely continuous function defined in an unbounded interval may not have bounded pointwise variation. As a corollary of the previous theorem we can now characterize the absolutely continuous functions that have bounded pointwise variation. Corollary 3.33. Let I ⊆ R be an interval and let u ∈ ACloc (I; RM ). Then u belongs to BP V (I; RM ) if and only if u belongs to L1 (I; RM ). In this case u belongs to AC(I; RM ). Proof. In view of Corollary 2.31, it remains to show that if u is Lebesgue integrable, then u belongs to BP V (I; RM ). By Proposition 3.9 we have that u ∈ BP Vloc (I; RM ). Hence by Theorem 3.25, for every [a, b] ⊆ I, b u dt = Var[a,b] u. a
Taking an and bn as in Step 2 of the proof of Theorem 3.25 and using Proposition 2.15 (note that in that proposition the function u is completely arbitrary), Exercise 2.22, and the Lebesgue monotone convergence theorem, we have that u dt = VarI u. I
Since u is Lebesgue integrable, it follows that VarI u < ∞. The last statement follows from Corollary 3.31.
3.3. Lusin (N ) Property In this section we show that absolutely continuous functions map sets of measure zero into sets of measure zero and that this property characterizes them among functions of bounded variation. We recall that H1 is the onedimensional Hausdorff measure (see Appendix C). Definition 3.34. Let E ⊆ R, let (Y, d) be a metric space, and let u : E → Y . We say that u satisfies the Lusin (N ) property if H1 (u(F )) = 0 for every set F ⊆ E with L1 (F ) = 0. In particular, when Y = R, then H1 = L1 and so u satisfies the Lusin (N ) property if L1 (u(F )) = 0 for every set F ⊆ E with L1 (F ) = 0. Example 3.35. The Cantor function u : [0, 1] → R does not satisfy the Lusin (N ) property since L1 (D) = 0 but L1 (u(D)) = 1.
3.3. Lusin (N ) Property
85
The next result shows that everywhere differentiable functions with bounded derivative satisfy the Lusin (N ) property. Theorem 3.36. Let I ⊆ R be an interval, let (Y, · ) be a normed space, and let u : I → Y . Assume that there exist a set E ⊆ I (not necessarily measurable) and c ≥ 0 such that u is differentiable for all x ∈ E, with u (x) ≤ c
(3.11)
for all x ∈ E.
Then Ho1 (u(E)) ≤ cL1o (E). Proof. Without loss of generality, we may assume that E ⊆ I ◦ . Fix ε > 0. By the properties of the Lebesgue outer measure we may find an open set U such that U ⊇ E and L1 (U ) ≤ L1o (E) + ε. By replacing U with U ∩ I ◦ , if necessary, we may suppose that U ⊆ I ◦ . Take E0 := ∅ and inductively for each n ∈ N let En be the set of points x ∈ U \ En−1 such that (x − 1/n, x + 1/n) ⊆ U and u(x + h) − u(x) ≤ (c + ε)|h|
(3.12)
for every 0 < |h| < 1/n. The sets En are Borel sets (exercise). We claim that ∞ En . (3.13) E⊆ n=1
To see this, fix x ∈ E. Since u (x) ≤ c, there exists δ > 0 such that u(x + h) − u(x) ≤ (c + ε)|h| for all x + h ∈ I, with |h| < δ. This implies that x ∈ En for every integer n > 1/δ so large that (x − 1/n, x + 1/n) ⊆ U and so the claim is proved. Decompose U as a countable family {Jk }k of pairwise disjoint intervals with 0 < length Jk < 1/n. Define Ek,n := En ∩ Jk . Note that the sets Ek,n are disjoint Borel sets contained in U . Hence, 1 L (Ek,n ) = L1 Ek,n ≤ L1 (U ). (3.14) k,n
k,n
It follows from (3.12) and the fact that length Jk < 1/n that if Ek,n is nonempty, the function u : Ek,n → Y is Lipschitz continuous with Lipschitz constant at most c + ε. Hence, by Proposition C.44, Ho1 (u(Ek,n )) ≤ (c + ε)L1o (Ek,n ).
86
3. Absolutely Continuous Functions
Then by (3.13) and (3.14), Ho1 (u(E)) ≤ Ho1 u(Ek,n ) ≤ Ho1 (u(Ek,n )) k,n
k,n
≤ (c + ε)
L1 (Ek,n ) ≤ (c + ε)L1 (U )
k,n
≤ (c + ε)(L1o (E) + ε).
It now suffices to let ε → 0+ .
As a consequence of the previous theorem, we have the following result. Corollary 3.37. Let I ⊆ R be an interval, let (Y, · ) be a normed space, and let u : I → Y . Assume that there exists a set E ⊆ I such that u is differentiable for all x ∈ E. If the set E has H1 measure zero or if u = 0 in E, then H1 (u(E)) = 0. Proof. Assume that E has Lebesgue measure zero. For every n ∈ N write En :={x ∈ E : u (x) ≤ n}. Since u is differentiable in E, it follows that E= ∞ n=1 En , while by the previous theorem, Ho1 (u(En )) ≤ nL1o (En ) = 0. By the countable subadditivity of Ho1 we obtain that Ho1 (u(E)) = 0. If u = 0 in E, then we may take c := 0 in the previous theorem.
Remark 3.38. The Cantor function shows that the previous corollary does not hold if we replace everywhere differentiability in E with L1 -a.e. differentiability in E. Another important consequence of Theorem 3.36 is the following. Theorem 3.39. Let I ⊆ R be an interval, let (Y, · ) be a normed space, let u : I → Y , and let E ⊆ I be a Lebesgue measurable set on which u is differentiable. Then u(E) is H1 measurable and 1 u (x) dx. (3.15) H (u(E)) ≤ E
Proof. Step 1: By the properties of the Lebesgue measure we may write E as the union of a set of Lebesgue measure zero and countably many compact sets, precisely, E = E0 ∪ n Kn , where Kn is compact and L1 (E0 ) = 0. By the previous corollary H1 (u(E0 )) = 0, while u(Kn ) is compact since u : Kn → R is continuous (since differentiable). Hence, u(E) is H1 measurable. Step 2: Assume that L1 (E) < ∞. Fix n ∈ N. For every k ∈ N write k ≤ u (x) < . Ek,n := x ∈ E : k−1 n n 2 2
3.3. Lusin (N ) Property
Then E =
87
∞
k=1 Ek,n ,
and so by Theorem 3.36, H1 (u(E)) = H1 u(Ek,n ) ≤ H1 (u(Ek,n )) ≤
k
k
k 1 2n L (Ek,n )
=
k
≤
k−1 1 2n L (Ek,n )
k
k
u (x) dx + Ek,n
+
L1 (Ek,n )
k
1 1 2n L (E)
1 2n
u (x) dx +
≤ E
1 1 2n L (E),
and it suffices to let n → ∞. Step 3: If L1 (E) = ∞, for every k ∈ N write Ek := E ∩ ([−k − 1, −k) ∪ (k, k + 1]). Then by the previous step applied to Ek we have Ek H1 (u(Ek )) ≤ H1 (u(E)) = H1 u ≤
k
k
k
u (x) dx =
Ek
u (x) dx, E
where we used the fact that the sets Ek are disjoint.
Remark 3.40. Let I ⊆ R be an interval, let (Y, · ) be a normed space, and let u : I → Y . If u is differentiable on the interval [a, b] ⊆ I, then, in particular, it is continuous on [a, b], and so by Lemma 2.63 and Theorem 3.39, b 1 u(b) − u(a) ≤ H (u([a, b])) ≤ u (x) dx. a
We are now ready to present the main theorem of this section. Theorem 3.41 (Lusin (N ) property). Let I ⊆ R be an interval. A function u : I → RM belongs to ACloc (I; RM ) if and only if (i) u is continuous on I, (ii) u is differentiable L1 -a.e. in I, and u ∈ L1loc (I; RM ), (iii) u satisfies the Lusin (N ) property. Proof. Step 1: Assume that u satisfies (i)–(iii) and fix [a, b] ⊆ I. We claim that u belongs to AC([a, b]; RM ). Let {(ak , bk )}nk=1 be a finite number of nonoverlapping intervals of [a, b] and let Ek := {x ∈ [ak , bk ] : u (x) exists}.
88
3. Absolutely Continuous Functions
By (ii), L1 ([ak , bk ] \ Ek ) = 0, and so by (iii), H1 (u([ak , bk ] \ Ek )) = 0. Since u is continuous, by Theorem 3.39 and by Lemma 2.63, n
u(bk ) − u(ak ) ≤
k=1
n
H1 (u([ak , bk ])) =
k=1
≤
H1 (u(Ek ))
k=1
n k=1
n
u (x) dx = Ek
n k=1
bk
u (x) dx.
ak
Since, by (ii), u is Lebesgue integrable in [a, b], in view of Exercise 3.27, u is equi-integrable in [a, b]. Hence, given ε > 0, we may find δ > 0 such that if n bk n k=1 (bk − ak ) ≤ δ, then k=1 ak u (x) dx ≤ ε. The absolute continuity of u in [a, b] follows from the previous inequality. Step 2: Conversely, assume that u ∈ ACloc (I; RM ). In view of Theorem 3.20, it remains to show that u satisfies property (iii). Fix [a, b] ⊆ I and a Lebesgue measurable set E ⊂ [a, b] with L1 (E) = 0. Without loss of generality, we may assume that E ⊆ (a, b). Fix ε > 0 and let δ > 0 be as in Definition 3.1. By the properties of the outer measure we may find an open set U such that U ⊇ E and L1 (U ) ≤ δ. By replacing U with U ∩ (a, b), we can assume that U ⊆ (a, b). Decompose U into a countable family {Jk }k of pairwise disjoint intervals. Since u is continuous, for every k we may find ak , bk ∈ J k such that diam u(J k ) = u(bk ) − u(ak ). Using the fact that L1 (U ) ≤ δ, we have that k |bk − ak | ≤ δ, and so, by the absolute continuity of u and Remark 3.2, u(bk ) − u(ak ) ≤ ε. k
Hence, by (C.28), Hε1 (u(E)) ≤ Hε1 (u(U )) ≤ diam u(J k ) = u(bk ) − u(ak ) ≤ ε. k
Letting ε →
0+
k
and using (C.29) we conclude that Ho1 (u(E)) = 0.
Remark 3.42. Note that since the sets Ek defined in the first part of the proof are Borel sets (why?), the previous theorem continues to hold if in place of the (N ) property we only require that u maps Borel sets of Lebesgue measure zero into sets of H1 measure zero. Corollary 3.43. Let I ⊆ R be an interval. A function u : I → RM belongs to AC(I; RM ) if and only if (i) u is continuous on I,
3.3. Lusin (N ) Property
89
(ii) u is differentiable L1 -a.e. in I, and u belongs to L1loc (I; RM ) and is equi-integrable, (iii) u satisfies the Lusin (N ) property. Proof. It is enough to repeat Step 1 of the proof of Theorem 3.41 word for word, with the only differences being that [a, b] should be replaced by I and that Exercise 3.27 is no longer needed, since, by (ii), u is assumed to be equi-integrable. Remark 3.44. Note that if I is a bounded interval, then u : I → RM belongs to AC(I; RM ) if and only if (i)–(iii) of the previous corollary hold. Indeed, if u ∈ AC(I; RM ), then by Corollary 3.10, u is integrable, and so equi-integrable by Exercise 3.27. Thus, property (ii) of the previous corollary holds. Properties (i) and (iii) follow from Theorem 3.41. Exercise 3.45. Let u : [a, b] → R be continuous and strictly increasing. (i) Prove that u is absolutely continuous if and only if it maps the set E := {x ∈ [a, b] : u (x) = ∞} into a set of Lebesgue measure zero. (ii) Is part (i) still true if u is only assumed to be continuous and increasing instead of strictly increasing? Exercise 3.46. Let u : [a, b] → R be continuous and strictly increasing. Prove that its inverse u−1 : [u(a), u(b)] → R is absolutely continuous if and only if the set E := {x ∈ [a, b] : u (x) = 0} has Lebesgue measure zero. In view of Remark 3.40 and Step 1 of the proof of Theorem 3.41, we have the following. Corollary 3.47. Let I ⊆ R be an interval. If u : I → RM is everywhere differentiable in I and u ∈ L1loc (I; RM ), then u belongs to ACloc (I; RM ). Exercise 3.48. Prove that if u : [a, b] → RM is continuous, differentiable on [a, b] except for at most a countable number of points, and if u is Lebesgue integrable, then u belongs to AC([a, b]; RM ). As a consequence of Theorem 3.41 we can characterize those functions with locally bounded pointwise variation that are locally absolutely continuous functions. Precisely, we have the following result. Corollary 3.49. Let I ⊆ R be an interval. A function u : I → RM belongs to ACloc (I; RM ) if and only if
90
3. Absolutely Continuous Functions
(i) u is continuous on I, (ii) u ∈ BP Vloc (I; RM ), (iii) u satisfies the Lusin (N ) property. Proof. In view of Corollary 2.31 we are in a position to apply Theorem 3.41. The next exercise gives an example of a function that satisfies the Lusin (N ) property, but it is not of bounded pointwise variation in any interval of (0, 1). Exercise 3.50. Let {(an , bn )}n be a base for the topology ∞ of (0, 1) and let {rn }n be a sequence of positive numbers such that n=1 rn < ∞. We construct inductively a sequence of functions un : [0, 1] → R and a sequence of intervals (cn , dn ) as follows. Let u0 (x) := x. Assume that un−1 : [0, 1] → R has been defined and that un−1 is a continuous piecewise affine function. Let (cn , dn ) ⊆ (an , bn ) be an interval such that un−1 restricted to (cn , dn ) is linear and L1 (un−1 ((cn , dn ))) < rn . We define the continuous function un to be un−1 outside (cn , dn ), while in (cn , dn ) we define it as a continuous piecewise affine function such that L1 (un ((cn , dn ))) < rn and Var(cn ,dn ) un > n. (i) Prove that {un }n converges uniformly to a continuous function u : [0, 1] → R. (ii) Prove that for every interval [a, b] ⊂ (0, 1) with a < b, Var[a,b] u = ∞. (iii) Prove that u satisfies the Lusin (N ) property. Hint: Let E0 := [0, 1] \
∞
∞
En := (cn , dn ) \
(ci , di ),
i=1
(ci , di )
i=n+1
and E∞ := For every set E ⊂ [0, 1],
∞ ∞
(ci , di ). n=1 i=n with L1 (E) = 0,
E = (E ∩ E∞ ) ∪
∞
write
(E ∩ En ).
n=0
Exercise 3.51 (The Cantor set and the Lusin (N ) property). Let D be the Cantor set. (i) Prove that every number x ∈ [0, 2] can be written as x=2
∞ cn (x) n=1
3n
,
3.4. Superposition in AC(I; Y )
91
where cn (x) ∈ {0, 1, 2}, and deduce that every element in [0, 2] can be written as the sum of two elements of D. (ii) Prove that if v : D → R is continuous and L1 (v(D)) = 0, then v can be extended to a continuous function on [0, 1] that has the Lusin (N ) property. Hint: Make v differentiable outside D. (iii) For each x ∈ D write x=2
∞ cn (x) n=1
3n
,
where cn (x) ∈ {0, 1} and prove that there exist two continuous functions u1 : [0, 1] → R and u2 : [0, 1] → R with the Lusin (N ) property and such that for all x ∈ D, u1 (x) =
∞ c2n (x) n=1
3n
,
u2 (x) =
∞ c2n+1 (x) n=1
3n
.
(iv) Prove that u1 + u2 does not have the Lusin (N ) property. (v) Why is this example important for absolute continuity? Exercise 3.52. Given an absolutely continuous function u : [a, b] → R, define v(x) := max u(y), x ∈ [a, b]. a≤y≤x
Prove that v is absolutely continuous and that v (x) = u (x)χ{v=u} (x) for L1 -a.e. x ∈ [a, b].
3.4. Superposition in AC(I; Y ) In this section we discuss the analog of Theorem 2.55. The following exercise (see Exercise 2.53) shows that the composition of absolutely continuous functions is not absolutely continuous (see however Exercise 3.67). Exercise 3.53. Consider the functions f : [0, 1] → R and u : [0, 1] → [0, 1], defined by 2 21 √ x sin x if 0 < x ≤ 1, f (z) := z, u(x) := 0 if x = 0. Prove that f and u are absolutely continuous but their composition f ◦ u is not. Remark 3.54. In Theorem 3.83 we will see that given f and u absolutely continuous, f ◦u is absolutely continuous if and only if (f ◦u)u is integrable, where f (u(x))u (x) is interpreted to be zero whenever u (x) = 0.
92
3. Absolutely Continuous Functions
The next result gives necessary and sufficient conditions on f : R → R for f ◦ u to be absolutely continuous for all absolutely continuous functions u : [a, b] → R. Theorem 3.55 (Superposition). Let I ⊆ R be an interval and let f : RM → R. Then f ◦ u ∈ ACloc (I) for all functions u ∈ ACloc (I; RM ) if and only if f is locally Lipschitz continuous. In particular, if M = 1, f is locally Lipschitz continuous and u ∈ ACloc (I), then the chain rule (3.24) holds. Proof. Step 1: Assume that f is locally Lipschitz continuous and let u ∈ ACloc (I). Fix an interval [a, b]. Since u is continuous, m := max u(x) < ∞, x∈[a,b]
and so there exists L > 0 such that |f (z1 ) − f (z2 ) ≤ Lz1 − z2
(3.16)
for all z1 , z2 ∈ B(0, m). We claim that f ◦ u is absolutely continuous in [a, b]. M Indeed, since u ∈ AC([a, b]; R ), for every ε > 0 there exists δ > 0 such that k u(bk ) − u(ak ) ≤ ε/L for every finite number of nonoverlapping intervals (ak , bk ) ⊆ [a, b], with k (bk − ak ) ≤ δ. Hence, by (3.16), |(f ◦ u)(bk ) − (f ◦ u)(ak )| ≤ L u(bk ) − u(ak ) ≤ ε, k
k
which proves the claim. The validity of the chain rule follows from Corollary 3.68. Step 2: Assume that f ◦ u ∈ ACloc (I) for all functions u ∈ ACloc (I; RM ). We claim that f is locally Lipschitz continuous. The proof follows closely that of Theorem 2.55, with the only difference that instead of discontinuous functions u (see (2.26), (2.31)) we will use piecewise affine functions. Fix [a, b] ⊆ I. We begin by showing that f is locally bounded. Fix r > 0. We claim that f is bounded in B(0, r). Indeed, assume by contradiction that there exists {zn }n in B(0, r) such that |f (zn )| → ∞ as n → ∞. By compactness we can assume that zn → z∞ , and by taking a further subsequence, also that zn − z∞ ≤ 1/n2 b−a for every n. For every n ∈ N, set In := [a + n+1 , a + b−a n ) and let xn be the middle point of In and n := (diam In )/2 = (b − a)/(2n(n + 1)). Consider the function ⎧ ⎨ zn + z∞ − zn (xn − x)/n if x ∈ In◦ , x ≤ xn , n ∈ N, zn + z∞ − zn (x − xn )/n if x ∈ In◦ , x ≥ xn , n ∈ N, (3.17) u(x) := ⎩ otherwise in I. z∞
3.4. Superposition in AC(I; Y )
93
We claim that u ∈ AC(I; RM ). Since zn → z∞ , we have that u is continuous at x = a. Hence, the function u is continuous and differentiable except for a countable number of points, and so, in view of Exercise 3.48, to prove that it is locally absolutely continuous in I, it remains to show that u is integrable. By (3.17) we have b ∞ ∞ u dx = u dx ≤ zn − z∞ ≤ 1/n2 < ∞. I
a
n=1
n=1
AC(I; RM ),
and so by hypothesis the function It follows that u belongs to f ◦ u belongs to ACloc (I). On the other hand, Var[a,b] (f ◦ u) ≥ VarI n (f ◦ u) ≥ |f (zn ) − f (z0 )| → ∞ as n → ∞, which is a contradiction. Step 3: Next we claim that f is Lipschitz continuous in B(0, r). Indeed, assume by contradiction that this is not the case. Then we may find two sequences {zn }n , {yn }n in B(0, r) such that zn = yn and (3.18)
|f (zn ) − f (yn )| > 2(n2 + n) zn − yn
for all n ∈ N. Since {zn }n is bounded, we may extract a subsequence (not relabeled) such that zn → z∞ . Take a further subsequence (not relabeled) such that (3.18) continues to hold and zn − z∞ < 1/(n + 1)2 .
(3.19)
By Step 2 f is bounded in B(0, r) by some constant cr > 0. Hence, by (3.18) for all n ∈ N we have (3.20)
2cr ≥ |f (zn ) − f (yn )| > 2(n2 + n)zn − yn ,
which gives 0 < δn :=
zn − yn (b − a) b−a → 0. < 2cr 2(n2 + n)
b−a , a + b−a For every n ∈ N, we divide the interval In := [a + n+1 n ] into subintervals of length δn . Thus, let 2cr diam In b−a = >2 = (3.21) n := 2 δn δn (n + n) zn − yn (n2 + n)
and set mn := max{j ∈ N0 : j < n }. Since n > 2, we have (3.22)
n /2 ≤ mn < n .
Consider the partition Pn of In given by b−a + 12 jδn : j = 0, . . . , 2mn ∪ a + Pn := a + n+1 = : {x0,n , . . . , x2mn +1,n }.
b−a n
94
3. Absolutely Continuous Functions
Let Ii,n := [xi,n , xi+1,n ], i = 0, . . . , 2mn . We now define the piecewise affine function u : I → R in the following way. Set z0 := z1 . Define u(x) := z∞ if x ≤ a, u(x) := z0 if x ≥ b, while in each interval In , n ∈ N, set (3.23) ⎧ if x ∈ I2i−1,n , 1 ≤ i ≤ mn − 1, ⎨2(zn − yn )(x − x2i−1,n )/δn + yn if x ∈ I2i,n , 0 ≤ i ≤ mn − 1, u(x) := 2(yn − zn )(x − x2i,n )/δn + zn ⎩ (zn−1 − yn )(x − x2mn −1,n )/dn + yn if x ∈ I2mn −1,n , where dn := x2mn +1,n − x2mn −1,n . Note that δn /2 ≤ dn ≤ 2δn . We claim that u ∈ AC(I; RM ). Since zn → z∞ and yn → z∞ , we have that u is continuous at x = a. Hence, the function u is continuous and differentiable except for a countable number of points, and so, in view of Exercise 3.48, to prove that it is locally absolutely continuous in I, it remains to show that u is integrable. By (3.19), (3.20), and (3.22) we have b ∞ u dx = u dx ≤ (2mn zn − yn + yn − zn−1 ) I
a
≤ ≤
∞
n=1
(2n zn − yn + yn − zn + zn − z∞ + z∞ − zn−1 )
n=1 ∞
4cr +2 n2
< ∞.
n=1
On the other hand, by (3.20)–(3.22), 2m n +1
|f (u(xi,n )) − f (u(xi−1,n ))| ≥ 2mn |f (zn ) − f (yn )|
i=1
> 2n (n2 + n)zn − yn = 4cr , and so for every n ∈ N by Proposition 2.15 we obtain that n VarI k (f ◦ u) ≥ 4cr n → ∞ Var[a,b] (f ◦ u) ≥ k=1
as n → ∞. Hence, we have obtained a contradiction.
Remark 3.56. Note that in the necessity part of the theorem we have actually proved a much stronger result, namely that if f : RM → R is such that f ◦ u ∈ BP Vloc (I) for all functions u ∈ AC(I; RM ) ⊆ ACloc (I; RM ), then f is locally Lipschitz continuous. Remark 3.57. The previous proof continues to hold if f : RM → Z, where (Z, dZ ) is a metric space. On the other hand, reasoning as in Step 1 we have that if (Y, dY ) and (Z, dZ ) are metric spaces and f : Y → Z is locally Lipschitz continuous, then f ◦ u belongs to BP Vloc (I; Z) (respectively, BP V (I; Z)) for all functions u ∈ BP Vloc (I; Y ) (respectively, BP V (I; Y )).
3.5. Chain Rule
95
Exercise 3.58. Let I ⊆ R be an interval, let f : RM → R, and let 1 ≤ p < ∞. Modify the proof of Theorem 3.55 to show that f ◦ u ∈ ACloc (I) for all functions u ∈ AC(I; RM ) with u ∈ Lp (I; RM ) if and only if f is locally Lipschitz continuous.
3.5. Chain Rule Next we discuss the validity of the chain rule for absolutely continuous functions. The next result establishes the validity of the chain rule under very weak hypotheses. Theorem 3.59 (Chain rule). Let I, J ⊆ R be two intervals, let (Y, · ) be a normed space, and let f : J → Y and u : I → J be such that f, u, and f ◦ u are differentiable L1 -a.e. in their respective domains. If f satisfies the Lusin (N ) property, then for L1 -a.e. x ∈ I, (3.24)
(f ◦ u) (x) = f (u(x))u (x),
where f (u(x))u (x) is interpreted to be zero whenever u (x) = 0 (even if f is not differentiable at u(x)). In the proof we will show that (f ◦ u) (x) = 0 and u (x) = 0 for L1 -a.e. x ∈ I such that f is not differentiable at u(x). To prove Theorem 3.59, we need an auxiliary result, which is a converse of Corollary 3.37. Lemma 3.60. Let I ⊆ R be an interval, let (Y, · ) be a normed space, and let u : I → Y . Assume that u is differentiable on a set E ⊆ I (not necessarily measurable), with H1 (u(E)) = 0. Then u (x) = 0 for L1 -a.e. x ∈ E. Proof. Let E0 := {x ∈ E : u (x) > 0}. We claim that L1 (E0 ) = 0. For every integer k ∈ N let Ek := {x ∈ E0 : u(x) − u(t) ≥ |x − t|/k for all t ∈ (x − 1/k, x + 1/k) ∩ I}. Noting that E0 = k Ek , we fix k and we let F := J ∩ Ek , where J is an interval of length less than 1/k. To prove that L1 (E0 ) = 0, it suffices to show that L1 (F ) = 0. Since H1 (u(E)) = 0 and F ⊆ E, for every ε > 0 we may find a sequence {Gn }n of subsets of Y such that Gn , diam Gn ≤ ε. u(F ) ⊆ n
n
96
3. Absolutely Continuous Functions
Let Fn := u−1 (Gn ) ∩ F . Since {Fn }n covers F , we have L1o (F ) ≤ L1o (Fn ) ≤ sup |x − t| n x,t∈Fn
n
≤
n
k sup u(x) − u(t) ≤ k
x,t∈Fn
diam Gn ≤ kε,
n
where we have used definition of Fn . It now suffices to let ε → 0+ .
Remark 3.61. In the case Y = R the previous lemma holds even if we assume in place of differentiability that u has a derivative at every x ∈ E. Corollary 3.62. Let I ⊆ R be an interval, let (Y, · ) be a normed space, and let u : I → Y , v : I → Y . Assume that there exists a set E ⊆ I such that u and v are differentiable for all x ∈ E and u(x) = v(x) for all x ∈ E. Then u (x) = v (x) for L1 -a.e. x ∈ E. Proof. Let w := u − v. Then w(E) = {0}, and so H1 (w(E)) = 0. By the previous lemma, u (x) − v (x) = w (x) = 0 for L1 -a.e. x ∈ E. We turn to the proof of Theorem 3.59. Proof of Theorem 3.59. Let G := {z ∈ J : f is not differentiable at z}, F := {x ∈ I : u or f ◦ u is not differentiable at x}. By hypothesis L1 (G) = L1 (F ) = 0. Let E := {x ∈ I ◦ \ F : u(x) ∈ G}. Since u(E) ⊆ G, we have L1 (u(E)) = 0, and since f satisfies the Lusin (N ) property, we obtain that H1 ((f ◦ u)(E)) = 0. By Lemma 3.60 applied to u and to f ◦ u, we conclude that u (x) = (f ◦ u) (x) = 0 for L1 -a.e. x ∈ E. / G, then we may apply On the other hand, if x ∈ I ◦ \ F and u(x) ∈ the standard chain rule to conclude that f ◦ u is differentiable at x with (f ◦ u) (x) = f (u(x))u (x). The next example shows the importance of the Lusin (N ) property. Example 3.63. Let u : [0, 1] → R be a strictly increasing function such that u (x) = 0 for L1 -a.e. x ∈ [0, 1] (see Theorem 1.37) and let f := u−1 . Note that f is strictly increasing, and so by Lebesgue’s differentiation theorem (see Theorem 1.18) it is differentiable for L1 -a.e. x ∈ [0, 1], despite the
3.5. Chain Rule
97
fact that u (x) = 0 for L1 -a.e. x ∈ [0, 1]. Moreover, (f ◦ u)(x) = x for all x ∈ [0, 1], and so (f ◦ u) (x) = 1 for all x ∈ [0, 1], while f (u(x))u (x) = 0 for L1 -a.e. x ∈ [0, 1], since u (x) = 0 for L1 -a.e. x ∈ [0, 1]. Corollary 3.64. Let I, J ⊆ R be two intervals, let (Y, · ) be a normed space, and let f : J → Y and u : I → J be such that f and u are differentiable L1 -a.e. in their respective domains. Suppose that u is zero at most on a set of Lebesgue measure zero. Then f ◦ u is differentiable L1 -a.e. in I and the chain rule (3.24) holds. Proof. Let G and F be as in the proof of Theorem 3.59 and let E := {x ∈ I ◦ : u(x) ∈ G}. Since u(E) ⊆ G, we have L1 (u(E)) = 0, and so by Lemma 3.60 we conclude that u (x) = 0 for L1 -a.e. x ∈ E. But then, according to our assumption that u is zero at most on a set of Lebesgue measure zero, E must have Lebesgue measure zero. If x ∈ I ◦ \ E, then f is differentiable at u(x), and so the chain rule (3.24) holds L1 -a.e. in I ◦ \ E. Corollary 3.65. Let I, J ⊆ R be two intervals, let f : J → RM and u : I → J be such that u and f ◦ u are differentiable L1 -a.e. in their respective domains. If f ∈ ACloc (J; RM ), then the chain rule (3.24) holds. Proof. In view of Theorem 3.41, we have that f is differentiable L1 -a.e. in I and satisfies the Lusin (N ) property and so all the hypotheses of Theorem 3.59 are satisfied. A less trivial consequence of Theorem 3.59 is the following result. Corollary 3.66. Let I, J ⊆ R be two intervals, let f ∈ ACloc (J; RM ), and let u : I → J be monotone. Then f ◦ u is differentiable L1 -a.e. in I and the chain rule (3.24) holds. Proof. Note that the composite f ◦u belongs to BP Vloc (I; RM ) by Exercise 2.54, and so by Corollary 2.31, f ◦ u is differentiable L1 -a.e. in I. We are now in a position to apply the previous corollary. Exercise 3.67. Let I, J ⊆ R be two intervals, let f ∈ ACloc (J), and let u : I → J be monotone and ACloc (I). Prove that f ◦ u belongs to ACloc (I). Corollary 3.68. Let I, J ⊆ R be two intervals, let f : J → R be locally Lipschitz continuous, and let u ∈ BP Vloc (I). Then f ◦ u is differentiable L1 -a.e. in I and the chain rule (3.24) holds. Proof. By Theorem 2.55, we have that f ◦ u ∈ BP Vloc (I). Hence we can apply Corollary 2.31 to conclude that f ◦ u is differentiable L1 -a.e. in I.
98
3. Absolutely Continuous Functions
Since f is locally Lipschitz continuous, it is locally absolutely continuous, and so the result follows from Corollary 3.65. To conclude this section, we observe that if f : RM → R is a Lipschitz continuous function with M ≥ 2 and if u : I → RM is absolutely continuous, then f ◦u ∈ AC(I) (see Step 1 of the proof of Theorem 3.55), but the analog of (3.24), which is M ∂f (u(x))ui (x), (f ◦ u) (x) = ∂zi
(3.25)
i=1
∂f ∂zi (u(x))ui (x)
where is interpreted to be zero whenever ui (x) = 0, may fail. This is illustrated by the next example. Example 3.69. Let M = 2, and consider the functions f (z) = f (z1 , z2 ) := max{z1 , z2 },
z ∈ R2 ,
and u(x) := (x, x) for x ∈ R. Then v(x) := (f ◦ u)(x) = x so that v (x) = 1, while the right-hand side of (3.25) is nowhere defined, since u (x) = (1, 1). The problem lies in the fact that for a Lipschitz continuous function f : RM → R, the set Σf := {z ∈ RM : f is not differentiable at z} has LM -measure zero by Rademacher’s theorem (see Theorem 9.14), but for M ≥ 2 it is easy to construct a set of LM -measure zero that contains the range of an absolutely continuous curve (see Example 3.69). Thus, to recover (3.25), stronger restrictions on the set Σf are needed. Definition 3.70. Let (Y, d) be a metric space. A set E ⊆ Y is called H1 -rectifiable if there exists a sequence of Lipschitz continuous functions un : R → Y , n ∈ N, such that un (R) = 0. H1 E \ n
A set E ⊆ Y is purely
H1 -unrectifiable
if
H (E ∩ u(R)) = 0 1
for every Lipschitz continuous function u : R → Y . Exercise 3.71. Let E ⊆ RM be purely H1 -unrectifiable. Prove that H1 (E ∩ u(I)) = 0 for every interval I ⊆ R and for all u ∈ AC(I; RM ). A useful characterization of purely H1 -unrectifiable sets is given by the following result. Theorem 3.72. Let E ⊆ RM . Then the following properties are equivalent:
3.5. Chain Rule
99
(i) E is purely H1 -unrectifiable. (ii) For every interval I ⊆ R and for all u ∈ AC(I; RM ), (3.26)
u (x) = 0
for L1 -a.e. x ∈ u−1 (E).
Proof. Step 1: Assume that E is purely H1 -unrectifiable, let I ⊆ R be an interval, and consider a function u ∈ AC(I; RM ). By the previous exercise, H1 (E ∩ u(I)) = 0. Let F := u−1 (E ∩ u(I)). By Lemma 3.60, u (x) = 0 for L1 -a.e. x ∈ F , which gives (ii). Step 2: Assume that (ii) holds. We claim that E is purely H1 -unrectifiable. Indeed, assume by contradiction that there exists a Lipschitz continuous function w : R → RM such that H1 (E ∩ w(R)) > 0.
(3.27)
Then H1 (E ∩ w([−n, n])) → H1 (E ∩ w(R)) > 0 as n → ∞. Fix n ∈ N so large that H1 (E ∩ w([−n, n])) > 0. Since w is Lipschitz continuous, H1 (w([−n, n])) < ∞. By property (ii) there exists a set G ⊂ [−n, n] with L1 (G) = 0 such that w (x) = 0 for all x ∈ [−n, n] \ G. Since w is Lipschitz continuous, w satisfies the Lusin (N ) property. Hence, H1 (w(G)) = 0. On the other hand, by Corollary 3.37, H1 (w([−n, n] \ G)) = 0. Thus, H1 (w([−n, n])) = 0, which is a contradiction. The following theorem is due to Marcus and Mizel [159]. Theorem 3.73 (Chain rule, II). Let f : RM → R be a locally Lipschitz continuous function and let I ⊆ R be an interval. Then f ◦ u ∈ ACloc (I) for every u ∈ ACloc (I; RM ). Moreover, if the set Σf is purely H1 -unrectifiable, then (3.28)
(f ◦ u) (x) =
M ∂f (u(x))ui (x) ∂zi i=1
for L1 -a.e. x ∈ I, where ui (x) = 0.
∂f ∂zi (u(x))ui (x)
is interpreted to be zero whenever
Proof. The fact that f ◦u ∈ ACloc (I) follows by Theorem 3.55. By Theorem 3.20, we have that u and f ◦ u are differentiable for L1 -a.e. x ∈ I. Thus, it suffices to prove (3.28) for L1 -a.e. x ∈ I such that f ◦ u and u are differentiable at x. If at any of these points x ∈ I, u(x) ∈ RM \ Σf , then f is differentiable at u(x), and so (3.28) follows by the classical chain rule.
100
3. Absolutely Continuous Functions
Thus, it remains to consider those x ∈ I such that x ∈ u−1 (Σf ) and f ◦ u and u are differentiable at x. By Theorem 3.72, we have that u (x) = 0
for L1 -a.e. x ∈ u−1 (Σf ).
Fix x ∈ I ◦ such that x ∈ u−1 (Σf ), u (x) = 0, and f ◦ u and u are differentiable at x. To conclude the proof, we need to show that (f ◦ u) (x) = 0. Since u is bounded in [x − δ, x + δ] ⊆ I, let L1 be the Lipschitz constant of f in the ball containing u([x − δ, x + δ]). Then, for all 0 < |h| < δ, |u(x + h) − u(x)| |(f ◦ u)(x + h) − (f ◦ u)(x)| ≤ L1 → L1 |u (x)| = 0 |h| |h| as h → 0. This concludes the proof.
Remark 3.74. Necessary and sufficient conditions for the validity of the chain rule (3.28) have been proved in [140] and [175]. If f : RM → R is only assumed to be a locally Lipschitz continuous function, then in place of (3.28) it is still possible to prove a weaker form of the chain rule. This has been established by Ambrosio and Dal Maso in [9].
3.6. Change of Variables As a corollary of Theorem 3.59 we have the following change of variables formula. Theorem 3.75 (Change of variables). Let g : [p, q] → R be an integrable function and let u : [a, b] → [p, q] be differentiable L1 -a.e. in [a, b]. Then (g ◦ u)u is integrable and the change of variables β u(β) g(t) dt = g(u(x))u (x) dx (3.29) u(α)
α
holds for all α, β ∈ [a, b] if and only if the function f ◦u belongs to AC([a, b]), where z f (z) := g(t) dt, z ∈ [p, q]. p
Proof. If f ◦ u ∈ AC([a, b]), then since f is absolutely continuous (see Theorem 3.16), we can apply Corollary 3.65 to obtain the chain rule formula (3.30)
(f ◦ u) (x) = g(u(x))u (x)
for L1 -a.e. x ∈ [a, b]. Since f ◦ u ∈ AC([a, b]), it follows from Corollary 3.10 and (3.30) that (g ◦ u)u is integrable, and by the fundamental theorem of
3.6. Change of Variables
101
calculus (see Theorem 3.20), for all α, β ∈ [a, b], u(β) g(t) dt = (f ◦ u)(β) − (f ◦ u)(α) u(α)
β
=
(f ◦ u) (x) dx =
α
β
g(u(x))u (x) dx.
α
Conversely, if (g ◦ u)u is integrable and the identity β g(u(x))u (x) dx (f ◦ u)(β) − (f ◦ u)(α) = α
holds for all α, β ∈ I, then, since the right-hand side is absolutely continuous by Theorem 3.16, it follows that f ◦ u is absolutely continuous. Remark 3.76. The previous proof shows, in particular, that the function x ∈ I → g(u(x))u (x) is measurable (see also Exercise 1.29), since it is the derivative of the function f ◦ u, but this does not imply that the function x ∈ I → g(u(x)) is measurable (see the next exercise). Exercise 3.77. Let I, J ⊆ R be intervals. Prove that there exist a continuous increasing function u : I → J and a Lebesgue measurable function g : J → R such that g ◦ u : I → R is not Lebesgue measurable. Hint: See Exercise 1.33. Corollary 3.78. Assume that g : [p, q] → R is an integrable function and that u : [a, b] → [p, q] is monotone and absolutely continuous. Then (g ◦ u)u is integrable and the change of variables formula (3.29) holds. Exercise 3.79. Prove that under the hypotheses of the previous corollary the function f ◦ u is absolutely continuous and then prove the corollary. Corollary 3.80. Assume that g : [p, q] → R is a measurable, bounded function and that u : [a, b] → [p, q] is absolutely continuous. Then (g ◦ u)u is integrable and the change of variables formula (3.29) holds. Exercise 3.81. Prove that under the hypotheses of the previous corollary the function f ◦ u is absolutely continuous and then prove the corollary. Corollary 3.82. Assume that g : [p, q] → R is an integrable function, that u : [a, b] → [p, q] is absolutely continuous, and that (g ◦ u)u is integrable. Then the change of variables formula (3.29) holds. Proof. Let
⎧ ⎨ gn (z) :=
n if g(z) > n, g(z) if − n ≤ g(z) ≤ n, ⎩ −n if g(z) < −n.
102
3. Absolutely Continuous Functions
Applying the Lebesgue dominated convergence theorem and the previous corollary, we obtain u(β) u(β) g(z) dz = lim gn (z) dz n→∞ u(α) β
u(α)
= lim
n→∞ α
β
gn (u(x))u (x) dx =
g(u(x))u (x) dx.
α
Theorem 3.83. Assume that f : [p, q] → R and u : [a, b] → [p, q] are absolutely continuous. Then f ◦u belongs to AC([a, b]) if and only if (f ◦u)u is integrable, where f (u(x))u (x) is interpreted to be zero whenever u (x) = 0. Proof. If f ◦ u belongs to AC([a, b]), then by Proposition 3.9, f ◦ u is differentiable L1 -a.e. in [a, b] and so by Theorem 3.59, (3.24) holds. In turn, (f ◦ u)u is integrable again by Proposition 3.9. Conversely, assume that (f ◦ u)u is integrable. By replacing f with f − f (p), without loss of generality, we may assume that f (p) = 0. Since f is absolutely continuous, by Theorem 3.20, z f (t) dt f (z) = p
for all z ∈ [p, q]. Extend f to be zero at points where f is not differentiable and let g be the resulting function. Then we are in a position to apply the previous corollary to obtain that the change of variables formula (3.29) holds. In turn, by Theorem 3.75, f ◦ u belongs to AC([a, b]). To extend the previous results to arbitrary intervals, we consider two functions g : J → R and u : I → J and we assume that there exist in R the limits (3.31)
lim
x→(inf I)+
u(x) = ,
lim
x→(sup I)−
u(x) = L.
In this case, the analog of (3.29) becomes sup I L g(z) dz = g(u(x))u (x) dx. (3.32)
inf I
Indeed, we have the following result. Theorem 3.84. Let I, J ⊆ R be two intervals, let g : J → R be an integrable function, and let u : I → J be differentiable L1 -a.e. in I and such that there exist in R the limits (3.31). Then (g ◦u)u is integrable on I and the changes
3.7. Singular Functions
103
of variables (3.29) and (3.32) hold for all [α, β] ⊆ I if and only if the function f ◦ u belongs to AC(I) ∩ BP V (I), where z f (z) := g(t) dt, z ∈ J. inf J
Proof. Assume that f ◦ u ∈ AC(I) ∩ BP V (I). Then, we can proceed as in the proof of Theorem 3.75 to show that (3.29) holds for all [α, β] ⊆ I. Since f ◦ u ∈ BP V (I), by Corollary 2.31 and (3.30) its derivative (g ◦ u)u is integrable on I. Hence, by taking limits as α and β approach the endpoints of I and using (3.31) and the Lebesgue dominated convergence theorem, we conclude that (3.32) holds. Conversely, if (g ◦ u)u is integrable and (3.29) and (3.32) hold for all [α, β] ⊆ I, then β g(u(x))u (x) dx (f ◦ u)(β) − (f ◦ u)(α) = α
for all α, β ∈ I. As in Theorem 3.75 we deduce that f ◦ u ∈ ACloc (I). In turn, by Corollary 3.33, it follows that f ◦ u ∈ AC(I) ∩ BP V (I). Corollary 3.85. Let I, J ⊆ R be two intervals, let f ∈ ACloc (J; R), and let u : I → J be ACloc (I). Then f ◦ u belongs to ACloc (I; R) if and only if (f ◦ u)u is locally integrable, where f (u(x))u (x) is interpreted to be zero whenever u (x) = 0. Exercise 3.86. Prove the previous corollary.
3.7. Singular Functions In this section we prove that every function of bounded pointwise variation may be decomposed into the sum of an absolutely continuous function and a singular function. Definition 3.87. Let I ⊆ R be an interval and let (Y, · ) be a normed space. A nonconstant function u : I → Y is said to be singular if it is differentiable at L1 -a.e. x ∈ I with u (x) = 0 for L1 -a.e. x ∈ I. The jump function uJ of a function u ∈ BP Vloc (I) is an example of a singular function. Another example is the Cantor function or the function given in Theorem 1.37. The following theorem provides a characterization of singular functions. Theorem 3.88 (Singular functions). Let I ⊆ R be an interval, let (Y, · ) be a normed space, and let u : I → Y be a nonconstant function such that u (x) exists for L1 -a.e. x ∈ I. Then u is a singular function if and only if there exists a Lebesgue measurable set E ⊆ I such that L1 (I \ E) = 0 and H1 (u(E)) = 0.
104
3. Absolutely Continuous Functions
Proof. Assume that u is singular and let E := {x ∈ I : u (x) = 0}. Then L1 (I \ E) = 0. By Corollary 3.37 we have that H1 (u(E)) = 0. Conversely, assume that there exists a Lebesgue measurable set E ⊆ I such that L1 (I \ E) = 0 and H1 (u(E)) = 0. Then by Lemma 3.60, u (x) = 0 for L1 -a.e. x ∈ E. Since L1 (I \ E) = 0, we have that u (x) = 0 for L1 -a.e. x ∈ I. As an application of Theorem 3.16 we obtain the standard decomposition of a monotone function into an absolutely continuous monotone function and a singular monotone function. Theorem 3.89. Let I ⊆ R be an interval and let u : I → R be an increasing function. Then u may be decomposed as the sum of three increasing functions u = uAC + uC + uJ ,
(3.33)
where uAC ∈ ACloc (I), uC is continuous and singular, and uJ is the jump function of u. Proof. Define v := u − uJ . By Exercises 1.6 and 1.36 we have that v is increasing, continuous, and v (x) = u (x) for L1 -a.e. x ∈ I. Fix x0 ∈ I and for every x ∈ I define x x (3.34) uAC (x) := v (t) dt = u (t) dt, uC (x) := v(x) − uAC (x). x0
x0
Then the decomposition (3.33) holds. Moreover, by Theorem 3.16, we have uC (x) = 0 for L1 -a.e. x ∈ I. It remains to show that uC is increasing. Let s, x ∈ I, with s < x. By Corollary 1.25, x v (t) dt ≤ v(x) − v(s), uAC (x) − uAC (s) = s
and so uC (x) ≥ uC (s) by (3.34)2 .
The function uC is called the Cantor part of u. Since every function with bounded pointwise variation may be written as a difference of two increasing functions, an analogous result holds for functions of bounded pointwise variation. Corollary 3.90. Let I ⊆ R be an interval and let u ∈ BP Vloc (I; RM ). Then u may be decomposed as the sum of three functions in BP Vloc (I; RM ), i.e., (3.35)
u = uAC + uC + uJ ,
where uAC ∈ ACloc (I; RM ), uC is continuous and singular, and (3.36) (u (t) − u− (t)) + u(x) − u− (x) if x ≥ t0 , t∈I, t0 ≤t 0 be as in Definition 3.1 for the absolutely continuous function uAC . Using the definition of differentiability, for every x ∈ E we may find hx > 0 such that x + hx < β and (3.42)
uC (x + h) − uC (x) ≤ εh/(β − α)
for all 0 < h ≤ hx . Let F be the family of all intervals (x, x + h), x ∈ E, 0 < h ≤ hx . Since, L1 (E) = β − α, by Lemma 1.20 there exist disjoint
106
3. Absolutely Continuous Functions
intervals (ai , bi ) ∈ F , i = 1, . . . , n such that L
1
(3.43)
n
(ai , bi ) ≥ β − α − δ.
i=1
By relabeling the points, if necessary, we may assume that bi−1 ≤ ai for all i = 2, . . . , n. Set b0 := α, an+1 := β and consider the partition P = {b0 , a1 , b1 , . . . , an , bn , an+1 }. By Proposition 2.15, (3.44)
Var[α,β] u =
n
Var[ai ,bi ] u +
n+1
i=1
Var[bi−1 ,ai ] u.
i=1
By Corollary 2.31 and the fact that uAC = u L1 -a.e. in I, n n bi (3.45) Var[ai ,bi ] u ≥ u dx i=1
=
i=1 ai n bi i=1
ai
uAC dx =
n
Var[ai ,bi ] uAC .
i=1
Using (2.17), we obtain that Var[bi−1 ,ai ] u ≥ Var[bi−1 ,ai ] uC − Var[bi−1 ,ai ] uAC ≥ uC (ai ) − uC (bi−1 ) − Var[bi−1 ,ai ] uAC , which, together with (3.44) and (3.45), yields (3.46)
Var[α,β] u ≥
n
Var[ai ,bi ] uAC +
i=1 n+1
−
n+1
uC (ai ) − uC (bi−1 )
i=1
Var[bi−1 ,ai ] uAC .
i=1
By (3.43) we have n+1
n
i=1
i=1
(ai − bi−1 ) = β − α −
(bi − ai ) ≤ δ.
Hence (see the proof of Theorem 3.29), (3.47)
n+1 i=1
Var[bi−1 ,ai ] uAC ≤ ε.
3.7. Singular Functions
107
On the other hand, by (3.42), n+1
uC (ai ) − uC (bi−1 ) ≥ uC (β) − uC (α) −
i=1
n
uC (bi ) − uC (ai )
i=1
ε (bi − ai ) β−α n
≥ uC (β) − uC (α) −
(3.48)
i=1
≥ uC (β) − uC (α) − ε. Combining (3.46), (3.47), and (3.48) and using Proposition 2.15 for uAC , we obtain Var[α,β] u ≥ Var[α,β] uAC + uC (β) − uC (α) − 3ε. By letting ε → 0+ , we obtain (3.41). Step 2: Assume that u is continuous. Fix an interval [a, b] ⊆ I and consider a partition P of [a, b], with a = x0 < · · · < xm = b. Applying (3.41) in each interval [xi−1 , xi ] and using Proposition 2.15 yields
Var[a,b] u = ≥
m i=1 m
Var[xi−1 ,xi ] u Var[xi−1 ,xi ] uAC +
i=1
= Var[a,b] uAC +
m
uC (xi ) − uC (xi−1 )
i=1 m
uC (xi ) − uC (xi−1 ).
i=1
Taking the supremum over all partitions of [a, b] gives Var[a,b] u ≥ Var[a,b] uAC + Var[a,b] uC . Thus, we have proved that (3.40) holds if u is continuous. Step 3: Assume that u has a finite number of discontinuity points in (a, b), say, a < x1 < · · · < x < b. Let x0 := a, x+1 := b and fix δ > 0 so small that 0 < δ < min{xi − xi−1 : i = 1, . . . , + 1}.
108
3. Absolutely Continuous Functions
Using Proposition 2.15, we have
Var[a,b] u =
+1
Var[xi−1 +δ,xi −δ] u +
i=1
≥
+1
+1
Var[xi −δ,xi +δ]∩[a,b] u
i=0
Var[xi−1 +δ,xi −δ] u
i=1
+
(u(xi + δ) − u(xi ) + u(xi ) − u(xi − δ))
i=1
+ u(a + δ) − u(a) + u(b) − u(b − δ) +1 (Var[xi−1 +δ,xi −δ] uAC + Var[xi−1 +δ,xi −δ] uC ) ≥ i=1
+
(u(xi + δ) − u(xi ) + u(xi ) − u(xi − δ))
i=1
+ u(a + δ) − u(a) + u(b) − u(b − δ), where we have used the previous step together with the fact that in each interval [xi−1 +δ, xi −δ] the jump function uJ is constant, and so the variation of u in those intervals reduces to the one of uC + uAC . Letting δ → 0+ in the previous inequality and using Proposition 2.15 and Exercise 2.22 yields
Var[a,b] u ≥
+1
(Var[xi−1 ,xi ] uAC + Var[xi−1 ,xi ] uC )
i=1
+
(u+ (xi ) − u(xi ) + u(xi ) − u− (xi ))
i=1
+ u+ (a) − u(a) + u(b) − u− (b) = Var[a,b] uAC + Var[a,b] uC + Var[a,b] uJ . Step 4: Finally, if u has an infinite number of discontinuity points in (a, b), say {xi }i , consider the saltus function uJ,k corresponding to the points xi with i < k. For each k ∈ N and x ∈ [a, b] define uk (x) := uAC (x) + uC (x) + uJ,k .
3.7. Singular Functions
109
Since the discontinuity points of uk in (a, b) are {x1 , . . . , xk }, by the previous step Var[a,b] uk = Var[a,b] uAC + Var[a,b] uC +
k (u+ (xi ) − u(xi ) + u(xi ) − u− (xi )) i=1
+ u+ (a) − u(a) + u(b) − u− (b). Since Var[a,b] uk = Var[a,b] (uk − u + u) ≤ Var[a,b] (uk − u) + Var[a,b] u ≤
∞
(u+ (xi ) − u(xi ) + u(xi ) − u− (xi )) + Var[a,b] u,
i=k
by the previous equality we have Var[a,b] uAC + Var[a,b] uC +
k
(u+ (xi ) − u(xi ) + u(xi ) − u− (xi ))
i=1
+ u+ (a) − u(a) + u(b) − u− (b) ≤
∞ (u+ (xi ) − u(xi ) + u(xi ) − u− (xi )) + Var[a,b] u. i=k
Letting k → ∞, we obtain (3.40). Step 5: If, in addition, u ∈ BP V (I; RM ), taking an and bn as in Step 2 of the proof of Theorem 3.25, we apply (3.37) in [an , bn ] and use Proposition 2.15 and the Lebesgue monotone convergence theorem. Reasoning as in Steps 3–5 of Corollary 3.90 we have the following result. Corollary 3.91. Let I ⊆ R be an interval, let (Y, · ) be a normed space and let u : I → Y be a jump function. Then for every [a, b] ⊆ I, (u+ (x) − u(x) + u(x) − u− (x)) Var[a,b] u = x∈(a,b)
+ u(a) − u+ (a) + u(b) − u− (b). Remark 3.92. We refer to the paper [41] of Cater for necessary and sufficient conditions on two continuous functions u : [a, b] → R and v : [a, b] → R with finite pointwise variation for the equality Var[a,b] (u + v) = Var[a,b] u + Var[a,b] v to hold.
Chapter 4
Decreasing Rearrangement Undergradese, IV: “Can I get an extension?” Translation: “Can you re-arrange your life around mine?” — Jorge Cham, www.phdcomics.com
In this chapter we introduce the notion of decreasing rearrangement u∗ of a Lebesgue measurable function u : E → [0, ∞], where E ⊆ R is a Lebesgue measurable set. Roughly speaking, u∗ is a function defined on the interval [0, L1 (E)), which is obtained from u by rearranging the superlevel sets {x ∈ E : u(x) > s} in such a way that u∗ becomes a decreasing function. For more information on this topics we refer to the monographs of Kawohl [130] and Kesavan [131]. For simplicity, in this chapter we deal only with nonnegative functions, although the theory can also be carried out for realvalued functions. Except for the last section, most of the results of this chapter continue to hold if we consider a measure space (X, M, μ) and replace the Lebesgue measurable set E ⊆ R and the Lebesgue measure L1 by a set E ∈ M of X and by μ, respectively. We leave the details as an exercise.
4.1. Definition and First Properties Let E ⊆ R be a Lebesgue measurable set and let u : E → [0, ∞] be a Lebesgue measurable function. The distribution function of u is the function
u : [0, ∞) → [0, L1 (E)], defined by (4.1)
u (s) := L1 ({x ∈ E : u(x) > s}),
s ≥ 0.
The function u is said to vanish at infinity if it is Lebesgue measurable and
u (s) < ∞ for every s > 0. Note that for a function u vanishing at infinity 111
112
4. Decreasing Rearrangement
the value u (0) could be infinite. For example, if E has infinite measure and u is everywhere positive, then u (0) = ∞. Moreover, for all s ≥ ess sup u,
(4.2)
u (s) = 0
(4.3)
u (s) = L (E) for all s < ess inf u,
E
1
E
where we recall that the essential supremum and essential infimum of u are defined as (4.4)
ess sup u := inf{s : u(x) ≤ s for L1 -a.e. x ∈ E}, E
(4.5)
ess inf u := sup{s : s ≤ u(x) for L1 -a.e. x ∈ E}. E
Note that if E has infinite measure and u : E → [0, ∞] vanishes at infinity, then ess inf E u = 0. Some important properties of u are summarized in the next proposition. Proposition 4.1. Let E ⊆ R be a Lebesgue measurable set and let u, v, un : E → [0, ∞], n ∈ N, be Lebesgue measurable functions. Then the following properties hold: (i) The function u : [0, ∞) → [0, L1 (E)] is decreasing and right continuous. (ii) If u vanishes at infinity, then lim u (s) = L1 ({x ∈ E : u(x) = ∞})
s→∞
and u is continuous at s > 0 if and only if L1 ({x ∈ E : u(x) = s}) = 0. (iii) If u(x) ≤ v(x) for L1 -a.e. x ∈ E, then u ≤ v . In particular, if u(x) = v(x) for L1 -a.e. x ∈ E, then u = v . (iv) If un (x) u(x) for L1 -a.e. x ∈ E, then un u . Proof. In what follows, for s ≥ 0 we set Es := {x ∈ E : u(x) > s}.
(4.6)
(i) If 0 ≤ s ≤ r, then Er ⊆ Es , and so u (r) ≤ u (s), which shows that u is decreasing. Since u is decreasing, to prove that u is right continuous at s ≥ 0 it is enough to consider a decreasing sequence sn → s+ . Then Esn ⊆ Esn+1 and ∞ n=1 Esn = Es , and so by Proposition B.9(i) we have that ∞ Esn = u (s). ( u )+ (s) = lim u (sn ) = lim L1 (Esn ) = L1 n→∞
n→∞
n=1
4.1. Definition and First Properties
113
(ii) Assume that u vanishes at infinity. Since u is decreasing, there exists lim u (s) = inf u (s),
s→∞
s>0
and thus, to evaluate it, it suffices to consider the sequence s = n → ∞. Then En ⊇ En+1 and ∞
En = {x ∈ E : u(x) = ∞} =: E∞ .
n=1
Since L1 (En ) < ∞, it follows by Proposition B.9(ii) that lim u (sn ) = lim L1 (En ) = L1 (E∞ ).
n→∞
n→∞
Similarly, if s > 0, consider an increasing sequence {sn }n such that sn → s− . Then Esn ⊇ Esn+1 and ∞
Esn = {x ∈ E : u(x) ≥ s}.
n=1
Since (4.7)
L1 (Esn )
< ∞, we conclude as before that
( u )− (s) = lim u (sn ) = L1 ({x ∈ E : u(x) ≥ s}) n→∞
= u (s) + L1 ({x ∈ E : u(x) = s}). Hence, u is continuous at s if and only if L1 ({x ∈ E : u(x) = s}) = 0. (iii) For every s ≥ 0, {x ∈ E : u(x) > s} ⊆ {x ∈ E : v(x) > s} ∪ F, where L1 (F ) = 0, and so u (s) ≤ v (s). (iv) In view of part (iii), by modifying each un on a set of measure zero, we can assume that un (x) u(x) for all x ∈ E. For every s ≥ 0 and n ∈ N set Es,n := {x ∈ E : un (x) > s}. Since un ≤ un+1 , we have that Es,n+1 for all n ∈ N, and since un (x) → u(x) for all x ∈ E, it follows Es,n ⊆ ∞ that n=1 Es,n = Es . Hence, again by Proposition B.9(i), (4.8)
lim un (s) = lim L1 (Es,n ) = L1 (Es ) = u (s).
n→∞
n→∞
Exercise 4.2. Give an example of a function u not vanishing at infinity for which u is not left continuous. Let E ⊆ R be a Lebesgue measurable set and let u : E → [0, ∞] be a Lebesgue measurable function. Since by the previous proposition the distribution function u : [0, ∞) → [0, ∞] is decreasing, it admits a left
114
4. Decreasing Rearrangement
inverse u∗ : [0, ∞) → [0, ∞] (see Theorem 1.8 and Remark 1.11), called the decreasing rearrangement of u. Precisely, for t ≥ 0 we set u∗ (t) := inf{s ∈ [0, ∞) : u (s) ≤ t}.
(4.9)
Note that if u vanishes at infinity, then by the previous proposition u (s) → t∞ as s → ∞, where t∞ := L1 ({x ∈ E : u(x) = ∞}), and so for t > t∞ the set {s ∈ [0, ∞) : u (s) ≤ t} is nonempty. Thus, u∗ (t) < ∞ for all t > t∞ . Observe that if L1 (E) < ∞, then it follows from the definition of u∗ that u∗ (t) = 0
(4.10)
for all t ≥ L1 (E).
In what follows, for every Lebesgue measurable set F ⊆ R we define F ∗ := [0, L1 (F )).
(4.11)
In particular, if L1 (F ) = ∞, then F ∗ = [0, ∞). In view of Proposition 4.1 we have the following result. Proposition 4.3. Let E ⊆ R be a Lebesgue measurable set and let u : E → [0, ∞] be a Lebesgue measurable function. Then the following properties hold: (i) The function u∗ : [0, ∞) → [0, ∞] is decreasing and right continuous, u∗ (0) = ess sup u, E
and u∗ (t) ≥ ess inf u E
for all t < L1 (E).
(ii) Assume that u vanishes at infinity and let t∞ := L1 ({x ∈ E : u(x) = ∞}). Then u∗ jumps at some point t0 ∈ (t∞ , L1 (E)) if and only if u (s) ≡ t0 for all s in some interval (s1 , s2 ) ⊂ [0, ∞), with s1 < s2 . Moreover, u∗ ( u (s)) ≤ s
for every s > 0,
with strict inequality holding if and only if u is constant in some interval [s , s] ⊂ [0, ∞), with s < s, while u∗ (t) ≡ s0 for all t in some interval (t1 , t2 ) ⊆ J, with t1 < t2 , and for some s0 > 0 if and only if L1 ({x ∈ E : u(x) = s0 }) > 0 and (t1 , t2 ) ⊆ ( u (s0 ), ( u )− (s0 )). (iii) For all s, t ≥ 0, u∗ (t) > s if and only if u (s) > t.
4.1. Definition and First Properties
115
(iv) The functions u and u∗ are equi-measurable, that is, for all s ≥ 0, L1 ({t ∈ E ∗ : u∗ (t) > s}) = L1 ({x ∈ E : u(x) > s}). In particular, if u vanishes at infinity, then so does u∗ . (v) L1 ({t ∈ E ∗ : u∗ (t) = 0}) ≤ L1 ({x ∈ E : u(x) = 0}) with equality holding if and only if either L1 ({x ∈ E : u(x) > 0}) < ∞ or L ({x ∈ E : u(x) > 0}) = ∞ and L1 ({x ∈ E : u(x) = 0}) = 0. 1
If L1 ({x ∈ E : u(x) > 0}) = ∞, then {t ∈ E ∗ : u∗ (t) = 0} is empty. Proof. (i) If 0 ≤ t1 ≤ t2 , then {s ∈ [0, ∞) : u (s) ≤ t1 } ⊆ {s ∈ [0, ∞) : u (s) ≤ t2 }, and so u∗ (t1 ) = inf{s ∈ [0, ∞) : u (s) ≤ t1 } ≥ inf{s ∈ [0, ∞) : u (s) ≤ t2 } = u∗ (t2 ). Thus, u∗ is decreasing. To prove that u∗ is right continuous, fix t0 ≥ 0. If u∗ (t0 ) = 0, then, since is decreasing, we have that u∗ (t) = 0 for all t ≥ t0 and there is nothing to prove. Thus, assume that u∗ (t0 ) > 0 and fix 0 < s0 < u∗ (t0 ). It follows from (4.9) that u (s0 ) > t0 . Hence, we can find δ > 0 such that u (s0 ) > t0 + δ. Using (4.9) once more, we have that u∗ (t) ≥ s0 for all t ∈ [t0 , t0 + δ], and so u∗
s0 ≤ u∗+ (t0 ) ≤ u∗ (t0 ). Letting s0 → u∗ (t0 ) proves that u∗ is right continuous. Note that if u were finite, then we could have applied Remark 1.11. Next we claim that (4.12)
u∗ (0) = u∗+ (0) = ess sup u. E
We begin by showing that (4.13)
0 ≤ u∗ (t) ≤ ess sup u E
for all t ≥ 0.
If ess supE u = ∞, then there is nothing to prove. Thus, assume that ess supE u < ∞. Then u (ess supE u) = 0 by (4.2). Hence, (4.13) holds by (4.9). It follows from (4.13) that if ess supE u = 0, then u∗ = 0, and so (4.12) follows. Thus, assume that ess supE u > 0 and fix any 0 < s0 < ess supE u. By the definition of essential supremum (see (4.4)) there exists a Lebesgue
116
4. Decreasing Rearrangement
measurable subset F ⊆ E, with t0 := L1 (F ) > 0, such that u(x) > s0 for all x ∈ F . Hence u (s0 ) ≥ t0 . In turn, by (4.9), for all 0 ≤ t < t0 we have that u∗ (t) ≥ s0 , which shows that s0 ≤ u∗+ (0) ≤ u∗ (0) ≤ ess sup u. E
Letting s0 → ess supE u, we conclude that (4.12) holds. On the other hand, if ess inf E u > 0, then u (s) = L1 (E) for all s < ess inf E u by (4.3), and so if t < L1 (E), then u∗ (t) ≥ ess inf E u by (4.9). (ii) The first two statements follow from Remark 1.11 and the last from Remark 1.11 and Proposition 4.1. (iii) Assume that u∗ (t) > s for some s, t ≥ 0. Then u (s) > t by (4.9). Conversely, assume that u (s) > t. Since u is right continuous, we have that u (r) > t for all r ∈ [s, s + δ] for some δ > 0. Hence, u∗ (t) > s by (4.9). (iv) By part (iii) for s ≥ 0, we have that u∗ (t) > s for t ≥ 0 if and only if u (s) > t, and so {t ≥ 0 : u∗ (t) > s} = {t ≥ 0 : u (s) > t} = [0, u (s)). Hence, L1 ({t ≥ 0 : u∗ (t) > s}) = u (s) = L1 ({x ∈ E : u(x) > s}) by the definition of u . The result now follows from (4.10) and (4.11). (v) If L1 ({x ∈ E : u(x) > 0}) < ∞, then by part (iv) and (4.11), L1 ({x ∈ E : u(x) = 0}) = L1 (E) − L1 ({x ∈ E : u(x) > 0}) = L1 (E ∗ ) − L1 ({s ∈ E ∗ : u∗ (s) > 0}) = L1 ({s ∈ E ∗ : u∗ (s) = 0}). On the other hand, if L1 ({x ∈ E : u(x) > 0}) = ∞, then by part (iv), L1 ({t ∈ E ∗ : u∗ (t) > 0}) = L1 ({x ∈ E : u(x) > 0}) = ∞. Since u∗ is decreasing, it follows that u∗ (t) > 0 for all t ≥ 0. Hence, the set {t ∈ E ∗ : u∗ (t) = 0} is empty. Remark 4.4. Note that parts (iv) and (v) of the previous proposition imply, in particular, that (4.10) can be strengthened as follows. Let E ⊆ R be a Lebesgue measurable set and let u : E → [0, ∞] be a Lebesgue measurable function. Recall that
u (0) = L1 ({x ∈ E : u(x) > 0}). If u (0) = ∞, then u∗ (t) > 0 for all t ≥ 0, while if u (0) < ∞, then u∗ (t) > 0 for all t ∈ [0, u (0)) and u∗ (t) = 0 for all t ≥ u (0).
4.1. Definition and First Properties
117
Exercise 4.5. What happens to part (ii) of the previous proposition if we remove the assumption that u vanishes at infinity? Corollary 4.6. Let E ⊆ R be a Lebesgue measurable set and let u : E → [0, ∞) be a function vanishing at infinity. Then (4.14)
L1 ({x ∈ E : u(x) ∈ B}) = L1 ({t ∈ E ∗ : u∗ (t) ∈ B})
for all Borel sets B ⊆ (0, ∞). Moreover, if either L1 ({x ∈ E : u(x) > 0}) < ∞ or L1 ({x ∈ E : u(x) > 0}) = ∞ and L1 ({x ∈ E : u(x) = 0}) = 0, then (4.14) holds for all Borel sets B ⊆ [0, ∞). Proof. Define the measures μ(B) := L1 ({x ∈ E : u(x) ∈ B}), ν(B) := L1 ({t ∈ E ∗ : u∗ (t) ∈ B}), where B belongs to the Borel σ-algebra B((0, ∞)). Since {x ∈ E : u(x) ∈ (0, ∞)} =
∞
{x ∈ E : u(x) > n1 },
n=1 ∞
{t ∈ E ∗ : u∗ (t) ∈ (0, ∞)} =
{t ∈ E ∗ : u∗ (t) > n1 },
n=1
u∗
and both u and vanish at infinity (see Proposition 4.3(iv)), it follows that μ and ν are σ-finite. By Proposition 4.3(iv), μ and ν coincide on all intervals (a, ∞), where a > 0. Since the family of these intervals generates the Borel σ-algebra B((0, ∞)), it follows by Corollary B.16 and Remark B.17 that μ and ν coincide on B((0, ∞)). The last part of the statement follows from Proposition 4.3(v).
Corollary 4.7. Let E ⊆ R be a Lebesgue measurable set and let u ∈ L∞ (E) be nonnegative. Then u∗ belongs to L∞ (E ∗ ) and u∗ L∞ (E ∗ ) = uL∞ (E) . Proof. By Proposition 4.3 we have that u∗ L∞ (E ∗ ) = u∗ (0) = uL∞ (E) < ∞. Exercise 4.8. Let E ⊆ R be a Lebesgue measurable set, let u : E → [0, ∞) be a function vanishing at infinity, and let Ψ : [0, ∞) → [0, ∞) be an increasing function. Prove that (Ψ◦u)∗ (t) = (Ψ◦u∗ )(t) for all but countably many t ∈ E ∗ .
118
4. Decreasing Rearrangement
Exercise 4.9. Let a > 0 and b ∈ R and consider the function u : R → R, defined by 1 if x ≤ b, (b−x+a)2 u(x) := 0 otherwise. Find u∗ . Corollary 4.10. Let I ⊆ R be an interval and let u : I → [0, ∞) be a continuous function vanishing at infinity. Then u∗ is continuous in I ∗ . Proof. By part (i) of Proposition 4.3, u∗ jumps at some point t0 ∈ (0, L1 (I)) if and only if u (s) ≡ t0 for all s in some interval (s1 , s2 ) ⊂ [0, ∞), with s1 < s2 . In view of (4.2) we have that ess inf I u ≤ s1 < s2 ≤ ess supI u. By replacing the interval (s1 , s2 ) with a smaller one, we can assume that
u (s1 ) = u (s2 ) = t0 . Hence, L1 ({y ∈ I : s1 < u(y) ≤ s2 }) = 0. On the other hand, since u is continuous, we have that u(I) is an interval containing (ess inf I u, ess supI u). In particular, u(I) contains (s1 , s2 ), which is a contradiction. The next proposition shows that decreasing rearrangement preserves inequalities between functions and behaves well with respect to monotone convergence. Proposition 4.11. Let E ⊆ R be a Lebesgue measurable set and let u, v, un : E → [0, ∞], n ∈ N, be Lebesgue measurable functions. (i) If u(x) ≤ v(x) for L1 -a.e. x ∈ E, then u∗ ≤ v ∗ . In particular, if u(x) = v(x) for L1 -a.e. x ∈ E, then u∗ = v ∗ . (ii) If un (x) u(x) for L1 -a.e. x ∈ E, then u∗n u∗ . Proof. (i) By Proposition 4.1(iii), for every t ≥ 0, {s ∈ [0, ∞) : v (s) ≤ t} ⊆ {s ∈ [0, ∞) : u (s) ≤ t} and hence
u∗ (t)
≤ v ∗ (t).
(ii) By part (i) we have that u∗n ≤ u∗n+1 ≤ u∗ for all n ∈ N. Hence, (4.15)
lim u∗ (t) ≤ u∗ (t) n→∞ n u∗ (t) > 0 for some t ≥
0 and fix s ∈ [0, u∗ (t)). By for all t ≥ 0. Suppose that Proposition 4.3(iii), u (s) > t. Hence, by Proposition 4.1(iv) there exists k ∈ N such that uk (s) > t. Again by Proposition 4.3(iii) we have that u∗k (t) > s and since u∗n ≤ u∗n+1 ≤ u∗ for all n ∈ N, it follows that lim u∗n (t) > s.
n→∞
4.1. Definition and First Properties
119
Letting s u∗ (t), we conclude that lim u∗n (t) ≥ u∗ (t), which, together n→∞
with (4.15), implies that lim u∗n (t) = u∗ (t)
n→∞
for all t ≥ 0.
Remark 4.12. It follows from part (i) of Proposition 4.11 that modifying a nonnegative Lebesgue measurable function on a set of measure zero does not change its decreasing rearrangement. In what follows, we will use this fact without further mention. For the same reason, we can replace the set E with a measurable subset F ⊆ E such that L1 (E \ F ) = 0. The next exercise shows that the decreasing rearrangement of a simple function is still a simple function. Exercise 4.13. Let E ⊆ R be a Lebesgue measurable set and let u : E → [0, ∞) be a simple function vanishing at infinity (see Figure 1), that is, u=
k
ci χEi ,
i=0
where c0 > · · · > ck > 0, Ei ⊆ E are pairwise disjoint Lebesgue measurable sets with 0 < L1 (Ei ) < ∞, i = 1, . . . , k. For every i = 1, . . . , k set Fi :=
i
Ej .
j=0
(i) Let ri := L1 (Fi ), i = 1, . . . , k. Prove that " k ! i 1 L (Ej ) χ[ci+1 ,ci ) ,
u = i=0 j=0
where ck+1 := 0, and that ∗
u =
k
ci χ[ri−1 ,ri ) ,
i=0
where r−1 := 0. (ii) Let {p0 , . . . , pk }, {q0 , . . . , qk } ⊂ R and prove that k i=0
pi qi =
k
Pi Q i ,
i=0
where Pi := pi − pi+1 for i = 0, . . . , k − 1, Pk := pk , and Qi = i j=0 qj for i = 0, . . . , k.
120
4. Decreasing Rearrangement
c0 c1 c2 E1
E0
x
E2
c0 c1 c2 r0
r1
r2
t
Figure 1. The graphs of u and u∗ .
(iii) Prove that (4.16)
u=
k
αi χFi ,
i=0
where αi := ci − ci+1 , i = 1, . . . , k, and that u∗ =
k
αi χFi∗ .
i=0
Using the previous exercise, it is possible to give a simple proof of a classical result of Hardy and Littlewood. Theorem 4.14 (Hardy–Littlewood’s inequality). Let E ⊆ R be a Lebesgue measurable set and let u, v : E → [0, ∞) be Lebesgue measurable functions. Then u(x)v(x) dx ≤ u∗ (t)v ∗ (t) dt. E
E∗
4.1. Definition and First Properties
121
Proof. Step 1: Assume first that u, v : E → [0, ∞) are simple functions vanishing at infinity and, using the notation (4.16), write u=
k
αi χFi ,
v=
i=0
l
β j χ Gj ,
j=0
so that by part (iii) of the previous exercise, ∗
u =
k
αi χ
Fi∗
,
l
∗
v =
i=0
βj χG∗j .
j=0
Then k l uv dx = αi βj χFi χGj dx E
E
i=0 j=0
=
k l
αi βj L1 (Fi ∩ Gj ) ≤
i=0 j=0
k l
αi βj min{L1 (Fi ), L1 (Gj )},
i=0 j=0
while
∗ ∗
E∗
u v dt =
l k i=0 j=0
=
k l
αi βj
E∗
χFi∗ χG∗j dt
αi βj L1 (Fi∗ ∩ G∗j )
i=0 j=0
=
k l
αi βj min{L1 (Fi∗ ), L1 (G∗i )}
i=0 j=0
=
k l
αi βj min{L1 (Fi ), L1 (Gj )},
i=0 j=0
where we have used (4.11) and the fact that L1 (Fi∗ ∩ G∗j ) = min{L1 (Fi∗ ), L1 (G∗j )}, since both sets are half-open intervals of the form [0, a) for some a > 0. Step 2: Now let u, v : E → [0, ∞) be two Lebesgue measurable functions and construct two increasing sequences {un }n and {vn }n of nonnegative simple functions vanishing at infinity, below u and v, and converging pointwise to u and v, respectively (see Corollary B.37). By Proposition 4.11 we have that ∗ for all n ∈ N and that {u∗n vn∗ }n converges pointwise to u∗n vn∗ ≤ u∗n+1 vn+1 ∗ ∗ u v . By Step 1, for each n ∈ N we have un vn dx ≤ u∗n vn∗ dt E
E∗
and the result follows by the Lebesgue monotone convergence theorem.
122
4. Decreasing Rearrangement
Exercise 4.15. Let E ⊆ R be a Lebesgue measurable set and let u, v : E → [0, ∞) be Lebesgue measurable functions. Prove that for every r ≥ 0, u(x)χ{v≤r} (x) dx ≥ u∗ (t)χ{v∗ ≤r} (t) dt. E∗
E
Hint: Assume first that u is integrable. We now turn to one of the most important properties of decreasing rearrangement. Theorem 4.16. Let E ⊆ R be a Lebesgue measurable set, let u : E → [0, ∞) be a function vanishing at infinity, and let f : [0, ∞) → [0, ∞) be a Borel function. Then f (u∗ (t)) dt ≤ f (u(x)) dx, (4.17) E∗
with equality holding if f (0) = 0 or (4.18)
E 1 L ({x ∈
E : u(x) > 0}) < ∞ or
L ({x ∈ E : u(x) > 0}) = ∞ and L1 ({x ∈ E : u(x) = 0}) = 0. 1
In particular, for any p > 0, ∗ p (u (t)) dt = (u(x))p dx. E∗
E
Proof. Step 1: We claim that f (u(x)) dx = {u>0}
{u∗ >0}
f (u∗ (t)) dt.
Let f˜ be the restriction of f to (0, ∞) and construct an increasing sequence {fn }n of nonnegative Borel simple functions defined on (0, ∞) and converging pointwise to f˜. Since each fn may be written as fn =
kn
ci,n χBi,n ,
i=0
where ci,n = cj,n for i = j and the Borel sets Bi,n ⊆ (0, ∞) are pairwise disjoint, by applying Corollary 4.6, we conclude that kn fn (u(x)) dx = ci,n L1 ({x ∈ E : u(x) ∈ Bi,n }) {u>0}
i=0
=
kn
ci,n L1 ({t ∈ E ∗ : u∗ (t) ∈ Bi,n })
i=0
=
{u∗ >0}
fn (u∗ (t)) dt.
The claim now follows by the Lebesgue monotone convergence theorem.
4.1. Definition and First Properties
123
Step 2: Write (4.19) f (u(x)) dx E f (u(x)) dx + f (0)L1 ({x ∈ E : u(x) = 0}), = {u>0}
where f (0)L1 ({x ∈ E : u(x) = 0}) is understood to be zero if f (0) = 0, independently of the value of L1 ({x ∈ E : u(x) = 0}). Similarly, (4.20) f (u∗ (t)) dt E∗ f (u∗ (t)) dt + f (0)L1 ({t ∈ E ∗ : u∗ (t) = 0}). = {u∗ >0}
Hence, it follows from the previous step, (4.19), (4.20), and Proposition 4.3(v) that (4.17) holds. Moreover, if f (0) = 0 or f (0) > 0 and L1 ({t ∈ E ∗ : u∗ (t) = 0}) = L1 ({x ∈ E : u(x) = 0}), then
f (u(x)) dx =
(4.21) E
E∗
f (u∗ (t)) dt.
Thus, the result now follows by Proposition 4.3(v) once more.
Remark 4.17. It follows from the previous proof that equality holds in (4.17) if and only if one of the following conditions holds: (i) {u>0} f (u(x)) dx = ∞. (ii) {u>0} f (u(x)) dx < ∞ and f (0) = 0. (iii) {u>0} f (u(x)) dx < ∞, f (0) > 0, and L1 ({x ∈ E : u(x) > 0}) < ∞. (iv) {u>0} f (u(x)) dx < ∞, f (0) > 0, and (4.18) holds. Exercise 4.18. Let u(x) = e−x χ[0,∞) (x) and let f = χ{0} . Find u∗ and prove directly that ∞ f (u∗ (t)) dt < f (u(x)) dx = ∞. 0= 0
R
The previous theorem shows that the operator u → u∗ preserves the norm in Lp for 1 ≤ p < ∞. Next we show that it is a continuous operator from Lp (E) into Lp (E ∗ ). More generally, we have the following result.
124
4. Decreasing Rearrangement
Theorem 4.19. Let Ψ : R → [0, ∞) be a convex function such that Ψ(0) = 0, let E ⊆ R be a Lebesgue measurable set, and let u, v : E → [0, ∞) be two functions vanishing at infinity. Then (4.22) Ψ(u∗ (t) − v ∗ (t)) dt ≤ Ψ(u(x) − v(x)) dx. E∗
In particular,
E
∗
E∗
∗
|u (t) − v (t)| dt ≤
|u(x) − v(x)|p dx
p
E
for all 1 ≤ p < ∞ and the operator u → u∗ is continuous from Lp (E) into Lp (E ∗ ). Proof. Define Ψ+ (s) :=
Ψ(s) if s ≥ 0, 0 otherwise,
Ψ− (s) :=
Ψ(s) if s < 0, 0 otherwise.
Then Ψ+ and Ψ− are still convex (why?) and Ψ = Ψ+ + Ψ− . Thus, to prove (4.22), it suffices to prove it for Ψ+ and Ψ− separately. We only prove (4.22) for Ψ+ , since the proof of (4.22) for Ψ− is very similar. By Exercise 3.21, the right derivative (Ψ+ )+ of Ψ+ exists in R and is increasing and for every s ∈ R, s Ψ+ (s) = (Ψ+ )+ (τ ) dτ, 0
where we have used the fact that Ψ+ (0) = 0. Hence, for x ∈ E, u(x)−v(x) (Ψ+ )+ (τ ) dτ Ψ+ (u(x) − v(x)) =
0
u(x)
= v(x) ∞
= 0
(Ψ+ )+ (u(x) − r) dr
(Ψ+ )+ (u(x) − r)χ{v≤r} (x) dr,
where in the second equality we have used the change of variable τ = u(x)−r and in the third equality we have used the facts that (Ψ+ )+ (s) = 0 for s < 0 and that v ≥ 0. Integrating both sides over E and using Tonelli’s theorem, we get ∞ Ψ+ (u(x) − v(x)) dx = (Ψ+ )+ (u(x) − r)χ{v≤r} (x) dxdr. (4.23) E
0
Similarly, Ψ+ (u∗ (t) − v ∗ (t)) dt = (4.24) E∗
E
∞ 0
E∗
(Ψ+ )+ (u∗ (t) − r)χ{v∗ ≤r} (t) dtdr.
4.1. Definition and First Properties
125
Since for every fixed t the function g(s) := (Ψ+ )+ (s − t) is increasing, by Exercise 4.8 we have that (g ◦ u)∗ (t) = (g ◦ u∗ )(t) for all but countably many t ∈ E ∗ , and so, also by Exercise 4.15, (Ψ+ )+ (u(x)−t)χ{v≤r}(x) dx ≥ (Ψ+ )+ (u∗ (t)−r)χ{v∗≤r} (t) dt. (4.25) E∗
E
The result now follows by combining (4.23), (4.24), and (4.25).
In the next exercise we present an alternative proof in the case in which Ψ is even. Exercise 4.20. Let Ψ : [0, ∞) → [0, ∞) be a convex function such that Ψ(0) = 0, let E ⊆ R be a Lebesgue measurable set, and let u, v : E → [0, ∞) be two functions vanishing at infinity. (i) Prove that if r1 ≥ r2 ≥ 0 and s1 ≥ s2 ≥ 0, then Ψ(|r1 − s1 |) + Ψ(|r2 − s2 |) ≤ Ψ(|r1 − s2 |) + Ψ(|r2 − s1 |). (ii) Prove that if r1 ≥ r2 ≥ · · · ≥ rn ≥ 0, s1 ≥ s2 ≥ · · · ≥ sn ≥ 0, n ∈ N, and if f : {1, . . . , n} → {1, . . . , n},
g : {1, . . . , n} → {1, . . . , n}
are two bijections, then n
Ψ(|ri − si |) ≤
i=1
n
Ψ(|rf (i) − sg(i) |).
i=1
(iii) Prove that if u, v : E → [0, ∞) are simple functions of the form (4.26)
u=
n
αi χEi ,
v=
i=0
(4.27)
n
βi χEi ,
i=0
where L1 (Ei ) = k for all i = 1, . . . , n, then ∗ ∗ Ψ(|u (t) − v (t)|) dt ≤ Ψ(|u(x) − v(x)|) dx. E∗
E
(iv) Prove that if u, v : E → [0, ∞) are simple functions of the form (4.26), where L1 (Ei ) = ri ∈ Q ∩ [0, ∞) for all i = 1, . . . , n, then (4.27) holds. (v) Prove that (4.27) holds for arbitrary u, v : E → [0, ∞) vanishing at infinity.
126
4. Decreasing Rearrangement
4.2. Function Spaces and Decreasing Rearrangement In this section we prove that if u belongs to some of the function spaces studied in the previous chapters, then so does u∗ and its corresponding norm decreases. We begin by showing that the pointwise variation of u∗ is less than the pointwise variation of u. Theorem 4.21 (Bounded variation). Let I ⊆ R be an interval and let u : I → [0, ∞) be a function vanishing at infinity. Then for all t1 , t2 ∈ I ∗ , with t1 ≤ t2 , (4.28)
u∗ (t1 ) − u∗ (t2 ) ≤ VarI u.
In particular, (4.29)
u∗ ≤ VarI u. VarI ∗ u∗ = sup u∗ − inf ∗ I∗
I
Proof. If either VarI u = ∞ or u is a constant function, then (4.28) holds. Thus assume that VarI u < ∞ and that u is not a constant function. It follows by Proposition 2.11 that u is bounded with sup u − inf u ≤ VarI u. I
I
On the other hand, by Proposition 4.3, if t1 , t2 ∈ I ∗ , with t1 ≤ t2 , then inf I u ≤ u∗ (t2 ) ≤ u∗ (t1 ) ≤ u∗ (0) = supI u, and so (4.28) holds. The inequal ity (4.29) follows by the arbitrariness of t1 and t2 . Next we show that if u vanishes at infinity and is absolutely continuous, then so is u∗ . Theorem 4.22 (Absolute continuity). Let I ⊆ R be an interval and let u : I → [0, ∞) be a function vanishing at infinity. If u belongs to AC(I), then u∗ belongs to AC(I ∗ ) and for every 1 ≤ p ≤ ∞, (4.30)
(u∗ ) Lp (I ∗ ) ≤ u Lp (I) .
Proof. Step 1: Since u is uniformly continuous (see Remark 3.4) and vanishes at infinity, if sup I = ∞, then there exists limx→∞ u(x) = 0 (why?). Similarly, if inf I = −∞, then limx→−∞ u(x) = 0. On the other hand, if one of the endpoints of I is finite, then by uniform continuity u can be extended to I as a uniform continuous function. Thus, in all cases, u is bounded. In turn, by Proposition 4.3, u∗ (0) = supI u < ∞ and u∗ is right continuous at 0.
4.2. Function Spaces and Decreasing Rearrangement
127
By Theorem 3.29, we have that u is equi-integrable in I and thus for every ε > 0 there exists δ > 0 such that |u (x)| dx ≤ ε (4.31) E
for every Lebesgue measurable set E ⊆ I with L1 (E) ≤ δ. Let a, b ∈ I ∗ with 0 < a < b and for x ∈ I define ⎧ ∗ ⎨ u (b) if u(x) < u∗ (b), u(x) if u∗ (b) ≤ u(x) ≤ u∗ (a), v(x) := ⎩ ∗ u (a) if u(x) ≥ u∗ (a). Since u(I) is an interval and inf I u ≤ u∗ (b) ≤ u∗ (a) ≤ u∗ (0) = supI u by Proposition 4.3, by Proposition 2.11, Theorems 3.25 and 3.59, we have ∗ ∗ |u (x)| dx. (4.32) u (a) − u (b) ≤ VarI v = |v (x)| dx = {u∗ (b) s0 })
n=1
= L1 ({t ∈ I ∗ : u∗ (t) = s0 }). Note that we could also have used Corollary 4.6. In turn, by (4.7), ( u )− (s0 ) = u (s0 ) + L1 ({x ∈ I : u(x) = s0 }) = u (s0 ) + L1 ({t ∈ I ∗ : u∗ (t) = s0 }) for s0 > 0 and so taking s0 = u∗ (t0 + h) we get ( u )− (u∗ (t0 + h)) = u (u∗ (t0 + h)) + L1 ({t ∈ I ∗ : u∗ (t) = u∗ (t0 + h)}). Since t0 is a Lebesgue point of (u∗ ) , we have t0 +r 1 |(u∗ ) (t) − (u∗ ) (t0 )| = 0, lim r→0+ 2r t0 −r and since (u∗ ) (t0 ) < 0, we can find a sequence hn → 0+ such that (u∗ ) (t0 + hn ) < 0 for every n. In turn, {t ∈ I ∗ : u∗ (t) = u∗ (t0 + hn )} is a singleton. Thus, (4.37)
( u )− (u∗ (t0 + hn )) = u (u∗ (t0 + hn )).
4.2. Function Spaces and Decreasing Rearrangement
129
By Proposition 4.3(iii), we have u (u∗ (t0 + hn )) ≤ t0 + hn and ( u (u∗ (t0 + hn )))− ≥ t0 + hn . It follows from (4.37) that u (u∗ (t0 + hn )) = t0 + hn and similarly, u (u∗ (t0 )) = t0 . Thus (4.35) holds by (4.36). In turn, by Jensen’s inequality (see Theorem B.50) and (4.32), p g(t0 + hn ) − g(t0 ) 1 ≥ |u (x)| dx hn hn {u∗ (t0 +hn ) α if either x ∈ E, with x > x0 , and s = s0 , or x ∈ E is arbitrary and 0 ≤ s < s0 . (iii) Prove that {x ∈ E : ϕ(x) ≤ α} = {x ∈ E : u(x) > s0 } ∪ {x ∈ E : x ≤ x0 , u(x) = s0 }.
4.2. Function Spaces and Decreasing Rearrangement
131
(iv) Consider the measure μ : B([0, L1 (E)]) → [0, ∞], defined by μ(B) := L1 (ϕ−1 (B)),
B ∈ B([0, L1 (E)]).
Prove that for every 0 ≤ α ≤ L1 (E) we have that μ([0, α]) = L1 (ϕ−1 ([0, α])) = L1 ({x ∈ E : ϕ(x) ≤ α}) = α and deduce that ϕ is measure-preserving. (v) Prove that for L1 -a.e. x ∈ E, u∗ (ϕ(x)) = u(x). (vi) Remove the assumption that E has finite measure. Exercise 4.31. Let E ⊆ R be a Lebesgue measurable set, let u : E → [0, ∞) be a function vanishing at infinity, and let (4.39)
E> := {x ∈ E : u(x) > 0},
F> := {t ∈ E ∗ : u∗ (t) > 0}.
Prove that the function ϕ : E> → F> given in (4.38) is measure-preserving and that u∗ (ϕ(x)) = u(x) for L1 -a.e. x ∈ E> . Exercise 4.32 (Lusin (N ) property). Let I ⊆ R be an interval, let u : I → [0, ∞) be a function vanishing at infinity. Prove that if u maps sets of Lebesgue measure zero into sets of Lebesgue measure zero, then so does u∗ . Exercise 4.33. Let I ⊆ R be an interval and let u : I → [0, ∞) be a function vanishing at infinity and differentiable L1 -a.e. in I. Let I> and F> be defined as in (4.39) and let ϕ be the function in (4.38) with E = I. (i) Prove that if u is differentiable at x0 ∈ I> and u (x0 ) = 0, then there exists δ > 0 such that ϕ(x) = ϕ(x0 ) for all x ∈ I> ∩ (x0 − δ, x0 + δ). (ii) Let Idiff ⊆ I> be the set of all x ∈ I> such that u is differentiable at x and u∗ is differentiable at ϕ(x). Prove that if ϕ|Idiff is differentiable at x0 ∈ Idiff , that is, if there exists in R the limit :=
lim
x∈Idiff , x→x0
ϕ(x) − ϕ(x0 ) , x − x0
then || ≥ 1. (iii) Prove that if x0 ∈ Idiff is such that (u∗ ) (ϕ(x0 )) = 0, then u (x0 ) = 0 and ϕ|Idiff is continuous at x0 . (iv) Prove that if x0 ∈ Idiff is such that (u∗ ) (ϕ(x0 )) = 0, then ϕ|Idiff is differentiable at x0 ∈ Idiff with (u∗ ) (ϕ(x0 ))(ϕ|Idiff ) (x0 ) = u (x0 ).
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4. Decreasing Rearrangement
(v) Prove that for any p > 0, ∗ p |(u ) (t)| dt ≤ |u (x)|p dx. I∗
I
Chapter 5
Curves Undergradese, V: “Hmmm, what do you mean by that?” Translation: “What’s the answer so we can all go home?” — Jorge Cham, www.phdcomics.com
In this chapter we will study rectifiable curves and their relation with functions of bounded pointwise variation.
5.1. Rectifiable Curves In the literature there are many different definitions of curves. Let I ⊆ R be an interval and let u : I → RM be a (possibly discontinuous) function. As the parameter t traverses I, u(t) traverses a curve in RM . Rather than calling u a curve, it is better to regard any vector function v obtained from u by a suitable change of parameter as representing the same curve as u. Thus, one should define a curve as an equivalence class of equivalent parametric representations. Definition 5.1. Given two intervals I, J ⊆ R, a metric space (Y, d), and two functions u : I → Y and v : J → Y , we say that they are (Lebesgue) equivalent if there exists a continuous, bijective function φ : I → J such that (5.1)
u(t) = v(φ(t))
for all t ∈ I. We write u ∼ v and we call u and v parametric representations and the function φ a parameter change. Note that in view of Theorem 1.8, φ−1 : J → I is also continuous. Exercise 5.2. Prove that ∼ is an equivalence relation. Definition 5.3. A curve γ is an equivalence class of parametric representations. The curve γ is said to be continuous if one (and so all) of its parametric representations is continuous. 133
134
5. Curves
When X is a normed space, by restricting parametric representations, parameter changes and their inverses in Definitions 5.1 and 5.3 to be differentiable L1 -a.e., or Lipschitz continuous, or of class C n , n ∈ N0 , etc., we may define curves γ that are differentiable L1 -a.e., or Lipschitz continuous, or of class C n , n ∈ N0 , respectively. Given a metric space (Y, d) and a curve γ with parametric representation u : I → Y , where I ⊆ R is an interval, the multiplicity of a point y ∈ Y is the (possibly infinite) number of points t ∈ I such that u(t) = y. Note that the multiplicity of a curve can also be expressed in terms of the Banach indicatrix (see Definition 2.58). Since every parameter change φ : I → J is bijective, the multiplicity of a point does not depend on the particular parametric representation. The range of γ is the set of points of Y with positive multiplicity, that is, u(I). A point in the range of γ with multiplicity one is called a simple point. If every point of the range is simple, then γ is called a simple arc. If I = [a, b] and u(a) = u(b), then a continuous curve γ is called a closed curve. A closed curve is called simple if every point of the range is simple, with the exception of u(a), which has multiplicity two. Example 5.4. The curves γn with parametric representations un : [0, π2 ] → R2 , given by un (t) := (sin2 nt, 0), t ∈ [0, π2 ], n ∈ N, all have the same range, but are different curves. Indeed, γn covers n times the segment of endpoints (0, 0) and (1, 0). Exercise 5.5. Given two intervals I, J ⊆ R and two functions u : I → RM and v : J → RM , we say that they are Fr´echet equivalent if for every ε > 0 there exists a continuous, bijective function φ : I → J such that |u(t) − v(φ(t))| < ε F
F
for all t ∈ I. We write u ∼ v. Prove that ∼ is an equivalence relation. Exercise 5.6. A Fr´echet curve γ is an equivalence class (with respect to F
∼) of parametric representations. Prove that if a Fr´echet curve γ has a continuous representative, then all of its representatives are continuous. The next exercise shows that two functions u : I → RM and v : J → RM may be Fr´echet equivalent but not (Lebesgue) equivalent. Exercise 5.7. For t, τ ∈ [0, 1] let u(t) := (t, 0) and ⎧ 3 if 0 ≤ τ ≤ 13 , ⎪ ⎨ ( 2 τ, 0) if 13 ≤ τ ≤ 23 , ( 12 , 0) v(τ ) := ⎪ ⎩ (1 + 32 (τ − 1), 0) if 23 ≤ τ ≤ 1.
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135
Figure 1. Peano’s curve.
Prove that u and v are Fr´echet equivalent but not (Lebesgue) equivalent. The next result shows that the class of continuous curves is somehow too large for our intuitive idea of curve. Indeed, we construct a continuous curve that fills the unit square in R2 . The first example of this type was given by Peano in 1890 [188]. We present here another example given by Hilbert [119]. The proof is taken from [174]. Theorem 5.8 (Hilbert). There exists a continuous function u : [0, 1] → R2 such that u([0, 1]) = [0, 1]2 . Proof. For every n ∈ N divide the interval [0, 1] into 4n closed intervals Ik,n , k = 1, . . . , 4n , of length 41n and the square [0, 1]2 into 4n closed squares Qk,n , k = 1, . . . , 4n , of side length 21n . Construct a bijective correspondence between the 4n intervals Ik,n and the 4n squares Qk,n in such a way that (see Figure 1) (i) to two adjacent intervals there correspond adjacent squares, 1 contained in some interval Ik,n (ii) to the four intervals of length 4n+1 1 there correspond the four squares of side length 2n+1 contained in the square corresponding to Ik,n .
By relabeling the squares, if necessary, we will assume that the square Qk,n corresponds to the interval Ik,n . Let F := {Ik,n : n ∈ N, k = 1, . . . , 4n }, G := {Qk,n : n ∈ N, k = 1, . . . , 4n }. By the axiom of continuity of the reals, if {Ji }i is any infinite sequence of intervals in F such that Ji+1 ⊆ Ji for all i ∈ N and {Ri }i is the corresponding sequence of squares in G, then there exist a unique t ∈ [0, 1] and (x, y) ∈ [0, 1]2 such that Ji , {(x, y)} = Ri . {t} = i
i
We set the point t and the point (x, y) in correspondence.
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5. Curves
We claim that this correspondence defines a continuous function u : [0, 1] → R2 with all the desired properties. Indeed, a point t ∈ [0, 1] that is not an endpoint of any interval determines uniquely a sequence {Ji }i to which it belongs and hence a point (x, y) belonging to the corresponding sequence {Ri }i . The same is true for t = 0 and t = 1. A point t that is common to two different intervals Ik1 ,m and Ik2, m for some m ∈ N is also common to two different intervals Ik,n , k = 1, . . . , 4n , for all n ≥ m. Hence, it belongs to two different sequences {Ji }i , {Ji }i . Since the squares Ri and Ri , corresponding to Ji and Ji , respectively, are adjacent by property (i), it follows that Ri = Ri . i
i
Thus, to every t ∈ [0, 1] there corresponds a unique (x, y) ∈ [0, 1]2 that we denote u(t). Since every (x, y) ∈ [0, 1]2 belongs to one, two, three, or four sequences {Ri }i , there exists one, two, three, or four t ∈ [0, 1] such that u(t) = (x, y). Hence, u([0, 1]) = [0, 1]2 . To prove that u is continuous, write u(t) = (x(t), y(t)), t ∈ [0, 1]. By conditions (i) and (ii) we have that |x(t1 ) − x(t2 )| ≤ 21−n ,
|y(t1 ) − y(t2 )| ≤ 21−n
for all t1 , t2 ∈ [0, 1], with |t1 −t2 | ≤ 1/4n . This proves the uniform continuity of u. In view of the previous result, to recover the intuitive idea of a curve, we restrict the class of continuous curves to those with finite or σ-finite length. Definition 5.9. Given a metric space (Y, d) and a curve γ in Y , let u : I → Y be a parametric representation of γ, where I ⊆ R is an interval. We define the length of γ as n d(u(ti ), u(ti−1 )) , L(γ) := Var u = sup i=1
where the supremum is taken over all partitions P = {t0 , . . . , tn } of I, n ∈ N. We say that the curve γ is rectifiable if L(γ) < ∞. Remark 5.10. Many authors define the notion of length only for continuous curves. This is motivated by the fact that if γ is a curve with parametric representation u : I → Y , where I ⊆ R is an interval and u is discontinuous at t ∈ I, then the length of γ also measures the length of the segment joining the points u(t) and u− (t), as well as the length of the segment joining the points u(t) and u+ (t) (see Corollary 5.14 below), although these segments may not belong to the curve. However, see Theorem 5.28.
5.1. Rectifiable Curves
137
Given a metric space (Y, d) and a curve γ in Y , with parametric representation u : I → Y , where I ⊆ R is an interval, we say that γ is locally rectifiable, or that it has σ-finite length, if Var[a,b] u < ∞ for every interval [a, b] ⊆ I. Exercise 5.11. Given a metric space (Y, d), prove that if two parametric representations u : I → Y and v : J → Y are equivalent, then VarI u = VarJ v. Hence, the definitions of rectifiability and of length of a curve do not depend on the particular parametric representation. In view of Theorem 3.25 we have Theorem 5.12. Given a rectifiable curve γ in RM with parametric representation u : I → RM , where I ⊆ R is an interval, then u (t) dt ≤ L(γ). (5.2) I
The equality holds in (5.2) if and only if u ∈ AC(I; RM ). Example 5.13. In view of the previous theorem, if f ∈ AC([a, b]; RN ), then its graph is given by the curve γ with parametric representation u : [a, b] → RN +1 , t → (t, f (t)), and L(Gr f ) = L(γ) =
b#
1 + f (t)2 dt.
a
When there is strict inequality in (5.2), given a rectifiable curve γ in RM with parametric representation u : I → RM , as in Corollary 3.90, we may decompose u : I → RM into an absolutely continuous part, a Cantor part, and a jump part. Thus, we obtain u = uAC + uC + uJ , where for a fixed z0 ∈ I and for all t ∈ I, t u (s) ds, uAC (t) := z0
uJ (t) :=
(u (z) − u− (z)) + u(t) − u− (t) if t ≥ z0 , z∈I, z0 ≤z 0, which contradicts the fact that gni (t0 ) → 0. Thus, assume that fn i (t0 ) → ∞. Fix y, z ∈ RN , and consider the function # g(s) := 1 + y + sz2 , s ∈ (0, ∞). Since g is convex and differentiable, the function s ∈ (0, ∞) → g(s)−2g( 2s ) is increasing. To see this, it suffices to differentiate and use the fact that g is an increasing function. Using this fact with y = f (t0 ) and z = fn i (t0 ) − f (t0 ), we have that for all 0 ≤ s < 1, # gni (t0 ) = 1 + f (t0 )2 + g(1) − 2g( 12 ) # ≥ 1 + f (t0 )2 + g(s) − 2g( 2s ) $ # = 1 + f (t0 )2 + 1 + f (t0 ) + s(fn i (t0 ) − f (t0 ))2 $ − 2 1 + f (t0 ) + 2s (fn i (t0 ) − f (t0 ))2 . Taking s := fn i (t0 ) − f (t0 )−1 ∈ (0, 1) for all i sufficiently large, we have that $ $ # gni (t0 ) ≥ 1 + f (t0 )2 + 1 + f (t0 ) + ξni 2 − 2 1 + f (t0 ) + 12 ξni 2 ,
5.1. Rectifiable Curves
where ξni :=
141
fn i (t0 )−f (t0 )
fn i (t0 )−f (t0 ) .
Since ξni = 1, we may find a subsequence
(not relabeled) such that ξni → ξ, with ξ = 1. Letting i → ∞ in the previous inequality yields # # 0 = lim gni (t0 ) ≥ 1 + f (t0 )2 + 1 + f (t0 ) + ξ2 i→∞ $ − 2 1 + f (t0 ) + 12 ξ2 > 0, where the last inequality follows from the fact that ξ = 0. Thus, we have reached a contradiction even in this case, and so the claim is proved. Next we show that b $ # 2 2 (5.7) lim 1 + f − 1 + f dt = 0. ni i→∞ a
Using the fact that ||x| − |x − y|| ≤ |y| for all x, y ∈ R, we have $ # $ # 1 + fn i 2 − 1 + fn i 2 − 1 + f 2 ≤ 1 + f 2 , and thus, by the Lebesgue dominated convergence theorem and the fact that fn i (t) → f (t) for L1 -a.e. t ∈ [a, b], b $ $ # 1 + fn i 2 − 1 + fn i 2 − 1 + f 2 dt lim i→∞ a
b# 1 + f 2 dt, = a
which, together with (5.6), implies (5.7). Finally, we prove that (5.8)
b
lim
i→∞ a
By Fatou’s lemma,
b
lim inf i→∞
fn i dt
a
b
=
fn i dt
f dt.
a
b
≥
f dt.
a
Thus, if (5.8) fails, then there is a subsequence (not relabeled) and ε > 0 such that b b f dt + 3ε(b − a) ≤ fn i dt (5.9) a
a
b# for all i ∈ N. Since a 1 + f 2 dt < ∞, we may find δ = δ(ε) > 0 such that # 1 + f 2 dt < ε(b − a) (5.10) F
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5. Curves
for every Lebesgue measurable set F ⊂ I, with L1 (F ) < δ. By Egoroff’s theorem there is a Lebesgue measurable set Eε ⊂ [a, b], with L1 (Eε ) < δ, such that {fn i }i converges uniformly in [a, b] \ Eε . Hence, we may find an integer iε ∈ N such that fn i (t) ≤ f (t) + ε for all t ∈ [a, b] \ Eε and for all i ≥ iε . In turn, by (5.9) for all i ≥ iε , b b f dt + 3ε(b − a) ≤ fni dt = fni dt + fn i dt a
a
[a,b]\Eε
b
≤
f dt + ε(b − a) +
a
It follows that
$
2ε(b − a) ≤
Eε
#
≤
$ Eε
Eε
1 + fn i 2 dt.
1 + fn i 2 dt 1 + f 2 dt +
b $ # 2 2 1 + f − 1 + f dt ni a
Eε
b $ # 2 2 ≤ ε(b − a) + 1 + fni − 1 + f dt a
for all i ≥ iε , where we have used the triangle inequality and (5.10). Letting i → ∞ and using (5.7), we obtain a contradiction. Hence, (5.8) holds, and in turn, reasoning as in the proof of (5.7), we have that b fn i − f dt = 0. (5.11) lim i→∞ a
To conclude, we observe that, by the uniqueness of the limit, (5.7) and (5.11) actually hold for the whole sequence (why?). Exercise 5.20. Let g : RN → [0, ∞) be strictly convex. (i) Prove that if g(s) → ∞ as s → ∞, then there exist c > 0 and S > 0 such that g(s) ≥ cs for all s ≥ S. (ii) Prove that if f, fn : [a, b] → RN , n ∈ N, are absolutely continuous functions such that f (t) = lim fn (t) n→∞
for all t ∈ [a, b] and b b g(fn (t)) dt = g(f (t)) dt < ∞, lim n→∞ a
then
a
lim
n→∞ a
b
|g(fn (t)) − g(f (t))| dt = 0.
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143
(You may use the fact that by (5.5), b b lim inf g(fn (t)) dt ≥ g(f (t)) dt; n→∞
a
a
see [8]). (iii) Prove that if in part (ii) we also assume that g(s) → ∞ as s → ∞, then b
lim
n→∞ a
fn − f dt = 0.
5.2. Arclength Next we introduce the notion of arclength of a locally rectifiable curve. Definition 5.21. Let (Y, d) be a metric space. Given a locally rectifiable curve γ in Y , we say that γ is parametrized by arclength if it admits a parametric representation w : J → Y such that Var[s1 ,s2 ] w = s2 − s1
(5.12)
for all s1 , s2 ∈ J, with s1 < s2 . Remark 5.22. Note that the function w is Lipschitz continuous with Lipschitz constant less than or equal to 1. Indeed, by (2.4), (5.13)
d(w(s1 ), w(s2 )) ≤ Var[s1 ,s2 ] w = s2 − s1
for all s1 , s2 ∈ J, with s1 < s2 . Proposition 5.23. Let γ be a curve in RM . Assume that γ can be parametrized by arclength and let w : J → RM be a parametric representation satisfying (5.12). Then w (s) = 1 for L1 -a.e. s ∈ J. Proof. Since w is Lipschitz continuous, it is differentiable for L1 -a.e. s ∈ J by Corollary 2.31. It follows from (5.13) that w (s) ≤ 1 for L1 -a.e. s ∈ J. On the other hand, by Theorem 5.12 and (5.12) for every interval J1 ⊆ J, L1 (J1 ) = VarJ1 w = w (s) ds. J1
Hence,
0 = L (J1 ) − 1
(1 − w (s)) ds
w (s) ds =
J1 1 L -a.e.
J1
s ∈ J1 , necessarily w (s) = 1 for L1 -a.e. and since w (s) ≤ 1 for s ∈ J1 and the proof is complete. The main theorem of this section is the following.
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5. Curves
Theorem 5.24 (Arclength). Let (Y, d) be a metric space and let γ be a continuous locally rectifiable curve. Assume that γ admits a parametric representation u : I → Y such that u is not constant on any proper subinterval of I. Then γ can be parametrized by arclength. In particular, if γ is a continuous simple locally rectifiable arc in Y , then γ can be parametrized by arclength. Proof. By Theorem 2.17 and Proposition 2.21 the indefinite pointwise variation V : I → V (I) of u, given by Var[t0 ,t] u if t ≥ t0 , (5.14) V (t) := − Var[t,t0 ] u if t < t0 , is continuous, increasing, and onto. Moreover, J := V (I) is an interval. We claim that V is strictly increasing. Indeed, if there exist t1 < t2 such that V (t1 ) = V (t2 ), then either t1 , t2 ≥ t0 or t1 , t2 ≤ t0 and so by Proposition 2.15, Var[t1 ,t2 ] u = 0, which implies that u is constant in [t1 , t2 ], which contradicts the hypothesis on u. Hence, V : I → V (I) is invertible, and so V −1 : J → I is continuous. Thus, the function w : J → X, defined by (5.15)
w(s) := u(V −1 (s)),
s ∈ J,
is equivalent to u. Moreover, by Theorem 2.17, for every s1 , s2 ∈ J, with s1 < s2 , we have that Var[s1 ,s2 ] w = Var[V −1 (s1 ),V −1 (s2 )] u = V (V −1 (s2 )) − V (V −1 (s1 )) = s2 − s1 . This concludes the proof.
Remark 5.25. If γ is rectifiable, then in place of the function V one can use the function V ∞ (t) := VarI∩(−∞,t] u so that J becomes the interval of endpoints 0 and L(γ). The next exercise shows that unlike (Lebesgue) curves, Fr´echet curves can always be parametrized by arclength. Exercise 5.26. Prove that if γ is a continuous rectifiable Fr´echet curve in RM with L(γ) > 0, then γ can be parametrized by arclength Exercise 5.27. Given two intervals I, J ⊆ R and two continuous functions u : I → RM and v : J → RM , we say that u ≈ v if there exists an increasing function (possibly nonstrictly increasing) φ : I → J such that φ(I) = J and for every discontinuity point of t ∈ I of φ, the interval [φ− (t), φ+ (t)] is contained in an interval [τ1 , τ1 ] ⊆ J on which v is constant. (i) Prove that ≈ is an equivalence relation.
5.2. Arclength
145
(ii) Given a continuous function u : I → RM of bounded pointwise variation, prove that the inverse function V −1 : J → I (properly defined) satisfies the conditions for an admissible parameter change and that the function w(s) := u(V −1 (s)),
s ∈ J,
is Lipschitz continuous. By modifying the proof of Theorem 5.24, we can show that if a simple arc γ is not continuous in a normed space, then the range of the curve γ is contained in the range of a (different) curve parametrized by arclength and with the same length. Theorem 5.28. Let (Y, · ) be a normed space and let γ be a rectifiable simple arc γ in Y with L(γ) > 0. Then there exists a curve parametrized by arclength whose range contains the range of Γ and whose length is L(γ). Proof. Let u : I → RM be a parametric representation of γ, where I ⊆ R is an interval. Since γ may be discontinuous, the indefinite pointwise variation V : I → V (I) given in (5.14) is still strictly increasing and invertible, but it may be discontinuous. To be precise, in view of Proposition 2.21, u and V have the same set of discontinuity points. In particular V (I) may not be an interval. Let w be the function defined in (5.15). Then (5.13) continues to hold. Hence, the function w is continuous in V (I), and so it may be uniquely extended to V (I) by continuity. Let J be the smallest interval containing V (I) and let {(an , bn )}n be the family of connected components of J ◦ \ V (I). Let t0 ∈ I be a discontinuity point of u. Then by Corollary 3.90, (5.16)
V (t0 ) − V− (t0 ) = u(t0 ) − u− (t0 ),
(5.17)
V + (t0 ) − V(t0 ) = u(t0 ) − u+ (t0 ).
If V (t0 ) − V − (t0 ) > 0, then there exists an interval (an , bn ) such that an = V − (t0 ),
(5.18)
bn = V (t0 ).
Hence, w(an ) = u− (t0 ) and w(bn ) = u(t0 ). In this case, we extend the function w to be affine in the interval (an , bn ). To be precise, we set w(s) :=
u(t0 ) − u− (t0 ) (s − an ) + u− (t0 ), bn − an
s ∈ (an , bn ).
Then w is continuous in [an , bn ] and Var[s1 ,s2 ] w = s2 − s1 for all s1 , s2 ∈ (an , bn ), with s1 < s2 , by (5.16) and (5.18).
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Similarly, if V + (t0 ) − V (t0 ) > 0, then there exists an interval (am , bm ) such that am = V (t0 ) and bm = V + (t0 ), and we define w(s) :=
u+ (t0 ) − u(t0 ) (s − am ) + u(t0 ), bm − am
s ∈ (am , bm ).
Thus, we have extended w to J in such a way that w satisfies (5.12). The curve γ˜ with parametric representation w : J → RM has length L(γ), and its range contains the range of γ.
5.3. Length Distance Let (Y, d) be a metric space. Given a set E ⊆ Y and two points y1 , y2 ∈ E, consider the family Fy1 ,y2 of all continuous curves with range in E and endpoints y1 and y2 . We are interested in finding a curve of minimal length, that is, a solution of the problem (5.19)
inf{L(γ) : γ ∈ Fy1 ,y2 }.
Any solution (if it exists) of the previous minimization problem is called a shortest path joining y1 and y2 . Remark 5.29. Let (Y, d) be a metric space and let E ⊆ Y . For every y1 , y2 ∈ E let dL (x1 , x2 ) := inf{L(γ) : γ ∈ Fy1 ,y2 }. Note that 0 ≤ dL (y1 , y2 ) ≤ ∞, dL (y1 , y2 ) = dL (y2 , y1 ). Moreover, since d(y1 , y2 ) ≤ dL (y1 , y2 ), we have that dL (y1 , y2 ) = 0 if and only if y1 = y2 . Finally, if y1 , y2 , y3 ∈ E, and dL (y1 , y3 ) + dL (y3 , y2 ) < ∞, let γ1 ∈ Fy1 ,y3 and γ2 ∈ Fy3 ,y2 . Then we can join γ1 to γ2 to get a curve γ with range in E and joining y1 to y2 . Hence, dL (y1 , y2 ) ≤ L(γ) ≤ L(γ1 ) + L(γ2 ). Taking first the infimum over all curves γ1 and then over all curves γ2 gives dL (y1 , y2 ) ≤ dL (y1 , y3 ) + dL (y3 , y2 ). Thus, dL : Y × Y → [0, ∞] has all the properties of a distance but it can be infinite. It is called the intrinsic or length distance with respect to E. A metric space (Y, d) is called a length space if dL = d, where dL is the length distance with respect to Y . The Euclidean space RM is a length space, while the unit sphere ∂B(0, 1) in RM is not. Exercise 5.30. Find the length distance of the metric space (R2 , d), where # d((x1 , y1 ), (x2 , y2 )) := |x1 − x2 | + |y1 − y2 |.
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147
We recall the following definitions. Definition 5.31. Given a metric space (Y, d), a set E ⊆ Y is said to be (i) disconnected if there exist two open sets U and V in Y such that U ∩ E and V ∩ E are nonempty, U ∩ V = ∅, and E ⊆ U ∪ V , (ii) connected if E is not disconnected, (iii) pathwise connected if for all z1 , z2 ∈ E there exists a continuous curve joining z1 and z2 and with range contained in E. Given a metric space (Y, d) and a set E ⊆ Y , a connected component of E is a maximal connected subset of E, that is, a connected subset of E that is properly contained in no other connected subset of E. Exercise 5.32. Let (Y, d) be a metric space and let E ⊆ Y . Prove that if E is pathwise connected, then E is connected. Exercise 5.33. Let (Y, · ) be a normed space and let E ⊆ Y be an open connected set. Consider the sets U :={y ∈ E : there exists a continuous curve joining y and y0 and with range contained in E} and B := E \ A. Prove that U and V are open and conclude that E is pathwise connected. Exercise 5.34. Let (Y, d) be a metric space and let E1 , E2 ⊆ Y be two connected sets. Prove that if E1 ∩E2 is nonempty, then E1 ∪E2 is connected. Exercise 5.35. Prove that the set E = E1 ∪ E2 of R2 , where E1 := {(0, x2 ) : −1 ≤ x2 ≤ 1},
E2 := {(x1 , sin 1/x1 ) : x1 > 0},
is connected but not pathwise connected. Exercise 5.36. Prove that in a length space every ball B(y, r) contains a ball B(y, ε) which is path connected. To prove the existence of shortest paths we will rely on the following compactness theorem. Theorem 5.37 (Ascoli–Arzel` a). Let I ⊂ R be a closed bounded interval and let (Y, · Y ) be a Banach space. Then U ⊆ C(I; Y ) is relatively compact in C(I; Y ) if and only (i) for every x ∈ I ◦ , the set {u(x) : u ∈ U } is relatively compact in Y , (ii) U is uniformly equicontinuous, that is, for every ε > 0 there exists δ > 0 such that u(x1 ) − u(x2 )Y ≤ ε for all x1 , x2 ∈ I with |x1 − x2 | ≤ δ and for all u ∈ U .
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5. Curves
Proof. We only prove that (i) and (ii) implies that U ⊆ C(I; Y ) is relatively compact in C(I; Y ) and leave the other implication as an exercise. Let I = [a, b]. We claim that the set {u(a) : u ∈ U } is relatively compact in Y . Indeed, fix ε > 0 and let 0 < δ < b − a be as in (ii). Since the set {u(a + δ) : u ∈ U } is relatively compact in Y , there exist y1 , . . . , ym ∈ Y such that for every u ∈ U there is i ∈ {1, . . . , m} with u(a + δ) − yi Y ≤ ε. In turn, by (ii) and the triangle inequality, u(a) − yi Y ≤ 2ε, which proves the claim. Similarly, {u(b) : u ∈ U } is relatively compact in Y . Next, given ε > 0 let δ > 0 be the number given in (ii). Consider a partition {x0 , . . . , xn } of [a, b] into subintervals of length less than or equal to δ. Since for each i the set {u(xi ) : u ∈ U } is relatively compact, there exist y1,i , . . . , ymi ,i ∈ Y such that for every u ∈ U there is j ∈ {1, . . . , mi } with u(xi ) − yj,i Y ≤ ε. For every u ∈ U let un be a continuous function such that un (xi ) = u(xi ) for all i = 1, . . . , n and un is affine in [xi−1 , xi ]. If x ∈ [xi−1 , xi ], then by (ii) ||u(x) − un (x)Y ≤ ||u(x) − u(xi )Y + ||u(xi ) − un (x)Y ≤ ε + ||u(xi ) − u(xi−1 )Y ≤ 2ε. Hence, u − un C(I;Y ) ≤ 2ε. Consider the finite family V ⊂ C(I; Y ) of all piecewise affine functions v such that for every i = 1, . . . , n, v(xi ) ∈ {y1,i , . . . , ymi ,i }, and v is affine in [xi−1 , xi ]. For every u ∈ U there exists v ∈ V such that ||u(xi ) − v(xi )Y ≤ ε for all i = 1, . . . , n. In turn, by (ii), for every i = 1, . . . , n, if x ∈ [xi−1 , xi ], ||u(x) − v(x)Y ≤ ||u(x) − un (x)Y + ||un (x) − v(x)Y ≤ 4ε.
This completes the proof.
Theorem 5.38 (Existence of shortest paths). Let C ⊆ RM be a closed set, let y1 , y2 ∈ C, and let Fy1 ,y2 be the family of all continuous curves with range in C and endpoints y1 and y2 . If the family Fy1 ,y2 is nonempty, then problem (5.19) admits a solution. Proof. Let L := inf{L(γ) : γ ∈ Fy1 ,y2 }. If L = ∞, then any γ ∈ Fy1 ,y2 will do, while if L = 0, then y1 = y2 and the problem becomes trivial. Thus, assume that 0 < L < ∞ and let {γn }n be a sequence of curves in Fy1 ,y2 such that lim L(γn ) = L.
n→∞
Let un : [an , bn ] → RM be a parametric representation of γn with un (an ) = y1 and un (bn ) = y2 . Note that in general we are not in a position to apply
5.4. Curves and Hausdorff Measure
149
Theorem 5.24 to γn , since un may be constant on a proper interval. However, by Corollary 2.24 and the fact that L(γn ) < ∞, we can write un = wn ◦ V n , where V n (t) := Var[an ,t] u, wn : [0, L(γn )] → RM is Lipschitz continuous with Lipschitz constant at most 1, and L(γn ) = VarIn un . Define vn (s) := wn (sL(γn )),
s ∈ [0, 1].
Then vn is Lipschitz continuous with Lipschitz constant at most L(γn ), vn (0) = y1 , vn (1) = y2 and Var[0,1] vn = L(γn ) ≤ L + 1 for all n sufficiently large. By the Ascoli–Arzel` a theorem there exists a function v : [0, 1] → RM and a subsequence (not relabeled) such that vn → w uniformly in [0, 1] as n → ∞. Using the facts that vn (s1 ) − vn (s2 ) ≤ L(γn )|s1 − s2 | for all s1 , s2 ∈ [0, 1] and that L(γn ) → L, letting n → ∞, we obtain that v is Lipschitz continuous with Lipschitz constant at most L. Moreover, v(0) = y1 , v(1) = y2 , and v([0, 1]) ⊆ C, since C is closed. Thus, the curve γ parametrized by w belongs to Fy1 ,y2 , and so L(γ) ≥ L. On the other hand, by Proposition 5.16, L(γ) ≤ lim inf L(γn ) = lim L(γn ) = L, n→∞
n→∞
and so L(γ) = L and the proof is concluded.
Exercise 5.39. Prove that the previous theorem continues to hold if RM is replaced by a proper metric space (Y, d), that is, a metric space in which {y ∈ Y : d(y, y0 ) ≤ r} is compact for every y0 ∈ Y and r > 0.
5.4. Curves and Hausdorff Measure In this section we study the relation between curves and the Hausdorff measure H1 . Theorem 5.40. Let γ be a locally rectifiable continuous simple arc in RM with range Γ and let f : Γ → [0, ∞] be an H1 measurable function. Then 1 f (y) dH (y) = f (u(t))u (t) dt, (5.20) Γ
I
where u : I → RM is any Lipschitz continuous parametric representation of γ. Note that γ admits a Lipschitz continuous parametric representation in view of Theorem 5.24.
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5. Curves
Proof. Step 1: Assume first that γ is rectifiable and that f = χW , where W ⊆ Γ is a relatively open set in Γ. Then u−1 (W ) is relatively open in I and thus we can find countably many intervals Ik such that (5.21) u−1 (W ) = Ik . k
We claim that
u (t) dt.
H (u(Ik )) = 1
(5.22)
Ik
To see this, fix a partition P := {t0 , . . . , tn } of Ik . Since γ is a simple arc, u is injective. Moreover, H1 ({y}) = 0 for all y ∈ Y . Hence, by Lemma 2.63, we have that n u([ti−1 , ti ]) H1 (u(Ik )) ≥ H1 (u([t0 , tn ])) = H1 =
n
i=1 n
H1 (u([ti−1 , ti ])) ≥
i=1
u(ti ) − u(ti−1 ).
i=1
Taking the supremum over all partitions P of Ik gives H1 (u(Ik )) ≥ VarIk u. On the other hand, reasoning exactly as in the second part of the proof of Lemma 2.63, we have that H1 (u(Ik )) ≤ VarIk u. Thus, H1 (u(Ik )) = VarIk u. The claim (5.22) now follows from Theorem 5.12. Summing both sides of (5.22) and using (5.21) and the fact that u is injective gives f (y) dH1 (y) = H1 (W ) = H1 u(Ik ) Γ
=
k
H (u(Ik )) = 1
k
=
u−1 (W )
k
u (t) dt =
u (t) dt Ik
f (u(t))u (t) dt. I
Step 2: Assume next that γ is rectifiable and that f = χE , where E ⊆ Γ is an H1 measurable set. Then for every k ∈ N there exists a relatively open set Wk ⊆ Γ containing E such that H1 (Wk \ E) ≤ 1/k.
Setting B := k Wk , we have that H1 (B \ E) = 0. Since Vn := nk=1 Wk is relatively open in Γ, by Step 1, χVn (y) dH1 (y) = χVn (u(t))u (t) dt. Γ
I
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151
Since γ is rectifiable, by Step 1, H1 (Γ) = VarI u < ∞. Letting n → ∞ it follows by the Lebesgue dominated convergence theorem applied to both sides that 1 (5.23) χB (y) dH (y) = χB (u(t))u (t) dt. Γ
I
Let F := {t ∈ I : u is differentiable at t}. Then L1 (I \ F ) = 0 and so by Theorem 3.41, H1 (u(I \ F )) = 0. On the other hand, since H1 (B \ E) = 0, by Lemma 3.60, u (t) = 0 for t ∈ u−1 (B \ E) ∩ F . Hence, χB\E (u(t))u (t) dt = χB\E (u(t))u (t) dt = 0.
L1 -a.e.
Since (5.24)
I 1 H (B
I∩F
\ E) = 0 it follows from the previous equality and (5.23) that 1 χE (y) dH (y) = χE (u(t))u (t) dt. Γ
I
Step 3: The case in which Γ is rectifiable and f is a simple function follows by (5.23) and the additivity of the integrals. If f is an H1 measurable function, find a increasing sequence of simple functions converging pointwise to f and apply the Lebesgue monotone convergence theorem to both sides. Finally, if γ is locally rectifiable, consider an increasing sequence of closed intervals [an , bn ] ⊆ I whose union gives I, by applying (5.20) to the rectifiable curve γn parametrized by u : [an , bn ] → RM and then applying the Lebesgue monotone convergence theorem to both sides. Remark 5.41. Since for a simple arc γ we have that Nu (y; I) = 1 for all y in the range of γ, the equality H1 (u(Ik )) = VarIk u is just a special case of Banach’s theorem, Theorem 2.60. Remark 5.42. It follows from the previous result with f = 1 that the Hausdorff dimension of a rectifiable continuous simple arc is one. This is in sharp contrast to the Peano curve whose Hausdorff dimension is two. Remark 5.43. If γ is parametrized by arclength v : J → RM , then by Proposition 5.23, v (s) = 1 for L1 -a.e. s ∈ J and so (5.20) takes the simpler form f (y) dH1 (y) = Γ
f (v(s)) ds. J
The right-hand side is called curve or line integral and is often denoted f ds. γ Remark 5.44. As usual, if f : Γ → R is an H1 integrable function, then by writing f = f + − f − , where s+ and s− are the positive and negative parts of s ∈ R, and applying (5.20) to f + and f − , we conclude that (5.20) holds for f .
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5.5. Jordan’s Curve Theorem We conclude this chapter with the Jordan curve theorem. We recall that a pathwise connected set is connected, but the converse is not true in general (see Exercise 5.35). However, if a set is open, then it is connected if and only if it is pathwise connected. In what follows, we will use this fact without further notice. In this section we use the notation z = (x, y) for a generic → point of R2 . Given two points z1 , z2 ∈ R2 , we denote by z−− 1 z2 the segment joining them, that is, → z−− 1 z2 := {tz1 + (1 − t)z2 : t ∈ [0, 1]}. We recall that a closed curve is continuous. Theorem 5.45 (Jordan’s curve theorem). Given a (continuous) closed simple curve γ in R2 with range Γ, the set R2 \ Γ consists of two connected components. We call the unbounded connected component the exterior component of γ and the other one the interior component of γ. We begin with some preliminary results. Lemma 5.46. Let a < b, c < d and let u : [−1, 1] → [a, b] × [c, d] and v : [−1, 1] → [a, b] × [c, d] be two continuous functions such that u1 (−1) = a, u1 (1) = b, and v2 (−1) = c, v2 (1) = d. Then there exists t0 , s0 ∈ [−1, 1] such that u(s0 ) = v(t0 ). Proof. Assume by contradiction that u(s) = v(t) for all s, t ∈ [−1, 1]. Let Q := [−1, 1] × [−1, 1] and consider the continuous function h : Q → Q defined by v (t) − u (s) u2 (s) − v2 (t) 1 1 , , h(s, t) := u(s) − v(t)∞ u(s) − v(t)∞ where z∞ := max{|x|, |y|}. Note that h(s, t) ∈ ∂Q. By the Brouwer fixed point theorem (see Theorem 9.36 below) there exists (s0 , t0 ) ∈ Q such that h(s0 , t0 ) = (s0 , t0 ). But since h(Q) ⊆ ∂Q, necessarily, (s0 , t0 ) ∈ ∂Q. Hence, s0 = ±1 or t0 = ±1. If s0 = 1, then 1=
v1 (t0 ) − b v1 (t0 ) − u1 (1) = ≤ 0, u(1) − v(t0 )∞ u(1) − v(t0 )∞
while if s0 = −1, then −1 =
v1 (t0 ) − a v1 (t0 ) − u1 (−1) = ≥ 0, u(−1) − v(t0 )∞ u(−1) − v(t0 )∞
which gives a contradiction.
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153
On the other hand, if t0 = 1, then 1=
u2 (s0 ) − v2 (1) u2 (s0 ) − d = ≤ 0, u(s0 ) − v(1)∞ u(s0 ) − v(1)∞
while if t0 = −1, then −1 =
u2 (s0 ) − c u2 (s0 ) − v2 (−1) = ≥ 0, u(s0 ) − v(−1)∞ u(s0 ) − v(1)∞
which gives again a contradiction. This completes the proof.
Remark 5.47. Since Γ is bounded, the set R2 \ Γ has only one unbounded component. Indeed, let Γ ⊆ B(0, r). The set R2 \ B(0, r) is open and pathwise connected and so it is connected. Moreover, R2 \ B(0, r) ⊆ R2 \ Γ, and so there is a connected component that contains R2 \ B(0, r). Lemma 5.48. Given a (continuous) closed simple curve γ in R2 with range Γ, if U is a connected component of R2 \ Γ, then ∂U = Γ. Proof. If V is any other connected component of R2 \ Γ, then by Exercise 5.34, U ∩ V = ∅, since otherwise U ∪ V would be connected, and this would contradict the maximality of U . Thus ∂U ⊆ Γ. Assume by contradiction that ∂U ⊂ Γ. Then there exists a simple arc whose range C ⊂ Γ is closed and contains ∂U . Moreover, since ∂U ⊂ Γ, there must be at least another connected component. Since there is only one unbounded connected component, there is a bounded connected component. Take p to be a point in this bounded connected component. If U is bounded, take p in U . Let r > 0 be so large that B(p, r) ⊃ Γ. Then ∂B(p, r) is contained in the unbounded connected component of R2 \ Γ. Define u(z) := z for all z ∈ C. Since C is homeomorphic to [0, 1], by Tietze’s extension theorem (see Theorem A.10 in the appendix) we can extend u to a continuous function u : B(p, r) → C. If U is bounded, define u(z) if z ∈ U , v(z) := z if z ∈ R2 \ U, while if U is unbounded, define z if z ∈ U , v(z) := u(z) if z ∈ R2 \ U. Then v : B(p, r) → B(p, r) \ {p}. Indeed, if U is bounded, then p belongs to U and v is mapped into C ⊂ Γ, while if U is unbounded, then p belongs to R2 \ U and v is mapped into C ⊂ Γ. Moreover, since ∂U ⊆ C and v(z) = z for all z ∈ C, we have that v is continuous. Note that v(z) = z for all z ∈ ∂B(p, r). Consider the projection π : B(p, r) \ {p} → ∂B(p, r) given
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5. Curves
Figure 2. Figure from R. Maehara, The Jordan curve theorem via the Brouwer fixed point theorem, Amer. Math. Monthly 91 (1984), no. 10, 641–643. Copyright 1984 Mathematical Association of America. All Rights Reserved. Reprinted with permission.
by π(z) := p + r z−p
z and let h : ∂B(p, r) → ∂B(p, r) be a rotation by 180 degrees. Then the continuous function h◦π◦v : B(p, r) → ∂B(p, r) ⊂ B(p, r) has no fixed point. This contradicts the Brouwer fixed point theorem (see Theorem 9.36 below). We now turn to the proof of Jordan’s curve theorem. Since γ is a simple curve, in what follows, with an abuse of notation, we will often confuse γ with its range Γ. Proof of Jordan’s curve theorem. Since Γ is compact, there exist a, b ∈ Γ such thata − b = diam Γ. By changing coordinates, without loss of generality we may assume that a = (−1, 0) and b = (1, 0). Then the rectangle R = [−1, 1] × [−2, 2] contains Γ and has only a and b on its boundary (see → Figure 2). Let n = (0, 2) and s = (0, −2). By Lemma 5.46 the segment − ns → with maximal y-component. intersects Γ. Let l be the point in Γ ∩ − ns The points a and b divide Γ in two arcs, let Γn be the one containing l → with minimal ns and let Γs be the other one. Let m be the point in Γn ∩ − y-component.
5.5. Jordan’s Curve Theorem
155
→ intersects Γ , since otherwise, denoting by lm % Then the segment − ms s the subarc contained in Γn with endpoints l and m, the curve given by → % −→ − nl + lm + ms would not intersect the curve Γs , contradicting Lemma 5.46. → be the points with maximal and minimal ms Let p and q be the points in Γs ∩ − y-component, respectively. Finally, let z0 be the middle point of the segment − → Note that m = p since γ is simple. Hence, z does not belong to Γ. Let mp. 0 U be the connected component of R2 \ Γ which contains z0 . We claim that U is bounded. Assume by contradiction that U is unbounded. Since U is open and connected, it is path-connected and so we can find a polygonal path in U joining z0 to a point outside R. Let w be the point at which this polygonal path first intersects ∂R and denote with γ1 the portion of this polygonal % be the subarc arc joining z0 and w. If w is on the lower half of R, let ws contained in ∂R with endpoints w and s and not intersecting a, b. Then the → % −−→ − % would not intersect the curve Γs , curve given by nl + lm + mz0 + γ1 + ws contradicting Lemma 5.46. On the other hand, if w is on the upper half of R, let wn % be the subarc contained in ∂R with endpoints w and n and not intersecting a, b. Then the → + γ + wn curve − sz % would not intersect the curve Γn , contradicting again 0 1 Lemma 5.46. This shows that U is bounded. It remains to show that U is the only bounded connected component of R2 \Γ. Assume by contradiction that there is another one, say, V . Then V ⊂ R, since R2 \ R is pathwise connected and thus contained in the unbounded connected component of R2 \ Γ. Since the → − → segments nl \ {l} and − qs \ {q} are contained in the unbounded connected component of R2 \ Γ, they do not intersect V . Similarly, since z0 ∈ U , the → \ {m, p} is contained in U , and thus it does not intersect V . It segment − mp → % −→ − → mp+ pq & +− qs, where pq & the subarc follows that the curve γ2 given by nl + lm+ contained in Γs with endpoints p and q, does not intersects V . Since a and b are not in γ2 , there are balls B(a, r) and B(b, r) which do not intersect γ2 . By Lemma 5.48, ∂V = Γ and so a and b belong to ∂V . Hence, there exist a1 ∈ V ∩ B(a, r) and b1 ∈ V ∩ B(b, r). Let a' 1 b1 be a polygonal path in V −→ − → ' joining a1 and b1 . Then the curve aa1 + a1 b1 + b1 b does not intersect γ2 , contradicting again Lemma 5.46. This concludes the proof.
Chapter 6
Lebesgue–Stieltjes Measures Undergradese, VI: “Are you going to have office hours today?” Translation: “Can I do my homework in your office?” — Jorge Cham, www.phdcomics.com
In the literature there have been several attempts to extend the notion of pointwise variation to functions u : R → RM , where now R is a rectangle of RN , but they have never been quite satisfactory. As we will see in the next chapters, a more successful approach is to give a different characterization of functions of bounded pointwise variation, namely as the class of functions whose “derivative” is a finite measure. We will make this concept more precise in Chapter 7. In this chapter we describe the correspondence between functions of bounded pointwise variation and Radon measures. We will start with increasing functions. We will show that there is a one-to-one correspondence between increasing, right continuous functions and (positive) Radon measures. In this chapter I ⊆ R is an open interval and L1 stands for the Lebesgue measure restricted to Lebesgue measurable subsets of I. Given an interval J ⊆ R we denote by S(J) the semiring of all intervals (a, b], where a, b ∈ J, with a ≤ b, and by R(J) the ring generated by S(J). The smallest σ-algebra containing R(J) is the Borel σ-algebra B(J). Finally, we recall that a Radon measure μ : B(J) → [0, ∞] is a measure finite on compact subsets of J.
6.1. Measures Versus Increasing Functions The next result shows that every (positive) Radon measure gives rise to a right (or left) continuous increasing function. 157
158
6. Lebesgue–Stieltjes Measures
Theorem 6.1. Let I ⊆ R be an open interval and let μ : B(I) → [0, ∞] be a Radon measure. Then for a fixed t0 ∈ I the function uμ : I → R, defined by μ((t0 , x]) if x ≥ t0 , (6.1) uμ (x) := −μ((x, t0 ]) if x < t0 , is increasing and right continuous and uμ (b) − uμ (a) = μ((a, b])
(6.2)
for all a, b ∈ I, with a ≤ b. Proof. By the properties of measures, if E, F ∈ B(I), with E ⊆ F , we have that μ(E) ≤ μ(F ), and so, if t0 ≤ x1 ≤ x2 , then uμ (x1 ) = μ((t0 , x1 ]) ≤ μ((t0 , x2 ]) = uμ (x2 ), while if x1 ≤ x2 ≤ t0 , then μ((x1 , t0 ]) ≥ μ((x2 , t0 ]), and so uμ (x1 ) = −μ((x1 , t0 ]) ≤ −μ((x2 , t0 ]) = uμ (x2 ). Finally, if x1 ≤ t0 ≤ x2 , then uμ (x1 ) ≤ 0 ≤ uμ (x2 ). Hence, uμ is increasing. To prove that uμ is right continuous, fix x ∈ I and let xn → x+ . Since uμ is increasing, by extracting a subsequence it suffices to assume that {xn }n is decreasing. There are two cases. If x ≥ t0 , then, since μ((t0 , x]) < ∞, by Proposition B.9(ii) we have that ∞ (t0 , xn ] = μ((t0 , x]), lim μ((t0 , xn ]) = μ n→∞
n=1
while if x < t0 , then by Proposition B.9(i), ∞ (xn , t0 ] = μ((x, t0 ]). lim μ((xn , t0 ]) = μ n→∞
n=1
Thus, uμ is right continuous. Property (6.2) follows from the definition of uμ .
Remark 6.2. (i) Note that if we work with intervals of the type [a, b), then the function uμ defined correspondingly is left continuous. (ii) If μ : B(I) → [0, ∞) is a finite measure, then we can work with the simpler function u(x) := μ(I ∩ (−∞, x]),
x ∈ I.
Next we show the relation between the properties of the measure μ and of the increasing function uμ . Set Idis := {x ∈ I : uμ is discontinuous at x}, Ider0 := {x ∈ I : uμ (x) = 0}.
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159
Theorem 6.3. Let I ⊆ R be an open interval and let μ : B(I) → [0, ∞] be a Radon measure and let uμ be the function given in (6.1). Then (i) The measure μ is purely atomic if and only if uμ is a jump function. Moreover, in this case, for every Borel set E ⊆ I, μ(E ∩ {x}). (6.3) μ(E) = x∈Idis
(ii) The measure μ is absolutely continuous with respect to L1 if and only uμ is locally absolutely continuous. Moreover, in this case, for every Borel set E ⊆ I, uμ dx. (6.4) μ(E) = E
(iii) The measure μ is nonatomic and μ ⊥ L1 if and only if uμ is a singular function. Moreover, in this case, (6.5)
μ(E) = μ(E \ Ider0 ) for every Borel set E ⊆ I.
Lemma 6.4. Let I ⊆ R be an open interval and let μ : B(I) → [0, ∞] be a Radon measure. Then E ∈ B(I) is an atom if and only if, up to a set of measure zero, E = {x} with x ∈ I a discontinuity point of uμ . Proof. Let E ∈ B(I) be an atom. In what follows we will neglect sets of measure zero. Then := μ(E) > 0 and for every Borel set F ⊆ E, either μ(F ) = 0 or μ(F ) = . Partition I into intervals Ii,1 of the form (a, b] with endpoints in I and of length less than or equal to 1. Since μ(Ii,1 ) < ∞ for every i and Ii,1 ∩ E has either measure zero or , it follows that < ∞. In turn, since the sets Ii,1 ∩ E, i ∈ N, are disjoint, there exists only one i1 ∈ N such that μ(Ii1 ,1 ∩ E) = , while μ(Ii,1 ∩ E) = 0 for all i = i1 . This implies that E ⊆ Ii1 ,1 =: I1 . Inductively, assume that E ⊆ In ⊆ · · · ⊆ I1 have been selected with diam In ≤ n1 and μ(In ) = . Partition In into intervals Ii,n+1 of the form 1 and let in+1 be the only integer (a, b] and of length less than or equal to n+1 such that μ(Iin+1 ,n+1 ∩ E) = , while μ(Ii,n+1 ∩ E) = 0 for i = in+1 . Then E ⊆ Iin+1 ,n+1 := In+1 . Thus we have constructed a decreasing sequence of intervals In containing E, with μ(In ) = and diam In ≤ n1 . But then ∞ E ⊆ n=1 In . Since diam In → 0, it follows that E must be a singleton, say, E = {x}. Moreover, by (6.2) we have that uμ (x) − uμ (x − h) = μ((x − h, x]) for all h > 0 such that x − h ∈ I. Letting h → 0+ gives (6.6)
uμ (x) − (uμ )− (x) = μ({x}).
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6. Lebesgue–Stieltjes Measures
Thus, {x} is an atom of μ if and only if x is a discontinuity point of uμ .
Lemma 6.5. Let I ⊆ R be an open interval and let μ : B(I) → [0, ∞] be a Radon measure. Then for every Borel set E ⊆ I, L1o (uμ (E)) ≤ μ(E).
(6.7)
Moreover, if uμ is continuous in I, then equality holds in (6.7). Proof. Given ε > 0 we can find an open set U ⊆ I containing E such that μ(U \ E) ≤ ε. DecomposeU into countably many disjoint intervals of the form (ai , bi ]. Then the set i [uμ (ai ), uμ (bi )] covers uμ (E), and so
1 1 Lo (uμ (E)) ≤ L [uμ (ai ), uμ (bi )] ≤ L1 ([uμ (ai ), uμ (bi )]) i
i
(uμ (bi ) − uμ (ai )) = μ((ai , bi ]) = i
i
= μ(U ) = μ(E) + μ(U \ E) ≤ μ(E) + ε, where we have used (6.2). Letting ε → 0 concludes the first part of the proof. Assume next that uμ is continuous in I, let E ⊆ I be a Borel set, and let V ⊆ R be an open set such that uμ (E) ⊆ V . We claim that μ(E) ≤ L1 (V ). Since I is open and uμ is continuous, the set U := u−1 μ (V ) is open and thus we can decompose it into countably many disjoint intervals of the form (ai , bi ]. It follows from (6.2) and the fact that E ⊆ U that μ((ai , bi ]) = (uμ (bi ) − uμ (ai )) μ(E) ≤ μ(U ) = =
i
i
L (uμ ((ai , bi ))) ≤ L (V ), 1
1
i
where we used the facts that uμ ((ai , bi )) is an interval of endpoints uμ (ai ) and uμ (bi ) because uμ is continuous and increasing, and that the sets (uμ (ai ), uμ (bi )) are disjoint and contained in V . This proves the claim. By taking the infimum over all open sets V containing uμ (E) we get μ(E) ≤ L1 (uμ (E)). Remark 6.6. Note that in general the inequality (6.7) is strict. Indeed, let x be a discontinuity point of uμ . Take E = {x}. Then L1o ({uμ (x)}) = 0 < uμ (x) − (uμ )− (x) = μ({x}), where we have used (6.6). We now turn to the proof of Theorem 6.3.
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161
Proof of Theorem 6.3. (i) Assume that μ is purely atomic. Then every set of positive measure contains an atom. Since Idis is countable, we can write Idis = {an : n ∈ G} for some G ⊆ N. Hence, for every Borel set E ⊆ I, μ(E ∩ {an }), μ(E) = μ(E \ Idis ) + μ(E ∩ Idis ) = 0 + n
where we have used Lemma 6.4. Hence, if x ∈ I is such that x ≥ t0 , by (6.1), μ((t0 , x] ∩ {an }) = μ({an }), (6.8) uμ (x) = μ((t0 , x]) = t0 nε
By relabeling an we can assume that an ≤ an+1 for all n < nε . Then by (6.9) we have μ((a, b] \ Idis ) ≤ μ((a, a1 )) +
n ε −1 n=1
μ((an , an+1 )) + μ((anε , b]) ≤
bn ≤ ε.
n>nε
Letting ε → 0+ , we have that μ((a, b] \ Idis ) = 0. By the arbitrariness of [a, b], it follows that μ(I \ Idis ) = 0, which shows that every set of positive measure contains an atom. Thus μ is purely atomic. (ii) Assume that μ L1 . Since singletons have Lebesgue measure zero, it follows that μ({x}) = 0 for all x ∈ I. Hence, by Lemma 6.4, μ has no atoms and uμ is continuous in I. In view of Theorem 3.41 to prove that uμ is locally absolutely continuous, it suffices to show that it satisfies the Lusin (N ) property. Let E ⊂ I be such that L1 (E) = 0. Find a Borel set F containing E such that L1 (F ) = 0. Since μ L1 , it follows that
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6. Lebesgue–Stieltjes Measures
μ(F ) = 0. For every ε > 0 we can find an open set U ⊆ I such that F ⊆ U and μ(U ) ≤ ε. By decomposing U into countably many disjoint intervals of the form (ai , bi ], it follows from (6.2) and the fact that uμ is increasing and continuous that L1o (uμ (E)) ≤ L1o (uμ (U )) ≤ L1o (uμ ((ai , bi ])) i
(uμ (bi ) − uμ (ai )) = μ((ai , bi ]) = μ(U ) ≤ ε. = i
Letting ε →
0+
i
shows that
L1o (uμ (E))
= 0.
Conversely, assume that uμ is locally absolutely continuous. Then by (6.2) and Theorem 3.20 we have b μ((a, b]) = uμ (b) − uμ (a) = uμ dx. a
Given an open set U ⊆ I, by decomposing U into countably many disjoint intervals of the form (ai , bi ], it follows that uμ dx. (6.11) μ(U ) = U
Since uμ is locally integrable, for every Borel set E ⊂ I with L1 (E) = 0, we have that E uμ dx = 0. In turn, given ε > 0 we can find an open set U ⊆ I containing E such that U uμ dx ≤ ε. It follows that μ(E) ≤ μ(U ) ≤ ε and by letting ε → 0+ we conclude that μ L1 . Now, let E ⊆ I be a bounded Borel set, with E ⊂ I. Then we can find a sequence of decreasing open sets Un and a set F ⊂ I with L1 (F ) = 0 such that ∞ Un . E∪F = n=1
Since E ⊂ I is compact and I open, we can assume that the sets Un are compactly contained in I so that μ(Un ) < ∞. Using the fact that μ L1 , we have that μ(F ) = 0. Taking U to be Un in (6.11) and using Proposition B.9(ii) and the Lebesgue dominated convergence theorem, we get (6.12) uμ dx μ(E) = μ(E ∪ F ) = lim μ(Un ) = lim n→∞ n→∞ U n uμ dx = uμ dx, = E∪F
E
where we used the fact that μ(F ) = L1 (F ) = 0. The case of an arbitrary Borel set E is left as an exercise.
6.1. Measures Versus Increasing Functions
163
(iii) Assume that μ is nonatomic and μ ⊥ L1 . As in part (ii) it follows that uμ is continuous. Since μ ⊥ L1 , there exists a Borel set Eμ ⊆ I such that for every Borel set E ⊆ I we have L1 (E) = L1 (E \ Eμ ).
μ(E) = μ(E ∩ Eμ ),
In particular, μ(I \ Eμ ) = 0 and L1 (Eμ ) = 0. Hence, L1 (uμ (I \ Eμ )) = 0 by Lemma 6.5. Since uμ is increasing, by Lebesgue’s differentiation theorem (see Theorem 1.18), the set F := {x ∈ I : uμ is not differentiable at x} satisfies L1 (F ) = 0. By Lemma 3.60 and the fact that L1o (uμ (I \ (Eμ ∪ F ))) = 0 we have that that uμ (x) = 0 for L1 -a.e. x ∈ I \ (Eμ ∪ F ). But L1 (Eμ ∪ F ) = 0, and so uμ is a singular function. Conversely, assume that uμ is a singular function. Then uμ is continuous, and so by Lemma 6.4, μ is nonatomic. Since uμ is singular, we have that L1 (I \ Ider0 ) = 0. By Corollary 3.37 and Lemma 6.5, μ(Ider0 ) = L1 (u(Ider0 )) = 0 and thus for every Borel set E ⊆ I, μ(E) = μ(E \ Ider0 ),
which shows that μ ⊥ L1 .
Next we prove the converse of Theorem 6.1; that is, we show that to every increasing function we may associate a unique Radon measure. Theorem 6.7. Let I ⊆ R be an open interval and let u : I → R be an increasing function. Then there exists a unique Radon measure μu : B(I) → [0, ∞] such that (6.13)
μu ((a, b]) = u+ (b) − u+ (a)
for all a, b ∈ I, with a ≤ b. Moreover, (6.14)
μu (I) = sup u − inf u. I
I
In particular, μu is finite if and only if u is bounded. The measure μu given by the previous theorem is called the Lebesgue– Stieltjes measure generated by u. To construct μu , we begin by defining μu on the semiring S(I) of all intervals (a, b], where a, b ∈ I, with a ≤ b. In view of Theorem 6.1, we set (6.15)
ρu ((a, b]) := u+ (b) − u+ (a).
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6. Lebesgue–Stieltjes Measures
Lemma 6.8. Let I ⊆ R be an open interval and let u : I → R be an increasing function. Then the set function ρu : S(I) → [0, ∞) is countably subadditive, that is, ∞ ρu ((an , bn ]) ρu ((a, b]) ≤ n=1
for every (a, b], (an , bn ] ∈ S(I), n ∈ N, such that (a, b] ⊆
∞
(an , bn ].
n=1
Proof. Let (a, b], (an , bn ] ∈ S(I) be such that (a, b] ⊆ ∞ n=1 (an , bn ]. If a = b there is nothing to prove, thus assume a < b. Fix ε > 0. Since u+ is rightcontinuous, there exist 0 < δ < b − a and 0 < δn < dist(bn , R \ I) such that u+ (x) − u+ (a) ≤ ε for all a ≤ x ≤ a + δ and u+ (x) − u+ (bn ) ≤ 2εn for all bn ≤ x ≤ bn + δn . Let a < α ≤ a + δ and bn < βn ≤ bn + δn . Then [α, b] ⊂ (a, b] ⊆
∞
∞
(an , bn ] ⊆
n=1
(an , βn )
n=1
and so by compactness we can find such that [α, b] ⊂
(an , βn ).
n=1
Then ρu ((a, b]) = u+ (b) − u+ (a) ≤ u+ (b) − u+ (α) + ε ≤
(u+ (βn ) − u+ (an )) + ε
n=1
≤
(u+ (bn ) − u+ (an )) + 2ε ≤
n=1
∞
ρu ((an , bn ]) + 2ε.
n=1
It suffices to let ε → 0+ . The outer measure μ∗u : P(I) → [0, ∞], given by ∞ ∞ ∗ ρu ((an , bn ]) : (an , bn ] ∈ S(I), E ⊆ (an , bn ] , (6.16) μu (E) := inf n=1
n=1
for E ⊆ I, is called the Lebesgue–Stieltjes outer measure generated by u. Exercise 6.9. Let I ⊆ R be an open interval and let u : I → R be an increasing function. Prove that μ∗u is a metric outer measure. We are now ready to prove Theorem 6.7.
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165
Proof of Theorem 6.7. It follows from Carath´eodory’s theorem (see Theorem B.13) that the restriction of μ∗u to the σ-algebra of all μ∗u -measurable sets is a complete measure. Moreover, by the previous exercise and Proposition B.19 every Borel set in I is μ∗u -measurable. In view of Lemma 6.8 and Remark B.3, (6.13) holds. Note that if [a, b] ⊂ I, then, since I is open, we can find a1 ∈ I with a1 < a. Then [a, b] ⊂ (a1 , b], and so (6.13) and the monotonicity of μu with respect to inclusion imply that μu is finite on [a, b]. Hence, μu is a Radon measure. To prove (6.14), it is enough to let a inf I and b sup I in (6.13) and to use Proposition B.9(i). Uniqueness follows from Corollary B.16. This concludes the proof.
Remark 6.10. (i) If I ⊆ R is an open interval and u, v : I → R are two increasing functions, then by the uniqueness established in the previous theorem, μu+v = μu + μv . (ii) If the interval I is not open, then it contains one of its endpoints, say a := inf I. In this case, if u : I → R is increasing, then we can extend u to (−∞, a) by setting u(x) := u(a) for all x < a. Thus, we can construct the Lebesgue–Stieltjes measure of the extension of u defined on the Borel σ-algebra of (−∞, sup I) and then consider its restriction to B(I). (iii) If μ : B(I) → [0, ∞] is a Radon measure and uμ : I → R is the function defined by (6.1), then by the uniqueness proved in the previous theorem, it follows that μ is the Lebesgue–Stieltjes measure generated by uμ , that is, μuμ = μ. Remark 6.11. Note that under the hypotheses of Theorem 6.7, if b ∈ I taking a ∈ I with a < b in (6.13) and letting a b− along a sequence yields (6.17)
μu ({b}) = u+ (b) − u− (b).
Hence, again by (6.13), if a, b ∈ I, with a ≤ b, then (6.18)
μu ([a, b]) = μu ((a, b]) + μu ({a}) = u+ (b) − u− (a),
(6.19)
μu ((a, b)) = μu ((a, b]) − μu ({b}) = u− (b) − u+ (a).
Exercise 6.12. Construct the Lebesgue–Stieltjes outer measures corresponding to
x for x ≥ 0, 1 for x ≥ 0, u2 (x) = u1 (x) = 0 for x < 0, −1 for x < 0, where x is the integer part of x.
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6. Lebesgue–Stieltjes Measures
Exercise 6.13. Let I ⊆ R be an open interval and let u : I → R be an increasing function. Prove that if in place of (6.15), we define
u ((a, b]) := u(b) − u(a),
(a, b] ∈ S(I),
and in (6.16) we use u instead of ρu , then (u+ (x) − u(x)) μu ((a, b]) = u(b) − u+ (a) − x∈(a,b)
for every (a, b] ∈ S(I). The following change of variables formula transforms a Lebesgue–Stieltjes integral into an integral with respect to the Lebesgue measure. Theorem 6.14. Let I ⊂ R be an open interval bounded from below, let u : I → R be a right continuous increasing function, and let f : I → [0, ∞) be a Borel function. Then for every a, b ∈ I, with a < b, u(b) f (x) dμu (x) = f (v(t)) dt, (6.20) (a,b]
u(a)
where μu is the Lebesgue–Stieltjes measure generated by u and v : J → R is the inverse of u defined as in Theorem 1.8. Proof. Step 1: We claim that if α, β ∈ I, with α < β, then α < v(t) ≤ β if and only if u(α) < t ≤ u(β). Indeed, if α < v(t), then by the definition of v (see (1.1)) we have that u(α) < t. On the other hand, if v(t) ≤ β, let ε > 0 be so small that β + ε ∈ I. Again by the definition of v and the fact that u is increasing we have that u(β + ε) ≥ t. Letting ε → 0+ and using the fact that u is right continuous gives u(β) ≥ t. Thus, if α < v(t) ≤ β, then u(α) < t ≤ u(β). Conversely, if u(α) < t, using the fact that u is right continuous, we may find x > α such that u(x) < t. Hence, v(t) ≥ x > α by the definition of v. Finally, if u(β) ≥ t, then v(t) ≤ β, again by the definition of v. This shows that if u(α) < t ≤ u(β), then α < v(t) ≤ β. Thus, the claim holds. Step 2: Assume that f = χ(α,β] , where α, β ∈ I, with α < β. There are now two cases. If (α, β] ∩ (a, b] = ∅, then the integral on the left-hand side of (6.20) is zero. On the other hand, by Step 1, if t ∈ (u(a), u(b)], then v(t) ∈ (a, b], and so χ(α,β] (v(t)) = 0, since (α, β] ∩ (a, b] = ∅. Thus, the integral on the right-hand side of (6.20) is zero also. If (α, β] ∩ (a, b] = ∅, then (α, β] ∩ (a, b] = (s, t], where s := max{α, a} and t := min{β, b}. Thus, χ(α,β] (x) dμu (x) = μu ((s, t]) = u(t) − u(s), (a,b]
6.1. Measures Versus Increasing Functions
167
where we have used (6.13) and the fact that u is right continuous. On the other hand, again by Step 1, v(t) ∈ (α, β] if and only if t ∈ (u(α), u(β)]. Thus, u(b) χ(α,β] (v(t)) dt = χ(u(a),u(b)] (t)χ(u(α),u(β)] (t) dt u(a)
J
= min{u(β), u(b)} − max{u(α), u(a)} = u(t) − u(s), where in the last equality we have used the fact that u is increasing. This proves (6.20) in the case that f = χ(α,β] . Step 3: For every Borel set B ⊆ I, define χB (x) dμu (x), μ2 (B) := μ1 (B) :=
u(b)
χB (v(t)) dt. u(a)
(a,b]
Note that if an inf I and bn sup I, then for all n sufficiently large, (a, b] ⊆ (an , bn ], and so by the previous step μ1 ((an , bn ]) = μ2 ((an , bn ]) = u(b) − u(a) < ∞. Then μ1 : B(I) → [0, ∞] and μ2 : B(I) → [0, ∞] are two Radon measures that coincide on intervals of the form (α, β]. It follows from Corollary B.16 that μ1 = μ2 . This shows that (6.20) holds in the case that f = χB , where B ⊆ I is a Borel set. A standard argument (using simple functions and the Lebesgue monotone convergence theorem) shows that (6.20) holds for a Borel function. Remark 6.15. A similar result holds if u is right continuous and decreasing. The following exercises discuss integration by parts for Lebesgue–Stieltjes measures. Exercise 6.16. Let μ : B((a, b)) → [0, ∞) and ν : B((a, b)) → [0, ∞) be two finite Borel measures. Prove that (i) (a,b) μ([t, b)) dν(t) = (a,b) ν((a, x]) dμ(x). (ii) (a,b) μ((a, t)) dν(t) + (a,b) ν((a, x]) dμ(x) = μ((a, b))ν((a, b)). (iii) Let f (x) := 12 (μ((a, x]) + μ((a, x))) and g(x) := ν((a, x))), x ∈ (a, b). Prove that f dν + g dμ = μ((a, b))ν((a, b)). (a,b)
1 2 (ν((a, x])
+
(a,b)
Exercise 6.17. Let I ⊆ R be an open interval and let u, v : I → R be increasing functions. Prove that for every [a, b] ⊂ I, 1 1 (u+ + u− ) dμv + 2 (v+ + v− ) dμu = u− (b)v− (b) − v+ (a)u+ (a). 2 (a,b)
(a,b)
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6. Lebesgue–Stieltjes Measures
Exercise 6.18. Let I ⊆ R be an open interval and let u : I → R be an increasing function. Prove that for every ϕ ∈ Cc1 (I), uϕ dx = − ϕ dμu . I
I
6.2. Vector-valued Measures Versus BP V (I; Y ) Next we extend Theorem 6.1 to vector-valued measures. As we have seen in Theorem 6.7 an increasing function u : I → R uniquely determines a Radon measure μu : B(I) → [0, ∞]. The measures μu is finite if and only if u is bounded. Hence, if now u : I → RM belongs to BP Vloc (I; RM ) we cannot expect u to determine a vector-valued measure λu : B(I) → RM unless u belongs to BP V (I; RM ). To overcome this difficulty we restrict our attention to vector-valued measures defined on the σ-field (6.21)
Bc (I) := {E ∈ B(I) : E is compactly contained in I},
that is, Bc (I) consists of all Borel sets whose closure is a compact subset of I. In particular, I does not belong to Bc (I) since I is open by assumption. Given an open interval I ⊆ R, a normed space (Y, · ), and a vectorvalued measure λ : Bc (I) → Y , we recall that the total variation of λ is the measure λ : Bc (I) → [0, ∞] given by n λ(Ei ) , (6.22) λ(E) := i=1
where the supremum is taken over all finite families of disjoint sets Ei ∈ Bc (I), i = 1, . . . , n, such that E = ni=1 Ei and over all n ∈ N. In particular, λ({x}) = λ({x})
(6.23) for every x ∈ I.
Theorem 6.19. Let I ⊆ R be an open interval, let (Y, · ) be a normed space and let λ : Bc (I) → Y be a vector-valued measure. Then for every fixed t0 ∈ I the function uλ : I → Y , defined by λ((t0 , x]) if x ≥ t0 , (6.24) uλ (x) := −λ((x, t0 ]) if x < t0 , is right continuous, (6.25)
uλ (b) − uλ (a) = λ((a, b])
for all a, b ∈ I, with a ≤ b, and for every interval [a, b] ⊆ I, (6.26)
Var[a,b] uλ = λ([a, b]),
Var(a,b] uλ = λ((a, b]).
In particular, λ : Bc (I) → [0, ∞] is a Radon measure if and only if uλ ∈ BP Vloc (I; Y ) and it is a finite measure if and only if uλ ∈ BP V (I; Y ).
6.2. Vector-valued Measures Versus BP V (I; Y )
169
Proof. Step 1: The fact that uλ is right continuous and (6.25) follow as in the proof of Theorem 6.1. We only prove (6.26)1 . To prove (6.26)1 , consider a partition x0 < x1 < · · · < xn of [a, b]. By (6.25) we have n
uλ (xi ) − uλ (xi−1 ) =
i=1
n
λ((xi−1 , xi ]) ≤ λ([a, b]).
i=1
Taking the supremum over all partitions gives Var[a,b] uλ ≤ λ([a, b]). Step 2: To prove the opposite inequality it suffices to assume that Var[a,b] uλ < ∞. Then the function V(x) := Var[a,x] uλ is finite in [a, b]. Since uλ is right continuous, by Proposition 2.21, so is V. Extend V to be V(a) for x < a. We claim that λ(E) ≤ μV (E)
(6.27)
for every E ∈ R([a, b]), where μV is the measure given in Theorem 6.7 (see also Remark 6.10(ii)). To see this, let E ∈ R([a, b]). Write E = ni=1 (ai , bi ], where (ai , bi ] ∈ S([a, b]), i = 1, . . . , n are pairwise disjoint intervals. By the additivity of λ, the right continuity of V, (2.4) and (6.25), we get λ(E) ≤
n
λ((ai , bi ]) =
i=1
=
n
n
uλ (bi ) − uλ (ai ) ≤
i=1
V+ (bi ) − V+ (ai ) =
i=1
n
V(bi ) − V(ai )
i=1 n
μV ((ai , bi ]) = μV (E).
i=1
Thus (6.27) holds. Next we show that (6.27) holds for all E ∈ B([a, b]). For E, F ∈ B([a, b]) define d(E, F ) := μV (EΔF ), where EΔF := (E \ F ) ∪ (F \ E) is the symmetric difference between the sets E and F . It can be checked that d is a semimetric in B([a, b]), that is, d(E, E) = 0, d(E, F ) = d(F, E) and d(E, F ) ≤ d(E, G) + d(G, F ) for all E, F, G ∈ B([a, b]). For every E ∈ B([a, b]) and ε > 0 there exists F ∈ R([a, b]) such that μV (EΔF ) ≤ ε. Hence, R([a, b]) is dense in B([a, b]). Moreover, for E, F ∈ R([a, b]), by the finite additivity of λ, λ(E) − λ(F ) = λ(E \ (E ∩ F )) − λ(F \ (E ∩ F )), and so by (6.27), λ(E) − λ(F ) ≤ μV (E \ (E ∩ F )) + μV (F \ (E ∩ F )) = d(E, F ).
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6. Lebesgue–Stieltjes Measures
It follows that λ : R([a, b]) → Y is Lipschitz continuous with respect to the semimetric d. Thus, by the density of R([a, b]) in B([a, b]) we have that λ(E) ≤ μV (E) for all E ∈ B([a, b]). Now, let E ∈B([a, b]) and let Ei ∈ B([a, b]), i = 1, . . . , n, be disjoint sets such that E = ni=1 Ei . By what we just proved and by the additivity of μV , n n λ(Ei ) ≤ μV (Ei ) = μV (E). i=1
i=1
Taking the supremum over all such partitions of E (see (6.22)) gives λ(E) ≤ μV (E). In particular, by (6.18), λ([a, b]) ≤ μV ([a, b]) = Var[a,b] uλ , which, together with Step 1, proves (6.26)1 . The last part of the statement follows from (6.26).
It follows from the previous theorem that if λ : Bc (I) → [0, ∞), then uλ ∈ BP Vloc (I; Y ). Next we show that the function u λ defined in (6.1) with μ replaced by λ coincides with the indefinite variation Vuλ of uλ (see (2.3). Corollary 6.20. Let I ⊆ R be an open interval, let (Y, · ) be a normed space, and let λ : Bc (I) → Y be a vector-valued measure such that λ : Bc (I) → [0, ∞). Then u λ = Vuλ . Proof. In view of the previous theorem uλ is right continuous, but then by Proposition 2.21, so is Vuλ . Hence, (Vuλ )+ (t0 ) = Vuλ (t0 ) = 0 by Remark 2.2 and (2.3). Let x ∈ I with t0 ≤ x. By Exercise 2.16, (2.4), (6.2), and (6.26)2 for λ and the fact that u λ (t0 ) = 0, in this order, we have Vuλ (x) = Var(t0 ,x] uλ = λ((t0 , x]) = u λ (x).
