This book provides a gentle introduction to fractional Sobolev spaces which play a central role in the calculus of varia
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Table of contents :
Cover
Title page
Copyright
Dedication
Contents
Preface
Fractional Sobolev spaces in one dimension
Fractional Sobolev spaces in one dimension
Embeddings and interpolation
A bit of wavelets
Rearrangements
Higher order fractional Sobolev spaces in one dimension
Fractional Sobolev spaces
Fractional Sobolev spaces
Embeddings and interpolation
Further properties
Trace theory
Symmetrization
Higher order fractional Sobolev spaces
Some equivalent seminorms
Applications
Interior regularity for the Poisson problem
The fractional Laplacian
Bibliography
Index
Back Cover
A First Course in Fractional Sobolev Spaces
GRADUATE STUDIES I N M AT H E M AT I C S
229
A First Course in Fractional Sobolev Spaces Giovanni Leoni
EDITORIAL COMMITTEE Matthew Baker Marco Gualtieri Gigliola Staﬃlani (Chair) Jeﬀ A. Viaclovsky Rachel Ward 2020 Mathematics Subject Classiﬁcation. Primary 46E35; Secondary 26A33, 26A46, 30H25, 35R11.
For additional information and updates on this book, visit www.ams.org/bookpages/gsm229
Library of Congress CataloginginPublication Data Names: Leoni, Giovanni, 1967– author. Title: A ﬁrst course in fractional Sobolev spaces / Giovanni Leoni. Description: Providence, Rhode Island : American Mathematical Society, [2023]  Series: Graduate studies in mathematics, 10657339 ; 229  Includes bibliographical references and index. Identiﬁers: LCCN 2022040942  ISBN 9781470468989 (hardcover ; acidfree paper)  ISBN 9781470472535 (paperback acidfree paper)  ISBN 9781470472528 (ebook) Subjects: LCSH: Sobolev spaces.  Function spaces.  AMS: Functional analysis – Linear function spaces and their duals – Sobolev spaces and other spaces of ”smooth’” functions, embedding theorems, trace theorems.  Real functions – Functions of one variable – Fractional derivatives and integrals.  Real functions – Functions of one variable – Absolutely continuous functions.  Functions of a complex variable – Spaces and algebras of analytic functions – Besov spaces and Qp spaces.  Partial diﬀerential equations – Miscellaneous topics – Fractional partial diﬀerential equations. Classiﬁcation: LCC QA323 .L458 2023  DDC 515/.782–dc23/eng20221110 LC record available at https://lccn.loc.gov/2022040942
Copying and reprinting. Individual readers of this publication, and nonproﬁt libraries acting for them, are permitted to make fair use of the material, such as to copy select pages for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of any material in this publication is permitted only under license from the American Mathematical Society. Requests for permission to reuse portions of AMS publication content are handled by the Copyright Clearance Center. For more information, please visit www.ams.org/publications/pubpermissions. Send requests for translation rights and licensed reprints to [email protected]. c 2023 by the American Mathematical Society. All rights reserved. The American Mathematical Society retains all rights except those granted to the United States Government. Printed in the United States of America. ∞ The paper used in this book is acidfree and falls within the guidelines
established to ensure permanence and durability. Visit the AMS home page at https://www.ams.org/ 10 9 8 7 6 5 4 3 2 1
28 27 26 25 24 23
Dedicated to Irene, with friendship, and to the memory of all the victims of the pandemic
Contents
Preface
xi
Part 1. Fractional Sobolev Spaces in One Dimension Chapter 1.
Fractional Sobolev Spaces in One Dimension
3
§1.1.
Some Useful Inequalities and Identities
§1.2.
Fractional Sobolev Spaces
12
§1.3.
Extensions
22
§1.4.
Some Equivalent Seminorms
27
§1.5.
Density
33
4
§1.6.
The Space
W0s,p (I)
38
§1.7.
Hardy’s Inequality
41
§1.8.
The Space
§1.9.
Notes
Chapter 2.
s,p W00
46 48
Embeddings and Interpolation
49
§2.1.
Embeddings: The Endpoints
50
§2.2.
Embeddings: The General Case
64
§2.3.
Interpolation Inequalities
74
§2.4.
Notes
86
Chapter 3.
A Bit of Wavelets `p
87
§3.1.
Weighted
§3.2.
Wavelets in Lp
89
§3.3.
L1
92
Wavelets in
Spaces
88
vii
viii
Contents
§3.4.
Wavelets in W 1,1
93
§3.5.
Wavelets in
W s,p
96
§3.6.
Back to Interpolation Inequalities
101
§3.7.
Notes
107
Chapter 4. §4.1.
Rearrangements
109
Polarization
109 W s,p (R)
§4.2.
Polarization in
§4.3.
Symmetric Decreasing Rearrangement
§4.4.
Symmetric Decreasing Rearrangement in
§4.5.
Notes
Chapter 5.
124 127 W s,p (R)
142 144
Higher Order Fractional Sobolev Spaces in One Dimension145
§5.1.
Higher Order Fractional Sobolev Spaces
145
§5.2.
Extensions
151
§5.3.
Density
158
§5.4.
Some Equivalent Seminorms
159
§5.5.
Hardy’s Inequality
165
§5.6.
Truncation
168
§5.7.
Embeddings
170
§5.8.
Interpolation Inequalities
174
§5.9.
Notes
181
Part 2. Fractional Sobolev Spaces Chapter 6.
Fractional Sobolev Spaces
185
§6.1.
Definition and Main Properties
185
§6.2.
Slicing
201
§6.3.
Some Equivalent Seminorms
214
§6.4.
Density of Smooth Functions
222
W0s,p
§6.5.
The Space
§6.6.
Hardy’s Inequality
§6.7.
The Space
§6.8.
Density in
§6.9.
Notes
Chapter 7. §7.1.
and Its Dual
s,p W00 ˙ s,p W
Embeddings and Interpolation
Campanato Spaces
228 234 244 251 255 257 258
Contents
§7.2. §7.3. §7.4. §7.5. §7.6. §7.7. §7.8.
ix
Embeddings: The Subcritical Case Embeddings: The Critical Case Embeddings: The Supercritical Case Embeddings: The General Case The Limit of W s,p as s → 1− Interpolation Inequalities Notes
261 264 275 277 296 300 308
Chapter §8.1. §8.2. §8.3. §8.4. §8.5.
8. Further Properties Extension: Lipschitz Domains Extension: The General Case Derivatives Embeddings and Interpolation Inequalities Notes
309 309 313 329 331 335
Chapter §9.1. §9.2. §9.3. §9.4. §9.5. §9.6. §9.7.
9. Trace Theory Traces of Weighted Sobolev Spaces The Trace Operator in W s,p (Ω) HalfSpaces Special Lipschitz Domains Bounded Lipschitz Domains Unbounded Lipschitz Domains Notes
337 337 349 355 361 366 372 384
Chapter 10. Symmetrization §10.1. Polarization §10.2. Polarization in W s,p (RN ) §10.3. Spherically Symmetric Rearrangement §10.4. Spherical Symmetric Rearrangement in W s,p (RN ) §10.5. Notes
387 387 391 394 402 405
Chapter 11. Higher Order Fractional Sobolev Spaces §11.1. Definition §11.2. Some Equivalent Seminorms §11.3. Slicing §11.4. Embeddings §11.5. Interpolation Inequalities §11.6. Superposition
407 407 415 422 426 436 448
x
Contents
§11.7.
Extension
451
§11.8.
Trace Theory
454
§11.9.
Notes
479
Chapter 12.
Some Equivalent Seminorms
481
§12.1.
The Interpolation Seminorm
481
§12.2.
The Littlewood–Paley Seminorm
490
§12.3.
Notes
518
Part 3. Applications Chapter 13.
Interior Regularity for the Poisson Problem
521
§13.1.
Cacciopoli’s Inequality
521
§13.2.
H2
525
§13.3.
W 2,p
Regularity
532
§13.4.
W s,p
Regularity
541
§13.5.
Notes
Chapter 14.
Regularity
545
The Fractional Laplacian
547
§14.1.
Definition and Main Properties
547
§14.2.
The Riesz Potential
556
§14.3.
Existence of Weak Solutions
561
§14.4.
An Extension Result
567
§14.5.
Notes
575
Bibliography
577
Index
585
Preface
What are fractional Sobolev spaces and why are they important? In his blog, Terence Tao gives a beautiful answer to these questions: If you consider a simple function u of the form u = hχE , where E ⊆ RN is a Lebesgue measurable set and h is a positive constant, then the Lp norm of u is given by the number h(LN (E))1/p , where LN (E) is the volume of the set E. Hence, you measure the height h of the function u and its width. Sobolev spaces also measure the regularity of the function and how quickly it oscillates (the frequency scale of u). A function u belongs to the Sobolev space W 1,p (RN ) if u belongs to Lp (RN ), and its (weak) gradient ∇u belongs to Lp (RN ; RN ). When p = 2, it is possible to give an equivalent definition using Fourier transforms. Indeed, you can check that u belongs to W 1,2 (RN ) if and only if its Fourier transform u ˆ and kξkˆ u belong to L2 (RN ), with Z Z 2 u(x) dx = ˆ u(ξ)2 dξ, N N Z R ZR k∇u(x)k2 dx = (2πkξk)2 ˆ u(ξ)2 dξ. RN
RN
To define the fractional Sobolev space W s,2 (RN ), replace kξk with kξks , where 0 < s < 1. Intuitively, this is saying that instead of having a full derivative, you only have a fraction of a derivative. After some computations, you find the following relations Z Z 2 u(x) dx = ˆ u(ξ)2 dξ, RN RN Z Z Z u(x) − u(y)2 dx = C kξk2s ˆ u(ξ)2 dξ, N +s2 RN RN kx − yk RN xi
xii
Preface
where C = C(N, s) > 0. You then extend this definition to the case p 6= 2. You say that a function u ∈ Lp (RN ) belongs to the fractional Sobolev space W s,p (RN ) if Z Z u(x) − u(y)p dx < ∞. (0.1) N +sp RN RN kx − yk All this is quite neat, but it still does not answer why you should care. It turns out that fractional Sobolev spaces play a central role in the calculus of variations, partial differential equations, and harmonic analysis. Let me try to explain. In 1957 Gagliardo [Gag57] proved that, when 1 < p < ∞, the fractional Sobolev space W 1−1/p,p (RN −1 ) is the trace space of the Sobolev space N W 1,p (RN + ), with R+ the open halfspace {x = (x1 , . . . , xN ) : xN > 0}. The case p = 2 had been studied by Aronszajn [Aro55], Prodi [Pro56], and Slobodecki˘ı [Slo58]. For a regular function, the trace is the restriction of the function to the boundary. This result is significant because if you want to establish the existence of a weak solution to partial differential equations with non homogeneous boundary conditions, you need to consider fractional Sobolev spaces. Important examples are the non homogeneous Dirichlet or Neumann problem for the Poisson equation or the heat equation. The same happens if you want to find a minimizer or a saddle point of a functional defined on a class of functions having the same inhomogeneous trace on the boundary. This problem is central in the calculus of variations. Besides existence, fractional Sobolev spaces are essential when studying the regularity of solutions of partial differential equations. Imagine that you proved a differential function satisfies the differential equation u0 = f (u), where f is a smooth function. But then the righthand side f (u) is a differentiable function (by the chain rule), and so you learn that u0 is differentiable itself. You have gained one extra derivative. In turn, f (u) is twice differentiable and, using the equation, u0 becomes twice differentiable. You can continue in this way, depending on how smooth the function f is. This procedure is a bootstrap method. Now, if you replace the righthand side f (u) with something more complicated, or if instead of an ordinary differential equation, you have a partial differential equation, then you might not be able to gain a full derivative. But in some cases, you might be able to get a fraction of a derivative. So again, this brings you to fractional Sobolev spaces. To give an example, the existence and smoothness of solutions to Navier– Stokes equations in R3 are among the Millenium Problems of the Clay Institute. While we are still far from finding a solution, fractional Sobolev spaces have played a significant role in the theory of the Navier–Stokes equations [BCD11, Chapter 5], [CDGG06, Theorem 3.5], [LR02, Part 5], [Soh01].
Preface
xiii
Finally, fractional Sobolev spaces and Besov spaces are ubiquitous in harmonic analysis and singular integrals. Indeed, it is possible to characterize these spaces using the Littlewood–Paley decomposition. In a way, Besov cases Bqs,p are the grownup version of fractional Sobolev spaces since they have an extra parameter q. When 0 < s < 1 and p = q, Bps,p (RN ) = W s,p (RN ). Hence, fractional Sobolev spaces are a subclass of Besov spaces. And so now we come to the sticky point. Why do we need another book on fractional Sobolev spaces? There are several ways to characterize Besov spaces and fractional Sobolev spaces. Besides the intrinsic formulation given in (0.1), you can use abstract interpolation theory to represent W s,p (RN ) as an intermediate space between Lp (RN ) and W 1,p (RN ). This method has the advantage that abstract interpolation theory lets you deduce several theorems (for example, embedding theorems) from the corresponding ones for W 1,p (RN ). And then there is also the Littlewood–Paley decomposition, which carries with it all the power of Fourier analysis (for example, Fourier multipliers). Leading experts navigate seamlessly from one formulation to the next to get the most elegant proof. In a way, this is great. I don’t know of any other area where such different disciplines all converge. But at the same time, if you are an advanced undergraduate student or a beginning graduate student, reading Russian literature [BIN79] or Triebel’s monographs [Tri95], [Tri10], [Tri92], [Tri06] can be challenging. I wrote this book as a gentle introduction to fractional Sobolev spaces for advanced undergraduate students (Part I) and graduate students (Part II). In a way, my book is like The Hobbit and [BIN79], [Tri95], [Tri10], [Tri92], [Tri06] are The Lord of the Rings. With the exception of Chapter 12, all the proofs in this book rely on the intrinsic seminorm (0.1). Since the “intrinsic” approach bypasses the need for harmonic analysis, it’s more readily amenable to the situations we encounter in studying boundary value problems. To my knowledge, no one has assembled the intrinsic techniques until now. As for why this book now? In 2007 Caffarelli and Silvestre published their paper [CS07], in which they studied properties of solutions to the fractional Laplacian. Their article, which has more than two thousand citations in Google Scholar, started a ripple in mathematics. Mathematicians extended several results from the theory of elliptic partial differential equations (the theory of viscosity solutions, critical point theory, etc.) and the calculus of variations to nonlocal operators.
xiv
Preface
I should make clear that this book is not on the fractional Laplacian. I wrote one chapter about it because if you consider the Euler–Lagrange equations of the fractional seminorm (0.1), you do get the fractional Laplacian. But this is as far as I went. As a byproduct of this new interest on the fractional Laplacian, new proofs of classical results on fractional Sobolev spaces appeared in the recent literature. When the pandemic hit (in March 2020), I suddenly found myself stuck at home with two cats (Gauss and Hilbert). While they kept me sane during the months of lockdown, I was unable to concentrate on my research. Try focusing when you have two cats squabbling (for hours!) on who gets to sleep on your lap. So instead, I devoted myself to online shopping (alas) and navigating the internet. Early enough, I stumbled on the paper of Nguyen, Diaz, and Nguyen [NDN20], in which they gave a new proof of a classical embedding for fractional Sobolev spaces using only maximal functions. When I was writing the chapter on Besov spaces for the first edition of my book [Leo09], I had spent months trying to find in the literature precisely these types of proofs. For the most part, I was not satisfied with the final result. In the second edition [Leo17], I changed that chapter completely. I caved in and used abstract interpolation and the Littlewood–Paley theory. But I was still not happy about it. So when I found the paper [NDN20], I decided to type that embedding theorem and post it on the AMS website of my book, where I put corrections and related results that I find interesting. The paper [NDN20] led me to another article [BN20], and I decided to type that up as well. After that I discovered the beautiful review paper of Mironescu [Mir18]. And so here we are several hundred pages later. References: The rule of thumb here is simple: I only quoted papers and books that I read. I believe that misquoting a paper is worse than not quoting it. Hence, if a meaningful and relevant article is not listed in the references, 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 specific topic. I claim no expertise in fractional Sobolev spaces. Important: Throughout the book the expression (0.2)
AB
means A ≤ CB
for some constant C > 0 that depends on the parameters quantified in the statement of the result (usually N , p and s), but not on the
Preface
xv
functions and on their domain of integration. Also, (0.3)
A≈B
means A B and B A.
When it is important to stress the explicit dependence of the constant C on some parameters, I will add a subscript on , for example, p,s will mean ≤ C, where C = C(p, s) > 0. Finally, when the constant C > 0 depends on the domain of integration I ⊆ R or Ω ⊆ RN , I write I or Ω , respectively. Web page for appendices, mistakes, comments, and exercises: I wrote some appendices for the book [Leo22b], [Leo22c], [Leo22d], but, due to the page limitation (600 pp!), I am posting them on the AMS website for this book. Also, typos and errors are inevitable in a book of this length and with an author who is a bit absent minded. I will be very grateful to those readers who write to [email protected] to indicate any errors they find. The AMS is hosting a webpage for this book at http://www.ams.org/bookpages/gsm229/ where updates, corrections, and other material may be found. Acknowledgments: I am profoundly indebted to Ian Tice, who was my goto person, whenever I got stuck on a proof. Unfortunately for him, I got stuck a lot! I would also like to thank Kerrek Stinson for reading parts of the book and for many valuable corrections. Maria Giovanna Mora and Massimiliano Morini proofread some of the more delicate proofs. Thanks! I worked with some undergraduate students on parts of this book: Andrew Chen, Jiewen Hu, Khunpob Sereesuchart, Braden Yates, and Grant Yu. I hope they had as much fun as I did! I want to thank Xavier RosOton and the anonymous referees for their feedback and Filippo Cagnetti and Tomasz Tkocz for useful conversations. I am grateful to Ina Mette, AMS publisher, Jennifer Wright Sharp, the production editor, and all the AMS staff I interacted with for their constant help and technical support during the preparation of this book. The National Science Foundation partially supported this research under Grants No. DMS1714098 and 2108784. Also, many thanks must go to all the people at Carnegie Mellon University’s interlibrary loan for finding the articles I needed quickly. The picture on the back cover of the book was taken by Kevin Lorenzi. Giovanni Leoni
Part 1
Fractional Sobolev Spaces in One Dimension
Chapter 1
Fractional Sobolev Spaces in One Dimension The only teaching that a professor can give, in my opinion, is that of thinking in front of his students. — Henri Lebesgue
If you are an advanced undergraduate or beginning graduate student, then the first part of the book is for you. However, before starting this chapter, I suggest you read the file on the Lebesgue measure ([Leo22c]) and the appendix on Sobolev spaces [Leo22d] (just the parts about Sobolev functions of one variable). You will also need a bit of functional analysis, so you might want to read the file ([Leo22b]). In this chapter, I will introduce fractional Sobolev spaces and study their main properties. An important reminder is that the definition of Sobolev spaces is discontinuous at s = 0 and s = 1, and more generally, at every s integer. Indeed, when s is an integer, the Sobolev space W s,p is the space defined in the appendix [Leo22d] using functions and weak derivatives. However, when s is not an integer, the fractional Sobolev space W s,p is defined using double integrals and finite differences. Another important observation is that elements of Lp are not functions but equivalence classes of functions. Everybody identifies an equivalence class of functions with a function, and I will do the same. However, this can be dangerous. When the distinction is needed, I will try to spell out that we are working with a representative in the class, but I will not always remember to do it, so please keep an eye on this fact.
3
4
1. Fractional Sobolev Spaces in One Dimension
I will denote by L1 the Lebesgue measure on the real line. The natural numbers N start with 1 because I hate dividing by zero when I write 1/n. So I will use N0 to denote the natural numbers starting from zero. I will assume that you are familiar with the various spaces of smooth functions C(I), C m (I), C ∞ (I), Cc∞ (I). OK, let’s start now: I will switch to what one of the reviewers called “the encyclopedic style”; otherwise, you will think this book is not serious. I will start using “we”. In what follows, “we” refers to either you and me or to me and my two cats, Gauss and Hilbert. If what follows is something dumb or a mistake, it is definitely the latter.
1.1. Some Useful Inequalities and Identities In this section we will prove some important inequalities and identities, which we will often use in what follows. Given 1 ≤ p < ∞, using the fact that the function f (t) = tp , t ∈ R, is convex, we have 1 1 1 1 t1 + t2 ≤ f (t1 ) + f (t2 ) for all t1 , t2 ∈ R, f 2 2 2 2 or, equivalently, taking a = 2t1 and b = 2t2 , (1.1)
a + bp ≤ 2p−1 ap + 2p−1 bp
for all a, b ∈ R.
After a few occurrences, we will use this inequality without further mention. We recall that a set I ⊆ R is an interval if for every x, y ∈ I with x ≤ y, we have that z ∈ I for all x ≤ z ≤ y. Given an interval I ⊆ R and ` ∈ R, we define (1.2)
I` := {x ∈ I : x + ` ∈ I}.
If ` > 0 and I = (a, b), then I` = (a, b − `) if ` < b − a and I` = ∅ otherwise, while if I = (−∞, b), then I` = (−∞, b − `); finally, I` = I if I = R or I = (a, ∞). Similarly, if ` < 0 and I = (a, b), then I` = (a − `, b) when ` < b − a and I` = ∅ otherwise, while if I = (a, ∞), then I` = (a − `, ∞); finally, I` = I if I = R or I = (−∞, b). We continue with a version of Young’s inequality. Theorem 1.1 (Young’s inequality). Let I ⊆ R be an interval, ` > 0, and 1 ≤ p, q < ∞, with 1/p + 1/q ≥ 1. Given u ∈ Lp (I) and v ∈ Lq ((0, `)), consider the function Z ` w(x) := u(x + t)v(t) dt, x ∈ I` . 0
1.1. Some Useful Inequalities and Identities
5
Then w ∈ Lr (I` ), where 1/r = 1/p + 1/q − 1, with kwkLr (I` ) ≤ kukLp (I) kvkLq ((0,`)) . Proof. For x ∈ I` we write Z ` w(x) ≤ u(x + t)v(t) dt 0 `
Z
(u(x + t)p/r v(t)q/r )u(x + t)1−p/r v(t)1−q/r dt.
= 0
Assume first that r < ∞. Since 1/r + (1/p − 1/r) + (1/q − 1/r) = 1, by the generalized H¨older’s inequality (see [Leo22c]), w(x)r Z ` r/p−1 Z ` r/q−1 Z ` p q p q ≤ (u(x + t) v(t) )dt u(x + t) dt v(t) dt 0
Z
0 ` p
≤
0
Z
q
(u(x + t) v(t) )dt
r/p−1 Z
p
u(y) dy
0
I
`
r/q−1 v(t) dt , q
0
where in the second inequality we use the change of variables x + t = y, so that dt = dy, and the fact that x < x + t < x + ` ∈ I for t ∈ (0, `) and x ∈ I` . Integrating both sides in x over I` and using Tonelli’s theorem gives Z w(x)r dx I`
Z
p
r/p−1 Z
`
u(y) dy
≤
0
I
Z
u(y)p dy
≤
r/p−1 Z
I
Z
r/p Z
I
I`
0 `
r/q−1 Z ` Z q q v(t) dt v(t) dt u(y)p dy
0
u(y)p dy
=
r/q−1 Z ` Z q u(x + t)p dxdt v(t) dt v(t) q
I
0
`
v(t)q dt
r/q ,
0
where we use the change of variables x + t = y, so that dx = dy, and the fact that x < x + t < x + ` ∈ I for x ∈ I` . If r = ∞, then 1 = 1/p + 1/q. Hence, by H¨older’s inequality, Z
`
w(x) ≤
Z u(x + t)v(t) dt ≤
0
`
1/p Z ` 1/q q u(x + t) dt v(t) dt
0
Z ≤ I
u(y)p dy
1/p Z 0
`
1/q q v(t) dt ,
p
0
6
1. Fractional Sobolev Spaces in One Dimension
where again we use the change of variables x + t = y, so that dt = dy, and the fact that x < x + t < x + ` ∈ I for t ∈ (0, `). It suffices to take the essential supremum in x over I` . Remark 1.2. Young’s inequality continues to hold if we assume that u belongs to Lp (I; RM ) and v belongs to Lq ((0, `)). Next we prove Hardy’s inequality. Theorem 1.3 (Hardy’s inequality). Let 0 < ` ≤ ∞, u : [0, `) → [0, ∞] be a Lebesgue measurable function, 1 ≤ p < ∞, and s > 0. Then Z ` Z x 1/p Z ` 1/p dt p dx 1 p dx (1.3) u(t) ≤ (u(x)) 1+sp . t x1+sp s x 0 0 0 Proof. Suppose that the righthand side is finite, since otherwise there is nothing to prove. Assume that u = 0 near 0, say, in [0, δ] and define Z x dt v(x) := u(t) , x ∈ [0, `). t 0 Then by H¨older’s inequality Z x t1/p−1+s u(t) v(x) = δ x
Z ≤
(1/p−1+s)p0
t
dt t1/p+s 1/p0 Z dt
δ
`
(u(t)) 0
p
dx
1/p
t1+sp
< ∞.
Thus, v is locally absolutely continuous. We have 1 d [(x−s v(x))p ] = −sx−sp−1 (v(x))p + x−sp−1 (v(x))p−1 u(x). p dx Since v = 0 in [0, δ], we can integrate both sides in [0, b], where b < `, and use the fundamental theorem of calculus ([Leo22d]) to get Z b Z b 1 −s p −sp−1 p 0 ≤ (b v(b)) = −s x (v(x)) dx + x−sp−1 (v(x))p−1 u(x) dx. p 0 0 Hence, by H¨older’s inequality Z b Z b s x−sp−1 (v(x))p dx ≤ x−sp−1 (v(x))p−1 u(x) dx 0
0
Z
b
=
[x−sp−1 (v(x))p ](p−1)/p [x−sp−1 (u(x))p ]1/p dx
0
Z ≤
b
x 0
−sp−1
p
(p−1)/p Z
(v(x)) dx
b
x 0
−sp−1
p
(u(x)) dx
1/p .
1.1. Some Useful Inequalities and Identities
7
Rb Since v = 0 in [0, δ] and is continuous in [δ, b], the integral 0 x−sp−1 (v(x))p dx is finite, and thus, Z b 1/p Z b 1/p −sp−1 p −sp−1 p s x (v(x)) dx ≤ x (u(x)) dx . 0
It suffices to let b → rem.
0
`−
and use the Lebesgue monotone convergence theo
To remove the additional hypothesis that u = 0 near 0, we apply (1.3) to the function uχ[1/n,`) and then use the Lebesgue monotone convergence theorem to let n → ∞. Exercise 1.4. Let 0 < ` ≤ ∞, u : [0, `) → [0, ∞] be a Lebesgue measurable function, 1 ≤ p < ∞, and s > 0. Prove that !1/p p Z ` 1/p Z ` Z ` dt 1 sp−1 p sp−1 u(t) x dx ≤ (u(x)) x dx . t s 0 x 0 As a corollary of Theorem 1.3, we obtain the following inequality, which is referred to as Hardy’s inequality in W 1,p (R+ ). Corollary 1.5 (Hardy’s inequality in W 1,p ). Let 1 < p < ∞, v ∈ W 1,p (R+ ), and v¯ be the absolutely continuous representative of v. Then 1/p Z ∞ 1/p Z ∞ p dx 0 0 p ≤p v (x) dx . ¯ v (x) − v¯(0) p x 0 0 Proof. By the fundamental theorem of calculus (see [Leo22d]) Z x ¯ v (x) − v¯(0) ≤ v 0 (t) dt. 0
Applying Theorem 1.76, with u(t) = v 0 (t)t and s = Z
∞
¯ v (x) − 0
dx v¯(0)p p x
1/p
1 p0 ,
we get
p 1/p dx ≤ v (t) dt xp 0 0 Z ∞ 1/p ≤ p0 v 0 (x)p dx . Z
∞ Z x
0
0
In what follows, we shall present some useful identities. Recall (1.2). Theorem 1.6. Let I ⊆ R be an interval and u ∈ C 1 (I). Then for all 0 < ` < L1 (I) and x ∈ I` , Z Z ` Z y 1 ` 1 (1.4) u(x) = u(x + t) dt − (u(x + y) − u(x + t)) dtdy. 2 ` 0 0 y 0
8
1. Fractional Sobolev Spaces in One Dimension
Proof. By translation it is enough to prove identity (1.4) at x = 0. Let 0 < ε < `. Using integration by parts, we have Z ` Z ` Z y Z ` Z y 1 u(y) 1 (u(y) − u(t)) dtdy = u(t) dtdy dy − 2 2 y ε y ε 0 ε y 0 y=` Z y Z ` Z ` u(y) u(y) 1 = u(t) dt dy − dy + y y y 0 ε ε y=ε Z ` Z ε Z ` 1 1 1 = u(t) dt − u(t) dt → u(t) dt − u(0) ` 0 ε 0 ` 0 as ε → 0+ since u is continuous at 0. R yOn the other hand, by the fundamental theorem of calculus u(y) − u(t) = t u0 (τ ) dτ . Hence, Z y Z yZ y Z 1 1 y 0 1 0 (u(y) − u(t)) dt ≤ 2 u (τ ) dτ dt ≤ u (τ ) dτ. y2 y 0 t y 0 0 ≤ ku0 kL∞ ((0,`)) . It follows by the Lebesgue dominated convergence theorem that Z y Z ` Z y Z ` 1 1 (u(y) − u(t)) dtdy → (u(y) − u(t)) dtdy, 2 2 0 0 y 0 ε y which concludes the proof.
Remark 1.7. Under the hypotheses of Theorem 1.6, we can apply Fubini’s theorem, the change of variables y = t + r, so that dy = dr, and Fubini’s theorem again, to write Z ` Z `Z ` Z y u(x + y) − u(x + t) 1 (u(x + y) − u(x + t)) dtdy = dydt 2 y2 t 0 y 0 0 Z ` Z `−t u(x + t + r) − u(x + t) = drdt (t + r)2 0 0 Z ` Z `−r u(x + t + r) − u(x + t) = dtdr. (t + r)2 0 0 Hence, u(x) =
1 `
Z
`
Z `Z u(x + t) dt −
0
0
0
`−r
u(x + t + r) − u(x + t) dtdr. (t + r)2
Remark 1.8. In what follows, for u : [0, ∞) → R locally integrable, we will also use a simpler identity, namely, Z Z 1 ` 1 ` u(x + t) dt + [u(x) − u(x + t)] dt, ` > 0. u(x) = ` 0 ` 0 In the next theorem, we write a function in terms of its mollification and higher order forward differences. We will only use this result in Chapter 11
1.1. Some Useful Inequalities and Identities
9
(see Corollaries 2.29 and 7.37 and Theorem 11.41), so feel free to skip it at a first reading. Given an interval I ⊆ R, a function u : I → R, and m ∈ N, with m ≥ 2, for h ∈ R and x ∈ Imh , we define inductively, ∆nh u(x) = ∆1h (∆n−1 h u(x)),
∆1h u(x) := u(x + h) − u(x),
n = 2, . . . , m,
where we recall that I` := {x ∈ I : x + ` ∈ I}. Exercise 1.9. Let I ⊆ R be an open interval and u : I → R be a function of class C m , m ∈ N. (i) Prove that for every h > 0 and x ∈ Imh , Z h Z h u(x) = · · · u(m) (x + t1 + · · · + tm ) dt1 · · · dtm . ∆m h 0
0
(ii) Prove that for every h > 0 and x ∈ Imh , Z m m m u(m) (x + ht) dt. (1.5) ∆h u(x) ≤ h 0
Exercise 1.10. Let I ⊆ R be an open interval and u, v : I → R. Prove that for every m ∈ N there exist numbers an , n = 0, . . . , m such that for all h > 0 and x ∈ Imh , ∆m h (uv)(x)
=
m X
an ∆m−n u(x + nh)∆nh v(x). h
n=0
In the proof of the following theorem, we will use the theory of mollifiers (see [Leo22c]). We recall that given a function ω : R → R and h > 0, we set ωh (x) := h1 ω hx , x ∈ R. Theorem 1.11. Let u : R → R be a function of class C m , m ∈ N. Then for every h > 0 and x ∈ R, Z u(x) = u (x + y) ωh (y) dy R
Z + where ω, φ, ψ ∈ and 0 < δ < 1.
h
1 ξ
Z
Z φξ (t)
0 R ∞ Cc (R), with
∆m δz u(x + t + z)ψξ (z) dzdtdξ,
R
support contained in [0, 1],
R
R ω(x) dx
= 1,
Proof. Let ϕ ∈ CcR∞ (R) be a nonnegative function such that supp ϕ ⊆ [0, 1/(2m + 2)] and R ϕ(x) dx = 1. Write (1.6)
∆m h u(x) =
m X j=0
(−1)m−j cj u(x + jh),
10
1. Fractional Sobolev Spaces in One Dimension
and define m
1 X (−1)m−j cj ω ˜ (x) = ϕ A (1 + δj)2
(1.7)
j=0
x 1 + δj
,
R P (−1)m−j cj m 1 (1 − xδ )m dx (exercise). Note that where A = m = (−1) j=0 1+δj 0 x supp ω ˜ ⊆ [0, 1/2]. Moreover, by the change of variables y = 1+δj , Z Z m x 1 X (−1)m−j cj ϕ dx ω ˜ (x) dx = A (1 + δj)2 R 1 + δj R j=0 Z m 1 X (−1)m−j cj = ϕ(y) dy = 1. A 1 + δj R j=0
Set Z (1.8)
ω ˜ (x − y)˜ ω (y) dy,
ω(x) :=
x ∈ R.
R
Then ω ∈ Cc∞ (R), supp ω ⊆ [0, 1], and, by Fubini’s theorem,
R
R ω(x) dx
= 1.
For x ∈ R and h > 0 define the function Z U (x, h) := u (x + y) ωh (y) dy. R
By differentiating with respect to a parameter and the properties of mollifiers (see [Leo22c]), we have that U ∈ C ∞ (R × R+ ). Hence, by the fundamental theorem of calculus, for 0 < ε < h we get Z h ∂U (x, ξ) (1.9) U (x, ε) = U (x, h) − dξ. ∂ξ ε To compute obtain (1.10)
∂U ∂ξ ,
we differentiate under the integral sign (see [Leo22c]) to ∂U (x, ξ) = ∂ξ
Z u (x + y) ∂ξ ωξ (y) dy. R
By (1.8) and the changes of variables t = z/ξ and y − z = w, Z y 1 ∂ξ ωξ (y) = ∂ξ ω ˜ −t ω ˜ (t) dt ξ R ξ Z 1 y−z 1 z = ∂ξ ω ˜ ω ˜ dz ξ ξ ξ R ξ Z 1 y−z 1 z =2 ω ˜ ∂ξ ω ˜ dz. ξ ξ ξ ξ R Define g(z) = z ω ˜ (z) and observe that z z 1 1 z 0 z 1 0 z ω ˜ = − 2ω ˜ − 3ω ˜ = − 2g . ∂ξ ξ ξ ξ ξ ξ ξ ξ ξ
1.1. Some Useful Inequalities and Identities
11
Hence, by the change of variables t = z/ξ and integration by parts, Z Z 2 2 y−z y 0 z ∂ξ ωξ (y) = − 3 g dz = − 2 ω ˜ ω ˜ − t g 0 (t) dt ξ R ξ ξ ξ R ξ Z Z 2 2 0 y 0 y =− 2 ω ˜ − t g(t) dt = − 2 ω ˜ − t t˜ ω (t) dt ξ R ξ ξ R ξ Z y−z z 2 z ω ˜0 dz. =− 3 ω ˜ ξ R ξ ξ ξ By plugging this expression into (1.10), we have ∂U (x, ξ) = ∂ξ
Z
u (x + y) ∂ξ ωξ (y) dy Z Z 2 z z 0 y−z =− 3 u (x + y) ω ˜ ω ˜ dzdy ξ R R ξ ξ ξ Z Z 2 t z z 0 =− 3 ω ˜ u (x + t + z) ω ˜ dzdt, ξ R ξ ξ ξ R R
where in the last equality we use Fubini’s theorem and the change of variables t = y − z, so that dt = dy. We now use the definition of ω ˜ (see (1.7)) to write Z t ∂U (x, ξ) 2 1 0 ω ˜ =− 3 ∂ξ ξ A R ξ Z m X (−1)m−j cj z z × u (x + t + z) ϕ dzdt (1 + δj)2 R ξ ξ(1 + δj) j=0 Z X Z m 1 2 t 0 =− 2 ω ˜ (−1)m−j cj u (x + t + ξ(1 + δj)y) yϕ(y) dydt ξ A R ξ R j=0 Z Z 1 2 t 0 =− 2 ω ˜ ∆m δξy u(x + t + ξy)yϕ(y) dydt, ξ A R ξ R z where we make the change of variables ξ(1+δj) = y and use formula (1.6). Setting z = ξy, so that dz = ξdy, we have
∂U (x, ξ) 1 2 =− 3 ∂ξ ξ A
Z t z z m ω ˜ ∆δz u(x + t + z) ϕ dzdt. ξ ξ ξ R R
Z
0
12
1. Fractional Sobolev Spaces in One Dimension
Hence, by (1.9), (1.11) Z U (x, ε) =
u (x + y) ωh (y) dy Z Z Z t 2 h 1 z z 0 m − ω ˜ ∆ u(x + t + z) dzdtdξ ϕ δz A ε ξ3 R ξ ξ ξ R =: A + B. R
R Since R ω(x) dx = 1, by standard properties of mollifiers (see [Leo22c]), we obtain Z (1.12) lim U (x, ε) = lim u (x + y) ωε (y) dy = u (x) . ε→0+
ε→0+
R
On the other hand, by (1.5), for z ≤ ξ, Z m m m u(m) (x + t + δzτ ) dτ. ∆δz u(x + t + z) ξ 0
Hence, since supp ϕ ⊆ [0, 1] and supp ω ˜ ⊆ [0, 1], by Tonelli’s theorem and a change of variables Z Z Z h 0 t m 1 z z ω ˜ dzdtdξ ∆δz u(x + t + z) ϕ 3 ξ ξ ξ 0 ξ R R Z Z Z h 1 ξ ξ m ∆δz u(x + t + z) dzdtdξ ϕ 3 0 0 0 ξ Z h m Z ξZ ξZ m ξ ϕ u(m) (x + t + δzτ ) dτ dzdtdξ 3 0 ξ 0 0 0 Z h Z (m+1)h m−1 ϕ ξ dξ u(m) (x + y) dy. 0
0
Thus, we can let ε → 0+ in (1.11), using the Lebesgue dominated convergence theorem on the righthand side and (1.12) on the lefthand side to get Z u(x) = u (x + y) ωh (y) dy R Z Z Z 2 h 1 t z z 0 m − ω ˜ ∆δz u(x + t + z) ϕ dzdtdξ. 3 A 0 ξ R ξ ξ ξ R
1.2. Fractional Sobolev Spaces In this section, we will introduce the main characters of this book, fractional Sobolev spaces, and begin studying their main properties.
1.2. Fractional Sobolev Spaces
13
Definition 1.12. Let I ⊆ R be an open interval, 1 ≤ p ≤ ∞, and 0 < s < 1. A function u ∈ Lp (I) belongs to the fractional Sobolev space W s,p (I) if kukW s,p (I) := kukLp (I) + uW s,p (I) < ∞, where Z Z (1.13)
uW s,p (I) :=
I
I
u(x) − u (y) p dxdy x − y1+sp
1/p
if 1 ≤ p < ∞ and uW s,∞ (I) := esssup
x,y∈I, x6=y
u(x) − u (y)  x − ys
if p = ∞. When p = 2 we write H s (I) := W s,2 (I). Definition 1.13. Let I ⊆ R be an open interval, 1 ≤ p ≤ ∞, and 0 < s < 1. A function u ∈ Lploc (I) belongs to the homogeneous fractional Sobolev space ˙ s,p (I) if u s,p < ∞. When p = 2 we write H˙ s (I) := W ˙ s,2 (I). W W (I) For brevity, we will sometimes use the notation ∆h u(x) = u(x + h) − u(x).
(1.14)
Remark 1.14. By symmetry, the change of variables x = y + h, so that dx = dh, and Tonelli’s theorem, we have Z Z u(x) − u (y) p upW s,p (I) = 2 dxdy x − y1+sp I {x∈I: x>y} Z L1 (I) Z u(y + h) − u (y) p dydh, =2 h1+sp 0 Ih where Ih = {x ∈ I : xR+ h ∈ I} (see (1.2)). Using the fact that Ih = ∅ if h > L1 (I) and setting ∅ f (x) dx := 0, we have Z L1 (I) Z Z ∞Z ∆h u(y)p ∆h u(y)p p (1.15) uW s,p (I) = 2 dydh = 2 dydh. h1+sp h1+sp 0 Ih 0 Ih Exercise 1.15. Let I ⊆ R be an open interval and 0 < s < 1. Prove that ˙ s,∞ (I), then u admits a representative u if u ∈ W ¯ that is H¨older continuous with exponent s and such that uW s,∞ (I) = Prove also that if u ∈
sup x,y∈I, x6=y
W s,∞ (I),
¯ u(x) − u ¯ (y)  = ¯ uC 0,s (I) . x − ys
then kukL∞ (I) = supx∈I ¯ u(x).
Remark 1.16. In view of Exercise 1.15, we can identify W s,∞ (I) with the space C 0,s (I). For this reason, in most of what follows, we will restrict our attention to the case 1 ≤ p < ∞.
14
1. Fractional Sobolev Spaces in One Dimension
˙ s,p (I) belongs to Next we prove that if I is bounded, a function u ∈ W This is no longer the case if I is unbounded, since any nonzero ˙ s,p (I) but not to W s,p (I). If E ⊂ R is a constant function belongs to W Lebesgue measurable set with positive and finite measure R and u : E → R is a Lebesgue integrable function, we define uE := L11(E) E u(x) dx. W s,p (I).
Theorem 1.17 (Poincar´e’s inequality). Let I = (a, b), E ⊆ I be a Lebesgue measurable set with positive measure, 1 ≤ p < ∞, and 0 < s < 1. Then Z Z Z b (b − a)1+sp b b u(x) − u(y)p p dxdy u(x) − uE  dx ≤ L1 (E) x − y1+sp a a a for all u ∈ W s,p (I). Proof. By Minkowski’s inequality for integrals (see [Leo22c]), we have p 1/p Z b Z Z b 1/p 1 p dx (u(x) − u(y)) dy u(x) − uE  dx = 1 L (E) E a a Z Z b 1/p 1 ≤ 1 u(x) − u(y)p dy dx. L (E) E a Hence, by H¨older’s inequality Z b 0 Z bZ b (L1 (E))p/p u(x) − uE p dx ≤ u(x) − u(y)p dydx 1 (E))p (L a a a Z Z (b − a)1+sp b b u(x) − u(y)p dxdy, ≤ L1 (E) x − y1+sp a a where in the last inequality we use the fact that if x, y ∈ I, then x − y ≤ b − a. Remark 1.18. If we take E b (a, b), then Poincar´e’s inequality continues to ˙ s,p (I). Indeed, consider E ⊆ [c, d] ⊂ (a, b). Since u ∈ W s,p ((c, d)), hold in W by applying Poincar´e’s inequality in (c, d) we get Z d Z Z (d − c)1+sp d d u(x) − u(y)p p dxdy u(x) − uE  dx ≤ L1 (E) x − y1+sp c c c Z Z (b − a)1+sp b b u(x) − u(y)p ≤ dxdy. L1 (E) x − y1+sp a a It suffices to let c → a+ , d → b− and use the Lebesgue monotone convergence theorem. ˙ s,p (I), Corollary 1.19. Let I = (a, b), 1 ≤ p < ∞, and 0 < s < 1. If u ∈ W s,p then u ∈ W (I).
1.2. Fractional Sobolev Spaces
15
Proof. Let [c, d] ⊂ (a, b). By Remark 1.18, Z b Z Z (b − a)1+sp b b u(x) − u(y)p p u(x) − u(c,d)  dx ≤ dxdy < ∞, d−c x − y1+sp a a a which implies that u − u(c,d) ∈ Lp (I). In turn, by Minkowski’s inequality, u ∈ Lp (I), with Z b 1/p Z b 1/p p p 1/p u(x) − u(c,d)  dx u(c,d) + u(x) dx ≤ (b − a) a
a
Z p b − a d ≤ u(y) dy (d − c)p c Z b Z b 1/p (b − a)1/p+s u(x) − u(y)p + dxdy . x − y1+sp (d − c)1/p a a If you don’t have a solid background in functional analysis, you can skip the part about reflexivity in the following theorem. We will only use it in the proof of Theorem 1.75 (which you can also skip). Theorem 1.20. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 0 < s < 1. The function u ∈ W s,p (I) 7→ kukW s,p (I) is a norm and (W s,p (I), k · kW s,p (I) ) is a Banach space. Moreover, W s,p (I) is reflexive for 1 < p < ∞. Proof. Let {un }n be a Cauchy sequence in W s,p (I). Then {un }n is a Cauchy sequence of Lp (I) and the sequence of functions vn : I × I → R, defined by un (x) − un (y) , vn (x, y) := kx − ykN/p+s is a Cauchy sequence in Lp (I × I). Since Lp is complete, we may find u ∈ Lp (I) and v ∈ Lp (I × I) such that un → u in Lp (I) and vn → v in Lp (I × I). Let u ¯n , u ¯, v¯ be representatives of un , u, v, respectively. Then u ¯n (x)−¯ un (y) v¯n (x, y) := kx−ykN/p+s is a representative of vn . Extract a subsequence {¯ unk }k of {¯ un }n such that u ¯nk (x) → u ¯(x) for all x ∈ I \ E and v¯nk (x, y) → v¯(x, y) for all (x, y) ∈ (I × I) \ F , where L1 (E) = 0 and L2 (F ) = 0. Consider the set G := (E × I) ∪ (I × E) ∪ F . By Tonelli’s theorem, L2 (G) = 0. If (x, y) ∈ (I × I) \ G, then x ∈ / E, y ∈ / E, and (x, y) ∈ / F . Hence, v¯nk (x, y) →
u ¯(x) − u ¯ (y) , N/p+s kx − yk
v¯nk (x, y) → v¯(x, y).
u ¯(x)−¯ u(y) By the uniqueness of limits, we have that v¯(x, y) = kx−yk N/p+s for every (x, y) ∈ (I × I) \ F . Hence, u ∈ W s,p (I) and kun − ukW s,p (I) → 0 as n → ∞.
16
1. Fractional Sobolev Spaces in One Dimension
Next assume that 1 < p < ∞. Then the space Lp (I)×Lp (I ×I), endowed with the norm (u, v) 7→ kukLp (I) + kvkLp (I×I) , is a reflexive Banach space. Consider the application T : W s,p (I) → Lp (I) × Lp (I × I), u 7→ (u, v), u(x)−u(y) where v(x, y) := x−y 1/p+s . The operator T is onetoone and continuous, and it preserves the norm, that is,
kT (u)kLp (I)×Lp (I×I) = kukW s,p (I) for all u ∈ W s,p (I). Hence, also by what we proved above, the subspace T (W s,p (I)) is closed in Lp (I) × Lp (I × I). Since Lp (I) × Lp (I × I) is reflexive (see [Leo22c]), so is its closed subspace T (W s,p (I)) (see [Leo22b]). In turn, W s,p (I) is is reflexive. Remark 1.21. The space H s (I) is a Hilbert space with the inner product Z Z Z (u(x) − u(y))(v(x) − v(y)) dxdy. (u, v)H s (I) := u(x)v(x) dx + x − y1+2s I I I Next we study what type of functions belong to W s,p (I). We begin by ¯ belong to W s,p (I). showing that functions in Cc1 (I) Theorem 1.22. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 0 < s < 1. ¯ then u ∈ W s,p (I). If u ∈ Cc1 (I), Proof. Step 1: Assume first that I = (a, b) and let u ∈ C 1 ([a, b]). By the mean value theorem, u(x) − u(y) ≤ ku0 k∞ x − y for every x, y ∈ I. Since Z bZ a
a
b
x − yp 1 dxdy = 1+ps x − y p(1 − s)
Z
b
[(y − a)p−sp + (b − y)p−sp ]dy
a p(1−s)+1
(b − a)
,
we have Z bZ a
a
b
u(x) − u(y)p dxdy (b − a)p(1−s)+1 ku0 kp∞ . x − y1+sp
¯ there exists b > a Step 2: Assume next that I = (a, ∞). Since u ∈ Cc1 (I) such that u(x) = 0 for all x ≥ b. By Tonelli’s theorem and relabeling the
1.2. Fractional Sobolev Spaces
17
variables of integration, we can write Z b+1 Z b+1 Z ∞Z ∞ u(x) − u(y)p u(x) − u(y)p dxdy = dxdy x − y1+sp x − y1+sp a a a a Z ∞ Z b+1 u(x)p +2 dxdy x − y1+sp b+1 a Z b+1 Z b+1 Z ∞Z b u(x) − u(y)p u(x)p = dxdy + 2 dxdy 1+sp x − y1+sp a a b+1 a x − y = A + B. The term A can be treated as in Step 1, while by Tonelli’s theorem we can write Z b Z b Z ∞ 1 1 1 p dydx = u(x)p dx B u(x) 1+sp sp a (b + 1 − x)sp a b+1 x − y Z b 1 (b − a) ≤ u(x)p dx ≤ kukp∞ . sp a sp This completes the proof in this case. The cases I = (−∞, b) and I = R are similar. Exercise 1.23. Let I = (a, b) and u : I → R be a H¨ older continuous function with exponent α > 0, that is, u(x) − u(y) ≤ Cx − yα for all x, y ∈ I and for some constant C > 0. Determine for which 1 ≤ p < ∞ and s ∈ (0, 1) the function u belongs to W s,p (I). Exercise 1.24. Let I = (0, 1) and u = χ(0,1/2) . Determine for which 1 ≤ p < ∞ and s ∈ (0, 1) the function u belongs to W s,p (I). We now show that Sobolev functions in W 1,p (I) belong to W s,p (I) (see [Leo22d]). Theorem 1.25. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 0 < s < 1. Then for all u ∈ W 1,p (I) and 0 < ` < L1 (I), uW s,p (I) `−s kukLp (I) + (1 − s)−1/p `1−s ku0 kLp (I) . In particular, W 1,p (I) ,→ W s,p (I). Proof. Write Z Z Z Z u(x) − u(y)p u(x) − u(y)p dxdy = dxdy 1+sp 1+sp I I x − y I B(y,`)∩I x − y Z Z u(x) − u(y)p + dxdy =: A + B. 1+sp I I\B(y,`) x − y
18
1. Fractional Sobolev Spaces in One Dimension
By a characterization of W 1,p (I) (see [Leo22d]), we can assume that u has a representative u ¯ that is absolutely continuous, with u ¯0 = u0 for L1 a.e. in I. Hence, by the fundamental theorem of calculus (see [Leo22d]), Z 1 u ¯(x) − u ¯(y) = u0 (tx + (1 − t)y)(x − y) dt, 0
and so, by H¨ older’s inequality, p
p
Z
1
¯ u(x) − u ¯(y) ≤ x − y
u0 (tx + (1 − t)y)p dt.
0
Define v(x) :=
u0 (x) if x ∈ I, 0 if x ∈ / I.
By Tonelli’s theorem and by making first the change of variables z = x − y, so that dz = dx, and then τ = y + tz, so that dτ = dy, Z 1Z Z u0 (tx + (1 − t)y)p A≤ dxdydt x − y1−(1−s)p 0 I B(y,`)∩I Z 1Z Z v(tx + (1 − t)y)p ≤ dxdydt x − y1−(1−s)p 0 R B(y,`) Z 1Z Z v(y + tz)p dzdydt = 1−(1−s)p 0 R B(0,`) z Z Z 1Z (1−s)p−1 v(y + tz)p dydzdt = z R 0 B(0,`) Z Z Z = v(τ )p dτ z(1−s)p−1 dz `(1−s)p u0 (τ )p dτ. R
I
B(0,`)
On the other hand, by Tonelli’s theorem and inequality (1.1), Z Z 1 B ≤ 2p−1 u(x)p dydx 1+sp x − y I I\B(x,`) Z Z Z 1 1 + 2p−1 u(y)p dxdy u(x)p . 1+sp sp x − y ` I I\B(y,`) I Exercise 1.26. Consider the function g(t) = At−a + Btb ,
t ∈ (0, t0 ),
where A, B ≥ 0, a, b > 0, and 0 < t0 ≤ ∞. (i) Prove that if t0 = ∞, then inf g(t) ≤ CAb/(a+b) B a/(a+b) ,
0 0 is independent of A and B.
1.2. Fractional Sobolev Spaces
19
(ii) Prove that if t0 < ∞, then b/(a+b) a/(a+b) inf g(t) ≤ CAt−a B , 0 + CA
0 0 is independent of A and B. Corollary 1.27. Let I ⊆ R be an open interval, 1 ≤ p < ∞, 0 < s < 1, and u ∈ W 1,p (I). If L1 (I) = ∞, then 0 s uW s,p (I) kuk1−s Lp (I) ku kLp (I) ,
while if L1 (I) < ∞, then (1.16)
uW s,p (I)
1 (L1 (I))s
0 s kukLp (I) + kuk1−s Lp (I) ku kLp (I) .
Proof. By Theorem 1.25 we have that (1.17)
uW s,p (I) ≤ inf g(t), 0 0, define ut (x) := u(x + t), x ∈ I. Prove that ut ∈ W s,p (I) and that lim kut − ukW s,p (I) = 0.
t→0+
Exercise 1.33. Let I = (0, 1), 1 ≤ p < ∞, 0 < s < 1, and u(x) = xa , a > 0. Determine for which p and s the function u belongs to W s,p (I). Exercise 1.34. Let I = (0, 1), 1 ≤ p < ∞, 0 < s < 1, and u(x) = log x. Determine for which p and s the function u belongs to W s,p (I). Hint: Make the change of variables z = x/y and split the integrals in z as R1 R∞ 0 dz and 1 dz. Exercise 1.35. Let I = (0, 1), 1 ≤ p < ∞, 0 < s < 1, and u(x) = xa log x, a > 0. Determine for which p and s the function u belongs to W s,p (I).
1.2. Fractional Sobolev Spaces
21
√ Exercise 1.36. Let I = (0, 1), 1 ≤ p < ∞, 0 < s < 1, and u(x) = n x, n ∈ N. Determine for which p and s the function u belongs to W s,p (I). Exercise 1.37. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and s > 1. (i) Assume that u ∈ C 1 (I) and let x ∈ I. Prove that if Z u(x) − u(y)p dy < ∞, 1+sp I x − y then u0 (x) = 0. (ii) Prove that if u ∈ C 1 (I) is such that then u is constant.
R R I
I
u(x)−u(y)p dxdy x−y1+sp
< ∞,
We conclude this section by studying products. Theorem 1.38 (Product). Let I ⊆ R be an open interval, 1 ≤ p < ∞, and ˙ s,p (I) ∩ L∞ (I), then uv ∈ W ˙ s,p (I), with 0 < s < 1. If u, v ∈ W uvW s,p (I) ≤ uW s,p (I) kvkL∞ (I) + kukL∞ (I) vW s,p (I) . Moreover, if u, v ∈ W s,p (I) ∩ L∞ (I), then kuvkLp (I) ≤ kukLp (I) kvkL∞ (I) . Proof. Since u(x)v(x) − u(y)v(y) = v(x)(u(x) − u(y)) + (v(x) − v(y))u(y), by Minkowski’s inequality’s we have 1/p u(x)v(x) − u(y)v(y)p dxdy x − y1+sp I Z Z 1/p v(x)(u(x) − u(y))p ≤ dxdy x − y1+sp I I Z Z 1/p (v(x) − v(y))u(y)p dxdy + x − y1+sp I I ≤ uW s,p (I) kvkL∞ (I) + kukL∞ (I) vW s,p (I) .
Z Z I
The last statement follows from the fact that u(x)v(x) ≤ u(x)kvkL∞ (I) for L1 a.e. x ∈ I. Remark 1.39. In Chapter 2, we will see that if sp > 1, then the hypothesis that u, v ∈ L∞ (I) is unnecessary provided u, v ∈ W s,p (I) (see Corollary 2.12).
22
1. Fractional Sobolev Spaces in One Dimension
1.3. Extensions In this section we show how to extend a function in W s,p (I) to a function in W s,p (R). We begin with the simpler case in which I is unbounded. Theorem 1.40. Let I ⊂ R be an unbounded open interval, 1 ≤ p < ∞, and 0 < s < 1. Given u ∈ W s,p (I), there exists a function v ∈ W s,p (R) such that v = u in I and kvkLp (R) ≤ 21/p kukLp (I) ,
vW s,p (R) ≤ 41/p uW s,p (I) .
Moreover, the map u ∈ W s,p (I) 7→ v ∈ W s,p (R) is linear. Proof. Since I 6= R, the interval I takes the form I = (a, ∞) or I = (−∞, b). We only treat the first case since the other one is similar. Define v : R → R by u(−x + 2a) if x ≤ a, (1.18) v(x) := u(x) if x > a. Then by Tonelli’s theorem, vpW s,p (R)
=
a u(−x + 2a) − u(−y + 2a)p + dxdy x − y1+sp −∞ −∞ Z a Z ∞ u(x) − u(−y + 2a)p +2 dxdy := A + B + C. x − y1+sp −∞ a
upW s,p (I)
Z
a
Z
Using the changes of variables t = −x + 2a and τ = −y + 2a, we get Z ∞Z ∞ u(t) − u(τ )p dtdτ. B= t − τ 1+sp a a On the other hand, if x > a and y < a, then x − 2a + y = x − y − 2(a − y) ≤ x − y ≤ x − y, 2a − y − x ≤ 2x − y − x ≤ x − y. Hence, x − 2a + y ≤ x − y, and so, Z a Z ∞ Z ∞Z ∞ u(x) − u(−y + 2a)p u(x) − u(τ )p C≤2 dxdy = 2 dxdy. x − 2a + y1+sp x − τ 1+sp −∞ a a a Combining the estimates for A, B, and C gives vpW s,p (R) ≤ 4upW s,p (I) . On the other hand, by the change of variables t = −x + 2a, Z a kvkpLp (R) = kukpLp (I) + u(−x + 2a)p dx = 2kukpLp (I) . −∞
˙ s,p (I), then Remark 1.41. Note that if in Theorem 1.40 we assume u ∈ W s,p 1/p ˙ (R) with vW s,p (R) ≤ 4 uW s,p (I) . the function v belongs to W
1.3. Extensions
23
Exercise 1.42. Let I = (a, ∞) and 1 ≤ p < ∞. Prove that if u ∈ W 1,p (I) ˙ 1,p (I)), then the function v defined in (1.18) belongs to (respectively, u ∈ W ˙ 1,p (R)), with kvkLp (R) ≤ 2kukLp (I) , kv 0 kLp (R) ≤ W 1,p (R) (respectively, u ∈ W 0 0 2ku kLp (I) (respectively, kv kLp (R) ≤ 2ku0 kLp (I) ). Next we consider the case in which I is bounded. Theorem 1.43. Let I = (a, b), 1 ≤ p < ∞, and 0 < s < 1. Given a function u ∈ W s,p (I), there exists a function v ∈ W s,p (R) such that v = u in (a, b) and kvkLp (R) kukLp (I) ,
vW s,p (R) (b − a)−s kukLp (I) + uW s,p (I) .
Moreover, the map u ∈ W s,p (I) 7→ v ∈ W s,p (R) is linear. Proof. By translation, without loss of generality, we may assume that I = (−r, r) for some r > 0. Step 1: Define fr : (−3r, 3r) → (−r, r) by −x − 2r if − 3r < x ≤ −r, x if − r < x < r. fr (x) := −x + 2r if r ≤ x < 3r. The function fr is Lipschitz continuous with Lipschitz constant 1. Hence, fr (x) − fr (y) ≤ x − y
(1.19)
for all x, y ∈ (−3r, 3r). Define w : (−3r, 3r) → R u(−x − 2r) u(x) (1.20) w(x) := u(fr (x)) = u(−x + 2r)
by if − 3r < x ≤ r, if − r < x < r. if r ≤ x < 3r.
Then wpW s,p ((−3r,3r)) Z p = uW s,p (I) +
Z
(−3r,3r)\I
Z
Z
+2 (−3r,3r)\I
I
(−3r,3r)\I
u(fr (x)) − u(fr (y))p dxdy x − y1+sp
u(fr (x)) − u(fr (y))p dxdy x − y1+sp
:= A + B + C. Using (1.19) and the changes of variables t = fr (x) and τ = fr (y), so that dt = fr0 (x) dx = dx and dτ = fr0 (y) dy = dy, we get Z Z u(fr (x)) − u(fr (y))p B≤ dxdy 1+sp (−3r,3r)\I (−3r,3r)\I fr (x) − fr (y) Z Z u(t) − u(τ )p =4 dtdτ. 1+sp I I t − τ 
24
1. Fractional Sobolev Spaces in One Dimension
Similarly, Z
Z
C≤2 (−3r,3r)\I
Z Z =4 I
I
I
u(fr (x)) − u(fr (y))p dxdy fr (x) − fr (y)1+sp
u(x) − u(τ )p dxdτ. x − τ 1+sp
Hence, (1.21)
wpW s,p ((−3r,3r)) ≤ 9upW s,p (I) .
Similarly, by the changes of variables t = fr (x), Z p p (1.22) kwkLp ((−3r,3r)) = kukLp (I) + u(fr (x))p dx = 3kukpLp (I) . (−3r,3r)\I
Cc∞ (R)
such that 0 ≤ φ ≤ 1, φ = 1 in (−1, 1), and Step 2: Consider φ ∈ φ = 0 outside (−2, 2). Define φr (x) = φ(x/r) for x ∈ R, and v := φr w, ¯ where w ¯ is the extension of w by zero outside (−3r, 3r). We write v(x) − v(y) ≤ w(x)φ ¯ ¯ − w(y). ¯ r (x) − φr (y) + φr (y)w(x) Hence, by inequality (1.1) and Tonelli’s theorem, Z 3r Z φr (x) − φr (y)p dydx vpW s,p (R) w(x)p x − y1+sp −3r R Z 2r Z 3r w(x) − w(y)p + dxdy x − y1+sp −2r −3r Z Z 2r 1 + w(y)p dxdy := D + E + F, 1+sp x − y R\(3r,3r) −2r where we use the fact that φr = 0 outside (−2r, 2r) and that 0 ≤ φr ≤ 1. Write Z Z φr (x) − φr (y)p φr (x) − φr (y)p dy = dy x − y1+sp x − y1+sp R B(y,4r) Z φr (x)p + dy =: D1 + D2 . 1+sp R\B(y,4r) x − y By the mean value theorem φr (x) − φr (y) ≤ Cr−1 x − y. Hence, Z rp−sp 1 D1 p x − yp−sp−1 dy , r B(y,4r) rp while
Z D2 ≤ R\B(y,4r)
1 1 dy sp . x − y1+sp r
Thus, using also (1.22), 1 D sp r
Z
3r
−3r
w(x)p dx
1 kukpLp (I) . sp r
1.3. Extensions
25
The term E is estimated by (1.21), while for F we observe that for y ∈ (−2r, 2r), Z Z 1 1 1 dx ≤ dx sp , 1+sp 1+sp r R\B(y,r) x − y R\(3r,3r) x − y and so, Z 2r 1 1 w(y)p dy sp kukpLp (I) , rsp −2r r again by (1.22). Combining the estimates for D, E, and F gives 1 vpW s,p (R) sp kukpLp (I) + upW s,p (I) . r Finally, since φr = 0 outside (−2r, 2r) and 0 ≤ φr ≤ 1, F
kvkpLp (R) ≤ kwkpLp ((−3r,3r)) ≤ 3kukpLp (I) by (1.22).
Remark 1.44. Note that if the function u ∈ W s,p (I) in Theorem 1.43 belongs to Lq (I) for some 1 ≤ q < ∞, then kvkqLq (R) ≤ kwkqLq ((−3r,3r)) ≤ 3kukqLq (I) . Exercise 1.45. Prove that in Step 1 we could have used the function fr : (−2r, 2r) → (−r, r) given by x if x ∈ (−r, r), fr (x) := r2 /x if x ∈ (−2r, 2r) \ (−r, r). Exercise 1.46. Let I = (a, b), 1 ≤ p < ∞, and u ∈ W 1,p (I). Prove that the function v defined in Step 2 of the proof of Theorem 1.43 belongs to W 1,p (R), with kvkLp (R) kukLp (I) , kv 0 kLp (R) (b − a)−1 kukLp (I) + ku0 kLp (I) . Corollary 1.47. Let I = (a, b), 1 ≤ p < ∞, and 0 < s < 1. Given a ˙ s,p (I), for every k ∈ N there exists a function v ∈ W ˙ s,p (Jk ), function u ∈ W where Jk = (a − k(b − a), b + k(b − a)), such that v = u in (a, b) and vpW s,p (Jk ) ≤ 9k 2 upW s,p (I) . Proof. By a translation we may assume that I = (−r, r). By repeated applications of Step 1 of the proof of Theorem 1.43, for every n we can extend p n ˙ ((−3n r, 3n r)), with wp s,p u to a function w in W W ((−3n r,3n r)) ≤ 9 uW s,p (I) . It suffices to take n so large that 3n ≥ k, that is, n = blog k/ log 3c + 1. Rd In the special case in which c u(x) dx = 0 for some [c, d] ⊂ (a, b), we can replace 9k 2 with a constant independent of k.
26
1. Fractional Sobolev Spaces in One Dimension
Theorem 1.48. Let I = (a, b), 1 ≤ p < ∞, and 0 < s < 1. Given a Rd function u ∈ W s,p (I) such that c u(x) dx = 0 for a ≤ c ≤ d ≤ b with c − d ≥ 12 (b − a). Then there exists a function v ∈ W s,p (R) such that v = u in (a, b) and (1.23)
kvkLp (R) (b − a)s uW s,p (I) ,
Proof. Since have that Z b
Rd
a
c
vW s,p (R) uW s,p (I) .
u(x) dx = 0, by Poincar´e’s inequality (Theorem 1.17), we
Z Z (b − a)1+sp b b u(x) − u(y)p u(x) dx dxdy (d − c) x − y1+sp a a Z bZ b u(x) − u(y)p sp dxdy, (b − a) x − y1+sp a a p
where we use the fact that d − c ≥ 12 (b − a). In turn, by Theorem 1.43, kvkLp (R) kukLp (I) (b − a)s uW s,p (I) , and vW s,p (R) (b − a)−s kukLp (I) + uW s,p (I) uW s,p (I) .
Remark 1.49. There is nothing special with 12 in the inequality c − d ≥ 1 2 (b − a). If we assume that c − d ≥ θ(b − a) for some θ ∈ (0, 1], then the constant C in (1.23) will depend on θ. The proof of Theorem 1.48 also shows the following result. Corollary 1.50. Let I = (a, b), 1 ≤ p < ∞, and 0 < s < 1. Given a R ˙ s,p (I) such that d u(x) dx = 0 for a < c < d < b, with function u ∈ W c c − d ≥ 12 (b − a), then there exists a function v ∈ W s,p (R) such that v = u in I and kvkLp (R) (b − a)s uW s,p (I) ,
vW s,p (R) uW s,p (I) .
˙ 1,p (I) be such that Exercise 1.51. Let I = (a, b), 1 ≤ p < ∞, and u ∈ W Rd 1 c u(x) dx = 0 for a < c < d < b, with c − d ≥ 2 (b − a). Prove that kukLp (I) (b − a)ku0 kLp (I) . Hint: Use the fundamental theorem of calculus. ˙ 1,p (I) be such that Exercise 1.52. Let I = (a, b), 1 ≤ p < ∞, and u ∈ W Rd 1 c u(x) dx = 0 for a < c < d < b, with c − d ≥ 2 (b − a). Prove that the function v defined in Step 2 of the proof of Theorem 1.43 belongs to W 1,p (R), with kvkLp (R) (b − a)ku0 kLp (I) ,
kv 0 kLp (R) ku0 kLp (I) .
1.4. Some Equivalent Seminorms
27
1.4. Some Equivalent Seminorms In this section we discuss some equivalent seminorms that will be used in what follows. We recall that given an open interval I ⊆ R and a function u : I → R, for h > 0 and x ∈ Ih , we define ∆h u(x) := u(x + h) − u(x), where Ih := {x ∈ I : x + h ∈ I}. Theorem 1.53. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 0 < s < 1. Then for every u ∈ Lploc (I), Z Z dh sp ∞ ∆t u(x)p dx 1+ps (1.24) sup 4 0 0≤t≤h It h Z ∞ Z dh p ≤ uW s,p (I) ≤ sup ∆t u(x)p dx 1+ps . h 0 0≤t≤h It Proof. For t/2 ≤ ζ ≤ t we have ∆t u(x) = u(x + t) − u(x) = u(x + ζ) − u(x) + u(x + t) − u(x + ζ) = ∆ζ u(x) + ∆t−ζ u(x + ζ). Hence, by Minkowski’s inequality, Z 1/p Z 1/p Z 1/p p p p ∆t u(x) dx ≤ ∆ζ u(x) dx + ∆t−ζ u(x + ζ) dx It
It
It
!1/p
Z ≤
!1/p
Z
p
∆ζ u(x) dx
p
∆t−ζ u(y) dy
+
Iζ
,
It−ζ
where we use the fact that It ⊆ Iζ and the change of variables y = x + ζ. By averaging in ζ over (t/2, t) we get (1.25) Z
∆t u(x)p dx
1/p
It
2 + t
Z
Z
t
2 ≤ t
t
2 ≤ t
Z
t/2
Z
t
t/2
Z
!1/p ∆t−ζ u(y)p dy !1/p p
Iζ
Z
≤4 0
dζ
Iζ
∆ζ u(x) dx t
∆ζ u(x)p dx
dζ
It−ζ
0
Z
!1/p
Z
Iζ
!1/p ∆ζ u(x)p dx
2 dζ + t dζ , ζ
Z
t
Z
p
∆τ u(y) dy t/2
Iτ
1/p dτ
28
1. Fractional Sobolev Spaces in One Dimension
where we make the change of variables τ = t − ζ and use the fact that 0 < ζ < t. In turn, !1/p Z 1/p Z h Z dζ ∆ζ u(x)p dx (1.26) sup ∆t u(x)p dx ≤4 . ζ 0 0. Then for every u ∈ W s,p (I), lim ku − φn ukW s,p (I) = 0.
n→∞
34
1. Fractional Sobolev Spaces in One Dimension
Proof. Define ψn := 1 − φn and un := ψn u. By inequality (1.1), we have Z Z (ψn (x) − ψn (y))u(y)p p p u − un W s,p (I) = ψn uW s,p (I) dxdy x − y1+ps I I Z Z u(x) − u(y)p + ψn (x)p dxdy =: A + B. x − y1+ps I I By the mean value theorem ψn (x) − ψn (y) ≤ kψn0 k∞ x − y ≤ Cx − y. Hence, since 0 ≤ ψn ≤ 1, we have ψn (x) − ψn (y) ≤ C min{1, x − y} for every x, y ∈ I. By Lemma 1.61 and Tonelli’s theorem, Z Z Z min{1, x − y}p u(y)p dxdy ≤ C u(y)p dy < ∞. 1+ps x − y I I I Since ψn (x) − ψn (y) → 0 as n → ∞ for every x, y ∈ I, it follows by the Lebesgue dominated convergence theorem that A → 0 as n → ∞. Similarly, since ψn (x) ≤ 1 and ψn (x) → 0 as n → ∞ for every x ∈ I, by the Lebesgue dominated convergence theorem, B → 0 as n → ∞. We turn to the proof of Theorem 1.62. Proof. In view of Lemma 1.63, we may assume that u is zero outside a compact set. By (1.15), Z Z Z ∞Z ∆h u(x)p u(x) − u(y)p (1.36) dxdy = 2 dxdh < ∞. 1+sp h1+sp R R x − y 0 R We will use mollifiers (see [Leo22c]). Let ϕε be a standard mollifier and let uε := u ∗ ϕε . Since ∆h uε = ϕε ∗ ∆h u, for every h > 0, k∆h uε kLp (R) ≤ k∆h ukLp (R)
(1.37) and (1.38)
lim kuε − ukLp (R) = 0,
ε→0+
lim k∆h uε − ∆h ukLp (R) = 0.
ε→0+
For h > 0 and ε > 0 define the functions 1 1 gε (h) := 1+sp k∆h uε − ∆h ukpLp (R) , g(h) := 1+sp k∆h ukpLp (R) . h h 1 By (1.36), g ∈ L (R+ ), and by (1.37), Minkowski’s inequality, and the convexity of the function tp , we have that gε (h) ≤ 2p g(h) for all h. Since gε (h) → 0 for all h by (1.38), we are in a position to apply the Lebesgue dominated convergence theorem to conclude that Z ∞Z Z ∞ ∆h uε (x) − ∆h u(x)p lim dxdh = lim gε (h) dh = 0. h1+sp ε→0+ 0 ε→0+ 0 R
1.5. Density
35
Next we consider the general case. Theorem 1.64. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 0 < s < 1. ¯ Then for every u ∈ W s,p (I) there exists a sequence {un }n in W s,p (I)∩Cc∞ (I) s,p such that un → u in W (I). We begin with a preliminary lemma. Lemma 1.65. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 0 < s < 1. Consider u ∈ L1loc (I), a, b ∈ I, with a < b, and ε > 0 so small that [a − ε, b + ε] ⊆ I. Then for every 0 < h < b − a, Z b−h Z b−h+ε p uε (x + h) − uε (x) dx ≤ u(x + h) − u(x)p dx. a
a−ε
Proof. Note that ε
Z uε (x + h) − uε (x) =
ϕε (y)[u(x + h − y) − u(x − y)] dy, −ε 0
1/p 1/p and using H¨ and thus, writing ϕε (y) older’s inequalR ε = (ϕε (y)) (ϕε (y)) ity and the fact that −ε ϕε (y) dy = 1, we obtain Z ε p uε (x + h) − uε (x) ≤ ϕε (y)u(x + h − y) − u(x − y)p dy. −ε
In turn, by Tonelli’s theorem, Z b−h uε (x + h) − uε (x)p dx a
Z
ε
≤ −ε ε
Z (1.39)
= −ε ε
Z ≤
b−h
Z ϕε (y)
u(x + h − y) − u(x − y)p dxdy
a
Z ϕε (y)
b−h−y
a−y b−h+ε
Z ϕε (y)
−ε Z b−h+ε
=
u(z + h) − u(z)p dzdy u(z + h) − u(z)p dzdy
a−ε
u(z + h) − u(z)p dz,
a−ε
where we make the Rchange of variables z = x − y, so that dz = dx, and use ε again the fact that −ε ϕε (y) dy = 1. Exercise 1.66. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 0 < s < 1. Consider u ∈ L1loc (I), a, b ∈ I with a < b, and ε > 0 so small that [a − ε, b + ε] ⊆ I. Prove that Z bZ b Z b+ε Z b+ε uε (x) − uε (y)p u(x) − u(y)p dxdy ≤ dxdy. x − y1+sp x − y1+sp a a a−ε a−ε
36
1. Fractional Sobolev Spaces in One Dimension
We turn to the proof of Theorem 1.64. Proof. Step 1: Assume that I = (a, ∞) (the case I = (−∞, b) is similar). Given u ∈ W s,p (I), by Lemma 1.63 we may assume that u = 0 outside a bounded set. Consider the function v in (1.18) obtained by a reflection. Since v ∈ W s,p (R), we can consider vε = v ∗ ϕε , where ϕε is a standard ¯ Then mollifier. Then vε ∈ Cc∞ (R). Let u ˜ε be the restriction of vε to I. ∞ ¯ u ˜ε ∈ Cc (I) and by Theorem 1.62, ˜ uε − uW s,p (I) ≤ vε − vW s,p (R) → 0
as ε → 0+
and k˜ uε − ukLp (I) ≤ kvε − vkLp (R) → 0
as ε → 0+ .
Step 2: Assume that I = (a, b). Using the full force of Theorem 1.43, we can extend u to a function v ∈ W s,p (R) and then proceed as in Step 1. Alternatively, we can use Corollary 1.47 to extend u to a function w ∈ W s,p (J), where J = (2a − b, 2b − a). Let wε be the mollification of w, where ¯ Then u 0 < ε < b − a, and let u ¯ε be the restriction of wε to I. ¯ε ∈ C ∞ (I) and kwε − ukLp (I) = kwε − wkLp (I) → 0 (see [Leo22c]). Moreover, for 0 < h < b − a, by Lemma 1.65, Z Z b−h+ε Z b−h p p ∆h w(x) dx ≤ ∆h w(x)p dx. ∆h u ¯ε (x) dx ≤ (1.40) a
a−ε
J
Hence, as in the proof of Theorem 1.62, if we define gε (h) :=
1 k∆h u ¯ε − ∆h ukpLp ((a,b−h)) , h1+sq
g(h) :=
1 k∆h wkpLp (J) , h1+sq
we have that g ∈ L1 ((0, b − a)) since w ∈ W s,p (J). By inequality (1.40), Minkowski’s inequality, and the convexity of the function tp , we have that gε (h) ≤ 2p g(h) for all 0 < h < b − a. Again, as in the proof of Theorem 1.62, using the fact that ∆h wε = ϕε ∗ ∆h w, we have that gε (h) → 0 for all 0 < h < b − a (see [Leo22c]). Hence, we are in a position to apply the Lebesgue dominated convergence theorem to conclude that Z b−a Z b−h Z b−a ∆h u ¯ε (x) − ∆h u(x)p lim dxdh = lim gε (h) dh = 0. h1+sp ε→0+ 0 ε→0+ 0 a It follows from (1.15) that ¯ uε − uW s,p (I) → 0.
In the next two exercises, we provide an alternative proof of Theorem 1.64. Exercise 1.67. Let I = (a, ∞), 1 ≤ p < ∞, 0 < s < 1, and u ∈ W s,p (I). (i) For every h > 0, consider the function uh (x) := u(x + h), x ∈ I. Prove that uh ∈ W s,p (I) and that kuh − ukW s,p (I) → 0 as h → 0+ .
1.5. Density
37
¯ such (ii) For every h > 0, prove that there is {vn }n in W s,p (I) ∩ Cc∞ (I) s,p that vn → uh in W (I). (iii) Use items (i) and (ii) to prove that there exists a sequence {un }n ¯ such that un → u in W s,p (I). in W s,p (I) ∩ Cc∞ (I) Exercise 1.68. Let I = (a, b), 1 ≤ p < ∞, 0 < s < 1, and u ∈ W s,p (I). (i) Construct two functions φ, ψ ∈ C ∞ (R) such that φ+ψ = 1 in [a, b], φ = 0 near a, ψ = 0 near b. (ii) Write u = φu + ψu and use Exercise 1.67 to construct a sequence ¯ such that un → u in W s,p (I). {un }n in W s,p (I) ∩ Cc∞ (I) Corollary 1.69. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 0 < s < 1. ˙ s,p (I) there exists a sequence {un }n in W ˙ s,p (I)∩C ∞ (I) Then for every u ∈ W p such that un − uW s,p (I) → 0 as n → ∞ and un → u in Lloc (I). Proof. If I is unbounded, by Remark 1.41 we can extend u to a function ˙ s,p (R), with vW s,p (R) uW s,p (I) . We can now consider vε = v ∗ ϕε . v∈W On the other hand, if I = (a, b), then u ∈ W s,p (I) by Corollary 1.19. Thus, we can apply Theorem 1.64. Corollary 1.70. Let I ⊂ R be an open interval, 1 ≤ p < ∞, and 0 < s < 1. Let u ∈ W s,p (I)∩C(I) be such that u(x) = 0 for all x ∈ ∂I. Then there there ¯ such that un → u in W s,p (I) exists a sequence {un }n in W s,p (I) ∩ Cc∞ (I) and un (x) = 0 for all x ∈ ∂I. Proof. Step 1: Assume first that I = (a, ∞). Given u ∈ W s,p (I) ∩ C(I) with u(a) = 0, by Lemma 1.63 we can assume that u(x) = 0 for all x ≥ b for some b > a. Fix δ ∈ (0, 1) and let ψ ∈ Cc∞ (R) be such that ψ = 1 in (a, b) and ψ(x) = 0 for x ≥ b + δ. Consider the function v in (1.18) obtained by a reflection of u and let vε = v ∗ ϕε , where ϕε is a standard mollifier. Since v is continuous and has compact support, we have that vε → v uniformly. In particular, vε (a) → u(a) = 0. Consider the functions uε (x) := ψ(x)(vε (x) − vε (a)),
¯ x ∈ I.
Then uε (0) = 0. Moreover, since ψ = 1 in (a, b) and u(x) = 0 for all x ≥ b, we have that u = ψu in I. Hence, by Minkowski’s inequality, kuε − ukLp (I) = kψ(vε − vε (a) − u)kLp (I) ≤ kvε − vε (a) − ukLp ((a,b+1)) ≤ kvε − ukLp ((a,b+1)) + vε (a)(b + 1 − a)1/p → 0.
38
1. Fractional Sobolev Spaces in One Dimension
On the other hand, as in the proof of Lemma 1.63, uε − uW s,p (I) = ψ(vε − vε (a) − u)W s,p (I) ≤ ψ(vε − u)W s,p (I) + vε (a)ψW s,p (I) kψ 0 k∞ kvε − ukLp (I) + kψk∞ vε − uW s,p (I) + vε (a)ψW s,p (I) . The righthand side goes to zero as ε → 0+ . Step 2: Assume that I = (a, b) and let u ∈ W s,p (I) ∩ C(I) with u(a) = u(b) = 0. By Theorem 1.43 we can extend u to a continuous function v ∈ W s,p (R). Consider a function φ ∈ C ∞ (R) such that 0 ≤ φ ≤ 1, φ = 1 near a, and φ = 0 near b. Then v = φv + (1 − φ)v =: v1 + v2 . Since v1 (a) = 0 and v1 (x) = 0 for all x ≥ b − 2δ for some 0 < δ < b−a 2 , by Step 1 we can s,p ∞ construct a sequence vε ∈ W (I) ∩ C ([a, b]) with vε (a) = 0 and vε = 0 for all x ∈ [b − δ, b] such that vε → v1 in W s,p (I). Similarly, since v2 (b) = 0 and v2 (x) = 0 for all x ≤ a + 2δ, by Step 1 (applied with (a, ∞) replaced by the interval (−∞, b)) we can construct a sequence wε ∈ W s,p (I) ∩ C ∞ ([a, b]), with wε (b) = 0 and wε = 0 for all x ∈ [a, a+δ] such that wε → v1 in W s,p (I). Define uε := vε + wε . Then uε (a) = uε (b) = 0 and uε → v1 + v2 = u in W s,p (I). Remark 1.71. We will see in the next section that the hypotheses of Corollary 1.70 imply that u ∈ W0s,p (I) (see Corollary 1.74).
1.6. The Space W0s,p (I) In the theory of differential equations, boundary value problems play a central role. Thus, it is important to understand functions that take value zero at the boundary of I. This leads us to the study of the (sub) spaces W0s,p (I). Definition 1.72. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 0 < s < 1. We define the space W0s,p (I) as the closure of Cc∞ (I) in W s,p (I) with respect to the norm k · kW s,p (I) . Thus, a function u ∈ W s,p (I) belongs to W0s,p (I) if and only if there exists a sequence of functions un ∈ Cc∞ (I) such that kun − ukW s,p (I) → 0. Theorem 1.73. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 0 < s < 1 with sp < 1. Then W0s,p (I) = W s,p (I). Proof. Let u ∈ W s,p (I). In view of Theorem 1.64, we can assume that ¯ u ∈ Cc∞ (I). Step 1: Assume that I = (a, ∞) (the case I = (−∞, b) is similar). Let φ ∈ C ∞ ([0, ∞)) be such that 0 ≤ φ ≤ 1, φ(t) = 1 for t ≥ 1, and φ(t) = 0 for
1.6. The Space W0s,p (I)
39
t ≤ 12 . Given ε > 0, define φε (x) := φ((x − a)/ε), x ∈ I. Then φε u ∈ Cc∞ (I). We estimate the W 1,p (I) norm of u − φε u. We have Z ∞ Z ∞ p p 1 − φ((x − a)/ε)p dx u(x) − φε (x)u(x) dx ≤ kuk∞ a a Z 1 p 1 − φ(y)p dy, = εkuk∞ 0
where in the second equality we make the change of variables x − a = εy. On the other hand, since (φε u)0 = φ0ε u + φε u0 , by inequality (1.1), Z ∞ Z ∞ Z ∞ 0 0 p 0 p u − (φε u)  dx φε u dx + u0 − φε u0 p dx a a a Z ∞ Z 1 (1.41) ε−p kukp∞ φ0 ((x − a)/ε)p dx + εku0 kp∞ 1 − φ(y)p dy a 0 Z 1 Z 1 ε−p+1 kukp∞ φ0 (y)p dy + εku0 kp∞ 1 − φ(y)p dy. 0
0
Hence, (1.42) (1.43)
ku − φε ukLp (I) ε1/p kuk∞ , ku0 − (φε u)0 kLp (I) ε−(1−1/p) kuk∞ + ε1/p ku0 k∞ .
In turn, by Corollary 1.27, (1.44)
0 0 s u − φε uW s,p (I) ku − φε uk1−s Lp (I) ku − (φε u) kLp (I) 0 s ε1/p−s kuk∞ + ε1/p kuk1−s L∞ (I) ku k∞ .
Since p < 1/s, then ku − φε ukW s,p (I) → 0 as ε → 0+ . Since φε u ∈ Cc∞ (I), this shows that u ∈ W0s,p (I). Step 2: Assume that I = (a, b). Consider a function ψ ∈ C ∞ (R) such that 0 ≤ ψ ≤ 1, ψ = 1 near a and ψ = 0 near b. Then v = ψv + (1 − ψ)u =: v1 + v2 . Let φε be as in Step 1. Then φε v1 ∈ Cc∞ (I) and, by Step 1, kv1 − φε v1 kW s,p ((a,∞)) → 0. Similarly, by Step 1, with (−∞, b) in place of (a, ∞), we can define ϕε (x) := φ((b − x)/ε), where φ is as in Step 1 and show that ϕε v2 ∈ Cc∞ (I) and kv2 − ϕε v2 kW s,p ((−∞,b)) → 0. The sequence uε := φε v1 + ϕε v2 belongs to Cc∞ (I) and satisfies kuε − ukW s,p (I) → 0. Corollary 1.74. Let I ⊂ R be an open interval, 1 ≤ p < ∞, and 0 < s < 1. Let u ∈ W s,p (I) ∩ C(I) be such that u(x) = 0 for all x ∈ ∂I. Then u ∈ W0s.p (I). Proof. By Corollary 1.70 we can assume that u ∈ W s,p (I) ∩ Cc∞ (I). Consider first the case I = (a, ∞). We proceed as in the proof of Step 1 of Theorem 1.73 until (1.41). Since u(a) = 0 we can give a better estimate
40
1. Fractional Sobolev Spaces in One Dimension
R∞ for a φ0ε up dx. Since φ0ε (x) = 1ε φ0 ((x − a)/ε), we have that φ0ε (x) = 0 for x ≥ a + ε. By the fundamental theorem of calculus, for x ∈ (a, a + ε), Z x Z x 0 u(x) = u(a) + u (t) dt = u0 (t) dt, a
a
and so, Z u(x) ≤
x
u0 (t) dt ≤ ku0 k∞ (x − a) ≤ ku0 k∞ ε.
a
In turn, ∞
Z a
φ0ε up dx
Z 1 a+ε 0 φ ((x − a)/ε)u(x)p dx ≤ p ε a 1 ≤ p kφ0 kp∞ ku0 kp∞ εp+1 = εkφ0 kp∞ ku0 kp∞ . ε
The estimate (1.43) should be replaced by ku0 − (φε u)0 kLp (I) ε1/p ku0 k∞ , and, consequently, (1.44) simplifies to 0 0 s u − φε uW s,p (I) ku − φε uk1−s Lp (I) ku − (φε u) kLp (I) 0 s ε1/p kuk1−s ∞ ku k∞ .
Together with (1.42), this proves that ku − φε ukW s,p (I) → 0 as ε → 0+ .
Next we consider the case sp = 1. The proof requires readers to be familiar with functional analysis, in particular, with the fact that in a reflexive Banach space a closed subspace is weakly closed. In [Leo22a] we present a much longer proof, but which does not rely on any deep knowledge. Theorem 1.75. Let I ⊆ R be an open interval and 1 < p < ∞. Then 1/p,p W0 (I) = W 1/p,p (I). Proof. As in the proof of Step 1 of Theorem 1.73, we assume that I = (a, ∞) ¯ We proceed exactly as in the proof of Theorem 1.73 and that u ∈ Cc∞ (I). to obtain (1.42) and (1.44), which reduce to (1.45) (1.46)
ku − φε ukLp (I) ε1/p kukL∞ (I) , u − φε uW 1/p,p (I) kuk∞ + ε1/p kuk1−1/p ku0 k1/p ∞ ∞ . 1/p,p
1/p,p
It follows that the family {φε u}ε is bounded in W0 (I). Since W0 (I) 1/p,p is a closed subspace of the reflexive space W (I) (see Theorem 1.20), 1/p,p we have that W0 (I) is reflexive (see [Leo22b]). Therefore, there exists
1.7. Hardy’s Inequality
41
1/p,p
a subsequence {φεn u}εn such that φεn u * v in W0 (I). In particular, 0 p p φεn u * v in L (I). Indeed, given g ∈ L (I), the function Z 1/p,p (I), Tg (w) := g(x)w(x) dx, w ∈ W0 I
is linear and continuous by H¨older’s inequality. Hence, Tg belongs to the 1/p,p dual of W0 (I), which implies that Tg (φεn u) → Tg (v). In turn, this proves that φεn u * v in Lp (I). On the other hand, φεn u → u in Lp (I) by (1.45), 1/p,p so u = v ∈ W0 (I).
1.7. Hardy’s Inequality In Corollary 1.5 we have proved that if u ∈ W 1,p (R+ ), 1 < p < ∞, then the following Hardy inequality holds: Z ∞ Z ∞ 1/p dx 1/p ¯ u(x) − u ¯(0)p p ≤ p0 u0 (x)p dx . x 0 0 Here u ¯ is the representative of u that is absolutely continuous. In this section we study the fractional version of this inequality. Theorem 1.76 (Hardy’s inequality in W s,p ). Let I = (a, ∞), 1 ≤ p < ∞, and 0 < s < 1, with sp 6= 1. If sp < 1, then for all u ∈ W s,p (I), Z ∞ u(x)p dx upW s,p (I) , sp (x − a) a ¯ while if sp > 1, then for all u ∈ W s,p (I) ∩ C(I), Z ∞ u(x) − u(a)p (1.47) dx upW s,p (I) . sp (x − a) a Remark 1.77. In Theorem 2.8 we will see that every function u in W s,p (I) admits a representative u ¯ that is H¨older continuous, and thus (1.47) holds for every u ∈ W s,p (I), provided we take the representative u ¯ in (1.47). We begin with a preliminary lemma. Lemma 1.78. Let λ > 0, with λ 6= 1, 1 ≤ p < ∞, a ∈ R, and u : (a, ∞) → R be a locally integrable function such that Z ∞ u(x)p dx < ∞. (1.48) (x − a)λ a Then Z a
∞
u(x)p dx (x − a)λ
Z a
∞Z ∞ a
u(x) − u(y)p dxdy. x − y1+λ
42
1. Fractional Sobolev Spaces in One Dimension
First proof. By defining v(x) := u(x + a), x > 0, and changing variables, it suffices to consider the case a = 0. For x, y ∈ R+ , by inequality (1.1), u(x)p ≤ 2p−1 u(y)p + 2p−1 u(x) − u(y)p . 1 Fix ρ > 0, divide by ρx1+λ , integrate first in y over (ρx, 2ρx) and then in x over R+ , and use Tonelli’s theorem to find Z ∞ Z 2ρx Z ∞ Z 2ρx Z ∞ u(x)p u(y)p u(x)p p−1 dx = dydx ≤ 2 dydx xλ ρx1+λ ρx1+λ 0 ρx 0 ρx 0 Z ∞ Z 2ρx u(x) − u(y)p + 2p−1 (1.49) dydx. ρx1+λ 0 ρx
By Tonelli’s theorem Z ∞ Z 2ρx Z Z y/ρ u(y)p 1 1 ∞ p u(y) dydx = dxdy 1+λ 1+λ ρ ρx x 0 ρx 0 y/(2ρ) Z ∞ λ u(y)p λ−1 2 − 1 dy, =ρ λ yλ 0 since Z
y/ρ
y/(2ρ)
1
x
dx = 1+λ
ρλ (2λ − 1) 1 . λ yλ
Hence, the righthand side of inequality (1.49) is majorized by Z ∞ Z ∞Z ∞ λ u(y)p u(x) − u(y)p λ−1 2 − 1 ρ dy + C dxdy, ρ λ yλ x − y1+λ 0 0 0 where Cρ := 2p−1
x − y1+λ 1 sup . ρ ρx 1 we can take ρ so small that ρλ−1 2 λ−1 ≤ 21 , while if λ λ < 1 we can take ρ so large that ρλ−1 2 λ−1 ≤ 12 . Thus, Z ∞ Z Z ∞Z ∞ u(x)p 1 ∞ u(x)p u(x) − u(y)p dx ≤ dx + C dxdy. ρ 2 0 xλ xλ x − y1+λ 0 0 0
1.7. Hardy’s Inequality
43
In view of assumption (1.48) we have Z Z ∞Z ∞ u(x) − u(y)p 1 ∞ u(x)p dx ≤ C dxdy. ρ 2 0 xλ x − y1+λ 0 0
We present a second proof of Lemma 1.78 in the case 0 < λ < 1. Second proof. Assume that 0 < λ < 1 and as before take a = 0. For ρ > 1 and x > 0 we can write Z ρx Z ρx 1 1 u(x) = u(y) dy + [u(x) − u(y)] dy. ρx 0 ρx 0 By H¨older’s inequality, and writing 1 = x − y(1+λ)/p x − y−(1+λ)/p , 1/p 0 Z ρx (ρx)1/p p u(x) ≤ u(y) dy ρx 0 Z ρx 1/p0 Z ρx 1/p 1 u(x) − u(y)p 0 + x − y(1+λ)p /p dy dy ρx x − y1+λ 0 0 Z ρx 1/p Z ρx 1/p 1 2 u(x) − u(y)p p ≤ u(y) dy + dy ρx 0 x − y1+λ (ρx)λ/p 0 since Z ρx Z x Z (1+λ)p0 /p (1+λ)/(p−1) x − y dy = (x − y) dy + 0
0 (1+λ)/(p−1)+1
ρx
(y − x)(1+λ)/(p−1) dy
x
≤ [1 + (ρ − 1) ]x(1+λ)/(p−1)+1 ≤ 2(ρx)(1+λ)/(p−1)+1 p−1 λ λ and 1+λ + 1 p−1 p − 1 = p . Raising both sides to power p, dividing by x , and integrating in x over R+ gives Z ∞ u(x)p dx xλ 0 Z Z ρx Z Z 2p−1 ∞ 1 2p+1 ∞ ρx u(x) − u(y)p p ≤ u(y) dydx + dydx ρ x1+λ 0 ρλ 0 x − y1+λ 0 0 Z ∞ Z Z 2p−1 u(y)p 2p+1 ∞ ρx u(x) − u(y)p = 1−λ dy + λ dydx, λρ yλ ρ x − y1+λ 0 0 0 where we use Tonelli’s theorem. Since 1 − λ > 0, we can choose ρ > 1 so 1 2p−1 large that λρ 1−λ ≤ 2 . Exercise 1.79. Prove that when 1 < λ < p + 1, condition (1.48) holds if u ∈ Cc∞ ([a, ∞)) with u(a) = 0, while when 0 < λ < 1, condition (1.48) holds for functions u ∈ Cc∞ ([a, ∞)).
44
1. Fractional Sobolev Spaces in One Dimension
Corollary 1.80. Let λ > 1, 1 ≤ p < ∞, a, b ∈ R, and u : (a, b) → R be a R b u(x)p locally integrable function such that a (x−a) λ dx < ∞. Then Z a
b
u(x)p dx (x − a)λ
Z bZ a
a
b
u(x) − u(y)p dxdy. x − y1+λ
Proof. As in the first proof of Lemma 1.78, we can assume that a = 0. Taking ρ ≤ 21 , we have that 2ρx ≤ x and so (ρx, 2ρx) ⊂ (0, b) for every x ∈ (0, b). Hence, we can now proceed as in the first proof of Lemma 1.78, R∞ Rb with the only change being that 0 should be replaced by 0 . We turn to the proof of Theorem 1.76. Proof of Theorem 1.76. By defining v(x) := u(x + a), x > 0, and changing variables, it suffices to consider the case a = 0. Step 1: Assume sp < 1. Let u ¯ be a representative of u. Since Z ∞Z ∞ ¯ u(x) − u ¯(y)p dxdy < ∞, x − y1+sp 0 0 by Fubini’s theorem there exists a Lebesgue measurable set E ⊆ R+ with L1 (R+ \ E) = 0 such that Z ∞ ¯ u(x) − u ¯(y)p dx < ∞ (1.50) x − y1+sp 0 for all y ∈ E and every y ∈ E is a Lebesgue point of u ¯. It follows that for y ∈ E, Z y+1 ¯ u(x) − u ¯(y)p dx < ∞. x − y1+sp y In turn, by inequality (1.1), Z y+1 Z y+1 Z y+1 ¯ u(x)p ¯ u(x) − u ¯(y)p ¯ u(y)p dx dx + dx x − ysp x − ysp x − ysp y y y Z y+1 ¯ u(x) − u ¯(y)p dx + ¯ u(y)p < ∞, 1+sp x − y y where in the last inequality we use the fact that x − y1+sp ≤ x − ysp and that sp < 1. On the other hand, since u ∈ Lp (R+ ), Z ∞ Z ∞ ¯ u(x)p dx ≤ ¯ u(x)p dx < ∞. sp y+1 x − y y+1 R ∞ ¯u(x)p Summing these two inequalities gives y x−y sp dx < ∞ for every y ∈ E.
1.7. Hardy’s Inequality
45
Let yn → 0+ with yn ∈ E. By Lemma 1.78, Z ∞ Z ∞Z ∞ ¯ u(x)p ¯ u(x) − u ¯(y)p dx dxdy sp x − y1+sp yn x − yn  yn yn Z ∞Z ∞ ¯ u(x) − u ¯(y)p dxdy. x − y1+sp 0 0 Letting n → ∞ and using Fatou’s lemma, we obtain Z ∞ Z ∞Z ∞ ¯ u(x)p u(x) − u(y)p dx dxdy. sp x x − y1+sp 0 0 0 Step 2: Assume sp > 1. By Theorem 1.64, without loss of generality, we s,p ∞ ¯ may assume that u ∈ W R x 0(I) ∩ C (I). By the fundamental theorem of calculus, u(x) − u(0) = 0 u (t) dt. Therefore, for every b > 0, Z p Z b Z b 1 x 0 u(x) − u(0)p dx ≤ u (t) dt dx sp sp x 0 x 0 0 Z b ≤ ku0 kp∞ xp(1−s) dx < ∞. 0
Hence, we may apply Corollary 1.80 to the function u − u(0) to obtain Z bZ b Z b u(x) − u(y)p u(x) − u(0)p dx dxdy. sp x x − y1+sp 0 0 0 Letting b → ∞ and using the Lebesgue monotone convergence theorem, we obtain Z ∞ Z ∞Z ∞ u(x) − u(0)p u(x) − u(y)p dx dxdy. xsp x − y1+sp 0 0 0 Note that in Step 2 of the proof of Theorem 1.76 we actually proved the following result: Corollary 1.81. Let I = (a, b), where a < b ≤ ∞, 1 ≤ p < ∞, and 0 < s < 1, with sp > 1. Then Z bZ b Z b u(x) − u(a)p u(x) − u(y)p dx dxdy (x − a)sp x − y1+sp a a a ¯ for all u ∈ W s,p (I) ∩ C(I). Exercise 1.82. Let 1 ≤ p < ∞ and 0 < s < 1 be such that sp ≥ 1. Prove R∞ p p s,p (R ). that the Hardy inequality 0 u(x) + xsp dx uW s,p (R+ ) fails in W
46
1. Fractional Sobolev Spaces in One Dimension
Exercise 1.83. Let I = (a, b), where a < b ≤ ∞, 1 ≤ p < ∞, and 0 < s < 1, with sp < 1. Prove that for all u ∈ W s,p (I), Z b u(x)p dx sp a min{x − a, b − x} Z b Z bZ b 1 u(x) − u(y)p p u(x) dx + dxdy. sp (b − a) x − y1+sp a a a When sp > 1 we have Hardy’s inequality in W0s,p (R+ ). Theorem 1.84 (Hardy’s inequality in W0s.p ). Let I = (a, ∞), 1 ≤ p < ∞, and 0 < s < 1 with sp > 1. Then for all u ∈ W0s,p (I), Z ∞Z ∞ Z ∞ u(x) − u(y)p u(x)p dx dxdy. (x − a)sp x − y1+sp a a a Proof. If u ∈ Cc∞ (I), by Theorem 1.76, Z ∞ Z ∞Z ∞ u(x)p u(x) − u(y)p dx dxdy. (x − a)sp x − y1+sp a a a If u ∈ W0s,p (I), construct a sequence of functions un ∈ Cc∞ (I) such that un → u in W s,p (I). By extracting a subsequence, we can assume that un → u L1 a.e. in I. Then Z ∞ Z ∞Z ∞ un (x)p un (x) − un (y)p dx dxdy. (x − a)sp x − y1+sp a a a Letting n → ∞ and using Fatou’s lemma on the left and the fact that un → u in W s,p (I) gives the desired result. s,p 1.8. The Space W00
The space W01,p (I) can be characterized as the subspace of all functions u ∈ W 1,p (I) such that u ¯ = 0 at ∂I, where u ¯ is the absolutely continuous representative of u. Thus, if we extend u ¯ to be zero in R\I, we have that the resulting function belongs to W 1,p (R). Conversely, if u ∈ W 1,p (I) and the function u ˜ obtained by extending u by zero in R\I belongs to W 1,p (R), then, necessarily, u ¯ = 0 at ∂I, where u ¯ is the absolutely continuous representative 1,p of u ˜. In turn, u ∈ W0 (I). This shows that W01,p (I) can be characterized as the space of all functions u ∈ W 1,p (I) whose extension by zero in R \ I belongs to W 1,p (R). In this section, we show that this characterization fails for fractional Sobolev spaces if sp = 1. Definition 1.85. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 0 < s < 1. s,p We define the space W00 (I) as the space of all functions u ∈ W s,p (I) such
s,p 1.8. The Space W00
47
that the function u ˜ : R → R, given by u(x) if x ∈ I, u ˜(x) := 0 if x ∈ R \ I, belongs to W s,p (R). s,p Exercise 1.86. Let 1 ≤ p < ∞ and 0 < s < 1. Prove that W00 (R+ ) ⊆ s,p W0 (R+ ).
Hint: Use translations. Theorem 1.87. Let 1 ≤ p < ∞ and 0 < s < 1. Then u ∈ W s,p (R+ ) belongs s,p to u ∈ W00 (R+ ) if and only if Z ∞ u(x)p dx < ∞. xsp 0 Moreover, (i) If sp < 1, then s,p W00 (R+ ) = W0s,p (R+ ) = W s,p (R+ ).
(ii) If sp = 1, then s,p (R+ ) ⊂ W0s,p (R+ ) = W s,p (R+ ), W00
where the inclusion is strict. (iii) If sp > 1, then s,p (R+ ) = W0s,p (R+ ) ⊂ W s,p (R+ ), W00
where the inclusion is strict. Proof. Step 1: In this step we show that u ∈ W s,p (R+ ) belongs to u ∈ R∞ p s,p W00 (R+ ) if and only if 0 u(x) xsp dx < ∞. Define u(x) if x ∈ R+ , u ˜(x) := 0 if x ∈ R− . Then by Tonelli’s theorem and the symmetry of the integral, Z Z Z Z ˜ u(x) − u ˜(y)p u(y)p dxdy = 2 dxdy 1+sp 1+sp R R x − y R+ R− x − y Z Z u(x) − u(y)p dxdy =: A + B. + 1+sp R+ R+ x − y R For y > 0, R− x−y11+sp dx ≈ y1sp , and so, by Tonelli’s theorem, Z Z Z 1 u(y)p p A=2 u(y) dxdy ≈ dy. 1+sp y sp R+ R− x − y R+ R∞ p This shows that A < ∞ if and only if 0 u(x) xsp dx < ∞.
48
1. Fractional Sobolev Spaces in One Dimension
Step 2: If sp < 1, then by Theorem 1.76, Hardy’s inequality holds in W s,p (R+ ), and thus, by Step 1 and Exercise 1.86 (or Theorem 1.73), item (i) holds. If sp = 1, item (ii) follows from Theorem 1.75, Step 1, and Exercises 1.82 and 1.86. If sp > 1, by Theorem 1.84, Hardy’s inequality holds in W0s,p (R+ ) and s,p thus, by Step 1 and Exercise 1.86, W00 (R+ ) = W s,p (R+ ). Item (iii) is then a consequence of Step 1 and Exercise 1.82. Exercise 1.88. Let I = (a, b), 1 ≤ p < ∞, and 0 < s < 1. Prove that Rb u(x)p s,p u ∈ W00 (I) if and only if a min{x−a,b−x} sp dx < ∞.
1.9. Notes For more information about Hardy’s inequality (Theorem 1.3), we refer to the books of Kufner and Persson [KP03], of Kufner, Maligranda, and Persson [KMP07] (see also [KMP06]), and the review paper of Mironescu [Mir18]. Theorem 1.6 is due to Il’in [Il’59] (see also Simon [Sim90]). Theorem 1.28 is adapted from [KP03, Theorem 5.3]. Exercises 1.33, 1.34, 1.35, and 1.36 are drawn from the book of F. Demengel and G. Demengel [DD12]. Theorem 1.48 is adapted from [Mir18]. The first proof of Lemma 1.78 appears in [Mir18], while the second is from the paper of Solonnikov [Sol67]. We refer to the paper of Krugljak, Maligranda, and Persson [KMP00] for s,p another proof. To my knowledge, the spaces W00 (I) were discovered by Lions and Magenes (see [LM61] and [LM72a, Section 11.3]).
Chapter 2
Embeddings and Interpolation Don’t always blindly follow guidance and stepbystep instructions; you might run into something interesting. — Georg Cantor
In this chapter you will learn the relationship between different fractional Sobolev spaces, and between fractional Sobolev spaces and other spaces. To be precise, we are interested in understanding continuous inclusions of the type W s2 ,p2 (I) ,→ W s1 ,p1 (I). As a starting point let’s take I = R and consider the inequality uW s1 ,p1 (R) uW s2 ,p2 (R) ,
(2.1)
where 1 ≤ p1 , p2 ≤ ∞ and 0 ≤ s1 , s2 ≤ 1. Here, if s1 = 1, we set uW 1,p1 (R) := ku0 kLp1 (R) , while if s1 = 0, we define W 0,p1 (R) := Lp1 (R) and uW 0,p1 (R) := kukLp1 (R) . We also recall that if p1 = ∞, then W s1 ,∞ (R) can be identified with the space C 0,s1 (R). Let’s assume that (2.1) holds for every u ∈ W s2 ,p2 (R), where 1 ≤ p2 < ∞, 0 < s2 < 1. Fix u ∈ W s2 ,p2 (R) and for λ > 0, consider the rescaled function uλ (x) := u(λx), x ∈ R. By the changes of variables λx = X and λy = Y , we get Z Z uλ 
W s2 ,p2 (R)
= =λ
R R s2 −1/p2
u(λx) − u(λy)p2 dxdy x − y1+s2 p2
1/p2
uW s2 ,p2 (R) .
The cases s2 = 1, s2 = 0, and p2 = ∞ are analogous. Similarly, we have that uλ W s1 ,p1 (R) = λs1 −1/p1 uW s1 ,p1 (R) . Since (2.1) holds for uλ , for every 49
50
2. Embeddings and Interpolation
λ > 0 we get (2.2)
uW s1 ,p1 (R) λs2 −1/p2 −s1 +1/p1 uW s2 ,p2 (R) .
If s2 −1/p2 −s1 +1/p1 > 0, we send λ → 0+ , while if s2 −1/p2 −s1 +1/p1 < 0, we send λ → ∞ to get uW s2 ,p2 (R) = 0, which is a contradiction, unless u is constant. Hence, a necessary condition for (2.1) to hold is 1 1 = s2 − . (2.3) s1 − p2 p1 In the first section, we are going to study the endpoints, namely the cases s1 = 0 and p1 = ∞. Note that if s1 = 0 and 1 ≤ p1 < ∞, necessarily, s2 − p12 < 0. This is called the subcritical case. If s1 = 0 and p1 = ∞, then s2 − p12 = 0, which is the critical case. We will show that (2.1) fails in this case. Finally, if s1 > 0 and p1 = ∞, then s2 − p12 > 0, which is the supercritical case.
2.1. Embeddings: The Endpoints Since in this section we will deal with only one fractional Sobolev space, we will replace p2 , s2 with p, s. In the subcritical case 1 ≤ p < 1s , we will prove ∗ that W s,p (R) ,→ Lps (R), where p p∗s := 1 − sp is called the Sobolev critical exponent for W s,p (R). Note that p1 − s = p1∗ , so s that (2.1) holds in this case. In the critical case p = 1/s, we will prove that W 1/p,p (R) ,→ Lq (R) for all p ≤ q < ∞. We will prove that the embedding W 1/p,p (R) ,→ L∞ (R) fails for 1 < p < ∞, so that (2.1) does not hold. Finally, in the supercritical case p > 1/s, we will show that W s,p (R) ,→ C 0,s−1/p (R). Since C 0,s−1/p (R) can be identified with W s−1/p,∞ (R), (2.1) holds in this case. We recall the following definition. Definition 2.1. Let E ⊆ R 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, (2.4)
L1 ({x ∈ E : u(x) > t}) < ∞.
We begin with the subcritical case p < 1/s. Theorem 2.2 (Sobolev–Gagliardo–Nirenberg’s embedding in W s,p ). Let ˙ s,p (R) vanishing at 0 < s < 1 and 1 ≤ p < 1/s. Then for every u ∈ W infinity, (2.5)
kukLp∗s (R) uW s,p (R) .
Moreover, W s,p (R) ,→ Lq (R) for all p ≤ q ≤ p∗s .
2.1. Embeddings: The Endpoints
51
∗
˙ s,p (R) ∩ Lps (R). For x ∈ R and r > 0 Proof. Step 1: Assume that u ∈ W we have Z Z 1 r 2p−1 r p p u(x) dh ≤ u(x + h) − u(x)p dh u(x) = r 0 r 0 Z 2p−1 r + u(x + h)p dh. r 0 Using H¨older’s inequality with exponents p∗s /p and (p∗s /p)0 = 1/(sp), we obtain Z x+r Z r u(y)p dy ≤ rsp kukpLp∗s (R) . u(x + h)p dh = 0
x
Combining these two inequalities gives 2p−1 u(x) − 1−sp kukpLp∗s (R) ≤ 2p−1 rsp r p
(2.6)
r
u(x + h) − u(x)p dh r1+sp 0 Z ∞ u(x + h) − u(x)p p−1 sp ≤2 r dh, h1+sp 0 Z
where in the last inequality we use the fact that h < r. If u(x) 6= 0 and kukLp∗s (R) > 0, we take r so that !p kukLp∗s (R) 1−sp p , r =2 u(x) so that inequality (2.6) becomes !sp2 /(1−sp) Z ∞ kukLp∗s (R) u(x + h) − u(x)p 1 p u(x) dh, 2 u(x) h1+sp 0 or, equivalently, since p + sp2 /(1 − sp) = p/(1 − sp) = p∗s , Z ∞ u(x + h) − u(x)p ∗ s −p u(x)ps kukp∗ dh. ∗ p s L (R) h1+sp 0 Note that the last inequality continues to hold if u(x) = 0 or kukLp∗s (R) = 0. s −p Integrating in x over R and dividing both sides by kukp∗ gives ∗ Lps (R) Z Z ∞ u(x + h) − u(x)p kukpLp∗s (R) dhdx upW s,p (R) . 1+sp h R 0 ∗
Thus, we have proved (2.5) under the additional hypothesis that u ∈ Lps (R). Step 2: For n ∈ N define (t − 1/n) sgn t if 1/n ≤ t ≤ n, 0 if t < 1/n, fn (t) := (n − 1/n) sgn t if t > n,
52
2. Embeddings and Interpolation
and vn := fn ◦ u. Then Z Z p∗s vn  dx = R
∗
vn ps dx
{u>1/n} ∗
≤ (n − 1/n)ps L1 ({x ∈ R : u(x) > 1/n}) < ∞, since u is vanishing at infinity. Moreover, since fn is Lipschitz continuous with Lipschitz constant 1, vn (x + h) − vn (x) = fn (u(x + h)) − fn (u(x)) ≤ u(x + h) − u(x), ∗
˙ s,p (R) ∩ Lps (R). Hence, we may apply Step 1 to vn and so vn belongs to W to obtain kvn kLp∗s (R) vn W s,p (R) uW s,p (R) . Letting n → ∞ and using Fatou’s lemma on the lefthand side gives (2.5). The proof of the last statement is left as an exercise.
Corollary 2.3. Let I ⊆ R be an open interval, 0 < s < 1 and 1 ≤ p < 1/s. ˙ s,p (I) vanishing at infinity, If I has infinite measure, then for every u ∈ W kukLp∗s (I) uW s,p (I) . ˙ s,p (I), On the other hand, if I has finite measure, then for every u ∈ W Z 1 + uW s,p (I) , ∗ kukLps (I) 1 u(x) dx ∗ (L (I))1−1/ps J where J b I is an open interval with L1 (I) ≤ 2L1 (J). Moreover, W s,p (I) ,→ Lq (I) for all p ≤ q ≤ p∗s . Proof. If I = (a, ∞) or I = (−∞, b), we can reflect u (see Theorem 1.40 ˙ s,p (R) such that vW s,p (R) and Remark 1.41), to obtain a function v ∈ W uW s,p (I) . It is now enough to apply Theorem 2.2 to v. If I = (a, b), let [c, d] ⊂ (a, b), with c − d ≥ 12 (b − a). Then the function R ˙ s,p (I) and d u u ˜ := u − u(c,d) belongs to W c ˜ dx = 0. Hence, by Corollary 1.50, we can extend u ˜ to a function v ∈ W s,p (R) such that kvkLp (R) (b − a)s k˜ ukLp (I) ,
vW s,p (R) uW s,p (I) .
By applying Theorem 2.2 to v, we obtain that k˜ ukLp∗s (I) ≤ kvkLp∗s (R) vW s,p (R) uW s,p (I) . In turn, ∗
kukLp∗s (I) ≤ (b − a)1/ps u(c,d)  + k˜ ukLp∗s (I) ∗
(b − a)1/ps u(c,d)  + uW s,p (I) .
2.1. Embeddings: The Endpoints
53
Next we study the critical case s = 1/p. Observe that as s → (1/p)− , p = 1−sp → ∞. This fact would suggest that in the critical case s = 1/p we could expect the embedding L∞ (R) ,→ W 1/p,p (R). The following exercise shows that this is not the case. p∗s
Exercise 2.4. Let f be a smooth function such that f (x) = (x log x)−1 for x ≥ 2 and f = 0 for x ≤ −1. Let v be its inverse Fourier transform (see [Leo22c]). (i) Prove that v ∈ H 1/2 (R). (ii) Let ϕ be a standard mollifier and vε := v∗ϕε . Prove that vε (0) → ∞ as ε → 0+ . (iii) Deduce that there is no constant C = C(p) > 0 such that kukL∞ (R) ≤ CkukH 1/2 (R)
for every u ∈ H 1/2 (R).
Theorem 2.5. Let 1 < p < ∞ and s = 1/p. Then for every p ≤ q < ∞ and for every u ∈ W 1/p,p (R), kukLq (R) kukW 1/p,p (R) . We begin with a preliminary embedding, which is of interest in itself. Lemma 2.6. Let I ⊆ R be an interval, 1 ≤ p < ∞, 0 < s1 < s2 < 1, and u ∈ W s2 ,p (I). If I is bounded, then uW s1 ,p (I) (L1 (I))s2 −s1 uW s2 ,p (I) , while if I is unbounded, then 1 kukLp (I) + `s2 −s1 uW s2 ,p (I) `s1 for every ` > 0. Moreover, W s2 ,p (I) ,→ W s1 ,p (I). uW s1 ,p (I)
Proof. Write Z Z Z Z u(x) − u(y)p u(x) − u(y)p dxdy = dxdy 1+s p 1+s1 p 1 I I x − y I I∩B(y,`) x − y Z Z u(x) − u(y)p + dxdy =: A + B. 1+s1 p I I\B(y,`) x − y Since s1 < s2 , if x − y ≤ `, we have that `(s2 −s1 )p x − y1+s1 p ≥ x − y1+s2 p , so that Z Z u(x) − u(y)p (s2 −s1 )p A≤` dxdy. 1+s2 p I I∩B(y,`) x − y If I is bounded, then by taking ` = 2L1 (I), we have that B = 0, and the proof is complete. Thus, in what follows we assume that I is unbounded.
54
2. Embeddings and Interpolation
By Tonelli’s theorem Z Z 1 p B u(x) dydx 1+s1 p I\B(x,`) x − y I Z Z Z 1 1 p + u(y) dxdy s1 p u(x)p dx. 1+s1 p x − y ` I\B(y,`) I I
We turn to the proof of Theorem 2.5. Proof. If 0 < s1 < s2 := 1/p, then s1 is subcritical and so we may apply Theorem 2.2 to obtain kukLp∗s1 (R) uW s1 ,p (R) , where p∗s1 = 1−sp 1 p . It follows by Lemma 2.6 that kukLp∗s1 (R) uW s1 ,p (R) kukW 1/p,p (R) . To conclude the proof, observe that for every p < q < ∞, we can find s1 ∈ (0, 1/p) such that q = p∗s1 . To be precise, take s1 = (q − p)/(pq). Exercise 2.7. Let I ⊆ R be an open interval, 1 < p < ∞, and s = 1/p. Prove that for every p ≤ q < ∞ and for every u ∈ W 1/p,p (I), kukLq (I) I kukW 1/p,p (I) . Next we prove that in the supercritical case p > 1/s, every function u in W s,p (I) admits a H¨older continuous representative. Theorem 2.8 (Morrey’s embedding in W s,p ). Let I ⊆ R be an open interval, ˙ s,p (I) admits a 0 < s < 1, and 1/s < p < ∞. Then every function u ∈ W H¨ older continuous representative u ¯ with exponent s − 1/p such that ¯ uC 0,s−1/p (I) uW s,p (I) . Moreover, if u ∈ W s,p (I), then for every 0 < ` < L1 (I), 1 (2.7) k¯ uk∞ 1/p kukLp (I) + `s−1/p uW s,p (I) . ` s,p 0,β In addition, W (I) ,→ C (I) for every 0 < β ≤ s − 1/p. We will present two proofs. ˙ s,p (I) ∩ C ∞ (I). By Corollary First proof. Step 1: Assume that u ∈ W 1.81, for every a, b ∈ I, with a < b, Z bZ b Z b u(x) − u(a)p u(x) − u(y)p (2.8) dx dxdy. sp (x − a) x − y1+sp a a a Similarly, Z (2.9) a
b
u(b) − u(x)p dx (b − x)sp
Z bZ a
a
b
u(x) − u(y)p dxdy. x − y1+sp
2.1. Embeddings: The Endpoints
55
Subdivide (a, b) into three equal intervals and let J be the middle one. For x ∈ J, p p u(b) − u(a)p p−1 u(b) − u(x) p−1 u(x) − u(a) ≤ 2 + 2 (b − a)sp (b − a)sp (b − a)sp p p u(b) − u(x) p−1 u(x) − u(a) + 2 , ≤ 2p−1 (b − x)sp (x − a)sp
where we use the fact that x − a ≤ b − a and b − x ≤ b − a. Integrating both sides over J gives Z Z u(b) − u(x)p u(x) − u(a)p u(b) − u(a)p p−1 p−1 ≤ 2 dx + 2 dx 3 sp−1 sp (b − a) (b − x) (x − a)sp J J Z bZ b u(x) − u(y)p dxdy, x − y1+sp a a where in the last inequality we use (2.8) and (2.9). Step 2: In this step we remove the additional hypothesis that u is of class ˙ s,p (I) ∩ C 1 . In view of Corollary 1.69, there exists a sequence {un }n in W p C ∞ (I) such that un − uW s,p (I) → 0 as n → ∞ and un → u in Lloc (I). Fix a representative u ¯ of u. By extracting a subsequence, we may assume that there exists E0 ⊂ I, with L1 (E0 ) = 0, such that un (x) → u ¯(x) as n → ∞ for all x ∈ I \ E0 . By Step 1 for every a, b ∈ I, with a < b, and for every n ∈ N, Z bZ b un (b) − un (a)p un (x) − un (y)p (2.10) dxdy. (b − a)sp−1 x − y1+sp a a In particular, if a, b ∈ I \ E0 , then un (b) − un (a) (b − a)s−1/p un W s,p (I) . Letting n → ∞ gives (2.11)
¯ u(b) − u ¯(a) (b − a)s−1/p uW s,p (I)
for all a, b ∈ I \ E0 . Now given x ∈ E0 , since L1 (E0 ) = 0, we can find ak ∈ I \ E0 such that ak → x. By (2.11) applied to am and ak , we get ¯ u(am ) − u ¯(ak ) am − ak s−1/p uW s,p (I) , which implies that {¯ u(ak )}k is a Cauchy sequence in R. Thus, u ¯(ak ) → `x as k → ∞. If bk ∈ I \ E0 is another sequence converging to x, then by (2.11) applied with bk and ak , we get u ¯(bk ) → `x . Thus `x does not depend on the particular sequence approaching x. Define v(x) := `x and observe that again by (2.11) with ak and b ∈ I \ E0 , ¯ u(b) − u ¯(an ) b − ak s−1/p uW s,p (I) .
56
2. Embeddings and Interpolation
Letting k → ∞ gives ¯ u(b) − `x  b − xs−1/p uW s,p (I) for all x ∈ E0 and all b ∈ I \ E0 . Thus, if we define v(x) := `x if x ∈ E0 and v(x) := u ¯(x) if x ∈ I \ E0 , we have that v(x) − v(y) x − ys−1/p uW s,p (I)
(2.12) for all x, y ∈ I.
Step 3: In this step we assume that u ∈ W s,p (I) and prove (2.7). v be the continuous representative of u constructed in Step 2. Let ` < 12 L1 (I). Given x ∈ I, assume that x + ` ∈ I (the case x − ` is similar). Since v is continuous, there exists yx ∈ (x, x + `) such R 1 x+` v(y) dy = v(yx ). By (2.12) and H¨older’s inequality, ` x
v(x) ≤ v(yx ) + v(x) − v(yx )
(2.13)
0 `1/p
`
Z
x+`
v(y)p dy
1/p
1 `
Z
Let 0 < ∈ I that
x+`
v(y) dy + `s−1/p uW s,p (I)
x
+ `s−1/p uW s,p (I) .
x
If L1 (I) < ∞ and 12 L1 (I) ≤ ` < L1 (I), we apply inequality (2.13) with ` 2 in place of `. Hence, (2.7) holds. Together with (2.12), this shows that W s,p (I) ,→ C 0,s−1/p (I). The embedding W s,p (I) ,→ C 0,β (I) follows from the embedding C 0,s−1/p (I) ,→ C 0,β (I) for all 0 < β ≤ s − 1/p, the latter being left as an exercise.
The second proof of Theorem 2.8 relies on the following lemma. We recall that Ih = {x ∈ I : x + h ∈ I}. Lemma 2.9. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 0 < s < 1. ˙ s,p (I), Then for all u ∈ W
sup h>0
1 hsp
Z Ih
u(x + h) − u(x)p dx
Z
∞
Z sup
0
0≤t≤h It
u(x + t) − u(x)p dx
dh . h1+ps
2.1. Embeddings: The Endpoints
57
R Proof. Since the function h 7→ sup0≤t≤h It u(x+t)−u(x)p dx is increasing, for r > 0 we have Z ∞ Z dh sup u(x + t) − u(x)p dx 1+ps h 0 0≤t≤h It Z ∞ Z dh sup u(x + t) − u(x)p dx 1+ps ≥ h r 0≤t≤h It Z ∞ Z 1 dh ≥ sup u(x + t) − u(x)p dx 1+ps h 0≤t≤r It r Z 1 1 = sup u(x + t) − u(x)p dx ps rps 0≤t≤r It Z 1 1 ≥ u(x + r) − u(x)p dx. ps rps Ir Taking the supremum over all r > 0 proves the lemma.
We turn to the second proof of Theorem 2.8. Second proof. Step 1: Assume that I is unbounded. We will consider only the cases in which I = R or I = (a, ∞) and leave the case I = (−∞, b) ˙ s,p (I)∩C 1 (I). as an exercise. Then Ih = I for all h > 0 (see (1.2)). Let u ∈ W We claim that Z 1/p u(x + h) − u(x) 1 p (2.14) sup sup sup s u(y + h) − u(y) dy . hs−1/p h>0 h h>0 x∈I I Let v ∈ C 1 (I). Then by Theorem 1.6 and Remark 1.7, for all h > 0 and x ∈ I, (2.15) Z Z h Z h−r 1 h 1 v(x) = v(x + t) dt − (v(x + t + r) − v(x + t)) dtdr. h 0 (t + r)2 0 0 In turn, by H¨ older’s inequality, Z h Z h Z h−r 1 1 v(x) ≤ v(x + t) dt + ∆r v(x + t) dtdr h 0 (t + r)2 0 0 Z h 1/p Z h Z h−r 1/p dr 1 p p ≤ 1−1/p0 v(x + t) dt + ∆r v(x + t) dt 0 2−1/p h r 0 0 0 Z x+h 1/p Z h Z x+h−r 1/p dr 1 p p = 1−1/p0 v(y) dy + ∆r v(y) dy 0, 2−1/p h r x 0 x where we use the fact that Z h−r 1 1 1 1 1 1 ≤ 0 0 dt = 0 −1 − 0 −1 0 2p 2p 2p 2p 2p − 1 r 2p − 1 r 0 −1 (t + r) h 0
58
2. Embeddings and Interpolation
and the change of variables x + t = y. Taking v(x) = u(x + h) − u(x) = ∆h u(x), we get u(x + h) − u(x) Z 1/p Z h Z 1/p 1 dr p p ≤ 1/p ∆h u(y) dy + ∆r u(y) dy 1+1/p h r I 0 I 1/p Z h Z dr ∆r u(y + h)p dy (2.16) + 1+1/p r 0 I Z 1/p 1/p Z h Z 1 dr p p ∆h u(y) dy , ≤ 1/p +2 ∆r u(y) dy 1+1/p h r I 0 I where in the third integral we use the change of variables ρ = y + h. In turn, Z 1/p u(x + h) − u(x) 1 p sup sup ∆h u(y) dy sup s hs−1/p h>0 x∈I h>0 h I 1/p Z h Z 1 dr + sup s−1/p ∆r u(y)p dy =: A + B. 1+1/p h r h>0 0 I Since 1 B ≤ sup s τ >0 τ
Z
p
∆τ u(y) dy
1/p sup
1
hs−1/p 1/p Z 1 1 ∆τ u(y)p dy , = sup s s − 1/p τ >0 τ I I
h>0
Z 0
h
rs r1+1/p
dr
we have shown (2.14). By (2.14), Lemma 2.9, and Theorem 1.53 we have Z 1/p u(x + h) − u(x) 1 p sup sup sup s u(y + h) − u(y) dy hs−1/p h>0 x∈I h>0 h I !1/p Z ∞ Z dh sup u(x + t) − u(x)p dx 1+ps uW s,p (I) . h 0 0≤t≤h I To remove the hypothesis that u ∈ C 1 (I), we proceed as in Step 2 of the first proof. Step 2: For the case in which I = (a, b), we use Corollary 1.47 to extend u to a function w ∈ W s,p (J), where J = (a − (b − a), b + (b − a)), with wW s,p (J) uW s,p (I) . Given x, y ∈ (a, b), with y > x, write y = x + h, then h < b − a and x ∈ Ih = (a, b − h) (see (1.2)). Thus, sup x,y∈I, x6=y
u(x) − u(y) u(x + h) − u(x) = sup sup . x − ys−1/p hs−1/p h>0 x∈Ih
2.1. Embeddings: The Endpoints
59
Hence, in (2.15) we take 0 < h < b − a and x ∈ Ih . In turn, when we define v(x) = ∆h u(x) just before (2.16), ∆h u should be replaced by ∆h w and I with (a, b + (b − a)) ⊆ J. We can now continue as in Step 1 with ∆h u, I, h u(x) h u(x) and suph>0 supx∈I ∆ replaced by ∆h w, J, and suph>0 supx∈Ih ∆ , hs−1/p hs−1/p R∞ R b−a respectively. We also replace 0 dh with 0 dh. Exercise 2.10. Let I ⊂ R be an open interval, with a = inf I ∈ R, and let u : I → R be a uniformly continuous function. Prove that there exists lim u(x) = ` ∈ R.
x→a+
Corollary 2.11. Let I ⊂ R be an open interval, 0 < s < 1, and 1/s < p < ∞. Given u ∈ W s,p (I), let u ¯ : I¯ → R be its H¨ older continuous representas,p tive. Then u ∈ W0 (I) if and only if u ¯(x) = 0 for all x ∈ ∂I. Proof. Assume that u ∈ W0s,p (I). Then there exists a sequence of functions un ∈ Cc∞ (I) such that un → u in W s,p (I). It follows from Theorem 2.8 that 1 k¯ u − un k∞ 1/p ku − un kLp (I) + `s−1/p u − un W s,p (I) → 0, ` and so un → u ¯ uniformly. If a = inf I ∈ R, then un (a) = 0 → u ¯(a) so that u ¯(a) = 0. Similarly, if sup I = b ∈ R, then u ¯(b) = 0. Conversely, assume that u ¯(x) = 0 for all x ∈ ∂I. Suppose that I = (a, ∞). Define u ¯(x) if x > a, u ˜(x) = 0 if x < a. Then by Tonelli’s theorem, Z ∞Z ∞ Z ∞Z a ˜ u(x) − u ˜(y)p ¯ u(y)p dxdy = 2 dxdy 1+sp x − y1+sp a a a −∞ x − y Z ∞Z ∞ ¯ u(x) − u ¯(y)p + dxdy x − y1+sp a a and Z a
(2.17)
∞Z a −∞
¯ u(y)p dxdy = x − y1+sp ≤
Z
∞ p
Z
a
¯ u(y) a
1 sp
Z a
∞
−∞ ¯ u(y)p
(y − a)sp
1 dxdy x − y1+sp dy.
By Hardy’s inequality (Corollary 1.81) and the fact that u ¯(a) = 0, the righthand side of inequality (2.17) is finite. Hence, u ˜ ∈ W s,p (R). Since translation is continuous in Lp , the functions un (x) := u ˜(x − 1/n), x ∈ I, s,p converge to u ¯ in W (I) and are zero in (a, a + 1/n). Hence, by mollifying un , we obtain a sequence of functions in Cc∞ (I) converging to u ¯ in W s,p (I). s,p This shows that u ¯ ∈ W0 (I).
60
2. Embeddings and Interpolation
The case I = (a, b) is left as an exercise (see Step 2 of the proof of Theorem 1.73). Another consequence of Morrey’s embedding theorem is the property that W s,p (I) is an algebra for sp > 1. Corollary 2.12 (Product). Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 0 < s < 1, with sp > 1. If u, v ∈ W s,p (I), then uv ∈ W s,p (I), with kuvkW s,p (I) I kukW s,p (I) kvkW s,p (I) . Proof. By Morrey’s embedding theorem (Theorem 2.8), u, v ∈ L∞ (I), with 1 kukL∞ (I) 1/p kukLp (I) + `s−1/p uW s,p (I) , ` 1 kvkL∞ (I) 1/p kvkLp (I) + `s−1/p vW s,p (I) , ` where 0 < ` < L1 (I). We can now apply Theorem 1.38, to obtain uvW s,p (I) ≤ uW s,p (I) kvkL∞ (I) + kukL∞ (I) vW s,p (I) 1 1 1/p uW s,p (I) kvkLp (I) + 1/p vW s,p (I) kukLp (I) ` ` + `s−1/p uW s,p (I) vW s,p (I) . Similarly, kuvkLp (I) ≤ kukLp (I) kvkL∞ (I) + kukL∞ (I) kvkLp (I) 1 1/p kukLp (I) kvkLp (I) + `s−1/p kvkLp (I) uW s,p (I) ` + `s−1/p kukLp (I) vW s,p (I) . Exercise 2.13. Let I = (−1, 1), 1 ≤ p < ∞, and 0 < s < 1. Define u(x) := xα , where α ∈ R. Prove that u ∈
W s,p (I)
x ∈ I \ {0}, if and only if s < α + 1/p.
Exercise 2.14. Let I = (−1/2, 1/2), 1 ≤ p < ∞, and 0 < s < 1. Define u(x) :=
xα , (− log x)β
x ∈ I,
where α ∈ R and β > 0. (i) Let m ∈ N. Prove that for 0 < h < 1/(6m), Z 2mh (m) ∆h u(x)p dx α,β,m hαp+1 (− log h)−βp . 0
(ii) Prove that u(m) (x) α,β,m xα−m (− log x)−β ,
0 < x < 1/2.
2.1. Embeddings: The Endpoints
61
(iii) Prove that Z 1/2−mh n0 X (m) p αp+1 ∆h u(x) dx α,β,m h n(α−m)p  log(nmh)−βp , 2mh
n=1
where n0 is the smallest integer such that 2(n0 + 1)hm > 1/2. (iv) Prove that if ε > 0 is sufficiently small and m sufficiently large, then for 0 < h < ε, Z 1/2−mh (m) ∆h u(x)p dx α,β,m hαp+1 (− log h)−βp . 2mh
(v) Let ε and m be as in the previous item. Prove that Z ε Z 1/2−mh dh (m) ∆h u(x)p dx 1+sp < ∞, h 0 0 provided either s < α + 1/p or s = α + 1/p and βp > 1. (vi) Prove that u ∈ W s,p (I) provided either s < α + 1/p or s = α + 1/p and βp > 1. Hint: Use Exercise 1.59. Exercise 2.15. Let I = (−1/2, 1/2), 1 ≤ p < ∞, and 0 < s < 1. Define u(x) :=
xα , (− log x)β
x ∈ I,
where α ∈ R and β > 0. (i) Let s = α + 1/p and βp = 1. Prove that hα (− log h)−β α,β,n xα (− log x)−β − (x + h)α (− log(x + h))−β  for 0 < x < h/n ≤ 1/2 and for some n ∈ N. (ii) Let s = α + 1/p and βp = 1. Prove that u ∈ / W s,p (I). Exercise 2.16. Let u ∈ L2 ((−π, π)) be such that ∞ X u(x) = (ak cos(kx) + bk sin(kx)), x ∈ (−π, π). k=−∞
(i) Prove that for k, l ∈ Z,with k 6= l, Z π Z π cos (kx) cos (lx) dx = 0, cos (kx) cos (lx) dx = 0, −π −π Z π Z π cos (kx) sin (lx) dx = 0, cos (kx) sin (kx) dx = 0. −π
−π
(ii) Prove that Z
π 2
(u(x)) dx ≈ −π
∞ X k=−∞
(a2k + b2k ).
62
2. Embeddings and Interpolation
Exercise 2.17. Let I = (−π, π), 0 < s < 1, and 0 ≤ δ ≤ 12 . Define ∞ X 1 1 cos(2n x), u(x) := 2ns nδ
x ∈ I.
n=1
(i) Prove that u ∈ / W s,2 (I). Hint: Use Exercise 2.16 and prove that for 0 < h < 1 small and for 1 ` ∈ N such that 2`+1 < h ≤ 21` , hs logδ (1/h)
` X 1 1 2n 2 2 h 2ns 2 n2δ
!1/2 k∆h ukL2 (Ih ) .
n=1
(ii) Prove that u ∈ W s,∞ (I). P∞ P` P Hint: Write ∞ n=`+1 , where ` is as in item (i). n=1 + n=1 = We conclude this section by discussing how to glue two functions. Given 1 ≤ p < ∞, 0 < s < 1, and two functions v ∈ W s,p (R+ ) and w ∈ W s,p (R− ), we are interested in finding necessary and sufficient conditions on v and w for the existence of a function u ∈ W s,p (R) such that u(x) = v(x) for x ∈ R+ and u(x) = w(x) for x ∈ R− . We will see that there are no restrictions when sp < 1, while when sp ≥ 1, some compatibility conditions are needed. Theorem 2.18 (Gluing two functions). Let 1 ≤ p < ∞, 0 < s < 1, v ∈ W s,p (R+ ), w ∈ W s,p (R− ), and (2.18)
u(x) :=
v(x) if x > 0, w(x) if x < 0.
Then u ∈ W s,p (R) if and only if the following conditions hold: (i) If s = 1/p, then Z ∞ dx (2.19) v(x) − w(−x)p < ∞. x 0 (ii) If s > 1/p, then v¯(0) = w(0), ¯ where v¯ and w ¯ are the H¨ older continuous representative of v and w. Proof. Step 1: Let 1 ≤ p < ∞, 0 < s < 1, and u ∈ W s,p (R). We claim that Z ∞ dx (2.20) u(x) − u(−x)p sp upW s,p (R) . x 0
2.1. Embeddings: The Endpoints
63
By the change of variables y = −z, we can write Z ∞Z ∞ u(x) − u(y)p upW s,p (R) = (2.21) dxdy x − y1+sp 0 0 Z 0 Z 0 u(x) − u(y)p + dxdy 1+sp −∞ −∞ x − y Z ∞Z ∞ u(x) − u(−z)p +2 dxdz =: A + B + C. (x + z)1+sp 0 0 Using the facts that x − y ≤ x + y for x, y > 0, Z ∞ 1 1 1 dy = 1+sp (x + y) sp xsp 0 for x > 0, (2.21), the changes of variables −x = τ and −y = t, and Tonelli’s theorem, we have Z ∞Z ∞ Z ∞ u(x) − u(−x)p p dx dxdy u(x) − u(−x) sp x (x + y)1+sp 0 0 0 Z ∞Z ∞ Z ∞Z ∞ u(x) − u(−y)p u(−y) − u(−x)p dxdy + dxdy (x + y)1+sp x − y1+sp 0 0 0 0 C + B upW s,p (R) . Step 2: Assume that v ∈ W s,p (R+ ), w ∈ W s,p (R− ) satisfy the hypotheses in items (i) and (ii), and let u be defined as in (2.18). We claim that u belongs to W s,p (R). Let A, B, and C be defined as in (2.21). Then A = vpW s,p (R+ ) and B = wpW s,p (R− ) . To estimate C, we distinguish the cases sp < 1, sp = 1, and sp > 1. When sp < 1, by Tonelli’s theorem, the change of variables −y = z, and Hardy’s inequality (Theorem 1.76), Z ∞Z ∞ v(x, 0) − w(−y)p dxdy C (x + y)1+sp 0 0 Z ∞ Z ∞ Z ∞ Z ∞ dy dx p dx + dy v(x)p w(−y) 1+sp (x + y) (x + y)1+sp 0 0 0 0 Z ∞ Z ∞ v(x)p w(−y)p dx + dy vpW s,p (R+ ) + wpW s,p (R− ) . xsp y sp 0 0 When sp ≥ 1, we use Tonelli’s theorem and the fact that x − y ≤ x + y for x, y > 0 to estimate Z ∞Z ∞ Z ∞Z ∞ v(x) − w(−x)p w(−x) − w(−y)p dxdy + dxdy C (x + y)1+sp (x + y)1+sp 0 0 0 0 Z ∞ Z ∞Z ∞ v(x) − w(−x)p w(−x) − w(−y)p dx + dxdy xsp x − y1+sp 0 0 0 =: D + E.
64
2. Embeddings and Interpolation
Then E ≤ wpW s,p (R− ) . On the other hand, if sp = 1, then D is finite by (2.19). If sp > 1, let v¯ and w ¯ be the H¨ older continuous representatives of v and w. Since v¯(0) = w(0), ¯ the function h(x1 ) := v¯(x) − w(−x), ¯
x ∈ R+ ,
W0s,p (R+ )
belongs to by Corollary 1.74. Hence, Hardy’s inequality (Theorem 1.76) holds for h, and so, D hpW s,p (R+ ) vpW s,p (R+ ) + wpW s,p (R− ) .
Exercise 2.19. Let I = (−`, `), where ` > 0, 1 ≤ p < ∞, and 0 < s < 1. Prove that if u ∈ W s,p (I), then Z ` u(x) − u(−x)p dx ` kukpW s,p (I) . sp x 0 Exercise 2.20. Let I = (−`, `), I− = (−`, 0), I+ = (0, `), where ` > 0, 1 ≤ p < ∞, 0 < s < 1, w ∈ W s,p (I− ), and v ∈ W s,p (I+ ). Define u(x) := w(x) for x ∈ I− , and u(x) := v(x) for x ∈ I+ . Prove that v ∈ W s,p (I) if and only if Z ` dx v(x) − w(−x)p 1, where v¯ and w ¯ are the H¨older continuous representative of v and w.
2.2. Embeddings: The General Case In this section we prove inequality (2.1) in the cases s2 = 1 and 0 < s1 < s2 < 1. The following theorem shows that, by decreasing the regularity parameter s, we can increase the integrability parameter p. In the proof, we will use Hardy–Littlewood maximal function (see [Leo22c]). Let v ∈ L1loc (R). The (Hardy–Littlewood) maximal function of v is defined by Z x+r 1 (2.22) M(v)(x) := sup v(y) dy, x ∈ R. r>0 2r x−r Theorem 2.21. Let I ⊆ R be an open interval, 1 ≤ p1 < ∞, 1 < p2 < ∞, and 0 < s1 < s2 = 1 be such that 1 1 (2.23) 1− = s1 − . p2 p1 ˙ 1,p2 (I), Then for every u ∈ W (2.24)
uW s1 ,p1 (I) ku0 kLp2 (I) .
Moreover, W 1,p2 (I) ,→ W s1 ,p1 (I).
2.2. Embeddings: The General Case
65
First proof. Step 1: Assume that I = R. By Tonelli’s theorem, Z ∞Z Z Z ∞ u(x + h) − u(x)p1 u(x + h) − u(x)p1 dxdh = dhdx. h1+s1 p1 h1+s1 p1 0 R R 0 Let u ¯ be the locally absolutely continuous representative of u. By the fundamental theorem of calculus (see [Leo22d]), for x ∈ R and h > 0, Z x+h u0 (y) dy. ¯ u(x + h) − u ¯(x) ≤ x
It follows that Z x+h p1 Z ∞ Z ∞ u(x + h) − u(x)p1 1 0 dh. (2.25) dh ≤ u (y) dy h1+s1 p1 h1+s1 p1 0 0 x We claim that Z ∞ (2.26) 0
Z
1
x+h
p1
0
u (y) dy
h1+s1 p1
x
2 dh ku0 kpL1p−p (M(u0 )(x))p2 , 2 (R)
where M(u0 ) is the Hardy–Littlewood maximal function of u0 . For ρ > 0 we write (2.27) Z ∞ 0
1
h1+s1 p1 Z ∞ + ρ
Z
x+h
p1
0
u (y) dy
dh =
x
1 h1+s1 p1
ρ
Z 0
Z
x+h
0
u (y) dy
1
Z
h1+s1 p1
x+h
0
p1
u (y) dy x
p1 dh =: A + B.
x
To estimate A observe that by (2.22), Z ρ hp1 ρp1 (1−s1 ) 0 p1 (M(u0 )(x))p1 . (2.28) A (M(u )(x)) dh = 1+s1 p1 p1 (1 − s1 ) 0 h On the other hand, by H¨older’s inequality and (2.23), Z ∞ p1 /p0 2 h 1 dhku0 kpL1p2 (R) = ku0 kpL1p2 (R) . B≤ 1+s p 1 1 h ρ ρ Combining the estimates for A and B gives (2.29)
1 A + B ρp1 (1−s1 ) (M(u0 )(x))p1 + ku0 kpL1p2 (R) . ρ
By Exercise 1.26 and (2.23), 2 A + B ku0 kpL1p−p (M(u0 )(x))p2 . 2 (R)
Hence, in view of (2.29), we have proved (2.26).
dh
66
2. Embeddings and Interpolation
It now follows from (2.25) and (2.26) that Z ∞Z Z u(x) − u(x + h)p1 0 p1 −p2 dxdh ku k (M(u0 )(x))p2 dx Lp2 (R) 1+s1 p1 h 0 R R ku0 kpL1p2 (R) , where last inequality is due to the fact that M : Lp2 (R) → Lp2 (R) is a bounded operator for p2 > 1 (see [Leo22c]). To prove the last part of the statement, observe that if u ∈ W 1,p2 (R), then, by Morrey’s embedding theorem (see [Leo22d]), we have that u ∈ Lp2 (R) ∩ L∞ (R). Since s1 < 1 by (2.23), we have that p2 < p1 < ∞, and so, 1−θ θ kukLp1 (R) ≤ kukθLp2 (R) kuk1−θ L∞ (R) ≤ kukLp2 (R) kukW 1,p2 (R) ,
where
1 p1
=
θ p2 .
Step 2: If I = (a, ∞) or I = (−∞, b), we first reflect u to obtain a function ˙ 1,p2 (R), with kv 0 kLp2 (R) ≤ 2ku0 kLp2 (I) , and then apply Step 1 to v to v∈W obtain uW s1 ,p1 (I) ≤ vW s1 ,p1 (R) kv 0 kLp2 (R) ku0 kLp2 (I) . If I = (a, b), let a < c < d < b, with d − c ≥ 12 (b − a). Since inequality (2.24) Rd 1 does not change if we replace u with u−u(c,d) , where u(c,d) := d−c c u(x) dx, without loss of generality, we may assume that u(c,d) = 0. By Exercise 1.52, we can extend u to a function v in W 1,p2 (R), with kvkLp2 (R) (b − a)ku0 kLp2 (I) ,
kv 0 kLp2 (R) ku0 kLp2 (I) .
Once again, we can apply Step 1 to v to obtain uW s1 ,p1 (I) ku0 kLp2 (I) . We now present a second proof in the case I = (a, b), which does not rely on extension but on the following lemma. Lemma 2.22. Let g : (0, T ] → [0, ∞) be a continuous function such that g(t) ≤ At−a + Btb for all t ∈ (0, T ) and g(T ) ≤ BT b , where A, B ≥ 0, a, b > 0. Then inf g(t) ≤ CAb/(a+b) B a/(a+b) , t∈(0,T ]
where C = C(a, b) > 0. Proof. If A = 0, then g(t) ≤ Btb for all t ∈ (0, T ]. Letting t → 0+ , it follows that inf t∈(0,T ] g(t) = 0. On the other hand, if B = 0, then g(T ) = 0 and so again, inf t∈(0,T ] g(t) = 0. Thus, in what follows, assume A, B > 0. Define h(t) = At−a + Btb . Then h0 (t) = −aAt−a−1 + bBtb−1 ≥ 0
2.2. Embeddings: The General Case
67
for t ≥ [aA/(bB)]1/(a+b) =: T0 . If T0 ≤ T , then inf g(t) ≤ inf h(t) = h(T0 )
t∈(0,T ]
t∈(0,T ]
= [(a/b)−a/(a+b) + (a/b)b/(a+b) ]Ab/(a+b) B a/(a+b) , while if T0 > T , then from the definition of T0 , aAT −(a+b) /b > B. In turn, inf g(t) ≤ g(T ) ≤ BT b = B b/(a+b) B a/(a+b) T b t∈(0,T ]
≤ (aAT −(a+b) /b)b/(a+b) B a/(a+b) T b = (a/b)b/(a+b) Ab/(a+b) B a/(a+b) .
We turn to the second proof of Theorem 2.21. Second proof of of Theorem 2.21. By Tonelli’s theorem Z ∞Z Z Z u(x + h) − u(x)p1 u(x + h) − u(x)p1 dxdh = dhdx, h1+s1 p1 h1+s1 p1 0 Ih I Ix where I x := {h > 0 : x + h ∈ I}. We proceed as in Step 1 of the first proof of Theorem 2.21, taking x ∈ I and h ∈ I x . The only differences areR thatR M(u0 ) Rshould be replaced by ∞ ρ ∞ M(u0 χI ), and in (2.25), (2.26), (2.27), 0 , 0 , and ρ should be replaced R R R by I x , I x ∩(0,ρ) , and I x \(0,ρ) , respectively. If I = (a, ∞), we can still apply item (i) of Exercise 1.26. On the other hand, if I = (a, b), fix x ∈ I and let h ∈ I x = (0, b−x). Taking 0 < ρ < b−x, we apply Lemma 2.22 with g the constant function given by the lefthand side of (2.27) and T = b − x. Note that if we take ρ = b − x, then B = 0. Hence, we obtain p1 Z x+h Z 1 0 2 u (y) dy dh ku0 kpL1p−p (M(u0 χI )(x))p2 . 2 (I) 1+s1 p1 h x I x Continuing as before, we get Z ∞Z u(x) − u(x + h)p1 dxdh ku0 kpL1p2 (I) . 1+s1 p1 h 0 Ih The following exercise shows why we required p2 > 1 in Theorem 2.21. Exercise 2.23. Prove that when p2 = 1 and 1 < p1 < ∞, then the embedding W 1,1 (R) ,→ W 1/p1 ,p1 (R) fails. Hint: See Exercise 1.24. Theorem 2.24. Let I ⊆ R be an open interval, 1 ≤ p2 < p1 < ∞, and ˙ s2 ,p2 (I), 0 < s1 < s2 < 1 be such that s2 − p12 = s1 − p11 . Then for every u ∈ W uW s1 ,p1 (I) uW s2 ,p2 (I) .
68
2. Embeddings and Interpolation
Moreover, W s2 ,p2 (I) ,→ W s1 ,p1 (I). We begin with a few preliminary results. 2 (R) and for Lemma 2.25. Let 1 ≤ p2 < p1 < ∞. Then for every u ∈ Lploc every ` > 0,
(2.30) Z 1/p1 p1 u(x) dx R
Z
1
p2
1/p2
u(x) dx `1/p2 −1/p1 R 1/p2 Z ` Z u(x + h) − u(x)p2 dx + 0
R
dh . h1+1/p2 −1/p1
Proof. Step 1: Assume that u ∈ C 1 (R) and that the righthand side of (2.30) is finite since otherwise there is nothing to prove. By Theorem 1.6 and Remark 1.7, for every ` > 0, Z Z ` Z `−y u(x + t + y) − u(x + t) 1 ` u(x + t) dt − dtdy u(x) = ` 0 (t + y)2 0 0 =: `−1 A + B. Set s := p12 − p11 ∈ (0, 1). Writing 1/p1 = 1/p2 − s = 1/p2 + 1 − s − 1, we apply Young’s inequality (see Theorem 1.1) with v = 1 to get `−1 kAkLp1 (R) ≤ `−1 kukLp2 (R) k1kL1/(1−s) ((0,`)) = `−s kukLp2 (R) . Again by Young’s inequality, with `, u, and v replaced by ` − y, x ∈ R 7→ 1 u(x + y) − u(x), and t ∈ (0, ` − y) 7→ (t+y) 2 , respectively, we get p1 !1/p1 u(x + t + y) − u(x + t) dt dx (t + y)2 R 0 1−s Z 1/p2 Z `−y 1 p2 ≤ dt u(x + y) − u(x) dx (t + y)2/(1−s) R 0 Z 1/p2 1 u(x + y) − u(x)p2 dx . 1+s y R Z Z
`−y
Using Minkowski’s inequality for integrals (see [Leo22c]), p1 !1/p1 Z ` Z Z `−y u(x + t + y) − u(x + t) kBkLp1 (R) ≤ dt dx dy 2 (t + y) 0 R 0 1/p2 Z ` Z dy p2 u(x + y) − u(x) dx . y 1+s 0 R The result now follows by combining the estimates for A and B.
2.2. Embeddings: The General Case
69
Step 2: To remove the additional assumption that u ∈ C 1 (R), we apply Step 1 to uε = u ∗ ϕε , where ϕε is a standard mollifier. We leave the details as an exercise. Lemma 2.26. Let 1 ≤ p2 < p1 < ∞ and 0 < s1 < s2 < 1 be such that ˙ s2 ,p2 (R), s2 − p12 = s1 − p11 . Then for every u ∈ W p2 /p1 Z ∞ Z dh p1 u(x + h) − u(x) dx 1+s h 1 p2 0 ZR ∞ Z dh u(x + h) − u(x)p2 dx 1+s2 p2 . h 0 R Proof. We apply Lemma 2.25 to the function u(· + t) − u(·), with t > 0 in place of ` > 0, to get Z Z 1/p1 1/p2 1 p1 p2 u(x + t) − u(x) dx u(x + t) − u(x) dx s2 −s1 t R R 1/p2 Z t Z dh p2 + u(x + t + h) − u(x + t) dx 1+s2 −s1 h 0 R Z 1/p2 1 p2 = s2 −s1 u(x + t) − u(x) dx t R 1/p2 Z t Z dh p2 , + u(y + h) − u(y) dy 1+s h 2 −s1 0 R where we make the change of variables y = x + t. Raising both sides to power p2 , dividing by t1+s1 p2 , and integrating in t over R+ gives p2 /p1 Z ∞ Z dt p1 u(x + t) − u(x) dx 1+s t 1 p2 0 Z ∞ ZR dt u(x + t) − u(x)p2 dx 1+s2 p2 t 0 R !p2 1/p2 Z ∞ Z t Z dh dt + u(y + h) − u(y)p2 dy =: A + B. 1+s −s 1+s 2 1 h t 1 p2 0 0 R Define Z
p2
u(y + h) − u(y) dy
g(h) = R
1/p2
1 hs2 −s1
.
By Hardy’s inequality (see Theorem 1.3), p Z ∞ Z t Z ∞ dh dh 2 dt B= g(h) (g(h))p2 1+s1 p2 1+s1 p2 h t h 0 Z0 ∞ Z 0 dh u(y + h) − u(y)p2 dy 1+s2 p2 . h 0 R
70
2. Embeddings and Interpolation
Remark 2.27. Lemma 2.26 is really an embedding between Besov spaces (see Simon’s paper [Sim90]). Exercise 2.28. Let 1 ≤ p2 < p1 < ∞. (i) Given a sequence {an }n of nonnegative numbers, prove that !1 !1 ∞ ∞ p1 p2 X X apn1 ≤ apn2 . n=1
n=1
(ii) Given an increasing function f : [0, ∞) → [0, ∞) and s > 0, prove that there exists a constant C > 0 independent of p1 such that 1 Z ∞ 1 Z ∞ f (x) p2 dx p2 f (x) p1 dx p1 ≤C . xs x xs x 0 0 (iii) Using part (ii), prove that Z ∞ 1 f (x) p2 dx p2 f (x) ≤C . sup s xs x 0 0, Z 1/p1 Z 1/p2 1 (2.33) u(x)p1 dx 1/p −1/p u(x)p2 dx 1 ` 2 R R 1/p2 Z ` Z dh m p2 + ∆δh u(y) dy . 1+1/p 2 −1/p1 h 0 R Proof. Step 1: Assume that u ∈ C m (R) and that the righthand side of (2.33) is finite since otherwise there is nothing to prove. By Theorem 1.11,
72
2. Embeddings and Interpolation
Fubini’s theorem and the facts that ω, φ, and ψ are supported on [0, 1], for every ` > 0 and x ∈ R, we have Z 1 ` u(x) u (x + y)  dy ` 0 Z ` Z Z 1 ξ ξ m 1 + ∆δz u(x + t + z) dtdzdξ =: A + B. 3 ` 0 ξ 0 0 Set s := p12 − p11 ∈ (0, 1). Writing 1/p1 = 1/p2 − s = 1/p2 + 1 − s − 1, we apply Young’s inequality (see Theorem 1.1) with v = 1 to get `−1 kAkLp1 (R) ≤ `−1 kukLp2 (R) k1kL1/(1−s) ((0,`)) = `−s kukLp2 (R) . Again by Young’s inequality, with u and ` replaced by x ∈ R 7→ ∆m δz u(x + z) and ξ, respectively, and with v = 1, we get p1 !1/p1 1/p2 Z Z Z ξ p2 m 1−s m ∆δz u(x + t + z) dt dx ≤ξ ∆δz u(x + z) dx R
(2.34)
0
R
=ξ
1−s
Z
p2 ∆m δz u(y) dy
1/p2 ,
R
where we make the change of variables y = x+z. Using Minkowski’s inequality for integrals (see [Leo22c]), inequality (2.34), and Tonelli’s theorem, we derive p1 !1/p1 Z Z ξ Z Z ` 1 ξ m ∆δz u(x + t + z) dt dx dzdξ kBkLp1 (R) ≤ 3 R 0 0 0 ξ 1/p2 Z ` 1−s Z ξ Z ξ p2 m u(y) dy ∆ dzdξ δz ξ3 0 0 R 1/p2 Z ` Z ` Z 1 m p2 = ∆δz u(y) dy dξdz 2+s 0 R z ξ 1/p2 Z ` Z dz m p2 ∆δz u(y) dy . z 1+s 0 R Inequality (2.33) now follows by combining the estimates for A and B. Step 2: To remove the additional assumption that u ∈ C m (R), we apply Step 1 to uε = u ∗ ϕε , where ϕε is a standard mollifier. We leave the details as an exercise. Next we show that if we use (non homogenous) fractional Sobolev spaces rather than homogeneous ones, then we can replace equality (2.3) with an inequality.
2.2. Embeddings: The General Case
73
Corollary 2.30. Let I ⊆ R be an open interval, 1 ≤ p2 < p1 ≤ ∞, and 0 ≤ s1 < s2 ≤ 1 be such that s2 − p12 ≥ s1 − p11 . Then kukW s1 ,p1 (I) kukW s2 ,p2 (I) for all u ∈ W s2 ,p2 (I), with the following exceptions: (i) if
1 p2
= s2 ∈ (0, 1) and s1 = 0, we require p1 < ∞;
(ii) if s2 = 1 and s1 > 0, we require p2 > 1. Here W 0,p1 (I) := Lp1 (I). Proof. Step 1: Assume that s1 = 0 < s2 < 1. In the subcritical case s2 p2 < 1, we have 1 1 1 ≤ . − s2 = p2 (p2 )∗s2 p1 Hence p2 ≤ p1 ≤ (p2 )∗s2 . By the Sobolev–Gagliardo–Nirenberg embedding theorem (see Theorem 2.2 and Corollary 2.3), W s2 ,p2 (I) ,→ Lq (I) for all p2 ≤ q ≤ (p2 )∗s2 . In the critical case, s2 p2 = 1, W s2 ,p2 (I) ,→ Lq (I) for all p2 ≤ q < ∞ by Exercise 2.7. Note that this is the exception in Corollary 2.30(i). Finally, in the subcritical case, s2 p2 > 1, by Morrey’s embedding theorem (Theorem 2.8), W s2 ,p2 (I) ,→ C 0,β (I) for every 0 < β ≤ s2 − 1/p2 . In particular, W s,p (I) ,→ L∞ (I), and so, by interpolation, W s2 ,p2 (I) ,→ Lq (I) for all p2 ≤ q ≤ ∞. Step 2: Assume that s1 = 0 < s2 = 1. Then W 1,p2 (I) ,→ Lp1 (I) for all p2 ≤ p1 ≤ ∞ (see [Leo22d]). Step 3: Assume that 0 < s1 < s2 = 1 and 1 < p2 < p1 < ∞ (we are in the exception in Corollary 2.30(ii)). Since p1 > p2 , 0 < s1 ≤ σ1 :=
1 1 − + 1 < 1. p1 p2
Then by Theorem 2.21, W 1,p2 (I) ,→ W σ1 ,p1 (I). If s1 < σ1 , we use Lemma 2.6 to obtain W σ1 ,p1 (I) ,→ W s1 ,p1 (I). On the other hand, if 0 < s1 < s2 = 1 and 1 < p2 < p1 = ∞, then by Morrey’s embedding theorem ([Leo22d]), W 1,p2 (I) ,→ C 0,σ (I) for every 0 < σ ≤ 1 − 1/p2 . We recall that C 0,s1 (I) can be identified with W s1 ,∞ (I). Step 4: Assume that 0 < s1 < s2 < 1 and 1 ≤ p2 < p1 < ∞. Since p1 > p2 , 0 < s1 ≤ σ1 :=
1 1 − + s2 < s2 . p1 p2
By Theorem 2.24, W s2 ,p2 (I) ,→ W σ1 ,p1 (I). If s1 < σ1 , we use Lemma 2.6 to obtain W σ1 ,p1 (I) ,→ W s1 ,p1 (I).
74
2. Embeddings and Interpolation
Finally, if 0 < s1 < s2 < 1 and 1 ≤ p2 < p1 = ∞, then s2 − p12 ≥ s1 > 0, which implies that s2 p2 > 1. Thus, by Morrey’s embedding theorem (Theorem 2.8), W s2 ,p2 (I) ,→ C 0,s (I) for every 0 < s ≤ s2 − 1/p2 . Remark 2.31. Exercises 2.4 and 2.23 provide the counterexample for Corollary 2.30 (i) and (ii), respectively.
2.3. Interpolation Inequalities In this section we study interpolation inequalities of the type (2.35)
uW s,p (R) uθW s1 ,p1 (R) u1−θ W s2 ,p2 (R)
˙ s2 ,p2 (R). Here, 0 < θ < 1, 1 ≤ p, p1 , p2 ≤ ∞, and ˙ s1 ,p1 (R) ∩ W for all u ∈ W ˙ 0,p (R) := Lp (R), and 0 ≤ s, s1 , s2 ≤ 1. We set W (2.36)
uW 0,p (R) := kukLp (R) ,
uW 1,p (R) := ku0 kLp (R) .
A scaling argument as in (2.2) shows that a necessary condition for (2.35) ˙ s1 ,p1 (R) ∩ W ˙ s2 ,p2 (R) is to hold for all u ∈ W 1 1 1 (2.37) s − = θ s1 − + (1 − θ) s2 − . p p1 p2 We will prove the following theorem. Theorem 2.32 (Gagliardo–Nirenberg). Let I ⊆ R be an open interval, 1 ≤ p1 , p2 < ∞, 0 ≤ s1 < s2 ≤ 1, and 0 < θ < 1 be such that (2.38)
θ 1−θ 1 = + , p p1 p2
s = θs1 + (1 − θ)s2 .
˙ s1 ,p1 (I) ∩ W ˙ s2 ,p2 (I), Then for all u ∈ W (2.39)
uW s,p (I) uθW s1 ,p1 (I) u1−θ W s2 ,p2 (I) ,
with the following exceptions: (i) if s1 = 0, we assume that p1 > 1; (ii) if s2 = 1, we assume that p2 > 1. We divide the proof of this theorem into several cases: • 0 < s1 < s2 < 1, • 0 = s1 < s2 < 1, • 0 < s1 < s2 = 1, • 0 = s1 < s2 = 1. We begin with the case 0 < s1 < s2 < 1.
2.3. Interpolation Inequalities
75
Theorem 2.33. Let I ⊆ R be an open interval, 1 ≤ p1 , p2 < ∞, 0 < s1 < s2 < 1, and 0 < θ < 1. Then uW s,p (I) uθW s1 ,p1 (I) u1−θ W s2 ,p2 (I) ˙ s1 ,p1 (I) ∩ W ˙ s2 ,p2 (I), where for all u ∈ W 1 1−θ θ + , = p p1 p2
(2.40)
s = θs1 + (1 − θ)s2 .
Proof. For 0 < δ < 1 and 1 < q < ∞, write Z Z Z Z u(x) − u(y)θp u(x) − u(y)(1−θ)p u(x) − u(y)p dxdy. dxdy = 1+sp 1/q+δp x − y1/q0 +(s−δ)p I I x − y I I x − y By H¨older’s inequality the righthand side can be majorized by Z Z I
Since
I p1 θp
u(x) − u(y)θpq dxdy x − y1+δpq =1+
(1−θ)p1 θp2
1/q
Z Z I
I
0
u(x) − u(y)(1−θ)pq dxdy x − y1+(s−δ)pq0
!1/q0 .
> 1, we can take q = p1 /(θp), so that by (2.40),
(1 − θ)pq 0 = (1 − θ)pp1 /(p1 − pθ) = p2 . On the other hand, with the choice δ = θs1 , we have δpq = s1 p1 , while (s − δ)pq 0 = (s − θs1 )p2 /(1 − θ) = p2 s2 .
Theorem 2.34. Let 1 < p1 < ∞, 1 ≤ p2 < ∞, and 0 = s1 < s < s2 < 1 be such that (2.41)
1 θ 1−θ = + , p p1 p2
s = (1 − θ)s2 .
˙ s2 ,p2 (R), Then for all u ∈ Lp1 (R) ∩ W (2.42)
uW s,p (R) kukθLp1 (R) u1−θ W s2 ,p2 (R) .
Proof. We claim that (2.43)
uW s,p (R) k M(u)kθLp1 (R) u1−θ W s2 ,p2 (R) .
For h > 0 and x ∈ R we can write Z Z 1 h 1 h (u(x + t) − u(x)) dt − (u(x + t) − u(x + h)) dt. u(x + h) − u(x) = h 0 h 0
76
2. Embeddings and Interpolation
Hence, Z 0
(2.44)
∞Z
u(x + h) − u(x)p dxdh h1+sp R p Z ∞Z Z h dh 1 u(x + t) − u(x) dt dx 1+sp h 0 R h 0 p Z ∞Z Z h 1 dh + u(x + t) − u(x + h) dt dx 1+sp h 0 R h 0 =: A + B.
By Tonelli’s theorem Z Z (2.45) A= R
0
∞
1 h
p
h
Z
u(x + t) − u(x) dt 0
We claim that p Z ∞ Z h 1 dh (2.46) u(x + t) − u(x) dt 1+sp h h 0 0 Z ∞ θp (M(u)(x)) u(x + t) − u(x)p2 0
dh dx. h1+sp
dt
(1−θ)p/p2
t1+s2 p2
for L1 a.e. x ∈ R. Using the fact that uW s2 ,p2 (R) < ∞, by Tonelli’s theorem, we can assume that Z ∞ dt (2.47) u(x + t) − u(x)p2 1+s2 p2 < ∞ t 0 for L1 a.e. x ∈ R. Moreover (see [Leo22c]), (2.48)
u(x) ≤ M(u)(x) < ∞
for L1 a.e. x ∈ R.
Fix x ∈ R such that (2.47) and (2.48) hold. By H¨ older’s inequality and the fact that 0 < t < h, Z h Z h dt 1/p2 +s2 u(x + t) − u(x) dt ≤ h u(x + t) − u(x) 1/p +s t 2 2 0 0 Z h 1/p2 dt ≤ h1+s2 u(x + t) − u(x)p2 1+s2 p2 . t 0 Hence, for ρ > 0, by (2.41), p Z ρ Z h dh 1 u(x + t) − u(x) dt 1+sp h 0 h 0 Z ρ p/p2 Z ρ 1+s2 p dt h dh (2.49) ≤ u(x + t) − u(x)p2 1+s2 p2 1+sp t h h 0 0 Z ρ p/p2 (s −s)p 2 ρ dt p2 = u(x + t) − u(x) 1+s2 p2 . (s2 − s)p t 0
2.3. Interpolation Inequalities
77
On the other hand, by the definition of maximal function (see (2.22)) and (2.48), 1 h
h
Z
1 u(x + t) − u(x) dt ≤ u(x) + h
0
x+h
Z
u(y) dy ≤ 3 M(u)(x). x
Hence, ∞
Z
1 h
Z
p
h
dh u(x + t) − u(x) dt 1+sp h ρ 0 Z ∞ dh 1 (M(u)(x))p (M(u)(x))p sp . 1+sp h ρ ρ
Summing this inequality to (2.49) gives Z
∞
0
1 h
u(x + t) − u(x) dt (s2 −s)p
Z
dh h1+sp
−h
+ρ
(2.50)
p
h
Z
(M(u)(x))p
∞
u(x + t) − u(x) 0
p/p2
dt
p2
1 ρsp
t1+s2 p2
for every ρ > 0. Using Exercise 1.26, (2.41), (2.47), and (2.48) gives (2.46). In turn, by (2.45), (2.46), Tonelli’s theorem, and H¨older’s inequality with exponents p2 /(p2 − (1 − θ)p) and p2 /((1 − θ)p), Z A
(M(u)(x))
Z
Z
∞ p2
u(x + t) − u(x) 0
R
(2.51)
θp
(M(u)(x))
θpp2 /(p2 −(1−θ)p)
dt
(1−θ)p/p2 dx
t1+s2 p2
(p2 −(1−θ)p)/p2 dx
R
Z Z
∞
×
p2
u(x + t) − u(x) R
0
dt t1+s2 p2
(1−θ)p/p2 dx
.
Note that by (2.41), (p2 − (1 − θ)p)/p2 = θp/p1 . To estimate B, we make the changes of variables t = h − τ , so that dt = dτ , and x + h = y, so that dx = dy, to write p Z h dh 1 B= u(y − τ ) − u(y) dτ dy 1+sp h 0 R h 0 p Z Z ∞ Z h 1 dh = u(y − τ ) − u(y) dτ dy, 1+sp h 0 h R 0 Z
∞Z
78
2. Embeddings and Interpolation
where the second equality follows from Tonelli’s theorem. Reasoning as for A, we can show Z θp/p1 (2.52) B (M(u)(x))p1 dx R
Z Z ×
∞
u(y − τ ) − u(y) R
dτ
p2
0
τ 1+s2 p2
(1−θ)p/p2 dy
.
It follows from (2.44), (2.51), (2.52), and Tonelli’s theorem that (2.43) holds. In turn, 1−θ θ uW s,p (R) k M(u)kθLp1 (R) u1−θ W s2 ,p2 (R) kukLp1 (R) uW s2 ,p2 (R) ,
where in the second inequality we use the continuity of the maximal operator M in Lp1 (R) for p1 > 1 (see [Leo22c]). Thirdly, we consider the case 0 < s1 < s2 = 1. Theorem 2.35. Let 1 ≤ p1 < ∞, 1 < p2 < ∞, 0 < s1 < s2 = 1, 0 < θ < 1 be such that θ 1−θ 1 = + , s = θs1 + (1 − θ)1. (2.53) p p1 p2 ˙ s1 ,p1 (R) ∩ W ˙ 1,p2 (R), Then for all u ∈ W (2.54)
uW s,p (R) uθW s1 ,p1 (R) ku0 k1−θ Lp2 (R) .
Proof. We proceed as in the proof of Theorem 2.34 up to (2.45). In place of (2.46), we now claim that for L1 a.e. x ∈ R, p Z ∞ Z h 1 dh (2.55) u(x + t) − u(x) dt 1+sp h 0 h 0 θp/p1 Z ∞ dt p1 u(x + t) − u(x) 1+s1 p1 (M(u0 )(x))(1−θ)p . t 0 Using the fact that uW s1 ,p1 (R) < ∞, by Tonelli’s theorem, we can assume that Z ∞ dt (2.56) u(x + t) − u(x)p1 1+s1 p1 < ∞ t 0 for L1 a.e. x ∈ R. Moreover, by the properties of the maximal function (see [Leo22c]), (2.57)
M(u0 )(x) < ∞
for L1 a.e. x ∈ R.
Fix x ∈ R such that (2.56) and (2.57) hold. Let u ¯ be the absolutely continuous representative of u (see [Leo22d]). By the fundamental theorem of
2.3. Interpolation Inequalities
79
calculus (see [Leo22d]) and the definition of maximal function (see (2.22)), Z x+t (2.58) ¯ u(x + t) − u ¯(x) ≤ u0 (y) dy ≤ 2t M(u0 )(x). x
Using inequality (2.58), for ρ > 0, we have p Z ρ Z h 1 dh ¯ u(x + t) − u ¯(x) dt 1+sp h h 0 0 p Z ρ Z h 1 dh (M(u0 )(x))p (2.59) t dt ρ(1−s)p (M(u0 )(x))p . 1+sp h h 0 0 On the other hand, by H¨older’s inequality and the fact that 0 < t < h, Z h Z h dt 1/p1 +s1 u(x + t) − u(x) dt ≤ h u(x + t) − u(x) 1/p +s t 1 1 0 0 1/p1 Z h dt 1+s1 p1 . ≤h u(x + t) − u(x) 1+s1 p1 t 0 Hence, Z ρ
(2.60)
p Z 1 h dh u(x + t) − u(x) dt h 0 h1+sp Z ∞ p/p1 Z ∞ 1+s1 p dt dh h p1 ≤ u(x + t) − u(x) 1+s1 p1 1+sp t h h 0 ρ Z ∞ p/p1 1 dt (s−s )p u(x + t) − u(x)p1 1+s1 p1 . 1 t ρ 0
∞
Summing inequalities (2.59) and (2.60) gives p Z ∞ Z h 1 dh u(x + t) − u(x) dt ρ(1−s)p (M(u0 )(x))p 1+sp h −h h 0 Z ∞ p/p1 dt 1 p1 u(x + t) − u(x) 1+s1 p1 + (s−s )p . 1 t ρ 0 Using Exercise 1.26, (2.53), (2.56), and (2.57) gives (2.55). In turn, by (2.45), (2.53), Tonelli’s theorem, and H¨ older’s inequality with exponents p1 /(θp) and p1 /(p1 − θp) = p2 /[(1 − θ)p], we obtain θp/p1 Z Z ∞ dt p1 (M(u0 )(x))(1−θ)p dx A u(x + t) − u(x) 1+s1 p1 t R 0 Z Z ∞ θp/p1 dt (2.61) ≤ u(x + t) − u(x)p1 1+s1 p1 dx t R 0 Z (1−θ)p/p2 × (M(u0 )(x))p2 dx . R
80
2. Embeddings and Interpolation
The term B in (2.44) can be estimated in a similar way. We leave the details as an exercise. It follows from (2.44) and (2.61) that uW s,p (R) uθW s1 ,p1 (R) k M(u0 )k1−θ Lp2 (R) uθW s1 ,p1 (R) ku0 k1−θ Lp2 (R) , where in the second inequality we use the continuity of the maximal operator M in Lp2 (R) for p2 > 1 (see [Leo22c]). Exercise 2.36. Let 1 ≤ p1 , p2 < ∞ and u ∈ Lp1 (R) ∩ W 1,p2 (R). Let ϕ ∈ Cc∞ (R) be such that ϕ = 1 in [−1, 1], ϕ = 0 outside [−2, 2], and let 0 ≤ ϕ ≤ 1. Define un := ϕn u, where ϕn (x) = ϕ(x/n), x ∈ R. Prove that un → u in Lp1 (R) and un → u in W 1,p2 (R). Finally, we consider the case 0 = s1 < s2 = 1. Theorem 2.37. Let 1 < p1 , p2 < ∞, and 0 < θ < 1. Then (2.62)
uW s,p (R) kukθLp1 (R) ku0 k1−θ Lp2 (R)
˙ 1,p2 (R), where for all u ∈ Lp1 (R) ∩ W 1 θ 1−θ = + , p p1 p2
(2.63)
s = 1 − θ.
Proof. Step 1: Assume first that u ∈ Cc∞ (R). If u is zero, there is nothing to prove, so we can suppose that u 6= 0. Choose s < σ < 1 and 1 < q < ∞ ¯ ¯ ¯ Then by Theorem 2.34, such that p1 = pθ1 + 1−q θ and σ = (1 − θ)s. ¯
¯
θ uW s,p (R) kukθLp1 (R) u1− W σ,q (R) .
On the other hand, writing 2.35, we have
1 q
=
θ˜ p
+ ˜
1−θ˜ p2
˜ + 1 − θ, ˜ by Theorem and σ = θs ˜
θ uW σ,q (R) uθW s,p (R) ku0 k1− Lp2 (R) .
Combining these two inequalities, we obtain 1−θ¯ ¯ ˜ θ˜ uW s,p (R) kukθLp1 (R) uθW s,p (R) ku0 k1− . p L 2 (R) Since u ∈ Cc∞ (R) and u 6= 0, necessarily, 0 < uW s,p (R) < ∞, and so, upon division, ¯ ˜ ¯ θ/[1− θ(1− θ)]
uW s,p (R) kukLp1 (R)
˜ ¯ ˜ ¯ (1−θ)(1− θ)//[1− θ(1− θ)]
ku0 kLp2 (R)
= kukθLp1 (R) ku0 k1−θ Lp2 (R) .
Step 2. If now u ∈ Lp1 (R) ∩ W 1,p2 (R), then by Exercise 2.36, we can find un ∈ Cc∞ (R) such that un → u in Lp1 (R) and u0n → u0 in Lp2 (R). By extracting a subsequence, not relabeled, we can suppose that un → u
2.3. Interpolation Inequalities
81
pointwise L1 a.e. in R. By applying what we just proved to each un , we obtain un W s,p (R) kun kθLp1 (R) ku0n k1−θ Lp2 (R) . Letting n → ∞ and using Fatou’s lemma on the lefthand side gives uW s,p (R) kukθLp1 (R) ku0 k1−θ Lp2 (R) . ˙ 1,p2 (R). For n ∈ N define Step 3: Finally, assume that u ∈ Lp2 (R) ∩ W (t − 1/n) sgn t if 1/n ≤ t ≤ n, 0 if t < 1/n, fn (t) := (n − 1/n) sgn t if t > n, and vn = fn ◦ u. Then Z Z p2 vn  dx = R
vn p2 dx
{u>1/n}
≤ (n − 1/n)p2 L1 ({x ∈ R : u(x) > 1/n}) < ∞, since u ∈ Lp1 (R). Moreover, since fn is Lipschitz continuous with Lipschitz constant 1, with fn (0) = 0, by the chain rule in Sobolev spaces (see [Leo22d]), vn (x) ≤ u(x) and vn0 (x) ≤ u0 (x) for L1 a.e. x ∈ R. It follows that vn belongs to Lp1 (R) ∩ W 1,p2 (R). Hence, we may apply the previous step to vn to obtain θ 0 1−θ vn W s,p (R) kvn kθLp1 (R) kvn0 k1−θ Lp2 (R) kukLp1 (R) ku kLp2 (R) .
Letting n → ∞ and using Fatou’s lemma on the lefthand side concludes the proof. We are finally ready to prove Theorem 2.32. Proof of Theorem 2.32. Step 1: Let 1 ≤ p1 , p2 < ∞, 0 ≤ s1 < s2 ≤ 1, 0 < θ < 1, and let p and s be given in (2.38). Assume that I = R. If 0 < s1 < s2 < 1, we apply Theorem 2.33. If 0 = s1 < s2 < 1, we use Theorem 2.34. If 0 < s1 < s2 = 1, we use Theorem 2.35. Finally, if 0 = s1 < s2 = 1, we can apply Theorem 2.37. Step 2: It remains to remove the additional hypothesis that I = R. Assume that I = (a, ∞) (the case I = (−∞, b) is similar). By Theorem 1.40 and Remark 1.41 if 0 < s1 < 1 or 0 < s2 < 1, and Exercise 1.42 if s2 = 1 (the case s1 = 0 is simpler), we can reflect u to obtain a function v ∈ ˙ s2 ,p2 (R), with ˙ s1 ,p1 (R) ∩ W W vW s1 ,p1 (R) uW s1 ,p1 (I) ,
vW s2 ,p2 (R) uW s2 ,p2 (I) ,
82
2. Embeddings and Interpolation
where we are using notation (2.36). Applying Step 1 to v gives uW s,p (I) ≤ vW s,p (R) vθW s1 ,p1 (R) v1−θ W s2 ,p2 (R) uθW s1 ,p1 (I) u1−θ W s2 ,p2 (I) . Step 3: Assume that I = (a, b), and let a < c < d < b, with d − c ≥ 1 2 (b − a). We claim that we can extend the function u − u(c,d) to a function v ∈ W s1 ,p1 (R) ∩ W s2 ,p2 (R), with (2.64)
vW s1 ,p1 (R) uW s1 ,p1 (I) ,
vW s2 ,p2 (R) uW s2 ,p2 (I) ,
where we are using notation (2.36). Observe that if the claim holds, then applying Step 1 to v and using (2.64) gives uW s,p (I) = u − u(c,d) W s,p (I) ≤ vW s,p (R) vθW s1 ,p1 (R) v1−θ W s2 ,p2 (R) uθW s1 ,p1 (I) u1−θ W s2 ,p2 (I) . To prove the claim, we first observe that if 1 ≤ q < ∞, then by Minkowski’s and H¨older’ inequalities Z (b − a)1/q d ku − u(c,d) kLq (I) ≤ kukLq (I) + u(x) dx d−c c ≤ kukLq (I) +
(b − a)1/q 0 (d − c)1/q kukLq (I) kukLq (I) , d−c
where we use the fact that c − d ≥ 12 (b − a). If 0 = s1 < s2 < 1, then the function u − u(c,d) belongs to Lp1 (I) ∩ R ˙ s2 ,p2 (I) and d (u − u(c,d) ) dx = 0. Hence, by Remark 1.44 and Corollary W c 1.50, we can extend u − u(c,d) to a function v ∈ Lp1 (R) ∩ W s2 ,p2 (R) such that kvkLp1 (R) ku − u(c,d) kLp1 (I) kukLp1 (I) , vW s2 ,p2 (R) uW s2 ,p2 (I) . The case 0 < s1 < s2 < 1 is similar. We omit the details. Similarly, if 0 < s1 < s2 = 1, by Corollary 1.50 and Exercise 1.52, we can extend u − u(c,d) to a function in W s1 ,p1 (R) ∩ W 1,p2 (R), with vW s1 ,p1 (R) u − u(c,d) W s1 ,p1 (I) uW s1 ,p1 (I) ,
kv 0 kLp2 (R) ku0 kLp2 (I) .
Finally, if 0 = s1 < s2 = 1, by Exercise 1.52, we can extend u − u(a,b) to a function v ∈ Lp1 (R) ∩ W 1,p2 (R), with kvkLp1 (R) ≤ ku − u(a,b) kLp1 (I) kukLp1 (I) , kv 0 kLp2 (R) (b − a)−1 ku − u(a,b) kLp2 (I) + ku0 kLp2 (I) ku0 kLp2 (I) .
Exercise 2.38. Study the case in which p1 = ∞ or p2 = ∞ in Theorem 2.32.
2.3. Interpolation Inequalities
83
In Chapter 3 we will use wavelets to remove exceptions (i) and (ii) in the statement of Theorem 2.32 and allow p1 (respectively, p2 ) to be 1 when s1 = 0 (respectively, s2 = 1). In what follows, we present some partial results. Theorem 2.39. Let 1 < p1 < ∞ and 0 < s1 < 1/p1 . Then uW s1 ,p1 (R) kukθLp1 (R) ku0 k1−θ L1 (R) ˙ 1,1 (R), where θ = 1 − s1 p1 . for all u ∈ Lp1 (R) ∩ W Proof. Let u ¯ be the absolutely continuous representative of u (see [Leo22d]). ¯(xn ) → 0. Since u ∈ Lp1 (R), there exists a sequence xn → ∞ such that u Hence, by the fundamental theorem of calculus (see [Leo22d]), Z xn u0 (t) dt. u ¯(x) = u ¯(xn ) − x R In turn, ¯ u(x) ≤ ¯ u(xn ) + R u0 (t) dt. Letting n → ∞ gives Z (2.65) ¯ u(x) ≤ u0 (t) dt for every x ∈ R. R
Write Z Z ¯ u(x) − u ¯(y)p1 ¯ u(x) − u ¯(y)p1 dxdy = dxdy 1+s1 p1 1+s1 p1 R x − y R B(y,ρ) x − y Z Z ¯ u(x) − u ¯(y)p1 dxdy =: A + B. + 1+s1 p1 R R\B(y,ρ) x − y
Z Z (2.66) R
By the fundamental theorem of calculus, Z 1 u ¯(x) − u ¯(y) = u0 (tx + (1 − t)y)(x − y) dt, 0
and so, ¯ u(x) − u ¯(y)p1 = ¯ u(x) − u ¯(y)p1 −1 ¯ u(x) − u ¯(y) Z 1 sup ¯ up1 −1 x − y u0 (tx + (1 − t)y) dt. 0
R
In turn, by Tonelli’s theorem and by making first the change of variables z = x − y, so that dz = dx, and then w = y + tz, so that dw = dy, Z Z Z 1 0 u (tx + (1 − t)y) A sup ¯ up1 −1 dtdxdy x − ys1 p1 R R B(y,ρ) 0 Z Z 1 p1 −1 0 sup ¯ u u (w) dw dz s1 p1 R R B(0,ρ) z Z 1−s1 p1 p1 −1 ρ sup ¯ u u0 (w) dw ρ1−s1 p1 ku0 kpL11 (R) , R
R
84
2. Embeddings and Interpolation
where in the last inequality we use (2.65). On the other hand, by Tonelli’s theorem and inequality (1.1), Z Z 1 B ¯ u(x)p1 dydx 1+s1 p1 R\B(x,ρ) x − y R Z Z 1 1 p1 + ¯ u(y) dxdy s1 p1 kukpL1p1 (R) . 1+s p 1 1 ρ R\B(y,ρ) x − y R Combining the last two estimates with (2.66) gives uW s1 ,p1 (R) kukLp1 (R) ρ−s1 + ku0 kL1 (R) ρ(1−s1 p1 )/p1 . By Exercise 1.26, 0 s1 p 1 1 p1 uW s1 ,p1 (R) kuk1−s Lp1 (R) ku kL1 (R) .
Exercise 2.40. Prove that Theorem 2.39 continues to hold if we replace the domain R with an open interval I ⊆ R. ˙ 1,∞ (R), Theorem 2.41. Let 1 < p < ∞. Then for all u ∈ L1 (R) ∩ W 1/p
1−1/p
uW 1−1/p,p (R) kukL1 (R) ku0 kL∞ (R) . Proof. Assume that u 6= 0, since otherwise there is nothing to prove. By replacing u with u/ku0 kL∞ (R) , without loss of generality, we may assume that ku0 kL∞ (R) = 1. Let u ¯ be the absolutely continuous representative of u (see [Leo22d]). Given h > 0, consider the sets Fh := {x ∈ R : ¯ u(x) ≤ h, ¯ u(x + h) ≤ h}, Eh := {x ∈ R : ¯ u(x) > h}. Write Z (2.67)
¯ u(x + h) − u ¯(x)p dx =
Z
¯ u(x + h) − u ¯(x)p dx
Fh
R
Z +
¯ u(x + h) − u ¯(x)p dx =: A + B.
R\Fh
By the layercake representation (see [Leo22c]), for every Lebesgue measurable set G ⊆ R and 1 ≤ q < ∞, Z Z ∞ q (2.68) ¯ u(x) dx = qrq−1 L1 ({x ∈ G : ¯ u(x) > r}) dr. G
0
Taking G = {x ∈ R : ¯ u(x) ≤ h} and q = p, we have that Z Z h p ¯ u(x) dx = prp−1 L1 ({x ∈ G : ¯ u(x) > r}) dr {¯ u≤h}
0
Z = 0
h
prp−1 [L1 (Er ) − L1 (Eh )] dr.
2.3. Interpolation Inequalities
85
Hence, by inequality (1.1) and the change of variables y = x + h, Z Z p p−1 p−1 (2.69) ¯ u(x)p dx ¯ u(x + h) dx + 2 A≤2 Fh
Fh
≤ 2p
Z
= 2p p
¯ u(x)p dx = 2p p
{¯ u≤h} h p−1
Z
r
Z
h
rp−1 [L1 (Er ) − L1 (Eh )] dr
0
L1 (Er ) dr − 2p hp L1 (Eh ).
0
On the other hand, by the fundamental theorem of calculus (see [Leo22d]) Z x+h ¯ u(x + h) − u ¯(x) ≤ u0 (t) dt ≤ hku0 kL∞ (R) = h, x
and so, B ≤ hp L1 (R \ Fh ) ≤ 2hp L1 (Eh ).
(2.70)
Combining estimates (2.67), (2.69), (2.70), we obtain Z ¯ u(x + h) − u ¯(x)p dx R p
Z
h
rp−1 L1 (Er ) dr − 2p hp L1 (Eh ) + 2hp L1 (Eh )
≤2 p 0 p
Z
=2 p
h
rp−1 L1 (Er ) dr.
0
In turn, by Tonelli’s theorem Z ∞Z Z Z h ¯ u(x + h) − u ¯(x)p1 dh p dxdh ≤ 2 p rp−1 L1 (Er ) dr p p h h 0 R ZR∞ 0 Z ∞ dh = 2p p rp−1 L1 (Er ) dr hp 0 r Z ∞ 2p p 2p p = L1 (Er ) dr = kukL1 (R) , p−1 0 p−1 where in the last equality we use (2.68) with q = 1. Hence, 1/p
uW 1−1/p,p (R) kukL1 (R) . Recalling that ku0 kL∞ (R) = 1, this concludes the proof.
Exercise 2.42. Prove that Theorem 2.41 continues to hold if we replace the domain R with an open interval I ⊆ R.
86
2. Embeddings and Interpolation
2.4. Notes Step 1 of the Sobolev–Gagliardo–Nirenberg embedding theorem (Theorem 2.2) draws upon the paper of Bru´e and Nguyen [BN20] and unpublished work of Brezis (see [Mir18]). The first proof of Morrey’s embedding theorem (Theorem 2.8) is inspired by the review paper of Mironescu [Mir18], and the second by Simon’s paper [Sim90]. Exercise 2.13 is proved in the book of F. Demengel and G. Demengel [DD12, Section 3.1.1], Exercises 2.14 and 2.15 are drawn from the book of Runst and Sickel [RS96, Section 2.3.1]. Exercise 2.17 is due to Taibleson [Tai64]. Theorem 2.18 is inspired by the book of Grisvard [Gri11]. The proof of Theorem 2.21 is adapted from the paper of Nguyen, Diaz, and Nguyen [NDN20]. The first time I have seen the use of the Hardy– Littlewood maximal function to prove Sobolevtype inequalities is in the paper of Hedberg [Hed72]; see also the paper of Maz’ya and Shaposhnikova [MS99]. Theorem 2.24 can be obtained as a special case of a general embedding result for Besov spaces contained in the paper of Simon [Sim90]. I worked on Section 2.3 with two undergraduate students, Andrew Chen and Grant Yu. The proof of Theorem 2.35 is in the same spirit of the proof of Theorem 2.21. I refer to the papers of Brezis and Mironescu [BM18], [BM19] for other proofs, extensive references, and necessary conditions. Theorem 2.39 is due to Bourdain, Brezis, and Mironescu [BBM04], while Theorem 2.41 is due to Besov (see [Kas78]).
Chapter 3
A Bit of Wavelets If only I had the Theorems! Then I should find the proofs easily enough. — Bernhard Riemann
Wavelet theory lies at the boundary between mathematics and engineering. It plays a crucial role in image compression and the numerical solution of differential equations [Coh03], [DL93], [Fra99], [Mey92]. Given a bounded function ϕ : R → R, we consider its dyadic dilates ϕ(2k x), k ∈ Z, and their translates ϕJ (x) = ϕ(2k x − j), j ∈ Z, where the index J corresponds to the dyadic interval (3.1)
J = Jj,k := [j/2k , (j + 1)/2k ).
We denote by D the collection of all dyadic intervals Jj,k , j, k ∈ Z. Note that if ϕ is concentrated around 0, then ϕJ is concentrated around 2−k j, and if ϕ is supported on an interval of length ` > 0, then ϕJ is concentrated on an interval of length 2−k `. Given a space X of functions u : R → R, we are interested in determining conditions on the function ϕ which guarantee that u can be decomposed in terms of the functions ϕJ , J ∈ D, that is, (3.2)
u=
X
aJ (u)ϕJ ,
J∈D
where aJ (u) are suitable coefficients and the convergence takes place in the space X. If (3.2) holds for all u ∈ X, we call ϕ the mother wavelet and (3.2) 1 a wavelet decomposition. If X = L2 (R) and the family { (L1 (J)) 1/2 ϕJ }J∈D is a complete orthonormal system in L2 (R), then ϕ is called an orthogonal wavelet. An example of orthogonal wavelet is given by the Haar function H := χ[0,1/2) − χ[1/2,1) . In this chapter, we will restrict our attention to this 87
88
3. A Bit of Wavelets
wavelet. The advantage is that H is a step function and has compact support. The disadvantage is that it is discontinuous. We have seen (Exercise 1.24) that the characteristic functions χ(a,b) belong to W s,p (R) only when sp < 1. Thus, throughout the chapter, we will confine ourselves to the case sp < 1. Daubechies constructed smooth wavelets with compact support (see, e.g., the book of Daubechies [Dau92] or the undergraduate textbook of Frazier [Fra99] for more details), but this construction goes beyond the purposes of this book. We will give a characterization of fractional Sobolev spaces in terms of Haar functions. Using this characterization, we will prove some interpolation inequalities (see Theorems 3.15, 3.17, and 3.19) in the borderline cases when ˙ 1,1 (R). one of the spaces involved is either L1 (R) or W
3.1. Weighted `p Spaces In view of the wavelet decomposition (3.2), we will be interested in studying sequences {aJ }J∈D . We will consider two type of spaces. Given γ ∈ R and 1 ≤ p < ∞, we define the weighted little ` p space ` pγ (D) as the space of all sequences {aJ }J∈D such that X (L1 (J))−γ(p−1) aJ p < ∞, J∈D
endowed with the norm !1/p (3.3)
k{aJ }J∈D k` pγ (D) :=
X
(L1 (J))−γ(p−1) aJ p
.
J∈D
We define the weak weighted little ` p space w` pγ (D) as the space of all sequences {aJ }J∈D for which there exists a constant C > 0 such that for every t > 0, X (L1 (J))γ ≤ Ct−p , J∈D: aJ ≥t(L1 (J))γ
endowed with the norm X
k{aJ }J∈D kw` pγ (D) := sup tp t>0
(L1 (J))γ .
J∈D: aJ ≥t(L1 (J))γ
Observe that ` pγ (D) ⊆ w` pγ (D). In what follows we will use the following interpolation result. Theorem 3.1. Let γ ∈ R, 1 ≤ p1 < p2 < ∞, and θ ∈ (0, 1). Then k{aJ }J∈D k` pγ (D) k{aJ }J∈D kθw` p1 (D) k{aJ }J∈D k1−θ p w` 2 (D) γ
for every sequence {aJ }J∈D in
w` pγ1 (D) ∩ w` pγ2 (D),
γ
and where
1 p
=
θ p1
+ 1−θ p2 .
3.2. Wavelets in Lp
89
Proof. Given {aJ }J∈D in w` pγ1 (D) ∩ w` pγ2 (D), define bJ := aJ /(L1 (J))γ . Then X Ai := k{aJ }J∈D kw` pγi (D) = sup tpi (L1 (J))γ t>0
J∈D: bJ ≥t
for i = 1, 2. Hence, for every t > 0, X (3.4) (L1 (J))γ ≤ min{t−p1 A1 , t−p2 A2 }. J∈D: bJ ≥t
We are going to use the layercake representation ([Leo22c]). By the Lebesgue monotone convergence theorem and inequality (3.4), for every ` > 0 we have Z bJ X X p 1 γ p 1 γ k{aJ }J∈D k` p (D) = (L (J)) bJ = (L (J)) ptp−1 dt γ
J∈D
Z
∞
X
= 0
J∈D ∞
(L1 (J))γ ptp−1 dt ≤
Z
`
≤ A1 p
p−p1 −1
t
min{t−p1 A1 , t−p2 A2 }ptp−1 dt
0
J∈D:bJ ≥t
Z
0
Z
∞
dt + A2 p
0
tp−p2 −1 dt
`
p 1 p A1 `p−p1 + A2 p2 −p . = p − p1 p2 − p ` Since this holds for every ` > 0, by Exercise 1.26, (p2 −p)/(p2 −p1 )
k{aJ }J∈D kp` p (D) A1 γ
(p−p1 )/(p2 −p1 )
A2
.
3.2. Wavelets in Lp The function (3.5)
H := χ[0,1/2) − χ[1/2,1)
is called the Haar function. Given a dyadic interval Jj,k = [j/2k , (j + 1)/2k ) ∈ D,
j, k ∈ Z,
we define the rescaled Haar function supported in Jj,k as (3.6)
Hj,k (x) := H(2k x − j),
x ∈ R.
Note that k2k/p Hj,k kLp (R) = 1. For every k ∈ Z, let Dk be the collection of all dyadic intervals Jj,k of length 2−k , j ∈ Z. Given 1 ≤ p ≤ ∞ and a function u ∈ Lploc (R), define the piecewise constant function X (3.7) Pk (u) := uJj,k χJj,k (x) , j∈Z
90
3. A Bit of Wavelets
where uE :=
1 L1 (E)
R E
u(y) dy. Observe that
1 2k (u, Hj,k )L2 (R) Hj,k = [uJ2j,k+1 − uJ2j+1,k+1 ](χJ2j,k+1 − χJ2j+1,k+1 ) 2 = uJ2j,k+1 χJ2j,k+1 + uJ2j+1,k+1 χJ2j+1,k+1 1 − [uJ2j,k+1 + uJ2j+1,k+1 ](χJ2j,k+1 + χJ2j+1,k+1 ) 2 = uJ2j,k+1 χJ2j,k+1 + uJ2j+1,k+1 χJ2j+1,k+1 − uJj,k χJj,k , and so, Pk+1 (u) − Pk (u) =
X
uJj,k+1 χJj,k+1 −
j∈Z k
(3.8)
=2
X
uJl,k χJl,k (x)
l∈Z
X
(u, Hj,k )L2 (R) Hj,k .
j∈Z
Using this identity, we can show that if 1 < p < ∞, we can write a function u ∈ Lp (R) as X X u= (Pk+1 (u) − Pk (u)) = 2k (u, Hj,k )L2 (R) Hj,k , k∈Z
j,k∈Z
where the convergence of the series takes place in Lp (R), namely,
n
X
(Pk+1 (u) − Pk (u)) →0
u −
k=−m
Lp (R)
as m, n → ∞. Theorem 3.2. Let 1 < p < ∞ and u ∈ Lp (R). Then X (3.9) u= (Pk+1 (u) − Pk (u)) in Lp (R). k∈Z
Proof. Fix k ∈ Z. Since the sets Jj,k , j ∈ Z, defined in (3.1) are pairwise disjoint, it follows from (3.7) and H¨older’s inequality that for every x ∈ R, X X 2−k(p−1) Z p p Pk (u)(x) = uJj,k  χJj,k (x) ≤ u(y)p dyχJj,k (x) . 2−kp Jj,k j∈Z
j∈Z
Hence, by Tonelli’s theorem and the fact that Jj,k has length 2−k , Z Z XZ p p (3.10) Pk (u)(x) dx ≤ u(y) dy = u(y)p dy. R
j∈Z
Jj,k
R
Using telescopic series, we can write u−
n−1 X k=−m
(Pk+1 (u) − Pk (u)) = u − Pn (u) + P−m (u).
3.2. Wavelets in Lp
91
Thus, to prove (3.9), it suffices to show that P−m (u) → 0 in Lp (R) as m → ∞ and Pn (u) → u in Lp (R) as n → ∞. Assume that u ∈ Cc (R), with supp u ⊆ [−2m0 , 2m0 ]. Since u is uniformly continuous, given ε > 0 there exists δ > 0 such that u(x) − u(y) ≤ ε for all x, y ∈ R, with x − y ≤ δ. Let n be so large that 21n ≤ δ. Then Jj,n has length less than or equal to δ. For every x ∈ R, let j be such that x ∈ Jj,n . Then by (3.7), Z −n u(x) − Pn (u)(x) ≤ 2 u(x) − u(y) dy ≤ ε. Jj,n
Since supp Pn (u) ⊆ [−2m0 +1 , 2m0 +1 ] for all n large, by raising both sides to power p and integrating over R, we obtain Z Z 2m0 +1 p u(x) − Pn (u)(x) dx = u(x) − Pn (u)(x)p dx ≤ εp 22m0 +2 . −2m0 +1
R
This shows that Pn (u) → u in Lp (R) as n → ∞. On the other hand, for all m ≥ m0 , we have that supp u ⊆ J−1,−m ∪ J0,−m , and so, by (3.7), Z P−m (u)(x)p dx = uJ−1,−m p 2m + uJ0,−m p 2m R
≤
2 2m(p−1)
kukL∞ (R) → 0
as m → ∞.
We now remove the additional hypothesis that u ∈ Cc (R). By density of smooth functions in Lp (R) (see [Leo22c]), given ε > 0, we can find v ∈ Cc (R) such that ku − vkLp (R) ≤ ε. By the linearity of Pk and (3.10), ku − Pn (u)kLp (R) ≤ ku − vkLp (R) + kv − Pn (v)kLp (R) + kPn (u − v)kLp (R) ≤ 2ku − vkLp (R) + kv − Pn (v)kLp (R) ≤ 2ε + kv − Pn (v)kLp (R) . Since v ∈ Cc (R), by what we just proved, kv − Pn (v)kLp (R) → 0 as n → ∞. Thus, Pn (u) → u as n → ∞. Similarly, kP−m (u)kLp (R) ≤ kP−m (v)kLp (R) + kP−m (u − v)kLp (R) ≤ ku − vkLp (R) + kP−m (v)kLp (R) ≤ ε + kP−m (v)kLp (R) . Since kP−m (v)kLp (R) → 0 as m → ∞, it follows that kP−m (u)kLp (R) → 0 as m → ∞. Exercise 3.3. Prove that (3.9) fails for p = 1 and p = ∞. Exercise 3.4. Prove that the family {2k/2 Hj,k }j,k∈Z of rescaled Haar functions is orthonormal in L2 (R).
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3. A Bit of Wavelets
3.3. Wavelets in L1 In view of Exercise 3.3, Theorem 3.2 fails in L1 (R). However, we can still show that the sequence {(u, HJ )L2 (R) }J∈D belongs to the weak space w`1γ (D) when γ < 0 or γ > 1 (see Section 3.1). We recall that D is the collection of all dyadic intervals Jj,k , j, k ∈ Z. Given J = Jj,k ∈ D, we set HJ := Hi,j , where Hi,j is the rescaled Haar function defined in (3.6). Theorem 3.5. Let u ∈ L1 (R) and γ ∈ R be such that γ < 0 or γ > 1. Then for every t > 0, 1 (L1 (J))γ kukL1 (R) . t γ
X
(3.11)
J∈D: (u,HJ )L2 (R) ≥t(L1 (J))
In particular, k{(u, HJ )L2 (R) }J∈D kw`1γ (D) kukL1 (R) . Proof. Given t > 0, let Z 1 γ (3.12) Λt := J ∈ D : u(x) dx ≥ t(L (J)) . J
By (3.6), Z (u, HJ )L2 (R)  ≤ max HJ  J
Z u(x) dx =
J
u(x) dx, J
R and so, if (u, HJ )L2 (R)  ≥ t(L1 (J))γ , then J u(x) dx ≥ t(L1 (J))γ , which shows that J ∈ D : (u, HJ )L2 (R)  ≥ t(L1 (J))γ ⊆ Λt . Hence, to show (3.11), it is enough to prove that Z X 1 γ −1 (L (J)) t u(x) dx. R
J∈Λt
Step 1: Assume that γ > 1. Let Λmax be the set of all maximal intervals t of Λt , that is, all dyadic intervals J in Λt with the property that if R ∈ Λt and J ⊆ R, then R = J. Given R ∈ Λt , note that Z u(x) dx ≥ t(L1 (R))γ . R
Since γ > 1, it follows that Λt does not contain intervals of arbitrarily large length. In turn, for every R ∈ Λt , there exists a unique maximal interval
3.4. Wavelets in W 1,1
93
J ∈ Λmax that contains R. Thus, we can write t X
(L1 (R))γ ≤
X
X
J∈Λmax R∈Λt , R⊆J t
R∈Λt
≤
X
(L1 (J))γ
J∈Λmax t
J∈Λmax k=0 t ∞ X
X
2k(1−γ)
(L1 (R))γ
R∈Λt , R⊆J, L1 (R)=2−k L1 (J)
(L1 (J))γ
u(x) dx t−1
Z u(x) dx,
J
J∈Λmax t
X
J∈Λmax t
k=0
X Z
t−1
∞ X X
(L1 (R))γ =
R
where we use (3.12) and the fact that there are at most 2k intervals R ∈ Λt contained in J with length L1 (R) = 2−k L1 (J), and the last inequality follows from the fact that the maximal intervals in Λmax are pairwise disjoint. t be the set of all minimal intervals of Step 2: Assume that γ < 0. Let Λmin t Λt , that is, all dyadic intervals J in Λt with the property that if R ∈ Λt and R ⊆ J, then R = J. Given R ∈ Λt , note that Z u(x) dx ≥ t(L1 (R))γ . R
Since γ < 0, we have that the length of R is bounded away from zero by a positive constant. Hence, Λt does not contain intervals of arbitrarily small length. In turn, for every R ∈ Λt , there exists a unique minimal interval contained in R. Thus, we can write J ∈ Λmin t X
X
(L1 (R))γ ≤
X
(L1 (R))γ =
R∈Λt , J⊆R J∈Λmin t
R∈Λt
X
≤
(L1 (J))γ
J∈Λmin t
t
−1
k=0
X Z J∈Λmin t
∞ X
J
∞ X X k=0 J∈Λmin t
X
2kγ
X
(L1 (R))γ
R∈Λt , J⊆R, L1 (R)=2k L1 (J)
(L1 (J))γ
J∈Λmin t
u(x) dx t
−1
Z u(x) dx, R
where we use (3.12) and the fact that there is at most one interval R ∈ Λt containing J with length L1 (R) = 2k L1 (J), and the last inequality follows from the fact that the minimal intervals in Λmin are pairwise disjoint. t
3.4. Wavelets in W 1,1 In this section we study the space W 1,1 (R). We will show that the rescaled sequence {(L1 (J))−1 (u, HJ )L2 (R) }J∈D belongs to the weak space w`1γ (D) when γ < 0 or γ > 1.
94
3. A Bit of Wavelets
˙ 1,1 (R) and γ ∈ R be such that γ < 0 or γ > 1. Theorem 3.6. Let u ∈ W Then for every t > 0, X 1 (3.13) (L1 (J))γ ku0 kL1 (R) . t 1 γ+1 J∈D: (u,HJ )L2 (R) ≥t(L (J))
In particular, k{(L1 (J))−1 (u, HJ )L2 (R) }J∈D kw`1γ (D) ku0 kL1 (R) . Proof. Step 1: Let ¯ be the absolutely continuous representative of u (see R u [Leo22d]). Since J HJ (x) dx = 0 by (3.5) and (3.6), we can write Z (u, HJ )L2 (R) = (¯ u(x) − uJ )HJ (x) dx. J
In turn, Z (3.14)
(u, HJ )L2 (R)  ≤ max HJ  J
Z ¯ u(x) − uJ  dx =
¯ u(x) − uJ  dx.
J
J
By the mean value theorem for integrals there exists xJ ∈ J such that Z 1 uJ = 1 u ¯(y) dy = u ¯(xJ ). L (J) J Hence, we can use the fundamental theorem of calculus (see [Leo22d]) to write Z x u ¯(x) − uJ = u0 (t) dt. xJ
Together with inequality (3.14), this implies that Z Z (3.15) (u, HJ )L2 (R)  ≤ ¯ u(x) − uJ  dx ≤ L1 (J) u0 (t) dt. J
J
Given t > 0, let (3.16)
Λt :=
Z J ∈D:
0
1
γ
u (t) dt ≥ t(L (J))
.
J
By inequality (3.15), if (u, HJ )L2 (R)  ≥ t(L1 (J))γ+1 , then t(L1 (J))γ , which shows that J ∈ D : (u, HJ )L2 (R)  ≥ t(L1 (J))γ+1 ⊆ Λt . Hence, to show (3.13), it is enough to prove that Z X 1 γ −1 (L (J)) t u0 (x) dx. J∈Λt
R
R J
u0 (x) dx ≥
3.4. Wavelets in W 1,1
95
Step 2: Assume that γ > 1. Let Λmax denote the set of all maximal t intervals of Λt , that is, all dyadic intervals J in Λt with the property that if R ∈ Λt and J ⊆ R, then R = J. Given R ∈ Λt , note that
Z
u0 (x) dx ≥ t(L1 (R))γ .
R
Since γ > 1, it follows that Λt does not contain intervals of arbitrarily large length. In turn, for every R ∈ Λt , there exists a unique maximal interval J ∈ Λmax that contains R. Thus, we can write t
X
(L1 (R))γ ≤
X
X
(L1 (R))γ
J∈Λmax R∈Λt , R⊆J t
R∈Λt
=
∞ X X J∈Λmax k=0 t
≤
X
X
R∈Λt , R⊆J, L1 (R)=2−k L1 (J)
(L1 (J))γ
J∈Λmax t
J∈Λmax t
∞ X
X
2k(1−γ)
k=0
X Z
t−1
(L1 (R))γ
u0 (x) dx t−1
J
(L1 (J))γ
J∈Λmax t
Z
u0 (x) dx,
R
where we use (3.16) and the fact that there are at most 2k intervals R ∈ Λt contained in J with length L1 (R) = 2−k L1 (J), and the last inequality follows from the fact that the maximal intervals in Λmax are pairwise disjoint. t Step 3: Assume that γ < 0. Let Λmin be the set of all minimal intervals of t Λt , that is, all dyadic intervals J in Λt with the property that if R ∈ Λt and R ⊆ J, then R = J. Given R ∈ Λt , note that
Z
u0 (x) dx ≥ t(L1 (R))γ .
R
Since γ < 0, we have that the length of R is bounded away from zero by a positive constant. Hence, Λt does not contain intervals of arbitrarily small length. In turn, for every R ∈ Λt , there exists a unique minimal interval
96
3. A Bit of Wavelets
J ∈ Λmin contained in R. Thus, we can write t X X X (L1 (R))γ (L1 (R))γ ≤ R∈Λt , J⊆R J∈Λmin t
R∈Λt
∞ X X
=
k=0 J∈Λmin t
X
≤
X R∈Λt , J⊆R, L1 (R)=2k L1 (J)
(L1 (J))γ
J∈Λmin t
∞ X
2kγ
J∈Λmin t
X
(L1 (J))γ
J∈Λmin t
k=0
X Z
t−1
(L1 (R))γ
u0 (x) dx t−1
Z
J
u0 (x) dx,
R
where we use (3.16) and the fact that there is at most one interval R ∈ Λt containing J with length L1 (R) = 2k L1 (J), and the last inequality follows from the fact that the minimal intervals in Λmin are pairwise disjoint. t
3.5. Wavelets in W s,p ˙ s,p (R) in In this section we use Theorem 3.2 to characterize the space W terms of Haar functions. As explained at the beginning of the chapter, we will restrict our attention to the case sp < 1 and refer to [Coh03], [DL93], [DP88], [Mey92] for the case sp ≥ 1. Theorem 3.7. Let 1 < p < ∞, 0 < s < 1, and u ∈ Lploc (R). Then !1/p X p ksp (3.17) 2 kPk+1 (u) − Pk (u)kLp (R) uW s,p (R) . k∈Z
Moreover, if sp < 1 and u ∈ Lp (R), then !1/p (3.18)
uW s,p (R)
X
2ksp kPk+1 (u) −
Pk (u)kpLp (R)
.
k∈Z
Lemma 3.8. Let k ∈ Z, 1 < p < ∞, and u ∈ Lploc (R). Then ku − Pk (u)kLp (R) ≤ 41/p
sup 0 1, assume that sp < 1. Then for all u ∈ ˙ 1,p2 (I), Lp1 (I) ∩ W (3.39)
uW s,p (I) kukθLp1 (I) ku0 k1−θ Lp2 (I) .
Proof. If p1 > 1 and p2 > 1, inequality (3.39) follows from Theorem 2.32. On the other hand, if p1 = p2 = 1, then u ∈ W 1,1 (I), and we can apply Corollary 1.27. Thus, it suffices to consider the cases p1 = 1 and p2 > 1 or p1 > 1 and p2 = 1. Step 1: Assume that p1 = 1 and p2 > 1. Suppose that I = R and let u ∈ Cc∞ (R). If u is zero, there is nothing to prove, so we can suppose that ¯ ¯ u 6= 0. Choose s < σ < 1 and q > 1 such that p1 = θ¯ + 1−q θ , σ = (1 − θ)s. Then by Theorem 3.15, ¯
¯
θ uW s,p (R) kukθL1 (R) u1− W σ,q (R) .
On the other hand, writing we have
1 q
=
θ˜ p
+
1−θ˜ p2 , ˜
˜ + 1 − θ, ˜ by Theorem 2.35, σ = θs ˜
θ uW σ,q (R) uθW s,p (R) ku0 k1− Lp2 (R) .
Combining these two inequalities, we obtain 1−θ¯ ¯ ˜ 1−θ˜ . uW s,p (R) kukθL1 (R) uθW s,p (R) ku0 kL p2 (R)
106
3. A Bit of Wavelets
Since u ∈ Cc∞ (R) and u 6= 0, necessarily, 0 < uW s,p (R) < ∞, and so, upon division, ¯ ˜ ¯ θ/[1− θ(1− θ)]
uW s,p (R) kukL1 (R)
˜ ¯ ˜ ¯ (1−θ)(1− θ)/[1− θ(1− θ)]
ku0 kLp2 (R)
= kukθL1 (R) ku0 k1−θ Lp2 (R) . ˙ 1,p2 (R), then by Exercise 2.36, we can find un ∈ If now u ∈ L1 (R) ∩ W Cc∞ (R) such that un → u in L1 (R) and u0n → u0 in Lp2 (R). By extracting a subsequence, not relabeled, we can suppose that un → u pointwise L1 a.e. in R. By applying what we just proved to each un , we obtain un W s,p (R) kun kθL1 (R) ku0n k1−θ Lp2 (R) . Letting n → ∞ and using Fatou’s lemma on the lefthand side gives uW s,p (R) kukθL1 (R) ku0 k1−θ Lp2 (R) . Step 2: Assume that p1 > 1 and p2 = 1. Suppose that I = R and let u ∈ Cc∞ (R). Take 0 < σ < s so small that σp1 < 1 and let θ¯ ∈ (0, 1) be such ¯ ¯ + (1 − θ)1. ¯ ¯ that s = θσ Let q > 1 be given by 1 = θ + (1 − θ)1. Then by p
q
Theorem 3.17, ¯
¯
θ uW s,p (R) uθW σ,q (R) ku0 k1− . L1 (R)
On the other hand, writing we have
1 q
=
θ˜ p
+
1−θ˜ p2 ,
˜ + 1 − θ, ˜ by Theorem 2.34, σ = θs ˜
˜
θ uW σ,q (R) kukθLp1 (R) u1− W s,p (R) .
Combining these two inequalities, we obtain θ¯ ˜ θ˜ θ¯ uW s,p (R) kukθLp1 (R) u1− ku0 k1− . s,p W (R) L1 (R) Since u ∈ Cc∞ (R) and u 6= 0, necessarily, 0 < uW s,p (R) < ∞, and so, upon division, ¯ ˜ θ] ¯ θ¯θ/[1−(1− θ)
uW s,p (R) kukLp1 (R)
¯ ˜ θ] ¯ (1−θ)/[1−(1− θ)
ku0 kL1 (R)
= kukθLp1 (R) ku0 k1−θ . L1 (R) ˙ 1,1 (I) is similar to the last part of Step 1 The case in which u ∈ Lp1 (I) ∩ W and is left as an exercise. Step 3: We leave the case in which I is an open interval as an exercise.
3.7. Notes
107
3.7. Notes Theorems 3.5 and 3.15 are drawn from Cohen’s paper [Coh00], while Theorems 3.6 and 3.17 from the paper of Cohen, Dahmen, Daubechies, and De Vore [CDDD03]. We refer to the papers [DL92], [DP88], [DS93] and the books [Dau92], [DL93], [Fra99], [Mey92] for more details about wavelets and fractional Sobolev spaces.
Chapter 4
Rearrangements I see it, but I don’t believe it. — Richard Dedekind
In this chapter, you will learn a bit about rearrangements. The idea is to replace a function u : E → R with another function v : F → R that is more symmetric than u but has the same distribution function, that is, L1 ({x ∈ E : u(x) > t}) = L1 ({y ∈ F : v(y) > t}) for all t ∈ R. You will study two types of rearrangements: polarization and symmetric decreasing rearrangement.
4.1. Polarization Given a ∈ R, consider the two halflines (a, ∞) and (−∞, a). If a 6= 0, we denote by I+ the halfline that contains the origin and by I− the other one. If a = 0, take I− = (0, ∞) and I+ = (−∞, 0). Consider the reflection σ : R → R such that σ(I± ) = I∓ , that is, σ(x) := −x + 2a,
x ∈ R.
Given a function u : R → R, the twopoint rearrangement or polarization of u with respect to σ is the function uσ : R → R, given by max{u(x), u(σ(x))} if x ∈ I+ , min{u(x), u(σ(x))} if x ∈ I− , (4.1) uσ (x) := u(a) if x = a. This definition can be extended to functions u : R → [−∞, ∞]. Theorem 4.1. Let u : R → R be a Lebesgue measurable function and let σ be a reflection. Then for every t ∈ R, L1 ({x ∈ R : u(x) > t}) = L1 ({x ∈ R : uσ (x) > t)}. 109
110
4. Rearrangements
Proof. Define F := {x ∈ I+ : u(x) > t}, σ
σ
F := {x ∈ I+ : u (x) > t},
G := {x ∈ I− : u(x) > t}, Gσ := {x ∈ I− : uσ (x) > t}.
In view of (4.1), x ∈ F σ if and only if x ∈ I+ and max{u(x), u(σ(x))} > t. Since σ(I± ) = I∓ and σ(σ(x)) = x for every x ∈ R, it follows that F σ = F ∪ σ(G). Similarly, Gσ = σ(F ) ∩ G. Using the fact that the reflection σ is measure preserving, we have L1 ({x ∈ R : u(x) > t}) = L1 (F ) + L1 (G) = L1 (F \ σ(G)) + L1 (F ∩ σ(G)) + L1 (G) = L1 (F \ σ(G)) + L1 (σ(G)) + L1 (σ(F ∩ σ(G))) = L1 (F ∪ σ(G)) + L1 (σ(F ) ∩ G) = L1 (F σ ) + L1 (Gσ ) = L1 ({x ∈ R : uσ (x) > t)}.
Exercise 4.2. Given a Lebesgue measurable function u : R → R and a reflection σ, let M be the family of Borel sets B ⊆ R such that L1 ({x ∈ R : u(x) ∈ B}) = L1 ({x ∈ R : uσ (x) ∈ B}). (i) Prove that if Bn ∈ M, n ∈ N, are pairwise disjoint, then M. S (ii) Prove that if Bn ∈ M, n ∈ N, then ∞ n=1 Bn ∈ M.
S∞
n=1 Bn
∈
(iii) Prove that if B ∈ M, then R \ B ∈ M. (iv) Prove that every open set U ⊆ R belongs to M. (v) Prove that every Borel set B ⊆ R belongs to M. Exercise 4.3. Let u, v : R → R be two Lebesgue measurable functions with u ≤ v and let σ be a reflection. Prove that uσ ≤ v σ . Exercise 4.4. Let u : R → R be a Lebesgue measurable function and let σ be a reflection. Prove that χσ{u>t} = χ{uσ >t} for every t ∈ R. Let E ⊆ R be a Lebesgue measurable set and let u : E → [0, ∞] be a Lebesgue measurable function. For all t ≥ 0 let Et := {x ∈ E : u(x) > t}). The identity Z ∞ (4.2) u(x) = χEt (x) dt, x ∈ E, 0
is called the layercake representation of u. Using the layercake representation, we can prove that polarization preserves Lp norms. Theorem 4.5. Let u : R → measurable function and R [0, ∞) be a Lebesgue R let σ be a reflection. Then R (u(x))p dx = R (uσ (x))p dx for every p > 0.
4.1. Polarization
111
Proof. By the layercake representation (4.2) applied to (u(x))p , Tonelli’s theorem, and the change of variables t = rp , we have Z ∞Z Z Z ∞ Z p χ{up >t} (x) dxdt χ{up >t} (x) dtdx = (u(x)) dx = 0 R R ZR∞ 0 L1 ({x ∈ R : (u(x))p > t}) dt = 0 Z ∞ = prp−1 L1 ({x ∈ R : u(x) > r}) dr. 0
Similarly, Z
σ
∞
Z
p
prp−1 L1 ({x ∈ R : uσ (x) > r}) dr.
(u (x)) dx = 0
R
To conclude, it suffices to observe that the righthand sides of these two identities coincide by Theorem 4.1. More generally, we have the following result. Theorem 4.6. Let f : R → [0, ∞) be a Borel function such that f (0) = 0, let u : R → R be a Lebesgue measurable function vanishing at infinity, and let σ be a reflection. Then Z Z f (u(x)) dx = f (uσ (x)) dx. (4.3) R
R
Proof. Step 1: Assume that f = χB , where B ⊆ R+ is a Borel set. Then by Exercise 4.2, Z f (u(x)) dx = L1 ({x ∈ R : u(x) ∈ B}) R Z 1 σ = L ({x ∈ R : u (x) ∈ B}) = f (uσ (x)) dx. R
Pn
Step 2: Assume that f = i=1 ci χBi , where Bi ⊆ R+ are Borel sets and ci ≥ 0. Then by Step 1, Z Z n X f (u(x)) dx = ci χBi (u(x)) dx R
=
i=1 n X i=1
R
Z ci
σ
Z
χBi (u (x)) dx = R
f (uσ (x)) dx.
R
Step 3: Let f be as in the statement and construct a sequence of simple Borel functions fn such R that 0 ≤ fn (t) ≤ R fn+1 (t) and fn (t) → f (t) for every t ∈ R. By Step 2, R fn (u(x)) dx = R fn (uσ (x)) dx. We can apply the Lebesgue monotone convergence theorem to obtain (4.3).
112
4. Rearrangements
Corollary 4.7. Let f : R → R be a Borel function such that f (0) = 0, let u : R → R be a Lebesgue measurable function vanishing at infinity and such that f ◦ u is Lebesgue integrable, and let σ be a reflection. Then Z Z f (u(x)) dx = f (uσ (x)) dx. R
R
Proof. Write f = f+ − f− , where t+ = max{t, 0} and t− = max{−t, 0}. Since f ◦ u is Lebesgue integrable, Z Z [f+ (u(x)) + f− (u(x))] dx = [(f (u(x))+ + (f (u(x))− ] dx < ∞. R
R
Since f+ and f− are Borel functions, by Theorem 4.6, Z Z Z Z σ f+ (u(x)) dx = f+ (u (x)) dx, f− (u(x)) dx = f− (uσ (x)) dx. R
R
R
R
To conclude the proof, it suffices to subtract these two equalities.
Next, we show that if u is uniformly continuous, then so is uσ , and that the modulus of continuity of uσ is less than or equal to the modulus of continuity of u. We recall that, given E ⊆ R and u : E → R, the modulus of continuity of u is the function ωu : [0, ∞) → [0, ∞], defined by ωu (δ) := sup{u(x) − u(y) : x, y ∈ E, x − y ≤ δ},
δ ≥ 0.
We leave it as an exercise to check that u is uniformly continuous in E if and only if ω(δ) → 0 as δ → 0+ . Theorem 4.8. Let u : R → R and let σ be a reflection. Then ωuσ ≤ ωu . Proof. Let σ(x) = −x + 2a, a ∈ R, and x, y ∈ R. We claim that there exist w, z ∈ R such that (4.4)
uσ (x) − uσ (y) ≤ u(w) − u(z),
w − z ≤ x − y.
We consider several cases. Case 1: Assume that x, y ∈ I+ . Then uσ (x) = max{u(x), u(σ(x))} and uσ (y) = max{u(y), u(σ(y))}. If uσ (x) = u(x) and uσ (y) = u(y), we take w = x and z = y. If uσ (x) = u(σ(x)) and uσ (y) = u(σ(y)), take w = σ(x), z = σ(y), and observe that σ(x) − σ(y) = x − y. If uσ (x) = u(σ(x)) and uσ (y) = u(y), then we consider two subcases. Subcase 1a: Assume that uσ (x) ≥ uσ (y). Then uσ (x) − uσ (y) = uσ (x) − uσ (y) = u(σ(x)) − u(y) ≤ u(σ(x)) − u(σ(y)), where we use the fact that u(y) ≥ u(σ(y)). Take w = σ(x), z = σ(y), so that σ(x) − σ(y) = x − y.
4.1. Polarization
113
Subcase 1b: Assume that uσ (x) < uσ (y). Then uσ (x) − uσ (y) = uσ (y) − uσ (x) = u(y) − u(σ(x)) ≤ u(y) − u(x), where we use the fact that u(x) ≤ u(σ(x)). Take w = x, z = y. The case uσ (x) = u(x) and uσ (y) = u(σ(y)) is similar. Case 2: Assume that x, y ∈ I− . Then uσ (x) = min{u(x), u(σ(x))} and uσ (y) = min{u(y), u(σ(y))}. If uσ (x) = u(x) and uσ (y) = u(y), we take w = x and z = y. If uσ (x) = u(σ(x)) and uσ (y) = u(σ(y)), take w = σ(x), z = σ(y), and observe that σ(x) − σ(y) = x − y. If uσ (x) = u(σ(x)) and uσ (y) = u(y), then we consider two subcases. Subcase 2a: Assume that uσ (x) ≥ uσ (y). Then uσ (x) − uσ (y) = uσ (x) − uσ (y) = u(σ(x)) − u(y) ≤ u(x) − u(y), where we use the fact that u(σ(x)) ≤ u(x). Take w = x, z = y. Subcase 2b: Assume that uσ (x) < uσ (y). Then uσ (x) − uσ (y) = uσ (y) − uσ (x) = u(y) − u(σ(x)) ≤ u(σ(y)) − u(σ(x)), where we use the fact that u(y) ≤ u(σ(y)). Take w = σ(x), z = σ(y), so that σ(x) − σ(y) = x − y. The case uσ (x) = u(x) and uσ (y) = u(σ(y)) is similar. Case 3: Assume that x and y are on the opposite side of a, say, x ∈ I+ and y ∈ I− . Then uσ (x) = max{u(x), u(σ(x))} and uσ (y) = min{u(y), u(σ(y))}. If uσ (x) = u(x) and uσ (y) = u(y), we take w = x and z = y. If uσ (x) = u(σ(x)) and uσ (y) = u(σ(y)), take w = σ(x), z = σ(y), and observe that σ(x) − σ(y) = x − y. If uσ (x) = u(σ(x)) and uσ (y) = u(y), then uσ (x) − uσ (y) = u(σ(x)) − u(y). Take w = σ(x), z = y, and observe that w − z ≤ x − y, since x and y lie on opposite sides of a and σ(x) is the reflection of x with respect to a. The case uσ (x) = u(x) and uσ (y) = u(σ(y)) is similar. In conclusion, we have proved claim (4.4). In turn, if x, y ∈ R, with x − y ≤ δ, then w − z ≤ x − y ≤ δ, and so uσ (x) − uσ (y) ≤ u(w) − u(z) ≤ ωu (δ). Taking the supremum over all such x, y gives ωuσ (δ) ≤ ωu (δ).
Remark 4.9. Note that in Cases 1 and 2 we have shown that if x, y ∈ I+ or x, y ∈ I− , then uσ (x) − uσ (y) ≤ u(x) − u(y) or
uσ (x) − uσ (y) ≤ u(σ(x)) − u(σ(y)).
Exercise 4.10. Let u : R → R and let f : R → R be an increasing function. Prove that (f ◦ u)σ = (f ◦ uσ ) for every reflection σ.
114
4. Rearrangements
Exercise 4.11. Let (a, b, c, d) ∈ R4 . Prove that  max{a, b} − max{c, d} +  min{a, b} − min{c, d} ≤ a − c + b − d. Exercise 4.12. Let u : R → R be such that Var u < ∞, where n X Var u = sup u(xi ) − u(xi−1 ), i=1
where the supremum is taken over all partitions of R. Prove that Var uσ ≤ Var u for every reflection σ. Hint: Use Exercise 4.11. Given an interval I ⊆ R and a function u : I → R, the modulus of absolute continuity of u is the function ωu,ac : [0, ∞) → [0, ∞], defined by ωu,ac (δ) := sup
n X
u(bi ) − u(ai ),
δ ≥ 0,
i=1
where the supremum is taken over all finite collections {(ai ,P bi )}ni=1 , n ∈ N, of nonoverlapping intervals with endpoints in I and such that ni=1 (bi −ai ) < δ. We leave it as an exercise to check that u is absolutely continuous in I if and only if ωu,ac (δ) → 0 as δ → 0+ . Theorem 4.13. Let u : R → R and let σ be a reflection. Then ωuσ ,ac ≤ ωu,ac . In particular, if u is absolutely continuous, then so is uσ . Proof. Step 1: Given δ > P 0, let {(ai , bi )}ni=1 be a collection of nonovern lapping intervals such that i=1 (bi − ai ) < δ. Assume that for every i = 1, . . . , n, either (ai , bi ) ⊆ I+ or (ai , bi ) ⊆ I− , and that either (ai , bi ) = (σ(bk ), σ(ak )) for some k 6= i or (ai , bi ) ∩ (σ(bk ), σ(ak )) = ∅ for all k 6= i. Fix i = 1, . . . , n. If (ai , bi ) = (σ(bk ), σ(ak )) for some k 6= i, define (ci , di ) := (ai , bi ) and (ck , dk ) := (ak , bk ). By (4.1), Exercise 4.11 and the fact that (ci , di ) and (ck , dk ) are on opposite sides of a, (4.5) uσ (bi ) − uσ (ai ) + uσ (bk ) − uσ (ak ) ≤ u(di ) − u(ci ) + u(dk ) − u(ck ). On the other hand, if (ai , bi ) ∩ (σ(bk ), σ(ak )) = ∅ for all k 6= i, then by Remark 4.9, uσ (bi ) − uσ (ai ) ≤ u(bi ) − u(ai ) or uσ (bi ) − uσ (ai ) ≤ u(σ(bi )) − u(σ(ai )). Define (ci , di ) := (ai , bi ) if the first equality holds and (ci , di ) := (σ(bi ), σ(ai )) if the second holds. Then uσ (bi ) − uσ (ai ) ≤ u(di ) − u(ci ).
(4.6)
n Note that di − ci = bi − ai . Moreover, Pnthe {(ci , di )}i=1 is still a collection of nonoverlapping intervals such that i=1 (ci − di ) < δ. It follows from (4.5) and (4.6) that n n X X σ σ u (bi ) − u (ai ) ≤ u(di ) − u(ci ) ≤ ωu,ac (δ). i=1
i=1
4.1. Polarization
115
Step 2: Given δ > let {(ai , bi )}ni=1 be a collection of nonoverlapping P0, n intervals such that i=1 (bi − ai ) < δ. In this step we replace {(ai , bi )}ni=1 with a collection {(αi , βi )}`i=1 satisfying the additional properties in Step 1. First, if a ∈ (ai , bi ), we replace (ai , bi ) with the two intervals (ai , a) and (a, bi ). Thus, for every i either (ai , bi ) ⊆ I+ or (ai , bi ) ⊆ I− . Next, if (ai , bi ) does not overlap with any of the reflected intervals (σ(bk ), σ(ak )), k 6= i, we leave (ai , bi ) as is. Otherwise, let ai < y1,i < y2,i < · · · < yni ,i < bi , where the yj,i are all the endpoints of the form σ(bk ) and σ(ak ) that are contained in (ai , bi ). In this case, we replace (ai , bi ) with the intervals (ai , y1,i ), . . . , (yni ,i , bi ). Let {(αi , βi )}`i=1 be the resulting collection of intervals. Since this collection has been obtained by partitioning some of the intervals (ai , bi ), the intervals (αi , βi ) are still nonoverlapping and their total length is still less than δ. Hence, by Step 1, n X
σ
σ
u (bi ) − u (ai ) ≤
` X
uσ (βi ) − uσ (αi ) ≤ ωu,ac (δ).
i=1
i=1
Taking the supremum over all such collections {(ai , bi )}ni=1 gives ωuσ ,ac ≤ ωu,ac . The next result shows that the support of the polarization of a function does not increase. Theorem 4.14. Let u : R → R and let σ be a reflection. Then for every t ∈ R, diam{x ∈ R : uσ (x) > t} ≤ diam{x ∈ R : u(x) > t}. Proof. Let σ(x) = −x + 2a, a ∈ R, and x, y ∈ R be such that uσ (x) > t and uσ (y) > t. We claim that there exist w, z ∈ R such that (4.7)
u(w) > t,
u(z) > t,
w − z ≥ x − y.
We consider several cases. Case 1: Assume that x, y ∈ I+ . Then uσ (x) = max{u(x), u(σ(x))} > t and uσ (y) = max{u(y), u(σ(y))} > t. If u(x) > t and u(y) > t, we take w = x and z = y. If u(σ(x)) > t and u(σ(y)) > t, take w = σ(x), z = σ(y), and observe that σ(x) − σ(y) = x − y. If u(σ(x)) > t and u(y) > t, then we take w = σ(x) and z = y and observe that x − y ≤ σ(x) − y, since σ(x) and y lie on opposite sides of a and x = σ(σ(x)) is the reflection of σ(x) with respect to a. The case u(x) > t and u(σ(y)) > t is similar. Case 2: Assume that x, y ∈ I− . Then uσ (x) = min{u(x), u(σ(x))} > t and uσ (y) = min{u(y), u(σ(y))} > t. We proceed exactly as in Case 1.
116
4. Rearrangements
Case 3: Assume that x and y are on the opposite side of a, say, x ∈ I+ and y ∈ I− . Then uσ (x) = max{u(x), u(σ(x))} > t and uσ (y) = min{u(y), u(σ(y))} > t, which implies that u(y) > t and u(σ(y)) > t. If u(x) = uσ (x) > t, we take w = x and z = y. If uσ (x) = u(σ(x)), we take w = σ(x), z = σ(y), and use the fact that σ(x) − σ(y) = x − y. In conclusion, we have proved (4.7). In turn, if x, y ∈ R are such that uσ (x) > t, uσ (y) > t, let z, w be as in (4.7). Then x − y ≤ w − z ≤ diam{u > t}. Taking the supremum over all such x, y gives diam{uσ > t} ≤ diam{u > t}. Next we prove some important inequalities. We begin with the Hardy– Littlewood rearrangement inequality. Theorem 4.15 (Hardy–Littlewood inequality). Let u : R → [0, ∞) and v : R → [0, ∞) be Lebesgue measurable functions and let σ be a reflection. Then Z Z (4.8) u(x)v(x) dx ≤ uσ (x)v σ (x) dx. R
R
Moreover, if the righthand side is finite, then equality holds if and only if (4.9)
(u(x) − u(σ(x))(v(x) − v(σ(x)) ≥ 0
for L1 a.e. x ∈ R.
Proof. Step 1: Since σ(I± ) = I∓ and σ = σ −1 , by the change of variables y = σ(x), Z Z (4.10) u(x)v(x) dx = (u(x)v(x) + u(σ(x))v(σ(x))) dx I+
R
=
4 Z X i=1
(u(x)v(x) + u(σ(x))v(σ(x))) dx =:
Ii
4 X
Ai
i=1
and (4.11) Z Z σ σ u (x)v (x) dx = (uσ (x)v σ (x) + uσ (σ(x))v σ (σ(x))) dx I+
R
=
4 Z X i=1
(uσ (x)v σ (x) + uσ (σ(x))v σ (σ(x))) dx =:
Ii
where I1 = {x ∈ I+ : u(x) ≥ u(σ(x)) and v(x) ≥ v(σ(x))}, I2 = {x ∈ I+ : u(σ(x)) > u(x) and v(σ(x)) ≥ v(x)}, I3 = {x ∈ I+ : u(x) ≥ u(σ(x)) and v(σ(x)) > v(x)}, I4 = {x ∈ I+ : u(σ(x)) > u(x) and v(x) > v(σ(x))}.
4 X i=1
Bi ,
4.1. Polarization
117
Let g(x) := uσ (x)v σ (x) + uσ (σ(x))v σ (σ(x)) − u(x)v(x) − u(σ(x))v(σ(x)). If x ∈ I1 , then uσ (x) = u(x), uσ (σ(x)) = u(σ(x)), v σ (x) = v(x), v σ (σ(x)) = v(σ(x)). Hence, g(x) = 0 and inequality (4.9) holds since both terms between parentheses are nonnegative. If x ∈ I2 , then uσ (x) = u(σ(x)), uσ (σ(x)) = u(x), v σ (x) = v(σ(x)), v σ (σ(x)) = v(x). Hence, g(x) = 0, and (4.9) holds since both terms between parentheses are nonpositive. If x ∈ I3 , then uσ (x) = u(x), uσ (σ(x)) = u(σ(x)), v σ (x) = v(σ(x)), v σ (σ(x)) = v(x). Then (4.12)
g(x) = (u(x) − u(σ(x)))(v(σ(x)) − v(x)) ≥ 0.
Similarly, if x ∈ I4 , then uσ (x) = u(σ(x)), uσ (σ(x)) = u(x), v σ (x) = v(x), v σ (σ(x)) = v(σ(x)). Then (4.13)
g(x) = (u(σ(x)) − u(x))(v(x) − v(σ(x))) ≥ 0.
Thus, A1 = B1 , A2 = B2 , A3 ≤ B3 , and A4 ≤ B4 . Hence, in view of (4.10) and (4.11), we obtain (4.8). Step 2: Assume that Z Z u(x)v(x) dx = uσ (x)v σ (x) dx < ∞. R
R
Using Step 1 also, we have B3 = A3 < ∞ and A4 = B4 < ∞. Since the integrand in the integral B3 − A3 = 0 is nonnegative by (4.12), necessarily (u(x) − u(σ(x)))(v(σ(x)) − v(x)) = 0 for L1 a.e. x ∈ I3 . Similarly, from B4 − A4 = 0 and (4.13), (u(x) − u(σ(x)))(v(σ(x)) − v(x)) = 0 for L1 a.e. x ∈ I4 . This proves that (4.9) is an equality for L1 a.e. x ∈ I3 ∪I4 . We have seen in Step 1 that (4.9) holds in I1 ∪ I2 . Hence, (4.9) holds for L1 a.e. x ∈ I+ . Taking x = σ(y) for y ∈ I− in (4.9) and using the fact that σ(σ(y)) = y, shows that (4.9) holds for L1 a.e. y ∈ I− . Exercise 4.16. Given a, b ∈ R and n ∈ N, prove that (max{min{a, n}, −n} − max{min{b, n}, −n})+ ≤ (max{min{a, n + 1}, −n − 1} − max{min{b, n + 1}, −n − 1})+ , where t+ = max{t, 0} is the positive part of t ∈ R. Corollary 4.17. Let g : R → [0, ∞) be a convex function with g(0) = 0, let u : R → R and v : R → R be Lebesgue measurable functions vanishing at infinity, and let σ be a reflection. Then Z Z σ σ (4.14) g(u (x) − v (x)) dx ≤ g(u(x) − v(x)) dx. R
R
118
4. Rearrangements
In particular, for 1 ≤ p < ∞, Z Z σ σ p u (x) − v (x) dx ≤ u(x) − v(x)p dx. R
R
Proof. Define g+ (t) :=
g(t) if t ≥ 0, 0 otherwise,
g− (t) :=
g(t) if t < 0, 0 otherwise.