The same holds for x ≤ t0 . In the same spirit of Theorem 6.3 we have the following results.
Theorem 6.21. Let I ⊆ R be an open interval, let (Y, · ) be a normed space and let λ : Bc (I) → Y be a vector-valued measure such that λ : Bc (I) → [0, ∞). Then the measure λ is purely atomic if and only if uλ is a jump function. Moreover, in this case, for every Borel set E ∈ Bc (I), λ(E ∩ {an }), λ(E) = λ(E ∩ {an }), (6.28) λ(E) = n
n
where {an }n is the set of discontinuity points of uλ .
6.2. Vector-valued Measures Versus BP V (I; Y )
171
Lemma 6.22. Let I ⊆ R be an open interval, let (Y, · ) be a normed space and let λ : Bc (I) → Y be a vector-valued measure such that λ : Bc (I) → [0, ∞). Then λ and λ have the same atoms. Proof. Let E ∈ Bc (I) be an atom for λ. Then λ(E) = y = 0 and for every subset F ∈ Bc (I) of E either λ(F ) = 0 or λ(F ) = λ(E). Let F ∈ Bc (I) be a subset of E and let Fi ∈ Bc (I), i = 1, . . . , n, be a family of disjoint sets such that n Fi . F = i=1
If λ(F ) = 0, then since each Fi can only have measure either 0 or y, it follows that all the sets Fi have measure zero. Hence, λ(F ) = 0. If λ(F ) = y, then only one Fi can have measure y and all others must have measure zero. It follows that n λ(Fi ) = λ(E) i=1
and taking the supremum over all such families gives λ(F ) = λ(E) > 0. Thus, E is an atom for λ. It follows from Lemma 6.4 applied to λ restricted to B([a, b]) for some E ⊆ [a, b] ⊆ I, that E must be a singleton. Conversely, if E ∈ Bc (I) is an atom for λ, then again by Lemma 6.4, E must be a singleton, E = {x}. Since the only subsets of E are E and the empty set, necessarily λ(E) = 0, and so E is an atom for λ. We now turn to the proof of Theorem 6.21. Proof of Theorem 6.21. Assume that λ is purely atomic. In view of the previous lemma, λ and λ have the same atoms. By Lemma 6.4 the atoms of λ are given by the singletons corresponding to the discontinuity points of u λ , but since u λ = Vuλ by Corollary 6.20, it follows from Proposition 2.21 that the discontinuity points of uλ are the same as the ones of u λ . Reasoning as in the first part of the proof of Theorem 6.3 with λ in place of μ (see (6.8)) we see that uλ is a jump function and (6.28)1 holds. Let E ∈ Bc (I) be such that λ(E) > 0. Then by (6.22) n there ex∈ B (I), i = 1, . . . , n, such that E = ist disjoint sets E i c i=1 Ei and n λ(E ) > 0. Hence, λ(E ) = 0 for at least one i. Since λ is purely i i i=1 atomic, the set Ei contains an atom. In turn, so does E. Thus λ is purely atomic, and so by Theorem 6.3, u λ = Vuλ is a jump function and (6.28)2 holds (see also (6.23)). Conversely, assume that uλ is a jump function. Then by (2.4) and Corollary 3.91, Vuλ is a jump function. In turn, by Theorem 6.3 and again the fact that u λ = Vuλ we have that λ is purely atomic. Since λ(E) ≤ λ(E) for every E ∈ Bc (I), it follows that if λ(E) = 0, then λ(E) > 0 and so
172
6. Lebesgue–Stieltjes Measures
E contains an atom for λ, but again by Lemma 6.4, the atom is also an atom for λ. Thus, λ is also purely atomic. This concludes the proof. Theorem 6.23. Let I ⊆ R be an open interval, let (Y, · ) be a normed space and let λ : Bc (I) → Y be a vector-valued measure such that λ : Bc (I) → [0, ∞). Then λ L1 if and only uλ is locally absolutely continuous. Moreover, in this case, if Y = RM , then for every E ∈ Bc (I), uλ dx, λ(E) = uλ dx. (6.29) λ(E) = E
E
Proof. Assume that uλ ∈ ACloc (I; Y ). By Theorem 3.11 the indefinite variation Vuλ of uλ belongs to ACloc (I). Since u λ = Vuλ by Corollary 6.20, it follows from Theorem 6.3 that λ L1 . Since λ(E) ≤ λ(E) for every E ∈ Bc (I), it follows that λ L1 . Conversely, assume that λ L1 . Let E ∈ Bc (I) be such that L1 (E) = 0. Then for every F ⊆ E with F ∈ Bc (I) we have that L1 (F ) = 0 and so λ(F ) = 0, since λ L1 . This implies that λ(E) = 0. Hence, λ L1 and since u λ = Vuλ by the previous corollary, it follows from Theorem 6.3 that Vuλ ∈ ACloc (I). Again by Theorem 3.11 we obtain that uλ ∈ ACloc (I; Y ). Next, assume that Y = RM and that λ L1 , or, equivalently, uλ ∈ ACloc (I; RM ). Then Vuλ ∈ ACloc (I) and so by Theorem 6.3 (see also Remark 6.10), for every E ∈ Bc (I), u λ dx = Vu λ dx. λ(E) = E
E
Since Vu λ (x) = uλ (x) for L1 -a.e. x ∈ I by Theorem 2.28, (6.29)2 holds. Reasoning as in the proof of (6.11) we get that (6.29)1 holds for every open set U ∈ Bc (I). Since λ L1 we can proceed as in (6.12) to conclude that (6.29)1 holds for every E ∈ Bc (I). This completes the proof. Theorem 6.24. Let I ⊆ R be an open interval and let λ : Bc (I) → RM be a vector-valued measure such that λ : Bc (I) → [0, ∞). Then λ is nonatomic and λ ⊥ L1 if and only if uλ is a singular function. Proof. Assume that λ is nonatomic and λ ⊥ L1 . By Lemma 6.22, λ is nonatomic. Since u λ = Vuλ by Corollary 6.20, it follows from Theorem 6.3 that Vuλ is continuous. In turn, by Proposition 2.21, so is uλ . Since λ ⊥ L1 , there exists a Borel set Eλ ⊆ I such that for every E ∈ Bc (I) we have λ(E) = λ(E ∩ Eλ ),
L1 (E) = L1 (E \ Eλ ).
6.2. Vector-valued Measures Versus BP V (I; Y )
173
In turn, given E ∈ Bc (I), let Ei ∈ B([a, b]), i = 1, . . . , n, be disjoint sets such that E = ni=1 Ei . Then n i=1
λ(Ei ) =
n
λ(Ei ∩ Eλ )
i=1
and so taking the supremum over all such families of sets gives λ(E) = λ(E ∩ Eλ ), which shows that λ ⊥ L1 . It follows from Theorem 6.3 that Vuλ is a singular function, and so Vu λ (x) = 0 for L1 -a.e. x ∈ I. Since Vu λ (x) = uλ (x) for L1 -a.e. x ∈ I by Theorem 2.28, we have that uλ is also a singular function. Conversely, assume that uλ is a singular function. Then by Proposition 2.21 and Theorem 2.28, Vuλ is a singular function, and so by Theorem 6.3 we have that λ is nonatomic and λ ⊥ L1 . By Lemma 6.22, λ is nonatomic. Since λ ⊥ L1 there exists a Borel set Eλ ⊆ I such that for every E ∈ Bc (I) we have λ(E) = λ(E ∩ Eλ ),
L1 (E) = L1 (E \ Eλ ).
It follows that λ(E \ Eλ ) = 0 for every E ∈ Bc (I), and since λ(F ) ≤ λ(F ) for every F ∈ Bc (I), we have that λ(E) = λ(E ∩ Eλ ) for every E ∈ Bc (I). Thus, λ ⊥ L1 . Next we discuss the converse of Theorem 6.19, namely the analog of Theorem 6.7. Theorem 6.25. Let I ⊆ R be an open interval, let (Y, · ) be a Banach space, and let u ∈ BP Vloc (I; Y ). Then there exists a unique vector-valued measure λu : Bc (I) → Y such that (6.30)
λu ((a, b]) = u+ (b) − u+ (a)
for all a, b ∈ I, with a ≤ b. Moreover, λu ≤ μV and equality holds if u is right continuous. Finally, if u ∈ BP V (I; Y ), then λu can be uniquely extended to a vector-valued measure λu : B(I) → Y and λu (I) ≤ Var u. In the previous statement V is the indefinite variation of u (see (2.3)) and μV is the Lebesgue–Stieltjes measure generated by V (see Theorem 6.7). The vector-valued measure λu given by the previous theorem is called the Lebesgue–Stieltjes vector-valued measure generated by u. Lemma 6.26. Let I ⊆ R be an open interval, let S ⊆ P(I) be a semiring, let (Y, · ) be a normed space, and let ρ : S → Y be a finitely additive
174
6. Lebesgue–Stieltjes Measures
measure. Then ρ can be uniquely extended to a finitely additive measure λ : R → Y , where R is the ring generated by S. Proof. Given E ∈ R, let E1 , . . . , En ∈ S be pairwise disjoint sets such that (6.31)
E=
n
Ei .
i=1
We define (6.32)
λ(E) :=
n
ρ(Ei ).
i=1
S be pairwise To prove that this is a good mdefinition, let F1 , . . . , Fm ∈ disjoint sets such that E = k=1 Fk . Then for each i, Ei = m k=1 (Ei ∩ Fk ) and the sets Ei ∩ F1 , . . . , Ei ∩ Fm are pairwise disjoint. Since ρ is finitely additive, we have that n i=1
ρ(Ei ) =
n m
ρ(Ei ∩ Fk ) =
i=1 k=1
n m
ρ(Ei ∩ Fk ) =
k=1 i=1
m
ρ(Fk ),
k=1
which shows that the extension of ρ to R does not depend on the particular partition of E. In particular, if E ∈ S, then λ(E) = ρ(E). It follows from the finite additivity of ρ and the definition of λ, that λ is finitely additive. Now, let λ1 : R → Y be another finitely additive extension of ρ. Given E ∈ R, let Ei be as in (6.31). Then by the additivity of λ and λ1 , λ1 (E) =
n i=1
λ1 (Ei ) =
n
ρ(Ei ) =
i=1
which proves the uniqueness of the extension.
n
λ(Ei ) = λ(E),
i=1
We turn to the proof of Theorem 6.25. Proof of Theorem 6.25. Step 1: We apply the previous lemma with S(I) the semiring of all intervals (a, b], where a, b ∈ I, with a ≤ b, and with ρu ((a, b]) := u+ (b) − u+ (a), to extend ρu to a finitely additive measure ρu : R(I) → Y , where R(I) is the ring generated by S(I). We claim that ρu (E) ≤ μV (E) for all E ∈ R(I). To see this, let (a, b] ∈ S(I) with a < b, and let a < x < b < t < sup I. It follows from (2.4) that u(t) − u(x) ≤ V (t) − V (x). Letting x a and t b, by Corollary 2.31 and the fact that V is increasing, we get u+ (b) − u+ (a) ≤ V+ (b) − V+ (a),
6.2. Vector-valued Measures Versus BP V (I; Y )
175
or, equivalently, by (6.15), ρu ((a, b]) ≤ μV ((a, b]).
(6.33)
By (6.32) and the additivity of μV it follows that ρu (E) ≤ μV (E) for every E ∈ R(I), which proves the claim. For E, F ∈ Bc (I) define d(E, F ) := μV (EΔF ). Reasoning as in Step 2 of the proof of Theorem 6.19 with λ replaced by ρu we have that d is a semimetric in Bc (I), that R(I) is dense in Bc (I), and that ρu : R(I) → Y is Lipschitz continuous with respect to the semimetric d. Thus, by the density of R(I) in Bc (I) we can uniquely extend ρu to a function λu : Bc (I) → Y . Moreover, λu (E) ≤ μV (E) for all E ∈ Bc (I). In Bc (I) × Bc (I) consider the semimetric d2 ((E1 , F1 ), (E1 , F1 )) := d(E1 , E2 ) + d(F1 , F2 ). Since the function f : Bc (I) × Bc (I) → Bc (I) given by f (E, F ) := E ∪ F is uniformly continuous, again by the density of R(I) in Bc (I) and the finite additivity of ρu , we obtain that λu is finitely additive. (I) be pairwise To prove thatλu is countably additive, let En ∈ Bc E ∈ Bc (I). Define Fn := E \ ni=1 Ei . Then disjoint sets with n En = Fn+1 ⊆ Fn for every n and n Fn = ∅. By the linearity of λu , λu (E) −
n
λu (Ei ) = λu (Fn ).
i=1
Since λu (Fn ) ≤ μV (Fn ) → 0 as n → ∞ by Proposition B.9(ii) applied to μV , it follows that λu is countably additive. Step 2: We prove that λu ≤ μV . Given E ∈ Bc (I) let (ai , bi ] ∈ S(I), i = 1, . . . , n, be pairwise disjoint with ni=1 (ai , bi ] ⊆ E, by the additivity and monotonicity of μV we have n i=1
λu ((ai , bi ]) ≤
n
μV ((ai , bi ]) ≤ μV (E).
i=1
Taking the supremum over all sets gives λu (E) ≤ μV (E). In particular, if u ∈ BP V (I; Y ), then λu (I) ≤ μV (I) = sup V − inf V = Var u, I
I
where we have used Theorem 6.1 and (2.4). Step 3: Assume that u is right continuous. We claim that λu = μV . Fix a, b ∈ I, with a ≤ b, and consider a partition a = x0 < x1 < · · · < xn = b of
176
6. Lebesgue–Stieltjes Measures
I. By (6.30) and the fact that u is right continuous we have n
u(xi ) − u(xi−1 ) =
i=1
n
λu ((xi−1 , xi ]) ≤ λu ((a, b]),
i=1
and so Var[a,b] u ≤ λu ((a, b]). In view of Proposition 2.21 the function V is right continuous. Hence by (2.4) and (6.13) we obtain that μV ((a, b]) = V (b) − V (a) = Var[a,b] u ≤ λu ((a, b]).
(6.34)
Reasoning as in the previous step, we conclude that μV ≤ λu , which, together with Step 2, gives λu = μV . Step 4: Finally, if u ∈ BP V (I; Y ), then μV (I) < ∞ by (6.14) and so it suffices to repeat Steps 1–3 with B(I) in place of Bc (I). Remark 6.27. If I ⊆ R is an open interval, (Y, · ) is a Banach space, and u, v ∈ BP Vloc (I; Y ), then by the uniqueness established in the previous theorem, λu+v = λu + λv . Remark 6.28. If I ⊆ R is an open interval, (Y, · ) is a normed space and λ : Bc (I) → Y is a vector-valued measure such that λ : Bc (I) → [0, ∞), then by the uniqueness established in the previous theorem, λ is the Lebesgue–Stieltjes measure generated by uλ , that is, λuλ = λ. Exercise 6.29. Let a ∈ R and let u := χ{a} . Find λu and μV and show that in general the inequality λu ≤ μV may be strict. Exercise 6.30. Let I ⊆ R be an open interval and let u ∈ BP Vloc (I) be a right continuous function. Prove that (λu )+ = μ 1 (V +u) , 2
)+
where (λu and (λu sition B.70.
)−
(λu )− = μ 1 (V −u) , 2
are the (positive) Radon measures given in Propo-
Exercise 6.31. Let I ⊆ R be an open interval and let u, v ∈ BP Vloc (I). Prove that for every [a, b] ⊂ I, 1 1 (u+ + u− ) dλv + 2 (v+ + v− ) dλu = u− (b)v− (b) − v+ (a)u+ (a). 2 (a,b)
(a,b)
Exercise 6.32. Let I ⊆ R be an open interval and let u ∈ BP Vloc (I). Prove that for every ϕ ∈ Cc1 (I), uϕ dx = − ϕ dλu . (6.35) I
I
6.3. Decomposition of Measures
177
6.3. Decomposition of Measures In Corollary 3.90 we have shown that every function in BP Vloc (I; RM ) can be decomposed as the sum of three functions in BP Vloc (I; RM ), a locally absolutely continuous function, a continuous singular function, and a saltus function. In this section we prove that the corresponding Lebesgue–Stieltjes measures are mutually singular. Let I ⊆ R be an open interval and let u ∈ BP Vloc (I; RM ). By Corollary 3.90, u may be decomposed as the sum of three functions in BP Vloc (I; RM ), u = uAC + uC + uJ ,
(6.36) where
x
uAC (x) :=
u dt,
x ∈ I,
t0
belongs to ACloc (I; RM ), uC is a singular function, and uJ is the jump function of u, that is, (6.37) (u (t) − u− (t)) + u(x) − u− (x) if x ≥ t0 , t∈I, t0 ≤t 0, we conclude that Du(Ω) ≤ essVar u. Conversely, assume that u ∈ BV (Ω; RM ). Then by Theorem 7.3, u admits a right continuous representative u, which has bounded pointwise variation and Var u = Du(Ω). In particular, essVar u ≤ Var u = Du(Ω). This concludes the proof.
Exercise 7.10. Let I ⊆ R be an open interval and let u ∈ BV (I; RM ). Prove that n u(xi ) − u(xi−1 ) , essVar u = sup i=1
188
7. Functions of Bounded Variation and Sobolev Functions
where the supremum is taken over all partitions P := {x0 , . . . , xn } of I such that each xi is a Lebesgue point of u (see Corollary B.119), i = 1, . . . , n, n ∈ N. In the remainder of this section we extend some of the previous results to functions that have only locally bounded variation. The space of functions of locally bounded variation BVloc (Ω; RM ) is defined as the space of all functions u ∈ L1loc (Ω; RM ) such that u ∈ BV (U ; RM ) for all open sets U Ω. We recall that Bc (I) is the σ-field of all Borel subsets compactly contained in I (see (6.21)). Exercise 7.11. Let Ω ⊆ R be an open set and let u ∈ BVloc (Ω; RM ). Prove that there is a vector-valued measure λ : Bc (Ω) → RM such that for all ϕ ∈ Cc1 (Ω), uϕ dx = − ϕ dλ. Ω
Ω
Exercise 7.12. Let Ω ⊆ R be an open set and let u : Ω → RM . (i) Prove that if u : Ω → RM has locally bounded pointwise variation, then u has locally bounded variation. (ii) Prove that if u ∈ L1loc (Ω; RM ) has locally bounded variation, then u admits a right continuous representative that has locally bounded pointwise variation.
7.2. Sobolev Functions Versus Absolutely Continuous Functions In this section we introduce the notion of Sobolev functions and study their relation to absolutely continuous functions. Definition 7.13. Given an open set Ω ⊆ R, n ∈ N, and 1 ≤ p ≤ ∞, we say that a function u ∈ L1loc (Ω; RM ) admits a weak or distributional derivative of order n in Lp (Ω; RM ) if there exists a function v ∈ Lp (Ω; RM ) such that (n) n uϕ dx = (−1) vϕ dx Ω
Ω
for all ϕ ∈ Cc∞ (Ω). The function v is denoted u(n) . A similar definition can be given when Lp (Ω; RM ) is replaced by Lploc (Ω; RM ). Definition 7.14. Given an open set Ω ⊆ R , m ∈ N, and 1 ≤ p ≤ ∞, the Sobolev space W m,p (Ω; RM ) is the space of all functions u ∈ Lp (Ω; RM )
7.2. Sobolev Functions Versus Absolutely Continuous Functions
189
which admit weak derivatives of order n in Lp (Ω; RM ) for every n = 1, . . . , m. The space W m,p (Ω; RM ) is endowed with the norm uW m,p (Ω;RM ) := uLp (Ω;RM ) +
m
u(n) Lp (Ω;RM ) .
n=1 m,p (Ω; RM ) is defined as the space of all functions u ∈ The space Wloc which admit weak derivatives of order n in Lploc (Ω; RM ) for every n = 1, . . . , m.
Lploc (Ω; RM )
m,p (Ω) for W m,p (Ω; R) When M=1 for simplicity we write W m,p (Ω) and Wloc m,p and Wloc (Ω; R), respectively.
Exercise 7.15. Let Ω ⊆ R be an open set, let n, M ∈ N, and let u ∈ L1loc (Ω; RM ) be such that its weak derivative u(n) of order n exists in L1loc (Ω; RM ). Prove that u(n) is unique. Note that W 1,1 (Ω; RM ) ⊂ BV (Ω; RM ). Indeed, if u ∈ W 1,1 (Ω; RM ), then in the definition of BV (Ω; RM ), we can take the vector-valued measure u dx, E ⊆ B(Ω). λ(E) := E
The following theorem, together with Corollary 3.90, shows that the inclusion W 1,1 (Ω; RM ) ⊂ BV (Ω; RM ) is strict. Theorem 7.16. Let Ω ⊆ R be an open set and let 1 ≤ p ≤ ∞. Then a function u : Ω → RM belongs to W 1,p (Ω; RM ) if and only if it admits ¯ and its an absolutely continuous representative u ¯ : Ω → RM such that u p M ¯ is classical derivative u ¯ belong to L (Ω; R ). Moreover, if p > 1, then u H¨ older continuous of exponent 1/p . Proof. By working on each connected component of Ω, we may assume that Ω = I, with I ⊆ R an open interval. If u ¯ : I → RM is absolutely continuous ¯ ∈ Lp (I; RM ), then by Corollary 3.23, with u ¯ ∈ Lp (I; RM ) and u uϕ dx = − u ϕ dx I
for all ϕ ∈ Cc1 (I). u ¯ ∈ W 1,p (I; RM ).
It follows that
I
u ¯
is the weak derivative of u ¯ and that
Conversely, assume that u ∈ W 1,p (I; RM ) and let v be its weak derivative. Fix a Lebesgue point x0 ∈ I of u (see Corollary B.119) and define x v(t) dt, x ∈ I. (7.3) u ¯(x) := u(x0 ) + x0
Lp (Ω; RM ),
we have that v is locally integrable, and thus, by Since v ∈ applying Corollary 3.19, we have that u ¯ is locally absolutely continuous,
190
7. Functions of Bounded Variation and Sobolev Functions
with u ¯ (x) = v(x) for L1 -a.e. x ∈ I, and, in turn, by Corollary 3.31 we have that u ¯ is actually absolutely continuous in I (when p = ∞ observe that since v is bounded, by (7.3) the function u ¯ is Lipschitz continuous). Using Corollary 3.23, we get u ¯ϕ dx = − u ¯ ϕ dx = − vϕ dx I
for all ϕ ∈
Cc1 (I).
I
I
Hence, we have that (u − u ¯)ϕ dx = 0 I
for all ϕ ∈ Cc1 (I). By Lemma 7.4 it follows that u − u ¯ is constant L1 -a.e. ¯(x0 ), we have in I, and since x0 ∈ I is a Lebesgue point of u and u(x0 ) = u that u = u ¯ L1 -a.e. in I. This shows that u has an absolutely continuous representative. To prove that u ¯ is H¨older continuous of exponent 1/p , let x, y ∈ I, with x < y. By Theorem 3.20, x u (t) dt, u ¯(x) − u ¯(y) = y
and so, by H¨older’s inequality, x 1/p ¯ u(x) − u ¯(y) ≤ (7.4) u (t) dt ≤ (x − y) y
≤ (x − y)1/p
u (t)p dt
1/p
x
1/p
u (t) dt p
y
,
Ω
which shows that u ¯ is H¨older continuous of exponent 1/p .
Exercise 7.17. Let Ω ⊆ R be an open set, let u : Ω → RM , let m ∈ N, with m ≥ 2, and let 1 ≤ p < ∞. Prove that u ∈ W m,p (Ω; RM ) if and only if it admits a representative u ¯ : Ω → RM of class C m−1 (Ω; RM ) such that its classical derivative of order m − 1 is absolutely continuous and u ¯ and all its classical derivatives u ¯(n) , n = 1, . . . , m, belong to Lp (Ω; RM ). Exercise 7.18. Given an open set Ω ⊆ R , m, M ∈ N, and 1 ≤ p ≤ ∞, prove that W m,p (Ω; RM ) is a Banach space. Exercise 7.19. Given an open set Ω ⊆ R , m, M ∈ N, prove that H m (Ω; RM ) := W m,2 (Ω; RM ) is a Hilbert space. As usual, when M = 1 we write H m (Ω) for H m (Ω; RM ).
7.2. Sobolev Functions Versus Absolutely Continuous Functions
191
Corollary 7.20. Let Ω ⊆ R be an open bounded set and let u : Ω → RM . Then u ∈ W 1,1 (Ω; RM ) if and only if it admits an absolutely continuous representative u ¯ : Ω → RM . Proof. If u ¯ : Ω → RM is absolutely continuous, then by Corollary 3.10, u ¯ is bounded, and so integrable, and u ¯ is integrable. We are now in a position to apply the previous theorem. Corollary 7.21. Let Ω ⊆ R be an open bounded set, let u : Ω → RM , and let 1 < p < ∞. Then u ∈ W 1,p (Ω; RM ) if and only if it admits an absolutely ¯ continuous representative u ¯ : Ω → RM such that its classical derivative u p M belongs to L (Ω; R ). Proof. Since for sets of finite measure Lp (Ω; RM ) ⊂ L1 (Ω; RM ), we have that W 1,p (Ω; RM ) ⊂ W 1,1 (Ω; RM ). Thus, the result follows from Corollary 7.20. The next exercise gives an example of an absolutely continuous function in L1 (R) \ W 1,1 (R). Exercise 7.22. Let 1 n if x ∈ (n − 1 + g(x) := − n1 if x ∈ (n − 1 + and let
2k 2k+1 2n , n − 1 + 2n ], 2k+1 2k+2 2n , n − 1 + 2n ],
n ∈ N, 0 ≤ k < n, n ∈ N, 0 ≤ k < n,
x u(x) :=
0 g(t) dt if x ≥ 0, u(−x) if x < 0.
(i) Prove that u is absolutely continuous. (ii) Prove that u ∈ L1 (R). (iii) Prove that u ∈ ε>0 L1+ε (R) \ L1 (R). Theorem 7.23 (Superposition). Let Ω ⊆ R be an open set, let 1 ≤ p < ∞ and let f : R → R. Then f ◦ u ∈ W 1,p (Ω) for all functions u ∈ W 1,p (Ω) if and only if f is locally Lipschitz continuous and, when L1 (Ω) = ∞, f (0) = 0. Proof. The proof relies on Theorems 3.55, Exercise 3.58, and Theorem 7.16 and is left as an exercise. Exercise 7.24. Given m, M ∈ N, and 1 ≤ p ≤ ∞, let u ∈ W m,p ((a, b); RM ). Prove that there exist a1 , . . . , am ∈ R and b1 , . . . , bm ∈ (0, 1] such that the function ⎧ m ⎨ n=1 an u(a + bn (a − x)) if 2a − b < x < a, u(x) if a < x < b, v(x) := ⎩ m n=1 an u(a + bn (x − b)) if b < x < 2b − a,
192
7. Functions of Bounded Variation and Sobolev Functions
is well-defined and belongs to W m,p ((2a − b, 2b − a); RM ). Prove also that vW m,p ((2a−b,2b−a)) ≤ cuW m,p ((a,b)) for some constant c = c(m, M, p) > 0. Exercise 7.25. Let I ⊂ R be an open interval with infinite length, let m, M ∈ N, let 1 ≤ p ≤ ∞, and let u ∈ W m,p (I; RM ). Prove that u can be extended to a function v ∈ W m,p (R; RM ) with vW m,p (R) ≤ cuW m,p (I) for some constant c = c(m, M, p) > 0. Theorem 7.26. Let I ⊆ R be an open interval, let m, M ∈ N, and let 1 ≤ p < ∞. Then functions in C ∞ (I; RM ) ∩ W m,p (I; RM ) are dense in W m,p (I; RM ). Proof. Let u ∈ W m,p (I; RM ). By Exercises 7.24 and 7.25 we can extend u to a function v ∈ W m,p (J; RM ) with vW m,p (J) ≤ cuW m,p (I) , where J is an interval containing I and such that dist(I, ∂J) > 0. For ϕ ∈ Cc∞ (R) satisfies 0 < ε < dist(I, ∂J) let vε := v ∗ ϕε , where with supp ϕ ⊆ B(0, 1), R ϕ(x) dx = 1 and ϕε (x) := 1ε ϕ( xε ). By Theorem C.20, vε ∈ C ∞ (R; RM ), with n n d ϕε d ϕε n (x − t)v(t) dt = (−1) (x − t)v(t) dt vε(n) (x) = n n J dx J dt for x ∈ R and n ∈ N. In particular, if x ∈ I, then since ε < dist(I, ∂J) we have that supp ϕε (x − ·) ⊆ B(x, ε) ⊆ J, and so we can integrate by parts to get (n) ϕε (x − t)v (n) (t) dt = (v (n) ∗ ϕε )(x). vε (x) = J
It now follows from Theorem C.16 applied to v and to v (n) that vε − uW m,p (I) = vε − vW m,p (I) → 0 as ε →
0+ .
Exercise 7.27. Prove that the previous theorem continues to hold when the interval I is replaced by an open set Ω ⊆ R. Finally, we consider the case p = ∞. Theorem 7.28. Let I ⊆ R be an open interval and let u : I → RM . Then u ∈ W 1,∞ (I; RM ) if and only if it admits a bounded, Lipschitz continuous representative u ¯ : I → RM .
7.2. Sobolev Functions Versus Absolutely Continuous Functions
193
Proof. If u ¯ : I → RM is Lipschitz continuous, then it is absolutely continuous, and so, as in the proof of Theorem 7.16, we conclude that u ¯ ∈ W 1,∞ (I; RM ). ¯ as Conversely, assume that u ∈ W 1,∞ (I; RM ) and define the function u in (7.3). Since v ∈ L∞ (I; RM ), we have that v is Lipschitz continuous and locally integrable, and thus we may proceed as in the proof of Theorem 7.16 to conclude that u ¯ = u L1 -a.e. in I. We conclude this section by showing that in bounded domains, up to constants, the Lp norm of a Sobolev function can be estimated by the Lp norm of its weak derivative. Theorem 7.29 (Poincar´e’s inequality). Let I = (a, b) and let 1 ≤ p < ∞. Then b b p p u(x) − uI dx ≤ (b − a) u (x)p dx (7.5) for all u ∈
a 1,p W ((a, b); RM ),
a
where
uI :=
1 b−a
b
u(x) dx. a
Proof. Fix u ∈ W 1,p ((a, b); RM ). By Corollaries 7.20 and 7.21 there exists ¯ an absolutely continuous representative of u, u ¯ : (a, b) → RM , such that u p M belongs to L ((a, b); R ). By Exercise 3.8 we may extend u ¯ to [a, b] in such a way that the extension is still absolutely continuous. Since u ¯ is continuous, by the mean value theorem for every i = 1,. . . ,M there exists xi ∈ [a, b] such that b 1 u ¯i (x) dx = u ¯i (xi ). (uI )i = b−a a By Theorem 3.20, for all x ∈ (a, b), x u ¯ (t) dt, u ¯i (x) − (uI )i = xi
and so as in (7.4), ¯ u(x) − uI ≤ (b − a)
1/p
b
1/p
¯ u (t) dt p
.
a
Raising both sides to exponent p and integrating in x gives b b u(x) − uI p dx ≤ (b − a)p ¯ u (t)p dt, a
where we have used the fact that p/p = p − 1.
a
Exercise 7.30. State and prove the analog of Poincar´e’s inequality in W m,p ((a, b)).
194
7. Functions of Bounded Variation and Sobolev Functions
Exercise 7.31. Let I = (a, b), let J = [c, d] ⊂ (a, b) and let 1 ≤ p < ∞. Prove that there exists a constant κ > 0 such that b b p p |u(x) − uJ | dx ≤ κ(b − a) |u (x)|p dx for all u ∈
a 1,p Wloc ((a, b)).
a
Exercise 7.32. Let u ∈ W 1,1 ((a, b)). (i) Prove that for all x ∈ (a, b), ! x " b 1 (s − a)u (s) ds − (b − s)u (s) ds . u ¯(x) − uI = b−a a x (ii) Prove that max (s − a)(b − s) = 14 (b − a)2 .
a≤s≤b
(iii) Prove that b a
|u(x) − uI | dx ≤ 12 (b − a)
(iv) Prove that the constant
1 2 (b
b
|u (x)| dx.
a
− a) is sharp.
We conclude this section with a weighted Poincar´e inequality, which will be used to prove Poincar´e’s inequality in convex domains (see Theorem 13.36). Proposition 7.33 (Weighted Poincar´e’s inequality). Let g : [0, d] → [0, ∞), g = 0, be increasing in [0, c] and decreasing in [c, d] for some 0 ≤ c ≤ d and let 1 ≤ p < ∞. Then there exists a constant cp > 0 depending only on p such that d d p p u(x) − u(0,d) g(x) dx ≤ cp d u (x)p g(x) dx (7.6) 0
for all u ∈
0
W 1,p ((0, d); RM ),
where
u(0,d) := d 0
1 g(x) dx
d
u(x)g(x) dx. 0
Proof. If g = 0 on some interval [0, b] (respectively, [b, d]), then all the integrals involved reduce to integrals over [b, d] (respectively, [0, b]). Thus, we can assume that g is strictly positive in (0, d). Also, by a scaling argument, it is enough 1 to prove the result for d = 1. Finally, by dividing the inequality (7.6) by 0 g(x) dx, we may assume that 1 g(x) dx = 1. (7.7) 0
7.2. Sobolev Functions Versus Absolutely Continuous Functions
195
By Corollaries 7.20 and 7.21, without loss of generality, we may assume that u is absolutely continuous. By Theorem 3.20 and (7.7), 1 1 u(t)g(t) dt = [u(x) − u(t)]g(t) dt u(x) − u(0,1) = u(x) − 0 0 1 x u (s)g(t) dsdt = 0 t x x 1 t = u (s)g(t) dsdt − u (s)g(t) ds dt 0 t x x s 1 1 x u (s) g(t) dtds − u (s) g(t) dtds, = 0
0
x
s
where we have used Fubini’s theorem. Hence,
x
u(x) − u(0,1) ≤
u (s)
0
s
1
g(t) dtds + 0
1
u (s)
x
g(t) dtds. s
Raising both sides to the power p, multiplying by g(x), and integrating over [0, 1] gives
1
u(x) − u(0,1) g(x) dx ≤
0
p
1
x
s
u (s)
g(t) dtds
p 1 1 u (s) g(t) dtds g(x) dx + x s
p 1 x s p−1 u (s) g(t) dtds g(x) dx ≤2 0 0 0
p 1 1 1 p−1 u (s) g(t) dtds g(x) dx +2 0
0
0
0
x
s
=: A + B, where in the second inequality we have used the inequality
(7.8)
n i=1
p ai
≤ np−1
n
api ,
i=1
where ai ≥ 0, i = 1, . . . , n, which follows from the convexity of the function f (t) := |t|p , t ∈ R.
196
7. Functions of Bounded Variation and Sobolev Functions
We estimate A. If 0 ≤ s ≤ c, then by hypothesis g(t) ≤ g(s) for all 0 ≤ t ≤ s, so for 0 ≤ x ≤ c we have that
p x
p x s s 1/p 1/p u (s) g(t) dtds ≤ u (s)(g(s)) (g(t)) dtds 0 0 0 0 x
p 1/p ≤ u (s)(g(s)) ds 0 1
u (s)p g(s) ds,
≤ 0
where we have used H¨older’s inequality twice and (7.7). On the other hand, if c ≤ s ≤ 1, then by hypothesis g(s) ≥ g(x) for all s ≤ x ≤ 1, and so for c ≤ x ≤ 1 we have that
p x s u (s) g(t) dtds g(x) 0 0
p c s x s u (s) g(t) dtds + u (s) g(t) dtds g(x) = 0 0 c 0
p
p c x s u (s) g(t) dtds g(x) + 2p−1 u (s) ds g(x) ≤ 2p−1 0 0 c
p 1 x p−1 p p−1 1/p u (s) g(s) ds g(x) + 2 u (s)(g(s)) ds ≤2 0 c
1 1 p−1 p p−1 p ≤2 u (s) g(s) ds g(x) + 2 u (s) g(s) ds , 0
0
again by H¨older’s inequality and (7.7). Thus, for all 0 ≤ x ≤ 1, x
p s u (s) g(t) dsds g(x) 0 0 1
p−1 p u (s) g(s) ds (g(x) + 1), ≤2 0
and, in turn, using (7.7) once more, 1 1 u (s)p g(s) ds (g(x) + 1) dx = 2c(p) A ≤ c(p) 0
0
1
u (s)p g(s) ds.
0
A similar estimate holds for B, thus giving the desired result for p > 1. The proof in the case p = 1 is simpler, since there is no need to use H¨older’s inequality.
7.3. Interpolation Inequalities In this section we prove that if a function u belongs to some Lq space and its highest order weak derivative u(m) belongs some Lp space, with p and q
7.3. Interpolation Inequalities
197
possibly different, then we can obtain additional information on the function itself and on all the intermediate derivatives of u. For simplicity we consider only scalar-valued functions, that is, M = 1, although all the results hold for vector-valued functions. Also, when there is no possibility of confusion, we write uLp for uLp (I) . We begin with the case m = 1. Theorem 7.34. Let I ⊆ R be an open interval, let 1 ≤ p, q, r ≤ ∞ be such 1,1 (I). Then that r ≥ q, and let u ∈ Wloc (7.9)
uLr (I) ≤ 1/r−1/q uLq (I) + 1−1/p+1/r u Lp (I)
for every 0 < < L1 (I). Proof. It is enough to prove (7.9) in the case L1 (I) < ∞ and then use the Lebesgue monotone convergence theorem to obtain the case L1 (I) = ∞. If u Lp = ∞, then the right-hand side of (7.9) is infinite and so there is nothing to prove. Thus, without loss of generality we may assume that u Lp < ∞. By Exercise 7.31 we can assume that u ∈ W 1,p (I). In turn, by Corollaries 7.20 and 7.21 we have that (a representative of) u is absolutely continuous in I, and by Exercise 3.8 we may extend u to I in such a way that the extension is still absolutely continuous. Step 1: Assume that r = ∞. For every x ∈ I, let Ix := [x − /2, x + /2] ∩ I. By the fundamental theorem of calculus (see Theorem 3.20), x u(x) = u(x1 ) + u (t) dt, x1
where x1 ∈ I is taken such that |u(x1 )| = minIx |u|. Then by H¨older’s inequality, |u (t)| dt ≤ −1/q uLq (Ix ) + 1/p u Lp (I) |u(x)| ≤ min |u| + Ix
≤
−1/q
Ix
uLq (I) + 1−1/p u Lp (I) .
This proves (7.9) when r = ∞. Step 2: Assume next that r < ∞ and r > q (if r = q there is nothing to prove). Then by Young’s inequality (B.17) with exponent r/q we can write
1/r 1−q/r q/r r−q q |u| |u| dx ≤ uL∞ uLq uLr = ≤
I 1/r
uL∞ + 1/r−1/q uLq .
By Step 1,
uL∞ ≤ −1/q uLq + 1−1/p u Lp . Combining the last two inequalities gives (7.9).
198
7. Functions of Bounded Variation and Sobolev Functions
Exercise 7.35. Prove that if p = 1 and u ∈ BVloc (I), then inequality (7.9) continues to hold with u L1 (I) replaced by the total variation Du(I). Corollary 7.36. Let I = (a, b), let 1 ≤ p, q, r ≤ ∞ be such that 1 + 1/r ≥ 1,1 1/p and r ≥ q and let u ∈ Wloc (I) with u ∈ Lp (I). Let x0 ∈ [a, b] be such that |u(x0 )| = min[a,b] |u|. Then 1−α u − u(x0 )Lr (I) ≤ 8uαLq (I) u L p (I) ,
where α := 0 if r = q and 1 − 1/p + 1/r = 0 and otherwise α :=
1 − 1/p + 1/r . 1 − 1/p + 1/q
Note that as in the proof of Theorem 7.34 we can assume that u is absolutely continuous in [a, b] and so by the Weierstrass theorem x0 exists. Proof. Define v(x) := u(x) − u(x0 ), x ∈ (a, b). If u Lp = 0, then u is constant and so v = 0 and there is nothing to prove. Thus, assume that u Lp > 0 and that uLq > 0. Note that 0 ≤ α ≤ 1. Moreover, by the definition of x0 , vLq ≤ 2uLq . Let I1 := (−∞, b) and I2 := (a, ∞). Since v ∈ C 1 ([a, b]) and v(x0 ) = 0, the functions v(x) for x0 < x < b, v(x) for a < x < x0 , v2 (x) := v1 (x) := 0 for x ≤ x0 , 0 for x ≥ x0 , belong to W 1,1 (I1 ) and W 1,1 (I2 ), respectively, with vi Lq (Ii ) ≤ vLq (I) ≤ 2uLq (I) and vi Lp (Ii ) ≤ u Lp (I) , i = 1, 2. By Theorem 7.34 applied to vi , i = 1, 2, we get vLr (I) ≤ v1 Lr (I1 ) + v2 Lr (I2 ) ≤ 41/r−1/q uLq (I) + 21−1/p+1/r u Lp (I) for every 0 < < ∞. It suffices to take
1/(1−1/p+1/q) −1 := uLq (I) u Lp (I) in the previous inequality.
Next we consider the case m = 2 and k = 1. Theorem 7.37. Let I ⊆ R be an open interval, let 1 ≤ p, q, r ≤ ∞ be such that 1 1 1 + ≥ , (7.10) 2q 2p r 2,1 (I). Then there exists c = c(p, q, r) > 0 such that and let u ∈ Wloc
(7.11)
u Lr (I) ≤ c1/r−1−1/q uLq (I) + c1−1/p+1/r u Lp (I)
for every 0 < < L1 (I).
7.3. Interpolation Inequalities
199
We begin with an auxiliary result which is of interest in itself. Lemma 7.38. Let I = (a, b), let 1 ≤ p, q, r ≤ ∞ satisfy (7.10) and let 2,1 u ∈ Wloc (I) be such u ∈ Lp (I) and u has at least one zero in [a, b]. Then there exists a constant c = c(p, q, r) > 0 such that 1−θ u Lr (I) ≤ cuθLq (I) u L p (I) ,
(7.12) where (7.13)
θ=
1 − 1/p + 1/r . 2 − 1/p + 1/q
Note that as in the proof of Theorem 7.34, but with u replaced by u , we can assume that u is absolutely continuous in [a, b]. Proof. If u Lp = 0, then u is constant in I and since it has a zero, then u must be zero. Thus, assume that u Lp > 0. Moreover, reasoning as in the beginning of the proof of Theorem 7.34 we can assume that u is absolutely continuous in [a, b]. In turn, u ∈ C 1 ([a, b]). Step 1: Assume first that either u is nonnegative in I or nonpositive in I. Then by the fundamental theorem of calculus, for every x, x1 ∈ (a, b), b |u (t)| dt = |u(b) − u(a)| |u(x) − u(x1 )| ≤ a
≤ |u(b) − u(x1 )| + |u(a) − u(x1 )| ≤ 2u − u(x1 )L∞ , which shows that (7.14)
u − u(x1 )L∞ ≤ u L1 ≤ 2u − u(x1 )L∞ .
If r > 1 we can apply Corollary 7.36 to both u and u to obtain 1 u − u(x0 )L∞ ≤ 8uαL1q u 1−α Lr ,
(7.15)
2 u Lr ≤ 8u αL21 u 1−α Lp ,
(7.16)
where x0 ∈ [a, b] is such that |u(x0 )| = min[a,b] |u| and α1 =
1 − 1/r , 1 − 1/r + 1/q
α2 =
1 − 1/p + 1/r . 2 − 1/p
Combining (7.14)–(7.16) we get 2 2 ≤ cu − u(x0 )αL2∞ u 1−α u Lr ≤ 8u αL21 u 1−α Lp Lp
≤ cuαL1q α2 u Lr
(1−α1 )α2
2 u 1−α Lp .
Since u ∈ C 1 ([a, b]), it follows that u Lr ≤ cuL1q
α α2 /(1−(1−α1 )α2 )
which gives (7.12).
u Lp
(1−α2 )/(1−(1−α1 )α2 )
,
200
7. Functions of Bounded Variation and Sobolev Functions
On the other hand, if r = 1, then by (7.10) necessarily q = p = 1. In this case by (7.14) and (7.15) with r = ∞, u L1 ≤ 2u − u(x1 )L∞ ≤ 8uL1 u L∞ . 1/2
1/2
Since u has at least one zero in [a, b], by the fundamental theorem of calculus (see Theorem 3.20), u L∞ ≤ u L1 , which together with the previous inequality gives (7.12) even in this case. Step 2: In the general case, we write the open set {x ∈ I : u (x) = 0} as a union of countably many open sets In in which u is either positive or negative. By applying Step 1 in each interval In we obtain |u |r dx = |u |r dx = |u |r dx {u =0}
I
≤c
In n r(1−θ) urθ Lq (In ) u Lp (In )
n
≤c
uqLq (In )
rθ/q
n
≤
r(1−θ)/p
n
r(1−θ) curθ , Lq u Lp
where we used the inequality (7.17)
u pLp (In )
n
asn btn ≤
an
s
n
t bn
,
n
which holds for s + t ≥ 1 and an , bn > 0 (exercise). Note that by (7.10) and (7.13).
rθ q
+ r(1−θ) ≥1 p
Exercise 7.39. Let (Y, · Y ) be a Banach space and let P ⊂ Y be a finite dimensional subspace. (i) Prove that for every y ∈ Y there exists p ∈ P such that y − pY = dist(y, P ). (ii) Let Y = Lq ([a, b]), let P be the set of polynomials of degree one and let f : [a, b] → R be a continuous increasing function. Prove that the polynomial p given in part (i) with y given by f touches f at least twice. We are now ready to prove Theorem 7.37. Proof of Theorem 7.37. Reasoning as in the beginning of the proof of Theorem 7.34 we can assume that I = (a, b), that u Lp < ∞, and that u is absolutely continuous in [a, b]. In turn, u ∈ C 1 ([a, b]).
7.3. Interpolation Inequalities
201
By Exercise 7.39 with Y = Lq (I) and P the subspace of polynomials of degree one, we can find a polynomial p1 of degree one such that u − p1 Lq = min u − pLq . p∈P
Let v := u − p1 . Then vLq ≤ u − 0Lq ,
(7.18) (7.19)
v Lp = u Lp ,
p1 Lq ≤ u − p1 Lq + uLq ≤ 2uLq .
Assume for simplicity that a = 0 (the general case can be obtained by translation). Fix s ∈ (0, 13 b) and t ∈ ( 23 b, b). Then for every x ∈ (0, b), p1 (t) − p1 (s) . t−s By averaging first in s over (0, 13 b) and then in t over ( 23 b, b) and by H¨older’s inequality we get b b/3 −2 −2 |p1 (t)| dt + 9b |p1 (s)| ds ≤ cb−1−1/q p1 Lq . |p1 (x)| ≤ 9b p1 (x) =
Finally, taking the (7.20)
2b/3 Lr norm
0
in the variable x on both sides gives
p1 Lr ≤ cb1/r−1−1/q p1 Lq ≤ c1/r−1−1/q p1 Lq ,
where in the second inequality we used the fact that 1 + 1/q − 1/r ≥ 0 by (7.10). Since the function v has at least two zeros by Exercise 7.39, its derivative vanishes at least once. Hence, we are in a position to apply Lemma 7.38 to obtain 1−θ u Lr ≤ v Lr + p1 Lr ≤ cvθLq v L p + p1 Lr 1−θ 1/r−1−1/q ≤ cuθLq u L uLq p + c
≤ c1/r−1−1/q uLq + c1−1/p+1/r u Lp , where we have used (7.18), (7.19), and (7.20) and Young’s inequality. This completes the proof. We now consider the general case m ≥ 2. Theorem 7.40. Let I ⊆ R be an open interval, let 1 ≤ p, q, r ≤ ∞, m ∈ N, k ∈ N0 , with 0 ≤ k < m, be such that
k 1 1 k 1 + ≥ , (7.21) 1− m q mp r m,1 (I). Then there exists a constant c = c(k, m, p, q, r) > 0 and let u ∈ Wloc such that
(7.22)
u(k) Lr (I) ≤ c1/r−k−1/q uLq (I) + cm−k−1/p+1/r u(m) Lp (I)
202
7. Functions of Bounded Variation and Sobolev Functions
for every 0 < < L1 (I). In particular, for p = q = r, (7.23)
u(k) Lp (I) ≤ c−k uLp (I) + cm−k u(m) Lp (I) .
Proof. The proof is by induction on m. We consider the cases k = 0 and 1 ≤ k < m separately. For m = 2 and k = 1, (7.22) holds in view of Theorem 7.37. Let n ∈ N and assume that (7.22) holds for all m ≤ n and all 1 ≤ k < m and for all 1 ≤ p, q, r ≤ ∞ satisfying (7.21). We claim that (7.22) holds with m = n + 1. Let 1 ≤ p, q, r ≤ ∞ and 1 ≤ k < n + 1, be such that
1 k 1 1 k + ≥ , (7.24) 1− n+1 q n+1p r n+1,1 (I). We claim that there exists c = c(k, n + 1, p, q, r) > 0 and let u ∈ Wloc such that
(7.25)
u(k) Lr ≤ c1/r−k−1/q uLq + cn+1−k−1/p+1/r u(n+1) Lp
for every 0 < < L1 (I). It is enough to prove this inequality in the case in which I is bounded. The general case follows from the Lebesgue monotone convergence theorem. Moreover, we can assume that uLq < ∞ and that u(n+1) Lp < ∞, since otherwise there is nothing to prove. Reasoning as in the beginning of the proof of Theorem 7.34, with u replaced by u(n−1) we can assume that u(n−1) ∈ C 1 (I), which implies that u ∈ C n (I). Step 1: Assume that k ≥ 2. In view of (7.24) we may find r1 ≥ 1 such that
1 1 k−1 1 k−11 1 1 11 + = , ≥ . 1− + 1− k q kr r1 n r1 n p r Since k ≤ n, by the induction hypothesis, for 0 < δ < 1 and 0 < < L1 (I), u Lr1 ≤ c(δ)1/r1−1−1/q uLq + (δ)k−1−1/r+1/r1 u(k) Lr , while again by the induction hypothesis applied to the function u in place of u, u(k) Lr ≤ c1/r−k+1−1/r1 u Lr1 + n+1−k−1/p+1/r u(n+1) Lp . Combining the last two inequalities gives u(k) Lr ≤ cδ 1/r1 −1−1/q 1/r−k−1/q uLq + cδ k−1−1/r+1/r1 u(k) Lr + n+1−k−1/p+1/r u(n+1) Lp . Since u ∈ C n (I) and I is bounded, it suffices to take δ so small that cδ k−1−1/r+1/r1 ≤ 12 . In the case k = 1 we can assume that n ≥ 3 and proceed as in Step 1 estimating u in terms of u and u(n−1) and u(n−1) in terms of u and u(n) . The details are left as an exercise.
7.3. Interpolation Inequalities
203
Step 2: The case k = 0 is similar and is left as an exercise. Observe that for m = 1 and k = 0, (7.22) holds in view of Theorem 7.34. As a corollary of Theorem 7.40 we have the following. Theorem 7.41. Let I ⊆ R be an open interval, let 1 ≤ p, q, r ≤ ∞, m ∈ N, m,1 k ∈ N0 , with 0 ≤ k < m, satisfy (7.21) and let u ∈ Wloc (I) with u ∈ Lq (I) (m) p ∈ L (I). Then there exists a constant c = c(k, m, p, q, r) > 0 such and u that (7.26)
1−θ u(k) Lr (I) ≤ cuθLq (I) u(m) L p (I) ,
if I has infinite length and (7.27)
1−θ u(k) Lr (I) ≤ c1/r−k−1/q uLq (I) + cuθLq (I) u(m) L p (I)
for every 0 < < L1 (I) if I has finite length, where (7.28)
θ :=
m − k − 1/p + 1/r m − 1/p + 1/q
and θ = 1 and r = ∞ if m = 1, q = ∞, and p = 1. Proof. If u(m) Lp L1 (I) = ∞, then u calculations. Thus, uLq > 0. Consider
= 0, then u is a polynomial of degree m − 1. If = 0, while if L1 (I) < ∞, (7.27) follows from direct in what follows, we assume that u(m) Lp > 0 and the function g(t) = At−a + Btb ,
(7.29)
where a := k + 1/q − 1/r, A := uLq , b := m − k − 1/p + 1/r, B := u(m) Lp and 0 < t ≤ t0 := . By Theorem 7.40 we have that (7.30)
u(k) Lr ≤ inf g(t). 0 32 u(x).
Then tn is still a simple Bochner integrable function, with tn (x) ≤ 32 u(x) for all x ∈ X and all n. Moreover, by (8.3), if u(x) > 0, taking ε = 1 1 2 u(x), we have that sn (x) − u(x) ≤ ε = 2 u(x) for all n sufficiently large, which implies that sn (x) ≤ 32 u(x) for all n sufficiently large. Hence, tn (x) = sn (x) for all n large and so tn (x) → u(x) as n → ∞. On the other hand, if u(x) = 0, then tn (x) = 0 = u(x) for all n. This shows that limn→∞ tn (x) − u(x) = 0 for μ-a.e. x ∈ X. In turn, tn − u dμ = 0 lim n→∞ X
by the Lebesgue dominated convergence theorem. It follows that u is Bochner integrable. Step 2. Conversely, assume that u : X → Y is Bochner integrable and consider a sequence {sn }n of Bochner integrable simple functions such that lim sn (x) − u(x) = 0
n→∞
for μ-a.e. x ∈ X
210
8. The Infinite-Dimensional Case
and lim
n→∞ X
sn − u dμ = 0.
Then for x ∈ X and for , n ∈ N, 0 ≤ |sn (x) − s (x)| ≤ sn (x) − s (x) ≤ sn (x) − u(x) + s (x) − u(x), and so
|sn − s | dμ ≤
sn − u dμ +
X
X
s − u dμ → 0 X
as , n → ∞. Thus, {sn }n is a Cauchy sequence in L1 (X, M, μ), and so it converges in L1 (X, M, μ) (and up to a subsequence also pointwise μ-a.e. in X) to a function v : X → R. On the other hand, for μ-a.e. x ∈ X, 0 ≤ |sn (x) − u(x)| ≤ sn (x) − u(x) → 0 as n → ∞. Hence, v(x) = u(x) for μ-a.e. x ∈ X, which shows that u is integrable. Since (8.2) holds for simple functions, by the properties of the norm, the general case follows by (8.1), the continuity of the norm, and Step 2. The proofs of the next three theorems follow the corresponding ones for Lebesgue integration and are left as an exercise. Theorem 8.10 (Lebesgue dominated convergence theorem). Let (X, M, μ) be a measure space and let (Y, · ) be a Banach space. Let un : X → Y , n ∈ N, and u : X → Y be strongly measurable and assume that un (x) ≤ v(x) for all n ∈ N and for μ-a.e. x ∈ X, for some Lebesgue integrable function v : X → [0, ∞), and that un (x) → u(x) as n → ∞ for μ-a.e. x ∈ X. Then u is Bochner integrable and un − u dμ = 0. lim n→∞ X
In particular,
lim
n→∞ X
un dμ =
u dμ. X
Theorem 8.11 (Fatou’s lemma). Let (X, M, μ) be a measure space and let (Y, · ) be a Banach space. Let un : X → Y , n ∈ N, and u : X → Y be strongly measurable and assume that un (x) u(x) as n → ∞ for μ-a.e. x ∈ X. If un dμ < ∞,
sup n
X
then u is Bochner integrable and u dμ ≤ lim inf un dμ. X
n→∞
X
8.1. The Bochner Integral
211
Theorem 8.12 (Egoroff). Let (X, M, μ) be a measure space with μ finite, let (Y, · ) be a Banach space, and let u, un : X → Y , n ∈ N, be strongly measurable functions such that lim un (x) − u(x) = 0
n→∞
for μ-a.e. x ∈ X. Then for every ε > 0 there exists a measurable set E ∈ M, with μ(X \ E) ≤ ε, such that lim sup un (x) − u(x) = 0.
n→∞ x∈E
Next we prove that the Bochner integral commutes with elements of Y . Theorem 8.13. Let (X, M, μ) be a measure space and let (Y, · ) be a Banach space. Let u : X → Y be Bochner integrable. Then for every T ∈ Y , u dμ = T (u) dμ. T X
X
Proof. We proceed as in the first part of the proof of Theorem 8.9 to construct a sequence {tn }n of simple functions tn : X → Y such that tn (x) ≤ 32 u(x) for all x ∈ X and all n, limn→∞ tn (x) − u(x) = 0 for μ-a.e. x ∈ X, and tn − u dμ = 0. lim n→∞ X
Write tn =
mn
ci,n χEi,n .
i=1
Now, let T ∈Y . Then by the continuity and linearity of T , and the fact that limn→∞ X tn dμ = X u dμ, u(x) dμ = lim T tn (x) dμ T X
n→∞
= lim
n→∞
X mn
T (ci,n )μ(Ei,n ) = lim
i=1
n→∞ X
T (tn (x)) dμ.
Since |T (tn (x))| ≤ T Y tn (x) ≤ 32 T Y u(x), we can apply the Lebesgue dominated convergence theorem to conclude that T (tn (x)) dμ = lim T (tn (x)) dμ = T (u(x)) dμ. lim n→∞ X
This concludes the proof.
X n→∞
X
212
8. The Infinite-Dimensional Case
8.2. Lp Spaces on Banach Spaces Definition 8.14. Let (X, M, μ) be a measure space, let (Y, ·) be a Banach space, and let 1 ≤ p < ∞. Then Lp (X; Y ) := {u : X → Y : u strongly measurable, uLp (X;Y ) < ∞}, where uLp (X;Y ) :=
up dμ
1/p .
X
If p = ∞, then L∞ (X; Y ) := {u : X → Y : u strongly measurable, uL∞ (X;Y ) < ∞}, where uL∞ (X;Y ) = ess sup u(x) x∈X
:= inf{t ≥ 0 : u(x) ≤ t for μ-a.e. x ∈ X}. As in the case Y = R we identify functions with their equivalence classes. Theorem 8.15. Let (X, M, μ) be a measure space and let Y be a Banach space. (i) Lp (X; Y ) is a Banach space for 1 ≤ p ≤ ∞; (ii) the family of all Bochner integrable simple functions is dense in Lp (X; Y ) for 1 ≤ p < ∞; (iii) if X is a separable metric space, μ is a σ-finite Radon measure, and if Y is separable, then Lp (X; Y ) is separable for 1 ≤ p < ∞. A major problem in the theory of Lp spaces on Banach spaces is the identification of the dual of Lp (X; Y ). Here we will consider only the two important special cases in which Y is either separable or reflexive. Definition 8.16. Let (X, M, μ) be a measure space, let Y be a Banach space, and let 1 ≤ p ≤ ∞. Then the space Lpw (X; Y ) is the space of all (equivalence classes of) weakly star measurable functions u : X → Y such that uY ∈ Lp (X; R). The space Lpw (X; Y ) is endowed with the norm
1/p p uY dμ for 1 ≤ p < ∞, uLpw (X;Y ) := X
and := ess sup u(x)Y uL∞ w (X;Y )
x∈X
for p = ∞.
8.2. Lp Spaces on Banach Spaces
213
Theorem 8.17 (Riesz representation theorem in Lp ). Let (X, M, μ) be a measure space with μ σ-finite, let Y be a Banach space, let 1 ≤ p < ∞, and let q be its H¨ older conjugate exponent. (i) Assume that Y is separable. If T ∈ (Lp (X; Y )) , then there exists a unique v ∈ Lqw (X; Y ) such that v, u Y ,Y dμ (8.4) T (u) = X
Moreover, the norm of T coincides with for every u ∈ vLqw (X;Y ) . Conversely, every functional of the form (8.4), where v ∈ Lqw (X; Y ), is a bounded linear functional on Lp (X; Y ). Lp (X; Y ).
(ii) Assume that Y is reflexive. Then for T ∈ (Lp (X; Y )) there exists a unique v ∈ Lq (X; Y ) such that v, u Y ,Y dμ (8.5) T (u) = X
for every u ∈ Moreover, the norm of T coincides with vLq (X;Y ) . Conversely, every functional of the form (8.5), where v ∈ Lq (X; Y ), is a bounded linear functional on Lp (X; Y ). Lp (X; Y ).
Corollary 8.18. Let (X, M, μ) be a measure space with μ σ-finite, let Y be a reflexive Banach space, and let 1 < p < ∞. Then Lp (X; Y ) is reflexive. In what follows we restrict our attention to the case in which the measure space (X, M, μ) is given by an interval I ⊆ R with the Lebesgue measure. Theorem 8.19. Let I ⊆ R be an interval, let (Y, · ) be a Banach space, and let v : I → Y be locally Bochner integrable. Then for L1 -a.e. x ∈ I, 1 x+r v(t) − v(x) dt = 0. lim r→0+ r x−r Proof. By the Pettis theorem (see Theorem 8.3) there exists a set E0 ⊂ I with L1 (E0 ) = 0 such that v(I \ E0 ) is separable. Let {yn }n be dense in v(I \E0 ). Since the function wn (x) := v(x)−yn , x ∈ I, is locally Lebesgue integrable, by Theorem 3.16 there exists En ⊂ I with L1 (En ) = 0 such that x+r 1 v(t) − yn dt = v(x) − yn lim r→0+ 2r x−r 1 for all x ∈ I \ En . Let E := ∞ n=0 En . Then L (E) = 0 and for x ∈ I \ E and n ∈ N, x+r x+r 1 1 v(t) − v(x) dt ≤ lim v(t) − yn dt lim sup r→0+ 2r x−r r→0+ 2r x−r + v(x) − yn = 2v(x) − yn .
214
8. The Infinite-Dimensional Case
Since {yn }n is dense in v(I \ E0 ) and x ∈ / E0 , we can find a subsequence {ynk }k of {yn }n such that ynk → v(x). By taking n = nk in the previous inequality and letting k → ∞ we conclude the proof. As a consequence of this theorem, we can extend the mollification theory (see Section C.3 in Appendix C) to Banach-valued functions. Given a nonnegative bounded function ϕ ∈ L1 (R) with ϕ (x) dx = 1, (8.6) supp ϕ ⊆ B (0, 1), R 1 ε ϕ (x/ε),
x ∈ R. The functions ϕε for every ε > 0 we define ϕε (x) := are called mollifiers. Note that supp ϕε ⊆ B (0, ε). Hence, given an open interval I ⊆ R, a Banach space Y , and a function u ∈ L1loc (I; Y ), we may define uε (x) := (u ∗ ϕε ) (x) =
ϕε (x − t) u (t) dt I
for x ∈ Iε , where the open set Iε is given by Iε := {x ∈ I : dist (x, ∂I) > ε} . The function uε : Iε → Y is called a mollification of u. Note that if u is Bochner integrable, then uε (x) is well-defined for every x ∈ R. The following theorems can be proved exactly as in Theorems C.16 and C.20. Theorem 8.20. Let I ⊆ R be an open interval, let ϕ ∈ L1 (I) be a nonnegative bounded function satisfying (8.6), let Y be a Banach space, and let u ∈ L1loc (I; Y ). (i) For every Lebesgue point x ∈ I, uε (x) → u (x) as ε → 0+ . (ii) If u ∈ Lp (I; Y ), 1 ≤ p ≤ ∞, then uε Lp (R;Y ) ≤ uLp (I;Y ) for every ε > 0 and uε Lp (R;Y ) → uLp (I;Y ) as ε → 0+ . (iii) If u ∈ Lp (I; Y ), 1 ≤ p < ∞, then lim uε − uLp (I;Y ) = 0.
ε→0+
Theorem 8.21. Let I ⊆ R be an open interval, let ϕ ∈ Cc∞ (R) be a nonnegative bounded function satisfying (8.6), let Y be a Banach space, and let u ∈ L1loc (I; Y ). Then uε ∈ C ∞ (Iε ; Y ) for all 0 < ε < 1, and for every n ∈ N, (n) (x) = (u ∗ ϕ )(x) = ϕ(n) (8.7) u(n) ε ε ε (x − t)u(t) dt I
8.2. Lp Spaces on Banach Spaces
215
for all x ∈ Iε . Moreover, if u ∈ Lp (I; Y ), 1 ≤ p ≤ ∞, then uε ∈ C ∞ (R; Y ). Exercise 8.22. Let I ⊆ R be an open interval, let Y be a Banach space, and let v : I → Y be a locally Bochner integrable function such that ϕ(x)v(x) dx = 0 I
for every function ϕ ∈ Cc∞ (I). Prove that v(x) = 0 for L1 -a.e. x ∈ I. Exercise 8.23. Let I ⊆ R be an open bounded interval, let Y be a Banach space, and let 1 ≤ p < ∞. Prove that functions in C ∞ (I; Y ) are dense in Lp (I; Y ). Next we characterize compact sets in Lp (I; Y ). Theorem 8.24 (Simon). Let I ⊂ R be a bounded interval, let 1 ≤ p < ∞, and let (Y, · ) be a Banach space. Then U ⊆ Lp (I; Y ) is relatively compact in Lp (I; Y ) if and only (i) for every x1 , x2 ∈ I ◦ , with x1 < x2 , the set x2 u(x) dx : u ∈ U x1
is relatively compact in Y , (ii) τh (u) − uLp (I h ;Y ) → 0 as h → 0+ uniformly for u ∈ U , where τh (u)(x) := u(x + h) and I h := {x ∈ I : x + h ∈ I}. Proof. Step 1: Assume that U ⊆ Lp (I; Y ) is relatively compact in Lp (I; Y ). Since for every x1 , x2 ∈ I ◦ , with x1 < x2 , the function x2 p u(x) dx u ∈ L (I; Y ) → x1
is continuous from Lp (I; Y ) into Y and since continuous functions map compact sets into compact sets, it follows that (i) holds. On the other hand, by the compactness of U , given ε > 0, we may find u1 , . . . , un ∈ Lp (I; Y ) such that for every u ∈ U there exists i ∈ {1, . . . , n} with u − ui Lp (I;Y ) ≤ ε. By the density of C(I; Y ) into Lp (I; Y ) (see Exercise 8.23), without loss of generality, we may assume that the functions ui are in C(I; Y ). Since each function ui is uniformly continuous, there exists δi > 0 such that ui (x) − ui (s) ≤ ε for all s, x ∈ I with |s − x| ≤ δi . In turn, for 0 < h ≤ δi , τh (ui )(x) − ui (x)p dx = ui (x + h) − ui (x)p dx ≤ εp L1 (I h ), Ih
Ih
216
8. The Infinite-Dimensional Case
and so τh (ui ) − ui Lp (I h ;Y ) ≤ ε(L1 (I))1/p . Let δ := mini δi > 0. For 0 < h ≤ δ and u ∈ U , let i ∈ {1, . . . , n} be such that u − ui Lp (I;Y ) ≤ ε. Writing τh (u) − u = τh (u − ui ) + (τh (ui ) − ui ) + (ui − u), it follows that τh (u) − uLp (I h ;Y ) ≤ 2ε + ε(L1 (I))1/p , which proves (ii). Step 2: Assume that U ⊆ Lp (I; Y ) satisfies properties (i) and (ii). Let 0 < η < L1 (I) and for every u ∈ U define the right mean function 1 x+η u(s) ds, x ∈ I η . Mη (u)(x) := η x For every x1 , x2 ∈ I η , with x1 < x2 , by a change of variables and H¨older’s inequality we have 1 x1 +η τx2− x1 (u) − u ds Mη (u)(x2 ) − Mη (u)(x1 ) ≤ η x1 1 ≤ 1/p τx2 −x1 (u) − uLp (I x2 −x1 ;Y ) . η It follows from property (ii) that the family of functions {Mη (u) : u ∈ U } belongs to C(I η ; Y ) and is uniformly equicontinuous. Moreover, by property (i) for every x ∈ (I η )◦ we have that the set {Mη (u)(x) : u ∈ U } is relatively compact in Y . Thus, we are in a position to apply the Ascoli–Arzel`a theorem (see Theorem 5.37) to obtain that the family {Mη (u) : u ∈ U } is relatively compact in C(I η ; Y ). Since by property (ii) the function h → τh (u) − u is continuous in Lp (I η ; Y ), we can write η p 1 Mη (u) − up dx = (τ (u)(x) − u(x)) dh dx h η η η 0 I I τh (u)(x) − u(x)p dx. ≤ sup 0 0 we can find 0 < ηε < 12 L1 (I) (independent of u) such that for every 0 < η ≤ ηε , the right-hand side of the previous inequality is bounded from above by εp for every u ∈ U . Hence, Mη (u) − up dx ≤ εp (8.8) Iη
for every 0 < η ≤ ηε and for every u ∈ U . Fix 0 < η ≤ ηε . Since {Mη (u) : u ∈ U } is relatively compact in C(I η ; Y ) there exist u1 ,. . . , un ∈ C(I η ; Y ) such that for every u ∈ U there is i ∈ {1, . . . , n} with (8.9)
Mη (u) − ui C(I η ;Y ) ≤ ε.
8.2. Lp Spaces on Banach Spaces
217
In turn, by (8.8) and (8.9), u − ui Lp (I η ;Y ) ≤ u − Mη (u)Lp (I η ;Y ) + Mη (u) − ui Lp (I η ;Y ) ≤ Mη (u) − ui C(I η ;Y ) (L1 (I))1/p + ε ≤ (1 + (L1 (I))1/p )ε. In particular, since η < 12 L1 (I) we have that I η ⊃ J1 , where J1 is the interval of endpoints inf I and the middle point of I, and so the set U1 := {u : J1 → Y : u ∈ U } is relatively compact in Lp (J1 ; Y ). With a similar proof, working with h negative, we can show that the set U2 := {u : J2 → Y : u ∈ U } is relatively compact in to Lp (J2 ; Y ), where J2 is the interval of endpoints the middle point of I and sup I. It follows (why?) that U is relatively compact in Lp (I; Y ). Exercise 8.25. Let I ⊂ R be a bounded interval and let Y be a Banach space. Prove that U ⊆ L∞ (I; Y ) is relatively compact in C(I; Y ) if and only if properties (i) and (ii) (with p = ∞) are satisfied. The following corollary will be useful in Section 8.5 below. Corollary 8.26. Let I ⊂ R be a bounded interval, let (Y0 , ·Y0 ), (Y1 , ·Y1 ), and (Y2 , · Y2 ) be Banach spaces with Y0 → Y1 → Y2 . Assume that the embedding Y0 → Y1 is compact. Let 1 ≤ p < ∞ and let U ⊆ Lp (I; Y0 ) be a bounded set such that τh (u) − uLp (I h ;Y2 ) → 0 as h → 0+ uniformly for u ∈ U . Then U is relatively compact in Lp (I; Y1 ). Proof. Step 1: We claim that for every ε > 0 there exists cε > 0 such that yY1 ≤ εyY0 + cε yY2
(8.10)
for every y ∈ Y0 . By hypothesis there exist c1 , c2 > 0 such that yY1 ≤ c1 yY0
(8.11) for all y ∈ Y0 and (8.12)
yY2 ≤ c2 yY1
for all y ∈ Y1 . Assume that (8.10) fails. Then for every n there exists yn ∈ Y0 such that yn Y1 > εyn Y0 + nyn Y2 . Note that by (8.11) the previous inequality implies that yn Y0 > 0. Let wn := yn /yn Y0 . Then (8.13)
wn Y1 > εwn Y0 + nwn Y2 = ε + nwn Y2 .
Since the sequence {wn }n is bounded in Y0 , by (8.11), it is also bounded in Y1 . In turn, by (8.13), wn → 0 in Y2 . On the other hand, since the embedding Y0 → Y1 is compact, there exist a subsequence {wnk }k and
218
8. The Infinite-Dimensional Case
w0 ∈ Y1 such that wnk → w0 in Y1 . Necessarily, w0 = 0 by (8.12). Hence, taking n = nk in (8.13) and letting k → ∞ we get 0 = lim wnk Y1 ≥ ε > 0, k→∞
which is a contradiction. Step 2: Since U ⊆ Lp (I; Y0 ) is bounded, by H¨older’s inequality for every x1 , x2 ∈ I ◦ , with x1 < x2 , the set x2 u(x) dx : u ∈ U Yx1 ,x2 := x1
is bounded in Y0 . On the other hand, the embedding Y0 → Y1 is compact, and so the set Yx1 ,x2 is relatively compact in Y1 . By (8.12), Yx1 ,x2 is also relatively compact in Y2 . It follows by Theorem 8.24 that U is relatively compact in Lp (I; Y2 ). It remains to show that U is relatively compact in Lp (I; Y1 ). Let {un }n be a sequence in U . Since U is relatively compact in Lp (I; Y2 ), by sequential compactness, there exists a subsequence, not relabeled, such that un → u0 in Lp (I; Y2 ) for some u0 ∈ Lp (I; Y2 ). In particular, {un }n is a Cauchy sequence in Lp (I; Y2 ). Let ε > 0. By Step 1 and the Minkowski inequality for every n, m ∈ N, un − um Lp (I;Y1 ) ≤ εun − um Lp (I;Y0 ) + cε un − um Lp (I;Y2 ) ≤ 2εK + cε un − um Lp (I;Y2 ) , where K := supu∈U uLp (I;Y0 ) < ∞. Letting n, m → ∞, by the fact that {un }n is a Cauchy sequence in Lp (I; Y2 ) we get lim sup un − um Lp (I;Y1 ) ≤ 2εK. n,m→∞
By letting ε → 0+ we obtain that a subsequence of {un }n is a Cauchy sequence in Lp (I; Y1 ). This shows that the closure of U is sequentially compact in Lp (I; Y1 ) and concludes the proof. Exercise 8.27. Let I ⊂ R be a bounded interval, let (Y0 , ·Y0 ), (Y1 , ·Y1 ), and (Y2 , · Y2 ) be Banach spaces with Y0 → Y1 → Y2 . Assume that the embedding Y0 → Y1 is compact. Let U ⊆ L∞ (I; Y0 ) be a bounded set such that τh (u) − uL∞ (I h ;Y2 ) → 0 as h → 0+ uniformly for u ∈ U . Prove that U is relatively compact in C(I; Y1 ). We conclude this section with a useful result. Theorem 8.28. Let I ⊆ R be an open interval, let E ⊆ RN be a Lebesgue measurable set, and let 1 ≤ p < ∞. Then Lp (I; Lp (E)) can be identified with Lp (E × I).
8.2. Lp Spaces on Banach Spaces
219
u ∈ Lp (E × I), by Fubini’s theorem, the function t ∈ I → Proof. Given p E |u (x, t) | dx is Lebesgue measurable with |u (x, t) |p dx dt = upLp (E×I) < ∞.
I
E
It follows that E |u (x, t) |p dx < ∞ for L1 -a.e. t ∈ I. Hence, setting v (t) := u (·, t), by Theorem 8.9, we have that v : I → Lp (E) with p p |u (x, t) |p dxdt = upLp (E×I) < ∞, vLp (I;Lp (E)) = v (t) Lp (E) dt = I
I
which shows that v ∈ sider the linear operator
Lp (I; Lp (E))
E
with vLp (I;Lp (E)) = uLp (E×I) . Con-
T : Lp (E × I) → Lp (I; Lp (E)) u → v. Then T preserves the norms and so it is injective. By identifying u with T (u), this shows that Lp (E × I) ⊆ Lp (I; Lp (E)). Conversely, given v ∈ Lp (I; Lp (E)), there exists a sequence {sn } of Bochner integrable simple functions sn : I → Lp (E) such that lim sn (t) − v(t)Lp (E) = 0 for L1 -a.e. t ∈ I
n→∞
and lim
n→∞ I
sn (t) − v (t) pLp (E) dt = 0.
Write sn =
mn
ui,n χEi,n ,
i=1
where ui,n ∈ Lp (E) and Ei,n ⊆ I are Lebesgue measurable pairwise disjoint sets. Define mn ui,n (x) χEi,n (t) , x ∈ E, t ∈ I. gn (x, t) := i=1
Then gn : E × I → R is Lebesgue measurable, since it is given by sums and products of Lebesgue measurable functions. Moreover, sn pLp (I;Lp (E)) = sn (t) pLp (E) dt I
=
mn I
=
|ui,n (x) |p dx dt
χEi,n (t)
i=1 gn pLp (E×I) .
E
A similar computation shows that sn − sm Lp (I;Lp (E)) = gn − gm Lp (E×I) . Since sn → v in Lp (I; Lp (E)), we have that {sn }n is a Cauchy sequence in
220
8. The Infinite-Dimensional Case
Lp (I; Lp (E)). In turn, {gn }n is a Cauchy sequence in Lp (E × I). Hence, there exists u ∈ Lp (E ×I) such that gn → u in Lp (E ×I). However, from the first part of the theorem Lp (E × I) ⊆ Lp (I; Lp (E)). Moreover, T (gn ) = sn . Hence, T (u) = v, which shows that Lp (I; Lp (E)) ⊆ Lp (E × I).
8.3. Functions of Bounded Pointwise Variation In this section we discuss the differentiability of functions of bounded pointwise variation in the case in which Y is an infinite dimensional normed space. We begin by establishing the analog of Corollary 1.28. Theorem 8.29. Let I ⊆ R be an interval, let (Y, · ) be a normed space, and let u ∈ BVloc (I; Y ). Then u(I) is separable, u is strongly measurable, and for h > 0, 1 u(x + h) − u(x) dx ≤ Var u, (8.14) h Ih where I h := {x ∈ I : x + h ∈ I}. Proof. Let’s prove that u is weakly measurable. Let T ∈ Y and consider the function v(x) := T (u(x)), x ∈ I. Let [a, b] ⊆ I and let x0 < · · · < xn be a partition of [a, b]. Then by the linearity and continuity of T , n i=1
|v(xi ) − v(xi−1 )| =
n
|T (u(xi ) − u(xi−1 ))|
i=1
≤ T Y
n
u(xi ) − u(xi−1 ) ≤ T Y Var[a,b] u.
i=1
Taking the supremum over all partitions gives Var[a,b] v ≤ T Y Var[a,b] u < ∞. Thus, v ∈ BVloc (I), and so it is measurable. Given the arbitrariness of T ∈ Y , it follows that u is weakly measurable. Next we show that u(I) is separable. By Theorem 2.17 the set of jump points of u is countable. Let {xn }n be the sequence obtained by taking the union of the rational numbers in I and the set of jump points of u. Consider the countable set F := {u(xn ) : n ∈ N}. Given y ∈ u(I), if y ∈ F , then there is nothing to prove. If y ∈ u(I) \ F , then there exists x ∈ I such that u(x) = y and u is continuous at x. By the density of the rationals, there exists a subsequence {xnk }k of {xn }n such that xnk → x as k → ∞. By continuity u(xnk ) → y. Hence, u(I) is separable. It now follows by the Pettis Theorem (see Theorem 8.3) that u is strongly measurable. Next we prove (8.14). If Var u = ∞, then there is nothing to prove. Thus, assume that Var u < ∞. If diam I h > 0, let [a, b] ⊆ I be such that
8.3. Functions of Bounded Pointwise Variation
221
0 < h ≤ b − a. By (2.4) and Corollary 1.28 applied to V we have 1 b−h 1 b−h u(x + h) − u(x) dx ≤ (V (x + h) − V (x)) dx h a h a ≤ V (b) − V (a) = Var[a,b] u. Construct an increasing sequence of intervals [an , bn ] such that an inf I, bn sup I. If diam I h > 0, then for all n sufficiently large we have that 0 < h < bn − an , and so by the previous inequality we have that 1 bn −h u(x + h) − u(x) dx ≤ Var[an ,bn ] u ≤ Var u. h an Letting n → ∞ and using the Lebesgue monotone convergence theorem gives 1 u(x + h) − u(x) dx ≤ Var u. h Ih The previous inequality continues to hold if diam I h = 0, since in this case the integral on the left-hand side is zero. This completes the proof. The following two exercises show that when Y is an infinite-dimensional normed space, functions of bounded pointwise variation may be nowhere differentiable. Exercise 8.30. Let Y = L1 ([a, b]) and let u : [a, b] → Y be defined by 1 if a ≤ y < x, (u(x))(y) := 0 if x ≤ y ≤ b. Prove that u has bounded pointwise variation but it is nowhere differentiable. Exercise 8.31. Let Y = c0 and let u : [0, 1] → Y be defined by x u(x) := {an (x)}n , an (x) := sin(2nπs) ds. 0
Prove that u has bounded pointwise variation but it is nowhere differentiable. We recall that c0 is the space of all sequences {an }n of real numbers converging to zero, endowed with the norm supn |an |. Exercise 8.32. Let Y = L∞ ((0, 1)) and let u : [0, 1] → Y be defined by (u(x))(y) := y sin(x/y),
x ∈ [0, 1], y ∈ (0, 1).
Prove that u has bounded pointwise variation but it is nowhere differentiable. The following theorem shows that if Y is reflexive, then a function of bounded pointwise variation is differentiable L1 -a.e.
222
8. The Infinite-Dimensional Case
Theorem 8.33. Let I ⊆ R be an interval, let (Y, · ) be a reflexive Banach space, and let u ∈ BVloc (I; Y ). Then u is differentiable at L1 -a.e. x ∈ I, the function u is locally Bochner integrable and for every interval J ⊆ I, u (x) dx ≤ VarJ u. J
In particular, if u ∈ BP V (I; Y ), then u is Bochner integrable. The proof relies on the fact that if Y is a reflexive Banach space, Lipschitz continuous functions u : I → Y are differentiable L1 -a.e. We will prove this result in Theorem 8.38 below. Proof. Step 1: Assume I = [a, b]. By Theorem 8.29 we have that u(I) is separable. Hence, by replacing Y with the closure of the subspace generated by the set u(I), without loss of generality, we may assume that Y is reflexive and separable. (Exercise. Hint: Use Theorem A.64.) By Theorem 2.36, the indefinite pointwise variation V of u is increasing and u(x + h) − u(x) ≤ |V (x + h) − V (x)| for all x, x + h ∈ I. Dividing by |h| and letting h → 0, it follows by Lebesgue’s differentiation theorem (see Theorem 1.18) applied to V , that for L1 -a.e. x ∈ I, (8.15)
lim sup h→0
u(x + h) − u(x) ≤ V (x) < ∞. |h|
Hence, the set (8.16)
u(x + h) − u(x) E0 := x ∈ I : lim sup =∞ |h| h→0
satisfies L1 (E0 ) = 0. For every n ∈ N let Cn be the set of points x ∈ I such that (8.17)
u(x + h) − u(x) ≤ n|h|
for I \ E0 = all x ∈ I and for all |h| < 1/n with x + h ∈ I. Observe that 1 -a.e. x ∈ C . C . Hence, it suffices to prove that u is differentiable for L n n n Fix n. We claim that Cn is closed. To see this, let xk ∈ Cn be such that xk → x ∈ I. Let |h| < 1/n be such that x + h ∈ I (if no such h exists, then x belongs to Cn automatically). Fix 0 < ε < 1/n − |h| and find k so large that |xk − x| < ε. Then |x + h − xk | ≤ |h| + |x − xk | < |h| + ε < 1/n,
8.3. Functions of Bounded Pointwise Variation
223
and so, since xk ∈ Cn , u(x + h) − u(x) ≤ u(x + h) − u(xk ) + u(xk ) − u(x) ≤ n|x + h − xk | + n|xk − x| < n(|h| + 2ε). Letting ε → 0+ in the previous inequality shows that x ∈ Cn . Thus, Cn is closed. Hence, the set I \ Cn is relatively open in I. Thus we can write I \ Cn as the disjoint union of countably many intervals Ik . These intervals will all be open with the possible exception of the two intervals that contain an endpoint of I. Extend u : I \ Cn → Y in each interval Ik of endpoints ak < bk by setting un (x) := u(ak ) +
u(bk ) − u(ak ) (x − ak ). bk − ak
Note that if |bk − ak | < 1/n, then, since ak ∈ Cn , it follows from (8.17) that u(bk ) − u(ak ) ≤ n(bk − ak ). On the other hand, if |bk − ak | ≥ 1/n, then u(bk ) − u(ak ) ≤ 2nu∞ (bk − ak ) (we recall that functions in BP V (I; Y ) are bounded; see Proposition 2.12). Thus, un : Ik → Y is Lipschitz continuous with Lipschitz constant at most n(1 + 2u∞ ). It follows also from (8.17) that un : I → Y is Lipschitz continuous (exercise). By Theorem 8.38 the function un is differentiable at L1 -a.e. x ∈ I. Moreover, since the function vn (x) := dist(x, Cn ), x ∈ I, is Lipschitz continuous, by Proposition 2.6 and Corollary 2.31, vn is differentiable at L1 -a.e. x ∈ I. Let x ∈ Cn be such that both un and vn are differentiable at x and x is not an endpoint of any Ik . We claim that u is differentiable at x. Since x is not an isolated point of Cn , necessarily, vn (x) = 0 (why?). Fix 0 < ε < 1/2 and find 0 < δ < 1/n such that (8.18)
un (x + h) − un (x) − un (x)h ≤ ε|h|, dist(x + h, Cn ) = |vn (x + h) − vn (x) − vn (x)h| < ε|h|
for all x + h ∈ I with 0 < |h| ≤ δ. Fix 0 < |h| ≤ δ/2. Since dist(x + h, Cn ) < ε|h|, there exists t ∈ Cn such that |x + h − t| < ε|h| < 1/n. Note that |x − t| ≤ |x + h − t| + |h| ≤ δ. Using the fact that u = un on Cn , we have that u(x + h) − u(x) − un (x)h ≤ u(x + h) − u(t) + un (t) − un (x) − un (x)(t − x) + un (x)|x + h − t| ≤ n|x + h − t| + ε|h| + un (x)|x + h − t| ≤ (n + 1 + un (x))ε|h|, where we used (8.17) and (8.18). This shows that u is differentiable at x.
224
8. The Infinite-Dimensional Case
Step 2: For an arbitrary interval I, it suffices to take an increasing sequence of closed intervals whose union is I and to apply Step 1 in each closed interval. Remark 8.34. The previous theorem applies, in particular, for Y = Lp (E), 1 < p < ∞, where E ⊆ RN is a Lebesgue measurable set. The Banach spaces Y for which Theorem 8.33 holds, namely, for which every function u : I → Y of bounded pointwise variation is differentiable L1 -a.e. in I, are said to have the Radon–Nikodym property. Definition 8.35. A Banach space Y has the Radon–Nikodym property (RNP) if for every interval I ⊆ R and every function u : I → Y of bounded pointwise variation, u is differentiable L1 -a.e. in I. Exercises 8.30 and 8.32 show that Y = L1 (I) and Y = L∞ (I) do not have the RNP. For spaces without the RNP property one can use a weaker notion of derivative, namely, the metric derivative. Definition 8.36. Given E ⊆ R, an accumulation point x0 ∈ E of E and a mu (x0 ) of a function u : E → Y metric space (Y, d), the metric derivative ddx at x0 is defined as the limit (8.19)
d(u(x + h), u(x)) h→0 |h| lim
provided this limit exists in [0, ∞]. We say that u is metric differentiable at mu x0 if there exists ddx (x0 ) < ∞. Exercise 8.37. Let I ⊆ R be an interval and let Y be a Banach space. Using the fact that every Lipschitz continuous function v : I → Y admits a finite metric derivative L1 -a.e. in I,1 modify the proof of Theorem 8.33 to prove that if u ∈ BVloc (I; Y ), then for L1 -a.e. x ∈ I, there exists in R mu mu (x), that the function ddx is locally integrable and the metric derivative ddx that for every interval J ⊆ I, dm u (x) dx ≤ VarJ u. J dx
8.4. Absolute Continuous Functions In this section we extend some of the results of Section 3.2 in Chapter 3 to functions u : I → Y , where Y is a normed space. In view of Exercise 8.30 we cannot expect Theorem 3.20 to hold for arbitrary Banach spaces. However, we will show that it holds if Y is reflexive. 1 We
will prove this in Exercise 8.40.
8.4. Absolute Continuous Functions
225
Theorem 8.38. Let I ⊆ R be an interval and let (Y, · ) be a reflexive Banach space. A function u : I → Y belongs to ACloc (I; Y ) if and only if (i) u is continuous in I, (ii) u is differentiable L1 -a.e. in I, and u is locally Bochner integrable, (iii) the fundamental theorem of calculus is valid; that is, for all x, x0 ∈ I, x u (t) dt. u(x) = u(x0 ) + x0
Proof. Step 1: Assume that u ∈ ACloc (I; Y ). For T ∈ Y consider the function w := T ◦ u. We claim that w belongs to ACloc (I). To see this, let [a, b] ⊆ I. Since u ∈ AC([a, b]; Y ), for every ε > 0 there exists δ > 0 such that n u(bi ) − u(ai ) ≤ ε i=1
for every finite number of nonoverlapping intervals (ai , bi ), i = 1, . . . , n, with [ai , bi ] ⊆ [a, b] and ni=1 (bi − ai ) ≤ δ. In turn, by the linearity of T , n
|w(bi ) − w(ai )| =
i=1
n
|T (u(bi ) − u(ai ))|
i=1
≤ T Y
n
u(bi ) − u(ai ) ≤ εT Y .
i=1
This proves the claim. Step 2: Assume that u ∈ ACloc (I; Y ). As in the proof of Theorem 8.29 we may assume that Y is reflexive and separable. In turn, its dual Y is reflexive and separable. Let {Tk }k be dense in Y . Consider the function wk := Tk ◦ u. By Step 1, wk belongs to ACloc (I). Hence, by Proposition 3.9, it is differentiable for all x ∈ I \ Ek , where L1 (Ek ) = 0. By the linearity of Tk , this implies that there exists u(x + h) − u(x) = wk (x) ∈ R (8.20) lim Tk h→0 h for all ∈ I \ Ek . Set E := ∞ k=0 Ek , where u(x + h) − u(x) =∞ . E0 := x ∈ I : lim sup |h| h→0 As in the proof of Theorem 8.29 we have that L1 (E0 ) = 0. In turn, L1 (E) = 0. Let x ∈ I \ E. Since x ∈ / E0 , there exists L0 > 0 such that (8.21)
u(x + h) − u(x) ≤ L0 |h|
226
8. The Infinite-Dimensional Case
for all h sufficiently small. Using the fact that Y is reflexive, by Corollary A.65 we can find a sequence hj → 0 and v(x) ∈ Y such that u(x + hj ) − u(x) v(x). hj It follows, also from (8.20), that wk (x) = Tk (v(x)). We claim that u(x + h) − u(x) v(x) h as h → 0+ . Fix T ∈ Y and η > 0. By the density of {Tk }k in Y , there (x) = T (v(x)), there exists exists Tm such that T − Tm Y ≤ η. Since vm m h0 > 0 such that u(x + h) − u(x) − Tm (v(x)) ≤ η Tm h for all 0 < |h| ≤ h0 . In turn, also by (8.21), u(x + h) − u(x) − T (v(x)) ≤ T − Tm Y (L0 + v(x)) T h u(x + h) − u(x) − Tm (v(x)) ≤ (L0 + v(x) + 1)η + Tm h for all 0 < |h| ≤ h0 . This proves the claim. (8.22)
The function x ∈ I \ E → v(x) is weakly measurable, since each function x ∈ I → u(x+h)−u(x) is strongly h measurable. Since Y is separable, it follows by Pettis’ theorem (see Theorem 8.3) that it is strongly measurable. Moreover, by the lower semicontinuity of the norm with respect to weak convergence (see Proposition A.67) and (8.15), u(x + h) − u(x) ≤ V (x) v(x) ≤ lim inf h→0 |h| for L1 -a.e. x ∈ I. It follows from Fatou’s lemma that u(x + h) − u(x) dx v(x) dx ≤ lim inf + h h→0 J J V (x) dx ≤ VarJ u, ≤ J
where in the last inequality we have used Corollary 1.28. Hence, v is locally Bochner integrable. Since for T ∈ Y the function w := T ◦ u belongs to ACloc (I), by Step 1, by Proposition 3.9, w is differentiable for L1 -a.e. x ∈ I. It follows from (8.22) and the linearity of T that for L1 -a.e. x ∈ I, u(x + h) − u(x) w(x + h) − w(x) =T → T (v(x)), h h
8.4. Absolute Continuous Functions
227
which shows that w (x) = T (v(x)) for L1 -a.e. x ∈ I. Hence, by Theorem 3.20 applied to the function w, we have that x x T (u(x)) − T (u(x0 )) = w(x) − w(x0 ) = w (t) dt = T (v(t)) dt x0 x0 x v(t) dt , =T x0
where we have used Theorem 8.13. Hence, x v(t) dt = 0. T u(x) − u(x0 ) − x0
Taking the supremum over all T with T Y ≤ 1 and using the fact that for y ∈ Y , y = sup T Y ≤1 T (y), we get x v(t) dt = 0. u(x) − u(x0 ) − x0
It now follows from Theorem 8.19 that u is differentiable for L1 -a.e. x ∈ I with u (x) = v(x). Step 3. Conversely, assume that u satisfies (i)–(iii). Then u is Lebesgue integrable on every [a, b] ⊆ I and so equi-integrable in [a, b]. Reasoning as in the second part of the proof of Theorem 3.29 we can show that u is absolutely continuous in [a, b].
This concludes the proof.
Remark 8.39. In view of the previous theorem, Theorems 3.25, 3.29, and Corollaries 3.31 and 3.33 continue to hold with RM replaced by a reflexive Banach space. Exercise 8.40. Let I ⊆ R be an interval, let (Y, · ) be a Banach space, and let u : I → Y be a Lipschitz continuous function. (i) Let {yn }n ⊆ u(I) be dense in u(I) and define vn (x) := u(x) − yn ,
x ∈ I.
Prove that for x, x + h ∈ I, (8.23)
|vn (x + h) − vn (x)| ≤ u(x + h) − u(t) and deduce that sup |vn (x)| ≤ lim inf n
for L1 -a.e. x ∈ I.
h→0
u(x + h) − u(t) |h|
228
8. The Infinite-Dimensional Case
(ii) Prove that for x, x + h ∈ I,
u(x + h) − u(t) = sup |vn (x + h) − vn (x)| ≤ n
x+h x
sup |vn (t)| dt n
and deduce that lim sup h→0
u(x + h) − u(t) ≤ sup |vn (x)| |h| n
x ∈ I. Deduce that u is metrically differentiable at for 1 L -a.e. x ∈ I. L1 -a.e.
Exercise 8.41. Let I ⊆ R be an interval, let (Y, · ) be a normed space, let u : I → Y . Prove that Theorems 3.36 and 3.39 continue to hold if we assume metric differentiability in place of differentiability (see (8.19)) and mu (x) in (3.11) and (3.15). replace u (x) with ddx Theorem 3.41 continues to hold for functions with values in a reflexive Banach space. To be precise, we have Theorem 8.42 (Lusin (N ) property, II). Let I ⊆ R be an interval, let Y be a reflexive Banach space. A function u : I → Y belongs to ACloc (I; Y ) if and only if (i) u is continuous on I, (ii) u is differentiable L1 -a.e. in I, and u ∈ L1loc (I; Y ), (iii) u satisfies the Lusin (N ) property. Proof. The proof is exactly the same of Theorem 3.41, with the only change that we use Theorem 8.38 in place of Theorem 3.20. Remark 8.43. Let I ⊆ R be an interval and let Y be a normed space. If a function u : I → Y belongs to ACloc (I; Y ) and is differentiable L1 -a.e. in I, then we can still show that u satisfies the Lusin (N ) property. In turn, Corollaries 3.43 and 3.49 continue to hold for functions u : I → Y , with Y a reflexive Banach space. Corollary 8.44. Let I ⊆ R be an interval and let Y be a reflexive Banach space. A function u : I → Y belongs to AC(I; Y ) if and only if (i) u is continuous on I, (ii) u is differentiable L1 -a.e. in I, and u belongs to L1loc (I; Y ) and is equi-integrable, (iii) u satisfies the Lusin (N ) property. Corollary 8.45. Let I ⊆ R be an interval and let Y be a reflexive Banach space. A function u : I → Y belongs to ACloc (I; Y ) if and only if
8.5. Sobolev Functions
229
(i) u is continuous on I, (ii) u ∈ BP Vloc (I; Y ), (iii) u satisfies the Lusin (N ) property. Exercise 8.46. Let I ⊆ R be an interval, let Y be a Banach space, and let u : I → Y . Using Exercises 8.37 and 8.41 prove that Corollary 8.45 continues to hold. State and prove the analog of Theorems 8.42 and Corollaries 8.44. Remark 8.47. In view of Theorems 8.33 and 8.42, Corollaries 3.65, 3.66, 3.68 continue to hold if f : I → Y , where Y is a reflexive Banach space. Remark 8.48. In view of Theorems 8.33 and 8.42, Corollary 3.90 continues to hold for functions u ∈ BP Vloc (I; Y ), where Y is a reflexive Banach space.
8.5. Sobolev Functions In this section we extend the notion of Sobolev spaces to Banach-valued functions. Definition 8.49. Given an open set Ω ⊆ R, a Banach space Y , n ∈ N, and 1 ≤ p ≤ ∞, we say that a function u ∈ L1loc (Ω; Y ) admits a weak or distributional derivative of order n in Lp (Ω; Y ) if there exists a function v ∈ Lp (Ω; Y ) such that uϕ(n) dx = (−1)n vϕ dx Ω
for all ϕ ∈
Cc∞ (Ω).
Ω
The function v is denoted u(n) .
A similar definition can be given when Lp (Ω; Y ) is replaced by Lploc (Ω; Y ). Remark 8.50. Let Ω ⊆ R be an open set, let Y be a Banach space, and let u : Ω → Y be a locally Bochner integrable function. If Y → Z, where Z is a Banach space, with immersion i : Y → Z (see Definition A.36), and if the function i ◦ u : Ω → Z admits a weak derivative of order n in Lploc (Ω; Z), then in view of Remark 8.8, we have that (n) (n) n ϕ u dx = ϕ (i ◦ u) dx = (−1) ϕ(i ◦ u)(n) dx. i Ω Ω Ω Since we agreed (see Remark 8.8) to identify Ω ϕ(n) u dx with i Ω ϕ(n) u dx , with an abuse of notation, we can write ϕ(n) u dx = − ϕu(n) dx. Ω
Ω
Exercise 8.51. Let Ω ⊆ R be an open set, let Y be a Banach space, let n ∈ N, and let u ∈ L1loc (Ω; Y ) be such that its weak derivative u(n) of order n exists in L1loc (Ω; Y ). Prove that u(n) is unique.
230
8. The Infinite-Dimensional Case
We now define Sobolev spaces for Banach-valued functions. Definition 8.52. Given an open set Ω ⊆ R, a Banach space Y , m ∈ N, and 1 ≤ p ≤ ∞, the Sobolev space W m,p (Ω; Y ) is the space of all functions u ∈ Lp (Ω; Y ) which admit weak derivatives of order n in Lp (Ω; Y ) for every n = 1, . . . , m. The space W m,p (Ω; Y ) is endowed with the norm m u(n) Lp (Ω;Y ) . uW m,p (Ω;Y ) := uLp (Ω;Y ) + n=1 m,p (Ω; Y ) is defined as the space of all functions u ∈ The space Wloc ) which admit weak derivatives of order n in Lploc (Ω; Y ) for every n = 1, . . . , m.
Lploc (Ω; Y
Exercise 8.53. Given an open set Ω ⊆ R, a Banach space Y , m ∈ N, and 1 ≤ p ≤ ∞, prove that W m,p (Ω; Y ) is a Banach space. Exercise 8.54. Given an open set Ω ⊆ R, a Hilbert space Y , m ∈ N, prove that H m (Ω; Y ) := W m,2 (Ω; Y ) is a Hilbert space. We have seen in Theorem 7.16 that W 1,1 ((a, b); RM ) coincides with AC((a, b); RM ). In the infinite-dimensional case this is no longer true, unless Y is reflexive or, more generally, satisfies the RNP property (see Definition 8.35). Indeed, in the next theorem we will show that if a function has a weak derivative, then it is strongly differentiable and locally absolutely continuous. However, when the Banach space Y does not satisfy the RNP, then there exist absolutely continuous functions that are nowhere differentiable (see Exercises 8.30 and 8.32). Thus the inclusion W 1,1 ((a, b); Y ) ⊆ AC((a, b); Y ) is strict unless Y satisfies the RNP. Theorem 8.55. Let I ⊆ R be an open interval, let (Y, · ) be a Banach space, and let u ∈ L1loc (I; Y ). Then u has a weak derivative u ∈ L1loc (I; Y ) if and only if there exist a representative u ¯ of u and a function v ∈ L1loc (I; Y ) such that x v(s) ds (8.24) u ¯(x) = u ¯(x0 ) + x0
¯ is (strongly) differentiable at L1 -a.e. x ∈ I, for all x, x0 ∈ T . Moreover, u its strong derivative coincides with v, and u ¯ is locally absolutely continuous. Lemma 8.56. Let I ⊆ R be an open interval, let (Y, ·) be a Banach space, and let u : I → Y be a locally integrable function whose weak derivative is zero. Then u is equivalent to a constant function. The proof is identical to the one of Lemma 7.4 and we omit it.
8.5. Sobolev Functions
231
Proof of Theorem 8.55. Assume that u ¯ is given by the formula (8.24), then u ¯(x + h) − u ¯(x) 1 x+h = v(s) ds → v(x) h h x ¯ is strongly differenas h → 0 for L1 -a.e. x ∈ I by Theorem 8.19. Hence, u tiable for L1 -a.e. x ∈ I with (strong) derivative given by v. Let’s prove that v is also the weak derivative of u ¯. Given ϕ ∈ Cc1 (I), let (a, b) I be such that the support of ϕ is contained in (a, b). Let v(x) if x ∈ (a, b), w(x) := 0 otherwise. Then w ∈ L1 (R; Y ). Hence, by Theorem 8.20 (with the mollifier χ(−1,0) ), 1 x+h w(s) ds → w in L1 (R; Y ) as h → 0+ . (8.25) h x On the other hand, for x ∈ supp ϕ and h > 0 small, 1 x+h 1 x+h u ¯(x + h) − u ¯(x) = v(s) ds = w(s) ds. (8.26) h h x h x Let 0 < h < dist (supp ϕ, I \ {a, b}). Then b 1 b ϕ(x) − ϕ(x − h) (8.27) u(x) dx = ϕ(x) (¯ u(x) − u ¯(x + h)) dx. h h a a Since,
b u ¯(x) − u ¯(x + h) − v(x) dx ϕ(x) h a b 1 x+h ϕ(x) w(s) ds − w(x) dx = h a x b x+h 1 ≤ ϕ∞ w(s) ds − w(x) dx → 0 h x a
by (8.25), it follows from the Lebesgue dominated convergence theorem, (8.26), and (8.27), b ϕ (x)¯ u(x) dx = ϕ (x)¯ u(x) dx I
a
b
= lim
h→0+
a
ϕ(x − h) − ϕ(x) u ¯(x) dx −h b
= − lim
h→0+
ϕ(x) a
u ¯(x + h) − u ¯(x) dx = − h
b
ϕ(x)v(x) dx. a
Together with Exercise 8.51 this shows that v is the weak derivative of u ¯.
232
8. The Infinite-Dimensional Case
Conversely, assume that v is the weak derivative of u and define w(x) = x u(x0 ) + x0 v(s) ds. Then by the first part of the proof, v is the weak derivative of w. Hence, for every ϕ ∈ Cc1 (I), b b ϕ (x)u(x) dx = − ϕ(x)v(x) dx = − ϕ (x)w(x) dx, a
I
and so
a
ϕ (x) (u(x) − w(x)) dx = 0 I
for all ϕ ∈ Cc1 (I). It follows by Lemma 8.56 that u − w is constant. Since u(x0 ) − w(x0 ) = 0, it follows that u = w. As a consequence of the following theorem we can now characterize elements in the space W 1,p (Ω; Y ). Theorem 8.57. Let Ω ⊆ R be an open set, let Y be a Banach space, and let 1 ≤ p < ∞. A function u : I → Y belongs to W 1,p (Ω; Y ) if and only if it admits an absolutely continuous representative u ¯ : Ω → Y which is ¯ and its classical derivative u ¯ differentiable L1 -a.e. in Ω and such that u p belong to L (Ω; Y ).
Proof. Exercise. Exercise 8.58. State and prove the analogous result for W m,p (Ω; Y ).
Next we discuss evolution triples. Let (Y, · Y ) be a Banach space embedded in a Hilbert space (H, (·, ·)H ). Assume that Y is dense in H. We claim that H → Y .
(8.28)
To see this, given h ∈ H, consider the linear functional Th : Y → R defined by Th (y) := (h, y)H , y ∈ Y . Then |Th (y)| = |(h, y)H | ≤ hH yH ≤ chH yY . Hence, Th ∈ Y , with Th Y ≤ chH . Note that (8.29)
(h, y)H = Th , y Y ,Y
for all y ∈ Y . The mapping h ∈ H → Th is linear and continuous. It is also injective, since if Th = 0, then (h, y)H = 0 for all y ∈ Y and so, by the density of Y in H, we have that h = 0. The claim (8.28) follows by identifying h with Th . In particular, we can rewrite (8.29) as (8.30) for all h ∈ H and y ∈ Y .
(h, y)H = h, y Y ,Y
8.5. Sobolev Functions
233
By identifying H with its dual, we have that Y → H ∼ = H → Y .
(8.31)
This is called an evolution triple. Remark 8.59. If h ∈ Y → H, then we identify h with Th and so in Y the norm of h is Th Y =
sup y∈Y \{0}, y Y ≤1
|(h, y)H | ≤ chH ≤ c hY .
In particular, if Y = H 1 (Ω) and H = L2 (Ω), then given u ∈ H 1 (Ω), we have that u is identified with Tu ∈ (H 1 (Ω)) , where sup Tu (H 1 (Ω)) = uv dx ≤ uL2 (Ω) < uH 1 (Ω) v∈H 1 (Ω)\{0}, v H 1 (Ω) ≤1
Ω
provided u is not constant on each connected component of Ω. The following theorem strengthens Theorem 8.57. Theorem 8.60. Let I ⊂ R be an open bounded interval, let (Y, · Y ) be a Banach space embedded in a Hilbert space (H, (·, ·)H ), with Y dense in H, and let 1 ≤ p < ∞. If u ∈ Lp (I; Y ) has a weak derivative u ∈ Lp (I; Y ), then u has a continuous representative u ¯ ∈ C(I; H) with ¯ uC(I;H) ≤ c(uLp (I;Y ) + u Lp (I;Y ) ) for some constant c > 0 independent of u. Moreover, the function x ∈ I → u(x)2H is absolutely continuous, with d u(x)2H = 2u (x)(u(x)) = 2 u (x), u(x) Y ,Y dx for L1 -a.e. x ∈ I. We begin with a preliminary lemma. Lemma 8.61. Let I ⊆ R be an open interval, let (H, (·, ·)H ) be a Hilbert space, and let u, v ∈ C 1 (I; H). Then the function g(x) := (u(x), v(x))H is of class C 1 (I) and g (x) = (u (x), v(x))H + (u(x), v (x))H .
234
8. The Infinite-Dimensional Case
Proof. Let x ∈ I and h = 0 small. Then using the bilinearity of the inner product, (u(x + h), v(x + h))H − (u(x), v(x))H g(x + h) − g (h) = h h (u(x + h) − u(x) + u(x), v(x + h))H − (u(x), v(x))H = h u(x + h) − u(x) v(x + h) − v(x) = , v(x + h) + u(x), . h h H H Letting h → 0 and using the continuity of the inner product gives the desired result. We now turn to the proof of Theorem 8.60. Proof. Let I = (a, b). As in Exercise 7.24 we can extend u to J := (2a − b, 2b − a). Let uε := u ∗ ϕε , where ϕε is a standard mollifier, with ϕ ∈ Cc∞ (R) and 0 < ε < b − a. Reasoning as in the proof of Theorem 7.26 and using Theorem 8.21 we have that uε ∈ Cc∞ (R; Y ), uε → u in Lp (I; Y ) and uε → u in Lp (I; Y ). Since uε ∈ Cc∞ (R; Y ) and Y → H, we have that uε ∈ Cc∞ (R; H) and so by the previous lemma, the fundamental theorem of calculus, and property (8.30), for x, x0 ∈ I, x d 2 2 uε (x)H = uε (x0 )H + uε (s)2H ds x0 ds x 2 (8.32) (uε (s), uε (s))H ds = uε (x0 )H + 2 x 0x uε (s), uε (s) Y ,Y ds. = uε (x0 )2H + 2 x0
By H¨older’s inequality uε (x)2H
≤
uε (x0 )2H
+2 I
uε (s)Y uε (s)Y ds
≤ uε (x0 )2H + 2uε Lp (I;Y ) uε Lp (I;Y ) . In turn, uε (x) − uδ (x)2H ≤ uε (x0 ) − uδ (x0 )2H + 2uε − uδ Lp (I;Y ) uε − uδ Lp (I;Y ) . Since by Theorems 8.20 and 8.21, uε (x) → u(x) for every Lebesgue point of u, uε → u in Lp (I; Y ) and uε → u in Lp (I; Y ), letting x0 be a Lebesgue point of u, it follows from the previous inequality that sup uε (x) − uδ (x)2H → 0 x∈I
8.5. Sobolev Functions
235
as ε, δ → 0+ . Hence, {uε }ε is a Cauchy sequence in C(I; H) and so it converges uniformly to a function w ∈ C(I; H). On the other hand, since uε (x) → u(x) for every Lebesgue point of u, we have that u(x) = w(x) for L1 -a.e. x ∈ I. Letting ε → 0+ in (8.32) gives
x
w(x)2H = u(x0 )2H + 2
u (s), u(s) Y ,Y ds
x0
for all x, x0 ∈ I. In turn,
x
u(x)2H = u(x0 )2H + 2
u (s), u(s) Y ,Y ds
x0
L1 -a.e.
for x ∈ I. This implies that the function x ∈ [a, b] → u(x)2H is absolutely continuous and concludes the proof. We conclude this section with an important compactness theorem. Theorem 8.62 (Aubin–Lions–Simon). Let I ⊂ R be an open bounded interval, let (Y0 · Y 0 ), (Y1 , · Y1 ), and let (Y2 , · Y2 ) be Banach spaces with Y0 → Y1 → Y2 . Assume that the embedding Y0 → Y1 is compact. Let 1 ≤ p < ∞, 1 ≤ q ≤ ∞, and let V be the Banach space of all functions u ∈ Lp (I; Y0 ) whose distributional derivative u belongs to Lq (I; Y2 ) endowed with the norm (8.33)
uV := uLp (I;Y0 ) + u Lq (I;Y2 ) .
Then the embedding V → Lp (I; Y1 ) is compact. Proof. Let {un }n be a bounded sequence in V. Using (8.33) and the fact that q ≥ 1 and I is bounded, by H¨older’s inequality we obtain that (8.34)
{un }n is bounded in Lp (I; Y0 ),
(8.35)
{un }n is bounded in L1 (I; Y2 ).
In view of Corollary 8.26, it suffices to show that τh (un ) − un Lp (I h ;Y2 ) → 0 as h → 0+ uniformly for n ∈ N. Since un ∈ W 1,1 (I; Y2 ), by Theorem 8.55, for every 0 < h < L1 (I) and x ∈ I h , x+h un (s) ds. un (x + h) − un (x) = x
Hence,
x+h
τh (un )(x) − u(x)Y2 ≤ x
un (s)Y2 ds.
If q ≥ p, raising both sides to power p and using H¨older’s inequality gives x+h un (s)pY2 ds. τh (un )(x) − u(x)pY2 ≤ hp−1 x
236
8. The Infinite-Dimensional Case
Integrating in x over I h and using Fubini’s theorem yields x+h τh (un )(x) − u(x)pY2 dx ≤ hp−1 un (s)pY2 dsdx Ih
≤ hp
Ih x sup I
inf I+h
It follows that
un (s)pY2 ds.
Ih
τh (un )(x) − u(x)p dx ≤ M hp ,
where M := supn un Lp (I;Y2 ) . This shows that τh (un ) − un Lp (I h ;Y2 ) → 0 as h → 0+ uniformly for n ∈ N and completes the proof in the case q ≥ p. The case q < p is left as an exercise. (Hint: Use Young’s inequality, Theorem 10.48.) Exercise 8.63. Assume that all the hypotheses of the previous theorem are satisfied with p = ∞. Prove that the embedding V → C(I; Y1 ) is compact.
Part 2
Functions of Several Variables
Chapter 9
Change of Variables and the Divergence Theorem Prospective Grad Students, III: “Will your qualifying exams procedure utterly destroy my dignity and sense of self-respect?” — Jorge Cham, www.phdcomics.com
In this chapter we extend some of the concepts and results of Chapter 3 to functions of several variables, including the change of variables formulas, the fundamental theorem of calculus, and integration by parts. We will prove the area formula and the divergence theorem.
9.1. Directional Derivatives and Differentiability Let (X, · X ), (Y, · Y ) be normed spaces, let E ⊆ X, let u : E → Y , and let x0 ∈ E. Given a direction v ∈ X \ {0}, let Lv be the line through x0 in the direction v, that is, Lv := {x ∈ X : x = x0 + tv, t ∈ R}, and assume that x0 is an accumulation point of the set E ∩ Lv . The directional derivative of u at x0 in the direction v is defined as u(x0 + tv) − u(x0 ) ∂u (x0 ) := lim , t→0 ∂v t provided the limit exists in Y . If Y = R, then we allow the limit to exist in [−∞, ∞]. When X = RN and v = ei , where ei is a vector of the canonical basis, ∂u (x0 ), if it exists, is called the partial derivative the directional derivative ∂e i ∂u (x0 ) or ∂i u(x0 ). We recall that the of u with respect to xi and is denoted ∂x i 239
240
9. Change of Variables and the Divergence Theorem
standard, or canonical, orthonormal basis of RN is given by the unit vectors e1 := (1, 0, . . . , 0), e2 := (0, 1, 0, . . . , 0), . . . , eN := (0, . . . , 0, 1). Definition 9.1. Let (X, · X ), (Y, · Y ) be normed spaces, let E ⊆ X, let u : E → Y , and let x0 ∈ E be an accumulation point of E. The function u is differentiable at x0 if there exists a continuous linear function T : X → Y (depending on u and x0 ) such that u(x) − u(x0 ) − T (x − x0 ) = 0, x − x0 X provided the limit exists. The function T , if it exists, is called the differential of u at x0 and is denoted du(x0 ) or dux0 . (9.1)
lim
x→x0
Exercise 9.2. Let (X, · X ), (Y, · Y ) be normed spaces, let E ⊆ X, let u : E → Y , and let x0 ∈ E be an accumulation point of E. Prove that if u is differentiable at x0 , then u is continuous at x0 and for every direction v such that x0 is an accumulation point of the set E ∩ Lv (if any), there exists ∂u ∂v (x0 ) = dux0 (v). In particular, if Y = R and u : E → R is differentiable at an interior point of E, then u is continuous at x0 , all partial and directional derivatives exist at x0 and for every direction v, ∂u ∂u (x0 ) = (x0 )vi . ∂v ∂xi N
(9.2)
i=1
The following exercise shows that these conditions are not enough to guarantee differentiability. Exercise 9.3. Let u : R2 → R be defined by x if y = x2 , x = 0, u(x, y) := 0 otherwise. Prove that u is continuous at (0, 0), all partial and directional derivatives exist at (0, 0) and that (9.2) holds but that u is not differentiable at (0, 0). Exercise 9.4. Let E ⊆ RN , let u : E → R, let x0 ∈ E, and let i ∈ {1, . . . , N }. Assume that there exists r > 0 such that B(x0 , r) ⊆ E and for all j = i and for all x ∈ B(x0 , r) the partial derivative ∂j u exists in R at x and is continuous at x0 . Assume also that ∂i u(x0 ) exists in R. Prove that u is differentiable at x0 . Hint: Use the mean value theorem on segments parallel to the axes. To define higher-order derivatives, for simplicity, we restrict ourselves to the case X = RN and Y = R, which is all we will need in this book. Let E ⊆ RN , let u : E → R and let x0 ∈ E. Let i ∈ {1, . . . , N } and assume that there exists in R the partial derivative ∂i u(x) for all x ∈ E. If j ∈ {1, . . . , N }
9.1. Directional Derivatives and Differentiability
241
and x0 is an accumulation point of E ∩Lj , where Lj is the line through x0 in the direction ej , then we can consider the partial derivative of the function ∂i u with respect to xj , that is, ∂ ∂u ∂2u (x0 ). (x0 ) =: ∂xj ∂xi ∂xj ∂xi Inductively, one can define higher-order partial derivatives
∂mu ∂xim ···∂xi1 .
Note that in general the order in which we take derivatives is important. Exercise 9.5. Let u : R2 → R be defined by 3 x y−xy 3 if (x, y) = (0, 0), x2 +y 2 u(x, y) := 0 if (x, y) = (0, 0). Prove that
∂2u ∂x∂y (0, 0)
=
∂2u ∂y∂x (0, 0).
The Schwartz theorem shows that if the second-order mixed partial derivatives are continuous, then they coincide. Theorem 9.6 (Schwartz). Let E ⊆ RN , let u : E → R, let x0 ∈ E, and let i, j ∈ {1, . . . , N }. Assume that there exists r > 0 such that B(x0 , r) ⊆ E 2u ∂u ∂u (x), ∂x (x), and ∂x∂j ∂x (x) and for all x ∈ B(x0 , r), the partial derivatives ∂x i j i exist in R. Assume also that ∂2u
∂xi ∂xj (x0 )
∂2u ∂xj ∂xi
is continuous at x0 . Then there exists
and ∂ 2u ∂ 2u (x0 ) = (x0 ). ∂xi ∂xj ∂xj ∂xi
Exercise 9.7. Let f : ((−r, r) \ {0}) × ((−r, r) \ {0}) → R. Assume that the double limit lim(s,t)→(0,0) f (s, t) exists in R and that the limit limt→0 f (s, t) exists in R for all s ∈ (−r, r) \ {0}. Prove that the iterated limit lims→0 limt→0 f (s, t) exists and lim lim f (s, t) =
s→0 t→0
lim
(s,t)→(0,0)
f (s, t).
Exercise 9.8. Prove the Schwartz theorem. Hint: For N = 2 take the function f in the previous exercise to be an appropriate second-order difference quotient and apply repeatedly the mean value theorem on segments parallel to the axes. In view of Schwartz’s theorem, for smooth functions the order in which we take partial derivatives is not important, and thus we can use multiindices. Given N ∈ N, for a multi-index α = (α1 , . . . , αN ) ∈ NN 0 we set ∂ |α| ∂ |α| := , ∂xα ∂xα1 1 · · · ∂xαNN
|α| := α1 + · · · + αN ,
242
9. Change of Variables and the Divergence Theorem
where x = (x1 , . . . , xN ) ∈ RN . For simplicity we will often write ∂ α for ∂ |α| ∂n n when the variables of differentiation are clear. Given ∂xα and ∂i for ∂xn i two multi-indices α = (α1 , . . . , αN ) and β = (β1 , . . . , βN ), we write α ≤ β if αi ≤ βi for all i = 1, . . . , N . Also α! := α1 ! . . . αN ! and α α! . (9.3) := β!(α − β)! β αN α1 α Given x ∈ RN and α ∈ NN 0 we write x := x1 · · · xN .
Given an open set Ω ⊆ RN , for every nonnegative integer m ∈ N0 we denote by C m (Ω) the vector space of all functions that are continuous ∞ together ∞ with their partial derivatives up to order m. We set C (Ω) := m=0 C m (Ω) and we define Ccm (Ω) and Cc∞ (Ω) as the subspaces of C m (Ω) and C ∞ (Ω), respectively, consisting of all functions that have compact support in Ω. In the following chapters we will often use Taylor’s formula. Theorem 9.9 (Taylor’s formula with integral remainder). Let Ω ⊆ RN be an open set, let u ∈ C m (Ω), m ∈ N, and let x0 ∈ Ω. Then for every x ∈ B(x0 , r) ⊆ Ω, 1 ∂ α u(x0 )(x − x0 )α u(x) = u(x0 ) + α! 1≤|α|≤m 1 m (x − x0 )α + (1−t)m−1 (∂ α u(tx + (1−t)x0 ) − ∂ α u(x0 )) dt. α! 0 |α|=m
Exercise 9.10. Prove the previous theorem. Hint: Use the one-dimensional function g(t) := u(tx0 + (1 − t)x), 0 ≤ t ≤ 1. Exercise 9.11. Let Ω ⊆ RN be an open set, let u ∈ C m (Ω), m ∈ N, and let x0 ∈ Ω. Prove that for every direction v ∈ RN \ {0}, 1 ∂ nu ∂ α u(x0 )v α , (x0 ) = (9.4) n ∂v α! |α|=n
where v.
∂nu ∂v n
stands for the repeated directional derivative of u in the direction
For more information on differentiability of functions of several variables, we refer the reader to the books of Bartle [20] and Fleming [80].
9.2. Lipschitz Continuous Functions In this section we prove two important theorems for Lipschitz continuous functions. We recall that if E ⊆ RN and u : E → R, then u is Lipschitz
9.2. Lipschitz Continuous Functions
243
continuous if Lip(u; E) =
|u(x) − u(y)| < ∞. x − y
sup x, y∈E, x=y
We usually write Lip u in place of Lip(u; E), whenever the underlying set is clear. The next exercise shows that a Lipschitz continuous function can always be extended to RN with the same Lipschitz constant. Exercise 9.12. Let E ⊆ RN and let u : E → R be a Lipschitz continuous function. Define v(x) := inf{u(y) + Lip(u; E)x − y : y ∈ E},
x ∈ RN .
(i) Prove that v is Lipschitz continuous with Lipschitz constant Lip(u; E). (ii) Prove that v(x) = u(x) for all x ∈ E. (iii) Prove that there exists a function w : RN → R with Lipschitz constant Lip(u; E) such that w(x) = u(x) for all x ∈ E and inf w = inf u, RN
E
sup w = sup u. RN
E
Exercise 9.13. Let u : RN → R be a Lipschitz continuous function. Let D ⊆ S N −1 be dense in S N −1 and let x0 ∈ RN be such that there exist all ∂u (x0 ), i = 1, . . . , N , all directional derivatives ∂u partial derivatives ∂x ∂v (x0 ) i for v ∈ D, and N ∂u ∂u (x0 )vi (x0 ) = ∂v ∂xi i=1
for all v = (v1 , . . . , vN ) ∈ D. Prove that u is differentiable at x0 . We recall that S N −1 is the unit sphere ∂B(0, 1). Compare the previous exercise with Exercise 9.3. Theorem 9.14 (Rademacher). Let u : RN → R be a Lipschitz continuous function. Then u is differentiable at LN -a.e. x ∈ RN . Proof. If N = 1, then u is absolutely continuous and the result follows from Lebesgue’s differentiation theorem (Theorem 1.18) and Proposition 3.9. Thus, assume that N ≥ 2. For every v ∈ S N −1 let ∂u (x) ∈ R Ev := x ∈ RN : there exists ∂v and let Hv be the hyperplane orthogonal to v. For every x0 ∈ Hv , the function g(t) := u(x0 + tv), t ∈ R, is Lipschitz continuous and thus, by the result for N = 1, g is differentiable for L1 -a.e. t ∈ R. Hence, ∂u ∂v (x0 + tv)
244
9. Change of Variables and the Divergence Theorem
exists in R for L1 -a.e. t ∈ R. This shows that x0 + tv ∈ Ev for L1 -a.e. t ∈ R. Since the set Ev is a Borel set (why?), it follows by Tonelli’s theorem that LN (RN \ Ev ) = 0. Next consider a countable set D ⊂ S N −1 dense in S N −1 . By what we just proved, for LN -a.e. x ∈ RN there exist ∂u ∂v (x) for all v ∈ D ∂u and ∂xi (x0 ) for all i = 1, . . . , N . We claim that ∂u ∂u (x0 ) = (x0 )vi ∂v ∂xi N
(9.5)
i=1
for LN -a.e. x ∈ RN and for all v ∈ D. To see this, fix v ∈ D and let ϕ ∈ Cc1 (RN ). Since u is Lipschitz continuous, by the Lebesgue dominated convergence theorem, u(x + tv) − u(x) ∂u lim ϕ(x) dx = (x)ϕ(x) dx, t t→0+ RN RN ∂v ϕ(x − tv) − ϕ(x) ∂ϕ dx = − (x)u(x) dx. u(x) lim + t t→0 RN RN ∂v By using the change of variables y = x + tv on the left-hand side of the first equality, we see that the two left-hand sides coincide. Hence, ∂u ∂ϕ (x)ϕ(x) dx = − (x)u(x) dx. ∂v N N R R ∂v Similarly,
N N ∂u ∂ϕ (x)vi ϕ(x) dx = − (x)vi u(x) dx. ∂xi ∂xi RN
RN i=1
i=1
Since the two right-hand sides coincide, we obtain that RN
∂u (x)ϕ(x) dx = ∂v
N ∂u (x)vi ϕ(x) dx ∂xi
RN i=1
for all ϕ ∈ Cc1 (RN ). Given the arbitrariness of ϕ ∈ Cc1 (RN ), it follows that (9.5) holds for LN -a.e. x ∈ RN (why?). We are now in a position to apply Exercise 9.13 to obtain the desired result. Exercise 9.15. Let E ⊆ RN and let u : E → R be a Lipschitz continuous function. Prove that for every set F ⊆ R with L1 (F ) = 0, ∇u(x) = 0 for LN -a.e. x ∈ u−1 (F ). (Hint: Use Lemma 3.60 and Tonelli’s theorem). As a corollary of Rademacher’s theorem we can show the following result.
9.2. Lipschitz Continuous Functions
245
Theorem 9.16 (Stepanoff). Every function u : RN → R is differentiable at LN -a.e. x in the (possibly empty) set |u(y) − u(x)| (9.6) Eu := x ∈ RN : lim sup 0 and L > 0 such that (9.8)
|u(y) − u(x)| ≤ Ly − x
for all y ∈ B(x, r). By the density of the rational numbers, we can find / E, it follows that n ≥ L such that Bn ⊆ B(x, r) and x ∈ Bn . Since x ∈ x ∈ Fn . By definition of vn and wn and by (9.8) we have that u(x) − ny − x ≤ vn (y) ≤ u(y) ≤ wn (y) ≤ u(x) + ny − x for all y ∈ Bn . In particular, for y = x we obtain vn (x) = u(x) = wn (x). Moreover, since x ∈ / Gn , we have that ∇vn (x) = ∇wn (x) =: A. Hence,
246
9. Change of Variables and the Divergence Theorem
using the previous inequality, together with the facts that vn and wn are differentiable at x and vn (x) = u(x) = wn (x), it follows that vn (y) − u(x) − A(y − x) u(y) − u(x) − A(y − x) ≤ lim inf y→x y − x y − x u(y) − u(x) − A(y − x) wn (y) − u(x) − A(y − x) ≤ lim = 0, ≤ lim sup y→x y − x y − x y→x
0 = lim
y→x
which shows that u is differentiable at x. This concludes the proof.
Although it was not needed in the previous theorem, it turns out that the set Eu given in (9.6) is Lebesgue measurable and so are the partial derivatives of u when restricted to Eu . Theorem 9.17. Let Ω ⊆ RN be an open set and let u : Ω → R be a function. Then the set E := {x ∈ Ω : u is differentiable at x} is Lebesgue measurable and so are the partial derivatives ∂i u : E → R, i = 1, . . . , N. Proof. Let Eu be the set given in (9.6) with RN replaced by Ω and for k ∈ N write Ek := {x ∈ Ω : |u(x + h) − u(x)| ≤ kh for all h with h < 1/k, dist(x, ∂Ω) > 1/k}. Then Eu = k Ek . Note that the sets Ek are relatively closed in the open set {x ∈ Ω : dist(x, ∂Ω) > 1/k}, and so Eu is a Borel set. In turn, by Stepanoff’s theorem the set E is Lebesgue measurable. Fix k and subdivide Ek into sets Ek, of diameter 1/k. The function u : Ek, → R is Lipschitz continuous and thus by Exercise 9.12 it can be extended to RN to a Lipschitz continuous function uk, with the same Lipschitz constant. For i = 1, . . . , N and for n ≥ k consider the function gn,k, : RN → R defined by gn,k, (x) :=
uk, (x + ei /n) − uk, (x) . 1/n
Since uk, is Lipschitz continuous, it follows that gn,k, is Lebesgue measurable, and so are the functions ∂i+ uk, := lim sup gn,k, , n→∞
∂i− uk, := lim inf gn,k, . n→∞
In turn the set Fk, := {x ∈ RN : ∂i+ uk, = ∂i− uk, } is Lebesgue measurable and so is the function ∂i uk, : Fk, → R. It remains to show that ∂i uk, = ∂i u
9.2. Lipschitz Continuous Functions
247
LN -a.e. in Ek, ∩ E. Let x be a point of density one of Ek, ∩ E (see Corollary B.121). Then for yn ∈ Ek, , |uk, (x + n1 ei ) − uk, (x) − n1 ∂i u(x)| = |uk, (x + n1 ei ) − u(x) − n1 ∂i u(x)| (9.9)
≤ |uk, (x + n1 ei ) − uk, (yn )| + |u(yn ) − u(x) − ∇u(x) · (yn − x)| + |∇u(x) · (x + n1 ei − yn )| ≤ (k + ∇u(x))x + n1 ei − yn + |u(yn ) − u(x) − ∇u(x) · (yn − x)|,
where we have used the facts that u(yn ) = uk, (yn ) and u(x) = uk, (x) and that uk, is Lipschitz continuous with constant less than or equal to k. Fix 0 < ε < 1. Since u is differentiable at x, there is δ1 > 0 such that (9.10)
|u(y) − u(x) − ∇u(x) · (y − x)| ≤ εy − x
for all y with y − x < δ1 . On the other hand, since x is a point of density one of Ek, ∩ E, there exists δ2 < 1/(2k) such that LN (Ek, ∩ E ∩ B(x, r)) > (1 − ( 2ε )N )LN (B(x, r)) for all r < δ2 . This implies that (9.11)
N LN (B(x, r) \ (Ek, ∩ E)) < ( 2ε )N LN (B(x, r)) = αN ( εr 2) .
Let n be so large that 1/n < min{δ1 , δ2 }/2 and find yn ∈ Ek, ∩ E such that x+ei /n−yn < ε/n. Note that if no such yn exists, then B(x+ei /n, ε/n) ⊆ RN \ (Ek, ∩ E). But since B(x + ei /n, ε/n) ⊂ B(x, 2/n) and 1/n < δ2 /2, by (9.11) we would have that B(x + n1 ei , nε ) ⊆ B(x, n2 ) \ (Ek, ∩ E), so that αN ( nε )N ≤ LN (B(x, n2 ) \ (Ek, ∩ E)) < αN ( nε )N , which is a contradiction. Hence, yn exists and so from (9.9) and (9.10) we get |uk, (x + n1 ei ) − uk, (x) − n1 ∂i u(x)| ≤ (k + ∇u(x))x + n1 ei − yn + εyn − x ≤ (k + ∇u(x)) nε + εyn − x − n1 ei +
ε n
≤ (k + ∇u(x) + 2) nε , and so
u (x + e /n) − u (x) k, i k, − ∂i u(x) ≤ (k + ∇u(x) + 2)ε. 1/n
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9. Change of Variables and the Divergence Theorem
Letting n → ∞ and then ε → 0 shows that ∂i uk, (x) = ∂i u(x). Since this is true for every k and , we have that ∂i u : E → R is Lebesgue measurable. In contrast to the previous theorem, the following exercise shows that the set where one of the partial derivatives ∂i u exists may not be Lebesgue measurable. Exercise 9.18. Let G ⊂ R be a non-Lebesgue measurable set and let u(x, y) := χQ (x)χG (y),
(x, y) ∈ R2 .
(i) Prove that u is Lebesgue measurable. (ii) Prove that the set {(x, y) ∈ R2 : Lebesgue measurable.
∂1 u exists at (x, y)} is not
In what follows, given x = (x1 , . . . , xN ) ∈ RN , we denote by x the (N − 1)-dimensional vector obtained by removing the N th component from x and with an abuse of notation we write x = (x , xN ) ∈ RN −1 × R.
(9.12)
The next exercise provides a sufficient condition to recover measurability. Exercise 9.19. Let R = I1 × · · · × IN ⊆ RN , where Ii are open intervals, let u : R → R be a Lebesgue measurable function such that for LN −1 -a.e. x ∈ I1 × · · · × IN −1 , (9.13) min lim sup u(x , t), lim sup u(x , t) ≤ u(x , xN ) t→x+ N
t→x− N
≤ max lim inf u(x , t), lim inf u(x , t) t→x+ N
t→x− N
for every xN ∈ IN , and let EN := {x ∈ R : there exists ∂N u(x) in R}. (i) For x ∈ R define rat u(x) := ∂N
lim
r→0+ , r∈Q
u(x + reN ) − u(x) r
rat := {x ∈ R : there exists whenever the limit exists and let EN rat rat rat u : ∂N u(x) in R}. Prove that EN is Lebesgue measurable and ∂N rat EN → R is a Lebesgue measurable function. rat be such that (9.13) holds for x and let (ii) Let x = (x , xN ) ∈ EN hn → 0 with hn = 0 for every n ∈ N. Prove that there exist two sequences {rn }n and {sn }n of rational numbers such that hrnn → 1, sn hn → 1 and
u(x + rn eN ) > u(x + hn eN ) − |hn |/n, u(x + sn eN ) < u(x + hn eN ) + |hn |/n
9.3. The Area Formula: The C 1 Case
249
for every n. Deduce that u(x + hn eN ) − u(x) n→∞ hn
rat u(x) = lim ∂N
rat u(x). and that there exists ∂N u(x) = ∂N
(iii) Prove that the set EN is Lebesgue measurable and ∂N u : EN → R is a Lebesgue measurable function. Observe that (9.13) holds if for LN −1 -a.e. x ∈ I1 × · · · × IN −1 , the function u(x , ·) is right continuous or left continuous at xN or is monotone in a neighborhood of xN . Exercise 9.20. Let R = I1 × · · · × IN ⊆ RN , where Ii are open intervals, let u : R → R be a Lebesgue measurable function such that for LN −1 -a.e. x ∈ I1 × · · · × IN −1 the function u(x , ·) has bounded pointwise variation in IN . Prove that there exists a Lebesgue measurable function v : R → R such that v(x) = u(x) for LN -a.e. x ∈ R, the set EN := {x ∈ R : there exists ∂N v(x) in R} is Lebesgue measurable and ∂N v : EN → R is a Lebesgue measurable function.
9.3. The Area Formula: The C 1 Case In this section we prove a change of variables formula for surface integrals. We begin by studying linear transformations. Given N ∈ N we denote by · N the Euclidean norm in RN . Given k, N ∈ N and a linear function L : Rk → RN , the adjoint of L is the linear function Lt : RN → Rk such that (9.14)
y · Lt (x) = L(y) · x
for all y ∈ Rk and x ∈ RN . The matrix representing Lt is simply the transpose of the matrix representing L. Given k, N ∈ N and a linear function L : Rk → RN , we say that L is orthogonal if L(y1 ) · L(y2 ) = y1 · y2 for all y1 , y2 ∈ Rk . We recall that Hok stands for the k-dimensional Hausdorff outer measure and Hk is the k-dimensional Hausdorff measure obtained by restricting Hok to the σ-algebra of all Hok -measurable sets (see the Carath´eodory Theorem B.13). We refer to Section C.7 for more details on the Hausdorff measure. Exercise 9.21. Given k, N ∈ N with 1 ≤ k ≤ N and a linear function L : Rk → RN , prove that if E ⊆ Rk is Lebesgue measurable, then L(E) is Hok -measurable.
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9. Change of Variables and the Divergence Theorem
Remark 9.22. An orthogonal function L : Rk → RN preserves inner products and distances, since # L(y1 ) − L(y2 )N = L(y1 − y2 ) · L(y1 − y2 ) # = (y1 − y2 ) · (y1 − y2 ) = y1 − y2 k . Thus, L is Lipschitz continuous with Lipschitz constant one and it is injective with L−1 : L(Rk ) → Rk Lipschitz continuous with Lipschitz constant one. If E ⊆ Rk is Lebesgue measurable, then by the previous exercise L(E) is Hok -measurable. Hence, by Propositions C.44 and C.50, Hk (L(E)) ≤ Hk (E) = Lk (E). On the other hand, since L−1 : L(Rk ) → Rk is Lipschitz continuous with Lipschitz constant one, Lk (E) = Hk (E) = Hk (L−1 (L(E))) ≤ Hk (L(E)) and thus (9.15)
Hk (L(E)) = Lk (E).
Observe also that Lt ◦ L = Ik . Definition 9.23. Given N ∈ N and a linear function L : RN → RN , we say that L is (i) a rotation if it is orthogonal and the corresponding matrix has positive determinant, (ii) symmetric if L = Lt , (iii) diagonal if the corresponding matrix is diagonal, (iv) positive definite if L(x) · x > 0 for all x ∈ RN \ {0}. Theorem 9.24 (Decomposition). Let 1 ≤ k ≤ N and let L : Rk → RN be a linear function. Assume that the corresponding matrix has rank k. Then there exist an orthogonal linear function P : Rk → Rk , a positive definite, diagonal linear function D : Rk → Rk , and an orthogonal linear function Q : Rk → RN such that L = Q ◦ D ◦ P. Proof. We claim that the function Lt ◦ L : Rk → Rk is symmetric and positive definite. Indeed, by (9.14), (Lt ◦ L)(y1 ) · y2 = (Lt (L(y1 ))) · y2 = L(y1 ) · L(y2 ) = y1 · (Lt (L(y2 ))) = y1 · (Lt ◦ L)(y2 ) and (Lt ◦ L)(y) · y = (Lt (L(y))) · y = L(y) · L(y) = L(y)2N > 0
9.3. The Area Formula: The C 1 Case
251
for all y ∈ Rk \ {0}, since the matrix corresponding to L has rank k. It follows that the eigenvalues μi of Lt ◦ L are all positive and that there exists √ an orthonormal basis {b1 , . . . , bk } of eigenvectors. Let λi := μi . Then (Lt ◦ L)(bi ) = λ2i bi for all i = 1, . . . , k. Let {e1 , . . . , ek } be the canonical basis in Rk . Given any two vector spaces of dimension k each with a given basis, there is a linear function between these two vector spaces that maps one basis into the other. Let P : Rk → Rk be the linear function that maps {b1 , . . . , bk } into {e1 , . . . , ek }, let D : Rk → Rk be the linear function that maps {e1 , . . . , ek } into {λ1 e1 , . . . , λk ek }, and let Q : Rk → RN be the linear function that maps {λ1 e1 , . . . , λk ek } into {L(b1 ), . . . , L(bk )}. Note that since JL has rank k, the vector space L(Rk ) ⊆ RN has dimension k and {L(b1 ), . . . , L(bk )} is a basis in L(Rk ). Since P (bi ) = ei , the function P is orthogonal. To verify that Q is orthogonal, note that Q(ei ) · Q(ej ) =
λ2j 1 1 L(bi ) · L(bj ) = bi · (Lt (L(bj ))) = bi · bj = δi,j . λi λj λi λj λi λj
It remains to show that L = Q ◦ D ◦ P . We have (Q ◦ D ◦ P )(bi ) = (Q ◦ D)(ei ) = Q(λi ei ) = L(bi ), and thus the result follows by linearity.
Given E ⊆ Rk and Ψ : E → RN , assume that Ψ is differentiable at some point y ∈ E. We recall that the Jacobian matrix of Ψ at y is the N × k matrix given by ⎛ ⎞ ∇Ψ1 (y) ⎜ ⎟ .. (9.16) JΨ (y) = ∇Ψ(y) := ⎝ ⎠ . ∇ΨN (y) and the Jacobian of Ψ at y is the number $ t (y)J (y)), (9.17) |||JΨ (y)||| := det(JΨ Ψ t (y) is the transpose of J (y). Note that when k = N , where JΨ Ψ
(9.18)
|||JΨ (y)||| = | det JΨ (y)|.
Theorem 9.25 (Cauchy–Binet formula). Let 1 ≤ k ≤ N , let A be a N × k matrix, and let B be a k × N matrix. Then det(aαi ,j )ki,j=1 det(bi,αj )ki,j=1 , (9.19) det BA = α∈ΛN,k
where ΛN,k := {α ∈ Nk : 1 ≤ α1 < α2 < · · · < αk ≤ N }.
252
9. Change of Variables and the Divergence Theorem
In particular,
det At A =
(det(aαi ,j )ki,j=1 )2 .
α∈ΛN,k
Proof. We only give a sketch of the proof. Using the fact that for square matrices the determinant of the product of two matrices is given by the product of the determinants of the two matrices, we have
I A I −A I A I −A det(I + AB) = det = det det 0 I B I 0 I B I
I −A I A = det = det(I + BA), 0 I B I where I and 0 are identity matrices and zero matrices of whatever dimension is needed to make sense of the previous expressions. The identity det(IN + AB) = det(Ik + BA) is called the Sylvester determinant identity. If we now let t ∈ R and rescale everything, we obtain det(tIN + AB) = tN −k det(tIk + BA). Since both the left-hand and right-hand sides are polynomials of degree N in t, by equating the coefficients of the tN −k terms we get (9.19). We are now ready to prove the area formula for injective linear functions. Theorem 9.26 (Area formula for linear functions). Let 1 ≤ k ≤ N and let L : Rk → RN be a linear function. Assume that JL has rank k. Then for every Lebesgue measurable E ⊆ Rk , L(E) is Hok -measurable and |||JL ||| dy = |||JL |||Lk (E). Hk (L(E)) = E
Proof. Step 1: Assume first that k = N and that L is a positive definite diagonal linear function D : Rk → Rk . Then D(y) = (λ1 y1 , . . . , λk yk ). Consider a rectangle R = I1 ×· · ·×Ik . Then by Fubini’s theorem and (9.18), Lk (D(R)) = λ1 · · · λk Lk (R) = det JD Lk (R) = |||JD |||Lk (R). If V ⊆ Rk is an open set, then we can write V as a countable union of disjoint rectangles Rn , and since D is injective, the sets D(Rn ) are also disjoint and so Lk (D(V )) = |||JD |||Lk (V ). By approximating Lebesgue measurable sets with open sets we obtain that Lk (D(E)) = |||JD |||Lk (E) for every Lebesgue measurable set E ⊆ Rk . Step 2: Given now L : Rk → RN such that JL has rank k, by Theorem 9.24 there exist an orthogonal linear function P : Rk → Rk , a positive definite
9.3. The Area Formula: The C 1 Case
253
diagonal linear function D : Rk → Rk , and an orthogonal linear function Q : Rk → RN such that L = Q ◦ D ◦ P . For every Lebesgue measurable set E ⊆ Rk , by Exercise 9.21, Proposition C.50, (9.15), and Step 1, Hk (L(E)) = Hk (Q(D(P (E)))) = Hk (D(P (E)))
(9.20)
= Lk (D(P (E))) = |||JD |||Lk (P (E)) = |||JD |||Lk (E). Now since Qt ◦ Q = Ik , Lt ◦ L = (Q ◦ D ◦ P )t ◦ (Q ◦ D ◦ P ) = P t ◦ D t ◦ Qt ◦ Q ◦ D ◦ P = P t ◦ D t ◦ D ◦ P, and so det(Lt ◦ L) = det P t det(D t ◦ D) det P. Since det P t = det P = ±1 we have det(Lt ◦ L) = det(D t ◦ D). Hence, # # |||JL ||| = det(Lt ◦ L) = det(D t ◦ D) = |||JD |||, which concludes the proof of the formula.
Next we extend the area formula to C 1 functions. Theorem 9.27 (Area formula, C 1 case). Let 1 ≤ k ≤ N , let V ⊆ Rk be an open set and let Ψ : V → RN be a function of class C 1 . Let E ⊆ V be a Lebesgue measurable set and assume that Ψ is injective in E. Then Ψ(E) is Hok -measurable and k H (Ψ(E)) = |||JΨ (y)||| dy. E
We divide the proof in a few lemmas. Lemma 9.28. Let 1 ≤ k ≤ N , let V ⊆ Rk be an open set and let Ψ : V → RN be a function of class C 1 and let y0 ∈ V . Assume that JΨ (y0 ) has rank k. Then for every ε > 0 there exists δ > 0 such that (1 − ε)L(y1 − y2 )N ≤ Ψ(y1 ) − Ψ(y2 )N ≤ (1 + ε)L(y1 − y2 )N for all y1 , y2 ∈ Bk (y0 , δ) ⊆ V , where L(y) := JΨ (y0 )y t , y ∈ Rk . Proof. Since JΨ (y0 ) has rank k there exists c ∈ (0, 1) such that (9.21)
L(y)N ≥ cyk
for all y ∈ Rk .
Since Ψ is of class C 1 there exists δ > 0 such that (9.22)
JΨ (y) − JΨ (y0 )N ×k ≤ cε
for every y ∈ Bk (y0 , δ). Let y1 , y2 ∈ Bk (y0 , δ). By the fundamental theorem of calculus applied to the function h(t) := Ψ(y1 t + (1 − t)y2 ) − L(y1 t + (1 − t)y2 )
254
9. Change of Variables and the Divergence Theorem
we have
1
Ψ(y1 ) − Ψ(y2 ) − L(y1 − y2 ) =
(JΨ (y1 t + (1 − t)y2 ) − JΨ (y0 ))(y1 − y2 )t dt.
0
Hence, by (9.21) and (9.22), Ψ(y1 ) − Ψ(y2 )N ≤ L(y1 − y2 )N + Ψ(y1 ) − Ψ(y2 ) − L(y1 − y2 )N ≤ L(y1 − y2 )N + cεy1 − y2 k ≤ (1 + ε)L(y1 − y2 )N , while Ψ(y1 ) − Ψ(y2 )N ≥ L(y1 − y2 )N − Ψ(y1 ) − Ψ(y2 ) − L(y1 − y2 )N ≥ L(y1 − y2 )N − cεy1 − y2 k ≥ (1 − ε)L(y1 − y2 )N ,
which completes the proof.
Lemma 9.29. Let 1 ≤ k ≤ N , let V ⊆ Rk be an open set, let Ψ : V → RN be a function of class C 1 , and let y0 ∈ V . Assume that JΨ (y0 ) has rank k. Then for every 0 < ε < 1 there exists δ > 0 such that for every Lebesgue measurable set E ⊆ Bk (y0 , δ) ⊆ V , Ψ(E) is Hok -measurable and k+1 k k+1 |||JΨ (y)||| dy ≤ H (Ψ(E)) ≤ (1 + ε) |||JΨ (y)||| dy. (1 − ε) E
E
Proof. Let L be as in the previous lemma. Since JΨ (y0 ) has rank k, the linear function L : Rk → RN is injective, and so there exists L−1 : L(Rk ) → Rk . Given 0 < ε < 1, let δ > 0 be so small that the conclusions of the previous lemma hold and also so that (9.23)
(1 + ε)−1 |||JΨ (y)||| ≤ |||JΨ (y0 )||| ≤ (1 + ε)|||JΨ (y)|||
for all y ∈ Bk (y0 , δ), where we used the fact that Ψ is of class C 1 . Since by the previous lemma, (9.24) (1 − ε)L(y1 − y2 )N ≤ Ψ(y1 ) − Ψ(y2 )N ≤ (1 + ε)L(y1 − y2 )N for all y1 , y2 ∈ Bk (y0 , δ) ⊆ V , taking y1 = L−1 (x1 ) and y1 = L−1 (x2 ) we get (Ψ ◦ L−1 )(x1 ) − (Ψ ◦ L−1 )(x2 )N ≤ (1 + ε)x1 − x2 N for all x1 , x2 ∈ L(Bk (y0 , δ)). Thus, Ψ ◦ L−1 is Lipschitz continuous with Lipschitz constant less than or equal to 1+ε. It follows by Proposition C.44, the area formula for L, and (9.23), (9.25)
Hok (Ψ(E)) = Hok ((Ψ ◦ L−1 )(L(E))) ≤ (1 + ε)k Hok (L(E)) k k+1 |||JL ||| dy ≤ (1 + ε) |||JΨ (y)||| dy. = (1 + ε) E
E
Similarly, by (9.24) and the fact that L is injective it follows that Ψ is injective, and so taking y1 = Ψ−1 (x1 ) and y1 = Ψ−1 (x2 ) we get (L ◦ Ψ−1 )(x1 ) − (L ◦ Ψ−1 )(x2 )N ≤ (1 − ε)−1 x1 − x2 N
9.3. The Area Formula: The C 1 Case
255
for all x1 , x2 ∈ Ψ(Bk (y0 , δ)). Thus, L ◦ Ψ−1 is Lipschitz continuous with Lipschitz constant less than or equal to (1 − ε)−1 . It follows by Proposition C.44, the area formula for L, and (9.23), −1 (1 + ε) |||JΨ (y)||| dy ≤ |||JL ||| dy = Hok (L(E)) E
=
E Hok ((L
◦ Ψ−1 )(Ψ(E))) ≤ (1 − ε)−k Hok (Ψ(E)),
which gives the other inequality since (1 − ε) ≤ (1 + ε)−1 . We leave as an exercise to prove that Ψ(E) is Hok -measurable.
Next we prove the area formula under the additional assumption that JΨ has rank k everywhere in V . Lemma 9.30. Let 1 ≤ k ≤ N , let V ⊆ Rk be an open set and let Ψ : V → RN be a function of class C 1 such that JΨ (y) has rank k for every y ∈ V . Let E ⊆ V be a Lebesgue measurable set and assume that Ψ is injective in E. Then |||JΨ (y)||| dy. Hk (Ψ(E)) = E
Proof. Fix ε > 0 and cover V with countably many balls Bi ⊆ V such that for every E ⊆ Bi the previous lemma applies. Given a Lebesgue measurable set E ⊆ V , define inductively, E1 := E ∩ B1 , Ei := (E ∩ Bi ) \ i−1 j=1 Bj . Then the sets Ei are disjoint and their union is E. By the previous lemma applied to each Ei we get |||JΨ (y)||| dy ≤ Hk (Ψ(Ei )) ≤ (1 + ε)k+1 |||JΨ (y)||| dy. (1 − ε)k+1 Ei
Ei
Summing over i and using the fact that Ψ is injective in E gives k+1 k k+1 (1 − ε) |||JΨ (y)||| dy ≤ H (Ψ(E)) ≤ (1 + ε) |||JΨ (y)||| dy. E
We now let ε → 0+ .
E
To remove the condition that JΨ has rank k in V we use a variation of Sard’s theorem (see Theorem 14.48). Lemma 9.31. Let 1 ≤ k ≤ N , let V ⊆ Rk be an open set, and let Ψ : V → RN be a function of class C 1 . Let Σ := {y ∈ V : Ψ is differentiable at y and JΨ (y) has rank less than k}. Then Hk (Ψ(Σ)) = 0. Proof. Step 1: Assume that V is bounded and that JΨ is bounded in V , say, JΨ (y) ≤ M for all y ∈ V . For ε > 0 consider the function Ψε : V → RN × Rk given by Ψε (y) := (Ψ(y), εy). Then
JΨ (y) , JΨε (y) = εIk
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9. Change of Variables and the Divergence Theorem
and so JΨε (y) has rank k for every y ∈ V . Then by the Cauchy–Binet formula (see Theorem 9.25),
2 ∂(Ψε )α |||JΨε (y)|||2 = (y) det ∂y α∈ΛN,k
≤ |||JΨ (y)|||2 + c(1 + M 2 )ε2 for some constant c > 0. If particular, if y ∈ Σ, then |||JΨε (y)|||2 ≤ cε2 ,
(9.26)
where as usual the constant c changes from line to line. Since Ψ = Π ◦ Ψε , where Π : RN × Rd → RN is the projection operator given by Π(x, y) := x and since Π is Lipschitz continuous with Lipschitz constant one, it follows from Proposition C.44, the area formula, and (9.26), Hk (Ψ(Σ)) = Hk (Π(Ψε (Σ))) ≤ 1Hk (Ψε (Σ)) |||JΨε (y)||| dy ≤ cεLk (V ). = Σ
Letting ε → 0 gives
Hk (Ψ(Σ))
= 0.
Step 2: The general case follows by writing V as an increasing sequence of open bounded sets Vn with Vn ⊆ Vn+1 ⊆ V for all n and by applying Step 1 in each set Vn . We now turn to the proof of Theorem 9.27. Proof of Theorem 9.27. Let Σ be as in Lemma 9.31. Then Hk (Ψ(Σ)) = 0, and so by Lemma 9.30, applied in the open set V \ Σ, k k |||JΨ (y)||| dy = |||JΨ (y)||| dy, H (Ψ(E)) = H (Ψ(E \ Σ)) = E\Σ
E
where the last equality follows from the fact that |||JΨ (y)||| = 0 for y ∈ Σ. Exercise 9.32. Let 1 ≤ k ≤ N , let V ⊆ Rk be an open set and let Ψ : V → RN be a function of class C 1 . Let E ⊆ V be a Lebesgue measurable set. Prove that 0 −1 k H (E ∩ Ψ ({x})) dH (x) = |||JΨ (y)||| dy. RN
E
As a consequence of the area formula we have the following change of variables formula for surface integrals. Theorem 9.33. Let 1 ≤ k ≤ N , let V ⊆ Rk be an open set and let Ψ : V → RN be a function of class C 1 . Let H ⊆ Ψ(V ) be a Borel set and let
9.3. The Area Formula: The C 1 Case
257
u : H → [−∞, ∞] be a Borel function, which is either Hk integrable or has a sign. Assume that Ψ is injective in Ψ−1 (H). Then (9.27) u(x) dHk (x) = u(Ψ(y))|||JΨ(y)||| dy. Ψ−1 (H)
H
Proof. Assume first that u = χG , where G ⊆ H is a Borel set. Then Ψ−1 (G) is a Borel set and so by Theorem 9.27, u(x) dHk (x) = Hk (G) = Hk (Ψ(Ψ−1 (G))) H |||JΨ (y)||| dy = χG (Ψ(y))|||JΨ(y)||| dy. = Ψ−1 (G)
Ψ−1 (H)
n
Next take u to be a simple function, u = i=1 ci χGi , where the Borel sets Gi are disjoint. Then by what we just proved and the linearity of integrals n n u(x) dHk (x) = ci Hk (Gi ) = ci |||JΨ (y)||| dy H
i=1
i=1
=
Ψ−1 (H)
Ψ−1 (Gi )
u(Ψ(y))|||JΨ(y)||| dy.
For a nonnegative Borel function u, construct an increasing sequence of Borel simple functions converging pointwise to u and apply the Lebesgue monotone convergence theorem on both sides. Finally, if the Borel function u : H → [−∞, ∞] is Hk integrable, then as usual we can write u = u+ − u− , apply (9.27) to u+ and u− , and use the linearity of integrals to deduce (9.27) for u. Exercise 9.34. Prove that the previous theorem continues to hold if we assume that Ψ is injective in Ψ−1 (H) \ E, where Lk (E) = 0. Since HN = LN (see Proposition C.50) in the case k = N we obtain the classical change of variables for Lebesgue integration. Corollary 9.35 (Change of variables, the C 1 case). Let Ω ⊆ RN be an open set and let Ψ : Ω → RN be a function of class C 1 . Let H ⊆ Ψ(Ω) be a Borel set and let u : H → [−∞, ∞] be a Borel function, which is either Lebesgue integrable or has a sign. Assume that Ψ is injective in Ψ−1 (H). Then u(x) dx = u(Ψ(y))| det JΨ (y)| dy. H
Ψ−1 (H)
Using the change of variables formula we can prove Brouwer’s fixed point theorem [36].
258
9. Change of Variables and the Divergence Theorem
Theorem 9.36 (Brouwer’s fixed point theorem). Let K ⊂ RN be a nonempty compact convex set and let Ψ : K → K be a continuous transformation. Then there exists x ∈ K such that Ψ(x) = x. We begin with a preliminary lemma. Lemma 9.37. There is no function Ψ : B(0, 1) → S N −1 such that Ψ(x) = x for all x ∈ S N −1 and which is continuous together with all its partial derivatives Proof. Assume by contradiction that Ψ exists and for t ∈ [0, 1] define Ψt (x) := tΨ(x) + (1 − t)x. Then for every x ∈ B(0, 1), Ψt (x) ≤ tΨ(x) + (1 − t)x ≤ 1, thus Ψt : B(0, 1) → B(0, 1). Moreover, Ψt (x) = x for every x ∈ S N −1 . Define Φ(x) := Ψ(x) − x. Let M > 0 be such that JΦ (x)N ×N ≤ M for all x ∈ B(0, 1). Then by the mean value theorem applied to each component, we √ at most √ obtain that Φ is Lipschitz continuous with Lipschitz constant N M . We claim that Ψt is injective for every 0 < t < min{1, 1/( N M )} =: t0 . Indeed, assume by contradiction that there exist x1 , x2 ∈ B(0, 1) such that Ψt (x1 ) = Ψt (x2 ) and x1 = x2 . Since Ψt (x) = x + tΦ(x) it follows that √ x2 − x1 = t(Φ(x2 ) − Φ(x1 )) ≤ t N M x2 − x1 < x2 − x1 , which is a contradiction. Hence, the claim holds. Since JΨt = IN + tJΦ and JΦ is bounded in B(0, 1), by taking t0 smaller, if necessary, we can assume that det JΨt (x) > 0 for all x ∈ B(0, 1). It follows by the inverse function theorem that the set Ut := Ψt (B(0, 1)) is open for all 0 < t < t0 . We claim that Ut = B(0, 1) for every 0 < t < t0 . Indeed, assume that this is not the case. Since Ut ⊆ B(0, 1) by what we proved above, then ∂Ut ⊆ B(0, 1), and so if Ut = B(0, 1), then there must exist y0 ∈ ∂Ut such that y0 ∈ B(0, 1). Let yn ∈ Ut be such that yn → y0 and find xn ∈ B(0, 1) such that yn = Ψt (xn ). By compactness, up to a subsequence, we may assume that xn → x0 ∈ B(0, 1) with Ψt (x0 ) = y0 , by the continuity of Ψt . Note that x0 cannot belong to B(0, 1) as otherwise y0 = Ψt (x0 ) would belong to Ψt (B(0, 1)) = Ut , so, necessarily x0 ∈ S N −1 . But then Ψt (x0 ) = x0 and so x0 = y0 , which is again a contradiction since y0 ∈ B(0, 1). This proves that Ut = B(0, 1) for every 0 < t < t0 . For t ∈ [0, 1] define the function det JΨt (x) dx = g(t) := B(0,1)
det(IN + tJΦ (x)) dx. B(0,1)
9.3. The Area Formula: The C 1 Case
259
Since for every 0 < t < t0 the function Ψt : B(0, 1) → B(0, 1) is a bijection and det JΨt > 0, it follows by Corollary 9.35 that g(t) = LN (B(0, 1)) for all 0 < t < t0 . Since g is a polynomial of degree N , we have that g(t) = LN (B(0, 1)) for all 0 ≤ t ≤ 1. In particular, N (9.28) 0 < L (B(0, 1)) = g(1) = det JΨ (x) dx. B(0,1)
On the other hand, since by hypothesis Ψ(x)2 = Ψ(x) · Ψ(x) = 1 for every x ∈ B(0, 1), if x ∈ B(0, 1) and v ∈ RN , then by replacing x with x + sv differentiating with respect to s we get (JΨ (x)v) · Ψ(x) = 0, which shows that the range of JΨ (x) is orthogonal to the vector Ψ(x). In turn, JΨ (x) has rank less than or equal to N −1 and so det JΨ (x) = 0, which contradicts (9.28). This concludes the proof. We now turn to the proof of Brouwer’s fixed point theorem. Proof of Theorem 9.36. Step 1: Assume first that K = B(0, 1) and that Ψ : B(0, 1) → B(0, 1) is continuous. Extend Ψ to a continuous function outside B(0, 1). Using standard mollifiers, we may construct a sequence {Ψn }n of functions in C 1 (RN ; RN ) such that Ψn → Ψ uniformly on compact sets. By selecting a subsequence, not relabeled, we may assume that Ψn − ΨC(B(0,1)) ≤ 1/n. Then for x ∈ B(0, 1), Ψn (x) ≤ Ψ(x) + Ψn (x) − Ψ(x) ≤ 1 + 1/n, and so the function Φn := Ψn /(1+1/n) maps B(0, 1) into B(0, 1). We claim that Φn has a fixed point. Indeed, if not, then Φn (x) = x for all x ∈ B(0, 1). Define Φ : B(0, 1) → S N −1 as follows. For each x ∈ B(0, 1) let Θn (x) be the intersection with the sphere S N −1 of the ray from Φn (x) to x, precisely1 , Θn (x) := x + Fn (x)(x − Φn (x)), 1 To
obtain Fn , we consider the line sΦn (x) + (1 − s)x,
s ∈ R,
through the distinct points Φn (x) and x and then find s ∈ R such that 1 = sΦn (x) + (1 − s)x 2 = s2 x − Φn (x) 2 + 2s(x · Φn (x) − x 2 ) + x 2 . It suffices to solve for s.
260
9. Change of Variables and the Divergence Theorem
where −x · Φn (x) + x2 x − Φn (x)2 # (x · Φn (x) − x2 )2 + x − Φn (x)2 (1 − x2 ) + . x − Φn (x)2
Fn (x) :=
Then Θn : B(0, 1) → S N −1 is continuous and is the identity on the unit sphere, which contradicts the previous lemma. Hence, the claim is true and thus there exists xn ∈ B(0, 1) such that Φn (xn ) = xn . By compactness we may assume that xn → x0 ∈ B(0, 1) and so since Ψn (xn ) = (1 + 1/n)xn , letting n → ∞ and using uniform convergence we obtain that Ψ(x0 ) = x0 . Step 2: Let K = B(0, R) for some R > 0 and let Ψ : B(0, R) → B(0, R) be a continuous transformation. To obtain a fixed point, it suffices to apply the previous step to the rescaled function ΨR (x) := R−1 Ψ(Rx), x ∈ B(0, 1). Step 3: Let K ⊂ RN be a nonempty compact convex set and let Ψ : K → K be a continuous transformation. Find R > 0 such that K ⊆ B(0, R) and for each x ∈ B(0, R) consider the continuous transformation Φ(x) := Ψ(Π(x)), x ∈ B(0, R), where Π : RN → K is the projection onto the convex set K. Note that Φ(K) ⊆ K ⊆ B(0, R), and so by the continuity of Π we have that Φ : B(0, R) → B(0, R) is continuous. By the previous step there exists x ∈ B(0, R) such that x = Φ(x) = Ψ(Π(x)). On the other hand, since Φ(K) ⊆ K, we have that x ∈ K, and so Π(x) = x. Thus, the previous identity reduces to x = Ψ(x) and the proof is completed. Exercise 9.38. Let Ψ : RN → RN be of class C 2 and let M1 , . . . , MN be the cofactors of the first row of the Jacobian matrix JΨ .2 (i) Prove that N i=1 ∂i Mi can be written symbolically as ⎛ ⎞ ∇ ⎜ ∇Ψ2 ⎟ ⎜ ⎟ det ⎜ . ⎟ . ⎝ .. ⎠ ∇ΨN (ii) Let N = 2. Using part (i), prove that ∂1 M1 + ∂2 M2 = 0. 2 If
A = (aij )i,j=1,...,N is an N × N matrix, the cofactor of the entry aij is defined by Mij := (−1)i+j det Cij ,
i, j = 1, . . . , N,
where Cij is the (N − 1) × (N − 1) matrix obtained from A by removing the ith row and the jth column.
9.3. The Area Formula: The C 1 Case
261
(iii) Let N > 2 and prove that ⎛ ⎛ ⎞ ∇ ∇ N ⎜ ∇Ψ2 ⎟ ⎜ ∇Ψ2 ⎜ ⎜ ⎟ det ⎜ . ⎟ = det ⎜ . . ⎝ . ⎠ ⎝ .. k=2 ∇ΨN ∇ΨN
⎞ ⎟ ⎟ ⎟ , ⎠ k
where the subscript k means that the differential operator ∇ in the first row acts only on the kth row. (iv) Prove that
⎛
∇ ∇Ψ2 .. .
⎜ ⎜ det ⎜ ⎝
⎞ ⎟ ⎟ ⎟ =0 ⎠
∇ΨN
k
for all k = 2, . . . , N and conclude that
N
i=1 ∂i Mi
≡ 0.
Exercise 9.39. Let Ψ ∈ C 2 (RN ; RN ) be such that Ψ(x) = x for x ≥ 1 and let u ∈ Cc1 (RN ). Using the notation (E.2) in the appendix, y1 u(y1 , t) dt, y ∈ RN . v(y) := −∞
(i) Prove that
⎛
⎜ ⎜ (u ◦ Ψ) det JΨ = det ⎜ ⎝
∇(v ◦ Ψ) ∇Ψ2 .. .
⎞ ⎟ ⎟ ⎟. ⎠
∇ΨN (ii) Prove that (u ◦ Ψ) det JΨ = M1 ∂1 (v ◦ Ψ) + · · · + MN ∂N (v ◦ Ψ), where M1 , . . . , MN are the cofactors of the first row of the Jacobian matrix JΨ . (iii) Let r > 0 be such that u = 0 outside the cube Q(0, r). Prove that Mi (x)∂i (v ◦ Ψ)(x) dx = − (v ◦ Ψ)(x)∂i Mi (x) dx Q(0,r)
Q(0,r)
for all i = 2, . . . , N and
M1 ∂1 (v ◦ Ψ) dx = − Q(0,r)
(v ◦ Ψ)∂1 M1 dx + Q(0,r)
(iv) Prove that (9.29) RN
u(y) dy = RN
u(Ψ(x)) det JΨ (x) dx.
u dx. Q(0,r)
262
9. Change of Variables and the Divergence Theorem
Exercise 9.40. Using mollifiers (see Appendix C) prove that (9.29) continues to hold if Ψ ∈ C 1 (RN ; RN ) is the identity outside some ball and u ∈ Cc (RN ). Exercise 9.41. Let Ψ ∈ C 1 (RN ; RN ) be the identity outside the unit ball. Prove that Ψ is onto. Exercise 9.42. Use the previous exercises to give an alternative proof of the Brouwer fixed point theorem.