Then g = g+ + g− . We leave it as an exercise to show that g+ and g− are still convex. Step 1: Assume that u and v are bounded. Since g+ is convex, the right derivative (g+ )0+ of g+ exists in R and is increasing. By the fundamental theorem of calculus and the fact that g+ (0) = 0, for every t ∈ R we have Rt that g+ (t) = 0 (g+ )0+ (τ ) dτ . Hence, for x ∈ R, Z u(x) Z u(x)−v(x) 0 (g+ )0+ (u(x) − r) dr (g+ )+ (τ ) dτ = g+ (u(x) − v(x)) = v(x) 0 Z = (g+ )0+ (u(x) − r)χ{v≤r} (x) dr, R
where, in the second equality, we use the change of variable τ = u(x) − r and, in the third equality, we use the fact that (g+ )0+ (t) = 0 for t < 0. Integrating both sides over R and using Tonelli’s theorem, we get Z Z Z (4.15) g+ (u(x) − v(x)) dx = (g+ )0+ (u(x) − r)χ{v≤r} (x) dxdr. R
R
R
Similarly, Z Z Z σ σ (4.16) g+ (u (x) − v (x)) dx = (g+ )0+ (uσ (x) − r)χ{vσ ≤r} (x) dxdr. R
R
R
By Hardy–Littlewood’s inequality (Theorem 4.15), we have (4.17) Z Z 0 (g+ )+ (u(x) − r)χ{v>r} (x) dx ≤ ((g+ )0+ (u(·) − r))σ (x)χσ{v>r} (x) dx R ZR = (g+ )0+ (uσ (x) − r)χ{vσ >r} (x) dx, R
where in the last equality we use Exercises 4.4 and 4.10 and the fact that (g+ )0+ is increasing. On the other hand, by Theorem 4.6 applied to the Borel function f (t) := (g+ )0+ (t − r), Z Z 0 (4.18) (g+ )+ (u(x) − r) dx = (g+ )0+ (uσ (x) − r) dx. R
R
4.1. Polarization
119
Note that these integrals are finite because Z Z 0 (g+ )+ (u(x) − r) dx = (g+ )0+ (u(x) − r) dx R
≤
{u>r} (g+ )0+ (sup u
− r)L1 ({x ∈ R : u(x) > r}) < ∞
since u is bounded and vanishes at infinity. Hence, if we write χ{v≤r} = 1 − χ{v>r} , then Z (g+ )0+ (u(x) − r)χ{v≤r} (x) dx R Z Z 0 = (g+ )+ (u(x) − r) dx − (g+ )0+ (u(x) − r)χ{v>r} (x) dx R ZR Z ≥ (g+ )0+ (uσ (x) − r) dx − (g+ )0+ (uσ (x) − r)χ{vσ >r} (x) dx R ZR 0 σ = (g+ )+ (u (x) − r)χ{vσ ≤r} (x) dx, R
where we use (4.17) and (4.18). Integrating both sides in r over R and using (4.15) and (4.16), we obtain Z Z σ σ g+ (u (x) − v (x)) dx ≤ g+ (u(x) − v(x)) dx. R
R
Step 2: Define un := max{min{u, n}, −n} and vn := max{min{v, n}, −n}. Then by Exercise 4.10, we have uσn = max{min{uσ , n}, −n} and vnσ = max{min{v σ , n}, −n}. By applying Step 1 to un and vn , we get Z Z σ σ g+ (un (x) − vn (x)) dx ≤ g+ (un (x) − vn (x)) dx. R
R
Since g+ (t) = 0 for t ≤ 0, it follows from Exercise 4.16 that for every x ∈ R, g+ (un (x) − vn (x)) ≤ g+ (un+1 (x) − vn+1 (x)), σ g+ (uσn (x) − vnσ (x)) ≤ g+ (uσn+1 (x) − vn+1 (x)).
Moreover, by (4.1), uσn (x) → uσ (x) and vnσ (x) → v σ (x) as n → ∞ for every x ∈ R. Hence, we can apply the Lebesgue monotone convergence theorem to conclude that Z Z σ σ g+ (u (x) − v (x)) dx ≤ g+ (u(x) − v(x)) dx. R
R
To obtain the same inequality for g− , observe that the function h(t) := g(−t) is convex and h+ = g− . Thus, we can apply what we have done so far with g+ replaced by h+ = g− to obtain Z Z g− (uσ (x) − v σ (x)) dx ≤ g− (u(x) − v(x)) dx. R
R
Summing the last two inequalities gives (4.49).
120
4. Rearrangements
Next we prove a simplified version of Riesz’s rearrangement inequality. Theorem 4.18 (Riesz). Let u : R → [0, ∞) and v : R → [0, ∞) be Lebesgue measurable functions vanishing at infinity, and let k : [0, ∞) → [0, ∞) be decreasing. Then for every reflection σ, Z Z Z Z u(x)v(y)k(x − y) dxdy ≤ uσ (x)v σ (y)k(x − y) dxdy. R
R
R
R
Proof. Since σ(I± ) = I∓ and σ = σ −1 , by the change of variables X = σ(x) and Y = σ(y), and the facts that σ(x) − σ(y) = x − y and σ(x) − y = x − σ(y), Z Z u(x)v(y)k(x − y) dxdy R R Z Z = {[u(x)v(y) + u(σ(x))v(σ(y))]k(x − y) I+
I+
+ [u(x)v(σ(y)) + u(σ(x))v(y)]k(x − σ(y))} dxdy =: A and Z Z R
uσ (x)v σ (y)k(x − y) dxdy R Z Z = {[uσ (x)v σ (y) + uσ (σ(x))v σ (σ(y))]k(x − y) I+
I+
+ [uσ (x)v σ (σ(y)) + uσ (σ(x))v σ (y)]k(x − σ(y))} dxdy =: B. Let I1 = {x ∈ I+ : u(x) ≥ u(σ(x))},
I2 = {x ∈ I+ : u(σ(x)) > u(x)},
J1 = {y ∈ I+ : v(y) ≥ v(σ(y))},
J2 = {y ∈ I+ : v(σ(y)) > v(y)},
and g(x, y) := [uσ (x)v σ (y) + uσ (σ(x))v σ (σ(y)) − u(x)v(y) − u(σ(x))v(σ(y))]k(x − y) + [uσ (x)v σ (σ(y)) + uσ (σ(x))v σ (y) − u(x)v(σ(y)) − u(σ(x))v(y)]k(x − σ(y)). If x ∈ I1 and y ∈ J1 , then uσ (x) = u(x), uσ (σ(x)) = u(σ(x)), v σ (y) = v(y), v σ (σ(y)) = v(σ(y)). Hence, g(x, y) = 0. If x ∈ I2 and y ∈ J2 , then uσ (x) = u(σ(x)), uσ (σ(x)) = u(x), v σ (y) = v(σ(y)), v σ (σ(y)) = v(y). Hence, g(x, y) = 0. If x ∈ I1 and y ∈ J2 , then uσ (x) = u(x), uσ (σ(x)) = u(σ(x)), v σ (y) = v(σ(y)), v σ (σ(y)) = v(y). Then g(x, y) = (u(x) − u(σ(x)))(v(σ(y)) − v(y))[k(x − y) − k(x − σ(y))] ≥ 0
4.1. Polarization
121
since the two terms between parentheses are nonnegative, k is decreasing, and x − y ≤ x − σ(y) for x, y ∈ I+ . Similarly, if x ∈ I2 and y ∈ J1 , then uσ (x) = u(σ(x)), uσ (σ(x)) = u(x), v σ (y) = v(y), v σ (σ(y)) = v(σ(y)). In turn, g(x, y) = (u(σ(x)) − u(x))(v(y) − v(σ(y)))[k(x − y) − k(x − σ(y))] ≥ 0. Thus, we have shown that g(x, y) ≥ 0 for all x, y ∈ I+ . It follows that A ≤ B, completing the proof. The analogue of Corollary 4.17 is the following result. Corollary 4.19. Let g : R → [0, ∞) be a convex function with g(0) = 0, let u : R → R and v : R → R be Lebesgue measurable functions vanishing at infinity, let k : [0, ∞) → [0, ∞) be decreasing and Lebesgue integrable, and let σ be a reflection. Then (4.19) Z Z Z Z g(uσ (x) − v σ (y))k(x − y) dxdy ≤
R
R
g(u(x) − v(y))k(x − y) dxdy. R
R
In particular, for 1 ≤ p < ∞, Z Z Z Z σ σ p u (x) − v (y) k(x − y) dxdy ≤ u(x) − v(y)p k(x − y) dxdy. R
R
R
R
Proof. Define g+ (t) :=
g(t) if t ≥ 0, 0 otherwise,
g− (t) :=
g(t) if t < 0, 0 otherwise.
Then g = g+ + g− . We leave it as an exercise to show that g+ and g− are still convex. Step 1: Assume that u and v are bounded. Since g+ is convex, the right derivative (g+ )0+ of g+ exists in R and is increasing. By the fundamental theorem of calculus and using g+ (0) = 0, for every t ∈ R we have Z t g+ (t) = (g+ )0+ (τ ) dτ. 0
Hence, for x, y ∈ R, Z
u(x)−v(y)
g+ (u(x) − v(y)) = 0
Z =
(g+ )0+ (τ ) dτ
Z
u(x)
=
(g+ )0+ (u(x) − r) dr
v(y)
(g+ )0+ (u(x) − r)χ{v≤r} (y) dr,
R
where, in the second equality, we use the change of variable τ = u(x) − r and, in the third equality, we use the fact that (g+ )0+ (t) = 0 for t < 0.
122
4. Rearrangements
Multiplying both sides by k(x − y), integrating in x and in y over R, and using Tonelli’s theorem, we get Z Z (4.20) g+ (u(x) − v(y))k(x − y) dxdy R R Z Z Z (g+ )0+ (u(x) − r)k(x − y)χ{v≤r} (y) dxdydr. = R
R
R
Similarly, Z Z
g+ (uσ (x) − v σ (y))k(x − y) dxdy Z Z Z = (g+ )0+ (uσ (x) − r)k(x − y)χ{vσ ≤r} (y) dxdydr.
(4.21)
R
R
R
R
R
By the simple form of Riesz’s inequality (Theorem 4.18), we have Z Z
(g+ )0+ (u(x) − r)k(x − y)χ{v>r} (y) dxdy Z Z ((g+ )0+ (u(·) − r))σ (x)k(x − y)χσ{v>r} (y) dxdy ≤ ZR ZR = (g+ )0+ (uσ (x) − r)k(x − y)χ{vσ >r} (y) dxdy,
(4.22)
R
R
R
R
where in the last equality we use Exercises 4.4 and 4.10 and the fact that (g+ )0+ is increasing. On the other hand, by Tonelli’s theorem, the change of variables z = x − y, so that dz = dy, and Theorem 4.6 applied to the Borel function f (t) := (g+ )0+ (t − r), Z Z R
(4.23)
R
Z
(g+ )0+ (u(x)
Z
− r)k(x − y) dxdy = k(z)dz (g+ )0+ (u(x) − r) dx R R Z Z = k(z)dz (g+ )0+ (uσ (x) − r) dx R R Z Z = (g+ )0+ (uσ (x) − r)k(x − y) dxdy. R
R
Note that these integrals are finite because k is integrable and Z Z 0 (g+ )+ (u(x) − r) dx = (g+ )0+ (u(x) − r) dx R
≤
{u>r} (g+ )0+ (sup u
− r)L1 ({x ∈ R : u(x) > r}) < ∞
4.1. Polarization
123
since u is bounded and vanishes at infinity. Hence, if we write χ{v≤r} = 1 − χ{v>r} , then Z Z (g+ )0+ (u(x) − r)k(x − y)χ{v≤r} (y) dxdy R R Z Z (g+ )0+ (u(x) − r)k(x − y) dxdy = R R Z Z − (g+ )0+ (u(x) − r)k(x − y)χ{v>r} (y) dxdy Z ZR R (g+ )0+ (uσ (x) − r)k(x − y) dxdy ≥ R R Z Z (g+ )0+ (uσ (x) − r)k(x − y)χ{vσ >r} (y) dxdy − R R Z Z = (g+ )0+ (uσ (x) − r)k(x − y)χ{vσ ≤r} (y) dxdy, R
R
where we use (4.22) and (4.23). Integrating both sides in r over R and using (4.20) and (4.21), we obtain Z Z Z Z σ σ g+ (u (x)−v (y))k(x−y) dxdy ≤ g+ (u(x)−v(y))k(x−y) dxdy. R
R
R
R
Step 2: Define un := max{min{u, n}, −n} and vn := max{min{v, n}, −n}. Then by Exercise 4.10, we have uσn = max{min{uσ , n}, −n} and vnσ = max{min{v σ , n}, −n}. By applying Step 1 to un and vn , we get Z Z Z Z g+ (uσn (x)−vnσ (y))k(x−y) dxdy ≤ g+ (u(x)−v(y))k(x−y) dxdy. R
R
R
R
Since g+ (t) = 0 for t ≤ 0, it follows from Exercise 4.16 that for every x ∈ R, g+ (un (x) − vn (y)) ≤ g+ (un+1 (x) − vn+1 (y)), σ g+ (uσn (x) − vnσ (y)) ≤ g+ (uσn+1 (x) − vn+1 (y)).
Moreover, by (4.1), uσn (x) → uσ (x) and vnσ (y) → v σ (y) as n → ∞ for every x, y ∈ R. Hence, we can apply the Lebesgue monotone convergence theorem to conclude that Z Z Z Z σ σ g+ (u (x)−v (y))k(x−y) dxdy ≤ g+ (u(x)−v(y))k(x−y) dxdy. R
R
R
R
To obtain the same inequality for g− , observe that the function h(t) := g(−t) is convex and h+ = g− . Thus, we can apply what we have done so far with g+ replaced by h+ = g− to obtain Z Z Z Z σ σ g− (u (x)−v (y))k(x−y) dxdy ≤ g− (u(x)−v(y))k(x−y) dxdy. R
R
R
R
Summing the last two inequalities gives (4.19).
124
4. Rearrangements
4.2. Polarization in W s,p (R) In this section, we prove that if u ∈ W s,p (R), 1 ≤ p < ∞, 0 < s < 1, then so does uσ , and that its fractional Sobolev seminorm doesn’t increase. Theorem 4.20. Let 1 ≤ p < ∞, 0 < s < 1, u : R → R be a Lebesgue measurable function vanishing at infinity, and let σ be a reflection. Then Z Z Z Z uσ (x) − uσ (y)p u(x) − u(y)p (4.24) dxdy ≤ dxdy. 1+sp x − y1+sp R R R R x − y Moreover, if un → u in W s,p (R), then uσn → uσ in W s,p (R). Proof. Step 1: For x > 0, by the change of variables y = x2 t, so that dy = x2 dt, we have Z ∞ Γ((1 + sp)/2) , t(1+sp)/2−1 exp(−x2 t) dt = x1+sp 0 R∞ where Γ is the Gamma function Γ(y) := 0 τ y−1 e−τ dτ , y > 0. Hence, by Tonelli’s theorem Z Z u(x) − u(y)p dxdy 1+sp R R x − y Z Z Z ∞ p (4.25) =c u(x) − u(y) t(1+sp)/2−1 exp(−x − y2 t) dtdxdy R R 0 Z ∞ Z Z (1+sp)/2−1 =c t u(x) − u(y)p exp(−x − y2 t) dxdydt, 0
R
R
where c := 1/Γ((1 + sp)/2). For each fixed t > 0, the function kt (x) = exp(−x2 t), x ≥ 0, is decreasing in [0, ∞) and Lebesgue integrable. It follows by Corollary 4.19 that Z Z (4.26) uσ (x) − uσ (y)p exp(−x − y2 t) dxdy R R Z Z ≤ u(x) − u(y)p exp(−x − y2 t) dxdy. R
R
By multiplying both sides by ct(1+sp)/2−1 , integrating in t over R+ , and using (4.25), we obtain (4.24). Step 2: Let un , u ∈ W s,p (R) be such that un → u in W s,p (R). For t > 0 define Z Z Ft (u) := u(x) − u(y)p exp(−x − y2 t) dxdy. R
R
We claim that (4.27)
lim Ft (uσn − uσ ) = 0.
n→∞
4.2. Polarization in W s,p (R)
125
By inequality (1.1), Tonelli’s theorem, and Corollary 4.19, Z Z σ σ p σ σ exp(−x − y2 t) dydx un (x) − u (x) Ft (un − u ) R R Z Z σ σ p (4.28) exp(−x − y2 t) dxdy un (y) − u (y) + R Z Z R σ σ p un (x) − u(x)p dx, un (x) − u (x) dx t t R
R
where we use the fact that, by a change of variables, Z Z 2 exp(−z 2 t) dz. exp(−x − y t) dy = R
R
The righthand side of inequality (4.28) converges to zero as n → ∞ since un → u in Lp (R). This proves (4.27). By inequality (1.1) and (4.26), (4.29) Ft (uσn − uσ ) ≤ 2p−1 Ft (uσn ) + 2p−1 Ft (uσ ) ≤ 2p−1 Ft (un ) + 2p−1 Ft (u) ≤ 22p−2 Ft (un − u) + 22p−1 Ft (u). Using the inequality a ≤ b + (a − b)+ , we have Ft (uσn − uσ ) ≤ 22p−2 Ft (un − u) + (Ft (uσn − uσ ) − 22p−2 Ft (un − u))+ . Multiplying both sides by ct(1+sp)/2−1 , integrating in t over R+ , and using (4.25), we have Z Z (uσn − uσ )(x) − (uσn − uσ )(y)p dxdy x − y1+sp R R Z Z (un − u)(x) − (un − u)(y)p dxdy x − y1+sp R R Z ∞ + t(1+sp)/2−1 (Ft (uσn − uσ ) − 22p−2 Ft (un − u))+ dt =: A + B. 0
By (4.29), (Ft (uσn − uσ ) − 22p−2 Ft (un − u))+ Ft (u) and since Z ∞ Z Z u(x) − u(y)p (1+sp)/2−1 c dxdy < ∞, t Ft (u) dt = 1+sp 0 R R x − y in view of (4.27), we can apply the Lebesgue dominated convergence theorem to conclude that B → 0. Since un → u in W s,p (R), we also have that A → 0. Hence, we have shown that uσn → uσ in W s,p (R). Next we study the case s = 1. ˙ 1,p (R), let u Theorem 4.21. Let 1 ≤ p < ∞, u ∈ W ¯ be the continuous ˙ 1,p (R), with representative of u, and let σ be a reflection. Then u ¯σ ∈ W Z Z (4.30) (¯ uσ )0 (x)p dx = u0 (x)p dx. R
R
126
4. Rearrangements
Moreover, if u ∈ W 1,p (R), then so does u ¯σ . Proof. Define v¯(x) := u ¯(σ(x)) = u ¯(−x + 2a), x ∈ R. Since u ¯ is locally absolutely continuous, so is v, with v¯0 (x) = −¯ u0 (−x + 2a) for L1 a.e. x ∈ R. ˙ 1,p (R), Consider the Lipschitz function f (t) := max{t, 0}, t ∈ R. If w ∈ W ˙ 1.p (R) and for L1 a.e. then by the chain rule (see [Leo22d]), f ◦ w ∈ W x ∈ R, 0 w (x) if w(x) ≥ 0, 0 (f ◦ w) (x) = 0 if w(x) < 0. Writing max{¯ u, v¯} = max{¯ u − v¯, 0} + v¯ = f ◦ (¯ u − v¯) + v¯, min{¯ u, v¯} = − max{¯ u − v¯, 0} + u ¯ = −f ◦ (¯ u − v¯) + u ¯, ˙ 1,p (R) and that for L1 a.e. x ∈ R, we deduce that max{¯ u, v¯}, min{¯ u, v¯} ∈ W 0 u ¯ (x) if u ¯(x) ≥ v¯(x), 0 (max{¯ u, v¯}) (x) = −¯ u0 (−x + 2a) if u ¯(x) < v¯(x), 0 u ¯ (x) if u ¯(x) ≤ v¯(x), (min{¯ u, v¯})0 (x) = −¯ u0 (−x + 2a) if u ¯(x) > v¯(x). Since u ¯(a) = v¯(a), the function u ¯σ 0 u ¯ (x) −¯ u0 (σ(x)) (¯ uσ )0 (x) = 0 u ¯ (x) −¯ u0 (σ(x))
is still locally absolutely continuous with if if if if
x ∈ I+ x ∈ I+ x ∈ I− x ∈ I−
u ¯(x) ≥ u ¯(σ(x)), u ¯(x) < u ¯(σ(x)), u ¯(x) ≤ u ¯(σ(x)), u ¯(x) > u ¯(σ(x)).
and and and and
Hence, Z (4.31)
(¯ uσ )0 p dx =
Z
¯ u0 (x)p dx +
I1 ∪I3
R
Z
¯ u0 (σ(x))p dx,
I2 ∪I4
where I1 = {x ∈ I+ : u ¯(x) ≥ u ¯(σ(x))},
I2 = {x ∈ I+ : u ¯(σ(x)) > u ¯(x)},
I3 = {x ∈ I− : u ¯(x) ≤ u ¯(σ(x))},
I4 = {x ∈ I− : u ¯(x) > u ¯(σ(x))}.
By considering the change of variables y = σ(x) and using the fact that σ(I2 ) = I4 , we have that Z Z ¯ u0 (σ(x))p dx = ¯ u0 (y)p dy, I2 ∪I4
I2 ∪I4
which, together with (4.31), shows (4.30). The second part of the statement follows from Theorem 4.5.
4.3. Symmetric Decreasing Rearrangement
127
4.3. Symmetric Decreasing Rearrangement Given a Lebesgue measurable set E ⊆ R, we define E \ to be the interval with the same measure and centered at the origin. To be precise, (4.32) E \ := −L1 (E)/2, L1 (E)/2 . If we now consider the functions χE and χE \ , we see that χE \ is even, that is, χE \ (−x) = χE \ (x) for all x ∈ R, χE \ is decreasing in [0, ∞) since χE \ (x) = 1 if 0 ≤ x < L1 (E)/2, and L1 ({x ∈ R : χE (x) > t)} = L1 ({x ∈ R : χE \ (x) > t)} for every t ≥ 0. We will see that we can extend this procedure to the case when χE is replaced by a nonnegative measurable function u, that is, we can construct a function u\ , which is even, decreasing in [0, ∞), and L1 ({u(x) > t)} = L1 ({u\ (x) > t)} for every t ≥ 0. Definition 4.22. Let E ⊆ R be a Lebesgue measurable set and let u : E → [0, ∞] be a Lebesgue measurable function. The symmetric decreasing rearrangement of u is the function u\ : R → [0, ∞], defined by Z ∞ \ (4.33) u (x) := χE \ (x) dt, x ∈ R, 0
t
where Et := {x ∈ E : u(x) > t} and (4.34)
Et\ := (Et )\ = (−L1 (Et )/2, L1 (Et )/2).
Remark 4.23. Let F ⊆ E ⊆ R be Lebesgue measurable sets. Consider the function χF : E → {0, 1}. Then Et = {x ∈ E : χF (x) > t} = F if 0 ≤ t < 1 and Et = ∅ if t ≥ 1. In turn, Z ∞ Z 1 (χF )\ (x) = χE \ (x) dt = χF \ (x) dt = χF \ (x). 0
t
0
Theorem 4.24. Let E ⊆ R be a Lebesgue measurable set, let u : E → [0, ∞] be a Lebesgue measurable function, and let u\ : R → [0, ∞] be its symmetric decreasing rearrangement. Then (i) u\ (x) = u\ (−x) for every x ∈ R, (ii) if 0 ≤ x1 < x2 , then u\ (x1 ) ≥ u\ (x2 ), (iii) {x ∈ E : u(x) > t}\ = {x ∈ R : u\ (x) > t} for every t ≥ 0 and (4.35)
L1 ({x ∈ E : u(x) > t}) = L1 ({x ∈ R : u\ (x) > t)},
(iv) u\ is lower semicontinuous.
128
4. Rearrangements
Proof. Since Et\ is symmetric with respect to the origin, we have that x ∈ Et\ if and only if −x ∈ Et\ . Thus, χE \ (−x) = χE \ (x) for every x ∈ R. In turn, t t by (4.33), Z ∞ Z ∞ χE \ (x) dt = u\ (x), χE \ (−x) dt = u\ (−x) = t
0
t
0
which shows that u\ is even, so item (i) holds. To prove item (ii), observe that if x2 ∈ Et\ = (−L1 (Et )/2, L1 (Et )/2), then, since 0 ≤ x1 < x2 , we have that 0 ≤ x1 < x2 < L1 (Et )/2, and so, x1 ∈ Et\ . Thus, if χE \ (x2 ) = 1, then χE \ (x1 ) = 1, which implies that t
t
χE \ (x2 ) ≤ χE \ (x1 ). Integrating in t and using (4.33) gives u\ (x2 ) ≤ u\ (x1 ). t
t
To prove item (iii), consider 0 ≤ t1 < t2 . Then (4.36)
Et2 = {x ∈ E : u(x) > t2 } ⊆ {x ∈ E : u(x) > t1 } = Et1 .
This proves that the function g(t) := L1 (Et ), t ≥ 0, is decreasing. We claim that g is rightcontinuous. Fix t0 ≥ 0. Since g is decreasing, there exists lim g(t) ≥ g(t0 ).
t→t+ 0
To prove equality, consider a decreasing sequence tn → t+ 0 . Then Etn ⊆ S∞ Etn+1 and n=1 Etn = Et0 , and so we have that lim g(tn ) = lim L1 (Etn ) = L1
n→∞
n→∞
∞ [
Etn = g(t0 ).
n=1
This proves the claim. Let t ≥ 0 and x ∈ Et\ = (−L1 (Et )/2, L1 (Et )/2). Assume that x > 0. Then 2x < L1 (Et ) = g(t). Since g is rightcontinuous, there exists δ > 0 such that g(r) > 2x for all r ∈ [t, t + δ). Similarly, if x < 0, then −L1 (Et )/2 < x, and so g(t) = L1 (Et ) > −2x. Again by the the rightcontinuity of g, by taking δ smaller, we have that g(r) > −2x for all r ∈ [t, t + δ). Thus, we have shown that x ∈ (−L1 (Er )/2, L1 (Er )/2) = Er\ for all r ∈ [t, t + δ). On the other hand, by (4.36), we have that x ∈ Er\ for all 0 ≤ r ≤ t. Hence, Z ∞ Z t+δ Z ∞ \ u (x) = χE \ (x) dr = 1 dr + χE \ (x) dr ≥ t + δ > t. 0
r
0
t+δ
r
This shows that (4.37)
Et\ ⊆ {x ∈ R : u\ (x) > t}.
To prove the other inclusion, assume that x ∈ / Et\ = (−L1 (Et )/2, L1 (Et )/2). 1 1 Assume x ≥ L (Et )/2 (the case x ≤ −L (Et )/2 is similar). Since L1 (Er ) ≤
4.3. Symmetric Decreasing Rearrangement
129
L1 (Et ) for all r ≥ t by (4.36), we have that x ∈ / Er\ = (−L1 (Er )/2, L1 (Er )/2), and so, Z ∞ Z t \ χE \ (x) dr = χE \ (x) dr ≤ t, u (x) = 0
r
0
r
which implies that x ∈ / {x ∈ R : u\ (x) > t}. Together with (4.37), this \ proves that Et = {x ∈ R : u\ (x) > t}. Since L1 (Et ) = L1 (Et\ ) by (4.34), it follows that L1 ({x ∈ E : u(x) > t}) = L1 (Et\ ) = L1 ({x ∈ R : u\ (x) > t}), and so item (iii) holds. Finally, to prove item (iv), it suffices to show that for every t ∈ R, the set Ft := {x ∈ R : u\ (x) > t} is open. If t < 0, then Ft = R since u\ ≥ 0, while if t ≥ 0, then Ft = Et\ by item (iii), so that Ft is an open interval. Exercise 4.25. Let E ⊆ R be a Lebesgue measurable set and let u, v : E → [0, ∞] be two Lebesgue measurable functions with u ≤ v. Prove that u\ ≤ v \ . Exercise 4.26. Given E ⊆ R a Lebesgue measurable set and a Lebesgue measurable function u : E → [0, ∞], let M be the family of Borel sets B ⊆ R+ such that L1 ({x ∈ E : u(x) ∈ B}) = L1 ({t ∈ E \ : u\ (t) ∈ B}). (i) Prove that if Bn ∈ M, n ∈ N, are pairwise disjoint, then M. S (ii) Prove that if Bn ∈ M, n ∈ N, then ∞ n=1 Bn ∈ M.
S∞
n=1 Bn
∈
(iii) Prove that if B ∈ M, then R+ \ B ∈ M. Hint: Assume first that E has finite measure. (iv) Prove that every open set U ⊆ R+ belongs to M. (v) Prove that every Borel set B ⊆ R+ belongs to M. Exercise 4.27. Let E ⊆ R be a Lebesgue measurable set R and let u : E → [0, ∞] be a Lebesgue measurable function. Prove that E (u(x))p dx = R \ p R (u (x)) dx for every p > 0. Exercise 4.28. Let f : R → [0, ∞) be a Borel function such that f (0) = 0, let E ⊆ R be a Lebesgue measurable set, and let u : E → [0, ∞) be a Lebesgue measurable function vanishing at infinity. Prove that Z Z (4.38) f (u(x)) dx = f (u\ (x)) dx. E
R
Exercise 4.29. Let f : R → R be a Borel function such that f (0) = 0, let E ⊆ R be a Lebesgue measurable set, and let u : E → [0, ∞) be a Lebesgue
130
4. Rearrangements
measurable function vanishingRat infinity and Rsuch that f ◦ u is Lebesgue integrable over E. Prove that E f (u(x)) dx = R f (u\ (x)) dx. Next we present some convergence results for the symmetric decreasing rearrangement. Theorem 4.30. Let E ⊆ R be a Lebesgue measurable set and let un : E → [0, ∞) Lebesgue measurable functions vanishing at infinity such that un ≤ un+1 for all n. Then u\n (x) → u\ (x) for all x ∈ E \ , where u = supn un . Proof. Let gu : [0, ∞) → [0, ∞] and gun : [0, ∞) → [0, ∞] be the decreasing functions defined by gu (t) := L1 ({x ∈ E : u(x) > t}),
gun (t) := L1 ({x ∈ E : un (x) > t}).
By (4.34) and Theorem 4.24(iii), (4.39)
(−gu (t)/2, gu (t)/2) = {x ∈ R : u\ (x) > t}, (−gun (t)/2, gun (t)/2) = {x ∈ R : u\n (x) > t}.
Since un ≤ un+1 ≤ u, by Exercise 4.25, u\n ≤ u\n+1 ≤ u\ , and so v(x) := sup u\n (x) ≤ u\ (x) n
for every x ∈ R.
To prove the opposite inequality, let 0 < x0 < L1 (E)/2 and δ > 0, and assume that u\ (x0 ) > v(x0 ) + δ. Let u\ (x0 ) − δ < t < u\ (x0 ). Then u\n (x0 ) < t < u\ (x0 ) for every n. By (4.39), gun (t)/2 ≤ x0 < gu (t)/2, which is a contradiction, since ∞ [ {x ∈ E : u(x) > t} = {x ∈ E : un (x) > t}, n=1
and {x ∈ E : un (x) > t} ⊆ {x ∈ E : un+1 (x) > t}, which implies that gun (t) → gu (t). Exercise 4.31. Given a Lebesgue measurable set E ⊆ R and Lebesgue measurable functions u : E → [0, ∞) and un : E → [0, ∞), let gu : [0, ∞) → [0, ∞] and gun : [0, ∞) → [0, ∞] be the decreasing functions defined by gu (t) := L1 ({x ∈ E : u(x) > t}),
gun (t) := L1 ({x ∈ E : un (x) > t}).
Prove that if un → u in measure as n → ∞, then for every t > 0, gu (t) ≤ lim inf gun (t) ≤ lim sup gun (t) ≤ lim gu (s). n→∞
n→∞
s→t−
Exercise 4.32. Let E ⊆ R be a Lebesgue measurable set and let un : E → [0, ∞) be Lebesgue measurable functions vanishing at infinity such that un → u in measure as n → ∞, for some function u : E → [0, ∞]. Prove that u vanishes at infinity.
4.3. Symmetric Decreasing Rearrangement
131
Theorem 4.33. Let E ⊆ R be a Lebesgue measurable set and let un : E → [0, ∞) be Lebesgue measurable functions vanishing at infinity such that un → u in measure as n → ∞, for some function u : E → [0, ∞). Then for every x ∈ (0, L1 (E)/2), (4.40)
u\ (x) ≤ lim inf u\n (x) ≤ lim sup u\n (x) ≤ lim u\ (y). n→∞
y→x−
n→∞
In particular, u\n (x) → u\ (x) for all but countably many x ∈ E \ . Proof. Define gu and gun as in Exercise 4.31. By (4.34) and Theorem 4.24(iii), (−gu (t)/2, gu (t)/2) = {x ∈ R : u\ (x) > t},
(4.41)
(−gun (t)/2, gun (t)/2) = {x ∈ R : u\n (x) > t}. Let 0 < x < L1 (E)/2 and δ > 0, and assume that u\ (x) > u\n (x) + δ for infinitely many n. Let u\ (x) − δ < t < u\ (x). Then u\n (x) < t < u\ (x) for infinitely many n. It follows from (4.41) that gu (t)/2 > x and gun (t)/2 ≤ x. In turn, lim inf n→∞ gun (t) ≤ 2x < gu (t), which contradicts Exercise 4.31. This shows that u\ (x) − δ ≤ u\n (x) for all but finitely many n. Hence, u\ (x) − δ ≤ lim inf u\n (x). n→∞
Letting δ →
0+
proves the first inequality in (4.40).
If limy→x− u\ (y) < lim supn→∞ u\n (x), then there exists δ > 0 so small that lim u\ (y) < lim sup u\n (x) − δ. y→x−
n→∞
Using the definition of limit, we can find 0 < y < x such that u\ (y) < lim sup u\n (x) − δ. n→∞
In turn, there exist infinitely many n such that u\ (y) < u\n (x) − δ. Let u\ (y) < s < t < u\ (y) + δ. Then t < u\n (x) and u\ (y) < s, and so, by (4.41), gun (t)/2 > x and gu (s)/2 ≤ y. Letting s → t− gives gun (t) > 2x > 2y ≥ lim gu (s). s→t−
Taking the limit as n → ∞, we obtain lim sup gun (t) > lim gu (s), n→∞
s→t−
which contradicts Exercise 4.31. This proves the last inequality in (4.40). The last part of the statement follows from the fact that u\ is even and decreasing in R+ , and so, it is continuous at all but countably many points.
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4. Rearrangements
Exercise 4.34. Let E ⊆ R be a Lebesgue measurable set and let un : E → [0, ∞) be Lebesgue measurable functions vanishing at infinity such that un → u in measure as n → ∞ for some function u : E → [0, ∞). (i) Prove that u\n → u\ in measure in [0, L1 (E)/2) as n → ∞. Hint: Prove that for 0 < x1 < x2 < L1 (E)/2, u\n → u\ in L1 ([x1 , x2 ]). (ii) Prove that if L1 (E) = ∞, then u\n (∞) → u\ (∞), where v \ (∞) := limx→∞ v \ (x). The following theorem shows the relationship between polarization and symmetric decreasing rearrangement. Theorem 4.35. Let u : R → [0, ∞) be a lower semicontinuous function vanishing at infinity. Then u = u\ if and only if u = uσ for every reflection σ with a 6= 0. Proof. Step 1: Assume that u = u\ . Then u is even and decreasing in R+ . Let a ∈ R \ {0} and assume first that a > 0. Then I+ = (−∞, a) and I− = (a, ∞). Let x ∈ I+ . If 0 ≤ x < a, then σ(x) = −x + 2a ≥ a > x, and so, since u is decreasing in R+ , u(x) ≥ u(σ(x)). If x < 0, then σ(x) = −x + 2a ≥ −x > 0, which implies that u(x) = u(−x) ≥ u(σ(x)). This shows that u = uσ in I+ . Next, let x ∈ I− . If a ≤ x ≤ 2a, then x ≥ −x + 2a = σ(x) ≥ 0, and thus u(σ(x)) ≥ u(x). If 2a < x, then 0 ≤ x − 2a ≤ x, which gives u(σ(x)) = u(−σ(x)) ≥ u(x). This proves that u = uσ in I− . In the case a < 0, we have I+ = (a, ∞) and I− = (−∞, a). Let x ∈ I+ . If a < x ≤ 0, then −x + 2a < x ≤ 0, so that u(x) ≥ u(σ(x)). If x ≥ 0, then x − 2a ≥ x > 0, and so, u(x) ≥ u(−σ(x)) = u(σ(x)). Hence, u = uσ in I+ . Next, let x ∈ I− . If 2a ≤ x < a, then x < −x + 2a ≤ 0, and hence, u(σ(x)) ≥ u(x). If x < 2a, then 0 < −x + 2a < −x, and so, u(σ(x)) ≥ u(−x) = u(x). It follows that u = uσ in I− . Thus, we have shown that u = uσ for every reflection σ. Step 2: Assume that u = uσ for every reflection σ with a 6= 0. If u is not even, then there exists x0 ∈ R such that u(x0 ) > u(−x0 ). Assume first that x0 > 0. Since u is lower semicontinuous, lim inf u(x) ≥ u(x0 ) > u(−x0 ). x→x0
Hence, we can find x1 > x0 such that u(x1 ) > u(−x0 ). Let σ be the 0 reflection such that σ(x1 ) = −x0 . To be precise, we take a = x1 −x > 0. 2 Then −x0 < 0 < a < x1 . Hence, −x0 ∈ I+ = (−∞, a) and x1 ∈ I− = (a, ∞). On the other hand, uσ (−x0 ) = max{u(−x0 ), u(x1 )} = u(x1 ) and
4.3. Symmetric Decreasing Rearrangement
133
uσ (x1 ) = min{u(−x0 ), u(x1 )} = u(−x0 ), which contradicts the fact that u = uσ by hypothesis. On the other hand, if x0 < 0, we take x1 < x0 such that u(x1 ) > u(−x0 ). 0 Then a = x1 −x < 0 and x1 < a < 0 < −x0 . Hence, −x0 ∈ I+ = (a, ∞) 2 and x1 ∈ I− = (−∞, a). But, uσ (−x0 ) = max{u(−x0 ), u(x1 )} = u(x1 ) and uσ (x1 ) = min{u(−x0 ), u(x1 )} = u(−x0 ), which contradicts again the fact that u = uσ by hypothesis. Thus, we have shown that u is even. Next we prove that u is decreasing in [0, ∞). Assume by contradiction that there exist 0 ≤ x1 < x2 such that u(x1 ) < u(x2 ). Let σ be the reflection 2 such that σ(x1 ) = x2 . To be precise, we take a = x1 +x > 0. Then 2 x2 > a > x1 . Hence, x1 ∈ I+ = (−∞, a) and x2 ∈ I− = (a, ∞). However, uσ (x1 ) = max{u(x1 ), u(x2 )} = u(x2 ) and uσ (x2 ) = min{u(x2 ), u(x1 )} = u(x1 ), which contradicts the fact that u = uσ by hypothesis. Therefore u is symmetric decreasing. Finally, observe that since u is lower semicontinuous, the set {x ∈ R : u(x) > t} is open for every t ≥ 0. But since u is symmetric decreasing, this set must be an open interval centered at 0. Hence, Et = Et\ . It follows from (4.32) and (4.33) that u = u\ . Exercise 4.36. Let u : R → [0, ∞) be a function vanishing at infinity and let f : [0, ∞) → [0, ∞) be an increasing function. Prove that (f ◦ u)σ = (f ◦ uσ ) for every reflection σ. Deduce that (f ◦ u)\ = f ◦ u\ . Using the Hardy–Littlewood inequality for polarizations, we can prove the following approximation result. In what follows, given n reflections σ1 , . . . , σn and u : R → R, we define inductively uσ1 σ2 := (uσ1 )σ2 ,
uσ1 ···σk := (uσ1 ···σk−1 )σk
for 3 ≤ k ≤ n.
Theorem 4.37. Given u ∈ Cc (R; [0, ∞)), let Pu := {uσ1 ···σn : σ1 , . . . , σn reflections, n ∈ N}. Then there exists un ∈ Pu such that un → u\ uniformly in R as n → ∞, and supp un ⊆ B(0, R) for all n and for some R > 0. Proof. Let f : [0, ∞) → [0, ∞) be a continuous, strictly decreasing, bounded, integrable function such that f (t) → 0 as t → ∞ and define g(x) := f (x), x ∈ R. Then g \ = g, and so, by Theorem 4.35, g = g σ for every reflection σ. Consider the functional Z w(x)g(x) dx, w ∈ Pu . F (w) := R
Note that for every reflection σ and w ∈ Pu , (4.42)
F (w) ≤ F (wσ ).
134
4. Rearrangements
Indeed, by the Hardy–Littlewood inequality for polarization (Theorem 4.15) Z Z F (w) = w(x)g(x) dx ≤ wσ (x)g(x) dx = F (wσ ), R
where we use the fact that
R
gσ
= g.
Step 1: We claim that there exists v ∈ Pu such that max F (w) = F (v).
w∈Pu
It follows by (4.1) that w ≤ kuk∞ for every w ∈ Pu , and so, Z F (w) ≤ kuk∞ g(x) dx < ∞. R
Thus, ` := supw∈Pu F (w) < ∞. Let wn ∈ Pu be such that ` − n1 ≤ F (wn ) ≤ `. Let supp u ⊆ B(0, r). Then by Theorem 4.14 with t = 0, supp wn ⊆ B(xn , r). If there exists yn  ≤ r such that wn (yn ) > 0, define vn := wn and observe that supp vn ⊆ B(0, 2r). Otherwise, if wn = 0 in B(0, r), let yn be such that wn (yn ) > 0 and consider a reflection σn which maps yn into 0. Then vn := wnσn ∈ Pu , and so F (vn ) ≤ `. Moreover, vn (0) = max{wn (0), wn (σn (0))} = wn (yn ) > 0. Since supp vn is contained in a closed segment of length 2r and supp wn ⊆ B(xn , r), it follows that supp vn ⊆ B(0, 2r). By (4.42), ` − n1 ≤ F (wn ) ≤ F (vn ) ≤ `. Hence, F (vn ) → ` as n → ∞. Since the modulus of continuity of vn is less than the modulus of continuity of u by Theorem 4.8, we have that the sequence {vn }n is equicontinuous with supp vn ⊆ B(0, 2r) for every n. Thus, by the Ascoli–Arzel´a theorem (see [Leo22b]), there exist a subsequence {vnk }k and v ∈ Cc (R) such that vnk → v uniformly. In turn, Z Z v(x)g(x) dx = lim vnk (x)g(x) dx = `. k→∞ R
R
Step 2: We claim that v = u\ . If σ is a reflection with a 6= 0, then by (4.1), vnσk → v σ pointwise in R. By Theorems 4.8 and 4.14, and the Ascoli– Arzel´a theorem (see [Leo22b]), necessarily, vnσk → v σ uniformly in R. By the Hardy–Littlewood inequality for polarization Z Z 1 ≤ vnk (x)g(x) dx ≤ vnσk (x)g(x) dx = F (vnσk ) ≤ `, `− nk R R where we use the fact that g σ = g. Letting k → ∞ and using the fact that vnσk → v σ uniformly in R gives Z Z σ v (x)g(x) dx = ` = v(x)g(x) dx. R
R
4.3. Symmetric Decreasing Rearrangement
135
In turn, by (4.9), (v(x) − v(σ(x)))(f (x) − f (σ(x)) ≥ 0
for L1 a.e. x ∈ R.
Since f is strictly decreasing, this implies that v = v σ . To see this, assume that a > 0. Then I+ = (−∞, a) and I− = (a, ∞). If x ∈ I+ , then σ(x) > x, and so, f (x) − f (σ(x)) > 0, which implies that v(x) − v(σ(x)) ≥ 0. Hence, v σ (x) = max{v(x), v(σ(x))} = v(x). On the other hand, if x ∈ I− , σ(x) < x, and so, f (x) − f (σ(x)) < 0, which implies that v(x) − v(σ(x)) ≤ 0. Hence, v σ (x) = min{v(x), v(σ(x))} = v(x). The case a < 0 is similar. Thus, we have shown that v = v σ for every reflection with a 6= 0. In turn, by Theorem 4.35, v = v \ . To conclude, observe that by Theorem 4.1, for every k and t ≥ 0, L1 ({x ∈ R : u(x) > t}) = L1 ({x ∈ R : vnk (x) > t}). On the other hand, since vnk → v uniformly, by Exercise 4.31, at all but countably many t, L1 ({x ∈ R : v(x) > t}) = lim L1 ({x ∈ R : vnk (x) > t}) k→∞ 1
= L ({x ∈ R : u(x) > t}). It follows from (4.32) and (4.33) that v = u\ .
Exercise 4.38. Given u ∈ Cc (R; [0, ∞)), prove that the modulus of continuity of u\ is less than or equal to the modulus of continuity of u, that is, ωu\ ≤ ωu . Exercise 4.39. Given u ∈ Cc (R; [0, ∞)), with supp u ⊆ B(0, R), prove that supp u\ ⊆ B(0, R). Exercise 4.40. Given u ∈ Cc (R; [0, ∞)), with supp u ⊆ B(0, R), consider the family F := {v ∈ Cc (R; [0, ∞)) : diam supp v ≤ R, ωv ≤ ωu , gv = gu in R+ }, where gv (t) := L1 ({x ∈ R : v(x) > t}). Let m := inf v∈F kv − u\ k2L2 (R) . (i) Prove that there exists v ∈ F such that kv − u\ k2L2 (R) = m. Hint: Follow the proof of Theorem 4.37. (ii) Prove that v = u\ . An important consequence of Theorem 4.37 is the fact that the symmetric decreasing rearrangement preserves absolute continuity. Corollary 4.41. Let u : R → [0, ∞) be an absolutely continuous function vanishing at infinity. Then u\ is absolutely continuous.
136
4. Rearrangements
Proof. Step 1: Assume that u has compact support. By Theorem 4.37 there exists un ∈ Pu such that un → u\ uniformly in R as n → ∞, and supp un ⊆ B(0, R) for all n and for some R > 0. In view of Theorem 4.13, each function un is absolutely continuous with ωun ,ac ≤ ωu,ac for every n. We claim that this implies that ωu\ ,ac ≤ ωu,ac . To see this, let δ > 0 and let {(ai , bi )}`i=1 be a collection of nonoverlapping intervals such that P` i=1 (bi − ai ) < δ. Then ` X
un (bi ) − un (ai ) ≤ ωun ,ac (δ) ≤ ωu,ac (δ).
i=1
Letting n → ∞ gives ` X
u\ (bi ) − u\ (ai ) ≤ ωu,ac (δ).
i=1
Taking the supremum over all such collection gives ωu\ ,ac ≤ ωu,ac . Step 2: Let u : R → [0, ∞) be an absolutely continuous function vanishing at infinity. Let ϕ ∈ Cc1 (R) be such that 0 ≤ ϕ ≤ 1, ϕ = 1 in (−1, 1) and ϕ = 0 outside (−2, 2). For every k ∈ R define ϕk (x) := ϕ(2−k x). The function ϕk u has compact support. Moreover, if δ > 0 and {(ai , bi )}`i=1 is a P collection of nonoverlapping intervals such that `i=1 (bi − ai ) < δ, then ` X
(ϕk u)(bi ) − (ϕk u)(ai ) ≤
i=1
` X
ϕk (bi )u(bi ) − u(ai )
i=1
+
` X
u(bi )ϕk (bi ) − ϕk (ai ) ≤ ωu,ac (δ) + 2−k kuk∞ kϕ0 k∞ δ.
i=1
Hence, ωϕk ◦u,ac (δ) ≤ ωu,ac (δ) + 2−k kuk∞ kϕ0 k∞ δ. Applying Step 1 to ϕk u, we obtain ω(ϕk u)\ ,ac (δ) ≤ ωϕk u,ac (δ) ≤ ωu,ac (δ) + 2−k kuk∞ kϕ0 k∞ δ for every k. Note that 0 ≤ ϕk ≤ ϕk+1 . Thus, ϕk u ≤ ϕk+1 u ≤ u and ϕk u → u pointwise as k → ∞. It follows from Theorem 4.30 that (ϕk u)\ → u\ . Reasoning as in the first step (with un replaced by (ϕk u)\ ), we obtain that ωu\ ,ac ≤ ωu,ac . By adapting the proof of Theorem 4.37, we can approximate two functions u, v ∈ Cc (R; [0, ∞)) using the same reflections. Corollary 4.42. Given u, v ∈ Cc (R; [0, ∞)), let Pu,v := {(uσ1 ···σn , v σ1 ···σn ) : σ1 , . . . , σn reflections, n ∈ N}.
4.3. Symmetric Decreasing Rearrangement
137
Then there exists (un , vn ) ∈ Pu,v such that un → u\ and vn → v \ uniformly in R as n → ∞, supp un ⊆ B(0, R), and supp vn ⊆ B(0, R) for all n and for some R > 0. Proof. Without loss of generality, we assume that u, v 6= 0, since if one of them is zero, say, v = 0, then v σ = 0 and so we can just consider the sequence un given in Theorem 4.37. Let f , g, and F be as in Theorem 4.37. Define (4.43)
` :=
sup
(F (w) + F (z))
(w,z)∈Pu,v
and let (wn , zn ) ∈ Pu,v be such that 1 (4.44) ` − ≤ F (wn ) + F (zn ) ≤ `. n Step 1: Let r > 0 be so large that supp u ⊆ B(0, r) and supp v ⊆ B(0, r). We claim that we can replace {(wn , zn )}n with a new maximizing sequence in Pu,v in such a way that supp wn ⊆ B(0, 6r) and supp zn ⊆ B(0, 6r). By Theorem 4.14 with t = 0, we have supp wn ⊆ B(xn , r) and supp zn ⊆ B(yn , r) for some xn , yn ∈ R. We consider three cases. Case 1: If B(xn , r) ∩ B(yn , r) 6= ∅ and (B(xn , r) ∪ B(yn , r)) ∩ B(0, 2r) 6= ∅, then we leave wn and zn unchanged. Case 2: If B(xn , r) ∩ B(yn , r) 6= ∅ and (B(xn , r) ∪ B(yn , r)) ∩ B(0, 2r) = ∅, we let ρn be the point of B(xn , r) ∪ B(yn , r) that is closest to the origin. Let an := 12 ρn be the middle point between the origin and ρn and let σn (x) = −x+2an , x ∈ R, be the corresponding reflection. Then 0 ∈ I+ and B(xn , r)∪ B(yn , r) ⊆ I− . If x ∈ I− , then σn (x) ∈ I+ and so wn (σn (x)) = 0 and zn (σn (x)) = 0. In turn, wnσn (x) = min{wn (x), wn (σn (x))} = 0. Let x ∈ B(xn , r) ⊆ I− be such that wn (x) > 0. Then σn (x) ∈ I+ . Moreover, since ρn ∈ B(xn , r) ∪ B(yn , r) and the two balls intersect, we have that x − ρn  < 4r. Using the fact that σn (ρn ) = 0, we obtain σn (x) = σn (x) − σn (ρn ) = x − ρn  < 4r. Since σn (x) ∈ I+ and σn (σn (x)) = x, by (4.1) we have wnσn (σn (x)) = max{wn (x), wn (σn (x))} = wn (x) > 0. Thus, σn (x) ∈ supp wnσn . By Theorem 4.14, the support of wnσn is contained in a ball of radius r. This implies that supp wnσn ⊆ B(0, 6r). Similarly, we have that supp znσn ⊆ B(0, 6r). n Case 3: If B(xn , r) ∩ B(yn , r) = ∅, let an := xn +y and let σn be the 2 corresponding reflection. Then B(xn , r) and B(yn , r) lie on different sides of an , say B(xn , r) ⊂ I+ . We claim that the support of wnσn and znσn is contained in B(xn , r). Note that σn maps B(yn , r) into B(xn , r). Hence, if x ∈ B(xn , r), wnσn (x) = max{wn (x), wn (σn (x))} = wn (x), since wn = 0
138
4. Rearrangements
in I− , while znσn (x) = max{zn (x), zn (σn (x))} = zn (σn (x)), where σn (x) ∈ B(yn , r), which contains the support of zn . This proves that the support of wnσn and znσn is contained in B(xn , r). Observe that if B(xn , r) ∩ B(0, 2r) 6= ∅, then wnσn and znσn satisfy the conditions of Case 1, while if B(xn , r) ∩ B(0, 2r) = ∅, we can apply Step 2 to wnσn and znσn . Step 2: Thanks to Step 1, without loss of generality, we can suppose that supp wn ⊆ B(0, 6r) and supp zn ⊆ B(0, 6r). Since the modulus of continuity of wn and zn is less than the modulus of continuity of u and v, respectively, by Theorem 4.8, we have that the sequence {wn }n and {zn }n are equicontinuous with support contained in B(0, 6r) for every n. By the Ascoli–Arzel´a theorem (see [Leo22b]), there exist a subsequence {wnk }k and w ∈ Cc (R) such that wnk → w uniformly in R. By extracting a further subsequence, again by the Ascoli–Arzel´ a theorem, we can find z ∈ Cc (R) such that zn → z uniformly in R. In turn, also by (4.44), Z Z w(x)g(x) dx + z(x)g(x) dx = lim (F (wnk ) + F (znk )) = `. R
k→∞
R
Reasoning as in Step 2 of the proof of Theorem 4.37, if σ is a reflection, we have that wnσk → wσ and znσk → z σ uniformly in R. By the Hardy– Littlewood inequality for polarization (Theorem 4.15), 1 ≤ F (wnk ) + F (znk ) ≤ F (wnσk ) + F (znσk ) ≤ `, `− nk where we use the fact that g σ = g. Letting k → ∞ and using the fact that wnσk → wσ and znσk → z σ uniformly in R gives Z Z Z Z σ σ w (x)g(x) dx + z (x)g(x) dx = ` = w(x)g(x) dx + z(x)g(x) dx. R
R
R
R
We now use the fact that a + b = c + d with a ≥ c and b ≥ d implies that a = c and b = d to conclude that Z Z Z Z σ σ z(x)g(x) dx. w(x)g(x) dx, z (x)g(x) dx = w (x)g(x) dx = R
R
R
R
We can now continue as in Step 2 of the proof of Theorem 4.37 to obtain that w = wσ and z = z σ for every reflection σ, that w = w\ and z = z \ , and finally, that w = u\ and z = v \ . This concludes the proof. Next we prove the Hardy–Littlewood rearrangement inequality and some of its generalizations. Theorem 4.43 (Hardy–Littlewood inequality). Let E ⊆ R be a Lebesgue measurable set and let u : E → [0, ∞) and v : E → [0, ∞) be Lebesgue measurable functions. Then Z Z (4.45) u(x)v(x) dx ≤ u\ (x)v \ (x) dx. E
R
4.3. Symmetric Decreasing Rearrangement
139
First proof. Given r, s ≥ 0, we have that L1 ({x ∈ E : u(x) > r} ∩ {y ∈ E : v(y) > t}) ≤ min{L1 ({x ∈ E : u(x) > r}), L1 ({y ∈ E : v(y) > t})} = min{L1 ({x ∈ R : u\ (x) > r}), L1 ({y ∈ R : v \ (y) > t})}, where in the last equality we use (4.35). Since u\ and v \ are symmetrically decreasing, the sets {x ∈ R : u\ (x) > r} and {y ∈ R : v \ (y) > t} are either R or balls centered at the origin. Hence, their intersection is the smallest ball, or, equivalently, min{L1 ({x ∈ R : u\ (x) > r}), L1 ({y ∈ R : v \ (y) > t})} = L1 ({x ∈ R : u\ (x) > r} ∩ {y ∈ R : v \ (y) > t}). Thus, (4.46)
L1 ({x ∈ E : u(x) > r} ∩ {y ∈ E : v(y) > t}) ≤ L1 ({x ∈ R : u\ (x) > r} ∩ {y ∈ R : v \ (y) > t}).
By the layercake representation (4.2) and Tonelli’s theorem, Z Z Z ∞Z ∞ u(x)v(x) dx = χ{u>r} (x)χ{v>t} (x) drdtdx E E 0 0 Z ∞Z ∞Z = (4.47) χ{u>r} (x)χ{v>t} (x) dxdrdt 0 0 E Z ∞Z ∞ = L1 ({x ∈ E : u(x) > r} ∩ {y ∈ E : v(y) > t}) drdt. 0
0
Similarly, Z (4.48) u\ (x)v \ (x) dx R Z ∞Z ∞ = L1 ({x ∈ R : u\ (x) > r} ∩ {y ∈ R : v \ (y) > t}) drdt. 0
0
In view of (4.46), (4.47), and (4.48), we have that (4.45) holds.
Next we present a second proof, which relies on polarization. Second proof of Theorem 4.43. Assume that u, v ∈ Cc (R; [0, ∞)), let Pu,v := {(uσ1 ···σn , v σ1 ···σn ) : σ1 , . . . , σn reflections, n ∈ N}, and use Corollary 4.42 to find (un , vn ) ∈ Pu,v such that un → u\ and vn → u\ uniformly in R as n → ∞, supp un ⊆ B(0, R), and supp vn ⊆ B(0, R) for all
140
4. Rearrangements
n and for some R > 0. By repeated applications of the Hardy–Littlewood inequality for polarization (Theorem 4.15) we have Z Z u(x)v(x) dx ≤ un (x)vn (x) dx. R
R
Letting n → ∞ and using the facts that un → u\ and vn → u\ uniformly in R as n → ∞, supp un ⊆ B(0, R), and supp vn ⊆ B(0, R) for all n, we obtain that Z Z u\ (x)v \ (x) dx.
u(x)v(x) dx ≤ R
R
The general case follows by assuming first that u and v are bounded Lebesgue integrable and using the density of Cc (R) in L1 (R) and Theorem 4.33. We leave the details as an exercise. An important consequence of Theorem 4.43 is the following. Corollary 4.44. Let g : R → [0, ∞) be a convex function with g(0) = 0, and let u : R → [0, ∞) and v : R → [0, ∞) be Lebesgue measurable functions vanishing at infinity. Then Z Z (4.49) g(u\ (x) − v \ (x)) dx ≤ g(u(x) − v(x)) dx. R
R
In particular, for 1 ≤ p < ∞, Z Z \ \ p u (x) − v (x) dx ≤ u(x) − v(x)p dx. R
R
Proof. The proof is very similar to that of Corollary 4.17 and is left as an exercise. The inequality Z Z u(x)(v ∗ w)(x) dx ≤ u\ (x)(v \ ∗ w\ )(x) dx, R
R
where u, v, w : R → [0, ∞) are measurable functions vanishing at infinity, is known as the Riesz rearrangement inequality (see [Bae19], [LL01]). Here, we prove a simpler version in which we assume that w = w\ . Theorem 4.45 (Riesz). Let u : R → [0, ∞) and v : R → [0, ∞) be Lebesgue measurable functions vanishing at infinity, and let k : [0, ∞) → [0, ∞) be decreasing. Then Z Z Z Z u(x)v(y)k(x − y) dxdy ≤ u\ (x)v \ (y)k(x − y) dxdy. R
R
R
R
4.3. Symmetric Decreasing Rearrangement
141
Proof. Step 1: Assume that u, v ∈ Cc (R; [0, ∞)), let Pu,v := {(uσ1 ···σn , v σ1 ···σn ) : σ1 , . . . , σn reflections, n ∈ N}, and use Corollary 4.42 to find (un , vn ) ∈ Pu,v such that un → u\ and vn → u\ uniformly in R as n → ∞, supp un ⊆ B(0, R), and supp vn ⊆ B(0, R) for all n and for some R > 0. By repeated applications of Theorem 4.18, we have Z Z Z Z un (x)vn (y)k(x − y) dxdy. u(x)v(y)k(x − y) dxdy ≤ R
R
R
R
Letting n → ∞ and using the facts that un → u\ and vn → u\ uniformly in R as n → ∞ and that supp un ⊆ B(0, R) and supp vn ⊆ B(0, R) for all n, we obtain that Z Z Z Z u(x)v(y)k(x − y) dxdy ≤ u\ (x)v \ (y)k(x − y) dxdy. R
R
R
R
Step 2: Assume next that u : R → [0, ∞), v : R → [0, ∞) are bounded Lebesgue measurable functions which vanish outside some ball B(0, R). By the density of Cc (R) in L1 (R) we can find un , vn ∈ Cc (R) such that un → u and vn → u in L1 (R) and pointwise L1 a.e. in R as n → ∞. We leave it as an exercise to prove that we can assume that supp un ⊆ B(0, R) and supp vn ⊆ B(0, R) for all n and that 0 ≤ un , vn ≤ M . By Exercises 4.25 and 4.39, this implies that supp u\n ⊆ B(0, R) and supp vn\ ⊆ B(0, R) and that 0 ≤ u\n , vn\ ≤ M for all n. By Step 1 applied to un and vn , Z Z Z Z un (x)vn (y)k(x − y) dxdy ≤ u\n (x)vn\ (y)k(x − y) dxdy. R
R
R
R
Since un → u and vn → v in L1 (R), we have that un → u and vn → v in measure. Hence, by Theorem 4.33, u\n → u\ and vn\ → v \ pointwise L1 a.e. in (−R, R). Using the fact that 0 ≤ un , vn , u\n , vn\ ≤ M , it follows by the Lebesgue dominated convergence theorem that Z Z Z Z u(x)v(y)k(x − y) dxdy ≤ u\ (x)v \ (y)k(x − y) dxdy. R
R
R
R
Step 3: Let u : R → [0, ∞), v : R → [0, ∞) be Lebesgue measurable functions vanishing at infinity. Define (4.50)
un := χB(0,n) min{u, n},
vn := χB(0,n) min{v, n}.
Then 0 ≤ un ≤ un+1 ≤ u, 0 ≤ vn ≤ vn+1 ≤ v, un % u, and vn % v. By Exercise 4.25 and Theorem 4.30, u\n % u\ , and vn\ % v \ . By Step 2, Z Z Z Z (4.51) un (x)vn (y)k(x − y) dxdy ≤ u\n (x)vn\ (y)k(x − y) dxdy. R
R
R
R
142
4. Rearrangements
Letting n → ∞ in (4.51) and using the Lebesgue monotone convergence theorem and Theorem 4.33, we obtain Z Z Z Z u\ (x)v \ (y)k(x − y) dxdy. u(x)v(y)k(x − y) dxdy ≤ R
R
R
R
The analogue of Corollary 4.44 is the following result. Corollary 4.46. Let g : R → [0, ∞) be a convex function with g(0) = 0, let u : R → [0, ∞) and v : R → [0, ∞) be Lebesgue measurable functions vanishing at infinity, and let k : [0, ∞) → [0, ∞) be decreasing and Lebesgue integrable. Then Z Z Z Z \ \ g(u (x) − v (y))k(x − y) dxdy ≤ g(u(x) − v(y))k(x − y) dxdy. R
R
R
R
In particular, for 1 ≤ p < ∞, Z Z Z Z \ \ p u (x) − v (y) k(x − y) dxdy ≤ u(x) − v(y)p k(x − y) dxdy. R
R
R
R
Proof. The proof is very similar to that of Corollary 4.19 and is left as an exercise. We conclude this section by observing that we can define the symmetric decreasing rearrangement of functions which takes both positive and negative signs. Definition 4.47. Let E ⊆ R be a Lebesgue measurable set and let u : E → [−∞, ∞] be a Lebesgue measurable function. The symmetric decreasing rearrangement of u is the function u\ : R → [0, ∞], defined by u\ := u\ .
4.4. Symmetric Decreasing Rearrangement in W s,p (R) In this section we will study the symmetric decreasing rearrangements of functions in W s,p (R) for 1 ≤ p < ∞ and 0 < s ≤ 1. We recall that u\ := u\ . Theorem 4.48. Let 1 ≤ p < ∞, 0 < s < 1, and let u : R → R be a Lebesgue measurable function vanishing at infinity. Then Z Z Z Z u\ (x) − u\ (y)p u(x) − u(y)p (4.52) dxdy ≤ dxdy. 1+sp x − y1+sp R R R R x − y Moreover, if un → u in W s,p (R), then u\n → u\ in W s,p (R). Proof. The proof is the same of Theorem 4.20, using Corollary 4.46 in place of Corollary 4.19. Inequality (4.52) fails if we replace R with a bounded interval.
4.4. Symmetric Decreasing Rearrangement in W s,p (R)
143
Exercise 4.49. Let f : R → [0, 1] be a symmetrically decreasing function of class C 1 such that f = 1 in [0, 1/2] and f = 0 in [1, ∞), and let 1 ≤ p < ∞ and 0 < s < 1. For 0 < ε < 1, define x − (1 − ε) uε (x) := f , x ∈ (−1, 1). ε (i) Prove that Z Z 1 (uε (x))p −1
R\(−1,1)
1 dydx ≥ Cε1−sp x − y1+sp
Z
Z
1
(ii) Prove that Z Z 1 (u\ε (x))p −1
R\(−1,1)
1 dydx ≤ Cε x − y1+sp
1
(f (x))p dx.
−1
(f (x))p dx.
−1
(iii) Prove that for all ε > 0 sufficiently small, Z 1Z 1 Z 1Z 1 \ u(x) − u(y)p u (x) − u\ (y)p dxdy > dxdy. 1+sp 1+sp x − y −1 −1 x − y −1 −1 Next we consider the case s = 1. Theorem 4.50. Let 1 ≤ p < ∞, and let u ∈ W 1,p (R) be a nonnegative function. Then u\ ∈ W 1,p (R) and Z Z \ 0 p (u ) (x) dx ≤ u0 (x)p dx. R
R
Proof. Step 1: Assume that u is zero outside a compact set. Let u ¯ be the absolutely continuous representative of u (see [Leo22d]). By Theorem 4.37, there exists a sequence un ∈ Pu such that un → u\ uniformly in R as n → ∞ and supp un ⊆ B(0, R R) for all nR and for some R > 0. By Theorem 4.21, un ∈ W 1,p (R) and R u0n p dx = R u0 p dx for every n. We now distinguish two cases. If p > 1, we use the fact (exercise) that for v ∈ Lp (R), Z 1 kvkLp (R) = sup vψ dx : ψ ∈ Cc (R), kψkLp0 (R) ≤ 1 , R
together with integration by parts, to get Z Z 0 0 0 ku kLp (R) = kun kLp (R) ≥ un ψ dx = − un ψ 0 dx R
R
for every ψ ∈ Cc1 (R) with kψkLp0 (R) ≤ 1. Letting n → ∞ and using the facts that un → u\ uniformly in R and that ψ has compact support gives Z 0 ku kLp (R) ≥ − u\ ψ 0 dx. R
144
4. Rearrangements
Since u\ is absolutely continuous by Corollary 4.41, we can integrate by R parts to obtain ku0 kLp (R) ≥ R (u\ )0 ψ dx. We leave as an exercise to show that this implies that (u\ )0 ∈ Lp (R) with ku0 kLp (R) ≥ k(u\ )0 kLp (R) . Rb If p = 1, we use the fact that for v ∈ AC([a, b]), a v 0 (x) dx = Var[a,b] v (see [Leo17, Theorem 3.25]) to estimate ku0 kL1 (R) = ku0n kL1 (R) ≥ Var[a,b] un . Letting n → ∞ and using the fact that un → u\ uniformly in R gives (exercise) ku0 kL1 (R) ≥ Var[a,b] u\ . Since u\ is absolutely continuous by Corollary Rb 4.41, we have that Var[a,b] u\ = a (u\ )0 (x) dx (see [Leo17, Theorem 3.25]). Take (a, b) = (−k, k) and use the Lebesgue monotone convergence theorem to get ku0 kL1 (R) ≥ k(u\ )0 kL1 (R) . Step 2: The general case is left as an exercise. Hint: See the proof of Step 2 of Corollary 4.41.
Remark 4.51. For p > 1 in Step 1, we could have used the fact that Lp (R) is reflexive and the lower semicontinuity of the norm in Lp with respect to weak convergence (see [Leo22b]).
4.5. Notes For more material on polarization, we refer to the book of Baernstein [Bae19], the lecture notes of Burchard [Bur09], as well as the papers [BS00], [BS01], [vS06] and the references therein. For more information on the symmetric decreasing rearrangement, we refer to the books of Baernstein [Bae19], Kawohl [Kaw85], Kesavan [Kes06], Leoni [Leo17], Lieb and Loss [LL01]. The proofs of Theorems 4.20 and 4.48 are due to Almgren and Lieb [AL89]. Exercise 4.49 is drawn from the paper of Li and Wang [LW19]. I worked on this chapter with the undergraduate students Braden Yates and Jiewen Hu.
Chapter 5
Higher Order Fractional Sobolev Spaces in One Dimension The reading of books, what is it but conversing with the wisest men of all ages and all countries. — Isaac Barrow
While most of the book focuses on fractional Sobolev spaces W s,p , with 0 < s < 1, in this chapter you will see a bit of higher order fractional Sobolev spaces. As a prerequisite, you should first read the section on Sobolev functions of one variable [Leo22d] and make sure you are comfortable with the notions of weak derivatives and absolute continuity. We will make heavy use of both in this chapter. For simplicity, we will often present the proofs in the case 1 < s < 2 and leave the remaining cases as an exercise.
5.1. Higher Order Fractional Sobolev Spaces In this section we introduce the fractional Sobolev spaces W s,p (I) for s > 1. Definition 5.1. Let I ⊆ R be an open interval, 1 ≤ p < ∞, 1 < s < ∞, with s ∈ / N, and m = bsc. A function u ∈ W m,p (I) belongs to the fractional Sobolev space W s,p (I) if kukW s,p (I) := kukW m,p (I) + u(m) W s−m,p (I) < ∞. When p = 2, we write H s (I) := W s,2 (I). Here bsc is the integer part of s. Definition 5.2. Let I ⊆ R be an open interval, 1 ≤ p < ∞, 1 < s < ∞, with m,p s∈ / N, and m = bsc. A function u ∈ Wloc (I) belongs to the homogeneous 145
146
5. Higher Order Fractional Sobolev Spaces in One Dimension
˙ s,p (I) if u(m) W s−m,p (I) < ∞. When p = 2, we fractional Sobolev space W ˙ s,2 (I). write H˙ s (I) := W Exercise 5.3. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 1 < s < ∞, with s ∈ / N. Prove that the function u ∈ W s,p (I) 7→ kukW s,p (I) is a norm and that (W s,p (I), k · kW s,p (I) ) is a Banach space. Exercise 5.4. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 1 < s < ∞, with s ∈ / N. Prove that if u ∈ Ccn (I), with n = bsc + 1, then u ∈ W s,p (I). Exercise 5.5. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 1 < s < ∞, with s ∈ / N. Prove that u(m) W s−m,p (I) `−s ku(m−1) kLp (I) + `1−s ku(m+1) kLp (I) for all u ∈ W m+1,p (I) and for every 0 < ` < L1 (I), where m = bsc, and u(0) := u. Conclude that W m+1,p (I) ,→ W s,p (I). ˙ s,p (I) coincides with In the following theorem, we prove that Lp (I) ∩ W for 1 < s < 2.
W s,p (I)
Theorem 5.6. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 1 < s < 2. ˙ s,p (I) and for every 0 < ` < L1 (I), Then for all u ∈ Lp (I) ∩ W ku0 kLp (I) `−1 kukLp (I) + `s−1 u0 W s−1,p (I) . ˙ s,p (I) = W s,p (I). In particular, Lp (I) ∩ W First proof. Step 1: Assume I = (a, b). Subdivide I into three disjoint intervals I1 , I2 , I3 of equal length, with a the left endpoint of I1 and b the right endpoint of I3 . By Poincar´e’s inequality (see Theorem 1.17 and Remark 1.18) applied to u0 and with E = (c, d), where c ∈ I1 and d ∈ I3 , Z b Z Z (b − a)1+(s−1)p b b u0 (x) − u0 (y)p 0 0 p dxdy. u (x) − (u )(c,d)  dx 1+(s−1)p d−c a a x − y a Therefore, Z b Z b 0 p 0 p u (x) dx (b − a)(u )(c,d)  + u0 (x) − (u0 )(c,d) p dx a
a 0
p
(b − a)(u )(c,d)  + 1,p Wloc (I),
(b −
a)1+(s−1)p d−c
Z bZ a
a
b
u0 (x) − u0 (y)p dxdy. x − y1+(s−1)p
Since u ∈ we can choose a representative u ¯ of u that is locally 0 0 1 absolutely continuous (see [Leo22d]) with u ¯ = u L a.e. in I. Hence, by the fundamental theorem of calculus (see [Leo22d]), Z d u ¯(d) − u ¯(c) 1 0 u0 (t) dt = . (u )(c,d) = d−c c d−c
5.1. Higher Order Fractional Sobolev Spaces
147
Note that d − c ≥ 13 (b − a), since c ∈ I1 and d ∈ I3 . Hence, Z b 1 1 (5.1) u0 (x)p dx ¯ u(c)p + ¯ u(d)p p−1 p−1 (b − a) (b − a) a Z bZ b 0 u (x) − u0 (y)p (s−1)p + (b − a) dxdy. 1+(s−1)p a a x − y We now average both sides of this inequality in the variable c over I1 and then in the variable d over I2 to get Z b Z b 1 0 p u (x) dx u(x)p dx (b − a)p a a Z bZ b 0 u (x) − u0 (y)p (s−1)p + (b − a) dxdy. 1+(s−1)p a a x − y Step 2: If I = (α, β) has finite length and 0 < ` < β − α, let m ∈ N be such that α + (m − 1)` ≤ β < α + m` and consider the subinterval In := (α + (n − 1)`, α + n`), n = 1, . . . , m − 1 and Im = (β − `, β). Then Z Z m−1 XZ 0 p 0 p u0 (x)p dx. u (x) dx + u (x) dx ≤ I
n=1
Im
In
We now apply Step 1 in each interval In . Step 3: If I has infinite length and ` > 0, we subdivide I into countably many disjoint intervals In of length ` and apply Step 1 in each one to get Z Z Z Z 1 u0 (x) − u0 (y)p 0 p p (s−1)p u (x) dx p dxdy. u(x) dx + ` 1+(s−1)p ` In In In I x − y Summing over n gives Z Z Z Z u0 (x) − u0 (y)p 1 p (s−1)p 0 p u (x) dx p u(x) dx + ` dxdy. 1+(s−1)p ` I I I x − y I
We present a second proof in the case p > 1 and I = R. We recall that for t ∈ R, It = {x ∈ I : x + t ∈ I}. The proof relies on the following identity. 1,1 Lemma 5.7. Let I ⊆ R be an open interval and let u ∈ Wloc (I). Then Z h Z h 1 1 u(x + t) sgn t dt + 2 (h − t)(u0 (x) − u0 (x + t)) dt u0 (x) = 2 h −h h −h
for all h > 0 and x ∈ Ih ∩ I−h . Proof. Integrating by parts, Z h Z 0 (h − t)u (x + t) dt = −h
h
−h
u(x + t) sgn t dt.
148
5. Higher Order Fractional Sobolev Spaces in One Dimension
Since sides.
Rh
−h (h − t) dt
= h2 , it suffices to subtract u0 (x)
Rh
−h (h − t) dt
to both
We turn to the second proof of Theorem 5.6, which only works for p > 1. Second proof of Theorem 5.6. Assume that I = R and p > 1. Using Lemma 5.7 and H¨older’s inequality, Z Z h 1 1 h t1/p+s−1 0 0 (5.2) u (x) ≤ 2 u (x) − u0 (x + t) dt u(x + t) dt + h −h h −h t1/p+s−1 Z h 0 1/p 1 u (x) − u0 (x + t)p s−1 M(u)(x) + h dt , h t1+(s−1)p −h where M(u) is the maximal function of u (see [Leo22c]), and we use the 1/p0 R 0 h ≈ hs . Raising both sides to power p and fact that 0 t(1/p+s−1)p dt integrating in x over R gives Z Z Z Z h 0 1 u (x) − u0 (x + t)p 0 p p (s−1)p u (x) dx p (M(u)(x)) dx + h dtdx. h R t1+(s−1)p R R −h To conclude, we use the continuity of the maximal operator M in Lp (R) for p > 1 (see [Leo22c]). A similar result holds for 2 < s < 3. Theorem 5.8. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 2 < s < 3. ˙ s,p (I) and for every 0 < ` < L1 (I), Then for all u ∈ Lp (I) ∩ W ku0 kLp (I) `−1 kukLp (I) + `s−1 u00 W s−2,p (I) , ku00 kLp (I) `−2 kukLp (I) + `s−2 u00 W s−2,p (I) . ˙ s,p (I) = W s,p (I). In particular, Lp (I) ∩ W Proof. Fix 0 < ` < L1 (I) and find an open set J b I with 0 < ` < L1 (J). 2,p Since u ∈ Wloc (I), we can apply the Gagliardo–Nirenberg interpolation theorem (see [Leo22d]) to get (5.3)
ku0 kLp (J) ≤ C0 δ −1 kukLp (J) + C0 δku00 kLp (J) ,
where C0 = C0 (p) ≥ 1 and 0 < δ < L1 (J). Let C1 = C1 (p, s) ≥ 1 be the constant in Theorem 5.6 (with s replaced by s − 1) and take δ = 2C10 C1 ` < `. ˙ s−1,p (J), By (5.3) and Theorem 5.6 applied to u0 ∈ Lp (J) ∩ W ku00 kLp (J) ≤ C1 `−1 ku0 kLp (J) + C1 `s−2 u00 W s−2,p (J) ≤ C1 `−1 (C0 δ −1 kukLp (J) + C0 δku00 kLp (J) ) + C1 `s−2 u00 W s−2,p (J) 1 = C1 C0 `−1 δ −1 kukLp (J) + ku00 kLp (J) + C1 `s−2 u00 W s−2,p (J) . 2
5.1. Higher Order Fractional Sobolev Spaces
149
It follows that (5.4)
1 00 ku kLp (J) `−2 kukLp (I) + `s−2 u00 W s−2,p (I) . 2
Combining inequality (5.4) with (5.3) (now with δ = `) gives ku0 kLp (J) `−1 kukLp (J) + `ku00 kLp (J) `−1 kukLp (J) + `(`−2 kukLp (I) + `s−2 u00 W s−2,p (I) ) (5.5)
`−1 kukLp (I) + `s−1 u00 W s−2,p (I) .
We now let J % I in (5.4) and (5.5) and apply Lebesgue’s monotone convergence theorem. Exercise 5.9. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and s > 1 with s∈ / N. Prove that ku(k) kLp (I) `−k kukLp (I) + `s−k u(m) W s−m,p (I) ˙ s,p (I), k = 1, . . . , m = bsc, and 0 < ` < L1 (I). for all u ∈ Lp (I) ∩ W Theorem 5.10 (Product). Let I ⊆ R be an open interval and 1 < s < 2. If u, v ∈ W s,p (I) ∩ W 1,∞ (I), then uv ∈ W s,p (I), with kuvkLp (I) ≤ kukLp (I) kvkL∞ (I) , k(uv)0 kLp (I) ≤ ku0 kLp (I) kvkL∞ (I) + kukL∞ (I) kv 0 kLp (I) , and (5.6)
(uv)0 W s−1,p (I) kukL∞ (I) v 0 W s−1,p (I) + uW s−1,p (I) kv 0 kL∞ (I) + ku0 kL∞ (I) vW s−1,p (I) + u0 W s−1,p (I) kvkL∞ (I) .
Proof. Since u, v ∈ W 1,p (I), they admit representatives u ¯ and v¯ that are absolutely continuous (see [Leo22d]). Moreover, their (classical) derivatives coincide with u0 and v 0 L1 a.e. in I. Hence, we can apply the product rule to obtain that k(uv)0 kLp (I) ≤ ku0 kLp (I) kvkL∞ (I) + kukL∞ (I) kv 0 kLp (I) , while kuvkLp (I) ≤ kukLp (I) kvkL∞ (I) . By the product rule we can write (¯ uv¯)0 (x + h) − (¯ uv¯)0 (x) = u0 (x + h)¯ v (x + h) + u ¯(x + h)v 0 (x + h) − u0 (x)¯ v (x) − u ¯(x)v 0 (x) = v¯(x + h)∆h u0 (x) + u ¯(x + h)∆h v 0 (x) + v 0 (x)∆h u ¯(x) + u0 (x)∆h v¯(x).
150
5. Higher Order Fractional Sobolev Spaces in One Dimension
Hence, by Minkowski’s inequality’s we have !1/p 0 (x)p ¯ v (x + h)∆ u h dxdh (uv)0 W s−1,p (I) ≤ h1+(s−1)p 0 Ih !1/p Z L1 (I) Z ¯ u(x + h)∆h v 0 (x)p dxdh + h1+(s−1)p 0 Ih !1/p Z L1 (I) Z v 0 (x)∆h u(x)p + dxdh h1+(s−1)p 0 Ih !1/p Z L1 (I) Z u0 (x)∆h v(x)p =: A + B + C + D. dxdh + h1+(s−1)p 0 Ih Z
(5.7)
L1 (I) Z
Then A + B ≤ u0 W s−1,p (I) kvkL∞ (I) + kukL∞ (I) v 0 W s−1,p (I) , while, C + D uW s−1,p (I) kv 0 kL∞ (I) + ku0 kL∞ (I) vW s−1,p (I) . Combining the estimates for A, B, C, and D concludes the proof.
Remark 5.11. Observe that if w ∈ W 1,p (I), by Theorem 1.25, wW s−1,p (I) `−(s−1) kwkLp (I) + `2−s kw0 kLp (I) . Hence, the righthand side of (5.6) can be bounded from above by (5.8) (uv)0 W s−1,p (I) `−(s−1) kukLp (I) kv 0 kL∞ (I) + kukL∞ (I) v 0 W s−1,p (I) + `2−s ku0 kLp (I) kv 0 kL∞ (I) + `−(s−1) ku0 kL∞ (I) kvkLp (I) + u0 W s−1,p (I) kvkL∞ (I) + `2−s ku0 kL∞ (I) kv 0 kLp (I) . Exercise 5.12. Let I ⊆ RN be an open interval, 1 ≤ p < ∞, 1 < s < ∞, with s ∈ / N, and m = bsc. Prove that if u, v ∈ W s,p (I) ∩ W m,∞ (I), then uv ∈ W s,p (I). We recall that given a Lebesgue measurable set E ⊆ R with 0 < L1 (E) < ∞R and a Lebesgue integrable function u : E → R, we define uE := L11(E) E u dx. Exercise 5.13 (Poincar´e’s inequality). Let I = (a, b), 1 ≤ p < ∞, and 1 < s < 2. Given u ∈ W s,p (I), let pI (x) := (u0 )I x − b+a + uI , x ∈ I. 2
5.2. Extensions
151
Prove that Z
u(x) − pI (x)p dx (b − a)sp u0 pW s−1,p (I) ,
I
Z I
u0 (x) − p0I (x)p dx (b − a)(s−1)p u0 pW s−1,p (I) .
Exercise 5.14 (Poincar´e’s inequality). Let I = (a, b), 1 ≤ p < ∞, and 1 < s < 2. ˙ 1,p (I), prove that (i) Given a < c < d < b and v ∈ W Z Z v(x) − v(c,d) (x)p dx (b − a)p v 0 (x)p dx. I
I
˙ s,p (I), prove that (ii) Given a < c < d < b and u ∈ W Z (b − a)1+sp 0 p u W s−1,p (I) , u(x) − p(c,d) (x)p dx d−c I Z (b − a)1+(s−1)p 0 p u0 (x) − p0(c,d) (x)p dx u W s−1,p (I) , d−c I where p(c,d) (x) := (u0 )(c,d) x − d+c + u(c,d) . 2
5.2. Extensions In this section we prove some extension results. We begin with the simpler case in which I is unbounded. Theorem 5.15. Let I ⊂ R be an unbounded open interval, 1 ≤ p < ∞, and 1 < s < 2. Given u ∈ W s,p (I), there exists a function v ∈ W s,p (R) such that v = u in I, and kvkLp (R) kukLp (I) ,
kv 0 kLp (R) ku0 kLp (I) ,
v 0 W s−1,p (R) u0 W s−1,p (I) .
Moreover, the map u ∈ W s,p (I) 7→ v ∈ W s,p (R) is linear. Proof. We only consider the case I = (a, ∞) since the case I = (−∞, b) is similar. By a translation we can take a = 0. Define 3u(−x) − 2u(−2x) if x < 0, v(x) := u(x) if x > 0. We leave it as an exercise to show that v ∈ W 1,p (R), with −3u0 (−x) + 4u0 (−2x) if x < 0, 0 v (x) = u(x) if x > 0,
152
5. Higher Order Fractional Sobolev Spaces in One Dimension
and that kvkLp (R) kukLp (R+ ) and kv 0 kLp (R) ku0 kLp (R+ ) . Write Z Z Z Z u0 (x) − u0 (y)p v 0 (x) − v 0 (y)p dxdy = dxdy 1+(s−1)p 1+(s−1)p R R x − y R+ R+ x − y Z Z Z Z v 0 (x) − v 0 (y)p u0 (x) − v 0 (y)p dxdy + dxdy +2 1+(s−1)p 1+(s−1)p R− R− x − y R− R+ x − y =: A + B + C. Writing u0 (x) = −3u0 (x) + 4u0 (x), for x ∈ R+ and y ∈ R− , we have 0 u (x) − v 0 (y) p u0 (x) − u0 (−y) p + u0 (x) − u0 (−2y) p . In turn, Z B R−
Z R+
u0 (x) − u0 (−y)p dxdy + x − y1+(s−1)p
Z R−
Z R+
u0 (x) − u0 (−2y)p dxdy. x − y1+(s−1)p
We now make the change of variables −y = w in the first integral on the right and −2y = z in the second to obtain Z Z Z Z u0 (x) − u0 (w)p u0 (x) − u0 (z)p B dxdw + dxdz, 1+(s−1)p 1+(s−1)p R+ R+ x + w R+ R+ x + z/2 and observe that x + w = x + w ≥ x − w and x + z/2 = x + z/2 ≥ (x + z)/2 ≥ x − z/2, so that B u0 pW s−1,p (R+ ) . For x, y ∈ R− , 0 v (x) − v 0 (y) p u0 (−x) − u0 (−y) p + u0 (−2x) − u0 (−2y) p . Hence, Z
Z
C R−
R−
Z
Z
u0 (−x) − u0 (−y)p dxdy x − y1+(s−1)p
+ R−
R−
u0 (−2x) − u0 (−2y)p dxdy. x − y1+(s−1)p
We now make the change of variables −x = z and −y = w in the first integral on the right and −2x = z and −2y = w in the second to obtain C u0 pW s−1,p (R+ ) . Combining the estimates for A, B, and C completes the proof of the theorem. Remark 5.16. It follows from Step 1 in the proof that if I ⊂ R is an unbounded open interval, 1 ≤ p < ∞, and 1 < s < 2, then for every ˙ s,p (I), there exists a function v ∈ W ˙ s,p (R) such that v = u in I, and u∈W 0 0 v W s−1,p (R) u W s−1,p (I) . Moreover, if u ∈ Lq (I) for some 1 ≤ q ≤ ∞, then v ∈ Lq (R), with kvkLq (R) kukLq (I) .
5.2. Extensions
153
Exercise 5.17. Given 1 ≤ p < ∞, 1 < s < ∞, with s ∈ / N, u ∈ W s,p (R+ ), and m = bsc, prove that there exist c1 , . . . , cm+1 ∈ R such that the function Pm+1 n=1 cn u(−nx) if x < 0, v(x) := u(x) if x > 0, is well defined and belongs to W s,p (R). Prove also that for every 0 ≤ k ≤ m, kv (k) kLp (R) ku(k) kLp (R+ ) and that v (m) W s−m,p (R) u(m) W s−m,p (R+ ) . The case of bounded intervals is more involved. Theorem 5.18. Let I = (a, b), 1 ≤ p < ∞, and 1 < s < 2. Given a function u ∈ W s,p (I), there exists a function v ∈ W s,p (R) such that v = u in (a, b) and kvkLp (R) kukLp (I) ,
kv 0 kLp (R) (b − a)−1 kukLp (I) + ku0 kLp (I) ,
v 0 W s−1,p (R) (b − a)−s kukLp (I) + (b − a)−(s−1) ku0 kLp (I) + u0 W s−1,p (I) . Moreover, the map u ∈ W s,p (I) 7→ v ∈ W s,p (R) is linear. Proof. After a translation, without loss of generality, we may assume that I = (−r, r) for some r > 0. Step 1: Define fr : (−2r, 2r) → (−r, r) and gr : (−2r, 2r) → (−r, r) by −x − 2r if − 2r < x ≤ −r, x if − r < x < r, fr (x) := −x + 2r if r ≤ x < 2r and −2x − 3r if − 2r < x ≤ −r, x if − r < x < r, gr (x) := −2x + 3r if r ≤ x < 2r. The functions fr and gr are Lipschitz continuous with Lipschitz constant 1. Hence, for all x, y ∈ (−2r, 2r), (5.9)
fr (x) − fr (y) ≤ x − y,
gr (x) − gr (y) ≤ x − y.
Define w : (−2r, 2r) → R by (5.10)
w(x) := 3u(fr (x)) − 2u(gr (x)) 3u(−x − 2r) − 2u(−2x − 3r) if − 2r < x ≤ −r, u(x) if − r < x < r. = 3u(−x + 2r) − 2u(−2x + 3r) if r ≤ x < 2r.
Let u ¯ be the absolutely continuous representative of u (see [Leo22d]). The composition of u ¯ with the piecewise affine functions fr and gr is absolutely
154
5. Higher Order Fractional Sobolev Spaces in One Dimension
continuous with w0 (x) = 3u0 (fr (x))fr0 (x) − 2u0 (gr (x))gr0 (x) −3u0 (−x − 2r) + 4u0 (−2x − 3r) if − 2r < x ≤ −r, u0 (x) if − r < x < r. = 0 0 −3u (−x + 2r) + 4u (−2x + 3r) if r ≤ x < 2r. Then w0 pW s−1,p ((−2r,2r)) Z ≤ u0 pW s−1,p (I) + Z
Z
+2 (−2r,2r)\I
Z
(−2r,2r)\I (−2r,2r)\I w0 (x) − w0 (y)p
I
x − y1+(s−1)p
w0 (x) − w0 (y)p dxdy x − y1+(s−1)p
dxdy := A + B + C.
Using (5.9) and the changes of variables t = fr (x), t = gr (x), τ = fr (y), and τ = gr (y), we get Z Z u0 (fr (x)) − u0 (fr (y))p B dxdy 1+(s−1)p (−2r,2r)\I (−2r,2r)\I fr (x) − fr (y) Z Z u0 (gr (x)) − u0 (gr (y))p + dxdy 1+(s−1)p (−2r,2r)\I (−2r,2r)\I gr (x) − gr (y) Z Z u0 (t) − u0 (τ )p dtdτ. 1+(s−1)p I I t − τ  Similarly, u0 (fr (x)) − u0 (fr (y))p dxdy 1+(s−1)p (−2r,2r)\I I fr (x) − fr (y) Z Z u0 (gr (x)) − u0 (gr (y))p + dxdy 1+(s−1)p (−2r,2r)\I I gr (x) − gr (y) Z Z u0 (x) − u0 (τ )p dxdτ. 1+(s−1)p I I x − τ  Z
Z
C
Hence, (5.11)
w0 pW s−1,p ((−2r,2r)) u0 pW s−1,p (I) .
Similarly, by the changes of variables t = fr (x) and τ = gr (x), Z p p (5.12) kwkLp ((−2r,2r)) kukLp (I) + u(fr (x))p dx (−2r,2r)\I Z + u(gr (x))p dx kukpLp (I) (−2r,2r)\I
5.2. Extensions
155
and (5.13)
Z u0 (fr (x))p dx kw0 kpLp ((−2r,2r)) ku0 kpLp (I) + (−2r,2r)\I Z u0 (gr (x))p dx ku0 kpLp (I) . + (−2r,2r)\I
This shows that w ∈ W s,p ((−2r, 2r)). Step 2: Consider φ ∈ Cc∞ (R) such that 0 ≤ φ ≤ 1, φ = 1 in (−1, 1), and φ = 0 outside (−3/2, 3/2). Define φr (x) = φ(x/r) for x ∈ R, and v := φr w, ˜ where w ˜ is the extension of w by zero outside (−2r, 2r). Then 0 0 0 v = φr w ˜ + φr w ˜ and (5.14)
v 0 W s−1,p (R) ≤ φr w ˜ 0 W s−1,p (R) + φ0r w ˜ W s−1,p (R) .
We estimate (φr w ˜ 0 )(x) − (φr w ˜ 0 )(y) ≤ w ˜ 0 (x)φr (x) − φr (y) + φr (y)w ˜ 0 (x) − w ˜ 0 (y). Hence, by inequality (1.1) and Tonelli’s theorem, Z Z 2r φr (x) − φr (y)p 0 p 0 p φr w ˜ W s−1,p (R) w (x) dydx 1+(s−1)p R x − y −2r Z 3r/2 Z 2r w0 (x) − w0 (y)p + dxdy 1+(s−1)p −3r/2 −2r x − y Z Z 3r/2 1 0 p dxdy =: D + E + F, + w (y) 1+(s−1)p R\(2r,2r) x − y −3r/2 where we use the fact that φr = 0 outside (−3r/2, 3r/2) and that 0 ≤ φr ≤ 1. Write Z Z φr (x) − φr (y)p φr (x) − φr (y)p dy = dy 1+(s−1)p 1+(s−1)p R x − y B(y,4r) x − y Z φr (x)p dy =: D1 + D2 . + 1+(s−1)p R\B(y,4r) x − y By the mean value theorem φr (x) − φr (y) r−1 x − y. Hence, Z 1 1 D1 p x − yp−(s−1)p−1 dy ≈ (s−1)p , r B(y,4r) r while
Z
1
D2 ≤ R\B(y,4r)
x −
y1+(s−1)p
dy ≈
1 r(s−1)p
.
Thus, D
1 r(s−1)p
Z
3r
w0 (x)p dx
−3r
where in the last inequality we use (5.13).
1 r(s−1)p
ku0 kpLp (I) ,
156
5. Higher Order Fractional Sobolev Spaces in One Dimension
The term E is estimated by (5.11), while for F we observe that for y ∈ (−2r, 2r), Z Z 1 1 1 dx ≤ dx ≈ (s−1)p . 1+(s−1)p 1+(s−1)p r R\(3r,3r) x − y R\B(y,r) x − y Therefore, F
Z
1 r(s−1)p
2r
w0 (y)p dy
−2r
1 r(s−1)p
ku0 kpLp (I)
again by (5.13). Combining the estimates for D, E, and F gives (5.15)
φr w ˜ 0 W s−1,p (R)
1 rs−1
ku0 kLp (I) + u0 W s−1,p (I) .
On the other hand, writing φ0r (x) = φ0 (x/r)/r =: ψr (x)/r, we have that the term ψr w ˜ W s−1,p (R) can be estimated as in (5.15) (replacing φr and w ˜ 0 with ψr and w, ˜ respectively). Hence, 1 ˜ s−1,p (R) φ0r w ˜ W s−1,p (R) = ψr w r W 1 1 kukLp (I) + uW s−1,p (I) . r rs−1 By Theorem 1.25, uW s−1,p (I)
1 rs−1
kukLp (I) + r2−s ku0 kLp (I) .
Combining these two inequalities gives (5.16)
φ0r w ˜ W s−1,p (R)
1 1 kukLp (I) + s−1 ku0 kLp (I) . s r r
By (5.14), (5.15), and (5.16), v 0 W s−1,p (R)
1 1 kukLp (I) + s−1 ku0 kLp (I) + u0 W s−1,p (I) . s r r
Since v 0 = φr w ˜ 0 + φ0r w ˜ = φr w ˜ 0 + 1r ψr w, ˜ with φr and ψr = 0 outside (−2r, 2r) 0 and 0 ≤ φr ≤ 1, ψr  ≤ kφ k∞ , by (5.12) and (5.13), kvkLp (R) ≤ kwkLp (−2r,2r) kukLp (I) , 1 1 kv 0 kLp (R) kwkLp (−2r,2r) + kw0 kLp (−2r,2r) kukLp (I) + ku0 kLp (I) . r r Corollary 5.19. Let I = (a, b), 1 ≤ p < ∞, and 1 < s < 2. Given a ˙ s,p (I), for every k ∈ N, there exists a function v ∈ W ˙ s,p (Jk ), function u ∈ W k k where Jk = (a − 2 (b − a), b + 2 (b − a)), such that v = u in (a, b) and vW s,p (Jk ) k uW s,p (I) .
5.2. Extensions
157
Proof. By a translation we may assume that I = (−r, r). By repeated applications of Step 1 of the proof of Theorem 5.18, for every k we can extend ˙ ((−2k r, 2k r)), with wW s,p ((−2k r,2k r)) k uW s,p (I) . u to a function w in W ˙ s,p (I) be Corollary 5.20. Let I = (a, b), 1 ≤ p < ∞, 1 < s < 2, and u ∈ W such that Z d Z d u0 (x) dx = 0 u(x) dx = (5.17) c
c
for a < c < d < b, with c − d ≥ 12 (b − a). Then there exists a function v ∈ W s,p (R) such that v = u in I and kvkLp (R) (b − a)s uW s,p (I) ,
kv 0 kLp (R) (b − a)s−1 uW s,p (I) ,
vW s,p (R) uW s,p (I) . Moreover, if u ∈
Lq (I)
for some 1 ≤ q ≤ ∞, then kvkLq (R) kukLq (I) .
Proof. By Poincar´e’s inequality (Exercise 5.14) and (5.17), kukLp (I) (b − a)s u0 W s−1,p (I) ,
ku0 kLp (I) (b − a)s−1 u0 W s−1,p (I) .
It follows that u ∈ W s,p (I), and so we can apply Theorem 5.18 to find a function v ∈ W s,p (R) such that v = u in (a, b) and kvkLp (R) kukLp (I) (b − a)s u0 W s−1,p (I) kv 0 kLp (R) (b − a)−1 kukLp (I) + ku0 kLp (I) (b − a)s−1 u0 W s−1,p (I) , and v 0 W s−1,p (R) (b − a)−s kukLp (I) + (b − a)−(s−1) ku0 kLp (I) + u0 W s−1,p (I) u0 W s−1,p (I) . The last statement follows from (5.12), with p replaced by q, and the facts that φr = 0 outside (−2r, 2r) and 0 ≤ φr ≤ 1. ˙ 2,p (I) be such that Exercise 5.21. Let I = (a, b), 1 ≤ p < ∞, and u ∈ W Rd Rd 0 1 c u(x) dx = c u (x) dx = 0 for a < c < d < b, with c − d ≥ 2 (b − a). Prove 2 00 0 00 that kukLp (I) (b − a) ku kLp (I) , ku kLp (I) (b − a)ku kLp (I) . ˙ 2,p (I) be such that Exercise 5.22. Let I = (a, b), 1 ≤ p < ∞, and u ∈ W Rd Rd 0 1 c u(x) dx = c u (x) dx = 0 for a < c < d < b with c − d ≥ 2 (b − a). Prove that the function v defined in Step 2 of the proof of Theorem 5.18 belongs to W 2,p (R), with kvkLp (R) (b − a)2 ku00 kLp (I) , kv 0 kLp (R) (b − a)ku00 kLp (I) , and kv 00 kLp (R) ku00 kLp (I) . Exercise 5.23. Let I = (a, b), 1 ≤ p < ∞, s > 1, s ∈ / N. Given a function u ∈ W s,p (I) , prove that there exists a function v ∈ W s,p (R) such that v = u in (a, b) and kvkW s,p (R) I kukW s,p (I) .
158
5. Higher Order Fractional Sobolev Spaces in One Dimension
5.3. Density In this section we show that smooth functions with compact support are dense in W s,p (I). Theorem 5.24. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 1 < s < 2. Then for every u ∈ W s,p (I) there exists a sequence {un }n in W s,p (I) ∩ Cc∞ (R) such that un → u in W s,p (I). We begin with a useful lemma. Lemma 5.25. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 1 < s < 2. Consider a sequence of functions φn ∈ Cc∞ (R) such that 0 ≤ φn ≤ 1, φn = 1 (k) in (−n, n), φn = 0 outside (−n−1, n+1), and kφn k∞ ≤ Ck for all n, k ∈ N and for some Ck > 0. Then φn u → u in W s,p (I) for every u ∈ W s,p (I). Proof. Define un := φn u and ψn = 1−φn . Then u−un = uψn . We proceed as in the proof of Theorem 5.10 to write 1/p Z Z (¯ uψn )0 (x) − (¯ uψn )0 (y)p dxdy x − y1+(s−1)p I I Z Z 1/p (u0 (x) − u0 (y))ψn (x)p ≤ dxdy x − y1+(s−1)p I I Z Z 1/p ¯ u(y)(ψn0 (x) − ψn0 (y))p + dxdy x − y1+(s−1)p I I 1/p Z Z (¯ u(x) − u ¯(y))ψn0 (x)p dxdy + x − y1+(s−1)p I I Z Z 1/p u0 (y)(ψn (x) − ψn (y))p + dxdy =: A + B + C + D, x − y1+(s−1)p I I where u ¯ is the absolutely continuous representative of u. Since 0 ≤ ψn ≤ 1, ψn (x) → 0 as n → ∞ for every x ∈ I, and u0 W s−1,p (I) < ∞, it follows from the Lebesgue dominated convergence theorem that A → 0 as n → ∞. By the mean value theorem ψn0 (x) − ψn0 (y) ≤ kψn00 k∞ x − y ≤ C2 x − y. Since kφ0n k∞ ≤ C1 , for every x, y ∈ R, we have ψn0 (x) − ψn0 (y) ≤ C min{1, x − y}. By Lemma 1.61 and Tonelli’s theorem, Z Z Z min{1, x − y}p ¯ u(u)p dxdy u(y)p dy < ∞. x − y1+(s−1)p I I I
5.4. Some Equivalent Seminorms
159
Since ψn0 (x) − ψn0 (y) → 0 as n → ∞ for every x, y ∈ I, it follows by the Lebesgue dominated convergence theorem that B → 0 as n → ∞. The term D can be treated in a similar way. To estimate C, observe that u ∈ W s−1,p (I) by Theorem 1.25 and the fact that u ∈ W 1,p (I). Since, kψn0 k∞ ≤ C1 and ψn0 (x) → 0 as n → ∞ for every x ∈ I, by the Lebesgue dominated convergence theorem, we have that C → 0 as n → ∞. We turn to the proof of Theorem 5.24. Proof. Step 1: Assume that I = R. In view of Lemma 5.25 we may assume that u is zero outside a compact set. We will use mollifiers (see [Leo22c]). Let ϕε be a standard mollifier and let uε := u ∗ ϕε . By integrating by parts, we have that u0ε = u0 ∗ ϕε . Since ∆h u0ε = ϕε ∗ ∆h u0 , for every h > 0, k∆h u0ε kLp (R) ≤ k∆h u0 kLp (R) ,
(5.18) and (5.19)
lim kuε − ukW 1,p (R) = 0,
ε→0+
lim k∆h u0ε − ∆h u0 kLp (R) = 0
ε→0+
(see [Leo22c]). For h > 0 and ε > 0 define the functions 1 1 gε (h) := 1+(s−1)p k∆h u0ε − ∆h u0 kpLp (R) , g(h) := 1+(s−1)p k∆h ukpLp (R) . h h 1 By hypothesis, g ∈ L (R+ ), and by (5.18), Minkowski’s inequality, and the convexity of the function tp , we have that gε (h) ≤ 2p g(h) for all h. Since gε (h) → 0 for all h by (5.19), we are in a position to apply the Lebesgue dominated convergence theorem to conclude that Z ∞ Z ∞Z ∆h u0ε (x) − ∆h u0 (x)p dxdh = lim gε (h) dh = 0. lim ε→0+ 0 ε→0+ 0 h1+(s−1)p R Step 2: If I ⊆ R is an open interval, we use Theorems 5.15 and 5.18 to extend u to a function v ∈ W s,p (R), with kvkW s,p (R) I kukW s,p (I) . By Step 1, we can find vn ∈ Cc∞ (R) such that vn → v in W s,p (R). In particular, vn restricted to I converges to u in W s,p (I). Exercise 5.26. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and s > 1, s∈ / N. Prove that for every u ∈ W s,p (I) there exists a sequence {un }n in W s,p (I) ∩ Cc∞ (R) such that un → u in W s,p (I).
5.4. Some Equivalent Seminorms In this section we show that instead of using higher order derivatives, we can use higher order differences. We recall that given an open interval I ⊆ R and h ∈ R, (5.20)
Ih := {x ∈ I : x + h ∈ I},
160
5. Higher Order Fractional Sobolev Spaces in One Dimension
and that, given u : I → R, for x ∈ I2h and 0 < h < L1 (I)/2, ∆2h u(x) = ∆h (∆h u)(x) = u(x + 2h) − 2u(x + h) + u(x). Exercise 5.27. Let 1 ≤ p < ∞, ak ≥ 0, k = 1, . . . n, n ∈ N, and a, b ≥ 0. Prove that !p n n X X p−1 ak ≤n apk , k=1 n X
k=1
!p ≤ (3n)p−1
ak + a + b
k=1
n X
apk + 3p−1 ap + 3p−1 bp .
k=1
Theorem 5.28. Let I = R or I = (a, ∞), 1 ≤ p < ∞, and 1 < s < 2. ˙ s,p (I), Then for every u ∈ W ∞Z
Z 0
I
∆2h u(x)p dxdh ≈ h1+sp
∞Z
Z 0
I
∆h u0 (x)p dxdh. h1+(s−1)p
˙ s,p (I), we have that u ∈ W 1,p (I). Therefore, Proof. Step 1: Since u ∈ W loc u admits a representative u ¯ : I → R that is locally absolutely continuous (see [Leo22d]). By the fundamental theorem of calculus (see [Leo22d]), for x ∈ I and h > 0, ¯(x) = (¯ u(x + 2h) − u ¯(x + h)) − (¯ u(x + h) − u ¯(x)) ∆2h u Z h Z h (5.21) = [u0 (x + h + t) − u0 (x + t)] dt = ∆h u0 (x + t) dt. 0
0
¯(x)p ≤ hp−1 By H¨ older’s inequality, ∆2h u Tonelli’s theorem, Z
∆2h u(x)p dx ≤ hp−1
I
Z Z I
= hp−1
Z 0
h
0 hZ
Rh 0
∆h u0 (x + t)p dt, and so, by
∆h u0 (x + t)p dtdx ∆h u0 (x + t)p dxdt ≤ hp
I
Z
∆h u0 (y)p dy,
I
where we make the change of variables y = x + t. In turn, Z 0
∞Z I
∆2h u(x)p dxdh ≤ h1+sp
Z 0
∞Z I
∆h u0 (y)p dydh. h1+(s−1)p
Step 2: As Step 1, let u ¯ : I → R be the locally absolutely continuous representative of u. Given n ∈ N and h > 0, for x ∈ I by (5.21), using
5.4. Some Equivalent Seminorms
161
telescopic sums, n−1 X
∆2h u ¯(x
+ kh) =
k=0
n−1 XZ h k=0 h
Z =
[u0 (x + (k + 1)h + t) − u0 (x + kh + t)] dt
0
[u0 (x + nh + t) − u0 (x + nh)] dt
0
Z
h
−
[u0 (x + t) − u0 (x)] dt + h[u0 (x + nh) − u0 (x)].
0
Hence, n−1
∆nh u0 (x) =
1X 2 ∆h u ¯(x + kh) h k=0 Z Z 1 h 1 h 0 ∆t u (x + nh) dt + ∆t u0 (x) dt. − h 0 h 0
Raising both sides to power p and using Exercise 5.27 gives n−1
(3n)p−1 X 2 ¯(x + kh)p ∆h u ∆nh u (x) ≤ hp k=0 Z h p Z h p p−1 3 3p−1 0 0 + p ∆t u (x + nh) dt + p ∆t u (x) dt . h h 0 0 0
p
We now divide both sides by h1+(s−1)p and then integrate first in x over I and then in h over R+ to obtain Z ∞Z dh A := ∆nh u0 (x)p dx 1+(s−1)p h 0 I Z Z n−1 X ∞ dh ≤ (3n)p−1 ∆2h u ¯(x + kh)p dx 1+sp h I k=0 0 p Z ∞ Z Z h dh p−1 0 +3 (∆t u (x + nh) dt dx 1+sp h 0 I 0 p Z ∞ Z Z h dh + 3p−1 ∆t u0 (x) dt dx 1+sp h 0 I 0 =: (3n)p−1
n−1 X
Bk + 3p−1 C + 3p−1 D.
k=0
By the change of variables ρ = nh, Z ∞Z (s−1)p A=n ∆ρ u0 (x)p dx 0
I
dρ ρ1+(s−1)p
.
162
5. Higher Order Fractional Sobolev Spaces in One Dimension
By the change of variables x + kh = y, so that dx = dy, we have Z ∞Z dh ∆2h u ¯(y)p dy 1+sp . Bk ≤ h 0 I By the change of variables y = x + nh, so that dy = dx, we obtain p Z ∞ Z Z h dh 0 ∆t u (y) dt dy 1+sp C≤ h 0 I 0 p Z Z ∞ Z h dh = ∆t u0 (y) dt dy, 1+sp h I 0 0 where the last equality follows from Tonelli’s theorem. By Hardy’s inequality (see Theorem 1.3), for every y ∈ I, p Z Z ∞ Z h dh dh 1 ∞ 0 ∆t u (y) dt ∆h u0 (y)p 1+(s−1)p , ≤ p 1+sp h s 0 h 0 0 R∞R dh and so, again by Tonelli’s theorem, C ≤ s1p 0 I ∆h u0 (y)p dy h1+(s−1)p . The term D can be treated in a similar way. Combining the estimates for A, Bk , C, and D gives Z ∞Z Z ∞Z dh dh p (s−1)p 0 p ¯(x)p dx 1+sp ∆2h u n ∆h u (x) dx 1+(s−1)p ≤ (3n) h h 0 I 0 ZI Z 3p ∞ dh + p ∆h u0 (x)p dx 1+(s−1)p . s 0 h I ˙ s,p (I), we have Since u ∈ W Z ∞Z dh 3p (s−1)p ∆h u0 (x)p dx 1+(s−1)p n − p s h 0 I Z ∞Z dh ∆2h u ¯(x)p dx 1+sp . ≤ (3n)p h 0 I It suffices to take n so large that n(s−1)p −
3p sp
≥ 1.
Remark 5.29. If I = (−∞, b), the proof of Theorem 5.28 continues to work by considering ∆2−h u and ∆−h u0 . A drawback of Theorem 5.28 is that it was proved under the additional ˙ s,p (I) (see Step 2 in the proof). When u ∈ Lp (I), we hypothesis that u ∈ W can remove this extra condition. Theorem 5.30. Let I = R or I = (a, ∞), 1 ≤ p < ∞, and 1 < s < 2. If R ∞ R ∆2 u(x)p u ∈ Lp (I) is such that 0 I hh1+sp dxdh < ∞, then u ∈ W s,p (I) and for
5.4. Some Equivalent Seminorms
163
every ` > 0, Z ∞Z Z Z ∆2h u(x)p p (s−1)p 0 p −p u(x) dx + ` u (x) dx ` dxdh, h1+sp 0 I I I Z ∞Z Z ∞Z ∆2h u(x)p ∆h u0 (x)p dxdh dxdh. 1+(s−1)p h1+sp 0 I h 0 I Proof. Assume that I = (a, ∞) (the case I = R is simpler), and let δ > 0. Define I δ = (a + δ, ∞). Let 0 < ε < δ and for x ∈ I δ consider the mollification uε = u ∗ ϕε . We claim that uε ∈ W s,p (I δ ). Since u0ε = ϕ0ε ∗ u and u00ε = ϕ00ε ∗ u, it follows that u0ε and u00ε belongs to Lp (I δ ) (see [Leo22c]), and so uε ∈ W 2,p (I δ ). In turn, uε ∈ W s,p (I δ ) by Theorem 1.25 applied to u0ε . We can now apply Theorem 5.28 to uε in I δ to obtain Z ∞Z Z ∞Z ∆2h uε (y)p ∆h u0ε (y)p dydh (5.22) dydh 1+(s−1)p h1+sp 0 Iδ 0 Iδ h Z ∞Z ∆2h u(y)p dydh =: M p , 1+sp h 0 I where the last inequality follows from the fact that by linearity ∆2h uε = (∆2h u) ∗ ϕε and that kv ∗ ϕε kLp (I δ ) ≤ kvkLp (I) for all v ∈ Lp (I). It follows from Theorem 5.6 and inequality (5.22) that ku0ε − u0ρ kLp (I δ ) `−1 kuε − uρ kLp (I δ ) + `s−1 u0ε − u0ρ W s−1,p (I δ ) `−1 kuε − uρ kLp (I δ ) + `s−1 M for every 0 < ε, ρ < δ and ` > 0. Take ε = 1/n and ρ = 1/k. Letting n, k → ∞ and using the fact that uε → u in Lp (I δ ) shows that lim sup ku01/n − u01/k kLp (I δ ) `s−1 M. n,k→∞
It turn, letting ` → 0+ shows that {u01/n }n is a Cauchy sequence in Lp (I δ ). Therefore, it converges to a function v ∈ Lp (I δ ). This implies that v is the weak derivative of u (Why?) and shows that u ∈ W 1,p (I δ ). By selecting a subsequence, not relabeled, we may assume that u01/n → u0 pointwise L1 a.e. in I δ . Letting ε = 1/n → 0+ in (5.22) and using Fatou’s lemma on the left gives Z ∞Z Z ∞Z ∆2h u(y)p ∆h u0 (y)p dydh dydh. 1+(s−1)p h1+sp 0 Iδ h 0 I We now let δ → 0+ and use the Lebesgue dominated convergence theorem ˙ s,p (I) with to conclude that u ∈ W Z ∞Z Z ∞Z ∆2h u(y)p ∆h u0 (y)p dydh dydh. 1+(s−1)p h1+sp 0 I h 0 I
164
5. Higher Order Fractional Sobolev Spaces in One Dimension
By applying Theorem 5.6, we obtain that u ∈ W s,p (I) with Z ∞Z Z Z ∆2h u(x)p u0 (x)p dx `−p u(x)p dx + `(s−1)p dxdh h1+sp 0 I I I for every ` > 0.
Remark 5.31. An alternative to using Theorem 5.6 would have been to use a compactness argument (see Theorem 6.13). As an application of the last two theorems, we can prove that functions in a weighted Sobolev space belong to W s,p (R+ ). Theorem 5.32. Let 1 < p < ∞ and 1 < s < 2. Then for every u ∈ 2,p Wloc (R+ ), Z ∞Z ∞ 2 Z ∞ ∆h u(x)p x(2−s)p u00 (x)p dx. dxdh 1+sp h 0 0 0 2,p (R+ ), it is possible to find a representative u ¯ of u Proof. Since u ∈ Wloc 1 0 that is C and such that u ¯ is locally absolutely continuous (exercise). By the fundamental theorem of calculus (see [Leo22d]), Z h u ¯(x + 2h) − 2¯ u(x + h) + u ¯(x) = [¯ u0 (x + h + t) − u ¯0 (x + t)] dt 0
Z
hZ h
= 0
u00 (x + τ + t) dτ dt,
0
so by the change of variable ρ = τ + t, with dρ = dτ , Z 2h Z hZ h 00 2 u (x + τ + t) dτ dt ≤ h u00 (x + ρ) dρ. ¯(x) ≤ ∆h u 0
0
0
Hence, by the changes of variables 2h = η, Hardy’s inequality (Theorem 1.3), and the change of variables x + ρ = y, in this order, p Z ∞ 2 Z ∞ Z 2h ∆h u ¯(x)p dh 00 dh ≤ u (x + ρ) dρ 1+sp 1+(s−1)p h h 0 0 0 Z ∞ Z ∞ p ρ u00 (x + ρ)p dρ = (y − x)(2−s)p−1 u00 (y)p dy. 1+(s−1)p ρ 0 x Therefore, again using Tonelli’s theorem Z ∞Z ∞ 2 Z ∞Z ∞ ∆h u ¯(x)p dxdh (y − x)(2−s)p−1 u00 (y)p dydx. 1+sp h 0 0 Z0 ∞ x Z y 00 p = u (y) (y − x)(2−s)p−1 dxdy 0 Z ∞ 0 1 y (2−s)p u00 (y)p dy. = (2 − s)p 0
5.5. Hardy’s Inequality
165
5.5. Hardy’s Inequality In view of the results of Section 5.4, we prove a form of Hardy’s inequality for second order differences. We recall that if u ∈ W s,p (R+ ) with sp > 1, then u admits a continuous representative. If 0 < s < 1, this fact follows from Morrey’s embedding theorem (see Theorem 2.8), while if s ≥ 1, then u ∈ W 1,p (R+ ), so u admits an absolutely continuous representative (see [Leo22d]). Exercise 5.33. Let 1 ≤ p < ∞, 0 < s < 2, and u ∈ Cc2 (R). Prove that R ∞ u(2x)−2u(x)+u(0)p dx < ∞. xsp 0 Theorem 5.34 (Hardy’s inequality in W s,p ). Let 1 ≤ p < ∞, 0 < s < 2, s 6= 1, and u ∈ W s,p (R+ ). If 0 < s < 1/p, then Z ∞ Z ∞Z ∞ u(x)p ∆h u(x)p (5.23) dx dxdh. xsp h1+sp 0 0 0 If 1/p < s < 1 + 1/p, then Z ∞ Z ∞Z ∞ 2 ∆h u(x)p ¯ u(x) − u ¯(0)p (5.24) dx dxdh. xsp h1+sp 0 0 0 If 1/p < s < 2, then Z ∞ Z ∞Z ∞ 2 ∆h u(x)p ¯ u(2x) − 2¯ u(x) + u ¯(0)p (5.25) dx dxdh, xsp h1+sp 0 0 0 where u ¯ is the continuous representative of u. Proof. Step 1: Inequality (5.23) follows from Theorem 1.76. We now prove inequality (5.25). Assume that u ∈ Cc∞ (R+ ). Write u(2x) − 2u(x) + u(0) = [u(2y) − 2u(y) + u(0)] + [u(2x) − 2u(x + y) + u(2y)] + 2 [u(x + y) − 2u(y + x/2) + u(y)] − 2[u(2y) − 2u(y + x/2) + u(x)], or, equivalently, ∆2x u(0) = ∆2y u(0) + ∆2x−y u(2y) + 2∆2x/2 u(y) − 2∆2y−x/2 u(x). Averaging in y over (0, x) gives Z Z 1 x 2 1 x 2 2 ∆y u(0) dy + ∆x−y u(2y) dy ∆x u(0) = x 0 x 0 Z x Z 2 2 x 2 2 + ∆x/2 u(y) dy − ∆y−x/2 u(x) dy. x 0 x 0
166
5. Higher Order Fractional Sobolev Spaces in One Dimension
By H¨older’s inequality, and writing 1 = y − x/2(1+sp)/p y − x/2−(1+sp)/p , Z 1 x 2 ∆y−x/2 u(x) dy x 0 !1/p Z x 1/p0 Z x ∆2 p 1 y−x/2 u(x) (1+sp)p0 /p ≤ y − x/2 dy dy 1+sp x 0 0 y − x/2 !1/p Z x ∆2 p y−x/2 u(x) s ≤ Cx dy 1+sp 0 y − x/2 Rx 0 p−1 + 1 since 0 y − x/2(1+sp)p /p dy ≤ x(1+sp)/(p−1)+1 and 1+sp p−1 p − 1 = s. Similarly, 1 x
x
Z
∆2x−y u(2y) dy
∆2x−y u(2y)p dy x − y1+sp
x
Z
s
≤ Cx
0
0
∆2x−z/2 u(z)p
2x
Z
s
= Cx
x − z/21+sp
0
!1/p
!1/p dz
,
where we make the change of variables 2y = z. Hence, ∆2x u(0) x1/p ≤ x
0
x
Z
∆2y u(0)p dy
1/p
+ Cx
0
x − z/21+sp
0
0
2x1/p + x
∆2x−z/2 u(z)p
2x
Z
s
Z 0
x
∆2x/2 u(y)p dy
1/p
x
Z
s
+ Cx
0
!1/p dz
∆2y−x/2 u(x)p y − x/21+sp
!1/p dy
.