9.4. The Area Formula: The Differentiable Case In this section we will prove that the area formula and the change of variables continue to hold for differentiable transformations. The proof relies on an important extension theorem due to Whitney [244]. Theorem 9.43 (Whitney extension). Let C ⊆ RN be a closed set and let u : C → R and w : C → RN be two continuous functions. Assume that for every x0 ∈ C and every ε > 0 there exists δ = δ(x0 , ε) > 0 such that (9.30)
|u(x) − u(y) − w(y) · (x − y)| ≤ εx − y
for all x, y ∈ C ∩ B(x0 , δ). Then there exists a function v ∈ C 1 (RN ) such that v = u and ∇v = w in C. Whitney’s extension theorem is based on the following decomposition result, which turned out to be quite useful in other extension theorems (see Exercise 13.7; see also the paper of Jones [122] and the monograph of Stein [220]). Theorem 9.44 (Whitney decomposition). Given a nonempty closed set C ⊂ RN , there exists a countable family F = {Qn : n ∈ N} of closed cubes such that (i) RN \ C = n Qn , (ii) Q◦n ∩ Q◦m = ∅ for all n = m, (iii) diam Qn ≤ dist(Qn , C) ≤ 4 diam Qn , (iv) if Qn and Qm touch, then
1 4
diam Qn ≤ diam Qm ≤ 4 diam Qn ,
(v) for every cube Qn in F there are at most (12)N cubes in F that touch Qn , (vi) for every fixed 0 < ε < 14 and every n let Q∗n be the closed cube with the same center of Qn and diameter (1 + ε) diam Qn , then for every x ∈ RN \ C there exist at most (12)N cubes Q∗n that contain x.
9.4. The Area Formula: The Differentiable Case
263
Proof. Let G0 := {Q(x, 1) : x ∈ ZN }. The family G0 leads to a family {Gk }k∈Z of collections of cubes with the property that each cube in the family Gk gives rise to 2N cubes in the family Gk+1 by bisecting the sides. The in the family Gk have side length 2−k and, in turn, diameter √ cubes N 2−k . For k ∈ Z define Uk := {x ∈ RN \ C :
√ −k+1 √ N2 < dist(x, C) ≤ N 2−k+2 }.
Then RN \ C =
(9.31)
Uk .
k∈Z
Consider the family F0 :=
{Q ∈ Gk : Q ∩ Uk = ∅}.
k∈Z
If x ∈ RN \ C, then by (9.31), x ∈ Uk for some k∈ Z. Since the family Gk is a partition of RN , there exists Q ∈ Gk such that x ∈ Q. Thus, Q ∈ F0 . This shows that Q. (9.32) RN \ C ⊆ Q∈F0
Next we claim that for every Q ∈ F0 , (9.33)
diam Q < dist(Q, C) ≤ 4 diam Q.
√ Indeed, if Q ∈ F0 , then Q ∈ Gk for some k∈ Z, so that diam Q = N 2−k . Moreover, since Q ∩ Uk = ∅, there exists x ∈ Q ∩ Uk . Thus, also by the definition of Uk , √ dist(Q, C) ≤ dist(x, C) ≤ 4 N 2−k . On the other hand, dist(Q, C) ≥ dist(x, C)−diam Q >
√
√ √ N 2−k+1 − N 2−k = N 2−k = diam Q,
which proves (9.33). Note that the first inequality in (9.33) shows, in particular, that every Q ∈ F0 is contained in RN \ C, so that also, by (9.32), Q. RN \ C = Q∈F0
Thus, properties (i) and (iii) are satisfied. To obtain (ii), we construct an appropriate subfamily of F0 . We begin by observing that if Q1 ∈ Gk1 and Q2 ∈ Gk2 intersect, with k1 < k2 , then, necessarily, Q1 ⊃ Q2 . Start from any cube Q ∈ F0 and consider the maximal cube Q (with respect to inclusion) in F0 that contains Q. Such a cube exists, since, by
264
9. Change of Variables and the Divergence Theorem
(9.33), the diameter of any cube in F0 containing Q cannot exceed 4 diam Q. Note that by the previous observation there is only one such maximal cube Q . Let F be the subfamily of maximal cubes in F0 . Then (i)–(iii) hold. To prove (iv), assume that Q1 , Q2 ∈ F touch. Then diam Q2 ≤ dist(Q2 , C) ≤ dist(Q1 , C) + diam Q1 ≤ 5 diam Q1 , where we have used (9.33). On the other hand, since diam Q2 = 2k diam Q1 for some integer k ∈ Z, then, necessarily, diam Q2 ≤ 4 diam Q1 . By reversing the roles of Q1 and Q2 , we obtain (iv). Next we show that (v) holds. Fix a cube Q ∈ F and let k∈ Z be such that Q ∈ Gk . In the family Gk there are only 3N − 1 cubes that touch Q. Each cube in Gk can contain at most 4N cubes of F with diameter greater than or equal to 14 diam Q. Hence (v) follows from (iv). Finally, we prove (vi). For every x ∈ RN \ C let Q ∈ F be such that x ∈ Q. We claim that if Qn ∈ F , then Q∗n intersects Q only if Qn touches Q. Indeed, consider the union of all the cubes in F that touch Q. By (iv), the diameter of each of these cubes is greater than or equal to 14 diam Q. Since 0 < ε < 14 , the union of these cubes contains Q∗n . By the maximality of the family F , it follows that Qn must be one of these cubes and the claim is proved. Property (vi) now follows from (v). The family F is called a Whitney decomposition of RN \ C. Lemma 9.45. Let C ⊂ RN be a closed set. Thenthere exists a locally finite N family of functions ψn ∈ Cc∞ (RN \ C) such that ∞ n=1 ψ = 1 in R \ C, for every n, (9.34) (9.35)
diam supp ψn ≤ 2 dist(supp ψn , C), c for all x ∈ RN \ C ∇ψn (x) ≤ dist(x, C)
and for some constant c = c(N ) > 0, and for every x ∈ RN \ C there are at most (12)N functions ψn such that x belongs to supp ψn . Proof. Let {Qn }n be the family given in the previous theorem. Construct a nonnegative function ϕ ∈ C ∞ (RN ) such that ϕ = 0 outside (−3/4, 3/4)N and ϕ = 1 in (−1/2, 1/2)N . If Qn is centered at yn and has side-length n , define ϕn (x) := ϕ((x − yn )/n ). that Φ ≥ 1 Then ϕn = 1 in Qn . Hence, if we define Φ := ∞ n=1 ϕn we have in RN \ C and so the function ψn := ϕn /Φ is well-defined and ∞ n=1 ψn = 1 N in R \ C.
(9.36)
9.4. The Area Formula: The Differentiable Case
265
√ Moreover, by the properties of ϕ, (9.36), and the fact that dist(Qn , C) ≥ N n , 3√ 1 N n ≥ diam supp ψn , dist(supp ψn , C) ≥ 4 2 which proves (9.34). By property (vi), for every x ∈ RN \ C there are at most (12)N functions ϕn whose support intersect a small neighborhood of x. Hence, Φ ∈ C ∞ (RN \ C) and for x ∈ RN \ C, ∇ψn (x) =
∇ϕn (x) ϕn (x)∇Φ(x) − . Φ(x) Φ2 (x)
By (9.36), ∇ϕk (x) = ∇ϕ((x − xk )/k )/k , and so
∇Φ(x) ≤ ∇ϕ∞
x∈supp ϕk
√ (12)N 4 N ∇ϕ∞ 1 , ≤ k dist(x, C)
√ where we used the inequality dist(x, C) ≤ dist(Qk , C) ≤ 4 N k , which follows from property (iii) and the fact that x ∈ supp ψk . Since for x ∈ RN \ C, 1 ≤ Φ(x) ≤ (12)N , (9.35) holds and the proof is complete. We now turn to the proof of Whitney’s theorem. Proof. For every x0 ∈ C and δ > 0 let |u(x) − u(y) − w(y) · (x − y)| : x, y ∈ C ∩B(x0 , δ) . (9.37) ρx0 (δ) := sup x − y Then by (9.30), (9.38)
lim ρx0 (δ) = 0
δ→0+
for every x0 ∈ C.
Step 1: For each n let xn ∈ C be a point with closest distance from supp ψn , and for x ∈ RN \ C define (9.39)
v(x) :=
∞
(u(xn ) + w(xn ) · (x − xn ))ψn (x),
n=0
while for x ∈ C set v(x) := u(x). We claim that if x0 ∈ C, then (9.40)
lim
x→x0
v(x) − u(x0 ) − w(x0 ) · (x − x0 ) = 0. x − x0
266
9. Change of Variables and the Divergence Theorem
In view of (9.38) it suffices toverify (9.40) for all x ∈ RN \ C with x → x0 . N N By (9.39) and the fact that ∞ n=0 ψ = 1 in R \ C, for x ∈ R \ C, with x − x0 ≤ 1, we have v(x) − u(x0 ) − w(x0 ) · (x − x0 ) =
∞
(u(xn ) − u(x0 ) − w(x0 ) · (xn − x0 ))ψn (x)
n=0 ∞
+
(w(xn ) − w(x0 )) · (x − xn )ψn (x) =: I + II.
n=0
For every n with x ∈ supp ψn , let yn ∈ supp ψn be such that dist(supp ψn , C) = xn − yn . Then by (9.34), dist(x, C) ≤ x − xn ≤ x − yn + yn − xn ≤ diam supp ψn
(9.41)
+ dist(supp ψn , C) ≤ 3 dist(supp ψn , C) ≤ 3 dist(x, C). Since dist(x, C) ≤ x − x0 , it follows that xn − x0 ≤ x − xn + x − x0 ≤ 4x − x0 .
(9.42) In turn, |I| ≤ ≤
∞ |u(xn ) − u(x0 ) − w(x0 ) · (x0 − xn ))| n=0 ∞
xn − x0
xn − x0 ψn (x)
ρx0 (xn − x0 )xn − x0 ψn (x)
n=0
≤ ρx0 (4x − x0 )4x − x0
∞
ψn (x)
n=0
= ρx0 (4x − x0 )4x − x0 . On the other hand, let ρw be a modulus of continuity for w. Then by (9.40), |II| ≤
∞
ρw (xn − x0 )x − xn ψn (x) ≤ ρw (4x − x0 )3x − x0 .
n=0
Combining the last two inequalities and letting x → x0 gives (9.40). Hence, we have shown that v is differentiable in C with ∇v = w. On the other hand, since each Q∗n intersects at most (12)N cubes Q∗n , for every x ∈ RN \ C there exists a neighborhood V of x such that in V only (12)N terms in the series (9.39) are nonzero. Hence, the function v belongs to
9.4. The Area Formula: The Differentiable Case
267
C ∞ (RN \ C) with (9.43)
∇v(x) =
for x ∈
RN
∞
(u(xn ) + w(xn ) · (x − xn ))∇ψn (x) +
n=0
∞
w(xn )ψn (x)
n=0
\ C.
Step 2: In view of Step 1, to prove that v ∈ C 1 (RN ), it remains to show that for x0 ∈ C, lim ∇v(x) = w(x0 ).
(9.44)
x→x0
By the continuity of w in C and Step 1 again, it suffices to verify (9.44) for all x ∈ RN \ C with x → x0 . For every y ∈ C we have that w(y) = ∇x (u(y) + w(y) · (x − y)) = ∇x
∞
(u(y) + w(y) · (x − y))ψn (x)
n=0
=
∞
(u(y) + w(y) · (x − y))∇ψn (x) +
n=0
∞
w(y)ψn (x).
n=0
Given x ∈ RN \ C, with x − x0 ≤ 1, let y ∈ C be such that dist(x, C) = x − y. Then by (9.43), ∇v(x) − w(y) =
∞
(u(xn ) − u(y) − w(y) · (xn − y))∇ψn (x)
n=0 ∞
+
+
(w(xn ) − w(y)) · (x − xn )∇ψn (x)
n=0 ∞
(w(xn ) − w(y))ψn (x) = I + II + III.
n=0
If n is such that x ∈ supp ψn , then by (9.41), xn − y ≤ x − xn + x − y ≤ 4 dist(x, C) ≤ 4x − x0 .
(9.45)
Hence, by (9.35), (9.37), and (9.45), I ≤
∞ |u(xn ) − u(y) − w(y) · (xn − y)| n=0
xn − y
xn − y∇ψn (x)
≤ (12)N 4cρx0 (4x − x0 ). On the other hand, by (9.35) and (9.45), II ≤
∞ n=0
ρw (xn − y)x − xn ∇ψn (x) ≤ (12)N 3cρw (4x − x0 ),
268
9. Change of Variables and the Divergence Theorem
while III ≤
∞
ρw (xn − y)ψn (x) ≤ (12)N ρw (4x − x0 ).
n=0
By combining the last three inequalities we have ∇v(x) − w(x0 ) ≤ ∇v(x) − w(y) + w(y) − w(x0 ) ≤ cρx0 (4x − x0 ) + cρw (4x − x0 ) → 0 as x → x0 . This concludes the proof.
Whitney’s theorem continues to hold for C m and C ∞ functions. Theorem 9.46 (Whitney extension, II). Let C ⊆ RN be a closed set, let m ∈ N0 and let u : C → R and uα : C → RN , α ∈ NN 0 , 0 < |α| ≤ m, be continuous functions. Assume that for every multi-index α, with 0 ≤ |α| ≤ m, for every x0 ∈ C, and for every ε > 0, there exists δ = δ(x0 , ε, α) > 0 such that uβ (y) β (x − y) ≤ εx − ym−|α| (9.46) uα (x) − β! |β|≤m−|α|
for all x, y ∈ C ∩ B(x0 , δ), and where u0 := u. Then there exists a function v ∈ C m (RN ) such that v = u in C, ∂ α v = uα in C for every multi-index α, with 0 < |α| ≤ m, and v is analytic in RN \ C. Proof. The proof is similar to the previous one. In (9.39) one should replace the polynomial of degree one u(xn ) + v(xn ) · (x − xn ) with uα (xn ) (x − xn )α . u(xn ) + α! |α|≤m
The details are left as an exercise.
Remark 9.47. The long-standing problem of determining whether a function u : C → R defined on a compact set C ⊆ RN can be extended to a function of class C m (RN ) has recently been solved by C. Fefferman (see [77]). In view of Whitney’s theorem, if Ω ⊆ RN is an open set and m ∈ N0 ∪{∞}, we define the space C m Ω as the space of all functions u ∈ C m (Ω) that can be extended to C m (RN ). Using Whitney’s extension theorem we can prove the following approximation result. Theorem 9.48. Let Ω ⊆ RN be an open set and let u : Ω → R be such that the set E := {x ∈ Ω : u is differentiable at x}
9.4. The Area Formula: The Differentiable Case
269
is nonempty. Then there exists a sequence of disjoint closed sets Cn ⊆ E and R such that gn = u and ∇gn = ∇u a sequence of C 1 functions gn : RN → in Cn and LN (E \ C) = 0, where C := ∞ n=1 Cn . Proof. For k ∈ N and x ∈ E set |u(y) − u(x) − ∇u(x) · (y − x)| 1 : y ∈ Ω, 0 < x − y ≤ . vk (x) := sup x − y k Then vk (x) → 0 as k → ∞. Hence, by Lusin’s and Egoroff’s theorems there ∂u is continuous in Cn exists a sequence of disjoint closed sets Cn such that ∂x i for every i =1, . . . , N , vk → 0 uniformly in each Cn , and LN (E \ C) = 0, where C := ∞ n=1 Cn . We can now apply Whitney’s extension theorem to find a C 1 function gn such that gn = u and ∇gn = ∇u in Cn . We can now extend Theorem 9.33 to differentiable functions. Theorem 9.49 (Area formula, the differentiable case). Let 1 ≤ k ≤ N , let V ⊆ Rk be an open set and let Ψ : V → RN be a differentiable function. Let H ⊆ Ψ(V ) be a Borel set and let u : H → [−∞, ∞] be a Borel function, which is either Hk integrable or has a sign. Assume that Ψ is injective in Ψ−1 (H). Then k u(x) dH (x) = u(Ψ(y))|||JΨ(y)||| dy. Ψ−1 (H)
H
Proof. In view of Theorem 9.48 applied to each component of Ψ we can find a sequence of disjoint closed sets Cn ⊆ V and a sequence of C 1 functions Ψn : Rk → RN such that Ψn = Ψ and JΨn = JΨ in Cn and where C :=
∞
Lk (V \ C) = 0,
n=1 Cn .
We claim that Hk (Ψ(V \ C)) = 0. To see this, write V \C =
∞
El ,
l=1
where
Ψ(y) − Ψ(z)N 1 ≤ l for all z ∈ V with 0 < z−yk ≤ . El := y ∈ V \C : y − zk l Then Ψ : El → RN is locally Lipschitz continuous. Indeed, if z, y belong to a compact subset K ⊆ El and z −yk ≤ 1/l, then Ψ(y)−Ψ(z)N ≤ ly−zk by the definition of El . On the other hand, if z − yk > 1/l, then Ψ(y) − Ψ(z)N ≤ 2 max ΨN ≤ 2l max ΨN y − zk . K
K
Hence, by Proposition C.44, Hk (Ψ(K)) ≤ LK Hk (K) = LK Lk (K) = 0,
270
9. Change of Variables and the Divergence Theorem
where LK := l(1 + 2 maxK ΨN ). By letting K El we obtain that Hk (Ψ(El )) = 0 and since this is true for every l, it follows that Hk (Ψ(V \ C)) = 0. Hence, if u ≥ 0, k u(x) dH (x) =
H
= =
u(x) dH (x) = H∩Ψ(C)
u(x) dHk (x) =
n
H∩Ψn (Cn )
n
Ψ−1 (H)∩Cn
=
k
Ψ−1 (H)
n
n
u(x) dHk (x) H∩Ψ(Cn )
Ψ−1 n (H)∩Cn
u(Ψ(y))|||JΨ(y)||| dy =
u(Ψn (y))|||JΨn (y)||| dy
Ψ−1 (H)∩C
u(Ψ(y))|||JΨ (y)||| dy
u(Ψ(y))|||JΨ (y)||| dy,
where we have used the fact that the sets H ∩ Ψ(Cn ) are disjoint since Ψ is injective in Ψ−1 (H) and Theorem 9.33 for Ψn . If u : H → [−∞, ∞] is Hk integrable, write u = u+ − u− , apply the previous identity to u+ and u− , and use the linearity of integrals to conclude the proof. Exercise 9.50. Prove that the previous theorem continues to hold if in place of everywhere differentiability of Ψ we assume that Ψ : V → RN is continuous and there exists a set F ⊆ V , with Lk (V \ F ) = 0 and Hk (Ψ(V \ F )) = 0, such that Ψ is differentiable in F . Corollary 9.51 (Change of variables, II). Let V ⊆ RN be an open set and let Ψ : V → RN be a differentiable function. Let H ⊆ Ψ(V ) be a Borel set and let u : H → [−∞, ∞] be a Borel function, which is either Lebesgue integrable or has a sign. Assume that Ψ is injective in Ψ−1 (H). Then u(x) dx = u(Ψ(y))| det JΨ (y)| dy. H
Ψ−1 (H)
Using Exercise 9.50 we obtain the standard change of variables formula for multiple integrals. Theorem 9.52 (Change of variables for multiple integrals). Let Ω ⊆ RN be an open set and let Ψ : Ω → RN be continuous. Assume that there exist two Lebesgue measurable sets F, G ⊆ Ω on which Ψ is differentiable and one-to-one, respectively. If (i) LN (Ω \ F ) = 0, (ii) LN (Ψ(Ω \ F )) = 0, (iii) LN (Ψ(Ω \ G)) = 0,
9.4. The Area Formula: The Differentiable Case
271
then the change of variables formula
u(y) dy = Ψ(E)
u(Ψ(x))| det JΨ (x)| dx E
holds for every Borel set E ⊆ Ω and for every Borel function u : Ψ(E) → [−∞, ∞], which is either Lebesgue integrable or has a sign. Remark 9.53. (i) The hypothesis that Ω is open may be replaced by the hypothesis that Ω is any set whose boundary has Lebesgue measure zero, provided we assume LN (Ψ(Ω\Ω◦ )) = 0. Then LN (Ω\Ω◦ ) = 0 and it is enough to apply the theorem to Ω◦ . (ii) Note that under the hypotheses of the previous theorem we obtain, as a by-product, that | det JΨ | is a locally integrable function such that | det JΨ (x)| dx < ∞ E
for every Borel set E ⊆ Ω for which LN (Ψ(E)) < ∞. Exercise 9.54. Let Ω ⊆ RN be an open set and let Ψ : Ω → RN be a Lipschitz continuous function. Prove that Ψ has the (N ) property, that is, that if E ⊆ Ω has Lebesgue measure zero, then so does Ψ(E). Deduce that if Ψ : Ω → RN is a Lipschitz continuous function, which is one-to-one except on a set of measure zero, then Theorem 9.52 holds. We will see in Chapter 12 (see Exercise 12.54), under suitable regularity hypotheses on the open set Ω, an important class of transformations satisfying the hypotheses of Theorem 9.52 is given by transformations Ψ : Ω → RN belonging to the Sobolev space W 1,p (Ω; RN ), p > N , and that are one-to-one except at most on a set of Lebesgue measure zero. Another extremely useful class of transformations is given by (locally) Lipschitz continuous functions Ψ : Ω → RN , which are again one-to-one except at most on a set of Lebesgue measure zero. Perhaps one of the most important applications of Theorem 9.52 and Remark 9.53 is given by spherical coordinates in RN . We proceed by induction. For N = 2 we consider the standard polar coordinate system. Given a point x = (x1 , . . . , xN ) ∈ RN , let r := x, let θ1 be the angle from the positive x1 -axis to x, precisely, θ1 := cos−1 (x1 /r),
0 < θ1 < π,
272
9. Change of Variables and the Divergence Theorem
and let (ρ, θ2 , . . . , θN −1 ) be the spherical coordinates for x1 = (x2 , . . . , xN ) ∈ RN −1 , where ρ := x1 N −1 = r sin θ1 . The coordinates of x are x1 = r cos θ1 , x2 = r sin θ1 cos θ2 , .. . xN −1 = r sin θ1 sin θ2 · · · sin θN −2 cos θN −1 , xN = r sin θ1 sin θ2 · · · sin θN −2 sin θN −1 . This defines a coordinate system (r, θ1 , . . . , θN −1 ), which is invertible except on a set of LN -measure zero. The Jacobian is det JΨ (r, θ1 , . . . , θN −1 ) = rN −1 sinN −2 θ1 sinN −3 θ2 · · · sin θN −2 . Example 9.55. Let u(x) = g(x), where g : [r1 , r2 ] → R is continuous and 0 ≤ r1 < r2 . Then r2 u(x) dx = βN g(r)rN −1 dr. r1 < x 0 such that, setting y := T (x), we have (9.49)
T (Ω ∩ B(x0 , r)) = {y ∈ B(0, r) : yN > f (y )}.
We say that ∂Ω is of class C m , m ∈ N0 , if the functions f are of class C m . Observe that R, f , and r depend on x0 . The coordinates x are called background coordinates while the coordinates y are called local coordinates. Remark 9.58. Without loss of generality, in the previous definition one can replace the ball B(x0 , r) with any small (open) neighborhood of x0 . We will use this fact without further notice. We remark that if ∂Ω is of class C m for m ∈ N, then ∂Ω is an (N − 1)dimensional surface of class C m . Remark 9.59. The previous definition implies in particular that ∂Ω = ∂(Ω), so that the open set Ω cannot lie on both sides of its boundary. Sets like RN \ {x ∈ RN : xN = 0} or B(0, 1) \ {x = (x1 , x2 ) ∈ R2 : 0 ≤ x1 < 1, x2 = 0} are excluded. Exercise 9.60. Let Ω ⊂ R2 be a bounded open set with the property that for each point x0 ∈ ∂Ω there exist local coordinates y = (y1 , y2 ) ∈ R × R, with y = 0 at x = x0 , a Lipschitz continuous function f : R →R, and r > 0 such that ∂Ω ∩ B(x0 , r) in local coordinates becomes {y ∈ B(0, r) : y2 = f (y1 )}.
274
9. Change of Variables and the Divergence Theorem
Prove that ∂Ω = ∂(Ω). Hint: Prove that if ∂Ω = ∂(Ω), then there exists a component Γ of ∂Ω that is a closed curve and such that R2 \ Γ is connected. Prove that this violates the Jordan curve theorem (see Theorem 5.45). The following example shows that domains with continuous boundary may have exponential cusps. Exercise 9.61. Prove that the domain {x = (x1 , x2 ) ∈ R2 : 0 < x1 < 1, − exp(−1/x1 ) < x2 < exp(−1/x1 )} is of class C 0 and that the function f in Definition 9.57 cannot be taken of class C 0,α for any α > 0. Definition 9.62. Given an open set Ω ⊆ RN with Lipschitz continuous boundary and x0 ∈ ∂Ω, a vector τ ∈ RN is called a tangent vector to ∂Ω at the point x0 if there exists a function ψ : (−δ, δ) → RN differentiable at 0 such that ψ((−δ, δ)) ⊆ ∂Ω, ψ(0) = x0 and ψ (0) = τ . The set of all tangent vectors to ∂Ω at x0 is called the tangent space to ∂Ω at x0 and is denoted T∂Ω (x0 ). Given an open set with Lipschitz continuous boundary, in a neighborhood of x0 ∈ ∂Ω we have that ∂Ω ∩ B(x0 , r) in local coordinates becomes {y ∈ B(0, r) : yN = f (y )}. In view of Rademacher’s theorem (see Theorem 9.14), the function f is differentiable for LN −1 -a.e. y ∈ RN −1 . At a point x ∈ ∂Ω at which f is differentiable, there exists the tangent space T∂Ω (x). Note that T∂Ω (x) is given by the (N − 1)-dimensional vector space given by ker ∇g, where g(y) = yN − f (y ) (exercise). Definition 9.63. Given an open set Ω ⊆ RN with Lipschitz continuous boundary and a point x0 ∈ ∂Ω for which there exists a tangent space T∂Ω (x0 ) of dimension N − 1, a vector ν ∈ RN \ {0} is called a normal vector to ∂Ω at x0 if it is orthogonal to all vectors in T∂Ω (x0 ), that is, ν · τ = 0 for all τ ∈ T∂Ω (x0 ). A unit normal vector ν to ∂Ω at x0 is called a unit outward normal to Ω at x0 if there exists δ > 0 such that x0 − tν ∈ Ω and x0 + tν ∈ RN \ Ω for all 0 < t < δ. We are ready to prove the divergence theorem. Theorem 9.64 (Divergence theorem). Let Ω ⊂ RN be an open, bounded set with Lipschitz continuous boundary and let u : Ω → RN be a Lipschitz continuous function. Then div u(x) dx = u(x) · ν(x) dHN −1 (x), (9.50) Ω
∂Ω
where ν(x) is the outward unit normal to ∂Ω at x and div u :=
N
i=1 ∂i ui .
9.5. The Divergence Theorem
275
Note that in view of Rademacher’s theorem (see Theorem 9.14) we have that u is differentiable at LN -a.e. x ∈ Ω. Thus both integrals in (9.50) are well-defined. Proof. Step 1: We first prove that the divergence theorem holds when Ω is given by a rectangle, R := (a1 , b1 ) × · · · × (aN , bN ). −1 N −1 . Then by Fubini’s theorem, Write R := ΠN i=1 (ai , bi ) ⊂ R
∂N uN (x) dx = R
R
bN
∂N uN (x , xN ) dxN dx
aN
(uN (x , bN ) − uN (x , aN )) dx = u(x , bN ) · eN dx + u(x , aN ) · (−eN ) dx R R N −1 = u(x) · ν dH (x)+ u(x) · ν dHN −1 (x), =
R
R ×{bN }
R ×{aN }
where in the third equality we have used the fundamental theorem of calculus (see Theorem 3.20) applied to the Lipschitz function of one variable xN ∈ [aN , bN ] → uN (x , xN ) with x fixed, and in the last equality we have used formula (9.48). Summing the resulting identities gives the desired result. Step 2: Next we prove that the divergence theorem holds when Ω is a set of the form Ω := {(x , xN ) ∈ R × R : f (x ) < xN < bN }, where R is as before and f : RN −1 → R is a Lipschitz continuous function and u = 0 in an open neighborhood of ∂Ω \ graph f . The change of variables Φ : RN → RN given by (9.51)
y = Φ(x) = (x , xN − f (x )),
maps the set Ω into the set W := {(y , yN ) ∈ R × R : 0 < yN < bN − f (y )}. Let R := R × (0, bN ) and consider the function u(y , yN + f (y )) if y ∈ W, w(y) := 0 if y ∈ R \ W.
276
9. Change of Variables and the Divergence Theorem
Since u = 0 near xN = bN , it follows that w is Lipschitz continuous in R. Hence, by Step 1, div w(y) dy = div w(y) dy = w(y) · ν(y) dHN −1 (y) W R ∂R =− wN (0, y ) dy . R
By the chain rule, for LN -a.e. y ∈ W , div w(y) =
∂uN (y , yN + f (y )) ∂xN N −1 ∂ui ∂f ∂ui (y , yN + f (y )) (y ) + (y , yN + f (y )) . + ∂xN ∂yi ∂xi i=1
Hence,
N ∂ui W i=1
∂xi
(y , yN + f (y )) dy = − −
R
uN (y , f (y )) dy
N −1 W
i=1
∂f ∂ui (y , yN + f (y )) (y ) dy. ∂xN ∂yi
Note that the change of variables Φ given in (9.51) is invertible, with inverse Ψ := Φ−1 given by x = Ψ(y) = (y , yN + f (y )). Moreover, we have
JΨ (y) =
IN −1 0 −∇y f (y ) 1
,
which implies that det JΨ (y) = 1. Hence, by changing variables (see Corollary 9.51),
div u(x) dx = −
(9.52) Ω
R
uN (x , f (x )) dx −
N −1 i=1
Ω
∂f ∂ui (x) (x ) dx. ∂xN ∂xi
By Fubini’s theorem, the fundamental theorem of calculus (see Theorem 3.20), and the fact that u = 0 near xN = bN , bN ∂f ∂f ∂ui ∂ui (x) (x ) dx = (x , xN ) dxN (x ) dx ∂x ∂x ∂x ∂x i i N N Ω R f (x ) ∂f = (0 − ui (x , f (x ))) (x ) dx , ∂x i R
9.5. The Divergence Theorem
and so from (9.52), div u(x) dx = − Ω
R
277
uN (x , f (x )) dxN +
N −1 i=1
R
ui (x , f (x ))
$ 1 + ∇f (x )2N −1 = u(x , f (x )) · (−1, ∇f (x )) $ dx 2 R 1 + ∇f (x )N −1 = u(x) · ν(x) dHN −1 (x),
∂f (x ) dx ∂xi
∂Ω
where in the last equality we have used formula (9.47). This concludes the proof in this case. Step 3: Assume that there exists a rigid motion T such that setting y =: T (x) = Ax + c, where A is an N × N rotation matrix and c ∈ RN , we can write T (Ω) = {(y , yN ) ∈ R × R : f (y ) < yN < bN }. Note that div u = trace Ju . Hence, if we define v(y) := Au(T −1 (y)), we have that Jv (y) = AJu (T −1 (y))At and so divy v(y) = trace Jv (y) = trace Ju (T −1 (y) = divx u(T −1 (y). By applying Step 2 to v we get −1 divx u(T (y)) dy = divy v(y) dy = v(y) · ν(y) dHN −1 (y) T (Ω) T (Ω) ∂T (Ω) = Au(T −1 (y)) · ν(y) dHN −1 (y) ∂T (Ω) u(T −1 (y)) · At ν(y) dHN −1 (y). = ∂T (Ω)
Since get
det At
= det A = 1, we can change variables (see Corollary 9.51) to div u(T T (Ω)
−1
(y)) dy =
div u(x) dx. Ω
On the other hand, since the Hausdorff measure is invariant by rotation (why?) we get −1 t N −1 u(T (y)) · A ν(y) dH (y) = u(x) · ν(x) dHN −1 (x). ∂T (Ω)
∂Ω
Step 4: We finally consider the general case. By definition of Lipschitz continuous boundary and Remark 9.58 for every point x0 ∈ ∂Ω there exist a rigid motion Tx0 : RN → RN with Tx0 (x0 ) = 0, a Lipschitz continuous
278
9. Change of Variables and the Divergence Theorem
function fx0 : RN −1 → R and rx0 > 0 such that, setting y := Tx0 (x), we have Tx0 (Ω ∩ QTx0 (x0 , rx0 )) = {y ∈ Q(0, rx0 ) : yN > fx0 (y )},
(9.53)
where QTx0 (x0 , rx0 ) is an open cube centered at x0 and of side-length rx0 > 0. On the other hand, if x0 ∈ Ω, which is open, then there exists Q(x0 , rx0 ) ⊆ Ω. Since the union of all these open cubes as x0 varies in Ω covers the compact set Ω, we can find Q1 , . . . , Qn which still cover Ω. Let ψ1 , . . . , ψ be a partition of unity subordinated to the family of open sets Q1 , . . . , Qn (see Theorem C.21). For every k = 1, . . . , , we have that the function ψk u is zero outside some Qj and since Qj is either contained in Ω or Ω∩Qk satisfies (9.53), we can apply either Step 1 or Step 3 to conclude that (9.54) div(ψk u) dx= div(ψk u) dx = ψk u · ν dHN −1 Ω
Ω∩Qj
∂(Ω∩Qj )
ψk u · ν dHN −1 ,
= ∂Ω
where we have used the fact that ψk u is zero outside Qj . Summing (9.54) over k and using the fact that k=1 ψk = 1 in Ω, we have
div u dx =
div
Ω
Ω
=
ψk u dx =
k=1
ψk u · ν dH
k=1
N −1
=
∂Ω
k=1
div(ψk u) dx Ω
ψk u · ν dHN −1
∂Ω k=1
u · ν dHN −1 ,
= ∂Ω
which is what we wanted.
Remark 9.65. In physics ∂Ω u(x) · ν(x) dHN −1 (x) represents the outward flux of a vector field u across the boundary of a region Ω. Corollary 9.66 (Integration by parts). Let Ω ⊂ RN be an open, bounded set with Lipschitz continuous boundary, let u : Ω → R, v : Ω → R be Lipschitz continuous. Then for every i = 1, . . . , N , ∂i u(x)v(x) dx = − ∂i v(x)u(x) dx + u(x)v(x)νi (x) dHN −1 (x), Ω
Ω
∂Ω
where ν(x) is the outward unit normal to ∂Ω at x. Proof. Apply the divergence theorem to the function w : Ω → RN given by wj := 0 for j = i and wi := uv.
9.5. The Divergence Theorem
279
Exercise 9.67. Let m ∈ N, let Ω ⊂ RN be an open, bounded set with Lipschitz continuous boundary, let u : Ω → R, v : Ω → R be such that u and v are bounded and continuous in Ω and there exist ∂ α u and ∂ α v in Ω for every multi-index α with 1 ≤ |α| ≤ m and they are continuous and bounded in Ω. (i) Prove that for every multi-index α with 1 ≤ |α| ≤ m, βi + 1 = 1. |β| + 1 β+ei =α
(i) Prove that for every multi-index α with 1 ≤ |α| ≤ m, (9.55) ∂ α u(x)v(x) dx = (−1)|α| ∂ α v(x)u(x) dx Ω Ω α! |β|!|γ|! |γ| (−1) ∂ β u(x)∂ γ v(x)νi (x) dHN −1 (x). + |α|! β!γ! ∂Ω β+γ+ei =α
Chapter 10
Distributions Prospective Grad Students, IV: “Can you really live comfortably in this major metropolitan area with that stipend, or will I find myself living out of a closet working part time as a shoe salesman?” — Jorge Cham, www.phdcomics.com
Throughout this chapter the implicit space is the Euclidean space RN and Ω ⊆ RN is an open set, not necessarily bounded.
10.1. The Spaces DK (Ω), D(Ω), and D (Ω) In this section, given an open set Ω ⊆ RN , we construct a topology on the space Cc∞ (Ω). We begin by considering the space of all functions with support in a given compact set K ⊂ Ω. In this subspace the topology should be consistent with the natural notion of convergence, which is uniform convergence of the functions and of all their partial derivatives of any order. Consider an open set Ω ⊆ RN and fix a compact set K ⊂ Ω. Let DK (Ω) be the set of all functions in Cc∞ (Ω) whose support is contained in K, that is, DK (Ω) := {φ ∈ Cc∞ (Ω) : supp φ ⊆ K}. For each j ∈ N0 define the norm · K,j on DK (Ω) by (10.1)
φK,j := sup{|∂ α φ(x)| : x ∈ K, α ∈ NN 0 with |α| ≤ j}.
By Theorem A.27, the family of norms { · K,j }j turns DK (Ω) into a locally convex space and a base for the topology τK is given by all sets of the form {φ ∈ DK (Ω) : φK,j1 < 1/1 , . . . , φK,jk < 1/k }, 281
282
10. Distributions
where j1 , . . . , jk ∈ N0 and 1 , . . . , k ∈ N, k ∈ N. Taking j := max{j1 , . . . , jk } and := max{1 , . . . , k }, it follows that (10.2)
VK,j, := {φ ∈ DK (Ω) : φK,j < 1/} ⊆ {φ ∈ DK (Ω) : φK,j1 < 1/1 , . . . , φK,jk < 1/k },
and so it suffices to consider as a local base for the topology τK the family of sets VK,j, , where j ∈ N0 and ∈ N. Exercise 10.1. Let Ω ⊆ RN be an open set and let K ⊂ Ω be a compact set. Define 1 φ − ψK,j , φ, ψ ∈ DK (Ω). (10.3) dK (φ, ψ) := max j j∈N0 2 1 + φ − ψK,j (i) Prove that dK is a metric. (ii) Prove that the topology τK is determined by the metric dK . (iii) Prove that DK (Ω) is complete. To construct a topology on Cc∞ (Ω), let B0 be the collection of all balanced,1 convex sets U ⊆ Cc∞ (Ω) such that (10.4)
U ∩ DK (Ω) ∈ τK
for every compact set K ⊂ Ω. Theorem 10.2. Let Ω ⊆ RN be an open set. The family B := {φ + V : φ ∈ Cc∞ (Ω), V ∈ B0 } is a base for a locally convex Hausdorff topology τ on Cc∞ (Ω) that turns Cc∞ (Ω) into a topological vector space. Proof. Step 1: We first prove that B0 is nonempty. Indeed, for every j ∈ N0 and ∈ N define the norms (10.5)
φj := sup{|∂ α φ(x)| : x ∈ Ω, α ∈ NN 0 with |α| ≤ j}
and (10.6)
Vj, := {φ ∈ Cc∞ (Ω) : φj < 1/}.
Then Vj, is balanced, convex, and (10.7)
Vj, ∩ DK (Ω) = {φ ∈ DK (Ω) : φK,j < 1/} = VK,j, ∈ τK
by (10.2), and so Vj, belongs to B0 . Step 2: To prove that B is a base for a topology, it suffices to verify the following two conditions: (i) for every φ ∈ Cc∞ (Ω) there exists U ∈ B such that φ ∈ U ; 1 We recall that a subset E of a vector space X is balanced if tx ∈ E for all x ∈ E and t ∈ [−1, 1].
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283
(ii) for every U1 , U2 ∈ B, with U1 ∩ U2 = ∅, and for every φ ∈ U1 ∩ U2 there exists U3 ∈ B such that φ ∈ U3 and U3 ⊆ U1 ∩ U2 . To prove (i), let φ0 ∈ Cc∞ (Ω). Fix j ∈ N0 and ∈ N and consider the set Vj, defined in (10.6). Then Vj, ∈ B0 , and so U := φ0 + Vj, ∈ B. To verify property (ii), let φ1 , φ2 ∈ Cc∞ (Ω) and V1 , V2 ∈ B0 be such that (φ1 + V1 ) ∩ (φ2 + V2 ) = ∅ and fix φ ∈ (φ1 + V1 ) ∩ (φ2 + V2 ). Since the supports of φ1 , φ2 , and φ are compact sets contained in Ω, we may find a compact set K ⊂ Ω such that K ⊇ supp φ1 ∪ supp φ2 ∪ supp φ. Note that if (φ − φi )(x) = 0 for some x ∈ Ω and for some i ∈ {1, 2}, then necessarily x ∈ supp φi ∪ supp φ, and so φ − φi ∈ DK (Ω), i = 1, 2. Since φ − φi ∈ Vi ∩ DK (Ω) ∈ τK , i = 1, 2, using the continuity of scalar multiplication in DK (Ω), we may find θ ∈ (0, 1) such that φ − φi ∈ (1 − θ)(Vi ∩ DK (Ω)) ⊆ (1 − θ)Vi for i = 1, 2. By the convexity of the sets Vi we have that φ − φi + θVi ⊆ (1 − θ)Vi + θVi = Vi for i = 1, 2, so that φ + θ(V1 ∩ V2 ) ⊆ (φ1 + V1 ) ∩ (φ2 + V2 ). Thus, B is a base for a topology τ given by all unions of members of B. Step 3: Next we show that (Cc∞ (Ω), τ ) is a topological vector space. To prove the continuity of scalar multiplication at a point (t0 , φ0 ) ∈ R × Cc∞ (Ω), consider an (open) neighborhood U ∈ τ of t0 φ0 . Since B is a base for the topology, we may find V ∈ B0 such that t0 φ0 + V ⊆ U . Let K := supp φ0 . Then φ0 ∈ DK (Ω) and since V ∩ DK (Ω) ∈ τK , by the continuity of scalar multiplication in DK (Ω) we may find δ > 0 so small that δφ0 ∈ 12 (V ∩ DK (Ω)) ⊆ 12 V. Let s := 1/[2(|t0 | + δ)]. Then for every |t − t0 | < δ and φ ∈ φ0 + sV we have that tφ − t0 φ0 = t(φ − φ0 ) + (t − t0 )φ0 ∈ tsV + 12 V ⊆ 12 V + 12 V = V, where we have used the fact that V is convex and balanced. Hence, tφ ∈ t0 φ0 + V ⊆ U for every |t − t0 | < δ and every φ ∈ φ0 + sV , which proves continuity of scalar multiplication at the point (t0 , φ0 ). To prove the continuity of addition at a point (φ1 , φ2 ) ∈ Cc∞ (Ω)×Cc∞ (Ω), consider a neighborhood U ∈ τ of φ1 +φ2 . Since B is a base for the topology,
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we may find V ∈ B0 such that φ1 + φ2 + V ⊆ U . The convexity of V ∈ B0 implies that (φ1 + 12 V ) + (φ2 + 12 V ) = φ1 + φ2 + V. By (10.4) for every compact set K ⊂ Ω we have that V ∩ DK (Ω) ∈ τK , and since the topology τK turns DK (Ω) into a topological vector space, it also follows that 12 V ∩ DK (Ω) ∈ τK . Thus, 12 V ∈ B0 and, in turn, φ1 + 12 V , φ2 + 12 V ∈ B.
Step 4: Finally, to prove that (Cc∞ (Ω), τ ) is a Hausdorff topological vector space, it suffices to show that singletons are closed. Let φ1 , φ2 ∈ Cc∞ (Ω) be two distinct elements and define V := {φ ∈ Cc∞ (Ω) : sup |φ(x)| < sup |φ1 (x) − φ2 (x)|}. x∈Ω
x∈Ω
/ φ2 + V . Hence, {φ1 } is In view of (10.6) the set V belongs to B0 and φ1 ∈ closed and the proof is completed. The space Cc∞ (Ω) endowed with the topology τ is denoted D(Ω) and its elements are called testing functions. A natural question is why we define τ in such a convoluted way, instead of directly considering the locally convex topology generated by the family of norms { · j }j∈N0 defined in (10.5). The problem is that this topology would not be complete. Exercise 10.3. Take N = 1, Ω = R, and consider a function φ ∈ Cc∞ (R) with support in [0, 1] such that φ > 0 in (0, 1). Prove that the sequence 1 1 φn (x) := φ(x − 1) + φ(x − 2) + · · · + φ(x − n), x ∈ R, 2 n is a Cauchy sequence in the topology generated by the family of norms { · j }j∈N0 defined in (10.5), but its limit does not have compact support, and so it does not belong to Cc∞ (R). Exercise 10.4. Let Ω ⊆ RN be an open set. Let {Kn }n∈N0 be an increasing sequence of compact sets of Ω, with K0 := ∅, such that dist(Kn , ∂Kn+1 ) > 0 and Ω = ∞ n=1 Kn . Given two sequences m := {mn }n∈N0 in N0 and a := {an }n∈N0 in (0, ∞), with mn → ∞ and an → 0, for every φ ∈ D(Ω) define 1 |∂ α φ(x)|. pm,a (φ) := sup sup n∈N0 x∈Ω\Kn an |α|≤mn
(i) Prove that pm,a is a seminorm. (ii) Prove that the family of seminorms {pm,a }m,a , where m and a vary among all sequences as above, generates the topology τ defined in Theorem 10.2.
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285
We now show that the topology τ , when restricted to DK (Ω), for some compact set K ⊂ Ω, does not produce more open sets than the ones in τK . Theorem 10.5. Let Ω ⊆ RN be an open set. Then for every compact set K ⊂ Ω the topology τK coincides with the relative topology of DK (Ω) as a subset of D(Ω). Proof. Fix a compact set K ⊂ Ω and let U ∈ τ . We claim that U ∩ DK (Ω) belongs to τK . To see this, it suffices to consider the case in which U ∩DK (Ω) is nonempty. Let φ ∈ U ∩DK (Ω). Since B is a base for τ , there exists V ∈ B0 such that φ + V ⊆ U . Hence, φ + (V ∩ DK (Ω)) ⊆ U ∩ DK (Ω). Since φ ∈ DK (Ω) and V ∩ DK (Ω) ∈ τK , we have that φ + (V ∩ DK (Ω)) ∈ τK . This shows that every point of U ∩ DK (Ω) is an interior point with respect to τK , and so U ∩ DK (Ω) ∈ τK . Conversely, let U ∈ τK . We claim that U = V ∩ DK (Ω)
(10.8)
for some V ∈ τ . Since the family of sets VK,j, , where j ∈ N0 and ∈ N, is a local base for the topology τK (see (10.2)), for every φ ∈ U we may find jφ ∈ N0 and φ ∈ N such that φ + VK,jφ ,φ ⊆ U . Let Vjφ ,φ be defined as in (10.6). By Step 1 of the proof of Theorem 10.2, (φ + Vjφ ,φ ) ∩ DK (Ω) = φ + VK,jφ ,φ ⊆ U and φ + Vjφ ,φ ∈ B. In turn, the set V := φ∈U (φ + Vjφ ,φ ) belongs to τ and (10.8) holds. Exercise 10.6. Let Ω ⊆ RN be an open set. Prove that for every x0 ∈ Ω, for every compact set K ⊂ Ω, and for every r > 0, the set U := {φ ∈ DK (Ω) : |φ(x0 )| < r} is open with respect to τK . Next we study topologically bounded sets in D(Ω) (see Definition A.20). The following result will be used to study convergence with respect to the topology τ . Theorem 10.7. Let Ω ⊆ RN be an open set and let W ⊆ D(Ω) be a topologically bounded set. Then there exists a compact set K ⊂ Ω such that W ⊆ DK (Ω). Moreover, for every j ∈ N0 there exists a constant Mj > 0 such that φj ≤ Mj for all φ ∈ W , where · j is defined in (10.5). Proof. Assume by contradiction that W is not contained in DK (Ω) for any compact set K ⊂ Ω. Let {Kn }n be an increasing ∞ sequence of compact sets of Ω such that dist(Kn , ∂Kn+1 ) > 0 and Ω = n=1 Kn . Then for each n ∈ N
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we may find a function φn ∈ W and a point xn ∈ Kn+1 \ Kn such that φn (xn ) = 0. Define U := {φ ∈ D(Ω) : |φ(xn )| < |φn (xn )|/n for all n ∈ N}. Since each compact set K ⊂ Ω contains only finitely many xn , by the previous exercise we have that U ∩DK (Ω) ∈ τK , and so U ∈ τ . Using the fact that the set W is topologically bounded, we may find t > 0 such that W ⊆ tU . Consider an integer n ≥ t. Then φn (xn ) = 0 and |φn (xn )|/t ≥ |φn (xn )|/n, / U , or, equivalently, that φn ∈ / tU , which is a which implies that t−1 φn ∈ contradiction. This shows that W ⊆ DK (Ω) for some compact set K ⊂ Ω. To prove the final part of the statement, note that by Theorem 10.5 the set W = W ∩ DK (Ω) is bounded with respect to the topology τK , and so, by Corollary A.28, for each j ∈ N0 the set {φj : φ ∈ W } is bounded in R. We are now ready to characterize convergence with respect to the topology τ and to prove the completeness of D(Ω) (recall Exercise 10.3). Theorem 10.8. Let Ω ⊆ RN be an open set. Then the space D(Ω) is complete. Moreover, a sequence {φn }n in D(Ω) converges to φ ∈ D(Ω) with respect to τ if and only if (i) there exists a compact set K ⊂ Ω such that the support of every φn is contained in K, (ii) limn→∞ ∂ α φn = ∂ α φ uniformly on K for every multi-index α. Proof. Let {φn }n be a Cauchy sequence in D(Ω). By Proposition A.22, the set {φn : n ∈ N} is topologically bounded. Hence, we may apply Theorem 10.7 to find a compact set K ⊂ Ω such that {φn : n ∈ N} is contained in DK (Ω). In turn, by Theorem 10.5, we have that {φn }n is a Cauchy sequence in DK (Ω). Hence, for all integers j ∈ N0 and ∈ N there exists an integer n ∈ N such that φn − φk ∈ VK,j, = {φ ∈ DK (Ω) : φK,j < 1/} for all k, n ≥ n; that is, for every multi-index α, with |α| ≤ j, we have that sup |∂ α φn (x) − ∂ α φk (x)| ≤ 1/.
x∈K α {∂ φn }n
is a Cauchy sequence in the space of continuous This implies that bounded functions, and so it converges uniformly in Ω to a function ψα with support in K. In particular, taking α = 0, we have that φn converges uniformly in Ω to a function ψ0 with support in K and ∂ α ψ0 = ψα for every multi-index α, with |α| ≤ j (why?). Given the arbitrariness of j ∈ N0 , we conclude that ψ0 ∈ DK (Ω) and that the sequence {φn }n converges to ψ0 ∈ D(Ω) with respect to τ . Thus, D(Ω) is complete.
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287
The proof of the second part of the statement is left as an exercise.
Exercise 10.9. Let Ω ⊆ RN be an open set. (i) Prove that for every compact K ⊂ Ω, the space DK (Ω) is closed and has empty interior in D(Ω). (ii) Prove that D(Ω) is not metrizable. In the previous exercise we have proved that D(Ω) is not a metrizable space. Despite this fact, we can still prove that linear functionals defined on D(Ω) are continuous if and only if they are sequentially continuous. Precisely, the following result holds. Theorem 10.10. Let Ω ⊆ RN be an open set and let T : D(Ω) → R be linear. Then the following properties are equivalent: (i) T is continuous. (ii) T is bounded. (iii) If {φn }n in D(Ω) converges to φ ∈ D(Ω) with respect to τ , then lim T (φn ) = T (φ).
n→∞
(iv) The restriction of T to DK (Ω) is continuous for every compact set K ⊂ Ω. (v) For every compact set K ⊂ Ω there exist an integer j ∈ N0 and a constant cK > 0 such that (10.9)
|T (φ)| ≤ cK φK,j
for all φ ∈ DK (Ω).
Proof. (i)⇒(ii) If T is continuous, then T is bounded by Theorem A.31. (ii)⇒(iii) Assume that T is bounded and let {φn }n in D(Ω) converge to φ ∈ D(Ω) with respect to τ . Since D(Ω) is a topological vector space, by replacing {φn } with {φn − φ}, without loss of generality, we may assume that {φn }n converges to 0 with respect to τ . By Proposition A.22, the set {φn : n ∈ N} is topologically bounded. Since T is bounded, it follows that the set {T (φn ) : n ∈ N} is bounded. By Exercise 10.1 and Theorem 10.8 there is a compact set K ⊂ Ω such that dK (φn , 0) → 0 as n → ∞. Since DK (Ω) is metrizable, it follows by Theorem A.31 that T (φn ) → 0. (iii)⇒(iv) Fix a compact set K ⊂ Ω and assume that {φn } is in DK (Ω) and such that dK (φn , 0) → 0 as n → ∞. By Theorem 10.8 we have that {φn } converges to φ with respect to τ . Hence, by property (iii), T (φn ) → 0 as n → ∞. Using Theorem A.31 once more, we get that the restriction of T to DK (Ω) is continuous.
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(iv)⇒(i) For every ε > 0 and for every compact set K ⊂ Ω the restriction of T to DK (Ω) is continuous at zero, and so T −1 ((−ε, ε)) ∩ DK (Ω) ∈ τK . Hence, T −1 ((−ε, ε)) ∈ τ , which shows that T is continuous at zero and, by linearity, everywhere. (iv)⇔(v) Assume that (iv) holds and fix a compact set K ⊂ Ω. Since T restricted to DK (Ω) is continuous at the origin, given ε = 1 there exist j ∈ N0 and ∈ N such that VK,j, ⊆ T −1 ((−1, 1)), that is, |T (φ)| ≤ 1 for all φ ∈ DK (Ω) with φK,j < 1/. If φ ∈ DK (Ω) and φ = 0, then φK,j = 0 and 1 φ 1 < . 2 φK,j K,j By the linearity of T it follows that |T (φ)| ≤ 2φK,j , which gives (iv). Conversely, if (v) holds, then by taking sufficiently large, we have that |T (φ)| ≤ ε for all φ ∈ VK,j, , which shows the continuity of T restricted to DK (Ω). The dual of D(Ω) is denoted D (Ω) and its elements are called distributions. We often use the duality notation T, φ to denote T (φ). The space D (Ω) is given the weak star topology (see Section A.5 in Appendix A), so that a sequence {Tn }n in D (Ω) converges to T ∈ D (Ω) if Tn (φ) → T (φ) for every φ ∈ D(Ω). In this case we say that {Tn }n converges to T in the sense of distributions. Exercise 10.11. Let Ω ⊆ RN be an open set and let T ∈ D (Ω). (i) Prove that if φ, ψ ∈ C ∞ (Ω), then for every multi-index β the Leibnitz formula cαβ ∂ β−α φ∂ α ψ ∂ α (φψ) = α≤β
holds for some cαβ ∈ R. (ii) Prove that if ψ ∈ C ∞ (Ω), then the linear functional ψT : D(Ω) → R, defined by (ψT )(φ) := T (ψφ),
φ ∈ D(Ω),
is a distribution.
10.2. Order of a Distribution In this section we define the order of a distribution. We begin by observing that the integer j ∈ N0 in (10.9) may change with the compact set K ⊂ Ω. If the same integer will do for all compact sets K ⊂ Ω, then the smallest
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289
integer j ∈ N0 for which (10.9) holds for all compact sets K ⊂ Ω is called the order of the distribution T . If no such integer exists, then the distribution T is said to have infinite order . Example 10.12. Let Ω ⊆ RN be an open set. (i) Let λ be a signed Radon measure on Ω. The functional φ dλ, φ ∈ D(Ω), Tλ (φ) := Ω
is a distribution of order zero. (ii) Fix x0 ∈ Ω. The functional δx0 , defined by δx0 (φ) := φ(x0 ), φ ∈ D(Ω), is a distribution and is called the delta Dirac with mass at x0 . The distribution δx0 has order zero. (iii) Let u ∈ L1loc (Ω). The functional φ(x)u(x) dx, Tu (φ) :=
φ ∈ D(Ω),
Ω
is a distribution of order zero. Actually, it turns out that all distributions of order zero may be identified with measures. Theorem 10.13. Let Ω ⊆ RN be an open set and let T ∈ D (Ω). (i) If T is positive, that is, if T (φ) ≥ 0 for all nonnegative functions φ ∈ D(Ω), then there exists a unique Radon measure μ : B(Ω) → [0, ∞] such that φ dμ for all φ ∈ D(Ω). T (φ) = Ω
(ii) If T has order zero, then there exist two Radon measure μ1 , μ1 : B(Ω) → [0, ∞] such that φ dμ1 − φ dμ2 for all φ ∈ D(Ω). T (φ) = Ω
Ω
Proof. (i) We claim that T has order zero. Fix a compact set K ⊂ Ω and find an open set U such that K ⊂ U Ω. Construct a smooth cut-off function ϕ ∈ C ∞ (Ω) such that ϕ ≡ 1 on K, supp ϕ ⊂ U , and 0 ≤ ϕ ≤ 1 (see Exercise C.23). In particular, ϕ ∈ D(Ω). Since ϕ ≡ 1 on K and ϕ ≥ 0, for every φ ∈ DK (Ω) we have that |φ(x)| ≤ φK,0 ϕ(x) for all x ∈ Ω, and so T (φK,0 ϕ − φ) ≥ 0,
T (φ + φK,0 ϕ) ≥ 0;
that is, by the linearity of T , |T (φ)| ≤ φK,0 T (ϕ) which shows that T has order zero.
for all φ ∈ DK (Ω),
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Let {Ωi }i be an increasing sequence of bounded open sets of Ω such that ∞ Ωi Ωi+1 and Ω = i=1 Ωi . We start by showing that for every fixed i ∈ N the distribution T can be extended in a unique way as a linear continuous map on Cc (Ωi ). Since Ki := Ωi is a compact set contained in Ω and T has order zero, by what we just proved there exists a constant ci > 0 such that (10.10)
|T (φ)| ≤ ci φKi ,0
for all φ ∈ DKi (Ω). If φ ∈ Cc (Ωi ), then dist(supp φ, ∂Ωi ) > 0, and thus if we consider φn := ϕ1/n ∗ φ, where ϕ1/n are standard mollifiers (with ε := 1/n) and 1/n < dist(supp φ, ∂Ωi ), we have that {φn }n is contained in DKi (Ω) and φn → φ uniformly on Ki . It follows by (10.10) that |T (φn − φl )| ≤ ci φn − φl Ki ,0 → 0 as l, n → ∞. Hence, {T (φn )}n is a Cauchy sequence, and therefore it converges to a limit that we denote by Ti (φ). Moreover, if φ ≥ 0, then φn ≥ 0 also, and so Ti (φ) ≥ 0. Note that, again by (10.10), Ti (φ) is independent of the choice of the approximating sequence {φn }n . By the linearity of T it follows that Ti : Cc (Ωi ) → R is linear and positive, while by (10.10) we have that |Ti (φ)| ≤ ci φCc (Ωi ) for all φ ∈ Cc (Ωi ). Since Ωi ⊂ Ωi+1 , it follows that Ti+1 (φ) = Ti (φ) for all φ ∈ Cc (Ωi ). Thus {Ti }i defines a unique linear positive extension of T to the union of all Cc (Ωi ), which coincides with Cc (Ω). The result now follows from the Riesz representation theorem in Cc (Ω) for positive linear functionals (see Theorem B.113). (ii) The second part of the proof of (i) continues to hold in this case. By (10.10) and the fact that {Ωi }i covers Ω, we have that the extended functional is locally bounded, and so the result now follows from the Riesz representation theorem in Cc (Ω) for locally bounded linear functionals (see Theorem B.113).
10.3. Derivatives of Distributions and Distributions as Derivatives We now define the notion of a derivative of a distribution. Definition 10.14. Let Ω ⊆ RN be an open set and let T ∈ D (Ω). Given a multi-index α ∈ NN 0 \ {0}, we define the αth derivative of T as |α| ∂ |α| T |α| ∂ φ (φ) := (−1) , φ ∈ D(Ω). ∂xα ∂xα For j ∈ N the symbol ∇j T stands for the collection of all αth distributional derivatives of T with |α| = j. For simplicity we often write ∂ α for
∂ |α| ∂xα
and ∂in for
∂n . ∂xn i
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291
Remark 10.15. It can be verified that ∂ α T is still a distribution. Indeed, let K ⊂ Ω be a compact set. By Theorem 10.10 there exist an integer j ∈ N0 and a constant cK > 0 such that |T (φ)| ≤ cK φK,j for all φ ∈ DK (Ω). It follows that |∂ α T (φ)| = |T (∂ α φ)| ≤ cK ∂ α φK,j ≤ cK φK,j+|α| for all φ ∈ DK (Ω), which shows that ∂ α T ∈ D (Ω), again by Theorem 10.10. In particular, if u ∈ L1loc (Ω) and α is a multi-index, then the αth weak, or distributional, derivative of u is the distribution ∂ α Tu . In a similar way we may define directional derivatives of a distribution (see Section 9.1). Definition 10.16. Let Ω ⊆ RN be an open set and let T ∈ D (Ω). Given a unit vector ν ∈ RN \ {0} and n ∈ N, we define the nth directional derivative of T in the direction ν as ∂nφ ∂ nT n (φ) := (−1) T , φ ∈ D(Ω). ∂ν n ∂ν n In particular, if u ∈ L1loc (Ω), ν ∈ RN \ {0} and n ∈ N, then the nth weak, or distributional, directional derivative of u in the direction ν is the n distribution ∂∂νTnu . Example 10.17. Let Ω ⊆ RN be an open set and let u ∈ L1loc (Ω). The Laplacian of u in the sense of distribution is the distribution TΔu := N 2 i=1 ∂i Tu , that is, TΔu (φ) =
=
N
∂i2 Tu (φ) =
i=1 N i=1
N
Tu (∂i2 φ)
i=1
u∂i2 φ dx Ω
=
uΔφ dx Ω
for all φ ∈ D(Ω). So a function u ∈ L1loc (Ω) is subharmonic if “Δu ≥ 0”, that is, uΔφ dx ≥ 0 TΔu (φ) = Ω
for all φ ∈ D(Ω) with φ ≥ 0. By Theorem 10.13 there exists a unique (positive) Radon measure μ : B(Ω) → [0, ∞] such that uΔφ dx = φ dμ for all φ ∈ D(Ω). TΔu (φ) = Ω
Ω
Thus, with an abuse of notation, we may write Δu = μ.
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Similarly, for u ∈ L1loc (Ω; RN ) one can define the divergence of u in the sense of distributions. Definition 10.18. Let Ω ⊆ RN be an open set, let u ∈ L1loc (Ω), and let 1 α ∈ NN 0 \ {0} be a multi-index. If there exists a function vα ∈ Lloc (Ω) such that Tvα (φ) = ∂ α Tu (φ) for all φ ∈ D(Ω), then vα is called the αth weak, or distributional, derivative of Tu . We write |α| ∂ α u = ∂∂xαu := vα . Thus, a function vα ∈ L1loc (Ω) is the αth weak derivative of u ∈ L1loc (Ω) if
φvα dx = (−1)
(10.11) Ω
|α|
u∂ α φ dx Ω
for all φ ∈ Cc∞ (Ω). Note that this is consistent with the integration by parts (9.55). Exercise 10.19. Let u : R → R be defined as follows: ⎧ ⎪ ⎨ cos xx if − π ≤ x ≤ 0, if 0 < x ≤ π, 1− u(x) := π ⎪ ⎩ 0 otherwise. (i) Calculate the derivative of u in the sense of distributions and find its order. (ii) Let v be the restriction of u in the interval (−π, π). Calculate the first and second derivative of v in the sense of distributions in (−π, π) and find their orders. (iii) Let w : R → R be differentiable in R \ {a} and assume that there exist lim w(x) ∈ R. lim w(x) ∈ R, x→a−
x→a+
Calculate the derivative of w in the sense of distributions. What can you say about its order? Under what conditions can you conclude that it has order zero? Exercise 10.20. Assume that u ∈ L1 ((−∞, δ) ∪ (δ, ∞)) for every δ > 0 and define the principal value integral −δ ∞ ∞ u(x) dx := lim u(x) dx + u(x) dx PV −∞
δ→0+
δ
whenever the limit exists. For φ ∈ D(R) define ∞ φ(x) log |x| dx. T (φ) := −∞
−∞
10.3. Derivatives of Distributions and Distributions as Derivatives
Prove that
∞
1 T (φ) = PV φ(x) dx, x −∞
T (φ) = − PV
∞ −∞
293
φ(x) − φ(0) dx. x2
Exercise 10.21. Let Ω ⊆ RN be an open set and let Tn ∈ D (Ω) be such that the limit T (φ) := lim Tn (φ) n→∞
exists in R for every φ ∈ D(Ω). (i) Prove that T ∈ D (Ω). (ii) Prove that for every multi-index α and for every φ ∈ D(Ω), ∂ α T (φ) = lim ∂ α Tn (φ). n→∞
In the next few theorems we characterize distributions as weak derivatives of continuous functions. Theorem 10.22. Let Ω ⊆ RN be an open set and let T ∈ D (Ω). Then for every compact set K ⊂ Ω there exist a continuous function u : Ω → R and a multi-index α (both depending on K) such that u∂ α φ dx for all φ ∈ DK (Ω). T (φ) = (−1)|α| Ω
Proof. In what follows, we use the notation (E.2) in Appendix E. Without loss of generality we may assume that K ⊆ Q := [0, 1]N . By the mean value theorem, for every ψ ∈ DQ (RN ), for all i = 1, . . . , N , and for all x ∈ Q we have |ψ(x)| = |ψ(xi , xi ) − ψ(xi , 0)|
(10.12)
= |∂i ψ(xi , t)(xi − 0)| ≤ max |∂i ψ|. Q
For x ∈ Q set Q(x) := {y ∈ Q : 0 ≤ yi ≤ xi , i = 1, . . . , N }. Then for ψ ∈ DQ (RN ) and x ∈ Q we have x1 ∂1 ψ(y1 , x2 , . . . , xN ) dy1 ψ(x) = ψ(0, x2 , . . . , xN ) + 0 x1 = ∂1 ψ(y1 , x2 , . . . , xN ) dy1 0 x1 x2 2 = ∂1,2 ψ(y1 , y2 , x3 , . . . , xN ) dy2 dy1 ∂1 ψ(y1 , 0, x3 , . . . , xN ) + 0 0 x1 x2 2 = ∂1,2 ψ(y1 , y2 , x3 , . . . , xN ) dy2 dy1 = ∂ β ψ(y) dy, 0
0
Q(x)
294
10. Distributions
where β := (1, . . . , 1), and so (10.13) ψ(x) =
for all x ∈ Q.
∂ β ψ(y) dy
Q(x)
Fix an integer j ∈ N0 (to be determined later) and let α be a multi-index with |α| ≤ j. By repeated applications of (10.12) we obtain that α jβ |∂ (j+1)β ψ(y)| dy, (10.14) max |∂ ψ| ≤ max |∂ ψ| ≤ Q
Q
Q
where jβ = (j, . . . , j) and where in the last inequality we have used (10.13). By (10.13) the linear operator D β : DK (Ω) → DK (Ω) φ → ∂ β φ is one-to-one, and hence so is the linear operator L := D jβ : DK (Ω) → DK (Ω) φ → ∂ (j+1)β φ, since D jβ = D β ◦ · · · ◦ D β , where the composition is done j times. Let Y := L(DK (Ω)) and define the linear functional T1 : Y → R as follows. Given ψ ∈ Y , there exists a unique φ ∈ DK (Ω) such that ψ = L(φ) = ∂ (j+1)β φ. Define T1 (ψ) := T (φ). Since T ∈ D (Ω), by Theorem 10.10 there exist an integer j ∈ N0 and a constant cK > 0 such that for all φ ∈ DK (Ω), |∂ (j+1)β φ| dx, |T (φ)| ≤ cK φK,j ≤ cK K
where in the last inequality we have used (10.14). It follows from the definition of T1 that |ψ| dx |T1 (ψ)| ≤ cK K
for all ψ ∈ Y , and thus we may apply the Hahn–Banach theorem (see Theorem A.32) to extend T1 as a continuous linear functional defined in L1 (K). By the Riesz representation theorem in L1 (K) (see Theorem B.93) there exists a function v ∈ L∞ (K) such that vψ dx T1 (ψ) = K
for all ψ ∈
In particular, if φ ∈ DK (Ω), then v∂ (j+1)β φ dx. T (φ) = T1 (∂ (j+1)β φ) =
L1 (K).
K
10.3. Derivatives of Distributions and Distributions as Derivatives
295
Extend v by zero outside K and define x1 xN u(x) := ··· v(y) dyN · · · dy1 . −∞
−∞
Then u is continuous and by integrating by parts N times we have that for all φ ∈ DK (Ω), u∂ (j+2)β φ dx. T (φ) = (−1)N K
To complete the proof, we may define α := (j + 2)β and, if needed, replace u with −u. The previous result is local in the sense that the function u and the multi-index α change with K. Next we show that if T has compact support, then the previous local result becomes global. Definition 10.23 (Support of a distribution). Let Ω ⊆ RN be an open set and let T ∈ D (Ω). If U ⊆ Ω is open, then we write that T = 0 in U if T (φ) = 0 for all φ ∈ D(U ). The support of T is the complement of V relative to Ω, where V is the union of all open subsets U ⊆ Ω in which T = 0. The support of T will be written as supp T . Exercise 10.24. Let Ω ⊆ RN be an open set and let T ∈ D (Ω). Show that T = 0 in the (possibly empty) open set Ω \ supp T . Prove also that ∂ α T = 0 in Ω \ supp T for every multi-index α. We now prove a global version of Theorem 10.22 for distributions with compact support. Theorem 10.25. Let Ω ⊆ RN be an open set and let T ∈ D (Ω) be such that supp T is a compact set of Ω. Then (i) there exist an integer j ∈ N0 and a constant c > 0 such that |T (φ)| ≤ cφj for all φ ∈ D(Ω) (in particular, T has finite order ≤ j), where · j is defined in (10.5), (ii) if U is an open set, with supp T ⊂ U ⊆ Ω, then for each multiindex α, with α ≤ β := ( + 2, . . . , + 2), there exists a function vα ∈ C(Ω), with supp vα ⊂ U , such that α |α| ∂ Tvα (φ) = (−1) vα ∂ α φ dx for all φ ∈ D(Ω). T (φ) = α≤β
α
Ω
Proof. (i) Consider an open set U , with supp T ⊂ U Ω, and (see Exercise C.23) construct a function ψ ∈ D(Ω) such that ψ = 1 on U . We claim that ψT = T . Indeed, if φ ∈ D(Ω), then since ψ = 1 on U , we have that φ − ψφ = 0 on U , and so supp(φ − ψφ) ∩ supp T = ∅, which implies that
296
10. Distributions
T (φ−ψφ) = 0, or, equivalently, T (φ) = T (ψφ) = (ψT )(φ). Hence, the claim holds. Since T ∈ D (Ω), by Theorem 10.10 there exist an integer j ∈ N0 and a constant cK > 0 such that |T (φ)| ≤ cK φK,j for all φ ∈ DK (Ω), where K := supp ψ. On the other hand, if φ ∈ D(Ω), then ψφ ∈ DK (Ω), and so, since ψT = T , |T (φ)| = |T (ψφ)| ≤ cK ψφK,j . By the Leibnitz formula (see Exercise 10.11), ψφK,j ≤ cψ φK,j , and so |T (φ)| ≤ cK cψ φK,j , which shows part (i). (ii) Consider an open set W , with supp T ⊂ W U . By Theorem 10.22 with W in place of K there exists a continuous function u : Ω → R such that |β| u∂ β φ dx T (φ) = (−1) Ω
for all φ ∈ DW (Ω), where β := ( + 2, . . . , + 2). Consider an open set V , with supp T ⊂ V W , and as in (i) construct a function ψ ∈ D(Ω) such that ψ = 1 on V and K := supp ψ ⊂ W . Then, by (i), for all φ ∈ D(Ω) we have that ψφ ∈ DW (Ω) and, by the Leibnitz formula (see Exercise 10.11) again, u∂ β (ψφ) dx T (φ) = T (ψφ) = (−1)|β| Ω cαβ (−1)|β| u∂ β−α ψ∂ α φ dx, = Ω
α≤β
and so it suffices to take vα := cαβ (−1)|β−α| u∂ β−α ψ. This concludes the proof. Finally, using partitions of unity, we have a similar representation for every distribution. Theorem 10.26. Let Ω ⊆ RN be an open set and let T ∈ D (Ω). Then for each multi-index α there exists a function vα ∈ C(Ω) such that (i) each compact set K ⊂ Ω intersects the support of finitely many vα , (ii) for all φ ∈ D(Ω), T (φ) =
α
α
∂ Tvα (φ) =
α
(−1)
|α|
vα ∂ α φ dx. Ω
If T has finite order, then only finitely many vα different from zero are needed.
10.3. Derivatives of Distributions and Distributions as Derivatives
297
Proof. (i) Let S be a countable dense set in Ω, e.g., S := {x ∈ QN ∩ Ω}, and consider the countable family F of open cubes F := {Q(x, r) : r ∈ Q, 0 < r < 1, x ∈ S, Q(x, r) ⊆ Ω}. Since F is countable, we may write F = {Q(xn , rn )}n∈N . By the density of S and of the rational numbers we have that Ω = ∞ n=1 Q(xn , rn /2). Note that every compact set K ⊂ Ω intersects only finitely many cubes Q(xn , rn ). For each n ∈ N construct a function φn ∈ D(Ω) such that φn = 1 on Q(xn , rn /2) and supp φn ⊂ Q(xn , rn ). Use this family to construct a partition of unity {ψn }n in D(Ω) with supp ψn ⊂ Q(xn , rn ) for each n ∈ N (see the proof of Theorem C.21). For each n ∈ N the distribution ψn T has support contained in Q(xn , rn ), and so by Theorem 10.25 it has finite order n and we may find finitely many functions {vα,n } in C(Ω), with α ≤ βn := (n + 2, . . . , n + 2) and supp vα,n ⊂ Q(xn , rn ), such that (−1)|α| vα,n ∂ α φ dx (ψn T )(φ) = Ω
α≤βn
for all φ ∈ D(Ω). For every multi-index α for which α ≤ βn fails define vα,n :≡ 0. Thus, |α| (−1) vα,n ∂ α φ dx (ψn T )(φ) = Ω
α
for all φ ∈ D(Ω).
Hence, for every multi-index α we may define vα := ∞ n=1 vα,n . Since each compact set intersects the support of only finitely many vα,n , the function vα is continuous and (i) holds. To prove (ii), let φ ∈ D(Ω). Since {ψn }n is a partition of unity, we have that φ = ∞ n=1 ψn φ, where the sum is actually finite since φ has compact support. Since the sum is finite and T is linear, we have that ∞ ∞ ∞ |α| T (ψn φ) = (ψn T )(φ) = (−1) vα,n ∂ α φ dx T (φ) = n=1
=
α
(−1)|α|
n=1
∞
Ω n=1
n=1 α
vα,n ∂ α φ dx =
α
(−1)|α|
Ω
vα ∂ α φ dx. Ω
This shows (ii). Finally, if T has finite order , then ψn T has finite order n ≤ , and so it suffices to consider only multi-indices α ≤ β := ( + 2, . . . , + 2). Exercise 10.27. Let Ω ⊆ RN be an open set. Prove that every continuous linear functional on C ∞ (Ω) is of the form φ → T (φ), where T is a distribution with compact support.
298
10. Distributions
Exercise 10.28. Let Ω = (a1 , b1 ) × · · · × (aN , bN ). (i) Prove that φ ∈ D(Ω) is such that Ω φ(x) dx = 0 if and only if φ= N i=1 ∂i φi for some φ1 , . . . , φN ∈ D(Ω). Hint: Use induction on N and look at Step 1 of the proof of Lemma 7.4. ∂T = 0 for all i = 1, . . . , N , (ii) Prove that if T ∈ D (Ω) is such that ∂x i then there exists a constant c ∈ R such that T (φ) = c Ω φ(x) dx for all φ ∈ D(Ω), i.e., T is constant.
(iii) Prove that if Ω ⊆ RN is an open connected set and T ∈ D (Ω) is such that ∂i T = 0 for all i = 1, . . . , N , then T is constant.
10.4. Rapidly Decreasing Functions and Tempered Distributions In view of the applications to Fourier transforms in this section we consider complex-valued functions. We recall that for a complex number z = x + iy, where x, y ∈ R, the complex conjugate of z is the number z := x − iy and the norm of z is # z := x2 + y 2 . Definition 10.29. The space of rapidly decreasing functions S(RN ; C) is the space of all functions φ : RN → C of class C ∞ such that for all multiindices α, β ∈ NN 0 , (10.15)
φα,β := sup xα ∂ β φ(x) < ∞. x∈RN
Thus, S(RN ; C) consists of all functions that, together with all their derivatives, decay to zero faster than any polynomial. Remark 10.30. The space Cc∞ (RN ; C) of all C ∞ functions φ : RN → C with compact support is contained in S(RN ; C). The function φ(x) := 2 e− x , x ∈ RN , is an example of a function in S(RN ; C) without compact support. Theorem 10.31. The space S(RN ; C) with the the topology induced by the family of seminorms · α,β is a Fr´echet space. Proof. Since there are countably many seminorms ·α,β , the space S(RN ; C) is metrizable, with metric ∞ 1 φ − ψαn ,βn . d(φ, ψ) := 2n 1 + φ − ψαn ,βn n=1
It remains to show that it is complete. Let {φn }n be a Cauchy sequence in S(RN ; C). Then {xα ∂ β φn }n is a Cauchy sequence in Cb (RN ; C) for every α, β ∈ N+ 0 and thus it converges uniformly to a function ψα,β . Let φ := ψ0,0 .
10.4. Rapidly Decreasing Functions and Tempered Distributions
299
By the fundamental theorem of calculus for x ∈ RN , t ∈ R and i = 1, . . . , N, t ∂φn φn (x + tei ) = φn (x) + (x + sei ) ds. 0 ∂xi Letting n → ∞ it follows by uniform convergence that t ψ0,ei (x + sei ) ds. φ(x + tei ) = φ(x) + 0 ∂φ ∂xi
= ψ0,ei . This proves that φ is of class C 1 . In a Hence, there exists similar way we can show that φ is of class C ∞ with ψα,β = xα ∂ β φ. Thus, φ ∈ S(RN ; C) and since φn − φα,β → 0 for all α, β ∈ N+ 0 , it follows that N S(R ; C) is complete. Definition 10.32. The dual of S(RN ; C) is called the space of tempered distributions and is denoted S (RN ; C). The following theorem is important for applications. For φ ∈ S(RN ; C) and m, n ∈ N0 we define φα,β . φm,n := |α|=m |β|=n
Theorem 10.33. A linear functional T : S(RN ; C) → C is continuous if and only if there exist a constant c > 0 and m, n ∈ N0 such that (10.16)
T (φ) ≤ cφm,n
for every φ ∈ S(RN ; C).
Proof. Exercise.
In view of the previous theorem, the restriction of a tempered distribution T : S(RN ; C) → C to D(RN ; C) defines a distribution. Corollary 10.34. Given T ∈ S (RN ; C), its restriction to D(RN ; C) is a distribution. Proof. By the previous theorem there exist a constant c > 0 and m, n ∈ N0 such that T (φ) ≤ cφm,n for every φ ∈ S(RN ; C). Given a compact set K ⊂ RN , let R > 0 be such that K ⊆ B(0, R). If φ ∈ DK (RN ; C), then supx∈RN xα ∂ β φ(x) ≤ (1 + Rm ) supx∈K ∂ β φ(x) and so |T (φ)| ≤ (1 + Rm )φK,n for all φ ∈ DK (RN ; C), where · K,n is defined in (10.1). It follows from Theorem 10.10 that T : D(RN ; C) → C is continuous. Next we show that S(RN ; C) is embedded in Lp for every p. Theorem 10.35. For all 1 ≤ p ≤ ∞, S(RN ; C) → Lp (RN ; C) → S (RN ; C).
300
10. Distributions
Proof. We only need to consider the case 1 ≤ p < ∞. For p = 1, write
RN
1 + xN +1 φ(x) dx N +1 RN 1 + x 1 ≤ cφN +1,0 dx. N +1 RN 1 + x
φ(x) dx =
For 1 < p < ∞ it is enough to observe that p p−1 φ(x) dx ≤ φ∞ φ(x) dx ≤ cφpN +1,0 . RN
RN
This shows that S(RN ; C) → Lp (RN ; C). Given ψ ∈ Lp (RN ; C), consider the linear functional Tψ : S(RN ; C) → C defined by (10.17)
Tψ (φ) :=
φψ dx, RN
φ ∈ S(RN ; C).
Then by H¨ older’s inequality Tψ (φ) ≤ φLp ψLp ≤ cφN +1,0 ψLp . Hence, by (10.16) the functional Tψ belongs to S (RN ; C) and the linear mapping ψ ∈ Lp (RN ; C) → Tψ is an embedding. Remark 10.36. In what follows we identify ψ with Tψ . Hence, Lp (RN ; C), and in particular S(RN ; C), can be thought as contained in S (RN ; C). Remark 10.37. Note that if f : RN → C is a polynomial, then Tf (φ) :=
φf dx, RN
φ ∈ S (RN ; C),
is well-defined and Tf (φ) ≤ cf φd+N +1,0 , where d is the degree of f and cf is the supremum of the norms of the coefficients of f . By (10.16) the functional Tf belongs to S (RN ; C) Example 10.38. Given a measure μ : B(RN ) → [0, ∞] with the property that μ(B(0, r)) ≤ c0 (1 + r)k for some c0 > 0, some k ∈ N, and for all r > 0, the linear functional Tμ : S(RN ; C) → C defined by Tμ (φ) := RN φ dμ is well-defined and continuous.
10.4. Rapidly Decreasing Functions and Tempered Distributions
Indeed, write RN
φ dμ =
φ dμ + B(0,1)
≤ 2c0 φ∞ +
∞
301
φ dμ
n=2 B(0,n)\B(0,n−1)
∞
n=2 B(0,n)
≤ 2c0 φ∞ + cc0 φ0,2k
(1 + x)2k φ dμ (1 + x)2k ∞ (1 + n)k . (1 + n)2k
n=1
Hence by (10.16), Tμ ∈
S (RN ; C).
Exercise 10.39. Prove that the linear mapping 1 φ(x) (φ) := lim dx, T (φ) := PV + x δ→0 R\[−δ,δ] x
φ ∈ S(R),
is well-defined and belongs to S (R; C). Exercise 10.40. Let ψ : RN → C be a function of class C ∞ such that for every multi-index α ∈ NN 0 there exist cα and nα ∈ N such that (10.18)
∂ α ψ(x) ≤ cα (1 + x2 )nα
for all x ∈ RN . (i) Prove that if φ ∈ S(RN ; C) then φψ ∈ S(RN ; C). (ii) Prove that if h : RN → C is a Lebesgue measurable function such that hφ ∈ S(RN ; C) for all φ ∈ S(RN ; C) and the mapping φ ∈ S(RN ; C) → hφ is continuous, then h must be of class C ∞ and satisfy (10.18). (iii) Given T ∈ S (RN ; C) prove that the linear functional ψT : S(RN ; C) → C defined by (ψT )(φ) := T (φψ), φ ∈ S(RN ; C), belongs to S (RN ; C). As in Definition 10.18 we can define the notion of a derivative of a tempered distribution. Definition 10.41. Given T ∈ S (RN ; C) and a multi-index α ∈ NN 0 \{0}, we α N define the αth derivative of T as the linear functional ∂ T : S(R ; C) → C defined by ∂ |α| φ ∂ |α| T |α| (φ) := (−1) T , φ ∈ S(RN ; C). ∂xα ∂xα For simplicity we often write ∂ α for
∂ |α| ∂xα
and ∂in for
∂n . ∂xn i
Theorem 10.42. For every T ∈ S (RN ; C) and every multi-index α, the functional ∂ α T belongs to S (RN ; C).
302
10. Distributions
Proof. Since T ∈ S (RN ; C), by Theorem 10.33 there exist a constant c > 0 and m, n ∈ N0 such that T (φ) ≤ cφm,n for every φ ∈ S(RN ; C). In turn, since for φ ∈ S(RN ; C), ∂ α φ still belongs to S(RN ; C), with (∂ α T )(φ) = T (∂ α φ) ≤ c∂ α φm,n ≤ cφm,n+|α| , and so, again by Theorem 10.33 it follows that ∂ α T belongs to S (RN ; C). Exercise 10.43. Prove that if T ∈ D (RN ; C) has compact support, then T ∈ S (RN ; C). Exercise 10.44. Prove that D(RN ; C) is dense in S(RN ; C) and that the inclusion D(RN ; C) → S(RN ; C) T → T is continuous. Exercise 10.45. Prove that if f is a polynomial, φ ∈ S(RN ; C), and T ∈ S (RN ; C), then f T and φT belong to S (RN ; C).
10.5. Convolutions Given two measurable functions φ : RN → C and ψ : RN → C, the convolution of φ and ψ is the function φ ∗ ψ defined by (10.19) (φ ∗ ψ)(x) := φ(x − y)ψ(y) dy RN
for all x ∈
RN
for which the right-hand side is well-defined.
Theorem 10.46. Given φ, ψ ∈ S(RN ; C), the function φ ∗ ψ belongs to S(RN ; C). Proof. Fix x ∈ RN . For m ∈ N with m > N , we can write φ(x − y)ψ(y) dy (φ ∗ ψ)(x) ≤ RN 1 1 ≤ cψ0,m φ0,m dy. m (1 + y) (1 + x − y)m RN We now split RN into the sets E := {y ∈ RN : 12 x ≤ x − y} and RN \ E. Then we have 1 1 dy m m E (1 + y) (1 + x − y) c 1 2m dy ≤ , ≤ m m (2 + x) (2 + x)m RN (1 + y)
10.5. Convolutions
303
while in RN \ E, y ≥ x − x − y ≥ x − 12 x = 12 x, and so 1 1 dy m m RN \E (1 + y) (1 + x − y) 1 2m c ≤ dy ≤ , (2 + x)m RN (1 + x − y)m (2 + x)m where c = c(m, N ) > 0. Hence, (2 + x)m (φ ∗ ψ)(x) ≤ cψ0,m φ0,m . This shows that φ ∗ ψ decays to zero faster than any power of x. On the other hand, by differentiating under the integral sign, for every multi-index α, ∂ |α| φ ∂ |α| φ ∂ |α| (φ ∗ ψ) (x) = (x − y)ψ(y) dy = ∗ ψ (x), α ∂xα ∂xα RN ∂x |α|
and so by repeating the same calculations above with φ replaced by ∂∂xαφ , we get that all derivatives of φ ∗ ψ decay to zero faster than any power of x, which shows that φ ∗ ψ ∈ S(RN ; C). Exercise 10.47. Prove that for every φ, ψ, ϕ ∈ S(RN ; C), (φ ∗ ψ) ∗ ϕ = φ ∗ (ψ ∗ ϕ). Theorem 10.48 (Young’s inequality). Let φ ∈ Lp (RN ; C), 1 ≤ p ≤ ∞, and ψ ∈ L1 (RN ; C). Then (φ ∗ ψ)(x) exists for LN -a.e. x ∈ RN and φ ∗ ψLp (RN ) ≤ φLp (RN ) ψL1 (RN ) . Proof. Consider two Borel functions φ0 and ψ0 such that φ0 (x) = φ(x) and ψ0 (x) = ψ(x) for LN -a.e. x ∈ RN . Since the integral in (10.19) is unchanged if we replace φ and ψ with φ0 and ψ0 , respectively, in what follows, without loss of generality, we may assume that φ and ψ are Borel functions. Let h : RN × RN → C be the function defined by h(x, y) := φ(x − y), (x, y) ∈ RN × RN . Then h is a Borel function, since it is the composition of the Borel function φ with the continuous function ψ : RN × RN → RN given by ψ(x, y) := x − y. In turn, the function (x, y) ∈ RN × RN → φ(x − y)ψ(y) is Borel measurable. We are now in a position to apply Minkowski’s inequality for integrals (Corollary B.83) and Tonelli’s theorem to conclude
304
10. Distributions
that
φ(· − y)ψ(y) dy p φ ∗ ψLp = N L R ≤ φ(· − y)ψ(y)Lp dy N R ψ(y)φ(· − y)Lp dy = φLp = RN
RN
ψ(y) dy,
where in the last equality we have used the fact that the Lebesgue measure is translation invariant. Hence, φ∗ψ belongs to Lp (RN ; C), and so it is finite LN -a.e. in RN . The following is the generalized form of the previous inequality. Theorem 10.49 (Young’s inequality, general form). Let 1 ≤ p ≤ q ≤ ∞ and let φ ∈ Lp (RN ; C) and ψ ∈ Lq (RN ; C). Then (φ ∗ ψ)(x) exists for LN -a.e. x ∈ RN and φ ∗ ψLr (RN ) ≤ φLp (RN ) ψLq (RN ) , where (10.20)
1/p + 1/q = 1 + 1/r.
Proof. If r = ∞, then q is the H¨ older conjugate exponent of p and the result follows from H¨ older’s inequality and the translation invariance of the Lebesgue measure, while if p = 1, then r = q and the result follows from the previous theorem (with φ and ψ interchanged). In the remaining cases, write φ(x − y)ψ(y) = (φ(x − y)p ψ(y)q )1/r φ(x − y)(r−p)/r ψ(y)(r−q)/r . Define p1 := r, p2 := pr/(r − p), p3 := qr/(r − q). Then 1/p1 + 1/p2 + 1/p3 = 1, and so by the extended H¨older inequality (see Exercise B.79) and the translation invariance of the Lebesgue measure, φ(x − y)ψ(y) dy (φ ∗ ψ)(x) ≤ RN 1/r 1−p/r 1−q/r φ(x − y)p ψ(y)q dy φLp ψLq . ≤ RN r L (RN )
Taking the norm in φ ∗ ψLr ≤ RN
RN
on both sides, we get
φ(x − y)p ψ(y)q dydx
1/r
1−p/r
φLp
1−q/r
ψLq
.
Applying the previous theorem (with p = 1), we get that the right-hand side of the previous inequality is less than or equal to 1−p/r
(φp L1 ψq L1 )1/r φLp
1−q/r
ψLq
= φLp ψLq .
10.6. Convolution of Distributions
305
This concludes the proof.
10.6. Convolution of Distributions In this section we define the convolution of a distribution (or a tempered distribution) T and a function ϕ. We begin with the case in which T = Tu for some function u ∈ L1loc (RN ), where we recall that Tu ∈ D (RN ) is defined by Tu (φ) :=
φ ∈ D(RN ).
ψ(x)φ(x) dx, RN
By Fubini’s theorem (u ∗ ϕ)(x)φ(x) dx = N RN R = N R = N R = RN
φ(x) R
N
u(ξ) R
N
u(ξ) RN
u(x − y)ϕ(y) dydx ϕ(x − ξ)φ(x) dxdξ ϕ(ξ . − x)φ(x) dxdξ
u(ξ)(ϕ . ∗ φ)(ξ) dξ,
where ξ := x − y and ϕ(x) . := ϕ(−x). Hence, we have shown that Tu∗ϕ (φ) = . ∗ φ) for all φ ∈ D(RN ). Motivated by this formula we define: Tu (ϕ Definition 10.50. If T ∈ D (RN ) and ϕ ∈ D(RN ) the convolution of T and . ϕ is the linear functional T ∗ϕ : D(RN ) → R defined by (T ∗ϕ)(φ) := T (ϕ∗φ), where ϕ(x) . := ϕ(−x),
(10.21)
x ∈ RN .
It turns out that T ∗ ϕ can be identified with a function. Given x ∈ RN and a function ϕ : RN → R we define the function (10.22)
ϕx (y) := ϕ(x − y),
y ∈ RN .
Exercise 10.51. Let φ, ϕ ∈ D(RN ). For h > 0 define uh (x) := hN ϕ(x − hy)φ(hy), x ∈ RN . y∈ZN
Prove that {uh }h converges uniformly to ϕ ∗ φ as h → 0+ . Hint: Use Riemann sums. Theorem 10.52. Let T ∈ D (RN ) and φ, ψ ∈ D(RN ). Then T ∗ ϕ = Tuϕ , where uϕ is the function given by uϕ (x) := T (ϕx ), x ∈ RN . Moreover, (i) uϕ ∈ C ∞ (RN ), (ii) supp uϕ ⊆ supp T + supp ϕ,
306
10. Distributions
α α α (iii) for every multi-index α ∈ NN 0 , ∂ uϕ = T ∗ ∂ ϕ = ∂ T ∗ ϕ,
(iv) (T ∗ ϕ) ∗ ψ = T ∗ (ϕ ∗ ψ). Proof. Step 1: Fix φ ∈ D(RN ) . We approximate the function ϕ . ∗ φ (see (10.21)) with the Riemann sum ϕ(x . − hy)φ(hy), x ∈ RN , uh (x) := hN y∈ZN
where h > 0. Note that for every multi-index α, ∂ |α| ϕ . ∂ |α| uh N (x) = h (x − hy)φ(hy), α α ∂x ∂x N
x ∈ RN .
y∈Z
Since and
∂ |α| uh ∂xα
h
converges uniformly to
∂ |α| ϕ ∂xα
∗ φ by the previous exercise
∂ |α| ∂ |α| ϕ . ∗ φ = (ϕ . ∗ φ) ∂xα ∂xα . ∗ φ as h → 0 (why?), by Theorem 10.8 we have that {uh }h converges to ϕ with respect to the topology τ in D(RN ). Hence, by the continuity and linearity of T and by Theorem 10.10 we have that T (ϕ(· . − hy))φ(hy) (T ∗ ϕ)(φ) = T (ϕ . ∗ φ) = lim T (uh ) = lim hN h→0
= lim h h→0
N
h→0
y∈ZN
T (ϕ(hy − ·))φ(hy) =
y∈ZN
T (ϕy )φ(y) dy, RN
where we have used the previous exercise. This shows that T ∗ ϕ = Tuϕ . In what follows we identify T ∗ ϕ with uϕ . Step 2: If xn → x in RN , then for every y ∈ RN , ϕxn (y) = ϕ(xn − y) → ϕ(x − y) = ϕx (y) and conditions (i) and (ii) of Theorem 10.8 are satisfied. Hence, {ϕxn }n converges to ϕx with respect to τ , and so by Theorem 10.10, (T ∗ ϕ)(xn ) = T (ϕxn ) → T (ϕx ) = (T ∗ ϕ)(x), which proves that T ∗ ϕ is a continuous function. To prove (ii), note that if x ∈ RN is such that supp ϕx ∩ supp T = ∅, then (T ∗ ϕ)(x) = 0. Thus, supp(T ∗ ϕ) ⊆ {x ∈ RN : supp ϕx ∩ supp T = ∅} = supp T + supp ϕ.
10.6. Convolution of Distributions
307
Next we prove (iii). Let ei be an element of the canonical basis of RN and for every x ∈ RN and h = 0 consider the function ϕx,h,i (y) :=
ϕ(x + hei − y) − ϕ(x − y) , h
y ∈ RN .
∂ϕ (x − y) for all y ∈ RN and conditions As h → 0, we have that ϕx,h,i (y) → ∂x i (i) and (ii) of Theorem 10.8 are satisfied (why?). Hence, {ϕx,h,i }h converges to (∂i ϕ)x with respect to τ as h → 0, and so, by the linearity of T and Theorem 10.10,
(T ∗ ϕ)(x + hei ) − (T ∗ ϕ)(x) = T (ϕx,h,i ) → T ((∂i ϕ)x ) h as h → 0, which proves that ∂i (T ∗ ϕ) = T ∗ ∂i ϕ. Moreover, since for all x, y ∈ RN , ∂ϕ x ∂ϕ ∂ϕ ∂ϕx (y) = (x − y) = − (x − y) = − (y), ∂xi ∂xi ∂yi ∂yi for all x ∈ RN we have ∂ϕx ∂T ∂T x ∗ ϕ (x) = (ϕ ) = −T ∂yi ∂yi ∂yi ∂ϕ x ∂ϕ =T − = T∗ (x), ∂yi ∂xi which, together with an induction argument, gives (iii). Finally, to prove (iv), we define ϕ(x − hy)ψ(hy), uh (x) := hN
x ∈ RN ,
y∈ZN
where h > 0. As before we have that {uh }h converges to ϕ ∗ ψ with respect to τ as h → 0. It follows that for every x ∈ RN , {(uh )x }h converges to (ϕ ∗ ψ)x with respect to τ as h → 0. By the linearity of T and by Theorem 10.10 we have that (T ∗ (ϕ ∗ ψ))(x) = T ((ϕ ∗ ψ)x ) = lim T ((uh )x ) = lim (T ∗ uh )(x) h→0 h→0 N (T ∗ ϕ)(x − hy)ψ(hy) = lim h h→0
y∈ZN
= ((T ∗ ϕ) ∗ ψ)(x), where we have used the previous exercise. This completes the proof.
In what follows we identify T ∗ ϕ with uϕ . As a consequence of the previous theorem, we can approximate distributions with C ∞ functions.
308
10. Distributions
Theorem 10.53. Let T ∈ D (RN ) and let {ϕε }ε , ε > 0, be a family of standard mollifiers2 . Then {T ∗ ϕε }ε converges to T in the sense of distributions as ε → 0+ ; that is, lim (T ∗ ϕε )(x)φ(x) dx = T (φ) ε→0+
for every φ ∈
RN
D(RN ).
Proof. By Theorems C.16(i) and C.20 and Theorem 10.8 we have that for every φ ∈ D(RN ) the sequence {ϕε ∗ φ} converges to φ with respect to τ as ε → 0+ . Then by Theorem 10.10, . 0) . . 0 ) = lim T ((ϕε ∗ φ) T (φ) = (T ∗ φ)(0) = T ((φ) ε→0+
. . = lim ((T ∗ ϕε ) ∗ φ)(0) = lim (T ∗ (ϕε ∗ φ))(0) ε→0+ ε→0+ . (T ∗ ϕε )(y)φ(0 − y) dy = lim (T ∗ ϕε )(y)φ(y) dy, = lim ε→0+
RN
ε→0+
RN
where we have used Theorem 10.52(iv).
Exercise 10.54. Let Ω ⊆ RN be an open set and let T ∈ D (Ω). (i) Prove that there exists a sequence {Tn }n in D (Ω) such that each Tn has support compactly contained in Ω and {Tn } converges to T in the sense of distributions. (ii) Prove that Cc∞ (Ω) is dense in D (Ω) with respect to the weak star topology of D (Ω). Similarly we can define the convolution of a tempered distribution and a function. We recall that ϕ(x) . := ϕ(−x), x ∈ RN (see (10.21)). Definition 10.55. If T ∈ S (RN ; C) and ϕ ∈ S(RN ; C) the convolution of T and ϕ is the continuous linear functional T ∗ ϕ : S(RN ; C) → C defined by (T ∗ ϕ)(φ) := T (ϕ . ∗ φ), φ ∈ S(RN ; C). The proof of the following theorem is similar to the one of Theorem 10.52 and is left as an exercise. We recall that ϕx (y) = ϕ(x − y), y ∈ RN . Theorem 10.56. If T ∈ S (RN ; C) and ϕ ∈ S(RN ; C), then T ∗ ϕ = Tuϕ , where uϕ is the function given by uϕ (x) := T (ϕx ), x ∈ RN . Moreover, uϕ ∈ C ∞ (RN ; C) and for every multi-index α there exist cα > 0 and nα ∈ N such that ∂ α uϕ (x) ≤ cα (1 + x2 )nα for all x ∈ RN . 2 See
Appendix C.
10.7. Fourier Transforms
309
In what follows we identify T ∗ ϕ with the function uϕ . Exercise 10.57. Let T ∈ S (RN ; C) and ϕ, ψ ∈ S(RN ; C). Prove that (T ∗ ϕ) ∗ ψ = (T ∗ ψ) ∗ ϕ.
10.7. Fourier Transforms Given φ ∈ S(RN ; C), the Fourier transform of φ is the function & e−2πix·y φ(y) dy, x ∈ RN , (10.23) φ(x) = F (φ)(x) := RN
while the inverse Fourier transform of φ is the function ∨ & (10.24) φ (x) := φ(−x) = e2πix·y φ(y) dy, x ∈ RN . RN
Since S(RN ; C) ⊂ L1 (RN ; C) (see Theorem 10.35), the functions φ& and φ∨ are well-defined. Theorem 10.58. The Fourier transform F maps S(RN ) into S(RN ). Moreover, for every φ ∈ S(RN ; C) and for every α, β ∈ NN 0 , & = ψ' xα ∂ β φ(x) α,β (x)
(10.25)
for every x ∈ RN ,
α φ(x). & where ψα,β (x) := ∂ α ((−2πix)β φ(x)). In particular, xα φ(x) = ∂'
Proof. By differentiating under the integral sign we have that ∂ |β| φ& (x) = ∂xβ
RN
∂ |β| −2πix·y (e )φ(y) dy = ∂xβ
(−2πiy)β e−2πix·y φ(y) dy. RN
Hence, xα
1 ∂ |β| φ& (x) = ∂xβ (−2πi)|α| =
1 (−2πi)|α|
(−2πiy)β (−2πix)α e−2πix·y φ(y) dy RN
(−2πiy)β φ(y) RN
∂ |α| −2πix·y (e ) dy. ∂y α
By integrating by parts (see (9.55)) and using the fact that φ and its derivatives decay to zero at infinity we get xα
∂ |β| φ& (x) = ∂xβ
e−2πix·y RN
∂ |α| ((−2πiy)β φ(y)) dy, ∂y α
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10. Distributions
which proves (10.25). It follows from Leibnitz rule (see Exercise 10.11) that |α| ∂ |β| & φα,β ≤ (2π) α ((−y)β φ(y)) dy RN ∂y 1 + yN +1 ∂ |α| |β| β = (2π) ((−y) φ(y)) dy N +1 ∂y α 1 + y N R ≤ cφN +1+|α|,|β| , where c = c(N, α, β) > 0, which shows that φ& ∈ S(RN ; C) and that the linear operator F : S(RN ; C) → S(RN ; C) is continuous. Exercise 10.59. Prove that the Fourier transform of the function φ(x) := 2 e−π x , x ∈ RN , is φ. Exercise 10.60. Prove that the Fourier transform of the function φε (x) := e2πix·x0 e−πε
2 x 2
,
x ∈ RN ,
1 2 φ&ε (x) = N e−π (x−x0 )/ε , ε
x ∈ RN .
where ε > 0 and x0 ∈ RN , is
Next we prove that F is invertible with inverse given by F −1 (φ) = φ∨ (see (10.24)). Proposition 10.61. For every φ, ψ ∈ S(RN ; C), we have & & dx = φ(x)ψ(x) dx. φ(x)ψ(x) (10.26) RN
RN
Proof. By Fubini’s theorem & φ(x)ψ(x) dx = φ(x) e−2πix·y ψ(y) dydx RN RN RN −2πix·y & dy, = ψ(y) e φ(x) dxdy = ψ(y)φ(y) RN
RN
RN
which shows (10.26). Theorem 10.62 (Fourier inversion theorem). For every φ ∈ S(RN ; C), ' ∨ ) = φ. & ∨ = (φ (φ)
In particular, the Fourier transform F is an isomorphism from S(RN ; C) to S(RN ; C) with inverse F −1 given by F −1 (φ) = φ∨ for every φ ∈ S(RN ; C). Proof. Fix x0 ∈ RN and ε > 0 and define ψε (x) := e2πix·x0 e−πε x , x ∈ %ε (x) = 1N e−π (x−x0 )/ε 2 and so, taking RN . By Exercise 10.60 we have that ψ ε ψ = ψε in (10.26), we get 1 2 2 2 & dy. φ(x) N e−π (x−x0 )/ε dx = e2πiy·x0 e−πε y φ(y) ε RN RN 2
2
10.7. Fourier Transforms
311
%ε is a mollifier. Hence, the left-hand side converges to φ(x0 ) Note that ψ by Exercise C.19 in Appendix C. On the other hand, by the Lebesgue & ∨ (x0 ). dominated convergence theorem the right-hand side converges to (φ) & ∨ (x0 ), which shows that (φ) & ∨ = φ. Similarly we can Hence, φ(x0 ) = (φ) ' ∨ ) = φ. show that, (φ & ∨ = 0∨ = 0, and so F is oneNext observe that if φ& = 0, then φ = (φ) ' ∨ ) = φ, it follows that F is onto and that the inverse of F to-one. Since (φ is F −1 (φ) = φ∨ .
We recall that for a complex number z = x + iy, where x, y ∈ R, the complex conjugate of z is the number z := x − iy. Corollary 10.63 (Plancherel). For every φ, h ∈ S(RN ; C), & & φ(x)h(x) dx = φ(x) h(x) dx (10.27) RN
and
RN
φ(x) dx = 2
(10.28) RN
RN
2 & φ(x) dx =
RN
φ∨ (x)2 dx.
In particular, F extends uniquely to an isomorphism of L2 (RN ; C) onto itself. The identity (10.27) is called Parseval identity, while the identity (10.28) is called Plancherel identity. h. Then, using the facts that the cosine is even and the Proof. Let ψ := & sine is odd, we have −2πiy·x& & e e2πiy·x& ψ(x) = h(y) dy = h(y) dy
RN
e2πiy·x& h(y) dy =
= RN
RN
e2πiy·x& h)∨ (x) = h(x), h(y) dy = (& RN
where in the last equality we have used the inversion theorem (Theorem 10.62). Hence, Parseval’s identity follows by (10.26). Taking h = φ and using the fact that φ(x)φ(x) = φ(x)2 gives the first equality in Plancherel’s identity. The second equality follows by replacing φ with φ∨ and using the inversion theorem. Since S(RN ; C) is dense in L2 (RN ; C), if {φn }n is a sequence in S(RN ; C) converging to φ in L2 (RN ; C), then by Plancherel’s identity the sequence %n }n is a Cauchy sequence in L2 (RN ; C) and so it converges to a function {φ ψ ∈ L2 (RN ; C). Again by Plancherel’s identity, the function ψ does not depend on the particular sequence {φn }n . We define φ& := ψ. Similarly we can extend uniquely the inverse Fourier transform to L2 (RN ; C) and
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10. Distributions
reasoning as in the last part of the proof of the inversion theorem (Theorem 10.62) we have that the Fourier transform F : L2 (RN ; C) → L2 (RN ; C) is an isomorphism with inverse given by the extension of F −1 to L2 (RN ; C). Remark 10.64 (Important). Note that the Fourier transform of a function φ in L2 (RN ; C) is obtained as a limit in L2 (RN ; C) of functions of the type (10.23), but in general we cannot say that φ& has the form (10.23), since the integral in (10.23) is well-defined for functions in L1 (RN ; C) but not for functions in L2 (RN ; C). On the other hand, if φ ∈ L1 (RN ; C), then (10.23) is well-defined. Hence, the Fourier transform of a function in L1 (RN ; C) is defined pointwise by (10.23), while the Fourier transform of a function in L2 (RN ; C) is defined as a limit in L2 (RN ; C). Theorem 10.65 (Riemann–Lebesgue lemma). F : L1 (RN ; C) → C0 (RN ; C) with (10.29)
sup F (φ)(x) ≤ φL1 (RN )
x∈RN
& In particular, lim x →∞ φ(x) = 0. Proof. By (10.23), for every φ ∈ L1 (RN ; C), & φ(x) ≤ φL1 for every x ∈ RN . Since S(RN ; C) is dense in L1 (RN ; C) let {φn }n in S(RN ; C) converge to φ in L1 (RN ; C). By the previous inequality %n (x) − φ(x) & sup φ ≤ φn − φL1 .
x∈RN
& On the other hand, %n }n converges uniformly to φ. Hence, the sequence {φ N %n ∈ S(R ; C) ⊂ C0 (RN ; C) and hence, by Theorem 10.58 we have that φ since C0 (RN ; C) is a closed under uniform convergence, it follows that φ& ∈ C0 (RN ; C). Exercise 10.66. Consider the Fourier transform F : L1 (R; C) → C0 (R; C). (i) Prove that if f ∈ L1 (R; C) is such that f& is odd, then for every b > 0, b f&(x) dx ≤ c x 1 for some constant c > 0 independent of b. (ii) Prove that there exists g ∈ C0 (R; C) such that g is not the Fourier transform of any function f in L1 (R; C). (iii) Prove that F (L1 (R; C)) is dense in C0 (R; C).
10.7. Fourier Transforms
313
Corollary 10.67 (Hausdorff–Young inequality). Let 1 < p < 2. Then F can be extended uniquely to Lp (RN ; C) and (10.30)
F (φ)Lp (RN ) ≤ φLp (RN )
for every φ ∈ Lp (RN ; C).
Proof. By Plancherel’s identity and (10.29), sup F (φ)(x) ≤ φL1 ,
F (φ)L2 = φL2
x∈RN
for every simple function φ vanishing at infinity. Hence, F is of strong type (1, ∞) and (2, 2) (see Definition B.95) and so we are in a position to apply the Riesz–Thorin theorem (see Theorem B.96) to obtain that F is of strong type (p, q) for every θ ∈ (0, 1), where 1−θ θ 1 1−θ θ 1 = + , = + . p 1 2 q ∞ 2 Hence, p =
2 2−θ
∈ (1, 2) and q = p = 2/θ. Moreover, (10.30) holds.
Exercise 10.68. Let φ ∈ S(RN ; C) and let L : RN → RN be an orthogonal transformation. & (i) Prove that (φ ◦ L)(x) = φ(L(x)) for every x ∈ RN . & (ii) Prove that if φ is radial, then so is φ. Exercise 10.69. Prove that if φ ∈ S(RN ; C) is real-valued and φ(x) = φ(−x) for every x ∈ RN , then φ& is real-valued. Exercise 10.70. Compute the Fourier transform of the function 2
φ(x) := e(−a+ib) x ,
x ∈ RN ,
where a > 0 and b ∈ R. & ∗ ψ = φ&ψ. Theorem 10.71. For every φ, ψ ∈ S(RN ; C), φ Proof. For x ∈ RN by Fubini’s theorem we have −2πix·y e (φ ∗ ψ)(y) dy = e−2πix·y φ(y − ξ)ψ(ξ) dξdy (φ ∗ ψ)(x) = RN RN RN = ψ(ξ) e−2πix·y φ(y − ξ) dydξ N N R R −2πix·ξ e ψ(ξ) e−2πix·(y−ξ) φ(y − ξ) dydξ = N N R R −2πix·ξ & φ(x), & e ψ(ξ) e−2πix·η φ(η) dηdξ = ψ(x) = RN
RN
where we have made the change of variables η := y − ξ.
Remark 10.72. The previous theorem continues to hold for φ ∈ L1 (RN ; C) and ψ ∈ S(RN ; C).
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10. Distributions
As a corollary of Theorem 10.71 we can show how the support of the Fourier transform relates to the regularity of a function. Theorem 10.73. Let φ ∈ S(RN ; C) be such that supp φ& ⊆ B(0, R) for some R > 0. Then for every 1 ≤ p ≤ q ≤ ∞ and for every multi-index α ∈ NN 0 , ∂ α φLq (RN ) ≤ cR|α|+N/p−N/q φLp (RN ) for some constant c = c(|α|, N, p, q) > 0. In particular, for α = 0, φLq (RN ) ≤ cRN/p−N/q φLp (RN ) . On the other hand, if φ ∈ S(RN ; C) is such that supp φ& ⊆ B(0, Rr2 ) \ B(0, Rr1 ) for some R > 0 and some 0 < r1 < r2 , then ∂ α φLp (RN ) ≤ cRm φLp (RN ) c−1 Rm φLp (RN ) ≤ |α|=m
for some constant c = c(N, m, p, r1 , r2 ) > 0. Proof. Using the change of variables x = Ry, we can assume R = 1. Step 1: Let φ ∈ S(RN ; C) be such that supp φ& ⊆ B(0, 1) and consider a function ϕ ∈ D(RN ) such that ϕ = 1 in B(0, 1). Then by Theorem 10.71, φ& = ϕφ& = ψ&φ& = ψ ∗ φ, where ψ := ϕ∨ , which, by Theorem 10.62, implies that φ = ψ ∗ φ. By differentiating inside the integral it follows from (10.19) that ∂ α φ = (∂ α ψ) ∗ φ, and so by Young’s inequality (see Theorem 10.49) ∂ α φLq = (∂ α ψ) ∗ φLq ≤ ∂ α ψLr φLp , where 1/p + 1/r = 1 + 1/q. Step 2: Let φ ∈ S(RN ; C) be such that supp φ& ⊆ B(0, r2 ) \ B(0, r1 ) and consider a function ϕ ∈ D(RN \{0}) such that ϕ = 1 in B(0, r2 ) \ B(0, r1 ). Writing y2m = cα (iy)α (−iy)α , |α|=m
we have that
& φ(y) =
& (iy)α φ(y)ψ α (y),
|α|=m
Note that since ϕ ∈ D(RN \{0}), where ψα (y) := ϕ(y)(cα N we have that ψα ∈ D(R \{0}). It follows from Theorems 10.58 and 10.71 that ∨ ψα ∗ ∂ α φ, φ= (−iy)α /y2m ).
|α|=m
and so by Young’s inequality (see Theorem 10.49) ∨ ∨ ψα ∗ ∂ α φLp ≤ ψα L1 ∂ α φLp ≤ c ∂ α φLp . φLp ≤ |α|=m
|α|=m
|α|=m
10.7. Fourier Transforms
315
The other inequality follows from Step 1. Exercise 10.74. For n, m ∈ N let fn = χ[−n,n] and gn,m :=
1 m fn
∗ fm .
(i) Compute the Fourier transform of fn . (ii) Compute the Fourier transform of gn,m . 2 (iii) Calculate the Lebesgue integral R sinx2 x dx. (iv) Compute the improper Riemann integral R
sin x x
dx.
We can also define the Fourier transform of tempered distributions. Given T ∈ S (RN ; C), the Fourier transform T& of T is the tempered distribution given by & T&(φ) := T (φ),
(10.31)
φ ∈ S(RN ; C).
Given ψ ∈ S(RN ; C), consider the linear functional T : S(RN ; C) → C defined by (10.17). By (10.26), for every φ ∈ S(RN ; C), we have & & =T & %ψ (φ). φ(x)ψ(x) dx = φ(x)ψ(x) dx = Tψ (φ) (10.32) Tψ(φ) = RN
RN
S (RN ; C),
this shows that the Fourier Since we are identifying ψ with Tψ in N transform defined on S (R ; C) extends the Fourier transform defined in S(RN ; C). Similarly, given T ∈ S (RN ; C), the inverse Fourier transform of T is the tempered distribution given by (10.33)
T ∨ (φ) := T (φ∨ ),
φ ∈ S(RN ; C).
Exercise 10.75. Let T ∈ S (RN ; C). (i) Prove that T& ∈ S (RN ; C). ∗
(ii) Prove that if {Tn }n is a sequence in S (RN ; C) such that Tn T ∗ & %n T. in S (RN ; C), then T (iii) Prove that Theorem 10.62 continues to hold in S (RN ; C). (iv) Prove that F : S (RN ; C) → S (RN ; C) is an isomorphism with inverse F −1 given by F −1 (T ) = T ∨ for every T ∈ S (RN ; C). In view of Theorem 10.35 and the previous exercise, for every function ψ ∈ Lp (RN ) with p > 2, the Fourier transform ψ& of ψ is the Fourier %ψ of the tempered distribution Tψ defined in (10.17). Hence, ψ& transform T belongs to S (RN ; C) but in general φ& cannot be identified with a function. A simple example is given by ψ = 1 ∈ L∞ (RN ). In this case φ1 dx, φ ∈ S(RN ; C), (10.34) T1 (φ) = RN
316
10. Distributions
and so by Theorem 10.62, & % & φ(x) dx = 1(φ) = T1 (φ) = RN
& dx = (φ) & ∨ (0) = φ(0), e2πi0·y φ(x) RN
which shows that & 1 is δ0 . Exercise 10.76. Given T ∈ S (RN ; C), prove that the following properties are equivalent: (i) supp T& = {0}, (ii) T = Tf for some polynomial f : RN → C, (iii) T = |α|≤n aα ∂ α δ0 for some n ∈ N and some aα ∈ C, α ∈ NN 0 , 0 ≤ |α| ≤ n. Exercise 10.77. Prove that Theorem 10.71 continues to hold, that is, that if T ∈ S (RN ; C) and ϕ ∈ S(RN ; C), then T ∗ϕ=ϕ &T&.
10.8. Littlewood–Paley Decomposition Let ω0 ∈ Cc∞ (RN ) be a nonnegative function such that (10.35)
supp ω0 = B(0, 1),
ω0 = 1 in B(0, 1/2),
and define (10.36)
ϕ0 (x) := ω0 (x/2) − ω0 (x),
x ∈ RN .
Given T ∈ S (RN ; C) and k ∈ Z we define the tempered distributions (10.37)
Sk (T ) := (ω0 (2−k ·)T&)∨ ,
(10.38)
Λ˙ k (T ) := (ϕ0 (2−k ·)T&)∨ .
The tempered distribution Λ˙ k (T ) is called the kth dyadic block of the Littlewood– Paley decomposition of T . Observe that by Exercise 10.40, (10.31), (10.33) and (10.37), for every φ ∈ S(RN ; C), (10.39)
Sk (T )(φ) = (ω0 (2−k ·)T&)∨ (φ) = T (F (ω0 (2−k ·)F −1 (φ))) = T (Sk (φ)),
where the last equality follows from Theorems 10.62 and 10.71, and where & ∨ . Similarly, for every φ ∈ S(RN ; C), Sk (φ) := (ω0 (2−k ·)φ) (10.40)
Λ˙ k (T )(φ) = T (Λ˙ k (φ)),
& ∨. where Λ˙ k (φ) := (ϕ0 (2−k ·)φ)
10.8. Littlewood–Paley Decomposition
317
Theorem 10.78 (Littlewood–Paley decomposition). Let T ∈ S (RN ; C) and let ω0 ∈ Cc∞ (RN ) be a nonnegative function satisfying (10.35). Then for every l ∈ Z, ∞ (10.41) T = Sl (T ) + Λ˙ k (T ). k=l
Moreover, if (10.42)
∗
Sk (T ) 0
as k → −∞ in S (RN ; C),
then (10.43)
∞
T =
Λ˙ k (T ).
k=−∞
The identity (10.43) is called the homogeneous Littlewood–Paley decomposition of T . Proof. By (10.36), (10.37), and (10.38) for n ∈ N we have n Λ˙ k (T ) = Sn+1 (T ) Sl (T ) + k=l
and so, by (10.39), for every φ ∈ S(RN ; C), Sn+1 (T )(φ) = T (Sn+1 (φ)). We leave as an exercise to check that since ω0 (2−n−1 x) = 1 for x ∈ B(0, 2n ) by (10.35), we have that ω0 (2−n−1 ·)φ& → φ& in S(RN ; C) as n → ∞. In turn, since F is an isomorphism from S(RN ; C) to S(RN ; C) (see Theorem 10.62), it follows that Sn+1 (φ) → φ in S(RN ; C), and so n Λ˙ k (T )(φ) = T (Sn+1 (φ)) → T (φ) Sl (T )(φ) + k=l
as n → ∞, which shows (10.41). If (10.42) holds, then (10.43) follows by letting l → −∞ in (10.41).
Remark 10.79. The previous theorem continues to hold if instead of (10.35) we take ω0 ∈ Cc∞ (RN ) to be a nonnegative function such that supp ω0 = B(0, r),
ω0 = 1 in B(0, r/2)
for some r > 0. The Littlewood–Paley decomposition (10.43) fails without (10.42). Indeed, by taking the functional T1 corresponding to the constant function 1 (see (10.34)), we have that Λ˙ k (T1 ) = 0 for all k, so (10.43) cannot hold.
Chapter 11
Sobolev Spaces Newton’s first law of graduation: A grad student in procrastination tends to stay in procrastination unless an external force is applied to it. — Jorge Cham, www.phdcomics.com
In this chapter we define Sobolev functions on domains of RN , N ≥ 1, and begin to study their properties.
11.1. Definition and Main Properties In view of the integration by parts formula (9.55) we can give the following definition. Definition 11.1. Given an open set Ω ⊆ RN , a multi-index α ∈ NN 0 \ {0}, and 1 ≤ p ≤ ∞, we say that a function u ∈ L1loc (Ω; RM ) admits a weak or distributional αth derivative in Lp (Ω; RM ) if there exists a function vα ∈ Lp (Ω; RM ) such that α |α| (11.1) u∂ φ dx = (−1) vα φ dx Ω
Ω
for all φ ∈ Cc∞ (Ω). The function vα is denoted ∂ α u or
∂ |α| u ∂xα .
A similar definition can be given when Lp (Ω; RM ) is replaced by Lploc (Ω; RM ). In terms of distributions the previous definition means that ∂ α Tu = Tvα (see Definition 10.18). Remark 11.2. We leave as an exercise to check that weak derivatives are unique. Similarly we can define weak directional derivatives in Lp (see Section 9.1). 319
320
11. Sobolev Spaces
Definition 11.3. Given an open set Ω ⊆ RN , a vector ν ∈ RN \ {0}, n ∈ N, and 1 ≤ p ≤ ∞, we say that a function u ∈ L1loc (Ω; RM ) admits a weak or distributional nth directional derivative in Lp (Ω; RM ) in the direction ν if there exists a function vν,n ∈ Lp (Ω; RM ) such that ∂nφ n u n dx = (−1) vν,n φ dx Ω ∂ν Ω for all φ ∈ Cc∞ (Ω). The function vν,n is denoted
∂nu ∂ν n .
Definition 11.4. Given an open set Ω ⊆ RN , m, M ∈ N, and 1 ≤ p ≤ ∞, the Sobolev space W m,p (Ω; RM ) is the space of all functions u ∈ Lp (Ω; RM ) which admit weak derivatives ∂ α u in Lp (Ω; RM ) for every α ∈ NN 0 with 1 ≤ |α| ≤ m. The space W m,p (Ω; RM ) is endowed with the norm uW m,p (Ω;RM ) := uLp (Ω;RM ) +
∂ α uLp (Ω;RM ) .
1≤|α|≤m m,p The space Wloc (Ω; RM ) is defined as the space of all functions u ∈ Lploc (Ω; RM ) which admit weak derivatives ∂ α u in Lploc (Ω; RM ) for every α ∈ NN 0 with 1 ≤ |α| ≤ m. m,p (Ω) for W m,p (Ω; R) and When M = 1 we write W m,p (Ω) and Wloc m,p (Ω; R), respectively. For simplicity, unless otherwise specified, in what Wloc follows we will focus on the case M = 1. Also, when there is no possibility of confusion, we write uLp and uW m,p for uLp (Ω) and uW m,p (Ω) , respectively.
Remark 11.5 (Important). Following the literature, we use the same notation to indicate the weak and the classical derivatives of a function. Unfortunately, this results in endless confusion for students. In the remainder of this book, when u ∈ W m,p (Ω), unless otherwise specified, ∂ α u is the αth weak derivative of u. Remark 11.6. Given Ω ⊆ RN open, m ∈ N, with m ≥ 2, and 1 ≤ p ≤ ∞, u ∈ W m,p (Ω), in view of (11.1) and Schwartz theorem on the order of differentiation for smooth functions (Theorem 9.6) applied to φ ∈ Cc∞ (Ω), Ω
∂ 2u φ dx = ∂xi ∂xj
2
∂ 2φ u dx = Ω ∂xi ∂xj 2
∂2φ u dx = Ω ∂xj ∂xi
Ω
∂2u φ dx, ∂xj ∂xi
u u = ∂x∂j ∂x . Thus for distributional derivatives the which shows that ∂x∂i ∂x j i order in which one differentiates is not important. This is in sharp contrast to classical derivatives.
11.1. Definition and Main Properties
321
For u ∈ W m,p (Ω) and 1 ≤ k ≤ m, we set ∇k u to be the vector of components ∂ α u, |α| = k. Then an equivalent norm in W m,p (Ω) is given by uLp (Ω) +
(11.2)
m
∇k uLp (Ω;RMk ) ,
k=1
where Mk is the number of multi-indices α ∈ NN 0 with |α| = k. For simplicity in what follows we will write ∇k uLp (Ω) for ∇k uLp (Ω;RMk ) . Another equivalent norm which we will sometimes use is given by 1/p ∂ α upLp (Ω) uW m,p (Ω) := upLp (Ω) + 1≤|α|≤m
for 1 ≤ p < ∞, and uW m,∞ (Ω) := max{∂ α uL∞ (Ω) : 0 ≤ |α| ≤ m}, where ∂ 0 u := u, for p = ∞. Exercise 11.7. Let Ω ⊆ RN be an open set, let m ∈ N, with m ≥ 2, and let 1 ≤ p ≤ ∞. Prove that u ∈ Lp (Ω) belongs to W m,p (Ω) if and only if ∇u ∈ W m−1,p (Ω; RN ). Exercise 11.8. Show that the function u : B(0, 1) → R, defined by 1 if xN > 0, u(x) = u(x1 , . . . , xN ) := 0 if xN < 0, does not belong to W 1,p (B(0, 1)) for any 1 ≤ p ≤ ∞. Exercise 11.9. Let Ω ⊂ RN be an open bounded set, let x0 ∈ Ω, and let 1 ≤ p < ∞. Prove that if u ∈ C(Ω)∩C 1 (Ω\{x0 }) is such that the (classical) gradient ∇u belongs to Lp (Ω; RN ), then u ∈ W 1,p (Ω). Exercise 11.10. Let Ω ⊂ RN be an open set and let C ⊂ Ω be a closed set with HN −1 (C) = 0, where we recall that HN −1 is the (N − 1)-dimensional Hausdorff measure (see Section C.7). Let m ∈ N and 1 ≤ p ≤ ∞ and let u ∈ W m,p (Ω \ C). Prove that u ∈ W m,p (Ω). Hint: Consider first the case m = 1 and for i = 1,. . . , N consider the projection Πi into the hyperplane orthogonal to the xi -axis. Prove that HN −1 (Πi (C)) = 0 and use Fubini’s theorem. Exercise 11.11. Let Ω ⊆ RN be an open set, let m ∈ N, and let 1 ≤ p ≤ ∞. Prove that if u ∈ W m,p (Ω) and T : RN → RN is a rigid motion, then u ◦ T ∈ W m,p (U ; RN ), where U := T −1 (Ω) and compute u ◦ T W m,p (U ) . We now show that W m,p (Ω) is a Banach space. Theorem 11.12. Let Ω ⊆ RN be an open set, let m ∈ N, and let1 ≤ p ≤ ∞.
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11. Sobolev Spaces
Then (i) the space W m,p (Ω) is a Banach space, (ii) the space H m (Ω) := W m,2 (Ω) is a Hilbert space with the inner product uv dx + ∂ α u∂ α v dx. u, v H 1 (Ω) := Ω
1≤|α|≤m Ω
Proof. We only prove part (i). Let {un }n be a Cauchy sequence in W m,p (Ω); that is, 0 = lim ul − un W m,p . l,n→∞ = lim ul − un Lp + l,n→∞
∂ α ul − ∂ α un Lp .
1≤|α|≤m
Since Lp (Ω) is a Banach space, there exist u, vα ∈ Lp (Ω) such that lim un − uLp = 0,
n→∞
lim ∂ α un − vα Lp = 0
n→∞
for every multi-index α with 1 ≤ |α| ≤ m. Fix a multi-index α with 1 ≤ |α| ≤ m. We claim that ∂ α u = vα . To see this, let φ ∈ Cc∞ (Ω) and note that α |α| φ∂ un dx = (−1) un ∂ α φ dx. Ω
Ω
Letting n → ∞ in the previous equality yields |α| φvα dx = (−1) u∂ α φ dx Ω
Ω
Cc∞ (Ω),
for all φ ∈ which proves the claim. Thus, u ∈ W m,p (Ω). This completes the proof. Exercise 11.13. Let Ω ⊆ RN be an open set and let 1 ≤ p < ∞. (i) Prove that a subset of a separable metric space is separable. (ii) Prove that W m,p (Ω) is separable. Hint: Consider the mapping W m,p (Ω) → Lp (Ω) × · · · × Lp (Ω) u → (u, ∂ α1 u, . . . , ∂ αM u), where α1 , . . . , αM are all the multi-indices α with 1 ≤ |α| ≤ m. Exercise 11.14. Let Ω ⊆ RN be an open set. Prove that W m,∞ (Ω) is not separable. Let Ω ⊆ RN be an open set, let m ∈ N, and let 1 ≤ p ≤ ∞. The space is defined as the closure of the space Cc∞ (Ω) in W m,p (Ω) (with respect to the topology of W m,p (Ω)).
W0m,p (Ω)
11.1. Definition and Main Properties
323
Remark 11.15. It is important to observe that since uniform convergence preserves continuity, functions in W0m,∞ (Ω) are necessarily of class C m (Ω). In particular, piecewise affine functions do not belong to W01,∞ (Ω). This is the reason why some authors refer to W01,∞ (Ω) as the closure of the space Cc∞ (Ω) in W 1,∞ (Ω) with respect to the weak star topology of W 1,∞ (Ω), rather than the strong topology. These two definitions are not equivalent. Thus, caution is needed when applying results for W01,∞ (Ω). We conclude this section by observing that in several applications in partial differential equations and in the calculus of variations, when one works with domains Ω of infinite measure, the Sobolev space W m,p (Ω), m ∈ N, 1 ≤ p ≤ ∞, may not be the right space to work with. Indeed, as the following exercise shows, there exist solutions of the Dirichlet problem for the Laplacian with the property that u ∈ Lploc (Ω), ∇u ∈ Lp (Ω; RN ), but u does not belong to W 1,p (Ω). Exercise 11.16. Let Ω := {x ∈ RN : x > 1} and let u(x) := 1 − x2−N , x ∈ Ω, if N ≥ 3, and u(x) := log x, x ∈ Ω, if N = 2. (i) Prove that u is a classical solution of the Dirichlet problem Δu = 0 in Ω, u=0 on ∂Ω. / Lq (Ω) for any (ii) Prove that u ∈ Lqloc (Ω) for all 1 ≤ q < ∞, but u ∈ 1 ≤ q < ∞. (iii) Prove that ∇u ∈ Lp (Ω; RN ) for all (iv) Prove that
∂2u ∂xi ∂xj
N N −1
< p < ∞.
∈ Lp (Ω) for all 1 < p < ∞.
To provide a functional setting for Dirichlet’s problems as in the previous ˙ m,p (Ω). exercise, we introduce the space W Definition 11.17. Let Ω ⊆ RN be an open set, let m ∈ N, and let 1 ≤ p < ˙ m,p (Ω) is the space of all functions ∞. The homogeneous Sobolev space W 1 u ∈ Lloc (Ω) whose αth weak derivative ∂ α u belongs to Lp (Ω) for every α ∈ NN 0 with |α| = m. Note that the inclusion ˙ m,p (Ω) W m,p (Ω) ⊆ W holds. We will see later on that as a consequence of Poincar´e’s inequality ˙ m,p (Ω) and for sufficiently regular domains of finite measure the spaces W m,p W (Ω) actually coincide. However, for domains of infinite measure (see Exercise 11.16) or irregular domains with finite measure (see Exercise 12.15) this is not the case.
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11. Sobolev Spaces
˙ m,p (Ω) is equipped with the seminorm The space W |u|W˙ m,p (Ω) := ∇m uLp (Ω) .
(11.3)
Sometimes we will also use the equivalent seminorm u → |α|=m ∂ α uLp (Ω) . Observe that |u|W˙ m,p (Ω) = 0 if and only if u is a polynomial of degree less than or equal to m − 1 in each connected component of Ω. When Ω has finitely many connected components and, in particular, when it is connected, ˙ m,p (Ω). then we can define a norm in W Exercise 11.18. Let Ω ⊆ RN be an open set with finitely many connected components Ω1 , . . . , Ω , let m ∈ N, and let 1 ≤ p < ∞. For each i = 1, . . . , ˙ m,p (Ω) is a consider a ball Bi Ωi and let U := i=1 Bi . Prove that W Banach space with the norm uW˙ m,p (Ω) :=
(11.4)
m−1
∇k uL1 (U ) + ∇m uLp (Ω) ,
k=0
where
∇0 u
:= u.
The following exercise shows that when Ω has infinitely many connected ˙ m,p (Ω) which is compatible with components, then there is no norm in W m,p ˙ the natural convergence in W (Ω), which is strong convergence in Lp (Ω) for the partial derivatives of order m and convergence in L1loc (Ω) for the functions and their derivatives of order less than m. Exercise 11.19. Let Ω ⊆ RN be an open set with infinitely many connected components Ωn , n ∈ N, and let 1 ≤ p < ∞. Prove that there cannot exist ˙ 1,p (Ω) compatible with the convergences un → u in a norm · W˙ 1,p in W L1loc (Ω) and ∇un → ∇u in Lp (Ω; RN ). Hint: Reason by contradiction and assume that · W˙ 1,p exists. For each n ∈ N consider a ball Bn Ωn . Prove that there exist ε > 0 and ∈ N such that |u| dx ≥ ε n=1
Bn
˙ 1,p (Ω) with norm u ˙ 1,p = 1. for all piecewise constant functions u ∈ W W We recall that S N −1 is the unit sphere ∂B(0, 1). Exercise 11.20. Let Ω ⊆ RN be an open set and let m ∈ N. 2 (i) Let ν1 , ν2 ∈ S N −1 and let ν := νν11 +tν +tν2 for t ∈ R, with ν1 +tν2 = 0. Prove that m m−k k m ∂ u k m ∂ m∂ u = t ν1 + tν2 ∂ν m k ∂ν1m−k ∂ν2k k=0
for every u ∈ C m (Ω).
11.2. Density of Smooth Functions
325
(ii) By taking a finite number of different values of t ∈ R show that ∂ m−k ∂ k u for each k = 1, . . . , m, m−k ∂ν k can be written as a linear ∂ν1
2
combination of a finite number of directional derivatives ν as in part (i).
∂mu ∂ν m ,
with
(iii) Prove that there exists a finite set Sm ⊂ S N −1 such that for every u ∈ C m (Ω), ∂ mu cα,ν m ∂αu = ∂ν ν∈Sm
for every multi-index α ∈ NN 0 , with |α| = m, and for some constants cα,ν ∈ R independent of u. Hint: Use induction on m and part (ii). Exercise 11.21. Given n ∈ N, let V be the vector space of all polynomials P : [0, 1] → R of degree less than or equal to n. (i) Given P ∈ V , let P := max{|a0 |, . . . , |an |}, where P (t) = a0 + · · · + an tn , t ∈ [0, 1]. Prove that · is a norm in V . (ii) Let 1 ≤ p ≤ ∞ and P ∈ V be such that P Lp ([0,1]) ≤ 1. Prove that there exists a constant c = c(n) > 0 such that |ak | ≤ c for all k = 1, . . . , n. Hint: The vector space V has finite dimension n. Exercise 11.22. Let Ω ⊆ RN be an open set, let m ∈ N, and let 1 ≤ p ≤ ∞. ˙ m,p (RN ): Prove that the following are equivalent seminorms in W
1/p ∂ m u p ∂mu N −1 (ν) , sup m p , m p dH L L S N −1 ∂ν ν∈S N −1 ∂ν m ∂ u m p, ∂ν L ν∈Sm
where Sm is the finite set of S N −1 given in the previous exercise. Hint: Use the previous exercises.
11.2. Density of Smooth Functions The next exercise shows that Sobolev functions can behave quite badly. Exercise 11.23. Let Ω := B(0, 1) ⊂ RN . (i) Consider the function u(x) := 1/xa , x = 0, where a > 0. (a) Prove that the (classical) gradient ∇u belongs to L1loc (Ω; RN ) if and only if a + 1 < N . (b) Let a + 1 < N and 1 ≤ p < ∞ and show that u ∈ W 1,p (Ω) if and only if (a + 1)p < N (in particular, u ∈ / W 1,p (Ω) for all p ≥ N ).
326
11. Sobolev Spaces
(ii) Let {xn : n ∈ N} ⊂ Ω be a countable dense set and define u(x) :=
∞ 1 1 , n 2 x − xn a
n=1
x∈Ω\
∞
{xn },
n=1
where 0 < a < N − 1. Prove that u ∈ W 1,p (Ω) if and only if (a + 1)p < N , but u is unbounded in each open subset of Ω. In view of the previous exercise, it is important to understand density of well-behaved functions in W m,p (Ω). Theorem 11.24 (Meyers–Serrin). Let Ω ⊆ RN be an open set, let m ∈ N, and let 1 ≤ p < ∞. Then the space C ∞ (Ω) ∩ W m,p (Ω) is dense in W m,p (Ω). We begin with an auxiliary result. We use standard mollifiers (see Appendix C). Lemma 11.25. Let Ω ⊆ RN be an open set, let m ∈ N, let 1 ≤ p < ∞, and let u ∈ W m,p (Ω). For every ε > 0 define uε := ϕε ∗ u, where ϕε is a standard mollifier, and let (11.5)
Ωε := {x ∈ Ω : dist(x, ∂Ω) > ε}.
Then uε − uW m,p (Ωε ) → 0 as ε → 0+ . In particular, if U ⊂ Ω, with dist(U, ∂Ω) > 0, then uε − uW m,p (U ) → 0 as ε → 0+ . Proof. By Theorem C.20 we have that uε ∈ C ∞ (RN ) and for x ∈ Ωε and α ∈ NN 0 , with 1 ≤ |α| ≤ m, |α| |α| ∂ ϕε ∂ ϕε ∂ |α| uε |α| (x) = (x − y)u(y) dy = (−1) (x − y)u(y) dy α α α ∂x Ω ∂x Ω ∂y ∂ |α| u ∂ |α| u ϕε (x − y) α (y) dy = ϕε ∗ (x), = ∂y ∂xα Ω where we have used (11.1) and the fact that for each x ∈ Ωε the function ϕε (x − ·) belongs to Cc∞ (Ω), since supp ϕε (x − ·) ⊆ B(x, ε) ⊂ Ω. The result |α| now follows from Theorem C.16 applied to the functions u and ∂∂xαu . Remark 11.26. Note that if Ω = RN , then Ωε = RN . Hence, uε → u in W m,p (RN ). Exercise 11.27. Let Ω ⊆ RN be an open set, let m ∈ N, and let 1 ≤ p ≤ ∞. Prove that if u ∈ W m,p (Ω) and ϕ ∈ Cc∞ (RN ), then ϕu ∈ W m,p (Ω). We now turn to the proof of the Meyers–Serrin theorem. Proof of Theorem 11.24. Let Ωi Ωi+1 be such that Ω = ∞ i=1 Ωi and use Theorem C.21 to construct a smooth partition of unity F subordinated
11.2. Density of Smooth Functions
327
to the open cover {Ωi+1 \ Ωi−1 }i , where Ω−1 = Ω0 := ∅.1 For each i ∈ N let ψi be the sum of all the finitely many ψ ∈ F such that supp ψ ⊂ Ωi+1 \ Ωi−1 and which have not already been selected at previous steps j < i. Then ψi ∈ Cc∞ (Ωi+1 \ Ωi−1 ) and ∞
(11.6)
ψi = 1
in Ω.
i=1
Fix η > 0. For each i ∈ N we have that supp(ψi u) ⊂ Ωi+1 \ Ωi−1 ,
(11.7)
and so, by the previous lemma, we may find εi > 0 so small that supp(ψi u)εi ⊂ Ωi+1 \ Ωi−1
(11.8) and
(ψi u)εi − ψi uW m,p (Ω) ≤
η , 2i
where we have used the previous exercise. Note that in view of (11.8), for every Ωi+1 \ U Ω only finitely many ∞ (Ω). In Ωi−1 cover U , and so the function v := ∞ (ψ u) belongs to C i ε i i=1 m,p particular, v ∈ Wloc (Ω). For x ∈ Ω , by (11.6), (11.7), and (11.8), (11.9)
u(x) =
(ψi u)(x),
i=1
v(x) =
(ψi u)εi (x).
i=1
Hence, (11.10)
u − vW m,p (Ω ) ≤
i=1
(ψi u)εi − ψi uW m,p (Ω) ≤
η ≤ η. 2i i=1
Letting → ∞, it follows from the Lebesgue monotone convergence theorem that u−vW m,p (Ω) ≤ η. This also implies that u−v (and, in turn, v) belongs to the space W m,p (Ω). ˙ m,p (Ω) (see Definition Remark 11.28. We will see later on that if u ∈ W p 11.17), then u ∈ Lloc (Ω). Hence, we can adapt the proof of the Meyers– ˙ m,p (Ω), then for every ε > 0 there Serrin theorem to show that if u ∈ W ∞ m,p ˙ exists a function v ∈ C (Ω) ∩ W (Ω) such that u − vW m,p (Ω) ≤ ε, despite the fact that neither u nor v need belong to W m,p (Ω). 1 Note
covered.
that we do not work simply with {Ωi+1 \ Ωi }, since, otherwise, ∂Ωi would not be
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11. Sobolev Spaces
Exercise 11.29. Let Ω ⊆ RN be an open set, let m ∈ N, let 1 ≤ p, q < ∞, ˙ m,q (Ω). Prove that for every ε > 0 there exists and let u ∈ Lq (Ω) ∩ W ∞ q ˙ v ∈ C (Ω) ∩ L (Ω) ∩ W m,p (Ω) such that u − vLq (Ω) + ∇m u − ∇m vLp (Ω) ≤ ε. Exercise 11.30. Prove that the function u(x) := |x|, x ∈ (−1, 1), belongs to W 1,∞ ((−1, 1)) but not to the closure of C ∞ ((−1, 1)) ∩ W 1,∞ ((−1, 1)). The previous exercise shows that the Meyers–Serrin theorem is false for p = ∞. This is intuitively clear, since if Ω ⊆ RN is an open set and {un }n is a sequence in C ∞ (Ω) ∩ W 1,∞ (Ω) such that un − uW 1,∞ (Ω) → 0, then u ∈ C 1 (Ω) (why?). However, the following weaker version of the Meyers– Serrin theorem holds. Exercise 11.31. Let Ω ⊂ RN be an open set with finite measure and let u ∈ W 1,∞ (Ω). Modify the proof of the Meyers–Serrin theorem to show that there exists a sequence {un }n in C ∞ (Ω) ∩ W 1,∞ (Ω) such that un − uL∞ (Ω) → 0, ∇un L∞ (Ω) → ∇uL∞ (Ω) , and ∇un (x) → ∇u(x) for LN -a.e. x ∈ Ω as n → ∞. We recall that given an open set Ω ⊆ RN , the space C ∞ (Ω) is defined as the space of all functions in C ∞ (Ω) that can be extended to functions in C ∞ (RN ). Exercise 11.32. Let Ω = B(0, 1) \ {x ∈ RN : xN = 0}. Show that the function u : Ω → R, defined by 1 if xN > 0, u(x) = u(x1 , . . . , xN ) := 0 if xN < 0, belongs to W 1,p (Ω) for all 1 ≤ p ≤ ∞, but it cannot be approximated by functions in C ∞ (Ω). The previous exercise shows that in the Meyers–Serrin theorem for general open sets Ω we may not replace C ∞ (Ω) with C ∞ (Ω). The problem is that the domain lies on both sides of its boundary. This motivates the next definition. Definition 11.33. An open set Ω ⊆ RN satisfies the segment property if for every x0 ∈ ∂Ω there exist r > 0 and a vector ν ∈ RN \ {0} such that if x ∈ Ω ∩ B(x0 , r), then x + θν ∈ Ω for all 0 < θ < 1. It turns out that the segment property is related to the regularity of the domain (see Definition 9.57). Indeed, we have the following result. Theorem 11.34. An open set Ω ⊆ RN satisfies the segment property if and only if ∂Ω is of class C.
11.2. Density of Smooth Functions
329
Proof. Assume that ∂Ω is of class C. Then for every x0 ∈ ∂Ω there exist local coordinates y = (y , yN ) ∈ RN −1 × R, with y = 0 at x = x0 , a continuous function f : RN −1 → R, and rx0 > 0 such that Ω ∩ B(x0 , rx0 ) is given in local coordinates by the set {(y , yN ) ∈ B(0, rx0 ) : yN > f (y )}. Note that y = R(x − x0 ), where R : RN → RN is an orthogonal transformation. Define the sets (11.11) V (x0 , r, t) := {(y , yN ) : y ∈ QN −1 (0, r), −t + f (y ) < yN < f (y ) + t}, (11.12) U (x0 , r, t) := R−1 (V (x0 , r, t)). Here, QN −1 (0, r) := (−r/2, r/2)N −1 . By taking r > 0 smaller, if necessary, and t > 0 sufficiently small, we have that the open set U (x0 , r, t) defined in (11.12) is contained in B(x0 , rx0 ). Then the neighborhood U (x0 , r, t/2) of x0 and the vector ν := R−1 ((0, . . . , 0, t/2)) satisfy the definition of segment property. Conversely, assume that Ω ⊆ RN satisfies the segment property. Fix x0 ∈ ∂Ω and find a neighborhood W of x0 and a vector ν ∈ RN \ {0} such that if x ∈ Ω ∩ W , then x + θν ∈ Ω for all 0 < θ < 1. Take local coordinates y = (y , yN ) ∈ RN −1 × R, with y = 0 at x = x0 and with y = eN at ν . For x ∈ Ω ∩ W , let y := L(x). The segment x = x0 + ν L({x + θν : 0 < θ < 1}) = y + {(0, θν) : 0 < θ < 1} will be called the arrow from y. From now on, for simplicity, we omit mention of L, and thus we write 0 ∈ ∂Ω, for example, in place of 0 ∈ L(∂Ω). Consider the closed segment S := {0} × [−η, η], where 0 < η < ν is chosen so small that S ⊆ W . Note that the points of S with yN > 0 are in Ω because they are on the arrow from 0 ∈ ∂Ω, while the points of S with yN < 0 are not in Ω because the arrows from them contain 0 that is not in Ω. Let Sl := {0} × [−η, − 12 η], Sm := {0} × [− 12 η, 12 η], and Su := {0} × [ 12 η, η] (here the subscripts stand for lower, middle, upper). Since Sl , Sm , and Su are compact sets, we may find finite open covers of cubes centered in S and contained in (RN \ Ω) ∩ W , W , and Ω ∩ W , respectively. By taking r > 0 sufficiently small, we obtain that the union of all these cubes contains the open set V := QN −1 (0, r) × (−η, η),
330
11. Sobolev Spaces
that Vl := QN −1 (0, r) × (−η, − 12 η) ⊂ RN \ Ω, and that Vu := QN −1 (0, r) × ( 12 η, η) ⊆ Ω. Define f ∈ QN −1 (0, r) → R by f (y ) := inf{yN : (y , yN ) ∈ Ω ∩ V }. Note that − 12 η ≤ f ≤ 12 η because Vl ⊂ RN \ Ω and Vu ⊂ Ω, and the infimum is attained because Ω is closed. Moreover, for every y = (y , yN ) ∈ V , if yN > f (y ), then y ∈ Ω, since y is on the arrow from (y , f (y )) ∈ Ω, while if yN < f (y ), then y ∈ RN \ Ω by the definition of f (y ). Thus, for every y ∈ QN −1 (0, r) the point (y , f (y )) belongs to ∂Ω and is the only such point in V . It remains to prove that f is continuous. Let y ∈ QN −1 (0, r) and let 0 < ε < 12 η. Define y+ := (y , f (y ) + ε),
y− := (y , f (y ) − ε).
By what we just proved and the bound |f | ≤ 12 η, we have that y+ ∈ Ω ∩ V and y− ∈ (RN \ Ω) ∩ V . Since Ω ∩ V and (RN \ Ω) ∩ V are open, there exist cubes QN (y+ , δ) ⊆ Ω ∩ V and QN (y− , δ) ⊆ (RN \ Ω) ∩ V . We claim that |f (y ) − f (z )| ≤ ε for all z ∈ QN (y , δ). Indeed, if f (z ) > f (y ) + ε for some z ∈ QN (y , δ), then (z , t) ∈ QN (y+ , δ) ⊆ Ω∩V for every f (y )+ε < t < f (z ) and this would contradict the minimum property of f (z ), while if f (z ) < f (y ) − ε, then the arrow from (z , f (z )) ∈ Ω would contain points of QN (y− , δ) ⊂ RN \ Ω, which is again a contradiction. This completes the proof. We now prove the density of functions Cc∞ (RN ) in W m,p (Ω) for domains with continuous boundary. Theorem 11.35. Let Ω ⊆ RN be an open set whose boundary is of class C, let m ∈ N, and let 1 ≤ p < ∞. Then the restriction to Ω of functions in Cc∞ (RN ) is dense in W m,p (Ω). To prove the theorem, we need an auxiliary result. Lemma 11.36. Let Ω ⊆ RN be an open set, let 1 ≤ p < ∞, and let u ∈ Lp (Ω). Extend u by zero outside Ω. Then for every ε > 0 there exists δ > 0 such that |u(x + h) − u(x)|p dx ≤ ε Ω
for all h ∈
RN ,
with h ≤ δ.
Proof. Exercise.
11.2. Density of Smooth Functions
331
Exercise 11.37. Let Ω ⊆ RN be an open set and let u, v ∈ C m (Ω), m ∈ N. Prove that for every α ∈ NN 0 with 1 ≤ |α| ≤ m, α α ∂ (uv) = ∂ α−β u∂ β v. β β≤α
Hint: Use induction on N . We now turn to the proof of Theorem 11.35. Proof. Fix u ∈ W m,p (Ω). By the Meyers–Serrin theorem (Theorem 11.24), without loss of generality, we may assume that u ∈ C ∞ (Ω) ∩ W m,p (Ω). Step 1: We first approximate u with a function with compact support. Consider a cut-off function ϕ ∈ Cc∞ (RN ) such that supp ϕ ⊆ B(0, 2), ϕ = 1 in B(0, 1) and 0 ≤ ϕ ≤ 1. For n ∈ N, define un (x) := ϕn (x)u(x),
ϕn (x) := ϕ(x/n),
x ∈ Ω.
By the Lebesgue dominated convergence theorem, we have that un → u in Lp (Ω) as n → ∞, while by Exercise 11.37, x α α 1 α β α−β β u(x) = ∂ ϕ ∂ ϕn (x)∂ ∂ un (x) = ∂ α−β u(x). β β n|β| n β≤α
β≤α
If β = 0, then again by the Lebesgue dominated convergence theorem, ϕn ∂ α u → ∂ α u in Lp (Ω) as n → ∞, while for β = 0, 1 c β α−β p |∂ ϕ(x/n)∂ u(x)| dx ≤ |∂ α−β u(x)|p dx → 0 n|β|p Ω n|β|p Ω as n → ∞. This concludes the proof. Step 2: By Step 1, we can assume that u = 0 outside a compact set. For every x0 ∈ ∂Ω there exist local coordinates y = (y , yN ) ∈ RN −1 × R, with y = 0 at x = x0 , a continuous function f : RN −1 → R, and rx0 > 0 such that Ω ∩ B(x0 , rx0 ) in local coordinates is given by (11.13)
{(y , yN ) ∈ B(0, rx0 ) : yN > f (y )}.
If the set Ω\ x∈∂Ω B(x, rx /2) is nonempty, for every x0 ∈ Ω\ x∈∂Ω B(x, rx /2) let B(x0 , rx0 ) be any open ball contained in Ω. The family {B(x, rx /2)}x∈Ω is an open cover of Ω. Since u = 0 outside a compact set K, we have that Ω ∩ K is compact. Hence, there is a finite number of balls B1 , . . . , B , where Bn := B(xn , rxn /2), that covers Ω ∩ K. Let {ψn }n=1 be a smooth partition of unity subordinated to B1 , . . . , B (see Exercise C.22). Fix n ∈ {1, . . . , } and define un := uψn ∈ W m,p (Ω) (see Exercise 11.27), where we extend un to be zero outside supp ψn . There are two cases. If supp ψn is contained in Ω, then we set vn := φn u ∈ Cc∞ (RN ). If supp ψn is not contained in Ω, let xn ∈ ∂Ω be such that supp ψn ⊆ Bn . For
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11. Sobolev Spaces
t > 0, using local coordinates in Bn = B(xn , rxn /2) (see Exercise 11.11) define the function un,t : Unt → R by un,t (y , yN ) := un (y , yN + t), where Unt := {(y , yN ) ∈ B(0, rxn ) : yN > f (y ) − t}. Note that for t > 0 sufficiently small Unt contains the set Ω ∩ Bn (written in local coordinates). Fix η > 0. By the previous lemma there exists tn > 0 so small that supp ψn + B(0, tn ) Untn and (11.14)
un,tn − un W m,p (Ω) ≤ η/2n .
Construct a function φn ∈ C ∞ (R) such that φn (yN ) = 1 if yN > f (y )−tn /2 and φn (yN ) = 0 if yN ≤ f (y ) − 3tn /4 and define vn := φn un,tn . Then vn ∈ Cc∞ (RN ) (provided we define un,tn to be zero, whenever φn is zero) and supp vn ⊂ Untn . Note that vn = un,tn in Ω, and so (11.14) may be rewritten as vn − uψn W m,p (Ω) ≤ η/2n . ∞ N Define the function v := n=1 vn . Then v ∈ Cc (R ) and so, v ∈ W m,p (Ω). Moreover, (11.15)
u − vW m,p (Ω) ≤
ψn u − vn W m,p (Ω) ≤ η
i=1
2−i ≤ η.
i=1
This concludes the proof.
Remark 11.38. Note that in view of Theorem 11.35 it follows that for all m ∈ N and 1 ≤ p < ∞, W0m,p (RN ) = W m,p (RN ). In Exercise 11.48 below we will show that there exists an open bounded domain Ω ⊂ R2 , with ∂Ω = ∂(Ω), such that W 1,p (Ω) ∩ C(Ω) is not dense in W 1,p (Ω), 1 ≤ p < ∞. As a corollary of Theorem 11.35, we can prove that piecewise affine functions are dense in W 1,p (Ω), whenever Ω ⊆ RN has a boundary of class C. Definition 11.39. An N -simplex Δ is the convex hull of N + 1 points xi ∈ RN (the vertices of Δ) that are not contained in a hyperplane, namely, N +1 N +1 θj xj , 0 ≤ θj ≤ 1, θj = 1 , Δ := x ∈ RN : x = j=1
j=1
and for all i ∈ {1, . . . , N + 1}, {xj − xi : j ∈ {1, . . . , N + 1} \ {i}} is a basis of RN .
11.2. Density of Smooth Functions
333
Let PA be the family of all continuous functions u : RN → R for which there exists a finite number of N -simplexes Δ1 , . . . , Δ with pairwise disjoint interiors such that the restriction of u to each Δi is affine and u = 0 outside Δ . i=1 i Theorem 11.40. Let Ω ⊆ RN be an open set whose boundary is of class C. Then the restriction to Ω of functions in PA is dense in W 1,p (Ω) for 1 ≤ p < ∞. Proof. Let u ∈ W 1,p (Ω). In view of Theorem 11.35 we may assume, without loss of generality, that u ∈ Cc∞ (RN ). Let K := supp u and consider B(0, R) ⊃ K. Fix α > 0 and construct N -simplexes Δ1 , . . . , Δ with pair wise disjoint interiors such that K ⊂ i=1 Δi ⊂ B(0, R) and diam Δi ≥ αri for all i = 1, . . . , , where ri is the radius of the inscribed ball in Δi . Let v be the continuous piecewise affine function that coincides with u in all the vertices of the simplexes. Then by Taylor’s formula (see Theorem 9.9), sup |u(x) − v(x)| ≤ ch2 sup ∇2 u(x) ≤ ch2 sup ∇2 u(x),
x∈Δi
x∈Δi
x∈RN
sup ∇u(x) − ∇v(x) ≤ ch sup ∇ u(x) ≤ ch sup ∇2 u(x) 2
x∈Δi
x∈Δi
x∈RN
for all i = 1, . . . , , where h := max1≤i≤ diam Δi . Thus,
∇u − ∇v dx ≤ p
Ω
∇u − ∇v dx = p
RN
≤ c u hp
i=1
∇u − ∇vp dx Δi
LN (Δi ) ≤ cu hp LN (B(0, R)),
i=1
where cu := c supx∈RN ∇2 u(x). A similar estimate holds for Ω |u − v|p dx. By choosing h sufficiently small (and so increasing the number of N simplexes), we have the desired approximation. Remark 11.41. If Ω ⊆ RN is an open set and u ∈ W 1,p (Ω) for 1 ≤ p < ∞, then for any U Ω we can construct a sequence {un }n in PA such that un → u in W 1,p (U ). Indeed, by the Meyers–Serrin theorem we may find vn ∈ C ∞ (Ω) ∩ W 1,p (Ω) such that vn → u in W 1,p (Ω). Construct a cut-off function ψ ∈ Cc∞ (Ω) such that ψ = 1 in U . Since ψvn ∈ Cc∞ (Ω), we may extend ψvn to be zero outside Ω and then apply the previous theorem to find un ∈ PA such that ψvn − un W 1,p (Ω) ≤ 1/n. Using the fact that ψ = 1 in U , it follows that u − un W 1,p (U ) ≤ u − vn W 1,p (Ω) + vn − un W 1,p (U ) ≤ u − vn W 1,p (Ω) + ψvn − un W 1,p (Ω) . It suffices to let n → ∞.
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11. Sobolev Spaces
We conclude this section by discussing the density of Cc∞ (RN ) functions ˙ m,p (RN ) (see Definition 11.17). in the homogeneous Sobolev space W x Exercise 11.42. Prove that if f0 ∈ Lp (R), p > 1, and fn+1 (x) := 0 fn (t) dt, for n ∈ N0 , then the function gn (x) := fn (x)|x|−n , x ∈ R, belongs to Lp (R) for every n ∈ N0 . Hint: Use Hardy’s inequality (see Theorem C.41). ˙ m,p (RN ), where m ∈ N and 1 ≤ p < ∞. Then Theorem 11.43. Let u ∈ W there exists a sequence {un }n of functions in Cc∞ (RN ) such that ∂ α un → ∂ α u in Lp (RN ) for every multi-index α with |α| = m if and only if N ≥ 2 or p > 1. Proof. Step p = 1. Let φ ∈ Cc∞ (R) be 1: Assume first that N = 1 and (m) = φ. Then u belongs to such that R φ(t) dt = 0 and let u be such that u ˙ m,1 (R). Assume, by contradiction, that there exists a sequence {un }n in W (m) Cc∞ (R) such that un → u(m) in L1 (R). Then by the fundamental theorem of calculus, 0= R
u(m) n (t) dt →
R
φ(t) dt = 0,
which is a contradiction. ˙ m,p (R). Step 2: We consider the case N = 1 and 1 < p < ∞. Fix u ∈ W ˙ m,p (R) with Using standard mollifiers, we have that C ∞ (R) is dense in W respect to the seminorm (11.3). Thus, without loss of generality, we may ˙ m,p (R). Consider a cut-off function ϕ ∈ C ∞ (R) assume that u ∈ C ∞ (R) ∩ W c such that supp ϕ ⊆ B(0, 2), ϕ = 1 in B(0, 1) and 0 ≤ ϕ ≤ 1. For n ∈ N, x define f0 := u(m) , fn+1 (x) := 0 fn (t) dt, for n ∈ N0 , and un (x) := ϕn (x)fm (x),
ϕn (x) := ϕ(x/n),
x ∈ R.
Then un ∈ Cc∞ (R). Thus, it remains to show that un → u(m) in Lp (R). By Exercise 11.37 and the definition of fm , we have m m m (k) m 1 (k) x (m) (m−k) un (x) = fk (x). (x) = ϕ ϕn (x)fm n k k nk (m)
k=0
k=0
For k = 0 we have that ϕn f0 = ϕn u(m) → u(m) in Lp (R) by the Lebesgue dominated convergence theorem. On the other hand, for 1 ≤ k ≤ m, c 1 (k) p |ϕ (x/n)f (x)| dx ≤ |fk (x)|p dx k nkp R nkp B(0,2n)\B(0,n) 1 |f (x)|p dx → 0 ≤c kp k |x| B(0,2n)\B(0,n) as n → ∞, since the function gk (x) := fk (x)|x|−k , x ∈ R, belongs to Lp (R) by Exercise 11.42.
11.2. Density of Smooth Functions
335
Step 3: Finally, we study the case N ≥ 2 and 1 ≤ p < ∞. Fix u ∈ ˙ m,p (RN ). Con˙ m,p (RN ). As before we can assume that u ∈ C ∞ (RN ) ∩ W W ∞ N sider a cut-off function ϕ ∈ Cc (R ) such that supp ϕ ⊂ B(0, 2), ϕ = 1 in B(0, 1) and 0 ≤ ϕ ≤ 1. For r > 0, define the annulus Ar := B(0, 2r)\B(0, r). By Poincar´e’s inequality (see Theorem 13.27) applied to the annulus A1 , there exists a constant c = c(m, p, N ) > 0 such that for every v ∈ W m,p (A1 ) and every k = 0, . . . ,m − 1, k k p ∇ v(x) − ∇ pA1 (v)(x) dx ≤ c ∇m v(x)p dx, A1
A1
where for every measurable set with positive finite measure E ⊂ RN , pE (v) is the polynomial of degree m − 1 such that 1 (∂ α v(x) − ∂ α pE (v)(x)) dx = 0 (11.16) LN (E) E for every multi-index α ∈ NN 0 with 0 ≤ |α| ≤ m − 1. A rescaling argument (exercise) shows that for every v ∈ W m,p (Ar ), r > 0, k k p p(m−k) ∇ v(x) − ∇ pAr (v)(x) dx ≤ cr ∇m v(x)p dx. (11.17) Ar
Ar
For n ∈ N, define (11.18) un (x) := ϕn (x)(u(x) − pAn (u)(x)),
ϕn (x) := ϕ(x/n),
x ∈ RN .
By Exercise 11.37, for every multi-index α with |α| = m, α α ∂ β ϕn (x)∂ α−β (u(x) − pAn (u)(x)) ∂ un (x) = β β≤α x α 1 β = ∂ α−β (u(x) − pAn (u)(x)). ∂ ϕ β n|β| n β≤α
If β = 0, then by the Lebesgue dominated convergence theorem and the fact that pAn (u) has degree m − 1, ϕn ∂ α (u − pAn (u)) = ϕn ∂ α u → ∂ α u in Lp (RN ) as n → ∞, while for β = 0, by (11.17), 1 |∂ β ϕ(x/n)∂ α−β (u(x) − pAn (u)(x))|p dx n|β|p An c |∂ α−β (u(x) − pAn (u)(x))|p dx ≤ |β|p n An cnp(m−|α−β|) m p ∇ u(x) dx = c ∇m u(x)p dx → 0 ≤ np|β| An An as n → ∞. This concludes the proof.
Exercise 11.44. Let m ∈ N and let v ∈ L1loc (RN ) be such that the function w(x) := (1 + x2 )m/2 v(x), x ∈ RN , belongs to L2 (RN ).
336
11. Sobolev Spaces
(i) Prove that there exists a function u ∈ W m,2 (RN ) whose Fourier transform is v. (ii) State and prove a similar result for 1 < p < ∞.
11.3. Absolute Continuity on Lines The next theorem relates weak partial derivatives with the (classical) partial derivatives. This result is the analog of Theorem 7.16. In what follows, we use the notation (E.2) in Appendix E. Given xi ∈ and a set E ⊆ RN , we write
RN −1
Exi := {xi ∈ R : (xi , xi ) ∈ E}.
(11.19)
Moreover, if v : Ω → R is Lebesgue integrable, with a slight abuse of notation, if Ωxi is empty, we set v(xi , xi ) dxi := 0, Ωx
i
so that by Fubini’s theorem v(x) dx = (11.20) Ω
RN −1
v(xi , xi ) dxi dxi . Ωx
i
Theorem 11.45 (Absolute continuity on lines). Let Ω ⊆ RN be an open set and let 1 ≤ p < ∞. A function u ∈ Lp (Ω) belongs to the space W 1,p (Ω) if and only if it has a representative u that is absolutely continuous on LN −1 -a.e. line segments of Ω that are parallel to the coordinate axes and whose firstorder (classical) partial derivatives belong to Lp (Ω). Moreover, the (classical) partial derivatives of u agree LN -a.e. with the weak derivatives of u. Proof. Step 1: Assume that u ∈ W 1,p (Ω). Consider a sequence of standard mollifiers {ϕε }ε>0 and for every ε > 0 define uε := u ∗ ϕε in Ωε := {x ∈ Ω : dist(x, ∂Ω) > ε}. By Lemma 11.25, ∇uε (x) − ∇u(x)p dx = 0. lim ε→0+
Ωε
It follows by Fubini’s theorem and (11.20) that for all i = 1, . . . , N , ∇uε (xi , xi ) − ∇u(xi , xi )p dxi dxi = 0, lim ε→0+
RN −1
(Ωε )x
i
where (Ωε )xi := {xi ∈ R : (xi , xi ) ∈ Ωε }, and so we may find a subsequence {εn }n such that for all i = 1, . . . , N and for LN −1 a.e. xi ∈ RN −1 , ∇uεn (xi , xi ) − ∇u(xi , xi )p dxi = 0. (11.21) lim n→∞ (Ω ) εn x
i
11.3. Absolute Continuity on Lines
337
Set un := uεn and E := {x ∈ Ω : limn→∞ un (x) exists in R}. Define lim un (x) if x ∈ E, n→∞ u(x) := 0 otherwise. Since by Theorem C.16, {un }n converges pointwise to u at every Lebesgue point of u, the set E contains every Lebesgue point of u. It follows from Corollary B.119 that LN (Ω \ E) = 0, and so u is a representative of u. It remains to prove that u has the desired properties. By Fubini’s theorem and (11.20) for every i = 1, . . . , N we have that
∇u(xi , xi )p dxi dxi < ∞ RN −1
and
RN −1
Ωx
i
L1 ({xi ∈ Ωxi : (xi , xi ) ∈ / E}) dxi = 0,
and so we may find a set Ni ⊂ RN −1 , with LN −1 (Ni ) = 0, such that for all xi ∈ RN −1 \ Ni for which Ωxi is nonempty we have that ∇u(xi , xi )p dxi < ∞, (11.22) Ωx
i
(11.21)
holds and (xi , xi ) ∈ any such xi and let
E for L1 -a.e. xi ∈ Ωxi .
I ⊆ Ωxi be a maximal interval. Fix t0 ∈ I such Fix that (xi , t0 ) ∈ E and let t ∈ I. For all n large, the interval of endpoints t and t0 is contained in (Ωεn )xi and so, since un ∈ C ∞ (Ωεn ), by the fundamental theorem of calculus, t ∂i un (xi , s) ds. un (xi , t) = un (xi , t0 ) + t0
(xi , t0 )
Since by (11.21),
∈ E, we have
(11.23)
lim
t
n→∞ t 0
un (xi , t0 )
→ u(xi , t0 ) ∈ R. On the other hand,
|∂i un (xi , s) − ∂i u(xi , s)| ds = 0.
Hence as n → ∞, un (xi , t)
→
u(xi , t0 )
t
+
∂i u(xi , s) ds.
t0
Note that by the definition of E and u, this implies, in particular, that t ∂i u(xi , s) ds (11.24) (xi , t) ∈ E and u(xi , t) = u(xi , t0 ) + t0
for all t ∈ I. Hence, by Theorem 3.16, the function u(xi , ·) is absolutely continuous in I and ∂i u(xi , t) = ∂i u(xi , t) for L1 -a.e. t ∈ I.