By Minkowski’s inequality, Z 0
∞
∆2x u(0)p dx xsp
1/p
Z
∞Z x
dx ∆2y u(0)p dy 1+sp x
≤ 0
0
∞Z x
Z +2 0
0
Z
∞Z
dx ∆2x/2 u(y)p dy 1+sp x
2x
+C 0
Z
0 ∞Z x
+C 0
0
1/p
∆2x−z/2 u(z)p x − z/21+sp
∆2y−x/2 u(x)p y − x/21+sp
1/p !1/p
dzdx !1/p
dydx
.
5.5. Hardy’s Inequality
167
By Tonelli’s theorem Z ∞ Z ∞ Z ∞ Z x 1 1 2 p 2 p ∆y u(0) ∆y u(0) dydx = dxdy 1+sp 1+sp x x y 0 0 0 Z ∞ 2 ∆y u(0)p 1 = dy, sp 0 y sp 1 and since sp < 1, by Exercise 5.33, Z ∞ 2 1/p ∆x u(0)p 1 dx 1− xsp (sp)1/p 0 !1/p !1/p Z ∞ Z ∞ ∆2 u(y)p Z ∞ Z ∞ ∆2 p x/2 x−z/2 u(z) ≤2 +C dydx dzdx x1+sp x − z/21+sp 0 0 0 0 !1/p Z ∞ Z ∞ ∆2 p y−x/2 u(x) +C dydx . y − x/21+sp 0 0
To conclude the proof in this case, it remains to use Tonelli’s theorem on each of the double integrals on the righthand side and make the appropriate changes of variables. We leave the details as an exercise. Hint: If h < 0, write ∆2h u(x) = ∆2−h u(x + 2h). Step 2: If 1/p < s < 1, inequality (5.24) follows from Theorems 1.57 and 1.76. Assume that 1 < s < 1 + 1/p and u ∈ Cc∞ (R+ ). For x > 0, write 1 1 u(x) − u(0) = [u(2x) − u(0)] − [u(2x) − 2u(x) + u(0)]. 2 2 Then by Minkowski’s inequality, Z ∞ 1/p Z ∞ 1/p ∆x u(0)p 1 ∆2x u(0)p dx ≤ dx xsp 2 xsp 0 0 Z ∞ Z ∞ 2 1/p 1/p ∆y u(0)p 1 ∆x u(0)p 2s + dx = 1+1/p dy 2 xsp y sp 2 0 0 Z ∞ 2 1/p 1 ∆x u(0)p + dx , 2 xsp 0 where we make the change of variables 2x = y. Hence, Z ∞ 1/p Z ∞ 2 1/p ∆x u(0)p 1 ∆x u(0)p 1 dx ≤ dx . 1 − 1+1/p−s xsp 2 xsp 2 0 0 We now use the Step 2 (note that we only used the fact that sp > 1) to R ∞ R ∞ ∆2h u(x)p bound the righthand side from above with C 0 0 dxdh. h1+sp Step 3: To remove the additional hypothesis that u ∈ Cc∞ (R+ ), we use Theorem 5.24. We leave the details as an exercise.
168
5. Higher Order Fractional Sobolev Spaces in One Dimension
Remark 5.35. In the case s = 1, the proof of Theorem 5.34 shows that Hardy’s inequality holds in the Besov space B 1,p (R+ ), which is the space of R ∞ R ∞ ∆2h u(x)p all functions u in Lp (R+ ) such that 0 0 dxdh < ∞. Functions h1+sp ∞ 1,p in Cc (R+ ) are dense in B (R+ ). For a proof we refer to [Leo17, Theorem 17.37], observing that Theorems 17.24 and 17.37 therein continue to hold when the domain R is replaced by R+ . For Hardy’s inequality in W 1,p (R+ ), see Corollary 1.5. Exercise 5.36. Let 1 < p < ∞ and 1/p < s < m + 1/p, s ∈ / N. Prove that for all u ∈ W s,p (R+ ), Z
∞
0
p ∆m x u(0) dx xsp
Z 0
∞Z ∞ 0
p ∆m h u(y) dydh. h1+sp
5.6. Truncation Since the absolute value is a Lipschitz continuous function, by the chain rule in Sobolev spaces (see [Leo22d]), if u ∈ W 1,p (I), then u ∈ W 1,p (I). However, if u ∈ W 2,p (I), then, in general, u ∈ / W 2,p (I). In this section, we show that if u ∈ W s,p (I), then u ∈ W s,p (I) for 0 < s < 1 + 1/p and that u ∈ / W s,p (I) for s ≥ 1 + 1/p. Theorem 5.37. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 0 < s < 1 + 1/p. If u ∈ W s,p (I), then u ∈ W s,p (I), with (5.26)
kukW s,p (I) kukW s,p (I) .
Proof. By Theorems 5.15 and 5.18 we can extend u to W s,p (R). Since u ∈ W 1,p (R), there exists a representative u ¯ of u which is absolutely continuous (see [Leo22d]). In turn, by the chain rule in Sobolev spaces, ¯ u0 (x) =
sgn u ¯(x)u0 (x) if u ¯(x) 6= 0, 0 if u ¯(x) = 0
for L1 a.e. x ∈ R (see [Leo22d]), which implies that u ∈ W 1,p (R). Consider the sets I+ := {x ∈ R : u ¯(x) > 0},
I− := {x ∈ R : u ¯(x) < 0},
I0 := {x ∈ R : u ¯(x) = 0}.
5.6. Truncation
169
Note that I+ and I− are open since u ¯ is continuous. By Tonelli’s theorem and by relabeling the variables, (5.27) Z Z R
Z Z u0 (x) − u0 (y)p ¯ u0 (x) − ¯ u0 (y)p dxdy = dxdy 1+(s−1)p 1+(s−1)p R x − y I+ I+ x − y Z Z u0 (x) − u0 (y)p + dxdy 1+(s−1)p I− I− x − y Z Z Z Z u0 (x) + u0 (y)p u0 (x)p +2 dxdy + 2 dxdy 1+(s−1)p 1+(s−1)p I− I+ x − y I0 I+ ∪I− x − y =: A + B + D + E.
We need to estimate D and E. We have Z Z u0 (x)p (5.28) dxdy D+E 1+(s−1)p I− ∪I0 I+ x − y Z Z u0 (y)p + dxdy =: A1 + A2 . 1+(s−1)p I− I+ ∪I0 x − y We will S only estimate A1 since A2 can be treated in a similar way. Write I+ = k Jk , where Jk are disjoint open intervals. If x ∈ I+ , then x ∈ Jk for some k, and so, Z Z 1 1 dy ≤ dy. 1+(s−1)p 1+(s−1)p I− ∪I0 x − y R\Jk x − y Case 1: If Jk has the form Jk = (ak , bk ), then Z Z ∞ Z ak dy dy dy (5.29) = + 1+(s−1)p 1+(s−1)p 1+(s−1)p R\Jk x − y bk (y − x) −∞ (x − y) 1 1 ≈ + . (bk − x)(s−1)p (x − ak )(s−1)p R Since u ¯(ak ) = u ¯(bk ) = 0 we have that Jk u0 (x) dx = 0. Hence, by Theorem 1.48 applied to u0 , we can find a function v ∈ W s−1,p (R) such that v = u0 in Jk and vW s−1,p (R) ≤ Cu0 W s−1,p (Jk ) , where C = C(p, s) > 0 is independent of Jk . It follows that Z Z I− ∪I0
Z u0 (x)p u0 (x)p dxdy dx 1+(s−1)p (s−1)p Jk x − y Jk (bk − x) Z u0 (x)p + dx (s−1)p Jk (x − ak ) Z bk Z ∞ v(x)p v(x)p dx + dx =: F + G. (s−1)p (s−1)p −∞ (bk − x) ak (x − ak )
170
5. Higher Order Fractional Sobolev Spaces in One Dimension
Since (s−1)p < 1, after the change of variables bk −x = t, we are in a position to apply Hardy’s inequality (Theorem 1.76) to the function w(t) = v(bk −x), t ∈ R+ , to obtain Z ∞ Z ∞Z ∞ w(t)p w(t) − w(τ )p F= dt dtdτ t(s−1)p t − τ 1+(s−1)p 0 0 0 Z Z Z bk Z bk v(x) − v(y)p u0 (x) − u0 (y)p dxdy dxdy. ≈ 1+(s−1)p 1+(s−1)p −∞ −∞ x − y Jk Jk x − y The integral G can be estimated in a similar way. Hence, Z Z Z Z u0 (x)p u0 (x) − u0 (y)p (5.30) dxdy dxdy. 1+(s−1)p 1+(s−1)p I− ∪I0 Jk x − y Jk Jk x − y Case 2: If Jk = (bk , ∞) or Jk = (−∞, ak ), then in (5.29) there is only one term on the right. This case is simpler because we can apply Hardy’s inequality directly to u0 instead of introducing the function v. Combining Cases S 1 and 2 gives (5.30) in all cases. Summing over k and recalling that I+ = k Jk , we get (5.31) Z Z Z Z u0 (x)p u0 (x) − u0 (y)p A1 = dxdy dxdy. 1+(s−1)p 1+(s−1)p I− ∪I0 I+ x − y I+ I+ x − y A similar estimate holds for A2 . Hence, by (5.27), (5.28), (5.30), and (5.31), we obtain that ¯ u0 W s−1,p (R) ≤ Cu0 W s−1,p (R) . This concludes the proof. Remark 5.38. Note that if I = R, then the proof of Theorem 5.37 shows that ˙ s,p (R). u0 W s−1,p (R) u0 W s−1,p (R) for all u ∈ W Corollary 5.39. Let I ⊆ R be an open interval, 1 ≤ p < ∞, and 0 < s < 1 + 1/p. If u ∈ W s,p (I), then u+ , u− ∈ W s,p (I), with ku± kW s,p (I) kukW s,p (I) . Here t+ = max{t, 0} and t− = max{−t, 0}. The following exercise shows that for s ≥ 1 + 1/p, Theorem 5.37 may fail. Exercise 5.40. Let 1 ≤ p < ∞ and u(x) = x, x ∈ (0, 1). Prove that u∈ / W 1+1/p,p ((0, 1)).
5.7. Embeddings As in Chapter 2, we are interested in understanding continuous inclusions of the type W s2 ,p2 (I) ,→ W s1 ,p1 (I), where now 1 ≤ p1 , p2 ≤ ∞ and 0 ≤ ˙ 0,p (R) := Lp (R) and, for m ∈ N, s1 , s2 < ∞. Here, when s = 0, we set W (5.32)
uW 0,p (R) := kukLp (R) ,
uW m,p (R) := ku(m) kLp (R) .
5.7. Embeddings
171
A scaling argument similar to the one at the beginning of Chapter 2 shows that if I = R, a necessary condition for the inequality uW s1 ,p1 (R) uW s2 ,p2 (R) to hold for all u ∈ W s2 ,p2 (R) is that s1 − p11 = s2 − p12 , as before. However, if we replace seminorms with full norms, then we can replace the equality s1 − p11 = s2 − p12 with the inequality s2 − p12 ≥ s1 − p11 . As already done throughout this chapter, for simplicity, we consider only the case s1 , s2 < 2. Theorem 5.41. Let I ⊆ R be an open interval, and let 1 ≤ p2 < p1 ≤ ∞ and 0 ≤ s1 < s2 < 2 be such that 1 1 (5.33) s2 − ≥ s1 − . p2 p1 s ,p 2 2 Then for all u ∈ W (I), kukW s1 ,p1 (I) I kukW s2 ,p2 (I) ,
(5.34)
with the following exceptions: (i) if (s2 − m2 )p2 = 1 and s1 = m2 , where m2 := bs2 c, then we require p1 < ∞; (ii) if s2 = m2 + 1 and s1 − m2 > 0, we require p2 > 1. Proof. Step 1: If s2 ≤ 1, then m2 = 0, and the result follows from Corollary 2.30 applied to u. Step 2: Assume 1 < s2 < 2. Since u ∈ W 1,p2 (I), for every p2 ≤ q ≤ ∞ we have (see [Leo22d]), kukLq (I) I kukW 1,p2 (I) .
(5.35)
If s1 ≥ 1, then s2 − 1 − p12 ≥ s1 − 1 − p11 , and so we can apply Corollary 2.30 to u0 , to obtain ku0 kW s1 −1,p1 (I) I ku0 kW s2 −1,p2 (I) . Assume next that s1 < 1. We consider three cases. If (s2 − 1)p2 < 1, we can apply the Sobolev–Gagliardo–Nirenberg embedding theorem (Corollary 2.3) to u0 to get ku0 k (p2 )∗s −1 I ku0 kW s2 −1,p2 (I) , where (p2 )∗s2 −1 = p2 /(1−(s2 −1)p2 ) > L
2
(I)
∗
1. Together with (5.35), this gives that u ∈ W 1,(p2 )s2 −1 (I). Since 1 − 1/(p2 )∗s2 −1 = s2 − 1/p2 ≥ s1 − 1/p1 by (5.33), we can apply Corollary 2.30, with s2 and p2 replaced by 1 and (p2 )∗s2 −1 , to obtain kukW s1 ,p1 (I) I kuk
W
1,(p2 )∗ s2 −1
(I)
I kukW s2 ,p2 (I) .
If (s2 − 1)p2 = 1, then by Theorem 2.5 and Exercise 2.7 applied to u0 , for every p2 ≤ q < ∞, ku0 kLq (I) I ku0 kW s2 −1,p2 (I) , and so, also by (5.35), u ∈ W 1,q (I). Since s1 < 1, by taking q sufficiently large, we have that
172
5. Higher Order Fractional Sobolev Spaces in One Dimension
1 − 1/q ≥ s1 − 1/p1 . Therefore, we can apply Corollary 2.30, with s2 and p2 replaced by 1 and q, to obtain kukW s1 ,p1 (I) I kukW 1,q (I) I kukW s2 ,p2 (I) . Finally, if (s2 − 1)p2 > 1, then by Morrey’s embedding theorem (Theorem 2.8), u0 ∈ L∞ (I), with ku0 kL∞ (I) I ku0 kW s2 −1,p2 (I) , and so u0 ∈ Lp1 (I) ∩ L∞ (I). In turn, u0 ∈ Lq (I), for every p2 ≤ q ≤ ∞, and so, we can proceed as in the case (s2 − 1)p2 = 1. Remark 5.42. Exercises 2.4 and 2.23 provide the counterexample for items (i) and (ii), respectively. As a consequence of Theorem 5.41, we have that W s,p (I) is an algebra for p > 1. Corollary 5.43 (Product). Let I ⊆ R be an open interval, 1 < p < ∞, and 1 < s < 2. If u, v ∈ W s,p (I), then uv ∈ W s,p (I), with kuvkW s,p (I) I kukW s,p (I) kvkW s,p (I) . Proof. By extending u and v (see Theorems 5.15 and 5.18), without loss of generality, we can assume that I = R. Since u, v ∈ W 1,p (R), we have kukL∞ (R) I kukW 1,p (R) and kvkL∞ (R) I kvkW 1,p (R) (see [Leo22d]). By the product rule we can write (¯ uv¯)0 (x + h) − (¯ uv¯)0 (x) = u0 (x + h)¯ v (x + h) + u ¯(x + h)v 0 (x + h) − u0 (x)¯ v (x) − u ¯(x)v 0 (x) = v¯(x + h)∆h u0 (x) + u ¯(x + h)∆h v 0 (x) + v 0 (x)∆h u ¯(x) + u0 (x)∆h v¯(x). Hence, by Minkowski’s inequality’s we have 1/p Z ∞ Z ¯ v (x + h)∆h u0 (x)p 0 dxdh (uv) W s−1,p (R) ≤ h1+(s−1)p 0 R Z ∞ Z 1/p ¯ u(x + h)∆h v 0 (x)p + dxdy h1+(s−1)p 0 R Z ∞ Z 1/p v 0 (x)∆h u ¯(x)p + dxdh h1+(s−1)p 0 R Z ∞ Z 1/p u0 (x)∆h v¯(x)p + =: A + B + C + D. dxdy h1+(s−1)p 0 R Observe that A ≤ u0 W s−1,p (R) kvkL∞ (R) kukW s,p (R) kvkW s,p (R) and similarly, B kukW s,p (R) kvkW s,p (R) . If (s − 1)p > 1, then by Morrey’s embedding theorem (Theorem 2.8), ku0 kL∞ (R) ku0 kW s−1,p (R) and
5.7. Embeddings
173
kv 0 kL∞ (R) kv 0 kW s−1,p (R) , and so, by Theorem 1.25, C ≤ uW s−1,p (R) kv 0 kL∞ (R) kukW 1,p (R) kv 0 kL∞ (R) kukW s,p (R) kvkW s,p (R) and, similarly, D kukW s,p (R) kvkW s,p (R) . When (s − 1)p ≤ 1, the estimates for C and D need to be changed since we no longer know that u0 and v 0 are bounded. If (s − 1)p < 1, we use H¨ older’s inequality to bound ∞ Z
Z C≤ 0
pq 0
v 0 (x)
R
1/q0 Z dx
1/q pq ∆h u(x)) dx
R
dh
!1/p .
h1+(s−1)p ∗
p 1 , we get pq 0 = p∗s−1 = 1−(s−1)p . Therefore, v 0 ∈ Lps−1 (R) Taking q = (s−1)p by the Sobolev–Gagliardo–Nirenberg embedding theorem (Theorem 2.2) applied to v 0 , with kv 0 kLpq0 (R) kvkW s,p (R) . On the other hand, by Lemma 2.26, with p2 = p, s2 = 1/p, p1 = pq = 1/(s − 1), s1 = s − 1, we have 1/q Z ∞ Z Z ∞Z dh dh ∆h u(x)pq dx ∆h u(x)p dx 2 . 1+(s−1)p h h 0 R 0 R
Hence, C kvkW 1,p (R) uW 1/p,p (R) kvkW 1,p (R) kukW 1,p (R) , where the last inequality follows by Theorem 1.25. If (s − 1)p = 1, by Theorem 2.5 applied to v 0 , kv 0 kLr (R) kvkW s,p (R) for every p ≤ r < ∞. Taking q = 1/(1 − ε), where 0 < ε < 1 is small, as in the previous case we can apply Lemma 2.26, with p2 = p, s2 = (1 + ε)/p, p1 = pq, and s1 = s − 1, to get 1/q Z ∞ Z Z ∞Z dh dh pq ∆h u(x) dx ∆h u(x)p dx 2+ε . 1+(s−1)p h h 0 R 0 R As before, uW (1+ε)/p,p (R) kukW 1,p (R) by Theorem 1.25, and so C ≤ kvkW 1,p (R) kukW 1,p (R) .
Exercise 5.44. Let I ⊆ R be an open interval, 1 < s < ∞, with s ∈ / N, ˙ s,p (I) m := bsc, and 1/(s − m) < p < ∞. Prove that every function u ∈ W admits a representative u ¯ which is of class C m with u ¯(m) H¨older continuous with exponent α = s − m − 1/p and such that ¯ u(m) C 0,α (I) u(m) W s−m,p (I) . s,p Prove also that if u ∈ W (I), then for every 0 < ` < L1 (I), 1 0 ku(k) kLp (I) + `1/p ku(k+1) kLp (I) , `1/p 1 1/p ku(m) kLp (I) + `s−m−1/p u(m) W s−m,p (I) `
k¯ u(k) kC 0 (I) k¯ u(m) kC 0 (I)
for every k = 0, . . . , m − 1, where u(0) := 0.
174
5. Higher Order Fractional Sobolev Spaces in One Dimension
Exercise 5.45. Let I ⊆ R be an open interval, 1 < s < ∞, with s ∈ / N, m := bsc, and 1/(s − m) < p < ∞. Prove that if u, v ∈ W s,p (I), then uv ∈ W s,p (I).
5.8. Interpolation Inequalities In this section we prove the Gagliardo–Nirenberg type inequality (5.36)
uW s,p (R) uθW s1 ,p1 (R) u1−θ W s2 ,p2 (R)
˙ s1 ,p1 (RN ) ∩ W ˙ s2 ,p2 (RN ). Here, 0 < θ < 1, 1 ≤ p, p1 , p2 ≤ ∞, for all u ∈ W and s, s1 , s2 ≥ 0, and we are using notation (5.32). We leave it as an exercise to check (see Section 2.3) that a necessary condition for (5.36) to hold for ˙ s1 ,p1 (R) ∩ W ˙ s2 ,p2 (R) is all u ∈ W 1 1 1 (5.37) s − = θ s1 − + (1 − θ) s2 − . p p1 p2 We will prove the following theorem. Theorem 5.46 (Gagliardo–Nirenberg). Let I ⊆ R be an unbounded open interval, 1 ≤ p1 , p2 < ∞, 0 ≤ s1 < s2 ≤ 2, and 0 < θ < 1 be such that (5.38)
θ 1−θ 1 = + , p p1 p2
s = θs1 + (1 − θ)s2 .
˙ s1 ,p1 (I) ∩ W ˙ s2 ,p2 (I), Then for all u ∈ W (5.39)
uW s,p (I) uθW s1 ,p1 (I) u1−θ W s2 ,p2 (I) ,
with the following exceptions: (i) if s1 = 0 or s1 = 1, we assume that p1 > 1; (ii) if s2 = 1 or s2 = 2, we assume that p2 > 1. We begin with the case s1 = 0 < s = 1 < s2 < 2. Theorem 5.47. Let I ⊆ R be an unbounded open interval, 1 < p1 < ∞, 1 ≤ p2 < ∞, 1 < s2 < 2, and 0 < θ < 1. Then (5.40)
ku0 kLp (I) kukθLp1 (I) u1−θ W s2 ,p2 (I)
˙ s2 ,p2 (I), where for all u ∈ Lp1 (I) ∩ W (5.41)
1 θ 1−θ = + , p p1 p2
1 = (1 − θ)s2 .
Proof. If I = (a, ∞) or I = (−∞, b), we use Remark 5.16 to extend u to a ˙ s2 ,p2 (R), with kukLp1 (R) kukLp1 (I) function, still denoted u, in Lp1 (R) ∩ W and uW s2 ,p2 (R) uW s2 ,p2 (I) .
5.8. Interpolation Inequalities
175
Step 1: Using Lemma 5.7 and H¨ older’s inequality (if p2 > 1), for every x ∈ R and h > 0, we get (5.42) Z 1 h t1/p2 +s2 −1 0 u (x) − u0 (x + t) dt u(x + t) dt + h −h t1/p2 +s2 −1 −h 1/p2 Z 1 u0 (x) − u0 (x + t)p2 s2 −1 M(u)(x) + h dt . h t1+(s2 −1)p2 R
1 u (x) ≤ 2 h 0
Z
h
By Exercise 1.26, we have 0
u (x) (M(u)(x))
1−1/s2
Z R
1/(s2 p2 ) u0 (x) − u0 (x + t)p2 . dt t1+(s2 −1)p2
Using the fact that 1 = (1 − θ)s2 , raising both sides to power p, and integrating in x over R gives Z (1−θ)p/p2 Z Z u0 (x) − u0 (x + t)p2 0 p θp u (x) dx (M(u)(x)) dt dx. t1+(s2 −1)p2 R R R (1−θ)p By (5.41), 1 = θp older’s p1 + p2 , which implies that θp < p1 . Hence, by H¨ inequality with exponents p1 /(θp) and p2 /[(1 − θ)p], the righthand side can be majorized by (1−θ)p/p2 Z θp/p1 Z Z u0 (x) − u0 (x + t)p2 p1 dtdx C (M(u)(x)) dx . t1+(s2 −1)p2 R R R
To conclude, we use the continuity of the maximal operator M in Lp1 (R) for p1 > 1 (see [Leo22c]). When I = (a, b) Theorem 5.47 fails. Indeed, by taking u(x) = mx + q, we have that the righthand side of (5.40) is zero. Corollary 5.48. Let I = (a, b), 1 ≤ p2 < p1 < ∞, 1 < s2 < 2, and 0 < θ < 1. Then ku0 kLp (I) ` kukLp1 (I) + kukθLp1 (I) u1−θ W s2 ,p2 (I) ˙ s2 ,p2 (I), where ` = b − a and p is given in (5.41). for all u ∈ Lp1 (I) ∩ W Proof. By Corollary 5.19 we can extend u to a function in the space Lp1 (J)∩ ˙ s2 ,p2 (J), where J = (a − (b − a), b + (b − a)), with W kukLp1 (J) kukLp1 (I) ,
uW s2 ,p2 (J) uW s2 ,p2 (I) .
We proceed as in the proof of Theorem 5.47 to obtain (5.42), except that we take x ∈ I and 0 < h < b − a, M(u) should be replaced by M(uχJ ), and R R` R by −` , where ` := b − a. Note that if x ∈ I and 0 < h < b − a, then
176
5. Higher Order Fractional Sobolev Spaces in One Dimension
x + t ∈ J for all −h < t < h. Hence, u(x + t) and u0 (x + t) are well defined. Using Exercise 1.26(ii), we get u0 (x)
1 M(uχJ )(x) ` + (M(uχJ )(x))
Z
1−1/s2
`
−`
1/(s2 p2 ) u0 (x) − u0 (x + t)p2 dt . t1+(s2 −1)p2
We now raise both sides to power p and integrate in x over I. By H¨ older’s inequality and the continuity of the maximal operator M in Lp1 (R) for p1 > 1, Z p/p1 Z 0 (M(uχJ ))p dx ≤ `1/(p1 /p) (M(uχJ ))p1 dx I
I
≤`
1/(p1 /p)0
Z
p1
p/p1
u dx
.
J
We can now continue as before to obtain Z p/p1 Z 0 p p1 u  dx ` u dx J
I
Z
θp/p1Z Z
p1
`
u dx
+ J
I
−`
(1−θ)p/p2 u0 (x) − u0 (x + t)p2 dtdx . t1+(s2 −1)p2
Exercise 5.49. Let I = (a, b), 1 ≤ p1 ≤ p2 < ∞, 1 < s2 < 2, and 0 < θ < 1. Prove that ku0 kLp2 (I) ` kukLp1 (I) + uW s2 ,p2 (I) . Hint: Start from (5.1) with p and s replaced by p2 and s2 , and raise both sides to power p1 /p2 . Next we consider the case 0 < s1 < s = 1 < s2 < 2. Theorem 5.50. Let I ⊆ R be an unbounded open interval, 1 ≤ p1 < ∞, 1 ≤ p2 < ∞, 0 < s1 < 1, 1 < s2 < 2, and 0 < θ < 1. Then for all ˙ s1 ,p1 (I) ∩ W ˙ s2 ,p2 (I), ku0 kLp (I) uθ s ,p u1−θ u∈W W 1 1 (I) W s2 ,p2 (I) , where (5.43)
1 θ 1−θ = + , p p1 p2
1 = θs1 + (1 − θ)s2 .
Proof. Reasoning as in Step 1 of the proof of Theorem 5.15, we can extend ˙ s2 ,p2 (R), with uW s1 ,p1 (R) ˙ s1 ,p1 (R)∩ W u to a function, still denoted u, in W uW s1 ,p1 (I) and uW s2 ,p2 (R) uW s2 ,p2 (I) . For x ∈ R and h > 0, write Z
h
Z u(x+t) sgn t dt =
−h
0
h
Z 0 Z h u(x+t) dt− u(x+t) dt = (u(x+t)−u(x−t)) dt, −h
0
5.8. Interpolation Inequalities
177
where in the last equality we made the change of variables τ = −t. Using this identity, it follows from Lemma 5.7 that (5.44) Z h Z h 1 1 0 u (x) = 2 (u(x + t) − u(x − t)) dt + 2 (h − t)(u0 (x) − u0 (x + t)) dt h 0 h −h 1/p1 +s1
1/p2 +s2 −1
t t older’s for x ∈ R and h > 0. Writing 1 = t 1/p1 +s1 and 1 = t1/p2 +s2 −1 , by H¨ inequality, Z Z h 1 1 h 0 u (x) − u0 (x + t) dt u0 (x) ≤ 2 u(x + t) − u(x − t) dt + h −h h −h Z 1/p1 1 u(x + t) − u(x − t)p1 1−s1 dt h t1+s1 p1 R 1/p2 Z u0 (x) − u0 (x + t)p2 s2 −1 dt , +h t1+(s2 −1)p2 R
where we use the fact that, for 0 < σ < 1 and 1 < q < ∞, Z h 0 0 (5.45) t(1/q+σ)q dt ≈ h(1+σ)q . 0
Using Exercise 1.26 we get Z (s2 −1)/[(s2 −s1 )p1 ] u(x + t) − u(x − t)p1 0 u (x) dt t1+s1 p1 R Z (1−s1 )/[(s2 −s1 )p2 ] u0 (x) − u0 (x + t)p2 × dt . t1+(s2 −1)p R Using the fact that s2 − 1 = θ(s2 − s1 ), raising both sides to power p, and integrating in x over R gives θp/p1 Z Z Z u(x + t) − u(x − t)p1 u0 (x)p dx dt t1+s1 p1 R R R Z (1−θ)p/p2 u0 (x) − u0 (x + t)p2 dx. × dt t1+(s2 −1)p2 R (1−θ)p By (5.41), 1 = θp p1 + p2 , which implies that θp < p1 . Hence, we can use H¨older’s inequality with exponents p1 /(θp) and p2 /[(1 − θ)p] to bound from above the righthand side of the last inequality with Z Z θp/p1 u(x + t) − u(x − t)p1 C dtdx t1+s1 p1 R R Z Z (1−θ)p/p2 u0 (x) − u0 (x + t)p2 × dtdx . t1+(s2 −1)p2 R R
178
5. Higher Order Fractional Sobolev Spaces in One Dimension
Exercise 5.51. Let I = (a, b), 1 ≤ p1 < ∞, 1 ≤ p2 < ∞, 0 < s1 < 1, 1 < s2 < 2, and 0 < θ < 1. Prove that, ku0 kLp (I) ` uW s1 ,p1 (I) + uθW s1 ,p1 (I) u1−θ W s2 ,p2 (I) ˙ s1 ,p1 (I) ∩ W ˙ s2 ,p2 (I), where ` = b − a and p is given in (5.43). for all u ∈ W Hint: Reason as in the proof of Corollary 5.48. Lastly, we treat the case 0 < s1 < s = 1 < s2 = 2. Exercise 5.52. Let I = (−r, r), J = (−2r, 2r), 1 ≤ p < ∞, and u ∈ ˙ 2,p (I). W ˙ 2,p (J) such that v = u in I and (i) Prove that there exists v ∈ W 00 00 kv kLp (J) ku kLp (I) . Hint: See Exercise 5.17 and Corollary 5.19. ˙ 2,p (I)∩ W ˙ s,q (I), where 1 ≤ q < ∞, and 0 < s < (ii) Prove that if u ∈ W ˙ s,q (J), 1, then the function v constructed in item (i) belongs to W with vW s,q (J) uW s,q (I) . (ii) Repeat items (i) and (ii) in the case in which I is an open interval of infinite length and J = R. Theorem 5.53. Let I ⊆ R be an unbounded open interval, 1 ≤ p1 < ∞, 1 < ˙ s1 ,p1 (I) ∩ W ˙ 2,p2 (I), p2 < ∞, 0 < s1 < 1, and 0 < θ < 1. Then for all u ∈ W ku0 kLp (I) uθW s1 ,p1 (I) ku00 k1−θ Lp2 (I) , where (5.46)
1 θ 1−θ = + , p p1 p2
1 = θs1 + (1 − θ)2.
˙ s1 ,p1 (R)∩ Proof. In view of Exercise 5.52, we can extend u to a function in W ˙ 2,p2 (R), with W (5.47)
uW s1 ,p1 (R) uW s1 ,p1 (I) ,
ku00 kLp2 (R) ku00 kLp2 (I) .
Using (5.44) for every x ∈ R and h > 0, we get Z h Z 1 1 h 0 0 u (x) ≤ 2 u(x + t) − u(x − t) dt + u (x) − u0 (x + t) dt. h −h h −h ˙ 1,p2 (R), u0 admits an absolute continuous representative v¯ (see Since u0 ∈ W [Leo22d]). Thus, by the fundamental theorem of calculus (see [Leo22d]), and (2.22), Z x+t Z t x+t 00 ¯ v (x + t) − v¯(x) ≤ u00 (y) dy = u (y) dy t M(u00 )(x). t x−t x−t
5.8. Interpolation Inequalities
Hence, writing 1 =
by H¨older’s inequality and (5.45),
Z 1 h ¯ v (x + t) − v¯(x) dt u(x + t) − u(x − t) dt + h −h −h Z 1/p1 1 u(x + t) − u(x − t)p1 1−s1 dt + h M(u00 )(x). 1+s1 p1 h t R
1 u (x) ≤ 2 h 0
t1/p1 +s1 , t1/p1 +s1
179
Z
h
By Exercise 1.26 we get Z 1/[(2−s1 )p1 ] u(x + t) − u(x − t)p1 0 u (x) dt (M(u00 )(x))(1−s1 )/(2−s1 ) . 1+s1 p1 t R Using the fact that (2 − s1 )θ = 1, raising both sides to power p, and integrating in x over R gives θp/p1 Z Z Z u(x + t) − u(x − t)p1 0 p dt (M(u00 )(x))(1−θ)p dx. u (x) dx 1+s1 p1 t R R R (1−θ)p By (5.48), 1 = θp p1 + p2 , which implies that θp < p1 . Hence, we can use H¨older’s inequality with exponents p1 /(θp) and p2 /[(1 − θ)p] to bound from above the righthand side of the last inequality with θp/p1 Z (1−θ)p/p1 Z Z u(x + t) − u(x − t)p1 00 p2 dtdx (M(u )) dx . C t1+s1 p1 R R R
To conclude, we use (5.47) and the continuity of the maximal operator M in Lp2 (R) for p2 > 1 (see [Leo22c]). Exercise 5.54. Let I = (a, b), 1 ≤ p1 < ∞, 1 < p2 < ∞, 0 < s1 < 1, and 0 < θ < 1. Prove that ku0 kLp (I) ` uW s1 ,p1 (I) + uθW s1 ,p1 (I) ku00 k1−θ Lp2 (I) ˙ s1 ,p1 (I) ∩ W ˙ 2,p2 (I), where ` = b − a and p is given in (5.48). for all u ∈ W We are finally ready to prove Theorem 5.46. Proof of Theorem 5.46. Let 1 < p1 , p2 < ∞, 0 ≤ s1 < s2 ≤ 2, and 0 < θ < 1 satisfy (5.38). If s2 ≤ 1, then we can use Theorem 2.32. Thus, assume that 1 < s2 ≤ 2. If s1 ≥ 1, then s − 1 = θ(s1 − 1) + (1 − θ)(s2 − 1) by (5.38), and so we can apply Theorem 2.32 to u0 to obtain uW s,p (I) = u0 W s−1,p (I) u0 θW s1 −1,p1 (I) u0 1−θ W s2 −1,p2 (I) = uθW s1 ,p1 (I) u1−θ W s2 ,p2 (I) . It remains to study the case 0 ≤ s1 < 1 and 1 < s2 ≤ 2.
180
5. Higher Order Fractional Sobolev Spaces in One Dimension
¯
¯
θ Let θ¯ ∈ (0, 1) and 1 < p¯ < ∞ be such that p1¯ = pθ1 + 1− p2 and 1 = ¯ 1 +(1−θ)s ¯ 2 . By Theorems 5.47, 5.50, and 5.53, or the Gagliardo–Nirenberg θs interpolation theorem in Sobolev spaces (see [Leo22d]), we have ¯
¯
θ ku0 kLp¯(I) uθW s1 ,p1 (I) u1− W s2 ,p2 (I) .
(5.48)
If s = 1, then p¯ = p and θ¯ = θ, and the proof is complete. If 0 < s < 1, let ˆ 1 + (1 − θ)1. ˆ θˆ ∈ (0, 1) be such that s = θs Then ˆ 1 + (1 − θ)1 ˆ = θs ˆ 1 + (1 − θ)( ˆ θs ¯ 1 + (1 − θ)s ¯ 2) s = θs ˆ θ]s ¯ 1 + (1 − θ)(1 ˆ − θ)s ¯ 2, = [θˆ + (1 − θ) ˆ θ. ¯ In turn, which implies that θ = θˆ + (1 − θ) ˆ θ¯ ¯ ˆ ˆ 1 θ 1−θ θˆ + (1 − θ) ˆ 1 − θ = θ + 1 − θ. = + = + (1 − θ) p p1 p2 p1 p¯ p1 p1 p¯ By Theorem 2.32 and (5.48), ˆ
ˆ
θ uW s,p (I) uθW s1 ,p1 (I) ku0 k1− Lp¯(I) ˆ
¯
¯
ˆ
1−θ 1−θ uθW s1 ,p1 (I) (uθW s1 ,p1 (I) uW s2 ,p2 (I) )
uθW s1 ,p1 (I) u1−θ W s2 ,p2 (I) . ˆ ˆ 2. On the other hand, if 1 < s < 2, let θˆ ∈ (0, 1) be such that s = θ1+(1− θ)s Then ˆ + (1 − θ)s ˆ 2 = θ( ˆ θs ¯ 1 + (1 − θ)s ¯ 2 ) + (1 − θ)s ˆ 2 s = θ1 ¯ 1 + [θ(1 ˆ − θ) ¯ + (1 − θ)]s ˆ 2, = θˆθs ¯ In turn, which implies that θ = θˆθ. 1 θ 1−θ 1 1 − θ¯ 1 − θˆθ¯ θˆ 1 − θˆ ˆ = + =θ − = + . + p p1 p2 p¯ p2 p2 p¯ p2 By Theorem 2.32 applied to u0 and (5.48), ˆ
ˆ
ˆ
ˆ
θ θ uW s,p (I) = u0 W s−1,p (I) ku0 kθLp¯(I) u0 1− = ku0 kθLp¯(I) u1− W s2 ,p2 (I) W s2 −1,p2 (I) ¯
¯
ˆ
ˆ
θ 1−θ 1−θ θ θ (uθW s1 ,p1 (I) u1− W s2 ,p2 (I) ) uW s2 ,p2 (I) = uW s1 ,p1 (I) uW s2 ,p2 (I) .
Remark 5.55. Using compactly supported smooth wavelets in place of Haar functions, it is possible to extend Theorem 3.15 to the case s2 > 1 (see [Coh00]) and Theorem 3.17 to the case s1 > 1. In turn, one can allow p1 = 1 or p2 = 1 in Theorem 5.46. To be precise, the exception (i) can be removed, while in exception (ii) if p2 = 1 and s2 = 1 or s2 = 2, then we require s2 − s1 > 1 − p11 (see [BM18]).
5.9. Notes
181
5.9. Notes Theorem 5.37 has been adapted from the review paper of Mironescu [Mir18]. I worked on Section 5.8 with the undergraduate student Grant Yu.
Part 2
Fractional Sobolev Spaces
Chapter 6
Fractional Sobolev Spaces I have discovered a truly marvelous proof of this, which however the margin is not large enough to contain. — Pierre de Fermat
In this chapter we introduce fractional Sobolev spaces W s,p (Ω), where Ω ⊆ RN is an open set, 1 ≤ p ≤ ∞, and 0 < s < 1. In what follows, k · k stands for the Euclidean norm in RN , LN is the N th dimensional Lebesgue measure, and HN −1 is the (N − 1)th dimensional surface measure. We write SN −1 := ∂B(0, 1), αN = LN (B(0, 1)) and βN = HN −1 (SN −1 ) for the measure of the unit ball and the surface area of the unit sphere, respectively. We recall that βN = N αN . Given x = (x1 , . . . , xN ) ∈ RN and i = 1, . . . , N , we denote by x0i ∈ RN −1 the vector obtained from x by removing the ith component xi . With an abuse of notation, we write (6.1)
x = (x0i , xi ) ∈ RN −1 × R.
When i = N , we will also use the notation (6.2)
x = (x0 , xN ) ∈ RN −1 × R.
We recall that we are using the notation from (0.2) and (0.3).
6.1. Definition and Main Properties Definition 6.1. Let Ω ⊆ RN be an open set, 1 ≤ p ≤ ∞, and 0 < s < 1. A function u ∈ Lp (Ω) belongs to the fractional Sobolev space W s,p (Ω) if kukW s,p (Ω) := kukLp (Ω) + uW s,p (Ω) < ∞, 185
186
6. Fractional Sobolev Spaces
where Z Z (6.3)
uW s,p (Ω) :=
Ω
Ω
u(x) − u(y)p dxdy kx − ykN +sp
1/p
if 1 ≤ p < ∞, and (6.4)
uW s,∞ (Ω) := esssup
x,y∈Ω, x6=y
u(x) − u(y) kx − yks
if p = ∞. Definition 6.2. Let Ω ⊆ RN be an open set, 1 ≤ p ≤ ∞, and 0 < s < 1. A function u ∈ Lploc (Ω) belongs to the homogeneous fractional Sobolev space ˙ s,p (Ω) if u s,p W W (Ω) < ∞. Definition 6.3. Let Ω ⊆ RN be an open set, 1 ≤ p < ∞, and 0 < s < 1. s,p We define the space Wloc (Ω) as the space of all functions u ∈ Lploc (Ω) such that for every open set U b Ω, the restriction of u to U belongs to W s,p (U ). For vectorvalued functions u : Ω → RM , we can give analogous defini˙ s,p (Ω; RM ), and W s,p (Ω; RM ) by replactions of the spaces W s,p (Ω; RM ), W loc ing the absolute value in (6.3) and (6.4) with the norm in RM . ˙ s,∞ (Ω), Remark 6.4. As in Exercise 1.15, it is possible to show that if u ∈ W 0 < s < 1, then u admits a representative u ¯ that is H¨ older continuous with exponent s and such that uW s,∞ (Ω) =
sup x,y∈Ω, x6=y
¯ u(x) − u ¯(y) = ¯ uC 0,s (Ω) . kx − yks
If in addition, u ∈ W s,∞ (Ω), then kukL∞ (Ω) = supx∈Ω ¯ u(x). Thus, we can s,∞ 0,s identify W (Ω) with the space C (Ω). For this reason, in most of what follows, we will restrict our attention to the case 1 ≤ p < ∞. Remark 6.5. When Ω = RN , by the change of variables y = x + h, so that dy = dh we can write Z Z Z Z u(x) − u(y)p ∆h u(x)p dxdy = dxdh, (6.5) N +sp N +sp RN RN kx − yk RN RN khk where ∆h u(x) := u(x + h) − u(x). Theorem 6.6. Let Ω ⊆ RN be an open set, 1 ≤ p < ∞, and 0 < s < 1. Then the space W s,p (Ω) is a Banach space. Proof. The proof is the same as the one of Theorem 1.20.
6.1. Definition and Main Properties
187
Remark 6.7. When 0 < s < 1 and p = 2, the space W s,2 (Ω) is a Hilbert space with the inner product Z Z Z (u(x) − u(y))(v(x) − v(x)) u(x)v(x) dx + (u, v) 7→ dxdy. kx − ykN +sp Ω Ω Ω ˙ s,2 (Ω), and H s (Ω) := In this case we write H s (Ω) := W s,2 (Ω), H˙ s (Ω) := W loc s,2 Wloc (Ω). In the following exercises we use the Fourier transform F to show that in the case p = 2 the fractional seminorm takes a simple form. We recall that S(RN ) is the space of rapidly decreasing realvalued functions (see [Leo22c]). Exercise 6.8. Let 0 < s < 1, and for every x ∈ RN consider the integral Z 1 − cos[2π(x · y)] I(x) := dy. kykN +2s RN (i) Prove that the integral is well defined. (ii) Prove that I(x) = I(kxke1 ), where e1 = (1, 0, . . . , 0). (iii) Prove that there exists a constant C = C(N, s) > 0 such that I(x) = Ckxk2s . Exercise 6.9. Let 0 < s < 1 and let u ∈ S(RN ). (i) Prove that Z Z RN
u(x + h) − u(x)2 dxdh khkN +2s RN 2 Z Z u(· + h) − u(·) dxdh. = (x) F N N khkN/2+s R
R
(ii) Prove that 2 Z Z F u(· + h) − u(·) (x) dxdh N/2+s khk RN RN Z Z 1 − cos[2π(h · x)] =2 F(u)(x)2 dxdh. N +2s khk N N R R (iii) Prove that there exists a constant C = C(N, s) > 0 such that Z Z Z u(x + h) − u(x)2 dxdh = C kxk2s F(u)(x)2 dx. N +2s khk N N N R R R Exercise 6.10 (The Besov space B 1,1 ). Let B 1,1 (RN ) be the set of all functions u ∈ L1 (RN ) such that (6.6)
kukB 1,1 (RN ) := kukL1 (RN ) + uB 1,1 (RN ) < ∞,
188
6. Fractional Sobolev Spaces
where Z (6.7)
uB 1,1 (RN ) :=
u(x + 2h) − 2u(x + h) + u(x) RN
dh . khkN +1
Prove that B 1,1 (RN ) is a Banach space. Theorem 6.11 (Reflexivity). Let Ω ⊆ RN be an open set, 1 < p < ∞, and 0 < s < 1. Then the space W s,p (Ω) is reflexive. Proof. Since 1 < p < ∞, the space Lp (Ω) × Lp (Ω × Ω), endowed with the norm (u, v) 7→ kukLp (Ω) + kvkLp (Ω×Ω) , is a reflexive Banach space. Consider the application T : W s,p (Ω) → Lp (Ω) × Lp (Ω × Ω),
(6.8)
u 7→ (u, v), u(x)−u(y) where v(x, y) := kx−yk N/p+s . The operator T is onetoone and continuous, and it preserves the norm; that is,
kT (u)kLp (Ω)×Lp (Ω×Ω) = kukW s,p (Ω) W s,p (Ω).
for all u ∈ Hence, also by Theorem 6.6, the subspace T (W s,p (Ω)) is closed in Lp (Ω) × Lp (Ω × Ω). Since a closed subspace of a reflexive space is reflexive (see [Leo22b]), T (W s,p (Ω)) is reflexive, and, in turn, so is W s,p (Ω). Corollary 6.12. Let Ω ⊆ RN be an open set, 1 < p < ∞, and 0 < s < 1. If un * u in W s,p (Ω), then un * u in Lp (Ω) and Z Z Z Z u(x) − u(y)p un (x) − un (y)p dxdy ≥ dxdy. lim inf N +sp n→∞ kx − ykN +sp Ω Ω kx − yk Ω Ω 0
Proof. To prove the first statement, it suffices to observe that if f ∈ Lp (Ω), then the linear function Tf : W s,p (Ω) → R, given by Z Tf (u) := uf dx, u ∈ W s,p (Ω), Ω
is well defined, and, by H¨ older’s inequality, Tf (u) ≤ kf kLp0 (Ω) kukLp (Ω) ≤ kf kLp0 (Ω) kukW s,p (Ω) , which implies that Tf belongs to the dual of W s,p (Ω). Hence, if un * u in 0 W s,p (Ω), Tf (un ) → Tf (u) for every f ∈ Lp (Ω). This implies that un * u in Lp (Ω). To prove the second part, let L ∈ (W s,p (Ω))0 and define ¯ L(w, z) := L(w), w ∈ W s,p (Ω),
6.1. Definition and Main Properties
189
w(x)−w(y) s,p (Ω)))0 , where T is defined ¯ where z(x, y) := kx−yk N/p+s . Then L ∈ (T (W in (6.8). By the Hahn–Banach theorem (see [Leo22b]), there exists a linear ˜ : Lp (Ω) × Lp (Ω × Ω) → R such that L ˜ = L ¯ on continuous function L s,p T (W (Ω)) and
˜ (Lp (Ω)×Lp (Ω×Ω))0 = kLk ¯ T (W s,p (Ω)) . kLk n (x)−un (y) Let vn (x, y) := ukx−yk N/p+s and v(x, y) := s,p W (Ω), we have that
u(x)−u(y) . kx−ykN/p+s
Since un * u in
˜ n , vn ) = L(u ¯ n , vn ) = L(un ) → L(u) = L(u, ˜ v). L(u Hence (un , vn ) * Lp (Ω)×Lp (Ω × Ω), and so, by the lower semicontinuity of the norms with respect to weak convergence, lim inf kvn kLp (Ω)×Lp (Ω) ≥ kvkLp (Ω)×Lp (Ω) , n→∞
which proves the second part of the statement.
The following theorem is important in applications. It is the analogue of the Rellich–Kondrachov compactness theorem in W 1,p (see [Leo22d]). Theorem 6.13 (Compactness). Let 1 ≤ p < ∞, 0 < s < 1, and {un }n be bounded in W s,p (RN ). Then there exist a subsequence {unk }k and a function u ∈ W s,p (RN ) such that unk → u in Lploc (RN ). We begin with a preliminary result, which is of interest in itself. Lemma 6.14. Let 1 ≤ p < ∞, 0 < s < 1, and h ∈ RN , with 0 < khk < 21 . Then kτh u − ukLp (RN ) khks uW s,p (RN ) ˙ s,p (RN ), where τh u(x) := u(x + h), x ∈ RN . for every u ∈ W Proof. Let x ∈ RN and y ∈ B(x, khk). Then u(x + h) − u(x)p ≤ 2p−1 u(x + h) − u(y)p + 2p−1 u(y) − u(x)p . Averaging in y over B(x, khk), we get Z 1 p u(x + h) − u(x) u(x + h) − u(y)p dy khkN B(x,khk) Z 1 + u(y) − u(x)p dy. khkN B(x,khk)
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6. Fractional Sobolev Spaces
In turn, (6.9) Z
Z Z 1 u(x + h) − u(x)p dx u(x + h) − u(y)p dydx N khk N N R R B(x,khk) Z Z 1 + u(y) − u(x)p dydx =: A + B. khkN RN B(x,khk)
Observe that for x ∈ RN and y ∈ B(x, khk), kx + h − yk < 2khk. Hence, Z Z 1 u(x + h) − u(y)p (6.10) A = kx + h − ykN +sp dydx khkN RN B(x,khk) kx + h − ykN +sp Z Z u(x + h) − u(y)p khksp dydx = khksp upW s,p (RN ) , N +sp kx + h − yk N N R R where in the last equality we have made the change of variables x + h = z. Similarly, B khksp upW s,p (RN ) .
(6.11)
Combining (6.9), (6.10), and (6.11) proves the result.
Lemma 6.15. Let 1 ≤ p < ∞ and {un }n be a sequence bounded in Lp (RN ) such that for every ε > 0 there exists δ > 0 such that Z un (x + h) − un (x)p dx ≤ ε RN
for all n ∈ N and for every h ∈ RN with khk ≤ δ. Then there exist a subsequence {unk }k of {un }n and u ∈ Lp (RN ) such that un → u in Lploc (RN ) and pointwise LN a.e. in RN . Proof. In view of the Kolmogorov–Riesz–Fr´echet compactness theorem (see [Leo22c]) for every Lebesgue measurable set E ⊂ RN of finite measure, the sequence {un }n restricted to E is relatively compact in Lp (E). We now use a diagonal argument. Take E = B(0, 1). Since {un }n restricted to B(0, 1) is relatively compact in Lp (B(0, 1)), we can find a subsequence {un,1 }n of {un }n and a function v1 ∈ Lp (B(0, 1)) such that un,1 → v1 in Lp (B(0, 1)) and pointwise LN a.e. in B(0, 1) as n → ∞. Next, take E = B(0, 2). Since {un,1 }n restricted to B(0, 2) is relatively compact in Lp (B(0, 2)), we can find a subsequence {un,2 }n of {un,1 }n and a function v2 ∈ Lp (B(0, 2)) such that un,2 → v2 in Lp (B(0, 2)) and pointwise LN a.e. in B(0, 2) as n → ∞. By the uniqueness of limits, we have that v2 = v1 in B(0, 1). Inductively, assume we have found a subsequence {un,k }n of {un,k−1 }n and a function vk ∈ Lp (B(0, k)) such that un,k → vk in Lp (B(0, k)) and pointwise LN a.e. in B(0, k) as n → ∞. Consider E = B(0, k + 1). Since {un,k }n restricted to B(0, k+1) is relatively compact in Lp (B(0, k+1)), we can find a
6.1. Definition and Main Properties
191
subsequence {un,k+1 }n of {un,k }n and a function vk+1 ∈ Lp (B(0, k +1)) such that un,k+1 → vk+1 in Lp (B(0, k + 1)) and pointwise LN a.e. in B(0, k + 1) as n → ∞. By the uniqueness of limits, we have that vk+1 = vk in B(0, k). Let unk := uk,k and define the function u : RN → R as follows. Given x ∈ RN find k so large that x ∈ B(0, k) and set u(x) := vk (x). Then u ∈ Lploc (RN ). Moreover, for every j ∈ N, we have that {unk }k≥j is a subsequence of {un,j }n , and so, unk → vj = u in Lp (B(0, j)). By the arbitrariness of j, this shows that unk → u in Lploc (RN ) and pointwise LN a.e. in RN as k → ∞. Since {unk }k is bounded in Lp (RN ), there exists C > 0 such that kunk kLp (RN ) ≤ C for every k. Using Fatou’s lemma, it follows that for every j, kukLp (B(0,j)) ≤ lim inf kunk kLp (B(0,j)) ≤ C. k→∞
Letting j → ∞, by the Lebesgue monotone convergence theorem we find that kukLp (RN ) ≤ C. We turn to the proof of Theorem 6.13. Proof of Theorem 6.13. Let {un }n be a sequence bounded in W s,p (RN ), with kun kW s,p (RN ) ≤ M for every n. By Lemma 6.14, kτh un − un kLp (RN ) khks un W s,p (RN ) M khks . Hence, we may apply Lemma 6.15 to find a subsequence {unk }k of {un }n and u ∈ Lp (RN ) such that unk → u in Lploc (RN ) as k → ∞. It remains to show that u ∈ W s,p (RN ). Let u ¯n be a representative of un . Fix j ∈ N. Since unk → u in Lp (B(0, j)) as k → ∞, there exists a subsequence, not relabeled, such that u ¯nk → u ¯ pointwise LN a.e. in B(0, j). Hence, by Fatou’s lemma we have that uW s,p (B(0,j)) ≤ lim inf unk W s,p (B(0,j)) ≤ M. k→∞
Letting j → ∞, by the Lebesgue monotone convergence theorem we find that uW s,p (RN ) ≤ M . Hence, u ∈ W s,p (RN ), and the proof is complete. Theorem 6.13 can be improved. Corollary 6.16. Let 1 ≤ p < ∞, 0 < s < 1, and {un }n be a sequence bounded in W s,p (RN ). Then there exist a subsequence {unk }k and a function σ,p u ∈ W s,p (RN ) such that unk → u in Wloc (RN ) for every 0 < σ < s. We begin with a preliminary embedding, which is of interest in itself.
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6. Fractional Sobolev Spaces
Lemma 6.17. Let Ω ⊆ RN be an open set, 1 ≤ p < ∞, and 0 < s1 < s2 < 1. Then for all u ∈ W s2 ,p (Ω) and ` > 0, uW s1 ,p (Ω)
1 kukLp (Ω) + `s2 −s1 uW s2 ,p (Ω) . `s 1
In particular, W s2 ,p (Ω) ,→ W s1 ,p (Ω). Proof. Given ` > 0, write Z Z Z Z u(x) − u(y)p u(x) − u(y)p dxdy = dxdy N +s1 p N +s1 p Ω Ω∩B(y,`) kx − yk Ω Ω kx − yk Z Z u(x) − u(y)p + dxdy =: A + B. N +s1 p Ω Ω\B(y,`) kx − yk Since s1 < s2 , if kx − yk ≤ `, we have that kx − ykN +s2 p ≤ `(s2 −s1 ) kx − ykN +s1 p , and so, A ≤ `(s2 −s1 )
Z Z Ω
Ω∩B(y,`)
u(x) − u(y)p dxdy. kx − ykN +s2 p
On the other hand, if Ω \ B(y, `) is nonempty, by Tonelli’s theorem, Z Z 1 p B u(x) dydx N +s1 p Ω RN \B(x,`) kx − yk Z Z Z 1 1 + u(y)p dxdy u(x)p dx. N +s1 p s1 p kx − yk ` N Ω R \B(y,`) Ω
Remark 6.18. If Ω is bounded, then by taking ` = 2 diam Ω in the proof of Lemma 6.17, we get that B = 0 and uW s1 ,p (Ω) (diam Ω)s2 −s1 uW s2 ,p (Ω) . Corollary 6.19. Let Ω ⊆ RN be an open set, 1 ≤ p < ∞, and 0 < s1 < s2 < 1. Then for all u ∈ W s2 ,p (Ω), 1−s /s
s /s
1 2 uW s1 ,p (Ω) kukLp (Ω) uW1 s22,p (Ω) .
Proof. The proof follows from Exercise 1.26.
We turn to the proof of Corollary 6.16. Proof of Corollary 6.16. Let {un }n be a sequence bounded in W s,p (RN ), with kun kW s,p (RN ) ≤ M for every n. By Theorem 6.13 there exist a subsequence {unk }k and a function u ∈ W s,p (RN ) such that unk → u in Lploc (RN ).
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193
Let Ω ⊂ RN be an open bounded set. Then unk → u in Lp (Ω). Applying Corollary 6.19 to unk − u, we get σ/s
1−σ/s
unk − uW σ,p (Ω) kunk − ukLp (Ω) unk − uW s,p (Ω) 1−σ/s
M σ/s kunk − ukLp (Ω) → 0 as k → ∞.
We now show that Sobolev functions in W 1,p (Ω) belong to W s,p (Ω) when Ω is sufficiently smooth. We recall the following definition. Definition 6.20. Given 1 ≤ p ≤ ∞, an open set Ω ⊆ RN is called an extension domain for the Sobolev space W 1,p (Ω) if there exists a continuous linear operator E : W 1,p (Ω) → W 1,p (RN ) with the property that for all u ∈ W 1,p (Ω), E(u)(x) = u(x) for LN a.e. x ∈ Ω. Theorem 6.21. Let 1 ≤ p < ∞, 0 < s < 1, and let Ω ⊆ RN be a W 1,p extension domain. Then for all u ∈ W 1,p (Ω), uW s,p (Ω) Ω kukW 1,p (Ω) . In particular, W 1,p (Ω) ,→ W s,p (Ω). Proof. Step 1: Assume that Ω = RN and let u ∈ W 1,p (RN ) ∩ C ∞ (RN ). For ` > 0, write Z Z Z Z u(x) − u(y)p u(x) − u(y)p dxdy = dxdy N +sp N +sp RN RN kx − yk RN B(y,`) kx − yk Z Z u(x) − u(y)p + dxdy =: A + B. N +sp RN RN \B(y,`) kx − yk By the fundamental theorem of calculus, for x, y ∈ RN , Z 1 u(x) − u(y) = ∇u(tx + (1 − t)y)(x − y) dt, 0
and so, by H¨ older’s inequality, p
p
Z
u(x) − u(y) ≤ kx − yk
0
1
k∇u(tx + (1 − t)y)kp dt.
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6. Fractional Sobolev Spaces
In turn, by Tonelli’s theorem and by making first the change of variables z = x − y, so that dz = dx, and then ξ = y + tz so that dξ = dy, Z 1Z Z k∇u(tx + (1 − t)y)kp A≤ dxdydt kx − ykN −(1−s)p 0 RN B(y,`) Z 1Z Z k∇u(y + tz)kp = dzdydt N −(1−s)p 0 RN B(0,`) kzk Z 1Z Z 1 k∇u(y + tz)kp dydzdt = N −(1−s)p 0 B(0,`) kzk RN Z Z Z 1 βN `(1−s)p p = k∇u(ξ)k dξ k∇ukp dξ. dz = N −(1−s)p (1 − s)p N N kzk B(0,`) R R On the other hand, by Tonelli’s theorem, Z Z 1 p B≤ u(x) dydx N +sp RN \B(x,`) kx − yk RN Z Z Z 1 2βN p + u(y) dxdy = u(x)p dx. N +sp sp`sp RN RN RN \B(y,`) kx − yk Combining the estimates for A and B gives Z Z Z u(x) − u(y)p 2βN (6.12) dxdy ≤ u(x)p N +sp sp kx − yk sp` N N N R R R Z (1−s)p βN ` + k∇u(x)kp dx. (1 − s)p RN If u ∈ W 1,p (RN ), we can use mollification to construct a sequence {un }n of functions in W 1,p (RN ) ∩ C ∞ (RN ) such that un → u in W 1,p (RN ) and un → u for LN a.e. x ∈ RN . Applying (6.12) to un , we obtain Z Z un (x) − un (y)p dxdy N +sp RN RN kx − yk Z Z 1 sp un (x)p + `(1−s)p k∇un (x)kp dx. ` N N R R Letting n → ∞ and using Fatou’s lemma on the lefthand side, we obtain that (6.12) holds for every u ∈ W s,p (RN ). Step 2: Let Ω be a W 1,p (Ω) extension domain. Given u ∈ W 1,p (Ω), we can find v ∈ W 1,p (RN ) such that v = u in Ω and kvkW 1,p (RN ) Ω kukW 1,p (Ω) . It suffices to apply (6.12) to v and use the fact that v = u in Ω to obtain uW s,p (Ω) ≤ vW s,p (RN ) kvkW 1,p (RN ) Ω kukW 1,p (Ω) . Theorem 6.21 can fail if Ω is not regular enough.
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195
Exercise 6.22. Let 1 ≤ p < ∞ and 0 < s < 1, with sp > 1. Let Ω := B(0, 1) \ C, where C := {(x1 , x2 ) ∈ R2 : x1 ≤ 0, x2  ≤ x1 a }, p+1 where a > sp−1 > 1. Let u be the function given by rθ in polar coordinates (r, θ), where r > 0, −π < θ < π.
(i) Prove that u ∈ W 1,p (Ω). (ii) Let 0 < r < 2−1/(a−1) , and define inductively r0 := r, rn := rn−1 − a , n ∈ N. For n ∈ N , let rn−1 0 Dn : = {(x, y) ∈ R2 × R2 : x1 , y1 ∈ (−rn , −rn+1 ), x2 ∈ (x1 a , 2x1 a ), y2 ∈ (−2y1 a , −y1 a )}. Prove that if (x, y) ∈ Dn , then kx − yk x1 a and u(x) − u(y) ≥ π 2 x1 . (iii) Prove that for every n ∈ N0 , r−a(sp−1)+p rna
rnp−asp+2a
Z
Z
Dn
Dn
u(x) − u(y)p dxdy. kx − yk2+sp
Deduce that for every 0 < r < 2−1/(a−1) , Z Z u(x) − u(y)p −a(sp−1)+p+1 dxdy, r 2+sp Ω Ω kx − yk and that u ∈ / W s,p (Ω). The following theorem will be important when localizing a function or when using partitions of unity. Theorem 6.23. Let Ω ⊆ RN be an open set, 1 ≤ p < ∞, and 0 < s < 1. If u ∈ W s,p (Ω) and ψ ∈ C 0,1 (Ω), then ψu ∈ W s,p (Ω), with (6.13)
kψukLp (Ω) ≤ kψk∞ kukLp (Ω) , s ψuW s,p (Ω) kψk1−s ∞ k∇ψk∞ kukLp (Ω) + kψk∞ uW s,p (Ω) .
Moreover, if ψ has support contained in Ω, with dist(supp ψ, ∂Ω) ≥ δ > 0, then the function ψu, extended to be zero outside Ω, belongs to W s,p (RN ), with 1 (6.14) ψuW s,p (RN ) s kψukLp (Ω) + ψuW s,p (Ω) . δ We begin with a preliminary computation, which we will use repeatedly. Lemma 6.24. Let 1 ≤ p < ∞ and 0 < s < 1. Then for every y ∈ RN and a, b > 0, Z min{a, bkx − yk}p dx a(1−s)p bsp . kx − ykN +sp RN
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6. Fractional Sobolev Spaces
Proof. Set ` = a/b. Using spherical coordinates, we have Z Z min{`, kx − yk}p p kx − ykp(1−s)−N dx bp dx ≤ b N +sp kx − yk N R B(y,`) Z `p p +b dx bp `(1−s)p . N +sp kx − yk N R \B(y,`) We turn to the proof of the theorem. Proof. By the inequality (a + b)p ≤ 2p−1 ap + 2p−1 bp for all a, b ≥ 0, we have Z Z (ψ(x) − ψ(y))u(y)p ψupW s,p (Ω) dxdy kx − ykN +sp Ω Ω Z Z u(x) − u(y)p + ψ(x)p dxdy =: A + B. kx − ykN +sp Ω Ω By the mean value theorem, for x, y ∈ RN , ψ(x) − ψ(y) ≤ k∇ψk∞ kx − yk. Then, by Tonelli’s theorem and Lemma 6.24, Z Z min{k∇ψkp∞ kx − ykp , 2p kψkp∞ } p A≤ u(y) dxdy kx − ykN +sp Ω Ω Z kψk(1−s)p k∇ψksp u(y)p dy. ∞ ∞ Ω
On the other hand, B ≤ kψkp∞
Z Z Ω
Ω
u(x) − u(y)p dxdy. kx − ykN +sp
To prove the second part of the statement, assume that F := supp ψ ⊂ Ω, with dist(F, ∂Ω) ≥ δ > 0 for some δ > 0. Extend ψu to be zero outside Ω. Then we have Z Z Z Z ψ(x)u(x) − ψ(y)u(y)p ψ(y)u(y)p dxdy = 2 dxdy N +sp kx − ykN +sp RN RN F RN \Ω kx − yk Z Z ψ(x)u(x) − ψ(y)u(y)p + dxdy =: C + ψupW s,p (Ω) . kx − ykN +sp Ω Ω If y ∈ F and x ∈ RN \ Ω, then kx − yk ≥ δ. Therefore, Z Z 1 1 1 dx ≤ dx ≈ sp . N +sp N +sp δ RN \Ω kx − yk RN \B(y,δ) kx − yk Hence, by Tonelli’s theorem, Z Z C ψ(y)u(y)p F
RN \Ω
1 1 dxdy sp kx − ykN +sp δ
Z F
ψ(y)u(y)p dy.
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197
Remark 6.25. If Ω is bounded, then (6.13) can be replaced by ψuW s,p (Ω) diam Ω k∇ψk∞ kukLp (Ω) + kψk∞ uW s,p (Ω) . Indeed, letting R = 2 diam Ω, for y ∈ Ω, we have Ω ⊆ B(y, 2R), and so, Z Z ψ(x) − ψ(y)p ψ(x) − ψ(y)p dx ≤ dx k∇ψkp∞ R(1−s)p . N +sp N +sp Ω kx − yk B(y,2R) kx − yk In turn, Z
p
A=
Z
u(y) Ω
Ω
ψ(x) − ψ(y)p dxdy k∇ψkp∞ R(1−s)p kx − ykN +sp
Z
u(y)p dy.
Ω
What happens if ψ has compact support (exercise)? Theorem 6.26 (Product). Let Ω ⊆ RN be an open set, 1 ≤ p < ∞, and ˙ s,p (Ω) ∩ L∞ (Ω), then uv ∈ W ˙ s,p (Ω), with 0 < s < 1. If u, v ∈ W uvW s,p (Ω) ≤ uW s,p (Ω) kvkL∞ (Ω) + kukL∞ (Ω) vW s,p (Ω) . Moreover, if u, v ∈ W s,p (Ω) ∩ L∞ (Ω), then uv ∈ W s,p (Ω), with kuvkLp (Ω) ≤ kukLp (Ω) kvkL∞ (Ω) . The proof of this theorem is simpler than that of Theorem 6.23 and is left as an exercise. Next, we study superpositions. Theorem 6.27. Let Ω ⊆ RN be an open set, 1 ≤ p < ∞, and 0 < s < 1. ˙ s,p (Ω) for every If f : R → R is Lipschitz continuous, then f ◦ u ∈ W ˙ s,p (Ω) with u∈W f ◦ uW s,p (Ω) ≤ Lip f uW s,p (Ω) . Moreover, if we assume that f (0) = 0 when Ω has infinite measure, then f ◦ u ∈ W s,p (Ω) for every u ∈ W s,p (Ω). Proof. For x, y ∈ Ω, we have f (u(x)) − f (u(y)) ≤ Lip f u(x) − u(y), and thus, Z Z Ω
Ω
f (u(x)) − f (u(y))p dxdy ≤ (Lip f )p kx − ykN +sp
Z Z Ω
Ω
u(x) − u(y)p dxdy. kx − ykN +sp
This completes the first part of the proof. Next, assume that u ∈ W s,p (Ω). Then f (u(x)) ≤ f (u(x)) − f (0) + f (0) ≤ Lip f u(x) + f (0). Raising both sides to power p, integrating over Ω, and using the fact that f (0) = 0 if Ω has infinite measure shows that f ◦ u ∈ Lp (Ω).
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6. Fractional Sobolev Spaces
Remark 6.28. By taking f (t) = t or f (t) = max{t, 0} = t+ , it follows that u ∈ W s,p (Ω) and u+ ∈ W s,p (Ω) for every u ∈ W s,p (Ω). Theorem 6.29 (Change of variables). Let Ω, U ⊆ RN be open sets, let Ψ : U → Ω be invertible with Ψ and Ψ−1 Lipschitz continuous functions, ˙ s,p (Ω), we have and let 1 ≤ p < ∞, and 0 < s < 1. Then for every u ∈ W ˙ s,p (U ), with that u ◦ Ψ ∈ W u ◦ ΨW s,p (U ) (Lip Ψ)s+N/p (Lip Ψ−1 )2N/p uW s,p (Ω) . Moreover, if u ∈ W s,p (Ω), then u ◦ Ψ ∈ W s,p (U ) and ku ◦ ΨkLp (U ) (Lip Ψ−1 )N/p kukLp (Ω) . Proof. If x, y ∈ Ω, (6.15) kx − yk = kΨ(Ψ−1 (x)) − Ψ(Ψ−1 (y))k ≤ Lip ΨkΨ−1 (x) − Ψ−1 (y)k. Moreover, since Ψ−1 is Lipschitz, for every y ∈ Ω,  det JΨ−1 (y) (Lip Ψ−1 )N . Therefore, for every X ∈ U , 1 1 =  det JΨ (X). −1 N (Lip Ψ )  det JΨ−1 (Ψ(X)) Using this inequality, we obtain Z Z u(Ψ(X)) − u(Ψ(Y ))p p u ◦ ΨW s,p (U ) = dXdY kX − Y kN +sp U U Z Z u(Ψ(X)) − u(Ψ(Y ))p (Lip Ψ−1 )2N kX − Y kN +sp Ψ−1 (Ω) Ψ−1 (Ω) ×  det JΨ (X)dX  det JΨ (Y )dY Z Z u(x) − u(y)p −1 2N = (Lip Ψ ) dxdy −1 −1 N +sp Ω Ω kΨ (x) − Ψ (y)k Z Z u(x) − u(y)p (Lip Ψ)N +sp (Lip Ψ−1 )2N dxdy, N +sp Ω Ω kx − yk where we use the change of variable theorem [Leo17, Theorem 9.52] and (6.15). Similarly, if u ∈ W s,p (Ω), then Z p ku ◦ ΨkLp (U ) = u(Ψ(X))p dX U Z −1 N (Lip Ψ ) u(Ψ(X))p  det JΨ (X)dX −1 Ψ (Ω) Z = (Lip Ψ−1 )N u(x)p dx. Ω
6.1. Definition and Main Properties
199
An N ×N matrix R is orthogonal if RT = R−1 , where RT is the transpose matrix of R. A rigid motion T : RN → RN is an affine function given by T (x) = Rx + c, x ∈ RN , where R is an orthogonal matrix and c ∈ RN . Exercise 6.30. Let Ω ⊆ RN be an open set, let 1 ≤ p < ∞, 0 < s < 1, and let T be a rigid motion. (i) Prove that det JR = ±1 and that for every x ∈ RN , kRxk = kxk. (ii) Prove that if u ∈ W s,p (Ω), then u ◦ T ∈ W s,p (T −1 (Ω)), with ku ◦ T kLp (T −1 (Ω)) = kukLp (Ω) ,
u ◦ T W s,p (T −1 (Ω)) = uW s,p (Ω) .
The following corollary will be important when dealing with domains Ω with Lipschitz continuous boundary. We will use notation (6.2). Corollary 6.31 (Flattening the boundary). Let f : RN −1 → R be a Lipschitz continuous function, let Ω = {x = (x0 , xN ) ∈ RN −1 × R : xN > f (x0 )}, ˙ s,p (Ω), 1 ≤ p < ∞, 0 < s < 1. Then the function v : RN and let u ∈ W + → R, given by v(y) := u(y 0 , yN + f (y 0 )), y ∈ RN +, ˙ s,p (RN ) and belongs to W +
vW s,p (RN ) (1 + Lip f )s+N/p uW s,p (Ω) . +
Moreover, if u ∈
W s,p (Ω),
then v ∈ W s,p (RN + ) and kvkLp (RN ) = kukLp (Ω) . +
Proof. Consider the transformation Ψ : RN → RN , given by Ψ(y) := (y 0 , yN + f (y 0 )). Note that Ψ is invertible, with inverse given by Ψ−1 (x) = (x0 , xN − f (x0 )). Moreover, for all y, z ∈ RN , kΨ(y) − Ψ(z)k = k(y 0 − z 0 , f (y 0 ) − f (z 0 ) + yN − zN )k q ≤ ky 0 − z 0 k2N −1 + (Lip f ky 0 − z 0 kN −1 + yN − zN )2 (1 + Lip f )ky − zk, which shows that Ψ (and similarly Ψ−1 ) is Lipschitz continuous. Since f is Lipschitz continuous, by Rademacher’s theorem (see [Leo17, Theorem 9.14]) f is differentiable for LN −1 a.e. y 0 ∈ RN −1 , and so, for any such y 0 ∈ RN −1 and for all yN ∈ R we have IN −1 0 JΨ (y) = , ∇y0 f (y 0 ) 1 which implies that det JΨ (y) = 1. Note that Ψ(RN + ) = Ω. We can now apply Theorem 6.29.
200
6. Fractional Sobolev Spaces
Exercise 6.32. Let 1 ≤ p < ∞, 0 < s < 1, and u ∈ W s,p (RN + ). Prove that the function v : RN → R, defined by u(x0 , −xN ) if xN < 0, v(x) := u(x0 , xN ) if xN > 0, belongs to W s,p (RN ), with vW s,p (RN ) uW s,p (RN ) and kvkLp (RN ) = + 2kukLp (RN ) . +
We end this section by showing that in domains of finite measure the ˙ s,p (Ω) coincides with W s,p (Ω). The homogeneous fractional Sobolev space W proof relies on Poincar´e’s inequality. As usual, if E ⊂ RN is a Lebesgue measurable set with positive and finite measure R and u : E → R is a Lebesgue 1 integrable function, we define uE := LN (E) E u(x) dx. Theorem 6.33 (Poincar´e’s inequality). Let Ω ⊂ RN be an open bounded set, let E ⊆ Ω be a Lebesgue measurable set with positive measure, and let 1 ≤ p < ∞ and 0 < s < 1. Then for all u ∈ W s,p (Ω), Z Z Z (diam Ω)N +sp u(x) − u(y)p p u(x) − uE  dx ≤ dxdy. N +sp LN (E) Ω Ω Ω kx − yk Proof. By Minkowski’s inequality for integrals ([Leo22c]), we have p 1/p Z 1/p Z Z 1 p dx u(x) − uE  dx = N (u(x) − u(y)) dy L (E) Ω Ω E 1/p Z Z 1 u(x) − u(y)p dy dx. ≤ N L (E) E Ω Hence, by H¨older’s inequality Z 0 Z Z (LN (E))p/p u(x) − uE p dx ≤ u(x) − u(y)p dydx N (E))p (L Ω Ω Ω Z Z u(x) − u(y)p (diam Ω)N +sp ≤ dxdy, N +sp LN (E) Ω Ω kx − yk where in the last inequality we use the fact that if x, y ∈ Ω, then kx − yk ≤ diam Ω. Remark 6.34. If we take E b Ω, then the Poincar´e inequality continues ˙ s,p (Ω). Indeed, consider an open set E ⊆ U b Ω. Since to hold in W u ∈ W s,p (U ), applying the Poincar´e inequality in U , we get Z Z Z u(x) − u(y)p (diam U )N +sp p u(x) − uE  dx ≤ dxdy N +sp LN (E) U U U kx − yk Z Z (diam Ω)N +sp u(x) − u(y)p ≤ dxdy. N +sp LN (E) Ω Ω kx − yk
6.2. Slicing
201
It suffices to let U % Ω and use the Lebesgue monotone convergence theorem.
6.2. Slicing In this section we provide a characterization of fractional Sobolev spaces, which is useful in applications. We show that a function u belongs to W s,p (RN ) if its onedimensional slices belong to W s,p (R) (we will make this precise in the statement). This fact allows us to prove results first in the simpler case N = 1. For simplicity, we begin with the case Ω = RN and then consider more general rectangles. In what follows we will use the notation x = (x0i , xi ) ∈ RN −1 × R (see (6.1)) and x = (x0 , xN ) ∈ RN −1 × R (see (6.2)). Theorem 6.35 (Slicing). Let 1 ≤ p < ∞ and 0 < s < 1. Then for every u ∈ Lploc (RN ), upW s,p (RN )
≈
N Z X
Z Z
RN −1
i=1
R
R
u(x0i , xi ) − u(x0i , yi )p dyi dxi dx0i . xi − yi 1+sp
In particular, a function u ∈ Lp (RN ) (respectively, u ∈ Lploc (RN )) belongs to ˙ s,p (RN )) if and only if it admits a representative u W s,p (RN ) (respectively, W ¯ 0 s,p 0 s,p N −1 ˙ such that u ¯(xi , ·) ∈ W (R) (respectively, u ¯(xi , ·) ∈ W (R)) for L a.e. x0i ∈ RN −1 and for all i = 1, . . . , N , with N Z X i=1
RN −1
Z Z R
R
¯ u(x0i , xi ) − u ¯(x0i , yi )p dyi dxi dx0i < ∞. xi − yi 1+sp
The following computation will be used in what follows. Lemma 6.36. Let 1 ≤ p < ∞ and λ > 0. Then for every y = (y 0 , yN ) ∈ RN −1 × R and every xN ∈ R, with xN 6= yN , Z 1 1 (6.16) dx0 ≈ . N +λ xN − yN 1+λ RN −1 kx − yk Moreover, for every y = (y 0 , yN ) ∈ RN −1 × R and every x0 ∈ RN −1 , with x0 6= y 0 , Z 1 1 (6.17) dxN ≈ . N +λ −1+λ kx0 − y 0 kN R kx − yk N −1
202
6. Fractional Sobolev Spaces
Proof. We only prove (6.16) since the second estimate is similar. Using the change of variables x0 − y 0 = xN − yN w0 , we have Z 1 dx0 N +λ RN −1 kx − yk Z xN − yN N −1 = dw0 2 kw 0 k2 2 )(N +λ)/2 N −1 (x − y  + x − y  N N N N R N −1 Z 1 1 = dw0 . xN − yN 1+λ RN −1 (kw0 k2N −1 + 1)(N +λ)/2 We turn to the proof of Theorem 6.35. Proof of Theorem 6.35. Step 1: Given x, y ∈ RN , with x 6= y, define X 0 := x, X 1 := (y1 , x2 , . . . , xN ), . . . , X N −1:= (y1 , . . . , yN −1 , xN ), X N := y. Then p
u(x) − u(y)
N X
u(X i ) − u(X i−1 )p ,
i=1
and so, upW s,p (RN )
N Z X i=1
RN
Z RN
N
X u(X i ) − u(X i−1 )p dxdy =: Ai . N +sp kx − yk i=1
We study AN (the other terms are similar) and use notation (6.2). By Tonelli’s theorem we can write Z Z Z Z 1 0 0 p dx0 dxN dyN dy 0 . AN = u(y , xN )−u(y , yN ) N +sp RN −1 R R RN −1 kx − yk We now use Lemma 6.36 to obtain Z Z Z u(y 0 , xN ) − u(y 0 , yN )p AN dxN dyN dy 0 . 1+sp x − y  N −1 N N R R R Step 2: Fix i ∈ {1, . . . , N }. We claim that Z Z Z u(x0i , xi ) − u(x0i , yi )p A := dyi dxi dx0i upW s,p (RN ) . 1+sp x − y  N −1 i i R R R We will show it for i = N and use notation (6.2). Let x0 ∈ RN −1 and xN , N yN ∈ R, with xN 6= yN . Take X := (x0 , xN +y ) and r := 14 xN − yN . Then 2 for z ∈ B(X, r), u(x0 , xN ) − u(x0 , yN )p ≤ 2p−1 u(x0 , xN ) − u(z)p + 2p−1 u(z) − u(x0 , yN )p . Dividing both sides by xN −yN 1+sp , averaging in z over B(X, r), integrating in x0 over RN −1 and in xN and yN over R, and using the fact that r =
6.2. Slicing
1 4 xN
203
− yN  gives Z
Z Z Z
A RN −1
Z
R
R
B(X,r)
u(x0 , xN ) − u(z)p dzdyN dxN dx0 xN − yN N +1+sp
u(x0 , yN ) − u(z)p dzdyN dxN dx0 N +1+sp R R B(X,r) xN − yN  Z 1 p χB(X,r) (z) u(x) − u(z) dyN dxdz, xN − yN N +1+sp R RN Z Z Z
+ RN −1 Z Z =2 RN
where in the last equality we use used Tonelli’s theorem and interchange the roles of xN and yN in one of the terms. It remains to show that for every x, z ∈ RN with x 6= z, Z 1 1 dyN . B := χB(X,r) (z) N +1+sp N +sp x − y  kx − zk N N R We have χB(X,r) (z) = 1 if and only if kz − Xk < xN − yN /4, which implies that xN − yN /4 > kz − Xk ≥ kx − zk − kx − Xk = kx − zk − xN − yN /2, that is, xN − yN  ≥ 43 kx − zk. Hence, Z
1
B≤ R\B1 (xN , 43 kx−zk) Z ∞
1
4 kx−zk 3
tN +1+sp
xN − yN N +1+sp dt
dyN
1 . kx − zkN +sp
Step 3: The last part of the statement follows from Tonelli’s theorem applied to a representative u ¯ of u. We leave the details as an exercise. Remark 6.37. We will see in Theorems 9.6 and 9.7 that W s,p (RN ) can be characterized as the trace space of a weighted Sobolev space. Thus, given u ∈ W s,p (RN ), an alternative approach to Theorem 6.35 would be to find a function v in the weighted space that has trace u, use a slicing theorem for v, and obtain as a byproduct a corresponding slicing result for u. We will not pursue this approach. Next we consider the case in which Ω is given by the Cartesian product of intervals of infinite length. Theorem 6.38 (Slicing). Let I1 , . . . , IN be open intervals of infinite length, and let R := I1 × · · · × IN , 1 ≤ p < ∞, and 0 < s < 1. Then for every
204
6. Fractional Sobolev Spaces
u ∈ Lploc (R), u(x) − u(y)p dxdy N +sp R kx − yk Z Z N Z X u(x0i , xi ) − u(x0i , yi )p dyi dxi dx0i . ≈ Q 1+sp x − y  i i Ii Ii j6=i Ij
Z Z (6.18) R
i=1
In particular, a function u ∈ Lp (R) (respectively, u ∈ Lploc (R)) belongs to ˙ s,p (R)) if and only if it admits a representative u W s,p (R) (respectively, W ¯ 0 s,p 0 s,p N −1 ˙ such Q that u ¯(xi , ·) ∈ W (Ii ) (respectively, u ¯(xi , ·) ∈ W (Ii )) for L a.e. x0i ∈ j6=i Ij , and for all i = 1, . . . , N , with N Z X i=1
Q
Z Z j6=i Ij
Ii
Ii
¯ u(x0i , xi ) − u ¯(x0i , yi )p dyi dxi dx0i < ∞. xi − yi 1+sp
Lemma 6.39. Let I ⊆ R be an open interval with infinite measure and λ > 0. Then for every ρ > 0 and y ∈ I, Z 1 1 dx ≈ λ . (1+λ)/2 2 2 ρ I (ρ + x − y ) Proof. Assume that I = (a, ∞) (the cases I = (−∞, b) and I = R are similar). Then Z 1 dx 2 2 (1+λ)/2 I (ρ + x − y ) Z y 1 1 dx = 1+λ 2 2 (1+λ)/2 ρ a (ρ + x − y ) Z ∞ 1 1 + 1+λ dx 2 ρ (ρ + x − y2 )(1+λ)/2 y Z (y−a)/ρ Z ∞ 1 1 1 1 = λ dt + dτ, ρ 0 ρλ 0 (1 + τ 2 )(1+λ)/2 (1 + t2 )(1+λ)/2 where we make the change of variables t = (y − x)/ρ and τ = (x − y)/ρ. Hence, Z ∞ Z 1 1 1 dt ≤ dx λ (1+λ)/2 2 2 2 (1+λ)/2 ρ 0 (1 + t ) I (ρ + x − y ) Z ∞ 2 1 ≤ λ dt. 2 ρ 0 (1 + t )(1+λ)/2
6.2. Slicing
205
Remark 6.40. If I = (a, b) and y ∈ I, a similar computation gives Z Z (y−a)/ρ 1 1 1 dx = dt λ 2 2 (1+λ)/2 ρ 0 (1 + t2 )(1+λ)/2 I (ρ + x − y ) Z (b−y)/ρ 1 1 + λ dτ. ρ 0 (1 + τ 2 )(1+λ)/2 b+a Depending on whether y ≥ b+a 2 or y ≤ 2 , we have that one of the two integrals on the right is greater than or equal to Z (b−a)/(2ρ) 1 1 dt, ρλ 0 (1 + t2 )(1+λ)/2
and so, we get Z (b−a)/(2ρ) Z 1 1 1 dt ≤ dx (1+λ)/2 2 2 2 (1+λ)/2 ρλ 0 (1 + t ) I (ρ + x − y ) Z ∞ 2 1 ≤ λ dt. 2 ρ 0 (1 + t )(1+λ)/2 In particular, if b − a ≥ ` and 0 < ρ ≤ M , then Z `/(2M ) Z (b−a)/(2ρ) 1 1 dt ≥ dt := C0 (`, M ). (1+λ)/2 2 2 (1 + t ) (1 + t )(1+λ)/2 0 0 Hence, 1 `,M ρλ
Z I
1 1 dx λ . ρ (ρ2 + x − y2 )(1+λ)/2
Corollary 6.41. Let I1 , . . . , IN be open intervals of infinite length, R := I1 × · · · × IN , and λ > 0. Then for every y = (yi0 , yi ) ∈ R and every xi ∈ Ii , with xi 6= yi , Z 1 1 (6.19) dx0i ≈ . Q N +λ kx − yk xi − yi 1+λ j6=i Ij Q Similarly, for every y = (y 0 , yN ) ∈ R and every x0i ∈ j6=i Ij , with x0i 6= yi0 , Z 1 1 (6.20) . dxi ≈ 0 N +λ −1+λ kxi − yi0 kN Ii kx − yk N −1 Proof. We only prove (6.19). The proof is by induction on N . If N = 2 and, say, i = 2, then by applying Lemma 6.39 in I1 , with ρ = x2 − y2 , we get Z 1 1 dx1 ≈ . 2+λ x2 − y2 1+λ I1 kx − yk
206
6. Fractional Sobolev Spaces
Next assume that (6.19) holds for N − 1 and let’s prove it for N . Without loss of generality, assume that i = 1. By applying Lemma 6.39 in IN , with ρ = kx0 − y 0 kN −1 and λ replaced by N − 1 + λ, we get Z 1 1 dxN ≈ . N +λ −1+λ 0 kx − y 0 kN IN kx − yk N −1 Integrating both sides in x2 , . . . , xN −1 gives Z Z dx2 · · · dxN dx2 · · · dxN −1 . (6.21) ≈ Q QN −1 N −1+λ 0 0 N kx − ykN +λ kx − y kN −1 j=2 Ij j=2 Ij Using the induction hypothesis in dimension N − 1 gives Z dx2 · · · dxN −1 1 ≈ (6.22) . QN −1 N −1+λ 0 0 x1 − y1 1+λ kx − y kN −1 j=2 Ij To prove (6.19), it is enough to combine (6.21) and (6.22).
Remark 6.42. If we assume that all the open intervals I1 , . . . , IN are bounded and set ` := mini L1 (Ii ) and M := maxi L1 (Ii ), then (6.19) and (6.20) continue to hold provided we allow the constants to depend on ` and M . This follows from Remark 6.40. We turn to the proof of Theorem 6.38. Proof of Theorem 6.38. Step 1: Given x, y ∈ R, with x 6= y, define X 0 := x, X 1 := (y1 , x2 , . . . , xN ), . . . , X N −1:= (y1 , . . . , yN −1 , xN ), X N := y. Then u(x) − u(y)p
N X
u(X i ) − u(X i−1 )p ,
i=1
and so, Z Z R
R
N
Z Z
i=1
R
X u(x) − u(y)p dxdy N +sp kx − yk
R
N
X u(X i ) − u(X i−1 )p dxdy =: Ai . N +sp kx − yk i=1
We study AN and use notation (6.2). By Tonelli’s theorem, we can write Z Z Z Z dx0 dxN dyN dy 0 0 0 p AN = Q u(y , xN ) − u(y , yN ) Q . kx − ykN +sp IN IN j6=N Ij j6=N Ij By Corollary 6.41, Z AN Q
j6=N
Z Ij
IN
Z IN
u(y 0 , xN ) − u(y 0 , yN )p dxN dyN dy 0 . xN − yN 1+sp
Similar estimates hold for the other Ai . Thus, we obtained the first inequality in (6.18).
6.2. Slicing
207
Step 2: Fix i ∈ {1, . . . , N }. We claim that Z Z Z u(x0i , xi ) − u(x0i , yi )p Ai := Q dyi dxi dx0i 1+sp x − y  i i Ii Ii j6=i Ij Z Z p u(x) − u(y) dxdy. N +sp R R kx − yk Q We will show it for i = N and use notation (6.2). Let x0 ∈ j6=N Ij and N ) and r := θxN − yN , xN , yN ∈ IN , with xN 6= yN . Take X := (x0 , xN +y 2 1 where θ = 4√N . For each j = 1, . . . , N , if [xj , ∞) ⊆ Ij consider the interval (xj , xj + r), while if (−∞, xj ] ⊆ Ij consider the interval (xj − r, xj ). Let Q(X, r) be the Cartesian product of these intervals. Then for z ∈ Q(X, r) ⊂ R, u(x0 , xN ) − u(x0 , yN )p ≤ 2p−1 u(x0 , xN ) − u(z)p + 2p−1 u(z) − u(x0 , yN )p . 1+sp , averaging in z over Q(X, r), integrating Dividing both Q sides by xN −yN  0 in x over j6=N Ij and in xN and yN over IN , and using the fact that Q(X, r) ⊂ R gives Z Z Z Z u(x0 , xN ) − u(z)p AN Q dzdyN dxN dx0 N +1+sp x − y  N N Q(X,r) I I I j N N j6=N Z Z Z Z u(x0 , yN ) − u(z)p dzdyN dxN dx0 + Q N +1+sp x − y  N N IN IN Q(X,r) j6=N Ij Z Z Z 1 χQ(X,r) (z) u(x) − u(z)p dyN dxdz, xN − yN N +1+sp R R R
where we use Tonelli’s theorem and interchange the roles of xN and yN in one of the terms. It remains to show that for every x, z ∈ R, with x 6= z, Z 1 1 D := χQ(X,r) (z) dyN . N +1+sp xN − yN  kx − zkN +sp R √ Since √ zj − Xj  < r for every j, kz − Xk < N r, and so, Q(X, r) ⊂ B(X, N r). Using the fact that r = θxN − yN , we have χB(X,√N r) (z) = 1 √ if and only if kz − Xk < N θxN − yN , which implies that √ N θxN − yN  > kz − Xk ≥ kx − zk − kx − Xk = kx − zk − xN − yN /2, √ that is, xN − yN  ≥ ( N θ + 1/2)−1 kx − zk. Recalling that θ = 4√1N , we √ have that ( N θ + 1/2)−1 = 43 . Hence, Z Z ∞ 1 1 D≤ dyN = 2 dt N +1+sp N +1+sp 4 R\B1 (xN , 4 kx−zk) xN − yN  kx−zk t 3
1 . kx − zkN +sp
3
208
6. Fractional Sobolev Spaces
Step 3: The last part of the statement follows from Tonelli’s theorem applied to a representative u ¯ of u. We leave the details as an exercise. Remark 6.43. An alternative proof would have been to use a reflection in each variable (see Theorem 1.40) to extend u to a function in RN , and then apply Theorem 6.35. We leave the details as an exercise. Remark 6.44. If all the intervals Ii have finite length, set ` := mini L1 (Ii ) and M := maxi L1 (Ii ). In Step 1 of the proof of Theorem 6.38 there are no changes except that, in view of Remark 6.42, √ the constant C depends on ` and √ M . In Step 2, we consider θ = `/(4 N M ). Then r = θxN − yN  < `/(4 N ). For each j = 1, . . . , N , let Ij = (aj , bj ) and consider the interval a +b a +b (xj , xj + r) if xj < j 2 j or the interval (xj − r, xj ) if xj > j 2 j . Since √ r < `/(4 N ) ≤ (bj − aj )/4, as before √ we have that Q(X, r) ⊂ R. In the last part of the proof, we replace 34 with ( N θ + 1/2)−1 = (`/(4M ) + 1/2)−1 . Q Exercise 6.45. Let 1 ≤ p < ∞, 0 < s < 1/p, and R = N i=1 [ai , ai + ri ], s,p N where ai ∈ R and ri > 0. Prove that χR ∈ W (R ), with !1/p N N Y X χR W s,p (RN ) ri 1/ris . i=1
i=1
Hint: Use Theorem 6.38. The slicing technique in Theorem 6.38 works well for RN , halfspaces of the type RN + , and bounded rectangles, but not for arbitrary open sets. Next, we consider a different slicing technique, which works for any open set Ω ⊆ RN . Given ν ∈ ∂B(0, 1) =: SN −1 , we denote by ν ⊥ the hyperplane orthogonal to ν, that is, ν ⊥ := {x ∈ RN : x · ν = 0}. In what follows, for brevity, we will write Z Z (6.23) f dΣ := f dHN −1 , where we recall that HN −1 is the (N − 1)dimensional Hausdorff measure. For simplicity we begin with the case Ω = RN . Theorem 6.46 (Slicing). Let 1 ≤ p < ∞ and 0 < s < 1. Then for every u ∈ Lploc (RN ), upW s,p (RN ) Z Z Z Z u(x + tν) − u(x + τ ν)p 1 = dtdτ dΣ(x)dΣ(ν). 2 SN −1 ν ⊥ R R t − τ 1+sp
6.2. Slicing
209
In particular, a function u ∈ Lp (RN ) (respectively, u ∈ Lploc (RN )) belongs to ˙ s,p (RN )) if and only if it admits a representative W s,p (RN ) (respectively, W ˙ s,p (R)) for u ¯ such that u ¯(x + ·ν) ∈ W s,p (R) (respectively, u ¯(x + ·ν) ∈ W N −1 N −1 N −1 ⊥ H a.e. ν ∈ S and H a.e. x ∈ ν , with Z Z Z Z u(x + tν) − u(x + τ ν)p dtdτ dΣ(x)dΣ(ν) < ∞. t − τ 1+sp SN −1 ν ⊥ R R Proof. For every x, h ∈ RN , using spherical coordinates centered at x, we can write x + h = x + rν, where r ≥ 0 and ν ∈ SN −1 . Then by Tonelli’s theorem for u ∈ Lploc (RN ), Z Z u(x + h) − u(x)p p uW s,p (RN ) = dhdx khkN +sp RN RN Z Z Z ∞ u(x + rν) − u(x)p N −1 r drdxdΣ(ν) =: A. = rN +sp SN −1 RN 0 For every fixed ν ∈ SN −1 and for every x ∈ RN we write x = y + τ ν, where y ∈ ν ⊥ and τ ∈ R. Since the Jacobian matrix of this change of variables has determinant 1, Z Z Z Z ∞ u(y + (τ + r)ν) − u(y + τ ν)p A= drdτ dΣ(y)dΣ(ν) r1+sp SN −1 ν ⊥ R 0 Z Z Z Z ∞ u(y + tν) − u(y + τ ν)p = dtdτ dΣ(y)dΣ(ν) (t − τ )1+sp SN −1 ν ⊥ R τ Z Z Z Z u(y + tν) − u(y + τ ν)p 1 drdtdΣ(y)dΣ(ν), = 2 SN −1 ν ⊥ R R t − τ 1+sp where we make the change of variables t = τ + r, so that dt = dr, and then relabel the variables. It follows that uW s,p (RN ) < ∞ if and only if the righthand side is finite. Next, we consider an arbitrary open set Ω ⊆ RN . Theorem 6.47 (Slicing). Let Ω ⊆ RN be an open set, 1 ≤ p < ∞, and 0 < s < 1. Then for every u ∈ Lploc (Ω), upW s,p (Ω) Z Z Z Z 1 u(x + tν) − u(x + τ ν)p dtdτ dΣ(x)dΣ(ν), = 2 SN −1 ν ⊥ Ω(x,ν) Ω(x,ν) t − τ 1+sp where Ω(x, ν) := {t ∈ R : x + tν ∈ Ω}. In particular, a function u ∈ Lp (Ω) ˙ s,p (Ω)) if and (respectively, u ∈ Lploc (Ω)) belongs to W s,p (Ω) (respectively, W only if it admits a representative u ¯ such that u ¯(x + ·ν) ∈ W s,p (Ω(x, ν))
210
6. Fractional Sobolev Spaces
˙ s,p (Ω(x, ν))) for HN −1 a.e. ν ∈ SN −1 and HN −1 (respectively, u ¯(x+·ν) ∈ W ⊥ a.e. x ∈ ν for which Ω(x, ν) 6= ∅, with Z Z Z Z u(x + tν) − u(x + τ ν)p dtdτ dΣ(x)dΣ(ν) < ∞. t − τ 1+sp SN −1 ν ⊥ Ω(x,ν) Ω(x,ν) We begin with a preliminary lemma, which is particular case of the affine Blaschke–Petkantschin formula (see [SW08, Theorem 7.2.7]). Lemma 6.48. Let E ⊆ RN be a Lebesgue measurable set and let f : E×E → [0, ∞] be a Lebesgue measurable function. Then Z Z Z Z f (x, y) dxdy = f (x, x + h) dhdx E E E Ex Z Z Z Z 1 = f (z + tν, z + τ ν)t − τ N −1 dtdτ dΣ(z)dΣ(ν), 2 SN −1 ν ⊥ E(z,ν) E(z,ν) where Ex := {h ∈ RN : x + h ∈ E} and E(x, ν) := {t ∈ R : x + tν ∈ E}. Proof. The first equality follows from Tonelli’s theorem and the change of variables y = x + h. To prove the second, we use spherical coordinates centered at x. For h ∈ Ex we write x + h = x + rν, where r ≥ 0 and ν ∈ SN −1 . Then, also by Tonelli’s theorem, we have Z Z Z Z Z f (x, x + h)dhdx = f (x, x + rν)rN −1 drdxdΣ(ν), E
Ex
SN −1
E
E+ (x,ν)
where E+ (x, ν) = {r ≥ 0 : x + rν ∈ E}. For every fixed ν ∈ SN −1 and for every x ∈ E, we write x = z + tν, where z ∈ ν ⊥ and t ∈ R. Since the Jacobian matrix of this change of variables has determinant 1, the righthand side equals Z Z Z Z f (z + tν, y + (t + r)ν)rN −1 drdtdΣ(z)dΣ(ν) N −1 ⊥ S ν E(z,ν) E+ (z+tν,ν) Z Z Z Z = f (z + tν, y + τ ν)(τ − t)N −1 dτ dtΣ(z)dΣ(ν) SN −1 ν ⊥ E(z,ν) E(z,ν)∩(t,∞) Z Z Z Z 1 = f (z + tν, y + τ ν)t − τ N −1 dτ dtdΣ(y)dΣ(ν), 2 SN −1 ν ⊥ E(z,ν) E(z,ν) where we make the change of variables τ = t + r, so that dτ = dr, used Tonelli’s theorem, and relabel the variables. We turn to the proof of Theorem 6.47.
6.2. Slicing
211
Proof of Theorem 6.47. By Lemma 6.48 applied to f (x, y) = u(x + h) − u(x)p , we have upW s,p (Ω) Z Z Z Z 1 u(y + tν) − u(y + τ ν)p = drdtdΣ(y)dΣ(ν). 2 SN −1 ν ⊥ Ω(y,ν) Ω(y,ν) t − τ 1+sp Hence, uW s,p (Ω) < ∞ if and only if the righthand side is finite. The result now follows from Tonelli’s theorem. We also consider the case s = 1. Theorem 6.49. Let Ω ⊆ RN be an open set and 1 ≤ p < ∞. (i) If u ∈ W 1,p (Ω), then there exists a representative u ¯ of u such that 1,p the function u(y + ·ν) belongs to W (Ω(y, ν)) for every ν ∈ SN −1 and for HN −1 a.e. y ∈ ν ⊥ , with Z Z (u(y + tν)p + ∂ν u(y + tν)p )dtdΣ(y) ≤ kukpW 1,p (Ω) . ν⊥
Ω(y,ν)
(ii) Given u ∈ Lploc (Ω), assume that there exist N linearly independent vectors νi ∈ SN −1 , i = 1, . . . , N , and a representative u ¯ of u such that the function u(y + ·νi ) belongs to W 1,p (Ω(y, νi )) for HN −1 a.e. y ∈ νi⊥ , with Z Z (u(y + tνi )p + ∂νi u(y + tνi )p ) dtdΣ(y) < ∞. (6.24) νi⊥
Ω(y,νi )
Then u ∈ W 1,p (Ω) with Z N Z X p kukW 1,p (Ω) ≤ C (u(y + tνi )p + ∂νi u(y + tνi )p ) dtdΣ(y) i=1
νi⊥
Ω(y,νi )
for some constant C = C(N, p, ν1 , . . . , νN ) > 0. Proof. Step 1: Assume that u ∈ W 1,p (Ω). We claim that for every ν ∈ SN −1 there exists the weak directional derivative of u in the direction ν and that ∂ν u = ∇u · ν. To see this, let ψ ∈ Cc1 (Ω). By the definition of weak partial derivatives, Z Z ∂i uψ dx = − Ω
u∂i ψ dx. Ω
Multiply this equation by νi and sum over i = 1, . . . , N . Then Z Z Z (∇u · ν)ψ dx = − u(∇ψ · ν) dx = − u∂ν ψ dx, Ω
Ω
Ω
where in the last equality we use the fact that ∇ψ · ν = ∂ν ψ, which follows since ψ ∈ Cc1 (Ω). This shows that the weak directional derivative of u in the direction ν is given by ∇u · ν.
212
6. Fractional Sobolev Spaces
Step 2: In this step we will prove item (i). Assume that u ∈ W 1,p (Ω). Let E := {x ∈ Ω : x is a pLebesgue point of u}. Then LN (Ω \ E) = 0 (see [Leo22c]). Moreover, if x ∈ E, then there exists `x ∈ R such that Z 1 lim u(y) − `x p = 0. r→0+ r N B(x,r)∩Ω Therefore, if we set u ¯(x) := `x for x ∈ E and u ¯(x) := 0 for x ∈ Ω \ E, then u ¯ is a representative of u (see [Leo22c]). 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 standard properties of mollifiers (see [Leo22d]), uε (x) → u ¯(x) for every x ∈ E and Z (uε (x) − u ¯(x)p + k∇uε (x) − ∇¯ u(x)kp ) dx = 0. lim ε→0+
Fix ν ∈
SN −1 . Z
lim
ε→0+
Ωε
By Tonelli’s theorem, Z (uε (y + tν) − u ¯(y + tν)p
ν⊥
Ωε (y,ν)
+ ∂ν uε (y + tν) − ∂ν u ¯(y + tν)p ) dtdΣ(y) = 0. Hence, we may find a subsequence {εn }n such that for HN −1 a.e. y ∈ ν ⊥ , (6.25)Z lim
n→∞ Ω (y,ν) εn
(uεn (y+tν)− u ¯(y+tν)p +∂ν uεn (y+tν)−∂ν u ¯(y+tν)p ) dt = 0.
Moreover, by Tonelli’s theorem Z Z (¯ u(y + tν)p + ∂ν u ¯(y + tν)p ) dtdHN −1 (y) ≤ kukpW 1,p (Ω) < ∞ ν⊥
Ω(y,ν)
and Z
N
0 = L (Ω \ E) =
L1 ({t ∈ Ω(y, ν) : y + tν ∈ / E}) dΣ(y),
ν⊥
and so, we may find a set N0 ⊂ ν ⊥ , with HN −1 (N0 ) = 0, such that for all y ∈ ν ⊥ \ N0 for which Ω(y, ν) is nonempty, we have that Z (6.26) (¯ u(y + tν)p + ∂ν u ¯(y + tν)p ) dt < ∞, Ω(y,ν)
(6.25) holds, and y + tν ∈ E for L1 a.e. t ∈ Ω(y, ν). Fix y ∈ ν ⊥ \ N0 and let I be a maximal interval of Ω(y, ν). Fix t0 ∈ I such that y + t0 ν ∈ E and let t ∈ I. For all n large, the interval of endpoints t and t0 is contained in
6.2. Slicing
213
Ωεn (y, ν). Therefore, since un ∈ C ∞ (Ωεn ), by the fundamental theorem of calculus, Z t un (y + tν) = un (y + t0 ν) + ∂ν un (y + rν) dr. t0
Since y + t0 ν ∈ E, we have un (y + t0 ν) → u(y + t0 ν) ∈ R. On the other hand, by (6.25), it follows that as n → ∞, Z t ∂ν u ¯(y + rν) dr =: v(y + tν). un (y + tν) → u(y + t0 ν) + t0
Since y + tν ∈ E for L1 a.e. t ∈ I, we have that un (y + tν) → u(y + tν). Hence, Z t (6.27) u(y + tν) = u(y + t0 ν) + ∂ν u ¯(y + rν) dr = v(y + tν) t0
for L1 a.e. t ∈ I. The function v(y + ·ν) is absolutely continuous and its (classical) derivative is ∂ν u ¯(y + tν) for L1 a.e. t ∈ I (see [Leo22d]). It follows from (6.27) that u(y + ·ν) is a representative of v(y + ·ν). Hence, by (6.26), u(y + ·ν) ∈ W 1,p (I) (see [Leo22d]). Since I is an arbitrary maximal interval of Ω(y, ν), again by (6.26), we obtain that u(y +·ν) ∈ W 1,p (Ω(y, ν)). Step 3: In this step we will prove item (ii). Let u ∈ Lploc (Ω) and ν1 , . . . , νN ∈ SN −1 be as in the statement of item (ii). For every function ϕ ∈ Cc∞ (Ω) and for HN −1 a.e. y ∈ νi⊥ such that Ω(y, νi ) 6= ∅, we can integrate by parts to get Z Z u(y + tνi )∂νi ϕ(y + tνi ) dt = − ∂νi u(y + tνi )ϕ(y + tνi ) dt. Ω(y,ν)
Ω(y,ν)
Since this holds for HN −1 a.e. y ∈ νi⊥ , integrating over νi⊥ and using Fubini’s theorem yields Z Z (6.28) u(x)∂νi ϕ(x) dx = − ∂νi u(x)ϕ(x) dx, Ω
Ω
Lp (Ω)
which implies that ∂νi u ∈ is the weak directional derivative of u in the direction νi . Since ν1 , . . . , νN are linearlyPindependent, for every k = 1, . . . , N there exist αk,i ∈ R such that ek = N i=1 αk,i νi . Multiplying (6.28) by αk,i , summing over i, and using the fact that ∇ϕ · ν = ∂ν ϕ, we obtain Z Z Z N X u∂k ϕ dx = u∇ϕ · ek dx = αk,i u∇ϕ · νi dx Ω
Ω
=−
i=1
Z X N Ω i=1
αk,i ∂νi uϕ dx.
Ω
214
6. Fractional Sobolev Spaces
This of u ¯ with respect to xk is given Pproves that the weak partial derivative 1,p by N α ∂ u. In turn, by (6.24), u ∈ W (Ω). i=1 k,i νi Exercise 6.50. Let Ω ⊆ RN be an open set and 1 ≤ p < ∞. Prove that if ˙ 1,p (Ω), then there exists a representative u u∈W ¯ of u such that the function 1,p ˙ u(y + ·ν) belongs to W (Ω(y, ν)) for every ν ∈ SN −1 and for HN −1 a.e. y ∈ ν ⊥ , with Z Z Z k∇u(x)kp dx. ∂ν u(y + tν)p dtdΣ(y) ≤ ν⊥
Ω
Ω(y,ν)
6.3. Some Equivalent Seminorms In this section we discuss some equivalent seminorms that will be used in what follows. Given a function u : RN → R, for every h ∈ RN and m ∈ N, we define inductively the forward difference operator (6.29)
∆1h u(x) := ∆h u(x) := u(x + h) − u(x),
x ∈ RN ,
for m = 1 and (6.30)
m−1 ∆m u(x)), h u(x) := ∆h (∆h
x ∈ RN ,
for m ≥ 2. Note that when m = 2, (6.31)
∆2h u(x) = u(x + 2h) − 2u(x + h) + u(x).
Remark 6.51. If u ∈ Lp (RN ), 1 ≤ p ≤ ∞, then for every h ∈ RN and m ∈ N, (6.32)
m k∆m h ukLp (RN ) ≤ 2 kukLp (RN ) .
This follows by induction on m from (6.29), (6.30), Minkowski’s inequality, and the change of variables y = x + h. Given a function u : RN → R and h ∈ RN , we define the translation operator Th as (6.33)
Th (u)(x) := u(x + h),
x ∈ RN .
Observe that Th ◦ Tξ = Tξ ◦ Th for every h, ξ ∈ RN . In addition, ∆h = Th − I and for m ∈ N, with m ≥ 2, (6.34)
m ∆m h = (Th − I) := (Th − I) ◦ · · · ◦ (Th − I), m times
where I is the identity operator. In most of the following results, the particular value of s plays no role. In particular, the restriction s < 1 is unnecessary.
6.3. Some Equivalent Seminorms
215
Theorem 6.52. Let m ∈ N, 1 ≤ p < ∞, and s > 0. Then Z Z ∞ Z dh dt p m p k∆h ukLp (RN ) sup ∆m (6.35) ≈ h u(x) dx 1+sp . N +sp khk t RN 0 khk≤t RN for all Lebesgue measurable functions u : RN → R. We begin with a preliminary result. Lemma 6.53. Let 1 ≤ p < ∞ and m ∈ N. Then for every Lebesgue measurable function u : RN → R and every t > 0, Z dζ m sup k∆h ukLp (RN ) k∆m . ζ ukLp (RN ) kζkN B(0,t) khk≤t Proof. By the binomial theorem, for y, z ∈ R, we have m m X m X X k m jk m−k m m−k (y − 1) z = (−1)j+k m (−1)m−k m k k j y z k=0
=
m X
(−1)m−j
m j
k=0 j=0 m X
(y j − z)m =
(−1)m−j
m j
(y j − z)m + (z − 1)m .
j=1
j=0
y0
Since = 1, after relabeling the summation index in the last sum, this identity can be written as m X k m m−k (−1)m−k m − (−1)m (z − y k )m ]. (z − 1)m = k [(y − 1) z k=1
By replacing y and z with the operators T(1/m)ξ and Th (see (6.33)), respectively, where h, ξ ∈ RN , and using (6.34), we get m X m (−1)m−k m u(x) = ∆m h k [∆(k/m)ξ u(x + (m − k)h) k=1
− (−1)m ∆m h−(k/m)ξ u(x + kξ/m)]. In turn, k∆m h ukLp (RN )
m X
m [k∆m (k/m)ξ ukLp (RN ) + k∆h−(k/m)ξ ukLp (RN ) ].
k=1
Assume that h 6= 0 and ξ ∈ B(h/2, khk/2). Then (k/m)ξ, h − (k/m)ξ ∈ B(0, khk), and so, by averaging in the variable ξ over B(h/2, khk/2), we obtain Z m X 1 m k∆h ukLp (RN ) [k∆m (k/m)ξ ukLp (RN ) khkN B(h/2,khk/2) k=1 Z 1 k∆m + k∆m uk ] dξ Lp (RN ) ζ ukLp (RN ) dζ, h−(k/m)ξ khkN B(0,khk)
216
6. Fractional Sobolev Spaces
where in the last inequality we use the changes of variables ζ = (k/m)ξ and ζ = h − (k/m)ξ. It follows that Z dζ m k∆m sup k∆h ukLp (RN ) . ζ ukLp (RN ) kζkN B(0,t) khk≤t We turn to the proof of Theorem 6.52. Proof of Theorem 6.52. Using Lemma 6.53 and spherical coordinates, we get !p Z ∞ Z ∞ Z dt dζ dt p m m sup k∆h ukLp 1+sp k∆ζ ukLp N 1+sp t kζk t 0 khk≤t 0 B(0,t) p Z ∞ Z t Z dt dρ m N −1 . ≈ k∆ρσ ukLp dH (σ) 1+sp ρ t 0 0 SN −1 By Hardy’s inequality (see Theorem 1.3) and H¨older’s inequality, we can bound from above the righthand side by p Z ∞ Z dt m N −1 p k∆tσ ukL dH (σ) 1+sp t N −1 0 ZS ∞ Z Z dt dy p p m N −1 k∆tσ ukLp dH (σ) 1+sp ≈ k∆m . y ukLp t kykN +sp 0 SN −1 RN This proves the second inequality in (6.35). To prove the first inequality in (6.35), we use again spherical coordinates to write Z Z ∞Z dh dt p m N −1 k∆h ukLp = k∆m (σ) 1+sp tσ ukLp dH N +sp khk t N N −1 R Z0 ∞ ZS dt N −1 (σ) 1+sp ≤ sup k∆m h ukLp dH t 0 SN −1 khk≤t Z ∞ dt = βN sup k∆m h ukLp 1+sp , t 0 khk≤t where we recall that βN = HN −1 (SN −1 ).
Exercise 6.54. Prove that Z Z ∞ Z dh dt 2 k∆h ukL1 (RN ) ≈ sup ∆2h u(x) dx 2 N +1 khk t RN 0 khk≤t RN for all Lebesgue measurable functions u : RN → R (see Exercise 6.10). Next we show that in the definition of W s,p we can replace firstorder differences ∆h with ∆m h . In the following theorems we do not restrict s to be in (0, 1).
6.3. Some Equivalent Seminorms
217
Theorem 6.55. Let 1 ≤ p < ∞, s > 0, s ∈ / N, and let m ∈ N be such that m ≥ bsc + 1. Then for all u ∈ Lploc (RN ) and for every n ∈ N, with bsc + 1 ≤ n < m, Z Z Z Z p ∆m ∆nh u(x)p (m−n)p h u(x) (6.36) dxdh ≤ 2 dxdh. N +sp N +sp RN RN khk RN RN khk Moreover, if u ∈ Lploc (RN ) is such that Z Z ∆nh u(x)p dxdh < ∞, (6.37) N +sp RN RN khk then Z (6.38) RN
∆nh u(x)p dxdh N +sp RN khk
Z
Z RN
p ∆m h u(x) dxdh. N +sp RN khk
Z
We begin with a preliminary result. Lemma 6.56. Given u : RN → R, h ∈ RN \ {0}, and m ∈ N, the following identity holds (6.39)
−m m ∆m ∆2h u + Pm−1 (Th ) ◦ ∆m+1 u, hu=2 h
where Th is the translation operator given in (6.33) and Pm−1 is a polynomial of degree m − 1. Proof. We use the following identity for polynomials (t − 1)m = 2−m (t2 − 1)m + (t − 1)m − 2−m (t2 − 1)m = 2−m (t2 − 1)m + Pm−1 (t)(t − 1)m+1 ,
t ∈ R,
where, by the binomial theorem, (t + 1)m − 2m ((t − 1) + 2)m − 2m = −2−m t−1 t−1 m X j−1 =− , t ∈ R. 2−j m j (t − 1)
Pm−1 (t) = −2−m
j=1
By replacing t with the translation operator Th given in (6.33) and using the fact that ∆h = Th − I, we get (6.39). We turn to the proof of Theorem 6.55. Proof of Theorem 6.55. By (6.32) applied with ∆nh u in place of u, we have that m−n k∆m (∆nh u)kLp (RN ) ≤ 2m−n k∆nh ukLp (RN ) . h ukLp (RN ) = k∆h
Raising both sides to power p, dividing by khkN +sp , and integrating in h over RN gives (6.36).
218
6. Fractional Sobolev Spaces
Thus, it remains to prove the opposite inequality (6.38). Using an induction argument, it is enough to prove that Z
∆nh u(x)p dxdh N +sp RN khk
Z
(6.40) RN
Z
p ∆n+1 h u(x) dxdh N +sp RN khk
Z
RN
for all u ∈ Lploc (RN ) such that Z
∆nh u(x)p dxdh < ∞ N +sp RN khk
Z
(6.41) RN
and for all n ≥ bsc + 1. Step 1: We consider first the case n = 1, which implies that 0 < s < 1. Let u ∈ Lploc (RN ) be such that (6.41) holds. 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)), or, equivalently, 2∆1h u(x) = ∆12h u(x) − ∆2h u(x).
(6.42)
It follows by Minkowski’s inequality that Z 2 RN
k∆1h ukpLp Z
+ RN
Z + RN
dh khkN +sp
k∆2h ukpLp k∆2h ukpLp
1/p
Z ≤ RN
dh khkN +sp
1/p
dh khkN +sp
1/p
k∆12h ukpLp
= 2s
Z RN
dh khkN +sp
k∆1η ukpLp
1/p
dη kηkN +sp
1/p
,
where we make the change of variables h = η/2. Using (6.41), we have (2 − 2s )
Z RN
k∆1h ukpLp
dh khkN +sp
1/p
Z ≤ RN
k∆2h ukpLp
dh khkN +sp
1/p .