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11. Sobolev Spaces
Step 2: Assume that u admits a representative u that is absolutely continuous on LN −1 -a.e. line segments of Ω that are parallel to the coordinate axes, and whose first-order (classical) partial derivatives belong to Lp (Ω). Fix i = 1, . . . , N and let xi ∈ RN −1 be such that u(xi , ·) is absolutely continuous on every connected component of the open set Ωxi . Then for every function ϕ ∈ Cc∞ (Ω), by the integration by parts formula for absolutely continuous functions, we have u(xi , t)∂i ϕ(xi , t) dt = − ∂i u(xi , t)ϕ(xi , t) dt. Ωx
Ωx
i
i
LN −1 -a.e.
xi
RN −1 ,
∈ integrating over RN −1 and using Since this holds for Fubini’s theorem yields u(x)∂i ϕ(x) dx = − ∂i u(x)ϕ(x) dx, Ω
Ω
which implies that ∈ is the weak partial derivative of u with respect to xi . This shows that u ∈ W 1,p (Ω). ∂u ∂xi
Lp (Ω)
Remark 11.46. If instead of assuming that u ∈ W 1,p (Ω) we only assume ˙ 1,p (Ω), then Step 1 of the previous proof still proves that u has that u ∈ W a representative u that is absolutely continuous on LN −1 -a.e. line segments of Ω that are parallel to the coordinate axes and whose first-order (classical) partial derivatives belong to Lp (Ω). Exercise 11.47. Let Ω ⊆ RN be an open set, let m ∈ N and let 1 ≤ p < ∞. Prove that if u belongs to W m,p (Ω), then u has a representative u such that for LN −1 -a.e. xi ∈ RN −1 the function u(xi , ·) is absolutely continuous in Ωxi together with all its derivatives of order up to m − 1. The next exercise shows that there exists an open bounded domain Ω ⊂ with ∂Ω = ∂(Ω), such that W 1,p (Ω) ∩ C(Ω) is not dense in W 1,p (Ω), 1 ≤ p < ∞ (see Theorem 11.35). R2 ,
Exercise 11.48. (i) Constructa sequence of open balls B((tn , 0),rn ) ⊂ R2 , ∞ 1 with 0 < tn < 1 and ∞ n=1 rn < 2 , in such a way that n=1 (tn − rn , tn + rn ) is dense in (0, 1) and ∂Ω = ∂(Ω). (ii) Let Ω := B((1, 0), 1) \
∞
B((tn , 0), rn )
n=1
and for (x1 , x2 ) ∈ Ω define ⎧ for 0 < x1 < 1/2, x2 > 0, ⎪ ⎨ 1 0 for 0 < x1 < 1/2, x2 < 0, u(x1 , x2 ) := ⎪ ⎩ anything convenient otherwise.
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339
Prove that u ∈ W 1,p (Ω) for all 1 ≤ p < ∞. (iii) Let Γ be the set of x1 ∈ ( 14 , 12 ) with (x1 , 0) ∈ ∂Ω. Assume by contradiction that there exists a sequence {un }n in W 1,p (Ω) ∩ C(Ω) that converges to u in W 1,p (Ω) and prove that for all n ∈ N and for all x2 ∈ (0, 12 ), ∂u n |un (x1 , x2 ) − un (x1 , −x2 )| dx1 ≤ (x , s) dx1 ds. 1 ∂x 2 Γ Γ×(−x2 ,x2 ) (iv) Using Fubini’s theorem, prove that there is a subsequence {unk }k of {un }n such that for L1 -a.e. x2 ∈ (0, 12 ), lim |unk (x1 , x2 ) − unk (x1 , −x2 )| dx1 = |u(x1 , x2 ) − u(x1 , −x2 )| dx1
k→∞ Γ
Γ
and show that this is in contradiction to part (iii). In the same spirit of Theorem 11.45, we discuss next Lipschitz functions and their relation with W 1,∞ . The next exercise shows that there exist functions that are in W 1,∞ (Ω) but are not Lipschitz continuous. Exercise 11.49. Let Ω ⊂ R2 be the open set defined in polar coordinates by −π < θ < π and r > 1. Prove that the function u(r, θ) = θ belongs to W 1,∞ (Ω) but that it is not Lipschitz continuous. The relation between Lipschitz functions and functions in W 1,∞ is discussed in the next exercise. Exercise 11.50 (Lipschitz functions and W 1,∞ ). (i) Given a function u ∈ ¯ that is Lipschitz conL1loc (RN ) prove that u has a representative u tinuous if and only if its distributional gradient ∇u belongs to L∞ (RN ; RN ). ¯ that is (ii) Prove that a function u ∈ L1loc (RN ) has a representative u bounded and Lipschitz continuous if and only if u ∈ W 1,∞ (RN ). (iii) Let Ω ⊆ RN be an open set and let u : Ω → R be bounded and Lipschitz continuous. Prove that u ∈ W 1,∞ (Ω). (iv) Let Ω ⊆ RN be an extension domain (see Definition 12.16) for W 1,∞ (Ω) and let u ∈ W 1,∞ (Ω). Prove that u has a representative u ¯ that is bounded and Lipschitz continuous. As a consequence of Theorem 11.45, of the properties of absolutely continuous functions, and of the previous exercises, we have the following results. Exercise 11.51. Let Ω ⊆ RN be an open set and let 1 ≤ p ≤ ∞.
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(i) (Chain rule) Let f : R → R be Lipschitz continuous and let u ∈ W 1,p (Ω). Assume that f (0) = 0 if Ω has infinite measure. Prove that f ◦ u ∈ W 1,p (Ω) and that for all i = 1, . . . , N and for LN -a.e. x ∈ Ω, ∂i (f ◦ u)(x) = f (u(x))∂i u(x), where f (u(x))∂i u(x) is interpreted to be zero whenever ∂i u(x) = 0. (ii) (Product rule) Let u, v ∈ W 1,p (Ω) ∩ L∞ (Ω). Prove that uv ∈ W 1,p (Ω) ∩ L∞ (Ω) for all i = 1, . . . , N and that for LN -a.e. x ∈ Ω, ∂i (uv)(x) = v(x)∂i u(x) + u(x)∂i v(x). N −1 × R : x > 0} and (iii) (Reflection) Let Ω = RN N + := {(x , xN ) ∈ R ). Prove that the function let u ∈ W 1,p (RN + u(x) if xN > 0, v(x) := u(x , −xN ) if xN < 0
belongs to W 1,p (RN ) and that for all i = 1, . . . , N and for LN -a.e. x ∈ RN , if xN > 0, ∂i u(x) ∂i v(x) = δ iN (−1) ∂i u(x , −xN ) if xN < 0, where δiN is the Kronecker delta, that is, δiN = 1 if i = N and δiN = 0 otherwise. 1,1 (iv) Let E ⊂ R be such that L1 (E) = 0, let u ∈ Wloc (Ω), and let u be its precise representative given in Theorem 11.45. Prove that ∇u(x) = 0 for LN -a.e. x ∈ (u)−1 (E).
(v) Prove Exercise 11.23 using Theorem 11.45. (vi) Prove analogous versions of parts (i)–(v) in the case in which the ˙ 1,p (Ω). functions u and v are in W Exercise 11.52. Let Ω ⊆ RN be an open set and let 1 ≤ p ≤ ∞. Let f : Rd → R be a Lipschitz continuous function such that the set Σf := {u ∈ Rd : f is not differentiable at u} is purely H1 -unrectifiable (see Definition 3.70). Assume that f (0) = 0 if Ω has infinite measure. Prove that f ◦ u ∈ W 1,p (Ω) for all u ∈ W 1,p (Ω; Rd ) and that for all i = 1, . . . , N and for LN -a.e. x ∈ Ω, ∂f ∂uj ∂(f ◦ u) (x) = (u(x)) (x), ∂xi ∂uj ∂xi d
j=1
where
∂uj ∂f ∂uj (u(x)) ∂x (x) i
is interpreted to be zero whenever
∂uj (x) = 0. ∂xi
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Theorem 11.53 (Change of variables). Let Ω, U ⊆ RN be open sets, let Ψ : U → Ω be invertible, with Ψ and Ψ−1 Lipschitz continuous functions, and let u ∈ W 1,p (Ω), 1 ≤ p ≤ ∞. Then u ◦ Ψ ∈ W 1,p (U ) and for all i = 1, . . . , N and for LN -a.e. y ∈ U , ∂u ∂Ψj ∂(u ◦ Ψ) (y) = (Ψ(y)) (y). ∂yi ∂xj ∂yi N
j=1
Proof. We only prove the case 1 ≤ p < ∞. Let Ωn Ωn+1 be such that Ω= ∞ n=1 Ωn . Fix n ∈ N and for 0 < ε < dist(Ωn , ∂Ω) define uε := u ∗ ϕε in Ωn . Let Un := Ψ−1 (Ωn ) and for y ∈ Un set vε (y) := uε (Ψ(y)).
(11.25)
By Rademacher’s theorem the Lipschitz continuous function Ψ is differentiable LN -a.e. in Un . Since uε ∈ C ∞ (Ωn ), we conclude that for all i = 1, . . . , N and for LN -a.e. y ∈ Un , ∂uε ∂Ψj ∂vε (y) = (Ψ(y)) (y). ∂yi ∂xj ∂yi j=1 ∂uε ε N (y) ≤ Lip Ψ (Ψ(y)) Hence, ∂v , and so j j=1 ∂yi ∂xj ∇vε (y)p dy ≤ c ∇uε (Ψ(y))p dy Un Ψ−1 (Ωn ) ∇uε (Ψ(y))p | det JΨ (y)| dy ≤c Ψ−1 (Ωn ) =c ∇uε (x)p dx, N
(11.26)
Ωn
where we have used the fact that | det JΨ | is bounded from below by a positive constant (since Ψ−1 is Lipschitz continuous) and Theorem 9.52. Taking a sequence εj → 0+ , the previous inequality implies that p ∇vεj (y) − ∇vεl (y) dy ≤ c ∇uεj (y) − ∇uεl (y)p dx Un
Ωn
for all j, l ∈ N. Similarly, p |vεj (y) − vεl (y)| dy ≤ c Un
|uεj (y) − uεl (y)|p dx Ωn
for all j, l ∈ N. Since uεj → u in W 1,p (Ωn ), it follows that {vεj }j is a Cauchy sequence in W 1,p (Un ), and so it converges to a function v (n) ∈ W 1,p (Un ). By extracting a subsequence, we can assume that vεj and ∇vεj converge, respectively, to v (n) and ∇v (n) pointwise LN -a.e. in Un . Since Ψ−1 has the (N ) property (see Exercise 9.54) and uεj → u and ∇uεj → ∇u pointwise
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11. Sobolev Spaces
LN -a.e. in Ωn , it follows that uεj ◦ Ψ → u ◦ Ψ and (∇uεj ) ◦ Ψ → ∇u ◦ Ψ pointwise LN -a.e. in Un . In view of (11.25) and (11.26), we conclude that v (n) (y) = u(Ψ(y)) for LN -a.e. y ∈ Ωn and that for all i = 1, . . . , N and for LN -a.e. y ∈ Un , ∂u ∂Ψj ∂(u ◦ Ψ) (y) = (Ψ(y)) (y). ∂yi ∂xj ∂yi N
(11.27)
j=1
Since u ◦ Ψ = v (n) in Un , it follows that u ◦ Ψ ∈ W 1,p (Un ) for all n ∈ N. Reasoning as in the first part of the proof, it follows from (11.27) and Theorem 9.52 that p ∇(u ◦ Ψ)(y) dy ≤ c ∇u(Ψ(y))p | det JΨ (y)| dy U Ψ−1 (Ω) = c ∇u(x)p dx, Ω
and similarly
|(u ◦ Ψ)(y)|p dy ≤ c U 1,p W (Un )
|u(x)|p dx. Ω
for all n ∈ N, applying either Theorem 11.45 or the Since u ◦ Ψ ∈ definition of weak derivatives yields u ◦ Ψ ∈ W 1,p (U ). To extend the previous theorem to W m,p (Ω) we need the extension of Faa di Bruno’s formula (see [123], [219]) to calculate higher-order derivatives of a composition. Theorem 11.54 (Faa di Bruno). Let Ω ⊆ RN and U ⊆ RM be open sets, let u : Ω → R be of class C m and let Ψ : U → Ω be of class C m . Then u ◦ Ψ belongs to C m (U ) and for every multi-index α ∈ NM 0 , with 1 ≤ |α| ≤ m, |β|
/ ∂ |γi | Ψl ∂ |β| u ∂ |α| i (u ◦ Ψ)(y) = c (Ψ(y)) (y), α,β,γ,l ∂y α ∂xβ ∂y γi i=1
where cα,β,γ,l ∈ R, the sum is done over all β ∈ NN 0 with 1 ≤ |β| ≤ |α|, γ = |β| M (γ1 , . . . , γ|β| ), γi ∈ N0 , with |γi | > 0 and i=1 γi = α, and l = (l1 , . . . , l|β| ), li ∈ {1, . . . , N }, i = 1, . . . , |β|. Proof. The proof is by induction on m and is left as an exercise.
We remark that the multi-indices γi , i = 1, . . . , |β|, are not pairwise ∂ |γi | Ψ
distinct so some of the derivatives ∂yγi li are repeated. The constants cα,β,γ,l can be computed explicitly (see [70]).
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Definition 11.55. Given an open set Ω ⊆ RN , m ∈ N0 , M ∈ N, and 0 < α ≤ 1, we define the space C m,α (Ω; RM ) as the space of all functions in C m (Ω; RM ) such that |∂ α u|C 0,α (Ω;RM ) < ∞. uC m,α (Ω;RM ) := uC m (Ω;RM ) + |α|=m
We recall that for a function v : Ω → RM , v(x) − v(y)M |v|C 0,α (Ω;RM ) := sup . x − yαN x, y∈Ω, x=y As usual, when M = 1 we write C m,α (Ω) for C m,α (Ω; R). Exercise 11.56. Given an open set Ω ⊆ RN , m ∈ N0 and 0 < α ≤ 1. Prove that the space C m,α (Ω), 0 < α ≤ 1, is a Banach space. Theorem 11.57. Let Ω, U ⊆ RN be open sets, let m ∈ N, let Ψ : U → Ω be invertible, with Ψ ∈ C m−1,1 (U ; RN ) and Ψ−1 Lipschitz continuous, and let u ∈ W m,p (Ω), 1 ≤ p ≤ ∞. Then u ◦ Ψ belongs to W m,p (U ) and for every N multi-index α ∈ NM 0 , with 0 < |α| ≤ m, and for L -a.e. y ∈ U , |β|
/ ∂ |γi | Ψl ∂ |α| ∂ |β| u i (u ◦ Ψ)(y) = c (Ψ(y)) (y), α,β,γ,l ∂y α ∂xβ ∂y γi i=1
where the summation is done as in Theorem 11.54. Proof. The proof is similar to the one of Theorem 11.53 and we omit it.
be an open set. Given n ∈ N, let u1 , . . . , Exercise 11.58. Let Ω ⊆ 1,p un ∈ W (Ω), 1 ≤ p ≤ ∞. Prove that vn := max1≤k≤n uk ∈ W 1,p (Ω) and that |∂i vn (x)| ≤ max |∂i uk (x)| RN
1≤k≤n
for
LN -a.e.
x ∈ Ω and all i = 1, . . . , N .
Exercise 11.59. Let u ∈ W 1,p (RN ), 1 ≤ p < ∞, and for r > 0 let 1 χ , χr := αN rN B(0,r) where we recall that αN = LN (B(0, 1)). Prove that |u| ∗ χr belongs to W 1,p (RN ) and that ∂i (|u| ∗ χr ) = χr ∗ ∂i |u| for i = 1, . . . , N . Exercise 11.60. Let u ∈ W 1,p (RN ), 1 < p < ∞, and let M(u) be the maximal function of u (see Definition C.25). (i) Let {rn }n be an enumeration of the positive rationals and define vn := max1≤k≤n |u| ∗ χrk . Prove that vn ∈ W 1,p (RN ) and that |∂i vn (x)| ≤ M(∂i u)(x) for LN -a.e. x ∈ RN . Deduce that the sequence {vn }n is bounded in W 1,p (RN ).
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11. Sobolev Spaces
(ii) Prove that M(u) belongs to W 1,p (RN ) and that ∂i (M(u)) = M(∂i u) for i = 1, . . . , N .
11.4. Duals and Weak Convergence In this section we study the dual spaces of W m,p (Ω) and W0m,p (Ω). Let Ω ⊆ RN be an open set, let m ∈ N. In this section it will be convenient to consider in W m,p (Ω) the equivalent norm 1/p ∂ α upLp (Ω) (11.28) uW m,p (Ω) := 0≤|α|≤m
for 1 ≤ p < ∞ and uW m,∞ (Ω) := max ∂ α uL∞ (Ω)
(11.29)
0≤|α|≤m
if p = ∞, where we recall that ∂ 0 u = u. For every 1 ≤ q ≤ ∞ we consider the space Lq (Ω; RM ), where M is the number of multi-indices α with 0 ≤ |α| ≤ m. Given f ∈ Lq (Ω; RM ) we label the components of the vector-valued function f as fα , with 0 ≤ |α| ≤ m, and we consider in Lq (Ω; RM ) the equivalent norm 1/q fα qLq (Ω) f Lq (Ω;RM ) := 0≤|α|≤m
if 1 ≤ q < ∞ and f L∞ (Ω;RM ) := max fα L∞ (Ω) 0≤|α|≤m
if q = ∞. The idea is to consider W m,p (Ω) as a closed subspace of Lp (Ω; RM ). In this way any element of L ∈ (W m,p (Ω)) can be extended by the Hahn– Banach theorem (see Theorem A.32) to an element of (Lp (Ω; RM )) . The Riesz representation theorem in Lp (Ω; RM ) (see Theorem B.92) will give a representation for the extension of L and, in turn, for L. We begin with the case 1 ≤ p < ∞. Observe that if p is the H¨ older p M conjugate exponent of p and f ∈ L (Ω; R ), then the functional fα (x)∂ α u(x) dx, u ∈ W m,p (Ω), (11.30) L(u) := 0≤|α|≤m Ω
belongs to (W m,p (Ω)) . Indeed, by H¨older’s inequality (for functions and for finite sums) we have that fα Lp (Ω) ∂ α uLp (Ω) ≤ f Lp (Ω;RM ) uW m,p (Ω) |L(u)| ≤ 0≤|α|≤m
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345
for all u ∈ W m,p (Ω), and thus by (11.28), (11.31)
L(W m,p (Ω)) =
|L(u)| ≤ f Lp (Ω;RM ) . u∈W m,p (Ω)\{0} uW m,p (Ω) sup
Note that in general we do not have equality in the previous inequality (see Remark 11.63 below). Exercise 11.61. Let Ω ⊆ RN be an open set, and let 1 ≤ p < ∞. Using the Riesz representation theorem in Lp (Ω), prove that for every L ∈ (Lp (Ω; RM )) there exists a unique f ∈ Lp (Ω; RM ) such that fα (x)gα (x) dx for all g ∈ Lp (Ω; RM ). L(g) = 0≤|α|≤m Ω
Prove also that L(Lp (Ω;RM )) = f Lp (Ω;RM ) . Theorem 11.62 (Riesz’s representation theorem in W m,p ). Let Ω ⊆ RN be older conjugate an open set, let m ∈ N, let 1 ≤ p < ∞, and let p be its H¨ exponent. Then for every bounded linear functional L ∈ (W m,p (Ω)) there exist f ∈ Lp (Ω; RM ) such that fα (x)∂ α u(x) dx (11.32) L(u) = 0≤|α|≤m Ω
for all u ∈ W m,p (Ω) and (11.33)
L(W m,p (Ω)) = f Lp (Ω;RM ) .
Proof. Consider the application (11.34)
T : W m,p (Ω) → Lp (Ω; RM ) u → (∂ 0 u, ∂ α1 u, . . . , ∂ αM u),
where α1 , . . . , αM are all the multi-indices α with 1 ≤ |α| ≤ m. The operator T is one-to-one and continuous and it preserves the norm; that is, T (u)Lp (Ω;RM ) = uW m,p (Ω) for all u ∈ W m,p (Ω). Hence, also by Theorem 11.12, the subspace Y := T (W m,p (Ω)) is closed in Lp (Ω; RM ). Given L ∈ (W m,p (Ω)) , define the functional L1 : Y → R g = (g0 , . . . , gαM ) → L(T −1 ((g0 , . . . , gαM ))). Since T preserves the norm, we have that L1 is linear and continuous, with L1 Y = L(W m,p (Ω)) . By the Hahn–Banach theorem (see Theorem A.32) we may extend L1 as a continuous linear operator L2 : Lp (Ω; RM ) → R in such a way that L2 (Lp (Ω;RM )) = L1 Y = L(W m,p (Ω)) . Note that since Y is closed, it is not dense in Lp (Ω; RM ), and so the extension will not be unique.
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By the previous exercise there exist a unique function f ∈ Lp (Ω; RM ) such that L1 (g) = fα (x)gα (x) dx 0≤|α|≤m Ω
for all g ∈ Lp (Ω; RM ) and L(W m,p (Ω)) = L2 (Lp (Ω;RM )) = f Lp (Ω;RM ) . It follows that (11.32) holds. Remark 11.63. Note that the previous theorem does not imply that the dual of W m,p (Ω) is Lp (Ω; RM ). Indeed, if f belongs to Lp (Ω; RM ), then we have shown that the functional L defined in (11.30) belongs to (W m,p (Ω)) and that (11.31) holds. On the other hand, by Theorem 11.62 there ex ists h in Lp (Ω; RM ) (possibly different from f ) such that L(W m,p (Ω)) = hLp (Ω;RM ) . Thus, we have that
L(W m,p (Ω)) = min{hLp (Ω) : h ∈ Lp (Ω; RM ) such that (11.32) holds}.
To explore the possible lack of uniqueness, assume that h ∈ Lp (Ω; RM ) is such that α fα ∂ u dx = hα ∂ α u dx 0≤|α|≤m Ω
0≤|α|≤m Ω
for all u ∈ W m,p (Ω). In particular, we obtain that fα ∂ α φ dx = hα ∂ α φ dx 0≤|α|≤m Ω
0≤|α|≤m Ω
for all φ ∈ D(RN ). If we rewrite this identity in the sense of distributions (see Definition 10.14), we get (−1)|α| ∂α Tfα χΩ (φ) = (−1)|α| ∂α Thα χΩ (φ) 0≤|α|≤m
0≤|α|≤m
for all φ ∈ D(RN ), or, equivalently, (−1)|α| ∂α (fα χΩ ) = (11.35) 0≤|α|≤m
(−1)|α| ∂α (hα χΩ )
0≤|α|≤m
for all φ ∈ D(RN ). Hence, we have shown that given f ∈ Lp (Ω; RM ), any solution h ∈ Lp (Ω; RM ) (in the sense of distributions) of the partial differential equation (11.35) in RN will give rise to a different representation in (11.32). In particular, when m = 1 and Ω and fi ∈ Lp (Ω), i = 1, . . . , N are sufficiently smooth, then any solution of the partial differential equation div h = div f in Ω, h · ν = f · ν on ∂Ω, where f := (f1 , . . . , fN ) will do.
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347
In view of Theorem 11.62, we can characterize weak convergence in W m,p (Ω). Exercise 11.64. Let Ω ⊆ RN be an open set, let m ∈ N, let 1 ≤ p < ∞, and let {un }n be a sequence in W m,p (Ω). Prove that un u in W m,p (Ω) if and only if un u in Lp (Ω) and ∂ α un ∂ α u in Lp (Ω) for every 1 ≤ |α| ≤ m. If Ω has additional properties, in the previous exercise one can replace un u in Lp (Ω) and ∂ α un ∂ α u in Lp (Ω) with un → u in Lp (Ω) and ∂ α un ∂ α u in Lp (Ω) for every 1 ≤ |α| ≤ m − 1 (see Exercise 12.22). Theorem 11.65 (Compactness). Let Ω ⊆ RN be an open set, let m ∈ N, and let 1 < p < ∞. Assume that {un } is bounded in W m,p (Ω). Then there exist a subsequence {unk }k of {un }n and u ∈ W m,p (Ω) such that unk u in W m,p (Ω). Proof. Since {un }n and {∂ α un }n are bounded in the reflexive Banach space Lp (Ω), we may select a subsequence {unk } such that unk u in Lp (Ω) and ∂ α unk vα in Lp (Ω) for every 1 ≤ |α| ≤ m and for some functions u, vα ∈ Lp (Ω), 1 ≤ |α| ≤ m. It remains to show that u ∈ W m,p (Ω). For every φ ∈ Cc∞ (Ω), 1 ≤ |α| ≤ m, and k ∈ N we have α |α| unk ∂ φ dx = (−1) ∂ α unk φ dx. Ω
Ω
Letting k → ∞ in the previous equality and using weak convergence in Lp (Ω) yields α |α| u∂ φ dx = (−1) vα φ dx, Ω
Ω
which shows that ∂ α u = vα . Hence, u ∈ W m,p (Ω).
Remark 11.66. The previous result fails for p = 1. Indeed, in this case, L1 (Ω) is not reflexive (so we do not expect weak sequential compactness of bounded sequences) and it is not the dual of a separable space (so we do not expect weak star sequential compactness of bounded sequences). To recover some compactness, we use the embedding L1 (Ω) → Mb (Ω)
u → λu (E)
where λu (E) := E u(x) dx, E ∈ B(Ω). Note that (why?) Ω |u| dx = λu Mb (Ω) . Thus, given a bounded sequence {un }n in L1 (Ω), one can only conclude that there exist a subsequence {unk }k and λ ∈ Mb (Ω) such that ∗ λunk λ in Mb (Ω). In particular, given a bounded sequence {un }n in W 1,1 (Ω), one will only recover some compactness in the space BV (Ω). See Theorem 14.39.
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Exercise 11.67. Let Ω = B(0, 1). Construct a bounded sequence {un }n in W 1,1 (Ω) converging strongly in L1 (Ω) to the function given in Exercise 11.8. Every functional L ∈ (W m,p (Ω)) is the extension of a distribution T ∈ More precisely, if L has the form (11.32), then its restriction to D(Ω) is given by
D (Ω).
(11.36)
T (φ) :=
(−1)|α| ∂ α Tfα (φ)
for all φ ∈ D(Ω).
0≤|α|≤m
Conversely, a distribution T ∈ D (Ω) of the form (11.36), where f ∈ Lp (Ω; RM ), can be extended to an element of (W m,p (Ω)) , but this extension may not be unique, unless W m,p (Ω) = W0m,p (Ω). Definition 11.68. Let Ω ⊆ RN be an open set, let m ∈ N, let 1 ≤ p ≤ ∞, and let p be its H¨older conjugate exponent. The dual space of W0m,p (Ω) is denoted by W −m,p (Ω). Exercise 11.69. Let Ω ⊆ RN be an open set, let m ∈ N, let 1 ≤ p < ∞, and let T ∈ D (Ω) be of the form (11.36), where f ∈ Lp (Ω; RM ). Prove that T may be uniquely extended to a functional L ∈ W −m,p (Ω). In view of the previous exercise we have the following characterization of the dual of W0m,p (Ω). Corollary 11.70 (Riesz’s representation theorem in W0m,p ). Let Ω ⊆ RN be an open set, let m ∈ N, and let 1 ≤ p < ∞. Then W −m,p (Ω) can be identified with the subspace of distributions T of the form T =
(−1)|α| ∂ α Tfα ,
0≤|α|≤m
where fα ∈ Lp (Ω; RM ). Next we study the case p = ∞. By Theorem B.94 the dual of L∞ (Ω) can be identified with the space of all bounded finitely additive signed measures that are absolutely continuous with respect to the Lebesgue measure restricted to Ω. Thus, if λα , 0 ≤ |α| ≤ m, are any such measures, then the functional ∂ α u dλα , u ∈ W m,∞ (Ω), L(u) := 0≤|α|≤m Ω
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349
belongs to (W m,∞ (Ω)) . Indeed, since each λα belongs to (L∞ (Ω)) , 0 ≤ |α| ≤ m, we have that |L(u)| ≤ λα (L∞ (Ω)) ∂ α uL∞ (Ω) 0≤|α|≤m
≤
λα (L∞ (Ω)) max ∂ α uL∞ (Ω) 0≤|α|≤m
0≤|α|≤m
for all u ∈ W m,∞ (Ω), and thus, L(W m,∞ (Ω)) =
|L(u)| ≤ u m,∞ W m,∞ (Ω) u∈W (Ω)\{0} sup
λα (L∞ (Ω)) .
0≤|α|≤m
Reasoning exactly as in the proof of Theorem 11.62, we have the following result. Theorem 11.71 (Riesz’s representation theorem in W m,∞ ). Let Ω ⊆ RN be an open set and let m ∈ N. For every bounded linear functional L ∈ (W m,∞ (Ω)) there exist bounded finitely additive signed measures λα , 0 ≤ |α| ≤ m, that are absolutely continuous with respect to the Lebesgue measure LN restricted to Ω, such that ∂ α u dλα (11.37) L(u) = 0≤|α|≤m Ω
for all u ∈ W m,∞ (Ω). The proofs of the next results are left as an exercise. Corollary 11.72 (Riesz’s representation theorem in W0m,∞ ). Let Ω ⊆ RN be an open set and let m ∈ N. The space W −m,1 (Ω) can be identified with the space of all distributions T of the form T = 0≤|α|≤m (−1)|α| ∂α Tλα , where λα , 0 ≤ |α| ≤ m, are bounded finitely additive measures that are absolutely continuous with respect to the Lebesgue measure LN restricted to Ω. Theorem 11.73 (Compactness). Let Ω ⊆ RN be an open set and let m ∈ N. Assume that {un }n is bounded in W m,∞ (Ω). Then there exist a subsequence ∗ {unk }k of {un }n and u ∈ W m,∞ (Ω) such that unk u in L∞ (Ω) and ∗ ∂ α unk ∂ α u in L∞ (Ω) for every 1 ≤ |α| ≤ m.
11.5. A Characterization of W 1,p (Ω) ˙ 1,p (Ω) in terms In this section we give a characterization of W 1,p (Ω) and W of difference quotients. An analogous result has been given in Corollary 2.51 for functions of bounded pointwise variation in one variable. As we already mentioned, results of this type are often useful in the regularity theory for partial differential equations (see, e.g., [33] and [71]). Moreover,
350
11. Sobolev Spaces
they provide characterizations that do not involve derivatives and thus they can be used to extend the definition of Sobolev spaces to more abstract settings (see, e.g., [12] and [107]). Let Ω ⊆ RN be an open set and for every h ∈ RN \ {0}, let Ωh := {x ∈ Ω : x + th ∈ Ω for all t ∈ [0, 1]}. Exercise 11.74. (i) Prove that if 0 < p < 1, then (a + b)p ≤ ap + bp for all a, b ≥ 0. (ii) Prove that if p ≥ 1 and ε > 0, then there exists a constant c = c(ε, p) > 0 such that (a + b)p ≤ (1 + ε)ap + cbp for all a, b ≥ 0. Hint: Use the convexity of the function g(t) = |t|p , t ∈ R. We recall that S N −1 is the unit sphere ∂B(0, 1) and that βN = HN −1 (S N −1 ), where we recall that HN −1 is the (N − 1)-dimensional Hausdorff measure (see Section C.7 in Appendix C). ˙ 1,p (Ω), 1 ≤ p < Theorem 11.75. Let Ω ⊆ RN be an open set and let u ∈ W N ∞. Then for every h ∈ R \ {0}, |u(x + h) − u(x)|p dx ≤ ∇u(x)p dx, (11.38) hp Ωh Ω while (11.39)
∇u(x) dx ≤ lim sup p
κN,p
h→0
Ω
where (11.40)
κN,p
1 := βN
S N −1
Ωh
|u(x + h) − u(x)|p dx, hp
|e1 · ξ|p dHN −1 (ξ).
Conversely, if 1 < p < ∞ and u ∈ L1loc (Ω) is such that |u(x + h) − u(x)|p dx < ∞, (11.41) lim sup hp h→0 Ωh ˙ 1,p (Ω). then u ∈ W ˙ 1,p (Ω) and h ∈ RN \ {0}. Let U Ω and for Proof. Step 1: Let u ∈ W 0 < ε < dist(U, ∂Ω) define uε := ϕε ∗ u, where ϕε is a standard mollifier. By
11.5. A Characterization of W 1,p (Ω)
351
the fundamental theorem of calculus for x ∈ Uh we have 1 |uε (x + h) − uε (x)| = ∇uε (x + th) · h dt 0 1 1 |∇uε (x + th) · h| dt ≤ h ∇uε (x + th) dt. ≤ 0
0
Raising to the power p and integrating over Uh , by H¨older’s inequality we get |uε (x + h) − uε (x)|p dx Uh
≤ h
≤ h
1
p Uh
0 1
p Uh
p ∇uε (x + th) dt
dx
∇uε (x + th) dtdx ≤ h p
0
∇uε (y)p dy,
p U
where we have used Tonelli’s theorem and the change of variables y = x+th. Since ∂i uε → ∂i u in Lp (U ) as ε → 0+ by Lemma 11.25 and uε → u pointwise for LN -a.e. in Ω, letting ε → 0+ and using Fatou’s lemma gives ||u(x + h) − u(x)||p dx ≤ hp ∇u(y)p dy ≤ hp ∇u(y)p dy. Uh
U
Ω
To prove (11.38) it suffices to let U Ω and use the Lebesgue monotone convergence theorem. ˙ 1,p (Ω). Let Step 2: We will prove (11.39) for u ∈ C ∞ (Ω) ∩ W |u(x + h) − u(x)|p dx. M := lim sup hp h→0 Ωh If M = ∞, then there is nothing to prove. Thus, in what follows we assume that M < ∞. Given δ > 0 let dδ > 0 be so small that |u(x + h) − u(x)|p dx ≤ M + δ hp Ωh for every 0 < h < dδ . Fix a compact set K ⊂ Ω. Let 1 min{dist(K, ∂Ω), 1}. 2 By Taylor’s formula there exists a constant cK > 0 (depending on u) such that |u(x + h) − u(x) − ∇u(x) · h| ≤ cK h2 d :=
for all x ∈ K and 0 < h < d. Hence, |∇u(x) · h| ≤ |u(x + h) − u(x)| + cK h2
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11. Sobolev Spaces
for all x ∈ K and 0 < h < d. By the previous exercise, for every δ > 0 we have |∇u(x) · h|p ≤ (1 + δ)|u(x + he) − u(x)|p + cK,δ h2p for all x ∈ K and 0 < h < d. Fix 0 < η < min{d, dδ }. Dividing by hp and integrating over K yields |u(x + h) − u(x)|p |∇u(x) · ξ|p dx ≤ (1 + δ) dx + cK,δ LN (K)hp p h K K ≤ (1 + δ)(M + δ) + cK,δ LN (K)η p for all x ∈ K and 0 < h < η, and where ξ := h/h. Integrating in ξ over the unit sphere S N −1 and using the fact that for every y ∈ RN \ {0}, by a rotation, p y · ξ |y · ξ|p dHN −1 (ξ) = yp dHN −1 (ξ) y N −1 N −1 S S |e1 · ξ|p dHN −1 (ξ) =: yp cN,p = yp S N −1
gives
∇u(x)p dx ≤ (1 + δ)(M + δ)βN + cK,δ βN LN (K)η p .
cN,p K
Letting η → 0+ , we obtain ∇u(x)p dx ≤ (1 + δ)(M + δ)βN . cN,p K
If δ → 0+ and K Ω, by the Lebesgue monotone convergence theorem we get |u(x + h) − u(x)|p ∇u(x)p dx ≤ βN M = βN lim sup dx. cN,p hp h→0 Ω Ωh ˙ 1,p (Ω). Let U Ω and for Step 3: We will prove (11.39) for u ∈ W 0 < ε < dist(U, ∂Ω) define uε := ϕε ∗ u, where ϕε is a standard mollifier. By Theorem C.16(iii) applied to the function x ∈ Ωh → u(x + h) − u(x), we have |uε (x + h) − uε (x)|p dx ≤ |u(x + he) − u(x)|p dx. Uh
Ωh
Dividing by hp , letting h → 0 and using the previous step applied to uε gives |u(x + h) − u(x)|p ∇uε (x)p dx ≤ lim sup dx. (11.42) κN,p hp h→0 U Ωh Since ∂i uε → ∂i u in Lp (U ) as ε → 0+ by Lemma 11.25, we obtain |u(x + h) − u(x)|p p ∇u(x) dx ≤ lim sup dx. (11.43) κN,p hp h→0 U Ωh
11.5. A Characterization of W 1,p (Ω)
353
By letting U Ω and using the Lebesgue monotone convergence theorem, we obtain (11.39). Step 4: To prove the final statement of the theorem, let 1 < p < ∞ and ˙ 1,p (Ω). To let u ∈ L1loc (Ω) be such that (11.41) holds. We claim that u ∈ W see this, let U Ω. Then reasoning as in Step 3, by (11.42) we get |u(x + h) − u(x)|p sup ∇uε (x)p dx ≤ lim sup dx < ∞. hp h→0 0 0, |u(x + hei ) − u(x)|p dx ≤ |∂i u(x)|p dx p h Ωh,i Ω and
lim
h→0+
Ωh,i
|u(x + hei ) − u(x)|p dx = hp
(ii) Prove that if 1 < p lim inf h→0+
|∂i u(x)|p dx. Ω
< ∞ and u ∈ L1loc (Ω) is such that |u(x + hei ) − u(x)|p dx < ∞ hp
Ωh,i
˙ 1,p (Ω). for every i = 1, . . . , N , then u ∈ W ˙ m,p (Ω) is given in Exercise 17.62. The extension of Theorem 11.75 to W
Chapter 12
Sobolev Spaces: Embeddings Newton’s second law of graduation: The age, a, of a doctoral process is directly proportional to the flexibility, f, given by the advisor and inversely proportional to the student’s motivation, m. — Jorge Cham, www.phdcomics.com
In this chapter we are interested in finding an exponent q such that uLq (RN ) ≤ c∇uLp (RN )
(12.1)
˙ 1,p (RN ). for all u in an appropriate subfamily of W 1,p (RN ) or W Assume for simplicity that u ∈ Cc1 (RN ) and for r > 0 define the rescaled function ur (x) := u(rx), x ∈ RN . If (12.1) holds for ur , we get
1/q
1/q
1/p q q p |u(rx)| dx = |ur (x)| dx ≤c ∇ur (x) dx RN
RN
= c rp
1/p RN
∇u(rx)p dx
RN
,
or, equivalently, after the change of variables y := rx,
1/q p
1/p r 1 q p |u(y)| dy ≤c N ∇u(y) dy , r N RN r RN that is,
1/q
|u(y)| dy q
RN
1/p
≤ cr
1−N/p+N/q
∇u(y) dy p
RN
.
If 1 − N/p + N/q > 0, let r → 0+ to conclude that u ≡ 0, while if 1 − N/p + N/q < 0, let r → ∞ to conclude again that u ≡ 0. Hence, the only possible 355
356
12. Sobolev Spaces: Embeddings
case is when 1 − N/p + N/q = 0. So in order for q to be positive, we need p < N , in which case, Np . (12.2) q = p∗ := N −p The number p∗ is called the Sobolev critical exponent. In what follows, when there is no possibility of confusion, we write uLp and uW m,p , for uLp (Ω) and uW m,p (Ω) , respectively. We also write ∇k uLp (Ω) or ∇k uLp for ∇k uLp (Ω;RMk ) . We recall, that βN is the surface area of the unit sphere, that is, (12.3)
βN := HN −1 (S N −1 ) =
2π N/2 . Γ(N/2)
12.1. Embeddings: mp < N In this section we prove inequality (12.1). We recall the following definition. Definition 12.1. Let E ⊆ RN be a Lebesgue measurable set and let u : E → R be a Lebesgue measurable function. The function u is said to vanish at infinity if for every t > 0, (12.4)
LN ({x ∈ E : |u(x)| > t}) < ∞.
Note that the previous definition is automatically satisfied if E has finite measure, while if E has infinite measure and u does not satisfy (12.4), then it cannot belong to any Lq (E) for 1 ≤ q < ∞. Exercise 12.2. Let E ⊆ RN be a Lebesgue measurable set and let u : E → R and v : E → R be two functions vanishing at infinity. Prove that u + v vanishes at infinity. Exercise 12.3. Prove that a nonzero polynomial P : RN → R cannot vanish at infinity. We recall (see Definition A.36 and (A.2)) that given two normed spaces (X, · X ) and (Y, · Y ), the normed space X is embedded in the normed space Y , written X → Y , if X is a subspace of Y and there exists a constant c > 0 such that xY ≤ cxX for all x ∈ X. Theorem 12.4 (Sobolev–Gagliardo–Nirenberg’s embedding in W 1,p ). Let 1 ≤ p < N . Then there exists a constant c = c(N, p) > 0 such that for every ˙ 1,p (RN ) vanishing at infinity, function u ∈ W (12.5)
uLp∗ (RN ) ≤ c∇uLp (RN ) .
In particular, W 1,p (RN ) → Lq (RN ) for all p ≤ q ≤ p∗ .
12.1. Embeddings: mp < N
357
The proof makes use of the following result. Exercise 12.5. Prove that if u ∈ Lp (R) for some 1 ≤ p < ∞, then for every representative v of u, lim inf x→−∞ |v(x)| = 0, lim inf x→∞ |v(x)| = 0. Prove that in general one cannot replace the limit inferiors with actual limits. In what follows, we use the notation (E.2) in Appendix E. To be precise, given x = (x1 , . . . , xN ) ∈ RN , for every i = 1, . . . , N we denote by xi the (N − 1)-dimensional vector obtained by removing the ith component from x and with an abuse of notation we write x = (xi , xi ) ∈ RN −1 × R. Lemma 12.6. Let N ≥ 2 and let ui ∈ LN −1 (RN −1 ), i = 1, . . . , N . Then the function u(x) := u1 (x1 )u2 (x2 ) · · · uN (xN ),
x ∈ RN ,
belongs to L1 (RN ) and uL1 (RN ) ≤
N /
ui LN −1 (RN −1 ) .
i=1
Proof. The proof is by induction on N . If N = 2, then u(x) := u1 (x2 )u2 (x1 ), x = (x1 , x2 ) ∈ R2 . Integrating both sides with respect to x and using Tonelli’s theorem, we get |u(x)| dx = |u1 (x2 )| dx2 |u2 (x1 )| dx1 . R2
R
R
Assume next that the result is true for N and let us prove it for N + 1. Let u(x) := u1 (x1 )u2 (x2 ) · · · uN +1 (xN +1 ),
x ∈ RN +1 ,
where ui ∈ LN (RN ), i = 1, . . . , N + 1. Fix xN +1 ∈ R. Integrating both sides of the previous identity with respect to x1 , . . . , xN and using H¨older’s inequality, we get (12.6) RN
|u(x)| dx1 · · · dxN ≤ uN +1 LN (RN )
N / RN i=1
|ui (xi )|N/(N −1) dx1 · · · dxN
(N −1)/N .
For every i = 1, . . . , N we denote by xi the (N − 1)-dimensional vector obtained by removing the last component from xi and with an abuse of notation we write xi = (xi , xN +1 ) ∈ RN −1 × R. Since xN +1 is fixed, by the induction hypothesis applied to the functions vi (xi ) := |ui (xi , xN +1 )|N/(N −1) ,
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12. Sobolev Spaces: Embeddings
xi ∈ RN −1 , i = 1, . . . , N , we obtain that / N N / |ui (xi )|N/(N −1) dx1 · · · dxN ≤ vi LN −1 (RN −1 ) RN i=1
=
i=1
N / i=1
RN −1
|ui (xi , xN +1 )|N dxi
1/(N −1) ,
and so from (12.6), |u(x)| dx1 · · · dxN RN
≤ uN +1 LN (RN )
N / i=1
RN −1
|ui (xi , xN +1 )|N dxi
1/N .
Integrating both sides with respect to xN +1 and using the extended H¨older inequality (see the Exercise B.79(i)), with 1 = 1/N + · · · + 1/N where the sum has N addends, and Tonelli’s theorem, we get N +1 / |u(x)| dx1 · · · dxN +1 ≤ ui LN (RN ) , RN +1
i=1
which concludes the proof.
We now turn to the proof of the Sobolev–Gagliardo–Nirenberg embedding theorem. Proof of Theorem 12.4. Step 1: We prove (12.5) in the case p = 1 and N under the additional hypothesis that u ∈ L N −1 (RN ) ∩ C 1 (RN ) with ∇u ∈ L1 (RN ; RN ). Fix i = 1, . . . , N . By Fubini’s theorem for LN −1 -a.e. xi ∈ N
RN −1 the function v(t) := u(xi , t), t ∈ R, belongs to L N −1 (R) ∩ C 1 (R) and v ∈ L1 (R). Fix any such xi ∈ RN −1 . By Exercise 12.5, lim inf t→−∞ |v(t)| = 0, and so we may find a sequence tn → −∞ such that v(tn ) → 0. Hence, for every t ∈ R we have that t v (s) ds. v(t) = v(tn ) + tn
In turn,
|v(t)| ≤ |v(tn )| +
R
|v (s)| ds
for all t ∈ R and n ∈ N. Letting n → ∞, we conclude that for each xi ∈ R we have ∂u (xi , yi ) dyi . |u(xi , xi )| ≤ ∂x i R
12.1. Embeddings: mp < N
359
Hence, we have shown that for all i = 1, . . . , N and for LN -a.e. x ∈ RN we have ∂u |u(x)| ≤ (xi , yi ) dyi . R ∂xi Multiplying these N inequalities and raising to the power N 1−1 , we get N N 1/(N −1) / / ∂u N/(N −1) ≤ (xi , yi ) dyi =: wi (xi ) |u(x)| R ∂xi i=1
LN -a.e.
i=1
RN .
for We now apply the previous lemma to the function 0N x ∈ w(x) := i=1 wi (xi ), x ∈ RN , to obtain that N / N/(N −1) |u(x)| dx ≤ |w(x)| dx ≤ wi LN −1 (RN −1 ) RN
RN
=
N / i=1
∂u (x) dx RN ∂xi
1/(N −1)
i=1
≤
RN
∇u(x) dx
N/(N −1) ,
where we have used Tonelli’s theorem. This gives the desired inequality for p = 1. ∗
Step 2: Assume next that 1 < p < N and that u ∈ Lp (RN ) ∩ C 1 (RN ) with ∇u ∈ Lp (RN ; RN ). Define v := |u|q , q := p(N − 1)/(N − p). Note that since q > 1, we have that v ∈ C 1 (RN ). Applying Step 1 to the function v, we get
(N −1)/N
(N −1)/N |u|p∗ dx = |v|N/(N −1) dx RN RN ≤ ∇v dx ≤ q |u|q−1 ∇u dx RN
RN
≤q
RN
|u|
(q−1)p
1/p
1/p ∇u dx p
dx RN
,
where in the last inequality we have used H¨ older’s inequality. Since (q − 1)p = p∗ , if u = 0, we obtain
(N −1)/N −(p−1)/p |u|p∗ dx = RN
1/p∗ RN
|u|p∗ dx
≤q
1/p ∇u dx p
RN
,
which proves the result. ∗
Step 3: Next consider the case u ∈ Lp (RN ) with distributional gradient ∇u ∈ Lp (RN ; RN ) for 1 ≤ p < N . Define uε := u ∗ ϕε , where ϕε is a standard mollifier. Then by Theorems C.16 and C.20 we have that uε ∈
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12. Sobolev Spaces: Embeddings
∗
Lp (RN )∩C ∞ (RN ), while, reasoning as in Lemma 11.25 and using Theorem C.16 once more, we obtain that ∇uε ∈ Lp (RN ; RN ). Hence, by the previous two steps uε Lp∗ ≤ q∇uε Lp . Letting ε → 0+ and using Theorem C.16 and Lemma 11.25 one more time, we obtain the same inequality for u. Step 4: Finally, we prove (12.5) in the general case. Assume that u ∈ ˙ 1,p (RN ) vanishes at infinity. For n ∈ N and x ∈ RN define W ⎧ ⎪ ⎨ |u(x)| − 1/n if 1/n ≤ |u(x)| ≤ n, 0 if |u(x)| < 1/n, vn (x) := ⎪ ⎩ n − 1/n if |u(x)| > n. By the chain rule (see Exercise 11.51(i) and (vi)) for LN -a.e. x ∈ RN , ∇u(x) if 1/n < |u(x)| < n, ∇vn (x) = 0 otherwise, and so ∇vn ∈ Lp (RN ; RN ), while p∗ |vn | dx = |vn |p∗ dx RN
{|u|>1/n}
≤ (n − 1/n)p∗ LN ({x ∈ RN : |u(x)| > 1/n}) < ∞, since u is vanishing at infinity. Hence, by the previous step applied to vn ,
1/p∗
1/p∗ p∗ p∗ (|u(x)| − 1/n) dx ≤ |vn | dx {1/n≤|u|≤n}
≤q
RN
1/p ∇vn dx p
RN
≤q
1/p ∇u dx p
RN
.
Letting n → ∞ and using Fatou’s lemma, we obtain the desired result. Step 5: To prove the last part of the theorem, assume that u ∈ W 1,p (RN ). ∗ Then by the previous steps we know that u ∈ Lp (RN ), and so we can now use Exercise B.79(ii) to conclude that u ∈ Lq (RN ) for all p ≤ q ≤ p∗ . Indeed, assume that p < q < p∗ and write 1/q = θ/p + (1 − θ)/p∗ for some 0 < θ < 1. Then 1−θ ∗ p uLq ≤ uθLp uL p∗ ≤ uL + uLp ,
where we have used Young’s inequality (see (B.17)) with exponents 1/θ and (1/θ) . Hence, uLq ≤ uLp + uLp∗ ≤ uLp + c∇uLp , which shows that W 1,p (RN ) → Lq (RN ).
12.1. Embeddings: mp < N
361
Remark 12.7. (i) If 1 < p < N , then the best constant in (12.5) is given by 1/N 1 1 p − 1 1−1/p Γ(1 + N/2)Γ(N ) c = 1/2 1/p N −p Γ(N/p)Γ(1 + N − N/p) π N and equality holds in (12.5) if u(x) = (a + bxp/(p−1) )1−N/p , x ∈ RN , where a, b are positive constants (see [15], [228]). If p = 1 < N , then the best constant in (12.5) is given by 1 1 c = 1/2 (Γ(1 + N/2))1/N . π N To see that this inequality is sharp, it suffices to take ⎧ if x ≤ 1, ⎨ 1 1 + n − nx if 1 < x ≤ 1 + 1/n, un (x) := ⎩ 0 if x > 1 + 1/n, which is a sequence of Lipschitz continuous functions that converges to the characteristic function of the unit ball. It is possible to show that equality does not hold in (12.5) for functions in W 1,1 (RN ) but that (12.5) still holds in BV (RN ) (see Chapter 14), with the norm of ∇u in L1 replaced by the total variation of the vectorial Radon measure Du, and that equality holds for the characteristic function of the unit ball. (ii) In Steps 1 and 2 of the previous proof we could have used Theorem 11.45, and so avoid Step 3. To extend the Sobolev–Gagliardo–Nirenberg embedding theorem to ˙ m,p (RN ), where m ≥ 2 and 1 ≤ p < ∞ are such that functions in W mp < N , for every k = 0, . . . , m we define the Sobolev critical exponent p∗m,k :=
(12.7)
Note that p∗m,m−1 = p∗ and p∗m,m
Np . N − (m − k)p = p.
Corollary 12.8 (Sobolev–Gagliardo–Nirenberg’s embedding in W m,p ). Let m ∈ N and 1 ≤ p < ∞ be such mp < N . Then there exists a constant c = c(N, m, p) > 0 such that for every function u ∈ W m,p (RN ) and for every k = 0, . . . , m − 1 and ∇k u
(12.8)
L
p∗ m,k (RN )
≤ c∇m uLp (RN ) .
In particular, W m,p (RN ) → Lq0 (RN ) ∩ W 1,q1 (RN ) ∩ · · · ∩ W m−1,qm−1 (RN ) for all p ≤ qk ≤ p∗m,k , k = 0, . . . , m − 1.
362
12. Sobolev Spaces: Embeddings
Proof. The proof is by induction and is left as an exercise.
˙ m,p (RN ) that do not vanish Next we consider the case of functions in W at infinity. As a corollary of Theorem 11.43, we have the following result. ˙ m,p ). Let Theorem 12.9 (Sobolev–Gagliardo–Nirenberg’s embedding in W m ∈ N and 1 ≤ p < ∞ be such mp < N . Then for every function u ∈ ˙ m,p (RN ) there exists a polynomial Pu of degree m − 1 such that W (12.9)
∇k (u − Pu )
L
p∗ m,k (RN )
≤ c∇m uLp (RN )
for every k = 0, . . . , m − 1, where c = c(N, m, p) > 0. Moreover, (Kα ∗ ∂ α u), (12.10) Pu = u − |α|=m
where Kα (x) :=
xα
m βN α! x N ,
x ∈ RN \ {0}.
We recall that βN is the surface area of the unit sphere (see (12.3)). We begin with a preliminary lemma. Lemma 12.10. Let u ∈ Ccm (RN ), m ∈ N. Then Kα ∗ ∂ α u. u= |α|=m
Proof. Fix x ∈ RN and θ ∈ S N −1 and consider the function g(r) := u(x + rθ), r ∈ R. Then by the fundamental theorem of calculus and by integrating by parts m − 1 times, ∞ ∞ (−1)m g(0) = − g (r) dr = rm−1 g (m) (r) dr. (m − 1)! 0 0 α α By the chain rule g (m) (r) = |α|=m m! α! θ ∂ u(x + rθ) and so 1 ∞ m rm−1 θα ∂ α u(x + rθ) dr. u(x) = m(−1) α! 0 |α|=m
Averaging both sides with respect to θ ∈ S N −1 and changing variables gives ∞ m(−1)m 1 rm−1 θα ∂ α u(x + rθ) drdHN −1 (θ) u(x) = βN α! S N −1 0 |α|=m m 1 (y − x)α α m(−1) 1 ∂ u(y) dy, = βN α! RN y − xN −m y − x|α| |α|=m
which gives the desired result. We turn to the proof of Theorem 12.9.
12.1. Embeddings: mp < N
363
˙ m,p (RN ), by Theorem 11.43, there Proof of Theorem 12.9. Given u ∈ W exists a sequence {un }n in Cc∞ (RN ) such that ∂ α un → ∂ α u in Lp (RN ) for every multi-index α with |α| = m. Since for every , n ∈ N, un − u ∈ W m,p (RN ), by the Sobolev–Gagliardo–Nirenberg theorem (see Theorem 12.8), ∇k (un − u ) p∗m,k ≤ c∇m (un − u )Lp L
for every k = 0, . . . , m − 1. This shows that {∇k un }n is a Cauchy ∗ ∗ sequence in Lpm,k (RN ; RMk ). Thus there exists v ∈ Lpm,0 (RN ) ∩ · · · ∩ ∗ ∗ Lm,pm,m (RN ) such that ∇k un → ∇k v in Lpm,k (RN ; RMk ). Since ∇m un → ∇m u in Lp (RN ; RMm ), it follows that ∇m u(x) = ∇m v(x) for LN -a.e. x ∈ RN . Hence, there exists a polynomial Pu of degree m − 1 such that u(x) = v(x) + Pu (x) for LN -a.e. x ∈ RN . Again by the Sobolev–Gagliardo– Nirenberg theorem (see Theorem 12.8), this time applied to un ∈ W m,p (RN ), n ∈ N, we have ∇k un p∗m,k ≤ c∇m un Lp for every k = 0, . . . , m − 1. L Letting n → ∞ gives (12.9). It remains to show (12.10). By Lemma 12.10, un = |α|=m Kα ∗ ∂ α un . Since |Kα (x)| ≤ x Nc −m and 1 ≤ p < N/m, applying Proposition C.29 (where α and p there are here m and pN/[pm + N (p − 1)] ≤ N/m, re˜ α ∗ ψ ∈ Lp (RN ) for every ψ ∈ C ∞ (RN ), where spectively) we get that K c ˜ α (x) := Kα (−x). Note that when p = 1, one should use (C.21) and the K ˜ α ∗ ψ ∈ L∞ (RN ). Hence, by Fubini’s fact that ψ ∈ Cc∞ (RN ) to obtain that K theorem un (x)ψ(x) dx = (Kα ∗ ∂ α un )(x)ψ(x) dx RN
N |α|=m R
=
|α|=m
RN
˜ α ∗ ψ)(y) dy. ∂ α un (y)(K
Letting n → ∞ and using Fubini’s theorem we get ˜ α ∗ ψ)(y) dy (u(x) − Pu (x))ψ(x) dx = ∂ α u(y)(K RN
N |α|=m R
=
N |α|=m R
(Kα ∗ ∂ α u)(x)ψ(x) dx.
Since this holds for every ψ ∈ Cc∞ (RN ), we obtain (12.10). This completes the proof.
Next we show that if u vanishes at infinity, then Pu = 0. Corollary 12.11. Let m ∈ N and 1 ≤ p < ∞ be such mp < N . Then there exists a constant c = c(N, m, p) > 0 such that for every function
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12. Sobolev Spaces: Embeddings
˙ m,p (RN ) vanishing at infinity, for every k = 0, . . . , m − 1, u∈W ∇k u
L
p∗ m,k (RN )
≤ c∇m uLp (RN ) . ∗
Proof. By the previous theorem we have that u − Pu ∈ Lpm,0 (RN ), which implies that u − Pu vanishes at infinity. But since u vanishes at infinity, then by Exercise 12.2 so does Pu , which contradicts Exercise 12.3 unless Pu = 0. The next exercises show that a similar result fails if p ≥ N . Exercise 12.12. Assume that p > N and consider a function u ∈ C ∞ (RN ) such that u(x) = xε for all x ∈ RN \ B(0, 1), where 0 < ε < 1. Prove that ˙ 1,p (RN ), but there is no if 0 < ε < 1 is chosen appropriately, then u ∈ W q N constant a ∈ R such that u − a belongs to L (R ) for some 1 ≤ q ≤ ∞. Exercise 12.13. Assume that p = N > 1 and consider a function u ∈ C ∞ (RN ) such that u(x) = log log x for all x ∈ RN \ B(0, e). Prove that ˙ 1,N (RN ), but there is no constant a ∈ R such that u − a belongs to u∈W q N L (R ) for some 1 ≤ q ≤ ∞. ˙ 1,1 (R), then there is a constant a ∈ R Exercise 12.14. Prove that if u ∈ W such that u − aL∞ (R) ≤ 2 |u (x)| dx. R
Next we discuss the validity of the Sobolev–Gagliardo–Nirenberg embedding theorem for arbitrary domains. Exercise 12.15 (Rooms and passages). Let {hn }n and {δ2n }n be two ∞ ∞, 0 < const. sequences of positive numbers such that n=1 hn = < ≤ hn+1 /hn ≤ 1, 0 < δ2n ≤ h2n+1 , and for n ∈ N let cn := ni=1 hi . Define Ω ⊂ R2 to be the union of all sets of the form Rj := (cj − hj , cj ) × (− 12 hj , 12 hj ), Pj+1 := [cj , cj + hj+1 ] × (− 12 δj+1 , 12 δj+1 ), for j = 1, 3, 5, . . . (see Figure 1). (i) Prove that ∂Ω is the range of a rectifiable curve but that it is not of class C. (ii) Let hn := 1/n3/2 , δ2n := 1/(2n)5/2 and for j = 1, 3, 5, . . . define ⎧ j ⎪ ⎨ =: Kj in Rj , log 2j u(x1 , x2 ) := x − cj ⎪ ⎩ Kj + (Kj+2 − Kj ) 1 in Pj+1 . hj+1 / Lq (Ω) for any q > 2. Prove that u ∈ W 1,2 (Ω) but u ∈
12.1. Embeddings: mp < N
R1
P2
365
R3
P4
R5
R7
Figure 1. Rooms and passages.
(iii) Let p > 1, q ≥ 12 (2p−1), hn := 1/np , δ2n := c/n2q+p , for some c > 0, and for n ∈ N define u(x1 , x2 ) := nq in R2n−1 and ∇u(x1 , x2 ) := q −nq , 0) in P2n . Prove that ∇u ∈ L2 (Ω; R2×2 ) but u ∈ / L2 (Ω). ( (n+1) 1/np The previous exercise shows that Theorem 12.4 fails for general domains. The problem is the regularity of the boundary. Definition 12.16. Given m ∈ N and 1 ≤ p ≤ ∞, an open set Ω ⊆ RN is called an extension domain for the Sobolev space W m,p (Ω) if there exists a continuous linear operator E : W m,p (Ω) → W m,p (RN ) with the property that for all u ∈ W m,p (Ω), E(u)(x) = u(x) for LN -a.e. x ∈ Ω. Note that the extension operator E strongly depends on m and p. In the next chapter we will show that “nice” Lipschitz continuous domains are extension domains for all m ∈ N and all 1 ≤ p ≤ ∞. Proposition 12.17. Let m ∈ N, 1 ≤ p < ∞ be such mp < N and let Ω ⊆ RN be an extension domain for W m,p (Ω). Then there exists a constant c = c(N, m, p, Ω) > 0 such that for every u ∈ W m,p (Ω), ∂ β uLq (Ω) ≤ cuW m,p (Ω) for every k = 0, . . . , m − 1, for every multi-index β with |β| = k and for all p ≤ q ≤ p∗m,k . Proof. Since Ω ⊆ RN is an extension domain for W m,p (Ω), there exists a constant c = c(N, m, p, Ω) > 0 such that E(u)W m,p (RN ) ≤ cuW m,p (Ω) for all u ∈ W m,p (Ω). On the other hand, by Theorem 12.4, for every k = 0, . . . , m − 1 and for all p ≤ q ≤ p∗m,k we have that ∇k uLq (Ω) = ∇k E(u)Lq (Ω) ≤ ∇k E(u)Lq (RN ) ≤ cE(u)W m,p (RN ) ≤ cuW m,p (Ω) ,
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12. Sobolev Spaces: Embeddings
where as usual c = c(N, m, p) > 0 changed from line to line and we have used the fact that E(u)(x) = u(x) for LN -a.e. x ∈ Ω. Next we show that if p ≤ q < p∗ and Ω is an extension domain for with finite measure, then the embedding W 1,p (Ω) → Lq (Ω) is actually compact. W 1,p (Ω)
Theorem 12.18 (Rellich–Kondrachov’s compactness). Let 1 ≤ p < N and let Ω ⊂ RN be an extension domain for W 1,p (Ω) with finite measure. Let {un }n be a bounded sequence in W 1,p (Ω). Then there exist a subsequence ∗ {unk }k of {un }n and a function u ∈ Lp (Ω) such that unk → u in Lq (Ω) for all 1 ≤ q < p∗ . The proof makes use of the following auxiliary results. Lemma 12.19. Let 1 ≤ p < ∞ and let u ∈ W 1,p (RN ). Then for all h ∈ RN \ {0}, p p |u(x + h) − u(x)| dx ≤ h ∇u(x)p dx. RN
RN
Proof. Assume that u ∈ W 1,p (RN )∩C ∞ (RN ). For x ∈ RN and h ∈ RN \{0} by the fundamental theorem of calculus we have that 1 1 d (u(x + th)) dt ≤ h ∇u(x + th) dt. |u(x + h) − u(x)| = 0 dt 0 Raising both sides to the power p and integrating over RN , by H¨older’s inequality we get
p 1 p p |u(x + h) − u(x)| dx ≤ h ∇u(x + th) dt dx RN
≤ h
1
p
= h
RN
RN
p RN
0
∇u(x + th) dtdx = h
0
p
1
p 0
RN
∇u(x + th)p dxdt
∇u(y)p dy,
where we have used Tonelli’s theorem and the change of variables y = x+th. To remove the additional hypothesis that u ∈ C ∞ (RN ), it suffices to apply the previous inequality to uε := ϕε ∗u, where ϕε is a standard mollifier, and to let ε → 0+ (see Theorem C.16 and Lemma 11.25). Lemma 12.20. Let 1 ≤ p < ∞ and let u ∈ W 1,p (RN ). Consider a standard mollifier ϕ as in (C.7). Then there exists a constant c = c(N, p) > 0 such that c |(u ∗ ϕ1/k )(x) − u(x)|p dx ≤ p ∇u(x)p dx. k N N R R
12.1. Embeddings: mp < N
367
Proof. By H¨older’s inequality, (C.4), and the change of variables h = y − x, we have p (12.11) ϕ1/k (x − y)|u(y) − u(x)|p dy |(u ∗ ϕ1/k )(x) − u(x)| ≤ N R ≤ ck N |u(x + h) − u(x)|p dh. B(0,1/k)
Hence,
(12.12) RN
|(u ∗ ϕ1/k )(x) − u(x)|p dx N |u(x + h) − u(x)|p dxdh. ≤ ck B(0,1/k)
RN
In turn, by the previous lemma and Tonelli’s theorem we get |(u ∗ ϕ1/k )(x) − u(x)|p dx ≤ ck N ∇u(x)p dx hp dh N N R R B(0,1/k) c ∇u(x)p dx, = p k RN
which concludes the proof. We now turn to the proof of the Rellich–Kondrachov theorem.
Proof of Theorem 12.18. Since Ω is an extension domain for W 1,p (Ω), we can extend each un to a function un ∈ W 1,p (RN ) in such a way that the sequence {un }n remains bounded in W 1,p (RN ). It follows by Theorem ∗ 12.4 that the sequence {un }n is bounded in Lp (RN ). Since p∗ > 1, by the ∗ reflexivity of Lp (RN ) (see Theorem B.92 and Corollary A.65) we may find ∗ a subsequence of {un }n (not relabeled) such that un u in Lp (RN ). We claim that un → u in Lp (Ω). For simplicity, for every v ∈ Lp (RN ) we set v (k) := v ∗ ϕ1/k . By the previous lemma and the fact that {un }n is bounded in W 1,p (RN ), we get c c (k) p |(un ) − un | dx ≤ p sup ∇un p dx ≤ p , sup k n∈N RN k n∈N RN and so (12.13)
lim sup
k→∞ n∈N RN
|(un )(k) − un |p dx = 0.
By Minkowski’s inequality, un − uLp (Ω) ≤ (un )(k) − un Lp (Ω) + (un )(k) − u(k) Lp (Ω) + u(k) − uLp (Ω) .
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12. Sobolev Spaces: Embeddings
Fix > 0. By (12.13) and Theorem C.16 there exists k¯ depending only on such that for all k ≥ k¯ and all n ∈ N the first and last terms in the previous inequality are both bounded by , and so (12.14)
un − uLp (Ω) ≤ (un )(k) − u(k) Lp (Ω) + 2
for all k ≥ k¯ and all n ∈ N. Hence, to complete the proof, it suffices to show that ¯
¯
lim (un )(k) − u(k) Lp (Ω) = 0.
(12.15)
n→∞ ∗ Lp (RN ),
it follows that for all x ∈ RN , Since un u in ¯ ¯ (k) ϕ1/k¯ (x − y)un (y) dy → ϕ1/k¯ (x − y)u(y) dy = u(k) (x) (un ) (x) = RN
RN
as n → ∞. Moreover, reasoning somewhat as in (12.11) (with u replaced by un ) and using the fact that {un }n is bounded in Lp (RN ), we get ¯ ¯ (k) (k) p N ¯ |un (x + h) − u(x + h)|p dh ≤ ck¯N |(un ) (x) − u (x)| ≤ ck ¯ B(0,1/k)
for all x ∈ RN and all n ∈ N. Since Ω has finite measure, we are in a position to apply the Lebesgue dominated convergence theorem to conclude that (12.15) holds. Hence, we have shown that un → u in Lp (Ω). Since {un }n is bounded in by Vitali’s convergence theorem (see Theorem B.101) this implies that un → u in Lq (Ω) for all 1 ≤ q < p∗ .
∗ Lp (RN ),
Remark 12.21. Note that when p > 1 the function u belongs to W 1,p (Ω). This follows from Theorem 11.65. On the other hand, when p = 1, we can only conclude that u ∈ BV (Ω) (see Theorem 14.39). Exercise 12.22. Prove that if Ω is an extension domain for W 1,p (Ω), 1 ≤ p < N , with finite measure, then un u in W 1,p (Ω) if and only if un → u in Lp (Ω) and ∇un ∇u in Lp (Ω; RN ). Theorem 12.23 (Rellich–Kondrachov’s compactness). Let m ∈ N, 1 ≤ p < ∞ be such mp < N , and let Ω ⊆ RN be an extension domain for W m,p (Ω) with finite measure. Let {un }n be a bounded sequence in W m,p (Ω). Then there exist a subsequence {unk }k of {un }n and a function ∗
∗
∗
u ∈ Lpm,0 (Ω) ∩ W 1,pm,1 (Ω) ∩ · · · ∩ W m−1,pm,m−1 (Ω) such that ∇j unk → ∇j u in Lq (Ω; RMj ) for every j = 0, . . . , m − 1, and for all p ≤ q < p∗m,j . We recall that ∇0 u := u. The proof is left as an exercise.
12.1. Embeddings: mp < N
369
The following exercises show that compactness fails for q = p∗ even for nice domains and that for general domains even the embedding W 1,p (Ω) → Lp (Ω) may fail to be compact. Exercise 12.24. Let 1 ≤ p < N , and consider the sequence of functions un : B(0, 1) → R defined by nN/p−1 (1 − nx) if x < 1/n, un (x) := 0 if x ≥ 1/n. Prove that {un }n is bounded in W 1,p (B(0, 1)) but that it does not admit ∗ any subsequence strongly convergent in Lp (B(0, 1)). Exercise 12.25. In Exercise 12.15 take δ2n := (h2n )a , for some a ≥ 3, and for j = 1, 3, 5, . . ., consider functions uj : Ω → R such that 1/hj in Rj , uj (x1 , x2 ) = 0 in Ω \ (Pj−1 ∪ Rj ∪ Pj+1 ) and
∇uj (x1 , x2 ) =
(1/(hj hj−1 ), 0) in Pj−1 , −(1/(hj hj−1 ), 0) in Pj+1 .
Prove that the sequence {uj }j is bounded in W 1,2 (Ω) but that it does not admit any subsequence strongly convergent in L2 (Ω). The previous exercise shows that the embedding W 1,p (Ω) → Lp (Ω) fails to be compact for arbitrary open bounded sets. The following exercise gives the only possible type of compactness that remains. Exercise 12.26. Prove that if Ω ⊂ RN is an open set with finite measure and 1 < p < ∞, then the embedding W 1,p (Ω) → Lq (Ω) is compact for all 1 ≤ q < p. Exercise 12.27. Let Ω ⊆ RN be an open set and let 1 ≤ p < ∞. Prove that the embedding W 1,p (Ω) → Lp (Ω) is compact if and only if p |u| dx : uW 1,p (Ω) ≤ 1 = 0, lim sup n→∞
Ω\Ωn
where Ωn := {x ∈ Ω : dist(x, ∂Ω) > 1/n and x < n}. The Rellich–Kondrachov theorem and its variations hold for special domains with finite measure. The next two exercises show that if we restrict our attention to the class of radial functions, then we have compactness in the entire space RN for all N < q < p∗ , but not for q = N or q = p∗ .
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12. Sobolev Spaces: Embeddings
Exercise 12.28 (Radial functions, I). Let N ≥ 2, let f ∈ C 1 ([0, ∞)), and let (12.16)
u(x) := f (x),
x ∈ RN .
(i) Let 1 ≤ p < ∞. Find necessary and sufficient conditions on f for u to be in W 1,p (RN ). (ii) Let a > 0. Prove that for r > 0, (r2a f 2 (r)) ≤ [(ra f (r)) ]2 + (ra f (r))2 = r2a [(f (r))2 + f 2 (r)] + a(r2a−1 f 2 (r)) − a(a − 1)r2a−2 f 2 (r). (iii) Prove that for every N ∈ N with N ≥ 3 and for all r > N − 1, r N −1 2 r f (r) ≤ 2 tN −1 [(f (t))2 + f 2 (t)] dt. 0
(iv) Prove that for all r > 1, ∞ t[(f (t))2 + f 2 (t)] dt. rf 2 (r) ≤ 2 r
(v) Prove that if the function u defined in (12.16) belongs to W 1,2 (RN ), N ≥ 2, then |u(x)| ≤ cx−(N −1)/2 uW 1,2 (RN )
for all x ∈ RN , with x > N − 1.
Exercise 12.29 (Radial functions, II). Let {un }n be a sequence of radial functions in W 1,2 (RN ) ∩ C 1 (RN ) with supn uW 1,2 (RN ) < ∞. (i) Prove that
lim sup
R→∞ n
RN \B(0,R)
|un (x)|2N/(N −2) dx = 0.
(ii) Prove that there exist a subsequence {unk }k of {un }n and a function u ∈ W 1,2 (RN ) such that unk → u in Lq (RN ) for all 2 < q < N2N −2 . (iii) Let ϕ ∈ Cc1 (R) be such that supp ϕ ⊆ [0, 1], ϕ = 0, and define un (x) := an ϕ(x − n), x ∈ RN . Find an in such a way that {un }n is bounded in W 1,2 (RN ) but does not converge in L2 (RN ). (iv) Let ϕ be as in part (iii) and define un (x) = an ϕ(2n x−1), x ∈ RN . Find an in such a way that {un }n is bounded in W 1,2 (RN ) but does ∗ not converge in L2 (RN ). We conclude this section by showing that the Rellich–Kondrachov theorem continues to hold for bounded domains with boundary of class C (see Definition 9.57).
12.1. Embeddings: mp < N
371
Theorem 12.30. Let 1 ≤ p < ∞ and let Ω ⊂ RN be a bounded domain whose boundary is of class C. Then the embedding W 1,p (Ω) → Lp (Ω) is compact. Proof. For every x0 ∈ ∂Ω there exist local coordinates y = (y , yN ) ∈ RN −1 × R, with y = 0 at x = x0 , a continuous function f : RN −1 → R, and rx0 > 0 such that Ω ∩ B(x0 , rx0 ) in local coordinates takes the form {(y , yN ) ∈ B(0, rx0 ) : yN > f (y )}. By taking r > 0 and t > 0 sufficiently small, we have that the open set U (x0 , r, t) defined in (11.12) is contained in B(x0 , rx0 ). Hence, Ω∩U (x0 , r, t) in local coordinates becomes {(y , yN ) : y ∈ QN −1 (0, r), f (y ) < yN < f (y ) + t}.
(12.17)
Assume that u ∈ Cc∞ (RN ). Fix y ∈ QN −1 (0, r) and let f (y ) < yN < f (y ) + t,
f (y ) + t/2 < τ < f (y ) + t.
By the fundamental theorem of calculus we have that τ ∂N u(y , s) ds, u(y , yN ) = u(y , τ ) − yN
t ∈ R, and H¨older’s inequality,
p f (y )+t p p−1 p p−1 |∂N u(y , s)| ds |u(y , yN )| ≤ 2 |u(y , τ )| + 2
and so by the convexity of g(t) =
|t|p ,
yN
≤2
p−1
|u(y , τ )| + 2 p
p−1 p/p
t
f (y )+t f (y )
|∂N u(y , s)|p ds.
By averaging in τ over (f (y ) + t/2, f (y ) + t), we obtain f (y )+t 2p f (y )+t |u(y , τ )|p dτ + 2p−1 tp/p |∂N u(y , s)|p ds. |u(y , yN )|p ≤ t f (y )+t/2 f (y ) Now we integrate in (y , yN ) over QN −1 (0, r) × (f (y ), f (y ) + δ), where 0 < δ ≤ t, to get f (y )+δ |u(y , yN )|p dyN dy QN −1 (0,r)
f (y )
2p δ ≤ t
(12.18)
+2
QN −1 (0,r)
p−1 p/p
t
f (y )+t f (y )+t/2
δ QN −1 (0,r)
M
|u(y , τ )|p dτ dy f (y )+t f (y )
|∂N u(y , s)|p dsdy .
is compact, ∂Ω ⊂ i=1 U (xi , ri , ti ) for some x1 , . . . , xM ∈ Since ∂Ω ⊂ ∂Ω. For every ε > 0 let Ωε := {x ∈ Ω : dist(x, ∂Ω) > ε}. By taking RN
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12. Sobolev Spaces: Embeddings
0 < δ < min1≤i≤M ti and εδ > 0 sufficiently small, we have that Ω \ Ωεδ ⊆ M i=1 U (xi , ri , δ). Hence, by (12.18) we get that 2p M δ p |u(x)| dx ≤ |u(x)|p dx t− Ω\Ωεδ Ω p/p p−1 + 2 M t+ δ ∇u(x)p dx, Ω
where t− := min1≤i≤M ti and t+ := max1≤i≤M ti . Since the restriction to Ω of functions in Cc∞ (RN ) is dense in W 1,p (Ω) by Theorem 11.35, the previous inequality continues to hold for all u ∈ W 1,p (Ω). In particular, 2p M δ p/p |u(x)|p dx ≤ + 2p−1 M t+ δ → 0 sup t−
u W 1,p (Ω) ≤1 Ω\Ωε δ
as δ → 0, and so the result follows from Exercise 12.27.
12.2. Embeddings: mp = N ∗
In the previous section we have seen that W 1,p (RN ) → Lp (RN ), where p∗ = N p/(N − p). Since p∗ → ∞ as p → N − , one would be tempted to say that if u ∈ W 1,N (RN ), then u ∈ L∞ (RN ). For N = 1 this is true since if u ∈ W 1,1 (R), then a representative u is absolutely continuous in R so that x u(x) = u(0) + u (s) ds, 0
u
u
and since = ∈ we have that u is bounded and continuous. For N > 1 this is not the case, as Exercise 12.13 shows. L1 (R),
For N ≥ 2, L∞ (RN ) should be replaced by BMO(RN ) (see Section C.5 in Appendix C). Indeed, we have the following result. Theorem 12.31. There exists a constant c = c(N ) > 0 such that for every ˙ 1,N (RN ), function u ∈ W (12.19)
|u|BMO(RN ) ≤ c∇uLN (RN ) .
In particular, W 1,N (RN ) → BMO(RN ). ˙ 1,N (RN ) and let Qr be a cube of side-length r > 0 Proof. Let u ∈ W and sides parallel to the axes. By Poincar´e’s inequality for rectangles (see Theorem 13.34) we have u − uQr LN (Qr ) ≤ N 1−1/N (LN (Qr ))1/N ∇uLN (RN ) ,
12.2. Embeddings: mp = N
where uQr :=
1 LN (Qr )
Qr
373
u(x) dx, and so by H¨older’s inequality,
|u(x) − uQr |dx ≤ u − uQr LN (Qr ) (LN (Qr ))1/N
≤ N 1−1/N LN (Qr )∇uLN (RN ) . By dividing by LN (Qr ) and taking the supremum over all cubes, we get (12.19). Next we extend the previous result to higher derivatives. ˙ m,N/m (RN ) Theorem 12.32. Let m ∈ N, m ≥ 2. Then for every u ∈ W there exists a polynomial Pu of degree m − 1 such that (12.20)
|u − Pu |BMO(RN ) ≤ c∇m uLN/m (RN )
and (12.21)
∇k (u − Pu )LN/k (RN ) ≤ c∇m uLN/m (RN )
for k = 1, . . . , m − 1 and for some constant c = c(N, m) > 0. In particular, W m,N/m (RN ) → BMO(RN ) ∩ W 1,q1 (RN ) ∩ · · · ∩ W m−1,qm−1 (RN ) for all N/m ≤ qk ≤ N/k, k = 1, . . . , m − 1. Observe that N/k is the critical exponent p∗m,k given in (12.7) for p = N/m. ˙ m,N/m (RN ), for every i = 1, . . . , N we have that ∂i u ∈ Proof. Given u ∈ W ˙ m−1,N/m (RN ) and (m − 1)N/m < N . Thus by the Sobolev–Gagliardo– W ˙ m−1,N/m (RN ) (see Theorem 12.9) we can find Nirenberg’s embedding in W a polynomial Pi of degree m − 2 such that (12.22)
∂i u − Pi LN ≤ c∇m uLN/m
and ∇k (∂i u − Pi )LN/k ≤ c∇m uLN/m for every k = 1, . . . , m − 2, where Pi = ∂i u − |α|=m−1 Kα ∗ ∂ α (∂i u). Using Remark 11.6 one can show (exercise) that ∂j Pi = ∂i Pj and so we can find a polynomial P of degree m−1 such that ∂i Pi = P . Since ∂i (u−P ) ∈ LN (RN ), by the previous theorem and (12.22) we have that ∂i u − Pi LN ≤ c∇m uLN/m , |u − P |BMO ≤ c (12.23)
i
while (12.21) follows from (12.23) and the fact that ∂i Pi = P . Next we study embeddings of the type W 1,N (RN ) → Lq (RN ).
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12. Sobolev Spaces: Embeddings
Theorem 12.33. Let m, N ∈ N be such N > m. Then there exists a constant c = c(N, m) > 0 such that for every function u ∈ W m,N/m (RN ), (12.24)
uLq (RN ) ≤ cq 1−m/N +1/q uW m,N/m (RN )
for every N/m < q < ∞ and (12.25)
∇k uLN/k (RN ) ≤ c∇m uLN/m (RN )
for every k = 1, . . . , m − 1. In particular, W m,N/m (RN ) → Lq0 (RN ) ∩ W 1,q1 (RN ) ∩ · · · ∩ W m−1,qm−1 (RN ) for all N/m < q0 < ∞ and all N/m ≤ qk ≤ N/k, k = 1, . . . , m − 1. The proof makes use of the following exercise. Exercise 12.34. Let wn : RN → R, n ∈ N, be Lebesgue measurable func∞ tions and let w := n=1 wn . Assume that there exists an integer M ∈ N such that for every x ∈ RN at most M terms wn (x) are nonzero. Prove that for every 1 ≤ p < ∞,
1/p ∞ wn pLp (RN ) , wLp (RN ) ≤ M 1/p n=1
while wL∞ (RN ) ≤ M supn wn L∞ (RN ) . We turn to the proof of Theorem 12.33. Proof of Theorem 12.33. It suffices to prove (12.24), since (12.25) follows from Theorem 12.8 and the fact that ∇k u ∈ W m−k,N/m (RN ; RMk ) for every k = 1, . . . , m − 1 and (m − k)N/m < N . Step 1: Assume first that u ∈ Cc∞ (RN ). By Lemma 12.10, c0 |∂ α u(y)| dy, |u(x)| ≤ N −m βN B(x,d) x − y |α|=m
where d is the diameter of the support of u and c0 = c0 (N, m). Let N/m < q < ∞. By Young’s inequality (see Theorem 10.49), with m/N + 1/s = 1 + 1/q,
1/s c0 1 dw ∇m uLN/m uLq ≤ (N −m)s βN B(0,d) w 1 = c0 1−1/s d[N −(N −m)s]/s ∇m uLN/m 1/s βN (N − (N − m)s) ≤ c0 q 1/s dN/q ∇m uLN/m , where in the last inequality we used the facts that m/N + 1/s = 1 + 1/q, q > N/m > 1, βN > 1, s > 1.
12.2. Embeddings: mp = N
375
Step 2: For every z ∈ ZN consider the open cube Q(z, 2) centered N at k and of side-length 2. Then R = z∈ZN Q(z, 2), and so reasoning as in the proof of Theorem C.21 we can construct a smooth partition of unity {ψz }z∈ZN subordinated to {Q(z, 2)}z∈ZN with the property that ∂ α ψz L∞ ≤ c(m, N ) for every |α| ≤ m and every z. Since for every x ∈ RN , there are at most 2N cubes Q(z, 2) that contain x, it follows from Exercise 12.34 that 1/q (uψz ) q ≤ 2N/q uψz qLq uLq = L
z∈ZN
√ ≤ 2N/q c0 q 1/s (2 N )N/q
z∈ZN
∇m (uψz )qLN/m
1/q ,
z∈ZN
√ where in the last inequality we have used Step 1 with d = 2 N . Using the fact that
aqz
1/q
z∈ZN
≤
aN/m z
m/N
z∈ZN
for az ≥ 0, we have that N/m m/N uLq ≤ cq 1/s ∇m (uψz )LN/m z∈ZN
≤ cq 1/s ≤ cq 1/s
z∈ZN m k=0
∇m (uψz )N/m dx
m/N
Q(z,2)
RN
∇k uN/m dx
m/N
= cq 1/s uW m,N/m ,
where we used the facts that z∈ZN χQ(z,2) ≤ 2N and that ∂ α ψz L∞ ≤ c for every multi-index α with |α| ≤ m and every z, and where the constant c = c(m, N ) has changed from line to line. A standard mollification argument concludes the proof.
Remark 12.35. Unlike the case mp < N , Step 2 of the proof of (12.24) uses strongly the fact that both u and its derivatives belong to Lp (RN ). Exercise 12.36. Using (12.25) in Step 2 of the previous proof shows that (12.24) can be replaced by the better estimate uLq (RN ) ≤ cq 1−m/N +1/q uLN/m (RN ) + ∇m uLN/m (RN ) . The following exercise provides another proof of Theorem 12.33 when m = 1 but with worse constants.
376
12. Sobolev Spaces: Embeddings
˙ 1,N (RN ), N ≥ 2, with u ∈ Lq (RN ) for some Exercise 12.37. Let u ∈ W 1 ≤ q < ∞. Let v := |u|t , where t > 1 is such that v ∈ W 1,1 (RN ). (i) Prove that (t−1)/t
1/t
uLtN/(N −1) ≤ cuL(t−1)N/(N −1) ∇uLN . (ii) Taking inductively t = n + q(N − 1)/N , n ∈ N, prove that for all q ≤ r < ∞, q/r
1−q/r
uLr ≤ cuLq ∇uLN
.
Note that, in general, if u ∈ W m,N/m (RN ), we cannot conclude that u ∈ Lq (RN ) for 1 ≤ q < N/m. However, since u ∈ Lq (RN ) for all q ≥ N/m, the function u must decay faster than algebraically at infinity. Indeed, we will show that it must have exponential decay. For every ∈ N consider the function (12.26)
∞ −2 1 n 1 n s = exp(s) − s , exp (s) := n! n!
s ∈ R.
n=0
n=−1
The next theorem gives an embedding of the space W m,N/m (RN ) into the Orlicz space generated by the function expN/m+1 (s(N/m) ). We introduce the notion of Orlicz space. Definition 12.38. Let E ⊆ RN be a Lebesgue measurable set and let Φ : [0, ∞) → [0, ∞] be a convex, lower semicontinuous function such that Φ(0) = 0 and Φ is not identically zero or infinity. The Orlicz space LΦ (E) generated by the Orlicz function Φ is the space of all Lebesgue measurable functions u : E → R such that E Φ(|u(x)|/s) dx < ∞ for some s > 0 (depending on u). Exercise 12.39. Let E and Φ be as in the previous definition. (i) Prove that LΦ (E) is a normed space with the norm Φ(|u(x)|/s) dx < ∞ . (12.27) uΦ := inf s > 0 : E
(ii) Prove that LΦ (E) is a Banach space. The following theorem is the main theorem of this section. Results of this kind in bounded domains are due to Pokhozhaev [189], Trudinger [234] and Yudovich [250] when m = 1 and to Strichartz [225] for m ≥ 2 (see also [179] and see [141] for some recent results).
12.2. Embeddings: mp = N
377
Theorem 12.40. Let m, N ∈ N be such N > m. Then there exist two constants c1 , c2 > 0 depending only on m and N such that
|u(x)|(N/m) (12.28) expN/m+1 c1 dx ≤ c2 (N/m) RN uW m,N/m (RN ) for all u ∈ W m,N/m (RN ) \ {0}. In particular, if
Φ(s) := expN/m+1 (c1 s(N/m) ), s ≥ 0, then W m,N/m (RN ) → LΦ (RN ). Proof. Let p := N/m. By replacing u with u/uW m,p , we can assume that uW m,p = 1. Then by Theorem 12.33 with q = np for a > 0, ∞ an p expp+1 (a|u(x)| ) dx = |u(x)|np dx n! RN RN n=p
∞ an cnp (np )n+1 . ≤ n!
n=p
By the ratio test,
an+1 c(n+1)p (n+1)!
an cnp n!
((n + 1)p )n+2 (np )n+1
= acp p (1 + 1/n)n+1 → acp p e < 1,
provided a < 1/(cp p e). Hence, the series converges.
Remark 12.41. When m = 1 we can take expN instead of expN +1 . Moreover, using symmetric rearrangement one can show (see Theorem 15.35) that (12.28) holds for any c1 ∈ (0, γN ), where (12.29)
1/(N −1)
γN := N βN
,
and where βN is the surface area of the unit sphere (see (12.3)). Moreover, the constant γN is sharp (see Exercise 15.37). Remark 12.42. The analog of this result for bounded domain can be found in Exercise 13.23. See also Remark 13.24 for more references on this topic. Exercise 12.43. Prove that for every function u ∈ W N,1 (RN ), ∂N u . uL∞ (RN ) ≤ ∂x1 · · · ∂xN 1 N L (R )
˙ N,1 (RN ) there exists a polynomial Prove also that for every function u ∈ W Pu of degree N − 1 such that ∂N u u − Pu L∞ (RN ) ≤ ∂x1 · · · ∂xN 1 N . L (R )
378
12. Sobolev Spaces: Embeddings
Theorem 12.44 (Rellich–Kondrachov). Let m, N ∈ N be such N > m and let Ω ⊂ RN be an extension domain for W m,N/m (Ω) with finite measure. Let {un }n be a bounded sequence in W m,N/m (Ω). Then there exist a subsequence {unk }k of {un }n and a function u ∈ LN/m (Ω) ∩ W 1,N (Ω) ∩ · · · ∩ W m,N/m (Ω) such that ∂ β unk → ∂ β u in Lq (Ω) for every multi-index β with |β| = k, k = 0, . . . , m − 1, and for all N/m ≤ q < N/k, where N/0 := ∞. The proof is left as an exercise. The following exercise shows that if u ∈ W 1,N (Ω), where Ω ⊆ RN is an open set, then one cannot have differentiability LN -a.e. in Ω (if u ∈ W 1,p (Ω), with p > N , then this is true by Corollary 12.50). Exercise 12.45 (Indian mystic’s bed of nails). Let Q := (0, 1)2 be the unit cube in R2 , subdivide it into 4k subcubes each having side length 2−k , and k let Ck := {x(1) , . . . , x(4 ) } be the set of centers of the subcubes. Define t + e−2k ln −2k gk (t) := 1 + e k , 0 ≤ t ≤ 1, e−2 ln k 1 + e−2 and (n) ) if x − x(n) ≤ ε for some x(n) ∈ C , gk (ε−1 k k k x − x fk (x) := 0 otherwise, k where 0 < εk < 4−k are chosen sufficiently small. Define u(x) := ∞ k=1 2k fk (x), x ∈ Q. (i) Prove that u ∈ W 1,2 (Q) ∩ C(Q). (ii) Let x ∈ Q be such that either fk (x) = 0 for all k or fk (x) = 0 for all k sufficiently large. Prove that u is not differentiable at x. (iii) Prove that if εk is chosen sufficiently small, then u is not differentiable on a set of positive measure.
12.3. Embeddings: mp > N In this section we study the case mp > N . We begin by showing that W m,p (RN ) is embedded in L∞ (RN ). Theorem 12.46. Let m ∈ N and 1 ≤ p < ∞ be such mp > N . Then W m,p (RN ) → L∞ (RN ). We begin with an auxiliary lemma, which will also be used in the sequel.
12.3. Embeddings: mp > N
379
Lemma 12.47. Let m ∈ N and 1 ≤ p < ∞ be such mp > N , let Qr be any cube with sides of length r parallel to the axes. Then for every u ∈ W m,p (Qr ) ∩ C ∞ (Qr ), |u(x) − u(y)| ≤ c
m−1
rk−N/p ∇k uLp (Qr ) + crm−N/p ∇m uLp (Qr )
k=1
for every x, y ∈ Qr , and where c > 0 is a constant depending on m, N , and p. In particular, if m = 1, then |u(x) − u(y)| ≤ cr1−N/p ∇uLp (Qr ) . Proof. Fix x, y ∈ Qr . By Taylor’s formula (see Theorem 9.9), u(x) − u(y) =
1 α ∂ u(y)(x − y)α α! 1≤|α|≤m−1 1 m (x − y)α + (1 − t)m−1 ∂ α u(tx + (1 − t)y) dt. α! 0 |α|=m
Averaging in the y variable over Qr yields 1 1 ∂ α u(y)(x − y)α dy u(x) − uQr = α! rN Qr 1≤|α|≤m−1 m 1 1 (x − y)α (1 − t)m−1 ∂ α u(tx + (1 − t)y) dtdy, + α! rN Qr 0 |α|=m
where we recall that uQr := |u(x) − uQr | ≤ c + cr−N ≤c
m−1
r
+ cr
r
−N
Qr
u(y) dy. Hence,
∇k u(y) dy
k−N Qr
k=1 1
x − ym (1 − t)m−1 ∇m u(tx + (1 − t)y) dydt Qr
∇k u(y) dy
k−N
k=1
m−1
0
1 rN
Qr
1 0
z − xm (1 − t)−N −1 ∇m u(z) dzdt, tx+Q(1−t)r
√ where we have used the fact that x − y ≤ N r in Qr , Tonelli’s theorem, the change of variables z = tx + (1 − t)y (so that z − x = (1 − t)(y − x) and dz = (1 − t)N dy). Using Tonelli’s theorem once more, we have that the
380
12. Sobolev Spaces: Embeddings
integral in the second term on the right-hand side of the previous inequality can be bound from above by 1− z−x /(r√N ) m m z − x ∇ u(z) (1 − t)−N −1 dtdz 0 Qr ≤ crN z − xm−N ∇m u(z) dz, Qr
and so |u(x) − uQr | ≤ c
m−1
r
∇k u(y) dy
k−N Qr
k=1
y − xm−N ∇m u(z) dy.
+c Qr
By H¨older’s inequality we now have |u(x) − uQr | ≤ c
m−1
rk−N +N/p ∇k uLp (Qr )
k=1
y − x
+c
(m−N )p
1/p dy
Qr
≤c
m−1
∇m uLp (Qr )
rk−N/p ∇k uLp (Qr ) + crm−N/p ∇m uLp (Qr ) ,
k=1
√ where we used polar coordinates and the fact that y−x ≤ N r to estimate y − x(m−N )p dy ≤ y − x(m−N )p dy √ Qr
B(x, N r)
βN (p − 1) √ ( N r)(mp−N )/(p−1) . = mp − N Since this is true for all y ∈ Qr , if x, y ∈ Qr , then |u(x) − u(y)| ≤ |u(x) − uQr | + |u(y) − uQr | ≤c
m−1
rk−N/p ∇k uLp (Qr ) + crm−N/p ∇m uLp (Qr ) ,
k=1
which completes the proof.
We turn to the proof of Theorem 12.46. Proof of Theorem 12.46. Assume that u ∈ W m,p (RN ) ∩ C ∞ (RN ). Fix x ∈ RN and let Q1 be a cube with sides of length 1 parallel to the axes containing x. By Lemma 12.47 with r = 1, |u(x)| ≤ |u(y)| + cuW m,p (Q1 )
12.3. Embeddings: mp > N
381
for every y ∈ Q1 . By averaging in y over Q1 and using H¨older’s inequality, we obtain |u(x)| ≤ cuW m,p (Q1 ) ≤ cuW m,p (RN ) . Now, if u ∈ W m,p (RN ), let uε := u ∗ ϕε , where ϕε is a standard mollifier. By the previous inequality applied to uε we get |uε (x)| ≤ cuε W m,p (RN ) . Since {uε } converge pointwise at every Lebesgue point by Theorem C.16 and in W m,p (RN ) by Lemma 11.25, it follows that |u(x)| ≤ cuW m,p (RN ) for LN -a.e. x ∈ RN . The next theorem shows that if p > N , a function u ∈ W 1,p (RN ) has a representative in the space C 0,1−N/p (RN ) of bounded functions that are H¨older continuous with exponent 1 − N/p. Theorem 12.48 (Morrey’s embedding in W 1,p ). Let N < p < ∞. Then W 1,p (RN ) → C 0,1−N/p (RN ). Moreover, if u ∈ W 1,p (RN ) and u ¯ is its representative in C 0,1−N/p (RN ), then ¯(x) = 0. lim u
x →∞
Proof. Let u ∈ W 1,p (RN ) ∩ C ∞ (RN ). If x, y ∈ RN , consider a cube Qr containing x and y and of side length r := 2x − y. By Lemma 12.47, (12.30)
|u(x) − u(y)| ≤ cx − y1−N/p ∇uLp (Qr ) .
Hence, u is H¨older continuous of exponent 1−N/p, while by Theorem 12.46, we get (12.31)
|u(x)| ≤ cuW 1,p (RN ) .
Next we remove the extra hypothesis that u ∈ C ∞ (RN ). Given any u ∈ W 1,p (RN ), let x, y ∈ RN be two Lebesgue points of u and let uε := u ∗ ϕε , where ϕε is a standard mollifier. By (12.30) we have that |uε (x) − uε (y)| ≤ cx − y1−N/p ∇uε Lp (Qr ) . Since {uε }ε converge at every Lebesgue point of u as ε → 0+ by Theorem C.16 and ∇uε = (∇u)ε → ∇u in Lp (RN ; RN ) by Theorems C.16 and C.20, letting ε → 0+ , we get (12.32)
|u(x) − u(y)| ≤ cx − y1−N/p ∇uLp (Qr )
for all Lebesgue points x, y ∈ RN of u. This implies that u : {Lebesgue points ¯ of u} → R can be uniquely extended to RN as a H¨older continuous function u N of exponent 1−N/p in such a way that (12.32) holds for all x, y ∈ R . With
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12. Sobolev Spaces: Embeddings
a similar argument from (12.31) we conclude that |¯ u(x)| ≤ cuW 1,p (RN ) for N all x ∈ R . Hence, ¯ uC 0,1−N/p (RN ) = sup |¯ u(x)| + x∈RN
sup x, y∈RN , x=y
|¯ u(x) − u ¯(y)| 1−N/p x − y
≤ cuW 1,p (RN ) . Finally, we prove that u ¯(x) → 0 as x → ∞. Let {un }n be any sequence in Cc∞ (RN ) that converges to u in W 1,p (RN ). By Theorem 12.46, u − un L∞ (RN ) ≤ cu − un W 1,p (RN ) → 0 as n → ∞. ¯ . Since Fix ε > 0 and find n ¯ ∈ N such that u − un L∞ (RN ) ≤ ε for all n ≥ n ∞ N un¯ ∈ Cc (R ), there exists Rn¯ > 0 such that un¯ (x) = 0 for all x ≥ Rn¯ . Hence, for LN -a.e. x ∈ RN with x ≥ Rn¯ we get |¯ u(x)| = |¯ u(x) − un¯ (x)| ≤ u − un L∞ (RN ) ≤ ε, and, since u ¯ is continuous, we get that the previous inequality actually holds for all x ∈ RN with x ≥ Rn¯ . Remark 12.49. The first part of the proof actually shows that if u ∈ ˙ 1,p (RN ) for some N < p < ∞, then a representative u W ¯ of u is H¨older continuous with exponent 1 − N/p and |¯ u(x) − u ¯(y)| ≤ cx − y1−N/p ∇uLp (RN ) for all x, y ∈ RN , where c = c(N, p) > 0. As a consequence of Theorem 12.48 we obtain the following result. ¯ is its representative Corollary 12.50. If u ∈ W 1,p (RN ), N < p < ∞, and u in C 0,1−N/p (RN ), then u ¯ is differentiable at LN -a.e. x ∈ RN and the weak partial derivatives of u coincide with the (classical) partial derivatives of u ¯ N N L -a.e. in R . Proof. Let x0 ∈ RN be a p-Lebesgue point for ∇u (see Corollary B.120), that is, 1 ∇u(x) − ∇u(x0 )p dx = 0, lim r→0+ r N Q(x0 ,r) where as usual Q(x0 , r) := x0 + (− 2r , 2r )N , and define v(x) := u ¯(x) − u ¯(x0 ) − ∇u(x0 ) · (x − x0 ), x ∈ RN . Reasoning as in the first part of the proof of Theorem 12.48, if x ∈ RN and r := 2x − x0 , we obtain that |v(x) − v(x0 )| ≤ cr1−N/p ∇vLp (Q(x0 ,r)) ,
12.3. Embeddings: mp > N
383
or, equivalently, |¯ u(x) − u ¯(x0 ) − ∇u(x0 ) · (x − x0 )| x − x0 1/p 1 ∇u(y) − ∇u(x0 )p dy →0 ≤c N r Q(x0 ,r) as x → x0 . This completes the proof.
In view of Exercise 11.50 and Corollary 12.50 we can give another proof of Rademacher’s theorem. Exercise 12.51 (Rademacher). Let Ω ⊆ RN be an open set. Prove that every Lipschitz continuous function u : Ω → R is differentiable at LN -a.e. x ∈ Ω. The following result is another important consequence of Theorem 12.48. Corollary 12.52. Let Ψ ∈ W 1,p (RN ; RN ), N < p < ∞, and let Ψ be its representative in C 0,1−N/p (RN ; RN ). Then there exists a constant c = c(N, p) > 0 such that for every Lebesgue measurable set E ⊆ RN the set Ψ(E) is Lebesgue measurable and LN (Ψ(E)) ≤ c(LN (E))1−N/p JΨ N Lp (E) . In particular, Ψ satisfies the (N ) property; that is, it maps sets of LN measure zero into sets of LN -measure zero. Proof. In the proof of Theorem 12.48, we have seen that for every cube Q with sides parallel to the axes and for every x, y ∈ Q, |Ψi (x) − Ψi (y)| ≤ c(diam Q)1−N/p ∇Ψi Lp (Q) for all i = 1, . . . , N . Hence, (12.33)
LN (Ψ(Q)) ≤ 2N (diam Ψ(Q))N ≤ c(diam Q)(1−N/p)N JΨ N Lp (Q) .
Next, if E is an arbitrary Lebesgue measurable set, for every ε >0 find a countable family {Qn }n of pairwise disjoint cubes such that E ⊆ n Qn := E0 and LN (Qn ) ≤ LN (E) + ε. n
384
12. Sobolev Spaces: Embeddings
Then by (12.33) and H¨older’s inequality (which one?) we have LN (Ψ(Qn )) LN o (Ψ(E)) ≤ n
≤c
(diam Qn )(1−N/p)N JΨ N Lp (Qn ) n
≤c
(diam Qn )N
n
≤c
LN (Qn )
1−N/p
1−N/p
JΨ pLp (Qn )
N/p
n
JΨ N Lp (E0 )
n N
≤ c(L (E) + ε)1−N/p JΨ N Lp (E0 ) . Letting ε → 0+ yields N 1−N/p LN JΨ N o (Ψ(E)) ≤ c(L (E)) Lp (E) .
This inequality implies, in particular, that Ψ has the (N ) property. It remains to show that Ψ(E) is Lebesgue measurable. Assume first that E has finite measure and write E = E∞ ∪ E1 , where E∞ is an Fσ set and E1 is a set of Lebesgue measure zero. Then Ψ(E) = Ψ(E∞ ) ∪ Ψ(E0 ). By the (N ) property the set Ψ(E1 ) has Lebesgue measure zero, and so it is Lebesgue measurable by the completeness of the Lebesgue measure, while by the continuity of Ψ the set Ψ(E∞ ) is Lebesgue measurable. If E is not bounded, we may write it as the countable union of bounded Lebesgue measurable sets, and so by what we have just proved, the set Ψ(E) is Lebesgue measurable, since it may be written as a countable union of Lebesgue measurable sets. ˙ 1,p (RN ; RN ) for Remark 12.53. The previous corollary still holds if Ψ ∈ W some N < p < ∞. Thus, from the previous corollary any such mapping can be used as a change of variables. More precisely, we have the following. Exercise 12.54. Let Ω ⊆ RN be an open set and let Ψ : Ω → RN be oneto-one on a set whose complement in Ω has Lebesgue measure zero. Prove that if Ψ ∈ W 1,p (Ω; RN ), N < p ≤ ∞, then it has a representative Ψ for which all the hypotheses of Theorem 9.52 are satisfied. Next we extend Morrey’s theorem to W m,p (RN ), m ≥ 2. Theorem 12.55 (Morrey’s embedding in W m,p ). Let m ∈ N, m ≥ 2, and 1 ≤ p < ∞ be such mp > N . Then W m,p (RN ) → C ,θ (RN ), where if m − N/p is not an integer, (12.34)
:= m − N/p ,
θ := m − − N/p,
12.3. Embeddings: mp > N
385
while if m − N/p is an integer, := m − 1 − N/p,
(12.35)
θ := any number less than 1.
Proof. Step 1: Assume that m−N/p is not an integer. Then = m−N/p and m − N/p = + θ, where θ ∈ (0, 1). Hence, (m − − 1)p < N < (m − )p. We now distinguish two cases. If = m − 1, then p > N and θ = 1 − N/p. Given u ∈ W m,p (RN ), since ∂ α u ∈ W 1,p (RN ) for every |α| ≤ , we can apply Morrey’s embedding theorem (Theorem 12.48) to ∂ α u to obtain that ∂ α u ∈ C 0,1−N/p (RN ). It follows that W m,p (RN ) → C ,θ (RN ). If < m − 1, then p < N . For every n = 0, . . . , and every β with |β| = − n, we have that ∂ β u ∈ W m−+n,p (RN ), and since (m − + n)p > N , by Theorem 12.46 we have that ∂ β u belongs to L∞ (RN ). Hence, u ∈ C −1,1 (RN ) if ≥ 1 and u ∈ L∞ (RN ) if = 0. In particular, for |β| = we have that ∂ β u ∈ L∞ (RN ) ∩ Lp (RN ) and, in turn, by Exercise B.79(ii), to all Lq (RN ) for p ≤ q ≤ ∞. On the other hand, since (m − − 1)p < N , using Sobolev–Gagliardo– Nirenberg’s embedding theorem (Theorem 12.4) we have that ∂ α u ∈ ∗ Np > N Lpm,+1 (RN ) for every |α| = + 1, where p∗m,+1 = N −(m−−1)p ∗
(see (12.7)). It follows that ∂ β u ∈ W 1,pm,+1 (RN ) for |β| = , and since p∗m,+1 > N , by Morrey’s embedding theorem (Theorem 12.48), ∂ β u ∈ ∗
C 0,1−N/pm,+1 (RN ). Note that 1 − N/p∗m,+1 = m − − N/p = θ. This shows that W m,p (RN ) → C ,θ (RN ). Step 2: Assume next that m − N/p = n ∈ N. If n = m − 1, then p = N and so by the critical embedding theorem (Theorem 12.33) applied to ∂ α u where |α| ≤ m − 1 we have that ∂ α u ∈ Lq (RN ) for every N ≤ q < ∞. In particular, u ∈ W m−1,q (RN ) for every N ≤ q < ∞. Hence, for every q > N we have that m − 2 < m − 1 − (N/q) < m − 1 and so by Step 1, W m,p (RN ) → C ,θ (RN ), where θ := m − N/p − (N/q). If n < m−1, then p < N . Since p∗m,n+1 = N (see (12.7)), using Sobolev– Gagliardo–Nirenberg’s embedding theorem (Theorem 12.4) we have that ∂ α u ∈ LN (RN ) for every |α| = n + 1. In turn, by repeated applications of Exercise 12.37, ∂ β u ∈ Lq (RN ) for 0 ≤ |β| ≤ n for all p ≤ q < ∞. Hence, u ∈ W n,q (RN ) for every p ≤ q < ∞. Taking q > N , by Step 1, W m,p (RN ) → C ,θ (RN ), where θ := m − N/p − (N/q). Remark 12.56. Let m ∈ N, m ≥ 2, and 1 ≤ p < ∞ be such mp > N ˙ m,p (RN ). As in Step 1 of and m − N/p is not an integer, and let u ∈ W the previous proof if = m − 1, by Remark 12.49, we have that ∂ α u is H¨older continuous with exponent θ for every |α| = . If < m − 1, consider v := ∂ β u for |β| = . Using Theorem 12.9 (with m replaced by m − − 1 and u by ∂i v) we can find a polynomial Pi,β of degree m − − 2 such that
386
12. Sobolev Spaces: Embeddings
∗ ∂i v − Pi,β ∈ Lpm,+1 (RN ), where Pi,β = ∂i v − |α|=m−−1 Kα ∗ ∂ α (∂i v). Using Remark 11.6 one can show (exercise) that ∂j Pi,β = ∂i Pj,β and so we can find a polynomial Pβ of degree m − − 1 such that Pi,β = ∂i Pβ . Since ∗ Np > N , by Remark 12.49, ∂i (v − Pβ ) ∈ Lpm,+1 (RN ) and p∗m,+1 = N −(m−−1)p we have that ∂ β u − Pβ is H¨older continuous with exponent θ.
If u ∈ Lq (RN ) for some 1 ≤ q < ∞, then Pβ = 0 (why?) and so ∂ β u is H¨older continuous with exponent θ. The next exercise provides an alternative proof of Step 1 in the case = 0 and 1 < p < ∞. Exercise 12.57. Let m ∈ N and 1 < p < ∞ be such mp > N and m−N/p ∈ (0, 1). Let u ∈ Cc∞ (RN ). (i) Use Lemma 12.10 to show that for x, h ∈ RN , with h = 0, |Kα (y − x − h) − Kα (y − x)||∂ α u(y)| dy |u(x + h) − u(x)| ≤ |α|=m B(x,3 h )
+
|α|=m
=:
|α|=m
RN \B(x,3 h )
Iα +
|Kα (y − x − h) − Kα (y − x)||∂ α u(y)| dy
IIα .
|α|=m
(ii) Prove that Iα ≤ chm−N/p ∂ α uLp and IIα ≤ chm−N/p ∂ α uLp . (iii) Use parts (i) and (ii) to show that |u|C 0,m−N/p ≤ c∇m uLp . Exercise 12.58. Prove that W N +1,1 (RN ) → C 0,1 (RN ). Exercise 12.59. Let 1 ≤ p < ∞ be such that p > N and let Ω ⊆ RN be an extension domain for W 1,p (Ω). Prove that W 1,p (Ω) → C 0,1−N/p (Ω) ¯ is its representative in C 0,1−N/p (Ω), then and that if u ∈ W 1,p (Ω) and u N ¯(x) → 0 as u ¯ is differentiable at L -a.e. x ∈ Ω and, if Ω is unbounded, u x → ∞. Exercise 12.60. Let m ∈ N, m ≥ 2, and 1 ≤ p < ∞ be such mp > N , and let Ω ⊆ RN be an extension domain for W m,p (Ω). Prove that W m,p (Ω) → C ,θ (Ω), where and θ are given in (12.34) and (12.35), respectively. Theorem 12.61 (Rellich–Kondrachov). Let 1 ≤ p < ∞ be such that p > N and let Ω ⊆ RN be a bounded extension domain for W 1,p (Ω). Then for all 0 < α < 1 − N/p the embedding W 1,p (Ω) → C 0,α (Ω) is compact. Proof. Exercise.
12.4. Superposition
387
Exercise 12.62. Let 1 ≤ p < ∞ be such that p > N and let Ω ⊆ RN be an extension domain for W 1,p (Ω) with finite measure. Prove that the embedding W 1,p (Ω) → Lp (Ω) is compact. Exercise 12.63. Let m ∈ N, m ≥ 2, and 1 ≤ p < ∞ be such mp > N , and let Ω ⊆ RN be a bounded extension domain for W m,p (Ω). Prove that for all 0 < α < θ the embedding W m,p (Ω) → C ,α (Ω) is compact. Here, and θ are given in (12.34) and (12.35), respectively. Exercise 12.64. Let m ∈ N and 1 ≤ p < ∞ be such mp > N or m ≥ N and p = 1 and let Ω ⊆ RN be an extension domain for W m,p (Ω). Prove that if u, v ∈ W m,p (Ω), then uv belongs to W m,p (Ω) and uvW m,p (Ω) ≤ cuW m,p (Ω) vW m,p (Ω) , where c = c(m, N, p, Ω) > 0. Exercise 12.65. Let 1 ≤ p < ∞ and let u ∈ C ∞ (RN ) be such that u ∈ W m,p (RN ) for every m ∈ N. Prove that u(x) → 0 as x → ∞. Exercise 12.66. Let u ∈ C(RN ) ∩ W 1,2 (RN ). Is it true that u(x) → 0 as x → ∞?
12.4. Superposition In Theorem 3.55 we have seen that given a function f : R → R, a necessary and sufficient condition for f ◦ u to be locally absolutely continuous for all u ∈ ACloc (I) is that f is locally Lipschitz continuous. In this section we establish similar results for u ∈ W m,p (RN ). The following proposition will be used to construct counterexamples in the proofs of the necessary conditions. Proposition 12.67. Let m ∈ N, 1 ≤ p < ∞, and let f : R → R be such that f (0) = 0 and f ◦ u belongs to W m,p (RN ) for every u ∈ W m,p (RN ). Then for every cube Q ⊂ RN there exist δ > 0 and M > 0 such that if u ∈ W m,p (RN ) has support in Q and uW m,p (RN ) ≤ δ, then f ◦ uW m,p (RN ) ≤ M . Proof. Assume for simplicity that Q = Q(0, 1). Arguing by contradiction, if the statement does not hold, for every n ∈ N we can find un ∈ W m,p (RN ), with supp un ⊂ Q, such that un W m,p ≤ 1/2n and f ◦ un W m,p ≥ n. Write ZN = {yn : n ∈ N}. Then the function vn (x) := un (x + yn ), x ∈ RN , of variables vn W m,p ≤ 1/2n is supported in Q(yn , 1), and by a change and f ◦ vn W m,p ≥ n. Define u := n vn . Then u ∈ W m,p (RN ) and so f ◦ uW m,p < ∞. On the other hand, since the cubes Q(yn , 1) are pairwise
388
12. Sobolev Spaces: Embeddings
disjoint, by a change of variables, m m k p ∞> ∇ (f ◦ u) dx = k=0
=
RN
n k=0
m n k=0
∇k (f ◦ un )p dx ≥ Q
∇k (f ◦ vn )p dx Q(yn ,1)
n = ∞,
n
which is a contradiction and completes the proof.
Exercise 12.68. Let m ∈ N and N ≥ 2, let 0 < ε < 1/2, and let vε (x) := log x/ log ε, x ∈ B(0, 1) \ B(0, ε). Let g ∈ C ∞ ([0, 1]) be zero near t = 0 and one near t = 1, and let uε (x) := g(vε (x)) for x ∈ B(0, 1) \ B(0, ε), u := 1 in B(0, ε), and u := 0 outside B(0, 1). Prove that uε ∈ W m,N/m (RN ) with uε W m,N/m (RN ) ≤ c| log ε|m/N −1 . We begin with the case m = 1. The following theorem is due to Marcus and Mizel [161] and [162]. The proof is taken from a paper of Bourdaud [31]. Theorem 12.69. Let 1 ≤ p < ∞ and let f : R → R. (i) If p > N or p = N = 1, then f ◦ u ∈ W 1,p (RN ) for all functions u ∈ W 1,p (RN ) if and only if f (0) = 0 and f is locally Lipschitz continuous. (ii) If p < N or p = N > 1, then f ◦ u ∈ W 1,p (RN ) for all functions u ∈ W 1,p (RN ) if and only if f (0) = 0 and f is Lipschitz continuous. Proof. We first prove the sufficient conditions in (i) and (ii). Step 1: Assume that p > N or p = N = 1 and that f (0) = 0 and f is locally Lipschitz continuous. Given u ∈ W 1,p (RN ), by Morrey’s theorem (Theorem 12.48) if p > N or by Theorem 7.16 and the fundamental theorem of calculus for functions in ACloc (R), if p = N = 1, we have that u is continuous and bounded. Using the notation (E.2), given u ∈ W 1,p (RN ) and i = 1, . . . , N , by Theorem 11.45, u has a representative u such that u(xi , ·) is absolutely continuous in R for LN −1 -a.e. xi ∈ RN −1 and whose first-order (classical) partial derivatives belong to Lp (RN ). Moreover, the (classical) partial derivatives of u agree LN -a.e. with the weak derivatives of u. Let xi ∈ RN −1 be such that u(xi , ·) is absolutely continuous in R and u(xi , ·), ∂i u(xi , ·) ∈ Lp (R). Since f is locally Lipschitz continuous, by Theorem 3.55, setting v := u(xi , ·), we have that f ◦ v is locally absolutely continuous and (12.36)
(f ◦ v) (xi ) = f (v(xi ))v (xi ) = f (u(xi , xi ))∂i u(xi , xi )
for L1 -a.e. xi ∈ R, where f (v(xi ))v (xi ) is interpreted to be zero whenever v (xi ) = 0 (even if f is not differentiable at v(xi ).
12.4. Superposition
389
Since u is bounded, so is u, and so, since f is locally Lipschitz continuous, by (12.36), (f ◦ v) ∈ Lp (R). Moreover, since f (0) = 0, |(f ◦ v)(xi )| = |f (v(xi )) − f (0)| ≤ Lu |v(xi ) − 0|, where Lu is the Lipschitz constant of f in [−uL∞ , uL∞ ]. Since v ∈ Lp (R), it follows that f ◦ v belongs to Lp (R). Using Theorem 11.45 once more, we conclude that f ◦ u belongs to W 1,p (RN ). Step 2: Assume that p < N or p = N > 1 and that f (0) = 0 and f is Lipschitz continuous. We can now proceed as in the previous step to show that f ◦ u ∈ W 1,p (RN ) for every function u ∈ W 1,p (RN ). Step 3: Assume that p > N or p = N = 1 and that f ◦ u ∈ W 1,p (RN ) for all functions u ∈ W 1,p (RN ). Since 0 ∈ W 1,p (RN ), we have that f (0) ∈ W 1,p (RN ), which implies that f (0) = 0. Given v ∈ W 1,p ((−1/4, 1/4)), by Theorem 7.16 we can assume that v is absolutely continuous in [−1/4, 1/4]. Extend v to [−1/2, 1/2] by setting v(t) := v(−1/4) for t ≤ −1/4 and v(t) := v(1/4) for t ≥ 1/4. Let ϕ ∈ Cc∞ (RN ) be such that ϕ(x) = 1 for all x ∈ Q(0, 1/2) and ϕ = 0 outside Q(0, 1). Then the function u(x) := ϕ(x)v(x1 ), x ∈ RN , belongs to W 1,p (RN ), and so f ◦ u ∈ W 1,p (RN ). Since for x ∈ Q(0, 1/2), f (u(x)) = f (v(x1 )), by Theorem 11.45, we have that f ◦ v ∈ W 1,p ((−1/4, 1/4)). Since this holds for every v ∈ W 1,p ((−1/4, 1/4)), it follows by Theorem 7.23 that f is locally Lipschitz continuous. Step 4: Assume that p < N and that f ◦ u ∈ W 1,p (RN ) for all functions u ∈ W 1,p (RN ). As in the previous step we have that f (0) = 0. Let ϕ, ψ ∈ Cc∞ (RN ) be such that ϕ(x) = 1 and ψ(x) = x1 for all x ∈ Q(0, 1/2) and supp ϕ, supp ψ ⊂ Q(0, 1). Given 0 < r, R < 1, t0 ∈ R, λ > 0, and ∈ N, define ψ(x/r − k) + t0 ϕ(x/R), x ∈ RN , (12.37) u(x) := λ k
where the sum is done over all k ∈ ZN such that |ki | ≤ for all i = 1, . . . , N . By the change of variables y = x/R and the facts that R < 1 and p < N , (12.38) Similarly, (12.39)
ϕ(·/R)W 1,p ≤ R(N −p)/p ϕW 1,p . ψ(·/r − k) k
W 1,p
≤ c1 N/p r(N −p)/p ,
where c1 depends on ψW 1,p . Let M , δ > 0 be the numbers given in Proposition 12.67 corresponding to Q(0, 1). Take R so small that |t0 |R(N −p)/p ϕW 1,p ≤ δ/2
390
12. Sobolev Spaces: Embeddings
and let r = r() > 0 be such that λc1 N/p r(N −p)/p = δ/2.
(12.40)
Then uW 1,p ≤ δ. Writing r = (δ/(2c1 λ))p/(N −p) −p/(N −p) and using the fact that p < N , we can take so large that 8r ≤ R. In turn, all the cubes Q(rk, r) are contained in Q(0, R/2), and thus supp u ⊂ Q(0, 1). It follows from Proposition 12.67 that f ◦ uW 1,p ≤ M . Note that the cubes Q(rk, r) are pairwise disjoint. Indeed, if j, k ∈ ZN are two distinct numbers such that |ji |, |ki | ≤ for all i = 1, . . . , N , then there exists l ∈ {1, . . . , N } such that jl = kl , say, jl < kl . If x ∈ Q(rj, r), then xl < rjl + r/2 = r(jl + 1) − r/2 ≤ rkl − r/2, which shows that x cannot belong to Q(rk, r). Since ϕ(x) = 1 and ψ(x) = x1 for all x ∈ Q(0, 1/2), it follows that on the cube Q(rk, r/2) we have that (f ◦ u)(x) = f (λx1 /r − λk1 + t0 ). Since f ◦ u ∈ W 1,p (RN ), for LN -a.e. x ∈ Q(rk, r/2) there exists (λ/r)f (λx1 /r − λk1 + t0 ) = ∂1 (f ◦ u)(x). In turn, by Fubini’s theorem, the change of variables t = λx1 /r − λk1 , and (12.40), p |∂1 (f ◦ u)(x)|p dx M ≥ RN p (λ/r) |f (λx1 /r − λk1 + t0 )|p dx ≥ Q(rk,r/2)
k
≥ cN λp−1 rN −p
λ/4 −λ/4
|f (t + t0 )|p dt = |f (t
cδ p (2c1 )−p λ
λ/4 −λ/4
|f (t + t0 )|p dt.
we obtain c2 ≥ for all t0 ∈ R which are Lebesgue Letting λ → 0 points of f . It follows that f is Lipschitz continuous. 0+
)|p
Step 5: Assume that p = N > 1 and that f ◦u ∈ W 1,N (RN ) for all functions u ∈ W 1,N (RN ). In this case we need to modify the function in (12.37) since in (12.38) we cannot make ϕ(·/R)W 1,N small. We consider instead the function ψ(x/r − k) + t0 uε (x), x ∈ RN , (12.41) u(x) := λ k
where uε is given in Exercise 12.68. Take 0 < ε < 1 so small that |t0 |uε W 1,N ≤ c|t0 || log ε|1/N −1 ≤ δ/2 and for 0 < λ < δ(4c1 )−1 , where c1 is the constant in (12.39), define :=
δ(2c1 λ)−1 > 1. Then δ(4c1 )−1 ≤ λ ≤ δ(2c1 )−1 . Finally, by taking r
12.4. Superposition
391
√ so small that r N ( + 1/2) ≤ ε, we have that all the cubes Q(rk, r) are contained in B(0, ε), and thus supp u ⊂ Q(0, 1). It follows from Proposition 12.67 that f ◦ uW 1,N ≤ M . Since ψ(x) = x1 for all x ∈ Q(0, 1/2) and uε = 1 in B(0, ε), we can proceed as in the previous step to obtain MN ≥ |∂1 (f ◦ u)(x)|N dx N R N (λ/r) |f (λx1 /r − λk1 + t0 )|N dx ≥ k N N −1
= c λ
Q(rk,r/2) λ/4
c2 |f (t + t0 )| dt ≥ λ −λ/4
λ/4
N
−λ/4
|f (t + t0 )|N dt,
where in the last inequality we used the fact that δ(4c1 )−1 ≤ λ. Letting λ → 0+ we obtain M N /c2 ≥ |f (t0 )|N for all t0 ∈ R which are Lebesgue points of f . It follows that f is Lipschitz continuous. Next we consider the case m ≥ 2. We begin with the subcritical case, mp < N . The following theorem is due to Dahlberg [55]. The proof is taken from a paper of Bourdaud [31]. Theorem 12.70. Let m ∈ N and 1 ≤ p < ∞ be such that 1 + 1/p < m < N/p and let f : R → R. Then f ◦ u ∈ W m,p (RN ) for all functions u ∈ W m,p (RN ) if and only if f is a linear function. Proof. If f (t) = ct for some c ∈ R and t ∈ R, then f ◦ u ∈ W m,p (RN ) for all functions u ∈ W m,p (RN ). Conversely, assume that f ◦u ∈ W m,p (RN ) for all functions u ∈ W m,p (RN ). We proceed as in Step 4 of the proof of Theorem 12.69. Let ψ ∈ Cc∞ (RN ) be such that ψ(x) = x1 for all x ∈ Q(0, 1/2) and supp ψ ⊂ Q(0, 1). Given 0 < r < 1, λ > 0, and ∈ N, define u(x) := λ ψ(x/r − k), x ∈ RN , k
where the sum is done over all k ∈ ZN such that |ki | ≤ for all i = 1, . . . , N . By the change of variables y = x/r − k and the facts that r < 1 and m < N/p, ψ(·/r − k) m,p ≤ c1 N/p r(N −mp)/p , W
k
where c1 depends on ψW m,p . Let M , δ > 0 be the numbers given in Proposition 12.67 corresponding to Q(0, 1). Take r = r() > 0 such that (12.42)
λc1 N/p r(N −mp)/p = δ.
Then uW m,p ≤ δ. Writing r = (δ/(c1 λ))p/(N −mp) −mp/(N −mp)
392
12. Sobolev Spaces: Embeddings
and using the fact that pm < N , we can take so large that 8r ≤ 1. In turn, all the cubes Q(rk, r) are contained in Q(0, 1), and thus supp u ⊂ Q(0, 1). It follows from Proposition 12.67 that f ◦ uW 1,p ≤ M . As in Step 4 of the proof of Theorem 12.69, we have that the cubes Q(rk, r) are pairwise disjoint. Since ψ(x) = x1 for all x ∈ Q(0, 1/2), it follows that on the cube Q(rk, r/2) we have that (f ◦u)(x) = f (λx1 /r−λk1 ). Since f ◦u ∈ W m,p (RN ), for LN -a.e. x ∈ Q(rk, r/2) there exists (λ/r)m f (m) (λx1 /r − λk1 ) = ∂1m (f ◦ u)(x). In turn, by Fubini’s theorem, the change of variables t = λx1 /r − λk1 , and (12.42), p |∂1m (f ◦ u)(x)|p dx M ≥ RN ≥ (λ/r)mp |f (m) (λx1 /r − λk1 )|p dx Q(rk,r/2)
k
≥ cN λmp−1 rN −mp mp−p−1 = cδ p c−p 1 λ
λ/4 −λ/4
λ/4 −λ/4
|f (m) (t)|p dt
|f (m) (t)|p dt.
Hence, (12.43)
c2 λmp−p−1
≥
λ/4 −λ/4
|f (m) (t)|p dt,
where c2 > 0 is independent ofλ. Letting λ → ∞ and using the fact that 1 + 1/p < m we obtain that R |f (m) (t)|p dt = 0. It follows that f is a polynomial of degree less than m, say f (t) = c1 t + · · · + ck tk , where ck = 0. Assume by contradiction that k > 1. Since m < N/p we can find θ > 0 such that θ + m < N/p < 2θ + m. Let ψ ∈ Cc∞ (RN ) be such that ψ(x) = 1 for all x ∈ B(0, 1) and define v(x) := x−θ ψ(x). Since θ + m < N/p, it follows that v ∈ W m,p (RN ). On the other hand, for x ∈ B(0, 1), f (v(x)) = c1 x−θ + · · · + ck c1 x−kθ , and since N/p < 2θ + m, passing to spherical coordinates, it can be checked that f ◦ v does not belong to W m,p (RN ). The details are left as an exercise. Thus, k = 1, which completes the proof. Exercise 12.71. Let m ∈ N and 1 ≤ p < ∞ be such that 1 + 1/p < m < N/p and let f : R → R be such that f ◦ u ∈ W m,p (RN ) for all functions u ∈ W m,p (RN ).
12.4. Superposition
393
(i) Construct two sequences {an }n and {εn }n of positive numbers such that εn < 1 < an for every n, εn → 0+ and an → ∞ as n → ∞, and ∞ ∞ −mp −mp apn εN < ∞, amp−1 εN = ∞. n n n n=1
n=1
(ii) Let {yn }n be a sequence of points in RN such that yn − y ≥ 10 for n = and let ψ ∈ Cc∞ (RN ) be such that ψ(x) = x1 for all x ∈ B(0, 1) and ψ = 0 outside B(0, 2). Let ∞ an ψ((x − yn )/εn ), x ∈ RN . u(x) := n=1
Prove that u ∈ W m,p (RN ). (iii) Assume by contradiction that ess inf a 0 such that LN (En ) ≥ cεN n /an for all n sufficiently large and deduce that f ◦ u cannot belong to W m,p (RN ). Next we consider the supercritical case mp > N . The remaining theorems of this section are due to Bourdaud [31]. Theorem 12.72. Let m ∈ N, m ≥ 2, and 1 ≤ p < ∞ be such either mp > N or m = N and p = 1, and let f : R → R. Then f ◦ u ∈ W m,p (RN ) m,p (R). for all functions u ∈ W m,p (RN ) if and only if f (0) = 0 and f ∈ Wloc We begin with some auxiliary results. 2,p (R) be such that f (0) = 0. Lemma 12.73. Let 1 ≤ p < ∞, let f ∈ Wloc Then p−1 p |f ◦ u|p |u |2p dx ≤ cf pLp ([−u ,u ]) uL ∞ (R) u Lp (R) R
for every u ∈ Cc∞ (R), where u := uL∞ (R) . Proof. Let
Mu :=
u −u
|f (s)| ds, p
u
g(t) :=
|f (s)|p ds.
t
Integrating by parts and using H¨ older’s inequality if p > 1, we have p 2p |f ◦ u| |u | dx = (2p − 1) (g ◦ u)|u |2p−2 u dx R
R
≤ (2p − 1)Mu u L2p (R) u Lp (R) . 2(p−1)
394
12. Sobolev Spaces: Embeddings
By Theorem 7.41 with k = 1, m = 2, r = 2p, q = ∞, we have u L2p (R) ≤ cuL∞ (R) u Lp (R) . 1/2
1/2
The result follows by combining these inequalities. We now turn to the proof of Theorem 12.72.
Proof of Theorem 12.72. Step 1: We only prove the “only if” part of the statement for m = 2. Assume that m = 2 and that either 2p > N 2,p (R) be such that f (0) = 0. Given or N = 2 and p = 1. Let f ∈ Wloc ∞ N u ∈ Cc (R ), by Exercise 12.43 or Theorem 12.46, u := uL∞ ≤ cuW 2,p . Define Lu := f L∞ ([−u ,u ]) and Mu := f pLp ([−u ,u ]) . By Theorem 3.59, ∂i (f ◦ u) = (f ◦ u)∂i u, ∂j ∂i (f ◦ u) = (f ◦ u)∂i u∂j u + (f ◦ u)∂j ∂i u, LN -a.e. in RN , where f (u(x))∂j u(x) is interpreted to be zero whenever ∂j u(x) = 0 (even if f is not differentiable at u(x). Hence, since f (0) = 0, |f ◦ u|p = |f ◦ u − f (0)|p ≤ Lpu |u|p ,
|∂i (f ◦ u)|p ≤ Lpu |∂i u|p ,
|∂j ∂i (f ◦ u)|p ≤ c|(f ◦ u)|p |∂i u|2p + c|(f ◦ u)|p |∂j u|2p + cLpu |∂j ∂i u|p , where c = c(p) > 0. Using the notation (E.2) in Appendix E and the previous lemma, p−1 p 2p p |f (u(xi , xi ))| |∂i u(xi , xi )| dxi ≤ cMu u(xi , ·)L∞ (R) |∂i2 u(xi , xi )|p dxi . R
Integrating in
R
xi RN
over
RN −1
and using Tonelli’s theorem gives p−1 p 2p p |f ◦ u| |∂i u| dx ≤ cMu uL∞ |∂i2 u|p dx. RN
Note that when p = 1 the previous inequality reduces to 2 |f ◦ u||∂i u| dx ≤ cMu |∂i2 u| dx. (12.44) RN
RN
This implies that f ◦ u belongs to
W 2,p (RN ).
Step 2: Assume that f ◦ u ∈ W m,p (RN ) for all functions u ∈ W m,p (RN ). Since 0 ∈ W m,p (RN ), we have that f (0) ∈ W m,p (RN ), which implies that f (0) = 0. Next fix a < b and consider a function u ∈ Cc∞ (RN ) such that u(x) = x1 for all x ∈ RN with x1 ∈ [a, b] and |xi | ≤ 1 for i = 2, . . . , N . Then (f ◦ u)(t, x2 , . . . , xN ) = f (t) for all t ∈ [a, b] and |xi | ≤ 1 for i = 2, . . . , N . Since f ◦ u ∈ W m,p (RN ), by Theorem 11.45 and an induction argument, it follows that f ∈ W m ((a, b)) with f (n) (t) = ∂1n (f ◦ u)(t, x2 , . . . , xN )
12.4. Superposition
395
for L1 -a.e. t ∈ [a, b] and for LN −1 -a.e. (x2 , . . . , xN ) ∈ RN −1 with |xi | ≤ 1 for i = 2, . . . , N . Exercise 12.74. Prove that in Step 1 of the previous proof the additional hypothesis that u ∈ Cc∞ (RN ) can be removed. , am and m Example 12.75. Prove that m positive numbers given 0n a1 , . . . pi m exponents p1 , . . . , pm , with i=1 1/pi = 1, we have i=1 ai ≤ m i=1 ai /pi . Exercise 12.76. Use Faa di Bruno’s theorem (see Theorem 11.54) and Exercise 12.75 to prove Step 1 of the previous proof for m ≥ 2. Theorem 12.77. Let N > 2 and let f : R → R. Then f ◦ u ∈ W 2,1 (RN ) ˙ 2,1 (R). for all functions u ∈ W 2,1 (RN ) if and only if f (0) = 0 and f ∈ W ˙ 2,1 (R). Since f ∈ L1 (R), Proof. Step 1: Assume that f (0) = 0 and f ∈ W ∞ by the fundamental theorem of calculus, f ∈ L (R). Given u ∈ Cc∞ (RN ), by (12.44), 2 |f ◦ u||∂i u| dx ≤ c |f (s)| ds |∂i2 u|2 dx. RN
R
RN
Thus we may continue as in Step 1 of Theorem 12.72 to conclude that f ◦ u belongs to W 2,1 (RN ). Step 2: Assume that f ◦ u ∈ W 2,1 (RN ) for all functions u ∈ W 2,1 (RN ). Since 0 ∈ W 2,1 (RN ), as usual we deduce that f (0) = 0. We proceed as in the proof of Theorem 12.70 until (12.43), which now becomes c2 ≥ λ/4 1 −λ/4 |f (t)|dt. Letting λ → ∞ shows that f ∈ L (R). Finally, we study the critical case mp = N . Theorem 12.78. Let m ∈ N be such that N > m ≥ 2, and let f : R → R. Then f ◦ u ∈ W m,N/m (RN ) for all functions u ∈ W m,m/N (RN ) if and only m,N/m (R) with f ∈ L∞ (R) and if f (0) = 0, f ∈ Wloc t+1 |f (m) (s)|N/m ds < ∞. (12.45) sup t∈R
t−1
We begin with an auxiliary result. m,p (R) be Lemma 12.79. Let m ∈ N, m ≥ 2, and 1 ≤ p < ∞, let g ∈ Wloc ∞ such that g ∈ L (R) and t+1 |g (m) (s)| ds < ∞. (12.46) M := sup t∈R
g (n)
t−1
Then L∞ (R) ≤ c(2gL∞ (R) + M ) for all 1 ≤ n ≤ m − 1 and for some constant c > 0 depending only on m.
396
12. Sobolev Spaces: Embeddings
Proof. Fix t0 ∈ R. By Taylor’s formula with integral remainder for t ∈ [0, 1], 1 m−1 1 tm (n) n g (t0 )t + (1 − s)m−1 g (m) (t0 + st) ds. g(t + t0 ) = g(t0 ) + n! (m − 1)! 0 n=1
Hence, by the change of variables y = t0 + st and (12.46), m−1 1 1 (n) tm n g (t0 )t ≤ |g(t + t0 )| + |g(t0 )| + |g (m) (t0 + st)| ds n! (m − 1)! 0
n=1
≤ 2gL∞ (R) +
tm−1 M (m − 1)!
.
It follows from Exercise 11.21 that there exists a constant cm > 0 depending only on m such that |g (n) (t0 )| ≤ cm n!(2gL∞ (R) + M ) for all n = 1, . . . , m − 1 and for all t0 ∈ R.
We are now ready to prove Theorem 12.78. Proof of Theorem 12.78. Step 1: We only prove the “only if” part of 2,N/2 the statement for m = 2. Let m = 2, let f ∈ Wloc (R) be such that f (0) = 0, f ∈ L∞ (R), and (12.45) holds, and let u ∈ Cc∞ (RN ). We follow the proof of Step 1 of Theorem 12.72. The only change is in the estimate for the term |f ◦ u|N/2 |∂i u|N . Let ψ ∈ C ∞ ([−1, 0]) be a nonnegative function such that ψ = 0 in [−1, −3/4], ψ = 1 in [−1/4, 0], and let ψ(t) := 1−ψ(t−1) √ for t ∈ [0, 1]. Extend ψ to be 0 outside [−1, 1]. Then the function ϕ := ψ belongs to Cc∞ (R), supp ϕ ⊆ [−1, 1], and ϕ2 (t)+ϕ2 (t−1) = 1 for all t ∈ [0, 1]. In turn, ϕ2 (t − k) = 1 (12.47) k∈Z
for every t ∈ R. Indeed, given t ∈ R, ϕ2 (t − k) = 0 for all k = t − 1, t , and so the series reduces to ϕ2 (t − 1 − t ) + ϕ2 (t − t ) = 1. Similarly, we have that ϕ(t − k) ≤ c, |ϕ (t − k)| ≤ c (12.48) k∈Z
k∈Z
for every t ∈ R and for some constant c > 0.
Using (12.47) by Tonelli’s theorem we can write N/2 N |f (u(x))| |∂i u(x)| dx = hk (u(x))ϕ(u(x) − k)|∂i u(x)|N dx,
RN
k∈Z
RN
12.4. Superposition
397
where hk (t) := |f (t)|N/2 ϕ(t − k), t ∈ R. Define gk (t) := Integrating by parts we have (hk ◦ u)ϕ ◦ (u − k)|∂i u|N dx
∞ t
hk (s) ds, t ∈ R.
N k∈Z R
=
N k∈Z R
=
k∈Z
RN
(gk ◦ u)∂i (ϕ ◦ (u − k)|∂i u|N −2 ∂i u) dx (gk ◦ u)ϕ ◦ (u − k)|∂i u|N dx
+ (N − 1)
k∈Z
RN
(gk ◦ u)ϕ ◦ (u − k)|∂i u|N −2 ∂i2 u dx =: I + II.
By (12.45) and the fact that supp ϕ ⊆ [−1, 1], we have 0 ≤ gk (t) ≤ hk L1 (R) ≤ c for all t ∈ R and k ∈ Z and for some constant c > 0. In turn, using (12.48) we obtain (12.49) |∂i u|N dx, |I| ≤ c N R 2(N/2−1) |∂i u|N −2 |∂i2 u| dx ≤ c∂i uLN ∂i2 uLN/2 , |II| ≤ c RN
where we used H¨older’s inequality if N > 2. Since (N/2)∗ = N , it follows by the Sobolev–Gagliardo–Nirenberg embedding theorem (Theorem 12.4) applied to ∂i u that ∂i uLN < ∞. (R) be such that f (0) = 0, f ∈ Step 2: Let m ≥ 2 and let f ∈ Wloc L∞ (R), and (12.45) holds and let u ∈ Cc∞ (RN ). By Faa di Bruno’s theorem (see Theorem 11.54) for every multi-index α ∈ NN 0 , with 1 ≤ |α| ≤ m, m,N/m
/ ∂ |γi | u ∂ |α| (n) (f ◦ u) = c (f ◦ u) , α,n,γ ∂xα ∂xγi n
i=1
where cα,n,γ,l ∈ R, the sum is done over all n ∈ N with 1 ≤ n ≤ |α|, |β| γ = (γ1 , . . . , γn ), γi ∈ N, with i=1 γi = α, i = 1, . . . , n. Using exercise 12.75 with pi = |α|/|γi | we get |∂ α (f ◦ u)(x)|N/m ≤
dα,n,γ,m |f (n) ◦ u|N/m f
n
|∂ γi u|N |α|/(|γi |m) .
i=1 ∞ L (R)
and with m replaced It follows from Lemma 12.79 applied to g = ∈ by m − 1, that f (n) ∈ L∞ (R) for every n = 2, . . . , m − 1, and so if 1 ≤ n ≤ |α| ≤ m − 1, we need to estimate terms of the form (12.50)
f (n) L∞ |∂ γi u|N |α|/(|γi |m) , N/m
398
12. Sobolev Spaces: Embeddings
while if |α| = m, then |γi | = 1 for all i = 1, . . . , m, and so we have terms of the form |f (m) ◦ u|N/m |∂i u|N . For the latter, we proceed as in Step 1 with the only change that (12.49) now becomes N |∂i u| dx, |II| ≤ c |∂i u|N −2 |∂i2 u| dx. (12.51) |I| ≤ c RN
RN
By the Sobolev–Gagliardo–Nirenberg embedding theorem (Theorem 12.8), for every multi-index β with |β| = 2, ∂ β u ∈ LN/2 (RN ) and ∂i u ∈ LN (RN ). Since (N − 2)(N/2) = N , it follows that the right-hand sides of both inequalities in (12.51) are finite. To estimate (12.50), again by Sobolev–Gagliardo–Nirenberg embedding theorem (Theorem 12.8), ∂ γi u ∈ LN/|γi | (RN ), and since N/m ≤ N |α|/(|γi |m) < N/|γi |, by Exercise B.79(ii) we have that ∂ γi u ∈ LN |α|/(|γi |m) (RN ). Step 3: The proof of the necessity part of the theorem is similar to the previous ones, using functions of the type (12.37) and is given in the following exercises. Exercise 12.80. Let m ∈ N be such that N > m ≥ 2 and let f : R → R be such that f ◦ u ∈ W m,N/m (RN ) for all functions u ∈ W m,m/N (RN ). Let ψ ∈ Cc∞ (RN ) be such that ψ(x) = x1 for all x ∈ Q(0, 1/2) and supp ψ ⊂ Q(0, 1). Given 0 < r < 1, t0 ∈ R, λ > 0, and 0 < ε < 1/2, define (12.52)
u(x) := λψ(x/r) + t0 uε (x),
x ∈ RN ,
where uε is the function given in Exercise 12.68. Reasoning as in the proofs of Theorems 12.69 and 12.70 prove that λ/4 |f (m) (t + t0 )|N/m dt M N/m ≥ cλN −1 −λ/4
and deduce that (12.45) holds. Exercise 12.81. Let m ∈ N be such that N > m ≥ 2 and let f : R → R be such that f ◦ u ∈ W m,N/m (RN ) for all functions u ∈ W m,m/N (RN ). Prove inductively that t+1 |f (n) (s)|N/m ds < ∞ (12.53) sup t∈R
t−1
for all n = 2, . . . , m − 1 as follows: (i) Assume that (12.45) holds for all k = n + 1, . . . , m and take u as in (12.52), where now ψ(x) = x1 x2 for all x ∈ Q(0, 1/2), where ∈ N is to be determined. Let v(x) := f (λx1 x2 + a) and prove that for
12.5. Interpolation Inequalities in RN
399
≥ m/n − 1, ∂ ne1 +(m−n)e2 v(x) = cn λn f (n) (λx1 x2 + a)x2n+n−m +
m
λk f (k) (λx1 x2 + a)Pk (x1 , x2 ),
k=n+1
where cn = 0 and Pk are polynomials in x1 and x2 . K
(ii) Let K = [−1/4, 1/4] ×[1/8, 1/4]. Prove that for k ∈ {n+1, . . . , m}, t+1 |λk f (k) (λx1 x2 + a)Pk (x1 , x2 )|N/m dx1 dx2 ≤ c sup |f (k) (s)|N/m ds. t∈R
t−1
Exercise 12.82. Let m ∈ N be such that N > m ≥ 2 and let f : R → R be such that f ◦ u ∈ W m,N/m (RN ) for all functions u ∈ W m,m/N (RN ). Let u be as in (12.41), where now ψ(x) = xm 1 for all x ∈ Q(0, 1/2) and prove that f is Lipschitz continuous. Hint: Compute ∂ me1 v, where v(x) := f (λxm 1 + a), and use the previous two exercises.
12.5. Interpolation Inequalities in RN This section is devoted to interpolation inequalities. These are particularly useful in partial differential equations or in calculus of variations in unbounded domains (see, e.g., [33], [71], [94], [142]). In those contexts one gains control of the Lp norm of the highest derivatives ∇m u of a function u (e.g., the solution of a PDE or a critical point of an energy functional) and has some additional information on u (e.g., boundary values, constraints) that allows them to conclude that u belongs to Lq . Then interpolation inequalities give bounds on the intermediate derivatives ∇k u, 1 ≤ k ≤ m − 1 and additional information on the function u itself. Thus the problem we consider in this section is the following: Assume that u ∈ Lq (RN ) and all its distributional derivatives of order m belong to Lp (RN ), for some 1 ≤ p, q ≤ ∞. What can we say about the intermediate derivatives ∇k u, 1 ≤ k ≤ m − 1? To give a unified treatment, only in this section, we use a different notation for Lr norms. To be precise, given r ∈ [−∞, ∞], r = 0, and a function u : RN → R, we define ⎧ if r > 0, ⎨ uLr (RN ) n ∇ uL∞ (RN ) if r < 0 and a = 0, (12.54) |u|r := ⎩ |∇n u|C 0,a (RN ) if r < 0 and 0 < a < 1, where if r < 0 we set n := −N/r and a := −n − N/r ∈ [0, 1), provided the right-hand sides are well-defined. We begin with the case m = 1.
400
12. Sobolev Spaces: Embeddings
Theorem 12.83 (Gagliardo–Nirenberg interpolation, m = 1). Let 1 ≤ p, q ≤ ∞, let 0 ≤ θ ≤ 1, and let r be such that 1 1 θ 1 − + = ∈ (−∞, 1]. (12.55) (1 − θ) p N q r Then there exists a constant c = c(N, p, q, θ) > 0 such that (12.56)
1−θ |u|r ≤ cuθLq (RN ) ∇uL p (RN )
˙ 1,p (RN ), with the exceptions that if p < N and for every u ∈ Lq (RN ) ∩ W q = ∞ we assume that u vanishes at infinity, while if p = N > 1 we take 0 < θ ≤ 1. The proof relies on the following interpolation result. older continuous with exponent Proposition 12.84. Let u : RN → R be H¨ 0 < α < 1 and such that u ∈ Lq (RN ) for some 1 ≤ q ≤ ∞. Then u belongs to Lr (RN ) for all q ≤ r ≤ ∞ and to C 0,β (RN ) for all 0 < β ≤ α, with (12.57)
1 , uLr (RN ) ≤ cuθL1q (RN ) |u|1−θ C 0,α (RN ) 2 , |u|C 0,β (RN ) ≤ cuθL2q (RN ) |u|1−θ C 0,α (RN )
for some constant c = c(α, N, q) > 0, where θ1 :=
α+N/r α+N/q
and θ2 :=
α−β α+N/q .
Proof. We begin by showing that u is bounded. If q = ∞, there is nothing to prove, thus assume that q < ∞. If |u|C 0,α = 0, then u is a constant and so it must be zero since uLq < ∞. Thus assume |u|C 0,α > 0. Given R > 0 and x ∈ RN , let y ∈ B(x, R) \ {x}. Then |u(x)| ≤ |u(y)| + |u(x) − u(y)| ≤ |u(y)| + Rα |u|C 0,α . Averaging both sides in y over B(x, R) (and using H¨older’s inequality if q > 1) we get 1 |u(y)| dy + Rα |u|C 0,α |u(x)| ≤ αN RN B(x,R) ≤ (αN RN )−1/q uLq + Rα |u|C 0,α . −1/q −N/q t u
Minimizing the function g(t) := αN t > 0 gives (12.58)
1 , uL∞ ≤ cuθL1q |u|1−θ C 0,α
Lq
+ tα |u|C 0,α over all
θ1 := α/(α + N/q).
In turn, by Exercise B.79(ii), for all q < r < ∞, 3 1 θ3 3 ≤ c(uθL1q |u|1−θ ) u1−θ uLr ≤ cuθL3∞ u1−θ Lq Lq C 0,α
(1−θ3 )+θ1 θ3
= cuLq
(1−θ )θ3
|u|C 0,α 1
where θ3 := 1 − q/r. This proves (12.57)1 .
,
12.5. Interpolation Inequalities in RN
401
To prove (12.57)2 observe that if 0 < β < α, then for x, y ∈ RN , with x = y, 1−β/α
|u(y) − u(x)|β/α
1−β/α
|u|C 0,α x − yβ .
|u(y) − u(x)| ≤ 21−β/α uL∞ ≤ 21−β/α uL∞
β/α
Hence, by (12.58), 1−β/α
|u|C 0,β ≤ cuL∞
β/α
θ (1−β/α)
= cuL1q
β/α
1 1−β/α |u|C 0,α ≤ c(uθL1q |u|1−θ ) |u|C 0,α C 0,α
β/α+(1−θ1 )(1−β/α)
|u|C 0,α
,
which completes the proof. We now turn to the proof of Theorem 12.83.
Proof of Theorem 12.83. If p < N and q < ∞, then since u ∈ Lq (RN ) we have that u vanishes at infinity, while if q = ∞, then u vanishes at infinity by hypothesis. Thus, in both cases we are in a position to apply the Sobolev–Gagliardo–Nirenberg embedding theorem (Theorem 12.4) to obtain (12.59)
uLp∗ ≤ c∇uLp ,
where p∗ = pN/(N − p). Note that when θ = 0 in (12.55) we get that r is exactly p∗ . Since u ∈ Lp∗ (RN )∩Lq (RN ), we can now apply Exercise B.79(ii) to obtain 1−θ 1−θ θ uLr ≤ cuθLq uL p∗ ≤ cuLq ∇uLp , where we used (12.59). In view of (12.54), this proves (12.56) when p < N . If p = N ≥ 2, then by Exercise 12.37, q/r
1−q/r
uLr ≤ cuLq ∇uLN
for all q < r < ∞, which, again by (12.54), gives (12.56) since (12.55) reduces to θ/q = 1/r for all 0 < θ ≤ 1. The case p = N = 1 is left as an exercise. If p > N , then by Remark 12.49, a representative u ¯ of u is H¨older continuous with exponent 1 − N/p and |u|C 0,1−N/p ≤ c∇uLp . We are now in a position to apply Proposition 12.84 to obtain (12.57), which by (12.54) gives 1−θ 1−θ θ |u|r ≤ cuθLq |u|C 0,1−N/p ≤ uLq ∇uLp , and so (12.56) holds even in this case.
Next we prove an interpolation result to estimate the Lr norm of lower order derivatives with the Lq norm of u and the Lp norm of the derivatives of order m of u.
402
12. Sobolev Spaces: Embeddings
Theorem 12.85 (Gagliardo–Nirenberg interpolation, I). Let 1 ≤ p, q ≤ ∞, m ∈ N, k ∈ N, with 1 ≤ k < m, r be such that k 1 1 k 1 + 1− = . (12.60) mp m q r Then there exists a constant c = c(k, m, N, p, q) > 0 such that (12.61)
1−k/m
k/m
∇k uLr (RN ) ≤ cuLq (RN ) ∇m uLp (RN )
˙ m,p (RN ). In particular, if 1 ≤ p, q ≤ ∞, and r is for every u ∈ Lq (RN ) ∩ W given by 1 1 1 + = , 2p 2q r
(12.62)
then there exists a constant c = c(N, p, q) > 0 such that (12.63)
1/2
1/2
∇uLr (RN ) ≤ cuLq (RN ) ∇2 uLp (RN )
˙ 2,p (RN ). for every u ∈ Lq (RN ) ∩ W In the proof we use the notation (E.2) in Appendix E. Proof. Step 1: For simplicity we present the proof first for k = 1 and m = 2. We treat the case p, q, r < ∞ and leave the other cases as an exercise. Assume first that u ∈ C ∞ (RN ). Let i ∈ {1, . . . , N }. For xi ∈ RN −1 , we have that the one-dimensional function u(xi , ·) satisfies the hypotheses of Theorem 7.41 and so r/(2q) r/(2p) r q |∂i u(xi , t)| dt ≤ c |u(xi , t)| dt |∂i2 u(xi , t)|p dt . R
R
R
Integrate the previous inequality in xi over RN −1 and use Tonelli’s theorem to get r/(2q) r/(2p) r q |∂i u| dx ≤ c |u(xi , t)| dt |∂i2 u(xi , t)|p dt dxi RN
≤
RN −1 r/2 r/2 2 cuLq ∂i uLp ,
R
R
where in the last inequality we have used H¨ older’s inequality and (12.62). q N 2,p N ˙ The general case u ∈ L (R )∩W (R ) follows by a mollification argument. We omit the details. Step 2: To prove the general case, fix a direction ν ∈ S N −1 and consider an orthonormal basis {b1 , . . . , bN −1 , ν}. Given u ∈ C ∞ (RN ), define v(y) = u(R(y)), y ∈ RN , where R is an orthogonal transformation that k ∂k v maps {e1 , . . . , eN −1 , eN } into {b1 , . . . , bN −1 , ν}. Then ∂∂νuk (R(y)) = ∂y k (y). N
12.5. Interpolation Inequalities in RN
403
Fix y ∈ RN −1 and apply Theorem 7.41 to the one-dimensional function v(y , ·) to get k p rk/(mp) r(m−k)/(mq) ∂ m v ∂ v r |v(y , t)|q dt . k (y , t) dt ≤ c m (y , t) dt ∂y ∂y R R R N N Integrating the previous inequality in y over RN −1 and reasoning exactly as in Step 1, using H¨ older’s inequality and (12.60), we obtain, after a change of variables, m ∂ku 1−k/m ∂ u k/m m p . k r ≤ cuLq ∂ν L ∂ν L The result now follows from Exercise 11.20. As a corollary of the previous theorem we obtain the following important result. Corollary 12.86. Let m ∈ N and 1 ≤ p ≤ ∞. Then ˙ m,p (RN ) W m,p (RN ) = Lp (RN ) ∩ W with equivalence of the norms uLp (RN ) +
m
∇k uLp (RN )
and
uLp (RN ) + ∇m uLp (RN ) .
k=1
Proof. It is enough to take p = q in (12.60).
Next we consider the general case. The following theorem is due to Nirenberg [184] in this generality and to Gagliardo [91] in the case r ≥ 1. Theorem 12.87 (Gagliardo–Nirenberg interpolation, general case). Let 1 ≤ p, q ≤ ∞, m ∈ N, k ∈ N0 , with 0 ≤ k < m, and let θ, r be such that 0 ≤ θ ≤ 1 − k/m
(12.64) and (12.65)
(1 − θ)
1 p
−
1 m − k k 1 +θ + = ∈ (−∞, 1]. N q N r
Then there exists a constant c = c(m, N, p, q, θ, k) > 0 such that (12.66)
1−θ |∇k u|r ≤ cuθLq (RN ) ∇m uL p (RN )
˙ m,p (RN ), with the following exceptional cases: for every u ∈ Lq (RN ) ∩ W (i) If k = 0, mp < N , and q = ∞, we assume that u vanishes at infinity. (ii) If 1 < p < ∞ and m − k − N/p is a nonnegative integer, then (12.66) only holds for 0 < θ ≤ 1 − k/m.
404
12. Sobolev Spaces: Embeddings
The strategy of the proof is to treat first the cases θ = 0 and θ = 1−k/m and then to use the following interpolation proposition, whose proof is due to Hormander [121], to treat the values 0 < θ < 1 − k/m. Proposition 12.88. Let −∞ < λ < μ < ν < ∞. Then there exists a constant c = c(λ, μ, ν, N ) > 0 such that 1−θ |u|θ1/λ |u|1/μ ≤ c|u|1/ν
(12.67)
for all u ∈ C ∞ (RN ), with |u|1/λ < ∞ and |u|1/ν < ∞, where θ := (ν − μ)/(ν − λ). Proof. We only give a sketch of the proof. Step 1: We begin with the case λ < μ < ν < 0 and −N μ ∈ N. We claim that if |u|1/λ ≤ 1 and |u|1/ν ≤ 1, then there exists a constant c = c(λ, μ, ν, N ) > 0 such that |u|1/μ ≤ c. Assume that 0 ≤ −N ν < 1. Let a := −N ν, let m := −N λ , and let b := −N λ − m ∈ [0, 1). Assume first that 0 < a, b < 1. Then by (12.54), |u|1/ν = |u|C 0,a and |u|1/λ = |∇m u|C 0,b . Thus, we need to show that if |u|C 0,a ≤ 1 and |∇m u|C 0,b ≤ 1, then there exists a constant c = c(a, b, m, N ) > 0 such that ∇n uL∞ ≤ c for every n = 1, . . . , m. Fix x0 ∈ RN and let B := B(x0 , 1). Then |u(x) − u(y)| ≤ |u|C 0,a x − ya ≤ 1
(12.68)
for every x, y ∈ B, while by Taylor’s formula (see Theorem 9.9), u(x) = u(y) + Py (x) + Ry (x), where (12.69)
Py (x) :=
1≤|α|≤m
1 α ∂ u(y)(x − y)α , α!
(12.70)
1 m α (x − y) (1 − t)m−1 (∂ α u(tx + (1 − t)y) − ∂ α u(y)) dt. Ry (x) := α! 0 |α|=m
Since
|∇m u|
(12.71)
C 0,b α
≤ 1,
|∂ u(tx + (1 − t)y) − ∂ α u(y)| ≤ |∇m u|C 0,b tx + (1 − t)y − yb ≤ tb x − yb ≤ tb (x − x0 + y − x0 )b ≤ 2b ,
and so, also by (12.68), |Py (x)| ≤ |u(x) − u(y)| 1 m 1 x − ym (1 − t)m−1 dt ≤ 1 + 2b + 2b α! α! 0 |α|=m
|α|=m
12.5. Interpolation Inequalities in RN
405
for all x, y ∈ B. Fix y ∈ B. Since the previous inequality holds for all x ∈ B, we leave as an exercise to show that this implies that there exists a constant c = c(b, m, N ) > 0 (independent of y) such that |∂ α u(y)| ≤ c for every 1 ≤ |α| ≤ m (see Exercise 11.21). Taking y = x0 , this concludes the proof of the claim in the case 0 ≤ −N ν < 1 and 0 < a, b < 1, since x0 is arbitrary. If a = 0, then |u|1/ν = uL∞ by (12.54), and so in place of (12.68) we use the estimate |u(x) − u(y)| ≤ 2uL∞ ≤ 2. Similarly, if b = 0, then the Taylor polynomial Py (see (12.69)) |u|1/λ = ∇m uL∞ .In this case in one should replace 1≤|α|≤m with 1≤|α|≤m−1 and replace the remainder term Ry with 1 m α ¯ Ry (x) = (x − y) (1 − t)m−1 (∂ α u(tx + (1 − t)y) dt. α! 0 |α|=m
1 1 m m ¯ y (x)| ≤ Then |R |α|=m α! x − y ∇ uL∞ ≤ |α|=m α! and we obtain the same bound on |Py (x)|. We can then continue as before to prove that ∇n uL∞ ≤ c for every n = 1, . . . , m − 1. Finally, if −N ν ≥ 1, it is enough to apply what we just proved to the function v = ∂ α u, where |α| = −N ν . Step 2: We will prove (12.67) in the case λ < μ < ν < 0. We will give a full proof in the case −N μ ∈ N and leave the other cases as an exercise. As in Step 1 assume first that 0 ≤ −N ν < 1, and that 0 < a, b < 1. If |u|C 0,a = 0, then u is a constant, and so ∇n u = 0 for all n = 1, . . . , m and there is nothing to prove. Thus, assume |u|C 0,a > 0. If |∇m u|C 0,b = 0, then u is a polynomial of degree less than or equal to m, but since |u|C 0,a < ∞, necessarily u must be a constant. Once more we have that ∇n u = 0 for all n = 1, . . . , m and there is nothing to prove. Thus, in what follows, we assume that |u|C 0,a > 0 and |∇m u|C 0,b > 0. Fix x0 ∈ RN and R > 0 and consider the function w(x) := u(x0 + R(x − x0 )), x ∈ RN . By the chain rule we have that |w|C 0,a ≤ Ra |u|C 0,a ,
|∇m w|C 0,b ≤ Rm+b |∇m u|C 0,b .
Since |u|C 0,a > 0 and |∇m u|C 0,b > 0, we can choose R > 0 in such a way that Rm+b |∇m u|C 0,b = Ra |u|C 0,a =: κ. Then |w|C 0,a ≤ κ and |∇m w|C 0,b ≤ κ and so by Step 1 applied to the function w/κ, for every multi-index α ∈ NN 0 with |α| = −N μ, R−N μ |∂ α u(x0 )| = |∂ α w(x0 )| ≤ cκ = c(Ra |u|C 0,a )1−θ (Rm+b |∇m u|C 0,b )θ . Since −N μ = −(1 − θ)N ν − θN λ = −(1 − θ)a − θ(m + b) the powers of R cancel to give 1−θ m θ |∂ α u(x0 )| ≤ c|u|C 0,a |∇ u|C 0,b ,
406
12. Sobolev Spaces: Embeddings
which completes the proof of the claim in the case 0 < a, b < 1. The cases a = 0 and b = 0 are similar and we omit them. Step 3: Next we study the case λ < μ = 0 < ν ≤ 1. Let m := −N λ , and let b := −N λ − m ∈ [0, 1). Assume first that m ≥ 1 and 0 < b < 1. Thus, setting q := 1/ν, by (12.54), we need to show that there exists a constant c = c(b, N, q) > 0 such that 1−θ m θ uL∞ ≤ cuL q |∇ u|C 0,b .
If uLq = 0, then u = 0. Thus, assume uLq > 0. If |∇m u|C 0,b = 0, then u is a polynomial of degree less than or equal to m, but since uLq < ∞, necessarily u must be a zero. Thus, in what follows, we assume that uLq > 0 and |∇m u|C 0,b > 0. Fix x0 ∈ RN and let B := B(x0 , 1) as before. Consider the vector space V of all polynomials P : RN → R of degree less than or equal to m. Then V is a finite-dimensional space. Hence, all norms in V are equivalent (exercise). It follows that there exist two constants c1 > 0 and c2 > 0 such that P L∞ (B) ≤ c1 P L1 (B) ≤ c2 P Lq (B)
(12.72) for all P ∈ V .
By rescaling, we can assume that |∇m u|C 0,b = 1. Fix R > 0 and let w be the function defined in Step 2. Reasoning somewhat as in Step 1 with w in place of u we have that for all x, y ∈ B, |w(x) − Qy (x)| ≤ c|∇m w|C 0,b ≤ cRm+b |∇m u|C 0,b = cRm+b , 1 α α where Qy (x) := 0≤|α|≤m α! ∂ w(y)(x − y) . Hence, by Minkowski’s inequality and a change of variables
(12.73)
Qy Lq (B) ≤ wLq (B) + cRm+b (LN (B))1/q ≤ R−N/q uLq + cRm+b αN . 1/q
Combining (12.72) applied to Qy , (12.73), and the previous inequality we obtain |u(x0 )| = |w(x0 )| ≤ |Qy (x0 )| + |w(x0 ) − Qy (x0 )| ≤ c2 Qy Lq (B) + cRm+b ≤ c2 R−N/q uLq + cRm+b . It suffices to choose R so that uLq = Rm+b+N/q . The cases m = 0 and b = 0 are similar and are left as an exercise (see Proposition 12.84). Step 4: In Step 3 we proved (12.67) for μ = 0. To prove (12.67) for 0 < −N μ < 1 it suffices to use Step 2 to estimate |u|C 0,−N μ in terms of uL∞ and |∇m u|C 0,b and Step 3 to estimate uL∞ in terms of uLq and |∇m u|C 0,b . We leave the remaining cases as an exercise. We are now ready to prove Theorem 12.87.
12.5. Interpolation Inequalities in RN
407
Proof of Theorem 12.87. Step 1: We begin by considering the case θ = 1 0. Then (12.65) reduces to 1p − m−k N = r . We claim that |∇k u|r ≤ c∇m uLp
(12.74)
˙ m,p (RN ) with the exception of the case 1 < p < ∞ for every u ∈ Lq (RN ) ∩ W and m − k − N/p is a nonnegative integer. If (m − k)p < N , then r = p∗m,k (see (12.7)) and so (12.74) follows from the Sobolev–Gagliardo–Nirenberg embedding theorem (see Corollary 12.11) (using hypothesis (i)). If (m − k)p = N and p = 1, then r = ∞ and (12.74) is a consequence of Exercise 12.43. The case (m − k)p = N and p > 1 is excluded by (ii). If (m − k)p > N and m − k − N/p is not a nonnegative integer, then 1/r < 0 and so by (12.54), |∇k u|r = |∇k+n u|C 0,a = |∇m−N/p u|C 0,a , where n := −N/r = m − k − N/p and a := −n − N/r = m − N/p −
m − N/p ∈ (0, 1). We can now use Remark 12.56 to obtain (12.74). Step 2: Next we prove (12.66) in the case in which 0 < θ < 1 − k/m but we are not in case (ii), that is, either m − k − N/p is not a nonnegative integer or if it is, then p = 1. By Step 1 and Theorem 12.85 for θ = 0 and θ = 1 − k/m, respectively, we have |∇k u|r0 ≤ c∇m uLp ,
1−k/m
|∇k u|r1 ≤ cuLq
k/m
∇m uLp ,
1 m−k 1 k 1 1 k 1 − = , + 1− = . p N r0 mp m q r1 Given 0 < θ < 1 − k/m we have that 1/r is in between 1/r0 and 1/r1 and thus by Proposition 12.84, where
1 |∇k u|r ≤ c|∇k u|θr01 |∇k u|1−θ r1
(1−k/m)θ1
≤ cuLq
θ +(k/m)(1−θ1 )
∇m uL1p
1−θ = cuθLq ∇m uL p ,
where θ1 := (1/r1 − 1/r)/(1/r1 − 1/r0 ) and θ := (1 − k/m)θ1 . Step 3: Finally, we prove (12.66) in the case in which 0 < θ < 1 − k/m, 1 < p < ∞, and m − k − N/p is a nonnegative integer. In view of Theorem 12.83 we can assume that m ≥ 2. If p = N , then by Theorem 12.85, 1−(m−1)/m
∇m−1 uLrm ≤ cuLq where
m−1 1 m N
+
1 1 mq
=
1 rm .