Step 2: Assume next that n ≥ 2. Let u ∈ Lploc (RN ) be such that (6.41) holds. By (6.39) we get p k∆nh ukLp ≤ 2−n k∆n2h ukLp + kPn−1 (Th ) ◦ ∆n+1 h ukL p ≤ 2−n k∆n2h ukLp + Ck∆n+1 h ukL ,
6.3. Some Equivalent Seminorms
219
where we use the fact that kTh (v)kLp = kvkLp . It follows by Minkowski’s inequality that Z Z 1/p 1/p dh dh p p n n −n k∆2h ukLp k∆h ukLp ≤2 khkN +sp khkN +sp RN RN Z 1/p Z 1/p dh dη p p n+1 n s−n +C k∆h ukLp k∆η ukLp =2 khkN +sp kηkN +sp RN RN 1/p Z dh p n+1 , +C k∆h ukLp khkN +sp RN where we make the change of variables h = η/2. Therefore, by (6.41), we get Z 1/p Z 1/p dh dh p p n+1 s−n n (1 − 2 ) k∆h ukLp k∆h ukLp . khkN +sp khkN +sp RN RN Note that here it is important that n ≥ bsc + 1 > s.
Remark 6.57. The restriction is crucial. Indeed, if u is a polynomial Z (6.37) Z of degree d > bsc + 1, then for all x ∈ RN if m > d.
RN RN
∆h u(x)p dxdh khkN +sp
= ∞, while ∆m h u(x) = 0
Exercise 6.58. Let 1 ≤ p, q < ∞, s > 0, s ∈ / N, and m, n ∈ N be such that m > n ≥ bsc + 1. (i) Prove that for all u ∈ Lploc (RN ), Z Z dh dh q m k∆h ukLp (RN ) k∆nh ukqLp (RN ) . N +sq khk khkN +sq RN RN (ii) Prove that if u ∈ Lploc (RN ) is such that Z dh k∆nh ukqLp (RN ) < ∞, khkN +sq RN then Z RN
k∆nh ukqLp (RN )
dh khkN +sq
Z RN
q k∆m h ukLp (RN )
dh . khkN +sq
Given i = 1, . . . , N , we recall that ∆1h,i u(x) := u(x + hei ) − u(x) and N := ∆1h,i (∆m−1 h,i u)(x) for x ∈ R and h ∈ R.
∆m h,i u(x)
Exercise 6.59. Let i = 1, . . . , N , 1 ≤ p, q < ∞, s > 0, s ∈ / N, and m, n ∈ N be such that m > n ≥ bsc + 1. (i) Prove that for all u ∈ Lploc (RN ), Z ∞ Z ∞ dh dh q m k∆nh,i ukqLp (RN ) 1+sq . k∆h,i ukLp (RN ) 1+sq h h 0 0
220
6. Fractional Sobolev Spaces
(ii) Prove that if u ∈ Lploc (RN ) is such that Z ∞ dh k∆nh,i ukqLp (RN ) 1+sq < ∞, h 0 then Z ∞ 0
k∆nh,i ukqLp (RN )
dh h1+sq
Z 0
∞
q k∆m h,i ukLp (RN )
dh h1+sq
.
Next we show that for u ∈ Lp (RN ), we can remove hypothesis (6.37). Corollary 6.60. Let 1 ≤ p < ∞, 0 < s < 1, m, n ∈ N, with n < m, and u ∈ Lp (RN ). Then Z Z Z Z p ∆nh u(x)p ∆m h u(x) dxdh dxdh. N +sp N +sp RN RN khk RN RN khk Proof. If the righthand side is infinite, there is nothing to prove, so assume that Z Z p ∆m h u(x) (6.43) dxdh < ∞. N +sp RN RN khk Let ε > 0 and consider the mollification uε = u ∗ ϕε ([Leo22c]). We claim that uε ∈ W s,p (RN ). Since ∂i uε = ∂i ϕε ∗ u ([Leo22c]), it follows that ∂uε belongs to Lp (RN ), and so uε ∈ W 1,p (RN ). In turn, uε ∈ W s,p (RN ) by Theorem 6.21. We can now apply Theorem 6.55 to obtain Z Z Z Z p ∆m ∆h uε (x)p h uε (x) dxdh dxdh N +sp N +sp RN RN khk RN RN khk Z Z p ∆m h u(x) dxdh, N +sp RN RN khk where the last inequality follows from the facts that, by linearity, ∆m h uε = p (RN ) (see (∆m u) ∗ ϕ and kv ∗ ϕ k ≤ kvk for every v ∈ L p N p N ε ε L (R ) L (R ) h [Leo22c]). Since uε → u pointwise LN a.e. in RN (see [Leo22c]), letting ε → 0+ and using Fatou’s lemma on the left gives Z Z Z Z p ∆m ∆h u(x)p h u(x) dxdh dxdh. N +sp N +sp RN RN khk RN RN khk If n = 1, we are done. Otherwise we apply (6.36), with m and n replaced by n and 1, respectively, to obtain Z Z Z Z ∆nh u(x)p ∆h u(x)p (n−1)p dxdh ≤ 2 dxdh N +sp N +sp RN RN khk RN RN khk Z Z p ∆m h u(x) dxdh. N +sp RN RN khk
6.3. Some Equivalent Seminorms
221
Theorem 6.61. Let 1 ≤ p < ∞, s > 0, s ∈ / N, and let m ∈ N be such that m ≥ bsc + 1. Then for all u ∈ Lploc (RN ), Z (6.44) RN
N Z Z p p X ∆m ∆mN t,i u(x) h u(x) dxdh dxdt N +sp t1+sp N RN khk R R i=1 Z Z p ∆m h u(x) dxdh. N +sp RN RN khk
Z
Proof. Step 1: Let h = (h1 , . . . , hN ) ∈ RN and consider the translation operator Th defined in (6.33). Then we can write Th − I = (Th1 e1 − I) + Th1 e1 ◦ (Th2 e2 − I) + · · · + Th1 e1 ◦ · · · ◦ ThN −1 eN −1 (ThN eN − I). Raising both sides to the power mN and using the multinomial theorem, which continues to work since Th ◦ Tξ = Tξ ◦ Th , we obtain X mN mN (Th −I) = Tα1 bh1 ◦(Th1 e1 −I)α1 ◦· · ·◦TαN bhN ◦(ThN eN −I)αN , α α=mN
)! mN , b h1 := 0, and b hk := where α ∈ NN = α(mN 0 is a multiindex, α 1 !···αN ! (h1 , . . . , hk−1 , 0, . . . , 0) for k ≥ 2. Note that since α = α1 +· · ·+αN = mN , there is at least one k ∈ {1, . . . , N } such that αk ≥ m. We denote that index by `α . It follows that for every function u : RN → RN and every x ∈ RN , X mN mN Tα1 bh1 ◦ ∆αh11,1 ◦ · · · ◦ TαN bhN ◦ ∆αhNN ,N u(x). (6.45) ∆h u(x) = α α=mN
¯ be a representative of u. We claim Step 2: Let u ∈ Lploc (RN ) and let u that the first inequality in (6.44) holds. By taking the Lp norm on both sides of (6.45) and using Minkowski’s inequality, (6.32), and some changes of variables, we have that k∆mN ¯kLp (RN ) h u
α
X
k∆h``α ,`α u ¯kLp (RN )
N X
α
k∆m ¯kLp (RN ) , hi ,i u
i=1
α=mN
where we recall that α`α ≥ m. Hence, by Tonelli’s theorem Z RN
N
X ∆mN ¯(x)p h u dxdh N +sp RN khk
Z
i=1
=
Z
∆m ¯(x)p hi ,i u
RN RN
khkN +sp
Z
N Z Z X i=1
R
RN
∆m ¯(x)p hi ,i u
dxdh
Z R
1 N +sp N −1 khk
dh0i dxdhi .
222
6. Fractional Sobolev Spaces
By Corollary 6.41, Z
1
R
N +sp N −1 khk
It follows that Z Z RN
dh0i ≈
N
X ∆mN ¯(x)p h u dxdh N +sp RN khk
1 . hi 1+sp
Z Z R
i=1
RN
∆m ¯(x)p dx hi ,i u
dhi . hi 1+sp
This shows the first inequality in (6.44). Step 3: Let u ∈ Lploc (RN ). Then for every i = 1, . . . , N and t ∈ R \ {0}, Z Z p p ∆m ∆m u(x) dx ≤ sup t,i h u(x) dx, RN
khk≤t RN
and so, also by Theorem 6.52, Z Z Z Z ∞ dt dt p m p ∆m ∆t,i u(x) dx 1+sp sup h u(x) dx 1+sp t t R RN 0 khk≤t RN Z Z p m ∆h u(x) dxdh. N +sp RN RN khk Summing over all i proves that the second inequality in (6.44) holds.
6.4. Density of Smooth Functions In this subsection we prove some density results for W s,p (Ω). We consider first the case in which Ω is the entire space. ˙ s,p (RN ), let Theorem 6.62. Let 1 ≤ p < ∞ and 0 < s < 1. Given u ∈ W uε := ϕε ∗ u, where ϕε is a standard mollifier. Then kuε kLp (RN ) ≤ kukLp (RN ) ,
uε W s,p (RN ) ≤ uW s,p (RN ) ,
and lim uε − uW s,p (RN ) = 0.
(6.46)
ε→0+
Moreover, W s,p (RN ) ∩ C ∞ (RN ) is dense in W s,p (RN ). Proof. By (6.5), Z Z RN
u(x) − u(y)p dxdy = N +sp RN kx − yk
Since ∆h uε = ϕε ∗ ∆h u, for every h ∈
Z
RN N R ,
∆h u(x)p dxdh. N +sp RN khk
Z
k∆h uε kLp (RN ) ≤ k∆h ukLp (RN )
(6.47) and (6.48)
lim kuε − ukLp (RN ) = 0,
ε→0+
lim k∆h uε − ∆h ukLp (RN ) = 0.
ε→0+
6.4. Density of Smooth Functions
223
For every h ∈ RN \ {0} and ε > 0 define gε (h) :=
1 khkN +sp
k∆h uε − ∆h ukpLp (RN ) ,
g(h) :=
1 khkN +sp
k∆h ukpLp (RN ) .
˙ s,p (RN ), we have that g ∈ L1 (RN ). Moreover, by (6.47), Since u ∈ W Minkowski’s inequality, and the convexity of the function tp , we have that gε (h) ≤ 2p g(h) for all h. Since gε (h) → 0 as ε → 0+ for all h by (6.48), we are in a position to apply the Lebesgue dominated convergence theorem to conclude that Z Z Z ∆h uε (x) − ∆h u(x)p lim dxdh = lim gε (h) dh = 0. khkN +sp ε→0+ RN RN ε→0+ RN This concludes the proof.
˙ s,p (RN ) has the property that u = 0 Remark 6.63. If the function u ∈ W outside some ball B(0, R), then for every δ > 0, we have that uε = 0 outside B(0, R + δ) for every 0 < ε < δ. Exercise 6.64. Let 1 ≤ p < ∞, 0 < s < 1, and u ∈ W s,p (RN ). Prove that kuε − ukLp (RN ) εs uW s,p (RN ) for all 0 < ε < 1. Here uε := ϕε ∗ u, where ϕε is a standard mollifier. Hint: Use Lemma 6.14. Next, we extend the Meyers–Serrin theorem (see [Leo22d]) to W s,p (Ω). Theorem 6.65 (Meyers–Serrin in W s,p ). Let Ω ⊆ RN be an open set, 1 ≤ p < ∞, and 0 < s < 1. For every u ∈ W s,p (Ω) there exists a sequence {un }n in W s,p (Ω) ∩ C ∞ (Ω) such that ku − un kW s,p (Ω) → 0 as n → ∞. In particular, W s,p (Ω) ∩ C ∞ (Ω) is dense in W s,p (Ω). S Proof. Let Ωi b Ωi+1 be such that Ω = ∞ i=1 Ωi and construct a smooth partition of unity F subordinated to the open cover {Ωi+1 \ Ωi−1 }i , where Ω−1 = Ω0 := ∅ (see [Leo17, Theorem C.21]). For each i ∈ N, let ψi be the sum of 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 (6.49)
∞ X
ψi = 1
in Ω.
i=1
Fix η > 0. For each i ∈ N we have that (6.50)
supp(ψi u) ⊂ Ωi+1 \ Ωi−1 .
224
6. Fractional Sobolev Spaces
Extend ψi u to be zero outside Ωi+1 \ Ωi−1 . Then by Theorem 6.23, ψi u ∈ W s,p (RN ), and so, by Theorem 6.62, we may find εi > 0 so small that supp(ψi u)εi ⊂ Ωi+1 \ Ωi−1
(6.51) and
η . 2i Note that in view of (6.51) for every Ωi+1 \ Ωi−1 P U b Ω only finitely many ∞ (Ω). cover U , and so, the function v := ∞ (ψ u) belongs to C i=1 i εi k(ψi u)εi − ψi ukW s,p (Ω) ≤ k(ψi u)εi − ψi ukW s,p (RN ) ≤
For x ∈ Ω` , by (6.49), (6.50), and (6.51), (6.52)
u(x) =
` X
(ψi u)(x),
v(x) =
i=1
` X
(ψi u)εi (x).
i=1
Hence, (6.53)
ku − vkW s,p (Ω` ) ≤
` X
k(ψi u)εi − ψi ukW s,p (Ω) ≤
i=1
` X η ≤ η. 2i i=1
Letting ` → ∞, it follows from the Lebesgue monotone convergence theorem that ku−vkW s,p (Ω) ≤ η. This also implies that u−v (and, in turn, v) belongs to the space W s,p (Ω). We consider now the density of functions W s,p (Ω)∩C ∞ (RN ) in W s,p (Ω). Here, with an abuse of notation, W s,p (Ω) ∩ C ∞ (RN ) stands for the space of functions u ∈ C ∞ (RN ) whose restriction to Ω belongs to W s,p (Ω). Theorem 6.66. Let 1 ≤ p < ∞ and 0 < s < 1. For every u ∈ W s,p (RN ), there exists a sequence {un }n in Cc∞ (RN ) such that ku − un kW s,p (RN ) → 0
as n → ∞.
We begin with a preliminary result, which is useful in itself. Lemma 6.67. Let Ω ⊆ RN be an open set, 1 ≤ p < ∞, and 0 < s < 1. Given u ∈ W s,p (Ω), there exists a sequence {un }n in W s,p (Ω) such that un → u in W s,p (Ω) as n → ∞ and for every n there exists rn > 0 such that un = 0 in Ω \ B(0, rn ). Proof. Consider a sequence of functions φn ∈ Cc∞ (RN ) such that 0 ≤ φn ≤ 1, φn = 1 in B(0, n), φn = 0 outside B(0, n + 1), and k∇φk∞ ≤ C for all n ∈ N. Define ψn := 1 − φn and un := ψn u. By the inequality (a + b)p ≤ 2p−1 ap + 2p−1 bp we have Z Z (ψn (x) − ψn (y))u(y)p u − un pW s,p (RN ) = ψn upW s,p (RN ) dxdy kx − ykN +sp RN RN Z Z u(x) − u(y)p + ψn (x)p dxdy =: A + B. kx − ykN +sp RN R N
6.4. Density of Smooth Functions
225
By the mean value theorem, for every x, y ∈ RN , ψn (x) − ψn (y) ≤ k∇ψn k∞ kx − yk ≤ Ckx − yk. Hence, since 0 ≤ ψn ≤ 1, we have ψn (x) − ψn (y) ≤ C min{1, kx − yk} for every x, y ∈ RN . Since by Lemma 6.24, Z Z Z min{1, kx − yk}p u(y)p dxdy u(y)p dy < ∞ kx − ykN +sp RN RN RN and ψn (x) → 0 as n → ∞ for every x ∈ RN , it follows by the Lebesgue dominated convergence theorem that A → 0 as n → ∞. Similarly, ψn (x) ≤ 1 and ψn (x) → 0 as n → ∞ for every x ∈ RN , and so, B → 0 as n → ∞ by the Lebesgue dominated convergence theorem. We turn to the proof of Theorem 6.66. Proof of Theorem 6.66. In view of Theorem 6.62, we can assume that u ∈ W s,p (RN ) ∩ C ∞ (RN ). Since the functions un constructed in Lemma 6.67 are obtained by multiplying u with a function φn ∈ Cc∞ (RN ), we have that un ∈ Cc∞ (RN ). Therefore, the conclusion follows from Lemma 6.67. Remark 6.68. If the function u ∈ W s,p (RN ) in Theorem 6.66 has the property that u = 0 outside some ball B(0, R), then for every δ > 0, we can construct each function un in such a way that un = 0 outside B(0, R + δ). Indeed, in view of Theorem 6.62 and Remark 6.63 we have that ϕε ∗ u → u in W s,p (RN ) as ε → 0+ and ϕε ∗ u = 0 outside B(0, R + δ) for all 0 < ε < δ. The functions un constructed in Lemma 6.67 are obtained by multiplying ϕε ∗ u by ψn . Hence, ψn (ϕε ∗ u) = 0 outside B(0, R + δ) for all 0 < ε < δ. We recall that a rigid motion T : RN → RN is an affine function given by T (x) = Rx + c, x ∈ RN , where R is an orthogonal matrix and c ∈ RN . Definition 6.69. Given an open set Ω ⊆ RN , we say that its boundary ∂Ω is Lipschitz continuous if for all x0 ∈ ∂Ω there exist a rigid motion T : RN → RN , with T (x0 ) = 0, a Lipschitz continuous function f : RN −1 → R, with f (0) = 0, and r > 0 such that, setting y := T (x), we have T (Ω ∩ B(x0 , r)) = {y ∈ B(0, r) : yN > f (y 0 )}. We say that ∂Ω is of class C m , m ∈ N0 , if the functions f are of class C m . Theorem 6.70. Let 1 ≤ p < ∞, 0 < s < 1, and Ω ⊆ RN be an open set with continuous boundary. For every u ∈ W s,p (Ω) there exists a sequence {un }n in W s,p (Ω) ∩ Cc∞ (RN ) such that ku − un kW s,p (Ω) → 0
as n → ∞.
226
6. Fractional Sobolev Spaces
We begin with an auxiliary result. Lemma 6.71. Let r > 0 and f : RN −1 → R be a continuous function such that f (0) = 0 and Ω = {x = (x0 , xN ) ∈ B(0, 2r) : xN > f (x0 )}. Let 1 ≤ p < ∞, 0 < s < 1, and u ∈ W s,p (Ω) be such that u = 0 in Ω\B(0, r). Define U := Ω ∪ (RN \ B(0, 2r)), and v : U → R by (6.54)
v(x) =
u(x) if x ∈ Ω, 0 if x ∈ RN \ B(0, 2r).
Then u ¯ ∈ W s,p (U ), with Z Z Z Z Z 1 v(x) − v(y)p u(x) − u(y)p p dxdy u(x) dx + dxdy. N +sp N +sp rsp Ω U U kx − yk Ω Ω kx − yk Proof. Write Z Z Z Z v(x) − v(y)p u(y)p dxdy = 2 dxdy N +sp N +sp U U kx − yk B(0,r)∩Ω RN \B(0,2r) kx − yk Z Z u(x) − u(y)p dxdy =: A + B. + N +sp Ω Ω kx − yk If y ∈ B(0, r) and x ∈ RN \ B(0, 2r), then kx − yk ≥ r. Hence, Z Z 1 1 1 dx ≤ dx ≈ sp . N +sp N +sp r RN \B(0,2r) kx − yk RN \B(y,r) kx − yk Hence, by Tonelli’s theorem, A
1 rsp
Z
u(y)p dy.
Ω
We turn to the proof of Theorem 6.70. Proof of Theorem 6.70. In view of Theorem 6.65, we can assume that u ∈ W s,p (Ω) ∩ C ∞ (Ω). Step 1: Assume first that there exist r > 0 and a continuous function f : RN −1 → R, with f (0) = 0, such that Ω = {x = (x0 , xN ) ∈ B(0, 2r) : xN > f (x0 )}. Let u ∈ W s,p (Ω) ∩ C ∞ (Ω), with u = 0 in Ω \ B(0, r). Consider the set U := Ω ∪ (RN \ B(0, 2r)) and let u ˜ : U → R be the extension of u to zero in U \ Ω; see (6.54). Since translation is continuous in Lp spaces, the function uδ (x) = u ˜(x0 , xN + δ),
x ∈ U,
0 < δ < r/2,
6.4. Density of Smooth Functions
227
belongs to W s,p (U ) and kuδ − u ˜kW s,p (U ) → 0 as δ → 0+ (exercise). Observe that uδ = 0 in U \ B(0, r + δ). Moreover, uδ is C ∞ in the larger open set Uδ = {x = (x0 , xN ) ∈ B(0, 2r) : xN > f (x0 ) − δ} ∪ (RN \ B(0, 2r)), which contains U . Construct a ϕδ ∈ C ∞ (RN ; [0, 1]) such that ϕδ (x) = 1 if x ∈ B(0, 2r) and xN > f (x0 ) and ϕδ (x) = 0 if x ∈ B(0, 2r) and xN < f (x0 ) − δ/2. Define vδ := ϕδ uδ in Uδ and vδ := 0 in RN \ Uδ . Then vδ ∈ Cc∞ (RN ). We claim that vδ → u ˜ in W s,p (U ). To see this, observe that vδ = uδ in Ω. On the other hand, u ˜(x) = vδ (x) = 0 if x ∈ U \ B(0, r + δ). Hence, we have vδ − u ˜pW s,p (U ) ≤ uδ − u ˜pW s,p (U ) Z Z uδ (y) − u ˜(y)p +2 dxdy =: A + B. N +sp B(0,r+δ)∩Ω RN \B(0,2r) kx − yk Then A → 0 as δ → 0 because kuδ − u ˜kW s,p (U ) → 0. Since r + δ < 3r/2, as in the proof of Lemma 6.71, for y ∈ B(0, r + δ) we have that R 1 1 RN \B(0,2r) kx−ykN +sp dx rsp , and so, Z 1 B sp uδ (y) − u ˜(y)p dy → 0 r B(0,r+δ)∩Ω as δ → 0. Finally, kvδ − u ˜kLp (U ) = kuδ − u ˜kLp (U ) → 0 as δ → 0. Step 2: We are ready to prove the general case. Let u ∈ W s,p (Ω) ∩ C ∞ (Ω). In view of Lemma 6.67, we can assume that there is a compact set K ⊂ RN such that u = 0 in Ω \ K. For every x0 ∈ ∂Ω there exist a rigid motion Tx0 : RN → RN , with Tx0 (x0 ) = 0, a continuous function fx0 : RN −1 → R, with fx0 (0) = 0, and rx0 > 0 such that Tx0 (Ω ∩ B(x0 , 2rx0 )) = {(y 0 , yN ) ∈ B(0, 2rx0 ) : yN > fx0 (y 0 )}. S S If the set Ω \ x∈∂Ω B(x, rx ) is nonempty, for every x0 ∈ Ω \ x∈∂Ω B(x, rx ), let B(x0 , rx0 ) ⊆ Ω. The family {B(x, rx )}x∈Ω is an open cover of the compact set Ω ∩ K. Hence, there is a finite number of balls B1 , . . . , B` that covers Ω ∩ K. Let {ψn }`n=1 be a smooth partition of unity subordinated to B1 , . . . , B` with supp ψn ⊆ Bn . Fix n ∈ {1, . . . , `} and define un := uψn ∈ W s,p (Ω) (see Theorem 6.23). There are two cases. If supp ψn is contained in Ω, then we set vn := ψn u ∈ Cc∞ (RN ) (see Theorem 6.23). If supp ψn is not contained in Ω, let xn ∈ ∂Ω be such Bn = B(xn , rn ). Then Tn (Ω ∩ B(xn , 2rn )) = {(y 0 , yN ) ∈ B(0, 2rn ) : yN > fn (y 0 )} =: Ωn , where fn : RN −1 → R is continuous and fn (0) = 0. Since ψn ∈ Cc∞ (RN ) and Tn is a rigid motion, the function (uψn ) ◦ Tn−1 belongs to W s,p (Ωn ) by Theorem 6.23 and Exercise 6.30, and it is zero in Ωn \ B(0, rn ). Hence,
228
6. Fractional Sobolev Spaces
given η > 0, we are in a position to apply Step 1 to (uψn ) ◦ Tn−1 to find v˜n ∈ Cc∞ (RN ) such that k˜ un − v˜n kW s,p (Un ) ≤ η/2n , where Un := Ωn ∪ (RN \ B(0, 2rn )) and u ˜n is obtained by extending u ◦ Tn−1 to be zero in Un \ Ωn . Define vn := v˜n ◦ Tn ∈ Cc∞ (RN ). By Exercise 6.30 and the fact that Tn (Ω) ⊆ Un , we have that kψn u − vn kW s,p (Ω) k˜ un − v˜n kW s,p (Un ) ≤ η/2n . P` ∞ N Define the function v := n . Then v ∈ Cc (R ), and so, v ∈ n=1 vP W s,p (Ω). Since u = 0 in Ω \ K and `n=1 ψn = 1 in Ω ∩ K, we have that P u = `n=1 ψn u in Ω. Hence, ku − vkW s,p (Ω) ≤
` X
kψn u − vn kW s,p (Ω) η
n=1
` X
2−n ≤ η.
n=1
6.5. The Space W0s,p and Its Dual We recall that the space W01,p (Ω) is the closure of the subspace Cc∞ (Ω) with respect to the norm in W 1,p (Ω). The space W01,p (Ω) plays an important role in the study of Dirichlet boundary value problems. In this section we define and study the space W0s,p (Ω) for 0 < s < 1. Definition 6.72. Let Ω ⊆ RN be an open set, 1 ≤ p < ∞, and 0 < s < 1. We define the fractional Sobolev space W0s,p (Ω) as the closure of Cc∞ (Ω) with respect to the norm in W s,p (Ω). The dual space of W0s,p (Ω) is denoted 0 by W −s,p (Ω). When p = 2 we write H0s (Ω) := W0s,2 (Ω) and H −s (Ω) := W −s,2 (Ω). 0
Next we study some properties of the dual space W −s,p (Ω) of W0s,p (Ω). Theorem 6.73. Let Ω ⊆ RN be an open set, 1 < p < ∞, and 0 < s < 1. 0 Then W0s,p (Ω) and W −s,p (Ω) are reflexive. Proof. Since W0s,p (Ω) is a closed subspace of the reflexive space W s,p (Ω) (see Theorem 6.11), it is reflexive (see [Leo22b]). In turn, since the dual 0 of a reflexive space is reflexive (see [Leo22b]), W −s,p (Ω) = (W0s,p (Ω))0 is reflexive. Theorem 6.74. Let Ω ⊆ RN be an open set, 1 < p < ∞, and 0 < s < 1. 0 0 Then Cc∞ (Ω) and Lp (Ω) are dense in W −s,p (Ω).
6.5. The Space W0s,p and Its Dual
229
0
Proof. Let v ∈ Lp (Ω) and consider the linear function Tv : W0s,p (Ω) → R, defined by Z v(x)u(x) dx, u ∈ W0s,p (Ω). Tv (u) := Ω
By H¨older’s inequality, Tv (u) ≤ kvkLp0 (Ω) kukLp (Ω) ≤ kvkLp0 (Ω) kukW s,p (Ω) , 0
which implies that Tv is linear and continuous, and so, Tv ∈ W −s,p (Ω). 0 Consider the subspace Y := {Tv : v ∈ Lp (Ω)}. We claim that Y is dense in 0 0 W −s,p (Ω). To prove the claim, it suffices to show that if L ∈ (W −s,p (Ω))0 0 is such that L(Tv ) = 0 for all v ∈ Lp (Ω), then L = 0. 0
Consider the linear function J : W0s,p (Ω) → (W −s,p (Ω))0 defined by 0 J(u)(T ) := T (u), u ∈ W0s,p (Ω), T ∈ W −s,p (Ω). Since W0s,p (Ω) is reflexive 0 by Theorem 6.73, we have that J(W0s,p (Ω)) = (W −s,p (Ω))0 , and so, since 0 L ∈ (W −s,p (Ω))0 , there exists u ∈ W0s,p (Ω) such that J(u) = L, that is, 0 L(T ) = T (u) for all T ∈ W −s,p (Ω). It follows that Z 0 = L(Tv ) = Tv (u) = v(x)u(x) dx Ω 0 Lp (Ω).
for all v ∈ This implies that u = 0, and, in turn, that L = 0. Hence, 0 we have shown that Y is dense in W −s,p (Ω). With a similar proof we can show that Z := {Tv : v ∈ Cc∞ (Ω)} is dense 0 in W −s,p (Ω). Note that in view of Theorem 6.66, for 1 ≤ p < ∞ and 0 < s < 1, W0s,p (RN ) = W s,p (RN ). In what follows, we show that if Ω is sufficiently regular and sp ≤ 1, then W0s,p (Ω) = W s,p (Ω). We begin with the case Ω = RN +. Theorem 6.75. Let 1 ≤ p < ∞ and 0 < s < 1, with sp ≤ 1. Then s,p (RN ). W0s,p (RN +) = W + Proof. Let u ∈ W s,p (RN + ), N ≥ 2. By Exercise 6.32, we can extend u to a s,p N function v ∈ W (R ). In turn, using Theorem 6.66, we can assume that u ∈ Cc∞ (RN ). Let R > 0 be so large that u(x) = 0 for all x = (x0 , xN ) ∈ RN with kx0 kN −1 ≥ R. Let φ ∈ C ∞ ([0, ∞)) be such that 0 ≤ φ ≤ 1, φ(t) = 1 for t ≥ 1, and φ(t) = 0 for t ≤ 21 . Given ε > 0, define φε (xN ) := φ(xN /ε) and ψε (xN ) := 1 − φε (xN ), x ∈ RN + . Then u(x) − φε (xN )u(x) = ψε (xN )u(x). 0 N −1 For every fixed x ∈ R , with kx0 kN −1 < R, we apply Theorem 1.73 (see (1.44)) to the function u(x0 , ·) to get that Z ∞ (ψε u)(x0 , xN )p dxN εku(x0 , ·)kp∞ εkukp∞ 0
230
6. Fractional Sobolev Spaces
and ∞Z ∞
Z 0
(ψε u)(x0 , xN ) − (ψε u)(x0 , yN )p dyN dxN xN − yN 1+sp
0
k∂N u(x0 , ·)ksp ε1−sp ku(x0 , ·)kp∞ + εku(x0 , ·)k(1−s)p ∞ ∞ ε1−sp kukp∞ + εkuk(1−s)p k∂N uksp ∞ ∞, where we recall that we are assuming that u ∈ Cc∞ (RN ). On the other hand, if kx0 kN −1 ≥ R, then (ψε u)(x0 , ·) = 0, and so, the lefthand sides of the last two inequalities are zero. Thus, if we integrate in x0 over BN −1 (0, R), we get Z Z ∞
RN −1
(ψε u)(x0 , xN )p dxN dx0 RN −1 εkukp∞
0
and Z
∞Z ∞
Z
RN −1 0 0 N −1 1−sp
R
ε
(ψε u)(x0 , xN ) − (ψε u)(x0 , yN )p dyN dxN dx0 xN − yN 1+sp
kukp∞ + RN −1 εkuk(1−s)p k∂N uksp ∞ ∞.
Next, we consider i = 1, . . . , N − 1. Since ψ depends only on xN , Z Z (ψε u)(x0i , xi ) − (ψε u)(x0i , yi )p dyi dxi xi − yi 1+sp R R Z Z u(x0i , xi ) − u(x0i , yi )p dyi dxi . = ψε (xN ) xi − yi 1+sp R R Since 0 ≤ ψε (xN ) ≤ 1 and ψε (xN ) = 0 for xN ≥ ε, it follows by the Lebesgue dominated convergence theorem that Z Z Z (ψε u)(x0i , xi ) − (ψε u)(x0i , yi )p (6.55) dyi dxi dx0i → 0. 1+sp N −1 x − y  i i R+ R R Hence, by Theorem 6.38, u − φε upW s,p (RN ) +
N −1 Z X
Z Z
i=1
−1 RN +
Z
Z
R
R
∞Z
∞
+
RN −1 N −1 Z X i=1
+R
0
−1 RN +
0
Z Z R
N −1 1−sp
ε
(ψε u)(x0i , xi ) − (ψε u)(x0i , yi )p dyi dxi dx0i xi − yi 1+sp
R
(ψε u)(x0 , xN ) − (ψε u)(x0 , yN )p dyN dxN dx0 xN − yN 1+sp
(ψε u)(x0i , xi ) − (ψε u)(x0i , yi )p dyi dxi dx0i xi − yi 1+sp
kukp∞ + RN −1 εkuk(1−s)p k∂N uksp ∞ ∞.
By (6.55), it follows that if p < 1/s, then ku − φε ukW s,p (RN ) → 0 as ε → 0+ . s,p N Since φε u ∈ W0s,p (RN + ), this shows that u ∈ W0 (R+ ).
+
6.5. The Space W0s,p and Its Dual
231
If p = 1/s, then ku − φε ukLp (RN ) → 0 and φε uW 1/p,p (RN ) ≤ C. Taking + + εn → 0 as n → ∞, we have that the sequence {φεn u}n is bounded in 1/p,p 1/p,p W0 (RN (RN + ), and so, since W0 + ) is reflexive (see Theorem 6.73), there 1/p,p (RN exist a subsequence {φεnk u}k of {φεn u}n and v ∈ W0 + ) such that 1/p,p
φεnk u * v in W0
(RN + ). In turn, by Corollary 6.12, which continues to
1/p,p
p N hold for W0 (RN + ), we have that φεnk u * v in L (R+ ). On the other hand, since ku − φε ukLp (RN ) → 0, necessarily, v = u, which implies that +
u∈
1/p,p W0 (RN + ).
Remark 6.76 (Important). Theorem 6.75 implies in particular that we cannot define a continuous trace operator in W s,p (RN + ) when sp ≤ 1 (see Chapter 9 for the definition of the trace). Indeed, given a smooth function s,p (RN ) with u(·, 0) 6= 0, there exists a sequence in {u } u ∈ Cc∞ (RN n n + + ) in W N ∞ in Cc (R+ ) such that ku − un kW s,p (RN ) → 0 and un (·, 0) = 0. +
Remark 6.77. If the function u ∈ W s,p (RN + ) has the property that u(x) = 0 \ B(0, R), then for every δ > 0, we can assume that each for all x ∈ RN + function φε u has the property that (φε u)(x) = 0 for all x ∈ RN + \ B(0, R + δ). Indeed, at the beginning of the proof of Theorem 6.75, we extend u to a function v in W s,p (RN ) and then use Theorem 6.66 to replace v with a function w in Cc∞ (RN ). In view of Remark 6.68, we can assume that w = 0 outside B(0, R + δ). In turn, φε w = 0 outside B(0, R + δ). Using a partition of unity we can now prove the following result. Theorem 6.78. Let 1 ≤ p < ∞, 0 < s < 1 with sp ≤ 1, and Ω ⊆ RN be an open set with Lipschitz continuous boundary. For every u ∈ W s,p (Ω) there exists a sequence {un }n in Cc∞ (Ω) such that ku − un kW s,p (Ω) → 0 as n → ∞. In particular, W0s,p (Ω) = W s,p (Ω). Proof. Step 1: Assume first that there exists a Lipschitz continuous function f : RN −1 → R such that Ω = {x = (x0 , xN ) ∈ RN −1 × R : xN > f (x0 )}, and let u ∈ W s,p (Ω). Set L := Lip f . By Corollary 6.31, the function s,p (RN ) and v v(y) := u(y 0 , yN + f (y 0 )), y ∈ RN + , belongs to W + W s,p (RN ) L +
uW s,p (Ω) . By Theorem 6.75 there exists {vn }n in Cc∞ (RN + ) such that kv − 0 vn kW s,p (RN ) → 0 as n → ∞. Define un (x) := vn (x , xN − f (x0 )), x ∈ Ω. + Then un is Lipschitz continuous with compact support contained in Ω and
232
6. Fractional Sobolev Spaces
by Corollary 6.31, un ∈ W s,p (Ω), with ku − un kW s,p (Ω) L kv − vn kW s,p (RN ) → 0. +
Since un has compact support contained in Ω, given a standard mollifier ϕε , we have that ϕε ∗ un has compact support contained in Ω for all ε > 0 sufficiently small (depending on n), and by Theorem 6.62, we have kϕε ∗ un − un kW s,p (Ω) → 0 as ε → 0+ . Hence, by a diagonal argument (see the proof of Lemma 6.15), we obtain the result in this case. Step 2: Let f , L, Ω be as in Step 1, with the additional hypothesis that f (0) = 0. Let u ∈ W s,p (Ω ∩ B(0, 2r)) be such that u = 0 in Ω ∩ B(0, 2r) \ B(0, r) for some r > 0. Define U := (Ω ∩ B(0, 2r)) ∪ (RN \ B(0, 2r)) and u ¯ : U → R by u ˜(x) :=
u(x) if x ∈ Ω ∩ B(0, 2r), 0 if x ∈ U \ (Ω ∩ B(0, 2r)).
By Lemma 6.71, u ˜ ∈ W s,p (U ). Define v(y) := u ˜(y 0 , yN + f (y 0 )), y ∈ RN +. N s,p N As in Step 1, v ∈ W (R+ ). We claim that v = 0 in R+ \ B(0, (2 + L)r). 0 To see this, let y ∈ RN + \ B(0, (2 + L)r). If ky kN −1 ≥ r, then we have that k(y 0 , yN + f (y 0 ))k ≥ r, and so, v(y) = 0. On the other hand, if ky 0 kN −1 < r, then using the facts that f is Lipschitz continuous and f (0) = 0, q k(y 0 , f (y 0 ))k = ky 0 k2N −1 + (f (y 0 ) − f (0))2 ≤ (1 + L)ky 0 kN −1 . Therefore, k(y 0 , yN + f (y 0 ))k ≥ k(y 0 , yN )k − k(y 0 , f (y 0 ))k > (2 + L)r − (1 + L)r = r, and so, again, v(y) = 0. This proves the claim. In turn, by Remark 6.77 with δ = Lr, the functions vn in Step 1 can be constructed in such a way 0 0 that vn = 0 in RN + \ B(0, (1 + L)2r). Let un (x) := vn (x , xN − f (x )), x ∈ Ω. ∞ Then un ∈ Cc (Ω) and k˜ u − un kW s,p (Ω) → 0 as n → ∞. Reasoning as above, we have that un = 0 in Ω \ B(0, R), where R := (1 + L)2 4r. Define V := Ω ∪ (RN \ B(0, 2R)), and u ˜n : V → R, by u ˜n (x) :=
un (x) if x ∈ Ω, 0 if x ∈ V \ Ω.
Reasoning as in Lemma 6.71, we have that u ˜n ∈ W s,p (V ), with ˜ u−u ˜n W s,p (V )
1 k˜ u − un kLp (Ω) + ˜ u − un W s,p (Ω) → 0 Rs
6.5. The Space W0s,p and Its Dual
233
as n → ∞. Also, k˜ u−u ˜n kLp (V ) = k˜ u − un kLp (Ω) → 0. Moreover, since un ∈ Cc∞ (Ω) we have that u ˜n ∈ Cc∞ (U ). Step 3: By Theorem 6.65, we can assume that u ∈ W s,p (Ω) ∩ C ∞ (Ω). Moreover, by Lemma 6.67, we can assume that u = 0 outside a compact set K ⊂ RN . For every x0 ∈ ∂Ω there exist a rigid motion Tx0 : RN → RN , with Tx0 (x0 ) = 0, a Lipschitz continuous function fx0 : RN −1 → R, with fx0 (0) = 0, and Rx0 > 0 such that (6.56)
Tx0 (Ω ∩ B(x0 , 2Rx0 )) = {(y 0 , yN ) ∈ B(0, 2Rx0 ) : yN > fx0 (y 0 )}.
Let Lx0 := Lip fx0 and rx0 := Rx0 /[4(1 + Lx0 )2 ]. S S If Ω \ x∈∂Ω B(x, rx ) is nonempty, for every x0 ∈ Ω \ x∈∂Ω B(x, rx ), let B(x0 , rx0 ) ⊆ Ω. The family {B(x, rx )}x∈Ω is an open cover of the compact set Ω ∩ K. Hence, there is a finite number of balls B1 , . . . , B` , where Bn := B(xn , rn ), that covers Ω ∩ K. Let {ψn }`n=1 be a smooth partition of unity subordinated to B1 , . . . , B` , with supp ψn ⊆ Bn for every n. Fix n ∈ {1, . . . , `} and define un := uψn ∈ W s,p (Ω) (see Theorem 6.23). There are two cases. If supp ψn is contained in Ω, then we set vn := ψn u ∈ Cc∞ (RN ). If supp ψn is not contained in Ω, let xn ∈ ∂Ω be such that supp ψn ⊆ Bn = B(xn , rn ). By Exercise 6.30, (uψn ) ◦ Tn−1 ∈ W s,p (Ωn ∩ B(0, 2rn )), where Ωn := {(y 0 , yN ) ∈ RN −1 × R : yN > fn (y 0 )}. Hence, we can apply Step 2 to find u ˜n ∈ Cc∞ (RN ) such that u ˜n (y) = 0 in Ωn \ B(0, Rn ) and k(uψn ) ◦ Tn−1 − u ˜n kW s,p (Vn ) ≤ η/2n , where Vn := Ωn ∪ (RN \ B(0, 2Rn )) and we extended (uψn ) ◦ Tn−1 to be zero in Vn \ B(0, rn ). Define un := u ˜n ◦ Tn . By Exercise 6.30 and the fact that Ω ⊆ Tn−1 (Vn ) (see (6.56)), kuψn − un kW s,p (Ω) ≤ k(uψn ) ◦ Tn−1 − u ˜n kW s,p (Vn ) ≤ η/2n . P Define the function v := `n=1 vn . Then v ∈ Cc∞ (RN ), and so, v ∈ W s,p (Ω). P P Since u = 0 in Ω\K and `n=1 ψn = 1 in Ω∩K, we have that u = `n=1 ψn u in Ω. In turn, ku − vkW s,p (Ω) ≤
` X n=1
kψn u − un kW s,p (Ω) ≤ η
` X n=1
2−n ≤ η.
234
6. Fractional Sobolev Spaces
6.6. Hardy’s Inequality In this section we study the validity of the following Hardy’s inequality Z Z Z u(x) − u(y)p u(x)p dx ≤ C dxdy. sp N +sp Ω Ω kx − yk Ω (dist(x, ∂Ω)) We begin with the case Ω = RN +. Theorem 6.79 (Hardy’s inequality in W s,p (RN + )). Let 1 ≤ p < ∞ and 0 < s < 1, with sp 6= 1. If sp < 1, then for all u ∈ W s,p (RN + ), Z Z Z u(x)p u(x) − u(y)p (6.57) dxdy, sp dx kx − ykN +sp xN RN RN RN + + + N while if sp > 1, then for all u ∈ W s,p (RN + ) ∩ C(R+ ), Z Z Z u(x) − u(y)p u(x) − u(x0 , 0)p (6.58) dx dxdy. kx − ykN +sp xsp RN RN RN N + + +
Proof. We only prove the case sp < 1 and leave the case sp > 1 as an exercise. Assume that N ≥ 2. Let u ¯ be a representative of u. By Theorem 6.38, Z Z ∞Z ∞ ¯ u(x0 , xN ) − u ¯(x0 , yN )p dyN dxN dx0 < ∞, 1+sp x − y  N −1 N N R 0 0 and thus, by Fubini’s theorem, Z ∞Z ∞ ¯ u(x0 , xN ) − u ¯(x0 , yN )p dyN dxN < ∞ xN − yN 1+sp 0 0 for LN −1 a.e. x0 ∈ RN −1 . Moreover, again by Fubini’s theorem, u ¯(x0 , ·) ∈ p N −1 0 N −1 0 N −1 L (R+ ) for L a.e. x ∈ R . Let x ∈ L be such that both are 0 s,p satisfied, so that u ¯(x , ·) ∈ W (R+ ). By Theorem 1.76 applied to u ¯(x0 , ·), Z ∞Z ∞ Z ∞ ¯ u(x0 , xN ) − u ¯(x0 , yN )p ¯ u(x0 , xN )p dxN dyN dxN sp xN − yN 1+sp xN 0 0 0 whenever the righthand side is finite. Integrating in x0 over RN −1 and using Tonelli’s theorem gives Z Z ∞Z ∞ Z u(x)p u(x0 , xN ) − u(x0 , yN )p dx dyN dxN dx0 . sp 1+sp N x − y  x N −1 N N R 0 0 R+ N The result now follows from Theorem 6.38.
Remark 6.80. In Theorem 9.14, we will see that if sp > 1 there exists a p N linear continuous operator Tr : W s,p (RN + ) → Lloc (∂R+ ) such that Tr(u) =
6.6. Hardy’s Inequality
235
N u∂RN for every u ∈ W s,p (RN + ) ∩ C(R+ ). Hence, inequality (6.58) can be + replaced by Z Z Z u(x) − Tr(u)(x0 , 0)p u(x) − u(y)p dxdy dx sp kx − ykN +sp xN RN RN RN + + +
for all W s,p (RN + ). Example 6.81. Inequality (6.57) fails when sp ≥ 1. To see this, take ϕ ∈ Cc∞ ([0, ∞)) with ϕ(0) = 1 and ψ ∈ Cc∞ (RN −1 ) \ {0}, and define u(x) := ϕ(xN )ψ(x0 ). Then by the mean value theorem and the fact that ϕ and ψ have compact support, one can show that u ∈ W s,p (RN + ) (exercise) but Z Z Z ∞ u(x)p ϕ(xN )p ψ(x0 )p dx0 dxN = ∞ sp dy = xN xsp RN RN −1 0 N + Rδ since ϕ(xN ) ≥ 12 for xN ∈ [0, δ) and 0 x1sp dxN = ∞. N
Exercise 6.82. Let λ > 0, with λ 6= 1, N ≥ 2, 1 ≤ p < ∞, and u ∈ Lploc (RN + ) be such that Z u(x)p dx < ∞. xλN RN + Prove that Z Z Z ∞Z ∞ u(x)p u(x0 , xN ) − u(x0 , yN )p dx dyN dxN dx0 . λ 1+λ N x − y  N −1 xN N N R+ R 0 0 Exercise 6.83. Let Ω = {x ∈ RN : xN > f (x0 )}, where f : RN −1 → R is a Lipschitz continuous function. Prove that for every x = (x0 , xN ) ∈ RN \ ∂Ω, 1 p
1 + (Lip f )2
f (x0 ) − xN  ≤ dist(x, ∂Ω) ≤ f (x0 ) − xN .
Next we consider Lipschitz continuous domains. Theorem 6.84. Let Ω = {x ∈ RN : xN > f (x0 )}, where f : RN −1 → R is a Lipschitz continuous function, 1 ≤ p < ∞, and 0 < s < 1, with sp < 1. If sp < 1, then for all u ∈ W s,p (Ω), Z Z Z u(x)p u(x) − u(y)p dx dxdy, Lip f sp N +sp Ω (dist(x, ∂Ω)) Ω Ω kx − yk while if sp > 1, then for all u ∈ W s,p (Ω) ∩ C(Ω), Z Z Z u(x) − u(x0 , f (x0 ))p u(x) − u(y)p dx dxdy. Lip f N +sp (dist(x, ∂Ω))sp Ω Ω Ω kx − yk
236
6. Fractional Sobolev Spaces
Proof. We only prove the case sp < 1 since the other one is similar. Define s,p (RN ) with v(y) = u(y 0 , yN + f (y 0 )), y ∈ RN + . By Corollary 6.31, v ∈ W + vW s,p (RN ) Lip f uW s,p (Ω) . +
Hence, by Theorem 6.79, Z Z Z Z u(y 0 , yN + f (y 0 ))p v(y)p v(x) − v(y)p dy = dy dxdy sp sp kx − ykN +sp yN yN RN RN RN RN + + + + Z Z u(x) − u(y)p Lip f dxdy. N +sp Ω Ω kx − yk By considering the change of variables x = (y 0 , yN + f (y 0 )) (see Corollary 6.31), it follows that Z Z u(y 0 , yN + f (y 0 ))p u(x)p dx = dy sp 0 ))sp N (x − f (x y N R RN N + + Z Z u(x) − u(y)p Lip f dxdy. N +sp Ω Ω kx − yk It now suffices to apply Exercise 6.83.
Exercise 6.85 shows that the inequality Z Z Z u(x)p u(x) − u(y)p dx dxdy (6.59) Ω sp N +sp Ω (dist(x, ∂Ω)) Ω Ω kx − yk in W0s,p (Ω) = W s,p (Ω) fails for bounded domains with Lipschitz continuous boundary when sp ≤ 1. Exercise 6.85. Let 1 ≤ p < ∞ and 0 < s < 1, with sp ≤ 1. Let Ω ⊂ RN be an open bounded set and assume that there exist ε ≥ 0 and c > 0 such that for every n ∈ N there exists a finite set En ⊂ ∂Ω with cardinality less than or equal to cnN −sp−ε such that [ B(x, 1/(2n)). ∂Ω ⊆ x∈En
(i) Construct un ∈ Cc∞ (RN ) such that 0 ≤ un ≤ 1, un = 1 on Ωn := {x ∈ Ω : dist(x, ∂Ω) > 1/(2n)}, un = 0 outside Ω2n , and k∇un kL∞ (Ω) ≤ Cn, for some constant C > 0 independent of n. (ii) Prove that for x0 ∈ En , Z Z un (x) − un (y)p dxdy ≤ Cnsp−N N +sp kx − yk B(x0 ,2/n) B(x0 ,1/n) Z
and that for k ∈ N, Z
B(x0 ,(2k+2)/n)\B(x0 ,2k/n)
B(x0 ,1/n)
un (x) − un (y)p nsp−N dxdy ≤ C kx − ykN +sp (k + 1)1+sp
6.6. Hardy’s Inequality
237
for some constant C > 0 independent of n and k. (iii) Prove that Z Z Ω
Ω
un (x) − un (y)p 1 dxdy ε N +sp kx − yk n
for some constant C > 0 independent of n. (iv) Prove that Z
un (x)p dx → sp Ω (dist(x, ∂Ω)) as n → ∞.
Z Ω
1 dx (dist(x, ∂Ω))sp
(v) Assume that Z Ω
1 dx = ∞ (dist(x, ∂Ω))sp
when ε = 0. Prove that Hardy’s inequality Z Z Z u(x) − u(y)p u(x)p dx ≤ C dxdy sp N +sp Ω Ω kx − yk Ω (dist(x, ∂Ω)) fails in W0s,p (Ω) in both cases ε > 0 and ε = 0. Exercise 6.86. Let 1 ≤ p < ∞ and 0 < s < 1 with sp ≤ 1. Let Ω ⊂ RN be an open bounded set with Lipschitz continuous boundary. Prove that that there exist ε ≥ 0 and c > 0 such that for every n ∈ N there exists a finite set En ⊂ ∂Ω with cardinality less than or equal to cnN −sp−ε such that [ B(x, 1/(2n)). ∂Ω ⊆ x∈En
Exercise 6.87. Let Ω ⊂ RN be an open set and let T : RN → RN be a rigid motion. Prove that for every x ∈ RN , dist(T (x), ∂Ω) = dist(x, ∂T −1 (Ω)). Theorem 6.88. Let Ω ⊂ RN be an open bounded set with Lipschitz continuous boundary, 1 ≤ p < ∞, and let 0 < s < 1 be such that sp 6= 1. Then Z Z Z Z u(x)p u(x) − u(y)p p dx u(x) dx + dxdy Ω sp N +sp Ω (dist(x, ∂Ω)) Ω Ω Ω kx − yk for all u ∈ W s,p (Ω) if sp < 1 and for all u ∈ W s,p (Ω) ∩ C(Ω) with u = 0 on ∂Ω if sp > 1. Proof. We only prove the case sp < 1 and leave the case sp > 1 as an exercise.
238
6. Fractional Sobolev Spaces
Step 1: Assume first that there exists a Lipschitz continuous function f : RN −1 → R such that f (0) = 0 and Ω = {x = (x0 , xN ) ∈ RN −1 × R : xN > f (x0 )}, and let u ∈ W s,p (Ω ∩ B(0, 2r)) be such that u = 0 in Ω ∩ (B(0, 2r) \ B(0, r)). Define u ˜ : Ω → R by u(x) if x ∈ Ω ∩ B(0, 2r), u ˜(x) := 0 if x ∈ Ω \ B(0, 2r). By Lemma 6.71, u ˜ ∈ W s,p (Ω), with Z Z Z 1 ˜ u(x) − u ˜(y)p dxdy sp u(x)p dx N +sp kx − yk r Ω Ω Ω∩B(0,r) Z Z u(x) − u(y)p + dxdy. N +sp Ω∩B(0,2r) Ω∩B(0,2r) kx − yk Hence, we can apply Theorem 6.84 to obtain Z Z u(x)p ˜ u(x)p dx = dx sp sp Ω∩B(0,r) (dist(x, ∂Ω)) Ω (dist(x, ∂Ω)) Z Z ˜ u(x) − u ˜(y)p Lip f dxdy N +sp Ω Ω kx − yk Z Z Z u(x) − u(y)p 1 p u(x) dx + dxdy. Lip f sp N +sp r Ω∩B(0,2r) Ω∩B(0,2r) kx − yk Ω∩B(0,r) Step 2: Let Ω ⊂ RN be an open bounded set with Lipschitz continuous boundary and let u ∈ W s,p (Ω). For every x0 ∈ ∂Ω there exist a rigid motion Tx0 : RN → RN , with Tx0 (x0 ) = 0, a Lipschitz continuous function fx0 : RN −1 → R, with fx0 (0) = 0, and rx0 > 0 such that Tx0 (Ω ∩ B(x0 , 2rx0 )) = {(y 0 , yN ) ∈ B(0, 2rx0 ) : yN > fx0 (y 0 )}. S S If the set Ω \ x∈∂Ω B(x, rx ) is nonempty, for every x0 ∈ Ω \ x∈∂Ω B(x, rx ), let B(x0 , rx0 ) ⊆ Ω. The family {B(x, rx )}x∈Ω is an open cover of the compact set Ω. Hence, there is a finite number of balls B1 , . . . , B` that covers Ω. Write Bn = B(xn , rn ) and set Ln := Lip fxn . Let {ψn }`n=1 be a smooth partition of unity subordinated to B1 , . . . , B` , with supp ψn ⊆ Bn for every n.
(6.60)
Fix n ∈ {1, . . . , `} and define un := uψn ∈ W s,p (Ω) (see Theorem 6.23). There are two cases. If supp ψn is contained in Ω, then δn := dist(supp ψn , ∂Ω) > 0, and so, Z (6.61) Ω
(uψn )(x)p 1 dx ≤ sp sp (dist(x, ∂Ω)) δn
Z
1 (uψn )(x) dx ≤ sp δn Ω p
Z Ω
u(x)p dx.
6.6. Hardy’s Inequality
239
If supp ψn is not contained in Ω, let xn ∈ ∂Ω be such Bn = B(xn , rn ). Then Tn (Ω ∩ B(xn , 2rn )) = {(y 0 , yN ) ∈ B(0, 2rn ) : yN > fn (y 0 )} =: Ωn , where fn : RN −1 → R is Lipschitz continuous and fn (0) = 0. Since ψn ∈ Cc∞ (RN ) and Tn is a rigid motion, the function (uψn ) ◦ Tn−1 belongs to W s,p (Ωn ) by Theorem 6.23 and Exercise 6.30, and it is zero in Ωn \ B(0, rn ). Hence, we are in a position to apply Step 1 to (uψn ) ◦ Tn−1 to find Z Z (uψn )(Tn−1 (y))p (uψn )(Tn−1 (y))p dy dy n sp (dist(y, ∂Ω )) n Ωn ∩B(0,rn ) Ωn ∩B(0,rn ) Z Z (uψn )(Tn−1 (w)) − (uψn )(Tn−1 (z))p + dwdz. kw − zkN +sp Ωn ∩B(0,2rn ) Ωn ∩B(0,2rn ) By Exercises 6.30 and 6.87, (6.60), and the change of variables x = we have Z Z (uψn )(x)p dy n (uψn )(x)p dx sp Ω∩B(xn ,rn ) (dist(x, ∂Ω)) Ω∩B(xn ,rn ) Z Z (uψn )(x) − (uψn )(y)p (6.62) dxdy + kx − ykN +sp Ω∩B(xn ,2rn ) Ω∩B(xn ,2rn ) Tx−1 (y), n
n kukpW s,p (Ω) , P where in the last inequality we use Theorem 6.23. Since `n=1 ψn = 1 in Ω, P we have that u = `n=1 ψn u in Ω. Hence, combining (6.61) and (6.62), we obtain Z ` Z X u(x)p (uψn )(x)p dx dx Ω kukpW s,p (Ω) . Ω sp sp Ω (dist(x, ∂Ω)) Ω (dist(x, ∂Ω)) n=1
Remark 6.89. When sp > 1, the stronger inequality Z Z Z u(x) − u(y)p u(x)p dx dxdy Ω sp N +sp Ω (dist(x, ∂Ω)) Ω Ω kx − yk holds for all u ∈ W s,p (Ω) ∩ Cc (Ω) (see [Dyd04]). Theorem 6.88 continues to hold for uniformly bounded domains. Definition 6.90. 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;
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6. Fractional Sobolev Spaces
(iii) for each n there exist a rigid motion Tn : RN → RN and a Lipschitz continuous function fn : RN −1 → R, with Lip fn ≤ L, such that Ωn ∩ Ω = Ωn ∩ Vn , where Vn := Tn−1 ({(y 0 , yN ) ∈ RN −1 × R : yN > fn (y 0 )}). Remark 6.91. Similarly, given m ∈ N0 and 0 < α ≤ 1, we can define open sets Ω ⊆ RN whose boundary is uniformly of class C m (respectively, uniformly of class C m,α ), with parameters ε, L > 0, M , provided items (i), (ii), and (iii) hold with fn of class C m (respectively, of class C m,α ) and with kfn kC m (RN −1 ) ≤ L (respectively, kfn kC m,α (RN −1 ) ≤ L). Exercise 6.92. 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. Theorem 6.93. Let Ω ⊂ RN be an open set with uniformly Lipschitz continuous boundary, 1 ≤ p < ∞, and let 0 < s < 1 be such that sp 6= 1. Then Z Z Z Z u(x)p u(x) − u(y)p p dx u(x) dx + dxdy ε,L sp N +sp Ω (dist(x, ∂Ω)) Ω Ω Ω kx − yk for all u ∈ W s,p (Ω) if sp < 1 and for all u ∈ W s,p (Ω) ∩ C(Ω) with u = 0 on ∂Ω if sp > 1, where ε, L are the parameters given in Definition 6.90. We begin with some preliminary estimates on dist(·, ∂Ω). Lemma 6.94. Let f : RN −1 → R be a Lipschitz continuous function such that f (0) = 0, ε > 0, and (6.63)
U := {(x0 , xN ) ∈ B(0, ε) : xN > f (x0 )}.
Then for every x ∈ U ∩ B(0, ε/[2(1 + Lip f )]), 1 (6.64) p (xN − f (x0 )) ≤ dist(x, B(0, ε) ∩ ∂U ) ≤ xN − f (x0 ). 1 + (Lip f )2 Proof. Step 1: Set L := Lip f . Given x ∈ U ∩ B(0, ε/[2(1 + L)]) and y = (y 0 , f (y 0 )) ∈ B(0, ε) ∩ ∂U , we have that 0 < xN − f (x0 ) = xN − f (y 0 ) + f (y 0 ) − f (x0 ) ≤ xN − f (y 0 ) + Lkx0 − y 0 kN −1 . In turn, (xN − f (x0 ))2 ≤ (xN − f (y 0 ))2 + L2 kx0 − y 0 k2N −1 + 2Lkx0 − y 0 kN −1 (xN − f (y 0 )) ≤ (xN − f (y 0 ))2 + L2 kx0 − y 0 k2N −1 + kx0 − y 0 k2N −1 + L2 (xN − f (y 0 ))2 = (1 + L2 )(kx0 − y 0 k2N −1 + (xN − f (y 0 ))2 ) = (1 + L2 )kx − yk2 ,
6.6. Hardy’s Inequality
241
which shows that √
1 (xN − f (x0 )) ≤ kx − yk 2 1+L
for all y ∈ B(0, ε) ∩ ∂U . Hence, the first inequality in (6.64) holds. To prove the second inequality, observe that if x ∈ U ∩B(0, ε/[2(1+L)]), then (x0 , f (x0 ) ∈ ∂U and k(x0 , f (x0 ))k2 = kx0 k2N −1 + (f (x0 ) − f (0))2 ≤ (1 + L2 )kx0 k2N −1 < (1 + L2 )
ε2 ε2 ≤ . 4(1 + L)2 4
Hence, (x0 , f (x0 )) ∈ B(0, ε) ∩ ∂U . Thus, dist(x, B(0, ε) ∩ ∂U ) ≤ kx − (x0 , f (x0 ))k = xN − f (x0 ).
Lemma 6.95. Let f , ε, and U be as in the Lemma 6.94 and Ξ := {ν = (ν 0 , νN ) ∈ SN −1 : νN ≥ Lip f kν 0 kN −1 + 1/2}. Then for every ν ∈ Ξ and x ∈ U ∩ B(0, ε/[8(1 + Lip f )]), (6.65)
1 2
p
1 + (Lip f )2
dν (x) ≤ dist(x, B(0, ε) ∩ ∂U ) ≤ dν (x),
where dν (x) := inf{t : x + tν ∈ ∂U }. Proof. Let L := Lip f , and fix ν ∈ Ξ and x ∈ U ∩ B(0, ε/[8(1 + L)]). Since f (0) = 0 and f is Lipschitz continuous, 0 < xN − f (x0 ) = xN − f (x0 ) + f (0) ≤ kxk + Lkxk < ε/8. Hence, if t < 0 is such that x + tν ∈ U , then xN + tνN ≥ f (x0 + tν 0 ), and so, (6.66)
ε/8 > xN − f (x0 ) ≥ tνN + f (x0 + tν 0 ) − f (x0 ) ≥ tLkν 0 kN −1 + t/2 − tLkν 0 kN −1 = t/2,
where we use the facts that ν ∈ Ξ and f is Lipschitz continuous. It follows that if t ≤ −ε/4, then xN + tνN < f (x0 + tν 0 ). By the intermediate value theorem, we can find 0 < t1 < −ε/4 such that xN + t1 νN = f (x0 + t1 ν 0 ). Moreover, kx + t1 νk < kxk + t1  < ε/2. On the other hand, if t is such that x + tν ∈ ∂B(0, ε), then t = ktνk ≥ kx + tνk − kxk > ε − ε/[8(1 + L)] > 7ε/8, while if t > 0, then, since f is Lipschitz continuous, ν ∈ Ξ, and x ∈ U , xN + tνN ≥ xN + tLkν 0 kN −1 ≥ xN + f (x0 + tν 0 ) − f (x0 ) > f (x0 + tν 0 ).
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6. Fractional Sobolev Spaces
Hence, we have proved that dν (x) = t, where t < 0 and x + tν ∈ B(0, ε) ∩ ∂U . In turn, dist(x, B(0, ε) ∩ ∂U ) ≤ kx + tν − xk = dν (x). This proves the second inequality in (6.65). To prove the second inequality, observe that since x + tν ∈ B(0, ε) and xN + tνN = f (x0 + tν 0 ), from (6.64), and (6.66), it follows that (6.67)
p 1 1 dν (x) = t ≤ xN − f (x0 ) ≤ 1 + L2 dist(x, B(0, ε) ∩ ∂U ). 2 2
We turn to the proof of Theorem 6.93. Proof of Theorem 6.93. Step 1: If sp < 1, assume that u ∈ W s,p (Ω) ∩ Cc∞ (RN ). Given 0 < ` < ε/[8(1 + L)] and ν ∈ S N −1 , we define Ω(ν, `) as the set of all points x ∈ Ω belonging to a segment parallel to ν, contained in Ω, and with length greater than `. Consider the hyperplane ν ⊥ := {x ∈ RN : x · ν = 0} and for x ∈ ν ⊥ define the slice Ω(x, ν, `) := {t ∈ R : x + tν ∈ Ω(ν, `)}. Note that if x ∈ ν ⊥ is such that Ω(x, ν, `) is nonempty, then Ω(x, ν, `) is the union of a family of pairwise disjoint open intervals with length greater than ` and we may define the function ux,ν : Ω(x, ν, `) → R by ux,ν (t) := u(x + tν). Let Ii be one such interval. If sp < 1, we apply Theorem 1.76 (if Ii has infinite measure) or Exercise 1.83 (if Ii is bounded) to the function ux,ν in Ii . If sp > 1, we apply Theorem 1.76 (if Ii has infinite measure) or Corollary 1.81 (if Ii is bounded) to the function ux,ν in Ii . In both cases we obtain Z Z Z Z ux,ν (t)p ux,ν (t) − ux,ν (τ )p 1 p dt u (t) dt + dtdτ, x,ν sp `sp Ii t − τ 1+sp Ii (dist(t, ∂Ii )) Ii Ii where we use the fact that length Ii ≥ `. Summing over all Ii in Ω(x, ν, `) gives (6.68) Z
Z 1 u(x + tν)p dt sp u(x + tν)p dt sp (dist(t, ∂Ω(x, ν, `))) ` Ω(x,ν,`) Ω(x,ν,`) Z Z u(x + tν) − u(x + τ ν)p + dtdτ t − τ 1+sp Ω(x,ν,`) Ω(x,ν,`) Z Z Z 1 u(x + tν) − u(x + τ ν)p sp u(x + tν)p dt + dtdτ, ` t − τ 1+sp Ω(x,ν) Ω(x,ν) Ω(x,ν)
6.6. Hardy’s Inequality
243
where Ω(x, ν) := {t ∈ R : x + tν ∈ Ω}. Define E(`, ν) := {x ∈ ν ⊥ : Ω(x, ν, `) 6= ∅} and observe that if x ∈ E(`, ν), then dist(t, ∂Ω(x, ν, `)) = inf{τ − t : τ ∈ ∂Ω(x, ν, `)} = inf{τ − t : x + τ ν ∈ ∂Ω(ν, `)} = dν,` (x + tν), where dν,` (x) := inf{r : x + rν ∈ ∂Ω(ν, `)}. Hence, if we integrate both sides of (6.68) over the set E(`, ν) and use Tonelli’s theorem (which we can apply since x · ν = 0 for x ∈ E(`, Ω)), we have Z Z u(y)p 1 u(y)p dy dy sp sp ` Ω(ν,`) (dν,` (y)) Ω Z Z Z u(x + tν) − u(x + τ ν)p + dtdτ dHN −1 (x). t − τ 1+sp ν ⊥ Ω(x,ν) Ω(x,ν) Finally, we integrate over ν in SN −1 and use Theorem 6.47 to get Z Z Z u(y)p 1 N −1 dydH (ν) sp u(x)p dx sp (d (y)) ` N −1 ν,` S Ω(ν,`) Ω Z Z u(x) − u(y)p + dxdy. N +sp Ω Ω kx − yk Step 2: To conclude the proof, it remains to show that Z Z Z u(x)p u(x)p dx dxdHN −1 (ν) ε,L sp sp (dist(x, ∂Ω)) (d (x)) N −1 ν,` Ω S Ω(ν,`) Z u(x)p dx. + Ω
Define Ω− := {x ∈ Ω : dist(x, ∂Ω) < `}, Then Z Ω+
Ω+ := {x ∈ Ω : dist(x, ∂Ω) ≥ `}.
u(x)p 1 dx ≤ sp (dist(x, ∂Ω))sp `
Z
u(x)p dx.
Ω+
On the other hand, if x ∈ Ω− , then dist(x, ∂Ω) < ` and so there exists x0 ∈ ∂Ω such that kx − x0 k < `. By items (i) and (iii) of Definition 6.90, there exist a rigid motion Tn : RN → RN , Tn (z) = Rn z + cn , z ∈ RN , where Rn is an orthogonal matrix and cn ∈ RN , and a Lipschitz continuous function fn : RN −1 → R, with Lip fn ≤ L, such that B(x0 , ε) ⊆ Ωn and Ωn ∩ Ω = Ωn ∩ Vn , where Vn := Tn−1 ({(y 0 , yN ) ∈ RN −1 × R : yN > fn (y 0 )}).
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6. Fractional Sobolev Spaces
In particular, B(x0 , ε) ∩ Ω = B(x0 , ε) ∩ Vn . Without loss of generality, we may assume that fn (0) = 0. Since ` < ε/(8(1 + L)), we have that x ∈ B(x0 , ε/[8(1 + L)]). Let Tn (x) = y¯ = (¯ y 0 , y¯N ). Since fn is Lipschitz 0 N −1 continuous, with Lip f ≤ L, for y ∈ R , we have that f (y 0 ) ≤ f (¯ y 0 ) + Lky 0 − y¯0 kN −1 < y¯N + Lky 0 − y¯0 kN −1 . Hence, the cone Kx = {(y 0 , yN ) ∈BN −1 (¯ y 0 , ε/[8(1 + L)]) × R : y¯N + Lky 0 − y¯0 kN −1 < yN < y¯N + ε/8} is contained in Tn (B(x0 , ε) ∩ Ω). Consider the sector Ξx := {ν = (ν 0 , νN ) ∈ S N −1 : νN > Lkν 0 kN −1 + 1/2}. Since ` < ε/(8(1 + L)), if t ∈ (0, `) and ν ∈ Ξx , then the point y¯ + tν belongs to Kx ⊆ Tn (Ω). Hence, Rn−1 (Ξx ) ⊆ G(x, `) := {ν ∈ SN −1 : x ∈ Ω(ν, `)}, and so, by Tonelli’s theorem, Z Z u(x)p dxdHN −1 (ν) sp (d (x)) N −1 S Ω(ν,`) ν,` Z Z 1 p u(x) dHN −1 (ν)dx = sp (d (x)) ν,` G(x,`) Ω Z Z 1 ≥ u(x)p dHN −1 (ν)dx sp −1 Ω− Rn (Ξx ) (dν,` (x)) Z u(x)p ε,L HN −1 (Rn−1 (Ξx )) dx, sp Ω− (dist(x, ∂Ω)) where in the last inequality we use Lemma 6.95. Since HN −1 (Rn−1 (Ξx )) is independent of x, we have shown that Z Z Z u(x)p u(x)p dx C dxdHN −1 (ν). ε,L sp sp Ω− (dist(x, ∂Ω)) SN −1 Ω(ν,`) (dν,` (x)) This concludes the proof under the additional hypothesis that u ∈ W s,p (Ω)∩ Cc∞ (RN ) when sp < 1. We leave the case u ∈ W s,p (Ω) as an exercise. s,p 6.7. The Space W00
If Ω ⊆ RN is an open set with Lipschitz continuous boundary, then the space W01,p (Ω) can be characterized as the subspace of all functions u ∈ W 1,p (Ω) such that Tr(u) = 0 on ∂Ω, where Tr(u) is the trace of u. Thus, if we extend u to be zero in RN \ Ω the resulting function belongs to W 1,p (RN ). Conversely, if u ∈ W 1,p (Ω) and the function u ˜ obtained by extending u by zero in RN \ Ω belongs to W 1,p (RN ), then, necessarily, Tr(u) = 0 at ∂Ω, and so u ∈ W01,p (Ω) [Leo22d]. This shows that W01,p (Ω) can be characterized as the space of all functions u ∈ W 1,p (Ω) whose extension by zero in RN \ Ω
s,p 6.7. The Space W00
245
belongs to W 1,p (RN ). In this section, we show that this characterization fails for fractional Sobolev spaces if sp = 1. Definition 6.96. Let Ω ⊆ RN be an open set, 1 ≤ p < ∞, and 0 < s < 1. s,p The space W00 (Ω) is the space of all functions u ∈ W s,p (Ω) such that the N function u ˜ : R → R, given by u(x) if x ∈ Ω, u ˜(x) := 0 if x ∈ RN \ Ω, s (Ω) := W s,2 (Ω). belongs to W s,p (RN ). When s = 2, we write H00 00
In the next theorem we characterize the functions u ∈ W s,p (Ω) that s,p (Ω). belong to W00 Theorem 6.97. Let Ω ⊆ RN be an open set, 1 ≤ p < ∞, and 0 < s < 1. s,p (Ω) if and only if Then u ∈ W s,p (Ω) belongs to W00 Z Z 1 (6.69) u(x)p dydx < ∞. N +sp Ω RN \Ω kx − yk Moreover, for all x ∈ Ω, Z (6.70) RN \Ω
1 1 dy . kx − ykN +sp (dist(x, ∂Ω))sp
In particular, if Z (6.71) Ω
u(x)p dx < ∞, (dist(x, ∂Ω))sp
then (6.69) holds. Proof. Given u ∈ W s,p (Ω), define u(x) if x ∈ Ω, u ˜(x) := 0 if x ∈ RN \ Ω. Then by Tonelli’s theorem and by symmetry, Z Z Z Z ˜ u(x) − u ˜(y)p 1 p dxdy = 2 u(x) dydx N +sp N +sp RN RN kx − yk Ω RN \Ω kx − yk Z Z u(x) − u(y)p + dxdy. N +sp Ω Ω kx − yk Thus, u ˜ belongs to W s,p (RN ) if and only if (6.69) holds. Next assume that (6.71) holds. Fix x ∈ Ω and let dx := dist(x, ∂Ω). Then RN \ Ω ⊆ RN \ B(x, dx ), and so, using spherical coordinates, Z Z 1 1 1 dy ≤ dy ≈ sp . N +sp N +sp kx − yk kx − yk d N N x R \Ω R \B(x,dx )
246
6. Fractional Sobolev Spaces
Hence, Z
p
Z
u(x)
RN \Ω
Ω
1 dydx kx − ykN +sp
Z Ω
u(x)p dx. (dist(x, ∂Ω))sp
s,p In view of Theorem 6.97, we can endow the space W00 (Ω) with a norm, which makes it into a Banach space.
Corollary 6.98. Let Ω ⊆ RN be an open set, 1 ≤ p < ∞, and 0 < s < 1. s,p Then the space W00 (Ω) is a Banach space with the norm (6.72) !1/p Z Z 1 p s,p dydx . kukW00 u(x) (Ω) := kukW s,p (Ω) + N +sp RN \Ω kx − yk Ω s,p (Ω). Then there exists a Proof. Let {un }n be a Cauchy sequence in W00 s,p constant M > 0 such that kun kW00 (Ω) ≤ M for all n. Moreover, {un }n is a Cauchy sequence in W s,p (Ω), and so, it converges in W s,p (Ω) to some function u. Consider a subsequence {unk }k of {un }n such that unk → u pointwise LN a.e. in Ω as k → ∞. By Fatou’s lemma, Z Z 1 p M ≥ lim inf unk (x) dydx N +sp k→∞ Ω RN \Ω kx − yk Z Z 1 p ≥ u(x) dydx, N +sp kx − yk N R \Ω Ω s,p (Ω) by Theorem 6.97. and so u ∈ W00 s,p (Ω). Let ε > 0 and find nε ∈ N It remains to show that un → u in W00 such that s,p kun − um kW00 (Ω) ≤ ε for all m, n ≥ nε . Taking m = nk and letting k → ∞, again by Fatou’s lemma, we have that s,p s,p kun − ukW00 (Ω) ≤ lim inf kun − unk kW00 (Ω) ≤ ε
k→∞
for all n ≥ nε .
In the following theorem we show that in halfspaces, (6.69) implies (6.71). Theorem 6.99. Let 1 ≤ p < ∞ and 0 < s < 1. Then for all x ∈ RN +, Z 1 1 dy ≈ sp . N +sp kx − yk xN RN − s,p N In particular, u ∈ W s,p (RN + ) belongs to W00 (R+ ) if and only if Z u(x)p dx < ∞. xsp RN N +
s,p 6.7. The Space W00
247
Proof. By (6.19), for xN > 0, Z 0 Z Z 0 1 1 1 dyN ≈ dy 0 dyN . sp ≈ 1+sp N +sp xN −∞ RN −1 kx − yk −∞ (xN − yN ) Hence, by Tonelli’s theorem, Z Z Z 1 u(x)p u(x)p dydx ≈ dx. kx − ykN +sp xsp RN RN RN N + − + The second part of the statement follows from Theorem 6.97.
Exercise 6.100. Let 1 ≤ p < ∞ and 0 < s < 1, with sp < 1. (i) Given u ∈ W s,p (RN + ), use Theorem 6.99 to show that there exists a N −1 ×(0, δ ), sequence of functions un ∈ W s,p (RN n + ), with un = 0 in R + where δn → 0 , such that kun − ukW s,p (RN ) → 0 as n → ∞. +
(ii) Prove that if there exists a ball B(0, r) such that u = 0 in RN + \ N B(0, r), then we can construct un such that supp un ⊂ R+ ∩ B(0, 2r). (iii) Use item (i) to give an alternative proof of the fact that W s,p (RN +) = s,p N W0 (R+ ) when sp < 1 (see Theorem 6.75). In view of Theorem 6.99 we have the following result. Corollary 6.101. Let 1 ≤ p < ∞ and 0 < s < 1. s,p s,p N s,p (RN ). (RN (i) If sp < 1, then W00 + + ) = W0 (R+ ) = W s,p s,p s,p (RN ), where the N (ii) If sp = 1, then W00 (RN + + ) ⊂ W0 (R+ ) = W inclusion is strict. s,p s,p N s,p (RN ), where the (iii) If sp > 1, then W00 (RN + ) = W0 (R+ ) ⊂ W + inclusion is strict.
Proof. This is left as an exercise.
More generally, we can show that for convex domains, (6.69) implies (6.71). Theorem 6.102. Let Ω ⊂ RN be an open convex set, 1 ≤ p < ∞, and 0 < s < 1. Then for all x ∈ Ω, Z 1 1 (6.73) dy ≈ . N +sp kx − yk (dist(x, ∂Ω))sp N R \Ω s,p In particular, u ∈ W s,p (Ω) belongs to W00 (Ω) if and only if Z u(x)p dx < ∞. sp Ω (dist(x, ∂Ω))
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6. Fractional Sobolev Spaces
Proof. Fix x ∈ Ω and let dx := dist(x, ∂Ω). By the convexity of Ω, if y0 ∈ ∂Ω is such that kx − y0 k = dx , then the halfline L+ = {ty0 + (1 − t)x : t ≥ 1} belongs to RN \ Ω, while B(x, dx ) ⊆ Ω. In particular, we can find x0 ∈ RN \Ω such that kx−x0 k = 2dx . We claim that B(x0 , dx ) ⊂ RN \Ω. To see this, let H be the hyperplane orthogonal to x−x0 and passing trough y0 , and let H+ be the closed halfspace that contains L+ and whose boundary is ◦ ∩ Ω. Consider the line H. Assume by contradiction that there exists z ∈ H+ through z and y0 and let w 6= y0 be the point where this line intersects the boundary of B(x, dx ). The twodimensional plane spanned by the vectors x−x0 and z−y0 intersects the boundary of B(x, dx ) into a circle C containing y0 and w. Also, w will be on the opposite side of z with respect to the line containing L+ . If we take a point ξ on the shortest arc of C connecting w and y0 , the line through z and ξ will be contained in Ω except possibly for ξ, since B(x, dx ) ⊆ Ω. However, this segment will also intersect L+ , since ξ and z are on opposite sides of the line containing L+ . This contradicts the ◦ . Hence, fact that L+ is contained in RN \ Ω and proves that Ω ⊆ RN \ H+ H is a supporting hyperplane for Ω. In turn, B(x0 , dx ) will be contained in H+ . Therefore B(x0 , dx ) ⊂ RN \ Ω, and the claim holds. For y ∈ RN \ B(x0 , dx ) we have (6.74)
kx − yk ≤ kx − x0 k + kx0 − yk = 2dx + kx0 − yk ≤ 3kx0 − yk.
We claim that Dx := {y ∈ RN \ B(x0 , dx ) : (y − x0 ) · (x − x0 ) < 0} ⊆ RN \ (Ω ∪ B(x0 , dx )). x−x0 Indeed, since kx−x0 k = 2dx , the point x0 +dx kx−x belongs to the boundary 0k of the ball B(x0 , dx ) and to ∂Ω. Given y ∈ Dx , observe that
(y − x0 ) · (x − x0 ) = ky − x0 kkx − x0 k cos θ < 0, where θ is the angle between the two vectors y − x0 and x − x0 . If y ∈ Ω, x−x0 then, since cos θ < 0, the segment joining x0 + dx kx−x and y would cross 0k the ball B(x0 , dx ), which is a contradiction since B(x0 , dx ) ⊂ RN \ Ω. This proves the claim. In turn, Z Z Z 1 1 1 dy ≥ dy dy N +sp N +sp N +sp Dx kx − yk Dx kx0 − yk RN \Ω kx − yk Z ∞ 1 1 N −1 =H (Σ) dρ ≈ HN −1 (Σ) sp , 1+sp dx dx ρ where in the second inequality we use (6.74) in first equality we use and N −1 : ν · e < 0 , which spherical coordinates. Here, we can take Σ = ν ∈ S 1 has the same surface measure of ν ∈ SN −1 : ν · (x − x0 ) < 0 by the symmetry of the sphere. This proves the first inequality in (6.73). The second was proved in Theorem 6.97.
s,p 6.7. The Space W00
249
Remark 6.103. It follows from Theorem 6.102 that in convex domains, the map Z 1/p u(x)p u 7→ kukW s,p (Ω) + dx sp Ω (dist(x, ∂Ω)) s,p s,p is an equivalent norm to k · kW00 (Ω) in W00 (Ω). Another important class of domains for which (6.73) holds is given by special Lipschitz continuous domains. Exercise 6.104. Let Ω = {x ∈ RN : xN > f (x0 )}, where f : RN −1 → R is a Lipschitz continuous function, 1 ≤ p < ∞, and 0 < s < 1. Prove that for all x ∈ Ω, Z 1 1 dy ≈Lip f . N +sp kx − yk (dist(x, ∂Ω))sp N R \Ω Theorem 6.105. Let Ω ⊂ RN be an open set with uniformly Lipschitz continuous boundary, 1 ≤ p < ∞, and 0 < s < 1. s,p (Ω) = W0s,p (Ω) = W s,p (Ω). (i) If sp < 1, then W00 s,p (Ω) ⊆ W0s,p (Ω) = W s,p (Ω), where in general (ii) If sp = 1, then W00 the inclusion is strict. s,p (iii) If sp > 1, then W00 (Ω) = W0s,p (Ω) ⊆ W s,p (Ω), where in general the inclusion is strict.
Proof. (i) Given u ∈ W s,p (Ω), by Theorem 6.93 we have that Z u(x)p dx < ∞. sp Ω (dist(x, ∂Ω)) s,p (Ω). In turn, by Theorem 6.97, u belongs to W00
(ii) This item follows from Theorem 6.78. (iii) Given u ∈ W0s,p (Ω), by Theorem 6.93 we have that Z u(x)p dx < ∞. sp Ω (dist(x, ∂Ω)) s,p In turn, by Theorem 6.97, u belongs to W00 (Ω). This shows that W0s,p (Ω) ⊆ s,p s,p W00 (Ω). Conversely, let u ∈ W00 (Ω). Define u(x) if x ∈ Ω, (6.75) u ˜(x) := 0 if x ∈ RN \ Ω.
Then u ˜ ∈ W s,p (RN ). By Lemma 6.67 there exists a sequence u ˜n ∈ W s,p (RN ) such that k˜ u−u ˜n kW s,p (RN ) → 0 and u ˜n = 0 outside some ball B(0, Rn ). Note that u ˜n is obtained by multiplying u ˜ by a smooth cutoff function.
250
6. Fractional Sobolev Spaces
Therefore, u ˜n = 0 in RN \ Ω. Since W0s,p (Ω) is closed, it suffices to show that the restriction of u ˜n to Ω belongs to W0s,p (Ω) for every n. Thus, in what follows, we assume that u = 0 outside B(0, R) for some R > 0. For every x0 ∈ ∂Ω there exist a rigid motion Tx0 : RN → RN , with Tx0 (x0 ) = 0, a Lipschitz continuous function fx0 : RN −1 → R, with fx0 (0) = 0, and rx0 > 0 such that Tx0 (Ω ∩ B(x0 , 2rx0 )) = {(y 0 , yN ) ∈ B(0, 2rx0 ) : yN > fx0 (y 0 )}. S S If Ω \ x∈∂Ω B(x, rx ) is nonempty, for every x0 ∈ Ω \ x∈∂Ω B(x, rx ), let B(x0 , rx0 ) ⊆ Ω. The family {B(x, rx )}x∈Ω is an open cover of the compact set Ω ∩ B(0, R). Hence, there is a finite number of balls B1 , . . . , B` , where Bk := B(xk , rxk ), that covers Ω∩B(0, R). Let {ψk }`k=1 be a smooth partition of unity subordinated to B1 , . . . , B` , with supp ψk ⊂ Bk . Fix k ∈ {1, . . . , `}. Then u ˜k := ψk u ˜ ∈ W s,p (RN ) by Theorem 6.23. There are two cases. If supp ψk is contained in Ω, then we set vk := ψk u. In this case, supp vk ⊆ supp ψk ⊂ Ω. If supp ψk is not contained in Ω, let xk ∈ ∂Ω be such that supp ψk ⊆ Bk = B(xk , rk ). Then (6.76)
Tk (Ω ∩ B(xk , 2rk )) = {(y 0 , yN ) ∈ B(0, 2rk ) : yN > fk (y 0 )} := Ωk ,
where fk : RN −1 → R is continuous and fk (0) = 0. By Exercise 6.87, (ψk u ˜) ◦ Tk−1 ∈ W s,p (RN ). Since translation is continuous in Lp spaces, the function u ˆk,δ (y) = ((ψk u ˜) ◦ Tk−1 )(y 0 , yN + δ),
y ∈ RN ,
δ > 0,
kˆ uk,δ −(ψk u ˜)◦Tk−1 kW s,p (RN )
→ 0 as δ → 0+ . Since belongs to W s,p (RN ) and ψk = 0 outside Bk , by (6.75) and (6.76), we have that (ψk u ˜)◦Tk−1 = 0 outside Ωk ∩ Bk . In turn, taking δk sufficiently small, we have that supp u ˆk,δk ⊂ Ωk and kˆ uk,δk − (ψk u ˜) ◦ Tk−1 kW s,p (RN ) ≤ ε/2k , where ε > 0 is fixed. Let v˜k := u ˆk,δk ◦Tk and let vk be its restriction to Ω. By Exercise 6.87, kvk − ψk ukW s,p (Ω) ≤ ε/2k and supp vk ⊂ Ω ∩ B(xk , rk ). Define P the function v := `k=1 vk . Then v ∈ W s,p (Ω). Since u = 0 in Ω \ B(0, R) P P and `k=1 ψk = 1 in Ω ∩ B(0, R), we have that u = `k=1 ψk u in Ω. Hence, ku − vkW s,p (Ω) ≤
` X k=1
kψk u − vk kW s,p (Ω) ≤ ε
` X
2−k ≤ ε.
k=1
Since dist(supp v, ∂Ω) > 0, by mollifying v, we obtain a function in Cc∞ (Ω). Remark 6.106. We remark that the inclusions and equalities are intended s,p as inclusions and equalities of sets. If we endow W00 (Ω) with the norm s,p (6.72), then, in the case sp = 1, we have that W00 (Ω) is embedded into
˙ s,p 6.8. Density in W
251
s,p W s,p (Ω), that is, W00 (Ω) ,→ W s,p (Ω). On the other hand, when sp 6= 1, by s,p Theorems 6.93 and 6.105, and by inequality (6.70), if u ∈ W00 (Ω), Z 1/p u(x)p s,p s,p s,p kukW (Ω) ≤ kukW00 (Ω) Ω kukW (Ω) + dx sp Ω (dist(x, ∂Ω)) Ω kukW s,p (Ω) . s,p This shows that the norm k·kW s,p (Ω) is equivalent to k·kW00 (Ω) when sp 6= 1.
˙ s,p 6.8. Density in W ˙ s,p N In this section R we prove a densitytype result for W (R ). We recall that 1 uE := LN (E) E u(x) dx whenever the righthand side is welldefined. ˙ s,p (RN ) Theorem 6.107. Let 1 ≤ p < ∞ and 0 < s < 1. For every u ∈ W there exists a sequence {un }n in Cc∞ (RN ) such that u − un W s,p (RN ) → 0
as n → ∞.
˙ s,p (RN ) ∩ Proof. In view of Theorem 6.62, we can assume that u ∈ W C ∞ (RN ). Let ϕ ∈ Cc∞ (RN ) be such that 0 ≤ ϕ ≤ 1, ϕ = 1 in B(0, 1), and ϕ = 0 outside B(0, 2). Define un (x) := ϕn (x)(u(x) − uAn ),
ϕn (x) := ϕ(x/n),
where An := B(0, 2n) \ B(0, n). Since u − un W s,p (RN ) = u − uAn − un W s,p (RN ) , it suffices to prove that vn W s,p (RN ) → 0 as n → ∞, where vn := ψn (u−uAn ) and ψn := 1−ϕn . Using the facts that ψn = 0 in B(0, n) and ψn = 1 outside B(0, 2n), by Tonelli’s theorem, we can write Z Z ψn (x)(u(x) − uAn )p p vn W s,p (RN ) = 2 dxdy kx − ykN +sp B(0,n) An Z Z Z Z u(x) − uAn p vn (x) − vn (y)p +2 dxdy + dxdy N +sp kx − ykN +sp B(0,n) RN \B(0,2n) kx − yk An An Z Z u(x) − uAn − ψn (y)(u(y) − uAn )p dxdy +2 kx − ykN +sp An RN \B(0,2n) Z Z u(x) − u(y)p + dxdy =: A + B + C + D + E. N +sp RN \B(0,2n) RN \B(0,2n) kx − yk Since ψn (y) = 0 for y ∈ B(0, n), by the mean value theorem, for every x ∈ An we have z 1 · (x − y) (6.77) ψn (x) = ψn (x) − ψn (y) = ∇ψn (z) · (x − y) = ∇ϕ n n
252
6. Fractional Sobolev Spaces
for some z in the segment of endpoints x and y. Hence, by Tonelli’s theorem Z Z 1 1 u(x) − uAn p (6.78) A ≤ p k∇ϕk∞ dydx. N +sp−p n B(0,n) kx − yk An If x ∈ An and y ∈ B(0, n), kx − yk ≤ kxk + kyk < 2n + n, and so, B(0, n) ⊂ B(x, 3n). In turn, Z Z 1 1 (6.79) dy ≤ dy N +sp−p N +sp−p kx − yk kx − yk B(0,n) B(x,3n) Z 3n N −1 r dr n(1−s)p . N r +sp−p 0 Hence, by Poincar´e’s inequality (see Theorem 6.33) and the fact that (diam An )N +sp /LN (An ) nsp , it follows from (6.78) and (6.79) that Z Z Z 1 u(x) − u(y)p p A sp u(x) − uAn  dx dxdy → 0 N +sp n An An An kx − yk as n → ∞ by the Lebesgue dominated convergence theorem. To estimate B, we write p Z 1 (u(x) − u(z)) dz u(x) − uAn  = N p (L (An )) An 0 Z N p/p (L (An )) ≤ u(x) − u(z)p dz (LN (An ))p An p
by H¨ older’s inequality. If z ∈ An , y ∈ B(0, n), and x ∈ RN \ B(0, 2n), kx − zk ≤ kx − yk + ky − zk ≤ kx − yk + 3n ≤ 4kx − yk. Hence, 1 B N L (An )
Z
Z
B(0,n) Z N L (B(0, n))
LN (An )
Z
RN \B(0,2n)
RN \B(0,2n)
Z An
An
u(x) − u(z)p dzdxdy kx − zkN +sp
u(x) − u(z)p dzdx → 0 kx − zkN +sp
as n → ∞, by the Lebesgue dominated convergence theorem and the fact N (B(0,n)) = 2N1−1 . that L LN (An )
˙ s,p 6.8. Density in W
253
By the inequality (a + b)p ≤ 2p−1 ap + 2p−1 bp and (6.77), we can bound (ψn (x) − ψn (y))(u(x) − uAn )p dxdy kx − ykN +sp An An Z Z u(x) − u(y)p ψn (y)p + dxdy kx − ykN +sp An An Z Z Z Z u(x) − uAn p u(x) − u(y)p k∇ϕkp∞ dxdy + dxdy. N +sp−p N +sp np An An kx − yk An An kx − yk Z
Z
C
The first integral on the righthand side can be estimated as A, while the second goes to zero as n → ∞ by the Lebesgue dominated convergence theorem. To study D, we use the fact that ψn = 1 outside B(0, 2n) to bound Z
Z
D An
RN \B(0,2n)
Z
(ψn (x) − ψn (y))(u(y) − uAn )p dxdy kx − ykN +sp
Z
+ An
RN \B(0,2n)
u(x) − u(y)p dxdy =: D1 + D2 . kx − ykN +sp
The integral D2 goes to 0 as n → ∞, while by (6.77) and the fact that 0 ≤ ψn ≤ 1, Z Z (ψn (x) − ψn (y))(u(y) − uAn )p D1 ≤ dxdy kx − ykN +sp An RN \B(0,3n) Z Z (ψn (x) − ψn (y))(u(y) − uAn )p + dxdy kx − ykN +sp An B(0,3n)\B(0,2n) Z Z u(y) − uAn p dxdy N +sp An RN \B(0,3n) kx − yk Z Z k∇ϕkp∞ u(y) − uAn p + dxdy =: D11 + D12. N +sp−p np An B(0,3n)\B(0,2n) kx − yk To estimate D11 note that if x ∈ RN \B(0, 3n) and y ∈ An , then kx−yk ≥ n. Hence, by Poincar´e’s inequality (Theorem 6.33), Z An
u(y) − uAn p dxdy N +sp RN \B(0,3n) kx − yk Z Z p ≤ u(y) − uAn 
Z
1 dxdy N +sp An RN \B(y,n) kx − yk Z Z Z u(x) − u(y)p 1 p sp u(y) − uAn  dy dxdy → 0 N +sp n An An An kx − yk
254
6. Fractional Sobolev Spaces
as n → ∞ by the Lebesgue dominated convergence theorem, and where we use the fact that Z ∞ Z 1 1 1 dxdy dr sp . N +sp 1+sp kx − yk r n N R \B(y,n) n The integral D12 can be estimated as A. Finally E goes to zero as n → ∞ by the Lebesgue dominated convergence theorem. ˙ s,p (RN ). As a corollary Theorem 6.107, we obtain Hardy’s inequality in W + Theorem 6.108. Let 1 ≤ p < ∞ and 0 < s < 1, with sp < 1. Then for ˙ s,p (RN every u ∈ W + ) there exists a constant mu ∈ R such that Z u(x) − mu p dx upW s,p (RN ) . sp N + x R+ N In particular, the function u ˜(x) =
u(x) − mu if xN > 0, 0 if xN < 0,
˙ s,p (RN ) and ˜ belongs to W uW s,p (RN ) uW s,p (RN ) . +
We begin with a preliminary result. Lemma 6.109. Let 1 ≤ p < ∞ and 0 < s < 1. Then for every u ∈ ˙ s,p (RN ), the function v : RN → R, defined by W + u(x) if xN > 0, v(x) := u(x0 , −xN ) if xN < 0, ˙ s,p (RN ), with belongs to W vW s,p (RN ) uW s,p (RN ) . +
Proof. We have Z Z Z Z v(x) − v(y)p u(x0 , −xN ) − u(y 0 , −yN )p dxdy = dxdy N +sp kx − ykN +sp RN RN RN RN kx − yk − − Z Z Z Z u(x0 , −xN ) − u(y)p u(x) − u(y)p +2 dxdy + dxdy kx − ykN +sp kx − ykN +sp RN RN RN RN + − + + =: A + B + C. In A we make the change of variables (x0 , −xN ) = z and (y 0 , −yN ) = w to get A = upW s,p (RN ) . In B we make the the change of variables (x0 , −xN ) = z + to get Z Z u(z) − u(y)p B= dxdy, k(z 0 − y 0 , zN + yN )kN +sp RN RN + +
6.9. Notes
255
and observe that zN + yN  = zN + yN ≥ zN − yN , so that B ≤ upW s,p (RN ) . +
Thus vpW s,p (RN ) upW s,p (RN ) .
+
We turn to the proof of Theorem 6.108. Proof of Theorem 6.108. By Lemma 6.109, we can reflect u to obtain a ˙ s,p (RN ) with vW s,p (RN ) u s,p N . By Theorem 6.107, function v ∈ W W (R ) +
there exists a sequence {vn }n in Cc∞ (RN ) such that vn − vW s,p (RN ) → 0. s,p (RN ), and so by Let un be the restriction of vn to RN + . Then un ∈ W + Theorem 6.79, Z un − uk p dx un − uk pW s,p (RN ) . sp + xN RN + Letting n, k → ∞, we obtain that {un }n is a Cauchy sequence in the space dx p N dx Lp (RN + ; xsp ), and thus, it converges to a function w in L (R+ ; xsp ) and, by N
N
extracting a subsequence, pointwise LN a.e. in RN + . Again by Theorem 6.79, Z p un  p sp dx un W s,p (RN ) . N + x R+ N Letting n → ∞ gives Z RN +
wp dx upW s,p (RN ) . + xsp N
On the other hand, again by Fatou’s lemma w − uW s,p (RN ) ≤ lim inf un − uW s,p (RN ) = 0, +
n→∞
+
which implies that w − u is a constant. This concludes the proof of the first statement. The second statement follows from Theorem 6.99.
6.9. Notes For more information about Sobolev spaces, we refer to the books of Adams and Fournier [AF03], Ambrosio, Fusco, and Pallara [AFP00], Brezis [Bre11], Burenkov [Bur98], F. Demengel and G. Demengel [DD12], Di Benedetto [DiB16], Evans [Eva10], Evans and Gariepy [EG15], Kufner [Kuf85], Leoni [Leo17], [Leo09], Neˇcas [Nec12], Tartar [Tar07], and Ziemer [Zie89]. We also refer to the book of L¨ utzen [L¨ ut82] for an historical account of distributions and Sobolev spaces, as well as the interesting review of the book by Dieudonne [Die84], which justifies my choice in not trying to dwell upon the history of these spaces. I simply don’t know enough. For fractional Sobolev spaces and Besov spaces, we refer to the books of Bahouri, Chemin, and Danchin [BCD11], Bennett and Sharpley [BS88],
256
6. Fractional Sobolev Spaces
Bergh and L¨ofstr¨ om [BL76], Besov, Il’in, and Nikol’ski˘ı [BIN79], [BIN78], F. Demengel and G. Demengel [DD12], Grafakos [Gra14], Grisvard [Gri11], Leoni [Leo17], [Leo09], Lions and Magenes (mostly for the case p = 2) [LM72b], [LM72a], Maz’ya [Maz11], Peetre [Pee76], Runst and Sickel [RS96], Sawano [Saw18], Stein [Ste70], and Triebel [Tri10], [Tri06], [Tri01], [Tri95], [Tri92]. The proof of Theorem 6.13 draws upon the paper of Bonder and Salort [BS19]. Using interpolation theory, it can be shown that the constant C in Lemma 6.14 may be taken to depend only on N . We refer to the paper of del Teso, G´ omezCastro, and V´ azquez [dTGCV20] for a proof. Exercise 6.22 is adapted from the paper of Di Nezza, Palatucci, and Valdinoci [DNPV12]. For more sophisticated versions of Poincar´e’s inequality (Theorem 6.33), we refer to the paper of Drelichman and Dur´an [DD18], and of HurriSyrj¨anen and V¨ah¨ akangas [HSV13] and the references therein. One of my undergraduate students, Khunpob Sereesuchart, worked on slicing theorems for domains of the type RN −1 ×(a, b), which are not covered by Theorems 6.35 and 6.38 and Remark 6.44 (see [Ser22]). Slicing theorems fail for Besov spaces Bqs,p (RN ) when q > p. See the papers of Mironescu, Russ, and Sire [MRS20] and the references therein. The proof of the Meyers–Serrin theorem (Theorem 6.65) is adapted from the paper of Muramatu [Mur70]. The proof of Theorem 6.75 is inspired by the paper of Lions and Magenes [LM61] (see also the papers of Fiscella, Servadei, and Valdinoci [FSV15] and Dyda and Kijaczko, [DK21] and the references therein for more recent density results). The slicing technique used in the proof of Theorem 6.93 is taken from the paper of Chermisi, Dal Maso, Fonseca, and Leoni [CDMFL11]. To my knowledge, the spaces s,p W00 (Ω) were discovered by Lions and Magenes (see [LM61] and [LM72a, Section 11.3]). The proof of Theorem 6.102 is inspired by the paper of Brasco and Cinti [BC18]. Tomasz Tkocz suggested the proof of the inclusion B(x0 , dx ) ⊂ RN \ Ω. The idea of the proof of Theorem 6.107 follows the ˙ m,p (RN ), m ∈ N, which is due one for the homogeneous Sobolev spaces W to Hajlasz and Kalamajska (see [HK95] or [Leo17, Theorem 11.43]). We refer also to the papers of Brasco, G´omezCastro, and V´azquez [BGCV21] and Mironescu [Mir18].
Chapter 7
Embeddings and Interpolation It is not knowledge, but the act of learning, not possession but the act of getting there, which grants the greatest enjoyment. — Carl Friedrich Gauss
In this chapter, we will study embeddings of fractional Sobolev spaces into Lq spaces, H¨ older spaces, and other fractional Sobolev spaces. We will also study interpolation inequalities of the type established by Gagliardo [Gag58] and Nirenberg [Nir59] for Sobolev spaces. We will begin with the embedding (7.1)
W s2 ,p2 (RN ) ,→ W s1 ,p1 (RN ),
where 1 ≤ p2 < p1 ≤ ∞ and 0 ≤ s1 < s2 ≤ 1 are such that (7.2)
s1 −
N N = s2 − . p1 p2
We consider first the endpoints, that is, the cases s1 = 0, s1 = 1, and p1 = ∞. When s1 = 0, we set W 0,p1 (RN ) := Lp1 (RN ). Solving for p1 p2 in (7.2), we obtain p1 = N N −s2 p2 , which is well defined when s2 p2 ≤ N . The case s2 p2 < N is called the subcritical case. We will show that in the subcritical case, the embedding (7.1) holds. This is the extension to fractional Sobolev spaces of the Sobolev–Gagliardo–Nirenberg embedding theorem in W 1,p (see [Leo22d]). If s2 p2 = N , the critical case, then p1 = ∞, but the embedding (7.1), W p2 /N,p2 (RN ) ,→ L∞ (RN ), fails. However, we will prove the weaker embedding W p2 /N,p2 (RN ) ,→ Lq (RN ), p2 ≤ q < ∞. Next we consider the case p1 = ∞. We recall that when 0 < s1 < 1, W s1 ,∞ (RN ) can be identified with the space C s1 ,0 (RN ) of all bounded, H¨ older continuous functions with exponent s1 . Note that from (7.2) we get 0 < s1 = s2 − pN2 , that is, s2 p2 > N . This is called the supercritical case. The embedding 257
258
7. Embeddings and Interpolation
W s2 ,p2 (RN ) ,→ C s2 −N/p2 (RN ) is the extension to fractional Sobolev spaces of the Morrey embedding theorem in W 1,p (see [Leo22d]).
7.1. Campanato Spaces In this section we introduce Campanato’s spaces and study some of their properties. For simplicity we consider only the case Ω = RN and refer to [GM12] and [Giu03, Chapter 2] for the case of an arbitrary open set. For 1 ≤ p < ∞ and 0 ≤ λ ≤ N + p, we define the Campanato space Lp,λ (RN ) as the space of all functions u ∈ Lploc (RN ) such that
(7.3) uLp,λ (RN ) :=
sup
r
−λ/p
x0 ∈RN , r>0
We recall that uE :=
1 LN (E)
R E
!1/p
Z B(x0 ,r)
p
u − uB(x0 ,r)  dx
< ∞.
u(x) dx whenever the righthand side is de
fined. Note that when λ = N , the space Lp,N (RN ) coincides with the space BMO(RN ) of functions of bounded mean oscillation (see [Leo17, Appendix C.5]). Now we prove that when N < λ ≤ N + p, functions in Lp,λ (RN ) are H¨older continuous. Theorem 7.1. Let 1 ≤ p < ∞ and N < λ ≤ N +p. Then every function u ∈ Lp,λ (RN ) admits a representative u ¯ that is H¨ older continuous with exponent α = (λ − N )/p such that ¯ uC 0,α (RN ) ≈ uLp,λ (RN ) . We begin with a preliminary lemma. Lemma 7.2. Let λ > N , 1 ≤ p < ∞, and u ∈ Lploc (RN ). Then for every x0 ∈ RN and 0 < r < R, (7.4)
uB(x0 ,r) − uB(x0 ,R)  R(λ−N )/p uLp,λ (RN ) .
In particular, if u ¯ is a representative of u, then (7.5)
¯ u(x0 ) − uB(x0 ,R)  R(λ−N )/p uLp,λ (RN )
for every Lebesgue point x0 of u ¯.
7.1. Campanato Spaces
259
Proof. For simplicity we write uρ := uB(x0 ,ρ) . For 0 < η < ρ, by H¨older’s inequality we have Z 1 u(x) − uρ  dx uη − uρ  ≤ N L (B(x0 , η)) B(x0 ,η) !1/p 1/p0 Z LN (B(x0 , η)) p ≤ u(x) − uρ  dx LN (B(x0 , η)) B(x0 ,η) !1/p Z ρλ/p p −λ (7.6) u(x) − uρ  dx ≤ N ρ (L (B(x0 , η)))1/p B(x0 ,ρ) λ/p ρλ/p ρ N/p uLp,λ (RN ) = η (λ−N )/p uLp,λ (RN ) . η η Take ri = R2−i . By applying the previous inequality with η = ri and ρ = ri−1 , we get uri − uri−1  (R2−i )(λ−N )/p uLp,λ (RN ) , and so, uR − urk  ≤
k X
uri − uri−1  R(λ−N )/p uLp,λ (RN )
i=1
k X i=1
1 2(λ−N )i/p
.
Let k be such that rk−1 ≤ r < rk . By (7.6), r λ/p k ur − urk  r(λ−N )/p uLp,λ (RN ) r rk λ/p (λ−N )/p r uLp,λ (RN ) R(λ−N )/p uLp,λ (RN ) . rk−1 Hence, uR − ur  ≤ uR − urk  + ur − urk  R(λ−N )/p uLp,λ (RN ) , which concludes the first part of proof. For the second part, let u ¯ be a representative of u and let x0 be a Lebesgue point of u ¯. Then ur → u ¯(x0 ) as r → 0+ ([Leo22c]), and so (7.5) + follows by letting r → 0 in (7.4). We turn to the proof of Theorem 7.1. Proof of Theorem 7.1. Step 1: Let u ∈ Lp,λ (RN ), u ¯ be a representative of u, and let x, y ∈ RN be Lebesgue points of u ¯, with x 6= y. Set r := kx − yk > 0. Then by (7.5), ¯ u(x) − u ¯(y) ≤ ¯ u(x) − uB(x,2r)  + ¯ u(y) − uB(y,2r)  + uB(x,2r) − uB(y,2r)  r(λ−N )/p uLp,λ (RN ) + uB(x,2r) − uB(y,2r) . To complete the proof, it remains to show that (7.7)
uB(x,2r) − uB(y,2r)  r(λ−N )/p uLp,λ (RN ) .
260
7. Embeddings and Interpolation
For z ∈ B(x, 2r) ∩ B(y, 2r), we have uB(x,2r) − uB(y,2r)  ≤ uB(x,2r) − u(z) + u(z) − uB(y,2r) . Averaging in z over B(x, 2r) ∩ B(y, 2r), we get uB(x,2r) − uB(y,2r)  Z 1 ≤ N u − u(z) dz L (B(x, 2r) ∩ B(y, 2r)) B(x,2r) B(x,2r) Z 1 + N u − u(z) dz =: A + B. L (B(x, 2r) ∩ B(y, 2r)) B(y,2r) B(y,2r) Since B(x, r) ⊂ B(x, 2r) ∩ B(y, 2r), by H¨older’s inequality and (7.3), we have Z 1 A≤ N u − u(z) dz L (B(x, r)) B(x,2r) B(x,2r) !1/p Z 1 uB(x,2r) − u(z)p dz N (L (B(x, r)))1−1/p0 B(x,2r)
(2r)λ/p u p,λ N r(λ−N )/p uLp,λ (RN ) . (LN (B(x, r)))1−1/p0 L (R )
A similar estimate holds for B. Hence, we have shown that ¯ u(x) − u ¯(y) r(λ−N )/p uLp,λ (RN ) = kx − yk(λ−N )/p uLp,λ (RN ) for all Lebesgue points x, y ∈ RN of u ¯. Let E be the set of Lebesgue points N N of u ¯. Then L (R \ E) = 0, and so if we consider the restriction of u ¯ to E, we can uniquely extend u ¯ to RN in such a way that the previous inequality holds for every x, y ∈ RN . This shows that u admits a representative u ¯ that is H¨older continuous with exponent α = (λ − N )/p and is such that ¯ uC 0,α (RN ) uLp,λ (RN ) . Step 2: Let u ∈ Lp,λ (RN ) and let u ¯ be the H¨ older continuous representative constructed in the previous step. Given r > 0 and x0 ∈ RN , for all x, y ∈ B(x0 , r) we have ¯ u(x) − u ¯(y) ≤ kx − yk(λ−N )/p ¯ uC 0,α (RN ) r(λ−N )/p ¯ uC 0,α (RN ) , or, equivalently, −r(λ−N )/p ¯ uC 0,α (RN ) u ¯(x) − u ¯(y) r(λ−N )/p ¯ uC 0,α (RN ) . Averaging in y over B(x0 , r) gives ¯ u(x) − u ¯B(x0 ,r)  r(λ−N )/p ¯ uC 0,α (RN ) .
7.2. Embeddings: The Subcritical Case
261
Raising both sides to power p and integrating over B(x0 , r), we obtain Z u − uB(x0 ,r) p dx rλ−N ¯ upC 0,α (RN ) rN , B(x0 ,r)
and so, r−λ/p
!1/p
Z B(x0 ,r)
u − uB(x0 ,r) p dx
¯ uC 0,α (RN ) .
By taking the supremum over all r > 0 and x0 ∈ RN , we get uLp,λ (RN ) ¯ uC 0,α (RN ) , which concludes the proof. Exercise 7.3. Let Ω ⊆ RN be an open set such that there exists a constant C > 0 with the property that LN (B(x, r) ∩ Ω) ≥ CrN for every x ∈ Ω and every 0 < r < diam Ω. For 1 ≤ p < ∞ and 0 ≤ λ ≤ N + p, let Lp,λ (Ω) be the space of all functions u ∈ Lploc (Ω) such that !1/p Z uLp,λ (Ω) :=
sup x0 ∈Ω, r>0
r−λ/p
B(x0 ,r)∩Ω
u − uB(x0 ,r)∩Ω p dx
< ∞.
Prove that Theorem 7.1 continues to hold in Lp,λ (Ω). Next we show that W s,p (RN ) is embedded into Lp,sp (RN ). Theorem 7.4. Let 0 < s < 1 and 1 ≤ p < ∞. Then uLp,sp (RN ) ˙ s,p (RN ). uW s,p (RN ) for every u ∈ W Proof. Let x0 ∈ RN and r > 0. By Poincar´e’s inequality in B(x0 , r) (see Theorem 6.33), Z Z Z u(x) − u(y)p p sp u(x) − uB(x0 ,r)  dx r dxdy. N +sp B(x0 ,r) B(x0 ,r) kx − yk B(x0 ,r) Dividing by rsp and taking the supremum over all r > 0 and x0 ∈ RN proves the theorem.
7.2. Embeddings: The Subcritical Case In this section, we prove the embedding (7.1) in the subcritical case. Given 1 ≤ p < ∞ and 0 < s < 1, with sp < N , we define the Sobolev critical exponent Np p∗s := . N − sp We will show that kukLp∗s (RN ) uW s,p (RN ) for all u ∈ W s,p (RN ). More ˙ s,p (RN ) ∩ Lq (RN ), 1 ≤ q < generally, this inequality holds whenever u ∈ W s,p N ˙ (R ) vanishes at infinity. We recall the following definition. ∞, or if u ∈ W
262
7. Embeddings and Interpolation
Definition 7.5. 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 LN ({x ∈ E : u(x) > ε}) < ∞ for every ε > 0. Theorem 7.6 (Sobolev–Gagliardo–Nirenberg embedding in W s,p ). Let 0 < ˙ s,p (RN ) vanishing at infinity, s < 1 and 1 ≤ p < N/s. Then for every u ∈ W kukLp∗s (RN ) uW s,p (RN ) .
(7.8)
Moreover, W s,p (RN ) ,→ Lq (RN ) for all p ≤ q ≤ p∗s . ˙ s,p (RN ) ∩ Lp∗s (RN ). Using the inequalProof. Step 1: Assume that u ∈ W ity (a + b)p ≤ 2p−1 ap + 2p−1 bp , for x ∈ RN and r > 0 we have Z 1 u(x)p = N u(x)p dh L (B(0, r)) B(0,r) Z 2p−1 u(x + h) − u(x)p dh ≤ N L (B(0, r)) B(0,r) Z 2p−1 + N u(x + h)p dh. L (B(0, r)) B(0,r) By H¨older’s inequality with exponents p∗s /p and (p∗s /p)0 = N/(sp), we have Z Z u(x + h)p dh = u(y)p dy ≤ kukpLp∗s (RN ) (LN (B(0, r)))sp/N . B(0,r)
B(x,r)
Combining these two inequalities gives p
u(x) −
1 c0 kukpLp∗s (RN ) N −sp r
2p−1 sp r ≤ αN
Z
2p−1
Z
≤
(7.9)
αN
rsp
B(0,r)
RN
u(x + h) − u(x)p dh rN +sp
u(x + h) − u(x)p dh, khkN +sp
where in the last inequality we use the fact that khk < r and c0 := If u(x) 6= 0 and kukLp∗s (RN ) > 0, we take r to be r
N −sp
= 2c0
kukLp∗s (RN ) u(x)
2p−1 1−sp/N αN
!p ,
so that the lefthand side of (7.9) becomes !sp2 /(N −sp) Z kukLp∗s (RN ) 1 u(x + h) − u(x)p p u(x) dh. 2 u(x) khkN +sp RN Since p + sp2 /(N − sp) = p∗s , it follows that Z u(x + h) − u(x)p p∗s −p p∗s dh. u(x) kukLp∗s (RN ) khkN +sp RN
.
7.2. Embeddings: The Subcritical Case
263
Note that the last inequality continues to hold if u(x) = 0 or kukLp∗s (RN ) = 0. s −p Integrating in x over RN and dividing both sides by kukp∗ gives ∗ Lps (RN ) Z Z u(x + h) − u(x)p p kukLp∗s (RN ) dhdx. khkN +sp RN RN ∗
This proves (7.8) under the additional hypothesis that u ∈ Lps (RN ). Step 2: For n ∈ N define (t − 1/n) sgn t if 1/n ≤ t ≤ n, 0 if t < 1/n, fn (t) := (n − 1/n) sgn t if t > n, and vn = fn ◦ u. Then Z Z p∗s vn  dx = RN
∗
vn ps dx
{u>1/n} ∗
≤ (n − 1/n)ps LN ({x ∈ RN : u(x) > 1/n}) < ∞, since u is vanishing at infinity. Moreover, since fn is Lipschitz continuous with Lipschitz constant 1, vn (x + h) − vn (x) = fn (u(x + h)) − fn (u(x)) ≤ u(x + h) − u(x), ˙ s,p (RN ) ∩ Lp∗s (RN ). Hence, we may apply the preand so, vn belongs to W vious step to vn to obtain kvn kLp∗s (RN ) vn W s,p (RN ) uW s,p (RN ) . Letting n → ∞ and using Fatou’s lemma on the lefthand side gives (7.8). Remark 7.7. As pointed out in [BN20], Theorem 7.6 continues to hold even if 0 < p ≤ 1. The only change is that in place of the inequality (a + b)p ≤ 2p−1 ap + 2p−1 bp for a, b ≥ 0, we use (a + b)p ≤ ap + bp . We recall that F denotes the Fourier transform. Exercise 7.8. Let 0 < s < N/2 and u ∈ S(RN ), u 6= 0. Define Z g 2s 2 uH s (RN ) := kxk F(u)(x) dx . RN
Given R > 0 write u = gR + hR , where gR := F −1 (χB(0,R) F(u)),
hR := F −1 (χRN \B(0,R) F(u)).
(i) Prove that there is a constant C0 = C0 (N, s) > 0 such that kgR kL∞ (RN ) ≤ C0 RN/2−s ug H s (RN ) .
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7. Embeddings and Interpolation
(ii) For every t > 0 take R=
t 4C0 ug H s (RN )
!2∗s /N .
Prove that Z ∞ Z ∗ ∗ t2s −1 LN ({x ∈ RN : hR (x) > t/2}) dt u(x)2s dx N R Z Z0 ∞ 2∗s −3 hR (x)2 dxdt. t 0
(iii) Prove that kukL2∗s (RN )
RN ug . H s (RN )
˙ s,p (RN ), we can prove the Using the density of functions Cc∞ (RN ) in W following result. ˙ s,p (RN ) Corollary 7.9. Let 0 < s < 1 and 1 ≤ p < N/s. For every u ∈ W there exists a constant mu ∈ R such that (7.10)
ku − mu kLp∗s (RN ) uW s,p (RN ) .
Proof. By Theorem 6.107, there exists a sequence {un }n in Cc∞ (RN ) such that un − uW s,p (RN ) → 0. By Theorem 7.6, kun − u` kLp∗s un − u` W s,p (RN ) ∗
for every n, ` ∈ N. This shows that {un }n is a Cauchy sequence in Lps (RN ). ∗ ∗ Thus, there exists v ∈ Lps (RN ) such that un → v in Lps (RN ). By extracting a subsequence, not relabeled, without loss of generality, we may assume that un → v pointwise LN a.e. in RN . Hence, by Fatou’s lemma u − vW s,p (RN ) lim inf un − uW s,p (RN ) = 0, n→∞
which implies that u − v is a constant mu ∈ R. Again by Theorem 7.6, this time applied to un ∈ W m,p (RN ), n ∈ N, we have kun kLp∗s un W s,p (RN ) . Letting n → ∞ gives (7.10).
7.3. Embeddings: The Critical Case When s → (N/p)+ , Sobolev’s critical exponent p∗s = ps/(N −sp) approaches ∞. Indeed, we will show that W N/p,p (RN ) is embedded in every Lq (RN ) for p ≤ q < ∞. However, we cannot allow q = ∞, as the following exercise shows. Exercise 7.10. LetN ≥ 2 and u : RN \ {0} → R be defined by u(x) = 1 ϕ(x) log log 1 + kxk where ϕ ∈ Cc∞ (RN ), with ϕ = 1 in B(0, 1) and ϕ = 0 outside B(0, 2). Prove that u ∈ W 1,N (RN ). Deduce that u ∈ W s,N (RN ) for all 0 < s < 1 but u ∈ / L∞ (RN ).