(m−1)/m
∇m uLN
In turn, by Exercise 12.37, r /s
1−r /s
∇m−1 uLs ≤ c∇m−1 uLmrm ∇m uLN m for all rm < s < ∞, which implies (12.66) for k = m − 1.
,
408
12. Sobolev Spaces: Embeddings
If p < N , then necessarily k < m − 1, since m − 1 is not an exceptional case. By applying Step 2 to m − 1 we obtain 1 ∇m−1 uLs ≤ cuθL1q ∇m u1−θ Lp
(12.75) where (1 − θ1 )
1 1 1 m−1 = 1s for all 0 p − N + θ1 q + N Ls (RN ; RMm−1 ) for all s between p∗ and (N
≤ θ1 ≤
1 m.
Hence,
+ (m − 1)q)/(N q). ∇m−1 u ∈ To conclude the proof it suffices to observe that for k < m − 1 we can write (12.65) as 1 1 k m − 1 − k 1 = (1 − θ) ∗ − +θ + . r p N q N Since θ = 0 is not allowed, we can write 1r as a convex combination of 1/s − (m−1−k)/N and 1/q +k/N for some s between p∗ and (N +(m−1)q)/(N q) with the property that m − 1 − k − N/s is not a nonnegative integer. We can now apply Step 2 to estimate ∇k uLr in terms of uLq and ∇m−1 uLs , which, together with (12.75), completes the proof. Exercise 12.89. Let u(x) = sin(λx1 )ϕ(x), x ∈ RN , where ϕ ∈ Cc∞ (RN ) and λ > 0 is large. Prove that (12.64) is a necessary condition for (12.66) k < θ ≤ 1. to hold, namely, that (12.66) fails if 1 − m Exercise 12.90. Let u ∈ C ∞ (RN ) and let ϕR is a standard mollifier, R > 0. (i) Prove that for every x, y ∈ RN and m ∈ N, 1 ∂α |α| (−1) u(y) α (ϕR (y − x)(x − y)α ) dy u(x) = α! ∂y RN 0≤|α|≤m 1 m ϕR (y − x) (x − y)α (1 − t)m−1 (∂ α u(tx + (1 − t)y) + α! RN 0
|α|=m α
− ∂ u(y)) dtdy. (ii) Using part (i) prove that for every 0 < b < 1 there exists a constant c = c(b, m, N ) > 0 such that for every x ∈ RN , |u(x)| ≤ cR
−N
|u(y)| dy + cRm+b |∇m u|C 0,b (B(x,R)) . B(x,R)
Use this inequality to give an alternative proof to Step 3 of Proposition 12.88.
12.5. Interpolation Inequalities in RN
409
(iii) Prove that for every x, y ∈ RN , m ∈ N, and every multi-index β ∈ NN 0 with |β| ≥ 1, 1 ∂ α+β β |α|+|β| (−1) u(y) α+β (ϕR (y − x)(x − y)α ) dy ∂ u(x) = α! ∂y RN 0≤|α|≤m 1 m + ϕR (y − x) (x − y)α (1 − t)m−1 (∂ α+β u(tx + (1 − t)y) α! RN 0 |α|=m
− ∂ α+β u(y)) dtdy. (iv) Using part (iii) to give an alternative proof to Step 1 of Proposition ∂ α+β α 12.88. Hint: Note that B(x,R) ∂y α+β (ϕR (y − x)(x − y) ) dy = 0. ˙ 1,2 (RN ). Exercise 12.91. Let u ∈ L1 (RN ) ∩ W (i) Prove that for every R > 0, |ˆ u|2 dx ≤ R−2 ∇u2L2 (RN ) , RN \B(0,R)
where u ˆ is the Fourier transform of u. (ii) Prove that for every R > 0, |ˆ u|2 dx ≤ αN RN u2L1 (RN ) . B(0,R)
(iii) Prove that 2/(N +2)
N/(N +2)
uL2 (RN ) ≤ cuL1 (RN ) ∇uL2 (RN ) .
Chapter 13
Sobolev Spaces: Further Properties Newton’s third law of graduation: For every action towards graduation there is an equal and opposite distraction. — Jorge Cham, www.phdcomics.com
13.1. Extension Domains As we have seen in the previous chapter, several embeddings that are valid for the entire space RN continue to hold for extension domains. The next exercise shows that in general an arbitrary open set is not an extension domain for W 1,p . Thus, an important problem is to characterize the class of domains that are extension domains for W m,p . To the author’s knowledge this problem is still open, with the exception of the case m = 1, N = 2, and p = 2 (see the paper of Jones [122]). Exercise 13.1. Let 1 ≤ p < ∞, let Ω = {x = (x1 , x2 ) ∈ R2 : 0 < x1 < 1, 0 < x2 < xα1 }, where α > 1, and let u : Ω → R be defined by u(x) := x11−β , x ∈ Ω, where 1 < β < (α + 1)/p. Prove that u ∈ W 1,p (Ω) and that if β is sufficiently close to (α + 1)/p, then u cannot be extended to a function in W 1,p (R2 ). Hint: Consider the three cases 1 ≤ p < 2, p = 2, and p > 2. Next we show that extension domains strongly depend on p. Exercise 13.2. Let Ω be as in the previous exercise, and using polar coordinates (r, θ), let u : R2 \ Ω → R be defined as u(x1 , x2 ) := r1−β ψ(θ)ϕ(r cos θ, r sin θ), where β > 0, ψ ∈ C ∞ ([0, 2π]), with ψ = 1 for θ small and ψ = 0 for θ ∈ [π, 2π], and ϕ ∈ Cc∞ (R2 ) with ϕ = 1 for 0 ≤ r < 1. 411
412
13. Sobolev Spaces: Further Properties
(i) Let 1 < p < ∞ and find for which values of β the function u belongs to W 1,p (R2 \ Ω). (ii) Prove that if 0 < β < 2/p is sufficiently close to 2/p, then u cannot be extended to a function in W 1,p (R2 ). (iii) For every function v ∈ W 1,1 (R2 \ Ω), with v = 0 for x1 > 12 , define v(x) if x ∈ / Ω, x2 + E0 (v)(x) := − − v (x) + α (v (x) − v (x)) if x ∈ Ω, x1 where v − (x) := v(x1 , −x2 ) and v + (x) := v(x1 , 2xα1 − x2 ). Prove that E0 (v) ∈ W 1,1 (R2 ) and that E0 (v)W 1,1 (R2 ) ≤ cvW 1,1 (R2 \Ω) . (iv) Prove R2 \ Ω is an extension domain for W 1,1 . Hint: Use cutoff functions to reduce to part (iii) and away from the origin use Theorem 13.4 below. The previous exercise shows that R2 \ Ω is an extension domain for W 1,1 but not for W 1,p with p > 1. We now prove that if ∂Ω is sufficiently regular, then it is possible to construct an extension operator E that works for all Sobolev spaces W m,p (Ω). We begin with the case in which Ω is the half space RN + (see Exercise 7.24). Exercise 13.3. Given m ∈ N, and 1 ≤ p ≤ ∞, let u ∈ W m,p (RN + ). Prove that there exist c1 , . . . , cm+1 ∈ R such that the function m+1 n=1 cn u(x , −nxN ) if xN < 0, v(x) := u(x) if xN > 0, is well-defined and belongs to W m,p (RN ). Prove also that for every 0 ≤ k ≤ m, ∇k vLp (RN ) ≤ c∇k uLp (RN ) for some constant c = c(m, N, p) > 0. +
Next we consider the important special case in which Ω lies above the graph of a Lipschitz continuous function. Theorem 13.4. Let f : RN −1 → R be a Lipschitz continuous function and let (13.1)
Ω := {(x , xN ) ∈ RN −1 × R : xN > f (x )}.
Then for all 1 ≤ p ≤ ∞ there exists a continuous linear operator E : W 1,p (Ω) → W 1,p (RN ) such that for all u ∈ W 1,p (Ω), E(u)(x) = u(x) for LN -a.e. x ∈ Ω and (13.2) (13.3)
E(u)Lp (RN ) = 2uLp (Ω) ,
∂N E(u)Lp (RN ) = ∂N uLp (Ω) ,
∂i E(u)Lp (RN ) ≤ 2∂i uLp (Ω) + 2 Lip f ∂N uLp (Ω)
for all i = 1, . . . , N .
13.1. Extension Domains
413
Proof. The idea of the proof is to first flatten the boundary to reduce to the case in which Ω = RN + and then use a reflection argument (see Exercise 11.51). We only prove the case 1 ≤ p < ∞ and leave the easier case p = ∞ as an exercise. Consider the transformation Ψ : RN → RN given by Ψ(y) := (y , yN + f (y )). Note that Ψ is invertible, with inverse given by Ψ−1 (x) = (x , xN − f (x )). Moreover, for all y, z ∈ RN , Ψ(y) − Ψ(z) = (y − z , f (y ) − f (z ) + yN − zN ) # ≤ y − z 2 + (Lip f ) y − z + |yN − zN |)2 ≤ 2(1 + Lip f )y − z, which shows that Ψ (and similarly Ψ−1 ) is Lipschitz continuous. Since f is Lipschitz continuous, by Rademacher’s theorem (see Theorem 9.14) it is differentiable for LN −1 -a.e. y ∈ RN −1 , and so for any such y ∈ RN −1 and for all yN ∈ R we have
IN −1 0 JΨ (y) = , ∇y f (y ) 1 which implies that det JΨ (y) = 1. Note that Ψ(RN + ) = Ω. Given a function u ∈ W 1,p (Ω), 1 ≤ p < ∞, define the function w(y) := u(Ψ(y)) = u(y , yN + f (y )),
y ∈ RN +.
By Theorem 11.53 the function w belongs to W 1,p (RN + ) and the usual chain rule formula for the partial derivatives holds. By Exercise 11.51 the function w ˆ : RN → R, defined by w(y) if yN > 0, w(y) ˆ := w(y , −yN ) if yN < 0, belongs to W 1,p (RN ) and the usual chain rule formula for the partial derivatives holds. Define the function v : RN → R by u(x) if xN > f (x ), (13.4) v(x) := (w ˆ ◦ Ψ−1 )(x) = u(x , 2f (x ) − xN ) if xN < f (x ). Again by Theorem 11.53, we have that v ∈ W 1,p (RN ) and the usual chain rule formula for the partial derivatives holds. Using Theorem 9.52 and the fact that det ∇Ψ = det ∇Ψ−1 = 1, we have that p p |v(x)| dx = |u(x , 2f (x ) − xN )| dx = |u(y)|p dy. RN \Ω
RN \Ω
Since for all i = 1, . . . , N − 1 and for (13.5)
Ω
LN -a.e.
x∈
RN
\ Ω,
∂i v(x) = ∂i u(x , 2f (x ) − xN ) + 2∂N u(x , 2f (x ) − xN )∂i f (x ),
414
13. Sobolev Spaces: Further Properties
again by Theorem 9.52 we have that
1/p
1/p p p |∂i v(x)| dx ≤ |∂i u(x , 2f (x ) − xN )| dx RN \Ω
RN \Ω
+ 2 Lip f RN \Ω
|∂N u(x , 2f (x ) − xN )| dx
1/p
≤
1/p
|∂i u(y)| dy p
p
1/p |∂N u(y)| dy p
+2 Lip f
Ω
Ω
Similarly, using the fact that ∂N v(x) = −∂N u(x , 2f (x ) − xN ) for LN -a.e. x ∈ RN \ Ω, , we obtain p |∂N v(x)| dx = |∂N u(x , 2f (x ) − xN )|p dx N N R \Ω R \Ω = |∂N u(y)|p dy. Ω
Hence, the linear extension operator u ∈ W 1,p (Ω) → E(u) := v ∈ W 1,p (RN ) is continuous and satisfies (13.2) and (13.3). Remark 13.5. Note that the operator E defined in the previous theorem does not depend on p. However, it has the disadvantage that it cannot be used for higher-order Sobolev spaces, unless one assumes that f is more regular (see Theorem 11.57). In order to construct an extension operator that can be used for higherorder Sobolev spaces, we need to introduce the notion of regularized distance. The construction given in the following theorem is due to Lieberman [143]. An alternative construction was obtained earlier by Stein [220] and is sketched in Exercise 13.7 below. Theorem 13.6 (Regularized distance). Let V ⊂ RN be an open set with nonempty boundary and let dist(x, ∂V ) if x ∈ V, d(x) := − dist(x, ∂V ) if x ∈ / V. Then there exists a Lipschitz continuous function dreg : RN → R such that dreg ∈ C ∞ (RN \ ∂V ), d(x) 3 1 ≤ ≤ 2 dreg (x) 2
(13.6) for all x ∈ RN \ ∂V , and (13.7)
|∂ α dreg (x)| ≤
cα |dreg (x)||α|−1
.
13.1. Extension Domains
415
for all x ∈ RN \ ∂V , for every multi-index α, |α| ≥ 1, and for some constant cα > 0. Proof. Step 1: Consider a nonnegative function ϕ ∈ Cc∞ (RN ) such that ϕ(x) dx = 1 (13.8) supp ϕ ⊆ B(0, 1), RN
and define (13.9)
G(x, t) := RN
d(x − 2t y)ϕ(y) dy,
x ∈ RN ,
t ∈ R.
Since d is 1-Lipschitz continuous, for all x ∈ RN and for all t1 , t2 ∈ R, by (13.8) we have (13.10) |d(x − t21 y) − d(x − t22 y)|ϕ(y) dy |G(x, t1 ) − G(x, t2 )| ≤ B(0,1) 1 ϕ(y) dy = 12 |t1 − t2 |, ≤ 2 |t1 − t2 | B(0,1)
while for all t ∈ R and x1 , x2 ∈ RN , (13.11) |G(x1 , t) − G(x2 , t)| ≤ |d(x1 − 2t y) − d(x2 − 2t y)|ϕ(y) dy ≤ x1 − x2 . B(0,1)
Step 2: By the change of variables z := x − 2t y, for t = 0 we can rewrite G as (13.12) G(x, t) = 2N t−N ϕ(2t−1 (x − z))d(z) dz = (d ∗ ϕt/2 )(x). RN
Using Theorem C.20 and the Lebesgue dominated convergence theorem we have that G belongs to C ∞ (RN × (R \ {0})) with ∂ |a| ∂ |a| G N (x, t) = 2 (t−N ϕ(2t−1 (x − z)))d(z) dz (13.13) α ∂(x, t)α ∂(x, t) N R +1 for every multi-index α ∈ NN . Moreover, by Exercise 11.37, an induction 0 argument, and the fact that ϕ ∈ Cc∞ (RN ), ∂ |a| aα −N −1 (t ϕ(2t (x − z))) ≤ N +|α| (13.14) ∂(x, t)α |t|
for all x ∈ RN and t = 0 and for some constant aα depending on ϕ. Since 2N RN t−N ϕ(2t−1 (x − z)) dz = 1 by (13.8), by differentiating we obtain ∂ |a| (t−N ϕ(2t−1 (x − z))) dz = 0 α ∂(x, t) N R
416
13. Sobolev Spaces: Further Properties
and so ∂ |a| G ∂ |a| N (x, t) = 2 (t−N ϕ(2t−1 (x − z)))(d(z) − d(x)) dz. α ∂(x, t)α ∂(x, t) B(x,t/2) Together with (13.14) and the fact that d is 1-Lipschitz continuous, this gives ∂ |a| G bα 2N aα |t| (x, t) (13.15) ≤ N +|α| LN (B(x, t/2)) ≤ |α|−1 . α ∂(x, t) 2 |t| |t| Step 3: By (13.10) and Banach’s fixed point theorem (Theorem A.14) for every x ∈ RN there exists a unique tx ∈ R such that tx = G(x, tx ). Defining dreg (x) := tx , from (13.10) and (13.11) we obtain that |dreg (x1 ) − dreg (x2 )| ≤ 2x1 − x2 for all x1 , x2 ∈ RN . We claim that dreg is of class C ∞ (RN \ ∂V ). Consider the function H(x, t) = t − G(x, t) for (x, t) ∈ RN × (R \ {0}). Then ∂t H(x, t) = 1 − ∂t G(x, t) ≥ 1 − 12 , since |∂t G(x, t)| ≤ 12 by (13.10). By the implicit function theorem for every point (x0 , tx0 ) ∈ RN × (R \ {0}) there are a neighborhood of (x0 , tx0 ) and a unique function f of class C ∞ such that H(x, f (x)) = 0. But by the uniqueness of tx , necessarily, f = dreg , which proves the claim. Step 4: We will prove (13.6). By (13.8), G(x, 0) = RN d(x)ϕ(y) dy = d(x). Using the fact that dreg (x) = G(x, dreg (x)), (13.9), and (13.10), we have |dreg (x) − d(x)| = |G(x, dreg (x)) − G(x, 0)| ≤ 12 |dreg (x) − 0|, which proves (13.6). Step 5: In this step we will prove (13.7). By differentiating the equality dreg (x) = G(x, dreg (x)) with respect to xi , we have ∂i dreg (x) = ∂i G(x, dreg (x)) + ∂t G(x, dreg (x))∂i dreg (x). Hence, (13.16)
∂i dreg (x) =
∂i G(x, dreg (x)) . 1 − ∂t G(x, dreg (x))
From the inequalities |∂i G(x, t)| ≤ 1 and |∂t G(x, t)| ≤ 12 (see (13.10) and (13.11)), we derive that |∂i dreg (x)| ≤ 2. By differentiating (13.16) and using (13.15), again the fact that |∂t G| ≤ 12 , and an induction argument on |α| we obtain (13.7). We leave the details as an exercise. Another way to construct a regularized distance is to use Whitney’s decomposition theorem (see Theorem 9.44) and is due to Stein (see [220]). Exercise 13.7 (Regularized distance). Let V ⊆ RN be an open set, let {Q(xn , rn )}n be a Whitney decomposition of V (see Theorem 9.44), and let 0 < ε < 14 . Consider a nonnegative function ϕ ∈ Cc∞ (RN ) such that
13.1. Extension Domains
417
ϕ(x) = 1 for all x ∈ Q(0, 1) and ϕ(x) = 0 for all x ∈ RN \ Q(0, 1 + ε) and for each n set ϕn (x) := ϕ((x − xn )/rn ), x ∈ RN . Define the regularized distance dreg (x) :=
diam Q(xn , rn )ϕn (x),
x ∈ RN .
n
(i) Prove that c1 dist(x, RN \ V ) ≤ dreg (x) ≤ c2 dist(x, RN \ V ) for all x ∈ V and for some constants c1 , c2 > 0 depending only on N . (ii) Prove that dreg ∈ C ∞ (V ). (iii) Prove that for every multi-index α, cα |∂ α dreg (x)| ≤ |dreg (x)||α|−1 for all x ∈ V and for some constant cα > 0 depending only on N and α. Using the regularized distance we can show that special Lipschitz continuous domains of the form (13.1) are W m,p extension domains. The following results are due to Stein (see [220]). Theorem 13.8 (Stein). Let f : RN −1 → R be a Lipschitz continuous function and let Ω := {(x , xN ) ∈ RN −1 × R : xN > f (x )}. Then for all m ∈ N and 1 ≤ p ≤ ∞ there exists a continuous linear operator E : W m,p (Ω) → W m,p (RN ) such that for all u ∈ W m,p (Ω), E(u)(x) = u(x) for LN -a.e. x ∈ Ω, and (13.17) (13.18)
E(u)Lp (RN ) ≤ cuLp (Ω) , ∇k E(u)Lp (RN ) ≤ c(1 + (Lip f )k )∇k uLp (Ω)
for every 1 ≤ k ≤ m and for some constant c = c(m, N, p) > 0. For simplicity we give the proof for m = 2 and then indicate the changes in the general case m ≥ 2. Proof for m = 2. Step 1: Let L := Lip f . We claim that for every x = (x , xN ) ∈ RN \ Ω, 1 √ (f (x ) − xN ) ≤ dist(x, Ω) ≤ f (x ) − xN . (13.19) 1 + L2 Observe that dist(x, Ω) = dist(x, ∂Ω). Indeed, for every y ∈ Ω, the segment of endpoints x and y must intersect ∂Ω at some point z and so x − y ≥ x − z. Next, given y = (y , f (y )) ∈ ∂Ω, we have that 0 ≤ f (x ) − xN = f (x ) − f (y ) + f (y ) − xN ≤ Lx − y N −1 + f (y ) − xN .
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13. Sobolev Spaces: Further Properties
In turn, (f (x ) − xN )2 ≤ L2 x − y 2N −1 + (f (y ) − xN )2 + 2Lx − y N −1 (f (y ) − xN ) ≤ L2 x − y 2N −1 + (f (y ) − xN )2 + x − y 2N −1 + L2 (f (y ) − xN )2 = (1 + L2 )(x − y 2N −1 + (f (y ) − xN )2 ) = (1 + L2 )x − y2 , which shows that 1 (f (x ) − xN ) ≤ x − y 2 1+L for all y ∈ ∂Ω. Hence, the first inequality in (13.19) holds. The second inequality follows by observing that (x , f (x )) ∈ Ω. Thus, we have proved (13.19). √
Using (13.6) and (13.19) for all x ∈ RN \ Ω we obtain 2 (f (x ) − xN ) ≤ dreg (x) ≤ 2(f (x ) − xN ). (13.20) 0< √ 2 3 1+L Define the function # (13.21) d1 (x) := 3 1 + L2 dreg (x). Then by (13.21) and (13.20), (13.22)
# 0 < 2(f (x ) − xN ) ≤ d1 (x) ≤ 6 1 + L2 (f (x ) − xN ).
Moreover, by (13.7), |d1 (x)||α|−1 |∂ α d1 (x)| ≤ cα (1 + L|α| )
(13.23)
for all x ∈ RN \ Ω and for every multi-index α. Step 2: Let φ : [1, ∞) → R be a continuous function such that ∞ ∞ φ(t) dt = 1, tn φ(t) dt = 0, lim tn φ(t) = 0 (13.24) 1
1
t→∞
for all n ∈ N. Given u ∈ Cc∞ (RN ), let v : RN → R be the function defined by ∞ φ(t)u(g t (x)) dt, (13.25) v(x) := 1
where (13.26)
g t (x) := (x , xN + td1 (x)).
Note that in view of (13.22), if x ∈ RN \ Ω and t ≥ 1, then xN + td1 (x) ≥ xN + 2(f (x ) − xN ) > xN + f (x ) − xN = f (x ), and so xt ∈ Ω. Moreover, if x ∈ ∂Ω, then d1 (x) = 0 and so by (13.24)1 , v(x) = u(x).
13.1. Extension Domains
419
By Theorem B.52 and the facts that u ∈ Cc∞ (RN ) and dreg ∈ C ∞ (RN \ ∂Ω) (see Theorem 13.6), we have that v ∈ C ∞ (RN \ ∂Ω) with ∞ α φ(t)∂ α (u ◦ g t )(x) dt (13.27) ∂ v(x) = 1 N for every multi-index α ∈ NN 0 with |α| ≤ 2 and every x ∈ R \ ∂Ω. We claim that
(13.28)
lim ∂ α v(x) = ∂ α u(x0 )
x→x0
for every x0 ∈ ∂Ω.
By (13.26), for all i = 1, . . . , N and x ∈ RN \ ∂Ω, (13.29)
∂i (u ◦ g t )(x) = ∂i u(g t (x)) + ∂N u(g t (x))t∂i d1 (x).
By (13.23), |∂i d1 (x)| ≤ c(1 + L) and so by the Lebesgue dominated convergence theorem, the fact that dreg (x) → 0 as x → x0 ∈ ∂Ω, (13.24)1 and (13.24)2 with n = 1, 2, (13.27), and (13.29), ∂i v(x) → ∂i u(x0 ) as x → x0 . The situation is more complicated for multi-indices of length greater than one. For all i, j = 1, . . . , N and x ∈ RN \ ∂Ω, (13.30)
2 2 2 2 ∂j,i (u ◦ g t ) = ∂j,i u ◦ g t + (∂i,N u ◦ g t )t∂j d1 + (∂j,N u ◦ g t )t∂i d1 2 2 u ◦ g t )t2 ∂i d1 ∂j d1 + (∂N u ◦ g t )t∂j,i d1 . + (∂N,N
The first four terms can be treated as before using (13.24)2 with n = 1, 2 and the fact that d1 is a Lipschitz continuous function. However, the term 2 d is unbounded. By the fundamental theorem of calculus applied to the ∂j,i 1 function h(t) := ∂N u(g t (x)), we have (13.31) t t 2 ∂N,N u(x , xN + sd1 (x)) ds. ∂N u(g (x)) = ∂N u(x , xN + d1 (x)) + d1 (x) 1
Hence, by (13.24)2 with n = 1, ∞ 2 (13.32) φ(t)∂N u(g t (x))t∂j,i d1 (x) dt 1 t ∞ 2 2 φ(t)td1 (x)∂j,i d1 (x) ∂N,N u(x , xN + sd1 (x)) dsdt. = 1
1
2 d is bounded (see (13.23)), we can apply the Since u ∈ Cc∞ (RN ) and d1 ∂j,i 1 Lebesgue dominated convergence theorem to show that the right-hand side of the previous integral goes to zero by (13.24) as x → x0 ∈ ∂Ω. Thus, (13.28) holds. It follows that the function v(x) for x ∈ RN \ Ω, (13.33) E(u)(x) := u(x) for x ∈ Ω,
is of class C 2 (RN ) (why?). Step 3: We will prove (13.17). By (13.24)3 we have that |φ(t)| ≤ ct−1 for all t ≥ 1 and for some c > 0. Using the change of variables r = xN + d1 (x)t,
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13. Sobolev Spaces: Further Properties
from (13.25) we get (13.34)
∞
|v(x)| ≤ c 1
≤c
1 |u(x , xN + d1 (x)t)| dt = c t
∞
∞
1 |u(x , r)| dr r − x N xN +d1 (x)
1 |u(x , r)| dr, r − f (x )
2f (x )−xN
where in the last inequality we used (13.22) and the fact that r − xN > r − f (x ) for all x ∈ RN \ Ω. Raising both sides to the power p, changing variables, and using Hardy’s inequality (see Theorem C.41) yield
f (x ) −∞
f (x ) ∞
p 1 |u(x , r)| dr dxN ≤c −∞ 2f (x )−xN r − f (x )
p ∞ ∞ 1 =c |u(x , s + f (x ))| ds dt s 0 t ∞ ∞ p ≤c |u(x , t + f (x ))| dt = c |u(x , xN )|p dxN ,
|v(x , xN )| dxN p
f (x )
0
where c changed from line to line, as usual. Integrating with respect to x ∈ RN −1 gives (13.17). Step 4: We show that ∇k vLp (RN \Ω) ≤ c(1 + Lk )∇k uLp (Ω)
(13.35)
for k = 1, 2 and for some c > 0 depending only on φ and p. By (13.27), (13.29), and the fact that ∂i d1 L∞ (RN ) ≤ c(1 + L), for all i = 1, . . . , N , and x ∈ RN \ ∂Ω, we have that
∞
|∂i v(x)| ≤ 1
∞
t|φ(t)||∂iu(g t (x))| dt + c(1 + L)
t|φ(t)||∂N u(g t (x))| dt.
1
We may proceed as in the previous step with u replaced by ∂i u and ∂N u, to obtain ∂i vLp (RN \Ω) ≤ c∂i uLp (Ω) + c(1 + L)∂N uLp (Ω) . Concerning the second-order derivatives of v, the first four terms on the right-hand side of (13.30) can be treated as in Step 3, again using the fact that ∂i d1 L∞ (RN ) ≤ c(1 + L) for all i = 1, . . . , N . Note that the fourth term will contribute to a factor of (1 + L2 ). It remains to study the last term on the right-hand side of (13.30). By (13.23), (13.32) and the fact that
13.1. Extension Domains
421
|φ(t)| ≤ ct−3 for all t ≥ 1 (see (13.24)3 ),
∞ 1
2 |φ(t)∂N u(g t (x))t∂j,i d1 (x)| dt ∞ t 1 2 2 |∂N,N u(x , xN + sd1 (x))| dsdt ≤ c(1 + L ) 2 t 1 1 ∞ 1 2 2 |∂N,N u(x , xN + sd1 (x))| ds, = c(1 + L ) s 1
where in the last equality we used Tonelli’s theorem. Using the change of variables r = xN + sd1 (x) we can now continue as in (13.34). We now indicate the main changes in the case m ≥ 2. Proof for m ≥ 2. Step 1: We claim that (13.28) holds for every multiindex α, with |α| ≤ m. As in Step 2 of the previous proof,
∞
α
∂ v(x) =
(13.36)
φ(t)∂ α (u ◦ g t )(x) dt
1
for every x ∈ RN \ ∂Ω. By Faa di Bruno’s theorem (see Theorem 11.54),
(13.37)
∂ (u ◦ g )(x) = α
t
β
t
cα,β,γ,l ∂ u(g (x))
|β| /
∂ γi glti (x),
i=1
where cα,β,γ,l ∈ R, the sum is done over all β ∈ NN 0 with 1 ≤ |β| ≤ |α|, γ = |β| N (γ1 , . . . , γ|β| ), γi ∈ N0 , with |γi | ≥ 1 and i=1 γi = α, and l = (l1 , . . . , l|β| ), li ∈ {1, . . . , N }, i = 1, . . . , |β|. Since g t (x) = x+td1 (x)eN , if li < N , then ∂ γi glti (x) = 0 unless γi = ei in t (x) = ∂ γi (x ) + t∂ γi d (x). Moreover, which case ∂ γi glti (x) = 1, while ∂ γi gN 1 N if |β| = |α|, then |γi | = 1 for all i and thus by (13.23) and the fact that t ≥ 1, (13.38)
|α| / β t ∂ γi glti (x) ≤ c(1 + L|α| )t|α| |∂ β u(g t (x))|. ∂ u(g (x)) i=1
In particular, if β = α, li = i and γi = ei for every i = 1, . . . , N , then the corresponding coefficient cα,α,γ,l is one. This can be shown by induction on |α| (see (13.29) and (13.30) for |α| = 1 and |α| = 2).
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13. Sobolev Spaces: Further Properties
If |β| < |α| we apply Taylor’s formula at t = 1 (see Theorem 9.9) to the function h(t) := ∂ β u(g t (x)) to write β
t
∂ u(g (x)) =
β (d1 (x))n n=0
n!
(d1 (x))β + β !
∂ β+neN u(x , xN + d1 (x))(t − 1)n
t
(t − s)β ∂ β+β eN u(x , xN + sd1 (x)) ds,
1
where β := |α| − |β|. By the binomial theorem and (13.24),
∞
(t − 1) t φ(t) dt = n k
n n
1
i=0
i
(−1)
∞
n−i
tk+i φ(t) dt = 0,
1
and so (13.39) ∞ 1
tk φ(t)∂ β u(g t (x)) dt t (d1 (x))β ∞ k t φ(t) (t − s)β ∂ β+β eN u(x , xN + sd1 (x)) dsdt. = β ! 1 1
By (13.23) and the facts that β = |α| − |β| = (13.40)
(d1 )β
|β| /
|∂ γi glti | ≤
i=1
|β| /
|β|
i=1 (|γi |
− 1) and t ≥ 1,
ci t(d1 )|γi |−1 (1 + |∂ γi d1 |) ≤ c(1 + L|α| )t|β| .
i=1
Hence, by (13.24) and (13.36)–(13.40), as in the previous steps, we can apply the Lebesgue dominated convergence theorem to obtain ∞ φ(t)∂ α u ◦ g t (x) dt = ∂ α u(x0 ), lim ∂ α v(x) = lim x→x0
x→x0
1
which gives (13.28). Step 2: Next we prove (13.35) for every multi-index α, with |α| ≤ m. If |β| = |α|, using (13.38) and the fact that |φ(t)| ≤ ct−|α|−1 for all t ≥ 1 (see (13.24)3 ), we have
|β| / γi t φ(t)∂ β u(g t (x)) ∂ gli (x) dt i=1 ∞ 1 β |∂ u(x , xN + td1 (x))| dt, ≤ c(1 + L|α| ) t 1
∞ 1
13.1. Extension Domains
423
while if |β| < |α|, using (13.39), (13.40), and the fact that |φ(t)| ≤ ct−|α|−2 for all t ≥ 1 we have ∞ |β| / γi t dt φ(t)∂ β u(g t (x)) ∂ g (x) li 1
≤c(1 + L|α| )
i=1
∞
1
t
(t−s)|α|−|β| |∂ β+β eN u(x , xN +sd1 (x))| dsdt |α|+2−|β| t 1 1 ∞ t 1 |∂ β+β eN u(x , xN + sd1 (x))| dsdt ≤c(1 + L|α| ) 2 1 t 1 ∞ 1 β+β eN =c(1 + L|α| ) |∂ u(x , xN + sd1 (x))| ds, s 1 where the last equality follows from Tonelli’s theorem. Using the change of variables r = xN + sd1 (x) we can now continue as in (13.34). Step 3: We have proved that the linear operator E : Cc∞ (RN ) → W m,p (RN ) defined by v(x) for x ∈ RN \ Ω, E(u)(x) := u(x) for x ∈ Ω, satisfies (13.17) and (13.18). Since functions in Cc∞ (RN ) are dense in W m,p (Ω) (see Theorem 11.35) we can extend E to a linear operator defined on W m,p (Ω) and with the same properties. Remark 13.9. Note that the operator E does not depend on m and p. e Im(e−λ(t−1) ), t ≥ 1, Exercise 13.10. Prove that the function φ(t) := πt where λ := eiπ/4 and Im z stands for the imaginary part of a complex number z ∈ C, satisfies properties (13.24). 1/4
Next we extend the previous theorem to more general Lipschitz continuous domains. Definition 13.11. The boundary ∂Ω of an open set Ω ⊆ RN is uniformly Lipschitz continuous if there exist ε, L > 0, M ∈ N, and a locally finite countable open cover {Ωn }n of ∂Ω such that (i) if x ∈ ∂Ω, then B(x, ε) ⊆ Ωn for some n ∈ N, (ii) no point of RN is contained in more than M of the Ωn ’s, (iii) for each n there exist local coordinates y = (y , yN ) ∈ RN −1 ×R and a Lipschitz continuous function f : RN −1 → R (both depending on n), with Lip f ≤ L, such that Ωn ∩ Ω = Ωn ∩ Vn , where Vn is given in local coordinates by {(y , yN ) ∈ RN −1 × R : yN > f (y )}.
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13. Sobolev Spaces: Further Properties
Remark 13.12. Similarly, given m ∈ N0 and 0 < α ≤ 1 we can define open sets Ω ⊆ RN whose boundary is uniformly of class C m (respectively, of class C m,α , see Definition 11.55) with parameters ε, L > 0, M provided (i), (ii), and (iii) hold but with f of class C m (respectively of class C m,α ) and with f C m (RN −1 ) ≤ L (respectively, f C m,α (RN −1 ) ≤ L). Exercise 13.13. Let Ω ⊆ RN be an open set such that ∂Ω is bounded. Prove that ∂Ω is uniformly Lipschitz continuous if and only if it is Lipschitz continuous. Exercise 13.14. Let Ω ⊆ R be such that Ω = n In , where the In are open intervals such that length In ≥ δ for all n ∈ N and dist(In , Ik ) ≥ δ for all n, k ∈ N with n = k and for some δ > 0. (i) Prove that ∂Ω is uniformly Lipschitz continuous. (ii) Prove that the condition length In ≥ δ for all n ∈ N is necessary to have an extension operator from W 1,1 (Ω) to W 1,p (R). (iii) Prove that the condition dist(In , Ik ) ≥ δ for all n, k ∈ N is necessary to have an extension operator from W 1,∞ (Ω) to W 1,∞ (R). In the next exercise, by a finite cone having vertex v ∈ RN , length , and vertex angle 2α ∈ (0, π) we mean a set of the form Kv,R := v + RK, where R is an orthogonal N × N matrix and K := {(x , xN ) ∈ RN −1 × R : xN > x cot α} ∩ B(0, ). Exercise 13.15. Let Ω ⊆ RN be an open set whose boundary ∂Ω is uniformly Lipschitz continuous. (i) Prove that for every x ∈ ∂Ω there exists a finite cone Kx,R , with vertex x, length ε, and vertex angle depending on L, such that Kx,R \ {x} ⊆ Ω. (ii) Prove that if Ω is unbounded, then LN (Ω) = ∞. Exercise 13.16. Prove that for every x = (x1 , . . . , xN ) ∈ RN and for every n ∈ N, n! xα . (x1 + · · · + xn )n = α! |α|=n
The next theorem shows that domains with uniformly Lipschitz continuous boundary are extension domains for W m,p (Ω) for all 1 ≤ p ≤ ∞. Theorem 13.17 (Stein). Let Ω ⊆ RN be an open set with uniformly Lipschitz continuous boundary. Then for all 1 ≤ p ≤ ∞ and m ∈ N there exists a continuous linear operator E : W m,p (Ω) → W m,p (RN ) such that for all
13.1. Extension Domains
425
u ∈ W m,p (Ω), E(u)(x) = u(x) for LN -a.e. x ∈ Ω, and (13.41) (13.42)
E(u)Lp (RN ) ≤ c(1 + M )uLp (Ω) , ∇k E(u)Lp (RN ) ≤ c(1 + M (1 + Lk ))
k (M/ε)k−i ∇i uLp (Ω) i=0
for every multi-index α ∈
NN 0
with 1 ≤ |α| ≤ m.
For simplicity we first prove the case m = 1 and then treat the general case. Proof for m = 1. Step 1: We only prove the case 1 ≤ p < ∞ and leave the easier case p = ∞ as an exercise. For every set E ⊆ RN and every r > 0, observe that E r ⊆ E and that we define E r := {x ∈ RN : B(x, r) ⊆ E}. We condition (i) in Definition 13.11 reads ∂Ω ⊂ n Ωεn . Define the regularized functions (13.43)
φn := ϕε/4 ∗ χΩ3ε/4 , n
where ϕε/4 is a standard mollifier. Then (13.44)
supp φn ⊂ Ωn ,
φn = 1 in Ωε/2 n .
By Theorem C.20 for every multi-index α ∈ NN 0 we have that ∂ α φn = ∂ α ϕε/4 ∗ χΩ3ε/4 , n
and so for all x ∈ Ωn we have that α (13.45) |∂ φn (x)| ≤ |∂ α ϕε/4 (y − x)| dy ≤ cα ε−|α| . RN
Next consider the three open sets Ω0 := {x ∈ RN : dist(x, Ω) < ε/4}, (13.46)
Ω+ := {x ∈ RN : dist(x, ∂Ω) < 3ε/4}, Ω− := {x ∈ Ω : dist(x, ∂Ω) > ε/4},
and define the regularized functions (13.47)
φ0 := ϕε/4 ∗ χΩ0 ,
φ± := ϕε/4 ∗ χΩ± .
Then φ0 = 1 in Ω, φ+ (x) = 1 if x ∈ RN and dist(x, ∂Ω) ≤ ε/2, and φ− (x) = 1 if x ∈ Ω and dist(x, ∂Ω) ≥ ε/2. Moreover, the supports of φ0 , φ+ , and φ− are contained, respectively, in an ε/2 neighborhood of Ω, in an ε neighborhood of ∂Ω, and in Ω. Finally, reasoning as in (13.45), we have that (13.48)
∂ α φ0 ∞ ≤ cε−|α| ,
∂ α φ± ∞ ≤ cε−|α| .
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13. Sobolev Spaces: Further Properties
Note that (13.49)
supp φ0 ⊆ {x ∈ RN : φ+ (x) + φ− (x) ≥ 1}.
Thus, with a slight abuse of notation, we may define (13.50)
ψ+ := φ0
φ+ , φ+ + φ−
ψ− := φ0
φ− , φ+ + φ−
where we interpret the right-hand sides to be zero whenever φ0 = 0. By (13.48) and (13.49) we have that all the derivatives of order α of ψ± are bounded by c|ε|−|α| . Also, ψ+ + ψ− = 1 in Ω and ψ+ = ψ− = 0 outside an ε/2 neighborhood of Ω. We are finally ready to construct the linear extension operator. Given a function u ∈ Cc∞ (RN ), 1 ≤ p < ∞, since supp(φn u) ⊂ Ωn for each n by (13.44), by condition (i) in Definition 13.11 and Theorem 13.8 we can extend φn u to a function vn ∈ W m,p (RN ) in such a way that (13.51)
vn Lp (RN ) = 2φn uLp (Ω∩Ωn ) , ∇k vn Lp (RN ) ≤ c(1 + Lk )∇k (φn u)Lp (Ω∩Ωn )
for every multi-index k with 1 ≤ k ≤ m. Again with a slight abuse of notation we define φn (x)vn (x) n + ψ− (x)u(x), x ∈ RN . (13.52) E(u)(x) := ψ+ (x) 2 φk (x) k
Note that if x ∈ RN is such that dist(x, ∂Ω) ≤ ε/2, then there exists an n ε/2 such that y ∈ Ωn , and so φn (x) = 1 by (13.44). In particular, since all the functions φn are nonnegative, it follows that φ2k (x) ≥ 1, (13.53) if x ∈ supp ψ+ , then k
and thus the first term on the right-hand side of (13.52) is well-defined, provided we interpret it to be zero whenever ψ+ = 0. Similarly, since supp ψ− ⊆ supp φ− ⊂ Ω, the term ψ− (x)u(x) is well-defined, provided we set it to be zero outside Ω. It remains to show that the linear operator E(u) is an extension operator and that it is bounded. For the former, it suffices to observe that if x ∈ Ω, then vn (x) = φn (x)u(x), and so E(u)(x) = ψ+ (x)u(x) + ψ− (x)u(x) = u(x) by (13.50). To prove that E(u) is bounded in Lp (RN ), we observe that by (13.52), condition (ii) in Definition 13.11, Exercise 12.34, (13.53), the fact
13.1. Extension Domains
427
that 0 ≤ ψ± , φn ≤ 1, and (13.51), in this order, we have
1/p 1/p p E(u)Lp (RN ) ≤ M |vn | dx + uLp (Ω) Ωn
n
≤ cM
(13.54)
1/p
|u|
p
Ω
1/p |φn | dx
+ uLp (Ω)
p
n
≤ c(1 + M )uLp (Ω) , where in the last inequality we have used the fact that
n
|φn |p ≤ M .
Step 2: In this step we will estimate ∂i E(u). Since {Ωn }n is locally finite, any small neighborhood of every point x ∈ RN intersects only finitely many Ωn ’s. Hence, by (13.52) we can calculate φn vn (vn ∂i φn + φn ∂i vn ) n n 2 (13.55) ∂i E(u) = 2 ∂i ψ+ + ψ+ φk φk k k φn vn φl ∂i φl n l + u∂i ψ− + ψ− ∂i u. − 2ψ+ 2 φ2k k
Using (13.51)–(13.53), condition (ii) in Definition 13.11, Exercise 12.34, and the facts that 0 ≤ ψ± , φn ≤ 1, ∂i ψ± ∞ , ∂i φn ∞ ≤ cε−1 , and φl ∂i φl ≤ cM ε−1 , |φn |p ≤ M, |∂i φn |p ≤ cM ε−p , (13.56) n
n
we obtain ∇E(u)Lp (RN ) ≤ cM
1/p
n
+ cε
−1
+ cε
−1
n
(M p ε−p |vn |p + ∇vn p ) dx
1/p
Ωn
uLp (Ω) + c∇uLp (Ω) 1/p 1/p ≤ cM (1 + L) (M p ε−p |φn u|p + ∇(φn u)p ) dx n
Ω
uLp (Ω) + c∇uLp (Ω)
≤ c(1 + M (1 + L))(M ε−1 uLp (Ω) + ∇uLp (Ω) ). To conclude the proof, it suffices to observe that since {Ωn }n is locally finite, in a neighborhood of every point the infinite sum in (13.52) is finite. We now turn to the proof for m ≥ 2.
428
13. Sobolev Spaces: Further Properties
Proof for m ≥ 2. It remains to estimate ∇k E(u)Lp (RN ) for k ≥ 2. Let α α ∈ NN 0 be a multi-index with |α| ≥ 2. We begin with ∂ ψ± . By Exercise 11.37, α α α −1 ∂ ψ± = ∂ (φ0 φ± (φ+ + φ− ) ) = ∂ β (φ0 φ± )∂ α−β (φ+ + φ− )−1 β β≤α α β = ∂ δ φ0 ∂ β−δ φ± ∂ α−β (φ+ + φ− )−1 , β δ β≤α
δ≤β
while by Theorem 11.54 with u replaced by t → t−1 and Ψ by φ+ + φ− , ∂ α−β (φ+ + φ− )−1 =
aα−β,l,γ (φ+ + φ− )−l−1
l /
∂ |γi | (φ+ + φ− ),
i=1
where aα−β,l,γ ∈ R, the sum is done over all l ∈ N0 with 1 ≤ l ≤ |α| − |β|, , with |γi | ≥ 1 and li=1 γi = α − β. Using the γ = (γ1 , . . . , γl ), γi ∈ NN 0 facts that β ≤ α, δ ≤ β, li=1 γi = α − β, and (13.48), we have that ∂ α ψ± ∞ ≤ cε−|α| .
(13.57)
By (13.52) and Exercise 11.37, −1 α α ∂ β vn ∂ α−β ψ+ φn ∂ E(u) = φ2k (13.58) β n β≤α k α ∂ β u∂ α−β ψ− + β β≤α
and ∂
α−β
ψ+ φn
φ2k
k
α − β = δ δ≤α−β
−1 α − β φ2k = ∂ α−β−δ (ψ+ φn )∂ δ δ δ≤α−β k −1 α − β − δ φ2k , ∂ η ψ+ ∂ α−β−δ−η φn ∂ δ η
−1
η≤α−β−δ
k
while by Theorem 11.54 with u replaced by t → t−1 and Ψ by
k
∂δ
φ2k
−1
=
k
=
bδ,l,γ
φ2k
l −l−1 /
k
bδ,l,γ
k
i=1
∂ γi
φ2k
φ2k ,
k
l −l−1 / γi αi 2 φk ∂ φk ∂ γi −δi φk , αi i=1 k αi ≤γi
where bδ,l,γ ∈ R are constant, the sum is done over all l ∈ N0 with 1 ≤ l ≤ |δ|, l γ = (γ1 , . . . , γl ), γi ∈ NN 0 , with |γi | ≥ 1 and i=1 γi = δ.
13.1. Extension Domains
429
Using (13.45) and (13.53), the fact that for each x ∈ RN at most M terms in the sum are different from zero (see condition (ii) in Definition 13.11), −1 Exercise 12.34, and the facts that 0 ≤ ψ± , φn ≤ 1, ∂ δ φ2k ≤ cM |δ| ε−|δ| ,
∞
k
and so by (13.45) and (13.57), −1 α−β φ2k ψ+ φn ≤ cM |α|−|β| ε−(|α|−|β|) . ∂ ∞
k
Hence, by (13.58), ∇k E(u) ≤ c
k n
M k−i ε−(k−i) ∇i vn + c
i=0
k
ε−(k−i) ∇i u.
i=0
By Exercise 12.34 and the fact that for each x ∈ RN at most M terms in the sum are different from zero (see condition (ii) in Definition 13.11), we have
1/p k k 1/p (k−i)p −(k−i)p i p M ε ∇ vn dx ∇ E(u)Lp (RN ) ≤ cM n
+c
k
Ωn
i=0
ε−(k−i) ∇i uLp (Ω) .
i=0
Finally, by (13.51), if i = 0, p |vn | dx ≤ c n
Ωn
n
|φn u| dx ≤ c p
Ω∩Ωn
|φn u|p dx
Ω n
|u|p dx,
≤ cM Ω
where we have used the facts that supp φn ⊆ Ωn and that n |φn |p ≤ M . Similarly, if k ≥ 1, by (13.51), ∇k vn p dx ≤ c(1 + Lk )p ∇k (φn u)p dx n
Ωn
Ω n
≤ cM (1 + L )
k p
k
ε
−(k−i)p
i=1
∇i up dx, Ω
where we have used Exercise 11.37 and the fact that cM ε−(k−i)p to estimate n
∇k (φn u)p ≤ c
k n
i=1
∇k−i φn p ∇i up ≤ cM
k i=1
n ∇
k−i φ p n
ε−(k−i)p ∇i up .
≤
430
13. Sobolev Spaces: Further Properties
In turn, ∇k E(u)Lp (RN ) ≤ cM (1 + Lk )
k
M k−i ε−(k−i) ∇i uLp (Ω)
i=0
+c
k
ε−(k−i) ∇i uLp (Ω) .
i=0
Thus we have shown (13.41) and (13.42) for u ∈ Cc∞ (RN ). Since functions in Cc∞ (RN ) are dense in W m,p (Ω), (see Theorem 11.35), we can now extend E to a linear operator defined on W m,p (Ω) and with the properties (13.41) and (13.42). Remark 13.18. Using Poincar´e’s inequality (see the next section), in the case of bounded Lipschitz continuous domains and m = 1 one can show that in the inequality (13.42) it is possible to drop the term M ε−1 uLp (Ω) (see Exercise 13.33 below) at the price of replacing RN with any large ball containing Ω.
13.2. Poincar´ e Inequalities The next result provides an equivalent norm in W0m,p (Ω) for a large class of domains Ω. We recall that ∇0 u := u. Theorem 13.19 (Poincar´e’s inequality in W0m,p (Ω)). Let Ω ⊂ RN be an open set with finite width, that is, Ω lies between two parallel hyperplanes, and let m ∈ N and 1 ≤ p < ∞. Then there exists a constant c = c(m, N, p) > 0 such that for all u ∈ W0m,p (Ω) and 0 ≤ k ≤ m − 1, dp(m−k) k p ∇ u(x) dx ≤ c m−k ∇m u(x)p dx, p Ω Ω where d is the distance between the two hyperplanes. In particular, if m = 1, for all u ∈ W01,p (Ω), dp p |u(x)| dx ≤ c ∇u(x)p dx. p Ω Ω Proof. Without loss of generality, up to a rotation and translation, we may assume that Ω lies between the two parallel hyperplanes xN = 0 and xN = d > 0. For u ∈ Cc∞ (Ω), by the fundamental theorem of calculus and H¨ older’s inequality, we have xN ∂N u(x , t) dt |u(x , xN )| = |u(x , xN ) − u x , 0 | = 1/p
≤ xN
0
d 0
|∂N u(x , t)|p dt
1/p .
13.2. Poincar´e Inequalities
431
Extend u to be zero outside RN \ Ω. Raising to the power p and integrating over RN −1 × [0, d], by Tonelli’s theorem we get p |u(x)| dx = |u(x , xN )|p dx Ω
≤
RN −1 ×[0,d] d RN −1
0
p/p xN
d 0
d
|∂N u(y)| dy p
= Ω
|∂N u(x , t)|p dtdxN dx
0
p−1 xN dxN
dp = p
|∂N u(y)|p dy. Ω
This proves the case m = 1. If m ≥ 2 and α is a multi-index with 0 ≤ |α| < m − 1, by repeated applications of the case m = 1 with u replaced by ∂ α+neN u(x), 0 ≤ n ≤ m − |α| − 1, we get dp α p |∂ u(x)| dx ≤ |∂ α+eN u(x)|p dx p Ω Ω dp(m−|α|) |∂ α+eN u(x)|p dx. ≤ · · · ≤ m−|α| p Ω This concludes the proof.
Remark 13.20. It follows from the previous theorem that if an open set Ω ⊂ RN has finite width, then the seminorm ∇m uLp (Ω) is actually an equivalent norm in the space W0m,p (Ω). Exercise 13.21. Show that if the open set Ω ⊆ RN contains a sequence of balls B(xn , rn ), where xn ∈ RN and rn → ∞, then the previous Poincar´e inequality fails. Exercise 13.22. Let Ω ⊂ RN be an open bounded set and let u ∈ W01,p (Ω), 1 ≤ p ≤ ∞. Prove that pp p |u(x)| dx ≤ p inf sup y − x0 ∇u(x)p dx. N x0 ∈RN y∈Ω Ω Ω Hint: Write N = div(x − x0 ) and use the divergence theorem. Exercise 13.23. Let Ω ⊂ RN be an open bounded set, let m ∈ N and 1 < p < ∞ be such mp = N . (i) Prove that there exist two constants c1 , c2 > 0 depending only on m and N such that for all u ∈ W01,N (Ω) \ {0},
|u(x)|p exp c1 dx ≤ c2 LN (Ω). (13.59) p uW m,p (RN ) Ω (ii) Prove that when m = 1 the previous inequality fails if c1 > γN , where γN is the constant given in (12.29).
432
13. Sobolev Spaces: Further Properties
Remark 13.24. The inequality (13.59) was established by Pokhozhaev [189], Trudinger [234] and Yudovich [250] when m = 1 and by Strichartz [225] for m ≥ 2. In [179], Moser proved that for bounded domains the inequality (13.59) also holds for γ = γN . The proof is not immediate and we refer to [179] for more details. See also the recent paper of Li and Ruf [141] and the bibliography contained therein for some recent results on unbounded domains. Exercise 13.25. Assume that the open set Ω ⊂ RN has finite width, let m ∈ N, and let 1 ≤ p < ∞. Prove that for every functional L ∈ fα ∈ Lp (Ω), α ∈ NN W −m,p (Ω) 0 , with |α| = m, such that there exist α L(u) = Ω |α|=m ∂ u(x)fα (x) dx for all u ∈ W0m,p (Ω). Another form of Poincar´e’s inequality is p |u(x) − uE | dx ≤ c ∇u(x)p dx Ω
Ω
for u ∈ W 1,p (Ω), where E ⊆ Ω is a Lebesgue measurable set with finite positive measure and 1 u(x) dx. (13.60) uE := N L (E) E Exercise 13.26. Let Ω ⊂ RN be an open bounded set, let E ⊆ Ω be a Lebesgue measurable set with finite positive measure, let m ∈ N, and let 1 ≤ p ≤ ∞. Prove that given u ∈ W m,p (Ω) there exists a polynomial pE (u) of degree m−1 such that for every multi-index α ∈ NN 0 , with 0 ≤ |α| ≤ m−1, (∂ α u(x) − ∂ α pE (u)(x)) dx = 0. (13.61) E
Theorem 13.27 (Poincar´e’s inequality in W m,p (Ω)). Let m ∈ N, let 1 ≤ p < ∞, and let Ω ⊂ RN be a connected extension domain for W m,p (Ω) with finite measure. Let E ⊆ Ω be a Lebesgue measurable set with positive measure. Then there exists a constant c = c(m, N, p, Ω, E) > 0 such that for all u ∈ W m,p (Ω), m−1
∇k (u − pE (u))Lp (Ω) ≤ c∇m uLp (Ω) .
k=0
In particular, if m = 1, for all u ∈ W 1,p (Ω), u − uE Lp (Ω) ≤ c∇uLp (Ω) .
13.2. Poincar´e Inequalities
433
Proof. Assume by contradiction that the result is false. Then we may find a sequence {un }n in W m,p (Ω) such that m−1
∇k (un − pE (un ))Lp (Ω) > n∇m un Lp (Ω) .
k=0
Define
un − pE (un ) , un − pE (un )W m−1,p (Ω) := · Lp (Ω) . Then vn ∈ W m,p (Ω) and by (13.61), vn :=
where · W 0,p (Ω)
vn W m−1,p (Ω) = 1,
(13.62)
pE (vn ) = 0,
∇m vn Lp (Ω) ≤ 1/n.
By the Rellich–Kondrachov theorem (Theorems 12.18, 12.44 and Exercise 12.63) there exists a subsequence {vnk }k such that vnk → v in W m−1,p (Ω) for some function v ∈ W m−1,p (Ω) with vW m−1,p (Ω) = 1, pE (v) = 0. Moreover, for every ψ ∈ Cc∞ (Ω) and every multi-index β with |β| = m, by H¨older’s inequality and (13.62), β β β v∂ ψ dx = lim vn ∂ ψ dx = lim ψ∂ vn dx k k k→∞ k→∞ Ω
Ω
Ω
≤ lim ∂ β vn Lp (Ω) ψLp (Ω) = 0, k→∞
with ∂ β v = 0. Since Ω is connected, this implies that v and so v ∈ is a polynomial of degree m − 1, but since pE (v) = 0, by (13.61) necessarily v = 0. This contradicts the fact that vW m−1,p (Ω) = 1 and completes the proof. W m,p (Ω)
Remark 13.28. In view of Theorem 13.17, it follows from the previous theorem that Poincar´e’s inequality holds for uniformly Lipschitz domains with finite measure. Exercise 13.29. Let 1 ≤ p < N and let Ω ⊂ RN be a connected extension domain for W 1,p (Ω) with finite measure. Let E ⊆ Ω be a Lebesgue measurable set with positive measure. Prove that there exists a constant c = c(N, p, Ω, E) > 0 such that for all u ∈ W 1,p (Ω), u − uE Lp∗ (Ω) ≤ c∇uLp (Ω) . Exercise 13.30. Let Ω ⊂ RN be an open connected set with finite measure and let 1 ≤ p < ∞. Assume that the embedding W 1,p (Ω) → Lp (Ω) is compact and prove that there exists a constant c = c(N, p, Ω) > 0 such that for all u ∈ W 1,p (Ω), u − uΩ Lp (Ω) ≤ c∇uLp (Ω) . Exercise 13.31. Let Ω ⊂ R3 be a connected bounded extension 3 domain for 1,2 1,2 W (Ω) and let S be the subset of all u ∈ W (Ω) such that Ω u (x) dx = 0. (i) Prove there exists a constant c = c(Ω) > 0 such that for all u ∈ S, uL2 (Ω) ≤ c∇uL2 (Ω) .
434
13. Sobolev Spaces: Further Properties
(ii) What properties of the subset S did you use in the proof? Corollary 13.32. Let m ∈ N, let 1 ≤ p < ∞, and let Ω ⊂ RN be a connected bounded domain whose boundary is of class C. Let E ⊆ Ω be a Lebesgue measurable set with positive measure. Then there exists a constant c = c(m, N, p, Ω, E) > 0 such that for all u ∈ W m,p (Ω), (13.63)
m−1
∇k (u − pE (u))Lp (Ω) ≤ c∇m uLp (Ω) .
k=0
In particular, if m = 1, for all u ∈ W 1,p (Ω), (13.64)
u − uE Lp (Ω) ≤ c∇uLp (Ω) .
Proof. For m = 1, in view of Theorem 12.30 we are in a position to apply Exercise 13.30. This gives the proof for m = 1. If m ≥ 2 and α is a multi-index with 0 ≤ |α| < m − 1, in view of (13.61) we can apply repeatedly inequality (13.64) with u − uE replaced by ∂ α+nei (u − pE (u)), 0 ≤ n ≤ m − |α| − 1, to get (13.63). Exercise 13.33. Let Ω ⊂ RN be an open bounded domain with Lipschitz continuous boundary and let B be a ball with Ω B. Prove that there exists a continuous linear operator EB : W 1,p (Ω) → W 1,p (B) such that for all u ∈ W 1,p (Ω), EB (u)(x) = u(x) for LN -a.e. x ∈ Ω and EB (u)Lp (B) ≤ cuLp (Ω) ,
∇EB (u)Lp (B) ≤ c∇uLp (Ω)
for some constant c = c(m, N, Ω, B). Hint: Let E be the extension operator given in Theorem 13.17 and let E1 (u) be the restriction of E(u) to B. Define EB (u) := E1 (u − uΩ ) + uΩ . In many applications it is of interest to have an estimate of the constant c(m, N, p, Ω, E). In the remainder of the chapter we study important special classes of domains for which there are explicit bounds on the Poincar´e constant c(m, N, p, Ω, E). We begin by considering rectangles. Proposition 13.34 (Poincar´e’s inequality for rectangles). Let m ∈ N, let 1 ≤ p < ∞, and let R = (0, a1 ) × · · · × (0, aN ) ⊂ RN . Then there exists a constant c = c(m, N, p) > 0 such that for all u ∈ W m,p (R), and every 0 ≤ k ≤ m − 1, ∇k (u − pR (u))Lp (R) ≤ c(max{a1 , . . . , aN })(m−k) ∇m uLp (R) . In particular, if m = 1, for all u ∈ W 1,p (R), (13.65)
u − uR Lp (R) ≤ c(max{a1 , . . . , aN })∇uLp (R) .
13.2. Poincar´e Inequalities
435
Proof. Step 1: Assume m = 1. Without loss of generality, by Theorem 11.35 we may assume that u ∈ C ∞ (R), and, by eventually replacing u with v(y) := u(λy), y ∈ λ1 R for an appropriate λ > 0, we may assume that max{a1 , . . . , aN } = 1.
(13.66)
By the fundamental theorem of calculus, for all x, y ∈ R, |u(x) − u(y)| ≤ |u(x) − u(x1 , . . . , xN −1 , yN )| + · · · + |u(x1 , y2 , . . . , yN ) − u(y)| N ai |∂i u(x1 , . . . , xi−1 , t, yi+1 , . . . , yN )| dt. ≤ 0
i=1
Raising both sides to power p, by H¨older’s inequality, (13.66), and the convexity of the function g(t) = |t|p , N ai 1 p p p p |∂i u(x1 , . . . , xi−1 , t, yi+1 , . . . , yN )| dt |u(x) − u(y)| ≤ ≤ N p−1
i=1
N i=1
ai
0
|∂i u(x1 , . . . , xi−1 , t, yi+1 , . . . , yN )|p dt.
0
Hence, again by H¨ older’s inequality and the fact that u(x) − uR = (u(x) − u)R , p 1 p |u(x) − uR | dx ≤ N |u(x) − u(y)| dy dx (L (R))p R R R 1 ≤ N |u(x) − u(y)|p dydx (L (R))p−p/p R R N ai N p−1 |∂i u(x1 , . . . , xi−1 , t, yi+1 , . . . , yN )|p dtdydx ≤ N L (R) R R 0 =N
p−1
i=1 N
|∂i u(z)| dz ≤ N p
ai
i=1
R
p−1
N i=1
p−p/p
where we have used the fact that This concludes the proof for m = 1.
|∂i u(z)|p dz, R
= 1, Fubini’s theorem, and (13.66).
Step 2: If m ≥ 2 and α is a multi-index with 0 ≤ |α| < m − 1, in view of (13.61) we can apply repeatedly inequality (13.65) with u − uR replaced by ∂ α+nei (u − pR (u)), 0 ≤ n ≤ m − |α| − 1 to conclude the proof. ˙ m,p (R). Exercise 13.35. Let R be as in the previous theorem and let u ∈ W m,p Prove that u ∈ W (R). Hint: Prove first that pR (u) can be replaced by pR1 (u), where R1 is rectangle compactly contained in R.
436
13. Sobolev Spaces: Further Properties
Next we study convex domains. In the literature there are simpler proofs (see, e.g., [94]). The present one gives a sharper constant and the approach is quite interesting. Theorem 13.36 (Poincar´e’s inequality for convex sets). Let m ∈ N, let 1 ≤ p < ∞ and let Ω ⊂ RN be an open bounded convex set. Then there exists a constant c = c(m, N, p) > 0 such that for all u ∈ W m,p (Ω), and every 0 ≤ k ≤ m − 1, ∇k (u − pΩ (u))Lp (Ω) ≤ c(diam Ω)m−k ∇m uLp (Ω) . In particular, if m = 1, for all u ∈ W 1,p (Ω), (13.67)
u − uΩ Lp (Ω) ≤ c diam Ω∇uLp (Ω) .
We begin with some preliminary results. Exercise 13.37. Let D ⊆ RN be a convex set. (i) Prove that the interior and the closure of D are convex. 2 (ii) Let u : D → R be a continuous function such that u( x1 +x 2 ) ≥ u(x1 )+u(x2 ) for all x1 , x2 ∈ D. Prove that u is concave. 2
Lemma 13.38. Let Ω ⊂ RN be an open bounded convex set. Then for every small δ > 0, the set Ωδ := {x ∈ Ω : dist(x, ∂Ω) > δ} is a convex set with Ωδ ⊂ Ω. Proof. We begin by showing that the function d(x) := dist(x, ∂Ω), x ∈ Ω, is concave in Ω. To see this, consider x1 , x2 ∈ Ω and let d1 := d(x1 ) and d2 := d(x2 ). Fix any direction ξ ∈ S N −1 and consider the trapezoid with vertices x1 , x2 , y1 , y2 , where y1 := x1 +d1 ξ and y2 := x2 +d2 ξ. Since the open balls B(xi , di ), i = 1, 2, are contained in Ω by the definition of di , it follows that yi ∈ Ω, i = 1, 2, and so the trapezoid also lies in Ω by the convexity 2 2 and d0 := d1 +d of Ω (see Exercise 13.37). In particular, if x := x1 +x 2 2 , y1 +y2 then the point y := 2 = x + d0 ξ lies on the edge of the trapezoid and therefore also in Ω. As the vector ξ varies in S N −1 , the corresponding point y = x + d0 ξ fills the surface of the ball B(x, d0 ). Hence, B(x, d0 ) ⊂ Ω, and d(x1 )+d(x2 ) 2 . Since d is Lipschitz continuous so d(x) ≥ d0 , that is, d( x1 +x 2 )≥ 2 by the previous exercise, it follows from Exercise 13.37 that d is concave. Next we show that Ωδ is convex. Let x1 , x2 ∈ Ωδ and let θ ∈ (0, 1). By the concavity of d, d(θx1 + (1 − θ)x2 ) ≥ θd(x1 ) + (1 − θ)d(x2 ) > δ, and so θx1 + (1 − θ)x2 ∈ Ωδ .
13.2. Poincar´e Inequalities
437
Lemma 13.39. LetΩ ⊂ RN be an open bounded convex set and let u ∈ L1 (Ω) be such that Ω u(x) dx = 0. Then for every δ > 0 there exists a decomposition of Ω into a finite number of pairwise disjoint convex domains Ωn , n = 1, . . . , , such that
Ω=
n=1
Ωn ,
u(x) dx = 0,
n = 1, . . . , ,
Ωn
and each Ωn is thin in all but one direction, that is, in an appropriate coordinate system y = (y , yN ) ∈ RN −1 × R, Ωn ⊂ [0, δ] × · · · × [0, δ] × [0, diam Ω]. Proof. For each α ∈ [0, 2π] there is a unique hyperplane Hα ⊂ RN with normal (cos α, sin α, 0, . . . , 0) that divides Ω into two convex sets Uα and Vα of equal volume. Let I(α) := Uα u(x) dx. Since I(α) = −I(α + π), by continuity there exists α0 such that I(α0 ) = 0. Since Uα u(x) dx = 0 Vα0 u(x) dx = 0, we now apply the same reasoning to each of the sets Uα0 and Vα0 . Continuing in this way, we can subdivide Ω into convex sets Wn with the property that each of these sets is contained between two hyperplanes with normal of the form (cos βn , sin βn , 0, . . . , 0) and at distance δ and the average of u vanishes on each of these sets. Fix any Wn . By a rotation we can assume that the normal of the two hyperplanes is (1, 0, . . . , 0). In these new coordinates we apply the same reasoning using hyperplanes with normals of the form (0, cos α, sin α, 0, . . . , 0). We continue in this way to obtain the desired conclusion.
Proof of Theorem 13.36. As in Proposition 13.34 it is enough to prove the case m = 1. Step 1: We begin by showing that we may assume u ∈ C 1 (Ω). By Lemma 13.38 for every δ > 0, the set Ωδ is convex and dist(Ωδ , ∂Ω) = δ > 0. Consider a sequence of standard mollifiers {ϕε }ε>0 . For 0 < ε < δ define uε := u ∗ ϕε in Ωδ . Then uε ∈ C ∞ (Ωδ ) by Theorem C.20, and so, if (13.67) holds for the pair uε , Ωδ , we get |uε (x) − (uε )Ωδ | dx ≤ c diam Ωδ p
Ωδ
N i=1
|∂i uε |p dx. Ωδ
438
13. Sobolev Spaces: Further Properties
Letting ε → 0+ and using Theorems C.16 and C.20 we obtain Ωδ
|u(x) − uΩδ |p dx ≤ c diam Ωδ ≤ c diam Ω
N
|∂i u|p dx
i=1 Ωδ N
|∂i u|p dx.
i=1
Ω
It now suffices to let δ → 0+ . Step 2: By the previous step and by replacing u with u − uΩ , we may 1 assume that u ∈ C (Ω) and that Ω u(x) dx = 0. By uniform continuity, given 0 < ε < 1, there exists δ > 0 such that (13.68)
||u(x1 )|p − |u(x2 )|p | + ||∂i u(x1 )|p − |∂i u(x2 )|p | ≤ ε,
(13.69)
|u(x1 ) − u(x2 )| + |∂i u(x1 ) − ∂i u(x2 )| ≤ ε
for all x1 , x2 ∈ Ω with x1 − x2 ≤ δ and all i = 1,. . . , N . By Lemma 13.39 we can decompose Ω into a finite number of pairwise disjoint convex domains Ωn , n = 1, . . . , , such that (13.70)
Ω=
n=1
Ωn ,
u(x) dx = 0,
n = 1, . . . , ,
Ωn
and for every n there is an appropriate coordinate system y = (y , yN ) ∈ RN −1 × R such that the projection of Ωn into the yN -axis is (0, dn ), where dn ≤ diam Ω, and (13.71)
√ √ Ωn ⊂ [0, δ/ N − 1] × · · · × [0, δ/ N − 1] × [0, dn ].
In particular, (13.72)
(y , yN ) − (0, yN ) ≤ δ
for all (y , yN ) ∈ Ωn .
Fix one such Ωn . By (13.68), (13.69), (13.70), and (13.72), (13.73) (13.74) (13.75)
u(y) dy − u(0, yN ) dy ≤ εLN (Ωn ), Ωn Ωn p p |u(y)| dy − |u(0, y )| dy ≤ εLN (Ωn ), N Ωn Ωn p |∂N u(y)| dy − |∂N u(0, yN )|p dy ≤ εLN (Ωn ). Ωn
Ωn
13.2. Poincar´e Inequalities
Let g(yN ) :=
439
χΩn (y , yN ) dy . By Fubini’s theorem,
RN −1
(13.76)
dn
u(0, yN ) dy = Ωn
u(0, yN )g(yN ) dyN ,
0
0
dn
|u(0, yN )| dy = p
(13.77) Ωn
dn
|∂N u(0, yN )| dy = p
(13.78) Ωn
|u(0, yN )|p g(yN ) dyN , |∂N u(0, yN )|p g(yN ) dyN .
0
By Minkowski’s inequality with respect to the measure g(yN ) dyN we have that
dn
|u(0, yN )|p g(yN ) dyN
0
≤
(13.79)
dn
|u(0, yN ) − u(0,dn ) |p g(yN ) dyN
0
1/p
dn
+ 0
|u(0,dn ) |p g(yN ) dyN
1/p
1/p =: I + II,
where dn u(0,dn ) :=
0
u(0, yN )g(yN ) dyN . dn 0 g(yN ) dyN
1
Since the function g N −1 is concave by Remark C.9, we are in a position to apply Proposition 7.33 to conclude that I ≤ cdn
(13.80)
dn
|∂N u(0, yN )|p g(yN ) dyN
1/p
0
1/p |∂N u(0, yN )|p dy Ωn 1/p |∂N u(y)|p dy + (εLN (Ωn ))1/p , ≤ c diam Ω
= cdn
Ωn
where we have used (13.78), (13.75), the inequality (13.81)
(a + b)1/p ≤ a1/p + b1/p
for a, b ≥ 0,
and the fact that dn ≤ diam Ω, in this order.
440
13. Sobolev Spaces: Further Properties
To estimate II, we distinguish two cases. If then by (13.70) and (13.73), II = =
dn 0 dn
u(0, yN )g(yN ) dyN
g(yN ) dyN 0
u(0, yN )g(yN ) dyN −
≤ εLN (Ωn )
Ωn
dn
g(yN ) dyN
0
dn
0
dn
−1+1/p
u(y) dy
−1+1/p
g(yN ) dyN > εLN (Ωn ),
dn
g(yN ) dyN
−1+1/p
0
≤ (εLN (Ωn ))1/p ,
0
while if
dn 0
g(yN ) dyN ≤ εLN (Ωn ), then dn 1/p 0 u(0, yN )g(yN ) dyN dn g(y ) dy II = dn N N 0 0 g(yN ) dyN dn 1/p ≤M g(yN ) dyN ≤ M (εLN (Ωn ))1/p , 0
where M := maxΩ |u| + maxΩ |∂uN | + 1 < ∞, since u ∈ C 1 (Ω). Combining both cases, we have that II ≤ (1 + M )(εLN (Ωn ))1/p .
(13.82)
It now follows from (13.79)–(13.82) that
dn
|u(0, yN )| g(yN ) dyN p
1/p
≤ c diam Ω
0
|∂N u(y)|p dy
1/p
Ωn N
+ c(εL (Ωn ))1/p , where the constant c = c(M, N, p) changes from line to line. By (13.74), (13.81), and the previous inequality,
|u| dy p
1/p
Ωn
1/p |u(0, yN )|p g(yN ) dyN + (εLN (Ωn ))1/p 0 1/p ≤ c diam Ω |∂N u|p dy + c(εLN (Ωn ))1/p . ≤
dn
Ωn
Using a discrete Minkowski inequality in the form
(an + bn )p
n=1
1/p
≤
n=1
apn
1/p +
n=1
bpn
1/p ,
13.2. Poincar´e Inequalities
441
where an , bn ≥ 0, it follows by summing over all n that
1/p 1/p |u|p dx = |u|p dy Ω
n=1 Ωn
≤c
diam Ω
|∂N u| dy p
1/p
N
+ (εL (Ωn ))
1/p
p 1/p
Ωn
n=1
≤ c diam Ω
1/p |∂N u| dy p
+ε
1/p
n=1 Ωn
1/p L (Ωn ) N
n=1
1/p |∂N u|p dy + ε1/p (LN (Ω))1/p . ≤ c diam Ω Ω
Since c does not depend on ε, it suffices to let ε → 0+ .
Remark 13.40. By choosing an appropriate norm in W 1,p (Ω), it can be shown that for p = 1 and p = 2 one can take the constant c in (13.67) to be 1 1 2 and π 2 , respectively (see [22] and [2], respectively), and that there exist convex domains for which these constants are optimal. Next we consider star-shaped sets. We recall that a set E ⊆ RN is starshaped with respect to a point x0 ∈ E if θx + (1 − θ)x0 ∈ E for all θ ∈ (0, 1) and for all x ∈ E. Theorem 13.41 (Poincar´e’s inequality for star-shaped sets). Let m ∈ N, let 1 ≤ p < ∞, and let Ω ⊂ RN be an open star-shaped set with respect to x0 ∈ Ω and such that (13.83)
Q(x0 , 4r) ⊆ Ω ⊆ B(x0 , R)
for some r, R > 0. Then there exists a constant c = c(m, N, p) > 0 such that for all u ∈ W m,p (Ω), and every 0 ≤ k ≤ m − 1, ∇k (u − pΩ (u))Lp (Ω) ≤ cRm−k (R/r)(N −1)(m−k)/p ∇m uLp (Ω) . In particular, if m = 1, for all u ∈ W 1,p (Ω), (13.84)
u − uΩ Lp (Ω) ≤ cR(R/r)(N −1)/p ∇uLp (Ω) .
Proof. As in Proposition 13.34 it is enough to prove the case m = 1. By the Meyers–Serrin theorem (Theorem 11.24), without loss of generality, we may assume that u ∈ C ∞ (Ω) ∩ W 1,p (Ω) and that x0 = 0. Hence, we may write Ω = {ρy ∈ RN : y ∈ S N −1 , 0 ≤ ρ < f (y)}, where f : S N −1 → [0, ∞) is the radial function of Ω, that is, f (y) := sup{ρ ≥ 0 : ρy ∈ Ω}. Note that for all y ∈ S N −1 we have that f (y)y ∈ ∂Ω and 2r ≤ f (y) ≤ R by (13.83).
442
13. Sobolev Spaces: Further Properties
Step 1: Assume that u vanishes in B(0, r). Since u(ry) = 0 for all y ∈ S N −1 , using polar coordinates and the fundamental theorem of calculus, for all r ≤ ρ < f (y) we have that
ρ
u(ρy) = u(ρy) − u(ry) =
∂u (ty) dt, ∂ρ
r
and so by H¨older’s inequality, ρ p p ρ ∂u ∂u p/p |u(ρy)| = (ty) dt ≤ ρ (ty) dt ∂ρ ∂ρ r r ρ ∇u(ty)p dt. ≤ cρp/p p
r
Hence, by Tonelli’s theorem
f (y)
p N −1
|u(ρy)| ρ
f (y)
dρ ≤ c
0
p/p +N −1
ρ
ρ
0
∇u(ty)p dtdρ
r
f (y)
≤c
∇u(ty)
r
=c
Rp+N −1 p+N −1
≤ cR (R/r) p
p
R
ρp+N −2 dρ dt
0
f (y)
∇u(ty)p dt
r
N −1
f (y)
∇u(ty)p tN −1 dt.
r
Integrating with respect to y over S N −1 and using spherical coordinates yields p p N −1 |u(x)| dx ≤ cR (R/r) ∇u(x)p dx. Ω
Ω
Step 2: Assume next that u ∈ C ∞ (Ω) ∩ W 1,p (Ω) is such that uQ(0,4r) = 0. Then by Proposition 13.34,
|u(x)| dx ≤ cr p
(13.85) Q(0,4r)
∇u(x)p dx.
p Q(0,4r)
Consider the function ⎧ if x ∈ B(0, r), ⎨ 0 (x − r)/r if x ∈ B(0, 2r) \ B(0, r), ϕ(x) := ⎩ 1 if x ∈ / B(0, 2r)
13.2. Poincar´e Inequalities
443
and write u = (1 − ϕ)u + ϕu =: u1 + u2 . Since u2 vanishes in B(0, r), by the previous step |u2 |p dx ≤ cRp (R/r)N −1 ∇u2 p dx Ω Ω u∇ϕ + ϕ∇up dx = cRp (R/r)N −1 Ω p N −1 −p p p N −1 (13.86) ≤ cR (R/r) r |u| dx + cR (R/r) ∇up dx Q(0,4r) Ω p N −1 ∇up dx, ≤ cR (R/r) Ω
where we have used the facts that ∂i ϕ = 0 outside B(0, 2r) and ∂i ϕ∞ ≤ cr−1 , and (13.85). On the other hand, since u1 is zero outside Q(0, 4r), p |u1 | dx = |u1 |p dx Ω Q(0,4r) p |u| dx + c |u2 |p dx ≤c Q(0,4r) Q(0,4r) p p p N −1 ≤ cr ∇u dx + cR (R/r) ∇up dx Q(0,4r) Ω ∇up dx, ≤ cRp (R/r)N −1 Ω
where we have used (13.85) and (13.86). Hence, by Minkowski’s and H¨older’s inequalities p p p |u − uΩ | dx ≤ c |u| dx ≤ c |u1 | dx + c |u2 |p dx Ω Ω Ω Ω ∇up dx. ≤ cRp (R/r)N −1 Ω
Step 3: Finally, if u ∈ C ∞ (Ω) ∩ W 1,p (Ω) is such that uQ(0,4r) = 0, apply the previous step to the function u − uQ(0,4r) to conclude that |u − uΩ |p dx = |u − uQ(0,4r) − (u − uQ(0,4r) )Ω |p dx Ω Ω p N −1 ∇u(x)p dx. ≤ cR (R/r) Ω
This concludes the proof.
˙ m,p (Ω). Exercise 13.42. Let Ω be as in the previous theorem and let u ∈ W Prove that u ∈ W m,p (Ω).
444
13. Sobolev Spaces: Further Properties
For p > 1 the constant Rp (R/r)N −1 in (13.84) can be significantly improved using Muckenhoupt’s weighted norm inequality for the maximal function [180]. Theorem 13.43 (Muckenhoupt). Let J ⊆ R be an interval, let 1 < p < ∞, and let w : J → [0, ∞) be a Lebesgue measurable function. Then there is a constant c > 0 such that for all Lebesgue measurable functions u : J → R, p (13.87) |(M1 u)(x)| w(x) dx ≤ c |u(x)|p w(x) dx J
J
if and only if there is a constant K > 0 such that for all intervals I ⊆ J, p−1 w(x) dx (w(x))−1/(p−1) dx ≤ K(length I)p . I
I
Furthermore, if c and K are the least constants for which the previous inequalities are true, then K ≤ c ≤ 2p K. In the previous theorem, (13.88)
1 (M1 u)(x) := sup y − x y∈J\{x}
y
|u(t)| dt,
x ∈ J.
x
The proof of this result is rather involved and goes beyond the purposes of this book. We refer to [180] for more details. Exercise 13.44. Let 0 < r < R/2 and let (13.89)
w(t) := min{(r + |t|)N −1 , RN −1 },
t ∈ R.
Prove that for all 1 < p < ∞, (13.90) p−1 1 −1/(p−1) w(x) dx (w(x)) dx K(r, R, p, N ) := sup 1 p I interval (L (I)) I I ⎧ N −p if 1 < p < N, ⎨ (R/r) N −1 ≤ (1 + R/r) if p = N, log ⎩ 1 if p > N. Corollary 13.45. Let 1 ≤ p < ∞ and let Ω ⊂ RN be an open star-shaped set with respect to x0 ∈ Ω and such that Q(x0 , 4r) ⊆ Ω ⊆ B(x0 , R) for some r, R > 0. Then there exists a constant c = c(m, N, p) > 0 such that for all u ∈ W m,p (Ω), and every 0 ≤ k ≤ m − 1, ∇k (u − pΩ (u))Lp (Ω) ≤ cR(m−k) (K(r, R, p, N ))(m−k)/p ∇m uLp (Ω) , where K is defined in (13.90). In particular, if m = 1, for all u ∈ W 1,p (Ω), u − uΩ Lp (Ω) ≤ cR(K(r, R, p, N ))1/p ∇uLp (Ω) .
13.2. Poincar´e Inequalities
445
Proof. The only change is in Step 1 of the proof of Theorem 13.41, where instead of using H¨older’s inequality, we write p p ρ ∂u 1 ρ ∂u p p |u(ρy)| = (ty) dt ≤ R (ty) dt ρ − r r ∂ρ r ∂ρ ≤ Rp |(M1 g)(ρ)|p , where
∂u g(ρ) := (ρy) χ(r,f (y)) (ρ), ρ > 0, ∂ρ and M1 g is defined in (13.88). In turn, since u = 0 in B(0, r), f (y) f (y) p N −1 |u(ρy)| ρ dρ = |u(ρy)|p ρN −1 dρ 0
r
≤R
≤R
f (y)
p
|(M1 g)(ρ)|p ρN −1 dρ
r f (y)
p
|(M1 g)(ρ)|p w(ρ) dρ
r
p ∂u (ρy) w(ρ) dρ ∂ρ r f (y) p ∂u = Rp K(r, R, p, N ) (ρy) ρN −1 dρ, ∂ρ r ≤ R K(r, R, p, N ) p
f (y)
where w and K(r, R, p, N ) are defined in (13.89) and (13.90), respectively, and we have used (13.87). Integrating with respect to y over S N −1 yields p p |u(x)| dx ≤ cR K(r, R, p, N ) ∇u(x)p dx Ω
Ω
and completes the proof.
The proof of Theorem 13.41 can be adapted to the case in which Ω is the subgraph of a lower semicontinuous function. Theorem 13.46. Let m ∈ N, let 1 ≤ p < ∞, let 0 < δ ≤ r < R, let f : QN −1 (0, δ) → [r, R] be a lower semicontinuous function, and let (13.91)
Ω := {(x , xN ) ∈ QN −1 (0, δ) × R : −2r < xN < f (x )}.
Then there exists a constant c = c(m, N, p) > 0 such that for all u ∈ W m,p (Ω), and every 0 ≤ k ≤ m − 1, ∇k (u − pΩ (u))Lp (Ω) ≤ cR(m−k) ∇m uLp (Ω) . In particular, if m = 1, for all u ∈ W 1,p (Ω), u − uΩ Lp (Ω) ≤ cR∇uLp (Ω) .
446
13. Sobolev Spaces: Further Properties
Proof. The proof is similar to the one of Theorem 13.41 and thus we indicate only the main changes. As in that theorem it suffices to prove the case m = 1 and to assume that u ∈ C ∞ (Ω) ∩ W 1,p (Ω). Step 1: Assume that u vanishes in QN −1 (0, δ)×(−2r, −r). Since u(x , −r) = 0 for all x ∈ QN −1 (0, δ), using polar coordinates and the fundamental theorem of calculus, for all −r ≤ xN < f (x ) we have that u(x , xN ) = xN older’s inequality, −r ∂N u(x , t) dt, and so by H¨ xN p/p p |∂N u(x , t)|p dt. |u(x , xN )| ≤ xN −r
Hence, by Tonelli’s theorem and the fact that f ≤ R, f (x ) xN f (x ) p/p p |u(x , xN )| dxN ≤ xN |∂N u(x , t)|p dtdxN −2r
≤
−2r f (x ) −2r
≤ Rp
−r
|∂N u(x , t)|
p
f (x ) t
f (x ) −2r
p−1 xN dxN dt
|∂N u(x , t)|p dt.
Integrating with respect to x over QN −1 (0, δ) yields |u(x)|p dx ≤ Rp |∂N u(x)|p dx. Ω
Ω ∞ C (Ω)∩W 1,p (Ω)
is such that uT = 0, where Step 2: Assume next that u ∈ T := QN −1 (0, δ) × (−r, 0). Then by Proposition 13.34, p p |u(x)| dx ≤ cr ∇u(x)p dx. (13.92) T
T
Consider the function
⎧ if xN ≤ −r, ⎨ 0 (r + xN )/r if − r ≤ xN ≤ 0, ϕ(x) := ⎩ 1 if xN ≥ 0
and write u = (1 − ϕ)u + ϕu =: u1 + u2 . Since u2 vanishes in QN −1 (0, δ) × (−2r, −r), as in (13.86), p p |u2 | dx ≤ cR ∇up dx, Ω
Ω
where we have used Minkowski’s inequality, the fact that ∂N ϕ = 0 outside T , and (13.92). We can now continue as in Step 2 of the proof of Theorem 13.41 to conclude that |u1 |p dx ≤ cRp ∇up dx, Ω
Ω
13.2. Poincar´e Inequalities
and in turn,
447
|u − uΩ | dx ≤ cR p
Ω
∇up dx.
p Ω
Step 3: Finally, if u ∈ C ∞ (Ω) ∩ W 1,p (Ω) is such that uT = 0, then we can reason as in Step 3 of the proof of Theorem 13.41 to conclude the proof. ˙ m,p (Ω). Exercise 13.47. Let Ω be as in the previous theorem and let u ∈ W m,p Prove that u ∈ W (Ω). In Proposition 13.34 and Theorems 13.36, 13.41, 13.46 we have proved Poincar´e’s inequality for several important classes of special domains. We now consider the case in which Ω can be written as union of overlapping sets in which Poincar´e’s inequality holds. A typical example is the case in which Ω is an open bounded, connected set of class C or more regular (see Definition 9.57). Then we can write Ω as a union of sets of the type (13.91) and of balls or cubes. In each of these subdomains Poincar´e’s inequality holds. To obtain Poincar´e’s inequality in Ω we then use a chaining argument. Chaining techniques are very useful in treating more general domains such as s-John domains (see, e.g., [106]). Theorem 13.48. Let m ∈ N, let 1 ≤ p < ∞, and let Ω ⊂ RN be a bounded, connected, open set. Assume that Ω can be written as Ω = n=1 Ωn , where each Ωn is a bounded, connected, open set, such that for all u ∈ W m,p (Ωn ), and every 0 ≤ k ≤ m − 1, (13.93)
∇k (u − pΩn (u))Lp (Ωn ) ≤ cn ∇m uLp (Ωn )
for some constant cn = cn (m, N, p, Ωn ) > 0. Then there is a constant c = c(m, N, p, Ω) > 0 such that for all u ∈ W m,p (Ω), and every 0 ≤ k ≤ m − 1, (13.94)
∇k (u − pΩ (u))Lp (Ωn ) ≤ c∇m uLp (Ωn ) .
Proof. As in Proposition 13.34 it is enough to prove the case m = 1. For every n = 1, . . . , , consider a ball Bn ⊆ Ωn . By H¨older’s inequality and (13.81), u − uΩ Lp (Ω) ≤ u − uB1 Lp (Ω) + |uΩ − uB1 |(LN (Ω))1/p ≤ 2u − uB1 Lp (Ω) ≤ 2
u − uB1 Lp (Ωn ) .
n=1
Thus, it suffices to prove there is kn > 0 such that (13.95)
u − uB1 Lp (Ωn ) ≤ kn ∇uLp (Ωn ) .
Since Ω is pathwise connected, there is a polygonal path in Ω joining the center of B1 with the center of Bn . In turn, we can find a chain of balls in Ω joining B1 to Bn . To be precise, we can find balls Bk∗ , k = 1, . . . , n , such
448
13. Sobolev Spaces: Further Properties
∗ that B1∗ = B1 , B∗n = Bn , and Bk∗ ∩ Bk+1 = ∅ for every k ∈ {1, . . . , n − 1}. Then
u − uB1 Lp (Ωn ) ≤ u − uBn Lp (Ωn ) +
n
∗ |uBk∗ − uBk+1 |(LN (Ωn ))1/p .
k=1
For the first term on the right-hand side of the previous inequality we use H¨older’s inequality and (13.93) to conclude u − uBn Lp (Ωn ) ≤ u − uΩn Lp (Ωn ) + |uΩn − uBn |(LN (Ωn ))1/p ≤ 2u − uΩn Lp (Ωn ) ≤ 2cn ∇uLp (Ωn ) . ∗ . Then by H¨ On the other hand, fix a ball B(xk , rk ) ⊂ Bk∗ ∩ Bk+1 older’s inequality ∗ |uBk∗ − uBk+1 |(LN (Ωn ))1/p ∗ ≤ |uBk∗ − uB(xk ,rk ) |(LN (Ωn ))1/p + |uB(xk ,rk ) − uBk+1 |(LN (Ωn ))1/p 1/p ≤ u − uB(xk ,rk ) Lp (Bk∗ ) LN (Ωn )/LN (Bk∗ ) 1/p N N ∗ ∗ + u − uB(xk ,rk ) Lp (Bk+1 ) L (Ωn )/L (Bk+1 ) ! 1/p ≤ c Rk (Rk /rk )(N −1)/p LN (Ωn )/LN (B ∗ ) 1/p " ∗ + Rk+1 (Rk+1 /rk )(N −1)/p LN (Ωn )/LN (Bk+1 ) ∇uLp (Ω) ,
∗ where in the last inequality we used the fact that Bk∗ and Bk+1 are starshaped with respect to B(xk , rk ) and Theorem 13.41, and where Rk is the radius of Bk∗ . By combining the last three inequalities we obtain (13.95).
Remark 13.49. It follows from the proof of the previous theorem that for m = 1 the constant c in (13.94) can be bounded from above by c∗ := 2
n=1
cn + c1 (m, n, p)
n −1
Rk,n (Rk,n /rk,n )(N −1)/p ×
n=1 k=1
1/p ∗ ) , LN (Ωn )/LN (Bk,n
∗ joining B to where for each n, Rk,n are the radii of the chain of balls Bk,n 1 ∗ ∗ . Bn and rk,n are the radii of the balls B(xk,n , rk,n ) ⊂ Bk,n ∩ Bk+1,n
Remark 13.50. By combining Theorems 13.46 and 13.48 it follows that Poincar´e’s inequality holds for open, connected, bounded sets with continuous boundary.
13.3. Interpolation Inequalities in Domains
449
13.3. Interpolation Inequalities in Domains In this section, we extend some of the results of Section 12.5. We begin by extending Theorem 12.85 for m = 2 to a rectangle R. Note that if R is ˙ 2,p (R), with p ≥ q, then we can use Poincar´e’s bounded and u ∈ Lq (R) ∩ W inequality (see Proposition 13.34) to conclude that u and ∂i u belong to Lp (R). Thus the interesting case is p < q. Theorem 13.51. Let R := I1 × · · · × IN ⊆ RN , where each Ii ⊆ R is an open interval, let 1 ≤ p, q, r ≤ ∞ be such that (13.96)
1 1 1 = + , r 2p 2q
˙ 2,p (R). Then there is a constant c = c(N, p, q) > 0 and let u ∈ Lq (R) ∩ W such that (13.97)
1/2
1/2
∂i uLr (R) ≤ cuLq (R) ∂i2 uLp (R)
for all i = 1, . . . , N , provided all the intervals Ii have infinite length, while (13.98) ∂i uLr (R) ≤ c−1 (LN (R))1/r−1/q uLq (R) + cuLq (R) ∂i2 uLp (R) 1/2
1/2
if all the intervals Ii have finite length and p ≤ q, where := mini L1 (Ii ). We begin with a preliminary lemma. Lemma 13.52. Let Ω ⊂ RN be an open bounded set, let 1 ≤ p, q, r ≤ ∞ be such that p < q and (13.96) hold, and let > 0. Assume that u ∈ ˙ 2,p (Ω) satisfies the inequality Lq (Ω) ∩ W (13.99)
∇uLr (Ω) ≤ c1/r−1/q−1uLq (Ω) + c−1−1/q (LN (Ω))1/r q/r
1/2
1/2
+ cuLq (Ω) ∇2 uLp (Ω) for some constant c = c(N, p, q) > 0. Then ∇uLr (Ω) ≤ c1 −1 (LN (Ω))1/r−1/q uLq (Ω) 1/2
1/2
+ c1 uLq (Ω) ∇2 uLp (Ω) for some constant c1 = c1 (N, p, q) > 0. Proof. By replacing u with su in (13.99) we obtain q/r
∇uLr ≤ csq/r−1 1/r−1/q−1uLq
+ cs−1 −1−1/q (LN (Ω))1/r + cuLq ∇2 uLp . 1/2
1/2
450
13. Sobolev Spaces: Further Properties
We now distinguish two cases. If ∇2 uLp ≤ auLq , where a > 0, we take s := bu−1 Lq , with b > 0, in the previous inequality to obtain ∇uLr ≤ cbq/r−1 1/r−1/q−1 + cb−1 −1−1/q (LN (Ω))1/r + ca1/2 uLq . On the other hand, if ∇2 uLp > auLq (R) then we take (r−2q)/[2(q−r)]
s := duLq
r/[2(q−r)]
∇2 uLp
,
−(r−2q)/[2(q−r)]
∇2 uLp
where d > 0, to obtain ∇uLr ≤ cd−1 −1−1/q (LN (Ω))1/r uLq
1/2
−r/[2(q−r)]
1/2
+ c(dq/r−1 1/r−1/q−1 + 1)uLq ∇2 uLp ≤ cd−1 a−r/[2(q−r)] −1−1/q (LN (Ω))1/r uLq 1/2
1/2
+ c(dq/r−1 1/r−1/q−1 + 1)uLq ∇2 uLp . We now choose a := −2 (LN (Ω))2/r−2/q ,
b := −1/q (LN (Ω))1/q ,
d := (r+r/q−1)/(q−r)
to obtain the desired inequality. We turn to the proof of Theorem 13.51.
Proof of Theorem 13.51. We only give the proof in the case p, q, r < ∞ and leave the other cases as an exercise. Step 1: Assume that R is bounded, that u ∈ C ∞ (R), and that p ≤ q. In this case p ≤ r ≤ q. Let i ∈ {1, . . . , N } and let Ri ⊆ RN −1 be obtained from R by removing the ith set in the cartesian product. Then for xi ∈ Ri , the one-dimensional function u(xi , ·) satisfies the hypotheses of Theorem 7.41, and so,
r/q r 1−r−r/q q |∂i u(xi , xi )| dxi ≤ c |u(xi , xi )| dxi Ii
Ii
|u(xi , xi )|q dxi
+c
r/(2q)
Ii
|∂i2 u(xi , xi )|p dxi
r/(2p) ,
Ii
where we have used the notation (E.2). If r < q we use the inequality tr/q ≤ t + 1, which holds for all t ≥ 0, to obtain |∂i u(xi , xi )|r dxi ≤ c1−r−r/q |u(xi , xi )|q dxi + c1−r−r/q Ii
Ii
|u(xi , xi )|q dxi
+c Ii
r/(2q)
|∂i2 u(xi , xi )|p dxi Ii
r/(2p) .
13.3. Interpolation Inequalities in Domains
451
Integrate the previous inequality in xi over Ri and use Tonelli’s theorem to get r 1−r−r/q |∂i u| dx ≤ c |u|q dx + c−r−r/q LN (R) R
R
+c
Ri
≤ c
|u(xi , xi )|q dxi Ii
r/(2q)
|∂i2 u(xi , xi )|p dxi
r/(2p)
|u|q dx + c−r−r/q LN (R) + cuLq (R) ∂i2 uLp (R) , r/2
1−r−r/q R
dxi
Ii r/2
where in the last inequality we have used H¨ older’s inequality and (13.96). r/2 We are now in a position to apply Lemma 13.52 with ∇2 uLp replaced by r/2 ∂i2 uLp to get (13.96). The additional hypothesis that u ∈ C ∞ (R) can be removed using Exercise 11.29. The proof is significantly simpler in the case p = q or in the case of R unbounded and is left as an exercise. Next we consider uniformly Lipschitz continuous domains (see Definition 13.11). Exercise 13.53. Given L > 0, consider the sector ΞL := S N −1 ∩ {x = (x , xN ) ∈ RN −1 × R : Lx N −1 < xN }. Prove that there exists a constant c = c(L, N, p) > 0 such that |x · y|p dHN −1 (y) ≥ cxp for every x ∈ RN . ΞL
Theorem 13.54 (Gagliardo–Nirenberg interpolation, m = 2). Let Ω ⊆ RN be an open set with uniformly Lipschitz continuous boundary (with parameters ε, L, M ), let 0 < < ε/(4(1 + L)), and let 1 ≤ p, q, r ≤ ∞ be such that p ≤ q and (13.96) holds. If p < q assume further that Ω is bounded. Then there is a constant c = c(L, N, p, q) > 0 such that for every ˙ 2,p (Ω), if p < q, u ∈ Lq (Ω) ∩ W (13.100) ∇uLr (Ω) ≤ c−1 (LN (Ω))1/r−1/q uLq (Ω) + cuLq (Ω) ∇2 uLp (Ω) 1/2
1/2
if p < q, while if p = q, (13.101)
∇uLp (Ω) ≤ c−1 uLp (Ω) + cuLp (Ω) ∇2 uLp (Ω) . 1/2
1/2
Proof. Step 1: Assume that Ω is bounded. We only give the proof in the case p < q < ∞ and leave the case q = ∞ as an exercise. Assume first that u ∈ C ∞ (Ω). Given > 0 and ξ ∈ S N −1 , we define Ω(ξ, ) as the set of all points x ∈ Ω belonging to a segment parallel to ξ, contained in
452
13. Sobolev Spaces: Further Properties
Ω, and with length greater than . We claim that there exists a constant c = c(L, N, r) > 0 such that r ∇u dx ≤ c |∇u(x) · ξ|r dxdHN −1 (ξ). S N −1
Ω
Ω(ξ,)
To see this, we use Tonelli’s theorem to write (13.102) r N −1 |∇u(x) · ξ| dxdH (ξ) = S N −1
Ω(ξ,)
Ω
|∇u(x) · ξ|r dHN −1 (ξ)dx,
G(x,)
where G(x, ) := {ξ ∈ S N −1 : x ∈ Ω(ξ, )}. Fix x ∈ Ω. There are two cases: If dist(x, ∂Ω) ≥ , then B(x, ) ⊂ Ω, and so G(x, ) = S N −1 . In this case, |∇u(x) · ξ|r dHN −1 (ξ) = cN,r ∇u(x)r , (13.103) G(x,)
where we used the fact that for every z ∈ RN \ {0}, by a rotation, z r r N −1 r |z · ξ| dH (ξ) = z · ξ dHN −1 (ξ) z N −1 N −1 S S |e1 · ξ|r dHN −1 (ξ) =: zr cN,r . = zr S N −1
On the other hand, if dist(x, ∂Ω) < , then there exists x0 ∈ ∂Ω with x − x0 < . By parts (i) and (iii) of Definition 13.11, there exist n ∈ N, local coordinates y = (y , yN ) ∈ RN −1 × R, and a Lipschitz continuous function f : RN −1 → R (both depending on n), with Lip f ≤ L, such that B(x0 , ε) ⊆ Ωn and Ωn ∩ Ω = Ωn ∩ Vn , where Vn is given in local coordinates by {(y , yN ) ∈ RN −1 × R : yN > f (y )}. In particular, B(x0 , ε) ∩ Ω = B(x0 , ε) ∩ Vn . Since < ε/(4(1 + L)), we have that x ∈ B(x0 , ε/[4(1 + L)]). y , f (¯ y )). Let (¯ y , y¯N ) be the local coordinates of x and consider the point (¯ N −1 , we have Since f is Lipschitz continuous, with Lip f ≤ L, for y ∈ R that f (y ) ≤ f (¯ y ) + Ly − y¯ N −1 < y¯N + Ly − y¯ N −1 . Hence, the cone y , ε/[4(1 + L)]) × R : Kx = {(y , yN ) ∈BN −1 (¯ y¯N + Ly − y¯ N −1 < yN < y¯N + ε/4} is contained in B(x0 , ε) ∩ Ω. Using local coordinates, consider the sector Ξx := {ξ = (ξ , ξN ) ∈ S N −1 : Lξ N −1 < ξN }.
13.3. Interpolation Inequalities in Domains
453
Since < ε/(4(1 + L)), if t ∈ (0, ) and ξ ∈ Ξx , then the point x + tξ belongs to Kx ⊆ Ω. Hence, G(x, ) ⊇ Ξx , and so |∇u(x) · ξ|r dHN −1 (ξ) ≥ |∇u(x) · ξ|r dHN −1 (ξ) ≥ cN,r,L ∇u(x)r , G(x,)
Ξx
where we have used the previous exercise. Combining this inequality with (13.102) and (13.103) proves the claim. Step 2: Let Ω be bounded, p < q < ∞, and u ∈ C ∞ (Ω). Given ξ ∈ S N −1 and 0 < < ε/(4(1+L)), consider the hyperplane Πξ := {x ∈ RN : x·ξ = 0} and for y ∈ Πξ define the slice Ω(ξ, , y) := {t ∈ R : y + tξ ∈ Ω(ξ, )}. Note that if y ∈ Πξ is such that Ω(ξ, , y) is nonempty, then Ω(ξ, , y) is the union of a finite family of pairwise disjoint open intervals with length greater than and we may define the function uy,ξ : Ω(ξ, , y) → R by uy,ξ (t) := u(y + tξ). Let Ii be one such interval. Applying Theorem 7.41 to the function uy,ξ in Ii we obtain
r/q r 1−r−r/q q |∇u(y + tξ) · ξ| dt ≤ c |u(y + tξ)| dt Ii
+c
Ii
1/(2q) 2 p 1/(2p) ∂ u q |u(y + tξ)| dt . 2 (y + tξ) dt Ii Ii ∂ξ
Using the inequality tr/q ≤ t + 1, which holds for all t ≥ 0, and the fact that L1 (Ii ) ≥ , we get |∇u(y + tξ) · ξ|r dt Ii 1−r−r/q |u(y + tξ)|q dt + c−r−r/q L1 (Ii ) ≤ c Ii
1/(2q)
1/(2p)
|u(y + tξ)| dt
||∇ u(y + tξ) dt
q
+c Ii
2
p
.
Ii
Summing over all Ii in Ω(ξ, , y) and using the inequality (7.17) gives |∇u(y + tξ) · ξ|r dt Ω(ξ,,y) |u(y + tξ)|q dt + c−r−r/q L1 (Ω(ξ, , y)) ≤ c1−r−r/q Ω(ξ,,y)
1/(2q)
1/(2p)
|u(y + tξ)ξ · ξ| dt
||∇ u(y + tξ) dt
q
+c Ω(ξ,,y)
2
p
.
Ω(ξ,,y)
Integrating both sides of the previous inequality over all y ∈ Πξ for which Ω(ξ, , y) is nonempty, by Tonelli’s theorem, H¨older’s inequality and (13.96),
454
13. Sobolev Spaces: Further Properties
we get
|u(x)|q dx + c−r−r/q LN (Ω)
|∇u(x) · ξ|r dx ≤ c1−r−r/q Ω(ξ,)
Ω
+
r/2 r/2 cuLq (Ω) ∇2 uLp (Ω) .
Integrating both sides of the previous inequality in the variable ξ over S N −1 and using Step 1 gives ∇ur dx ≤ c1−r−r/q |u|q dx + c−r−r/q LN (Ω) Ω
+
Ω r/2 r/2 cuLq (Ω) ∇2 uLp (Ω) .
We are now in a position to apply Lemma 13.52 to obtain (13.100). The additional hypothesis that u ∈ C ∞ (Ω) can be removed using Exercise 11.29. We omit the details. Step 3: The case q = p and Ω not necessarily bounded is simpler. Indeed, we do not need to use the inequality tr/q ≤ t + 1 and so the term c−r−r/q LN (Ω) does not appear. The details are left as an exercise. Remark 13.55. Note that in the previous proof we only used a uniform cone property. Remark 13.56. We remark that when Ω ⊂ RN is an open, bounded, ˙ 2,p (Ω), where connected set with Lipschitz boundary and if u ∈ Lq (Ω) ∩ W 1 ≤ q ≤ p, then it follows from Poincar´e’s inequality that u and ∇u are in Lp (Ω) and so in all Lr (Ω) for 1 ≤ r ≤ p. This is why we only considered the interesting case p < q in the previous theorem. Remark 13.57. We recall that if Ω ⊂ RN is an an open bounded set with Lipschitz continuous boundary, then in view of Exercise 13.13 there exist ε > 0, L > 0, and M ∈ N such that Ω has uniformly Lipschitz continuous boundary with parameters ε, L, and M . Exercise 13.58. Let Ω ⊂ RN be an open bounded set with Lipschitz continuous boundary, let 0 < < ε/(4(1 + L)), let m ∈ N, with m ≥ 2, and let 1 ≤ p, q, r ≤ ∞ be such that p < q and
1 1 1 1 1 + 1− = . mp m q r Prove there exists a constant c = c(m, N, p, q) > 0 such that for every ˙ m,p (Ω), u ∈ Lq (Ω) ∩ W ∇uLr (Ω) ≤ c−1 (LN (Ω))1/r−1/q uLq (Ω) + cuLq (Ω) ∇m uLp (Ω) . 1−1/m
Hint: Repeat the proof of the previous theorem.
1/m
13.3. Interpolation Inequalities in Domains
455
An important consequence of the previous theorem is the following result. Corollary 13.59. Let Ω ⊆ RN be an open set with uniformly Lipschitz ˙ 2,p (Ω) continuous boundary and let 1 ≤ p ≤ ∞. Then W 2,p (Ω) = Lp (Ω) ∩ W 2 with equivalence of the norms k=0 ∇k uLp (Ω) and uLp (Ω) + ∇2 uLp (Ω) . Remark 13.60. Note that for sets of infinite measure the previous corollary cannot be obtained using the Poincar´e inequality. Next we consider the case m ≥ 2. Theorem 13.61 (Gagliardo–Nirenberg interpolation, m ≥ 2). Let Ω ⊆ RN be an open set with uniformly Lipschitz continuous boundary (with parameters ε, L, M ), let 0 < < ε/(4(1 + L)), let m, k ∈ N, with m ≥ 2 and 1 ≤ k < m, and let 1 ≤ p, q, r ≤ ∞ be such that p ≤ q and
k 1 1 k 1 + 1− = . (13.104) mp m q r If p < q, assume further that Ω is bounded. Then for every u ∈ Lq (Ω) ∩ ˙ m,p (Ω), W (13.105) 1−k/m k/m ∇k uLr (Ω) ≤ c−k (LN (Ω))1/r−1/q uLq (Ω) + cuLq (Ω) ∇m uLp (Ω) if p < q, while (13.106)
∇k uLp (Ω) ≤ c−k uLp (Ω) + cuLp (Ω) ∇m uLp (Ω) 1−k/m
k/m
if p = q. Here, c > 0 is a constant depending on m, N , p, q. Proof. We prove the case p < q. By hypothesis Ω is bounded. The proof is by induction on m. When m = 2 and k = 1, (13.105) follows from Theorem 14.47. Let n ∈ N, with n ≥ 2, and assume that (13.105) holds for all 1 ≤ p, q ≤ ∞ with p < q all m ≤ n, all 1 ≤ k < m, all r given by (13.104), ˙ m,p (Ω) and for every multi-index β ∈ NN , with |β| = k. all u ∈ Lq (Ω) ∩ W 0 Fix 1 ≤ p, q ≤ ∞ with p < q. We claim that (13.105) holds with m = n + 1. To be precise, let 1 ≤ k < n + 1 and let r be given by
k 1 1 k 1 + 1− = (13.107) n+1p n+1 q r ˙ n+1,p (Ω). We claim that there exists a constant and let u ∈ Lq (Ω) ∩ W c = c(k, n, N, p, q) > 0 such that (13.108) 1−k/(n+1) k/(n+1) ∇n+1 uLp . ∇k uLr ≤ c−k (LN (Ω))1/r−1/q uLq + cuLq
456
13. Sobolev Spaces: Further Properties
If k = 1, this follows from Exercise 13.58. Thus, assume that k ≥ 2. Since 1r belongs to the segment of endpoints 1q and 1p by (13.107), we can find r1 ≥ 1 such that
1 1 1 11 + 1− = . (13.109) kr k q r1 Using (13.107) and (13.109), one can verify that
k−1 1 1 k−11 + 1− = . (13.110) n p n r1 r By Exercise 13.58, ∇uLr1 ≤ c−1 (LN (Ω))1/r1 −1/q uLq + cuLq
1−1/k
1/k
∇k uLr .
Given a multi-index β ∈ NN 0 , with |β| = k, let i ∈ {1, . . . , N } be such that β = ei + δ, with |δ| = k − 1. Hence, by the induction hypothesis applied to the function ∂i u in place of u, 1−(k−1)/n
∂ β uLr ≤ c1−k (LN (Ω))1/r−1/r1 ∂i uLr1 +c∂i uLr1
(k−1)/n
∇n (∂i u)Lp
.
Combining the last two inequalities and summing over all β with |β| = k gives ∇k uLr ≤ c−k (LN (Ω))1/r−1/q uLq 1−1/k
+ c1−k (LN (Ω))1/r−1/r1 uLq
1/k
∇k uLr
+ c−1+(k−1)/n(LN (Ω))(1/r1 −1/q)(1−(k−1)/n)uLq
1−(k−1)/n
(1−(k−1)/n)(1−1/k)
+ cuLq
(1/k)(1−(k−1)/n)
∇k uLr
(k−1)/n
∇n+1 uLp
(k−1)/n
∇n+1 uLp
=: I + II + III + IV. We now use Young’s inequality in the form ab ≤ εat + cε bt for a, b ≥ 0, ε > 0, 1 < t < ∞. Taking ε =
1 3
and t = k we have
1 II ≤ ∇k uLr + c−k (LN (Ω))1/r−1/q uLq , 3 where we used (13.109). Similarly, with ε = that t = nk/[(n + 1)(k − 1)],
1 3
and t = nk/(n + 1 − k) so
1 1−k/(n+1) k/(n+1) ∇n+1 uLp . IV ≤ ∇k uLr + cuLq 3 Finally, with ε = 1 and t = kn/(n + 1 − k), so that t = kn/[(n + 1)(k − 1)], and writing 1−(k−1)/n
uLq
(n+1−k)/(nk)
= uLq
(1−1/k)(n+1−k)/n
uLq
,
13.3. Interpolation Inequalities in Domains
457
we get III ≤ c−k (LN (Ω))1/r−1/q uLq + cuLq
1−k/(n+1)
k/(n+1)
∇n+1 uLp
,
where we used (13.109). Combining the last four inequalities gives (13.105). The case p = q is simpler and is left as an exercise.
As in the case m = 2, the inequality (13.105) implies the following important consequence. Corollary 13.62. Let Ω ⊆ RN be an open set with uniformly Lipschitz continuous boundary, let m ∈ N with m ≥ 2 and let 1 ≤ p ≤ ∞. Then m,p p m,p k ˙ W (Ω) = L (Ω)∩W (Ω) with equivalence of the norms m k=0 ∇ uLp (Ω) m and uLp (Ω) + ∇ uLp (Ω) .
Chapter 14
Functions of Bounded Variation Living with P.Q.S, I: What is PQS? “Post-Quals-Slump” or PQS, affects 99% of all grad students. It is the #1 cause of delayed graduation dates and contributes to the Ph.D. degree drop rate. — Jorge Cham, www.phdcomics.com
In this chapter we introduce the space of functions of bounded variation in domains of RN and study some of their basic properties. What is covered here is just the tip of the iceberg. We refer the interested reader to the monographs [10], [72], [76], [95], [149], and [251] for more information on the subject.
14.1. Definition and Main Properties Definition 14.1. Let Ω ⊆ RN be an open set. We define the space of functions of bounded variation BV (Ω) as the space of all functions u ∈ L1 (Ω) whose distributional first-order partial derivatives are finite signed Radon measures; that is, for all i = 1, . . . , N there exists a finite signed measure λi : B(Ω) → R such that u∂i φ dx = − φ dλi (14.1) Ω
Ω
Cc∞ (Ω).
for all φ ∈ The measure λi is called the weak, or distributional, partial derivative of u with respect to xi and is denoted Di u. Here B(Ω) is the Borel σ-algebra in Ω. We define BVloc (Ω) := {u ∈ L1loc (Ω) : u ∈ BV (U ) for all open sets U Ω}. 459
460
14. Functions of Bounded Variation
For u ∈ BV (Ω) we set Du := (D1 u, . . . , DN u). Thus, if u ∈ BV (Ω), then Du ∈ Mb (Ω; RN ) (see (B.26)) and so the total variation measure of Du, defined by ∞ (14.2) Du(E) := sup Du(En ) , E ∈ B(Ω), n=1
where the supremum is taken over all partitions {En }n in B(Ω) of E, is a finite Radon measure (see Proposition B.74). Moreover, since Mb (Ω; RN ) may be identified with the dual of C0 (Ω; RN ) (see Theorem B.111), we have that Du(Ω) = DuMb (Ω;RN ) N N = sup Φi dDi u : Φ ∈ C0 (Ω; R ), ΦC0 (Ω) ≤ 1 < ∞. i=1
Ω
Definition 14.2. Let Ω ⊆ RN be an open set and let u ∈ L1loc (Ω). The variation of u in Ω is defined by N ∞ N ∂i Φi u dx : Φ ∈ Cc (Ω; R ), ΦC0 (Ω) ≤ 1 . V (u, Ω) := sup i=1
Ω
Exercise 14.3. Let Ω ⊆ RN be an open set and let u ∈ L1loc (Ω). Prove the following. (i) If the distributional gradient Du of u belongs to Mb (Ω; RN ), then Du(Ω) = V (u, Ω). (ii) If V (u, Ω) < ∞, then the distributional gradient Du of u belongs to Mb (Ω; RN ). In particular, if u ∈ L1 (Ω), then u belongs to BV (Ω) if and only if V (u, Ω) < ∞. Hint: Use the Riesz representation theorem in C0 (Ω; RN ). (iii) If {un }n is a sequence of functions in L1loc (Ω) converging to u in L1loc (Ω), then V (u, Ω) ≤ lim inf V (un , Ω). n→∞
Exercise 14.4. Let Ω ⊆ RN be an open set. Prove that the space BV (Ω) is a Banach space with the norm uBV (Ω) := uL1 (Ω) + Du(Ω). It follows from the definition of BV (Ω) that W 1,1 (Ω) ⊂ BV (Ω). We have already seen that for N = 1 the inclusion is strict. Exercise 14.5 shows that this is also the case in higher dimensions.
14.1. Definition and Main Properties
461
Exercise 14.5. Let Ω ⊆ RN be an open set and let E ⊂ RN , N ≥ 2, be a bounded set with C 2 boundary. (i) Prove that χE ∈ / W 1,1 (Ω). (ii) Prove that D(χE )(Ω) ≤ HN −1 (∂E ∩ Ω). Hint: Use Exercise 14.3(i). (iii) Prove that the normal ν to ∂E may be extended to a function C 1 (RN ; RN ). (iv) Prove that (14.3)
D(χE )(Ω) = HN −1 (∂E ∩ Ω).
The previous example shows that characteristic functions of smooth sets belong to BV (Ω). More generally, we have the following. Definition 14.6. Let E ⊆ RN be a Lebesgue measurable set and let Ω ⊆ RN be an open set. The perimeter of E in Ω, denoted P(E, Ω), is the variation of χE in Ω, that is, P(E, Ω) := V (χE , Ω) N ∞ N ∂i Φi dx : Φ ∈ Cc (Ω; R ), ΦC0 (Ω) ≤ 1 . = sup i=1
E
The set E is said to have finite perimeter in Ω if P(E, Ω) < ∞. If Ω = RN , we write P(E) := P(E, RN ).
(14.4)
Remark 14.7. In view of Exercise 14.3, if Ω ⊆ RN is an open set and E ⊆ RN is a Lebesgue measurable set with LN (E ∩ Ω) < ∞, then χE belongs to BV (Ω) if and only if P(E, Ω) < ∞. Note that Exercise 14.5 shows that for smooth sets E the perimeter of E in Ω is simply HN −1 (∂E ∩ Ω). The next exercise shows that this fact no longer holds if the boundary of the set E is not (sufficiently) smooth. Exercise 14.8. Let QN = {xn : n ∈ N} and let U ⊂ RN be the open set U :=
∞
B(xn , 1/2n ).
n=1
(i) Prove that
LN (U )
< ∞ but LN (∂U ) = ∞.
(ii) Prove that HN −1 (∂U ) = ∞. (iii) Prove that m U has finiten perimeter. Hint: Define um := χUm , where Um := n=1 B(xn , 1/2 ), and use Exercise 14.3(iii).
462
14. Functions of Bounded Variation
14.2. Approximation by Smooth Functions Since every function u ∈ C ∞ (Ω) ∩ BV (Ω) belongs to W 1,1 (Ω) (why?) and ∇u dx,
Du(Ω) = Ω
the closure of C ∞ (Ω) ∩ BV (Ω) in BV (Ω) is W 1,1 (Ω). Thus, we cannot expect the Meyers–Serrin density theorem (see Theorem 11.24) to hold in BV (Ω). However, the following weaker version holds. Theorem 14.9. Let Ω ⊆ RN be an open set and let u ∈ BV (Ω). Then there exists a sequence {un }n in C ∞ (Ω) ∩ W 1,1 (Ω) such that un → u in L1 (Ω) and ∇un dx = Du(Ω). lim n→∞ Ω
We begin with an auxiliary result. Lemma 14.10. Let Ω ⊆ RN be an open set and let u ∈ BV (Ω). For every ε > 0 define uε := ϕε ∗ u, where ϕε is a standard mollifier and let Ωε be defined as in (C.6). Then (14.5)
∇uε dx = Du(Ω).
lim
ε→0+
Ωε
Moreover, for every open set U ⊂ Ω, with dist(U, ∂Ω) > 0 and Du(∂U ) = 0, (14.6)
∇uε dx = Du(U ).
lim
ε→0+
U
Proof. Step 1: For every i = 1, . . . , N , define λi := Di u ∈ Mb (Ω). By Theorem C.20, for every x ∈ Ωε we have (14.7)
∂uε (x) = ∂xi
Ω
∂ i ϕε (x − y)u(y) dy = − ∂xi
Ω
∂ϕε (x − y)u(y) dy ∂yi
ϕε (x − y) dλi (y),
= Ω
where we have used (14.1) and the fact that the function ϕε (x − ·) belongs to Cc∞ (Ω), since supp ϕε (x − ·) ⊆ B(x, ε).
14.2. Approximation by Smooth Functions
463
Let Φ ∈ Cc∞ (Ωε ; RN ) be such that ΦC0 (Ωε ) ≤ 1 and extend Φ to be zero outside Ωε . By (14.7) and an integration by parts, we have ∂Φi ∂uε uε (x) (x) dx = − (x)Φi (x) dx ∂x ∂xi i Ωε Ωε Φi (x)ϕε (x − y) dλi (y)dx =− Ωε Ω ∂(Φi )ε = − (Φi )ε (y) dλi (y) = (y) u(y) dy, Ω Ω ∂yi where we have used Fubini’s theorem and (14.1). Hence, uε div Φ dx = u div Φε dx. (14.8) Ωε
Ω
Moreover, for each x ∈ Ωε , by Jensen’s inequality (see Theorem B.50), which can be applied since Ω ϕε (x − y) dy = 1, and by the fact that Φ ≤ 1, we have N N 2 2 ((Φi )ε ) (x) = Φi (y)ϕε (x − y) dy i=1
i=1
≤
(14.9)
Ω
N i=1
Φ2i (y)ϕε (x − y) dy Ω
Φ(y) ϕε (x − y) dy ≤
ϕε (x − y) dy = 1,
2
= Ω
Ω
and (since supp Φ ⊂ Ωε ) (14.10)
supp Φε ⊂ {x ∈ Ω : dist(x, supp Φ) < ε} ⊂ Ω.
Thus, by (14.8) and Exercise 14.3, uε div Φ dx = u div Φε dx ≤ Du(Ω). Ωε
Ω
Taking the supremum over all admissible Φ and using Exercise 14.3 once more, this time applied to Ωε , yields (14.11)
Duε (Ωε ) ≤ Du(Ω).
In particular, uε ∈ BV (Ωε ) ∩ C ∞ (Ωε ). Letting ε → 0+ , we conclude that (14.12)
lim sup Duε (Ωε ) ≤ Du(Ω). ε→0+
To prove the converse inequality, fix an open set U ⊂ Ω with the property that dist(U, ∂Ω) > 0. Since uε → u in L1 (U ) as ε → 0+ (see Theorem C.16), by Exercise 14.3, (14.13)
Du(U ) ≤ lim inf Duε (U ). ε→0+
464
14. Functions of Bounded Variation
Since Ωε ⊃ U for all 0 < ε < dist(U, ∂Ω), it follows that Du(U ) ≤ lim inf Duε (Ωε ). ε→0+
Letting U Ω and using Proposition B.9 gives Du(Ω) ≤ lim inf Duε (Ωε ),
(14.14)
ε→0+
which, together with (14.12), gives (14.5). Step 2: Fix an open set U ⊂ Ω, with dist(U, ∂Ω) > 0 and Du(∂Ω) = 0. In view of (14.13), it remains to show that lim sup Duε (U ) ≤ Du(U ). ε→0+
We proceed as in the previous step, with the only change that we now consider Φ ∈ Cc∞ (U ; RN ) such that ΦC0 (U ) ≤ 1. If 0 < ε < dist(U, ∂Ω), then by (14.10), supp Φε ⊂ U ε := {x ∈ Ω : dist(x, U ) < ε} ⊂ Ω, and so (14.11) should be replaced by Duε (U ) ≤ Du(U ε ). Letting ε → 0+ and using Proposition B.9, we conclude that lim sup Duε (U ) ≤ Du(U ) = Du(U ), ε→0+
since by hypothesis Du(∂U ) = 0.
Remark 14.11. If Ω = RN , then Ωε = RN , and so uε → u in L1 (RN ) and by (14.6), ∇uε dx = Du(RN ). (14.15) lim ε→0+
RN
Note that the previous proof continues to hold if u ∈ L1loc (RN ) and its distributional gradient Du belongs to Mb (RN ; RN ). The only difference is that now uε → u in L1loc (RN ). We observe that if u = χE , then (14.15) gives ∇(χE )ε dx = P(E). lim ε→0+
RN
This identity can be used to define the perimeter of a set. This approach was taken by De Giorgi [58], [59], who considered a somewhat different family of mollifiers.
14.2. Approximation by Smooth Functions
465
Exercise 14.12. Let Ω ⊆ RN be an open set, let u ∈ BV (Ω), and let {un }n be a sequence of functions in BV (Ω) converging in L1loc (Ω) to u and such that Du(Ω) = lim Dun (Ω). n→∞
∗
Prove that Di un Di u in Mb (Ω) for all i = 1, . . . , N and that Du(U ) = lim Dun (U ) n→∞
for every open set U ⊆ Ω, with Du(∂U ∩ Ω) = 0. Exercise 14.13. Let Ω ⊆ RN be an open set and let u ∈ BV (Ω). Fix 1 ≤ i ≤ N. (i) Prove that
∞ |Di u|(Ω) = sup φdDi u : φ ∈ Cc (Ω), φC0 (Ω) ≤ 1 . Ω
(ii) Prove that
|∂i uε | dx ≤ |Di u|(Ω) Ωε
for all ε > 0, that
|∂i uε | dx = |Di u|(Ω),
lim
ε→0+
Ωε
and that for every open set U ⊂ Ω, with dist(U, ∂Ω) > 0 and |Di u|(∂U ) = 0, |∂i uε | dx = |Di u|(U ). lim ε→0+
(iii)
U
Prove that if u ∈ L1loc (RN ) and longs to Mb (RN ; RN ), then
its distributional gradient Du be-
lim
ε→0+
RN
|∂i uε | dx = |Di u|(RN ).
Exercise 14.14. Let Ω ⊆ RN be an open set. Prove that if u ∈ BV (Ω) and φ ∈ Cc∞ (Ω), then φu ∈ BV (Ω) and D(φu)(Ω) ≤ φC0 (Ω) Du(U ) + ∇φC0 (Ω) uL1 (U ) , where U ⊆ Ω is any open set such that supp φ ⊂ U . We now turn to the proof of Theorem 14.9. Proof of Theorem 14.9. Since the measure Du is finite, lim Du(Ω \ {x ∈ Ω : dist(x, ∂Ω) > 1/j, x < j}) = 0.
j→∞
466
14. Functions of Bounded Variation
Fix η > 0 and let j0 ∈ N be so large that Du(Ω \ {x ∈ Ω : dist(x, ∂Ω) > 1/j, x < j}) ≤ η
(14.16)
for all j ≥ j0 . For i ∈ N define Ωi := {x ∈ Ω : dist(x, ∂Ω) > 1/(j0 + i), x < j0 + i}. As in the proof of the Meyers–Serrin density theorem (see Theorem 11.24), consider a smooth partition of unity F subordinated to the open cover {Ωi+1 \ Ωi−1 }, where Ω−1 = Ω0 := ∅. For each i ∈ N let ψi be the sum of all the finitely many ψ ∈ F such that supp ψ ⊂ Ωi+1 \ Ωi−1 and such that they have not already been selected at previous steps j < i. For every i ∈ N find εi > 0 so small that supp(ψi u)εi ⊂ Ωi+1 \ Ωi−1 and, using properties of mollifiers (see Theorems C.16 and C.20),
|(ψi u)εi − ψi u| dx ≤
(14.17) Ω
η , 2i
(u∇ψi )εi − u∇ψi dx ≤ Ω
η . 2i
Define vη := ∞ i=1 (ψi u)εi , where we have extended each (ψi u)εi to be zero outside its support. As in the Meyers–Serrin theorem we have that vη ∈ ∞ C (Ω) and (see the proof of (11.10)), Ω |u − vη | dx ≤ η. Thus, vη → u in L1 (Ω) as η → 0+ . Next we prove that lim sup Dvη (Ω) ≤ Du(Ω).
(14.18)
η→0+
Let Φ ∈ Cc∞ (Ω; RN ) be such that ΦC0 (Ω) ≤ 1. Since by the previous exercise ψi u ∈ BV (Ω) for every i ∈ N, by (14.8) with ψi u in place of u we have (ψi u)εi div Φ dx = ψi u div Φεi dx, Ω
Ω
and so, using the facts that the support of Φ is compact and that the partition of unity is locally finite, we have that (14.19)
vη div Φ dx = Ω
=
∞
(ψi u)εi div Φ dx =
i=1 Ω ∞ i=1
Ω
∞ i=1
ψi u div Φεi dx Ω
u[div(ψi Φεi ) − ∇ψi · Φεi ] dx =: I + II.
14.2. Approximation by Smooth Functions
467
Since ψi Φεi C0 (Ω) ≤ 1 by (14.9), it follows from Exercise 14.3 applied to Ω and to Ωi+1 \ Ωi−1 (see also (11.8)) that ∞ I= u div(ψ1 Φε1 ) dx + u div(ψi Φεi ) dx Ω
i=2
≤ Du(Ω) +
∞
Ω
Du(Ωi+1 \ Ωi−1 ).
i=2
Using telescopic series, the right-hand side of the previous inequality may be bounded by Du(Ω) + 3Du(Ω \ Ω1 ) ≤ Du(Ω) + 3η by (14.16), and so I ≤ Du(Ω) + 3η.
(14.20)
To estimate II, note that by Fubini’s theorem ∞ ∞ II = − (u∇ψi )εi · Φ dx = − [(u∇ψi )εi − u∇ψi ] · Φ dx, i=1
Ω
i=1
Ω
where in the second identity we have used the fact that ∞ i=1 ∇ψi = 0 by (11.6) and the local finiteness of the partition of unity. We now use (14.17) and the fact that ΦC0 (Ω) ≤ 1 to conclude that II ≤ η, which, together with (14.19) and (14.20), yields vη div Φ dx ≤ Du(Ω) + 3η. Ω
Taking the supremum over all admissible Φ and using Exercise 14.3 gives Dvη (Ω) ≤ Du(Ω) + 3η. It now suffices to let η → 0+ to obtain (14.18). To prove the converse inequality, note that, since vη → u in L1 (Ω) as η → 0+ , by Exercise 14.3, Du(Ω) ≤ lim inf Dvη (Ω). η→0+
This concludes the proof.
Remark 14.15. Note that the previous proof continues to hold if u ∈ L1loc (Ω) and its distributional gradient Du belongs to Mb (Ω; RN ). The only 1,1 (Ω) and uε → u in L1loc (Ω). difference is that now {un }n is in C ∞ (Ω) ∩ Wloc Exercise 14.16. Prove that the sequence constructed in Theorem 14.9 has the property that |∂i un | dx = |Di u|(Ω) lim ε→0+
for every i = 1, . . . , N .
Ω
468
14. Functions of Bounded Variation
Remark 14.17. If N ≥ 2 and u ∈ L1loc (RN ) is such that its distributional gradient Du belongs to Mb (Ω; RN ), then reasoning as in Step 3 of the proof of Theorem 11.43, one can find a sequence {un }n in Cc∞ (RN ) such that ∇un dx = Du(RN ). lim n→∞ RN
14.3. Bounded Pointwise Variation on Lines In Theorem 11.45 we have seen that if a function u belongs to W 1,p (Ω), then it admits a representative that is absolutely continuous on LN −1 -a.a. line segments of Ω that are parallel to the coordinate axes and whose first-order (classical) partial derivatives agree LN -a.e. with the weak derivatives of u. In this section we prove a similar result for a function in BV (Ω). In what follows, we use the notation (E.2) in Appendix E. Given an open rectangle R = (a1 , b1 ) × · · · × (aN , bN ) ⊂ RN , following (E.2), we consider the rectangle R to be the Cartesian product of a rectangle Ri ⊂ RN −1 of the variable xi and of an interval Ri ⊂ R in the xi variable and we write R = Ri × Ri . Given xi ∈ RN −1 and a set E ⊆ RN , we write Exi := {xi ∈ R : (xi , xi ) ∈ E}.
(14.21)
Moreover, if v is a real-valued function defined LN -a.e. on an open set Ω ⊆ RN , for all xi ∈ RN −1 for which Ωxi is nonempty we let essVarΩx v(xi , ·) i
be the essential pointwise variation of the function xi ∈ Ωxi → v(xi , ·) (see Definition 7.8 and Exercise 7.10). If v is Lebesgue integrable in Ω, with a slight abuse of notation, if Ωxi is empty, we set Ω v(xi , xi ) dxi := 0, so x
that by Fubini’s theorem v(x) dx = (14.22) Ω
RN −1
i
v(xi , xi ) dxi dxi . Ωx
i
Finally, for every set F in some Euclidean space Rm and for every ε > 0 we set (14.23)
Fε := {y ∈ F : dist(y, ∂F ) > ε}.
Let Ω ⊆ RN be an open set and let u ∈ BV (Ω). Setting λ := Du ∈ Mb (Ω; RN ), by Theorem B.76, the vectorial measure λ can be expressed in the form Ψ(x) dx + λs (E) (14.24) λ(E) = E
for every Borel set E ⊆ Ω, where Ψ ∈ L1 (Ω; RN ) and λs is a vector-valued measure singular with respect to the Lebesgue measure. Here Ψ is the
14.3. Bounded Pointwise Variation on Lines
dDu of the vectorial dLN Ω N L restricted to Ω.
Radon–Nikodym derivative to the Lebesgue measure
469
measure Du with respect
The following result shows the relation between the weak derivatives of a function in BV (Ω) and the (classical) partial derivatives of its representative (when they exist). Theorem 14.18. Let Ω ⊆ RN be an open set, let u ∈ BV (Ω), and let 1 ≤ i ≤ N . If v is any function equivalent to u for which the (classical) partial derivative ∂i v(x) exists in R for LN -a.e. x ∈ Ω, then ∂i v(x) = Ψi (x) for LN -a.e. x ∈ Ω, where Ψ is the vectorial function given in (14.24). Proof. Consider a convex function f : R → [0, ∞) such that f (s + t) ≤ f (s) + |t|
(14.25)
for all s, t ∈ R. We claim that for every rectangle R whose closure is contained in Ω, f (∂i v) dx ≤ lim inf f (∂i uε ) dx. (14.26) R
ε→0+
Rε
To begin with, for every function φ ∈ C ∞ (R), for x = (xi , xi ) ∈ R and for all h > 0 sufficiently small by Jensen’s inequality (see Theorem B.50) one has 1 xi +h φ(x , x + h) − φ(x , x ) i i i i =f f ∂i φ(xi , s) ds h h xi xi +h 1 f (∂i φ(xi , s)) ds. ≤ h xi Integrating this inequality along the interval (Ri )h (see (14.23)) and using Tonelli’s theorem yields φ(x , x + h) − φ(x , x ) i i i i dxi f h (Ri )h xi +h 1 ≤ f (∂i φ(xi , s)) dsdxi h (Ri )h xi s 1 ≤ f (∂i φ(xi , s)) dxi ds = f (∂i φ(xi , s)) ds. h Ri s−h Ri Integrating in xi and using Tonelli’s theorem, one then obtains φ(x , x + h) − φ(x , x ) i i i i dx ≤ f f (∂i φ(x)) dx. h Rh R
470
14. Functions of Bounded Variation
Setting φ = uε in the previous inequality, replacing R with Rε , and then letting ε → 0+ , we deduce from Fatou’s lemma that u(x , x + h) − u(x , x ) i i i i dx ≤ lim inf f f (∂i uε (x)) dx. h ε→0+ Rh Rε Since v is a representative of u, the previous inequality may be written as v(x , x + h) − v(x , x ) i i i i dx ≤ lim inf f f (∂uε (x)) dx. h ε→0+ Rh Rε If we now let h → 0+ and use Fatou’s lemma, we obtain (14.26). Next we prove that f (∂i uε ) dx ≤ f (Ψi ) dx + |λsi |(R), (14.27) Rε
R
where |λsi | is the total variation measure of λsi . By (14.7) and (14.24) for x ∈ Rε we have ∂i uε (x) = ϕε (x − y)Ψi (y) dy + ϕε (x − y) dλsi (y). Ω
Ω
Therefore by virtue of (14.25) and Jensen’s inequality (see Theorem B.50) ϕε (x − y)Ψi (y) dy + ϕε (x − y) dλsi (y) f (∂i uε (x)) ≤ f Ω Ω ϕε (x − y)f (Ψi (y)) dy + ϕε (x − y) d|λsi |(y). ≤ Ω
Ω
Integrating over Rε and applying Tonelli’s theorem yields (14.27). Combining (14.26) and (14.27), we obtain f (∂i v) dx ≤ f (Ψi ) dx + |λsi |(R). (14.28) R
R
By Besicovitch’s derivation theorem (see Theorem B.116) lim
r→0+
|λsi |(Q(x0 , r)) =0 rN
for LN -a.e. x0 ∈ Ω. Let x0 be any such point and assume that x0 is also a Lebesgue point for f (∂i v) and f (Ψi ). By replacing R with Q(x0 , r) in (14.28), dividing by rN , and letting r → 0+ , we obtain (14.29)
f (∂i v(x0 )) ≤ f (Ψi (x0 )).
Now choose f to be the function t e if t < 0, f1 (t) := t + 1 if t ≥ 0.
14.3. Bounded Pointwise Variation on Lines
471
Since f1 is monotone increasing, it follows from (14.29) that ∂i v(x) ≤ Ψi (x) for LN -a.e. x ∈ Ω. Similarly, setting f (t) := f1 (−t), we find that ∂i v(x) ≥ Ψi (x) for LN -a.e. x ∈ Ω. Next we prove that Theorem 11.45 continues to hold for BV (Ω). Definition 14.19. Let Ω ⊆ RN be an open set. A function u : Ω → R is called of bounded pointwise variation in the sense of Cesari if there exist an equivalent function v defined LN -a.e. in Ω and N nonnegative functions V1 , . . . , VN ∈ L1 (RN −1 ) such that essVarΩx v(xi , ·) ≤ Vi (xi )
(14.30)
i
for all xi ∈ RN −1 for which Ωxi is nonempty and for all 1 ≤ i ≤ N . Theorem 14.20 (Serrin). Let Ω ⊆ RN be an open set and let u ∈ L1 (Ω). Then u ∈ BV (Ω) if and only if u is of bounded pointwise variation in the sense of Cesari. Moreover, if u ∈ BV (Ω), then u has a representative u ¯ that N u(x) = Ψ(x) for admits (classical) partial derivatives L -a.e. in Ω and ∇¯ LN -a.e. x ∈ Ω, where Ψ is the vectorial function given in (14.24). We begin with some preliminary results, which are of interest in themselves. Lemma 14.21. Let R ⊂ RN be an open rectangle, let i ∈ {1, . . . , N }, and let u : R → R be a Lebesgue measurable function that is monotone on LN −1 a.e. line parallel to the xi axis. Then the (classical) partial derivative ∂i u exists and is finite LN -a.e. in R. Moreover, ∂i u is Lebesgue measurable. Proof. For x = (xi , xi ) ∈ R and n ∈ N we define the upper right (respectively, left) sequential derivative D ± u(x) := lim sup n→∞
u(xi , xi ± 1/n) − u(xi , xi ) ±1/n
and the lower right (respectively, left) sequential derivative D± u(x) := lim inf n→∞
u(xi , xi ± 1/n) − u(xi , xi ) . ±1/n
Then the four functions D ± u and D± u are Lebesgue measurable. Moreover, by Lebesgue’s differentiation theorem (Theorem 1.18), for LN −1 -a.e. xi ∈ Ri ∂u we have that ∂x (xi , xi ) exists and is finite for L1 -a.e. xi ∈ Ri . By Tonelli’s i theorem D ± u and D± u are finite LN -a.e. in R and |D ± u(x) − D± u(x)| dx R |D ± u(xi , xi ) − D± u(xi , xi )| dxi dxi = 0. = Ri
Ri
472
14. Functions of Bounded Variation
It follows that both limits u(xi , xi + 1/n) − u(xi , xi ) , n→∞ 1/n lim
u(xi , xi − 1/n) − u(xi , xi ) n→∞ −1/n lim
exist in R and are equal LN -a.e. in R. Let v(x) be the common limit where it exists; otherwise put v(x) := ∞. Let E be the set of points x = (xi , xi ) ∈ R such that v(x) is finite and u(xi , ·) is monotone. Note that LN (R \ E) = 0. If x = (xi , xi ) ∈ E, with, say, u(xi , ·) increasing, and 1/(n + 1) ≤ h ≤ 1/n, then u(xi , xi + h) − u(xi , xi ) u(xi , xi + 1/(n + 1)) − u(xi , xi ) ≤ 1/n h u(xi , xi + 1/n) − u(xi , xi ) . ≤ 1/(n + 1) Since both the right- and left-hand sides of the above inequality tend to v(x) as h → 0+ , we conclude that lim
h→0+
u(xi , xi + h) − u(xi , xi ) = v(x). h
Similarly, we have that lim
h→0−
u(xi , xi + h) − u(xi , xi ) = v(x) h
and so ∂i u(x) = v(x), which shows that ∂i u exists and is finite LN -a.e. in R. Moreover, ∂i u is Lebesgue measurable. Lemma 14.22. Let R ⊂ RN be an open rectangle and let u be a Lebesgue integrable real-valued function defined LN -a.e. in R and such that for some 1 ≤ i ≤ N and for all xi ∈ Ri , (14.31)
essVarRi u(xi , ·) ≤ Vi (xi )
for some nonnegative Lebesgue integrable function Vi defined in Ri . Then there exist two Lebesgue integrable functions u1 and u2 on R, each increasing in the variable xi , such that u(x) = u1 (x) − u2 (x) for LN -a.e. x ∈ R. In particular, the (classical) partial derivative ∂i u exists and is finite LN -a.e. in R. Proof. We construct a function v equivalent to u but with fewer discontinuities. Let Si be the set of points xi ∈ Ri such that u(xi , ·) is defined for L1 -a.e. xi ∈ Ri and Vi (xi ) < ∞. By hypothesis LN −1 (Ri \ Si ) = 0.
14.3. Bounded Pointwise Variation on Lines
473
For every xi ∈ Si , essVarRi u(xi , ·) < ∞ by (14.31) and the definition of Si . Hence, by Theorems 7.3 and 7.9 we may find a right continuous function w(xi , ·) such that (14.32)
w(xi , xi ) = u(xi , xi )
for L1 -a.e. xi ∈ Ri
and VarRi w(xi , ·) = essVarRi u(xi , ·) < ∞. For x = (xi , xi ) ∈ R define w(xi , xi ) if xi ∈ Si , (14.33) v(x) := 0 otherwise. Since LN −1 (Ri \ Si ) = 0, it follows by (14.32) and by Tonelli’s theorem that v(x) = u(x) for LN -a.e. x ∈ R. By the completeness of the Lebesgue measure, the function v is Lebesgue measurable. Now let a ∈ Ri be such that |v(xi , a)| dxi < ∞ Ri
(such a point exists by Fubini’s theorem), and for any fixed xi ∈ Ri let V (xi , ·) be the indefinite pointwise variation of the function xi ∈ Ri → v(xi , xi ), normalized so that V (xi , a) ≡ 0, that is (see (2.3)), Var[a,xi ] v(xi , xi ) if xi ≥ a, V (xi , xi ) := − Var[xi ,a] v(xi , xi ) if xi < a. We claim that V is Lebesgue measurable. To see this, denote by Pk the partition of the interval Ri by k + 1 equally spaced points t0 < · · · < tk and for x = (xi , xi ) ∈ R define |v(xi , tj ) − v(xi , tj−1 )| sgn(tj − a), V k (xi , xi ) := where the sum is extended over all partition points tj which lie in the interval of endpoints a and xi . Since v(xi , ·) is right continuous, it follows by Proposition 2.21 that V (xi , xi ) = lim V k (xi , xi ) k→∞
for all x = (xi , xi ) ∈ R. Thus, to prove the measurability of V , it remains to show that each V k is Lebesgue measurable. This follows from the fact that, for each xi fixed, the function V k (xi , ·) is a step function. More precisely, for xi between any two consecutive points of Pk we have that V k (xi , xi ) = finite sum of Lebesgue measurable functions of xi . Hence, we have shown that V is Lebesgue measurable. We can now define (14.34)
u1 (x) := 12 [V (x) + v(x)],
u2 (x) := 12 [V (x) − v(x)]
474
14. Functions of Bounded Variation
for x = (xi , xi ) ∈ R. The functions u1 and u2 are Lebesgue measurable, increasing in the variable xi , and u = u1 − u2 LN -a.e. in R. Furthermore, by Proposition 2.11, |u1 (x)| + |u2 (x)| ≤ (VarRi u1 (xi , ·) + 12 |v(xi , a)|) + (VarRi u2 (xi , ·) + 12 |v(xi , a)|) = VarRi v(xi , ·) + |v(xi , a)| ≤ Vi (xi ) + |v(xi , a)|, and so u1 and u2 are Lebesgue integrable. The last part of the statement follows by applying the previous lemma to the functions u1 and u2 . Remark 14.23. Note that by Proposition 2.21 for every xi ∈ Ri , for every xi ∈ Ri the functions u1 (xi , ·) and u2 (xi , ·) defined in (14.34) cannot both be discontinuous at xi . We now turn to the proof of Theorem 14.20. Proof of Theorem 14.20. Step 1: We prove that if u is of locally bounded pointwise variation in the sense of Cesari, then u ∈ BVloc (Ω). Let v, V1 , . . . , VN be as in Definition 14.19. For φ ∈ Cc1 (Ω) and for every fixed i, 1 ≤ i ≤ N , set u∂i φ dx.
L(φ) := Ω
The functional L is linear. We claim that it is continuous with respect to the topology of C0 (Ω). To see this, let us first observe that, since essVarΩx v(xi , ·) ≤ Vi (xi ) for all xi ∈ RN −1 for which Ωxi is nonempty, for i
LN −1 -a.e. xi ∈ RN −1 we have that Vi (xi ) (and, in turn, essVarRi v(xi , ·)) is finite. Thus, by Theorem 7.9 for LN −1 -a.e. xi ∈ RN −1 for which Ωxi is nonempty the function v(xi , ·) belongs to BV (Ωxi ), and so there exists a measure λxi ∈ Mb (Ωxi ) such that ∂i v(xi , ·) = λxi and |λxi |(Ωxi ) = essVarΩx v(xi , ·) ≤ Vi (xi ), i
again by Theorem 7.9. Using (14.22), it follows that u∂i φ dx = u∂i φ dxi dxi L(φ) = Ω
=−
RN −1
RN −1
Ωx
Ωx
i
φ dλxi dxi ,
i
and so
|L(φ)| ≤ max |φ| Ω
RN −1
Vi dxi .
14.3. Bounded Pointwise Variation on Lines
475
Since L is continuous on C0 (Ω), by the Hahn–Banach theorem (see Theorem A.32) it may be extended to a continuous linear functional on C0 (Ω). But then by the Riesz representation theorem in C0 (Ω) (see Theorem B.111) there exists a finite Radon measure λi ∈ Mb (Ω) such that L(φ) = − Ω φ dλi for all φ ∈ C0 (Ω) and, by the previous inequality, Vi dxi . |λi |(Ω) = L(C0 (Ω)) ≤ RN −1
This shows that u ∈ BV (Ω). Step 2: Assume that u ∈ BVloc (Ω). We prove that u is of bounded pointwise variation in the sense of Cesari. Let {ϕε }ε>0 be a sequence of standard mollifiers and define uε := u ∗ ϕε in Ωε . Set E := {x ∈ Ω : lim uε (x) exists in R}
(14.35)
ε→0+
and
(14.36)
lim uε (x) if x ∈ E,
ε→0+
v(x) :=
0
otherwise.
Since {uε }ε converges pointwise to u at every Lebesgue point of u by Theorem C.16, we have that E contains every Lebesgue point of u. Moreover, since by Corollary B.119 the complement in Ω of the set of Lebesgue points has Lebesgue measure zero, it follows that LN (Ω \ E) = 0. This shows that the function v is a representative of u. Let i be a fixed integer, 1 ≤ i ≤ N . By Exercise 14.13 we get |∂i uε | dx ≤ |Di u|(Ω). (14.37) Ωε
We now take εn := 1/n and for simplicity we set un := uεn , (Ωxi )n := (Ωxi )εn , and so on. For xi ∈ RN −1 define Vi (xi )
(14.38)
:= lim inf n→∞
(Ωx )n
|∂i un (xi , xi )| dxi
i
Vi (xi )
if Ωxi is nonempty and := 0 otherwise. The function Vi is Lebesgue N −1 . By Fatou’s lemma, Fubini’s theorem, and (14.37), measurable on R Vi (xi ) dxi ≤ lim inf |∂i un (xi , xi )| dxi dxi RN −1
n→∞
RN −1
i
|∂i un | dx ≤ |Di u|(Ω).
= lim inf n→∞
(Ωx )n
Ωn
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14. Functions of Bounded Variation
Hence,
(14.39)
RN −1
Vi (xi ) dxi ≤ |Di u|(Ω).
Since LN (Ω \ E) = 0, by Fubini’s theorem L1 (Ωxi \ Exi ) = 0
(14.40)
for LN −1 -a.e. xi ∈ RN −1 for which Ωxi is nonempty. Fix any such xi ∈ RN −1 and let Ri ⊂ R be any maximal open interval such that {xi } × Ri ⊂ Ωxi . We claim that essVarRi v(xi , ·) ≤ Vi (xi ).
(14.41)
By (14.40), (xi , xi ) ∈ E for L1 -a.e. xi ∈ Ri . Thus, to prove (14.41), it is enough to consider partitions contained in Exi (see Definition 7.8 and Exercise 7.10). Let inf Ri < t0 < t1 < · · · < tk < sup Ri , where tj ∈ Exi for all j = 0, . . . , k. Then [t0 , tk ] ⊂ (Ri )n for all n sufficiently large, and so, by the fundamental theorem of calculus, k k tj−1 |un (xi , tj ) − un (xi , tj−1 )| ≤ |∂i un (xi , xi )| dxi j=1
j=1
tj−1
|∂i un (xi , xi )| dxi .
≤ (Ri )n
Letting n → ∞, by (14.35), (14.36), and (14.38) we get k
|v(xi , tj ) − v(xi , tj−1 )| ≤ Vi (xi ).
j=1
Taking the supremum over all admissible partitions yields (14.41). By eventually redefining Vi to be infinite on a set of LN −1 -measure zero, we have proved that (14.30) holds. This shows that u is of bounded pointwise variation in the sense of Cesari. Step 3: We prove the last part of the theorem. Since the function v defined in (14.36) satisfies the hypotheses of Lemma 14.22, we may find two Lebesgue integrable functions v1 and v2 on R, each increasing in the variable xi , such that v(x) = v1 (x) − v2 (x) for LN -a.e. x ∈ R. Let Si be the set of points xi ∈ Ri such that v1 (xi , xi ) = lim (v1 )ε (xi , xi ), ε→0+
for
LN −1 (Ri
xi ∈ Ri . Note that (xi , xi ) ∈ Si × Ri . We
L1 -a.e.
v2 (xi , xi ) = lim (v2 )ε (xi , xi ) ε→0+
\
Si )
= 0.
assert that if the limit limε→0+ (v1 )ε (xi , xi ) Fix does not exist, then the limit limε→0+ (v2 )ε (xi , xi ) does exist. Indeed, since
14.3. Bounded Pointwise Variation on Lines
477
v1 (xi , ·) and (v1 )ε (xi , ·) are increasing, if the limit limε→0+ (v1 )ε (xi , xi ) does not exist, then (why?) (v1 )− (xi , xi ) ≤ lim inf (v1 )ε (xi , xi ) < lim sup(v1 )ε (xi , xi ) ≤ (v1 )+ (xi , xi ). ε→0+
ε→0+
Thus, v1 (xi , ·) is discontinuous at xi . But then by Remark 14.23, v2 (xi , ·) must be continuous at xi , and hence v2 (xi , xi ) = lim (v2 )ε (xi , xi ). ε→0+
We now define the function u ¯(x) := lim supε→0+ uε (x) for x = (xi , xi ) ∈ R. The function u ¯ is a precise representative of u and is independent of i. We ¯2 by means of also introduce the functions u ¯1 and u lim sup(v1 )ε (xi , xi ) if xi ∈ Si , ε→0+ u ¯1 (x) := v(x) otherwise, lim inf (v2 )ε (xi , xi ) if xi ∈ Si , ε→0+ u ¯2 (x) := 0 otherwise. ¯2 are equivalent to v1 and v2 , respectively, and so they are Then u ¯1 and u increasing on LN −1 -a.e. lines parallel to the xi -axis. Furthermore, from the identity uε = (v1 − v2 )ε = (v1 )ε − (v2 )ε and the assertion of the preceding paragraph it follows that1 u ¯(x) = u ¯1 (x) − u ¯2 (x) for all x ∈ R. By Lemma 14.21 the partial derivatives ∂∂xu¯1i and ∂∂xu¯2i ∂u ¯ exist and are finite LN -a.e. in R. The same is true for ∂x . The fact that i ∇¯ u(x) = Ψ(x) follows from Theorem 14.18. The following exercises are the analog of Exercise 11.51 and Theorem 11.53. 1,1 (RN ). Prove that the Exercise 14.24. Let u ∈ W 1,1 (RN + ) and v ∈ W − N function w : R → R, defined by u(x) if xN > 0, w(x) := v(x) if xN < 0,
belongs to BV (RN ). 1 The key point here is that given two sequences of real numbers {a } and {b } , in general n n n n one can only conclude that
lim sup(an + bn ) ≤ lim sup an + lim sup bn , n→∞
n→∞
n→∞
with the strict inequality possible. However, if one of the two sequences converges, say {an }n , then lim sup(an + bn ) = lim an + lim sup bn . n→∞
n→∞
n→∞
478
14. Functions of Bounded Variation
Exercise 14.25. Let Ω, U ⊆ RN be open sets, let Φ : U → Ω be invertible, with Φ and Φ−1 Lipschitz continuous functions, and let u ∈ BV (Ω). Prove that u ◦ Φ ∈ BV (U ). Exercise 14.26. Let Ω ⊆ RN be an open set, let u ∈ BVloc (Ω), and let f : R → R be a Lipschitz continuous function with f (0) = 0 if LN (Ω) = ∞. Prove that f ◦ u ∈ BVloc (Ω) and that D(f ◦ u)(U ) ≤ Lip f Du(U ) for every open set U ⊆ Ω. Remark 14.27. One can actually show that a chain rule formula holds in BV (Ω). We refer to the paper of Ambrosio and Dal Maso [9] (see also [140]).
14.4. Coarea Formula for BV Functions In this section we prove the coarea formula for a function u ∈ BV (Ω). This formula relates the total variation measure Du with the perimeter of its superlevel sets {x ∈ Ω : u(x) > t}. Theorem 14.28 (Coarea formula). Let Ω ⊆ RN be an open set and let u ∈ L1loc (Ω). Then P({x ∈ Ω : u(x) > t}, Ω) dt. V (u, Ω) = R
In particular, if u ∈ BV (Ω), then the set {x ∈ Ω : u(x) > t} has finite perimeter in Ω for L1 -a.e. t ∈ R and (14.42) Du(Ω) = P({x ∈ Ω : u(x) > t}, Ω) dt. R
We begin by proving the theorem for affine functions. Lemma 14.29. Let u : RN → R be an affine function of the form u(x) = a + b · x, x ∈ RN , where a ∈ R and b ∈ RN . Then for every Lebesgue measurable set E ⊆ RN the function t ∈ R → HN −1 (E ∩ u−1 ({t})) is Lebesgue measurable and N HN −1 (E ∩ u−1 ({t})) dt. (14.43) bL (E) = R
Proof. If b = 0, then there is nothing to prove, since both sides of (14.43) are zero. Thus, assume that b = 0. Hence, u−1 ({t}) = {x ∈ RN : a+b· x = t} is the translate of a hyperplane in RN , and so the measure HN −1 restricted to u−1 ({t}) coincides with the measure LN −1 . It follows by Tonelli’s theorem that the function t ∈ R → HN −1 (E ∩ u−1 ({t}))
14.4. Coarea Formula for BV Functions
479
is Lebesgue measurable. Since LN and HN −1 are both rotation invariant, b = e1 . By the change of without loss of generality, we may assume that b variables t = bs + a we obtain N −1 −1 H (E ∩ u ({t})) dt = b HN −1 (E ∩ u−1 ({bs + a})) ds R R = b HN −1 ({x ∈ E : x1 = s}) ds R = b LN −1 ({x ∈ E : x1 = s}) ds = bLN (E), R
where in the last inequality we have used Tonelli’s theorem.
Remark 14.30. Note that ∇u ≡ b. Exercise 14.31. Let Ω ⊆ RN be an open set and let u ∈ L1loc (Ω). Prove that the function t ∈ R → P({x ∈ Ω : u(x) > t}, Ω) is Lebesgue measurable. We are now ready to prove Theorem 14.28. In what follows, for t ∈ R we set Ωt := {x ∈ Ω : u(x) > t}. Proof of Theorem 14.28. Step 1: We claim that if {un }n is a sequence in L1loc (Ω) converging to u in L1loc (Ω), then there exists a subsequence {unk }k of {un }n such that P(Ωt , Ω) ≤ lim inf P(Ωnt k , Ω) k→∞
for
L1 -a.e.
t ∈ R and P(Ωt , Ω) dt ≤ lim inf P(Ωnt k , Ω) dt, k→∞
R
where
Ωnt
R
:= {x ∈ Ω : un (x) > t}.
Indeed, for every x ∈ Ω and n ∈ N, max{un (x),u(x)} |χΩnt (x) − χΩt (x)| dt = 1 dt = |un (x) − u(x)|. R
min{un (x),u(x)}
Hence, by Fubini’s theorem, for every U Ω, |χΩnt (x) − χΩt (x)| dxdt = |un (x) − u(x)| dx → 0. R
U
U
By considering an increasing sequence Uj Ω, with Uj Ω and using a diagonal argument, we may find a subsequence {unk }k of {un }n such that χΩnk → χΩt in L1loc (Ω) for L1 -a.e. t ∈ R. It follows by Exercise 14.3 that t
P(Ωt , Ω) ≤ lim inf P(Ωnt k , Ω) k→∞
480
14. Functions of Bounded Variation
for L1 -a.e. t ∈ R. In turn, by Fatou’s lemma, P(Ωt , Ω) dt ≤ lim inf P(Ωnt k , Ω) dt. k→∞
R
R
Step 2: We prove that if u is piecewise affine in the sense of Definition 11.39, then D(χΩt )(Ω) dt = ∇u dx. R
Ω
To see this, let Δ1 , . . . , Δ be N -simplexes with pairwise disjoint interiors such that the restriction of u to each Δi is affine, say u(x) = ai + bi · x for x ∈ Δi , where ai ∈ R and bi ∈ RN , and u = 0 outside i=1 Δi . Then by the previous lemma, applied to the affine function ui (x) = ai + bi · x, x ∈ RN , N ∇u dx = bi L (Ω ∩ Δi ) = HN −1 (Ω ∩ Δi ∩ u−1 ({t})) dt. Ω
i=1
i=1
R
Since by Exercise 14.5 (note that in Ω ∩ Δi the function u coincides with the smooth function v(x) := ai + bi · x, x ∈ RN ) D(χΩt )(Ω ∩ Δi ) = HN −1 (Ω ∩ Δi ∩ u−1 ({t})), we have
i=1
R
H
N −1
−1
(Ω ∩ Δi ∩ u
({t})) dt =
i=1
=
R
R
D(χΩt )(Ω ∩ Δi ) dt
D(χΩt )(Ω) dt.
1,1 (Ω), then Step 3: We prove that if u ∈ C ∞ (Ω) ∩ Wloc P(Ωt , Ω) dt ≤ ∇u dx. R
Ω
It suffices to assume that Ω ∇u dx < ∞. Fix an open set U Ω. By Remark 11.41 we may find a sequence {un }n in W 1,1 (U ) of piecewise affine functions such that un → u in W 1,1 (U ). By the previous two steps (with Ω replaced by U ) we may find a subsequence (not relabeled) of {un }n such that D(χUt )(U ) dt ≤ lim inf D(χUtn )(U ) dt n→∞ R R ∇un dx = ∇u dx ≤ ∇u dx. = lim n→∞ U
U
Ω
Since χUt = χΩt in U , we have that D(χUt )(U ) = D(χΩt )(U ). Hence, D(χΩt )(U ) dt ≤ ∇u dx. R
Ω
14.4. Coarea Formula for BV Functions
481
By considering an increasing sequence Uk Ω, with Uk Ω, in the previous inequality and using the Lebesgue monotone convergence theorem, we obtain the result. Step 4: We prove that for u ∈ L1loc (Ω), P(Ωt , Ω) dt ≤ V (u, Ω). R
It suffices to assume that V (u, Ω) < ∞. Hence, by Exercise 14.3 we have that the distributional gradient Du of u belongs to Mb (Ω; RN ). By Theorem 1,1 (Ω) 14.9 and Remark 14.15 there exists a sequence {un }n in C ∞ (Ω) ∩ Wloc such that un → u in L1loc (Ω) and ∇un dx = Du(Ω). lim n→∞ Ω
By Step 1 and Step 3 applied to each function un , D(χΩt )(Ω) dt ≤ lim inf D(χΩnt )(Ω) dt n→∞ R R ∇un dx = Du(Ω). ≤ lim n→∞ Ω
Step 5: We claim that for every Φ ∈ Cc∞ (Ω; RN ) with ΦC0 (Ω) ≤ 1, u div Φ dx = div Φ dxdt. R
Ω
Ωt
N To∞ see this, assume first that u ≥ 0. Then for L -a.e. x ∈ Ω, u(x) = 0 χΩt (x) dt, and so, by Fubini’s theorem ∞ u(x) div Φ(x) dx = χΩt (x) div Φ(x) dtdx Ω Ω∞ 0 χΩt (x) div Φ(x) dxdt = 0 Ω ∞ div Φ(x) dxdt. = 0
Ωt
0 Similarly, if u ≤ 0, then for LN -a.e. x ∈ Ω, u(x) = −∞ (χΩt (x) − 1) dt, and so 0 u(x) div Φ(x) dx = (χΩt (x) − 1) div Φ(x) dt dx Ω
Ω −∞ 0
= −∞ Ω 0
(χΩt (x) − 1) div Φ(x) dxdt div Φ(x) dxdt,
= −∞
Ωt
482
14. Functions of Bounded Variation
where we have used the fact that Ω div Φ dx = 0 by the divergence theorem and the fact that Φ ∈ Cc∞ (Ω; RN ). In the general case, write u = u+ − u− . Note that for t > 0, Ωt = {x ∈ Ω : u(x) > t} = {x ∈ Ω : u+ (x) > t}, while for t < 0, Ωt = {x ∈ Ω : u(x) > t} = {x ∈ Ω : −u− (x) > t}. Hence,
(u+ (x) − u− (x)) div Φ(x) dx Ω div Φ(x) dxdt, =
u(x) div Φ(x) dx = Ω
R
Ωt
which proves the claim. Step 6: We show that for u ∈ L1loc (Ω), V (u, Ω) ≤ P(Ωt , Ω) dt. R
Cc∞ (Ω; RN )
with ΦC0 (Ω) ≤ 1. By the previous step and the Let Φ ∈ definition of the variation of the perimeter of Ωt in Ω, u div Φ dx = div Φ dxdt ≤ P(Ωt , Ω) dt. R
Ω
Ωt
R
This concludes the proof.
Remark 14.32. We refer to [72] for a different proof of Step 3, which does not make use of piecewise affine functions.
14.5. Embeddings and Isoperimetric Inequalities In this section we prove that the Sobolev–Gagliardo–Nirenberg embedding theorem continues to hold for functions of bounded variation. We recall that for N ≥ 2, 1∗ = NN−1 . Theorem 14.33 (Sobolev–Gagliardo–Nirenberg embedding in BV ). Let u ∈ L1loc (RN ), N ≥ 2, be a function vanishing at infinity such that its distributional gradient Du belongs to Mb (RN ; RN ). Then there exists a constant c = c(N ) > 0 such that uL1∗ (RN ) ≤ cDu(RN ). In particular, BV (RN ) → Lq (RN ) for all 1 ≤ q ≤ 1∗ .
14.5. Embeddings and Isoperimetric Inequalities
483
∗
Proof. Step 1: Assume that u ∈ L1 (RN ). Consider a sequence of standard mollifiers {ϕε }ε>0 and define uε := u ∗ ϕε . By Lemma 14.10, Remark ∗ 14.11, and Theorem C.16, uε → u in L1 (RN ) and lim ∇uε L1 (RN ) = Du(RN ).
(14.44)
ε→0+
Applying the Sobolev–Gagliardo–Nirenberg embedding theorem (Theorem 12.4) to each uε , we obtain uε L1∗ (RN ) ≤ c∇uε L1 (RN ) . It now suffices to let ε → 0+ and to use (14.44) and Theorem C.16. ∗
Step 2: To remove the additional hypothesis that u ∈ L1 (RN ), we truncate u as in Step 4 of the proof of the Sobolev–Gagliardo–Nirenberg embedding theorem (Theorem 12.4) and use Exercise 14.26. Remark 14.34. If N ≥ 2 and u ∈ L1loc (RN ) is such that its distributional gradient Du belongs to Mb (Ω; RN ), then reasoning as in the proof of Theorem 12.9, one can show that there exists a unique constant cu ∈ R (depending on u) such that u − cu L1∗ (RN ) ≤ cDu(RN ), where c = c(N ) > 0. See Theorem 3.47 in [10] for a different proof of this result. Definition 14.35. An open set Ω ⊂ RN is called an extension domain for the space BV (Ω) if there exists a continuous linear operator E : BV (Ω) → BV (RN ) such that (i) for all u ∈ BV (Ω), E(u)(x) = u(x) for LN -a.e. x ∈ Ω, (ii) E(u) ∈ W 1,1 (RN ) whenever u ∈ W 1,1 (Ω). Next we prove that the Rellich–Kondrachov theorem continues to hold in BV (Ω). Theorem 14.36 (Rellich–Kondrachov). Let Ω ⊂ RN , N ≥ 2, be an extension domain for BV (Ω) with finite measure. Let {un }n be a bounded sequence in BV (Ω). Then there exist a subsequence {unk }n of {un }n and a ∗ function u ∈ BV (Ω)∩L1 (Ω) such that unk → u in Lq (Ω) for all 1 ≤ q < 1∗ . The proof is very similar to that of Theorem 12.18. The following result is the analog of Lemma 12.19. Lemma 14.37. Let u ∈ BV (RN ). Then for all h ∈ RN \ {0}, |u(x + h) − u(x)| dx ≤ hDu(RN ). RN
484
14. Functions of Bounded Variation
Proof. Assume first that u ∈ W 1,1 (RN ) ∩ C ∞ (RN ). Then by Lemma 12.19 we have that for all h ∈ RN \ {0}, |u(x + h) − u(x)| dx ≤ h ∇u(x) dx. RN
RN
To remove the additional hypothesis that u ∈ C ∞ (RN ), it suffices to apply the previous inequality to uε := ϕε ∗u, where ϕε is a standard mollifier, and to let ε → 0+ (see Theorem C.16 and Lemma 14.10). The following result is the analog of Lemma 12.20. Lemma 14.38. Let u ∈ BV (RN ). Consider a standard mollifier ϕ as in (C.7). Then |(u ∗ ϕ1/k )(x) − u(x)| dx ≤ ck −1 Du(RN ) RN
for some constant c = c(N ) > 0. Proof. By Lemma 12.20 (see (12.12)) we have |(u ∗ ϕ1/k )(x) − u(x)| dx RN |u(x + h) − u(x)| dxdh. ≤ ck N B(0,1/k)
RN
In turn, by the previous lemma we get N N |(u ∗ ϕ1/k )(x) − u(x)| dx ≤ ck Du(R ) RN
h dh B(0,1/k)
= ck −1 Du(RN ). We now turn to the proof of the Rellich–Kondrachov theorem. Proof of Theorem 14.36. The proof of the first statement is the same as that for Theorem 12.18 with the only difference that we use the previous lemma in place of Lemma 12.20. It remains to prove that u belongs to BV (Ω). For all i = 1, . . . , N and k ∈ N define λk,i := Di unk . Then supk∈N |λk,i |(Ω) < ∞ for all i = 1, . . . , N . Since Mb (Ω) is the dual of C0 (Ω) and C0 (Ω) is separable, by Corollary A.54 we may select a further subsequence, not relabeled, such that for each i = 1, . . . , N , ∗
λk,i λi
in Mb (Ω)
14.5. Embeddings and Isoperimetric Inequalities
485
for some finite signed Radon measures λ1 ,. . . , λN ∈ Mb (Ω). For every φ ∈ Cc∞ (Ω), i = 1, . . . , N , and k ∈ N we have unk ∂i φ dx = − ∂i unk φ dx. Ω
Ω
Letting k → ∞ in the previous equality yields u∂i φ dx = − φ dλi , Ω
Ω
which shows that Di u = λi . Hence, u ∈ BV (Ω) and the proof is complete. As a consequence of the previous theorem, we can prove a compactness result in BV (Ω) (see Remark 11.66). Theorem 14.39 (Compactness). Let Ω ⊆ RN be an open set. Assume that {un }n is bounded in BV (Ω). Then there exist a subsequence {unk }k of ∗ {un }n and u ∈ BV (Ω) such that unk → u in L1loc (Ω) and Di unk Di u in Mb (Ω) for all i = 1, . . . , N . Proof. Fix an open set Ω1 Ω. Construct a cut-off function φ ∈ Cc∞ (Ω) such that φ ≡ 1 in Ω1 and for every n ∈ N define φ(x)un (x) if x ∈ Ω, vn (x) := 0 otherwise. Since supp vn ⊆ supp φ and the latter set is compact, we are in a position to apply the Rellich–Kondrachov theorem to find a subsequence {vn,1 }n,1 of {vn } and a function u(1) ∈ BV (Ω1 ) such that vn,1 → u(1) in L1 (Ω). Using the fact that φ ≡ 1 in Ω1 , it follows that un,1 → u(1) in L1 (Ω1 ). By considering an increasing sequence of open sets Ωj Ωj+1 Ω such that ∞ Ω = j=1 Ωj and using a diagonal argument (see, e.g., the proof of the Helly theorem, Theorem 2.44), we may find a subsequence {unk }k and a function u ∈ BVloc (Ω) such that unk → u in L1 (Ωj ) as k → ∞ for every j ∈ N. By selecting a further subsequence, not relabeled, we may assume that unk (x) → u(x) for LN -a.e. x ∈ Ω. Hence, by Fatou’s lemma |u(x)| dx ≤ lim inf |unk (x)| dx < ∞, k→∞
Ω
which implies that u ∈ Exercise 14.3 that
L1 (Ω).
Ω
Since unk → u in L1loc (Ω), it follows by
V (u, Ω) ≤ lim inf V (unk , Ω) = lim inf Dunk (Ω) < ∞, k→∞
which implies that u ∈ BV (Ω).
k→∞
486
14. Functions of Bounded Variation
Remark 14.40. In particular, if Ω ⊆ RN is an open set and {un } is bounded in W 1,1 (Ω), then it follows from the previous theorem that there exist a subsequence {unk }k of {un }n and u ∈ BV (Ω) such that unk → u in L1loc (Ω) and ∗ λnk ,i Di u in Mb (Ω) for all i = 1, . . . , N , where λnk ,i (E) := E ∂i unk dx, E ∈ B(Ω). Exercise 14.41. Let Ω ⊆ RN be an open set with uniformly Lipschitz continuous boundary. Prove that Ω ⊆ RN is an extension domain for BV (Ω). Hint: Use Theorems 13.17 and 14.9. Exercise 14.42 (Poincar´e’s inequality). Let Ω ⊂ RN be a connected extension domain for BV (Ω) with finite measure. Let E ⊆ Ω be a Lebesgue measurable set with positive measure. Prove that there exists a constant c = c(Ω, E) > 0 such that for all u ∈ BV (Ω), u − uE L1∗ (Ω) ≤ cDu(Ω), where, we recall, uE := LN1(E) E u(x) dx. Exercise 14.43 (Poincar´e’s inequality for balls). Let x0 ∈ RN and R > 0. Prove that there exists a constant c = c(N ) > 0 (independent of R and x0 ) such that for all u ∈ BV (B(x0 , R)), u − uB(x0 ,R) L1∗ (B(x0 ,R)) ≤ cDu(B(x0 , R)). As a corollary of the Sobolev–Gagliardo–Nirenberg embedding theorem (see Theorem 14.33) we can prove the following isoperimetric inequality (see also Theorem C.12). Theorem 14.44 (Isoperimetric inequality). Let E ⊆ RN , N ≥ 2, be a set of finite perimeter. Then either E or RN \ E has finite Lebesgue measure and (14.45)
∗
min{LN (E), LN (RN \ E)}1/1 ≤ c P(E)
for some constant c = c(N ) > 0. Proof. If LN (E) < ∞, then χE ∈ BV (RN ), and so (14.45) follows from the Sobolev–Gagliardo–Nirenberg embedding theorem (Theorem 14.33) applied to χE . It remains to show that if E has finite perimeter, then either E or RN \ E has finite Lebesgue measure. Consider a ball BR := B(0, R) and set u := χE . We claim that ∗
min{LN (BR ∩ E), LN (BR \ E)}1/1 ≤ cD(χE )(BR ). To see this, note that uBR
1 = N L (BR )
u(x) dx = BR
LN (BR ∩ E) , LN (BR )
14.5. Embeddings and Isoperimetric Inequalities
and so
487
LN (BR ∩ E) 1∗ N |u − uBR |1∗ dx = 1 − L (BR ∩ E) LN (BR ) BR LN (B ∩ E) 1∗ R + LN (BR \ E). LN (BR )
If LN (BR ∩ E) ≤ LN (BR \ E), then 1/1∗ LN (BR \ E) N ∗ (L (BR ∩ E))1/1 |u − uBR |1∗ dx ≥ N L (BR ) BR ≥ 12 (LN (BR ∩ E))1/1 =
1 2
∗ ∗
min{LN (BR ∩ E), LN (BR \ E)}1/1 .
The other case is analogous. By applying Poincar´e’s inequality for balls (see the previous exercise), we get that the left-hand side of the previous inequality is bounded from above by cD(χE )(BR ), and so 1 2
∗
min{LN (BR ∩ E), LN (BR \ E)}1/1 ≤ cD(χE )(BR ) ≤ cD(χE )(RN ).
Hence, the claim is proved. By letting R → ∞ in the previous inequality and using Proposition B.9, it follows that either E or RN \ E has finite Lebesgue measure. Thus, we have shown that the Sobolev–Gagliardo–Nirenberg embedding theorem in BV implies the isoperimetric inequality. Next we show that the opposite is also true. Theorem 14.45. Assume that the isoperimetric inequality (14.45) holds for all sets with finite perimeter. Then there exists a constant c = c(N ) > 0 such that uL1∗ (RN ) ≤ cDu(RN ) for all u ∈ BV (RN ). Proof. Assume first that u ≥ 0 and that u ∈ C ∞ (RN ) ∩ W 1,1 (RN ). For t ∈ R, define At := {x ∈ RN : u(x) > t}. Then by the coarea formula (14.42) and the isoperimetric inequality (14.45), ∞ 1 ∞ N ∗ ∇u dx = P(At ) dt ≥ (L (At ))1/1 dt. (14.46) c 0 0 RN
488
14. Functions of Bounded Variation
For every t ≥ 0, consider the function ut := min{u, t} and let f (t) := ut L1∗ (RN ) . Note that f is finite, since ∗ 1∗ 1∗ −1 (ut (x)) dx ≤ t u(x) dx + t1 LN (At ) < ∞. RN
{u≤t}
Moreover, f is increasing in [0, ∞) and by the Lebesgue monotone convergence theorem, (14.47)
lim f (t) = uL1∗ (RN ) ,
t→∞
∗
∗
while by the fact that 0 ≤ (ut (x))1 ≤ (u1 (x))1 for all t ∈ [0, 1], it follows by the Lebesgue dominated convergence theorem that (14.48)
lim f (t) = 0 = f (0).
t→0+
Furthermore, for h, t > 0, by Minkowski’s inequality, ∗
0 ≤ f (t + h) − f (t) ≤ ut+h − ut L1∗ (RN ) ≤ h(LN (At ))1/1 , where in the last inequality we have used the facts that ut+h (x) = u(x) = ut (x) if u(x) ≤ t, while |ut+h (x) − ut (x)| = min{u(x), t + h} − t ≤ t + h − t = h if u(x) > t. Thus, the function f is locally Lipschitz continuous in (0, ∞), ∗ with f (t) ≤ (LN (At ))1/1 for L1 -a.e. t > 0. It follows by the fundamental theorem of calculus (see Theorem 3.20), (14.46), (14.47), and (14.48) that ∞ uL1∗ (RN ) = lim f (t) = f (s) ds t→∞ 0 ∞ ∗ (LN (As ))1/1 ds ≤ c ∇u dx. ≤ RN
0
The general case follows as in the proof of Theorem 14.33.
The proofs of the following theorems are left as an exercise. Theorem 14.46 (Gagliardo–Nirenberg interpolation, m = 2). Let 1 ≤ q ≤ ∞ and let r be given by 1 1 1 + = . (14.49) 2 2q r There exists c = c(N, q) > 0 such that 1/2
∇uLr (RN ) ≤ cuLq (RN ) (D(∇u)(RN ))1/2 1,1 for every u ∈ Lq (RN ) ∩ Wloc (RN ) with Dj (∂k u) ∈ Mb (RN ) for all j, k = 1, . . . , N.
Next we consider uniformly Lipschitz continuous domains (see Definition 13.11).
14.6. Density of Smooth Sets
489
Theorem 14.47 (Gagliardo–Nirenberg interpolation, m = 2). Let Ω ⊆ RN be an open set with uniformly Lipschitz continuous boundary (with parameters ε, L, M ), let 0 < < ε/(4(1 + L)), and let 1 ≤ q, r ≤ ∞ be such that (14.49) holds. If q > 1 assume further that Ω is bounded. Then there exists 1,1 a constant c = c(N, q) > 0 such that for every u ∈ Lq (Ω) ∩ Wloc (Ω) with Dj (∂k u) ∈ Mb (Ω) for all j, k = 1, . . . , N , if q > 1, ∇uLr (Ω) ≤ c−1 (LN (Ω))1/r−1/q uLq (Ω) + cuLq (Ω) (D(∇u)(Ω))1/2 1/2
while if q = 1, ∇uL1 (Ω) ≤ c−1 uL1 (Ω) + cuL1 (Ω) (D(∇u)(Ω))1/2 . 1/2
14.6. Density of Smooth Sets In this section we prove that sets of finite perimeter can be approximated by smooth sets. The proof makes use of a theorem of Sard (see [171], [204]). Theorem 14.48 (Sard). Let Ω ⊆ RN be an open set, let u ∈ C ∞ (Ω), and let Σ := {x ∈ Ω : ∇u(x) = 0}. Then L1 (u(Σ)) = 0. The proof uses the following covering result. Lemma 14.49 (Vitali). Let E ⊂ RN be the union of a finite number of balls . . , } such that B(xi , ri ), i = 1, . . . , . Then there exists a subset I ⊆ {1, . the balls B(xi , ri ) with i ∈ I are pairwise disjoint and E ⊆ i∈I B(xi , 3ri ). Proof. Without loss of generality, we may assume that r1 ≥ r2 ≥ . . . ≥ r . Put i1 := 1 and discard all the balls that intersect B(x1 , r1 ). Let i2 be the first integer, if it exists, such that B(xi2 , ri2 ) does not intersect B(x1 , r1 ). If i2 does not exist, then set I := {i1 }, while if i2 exists, discard all the balls B(xi , ri ), with i > i2 , that intersect B(xi2 , ri2 ). Let i3 > i2 be the first integer, if it exists, such that B(xi3 , ri3 ) does not intersect B(xi2 , ri2 ). If i3 does not exist, then set I := {i1 , i2 }; if i3 exists, continue the process. Since there are only a finite number of balls, the process stops after a finite number of steps and we obtain the desired set I ⊆ {1, . . . , }. By construction, the balls B(xi , ri ) with i ∈ I are pairwise disjoint. To prove the last statement, let x ∈ E. Then there exists i = 1, . . . , such that / I, then B(xi , ri ) x ∈ B(xi , ri ). If i ∈ I, then there is nothing to prove. If i ∈ is one of the balls that has been discarded and thus there exists j ∈ I such that rj ≥ ri and B(xj , rj ) ∩ B(xi , ri ) = ∅. Let z ∈ B(xj , rj ) ∩ B(xi , ri ). Then x − xj ≤ x − xi + xi − z + z − xj < ri + ri + rj ≤ 3rj , and so x ∈ B(xj , 3rj ) and the proof is complete.
490
14. Functions of Bounded Variation
We are now ready to prove Theorem 14.48. We use the notation (E.3) in Appendix E. Proof of Theorem 14.48. Step 1: Assume that Ω is bounded and let ΣN := {x ∈ Ω : ∇u(x) = 0, ∇2 u(x) = 0, . . . , ∇N u(x) = 0}. We claim that L1 (u(ΣN )) = 0. To see this, fix ε > 0 and let x0 ∈ ΣN . By Taylor’s formula centered at x0 (see Theorem 9.9) we have N 1 α u(x) = ∂ u(x0 )(x − x0 )α + o(x − x0 N ) α! |α|=0
= u(x0 ) + o(x − x0 N ) as x → x0 , and thus there exists rx0 > 0 such that |u(x) − u(x0 )| ≤ εx − x0 N for all x ∈ B(x0 , rx0 ) ⊂ Ω. In particular, for all 0 < r ≤ rx0 , osc u ≤ 2εrN .
(14.50)
B(x0 ,r)
Fix a compact set K ⊆ ΣN . Since {B(x, rx /3)}x∈ΣN is an open cover for K, we may find a finite subcover. By the previous lemma there exists a disjoint finite subfamily {B(xi , ri /3)}ni=1 , such that K ⊂ ni=1 B(xi , ri ). Hence, L1 (u(K)) ≤
n
L1 (u(K ∩ B(xi , ri )))
i=1
≤ 2ε
n i=1
riN
≤2·3
N
−1 αN ε
n
αN
r N i
3
≤ cεLN (Ω),
i=1
where c = c(N ) and we have used (14.50) and the fact that {B(xi , ri /3)}ni=1 is a disjoint family contained in Ω. Given the arbitrariness of ε > 0, we conclude that L1 (u(K)) = 0. that Now, let {Kn }n be an increasing sequence of compact sets such ∞ (Ω)), K . Then Σ ∩ K is a compact set (since u is of class C Ω= ∞ n n N n=1 and so, by what we just proved, L1 (u(ΣN ∩ Kn )) = 0 for every n ∈ N. Thus, L1 (u(ΣN )) = 0. The case in which Ω is unbounded follows by considering an increasing sequence of bounded open sets whose union is Ω and applying what we just proved in each open bounded set. Step 2: We now prove the theorem by induction on N . The previous step shows that the result is true for N = 1. Thus we assume that the result is true for dimension N − 1 and we prove it for dimension N . Again by the previous step it remains to show that L1 (u(Σ \ ΣN )) = 0. If x ∈ Σ \ ΣN ,
14.6. Density of Smooth Sets
491
then there exist a multi-index α, with |α| < N , and i ∈ {1, . . . , N } such that (14.51)
∂ α u(x) = 0,
∂i (∂ α u)(x) = 0.
Let Eα,i be the set of points x ∈ Σ \ ΣN for which (14.51) holds. Since Eα,i , Σ \ ΣN = |α| 0 define uε := ϕε ∗ χE , where ϕε is a standard mollifier. By Lemma 14.10, uε → χE in L1 (RN ) and ∇uε dx = D(χE )(RN ) = P(E). lim ε→0+
RN
Note that since 0 ≤ χE ≤ 1, then 0 ≤ uεn ≤ 1. By Step 1 of the proof of Theorem 14.28 there exists a sequence εn → 0+ such that lim inf P(Ant ) ≥ P(At ) n→∞
for
L1 -a.e.
t ∈ R, where for t ∈ R,
At := {x ∈ RN : χE (x) > t},
Since At =
Ant := {x ∈ RN : uεn (x) > t}.
E if t < 1, ∅ if t ≥ 1,
it follows that lim inf P(Ant ) ≥ P(E)
(14.52) for
L1 -a.e.
n→∞
t ∈ [0, 1].
On the other hand, by the coarea formula, Fatou’s lemma, and the fact that 0 ≤ uεn ≤ 1, 1 1 ∇uε dx = lim P(Ant ) dt ≥ lim inf P(Ant ) dt. P(E) = lim n→∞ RN
n→∞ 0
0
n→∞
Together with (14.52), this implies that lim inf P(Ant ) = P(E) n→∞
for
L1 -a.e.
t ∈ [0, 1].
Since uεn ∈ C ∞ (RN ), it follows by Sard’s theorem that the sets {x ∈ RN : uεn (x) = t} are C ∞ -manifolds for L1 -a.e. t ∈ [0, 1]. Hence, we may find t ∈ (0, 1) such that lim inf P(Ant ) = P(E) n→∞
and the sets {x ∈
RN
: uεn (x) = t} are C ∞ -manifolds for all n ∈ N.
14.7. A Characterization of BV (Ω)
493
Define En := Ant . It remains to show that LN (En ΔE) → 0. To see this, note that uεn (x) − χE (x) > t for x ∈ En \ E, while χE (x) − uεn (x) ≥ 1 − t for x ∈ E \ En . Hence, |uεn − χE | dx ≥ |uεn − χE | dx + |uεn − χE | dx RN
En \E N
E\En N
≥ tL (En \ E) + (1 − t)L (E \ En ), which implies that LN (En ΔE) → 0, since 0 < t < 1. This concludes the proof. Exercise 14.52. Prove that, in general, we cannot approximate E with smooth sets contained outside E (or inside). Hint: Consider the set A given in Exercise 14.8.
14.7. A Characterization of BV (Ω) In this section we give a characterization of BV (Ω) in terms of difference quotients. This is the extension of Corollary 2.51 in higher dimension. A similar characterization has been given in Section 11.5 for the Sobolev space W 1,p (Ω), 1 < p < ∞. Let Ω ⊆ RN be an open set and for every h ∈ RN \ {0}, let Ωh := {x ∈ Ω : x + th ∈ Ω for all t ∈ [0, 1]}. We recall that S N −1 is the unit sphere ∂B(0, 1) and that βN = HN −1 (S N −1 ). Theorem 14.53. Let Ω ⊆ RN be an open set and let u ∈ L1loc (Ω) be such that Du ∈ Mb (Ω; RN ). Then for every h ∈ RN \ {0}, |u(x + h) − u(x)| dx ≤ Du(Ω), (14.53) h Ωh while (14.54)
κN,1 Du(Ω) ≤ lim sup h→0
where (14.55)
κN,1
1 := βN
S N −1
Ωh
|u(x + h) − u(x)| dx, h
|e1 · ξ|dHN −1 (ξ).
Conversely, if u ∈ L1loc (Ω) is such that |u(x + h) − u(x)| dx < ∞, (14.56) lim sup h h→0 Ωh then Du ∈ Mb (Ω; RN ).
494
14. Functions of Bounded Variation
Proof. The proof is very similar to the one of Theorem 11.75 and we only indicate the main changes. In Step 1 we assume that U satisfies the additional hypothesis Du(∂U ) = 0. We proceed as before to get |uε (x + h) − uε (x)| dx ≤ h ∇uε (y) dy. Uh
U
We are now in a position to apply Lemma 14.10 together with the fact that uε → u pointwise for LN -a.e. in Ω, and Fatou’s lemma to obtain |u(x + h) − u(x)| dx ≤ hDu(U ) ≤ hDu(Ω). Uh
To prove (14.53) it suffices to let U Ω and use the Lebesgue monotone convergence theorem. Step 2 of the proof of Theorem 11.75 remains unchanged. In Step 3 we assume that U satisfies the additional hypothesis Du(∂U ) = 0. We proceed as before to obtain (11.42), which now reads |u(x + h) − u(x)| dx. ∇uε (x) dx ≤ lim sup (14.57) κN,1 h h→0 U Ωh We are now in a position to apply Lemma 14.10 to obtain |u(x + h) − u(x)| dx. (14.58) κN,1 Du(U ) ≤ lim sup h h→0 Ωh By letting U Ω along a sequence and using Proposition B.9, we have that |u(x + h) − u(x)| (14.59) κN,1 Du(Ω) ≤ lim sup dx, h h→0 Ωh which completes Step 3. Step 4: To prove the final statement of the theorem, let u ∈ L1loc (Ω) be such that (14.56) holds. We claim that Du ∈ Mb (Ω; RN ). To see this, let U Ω. Then reasoning as in Step 4, by (14.57) we get |u(x + h) − u(x)| sup dx < ∞. ∇uε (x) dx ≤ lim sup h h→0 0 0, |u(x + hei ) − u(x)| dx ≤ h|Di u|(Ω) Ωh,i
and
lim
h→0+
Ωh,i
|u(x + hei ) − u(x)| dx = |Di u|(Ω). h
(ii) Let u ∈ L1loc (Ω) be such that |u(x + hei ) − u(x)| dx < ∞ lim inf + h h→0 Ωh,i for every i = 1, . . . , N . Prove that Du ∈ Mb (Ω; RN ).
Chapter 15
Sobolev Spaces: Symmetrization Living with P.Q.S, II: What are the symptoms of PQS? Symptoms include: Inability to maintain interest in research area; Advisor Avoidance; Increased web-surfing; Healthy sleeping-pattern; Cynicism towards Academia. — Jorge Cham, www.phdcomics.com
In this chapter we introduce and study the notion of spherically symmetric rearrangement of a function. For more information on this topic and on its applications to partial differential equations we refer the reader to the monographs of Kawohl [130] and Kesavan [131] and the review paper of Talenti [230]. Except for the last two sections, most of the results of this chapter continue to hold if we consider a measure space (X, M, μ) and replace the Lebesgue measurable set E ⊆ RN and the Lebesgue measure LN by a set E ∈ M of X and by μ, respectively. We leave the details as an exercise. In what follows, given a set E ⊆ RN , a function u : E → [−∞, ∞], and s ∈ R, when there is no possibility of confusion we write {u > s}, {u ≥ s} for the sets {x ∈ E : u(x) > s}, {x ∈ E : u(x) ≥ s}, respectively.
15.1. Symmetrization in Lp Spaces As in Chapter 4, given a Lebesgue measurable set E ⊆ RN and a Lebesgue measurable function u : E → [0, ∞], the distribution function of u is the function u : [0, ∞) → [0, LN (E)], defined by (15.1)
u (s) := LN ({u > s}),
s ≥ 0.
ςu (s) := LN ({u ≥ s}),
s ≥ 0.
Similarly, we define (15.2)
497
498
15. Sobolev Spaces: Symmetrization
We recall that u : E → [0, ∞] vanishes at infinity if u is Lebesgue measurable and u (s) < ∞ for s > 0 ( u (0) can be infinite). The proof of the following proposition is very similar to that of Proposition 4.1 and is left as an exercise. Proposition 15.1. Let E ⊆ RN be a Lebesgue measurable set and let u, v, un : E → [0, ∞], n ∈ N, be Lebesgue measurable functions. Then the following properties hold: (i) The function u : [0, ∞) → [0, LN (E)] is decreasing and right continuous, while ςu : [0, ∞) → [0, LN (E)] is decreasing. (ii) If u vanishes at infinity, then ςu is left continuous, lim u (s) = lim ςu (s) = 0,
s→∞
s→∞
and u and ςu are continuous at s > 0 if and only if LN ({u = s}) = 0. In particular, ςu (s) = u (s) for all but countably many s ≥ 0. (iii) If u(x) ≤ v(x) for LN -a.e. x ∈ E, then u ≤ v . In particular, if u(x) = v(x) for LN -a.e. x ∈ E, then u = v . (iv) If un (x) u(x) for LN -a.e. x ∈ E, then un u . Let E ⊆ RN be a Lebesgue measurable set and let u : E → [0, ∞] be a Lebesgue measurable function. As in Chapter 4, the function u∗ : [0, ∞) → [0, ∞], defined by (15.3)
u∗ (t) := inf{s ∈ [0, ∞) : u (s) ≤ t},
t ≥ 0,
is the decreasing rearrangement of u. Proposition 15.2. Let E ⊆ RN be a Lebesgue measurable set and let u : E → [0, ∞] be a measurable function. Then the following properties hold: (i) The function u∗ is decreasing, right continuous, and u∗ (0) = ess supE u. (ii) For all s, t ≥ 0, u∗ (t) > s if and only if u (s) > t. (iii) If u vanishes at infinity, then for all 0 < t < u (0), t ≤ ςu (u∗ (t)). (iv) If u vanishes at infinity, then for all t > 0, u (u∗ (t)) ≤ t, and if
u (u∗ (t1 )) < t1 for some t1 > 0, then u∗ is constant on [t0 , t1 ], where t0 := u (u∗ (t1 )). (v) The functions u and u∗ are equimeasurable; that is, for all s ≥ 0, L1 ({t ∈ E ∗ : u∗ (t) > s}) = LN ({x ∈ E : u(x) > s}), where E ∗ := [0, L1 (E)).
15.1. Symmetrization in Lp Spaces
499
Proof. Properties (i), (ii), and (v) follow as in the proof of Proposition 4.3. (iii) If u (0) > t > 0, then u∗ (t) > 0 by property (ii). Fix 0 ≤ s < u∗ (t). Then u (s) > t by part (ii), and since u ≤ ςu , we have ςu (s) > t. Using the fact that ςu is left continuous, letting s u∗ (t), we get ςu (u∗ (t)) ≥ t. (iv) If u∗ (t) ≤ s, then by part (ii), u (s) ≤ t. Taking s = u∗ (t) gives
u (u∗ (t)) ≤ t. Assume next that t0 := u (u∗ (t1 )) < t1 for some t1 > 0. Then for t0 ≤ t ≤ t1 we have u (u∗ (t1 )) = t0 ≤ t and hence u∗ (t1 ) is an admissible s in the definition of u∗ (t), and so u∗ (t) ≤ u∗ (t1 ). But since u∗ is decreasing, this implies that u∗ (t) = u∗ (t1 ) for all t ∈ [t0 , t1 ]. Remark 15.3. If in the previous proposition we assume that u is also bounded, then by part (i), u∗ (0) < ∞. Hence, reasoning as in part (iv), we have that u (u∗ (0)) = 0. Exercise 15.4. State and prove the analogous statement of part (iv) of Proposition 15.2 for the function ςu . Definition 15.5. Let E ⊆ RN be a Lebesgue measurable set and let u : E → [0, ∞] be a Lebesgue measurable function. The function u : RN → [0, ∞], defined by (15.4)
u (y) := u∗ (αN yN ) = inf{s ∈ [0, ∞) : u (s) ≤ αN yN }
for y ∈ RN , is the spherically symmetric rearrangement of u. We recall that αN is the Lebesgue measure of the unit ball. The function u is also called the Schwarz symmetric rearrangement of u. Note that if u vanishes at infinity, then by Proposition 15.1(ii), u (y) < ∞ for all y = 0. Given a Lebesgue measurable set F ⊂ RN , with 0 < LN (F ) < ∞, we define the spherically symmetric rearrangement of the set F to be the open ball centered at the origin and with the same measure of F , precisely F := B(0, r), where (15.5)
r := (LN (F )/αN )1/N .
If F has measure zero, then we take F to be the empty set, while if F has infinite measure, then we take F to be RN . Note that F is always open even if F is not. Hence, in all cases, (15.6)
LN (F ) = LN (F ).
The next proposition collects some elementary properties of symmetrization. Proposition 15.6. Let E ⊆ RN be a Lebesgue measurable set and let u : E → [0, ∞] be a measurable function. Then the following properties hold: (i) For all s ≥ 0, {y ∈ RN : u (y) > s} = {x ∈ E : u(x) > s} .
500
15. Sobolev Spaces: Symmetrization
(ii) If u vanishes at infinity, then so does u . Moreover, LN ({y ∈ E : u (y) = 0}) ≤ LN ({x ∈ E : u(x) = 0}) with equality holding if and only if either LN ({u > 0}) < ∞ or LN ({u > 0}) = ∞ and LN ({u = 0}) = 0. (iii) The function u is lower semicontinuous and u (0) = ess supE u. (iv) If v : E → [0, ∞] is another Lebesgue measurable function with u ≤ v LN -a.e. on E, then u ≤ v on RN . (v) If {un }n is an increasing sequence of nonnegative Lebesgue measurable functions un : E → [0, ∞] such that un (x) → u(x) as n → ∞ for all x ∈ E, then un (y) → u (y) as n → ∞ for all y ∈ RN . Proof. (i) For s ≥ 0, {y ∈ RN : u (y) > s} = {y ∈ RN : u∗ (αN yN ) > s}. By Proposition 15.2 we have that u∗ (αN yN ) > s if and only if u (s) > αN yN , and so {y ∈ RN : u (y) > s} = {y ∈ RN : y < ( u (s)/αN )1/N } = {y ∈ RN : y < (LN ({x ∈ E : u(x) > s})/αN )1/N } = {x ∈ E : u(x) > s} , where in the last equality we have used (15.5). The proof of part (ii) is the same as the one of Proposition 4.3(v). (iii) To prove that u is lower semicontinuous, it suffices to show that the set {y ∈ RN : u (y) > s} is open for all s ∈ R. If s ≥ 0, then by part (i) the set {y ∈ RN : u (y) > s} is an open ball or empty or RN , while if s < 0, then {y ∈ RN : u (y) > s} = RN since u ≥ 0. Properties (iv) and (v) follow as in the proof of Proposition 4.11(i) and (ii), respectively. Corollary 15.7. Let E ⊆ RN be a Lebesgue measurable set and let u ∈ L∞ (E) be nonnegative. Then u belongs to L∞ (E ) and u L∞ (E ) = uL∞ (E) . Proof. By (15.4) and Propositions 15.2 and 15.6 we have that u L∞ (E ) = u (0) = u∗ (0) = uL∞ (E) < ∞.
Exercise 15.8 (Uniqueness of Schwarz symmetrization). (i) Let f1 , f2 : (0, ∞) → [0, ∞] be two decreasing right continuous functions such that LN ({x ∈ RN : f1 (x) > s}) = LN ({x ∈ RN : f2 (x) > s}) for all s > 0. Prove that f1 = f2 .
15.1. Symmetrization in Lp Spaces
501
(ii) Let u : RN → [0, ∞] be a function vanishing at infinity such that u(x) = f (x) for LN -a.e. x ∈ RN for some decreasing function f : (0, ∞) → [0, ∞). Prove that there exists a unique function v : RN → [0, ∞] vanishing at infinity such that (a) v(x) = f˜(x) for all x ∈ RN \ {0} for some decreasing right continuous function f˜ : (0, ∞) → [0, ∞), (b) v(0) = limt→0+ f˜(t), (c) u(x) = v(x) for LN -a.e. x ∈ RN . Prove also that the function v is lower semicontinuous. (iii) Let u : RN → [0, ∞] be as in part (ii). Prove that u(x) = u (x) for LN -a.e. x ∈ RN . The proofs of the following results are very similar to the analogous ones in Chapter 4 and are left as an exercise. Theorem 15.9 (Hardy–Littlewood’s inequality). Let E ⊆ RN be a Lebesgue measurable set and let u, v : E → [0, ∞) be two Lebesgue measurable functions. Then LN (E) u(x)v(x) dx ≤ u (y)v (y) dy = u∗ (t)v ∗ (t) dt. E
E
0
Theorem 15.10. Let E ⊆ RN be a Lebesgue measurable set, let u : E → [0, ∞) be a function vanishing at infinity, and let f : [0, ∞) → [0, ∞) be a Borel function. Then (15.7) f (u (y)) dy ≤ f (u(x)) dx, E
E
> 0}) < ∞ or LN ({u > 0}) = ∞ with equality holding if f (0) = 0 or N and L ({u = 0}) = 0. In particular, for any p > 0, (u (y))p dy = (u(x))p dx. LN ({u
E
E
Theorem 15.11. Let Ψ : R → [0, ∞) be a convex function such that Ψ(0) = 0, let E ⊆ RN be a Lebesgue measurable set, and let u, v : E → [0, ∞) be two functions vanishing at infinity. Then Ψ(u (y) − v (y)) dy ≤ Ψ(u(x) − v(x)) dx. E
In particular,
E
|u (y) − v (y)| dy ≤
E
|u(x) − v(x)|p dx
p
E
for all 1 ≤ p < ∞ and the operator u → u is a continuous operator from Lp (E) into Lp (E ).
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15. Sobolev Spaces: Symmetrization
Remark 15.12. It is important to observe that it is also possible to define the decreasing rearrangement and spherically symmetric rearrangement of a measurable function u : X → [0, ∞], where (X, M, μ) is a measure space. Remark 15.13. Given a Lebesgue measurable set E ⊆ RN and a Lebesgue measurable function u : E → [−∞, ∞], we can define the decreasing rearrangement u∗ of u to be the decreasing rearrangement of |u|, that is, u∗ := (|u|)∗ . In turn, we can define the spherically symmetric rearrangement u of u to be the spherically symmetric rearrangement of u.
15.2. Lorentz Spaces In this section we introduce the Lorentz spaces Lp,q . We will see in Section 16.1 of Chapter 16 that Lp,q is an interpolation space between L1 and L∞ . Definition 15.14. Let E ⊆ RN be a Lebesgue measurable set and let 0 < p < ∞ and 0 < q ≤ ∞. The Lorentz space Lp,q (E) is defined as the space of all measurable functions u : E → R such that ∞ dt 1/q (t1/p u∗ (t))q 0} < ∞
if q = ∞, where u∗ is defined in Remark 15.13. In view of Proposition 15.1, we can identify functions in Lp,q (E) that coincide up to a set of measure zero. Observe that Lp (E) = Lp,p (E). Indeed, if p < ∞, then by Theorem 15.10, uLp,p (E) = u∗ Lp ([0,∞)) = u∗ Lp (E ) = uLp (E) . On the other hand, if p = ∞, then since u∗ is decreasing by Proposition 15.2, uL∞,∞ (E) = sup{u∗ (t) : t > 0} = u∗ (0) = ess sup |u|. E
Exercise 15.15. Let E ⊆ q < ∞, L∞,q (E) = {0}.
RN
be a Lebesgue measurable set. Prove that if
Proposition 15.16. Let E ⊆ RN be a Lebesgue measurable set and let 0 < p < ∞ and 0 < q ≤ ∞. Then Lp,q (E) is a quasi-Banach space. Proof. We only prove that · Lp,q is a quasi-norm and leave completeness as an exercise. Let u, v ∈ Lp,q (E) and let t > 0. We claim that (15.10)
(u + v)∗ (t) ≤ u∗ (t/2) + v ∗ (t/2).
15.2. Lorentz Spaces
503
To see this, consider the three sets Eu+v := {s ∈ [0, ∞) : |u+v| (s) ≤ t}, Eu := {s ∈ [0, ∞) : |u| (s) ≤ t/2},
Ev := {s ∈ [0, ∞) : |v| (s) ≤ t/2}.
Let s1 ∈ Eu and s2 ∈ Ev . Then |u| (s1 ) = μ({|u| > s1 }) ≤ t/2 and
|v| (s2 ) = μ({|v| > s2 }) ≤ t/2. But since {|u + v| > s1 + s2 } ⊆ {|u| > s1 } ∪ {|v| > s2 }, it follows that
|u+v| (s1 + s2 ) = μ({|u + v| > s1 + s2 }) ≤ |u| (s1 ) + |v| (s2 ) ≤ t. Hence, s1 + s2 ∈ Eu+v , and so (u + v)∗ (t) = inf Eu+v ≤ s1 + s2 . By taking first the infimum over all s1 ∈ Eu and then the infimum over all s2 ∈ Ev , it follows that (u + v)∗ (t) ≤ inf Eu + inf Ev = u∗ (t/2) + v ∗ (t/2). Multiplying both sides by t1/p we get t1/p (u + v)∗ (t) ≤ t1/p u∗ (t/2) + t1/p v ∗ (t/2). If q ≥ 1, we can apply Minkowski’s inequality with respect to the measure 1 t dt to obtain ∞ dt 1/q ∞ 1/p ∗ dt 1/q (t1/p (u + v)∗ (t))q ≤ (t u (t/2) + t1/p v ∗ (t/2))q t t 0 0∞ ∞ 1/q 1/q dt dt ≤ (t1/p u∗ (t/2))q + (t1/p v ∗ (t/2))q . t t 0 0 To obtain the triangle inequality we use the change of variables s = t/2. On the other hand, if q ≤ 1, then ∞ ∞ dt 1/p ∗ q dt ≤ (t (u + v) (t)) (t1/p u∗ (t/2) + t1/p v ∗ (t/2))q t t 0 ∞ 0 ∞ dt dt (t1/p u∗ (t/2))q + (t1/p g ∗ (t/2))q . ≤ t t 0 0 We now make the change of variables s = t/2, raise everything to power 1/q ≥ 1, and then use the inequality (a+b)1/q ≤ 21/q−1 a1/q +21/q−1 b1/q .
Exercise 15.17. Let E ⊆ RN be a Lebesgue measurable set and let u : E → R and v : E → R be two Lebesgue measurable functions. Prove that for every 0 < ε < 1 and every t > 0, (u + v)∗ (t) ≤ u∗ ((1 − ε)t) + v ∗ (εt). Remark 15.18. We will see in Section 16.1 of Chapter 16 that for 1 < p < ∞ and 1 ≤ q ≤ ∞, Lp,q (E) is a Banach space.
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15. Sobolev Spaces: Symmetrization
Exercise 15.19. Let E ⊆ RN be a Lebesgue measurable set and let 0 < p ≤ ∞ and 0 < q ≤ r ≤ ∞. Prove that Lp,q (E) ⊆ Lp,r (E), with uLp,r (E) ≤ cuLp,q (E) for all u ∈ Lp,q (E) and for some constant c = c(p, q, r) > 0. In particular, Lp (E) ⊆ Lp,r (E) ⊆ Lp,∞ (E) for all p ≤ r ≤ ∞. Exercise 15.20. Let E ⊆ RN be a Lebesgue measurable set and let 0 < p < ∞. The weak space Lpw (E) is defined as the space of all measurable functions u : E → R such that uLpw (E) := sup{s( u (s))1/p : s > 0} < ∞. Prove that Lpw (E) = Lp,∞ (E). Exercise 15.21. Given a > 0, let u(x) := x−a , where x ∈ RN \ {0}. Find for which p and q (if any) u ∈ Lp,q (RN ). Exercise 15.22 (H¨ older inequality). Let E ⊆ RN be a Lebesgue measurable set and let 1 ≤ p, q ≤ ∞. Prove that if u ∈ Lp,q (E) and v ∈ Lp ,q (E), then uv ∈ L1 (E) and |u(x)v(x)| dx ≤ uLp,q (E) vLp ,q (E) . E
15.3. Symmetrization of W 1,p and BV Functions In this section we prove that if u ∈ W 1,p (RN ) is nonnegative, 1 ≤ p ≤ ∞, then u ∈ W 1,p (RN ) and (15.11)
u Lp (RN ) = uLp (RN ) ,
∇u Lp (RN ) ≤ ∇uLp (RN ) .
Theorem 15.23. Let u ∈ W 1,p (RN ), 1 ≤ p ≤ ∞, be a nonnegative function. If p = ∞ assume also that u vanishes at infinity. Then for all 0 < a < b we have 1/N ∗ ∗ (N −1)/N ≤ ∇u dx. αN (u (a) − u (b))a In particular,
u∗
{u∗ (b) 0 be such that LN (E) = LN (B) = αN rN , where B := B(y, r). Then LN (B \ E) = LN (E \ B) and so 1−N 1−N 1−N x − y dx ≤ r dx = r dx ≤ x − y1−N dx. E\B
E\B
B\E
B\E
Adding E∩B x − y1−N dx to both sides and using spherical coordinates and the fact that LN (E) = LN (B) gives 1−1/N x − y1−N dx ≤ x − y1−N dx = N αN r = N αN (LN (E))1/N , E
B
which proves the claim. Step 2: By the previous exercise and a change of variables, for LN -a.e. x ∈ RN , 1 1 ∇u(y) dy. |u(x)| ≤ N αN RN x − yN −1 Let K ⊆ {u ≥ s} be a compact set. Integrating the previous inequality over K and using Tonelli’s theorem yields 1 1 N |u(x)| dx ≤ ∇u(y) dxdy, sL (K) ≤ N −1 N αN RN K K x − y which, together with the previous step, gives −1/N N N 1/N sL (K) ≤ αN (L (K)) Since LN (K) < ∞, −1/N
s(LN (K))1−1/N ≤ αN
RN
∇u(y) dy.
RN
∇u(y) dy.
Since this holds for every compact set K ⊆ {u ≥ s}, by standard properties of the Lebesgue measure we get that the previous inequality holds with Es in place of K. It now suffices to take the supremum over all s > 0. Exercise 15.26. Prove Step 1 of the previous theorem using the Hardy– Littlewood inequality (see Theorem 15.9).
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15. Sobolev Spaces: Symmetrization
We are now ready to prove Theorem 15.23. Proof of Theorem 15.23. Step 1: Let 0 ≤ s1 < s2 be such that LN ({s1 < u < s2 )} < ∞. For 0 ≤ s1 < s2 define ⎧ if s ≤ s1 , ⎨ 0 s − s1 if s1 < s < s2 , f (s) := ⎩ s2 − s1 if s ≥ s2 . Since Lp (E) ⊂ L1 (E) for sets of finite measure, LN ({s1 < u < s2 )} < ∞, LN ({u ≥ s2 )} < ∞, and f is a Lipschitz continuous function, it follows by the chain rule (see Exercise 11.51) that the function v := f ◦ u belongs to W 1,1 (RN ) with ∂i u(x) if s1 < u(x) < s2 , ∂i v(x) = f (u(x))∂i u(x) = 0 otherwise, for all i = 1, . . . , N and for LN -a.e. x ∈ RN . Applying the previous proposition to v, we obtain 1/N N (N −1)/N ≤ ∇v dx = ∇u dx. αN sup s[L ({v ≥ s})] RN
s>0
{s1 0, (15.20)
αN s[LN ({v ≥ s})](N −1)/N ≤ ∇vL1 ({v>0}) 1/N
≤ ∇vLp ({v>0}) [LN ({v > 0})]1/p . If v ∈ W 1,1 (RN ) and p = 1, we only consider the first inequality in (15.20). Let 0 < s1 < s2 and for 0 < s1 < s2 define ⎧ if |s| ≤ s1 , ⎨ 0 |s| − s1 if s1 < |s| < s2 , f (s) := ⎩ s2 − s1 if |s| ≥ s2 . Since f is a Lipschitz continuous function, it follows by the chain rule (see Exercise 11.51) and the fact that u vanishes at infinity that the function ˙ 1,p (RN ) (see Step 4 of the proof of v := f ◦ u belongs to W 1,1 (RN ) ∩ W Theorem 12.4 for more details) with ∇u(x) if s1 < |u(x)| < s2 , ∇v(x) = 0 otherwise. Hence, by applying (15.20) with s = s2 − s1 we get αN (s2 − s1 )[LN ({u ≥ s2 })](N −1)/N 1/N
≤ ∇uLp ({s1 s1 })]1/p := 1 if p = 1. Using the fact that LN ({|u| > u∗ (tn )}) ≤ tn ≤ LN ({|u| ≥ u∗ (tn )}) (see Proposition 15.2(iii) and (iv))) and the previous inequality with s1 = u∗ (tn ) and s2 = u∗ (tn+1 ), we get (N −1)/N
αN (u∗ (tn+1 ) − u∗ (tn ))tn+1 1/N
≤ ∇uLp ({u∗ (tn ) 0 such that for all u ∈ Cc∞ (RN ) with ∇m u ∈ LN/m,1 (RN ; RMm ), uCb (RN ) ≤ c∇m uLN/m,1 (RN ) . Hint: Use Lemma 12.10 and Exercise 15.22. Next we can give the sharp version of the embedding Theorem 12.40 for 1 n m = 1. We recall that for every ∈ N, exp (s) := ∞ n=−1 n! s , s ∈ R, (see 1/(N −1)
(12.26)) and that γN := N βN measure of the unit sphere.
(see (12.29)), where βN is the surface
Theorem 15.35. Suppose N ≥ 2. Then for every γ ∈ (0, γN ) there exists a constant c = c(N, γ) > 0 such that uN |u(x)|N LN (RN ) dx ≤ c (15.21) expN γ ∇uN ∇uN RN LN (RN ;RN ) LN (RN ;RN )
for all u ∈ W 1,N (RN ) \ {0}. In particular, if Φγ (s) := expN (γsN ), s ≥ 0, then uΦγ ≤ c∇uLN (RN ;RN )
(15.22) for all u ∈ W 1,N (RN ).
Proof. Fix u ∈ W 1,N (RN )\{0} and define v(x) := |u(x)|/∇uLN , x ∈ RN . Then ∇vLN = 1, and (15.21) reduces to expN (γ(v(x))N ) dx ≤ cvN (15.23) LN . RN
v∗
be the spherically symmetric rearrangement of v. Then v ∗ (x) = Let w(x), x ∈ RN , where w is nonnegative, decreasing, and locally absolutely continuous by Theorem 15.29. Hence, ∇v ∗ (x) = w (x)x/x for LN -a.e. x ∈ RN . Using spherical coordinates and Theorem 15.29, it follows that ∞ N |w (r)|N rN −1 dr = ∇v ∗ N (15.24) βN LN ≤ ∇vLN = 1. 0
15.4. Sobolev Embeddings Revisited
513
Define r0 := inf{r ≥ 0 : w(r) ≤ 1}.
(15.25)
Since w(r) → 0 as r → ∞, we have that r0 must be finite. Using spherical coordinates and Theorem 15.10, we have that N expN (γ(v(x)) ) dx = expN (γ(v ∗ (x))N ) dx RN RN ∞ (15.26) expN (γ(w(r))N )rN −1 dr = βN 0 r 0 expN (γ(w(r))N )rN −1 dr = βN 0 ∞ expN (γ(w(r))N )rN −1 dr =: I + II. + βN r0
To estimate I, it is enough to consider the case r0 > 0, so that w(r0 ) = 1 by (15.25). Since w is locally absolutely continuous, by the fundamental theorem of calculus (see Theorem 3.20), H¨older’s inequality, (15.24), and (15.25), for 0 < r < r0 we have that r0 r0 τN w(r) = w(r0 ) − w (τ ) dτ ≤ 1 + |w (τ )| N dτ τ r r 1/N ∞ |w (τ )|N τ N −1 dr (log(r0 /r))1/N ≤1+ ≤1
0 1 −N + βN (log(r0 /r))1/N .
By the convexity of the function s → sN , for every ε > 0 we may find a constant cε = cε (N ) > 0 such that (1 + s)N ≤ (1 + ε)sN + cε for all s ≥ 0. Hence, −1/(N −1)
(w(r))N ≤ (1 + ε)βN
(15.27)
log(r0 /r) + cε
for all 0 < r < r0 . Since γ < γN , we may take ε so small that γ(1 + ε) < 1/(N −1) −1/(N −1) , and so γ(1 + ε)βN < N . Hence, also by (15.26) γN = N β N and (15.27), r0 −1/(N −1) γcε exp(γ(1 + ε)βN log(r0 /r))rN −1 dr I ≤ βN e 0 r0 −1/(N −1) −1/(N −1) γ(1+ε)β γcε N (15.28) rN −1−γ(1+ε)βN dr = βN e r0 0
=
βN eγcε N − γ(1
rN −1/(N −1) 0 + ε)βN
=: c1 (N, γ)r0N .
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15. Sobolev Spaces: Symmetrization
On the other hand, by the Lebesgue monotone convergence theorem and the fact that w(r) ≤ 1 for all r ≥ r0 by (15.25), we have that ∞ 1 n ∞ II = βN γ (w(r))nN/(N −1) rN −1 dr n! r0 n=N −1 ∞ ∞ 1 γn (w(r))N rN −1 dr ≤ βN n! r 0 n=N −1 ∞ (w(r))N rN −1 dr. = βN (expN γ) r0
Combining this estimate with (15.26) and (15.28), we get ∞ N N (15.29) expN (γ(v(x)) ) dx ≤ c1 r0 +(expN γ)βN (w(r))N rN −1 dr. RN
r0
Using spherical coordinates and Theorem 15.10, we have that ∞ (w(r))N rN −1 dr ≤ (v ∗ (x))N dx = (v(x))N dx. βN RN
r0
RN
Thus, to obtain (15.23), it remains to estimate r0N in the case that r0 > 0. By (15.25) and the fact that w is decreasing, we have that w(r) > w(r0 ) = 1 if and only if 0 ≤ r < r0 . Hence, {x ∈ RN : v ∗ (x) > 1} = {x ∈ RN : w(x) > 1} = B(0, r0 ). By Proposition 15.6, we have that αN r0N = LN ({x ∈ RN : v ∗ (x) > 1}) = L ({x ∈ R N
N
: v(x) > 1}) ≤
(v(x))N dx, {v>1}
and so we have proved (15.23) and, in turn, (15.21).
By (15.21), the number s = ∇uN is admissible in the definition of LN uΦγ (see (12.27)), and so (15.22) follows. Remark 15.36. Note that the inequality (15.21) is scale invariant; that is, if for r > 0 we define the rescaled function ur (x) := u(rx), x ∈ RN , then = r1N uN , ∇ur LN = ∇uLN , and ur N LN LN |ur (x)|N expN c1 dx ∇ur N RN LN |u(x)|N 1 expN c1 dx. = N r ∇uN RN LN The previous theorem is complemented by the following exercise, which shows that (15.21) fails for all γ ≥ γN .
15.4. Sobolev Embeddings Revisited
515
Exercise 15.37. Suppose N ≥ 2 and let r > 0. (i) Construct a sequence {un }n in W 1,N (RN ) of nonnegative radial functions such that supp un ⊆ B(0, r), ∇vn LN (RN ;RN ) = 1, and N RN expN (γN (vn (x)) ) dx → ∞. vn N LN (RN ) (ii) Prove that for every open set Ω ⊆ RN and for every c > 0, the inequality u N N |u(x)|N L (Ω) expN γN dx ≤ c ∇uLN (Ω;RN ) ∇uN Ω LN (Ω;RN ) fails for some u ∈ W01,N (Ω) \ {0}.
Chapter 16
Interpolation of Banach Spaces Living with P.Q.S, III: How do I treat PQS? Experimental treatment such as Advisor-Pressure (AP), Spousal-Income-Frustration (SIF), Lack-of-Savings-Realization (LSR) and “cocktails” of these have known to cause a remission of the disease long enough to at least defend a thesis. More severe treatments are required to actually finish writing the thesis. — Jorge Cham, www.phdcomics.com
In this chapter we give a brief overview of the real method of interpolation. We refer to [7], [24], [25], [40], [146], [232], and [233] for more details about interpolation theory. In Chapter 17 we will use this theory to interpolate between Lp (RN ) and W m,p (RN ) to obtain the Besov spaces Bqs,p (RN ).
16.1. Interpolation: K-Method Definition 16.1. We say that two normed spaces (X0 , · X0 ), (X1 , · X1 ) are an admissible pair if they are embedded into a common Hausdorff topological vector space X. It can be checked (exercise) that the vector space X0 ∩ X1 endowed with the norm (16.1)
xX0 ∩X1 := max{xX0 , xX1 }
is a normed space. The vector space X0 + X1 := {x ∈ X : x = x0 + x1 , x0 ∈ X0 , x1 ∈ X1 } is also a normed space when endowed with the norm (16.2)
xX0 +X1 := inf{x0 X0 + x1 X1 }, 517
518
16. Interpolation of Banach Spaces
where the infimum is taken over all possible decompositions x = x0 + x1 , x0 ∈ X0 , x1 ∈ X1 . Moreover, if X0 and X1 are Banach spaces, then so are X0 ∩ X1 and X0 + X1 . Exercise 16.2. Let (X0 , · X0 ), (X1 , · X1 ) be an admissible pair of Banach spaces. (i) Prove that the set D := {(x, −x) : x ∈ X0 ∩ X1 } is closed in the Banach space X0 × X1 , endowed with the norm (x0 , x1 )X0 ×X1 := x0 X0 + x1 X1 . (ii) Prove that X0 +X1 is isometric to the quotient space (X0 ×X1 )/D. (iii) Prove that X0 + X1 is a Banach space. Given t > 0, in X0 + X1 we can also consider the equivalent norm (16.3)
x ∈ X0 + X1 → K(x, t) := inf{x0 X0 + tx1 X1 },
where as before the infimum is taken over all possible decompositions x = x0 + x1 , x0 ∈ X0 , x1 ∈ X1 . To highlight the dependence of K on X0 and X1 , when needed, we write K(x, t; X0 , X1 ) := K(x, t).
(16.4)
Remark 16.3. The function K(·, t) can be extended to X \ (X0 + X1 ) by setting K(x, t) := ∞ if x ∈ X \ (X0 + X1 ), where X is the Hausdorff topological vector space in Definition 16.1. Exercise 16.4. Let (X0 , · X0 ), (X1 , · X1 ) be an admissible pair. Prove that for every x ∈ X0 + X1 , the function t → K(x, t) is an increasing, concave function and such that t−1 K(x, t) = K(x, t−1 ) and K(x, t) ≤ max{1, t/τ }K(x, τ ) for every t > 0 and τ > 0. For 0 < σ < 1 and 1 ≤ q ≤ ∞ we can define the real interpolation space (X0 , X1 )σ,q := {x ∈ X0 + X1 : xσ,q < ∞}, where if 1 ≤ q < ∞, (16.5)
xσ,q :=
∞
(K(x, t))q
0
dt 1/q t1+σq
,
while if q = ∞, (16.6)
xσ,∞ := sup t−σ K(x, t). t>0
Theorem 16.5. Let (X0 , · X0 ), (X1 , · X1 ) be an admissible pair and let 1 ≤ q ≤ ∞ and 0 < σ < 1. Then the following embeddings hold: X0 ∩ X1 → (X0 , X1 )σ,q → X0 + X1 .
16.1. Interpolation: K-Method
519
Moreover, if X0 and X1 are Banach spaces, then so is (X0 , X1 )σ,q . Proof. We leave as an exercise to check that · σ,q is a norm. In view of (16.1) and (16.3) for every x ∈ X0 ∩ X1 , K(x, t) ≤ xX0 ≤ xX0 ∩X1 ,
K(x, t) ≤ txX1 ≤ txX0 ∩X1 ,
and so K(x, t) ≤ min{1, t}xX0 ∩X1 . In turn, if 1 ≤ q < ∞, by (16.5), ∞ dt 1/q min{1, tq } 1+σq = cq xX0 ∩X1 , xσ,q ≤ xX0 ∩X1 t 0 where cq = 1/(q(1 − σ))1/q + 1/(σq)1/q . If q = ∞, then t−σ K(x, t) ≤ min{t−σ , t1−σ }xX0 ∩X1 ≤ xX0 ∩X1 , where we used the fact that supt>0 min{t−σ , t1−σ } = 1, and so by (16.6), xσ,∞ ≤ xX0 ∩X1 . Thus, in both cases, we have shown that X0 ∩ X1 → (X0 , X1 )σ,q . On the other hand, by (16.2) and (16.3), for every x ∈ (X0 , X1 )σ,q , (16.7)
min{1, t}xX0 +X1 ≤ K(x, t) ≤ max{1, t}xX0 +X1
and thus as before, for 1 ≤ q ≤ ∞, cq xX0 +X1 ≤ xσ,q ,
(16.8)
where c∞ := 1, which proves that (X0 , X1 )σ,q → X0 + X1 . Next we claim that (X0 , X1 )σ,q is a Banach space. Let {xn }n be a Cauchy sequence in (X0 , X1 )σ,q . In view of (16.8), {xn }n is a Cauchy sequence in X0 +X1 , and so, since X0 +X1 is a Banach space (see Exercise 16.2), xn → x in X0 + X1 for some x ∈ X0 + X1 . Given ε > 0 we can find nε ∈ N such that xm − xn σ,q ≤ ε for all m, n ≥ nε . Since K(·, t) is a norm in X0 + X1 , by the triangle inequality and (16.7)2 , (16.9) t−σ K(x−xn , t) ≤ t−σ K(xm −xn , t)+max{t−σ , t1−σ }x−xm X0 +X1 . In turn, if 1 ≤ q < ∞, for every ∈ N, dt 1/q (K(x − xn , t))q 1+σq ≤ xm − xn σ,q + c,q,σ x − xm X0 +X1 t 1/ ≤ ε + c,q,σ x − xm X0 +X1 , where
(max{t−σ , t1−σ })q
dt 1/q
< ∞. t1+σq Letting first m → ∞ and then → ∞ in the previous inequality gives x − xn σ,q ≤ ε for all n ≥ nε , which shows that (X0 , X1 )σ,q is a Banach space. c,q,σ :=
1/
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If q = ∞, then by (16.6) and (16.9), t−σ K(x − xn , t) ≤ ε + max{t−σ , t1−σ }x − xm X0 +X1 . Letting m → ∞ shows that t−σ K(x − xn , t) ≤ ε for all t > 0 and for all n ≥ nε . Hence, x−xn σ,∞ ≤ ε for all n ≥ nε , and the proof is complete. Remark 16.6. If X0 = X1 , then it follows from the previous theorem that X0 = X0 ∩ X1 = (X0 , X1 )σ,q = X0 + X1 = X0 . On the other hand, if X0 ∩ X1 = {0}, then for every x ∈ X0 + X1 there exist unique x0 ∈ X0 and x1 ∈ X1 such that x = x0 + x1 . In turn, by (16.3), K(x, t) = x0 X0 + tx1 X1 and so for 1 ≤ q < ∞, ∞ dt 1/q (x0 X0 + tx1 X1 )q 1+σq =∞ xσ,q = t 0 unless x0 = x1 = 0. Similarly, xσ,∞ = supt>0 t−σ (x0 X0 + tx1 X1 ) = ∞ unless x0 = x1 = 0. Thus, (X0 , X1 )σ,q = {0} for every 1 ≤ q ≤ ∞. Exercise 16.7. Let (X0 , · X0 ), (X1 , · X1 ) be an admissible pair, with X1 → X0 , and let 1 ≤ q ≤ ∞ and 0 < σ < 1. (i) Prove that · X0 is an equivalent norm in X0 + X1 . (ii) Prove that for every T > 0, T dt 1/q (K(x, t))q 1+σq x → xX0 + t 0 is an equivalent norm in (X0 , X1 )σ,q for 1 ≤ q < ∞, while x → xX0 + sup t−σ K(x, t) 0 0, τ −σ K(x, τ ) ≤ (σq)1/q xσ,q . Taking the supremum over all τ > 0 gives xσ,∞ ≤ (σq)1/q xσ,q .
(16.10)
This proves the embedding (X0 , X1 )σ,q → (X0 , X1 )σ,∞ . Now, if 1 ≤ q1 < q2 < ∞, then for x ∈ (X0 , X1 )σ,q1 , ∞ dt 1/q2 (K(x, t))q2 1+σq2 t 0 ∞ ≤ (sup τ −σ K(x, τ ))(q2 −q1 )/q2 (K(x, t))q1 τ >0
≤
0 (q2 −q1 )/(q1 q2 ) (q2 −q1 )/q2 +q1 /q2 (σq1 ) xσ,q1
dt 1/q2 t1+σq1
= cxσ,q1 ,
where in the last inequality we used (16.10) with q1 in place of q.
An important property of interpolation spaces is given by the following theorem. Theorem 16.12. Let (X0 , · X0 ), (X1 , · X1 ) and (Y0 , · Y0 ), (Y1 , · Y1 ) be two admissible pairs and let T : X0 + X1 → Y0 + Y1 be a linear operator such that T : X0 → Y0 and T : X1 → Y1 are continuous. Then for every 1 ≤ q ≤ ∞ and 0 < σ < 1, T : (X0 , X1 )σ,q → (Y0 , Y1 )σ,q with 1−σ T σL(X1 ;Y1 ) . T L((X0 ,X1 )σ,q ;(Y0 ,Y1 )σ,q ) ≤ T L(X 0 ;Y0 )
Proof. Let c0 , c1 > 0 be such that T (x0 )Y0 ≤ c0 x0 X0 ,
T (x1 )Y1 ≤ c1 x1 X1
for all x0 ∈ X0 and x1 ∈ X1 . If x ∈ (X0 , X1 )σ,q and x = x0 +x1 , with x0 ∈ X0 and x1 ∈ X1 , it follows by the linearity of T that T (x) = T (x0 ) + T (x1 ), with T (x0 ) ∈ Y0 and T (x1 ) ∈ Y1 . Hence, by (16.3) and (16.4), K(T (x), t; Y0 , Y1 ) ≤ T (x0 )Y0 + tT (x1 )Y1 ≤ c0 x0 X0 + tc1 x1 X1 = c0 (x0 X0 + tc1 c−1 0 x1 X1 ).
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It follows that K(T (x), t; Y0 , Y1 ) ≤ c0 K(x, tc1 c−1 0 ; X0 , X1 ), and so, if 1 ≤ q < ∞, by (16.5) and the change of variables τ = tc1 c−1 0 , ∞ 1/q q dt (K(x, tc1 c−1 ; X , X )) T (x)σ,q ≤ c0 0 1 0 t1+σq 0 ∞ 1/q σ q dτ = c0 (c1 c−1 ) (K(x, τ ; X , X )) 0 1 0 τ 1+σq 0 = c01−σ cσ1 xσ,q . Similarly, if q = ∞, T (x)σ,∞ ≤ c01−σ cσ1 xσ,∞ . It remains to let c0 T L(X0 ;Y0 ) and c1 T L(X1 ;Y1 ) . Exercise 16.13. Let (X0 , · X0 ), (X1 , · X1 ) be an admissible pair, let 1−σ xσX1 for every 1 ≤ q ≤ ∞ and 0 < σ < 1. Prove that xσ,q ≤ cxX 0 x ∈ X0 ∩ X1 and for some constant c = c(q, σ) > 0. Exercise 16.14. Let (X0 , · X0 ), (X1 , · X1 ) be an admissible pair of Banach spaces, let (Y, · Y ) be a Banach space and let T : X0 + X1 → Y be a linear operator such that T : X0 → Y is compact and T : X1 → Y is continuous. Given 1 ≤ q ≤ ∞ and 0 < σ < 1, prove that T : (X0 , X1 )σ,q → Y is compact. Hint: Given ε > 0 take t > 0 so large that tσ ≤ εt and use Exercise 16.8. The following theorem tells us that the interpolation of two interpolation spaces may be realized as an interpolation between the original spaces. This is one of the key results in interpolation theory. Theorem 16.15 (Reiteration). Let (X0 , · X0 ), (X1 , · X1 ) be an admissible pair, let 1 ≤ q ≤ ∞, and let 0 ≤ σ0 < σ1 ≤ 1. Then for every 0 < σ < 1, (Xσ0 , Xσ1 )σ,q = (X0 , X1 )θ,q , where θ := (1 − σ)σ0 + σσ1 , Xσ0 := (X0 , X1 )σ0 ,q0 if σ0 > 0 for some 1 ≤ q0 ≤ ∞, Xσ1 := (X0 , X1 )σ1 ,q1 if σ1 < 1 for some 1 ≤ q1 ≤ ∞. ¯ := K(·, ·; Xσ , Xσ ) and K := K(·, ·; X0 , X1 ). Proof. Step 1: We write K 0 1 Given x ∈ Xσ0 + Xσ1 , write x = x0 + x1 , where x0 ∈ Xσ0 and x1 ∈ Xσ1 . Since K(·, t) is a norm in X0 + X1 and Xσi → X0 + X1 for i = 0, 1 (see Theorem 16.5), K(x, t) ≤ K(x0 , t) + K(x1 , t) ≤ ctσ0 x0 Xσ0 + ctσ1 x1 Xσ1 = ctσ0 (x0 Xσ0 + tσ1 −σ0 x1 Xσ1 ), where we used (16.3) if σ0 = 0 or σ1 = 1, and Exercise 16.8 if σ0 > 0 and q0 < ∞ or σ1 < 1 and q1 < ∞. Taking the infimum over all such decompositions of x, gives ¯ tσ1 −σ0 ). K(x, t) ≤ ctσ0 K(x,
16.1. Interpolation: K-Method
523
Hence, if q < ∞, by the change of variables τ := tσ1 −σ0 and the fact that θ = (1 − σ)σ0 + σσ1 , ∞ dt 1/q ¯ tσ1 −σ0 ))q 1+θq xθ,q ≤ c (tσ0 K(x, t 0 ∞ 1/q dτ ¯ =c (K(x, τ ))q 1+σq = cx− σ,q , τ 0 where · − σ,q is the norm of (Xσ0 , Xσ1 )σ,q . The proof is simpler when q = ∞ and is left as an exercise. Step 2: Assume that 0 < σ0 < σ1 < 1. Given x ∈ X0 + X1 and t > 0, let x0 ∈ X0 and x1 ∈ X1 be such that x = x0 + x1 and (see (16.3)) (16.11)
x0 X0 + tx1 X1 ≤ 2K(x, t).
(if K(x, t) = 0 we take x0 = 0 and x1 = 0). Let t ∞ dτ dτ K(x, τ ) 1+σ , gi (x, t) := K(x, τ ) 1+σ , (16.12) fi (x, t) := i i τ τ 0 t i = 0, 1. We claim that if f0 (x, t) and g1 (x, t) are finite, then x ∈ Xσ0 + Xσ1 and (16.13)
¯ K(x, tσ1 −σ0 ) ≤ cf0 (x, t) + ctσ1 −σ0 g1 (x, t).
Since x0 ∈ X0 and x1 ∈ X1 , by (16.3) and (16.11), (16.14) K(x0 , τ ) ≤ x0 X0 ≤ 2K(x, t), K(x1 , τ ) ≤ τ x1 X1 ≤ 2τ t−1 K(x, t). Since K is increasing as a function of t, using the previous inequalities we get ∞ dτ 2 (16.15) K(x0 , τ ) 1+σ ≤ K(x, t) g0 (x0 , t) = σ 0 τ σ 0t 0 t t 2(1 − σ0 ) 2(1 − σ0 ) dτ K(x, t) ≤ f0 (x, t), = σ 0 σ0 t σ0 0 τ where in the last inequality we used (16.12) and the fact that K(x, t)/t ≤ K(x, τ )/τ (see Exercise 16.4). Moreover, again by (16.14) and the fact that K(x, t)/t ≤ K(x, τ )/τ , t t dτ τ dτ K(x1 , τ ) 1+σ0 ≤ 2 K(x, t) 1+σ0 ≤ 2f0 (x, t). f0 (x1 , t) = τ tτ 0 0 Since x0 = x − x1 and K(·, τ ) is a norm, it follows that (16.16)
f0 (x0 , t) ≤ f0 (x, t) + f0 (x1 , t) ≤ 3f0 (x, t).
Thus, by (16.15) and (16.16), x0 σ0 ,1 = f0 (x0 , t) + g0 (x0 , t) ≤ cf0 (x, t) < ∞.
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Similarly, one can show that x1 σ1 ,1 = f1 (x1 , t) + g1 (x1 , t) ≤ cg1 (x, t) < ∞. By Theorem 16.11 and the last two inequalities it follows that x0 σ0 ,q0 ≤ cx0 σ0 ,1 ≤ cf0 (x, t),
x1 σ1 ,q1 ≤ cx1 σ1 ,1 ≤ cg1 (x, t).
This proves that x0 ∈ Xσ0 and x1 ∈ Xσ1 . In turn, x ∈ Xσ0 + Xσ1 , and so ¯ K(x, tσ1 −σ0 ) ≤ x0 σ0 ,q0 + tσ1 −σ0 x1 σ1 ,q1 ≤ cf0 (x, t) + ctσ1 −σ0 g1 (x, t). Step 3: Assume that 0 < σ0 < σ1 < 1. Given x ∈ X0 + X1 , if q < ∞ we apply Hardy’s inequalities (see Theorem C.41) to obtain 1/q ∞ t dτ q dt K(x, τ ) 1+σ0 τ t1+σ(σ1 −σ0 )q 0 0 ∞ 1 dt 1/q ≤ (K(x, t))q 1+θq σ(σ1 − σ0 ) 0 t and
∞ ∞
K(x, τ )) 0
t
dτ q
dt
1/q
τ 1+σ1 t1−(1−σ)(σ1 −σ0 )q ∞ 1 dt 1/q ≤ (K(x, t))q 1+θq , (1 − σ)(σ1 − σ0 ) 0 t
where we used the fact that θ = (1 − σ)σ0 + σσ1 . It follows from (16.12) that if xθ,q < ∞, then for L1 -a.e. t > 0, f0 (x, t) and g1 (x, t) are finite and so we are in a position to apply Step 2 to get that x ∈ Xσ0 + Xσ1 and that (16.13) holds. Hence, with the change of variables r := tσ1 −σ0 , by (16.12), (16.13), and the previous inequalities, ∞ 1/q dt − ¯ (K(x, tσ1 −σ0 ))q 1+σ(σ −σ )q xσ,q = (σ1 − σ0 ) 1 0 t 0 ∞ 1/q dt σ1 −σ0 q ≤c (f0 (x, t) + ct g1 (x, t)) 1+σ(σ −σ )q ≤ cxθ,q , 1 0 t 0 which concludes the proof in the case 0 < σ0 < σ1 < 1. The cases σ0 = 0 and σ1 = 1 are similar and are left as an exercise. We conclude this section by studying the limits lim (σ(1 − σ)q)1/q xσ,q
σ→0+
and
lim (σ(1 − σ)q)1/q xσ,q .
σ→1−
The following result will play an important role in Section 17.5 in Chapter 17 to show that the norm in Lp (RN ) and a suitable seminorm in W m,p (RN ) can be regarded as limits of seminorms in the Besov space Bqs,p (RN ) as s → 0+ and s → m− , respectively.
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525
Theorem 16.16. Let (X0 , · X0 ), (X1 , · X1 ) be an admissible pair and let 1 ≤ q < ∞. Assume that x ∈ (X0 , X1 )σ0 ,q for some 0 < σ0 < 1. Then lim inf K(x, t) ≤ lim inf (σ(1 − σ)q)1/q xσ,q t→∞
σ→0+
≤ lim sup(σ(1 − σ)q)1/q xσ,q ≤ lim sup K(x, t) t→∞
σ→0+
and lim inf t→0+
K(x, t) ≤ lim inf (σ(1 − σ)q)1/q xσ,q t σ→1− ≤ lim sup(σ(1 − σ)q)1/q xσ,q ≤ lim sup σ→1−
t→0+
K(x, t) . t
Let 0 < q ≤ ∞ and σ ∈ (0, 1) and for every nonnegative Lebesgue measurable function g : [0, ∞) → [0, ∞) define for 0 < q < ∞, ∞ dt 1/q ∼ (16.17) gσ,q := (g(t))q 1+σq , t 0 while for q = ∞, g∼ σ,∞ := sup t>0
g(t) . tσ
Theorem 16.16 follows by applying the following lemma to the function g(t) = K(x, t), t ≥ 0. Lemma 16.17. Let 0 < q < ∞ and let g : [0, ∞) → [0, ∞) be a Lebesgue measurable function such that gσ0 ,q < ∞ for some σ0 ∈ (0, 1). Then (16.18)
lim inf g(t) ≤ lim inf (σ(1 − σ)q)1/q g∼ σ,q t→∞
σ→0+
≤ lim sup(σ(1 − σ)q)1/q g∼ σ,q ≤ lim sup g(t) t→∞
σ→0+
and (16.19)
lim inf t→0+
g(t) ≤ lim inf (σ(1 − σ)q)1/q g∼ σ,q t σ→1− ≤ lim sup(σ(1 − σ)q)1/q g∼ σ,q ≤ lim sup σ→1−
t→0+
g(t) . t
Proof. We prove only the first inequality in (16.18) and leave the second and (16.19) as exercises. Let := lim inf t→∞ g(t). Note that ∈ [0, ∞]. If = 0, then there is nothing to prove. Thus, assume that > 0 and fix 0 < 1 < . Then there exists T > 0 such that (16.20)
g(t) ≥ 1
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16. Interpolation of Banach Spaces
for all t ≥ T . Write
q σ(1−σ)q(g∼ σ,q ) = σ(1−σ)q
T 0
Then for all 0 < σ ≤ σ0 , I ≤ σ(1 − σ)qT (σ0 −σ)q
T 0
g q (t) dt+σ(1−σ)q t1+σq
∞ T
g q (t) dt =: I +II. t1+σq
g q (t) q dt ≤ σ(1 − σ)qT (σ0 −σ)q (g∼ σ0 ,q ) , t1+σ0 q
and so letting σ → 0+ , we obtain limσ→0+ I = 0. On the other hand, by (16.20), ∞ 1 1 q dt = (1 − σ) σq q1 . II ≥ σ(1 − σ)q1 1+σq t T T Hence, lim inf σ→0+ II ≥ q1 . Letting 1 gives the desired result. Exercise 16.18. Prove (16.18)2 and (16.19). Exercise 16.19. Let g : [0, ∞) → [0, ∞) be a Lebesgue measurable function such that g∼ σ0 ,∞ < ∞ for some σ0 ∈ (0, 1). (i) Prove that if g is increasing, then there exists lim g∼ σ,∞ = lim g(t).
σ→0+
t→∞
What happens if we remove the hypothesis that g is increasing? (ii) Prove that if g(t)/t is decreasing, then there exists lim g∼ σ,∞ = lim g(t)/t.
σ→1−
t→0+
What happens if we remove the hypothesis that g(t)/t is decreasing?
16.2. Interpolation: J-Method In this section we introduce an equivalent way to introduce the interpolation space (X0 , X1 )σ,q . Let (X0 , · X0 ), (X1 , · X1 ) be an admissible pair. We have seen that X0 ∩ X1 is a normed space with the norm given in (16.1). Given t > 0, we can also consider the equivalent norm (16.21)
x ∈ X0 ∩ X1 → J(x, t) := max{xX0 , txX1 }.
To highlight the dependence of J on X0 and X1 , when needed, we write (16.22)
J(x, t; X0 , X1 ) := J(x, t).
Exercise 16.20. Let (X0 , · X0 ), (X1 , · X1 ) be an admissible pair. Prove that for every x ∈ X0 ∩ X1 , the function t → J(x, t) is increasing, convex, and such that for every t > 0 and τ > 0, J(x, t) ≤ max{1, t/τ }J(x, τ ),
K(x, t) ≤ min{1, t/τ }J(x, τ ).
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For 0 < σ < 1 and 1 ≤ q ≤ ∞ we can define the real interpolation space (X0 , X1 )σ,q,J as the space of all x ∈ X0 + X1 such that xσ,q,J < ∞, where if 1 ≤ q < ∞, ∞ dt 1/q (J(u(t), t))q 1+σq , (16.23) xσ,q,J := inf t 0 while if q = ∞, (16.24)
xσ,∞,J := inf sup t−σ J(u(t), t), t>0
where the infimum in (16.23) and (16.24) is done over all Bochner integrable functions (see Section 8.1 in Chapter 8) u : (0, ∞) → X0 ∩ X1 satisfying ∞ dt u(t) , (16.25) x= t 0 where the Bochner integral is considered with respect to the normed space X0 + X1 . The next theorem shows that the spaces (X0 , X1 )σ,q and (X0 , X1 )σ,q,J coincide. Theorem 16.21. Let (X0 , · X0 ), (X1 , · X1 ) be an admissible pair, let 1 ≤ q ≤ ∞, and let 0 < σ < 1. Then there exists a constant c > 0 such that c−1 xσ,q ≤ xσ,q,J ≤ cxσ,q for every x ∈ X0 + X1 . In particular, (X0 , X1 )σ,q = (X0 , X1 )σ,q,J , with equivalent norms. Proof. We only prove the case 1 ≤ q < ∞ and leave the case q = ∞ as an exercise. Step 1: We claim that there exists a constant c > 0 such that xσ,q ≤ cxσ,q,J for every x ∈ X0 + X1 . Without loss of generality, we may assume that xσ,q,J < ∞, since otherwise there is nothing to prove. Hence, there exists a Bochner integrable function u : (0, ∞) → X0 ∩ X1 satisfying (16.25). Since K(·, t) is an equivalent norm in X0 + X1 , by Theorem 8.9 and Exercise 16.20, ∞ ∞ dτ dτ u(τ ) , t ≤ K(u(τ ), t) K(x, t) = K τ τ 0 ∞0 dτ min{1, t/τ }J(u(τ ), τ ) ≤ τ 0 ∞ t dτ dτ +t J(u(τ ), τ ) J(u(τ ), τ ) 2 . = τ τ 0 t
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In turn, by Hardy’s inequalities (see Theorem C.41),
∞
1/q ∞ t
1/q dτ q dt dt (K(x, t))q 1+σq ≤ J(u(τ ), τ ) t τ t1+σq 0 0 0
1/q ∞ ∞ dτ q dt + J(u(τ ), τ ) 2 τ t1−q+σq 0 t
1/q 1 ∞ 1 q dt + ≤ (J(u(t), t)) 1+σq . σ 1−σ t 0 By taking the infimum over all admissible functions u, the claim follows. Step 2: We claim that if x ∈ X0 + X1 is such that (16.26)
lim K(x, t) = 0,
t→0+
lim K(x, t)/t = 0,
t→∞
then there exists a strongly measurable function u : (0, ∞) → X0 ∩ X1 (see Definition 8.1) satisfying 1/ε dt →x u(t) (16.27) t ε as ε → 0+ (where the limit is in the space X0 + X1 ) and such that (16.28)
J(u(t), t) ≤ 4eK(x, t) for all t > 0.
To see this, by the definition of K(x, t) (see (16.3)) for every k ∈ Z we can find xk ∈ X0 and yk ∈ X1 such that (16.29)
x = xk + yk ,
xk X0 + ek yk X1 ≤ 2K(x, ek )
(if K(x, ek ) = 0 we take xk = 0 and yk = 0). In turn, by (16.26), (16.30)
lim xk X0 = lim yk X1 = 0.
k→−∞
k→∞
Define u(t) := xk+1 − xk = yk − yk+1 ∈ X0 ∩ X1 for t ∈ [ek , ek+1 ). Then u is strongly measurable and for every ∈ N, using telescopic sums, e −1 dt u(t) (xk+1 − xk ) = x − x− = x − y − x− → x = t e− k=−
as → ∞ by (16.2), (16.29), and (16.30). This shows (16.27). Moreover, for ek ≤ t < ek+1 , by the definition of J (see (16.21)) and of u(t), Exercise 16.4, (16.29), and J(u(t), t) ≤ max{xk X0 + xk+1 X0 , tyk X1 + tyk+1 X1 } ≤ e(xk X0 + ek yk X1 ) + xk+1 X0 + ek+1 yk+1 X1 ≤ 2eK(x, ek )+2K(x, ek+1 ) ≤ 2eK(x, t)+2 max{1, ek+1 /t}K(x, t) ≤ 4eK(x, t), which proves (16.28).
16.2. Interpolation: J-Method
529
Step 3: We claim that there exists a constant c > 0 such that xσ,q,J ≤ cxσ,q for every x ∈ X0 + X1 . Without loss of generality, we may assume that xσ,q < ∞, since otherwise there is nothing to prove. Then by Exercise 16.8, K(x, t) ≤ (σq)1/q tσ xσ,q , which implies that (16.26) holds. Thus, we are in a position to apply Step 2 to find a strongly measurable function u : (0, ∞) → X0 ∩ X1 satisfying (16.27) and (16.28). By (16.28), ∞ 1/q ∞ dt 1/q q dt (J(u(t), t)) 1+σq ≤ 4e (K(x, t))q 1+σq = 4exσ,q < ∞. t t 0 0 Moreover, since by the definition of J (see (16.21)) for y ∈ X0 ∩ X1 , yX0 +X1 ≤ min{yX0 , yX1 } ≤ min{1, 1/t}J(y, t), when 1 < q < ∞ it follows from H¨older’s inequality and the fact that 0 < σ < 1, ∞ ∞ dt dt ≤ u(t)X0 +X1 min{1, 1/t}J(u(t), t) t t 0 0 ∞ ∞ dt 1/q dt 1/q tσq (min{1, 1/t})q (J(u(t), t))q 1+σq < ∞. ≤ t t 0 0 On the other hand, if q = 1, then instead of H¨older’s inequality we use the fact that min{1, 1/t} ≤ 1/tσ for every t > 0. This shows that u is Bochner integrable, and so it is admissible in the definition of · σ,q,J (see (16.23)). Exercise 16.22. Let (X0 , · X0 ), (X1 , · X1 ) be an admissible pair of Banach spaces, let 1 ≤ q < ∞, and let 0 < σ < 1. Prove that X0 ∩ X1 is dense in (X0 , X1 )σ,q . Hint: Modify the proof of Steps 2 and 3 of the previous theorem. Exercise 16.23. Let (X0 , · X0 ), (X1 , · X1 ) be an admissible pair, with X1 → X0 , and let 1 ≤ q ≤ ∞ and 0 < σ < 1. Prove that for every T > 0, T dt 1/q (J(u(t), t))q 1+σq x → xX0 + inf t 0 is an equivalent norm in (X0 , X1 )σ,q for 1 ≤ q < ∞, while x → xX0 + inf sup t−σ J(u(t), t) 0 0 so small that εt ≤ tσ and use (16.23) and (16.24). Exercise 16.26. Let (X0 , · X0 ), (X1 , · X1 ) be an admissible pair of Banach spaces with X1 compactly embedded in X0 . Given 1 ≤ q1 , q2 ≤ ∞ and 0 < σ1 < σ2 < 1, prove that (X0 , X1 )σ2 ,q2 is compactly embedded in (X0 , X1 )σ1 ,q1 . Hint: Use Exercise 16.14 with the identity operator to prove that Xσ2 := (X0 , X1 )σ2 ,q2 is compactly embedded in X0 , then use the previous exercise to show that Xσ2 is compactly embedded in (X0 , Xσ2 )σ,q , and finally use the reiteration theorem (Theorem 16.15).
16.3. Duality In this section we study the dual space of (X0 , X1 )σ,q . Given an admissible pair (X0 , · X0 ), (X1 , · X1 ) of Banach spaces, in general their duals will not be an admissible pair. To overcome this problem we assume that X0 ∩X1 is dense in both X0 and X1 . Given L ∈ X0 , we denote by L∗ its restriction to X0 ∩ X1 . Then for every x ∈ X0 ∩ X1 , |L∗ (x)| ≤ LX0 xX0 ≤ LX0 xX0 ∩X1 , where in the last inequality we used (16.1). It follows that L∗ ∈ (X0 ∩ X1 ) . Moreover, using the first inequality, (16.31) L∗ X0∗ := sup{|L∗ (x)| : x ∈ X0 ∩ X1 , xX0 ≤ 1} ≤ LX0 < ∞. Conversely, given L∗ ∈ (X0 ∩ X1 ) such that L∗ X0∗ < ∞, since X0 ∩ X1 is dense in X0 , we can uniquely extend L∗ to a linear continuous function L : X0 → R with LX0 = L∗ X0∗ . Thus, we can identify X0 with the space X0∗ of all L∗ ∈ (X0 ∩ X1 ) for which L∗ X0∗ < ∞. Similarly, we can identify X1 with the space X1∗ , which is defined analogously to X0∗ with the norm in X1 in place of the one in X0 in the definition of L∗ X1∗ . With this identification, X0 and X1 can be embedded into (X0 ∩ X1 ) , and thus they are an admissible pair of Banach spaces.
16.3. Duality
531
Theorem 16.27 (Duality). Let (X0 , · X0 ), (X1 , · X1 ) be an admissible pair of Banach spaces and assume that X0 ∩ X1 is dense in both X0 and X1 . Then (X0 , · X0 ), (X1 , · X1 ) is an admissible pair and for 1 ≤ q < ∞ and 0 < σ < 1, the dual of (X0 , X1 )σ,q is given by (X0 , X1 )σ,q . We begin with some preliminary results. Exercise 16.28. Let (X0 , · X0 ), (X1 , · X1 ) be an admissible pair of Banach spaces and let t > 0. Prove that the dual space of X0 × X1 equipped with the norm (x0 , x1 )X0 ×X1 := max{x0 X0 , tx1 X1 }, x0 ∈ X0 , x1 ∈ X1 , is given by X0 × X1 , equipped with the norm L0 X0 + t−1 L1 X1 , L0 ∈ X0 , L1 ∈ X1 . Lemma 16.29. Let (X0 , · X0 ), (X1 , · X1 ) be an admissible pair of Banach spaces and assume that X0 ∩ X1 is dense in both X0 and X1 . If the space X0 + X1 is equipped with the norm K(·, t; X0 , X1 ), where t > 0, then its dual space is X0 ∩ X1 equipped with the norm J(·, t−1 ; X0 , X1 ), while if the space X0 ∩ X1 is equipped with the norm J(·, t; X0 , X1 ), where t > 0, then its dual space is X0 + X1 equipped with the norm K(·, t−1 ; X0 , X1 ). Proof. Step 1: We prove the first part of the statement. Assume that X0 + X1 is equipped with the norm K(·, t; X0 , X1 ) and X0 ∩ X1 with the norm J(·, t−1 ; X0 , X1 ). Given x ∈ X0 +X1 , write x = x0 +x1 , where x0 ∈ X0 and x1 ∈ X1 . Then for L ∈ X0 ∩X1 , we can extend L to X0 +X1 by defining L(x) := L(x0 ) + L(x1 ) (why is this a good definition?). It follows that |L(x)| ≤ |L(x0 )| + |L(x1 )| ≤ LX0 x0 X0 + tt−1 LX1 x1 X1 ≤ max{LX0 , t−1 LX1 }(x0 X0 + tx1 X1 ) = J(L, t−1 ; X0 , X1 )(x0 X0 + tx1 X1 ), where we used (16.22). Taking the infimum over all such decompositions of x and using (16.4) gives |L(x)| ≤ J(L, t−1 ; X0 , X1 )K(x, t; X0 , X1 ), which shows that L(X0 +X1 ) ≤ J(L, t−1 ; X0 , X1 ). Conversely, given L ∈ (X0 + X1 ) , define (16.32)
M := sup{|L(x)| : x ∈ X0 + X1 , K(x, t; X0 , X1 ) = 1}.
If x0 ∈ X0 , then |L(x0 )| ≤ M K(x0 , t; X0 , X1 ) ≤ M x0 X0 by (16.3) and, similarly, if x1 ∈ X1 , |L(x1 )| ≤ M K(x1 , t; X0 , X1 ) ≤ M tx1 X1 .
532
16. Interpolation of Banach Spaces
Thus, L ∈ X0 ∩ X1 . To estimate its norm, given ε > 0 find x0 ∈ X0 with x0 X0 = 1 and x1 ∈ X1 with x1 X1 = 1 such that LX0 ≤ (1 + ε)L(x0 ),
LX1 ≤ (1 + ε)L(x1 ).
Given a, b > 0, by replacing x0 with ax0 and x1 with bx1 , we have that x0 X0 = a, x1 X1 = b and aLX0 ≤ (1 + ε)L(x0 ),
bLX1 ≤ (1 + ε)L(x1 ).
Hence, by (16.3) the vector x = x0 + x1 satisfies K(x, t; X0 , X1 ) ≤ a + tb, while L(x) = L(x0 ) + L(x1 ) ≥ (1 + ε)−1 (aLX0 + bLX1 ). In turn, L(X0 +X1 ) ≥ (1 + ε)−1
aLX0 + bLX1
. a + tb Since ε, a and b are arbitrary, by letting ε → 0+ and then separately a → 0+ and b → 0+ we obtain L(X0 +X1 ) ≥ max{LX0 , t−1 LX1 } = J(L, t−1 ; X0 , X1 ). Step 2: Assume that X0 ∩ X1 is equipped with the norm J(·, t; X0 , X1 ) and X0 + X1 with the norm K(·, t−1 ; X0 , X1 ). Given L ∈ X0 + X1 , let L = L0 + L1 , where L0 ∈ X0 and L1 ∈ X1 . Then for x ∈ X0 ∩ X1 we have |L(x)| ≤ |L0 (x)| + |L1 (x)| ≤ (L0 X0 + t−1 L1 X1 ) max{xX0 , txX1 } = (L0 X0 + t−1 L1 X1 )J(x, t; X0 , X1 ). Taking the infimum over all such decompositions of L gives |L(x)| ≤ K(L, t−1 ; X0 , X1 )J(x, t; X0 , X1 ). In turn, L ∈ (X0 ∩ X1 ) with L(X0 ∩X1 ) ≤ K(L, t−1 ; X0 , X1 ). Conversely, let L ∈ (X0 ∩X1 ) and consider the linear function T : Y → R defined by T (x0 , x1 ) = L(x0 + x1 )/2, where Y is the diagonal Y := {(x0 , x1 ) ∈ X0 × X1 : x0 = x1 } and we endow X0 × X1 with the norm (x0 , x1 )X0 ×X1 := max{x0 X0 , tx1 X1 } (see Exercise 16.28). Since (x, x)X0 ×X1 = J(x, t; X0 , X1 ), we have that |T (x0 , x1 )| ≤ L(X0 ∩X1 ) (x0 , x1 )X0 ×X1 for all (x0 , x1 ) ∈ Y . Thus, by the Hahn–Banach theorem (see Theorem A.32) we can extend T to a continuous linear function T1 : X0 × X1 → R with T1 (X0 ×X1 ) ≤ L(X0 ∩X1 ) . By Exercise 16.28 we can find L0 ∈ X0 and
16.3. Duality
533
L1 ∈ X1 such that T1 (x0 , x1 ) = L0 (x0 ) + L1 (x1 ) for all (x0 , x1 ) ∈ X0 × X1 and T1 (X0 ×X1 ) = L0 X0 + t−1 L1 X1 . Hence, L(x) = L0 (x) + L1 (x) for every x ∈ X0 ∩ X1 and L0 X0 + t−1 L1 X1 ≤ L(X0 ∩X1 ) , which shows that K(L, t−1 ; X0 , X1 ) ≤ L(X0 ∩X1 ) . We are now ready to prove Theorem 16.27. Proof of Theorem 16.27. Step 1: We will show that (X0 , X1 )σ,q → (X0 , X1 )σ,q , where (X0 , X1 )σ,q is the dual of (X0 , X1 )σ,q . Let L ∈ (X0 , X1 )σ,q and let u : (0, ∞) → X0 ∩ X1 be such that ∞ dt u(t) , (16.33) L= t 0 where the Bochner integral is considered with respect to the normed space X0 + X1 . By the previous lemma we have that X0 + X1 coincides with (X0 ∩ X1 ) and for every t > 0, J(u(t), t; X0 , X1 ) = sup{|u(t)(x)| : x ∈ X0 + X1 , K(x, t−1 ; X0 , X1 ) = 1}. Since (X0 , X1 )σ,q → X0 + X1 (see Theorem 16.5), it follows that for every x ∈ (X0 , X1 )σ,q , |u(t)(x)| ≤ J(u(t), t; X0 , X1 )K(x, t−1 ; X0 , X1 ). In turn, by (16.33) and H¨older’s inequality, ∞ ∞ dt dt ≤ |u(t)(x)| J(u(t), t; X0 , X1 )K(x, t−1 ; X0 , X1 ) |L(x)| ≤ t t 0 0 ∞ ∞ dt dt ≤ (J(u(t), t; X0 , X1 ))q 1+σq 1/q (K(x, t−1 ; X0 , X1 ))q 1−σq 1/q . t t 0 0 By taking the infimum over all representations (16.33) (see (16.23)) and by the change of variables t = τ −1 in the last integral, we obtain |L(x)| ≤ Lσ,q ,J xσ,q for every x ∈ (X0 , X1 )σ,q . This shows that L belongs to the dual (X0 , X1 )σ,q of (X0 , X1 )σ,q with L(X0 ,X1 )σ,q ≤ Lσ,q ,J . It follows by Theorem 16.21 that L(X0 ,X1 )σ,q ≤ cLσ,q . Step 2: We will show that (X0 , X1 )σ,q → (X0 , X1 )σ,q . Let L ∈ (X0 , X1 )σ,q . Since X0 ∩X1 → (X0 , X1 )σ,q by Theorem 16.5, it follows that the restriction of L to X0 ∩ X1 is linear and continuous and thus belongs to (X0 ∩ X1 ) . Hence, by the previous lemma, for every t > 0, K(L, t; X0 , X1 ) = sup{|L(x)| : x ∈ X0 ∩ X1 , J(x, t−1 ; X0 , X1 ) = 1}.
534
16. Interpolation of Banach Spaces
In turn, for every ε > 0 and k ∈ Z we can find xk ∈ X0 ∩ X1 such that J(xk , 2k ; X0 , X1 ) = 1 and L(xk ) ≥ K(L, 2−k ; X0 , X1 ) − ε min{1, 2−k }.
(16.34)
Let a := {ak }k∈Z be a sequence of nonnegative numbers such that the seto Lq (Z, H0 ), where H0 is the counting meaquence {2−σk ak }k∈Z belongs k sure. Define za := k ak xk . By the properties of a norm, J(ak xk , 2 ; k X0 , X1 ) = ak J(xk , 2 ; X0 , X1 ) = ak and so {2−kσ J(ak xk , 2k )}k∈Z Lq (Z,H0 ) = {2−kσ ak }k∈Z Lq (Z,H0 ) . It follows from Exercise 16.24 (see also Step 3 of the proof of Theorem 16.21) that za belongs to (X0 , X1 )σ,q with za σ,q,J ≤ {2−kσ ak }k∈Z Lq (Z,H0 ) . Using the fact that L ∈ (X0 , X1 )σ,q , the linearity of L, (16.34), and the previous inequality, we get L(X0 ,X1 )σ,q {2−kσ ak }k∈Z Lq (Z,H0 ) ≥ L(X0 ,X1 )σ,q za σ,q,J ≥ L(za ) = ak L(xk ) ≥ Since
k
ak K(L, 2−k ; X0 , X1 ) − ε
k
ak min{1, 2−k }.
k
min{1, 2−k }
< ∞ (why?), letting ε → 0+ gives ak K(L, 2−k ; X0 , X1 ) L(X0 ,X1 )σ,q {2−kσ ak }k∈Z Lq (Z,H0 ) ≥ k
ak
k
for every sequence {ak }k∈Z of nonnegative numbers such that {2−σk ak }k∈Z belongs to Lq (Z, H0 ), or, equivalently, dk 2σk K(L, 2−k ; X0 , X1 ) L(X0 ,X1 )σ,q {dk }k∈Z Lq (Z,H0 ) ≥ k
for every sequence {dk }k∈Z of nonnegative numbers which belongs to Lq (Z, H0 ). It follows by Corollary B.81 that L(X0 ,X1 )σ,q ≥ {2kσ K(L, 2−k ; X0 , X1 )}k∈Z Lq (Z,H0 ) . Using Exercise 16.4 and (16.3) we obtain that cL(X0 ,X1 )σ,q ≥ L(X0 ,X1 )σ,q , which concludes the proof.
Next we show that if X0 and X1 are reflexive Banach spaces, then so is (X0 , X1 )σ,q for q < ∞. For simplicity, we consider here a very special case, which is all we need in the next chapter.
16.4. Lorentz Spaces as Interpolation Spaces
535
Theorem 16.30 (Reflexivity). Let (X0 , ·X0 ), (X1 , ·X1 ) be an admissible pair of Banach spaces with X0 and X1 reflexive and assume that X1 → X0 and that X1 is dense in X0 . Then for 1 ≤ q < ∞ and 0 < σ < 1, the space (X0 , X1 )σ,q is reflexive. Proof. Step 1: Let X0∗ and X1∗ be the spaces introduced before Theorem 16.27. Since X1 → X0 , we have that X1 ∩ X0 = X1 and so X1∗ = X1 . We claim that X0∗ is dense in X1 . In view of the second geometric form of the Hahn–Banach theorem (Theorem A.35), it is enough to show that if T ∈ X1 is such that T (L∗ ) = 0 for all L∗ ∈ X0∗ , then T = 0. Since X1 is reflexive, there exists x1 ∈ X1 such that T (L) = L(x1 ) for all L ∈ X1 . It follows by the hypothesis on T that L∗ (x1 ) = 0 for L∗ ∈ X0∗ . Using the fact that X1 → X0 , we have that x1 ∈ X0 . Moreover, every L∗ ∈ X0∗ can be uniquely extended to a linear continuous function L ∈ X0 with LX0 = L∗ X0∗ , where L∗ X0∗ is defined in (16.31). Thus, L(x1 ) = 0 for every L ∈ X0 . Again by the second geometric form of the Hahn–Banach theorem (Theorem A.35) this implies that x1 = 0, and in turn, that T = 0. Step 2: By Theorem 16.27, the dual of (X0 , X1 )σ,q is given by (X0∗ , X1∗ )σ,q . In turn, by Step 1, we are in a position to apply Theorem 16.27 once more, to conclude that the dual of (X0∗ , X1∗ )σ,q is given by ((X0∗ )∗ , (X1∗ )∗ )σ,q . By Step 1, (X0∗ )∗ and (X1∗ )∗ can be identified with (X0∗ ) and (X1∗ ) . On the other hand, by the density of X1 in X0 , we have that X0∗ and X1∗ can be identified with X0 and X1 , respectively. Hence, (X0∗ )∗ and (X1∗ )∗ can be identified with X0 and X1 , , respectively. Since X0 and X1 are reflexive, it follows that ((X0∗ )∗ , (X1∗ )∗ )σ,q can be identified with (X0 , X1 )σ,q . This completes the proof. Remark 16.31. More generally, if (X0 , · X0 ), (X1 , · X1 ) are an admissible pair of Banach spaces with X0 ∩ X1 dense in X0 and X1 , one can show that for 1 ≤ q < ∞, (X0 , X1 )σ,q is reflexive if and only if the the inclusion i : X0 ∩ X1 → X0 + X1 is weakly compact (see, e.g., [21] and [152]). We will not need this result.
16.4. Lorentz Spaces as Interpolation Spaces As an application of the interpolation theory, we prove that the interpolation space between L1 (E) and L∞ (E) is the Lorentz space Lσm,p (E) (see Section 15.2 in Chapter 15). Theorem 16.32. Let E ⊆ RN be a Lebesgue measurable set, let 1 < p < ∞, and let 1 ≤ q ≤ ∞. Then (L1 (E), L∞ (E))σ,q = Lp,q (E),
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16. Interpolation of Banach Spaces
where σ := 1 − 1/p. Moreover, a norm in Lp,q (E) is given by ∞ t q dt 1/q u∗ (τ ) dτ , uσ,q = t1+σq 0 0 where u∗ is the decreasing rearrangement of u. Proof. The fact that L1 (E) and L∞ (E) are an admissible pair follows from Exercise B.105. Given a Lebesgue measurable function u : E → R and t > 0, we claim that t u∗ (τ ) dτ. (16.35) K(u, t) = t
0
u∗ dτ
We begin by proving 0 ≤ K(u, t). Without loss of generality, we may assume that K(u, t) < ∞ (see Remark 16.3), so that u ∈ L1 (E) + L∞ (E). Write u = v +w, where v ∈ L1 (E) and w ∈ L∞ (E). Then by Exercise 15.17, a change of variables and the fact that w∗ is decreasing t t t ∗ ∗ u (τ ) dτ ≤ v ((1 − ε)τ ) dτ + w∗ (ετ ) dτ 0 0 0 ∞ −1 v ∗ (r) dr + tw∗ (0) ≤ (1 − ε) 0
= (1 − ε)
−1
vL1 (E) + twL∞ (E) ,
where the last equality follows from Corollary 4.7 and Theorem 4.16 (which continue to hold without changes when N ≥ 1). Letting ε → 0+ shows that t ∗ tw ∞ . Since this holds for every decomposition 0 u dτ ≤ vL1 (E) + t ∗ L (E) of u, it follows that 0 u dτ ≤ K(u, t). t To prove the converse inequality, assume that 0 u∗ dτ < ∞ and define v(x) := max{|u(x)| − u∗ (t), 0} sgn u(x),
w(x) := u(x) − v(x).
u∗ (t)}
and Ft := {τ > 0 : u∗ (τ ) > u∗ (t)}. By Let Et := {x ∈ E : |u(x)| > Proposition 15.2(iv), LN (Et ) ≤ t and u∗ is constant on [LN (Et ), t]. Hence, LN (Et ) ∗ (|u(x)| − u (t)) dx = (u∗ (τ ) − u∗ (t)) dτ vL1 (E) = ≤
0
Et t
(u∗ (τ ) − u∗ (t)) dτ,
0
where in the second equality we used Proposition 15.2(v) and Theorem B.60. On the other hand, |w(x)| = u∗ (t) in Et and |w(x)| ≤ u∗ (t) outside Et . It follows that t t ∗ ∗ ∗ (u (τ ) − u (t)) dτ + tu (t) = u∗ (τ ) dτ, vL1 (E) + twL∞ (E) ≤ 0
which shows that K(u, t) ≤
t 0
0
u∗ dτ .
This proves (16.35).
16.4. Lorentz Spaces as Interpolation Spaces
537
In turn, since u∗ is decreasing, K(u, t) ≥ tu∗ (t) and so if q < ∞, using the fact that σ = 1 − 1/p, ∞ ∞ dt dt uqσ,q = (K(x, t))q 1+σq ≥ (tu∗ (t))q 1+σq = uqLp,q (E) . t t 0 0 On the other hand, by Hardy’s inequality (see Theorem C.41), (16.35), and again the fact that σ = 1 − 1/p, ∞ q dt 1/q 1/q ∞ t q dt (K(x, t)) 1+σq = u∗ dτ uσ,q = t t1+σq 0 0 0 ∞ 1/q dt ≤σ (u∗ (t))q 1+σq−q = σuLp,q (E) . t 0 The case q = ∞ is similar and is left as an exercise.
Exercise 16.33. Let (X, M, μ) be a measure space. Extend the previous theorem to the case in which the Lebesgue measurable set E ⊆ RN and the Lebesgue measure LN are replaced by a set E ∈ M of X and by μ, respectively. Do you need any hypotheses on μ? Exercise 16.34. Let E ⊆ RN be a Lebesgue measurable set, let 1 ≤ q ≤ ∞, and let 0 < σ < 1. Prove that if if p0 = p1 , (Lp0 ,q0 (E), Lp1 ,q1 (E))σ,q = Lp,q (E), with equivalent norms, where
1 p
=
1−σ p0
+
σ p1
and deduce that
(Lp0 (E), Lp1 (E))σ,q = Lp,q (E). Exercise 16.35. Let E ⊆ RN be a Lebesgue measurable set, let 1 < p < r < s < ∞ and let 1 ≤ q ≤ ∞. Prove that if u ∈ Lp,∞ (E) ∩ Ls,∞ (E), then u ∈ Lr,q (E) and estimate its Lr,q (E) norm. Exercise 16.36. Let E ⊆ RN be a Lebesgue measurable set, let 1 < p < ∞ and let 1 ≤ q < ∞. Prove that L1 (E) ∩ L∞ (E) is dense in Lp,q (E). Deduce that simple functions are dense in Lp,q (E). Exercise 16.37. Let E ⊆ RN be a Lebesgue measurable set, let 1 < p < ∞ and let 1 ≤ q1 < q2 < ∞. Prove that Lp,q1 (E) → Lp,q2 (E). Exercise 16.38. Let 1 < p ≤ 2 and let 1 ≤ q ≤ ∞. Prove that for every u ∈ Lp,q (RN ), its Fourier transform u & (see (10.23)) belongs to Lp ,q (RN ) with & uLp ,q (RN ) ≤ cuLp,q (RN ) (compare this with the Hausdorff–Young inequality (10.30)). Exercise 16.39. Let 1 < r < ∞, let 1 < p < r , let 1 ≤ q ≤ ∞, and let 1 1 1 p,q N r N s = p + r − 1. Prove that if u ∈ L (R ) and v ∈ L (R ), then u ∗ v belongs to Ls,q (RN ) with u ∗ vLs,q (RN ) ≤ cuLp,q (RN ) vLr (RN ) (compare this inequality with Young’s inequality, Theorem 10.49).
Chapter 17
Besov Spaces Another principal concept in Quantum Gradnamics is the observation that graduate students do not move toward graduation in a steady and continuous manner. Rather they make progress through discrete bursts of random productivity called “wanta” (short for “want data”) whose energy is proportional to the frequency of meetings with their advisor. — Jorge Cham, www.phdcomics.com
17.1. Besov Spaces Bqs,p In this section we define the Besov spaces Bqs,p (RN ). Given a function u : RN → R, for every h ∈ RN , m ∈ N, and x ∈ RN , we define inductively the forward difference operator Δ1h u(x) := Δh u(x) := u(x + h) − u(x)
(17.1) for m = 1 and
m−1 u(x)) Δm h u(x) := Δh (Δh
(17.2)
for m ≥ 2. In particular, Δ2h u(x) = u(x + 2h) − 2u(x + h) + u(x).
(17.3)
We recall that the symbol s stands for the integer part of s. Definition 17.1. Let 1 ≤ p, q ≤ ∞ and let s > 0. A Lebesgue measurable function u : RN → R belongs to the Besov space Bqs,p (RN ) if (17.4)
uBqs,p (RN ) := uLp (RN ) + |u|Bqs,p (RN ) < ∞,
where for q < ∞, (17.5)
|u|Bqs,p (RN ) :=
RN
dh s+1 Δh uqLp (RN ) hN +sq
1/q , 539
540
17. Besov Spaces
while for q = ∞, (17.6)
s,p |u|B∞ (RN ) :=
1 s+1 Δh uLp (RN ) . s h N h∈R \{0} sup
Remark 17.2. (i) Given a Lebesgue measurable function u : RN → R, to see that the seminorm |u|Bqs,p (RN ) is well-defined, consider a representative u ¯ of u that is Borel measurable and let w : RN × RN → R be the function defined by w(x, h) := u ¯(x + h). Then w is a Borel function, since it is the composition of u ¯ with the continuous function g : RN × RN → RN given by g(x, h) := x + h. Thus, (x, h) → Δ1h u(x) is Lebesgue measurable and, in turn, so is (x, h) → Δm h u(x). (ii) Observe that given a function u ∈ L1loc (RN ), we have that Δh uLp (RN ) = 0 if and only if u is equivalent to a constant. More generally, Δm h uLp (RN ) = 0 if and only if u has a representative that is a polynomial of degree less than m (see (17.25) below). (iii) In the sequel we will often use the fact that if u ∈ Lp (RN ), 1 ≤ p ≤ ∞, then for every h ∈ RN , and m ∈ N, m Δm h uLp (RN ) ≤ 2 uLp (RN ) .
(17.7)
This follows by induction on m from (17.1), (17.2), Minkowski’s inequality, and the change of variables y = x + h. (iv) If u : RN → R is Lebesgue measurable, by a change of variables, we have that Δ2h uLp (RN ) = ∼ Δ2h uLp (RN ) , where for h ∈ RN and x ∈ RN , (17.8)
∼
Δ2h u(x) := u(x + h) − 2u(x) + u(x − h).
If q = p, we write (17.9)
B s,p (RN ) := Bps,p (RN ).
Remark 17.3. If p = q < ∞ and 0 < s < 1, then (17.5) takes the simpler form
1/p |u(x + h) − u(x)|p dxdh , |u|B s,p (RN ) = hN +sp RN RN
17.1. Besov Spaces Bqs,p
541
while if p = q < ∞ and s = 1,
1/p |u(x + h) − 2u(x) + u(x − h)|p dxdh . |u|B 1,p (RN ) = hN +p RN RN On the other hand, if p = q = ∞ and 0 < s < 1, then |u|B s,∞ (RN ) = sup ess sup h∈RN x∈RN
|u(x + h) − u(x)| , hs
and thus a Lebesgue measurable function u : RN → R is such that |u|B s,∞ (RN ) < ∞ if and only if it has a H¨ older continuous representative of exponent ∞ N s. In particular, u ∈ L (R ) belongs to B s,∞ (RN ) if and only if it has a representative in C 0,s (RN ). If p = q = 1, then |u|B 1,∞ (RN ) = sup ess sup h∈RN x∈RN
|u(x + h) − 2u(x) + u(x − h)| . h
The space B 1,∞ (RN ) is known as the Zygmund space Λ1 (RN ). The next exercise shows that a function in B 1,∞ (RN ) admits a representative which is H¨older continuous of exponent s for every 0 < s < 1. Exercise 17.4. Let u ∈ B 1,∞ (RN ). Prove that for a precise representative of u, (i) for every n ∈ N and every x, y ∈ RN , |2n Δ2−n y u(x) − 2n−1 Δ2−n+1 y u(x)| ≤ |u|B 1,∞ (RN ) y, (ii) for every l ∈ N and every x, h ∈ RN , with 2−l−1 ≤ h < 2−l , |Δh u(x)| ≤ 2−l+1 uL∞ (RN ) + l|u|B 1,∞ (RN ) h, (iii) there exists a constant c > 0 independent of u such that |u(x + h) − u(x)| ≤ cuB 1,∞ (RN ) h(1 + | log h|) for every x ∈ RN and h ∈ RN \ {0}. Deduce that B 1,∞ (RN ) → B s,∞ (RN ) for 0 < s < 1. Note that Λ1 (RN ) ⊂ C 0,1 (RN ) with strict inclusion. Indeed, Lipschitz continuous functions are differentiable LN -a.e. in their domain by Rademacher’s theorem (see Theorem 9.14), while the Weierstrass nowhere differentiable function (1.2) belongs to Λ1 (R) (exercise). The next exercise shows that without measurability, there are everywhere discontinuous functions with Δ2h u(x) ≡ 0.
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17. Besov Spaces
Exercise 17.5. Prove that there exists an everywhere discontinuous function u : R → R such that u(x + h) − 2u(x) + u(x − h) = 0 for every x, h ∈ R. Hint: Consider R as a vector space over the rationals and take a Hamel basis. Next we show that Bqs,p (RN ) is a Banach space. Proposition 17.6 (Completeness). Let 1 ≤ p, q ≤ ∞ and s > 0. Then the Besov space Bqs,p (RN ) is a Banach space. Proof. We prove the result for q < ∞ and leave the case q = ∞ as an exercise. Let {un }n be a Cauchy sequence in Bqs,p (RN ). Then by (17.4), {un }n is a Cauchy sequence of Lp (RN ) and so it converges to a function u in Lp (RN ). On the other hand, as in the proof of Theorem 8.28, the functions vn : RN × RN → R, defined by vn (x, h) :=
1 hN/q+s
Δm h un (x),
can be identified with functions in Lq (RN ; Lp (RN )). It follows from (17.4) that {vn }n is a Cauchy sequence in Lq (RN ; Lp (RN )). By Theorem 8.15 we may find v ∈ Lq (RN ; Lp (RN )) such that vn → v in Lq (RN ; Lp (RN )). Extract a subsequence {unk }k of {un }n such that unk (x) → u(x) for LN -a.e. x ∈ RN and vnk (x, h) → v(x, h) for LN -a.e. x ∈ RN and LN -a.e. h ∈ RN . It follows that for LN -a.e. x ∈ RN and LN -a.e. h ∈ RN the function v(x, h) coincides with the function 1 Δm u(x). w(x, h) := hN/q+s h Hence, u ∈ Bqs,p (RN ) and un − uBqs,p → 0.
Exercise 17.7. Let f : R → R be a Lipschitz continuous function and let u : RN → R be a Lebesgue measurable function such that |u|Bqs,p (RN ) < ∞ for some 1 ≤ p, q ≤ ∞ and 0 < s < 1. Prove that |f ◦ u|Bqs,p (RN ) ≤ Lip f |u|Bqs,p (RN ) . Remark 17.8. Given a function u : RN → R and h ∈ RN consider the translation operator Th defined by (17.10)
Th (u)(x) := u(x + h),
x ∈ RN .
Note that Th ◦ Tξ = Tξ ◦ Th for every h, ξ ∈ RN . Moreover, Δh := Th − I and for m ∈ N, with m ≥ 2, m Δm h = (Th − I) := (Th − I) ◦ · · · ◦ (Th − I). m times
17.1. Besov Spaces Bqs,p
543
Here, I is the identity operator. Hence, by the binomial theorem, m m (17.11) Δh u(x) = (−1)m−j m x ∈ RN . j u(x + jh), j=0
In particular, setting y := x + mh and k := m − j, we have (17.12) m m u(x) = (−1)k m−k u(y − kh) Δm h k=0
= (−1)m
m m m (−1)m−k m k u(y − kh) = (−1) Δ−h u(x + mh). k=0
Exercise 17.9. Let 1 ≤ p < ∞, 0 < s < N/p, and a > 0 and consider the function u(x) := xs−N/p | log x|−a if x ∈ RN with x < 1/2, with u extended to RN in a smooth way and with compact support. Find the values of q (if any) for which u ∈ Bqs,p (RN ). In the following exercises, given an open set Ω ⊆ RN , 1 ≤ p, q < ∞, and s > 0, for a Lebesgue measurable function u : Ω → R and 1 ≤ q < ∞ we define
1/q dh s+1 Δh uqLp (Ωh ) , (17.13) |u|Bqs,p (Ω) := hN +sq RN while for q = ∞, 1 s+1 s,p (17.14) |u|B∞ sup Δh uLp (Ωh ) , (Ω) := s h∈RN \{0} h where Ωh := {x ∈ Ω : x + h ∈ Ω}. Exercise 17.10. Let 0 < s < 1, let I = (0, 1), and let ∞ u(x) = 2−ns n−1/2 cos(2n x), x ∈ I. n=1
(i) Prove that |u|Bqs,2 (I) < ∞ if q > 2. Hint: Given h find m such that 2m |h| ≤ 1 and 2m+1 |h| > 1 and break the series in the sum from 1 to m and from m + 1 to infinity (see the proof of Theorem 1.15).
(ii) Prove that |u|Bqs,2 (I) = ∞ if q ≤ 2. Exercise 17.11. Let 1 ≤ p, q ≤ ∞ and let 0 < s < 1. Given a function < ∞ (see (17.13) and (17.14)), define u ∈ L1loc (RN + ) such that |u|Bqs,p (RN +) if xN > 0, u(x , xN ) v(x) := u(x , −xN ) if xN < 0.
544
17. Besov Spaces
Prove that |v|Bqs,p (RN ) ≤ c|u|Bqs,p (RN ) < ∞. Can you do something similar + for s ≥ 1? Hint: See Exercise 13.3. We conclude this section by discussing the density of functions in C ∞ (RN ). Proposition 17.12. Let 1 ≤ p, q ≤ ∞ and s > 0. For any u ∈ L1loc (RN ), let uε := ϕε ∗ u, where ϕε is a standard mollifier. Then |uε |Bqs,p (RN ) ≤ |u|Bqs,p (RN )
(17.15) for all ε > 0 and
lim |uε |Bqs,p (RN ) = |u|Bqs,p (RN ) .
(17.16)
ε→0+
Moreover, if 1 ≤ p, q < ∞ and u ∈ L1loc (RN ) is such that |u|Bqs,p (RN ) < ∞, then lim |uε − u|Bqs,p (RN ) = 0.
(17.17)
ε→0+
In particular, if 1 ≤ p, q < ∞, then C ∞ (RN ) ∩ Bqs,p (RN ) is dense in Bqs,p (RN ). Proof. Given u ∈ L1loc (RN ) observe that for m ∈ N, h ∈ RN , and ε > 0, m Δm h uε = ϕε ∗ Δh u. Hence, by Theorem C.16 we have that m Δm h uε Lp ≤ Δh uLp
(17.18) and
m Δm h uε Lp → Δh uLp
(17.19)
as ε → 0+ . It follows from (17.18) that (17.15) holds. In turn, lim sup |uε |Bqs,p ≤ |u|Bqs,p . ε→0+
To prove the opposite inequality, assume first that q < ∞. For h ∈ RN and ε > 0 we define 1 1 q q fε (h) := Δm f (h) := Δm h uε Lp , h uLp . hN +sq hN +sq Since fε (h) → f (h) for every h by (17.19), by Fatou’s lemma we have that f (h) dh = lim fε (h) dh ≤ lim inf fε (h) dh. RN
+ RN ε→0
ε→0+
RN
Thus, (17.16) holds. m N If q = ∞, then since Δm h uε Lp → Δh uLp for all h ∈ R by (17.19), we have that 1 1 1 Δm Δm sup Δm h uLp = lim+ h uε Lp ≤ lim inf h uε Lp s s + h ε→0 h ε→0 h∈RN hs
17.2. Some Equivalent Seminorms
545
for all h ∈ RN . It follows that 1 1 sup Δm sup Δm h uLp ≤ lim inf h uε Lp , s + ε→0 h∈RN hs h∈RN h and so (17.16) is satisfied. Assume next that p < ∞, q < ∞, and u ∈ L1loc (RN ) is such that |u|Bqs,p (RN ) < ∞. By Theorem C.16(iv), (17.20)
m lim Δm h uε − Δh uLp = 0.
ε→0+
For h > 0 and ε > 0 define the functions 1 q m gε (h) := Δm h uε − Δh uLp , hN +sq
g(h) :=
1 q Δm h uLp . hN +sq
By hypothesis g ∈ L1 (RN ), and by (17.18), Minkowski’s inequality, and the convexity of the function |y|q we have that gε (h) ≤ 2q g(h) for all h. Since gε (h) → 0 for all h by (17.20), we are in a position to apply the Lebesgue dominated convergence theorem to conclude that (17.17) holds. Remark 17.13. If u ∈ Bqs,p (RN ) with 1 ≤ p, q < ∞, in the previous theorem, instead of taking ϕ ∈ Cc∞ (RN ), it is enough to consider ϕ ∈ S(RN ). Exercise 17.14. Let u ∈ L1loc (R) be such that |u(x) − u(y)| dxdy < ∞. |x − y|2 2 R Prove that u is constant. Hint: Let g(t) be the integral over the set {(x, y) ∈ R2 : |x − y| ≤ t} and show that g(t) ≤ g(t/2). Exercise 17.15. Let 1 ≤ p, q ≤ ∞ and let 0 < s < 1. Prove that if u ∈ Bqs,p (RN ) and ψ ∈ C 0,1 (RN ), then ψu ∈ Bqs,p (RN ) and estimate ψuBqs,p (RN ) in terms of uBqs,p (RN ) and ψC 0,1 (RN ) . Exercise 17.16. Let 1 ≤ p, q ≤ ∞ and let s > 0. Prove that if u ∈ S(RN ), then u ∈ Bqs,p (RN ).
17.2. Some Equivalent Seminorms In this section we introduce some equivalent seminorms, which are useful in different applications. We begin by showing that in the definition of Bqs,p , s+1 with Δnh for any n ∈ N with n > s + 1. To be one can replace Δh precise, let 1 ≤ p, q ≤ ∞, let s > 0, and let n ∈ N be such that n ≥ s + 1. For every Lebesgue measurable function u : RN → R if q < ∞ define
1/q dh (n) q n Δh uLp (RN ) , (17.21) |u|B s,p (RN ) := q hN +sq RN
546
17. Besov Spaces
while for q = ∞, (n)
|u|B s,p (RN ) := sup ∞
h
1 Δn u p N . hs h L (R )
Proposition 17.17. Let 1 ≤ p, q ≤ ∞, let s > 0, and let n ∈ N be such that n ≥ s + 1. Then there exist two constants c1 , c2 > 0, depending only on n, N , p, q, and s, such that for every Lebesgue measurable function u : RN → R with |u|Bqs,p (RN ) < ∞, (n)
(n)
c1 |u|B s,p (RN ) ≤ |u|Bqs,p (RN ) ≤ c2 |u|B s,p (RN ) . q
q
We begin with a preliminary result. Lemma 17.18. Given u : RN → R, h ∈ RN \ {0}, and n ∈ N, the following identity holds: (17.22)
Δnh u = 2−n Δn2h u + Pn−1 (Th ) ◦ Δn+1 h u,
where Th is the translation operator given in (17.10) and Pn−1 is a polynomial of degree n − 1. Proof. We use the following identity for polynomials (t − 1)n = 2−n (t2 − 1)n + (t − 1)n − 2−n (t2 − 1)n = 2−n (t2 − 1)n + Pn−1 (t)(t − 1)n+1 ,
t ∈ R,
where by the binomial theorem, (t + 1)n − 2n ((t − 1) + 2)n − 2n = −2−n t−1 t−1 n 2−j nj (t − 1)j−1 , t ∈ R. =−
Pn−1 (t) = −2−n
j=1
By replacing t with the translation operator Th given in (17.10) we get (17.22). Exercise 17.19. Let u : RN → R, h ∈ RN \ {0}, and n ∈ N. (i) Prove that there exists a polynomial P such that for every t ∈ R, (t − 1)n = −2−2n−1 (t4 − 1)n + 2−n−1 3(t2 − 1)n + P (t)(t − 1)n+2 . (ii) Prove that Δnh u = −2−2n−1 Δn4h u + 2−n−1 3Δn2h u + P (Th ) ◦ Δn+2 h u. We turn to the proof of Proposition 17.17.
17.2. Some Equivalent Seminorms
547
Proof of Proposition 17.17. Using an induction argument, it is enough (n) (n+1) to prove that the seminorms |u|B s,p and |u|B s,p are equivalent for some q
n ≥ s + 1 whenever
(n) |u|B s,p q
q
< ∞. By (17.7) for m = 1 we have that
1 n n p p p Δn+1 h uL = Δh (Δh u)L ≤ 2Δh uL .
Thus, it remains to prove the opposite inequality. Step 1: We consider first the case 0 < s < 1 and n = 1. Fix a Lebesgue measurable function u : RN → R with |u|Bqs,p < ∞. For x ∈ RN and h ∈ RN , we have 2(u(x + h) − u(x)) = (u(x + 2h) − u(x)) − (u(x + 2h) − 2u(x + h) + u(x)), and so 2Δ1h u(x) = Δ12h u(x) − Δ2h u(x).
(17.23)
If q < ∞, it follows by Minkowski’s inequality that 2 RN
Δ1h uqLp
+
RN
+ RN
dh hN +sq
Δ2h uqLp Δ2h uqLp
1/q ≤
dh hN +sq dh hN +sq
RN
Δ12h uqLp
1/q
=2
s RN
1/q
dh hN +sq
Δ1η uqLp
1/q
dη ηN +sq
1/q
,
where we have made the change of variables h = η/2. The case q = ∞ is similar. Since |u|Bqs,p < ∞, we have (2)
(2 − 2s )|u|Bqs,p ≤ |u|B s,p . q
Step 2: Assume next that n ≥ 2. Fix a Lebesgue measurable function (n) u : RN → R with |u|B s,p < ∞. By (17.22) we get q
p Δnh uLp ≤ 2−n Δn2h uLp + Pn−1 (Th ) ◦ Δn+1 h uL p ≤ 2−n Δn2h uLp + cΔn+1 h uL ,
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17. Besov Spaces
where we have used the fact that Th (v)Lp = vLp . If q < ∞, it follows by Minkowski’s inequality that
1/q
1/q dh dh q −n n Δnh uqLp ≤ 2 Δ u 2h Lp hN +sq hN +sq RN RN
1/q dh q n+1 +c Δh uLp hN +sq RN
1/q dη q s−n n =2 Δη uLp ηN +sq RN
1/q dh q +c Δn+1 u , Lp h hN +sq RN where we have made the change of variables h = η/2. Therefore, since (n) |u|B s,p < ∞ we get q
(1 − 2
s−n
) RN
Δnh uqLp
≤c
RN
dh hN +sq
1/q
dh q Δn+1 h uLp hN +sq
1/q .
Note that here it is important that n ≥ s + 1 > s. The case q = ∞ is similar. Remark 17.20. It is important to observe that the restriction |u|Bqs,p (RN ) < ∞ is needed. Indeed, if u is a polynomial of degree d > s + 1, then |u|Bqs,p (RN ) = ∞, while Δnh u(x) = 0 for all x ∈ RN if n > d. Another useful seminorm is the following: Let 1 ≤ p, q ≤ ∞ and s > 0. For every Lebesgue measurable function u : RN → R and for 1 ≤ q < ∞ define ∞
dt 1/q s+1 q ∞ sup Δh uLp 1+qs , (17.24) |u|Bqs,p (RN ) := t 0 h ≤t while for q = ∞, s+1
−s s,p sup Δh |u|∞ (RN ) := sup t B∞ t>0
h ≤t
uLp (RN ) .
Proposition 17.21. Let 1 ≤ p, q ≤ ∞ and let s > 0. Then there exist two constants c1 , c2 > 0 depending only on N , p, q, and s such that for every Lebesgue measurable function u : RN → R, ∞ s,p c1 |u|∞ Bqs,p (RN ) ≤ |u|Bq (RN ) ≤ c2 |u|Bqs,p (RN ) .
We begin with a preliminary result.
17.2. Some Equivalent Seminorms
549
Lemma 17.22. Let 1 ≤ p ≤ ∞ and let m ∈ N. Then there exists a constant c = c(m, N, p) > 0 such that for every Lebesgue measurable function u : RN → R and every t > 0, dζ m Δm . sup Δh uLp ≤ c ζ uLp ζN B(0,t)
h ≤t Proof. By the binomial theorem for y, z ∈ R, we have m
m m k m jk m−k m m−k = (−1)j+k m (y − 1) z k j y z
(−1)m−k m k
k=0
=
k=0 j=0 m
(−1)m−j
m j m j (y − z) ,
j=0
which, after relabeling the summation index in the last sum, gives (z − 1)
m
m k m m−k = (−1)m−k m − (−1)m (z − y k )m ]. k [(y − 1) z k=1
By replacing y and z with the operators T(1/m)ξ and Th (see (17.10)), respectively, where h, ξ ∈ RN , we get Δm h u(x)
m m = (−1)m−k m k [Δ(k/m)ξ u(x + (m − k)h) k=1
− (−1)m Δm h−(k/m)ξ u(x + kξ)]. In turn, Δm h uLp ≤ c
m
m [Δm (k/m)ξ uLp + Δh−(k/m)ξ uLp ].
k=1
Now, let h = 0 and ξ ∈ B(h/2, h/2). Then (k/m)ξ, h − (k/m)ξ ∈ B(0, h), and so by averaging in the variable ξ over B(h/2, h/2) we obtain m c m m [Δm Δh uLp ≤ (k/m)ξ uLp + Δh−(k/m)ξ uLp ] dξ hN B(h/2, h /2) k=1 c Δm ≤ ζ uLp dζ, hN B(0, h ) where we have used the changes of variables ζ = (k/m)ξ and ζ = h−(k/m)ξ. It follows that dζ m Δm , sup Δh uLp ≤ c ζ uLp ζN
h ≤t B(0,t) which concludes the proof.
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17. Besov Spaces
We turn to the proof of Proposition 17.21. Proof of Proposition 17.21. We only prove the case 1 ≤ q < ∞ and leave the case q = ∞ as an exercise. Let m := s + 1. By Lemma 17.22 and using spherical coordinates we get ∞ ∞ dt dζ q dt q m sup Δh uLp 1+qs ≤ c Δm ζ uLp t ζN t1+qs 0 h ≤t 0 B(0,t) ∞ t dρ q dt N −1 p =c Δm u dH (σ) . L ρσ ρ t1+qs 0 0 S N −1 By Hardy’s inequality (see Theorem C.41) and H¨older’s inequality, up to a multiplicative constant, we can bound from above the right-hand side of the previous inequality by ∞ q dt N −1 p dH Δm u (σ) c L tσ t1+qs 0 S N −1 ∞ dt dy q q N −1 ≤c Δm u dH (σ) = c Δm . tσ y uLp Lp 1+qs t yN +qs 0 S N −1 RN This shows that |u|∞ B s,p ≤ c|u|B s,p . On the other hand, again by spherical coordinates, ∞ dh dt q q m N −1 Δh uLp = Δm (σ) 1+qs tσ uLp dH N +sq h t N N −1 R 0 ∞ S dt q N −1 ≤ sup Δm (σ) 1+qs h uLp dH t 0 S N −1 h ≤t ∞ dt q = βN sup Δm h uLp 1+qs , t 0 h ≤t where we recall that βN = HN −1 (S N −1 ). This concludes the proof. Exercise 17.23. Let 1 ≤ p, q ≤ ∞ and s > 0, and let 0 < R < ∞. (i) Prove that [R]
u → uLp (RN ) + |u|B s,p (RN ) q
is an equivalent norm in Bqs,p (RN ), where for q < ∞,
1/q dh [R] s+1 Δh uqLp (RN ) , |u|B s,p (RN ) := q hN +sq B(0,R) while for q = ∞, [R]
|u|B s,p (RN ) := ∞
1 s+1 Δh uLp (RN ) . s h h∈B(0,R)\{0} sup
17.3. Besov Spaces as Interpolation Spaces
551
(ii) Similarly, prove that given n ≥ s + 1, an equivalent norm in Bqs,p (RN ) for q < ∞ is given by R
dt 1/q q n u → uLp (RN ) + sup Δh uLp 1+qs t 0 h ≤t and for q = ∞ by 1 s+1 sup Δh uLp (RN ) . s t t∈(0,R)
h ≤t
u → uLp (RN ) + sup
17.3. Besov Spaces as Interpolation Spaces In this section we show that the interpolation space between Lp (RN ) and W m,p (RN ) is the Besov space Bqσm,p (RN ). We recall that interpolation spaces have been defined in Chapter 16. Theorem 17.24. Let m ∈ N, 1 ≤ p, q ≤ ∞, and 0 < σ < 1. Then (Lp (RN ), W m,p (RN ))σ,q = Bqσm,p (RN ). We begin with a preliminary result, which will be used also in the sequel. Lemma 17.25. Let u ∈ C ∞ (RN ) and let m ∈ N. Then for every h ∈ RN \ {0} and x ∈ RN , m m ∂ u m m (17.25) (x + th)χm (t) dt Δh u(x) = h m 0 ∂ν m 1 m α ν ∂ α u(x + th)χm (t) dt, = h α! 0 |α|=m
where ν := h/h, χ1 := χ(0,1) and χm := χ(0,1) ∗ · · · ∗ χ(0,1) (m times) for m ≥ 2. In particular, there is a constant c = c(m, N ) > 0 such that for every x ∈ RN , m m m tm−1 ∇m u(x + th) dt. (17.26) |Δh u(x)| ≤ ch 0
Proof. Fix x ∈ RN and h ∈ RN \{0} and define f (t) := u(x+th), t ∈ [0, m]. m Then Δm h u(x) = Δ1 f (0). By the fundamental theorem of calculus and induction on m we have that 1 1 m ··· f (m) (t1 + · · · + tm ) dt1 · · · dtm Δ1 f (0) = 0 0 = · · · χI (t1 ) · · · χI (tm )f (m) (t1 + · · · + tm ) dt1 · · · dtm R R = f (m) (t)χm (t) dt. R
552
17. Besov Spaces
Writing f (t) = u(x + htν), it follows by the chain rule and an induction argument that 1 ∂ mu ν α ∂ α u(x + th), f (m) (t) = hm m (x + th) = hm ∂ν α! |α|=m
which, together with the previous identity, shows (17.25). Inequality (17.26) now follows from the facts that χm (t) = 0 for t ≤ 0 and t ≥ m and χm (t) ≤ tm−1 for all 0 ≤ t ≤ m. We turn to the proof of Theorem 17.24. Proof of Theorem 17.24. Step 1: Since W m,p (RN ) ⊂ Lp (RN ), by Theorem 16.5 we have that (Lp (RN ), W m,p (RN ))σ,q → Lp (RN ) + W m,p (RN ) = Lp (RN ). Let u ∈ (Lp (RN ), W m,p (RN ))σ,q and let v ∈ Lp (RN ) and w ∈ W m,p (RN ) be such that u = v + w. Then for every h ∈ RN \ {0}, by (17.7), (17.26), Tonelli’s theorem and a change of variables, m m Δm h uLp ≤ Δh vLp + Δh wLp
≤ 2m vLp + chm ∇m wLp . Taking the infimum over all possible decompositions of u and using (16.3) gives (17.27)
m Δm h uLp ≤ cK(u, h ).
In turn, if 1 ≤ q < ∞, by (16.5) and (17.5), using spherical coordinates and the change of variables rm = t,
1/q dh m q (K(u, h )) |u|Bqσm,p ≤ c hN +σmq RN ∞
1/q dr =c (K(u, rm ))q 1+σmq = cuσ,q , r 0 where as usual the constant c changed from expression to expression. On the other hand, if q = ∞, then by multiplying both sides of (17.27) by h−mσ σm,p ≤ cuσ,q . it follows from (16.6) and (17.24) that |u|B∞ Step 2: For simplicity we present the proof of the opposite inequality first for m = 1. In view of Exercise 16.7, it suffices to consider 0 < t < 1. Given u ∈ Bqσ,p (RN ) and t > 0, for x ∈ RN write 1 1 Δy u(x) dy + N u(x + y) dy u(x) = − N t t Q(0,t) Q(0,t) =: vt (x) + wt (x).
17.3. Besov Spaces as Interpolation Spaces
553
By Minkowski’s inequality for integrals (see Corollary B.83), 1 vt Lp ≤ N Δy uLp dy ≤ sup Δh uLp . √ t Q(0,t)
h ≤ N t √ Hence, by (17.24) and the change of variables τ = N t, 1 1 dt dt q q vt Lp 1+σq ≤ sup Δh uqLp 1+σq ≤ c(|u|∞ Bqσ,p ) . √ t t 0 0 h ≤ N t To estimate wt , given i = 1, . . . , N , by a change of variables and using the notation (E.2) we have xi +t/2 t/2 u(xi + yi , xi + yi ) dyi = u(xi + yi , τi ) dτi . −t/2
xi −t/2
Hence, for LN −1 -a.e. xi ∈ RN −1 , t/2 ∂ u(xi + yi , xi + yi ) dyi = Δtei u(xi + yi , xi − t/2), ∂xi −t/2 which by Fubini’s theorem and differentiation under the integral sign implies that 1 Δtei u(xi + yi , xi − t/2) dyi . ∂i wt (x) = N t QN −1 (0,t) By Minkowski’s inequality for integrals (see Corollary B.83) and a change of variables, 1 1 sup Δh uLp . Δtei uLp dyi ≤ ∂i wt Lp ≤ N t t h ≤√N t QN −1 (0,t) In turn, ∞ t 0
q
dt ∂i wt qLp 1+σq t
∞
≤ 0
sup √
h ≤ N t
Δh uLp
dt t1+σq
q = c(|u|∞ Bqσ,p ) .
On the other hand, by (C.8), wt Lp ≤ uLp and so 1 1 1 dt q q q uqLp . t wt Lp 1+σq ≤ uLp tq(1−σ)−1dt = t q(1 − σ) 0 0 Note that this is the only estimate which uses heavily the fact that 0 < t < 1. In conclusion we have shown that vt ∈ Lp (RN ), wt ∈ W 1,p (RN ), and 1 1 dt dt q (K(u, t))q 1+σq ≤ (vt qLp + tq wt qW 1,p ) 1+σq ≤ cuqLp + c(|u|∞ Bqσ,p ) . t t 0 0 The case q = ∞ is simpler and is left as an exercise.
554
17. Besov Spaces
Step 3: Next we study the case m = 2. Given u ∈ Bq2σ,p (RN ) and t > 0, for x ∈ RN write 1 Δ2y+z u(x) dydz u(x) = 2N t Q(0,t) Q(0,t) 1 (17.28) (u(x + 2(y + z)) − 2u(x + y + z)) dydz − 2N t Q(0,t) Q(0,t) =: vt (x) + wt (x). By Minkowski’s inequality for integrals (see Corollary B.83), 1 Δ2y+z uLp dydz ≤ sup√ Δ2h uLp . vt Lp ≤ 2N t Q(0,t) Q(0,t)
h ≤2 N t √ Hence, by (17.24) and the change of variables τ = 2 N t,
1 0
dt vt qLp 1+2σq t
1
≤
sup√
0 h ≤2 N t
Δ2h uqLp
dt t1+2σq
≤ c(|u|∞ )q . B 2σ,p q
To estimate wt , define 1
ft (x) :=
u(x + 2(y + z)) dydz, t2N Q(0,t) Q(0,t) 1 gt (x) := 2N u(x + y + z) dydz. t Q(0,t) Q(0,t)
Given i = 1, . . . , N , by a change of variables and using the notation (E.2) and Fubini’s theorem, we have
t/2 −t/2
t/2 −t/2
u(xi + yi + zi , xi + yi + zi ) dyi dzi
xi +t/2 ξi +t/2
= xi −t/2
ξi −t/2
u(xi + yi + zi , τi ) dτi dξi .
Hence, for LN −1 -a.e. xi ∈ RN −1 , ∂2 ∂x2i
t/2 −t/2
t/2 −t/2
u(xi + yi + zi , xi + yi + zi , ) dyi dzi =∼Δ2tei u(xi + yi + zi , xi ),
which implies that (17.29)
∂i2 gt (x)
=
1 t2N
QN −1 (0,t)
QN −1 (0,t)
∼ 2 Δtei u(xi
+ yi + zi , xi ) dyi dzi .
17.3. Besov Spaces as Interpolation Spaces
555
Similarly, (17.30) 2 ∂i,j gt (x)
=
1 t2N
QN −1 (0,t)
QN −1 (0,t)
Δtej (Δtei u(xi,j + yi,j + zi,j ,
xi + zi − t/2, xj + zj − t/2)) dyi dzj ,
where for N > 2, xi,j ∈ RN −2 is the vector obtained from x by removing the ith and jth components and with a slight abuse of notation we write x = (xi,j , xi , xj ). Similar computations can be done for ft . In view of Exercise 17.26 for every i, j = 1, . . . , N , Δtej (Δtei u)Lp ≤ sup Δ2h uLp .
h ≤ct
By Minkowski’s inequality for integrals (see Corollary B.83), it follows from (17.29), (17.30), and the analogous identities for ft that, ∇2 wt Lp ≤ ct−2 sup Δ2h uLp .
h ≤ct
In turn, 1 0
t ∇ 2q
2
dt wt qLp 1+2σq t
≤c
1
sup Δ2h uqLp
0 h ≤ct
dt t1+2σq
≤ c(|u|∞ )q . B 2σ,p q
On the other hand, by (C.8), wt Lp ≤ uLp , and so 1 1 1 dt q q 2q uqLp . t wt Lp 1+2σq ≤ uLp t2q(1−σ)−1 dt = t 2q(1 − σ) 0 0 Note that this is the only estimate which uses heavily the fact that 0 < t < 1. In conclusion we have shown that vt ∈ Lp (RN ), wt ∈ W 2,p (RN ), and 1 1 dt q dr (K(u, r)) 1+σq = 2 (K(u, t2 ))q 1+2σq r t 0 0 1 dt ≤c (vt qLp + t2q wt qW 2,p ) 1+2σq ≤ cuqLp + c(|u|∞ )q . Bq2σ,p t 0 Step 4: The case m > 2 is similar to Step 3. The main difference is that in (17.28) to define the function vt we use m repeated integrals over Q(0, t), N for rescale by tmN , replace Δ2y+z u(x) with Δm h1 +···+hm u(x) where hi ∈ R i = 1, . . . , m, and use Exercise 17.26. We leave the details as an exercise. Exercise 17.26. Let h1 , . . . , hm ∈ RN and let Thi be the translation operator defined in (17.10). (i) Prove that Δ2h2 ◦ Δh1 = −Th1 ◦ Δ2−h1 − T2h2 ◦ Δ2−h2 + Th1 +2h2 ◦ Δ2−h1 −h2 .
556
17. Besov Spaces
(ii) Prove that for every 0 ≤ ≤ m, (T(m−)hm − I) ◦ · · · ◦ (T(1−)h1 − I) =
m (−1)m−card E Th∗E ◦ T−h , E E
∗ where the sum is taken over all subsets E ⊆ {1, . . . , m}, hE := k∈E khk and hE := k∈E hk .
(iii) Prove that Δmhm ◦ · · · ◦ Δh1 =
(−1)card E Th∗E ◦ Δm −hE .
E
Remark 17.27. Steps 2–4 of the previous proof show that for u ∈ L1loc (RN ) with |u|Bqσm,p (RN ) < ∞ it is always possible to decompose u = v + w with ˙ m,p (RN ). The previous proof actually shows that for v ∈ Lp (RN ), w ∈ W every u ∈ L1loc (RN ) and every t > 0, m ˙ ˙ tm ) ≤ sup Δm c1 K(u, h uLp (RN ) ≤ c2 K(u, t ),
h ≤ct
where ˙ K(u, t) := inf{vLp (RN ) + t|w|W m,p (RN ) }, where the infimum is taken over all possible decompositions u = v + w, ˙ m,p (RN ). In turn, by the change of variables rm = t, v ∈ Lp (RN ), w ∈ W c−1 uσ,q ≤ |u|∞ Bqσm,p (RN ) ≤ cuσ,q , where · σ,q is defined as in (16.5) with K˙ in place of K and | · |∞ Bqσm,p (RN ) is given in (17.24). Motivated by the previous remark we give the following definition. Definition 17.28. Let 1 ≤ p, q ≤ ∞ and let s > 0. A function u ∈ L1loc (RN ) belongs to the homogeneous Besov space B˙ qs,p (RN ) if |u|Bqs,p (RN ) < ∞, where | · |Bqs,p (RN ) is given in (17.5) and (17.6). Remark 17.29. By Remark 17.27 every function u ∈ B˙ qs,p (RN ), where 1 ≤ p, q ≤ ∞, and s > 0, can be written as the sum of a function v in ˙ m,p (RN ) where m > s can be taken arbitrarily Lp (RN ) and a function w in W large. In particular, if mp > N , then by Remark 12.56, the function w has polynomial growth. It follows that u can be identified with a tempered distribution (see Definition 10.32). In view of Remark 17.27 we have the following interpolation result. Theorem 17.30. Let m ∈ N, 1 ≤ p, q ≤ ∞, and 0 < σ < 1. Then ˙ m,p (RN ))σ,q = B˙ σm,p (RN ). (Lp (RN ), W q
17.3. Besov Spaces as Interpolation Spaces
557
Remark 17.31 (Important). We observe that strictly speaking we cannot ˙ m,p (RN ) since | · |W m,p (RN ) is apply the interpolation theory to the space W only a seminorm and not a norm. One could still define interpolation spaces starting from an admissible pair of seminormed spaces instead of normed spaces in the definition of the function K in (16.3). With this modified definition, Theorem 17.30 would be legitimate. However, many theorems in Chapter 16 rely on the fact that the admissible pairs of spaces are endowed ˙ m,p (RN ) with with a norm. This difficulty can be overcome by replacing W m,p N N N ˙ (R )/Pm−1 (R ), where Pm−1 (R ) is the family the quotient space W N of polynomials P : R → R of degree less than or equal to m − 1. In this quotient space | · |W m,p (RN ) becomes a norm. In the sequel we will sometimes ˙ m,p (RN )/Pm−1 (RN ), without further notice. ˙ m,p (RN ) with W identify W Exercise 17.32. Prove that for 0 < s < 1, (C(RN ), C 1 (RN ))s,∞ = C 0,s (RN ). Hint: Follow the proof of Theorem 17.24. In view of Theorems 16.5 and 17.24, the fact that W m,p (RN ) → Lp (RN ), and Exercise 16.7 we have the following embedding. Theorem 17.33. Let 1 ≤ p, q ≤ ∞, let s > 0, and let m ∈ N be such that m ≥ s + 1. Then there exists a constant c = c(m, N, p, q, s) > 0 such that (17.31)
|u|Bqs,p (RN ) ≤ cuLp (RN ) + c∇m uLp (RN ;RMm )
for all u ∈ W m,p (RN ). In particular, W m,p (RN ) → Bqs,p (RN ). Remark 17.34. Note that if we had used the seminorm in Exercise 17.23, then the term cuLp (RN ) would not be needed on the right-hand side of (17.31). Using Lemma 17.25 it is possible to give a direct proof of this theorem. Exercise 17.35. Let 1 ≤ p, q ≤ ∞, let s > 0, and let m ∈ N be such that m ≥ s + 1. (i) If 1 ≤ q < ∞, prove that for all u ∈ W m,p (RN ),
1/q dh q m (17.32) Δh uLp hN +sq RN
1/q 1/q ∂ m u q 2m β N 1 N −1 dH (ν) + uLp , ≤ [(m − s)q]1/q S N −1 ∂ν m Lp (sq)1/q where βN = HN −1 (S N −1 ). Hint: Write RN = B(0,1) + RN \B(0,1) and use Lemma 17.25.
558
17. Besov Spaces
(ii) If q = ∞, prove that for all u ∈ W m,p (RN ), ∂mu 1 m p sup Δ u ≤ sup m p + 2m uLp . L h s h ∂ν L N
ν =1 h∈R \{0} Exercise 17.36. Let 0 < s < 1 and 1 ≤ q ≤ ∞. Prove that there exists a constant c = c(N, q, s) > 0 such that |u|Bqs,1 (RN ) ≤ cuBV (RN ) for all
u ∈ BV (RN ).
As a corollary of Theorem 17.33 we can prove the density of Cc∞ (RN ) in Bqs,p (RN ) for 1 ≤ p, q < ∞. Theorem 17.37. Let 1 ≤ p, q < ∞ and s > 0. Then Cc∞ (RN ) is dense in Bqs,p (RN ) and B˙ qs,p (RN ). Proof. Step 1: By Theorem 17.24, (Lp (RN ), W m,p (RN ))σ,q = Bqs,p (RN ), where m > s, σ := s/m. In turn, by Exercise 16.22 we have that W m,p (RN ) is dense in Bqs,p (RN ). By the density of Cc∞ (RN ) in W m,p (RN ) (see Theorem 11.35) and (17.31), it follows that Cc∞ (RN ) is dense in Bqs,p (RN ). ˙ m,p (RN ) by Corollary 12.86, again Step 2: Since W m,p (RN ) = Lp (RN ) ∩ W by Exercise 16.22 and Theorem 17.30 we have that W m,p (RN ) is dense in B˙ qs,p (RN ). We can continue as in the previous step to conclude that Cc∞ (RN ) is dense in B˙ qs,p (RN ). Theorems 16.11 and 17.24 give the following embedding. Theorem 17.38. Let 1 ≤ p ≤ ∞, let 1 ≤ q1 < q2 ≤ ∞, and let s > 0. Then there exists a constant c = c(N, p, q1 , q2 , s) > 0 such that |u|Bqs,p (RN ) ≤ c|u|Bqs,p (RN ) 2
for all u ∈ N Bqs,p 2 (R ).
L1loc (RN ).
In particular,
1
N B˙ qs,p 1 (R )
s,p N N → B˙ qs,p 2 (R ) and Bq1 (R ) →
Using the seminorm | · |∞ B s,p defined in (17.24), it is possible to give an alternative proof of Theorem 17.38, as described in the next two exercises. Exercise 17.39. Let 1 ≤ q1 < q2 < ∞. (i) Given a sequence {ak }k∈Z of nonnegative numbers, prove that 1/q2 1/q1 aqk2 ≤ aqk1 . k∈Z
k∈Z
(ii) Given an increasing function g : [0, ∞) → [0, ∞) and r > 0, prove that there exists a constant c > 0 independent of g and q2 such that
∞ ∞ g(t) q2 dt 1/q2 g(t) q1 dt 1/q1 ≤c . tr t tr t 0 0
17.3. Besov Spaces as Interpolation Spaces
559
(iii) Using part (ii), prove that
∞ g(t) q1 dt 1/q1 g(t) . sup r ≤ c tr t t>0 t 0 Exercise 17.40. Use the previous exercise to give an alternative proof of Theorem 17.38 using the equivalent seminorm | · |∞ B s,p defined in (17.24). In view of the duality theorem for interpolation spaces (see Theorem 16.27) and of Theorem 16.30, we can identify the dual of Bqs,p (RN ). Theorem 17.41 (Duality). Let m ∈ N, 1 ≤ p ≤ ∞, 1 ≤ q < ∞, and 0 < σ < 1. Then the dual of Bqσm,p (RN ) can be identified with
(Lp (RN ), (W m,p (RN )) )σ,q . Moreover, if 1 < p < ∞, then Bqσm,p (RN ) is reflexive. As a consequence of the reiteration theorem (Theorem 16.15) and of Theorems 17.24 and 17.30 we have the following result. Corollary 17.42. Let m ∈ N, 1 ≤ p, q, q0 , q1 ≤ ∞, 0 < σ < 1, and 0 < s0 < s1 . Then (B˙ qs00 ,p (RN ), B˙ qs11 ,p (RN ))σ,q = B˙ qs,p (RN ), (Bqs00 ,p (RN ), Bqs11 ,p (RN ))σ,q = Bqs,p (RN ), where s := (1 − σ)s0 + σs1 . In view of the previous corollary and Exercise 16.13 we have the following inequality, which is similar in spirit to the Gagliardo–Nirenberg interpolation theorem (see Theorem 12.87). Corollary 17.43. Let m ∈ N, 1 ≤ p, q, q0 , q1 ≤ ∞, 0 < σ < 1, and 0 < s0 < s1 . Then there is a constant c = c(m, p, q, q0 , q1 , σ, s0 , s1 ) > 0 such that 1−σ |u|Bqs,p (RN ) ≤ c|u|B |u|σB s1 ,p (RN ) s0 ,p (RN ) q0
q1
for all u ∈ B˙ qs00 ,p (RN ) ∩ B˙ qs11 ,p (RN ), where s := (1 − σ)s0 + σs1 . A similar result holds for Bqs,p (RN ). We conclude this section with a few remarks on Besov spaces defined on open sets Ω ⊆ RN . In view of Theorem 17.24, given 1 ≤ p, q ≤ ∞ and s > 0, one can define the Besov spaces Bqs,p (Ω) and B˙ qs,p (Ω) as Bqs,p (Ω) := (Lp (Ω), W m,p (Ω))s/m,q , ˙ m,p (Ω))s/m,q , B˙ s,p (Ω) := (Lp (Ω), W q
560
17. Besov Spaces
where m ∈ N is such that m > s. If ∂Ω is uniformly Lipschitz continuous, then it follows from the proof of Theorems 13.8 and 13.17 (see (13.25)) that there exists a continuous linear operator E : Lp (Ω) → Lp (RN ) with the property that for all u ∈ Lp (Ω), E(u)(x) = u(x) for LN -a.e. x ∈ Ω. Moreover, E : W m,p (Ω) → W m,p (RN ) is linear and continuous. It follows by Theorem 16.12 that E : Bqs,p (Ω) → Bqs,p (RN )
(17.33) with (17.34)
1−σ σ EL(Bs,p (Ω);Bs,p (RN )) ≤ EL(L p (Ω);Lp (RN )) EL(W m,p (Ω);W m,p (RN )) . q
q
Since for all u ∈ E(u)(x) = u(x) for LN -a.e. x ∈ Ω, it follows in particular that for all u ∈ B s,p (Ω), E(u)(x) = u(x) for LN -a.e. x ∈ Ω. We remark that (17.33) can be used to obtain embedding theorems in Bqs,p (Ω). Indeed, one can first prove embeddings for Bqs,p (RN ) (see the following section) and then use (17.33) to obtain the same embeddings for Bqs,p (Ω). Lp (Ω),
Exercise 17.44 (Compactness). Let Ω ⊂ RN be an open bounded set with Lipschitz continuous boundary, let 1 ≤ p, q ≤ ∞, and let 0 < s < 1. Prove that the embedding Bqs,p (Ω) → Lp (Ω) is compact. Hint: Use the fact that the embedding W 1,p (Ω) → Lp (Ω) is compact (see Theorem 12.30) and Exercises 16.10 and 16.14. Exercise 17.45 (Compactness). Let Ω ⊂ RN be an open bounded set with Lipschitz continuous boundary, let 1 ≤ p, q ≤ ∞, and let 0 < s1 < s2 < 1. Prove that the embedding Bqs2 ,p (Ω) → Bqs1 ,p (Ω) is compact. Hint: Use the fact that the embedding W 1,p (Ω) → Lp (Ω) is compact (see Theorem 12.30) and Exercise 16.26. Exercise 17.46. Let Ω, U ⊆ RN be open sets, let Ψ : U → Ω be invertible, with Ψ and Ψ−1 Lipschitz continuous functions, let 1 ≤ p, q ≤ ∞, and let 0 < s < 1. Prove that u ∈ Bqs,p (Ω) if and only if u ◦ Ψ ∈ Bqs,p (U ) and estimate u ◦ ΨBqs,p (U ) . Hint: Use Theorems 11.53 and 16.12. Exercise 17.47. Let Ω ⊆ RN be an open set with bounded Lipschitz continuous boundary, let 1 ≤ p, q ≤ ∞, and let 0 < s < 1. Prove that u → uLp (Ω) + |u|Bqs,p (Ω) , where | · |Bqs,p (Ω) is defined in (17.13), is an equivalent norm in Bqs,p (Ω). Hint: Consider first the case in which Ω = RN + and use Exercise 17.11, then use a partition of unity together with Exercises 17.15 and 17.46. Exercise 17.48 (Poincar´e’s inequality). Let B be an open ball of radius r > 0, let 1 ≤ p < ∞, and 0 < s < 1. Prove that there exists a constant c = c(N, p, s) > 0 such that for every u ∈ B˙ s,p (B), u − uB Lp (B) ≤ crs |u|B s,p (B) ,
17.4. Sobolev Embeddings
561
where we recall that (see (13.60)) for a measurable set E ⊂ RN of positive finite measure, uE := LN1(E) E u(x) dx. Hint: Use Poincar´e’s inequality for balls (see, e.g. Theorem 13.41) and Theorem 16.12.
17.4. Sobolev Embeddings In this section we use interpolation theory to extend the embeddings for Sobolev spaces to Besov spaces. We begin with the Sobolev–Gagliardo– Nirenberg embedding theorem (Theorem 12.4). Given s > 0 and 1 ≤ p < N/s we define the Sobolev critical exponent. (17.35)
p∗s :=
Np . N − sp
Note that p∗1 = p∗ , which is the Sobolev critical exponent for W 1,p (RN ) (see (12.2)). In what follows Remark 17.31 will be of particular relevance. Theorem 17.49 (Sobolev–Gagliardo–Nirenberg embedding in Bqs,p ). Let s > 0, 1 ≤ p < N/s, and 1 ≤ q ≤ ∞. Then for every function u ∈ B˙ qs,p (RN ) there exists a polynomial Pu such that (17.36)
u − Pu Lp∗s ,q (RN ) ≤ c|u|Bqs,p (RN )
for some constant c = c(N, p, q, s) > 0. In particular, B˙ s,p (RN )/P(RN ) → ∗ Lps (RN ) and Bqs,p (RN ) → Lr (RN ) for all p ≤ r ≤ p∗s and 1 ≤ q ≤ p∗s . Proof. Step 1: Let m ∈ N be such that mp > N . By the Gagliardo– Nirenberg interpolation theorem (see Theorem 12.87) with k = 0, β = 0, q = p < r < ∞, and θ = 1 − σ, where σ := (1/p − 1/r)N/m ∈ (0, 1), 1−σ σ vLr ≤ cvL p |v|W m,p
˙ m,p (RN ). Since by (16.21) for every t > 0, J(v, ˙ t) := for every v ∈ Lp (RN )∩W max{vLp , t|v|W m,p }, from the previous inequality we deduce that ˙ t). vLr ≤ ct−σ J(v, Step 2: Let m ∈ N be such that mp > N and let 0 < σ < 1. Given ˙ m,p (RN ) be a Bochner integrable u ∈ B˙ 1σm,p (RN ) let v : (0, ∞) → Lp (RN )∩W function such that ∞ dt (v(t))(x) , x ∈ RN . u(x) = t 0 By Minkowski’s inequality for integrals (see Corollary B.83), ∞ ∞ ∞ dt dt dt |v(t)(·)| ≤ v(t)Lr ≤ c J˙(v(t), t) 1+σ , uLr ≤ t Lr t t 0 0 0
562
17. Besov Spaces
where in the last inequality we applied Step 1 to the function v(t) ∈ Lp (RN )∩ ˙ m,p (RN ) for every t > 0. By taking the infimum over all such functions v W and using (16.23) we get uLr ≤ cuσ,1,J ≤ cuσ,1 ≤ c|u|∞ B σm,p , 1
where in the last two inequalities we applied Theorems 16.21 and 17.24 and Remark 17.27. This shows that B˙ 1σm,p (RN ) → Lr (RN ). Step 3: By taking m ∈ N such that m > N/p > s we have that σ := s/m ∈ (0, 1) and thus, by the previous step (17.37)
B˙ 1s,p (RN ) → Lr (RN ),
where s = (1/p − 1/r)N , that is, r = ps , where ps is the critical exponent given in (17.35). To obtain (17.36), consider 0 < s0 < s < s1 < N/p and let σ ∈ (0, 1) be such that s = (1 − σ)s0 + σs1 . By (17.37) for s0 and s1 , B˙ 1s0 ,p (RN ) → Lps0 (RN ),
B˙ 1s1 ,p (RN ) → Lps1 (RN ).
In turn, by Exercise 16.34, Theorem 16.12 and Corollary 17.42, B˙ qs,p (RN ) = (B˙ 1s0 ,p (RN ), B˙ 1s1 ,p (RN ))σ,q → (Lps0 (RN ), Lps1 (RN ))σ,q = Lr,q (RN ), where 1/r = (1 − σ)/ps0 + σ/ps1 = 1/p − s/N .
N/p,p (RN ) Exercise 17.50. Let 1 ≤ p < ∞. Prove that for every u ∈ B˙ 1 there exists a polynomial Pu such that
u − Pu L∞ (RN ) ≤ c|u|B N/p,p (RN ) 1
for some constant c = c(N, p) > 0. Hint: Use (12.54) with r < 0 and a = 0, in Step 1. Exercise 17.51. Let Ω ⊂ RN be an open bounded set with Lipschitz continuous boundary, let 0 < s < 1, let 1 ≤ p < N/s, and let 1 ≤ q ≤ ∞. Prove that the embedding Bqs,p (Ω) → Lr (Ω) is compact for every 1 ≤ r < p∗s and 1 ≤ q ≤ p∗s . Hint: Use the previous theorem, (17.33), and Exercise 17.44. Next we extend Morrey’s embedding theorem (see Theorems 12.48 and 12.55) to Besov spaces. Theorem 17.52 (Morrey embedding in Bqs,p ). Let 1 ≤ p < ∞ and s > 0 be such that 0 < s − N/p < 1, and let 1 ≤ q ≤ ∞. Then for every function u ∈ B˙ qs,p (RN ) there exists a polynomial Pu such that (17.38)
|u − Pu |C 0,s−N/p (RN ) ≤ c|u|Bqs,p (RN )
17.4. Sobolev Embeddings
563
for some constant c = c(N, p, q, s) > 0. In particular, Bqs,p (RN ) → C 0,s−N/p (RN ). Proof. Step 1: Let m ∈ N be such that m > s. By the Gagliardo– Nirenberg interpolation theorem (see Theorem 12.87) with k = 0, β = 0, q = p, θ = 1 − s/m, and 1/r = 1/p − s/N < 0, 1−s/m
|v|C 0,s−N/p ≤ cvLp
s/m
|v|W m,p
˙ m,p (RN ) (see (12.54)). Since by (16.21) for every for every v ∈ Lp (RN ) ∩ W t > 0, J˙(v, t) := max{vLp , t|v|W m,p }, from the previous inequality we deduce that ˙ t). |v|C 0,s−N/p ≤ ct−s/m J(v, s,p ˙ m,p (RN ) be a Step 2: Given u ∈ B˙ 1 (RN ) let v : (0, ∞) → Lp (RN ) ∩ W Bochner integrable function such that ∞ dt (v(t))(x) , x ∈ RN . u(x) = t 0 Then for x, y ∈ RN ,
∞
dt |(v(t))(x) − (v(t))(y)| t 0 ∞ dt |v(t)|C 0,s−N/p ≤ x − ys−N/p t 0 ∞ dt ≤ cx − ys−N/p J˙(v(t), t) 1+s/m , t 0
|u(x) − u(y)| ≤
where in the last inequality we applied Step 1 to the function v(t) ∈ Lp (RN )∩ ˙ m,p (RN ) for every t > 0. By dividing by x − ys−N/p for x = y, taking W the supremum over all such x, y, then taking the infimum over all such functions v, and using (16.23) we get |u|C 0,s−N/p ≤ cus/m,1,J˙ ≤ cus/m,1 ≤ c|u|∞ B s,p , 1
where in the last two inequalities we applied Theorems 16.21 and 17.24 and Remark 17.27. This shows that (17.39) B˙ sm,p (RN ) → B˙ s−N/p,∞ (RN ). 1
To obtain (17.38), consider N/p < s0 < s < s1 < 1 + N/p and let σ ∈ (0, 1) be such that s = (1 − σ)s0 + σs1 . By (17.39) for s0 and s1 , B˙ s0 ,p (RN ) → B˙ s0 −N/p,∞ (RN ), B˙ s1 ,p (RN ) → B˙ s1 −N/p,∞ (RN ). 1
1
In turn, by Theorem 16.12 and Corollary 17.42, B˙ s,p (RN ) = (B˙ s0 ,p (RN ), B˙ s1 ,p (RN ))σ,q q
1
1
→ (B˙ s0 −N/p,∞ (RN ), B˙ s1 −N/p,∞ (RN ))σ,q = B˙ qs1 −N/p,∞ (RN ),
564
17. Besov Spaces
where 1/r = (1 − σ)/ps0 + σ/ps1 = 1/p − s/N .
Exercise 17.53. Let 1 ≤ p < ∞, let s = 1 + N/p, let θ ∈ (0, 1), and 1+N/p,p let 1 ≤ q ≤ ∞. Prove that for every u ∈ B˙ q (RN ) there exists a polynomial Pu such that |u − Pu |C 0,θ (RN ) ≤ c|u|B 1+N/p,p (RN ) q
for some constant c = c(N, p, q, s, θ) > 0. Hint: Use (12.54) to estimate the Lipschitz norm of v in Step 1. Exercise 17.54. Let Ω ⊂ RN be an open bounded set with Lipschitz continuous boundary, let 1 ≤ p < ∞ and 0 < s < 1 be such that 0 < s − N/p < 1, and let 1 ≤ q ≤ ∞. Prove that the embedding Bqs,p (Ω) → C 0,θ (Ω) is compact for every 0 < θ < s − N/p. Hint: Use the previous theorem, (17.33), and Proposition 12.84. See Exercises 17.70 and 17.71 for the case s − N/p > 1. Finally, we consider the critical case s = N/p. As in the Sobolev case (see, e.g., Theorems 12.33 or 12.40), we will need the full Besov norm and not just the seminorm. Indeed, we will make use of the equivalent norm given in Exercise 16.23. Theorem 17.55. Let 1 ≤ p < ∞ and 1 ≤ q ≤ ∞. Then for every p ≤ r < ∞ there exists a constant c = c(N, p, q, r) > 0 such that for every N/p,p (RN ), u ∈ Bq uLr (RN ) ≤ cuB N/p,p (RN ) . q
In particular,
N/p,p (RN ) Bq
→ Lr (RN ) for all p ≤ r < ∞.
Proof. We only prove the case q < ∞. Let m ∈ N be such that mp > N . As in Step 1 of the proof of Theorem 17.49, for every p < r < ∞ we have that (17.40)
vLr ≤ ct−σ J(v, t)
for every v ∈ W m,p (RN ), where σ := (1/p − 1/r)N/m ∈ (0, 1). N/p,p
(RN ) Let σ0 := N/(mp) ∈ (0, 1). Then (Lp (RN ), W m,p (RN ))σ0 ,q = Bq by Theorem 17.24. Moreover, since W m,p (RN ) → Lm (RN ), we can apply Exercise 16.23. Reasoning as in Steps 2 and 3 of the proof of Theorem N/p,p (RN ) we can find a Bochner integrable function 16.21, given u ∈ Bq m,p N v : (0, 1) → W (R ) such that 1 dt (v(t))(x) , x ∈ RN , u(x) = t 0 and J(v(t), t) ≤ cK(u, t) for all t ∈ (0, 1).
17.5. The Limit of Bqs,p as s → 0+ and s → m−
565
Reasoning as in Step 2 of the proof of Theorem 17.49, we get 1 1 dt dt r uL ≤ c J(v(t), t) 1+σ ≤ c K(u, t) 1+σ t t 0 0 1 1/q 1 dt 1/q ≤c t(σ0 −σ)q −1 dt (K(u, t))q 1+σ0 q t 0 0 c ≤ uB N/p,p , q (σ0 − σ)1/q where we have used H¨older’s inequality for 1 < q < ∞ and as usual c changed from line to line. Note that σ0 − σ = N/(rm) > 0. Exercise 17.56. Let Ω ⊂ RN be an open bounded set with Lipschitz continuous boundary, let 1 ≤ q ≤ ∞, and let 0 < s < 1. Prove that the s,N/s (Ω) → Lr (Ω) is compact for every 1 ≤ r < ∞. Hint: Use embedding Bq the previous theorem and (17.33).
17.5. The Limit of Bqs,p as s → 0+ and s → m− In this section we prove that the norm in Lp (RN ) and a suitable seminorm in W m,p (RN ) can be regarded as limits of seminorms in Bqs,p (RN ) as s → 0+ and s → m− , respectively. We begin with the case s → 0+ . For simplicity we take m = 1 and leave the other cases as an exercise. We recall that βN = HN −1 (S N −1 ). Theorem 17.57. Let 1 ≤ p, q < ∞, let 0 < s0 < 1 and let u ∈ B˙ qs0 ,p (RN ) vanish at infinity. Then u ∈ Lp (RN ) if and only if lim inf [s(1 − s)q]1/q |u|Bqs,p (RN ) < ∞. s→0+
Moreover, if u ∈ Lp (RN ), then (17.41)
1/q
lim [s(1 − s)q]1/q |u|Bqs,p (RN ) = 2βN uLp (RN ) .
s→0+
The idea of the proof is to use Lemma 16.17. Lemma 17.58. Let 1 ≤ p, q < ∞ and let u ∈ Lp (RN ). Then (17.42) lim Δtν uqLp (RN ) dHN −1 (ν) = 2q βN uqLp (RN ) . t→∞ S N −1
Proof. Since u ∈ Lp (RN ), by Theorem C.24 for every ε > 0 we may find a function v ∈ Cc (RN ) such that (17.43)
u − vLp ≤ ε.