7.3. Embeddings: The Critical Case
265
Theorem 7.11. Let N < p < ∞. Then uBMO(RN ) uW N/p,p (RN ) for ˙ N/p,p (RN ). every u ∈ W Proof. By Theorem 7.4, uLp,N (RN ) uW s,p (RN ) . As we already observed, Lp,N (RN ) = BMO(RN ). As a corollary of Theorem 7.6, we can prove that in the critical case s = N/p, W N/p,p (RN ) ,→ Lq (RN ) for every p ≤ q < ∞. Theorem 7.12. Let N < p < ∞. Then kukLq (RN ) kukW N/p,p (RN ) for every p ≤ q < ∞ and every u ∈ W N/p,p (RN ). Proof. If 0 < s1 < s2 := N/p, then s1 is subcritical, and so we may apply Theorem 7.6 to obtain kukLp∗s1 (RN ) uW s1 ,p (RN ) , where p∗s1 = N pN −s1 p . It follows by Lemma 6.17 that kukLp∗s1 (RN ) uW s1 ,p (RN ) kukW N/p,p (RN ) . To conclude the proof, observe that for every p < q < ∞, we can find s1 ∈ (0, N/p) such that q = p∗s1 . To be precise, take s1 = N (q − p)/(pq). We now present a second proof of Theorem 7.12, which is more involved, but provides a better estimate on the constants. This improvement will be crucial to prove exponential decay of functions in W N/p,p (RN ). Theorem 7.13. Let N < p < ∞. Then for every p ≤ q < ∞ there exists a constant C = C(N, p) > 0 such that for every u ∈ W N/p,p (RN ), 0
kukLq (RN ) ≤ Cq 1/p +1/q kukW N/p,p (RN ) . We refer to Theorem 14.10 below for a more general version of the following lemma but without such an explicit dependence on the relevant constants. Lemma 7.14. Let E ⊂ RN be a Lebesgue measurable set with finite measure, 0 < α < N , and let 1 ≤ p, q ≤ ∞ be such that (7.11)
0≤
1 1 α − < . p q N
Then for every nonnegative function v ∈ Lp (E), Z Z q 1/q v(x) dx dy N −α E kx − yk E 1/p0 +1/q 1/p0 + 1/q 1−α/N ≤ αN (LN (E))α/N +1/q−1/p kvkLp (E) . α/N + 1/q − 1/p
266
7. Embeddings and Interpolation
Proof. Step 1: Let 0 < β < N . We claim that for every y ∈ RN , Z
β/N
E
kx − yk−β dx ≤ N (N − β)−1 αN (LN (E))1−β/N .
To see this, let R > 0 be such that LN (E) = LN (B) = αN RN , where B := B(y, R). Then LN (B \ E) = LN (E \ B), and so, Z
kx − yk−β dx ≤
Z
E\B
R−β dx =
E\B
Z
R−β dx ≤
B\E
Z
kx − yk−β dx.
B\E
R Adding E∩B kx − yk−β dx to both sides and using spherical coordinates and the fact that LN (E) = αN RN gives Z
kx − yk−β dx ≤
Z
kx − yk−β dx = N (N − β)−1 αN RN −β
B
E
β/N
= N (N − β)−1 αN (LN (E))1−β/N , which proves the claim. Step 2: By (7.11) and the fact that α < N , the number 1/r = 1/p0 + 1/q satisfies 1 ≤ r < ∞. For every fixed y ∈ RN , let hy (x) := kx − ykα−N , x ∈ RN \ {y}. Write 0
0
p/q 1−r/q 1−p/q p/q r/p p/r hy v = (hr/q )hy v = (hr/q )hy v . y v y v
By the generalized H¨older’s inequality (see [Leo22c]) with exponents q, p0 , and r0 , Z
Z v(x) dx = hy (x)v(x) dx N −α E kx − yk E 1/q Z 1−1/p Z 1−1/r Z r p r p ≤ (hy (x)) (v(x)) dx (hy (x)) dx (v(x)) dx . E
E
E
Raising both sides to power q, integrating in y over E, and using Tonelli’s theorem gives q Z q−q/r v(x) p dx dy ≤ (v(x)) dx N −α E E kx − yk Z q−q/p Z Z (α−N )r p (α−N )r × kx − yk (v(x)) dx kx − yk dx dy.
Z Z E
E
E
E
7.3. Embeddings: The Critical Case
267
Since β := (N − α)r < N by (7.11), using Step 1 and Tonelli’s theorem, we have that Z q−q/p Z Z (α−N )r p (α−N )r kx − yk (v(x)) dx kx − yk dx dy E E E Z q−q/pZ β/N kx − yk(α−N )r dydx (v(x))p ≤ N (N − β)−1 αN (LN (E))1−β/N E E 1+q−q/p Z β/N (v(x))p dx. ≤ N (N − β)−1 αN (LN (E))1−β/N E
Combining the last two inequalities shows that q 1/q Z Z 1/r v(x) −1 β/N N 1−β/N dx dy ≤ N (N − β) α (L (E)) N N −α E E kx − yk Z 1+1/q−1/r × (v(x))p dx . E
To conclude, observe that 1/r − β/(N r) = α/N + 1/q − 1/p, N N 1/r 1/p0 + 1/q = = = , N −β N − (N − α)r 1/r − 1 + α/N α/N + 1/q − 1/p and 1 + 1/q − 1/r = 1/p.
In what follows, given a ball B = B(x0 , r) and n ∈ N, we denote by nB the ball B(x0 , nr). Lemma 7.15. Let N < p < q < ∞ and B = B(x0 , 2−k0 ), where x0 ∈ RN ˙ N/p,p (8B), and k0 ∈ Z. Then for every u ∈ W 0
ku − uB kLq (B) ≤ Cq 1/p +1/q (LN (2B))1/q uW N/p,p (8B) , where C = C(N, p) > 0. Proof. Define jN/p
f (x) := sup 2 j≥k0 −2
1 N L (B(x, 2−j ))
Z B(x,2−j )
u(y) − uB(x,2−j )  dy
for x ∈ 2B and f (x) := 0 for x ∈ RN \ 2B. Step 1: We claim that Z f (x) ≤ C
(7.12)
8B
for all x ∈
RN
u(y) − u(x)p dy kx − yk2N
1/p
and for some constant C = C(N, p) > 0. For x ∈ 2B, Z 1 u(x) − uB(x,2−j )  ≤ N u(y) − u(x) dy, L (B(x, 2−j )) B(x,2−j )
268
7. Embeddings and Interpolation
and so, Z B(x,2−j )
Z u(y) − uB(x,2−j )  dy ≤
u(y) − u(x) dy Z N −j + L (B(x, 2 ))u(x) − uB(x,2−j )  ≤ 2 B(x,2−j )
u(y) − u(x) dy
B(x,2−j )
≤ C2
−j2N/p
Z B(x,2−j )
u(y) − u(x) dy, ky − xk2N/p
where in the last inequality we use the fact that kx − yk ≤ 2−j . By H¨ older’s inequality we can bound the righthand side from above by !1/p Z p u(y) − u(x) 0 2−jN/p 2−j2N/p dy . kx − yk2N B(x,2−j ) Combining the last two inequalities gives u(y) − u(x)p f (x) ≤ C sup dy kx − yk2N j≥k0 −2 B(x,2−j ) Z 1/p u(y) − u(x)p ≤C dy , kx − yk2N 8B Z
!1/p
where we use the fact that B(x, 2−j ) ⊆ 8B. To see this, observe that if x ∈ 2B = B(x0 , 2−k0 +1 ) and y ∈ B(x, 2−j ), then ky − x0 k ≤ ky − xk + kx − x0 k < 2−j + 2−k0 +1 ≤ 2−k0 +2 + 2−k0 +1 ≤ 2−k0 +3 . This proves claim (7.12). Step 2: Let u ¯ be a representative of u. We claim that for every Lebesgue point x ∈ B of u ¯, Z f (z) (7.13) ¯ u(x) − uB  ≤ C dz N −N/p 2B kx − zk for some constant C = C(N, p) > 0. The proof is similar to the one of Lemma 7.2. We will use the fact that uB(x,2−j ) → u ¯(x) as j → ∞ (see [Leo22c]). Using telescopic sums, it follows that X (7.14) ¯ u(x) − uB(x,2−k0 −1 )  = (uB(x,2−j−1 ) − uB(x,2−j ) ) j≥k0 +1 X ≤ uB(x,2−j−1 ) − uB(x,2−j ) . j≥k0 +1
7.3. Embeddings: The Critical Case
269
Consider a ball B(xj , 2−j−2 ) ⊂ B(x, 2−j ) \ B(x, 2−j−1 ). Then for z ∈ B(xj , 2−j−2 ), we have that B(x, 2−j ) ⊂ B(z, 2−j+1 ) since for y ∈ B(x, 2−j ), ky − zk ≤ ky − xk + kx − zk < 2−j + 2−j = 2−j+1 . Hence, uB(x,2−j−1 ) − uB(x,2−j )  ≤ uB(x,2−j−1 ) − uB(z,2−j+1 )  Z jN u(y) − uB(z,2−j+1 )  dy + uB(x,2−j ) − uB(z,2−j+1 )  ≤ C2 B(x,2−j ) Z ≤ C2jN u(y) − uB(z,2−j+1 )  dy B(z,2−j+1 )
≤ C2−jN/p f (z). Averaging both sides in z over B(xj , 2−j−2 ) gives uB(x,2−j−1 ) − uB(x,2−j )  ≤ C2jN (1−1/p)
Z f (z) dz. B(xj ,2−j−2 )
Hence, by (7.14) and the fact that the balls B(xj , 2−j−2 ) ⊂ 2B are disjoint, X
¯ u(x) − uB(x,2−k0 −1 )  ≤ C
2
jN (1−1/p)
Z
j≥k0 +1
(7.15)
X Z
≤C
−j−2 ) j≥k0 +1 B(xj ,2
Z ≤C 2B
f (z) dz B(xj ,2−j−2 )
f (z) dz kz − xkN −N/p
f (z) dz. kz − xkN −N/p
Next we estimate uB − uB(x,2−k0 −1 ) . For z ∈ B(x, 2−k0 −2 ) we have that B ⊂ B(z, 2−k0 +2 ) since for y ∈ B = B(x0 , 2−k0 ) we have ky − zk ≤ ky − x0 k + kx0 − xk + kx − zk < 2−k0 + 2−k0 + 2−k0 −2 ≤ 2−k0 +2 . Similarly, B(x, 2−k0 −1 ) ⊂ B(z, 2−k0 ) since for y ∈ B(x, 2−k0 −1 ) we have ky − zk ≤ ky − xk + kx − zk < 2−k0 −1 + 2−k0 −2 < 2−k0 .
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7. Embeddings and Interpolation
Hence, uB − uB(x,2−k0 −1 )  ≤ uB − uB(z,2−k0 +2 ) 
(7.16)
+ uB(x,2−k0 −1 ) − uB(z,2−k0 +2 )  Z k0 N u(y) − uB(z,2−k0 +2 )  dy ≤ C2 B Z u(y) − uB(z,2−k0 +2 )  dy + C2k0 N B(x,2−k0 −1 ) Z k0 N u(y) − uB(z,2−k0 +2 )  dy ≤ C2−k0 N/p f (z). ≤ C2 B(z,2−k0 +2 )
Averaging both sides in z over B(x, 2−k0 −2 ) ⊂ 2B gives Z k0 N (1−1/p) f (z) dz uB − uB(x,2−k0 −1 )  ≤ C2 B(x,2−k0 −2 )
Z ≤C
(7.17)
B(x,2−k0 −2 )
Z ≤C 2B
f (z) dz kz − xkN −N/p
f (z) dz. kz − xkN −N/p
Combining this inequality with (7.15) shows (7.13). Step 3: By (7.13) and Lemma 7.14, with α = N/p, for every p < q < ∞, Z 1/q Z Z q 1/q f (z) q u − uB  dx ≤C dz dx N −N/p B 2B 2B kx − zk 0 1/p0 + 1/q 1/p +1/q N (L (2B))1/q kf kLp (2B) ≤C 1/q 0
≤ Cq 1/p +1/q (LN (2B))1/q uW N/p,p (8B) , where we use Step 1, which gives kf kLp (2B) ≤ CuW N/p,p (8B) .
Remark 7.16. In Lemma 7.15, we can replace uB with uB1 , where B1 ⊆ B = B(x0 , 2−k0 ) has radius 2−k0 −2 . It suffices to replace uB with uB1 in (7.16) and (7.17) and observe that the constant C changes by a factor of 2−2N . Hence, 0
ku − uB1 kLq (B) ≤ Cq 1/p +1/q (LN (2B))1/q uW N/p,p (8B) . N/p.p
Exercise 7.17. Let N < p < ∞ and u ∈ W00 (B(0, 1)). Prove that for 0 every p < q < ∞, kukLq (B) ≤ Cq 1/p +1/q ˜ uW N/p,p (RN ) , where C = C(N, p) > 0 and u ˜ is the function obtained by extending u to be zero outside B(0, 1). Hint: Use Remark 7.16.
7.3. Embeddings: The Critical Case
271
Exercise 7.18. LetPwn : RN → [0, ∞), n ∈ N, be Lebesgue measurable functions and 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 X ∞ p 1/p0 kwkLp (RN ) ≤ M kwn kLp (RN ) , n=1
while kwkL∞ (RN ) ≤ M supn kwn kL∞ (RN ) . We turn to the proof of Theorem 7.13. Proof of Theorem 7.13. Step 1: Assume first that u ∈ Cc∞ (RN ), with supp u ⊆ B(x0 , 2−k0 −1 ) for some x0 ∈ RN and k0 ∈ Z. Let B = B(x0 , 2−k0 ) and consider a ball B1 := B(x1 , 2−k0 −1 ) ⊆ B(x0 , 2−k0 )\B(x0 , 2−k0 −1 ). Then uB1 = 0, and so, by Remark 7.16, for p < q < ∞, 0
kukLq (RN ) ≤ Cq 1/p +1/q (2−k0 +1 )N/q uW N/p,p (8B) , where C = C(N, p) > 0. Step 2: For every z ∈ ZN consider the open cube Q(z, 2) centered at z and S N of sidelength 2. Then R = z∈ZN Q(z, 2). Construct a smooth partition of unity {ψz }z∈ZN subordinated to {Q(z, 2)}z∈ZN , with the property that k∇ψz k∞ ≤ C for every z. Since for every x ∈ RN , there are at most 2N cubes Q(z, 2) that contain x, it follows from Exercise 7.18 that
1/q
X
X
0 kukLq (RN ) = (uψz ) ≤ 2N/q kuψz kqLq (RN ) .
z∈ZN
q N z∈ZN L (R )
√ Let k0√ ∈ Z be such that 2−k0 −2 < 2 N ≤ 2−k0 −1 . Then Q(z, 2) ⊂ B(z, 2 N ) ⊆ B(z, 2−k0 −1 ). Hence we can apply Step 1 to the function uψz to obtain 1/q X 0 0 kukLq (RN ) ≤ C2N/q q 1/p +1/q (2−k0 +1 )N/q uψz qW N/p,p (8B ) , z
z∈ZN
where Bz := B(z, 2−k0 ). Since q > 1, N/q ≤ N , and so, we have that 0
C2N/q (2−k0 +1 )N/q ≤ C2(−k0 +1)N =: C0 , where C0 = C0 (N, p) > 0 does not depend on q. Hence, X 0 (7.18) kukLq (RN ) ≤ C0 q 1/p +1/q uψz q N/p,p W
z∈ZN
1/q (8Bz )
.
272
7. Embeddings and Interpolation
Using the fact that we have that X uψz q
aqz
P
z∈ZN
1/q
≤
1/q
W N/p,p (8Bz )
P
z∈ZN
apz
1/p
for az ≥ 0 (exercise), 1/p
≤
z∈ZN
X
uψz pW N/p,p (8B ) z
.
z∈ZN
Thus, in view of (7.18), to conclude the proof, it is enough to show that X uψz pW N/p,p (8B ) ≤ CkukpW N/p,p (RN ) . (7.19) z
z∈ZN
By Theorem 6.23 and Remark 6.25, uψz W N/p,p (8Bz ) ≤ Ck∇ψz k∞ kukLp (8Bz ) + Ckψz k∞ uW N/p,p (8Bz ) ≤ CkukW N/p,p (8Bz ) , where C = C(N, p) > 0. Hence, X
uψz pW N/p,p (8B
z)
≤C
X Z z∈ZN
z∈ZN
+C
X Z z∈ZN
8Bz
Z RN
up dx
8Bz
u(y) − u(x)p dxdy = A + B. kx − yk2N
By the Lebesgue monotone convergence theorem Z Z X p A= χ8Bz u dx ≤ C RN
up dx,
RN
z∈ZN
P where we use the fact that z∈ZN χ8Bz ≤ C for some constant depending on N . A similar inequality holds for B. This shows (7.19). Note that in general, if u ∈ W N/p,p (RN ), we cannot conclude that u ∈ for 1 ≤ q < p. However, since u ∈ Lq (RN ) for all q ≥ p, 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 Lq (RN )
(7.20)
∞ `−1 X X 1 n 1 n exp` (t) := t = exp(t) − t , n! n! n=`
t ∈ R.
n=0
Theorem 7.19. Let N < p < ∞. Then there exist two positive constants C1 , C2 , depending only on N and p, such that for all u ∈ W N/p,p (RN ) \ {0}, Z 0 u(x)p expbp−1c+1 C1 dx ≤ C2 . (7.21) 0 kukpW N/p,p (RN ) RN
7.3. Embeddings: The Critical Case
273
Proof. By Theorem 7.13 there exists C0 = C0 (N, p) > 0 such that for every u ∈ W N/p,p (RN ) and p < q < ∞, 0
kukLq (RN ) ≤ C0 q 1/p +1/q kukW N/p,p (RN ) . Taking q = np0 , where n > bp − 1c, n ∈ N, yields Z 0 0 0 unp dx ≤ C0np (np0 )n+1 kuknp . W N/p,p (RN ) RN
Let C1 > 0. Then X n>bp−1c
C1n n!
unp
Z
0
0
0
RN
kuknp W N/p,p (RN )
dx ≤
X n>bp−1c
C1n C0np (np0 )n+1 . n!
To use the ratio test, observe that (n+1)p0
lim
n→∞
C1n+1 C0 (n+1)!
0 C1n C0np
n!
((n + 1)p0 )n+2
0
= C1 C0p p0 lim (1 + 1/n)n+1 n→∞
(np0 )n+1
0
= C1 C0p p0 e < 1, 0
provided C1 < 1/(C0p p0 e). To complete the proof, it suffices to take 0
C2 :=
X n>bp−1c
C1n C0np (np0 )n+1 < ∞. n!
While the expression on the lefthand side of (7.21) looks unusual, it is natural in the context of Orlicz spaces (see [RR91]). Definition 7.20. 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 functionR 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). We define Z n o kukΦ := inf t > 0 : Φ(u(x)/t) dx < ∞ . E
We leave it as an exercise to check that k·kΦ is a norm and that LΦ (E) is a Banach space. Note that Theorem 7.19 shows that W N/p,p (RN ) ,→ LΦ (RN ), 0 where Φ(s) := expbp−1c+1 (C1 sp ), s ≥ 0. ˙ N/p/p . We now give a local version of Theorem 7.19 for W
274
7. Embeddings and Interpolation
Theorem 7.21. Let N < p < ∞. Then there exist two positive constants C1 , C2 , depending only on N and p, such that for every open ball B ⊂ RN ˙ N/p,p (8B) \ {0}, and every u ∈ W Z p0 u(x) − u  B dx ≤ C2 LN (B). exp C1 p0 uW N/p,p (8B) B Proof. By Lemma 7.15 for p < q < ∞, Z u − uB q dx ≤ C0q q q+1−q/p LN (2B)uqW N/p,p (8B) , B
where C0 = C0 (N, p) > 0. Taking q = np0 , where n > bp − 1c, n ∈ N, Z 0 0 0 . u − uB np dx ≤ C0np (np0 )n+1 LN (2B)unp W N/p,p (8B) B
Let C1 > 0. Summing over all n > bp − 1c gives X C n Z u − uB np0 1 (7.22) dx ≤ LN (2B) 0 n! B unpN/p,p n>bp−1c
W
(8B)
X n>bp−1c
C1n np0 C (np0 )n+1 . n! 0
Observe that C1n+1 (n+1)p0 ((n + 1)p0 )n+2 (n+1)! C0 lim C1n np0 n→∞ 0 n+1 n! C0 (np )
0
= C1 C0p p0 lim (1 + 1/n)n+1 n→∞
0
= C1 C0p p0 e < 1, P 0 0 Cn provided C1 < 1/(C0p p0 e). Hence, 2N n>bp−1c n!1 C0np (np0 )n+1 < ∞ by the ratio test. On the other hand, if n ≤ bp−1c, then np0 ≤ p, and so, by H¨ older’s 0 0 inequality with exponent q = p/(p n) = (p−1)/n and q = (p−1)/(p−1−n), and by Poincar´e’s inequality (Theorem 6.33), Z n/(p−1) Z np0 N (p−1−n)/(p−1) p u − uB  dx ≤ (L (B)) u − uB  dx B
B N
(p−1−n)/(p−1)
≤ C(L (B))
(diam B)2N LN (B)
upW N/p,p (B)
n/(p−1)
0
≤ CLN (B)unp . W N/p,p (8B) It follows that bp−1c n X C n Z u − uB np0 X C N 1 1 dx ≤ CL (B) . 0 np n! B u N/p,p n! n=0 n=0 W (8B)
bp−1c
(7.23)
7.4. Embeddings: The Supercritical Case
275
Summing inequalities (7.22) and (7.23) and using the Lebesgue monotone convergence theorem gives Z X 0 ∞ C1n u − uB np dx np0 B n=0 n! u N/p,p W (8B) bp−1c n X C n np0 X C 1 1 ≤ LN (B) C (np0 )n+1 + C . n! 0 n! n=0
n>bp−1c
N/p,p
Exercise 7.22. Let N < p < ∞ and u ∈ W00 (B(0, 1)) be such that ˜ uW N/p,p (RN ) ≤ 1, where u ˜ is the function obtained by extending u to be R 0 zero outside B(0, 1). Prove that B(0,1) exp(C1 u(x)p ) dx ≤ C2 , where C1 = C1 (N, p) > 0 and C2 = C2 (N, p) > 0. Hint: Use Exercise 7.17.
7.4. Embeddings: The Supercritical Case In this section we study the case p > N/s. The next result is the extension of the Morrey embedding theorem in W 1,p (see [Leo22d]) to fractional Sobolev spaces. Theorem 7.23 (Morrey’s embedding in W s,p ). Let 0 < s < 1 and N/s < ˙ s,p (RN ) admits a representative u p < ∞. Then every u ∈ W ¯ that is H¨ older continuous with exponent α = s − N/p such that ¯ uC 0,α (RN ) uW s,p (RN ) . Moreover, if u ∈ W s,p (RN ), then ¯ uC 0 (RN ) kukW s,p (RN )
and
lim u ¯(x) = 0.
kxk→∞
In particular, W s,p (RN ) ,→ C 0,α (RN ). ˙ s,p (RN ), uLp,sp (RN ) uW s,p (RN ) . Proof. By Theorem 7.4, for every u ∈ W The inequality ¯ uC 0,α (RN ) uW s,p (RN ) follows from Theorem 7.1. Next we prove that if u ∈ W s,p (RN ) and u ¯ is its H¨ older continuous representative, then (7.24)
k¯ uk∞ kukW s,p (RN ) .
By (7.5) with R = 1 and what we just proved, ¯ u(x) − uB(x,1)  uLp,sp (RN ) uW s,p (RN ) for every x ∈ RN . Hence, by H¨older’s inequality ¯ u(x) ≤ uB(x,1)  + ¯ u(x) − uB(x,1)  kukLp (B(x,1)) + uW s,p (RN ) .
276
7. Embeddings and Interpolation
Finally, we prove that u ¯(x) → 0 as kxk → ∞. Let {un }n be any sequence in Cc∞ (RN ) that converges to u in W s,p (RN ) (see Theorem 6.66). By (7.24), k¯ u − un k∞ ku − un kW s,p (RN ) → 0
as n → ∞.
Fix ε > 0 and find n ¯ ∈ N such that k¯ u − un k∞ ≤ ε for all n ≥ n ¯ . Since un¯ ∈ Cc∞ (RN ), there exists Rn¯ > 0 such that un¯ (x) = 0 for all kxk ≥ Rn¯ . Hence, for x ∈ RN , with kxk ≥ Rn¯ , we get ¯ u(x) = ¯ u(x) − un¯ (x) ≤ k¯ u − un k∞ ≤ ε.
It follows from Morrey’s embedding theorem that W s,p (RN ) is an algebra when sp > N . Corollary 7.24 (Product). Let 1 ≤ p < ∞ and 0 < s < 1, with sp > N . If u, v ∈ W s,p (RN ), then uv ∈ W s,p (RN ), with kuvkW s,p (RN ) kukW s,p (RN ) kvkW s,p (RN ) . Proof. By Morrey’s embedding theorem, kukL∞ (RN ) kukW s,p (RN ) ,
kvkL∞ (RN ) kvkW s,p (RN ) .
Hence, we are in a position to apply Theorem 6.26 to obtain kuvkLp (RN ) ≤ kukLp (RN ) kvkL∞ (RN ) kukLp (RN ) kvkW s,p (RN ) and uvW s,p (RN ) ≤ uW s,p (RN ) kvkL∞ (RN ) + kukL∞ (RN ) vW s,p (RN ) kukW s,p (RN ) kvkW s,p (RN ) .
Exercise 7.25. Let α ∈ R, 1 ≤ p < ∞, and 0 < s < 1. Consider a function u ∈ C ∞ (RN \ {0}) such that u(x) = kxkα for x ∈ B(0, 1) \ {0} and u(x) = 0 for x ∈ RN \ B(0, 2). Prove that u ∈ W s,p (RN ) if and only if s < α + N/p. Exercise 7.26. Let α ∈ R, β > 0, 1 ≤ p < ∞, and 0 < s < 1. Consider a function u ∈ C ∞ (RN \ {0}) such that u(x) :=
kxkα , (− log kxk)β
x ∈ B(0, 1/2) \ {0},
and u(x) = 0 for x ∈ RN \B(0, 1). Prove that u ∈ W s,p (RN ) provided either s < α + N/p or s = α + N/p and βp > 1. Hint: Follow the scheme in Exercise 2.14. Exercise 7.27. Let α ∈ R and 0 < β < 1 be such 0 < α+N β < 1. Consider a function u ∈ C ∞ (RN \ {0}) such that u(x) :=
kxkα , (− log kxk)β
x ∈ B(0, 1/2) \ {0},
7.5. Embeddings: The General Case
277
and u(x) = 0 for x ∈ RN \ B(0, 1). Let p := 1/β and s := α + N/p. Prove that u ∈ / W s,p (RN ). Hint: Follow the scheme in Exercise 2.15
7.5. Embeddings: The General Case We begin this section by studying the relationship between Sobolev spaces and fractional Sobolev spaces. Theorem 7.28. Let 1 ≤ p1 < ∞, 1 < p2 < ∞, and 0 < s1 < s2 = 1 be such that N N (7.25) 1− = s1 − . p2 p1 ˙ 1,p2 (RN ), Then for every u ∈ W uW s1 ,p1 (RN ) k∇ukLp2 (RN ) .
(7.26)
Moreover, W 1,p2 (RN ) ,→ W s1 ,p1 (RN ). We begin with a preliminary result. Lemma 7.29. Let Ω ⊂ RN be an open bounded convex set, let E ⊆ Ω be a Lebesgue measurable set with positive measure, and let u ∈ C 1 (Ω). Then Z (diam Ω)N k∇u(y)k u(x) − uE  ≤ dy N N L (E) Ω kx − ykN −1 for every x ∈ Ω. Proof. Given x, y ∈ Ω, with x 6= y, by the fundamental theorem of calculus, Z kx−yk y−x u(x) − u(y) = − ∇u(x + rω) · ω dr, ω := . ky − xk 0 Averaging in y over E, we obtain Z Z kx−yk 1 u(x) − uE = − N ∇u(x + rω) · ω drdy. L (E) E 0 Hence, (7.27)
1 u(x) − uE  ≤ N L (E)
Z Z
kx−yk
k∇u(x + rω)k drdy. E
0
Define f (x) :=
k∇u(x)k if x ∈ Ω, 0 otherwise.
278
7. Embeddings and Interpolation
Then, by (7.27) and the fact that E ⊆ B(x, d), d := diam Ω, Z Z ∞ 1 u(x) − uE  ≤ N f (x + rω) drdy L (E) B(x,d) 0 Z dZ Z ∞ 1 = N f (x + rω)ρN −1 drdHN −1 (ω)dρ L (E) 0 ∂B(0,1) 0 Z ∞Z dN = f (x + rω) dHN −1 (ω)dr =: A, N LN (E) 0 ∂B(0,1) where we use spherical coordinates and Tonelli’s theorem. Using spherical coordinates centered at x, we can rewrite A as Z ∞Z dN rN −1 A= f (x + rω) dHN −1 (ω)dr N −1 N LN (E) 0 r ∂B(0,1) Z Z N d f (y) dN k∇u(y)k = dy = dy. N LN (E) RN kx − ykN −1 N LN (E) Ω kx − ykN −1 Lemma 7.30. Let v ∈ L1loc (RN ), N ≥ 2. Then for every R > 0 and every x ∈ RN , Z Z 2R Z v(y) 1 (7.28) v(y) dydr. dy N −1 rN B(x,r) B(x,R) kx − yk 0 Proof. Using spherical coordinates and Tonelli’s theorem, we can write Z 2R Z Z 2R Z rZ 1 1 v(y) dydr = v(ρw) dHN −1 (w)dρdr N N r r 0 B(x,r) 0 0 ∂B(x,ρ) Z 2R Z Z 2R 1 = v(ρw) dHN −1 (w) drdρ rN 0 ∂B(x,ρ) ρ Z 2R Z 1 1 1 N −1 = v(ρw) dH (w) − dρ = A. N −1 0 ρN −1 (2R)N −1 ∂B(x,ρ) If 0 < ρ < R, then 1 ρN −1
−
1 1 1 C ≥ N −1 − ≥ N −1 . N −1 N −1 (2R) ρ (2ρ) ρ
It follows that Z R Z Z 1 v(y) N −1 A dy. v(ρw) dH (w)dr = N −1 N −1 0 ρ ∂B(x,ρ) B(x,R) kx − yk We recall that, given v ∈ L1loc (RN ), the (Hardy–Littlewood) maximal function of v is defined by Z 1 v(y) dy (7.29) M(v)(x) := sup N r>0 L (B(x, r)) B(x,r)
7.5. Embeddings: The General Case
279
for all x ∈ RN (see [Leo22c]). We turn to the proof of Theorem 7.28.
Proof of Theorem 7.28. If N = 1, then the result follows from Theorem 2.21. Thus, in what follows, we take N ≥ 2. Assume that u ∈ C 1 (RN ). Given x, h ∈ RN , with h 6= 0, let E = B(x, khk) ∩ B(x + h, khk). Note that LN (E) ≥ c0 khkN . Applying Lemma 7.29, with Ω replaced by B(x, khk) and B(x + h, khk), we have
u(x + h) − u(x) ≤ u(x + h) − uE  + u(x) − uE  Z khkN k∇u(y)k N dy L (E) B(x+h,khk) kx + h − ykN −1 Z k∇u(y)k khkN dy + N L (E) B(x,khk) kx − ykN −1 Z Z k∇u(y)k k∇u(y)k dy + dy. N −1 N −1 B(x+h,khk) kx + h − yk B(x,khk) kx − yk
It follows that Z RN
(7.30)
u(x + h) − u(x)p1 dhdx khkN +s1 p1 RN Z Z Z
Z
RN
RN
Z
Z
+ RN
RN
!p 1 k∇u(y)k dh dy dx N −1 N +s1 p1 kx + h − yk khk B(x+h,khk) !p 1 Z k∇u(y)k dh dy dx. N −1 N khk +s1 p1 B(x,khk) kx − yk
Using Tonelli’s theorem and making the change of variables x + h = z, so that dx = dz, the first integral on the righthand side reduces to the second. We claim that
Z
Z
(7.31) RN
B(x,khk)
k∇u(y)k dy kx − ykN −1
!p1
dh khkN +s1 p1 2 k∇ukpL1p−p (M(k∇uk)(x))p2 , 2 (RN )
280
7. Embeddings and Interpolation
where M(k∇uk) is the Hardy–Littlewood maximal function of k∇uk. Using (7.28) for ρ > 0 we can write !p1 Z Z k∇u(y)k dh dy N −1 N khk +s1 p1 RN B(x,khk) kx − yk !p1 Z Z 2khk Z 1 dh (7.32) k∇u(y)k dydr N N r khk +s1 p1 B(0,ρ) 0 B(x,r) !p1 Z Z Z 2khk dh 1 k∇u(y)k dydr + N N +s1 p1 r khk N B(x,r) R \B(0,ρ) 0 =: A + B. To estimate A observe that A (M(k∇uk)(x))p1
(M(k∇uk)(x)) ρ On the other hand, if 0 < α < H¨older’s inequality,
0
1 rN
p2 N
khkN +s1 p1
0
B(0,ρ)
RN \B(0,ρ)
dh
1 dr
p1 (1−s1 )p1
B (M(k∇uk)(x))(1−α)p1 Z 2khk Z ×
!p1
2khk
Z
Z
.
is such that s1 > 1 − αN/p2 , then by
!p1
!α
Z k∇u(y)kdy
dr
!α
!p1
B(x,r)
dh khkN +s1 p1
(1−α)p1
(M(k∇uk)(x)) Z Z × RN \B(0,ρ)
0
2khk
0
rN/p2 k∇ukLp2 (RN ) rN
1 k∇ukαp (M(k∇uk)(x))(1−α)p1 Lp2 (RN )
dr
dh khkN +s1 p1
1 , ρ(s1 −1+αN/p2 )p1
where we use the facts that s1 > 1 − αN/p2 and !p1 Z Z 2khk 1 1 dr dh N +s1 p1 rαN/p2 RN \B(0,ρ) khk 0 p 1 Z khk1−αN/p2 dh khkN +s1 p1 RN \B(0,ρ) Z ∞ 1 1 dη (s −1+αN/p )p . 1+(s −1+αN/p )p 1 2 1 1 2 1 η ρ ρ
7.5. Embeddings: The General Case
281
Combining the estimates for A and B gives A + B (M(k∇uk)(x))p1 ρ(1−s1 )p1 1 . ρ(s1 −1+αN/p2 )p1 Note that if M(k∇uk)(x) = 0, then the righthand side of the previous inequality is zero. Otherwise, taking k∇ukLp2 (RN ) p2 /N ρ= M(k∇uk)(x) gives 1 + k∇ukαp (M(k∇uk)(x))(1−α)p1 Lp2 (RN )
(1−s )p p2 /N
A + B k∇ukLp2 (R1 N 1)
(M(k∇uk)(x))p1 −(1−s1 )p1 p2 /N
αp −(s −1+αN/p2 )p1 p2 /N
+ k∇ukLp21 (RN1)
(M(k∇uk)(x))(1−α)p1 +(s1 −1+αN/p2 )p1 p2 /N
2 k∇ukpL1p−p (M(k∇uk)(x))p2 , 2 (RN )
where we use the facts that, by (7.25), αp1 − (s1 − 1 + αN/p2 )p1 p2 /N = (1 − s1 )p1 p2 /N = p1 − p2 , (1 − α)p1 + (s1 − 1 + αN/p2 )p1 p2 /N = p1 − (1 − s1 )p1 p2 /N = p2 . Hence, in view of (7.32), we have proved (7.31). It now follows from (7.30) and (7.31) that Z Z u(x) − u(x + h)p1 p1 −p2 (M(k∇uk)(x))p2 dx dxdh k∇uk N +s1 p1 Lp2 (RN ) khk N N N R R R k∇ukpL1p2 (RN ) ,
Z
where the last inequality is due to the fact that M : Lp2 (RN ) → Lp2 (RN ) is a bounded operator for p2 > 1 (see [Leo22c]). This proves (7.26) for ˙ 1,p2 (RN ) ∩ C 1 (RN ). In the case u ∈ W ˙ 1,p2 (RN ), we apply (7.26) to u∈W uε = ϕε ∗ u to get Z Z uε (x) − uε (x + h)p1 dxdh k∇uε kpL1p2 (RN ) k∇ukpL1p2 (RN ) , N +s p 1 1 khk RN RN where the last inequality follows from the fact that ∇uε = ϕε ∗ ∇u and from standard properties of mollifiers (see [Leo22c]). Letting ε → 0+ and using Fatou’s lemma on the lefthand side, we find that (7.26) holds for u. To prove the last part of the statement, observe that if u ∈ W 1,p2 (RN ), then by the embedding theorems for W 1,p2 (RN ) (see [Leo22d]), we have that u ∈ Lp2 (RN ) ∩ Lr (RN ), where ∗ if p2 < N, p2 any number in (p2 , ∞) if p2 = N, r= ∞ if p2 > N.
282
7. Embeddings and Interpolation
Note that if p2 < N , then by (7.25), p∗2 =
N p2
N = −1
N p1
N = (p1 )∗s1 > p1 . − s1
Since s1 < 1, by (7.25) we have that p2 < p1 < r, and so, kukLp1 (RN ) kukθLp2 (RN ) kuk1−θ kukθLp2 (RN ) kuk1−θ , Lr (RN ) W 1,p2 (RN ) where
1 p1
=
θ p2
+
1−θ r .
Exercise 2.23 shows that when p2 = 1 and 1 < p1 < ∞, the embedding W 1,1 (R) ,→ W 1/p1 ,p1 (R) fails. However, when N ≥ 2, Theorem 7.28 continues to hold for p2 = 1. Theorem 7.31. Let N ≥ 2, 1 < p1 < ∞, and 0 < s1 < s2 = 1 be such that (7.33)
1+
N = s1 + N. p1
˙ 1,1 (RN ), Then for every u ∈ W uW s1 ,p1 (RN ) k∇ukL1 (RN ) .
(7.34)
Moreover, W 1,1 (RN ) ,→ W s1 ,p1 (RN ). Proof. Step 1: Assume that N ≥ 3 and that u ∈ Cc∞ (RN ). Write x = (x0 , xN ) ∈ RN −1 × R. By (7.33), 1 s1 N − 1 1 − s1 = + = s1 + N . p1 N N N −1
(7.35)
It follows that for every x0 ∈ RN −1 , 1 ku(x0 , ·)kLp1 (R) ≤ ku(x0 , ·)ksL11 (R) ku(x0 , ·)k1−s . LN/(N −1) (R)
(7.36)
By Theorem 2.39 applied to u(x0 , ·) and (7.36), we obtain Z ∞Z u(x0 , xN + h) − u(x0 , xN )p1 (7.37) dxN dh h1+s1 p1 0 R Z 1−s1 p1 Z s1 p21 0 p1 0 u(x , xN ) dxN ∂N u(x , xN ) dxN R
Z
R
u(x0 , xN )dxN
s1 p1 (1−s1 p1 ) Z
R
∂N u(x0 , xN ) dxN
R
Z × R
u(x0 , xN )N/(N −1) dxN
(1−s1 )p1 (1−s1 p1 )(N −1)/N .
s1 p21
7.5. Embeddings: The General Case
283
Define N −1 1 , Q := , s1 p1 (1 − s1 p1 )(N − 2) s1 p21 N R := , (1 − s1 )p1 (1 − s1 p1 )(N − 1)
P := (7.38)
and observe that by (7.33), 1 1 1 1 − s1 s1 = 1. + + = p1 − (1 − s1 p1 )p1 + P Q R N −1 N We now integrate both sides of (7.37) in x0 over RN −1 and apply the generalized H¨older inequality (see [Leo22c]) and Tonelli’s theorem to obtain Z Z ∞Z u(x0 , xN + h) − u(x0 , xN )p1 dxN dhdx0 1+s1 p1 h N −1 R 0 R Z s1 p1 (1−s1 p1 ) Z s1 p21 Z 0 0 u(x , xN )dxN ∂N u(x , xN ) dxN RN −1
R
R
(7.39) Z ×
0
N/(N −1)
u(x , xN )
p1 (1−s1 )(1−s1 p1 )(N −1)/N dxN
dx0
R
Z
Z
u(x0 , xN )dxN
RN −1
(N −1)/(N −2)
!1/P dx0
R N/[(N −1)R]
1/Q
× k∇ukL1 (RN ) kukLN/(N −1) (RN ) . By Minkowski’s inequality for integrals (see [Leo22c]), the Sobolev–Gagliardo–Nirenberg embedding theorem (see [Leo22d]) applied to the function u(·, xN ), and Tonelli’s theorem !(N −2)/(N −1) Z (N −1)/(N −2) Z u(x0 , xN )dxN dx0 RN −1
R
Z Z
0
≤ RN −1
R
Z Z R
RN −1
(N −1)/(N −2)
u(x , xN )
dx
0
(N −2)/(N −1) dxN
k∇x0 u(x0 , xN )kdx0 dxN ≤ k∇ukL1 (RN ) .
Raising both sides to the power (N − 1)/[(N − 2)P ] and using the resulting inequality to bound the first integral on the righthand side of (7.39), we obtain Z Z ∞Z u(x0 , xN + h) − u(x0 , xN )p1 dxN dhdx0 1+s1 p1 h N −1 R 0 R (N −1)/[(N −2)P ]+1/Q+N/[(N −1)R]
k∇ukL1 (RN )
= k∇ukpL11 (RN ) ,
284
7. Embeddings and Interpolation
where we use (7.38) and the Sobolev–Gagliardo–Nirenberg embedding theorem (see [Leo22d]). Step 2: Assume that N = 2 and u ∈ Cc∞ (R2 ). Then the exponent P in (7.38) becomes ∞, and so, the inequality (7.39) should be replaced by Z Z ∞Z u(x1 , x2 + h) − u(x1 , x2 )p1 dx2 dhdx1 h1+s1 p1 R 0 R s1 p21 s1 p1 (1−s1 p1 ) Z Z Z ∂2 u(x1 , x2 ) dx2 u(x1 , x2 )dx2 R
×
(7.40)
R
R
Z
u(x1 , x2 )2 dx2
p1 (1−s1 )(1−s1 p1 )/2 dx2
R
s1 p1 (1−s1 p1 )
Z sup
u(x1 , x2 )dx2
1/Q
2/R
k∇ukL1 (R2 ) kukL2 (R2 ) .
x1 ∈R R
By the fundamental theorem of calculus and the fact that u(·, x2 ) ∈ Cc∞ (R), Z x1 u(x1 , x2 ) = ∂1 u(t, x2 ) dt. −∞
R
In turn, supx1 ∈R u(x1 , x2 ) ≤ R ∂1 u(t, x2 ) dt, and so, by Tonelli’s theorem, Z Z Z sup u(x1 , x2 ) dx2 ≤ sup u(x1 , x2 ) dx2 ≤ ∂1 u(x) dx. x1 ∈R R
R x1 ∈R
R2
Using this inequality to bound the first integral on the righthand side of (7.40), we obtain Z Z ∞Z u(x1 , x2 + h) − u(x1 , x2 )p1 dx2 dhdx01 1+s1 p1 h R 0 R s p (1−s1 p1 )+1/Q+2/R
1 k∇ukL11 (R 2)
= k∇ukpL11 (R2 ) ,
where we use (7.38) and the Sobolev–Gagliardo–Nirenberg embedding theorem (see [Leo22d]). Step 3: By the previous two steps and the Sobolev–Gagliardo–Nirenberg embedding theorem (see [Leo22d]), (7.41)
kwkLN/(N −1) (RN ) + wW s1 ,p1 (RN ) k∇wkL1 (RN )
˙ 1,1 (RN ). By the density of C ∞ (RN ) for all w ∈ Cc∞ (RN ), N ≥ 2. Let u ∈ W c 1,1 N ˙ (R ) (see [Leo22d]), there exists a sequence {un }n of functions in W in Cc∞ (RN ) such that ∂i un → ∂i u in L1 (RN ) for every i = 1, . . . , N . Then {un }n is a Cauchy sequence in LN/(N −1) (RN ), and so, there exists a function v ∈ LN/(N −1) (RN ) such that un → v in LN/(N −1) (RN ). It follows (Why?) ˙ 1,1 (RN ), with ∇v = ∇u. In turn, u = v + c for some constant that v ∈ W c ∈ R. By extracting a subsequence, not relabeled, we can assume that
7.5. Embeddings: The General Case
285
un → v LN a.e. in RN . Taking w = un in (7.41) and letting n → ∞, it follows from Fatou’s lemma that uW s1 ,p1 (RN ) = vW s1 ,p1 (RN ) k∇ukL1 (RN ) , which concludes the proof of (7.34). Step 4: It remains to show the embedding W 1,1 (RN ) ,→ W s1 ,p1 (RN ). Given u ∈ W 1,1 (RN ), by the Sobolev–Gagliardo–Nirenberg embedding theorem (see [Leo22d]) and (7.35), 1 kukLp1 (RN ) ≤ kuksL11 (RN ) kuk1−s LN/(N −1) (RN ) 1 kuksL11 (RN ) k∇uk1−s . L1 (RN )
Together with (7.34), this shows that W 1,1 (RN ) ,→ W s1 ,p1 (RN ).
Next we study the case s1 < 1. Theorem 7.32. Let 0 < s1 < s2 < 1, 1 ≤ p1 < that (7.42)
s2 −
N s1 ,
and 1 ≤ p2
0 r B(0,r) Let ρ > 0. Since kzk < 2khk, !q Z Z dh 1 u(x + z) − u(x) dz N khk khkN +σq B(0,2khk) B(0,ρ) !q Z Z khkσ1 dz dh (7.47) u(x + z) − u(x) N σ 1 khk kzk khkN +σq B(0,2khk) B(0,ρ) ! q Z 1 dz (σ1 −σ)q ρ sup N u(x + z) − u(x) , kzkσ1 r>0 r B(0,r)
7.5. Embeddings: The General Case
where we use the fact that Z B(0,ρ)
1
khkN +(σ−σ1 )q
dh =
287
βN ρ(σ1 −σ)q . (σ1 − σ)q
On the other hand, by (7.44) and the definition of maximal function, !q Z Z 1 dh u(x + z) − u(x) dz N khk khkN +σq RN \B(0,ρ) B(0,2khk) Z 1 1 dh (M(u)(x))q σq . (M(u)(x))q N +σq ρ RN \B(0,ρ) khk Summing this inequality to (7.47) gives (7.48) Z RN
1 khkN
!q
Z
dh 1 (M(u)(x))q σq N +σq khk ρ B(0,2khk) !q Z 1 dz . + ρ(σ1 −σ)q sup N u(x + z) − u(x) kzkσ1 r>0 r B(0,r) u(x + z) − u(x) dz
Since uW σ1 ,q1 (RN ) < ∞, by Tonelli’s theorem, Z dz (7.49) u(x + z) − u(x)q1 0 r B(0,r) then u is a constant, and so, there is nothing to prove. Thus, we can assume R 1 dz that 0 < rN B(0,r) u(x + z) − u(x) kzk σ1 < ∞. Hence, we can take !1/σ1 M(u)(x) R ρ= dz supr>0 r1N B(0,r) u(x + z) − u(x) kzk σ1 in (7.48). Observing that θ = σ/σ1 , we obtain (7.46).
288
7. Embeddings and Interpolation
In turn, by (7.45), (7.46), and H¨ older’s inequality with exponents q1 /(q1 − θq) and q1 /(qθ), (7.51) Z Z RN
RN
u(x + h) − u(x)q dxdh khkN +σq
Z
1 sup N r>0 r
Z
dz (M(u)(x))(1−θ)q u(x + z) − u(x) kzkσ1 RN B(0,r) Z (q1 −θq)/q1 (1−θ)qq1 /(q1 −θq) (M(u)(x)) dx
!θq dx
RN
Z
1 sup N r r>0
× RN
Z
dz u(x + z) − u(x) σ1 kzk B(0,r)
!θq/q1
!q1 dx
.
Hence, also by (7.50), the righthand side of the inequality (7.51) is bounded from above up to a multiplicative constant by Z (q1 −θq)/q1 (1−θ)qq1 /(q1 −θq) (M(u)(x)) dx RN
Z
Z
×
u(x + z) − u(x) RN
RN
θq/q1
dz
q1
kzkN +σ1 q1
dx
.
To conclude, we observe that by (7.43), (q1 − θq)/q1 = (1 − θ)q/q2 and (1 − θ)qq1 /(q1 − θq) = q2 . We turn to the proof of Theorem 7.32 Proof of Theorem 7.32. By Corollary 7.9, there exists a constant mu ∈ R ∗ 2N such that u − mu ∈ L(p2 )s2 (RN ), where (p2 )∗s2 = N p−s . Since the fractional 2 p2 seminorms do not see constants, without loss of generality, we may assume ∗ that u ∈ L(p2 )s2 (RN ). Setting q1 = p2 , q2 = (p2 )∗s2 , σ1 = s2 , σ = s1 , and q = p1 , we have s1 = θs2 , and so, by (7.42), θ 1−θ s1 s2 − s1 1 s2 + = + − q1 q2 s2 p2 s2 p2 N 1 1 1 1 = + (−s2 + s1 ) = = . p2 N p1 q Hence, we can apply Lemma 7.33 to obtain uW s1 ,p1 (RN ) uθW s2 ,p2 (RN ) k M(u)k1−θ (p2 )∗ s L
uθW s2 ,p2 (RN ) kuk1−θ (p2 )∗ s L
2 (RN )
2 (RN )
uW s2 ,p2 (RN ) ,
7.5. Embeddings: The General Case
289
where, in the second inequality, we use the continuity of the maximal operator M (see [Leo22c]) in Lq (RN ) for q > 1, and in the last, the Sobolev– Gagliardo–Nirenberg embedding theorem (Theorem 7.6). This concludes the proof of the first part of the statement. To prove the second part, assume that u ∈ W s2 ,p2 (RN ). Again by the Sobolev–Gagliardo–Nirenberg embedding theorem (Theorem 7.6), u ∈ ∗ Lp2 (RN ) ∩ L(p2 )s2 (RN ). Observe that by (7.42), (7.52)
(p2 )∗s2 =
N p2
N = − s2
N p1
N = (p1 )∗s1 . − s1
Since s1 < s2 < 1, by (7.42), we have that p2 < p1 < (p1 )∗s1 = (p2 )∗s2 , and so, kukLp1 (RN ) ≤ kukθLp2 (RN ) kuk1−θ (p2 )∗ s L
2 (RN )
. kukθLp2 (RN ) u1−θ W s2 ,p2 (RN )
Next we consider the supercritical case. Theorem 7.34. Let 0 < s1 < s2 < 1, sN1 < p1 < ∞, and sN2 < p2 < ∞ be ˙ s2 ,p2 (RN ), such that s2 − pN2 = s1 − pN1 . Then for every u ∈ W uW s1 ,p1 (RN ) uW s2 ,p2 (RN ) . Moreover, W s2 ,p2 (RN ) ,→ W s1 ,p1 (RN ). Proof. We proceed as in the proof of Lemma 7.33 up to (7.45) with q1 = p2 , σ1 = s2 , σ = s1 , and q = p1 . We claim that !p1 Z Z 1 dh u(x + z) − u(x) dz (7.53) N N +s1 p1 khk khk N R B(0,2khk) !p2 Z 1 dz −p2 upC10,α sup N u(x + z) − u(x) , (RN ) kzks2 r>0 r B(0,r) where α = s2 − N/p2 . As before we fix ρ > 0 and we obtain (7.47). On the other hand, by Morrey’s embedding theorem (Theorem 7.23) and the fact that N + s1 p1 − αp1 = 2N , !p1 Z Z 1 dh u(x + z) − u(x) dz N N khk khk +s1 p1 RN \B(0,ρ) B(0,2khk) Z khkαp1 1 ≤ upC10,α (RN ) dh upC10,α (RN ) N . N +s1 p1 khk ρ N R \B(0,ρ)
290
7. Embeddings and Interpolation
Summing (7.47) and the last inequality gives Z RN
1 khkN
u(x + z) − u(x) dz B(0,2khk)
upC10,α (RN ) +ρ
!p1
Z
dh khkN +s1 p1
1 ρN
(s2 −s1 )p1
1 sup N r>0 r
Z
dz u(x + z) − u(x) kzks2 B(0,r)
!p1 .
Taking ρ=
uC 0,α (RN ) supr>0
1 rN
and using the fact that s2 −
B(0,r) u(x
dz + z) − u(x) kzk s2
N p2
N p1
R
= s1 −
!p2 /N
gives (7.53).
In turn, by (7.45), (7.53), and H¨ older’s inequality, Z Z u(x + h) − u(x)p1 dxdh khkN +s1 p1 RN RN !p2 Z Z dz 1 p1 −p2 u(x + z) − u(x) sup N dx uC 0,α (RN ) kzks2 r>0 r B(0,r) RN Z Z 1 u(x + z) − u(x)p2 p1 −p2 uC 0,α (RN ) sup N dzdx kzks2 p2 RN r>0 r B(0,r) Z Z u(x + z) − u(x)p2 p1 −p2 uC 0,α (RN ) sup dzdx kzkN +s2 p2 RN r>0 B(0,r) p2 2 upW1 −p s2 ,p2 (RN ) uW s2 ,p2 (RN ) ,
where in the last inequality we use Morrey’s embedding theorem (Theorem 7.23). This concludes the first part of the proof. Next observe that if u ∈ W s2 ,p2 (RN ), then, by Morrey’s embedding theorem (Theorem 7.23), u ∈ Lp2 (RN ) ∩ L∞ (RN ). Since s1 < s2 < 1 by (7.42), we have that p2 < p1 < ∞, and so, kukLp1 (RN ) ≤ kukθLp2 (RN ) kuk1−θ kukθLp2 (RN ) u1−θ . L∞ (RN ) W s2 ,p2 (RN )
We now present a completely different proof, which includes the critical case. Theorem 7.35. Let 1 ≤ p1 , p2 < ∞ and 0 < s1 < s2 < 1 be such that (7.54)
s2 −
N N = s1 − . p2 p1
7.5. Embeddings: The General Case
291
˙ s2 ,p2 (RN ), Then for every u ∈ W uW s1 ,p1 (RN ) uW s2 ,p2 (RN ) .
(7.55)
Moreover, W s2 ,p2 (RN ) ,→ W s1 ,p1 (RN ). We recall that given n = 1, . . . , N , and t ∈ R, ∆t,n u(x) := u(x + ten ) − u(x), where en is the nth vector of the canonical basis. Lemma 7.36. Let j = 1, . . . , N , 1 ≤ p, q < ∞, and 0 < s < 1. Then for every u ∈ L1loc (RN ), Z ∞ Z ∞ dh dh q k∆h,j ukqLp (RN ) 1+qs . sup k∆t,j ukLp (RN ) 1+qs ≈ (7.56) h h 0 0 0≤t≤h Proof. Write x = (x0j , xj ) ∈ RN −1 × R. Fix x0j ∈ RN −1 . By applying (1.26) to the function u(x0j , ·), we obtain Z 1/p Z t Z 1/p dζ ∆t u(x0j , xj )p dxj . ∆ζ u(x0j , xj )p dxj ζ R 0 R By taking the Lp (RN −1 ) norm in the variable x0j on both sides and using Minkowski’s inequality for integrals (see [Leo22c]) and Tonelli’s theorem on the righthand side, we get Z 1/p Z t Z 1/p dζ . ∆t,j u(x)p dx ∆ζ,j u(x)p dx ζ N N R 0 R Raising both sides to power q gives sup 0≤t≤h
Hence, Z ∞ sup 0
0 0, then p and p differ only in the kth component. Hence, by Step 2, (7.62)
N Z X
∞
k∆h,j ukr p(k) L
0
j=1
N Z X
1/r
(RN ) h1+σk r
∞
0
i=1
dh
k∆h,i ukr p(k−1) L
1/r
dh
.
(RN ) h1+(σk +sk )r
On the other hand, if sk = 0, then p(k−1) = p(k) , and so (7.62) continues to hold. Since p(0) := p, p(N ) := q, σN = s, and σ1 = 1, (7.61) follows from repeated applications of (7.62). Step 4: Let 1 ≤ p2 < p1 < ∞ and 0 < s1 < s2 < 1 satisfy (7.54). Define p := (p2 , . . . , p2 ), q := (p1 , . . . , p1 ), r = p2 , σ := s1 , and s = s2 . Then by (7.54), relation (7.60) is satisfied, and so, we may apply Step 3 to obtain (7.63) Z ∞ 1/p2 X 1/p2 N Z ∞ dh dh p2 p2 k∆h,j ukLp1 (RN ) 1+s1 p2 k∆h,i ukLp2 (RN ) 1+s2 p2 . h h 0 0 i=1
Since k∆h,j ukLp1 (RN ) ≤ sup0≤t≤h k∆t,j ukLp1 (RN ) , we have Z 0
∞
dh k∆h,j ukpL1p1 (RN ) 1+s1 p1 h Z
≤ 0
∞
1/p1
sup k∆t,j ukpL1p1 (RN )
0≤t≤h
dh h1+s1 p1
!1/p1 .
7.5. Embeddings: The General Case
295
Since the function f (h) := sup0≤t≤h k∆t,j ukLp1 (RN ) is increasing and p2 < p1 , we can apply Exercise 2.28(ii) to the righthand side to obtain Z ∞ 1/p1 dh p1 k∆h,j ukLp1 (RN ) 1+s1 p1 h 0 !1/p2 Z ∞ dh sup k∆t,j ukpL2p1 (RN ) 1+s1 p2 h 0 0≤t≤h 1/p2 Z ∞ dh p2 , k∆h,j ukLp1 (RN ) 1+s1 p2 h 0 where in the second inequality we use Lemma 7.36. Combining this inequality with (7.63), we obtain Z ∞ 1/p1 X 1/p2 N Z ∞ dh dh p1 p2 k∆h,j ukLp1 (RN ) 1+s1 p1 k∆h,i ukLp2 (RN ) 1+s2 p2 h h 0 0 i=1
for every j = 1, . . . , N . To conclude the proof of (7.55), it suffices to apply the slicing theorem (Theorem 6.38) to both sides. The additional assumption that u ∈ C ∞ RN can be removed by mollification. Step 5: It remains to show that W s2 ,p2 (RN ) ,→ W s1 ,p1 (RN ). In view of Theorems 7.32 and 7.34, we can assume that s2 p2 = N = s1 p1 . Then by Theorem 7.12, we have that u ∈ Lp2 (RN ) ∩ Lq (RN ) for every p2 ≤ q < ∞, with kukLq (RN ) kukW N/p2 ,p2 (RN ) , and so, since p2 < p1 < ∞, kukLp1 (RN ) kukW N/p2 ,p2 (RN ) . For future applications in Chapter 11 (see Theorem 11.41), we will need a version of Step 3 that uses higher order differences. Corollary 7.37. Let m ∈ N, 1 ≤ p2 < p1 < ∞, 1 ≤ r < ∞, and 0 < s1 < s2 be such that s2 − pN2 = s1 − pN1 . Then for every u ∈ C ∞ (RN ) and j = 1, . . . , N , Z 0
∞ r k∆m h,j ukLp1 (RN )
dh h1+s1 r
1/r
N Z X i=1
0
∞ r k∆m h,i ukLp2 (RN )
dh h1+s2 r
1/r .
Proof. We proceed as in the proof of Theorem 7.35, with the only difference that we apply Corollary 2.29 in place of Lemma 2.25 in Step 1. In formula m (7.58), ∆η,k v should be replaced by ∆m δη,k v. In Step 2 we take v = ∆h,j u m m and obtain (7.59) with ∆h,i u and ∆h,j u replaced by ∆h,i u and ∆h,j u, respectively. Note that we never use the fact that σk < 1. Similarly, in Step m 3, we obtain (7.61) with ∆h,i u and ∆h,j u replaced by ∆m h,i u and ∆h,j u, respectively. Observe that we do not use the fact that s, σ < 1.
296
7. Embeddings and Interpolation
If we replace norms with seminorms, then we can replace the equality in (7.54) with s2 − pN2 ≥ s1 − pN1 . Corollary 7.38. Let 1 ≤ p2 < p1 < ∞ and 0 < s1 < s2 < 1 be such that s2 − pN2 ≥ s1 − pN1 . Then for every u ∈ W s2 ,p2 (RN ), kukW s1 ,p1 (RN ) kukW s2 ,p2 (RN ) . Proof. By the hypotheses on the exponents, 0 < s1 ≤ σ2 := p11 − p12 + s2 < s2 . Hence, by Theorem 7.35, W s2 ,p2 (RN ) ,→ W σ2 ,p1 (RN ). If s1 < σ2 , we use Lemma 6.17 to obtain W σ2 ,p1 (RN ) ,→ W s1 ,p1 (RN ).
7.6. The Limit of W s,p as s → 1− ˙ 1,p (RN ), then u ∈ W ˙ s,p (RN ) In Theorem 6.21 we have seen that if u ∈ W for every 0 < s < 1. In this section we study the limit of the fractional seminorms as s → 1− and prove that if u ∈ W s,p (RN ) for every 0 < s < 1 and Z Z u(x) − u(y)p dxdy < ∞, lim inf (1 − s) N +sp s→1− RN RN kx − yk then u ∈ W 1,p (RN ). Theorem 7.39. Let u ∈ Lp (RN ), 1 < p < ∞, be such that Z Z u(x) − u(y)p lim inf (1 − s) dxdy < ∞. N +sp s→1− RN RN kx − yk Then u ∈ W 1,p (RN ). Conversely, if u ∈ W 1,p (RN ), then Z Z Z u(x) − u(y)p lim (1 − s) dxdy = KN,p k∇u(x)kp dx, N +sp kx − yk N N N s→1− R R R where (7.64)
KN,p :=
1 p
Z SN −1
e1 · νp dHN −1 (ν).
Proof. Step 1: Assume that u ∈ C ∞ (RN ). We claim that Z Z Z u(x) − u(y)p p KN,p k∇u(x)k dx ≤ lim inf (1 − s) dxdy. N +sp s→1− RN RN RN kx − yk Let K ⊂ RN be a compact set. By Taylor’s formula, for x ∈ K and h ∈ RN , with khk < 1, u(x + h) − u(x) − ∇u(x) · h ≤ CK khk2 , where Ck := max{k∇2 u(y)k : dist(y, K) ≤ 1}. In turn, ∇u(x) · h ≤ u(x + h) − u(x) + u(x + h) − u(x) − ∇u(x) · h ≤ u(x + h) − u(x) + CK khk2 .
7.6. The Limit of W s,p as s → 1−
297
Raising both sides to the power p and using the inequality (a + b)p ≤ (1 + ε)ap + Cε b p , we obtain ∇u(x) · hp ≤ (1 + ε)u(x + h) − u(x)p + CK,ε khk2p . Multiplying both sides by (1 − s)/khkN +sp and integrating in x over K and in h over B(0, 1) gives Z Z ∇u(x) · hp (1 − s) dhdx N +sp K B(0,1) khk Z Z u(x + h) − u(x)p (7.65) ≤ (1 + ε)(1 − s) dhdx khkN +sp K B(0,1) 1−s + CK,ε LN (K)βN , p(2 − s) where we use the fact that Z (7.66) B(0,1)
khk2p 1 dh = βN . khkN +sp p(2 − s)
Since the ball is invariant by rotations, Z Z ∇u(x) · hp e1 · hp p dh = k∇u(x)k dh N +sp N +sp B(0,1) khk B(0,1) khk Z 1 N −1+p Z r (7.67) = k∇u(x)kp dr e1 · νp dHN −1 (ν) N +sp r N −1 S Z0 k∇u(x)kp k∇u(x)kp = e1 · νp dHN −1 (ν) = KN,p , p(1 − s) SN −1 1−s where KN,p is given in (7.64). Using this identity in (7.65), we obtain Z Z Z u(x + h) − u(x)p p KN,p k∇u(x)k dx ≤ (1 + ε)(1 − s) dhdx khkN +sp K RN R N 1−s + CK,ε LN (K)βN . p(2 − s) Now, we first take the liminf as s → 1− and then the limit as ε → 0+ to obtain Z Z Z u(x) − u(y)p KN,p k∇u(x)kp dx ≤ lim inf (1 − s) dxdy. N +sp s→1− K RN RN kx − yk To conclude the proof of the claim, it suffices to consider an increasing sequence of compact sets Kn % RN , and use the Lebesgue monotone convergence theorem on the lefthand side. Step 2: Let u ∈ Lp (RN ) be such that Z Z u(x) − u(y)p lim inf (1 − s) dxdy =: M < ∞. N +sp s→1− RN RN kx − yk
298
7. Embeddings and Interpolation
For ε > 0, let uε := ϕε ∗ u, where ϕε is a standard mollifier. By Theorem 6.62, uε W s,p (RN ) ≤ uW s,p (RN ) , and so, by the previous inequality, Z Z uε (x) − uε (y)p lim inf (1 − s) dxdy ≤ M. kx − ykN +sp s→1− RN RN Since uε ∈ C ∞ (RN ) (see [Leo22c]), we can apply Step 1 to uε to obtain Z k∇uε (x)kp dx ≤ M (7.68) KN,p K
for every ε > 0. Taking ε = εn → 0+ , we have that {uεn }n is bounded in W 1,p (RN ). Since 1 < p < ∞, W 1,p (RN ) is reflexive, and so there exist a subsequence, not relabeled, and u ∈ W 1,p (RN ) such that uεn * u in W 1,p (RN ). Taking ε = εn in (7.68) and letting n → ∞, by the lower semicontinuity of the norm with respect to weak convergence in Lp , we get Z (7.69) KN,p k∇u(x)kp dx ≤ M. K
We have proved the first statement in the theorem. Step 3: Assume that u ∈ Cc∞ (RN ). We claim that Z Z Z u(x + h) − u(x)p dhdx ≤ K k∇u(x)kp dx. lim sup(1 − s) N,p N +sp khk N N N − R R R s→1 By Taylor’s formula, for x ∈ RN and h ∈ RN , with khk < 1, u(x + h) − u(x) − ∇u(x) · h ≤ k∇2 uk∞ khk2 . In turn, u(x + h) − u(x) ≤ ∇u(x) · h + u(x + h) − u(x) − ∇u(x) · h ≤ ∇u(x) · h + k∇2 uk∞ khk2 . Raising both sides to the power p and using the inequality (a + b)p ≤ (1 + ε)ap + Cε b p , we obtain u(x + h) − u(x)p ≤ (1 + ε)∇u(x) · hp + Cε khk2p . Let supp u ⊂ B(0, R). Multiplying both sides by (1 − s)/khkN +sp and integrating in x over B(0, R + 1) and in h over B(0, 1) gives Z Z u(x + h) − u(x)p dhdx (1 − s) khkN +sp B(0,R+1) B(0,1) Z Z ∇u(x) · hp ≤ (1 + ε)(1 − s) dhdx N +sp B(0,R+1) B(0,1) khk 1−s , + Cε αN (R + 1)N βN p(2 − s)
7.6. The Limit of W s,p as s → 1−
299
where we use (7.66). Note that for x ∈ RN \ B(0, R + 1) and khk ≤ 1, kx + hk ≥ R. Since supp u ⊂ B(0, R RR), the integral on the lefthand side can be replaced with the integral RN B(0,1) . Hence, also by (7.67) we have Z Z u(x + h) − u(x)p (1 − s) dhdx khkN +sp RN B(0,1) Z 1−s k∇u(x)kp dx + Cε αN (R + 1)N βN ≤ (1 + ε)KN,p . p(2 − s) B(0,R) Letting first s → 1− and then ε → 0+ gives Z Z Z u(x + h) − u(x)p dhdx ≤ KN,p lim sup(1 − s) k∇u(x)kp dx. khkN +sp RN B(0,1) RN s→1− On the other hand, since supp u ⊆ B(0, R), by Tonelli’s theorem and the change of variables y = x + h, Z Z u(x + h) − u(x)p dhdx (1 − s) khkN +sp RN RN \B(0,1) Z Z 1 p p ≤ 2 (1 − s) u(x) dhdx N +sp RN RN \B(0,1) khk Z 2p βN = (1 − s) u(x)p dx → 0 sp N R as s → 1− . Adding the two limsup proves the claim. Step 4: Suppose that u ∈ W 1,p (RN ). By the density of Cc∞ (RN ) in W 1,p (RN ) (see [Leo22d]), we can find a sequence {un }n in Cc∞ (RN ) such that un → u in W 1,p (RN ) as n → ∞. Using inequality (6.12), we have uW s,p (RN ) ≤ un − uW s,p (RN ) + un W s,p (RN ) ≤ Cs−1/p kun − ukLp (RN ) + C(1 − s)−1/p k∇(un − u)kLp (RN ) + un W s,p (RN ) . By Step 3 applied to un , if we multiply both sides by (1 − s)1/p and let s → 1− , we obtain lim sup(1 − s)1/p uW s,p (RN ) ≤ Ck∇(un − u)kLp (RN ) s→1−
+ lim sup(1 − s)1/p un W s,p (RN ) s→1−
1/p
≤ Ck∇(un − u)kLp (RN ) + KN,p k∇un kLp (RN ) . Letting n → ∞ and using the fact that un → u in W 1,p (RN ) as n → ∞, we obtain 1/p
lim sup(1 − s)1/p uW s,p (RN ) ≤ KN,p k∇ukLp (RN ) . s→1−
300
7. Embeddings and Interpolation
Combining this inequality with (7.69) shows that there exists Z Z Z u(x + h) − u(x)p lim (1 − s) k∇u(x)kp dx. dhdx = KN,p N +sp khk N N N s→1− R R R Exercise 7.40. Prove that if u ∈ W 1,1 (RN ), then there exists Z Z Z u(x + h) − u(x) k∇u(x)k dx. dhdx = KN,1 lim (1 − s) khkN +s s→1− RN RN RN
7.7. Interpolation Inequalities In this section we study some interpolation inequalities of the type established by Gagliardo [Gag58] and Nirenberg [Nir59] for the Sobolev spaces W m,p (Ω), m ∈ N, (see [FFRS21], [Leo22d]). To be precise, we are interested in inequalities of the type (7.70)
uW s,p (RN ) uθW s1 ,p1 (RN ) u1−θ W s2 ,p2 (RN )
˙ s1 ,p1 (RN ) ∩ W ˙ s2 ,p2 (RN ). Here, 0 < θ < 1, 1 ≤ p, p1 , p2 ≤ ∞, for all u ∈ W ˙ 0,p (RN ) := Lp (RN ) and and 0 ≤ s, s1 , s2 ≤ 1. We set W (7.71)
uW 0,p (RN ) := kukLp (RN ) ,
uW 1,p (RN ) := k∇ukLp (RN ) .
As in the beginning of Chapter 2, a scaling argument shows that a necessary ˙ s1 ,p1 (RN ) ∩ W ˙ s2 ,p2 (RN ) is condition for (7.70) to hold for all u ∈ W N N N (7.72) s− = θ s1 − + (1 − θ) s2 − . p p1 p2 The main theorem of this section is the following. Theorem 7.41 (Gagliardo–Nirenberg). Let 1 < p1 , p2 < ∞, 0 ≤ s1 < s2 ≤ 1, and 0 < θ < 1. Then uW s,p (RN ) uθW s1 ,p1 (RN ) u1−θ W s2 ,p2 (RN ) ˙ s1 ,p1 (RN ) ∩ W ˙ s2 ,p2 (RN ), where for all u ∈ W (1 − θ)s2 , and we are using notation (7.71).
1 p
=
θ p1
+
1−θ p2
and s = θs1 +
We will divide the proof of this theorem into four subcases: 0 < s1 < s2 < 1, s1 = 0 < s2 < 1, s1 = 0, s2 = 1, and 0 < s1 < s2 = 1. We begin with the case 0 < s1 < s2 < 1. Theorem 7.42. Let 1 ≤ p1 < ∞, 1 ≤ p2 < ∞, 0 < s1 < s2 < 1, and 0 < θ < 1. Then uW s,p (RN ) uθW s1 ,p1 (RN ) u1−θ for all W s2 ,p2 (RN ) 1 θ 1−θ s ,p N s ,p N 1 1 2 2 ˙ ˙ u∈W (R ) ∩ W (R ), where = + and s = θs1 + (1 − θ)s2 . p
p1
p2
Proof. The proof is similar to the one of Theorem 2.33 and is left as an exercise.
7.7. Interpolation Inequalities
301
Next we study the case s1 = 0 < s2 < 1. Theorem 7.43. Let 1 < p1 < ∞, 1 ≤ p2 < ∞, 0 < s2 < 1, and 0 < θ < 1. Then uW s,p (RN ) kukθLp1 (RN ) u1−θ for all u ∈ Lp1 (RN ) ∩ W s2 ,p2 (RN ) ˙ s2 ,p2 (RN ), where 1 = θ + 1−θ , s = (1 − θ)s2 . W p
p1
p2
Proof. Setting q1 = p2 , q2 = p1 , σ1 = s2 , σ = s, and q = p, by Lemma 7.33, we have uW s,p (RN ) k M(u)kθLp1 (RN ) u1−θ kukθLp1 (RN ) u1−θ , W s2 ,p2 (RN ) W s2 ,p2 (RN ) where in the second inequality we use the continuity of the maximal operator M in Lp1 (RN ) for p1 > 1 (see [Leo22c]). Next, we consider the case 0 < s1 < s2 = 1. Theorem 7.44. Let 1 ≤ p1 < ∞, 1 < p2 < ∞, 0 < s1 < 1, and 0 < θ < 1. ˙ s1 ,p1 (RN ) ∩ W ˙ 1,p2 (RN ), Then for all u ∈ W uW s,p (RN ) uθW s1 ,p1 (RN ) k∇uk1−θ , Lp2 (RN )
(7.73) where
1 θ 1−θ = + , p p1 p2
(7.74)
s = θs1 + (1 − θ)1.
Proof. Assume that u ∈ C 1 (RN ). Given x, h ∈ RN , with h 6= 0, let E := B(x, khk) ∩ B(x + h, khk). Note that LN (E) ≥ c0 khkN . Then u(x + h) − u(x) ≤ u(x + h) − uE  + u(x) − uE . It follows that Z Z
Z Z u(x + h) − u(x)p u(x + h) − uE p dxdh dxdh khkN +sp khkN +sp RN RN RN RN Z Z Z Z u(z) − uF p u(x) − uE p (7.75) dxdh = dzdh + khkN +sp khkN +sp RN RN RN RN Z Z u(x) − uE p + dxdh =: A + B, khkN +sp RN RN where we make the change of variables x + h = z, which gives dx = dz, we use the fact that the Lebesgue measure is translation invariant, and where F := B(z − h, khk) ∩ B(z, khk). We will estimate B since the term A can be treated similarly. Given x ∈ RN and ρ > 0, we write Z Z u(x) − uE p u(x) − uE p (7.76) dh = dh khkN +sp khkN +sp RN B(0,ρ) Z u(x) − uE p + dh =: C + D. khkN +sp RN \B(0,ρ)
302
7. Embeddings and Interpolation
Using the facts that E = B(x, khk) ∩ B(x + h, khk) and LN (E) ≥ c0 khkN , by Lemmas 7.29 and 7.30, for x, h ∈ RN , with h 6= 0, we have Z Z k∇u(y)k khkN k∇u(y)k u(x) − uE  N dy dy N −1 N −1 L (E) B(x,khk) kx − yk B(x,khk) kx − yk Z 2khk Z 1 (7.77) k∇u(y)kdydr khk M(k∇uk)(x), rN B(x,r) 0 where M(k∇uk) is the Hardy–Littlewood maximal function of k∇uk. By (7.76) and (7.77), Z khkp p dh (M(k∇uk)(x))p ρ(1−s)p . (7.78) C (M(k∇uk)(x)) N +sp B(0,ρ) khk To estimate D, we use the facts that E = B(x, khk) ∩ B(x + h, khk) and LN (E) ≥ c0 khkN to obtain Z Z 1 c0 u(x) − uE  ≤ N u(y) − u(x) dy ≤ u(y) − u(x) dy L (E) E khkN B(x,khk) Z c0 = u(x + z) − u(x) dz. khkN B(0,khk) In turn, by (7.76) and since kzk < khk, !p Z Z dh 1 u(x + z) − u(x) dz D N khk khkN +sp B(0,khk) RN \B(0,ρ) !p Z Z khks1 dh dz (7.79) u(x + z) − u(x) N s1 N +sp khk kzk khk N R \B(0,ρ) B(0,khk) ! p Z 1 1 dz (s−s )p sup N u(x + z) − u(x) , 1 kzks1 ρ r>0 r B(0,r) where we use the facts that s1 < s and that Z 1 1 dh (s−s )p . N +(s−s )p 1 1 ρ RN \B(0,ρ) khk It follows from (7.78) and (7.79) that for every ρ > 0, (7.80)
C + D (M(k∇uk)(x))p ρ(1−s)p !p Z 1 1 dz + (s−s )p sup N u(x + z) − u(x) . 1 kzks1 ρ r>0 r B(0,r)
Since uW s1 ,p1 (RN ) < ∞, by Tonelli’s theorem, Z dz u(x + z) − u(x)p1 0 r B(0,r) (1−θ)p
(M(k∇uk)(x))
1 sup N r r>0
Z
dz u(x + z) − u(x) s1 kzk B(0,r)
!θp ,
where in the last inequality we use the facts that s − s1 = (1 − θ)(1 − s1 ) and 1 − s = θ(1 − s1 ) (see (7.74)). In turn, by (7.75) and (7.76) and H¨older’s inequality with exponents p1 /(p1 − θp) and p1 /(pθ), !θp Z Z 1 dz B (M(k∇uk)(x))(1−θ)p sup N u(x + z) − u(x) dx kzks1 r>0 r RN B(0,r) Z (p1 −θp)/p1 (1−θ)pp1 /(p1 −θp) (M(k∇uk)(x)) dx RN
Z × RN
1 sup N r>0 r
Z
dz u(x + z) − u(x) kzks1 B(0,r)
!θp/p1
!p1 dx
.
By (7.82), the righthand side of the previous inequality is bounded from above by Z (p1 −θp)/p1 C (M(k∇uk)(x))(1−θ)pp1 /(p1 −θp) dx RN
Z
Z
×
p1
u(x + z) − u(x) RN
RN
dz kzkN +s1 p1
θp/p1 dx
.
By (7.74), (p1 − θp)/p1 = (1 − θ)p/p2 and (1 − θ)pp1 /(p1 − θp) = p2 . Hence, Z (1−θ)p/p2 θp p2 B uW s1 ,p1 (RN ) (M(k∇uk)(x)) dx . RN
304
7. Embeddings and Interpolation
Since M is a bounded operator in Lq (RN ) for q > 1 (see [Leo22c]), we get Z (1−θ)p/p2 θp p2 k∇uk dx . B uW s1 ,p1 (RN ) RN
˙ 1,p1 (RN ) ∩ A similar estimate holds for A. This proves (7.73) for u ∈ W ˙ 1,p2 (RN ), we apply (7.73) to uε = ϕε ∗u C 1 (RN ). In the case u ∈ Lp1 (RN )∩W to get Z Z uε (x) − uε (x + h)p (1−θ)p dxdh uε θp s1 ,p1 (RN ) k∇uε kLp1 (RN ) N +sp W khk RN RN (1−θ)p
uθp k∇ukLp1 (RN ) , W s1 ,p1 (RN ) where the last inequality follows from the fact that ∇uε = ϕε ∗ ∇u and standard properties of mollifiers (see [Leo22c]). Letting ε → 0+ and applying Fatou’s lemma on the lefthand side, we find that (7.73) holds for u. Lastly, we study the case s1 = 0 < s2 = 1. Theorem 7.45. Let 1 < p1 , p2 < ∞ and 0 < θ < 1. Then for all u ∈ ˙ 1,p2 (RN ), uW s,p (RN ) kukθ p N k∇uk1−θ Lp1 (RN ) ∩ W , where p1 = L 1 (R ) Lp2 (RN ) θ p1
+
1−θ p2 ,
s = 1 − θ.
Proof. The proof is very similar to the one of Theorem 2.37, and so we omit it. Example 7.46. When N = 2, taking p1 = p2 = 2 and θ =
1 2
in Theorem
1/2 1/2 7.45 gives uW 1/2,2 (R2 ) kukL2 (R2 ) k∇ukL2 (R2 ) for all u ∈ W 1,2 (R2 ). In turn, by the embedding theorem in W 1/2,2 (R2 ) in the critical case (Theorem 7.12),
kukLq (R2 ) kukW 1/2,2 (R2 ) for every 2 ≤ q < ∞. In particular, kukL4 (R2 ) kukW 1/2,2 (R2 ) = kukL2 (R2 ) + uW 1/2,2 (R2 ) 1/2
1/2
kukL2 (R2 ) + kukL2 (RN ) k∇ukL2 (RN ) . This is a version of Ladyzhenskaya’s inequality (see [BM19]). Exercise 7.47. Let Ω ⊂ R2 be an open bounded set and let u ∈ W01,2 (Ω). 1/2 1/2 Prove that kukL4 (Ω) diam Ω kukL2 (Ω) k∇ukL2 (Ω) . In Theorem 7.41 we have excluded p1 = 1 and p2 = 1. We now use a slicing argument and the interpolation results obtained in Chapter 3 using wavelets to study these cases. We begin with p1 = 1. Theorem 7.48. Let p1 = 1, 1 ≤ p2 < ∞, and 0 < s < s2 < 1, with s2 p2 < 1, be such that 1−θ 1 =θ+ , s = (1 − θ)s2 . (7.83) p p2
7.7. Interpolation Inequalities
305
˙ s2 ,p2 (RN ), Then for all u ∈ L1 (RN ) ∩ W uW s,p (RN ) kukθL1 (RN ) u1−θ . W s2 ,p2 (RN ) ˙ s2 ,p2 (RN ), let u Proof. Given u ∈ L1 (RN ) ∩ W ¯ be a representative of u, and let i = 1, . . . , N . By Fubini’s theorem and Theorem 6.35, Z Z Z ¯ u(x) dx < ∞, ¯ u(x0i , xi ) dxi dx0i = RN RN −1 R Z Z Z ¯ u(x0i , xi ) − u ¯(x0i , yi )p2 dyi dxi dx0i upW2 s2 ,p2 (RN ) < ∞, 1+s p 2 2 xi − yi  RN −1 R R where we write x = (x0i , xi ) ∈ RN −1 × R. Hence, for LN −1 a.e. x0i ∈ RN −1 , Z Z Z ¯ u(x0i , xi ) − u ¯(x0i , yi )p2 0 ¯ u(xi , xi ) dxi < ∞, dyi dxi < ∞, xi − yi 1+s2 p2 R R R ˙ s2 ,p2 (R). It follows from Theorem which implies that u ¯(x0i , ·) ∈ L1 (R) ∩ W N −1 0 N −1 3.15 that for L a.e. xi ∈ R , Z θp Z Z ¯ u(x0i , xi ) − u ¯(x0i , yi )p 0 dyi dxi ¯ u(xi , xi ) dxi xi − yi 1+sp R R R Z Z (1−θ)p ¯ u(x0i , xi ) − u ¯(x0i , yi )p2 × dy dx . i i xi − yi 1+s2 p2 R R Integrating both sides in x0i over RN −1 and using H¨older’s inequality with exponents 1/θp and p2 /[(1 − θ)p] (see (7.83)) and Tonelli’s theorem, we get θp Z Z Z Z Z ¯ u(x0i , xi ) − u ¯(x0i , yi )p 0 0 ¯ u (x , x ) dx dy dx dx i i i i i i xi − yi 1+sp RN −1 R RN −1 R R Z Z (1−θ)p ¯ u(x0i , xi ) − u ¯(x0i , yi )p2 × dyi dxi dx0i xi − yi 1+s2 p2 R R Z θp ≤ ¯ u(x) dx RN
Z
Z Z
× RN −1
R
R
¯ u(x0i , xi ) − u ¯(x0i , yi )p2 dyi dxi dx0i xi − yi 1+s2 p2
(1−θ)p/p2 .
It follows from Theorem 6.35 that Z Z Z ¯ u(x0i , xi ) − u ¯(x0i , yi )p (1−θ)p dyi dxi dx0i kukθp uW s2 ,p2 (RN ) . 1+sp L1 (RN ) x − y  N −1 i i R R R Summing the previous inequality over all i = 1, . . . , N and using Theorem 6.35 once more completes the proof. Remark 7.49. The restriction s2 p2 < 1 is due to the fact that in Chapter 3 we only consider Haar functions. By using Daubechies’ smooth wavelets
306
7. Embeddings and Interpolation
with compact support, it is possible to remove this extra assumption (see [Coh00], [CDDD03], and [BM18]). Next we consider the case p2 = 1. Theorem 7.50. Let 1 < p1 < ∞, p2 = 1, 0 < s1 < s2 = 1, with s1 p1 < 1, ˙ s1 ,p1 (RN ) ∩ W ˙ 1,1 (RN ) vanishing at and θ ∈ (0, 1). Then for all u ∈ W infinity, uW s,p (RN ) uθW s1 ,p1 (RN ) k∇uk1−θ , L1 (RN ) where 1 θ = + 1 − θ, p p1
(7.84)
s = θs1 + 1 − θ.
˙ s1 ,p1 (RN ) ∩ W ˙ 1,1 (RN ), there exists a representative u Proof. Given u ∈ W ¯ N −1 of u that is locally absolutely continuous on L a.e. line parallel to the ˙ 1,1 (R) for LN −1 a.e. x0 ∈ RN −1 and i = 1, . . . , N . axes, with u ¯(x0i , ·) ∈ W i Moreover, the (classical) derivative of u ¯(x0i , ·) coincides L1 a.e. in R with the weak derivative ∂i u(x0i , ·) (see [Leo22d]). Let {rk }k be an enumeration of the positive rational numbers. Since u ¯ vanishes at infinity, by Fubini’s theorem and Theorem 6.35, Z L1 ({xi ∈ R : ¯ u(x0i , xi ) > rk }) dx0i RN −1
= LN ({x ∈ RN : ¯ u(x) > rk )} < ∞, and Z RN −1
Z Z R
R
¯ u(x0i , xi ) − u ¯(x0i , yi )p1 dyi dxi dx0i upW1 s1 ,p1 (RN ) < ∞, xi − yi 1+s1 p1
where we are writing x = (x0i , xi ) ∈ RN −1 × R. Hence, for LN −1 a.e. x0i ∈ RN −1 and all k ∈ N, L1 ({xi ∈ R : ¯ u(x0i , xi ) > rk }) < ∞, Z Z ¯(x0i , yi )p1 ¯ u(x0i , xi ) − u dyi dxi < ∞. xi − yi 1+s1 p1 R R ˙ s1 ,p1 (R) ∩ W ˙ 1,1 (R) and It follows that for LN −1 a.e. x0i ∈ RN −1 , u ¯(x0i , ·) ∈ W 0 u ¯(xi , ·) vanishes at infinity (Why?). In turn, by Theorem 3.17, for LN −1 a.e. x0i ∈ RN −1 , we have Z Z θp Z Z ¯ u(x0i , xi )− u ¯(x0i , yi )p ¯(x0i , yi )p1 ¯ u(x0i , xi )− u dy dx dy dx i i i i xi − yi 1+sp xi − yi 1+s1 p1 R R R R Z (1−θ)p 0 × ∂i u(xi , xi ) dxi . R
7.7. Interpolation Inequalities
307
Integrating both sides in x0i over RN −1 and using H¨older’s inequality with exponents p1 /(θp) and 1/[(1 − θ)p] (see (7.83)) and Tonelli’s theorem, we get Z Z Z ¯ u(x0i , xi ) − u ¯(x0i , yi )p dyi dxi dx0i 1+sp x − y  N −1 i i R R R θp Z Z Z 0 ¯ u(xi , xi ) − u ¯(x0i , yi )p1 dyi dxi xi − yi 1+s1 p1 RN −1 R R (1−θ)p Z 0 × ∂i u(xi , xi ) dxi dx0i R
Z
¯ u(x0i , xi ) − u ¯(x0i , yi )p1 dyi dxi dx0i 1+s1 p1 x − y  i i R (1−θ)p ∂i u(x) dx .
Z Z
≤ RN −1
R
Z ×
θp/p1
RN
It follows from Theorem 6.35 that Z Z Z ¯ u(x0i , xi ) − u ¯(x0i , yi )p (1−θ)p dyi dxi dx0i uθp k∂ ukL1 (RN ) . 1+sp W s1 ,p1 (RN ) i x − y  N −1 i i R R R Summing the previous inequality over all i = 1, . . . , N and using Theorem 6.35 once more completes the proof. Remark 7.51. In this case, the restriction s1 p1 < 1 is needed (see [BM18]). As a consequence of the previous two theorems, we can also treat the case in which p1 = 1, s1 = 0, s2 = 1, p2 > 1 and sp < 1. Theorem 7.52. Let p1 = 1 < p < p2 , s1 = 0 < s < s2 = 1, with sp < 1, be 1 N ˙ 1,p2 (RN ), such that p1 = θ + 1−θ p2 and s = 1 − θ. Then for all u ∈ L (R ) ∩ W . uW s,p (RN ) kukθL1 (RN ) k∇uk1−θ Lp2 (RN ) Proof. The proof is similar to the one of Theorem 3.19 and is left as an exercise. Exercise 7.53. Use the slicing argument of Theorems 7.48 and 7.50 to give an alternative proof of Theorem 7.41. Example 7.54. When N = 2, taking p1 = 1, p2 = 2, and θ = 1/2 1/2 kukL1 (R2 ) k∇ukL2 (R2 )
1 2
in The
L1 (R2 )
orem 7.52 gives uW 1/2,4/3 (R2 ) for all u ∈ ∩ 1,2 2 1/2,4/3 2 ˙ (R ). In turn, by the Sobolev–Gagliardo–Nirenberg in W W (R ) (Theorem 7.6), kukL2 (R2 ) uW 1/2,4/3 (R2 ) . This is a version of Nash’s inequality (see [BM19]).
308
7. Embeddings and Interpolation
7.8. Notes Step 1 of Theorem 7.6 draws upon the paper of Bru´e and Nguyen [BN20] and an unpublished work of Brezis (see [Mir18]). For an alternative proof, which gives a precise estimate on the dependence on s in the constant, we refer to the paper of Ponce and Spector [PS20]. Exercise 7.8 is adapted from the paper of Chemin and Xu [CX97]. Lemma 7.14 is adapted from the book of Gilbarg and Trudinger [GT01, Lemma 7.12] although Step 1 is due to Mal´ y and Pick [MP02]. Lemma 7.15 and Theorem 7.21 are adapted from the paper of Zhou [Zho15]. The proof of Theorem 7.23 draws upon the paper of Di Nezza, Palatucci, and Valdinoci [DNPV12]. The proof of Theorems 7.28 and 7.32 is due to A.D. Nguyen, J.I. D´ıaz, and Q.H. Nguyen [NDN20]. I was not able to adapt the proof in the critical case s 2 p2 = N = s 1 p1 . The proof of Theorem 7.35 is due to Solonnikov [Sol75]. Theorem 7.39 is due to Bourgain, Brezis, and Mironescu [BBM01] (see also [BBM02], [KMX05], [LS11], [LS14], [Maz11, Chapter 10]). The proof has been simplified by E. Stein [Bre02]. We refer to the papers of Brezis and Mironescu [BM18], [BM19] for more information on Gagliardo–Nirenberg interpolation inequalities in Section 7.7 (see also the paper of Maz’ya and Shaposhnikova [MS99]). Theorem 7.48 is due to Cohen [Coh00], while Theorem 7.50 is due to Cohen, Dahmen, Daubechies, and De Vore [CDDD03], although the proofs in these papers do no rely on slicing methods.
Chapter 8
Further Properties It matters little who first arrives at an idea, rather what is significant is how far that idea can go. — Sophie Germain
In Chapter 7, we have studied embeddings and interpolation inequalities for fractional Sobolev spaces in the case Ω = RN . In this chapter, we will establish analogous properties in the case of smooth domains Ω ⊆ RN . To accomplish this goal, we need to understand when it is possible to extend a function u ∈ W s,p (Ω) to a function in W s,p (RN ).
8.1. Extension: Lipschitz Domains Definition 8.1. Given 1 ≤ p < ∞ and 0 < s < 1, an open set Ω ⊆ RN is called an extension domain for the Sobolev space W s,p (Ω) if there exists a continuous linear operator E : W s,p (Ω) → W s,p (RN ) such that E(u)(x) = u(x) for LN a.e. x ∈ Ω and for all u ∈ W s,p (Ω). Theorem 8.2. Let f : RN −1 → R be a Lipschitz continuous function and (8.1)
Ω := {(x0 , xN ) ∈ RN −1 × R : xN > f (x0 )}.
Then for all 1 ≤ p < ∞ and 0 < s < 1, there exists a continuous linear ˙ s,p (Ω) → W ˙ s,p (RN ) such that E(u)(x) = u(x) for LN a.e. operator E : W s,p ˙ (Ω), with x ∈ Ω and for all u ∈ W (8.2)
E(u)W s,p (RN ) (1 + Lip f )N/p+s uW s,p (Ω) .
Proof. The idea is to first flatten the boundary, then reflect the resulting function, and, finally, to unflatten the boundary. This leads us to define the function v : RN → R by u(x) if xN > f (x0 ), (8.3) v(x) := 0 0 u(x , 2f (x ) − xN ) if xN < f (x0 ). 309
310
8. Further Properties
We have Z Z
Z Z v(x) − v(y)p u(x) − u(y)p dxdy = dxdy N +sp N +sp RN RN kx − yk Ω Ω kx − yk Z Z u(x0 , 2f (x0 ) − xN ) − u(y 0 , 2f (y 0 ) − yN )p + dxdy kx − ykN +sp RN \Ω RN \Ω Z Z u(x0 , 2f (x0 ) − xN ) − u(y)p dxdy =: A + B + C. +2 kx − ykN +sp Ω RN \Ω
Consider the transformation Ψ : RN → RN given by Ψ(x) := (x0 , 2f (x0 ) − xN ). Note that Ψ is invertible, with inverse given by Ψ−1 = Ψ. Since f is Lipschitz continuous, by Rademacher’s theorem (see [Leo17, Theorem 9.14]), it is differentiable for LN −1 a.e. x0 ∈ RN −1 , and so for any such x0 ∈ RN −1 and for all xN ∈ R we have IN −1 0 JΨ (x) = , 2∇x0 f (x0 ) −1 which implies that det JΨ (x) = −1. Note that Ψ(RN \ Ω) = Ω. Moreover, for all x, y ∈ RN , kΨ(x) − Ψ(y)k = k(x0 − y 0 , 2f (x0 ) − 2f (y 0 ) − xN + yN )k q ≤ kx0 − y 0 k2N −1 + (2Lkx0 − y 0 kN −1 + xN − yN )2 (1 + L)kx − yk, which shows that Ψ and Ψ−1 are Lipschitz continuous. Here L := Lip f . Hence, by a change of variables and the fact that det ∇Ψ = det ∇Ψ−1 = −1, Z Z u(Ψ(x))−u(Ψ(y))p det JΨ (x) dx det JΨ (y) dy B (1 + L)N +sp N +sp RN \Ω RN \Ω kΨ(x) − Ψ(y)k Z Z u(z) − u(w)p N +sp = (1 + L) dzdw. N +sp Ω Ω kz − wk On the other hand if x ∈ RN \Ω and y ∈ Ω, then f (x0 ) ≥ xN and f (y 0 ) < yN . Hence, yN − 2f (x0 ) + xN = yN − xN + 2(xN − f (x0 )) ≤ yN − xN ≤ yN − xN , −yN + 2f (x0 ) − xN = 2f (x0 ) − 2f (y 0 ) + 2f (y 0 ) − xN − yN ≤ 2Lkx0 − y 0 kN −1 + 2yN − xN − yN ≤ 2Lkx0 − y 0 kN −1 + yN − xN . Therefore, yN − 2f (x0 ) + xN  ≤ (1 + 2L)(kx0 − y 0 kN −1 + yN − xN ),
8.1. Extension: Lipschitz Domains
311
and so kΨ(x) − yk (1 + L)kx − yk. Reasoning as before, we have Z Z u(Ψ(x)) − u(y)p C (1 + L)N +sp  det JΨ (x) dxdy N +sp Ω RN \Ω kΨ(x) − yk Z Z u(z) − u(y)p dzdy. = (1 + L)N +sp N +sp Ω Ω kz − yk Combining the estimates for A, B, and C, we get E(u)pW s,p (RN ) (1 + L)N +sp upW s,p (Ω) .
Corollary 8.3. Let f : RN −1 → R be a Lipschitz continuous function and let Ω be as in (8.1). Then for all 1 ≤ p < ∞ and 0 < s < 1 there exists a continuous linear operator E : W s,p (Ω) → W s,p (RN ) such that for all u ∈ W s,p (Ω), E(u)(x) = u(x) for LN a.e. x ∈ Ω and (8.4) kE(u)kLp (RN ) = 2kukLp (Ω) , E(u)W s,p (RN ) (1 + Lip f )N/p+s uW s,p (Ω) . Proof. Define v and Ψ as in Theorem 8.2. Using a change of variables and the fact that det ∇Ψ = −1, we have that Z Z Z v(x)p dx = u(Ψ(x))p  det JΨ (x) dx = u(y)p dy. RN \Ω
RN \Ω
Ω
Hence, the linear extension operator u ∈ W s,p (Ω) 7→ E(u) := v ∈ W s,p (RN ) is continuous and satisfies (8.4). Next we consider the case of bounded Lipschitz domains. Theorem 8.4. Let Ω ⊂ RN be an open bounded set with Lipschitz continuous boundary, 1 ≤ p < ∞, and 0 < s < 1. Then there exists a continuous linear operator E : W s,p (Ω) → W s,p (RN ) such that E(u)(x) = u(x) for LN a.e. x ∈ Ω and for all u ∈ W s,p (Ω), with kE(u)kLp (RN ) Ω kukLp (Ω) ,
E(u)W s,p (RN ) Ω kukW s,p (Ω) .
Proof. Let u ∈ W s,p (Ω). For every x0 ∈ ∂Ω there exist a rigid motion Tx0 : RN → RN , with Tx0 (x0 ) = 0, a Lipschitz continuous function fx0 : RN −1 → R, with fx0 (0) = 0, and rx0 > 0 such that Tx0 (Ω ∩ B(x0 , 2rx0 )) = {(y 0 , yN ) ∈ B(0, 2rx0 ) : yN > fx0 (y 0 )}. S S If the set Ω \ x∈∂Ω B(x, rx ) is nonempty, for every x0 ∈ Ω \ x∈∂Ω B(x, rx ) let B(x0 , rx0 ) ⊆ Ω. The family {B(x, rx )}x∈Ω is an open cover of Ω. Since Ω is compact, there is a finite number of balls B1 , . . . , B` , where Bn := B(xn , rxn ), that covers Ω ∩ K. Let {ψn }`n=1 be a smooth partition of unity P subordinated to B1 , . . . , B` , with supp ψn ⊆ Bn . Then `n=1 ψn = 1 in Ω. (8.5)
312
8. Further Properties
Fix n ∈ {1, . . . , `} and define un := uψn ∈ W s,p (Ω) (see Theorem 6.23). There are two cases. If supp ψn is contained in Ω, then by the second part of the statement of Theorem 6.23, if we extend un by zero outside Ω, the resulting function, denoted by vn , belongs to W s,p (RN ), with kvn kW s,p (RN ) n kukW s,p (Ω) .
(8.6)
If supp ψn is not contained in Ω, let xn ∈ ∂Ω be such Bn = B(xn , rn ). Then Tn (Ω ∩ B(xn , 2rn )) = {(y 0 , yN ) ∈ B(0, 2rn ) : yN > fn (y 0 )} =: Ωn , where fn : RN −1 → R is Lipschitz continuous and fn (0) = 0. Since ψn ∈ Cc∞ (RN ) and Tn is a rigid motion, the function (uψn ) ◦ Tn−1 belongs to W s,p (Ωn ) by Theorem 6.23 and Exercise 6.30, and it is zero in Ωn \ B(0, rn ). Let Un := Ωn ∪ (RN \ B(0, 2rn )) and let u ˜n be the function obtained by extending (uψn )◦Tn−1 to be zero in Un \Ωn . By Lemma 6.71, u ˜n ∈ W s,p (Un ), with k˜ un kW s,p (Un ) n k(uψn ) ◦ Tn−1 kW s,p (Ωn ) n kuψn kW s,p (Ω) . Let Vn := {(y 0 , yN ) ∈ RN −1 × R : yN > fn (y 0 )}. By Theorem 8.2 we can extend u ˜n Vn to a function v˜n ∈ W s,p (RN ) with (8.7)
k˜ vn kW s,p (RN ) n k˜ un kW s,p (Un ) n kuψn kW s,p (Ω) n kukW s,p (Ω) ,
where in the last inequality we use Theorem 6.23. Define vn := v˜n ◦ Tn . By Exercise 6.30, kvn kW s,p (RN ) k˜ vn kW s,p (RN ) n kukW s,p (Ω) . Define v :=
Pn
n=1 vn . n X
v(x) =
If x ∈ Ω, then vn (x) =
n=1
n X n=1
un (x) = u(x)
n X
ψn (x) = u(x).
n=1
Moreover, since the mapping u ˜n Vn 7→ v˜n given by Theorem 8.2 is linear, so is the mapping u 7→ v. Finally, by (8.6) and (8.7), kvkW s,p (RN ) ≤
n X
kvn kW s,p (RN ) Ω kukW s,p (Ω) .
n=1
Theorem 8.5 (Compactness). Let Ω ⊂ RN be an open bounded set with Lipschitz continuous boundary, let 1 ≤ p < ∞ and 0 < s < 1, and let {un }n be a sequence bounded in W s,p (Ω). Then there exist a subsequence {unk }k and a function u ∈ W s,p (Ω) such that unk → u in W σ,p (Ω) for every 0 < σ < s. Proof. By Theorem 8.4 there exists a function vn ∈ W s,p (RN ) such that vn = un in Ω and kvn kW s,p (RN ) Ω kun kW s,p (Ω) . Since the sequence {un }n is bounded in W s,p (Ω), the sequence {vn }n is bounded in W s,p (RN ). By Corollary 6.16, there exist a subsequence {vnk }k and a function v ∈ W s,p (RN )
8.2. Extension: The General Case
313
σ,p (RN ) for every 0 < σ < s. Let u be the restriction such that vnk → v in Wloc of v to Ω. Since Ω is bounded, unk → u in W σ,p (Ω) for every 0 < σ < s.
8.2. Extension: The General Case Unlike the case s = 1 (see [Leo17, Section 13.1]), for 0 < s < 1, the class of extension domains for W s,p (Ω) has been completely characterized. Indeed, we have the following result. Theorem 8.6 (Extension). Let Ω ⊆ RN be an open connected set, 1 ≤ p ≤ ∞, and 0 < s < 1. Then Ω is an extension domain for W s,p (Ω) if and only if there exists a constant C > 0 such that LN (B(x, r) ∩ Ω) ≥ CrN
(8.8)
for all x ∈ Ω and all 0 < r ≤ 1. Exercise 8.7. Let Ω ⊂ RN be an open bounded connected set with Lipschitz continuous boundary. Prove that (8.8) holds. Exercise 8.8. Let Ω ⊂ RN be an open bounded connected set with uniformly Lipschitz continuous boundary (see Definition 6.90). Prove that (8.8) holds. Lemma 8.9. Let Ω ⊆ RN be an open set, 1 ≤ p < ∞, and 0 < s < 1. Given x0 ∈ Ω and 0 < r < R < 1, define if x ∈ B(x0 , r), 1 R−kx−x0 k (8.9) v(x) := if x ∈ B(x0 , R) \ B(x0 , r), R−r 0 if x ∈ RN \ B(x0 , R), and u := vχΩ . Then kukLp (Ω) (LN (B(x0 , R) ∩ Ω))1/p , uW s,p (Ω)
(LN (B(x0 , R) ∩ Ω))1/p . (R − r)s
Proof. Since 0 ≤ u ≤ 1 and u = 0 in Ω \ B(x0 , R), we have kukLp (Ω) ≤ (LN (B(x0 , R) ∩ Ω))1/p , while, by Tonelli’s theorem, Z Z Z Z u(x) − u(y)p u(y)p dxdy = 2 dxdy N +sp N +sp Ω Ω kx − yk B(x0 ,R)∩Ω Ω\B(x0 ,R) kx − yk Z Z u(x) − u(y)p + dxdy =: A + B. N +sp B(x0 ,R)∩Ω B(x0 ,R)∩Ω kx − yk
314
8. Further Properties
If y ∈ B(x0 , R) ∩ Ω, then Ω \ B(x0 , R) ⊂ RN \ B(y, R − ky − x0 k). Hence, using spherical coordinates, Z Z 1 1 dx ≤ dx N +sp N +sp Ω\B(x0 ,R) kx − yk RN \B(y,R−ky−x0 k) kx − yk 1 . (R − ky − x0 k)sp In turn, Z A B(x0 ,R)∩Ω
u(y)p dy (R − ky − x0 k)sp
Z + (B(x0 ,R)\B(x0 ,r))∩Ω
Z B(x0 ,r)∩Ω
1 dy (R − ky − x0 k)sp
LN (B(x0 , R) ∩ Ω) (R − ky − x0 k)p(1−s) dy , (R − r)p (R − r)sp
where we use the facts that R − ky − x0 k ≥ R − r for y ∈ B(x0 , r), while R − ky − x0 k ≤ R − r for y ∈ B(x0 , R) \ B(x0 , r). To estimate B, write Z Z B= B(x0 ,R)∩Ω
B(y,R−r)∩Ω
Z
u(x) − u(y)p dxdy kx − ykN +sp
Z
+ B(x0 ,R)∩Ω
(B(x0 ,R)\B(y,R−r))∩Ω
u(x) − u(y)p dxdy =: B1 + B2 . kx − ykN +sp
1 and u = vχΩ , by the mean value theorem applied to v, Since k∇vk∞ ≤ R−r for x, y ∈ Ω we have 1 u(x) − u(y) = v(x) − v(y) ≤ kx − yk. R−r Hence, for y ∈ B(x0 , R) ∩ Ω, Z Z u(x) − u(y)p 1 kx − ykp dx ≤ dx N +sp (R − r)p B(y,R−r) kx − ykN +sp B(y,R−r)∩Ω kx − yk 1 . (R − r)sp
In turn, LN (B(x0 , R) ∩ Ω) . (R − r)sp On the other hand, since for y ∈ B(x0 , R) ∩ Ω, Z 1 1 dx , N +sp kx − yk (R − r)sp N R \B(y,R−r) B1
using the fact that 0 ≤ u ≤ 1, we have that Z Z 1 LN (B(x0 , R) ∩ Ω) B2 dxdy . N +sp (R − r)sp B(x0 ,R)∩Ω RN \B(y,R−r) kx − yk
8.2. Extension: The General Case
315
Combining the estimates for A, B1 , and B2 gives the desired inequality.
We now turn to the proof of the necessity part of Theorem 8.6. Proof of necessity. Assume that Ω is a W s,p extension domain. Step 1: The subcritical case. Assume that sp < N . Let x0 ∈ Ω, 0 < r < R ≤ 1, and let u be the function given in Lemma 8.9. Since Ω is a W s,p extension domain, u can be extended to a function w ∈ W s,p (RN ), with kwkW s,p (RN ) ≤Ω kukW s,p (Ω) . By Theorem 7.6, kwkLp∗s (RN ) wW s,p (RN ) . Using the fact that u = 1 in B(x0 , r) ∩ Ω and Lemma 8.9, we have ∗
(8.10) (LN (B(x0 , r) ∩ Ω))1/ps ≤ kwkLp∗s (RN ) wW s,p (RN ) Ω kukW s,p (Ω) Ω
(LN (B(x0 , R) ∩ Ω))1/p . (R − r)s
Since the function θ ∈ [0, 1] → LN (B(x0 , θR) ∩ Ω) is continuous and takes every value between 0 and LN (B(x0 , R) ∩ Ω), there exists θ ∈ (0, 1) such that 1 LN (B(x0 , θR) ∩ Ω) = LN (B(x0 , R) ∩ Ω). 2 Taking r = θR in (8.10) gives ∗
(R − θR)s Ω (LN (B(x0 , R) ∩ Ω))1/p−1/ps ≈Ω (LN (B(x0 , R) ∩ Ω))s/N , where we use the fact that 1/p − 1/p∗s = s/N . Hence, R − θR Ω (LN (B(x0 , R) ∩ Ω))1/N , where the constant depends on Ω, N , p, and s but not on x0 , θ, and R. Taking θ0 := 1 and replacing R with θn−1 R in what we just did, we can define inductively θn ∈ (0, θn−1 ) such that (8.11) 1 1 LN (B(x0 , θn R) ∩ Ω) = LN (B(x0 , θn−1 R) ∩ Ω) = n LN (B(x0 , R) ∩ Ω) 2 2 and θn−1 R − θn R Ω (LN (B(x0 , θn−1 R) ∩ Ω))1/N .
(8.12)
Since θn → 0 as n → ∞ by (8.11), we can use telescopic series to write R=
∞ X
(θn−1 R − θn R) Ω
n=1
∞ X
(LN (B(x0 , θn−1 R) ∩ Ω))1/N
n=1 ∞ X 1/N
Ω (LN (B(x0 , R) ∩ Ω))
n=1
1 2n/N
,
where we use (8.11) and (8.12). This proves (8.8).
316
8. Further Properties
Step 2: The supercritical case. Assume that sp > N . Let x0 ∈ Ω, 0 < r < R ≤ 1, and u be the function given in Lemma 8.9. Since Ω is a W s,p extension domain, u can be extended to a function w ∈ W s,p (RN ), with kwkW s,p (RN ) Ω kukW s,p (Ω) . In turn, by Theorem 7.23 w admits a representative w ¯ that is H¨older continuous with exponent α = s − N/p, with w ¯ C 0,α (RN ) wW s,p (RN ) . Since u is continuous in Ω, it follows that w ¯=u in Ω, and so, u(x) − u(y) w ¯ C 0,α (RN ) kx − yks−N/p wW s,p (RN ) kx − yks−N/p Ω kukW s,p (Ω) kx − yks−N/p Ω
(LN (B(x0 , R) ∩ Ω))1/p kx − yks−N/p , (R − r)s
where in the last inequality we use Lemma 8.9. Taking 0 < r < R4 , we have that R − r ≥ 43 R. Let x ∈ B(x0 , r) ∩ Ω, so that u(x) = 1, and y ∈ (B(x0 , R + R/2) \ B(x0 , R)) ∩ Ω, so that u(y) = 0. Then kx − yk ≤ kx − x0 k + kx0 − yk ≤ R/4 + R + R/2 ≤ 2R. Thus, 1 Ω
(LN (B(x0 , R) ∩ Ω))1/p s−N/p R Ω (LN (B(x0 , R) ∩ Ω))1/p R−N/p , Rs
which proves (8.8) even in this case. Step 3: The critical case. Assume that sp = N . Let x0 ∈ Ω, 0 < R < min{1, diam Ω}, θ0 = 1, and θn ∈ (0, 1) satisfy (8.11). Consider the function u given in the Lemma 8.9, with r and R replaced by θ2 R and θ1 R, respectively. Since Ω is a W s,p extension domain, u can be extended to a function w ∈ W N/p,p (RN ), with kwkW N/p,p (RN ) ≤ C0 kukW N/p,p (Ω) , where C0 = C0 (N, p, Ω) > 0. By Theorem 7.21 applied to the ball B = B(x0 , R), Z Z 0 0 p p 0 u(x) − c w(x) − c dx ≤ inf inf expC1 C0−p expC1 p0 dx 0 p c∈R B∩Ω c∈R B kukW N/p,p (Ω) wW N/p,p (8B) Z 0 p w(x) − wB  ≤ exp C1 dx ≤ C2 LN (B), 0 wpW N/p,p (8B) B where 8B := B(x0 , 8R). In turn, by Lemma 8.9, ! 0 0 u(x) − cp ((θ1 − θ2 )R)N p /p inf exp C dx ≤ C2 LN (B(x0 , R)). c∈R B(x0 ,R)∩Ω (LN (B(x0 , θ1 R) ∩ Ω))p0 /p Z
In view of (8.9), if c ≥ 3/2 or c ≤ 1/2, then u(x) − c = 1 − c ≥ 1/2 in B(x0 , θ2 R) ∩ Ω, while if 1/2 ≤ c ≤ 3/2, then u(x) − c = c ≥ 1/2 in
8.2. Extension: The General Case
317
(B(x0 , R) \ B(x0 , θ1 R)) ∩ Ω. Since, by (8.11), 1 N L (B(x0 , R) ∩ Ω), 22 1 LN ((B(x0 , R) \ B(x0 , θ1 R)) ∩ Ω) = LN (B(x0 , R) ∩ Ω), 2 we have that ! 0 C 1 N ((θ1 − θ2 )R)N p /p L (B(x0 , R)∩Ω) exp Ω LN (B(x0 , R)). 22 2p0 (LN (B(x0 , θ1 R) ∩ Ω))p0 /p LN (B(x0 , θ2 R) ∩ Ω) =
This implies that N
(θ1 − θ2 )R Ω (L (B(x0 , R) ∩ Ω))
1/N
C4 LN (B(x0 , R)) log N L (B(x0 , θ1 R) ∩ Ω)
p/(p0 N ) .
By repeating the same argument, that is, letting u be the function given in Lemma 8.9, with r and R replaced by θn+1 R and θn R, respectively, and applying Theorem 7.21 in the ball B(x0 , θn−1 R), we obtain (θn − θn+1 )R p/(p0 N ) C4 LN (B(x0 , θn−1 R)) Ω (L (B(x0 , θn−1 R) ∩ Ω)) log N L (B(x0 , θn R) ∩ Ω) " #p/(p0 N ) N C α RN 2n θn−1 4 N −(n−1)/N N 1/N Ω 2 (L (B(x0 , R) ∩ Ω)) log N , L (B(x0 , R) ∩ Ω) N
1/N
where in the last inequality we use (8.11). Using telescoping sums, we get θ1 R =
X
(θn − θn+1 )R
n N
Ω (L (B(x0 , R) ∩ Ω))
1/N
X
2
−n/N
n
log
2n C4 RN LN (B(x0 , R) ∩ Ω)
p/(p0 N )
C4 R N Ω (L (B(x0 , R) ∩ Ω)) 2 LN (B(x0 , R) ∩ Ω) n p/(p0 N ) C4 R N N 1/N , Ω (L (B(x0 , R) ∩ Ω)) log N L (B(x0 , R) ∩ Ω) N
1/N
X
−n/N
log 2n + log
p/(p0 N )
N where we use the fact that θn−1 ≤ 1 and C4 > 1 changes from line to line, as usual. Equivalently, p−1 C4 R N N N N θ1 R Ω (L (B(x0 , R) ∩ Ω)) log N . L (B(x0 , R) ∩ Ω)
318
8. Further Properties
If θ1 ≥
1 10 ,
using the fact that t(log t−1 )p−1 ≥ δ implies t ≥ cδ > 0, gives LN (B(x0 , R) ∩ Ω) ≥ cδ C4 RN ,
1 which is (8.8). If θ1 < 10 , then since 0 < R < diam Ω, we can find a point y ∈ Ω such that kx0 − yk > θ1 R + R/5. Since Ω is open and connected, it is pathwise connected, and thus it contains a polygonal path joining x0 and y. Let y0 ∈ Ω be a point on this polygonal path such that kx0 −y0 k = θ1 R+R/5. Then (8.13) B(x0 , θ1 R) ⊂ B(y0 , 2R/5) ⊂ B(x0 , R), B(y0 , R/5) ∩ B(x0 , θ1 R) = ∅.
Reasoning as in Step 1, but with B(x0 , R) replaced by B(y0 , 2R/5), we can find 0 < δ1 < 1 such that 1 LN (B(y0 , δ1 2R/5) ∩ Ω) = LN (B(y0 , 2R/5) ∩ Ω). 2 By (8.11) and (8.13), 1 1 LN (B(x0 , θ1 R) ∩ Ω) = LN (B(x0 , R) ∩ Ω) ≥ LN (B(y0 , 2R/5) ∩ Ω). 2 2 Since B(y0 , R/5) ∩ B(x0 , θ1 R) = ∅ and B(x0 , θ1 R) ⊂ B(y0 , 2R/5), it follows that B(x0 , θ1 R) ⊂ B(y0 , 2R/5) \ B(y0 , R/5). Hence, δ1 ≥ 21 . Thus, we can apply the first part of this step, with B(x0 , R) replaced by B(y0 , 2R/5), to conclude that LN (B(y0 , 2R/5) ∩ Ω) ≥ C(2R/5)N . Since B(y0 , 2R/5) ⊂ B(x0 , R) by (8.13), this implies that LN (B(x0 , R) ∩ Ω) ≥ LN (B(y0 , 2R/5) ∩ Ω) ≥ C(2R/5)N , which shows (8.8) in the case θ1
0 and k ∈ N. Prove that there exists `k ∈ N depending only on N and k with the property that there cannot exist more that `k points xn in Br such that kxn −xm k ≥ 2rk for n 6= m.
8.2. Extension: The General Case
319
Lemma 8.11. Let U ⊂ RN be an open set and (8.14)
r(x) := dist(x, RN \ U )/10,
x ∈ U.
Then the family {B(x, r(x)/5)}x∈U admits a maximal countable subfamily {B(xn , rn /5)}n of pairwise disjoint balls, where rn := r(xn ). Moreover, the following properties hold: S (i) U = n B(xn , rn ); (ii) if x ∈ B(xn , 5rn ), then 5rn < dist(x, RN \ U ) < 15rn ; (iii) there exists yn ∈ RN \ U such that kxn − yn k < 15rn ; P (iv) there exists ` ∈ N such that n χB(xn ,5rn ) (x) ≤ ` for all x ∈ U . Proof. Let X be the set of all subfamilies of {B(x, r(x)/5)}x∈U with pairwise disjoint elements. Given F1 and F2 , we say that F1 ≤ F2 if every ball of F1 is contained in some ball of F2 . The relation ≤ is a partial order in X. Let Y ⊆ X be a totally ordered set. Consider the family F∞ := {B : B ∈ F for some F ∈ Y }. If B1 , B2 ∈ F∞ , then B1 ∈ F1 and B2 ∈ F2 for some F1 and F2 in Y . Since Y is totally ordered, F1 ≤ F2 or F2 ≤ F1 ; let’s say the first. Then B1 , B2 ∈ F2 , and so either B1 = B2 or B1 ∩ B2 = ∅. This shows that F∞ is a subfamily of {B(x, r(x)/5)}x∈U with pairwise disjoint elements. Thus, F∞ ∈ X. Since F ≤ F∞ for every F ∈ Y , we have that F∞ is an upper bound of Y . It follows by Zorn’s lemma that X admits a maximal element F0 = {Bα }α∈Λ . Each ball Bα is contained in U . Consider all the points of U with rational coordinates. Each of these points can be contained in at most one Bα . Moreover, since each ball is open, by density, it must contain one such point. It follows that the family F0 must be countable, and so we can write it as a sequence {B(xn , rn /5)}n . To prove item (i), let x ∈ U . If x = xn for some n, then there is nothing to prove. Otherwise, there exists n such that B(x, r(x)/5)∩B(xn , rn /5) 6= ∅, since otherwise we could add B(x, r(x)/5) to F0 , which would contradict the maximality of F0 . We claim that rn ≥ r(x)/2. Assume by contradiction that rn < r(x)/2. Let y ∈ B(x, r(x)/5) ∩ B(xn , rn /5). Then for every z ∈ B(xn , 15rn ) we have kz − xk ≤ kz − xn k + kxn − yk + ky − xk < 15rn + rn /5 + r(x)/5 < [15/2 + 1/10 + 1/5]r(x) < 10r(x) = dist(x, RN \ U ). This shows that B(xn , 15rn ) ⊆ U , which is a contradiction since 15rn = 3 N 2 dist(xn , R \ U ). Therefore, rn ≥ r(x)/2. In turn, kx − xn k ≤ kx − yk + ky − xn k < r(x)/5 + rn /5 ≤ 3rn /5 < rn .
320
8. Further Properties
Hence, x ∈ B(xn , rn ). This proves item (i). Item (ii) follows from the facts that if x ∈ B(xn , 5rn ), then dist(x, RN \ U ) ≤ dist(xn , RN \ U ) + kx − xn k < 10 rn + 5rn , dist(x, RN \ U ) ≥ dist(xn , RN \ U ) − kx − xn k ≥ 10 rn − 5rn . To show item (iii), observe that, by the definition of rn = r(xn ) (see (8.14)), for every 0 < ε < 5rn we can find yn ∈ RN \ U such that 10 rn ≤ kyn − xn k < 10 rn + ε. Finally to prove (iv), given x ∈ U , let Ix := {n : x ∈ B(xn , 5rn )}. If n ∈ Ix , then by item (ii), 1 1 dist(x, RN \ U ) < rn < dist(x, RN \ U ), 15 5 so that xn ∈ B(x, dist(x, RN \ U )). Moreover, for every n, m ∈ Ix with n 6= m, since B(xn , rn /5) and B(xm , rm /5) are disjoint, we have that rn 1 kxm − xn k ≥ > dist(x, RN \ U ). 5 75 It follows from Exercise 8.10 that the cardinality of Ix is at most `15 . Next we construct a partition of unity subordinated to the family of balls constructed in Lemma 8.9. Lemma 8.12. Let U ⊂ RN be an open set and let {B(xn , rn /5)}n be the countable family of pairwise disjoint balls constructed in Lemma 8.9. Let ψ ∈ Cc∞ (R) be such that 0 ≤ ψ ≤ 1, ψ = 1 in [0, 1], and ψ = 0 in [3/2, ∞). Define ψn (x) := ψ (kx − xn k/rn ) , x ∈ RN , and ( Pψn (x) if x ∈ U, k ψk (x) (8.15) ϕn (x) := 0 otherwise. Then ϕn is well defined, Lipschitz continuous, and satisfies the following properties: (i) supp ϕn ⊆ B(xn , 2rn ); (ii) ϕn (x) ≥ 1/` for x ∈ B(xn , rn ); (iii) there exists L > 1 such k∇ϕn k∞ ≤ L/rn ; P N (iv) n ϕn (x) = χU (x) for all x ∈ R . Proof. For x ∈ U , set dx := dist(x, RN \ U ). By item (i) of Lemma 8.11, if x ∈ U , then x ∈ B(xk , rk ) for some k, so ψk (x) = 1. Thus the denominator in (8.15) is always different from zero in U . This shows that ϕn is well defined.
8.2. Extension: The General Case
321
If kx − xn k/rn ≥ 2, then ψn (x) = 0. Thus item (i) follows. To prove item (ii), observe that if x ∈ B(xn , rn ), then ψn (x) = 1, while by item (i), X X X χB(xk ,5rk ) ≤ ` ψk (x) ≤ ψk (x) = k
k: x∈B(xk ,5rk )
k: x∈B(xk ,2rk )
by item (iv) of Lemma 8.11. To prove item (iii), we use the fact that supp ϕn ⊆ B(xn , 2rn ) ⊆ U , where the second inclusion follows from the definition of rn = r(xn ) (see (8.14)). Given x ∈ supp ϕn , let Jx := {k : x ∈ B(xk , 5rk )}. By item (iv) of Lemma 8.11, we have that the cardinality of Jx is at most `. Since supp ψk ⊆ B(xk , 2rk ), it follows that in a small neighborhood of x only the functions ψk with k ∈ Jx are different from zero. To be precise, let y ∈ B(x, rn ). Then xn − y < xn − x + x − y < 3rn , so 5rn < dy < 15rn by item (ii) of Lemma 8.11. If ψk (y) > 0 for some k, then y ∈ B(xk , 2rk ). By item (ii) of Lemma 8.11, 5rk < dy < 15rk . It follows that rk < 3rn and rn < 3rk . In turn, x − xk  < x − y + y − xk  < rn + rk < 4rk . This implies that k ∈ Jx . Thus, for all y ∈ B(x, rn ), ψn (y) . k∈Jx ψk (y)
ϕn (y) = P
Hence, we can differentiate ϕn at all y ∈ B(x, rn ), with y 6= xk , k ∈ Jx , to get ∂i ϕn (y) =
ψ 0 (ky − xn k/rn )∂i (ky − xn k) P rn k∈Jx ψk (y) P ψn (y) k∈Jx rk−1 ψ 0 (ky − xk k/rk )∂i (ky − xk k) . − 2 P ψ (y) k k∈Jx
Since rk < 3rn and rn < 3rk whenever ψk (y) > 0, k∂i ϕn k∞ ≤ Item (iv) follows from item (i).
C 0 rn ` kψ k∞ .
Exercise 8.13. Let Ω ⊂ RN be an open connected set satisfying (8.8). Prove that LN (∂Ω) = 0. Definition 8.14. Given a Lebesgue measurable set E ⊂ RN with finite measure and a Lebesgue measurable function u : E → R, the median of u over E is defined as med(u, E) := sup{t ∈ R : LN ({x ∈ E : u(x) < t}) ≤ 21 LN (E)}. Exercise 8.15. Let E ⊂ RN a Lebesgue measurable set with finite measure and let u : E → R be a Lebesgue measurable function. Prove that: (i) for every c ∈ R, med(u − c, E) = med(u, E) − c;
322
8. Further Properties
(ii)  med(u, E) ≤ med(u, E); (iii) for every c ∈ R and q > 0, 1/q Z 2 q u(x) − c dx  med(u − c, E) ≤ . LN (E) E We are ready to prove the sufficiency part of Theorem 8.6. Proof of sufficiency. Step 1: Assume that (8.8) holds. Then by Exercise 8.13, LN (∂Ω) = 0. Define U := RN \ Ω and let B(xn , rn /5) and yn be as in Lemma 8.11. Since yn ∈ RN \ U = Ω and kxn − yn k < 15rn , if yn ∈ ∂Ω, there exists zn ∈ Ω ∩ B(xn , 15rn − kxn − yn k). Hence, by replacing yn with zn , in what follows, we can assume that yn ∈ Ω. Let V := {x ∈ RN : dist(x, Ω) ≤ 8}. Fix x ∈ V \ Ω and let Ix := {n : x ∈ B(xn , 2rn )}.
(8.16) By Lemma 8.11, (8.17)
]Ix ≤ `,
and for n ∈ Ix , (8.18)
B(yn , rn ) ⊂ B(xn , 16rn ) ⊂ B(x, 5dx ),
where (8.19)
dx := dist(x, RN \ U ),
and we use the facts that kxn − yn k < 15rn , kx − xn k < 2rn , and 5rn < dx < 15rn . We claim that Ix ⊆ J := {n : rn ≤ 1}. Indeed, if rn > 1 for some n ∈ Ix , then for all z ∈ B(xn , 2rn ), dist(z, RN \ U ) ≥ dist(xn , RN \ U ) − kxn − zk > 10rn − 2rn = 8rn > 8, which implies that B(xn , 2rn ) ∩ V = ∅. This contradicts the fact that x ∈ B(xn , 2rn ). This shows that Ix ⊆ J. In turn, by (8.8) and the facts that yn ∈ Ω and 5rn < dx < 15rn , we get (8.20)
N N N C1 dN x /(15) ≤ C1 rn ≤ L (B(yn , rn ) ∩ Ω).
Step 2: Given u ∈ W s,p (Ω), define v : V → R by if x ∈ Ω, u(x) 0 if x ∈ Ω \ Ω, (8.21) v(x) := P ϕ (x) med(u, B(y , r ) ∩ Ω) if x ∈ V \ Ω, n n n n
8.2. Extension: The General Case
323
where med(u, B(yn , rn ) ∩ Ω) is the median of u in the set B(yn , rn ) ∩ Ω. In this step we prove that v ∈ Lp (V ), with kvkLp (V ) Ω kukLp (Ω) . Since supp ϕn ⊆ B(xn , 2rn ) by items (i) and (iv) of Lemma 8.12, for all x ∈ V \ Ω, X X ϕn (x) med(u, B(yn , rn ) ∩ Ω) = ϕn (x) med(u, B(yn , rn ) ∩ Ω), n
n∈Ix
where Ix = {n : x ∈ B(xn , 2rn )} (see (8.16)), and X X (8.22) 1= ϕn (x) = ϕn (x). n
n∈Ix
Hence, for 0 < q < p by Exercise 8.15, (8.18), (8.20), and (8.21), for x ∈ V \ Ω, X v(x) ≤ ϕn (x) med(u, B(yn , rn ) ∩ Ω) n∈Ix
≤
X
ϕn (x)
n∈Ix
X
Ω
n∈Ix
ϕn (x)
1 N L (B(yn , rn ) ∩ Ω) 1 dN x
Z
!1/q
Z
u(y)q dy
B(yn ,rn )∩Ω
!1/q q
u(y) dy
Ω (M(χΩ uq )(x))1/q ,
B(x,5dx )∩Ω
where in the last inequality we use (8.17). Raising both sides to power p and integrating in x over V , we get Z Z Z v(x)p dx Ω (M(χΩ uq )(x))p/q dx Ω u(x)p dx, V
RN
Ω
where we use the continuity of the maximal operator M in Lp/q and the fact that p/q > 1 (see [Leo22c]). Step 3: We claim that Z Z Z Z v(x) − v(y)p u(x) − u(y)p dxdy dxdy. Ω N +sp N +sp V V kx − yk Ω Ω kx − yk By Tonelli’s theorem, Exercise 8.13, and (8.21), we can write Z Z Z Z v(x) − v(y)p u(x) − u(y)p dxdy = dxdy N +sp N +sp V V kx − yk Ω Ω kx − yk Z Z v(x) − u(y)p (8.23) +2 dxdy N +sp Ω V \Ω kx − yk Z Z v(x) − v(y)p + dxdy =: A + B + C. N +sp V \Ω V \Ω kx − yk We will estimate the terms B and C in the following substeps.
324
8. Further Properties
SubstepP 3a: We estimate B. Given x ∈ V \ Ω ⊂ U and y ∈ Ω, by Lemma 8.12(iv), n∈Ix ϕn (x) = 1, and so, by (8.21), we can write X v(x) − u(y) = ϕn (x)[med(u, B(yn , rn ) ∩ Ω) − u(y)] n∈Ix X ≤ ϕn (x) med(u − u(y), B(yn , rn ) ∩ Ω) n∈Ix
≤
X
ϕn (x)
n∈Ix
X
Ω
ϕn (x)
n∈Ix
1 N L (B(yn , rn ) ∩ Ω) 1 dN x
!1/q
Z
u(z) − u(y)q dz
B(yn ,rn )∩Ω
!1/q
Z
q
u(z) − u(y) dz B(x,5dx )∩Ω
for 0 < q < p and where we use Exercise 8.15, (8.18), and (8.20). If z ∈ B(x, 5dx ) ∩ Ω, then since x ∈ / Ω , by (8.19) we have that kx − yk ≥ dx , so kz − yk ≤ kz − xk + kx − yk < 5dx + kx − yk ≤ 6kx − yk. Hence, X v(x) − u(y) Ω ϕn (x) N/p+s kx − yk n∈I x
Ω
1 dN x
1 dN x
Z B(x,5dx )∩Ω
u(z) − u(y)q χΩ (z) dz kz − yk(N/p+s)q B(x,5dx )
Z
u(z) − u(y)q dz kz − yk(N/p+s)q
!1/q
!1/q Ω (M(gyq )(x))1/q ,
where we use (8.17) and gy (x) := χΩ (x)
u(x) − u(y) , kx − ykN/p+s
x ∈ RN .
Raising both sides to power p and integrating in x over V \ Ω and in y over Ω, we get Z Z Z Z q p/q (M(gyq )(x))p/q dxdy B Ω (M(gy )(x)) dxdy Ω Ω
V \Ω
Z Z Ω Ω
RN
(gy (x))p dxdy =
Ω
Z Z Ω
Ω
RN
u(x) − u(y)p dxdy kx − ykN +sp
by the continuity of the maximal operator M in Lp/q and the fact that p/q > 1 (see [Leo22c]). Substep 3b: To estimate C in (8.23), for every y ∈ V \ Ω define (8.24)
C1,y := {x ∈ V \ Ω : kx − yk ≥
(8.25)
C2,y := {x ∈ V \ Ω : kx − yk
1 (see [Leo22c]). Substep 3c: Finally, we estimate C2 in (8.26). Fix y ∈ V \ Ω and x ∈ C2,y . We claim that 1 dy ≤ dx ≤ 3dy . 3
(8.29)
Let z ∈ Ω be such that kz − yk = dy . Then since x ∈ C2,y , by (8.25), dx ≤ kx − zk ≤ kx − yk + kz − yk ≤
1 1 1 max{dx , dy } + dy ≤ dx + dy + dy , 2 2 2
which implies that dx ≤ 3dy . By interchanging the roles of x and y we get that dy ≤ 3dx . Hence, we have shown (8.29). Since 5rn < dz < 15rn for all n ∈ Iz by Lemma 8.11(ii), (8.29) implies in particular, that rn ≈ dy
(8.30)
for all n ∈ Ix ∪ Iy .
Next we claim that B(yn , rn ) ⊆ B(y, 20dy )
(8.31)
for all n ∈ Ix ∪ Iy .
If n ∈ Iy , this follows from (8.18). If n ∈ Ix , then B(yn , rn ) ⊆ B(x, 5dx ) again by (8.18). On the other hand, since x ∈ C2,y , we have that kx − yk < 1 3 2 max{dx , dy } ≤ 2 dy by (8.29), and so, B(x, 5dx ) ⊆ B(y, 20dy ), again by (8.29). Observe that by (8.22), X X 1= ϕn (y) = ϕn (y) = n∈Iy
n∈Ix ∪Iy
X
ϕn (x) =
n∈Ix ∪Iy
so X n∈Ix ∪Iy
(ϕn (x) − ϕn (y)) = 0.
X n∈Iy
ϕn (y),
8.2. Extension: The General Case
327
Hence, by (8.21), we can write X v(x) − v(y) = (ϕn (x) − ϕn (y)) med(u, B(yn , rn ) ∩ Ω) n∈Ix ∪Iy
X
=
(ϕn (x) − ϕn (y))[med(u, B(yn , rn ) ∩ Ω) − med(u, B(y, 5dy ) ∩ Ω)].
n∈Ix ∪Iy
Since k∇ϕn k∞ ≤ L/rn by Lemma 8.12, (8.32) v(x) − v(y) X kx − yk ≤L  med(u, B(yn , rn ) ∩ Ω) − med(u, B(y, 5dy ) ∩ Ω). rn n∈Ix ∪Iy
Consider some m ∈ Iy . Then by (8.18) and (8.20), N N N N C1 dN y /(15) ≤ C1 rm ≤ L (B(ym , rm ) ∩ Ω) ≤ L (B(y, 5dy ) ∩ Ω),
while C1 rnN ≤ LN (B(yn , rn ) ∩ Ω) for every n ∈ Ix ∪ Iy by (8.8) since yn ∈ Ω. In turn, by Exercise 8.15 applied twice with for 0 < q < p, Tonelli’s theorem, and H¨older’s inequality with exponent p/q > 1,  med(u, B(yn , rn ) ∩ Ω) − med(u, B(y, 5dy ) ∩ Ω) !1/q Z 1 q (8.33) Ω u(z) − med(u, B(yn , rn ) ∩ Ω) dz dN B(y,5dy )∩Ω y !1/q Z Z 1 q Ω u(z) − u(ξ) dξdz rnN dN B(y,5dy )∩Ω B(yn ,rn )∩Ω y !q/p 1/q Z Z 1 1 Ω N dξ . u(z) − u(ξ)p dz rn B(yn ,rn )∩Ω dN B(y,5dy )∩Ω y Combining (8.32) and (8.33), using (8.30) and (8.31), and writing 1 = kz−ξkN +sp , we obtain kz−ξkN +sp v(x) − v(y) Ω
X n∈Ix ∪Iy
kx − yk dy
×
1 dN y
Z B(y,20dy )∩Ω
1 dN y
!q/p
Z
u(z) − u(ξ)p dz
1/q dξ
B(y,5dy )∩Ω
Z kx − yk 1 Ω dy dN B(y,20dy )∩Ω y
u(z) − u(ξ)p dsp dz y kz − ξkN +sp B(y,5dy )∩Ω
Z
!q/p
1/q dξ
,
328
8. Further Properties
where in the last inequality we use the facts that kz−ξk ≤ kz−yk+ky−ξk ≤ P 5dy + 20dy and that n∈Ix ∪Iy 1 ≤ 2` by Lemma 8.11(v). Define Z q/p u(z) − u(ξ)p h(ξ) := χΩ (ξ) dz . N +sp Ω kz − ξk Then
kx − ykdsy (M(h)(y))1/q . dy Raising both sides to power p, dividing by kx − ykN +sp and integrating in y over V \ Ω and in x over C2,y , by (8.23), we get Z Z Z kx − ykp−N −sp p/q C2 Ω (M(h)(y)) dxdy Ω (M(h)(y))p/q dy p(1−s) N V \Ω C2,y R dy Z Z Z p u(z) − u(y) Ω (h(y))p/q dy Ω dzdy, N +sp N R Ω Ω kz − yk where we use Tonelli’s theorem, the continuity of the maximal operator M in Lp/q , and the facts that p/q > 1 (see [Leo22c]), and that Z Z kx − ykp−N −sp kx − ykp−N −sp dx ≤ dx ≈ 1, p(1−s) p(1−s) C2,y B(y,3dy /2) dy dy v(x) − v(y) Ω
since C2,y ⊆ B(y, 3dy /2) by (8.25) and (8.29). Combining the estimates for B and C obtained in Substeps 3a, 3b, and 3c, it follows from (8.23) that Z Z Z Z u(x) − u(y)p v(x) − v(y)p dxdy dxdy. Ω N +sp N +sp V V kx − yk Ω Ω kx − yk Step 4. To extend v from V to RN we can use a cutoff function. We leave the details as an exercise. Remark 8.16 (Important). Note that if Ω ⊆ RN satisfies the hypotheses of Theorem 8.6, 1 ≤ p < ∞, and 0 < s < 1, then for every u ∈ W s,p (Ω) we can find v ∈ W s,p (RN ) such that v = u and kvkW s,p (RN ) Ω kukW s,p (Ω) . On the other hand, if v ∈ W s,p (RN ) is such that v = u, then kvkW s,p (RN ) ≥ kukW s,p (Ω) . Hence, kukW s,p (Ω) ≈Ω inf{kvkW s,p (RN ) : v ∈ W s,p (RN ), v = u in Ω}. We leave it as an exercise to check that (8.34)
u 7→ inf{kvkW s,p (RN ) : v ∈ W s,p (RN ), v = u in Ω}
is a norm. In the literature (see, e.g., [Mir18], [Tri10, Chapter 3]), often W s,p (Ω) is defined as the space of functions u ∈ Lp (Ω) for which that infimum in (8.34) is finite. This space is endowed with the norm (8.34). If Ω does not satisfy the hypotheses of Theorem 8.6, then I suspect that this
8.3. Derivatives
329
definition of fractional Sobolev spaces does not coincide with the one given in this book (Definition 6.1). Exercise 8.17. Let Ω = {x = (x1 , x2 ) ∈ R2 : 0 < x1 < 1, 0 < x2 < x21 }, 0 < s < 1, p > 2/s, and u(x) = kxkα , where s − 3/p < α < s − 2/p. • Prove that u ∈ W s,p (Ω). • Prove that Ω is not a W s,p extension domain without making use of Theorem 8.6. Theorem 8.18 (Compactness). Let Ω ⊆ RN be an open connected set satisfying (8.8). Given 1 ≤ p < ∞ and 0 < s < 1, let {un }n be bounded in W s,p (Ω). Then there exist a subsequence {unk }k and a function u ∈ W s,p (Ω) σ,p (Ω) for every 0 < σ < s. such that unk → u in Wloc Proof. By Theorem 8.6 there exists a function vn ∈ W s,p (RN ) such that vn = un in Ω and kvn kW s,p (RN ) Ω kun kW s,p (Ω) . Since the sequence {un }n is bounded in W s,p (Ω), the sequence {vn }n is bounded in W s,p (RN ). By Corollary 6.16, there exist a subsequence {vnk }k and a function v ∈ W s,p (RN ) σ,p such that vnk → v in Wloc (RN ) for every 0 < σ < s. Let u be the restriction σ,p (Ω) for every 0 < σ < s. of v to Ω. Then unk → u in Wloc
8.3. Derivatives Given an open set Ω ⊆ RN , 1 ≤ p < ∞, and m ∈ N, for every i = 1, . . . , N , ∂ the partial derivative ∂x is a linear continuous operator from W m,p (Ω) i m−1,p 0,p into W (Ω), where W (Ω) := Lp (Ω). In this section we discuss the analogue in W s,p (Ω). Given an open set Ω ⊆ RN , 1 ≤ p < ∞, 0 < s < 1, and a function u ∈ W s,p (Ω), we consider the distribution (see [Leo17, Chapter 10]) Z Tu (ϕ) :=
u(x)ϕ(x) dx,
ϕ ∈ D(Ω).
Ω u For every i = 1, . . . , N , the distributional derivative ∂T ∂xi : D(Ω) → R is the distribution defined by Z ∂Tu ∂ϕ (ϕ) := − u(x) (x) dx, ϕ ∈ D(Ω). ∂xi ∂xi Ω
∂u u As usual, with an abuse of notation, we can write ∂x to mean ∂T ∂xi . In i the next theorem we show that when sp 6= 1 and Ω is sufficiently regular, s−1,p (Ω). Note that u the distribution ∂T ∂xi can be extended continuously to W 0
s − 1 < 0, so W s−1,p (Ω) = (W01−s,p (Ω))0 . We begin with the case Ω = RN . Theorem 8.19. Let 1 < p < ∞ and 0 < s < 1. Then for every i = ∂ 1, . . . , N , the linear map ∂x : W s,p (RN ) → W s−1,p (RN ) is continuous. i
330
8. Further Properties
The proof of Theorem 8.19 relies on trace theory and will be given in s,p Chapter 9. Note that W s,p (RN ) = W00 (RN ). Theorem 8.20. Let Ω ⊆ RN be an open set with uniformly Lipschitz continuous boundary, 1 < p < ∞, and 0 < s < 1. Then for every i = 1, . . . , N , 0 ∂ s,p (Ω) → (W 1−s,p (Ω))0 is continuous. In particular, the linear map ∂x : W 00 i ∂ if sp 6= 1, then the linear map ∂x : W s,p (Ω) → W s−1,p (Ω) is continuous. i Proof. Step 1: Let Ω ⊂ RN be an open bounded set with Lipschitz continuous boundary. Given u ∈ W s,p (Ω), by Theorem 8.4 there exists a function v ∈ W s,p (RN ) such that v = u on Ω and kvkW s,p (RN ) Ω kukW s,p (Ω) . For every ϕ ∈ D(Ω), we extend ϕ to be zero outside Ω. By Theorem 8.19, Z Z ∂Tu ∂ϕ ∂ϕ ∂xi (ϕ) = u(x) ∂xi (x) dx = N v(x) ∂xi (x) dx Ω R
∂Tv
kϕkW 1−s,p0 (RN ) ∂xi W s−1,p (RN )
∂Tv
, 1−s,p0
∂xi s−1,p N kϕkW00 (Ω) W
(R )
where we recall that for 1 ≤ q < ∞, 0 < σ < 1, !1/q Z Z v(x)q σ,q dydx kvkW00 (Ω) := kvkW σ,q (Ω) + N +σq Ω RN \Ω kx − yk ∂Tu ∂xi : D(Ω) → R can 1−s,p0 u operator ∂T (Ω) → ∂xi : W00
(see Corollary 6.98). This inequality implies that
be
uniquely extended to a linear continuous
R.
1−s,p0 (Ω))0 . To conclude the proof of this step, observe Hence, ∈ (W00 0 1−s,p0 (Ω) = W01−s,p (Ω), with equivalent norms that by Theorem 6.105, W00 provided (1 − s)p0 6= 1, which is equivalent to sp 6= 1. This implies that ∂Tu ∂xi
∂Tu ∂xi
0
∈ (W01−s,p (Ω))0 = W s−1,p (Ω).
Step 2: Let Ω ⊆ RN be an open set with uniformly Lipschitz continuous boundary. We proceed as in Step 1. The only difference is that we use Theorem 8.6 and Exercise 8.8 in place of Theorem 8.4. Next we show that when sp = 1 in general into W 1/p−1,p (Ω).
∂ ∂xi
does not map W 1/p,p (Ω)
Theorem 8.21. Let I = (a, b) and 1 < p < ∞. Then the linear map d : W 1/p,p (I) ∩ D(I) 7→ W 1/p−1,p (I) dx cannot be extended continuously to W 1/p,p (I).
8.4. Embeddings and Interpolation Inequalities
331
d Proof. Assume by contraction that the linear operator dx can be extended 1/p,p continuously to W (I). Then Z b dTu 0 I kϕk 1−1/p,p0 kuk 1/p,p (ϕ) = ϕ (x)u(x) dx W (I) W (I) dx a
for all ϕ ∈ D(I) and u ∈ W 1/p,p (I). In particular, if u ∈ C 1 (I), then we can integrate by parts to get Z b 0 ϕ(x)u (x) dx I kϕkW 1−1/p,p0 (I) kukW 1/p,p (I) (8.35) a
0
for all ϕ ∈ D(I). Since (1 − 1/p)p0 = 1, by Theorem 1.75, W 1−1/p,p (I) = 0 1−1/p,p0 W0 (I), thus, for every ϕ ∈ W 1−1/p,p (I), we can find a sequence ϕn ∈ 0 D(I) such that ϕn → ϕ in W 1−1/p,p (I). Taking ϕn in (8.35) and letting 0 n → ∞ shows that (8.35) continues to hold for every ϕ ∈ W 1−1/p,p (I). In particular, (8.35) holds if ϕ ∈ C 1 (I). Take ϕ = 1 and assume that u(b) = 0. Then Z b 0 u(a) = u (x) dx I kukW 1/p,p (I) . a
1/p,p = W0 (I), by Theorem 1.75, so, we can find a sequence ∞ un ∈ Cc (I) such that un → u in W 1/p,p (I). Taking u − un in
W 1/p,p (I)
Since of functions the previous inequality gives
u(a) = u(a) − un (a) I ku − un kW 1/p,p (I) → 0, which gives a contradiction if u ∈ C 1 (I) is such that u(a) 6= 0.
8.4. Embeddings and Interpolation Inequalities In this section, we extend the results of Chapter 7 to smooth domains. Theorem 8.22. Let 1 ≤ p < ∞, 0 < s < 1, and let Ω ⊆ RN be an open connected set satisfying (8.8). (i) Sobolev–Gagliardo–Nirenberg embedding. If 1 ≤ p < N/s, then kukLp∗s (Ω) Ω kukW s,p (Ω) for every u ∈ W s,p (Ω). Moreover, W s,p (Ω) ,→ Lq (Ω) for all p ≤ q ≤ p∗s . (ii) Critical embedding. If s = N/p, then for every p ≤ q < ∞, kukLq (Ω) Ω kukW N/p,p (Ω) for every u ∈ W s,p (Ω). (iii) Morrey embedding. If s > N/p, then every function u ∈ W s,p (Ω) admits a representative u ¯ that is H¨ older continuous with exponent α = s−N/p, with k¯ ukC 0,α (Ω) Ω kukW s,p (Ω) . Moreover, W s,p (Ω) ,→ Lq (Ω) for every p ≤ q ≤ ∞.
332
8. Further Properties
Proof. Given u ∈ W s,p (Ω), by Theorem 8.6 there exists v ∈ W s,p (RN ) such that v = u in Ω and kvkW s,p (RN ) Ω kukW s,p (Ω) . It is now enough to apply the embedding theorems (Theorems 7.6, 7.12, and 7.23). In view of Corollary 7.24, we have the following result. Corollary 8.23 (Product). Let 1 ≤ p < ∞, 0 < s < 1, with sp > N , and let Ω ⊆ RN be an open connected set satisfying (8.8). If u, v ∈ W s,p (Ω), then uv ∈ W s,p (Ω), with kuvkW s,p (Ω) Ω kukW s,p (Ω) kvkW s,p (Ω) . In a similar way, using Theorems 7.31, 7.32, and 8.6, we obtain: Theorem 8.24. Let 1 < p1 < ∞, 1 ≤ p2 < ∞, and 0 < s2 < s1 = 1 be such that 1 + pN2 = s2 + pN1 , and let Ω ⊆ RN be a W 1,p1 (Ω) extension domain. Then kukW s2 ,p2 (Ω) Ω kukW 1,p1 (Ω) for every u ∈ W 1,p1 (Ω). Theorem 8.25. Let N ≥ 2, 1 < p2 < ∞, and 0 < s2 < s1 = 1 be such that 1 + pN2 = s2 + N , and let and Ω ⊆ RN be a W 1,1 (Ω) extension domain. Then kukW s2 ,p2 (Ω) Ω kukW 1,1 (Ω) for every u ∈ W 1,1 (Ω). Theorem 8.26. Let 0 < s2 < s1 < 1, 1 ≤ p2 < sN2 , and 1 ≤ p1 < sN1 be such that s2 + pN1 = s1 + pN2 , and let Ω ⊆ RN be an open connected set satisfying (8.8). Then kukW s2 ,p2 (Ω) Ω kukW s1 ,p1 (Ω) for every u ∈ W s1 ,p1 (Ω). Next we extend the interpolation inequalities of Section 7.7. We set := Lp1 (Ω). We begin with the case s2 = 1.
W 0,p1 (Ω)
Theorem 8.27. Let 1 < p1 , p2 < ∞, 0 ≤ s1 < s2 = 1, 0 < θ < 1, and let Ω ⊆ RN be a connected W 1,p2 (Ω) extension domain satisfying (8.8). Then kukW s,p (Ω) Ω kukθW s1 ,p1 (Ω) kuk1−θ W 1,p2 (Ω) for all u ∈ W s1 ,p1 (Ω) ∩ W 1,p2 (Ω), where
1 p
=
θ p1
+
1−θ p2
and s = θs1 + 1 − θ.
Theorem 8.28. Let 1 < p1 , p2 < ∞, 0 ≤ s1 < s2 < 1, and 0 < θ < 1, and let Ω ⊆ RN be an open connected set satisfying (8.8). Then kukW s,p (Ω) Ω kukθW s1 ,p1 (Ω) kuk1−θ W s2 ,p2 (Ω) for all u ∈ W s1 ,p1 (Ω)∩W s2 ,p2 (Ω), where
1 p
=
θ 1−θ p1 + p2
and s = θs1 +(1−θ)s2 .
The main difference between the last two theorems and the ones in Section 7.7 is that here we are using full norms, while there only seminorms. We now show that in convex domains we can use seminorms. We recall that, given ν ∈ SN −1 = ∂B(0, 1), we denote by ν ⊥ the hyperplane orthogonal to ν,
8.4. Embeddings and Interpolation Inequalities
333
˙ 0,p (Ω) := Lp (Ω), that is, ν ⊥ := {x ∈ RN : x · ν = 0}. When s = 0, we set W and
uW 0,p (Ω) := kukLp (Ω) ,
(8.36)
uW 1,p (Ω) := k∇ukLp (Ω) .
We will prove the following theorem. For brevity, we write dΣ = dHN −1 . Theorem 8.29 (Gagliardo–Nirenberg). Let Ω ⊆ RN be an open convex set, 1 < p1 , p2 < ∞, 0 ≤ s1 < s2 ≤ 1, and 0 < θ < 1. Then uW s,p (Ω) uθW s1 ,p1 (Ω) u1−θ W s2 ,p2 (Ω) ˙ s1 ,p1 (Ω) ∩ W ˙ s2 ,p2 (Ω), where for all u ∈ W and we are using notation (8.36).
1 p
=
θ p1
+
1−θ p2 ,
s = θs1 + (1 − θ)s2 ,
Proof. Step 1: Assume 0 = s1 < s2 < 1. In view of the slicing theorem (Theorem 6.47), u ¯ admits a representative u ¯ such that u ¯(y + ·ν) ∈ s ,p N −1 N −1 N −1 ⊥ 2 2 ˙ W (Ω(y, ν)) for H a.e. ν ∈ S and H a.e. y ∈ ν for which Ω(y, ν) := {t ∈ R : y + tν ∈ Ω} 6= ∅, with upW2 s2 ,p2 (Ω) Z Z Z Z 1 ¯ u(y + tν) − u ¯(y + τ ν)p2 = dtdτ dΣ(y)dΣ(ν). 2 SN −1 ν ⊥ Ω(y,ν) Ω(y,ν) t − τ 1+s2 p2 Let ν ∈ SN −1 and y ∈ ν ⊥ be such that the function u ¯(y + ·ν) belongs to ˙ s2 ,p2 (Ω(y, ν)). Since Ω is convex, Ω(y, ν) is an open interval, Lp1 (Ω(y, ν))∩ W so, by Theorem 2.32,
Z Ω(y,ν)
¯ u(y + tν) − u ¯(y + τ ν)p dtdτ t − τ 1+sp Ω(y,ν)
Z
Z
Z
× Ω(y,ν)
Ω(y,ν)
Z
!θp/p1 ¯ u(y + tν)p1 dt
Ω(y,ν)
¯ u(y + tν) − u ¯(y + τ ν)p2 dtdτ t − τ 1+s2 p2
!(1−θ)p/p2 .
334
8. Further Properties
Integrating both sides in y over ν ⊥ and in ν over SN −1 , and using H¨older’s inequality, we obtain Z Z Z Z ¯ u(y + tν) − u ¯(y + τ ν)p dtdτ dΣ(y)dΣ(ν) t − τ 1+sp SN −1 ν ⊥ Ω(y,ν) Ω(y,ν) !θp/p1 Z Z Z p1 ¯ u(y + tν) dt SN −1
ν⊥
Z
Ω(y,ν)
Z
× Ω(y,ν)
Z
Ω(y,ν)
Z
Z
!(1−θ)p/p2 ¯ u(y + tν) − u ¯(y + τ ν)p2 dtdτ dΣ(y)dΣ(ν) t − τ 1+s2 p2 !θp/p1 ¯ u(y + tν)p1 dtdΣ(y)dΣ(ν)
SN −1
Z
Z
ν⊥
Ω(y,ν)
Z
Z
× SN −1 ν ⊥ Ω(y,ν)
!(1−θ)p/p2 ¯ u(y+tν)− u ¯(y+τ ν)p2 . dtdτ dΣ(y)dΣ(ν) t − τ 1+s2 p2 Ω(y,ν)
Using the slicing theorem (Theorem 6.47) once more, together with Tonelli’s theorem, we can write the previous inequality as (1−θ)p
upW s,p (Ω) kukθp Lp1 (Ω) uW s2 ,p2 (Ω) . Step 2: Assume 0 < s1 < s2 < 1. Then the proof follows as in Theorem 7.42. Step 3: Assume 0 = s1 < s2 = 1. By a slicing argument (Exercise 6.50) and Tonelli’s theorem, there exists a representative u ¯ of u such that, for N −1 N −1 ⊥ every ν ∈ S and for H a.e. y ∈ ν , the function u(y + ·ν) belongs to ˙ 1,p2 (Ω(y, ν)), with Lp1 (Ω(y, ν)) ∩ W Z Z (8.37) u(y + tν)p1 dtdΣ(y) = kukpL1p1 (Ω) , ν ⊥ Ω(y,ν) Z Z ∂ν u(y + tν)p2 dtdΣ(y) ≤ k∇ukpL2p2 (Ω) , ν⊥
Ω(y,ν)
where Ω(y, ν) := {t ∈ R : y + tν ∈ Ω}. Let ν ∈ SN −1 and y ∈ ν ⊥ be such ˙ s2 ,p2 (Ω(y, ν)). Since Ω is that u ¯(y + ·ν) ∈ Lp1 (Ω(y, ν)) and u ¯(y + ·ν) ∈ W convex, Ω(y, ν) is an open interval, so, by Theorem 2.32, !θp/p1 Z Z Z ¯ u(y + tν) − u ¯(y + τ ν)p dtdτ ¯ u(y + tν)p1 dt t − τ 1+sp Ω(y,ν) Ω(y,ν) Ω(y,ν) !(1−θ)p/p2 Z ∂ν u(y + tν)p2 dt
× Ω(y,ν)
.
8.5. Notes
335
Integrating both sides in y over ν ⊥ and in ν over SN −1 , and using H¨older’s inequality, we obtain Z Z Z Z ¯ u(y + tν) − u ¯(y + τ ν)p dtdτ dΣ(y)dΣ(ν) t − τ 1+sp SN −1 ν ⊥ Ω(y,ν) Ω(y,ν) !θp/p1 Z Z Z ¯ u(y + tν)p1 dt SN −1
ν⊥
Ω(y,ν)
!(1−θ)p/p2
Z
∂ν u(y + tν)p2 dt
×
dΣ(y)dΣ(ν)
Ω(y,ν)
Z
Z
!θp/p1
Z
¯ u(y + tν)p1 dtdΣ(y)dΣ(ν)
SN −1
ν⊥
Z
Ω(y,ν)
Z
Z
∂ν u(y + tν) dtdΣ(y)dΣ(ν)
× SN −1
ν⊥
!(1−θ)p/p2 p2
.
Ω(y,ν)
Using a slicing theorem (Theorem 6.47), together with (8.37) and the previous inequality, we obtain (1−θ)p
upW s,p (Ω) kukθp Lp1 (Ω) k∇ukLp2 (Ω) . Step 4: The case 0 < s1 < s2 = 1 follows similarly, using Theorems 2.32, Theorem 6.47, and Exercise 6.50. We leave the details as an exercise. Remark 8.30. If p1 = 1, s1 = 0, and s2 p2 < 1, then Theorem 8.29 continues to hold in view of Theorem 3.15. If p1 = 1 and 0 < s1 < s2 < 1, then Theorem 8.29 continues to hold since Theorem 7.42 is proved also for p1 = 1. On the other hand, if p1 = 1, s1 = 0, and 1 < p2 < ∞, s2 = 1 and sp < 1, then Theorem 8.29 continues to hold in view of Theorem 3.19.
8.5. Notes Theorem 8.6 is due to Zhou [Zho15]. Theorem 8.21 was proved by Lions and Magenes [LM61].
Chapter 9
Trace Theory It is impossible to be a mathematician without being a poet in soul. — Sofia Kovalevskaya
As explained in the preface, one of the reasons why fractional Sobolev spaces are so important is that when 1 < p < ∞ and Ω ⊂ RN is an open bounded set with smooth boundary, then the fractional Sobolev space W 1−1/p,p (∂Ω) is the trace space of the Sobolev space W 1,p (Ω), where the trace operator Tr : W 1,p (Ω) → Lploc (∂Ω) is the continuous extension of the operator Tr : W 1,p (Ω) ∩ C(Ω) → Lploc (∂Ω) defined by Tr(u) = u ¯∂Ω and where u ¯ is the continuous representative of u (see [Leo22d]). In this chapter we will show that if Ω ⊂ RN is an open set with Lipschitz continuous boundary and sp > 1, then functions in W s,p (Ω) admit a trace. We will then characterize the trace space Tr(W s,p (Ω)) in the case of uniformly Lipschitz continuous domains Ω.
9.1. Traces of Weighted Sobolev Spaces In this section we will show that fractional Sobolev spaces can be regarded as the trace spaces of a class of weighted Sobolev spaces. The results of this section will play a central role in the study of the fractional Laplacian in Chapter 14. Definition 9.1. Given an open set Ω ⊆ RN , α ∈ R, 1 ≤ p < ∞, we define 1,p the weighted Sobolev space Wα1,p (Ω) as the space of all functions u ∈ Wloc (Ω) 337
338
9. Trace Theory
such that Z kukWα1,p (Ω) : = +
(dist(x, ∂Ω))α u(x)p dx
Ω N X Z
1/p
α
p
1/p
(dist(x, ∂Ω)) ∂i u(x) dx
< ∞.
Ω
i=1
˙ α1,p (Ω) as the Similarly, we define the homogeneous weighted Sobolev space W 1,p space of all functions u ∈ Wloc (Ω) such that 1/p N Z X uWα1,p (Ω) := (dist(x, ∂Ω))α ∂i u(x)p dx < ∞. i=1
Ω
In what follows we will consider only the case in which Ω = RN + , so N −1 dist(x, ∂Ω) = xN . Given R > 0, we define QR := (−R/2, R/2) × (0, R). Theorem 9.2. Let α ∈ R, N ≥ 2, and 1 ≤ p < ∞, with α < p − 1, and 1,1 (Q ) for every R > 0, with u ∈ Wα1,p (RN R + ). Then u ∈ W Z 1/p Z α p u(x) dx R xN u(x) dx , QR
QR
Z
Z k∇u(x)k dx R QR
xαN k∇u(x)kp dx
1/p .
QR
In particular, u admits a representative u that is locally absolutely continuous on LN −1 a.e. line segments of RN + that are parallel to the coordinate axes, 0 with u ¯(x , ·) absolutely continuous in (0, R) for LN −1 a.e. x0 ∈ RN −1 and every R > 0. Moreover the firstorder (classical) partial derivatives of u ¯ N agree L a.e. with the weak derivatives of u. 1,1 (Q ) for every R > Proof. Let u ∈ Wα1,p (RN R + ). We claim that u ∈ W 1,p N 0. Since u ∈ Wloc (R+ ), it suffices to prove that u and ∇u are Lebesgue integrable over QR . H¨older’s inequality gives Z 1/p Z Z −α/p α/p α p u(x) dx = xN xN u(x) dx R xN u(x) dx , QR
QR
QR
where we use the facts that α < p − 1 and Z Z R R1−α/(p−1) −αp0 /p −α/(p−1) N −1 . (9.1) xN dx = R xN dxN = RN −1 1 − α/(p − 1) QR 0 Similarly, Z
Z k∇u(x)k dx R QR
QR
xαN k∇u(x)kp dx
1/p .
9.1. Traces of Weighted Sobolev Spaces
339
This proves the claim. In turn, u has a representative uR that is absolutely continuous on LN −1 a.e. line segments of QR that are parallel to the coordinate axes and the firstorder (classical) partial derivatives of uR agree LN a.e. in QR with the weak derivatives of u (see [Leo22d]). In particular, uR = uR+1 for LN a.e. in QR . Since countable unions of sets of Lebesgue measure zero have Lebesgue measure zero, by taking R = n, where n ∈ N, we can find a representative u that is locally absolutely continuous on LN −1 a.e. line segments of RN + that are parallel to the coordinate axes and whose firstorder (classical) partial derivatives agree LN a.e. with the weak derivatives of u. Moreover, u ¯(x0 , ·) is absolutely continuous in N −1 0 N −1 (0, n) for L a.e. x ∈ R and every n ∈ N. Corollary 9.3. Let α ∈ R, N ≥ 2, and 1 ≤ p < ∞, with α < p − 1, and ˙ α1,p (RN ). Then u ∈ W 1,1 (QR ) for every R > 0, with u∈W + Z 1/p Z α p (9.2) k∇u(x)k dx R xN k∇u(x)k dx . QR
QR
In particular, u admits a representative u that is locally absolutely continuous on LN −1 a.e. line segments of RN + that are parallel to the coordinate axes, 0 with u ¯(x , ·) absolutely continuous in (0, R) for LN −1 a.e. x0 ∈ RN −1 and every R > 0. Moreover the firstorder (classical) partial derivatives of u ¯ N agree L a.e. with the weak derivatives of u. Proof. We proceed as in the proof of Theorem 9.2 to show that (9.2) holds. ˙ 1,1 (QR ). Consider a cube TR b QR of sidelength This implies that u ∈ W 1,p N R/4. Since u ∈ Wloc (R+ ), u ∈ L1 (QR ). Hence, by Poincar´e’s inequality in ˙ 1,1 (QR ) (see [Leo22d]), W Z Z u(x) − uTR  dx R k∇u(x)k dx. QR
QR
W 1,1 (QR ).
This implies that u ∈ as in the proof of Theorem 9.2.
The second part of the statement follows
Theorem 9.4. Let α ∈ R, N ≥ 2, and 1 ≤ p < ∞, with α < p − 1. Then there exists a unique linear operator p N −1 Tr : Wα1,p (RN ) + ) → L (R
such that (i) Tr(u)(x0 ) = u(x0 , 0) for all x0 ∈ RN −1 and all u ∈ Wα1,p (RN +) ∩ C(RN −1 × [0, ∞)); (ii) the integration by parts formula Z Z Z (9.3) u∂i ψ dx = − ψ∂i u dx + RN +
RN +
RN −1
ψ(x0 , 0) Tr(u)(x0 )δi,N dx0
340
9. Trace Theory
1 N holds for all u ∈ Wα1,p (RN + ), ψ ∈ Cc (R ), and i = 1, . . . , N , where δi,j = 1 if i = j and δi,j = 0 otherwise;
(iii) for every R > 0, Z Z (9.4) xαN u(x)p dx  Tr(u)(x0 )p dx0 R N −1 N −1 R ×(0,R) R Z xαN k∇u(x)kp dx. + RN −1 ×(0,R)
The function Tr(u) is called the trace of u on ∂Ω. Proof. Since u ∈ W 1,1 (QR ) for every R > 0 by Theorem 9.2, it follows from standard trace theory (see [Leo22d]) that u restricted to QR admits a unique trace TrR (u) on (−R/2, R/2)N −1 , with Z Z Z (9.5)  Tr(u)(x0 ) dx0 R u(x) dx + k∇u(x)k dx. (−R/2,R/2)N −1
QR
QR
By uniqueness, we have that TrR (u)(x0 ) = TrR+1 (u)(x0 ) for LN −1 a.e. x0 ∈ (−R/2, R/2)N −1 . Hence, taking R = n and letting n → ∞, we obtain a N −1 ) such that item (i) holds. Item 1 linear operator Tr : Wα1,p (RN + ) → Lloc (R 1 (ii) follows from the fact that if ψ ∈ Cc (RN ), then by taking R > 0 so large that supp ψ ⊆ (−R/2, R/2)N , by standard trace theory (see [Leo22d]) we have that Z Z Z u∂i ψ dx + ψ∂i u dx − ψ(x0 , 0) Tr(u)(x0 )δi,N dx0 RN +
RN +
RN −1
Z
Z
=
u∂i ψ dx + QR
Z − (−R/2,R/2)N −1
ψ∂i u dx QR
ψ(x0 , 0) Tr(u)(x0 )δi,N dx0 = 0.
It remains to prove item (iii). By Theorem 9.2, u has a representative u such that u ¯(x0 , ·) is absolutely continuous in (0, R) for every R > 0 and for LN −1 a.e. x0 ∈ RN −1 . Then u ¯ admits a continuous extension to xN = 0 and by the fundamental theorem of calculus, we can write Z xN 0 0 (9.6) ¯ u(x , 0) ≤ ¯ u(x , xN ) + ∂N u(x0 , s) ds 0
for all 0 < xN < R. Averaging both sides in xN over (0, R) and integrating in x0 over (−R/2, R/2)N −1 , we obtain Z Z Z ¯ u(x0 , 0) dx0 R u dx + k∇uk dx. (−R/2,R/2)N −1
QR
QR
9.1. Traces of Weighted Sobolev Spaces
341
It follows from (9.5) and the uniqueness of the trace operator that Tr(u)(x0 ) = u ¯(x0 , 0) for LN −1 a.e. x0 ∈ (−R/2, R/2)N −1 . Since this holds for every R > 0, we have that Tr(u)(x0 ) = u ¯(x0 , 0)
(9.7)
for LN −1 a.e. x0 ∈ RN −1 .
Using this identity in (9.6), averaging both sides of (9.6) in xN over (0, R), and using H¨ older’s inequality gives Z R Z R ∂N u(x0 , s) ds u(x0 , xN )dxN +  Tr(u)(x0 ) ≤ R−1 0
0
=R
Z
−1
R
0
Z R
R
−α/p α/p xN xN u(x0 , xN )dxN
1/p
xαN u(x0 , xN )p dxN
Z
R
+
s−α/p sα/p ∂N u(x0 , s) ds
0
Z
R
0
α
p
1/p
s ∂N u(x , s) ds
+
0
,
0
where we use (9.1). Raising both sides to power p, integrating in x0 over RN −1 , and using Tonelli’s theorem gives Z Z RZ  Tr(u)(x0 )p dx0 R xαN u(x0 , xN )p dx0 dxN RN −1
RN −1 RZ
0
Z +
RN −1
0
xαN ∂N u(x0 , xN )p dx0 dxN .
Remark 9.5. Averaging in xN over (R/2, R) instead of (0, R) in the proof of Theorem 9.4, we have that if α ∈ R, N ≥ 2, and 1 ≤ p < ∞, with α < p − 1, then there exists a unique linear operator p N −1 ˙ α1,p (RN ) Tr : W + ) → L (R loc
such that items (i)–(iii) in the statement of the theorem hold with Wα1,p (RN +) 1,p N ˙ replaced by Wα (R+ ) in (i), (ii), and (iii) replaced by Z  Tr(u)(x0 )p dx0 (−T /2,T /2)N −1
R
−(1+α)
Z
R
Z (−T /2,T /2)N −1
+ Rp−1−α
R/2 Z RZ 0
RN −1
xαN u(x0 , xN )p dx0 dxN
xαN ∂N u(x0 , xN )p dx0 dxN
for every R, T > 0. N σ,p (RN −1 ). ˙ 1,p Next we show that Tr(W (1−σ)p−1 (R+ )) ⊆ W
Theorem 9.6 (Gagliardo). Let N ≥ 2, 1 < p < ∞, and 0 < σ < 1. Then N ˙ 1,p for all u ∈ W (1−σ)p−1 (R+ ), (9.8)
 Tr(u)W σ,p (RN −1 ) k∇ukLp
(1−σ)p−1
. (RN +)
342
9. Trace Theory
N ˙ 1,p Proof. Let u ∈ W (1−σ)p−1 (R+ ). By Corollary 9.3, u has a representative u that is locally absolutely continuous on LN −1 a.e. line segments of RN + that are parallel to the coordinate axes, with u ¯(x0 , ·) absolutely continuous in (0, R) for LN −1 a.e. x0 ∈ RN −1 and every R > 0. Moreover the firstorder (classical) partial derivatives of u ¯ agree LN a.e. with the weak derivatives of u. Also, Tr(u)(x0 ) = u ¯(x0 , 0) for LN −1 a.e. x0 ∈ RN −1 (see (9.7)). For x0 , 0 N −1 h ∈R , with h 6= 0, set r := kh0 kN −1 . Then
Z 1 r 0 0 u ¯(x , t) dt ¯ u(x + h , 0) − u ¯(x , 0) ≤ u ¯(x , 0) − r 0 Z r 1 0 0 0 u ¯(x , t) dt + u ¯(x + h , 0) − r 0 Z Z 1 r 1 r 0 0 ≤ ¯ u(x , 0) − u ¯(x , t) dt + ¯ u(x0 + h0 , t) − u ¯(x0 , t) dt r 0 r 0 Z 1 r + ¯ u(x0 + h0 , 0) − u ¯(x0 + h0 , t) dt. r 0 0
0
0
Hence, by Young’s inequality and the change of variables x0 + h0 = z 0 , so that dx0 = dz 0 , Z
¯ u(x0 + h0 , 0) − u ¯(x0 , 0)p
Z
RN −1
Z (9.9)
dx0 dh0 −1+σp kh0 kN N −1 p R 1 r u(x0 , 0) − u ¯(x0 , t) dt r 0 ¯
RN −1
Z
RN −1
−1+σp kh0 kN N −1
RN −1
Z
Z
+ RN −1
1 r
Rr 0
dx0 dh0
p ¯ u(x0 + h0 , t) − u ¯(x0 , t) dt −1+σp kh0 kN N −1
RN −1
dx0 dh0 =: A + B.
Let x0 ∈ RN −1 be such that u ¯(x0 , ·) is absolutely continuous in (0, r). By the fundamental theorem of calculus, 1 r
Z
r
¯ u(x0 , 0)−¯ u(x0 , t) dt ≤
0
1 r
Z 0
r
Z t Z ∂N u(x0 , ρ) dρ dt ≤ 0
r
k∇u(x0 , ρ)k dρ.
0
Using Minkowski’s inequality for integrals (see [Leo22c]), (9.10) Z r p Z r p Z Z 1 0 0 0 0 ¯ u(x , 0) − u ¯(x , t) dt dx ≤ k∇u(x , ρ)k dρ dx0 r N −1 N −1 0 R 0 R Z r p ≤ k∇u(·, ρ)kLp (RN −1 ) dρ . 0
9.1. Traces of Weighted Sobolev Spaces
343
Fix 1/p0 − σ < ε < 1/p0 . By H¨older’s inequality and the identity 1 = ρ−ε ρε , the righthand side of the previous inequality is bounded from above by Z r p/p0 Z r −εp0 ρεp k∇u(·, ρ)kpLp (RN −1 ) dρ ≤ ρ dρ 0 0 Z r p−1−εp r = ρεp k∇u(·, ρ)kpLp (RN −1 ) dρ. (1 − εp0 )p−1 0 Recalling that r = kh0 kN −1 , by (9.10), (9.9), and Tonelli’s theorem, we have (9.11) Z A Z
ρ
N +(ε−1+σ)p kh0 kN −1
RN −1
∞
Z
=
kh0 kN −1
Z
1
Z
0
RN −1
k∇u(x0 , ρ)kp dx0 dρdh0
Z
ρεp k∇u(x0 , ρ)kp
∞
Z
Z RN −1
1
RN −1 \BN −1 (0,ρ) kh0 k
RN −1 0
εp
N +(ε−1+σ)p N −1
! dh0
dρdx0
ρ(1−σ)p−1 k∇u(x0 , ρ)kp dρdx0 ,
0
where we use the facts that 1/p0 − σ < ε and Z 1 1 dh0 (ε−1+σ)p+1 . N +(ε−1+σ)p 0 N −1 ρ R \BN −1 (0,ρ) kh k N −1 To estimate B in (9.10), define X00 := x0 ,
0 Xn0 := Xn−1 + hn en ,
n = 1, . . . , N − 1.
Then 0
0
0
u ¯(x + h , t) − u ¯(x , t) =
N −1 X
0 u ¯(Xn0 , t) − u(Xn−1 , t).
n=1
x0 ,
h0 ,
By fixing and t in such a way that u ¯ is absolutely continuous along 0 0 the segments (ηXn−1 + (1 − η)Xn−1 , t), η ∈ [0, 1] for every n = 1, . . . , N − 1, by the fundamental theorem of calculus, we have N −1 Z 1 X 0 0 u ¯(x0 + h0 , t) − u ¯(x0 , t) = ∂n u(ηXn−1 + (1 − η)Xn−1 , t)hn dη. n=1
0
Hence, 1 r
Z
r
¯ u(x0 + h0 , t) − u ¯(x0 , t) dt
0
≤
N −1 Z r X n=1
0
Z 0
1
0 0 ∂n u(ηXn−1 + (1 − η)Xn−1 , t) dηdt.
344
9. Trace Theory
Using Minkowski’s inequality for integrals (see [Leo22c]) and the change of 0 0 variables ηXn−1 + (1 − η)Xn−1 = z 0 , we get Z r p 1 0 0 0 ¯ u(x + h , t) − u ¯(x , t) dt dx0 r N −1 R 0 Z r Z 1 p N −1 Z X 0 0 ∂n u(ηXn−1 + (1 − η)Xn−1 , t) dηdt dx0
Z
RN −1
n=1
N −1 X
Z 0
n=1
Z 0
r
r
0
0
1 Z
Z
RN −1
0
0 0 ∂n u(ηXn−1 + (1 − η)Xn−1 , t)p dx0
!p
1/p
dηdt
p k∇u(·, t)kLp (RN −1 ) dt .
Since the righthand side is the same as the righthand side in (9.10) we can now continue as in the estimate (9.11) of A to obtain Z
Z
∞
B RN −1
ρ(1−σ)p−1 k∇u(x0 , ρ)kp dρdx0 .
0
Combining this inequality with (9.9) and (9.11) and using the fact that Tr(u)(x0 ) = u ¯(x0 , 0) for LN −1 a.e. x0 ∈ RN −1 , we have Z RN −1
Z
 Tr(u)(x0 + h0 ) − Tr(u)(x0 )p
dx0 dh0
−1+σp kh0 kN N −1
RN −1
Z
(1−σ)p−1
RN +
xN
k∇u(x)kp dx.
N σ,p (RN −1 ). ˙ 1,p We now prove that Tr(W (1−σ)p−1 (R+ )) = W
Theorem 9.7 (Gagliardo). Let N ≥ 2, 1 < p < ∞, 0 < σ < 1, and N ˙ σ,p (RN −1 ). Then there exists a function v ∈ W ˙ 1,p g ∈ W (1−σ)p−1 (R+ ) such that Tr(v) = g and vW 1,p gW σ,p (RN −1 ) . (RN ) +
(1−σ)p−1
R Proof. Let ϕ ∈ Cc∞ (RN −1 ) be such that RN −1 ϕ(x0 ) dx0 = 1 and supp ϕ ⊆ BN −1 (0, 1). For x0 ∈ RN −1 and xN > 0 define (9.12)
0
v(x) := (ϕxN ∗ g)(x ) =
1 −1 xN N
Z ϕ RN −1
x0 − y 0 xN
g(y 0 ) dy 0 .
9.1. Traces of Weighted Sobolev Spaces
345
By standard properties of mollifiers (see [Leo22c]), where xN plays the role of ε, for any i = 1, . . . , N − 1 we have that Z ∂v 1 ∂ϕ x0 − y 0 (x) = N g(y 0 ) dy 0 ∂xi ∂x x N −1 xN R i N 0 Z 1 ∂ϕ x − y 0 = N [g(y 0 ) − g(x0 )] dy 0 , xN xN RN −1 ∂xi where in the second equality we use the fact that ! 0 Z ∂ ∂ 1 x − y0 0= (1) = dy 0 ϕ −1 ∂xi ∂xi xN x N −1 N R N 0 Z 0 1 ∂ϕ x − y = N dy 0 . ∂x x xN RN −1 i N Since supp ϕ ⊆ BN −1 (0, 1), we have that Z 1 (9.13) ∂i v(x) N g(y 0 ) − g(x0 ) dy 0 . xN BN −1 (x0 ,xN ) (1−σ)p−1
Raising both sides to the power p, multiplying by xN older’s inequality, we get x over RN + , and using H¨ Z (1−σ)p−1 xN ∂i v(x)p dx
, integrating in
RN +
(1−σ)p−1
Z
xN
p xN N
RN +
!p
Z
g(y 0 ) − g(x0 ) dy 0
(1−σ)p−1+(N −1)(p−1)
Z
xN
p xN N
RN +
0
RN −1
Z
Z
g(y 0 ) − g(x0 )p dy 0 dx =: A. (x0 ,x
BN −1
By Tonelli’s theorem, we get that Z ∞Z Z A= Z
BN −1
(x0 ,x
0
1
N +σp N ) xN 0 p
Z
N)
g(y 0 ) − g(x0 )p dy 0 dx0 dxN ∞
1
g(y ) − g(x )
= RN −1
Z
RN −1
Z
RN −1
dx
BN −1 (x0 ,xN )
RN −1
kx0 −y 0 kN −1
g(y 0 ) − g(x0 )p −1+σp kx0 − y 0 kN N −1
Hence, we have shown that Z Z (1−σ)p−1 (9.14) xN ∂i v(x)p dx RN +
for all i = 1, . . . , N − 1.
+σp xN N
dxN dy 0 dx0
dx0 dy 0 .
RN −1
Z RN −1
g(y 0 ) − g(x0 )p kx0
−
−1+σp y 0 kN N −1
dx0 dy 0
346
9. Trace Theory
Similarly, by differentiating under the integral sign (see [Leo22c]), we obtain that ! 0 Z ∂ 1 x − y0 ∂N v(x) = g(y 0 ) dy 0 ϕ N −1 ∂x x N −1 x N N R N ! 0 Z ∂ 1 x − y0 = [g(y 0 ) − g(x0 )] dy 0 , ϕ N −1 ∂x x N −1 xN N N R where in the second equality we use the fact that ! 0 Z ∂ ∂ 1 x − y0 0= dy 0 (1) = ϕ N −1 ∂xN ∂xN xN RN −1 xN 0 ! Z ∂ x − y0 1 = ϕ dy 0 . N −1 ∂x x N −1 x N N R N Since supp ϕ ⊆ BN −1 (0, 1), we have that 0 ! Z x − y0 ∂ 1 ϕ [g(y 0 ) − g(x0 )] dy 0 . ∂N v(x) = −1 x xN N BN −1 (x0 ,xN ) ∂xN N Now for y 0 ∈ BN −1 (x0 , xN ), 0 0 0 0 ϕ((x − y )/x )) = ∂N (x1−N −(N − 1)x−N N N ϕ((x − y )/xN ) N −
x1−N N
x−N N
N −1 X
∂i ϕ((x0 − y 0 )/xN )(xi − yi )x−2 N
i=1 0
x−N + kx − y 0 kN −1 x−1−N N . N
In turn, 1 ∂N v(x) N xN
Z BN −1
g(y 0 ) − g(x0 ) dy 0 . (x0 ,x
N)
We can now continue as before (see (9.13)) to conclude that Z Z Z g(y 0 ) − g(x0 )p (1−σ)p−1 (9.15) xN ∂N v(x)p dx dx0 dy 0 . N −1+σp 0 0 N N −1 N −1 kx − y kN −1 R+ R R N ˙ 1,p Since v ∈ C 1 (RN + ), we have shown that v ∈ W(1−σ)p−1 (R+ ). 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 W σ,p (RN −1 ) < ∞ such that g − gn W σ,p (RN −1 ) → 0. Let vn be defined as in (9.12) with g replaced by gn . Then vn ∈ C(RN −1 × [0, ∞)), with vn (x0 , 0) = gn (x0 ). By (9.15) 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 (9.4), we obtain that Tr(u) = g.
9.1. Traces of Weighted Sobolev Spaces
347
Corollary 9.8. Let N ≥ 2, 1 < p < ∞, 0 < σ < 1, and g ∈ W σ,p (RN −1 ). 1,p Then for every 0 < ε ≤ 1 there exists a function u ∈ W(1−σ)p−1 (RN + ) such that Tr(u) = g, supp u ⊆ RN −1 × [−ε, ε] and kukW 1,p
(1−σ)p−1
(RN +)
ε kgkW σ,p (RN −1 ) .
Proof. Let v be as in Theorem 9.7. Since v(·, xN ) = (ϕxN ∗ g)(·), by standard properties of mollifiers (see [Leo22c]), we have that for all xN > 0, Z Z g(x0 )p dx0 . v(x0 , xN )p dx0 ≤ (9.16) RN −1
RN −1
(1−σ)p−1
Multiplying both sides by xN and integrating in xN in (0, ε) gives Z εZ Z (1−σ)p−1 0 p 0 (1−σ)p (9.17) xN v(x , xN ) dx dxN ε g(x0 )p dx0 . 0
RN −1
RN −1
Let ψ ∈ C ∞ ([0, ∞)) be a decreasing function such that ψ = 1 in [0, ε/2], ψ(xN ) = 0 for xN ≥ ε and kψ 0 k∞ ≤ Cε−1 . For x = (x0 , xN ) ∈ RN + we define u(x) := ψ(xN )v(x). By (9.17), Tonelli’s theorem, and the fact that ψ(xN ) = 0 for xN ≥ ε, we have (9.18) Z RN +
(1−σ)p−1 xN u(x)p dx
Z = 0
Z
ε
(1−σ)p−1 xN (ψ(xN ))p
εZ
(1−σ)p−1
≤ 0
ε
RN −1
(1−σ)p
xN
Z
Z RN −1
v(x)p dx0 dxN
v(x)p dx0 dxN
g(x0 )p dx0 ,
RN −1
while for i = 1, . . . , N − 1, ∂i u(x) = ψ(xN )∂i v(x) ∂i v(x). In turn, by (9.14), we obtain that Z Z (1−σ)p−1 p xN ∂i u(x) dx RN +
RN −1
Z RN −1
g(y 0 ) − g(x0 )p −1+σp kx0 − y 0 kN N −1
for all i = 1, . . . , N − 1. On the other hand, ∂N u(x) = −ψ 0 (xN )v(x) + ψ(xN )∂N v(x),
dx0 dy 0
348
9. Trace Theory
and so, by (9.15), (9.17), and the facts that ψ 0 (xN ) = 0 for xN ≥ ε, and kψ 0 k∞ ε−1 , Z (1−σ)p−1 xN ∂N up dx RN +
ε
−p
Z
εZ RN −1
0
ε
−σp
Z
(1−σ)p−1 xN vp dx0 dxN 0
p
Z RN −1
+ RN +
(1−σ)p−1
xN
g(y 0 ) − g(x0 )p
Z
g dx + RN −1
Z
RN −1
−1+σp kx0 − y 0 kN N −1
∂N vp dx dx0 dy 0 .
Reasoning as in the last part of Theorem 9.7 we can show that Tr(u) = g.
The following corollary will play an important role in the study of the fractional Laplacian (see Chapter 14). ˙ σ,p (RN −1 ). Corollary 9.9. Let N ≥ 2, 1 < p < ∞, 0 < σ < 1, and g ∈ W Then N ˙ 1,p : u∈W gW σ,p (RN −1 ) ≈ inf{uW 1,p (1−σ)p−1 (R+ ), Tr(u) = g}. (RN ) (1−σ)p−1
+
Proof. Let ` := inf{uW 1,p
(1−σ)p−1
(RN +)
N ˙ 1,p : u∈W (1−σ)p−1 (R+ ), Tr(u) = g}.
N ˙ 1,p By Theorem 9.6, if u ∈ W (1−σ)p−1 (R+ ) is such that Tr(u) = g, then
gW σ,p (RN −1 ) uW 1,p
(1−σ)p−1
(RN +)
.
Take the infimum over all such u gives gW σ,p (RN −1 ) `. On the other hand, N ˙ 1,p by Theorem 9.7, there exists v ∈ W (1−σ)p−1 (R+ ), such that Tr(v) = g and vW 1,p gW σ,p (RN −1 ) . It follows that (RN ) (1−σ)p−1
+
` ≤ vW 1,p
(1−σ)p−1
(RN +)
gW σ,p (RN −1 ) .
Remark 9.10. In Chapter 14, we will see that when p = 2, then C1 gW σ,2 (RN −1 ) = inf{uW 1,2
N 1−2σ (R+ )
˙ 1,2 (RN : u∈W + ), Tr(u) = g}. 1−2σ
As a corollary of the results in this section, we are able to prove Theorem 8.19. Theorem 9.11 (Derivative). Let 1 < p < ∞ and 0 < s < 1. Then for every i = 1, . . . , N , the linear map ∂ : W s,p (RN ) → W s−1,p (RN ) ∂xi is continuous.
9.2. The Trace Operator in W s,p (Ω)
349
Proof. Define the distribution Z u(x)φ(x) dx, Tu (φ) =
φ ∈ D(RN ).
RN
Then
∂Tu (φ) = − ∂xi
Z u(x) RN
∂φ (x) dx, ∂xi
φ ∈ D(RN ). 0
1,p 1,p +1 N +1 By Corollary 9.8, we can find U ∈ W(1−s)p−1 (RN ) and Ψ ∈ Wsp ) 0 −1 (R+ + such that Tr(U ) = u, Tr(Ψ) = φ,
kU kW 1,p
(1−s)p−1
+1 ) (RN +
kukW s,p (RN ) ,
kΨkW 1,p0
+1 ) (RN + sp0 −1
kφkW 1−s,p0 (RN ) ,
and U = Ψ = 0 for xN +1 ≥ 1. Note that the function Ψ is defined as Ψ(x, xN +1 ) = ψ(xN +1 )(ϕxN +1 ∗ φ)(x), where ψ ∈ Cc∞ (RN +1 ).
Cc∞ (R),
ϕ ∈ Cc∞ (RN ). Since φ ∈ D(RN ), we have that Ψ ∈
Using integration by parts (see item (ii) in Theorem 9.4), Z ∂Tu ∂Ψ (φ) = − Tr(U )(x) (x, 0) dx ∂xi ∂xi RN Z ∂2Ψ ∂U ∂Ψ =− U + dxdxN +1 +1 ∂x ∂x ∂x ∂x i i N +1 N +1 RN + Z ∂U ∂Ψ ∂U ∂Ψ + dxdxN +1 . =− − +1 ∂xi ∂xN +1 ∂xN +1 ∂xi RN + 1−s−1/p s−1/p0
Writing 1 = xN +1 xN +1 , it follows from H¨older’s inequality that ∂Tu 1,p +1 kΨk 1,p0 N +1 ∂xi (φ) kU kW(1−s)p−1 (RN ) Wsp ) + 0 −1 (R+ kukW s,p (RN ) kφkW 1−s,p0 (RN ) . u This inequality implies that ∂T ∂xi can be uniquely extended to a continuous 0 1−s,p (RN ) → R. Hence, ∂Tu ∈ (W 1−s,p0 (RN ))0 = u linear function ∂T ∂xi : W ∂xi
∂Tu s−1,p N W (R ), with ∂xi s−1,p N kukW s,p (RN ) .
W
(R )
9.2. The Trace Operator in W s,p (Ω) We have seen in Theorem 6.78 that when sp ≤ 1, W0s,p (Ω) = W s,p (Ω), which implies in particular that we cannot have a continuous trace in this case, since any function in W s,p (Ω) can be approximated by a sequence of functions in Cc∞ (Ω). In the following exercises, we present an alternative argument. Exercise 9.12. Let 1 < p < ∞ and 0 < s < 1, with sp < 1.
350
9. Trace Theory
(i) Assume N = 1 and consider the sequence of functions un (x) = e−nx , x ∈ (0, 1). Prove that un → 0 in W s,p ((0, 1)) while un (0) = 1. (ii) Assume that N ≥ 2, and let vn (x) = e−nxN , x ∈ (0, 1)N . Prove that the trace of vn on (0, 1)N −1 × {0} is the constant function 1 but that vn → 0 in W s,p ((0, 1)N ). Exercise 9.13. Let 1 < p < ∞ and s = 1/p. (i) Prove that 1
 log rp dr < ∞. 2 0 r − 1 (ii) Assume N = 1 and consider the sequence of functions Z
un (x) = −
(9.19)
log(x + n−1 ) , log n
x ∈ (0, 1), n ≥ 2.
Then un (0) = 1. Prove that un → 0 in W 1/p,p ((0, 1)). (iii) Assume that N ≥ 2, and let vn (x) = un (xN ), x ∈ (0, 1)N , where un is defined in (9.19). Prove that the trace of vn on (0, 1)N −1 × {0} is the constant function 1, but that vn → 0 in W s,p ((0, 1)N ). We will now see that when sp > 1, we can define the trace of functions in W s,p (Ω). Theorem 9.14. Let Ω ⊆ RN , N ≥ 2, be an open set whose boundary ∂Ω is Lipschitz continuous, and let 1 < p < ∞ and 0 < s < 1, with sp > 1. Then there exists a unique linear operator Tr : W s,p (Ω) → Lploc (∂Ω) such that (i) Tr(u) = u on ∂Ω for all u ∈ W s,p (Ω) ∩ C(Ω), (ii) for every R ≥ 1, 0 < ε < 1, and u ∈ W s,p (Ω), (9.20) Z ∂Ω∩B(0,R)
 Tr(u)p dHN −1 Ω,R ε−1 kukpLp (Ω∩B(0,2R)) + εsp−1 kukpW s,p (Ω) .
The function Tr(u) is called the trace of u on ∂Ω. s,p (RN ) ∩ C(RN ). Proof. Step 1: Assume that Ω = RN + and let u ∈ W + + Write Z ε Z ε 1 1 u(x0 , 0) = u(x0 , t) dt − [u(x0 , t) − u(x0 , 0)] dt. ε 0 ε 0 Then Z ε Z ε u(x0 , 0) ≤ 1 u x0 , t dt + 1 u(x0 , t) − u(x0 , 0) dt ε 0 ε 0 Z Z ε 1 ε 1 dt 0 ≤ u x , t dt + 1−s u(x0 , t) − u(x0 , 0) s , ε 0 ε t 0
9.2. The Trace Operator in W s,p (Ω)
351
where we use the fact that 0 < t < ε. Raising both sides to power p and using H¨older’s inequality, we get (9.21) p−1 Z ε p−1 Z ε dt u(x0 , 0) p ε u x0 , t p dt + ε u(x0 , t) − u(x0 , 0)p sp p p(1−s) ε t ε Z ε Z ε0 Z ε 0 0 0 p u(x , t) − u(x , τ ) 1 u x0 , t p dt + εsp−1 dtdτ, ε 0 t − τ 1+sp 0 0 where we use Hardy’s inequality (Corollary 1.81). Integrate the previous inequality in x0 over RN −1 to obtain Z Z ∞ Z u x0 , t p dtdx0 u(x0 , 0) p dx0 1 ε RN −1 0 RN −1 Z Z εZ ε u(x0 , t) − u(x0 , τ )p dtdτ dx0 + εsp−1 1+sp t − τ  N −1 0 0 R Z Z Z u(x)−u(y)p 1 p sp−1 u (x) dx + ε dxdy, ε RN kx − ykN +sp RN RN + + + where we use Theorem 6.38. Step 2: Assume that (9.22)
Ω = {(x0 , xN ) ∈ RN −1 × R : xN > f (x0 )},
where f : RN −1 → R is Lipschitz continuous. Given u ∈ W s,p (Ω) ∩ C(Ω), define v(x) := u(x0 , xN + f (x0 )), x ∈ RN +. In view of Corollary 6.31, we can apply Step 1 to v to get Z Z u(x0 , f (x0 )) p dx0 1 u x0 , xN + f (x0 ) p dx ε RN RN −1 + Z Z 0 , x + f (x0 )) − u(y 0 , y + f (y 0 ))p u (x N N dxdy. + εsp−1 N +sp N N kx − yk R+ R+ Using Corollary 6.31, it follows that Z Z 1 1 p N −1 u dH up dx 1 + Lip f ∂Ω ε Ω + (1 + Lip f )sp+N εsp−1 upW s,p (Ω) . Step 3: We now consider an open set Ω ⊆ RN with Lipschitz continuous boundary. Let u ∈ W s,p (Ω) ∩ C(Ω). For every x0 ∈ ∂Ω there exist a rigid motion Tx0 : RN → RN with Tx0 (x0 ) = 0, a Lipschitz continuous function fx0 : RN −1 → R, with fx0 (0) = 0, and 0 < rx0 < 1/2 such that Tx0 (Ω ∩ B(x0 , 2rx0 )) = {(y 0 , yN ) ∈ B(0, 2rx0 ) : yN > fx0 (y 0 )}.
352
9. Trace Theory
The family {B(x, rx )}x∈∂Ω is an open cover of ∂Ω. Since ∂Ω ∩ B(0, R) is compact, there exist B1 , . . . , B` that cover ∂Ω ∩ B(0, R). Without loss of generality, we discard the balls that don’t intersect ∂Ω ∩ B(0, R). Let {ψn }`n=1 be a smooth partition of unity subordinated to B1 , . . . , B` . Fix n ∈ {1, . . . , `} and let xn ∈ ∂Ω be such Bn = B(xn , rn ). Then Tn (Ω ∩ B(xn , 2rn )) = {(y 0 , yN ) ∈ B(0, 2rn ) : yN > fn (y 0 )} =: Ωn , where fn : RN −1 → R is Lipschitz continuous and fn (0) = 0. Since ψn ∈ Cc∞ (RN ) and Tn is a rigid motion, the function (uψn ) ◦ Tn−1 is zero in Ωn \ B(0, rn ) and belongs to W s,p (Ωn ) by Theorem 6.23 and Exercise 6.30, with k(uψn )◦Tn−1 kW s,p (Un ) n kukW s,p (Ω) . Hence, by Lemma 6.71, the function u ˜n , obtained by extending (uψn ) ◦ Tn−1 to be zero in Un \ Ωn , where Un := Ωn ∪ (RN \ B(0, 2rn )), belongs to W s,p (Un ), with (9.23)
k˜ un kW s,p (Un ) n k(uψn ) ◦ Tn−1 kW s,p (Ωn ) n kukW s,p (Ω) .
By Step 2 applied to u ˜n , the previous inequality, and the fact that u ˜n = 0 outside B(0, rn ), we have Z Z (uψn ) ◦ Tn−1 p dx (uψn ) ◦ Tn−1 p dHN −1 n ε−1 ∂Ωn ∩B(0,rn )
+
Ωn ∩B(0,rn ) sp−1 ε ˜ un pW s,p (Un ) .
Since the surface measure HN −1 is invariant by rotations (see [Leo22c]), by changing variables and (9.23), we have Z uψn p dHN −1 n ε−1 kukpLp (Ω∩B(0,2R)) + εsp−1 kukpW s,p (Ω) . ∂Ω∩B(xn ,rn )
Note that we used the fact that each ball Bn intersects ∂Ω ∩ B(0, R), so that Bn ⊆ B(0, 2R), since rn ≤ 1/2 and 1 ≤ R. Using the fact that P u = `n=1 uψn on B(0, R) ∩ ∂Ω, we get !1/p !1/p Z Z ` X up dHN −1 ≤ uψn p dHN −1 . ∂Ω∩B(0,R)
n=1
∂Ω∩B(xn ,rn )
Combining the last two inequalities gives (9.20). Step 4: Assume that Ω ⊂ RN is an open set with Lipschitz continuous boundary and let u ∈ W s,p (Ω). By Theorem 6.70 there exists a sequence {un }n in W s,p (Ω) ∩ C ∞ (RN ) such that un → u in W s,p (Ω). By applying the previous step to un − uk we get Z un − uk p dHN −1 Ω,R ε−1 kun − uk kpLp (Ω∩B(0,2R)) ∂Ω∩B(0,R)
+ εsp−1 kun − uk kpW s,p (Ω)
9.2. The Trace Operator in W s,p (Ω)
353
for every R ≥ 1 and every 0 < ε < 1. It follows that {un }n is a Cauchy sequence in Lp (∂Ω ∩ B(0, R)), so un ∂Ω∩B(0,R) → vR in Lp (∂Ω ∩ B(0, R)) for some function vR . Applying (9.20) to un and letting n → ∞, we obtain that (9.20) holds with vR in place of Tr(u). Taking R = j → ∞, it follows that vj+1 = vj HN −1 a.e. in ∂Ω ∩ B(0, j), so we have constructed a function v ∈ Lploc (∂Ω). We define Tr(u) := v. Note that if {vn }n is another sequence in W s,p (Ω) ∩ C ∞ (RN ) such that vn → u in W s,p (Ω), then by applying (9.20) to un − vn , it follows that vn ∂Ω∩B(0,R) → Tr(u)∂Ω∩B(0,R) in Lp (∂Ω ∩ B(0, R)). Thus, the definition of Tr(u) is independent of the particular sequence. We leave it as an exercise to check that Tr is linear and unique. N ˙ s,p (RN Remark 9.15. If Ω = RN + ) ∩ C(R+ ), then reasoning as + and u ∈ W in Step 1, we can still obtain estimate (9.21). Integrating in x0 over a cube Q0R of sidelength R > 0, we get Z Z Z ε u(x0 , 0) p dx0 1 u x0 , t p dtdx0 ε Q0R 0 Q0R Z Z εZ ε u(x0 , t) − u(x0 , τ )p + εsp−1 dtdτ dx0 t − τ 1+sp Q0R 0 0 Z Z Z 1 u (x) − u(y)p p sp−1 u (x) dx + ε dxdy. N +sp ε Q0R ×(0,ε) Q0R ×(0,ε) Q0R ×(0,ε) kx − yk
˙ s,p (RN ), then by Poincar´e’s inequality (see Theorem 6.33), we If u ∈ W + have that u ∈ W s,p (Q0R × (0, ε)), so we may extend u outside Q0R × (0, ε) to obtain a function v ∈ W s,p (RN ). Using mollifiers (see Theorem 6.62), we can use approximate v in W s,p (RN ) with a sequence of functions in W s,p (RN ) ∩ C ∞ (RN ). Reasoning as in Step 3, we obtain that v restricted to 0 RN + has a trace. In turn, u has a trace in QR × (0, ε). By the arbitrariness of R, we obtain that u admits a trace Tr(u) ∈ Lploc (∂Ω). Corollary 9.16 (Compactness of traces). Let Ω ⊆ RN , N ≥ 2, be an open set whose boundary ∂Ω is Lipschitz continuous, let 1 < p < ∞, 0 < s < 1, with sp > 1, and let {un }n be a bounded sequence in W s,p (Ω). Then there exist a subsequence {unk }k of {un }n and a function u ∈ W s,p (Ω) such that unk → u in Lploc (Ω) and Tr(unk ) → Tr(u) in Lploc (∂Ω). Proof. Assume that Ω is bounded. Then by (9.20) we have Z Z p N −1 −1  Tr(v) dH Ω ε vp dx + εsp−1 kvkpW s,p (Ω) ∂Ω
Ω
W s,p (Ω)
for every v ∈ and for every 0 < ε < 1. By Theorem 6.13, there exist a subsequence {unk }k of {un }n and a function u ∈ W s,p (Ω) such that
354
9. Trace Theory
unk → u in Lp (Ω). Hence, Z Z unk − up dx + εsp−1 ,  Tr(unk ) − Tr(u)p dHN −1 Ω,R ε−1 ∂Ω
Ω
where we use the fact that {un }n is bounded in W s,p (Ω). Letting first k → ∞ and then ε → 0+ completes the proof. In the case in which Ω is unbounded, it suffices to apply what we just did to an increasing sequence of bounded sets Ωl with Lipschitz continuous boundary whose union gives Ω and then use a diagonal argument. We leave the details as an exercise. Next we show that if the domain Ω is sufficiently regular, we may characterize W0s,p (Ω) as the subspace of functions in W s,p (Ω) with trace zero (see Theorem 6.78 for the case sp ≤ 1). Exercise 9.17. Let Ω ⊂ RN be an open bounded set with Lipschitz continuous boundary, and let 1 ≤ p < ∞ and 0 < s < 1 be such that sp > 1. Prove that for all u ∈ W s,p (Ω) with Tr(u) = 0 on ∂Ω, Z Z Z Z u(x)p u(x) − u(y)p p dx u(x) dx + dxdy. Ω sp N +sp Ω (dist(x, ∂Ω)) Ω Ω Ω kx − yk Hint: Use Theorem 6.84. Theorem 9.18 (Traces and W01,p ). Let Ω ⊂ RN , N ≥ 2, be an open set whose boundary ∂Ω is Lipschitz continuous, let 1 < p < ∞, 0 < s < 1, with sp > 1, and let u ∈ W s,p (Ω). Then Tr(u) = 0 if and only if u ∈ W0s,p (Ω). Proof. If u ∈ Cc∞ (Ω), then Tr(u) = 0, so, since W0s,p (Ω) is the closure of Cc∞ (Ω) with respect to the norm k · kW s,p (Ω) , it follows from Corollary 9.16 that Tr(u) = 0 for all u ∈ W0s,p (Ω). To prove the converse implication, let u ∈ W s,p (Ω) be such that Tr(u) = 0. As in the proof of Theorem 6.66, we can assume that there exists R > 0 such that u = 0 in Ω \ B(0, R). By Exercise 9.17 applied in Ω ∩ B(0, R) and the fact that u = 0 in Ω \ B(0, R), we have that Z Z Z Z u(x)p u(x) − u(y)p p dx u(x) dx + dxdy. Ω,R sp N +sp Ω (dist(x, ∂Ω)) Ω Ω Ω kx − yk s,p Hence, by Theorem 6.97, u ∈ W00 (Ω). We can reason as in the proof of Theorem 6.105 (noting that the hypothesis that Ω is uniformly Lipschitz s,p continuous was only used to prove that u ∈ W00 (Ω)) to conclude that s,p u ∈ W0 (Ω).
9.3. HalfSpaces
355
9.3. HalfSpaces In this section we assume that Ω is the halfspace RN + and prove that when s−1/p,p (RN −1 ). Note that this sp > 1, Tr(W s,p (RN )) can be identified with W + is consistent with the case s = 1, where Tr(W 1,p (RN + )) can be identified with N −1 ×{0}, in this section we W 1−1/p,p (RN −1 ) (see [Leo22d]). Since ∂RN = R + should be working with spaces Lp (RN −1 × {0}), which I find inelegant. For this reason, with an abuse of notation, we identify Tr(u) with the function Tr(u)(·, 0). In view of this identification, we will be able to write Tr(u)(x0 ) and Tr(u) ∈ Lp (RN −1 ) rather than Tr(u)(x0 , 0) and Tr(u) ∈ Lp (RN −1 ×{0}). I hope this is not too confusing. s−1/p,p (RN −1 ). We begin by showing that Tr(W s,p (RN + )) ⊆ W
Theorem 9.19. Let N ≥ 2, 1 < p < ∞, and 0 < s < 1, with sp > 1. Then for all u ∈ W s,p (RN + ), k Tr(u)kLp (RN −1 ) kukW s,p (RN ) , +
 Tr(u)W s−1/p,p (RN −1 ) uW s,p (RN ) . +
N Proof. Step 1: Assume that u ∈ W s,p (RN + ) ∩ C(R+ ). By Step 1 of the proof of Theorem 9.14, for every ε > 0, Z Z Z Z u (x) − u(y)p p sp−1 u(x0 , 0) p dx0 1 u (x) dx + ε dxdy. ε RN kx − ykN +sp RN RN −1 RN + + +
We recall that for n = 1, . . . , N and τ > 0, ∆τ,n u(x) := u(x + τ en ) − u(x). Write Z Z 1 τ 1 τ u(x0 , 0) = u(x0 , t) dt − [∆t,N u(x0 , 0)] dt. τ 0 τ 0 Then Z τ Z τ u x0 , t dt + 1 u(x0 , 0) ≤ 1 ∆t,N u(x0 , 0) dt. τ 0 τ 0 Raising both sides to power p and using H¨ older’s inequality, we get Z τ p−1 p−1 Z τ u(x0 , 0) p τ u x0 , t p dt + τ ∆t,N u(x0 , 0)p dt. τp 0 τp 0 Replacing u with ∆τ,n u, where n = 1, . . . , N − 1, we obtain Z τ Z τ ∆τ,n u(x0 , 0) p 1 ∆τ,n u x0 , t p dt + 1 ∆τ,n ∆t,N u(x0 , 0)p dt. τ 0 τ 0 Dividing by τ sp and integrating in τ over R+ and in x0 over RN −1 , it follows that Z Z ∞ Z Z ∞Z τ 0 p dτ 0 ∆τ,n u x0 , t p dt dτ dx0 ∆τ,n u(x , 0) sp dx τ τ 1+sp RN −1 0 RN −1 0 0 Z Z ∞Z τ dτ + ∆τ,n ∆t,N u(x0 , 0)p dt 1+sp dx0 =: A + B. τ RN −1 0 0
356
9. Trace Theory
Since k∆τ,n vkLp (RN −1 ) ≤ 2kvkLp (RN −1 ) , using Tonelli’s theorem, we can estimate Z Z ∞Z τ dτ B ∆t,N u(x0 , 0)p dt 1+sp dx0 τ N −1 Z ∞ ZR Z0 ∞ 0 1 0 p ∆t,N u(x , 0) dτ dtdx0 1+sp τ t RN −1 0 Z Z ∞ dt u(x0 , t) − u(x0 , 0)p sp dx0 . t RN −1 0 We now apply Hardy’s inequality (see (1.47)) to the function u(x0 , ·) to conclude that Z Z ∞Z ∞ u(x0 , t) − u(x0 , τ )p B dtdτ dx0 1+sp t − τ  N −1 R 0 0 Z Z ∞Z ∞ u(x0 , t) − u(x0 , τ )p dtdτ dx0 1+sp (t − τ ) N −1 R 0 τ Z Z ∞Z ∞ u(x0 , t) − u(x0 , τ )p dτ dtdx0 + (τ − t)1+sp RN −1 0 t Z Z ∞Z ∞ ∆η,N u(x0 , t)p dtdηdx0 , 1+sp η N −1 R 0 0 where we use Tonelli’s theorem and a change of variables. In conclusion we have shown that Z Z ∞ Z Z ∞Z ∞ ∆τ,n u(x0 , 0)p ∆τ,n u (x0 , t)p 0 dτ dx dtdτ dx0 τ sp τ 1+sp RN −1 0 RN −1 0 0 Z Z ∞Z ∞ ∆η,N u(x0 , t)p + dtdηdx0 . 1+sp η N −1 R 0 0 The result now follows from Theorem 6.38. Step 2: Assume next that u ∈ W s,p (RN + ). By Theorem 6.70, there exists s,p N a sequence of functions un ∈ W (R+ ) ∩ C(RN + ) such that un → u in W s,p (RN ). By the previous step applied to u − uk , n + kun (·, 0) − uk (·, 0)kW s−1/p,p (RN −1 ) kun − uk kW s,p (RN ) . +
Hence, the sequence {un (·, 0)}n is a Cauchy sequence in W s−1/p,p (RN −1 ). By Theorem 6.6, there exists v ∈ W s−1/p,p (RN −1 ) such that un (·, 0) → v in W s−1/p,p (RN −1 ). By applying Step 1 to un , we obtain kun (·, 0)kW s−1/p,p (RN −1 ) kun kW s,p (RN ) . +
Letting n → ∞ gives kvkW s−1/p,p (RN −1 ) kukW s,p (RN ) . By the uniqueness of + the trace operator established in Theorem 9.14, necessarily, Tr(u) = v.
9.3. HalfSpaces
357
Exercise 9.20. Let N ≥ 2, 1 < p < ∞, and 0 < s < 1, with sp > 1. Prove ˙ s,p (RN ˙ s−1.p,p (RN −1 ), with that if u ∈ W + ), then Tr(u) ∈ W  Tr(u)W s−1/p,p (RN −1 ) uW s,p (RN ) . +
Hint: See Remark 9.15. s−1/p,p (RN −1 ). In what follows, Next, we prove that Tr(W s,p (RN + )) = W given u ∈ W s,p (RN ), by its trace we mean the trace of the restriction of u to RN +.
Theorem 9.21. Let N ≥ 2, 1 < p < ∞, 0 < s < 1, with sp > 1, and let g ∈ W s−1/p,p (RN −1 ). Then there exists a function u ∈ W s,p (RN ) such that Tr(u) = g in RN −1 and kukLp (RN ) kgkLp (RN −1 ) ,
uW s,p (RN ) kgkW s−1/p,p (RN −1 ) .
Proof. Step 1: Let ϕ ∈ RCc∞ (RN −1 ) be a nonnegative function such that supp ϕ ⊆ BN −1 (0, 1) and RN −1 ϕ(x0 ) dx0 = 1. For x0 ∈ RN −1 and xN > 0, define 0 Z 1 x − y0 0 (9.24) v(x) := (ϕxN ∗ g)(x ) = N −1 ϕ g(y 0 ) dy 0 . x N −1 xN N R Note that, by the change of variables (x0 − y 0 )/xN = w0 , we can write Z (9.25) v(x) = ϕ(w0 )g(x0 − xN w0 ) dw0 . RN −1
We claim that Z Z (9.26) RN +
∞
0
∆t,i v(x)p dtdx t1+sp
Z RN −1
Z RN −1
g(y 0 ) − g(z 0 )p ky 0
−
−2+sp z 0 kN N −1
dz 0 dy 0
for every i = 1, . . . , N − 1. To see this, we use Tonelli’s theorem and write Z Z ∞ Z ∞Z tZ ∆t,i v(x)p ∆t,i v(x)p 0 dtdx = dx dxN dt t1+sp t1+sp RN 0 0 0 RN −1 + Z ∞Z ∞Z ∆t,i v(x)p 0 + dx dxN dt =: A + B. t1+sp 0 t RN −1 By (9.25), Z
ϕ(w0 )∆t,i g(x0 − xN w0 ) dw0 . R It follows from H¨older’s inequality and the fact that RN −1 ϕ(x0 ) dx0 = 1 that Z p ∆t,i v(x) ≤ ϕ(w0 )∆t,i g(x0 − xN w0 )p dw0 . ∆t,i v(x) =
RN −1
RN −1
358
9. Trace Theory
Integrating in x0 over RN −1 , using Tonelli’s theorem and the change of variables x0 − xN w0 = z 0 , so that dx0 = dz 0 , gives Z
Z
Z
ϕ(w0 )∆t,i g(x0 − xN w0 )p dw0 dx0 ∆t,i v(x) dx ≤ N −1 N −1 R Z R Z Z 0 p 0 0 0 ∆t,i g(z ) dz = ∆t,i g(z 0 )p dz 0 . = ϕ(w ) dw 0
p
RN −1
RN −1
RN −1
RN −1
Hence, ∞Z tZ
Z
0 p
∆t,i g(z ) dz dxN
A≤ 0
RN −1
0
Z
dt
0
∞Z
=
t1+sp
RN −1
0
∆t,i g(z 0 )p dz 0
dt . tsp
To estimate B, we use the fundamental theorem of calculus and Fubini’s theorem to write Z (9.27)
∆t,i v(x) =
RN −1 Z tZ
−1 xN N
= RN −1
0
Since
R
RN −1
∆t,i v(x) =
1
∆t,i ϕ
1 ∂i ϕ xN N
x0 − y 0 xN
g(y 0 ) dy 0
x0 − y 0 + rei xN
g(y 0 ) dy 0 dr.
∂i ϕ(x0 ) dx0 = 0, we can write
1 xN
Z tZ 0
1
RN −1
∂ϕ N −1 i
xN
x0 − y 0 + rei xN
[g(y 0 ) − g(x0 + rei )] dy 0 dr.
By the change of variables w0 := (x0 − y 0 + rei )/xN , we have Z
1
RN −1
−1 xN N
0 Z 0 ∂i ϕ x − y + rei dy 0 = xN
RN −1
∂i ϕ(w0 ) dw0 .
Hence, if we write 1/p0 1/p 1 1 1 N −1 ∂i ϕ , N −1 ∂i ϕ = N −1 ∂i ϕ xN xN xN it follows from H¨older’s inequality that p
∆t,i v(x)
tp−1 −1 xp+N N
Z tZ 0
R
0 0 ∂i ϕ x − y +rei g(y 0 )−g(x0 +rei )p dy 0 dr. xN N −1
9.3. HalfSpaces
359
Integrating in x0 over RN −1 , using Tonelli’s theorem and the change of variables x0 + rei = z 0 gives Z tp−1 ∆t,i v(x)p dx0 p+N −1 xN RN −1 0 Z tZ Z x − y 0 + rei 0 0 p 0 0 × g(y ) − g(x + rei ) dx dy dr ∂i ϕ x N −1 N −1 N 0 R R 0 Z Z tp z − y 0 0 0 p 0 0 p+N −1 ∂i ϕ g(y ) − g(z ) dz dy . xN x RN −1 RN −1 N
Since supp ϕ ⊆ BN −1 (0, 1), we have that Z Z Z tp p 0 g(y 0 ) − g(z 0 )p dz 0 dy. ∆t,i v(x) dx p+N −1 N −1 0 N −1 xN R BN −1 (y ,xN ) R In turn, again by Tonelli’s theorem, Z ∞ Z ∞ Z Z 1 p(1−s)−1 B t g(y 0 )−g(z 0 )p dz 0 dy 0 dxN dt p+N −1 t xN 0 RN −1 BN −1 (y 0 ,xN ) Z xN Z Z Z ∞ 1 p(1−s)−1 t g(y 0 )−g(z 0 )p dz 0 dy 0 dtdxN p+N −1 N −1 0 x 0 R BN −1 (y ,xN ) 0 N ! Z Z Z ∞ p(1−s) xN dxN g(y 0 ) − g(z 0 )p dz 0 dy 0 p+N −1 N −1 N −1 0 0 R R ky −z kN −1 xN Z Z g(y 0 ) − g(z 0 )p 0 0 dz dy , N −2+sp RN −1 RN −1 ky 0 − z 0 kN −1 where Z xN
p(1−s)−1
t 0
dt
p(1−s) xN ,
Z
∞
ky 0 −z 0 kN −1
1 −1+sp xN N
dxN
1 −2+sp ky 0 − z 0 kN N −1
.
Combining the estimates for A and B and using Theorem 6.61 gives (9.26) when i = 1, . . . , N − 1. Step 2: We claim that Z Z ∞ Z Z ∆t,N v(x)p g(y 0 ) − g(z 0 )p 0 0 dtdx dz dy . N −2+sp t1+sp RN 0 RN −1 RN −1 ky 0 − z 0 kN −1 + Fix x0 ∈ RN −1 . By Theorem 1.28 applied to v(x0 , ·), we get Z ∞Z ∞ Z ∞ ∆t,N v(x0 , xN )p (1−s)p dtdxN xN ∂N v(x0 , xN )p dxN . 1+sp t 0 0 0 Integrating both sides in x0 over RN −1 and using Tonelli’s theorem gives Z Z ∞ Z ∆t,N v(x)p (1−s)p dtdx xN ∂N v(x)p dx. 1+sp N N t R+ 0 R+
360
9. Trace Theory
By (9.15), with σ = s − 1/p, we get Z Z (1−s)p xN ∂N v(x)p dx RN +
RN −1
g(y 0 ) − g(x0 )p
Z RN −1
kx0
−
−2+sp y 0 kN N −1
dx0 dy 0 .
The claim follows by combining the last two inequalities. ˙ s,p (RN Step 3: The function v in (9.25) belongs to W + ). We now modify v to s,p N obtain a function in W (R+ ) with the same trace. Let ψ ∈ Cc∞ ([0, ∞)) be such that ψ(t) = 1 for t ∈ [0, 1), ψ(t) = 0 for t ≥ 2, and 0 ≤ ψ ≤ 1. Define u(x) := ψ(xN )v(x). If i = 1, . . . , N − 1, then ∆t,i u(x) = ψ(xN )∆t,i v(x), so, since 0 ≤ ψ ≤ 1, by Step 1, Z Z ∞ Z Z ∞ ∆t,i u(x)p ∆t,i v(x)p dtdx ≤ dtdx t1+sp t1+sp N RN 0 ZR 0Z g(y 0 ) − g(x0 )p dx0 dy 0 . N −2+sp 0 0 N −1 N −1 ky − x k R R N −1 On the other hand, if i = N , u(x0 , xN + t) − u(x) = ψ(xN + t)(v(x0 , xN + t) − v(x)) + (ψ(xN + t) − ψ(xN ))v(x), so, ∆t,N u(x)p ∆t,N v(x)p + ψ(xN + t) − ψ(xN )p v(x)p . Since v(x) = (ϕxN ∗ g)(x0 ), by the properties of mollifiers (see [Leo22c]), Z Z p 0 v(x) dx ≤ g(x0 )p dx0 , RN −1
RN −1
so, by Theorem 1.22 applied to ψ, and Step 2, Z Z ∞ Z Z ∞ ∆t,N u(x)p ∆t,N v(x)p dtdx dtdx t1+sp t1+sp RN 0 RN 0 Z Z ∞ Z ∆t,N ψ(xN )p + dtdxN g(x0 )p dx0 t1+sp R 0 RN −1 Z Z Z g(y 0 ) − g(x0 )p 0 0 dx dy + g(x0 )p dx0 . N −2+sp 0 0 N −1 N −1 N −1 ky − x kN −1 R R R Using the same inequality, we also have Z ∞ Z Z p p u(x) dx ≤ (ψ(xN )) RN +
RN −1
0
Z ≤
∞ p
(ψ(xN )) dxN 0
v(x)p dx0 dxN
Z
g(x0 )p dx0 .
RN −1
Step 4: In this step we show that g is the trace Rof u. Assume first that g ∈ Cc∞ (RN −1 ). Since supp ϕ ⊆ BN −1 (0, 1) and RN −1 ϕ(x0 ) dx0 = 1, for
9.4. Special Lipschitz Domains
361
x00 ∈ RN −1 and 0 < xN < 1, by (9.25) and the fact that ψ(xN ) = 1, we can write Z ϕ(w0 )[g(x0 − xN w0 ) − g(x00 )] dw0 . u(x) − g(x00 ) = BN −1 (0,1)
Since g is uniformly continuous, given ε > 0 there exists δ > 0 such that g(y 0 ) − g(z 0 ) ≤ ε for all y 0 , z 0 ∈ RN −1 with ky 0 − z 0 kN −1 ≤ δ. Hence, by taking 0 < xN < min{1, δ/2} and kx0 − x00 kN −1 ≤ δ/2, we have that kx0 − xN w0 − x00 kN −1 ≤ δ, so, Z 0 ϕ(w0 ) dw0 = ε. u(x) − g(x0 ) ≤ ε BN −1 (0,1) N Since u ∈ C ∞ (RN + ), this shows that the function w : R+ → R, given by v(x) if xN > 0, w(x) := g(x0 ) if xN = 0,
is continuous. Thus, g is the trace of u. To remove the additional assumption that g ∈ Cc∞ (RN −1 ), given g ∈ by Theorem 6.66, we can find a sequence {gn }n of functions in Cc∞ (RN −1 ) such that gn → g in W s−1/p,p (RN −1 ). By the linearity of the mapping g 7→ u given in (9.24) and the previous steps, we have W s−1/p,p (RN −1 ),
kun − ukW s,p (RN ) kgn − gkW s−1/p,p (RN −1 ) → 0. +
By the continuity of the trace (see Theorem 9.14), it follows that g is the trace of u. Step 5: Finally, by reflecting u (see Exercise 6.32), we can extend u to RN . Remark 9.22. Steps 1 and 2 of the proof of Theorem 9.21 show that if ˙ s−1/p,p (RN −1 ), then N ≥ 2, 1 < p < ∞, 0 < s < 1, with sp > 1, and g ∈ W s,p N ˙ the function v defined in (9.24) belongs to W (R ), with vW s,p (RN ) gW s−1/p,p (RN −1 ) . We leave it as an exercise to prove that g is the trace of v.
9.4. Special Lipschitz Domains In this and in the following sections, we characterize the image of the trace operator Tr : W s,p (Ω) → Lploc (∂Ω), in the case in which Ω ⊂ RN is sufficiently regular. We will begin with the case of the supergraph of a Lipschitz continuous function, then consider bounded Lipschitz continuous domains, and finally (unbounded) uniformly Lipschitz domains. With this in mind, we need to define fractional Sobolev spaces on the boundary ∂Ω of an open set. More generally, we will define fractional Sobolev spaces on a manifold. In this section we consider the simple case in which the manifold can be parametrized by a single chart.
362
9. Trace Theory
Definition 9.23. Given an integer 1 ≤ k < N , a nonempty set M ⊂ RN is called a kdimensional Lipschitz continuous parametrized surface or manifold if there exist an open set V ⊆ Rk and a function ϕ : V → RN such that (i) ϕ(V ) = M , (ii) ϕ is injective, (iii) ϕ : V → M is biLipschitz continuous, that is, ϕ : V → M and ϕ−1 : M → V are Lipschitz continuous. The function ϕ is called a chart or a parametrization or a system of coordinates of M . If ψ : W → M is another chart, then the function ψ −1 ◦ ϕ : V → W is biLipschitz continuous, that is, ψ −1 ◦ ϕ and ϕ−1 ◦ ψ are both Lipschitz continuous. The function ψ −1 ◦ ϕ is called a change of parameters or a change of coordinates. Example 9.24. Given a Lipschitz continuous function f : RN −1 → R, the graph of f , M = {(x0 , f (x0 )) : x0 ∈ RN −1 }, is an (N − 1)dimensional Lipschitz continuous parametrized surface. To see this, take ϕ : RN −1 → RN to be ϕ(x0 ) = (x0 , f (x0 )). Then ϕ is injective, since the first N − 1 coordinates are. Moreover, for x0 , y 0 ∈ RN −1 , (9.28) 0
0
kx − y kN −1
q ≤ kϕ(x ) − ϕ(y )k = kx0 − y 0 k2N −1 + (f (x0 ) − f (y 0 ))2 p ≤ 1 + (Lip f )2 kx0 − y 0 kN −1 , 0
0
which shows that ϕ is Lipschitz continuous. Finally, ϕ−1 (z 0 , zN ) = z 0 for z ∈ M , which is Lipschitz continuous. Hence, ϕ : RN −1 → M is biLipschitz continuous. Definition 9.25. Given a kdimensional Lipschitz continuous parametrized surface M ⊂ RN , where 1 ≤ k < N , 1 ≤ p < ∞, and 0 < σ < 1, we say that a function v : M → R belongs to the fractional Sobolev space W σ,p (M ) if the function v ◦ ϕ belongs to W σ,p (V ). We set kvkLp (M ) := kv ◦ ϕkLp (V ) ,
vW σ,p (M ) := v ◦ ϕW σ,p (V ) ,
kvkW σ,p (M ) := kv ◦ ϕkW σ,p (V ) . Here ϕ : V → RN is a chart of M . Remark 9.26. If ψ : W → M is another chart, since the function ψ −1 ◦ ϕ : V → W is biLipschitz continuous, we can apply Theorem 6.29 to obtain
9.4. Special Lipschitz Domains
363
that v ◦ ψ ∈ W σ,p (W ), with C1 kv ◦ ψkW σ,p (W ) ≤ kv ◦ ϕkW σ,p (V ) ≤ C2 kv ◦ ψkW σ,p (W ) , where the constants C1 and C2 depend on Lip ψ and Lip ϕ. Hence, if we change charts, we obtain an equivalent norm, but the space W σ,p (M ) is independent of the particular chart. The following theorem provides an intrinsic norm for W σ,p (M ). We will not use this result to characterize the trace space of W s,p (Ω) below. Theorem 9.27. Let M ⊂ RN be a kdimensional Lipschitz continuous parametrized surface, where 1 ≤ k < N , 1 ≤ p < ∞, and 0 < σ < 1. Then a Borel function v : M → R belongs to W σ,p (M ) if and only if Z 1/p ∗ p k kvkW σ,p (M ) : = v(x) dH (x) M
Z
Z
+ M
Moreover, k ·
k∗W σ,p (M )
M
1/p v(x) − v (y) p k k dH (x)dH (y) < ∞. kx − ykk+σp
is an equivalent norm.
Lemma 9.28. Let 1 ≤ k ≤ N , V ⊆ Rk be an open set, and let ϕ : V → RN be an injective function. Assume that ϕ : V → ϕ(V ) is biLipschitz continuous. Then for all y1 , y2 ∈ V , (9.29)
(Lip ϕ−1 )−1 ky1 − y2 kk ≤ kϕ(y1 ) − ϕ(y2 )k ≤ Lip ϕky1 − y2 kk ,
and for Lk a.e. y ∈ V , (9.30)
(Lip ϕ−1 )−k ≤ det(Jϕt (y)Jϕ (y)) ≤ (Lip ϕ)k .
Proof. Since ϕ : V → ϕ(V ) is biLipschitz continuous, there exist L1 , L2 > 0 such that kϕ(y1 ) − ϕ(y2 )k ≤ L1 ky1 − y2 kk , −1
kϕ
(x1 ) − ϕ−1 (x2 )kk ≤ L2 kx1 − x2 k
for all y1 , y2 ∈ V and all x1 , x2 ∈ ϕ−1 (V ). It follows that L−1 2 ky1 − y2 kk ≤ kϕ(y1 ) − ϕ(y2 )k ≤ L1 ky1 − y2 kk for all y1 , y2 ∈ V . By Rademacher’s theorem ([Leo17, Theorem 9.14]), ϕ is differentiable for Lk a.e. y ∈ V . Fix any such y ∈ V and consider a unit vector v ∈ Rk . Taking y1 = y + tv and y2 = y in the previous inequality and letting t → 0 gives L−1 2 ≤ k∂v ϕ(y)k ≤ L1 . In turn, (Jϕt (y)Jϕ (y))v · v = (Jϕ (y)v) · (Jϕ (y)v) 2 = ∂v ϕ(y) · ∂v ϕ(y) ∈ [L−2 2 , L1 ].
364
9. Trace Theory
Since Jϕt (y)Jϕ (y) is a symmetric matrix, the previous inequality implies that its smallest eigenvalue λ1 is bounded from below by L−1 2 and its largest eigenvalue λk is bounded from above by L1 . In turn, k t k k L−k 2 ≤ λ1 ≤ det(Jϕ (y)Jϕ (y)) ≤ λk ≤ L1 .
We turn to the proof of Theorem 9.27. Proof. Let M = ϕ(V ), where ϕ : V → RN is injective and biLipschitz continuous. By the area formula for Lipschitz functions ([Leo17, Theorem 9.49 and Exercise 9.50]), Z Z p k v(x) dH (x) = (v ◦ ϕ)(z)p Jϕ (z) dz, M
det(Jϕt (z)Jϕ (z)). Hence, by Lemma 9.28, Z Z −1 −k/2 p (Lip ϕ ) (v ◦ ϕ)(z) dz ≤ v(x)p dHk (x) V M Z ≤ (Lip ϕ)k/2 (v ◦ ϕ)(z)p dz.
where Jϕ (z) = (9.31)
V
q
V
Again by the area formula, Z Z v(x) − v (y) p dHk (x)dHk (y) k+σp kx − yk M M Z Z (v ◦ ϕ)(z) − (v ◦ ϕ)(w)p Jϕ (z) Jϕ (w) dzdw. = kϕ(z) − ϕ(w)kk+σp V V Since (Lip ϕ−1 )−1 kz − wkk ≤ kϕ(z) − ϕ(w)k ≤ Lip ϕkz − wkk for all z, w ∈ V , also by (9.30), we have (9.32) Z Z (Lip ϕ−1 )−k (v ◦ ϕ)(z) − (v ◦ ϕ)(w)p dzdw (Lip ϕ)k+σp V V kz − wkk+σp k Z Z v(x) − v (y) p ≤ dHk (x)dHk (y) k+σp kx − yk M M Z Z (Lip ϕ)k (v ◦ ϕ)(z) − (v ◦ ϕ)(w)p ≤ dzdw. (Lip ϕ−1 )−k−σp V V kz − wkk+σp k We now characterize the trace operator in the case in which Ω is the supergraph of a Lipschitz continuous function. In what follows, given u ∈ W s,p (RN ), by its trace on ∂Ω we mean the trace of the restriction of u to Ω.
9.4. Special Lipschitz Domains
365
Theorem 9.29. Let f : RN −1 → R be Lipschitz continuous, N ≥ 2, Ω := {(x0 , xN ) ∈ RN −1 × R : xN > f (x0 )},
(9.33)
1 < p < ∞, and 0 < s < 1, with sp > 1. Then for all u ∈ W s,p (Ω), k Tr(u)kLp (∂Ω) kukLp (Ω) + (1 + Lip f )s+N/p uW s,p (Ω) ,  Tr(u)W s−1/p,p (∂Ω) (1 + Lip f )s+N/p uW s,p (Ω) . Conversely, for every g ∈ W s−1/p,p (∂Ω) there is a function u ∈ W s,p (RN ) such that Tr(u) = g on ∂Ω and uW s,p (RN ) (1 + Lip f )s+N/p kgkW s−1/p,p (∂Ω) .
kukLp (RN ) kgkLp (∂Ω) ,
Proof. Set L := Lip f and ϕ(x0 ) := (x0 , f (x0 )), x0 ∈ RN −1 . Step 1: If u ∈ W s,p (Ω), by Corollary 6.31 the function v : RN + → R, given s,p (RN ), with by v(y) := u(y 0 , yN + f (y 0 )), y ∈ RN , belongs to W + + kvkLp (RN ) = kukLp (Ω) , +
vW s,p (RN ) (1 + L)s+N/p uW s,p (Ω) . +
Since Tr(u) ◦ ϕ = Tr(v), it follows from Theorem 9.19 and Definition 9.25 that  Tr(u)W s−1/p,p (∂Ω) =  Tr(u) ◦ ϕW s−1/p,p (RN −1 ) =  Tr(v)W s−1/p,p (RN −1 ) vW s,p (RN ) (1 + L)s+N/p uW s,p (Ω) +
and that (see Theorem 6.29 and Corollary 6.31), Z Z k Tr(u)kpLp (∂Ω) = (Tr(u) ◦ ϕ)(y 0 )p dy 0 =  Tr(v)(y 0 )p dy 0 N −1 N −1 R ZR p p v (y) dy + vW s,p (RN ) RN +
Z Ω
+
u(x)p dx + (1 + L)sp+N upW s,p (Ω) .
Step 2: Conversely, given g ∈ W s−1/p,p (∂Ω), by Definition 9.25, g ◦ ϕ ∈ W s−1/p,p (RN −1 ), so, by Theorem 9.21, there exists a function w ∈ W s,p (RN ) such that Tr(w) = g ◦ ϕ in RN −1 and kwkLp (RN ) kg ◦ ϕkLp (RN −1 ) ,
wW s,p (RN ) kg ◦ ϕkW s−1/p,p (RN −1 ) .
Setting u(x) := w(x0 , xN − f (x0 )), x ∈ RN , by Theorem 6.29 (see also the proof of Corollary 6.31), we have that u ∈ W s,p (RN ), with kukLp (RN ) = kwkLp (RN ) kg ◦ ϕkLp (RN −1 ) , uW s,p (RN ) (1 + L)s+N/p wW s,p (RN ) (1 + L)s+N/p kg ◦ ϕkW s−1/p,p (RN −1 ) .
366
9. Trace Theory
Exercise 9.30. Let f : RN −1 → R be Lipschitz continuous, with N ≥ 2, and let Ω be as in (9.33), 1 < p < ∞, 0 < σ < 1, with σp < N − 1, and ] −1)p v ∈ W σ,p (∂Ω). Prove that v ∈ Lpσ (∂Ω), where p]σ := N(N−1−σp is the Sobolev N −1 critical exponent in R , with kvk p]σ f kvkW σ,p (∂Ω) . L
(∂Ω)
RN −1
Exercise 9.31. Let f : → R be Lipschitz continuous, with N ≥ 2, and let Ω be as in (9.33), 1 < p < ∞, and v ∈ W (N −1)/p,p (∂Ω). Prove that v ∈ Lq (∂Ω) for every p ≤ q < ∞, with kvkLq (∂Ω) f kvkW (N −1)/p,p (∂Ω) . Exercise 9.32. Let f : RN −1 → R be Lipschitz continuous, with N ≥ 2, and let Ω be as in (9.33), 1 < p < ∞, 0 < σ < 1, with σp > N − 1, and v ∈ W σ,p (∂Ω). Prove that v admits a representative v¯ that is H¨older continuous with exponent α = σ − (N − 1)/p, with k¯ v kC 0,α (∂Ω) f kvkW σ,p (∂Ω) .
9.5. Bounded Lipschitz Domains In this section we characterize Tr(W s,p (Ω)) when Ω is an open bounded domain with Lipschitz continuous boundary. Note that all the results of this section are a subcase of the ones in the following section, for which we treat uniformly Lipschitz continuous boundaries. However, the proofs in that case are significantly more involved, so we present them first in this friendlier case. Definition 9.33. Given an integer 1 ≤ k < N , a nonempty set M ⊂ RN is called a kdimensional Lipschitz continuous surface or manifold if for every x0 ∈ M there exist an open set U containing x0 , an open set V ⊆ Rk , and a function ϕ : V → RN such that (i) ϕ(V ) = M ∩ U , (ii) ϕ is injective, (iii) ϕ : V → M ∩ U is biLipschitz continuous, that is, ϕ : V → M ∩ U and ϕ−1 : M ∩ U → V are Lipschitz continuous. The function ϕ is called a local chart or a local parametrization or a local system of coordinates of M . An atlas for S M is a family of local charts ϕα : Vα → RN , α ∈ Λ, with the property that α∈Λ ϕα (Vα ) = M . Remark 9.34. Given an atlas {ϕα }α∈Λ , for every α ∈ Λ, let Uα ⊆ RN be an open set such that ϕα (Vα ) = M ∩ Uα . If x ∈ M ∩ Uα , we can find yx ∈ Vα such that ϕα (yx ) = x. Since Vα is open, there exists Bk (yx , rx ) ⊆ Vα . Hence, if we replace each local chart ϕα with the restriction ϕα Bk (yx ,rx ) , x ∈ M ∩ Uα , we obtain a new atlas {ωβ }β∈∆ , where the domain of each ωβ is a ball. Moreover, since a ball Bk (y0 , r) in Rk is diffeomorphic to Rk via the diffeomorphism φ(y) = √ 2 y−y0 2 , we can construct an atlas, where r −ky−y0 kk
each local chart is defined in the entire space Rk .
9.5. Bounded Lipschitz Domains
367
Proposition 9.35. Let Ω ⊂ RN be an open set with a Lipschitz continuous boundary. Then ∂Ω is an (N − 1)dimensional Lipschitz continuous surface. Proof. For every x0 ∈ ∂Ω there exist a rigid motion Tx0 : RN → RN with Tx0 (x0 ) = 0, a Lipschitz continuous function fx0 : RN −1 → R with fx0 (0) = 0, and rx0 > 0 such that Tx0 (Ω ∩ B(x0 , rx0 )) = {(y 0 , yN ) ∈ B(0, rx0 ) : yN > fx0 (y 0 )}. ((y 0 , fx0 (y 0 ))), y 0 ∈ RN −1 . By (9.28) and the fact Define ϕx0 (y 0 ) = Tx−1 0 that Tx0 is a rigid motion, we have that ϕx0 : RN −1 → ϕx0 (RN −1 ) is biLipschitz continuous. Hence, if we define the open sets Ux0 := B(x0 , rx0 ) N −1 , then items (i)–(iii) of Definition 9.33 and Vx0 := ϕ−1 x0 (B(x0 , rx0 )) ⊆ R are satisfied. Definition 9.36. Given a bounded kdimensional Lipschitz continuous surface M ⊂ RN , where 1 ≤ k < N , 1 ≤ p < ∞, and 0 < σ < 1, let {ϕn }`n=1 be an atlas, with ϕn : Vn → RN , Un ⊆ RN open sets such that ϕn (Vn ) = M ∩ Un , and {ψn }`n=1 a smooth partition of unity subordinated to {Un }`n=1 , with supp ψn ⊂ Un . We say that a function v : M → R belongs to the fractional Sobolev space W σ,p (M ) if the function (ψn v) ◦ ϕn belongs to W σ,p (Vn ) for every n = 1, . . . , `, and we set !1/p ` X p kvkLp (M ) := k(ψn v) ◦ ϕn kLp (Vn ) , n=1
vW σ,p (M ) :=
` X
!1/p (ψn v) ◦
ϕn pW σ,p (Vn )
,
n=1
kvkW σ,p (M ) := kvkLp (M ) + vW σ,p (M ) . Exercise 9.37. Prove that the space W σ,p (M ) in Definition 9.36 does not depend on the particular atlas and partition of unity. Hint: Use Theorem 6.29. The following theorem provides an intrinsic norm for W σ,p (M ). We will not use this result to characterize the trace space of W s,p (Ω) below. Theorem 9.38. Let M ⊂ RN be a bounded kdimensional Lipschitz continuous surface, where 1 ≤ k < N , 1 ≤ p < ∞, and 0 < σ < 1. Then a Borel function v : M → R belongs to W σ,p (M ) if and only if 1/p Z (9.34) kvk∗W σ,p (M ) : = v(x)p dHk (x) M
Z
Z
+ M
M
v(x) − v (y) p dHk (x)dHk (y) kx − ykk+σp
1/p < ∞.