Writing Δtν v(x) = v(x + tν) − u(x + tν) + Δtν u(x) + u(x) − v(x),
566
17. Besov Spaces
from (17.43), a change of variables, and Minkowski’s inequality, it follows that Δtν vLp ≤ Δtν uLp + 2u − vLp ≤ Δtν uLp + 2ε,
(17.44) and, similarly,
Δtν uLp ≤ Δtν vLp + 2ε.
(17.45)
Using (17.44), (17.45), and Minkowski’s inequality we get
1/q 1/q q N −1 Δtν vLp dH (ν) − 2βN ε S N −1
≤ ≤
S N −1
S N −1
Δtν uqLp dHN −1 (ν) Δtν vqLp dHN −1 (ν)
1/q
1/q
1/q
+ 2βN ε.
Thus, also by (17.43), it suffices to prove (17.42) for the function v. Next, we write |Δtν v(x)|p dx RN p |Δtν v(x)| dx + =
Et,ν
RN \Et,ν
|Δtν v(x)| dx + p
= Et,ν
Ft,ν
|Δtν v(x)|p dx
|Δ−tν v(y)|p dy =: It,ν + IIt,ν ,
where Et,ν := {x ∈ RN : x + tν ≥ x} and Ft,ν := {y ∈ RN : y − tν > y}. Since v ∈ Cc (RN ), there exists r > 0 such that v(x) = 0 for all x ∈ RN with x ≥ r. Note that if x + tν ≥ x ≥ r, then v(x + tν) = 0. Hence we may write It,ν = B(0,r)
χEt,ν (x) |Δtν v(x)|p dx.
Since χEt,ν (x) Δtν v(x) → −v(x) as t → ∞ and |Δtν v(x)| ≤ 2v∞ for all t, ν and x, we may apply Lebesgue’s dominated convergence theorem (twice) to conclude that q/p It,ν dHN −1 (ν) lim t→∞ S N −1
lim χEt,ν (x) |Δtν v(x)| dx p
= =
q/p
S N −1
B(0,r) t→∞
q/p
|v(x)| dx p
S N −1
B(0,r)
dHN −1 (ν)
dHN −1 (ν) = βN vqLp .
The term IIt,ν can be treated in the same way. We omit the details.
17.5. The Limit of Bqs,p as s → 0+ and s → m−
567
Exercise 17.59. Let u : RN → R be a Lebesgue measurable function vanishing at infinity and let 1 ≤ p, q < ∞. Prove that lim inf Δtν uqLp (RN ) dHN −1 (ν) ≥ 2q βN uqLp (RN ) . t→∞
S N −1
Hint: Assume first that u ∈ Cc (RN ), then that u ∈ Lp (RN ), and then do the general case. We turn to the proof of Theorem 17.57. Proof of Theorem 17.57. Let u ∈ B˙ qs0 ,p (RN ) vanish at infinity. Using spherical coordinates we have ∞ dh dt q Δh uLp = Δtν uqLp dHN −1 (ν) 1+sq , N +sq h t 0 RN S N −1 and so by (16.17) and (17.5), |u|Bqs,p = g∼ s,q , where for t > 0,
1/q q N −1 Δtν uLp dH (ν) . g(t) := S N −1
Hence, if u ∈ Lp (RN ), in view of Lemma 17.58, we are in a position to apply Lemma 16.17 to obtain (17.41). On the other hand, if lim inf (s(1 − s)pq)1/q |u|Bqs,p < ∞, s→0+
then by Lemma 16.17 and Exercise 17.59, we obtain 1/q
lim inf [s(1 − s)q]1/q |u|Bqs,p ≥ 2βN uLp . s→0+
Thus, u ∈ Lp (RN ). This completes the proof.
Next we study the case s → m− . Theorem 17.60. Let m ∈ N, let 1 < p < ∞ and 1 ≤ q < ∞, let m − 1 < s0 < m, and let u ∈ B˙ qs0 ,p (RN ) vanish at infinity. Then u belongs to ˙ m,p (RN ) if and only if W lim inf [s(m − s)q]1/q |u|Bqs,p (RN ) < ∞. s→m−
˙ m,p (RN ), then Moreover, if u ∈ W (17.46) lim [s(m − s)q]
s→m−
1/q
|u|Bqs,p (RN )
= m
1/q ∂ m u q N −1 (ν) . m p N dH L (R ) S N −1 ∂ν
The overall strategy of the proof is similar to the one of Theorem 11.75. We begin with a preliminary lemma.
568
17. Besov Spaces
Lemma 17.61. Let m ∈ N and 1 ≤ p, q < ∞. Then for every u ∈ m,p (RN ), Wloc 1 q N −1 lim inf mq (17.47) Δm (ν) tν uLp dH N −1 t→0+ t S ∂ m u q ≥ m p dHN −1 (ν). ∂ν N −1 L S Moreover, if u ∈ W m,p (RN ), then ∂ m u q 1 q m N −1 Δtν uLp dH (ν) = (17.48) lim mq m p dHN −1 (ν). + L t→0 t S N −1 S N −1 ∂ν m m,p Proof. Step 1: Assume that u ∈ C ∞ (RN ) ∩ Wloc (RN ). Since 0 χm (t) dt = 1, by (17.25), m m ∂mu ∂ u ∂mu −m m (x) = t Δtν u(x) + (x) − m (x + tτ ν) χm (τ ) dτ. (17.49) ∂ν m ∂ν m ∂ν 0 Let Br := B(0, r), r > 0. Taking the Lp norm in the x variable over Br on both sides and using Minkowski’s inequality for integrals (see Corollary B.83) gives ∂mu (17.50) ≤ t−m Δm m p tν uLp ∂ν L (Br ) m m ∂ u ∂ mu − (· + tτ ν) dτ. +c p m m ∂ν ∂ν L (Br ) 0 m,p By mollification the previous inequality holds for u ∈ Wloc (RN ). Taking the Lq (S N −1 ; HN −1 ) norm in the ν variable on both sides of (17.50) and using Minkowski’s inequality and Minkowski’s inequality for integrals (see Corollary B.83) yields
1/q
1/q ∂ m u q q N −1 −m m N −1 dH (ν) ≤t Δtν uLp dH (ν) m p L (Br ) S N −1 ∂ν S N −1
1/q m q ∂mu ∂mu N −1 +c dH (ν) dτ. m − m (· + tτ ν) p ∂ν L (Br ) 0 S N −1 ∂ν
Letting t → 0+ and using the facts that the Lp norm is continuous with m,p (RN ), it follows from the Lebesgue respect to translations and that u ∈ Wloc dominated convergence theorem that
1/q ∂ m u q N −1 dH (ν) m p L (Br ) S N −1 ∂ν
1/q q −m m N −1 ≤ lim inf t Δtν uLp dH (ν) . t→0+
S N −1
17.5. The Limit of Bqs,p as s → 0+ and s → m−
569
Letting r → ∞ and using the Lebesgue monotone convergence theorem gives (17.47). Step 2: To complete the proof it is enough to assume that u ∈ W m,p (RN ) (why?). Then, reasoning in a similar way, from (17.49) we obtain m m ∂mu ∂mu 1 ∂ u m p ≤ Δ u + c (· + tτ ν) − dτ, L tν m m m m p t ∂ν L ∂ν ∂ν Lp 0 and in turn
1/q
1/q ∂ m u q 1 q N −1 N −1 Δ v dH (ν) ≤ dH (ν) m p tν Lp tm S N −1 L S N −1 ∂ν
1/q m ∂ mu ∂mu q N −1 +c (ν) dτ. m (· + tτ ν) − m p dH ∂ν L 0 S N −1 ∂ν Letting t → 0+ and using the fact that the Lp norm is continuous with respect to translations, it follows by the Lebesgue dominated convergence theorem that
1/q 1 q N −1 Δtν vLp dH (ν) lim sup m S N −1 t→0+ t
1/q ∂ m u q N −1 ≤ (ν) . m p dH L S N −1 ∂ν Combining this inequality with (17.47) completes the proof.
We turn to the proof of Theorem 17.60. Proof of Theorem 17.60. Let u ∈ B˙ qs0 ,p (RN ) vanish at infinity. Using spherical coordinates and the change of variables ρ = t1/m we have ∞ dρ dh q q N −1 Δm u = Δm (ν) 1+sq ρν uLp dH h Lp N +sq h ρ N N −1 0 R S 1 ∞ dρ q N −1 = Δm (ν) 1+sq/m 1/m ν uLp dH t m 0 t S N −1 and so by (16.17) and (17.5), |u|Bqs,p = g∼ s/m,q , where for t > 0, g(t) :=
S N −1
Δm uqLp dHN −1 (ν) t1/m ν
1/q .
˙ m,p (RN ), in view of Lemma 17.61, we are in a position to Hence, if u ∈ W apply Lemma 16.17 to obtain (17.46). On the other hand, if M := lim inf [s(m − s)q]1/q |u|Bqs,p < ∞, s→m−
570
17. Besov Spaces
let uε := ϕε ∗ u, where ϕε is a standard mollifier. By (17.18), Δm h uε Lp ≤ q m Δh uLp , and so by Lemma 16.17 and (17.47), we obtain
1/q ∂ m u q ε N −1 (ν) ≤ M. m m p dH L S N −1 ∂ν Since p > 1 and uε → u in L1loc (RN ), it follows from Exercise 11.22 that ˙ m,p (RN ). This completes the proof. u∈W By adapting the proofs of Theorem 11.75 and of the previous theorem ˙ m,p (Ω), we can show the following result, which extends Theorem 11.75 to W m ≥ 2. Exercise 17.62. Let Ω ⊆ RN be an open set and for every h ∈ RN \ {0}, let Ωh := {x ∈ Ω : x + th ∈ Ω for all t ∈ [0, 1]}. ˙ m.p (Ω). Prove that (i) Let m ∈ N, 1 ≤ p < ∞, and let u ∈ W |Δh u(x)|p dx ≤ hmp ∇m u(x)p dx, Ωh
Ω
and that there exists a constant κ = κ(m, N, p) > 0 such that |Δh u(x)|p m p κ ∇ u(x) dx ≤ lim sup dx. hmp h→0 Ω Ωh (ii) Let m ∈ N, 1 < p < ∞, and let u ∈ L1loc (Ω) be such that |Δh u(x)|p dx < ∞. lim sup hmp h→0 Ωh ˙ m.p (Ω). Prove that u ∈ W Next we consider the case p = 1. Theorem 17.63. Let m ∈ N, let 1 ≤ q < ∞, and let u ∈ L1loc (RN ) be a function vanishing at infinity such that |u|B s0 ,1 (RN ) < ∞ for some m − 1 < q
m−1,1 (RN ) and the distributional gradient s0 < m. Then u belongs to Wloc α α N N D(∂ u) of ∂ u belongs to Mb (R ; R ) for every multi-index α with |α| = m − 1 if and only if
lim inf [s(m − s)q]1/q |u|Bqs,1 (RN ) < ∞. s→m−
The proof is left as an exercise. Exercise 17.64. Given a function u ∈ L1loc (RN ), a multi-index α ∈ NN 0 , N and h ∈ R , define Δαh u(x) := ΔαhNN eN (· · · Δαh22e2 (Δαh11e1 u(x)))
17.6. Besov Spaces and Derivatives
571
and for 1 ≤ p ≤ ∞, 1 ≤ q < ∞, and s > 0,
|u|eBqs,p (RN ) :=
|α|=s+1
RN
Δαh uqLp (RN )
dh hN +sq
1/q .
(i) Prove that | · |eB s,p (RN ) is an equivalent seminorm. q
(ii) Given m ∈ N, 1 ≤ p < ∞, and m − 1 < s0 < m, let u ∈ B˙ qs0 ,p (RN ) ˙ m,p (RN ), then the limit vanish at infinity. Prove that if u ∈ W lim [s(m − s)q]1/q |u|eBqs,p (RN )
s→m−
exists, and compute it. Exercise 17.65. Given m ∈ N, 1 ≤ p < ∞, and m − 1 < s0 < m, let ˙ m,p (RN ), then the u ∈ B˙ qs0 ,p (RN ) vanish at infinity. Prove that if u ∈ W limit lim [s(m − s)q]1/q |u|∞ Bqs,p (RN )
s→m−
exists, and compute it. Here, | · |∞ is the seminorm defined in (17.24). B s,p (RN )
17.6. Besov Spaces and Derivatives In this section we study the regularity of functions in Besov spaces. We begin by considering the space Bqs,p (RN ) when s ∈ N. In this case we can show that B1s,p (RN ) → W s,p (RN ). Theorem 17.66. Let m ∈ N and let 1 ≤ p ≤ ∞. If u ∈ L1loc (RN ) is such that 1 dt sup Δm+1 uLp 1+m < ∞, (17.51) h t 0 h ≤t ˙ m,p (RN ). If, in addition, u ∈ B m,p (RN ), then then u ∈ W 1 (17.52)
∇m uLp (RN ) ≤ cuB1m,p (RN )
for some constant c = c(N, m) > 0. In particular, B1m,p (RN ) → W m,p (RN ). Proof. Step 1: We consider first the case m = 1. Assume that u satisfies (17.51) with m = 1. Then for every ν ∈ S N −1 and x ∈ RN , using telescopic
572
17. Besov Spaces
sums for k, l ∈ N we can write 2l+k Δ2−l−k ν u(x) − 2k Δ2−k ν u(x) =−
l+k−1
[2n (u(x + 2−n ν) − u(x)) − 2n+1 (u(x + 2−n−1 ν) − u(x))]
n=k
=−
l+k−1
2n Δ22−n−1 ν u(x).
n=k
Taking the Lp norm in x and making a change of variables we obtain 2
l+k
Δ2−l−k ν u − 2 Δ2−k ν uLp ≤ k
≤
∞
2n Δ22−n−1 ν uLp
n=k ∞ 2−n n=k
2−n−1 h ≤t
2−k
= 0
sup Δ2h uLp
sup Δ2h uLp
h ≤t
dt t2
dt . t2
By (17.51) the right-hand side of the previous inequality goes to zero as k → ∞. Hence, the sequence {2k Δ2−k ν u}k is a Cauchy sequence in Lp (RN ), and thus it converges to some function vν ∈ Lp (RN ). For every ϕ ∈ Cc∞ (RN ), by a change of variables we have that ϕ(y − 2−k ν) − ϕ(y) k 2 Δ2−k ν u(x)ϕ(x) dx = u(y) dx. 2−k RN RN Letting k → ∞ and using the facts that 2k Δ2−k ν u → vν in Lp (RN ) on the left-hand side and the Lebesgue dominated convergence theorem on the right-hand side, we obtain ∂ϕ vν (x)ϕ(x) dx = − u(y) (y) dy, ∂ν N N R R which shows that vν is the distributional directional derivative ˙ 1,p (RN ). u∈W
∂u ∂ν .
Step 2: If u ∈ B11,p (RN ) ∩ C ∞ (RN ), then u(x + 2−n ν) − u(x) ∂u (x) = lim , n→∞ ∂ν 2−n and so using telescopic sums as in the previous step we can write ∞
Δν u(x) −
∂u (x) = 2n Δ22−n−1 ν u(x). ∂ν n=0
Thus,
17.6. Besov Spaces and Derivatives
573
Reasoning as in the previous step we obtain ∞ ∂u 2n Δ22−n−1 ν uLp p ≤ Δν uLp + ∂ν L n=0 ∞ 2−n dt ≤ 2uLp + sup Δ2h uLp 2 t −n−1 2
h ≤t n=0
≤ 2uLp + |u|∞ , B 1,p 1
where we used (17.7) and (17.24). Using a mollification argument, it follows from Proposition 17.12 that the same inequality holds for u ∈ B11,p (RN ). Hence, u ∈ W 1,p (RN ). Step 3: Assume next that m ≥ 2 and that u satisfies (17.51). By Lemma 17.18 we can write −m m Δm Δ2h u + Pm−1 (Th )Δm+1 u, hu=2 h
where Th is the operator given in (17.10) and Pm−1 is a polynomial of degree m − 1. Hence, nm m Δ2−n ν u = 2(n+1)m Pm−1 (T2−n−1 ν )Δm+1 u, 2(n+1)m Δm 2−n−1 ν u − 2 2−n−1 ν
and so using telescopic sums for k, l ∈ N we can write km 2(l+k)m Δm Δ2−k ν u 2−l−k ν u − 2
=
l+k−1
2(n+1)m Pm−1 (T2−n−1 ν )Δm+1 u. 2−n−1 ν
n=k
Taking the Lp norm in x and making a change of variables we obtain km 2(l+k)m Δm Δ2−k ν uLp ≤ c 2−l−k ν u − 2
≤c
∞
2nm Δm+1 uLp 2−n−1 ν
n=k ∞ 2−n n=k
2−n−1
2−k
=c 0
sup Δm+1 uLp h
h ≤t
sup Δm+1 uLp h
h ≤t
dt t1+m
dt t1+m .
We can now continue as in Step 1 to conclude that if u ∈ L1loc (RN ) satisfies (17.51), then for every ν ∈ S N −1 there exists the distributional directional m derivative ∂∂ν mu ∈ Lp (RN ). In view of Exercise 11.22, we have that u ∈ ˙ m,p (RN ). W
574
17. Besov Spaces
Step 4: In the case m ≥ 2 and u ∈ B1m,p (RN ) ∩ C ∞ (RN ) we use (17.25) to show that for every ν ∈ S N −1 and x ∈ RN we have Δm ∂mu 2−n ν u(x) (x) = lim . m n→∞ ∂ν 2−nm We can now proceed as in Step 2 using Step 3 in place of Step 1. We omit the details. Step 5: The embedding B1m,p (RN ) → W m,p (RN ) follows from the Gagliardo–Nirenberg interpolation theorem (see Theorem 12.87) and (17.52). Exercise 17.67. Let m ∈ N, let 1 ≤ p ≤ ∞, and let u ∈ L1loc (RN ) be such that 1 dt sup Δm+2 uLp 1+m < ∞. h t 0 h ≤t Imitate the proof of Steps 1 and 3 to show that the sequence (k−2)m−1 m {2km Δm Δ2−k+2 ν u}k∈N 2−k ν u + 2
is a Cauchy sequence in Lp (RN ), and that for every ν ∈ S N −1 there exists m the distributional directional derivative ∂∂ν mu ∈ Lp (RN ). Hint: Use Exercise 17.19. As a consequence of the previous theorem we can prove the inclusion ⊆ W s,p (RN ), where s > 1 is not an integer.
Bqs,p (RN )
Corollary 17.68. Let 1 ≤ p, q ≤ ∞ and let s ∈ (1, ∞) \ N. If u ∈ L1loc (RN ) is such that 1 dt s+1 sup Δh uqLp 1+sq < ∞, t 0 h ≤t ˙ s,p (RN ). Moreover, if u ∈ Bqs,p (RN ), then there exists a conthen u ∈ W stant c = c(N, p, q, s) > 0 such that ∇s uLp (RN ) ≤ cuBqm,p (RN ) . In particular, Bqs,p (RN ) → W s,p (RN ). Proof. Reasoning as in Step 3 of the previous proof with m = s we get 2−k dt km p u − 2 Δ u ≤ c sup Δm+1 uLp 1+m 2(l+k)m Δm −k L 2 ν h 2−l−k ν t 0
h ≤t 2−k 2−k 1/q dt 1/q dt q ≤c sup Δm+1 u , Lp 1+sq h t t1−(s−m)q 0
h ≤t 0 where we used H¨older’s inequality. We can now continue as in the previous proof. We omit the details.
17.6. Besov Spaces and Derivatives
575
In conclusion, also by Theorem 17.33, we have shown that if s > 0 is not an integer, then W s+1,p (RN ) → Bqs,p (RN ) → W s,p (RN ), where W 0,p (RN ) := Lp (RN ), while if s = m ∈ N, with m ≥ 2, then B1m,p (RN ) → W m,p (RN ) → Bqm−1,p (RN ). Finally, when s = 1, B11,p (RN ) → W 1,p (RN ). In what follows we say that a function u ∈ L1loc (RN ; RM ) belongs to if all its components ui belong to B˙ qs−,p (RN ) for every i = 1, s,p N M s,p . . . , M and we set |u|Bqs,p (RN ) := M i=1 |ui |Bq (RN ) . We define Bq (R ; R ) analogously. B˙ qs,p (RN ; RM )
The next theorem is the main result of this section. Theorem 17.69. Given 1 ≤ p, q ≤ ∞ and s > 1, let ∈ N be such that ≤ max{n ∈ N : n < s}. Then u ∈ L1loc (RN ) belongs to B˙ qs,p (RN ) if and only if the weak derivative ∇ u belongs to B˙ qs−,p (RN ; RM ). Moreover, there exists a constant c = c(N, p, q, s) > 0 such that c−1 |u|Bqs,p (RN ) ≤ |∇ u|B s−,p (RN ) ≤ c|u|Bqs,p (RN ) . q
,p (RN ). Proof. Step 1: Let , m ∈ N be such that 1 ≤ < m and let u ∈ Wloc We claim that ∞ dt sup Δm (17.53) |u|W ,p ≤ c h uLp 1+ . t 0 h ≤t
Without loss of generality, we may assume that the right-hand side is finite. By Remark 17.27 for every k ∈ Z we can find vk ∈ Lp (RN ) and wk ∈ ˙ m,p (RN ) such that u = vk + wk and W vk Lp + 2km |wk |W m,p ≤ c1 sup Δm h uLp .
h ≤c2k
Multiplying both sides by 2−k and summing over k gives (2−k vk Lp + 2k(m−) |wk |W m,p ) ≤ c1 2−k sup Δm h uLp k∈Z
k∈Z ∞
≤ c1
0
h ≤c2k
sup Δm h uLp
h ≤ct
dt t1+
,
where the last inequality follows since the function t → sup h ≤ct Δm h uLp is increasing. Since the right-hand side of the previous inequality is finite and m > , we have that vk Lp → 0 as k → −∞ and |wk |W m,p → 0 as k → ∞. As u = vk + wk , we obtain that u − wk Lp → 0 as k → −∞.
576
17. Besov Spaces
Moreover, writing u = vk +wk = vk+1 +wk+1 , we have that fk := vk+1 −vk = wk − wk+1 ∈ W m,p (RN ) with fk Lp + 2km |fk |W m,p ≤ c1
sup
h ≤c2k+1
Δm h uLp .
In turn, by the Gagliardo–Nirenberg interpolation theorem (see Theorem 12.87), Young’s inequality with exponent m/, and the previous inequality, 1−/m
|fk |W ,p ≤ fk Lp ≤ c1 2−k Hence,
|fk |W m,p ≤ 2−k fk Lp + 2k(m−) |fk |W m,p /m
sup
h ≤c2k+1
|fk |W ,p ≤ c1
k∈Z
Δm h uLp .
2−k
k∈Z ∞
≤ c1
0
sup
h ≤c2k+1
Δm h uLp
sup Δm h uLp
h ≤ct
dt < ∞. t1+
Fix n ∈ N. Since sj :=
n−1
fk = w−j − wn → u − wn
in Lp (RN )
k=−j
as j → ∞ and for j > i, |sj − si |W ,p ≤ −i k=−j |fk |W ,p → 0 as i → ∞, it follows from Corollary 12.86 that the sequence {sj }j∈N is a Cauchy sequence in W ,p (RN ), but since it converges to u − wn in Lp (RN ), necessarily, sj → ,p (RN ) u − wn in W ,p (RN ) as j → ∞. In turn, n−1 k=−∞ fk = u − wn in W and ∞ n−1 dt |fk |W ,p ≤ c1 sup Δm (17.54) |u − wn |W ,p ≤ h uLp 1+ . t 0 h ≤ct k=−∞ ∞ For i < n we have |wi − wn |W ,p ≤ k=i |fk |W ,p → 0 as i → ∞. Hence, ˙ ,p (RN ), but since |wn |W m,p → 0 as {wn }n∈N is a Cauchy sequence in W ˙ ,p (RN ) to a polynomial Pu . n → ∞, we have that {wn }n∈N converges in W Since |Pu |W ,p < ∞, Pu must have degree less than , so that |Pu |W ,p = 0. In turn, letting n → ∞ in (17.54) we conclude that (17.53) holds. ,p Step 2: Let , m ∈ N be such that 1 ≤ < m and let u ∈ Wloc (RN ). We claim that for every α ∈ NN 0 with |α| = and every t > 0, t dτ m α sup Δm (17.55) sup Δξ (∂ u)Lp ≤ c h uLp 1+ . τ 0 h ≤2τ
ξ ≤t
By the previous step
∞
∂ α uLp ≤ c 0
sup Δm h uLp
h ≤τ
dτ . τ 1+
17.6. Besov Spaces and Derivatives
577
Replacing u with Δm ξ u, where ξ ≤ t, gives ∞ dτ m α α m m Δξ (∂ u)Lp = ∂ (Δξ u)Lp ≤ c sup Δm h (Δξ u)Lp 1+ . τ 0 h ≤τ By (17.7), if h ≤ τ and ξ ≤ t, m m sup Δm h (Δξ u)Lp ≤ 2
h ≤τ
and so α Δm ξ (∂ u)Lp
≤c
t
sup
0 h ≤τ t
=c ≤c
Δm h uLp
sup Δm h uLp
0 h ≤τ t
ζ ≤min{τ,t}
Δm ζ uLp ,
dτ + c sup Δm h uLp τ 1+
h ≤t
∞ t
dτ τ 1+
c dτ + sup Δm uLp 1+ τ t h ≤t h
sup Δm h uLp
0 h ≤2τ
sup
dτ , τ 1+
where in the last inequality we used the fact that for t > 0, t dτ 1 m p sup Δ u ≤ c sup Δm L h h uLp 1+ , t h ≤t τ t/2 h ≤2τ since the function τ → sup h ≤τ Δm h uLp is increasing. Step 3: Let u ∈ C ∞ (RN ) be such that |u|Bqs,p (RN ) < ∞ and let := max{n ∈ N : n < s} and m := s + 1. Assume that 1 ≤ q < ∞. For every α ∈ NN 0 with |α| = , by the previous step, ∞
1/q dt q α ∞ m α sup Δξ (∂ u)Lp 1+q(s−) |∂ u|B s−,p (RN ) = q t 0 ξ ≤t ∞ t
1/q dτ q dt m ≤c sup Δh uLp 1+ . τ t1+q(s−) 0 0 h ≤2τ By Hardy’s inequality (see Theorem C.41) the right-hand of the previous . The hypothesis that inequality side is bounded from above by c|u|∞ B s,p (RN ) q
u ∈ C ∞ (RN ) can be removed by mollification (see the proof of Proposition 17.12). ˙ s,p (RN ) Next, if u ∈ B˙ qs,p (RN ) and s is not an integer, then u ∈ W by Corollary 17.68. On the other hand, if s ∈ N, then = s − 1 and m = s+1 and again by Hardy’s inequality the right-hand side of the previous inequality is finite, and so t dτ p sup Δ+2 h uL τ 1+ < ∞ 0 h ≤2τ
578
17. Besov Spaces
˙ ,p (RN ). Hence, in for L1 -a.e. t > 0. It follows by Exercise 17.67 that u ∈ W . both cases by the first part of this step we get |∇ u|∞s−,p N ≤ c|u|∞ B s,p (RN ) Bq
Conversely, if u ∈ Lemma 17.25 applied
,p Wloc (RN ) is u, to Δm− h
such that
(R )
|∇ u|∞s−,p N Bq (R )
q
< ∞, then by
m− u)Lp ≤ ch ∇ (Δm− u)Lp Δm h uLp = Δh (Δh h
≤ ct sup Δm− (∇ u)Lp . h
h ≤t
Integrating both sides gives
1/q ∞ ∞ dt 1/q dt q q m− m sup Δh uLp 1+qs ≤c sup Δh (∇ u)Lp 1+(q−)s t t 0 h ≤t 0 h ≤t ≤ c|∇ u|∞ , B s−,p (RN ) q
which concludes the proof.
Exercise 17.70. Let s > 0 and 1 ≤ p ≤ ∞ be such that m < s − N/p < m + 1 for some m ∈ N. Prove that for every u ∈ B˙ qs,p (RN ) there exists a polynomial Pu such that |∇m (u − Pu )|C 0,s−m−N/p (RN ) ≤ c|u|Bqs,p (RN ) for some constant c = c(N, p, q, s) > 0. Exercise 17.71. Let s > 0 and 1 ≤ p ≤ ∞ be such that s − N/p = m + 1 for some m ∈ N, let θ ∈ (0, 1), and let 1 ≤ q ≤ ∞. Prove that for every u ∈ B˙ qs,p (RN ) there exists a polynomial Pu such that |∇m (u − Pu )|C 0,θ (RN ) ≤ c|u|Bqs,p (RN ) for some constant c = c(N, p, q, s, θ) > 0. Exercise 17.72. Given 1 ≤ p, q ≤ ∞ and s > 1, let ∈ N be such that ≤ max{n ∈ N : n < s}. Prove that u ∈ Lp (RN ) belongs to Bqs,p (RN ) if and only if ∇ u belongs to Bqs−,p (RN ; RM ).
17.7. Yet Another Equivalent Norm In this section we introduce an equivalent norm in Bqs,p (RN ), which relies on the Littlewood–Paley decomposition (see Theorem 10.78). Given a > 0, let ϕ0 ∈ Cc∞ (RN ) be a nonnegative radial function such that (17.56) supp ϕ0 = B(0, 2a) \ B(0, a/2), ϕ0 > 0 in B(0, 2a) \ B(0, a/2), ϕ0 (2−k x) = 1 for every x ∈ RN \ {0}. (17.57) k∈Z
17.7. Yet Another Equivalent Norm
579
Note that for every x ∈ RN \ {0} we can find a unique kx ∈ Z such that 2kx −1 a < x ≤ 2kx a, and since supp ϕ0 (2−k ·) = B(0, 2k+1 a) \ B(0, 2k−1 a)
(17.58)
for every k ∈ Z, we have that ϕ0 (2−k x) = ϕ0 (2−kx −1 x) + ϕ0 (2−kx x) + ϕ0 (2−kx +1 x). (17.59) 1= k∈Z
By Theorem 10.62 the inverse Fourier transform of the function ϕ0 (2−k ·) is given by (17.60) −k ∨ e2πix·y ϕ0 (2−k y) dy (ϕ0 (2 ·)) (x) = RN k kN k e2πix2 ·z ϕ0 (z) dz = 2kN ϕ∨ =2 0 (2 x) =: ψk (x), RN
where we used the change of variables z = 2−k y. Since ϕ0 is radial, it follows that ψk is real-valued (see Exercise 10.69). Note that ψk belongs to the space of real-valued rapidly decreasing functions S(RN ) (see Definition 10.29). By Young’s inequality (Theorem 10.48) for every u ∈ Lp (RN ), 1 ≤ p ≤ ∞, we have that ψk ∗ u ∈ Lp (RN ), and so if 1 ≤ q < ∞ and s ∈ R we may define
1/q q ∨ kqs (17.61) |u|Bqs,p (RN ) := 2 u ∗ ψk Lp (RN ) , k∈Z
while if q = ∞, (17.62)
ks s,p |u|∨ (RN ) := sup 2 u ∗ ψk Lp (RN ) . B∞ k∈Z
Exercise 17.73. Prove that there exists a radial function ϕ0 ∈ Cc∞ (RN ) satisfying (17.56)–(17.57). Hint: Let ϕ ∈ Cc∞ (RN ) satisfy (17.56) and take ϕ0 (x) :=
ϕ(x) , −k k∈Z ϕ(2 x)
x ∈ RN .
Remark 17.74. To relate (17.56)–(17.57) to the Littlewood–Paley decomposition of u (see Theorem 10.78), define ∞ ϕ0 (2−k x), x ∈ RN . (17.63) ω0 (x) := 1 − k=0
Then by (17.58), ω0 = 1 in B(0, 2−1 a), while by (17.56) and (17.59), supp ω0 = B(0, a). Moreover, ∞ ∞ ϕ0 (2−k x) − ϕ0 (2−k−1 x) = ϕ0 (x), ω0 (x/2) − ω0 (x) = k=0
k=0
580
17. Besov Spaces
and so (10.36) holds. Hence, (10.35) holds. In turn, given u ∈ Lp (RN ), by (10.37) and (10.38) and Theorems 10.71 and 10.62 and Exercises 10.75 and 10.77, we have that Sk (u) = (ω0 (2−k ·)& u)∨ = u ∗ (ω0 (2−k ·))∨ , u)∨ = u ∗ ψk . Λ˙ k (u) = (ϕ0 (2−k ·)&
(17.64) (17.65)
Observe that the seminorm 17.61 has been defined also for s ≤ 0. More generally, given a real-valued tempered distribution T ∈ S (RN ) (see Definition 10.32), by Theorem 10.56 we have that the tempered distribution T ∗ψk can be identified with the C ∞ function x → T (ψkx ), where ψkx (y) := ψk (x−y), y ∈ RN . Thus, with a slight abuse of notation, given 1 ≤ p, q ≤ ∞ and s ∈ R, for T ∈ S (RN ) and 1 ≤ q < ∞ we can define
1/q q ∨ kqs 2 T ∗ ψk Lp (RN ) , (17.66) |T |Bqs,p (RN ) := k∈Z
while for q = ∞, ks s,p |T |∨ (RN ) := sup 2 T ∗ ψk Lp (RN ) . B∞
(17.67)
k∈Z
Exercise 17.75. Let 1 ≤ p, q ≤ ∞ and s ∈ R, and let ϕ0 ∈ Cc∞ (RN ) satisfy = 0 if and only (17.56)–(17.57). Given T ∈ S (RN ), prove that |T |∨ Bqs,p (RN ) if T can be identified with a polynomial. We have seen in Remark 17.29 that a function u ∈ B˙ qs,p (RN ), s > 0, can always be identified with a tempered distribution Tu . In the next theorem we prove the opposite implication, that is, that if T ∈ S (RN ) is such that < ∞ for s > 0, then T can be identified with a function in |T |∨ B s,p (RN ) q
L1loc (RN ). Theorem 17.76. Let 1 ≤ p, q < ∞ and s > 0. If T ∈ S (RN ) is such that |T |∨ < ∞, then there exists a function u ∈ L1loc (RN ) such that Bqs,p (RN ) T (φ) = RN φu dx for every φ ∈ Cc∞ (RN ). Proof. In view of Theorem 10.78 and using the function ω0 in Remark 17.74 we can write ∞ T ∗ ψk . T = T ∗ ω0∨ + k=0
If q > 1, by H¨older’s and Minkowski’s inequalities, then
1/q
1/q ∞ ∞ ∞ s q −kq kqs (17.68) T ∗ ψk Lp ≤ 2 2 T ∗ ψk Lp k=0
k=0
≤ c|T |∨ Bqs,p .
k=0
17.7. Yet Another Equivalent Norm
581
The same inequality holds if q = 1. On the other hand, since ω0 (2−k ·) ∈ S(RN ), T ∗(ω0 (2−k ·))∨ can be identified with a C ∞ function with polynomial growth by Theorem 10.56. Hence, T is given by the sum of a C ∞ function with polynomial growth and an Lp function. Thus, it can be identified with a function u ∈ L1loc (RN ). This concludes the proof. is an equivalent In the following theorem we will prove that · ∨ B s,p (RN ) q
norm in Bqs,p (RN ). In the statement the norms are obtained by adding the Lp norm to the corresponding seminorm. Theorem 17.77. Let 1 ≤ p, q ≤ ∞ and s > 0, and let ϕ0 ∈ Cc∞ (RN ) satisfy (17.56)–(17.57). Then there exists a constant c = c(N, p, q, s) > 0 such that for every u ∈ Bqs,p (RN ), ∞ c−1 uBqs,p (RN ) ≤ u∨ Bqs,p (RN ) ≤ cuBqs,p (RN ) .
Proof. Step 1: Let u ∈ Bqs,p (RN ) and let m ∈ N be so large that mp > N and m > s. Write u = v + w, where v ∈ Lp (RN ) and w ∈ W m,p (RN ). By Young’s inequality (Theorem 10.48), (17.60), and the change of variables x = 2k y, v ∗ ψk Lp ≤ ψk L1 vLp = ϕ∨ 0 L1 vLp . %k = On the other hand, by Theorem 10.71 and (17.60), w ∗ ψk = w &ψ ∗ ψk ⊆ B(0, 2k+1 a) \ B(0, 2k−1 a) wϕ & 0 (2−k ·). It follows from (17.58) that supp w and thus we are in a position to apply Theorem 10.73 to get w ∗ ψk Lp ≤ c2−km ∇m (w ∗ ψk )Lp = c2−km ∇m w ∗ ψk Lp ≤ c2−km ψk L1 ∇m wLp = c2−km ϕ∨ 0 L1 |w|W m,p , where we used again Young’s inequality (Theorem 10.48), (17.60), and the change of variables x = 2k y. Combining the estimates for v and w and using Minkowski’s inequality we get u ∗ ψk Lp ≤ v ∗ ψk Lp + w ∗ ψk Lp −km ≤ ϕ∨ ϕ∨ 0 L1 vLp + c2 0 L1 |w|W m,p .
Taking the infimum over all such decomposition and using Remark 17.27 gives (17.69)
˙ 2−km ). u ∗ ψk Lp ≤ cK(u,
582
17. Besov Spaces
In turn, assuming q < ∞, ˙ 2kqs u ∗ ψk qLp ≤ c 2kqs (K(u, 2−km ))q k∈Z
k∈Z ∞
≤c
0
dτ ˙ (K(u, τ m ))q 1+qs , τ
˙ where we used Exercise 16.9. Since K(u, τ m ) ≤ c sup h ≤cτ Δm h uLp (RN ) (see Remark 17.27) it follows that ∞ dτ q q kqs 2 u ∗ ψk Lp ≤ c sup Δm h uLp (RN ) 1+qs , τ 0 h ≤cτ k∈Z
and so |u|∨ ≤ c|u|∞ . The case q = ∞ is left as an exercise. B s,p B s,p q
q
< ∞. In this step we will use Step 2: Let u ∈ Lp (RN ) be such that |u|∨ Bqs,p the J-method of interpolation (see Section 16.2 in Chapter 16), especially Exercise 16.23. Let m ∈ N be so large that mp > N and m > s. By Exercise u. Hence, by (17.58), 10.77 and (17.60), u ∗ ψk = ϕ0 (2−k ·)& supp u ∗ ψk ⊆ B(0, 2k+1 a) and thus we are in a position to apply Theorem 10.73 to conclude that u ∗ ψk ∈ W m,p (RN ), with (17.70)
∇m (u ∗ ψk )Lp ≤ c2km u ∗ ψk Lp .
Moreover, if k ∈ N0 , then u ∗ ψk Lp ≤ 2km u ∗ ψk Lp . Hence, for k ∈ N0 , J(u ∗ ψk , 2−km ) ≤ cu ∗ ψk Lp , where for v ∈ W m,p (RN ) and t > 0, J(v, t) := max{vLp , tvW m,p }. Similar estimates can be obtained for u ∗ ω0 (see Remark 17.74). Set g0 := ψ0 + ω0∨ and gk := ψk for k ∈ N. Define u(t) := (u ∗ gk )/ log 2 for t ∈ (2−(k+1)m , 2−km ], k ∈ N0 . Then 1 2−km dt dt (J(u(t), t))q 1+qs/m ≤ (J(u(t), t))q 1+qs/m −(k+1)m t t 0 k∈N0 2 2−km 1 dt (17.71) ≤ (J(u ∗ gk , 2−km ))q 1+qs/m q (log 2) −(k+1)m t k∈N0 2 2kqs u ∗ gk qLp < ∞, ≤c k∈N0
17.7. Yet Another Equivalent Norm
583
where we used the fact that J(·, t) is a norm and that J(v, ·) is increasing (see Exercise 16.20). Moreover, 1 −1 dt u(t) u ∗ gk . (17.72) = t 2−m k=0
In view of (17.68), ∞
u ∗ gk Lp ≤ cu∨ Bqs,p < ∞.
k=0
In turn, by (10.41), (17.72) (see also the Step 3 of the proof of Theorem 16.21), 1 dt u= u ∗ gk = u(t) , t 0 k∈N0
where the series converges in Lp (RN ) (and so also in Lp (RN ) + W m,p (RN )). It follows by Theorem 16.21, Exercise 16.23, Remark 17.27, and (17.71), ≤ cu∨ , which concludes the proof. that u∞ B s,p (RN ) B s,p q
q
Remark 17.78. We remark that Step 1 of the previous proof actually shows that (17.73)
s,p |u|∨ Bqs,p (RN ) ≤ c|u|Bq (RN )
for every u ∈ B˙ qs,p (RN ). Indeed, if we write u = v + w, where v ∈ Lp (RN ) ˙ m,p (RN ), since w has polynomial growth (see Remark 12.56), the and w ∈ W convolution of w (and, in turn, of u) with a function in S(RN ) is well-defined. Thus we can continue as in the proof of Step 1 to obtain (17.73). < ∞, consider Conversely, for every T ∈ S (RN ) such that |T |∨ Bqs,p (RN ) ˙ m,p (RN )/Pm−1 (RN ) endowed with the norm |·|W m,p (RN ) , the Banach space W where Pm−1 (RN ) is the family of polynomials of degree less than or equal to m − 1. Then in view of (17.70), for every k ∈ Z, ˙ ∗ ψk , 2−km ) ≤ cT ∗ ψk Lp , J(T where for v ∈ W m,p (RN ) and t > 0, J˙(v, t) := max{vLp , t|v|W m,p }. Define u(t) as in Step 2 but now with k ∈ Z. Then (17.71) becomes ∞ dt ˙ (J(u(t), t))q 1+qs/m ≤ c 2kqs T ∗ ψk qLp , t 0 k∈Z while (17.72) should be replaced by 2m −1 dt = u(t) T ∗ ψk . t 2−m k=−
584
17. Besov Spaces
In view of (17.68), ∞
T ∗ ψk Lp +W˙ m,p /Pm−1 ≤
k=0
∞
T ∗ ψk Lp ≤ c|T |∨ Bqs,p < ∞,
k=0
where · Lp +W˙ m,p /Pm−1 is defined as in (16.2). On the other hand, if q > 1, by (17.70), the fact that m > s and H¨older’s and Minkowski’s inequalities, we get ∞
∞
T ∗ ψ−n Lp +W˙ m,p /Pm−1 ≤
n=1
≤
∞
|T ∗ ψ−n |W m,p ≤ c
n=1
2−n(m−s)q
1/q ∞
n=1
∞
2−nm T ∗ ψ−n Lp
n=1
2−nqs T ∗ ψ−n qLp
1/q
≤ c|T |∨ Bqs,p .
n=1
If q = 1, the same inequality holds since 2−nm ≤ 2−ns . Hence, we have shown that T ∗ ψk Lp +W˙ m,p /Pm−1 < ∞. k∈Z
In turn, by (17.72) (see also the Step 3 of the proof of Theorem 16.21), ∞ dt T ∗ ψk = u(t) , v := t 0 k∈Z
˙ m,p (RN )/Pm−1 (RN ). It follows where the series converges in Lp (RN ) + W by Theorem 16.21, Remark 17.27, and (17.71), ∨ |v|∞ Bqs,p (RN ) ≤ c|T |Bqs,p .
The following exercises give an alternative proof of Step 1 in the cases 0 < s < 1 and s = 1. Exercise 17.79. Let 1 ≤ p ≤ ∞, 1 ≤ q < ∞, and 0 < s < 1. (i) Prove that (u∗ψk )(x) = RN ψk (y)Δ−y u(x) dy for every u ∈ B˙ qs,p (RN ) and x ∈ RN . ≤ c|u|Bqs,p . Hint: Write (ii) Prove that |u|∨ Bqs,p = + RN
B(0,2−k )
RN \B(0,2−k )
and use Minkowski’s inequality for integrals. Exercise 17.80. Let 1 ≤ p ≤ ∞, 1 ≤ q < ∞, and s = 1. (i) Prove that (u ∗ ψk )(x) = 12 RN ψk (y)∼ Δ2y u(x) dy for every u ∈ B˙ q1,p (RN ) and x ∈ RN , where ∼ Δ2y u is defined in (17.8). Hint: ψk is even.
17.8. And More Embeddings
585
(ii) Prove that |u|∨ 1,p ≤ c|u|Bq1,p . Bq
Exercise 17.81. Let u ∈ S(RN ) and let h ∈ RN and m ∈ N. Prove that 2πih·x ' − 1)& u(x), Δ h u(x) = (e
and that
x ∈ RN ,
m u(x) = (e2πih·x − 1)m u &(x), Δ h
x ∈ RN .
17.8. And More Embeddings In this section we show how we can use the seminorm (17.61) to give simple proofs of more embeddings for Besov spaces. Theorem 17.82. Let 1 ≤ p, q ≤ ∞ and s2 > s1 > 0. Then there exists a constant c = c(N, p, q, s1 , s2 ) > 0 such that s ,p s ,p ≤ |u|∨ + cuLp (RN ) |u|∨ B 1 (RN ) B 2 (RN ) q
for all u ∈
Bqs2 ,p (RN ).
q
In particular, Bqs2 ,p (RN ) → Bqs1 ,p (RN ).
Proof. Assume that 1 ≤ q < ∞. Given u ∈ Bqs2 ,p (RN ), by Young’s inequality (Theorem 10.48), (17.60), and the change of variables x = 2k y, we have u ∗ ψk Lp ≤ ψk L1 uLp = ϕ∨ 0 L1 uLp . Hence, (17.74)
0
2
kqs1
u ∗
ψk qLp
≤
q q ϕ∨ 0 L1 uLp
k=−∞
0
2kqs1 = cuqLp .
k=−∞
On the other hand, since s1 < s2 , we get ∞ ∞ 2kqs1 u ∗ ψk qLp ≤ 2kqs2 u ∗ ψk qLp . k=0
k=0
Combining the last two inequalities and using (17.61) completes the proof in the case 1 ≤ q < ∞. The case q = ∞ is similar and is left as an exercise. Exercise 17.83. Prove the previous theorem in the special case s1 = s2 using the seminorms (17.21). Next we show that by slightly lowering the regularity parameter s we can drastically change q. To be precise, we have the following result. Theorem 17.84. Let 1 ≤ p ≤ ∞, let 1 ≤ q1 , q2 ≤ ∞, and let s > 0. Then for every 0 < ε < s there exists a constant c = c(ε, N, p, q1 , q2 , s) > 0 such that ≤ c|u|∨ |u|∨ Bqs,p (RN ) + cuLp (RN ) B s−ε,p (RN ) q2
for all u ∈
N Bqs,p 1 (R ).
In particular,
1
N Bqs,p 1 (R )
→ Bqs−ε,p (RN ). 2
586
17. Besov Spaces
Observe that there is no relationship between q1 and q2 . Proof. If q1 ≤ q2 , then by Theorems 17.38 and 17.82, (RN ) → Bqs−ε,p (RN ) → Bqs−ε,p (RN ). Bqs,p 1 1 2 Thus it remains to study the case q1 > q2 . By (17.74), with q and s replaced here by q2 and s − ε, respectively, 0
2q2 k(s−ε) u ∗ ψk qL2p ≤ cuqL2p .
k=−∞
On the other hand, by H¨older’s inequality with exponents q1 /q2 , ∞ ∞ q2 /q1 2q2 k(s−ε) u ∗ ψk qL2p ≤ 2q1 ks u ∗ ψk qL1p k=0
k=0 ∞
×
2−εk(q1 /q2 )
1/(q1 /q2 )
.
k=0
This concludes the proof.
The next theorem shows that by lowering the regularity parameter s of functions in Besov spaces, one can increase their integrability parameter p. Theorem 17.85. Let 1 ≤ p2 < p1 ≤ ∞ and 0 < s1 < s2 be such that N N = s2 + (17.75) s1 + p2 p1 and let 1 ≤ q2 ≤ q1 ≤ ∞. Then there exists a constant c = c(N, p1 , p2 , q1 , q2 , s1 , s2 ) > 0 such that s ,p s ,p ≤ c|u|∨ |u|∨ B 1 1 (RN ) B 2 2 (RN ) q1
q2
for every u ∈ S(RN ). Proof. Assume that q1 < ∞. By Theorems 10.62 and 10.71, %k (x) = u &(x)ψ &(x)ϕ0 (2−k x). u ∗ ψk (x) = u Hence, by (17.56), supp u ∗ ψk ⊆ B(0, 2k+1 ). It follows from Theorem 10.73 and (17.75) that u ∗ ψk Lp1 ≤ c2kN (1/p2 −1/p1 ) u ∗ ψk Lp2 = c2k(s2 −s1 ) u ∗ ψk Lp2 . Hence, 2ks1 u ∗ ψk Lp1 ≤ c2ks u ∗ ψk Lp2 . Using Exercise 17.39 when q2 < q1 , we get
1/q1
1/q2 q1 q2 q1 ks1 q2 ks2 2 u ∗ ψk Lp1 ≤c 2 u ∗ ψk Lp , k∈Z
k∈Z
17.8. And More Embeddings
587
which completes the proof in this case. The case q1 = ∞ is left as an exercise. Exercise 17.86. Use Theorem 17.85 to give an alternative proof of Theorem 17.52. We conclude this section with an alternative proof of (a slightly weaker version of) Theorem 17.49. Theorem 17.87 (Sobolev–Gagliardo–Nirenberg embedding in Bqs,p ). Let s > 0, 1 ≤ p < N/s, and 1 ≤ q ≤ p. Then there exists a constant c = c(N, p, q, s) > 0 such that uLp∗s ≤ c|u|∨ Bqs,p (RN )
(17.76)
for every u ∈ S(RN ). In particular, Bqs,p (RN ) → Lr (RN ) for all p ≤ r ≤ p∗s . We begin with a preliminary result, which is of interest in itself. We recall that the seminorm (17.61) has been defined also for negative values of s. Lemma 17.88. Let 1 ≤ p < r < ∞ and s > 0. Then there exists a constant c = c(N, p, r, s) > 0 such that 1−p/r p/r ∨ |u| uLr (RN ) ≤ c |u|∨ B s,p (RN ) B −θ,∞ (RN ) for every u ∈ S(RN ), where θ := sp/(r − p). Proof. Given a function v : RN → R and t ∈ R for brevity we write {v > t} < ∞, by (17.62) we have for the set {x ∈ RN : v(x) > t}. Since |u|∨ B −θ,∞ u ∗ ψk L∞ ≤ 2kθ |u|∨ B −θ,∞ , and so for l ∈ Z, (17.77)
l
u ∗ ψk L∞ ≤ |u|∨ B −θ,∞
k=−∞
l k=−∞
|u|∨ B −θ,∞
By rescaling we may assume that 10.78, Remark 17.74 and (17.77) that (17.78)
lθ 2kθ = |u|∨ B −θ,∞ 2 .
u = Sl (u) + (u − Sl (u)) =
= 1. It follows from Theorem
l
u ∗ ψk +
k=−∞
∞
u ∗ ψk .
k=l+1
Hence, for every t > 0, (17.79)
{|u| > t} ⊆ {|Sl (u)| > t/2} ∪ {|u − Sl (u)| > t/2}.
Let lt ∈ Z be such that (17.80)
2−1 (t/2)1/θ ≤ 2lt ≤ (t/2)1/θ .
588
17. Besov Spaces
Then by (17.77), Slt (u)L∞ ≤ 2lt θ ≤ t/2, which implies that the set {|Slt (u)| > t/2} has Lebesgue measure zero. Thus, by (17.79), LN ({|u| > t}) ≤ LN ({|u − Slt (u)| > t/2}), and so by Tonelli’s theorem, ∞ urLr = r tr−1 LN ({|u| > t}) dt 0 ∞ (17.81) tr−1 LN ({|u − Slt (u)| > t/2}) dt ≤r 0 ∞ tr−p−1 u − Slt (u)pLp dt, ≤c 0
where we used the inequality τ L ({|v| > τ }) ≤ p
N
|v(x)| dx ≤ p
{|v|>τ }
RN
|v(x)|p dx.
To estimate u−Slt (u)Lp , observe by Minkowski’s and H¨older’s inequalities that ∞ ∞ u ∗ ψk Lp = 2−ka 2ka u ∗ ψk Lp u − Slt (u)Lp ≤ ≤
∞
k=lt +1
2−kap
k=lt +1
= 2−(lt +1)a
∞ 1/p
∞
k=lt +1
2kap u ∗ ψk pLp
k=lt +1
2kap u ∗ ψk pLp
1/p
1/p ,
k=lt +1
where 0 < a < θ(r − p)/p = s. Combining the previous inequality with (17.80) and (17.81) gives ∞ ∞ r r−p−1−ap/θ t 2kap u ∗ ψk pLp dt uLr ≤ c 0
≤c
∞
k=lt +1
2
kap
u ∗
ψk pLp
k=−∞
=
c2r−p−ap/θ r − p − ap/θ
∞
2kθ+1
tr−p−1−ap/θ dt
0
2kθ(r−p) u ∗ ψk pLp ,
k=−∞
where we used Fubini’s theorem and the fact that if k ≥ lt +1, then t ≤ 2kθ+1 . Since θ(r − p) = sp, the proof is concluded in view of (17.61). We turn to the proof of Theorem 17.87.
17.8. And More Embeddings
589
Proof of Theorem 17.87. Since q ≤ p it follows by Theorem 17.38 that ∨ |u|∨ B s,p ≤ c|u|Bqs,p .
Thus it suffices to show (17.76) for q = p. By the general form of Young’s inequality (see Theorem 10.49) with 1/p∗s + 1/(p∗s ) = 1 + 1/∞ (see (10.20)), ∗
u ∗ ψk L∞ ≤ ψk L(p∗s ) uLp∗s ≤ c2kN/ps uLp∗s , where we used (17.60), and a change of variables to estimate 1/(p∗s ) k (p∗s ) (ϕ∨ (2 y)) dy ψk L(p∗s ) = 2kN 0 RN 1/(p∗s ) kN (1−1/(p∗s ) ) (p∗s ) ≤2 (ϕ∨ (x)) dx . 0 RN
Hence, u ∈
∗ B −N/ps ,∞ (RN ),
with
≤ cuLp∗s . |u|∨ ∗ B −N/ps ,∞ We are now in a position to apply the previous lemma to conclude that 1−p/p∗s p/p∗s ∨ uLp∗s ≤ c |u|∨ ∗ |u| s,p B B −N/ps ,∞ ∗ p/p s 1−p/p∗ ≤ cuLp∗s s |u|∨ . B s,p 1−p/p∗s
Since uLp∗s < ∞, if u = 0, we can divide by uLp∗s
to obtain (17.76).
Chapter 18
Sobolev Spaces: Traces Grad students, or “p-ons” as Einstein called them, can only occupy a discrete number of energy states: working, thinking about working, sleeping. — Jorge Cham, www.phdcomics.com
Sobolev functions u ∈ W m,p (Ω) have been defined using integration by parts (see (11.1)) with test functions φ ∈ Cc∞ (Ω) vanishing on the boundary. To drop this requirement we need to make sense of the “restriction” of u (and of its normal derivatives when m ≥ 2) to the boundary ∂Ω (see (9.55)). Since Sobolev functions are Lp functions, they are equivalence classes of functions, and thus talking about their pointwise value does not make sense in general. One possibility would be to find a good representative. Indeed, in the supercritical case mp > N and if Ω is a W m,p extension domain, we can extend u to W m,p (RN ) and then apply Morrey’s theorem (see Theorem 12.55) to conclude that u has a H¨older continuous representative u ¯. Thus, the value of u on the boundary of Ω is well-defined. The situation is quite different in the subcritical and critical cases mp ≤ N (unless N = 1). In this case we will prove that if ∂Ω is sufficiently regular, say, Lipschitz continuous, we can introduce a linear operator Tr : W m,p (Ω) → Lploc (∂Ω, HN −1 ) such that Tr(u) = u on ∂Ω for all u ∈ W m,p (Ω) ∩ C(Ω) and for which integration by parts holds, that is,
u∂i ψ dx = −
(18.1) Ω
ψ Tr(u)νi dHN −1
ψ∂i u dx + Ω
∂Ω
for all u ∈ W m,p (Ω), ψ ∈ Cc1 (RN ), and i = 1, . . . , N , where ν is the outward unit normal to ∂Ω. We will study the continuity properties of this operator. 591
592
18. Sobolev Spaces: Traces
In what follows, we will use the abbreviations Lp (∂Ω) := Lp (∂Ω, HN −1 ),
(18.2)
Lploc (∂Ω) := Lploc (∂Ω, HN −1 ),
· Lp (∂Ω) := · Lp (∂Ω,HN −1 ) .
18.1. The Trace Operator In this section we establish the existence of a trace operator. For simplicity we will consider separately the cases m = 1 and m ≥ 2. The parameter ε > 0 in the estimate (18.3) will be used to prove compactness of the trace operator. Theorem 18.1. Let Ω ⊆ RN , N ≥ 2, be an open set whose boundary ∂Ω is Lipschitz continuous, let 1 ≤ p < ∞. There exists a unique linear operator Tr : W 1,p (Ω) → Lploc (∂Ω) such that (i) Tr(u) = u on ∂Ω for all u ∈ W 1,p (Ω) ∩ C(Ω), (ii) the integration by parts formula (18.1) holds for all u ∈ W 1,p (Ω), all ψ ∈ Cc1 (RN ), and all i = 1, . . . , N , (iii) for every R > 0 there exist two constants cR , εR > 0 depending on R, Ω and p such that p N −1 −1 (18.3) | Tr(u)| dH ≤ cR ε |u|p dx B(0,R)∩∂Ω B(0,R)∩(Ω\Ωε ) p−1 + cR ε ∇up dx B(0,R)∩(Ω\Ωε )
for every 0 < ε ≤ εR , where Ωε := {x ∈ Ω : dist(x, ∂Ω) > ε}. The function Tr(u) is called the trace of u on ∂Ω. Proof. Since ∂Ω is Lipschitz continuous, by Theorem 11.35 the restriction to Ω of functions in Cc∞ (RN ) is dense in W 1,p (Ω). Let u ∈ Cc∞ (RN ). We begin by proving (iii). Let x0 ∈ ∂Ω. Since ∂Ω is Lipschitz continuous (see Definition 9.57) there exist a rigid motion T : RN → RN with T (x0 ) = 0, a Lipschitz continuous function f : RN −1 → R, and rx0 > 0 such that, setting y := T (x), we have T (Ω ∩ QT (x0 , rx0 )) = {y ∈ Q(0, rx0 ) : yN > f (y )}, at x0 and of side-length rx0 > 0. where QT (x0 , rx0 ) is an open cube centered √ Fix 0 < ε < r ≤ rx0 /[2(Lip f + 1) N − 1]. Setting v(y) := u(T −1 (y)), by the fundamental theorem of calculus we can write yN |∂N v(y , s)| ds |v(y , f (y )| ≤ |v(y , yN )| + f (y )
18.1. The Trace Operator
593
for y ∈ QN −1 (0, r) and f (y ) < yN < f (y ) + ε. Raising both sides to the power p and using Young’s and H¨older’s inequalities if p > 1 gives yN p p−1 p p−1 p−1 |∂N v(y , s)|p ds |v(y , f (y )| ≤ 2 |v(y , yN )| + 2 (yN − f (y )) ≤ 2p−1 |v(y , yN )|p + 2p−1 εp−1 $ Multiplying both sides by
f (y )
f (y )+ε f (y )
|∂N v(y , s)|p ds.
1 + ∇y f (y )2N −1 , averaging in yN over the
interval (f (y ), f (y ) + ε), and integrating in y over QN −1 (0, r) gives $ p |v(y , f (y )| 1 + ∇y f (y )2N −1 dy QN −1 (0,r)
(18.4)
≤2
p−1
(1 + Lip f )ε
−1
QN −1 (0,r)
+2
p−1
(1 + Lip f )ε
f (y )+ε f (y )
p−1 QN −1 (0,r)
|v(y , yN )|p dyN dy
f (y )+ε f (y )
|∂N v(y , s)|p dsdy .
By a change of variables, we obtain that for every 0 < ε < r, |u|p dHN −1 ≤ 2p−1 (1 + Lip f )ε−1 |u|p dx QT (x0 ,r)∩∂Ω QT (x0 ,r)∩(Ω\Ωε ) p−1 p−1 ∇up dx. + 2 (1 + Lip f )ε QT (x0 ,r)∩(Ω\Ωε )
By a compactness argument, given R > 0 we can cover B(0, R) ∩ ∂Ω with a finite number of cubes as above, say Q1 , . . . , Q . Then p N −1 p−1 −1 |u| dH ≤ 2 (1 + L)ε |u|p dx B(0,R)∩∂Ω B(0,R)∩(Ω\Ωε ) p−1 p−1 ∇up dx, + 2 (1 + L)ε B(0,R)∩(Ω\Ωε )
where L is the maximum of the Lipschitz constants in each cube and ε is taken to be smaller than every side-length of the cubes. This shows that (iii) holds for every u ∈ Cc∞ (RN ). Now, if u ∈ W 1,p (Ω), consider a sequence {un }n in Cc∞ (RN ) such that un → u in W 1,p (Ω). Applying (18.3) to un − um , we obtain that the sequence of functions un restricted to B(0, R) ∩ ∂Ω is a Cauchy sequence in Lp (B(0, R)∩∂Ω) and thus it converges to a function wR ∈ Lp (B(0, R)∩∂Ω). By uniqueness, we have that wR (x) = wS (x) for HN −1 -a.e. x ∈ B(0, R)∩∂Ω for R < S and thus, by taking a sequence Rk → ∞ we obtain that the functions un restricted to ∂Ω converge on bounded sets of ∂Ω to a function in Lploc (∂Ω) that we denote Tr(u). Since (18.1) holds for each un and
594
18. Sobolev Spaces: Traces
ψ ∈ Cc∞ (RN ) in (see (11.1)), letting n → ∞ and using the fact that un → u in W 1,p (Ω) and {un }n restricted to ∂Ω converges to Tr(u) in Lp (supp ψ∩∂Ω), we obtain that (18.1) holds for u. Finally, the uniqueness of the trace operator follows from (18.3). We omit the details. Exercise 18.2. Let Ω ⊆ RN , N ≥ 2, be an open set whose boundary ∂Ω is Lipschitz continuous, and let 1 ≤ p < ∞. Prove that the previous theorem ˙ 1,p (Ω). Hint: Use Poincar´e’s inequality. continues to hold in W Exercise 18.3. Let Ω ⊂ R2 and u ∈ W 1,2 (Ω) be given as in Exercise 12.15. Prove that the trace of u on the line segment Γ := (0, ) × {0} is not in Lq (Γ, H1 ) for any q ≥ 1. An important corollary of the previous theorem is compactness of the trace operator for p > 1. Corollary 18.4 (Compactness of traces). Let Ω ⊆ RN , N ≥ 2, be an open bounded set whose boundary ∂Ω is Lipschitz continuous, let 1 < p < ∞, and let {un }n be a bounded sequence in W 1,p (Ω). Then there exist a subsequence {unk }k of {un }n and a function u ∈ W 1,p (Ω) such that unk → u in Lp (Ω) and Tr(unk ) → Tr(u) in Lp (∂Ω). Proof. Since Ω is bounded, by (18.3) we have p N −1 −1 p p−1 | Tr(v)| dH ≤ cε |v| dx + cε ∇vp dx ∂Ω
Ω
Ω
and for every ε > 0 sufficiently small (depending for every v ∈ only on Ω and p). By the Rellich–Kondrachov theorem (Theorems 12.18, 12.44 and 12.63) there exist a subsequence {unk }k of {un }n and a function u ∈ W 1,p (Ω) such that unk → u in Lp (Ω). Hence, p N −1 −1 | Tr(unk ) − Tr(u)| dH ≤ cε |unk − u|p dx + c1 εp−1 W 1,p (Ω)
∂Ω
Ω
for some constant c1 > 0, where we used the fact that {un }n is bounded in W 1,p (Ω). First, letting k → ∞ and then ε → 0+ completes the proof. Exercise 18.5. Prove that the previous corollary fails for p = 1. Exercise 18.6. Let Ω ⊆ RN , N ≥ 2, be an open bounded set whose boundary ∂Ω is Lipschitz continuous. Prove that if un u in W 1,1 (Ω), then Tr(un ) converges to Tr(u) in L1 (∂Ω). Hint: Use equi-integrability. Next we show that if the domain Ω is sufficiently regular, we may characterize W01,p (Ω) as the subspace of functions in W 1,p (Ω) with trace zero.
18.1. The Trace Operator
595
Theorem 18.7 (Traces and W01,p ). Let Ω ⊂ RN , N ≥ 2, be an open set whose boundary ∂Ω is Lipschitz continuous, let 1 ≤ p < ∞, and let u ∈ W 1,p (Ω). Then Tr(u) = 0 if and only if u ∈ W01,p (Ω). Proof. If u ∈ Cc∞ (Ω), then Tr(u) = 0, and so, since W01,p (Ω) is the closure of Cc∞ (Ω) with respect to the norm · W 1,p (Ω) it follows from (18.3) that Tr(u) = 0 for all u ∈ W01,p (Ω). To prove the converse implication, let u ∈ W 1,p (Ω) be such that Tr(u) = 0. Using a partition of unity and flattening out ∂Ω, we may assume that N N Ω = RN + and that u = 0 for all x ∈ R+ with x ≥ R (exercise). For x ∈ R define u(x) if xN > 0, (18.5) v(x) := 0 if xN ≤ 0. By Theorem 18.1 and the fact that Tr(u) = 0, for all ψ ∈ Cc1 (RN ) and i = 1, . . . , N we have that v∂i ψ dx = u∂i ψ dx = − ψ∂i u dx, RN
RN +
RN +
which shows that v ∈ W 1,p (RN ), with ∂i u(x) if xN > 0, ∂i v(x) = 0 if xN < 0, i = 1, . . . , N . We now translate v upwards. To be precise, for t > 0 and x ∈ RN define vt (x) := v(x , xN − t). Note that the support of vt is a compact set contained in RN −1 × (t/2, ∞). For each η > 0, by Lemma 11.36 we may find t so small that u − vt W 1,p (RN ) ≤ η. It now suffices to consider + a sequence of mollifications {(vt )ε }ε>0 of vt . Find 0 < ε < t/4 so small that (vt )ε − vε W 1,p (RN ) ≤ η. Since, for x ∈ RN , + ϕε (x − y) vt (y) dy = ϕε (x − y) vt (y) dy (vt )ε (x) = RN
RN −1 ×(t/2,∞)
and supp ϕε ⊆ B(0, ε), it follows that if 0 < xN < ε, then (vt )ε (x) = 0. 1,p N Thus vε ∈ Cc∞ (RN + ). This shows that u ∈ W0 (R+ ). We now consider the case m ≥ 2. Let Ω ⊆ RN , N ≥ 2, be an open set whose boundary ∂Ω is Lipschitz continuous. Given a function u ∈ W m,p (Ω), 1 ≤ p < ∞, and a multi-index α ∈ NN 0 , with 1 ≤ |α| ≤ m − 1, since α 1,p ∂ u ∈ W (Ω), by Theorem 18.1 there exists Tr(∂ α u) ∈ Lploc (∂Ω). Similarly, n for every direction ξ and 1 ≤ n ≤ m − 1 we have that there exists Tr( ∂∂ξnu ) ∈
596
18. Sobolev Spaces: Traces
Lploc (∂Ω). Moreover, since by (9.4) for u ∈ Cc∞ (RN ) and for every x ∈ RN we have that 1 ∂ nu ∂ α u(x)ξ α , (x) = ∂ξ n α! |α|=n
by Theorem 18.1(i) it follows that ∂nu 1 = (18.6) Tr Tr(∂ α u)ξ α . ∂ξ n α! |α|=n
We have seen in Chapter 9 that, in view of Rademacher’s theorem, at x0 ∈ ∂Ω there exist the tangent space T∂Ω (x0 ) to ∂Ω at x0 and the outward unit normal ν to ∂Ω at x0 (see Definitions 9.62 and 9.63). Thus for 1 ≤ n ≤ m − 1, in view of (18.6), we can define ∂nu 1 := Tr(∂ α u)ν α . (18.7) Tr ∂ν n α! HN −1 -a.e.
|α|=n
We will study the linear mapping (18.8)
∂ m−1 u ∂u , · · · , Tr . u ∈ W m,p (Ω) → Trm (u) := Tr(u), Tr ∂ν ∂ν m−1
The reason we are only interested in the traces of the normal derivatives is that all tangential derivatives depend only on Tr(u). To be precise, given a function v : ∂Ω → R differentiable at x0 and a tangent vector t ∈ T∂Ω (x0 ), there exists a function ψ : (−δ, δ) → RN differentiable at 0 such that ψ((−δ, δ)) ⊆ ∂Ω, ψ(0) = x0 and ψ (0) = t. Hence, by the chain rule, the function v ◦ ψ : (−δ, δ) → R is differentiable at 0, with (v ◦ ψ) (0) = dvx0 (t). In particular, if v : ∂Ω → R is of class C 1 , thenit can be extended to a N function u ∈ C 1 (RN ) and so dvx0 (t) = ∂u i=1 ∂i u(x0 )ti . It follows ∂t (x0 ) = 1 N that, given a function u ∈ C (R ), for every tangent vector t ∈ T∂Ω (x0 ), the directional derivative ∂u ∂t (x0 ) is completely determined by the restriction of u to ∂Ω, that is, by Tr(u). Since T∂Ω (x0 ) is a vector space of dimension N − 1, by taking N − 1 linearly independent tangent unit vectors b1 , . . . , bN −1 and adding to them the outward unit normal ν to ∂Ω at x0 , we obtain a basis in RN . In particular every partial and directional derivative of u at ∂u (x0 ), i = 1, . . . , N − 1, and x0 can be written as a linear combination of ∂b i ∂u of the normal derivative ∂ν (x0 ). To extend Theorem 18.7 we assume that ∂Ω is of class C m−1,1 (see Definition 11.55). Theorem 18.8 (Traces and W0m,p ). Let m ∈ N, m ≥ 2, let Ω ⊂ RN , N ≥ 2, be an open set whose boundary ∂Ω is of class C m−1,1 , let 1 ≤ p < ∞, and ∂ m−1 u let u ∈ W m,p (Ω). Then Tr(u) = 0, Tr( ∂u ∂ν ) = 0, . . . , Tr( ∂ν m−1 ) = 0 if and only if u ∈ W0m,p (Ω).
18.1. The Trace Operator
597
Proof. We only give the proof for m = 2 and leave the case m > 2 as an exercise. Step 1: Let u ∈ W 2,p (Ω). Let x0 ∈ ∂Ω. Since ∂Ω is of class C 1,1 there exist a rigid motion T : RN → RN with T (x0 ) = 0, a function f : RN −1 → R of class C 1,1 , and rx0 > 0 such that, setting y := T (x), we have T (Ω ∩ QT (x0 , rx0 )) = {y ∈ Q(0, rx0 ) : yN > f (y )}, at x0 and of side-length rx0 > 0. where QT (x0 , rx0 ) is an open cube centered √ Fix 0 < ε < r ≤ rx0 /[2(Lip f + 1) N − 1]. Setting w(y) := u(T −1 (y)) for y ∈ Q(0, r), by Theorem 11.53, w ∈ W 2,p (Q(0, r)). We claim that Tr(w) ∈ W 1,p (QN −1 (0, r)). If u ∈ Cc∞ (RN ), then g(y ) := Tr(w)(y , f (y )) = w(y , f (y )) and thus we can differentiate with respect to yi for i = 1, . . . , N − 1 to obtain that for LN −1 -a.e. y ∈ QN −1 (0, r), ∂w ∂w ∂f ∂g (y ) = (y , f (y )) + (y , f (y )) f (y ). ∂yi ∂yi ∂yN ∂yi Hence,
∂w ∂w ∂g (y ) ≤ (y , f (y )) + Lip f (y , f (y )) . ∂yi ∂yi ∂yN ∂w ∂w and by ∂y respectively, we obtain Using (18.4) with w replaced by ∂y i N ∂g p $ (y ) 1 + ∇y f (y )2N −1 dy ∂y i QN −1 (0,r) f (y )+ε p ∂w p ∂w ≤c (y) + (y) dyN dy ∂y ∂y i N QN −1 (0,r) f (y ) f (y )+ε 2 p ∂ 2 w p ∂ w +c (y) + 2 (y) dyN dy . ∂yN ∂yi ∂yN QN −1 (0,r) f (y )
It follows by (18.4) and the previous inequality that g ∈ W 1,p (QN −1 (0, r)). Now, if u ∈ W 2,p (Ω), then by Theorem 11.35 there exists a sequence {un } in Cc∞ (RN ) such that un restricted to Ω converges to u in W 2,p (Ω). Setting wn (y) := un (T −1 (y)) and gn (y ) := wn (y , f (y )), by (18.4) and the previous inequality applied to gn − g , it follows that {gn } is a Cauchy sequence in W 1,p (QN −1 (0, r)) and thus it converges to a function g in W 1,p (QN −1 (0, r)). This proves the claim. Step 2: If u ∈ Cc∞ (Ω), then Tr(u) = 0 and Tr(∂i u) = 0 for all i = 1, . . . , N 2,p ∞ and so Tr( ∂u ∂ν ) = 0 by (18.7). Since W0 (Ω) is the closure of Cc (Ω) with respect to the norm · W 2,p (Ω) it follows from (18.3) applied to u and to ∂i u that Tr(u) = Tr(∂i u) = 0 for all i = 1, . . . , N and for all u ∈ W02,p (Ω). In turn, Tr( ∂u ∂ν ) = 0 by (18.7). Conversely, assume that u ∈ W 2,p (Ω) is such that Tr(u) = 0, Tr( ∂u ∂ν ) = 0. Let f , r, and w be as in Step 1. Since Tr(u) = 0, we have that
598
18. Sobolev Spaces: Traces
Tr(w)(y , f (y )) = 0 for all y ∈ QN −1 (0, r). Since Tr(w) ∈ W 1,p (QN −1 (0, r)), by differentiating with respect to yi for i = 1, . . . , N − 1, for LN −1 -a.e. y ∈ QN −1 (0, r) we have that ∂w ∂w ∂f (y ) = 0. (y , f (y )) + Tr (y , f (y )) (18.9) Tr ∂yi ∂yN ∂yi On the other hand, since Tr( ∂u ∂ν )(y , f (y )) = 0 and the outward unit normal ν to ∂Ω at (y , f (y )) is given by
∇y f (y ) −1 (18.10) ν(y , f (y )) = $ ,$ , 1 + ∇y f (y )2N −1 1 + ∇y f (y )2N −1
by (18.7) we get (18.11)
N −1 i=1
Tr
∂w ∂yi
(y , f (y ))
∂w ∂f (y ) − Tr (y , f (y )) = 0. ∂yi ∂yN
The N − 1 equalities in (18.9) and (18.11) give a system of N equations ∂w )(y , f (y )), i = 1, . . . , N . Since the determinant is in the unknowns Tr( ∂y i ∂w )(y , f (y )) = 0, i = 1, different from zero (exercise), it follows that Tr( ∂y i . . . , N. ∂u ) = 0 on ∂Ω. We can now continue as in the proof It follows that Tr( ∂x i of Theorem 18.7 to prove that the function v defined in (18.5) belongs to W 2,p (RN ) and then proceed as in that proof to show that u ∈ W02,p (Ω).
In what follows we characterize the image of the trace operator. We will see that the cases p = 1 and p > 1 are quite different. We start with p = 1.
18.2. Traces of Functions in W 1,1 (Ω) In this section we show that if Ω is sufficiently regular, then Tr(W 1,1 (Ω)) = L1 (∂Ω). We begin with the simplest case in which Ω is the half-space N −1 × R : xN > 0}. RN + = {x = (x , xN ) ∈ R
˙ 1,1 (RN ) vanishing Theorem 18.9. Let N ≥ 2. Then for all functions u ∈ W + at infinity, (18.12)
Tr(u)(·, 0)L1 (RN −1 ) ≤ ∂N uL1 (RN ) . +
∗
Proof. Step 1: Assume first that u ∈ L1 (RN ) ∩ C 1 (RN ) with ∇u ∈ L1 (RN ; RN ). Reasoning as in the first step of the proof of Theorem 12.4,
18.2. Traces of Functions in W 1,1 (Ω) for every x ∈ RN −1 we have that |u(x , 0)| ≤
∞
599
|∂N u(x , xN )| dxN .
0
Integrate both sides with respect to x and use Tonelli’s theorem to conclude that |u(x , 0)| dx ≤ |∂N u(x)| dx. RN −1
RN +
∗
Step 2: To remove the additional assumption that u ∈ L1 (RN ) ∩ C 1 (RN ) ˙ 1,1 (RN ) vanishing with ∇u ∈ L1 (RN ; RN ), note that given a function u ∈ W + at infinity, using reflection (see Exercise 11.51), we can extend u to a function ˙ 1,1 (RN ) vanishing at infinity. In turn, by Theorem 12.4, we have that u∈W ∗ u ∈ L1 (RN ). Let uε := ϕε ∗ u, where ϕε is a standard mollifier. As in the proof of Lemma 11.25, we obtain that lim uε − uL1∗ (RN ) = 0,
ε→0+
Since, by Step 1,
RN −1
lim ∇uε − ∇uL1 (RN ) = 0.
ε→0+
|uε (x , 0)| dx ≤
RN
|∂N uε (x)| dx
for all ε > 0, by letting ε → 0+ we conclude that (18.12) holds.
Remark 18.10. In particular, it follows from the previous theorem that the linear operator 1 N −1 ) Tr : W 1,1 (RN + ) → L (R
u → Tr(u)(·, 0) is continuous. Exercise 18.11. Let u, v ∈ W 1,1 (RN + ). Prove that for all i = 1, . . . , N , u∂i v dx = − v∂i u dx + Tr(u) Tr(v)νi dx , RN +
RN +
RN −1
where ν = −eN . 1,1 (RN ), where Exercise 18.12. Let u ∈ W 1,1 (RN + ) and v ∈ W − N −1 × R : xN < 0}. RN − := {(x , xN ) ∈ R
(i) Prove that the function w : RN → R, defined by u(x) if x ∈ RN +, w(x) := v(x) if x ∈ RN −, belongs to BVloc (RN ). (ii) Prove that the function w belongs to W 1,1 (RN ) if and only if Tr(u) = Tr(v).
600
18. Sobolev Spaces: Traces
Next we prove that the operator Tr is onto. We use the abbreviation 18.2. The following theorem is due to Gagliardo [89]. The proof we present here is due to Mironescu [172]. Theorem 18.13 (Gagliardo). Let g ∈ L1 (RN −1 ), N ≥ 2. Then for every 0 < δ < 1 there exists a function u ∈ W 1,1 (RN + ) such that Tr(u) = g and uL1 (RN ) ≤ δgL1 (RN −1 ) , +
∇uL1 (RN ) ≤ (1 + δ)gL1 (RN −1 ) . +
Proof. Step 1: Assume that g ∈ Cc∞ (RN −1) \ {0}. Fix δ > 0 and let ϕ ∈ Cc∞ ([0, ∞)) be such that ϕ(0) = 1 and R+ |ϕ (t)| dt < 1 + δ. Note Lipschitz continuous function ϕ0 (t) = (1 − t)+ , t ≥ 0, satisfies that the R+ |ϕ0 (t)| dt = 1. Hence, to obtain ϕ it suffices to regularize ϕ0 . For n ∈ N and x = (x , xN ) ∈ RN −1 × [0, ∞) define un (x) := g(x )ϕ(nxN ). Then un ∈ C ∞ (RN −1 × [0, ∞)), un (x , 0) = g(x ) for every x ∈ RN −1 , while ∂i un (x) = ∂i g(x )ϕ(nxN ),
∂N un (x) = ng(x )ϕ (nxN )
for x ∈ RN + , i = 1, . . . , N − 1. Moreover, by Fubini’s theorem and the change of variables t = nxN , 1 |un (x)| dx = |g(x )| dx |ϕ(t)| dt → 0 n RN −1 RN R+ + as n → ∞. Similarly, for i = 1, . . . , N − 1, 1 |∂i un (x)| dx = |∂i g(x )| dx |ϕ(t)| dt → 0 n N −1 + RN R R + as n → ∞, while |∂N un (x)| dx = RN +
RN −1
|g(x )| dx
R+
|ϕ (t)| dt ≤ (1+δ)
RN −1
|g(x )| dx .
By taking n large enough we obtain the desired result in the case g ∈ Cc∞ (RN −1 ). Step 2: In the general case, it suffices to use the continuity of the trace operator and the density of smooth functions. We omit the details. Remark 18.14. (i) Note that from the proof we actually obtain a stronger estimate than the one in the statement of the theorem, to be precise, ∂i uL1 (RN ) ≤ δgL1 (RN −1 ) , +
∂N uL1 (RN ) ≤ (1 + δ)gL1 (RN −1 ) +
for i = 1, . . . , N − 1. Moreover, by taking n sufficiently large, given ε > 0 we can assume that supp u ⊆ supp g × [−ε, ε].
18.2. Traces of Functions in W 1,1 (Ω)
601
(ii) Peetre [193] has proved that there does not exist a bounded linear operator L :L1 (RN −1 ) → W 1,1 (RN +) g → L(g) with the property that Tr(L(g)) = g. Next we extend the previous theorem to special Lipschitz continuous domains. Theorem 18.15. Let f : RN −1 → R be a Lipschitz continuous function, N ≥ 2, and let Ω := {(x , xN ) ∈ RN −1 × R : xN > f (x )}.
(18.13)
Then for all u ∈ W 1,1 (Ω), Tr(u)L1 (∂Ω) ≤ (1 + Lip f )∂N uL1 (Ω) .
(18.14)
Moreover, for every g ∈ L1 (∂Ω) and 0 < δ < 1 there exists a function u ∈ W 1,1 (Ω) such that Tr(u) = g and uL1 (Ω) ≤ δgL1 (∂Ω) ,
∇uL1 (Ω;RN ) ≤ 2(1 + δ)(1 + Lip f )gL1 (∂Ω) .
Proof. We only prove the second part of the statement and leave the first as an exercise. Given g ∈ L1 (∂Ω), let h(x ) := g(x , f (x )), x ∈ RN −1 . Then (18.15) $ |h(y )| dy ≤ |g(x , f (x ))| 1 + ∇x f (x )2N −1 dx < ∞, RN −1
RN −1
and so by Theorem 18.13 for 0 < δ < 1 there exists w ∈ W 1,1 (RN + ) such that Tr(w) = h and (18.16) |w| dy ≤ δ |h| dy , |∂i w| dy ≤ δ |h| dy RN +
RN +
RN −1
|∂N w| dy ≤ (1 + δ)
RN −1
RN +
RN −1
|h| dy
for i = 1, . . . , N − 1. Set u(x) := w(Φ(x)) = w(x , xN − f (x )),
x ∈ Ω.
As in the proof of Theorem 13.4 we have that Φ is invertible, with inverse given by Φ−1 (y , yN ) = (y , yN + f (y )), and that Φ and Φ−1 are Lipschitz N −1 -a.e. x ∈ RN −1 continuous, Φ(Ω) = RN + , and det ∇Φ(x , xN ) = 1 for L
602
18. Sobolev Spaces: Traces
and for all xN ∈ R. Hence, by Exercise 11.51(iv), we have that u ∈ W 1,1 (Ω), with ∂u ∂w ∂w ∂f (18.17) (x) = (x , xN − f (x )) − (x , xN − f (x )) (y ), ∂xi ∂yi ∂yN ∂yi ∂w ∂u (x) = (x , xN − f (x )). ∂xN ∂yN In turn, by Theorems 9.52 and 11.53, (18.15), (18.16), |∂i u(x)| dx ≤ |∂i w(Φ(x))| dx + Lip f |∂N w(Φ(x))| dx Ω Ω Ω |∂i w(y)| dy + Lip f |∂N w(y)| dy = RN +
RN +
|g| dHN −1 ,
≤ (δ + (1 + δ) Lip f ) ∂Ω
where we have used the fact that det ∇Φ = det ∇Φ−1 = 1. Similarly, |∂N u(x)| dx = |∂N w(y)| dy ≤ (1 + δ) |g| dHN −1 RN +
Ω
and
∂Ω
|u(x)| dx = Ω
RN +
|g| dHN −1 ,
|w(y)| dy ≤ δ ∂Ω
which concludes the proof.
Remark 18.16. In view of Remark 18.14, given ε > 0 we can assume that (18.18)
supp u ⊆ {x ∈ RN : x ∈ supp g, −ε + f (x ) ≤ xN ≤ f (x ) + ε}.
The first part of the previous theorem continues to hold with W 1,1 (Ω) ˙ 1,1 (Ω). replaced with W Exercise 18.17. Let Ω ⊂ RN be as in (18.13) and let u ∈ W 1,1 (Ω) and φ ∈ Cc1 (RN ). Prove that for LN −1 -a.e. x ∈ RN −1 , Tr(uφ)(x , f (x )) = φ(x , f (x )) Tr(u)(x , f (x )). Finally, we extend the previous result to open sets Ω ⊂ RN with uniformly Lipschitz continuous boundary (see Definition 13.11). Theorem 18.18. Let Ω ⊆ RN , N ≥ 2, be an open set whose boundary ∂Ω is uniformly Lipschitz continuous. Then Tr(u)L1 (∂Ω) ≤ cM ε−1 (1 + L)uL1 (Ω) + c(1 + L)∇uL1 (Ω) for all u ∈ W 1,1 (Ω), where c = c(N ) > 0 and ε, L > 0, and M ∈ N are given in Definition 13.11.
18.2. Traces of Functions in W 1,1 (Ω)
603
Moreover, for every g ∈ L1 (∂Ω) there exists a function u ∈ W 1,1 (Ω) such that Tr(u) = g and uL1 (Ω) ≤ gL1 (∂Ω) ,
∇uL1 (Ω) ≤ 4(1 + L)gL1 (∂Ω) .
Proof. Step 1: Let u ∈ W 1,1 (Ω). Define φn , Ω0 , Ω± , φ0 , φ± , ψ± as in (13.43), (13.46), (13.47), (13.50), respectively. Let (18.19)
ψn (x) :=
ψ+ (x)φn (x) . φk (x) k
Using the facts that 0 ≤ ψ± , φn ≤ 1, ∇ψ± ∞ ≤ cε−1 , and (13.56), we get |ψn | ≤ M, ∇ψn ≤ M cε−1 . (18.20) n
n
Then in Ω we may write (see (13.52)) ψn u + ψ− u =: un + ψ− u. (18.21) u= n
n
Observe that since {Ωn }n is locally finite, any bounded neighborhood of every point x ∈ Ω intersects only finitely many Ωn ’s. By (13.50) the support of ψ− is contained in Ω. Hence, Tr(ψ− u) = 0 by Theorem 18.7. Next we study the functions un . Fix n. By property (iii) of Definition 13.11 there exist local coordinates y = (y , yN ) ∈ RN −1 × R and a Lipschitz continuous function f : RN −1 → R (both depending on n), with Lip f ≤ L, such that Ωn ∩ Ω = Ωn ∩ Vn ,
(18.22)
where Vn in local coordinates is given by {(y , yN ) ∈ RN −1 × R : yN > f (y )}. Since by (13.44) the support of un is contained in Ωn , we may extend un to be zero in Vn \ Ωn . Thus, we are in a position to apply Theorem 18.15 to find that Tr(un ) satisfies (18.14) with Vn in place of Ω. Since the support of ψn is contained in Ωn , it follows from (18.22) that ψn = 0 on ∂Vn \ ∂Ω. Hence, the same holds for Tr(un ). This shows that N −1 (18.23) | Tr(un )| dH ≤ (1 + L) ∇un dx ∂Ω∩∂Vn Vn ∇un dx. = (1 + L) Ω∩Ωn
Since by (18.21), in Ω ∩ Ωn , (18.24)
∇un = ∇ψn u + ψn ∇u,
604
18. Sobolev Spaces: Traces
using (18.20), we get that in Ω ∩ Ωn , ∇un ≤ cM ε−1 |u| + ∇u. Hence, by Exercise 18.17 and (18.23), N −1 | Tr(u)| dH ≤ | Tr(un )| dHN −1 ∂Ω
∂Ω∩∂Vn
n
≤ cM ε
−1
|u| dx + (1 + L)
(1 + L) Ω
∇u dx. Ω
Step 2: Given g ∈ L1 (∂Ω), since ψ+ = 1 on ∂Ω, we can write g =
ψn g.
n
By extending ψn g to be zero on ∂Vn \ ∂Ωn , we are in a position to apply Theorem 18.15 to find a function un ∈ W 1,1 (Vn ) such that Tr(un ) = ψn g on ∂Vn and N −1 |un | dx ≤ ψn |g| dH = ψn |g| dHN −1 , Vn ∂Vn ∂Ω ∇un dx ≤ 4(1 + L) ψn |g| dHN −1 , Vn
∂Ω
\ ∂Ω. Moreover, by (18.18) we where we used the fact that ψn = 0 in ∂Vn can assume that supp un ⊆ Ωn . Let u := n un . Since {Ωn }n is locally 1,1 (Ω) and Tr(u) = g. Moreover, finite we have that u ∈ Wloc N −1 |u| dx ≤ |un | dx ≤ ψn |g| dH = |g| dHN −1 , Ω
n
Vn
n
and similarly ∇u dx ≤ Ω
n
∂Ω
∂Ω
|g| dHN −1 .
∇un dx ≤ 4(1 + L) Vn
∂Ω
This completes the proof.
Remark 18.19. Note that by Exercise 13.15, if Ω ⊆ RN is an unbounded open set whose boundary ∂Ω is uniformly Lipschitz continuous, then Ω has infinite measure. On the other hand, if Ω ⊂ RN is an unbounded open set with finite measure, then ∂Ω is unbounded. In this case the function 1 belongs to W 1,1 (Ω), but its trace does not belong to L1 (∂Ω). Exercise 18.20. Let Ω ⊆ RN , N ≥ 2, be an open set whose boundary ∂Ω is uniformly Lipschitz continuous and let u ∈ W 1,1 (Ω) and v ∈ W 1,1 (RN \ Ω). (i) Prove that the function w : RN → R, defined by u(x) if x ∈ Ω, w(x) := v(x) if x ∈ RN \ Ω, belongs to BV (RN ).
18.3. Traces of Functions in BV (Ω)
605
(ii) Prove that the function w belongs to W 1,1 (RN ) if and only if Tr(u) = Tr(v).
18.3. Traces of Functions in BV (Ω) In this section we prove that every function in BV (Ω) has a trace in L1 (∂Ω), provided Ω is sufficiently regular. Theorem 18.21. Let Ω ⊆ RN , N ≥ 2, be an open set whose boundary ∂Ω is Lipschitz continuous. There exists a unique linear operator Tr : BV (Ω) → L1loc (∂Ω) such that (i) Tr(u) = u on ∂Ω for all u ∈ BV (Ω) ∩ C(Ω), (ii) for all ψ ∈ Cc1 (RN ), u ∈ BV (Ω), and i = 1, . . . , N , u∂i ψ dx = − ψDi u dx + ψ Tr(u)νi dHN −1 , (18.25) Ω
Ω
∂Ω
where ν is the outward unit normal to ∂Ω, (iii) for all but countably many R > 0 there exist two constants cR , εR > 0 depending on R and Ω such that | Tr(u)| dHN −1 ≤ cR ε−1 |u| dx B(0,R)∩∂Ω
B(0,R)∩(Ω\Ωε )
+ cR Du(B(0, R) ∩ (Ω \ Ωε )) for all but countably many 0 < ε ≤ εR , where Ωε := {x ∈ Ω : dist(x, ∂Ω) > ε}. Proof. By Theorem 14.9 for every u ∈ BV (Ω) there exists a sequence {un }n 1 ∗ in C ∞ (Ω) ∩ W 1,1 (Ω) such that un → u in L1 (Ω), ∇un LN Ω Du in the sense of measures and ∇un dx = Du(Ω). lim n→∞ Ω
It follows from Theorem 18.1, | Tr(un ) − Tr(u )| dHN −1 ≤ cR ε−1 |un − u | dx B(0,R)∩∂Ω B(0,R)∩(Ω\Ωε ) (∇un + ∇u ) dx + cR B(0,R)∩(Ω\Ωε )
for every , n ∈ N. By Exercise 14.12 and Proposition B.12, letting n, → ∞ first and then ε → 0+ , we obtain that {Tr(un )}n is a Cauchy sequence in L1 (B(0, R) ∩ ∂Ω) for all but countably many R > 0 and thus it converges in L1loc (∂Ω) to a function function Tr(u).
606
18. Sobolev Spaces: Traces
Since (18.1) holds for all un , all ψ ∈ Cc1 (RN ), and all i = 1, . . . , N , letting n → ∞ in (18.1) gives (ii). Moreover, since | Tr(un )| dHN −1 ≤ cR ε−1 |un | dx B(0,R)∩∂Ω B(0,R)∩(Ω\Ωε ) ∇un dx, + cR B(0,R)∩(Ω\Ωε )
if R and ε are such that Du(∂(B(0, R) ∩(Ω\Ωε )) ∩Ω) = 0, which happens for all but countably many ε and R, then by Exercise 14.12, ∇un dx = Du(B(0, R) ∩ (Ω \ Ωε )) lim n→∞ B(0,R)∩(Ω\Ω ) ε
and so, letting n → ∞ in the previous inequality we obtain (iii).
The analog of Theorem 18.18 is given by the following theorem. The proof is very similar to the one of Theorem 18.18 and is left as an exercise. Theorem 18.22. Let Ω ⊆ RN , N ≥ 2, be an open set whose boundary ∂Ω is uniformly Lipschitz continuous. Then Tr(u)L1 (∂Ω) ≤ cM ε−1 (1 + L)uL1 (Ω) + c(1 + L)Du(Ω) for all u ∈ BV (Ω), where ε, L > 0, M ∈ N are given as in Definition 13.11. Since W 1,1 (Ω) ⊂ BV (Ω), it follows from Theorem 18.18 that if Ω ⊆ RN , N ≥ 2, is an open set whose boundary ∂Ω is uniformly Lipschitz continuous, then Tr : BV (Ω) → L1 (Ω) is onto. Exercise 18.23 (Compactness of traces). Let Ω ⊂ RN be an open bounded set with Lipschitz continuous boundary. Prove that if {un }n is a sequence in BV (Ω) converging in L1 (Ω) to a function u ∈ BV (Ω) and such that 1 ∗ ∇un LN Ω Du in the sense of measures and Dun (Ω) → Du(Ω), then Tr(un ) → Tr(u) in L1 (∂Ω). Hint: See Exercise 18.6.
18.4. Traces of Functions in W 1,p (Ω), p > 1 In this section we study trace of functions in W 1,p (Ω) for 1 < p < N . We will see that the situation is quite different from the case p = 1. Theorem 18.24. Let 1 < p < N . Then there exists a constant c = ˙ 1,p (RN ) vanishing at infinc(N, p) > 0 such that for all functions u ∈ W + ity, (18.26)
Tr(u)(·, 0)Lp(N −1)/(N −p)(RN −1 ) ≤ c∇uLp (RN ) . +
18.4. Traces of Functions in W 1,p (Ω), p > 1
607 ∗
Proof. Step 1: Assume first that u ∈ Lp (RN ) ∩ C 1 (RN ) with ∇u ∈ Lp (RN ; RN ). Reasoning as in the first step of the proof of Theorem 12.4, for every x ∈ RN −1 and for r > 1 we have that ∞ r |u(x , 0)| ≤ r |u(x , xN )|r−1 |∂N u(x , xN )| dxN . 0
Integrate both sides with respect to x and use Tonelli’s theorem to conclude that (18.27) |u(x , 0)|r dx ≤ r |u(x)|r−1 |∂N u(x)| dx RN −1
RN +
(r−1)
≤ ruL(r−1)p (RN ) ∂N uLp (RN ) , +
+
where we have used H¨older’s inequality. Taking r := p(N − 1)/(N − p), we have that (r − 1)p = p∗ , and so, by Theorem 12.4, p∗ /p |u(x , 0)|p(N −1)/(N −p) dx ≤ cuLp∗ (RN ) ∂N uLp (RN ) RN −1
+
+
1+p∗ /p
≤ c∇uLp (RN ) . +
Note that 1 +
p∗ /p
= p(N − 1)/(N − p), so that (18.26) holds. ∗
Step 2: To remove the additional assumption that u ∈ Lp (RN ) ∩ C 1 (RN ) with ∇u ∈ Lp (RN ; RN ), we proceed as in the proof of Theorem 18.9 by first ˙ 1,p (RN ) vanishing at infinity to a function u ∈ extending a function u ∈ W + ˙ 1,p (RN ) vanishing at infinity and such that ∇uLp (RN ) ≤ c∇u p N W L (R+ ) and then using mollifiers. We omit the details. Remark 18.25. In particular, it follows from the previous theorem that the linear operator p(N −1)/(N −p) (RN −1 ) Tr : W 1,p (RN +) → L
is continuous. Moreover, taking r = p in (18.27) yields 1/p 1/p |u(x , 0)|p dx ≤ p |u(x)|p dx |∂N u(x)|p dx RN −1
≤p
RN +
RN +
RN +
|u(x)|p dx + p
RN +
|∂N u(x)|p dx,
where we have used Young’s inequality (see (B.17) in Appendix B). Note that this inequality actually holds for every 1 ≤ p < ∞. Hence, p p | Tr(u)(x , 0)| dx ≤ p |u(x)| dx + p |∂N u(x)|p dx, RN −1
for all u ∈ W 1,p (RN + ).
RN +
RN +
608
18. Sobolev Spaces: Traces
Exercise 18.26. Prove that if p = N , then for every N ≤ q < ∞ we have that q N −1 Tr : W 1,N (RN ) + ) → L (R is a continuous operator. Unlike the case p = 1, when 1 < p < ∞, the trace operator p N −1 ) Tr : W 1,p (RN + ) → L (R
is not onto. We now show that if u ∈ W 1,p (RN + ), 1 < p < ∞, then its trace Tr(u) belongs to the Besov space B 1−1/p,p (RN −1 ) (see Definition 17.1). The following theorem is due to Gagliardo [89]. Theorem 18.27 (Gagliardo). Let 1 < p < ∞ and let N ≥ 2. Then there ˙ 1,p (RN ), exists a constant c = c(p, N ) > 0 such that for all u ∈ W + (18.28)
| Tr(u)(·, 0)|B 1−1/p,p(RN −1 ) ≤ c∇uLp (RN ) . +
˙ 1,p (RN ) ∩ C ∞ (RN ). By Theorem 18.57 and Proof. Assume that u ∈ W + Remark 18.60 below we have that |u(·, 0)|B 1−1/p,p(RN −1 ) ≤ c∇uLp (RN ) . +
To remove the extra assumption that u ∈ C ∞ (RN ), we first use a re˙ 1,p (RN ) to a function flection argument to extend every function u ∈ W + 1,p N ˙ (R ) and then consider a sequence of {uε }ε , where uε := u ∗ ϕε u∈W and the ϕε are standard mollifiers. Since uε → u in L1loc (RN ), ∇uε → ∇u in Lp (RN ; RN ), and uε (·, 0) → Tr(u) in L1loc (RN −1 ), by selecting a subsequence, we may assume that uε (x , 0) → Tr(u)(x ) for LN −1 -a.e. x ∈ RN −1 . Hence, by Fatou’s lemma, | Tr(u)(x , 0) − Tr(u)(y , 0)|p dx dy p+N −2 x − y N RN −1 RN −1 −1 |uε (x , 0) − uε (y , 0)|p ≤ lim inf dx dy p+N −2 ε→0+ x − y N RN −1 RN −1 −1 ≤ c lim ∇uε (x)p dx = c ∇u(x)p dx. ε→0+
RN +
RN +
This concludes the proof.
The previous result shows in particular that 1−1/p,p (RN −1 ). Tr(W 1,p (RN + )) ⊆ B
To prove the opposite inclusion one could use Theorem 18.53 below, however, we present a different proof, which relies on the seminorm | · |B 1−1/p,p (RN −1 ) .
18.4. Traces of Functions in W 1,p (Ω), p > 1
609
Theorem 18.28 (Gagliardo). Let 1 < p < ∞, let N ≥ 2, and let g ∈ L1loc (RN −1 ) be such that |g|B 1−1/p,p (RN −1 ) < ∞. Then there exists a function ˙ 1,p (RN ) such that Tr(v)(·, 0) = g and v∈W + ∇vLp (RN ) ≤ c|g|B 1−1/p,p (RN −1 ) ,
(18.29)
+
where c = c(N, p) > 0. Proof. Let ϕ ∈ Cc∞ (RN −1 ) be such that supp ϕ ⊆ BN −1 (0, 1)
and
RN −1
ϕ(x ) dx = 1.
For x ∈ RN −1 and xN > 0 define 1 ϕ((x − y )/xN )g(y ) dy . (18.30) v(x) := N −1 xN RN −1 By Theorem C.20 (where xN we have that ∂v 1 (x) = N ∂xi xN RN −1 1 = N xN RN −1
plays the role of ε), for any i = 1, . . . , N − 1 ∂ϕ ((x − y )/xN )g(y ) dy ∂xi ∂ϕ ((x − y )/xN )[g(y ) − g(x )] dy , ∂xi
where in the second equality we used the fact that ∂ 1 ∂ (1) = ϕ((x − y )/x ) dy 0= N −1 ∂xi ∂xi xN RN −1 N ∂ϕ 1 = N ((x − y )/xN ) dy . xN RN −1 ∂xi Since supp ϕ ⊆ BN −1 (0, 1), we have that c |g(y ) − g(x )| dy . (18.31) |∂i v(x)| ≤ N xN BN −1 (x ,xN ) Raising both sides to the power p, integrating in x over RN + , and using H¨older’s inequality, we get |∂i v(x)|p dx RN +
1
≤c ≤c
RN +
p xN N
BN −1 (x ,xN )
(N −1)(p−1)
xN RN +
p xN N
|g(y ) − g(x )| dy
BN −1 (x ,xN )
p dx
|g(y ) − g(x )|p dy dx =: cI.
610
18. Sobolev Spaces: Traces
By Tonelli’s theorem, we get that ∞ I= =
RN −1
0
RN −1
RN −1
1 = N +p−2
1
|g(y ) − g(x )|p dy dx dxN
N +p−1 BN −1 (x ,xN ) xN ∞ p
1
|g(y ) − g(x )|
x −y N −1 |g(y ) − g(x )|p
RN −1
RN −1
p+N −2 x − y N −1
Hence, we have shown that p (18.32) |∂i v(x)| dx ≤ c RN +
RN −1
N +p−1 xN
dxN dy dx
dy dx .
|g(y ) − g(x )|p RN −1
x
−
p+N −2 y N −1
dy dx
for all i = 1, . . . , N − 1. Similarly, by Theorem B.52, we obtain that ∂ 1 ϕ((x − y )/x ) g(y ) dy ∂N v(x) = N N −1 ∂x N −1 x N R N ∂ 1 = ϕ((x − y )/x ) [g(y ) − g(x )] dy , N N −1 ∂x N −1 x N R N where in the second equality we used the fact that ∂ ∂ 1 0= (1) = ϕ((x − y )/x ) dy N −1 ∂xN ∂xN RN −1 xN N ∂ 1 = ϕ((x − y )/x ) dy . N N −1 ∂x N −1 N xN R Since supp ϕ ⊆ BN −1 (0, 1), we have that ∂ 1 ϕ((x − y )/x ) [g(y ) − g(x )] dy . ∂N v(x) = N N −1 BN −1 (x ,xN ) ∂xN xN Now for y ∈ BN −1 (x , xN ),
1−N ϕ((x − y )/xN ))| = −(N − 1)x−N |∂N (xN N ϕ((x − y )/xN ) 1−N − xN
N −1
∂i ϕ((x − y )/xN )(xi − yi )x−2 N
i=1
≤
cx−N N
In turn, c |∂N v(x)| ≤ N xN
−1−N + cx − y N −1 xN ≤ cx−N N .
BN −1 (x ,xN )
|g(y ) − g(x )| dy .
18.4. Traces of Functions in W 1,p (Ω), p > 1
611
We can now continue as before (see (18.31)) to conclude that |g(y ) − g(x )|p p (18.33) |∂N v(x)| dx ≤ c dy dx . p+N −2 RN RN −1 RN −1 x − y N −1 + ˙ 1,p (RN ) and that (18.29) Since v ∈ C 1 (RN + ), we have shown that v ∈ W + holds. To conclude the proof, it remains to prove that Tr(v) = g. Using standard mollifiers (exercise), we may find a sequence {gn }n in C ∞ (RN −1 ) with |gn |B 1−1/p,p (RN −1 ) < ∞ such that |g − gn |B 1−1/p,p (RN −1 ) → 0. Let vn be defined as in (18.30) with g replaced by gn . Reasoning as in the proof of Theorem C.16(i), we have that vn ∈ C(RN −1 × [0, ∞)), with vn (x , 0) = gn (x ). By (18.29) applied to vn − v and gn − g we have that ∂i vn → ∂i v in Lp (RN + ) for every i = 1, . . . , N . In turn, by (18.3) and Exercise 18.2, we obtain that Tr(u)(·, 0) = g. Corollary 18.29. Let 1 < p < ∞, let N ≥ 2, and let g ∈ B 1−1/p,p (RN −1 ). Then for every 0 < ε ≤ 1 there exists a function u ∈ W 1,p (RN + ) such that Tr(u)(·, 0) = g, supp u ⊆ RN −1 × [−ε, ε] and uLp (RN ) ≤ ε1/p gLp (RN −1 ) , +
|u|W 1,p (RN ) ≤ cε−1/p gLp (RN −1 ) + c|g|B 1−1/p,p(RN −1 ) , +
where c = c(N, p) > 0. Proof. Let v be as in the previous theorem. By Theorem C.16 (with xN in place of ε) we have that for all xN > 0, p |v(x , xN )| dx ≤ |g(x )|p dx . (18.34) RN −1
RN −1
Integrating in xN in (0, ε) gives ε p |v(x , xN )| dx dxN ≤ ε (18.35) 0
RN −1
RN −1
|g(x )|p dx .
Let ψ ∈ C ∞ ([0, ∞)) be a decreasing function such that ψ = 1 in [0, ε/2], ψ(xN ) = 0 for xN ≥ ε and ψ ∞ ≤ cε−1 . For x = (x , xN ) ∈ RN + we define u(x) := ψ(xN )v(x). By (18.35), Tonelli’s theorem, and the fact that ψ(xN ) = 0 for xN ≥ ε we have ε p p (18.36) |u(x)| dx = (ψ(xN )) |v(x)|p dx dxN RN +
0
≤ 0
RN −1
ε RN −1
|v(x)|p dx dxN ≤ ε
RN −1
|g(x )|p dx ,
612
18. Sobolev Spaces: Traces
while for i = 1, . . . , N − 1, |∂i u(x)| = |ψ(xN )∂i v(x)| ≤ |∂i v(x)|, where we used the fact that ψ∞ ≤ 1. In turn, by (18.32), we obtain that |g(y ) − g(x )|p p |∂i u(x)| dx ≤ c dy dx p+N −2 RN RN −1 RN −1 x − y N −1 + for all i = 1, . . . , N − 1. On the other hand, ∂N u(x) = ψ(xN )∂N v(x) + ψ (xN )v(x), and so, by (18.33), (18.35) and the fact that ψ (xN ) = 0 for xN ≥ ε, 1/p 1/p ε 1/p |∂N u|p dx ≤ |∂N v|p dx +c ε−p |v|p dx dxN RN +
≤c
RN +
RN −1
|g(y ) − g(x )|p RN −1
p+N −2 x − y N −1
dy dx
0
1/p
+cε
RN −1
−(p−1)/p
RN −1
|g|p dx
1/p .
Reasoning as in the last part of the previous theorem we can show that Tr(u) = g. Exercise 18.30. Prove that the previous corollary (without the condition supp u ⊆ RN −1 ×[−ε, ε]) continues to hold if we consider the function u(x) := e−xN /p v(x), x ∈ RN +. Exercise 18.31. Let N = 2 and x = (x1 , x2 ). ˙ 1,2 (R2 ), then it has a trace on the line R × {0} (i) Prove that if u ∈ W and |v(x1 ) − v(y1 )|2 dx1 dy1 ≤ π ∇u(x)2 dx, 2 |x − y | 2 1 1 R R R where v(x1 ) := Tr(u)(x1 , 0). Hint: Use Fourier transforms for v and u. ˙ 1,2 (B(0, r) ∩ R2 ), then (ii) Prove that if u ∈ W + r r |v(x1 ) − v(y1 )|2 dx1 dy1 ≤ 2π ∇u(x)2 dx, 2 |x − y | 1 1 −r −r R2 where v(x1 ) := Tr(u)(x1 , 0). (iii) Prove that the constant 2π in (ii) is sharp. Hint: Use polar coordinates. To characterize the traces of Sobolev functions in W 1,p (Ω) when p > 1, we need to extend the definition of Besov spaces (see Chapter 17) to ∂Ω.
18.4. Traces of Functions in W 1,p (Ω), p > 1
613
Definition 18.32. Let Ω ⊂ RN be as in (18.13). Given 1 ≤ p < ∞ and 0 < s < 1, for every function u ∈ L1loc (∂Ω, HN −1 ) we set 1/p |u(x) − u(y)|p N −1 N −1 |u|B s,p (∂Ω,HN −1 ) := dH (y)dH (x) N −1+sp ∂Ω ∂Ω x − y and we say that u belongs to the Besov space B s,p (∂Ω, HN −1 ) if (18.37)
uB s,p (∂Ω,HN −1 ) := uLp (∂Ω) + |u|B s,p (∂Ω,HN −1 ) < ∞.
In what follows we use the abbreviations (18.38) B s,p (∂Ω) := B s,p (∂Ω, HN −1 ), | · |B s,p (∂Ω) := | · |B s,p (∂Ω,HN −1 ) ,
· B s,p (∂Ω) := · B s,p (∂Ω,HN −1 ) .
Exercise 18.33. Given 1 ≤ p < ∞, 0 < s < 1, and Ω as in (18.13), prove that (1+Lip f )−((N +1)/p+s) |w|B s,p (RN −1 ) ≤ |u|B s,p (∂Ω) ≤ c(1+Lip f )2/p |w|B s,p (RN −1 ) , where w(x ) := u(x , f (x )), x ∈ RN −1 . Theorem 18.34. Let f : RN −1 → R be a Lipschitz continuous function, N ≥ 2, let Ω := {(x , xN ) ∈ RN −1 × R : xN > f (x )}, and let 1 < p < ∞. Then there exists a constant c = c(p, N ) > 0 such that Tr(u)Lp (∂Ω) ≤ c(1 + Lip f )1/p uW 1,p (Ω) , | Tr(u)|B 1−1/p,p (∂Ω) ≤ c(1 + Lip f )1+2/p uW 1,p (Ω) for all u ∈ W 1,p (Ω). Conversely, for every g ∈ B 1−1/p,p (∂Ω) there exists a function u ∈ W 1,p (RN + ) such that Tr(u) = g and uLp (Ω) ≤ cgLp (∂Ω) , ∇uLp (Ω) ≤ (1 + Lip f )gLp (∂Ω) + c(1 + Lip f )2+N/p |g|B 1−1/p,p (∂Ω) , where c = c(N, p) > 0. ˙ 1,p (Ω), by Theorems 9.52 and 11.53 the function Proof. Step 1: If u ∈ W w(y) := u(Ψ(y)) = u(y , yN + f (y )),
y ∈ RN +
˙ 1,p (RN ) with belongs to W + (18.39)
∂w (y) = ∂yi ∂w (y) = ∂yN
∂u ∂u ∂f (y , yN + f (y )) + (y , yN + f (y )) (y ), ∂xi ∂xN ∂xi ∂u (y , yN + f (y )). ∂xN
614
18. Sobolev Spaces: Traces
Hence, by Theorem 18.27, Exercise 18.33, (18.39), and again Theorem 11.53, | Tr(u)|B 1−1/p,p (∂Ω) ≤ (1 + Lip f )2/p | Tr(w)|B 1−1/p,p (RN −1 ) ≤ c(1 + Lip f )2/p ∇y wLp (RN ) +
≤ c(1 + Lip f )
1+2/p
∇uLp (Ω) ,
where c = c(N, p). Moreover, if u ∈ W 1,p (Ω), then by (9.47), Remark 18.25, (18.39), and again Theorem 11.53, | Tr(u)(x )|p dHN −1 ≤ (1 + Lip f ) | Tr(w)(y )|p dy N −1 ∂Ω R p ∂w p ≤ p(1 + Lip f ) |w(y)| dy + p(1 + Lip f ) (y) dy RN RN ∂yN + + p ∂u = p(1 + Lip f ) |u(x)|p dx + p(1 + Lip f ) (x) dx. Ω Ω ∂xN Step 2: Conversely, given g ∈ L1loc (∂Ω) with |g|B 1−1/p,p (∂Ω) < ∞, setting h(x ) := g(x , f (x )), by Theorem 18.28 there exists a function w ∈ ˙ 1,p (RN ) such that Tr(v) = h and W + ∇y wLp (RN ) ≤ c|h|B 1−1/p,p (RN −1 ) +
≤ c(1 + Lip f )1+N/p |g|B 1−1/p,p (∂Ω) , where c = c(N, p) > 0 and in the last inequality we used Exercise 18.33. Setting u(x) := w(x , xN − f (x )), x ∈ Ω, reasoning somewhat as in the first ˙ 1,p (Ω), with part of the proof we have that u ∈ W (18.40)
∇uLp (Ω) ≤ c(1 + Lip f )∇y wLp (RN ) +
≤ c(1 + Lip f )
2+N/p
|g|B 1−1/p,p(∂Ω) .
Moreover, by Corollary 18.29, if g ∈ Lp (∂Ω), then we can assume that N −1 × [−ε, ε] and w ∈ W 1,p (RN + ) with supp w ⊆ R wLp (RN ) ≤ cε1/p hLp (RN −1 ) ≤ cε1/p gLp (∂Ω) , +
|w|W 1,p (RN ) ≤ cε−1/p hLp (RN −1 ) + c|h|B 1−1/p,p (RN −1 ) +
≤ cε−1/p gLp (∂Ω) + c(1 + Lip f )1+N/p |g|B 1−1/p,p(∂Ω) , $ where we used the fact that 1 ≤ 1 + ∇x f (x )2N −1 (see (9.47)) and Exercise 18.33. In turn, by Theorem 11.53, (18.41)
uLp (Ω) = wLp (RN ) ≤ cε1/p gLp (∂Ω) , +
18.4. Traces of Functions in W 1,p (Ω), p > 1
615
while as in (18.40),
(18.42) |u|W 1,p (Ω) ≤ c(1 + Lip f )|w|W 1,p (RN ) ≤ c(1 + Lip f )ε−1/p gLp (∂Ω) +
+ c(1 + Lip f )
2+N/p
|g|B 1−1/p,p(∂Ω) .
This completes the proof. Remark 18.35. Note that in the second part of the proof we have that supp u ⊆ {x ∈ RN : x ∈ supp g, −ε + f (x ) ≤ xN ≤ f (x ) + ε}.
To extend the previous theorem to open sets with uniformly Lipschitz continuous boundary, we consider a different seminorm. Definition 18.36. Let Ω ⊆ RN , N ≥ 2, be an open set whose boundary ∂Ω is uniformly Lipschitz continuous. Given 1 ≤ p < ∞ and 0 < s < 1, for every function u ∈ L1loc (∂Ω) we set |u|B s,p (∂Ω,HN −1 ) := ∂Ω
∂Ω∩B(x,ε)
1/p |u(x) − u(y)|p N −1 N −1 dH (y) dH (x) x − yN −1+sp
and we say that u belongs to the Besov space B s,p (∂Ω, HN −1 ) if (18.43)
uB s,p (∂Ω,HN −1 ) := uLp (∂Ω) + |u|B s,p (∂Ω,HN −1 ) < ∞.
Here ε > 0 is the parameter given in Definition 13.11. In what follows we will use the abbreviations (18.38). Exercise 18.37. Prove that if ∂Ω is bounded, then the norm (18.43) is equivalent to the norm (18.10). Exercise 18.38. Let Ω ⊆ RN , N ≥ 2, be an open set of the form (18.13), let 1 ≤ p < ∞ and 0 < s < 1. Prove that for every ε > 0 and every u ∈ B s,p (∂Ω), |u|B s,p (∂Ω) ≤ c(1 + Lip f )s+1/p ε−s uLp (∂Ω) + |u|B s,p (∂Ω) , where c = c(N, p, s) > 0. Exercise 18.39. Let Ω ⊆ RN , N ≥ 2, be an open set whose boundary ∂Ω : ∂Ω → R, n ∈ N, be Lebesgue is uniformly Lipschitz continuous, let wn ∞ measurable functions, and let w(x) := n=1 wn (x), x ∈ ∂Ω, where we assume that there exists an integer M ∈ N such that for every x ∈ ∂Ω at most M terms wn (x) are nonzero. Prove that for every 1 ≤ p < ∞ and 0 < s < 1,
1/p ∞ p 1/p wn Lp (∂Ω) , wLp (∂Ω) ≤ M n=1
616
18. Sobolev Spaces: Traces
while |w|B s,p (∂Ω)
≤ (2M )
1/p
∞
(|wn |B s,p (∂Ω) )p n=1
1/p .
Theorem 18.40. Let Ω ⊆ RN , N ≥ 2, be an open set whose boundary ∂Ω is uniformly Lipschitz continuous and let 1 < p < ∞. Then Tr(u)Lp (∂Ω) ≤ cM (1 + L)1/p (ε−1 uLp (Ω) + ∇uLp (Ω;RN ) ), | Tr(u)|B 1−1/p,p (∂Ω) ≤ cM (1 + L)1+2/p ε−1 (ε−1 uLp (Ω) + ∇uLp (Ω;RN ) ) for all u ∈ W 1,p (Ω), where c = c(N, p) > 0 and ε, L > 0, and M ∈ N are given in Definition 13.11. Moreover, for every g ∈ B 1−1/p,p (∂Ω) there exist a constant c = c(N, p) and a function u ∈ W 1,p (Ω) such that Tr(u) = g,
uLp (Ω) ≤ M 1/p ε1/p gLp (∂Ω) and
∇uLp (Ω;RN ) ≤ cM (1 + L)3+N/p ε−1/p gLp (∂Ω) + cM (1 + L)2+(N +1)/p |g|B 1−1/p,p(∂Ω) . We begin with a preliminary result. Define φn , Ω0 , Ω± , φ0 , φ± , ψ± as in (13.43), (13.46), (13.47), (13.50), respectively, and let ψn be as in (18.19). Lemma 18.41. Let Ω ⊆ RN , N ≥ 2, be an open set whose boundary ∂Ω is uniformly Lipschitz continuous, let 1 ≤ p < ∞, and let 0 < s < 1. Then there exists a constant c = c(N, p, s) > 0 such that for every function u ∈ B s,p (∂Ω), 1/p (18.44) ψn upLp (∂Vn ) |u|B s,p (∂Ω) ≤ cM 1/p (1 + L)1/p ε−1/p−s + cM
1/p
n
(|ψn u|B s,p (∂Vn ) )p
1/p ,
n
while (18.45)
(|ψn u|B s,p (∂Vn ) )p
1/p
n
≤ cM 1/p (1 + L)(1−s−(N −1)/p)
+ +1/p
ε−s uLp (∂Ω)
+ cM 1/p |u|B s,p (∂Ω) . Proof. Step 1: We begin by proving that |ψn u|B s,p (∂Ω) ≤ c(1 + L)1/p ε−1/p−s ψn uLp (∂Ω∩∂Vn ) + c|ψn u|B s,p (∂Vn ) , |ψn u|B s,p (∂Vn ) ≤ c(1 + L)1/p ε−1/p−s ψn uLp (∂Ω∩∂Vn ) + c|ψn u|B s,p (∂Ω) ,
18.4. Traces of Functions in W 1,p (Ω), p > 1
617
where we recall that Ωn ∩ Ω = Ωn ∩ Vn , where Vn is (up to a rigid motion) a special Lipschitz domain. It is enough to show the first inequality, since the second is similar. If x, y ∈ / Ωn , then (ψn u)(x) = (ψn u)(y) = 0 since the support of ψn is contained in Ωn , while if x, y ∈ Ωn , then Ωn ∩ ∂Ω = Ωn ∩ ∂Vn . Hence, by interchanging x and y if needed, and using the fact that χB(x,ε) (y) = χB(y,ε) (x), we can write
∂Ω
=
∂Ω∩B(x,ε)
|(ψn u)(x) − (ψn u)(y)|p dHN −1 (y)dHN −1 (x) x − yN −1+sp
|(ψn u)(x) − (ψn u)(y)|p dHN −1 (y)dHN −1 (x) N −1+sp x − y ∂Vn ∩Ωn ∂Vn ∩Ωn ∩B(x,ε) dHN −1 (y) p |(ψn u)(x)| dHN −1 (x) +2 N −1+sp ∂Ω∩Ωn (∂Ω\Ωn )∩B(x,ε) x − y
=: I + II. It remains to estimate II. Note that we can replace ∂Ω∩Ωn with ∂Ω∩{ψn >0} . We recall that φn := ϕε/4 ∗ χΩ3ε/4 (see (13.43)), where ϕε/4 is a standard n 3ε/4 = x ∈ RN : B (x, 3ε/4) ⊆ Ωn . Hence, if we assume mollifier and Ωn that supp ϕ ⊆ B(0, 1/2), for every x ∈ ∂Ω ∩ {ψn > 0}, we have that dist(x, ∂Ωn ) ≥ 3ε/8. On the other hand, if y ∈ (∂Ω \ Ωn ) ∩ B(x, ε), we have that x − y ≥ 3ε/8. Moreover, by Definition 13.11(i), there exists k such that B(x, ε) ⊆ Ωk . Hence, up to a rotation, we can write (18.46)
dHN −1 (y) N −1+sp (∂Ω\Ωn )∩B(x,ε) x − y ≤ cε−N +1−sp ≤ c(1 + L)ε
$
BN −1 −1−sp
(x ,ε)
1 + ∇y fk (y )2N −1 dy
.
Thus, II ≤ c(1 + L)ε
−1−sp
|(ψn u)(x)|p dHN −1 (x). ∂Ω∩Ωn
618
18. Sobolev Spaces: Traces
Step 2: Since u = |u|B s,p (∂Ω)
n ψn u,
≤ (2M )
by Exercise 18.39 and Step 1,
1/p
∞
(|ψn u|B s,p (∂Ω) )p n=1
≤ cM 1/p (1 + L)1/p ε−1/p−s + cM
1/p
1/p
ψn upLp (∂Vn )
n
(|ψn u|B s,p (∂Vn ) )p
1/p
1/p ,
n
which proves (18.44). Step 3: To prove (18.45), we first apply Step 1 to obtain 1/p 1/p (|ψn u|B s,p (∂Vn ) )p ≤ c(1 + L)1/p ε−1/p−s ψn upLp (∂Ω∩∂Vn ) n
+c
n
(|ψn u|B s,p (∂Ω) )p
1/p .
n
We now estimate the second term on the right-hand side of the previous inequality. For x ∈ ∂Ω and y ∈ ∂Ω ∩ B(x, ε) we have |(ψn u)(x) − (ψn u)(y)|p ≤ 2p−1 (ψn (y))p |u(x) − u(y)|p + 2p−1 |u(x)|p |ψn (x) − ψn (y)|p ≤ 2p−1 (ψn (y))p |u(x) − u(y)|p 1 + 2p−1 |u(x)|p x − yp ∇ψn (tx + (1 − t)y)p dt, 0
where in the last term we used the fundamental theorem of calculus and H¨older’s inequality. By (13.56), (18.47) |ψn |p ≤ M, ∇ψn p ≤ cM ε−p . n
n
Hence,
|(ψn u)(x) − (ψn u)(y)|p ≤ 2p−1 M |u(x) − u(y)|p
n
+ cM ε−p |u(x)|p x − yp . In turn, (|ψn u|B s,p (∂Ω) )p ≤ cM (|u|B s,p (∂Ω) )p n
+ cM ε
−p
|u(x)| ∂Ω
x − yp(1−s)−N +1 dHN −1 (y)dHN −1 (x).
p ∂Ω∩B(x,ε)
18.4. Traces of Functions in W 1,p (Ω), p > 1
619
By Definition 13.11(i), if x ∈ ∂Ω there exists k such that B(x, ε) ⊆ Ωk . Hence, up to a rotation, we can write x − yp(1−s)−N +1 dHN −1 (y) ∂Ω∩B(x,ε) + p(1−s)−(N −1) ≤ (1 + L)(p(1−s)−(N −1)) x − y N −1 Jk (y ) dy BN −1 (x ,ε)
≤ c(1 + L) where Jk (y ) :=
(p(1−s)−(N −1))+ +1
$
εp(1−s) ,
1 + ∇y fk (y )2N −1 and we used the fact that x −
y N −1 ≤ x−y ≤ (1+L)x −y N −1 . Combining the last two inequalities gives (18.45). We turn to the proof of Theorem 18.40. Proof of Theorem 18.40. The proof is similar to the one of Theorem 18.18 and we indicate only the main changes. Step 1: Given u ∈ W 1,p (Ω), as in Theorem 18.18 we write u as in (18.21) and study the functions un . Since the support of ψn is contained in Ωn and Ωn ∩ Ω = Ωn ∩ Vn , where Vn is (up to a rigid motion) a special Lipschitz domain, by extending un to be zero in Vn \ Ωn , we are in a position to apply Theorem 18.34 to find that Tr(un )Lp (∂Vn ) ≤ c(1 + L)1/p un W 1,p (Vn ) ,
(18.48)
| Tr(un )|B 1−1/p,p (∂Vn ) ≤ c(1 + L)1+2/p un W 1,p (Vn ) . Since ψ+ = 1 on ∂Vn ∩ ∂Ω and ψn = 0 on ∂Vn \ ∂Ω, we have that Tr(un ) = Tr(ψn u) on ∂Vn , and so by (18.44) and (18.48), 1/p Tr(un )pLp (∂Vn ) | Tr(u)|B 1−1/p,p (∂Ω) ≤ cM 1/p ε−1 (1 + L)1/p + cM 1/p
n
(| Tr(un )|B 1−1/p,p (∂Vn ) )p
n
≤ cM
1/p
(1 + L)1+2/p ε−1
1/p
un pW 1,p (Ωn ∩Ω)
n
By (18.24), (18.47), the following pointwise inequalities hold in Ω, ψnp |u|p ≤ M |u|p , n
∇un p ≤ 2p−1
n
≤ cM ε
n −p
∇ψn p |u|p + 2p−1
n
|u| + cM ∇u , p
p
ψnp ∇up
1/p .
620
18. Sobolev Spaces: Traces
which, combined with the previous inequality, give | Tr(u)|B 1−1/p,p (∂Ω) ≤ cM (1 + L)1+2/p ε−2 uLp (Ω) + cM (1 + L)1+2/p ε−1 ∇uLp (Ω;RN ) . Similarly, by Exercise 18.39 and (18.48),
Tr(u)Lp (∂Ω) ≤ M 1/p
∞
1/p Tr(un )pLp (∂Ω)
n=1
≤M
1/p
(1 + L)1/p
un pW 1,p (Ωn ∩Ω)
1/p
n
≤ cM (1 + L)
1/p
(ε
−1
uLp (Ω) + ∇uLp (Ω;RN ) ).
Step 2: Given g ∈ B 1−1/p,p (∂Ω), since ψ+ = 1 on ∂Ω, we can write g = ψn g. By extending ψn g to be zero on ∂Vn \ ∂Ωn , we are in a position n
to apply Theorem 18.34 (see (18.41) and (18.42)) and Remark 18.35 to find a function un ∈ W 1,p (Vn ) such that Tr(un ) = ψn g on ∂Vn , supp un ⊂ Ωn , un Lp (Vn ) ≤ ε1/p ψn gLp (∂Vn ) , and
|un |W 1,p (Vn ) ≤ c(1 + L)ε−1/p ψn gLp (∂Vn ) + c(1 + L)2+N/p |ψn g|B 1−1/p,p (∂Vn )
≤ c(1 + L)3+N/p ε−1/p ψn gLp (∂Vn ) + c(1 + L)2+N/p |ψn g|B 1−1/p,p (∂Vn ) , where c = c(N, p) > 0 and in the second inequality we used Exercise 18.38. 1,p (Ω) Let u := n un . Since {Ωn }n is locally finite we have that u ∈ Wloc ⊂ Ω , Ω ∩Ω = and Tr(u) = g. By Exercise 12.34 and the facts that supp u n n n ψ = 1 on ∂Ω, Ωn ∩ Vn , ψn = 0 on ∂Vn \ ∂Ω, and ∞ n=1 n
uLp (Ω) ≤ M
1/p
∞
1/p un pLp (Ω)
n=1
≤M
1/p 1/p
ε
∞ n=1
1/p ψn gpLp (∂Ω)
= M 1/p ε1/p gLp (∂Ω) .
18.5. Traces of Functions in W m,1 (Ω)
621
Similarly, |u|W 1,p (Ω) ≤ M
1/p
∞
1/p |un |pW 1,p (Ω)
≤M
n=1
≤ cM
1/p
+M
(1 + L)
1/p
3+N/p −1/p
ε
(1 + L)
2+N/p
∞
∞
1/p
∞
1/p |un |pW 1,p (Vn )
n=1
1/p
ψn gpLp (∂Vn )
n=1
(|ψn g|B 1−1/p,p (∂Vn ) )p n=1
1/p
≤ cM (1 + L)3+N/p ε−1/p gLp (∂Ω) + cM (1 + L)2+(N +1)/p |g|B 1−1/p,p (∂Ω) , where the last inequality follows from (18.45) and the fact that (−(N − 2)/p)+ = 0. As in Remark 18.19, if Ω ⊆ RN is Lipschitz continuous but ∂Ω is unbounded, then the results in the previous theorem continue to hold locally. Exercise 18.42. Let Ω ⊂ RN be an open set whose boundary ∂Ω is uniformly Lipschitz continuous and let u ∈ W 1,p (Ω) and v ∈ W 1,p (RN \ Ω), 1 < p < ∞. Prove that the function u(x) if x ∈ Ω, w(x) := v(x) if x ∈ RN \ Ω belongs to W 1,p (RN ) if and only if Tr(u) = Tr(v).
18.5. Traces of Functions in W m,1 (Ω) In this section given an integer m ∈ N, m ≥ 2, we characterize the image of the trace operator Trm in (18.8) when p = 1. Theorem 18.43. Let N ≥ 2 and let m ∈ N with m ≥ 2. Then the image of the linear mapping (18.8) is given by Trm (W m,1 (RN + )) =
m−2 /
B m−k−1,1 (RN −1 ) × L1 (RN −1 ).
k=0
Proof. We only prove the case m = 2. ∞ N −1 × [0, ∞)), by Theorem 18.57 Step 1: Given u ∈ W 2,1 (RN + ) ∩ C (R below we have that
| Tr(u)(·, 0)|B 1,1(RN −1 ) ≤ c∇2 uL1 (RN ) . +
622
18. Sobolev Spaces: Traces
The general case follows by density (see Theorem 11.35) and is left as an exercise. Moreover, by Theorem 18.9, Tr(u)(·, 0)L1 (RN −1 ) ≤ ∂N uL1 (RN ) . +
Thus,
Tr(W 2,1 (RN + ))
⊆
B 1,1 (RN −1 ).
To prove the opposite inclusion, let g ∈ S(RN −1 ) and let ϕ ∈ S(RN −1 ) be & = 1 in BN −1 (0, 1/2), a radial function such that supp ϕ & ⊆ BN −1 (0, 1) and ϕ where ϕ & is the Fourier transform of ϕ. Then by Theorems 17.77 and 18.53 below the function x − y 1 ϕ g(y ) dy , x ∈ RN v(x) := N +, xN x N RN satisfies v(x , 0) = g(x ) for all x ∈ RN −1 , and (18.49)
∇2 vL1 (RN ) ≤ c|g|B 1,1 (RN −1 ) . +
Moreover, (18.35) continues to hold with p = 1. Hence, by Theorem 13.51 with p = q = 1 (see (13.98)), ∇vL1 (R) ≤ cε−1 vL1 (R) + cvL1 (R) ∇2 vL1 (R) 1/2
1/2
≤ cε−1 vL1 (R) + cε∇2 vL1 (R) , where R = I1 × · · · × IN is a rectangle with ε := mini L1 (Ii ) and we used Young’s inequality. By the Lebesgue monotone convergence theorem, letting L1 (Ii ) → ∞ we get (18.50)
∇vL1 (RN −1 ×(0,ε)) ≤ cε−1 vL1 (RN −1 ×(0,ε)) + cε∇2 vL1 (RN ) +
≤ cgL1 (RN −1 ) + cε|g|B 1,1(RN −1 ) , where the last inequality follows from (18.35) and (18.49). Let ψ ∈ C ∞ ([0, ∞)) be a decreasing function such that ψ = 1 in [0, ε/2], ψ(xN ) = 0 for xN ≥ ε, ψ ∞ ≤ cε−1 and ψ ∞ ≤ cε−2 for some constant c > 0. For x = (x , xN ) ∈ RN + we define u(x) := ψ(xN )v(x). Then for i, j = 1, . . . , N − 1, ∂i u(x) = ψ(xN )∂i v(x), 2 2 ∂i,j u(x) = ψ(xN )∂i,j v(x),
∂N u(x) = ψ(xN )∂N v(x) + ψ (xN )v(x), 2 2 ∂i,N u(x) = ψ(xN )∂i,N v(x) + ψ (xN )∂i v(x),
2 2 u(x) = ψ(xN )∂N v(x) + 2ψ (xN )∂N v(x) + ψ (xN )v(x). ∂N
Hence, also by (18.35), (18.36), (18.50), and the properties of ψ, we have that ∇uL1 (RN ) ≤ cgL1 (RN −1 ) + cε|g|B 1,1(RN −1 ) , +
while
∇2 uL1 (RN ) ≤ cε−1 gL1 (RN −1 ) + c|g|B 1,1 (RN −1 ) . +
The general case follows by density.
18.5. Traces of Functions in W m,1 (Ω)
623
Step 2: We prove that for every h ∈ L1 (RN −1 ) there exists a function w ∈ W 2,1 (RN + ) such that Tr(w) = 0 and Tr(∂N w) = h. Consider a sequence {hn }n of functions in Cc2 (RN −1 ) such that hn → h in L1 (RN −1 ). Since {hn }n is a Cauchy sequence in L1 (RN −1 ), for each k ∈ N there exists Nk such that for all n ≥ Nk and ≥ 1, hn+ − hn L1 (RN −1 ) ≤ 2−k hL1 (RN −1 ) . Let nk := max{Nk , Nk−1 + 1}. The subsequence {hnk }k has the property ∞
hnk+1 − hnk L1 (RN −1 ) ≤ 2hL1 (RN −1 ) .
k=1
Without loss of generality, we assume that the entire sequence {hn }n satisfies ∞
(18.51)
hn+1 − hn L1 (RN −1 ) ≤ 2hL1 (RN −1 ) .
n=1
Define h0 := 0 and construct a strictly decreasing sequence {tn }n in (0, 1) such that tn → 0+ and |tn − tn+1 | ≤
(18.52)
hL1 (RN −1 ) 1 . n 2 hn+1 W 2,1 (RN −1 ) + hn W 2,1 (RN −1 ) + 1
N −1 and x ≥ 0 define Note that t0 = ∞ N n=0 (tn − tn+1 ). For x ∈ R ⎧ ⎨ 0 if xN ≥ t0 , x − t t n − xN ω(x , xN ) := n+1 N hn+1 (x ) + hn (x ) if tn+1 ≤ xN ≤ tn , ⎩ tn − tn+1 tn − tn+1 and w(x , xN ) :=
xN
ω(x , s) ds.
0
Then lim ω(·, xN ) − hL1 (RN −1 ) = 0,
xN →0+
and so w(x , 0) = 0,
∂N w(x , 0) = h(x ).
Moreover, by Tonelli’s theorem 2 |∂N w| dx = |∂N ω| dx RN +
= ≤
RN +
∞
RN −1 n=0 ∞
h n+1 (x ) − hn (x ) dxN dx t − t n n+1 tn+1 tn
hn+1 − hn L1 (RN −1 ) ≤ 2hL1 (RN −1 )
n=0
624
18. Sobolev Spaces: Traces
by (18.51), while for i = 1, . . . , N − 1, 2 |∂N,i w| dx = |∂i ω| dx RN +
RN +
∞
= ≤
RN −1 n=0 ∞
t −x xN − tn+1 n N ∂i hn+1 (x ) + ∂i hn (x ) dxN dx tn − tn+1 tn+1 tn − tn+1 tn
∇x hn+1 L1 (RN −1 ) + ∇x hn L1 (RN −1 ) |tn+1 − tn |
n=0
≤ 2hL1 (RN −1 ) , by (18.52). Similarly, for i, j = 1, . . . , N − 1, xN 2 2 |∂i,j w| dx dxN = |∂i,j ω| dsdx RN +
∞ tn
≤ ≤
RN −1 ∞
RN +
0
2 2 |∂i,j hn+1 (x )| + |∂i,j hn (x )| dxN dx
n=0 tn+1
2 ∇x hn+1 L1 (RN −1 ) + ∇2x hn L1 (RN −1 ) |tn+1 − tn |
n=0
≤ 2hL1 (RN −1 ) by (18.52). Similar estimates hold for the first order derivatives of w and for w. Step 3: To prove that 1,1 N −1 ) × L1 (RN −1 ) Tr2 (W 2,1 (RN + )) = B (R
consider g ∈ B 1,1 (RN −1 ) and h ∈ L1 (RN −1 ). By Step 1 there exists v ∈ 1,1 (RN ), we have that W 2,1 (RN + ) such that Tr(v) = g. Since ∂N v ∈ W + 1 N −1 ) by Theorem 18.9. In turn, by Step 2 we can find a Tr(∂N v) ∈ L (R function w ∈ W 2,1 (RN + ) such that Tr(w) = 0 and Tr(∂N w) = h − Tr(∂N v). Taking u := v + w, we have that Tr(u) = g and Tr(∂N u) = h, as desired. Exercise 18.44. Prove the case m ≥ 2. Remark 18.45. Step 1 of the previous proof actually shows that ˙ 2,1 (RN ˙ 1,1 N −1 ). Tr(W + )) = B (R Remark 18.46. By taking t0 ≤ ε in Step 2 of the previous proof, we can assume that supp u ⊆ RN −1 × [−ε, ε] in Step 3. Unlike the first order difference operator Δh u(x) = u(x + h) − u(x), there is no simple way (see the papers of Jonsson and Wallin [124], [125]) to define in an intrinsic way the second-order difference operator Δ2h u(x) = u(x + 2h) − 2u(x + h) + u(x) or the higher-order ones Δm h u(x) for functions
18.5. Traces of Functions in W m,1 (Ω)
625
defined on the boundary of special Lipschitz domains or uniformly Lipschitz domains. Thus, we first flatten the boundary and then use the definition on RN −1 . However, this requires more regularity on the boundary. We recall Remark 13.12. Definition 18.47. Given 1 ≤ p < ∞ and s ≥ 1, let m := s + 1, let f ∈ C m−1,1 (RN −1 ), N ≥ 2, and let (18.53)
Ω := {(x , xN ) ∈ RN −1 × R : xN > f (x )}.
For every function u ∈ L1loc (∂Ω) we set (18.54)
|u|B s,p (∂Ω) := |v|B s,p (RN −1 ) ,
where v(x ) := u(x , f (x )), x ∈ RN −1 , and we say that u belongs to the Besov space B s,p (∂Ω) if the function v belongs to B s,p (RN −1 ). We define uB s,p (∂Ω) := uLp (∂Ω) + |u|B s,p (∂Ω) . Given 1 ≤ p < ∞ and s > 0, if Ω ⊂ RN , N ≥ 2, is an open set for which there exists a rigid motion T : RN → RN such that T (Ω) can be written in the form (18.53), for every function u ∈ L1loc (∂Ω) we set |u|B s,p (∂Ω) := |u ◦ T −1 |B s,p (∂T (Ω)) and we say that u belongs to the Besov space B s,p (∂Ω) if the function u◦T −1 belongs to B s,p (∂T (Ω)). We define uB s,p (∂Ω) := uLp (∂Ω) + |u|B s,p (∂Ω) . Using partitions of unity we are now ready to extend Definition 18.47 to domains with Lipschitz continuous boundary. Definition 18.48. Given 1 ≤ p < ∞ and s ≥ 1, let Ω ⊆ RN , N ≥ 2, be an open set whose boundary ∂Ω is of class C m−1,1 , where m := s + 1, and let {ψn }n be a partition of unity such that supp ψn ⊂ Ωn , with Ωn ⊂ RN an open set such that Ω ∩ Ωn = Ω ∩ Vn , where Vn is, up to a rigid motion, a domain of the type (18.53). We say that a function u ∈ Lp (∂Ω) belongs to the Besov space B s,p (∂Ω) if uBqs,p (∂Ω) := uLp (∂Ω) + |u|Bqs,p (∂Ω) < ∞, where (18.55)
|u|Bqs,p (∂Ω) :=
|ψn u|pB s,p (∂Vn )
1/p .
n
We leave as an exercise to check that if ∂Ω is bounded, then the space does not depend on the particular partition of unity and that a different partition of unity gives rise to an equivalent seminorm | · |Bqs,p (∂Ω) . If ∂Ω is uniformly Lipschitz continuous, we take the sets Ωn and Vn in the previous definition to be the ones in Definition 13.11 and ψn as in (18.19). B s,p (∂Ω)
626
18. Sobolev Spaces: Traces
Theorem 18.49. Let Ω ⊆ RN , N ≥ 2, be an open set whose boundary ∂Ω is uniformly of class C m−1,1 , where m ∈ N with m ≥ 2. Then the image of the linear mapping (18.8) is given by Trm (W m,1 (Ω)) =
m−2 /
B m−k−1,1 (∂Ω) × L1 (∂Ω).
k=0
The proof follows the lines of Theorems 18.34 and 18.40 using Theorem 18.43 in place of Theorems 18.27 and 18.28. I am out of time and out of space, so I am going to omit it.
18.6. Traces of Functions in W m,p (Ω), p > 1 In this section given an integer m ∈ N, we characterize the image of the trace operator Trm in (18.8) when 1 < p < ∞. Theorem 18.50. Let N ≥ 2, let m ∈ N, with m ≥ 2, and let 1 < p < ∞. Then the image of the linear mapping (18.8) is given by Trm (W
m,p
(RN + ))
=
m−1 /
B m−k−1/p,p (RN −1 ).
k=0
Proof. The proof follows by an induction argument of m using Theorems 17.69 and Theorems 18.27 and 18.28. We leave the details as an exercise. In turn, reasoning as in Theorems 18.34 and 18.40 using Theorem 18.50 in place of Theorems 18.27 and 18.28 we have the following result, whose proof we omit. Theorem 18.51. Let Ω ⊆ RN , N ≥ 2, be an open set whose boundary ∂Ω is uniformly of class C m−1,1 , where m ∈ N with m ≥ 2, and let 1 < p < ∞. Then the image of the linear mapping (18.8) is given by Trm (W
m,p
(Ω)) =
m−1 /
B m−k−1/p,1 (∂Ω).
k=0
Remark 18.52. The previous theorem was recently extended by Maz ja, Mitrea, and Shaposhnikova [166] to bounded open sets with Lipschitz continuous boundary.
18.7. Besov Spaces and Weighted Sobolev Spaces In this section we will show that a smooth function u belongs to Bqs,p (RN −1 ) if and only if it is the trace of a function v : RN + → R in some weighted N −1 ) be a radial function such that supp ϕ &⊂ Sobolev space. Let ϕ ∈ S(R
18.7. Besov Spaces and Weighted Sobolev Spaces
627
BN −1 (0, 1), and ϕ & = 1 in BN −1 (0, 1/2), where ϕ & is the Fourier transform of ϕ. In particular, ϕ(x ) dx = ϕ(0) & = 1. RN −1
Given u ∈
S(RN −1 ),
for x = (x , xN ) ∈ RN + define x − y 1 ϕ u(y ) dy . v(x) := N −1 x N −1 xN N R
(18.56)
Note that v(x) = uxN (x ) = (u ∗ ϕxN )(x ), that is, v is just the mollification of u with xN in place of ε. The proof of the following theorem is due to Mironescu and Russ [173]. Theorem 18.53. Let 1 ≤ p ≤ ∞, 1 ≤ q < ∞, and s > 0. Given u ∈ S(RN −1 ), let v be the function given in (18.56). Then for every α ∈ NN 0 , with |α| > s, ∞ 1/q dxN ∂ α v(·, xN )qLp (RN −1 ) 1−q(|α|−s) ≤ c|u|∨ Bqs,p (RN −1 ) , 0 xN where c = c (α, p, q, s) > 0. We begin with a preliminary result. Lemma 18.54. Let w ∈ C ∞ (RN + ) be such that ∞ ∂w (18.57) w(x) = − (x , r) dr ∂x N xN for every x ∈ RN + . Let 1 ≤ p ≤ ∞, 1 ≤ q < ∞, and a > −1. Then ∞ ∞ q q a+q ∂w a xN w(·, xN )Lp (RN −1 ) dxN ≤ c xN (·, xN ) p N −1 dxN . ∂x ) L (R N 0 0 Proof. We have
w(·, xN )Lp (RN −1 ) ≤
∞ xN
and so ∞ q a xN w(·, xN )Lp (RN −1 ) dxN ≤ 0
∞ 0
∂w (·, s) p N −1 ds ∂xN L (R )
xaN
q ∂w (·, r) p N −1 dr dxN . ∂xN L (R )
∞ xN
We now apply Hardy’s inequality (see Theorem C.41) to the function ∂w (·, xN ) p N −1 . f (xN ) := ∂xN L (R ) We turn to the proof of Theorem 18.53.
628
18. Sobolev Spaces: Traces
Proof of Theorem 18.53. Step 1: Given an integer ∈ N we consider ∂v and take the Fourier transform Fx in the variable x . By Theorem ∂x N
& N y ), for y ∈ RN −1 we have 10.71 and the fact that ϕ' xN (y ) = ϕ(x ∂v ∂ ∂ ) = ( ϕ ∗ u)(y ) = (ϕ' u(y )) (y Fx xN xN (y )& ∂xN ∂xN ∂xN
=
∂ (ϕ(x & N y )& u(y )). ∂xN
Consider two functions η ∈ S(R) and ψ ∈ S(RN −1 ) such that (18.58)
ϕ(y & ) = ψ(y N −1 )
η&(y ) = y N −1 ψ () (y N −1 )
and
for all y ∈ RN −1 . Then we can rewrite the previous equality as ∂v ∂ ) = (ψ(xN y N −1 )& u(y )) (y Fx ∂xN ∂xN u(y ) = y N −1 ψ () (xN y N −1 )& 1 1 u(y ) = η&(xN y )& u(y ). = (xN y N −1 ) ψ () (xN y N −1 )& xN xN Since supp ϕ & ⊂ BN −1 (0, 1), ϕ & = 1 in BN −1 (0, 1/2), by (18.58) we have that η&(y ) = 0 for y N −1 < 1/2 or y N −1 > 1. Hence, by (17.57) and (17.58), we can write ∂ v 1 1 u(y ) = η&(xN y )ϕ0 (2−k y )& u(y ) (y ) = η&(xN y )& Fx ∂xN xN xN k∈Z 1 η&(xN y )ϕ0 (2−k y )& u(y ), = xN k∈I xN
where (18.59)
IxN := {k ∈ Z : supp ϕ0 (2−k ·) ∩ supp η(xN ·) = ∅},
and so, by Theorem 10.71 again, ∂v 1 (x , x ) = (ηxN ∗ (u ∗ ψk ))(x ), N ∂xN xN k∈I xN
where ψk is the inverse Fourier transform of the function ϕ0 (2−k ·) (see 17.60) and we used Theorem 10.62. Now, if α ∈ NN 0 is a multi-index of the form α = (β, ), we have 1 ∂ |α| v (x , x ) = (ηxN ∗ ∂ β (u ∗ ψk ))(x ). N ∂(x )β ∂xN xN k∈I xN
18.7. Besov Spaces and Weighted Sobolev Spaces
629
Hence, 1 ηxN ∗ ∂ β (u ∗ ψk )Lp (RN −1 ) xN k∈I xN cη ≤ ∂ β (u ∗ ψk )Lp (RN −1 ) , xN k∈I
∂ α v(·, xN )Lp (RN −1 ) ≤
xN
where we used the fact that ηxN ∗∂ β (u∗ψk )Lp (RN −1 ) ≤ cη ∂ β (u∗ψk )Lp (RN −1 ) (see the proof of Theorem C.16(iii)). Since supp u ∗ ψk ⊆ BN −1 (0, 2k+1 a) we β can now apply Theorem 10.73 to get ∂ (u ∗ ψk )Lp (RN −1 ) ≤ c2(k+1)|β| u ∗ ψk Lp (RN −1 ) and so c (k+1)|β| 2 u ∗ ψk Lp (RN −1 ) xN k∈I xN c ≤ |α| u ∗ ψk Lp (RN −1 ) , xN k∈Ix
∂ α v(·, xN )Lp (RN −1 ) ≤
N
where in the last inequality we used the facts that |α| = |β| + and that by taking a = 1/4 in (17.56) we have that supp ϕ0 (2−k ·) = B(0, 2k−1 ) \ B(0, 2k−3 ) (see (17.58)), and so the set IxN in (18.59) is contained in the set of k ∈ Z such that 1/xN < 2k < 8/xN . Since for every xN > 0 there are at most three k such that 1/xN < 2k < 8/xN , we obtain
∞ 0
∂ α v(·, xN )qLp (RN −1 )
∞
≤c
x1+qs N
0
∞
≤c 0
≤c
1
k∈Z
q
u ∗ ψk Lp (RN −1 )
dxN
k∈IxN
1
x1+qs N
k∈IxN
u ∗
dxN 1−q(|α|−s) xN
u ∗ ψk qLp (RN −1 ) dxN
ψk qLp (RN −1 )
23−k 2−k
1
1+qs dxN = c
xN
2qsk u ∗ ψk qLp (RN −1 ) ,
k∈Z
where in the third inequality we used Tonelli’s theorem and as usual the constant c changed from line to line. In view of Theorem 17.77 we have proved the conclusion of the theorem for every multi-index α ∈ NN 0 with αN > 0. Step 2: It remains to study the case of a multi-index α ∈ NN 0 with αN = 0 N −1 is such that |β| = |α| > s. and |α| > s. Write α = (β, 0), where β ∈ N0
630
18. Sobolev Spaces: Traces
Then a := q (|β| − s)−1 > −1, and so by Lemma 18.54 applied to w := ∂ α v,
∞ 0
∂ α v(·, xN )qLp (RN −1 )
∞
≤ 0
dxN 1−q(|α|−s) xN
∂ (β,1) v(·, xN )qLp (RN −1 )
dxN . 1−q(|β|+1−s) xN
It now suffices to apply Step 1 to ∂ (β,1) v. This concludes the proof.
Exercise 18.55. Prove that the previous theorem continues to hold for q = ∞. Remark 18.56. Since the function v defined in (18.56) is given by v(x , xN ) = (u ∗ ϕxN )(x ), we can define v also for a tempered distribu tion T ∈ S (RN −1 ), by setting (see Theorem 10.56), v(x , xN ) = T (ϕxxN ), where we recall that ϕxxN (y ) := ϕxN (x − y ), y ∈ RN −1 . Reasoning as in the proof of Theorem 10.52), one can show that v ∈ C ∞ (RN + ). Reasoning as in Step 1 of the previous proof, we still have that 1/q ∞ dxN ∂ α v(·, xN )qLp (RN −1 ) 1−q(|α|−s) ≤ c|T |∨ Bqs,p (RN −1 ) 0 xN for every multi-index α ∈ NN 0 , with αN ≥ 1 and |α| > s. However, Lemma 18.54 in general fails in this case. As an example, taking the distribution Tp associated to a polynomial p (see Remark 10.37), one can show that v(x , xN ) = p(x ) for every xN > 0 and x ∈ RN −1 (exercise), and so w := ∂ α v = ∂ α p does not satisfy (18.57). We now prove the opposite of Theorem 18.53. Theorem 18.57. Let 1 ≤ p ≤ ∞, 1 ≤ q < ∞, and s > 0. Given v ∈ C ∞ (RN −1 ×[0, ∞)), let u(x ) := v(x , 0), x ∈ RN −1 . Then for every m ∈ N, with m > s, ∞ dxN q ∂ α v(·, xN )qLp (RN −1 ) 1−q(m−s) , |u|B s,p (RN −1 ) ≤ c q xN α∈Ξm 0 −1 , |β| = m} ∪ where c = c (m, p, q, s) > 0 and Ξm := {(β, 0) : β ∈ NN 0 {(0, m)}.
We begin with a preliminary result.
18.7. Besov Spaces and Weighted Sobolev Spaces
631
Lemma 18.58. Given v ∈ C ∞ (RN −1 × [0, ∞)), let u(x ) := v(x , 0), x ∈ RN −1 . Then for x , h ∈ RN −1 and m ∈ N, Δm h u(x ) =
m
(−1)k
m m k Δ h N −1 eN v(x + kh, 0)
k=0 m
(−1)k+1
+
m m k Δ(h,0) v(x , khN −1 ).
k=1
Proof. Set r := hN −1 . For every x ∈ RN −1 we can write u(x ) =
m
(−1)k
m m , kr) + (−1)k+1 m v(x k k v(x , kr).
k=0
k=1
By applying the previous identity to u(x + jh) it follows from (17.11) that Δm h u(x )
m = (−1)m−j m + jh u x j
=
j=0 m
(−1)m−j
m m (−1)k m j k v(x + jh, kr)
j=0 m
k=0 m k+1 m m−j m (−1) (−1) + k j v(x j=0 k=1 m m (−1)j m = j Δ h N −1 eN v(x + jh, 0) j=0 m m (−1)k+1 m + k Δ(h,0) v(x , kr), k=1
+ jh, kr)
which concludes the proof. We turn to the proof of Theorem 18.57.
Proof of Theorem 18.57. Reasoning as in the proof of Lemma 17.25 we can show that if ξ = (h, τ ) ∈ RN −1 × [0, ∞) with either h = 0 or τ = 0, then (18.60)
|Δm ξ v(x)|
≤
cξm N
α∈Ξm
m 0
xm−1 |∂ α v(x + xN ξ)| dxN . N
632
18. Sobolev Spaces: Traces
For h ∈ RN −1 set r := hN −1 . In view of Lemma 18.58 and (18.60) we have m m m m |ΔreN v(x + kh, 0)| + |Δm |Δh u(x )| ≤ c (h,0) v(x , kr)| k=0
k=1
m
≤ crm + cr
k=0 α∈Ξm m m
m 0
k=1 α∈Ξm
xm−1 |∂ α v(x + kh, xN r)| dxN N m
0
xm−1 |∂ α v(x + xN h, kr)| dxN . N
Taking the Lp norm in x on both sides and using Tonelli’s theorem and a change of variables gives m m m−1 m m m Δh uLp (RN −1 ) ≤ cr xN f (xN r) dxN + cr f (kr), 0
k=0
mr
≤c
sm−1 f (s) ds + crm 0
m
f (kr),
k=0
where f (xN ) := α∈Ξm ∂ α v(·, xN )Lp (RN −1 ) . In turn, dh q Δm h uLp (RN −1 ) N hN −1+sq RN −1 −1 m h N −1
q dh m−1 ≤c τ f (τ ) dτ N hN −1+sq RN −1 0 −1 m dh +c (f (khN −1 ))q =: I + II. N −1−(m−s)q N −1 hN −1 k=0 R Using spherical coordinates and Hardy’s inequality (see Theorem C.41) we have that
q ∞ mr dr m−1 I≤c τ f (τ ) dτ 1+sq r 0 ∞ 0 dxN ≤c (f (xN ))q 1−q(m−s) . 0 xN Similarly, again by spherical coordinates and a change of variables, ∞ m ∞ dr dxN q (f (kr)) 1−(m−s)q ≤ c (f (xN ))q 1−q(m−s) , II ≤ c r 0 x k=0 0 N
which concludes the proof.
Exercise 18.59. Prove that the previous theorem continues to hold for q = ∞.
18.7. Besov Spaces and Weighted Sobolev Spaces
633
Remark 18.60. If m ∈ N, 1 ≤ p = q < ∞, and s = m − 1/p > 0, we obtain ∞ p |u|B m−1/p,p (RN −1 ) ≤ c |∂ α v(x , xN )|p dx dxN . α∈Ξm
0
RN −1
Appendix A
Functional Analysis When probed under pressure, grad students will either blurt out what they are doing (but won’t know if it means anything), or they will blurt out what they plan to do (but won’t know how to do it). — Jorge Cham, www.phdcomics.com
In this chapter we present without proofs most of the results from functional analysis used in the text.
A.1. Topological Spaces Definition A.1. Let X be a nonempty set. A collection τ ⊆ P(X) is a topology if the following hold. (i) ∅, X ∈ τ . (ii) If Ui ∈ τ for i = 1, . . . , n, n ∈ N, then U1 ∩ · · · ∩ Un ∈ τ . (iii) If {Uα }α∈Λ is an arbitrary collection of elements of τ , then Uα ∈ τ. α∈Λ
Example A.2. Given a nonempty set X, the smallest topology consists of {∅, X}, while the largest topology contains all subsets as open sets. The pair (X, τ ) is called a topological space and the elements of τ are open sets. For simplicity, we often apply the term topological space only to X. A set C ⊆ X is closed if its complement X \ C is open. The closure E of a set E ⊆ X is the smallest closed set that contains E. The interior E ◦ of a set E ⊆ X is the union of all its open subsets. A point x0 ∈ X is an accumulation point for a set E ⊆ X if for every open set U that contains x0 there exists x ∈ E ∩ U , with x = x0 . 635
636
A. Functional Analysis
Given a set E ⊆ X we say that a set F ⊆ E is relatively open (respectively, relatively closed ) if there exists an open set U ⊆ X (respectively, a closed set C ⊆ X) such that F = E ∩ U (respectively, F = E ∩ C). A subset E of a topological space X is said to be dense if its closure is the entire space, i.e., E = X. We say that a topological space is separable if it contains a countable dense subset. Given a point x ∈ X, a neighborhood 1 of x is an open set U ∈ τ that contains x. Given a set E ⊆ X, a neighborhood of E is an open set U ∈ τ that contains E. A topological space is a Hausdorff space if for all x, y ∈ X with x = y there exist two disjoint neighborhoods of x and y. Given a topological space X and a sequence {xn }n in X, we say that {xn }n converges to a point x ∈ X if for every neighborhood U of x we have that xn ∈ U for all n sufficiently large. Note that unless the space is Hausdorff, the limit may not be unique. We now introduce the notion of a base for a topology. Let (X, τ ) be a topological space. A family β of open sets of X is a base for the topology τ if every open set U ∈ τ may be written as a union of elements of β. Given a point x ∈ X, a family βx of neighborhoods of x is a local base at x if every neighborhood of x contains an element of βx . Proposition A.3. Let X be a nonempty set and let β ⊆ P(X) be a family of sets such that (i) for every x ∈ X there exists B ∈ β such that x ∈ B, (ii) for every B1 , B2 ∈ β, with B1 ∩ B2 = ∅, and for every x ∈ B1 ∩ B2 there exists B3 ∈ β such that x ∈ B3 and B3 ⊆ B1 ∩ B2 . Then the collection τ ⊆ P(X) of arbitrary unions of members of β is a topology for which β is a base. As we will see later on, the metrizability and the normability of a given topology depend on the properties of a base (see Theorems A.15 and A.42 below). Definition A.4. Let (X, τ ) be a topological space. (i) The space X satisfies the first axiom of countability if every x ∈ X admits a countable local base at x. (ii) The space X satisfies the second axiom of countability if it has a countable base. In the text we use several notions of compactness. 1 The reader should be warned that in some texts (e.g., [200]) the definition of neighborhood is different.
A.1. Topological Spaces
637
Definition A.5. Let X be a topological space. (i) A set K ⊆ X is compact if for every open cover of K, i.e., for every collection {Uα }α∈Λ of elements of τ such that K ⊆ α∈Λ Uα , there exists a finite subcover (i.e., a finite subcollection of {Uα }α∈Λ whose union still contains K). (ii) A set E ⊆ X is relatively compact (or precompact) if its closure E is compact. (iii) A set E ⊆ X is σ-compact if it can be written as a countable union of compact sets. (iv) The topological space X is locally compact if every point x ∈ X has a neighborhood whose closure is compact. Remark A.6. A closed subset of a compact topological space is compact. On the other hand, a compact set of a Hausdorff space is closed. Definition A.7. Let X be a topological space and let F be a collection of subsets of X. Then (i) F is locally finite if every x ∈ X has a neighborhood meeting only finitely many U ∈ F , (ii) F is σ-locally finite if F = ∞ n=1 Fn , where each Fn is a locally finite collection in X. Definition A.8. Let X, Y be two topological spaces, let E ⊆ X and let u : E → Y . We say that u is continuous if u−1 (U ) is relatively open in E for every open set U ⊆ Y . The space of all continuous functions u : X → Y is denoted by C(X; Y ). If Y = R, we write C(X) in place of C(X; R). If X is a topological space, we denote by Cc (X) the space of all continuous functions u : X → R whose support is compact. A topological space is a normal space if for every pair of disjoint closed sets C1 , C2 ⊂ X there exist two disjoint neighborhoods of C1 and C2 . The next theorems give an important characterization of normal spaces. Theorem A.9 (Urysohn). A topological space X is normal if and only if for all disjoint nonempty closed sets C1 , C2 ⊂ X there exists a continuous function u : X → [0, 1] such that u ≡ 1 in C1 and u ≡ 0 in C2 . Theorem A.10 (Tietze’s extension theorem). A topological space X is normal if and only if for every closed set C ⊂ X and every continuous function u : C → R there exists a continuous function v : X → R such that v(x) = u(x) for all x ∈ C. Moreover, if u(C) ⊆ [a, b], then v may be constructed so that v(X) ⊆ [a, b].
638
A. Functional Analysis
A.2. Metric Spaces Definition A.11. A metric on a set X is a map d : X × X → [0, ∞) such that (i) d(x, y) ≤ d(x, z) + d(z, y) for all x, y, z ∈ X, (ii) d(x, y) = d(y, x) for all x, y ∈ X, (iii) d(x, y) = 0 if and only if x = y. A metric space (X, d) is a set X endowed with a metric d. When there is no possibility of confusion, we abbreviate by saying that X is a metric space. If r > 0, the (open) ball of center x0 ∈ X and radius r is the set B(x0 , r) := {x ∈ X : d(x0 , x) < r}. If x ∈ X and E ⊆ X, the distance of x from the set E is defined by dist(x, E) := inf{d(x, y) : y ∈ E}, while the distance between two sets E1 , E2 ⊆ X is defined by dist(E1 , E2 ) := inf{d(x, y) : x ∈ E1 , y ∈ E2 }. A sequence {xn }n in X converges (strongly) to x ∈ X if lim d(xn , x0 ) = 0.
n→∞
A Cauchy sequence in a metric space is a sequence {xn }n in X such that lim d(xn , xm ) = 0.
n,m→∞
A metric space X is said to be complete if every Cauchy sequence is convergent. Theorem A.12 (Baire category theorem). Let (X, d) be a complete metric space. Then the intersection of a countable family of open dense sets in X is still dense in X. Corollary A.13. Let (X, d) be a nonempty complete metric space. If X = n=1 Cn , where Cn is closed for every n ∈ N, then at least one Cn has nonempty interior. Theorem A.14 (Banach’s fixed point). Let (X, d) be a nonempty complete metric space and let u : X → X be a contraction, that is there exists 0 < L < 1 such that d(u(x), u(y)) ≤ Ld(x, y) for all x, y ∈ X. Then u has a unique fixed point; that is, there is a unique x ∈ X such that u(x) = x. A metric space (X, d) can always be turned into a topological space (X, τ ) by taking as a base for the topology τ the family of all open balls. We then say that τ is determined by d. Note that (X, τ ) is a Hausdorff normal space.
A.3. Topological Vector Spaces
639
A topological space X is metrizable if its topology can be determined by a metric. Theorem A.15. A topological space is metrizable if and only if the following properties hold: (i) Singletons are closed. (ii) For every closed set C ⊂ X and for every x ∈ / C there exist disjoint open neighborhoods of C and x. (iii) X has a σ-locally finite base.
A.3. Topological Vector Spaces Definition A.16. A vector space, or linear space, over R is a nonempty set X, whose elements are called vectors, together with two operations, addition and multiplication by scalars, X × X → X, (x, y) → x + y,
and
R × X → X, (t, x) → tx,
with the properties that (i) (X, +) is a commutative group, that is, (a) x + y = y + x for all x, y ∈ X (commutative property), (b) x+(y+z) = (x+y)+z for all x, y, z ∈ X (associative property), (c) there is a vector 0 ∈ X, called zero, such that x+0 = 0+x = x for all x ∈ X, (d) for every x ∈ X there exists a vector in X, called the opposite of x and denoted −x, such that x + (−x) = 0, (ii) for all x, y ∈ X and s, t ∈ R, (a) s(tx) = (st)x, (b) 1x = x, (c) s(x + y) = (sx) + (sy), (d) (s + t)x = (sx) + (tx). Let X be a vector space over R and let E ⊆ X. The set E is said to be balanced if tx ∈ E for all x ∈ E and t ∈ [−1, 1]. We say that E ⊆ X is absorbing if for every x ∈ X there exists t > 0 such that sx ∈ E for all 0 ≤ s ≤ t. Definition A.17. Given a vector space X over R endowed with a topology τ , the pair (X, τ ) is called a topological vector space if the functions X × X → X, (x, y) → x + y, are continuous.
and
R × X → X, (t, x) → tx,
640
A. Functional Analysis
Remark A.18. (i) In a topological vector space a set U is open if and only if x + U is open for all x ∈ X. Hence, to give a base, it is enough to give a local base at the origin. (ii) Using the continuity of addition and scalar multiplication, it is possible to show that each neighborhood U of the origin is absorbing and it contains a neighborhood W of zero such that W + W ⊆ U and W ⊆ U , as well as a balanced neighborhood of zero. As a corollary of Theorem A.15 we have the following: Corollary A.19. A topological vector space X is metrizable if and only if (i) singletons are closed, (ii) X has a countable local base at 0. Definition A.20. Let X be a topological vector space. A set E ⊆ X is said to be topologically bounded if for each neighborhood U of 0 there exists t > 0 such that E ⊆ tU . Note that when the topology τ is generated by a metric d, sets bounded in the topological sense and in the metric sense may be different. To see this, d generates the same topology it suffices to observe that the metric d1 := d+1 as d, but since d1 ≤ 1, every set in X is bounded with respect to d1 . We now define Cauchy sequences in a topological vector space. Definition A.21. Let X be a topological vector space. A sequence {xn }n in X is called a Cauchy sequence if for every neighborhood U of the origin there exists an integer n ∈ N such that xn − xm ∈ U for all n, m ≥ n. The space X is complete if every Cauchy sequence is convergent. Note that Cauchy (and hence convergent) sequences are bounded in the topological sense. Proposition A.22. Let X be a topological vector space and let {xn }n be a Cauchy sequence in X. Then the set {xn : n ∈ N} is topologically bounded. Topologically bounded sets play an important role in the normability of locally convex topological vector spaces (see Theorem A.42 below). Definition A.23. A topological vector space X is locally convex if every point x ∈ X has a neighborhood that is convex. Proposition A.24. A locally convex topological vector space admits a local base at the origin consisting of balanced convex neighborhoods of zero.
A.3. Topological Vector Spaces
641
Let X be a vector space over R and let E ⊆ X. The function pE : X → R defined by pE (x) := inf{t > 0 : x ∈ tE}, x ∈ X, is called the gauge or Minkowski functional of E. Definition A.25. Let X be a vector space over R. A function p : X → R is called a seminorm if p(x + y) ≤ p(x) + p(y) for all x, y ∈ X and p(tx) = |t|p(x) for all x ∈ X and t ∈ R. Remark A.26. Let X be a vector space over R and let E ⊆ X. The gauge pE of E is a seminorm if and only if E is balanced, absorbing, and convex. Theorem A.27. If F is a balanced, convex local base of 0 for a locally convex topological vector space X, then the family {pU : U ∈ F } is a family of continuous seminorms. Conversely, given a family P of seminorms on a vector space X, the collection of all finite intersections of sets of the form V (p, n) := {x ∈ X : p(x) < 1/n},
p ∈ P, n ∈ N,
is a balanced, convex local base of 0 for a topology τ that turns X into a locally convex topological vector space such that each p is continuous with respect to τ . We now give some necessary and sufficient conditions for the topology τ given in the previous theorem to be Hausdorff and for a set to be topologically bounded. Corollary A.28. Let P be a family of seminorms on a vector space X and let τ be the locally convex topology generated by P. Then (i) τ is Hausdorff if and only if p(x) = 0 for all p ∈ P implies that x = 0, (ii) a set E ⊆ X is topologically bounded if and only if the set p(E) is bounded in R for all p ∈ P. A topological vector space X is called seminormable if the topology of the space can be induced by a single seminorm p (see Theorem A.27). The pair (X, p) is called a seminormed space. We now introduce the notion of dual space. Definition A.29. Let X and Y be two vector spaces. A map L : X → Y is called a linear operator if (i) L(x1 + x2 ) = L(x1 ) + L(x2 ) for all x1 , x2 ∈ X, (ii) L(tx) = tL(x) for all x ∈ X and t ∈ R.
642
A. Functional Analysis
If X and Y are topological vector spaces, then the vector space of all continuous linear operators from X to Y is denoted by L(X; Y ). In the special case Y = R, the space L(X; R) is called the (topological) dual space of X and it is denoted by X . The elements of X are also called continuous linear functionals. The bilinear (i.e., linear in each variable) mapping ·, · X ,X : X × X → R
(A.1)
(L, x) → L(x) is called the duality pairing. The dual space L(X ; R) of X is called the bidual space of X and it is denoted by X . Definition A.30. Let X, Y be topological vector spaces. An operator L : X → Y is bounded if it sends topologically bounded sets of X into topologically bounded sets of Y . Theorem A.31. Let X, Y be topological vector spaces and let L : X → Y be a linear operator. Consider the following properties: (i) L is continuous. (ii) L is bounded. (iii) If {xn }n is a sequence in X and xn → 0, then {L(xn )}n is a topologically bounded set. (iv) If {xn }n is a sequence in X and xn → 0, then L(xn ) → 0. Then (i)⇒(ii)⇒(iii). Moreover, if X is metrizable, then (iii)⇒(iv)⇒(i) so that all four properties are equivalent. Theorem A.32 (Hahn–Banach theorem, analytic form). Let X be a vector space, let Y be a subspace of X, and let p : X → R be a convex function. Then for every linear functional L : Y → R such that L(x) ≤ p(x) for all x ∈ Y , there exists a linear functional L1 : X → R such that L1 (x) = L(x) for all x ∈ Y and L(x) ≤ p(x) for all x ∈ X. Theorem A.33 (Hahn–Banach theorem, first geometric form). Let X be a topological vector space and let E, F ⊂ X be nonempty disjoint convex sets. Assume that E has an interior point. Then there exist a continuous linear functional L : X → R, L = 0, and a number a ∈ R such that L(x) ≥ a for all x ∈ E and L(x) ≤ a for all x ∈ F . Corollary A.34. Let X be a topological vector space. Then the dual X of X is not the null space if and only if X has a convex neighborhood of the origin strictly contained in X.
A.4. Normed Spaces
643
In view of the previous corollary it is natural to restrict our attention to locally convex topological vector spaces. Theorem A.35 (Hahn–Banach theorem, second geometric form). Let X be a locally convex topological vector space and let C, K ⊂ X be nonempty disjoint convex sets, with C closed and K compact. Then there exist a continuous linear functional L : X → R and two numbers a ∈ R and ε > 0 such that L(x) ≤ a − ε for all x ∈ C and L(x) ≥ a + ε for all x ∈ K. Definition A.36. We say that a topological vector space (X, τX ) is embedded in a topological vector space (Y, τY ) and we write X → Y if X is a subspace of Y and the immersion i : X → Y , given by i(x) := x, is continuous. Remark A.37 (Important). The definition of embedding changes with books. In some books an embedding only means that X is a subspace of Y and our definition is called a continuous embedding.
A.4. Normed Spaces Definition A.38. A norm on a vector space X is a map · : X → [0, ∞) such that (i) x + y ≤ x + y for all x, y ∈ X, (ii) tx = |t| x for all x ∈ X and t ∈ R, (iii) x = 0 implies x = 0. A normed space (X, · ) is a vector space X endowed with a norm · . For simplicity, we often say that X is a normed space. If for every x, y ∈ X we define d(x, y) := x − y, then (X, d) is a metric space. We say that a normed space X is a Banach space if it is complete as a metric space. Example A.39 (Euclidean space). The most important example of Banach space in this book is the Euclidean space RN . Given x = (x1 , . . . , xN ) ∈ RN the Euclidean norm of x is the number $ x := x21 + · · · + x2N . When working with Euclidean spaces of different dimension, we will sometimes use the notation xN for x. Remark A.40. A seminormable space (X, p) induces a normed space on the quotient space X/Y , where Y := {x ∈ X : p(x) = 0}. The induced norm on X/Y is given by [x] := p(x), where [x] := x + Y .
644
A. Functional Analysis
Example A.41. Let X be a topological space and consider the space Cc (X) of all continuous functions u : X → R whose support is compact. For u ∈ Cc (X) define u∞ := maxx∈X |u(x)|. Then · ∞ is a norm. In general Cc (X) is not complete. We denote by C0 (X) the completion of Cc (X) relative to the supremum norm · ∞ . If X is compact, then C0 (X) = Cc (X) = C(X). If X is a locally compact Hausdorff space, then it can be shown that u ∈ C0 (X) if and only if u ∈ C(X) and for every ε > 0 there exists a compact set K ⊆ X such that |u(x)| < ε for all x ∈ X \ K. We also define Cc (X; Rm ) and C0 (X; Rm ) as the spaces of all functions u : X → Rm whose components belong to Cc (X) and C0 (X), respectively. A topological vector space is normable if its topology can be determined by a norm. Theorem A.42. A topological vector space X is normable if and only if it is Hausdorff, locally convex and it has a topologically bounded neighborhood of 0. Two norms · 1 and · 2 on the same vector space X are equivalent if there exists a positive constant c > 0 such that c−1 x1 ≤ x2 ≤ cx1
for all x ∈ X.
Equivalent norms induce the same topology on X. Proposition A.43. Let (X, · X ) and (Y, · Y ) be normed spaces. (i) A linear operator L : X → Y is continuous if and only if LL(X;Y ) :=
L(x)Y < ∞. x∈X\{0} xX sup
(ii) The mapping L ∈ L(X; Y ) → LL(X;Y ) is a norm. (iii) If Y is a Banach space, then so is L(X; Y ). Conversely, if X = {0} and L(X; Y ) is a Banach space, then so is Y . As a corollary of the Hahn–Banach theorem (Theorem A.32) one has the following result. Corollary A.44. Let X be a normed space. Then for all x ∈ X, xX =
|L(x)| . X ≤1 xX
max
L∈X , L
Definition A.45. Let (X, · X ) and (Y, · Y ) be normed spaces. A continuous linear operator L ∈ L(X; Y ) is said to be compact if it maps every bounded subset of X onto a relatively compact subset of Y .
A.5. Weak Topologies
645
In particular, if L is compact, then from every bounded sequence {xn }n in X we may extract a subsequence {xnk }k such that {L(xnk )}k converges in Y . Let (X, · X ) and (Y, · Y ) be normed spaces with X embedded in Y (see Definition A.36). Since the immersion is linear, in view of Proposition A.43 the continuity of i is equivalent to requiring the existence of a constant M > 0 such that (A.2)
xY ≤ M xX
for all x ∈ X.
We say that X is compactly embedded in Y if the immersion i is a compact operator. Given a vector space X, a quasi-norm is a map · : X → [0, ∞) such that it satisfies conditions (ii) and (iii) of Definition A.38 and such that x + y ≤ c(x + y) for all x, y ∈ X and for some constant c > 0. A quasi-normed space (X, · ) is a vector space X endowed with a quasi-norm · . For simplicity, we often say that X is a quasi-normed space. Setting d(x, y) := x − y, x, y ∈ X, we cannot conclude that d is a metric, since the triangle inequality is now replaced by d(x, y) ≤ c(d(x, z) + d(z, y)). However, we can carry on most of the analysis done for metric space. In particular we can define limits and Cauchy sequences and define completeness. A quasi-normed space X is a quasi-Banach space if every Cauchy sequence has a limit in X.
A.5. Weak Topologies Given a locally convex topological vector space X, for each L ∈ X the function pL : X → [0, ∞) defined by (A.3)
pL (x) := |L(x)|,
x ∈ X,
is a seminorm. In view of Theorem A.27, the family of seminorms {pL }L∈X generates a locally convex topology σ(X, X ) on the space X, called the weak topology, such that each pL is continuous with respect to σ(X, X ). In turn, this implies that every L ∈ X is σ(X, X ) continuous. Theorem A.46. Let X be a locally convex topological vector space and let E ⊆ X. Then (i) E is bounded with respect to the (strong) topology if and only if it is weakly bounded, (ii) if E is convex, then E is closed if and only if it is weakly closed. Definition A.47. Given a locally convex topological vector space X, a sequence {xn }n in X converges weakly to x ∈ X if it converges to x with respect to the weak topology σ(X, X ).
646
A. Functional Analysis
We write xn x. In view of Theorem A.27 and (A.3), we have the following result. Proposition A.48. Let X be a locally convex topological vector space. A sequence {xn }n in X converges weakly to x ∈ X if and only if limn→∞ L(xn ) = L(x) for every L ∈ X . Similarly, given a locally convex topological vector space X, for each x ∈ X the function px : X → [0, ∞) defined by px (L) := |L(x)|,
(A.4)
L ∈ X ,
is a seminorm. In view of Theorem A.27, the family of seminorms {px }x∈X generates a locally convex topology σ(X , X) on the space X , called the weak star topology, such that each px is continuous with respect to σ(X , X). Definition A.49. Let X be a locally convex topological vector space. A sequence {Ln }n in X is weakly star convergent to L in X if it converges to ∗ L with respect to the weak star topology σ(X , X). We write Ln L. In view of Theorem A.27 and (A.4), we have the following result. Proposition A.50. Let X be a locally convex topological vector space. A sequence {Ln }n in X converges weakly star to L ∈ X if and only if limn→∞ Ln (x) = L(x) for every x ∈ X. Theorem A.51 (Banach–Alaoglu). If U is a neighborhood of 0 in a locally convex topological vector space X, then K := {L ∈ X : |L(x)| ≤ 1 for every x ∈ U } is weak star compact. Corollary A.52. If X is a normed space, then the closed unit ball {L ∈ X : LX ≤ 1} of X is weak star compact. If X is separable, it actually turns out that weak star compact sets are metrizable, and thus one can work with the friendlier notion of sequential compactness. Theorem A.53. Let X be a separable locally convex topological vector space and let K ⊆ X be weak star compact. Then (K, σ(X , X)) is metrizable. Hence, also in view of the Banach–Alaoglu theorem, we have the following: Corollary A.54. Let U be a neighborhood of 0 in a separable locally convex topological vector space X and let {Ln }n be a sequence in X be such that sup sup |Ln (x)| < ∞. n x∈U
A.5. Weak Topologies
647
Then there exists a subsequence {Lnk }k that is weak star convergent. In particular, if X is a separable Banach space and {Ln }n is any bounded sequence in X , then there exists a subsequence that is weak star convergent. For Banach spaces the converse of Theorem A.53 holds: Theorem A.55. Let X be a Banach space. Then the unit ball B(0; 1) in X endowed with the weak star topology is metrizable if and only if X is separable. Proposition A.56. Let X be a Banach space. If a sequence {Ln }n in X converges weakly star to L ∈ X , then it is bounded and LX ≤ lim inf Ln X . n→∞
Proposition A.57. Let X be a Banach space. If X is separable, then so is X. The converse is false in general (take, for example, the separable space L1 (RN ) and its dual L∞ (RN )). We now study analogous results for the weak topology. An infinitedimensional Banach space when endowed with the weak topology is never metrizable. However, we have the following: Theorem A.58. Let X be a Banach space whose dual X is separable. Then the unit ball B(0; 1) endowed with the weak topology is metrizable. Definition A.59. Let X be a locally convex topological vector space. A set K ⊆ X is called sequentially weakly compact if every sequence {xn }n in K has a subsequence converging weakly to a point in K. Theorem A.60. Let X be a Banach space. If K ⊆ X is weakly compact, then it is weakly sequentially compact. Using Banach–Alaoglu’s theorem, one can prove the following theorem: ˇ Theorem A.61 (Eberlein–Smulian). Let E be a subset of a Banach space X. Then the weak closure of E is weakly compact if and only if for every sequence {xn }n in E there exists a subsequence weakly convergent to some element of X. As an immediate application of the Hahn–Banach theorem we have the following: Proposition A.62. Let X be a normed space and consider the linear operator mapping J : X → X defined by J(x)(L) := L(x), L ∈ X . Then J(x)X = xX for all x ∈ X. In particular, J is injective and continuous. Definition A.63. A normed space X is reflexive if J(X) = X .
648
A. Functional Analysis
In this case it is possible to identify X with its bidual X . Theorem A.64 (Kakutani). A Banach space is reflexive if and only if the closed unit ball {x ∈ X : x ≤ 1} is weakly compact. ˇ In view of the previous theorem and the Eberlein–Smulian theorem we have the following corollary: Corollary A.65. Let X be a reflexive Banach space and let {xn }n be a bounded sequence in X. Then there exists a subsequence that is weakly convergent. Proposition A.66. A normed space X is reflexive if and only if X is reflexive. The following proposition is used throughout the text, sometimes without mention. Proposition A.67. Let X be a Banach space. If a sequence {xn }n in X converges weakly to x ∈ X, then it is bounded and x ≤ lim inf xn . n→∞
A.6. Hilbert Spaces Definition A.68. An inner product on a vector space X is a map (·, ·) : X × X → R such that (i) (x, y) = (y, x) for all x, y ∈ X, (ii) (sx + ty, z) = s(x, z) + t(y, z) for all x, y, z ∈ X and s, t ∈ R, (iii) (x, x) ≥ 0 for every x ∈ X and (x, x) = 0 if and only if x = 0. If for every x ∈ X we define (A.5)
x :=
# (x, x),
then X becomes a normed space and the following inequality, called Cauchy’s inequality, holds |(x · y)| ≤ x y for all x, y ∈ X. We say that X is a Hilbert space if it is a Banach space when endowed with the norm (A.5). Theorem A.69. Let (X, · ) be a normed space. # Then there exists an inner product (·, ·) : X × X → R such that x = (x, x) for all x ∈ X if and only if · satisfies the parallelogram law x + y2 + x − y2 = 2x2 + 2y2 for all x, y ∈ X.
A.6. Hilbert Spaces
649
Remark A.70. If · satisfies the parallelogram law, then the inner product in the previous theorem is defined as (x, y) := 14 [x + y2 − x − y2 ] for all x, y ∈ X. Example A.71. The most important example of Hilbert space in this book is the Euclidean space RN . Given x = (x1 , . . . , xN ) ∈ RN and y = (y1 , . . . , yN ) ∈ RN , the Euclidean inner product of x and y is the real number defined by N xi yi . x · y := i=1
Appendix B
Measures ANOVA: Analysis of Value. Is your research worth anything? Significance is determined by comparing one’s research with the Dull Hypothesis: H0 : μ1 = μ2 ? where H0 : the Dull Hypothesis, μ1 : significance of your research, μ2 : significance of a monkey typing randomly on a typewriter in a forest where no one can hear it. — Jorge Cham, www.phdcomics.com
In this chapter we present, without proofs, all the results in abstract measure theory and Lebesgue integration that we used in the text. This chapter draws upon [83], to which we refer for the proofs of all the results that cannot be found in classical texts (such as [61], [72], [82], [199]). Here, we prove only a few results that are used in this text and that cannot be found in [83] or in classical textbooks.
B.1. Outer Measures and Measures Definition B.1. Let X be a nonempty set. A map μ∗ : P(X) → [0, ∞] is called an outer measure if (i) μ∗ (∅) = 0, (ii) μ∗ (E) ≤ μ∗ (F ) for all E ⊆ F ⊆ X, ∞ ∗ (iii) μ∗ ( ∞ n=1 En ) ≤ n=1 μ (En ) for every countable collection {En }n in P(X). To construct outer measures, we usually start with a family G of “elementary sets” (e.g., cubes in RN ) for which we have an “elementary notion of measure” ρ : G → [0, ∞] (i.e., in RN the volume of a cube Q(x, r) is rN ). Proposition B.2. Let X be a nonempty set and ∞let G ⊆ P(X) be such that ∅ ∈ G and there exists {Xn }n in G with X = n=1 Xn . Let ρ : G → [0, ∞] 651
652
B. Measures
be such that ρ(∅) = 0. Then the map μ∗ : P(X) → [0, ∞], defined by ∞ ∞ μ∗ (E) := inf ρ(En ) : {En }n in G, E ⊆ En , E ⊆ X, n=1
n=1
is an outer measure. Remark B.3. Note that if E ∈ G, then taking E1 := E, En := ∅ for all n ≥ 2, it follows from the definition of μ∗ that μ∗ (E) ≤ ρ(E), with the strict inequality possible. However, ∞ if ρ is countably subadditive, that is, if ρ(E) ≤ ∞ ρ(E ) for all E ⊆ n n=1 n=1 En with E ∈ G and {En }n in G, then μ∗ = ρ on G. The main problem with outer measures is that they are not additive on disjoint sets. To circumvent this problem, we restrict an outer measure μ∗ : P(X) → [0, ∞] to a smaller class of subsets for which additivity of disjoint sets holds. Definition B.4. Let X be a nonempty set and let μ∗ : P(X) → [0, ∞] be an outer measure. A set E ⊆ X is said to be μ∗ -measurable if μ∗ (F ) = μ∗ (F ∩ E) + μ∗ (F \ E) for all sets F ⊆ X. We will see below in Theorem B.13 that the restriction of μ∗ to the class M∗ := {E ⊆ X : E is μ∗ -measurable} is additive, actually countably additive, and that the class M∗ has some important properties; to be precise, it is a σ-algebra. Definition B.5. Let X be a nonempty set. A collection M ⊆ P(X) is an algebra if the following properties hold: (i) ∅ ∈ M. (ii) If E ∈ M, then X \ E ∈ M. (iii) If E1 , E2 ∈ M, then E1 ∪ E2 ∈ M. M is said to be a σ-algebra if it satisfies (i)–(ii) and (iii) if En ∈ M for every n ∈ N, then ∞ n=1 En ∈ M. If M is a σ-algebra, then the pair (X, M) is called a measurable space. Let X be a nonempty set. Given a subset F ⊆ P(X), the smallest (in the sense of inclusion) σ-algebra that contains F is given by the intersection of all σ-algebras on X that contain F . If X is a topological space, then the Borel σ-algebra B(X) is the smallest σ-algebra containing all open subsets of X.
B.1. Outer Measures and Measures
653
Definition B.6. Let X be a nonempty set. A collection M ⊆ P(X) is a semiring if the following properties hold: (i) ∅ ∈ M. (ii) If E, F ∈ M, with E ⊆ F , then there exist E0 , . . . , En in M, with E0 = E and En = F , such that Ei−i ⊆ Ei and Ei \ Ei−1 ∈ M for every i = 1, . . . , n. (iii) If E, F ∈ M, then E ∩ F ∈ M. M is said to be a ring if for every E, F ∈ M, we have that E ∪ F ∈ M and E \ F ∈ M. The family of intervals (a, b] is a semiring. Definition B.7. Let X be a nonempty set and let M ⊆ P(X) be an algebra. A map μ : M → [0, ∞] is called a (positive) finitely additive measure if μ(∅) = 0,
μ(E1 ∪ E2 ) = μ(E1 ) + μ(E2 )
for all E1 , E2 ∈ M with E1 ∩ E2 = ∅. Definition B.8. Let X be a nonempty set, let M ⊆ P(X) be a σ-algebra. A map μ : M → [0, ∞] is called a (positive) measure if ∞ ∞ En = μ(En ) μ(∅) = 0, μ n=1
n=1
for every countable collection {En }n of pairwise disjoint sets in M. The triple (X, M, μ) is said to be a measure space. Given a measure space (X, M, μ), where X is a topological space, the measure μ is called a Borel measure if M contains B(X). One of the most important properties of measures is given in the next proposition. Proposition B.9. Let (X, M, μ) be a measure space. (i) If {En }n is an increasing sequence of subsets of M, then ∞ En = lim μ(En ). μ n=1
n→∞
(ii) If {En }n is a decreasing sequence of subsets of M and μ(E1 ) < ∞, then ∞ En = lim μ(En ). μ n=1
n→∞
Definition B.10. Let (X, M, μ) be a measure space.
654
B. Measures
(i) The measure μ is said to be complete if for every E ∈ M with μ(E) = 0 it follows that every F ⊆ E belongs to M. (ii) A set E ∈ M has σ-finite μ-measure if it can be written as a countable union of measurable sets of finite measure; μ is said to be σ-finite if X has σ-finite μ-measure; μ is said to be finite if μ(X) < ∞. (iii) A set E ∈ M of positive measure is said to be an atom if for every F ∈ M, with F ⊆ E, either μ(F ) = 0 or μ(F ) = μ(E). The measure μ is said to be nonatomic if there are no atoms, that is, if for every set E ∈ M of positive measure there exists F ∈ M, with F ⊆ E, such that 0 < μ(F ) < μ(E). The measure μ is purely atomic, if every set of positive measure contains an atom. Analogous definitions can be given for finitely additive measures. Remark B.11. The Hausdorff measure Hs (see Appendix C) is not σ-finite. The following is a useful property of finite measures. Proposition B.12. Let (X, M, μ) be a measure space with μ finite and let {Eλ }λ∈Λ be an arbitrary family of pairwise disjoint subsets in M. Then μ(Eλ ) = 0 for all but at most countably many λ ∈ Λ. Theorem B.13 (Carath´eodory). Let X be a nonempty set and let μ∗ : P(X) → [0, ∞] be an outer measure. Then (B.1)
M∗ := {E ⊆ X : E is μ∗ -measurable}
is a σ-algebra and μ∗ : M∗ → [0, ∞] is a complete measure. From Proposition B.2 and Remark B.3 we have the following. Corollary B.14. Let X be a nonempty set, let M ⊆ P(X) be an algebra, and let μ : M → [0, ∞] be a finitely additive measure. Let μ∗ be the outer measure defined in Proposition B.2 (with G := M and ρ := μ). Then every element of M is μ∗ -measurable. Moreover, if μ is countably additive, then μ∗ = μ on M. Remark B.15. Note that the previous result implies that every countably additive measure μ : M → [0, ∞] defined on an algebra M may be extended as a measure to a σ-algebra that contains M. It actually turns out that when μ is σ-finite, this extension is unique. Corollary B.16. Let I ⊆ R be an open interval and let μ, ν : B(I) → [0, ∞] be two measures such that μ((a, b]) = ν((a, b]) < ∞ for all intervals (a, b] with a, b ∈ I. Then μ = ν.
B.2. Measurable and Integrable Functions
655
Remark B.17. In the previous corollary, we could have used any other family (e.g., open, closed) of intervals that generates B(I). Using Carath´eodory’s theorem, we have created a large class of complete measures. The next problem is to understand the class M∗ of the μ∗ -measurable sets. To do this, we consider a special class of outer measures. Definition B.18. Let X be a metric space and let μ∗ : P(X) → [0, ∞] be an outer measure. Then μ∗ is said to be a metric outer measure if μ∗ (E ∪ F ) = μ∗ (E) + μ∗ (F ) for all sets E, F ⊆ X, with dist(E, F ) > 0. Proposition B.19. Let X be a metric space and let μ∗ : P(X) → [0, ∞] be a metric outer measure. Then every Borel set is μ∗ -measurable.
B.2. Measurable and Integrable Functions In this section we introduce the notions of measurable and integrable functions. Definition B.20. Let X and Y be nonempty sets and let M and N be algebras on X and Y , respectively. A function u : X → Y is said to be measurable if u−1 (F ) ∈ M for every set F ∈ N. If X and Y are topological spaces, M := B(X) and N := B(Y ), then a measurable function u : X → Y will be called a Borel function, or Borel measurable. If in a topological space (usually, RN , R, or [−∞, ∞]) the σalgebra is not specified, it is understood that we take the Borel σ-algebra B(X). Proposition B.21. If M is a σ-algebra on a set X and N is the smallest σalgebra that contains a given family G of subsets of a set Y , then u : X → Y is measurable if and only if u−1 (F ) ∈ M for every set F ∈ G. Remark B.22. In particular, if in the previous proposition Y is a topological space and N = B(Y ), then it suffices to verify that u−1 (V ) ∈ M for every open set V ⊆ Y . Moreover, if Y = R (respectively, Y = [−∞, ∞]), then it suffices to check that u−1 ((a, ∞)) ∈ M (respectively, u−1 ((a, ∞]) ∈ M) for every a ∈ R. Remark B.23. If X and Y are topological spaces, then, in view of the previous remark, every continuous function u : X → Y is a Borel function. Proposition B.24. Let (X, M), (Y, N), (Z, O) be measurable spaces and let u : X → Y and v : Y → Z be two measurable functions. Then v◦u = X → Z is measurable.
656
B. Measures
Corollary B.25. Let (X, M) be a measurable space and let u : X → R (respectively, u : X → [−∞, ∞]) be a measurable function. Then u2 , |u|, u+ , u− , cu, where c ∈ R, are measurable. Remark B.26. If c = 0 and u : X → [−∞, ∞], the function cu is defined to be identically equal to zero. Given two measurable spaces (X, M) and (Y, N), we denote by M ⊗ N ⊆ P(X × Y ) the smallest σ-algebra that contains all sets of the form E × F , where E ∈ M, F ∈ N. Then M ⊗ N is called the product σ-algebra of M and N. Proposition B.27. Let (X, M), (Y1 , N1 ),. . ., (Yn , Nn ) be measurable spaces and consider (Y1 × · · · × Yn , N1 ⊗ · · · ⊗ Nn ). Then the vector-valued function u : X → Y1 × · · · × Yn is measurable if and only if its components ui : X → Yi are measurable functions for all i = 1, . . . , n. Corollary B.28. Let (X, M) be a measurable space and let u : X → R and v : X → R be two measurable functions. Then u + v, uv, min{u, v}, max{u, v} are measurable. Remark B.29. The previous corollary continues to hold if R is replaced by [−∞, ∞], provided u + v are well-defined, i.e., (u(x), v(x)) = (±∞, ∓∞) for all x ∈ X. Concerning uv, we define (uv)(x) := 0 whenever u(x) or v(x) is zero. Proposition B.30. Let (X, M) be a measurable space and let un : X → [−∞, ∞], n ∈ N, be measurable functions. Then the functions sup un , inf un , lim inf un , lim sup un n
n
n→∞
n→∞
are measurable. Remark B.31. The previous proposition uses in a crucial way the fact that M is a σ-algebra. Let (X, M, μ) be a measure space. We will see later on that the Lebesgue integration does not “see” sets of μ-measure zero. Also, several properties (existence of pointwise limit of a sequence of functions, convergence of a series of functions) do not hold at every point x ∈ X. Thus, it is important to work with functions u that are defined only on X \ E with μ(E) = 0. For this reason, we extend Definition B.20 to read as follows. Definition B.32. Let (X, M) and (Y, N) be two measurable spaces and let μ : M → [0, ∞] be a measure. Given a function u : X \ E → Y where
B.2. Measurable and Integrable Functions
657
μ(E) = 0, u is said to be measurable over X if u−1 (F ) ∈ M for every set F ∈ N. In general, measurability of u on X \ E does not entail the measurability of an arbitrary extension of u to X unless μ is complete. However, if we define u(x) if x ∈ X \ E, w(x) := y1 if x ∈ E, where y1 ∈ Y , then w is measurable. Hence, in general, there are extensions of u that are measurable and others that are not. The next result shows that if the measure μ is complete, then this cannot happen. In what follows, if μ is a measure, we write that a property holds μ-a.e. on a measurable set E if there exists a measurable set F ⊆ E such that μ(F ) = 0 and the property holds everywhere on the set E \ F . Proposition B.33. Let (X, M) and (Y, N) be two measurable spaces and let u : X → Y be a measurable function. Let μ : M → [0, ∞] be a complete measure. If v : X → Y is a function such that u(x) = v(x) for μ-a.e. x ∈ X, then v is measurable. Going back to the setting in which u : X \ E → Y with μ(E) = 0, since Lebesgue integration does not take into account sets of measure zero, we will see that integration of u depends mostly on its measurability on X \ E. Corollary B.34. Let (X, M) be a measurable space and let un : X → [−∞, ∞], n ∈ N, be measurable functions. Let μ : M → [0, ∞] be a complete measure. If there exists limn→∞ un (x) for μ-a.e. x ∈ X, then limn→∞ un is measurable. We are now in a position to introduce the notion of integral. Definition B.35. Let X be a nonempty set and let M be an algebra on X. A simple function is a measurable function s : X → R whose range consists of finitely many points. If c1 , . . . , c are the distinct values of s, then we write s=
cn χEn ,
n=1
where En := {x ∈ X : s(x) = cn } and for every set E ⊆ X, χE is the characteristic function of the set E, i.e., 1 if x ∈ E, χE (x) := 0 otherwise.
658
B. Measures
If μ is a finitely additive (positive) measure on X and s ≥ 0, then for every measurable set E ∈ M we define the Lebesgue integral of s over E as (B.2) s dμ := cn μ(En ∩ E), E
n=1
where if cn = 0 and μ(En ∩ E) = ∞, then we use the convention cn μ(En ∩ E) := 0. Theorem B.36. Let X be a nonempty set, let M be an algebra on X, and let u : X → [0, ∞] be a measurable function. Then there exists a sequence {sn }n of simple functions such that 0 ≤ s1 (x) ≤ s2 (x) ≤ · · · ≤ sn (x) → u(x) for every x ∈ X. The convergence is uniform on every set on which u is bounded from above. Corollary B.37. Let (X, M, μ) be a measure space and let u : X → [0, ∞] be a measurable function. If the set {x ∈ X : u(x) > 0} has σ-finite measure and u is finite μ-a.e., then there exists a sequence {sn }n of simple functions, each of them bounded and vanishing outside a set of finite measure (depending on n), such that 0 ≤ s1 (x) ≤ s2 (x) ≤ · · · ≤ sn (x) → u(x) for every x ∈ X. In the remainder of this section, M is a σ-algebra and μ a (countably additive) measure. In view of the previous theorem, if u : X → [0, ∞] is a measurable function, then we define its (Lebesgue) integral over a measurable set E as u dμ := sup s dμ : s simple, 0 ≤ s ≤ u . E
E
We list below some basic properties of Lebesgue integration for nonnegative functions. Proposition B.38. Let (X, M, μ) be a measure space and let u, v : X → [0, ∞] be two measurable functions. (i) If 0 ≤ u ≤ v, then E u dμ ≤ E v dμ for every measurable set E. (ii) If c ∈ [0, ∞], then E cu dμ = c E u dμ (here we set 0∞ := 0). (iii) If E ∈ M, then E u dμ = 0 if and only if u(x) = 0 for μ-a.e. x ∈ E (even if μ(E) = ∞). (iv) If E ∈ M and μ(E) = 0, then E u dμ = 0 (even if u ≡ ∞ in E). (v) E u dμ = X χE u dμ for every measurable set E.
B.2. Measurable and Integrable Functions
659
The next results are central to the theory of integration of nonnegative functions. Theorem B.39 (Lebesgue’s monotone convergence). Let (X, M, μ) be a measure space and let un : X → [0, ∞] be a sequence of measurable functions such that 0 ≤ u1 (x) ≤ u2 (x) ≤ · · · ≤ un (x) → u(x) for every x ∈ X. Then u is measurable and un dμ = u dμ. lim n→∞ X
X
Remark B.40. The previous theorem continues to hold if we assume that un (x) → u(x) for μ-a.e. x ∈ X. Indeed, in view of Proposition B.38(iv), it suffices to redefine un and u to be zero in the set of measure zero in which there is no pointwise convergence. Corollary B.41. Let (X, M, μ) be a measure space and let u, v : X → [0, ∞] be two measurable functions. Then (u + v) dμ = u dμ + v dμ. X
X
X
Corollary B.42. Let (X, M, μ) be a measure space and let un : X → [0, ∞] be a sequence of measurable functions. Then ∞ ∞ un dμ = un dμ. n=1 X
X n=1
Example B.43. Given a doubly indexed sequence {ak,n }k,n∈N , with ak,n ≥ 0 for all n, k ∈ N, we have ∞ ∞ n=1 k=1
ak,n =
∞ ∞
ak,n .
k=1 n=1
To see this, it suffices to consider X = N with the counting measure 1 and to define un : N → [0, ∞] by un (k) := ak,n . Then ∞ un dμ = ak,n , X
k=1
and the result now follows from the previous corollary. 1 Given
a set X, the counting measure μ : P(X) → [0, ∞] is defined by the number of elements of E if E is a finite set, μ(E) := ∞ otherwise
for every E ⊆ X.
660
B. Measures
Lemma B.44 (Fatou). Let (X, M, μ) be a measure space. If un : X → [0, ∞] is a sequence of measurable functions, then lim inf un dμ ≤ lim inf un dμ. X n→∞
n→∞
X
In order to extend the notion of integral to functions of arbitrary sign, consider u : X → [−∞, ∞]. Note that u = u+ − u− and |u| = u+ + u− and that u is measurable if and only if u+ and u− are measurable. Also, if u is bounded, then so are u+ and u− , and in view of Theorem B.36, u is then the uniform limit of a sequence of simple functions. Definition B.45. Let (X, M, μ) be a measure space and let u : X → [−∞, ∞] be a measurable function. Givena measurable set E ∈ M, if at least one of the two integrals E u+ dμ and E u− dμ is finite, then we define the (Lebesgue) integral of u over the measurable set E by + u dμ := u dμ − u− dμ.
E
u+ dμ
E
E
u− dμ
and E are finite, then u is said to be (Lebesgue) If both E integrable over the measurable set E. A measurable function u : X → [−∞, ∞] is Lebesgue integrable over the measurable set E if and only if E |u| dμ < ∞. Remark B.46. If (X, M, μ) is a measure space and u : X → [−∞, ∞] is Lebesgue integrable, then the set {x ∈ X : |u(x)| = ∞} has measure zero, while the set {x ∈ X : |u(x)| > 0} is σ-finite. If (X, M, μ) is a measure space, with X a topological space, and if M contains B(X), then u : X → [−∞, ∞] is said to be locally integrable if it is Lebesgue integrable over every compact set. Proposition B.47. Let (X, M, μ) be a measure space and let u, v : X → [−∞, ∞] be two integrable functions. (i) If α, β ∈ R, then αu + βv is integrable and (αu + βv) dμ = α u dμ + β v dμ. X
X
(ii) If u(x) = v(x) for μ-a.e. x ∈ X, then u dμ = v dμ. (B.3) X
(iii)
X
|u| dμ. u dμ ≤ X
X
X
B.2. Measurable and Integrable Functions
661
Part (ii) of the previous proposition motivates the next definition. Let (X, M, μ) be a measure space. If F ∈ M is such that μ(F ) = 0 and u : X \ F → [−∞, ∞] is a measurable function in the sense of Definition B.32, then we define the (Lebesgue) integral of u over the measurable set E as the Lebesgue integral of the function u(x) if x ∈ X \ F, v(x) := 0 otherwise, provided E v dμ is well-defined. Note that in this case E v dμ = E v˜ dμ, where u(x) if x ∈ X \ F, v˜(x) := w(x) otherwise, and w is an arbitrary measurable function defined on F . If the measure μ is complete, then E v dμ is well-defined if and only if E\F u dμ is well-defined. For functions of arbitrary sign we have the following convergence result. Theorem B.48 (Lebesgue’s dominated convergence). Let (X, M, μ) be a measure space and let un : X → [−∞, ∞] be a sequence of measurable functions such that limn→∞ un (x) = u(x) for μ-a.e. x ∈ X. If there exists a Lebesgue integrable function v such that |un (x)| ≤ v(x) for μ-a.e. x ∈ X and all n ∈ N, then u is Lebesgue integrable and |un − u| dμ = 0. lim n→∞ X
In particular,
lim
n→∞ X
un dμ =
u dμ. X
Corollary B.49. Let (X, M, μ) be a measure space and let un : X → ∞ [−∞, ∞] be a sequence of measurable functions. If n=1 X |un | dμ < ∞, ∞ u (x) converges for μ-a.e. x ∈ X, the function u(x) := then the series n n=1 ∞ u (x), defined for μ-a.e. x ∈ X, is integrable, and n=1 n ∞
un dμ =
n=1 X
∞
un dμ.
X n=1
Theorem B.50 (Jensen’s inequality). Let f : R → R be convex and let (X, M, μ) be a measure space, where μ(X) = 1. Then for every integrable function u : X → R, u dμ ≤ f ◦ u dμ. (B.4) f X
X
662
B. Measures
B.3. Integrals Depending on a Parameter In this section we study the continuity and the differentiability of functions of the type f (x, y) dμ(x), y ∈ Y. F (y) = X
We begin by studying continuity. Theorem B.51. Let (X, M, μ) be a measure space, let Y be a metric space, and let f : X × Y → R be a function. Assume that for each fixed y ∈ Y the function x ∈ X → f (x, y) is measurable and that there exists y0 ∈ Y such that lim f (x, y) = f (x, y0 ) y→y0
for every x ∈ X. Assume also that there exists an integrable function g : X → [0, ∞) such that |f (x, y)| ≤ g(x) for μ-a.e. x ∈ X and for all y ∈ Y . Then the function F : Y → R, defined by f (x, y) dμ(x), y ∈ Y, F (y) = X
is well-defined and is continuous at y0 . Next we study the differentiability of F . Theorem B.52. Let Y be an interval of R and assume that for each fixed x ∈ X the function y ∈ Y → f (x, y) is differentiable and that for each fixed y ∈ Y the functions x ∈ X → f (x, y) and x ∈ X → ∂f ∂y (x, y) are measurable. Assume also that for some y0 ∈ Y the function x ∈ X → f (x, y0 ) is integrable and that there exists an integrable function h : X → [0, ∞) such that | ∂f ∂y (x, y)| ≤ h(x) for μ-a.e. x ∈ X and for all y ∈ Y . Then the function F : Y → R, defined by f (x, y) dμ(x), y ∈ Y, F (y) = X
is well-defined and differentiable, with ∂f F (y) = (x, y) dμ(x). X ∂y The next exercise shows that, without the integrability of g, Theorem B.51 fails. Exercise B.53. Consider the function
|y|−|x| y2
f (x, y) = Prove that the function F (y) = not continuous at y = 0.
0
if |y| < |x|, if |y| ≥ |x|.
R f (x, y) dx,
y ∈ R, is well-defined and is
B.4. Product Spaces
663
B.4. Product Spaces We recall that, given two measurable spaces (X, M) and (Y, N), we denote by M ⊗ N ⊆ P(X × Y ) the smallest σ-algebra that contains all sets of the form E × F , where E ∈ M, F ∈ N. Then M ⊗ N is called the product σ-algebra of M and N. Exercise B.54. Let X and Y be topological spaces and let B(X) and B(Y ) be their respective Borel σ-algebras. Prove that B(X) ⊗ B(Y ) ⊆ B(X × Y ). Show also that if X and Y are separable metric spaces, then B(X) ⊗ B(Y ) = B(X × Y ). In particular, B(RN ) = B(R) ⊗ · · · ⊗ B(R). Let (X, M, μ) and (Y, N, ν) be two measure spaces. For every E ∈ X ×Y define ∞ ∗ (B.5) μ(Fn )ν(Gn ) : Fn ∈ M, Gn ∈ N, (μ × ν) (E) := inf n=1
E⊆
∞
(Fn × Gn ) .
n=1
ν)∗
: P(X) → [0, ∞] is an outer measure, called By Proposition B.2, (μ × the product outer measure of μ and ν. By Carath´eodory’s theorem, the restriction of (μ × ν)∗ to the σ-algebra M × N of (μ × ν)∗ -measurable sets is a complete measure, denoted by μ × ν and called the product measure of μ and ν. Note that M×N is, in general, larger than the product σ-algebra M⊗N. Theorem B.55. Let (X, M, μ) and (Y, N, ν) be two measure spaces. (i) If F ∈ M and G ∈ N, then F × G is (μ × ν)∗ -measurable and (B.6)
(μ × ν)(F × G) = μ(F )ν(G).
(ii) If μ and ν are complete and E has σ-finite (μ × ν)-measure, then for μ-a.e. x ∈ X the section Ex := {y ∈ Y : (x, y) ∈ E} belongs to the σ-algebra N and for ν-a.e. y ∈ Y the section Ey := {x ∈ X : (x, y) ∈ E} belongs to the σ-algebra M. Moreover, the functions y → μ(Ey ) and x → ν(Ex ) are measurable and μ(Ey ) dν(y) = ν(Ex ) dμ(x). (μ × ν)(E) = Y
X
The previous result is a particular case of Tonelli’s theorem in the case that u = χE .
664
B. Measures
Theorem B.56 (Tonelli). Let (X, M, μ) and (Y, N, ν) be two measure spaces. Assume that μ and ν are complete and σ-finite and let u : X × Y → [0, ∞] be an (M × N)-measurable function. Then for μ-a.e. x ∈ X the function u(x, ·) is measurable and the function Y u(·, y) dν(y) is measurable. Similarly, for ν-a.e. y ∈ Y the function u(·, y) is measurable and the function X u(x, ·) dμ(x) is measurable. Moreover, u(x, y) d(μ × ν)(x, y) = u(x, y) dν(y) dμ(x) X×Y X Y = u(x, y) dμ(x) dν(y). Y
X
Throughout this book, for simplicity, we write u(x, y) dν(y)dμ(x) := u(x, y) dν(y) dμ(x). X
Y
X
Y
Exercise B.57. Prove that in the case that u : X × Y → [0, ∞] is (M ⊗ N)measurable, then Tonelli’s theorem still holds even if the measures μ and ν are not complete, and the statements are satisfied for every x ∈ X and y ∈ Y (as opposed to for μ-a.e. x ∈ X and for ν-a.e. y ∈ Y ). Theorem B.58 (Fubini). Let (X, M, μ) and (Y, N, ν) be two measure spaces. Assume that μ and ν are complete and let u : X × Y → [−∞, ∞] be (μ × ν)integrable. Then for μ-a.e. x ∈ X the function u(x, ·) is ν-integrable and the function Y u(·, y) dν(y) is μ-integrable. Similarly, for ν-a.e. y ∈ Y the function u(·, y) is μ-integrable and the function X u(x, ·) dμ(x) is ν-integrable. Moreover, u(x, y) d(μ × ν)(x, y) = u(x, y) dν(y) dμ(x) X×Y X Y u(x, y) dμ(x) dν(y). = Y
X
Exercise B.59. Prove that in the case that u : X × Y → [−∞, ∞] is (M ⊗ N)-measurable, then Fubini’s theorem still holds even if the measures μ and ν are not complete. The following result is a consequence of Tonelli’s theorem. Theorem B.60. Let (X, M, μ) be a measure space and let u : X → [0, ∞] be a measurable function. Then ∞ u dμ = μ({x ∈ X : u(x) > t}) dt. (B.7) X
0
B.5. Radon–Nikodym’s and Lebesgue’s Decomposition Theorems
665
B.5. Radon–Nikodym’s and Lebesgue’s Decomposition Theorems Definition B.61. Let (X, M) be a measurable space and let μ, ν : M → [0, ∞] be two measures. The measure ν is said to be absolutely continuous with respect to μ, and we write ν μ, if for every E ∈ M with μ(E) = 0 we have ν(E) = 0. Proposition B.62. Let (X, M) be a measurable space and let μ, ν : M → [0, ∞] be two measures with ν finite. Then ν is absolutely continuous with respect to μ if and only if for every ε > 0 there exists δ > 0 such that ν(E) ≤ ε
(B.8)
for every measurable set E ⊆ X with μ(E) ≤ δ. Proposition B.63. Let (X, M) be a measurable space and let μ, ν : M → [0, ∞] be two measures. For every E ∈ M define (B.9) u dμ : u : X → [0, ∞] is measurable, νac (E) := sup E u dμ ≤ ν(F ) for all F ⊆ E, F ∈ M . F
Then νac is a measure, with νac μ, and for each E ∈ M the supremum in the definition of νac is actually attained by a function u admissible for νac (E). Moreover, if νac is σ-finite, then u may be chosen independently of the set E. Theorem B.64 (Radon–Nikodym). Let (X, M) be a measurable space and let μ, ν : M → [0, ∞] be two measures, with μ σ-finite and ν absolutely continuous with respect to μ. Then there exists a measurable function u : X → [0, ∞] such that ν(E) = E u dμ for every E ∈ M. The function u is unique up to a set of μ-measure zero. The function u is called the Radon–Nikodym derivative of ν with respect dν . to μ, and we write u = dμ Definition B.65. Let (X, M) be a measurable space. Two measures μ, ν : M → [0, ∞] are said to be mutually singular , and we write ν ⊥ μ, if there exist two disjoint sets Xμ , Xν ∈ M such that X = Xμ ∪ Xν and for every E ∈ M we have μ(E) = μ(E ∩ Xμ ),
ν(E) = ν(E ∩ Xν ).
Proposition B.66. Let (X, M) be a measurable space and let μ, ν : M → [0, ∞] be two measures. For every E ∈ M define (B.10)
νs (E) := sup{ν(F ) : F ⊆ E, F ∈ M, μ(F ) = 0}.
666
B. Measures
Then νs is a measure and for each E ∈ M the supremum in the definition of νs is actually attained by a measurable set. Moreover, if νs is σ-finite, then νs ⊥ μ. Theorem B.67 (Lebesgue’s decomposition). Let (X, M) be a measurable space and let μ, ν : M → [0, ∞] be two measures, with μ σ-finite. Then ν = νac + νs ,
(B.11)
where νac and νs are defined in (B.9) and (B.10), respectively, and νac μ. Moreover, if ν is σ-finite, then νs ⊥ μ and the decomposition (B.11) is unique, that is, if ν = ν ac + ν s , for some measures ν ac , ν s , with ν ac μ and ν s ⊥ μ, then νac = ν ac and νs = ν s .
B.6. Signed Measures Definition B.68. Let X be a nonempty set and let M be an algebra on X. A finitely additive signed measure is a function λ : M → [−∞, ∞] such that (i) λ(∅) = 0, (ii) λ takes at most one of the two values ∞ and −∞, that is, either λ : M → (−∞, ∞] or λ : M → [−∞, ∞), (iii) if E1 , E2 ∈ M are disjoint sets, then λ(E1 ∪ E1 ) = λ(E1 ) + λ(E2 ). Definition B.69. Let (X, M) be a measurable space. A signed measure is a function λ : M → [−∞, ∞] such that (i) λ(∅) = 0, (ii) λ takes at most one of the two values ∞ and −∞, that is, either λ : M → (−∞, ∞] or λ : M → [−∞, ∞), (iii) for every countable collection {En }n of pairwise disjoint sets in M, ∞ ∞ En = λ(En ). λ n=1
n=1
Proposition B.70. Let (X, M) be a measurable space and let λ : M → [−∞, ∞] be a signed measure. For every E ∈ M define (B.12)
λ+ (E) : = sup{λ(F ) : F ⊆ E, F ∈ M},
(B.13)
λ− (E) : = − inf{λ(F ) : F ⊆ E, F ∈ M}.
Then λ+ and λ− are measures. Moreover, if λ : M → [−∞, ∞), then for every E ∈ M we have (B.14)
λ+ (E) = sup{λ(F ) : F ⊆ E, F ∈ M, λ− (F ) = 0},
λ+ is finite, and λ = λ+ − λ− .
B.6. Signed Measures
667
Theorem B.71 (Jordan’s decomposition). Let (X, M) be a measurable space and let λ : M → [−∞, ∞] be a signed measure. Then there exists a unique pair λ+ , λ− of mutually singular (nonnegative) measures, one of which is finite, such that λ = λ+ − λ− . The measures λ+ , λ− are called, respectively, the upper and lower variation of λ, while the measure |λ| := λ+ + λ− is called the total variation of λ . We say that λ is bounded, or finite, if the measure |λ| is finite. Proposition B.72. Let (X, M) be a measurable space and let λ : M → [−∞, ∞] be a signed measure. Then for every E ∈ M, ∞ |λ|(E) = sup |λ(En )| , n=1
where the supremum is taken over all partitions {En }n in M of E. Definition B.73. Let (X, M) be a measurable space and let Y be a normed space. A set function λ : M → Y is a vector-valued measure if λ(∅) = 0 and for every countable collection {En }n of pairwise disjoint sets in M, ∞ ∞ En = λ(En ). λ n=1
n=1
Proposition B.74. Let (X, M) be a measurable space, let (Y, · Y ) be a normed space and let λ : M → Y be a vector-valued measure. Then the function λ : M → [0, ∞], defined by ∞ λ(En )Y , E ∈ M, λ(E) := sup n=1
where the supremum is taken over all partitions {En }n in M of E, is a measure. Definition B.75. Let (X, M) be a measurable space, let μ : M → [0, ∞] be a measure, and let λ : M → [−∞, ∞] and ρ : M → [−∞, ∞] be signed measures. (i) λ is said to be absolutely continuous with respect to μ, and we write λ μ, if λ(E) = 0 whenever E ∈ M and μ(E) = 0. (ii) λ and ρ are said to be mutually singular , and we write λ ⊥ ρ, if there exist two disjoint sets Xλ , Xρ ∈ M such that X = Xλ ∪ Xρ and for every E ∈ M we have λ(E) = λ(E ∩ Xλ ),
ρ(E) = ρ(E ∩ Xρ ).
Note that if λ μ, then λ+ μ and λ− μ.
668
B. Measures
Theorem B.76 (Lebesgue’s decomposition). Let (X, M) be a measurable space, let λ : M → [−∞, ∞] be a signed measure, and let μ : M → [0, ∞] exists be a σ-finite measure. Then λ = λac + λs , with λac μ, and there a measurable function u : X → [−∞, ∞] such that λac (E) = E u dμ for all E ∈ M. Moreover, if λ is σ-finite, then λs ⊥ μ and the decomposition is unique; that is, if λ = λac + λs , for some signed measures λac , λs , with λac μ and λs ⊥ μ, then λac = λac and λs = λs . It is implicit in the statement of the theorem that E u dμ is well-defined for every E ∈ M, which implies that u+ or u− is Lebesgue integrable. We call λac and λs , respectively, the absolutely continuous part and the singular part of λ with respect to μ, and often we write u = dλdμac .
B.7. Lp Spaces Let (X, M, μ) be a measure space. Given two measurable functions u, v : X → [−∞, ∞], we say that u is equivalent to v and we write u ∼ v if u(x) = v(x) for μ-a.e. x ∈ X.
(B.15)
Note that ∼ is an equivalence relation in the class of measurable functions. With an abuse of notation, from now on we identify a measurable function u : X → [−∞, ∞] with its equivalence class [u]. Definition B.77. Let (X, M, μ) be a measure space and let 0 < p < ∞. Then Lp (X, M, μ) := {u : X → [−∞, ∞] : u is measurable, uLp (X,M,μ) < ∞}, where
1/p |u| dμ
uLp (X,M,μ) :=
p
.
X
If p = ∞, then L∞ (X, M, μ) := {u : X → [−∞, ∞] : u is measurable, uL∞ (X,M,μ) < ∞}, where uL∞ (X,M,μ) is the essential supremum ess sup |u| of the function |u|, that is, ess sup |u| := inf{t ≥ 0 : |u(x)| ≤ t for μ-a.e. x ∈ X}. For simplicity, and when there is no possibility of confusion, we denote the spaces Lp (X, M, μ) simply by Lp (X, μ) or Lp (X) and the norms uLp (X,M,μ) by uLp (X) , or uLp . We denote by Lp (X, M, μ; Rm ) (or more simply by Lp (X; Rm )) the space of all functions u : X → Rm whose components are in Lp (X, M, μ). We will
B.7. Lp Spaces
669
endow Lp (X, M, μ; Rm ) with the norm uLp (X,M,μ;Rm ) :=
m
ui pLp (X,M,μ)
1/p
i=1
for 1 ≤ p < ∞ and uL∞ (X,M,μ;Rm ) := max ui L∞ (X,M,μ) i
if p = ∞. For 1 ≤ p < ∞ sometimes it will be more convenient to use the equivalent norm
1/p p uLp (X,M,μ;Rm ) := u dμ . X
Given 1 ≤ p ≤ ∞, the H¨ older conjugate exponent of p is the extended real number p ∈ [1, ∞], defined by ⎧ p ⎨ p−1 if 1 < p < ∞, p := ∞ if p = 1, ⎩ 1 if p = ∞. Note that, with an abuse of notation, we have 1 1 + = 1. p p Theorem B.78 (H¨ older’s inequality). Let (X, M, μ) be a measure space, older conjugate exponent. If u, v : X → let 1 ≤ p ≤ ∞, and let p be its H¨ [−∞, ∞] are measurable functions, then (B.16)
uvL1 ≤ uLp vLp .
In particular, if u ∈ Lp (X) and v ∈ Lp (X), then uv ∈ L1 (X). The proof of H¨older’s inequality for 1 ≤ p < ∞ is based on Young’s inequality (B.17)
ab ≤ 1p ap +
1 p p b ,
which holds for all a, b ≥ 0 and whose proof is left as an exercise. Exercise B.79. Let (X, M, μ) be a measure space. (i) Let 1 ≤ p1 , . . . , pn , p ≤ ∞, with p11 + · · · + p1n = 1p , and ui ∈ Lpi (X), i = 1, . . . , n. Prove that n n / / ui ≤ ui Lpi . i=1
Lp
i=1
670
B. Measures
(ii) Let u : X → R be a measurable function. Prove that 1−θ uLq ≤ uθLp uL r ,
where 0 < p ≤ q ≤ r ≤ ∞ and
1 q
=
θ p
+
1−θ r .
(iii) Let u : X → R be a measurable function. Prove that for every R > 0, uLq ≤ R−1/q+1/p uLp + R−1/q+1/r uLr , where 0 < p ≤ q ≤ r ≤ ∞. Proposition B.80. Let (X, M, μ) be a measure space, let 1 ≤ p ≤ ∞, and older conjugate exponent of p. If p = ∞, assume that μ is let p be the H¨ σ-finite. Then for every function u ∈ Lp (X), (B.18) ⎧ ⎪ |uv| dμ : v ∈ Lp (X), vLp ≤ 1 if p < ∞, ⎨ max X uLp = ⎪ ⎩ sup |uv| dμ : v ∈ L1 (X), vL1 ≤ 1 if p = ∞. X
Corollary B.81. Let (X, M, μ) be a measure space, let 1 ≤ p ≤ ∞, and let p be the H¨ older conjugate exponent of p. Assume that μ is σ-finite. Then for every measurable function u : X → R, |uv| dμ : v ∈ Lp (X), vLp ≤ 1 . (B.19) uLp = sup X
Exercise B.82. Prove the previous corollary. As a corollary of Proposition B.80 we have the following result. Corollary B.83 (Minkowski’s inequality for integrals). Let (X, M, μ) and (Y, N, ν) be two measure spaces. Assume that μ and ν are complete and σ-finite. Let u : X × Y → [0, ∞] be an (M × N)-measurable function and let 1 ≤ p ≤ ∞.Then |u(x, ·)| dμ(x) ≤ u(x, ·)Lp (Y,N,ν) dμ(x). p X
L (Y,N,ν)
X
Exercise B.84. Prove the previous corollary. Hint: Use Proposition B.80 and Tonelli’s theorem. We now turn to the relation between different Lp spaces. Theorem B.85. Let (X, M, μ) be a measure space. Suppose that 1 ≤ p < q < ∞. Then (i) Lp (X) is not contained in Lq (X) if and only if X contains measurable sets of arbitrarily small positive measure2 , 2 By
this we mean that for every ε > 0 there is a measurable set of positive measure less than ε.
B.7. Lp Spaces
671
(ii) Lq (X) is not contained in Lp (X) if and only if X contains measurable sets of arbitrarily large finite measure3 . Corollary B.86. Let (X, M, μ) be a measure space. Suppose that 1 ≤ p < q ≤ ∞. If μ(X) < ∞, then Lq (X) ⊆ Lp (X). Theorem B.87 (Minkowski’s inequality). Let (X, M, μ) be a measure space, let 1 ≤ p ≤ ∞, and let u, v : X → [−∞, ∞] be measurable functions. Then, u + vLp ≤ uLp + vLp . In particular, if u, v ∈ Lp (X), then u + v ∈ Lp (X). By identifying functions with their equivalence classes [u], it follows from Minkowski’s inequality that · Lp is a norm on Lp (X). Theorem B.88. Let (X, M, μ) be a measure space. Then Lp (X) is a Banach space for 1 ≤ p ≤ ∞. Next we study some density results for Lp (X) spaces. Theorem B.89. Let (X, M, μ) be a measure space. Then the family of all simple functions in Lp (X) is dense in Lp (X) for 1 ≤ p ≤ ∞. The next result gives conditions on X and μ that ensure the density of continuous functions in Lp (X). Theorem B.90. Let (X, M, μ) be a measure space, with X a normal space and μ a Borel measure such that μ(E) = sup{μ(C) : C is closed, C ⊆ E} = inf{μ(U ) : U is open, U ⊇ E} for every set E ∈ M with finite measure. Then Lp (X) ∩ Cb (X) is dense in Lp (X) for 1 ≤ p < ∞. Definition B.91. Let (X, M, μ) be a measure space, with X a topological space, μ : M → [0, ∞] a Borel measure, and 1 ≤ p ≤ ∞. A measurable function u : X → [−∞, ∞] is said to belong to Lploc (X) if u ∈ Lp (K) for every compact set K ⊆ X. A sequence {un }n in Lploc (X) is said to converge to u in Lploc (X) if un → u in Lp (K) for every compact set K ⊆ X. Theorem B.92 (Riesz’s representation theorem in Lp ). Let (X, M, μ) be a older conjugate exponent. measure space, let 1 < p < ∞, and let p be its H¨ Then every bounded linear functional L : Lp (X) → R is represented by a unique v ∈ Lp (X) in the sense that uv dμ for every u ∈ Lp (X). (B.20) L(u) = X 3 By this we mean that for every M > 0 there is a measurable set of finite positive measure greater than M .
672
B. Measures
Moreover, the norm of L coincides with vLp . Conversely, every func tional of the form (B.20), where v ∈ Lp (X), is a bounded linear functional on Lp (X). Thus, the dual of Lp (X) may be identified with Lp (X). In particular, Lp (X) is reflexive. Theorem B.93 (Riesz’s representation theorem in L1 ). Let (X, M, μ) be a measure space with μ σ-finite. Then the dual of L1 (X) may be identified with L∞ (X). The next result shows that L1 (X) is not reflexive, since in general its bidual is larger than L1 (X). Let (X, M, μ) be a measure space. The dual of L∞ (X) may be identified with the space ba(X, M, μ) of all bounded finitely additive signed measures absolutely continuous with respect to μ, that is, all maps λ : M → R such that (i) λ is a finitely additive signed measure, (ii) λ is bounded, that is, its total variation norm l |λ(En )| , λ := sup n=1
where the supremum is taken over all finite partitions {En }ln=1 ⊆ M of X, l ∈ N, is finite, (iii) λ(E) = 0 whenever E ∈ M and μ(E) = 0. Given a measure space (X, M, μ), λ ∈ ba(X, M, μ), and u ∈ L∞ (X), by ∞ Theorem B.36 we may find a sequence {sn }n of simple functions in L (X) that converges uniformly to u. For every E ∈ M, we define E sn dλ as in (B.2) with λ and sn in place of μ and s, respectively. Then (sm − sn ) dλ ≤ sm − sn L∞ (X) |λ|(E) E
for all m, n ∈ N, and so there exists the limit u dλ := lim sn dλ. (B.21) E
It may be verified that the integral approximating sequence {sn }n .
n→∞ E
E
u dλ does not depend on the particular
Theorem B.94 (Riesz’s representation theorem in L∞ ). Let (X, M, μ) be a measure space. Then every bounded linear functional L : L∞ (X) → R is represented by a unique λ ∈ ba(X, M, μ) in the sense that u dλ for every u ∈ L∞ (X). (B.22) L(u) = X
B.8. Modes of Convergence
673
Moreover, the norm of L coincides with λ. Conversely, every functional of the form (B.22), where λ ∈ ba(X, M, μ), is a bounded linear functional on L∞ (X). Definition B.95. Let (X, M, μ) and (Y, N, ν) be two measure spaces, let L0 (X) := {u : X → C measurable}, L0 (Y ) := {v : Y → C measurable}, let V ⊆ L0 (X) be a subspace. We say that T : V → L0 (Y ) is of strong type (p, q) if it is bounded from Lp (X, μ) ∩ V into Lq (Y, ν), that is, T (u)Lq (Y ) ≤ cp,q uLp (X) for all u ∈ Lp (X, μ) ∩ V and for some constant cp,q > 0. Let (X, M, μ) and (Y, N, ν) be two measure spaces, and let S be the family of simple functions in L0 (X) vanishing at infinity. Theorem B.96 (Riesz–Thorin). Let (X, M, μ) and (Y, N, ν) be two measure spaces, let 1 ≤ p1 , q1 , p2 , q2 ≤ ∞, let T : S → L0 (Y ) be a linear operator of strong type (p1 , q1 ) and (p2 , q2 ). Then for every θ ∈ [0, 1], T is of strong type (p, q), with cθ uLp (X) T (u)Lq (Y ) ≤ cp1−θ 1 ,q1 p2 ,q2 for all u ∈ S, with 1 1−θ 1−θ θ 1 θ = = + , + . p p1 p2 q q1 q2 p In particular, T can be uniquely extended to L (X) as a linear operator of strong type (p, q).
B.8. Modes of Convergence In this section we study different modes of convergence and their relation to one another. Definition B.97. Let (X, M, μ) be a measure space and let un , u : X → R be measurable functions. (i) {un }n is said to converge to u pointwise μ-a.e. if there exists a set E ∈ M such that μ(E) = 0 and limn→∞ un (x) = u(x) for all x ∈ X \ E. (ii) {un }n is said to converge to u almost uniformly if for every ε > 0 there exists a set E ∈ M such that μ(E) ≤ ε and {un }n converges to u uniformly in X \ E, that is, lim
sup |un (x) − u(x)| = 0.
n→∞ x∈X\E
(iii) {un }n is said to converge to u in measure if for every ε > 0, lim μ({x ∈ X : |un (x) − u(x)| > ε}) = 0.
n→∞
674
B. Measures
The next theorem relates the types of convergence introduced in Definition B.97 to convergence in Lp (X). Theorem B.98. Let (X, M, μ) be a measure space and let un , u : X → R be measurable functions. (i) If {un }n converges to u almost uniformly, then it converges to u in measure and pointwise μ-a.e. (ii) If {un }n converges to u in measure, then there exists a subsequence {unk }k such that {unk }k converges to u almost uniformly (and hence pointwise μ-a.e.). (iii) If {un }n converges to u in Lp (X), 1 ≤ p < ∞, then it converges to u in measure and there exist a subsequence {unk }k and an integrable function v such that {unk } converges to u almost uniformly (and hence pointwise μ-a.e.) and |unk (x)|p ≤ v(x) for μ-a.e. x ∈ X and for all k ∈ N. Theorem B.99 (Egoroff). Let (X, M, μ) be a measure space with μ finite and let un : X → R be measurable functions converging pointwise μ-a.e. Then {un }n converges almost uniformly (and hence in measure). In order to characterize convergence in Lp , we need to introduce the notion of p-equi-integrability. Definition B.100. Let (X, M, μ) be a measure space and let 1 ≤ p < ∞. A family F of measurable functions u : X → [−∞, ∞] is said to be p-equiintegrable if for every ε > 0 there exists δ > 0 such that |u|p dμ ≤ ε E
for all u ∈ F and for every measurable set E ⊆ X with μ(E) ≤ δ. When p = 1, we refer to 1-equi-integrability simply as equi-integrability. Theorem B.101 (Vitali’s convergence). Let (X, M, μ) be a measure space, let 1 ≤ p < ∞, and let un , u ∈ Lp (X), n ∈ N. Then {un }n converges to u in Lp (X) if and only if the following conditions hold: (i) {un }n converges to u in measure. (ii) {un }n is p-equi-integrable. (iii) For every ε > 0 there exists E ⊆ X with E ∈ M such that μ(E) < ∞ and |un |p dμ ≤ ε X\E
for all n.
B.8. Modes of Convergence
675
Remark B.102. Note that condition (iii) is automatically satisfied when X has finite measure. Theorem B.103 (Dunford–Pettis). Let(X, M, μ) be a measure space and and let un , u ∈ L1 (X), n ∈ N, with supn X |un | dμ < ∞. Then there exist a subsequence {unk }k of {un }n and an integrable function u such that unk u in L1 (X) if and only if {un }n is equi-integrable and satisfies property (iii) of Vitali’s convergence theorem (with p = 1). In view of Vitali’s convergence theorem it becomes important to understand equi-integrability. Theorem B.104. Let (X, M, μ) be a measure space and let F be a family of integrable functions u : X → [−∞, ∞]. Consider the following conditions: (i) F is equi-integrable. (ii)
lim sup
(B.23)
t→∞ u∈F
{|u|>t}
|u| dμ = 0.
(iii) (De la Vall´ ee Poussin) There exists an increasing function g : [0, ∞) → [0, ∞] with g(t) =∞ t→∞ t
(B.24)
lim
such that
(B.25)
g(|u|) dμ < ∞.
sup u∈F
X
Then (ii) and (iii) are equivalent and either one implies (i). If in addition we assume that |u| dμ < ∞, sup u∈F
X
then (i) implies (ii) (and so all three conditions are equivalent in this case). Exercise B.105. Let (X, M, μ) be a measure space and let L0 (X) be the space of all (equivalence classes) of measurable functions u : X → R. Given u, v ∈ L0 (X) define d0 (u, v) := min{d(u, v), 1}, where d(u, v) := inf [ε + μ({x ∈ X : |u(x) − v(x)| > ε})]. ε>0
(i) Prove that L0 (X) is a complete metric space and that a sequence {un }n of functions in L0 (X) converges if and only if it converges in measure. (ii) Prove that L1 (X) and L∞ (X) are embedded in L0 (X).
676
B. Measures
B.9. Radon Measures In this section we study regularity properties for measures. Definition B.106. Let X be a topological space and let μ : B(X) → [0, ∞] be a Borel measure. (i) A Borel set E ⊆ X is said to be inner regular if μ(E) = sup{μ(K) : K ⊆ E, K is compact}, and it is outer regular if μ(E) = inf{μ(U ) : U ⊇ E, U is open}. (ii) A Borel set E ⊆ X is said to be regular if it is both inner and outer regular. Definition B.107. Let (X, M, μ) be a measure space. If X is a topological space, then the following hold. (i) μ is a Borel regular measure if μ is a Borel measure and if for every set E ∈ M there exists a Borel set F such that E ⊆ F and μ(E) = μ(F ). (ii) A Borel measure μ : M → [0, ∞] is a Radon measure if (a) μ(K) < ∞ for every compact set K ⊆ X, (b) every open set U ⊆ X is inner regular, (c) every set E ∈ M is outer regular. Hausdorff measures Hs , s > 0, represent an important class of regular Borel measures that are not Radon measures. Proposition B.108. Let X be a locally compact Hausdorff space such that every open set is σ-compact. Let μ : B(X) → [0, ∞] be a measure finite on compact sets. Then μ is a Radon measure and every Borel set E is inner regular. Theorem B.109 (Lusin). Let (X, M, μ) be a measure space, with X a topological space and μ a Radon measure, and let u : X → R be a measurable function. Then for every set E ∈ M with finite measure and every ε > 0 there exists a closed set C ⊆ E with μ(E \ C) ≤ ε such that u : C → R is continuous. Definition B.110. Let (X, M) be a measurable space and let λ : M → [−∞, ∞] be a signed measure. If X is a topological space, then λ is a signed Radon measure if |λ| : M → [0, ∞] is a Radon measure. If X is a topological space, then Mb (X; R) is the space of all signed finite Radon measures λ : B(X) → R endowed with the total variation norm. It
B.9. Radon Measures
677
can be verified that Mb (X; R) is a Banach space with the norm λMb (X;R) := |λ|(X). Similarly, if (X, M) is a measurable space and X is a topological space, then λ : M → Rm is a vectorial Radon measure if each component λi : M → R is a signed Radon measure. The space Mb (X; Rm ) of all vectorial Radon measures λ : B(X) → Rm is a Banach space with the norm λMb (X;Rm ) := λ(X).
(B.26)
Theorem B.111 (Riesz’s representation theorem in C0 ). Let X be a locally compact Hausdorff space. Then for every bounded linear functional L : C0 (X; Rm ) → R there exists a unique λ ∈ Mb (X; Rm ) such that u dλ for every u ∈ C0 (X; Rm ). (B.27) L(u) = X
Moreover, the norm of L coincides with λMb (X;Rm ) . Conversely, every functional of the form (B.27), where λ ∈ Mb (X; Rm ), is a bounded linear functional on C0 (X; Rm ). Remark B.112. A similar result holds for bounded linear functionals L : C0 (X; C) → R. Theorem B.113 (Riesz’s representation theorem in Cc ). Let X be a locally compact Hausdorff space and let L : Cc (X) → R be a linear functional. Then (i) if L is positive, that is, L(v) ≥ 0 for all v ∈ Cc (X), then there exists a unique (positive) Radon measure μ : B(X) → [0, ∞] such that L(u) =
u dμ
for every u ∈ Cc (X),
X
(ii) if L is locally bounded, that is, for every compact set K ⊆ X there exists a constant cK > 0 such that |L(u)| ≤ cK vCc (X) for all v ∈ Cc (X) with supp v ⊆ K, then there exist two (positive) Radon measures μ1 , μ2 : B(X) → [0, ∞] such that u dμ1 − u dμ2 for every u ∈ Cc (X). L(u) = X
X
Note that since both μ1 and μ2 could have infinite measure, their difference is not defined in general, although on every compact set it is a well-defined finite signed measure.
678
B. Measures
B.10. Covering Theorems in RN Theorem B.114 (Besicovitch’s covering). There exists a constant , depending only on the dimension N of RN , such that for every collection F of (nondegenerate) closed balls with (B.28)
sup{diam B : B ∈ F } < ∞
there exist F1 , . . . , F ⊆ F such that each Fn , n = 1, . . . , , is a countable family of disjoint balls in F and E⊆
B,
n=1 B∈Fn
where E is the set of centers of balls in F . The following result is an important consequence of the Besicovitch covering theorem. Theorem B.115 (Vitali–Besicovitch’s covering). Let E ⊆ RN be a Borel set and let F be a family of closed balls such that each point of E is the center of arbitrarily small balls, that is, inf{r : B(x, r) ∈ F } = 0 for every x ∈ E. a countable Let μ : B(RN ) → [0, ∞] be a Radon measure. Then there exists family {Bn }n in F of disjoint closed balls such that μ E \ n Bn = 0. Theorem B.116 (Besicovitch’s derivation). Let μ, ν : B(RN ) → [0, ∞] be two Radon measures. Then there exists a Borel set E0 ⊆ RN , with μ(E0 ) = 0, such that for every x ∈ RN \ E0 , (B.29)
dνac ν(B(x, r)) ∈R (x) = lim dμ r→0+ μ(B(x, r))
and
lim
r→0+
νs (B(x, r)) μ(B(x, r))
= 0,
where ν = νac + νs , νac μ, νs ⊥ μ. Remark B.117. In the previous theorems one can use closed cubes instead of balls. Theorem B.118. Let μ : B(RN ) → [0, ∞] be a Radon measure, let u : RN → [−∞, ∞] be a locally integrable function. Then there exists a Borel set E0 ⊆ RN , with μ(E0 ) = 0, such that for every x ∈ RN \ E0 , 1 u(y) dμ(y) = u(x). lim r→0+ μ(B(x, r)) B(x,r) By enlarging the “bad” set E0 , we can strengthen the conclusion of the previous theorem.
B.10. Covering Theorems in RN
679
Corollary B.119. Let μ : B(RN ) → [0, ∞] be a Radon measure and let u : RN → [−∞, ∞] be a locally integrable function. Then there exists a Borel set E0 ⊆ RN , with μ(E0 ) = 0, such that for every x ∈ RN \ E0 , 1 (B.30) lim |u(y) − u(x)| dμ(y) = 0. r→0+ μ(B(x, r)) B(x,r) A point x ∈ RN for which (B.30) holds is called a Lebesgue point of u. Corollary B.120. Let μ : B(RN ) → [0, ∞] be a Radon measure, let 1 ≤ p < ∞, and let u ∈ Lp (RN ). Then there exists a Borel set E0 ⊆ RN , with μ(E0 ) = 0, such that for every x ∈ RN \ E0 , 1 (B.31) lim |u(y) − u(x)|p dμ(y) = 0. r→0+ μ(B(x, r)) B(x,r) A point x ∈ RN for which (B.31) holds is called a p-Lebesgue point of u. By applying Theorem B.118 to χE , we obtain the following result. Corollary B.121. Let μ : B(RN ) → [0, ∞] be a Radon measure and let E ⊆ RN be a Borel set. Then there exists a Borel set E0 ⊆ RN , with μ(E0 ) = 0, such that for every x ∈ RN \ E0 , lim
r→0+
μ(B(x, r) ∩ E) μ(B(x, r))
= χE (x).
A point x ∈ E for which the previous limit is one is called a point of density one for E. More generally, for any t ∈ [0, 1] a point x ∈ RN such that μ(B(x, r) ∩ E) =t (B.32) lim r→0+ μ(B(x, r)) is called a point of density t for E.
Appendix C
The Lebesgue and Hausdorff Measures The Analysis of Value must be used carefully to avoid the following two types of errors. Type I: You incorrectly believe your research is not dull. Type II: No conclusions can be made. Good luck graduating. — Jorge Cham, www.phdcomics.com
C.1. The Lebesgue Measure For each set E ⊆ RN define LN o (E) := inf
∞ n=1
rnN : E ⊆
∞
Q(xn , rn ) ,
n=1
where for x ∈ RN and r > 0, Q(x, r) := x + (− 2r , 2r )N . By Proposition B.2, LN o is an outer measure, called the N -dimensional Lebesgue outer measure. Using Remark B.3, it can be shown that (C.1)
N LN o (Q(x, r)) = r
N N N and that LN o is translation-invariant, i.e., Lo (x+E) = Lo (E) for all x ∈ R N N N and all E ⊆ R . The class of all Lo -measurable subsets of R is called the σ-algebra of Lebesgue measurable sets, and by Carath´eodory’s theorem (Theorem B.13), LN o restricted to this σ-algebra is a complete measure, called the N -dimensional Lebesgue measure and denoted by LN . Since N that LN LN (RN ) ≥ LN o (Q(x, r)) = r , by sending r → ∞, we conclude is not a finite measure. However, it is σ-finite, since RN = ∞ n=1 Q(0, n) and LN (Q(0, n)) = nN < ∞.
681
682
C. The Lebesgue and Hausdorff Measures
Exercise C.1. Prove that for every δ > 0 and for each set E ⊆ RN , ∞ ∞ N LN (E) = inf r : r < δ, E ⊆ Q(x , r ) . n n n o n n=1
Conclude that
LN o
n=1
is a metric outer measure.
It follows from the previous exercise and Proposition B.19 that every Borel subset of RN is Lebesgue measurable. It may be proved that there are Lebesgue measurable sets that are not Borel sets (see Exercise 1.33). Hence, LN : B(RN ) → [0, ∞] is not a complete measure. Using the axiom of choice, it is also possible to construct sets that are not Lebesgue measurable. Exercise C.2 (A non-Lebesgue measurable set). On the real line consider the equivalence relation x ∼ y if x − y ∈ Q. By the axiom of choice we may construct a set E ⊆ (0, 1) that contains exactly one element from each equivalence class. Let (r + E) ⊆ (−1, 2). F := r∈(−1,1)∩Q
(i) Prove that F ⊇ (0, 1). (ii) Prove that if r, q ∈ Q, with r = q, then (r + E) ∩ (q + E) = ∅. (iii) Prove that E is not Lebesgue measurable. We observe that the Lebesgue measure is a Radon measure. Indeed, outer regularity of arbitrary sets follows from (C.1) and the definition of LN o , while inner regularity of open sets is an immediate consequence of the fact that each open set can be written as a countable union of closed cubes with pairwise disjoint interiors. Proposition C.3. LN is a Radon measure. Moreover, every Lebesgue measurable set E is the union of a Borel set and a set of Lebesgue measure zero. In addition, for every set F ⊆ RN there exists a Lebesgue measurable set E containing F with LN (E) = LN o (F ). Using the previous proposition and Proposition B.9 we have: Proposition C.4. Let {En }n be an increasing sequence of subsets of RN , then ∞ N E LN n = lim Lo (En ). o n=1
n→∞
Using the notation introduced in Section B.4, we have
C.2. The Brunn–Minkowski Inequality
683
Proposition C.5. Let N = n + m, where N, n, m ∈ N. Then (Ln × Lm )∗ = LN o . If E ⊆ RN is a Lebesgue measurable set, M is the σ-algebra of all Lebesgue measurable subsets of E, Y is a nonempty set, and N is an algebra on Y , then a measurable function u : E → Y is called Lebesgue measurable. Proposition C.6. Let E ⊆ RN be a Lebesgue measurable set and let u : E → R be a Lebesgue measurable function. Then there exists a Borel function v : E → R such that v(x) = u(x) for LN -a.e. x ∈ E. If E ⊆ RN is a Lebesgue measurable set and u : E → [−∞, ∞] is a Lebesgue measurablefunction, then, whenever it is well-defined, we denote N simply by u dL u dx. If N = 1 and I is an interval of endpoints a E Eb and b, we also write a u dx for I u dx.
C.2. The Brunn–Minkowski Inequality We leave the following preliminary result as an exercise. Exercise C.7. Let E, F ⊂ RN be two compact sets. (i) Construct sequences of sets {En }n and {Fn }n such that decreasing ∞ E , F = E = ∞ n=1 n n=1 Fn , and all En and Fn consist of finite unions of rectangles with sides parallel to the axes. (ii) Prove that LN (En + Fn ) → LN (E + F ) as n → ∞. Theorem C.8 (Brunn–Minkowski’s inequality). Let E, F ⊂ RN be two Lebesgue measurable sets such that E + F := {x + y : x ∈ E, y ∈ F } is also Lebesgue measurable. Then (LN (E))1/N + (LN (F ))1/N ≤ (LN (E + F ))1/N . Proof. Step 1: Assume that E and F are rectangles whose sides are parallel to the axes and let xi and yi , i = 1, . . . , N , be their respective side lengths. Then LN (E) =
N /
xi ,
i=1
LN (F ) =
N /
yi ,
LN (E + F ) =
i=1
N / (xi + yi ). i=1
By the arithmetic-geometric mean inequality N / i=1
N N N 1 xi 1 yi xi 1/N / yi 1/N + ≤ + = 1, xi + yi xi + yi N xi + yi N xi + yi i=1
i=1
i=1
684
C. The Lebesgue and Hausdorff Measures
which gives the Brunn–Minkowski inequality for rectangles. Step 2: We now suppose that E and F are finite unions of rectangles as in Step 1. The proof is by induction on the sum of the numbers of rectangles in E and in F . By interchanging E with F , if necessary, we may assume that E has at least two rectangles. By translating E if necessary, we may also assume that a coordinate hyperplane, say {xN = 0}, separates two rectangles in E. Let E + and E − denote the union of the rectangles formed by intersecting the rectangles in E with the half spaces {xN ≥ 0} and N ±) LN (F ± ) = , where F + {xN ≤ 0}, respectively. Translate F so that LLN(E (E) LN (F ) and F − are defined analogously to E + and E − . Then E + + F + ⊆ {xN ≥ 0} and E − + F − ⊆ {xN ≤ 0}, and the numbers of rectangles in E + ∪ F + and in E − ∪ F − are both smaller than the number of rectangles in E ∪ F . By induction and Step 1 we obtain LN (E + F ) ≥ LN (E + + F + ) + LN (E − + F − ) ≥ ((LN (E + ))1/N + (LN (F + ))1/N )N + ((LN (E − ))1/N + (LN (F − ))1/N )N (LN (F ))1/N N (LN (F ))1/N N N − = LN (E + ) 1 + N + L (E ) 1 + (L (E))1/N (LN (E))1/N (LN (F ))1/N N = LN (E) 1 + N = ((LN (E))1/N + (LN (F ))1/N )N . (L (E))1/N Step 3: Assume next that E and F are compact sets. Let {En }n and {Fn }n be as in the previous exercise. Then for all n ∈ N, (LN (En ))1/N + (LN (Fn ))1/N ≤ (LN (En + Fn ))1/N . It now suffices to let n → ∞. Finally, if E and F are Lebesgue measurable sets, with E + F measurable, fix two compact sets K1 ⊆ E, K2 ⊆ F . Then K1 + K2 ⊆ E + F , and so (LN (K1 ))1/N + (LN (K2 ))1/N ≤ (LN (K1 + K2 ))1/N ≤ (LN (E + F ))1/N . Using the inner regularity of the Lebesgue measure, we obtain the desired result. Remark C.9. (i) Note that the hypothesis that E +F is measurable cannot be omitted. Indeed, Sierpi´ nsky [211] constructed an example of two Lebesgue measurable sets whose sum is not measurable. However, if E and F are Borel sets, then E + F is an analytic set, and so it is measurable. We refer to [83] for more information on this topic.
C.2. The Brunn–Minkowski Inequality
685
(ii) Fix θ ∈ (0, 1). By replacing E with θE and F with (1 − θ)F and using the N -homogeneity of the Lebesgue measure, we obtain that θ(LN (E))1/N + (1 − θ)(LN (F ))1/N = (LN (θE))1/N + (LN ((1 − θ)F ))1/N ≤ (LN (θE + (1 − θ)F ))1/N . Thus, the function f (t) := (LN (tE + (1 − t)F ))1/N is concave in [0, 1]. In particular, if C ⊆ RN +1 is a convex set and E and F are the intersections of C with the hyperplanes {x1 = 1} and {x1 = 0} (we write x = (x1 , x2 , . . . , xN +1 ) ∈ RN +1 ), then tE + (1 − t)F is the intersection of the convex hull of E and F with the hyperplane x1 = t and is therefore contained in the intersection of C with the hyperplane {x1 = t}. Hence, the function giving the nth root of the volume of parallel hyperplane sections of an (n+1)-dimensional convex set is concave (see [92]). We used this fact in the proof of Poincar´e’s inequality in convex domains (see Theorem 13.36). # Exercise C.10. Let C = {x ∈ R3 : x22 + x23 ≤ 12 (x1 + 1), 0 ≤ x1 ≤ 1}. Prove that C is convex but the function χE (x1 , x2 , x3 ) dx2 dx3 g(x1 ) := R2
is not concave (although its square root is by the previous remark). Definition C.11. Given a set E ⊆ RN and an integer 0 ≤ n ≤ N , we define the n-dimensional upper Minkowski content of E, LN x ∈ RN : dist(x, E) < ε ∗n , M (E) := lim sup αN −n εN −n ε→0+ and the n-dimensional lower Minkowski content of E, N N : dist(x, E) < ε L x ∈ R Mn∗ (E) := lim inf . αN −n εN −n ε→0+ When the n-dimensional upper and lower Minkowski contents of E coincide, their common value is called the n-dimensional Minkowski content of E and is denoted Mn (E). Next we prove the isoperimetric inequality (see also Theorem 14.44). Theorem C.12 (Isoperimetric inequality). Let E ⊆ RN be a Lebesgue measurable set with LN (E) < ∞. Then −(N −1)/N
αN
−1 −1 (LN (E))(N −1)/N ≤ βN MN (∂E). ∗
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C. The Lebesgue and Hausdorff Measures
Proof. Since {x ∈ RN : dist(x, E) < ε} = E + B(0, ε), by the Brunn–Minkowski inequality
N 1/N LN ({x ∈ RN : dist(x, E) < ε}) ≥ (LN (E))1/N + αN ε ≥ LN (E) + N (LN (E))(N −1)/N αN ε, 1/N
where we have used the inequality (a + b)N ≥ aN + N aN −1 b for a, b ≥ 0. Hence, (C.2)
LN ({x ∈ RN : 0 < dist(x, E) < ε}) = LN ({x ∈ RN : dist(x, E) < ε}) − LN (E) ≥ N (LN (E))(N −1)/N αN ε. 1/N
On the other hand,
E ◦ ⊇ {x ∈ RN : dist x, RN \ E ≥ ε} + B(0, ε),
and so, again by the Brunn–Minkowski inequality and reasoning as before, LN (E ◦ ) ≥ LN ({x ∈ RN : dist(x, RN \ E) ≥ ε}) + N (LN ({x ∈ RN : dist(x, RN \ E) ≥ ε}))(N −1)/N αN ε. 1/N
It follows that LN ({x ∈ RN : 0 < dist(x, RN \ E) < ε}) = LN (E ◦ ) − LN ({x ∈ RN : dist(x, RN \ E) ≥ ε})
(C.3)
≥ N (LN ({x ∈ RN : dist(x, RN \ E) ≥ ε}))(N −1)/N αN ε. 1/N
Since the sets {x ∈ RN : 0 < dist(x, E) < ε} and
{x ∈ RN : 0 < dist x, RN \ E < ε}
are disjoint subsets of {x ∈ RN : dist (x, ∂E) < ε}, combining the inequalities (C.2) and (C.3), we obtain LN ({x ∈ RN : dist (x, ∂E) < ε}) ε 1/N N ≥ N (L (E))(N −1)/N αN + N (LN ({x ∈ RN : dist(x, RN \ E) ≥ ε}))(N −1)/N αN . 1/N
Letting ε → 0+ and using the facts that α1 = 2 and LN ({x ∈ RN : dist(x, RN \ E) ≥ ε}) LN (E ◦ ),
C.3. Mollifiers
687
we deduce −1 (∂E) = lim inf 2MN ∗
≥
LN ({x ∈ RN : dist (x, ∂E) < ε}) ε
ε→0+ 1/N N αN [(LN (E))(N −1)/N
+ (LN (E ◦ ))(N −1)/N ].
−1 (∂E) < ∞, then, necessarily, To conclude, we now observe that if MN ∗ N L (∂E) = 0.
Exercise C.13. Let Q := (0, 1)N −1 , let f ∈ C 1 (Q ), and let E := {(x , xN ) ∈ Q × R : f (x ) < xN < f (x ) + 1}. −1 (∂E) < ∞ and give an explicit formula. Prove that MN ∗
C.3. Mollifiers Given a nonnegative bounded function ϕ ∈ L1 (RN ) with ϕ(x) dx = 1, (C.4) supp ϕ ⊆ B(0, 1), RN
for every ε > 0 we define
1 x , x ∈ RN . ϕ εN ε The functions ϕε are called mollifiers. Note that supp ϕε ⊆ B(0, ε). Hence, given an open set Ω ⊆ RN and a function u ∈ L1loc (Ω), we may define ϕε (x − y)u(y) dy (C.5) uε (x) := (u ∗ ϕε )(x) = ϕε (x) :=
Ω
for x ∈ Ωε , where the open set Ωε is given by (C.6)
Ωε := {x ∈ Ω : dist(x, ∂Ω) > ε}.
The function uε : Ωε → R is called a mollification of u. Note that if x ∈ Ω, then uε (x) is well-defined for all 0 < ε < dist(x, ∂Ω). Thus, it makes sense to talk about limε→0+ uε (x). We will use this fact without further mention. Moreover, if u ∈ L1 (Ω), then uε (x) is well-defined for every x ∈ RN . Remark C.14. Note that if Ω = RN , then Ωε = RN ; thus uε is defined in the entire space RN . Remark C.15. In the applications we will consider two important types of mollifiers: (i) ϕ is the (renormalized) characteristic function of the unit ball, that is, 1 χ (x), x ∈ RN , ϕ(x) := αN B(0,1)
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C. The Lebesgue and Hausdorff Measures
(ii) ϕ is the function of class Cc∞ defined by c exp x 12 −1 if x < 1, (C.7) ϕ(x) := 0 if x ≥ 1, where we choose c > 0 so that (C.4) is satisfied. In this case, the functions ϕε are called standard mollifiers. The following theorem is the first main result of this subsection. Theorem C.16. Let Ω ⊆ RN be an open set, let ϕ ∈ L1 (RN ) be a nonnegative bounded function satisfying (C.4), and let u ∈ L1loc (Ω). (i) If u ∈ C(Ω), then uε → u as ε → 0+ uniformly on compact subsets of Ω. (ii) For every Lebesgue point x ∈ Ω (and so for LN -a.e. x ∈ Ω), uε (x) → u(x) as ε → 0+ . Moreover, if u ∈ Lp (Ω), 1 ≤ p ≤ ∞, then uε (x) → 0 for every x ∈ RN \ Ω. (iii) If u ∈ Lp (Ω), 1 ≤ p ≤ ∞, then (C.8)
uε Lp (RN ) ≤ uLp (Ω) for every ε > 0 and
(C.9)
uε Lp (RN ) → uLp (Ω) as ε → 0+ .
(iv) If u ∈ Lp (Ω), 1 ≤ p < ∞, then lim uε − uLp (Ω) = 0.
ε→0+
Proof. (i) Let K ⊂ Ω be a compact set. For any fixed 0 < η < dist(K, ∂Ω), let Kη := {x ∈ RN : dist(x, K) ≤ η}, so that Kη ⊂ Ω. Note that for ε > 0 sufficiently small we have that Kη ⊂ Ωε . Since Kη is compact and u is uniformly continuous on Kη , for every ρ > 0 there exists δ = δ(η, K, ρ) > 0 such that (C.10)
|u(x) − u(y)| ≤ ρ
for all x, y ∈ Kη , with x − y ≤ δ. Let 0 < ε < min{δ, η}. Then for all x ∈ K, (C.11) |uε (x) − u(x)| = ϕε (x − y)u(y) dy − u(x) Ω x − y 1 [u(y) − u(x)] dy ϕ = N ε ε B(x,ε) 1 |u(y) − u(x)| dy, ≤ ϕ∞ N ε B(x,ε)
C.3. Mollifiers
689
where we have used (C.4) and the fact that supp ϕε (· − y) ⊆ B(x, ε). It follows by (C.10) that |uε (x) − u(x)| ≤ ραN ϕ∞ for all x ∈ K, and so uε − uC(K) ≤ ραN ϕ∞ . (ii) Let x ∈ Ω be a Lebesgue point of u, that is, 1 |u(y) − u(x)| dy = 0, lim ε→0+ εN B(x,ε) then from (C.11) it follows that uε (x) → u(x) as ε → 0+ . (iii) To prove (C.8) it is enough to assume that u ∈ Lp (Ω). If 1 ≤ p < ∞, then by H¨older’s inequality and (C.4) for all x ∈ RN , 1/p 1/p |uε (x)| = (ϕε (x − y)) (ϕε (x − y)) u(y) dy Ω 1/p 1/p (C.12) ϕε (x − y) dy ϕε (x − y)|u(y)|p dy ≤ Ω Ω 1/p ≤ ϕε (x − y)|u(y)|p dy , Ω
and so by Fubini’s theorem and (C.4) once more |uε (x)|p dx ≤ ϕε (x − y)|u(y)|pdydx N N R R Ω p |u(y)| ϕε (x − y) dx dy = |u(y)|p dy. = RN
Ω
Ω
RN ,
On the other hand, if p = ∞, then for every x ∈ ϕε (x − y)|u(y)| dy |uε (x)| ≤ Ω ϕε (x − y) dy ≤ uL∞ (Ω) ≤ uL∞ (Ω) Ω
again by (C.4), and so (C.8) holds for all 1 ≤ p ≤ ∞. In particular, lim sup uε Lp (RN ) ≤ uLp (Ω) . ε→0+
To prove the opposite inequality, assume first that 1 ≤ p < ∞. By part (ii), uε (x) → u(x) as ε → 0+ for LN a.e. x ∈ Ω, and so by Fatou’s lemma |u(x)|p dx = lim |uε (x)|p dx ≤ lim inf |uε (x)|p dx. Ω
Ω ε→0
+
ε→0+
RN
If p = ∞, then again by part (ii) uε (x) → u(x) as ε → 0+ for LN a.e. x ∈ Ω. Hence, |u(x)| = lim |uε (x)| ≤ lim inf uε L∞ (RN ) ε→0+
ε→0+
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C. The Lebesgue and Hausdorff Measures
for LN a.e. x ∈ Ω. It follows that uL∞ (Ω) ≤ lim inf uε L∞ (RN ) . ε→0+
Hence, (C.9) holds also in this case. (iv) Fix ρ > 0 and find a function v ∈ Cc (Ω) such that u − vLp (Ω) ≤ ρ. Since K := supp v is compact, it follows from part (i) that for every 0 < η < dist(K, ∂Ω), the mollification vε of v converges to v uniformly in the compact set Kη := {x ∈ RN : dist(x, K) ≤ η}. Since vε = v = 0 in Ω \ Kη for 0 < ε < η, we have that p |vε − v| dx = |vε − v|p dx ≤ vε − vpC(Kη ) LN (Kη ) ≤ ρ, Ω
Kη
provided ε > 0 is sufficiently small. By Minkowski’s inequality, uε − uLp (Ω) ≤ uε − vε Lp (Ω) + vε − vLp (Ω) + v − uLp (Ω) ≤ 2u − vLp (Ω) + vε − vLp (Ω) ≤ 3ρ, where we have used (C.8) for the function u − v.
Remark C.17. If we do not assume that u ∈ Lp (Ω) in part (iii), then we can still prove that uε Lp (Ωε ) ≤ uLp (Ω) and that uε Lp (Ωε ) → uLp (Ω) as ε → 0+ . Exercise C.18. Let (C.13)
& N ) := {φ& : φ ∈ C ∞ (RN ), 0 ∈ / supp φ} ⊂ S(RN ), O(R c
where we recall that φ& is the Fourier transform of φ (see (10.23)). Prove & N ) is dense in Lp (RN ) for 1 ≤ p < ∞. that O(R Exercise C.19. Let ϕ ∈ L1 (RN ) be a nonnegative function satisfying RN ϕ(x) dx = 1. (i) Prove that for every ρ > 0 there exist ψ ∈ S(RN ) and δ = δ(ρ) > 0 & & = 1 for x ≤ δ and ψ(x) =0 such that ϕ − ψL1 (RN ) ≤ ρ, ψ(x) for x > δ. (ii) Let Ω ⊆ RN be an open set and let u ∈ Lp (Ω), 1 ≤ p < ∞. Prove that uε → u in Lp (Ω) as ε → 0+ . Hint: Use part (i) and Exercise C.18. (iii) Let Ω, u and p be as in part (ii). Prove that uε (x) → u(x) as ε → 0+ for every Lebesgue point x ∈ Ω of u. More can be said about the regularity of uε if we restrict our attention to standard mollifiers.
C.3. Mollifiers
691
Theorem C.20. Let Ω ⊆ RN be an open set, let ϕ ∈ L1 (RN ) be defined as in (C.7), and let u ∈ L1loc (Ω). Then uε ∈ C ∞ (Ωε ) for all 0 < ε < 1, and for every multi-index α, ∂ |α| ϕε α α (C.14) ∂ uε (x) = (u ∗ ∂ ϕε )(x) = (x − y)u(y) dy α RN ∂x for all x ∈ Ωε . Moreover, if u ∈ Lp (RN ) for 1 ≤ p ≤ ∞, then uε ∈ C ∞ (RN ). Proof. Fix x ∈ Ωε and 0 < η < dist(x, ∂Ω) − ε. Let ei , i = 1, . . . , N , be an element of the canonical basis of RN and for every h ∈ R, with 0 < |h| ≤ η, consider ∂ϕε uε (x + hei ) − uε (x) − (x − y)u(y) dy h Ω ∂xi ϕε (x + hei − y) − ϕε (x − y) ∂ϕε = − (x − y) u(y) dy h ∂xi Ω h ∂ϕε ∂ϕε 1 (x − y + tei ) dt − (x − y) u(y) dy = ∂xi Ω h 0 ∂xi h ∂ϕε ∂ϕε 1 (x − y + tei ) − (x − y) u(y) dydt, = h 0 B(x,ε+η) ∂xi ∂xi where we have used Fubini’s theorem and the fact that supp ϕε ⊆ B(0, ε). Since ϕε ∈ Cc∞ (RN ), its partial derivatives are uniformly continuous. Hence, for every ρ > 0 there exists δ = δ(η, x, ρ, ε) > 0 such that ∂ϕ ∂ϕε ρ ε (z) − (w) ≤ ∂xi ∂xi 1 + uL1 (B(x,ε+η)) for all z, y ∈ B(x, ε + η), with z − w ≤ δ. Then for 0 < |h| ≤ min{η, δ} we have u (x + he ) − u (x) ∂ϕ ε i ε ε − (x − y) u(y) dy ≤ ρ, h Ω ∂xi which shows that ∂uε (x) = ∂xi
Ω
∂ϕε (x − y) u(y) dy. ∂xi
Note that the only properties that we have used on the function ϕε are that ϕε ∈ Cc∞ (RN ) with supp ϕε ⊆ B(0, ε). Hence, the same proof carries over ε if we replace ϕε with ψε := ∂ϕ ∂xi . Thus, by induction we may prove that for every multi-index α there holds |α| ∂ ϕε (x − y) u(y) dy. ∂ α uε (x) = (u ∗ ∂ α ϕε )(x) = α Ω ∂x
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C. The Lebesgue and Hausdorff Measures
This completes the first part of the proof. We leave as an exercise to modify the previous proof to show that if u ∈ Lp (RN ) for 1 ≤ p ≤ ∞, then uε ∈ C ∞ (RN ). An important application of the theory of mollifiers is the existence of smooth partitions of unity. Theorem C.21 (Smooth partition of unity). Let Ω ⊆ RN be an open set and let {Uα }α∈Λ be an open cover of Ω. Then there exists a sequence {ψn }n of nonnegative functions in Cc∞ (Ω) such that (i) each ψn has support in some Uα , ∞ (ii) n=1 ψn (x) = 1 for all x ∈ Ω, (iii) for every compact set K ⊂ Ω there exists an integer ∈ N and an open set U , with K ⊂ U ⊆ Ω, such that n=1 ψn (x) = 1 for all x ∈ U. Proof. Let S be a countable dense set in Ω, e.g., S := {x ∈ QN ∩ Ω}, and consider the countable family F of closed balls (C.15)
F :={B(x, r) : r ∈ Q, 0 < r < 1, x ∈ S, B(x, r) ⊆ Uα ∩ Ω for some α ∈ Λ}.
Since F is countable, we may write F = {B(xn , rn ) : n ∈ N}. Since {Uα }α∈Λ is an open cover of Ω, by the density of S and of the rational numbers we have that (C.16)
Ω=
∞
B(xn , rn /2).
n=1
For each n ∈ N consider φn := ϕrn /4 ∗ χB(xn ,3rn /4) , where the ϕrn /4 are standard mollifiers (with ε := rn /4). By Theorem C.20, φn ∈ C ∞ (RN ). Moreover, if x ∈ B(xn , rn /2), then ϕrn /4 (x − y) χB(xn ,3rn /4) (y) dy φn (x) = N R ϕrn /4 (x − y) χB(xn ,3rn /4) (y) dy = B(x,rn /4) ϕrn /4 (x − y) dy = 1, = B(x,rn /4)
where we have used (C.4) and the fact that if x ∈ B(xn , rn /2), then B(x, rn /4) ⊆ B(xn , 3rn /4). Since 0 ≤ χB(xn ,3rn /4) ≤ 1, a similar calculation shows that
C.3. Mollifiers
693
0 ≤ φn ≤ 1. On the other hand, if x ∈ / B(xn , rn ), then φn (x) = ϕrn /4 (x − y) χB(xn ,3rn /4) (y) dy RN ϕrn /4 (x − y) χB(xn ,3rn /4) (y) dy = 0, = B(x,rn /4)
where we have used the fact that if x ∈ / B(xn , rn ), then B(x, rn /4) ∩ ∞ B(xn , 3rn /4) = ∅. In particular, φn ∈ Cc (RN ) and supp φn ⊆ B(xn , rn ). Note that in view of the definition of F , supp φn ⊂ Uα ∩ Ω for some α ∈ Λ. Define ψ1 := φ1 and (C.17)
ψn := (1 − φ1 ) · · · (1 − φn−1 )φn
for n ≥ 2, n ∈ N. Since 0 ≤ φk ≤ 1 and supp φk ⊆ B(xk , rk ) for all k ∈ N, we have that 0 ≤ ψn ≤ 1 and supp ψn ⊆ B(xn , rn ). This gives (i). To prove (ii), we prove by induction that (C.18)
ψ1 + · · · + ψn = 1 − (1 − φ1 ) · · · (1 − φn )
for all n ∈ N. The relation (C.18) is true for n = 1, since ψ1 := φ1 . Assume that (C.18) holds for n. Then by (C.17), ψ1 + · · · + ψn + ψn+1 = 1 − (1 − φ1 ) · · · (1 − φn ) + ψn+1 = 1 − (1 − φ1 ) · · · (1 − φn ) + (1 − φ1 ) · · · (1 − φn )φn+1 = 1 − (1 − φ1 ) · · · (1 − φn+1 ). Hence, (C.18) holds for all n ∈ N. Since φk = 1 in B(xk , rk /2) for all k ∈ N, it follows from (C.18) that n B(xk , rk /2). (C.19) ψ1 (x) + · · · + ψn (x) = 1 for all x ∈ k=1
Thus, in view of (C.16), property (ii) holds. Finally, if K ⊂ Ω is compact, again by (C.16), we may find ∈ N so large that k=1 B(xk , rk /2) ⊃ K, and so (iii) follows by (C.19). Exercise C.22. Prove that if Λ = {1, . . . , }, then in the previous theorem we can construct a partition of unity consisting of functions. Exercise C.23 (Cut-off function). ; Let Ω ⊆ RN be an open set and let K ⊂ Ω be a compact set. Prove that there exists a function ϕ ∈ Cc∞ (Ω) such that 0 ≤ ϕ ≤ 1 and ϕ = 1 on K. The function ϕ constructed in the previous exercise is usually called a cut-off function. Finally, using mollifiers, it is possible to give the following density result in Lp spaces.
694
C. The Lebesgue and Hausdorff Measures
Theorem C.24. Let Ω ⊆ RN be an open set. Then the space Cc∞ (Ω) is dense in Lp (Ω) for 1 ≤ p < ∞.
C.4. Maximal Functions In this section we introduce the notion of maximal function and study its properties. Throughout this section we consider Lp spaces in the case that the underlying measure is the Lebesgue measure. Definition C.25. Let u ∈ L1loc (RN ). The (Hardy–Littlewood) maximal function of u is defined by 1 |u(y)| dy M(u)(x) := sup N r>0 L (B(x, r)) B(x,r) for all x ∈ RN . Exercise C.26. Prove that the set {x ∈ RN : M(u)(x) > t} is open. Theorem C.27. Let u ∈ Lp (RN ), 1 ≤ p ≤ ∞. Then (i) M(u)(x) < ∞ for LN -a.e. x ∈ RN , (ii) if p = 1, then for any t > 0, (C.20)
L ({x ∈ R N
N
3N : M(u)(x) > t}) ≤ t
RN
|u(x)| dx,
(iii) if 1 < p ≤ ∞, then M(u) ∈ Lp (RN ) and M(u)Lp (RN ) ≤ cuLp (RN ) for some constant c = c(N, p) > 0. Exercise C.28. Let u, v ∈ L1 (RN ). Assume that |u(x)| ≤ v(x) for LN a.e. x ∈ RN , where v ∈ L1 (RN ) has the form v(x) = v1 (x) for some nonnegative, decreasing function v1 . Prove that u(x)v(x − y) dy ≤ vL1 (RN ) M(u)(x) RN
for LN -a.e. x ∈ RN . Hint: Consider first the case in which v1 = χ[0,r] for some r > 0. Using the previous exercise and Theorem C.27, we can prove the following result. Proposition C.29. Let 0 < α < N , 1 < p < N/α. Then for every u ∈ Lp (RN ), q 1/q u(y) dy ≤ cuLp (RN ) dx N −α x − y N N R R for some constant c = c(α, p, N ) > 0, where q := pN/(N − αp).
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695
Proof. Define
v1 (t) :=
1 tN −α
0
if 0 ≤ t ≤ R, otherwise.
Then by the previous exercise, for every R > 0 and for LN -a.e. x ∈ RN we have that |u(y)| dy ≤ cRα M(u)(x), N −α x − y B(x,R) and so |u(y)| |u(y)| dy ≤ dy N −α N −α RN x − y B(x,R) x − y |u(y)| dy. + N −α RN \B(x,R) x − y (C.21)
≤ cRα M(u)(x) + uLp
RN \B(x,R)
1/p 1 dy x − y(N −α)p
= cRα M(u)(x) + cuLp Rα−N/p , where we have used H¨older’s inequality. Taking u p p/N L , R := M(u)(x) + ε where ε > 0, we get |u(y)| αp/N dy ≤ cuLp (M(u)(x) + ε)1−αp/N N −α RN x − y for LN -a.e. x ∈ RN . Letting ε → 0+ and taking the norm in Lq (RN ) on both sides yields q 1/q 1/q u(y) 1−p/q p dy dx ≤ cu (M(u)) dx p L N −α RN RN x − y RN ≤ cuLp , where we have used Theorem C.27.
Exercise C.30. What happens if p = 1 or p = N/α?
C.5. BMO Spaces Definition C.31. We say that a function u ∈ L1loc (RN ) has bounded mean oscillation, and we write u ∈ BMO(RN ), if 1 |u(x) − uQ | dx < ∞, (C.22) |u|BMO := sup N L (Q) Q where the supremum is taken over all cubes Q.
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C. The Lebesgue and Hausdorff Measures
Here we are using the notation (C.23)
uE :=
1 LN (E)
u(x) dx. E
The quantity LN1(Q) Q |u(x) − uQ | dx is called the mean oscillation of u over Q. It measures the average distance of u from uQ over the same cube. Remark C.32. The space BMO(RN ) is a vector space and | · |BMO is a seminorm. Since |u|BMO = 0 if and only if u ≡const, | · |BMO is a norm in the quotient space BMO(RN )/R. Remark C.33. In the sequel we will use the fact that u ∈ L1loc (RN ) belongs to BMO(RN ) if and only if for every cube Q there exists cQ,u ∈ R such that 1 |u(x) − cQ,u | dx < ∞. sup N L (Q) Q To see this assume that the latter holds. Then 1 u(x) − uQ = u(x) − cQ,u + N (u(y) − cQ,u ) dy L (Q) Q and so 1 |u(x) − uQ | ≤ |u(x) − cQ,u | + N L (Q)
|u(y) − cQ,u | dy. Q
By averaging over Q in x we get 1 1 |u(x) − uQ | dx ≤ 2 N |u(x) − cQ,u | dx. (C.24) LN (Q) Q L (Q) Q Note that if we define |u|∗BMO
1 inf := sup N L (Q) cQ ∈R
|u(x) − cQ | dx, Q
then by (C.24), 12 |u|BMO ≤ |u|∗BMO ≤ |u|BMO . Exercise C.34. Prove that | · |∗BMO is a seminorm. Using the previous property we can show that L∞ (RN ) BMO(RN ). Theorem C.35. The space L∞ (RN ) is strictly contained in BMO(RN ). Proof. Given u ∈ L∞ (RN ) take cQ,u := 0. Then 1 |u(x) − 0| dx ≤ uL∞ , LN (Q) Q which, by Remark C.33, implies that u ∈ BMO(RN ).
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697
Next we show that the unbounded function u(x) := log x belongs to BMO(RN ). Let Q = Q(x0 , r) for some x0 and r > 0. If Q ∩ B(0, 2r) = ∅, take cQ,u := log x0 . By the mean value theorem for x ∈ Q, log x − cQ,u = log x − log x0 =
y · (x − x0 ) y2
for some y between x and x0 , and so
√ √ r N x − x0 N ≤ = . | log x − cQ,u | ≤ y 2r 2
It follows that 1 N L (Q)
√ N . |u(x) − cQ,u | dx ≤ 2 Q
√ On the other hand, if Q ∩ B(0, 2r) = ∅, then Q ⊂ B(0, 2r + N r). In this case take cQ,u := log r. Then 1 1 |u(x) − cQ,u | dx = N | log x − log r| dx LN (Q) Q r Q √ αN 2r+ N r N −1 1 | log x/r| dx = N s | log s/r| ds ≤ N √ r r 0 B(0,2r+ N r) 2+√N = αN tN −1 | log t| dt < ∞, 0
where rt = s. Hence, u belongs to BMO(RN ).
Exercise C.36. Prove that for N = 1, the function log x if x > 0, v(x) = 0 if x ≤ 0 does not belong to BMO(R). So multiplying a BMO function by a characteristic function does not preserve BMO. Exercise C.37. Let u, v ∈ BMO(RN ). (i) Prove that the function h := |u| belongs to BMO(RN ) and |h|∗BMO ≤ |u|∗BMO . (ii) Prove that the functions min{u, v} and max{u, v} belong to BMO(RN ), with | min{u, v}|∗BMO ≤ |u|∗BMO + |v|∗BMO and | max{u, v}|∗BMO ≤ |u|∗BMO + |v|∗BMO . (iii) Prove that for every t > 0 the truncated function ut (x) := min{t, max{u, −t}} belongs to BMO(RN ) with |ut |∗BMO ≤ |u|∗BMO . Exercise C.38. Prove that BMO(RN )/R is a Banach space.
698
C. The Lebesgue and Hausdorff Measures
C.6. Hardy’s Inequality In this section we prove Hardy’s inequality. Theorem C.39. Let f : [0, ∞) → [0, ∞) be a convex, strictly increasing function and let v : [0, ∞) → [0, ∞) and g : [0, ∞) → [0, ∞] be measurable functions. Then
x
∞ ∞ ∞ 1 dx dy. g(x)f v(y) dy dx ≤ f (v(y)) g(x) x 0 x 0 0 y Proof. By Jensen’s inequality (see Theorem B.50) for every measurable function w : [0, ∞) → [0, ∞) and every x > 0,
x 1 x 1 (f −1 ◦ w)(y) dy ≤ (f ◦ (f −1 ◦ w))(y) dy f x 0 x 0 1 x w(y) dy. = x 0 Multiplying both sides by g(x), integrating with respect to x, and using Tonelli’s theorem gives x
x
∞ ∞ 1 1 −1 g(x)f (f ◦ w)(y) dy dx ≤ g(x) w(y) dy dx x 0 x 0 0 0
∞ ∞ dx dy. w(y) g(x) = x 0 y It suffices to replace w with f ◦ v.
Remark C.40. Taking g(x) = x1 and f (t) = tp , where 1 ≤ p < ∞ in the previous theorem gives x
p ∞ ∞ −p−1 x v(y) dy dx ≤ x−1 v p (x) dx. (C.25) 0
0
0
An important corollary of the previous theorem is Hardy’s inequality. Theorem C.41 (Hardy’s inequality). Let u : [0, ∞) → [0, ∞] be a measurable function, where 0 ≤ a < b ≤ ∞, and let 1 ≤ p < ∞ and 0 < s < ∞. Then ∞
p 1/p
1/p x ∞ p −s−1 −s+p−1 p x u(y) dy dx ≤ x u (x) dx (C.26) s 0 0 0 and
∞
x
(C.27) 0
p
∞
s−1
u(y) dy x
1/p dx
p ≤ s
1/p
∞
x 0
s+p−1 p
u (x) dx
.
C.7. Hausdorff Measures
699
Proof. Take v(x) := u(x1/a )x1/a−1 in (C.25), where a > 0. By the changes of variables t = y 1/a and z = x1/a we have x x
p
p ∞ ∞ −p−1 −p−1 1/a 1/a−1 x v(y) dy dx = x u(y )y dy dx 0
0
0
0
p x1/a −p−1 x u(t) dt dx a = 0 0
p z ∞ −pa−1 =a z u(t) dt dz, a
∞
0
0
while by the changes of variables z = x1/a , ∞ ∞ −1 p −1 p 1/a p/a−p x v (x) dx = x u (x )x dx = a 0
0
∞
z p−pa−1 up (z) dz.
0
Combining these two inequalities from (C.25) we get z
p ∞ ∞ −pa−1 a z u(t) dt dz ≤ z p−pa−1 up (z) dz. 0
0
0
It suffices to take a := s/p. Exercise C.42. Prove (C.27). Hint: Take u(x) = v(x−1 )x−2 in (C.26).
C.7. Hausdorff Measures In this section we introduce the Hausdorff measure in a metric space. In RN , loosely speaking, the Hausdorff measure is a measure that is adapted to measure sets of lower dimensions in RN , say, a curve in the plane or a surface in R3 . It is also used to measure fractals. Given a metric space (X, d), for 0 < δ ≤ ∞ and for each set E ⊆ X we define ∞ ∞ (C.28) Hδs (E) := inf cs (diam En )s : E ⊆ En , diam En < δ , n=1
n=1
where, when s = 0, we sum only over those En = ∅. Here, cs > 0 is a renormalization constant. In many books cs is taken to be 1. However, in RN it is more convenient to take cs as in (C.30) below. The advantage is that with this choice for s ∈ {1, . . . , N } the Hausdorff measure coincides with the surface measure on a k-dimensional manifold if s < N and with the Lebesgue measure LN when s = N . By Proposition B.2, Hδs is an outer measure. We next define the Hausdorff outer measure. Since for each set E ⊆ X the function δ → Hδs (E) is decreasing, there exists (C.29)
Hos (E) := lim Hδs (E) = sup Hδs (E). δ→0+
δ>0
700
C. The Lebesgue and Hausdorff Measures
Hos is called the s-dimensional Hausdorff outer measure of E. Exercise C.43. Prove that if X = RN in the definition (C.28) it is possible to restrict the class of admissible sets in the covers {En }n to closed and convex sets (open and convex, respectively) and that the condition diam En < δ can be replaced by diam En ≤ δ, without changing the value of Hos (E). If X is a normed space, it follows from (C.28) and (C.29) that for all E ⊆ X, x ∈ RN , and t > 0, we have Hos (x + E) = Hos (E),
Hos (tE) = ts Hos (E).
Another useful property of Hos is the following result whose proof is left as an exercise. Proposition C.44. Let (X, dX ) and (Y, dY ) be metric spaces, let E ⊆ X and let u : E → Y be a Lipschitz continuous function. Prove that for all s > 0,1 Hos (u(E)) ≤ (Lip u)s Hos (E). Proposition C.45. Let (X, d) be a metric space and let 0 ≤ s < ∞. Then Hos is an outer measure. By Carath´eodory’s theorem, Hos restricted to the σ-algebra of all Hos measurable subsets of X is a complete measure denoted Hs and called the s-dimensional Hausdorff measure. Using Proposition B.19 we have Proposition C.46. Let (X, d) be a metric space. For 0 ≤ s < ∞ the outer measure Hos is a metric outer measure, so that every Borel subset of X is Hos -measurable. Remark C.47. We remark that unlike Hos , Hδs is not a metric outer measure in general. As an example, consider in RN the outer measure Hδ0 , given by ∞ 1: E⊆ En , diam En < δ , E ⊆ RN . Hδ0 (E) := inf En =∅
n=1
If we take E1 = {x} and E2 = {y}, where 0 < x − y < δ, then to cover E1 ∪ E2 it is enough to consider one set, so Hδ0 (E1 ∪ E2 ) = 1, while Hδ0 (E1 ) = 1, Hδ0 (E2 ) = 1. Thus, Hδ0 (E1 ∪ E2 ) = 1 < 2 = Hδ0 (E1 ) + Hδ0 (E2 ). Proposition C.48. Let (X, d) be a metric space, let E ⊆ X, and let 0 ≤ s < t < ∞. 1 Hence, on the right-hand side we have the Hausdorff outer measure in Y and on the left-hand side we have the Hausdorff outer measure in X. We use the same notation.
C.7. Hausdorff Measures
701
(i) If Hδs (E) = 0 for some 0 < δ ≤ ∞, then Hos (E) = 0. (ii) If Hos (E) < ∞, then Hot (E) = 0. (iii) If Hot (E) > 0, then Hos (E) = ∞. In view of Proposition C.48 the following definition makes sense. Definition C.49. Let (X, d) be a metric space. The Hausdorff dimension of a set E ⊆ X is defined by dimH (E) := inf{0 ≤ s < ∞ : Hos (E) = 0}. Note that dimH may be any number in [0, ∞], not necessarily an integer number. Moreover, if t := dimH (E), then by the previous proposition we have that Hos (E) = 0 for all s > t, while if t > 0, then Hos (E) = ∞ for all 0 < s < t.‘ We now restrict our attention to the case X = RN (with the Euclidean norm). For 0 ≤ s < ∞ we define (C.30)
cs :=
π s/2 , 2s Γ(1 + s/2)
where Γ(t) is the Euler Gamma function ∞ Γ(t) := e−x xt−1 dx,
0 < t < ∞.
0
Note that Γ(n) = (n − 1)! for all n ∈ N. We remark that 2N cN = αN := π N/2 N Γ(1+N/2) , where αN is the Lebesgue measure of the unit ball in R , so that LN (B(x, r)) = αN rN for every ball B(x, r) in RN . Theorem C.50. Let X = RN and let 0 ≤ s < ∞. Then (i) Ho0 is the counting measure, (ii) HoN = LN o , (iii) Hos ≡ 0 if s > N . Remark C.51. In view of Theorem C.50, it follows that the Hausdorff outer measures Hos are of interest only when 0 < s < N . Proposition C.52. We remark that Proposition C.48(ii) and Theorem C.50(iii) imply that dimH (E) ≤ N for every set E ⊆ RN . Exercise C.53. Give a direct, simple proof of the fact that Ho1 = L1o . Hint: Show that Hδ1 = L1o for all δ > 0. As a consequence of Proposition C.48 we have the following result. Proposition C.54. Let X = RN and let 0 < s < N . Then the Hausdorff measure Hs is not σ-finite.
Appendix D
Notes Deciphering Academese, I: “To the best of the author’s knowledge” = “We were too lazy to do a real literature search.” — Jorge Cham, www.phdcomics.com
Chapters 1, 2, and 3 draw upon the book of van Rooij and Schikhof [237]1 . Chapter 1: For a proof of Lebesgue’s differentiation theorem which relies only on the definition of Lebesgue outer measure we refer to the paper of Faure [74] (see also [29] for another proof). For more information and extensions on the Weierstrass function (see (1.2)) we refer to the paper of Hardy [108], Krantz [135], and the recent paper of Pinkus [194]. We refer to the paper of Dovgoshey, Martio, Ryazanov, and Vuorinen [65] and the references contained therein for an extensive treatment of the Cantor function. For Exercise 1.32 see [12]. For Exercise 1.38 see [127]. Theorem 1.37 is due to Freilich [85], while Exercise 1.39 is due to Tak´ acs [227] (see also [202] and [118]). Exercise 1.40 is due to Pompeiu ([190]). Chapter 2: Exercises 2.25 and Exercise 2.26 are based on papers of Gehring [93] and Wiener [247], while Exercise 2.30 is due to Heuer [115]. For Exercise 2.33 see [23]. Theorem 2.40 is due to Weil [243] (see also Katznelson and Stromberg [129] for an explicit construction). The proof of the Helly selection theorem (Theorem 2.44) follows [181]. For Exercises 2.43 and 2.50 see [5]. Corollary 2.51 is proved in [252]. Theorem 2.55 and Exercise 2.57 are due to Josephy [126]. Theorem 2.60 is due to Banach [18] in the real-valued case, while Theorem 2.59, Lemma 2.61 and Theorem 2.60 in the general case are due to Federer [75].
1 This
is a really nice book that is not as well known as it should be. I highly recommend it.
703
704
D. Notes
Chapter 3: For Exercise 3.3 see [181]. Exercise 3.14 is due to Botsko [30]. The author would like to thank Jan Mal´ y for useful conversations on the proof of Theorem 3.29. Section 3.3 draws from papers of Goffman [96], Van Vleck [238], and Varberg [239]. Exercise 3.32 is based on a paper of Gehring [93]. For Exercises 3.45 and 3.46 see [181], while Exercise 3.50 is based on a paper of Lindner [144]. Theorem 3.59 was first proved by Krzy˙zewski [136] and then later independently by Serrin and Varberg [210] (see also [97] for some extensions). Theorem 3.55 is adapted from a paper of Josephy [126]. An alternative proof of Lemma 3.60 in the Sobolev setting due to A. Ancona may be found in [28]. Example 3.69 and Theorem 3.73 are due to Marcus and Mizel [159] (see also [9], [140], and [175]). Theorem 3.75 and all its corollaries were proved by Serrin and Varberg [210]. The proof of Corollary 3.90 uses some ideas of [203]. We refer also to the papers of Morse [176] and of Kober [133] for more information on singular functions. Chapter 4: Theorem 4.16 and Exercise 4.18 follow [102]. For Theorem 4.19 see [142]. The case in which Ψ is even was first proved independently by Chiti [43] and by Crandall and Tartar [53]. We refer to the paper of Hajaiej [103] for extensions to the case in which Ψ is real-valued. Exercise 4.20 is based on [43]. The proof of Theorem 4.22 is adapted from the review paper of Talenti [229] (see also the paper of Duff [66]). Exercise 4.24 is based on a paper of Cianchi [47]. We refer to the papers of Cianchi [47] and Dahlberg [56] for more information on the regularity of (u∗ ) . Corollary 4.25 is due to Novak [186]. Exercises 4.30–4.33 are based on a paper of Ryff [198], (see also [48] and [78]). For more information on the topics of this chapter and for an extensive bibliography we refer to the monographs of Kawohl [130] and Kesavan [131]; see also the book of Lieb and Loss [142] and the recent review paper of Talenti [230]. Chapter 5: The first two sections of Chapter 5 draw upon the review paper of Cesari [42]. Theorem 5.8 is due to Hilbert [119]. Our proof is due to Moore [174]. The first example of this type was given by Peano in 1890 [188]. Theorem 5.19 is due to Almgren and Lieb [8]. For Theorem 5.38 see [12].
D. Notes
705
Section 5.4 draws upon the books of Ambrosio and Tilli [12] and of Falconer [73]. Our proof of the Jordan curve theorem (Theorem 5.45) is due to Maehara [148] (see also the paper of Tverberg [235] for a different proof which does not rely on Brouwer fixed point theorem). Chapter 6: This chapter draws upon the books of Dinculeanu [63], Dunford and Schwartz [67], and Saks [201]. For Theorem 6.14 see the paper of Winter [248]. Exercises 6.16 and 6.17 were proved by Hewitt [117] to which we refer for extensions to the case of unbounded intervals. Chapter 7 For more information on functions in BV (Ω) we refer to the monographs of Ambrosio, Fusco, and Pallara [10], Evans and Gariepy [72], and Ziemer [251]. The proof of Lemma 7.4 is adapted from [33]. For Exercise 7.6 see the lecture notes of Dal Maso [57]. For more information on Sobolev functions of one variable we refer to the monographs of Brezis [33] and Burenkov [39]. The inequality (7.17) can be found in the book of Hardy, Littlewood, and P´olya [109, page 29]. Exercise 7.22 is based on a paper of Helmberg [113]. Exercise 7.32 is due to Lu and Wheeden [147]. Proposition 7.33 was proved by Chua and Wheeden [45]. Section 7.3 draws upon the paper of Gabushin [88]. Chapter 8: Sections 8.1 and 8.2 draw upon [62], [63], [67], [69], [83], and [206]. Theorem 8.24 is due to Simon [213]. For more information on the material in Sections 8.3, 8.4, and 8.5 and its applications to abstract evolution equations we refer to the monographs of Barbu [19], Brezis [34], and Lions [145]. For the notion of metric derivative (see Definition 8.36) and its applications we refer to the books of Ambrosio, Gigli and Savare [11] and Ambrosio and Tilli [12]. Theorem 8.62 was proved by Aubin [14] and Lions [145, Chapter 5], under the additional hypothesis that Y0 and Y2 are reflexive, and by Simon [213] in the general case. Chapter 9: Theorem 9.14 is due to Rademacher [196]. The proof presented here follows the lecture notes of Lang [137]. Theorem 9.16 is due to Stepanoff [223]. The proof presented here is due to Mal´ y [154]. Theorem 9.17 is essentially due to Haslam-Jones [110]. The proof presented here is based on an argument of Federer (see [220, page 268]) and its details are carried out in [164]. Exercise 9.19 is taken from the paper of Marcus and Mizel [160] (see also the paper [208] of Serrin). Sections 9.3 and 9.4 draw upon [155] (see also [10], [72], [149], [251]).
706
D. Notes
The proof of the Cauchy–Binet formula in Theorem 9.25 is due to Tao [231]. For a proof of Lemma 9.31 in the case k = N , which does not make use of Hausdorff measures, we refer to Flett [81]. Exercises 9.38–9.42 are taken from the paper of Lax [138] to which we refer for some historical background and references on the Brouwer fixed point theorem [36]. Theorems 9.43, 9.44, 9.46 are due to Whitney [244]. Theorem 9.52 is due to Varberg [241] (see also Rudin [199] and the paper of Schwartz [207]). We refer to the paper of Hajlasz [104] and the references therein for significant improvements. In recent years there has been a renewed interest in absolutely continuous functions of several variables and the Lusin (N ) property. In [153], Mal´ y proposed an alternative definition. We refer to the papers of Cs¨ornyei [54] and of Hencl and Mal´ y [114] for more information and references on this topic. Exercise 9.60 is taken from a paper of Fraenkel [84] to which we refer for more information about different definitions of regular domains (see also the monograph of Delfour and Zol´esio [60]). Chapter 10: The material on distributions is based on [69], [200], and [249]. For Exercise 10.4 see the lecture notes of Acquistapace [3]. The second part of the chapter, starting from Section 10.4 draws upon [16], [98], [99], [142], and [221]. Chapter 11: For more information on Sobolev functions we refer to the monographs [7], [39], [72], [165], and [251]. Exercise 11.16 is based on a paper of Simader [212]. Exercise 11.19 is due to Nicol´ as Garc´ıa Trillos and Matteo Rinaldi, Exercises 11.20 and 11.21 to Ian Tice. Theorem 11.24 is due to Meyers and Serrin [169]. Theorem 11.34 follows a paper of Fraenkel [84] (see also the monograph of Delfour and Zol´esio [60]). For the proof of Theorem 11.40 see Hild´en [120]. Theorem 11.43 is due to Hajlasz and Kalamajska [105]. Exercise 11.48 is due to Kolsrud [134]. For Exercise 11.49 see [232]. Exercises 11.58, 11.59, and 11.60 are based on a paper of Kinnunen [132]. The first step of the proof of Theorem 11.75 is due to Stein (see the paper of Brezis [35]). Theorem 11.75 is quite important for two different reasons. This characterization is often used to prove higher regularity of solutions of partial differential equations (see, e.g., [33], [71], and [94]). It is also important because it allows us to give a definition of Sobolev spaces
D. Notes
707
that does not involve derivatives. Such characterizations have been studied extensively in recent years, since they can be used to define Sobolev spaces on metric spaces. We refer to the books of Ambrosio and Tilli [12] and Hajlasz and Koskela [107] and to the recent survey paper of Heinonen [112] for more details on this subject. Chapter 12: Step 4 of the proof of the Sobolev–Gagliardo–Nirenberg embedding theorem follows the paper of Maggi and Villani [150]. For a proof of the results mentioned in Remark 12.7 we refer to the papers of Aubin [15] and Talenti [228]. For sharp forms of the Sobolev–Gagliardo–Nirenberg inequality we refer to the recent work of Cianchi, Fusco, Maggi, and Pratelli [50] and to the references contained therein. Exercises 12.15 (i) and (ii) and 12.25 are based on [84]. For Exercise 12.15(iii) see [13]. The proof of the Rellich–Kondrachov theorem is adapted from a paper of Serrin [209]. For Exercise 12.24 see [209]. Exercise 12.28 and the first part of Exercise 12.29 are based on a paper of Strauss [224], while the second part of Exercise 12.29 is based on a paper of Ebihara and Schonbek [68]. The author would like to thank Bill Hrusa for suggesting the reference [224]. The proof of Theorem 12.33 is adapted from [165]. Results similar to Theorem 12.40 in bounded domains are due to Pokhozhaev [189], Trudinger [234] and Yudovich [250] when m = 1 and to Strichartz [225] for m ≥ 2 (see also the work of Moser [179] and the recent paper of Li and Ruf [141] for some recent results). Corollary 12.50 and Exercise 12.45 follow a paper of Serrin [208]. Corollary 12.52 follows a paper of Marcus and Mizel [158]. In [156] Mal´ y and Martio have constructed a continuous transformation Ψ ∈ W 1,N (RN ; RN ) that does not satisfy the (N ) property and whose Jacobian is zero LN -a.e. in RN . Section 12.4 is based on paper of Bourdaud [31]. For Exercise 12.68 see [165]. Theorem 12.69 is due to Marcus and Mizel [161] and [162]. Theorem 12.70 and Exercise 12.71 are due to Dahlberg [55]. All the other theorems are due to Bourdaud [31]. Section 12.5 is based on the papers of Gagliardo [91] and Nirenberg [184]. The author would like to thank Ian Tice for his crucial help and endless discussions on this section. The proof of Proposition 12.88 is due to Hormander [121]. Chapter 13: A necessary and sufficient condition on the regularity of ∂Ω for the existence of an extension operator is still missing except in the twodimensional case (see the paper of Jones [122]). For Exercises 13.1 and 13.2
708
D. Notes
see the book of Maz ja [165]. The proof of Theorem 13.6 is adapted from a paper of Lieberman [143]. Exercises 13.7 and 13.14 and the proof of Theorem 13.17 follow the book of Stein [220]. Definition 13.11 is also from the book of Stein [220], but we added the condition that {Ωn }n is locally finite, since we were not able to show that conditions (i)–(iii) imply that {Ωn }n is locally finite and we used this property in the proof of Theorem 13.17. We refer to the book of Ziemer [251] for an extensive treatment of Poincar´e’s inequality. Exercise 13.30, Proposition 13.34, and Theorems 13.41 and 13.46 are due to Smith and Stegenga [214]. Exercises 13.22 and 13.31 were suggested by Ian Tice and Bill Hrusa, respectively. Theorem 13.36 follows a paper of Chua and Wheeden [46]. The idea of the proof comes from the paper of Bebendorf [22], who corrected a mistake in the original paper of Payne and Weinberger [187] in the case p = 2. See also the paper of Acosta and Dur´an [2] for the case p = 1. Section 13.3 is based on the papers of Gagliardo [91] and Nirenberg [184]. In particular, Lemma 13.52 is adapted from [91]. The author would like to thank Ian Tice for several useful discussions on this section. Chapter 14: As we already mentioned at the beginning of Chapter 14, what is covered here is just the tip of the iceberg. We refer the interested reader to the monographs [10], [72], [76], [95], [149], and [251] for more information on functions of bounded variation. For Exercises 14.5 and 14.8 see the book of Giusti [95]. Section 14.3 is based on a paper of Serrin [208]. Theorem 14.28 is due to Fleming and Rishel [79]. For sharp forms of the Sobolev–Gagliardo–Nirenberg inequality in BV (Theorem 14.33) we refer to the recent work of Fusco, Maggi, and Pratelli [87] and to the references contained therein. Theorem 14.48 is due to Sard [204], and the proof follows the book of Milnor [171]. Chapter 15: Exercise 15.8 is based on [102]. Theorems 15.23, 15.29 and 15.32 were proved independently by Hild´en [120] and Talenti [228]. We follow here the approach of Talenti [229]. Step 1 of the proof of Proposition 15.25 is due to Mal´ y and Pick [157], while Step 2 is due to Martin, Milman, and Pustylnik [163]. Theorem 15.32 has been significantly extended by Cianchi and Fusco [49]. The proof of Theorem 15.33 is due to Mal´ y and Pick [157]. Exercise 15.34 was suggested by Ian Tice.
D. Notes
709
Theorem 15.35 and Exercise 15.37 are due to Adachi and Tanaka [4]. The simplified proof of Theorem 15.35 is due to Francesco Maggi. For more information on the topics of this chapter and for an extensive bibliography we refer to the monographs of Kawohl [130] and Kesavan [131] and the review paper of Talenti [230]; see also the book of Lieb and Loss [142]. Chapter 16: This chapter draws upon the books [7], [24], [25], [40], [146], [232], and [233] to which we refer for more details about interpolation theory. Theorem 16.16 is due to Karadzhov, Milman, and Xiao [128] and its generalization to abstract settings of papers of Bourgain, Brezis, and Mironescu [32] and Maz ja and Shaposhnikova [167]. Some of the exercises at the end of Section 16.4 were suggested by Ian Tice. Exercises 16.38 and 16.39 are adapted from a paper of Maligranda and Persson [151]. Chapter 17: This chapter was written in collaboration with Ian Tice, whom I would like to thank for a wonderful summer spent discussing Besov spaces. Any mistake is purely due to me. What is written here is just the tip of the iceberg. We refer to the textbooks [7], [16], [24], [25], [26], [27], [39], [40], [146], [192], [220], [232], and [233] for more information on this subject. Although I spent many months rewriting this chapter, I am still not satisfied with the results. My initial hope was to pick the seminorm 17.5 and give simple proofs of all the results using only that seminorm. Needless to say, I failed miserably and ended up using more seminorms that I would have liked. Exercise 17.10 and similar counterexamples can be found in the paper of Taibleson [226] (see also [220]). Exercise 17.14 is a particular case of more general results contained in a paper of Brezis [35]. The suggestion for the proof came from one of the questions in http://math.stackexchange.com Theorem 17.24 is adapted from [24]. The embedding Theorems 17.49, 17.52, and 17.55 are based on a paper of Peetre [191]. Theorem 17.57 is due to Maz ja and Shaposhnikova [167]. The present proof is based on that of Karadzhov, Milman, and Xiao [128]. Theorem 17.60 was first proved by Bourgain, Brezis, and Mironescu [32] and by Ponce [195] (see also the paper of Brezis [35]). The present proof is based on that of Karadzhov, Milman, and Xiao [128]. The proofs of Theorem 17.66 and of Corollary 17.68 were suggested by Ian Tice. Theorem 17.69 is adapted from [24].
710
D. Notes
For refined inequalities of the type in Lemma 17.88 in Lorentz spaces we refer to the recent paper [17] and the references therein. Chapter 18: Theorems 18.13, 18.27, and 18.28 were first proved for bounded Lipschitz continuous domains by Gagliardo in [89]. The proof of Theorem 18.27 is due to Solonnikov [216]. The proof of Theorem 18.28 is adapted from a paper of Uspenski˘ı [236]. Exercise 18.3 is based on [84]. Appendix A: This chapter draws upon [83]. Appendix B: This chapter draws upon [83], to which we refer for the proofs of all the results that cannot be found in classical texts (such as [61], [72], [82], [199]). Appendix C: The proof of the Brunn–Minkowski inequality (see Theorem C.8) follows a survey paper of Gardner [92], to which we refer for more information and an extensive bibliography. Exercise C.10 is based on a paper of Bebendorf [22]. Exercise C.28 is due to L. Tartar. The proof of Proposition C.29 follows a paper of Hedberg [111] but is presented here in a simplified form due L. Tartar.
Appendix E
Notation and List of Symbols Deciphering Academese, II: “. . . remains an open question” = “we have no clue either.” — Jorge Cham, www.phdcomics.com
Since the number of letters and symbols (and the author’s imagination) are limited, sometimes, and when there is no possibility of confusion, we use the same letter or symbol for different objects (not in the same theorem, we hope). For example, the letter C is used for constants, but also for closed sets, while the letter α is used for multi-indices, but also as a real number. A subscript on an equation number refers to that expression in the display. For example, given (E.1)
Expression 1,
Expression 2,
(E.1)2 refers to Expression 2 in (E.1). Constants • C, c: arbitrary constants that can change from line to line and from expression to expression that can be computed in terms of known quantities. Sets • X, Y : sets or spaces; card X: the cardinality of a set X; P(X): the family of all subsets of a set X; F , G: family of sets or of functions. • ⊆: inclusion between two sets, with equality possible, ⊂: strict inclusion between two sets, • XΔY = (X \ Y ) ∪ (Y \ X): symmetric difference between the sets X and Y . 711
712
E. Notation and List of Symbols
Functional Analysis • Ω, U , V : open sets; K: a compact set; B: a Borel set; C: a closed or convex set. For two open sets U , V , U V means that the closure of U is compact and contained in V . • E, F, G: usually denote sets; ∂E: boundary of E; E ◦ : the interior of E; E: the closure of E. • χE : characteristic function of the set E. • τ : topology; · : norm. • dist: distance; diam: diameter. • ·, · X,X : duality pairing. ∗
• : weak convergence; : weak star convergence. • C(X): the space of all continuous functions; Cc (X): the space of all continuous functions whose support is compact; C0 (X): the closure of Cc (X) in the sup norm. Measure Theory • M, N: algebras or σ-algebras; M ⊗ N: product σ-algebra of M and N (not to be confused with M × N); B(X): Borel σ-algebra, Bb (X) the family of bounded Borel sets, Bc (X) the family of Borel sets compactly contained in X. • μ, ν: (positive) finitely additive measures or (positive) measures; ν ⊥ μ means that μ, ν are mutually singular measures; ν μ means that the measure ν is absolutely continuous with respect dν : the Radon–Nikodym derivative of ν with to the measure μ; dμ ∗ respect to μ; μ : outer measure. • λ: a finitely additive signed measure or a signed measure; λ+ , λ− , |λ|: the upper, lower, and total variation measures of λ. For a vector-valued measure λ, λ the total variation measure of λ. • v, and w usually denote functions or variables. • f , g, ϕ, ψ, φ, u usually denote functions; supp u: support of the function u; Lip u: Lipschitz constant of the function u; osc(u; E): oscillation of u on E; u|E the restriction of u to the set E; u ∗ v: convolution of the functions u and v, u & or F (u): the Fourier transform of u, u∨ the inverse Fourier transform; M(u) : maximal function of u; u : the distribution function of a function u; u∗ : the decreasing rearrangement of u; u : the spherically symmetric rearrangement (or Schwarz symmetric rearrangement) of a function u. • Mb (X; R): the space of all signed finite Radon measures.
E. Notation and List of Symbols
713
• ba(X, M, μ): space of all bounded finitely additive signed measures absolutely continuous with respect to μ. • Lp (X, M, μ), Lp (X, μ), Lp (X) are various notations for Lp spaces; p : H¨older conjugate exponent of p; · Lp (X,M,μ) , · Lp (X) , · Lp , or · p are various notations for the norm in Lp . Similarly, given Lp (X; Y ) we sometimes write · Lp (X) or · Lp for · Lp (X;Y ) . Lp,q (X) Lorentz spaces. LΦ (X): an Orlicz space. • Given a function u and a set E, uE = LN1(E) E u dμ is the integral average of u over a set E. Functions of One Variable • N: the set of positive integers; N0 := N ∪ {0}; Z: the integers; Q: the rational numbers; R: the real line; R := [−∞, ∞]: the extended real line; C: the complex numbers; if x ∈ R, then x+ := max{x, 0}, x− := max{−x, 0}, |x| := x+ + x− ; x : the integer part of x, sgn x the sign of x, sgn x := x/|x| for x = 0 and sgn 0 := 0. • For a complex number z = x + iy, where x, y ∈ R, the complex conjugate # of z is the number z := x − iy and the norm of z is z := x2 + y 2 . • I: an interval; D: the Cantor set. • L1o : the 1-dimensional Lebesgue outer measure; L1 : the 1-dimensional Lebesgue measure. • BP V (I): the space of all functions of bounded pointwise variation on I; AC(I): the space of all absolutely continuous functions defined on I. • Given a function u, u+ (x) and u− (x) are the right and left limits of u at a point x, uJ is the jump function of u, uC is the Cantor part of u, uAC is the absolutely continuous part of u, Var u is the pointwise variation of u, V (or V u ) is the indefinite pointwise variation of u, PVar u and NVar u are the positive and negative pointwise variations of u, Varp u is the p-variation of u, essVar u is the essential variation of u, Nu (·; E) is the Banach indicatrix (or counting function) of u over a set E, and D − u(x), D− u(x), D+ u(x), and D + u are Dini’s derivatives of u at x. • μu : the Lebesgue–Stieltjes measure generated by an increasing function u; λu : the Lebesgue–Stieltjes signed measure generated by a function u with bounded pointwise variation. • L(γ): the length of a curve γ.
714
E. Notation and List of Symbols
Functions of Several Variables • RN : the N -dimensional Euclidean space, N ≥ 1, for x = (x1 , . . . , xN ) and y = (y1 , . . . , yN ) ∈ RN in RN ,1 x · y = N i=1 xi yi the Euclidean √ inner product of x and y, x = x · x the Eucliden norm of x. When working with Euclidean spaces of different dimension, we will N sometimes use the notation xN for x. RN + = {x ∈ R : xN > N N 0}, R− = {x ∈ R : xN < 0}. Given x = (x1 , . . . , xN ) ∈ RN , for every i = 1, . . . , N we denote by xi the (N − 1)-dimensional vector obtained by removing the ith component from x and with an abuse of notation we write (E.2)
x = (xi , xi ) ∈ RN −1 × R. When i = N , we will also use the simpler notation
(E.3)
x = (x , xN ) ∈ RN −1 × R. • δij : the Kronecker delta; that is, δij := 1 if i = j and δiN := 0 otherwise. • ei , i = 1, . . . , N ,: the unit vectors of the standard (or canonical) orthonormal basis of RN . • det: determinant of a matrix or a linear mapping. • Ω: an open set of RN (not necessarily bounded); ν: the outward unit normal to ∂Ω; BN (x, r) (or simply B(x0 , r)): open ball in RN of center x0 and radius r; QN (x0 , r) := x0 + (− 2r , 2r )N , (or simply Q(x0 , r)); S N −1 : unit sphere ∂B(0, 1) in RN . N • LN o : the N -dimensional Lebesgue outer measure; L : the N N dimensional Lebesgue measure; αN := L (B(0, 1)),
• Hos : the s-dimensional Hausdorff outer measure; Hs : the s-dimensional Hausdorff measure. βN is the surface area of the unit sphere, 2π N/2 . Hence, βN = N αN . that is, βN := HN −1 (S N −1 ) = Γ(N/2) • Given E ⊆ RN and Ψ : E → RM , dΨ(x) is the differential of Ψ at x, ∂Ψ ∂Ψ ∂xi (x) is the partial derivative of Ψ at x0 with respect to xi , ∂v (x) is the directional derivative of Ψ at x in the direction v ∈ S N −1 , ≥ 2, JΨ (x) = ∇Ψ(x) ∇Ψ(x) is the gradient of Ψ at x0 , and for M # t (x)J (x)) is is the Jacobian matrix at x and |||JΨ (x)||| := det(JΨ Ψ the Jacobian of Ψ at x. • C m (Ω): the space of all functions that are continuous together with their partial derivatives up to order m ∈ N0 , C ∞ (Ω) := 1 When there is no possibility of confusion, we will also use x , x , etc., to denote different 1 2 points of RN . Thus, depending on the context, xi is either a point of RN or the ith coordinate of the point x ∈ RN .
E. Notation and List of Symbols
715
∞
m m ∞ m m=0 C (Ω); Cc (Ω) and Cc (Ω): the subspaces of C (Ω) C ∞ (Ω), respectively, consisting of all functions with compact
and supolder continuous functions port; C 0,a (Ω): the space of all bounded H¨ with exponent 0 < a ≤ 1.
α • For a multi-index α = (α1 , . . . , αN ) ∈ NN 0 ,∂ =
∂ |α| ∂xα
:=
∇m u
(E.4)
∂ |α| α α ∂x1 1 ...∂xNN
,
the vector of compo|α| := α1 + · · · + αN : the length of α; α nents ∂ u, Mm is the number of all multi-indices α with |α| = m; ∂in := ∂ nei . Given two multi-indices α = (α1 , . . . , αN ) and β = (β1 , . . . , βN ), we write α ≤ β if αi ≤ βi for all i = 1, . . . , N . Also, α! := α1 ! . . . αN ! and α α! . := β!(α − β)! β αN α1 α Given x ∈ RN and α ∈ NN 0 we write x := x1 · · · xN .
• ∇k u: to be the vector of components ∂ α u, |α| = k; Mk , ∇0 u := u. • DK (Ω): the set of all functions in Cc∞ (Ω) with support in the compact set K; D(Ω): the space Cc∞ (Ω) endowed with a particular topology τ ; D (Ω): the topological dual of D(Ω); S(RN ): the space of rapidly decreasing functions; S (RN ): the topological dual of S(RN ), called the space of tempered distributions, rapidly decreasing functions. • In Chapter 10, T : a distribution or a tempered distribution; δx0 : α the delta Dirac with mass at x0 ; ∂∂xTα or ∂ α T : the derivative of T with respect to the multi-index α; supp T : the support of T ; T ∗ φ: the convolution of T with the function φ. • W m,p (Ω): a Sobolev space; · W m,p (Ω) : the norm in W m,p (Ω); for |α|
u ∈ W m,p (Ω), ∂ α u or ∂∂xαu : the αth weak (or distributional) par∂u : the weak (or distributional) partial tial derivative of u, ∂i u or ∂x i derivative of u with respect to xi ; |u|W m,p (Ω) := ∇m uLp (Ω;RMm ) ; ˙ m,p (Ω): the homogeneous H m (Ω): the Hilbert space W m,2 (Ω); W Sobolev space; (W m,p (Ω)) : the topological dual of W m,p (Ω); W0m,p (Ω): the closure of Cc∞ (Ω) in W m,p (Ω); W −m,p (Ω): the topological dual of W0m,p (Ω); E : W m,p (Ω) → W m,p (RN ): an extenp , sion operator; Tr(u): the trace of a Sobolev function; p∗ := NN−p p∗m,k :=
Np N −(m−k)p :
Sobolev critical exponents.
• ϕε : a standard mollifier; uε := ϕε ∗ u: the mollification of u. • PA: a particular family of continuous piecewise affine functions. • dreg : the regularized distance.
716
E. Notation and List of Symbols
• BV (Ω): the space of functions of bounded variation. Given u ∈ BV (Ω), Di u is the weak (or distributional) partial derivative of u with respect to xi , Du is the weak (or distributional) gradient of u, Du is the total variation measure of the measure Du; V (u, Ω): the variation of a function u in Ω; P(E, Ω): the perimeter of a set E in Ω. • For x, h ∈ RN , Δh u(x) := u(x + h) − u(x), the forward differm−1 u(x)); ∼ Δ2h u(x) := u(x + h) − ence operator; Δm h u(x) := Δh (Δh s,p N 2u(x) + u(x − h); Bq (R ): a Besov space; B˙ qs,p (RN ): a homoge(n) [R] , | · |B s,p (RN ) , neous Besov space; | · |Bqs,p (RN ) , | · |B s,p (RN ) , | · |∞ B s,p (RN ) q
q
q
: various seminorms in Bqs,p (RN ); Λ1 (RN ): the Zygmund | · |∨ Bqs,p (RN ) space.
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Index
algebra, 652 atom, 654 ball open, 638 base for a topology, 636 local, 636 basis orthonormal, 240 boundary Lipschitz, 273 uniformly of class C m , 424 uniformly of class C m,α , 424 uniformly Lipschitz, 423 Cantor diagonal argument, 49 Cantor part of a function, 104 Cauchy–Binet formula, 251 change of variables, 166 for multiple integrals, 270 coarea formula, 478 cofactor, 260 compact embedding, 366 connected component, 147 exterior, 152 interior, 152 convergence almost everywhere, 673 almost uniform, 673 in measure, 673 in the sense of distributions, 288 weak, 645
weak star, 646 convolution, 302 of a distribution, 305, 308 coordinates background, 273 local, 273 countability first axiom, 636 second axiom, 636 curve, 133 arclength, 143 closed, 134 closed simple, 134 continuous, 133 equivalent, 133 length, 136 locally rectifiable, 137 parameter change, 133 parametric representation, 133 range, 134 rectifiable, 136 simple arc, 134 simple point, 134 cut-off function, 693 delta Dirac, 289 derivative, 9 ∇k , 321 Dini, 14 ∂ directional ∂ν , 239 distributional, 183, 188, 229, 291, 319 directional, 320 distributional partial , 459
729
730
dm , dx
224 of a distribution, 290 directional, 291 partial ∂i , 239 Radon–Nikodym, 665 weak, 183, 188, 229, 291, 319 directional, 320 weak partial, 459 differentiability, 9 distance, 638 regularized, 414 distribution, 288 infinite order, 289 order of, 289 dual spaces D (Ω), 288 Mb (X; R), Radon measures, 676 of W 1,p (Ω), 345 W −1,p (Ω), 348 S (RN ; C), 299 duality pairing, 642 embedding, 356, 643 compact, 645 equi-integrability, 674 essential infimum, 112 essential supremum, 112, 668 evolution triple, 233 extension domain for BV (Ω), 483 for W m,p (Ω), 365 extension operator, 365 finite cone, 424 Fourier transform, 309 inverse, 309 of a tempered distribution, 315 inverse, 315 function absolutely continuous, 67 Banach indicatrix, 59 Bochner integrable, 208 Borel, 655 Cantor, 22 characteristic, χ, 657 continuous, 637 counting, 59 cut-off, 693 decreasing, 3 decreasing rearrangement, 502 distribution, 111, 497 equi-integrable, 82
Index
equivalent, 668 Gamma, 701 H¨ older continuous, 10 increasing, 3 inverse of a monotone, 6 jump, 5, 6 Lebesgue integrable, 660 Lebesgue measurable, 683 linear adjoint, 249 diagonal, 250 orthogonal, 249 positive definite, 250 rotation, 250 symmetric, 250 Lipschitz continuous, 9, 242 locally absolutely continuous, 68 locally integrable, 660 maximal, 694 measurable, 655, 657 measure-preserving, 130 monotone, 3 of bounded pointwise variation, 30 in the sense of Cesari, 471 of bounded variation, 459 radial of a star-shaped domain, 441 rearrangement decreasing, 114, 498 Schwarz symmetric, 499, 502 spherically symmetric, 499 saltus, 5 simple, 205, 657 singular, 103 Sobolev, 188, 230, 320 strictly decreasing, 3 strictly increasing, 3 strongly measurable, 205 subharmonic, 291 testing, 284 vanishing at infinity, 111, 356, 498 weakly measurable, 205 weakly star measurable, 206 Weierstrass, 9 function space ACp ([a, b]), 83 AC(I), 68 ACloc (I), 68 ACloc (I; Y ), 68 AC(Ω; Y ), 68 AC(I; Y ), 68
Index
Bqs,p (RN ), 539 Besov B s,p (∂Ω, HN −1 ), 613, 615, 625 B˙ qs,p (RN ), 556 BMO, 695 BV P (I), 30 BV P (I; X), 30 BV P (Ω; X), 31 BV Ploc (I), 30 BV Ploc (I; X), 30 BV (Ω), 183, 459 BVloc (Ω), 188 C m,α (Ω), 343 C(X; Y ), 637 C0 (X), 644 c0 , 221 Cc (X), 644 C ∞ (Ω), 242 Cc∞ (Ω), 242 C m (Ω), 242 Ccm (Ω), 242 Cc (X), 637 D(Ω), 284 DK (Ω), 281 H m (Ω), 190 H m (Ω; RM ), 190, 230 H m (Ω), 322 L∞ (X), 668 Lp,q , Lorentz, 502 Lp (X), 668 Lploc , 671 Lpw , weak Lp , 504 LΦ (E), 376 PA, 333 S(RN ; C), 298 W m,p (Ω), 189, 320 W m,p (Ω; RM ), 188, 320 W m,p (Ω; Y ), 230 ˙ m,p (Ω), 323 W m,p (Ω), 189, 320 Wloc m,p (Ω; RM ), 189, 320 Wloc m,p Wloc (Ω; Y ), 230 W0m,p (Ω), 322 Zygmund Λ1 (RN ), 541 functional locally bounded, 677 positive, 677 gauge, 641 Hausdorff dimension, 701 H¨ older’s conjugate exponent, 669
731
identity Parseval, 311 Plancherel, 311 immersion, 643 inequality Brunn–Minkowski, 683 Cauchy, 648 Hardy, 698 Hardy–Littlewood, 120, 501 H¨ older, 669 isoperimetric, 486, 685 Jensen, 661 Minkowski, 671 for integrals, 670 Poincar´e, 193 for continuous domains, 434 for convex sets, 436 for rectangles, 434 for star-shaped sets, 441 in BV , 486 in W0m,p , 430 in W m,p , 432 weighted, 194 Young, 303, 669 Young, general form, 304 infinite sum, 4 inner product, 648 Euclidean, 649 integral Bochner, 208 Lebesgue of a nonnegative function, 658 of a real-valued function, 660 of a simple function, 658 integrals depending on a parameter, 662 integration by parts, 78, 278 Riemann, 73 interval, 3 partition, 29 Jacobian, 251 Jacobian matrix, 251 Laplacian, 291 Leibnitz formula, 288 lemma Fatou, 210, 660 Riemann–Lebesgue, 312 length distance, 146 length of a curve σ-finite, 137
732
line integral, 151 Lipschitz constant, 10 Littlewood–Paley decomposition dyadic block, Λ˙ k , 316 homogeneous, 317 locally finite, 637 Lusin (N ) property, 84, 383 measure, 653 σ-finite, 654 absolute continuous, 665 absolutely continuous part, 668 Borel, 653 Borel regular, 676 complete, 654 counting, 659 finite, 654 finitely additive, 653 Hausdorff, 700 Lebesgue, 681 Lebesgue–Stieltjes, 163 lower variation, 667 metric outer, 655 nonatomic, 654 outer, 651 Hausdorff, 700 Lebesgue, 12, 681 Lebesgue–Stieltjes, 164 product, 663 purely atomic, 654 Radon, 157, 676 signed, 666 absolutely continuous, 667 bounded, 667 signed finitely additive, 666 signed Radon, 676 singular part, 668 total variation, 460, 667 upper variation, 667 vector-valued, 667 Radon, 677 vector-valued Lebesgue–Stieltjes, 173 measure space, 653 measures mutually singular, 665, 667 metric, 638 metric space length space, 146 Minkowski content lower, 685 upper, 685
Index
Minkowski functional, 641 mollification, 687 mollifier, 687 standard, 688 multi-index, 241 multiplicity of a point, 134 N -simplex, 332 neighborhood, 636 norm, 643 equivalent, 644 Euclidean, 643 normal vector, 274 operator bounded, 642 compact, 644 linear, 641 outer measure product, 663 p-equi-integrability, 674 parallelogram law, 648 parameter of a curve, 133 partition of unity smooth, 692 perimeter of a set, 461 point accumulation, 635 Lebesgue, 679 of density one, 679 of density t, 679 p-Lebesgue, 679 pointwise variation, 29 essential, 187 indefinite, 36 locally bounded, 30 negative, 31 p-variation, 39 positive, 31 principal value integral, 292 quasi-norm, 645 Radon–Nikodym property, 224 regularized distance, 417 ring, 653 σ-algebra, 652 Borel, 652 product, 656, 663 σ-locally finite, 637 section, 663
Index
segment property, 328 seminorm, 641 semiring, 653 sequence Cauchy, 638, 640 convergent, 636, 638 sequentially weakly compact set, 647 set absorbing, 639 balanced, 282, 639 Cantor , 21 closed, 635 closure, 635 compact, 637 connected, 147 dense, 636 disconnected, 147 Fσ , 20 finite width, 430 Gδ , 20 Hk -rectifiable, 98 inner regular, 676 interior, 635 Lebesgue measurable, 681 μ∗ -measurable, 652 of finite perimeter, 461 open, 635 outer regular, 676 pathwise connected, 147 precompact, 637 purely Hk -unrectifiable , 98 regular, 676 relatively closed, 636 relatively compact , 637 relatively open, 636 σ-compact, 637 spherically symmetric rearrangement, 499 star-shaped, 441 symmetric difference, 169 topologically bounded, 640 shortest distance, 146 Sobolev critical exponent, 356, 361, 561 space Banach, 643 bidual, 642 complete, 638, 640 dual, 642 Hausdorff, 636 Hilbert, 648 locally compact, 637
733
locally convex, 640 measurable , 652 metric, 638 metrizable, 639 normable, 644 normal, 637 normed, 643 quasi-Banach, 645 quasi-normed, 645 (Y0 , Y1 )s,q,J real interpolation, 527 (Y0 , Y1 )s,q real interpolation, 518 reflexive, 647 seminormable, 641 seminormed, 641 separable, 636 topological, 635 topological vector, 639 vector, 639 spherical coordinates, 271 support of a distribution, 295 tangent line, 138 tangent space, 274 tangent vector, 138, 274 Taylor’s formula, 242 theorem area formula, C 1 case, 253 area formula, the differentiable case, 269 Ascoli–Arzel` a, 147 Aubin–Lions–Simon, 235 Baire category, 638 Banach, 61 Banach fixed point, 638 Banach–Alaoglu, 646 Besicovitch’s covering, 678 Besicovitch’s derivation, 678 Brouwer fixed point, 258 Carath´eodory, 654 chain rule, 95, 99 change of variables, 100, 341 change of variables, the C 1 case, 257 change of variables, the differentiable case, 270 De la Vall´ee Poussin, 180, 675 decomposition, 250 divergence, 274 Dunford–Pettis, 675 ˇ Eberlein–Smulian, 647 Egoroff, 211, 674 Faa di Bruno, 342 Fubini, 23, 664
734
fundamental theorem of calculus, 77, 82 Gagliardo, 600, 608, 609 Gagliardo–Nirenberg interpolation, 400, 403, 451, 455, 488, 489 Hahn–Banach analytic form, 642 first geometric form, 642 second geometric form, 643 Helly’s selection, 49 Hilbert, 135 integration by parts, 278 Jordan’s curve, 152 Jordan’s decomposition, 667 Josephy, 55 Kakutani, 648 Lebesgue differentiation, 11 Lebesgue dominated convergence, 210 Lebesgue’s decomposition, 666, 668 Lebesgue’s dominated convergence, 661 Lebesgue’s monotone convergence, 659 Littlewood–Paley decomposition, 317 Lusin, 676 Meyers–Serrin, 326 Morrey’s embedding in Bqσ,p , 562 in W 1,p , 381 in W m,p , 384 Muckenhoupt, 444 Peano, 135 Pettis, 206 Plancherel, 311 Rademacher, 243 Radon–Nikodym, 665 reiteration, 522 Rellich–Kondrachov, 483 Rellich–Kondrachov’s compactness, 366, 368, 378, 386 for continuous domains, 371 Riesz representation in C0 , 677 in Cc , 677 in L1 , 672 in L∞ , 672 in Lp , 671 in Lp (X; Y ), 213 in W m,∞ (Ω), 349 in W0m,∞ (Ω), 349 in W m,p (Ω), 345
Index
in W0m,p (Ω), 348 Riesz–Thorin, 673 Sard, 489, 491 Schwartz, 241 Serrin, 471 Simon, 215 Sobolev–Gagliardo–Nirenberg in Bqs,p , 561, 587 in BV , 482 in W 1,p , 356, 510 ˙ m,p , 362 in W in W m,p , 361 Stepanoff, 245 superposition, 388, 391, 393, 395 in ACloc , 92, 191 Tietze extension, 637 Tonelli, 80, 137, 664 Urysohn, 637 Vitali’s convergence , 674 Vitali’s covering, 489 Vitali–Besicovitch’s covering , 678 Weil, 45 Whitney extension, 262 Whitney extension, II, 268 Whitney’s decomposition, 262 topology, 635 weak, 645 weak star, 646 total variation norm, 672 trace of a function, 592 unit sphere S N −1 , 243 vanishing at infinity, 356 variation, 460 vertex of a symplex, 332 Whitney’s decomposition, 264
Selected Published Titles in This Series 181 180 179 178
Giovanni Leoni, A First Course in Sobolev Spaces, Second Edition, 2017 Joseph J. Rotman, Advanced Modern Algebra: Third Edition, Part 2, 2017 Henri Cohen and Fredrik Str¨ omberg, Modular Forms, 2017 Jeanne N. Clelland, From Frenet to Cartan: The Method of Moving Frames, 2017
177 Jacques Sauloy, Differential Galois Theory through Riemann-Hilbert Correspondence, 2016 176 Adam Clay and Dale Rolfsen, Ordered Groups and Topology, 2016 175 Thomas A. Ivey and Joseph M. Landsberg, Cartan for Beginners: Differential Geometry via Moving Frames and Exterior Differential Systems, Second Edition, 2016 174 Alexander Kirillov Jr., Quiver Representations and Quiver Varieties, 2016 173 Lan Wen, Differentiable Dynamical Systems, 2016 172 Jinho Baik, Percy Deift, and Toufic Suidan, Combinatorics and Random Matrix Theory, 2016 171 Qing Han, Nonlinear Elliptic Equations of the Second Order, 2016 170 169 168 167
Donald Yau, Colored Operads, 2016 Andr´ as Vasy, Partial Differential Equations, 2015 Michael Aizenman and Simone Warzel, Random Operators, 2015 John C. Neu, Singular Perturbation in the Physical Sciences, 2015
166 165 164 163
Alberto Torchinsky, Problems in Real and Functional Analysis, 2015 Joseph J. Rotman, Advanced Modern Algebra: Third Edition, Part 1, 2015 Terence Tao, Expansion in Finite Simple Groups of Lie Type, 2015 G´ erald Tenenbaum, Introduction to Analytic and Probabilistic Number Theory, Third Edition, 2015
162 Firas Rassoul-Agha and Timo Sepp¨ al¨ ainen, A Course on Large Deviations with an Introduction to Gibbs Measures, 2015 161 Diane Maclagan and Bernd Sturmfels, Introduction to Tropical Geometry, 2015 160 Marius Overholt, A Course in Analytic Number Theory, 2014 159 John R. Faulkner, The Role of Nonassociative Algebra in Projective Geometry, 2014 158 Fritz Colonius and Wolfgang Kliemann, Dynamical Systems and Linear Algebra, 2014 157 Gerald Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schr¨ odinger Operators, Second Edition, 2014 156 Markus Haase, Functional Analysis, 2014 155 154 153 152
Emmanuel Kowalski, An Introduction to the Representation Theory of Groups, 2014 Wilhelm Schlag, A Course in Complex Analysis and Riemann Surfaces, 2014 Terence Tao, Hilbert’s Fifth Problem and Related Topics, 2014 G´ abor Sz´ ekelyhidi, An Introduction to Extremal K¨ ahler Metrics, 2014
151 150 149 148
Jennifer Schultens, Introduction to 3-Manifolds, 2014 Joe Diestel and Angela Spalsbury, The Joys of Haar Measure, 2013 Daniel W. Stroock, Mathematics of Probability, 2013 Luis Barreira and Yakov Pesin, Introduction to Smooth Ergodic Theory, 2013
147 146 145 144
Xingzhi Zhan, Matrix Theory, 2013 Aaron N. Siegel, Combinatorial Game Theory, 2013 Charles A. Weibel, The K-book, 2013 Shun-Jen Cheng and Weiqiang Wang, Dualities and Representations of Lie Superalgebras, 2012
For a complete list of titles in this series, visit the AMS Bookstore at www.ams.org/bookstore/gsmseries/.
Photo by Adella Guo
This book is about differentiation of functions. It is divided into two parts, which can be used as different textbooks, one for an advanced undergraduate course in functions of one variable and one for a graduate course on Sobolev functions. The ½rst part develops the theory of monotone, absolutely continuous, and bounded variation functions of one variable and their relationship with Lebesgue– Stieltjes measures and Sobolev functions. It also studies decreasing rearrangement and curves. The second edition includes a chapter on functions mapping time into Banach spaces. The second part of the book studies functions of several variables. It begins with an overview of classical results such as Rademacher’s and Stepanoff’s differentiability theorems, ;hitney’s extension theorem, Brouwer’s ½xed point theorem, and the divergence theorem for Lipschitz domains. It then moves to distributions, Fourier transforms and tempered distributions. The remaining chapters are a treatise on Sobolev functions. The second edition focuses more on higher order derivatives and it includes the interpolation theorems of Gagliardo and Nirenberg. It studies embedding theorems, extension domains, chain rule, superposition, Poincaré’s inequalities and traces. A major change compared to the ½rst edition is the chapter on Besov spaces, which are now treated using interpolation theory.
For additional information and updates on this book, visit www.ams.org/bookpages/gsm-181
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