368
9. Trace Theory
Moreover, k · k∗W σ,p (M ) is an equivalent norm. Proof. Step 1: For every x ∈ M there exist an open set Ux containing x and a local chart ϕx : Wx → RN , where Wx ⊆ Rk is open and ϕx (Wx ) = M ∩ Ux . Let rx > 0 be such that B(x, 2rx ) ⊆ Ux and consider Vx := ϕ−1 x (B(x, 2rx )). Since M is compact, there exist x1 , . . . , x` ∈ M such that M⊆
` [
B(xn , rxn ).
n=1
Write ϕn := ϕxn , rn := rxn , Vn := Vxn , and Ln := Lip ϕn . Consider a smooth partition of unity {ψn }`n=1 subordinated to {B(xn , rn )}`n=1 , with supp ψn ⊂ B(xn , rn ). Writing v(x) − v (y) =
` X
[(ψn v)(x) − (ψn v)(y)],
n=1
we have v(x) − v (y) p dHk (x)dHk (y) k+σp kx − yk M M ` Z Z ` X X (ψn v)(x) − (ψn v) (y) p k k ` dH (x)dH (y) =: An . kx − ykk+σp M M
Z (9.35)
Z
n=1
n=1
Since supp ψn ⊂ B(xn , rn ), if x, y ∈ / B(xn , rn ), then (ψn v)(x) = (ψn v)(y) = 0. Hence, by interchanging x and y if needed, we can write (9.36) (ψn v)(x) − (ψn v) (y) p dHk (y)dHk (x) kx − ykk+σp M ∩B(xn ,2rn ) M ∩B(xn ,2rn ) Z Z dHk (y) p +2 (ψn v)(x) dHk (x) k+σp M ∩B(xn ,rn ) M \B(xn ,2rn ) kx − yk Z
An =
Z
=: A + B. By the area formula ([Leo17, Theorem 9.49 and Exercise 9.50]), Z Z ((ψn v) ◦ ϕn )(z) − ((ψn v) ◦ ϕn )(w)p A= kϕn (z) − ϕn (w)kk+σp Vn Vn × Jϕn (z) Jϕn (w) dzdw, q where Jϕn (z) = det(Jϕt n (z)Jϕn (z)) and Jϕt n (z) is the transpose of Jϕn (z). As in the proof of Theorem 9.27 (see (9.32)), we conclude that (9.37)
A n (ψn v) ◦ ϕn pW σ,p (Vn ) .
9.5. Bounded Lipschitz Domains
369
To estimate B, observe that if x ∈ M ∩ B(xn , rn ) and y ∈ M \ B(xn , 2rn ), then kx − yk ≥ rn . Hence, Z dHk (y) rn−k−σp Hk (M ). k+σp M \B(xn ,2rn ) kx − yk Thus, again by the area formula and Lemma 9.28, Z −k−σp (ψn v)(x)p dHk (x) B rn M ∩B(xn ,rn ) Z −k−σp ((ψn v) ◦ ϕn )(z)p Jϕn (z) dz rn Vn
n k(ψn v) ◦ ϕn kpLp (Vn ) . Combining this last inequality with (9.35), (9.36), and (9.37) gives Z M
Z M
` X v(x) − v (y) p k k dH (x)dH (y) M k(ψn v) ◦ ϕn kpW σ,p (Vn ) k+σp kx − yk
=
n=1 kvkpW σ,p (M ) .
A simpler argument, which we leave as an exercise, shows that Z
p
k
v(x) dH (x) M M
` X
k(ψn v) ◦ ϕn kpLp (Vn ) = kvkpLp (M ) .
n=1
Step 2: As in the proof of Theorem 9.27 (see (9.32)), Z Z ((ψn v) ◦ ϕn )(z) − ((ψn v) ◦ ϕn )(w)p p (ψn v) ◦ ϕn W σ,p (Vn ) n kϕn (z) − ϕn (w)kk+σp Vn Vn × Jϕn (x0 ) Jϕn (y 0 ) dx0 dy 0 Z Z (ψn v)(x) − (ψn v) (y) p ≤ dHk (y)dHk (x) kx − ykk+σp M M n (kvk∗W σ,p (M ) )p , where the last inequality follows as in the proof of Theorem 6.23. Summing over n gives vW σ,p (M ) M kvk∗W σ,p (M ) . A simpler argument shows that Z kvkLp (M ) M
1/p v(x) dH (x) . p
k
M
We now characterize the trace operator of W s,p (Ω). In what follows, given u ∈ W s,p (RN ), by its trace on ∂Ω we mean the trace of the restriction of u to Ω.
370
9. Trace Theory
Theorem 9.39. Let Ω ⊂ RN be an open set with bounded Lipschitz continuous boundary, with N ≥ 2, and let 1 < p < ∞ and 0 < s < 1, with sp > 1. Then for all u ∈ W s,p (Ω), k Tr(u)kW s−1/p,p (∂Ω) Ω kukW s,p (Ω) . Conversely, for every g ∈ W s−1/p,p (∂Ω), there is a function u ∈ W s,p (RN ) such that Tr(u) = g on ∂Ω and kukW s,p (RN ) Ω kgkW s−1/p,p (∂Ω) . Proof. Step 1: Let u ∈ W s,p (Ω). For every x ∈ ∂Ω there exist a rigid motion Tx : RN → RN with Tx (x) = 0, a Lipschitz continuous function fx : RN −1 → R with fx (0) = 0, and rx > 0 such that Tx (∂Ω ∩ B(x, 2rx )) = {(y 0 , yN ) ∈ B(0, 2rx ) : yN = fx (y 0 )} =: Wx . As in Proposition 9.35, we consider ϕx (y 0 ) = Tx−1 ((y 0 , fx (y 0 ))), y 0 ∈ RN −1 , and we define Vx := ϕ−1 x (B(x, 2rx )). Since ∂Ω is compact, there exist x1 , . . . , x` ∈ ∂Ω such that ∂Ω ⊆
` [
B(xn , rxn ).
n=1
Write fn := fxn , ϕn := ϕxn , rn := rxn , Tn := Txn , Vn := Vxn , and Wn := Wxn . Consider a smooth partition of unity {ψn }`n=1 subordinated to {B(xn , rn )}`n=1 , with supp ψn ⊂ B(xn , rn ). Fix n ∈ {1, . . . , `}. By Theorem 6.23, ψn u ∈ W s,p (Ω), so, by Theorem 6.29, (ψn u)◦Tn−1 ∈ W s,p (Tn (Ω)), with k(ψn u) ◦ Tn−1 kW s,p (Tn (Ω)) n kψn ukW s,p (Ω) . Define (9.38)
Ωn := {(y 0 , yN ) ∈ RN −1 × R : yN > fn (y 0 )}.
By Lemma 6.71, since supp ψn ⊂ B(xn , rn ), if we extend (ψn u) ◦ Tn−1 to be zero in Ωn \ Wn , we obtain that the extended function (ψn u) ◦ Tn−1 belongs to W s,p (Ωn ), with k(ψn u) ◦ Tn−1 kW s,p (Ωn ) n k(ψn u) ◦ Tn−1 kW s,p (Tn (Ω)) n kψn ukW s,p (Ω) . By Theorem 9.29, k Tr((ψn u) ◦ Tn−1 )(·, fn (·))kW s−1/p,p (RN −1 ) = k Tr((ψn u) ◦ Tn−1 )kW s−1/p,p (∂Ωn ) n k(ψn u) ◦ Tn−1 kW s,p (Ωn ) n kψn ukW s,p (Ω) . Since ϕn (y 0 ) = Tn−1 ((y 0 , fn (y 0 ))), the previous inequality implies k Tr((ψn u) ◦ ϕn )kW s−1/p,p (Vn ) n kψn ukW s,p (Ω) .
9.5. Bounded Lipschitz Domains
371
By Definition 9.36,
k Tr(u)kW s−1/p,p (∂Ω) =
` X
!1/p k(ψn Tr(u)) ◦
ϕn kpW s−1/p,p (V ) n
n=1
Ω
` X
!1/p kψn ukpW s,p (Ω)
Ω kukpW s,p (Ω) ,
n=1
where in the last inequality we use Theorem 6.23. Step 2: Let g ∈ W s−1/p,p (∂Ω). By Definition 9.36,
kgkW s−1/p,p (∂Ω) =
` X
!1/p k(ψn g) ◦
ϕn kpW s−1/p,p (V ) n
.
n=1
Since supp ψn ⊂ B(xn , rn ) and Vn = ϕ−1 n (B(xn , 2rn )), where we recall that ϕn is biLipschitz continuous, if we extend (ψn g) ◦ ϕn to be zero outside of Vn , as in the second part of the proof of Theorem 6.23, we have that k(ψn g) ◦ ϕn kW s−1/p,p (RN −1 ) n k(ψn g) ◦ ϕn kW s−1/p,p (Vn ) . Recalling that ϕn (y 0 ) = Tn−1 ((y 0 , fn (y 0 ))) and that ωn (y 0 ) := (y 0 , fn (y 0 )) is a chart for ∂Ωn (see (9.38) and Example 9.24), we have that (ψn g) ◦ Tn−1 ∈ W s−1/p,p (∂Ωn ), with k(ψn g) ◦ Tn−1 kW s−1/p,p (∂Ωn ) = k(ψn g) ◦ ϕn kW s−1/p,p (RN −1 ) n k(ψn g) ◦ ϕn kW s−1/p,p (Vn ) . Hence, by Theorem 9.29 there exists a function wn ∈ W s,p (RN ) such that Tr(wn ) = (ψn g) ◦ Tn−1 on ∂Ωn and kwn kW s,p (RN ) n k(ψn g) ◦ Tn−1 kW s−1/p,p (∂Ωn ) n k(ψn g) ◦ ϕn kW s−1/p,p (Vn ) . Setting vn = wn ◦ Tn , by Theorem 6.29, we have that vn ∈ W s,p (RN ), with kvn kW s,p (RN ) kwn kW s,p (RN ) n k(ψn g) ◦ ϕn kW s−1/p,p (Vn ) . Moreover, Tr(vn ) = ψn g on Tn (∂Ωn ) = ∂Tn (Ωn ). Since supp ψn ⊂ B(xn , rn ), we can find φn ∈ Cc∞ (RN ) such that supp φn ⊂ B(xn , rn ) and φn ψn = ψn on ∂Ω ∩ B(xn , rn ). Define un := φn vn . By Theorem 6.23, kun kW s,p (RN ) n kvn kW s,p (RN ) n k(ψn g) ◦ ϕn kW s−1/p,p (Vn ) .
372
9. Trace Theory
Note that if x ∈ ∂Ω ∩ B(xn , rn ), (φn ψn )(x) = ψn (x), so Tr(un )(x) = P Tr(vn )(x) = (ψn g)(x). Hence, if we define u := `n=1 un , then kukW s,p (RN ) ≤
` X
kun kW s,p (RN )
n=1
Ω
` X
k(ψn g) ◦ ϕn kW s−1/p,p (Vn ) Ω kgkW s−1/p,p (∂Ω) .
n=1
Since Tr is linear, Tr(u) =
` X
Tr(un ) =
n=1
where we us the fact that
` X
ψn g = g
on ∂Ω,
n=1
P`
n=1 ψn
= 1 in ∂Ω.
Exercise 9.40. Let Ω ⊂ RN be an open bounded set with Lipschitz continuous boundary, with N ≥ 2, and let 1 < p < ∞, 0 < σ < 1, with ] Ω σp < N −1, and v ∈ W σ,p (∂Ω). Prove that v ∈ Lpσ (∂Ω), with kvk p]σ L
kvkW σ,p (∂Ω) , where p]σ :=
(N −1)p N −1−σp
(∂Ω)
is the Sobolev critical exponent in RN −1 .
Exercise 9.41. Let Ω ⊂ RN be an open bounded set with Lipschitz continuous boundary, with N ≥ 2, and let 1 < p < ∞, and v ∈ W (N −1)/p,p (∂Ω). Prove that v ∈ Lq (∂Ω) for every 1 ≤ q < ∞, with kvkLq (∂Ω) Ω kvkW (N −1)/p,p (∂Ω) . Exercise 9.42. Let Ω ⊂ RN be an open bounded set with Lipschitz continuous boundary, with N ≥ 2, and let 1 < p < ∞, 0 < σ < 1, with σp > N − 1, and v ∈ W σ,p (∂Ω). Prove that v admits a representative v¯ that is H¨older continuous with exponent α = σ − (N − 1)/p, with k¯ v kC 0,α (∂Ω) Ω kvkW σ,p (∂Ω) .
9.6. Unbounded Lipschitz Domains The natural way to define fractional Sobolev spaces on an unbounded Lipschitz continuous manifold would be to allow ` = ∞ in Definition 9.33. The problem is that I don’t know if in this case the definition is independent of the particular atlas and partition of unity. To overcome this problem, I will introduce the notion of uniformly Lipschitz manifolds, which is consistent with Stein’s definition of open sets with uniformly Lipschitz continuous boundary. We recall the definition of uniformly Lipschitz continuous domains.
9.6. Unbounded Lipschitz Domains
373
Definition 9.43. 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 {Un }n of ∂Ω such that (i) if x ∈ ∂Ω, then B(x, ε) ⊆ Un for some n ∈ N, (ii) no point of RN is contained in more than m of the Un ’s, (iii) for each n there exist a rigid motion Tn : RN → RN and a Lipschitz continuous function fn : RN −1 → R, with Lip fn ≤ L, such that Un ∩ Ω = Un ∩ Vn , where Wn = Tn (Vn ) and Wn := {(y 0 , yN ) ∈ RN −1 × R : yN > fn (y 0 )}. Guided by Definition 9.43, we define uniformly Lipschitz continuous manifolds. Definition 9.44. Let 1 ≤ k < N and M ⊂ RN be a kdimensional Lipschitz continuous surface. We say that M is a uniformly Lipschitz continuous surface or manifold if there exist ε, L > 0, m ∈ N, an atlas {ϕn }n , ϕn : Vn → RN , where Vn ⊆ Rk is open, and a locally finite countable open cover {Un }n of M , where ϕn (Vn ) = M ∩ Un , such that (i) if x ∈ M , then B(x, ε) ⊆ Un for some n ∈ N, (ii) no point of RN is contained in more than m of the Un ’s, (iii) Lip ϕn ≤ L and Lip ϕ−1 n ≤ L for each n. We now construct a partition of unity subordinated to the family of open sets Un in Definition 9.44. Proposition 9.45. Given a kdimensional uniformly Lipschitz continuous surface M ⊂ RN , where 1 ≤ k < N , let {ϕn }n be an atlas satisfying items (i)–(iii) of Definition 9.44. Then there exist nonnegative functions ψn ∈ C ∞ (RN ) such that (9.39) supp ψn ⊂ {x ∈ Un : dist(x, ∂Un ) ≥ ε/4} ∩ {x ∈ RN : dist(x, M ) < ε}, (9.40) X
ψn = 1
on M,
n
and (9.41)
X n
ψn (x)q ≤ m,
X
ψn (x) − ψn (y)q mq+1 ε−q kx − ykq
n
for every 1 ≤ q < ∞ and x, y ∈ RN .
374
9. Trace Theory
Proof. For every set E ⊆ RN and r > 0, we define E r := {x ∈ RN : r B(x, r) ⊆ E}. Observe S ε that E ⊆ E and that item (i) in Definition 9.44 implies that M ⊂ n Un . Define φn := ϕε/8 ∗χU 3ε/8 , where ϕε/8 is a standard n mollifier. Then (9.42)
supp φn ⊂ {x ∈ Un : dist(x, ∂Un ) ≥ ε/4},
φn = 1 in Unε/2 .
By standard properties of mollifiers ([Leo22c]), for every i = 1, . . . , N , we have that ∂i φn = ∂i ϕε/8 ∗ χU 3ε/8 , so that n
Z (9.43)
∂i φn (x) ≤ RN
∂i ϕε/8 (y − x) dy ε−1 ,
x ∈ RN .
Next consider the open set U := {x ∈ RN : dist(x, M ) < ε/4},
(9.44)
and define φ+ := ϕε/4 ∗χU . Then φ+ (x) = 1 if x ∈ RN and dist(x, M ) ≤ ε/4. Moreover, the support of φ+ is contained in an ε/2 neighborhood of M . Finally, reasoning as in (9.43), we have that k∂i φ+ k∞ ε−1 . Let (9.45)
ψn (x) :=
φ+ (x)φn (x) P , φk (x)
x ∈ RN .
k
Note that if x ∈ RN is such that dist(x, M ) < ε/2, then there exists l such ε/2 that x ∈ Ul , so φl (x) = 1 by (9.42). In particular, since all the functions φk are nonnegative, it follows that X (9.46) if x ∈ supp φ+ , then φk (x) ≥ 1. k
This shows that ψn is well defined, provided we interpret it to be zero whenever φ+ = 0. In turn, using (9.42), (9.46), the fact that 0 ≤ φ+ , φn ≤ 1, and item (ii) in Definition 9.44, for every 1 ≤ q < ∞ and x ∈ RN , we get X X X ψn (x)q ≤ φn (x)q ≤ χUn (x) ≤ m. n
n
n
This proves the first inequality in (9.41). To prove the second, observe that since the family {Un }n is locally finite, we can differentiate ψn with respect to xi , to get P φ+ φn ∂i φl φn ∂i φ+ + φ+ ∂i φn P ∂i ψn = − P l 2 . φk φk k k
9.6. Unbounded Lipschitz Domains
375
Using (9.46), the facts that 0 ≤ φ+ , φn ≤ 1, k∇φn k∞ ε−1 , k∇φ+ k∞ ε−1 , and item (ii) in Definition 9.44, for x ∈ RN we obtain X k∇φl k∞ mε−1 , ∂i ψn (x) k∇φ+ k∞ + k∇φn k∞ + l∈Ix
where Ix = {l : x ∈ Ul }. Therefore, for every 1 ≤ q < ∞ and x, y ∈ RN , by the mean value theorem and the fact that and Ix and Iy have cardinality at most m, X X ψn (y) − ψn (x)q ≤ ψn (y) − ψn (x)q n
n∈Iy
+
X
ψn (y) − ψn (x)q mq+1 ε−q kx − ykq .
n∈Ix
Remark 9.46. Under the hypotheses of Proposition 9.45, define ωn := ϕε/16 ∗ χU 3ε/16 , where ϕε/16 is a standard mollifier. Then n
supp ωn ⊂ {x ∈ Un : dist(x, ∂Un ) ≥ ε/8}, ωn = 1 in Unε/4 . P As in the proof of (9.41), k∇ωn k∞ ε−1 , n ωn (x)q ≤ m, and X ωn (x) − ωn (y)q mε−q kx − ykq n
for every x, y ∈ RN and 1 ≤ q < ∞. Observe that since supp ψn ⊂ {x ∈ ε/4 Un : dist(x, ∂Un ) ≥ ε/4} and ωn = 1 in Un , we have that ωn ψn = ψn on M ∩ Un . Definition 9.47. Given a kdimensional uniformly Lipschitz continuous surface M ⊂ RN , where 1 ≤ k < N , 1 ≤ p < ∞, and 0 < σ < 1, let {ϕn }n be an atlas satisfying items (i)–(iii) of Definition 9.44, and let {ψn }n be a smooth partition of unity subordinated to {Un }n , with supp ψn ⊂ {x ∈ Un : dist(x, ∂Un ) ≥ ε/4}. We say that a function v : M → R belongs to the fractional Sobolev space W σ,p (M ) if the function (ψn v) ◦ ϕn belongs to W σ,p (Vn ) for every n, and we set !1/p X p kvkLp (M ) := k(ψn v) ◦ ϕn kLp (Vn ) , n
!1/p vW σ,p (M ) :=
X
(ψn v) ◦
ϕn pW σ,p (Vn )
,
n
kvkW σ,p (M ) := kvkLp (M ) + vW σ,p (M ) . The following theorem provides an intrinsic norm for W σ,p (M ). We will not use this result to characterize the trace space of W s,p (Ω) below.
376
9. Trace Theory
Theorem 9.48. Let M ⊂ RN be a kdimensional uniformly Lipschitz continuous surface, where 1 ≤ k < N , 1 ≤ p < ∞, and 0 < σ < 1. Then a Borel function v : M → R belongs to W σ,p (M ) if and only if Z 1/p p k v(x) dH (x) kvkW σ,p (M ) := + vW σ,p (M ) < ∞, M
where (9.47)
vW σ,p (M )
Moreover, k ·
:=
Z
Z
M
kW σ,p (∂Ω)
M ∩B(y,ε)
1/p v(x) − v(y)p k k . dH (x)dH (y) kx − ykk+σp
is an equivalent norm.
Here ε > 0 is the number given in Definition 9.44. Proof. Step 1: Let v ∈ Lp (M ). We claim that kvkW σ,p (M ) ε,L,m kvkW σ,p (M ) . P By (9.40), n ψn = 1 on M . Moreover, if we consider ωn := ϕε/16 ∗ χU 3ε/16 , n then by Remark 9.46, ω ψ = ψ on M . Define v := ψ v. Then v = n n n n n P ω v on M . By the discrete H¨older’s inequality and the fact that Pn n n p 0 (ω n n (x)) ≤ m (see Remark 9.46), we have that p Z Z X XZ p k k p−1 v dH = ωn vn dH ≤ m vn p dHk M M n M n XZ vn p dHk . = mp−1 M ∩Un
n
By Theorem 9.27 (see (9.31)), Z Z (9.48) vn (x)p dHk (x) L M ∩Un
(vn ◦ ϕn )(z)p dz.
Vn
Thus, Z
vp dHk L,m
M
On the other hand, since v =
XZ n
P
n ωn v n
(vn ◦ ϕn )(z)p dz.
Vn
on M , we have
(9.49) (vW σ,p (M ) )p Z Z
P  n ωn (x)(vn (x) − vn (y))p dHk (x)dHk (y) k+σp kx − yk M M ∩B(y,ε) P Z Z  n vn (y)(ωn (y) − ωn (x))p + dHk (x)dHk (y) =: A + B. k+σp kx − yk M M ∩B(y,ε)
9.6. Unbounded Lipschitz Domains
377
P 0 By the discrete H¨ older’s inequality and the fact that n (ωn (x))p ≤ m (see Remark 9.46), we have that XZ Z vn (x) − vn (y)]p (9.50) dHk (x)dHk (y) A ≤ mp−1 k+σp kx − yk M M ∩B(y,ε) n X An . =: mp−1 n
If x, y ∈ / Un , then (ψn v)(x) = (ψn v)(y) = 0 since the support of ψn is contained in Un by Definition 9.47. Hence, by interchanging x and y, if needed, we can write Z Z vn (x) − vn (y)p An = dHk (x)dHk (y) k+σp kx − yk M ∩Un M ∩Un ∩B(y,ε) Z Z dHk (x) p (9.51) +2 vn (y) dHk (y) k+σp kx − yk M ∩Un (M \Un )∩B(y,ε) =: C + D. Since M ∩ Un = ϕn (Vn ), by Theorem 9.27 (see (9.32)), Z Z (vn ◦ ϕn )(z) − (vn ◦ ϕn )(w)p dzdw. (9.52) C L kz − wkk+σp Vn Vn k R R We estimate D. Note that we can replace M ∩Un with M ∩{ψn >0} . Since supp ψn ⊂ {z ∈ Un : dist(z, ∂Un ) ≥ ε/4} by Definition 9.47, for every y ∈ M ∩{ψn > 0}, we have that dist(y, ∂Un ) ≥ ε/4. Hence, if x ∈ (M \Un )∩ B(y, ε), we have that kx − yk ≥ ε/4. Moreover, by item (i) in Definition 9.44, there exists j such that B(y, ε) ⊆ Ωj . Hence, by the area formula ([Leo17, Theorem 9.49 and Exercise 9.50]) and Lemma 9.28, (9.53) Z Z dHk (x) −k−σp Jϕj (w) dw L ε−σp , ε k+σp (M \Un )∩B(y,ε) kx − yk ϕ−1 (B(y,ε)) j where in the last inequality we use (9.30) and the fact that kz − wkk ≤ Lkϕj (z) − ϕj (w)k < 2Lε for all z, w ∈ ϕ−1 j (B(y, ε)), so that k Lk (ϕ−1 j (B(y, ε))) (Lε) .
Thus, also by (9.48), Z D ε,L
p
k
Z
vn (x) dH (x) ε,L
M ∩Un
(vn ◦ ϕn )(z)p dz.
Vn
Combining this inequality with (9.50), (9.51), and (9.52) gives X (9.54) A ε,L,m kvn ◦ ϕn kpW σ,p (Vn ) . n
378
9. Trace Theory
To estimate B in (9.49), note that by the discrete H¨ older’s inequality, p !p/p0 X X X p p0 vn (y)(ωn (y) − ωn (x)) ≤ vn (y) ωn (y) − ωn (x) . n n n If x ∈ B(y, ε), by Remark 9.46, X 0 0 0 ωn (y) − ωn (x)p mε−p kx − ykp . n
It follows that Z

M ∩B(y,ε)
ε,m
X
P
n vn (y)(ωn (y) − ωn (x)) kx − ykk+σp p
Z
vn (y)
M ∩B(y,ε)
n
p
dHk (x)
kx − ykp dHk (x). kx − ykk+σp
By item (i) in Definition 9.44, if y ∈ M there exists j such that B(y, ε) ⊆ Ωj . Hence, y = ϕj (z) for some z ∈ Vj and by the area formula ([Leo17, Theorem 9.49 and Exercise 9.50]) and Lemma 9.28, we can estimate Z kx − ykp(1−σ)−k dHk (x) M ∩B(y,ε) Z = kϕj (z) − ϕj (w)kp(1−σ)−k Jϕj (w) dw Vj ∩ϕ−1 j (B(y,ε))
Z (9.55)
p(1−σ)−k
L Bk (w,Lε)
kz − wkk
dz L εp(1−σ) ,
where we use the fact that L−1 kz − wkk ≤ kϕj (z) − ϕj (w)k ≤ Lkz − wkk . It follows from the last two inequalities, the fact that supp ψn ⊂ Un , (9.48), and (9.49) that Z XZ kx − ykp p vn (y) dHk (x)dHk (y) B ε,m k+σp kx − yk M ∩Un M ∩B(y,ε) n XZ XZ p N −1 ε,L,m vn (y) dH (y) ε,L,m (vn ◦ ϕn )(z)p dz. n
M
n
Vn
Combining this estimate with (9.49) and (9.54), we obtain X (vW σ,p (M ) )p ε,L,m kvn ◦ ϕn kpW σ,p (Vn ) . n
Step 2: Let v ∈ Lp (M ). We claim that kvkW σ,p (M ) ε,L,m kvkW σ,p (M ) .
9.6. Unbounded Lipschitz Domains
379
P By Theorem 9.27 (see (9.31)) and the fact that n (ψn (x))p ≤ m by (9.41), XZ XZ vn (x)p dHk (x) (vn ◦ ϕn )(z)p dz L Vn
n
n
Z
X
L M
p
p
M ∩Un k
(ψn (x)) v(x) dH (x) L,m
Vn
L
n
kz − wkk+σp k
Vn
XZ
Z
M ∩Un
n
v(x)p dHk (x).
M
Again by Theorem 9.27 (see (9.32)), X Z Z (vn ◦ ϕn )(z) − (vn ◦ ϕn )(w)p n
Z
M ∩Un
dzdw
vn (x) − vn (y) p dHk (x)dHk (y) =: A. kx − ykk+σp
For y ∈ M and x ∈ M ∩ B(y, ε), we have (ψn v)(x) − (ψn v)(y)p (ψn (x))p v(x) − v(y)p + v(y)p ψn (x) − ψn (y)p . Hence, by (9.41), X (ψn v)(x) − (ψn v)(y)p mv(x) − v(y)p n
+ mp+1 ε−p v(y)p kx − ykp . On the other hand, for y ∈ M and x ∈ M \ B(y, ε), by (9.41) we have that X (ψn v)(x) − (ψn v)(y)p m(v(x)p + v(y)p ). n
Hence, by Tonelli’s theorem Z Z A ≤ε,L,m M
M ∩B(y,ε)
v(x) − v (y) p dHk (x)dHk (y) kx − ykk+σp
kx − ykp dHk (x)dHk (y) k+σp M M ∩B(y,ε) kx − yk Z Z 1 p + v(y) dHk (x)dHk (y). k+σp kx − yk M M \B(y,ε) Z
+
p
Z
v(y)
Using (9.53) and (9.55), we obtain Z A ε,L,m v(y)p dHk (y) + (vW σ,p (M ) )p .
M
We now characterize the trace space of W s,p (Ω) for domains with uniformly Lipschitz continuous boundary. In what follows, given u ∈ W s,p (RN ), by its trace on ∂Ω we mean the trace of the restriction of u to Ω.
380
9. Trace Theory
Theorem 9.49. Let Ω ⊆ RN , N ≥ 2, be an open set whose boundary ∂Ω is uniformly Lipschitz continuous, and let 1 < p < ∞, and 0 < s < 1, with sp > 1. Then for all u ∈ W s,p (Ω), k Tr(u)kW s−1/p,p (∂Ω) ε,L,m kukW s,p (Ω) ,
(9.56)
where ε, L, m are given in Definition 9.43. Moreover, for every g ∈ W s−1/p,p (∂Ω), there exists a function u ∈ such that Tr(u) = g on ∂Ω and
W s.p (RN )
kukW s,p (Ω) ε,L,m kgkW s−1/p,p (∂Ω) . In the next lemma, given a function u : Ω → R and ψn as in Proposition 9.45, we extend ψn u to be zero in Vn \ Ω. Lemma 9.50. Let Ω ⊆ RN , N ≥ 2, be an open set whose boundary ∂Ω is uniformly Lipschitz continuous, and let 1 ≤ p < ∞, and 0 < s < 1. Then for every function u ∈ Lp (Ω), X 1/p (9.57) kψn ukpLp (Vn ) m kukLp (Ω) n
and X
(9.58)
ψn upW s,p (Vn )
1/p
ε,m kukW s,p (Ω) ,
n
where ψn and Vn are given in Definition 9.47. Proof. Set un := ψn u. Since the support of ψn is contained in Un , and Un ∩ Ω = Un ∩ Vn , by (9.41) and Tonelli’s theorem, XZ XZ un (x)p dx = ψn (x)u(x)p dx n
Vn
Zn
Un ∩Ω p
u(x)
= Ω
X n
p
Z
(ψn (x)) dx ≤ m
u(x)p dx.
Ω
This proves (9.57). Set Fn := supp ψn ⊂ Un . By Tonelli’s theorem X Z Z un (x) − un (y)p dxdy N +sp kx − yk V V n n n Z XZ dx p un (y) =2 dy kx − ykN +sp Fn ∩Vn Vn \Un n Z XZ un (x) − un (y)p (9.59) + dxdy =: A + B. kx − ykN +sp Un ∩Vn Un ∩Vn n
9.6. Unbounded Lipschitz Domains
381
By Definition 9.47, we have that dist(Fn , ∂Un ) ≥ ε/4. Hence, if y ∈ Fn ∩ Vn and x ∈ Vn \ Un , then kx − yk ≥ ε/4. It follows that Z Z 1 1 1 dx ≤ dx sp . N +sp N +sp ε Vn \Un kx − yk RN \B(y,ε/4) kx − yk In turn, by Tonelli’s theorem and (9.57), Z XZ (9.60) A ε u(x)p dx. un (y)p dy ε,m n
Un ∩Ω
Ω
For x, y ∈ Un ∩ Vn with kx − yk < ε, we have un (x) − un (y)p (ψn (y))p u(x) − u(y)p + u(x)p ψn (x) − ψn (y)p . Hence, by (9.41), X un (x) − un (y)p mu(x) − u(y)p + mp+1 ε−p u(x)p kx − ykp . n
In turn, XZ n
Un ∩Vn
Z Un ∩Vn ∩B(y,ε)
un (x) − un (y)p dydx kx − ykN +sp
u(x) − u(y)p ε,m dxdy N +sp Ω Ω kx − yk Z Z 1 + u(x)p dydx N +sp−p B(y,ε) kx − yk Ω Z Z Z u(x) − u(y)p dxdy + u(x)p dx, ε,m N +sp kx − yk Ω Ω Ω Z Z
(9.61)
since Z B(y,ε)
1 dx ε(1−s)p . kx − ykN +sp−p
On the other hand, if x, y ∈ Un ∩ Ω are such that kx − yk > ε, then by (9.41), X (ψn u)(x) − (ψn u)(y)p mu(x)p + mu(y)p . n
In turn, by Tonelli’s theorem and the symmetry of the integrals, Z XZ un (x) − un (y)p dxdy kx − ykN +sp Un ∩Vn (Un ∩Vn )\B(x,ε) n Z Z Z 1 −sp (9.62) m u(x)p dydx ε u(x)p dx. m N +sp kx − yk N Ω R \B(x,ε) Ω
382
9. Trace Theory
Combining the estimates in (9.61) and (9.62), we obtain Z Z Z u(x) − u(y)p u(x)p dx + dxdy. B ε,m N +sp Ω Ω kx − yk Ω By (9.59), (9.60), and the previous inequality, we obtain (9.58).
We turn to the proof of Theorem 9.49. Proof of Theorem 9.49. Step 1: Given u ∈ W s.p (Ω), consider ψn u. By Lemma 9.50, ψn u ∈ W s,p (Vn ), where Vn is given in Definition 9.43. Hence, by Theorem 6.29, (ψn u) ◦ Tn−1 ∈ W s,p (Wn ), with k(ψn u) ◦ Tn−1 kW s,p (Wn ) kψn ukW s,p (Vn ) , where Wn := {(y 0 , yN ) ∈ RN −1 × R : yN > fn (y 0 )}. Hence, we are in a position to apply Theorem 9.29 to the function (ψn u)◦Tn−1 to find that (9.63)
k Tr((ψn u) ◦ Tn−1 )kW s−1/p,p (∂Wn ) L kψn ukW s,p (Vn ) .
Note that in view of Definition 9.25 and the fact that $n (y 0 ) = (y 0 , fn (y 0 )), y 0 ∈ RN −1 is a chart for ∂Wn , k Tr((ψn u) ◦ Tn−1 )kW s−1/p,p (∂Wn ) = k Tr((ψn u) ◦ ϕn )kW s−1/p,p (RN −1 ) , where ϕn = Tn−1 ◦ ωn . By Definition 9.47, (9.57), (9.58), and (9.63) X 1/p k Tr(u)kW s−1/p,p (∂Ω) = k Tr((ψn u) ◦ ϕn )kpW s−1/p,p (RN −1 ) n
L
X
kψn ukpW s,p (Vn )
1/p
ε,L,m kukW s,p (Ω) .
n
Step 2: Let g ∈ W s−1/p,p (∂Ω). Then by Definition 9.47, !1/p kgkW s−1/p,p (∂Ω) =
X
k(ψn g) ◦
ϕn kpW s−1/p,p (RN −1 )
.
n
Recalling that Wn := {(y 0 , yN ) ∈ RN −1 × R : yN > fn (y 0 )}, that ϕn (y 0 ) = Tn−1 ((y 0 , fn (y 0 )), and that $n (y 0 ) = (y 0 , fn (y 0 )), y 0 ∈ RN −1 , is a chart for ∂Wn (see Example 9.24), we have that (ψn g) ◦ Tn−1 belongs to W s−1/p,p (∂Wn ), with k(ψn g) ◦ Tn−1 kW s−1/p,p (∂Wn ) = k(ψn g) ◦ ϕn kW s−1/p,p (RN −1 ) .
9.6. Unbounded Lipschitz Domains
383
Hence, by Theorem 9.29 there exists a function wn ∈ W s,p (RN ) such that Tr(wn ) = (ψn g) ◦ Tn−1 on ∂Wn and kwn kW s,p (RN ) L k(ψn g) ◦ ϕn kW s−1/p,p (RN −1 ) . Setting vn = wn ◦ Tn , by Theorem 6.29, we have that vn ∈ W s,p (RN ), with kvn kW s,p (RN ) kwn kW s,p (RN ) L k(ψn g) ◦ ϕn kW s−1/p,p (RN −1 ) . Moreover, Tr(vn ) = ψn g on Tn (∂Wn ) = ∂Tn (Wn ). Define ωn := ϕε/16 ∗ χU 3ε/16 , where ϕε/16 is a standard mollifier, and we n
recall that for every set E ⊆ RN and r > 0, we define E r := {x ∈ RN : B(x, r) ⊆ E}. Then by Remark 9.46, supp ωn ⊂ {x ∈ Un : dist(x, ∂Un ) ≥ ε/8},
ωn = 1 in Unε/4 ,
and k∇ωn k∞ ε−1 . Since supp ψn ⊂ {x ∈ Un : dist(x, ∂Un ) ≥ ε/4} and ε/4 ωn = 1 in Un , we have that ωn ψn = ψn on ∂Ω ∩ Un . P Let u := n ωn vn . Since supp ωn ⊂ Un and {Un }n is locally finite, we have that u ∈ Lploc (RN ). We claim that u ∈ W s,p (RN ). By the discrete P 0 H¨older’s inequality and the fact that n (ωn (x))p ≤ m by Remark 9.46, we have that Z XZ u(x)p dx ≤ mp−1 vn (x)p dx. RN
n
RN
On the other hand, Z Z u(x) − u(y)p dxdy N +sp RN RN kx − yk P Z Z  n ωn (x)(vn (x) − vn (y))p (9.64) dxdy kx − ykN +sp RN RN P Z Z  n vn (y)(ωn (y) − ωn (x))p + dxdy = A + B. kx − ykN +sp RN RN P 0 By the discrete H¨older’s inequality and the fact that n (ωn (x))p ≤ m by Remark 9.46, we have that X Z Z vn (x) − vn (y)p dxdy. (9.65) A ≤ mp−1 N +sp kx − yk N N R R n To estimate B, for every x ∈ RN let Ix := {n ∈ N : x ∈ Un }. Note that by item (i) in Definition 9.43, the cardinality of Ix is at most m. By the discrete H¨older’s inequality p !p/p0 X X X 0 vn (y)(ωn (y) − ωn (x)) ≤ vn (y)p ωn (y) − ωn (x)p . n n n
384
9. Trace Theory
If x ∈ RN \ B(y, ε), then X X X 0 0 0 ωn (y) − ωn (x)p ωn (y)p + ωn (x)p m, n
n∈Iy
n∈Ix
while if x ∈ B(y, ε), by Remark 9.46, X 0 0 0 ωn (y) − ωn (x)p mε−p kx − ykp . n
It follows that XZ
"Z
1 dx N +sp RN RN \B(y,ε) kx − yk n # Z XZ kx − ykp dy vn (y)p . dx + ε,m N +sp kx − yk N R B(y,ε) n
B ε,m
vn (y)p
Combining this estimate with (9.64) and (9.65), we obtain X X upW s,p (RN ) ε,m kvn kpW s,p (RN ) ε,L,m k(ψn g) ◦ ϕn kpW s−1/p,p (RN −1 ) . n
n
Note that if x ∈ ∂Ω ∩ Un , (ωn ψn )(x) = ψn (x), so, Tr(ωn vn ) = Tr(vn ) = ψn g on ∂Ω ∩ Un . Since supp ωn ⊂ Un and supp ψn ⊂ Un , it follows that Tr(ωn vn ) = ψn g = 0 in ∂Ω \ Un . Hence, Tr(ωn vn ) = ψn g on ∂Ω, so, by the linearity of the trace and the fact that {Un }n is locally finite, X X Tr(u) = Tr(ωn vn ) = ψn g = g on ∂Ω. n
n
RN
Exercise 9.51. Let Ω ⊂ be an open set whose boundary ∂Ω is uniformly Lipschitz continuous, and let 1 < p < ∞, 0 < s < 1, with sp > 1, u ∈ W s.p (Ω), and v ∈ W s.p (RN \ Ω). Prove that the function u(x) if x ∈ Ω, w(x) := v(x) if x ∈ RN \ Ω belongs to W s.p (RN ) if and only if Tr(u) = Tr(v) on ∂Ω. Exercise 9.52. Let Ω ⊂ RN be an open set whose boundary ∂Ω is uniformly Lipschitz continuous. Prove that if ∂Ω is bounded, then the norm (9.34) is equivalent to the norm (9.47).
9.7. Notes Versions of Theorems 9.4, 9.6, and 9.7 and Corollary 9.8 can be found in the papers of Grisvard [Gri63] and Nekvinda [Nek93]. The proof of Theorem 9.6 is adapted from the paper of A.D. Nguyen, J.I. D´ıaz, and Q.H. Nguyen [NDN20]. For some recent results on the trace spaces of homogeneous Sobolev spaces in striplike domains and new types of fractional Sobolev
9.7. Notes
385
spaces, we refer to the papers of Strichartz [Str16], Leoni and Tice [LT19], and Stevenson and Tice [ST20] (see also Sereesuchart [Ser22]). Exercises 9.12 and 9.13 are due to H¨ormander (see [Gru91]). The proof of Theorem 9.19 is adapted from the paper of Solonnikov [Sol67], while the one of Theorem 9.21 is adapted from the paper of Ilin and Solonnikov [IS69] (see also [Mar87]). I am grateful to one of my undergraduate students, Khunpob Sereesuchart, for catching a serious mistake in an earlier version of the proof of Theorem 9.21. I got the idea of using weighted Sobolev spaces in Step 2 of the proof of Theorem 9.21, in place of second order differences, from the paper of Kim [Kim07] (see also [Tri95, Section 2.9.2]). Theorem 9.11 is adapted from Grisvard [Gri11, Lemma 1.4.1.3]. We refer to the papers of Jonsson and Wallin [JW78], [JW80] for a generalization of Theorems 9.19 and 9.21 to the case when the Lipschitz continuous manifold ∂Ω is replaced by a closed “kset”, where 1 ≤ k ≤ N . For more recent results on fractional Sobolev spaces on manifolds, see the papers of Amann [Ama13] and Mugnolo and Vitillaro [MV22] and the references therein.
Chapter 10
Symmetrization If I had inherited a fortune I should probably not have cast my lot with mathematics. — JosephLouis Lagrange
In this chapter we will study two types of rearrangements: the twopoint rearrangement, or polarization, and the spherically symmetric rearrangement.
10.1. Polarization In this section, we extend the results of Section 4.1 to higher dimensions. Given an affine hyperplane H ⊂ RN , RN \ H is given by the union of two open halfspaces. If H does not contain the origin, let H+ be the open halfspace that contains the origin, and let H− be the other one. If H contains the origin, label H+ and H− arbitrarily. Consider the reflection σ : RN → RN that maps H+ into H− . Let ν ∈ SN −1 be a normal vector to H and denote by ν ⊥ the hyperplane orthogonal to ν, that is, ν ⊥ := {x ∈ RN : x · ν = 0}. For every x ∈ RN , we write x = y + tν, where y ∈ ν ⊥ , and t ∈ R. If x ∈ RN \ H, let a ∈ R be the unique real number such that y + aν ∈ H. Then (10.1)
σ(x) = y + (−t + 2a)ν.
Given a function u : RN → R, the twopoint rearrangement or polarization of u with respect to σ is the function uσ : RN → R, given by max{u(x), u(σ(x))} if x ∈ H+ , σ min{u(x), u(σ(x))} if x ∈ H− , (10.2) u (x) := u(x) if x ∈ H. The following theorem is the analogue of Theorem 4.1 for N ≥ 2. 387
388
10. Symmetrization
Theorem 10.1. Let u : RN → R be a Lebesgue measurable function and let σ be a reflection. Then for every t ∈ R, LN ({x ∈ RN : u(x) > t}) = LN ({x ∈ RN : uσ (x) > t)}. Proof. Let H be the affine hyperplane corresponding to σ and let ν be a unit normal to H. Assume first that ν = eN . Then H = {x ∈ RN : xN = a} for some a ∈ R. In turn, σ(x) = (x0 , −xN + 2a). Hence, for every fixed x0 ∈ RN −1 by (10.2), uσ (x0 , ·) is the polarization of the onedimensional function u(x0 , ·) with respect to the onedimensional reflection determined by a (see (4.1)). It follows from Tonelli’s theorem and Theorem 4.1, that Z N N L ({x ∈ R : u(x) > t}) = L1 ({xN ∈ R : u(x0 , xN ) > t}) dx0 RN −1 Z = L1 ({xN ∈ R : uσ (x0 , xN ) > t}) dx0 RN −1 N
= L ({x ∈ RN : uσ (x) > t)}. The case in which ν 6= eN follows in a similar way.
Exercise 10.2. Let u : RN → [0, ∞) be a Lebesgue measurable function and let σ be a reflection. Prove that for every p > 0, Z Z p (u(x)) dx = (uσ (x))p dx. RN
RN
Exercise 10.3. Given a Lebesgue measurable function u : RN → R and a reflection σ, let M be the family of Borel sets B ⊆ R such that LN ({x ∈ RN : u(x) ∈ B}) = LN ({y ∈ RN : uσ (y) ∈ B}). S (i) Prove that if Bn ∈ M, n ∈ N, are pairwise disjoint, then ∞ n=1 Bn ∈ M. S (ii) Prove that if Bn ∈ M, n ∈ N, then ∞ n=1 Bn ∈ M. (iii) Prove that if B ∈ M, then R \ B ∈ M. (iv) Prove that every open set U ⊆ R belongs to M. (v) Prove that every Borel set B ⊆ R belongs to M. Exercise 10.4. Let f : R → [0, ∞) be a Borel function such that f (0) = 0, let u : RN → R be a Lebesgue measurable function vanishing at infinity, and let σ be a reflection. Prove that Z Z f (u(x)) dx = f (uσ (x)) dx. RN
RN
Exercise 10.5. Let f : R → R be a Borel function such that f (0) = 0, let u : RN → R be a Lebesgue measurable function vanishing at infinity and be
10.1. Polarization
389
such that f ◦ u is Lebesgue integrable, and let σ be a reflection. Prove that Z Z f (uσ (x)) dx. f (u(x)) dx = RN
RN
Next, we extend Theorem 4.8. We recall that, given E ⊆ RN and u : E → R, the modulus of continuity of u is the function ωu : [0, ∞) → [0, ∞], defined by ωu (δ) := sup{u(x) − u(y) : x, y ∈ E, kx − yk ≤ δ},
δ ≥ 0.
Theorem 10.6. Let u : RN → R and let σ be a reflection. Then ωuσ ≤ ωu . Lemma 10.7. Let H be an affine hyperplane and let σ be the corresponding reflection. Then (i) If x, y ∈ RN , then kσ(x) − σ(y)k = kx − yk. (ii) If x ∈ H+ and y ∈ H− , then kσ(x) − yk ≤ kx − yk. (iii) If x ∈ H− and y ∈ H+ , then kx − σ(x)k ≤ kx − yk. Proof. Assume first that ν = eN . Then H = {x ∈ RN : xN = a} for some a ∈ R. In turn, σ(x) = (x0 , −xN + 2a). Hence, if x, y ∈ RN , kσ(x) − σ(y)k2 = kx0 − y 0 k2N −1 + (−xN + 2a) − (−yN + 2a)2 = kx − yk2 . This proves item (i). If x ∈ H+ and y ∈ H− , then kσ(x) − yk2 = kx0 − y 0 k2N −1 + (−xN + 2a) − yN 2 ≤ kx − yk2 since xN and yN lie on opposite sides of a, and −xN + 2a is the reflection of xN with respect to a. Hence, item (ii) holds. Item (iii) is similar. In the case in which ν 6= eN , we can consider a rotation T : RN → RN such that T (H) = {x ∈ RN : xN = a} and observe that T preserves distances. We leave the details as an exercise. Using Lemma 10.7, the proof of Theorem 10.6 is very similar to that of Theorem 4.14, and it is left as an exercise. Theorem 10.8. Let u : RN → R and let σ be a reflection. Then for every t ∈ R, diam{x ∈ RN : uσ (x) > t} ≤ diam{x ∈ RN : u(x) > t}. Lemma 10.7 allows us to adapt the proof of Theorem 4.14 to the N th dimensional case. The details are left as an exercise.
390
10. Symmetrization
The proofs of Theorems 10.9 and 10.11 and Corollaries 10.10 and 10.12 are similar to those of Theorems 4.15 and 4.18 and Corollaries 4.17 and 4.19, respectively. Theorem 10.9 (Hardy–Littlewood’s inequality for polarization). Let u : RN → [0, ∞) and v : RN → [0, ∞) be Lebesgue measurable functions and let σ be a reflection. Then Z Z uσ (x)v σ (x) dx. u(x)v(x) dx ≤ RN
RN
Moreover, if the righthand side is finite, then equality holds if and only if (u(x) − u(σ(x)))(v(x) − v(σ(x))) ≥ 0 for LN a.e. x ∈ RN . Corollary 10.10. Let g : R → [0, ∞) be a convex function with g(0) = 0, let u : RN → R and v : RN → R be Lebesgue measurable functions vanishing at infinity, and let σ be a reflection. Then Z Z σ σ g(u (x) − v (x)) dx ≤ g(u(x) − v(x)) dx. RN
RN
In particular, for 1 ≤ p < ∞, Z Z uσ (x) − v σ (x)p dx ≤ RN
u(x) − v(x)p dx.
RN
Theorem 10.11 (Riesz). Let u : RN → [0, ∞) and v : RN → [0, ∞) be Lebesgue measurable functions vanishing at infinity, let k : [0, ∞) → [0, ∞) be decreasing, and let σ be a reflection. Then Z Z u(x)v(y)k(kx − yk) dxdy RN RN Z Z ≤ uσ (x)v σ (y)k(kx − yk) dxdy. RN
RN
Corollary 10.12. Let g : R → [0, ∞) be a convex function with g(0) = 0, let u : RN → R and v : RN → R be Lebesgue measurable functions vanishing at infinity, let k : [0, ∞) → [0, ∞) be decreasing and Lebesgue integrable, and let σ be a reflection. Then Z Z g(uσ (x) − v σ (y))k(kx − yk) dxdy RN RN Z Z ≤ g(u(x) − v(y))k(kx − yk) dxdy. RN
RN
10.2. Polarization in W s,p (RN )
391
In particular, for 1 ≤ p < ∞, Z Z uσ (x) − v σ (y)p k(kx − yk) dxdy N N R R Z Z u(x) − v(y)p k(kx − yk) dxdy. ≤ RN
RN
10.2. Polarization in W s,p (RN ) In this section we prove that polarization decreases the fractional Sobolev seminorm and the Sobolev seminorm. Theorem 10.13. Let 1 ≤ p < ∞, 0 < s < 1, let u : RN → R be a Lebesgue measurable function vanishing at infinity, and let σ be a reflection. Then Z Z Z Z uσ (x) − uσ (y)p u(x) − u(y)p (10.3) dxdy ≤ dxdy. N +sp kx − ykN +sp RN RN RN RN kx − yk Moreover, if un → u in W s,p (RN ), then uσn → uσ in W s,p (RN ). Proof. Step 1: By the change of variables y = kxk2 t, so that dy = kxk2 dt, we have Z ∞ Z ∞ 1 (N +sp)/2−1 2 t exp(−kxk t) dt = y (N +sp)/2−1 exp(−y) dy kxkN +sp 0 0 Γ((N + sp)/2) = , kxkN +sp R∞ where Γ is the Gamma function Γ(y) := 0 τ y−1 e−τ dτ , y > 0. Hence, by Tonelli’s theorem (10.4) Z Z u(x) − u(y)p dxdy N +sp RN RN kx − yk Z Z Z ∞ p =c u(x) − u(y) t(N +sp)/2−1 exp(−kx − yk2 t) dtdxdy RN RN 0 Z ∞ Z Z =c t(N +sp)/2−1 u(x) − u(y)p exp(−kx − yk2 t) dxdydt, RN
0
RN
where c := 1/Γ((N + sp)/2). For each fixed t > 0, the function kt (y) = exp(−y 2 t), y ≥ 0, is decreasing and Lebesgue integrable. It follows by Corollary 10.12 that Z Z (10.5) uσ (x) − uσ (y)p exp(−kx − yk2 t) dxdy RN RN Z Z ≤ u(x) − u(y)p exp(−kx − yk2 t) dxdy. RN
RN
392
10. Symmetrization
By multiplying both sides by ct(N +sp)/2−1 , integrating in t over R+ , and using (10.4), we obtain (10.3). Step 2: Let un , u ∈ W s,p (RN ) be such that un → u in W s,p (RN ). For t > 0, define Z Z (10.6) Ft (u) := u(x) − u(y)p exp(−kx − yk2 t) dxdy. RN
RN
We claim that lim Ft (uσn − uσ ) = 0.
(10.7)
n→∞
By Tonelli’s theorem, Corollary 10.12, and the fact that, by a change of variables, Z Z 2 exp(−kx − yk t) dy = exp(−kzk2 t) dz, RN
RN
we have Ft (uσn − uσ )
Z RN
uσn (x) − uσ (x)p
Z + Z t RN
σ
exp(−kx − yk2 t) dydx
RN p
Z
exp(−kx − yk2 t) dxdy Z σ σ p un (x) − u (x) dx t un (x) − u(x)p dx → 0
RN
uσn (y)
Z
− u (y)
RN
RN
as n → ∞ since un → u in Lp (RN ). This proves (10.7). By the inequality (a + b)p ≤ 2p−1 ap + 2p−1 bp and (10.5), (10.8) Ft (uσn − uσ ) ≤ 2p−1 Ft (uσn ) + 2p−1 Ft (uσ ) ≤ 2p−1 Ft (un ) + 2p−1 Ft (u) ≤ 22p−2 Ft (un − u) + 22p−1 Ft (u). Using the inequality a ≤ b + (a − b)+ , we have Ft (uσn − uσ ) ≤ 22p−2 Ft (un − u) + (Ft (uσn − uσ ) − 22p−2 Ft (un − u))+ . By multiplying both sides by ct(N +sp)/2−1 , integrating in t over R+ , and using (10.4), we obtain Z Z (uσn − uσ )(x) − (uσn − uσ )(y)p dxdy kx − ykN +sp RN RN Z Z (un − u)(x) − (un − u)(y)p 2p−2 ≤2 dxdy kx − ykN +sp N N Z ∞R R +c t(N +sp)/2−1 (Ft (uσn − uσ ) − 22p−2 Ft (un − u))+ dt =: A + B. 0
By (10.8), (Ft (uσn − uσ ) − 22p−2 Ft (un − u))+ ≤ 22p−1 Ft (u)
10.2. Polarization in W s,p (RN )
and since Z c 0
∞ (N +sp)/2−1
t
393
Z
Z
Ft (u) dt = RN
RN
u(x) − u(y)p dxdy < ∞, kx − ykN +sp
in view of (10.7) and (10.8), we can apply the Lebesgue dominated convergence theorem to conclude that B → 0. Since un → u in W s,p (RN ), we also have that A → 0. Hence, we have shown that uσn → uσ in W s,p (RN ). ˙ 1,p (RN ), and let σ be a reflection. Theorem 10.14. Let 1 ≤ p < ∞, u ∈ W ˙ 1,p (RN ) and Consider a representative u ¯ of u. Then u ¯σ ∈ W (10.9)
k∇¯ uσ kLp (RN ) = k∇ukLp (RN ) .
Moreover, if u ∈ W 1,p (RN ), then so does u ¯σ , with k¯ uσ kLp (RN ) = kukLp (RN ) . Proof. For simplicity we assume that ν = eN . Then H = {x ∈ RN : xN = a} for some a ∈ R. In turn, σ(x) = (x0 , −xN + 2a). Define v¯(x) := u ¯(σ(x)), N 1,p N ˙ x ∈ R . Since σ is biLipschitz continuous, u ¯ ∈ W (R ) (see [Leo22d]), with ∂i v¯(x) = ∂i u ¯(σ(x)), ∂N v¯(x) = −∂N u ¯(σ(x)) for LN a.e. x ∈ RN . Consider the Lipschitz function f (t) := max{t, 0}, ˙ 1,p (RN ), then by the chain rule (see [Leo22d]), f ◦ w ∈ t ∈ R. If w ∈ W 1.p N ˙ (R ) and for LN a.e. x ∈ RN , W ∇w(x) if w(x) ≥ 0, ∇(f ◦ w)(x) = 0 if w(x) < 0. Writing max{¯ u, v¯} = max{¯ u − v¯, 0} + v¯ = f ◦ (¯ u − v¯) + v¯, min{¯ u, v¯} = − max{¯ u − v¯, 0} + u ¯ = −f ◦ (¯ u − v¯) + u ¯, ˙ 1,p (RN ) and that for LN a.e. we deduce that max{¯ u, v¯}, min{¯ u, v¯} ∈ W N x∈R , ∇¯ u(x) if u ¯(x) ≥ v¯(x), ∇(max{¯ u, v¯})(x) = ∇¯ u(σ(x)) if u ¯(x) < v¯(x), ∇¯ u(x) if u ¯(x) ≤ v¯(x), ∇(min{¯ u, v¯})(x) = ∇¯ u(σ(x)) if u ¯(x) > v¯(x). Since u ¯ and v¯ have the same trace on H = {x ∈ RN : xN = a} (see ˙ 1,p (RN ), with [Leo22d]), we leave it as an exercise to check that u ¯σ ∈ W ∇¯ u(x) if x ∈ H+ and u ¯(x) ≥ u ¯(σ(x)), −∇¯ u(σ(x)) if x ∈ H+ and u ¯(x) < u ¯(σ(x)), ∇(¯ uσ )(x) = ∇¯ u (x) if x ∈ H and u ¯ (x) ≤ u ¯(σ(x)), − −∇¯ u(σ(x)) if x ∈ H− and u ¯(x) > u ¯(σ(x)).
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10. Symmetrization
Hence, Z
σ p
Z
p
k∇¯ u(x)k dx +
k∇¯ u k dx =
(10.10) RN
Z
k∇¯ u(σ(x))kp dx,
E2 ∪E4
E1 ∪E3
where E1 = {x ∈ H+ : u ¯(x) ≥ u ¯(σ(x))},
E2 = {x ∈ H+ : u ¯(σ(x)) > u ¯(x)},
E3 = {x ∈ H− : u ¯(x) ≤ u ¯(σ(x))},
E4 = {x ∈ H− : u ¯(x) > u ¯(σ(x))}.
By considering the change of variables y = σ(x) and using the fact that σ(E2 ) = E4 , we have that Z Z k∇¯ u(σ(x))kp dx = k∇¯ u(y)kp dy, E2 ∪E4
E2 ∪E4
which, together with (10.10), shows (10.9). The second part of the statement follows from Exercise 10.4.
10.3. Spherically Symmetric Rearrangement In this section we extend the results of Section 4.3 to functions of several variables. Definition 10.15. Given a Lebesgue measurable set E ⊂ RN with 0 < LN (E) < ∞, the spherically symmetric rearrangement of E, or Schwarz symmetrization of E, is the open ball centered at the origin and with the same measure of E, precisely E ] := B(0, r), where (10.11)
r := (LN (E)/αN )1/N ,
where we recall that αN = LN (B(0, 1)). If E has measure zero, then we take E ] to be the empty set, while if E has infinite measure, then we take E ] to be RN . Note that E ] is always open even if E is not. Hence, E ] is Lebesgue measurable and (10.12)
LN (E) = LN (E ] ).
Definition 10.16. Let E ⊆ RN be a Lebesgue measurable set and let u : E → [0, ∞] be a Lebesgue measurable function. The spherically symmetric rearrangement of u is the function u] : RN → [0, ∞], defined by Z ∞ ] (10.13) u (x) := χE ] (x) dt, x ∈ RN , 0
t
where Et := {x ∈ E : u(x) > t} and Et] is the spherically symmetric rearrangement of the set Et , that is, Et] = B(0, r), with (10.14)
r := (LN (Et )/αN )1/N ,
if LN (Et ) < ∞, and Et] = RN if LN (Et ) = ∞.
10.3. Spherically Symmetric Rearrangement
395
The function u] is also called the Schwarz symmetric rearrangement of u. Theorem 10.17. Let E ⊆ RN be a Lebesgue measurable set, let u : E → [0, ∞] be a Lebesgue measurable function, and let u] : RN → [0, ∞] be its spherically symmetric rearrangement. Then (i) u] is radial, that is, u] (x) = u] (y) for every x, y ∈ RN with kxk = kyk; (ii) if 0 ≤ kxk < kyk, then u] (x) ≥ u] (y); (iii) for every t ≥ 0, {x ∈ E : u(x) > t}] = {x ∈ RN : u] (x) > t} and LN ({x ∈ E : u(x) > t}) = LN ({x ∈ RN : u] (x) > t)};
(10.15)
(iv) the function u] is lower semicontinuous. Proof. Since Et] is an open ball (or the entire space), we have that x ∈ Et] if and only if y ∈ Et] for every y ∈ RN with kyk = kxk. Thus, χE ] (y) = χE ] (x) for every x, y ∈ RN , with kyk = kxk. In turn, by (10.13), Z ∞ Z ∞ u] (y) = χE ] (y) dt = χE ] (x) dt = u] (x), 0
t
0
t
t
t
which shows that item (i) holds. To prove item (ii), observe that if y ∈ Et] , then, since Et] is an open ball (or the entire space), if 0 ≤ kxk < kyk, then x ∈ Et] . Thus, if χE ] (y) = 1, t then χE ] (x) = 1, which implies that χE ] (y) ≤ χE ] (x). Integrating in t and t
using (10.13) gives u] (y) ≤ u] (x).
t
t
To prove item (iii), consider 0 ≤ t1 < t2 . Then (10.16)
Et2 = {x ∈ E : u(x) > t2 } ⊆ {x ∈ E : u(x) > t1 } = Et1 .
This proves that the function g(t) := LN (Et ), t ≥ 0, is decreasing. We claim that g is rightcontinuous. Fix t0 ≥ 0. Since g is decreasing, there exists lim g(t) ≥ g(t0 ).
t→t+ 0
To prove equality, consider a decreasing sequence tn → t+ 0 . Then Etn ⊆ S∞ Etn+1 and n=1 Etn = Et0 , so, we have that ∞ [ lim g(tn ) = lim LN (Etn ) = LN Etn = g(t0 ). n→∞
n→∞
n=1
This proves the claim. Let t ≥ 0 and x ∈ Et] . By (10.14), kxk < (LN (Et )/αN )1/N = (g(t)/αN )1/N .
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10. Symmetrization
Since g is rightcontinuous, there exists δ > 0 such that kxk < (g(r)/αN )1/N for all r ∈ [t, t + δ). Thus, we have shown that x ∈ Er] for all r ∈ [t, t + δ). On the other hand, by (10.16), we have that x ∈ Er] for all 0 ≤ r ≤ t. Hence, Z ∞ Z t+δ Z ∞ ] u (x) = χE ] (x) dr = 1 dr + χE ] (x) dr ≥ t + δ > t. 0
r
0
r
t+δ
This shows that Et] ⊆ {x ∈ RN : u] (x) > t}.
(10.17)
To prove the other inclusion, assume that x ∈ / Et] . Then kxk ≥ (LN (Et )/αN )1/N . Since LN (Er ) ≤ LN (Et ) for all r ≥ t by (10.16), it follows that x ∈ / Er] , and so Z t Z ∞ χE ] (x) dr ≤ t, u] (x) = χE ] (x) dr = 0
r
RN
0
r
u] (y)
which implies that x ∈ / {y ∈ : > t}. Together with (10.17), this proves that Et] = {x ∈ RN : u] (x) > t}. Since LN (Et ) = LN (Et] ) by (10.12), it follows that LN ({x ∈ E : u(x) > t)} = LN (Et] ) = LN ({x ∈ R : u] (x) > t}), and so, item (iii) holds. Finally, to prove item (iv), it suffices to show that for every t ∈ R, the set Ft := {x ∈ RN : u\ (x) > t} is open. If t < 0, then Ft = RN since u\ ≥ 0, while if t ≥ 0, then Ft = Et\ by item (iii), so that Ft is an open ball or the entire space. Exercise 10.18. Let E ⊆ RN be a Lebesgue measurable set and let u, v : E → [0, ∞] be two Lebesgue measurable functions with u ≤ v. Prove that u] ≤ v ] . The proofs of the next theorems are left as exercises. Theorem 10.19. Let E ⊆ RN be a Lebesgue measurable set and let u : E → [0, ∞] be a Lebesgue measurable function. Then for every p > 0, Z Z p (u(x)) dx = (u] (x))p dx. E
RN
Theorem 10.20. Let f : R → [0, ∞) be a Borel function such that f (0) = 0, let E ⊆ RN be a Lebesgue measurable set, and let u : E → [0, ∞) be a Lebesgue measurable function vanishing at infinity. Then Z Z f (u(x)) dx = f (u] (x)) dx. E
RN
10.3. Spherically Symmetric Rearrangement
397
Corollary 10.21. Let f : R → R be a Borel function such that f (0) = 0, let E ⊆ RN be a Lebesgue measurable set, and let u : E → [0, ∞) be a Lebesgue measurable function vanishing at infinity and be such that f ◦ u is Lebesgue integrable over E. Then Z Z f (u\ (x)) dx. f (u(x)) dx = E
RN
Next we present some convergence results for the spherically symmetric rearrangement. Theorem 10.22. Let E ⊆ RN be a Lebesgue measurable set and let un : E → [0, ∞) be Lebesgue measurable functions vanishing at infinity such # # that un ≤ un+1 for all n. Then u# n (x) → u (x) for all x ∈ E , where u = supn un . Proof. Let gu : [0, ∞) → [0, ∞] and gun : [0, ∞) → [0, ∞] be the decreasing functions defined by gu (t) := LN ({x ∈ E : u(x) > t}),
gun (t) := LN ({x ∈ E : un (x) > t}).
By Definition 10.16 and Theorem 10.17(iii), (10.18)
B(0, (gu (t)/αN )1/N ) = {x ∈ RN : u# (x) > t}, B(0, (gun (t)/αN )1/N ) = {x ∈ RN : u# n (x) > t}.
# # Since un ≤ un+1 ≤ u, by Exercise 10.18, u# n ≤ un+1 ≤ u , and so # v(x) := lim u# n (x) ≤ u (x) n→∞
for every x ∈ RN .
To prove the opposite inequality, let x0 ∈ E # and assume that v(x0 ) < u# (x0 ). Let δ > 0 be so small that v(x0 ) < u# (x0 ) − δ and let u# (x0 ) − δ < # t < u# (x0 ). Then u# n (x0 ) < t < u (x0 ) for every n. In turn, by (10.18), (gun (t)/αN )1/N ≤ kx0 k < (gu (t)/αN )1/N , which is a contradiction since ∞ [ {x ∈ E : u(x) > t} = {x ∈ E : un (x) > t}, n=1
and {x ∈ E : un (x) > t} ⊆ {x ∈ E : un+1 (x) > t}, which implies that gun (t) → gu (t). Thus, v(x) = u# (x) for all x ∈ E # . Exercise 10.23. Given a Lebesgue measurable set E ⊆ RN and Lebesgue measurable functions u : E → [0, ∞] and un : E → [0, ∞], let gu : [0, ∞) → [0, ∞] and gun : [0, ∞) → [0, ∞] be the decreasing functions defined by gu (t) := LN ({x ∈ E : u(x) > t}),
gun (t) := LN ({x ∈ E : un (x) > t}).
Prove that if un → u in measure as n → ∞, then for every t > 0, gu (t) ≤ lim inf gun (t) ≤ lim sup gun (t) ≤ lim gu (s). n→∞
n→∞
s→t−
398
10. Symmetrization
Exercise 10.24. Let E ⊆ RN be a Lebesgue measurable set and let un : E → [0, ∞) be Lebesgue measurable functions vanishing at infinity such that un → u in measure as n → ∞ for some function u : E → [0, ∞]. Prove that u vanishes at infinity. Theorem 10.25. Let E ⊆ RN be a Lebesgue measurable set and let un : E → [0, ∞) be Lebesgue measurable functions vanishing at infinity such that un → u in measure as n → ∞ for some function u : E → [0, ∞]. Then # N # u# n (x) → u (x) as n → ∞ for all L a.e. x ∈ E . Proof. Define gu and gun as in Exercise 10.23. By Definition 10.16 and Theorem 10.17(iii), (10.19)
B(0, (gu (t)/αN )1/N ) = {x ∈ RN : u# (x) > t}, B(0, (gun (t)/αN )1/N ) = {x ∈ RN : u# n (x) > t}.
We claim that (10.20)
u# (x) ≤ lim inf u# n (x) n→∞
for all x ∈ E # \ {0}.
To see this, assume that (10.20) does not hold for some x ∈ E # \ {0}. Then we can find δ > 0 such that u# (x) > u# n (x) + δ for infinitely many n. Let # # # u (x) − δ < t < u (x). Then un (x) < t < u# (x) for infinitely many n. It follows from (10.19) that (gun (t)/αN )1/N ≤ kxk < (gu (t)/αN )1/N for infinitely many n. In turn, lim inf gun (t) ≤ αN kxkN < gu (t), n→∞
which contradicts Exercise 10.23. This shows that u# (x) − δ ≤ u# n (x) for all but finitely many n. Hence, u# (x) − δ ≤ lim inf u# n (x). n→∞
Letting δ → 0+ proves (10.20). Next, write u# (x) = v(kxk). We claim that (10.21)
lim sup u# n (x) ≤ n→∞
lim v(r)
r→kxk−
for every x ∈ E # \{0}. If the previous inequality fails for some x ∈ E # \{0}, then there exists δ > 0 so small that lim v(r) < lim sup u# n (x) − δ.
r→kxk−
n→∞
Using the definition of limit, we can find 0 < kyk < kxk such that u# (y) = v(kyk) < lim sup u# n (x) − δ. n→∞
10.3. Spherically Symmetric Rearrangement
399
In turn, there exist infinitely many n such that u# (y) < u# n (x) − δ. Let # # # # u (y) < s < t < u (y) + δ. Then t < un (x) and u (y) < s, and so, by (10.19), (gu (s)/αN )1/N ≤ kyk and kxk < (gun (t)/αN )1/N for infinitely many n. Letting s → t− gives gun (t) > αN kxkN > αN kykN ≥ lim gu (s). s→t−
Taking the limit as n → ∞, we obtain lim sup gun (t) > lim gu (s), n→∞
s→t−
which contradicts Exercise 10.23. This proves (10.21). Since v is decreasing in R+ , it is continuous at all but countably many points. If x ∈ E # \ {0} is such that v is continuous at kxk, then by (10.20) and (10.21), # u# (x) ≤ lim inf u# n (x) ≤ lim sup un (x) ≤ n→∞
n→∞
lim v(r) = v(kxk) = u# (x).
r→kxk−
This proves pointwise convergence for all x in E # except those on the boundary of countably many balls. Since the boundary of a ball has Lebesgue measure zero, the proof is concluded. We now show that Theorem 4.35 continues to hold for the spherically symmetric rearrangement. Theorem 10.26. Let u : RN → [0, ∞) be a lower semicontinuous function vanishing at infinity. Then u = u# if and only if u = uσ for every reflection σ such that H+ contains zero. Proof. Step 1: Assume that u = u# . Given a reflection σ, let H be the affine hyperplane corresponding to σ and let ν be a unit normal to H. Assume first that ν = eN . Then H = {x ∈ RN : xN = a} for some a ∈ R. In turn, σ(x) = (x0 , −xN + 2a). Hence, for every fixed x0 ∈ RN −1 by (10.2), uσ (x0 , ·) is the polarization of the onedimensional function u(x0 , ·) with respect to the onedimensional reflection determined by a (see (4.1)). Since u = u# , we have that u(x0 , ·) is even in R and decreasing in R+ . Hence, by Theorem 4.35 applied to u(x0 , ·), max{u(x0 , xN ), u(x0 , −xN + 2a)} if xN ∈ I+ , 0 min{u(x0 , xN ), u(x0 , −xN + 2a)} if xN ∈ I− , u(x , xN ) = u(x0 , xN ) if xN = a. This implies that u = uσ . In the case in which ν 6= eN , since u is radial, we can consider a rotation T : RN → RN such that T (H) = {x ∈ RN : xN = a}. We omit the details.
400
10. Symmetrization
Step 2: Assume that u = uσ for every reflection σ such that H+ contains zero. If u is not radial, then we can find x0 , y0 ∈ RN , with kx0 k = ky0 k, such that u(y0 ) < u(x0 ). Since u is lower semicontinuous, lim inf u(x) ≥ u(x0 ) > u(y0 ). x→x0
Hence, we can find x1 ∈ RN such that kx1 k > kx0 k and u(x1 ) > u(y0 ). Let σ be the reflection such that σ(x1 ) = y0 . To be precise, we consider the 0 hyperplane H passing through x1 +y and perpendicular to the line joining 2 x1 and y0 . Note that, since ky0 k = kx0 k < kx1 k, 0 does not belong to H. Hence, y0 belongs to halfspace H+ that contains the origin, and x1 belongs to H− . But then uσ (y0 ) = max{u(y0 ), u(x1 )} = u(x1 ) and uσ (x1 ) = min{u(y0 ), u(x1 )} = u(y0 ), which contradicts the fact that u = uσ . Thus, u is radial. Next we show that u is radially decreasing. To see this, assume by contradiction that there exist x, y ∈ RN , with kxk < kyk, such that u(x) < u(y). Let σ be the reflection such that σ(x1 ) = x2 . To be precise, we consider the hyperplane H passing through x+y 2 and which is perpendicular to the line joining x and y. Note that, since kxk < kyk, 0 does not belong to H. Hence, x belongs to halfspace H + that contains the origin, and y belongs to H − . But then uσ (x) = max{u(x), u(y)} = u(y) and uσ (y) = min{u(x), u(y)} = u(x), which contradicts the fact that u = uσ . Finally, observe that since u is lower semicontinuous, the set {x ∈ RN : u(x) > t} is open for every t ≥ 0. But since u is radially decreasing, this set must be an open ball centered at 0. Hence, Et = Et\ . It follows from (10.13) that u = u\ . Using the Hardy–Littlewood inequality for polarizations, we can prove the following approximation result. In what follows, given n reflections σ1 , . . . , σn and u : RN → R, we define inductively uσ1 σ2 := (uσ1 )σ2 ,
uσ1 ···σk := (uσ1 ···σk−1 )σk
for 3 ≤ k ≤ n.
Theorem 10.27. Given u ∈ Cc (RN ; [0, ∞)), let Pu := {uσ1 ···σn : σ1 , . . . , σn reflections, n ∈ N}. Then there exists un ∈ Pu such that un → u\ uniformly in RN as n → ∞, and supp un ⊆ B(0, R) for all n and for some R > 0. Proof. The proof is similar to that of Theorem 4.37, with the only difference being that we use Theorem 10.26 in place of Theorem 4.35. We leave the details as an exercise. Using Theorem 10.27, one can show the following approximation result.
10.3. Spherically Symmetric Rearrangement
401
Corollary 10.28. Given u, v ∈ Cc (RN ; [0, ∞)), let Pu,v := {(uσ1 ···σn , v σ1 ···σn ) : σ1 , . . . , σn reflections, n ∈ N}. Then there exists (un , vn ) ∈ Pu,v such that un → u# and vn → v # uniformly in RN as n → ∞ and supp un ⊆ B(0, R) and supp vn ⊆ B(0, R) for all n and for some R > 0. Proof. The proof is similar to that of Corollary 4.42, with the only differences being that we use Theorem 10.26 in place of Theorem 4.35 and that in Case 2, when we consider the reflection with respect to the middle point ρ2n , we consider the hyperplane through the middle point ρ2n that is orthogonal to the line joining 0 and ρn . Similarly, in Case 3, we consider the hyperplane n through the middle point xn +y that is orthogonal to the line joining xn and 2 yn . We leave the details as an exercise. Theorem 10.29 (Hardy–Littlewood). Let E ⊆ RN be a Lebesgue measurable set and let u : E → [0, ∞) and v : E → [0, ∞) be Lebesgue measurable functions. Then Z Z u(x)v(x) dx ≤ u] (x)v ] (x) dx. E
RN
To prove this theorem, one can proceed as in the second proof of the Hardy–Littlewood inequality for N = 1 (see Theorem 4.43) using Corollary 10.28. The proof of the following results follows closely that of their onedimensional versions in Chapter 4, and it consists in first proving these inequalities for polarization and then using Corollary 10.28 to extend them to spherically symmetric rearrangements. We refer to Corollary 4.44, Theorem 4.45, and Corollary 4.46, and leave the details as an exercise. Corollary 10.30. Let g : R → [0, ∞) be a convex function with g(0) = 0, and let u : RN → [0, ∞) and v : RN → [0, ∞) be Lebesgue measurable functions vanishing at infinity. Then Z Z ] ] g(u (x) − v (x)) dx ≤ g(u(x) − v(x)) dx. RN
RN
In particular, for 1 ≤ p < ∞, Z Z ] ] p u (x) − v (x) dx ≤ RN
u(x) − v(x)p dx.
RN
Theorem 10.31 (Riesz). Let u : RN → [0, ∞) and v : RN → [0, ∞) be Lebesgue measurable functions vanishing at infinity, and let k : [0, ∞) →
402
10. Symmetrization
[0, ∞) be decreasing. Then Z Z u(x)v(y)k(kx − yk) dxdy RN RN Z Z u# (x)v # (y)k(kx − yk) dxdy. ≤ RN
RN
Corollary 10.32. Let g : R → [0, ∞) be a convex function with g(0) = 0, let u : RN → [0, ∞) and v : RN → [0, ∞) Lebesgue measurable functions vanishing at infinity, and let k : [0, ∞) → [0, ∞) be decreasing and Lebesgue integrable. Then Z Z g(u# (x) − v # (y))k(kx − yk) dxdy N N R R Z Z ≤ g(u(x) − v(y))k(kx − yk) dxdy. RN
RN
In particular, for 1 ≤ p < ∞, Z Z u# (x) − v # (y)p k(kx − yk) dxdy N N R R Z Z ≤ u(x) − v(y)p k(kx − yk) dxdy. RN
RN
We conclude this section by observing that we can define the symmetric decreasing rearrangement of functions which take both positive and negative signs. Definition 10.33. Let E ⊆ RN be a Lebesgue measurable set and let u : E → [−∞, ∞] be a Lebesgue measurable function. The spherically symmetric rearrangement of u is the function u] : R → [0, ∞], defined by u] := u] .
10.4. Spherical Symmetric Rearrangement in W s,p (RN ) In this section we will study the spherical symmetric rearrangements of functions in W s,p (RN ) for 1 ≤ p < ∞ and 0 < s < 1. We recall that u# := u# . Theorem 10.34. Let 1 ≤ p < ∞, 0 < s < 1, and let u : RN → R be a Lebesgue measurable function vanishing at infinity. Then Z Z Z Z u# (x) − u# (y)p u(x) − u(y)p (10.22) dxdy ≤ dxdy. N +sp kx − ykN +sp RN RN RN RN kx − yk # s,p (RN ). Moreover, if un → u in W s,p (RN ), then u# n → u in W
Proof. The proof is very similar to that of Theorem 10.13 and it is left as an exercise.
10.4. Spherical Symmetric Rearrangement in W s,p (RN )
403
Next we consider the case s = 1. ˙ 1,p (RN ) be a Theorem 10.35. Let 1 < p < ∞, 0 < s < 1, and let u ∈ W # 1,p ˙ nonnegative function vanishing at infinity. Then u ∈ W (RN ) and Z Z # p k∇u(x)kp dx. k∇u k dx ≤ RN
RN
First proof. Step 1: Assume that u ∈ W 1,p (RN ) is nonnegative. Then, by Theorem 6.21, u ∈ W s,p (RN ) for every 0 < s < 1. In turn, by Theorem 10.34, u# ∈ W s,p (RN ), with Z Z Z Z u# (x) − u# (y)p u(x) − u(y)p dxdy ≤ dxdy. N +sp N +sp kx − yk RN RN RN RN kx − yk Multiplying both sides by 1 − s and applying Theorem 7.39 to u, we obtain Z Z u# (x) − u# (y)p lim sup(1 − s) dxdy kx − ykN +sp RN R N s→1− Z Z u(x) − u(y)p dxdy ≤ lim sup(1 − s) N +sp RN RN kx − yk s→1− Z = KN,p k∇u(x)kp dx < ∞. RN
Hence, the lefthand side is finite and since 1 < p < ∞, again by Theorem ˙ 1,p (RN ), with 7.39, u# ∈ W Z Z Z u# (x) − u# (y)p KN,p k∇u# (x)kp dx = lim sup(1 − s) dxdy kx − ykN +sp RN RN R N s→1− Z ≤ KN,p k∇u(x)kp dx. RN
By Theorem 10.19, u# ∈ Lp (RN ) with ku# kLp (RN ) = kukLp (RN ) . This proves that u# ∈ W 1,p (RN ). ˙ 1,p (RN ) is a nonnegative function vanishing at Step 2: Assume that u ∈ W infinity. For n ∈ N define t − 1/n if 1/n ≤ t ≤ n, 0 if 0 ≤ t < 1/n, fn (t) := n − 1/n if t > n, and un := fn ◦ u. Then Z Z upn dx ≤ (n − 1/n)p LN ({x ∈ RN : u(x) > 1/n}) < ∞, upn dx = RN
{u>1/n}
404
10. Symmetrization
since u is vanishing at infinity. Moreover, since fn is Lipschitz continuous with Lipschitz constant 1, un belongs to W 1,p (RN ) with ( ∇u(x) if 1/n ≤ u(x) ≤ n, ∇un (x) = 0 if 0 ≤ u(x) < 1/n or u(x) > n for LN a.e. x ∈ RN (see [Leo22d]). Hence, by the previous step Z Z Z p # p k∇u(x)kp dx. k∇un (x)k dx ≤ k∇un (x)k dx ≤ (10.23) RN
RN
RN
Since un ≤ un+1 and un → u pointwise, it follows from Theorem 10.22 # # # pointwise. Since {∇u# } is bounded in that u# n ≤ un+1 and un → u n n Lp (RN ; RN ) because of (10.23), and Lp (RN ) is reflexive for 1 < p < ∞, there exist a subsequence not relabeled and v ∈ Lp (RN ; RN ) such that ∇u# εn * v p N N # in L (R ; R ). It follows that v = ∇u (Why?). Letting n → ∞ in the (10.23) and using the lower semicontinuity of the Lp norm with respect to weak convergence, we obtain Z Z Z # p # p k∇u (x)k dx ≤ lim inf k∇un (x)k dx ≤ k∇u(x)kp dx. n→∞
RN
RN
RN
We present a second proof that relies on polarization. Second proof. Step 1: Assume that u ∈ Cc∞ (RN ). By Theorem 10.27, there exists a sequence un ∈ Pu such that un → u\ uniformly in RN as n → ∞ and supp un ⊆ B(0, R) for all n and for some R > 0. By Theorem 10.14, un ∈ W 1,p (RN ) and Z Z p k∇un (x)k dx = k∇u(x)kp dx RN
RN
for every n. Hence, the sequence {un }n is bounded in W 1,p (RN ). Since 1 < p < ∞, W 1,p (RN ) is reflexive. Thus, given any subsequence, we can find a further subsequence that converges weakly in W 1,p (RN ). Since un → u\ uniformly in RN , by the uniqueness of limits, it follows that un * u\ in W 1,p (RN ). In turn, Z Z Z k∇u\ (x)kp dx ≤ lim inf k∇un (x)kp dx = k∇u(x)kp dx RN
n→∞
RN
RN
by the lower semicontinuity of the norm in Lp with respect to weak convergence. Step 2: Assume that u ∈ W 1,p (RN ) is nonnegative. By the density of Cc∞ (RN ) in W 1,p (RN ) (see [Leo22d]) there exists a sequence un ∈ Cc∞ (RN ) such that un → u in W 1,p (RN ). It follows from Corollary 10.30 that u\n → u\
10.5. Notes
405
in Lp (RN ). Applying Step 1 to each un , we have that u\n ∈ W 1,p (RN ) with Z Z \ p k∇un (x)k dx ≤ k∇un (x)kp dx. RN
RN
Hence, the sequence {un }n is bounded in W 1,p (RN ). Reasoning as in the previous step, we conclude that u\ ∈ W 1,p (RN ). Moreover, Z Z Z k∇u\ (x)kp dx ≤ lim inf k∇u\n (x)kp dx ≤ k∇u(x)kp dx RN
n→∞
RN
by the lower semicontinuity of the norm in gence.
RN
Lp
with respect to weak conver
Remark 10.36. Theorem 4.50 continues to work also for p = 1 but the proof is more involved. Indeed, the proof of Theorem 4.50 would only lead to u# ∈ BV (RN ) whenever u ∈ W 1,1 (RN ). We refer to [Leo17, Theorem 15.23] for more details.
10.5. Notes For more material on polarization, we refer to the book of Baernstein [Bae19], the lecture notes of Burchard [Bur09], as well as the papers [BS00], [BS01], [vS06] and the references therein. For more information on symmetric decreasing rearrangement, we refer to the books of Baernstein [Bae19], Kawohl [Kaw85], Kesavan [Kes06], and [Leo17]. The proofs of Theorems 10.13 and 10.34 are due to Almgren and Lieb [AL89]. I would like to thank Filippo Cagnetti for many useful conversations on Steiner symmetrization, even if at the end I had to cut that part from the book. Sorry, Filippo!
Chapter 11
Higher Order Fractional Sobolev Spaces Philosophy consists mostly of kicking up a lot of dust and then complaining that you can’t see anything. — Gottfried Leibniz
The focus of this book is on fractional Sobolev spaces W s,p , with 0 < s < 1. However, I would feel remiss if I did not give a brief overview of the case s > 1. For simplicity, I will often consider the case 1 < s < 2, and leave the general case as an exercise.
11.1. Definition In this section we introduce the fractional Sobolev spaces W s,p (Ω) for s > 1, s∈ / N. Definition 11.1. Let Ω ⊆ RN be an open set, 1 ≤ p < ∞, and 1 < s < ∞, with s ∈ / N. A function u ∈ W bsc,p (Ω) belongs to the fractional Sobolev space W s,p (Ω) if kukW s,p (Ω) := kukW bsc,p (Ω) + uW s,p (Ω) < ∞, where uW s,p (Ω) := ∇bsc uW s−bsc,p (Ω) . Here bsc is the integer part of s. When p = 2, the space W s,2 (Ω) is a Hilbert space and is denoted H s (Ω). Definition 11.2. Let Ω ⊆ RN be an open set, 1 ≤ p < ∞, and s > 1, s ∈ / N. We define the fractional Sobolev space W0s,p (Ω) as the closure of Cc∞ (Ω) with respect to the norm in W s,p (Ω). When p = 2 we write H0s (Ω) := W0s,2 (Ω). 0 The dual space of W0s,p (Ω) is denoted by W −s,p (Ω). 407
408
11. Higher Order Fractional Sobolev Spaces
Definition 11.3. Let Ω ⊆ RN be an open set, 1 ≤ p < ∞, and 1 < s < ∞, bsc,p with s ∈ / N. A function u ∈ Wloc (Ω) belongs to the homogeneous fractional ˙ s,p (Ω) if ∇bsc u s−bsc,p Sobolev space W W (Ω) < ∞. When p = 2 we write s s,p ˙ (Ω). H˙ (Ω) := W Exercise 11.4. Let 1 ≤ p < ∞ and 1 < s < ∞, with s ∈ / N. Given s,p N ˙ u ∈ W (R ), let uε := ϕε ∗ u, where ϕε is a standard mollifier. Prove that k∇k uε kLp (RN ) ≤ k∇k ukLp (RN ) ,
uε W s,p (RN ) ≤ uW s,p (RN )
for all k = 0, . . . , bsc, and uε → u in W s,p (RN ). Prove also that W s,p (RN ) ∩ C ∞ (RN ) is dense in W s,p (RN ). Exercise 11.5. Let Ω ⊆ RN be an open set, 1 ≤ p < ∞, and 1 < s < ∞ with s ∈ / N. Prove that W s,p (Ω) is a Banach space. We recall that F denotes the Fourier transform. Exercise 11.6. Let 1 < s < ∞ with s ∈ / N and let u ∈ S(RN ). Prove that Z 1/2 2 2 s (1 + kxk ) F(u)(x) dx ∼ kukH s (RN ) . RN
Hint: See Exercise 6.9. Exercise 11.7. Given m ∈ N and a sufficiently smooth function u : RN → R, we define X 1 Pm u(x, y) := ∂ α u(x)y α , x, y ∈ RN , α! 0≤α≤m
where, as usual,
∂0u
:= u.
˙ s,p (RN ) ∩ C ∞ (RN ). (i) Let 1 ≤ p < ∞ and 1 < s < 2 and let u ∈ W 1 N Prove that for h ∈ R , with 0 < khk < 2 , Z 1 u(x + h) − P1 (u)(x, h)p ≤ khkp k∇u(x + th) − ∇u(x)kp dt. 0
˙ s,p (RN ) ∩ C ∞ (RN ). (ii) Let 1 ≤ p < ∞ and 1 < s < 2 and let u ∈ W 1 N Prove that for h ∈ R , with 0 < khk < 2 , Z u(x + h) − P1 (u)(x, h)p dx khksp ∇upW s−1,p (RN ) . RN
(iii) Let 1 ≤ p < ∞, s > 1, s ∈ / N, and h ∈ RN , with 0 < khk < Prove that ku(· + h) − Pbsc u(·, h)kLp (RN ) khks uW s,p (RN ) ˙ s,p (RN ). for every u ∈ W
1 2.
11.1. Definition
409
Theorem 11.8 (Compactness). Let 1 ≤ p < ∞, s > 1, s ∈ / N, and {un }n be bounded in W s,p (RN ). Then there exist a subsequence {unk }k and a function bsc,p u ∈ W s,p (RN ) such that unk → u in Wloc (RN ) as k → ∞. Proof. Let {un }n be a sequence bounded in W s,p (RN ), with kun kW s,p (RN ) ≤ M for every n. Then {un }n is bounded in W bsc,p (RN ), and so by the Rellich– Kondrachov compactness theorem (see [Leo22d]), there exist a subsequence bsc−1,p (RN ), where {unk }k and u ∈ W bsc,p (RN ) such that unk → u in Wloc 0,p p Wloc := Lloc . On the other hand, if α is a multiindex with α = bsc, since ∂ α unk ∈ W s−bsc,p (RN ), by Theorem 6.13 we can find further subsequences, not relabeled, and a function vα ∈ Lp (RN ) such that ∂ α unk → vα in Lploc (RN ) and pointwise LN a.e. in RN . It follows that vα is the αth weak partial derivative of u (Why?). By Fatou’s lemma, we have uW s,p (RN ) ≤ lim inf unk W s,p (RN ) ≤ M. k→∞
Hence, u ∈ W s,p (RN ), and the proof is complete.
Exercise 11.9. Let 1 ≤ p < ∞, s > 1, with s ∈ / N, and {un }n be bounded in W s,p (RN ). Prove that there exist a subsequence {unk }k and a function σ,p (RN ) for every bsc < σ < s. u ∈ W s,p (RN ) such that unk → u in Wloc Next we discuss the product of two functions. Exercise 11.10. Let Ω ⊆ RN be an open set, m ∈ N, 1 ≤ p < ∞, and ∞ u, v ∈ W m,p (Ω) ∩ W m−1,∞ (Ω), where W 0,∞ (Ω) Prove that for P := Lα (Ω). α N every α ∈ N0 with 1 ≤ α ≤ m, ∂ (uv) = β≤α β ∂ α−β u∂ β v, and that uv ∈ W m,p (Ω). Hint: Use induction. Theorem 11.11 (Product). Let 1 ≤ p < ∞, s > 1, s ∈ / N, u, v ∈ W s,p (RN ) ∩ W m−1,∞ (RN ), where m = bsc. Then uv ∈ W s,p (RN ), with kuvkW s,p (RN ) kukW s,p (RN ) kvkW m−1,∞ (RN ) + kukW m−1,∞ (RN ) kukW s,p (RN ) . Proof. Since u ∈ W m,p (RN ), for every multiindex β ∈ NN 0 , with β = 0, . . . , m − 1, we have that ∂ β u ∈ W 1,p (RN ), where ∂ 0 u := u. Hence, by Theorem 6.21, ∂ β u ∈ W s−m,p (RN ), with ∂ β uW s−m,p (RN ) k∂ β ukW 1,p (RN ) kukW β+1,p (RN ) .
410
11. Higher Order Fractional Sobolev Spaces
In turn, given α ∈ NN 0 , with α = m, by Exercise 11.10 and Theorems 6.21, 6.26, X ∂ α (uv)W s−m,p (RN ) ∂ α−β u∂ β vW s−m,p (RN ) β≤α
X
[∂
α−β
uW s−m,p (RN ) k∂ β vkL∞ (RN ) + k∂ α−β ukL∞ (RN ) ∂ β vW s−m,p (RN ) ]
β≤α
kukW s,p (RN ) kvkW m−1,∞ (RN ) + kukW m−1,∞ (RN ) kukW s,p (RN ) . Moreover, again by Exercise 11.10, X k∂ α (uv)kLp (RN ) k∂ α−β u∂ β vkLp (RN ) β≤α
X
k∂ α−β ukLp (RN ) k∂ β vkL∞ (RN ) .
β≤α
Remark 11.12. Theorem 11.11 continues to hold for an open connected set Ω ⊆ RN with the property that there exists a linear operator E : W s,p (Ω) ∩ W m−1,∞ (Ω) → W s,p (RN ) ∩ W m−1,∞ (RN ) such that E(u)(x) = u(x) for x ∈ Ω and kE(u)kW s,p (RN ) Ω kukW s,p (Ω) ,
kE(u)kW m−1,∞ (RN ) Ω kukW m−1,∞ (Ω)
for all u ∈ W s,p (Ω) ∩ W m−1,∞ (Ω). Indeed, it is enough to apply the previous theorem to E(u) and E(v). We will see in Section 11.7 that if Ω has sufficiently regular boundary, then such an extension operator E exists. Exercise 11.13. Let Ω ⊆ RN be an open set, 1 ≤ p < ∞, s > 1, s ∈ / N, s,p m+1,1 u ∈ W (Ω), and ψ ∈ C (Ω), where m = bsc. (i) Prove that ψu ∈ W s,p (Ω), with kψukW s,p (Ω) kψkC m+1,1 (Ω) kukW s,p (Ω) . (ii) Prove that if ψ has support contained in Ω, with dist(supp ψ, ∂Ω) ≥ δ0 > 0, then the function ψu, extended to be zero outside Ω, belongs to W s,p (RN ), with kψukW s,p (RN ) δ0 kψukW s,p (Ω) . Definition 11.14. Given 1 ≤ p < ∞ and s > 0, an open set Ω ⊆ RN is called an extension domain for the Sobolev space W s,p (Ω) if there exists a continuous linear operator E : W s,p (Ω) → W s,p (RN ) such that for all u ∈ W s,p (Ω), E(u)(x) = u(x) for LN a.e. x ∈ Ω. Theorem 11.15. Let 1 ≤ p < ∞, n ∈ N, and Ω ⊆ RN be a W 1,p extension domain. Then kukW s,p (Ω) Ω kukW n,p (Ω) for every 0 < s < n and for all u ∈ W n,p (Ω). In particular, W n,p (Ω) ,→ W s,p (Ω).
11.1. Definition
411
Proof. If s ∈ N, then there is nothing to prove. Assume that s ∈ / N and let m := bsc < s < n. If m = 0, then the result follows from Theorem 6.21. Assume that m ≥ 1 and let α ∈ NN 0 be a multiindex with α = m. Then ∂ α u ∈ W n−m,p (Ω), and so, by Theorem 6.21, ∂ α uW s−m,p (Ω) Ω k∂ α ukW 1,p (Ω) Ω kukW n,p (Ω) . It follows that uW s,p (Ω) Ω kukW n,p (Ω) .
Theorem 11.16. Let 1 ≤ p < ∞, s > 1, with s ∈ / N, and Ω ⊆ RN be a s,p W extension domain. Then kukW σ,p (Ω) Ω kukW s,p (Ω) for every 0 < σ < s and all u ∈ W s,p (Ω). Proof. We give the proof only in the case 1 < s < 2. Step 1: Assume that Ω = RN and let u ∈ W s,p (RN ). If σ = 1, there is nothing to prove, while if σ > 1, then we can apply Lemma 6.17 to the partial ∂u derivatives of u, to obtain that ∂x ∈ W σ−1,p (RN ), and so u ∈ W σ,p (RN ). i Finally, if σ < 1, we apply Theorem 6.21. Step 2: Let Ω be a W s,p extension domain and let u ∈ W s,p (Ω). Then there exists v ∈ W s,p (RN ) such that v = u on Ω and kvkW s,p (RN ) Ω kukW s,p (Ω) . It suffices to apply the previous step to v. Definition 11.17. 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 X kukC m,α (Ω;RM ) := kukC m (Ω;RM ) + ∂ α uC 0,α (Ω;RM ) < ∞. α=m
We recall that for a function v : Ω → RM , kv(x) − v(y)kM sup vC 0,α (Ω;RM ) := . kx − ykαN x, y∈Ω, x6=y As usual, when M = 1 we write C m,α (Ω) for C m,α (Ω; R). Theorem 11.18 (Change of variables). Let 1 ≤ p < ∞, s > 1, s ∈ / N, N m = bsc, let Ω, U ⊆ R be open sets, let Ψ : U → Ω be invertible, with Ψ ∈ C m,1 (U ; RN ) and Ψ−1 Lipschitz continuous, and let u ∈ W s,p (Ω). If m > 2, assume also that Ω is a W 1,p extension domain. Then u ◦ Ψ ∈ W s,p (U ), with ku ◦ ΨkW s,p (U ) Ψ kukW s,p (Ω) . Proof. Step 1: Assume that 1 < s < 2. Then u ∈ W 1,p (Ω) and so by the change of variables in Sobolev spaces (see [Leo22d]), u ◦ Ψ belongs to W 1,p (U ), and for all i = 1, . . . , N and for LN a.e. y ∈ U , N
X ∂u ∂Ψj ∂(u ◦ Ψ) (y) = (Ψ(y)) (y). ∂yi ∂xj ∂yi j=1
412
11. Higher Order Fractional Sobolev Spaces
We claim that
∂u ∂u ◦ Ψ ∈ W s−1,p (U ). For simplicity we write ∂j u := . ∂xj ∂xj
If x, y ∈ Ω, (11.1) kx − yk = kΨ(Ψ−1 (x)) − Ψ(Ψ−1 (y))k ≤ Lip ΨkΨ−1 (x) − Ψ−1 (y)k. Moreover, since Ψ−1 is Lipschitz,  det JΨ−1 (y) (Lip Ψ−1 )N for every y ∈ Ω, and so, for every X ∈ U , 1 1 =  det JΨ (X). −1 N (Lip Ψ )  det JΨ−1 (Ψ(X)) It follows that ∂j u(Ψ(X)) − ∂j u(Ψ(Y ))p dXdY kX − Y kN +(s−1)p U U Z Z ∂j u(Ψ(X)) − ∂j u(Ψ(Y ))p −1 2N (Lip Ψ ) kX − Y kN +(s−1)p Ψ−1 (Ω) Ψ−1 (Ω)
∂j u ◦
(11.2)
ΨpW s−1,p (U )
Z Z
=
×  det JΨ (X)dX  det JΨ (Y )dY Z Z ∂j u(x) − ∂j u(y)p −1 2N (Lip Ψ ) dxdy N +(s−1)p −1 −1 Ω Ω kΨ (x) − Ψ (y)k Z Z ∂j u(x) − ∂j u(y)p −1 2N N +(s−1)p (Lip Ψ ) (Lip Ψ) dxdy, N +(s−1)p Ω Ω kx − yk
where we use (11.1) and the changes of variables x = Ψ(X) and y = Ψ(Y ) (see [Leo17, Theorem 9.52]). Similarly, k∂j u ◦
ΨkpLp (U )
Z
∂j u(Ψ(X))p dX Z −1 N (Lip Ψ ) ∂j u(Ψ(X))p  det JΨ (X)dX −1 Ψ (Ω) Z (Lip Ψ−1 )N ∂j u(x)p dx. =
U
Ω
∂Ψj ∂u ◦ Ψ ∈ W s−1,p (U ). Since ∈ C 0,1 (U ), we can now ∂xj ∂y i ∂Ψj ∂u ◦Ψ ∈ W s−1,p (U ), with apply Theorem 6.23 to obtain that ∂xj ∂yi This shows that
∂u ∂Ψj
∂xj ◦ Ψ ∂yi
Lp (U )
∂u
≤ k∇Ψj kL∞ (U ) ◦ Ψ
∂xj
Lp (U )
∂u
Ψ ∂xj
Lp (Ω)
11.1. Definition
413
and
∂u
∂u ∂Ψj 2−s s−1 2
◦ Ψ k∇Ψj kL∞ (U ) k∇ Ψj kL∞ (U ) ∂xj ◦ Ψ ∂yi s−1,p
p ∂xj W (U ) L (U )
∂u ∂u
◦ Ψ Ψ + k∇Ψj kL∞ (U )
∂xj s−1,p . ∂xj s−1,p W (U ) W (Ω) Step 2: The situation is more involved in the case s > 2. Let m = bsc. Then u ◦ Ψ belongs to W m,p (U ) (see [Leo22d]), and for every multiindex N α ∈ NM 0 , with α = m, and for L a.e. y ∈ U , β
X Y ∂ γi  Ψl ∂ α ∂ β u i (u ◦ Ψ)(y) = c (Ψ(y)) (y), α,β,γ,l ∂y α ∂y γi ∂xβ i=1
where cα,β,γ,l ∈ R, the sum is done over all β ∈ NN 0 with 1 ≤ β ≤ α, γ = Pβ M (γ1 , . . . , γβ ), γi ∈ N0 , with γi  > 0 and i=1 γi = α, and l = (l1 , . . . , lβ ), li ∈ {1, . . . , N }, i = 1, . . . , β. There are now two cases. If β = m, then β we can proceed as in (11.2) to show that ∂∂xβu ◦ Ψ ∈ W s−m,p (U ), with p ∂ β u ∂ β u p Ψ β , β ◦ Ψ s−m,p ∂x ∂x s−m,p W (U ) W (Ω)
p
∂ β u
∂ β u p
Ψ β .
β ◦ Ψ
p
∂x p
∂x L (U )
By Theorem 6.23, we have that
! β
β
Y ∂ γi  Ψl
∂ u i
∂xβ ◦ Ψ ∂y γi
i=1
∂ β u ∂xβ
Lp (U )
L (Ω)
Q β ◦Ψ i=1
∂ γi  Ψli ∂y γi
∈ W s−m,p (U ), with
β γ 
∂ β u
Y ∂ i Ψli
≤ ◦ Ψ
γ β i
p ∂x
i=1 ∂y
∞ L (U ) L (U )
∂ β u
Ψ β ,
∂x p L (Ω)
414
11. Higher Order Fractional Sobolev Spaces
and ! β β γ  Y i ∂ u ∂ Ψli ∂xβ ◦ Ψ ∂y γi i=1 W s−m,p (U )
1−(s−m)
!
Y
s−m
Y
∂ β u γi  Ψ
β ∂ γi  Ψli
β ∂
li
∇ ◦ Ψ
γi γi β
p
∂y ∂y ∂x
i=1
∞
∞
i=1 L (U ) L (U ) L (U )
Y
∂ β u
∂ β u
β ∂ γi  Ψli
+ ◦ Ψ .
β Ψ
γ β i ∂x s−m,p
∂x s−m,p ∂y
i=1
If 1 ≤ β < m, then ∂ β u
W
L∞ (U ) ∂ β u ∂xβ
(U )
W
(Ω)
∈ W m−β,p (Ω). Since Ω is a W 1,p extension
domain, ∂xβ belongs to W s−m,p (Ω) (see Theorem 11.15). We can now proceed as in the case β = m. Corollary 11.19 will be important when dealing with domains Ω with smooth boundary. Corollary 11.19 (Flattening the boundary). Let 1 ≤ p < ∞, s > 1, s ∈ / N, m = bsc, f ∈ C m,1 (RN −1 ), Ω = {x = (x0 , xN ) ∈ RN −1 × R : xN > f (x0 )}, and u ∈ W s,p (Ω). Then the function v : RN + → R, given by v(y) := u(y 0 , yN + f (y 0 )),
y ∈ RN +,
belongs to W s,p (RN + ), with kvkW s,p (RN ) f kukW s,p (Ω) . +
Proof. Consider the transformation Ψ : RN → RN given by Ψ(y) := (y 0 , yN + f (y 0 )). Note that Ψ is of class C m,1 , invertible, with inverse given by Ψ−1 (x) = (x0 , xN − f (x0 )). Moreover, for all y, z ∈ RN , kΨ(y) − Ψ(z)k = k(y 0 − z 0 , f (y 0 ) − f (z 0 ) + yN − zN )k q ≤ ky 0 − z 0 k2N −1 + (Lip f ky 0 − z 0 kN −1 + yN − zN )2 (1 + Lip f )ky − zk, which shows that Ψ (and similarly Ψ−1 ) is Lipschitz continuous. Note that Ψ(RN + ) = Ω. We can now apply Theorem 11.18. Exercise 11.20. Let Ω ⊆ RN be an open set, and let 1 ≤ p < ∞, s > 1, s∈ / N, u ∈ W s,p (Ω). Consider ϕn ∈ Cc∞ (RN ) such that 0 ≤ ϕn ≤ 1, ϕn = 1 in B(0, n), ϕn = 0 outside B(0, n + 1), and k∇k ϕn k∞ ≤ C for all n ∈ N and all k = 1, . . . , bsc + 1. Prove that ϕn u → u in W s,p (Ω). Hint: Adapt the proof of Theorem 6.66 and use Exercise 11.10.
11.2. Some Equivalent Seminorms
415
Exercise 11.21. Let 1 ≤ p < ∞ and s > 1, s ∈ / N. Prove that Cc∞ (RN ) is s,p N dense in W (R ). Exercise 11.22. Let Ω ⊆ RN be an open set, and let 1 ≤ p < ∞, and 1 < s < ∞ with s ∈ / N. Prove that for every u ∈ W s,p (Ω) there exists a sequence {un }n in W s,p (Ω) ∩ C ∞ (Ω) such that un → u in W s,p (Ω) as n → ∞.
11.2. Some Equivalent Seminorms Given u : RN → R, m ∈ N, n = 1, . . . , N , and t ∈ R, we define inductively m−1 ∆m t,n u(x) := ∆t,n (∆t,n u)(x).
∆1t,n u(x) := ∆t,n u(x) := u(x + ten ) − u(x),
Theorem 11.23. Let 1 ≤ p < ∞, s > 1, s ∈ / N, and m = bsc. A function p N s,p N u ∈ L (R ) belongs to W (R ) if and only if Z ∞ Z dt u(x)p dx 1+sp < ∞. sup ∆m+1 h t 0 khk≤t RN Moreover, for all u ∈ W s,p (RN ), ∇
(11.3)
m
upW s−m,p (RN )
Z
∞
≈
Z sup
0
RN
khk≤t
∆m+1 u(x)p dx h
dt . t1+sp
Proof. We prove the case 1 < s < 2 and leave the general case s > 2 as an exercise (see Exercise 11.25). Step 1: Let 1 < s < 2 and u ∈ Lp (RN ) be such that Z ∞ Z dt (11.4) sup ∆2h u(x)p dx 1+sp < ∞. t N 0 khk≤t R We claim that u ∈ W 1,p (RN ) and that for every ` > 0, Z
1 k∂i ukLp (RN ) kukLp (RN ) + `
(11.5)
`
sup k∆2h ukLp (RN )
0 khk≤t
dt . t2
Note that the righthand side of the previous inequality is finite. To see this, 0 write t2 = t1/p+s t1+1/p −s and apply H¨ older’s inequality to get Z (11.6)
`
sup 0 khk≤t
dt k∆2h ukLp (RN ) 2 t
`
s−1
Z
`
sup k∆2h ukpLp (RN )
0 khk≤t
dt 1/p t1+sp
,
416
11. Higher Order Fractional Sobolev Spaces
R` 0 0 where we use the fact that 0 t(s−1)p −1 dt `(s−1)p . Using telescopic sums, for i = 1, . . . , N , k ∈ Z, l ∈ N, and x ∈ RN , we can write 2l+k ∆2−l−k ,i u(x) − 2k ∆2−k ,i u(x) =−
l+k−1 X
[2j (u(x + 2−j ei ) − u(x)) − 2j+1 (u(x + 2−j−1 ei ) − u(x))]
j=k
=−
l+k−1 X
2j ∆22−j−1 ,i u(x).
j=k
Taking the Lp norm in x, we obtain (11.7) ∞ X
k2l+k ∆2−l−k ,i u − 2k ∆2−k ,i ukLp (RN ) ≤
2j k∆22−j−1 ,i ukLp (RN )
j=k
≤2
∞ Z X
2−j
sup
2−j−1 khk≤t
j=k
Z
dt k∆2h ukLp (RN ) 2
2−k
=2
t
0
sup k∆2h ukLp (RN )
khk≤t
dt . t2
By (11.4) and (11.6), the righthand side of the previous inequality goes to zero as k → ∞. Hence, the sequence {2k ∆2−k ,i u}k is a Cauchy sequence in Lp (RN ), and thus, it converges to some function vi ∈ Lp (RN ). For every ϕ ∈ Cc∞ (RN ), by a change of variables we have that Z
Z
k
RN
2 ∆2−k ,i u(x)ϕ(x) dx =
u(y) RN
ϕ(y − 2−k ei ) − ϕ(y) dx. 2−k
Letting k → ∞ and using the facts that 2k ∆2−k ,i u → vi in Lp (RN ) on the lefthand side and the Lebesgue dominated convergence theorem on the righthand side, we obtain Z
Z vi (x)ϕ(x) dx = −
u(y)∂i ϕ(y) dy,
RN
RN
which shows that vi is the distributional partial derivative ∂i u. Thus, u ∈ W 1,p (RN ). To prove (11.5), let l → ∞ in (11.7) to get k
Z
k∂i u − 2 ∆2−k ,i ukLp (RN ) ≤ 2 0
2−k
sup k∆2h ukLp (RN )
khk≤t
dt . t2
11.2. Some Equivalent Seminorms
417
Given ` > 0, let k ∈ Z be such that 2−k ≤ ` < 2−k+1 . Then Z 2−k dt k sup k∆2h ukLp (RN ) 2 (11.8) k∂i ukLp (RN ) ≤ 2 k∆2−k ,i ukLp (RN ) + 2 t 0 khk≤t Z ` 1 dt kukLp (RN ) + sup k∆2h ukLp (RN ) 2 . ` t 0 khk≤t Step 2: Let 1 < s < 2 and u ∈ Lp (RN ) be such that (11.4) holds. We claim that every t > 0, Z t dτ 2 (11.9) sup k∆ξ (∂i u)kLp (RN ) sup k∆2h ukLp (RN ) 2 . τ 0 khk≤τ kξk≤t By Step 2, k∂i ukLp (RN )
1 kukLp (RN ) + t
Z
t
sup k∆2h ukLp (RN )
0 khk≤τ
dτ . τ2
Replacing u with ∆2ξ u, where kξk ≤ t, gives k∆2ξ (∂i u)kLp (RN ) = k∂i (∆2ξ u)kLp (RN ) 1 k∆2ξ ukLp (RN ) + t
Z
t
sup k∆2h (∆2ξ u)kLp (RN )
0 khk≤τ
dτ . τ2
Since k∆ζ vkLp (RN ) ≤ 2kvkLp (RN ) , we have Z t 1 2 dτ 2 k∆ξ (∂i u)kLp (RN ) k∆ξ ukLp (RN ) + sup k∆2h ukLp (RN ) 2 t τ 0 khk≤τ Z t dτ 1 sup k∆2h ukLp (RN ) 2 sup k∆2h ukLp (RN ) + t khk≤t τ 0 khk≤τ Z t dτ sup k∆2h ukLp (RN ) 2 , τ 0 khk≤2τ where in the last inequality we use the fact that for t > 0, Z t 1 dτ 2 sup k∆h ukLp (RN ) sup k∆2h ukLp (RN ) 2 , t khk≤t τ t/2 khk≤2τ since the function τ 7→ supkhk≤τ k∆2h ukLp (RN ) is increasing. Step 3: Let 1 < s < 2 and u ∈ Lp (RN ) be such that (11.4) holds. By Step 2, Z ∞ dt sup k∆2ξ (∂i u)kpLp (RN ) 1+(s−1)p t 0 kξk≤t Z ∞ Z t dτ p dt sup k∆2h ukLp (RN ) 2 . 1+(s−1)p τ t 0 0 khk≤τ
418
11. Higher Order Fractional Sobolev Spaces
By Hardy’s inequality (see Theorem 1.3) applied on the righthand side of the previous inequality and Theorem 6.52 applied on the lefthand side, we obtain Z ∞ Z Z Z ∆2h (∂i u)(x)p dt sup ∆2h u(x)p dx 1+sp . dxdh N +(s−1)p t RN RN khk 0 khk≤t RN Since u ∈ W 1,p (RN ), we can now apply Corollary 6.60 to ∂i u ∈ Lp (RN ) to obtain the first inequality in (11.3). Step 4: Let 1 < s < 2. To prove the second inequality in (11.3), let u ∈ W s,p (RN ) ∩ C ∞ (RN ). By the fundamental theorem of calculus, ∆2h u(x) = [u(x + 2h) − u(x + h)] − [u(x + h) − u(x)] Z 1 Z 1 = ∇u(x + h + th) · h dt − ∇u(x + th) · h dt, 0
0
R1
and so ≤ khk 0 k∇u(x + h + th) − ∇u(x + th)k dt. Taking the Lp norm in x on both sides and using Minkowski’s inequality for integrals (see [Leo22c]) gives Z 1 2 k∆h ukLp (RN ) ≤ khk k∆h ∇u(· + th)kLp (RN ) dt = khkk∆h ∇ukLp (RN ) , ∆2 u(x)
0
where in the last equality we make the change of variables x + th = y, so that dx = dy. Raising both sides to power p, dividing by khkN +sp , and integrating in h over RN gives Z Z Z Z ∆2h u(x)p k∆h ∇u(x)p dxdh ≤ dxdh. N +sp N +(s−1)p RN RN khk RN RN khk Now, if u ∈ W s,p (RN ), let uε = u ∗ ϕε , where ϕε is a standard mollifier. Then, by what we just proved, we have Z Z Z Z ∆2h uε (x)p k∆h ∇uε (x)p dxdh ≤ dxdh N +sp N +(s−1)p RN RN khk RN RN khk Z Z k∆h ∇u(x)p ≤ dxdh, N +(s−1)p RN RN khk where the last inequality follows from the fact that ∇uε = ∇u ∗ ϕε (see [Leo22c]). Let u ¯ be a representative of u. Letting ε → 0+ and using Fatou’s lemma on the left together with the fact that uε (x) → u ¯(x) at every Lebesgue point of u ¯ (see [Leo22c]), we obtain Z Z Z Z ∆2h u(x)p k∆h ∇u(x)p dxdh ≤ dxdh. N +sp N +(s−1)p RN RN khk RN RN khk Step 5: For the case s > 2, see Exercise 11.25.
11.2. Some Equivalent Seminorms
419
Remark 11.24. Note that Step 1 of the proof of Theorem 11.23 shows that B 1,1 (RN ) ,→ W 1,1 (RN ) (see Exercises 6.10 and 6.54). Exercise 11.25. Let 1 ≤ p < ∞, s > 1, s ∈ / N, and m = bsc. (i) Let u ∈ Lp (RN ) be such that Z ∞ Z dt sup ∆m+1 u(x)p dx 1+sp < ∞. h t 0 khk≤t RN Prove that for all ν ∈ SN −1 , k ∈ Z, and l ∈ N, km m 2(l+k)m ∆m ∆2−k ν u 2−l−k ν u − 2
=
l+k−1 X
2(n+1)m Pm−1 (T2−n−1 ν )∆m+1 u. 2−n−1 ν
n=k
Hint: Use Lemma 6.56. (ii) Let u be as in item (i). Prove that Z (l+k)m m km m k2 ∆2−l−k ν u − 2 ∆2−k ν ukLp (RN )
2−k
ukLp (RN ) sup k∆m+1 h
khk≤t
0
dt t1+m
Deduce that there is the distributional directional derivative m and that 2km ∆m u → ∂∂ν mu in Lp (RN ) as k → ∞. 2−k ν
.
∂mu ∂ν m
(iii) Let u be as in item (i). Prove that u belongs to W s,p (RN ) and that Z ∞ Z dt m p ∇ uW s−m,p (RN ) ≈ sup ∆m+1 u(x)p dx 1+sp h t 0 khk≤t RN for all u ∈ W s,p (RN ). Corollary 11.26. Let 1 ≤ p < ∞ and s > 1, s ∈ / N. Then for every p N s,p N N ˙ u ∈ L (R ) ∩ W (R ) and α ∈ N0 , with 1 ≤ α ≤ m = bsc, k∂ α ukLp (RN ) kukLp (RN ) + ∇m uW s−m,p (RN ) . Proof. Given ` > 0 and ν ∈ SN −1 , let k ∈ Z be such that 2−k ≤ ` < 2−k+1 . Then by item (ii) in Exercise 11.25, k2(l+k)m ∆m 2−l−k ν ukLp (RN ) 2
km
k∆m 2−k ν ukLp (RN )
2km+m kukLp (RN ) +
Z
2−k
+ 0
`−m kukLp (RN ) +
khk≤t
2−k
Z 0
Z
sup k∆m+1 ukLp (RN ) h
sup k∆m+1 ukLp (RN ) h
khk≤t
`
sup k∆m+1 ukLp (RN ) h
0 khk≤t
dt t1+m
dt t1+m
dt t1+m .
420
11. Higher Order Fractional Sobolev Spaces
m
Letting l → ∞ and using the fact that 2(l+k)m ∆m u → ∂∂ν mu in Lp (RN ) 2−l−k ν as l → ∞, and H¨ older’s inequality, we obtain !1/p
m Z `
∂ u dt p −m s−m
sup k∆m+1 ukLp (RN ) 1+sp , h
∂ν m p N ` kukLp (RN ) +` t 0 khk≤t L (R ) where we write
1 t1+m
1 = t1/p+s ts−m−1+1/p and use the fact that Z ` 1/p0 0 t(s−m)p −1 dt `s−m . 0
By applying the previous theorem, we obtain
m
∂ u −m s−m
∇m uW s−m,p (RN ) .
∂ν m p N ` kukLp (RN ) + ` L (R ) We leave it as an exercise to check that this implies that for every multiindex α with α = m, k∂ α ukLp (RN ) `−m kukLp (RN ) + `s−m ∇m uW s−m,p (RN ) . In turn, by the Gagliardo–Nirenberg interpolation theorem (see [Leo22d]), for 1 ≤ k < m and ` = 1, 1−k/m
k/m
k∇k ukLp (RN ) kukLp (RN ) k∇m ukLp (RN ) kukLp (RN ) + k∇m ukLp (RN ) kukLp (RN ) + ∇m uW s−m,p (RN ) .
Now we show that we can replace the first order difference of the highest derivatives with higher order differences. Theorem 11.27. Let m ∈ N, 1 ≤ p < ∞, and s > 1, s ∈ / N, with m > bsc. A function u ∈ Lp (RN ) belongs to W s,p (RN ) if and only if Z dh p < ∞. (11.10) k∆m h ukLp (RN ) khkN +sp RN Moreover, for all u ∈ W s,p (RN ), (11.11)
∇
bsc
upW s−bsc,p (RN )
Z ≈ RN
p k∆m h ukLp (RN )
dh . khkN +sp
Proof. Let u ∈ Lp (RN ) be such that (11.10) holds. By Theorem 6.52 (which was proved for every s), Z Z ∞ Z dh dt p m p (11.12) k∆h ukLp (RN ) ≈ sup ∆m h u(x) dx 1+sp . N +sp khk t N N R 0 khk≤t R If m = bsc+1, it follows from Theorem 11.23 that u ∈ W s,p (RN ). Moreover, by combining (11.3) and (11.12), we obtain (11.11). On the other hand, if m > bsc+1, let ε > 0 and consider the mollification m uε = u ∗ ϕε (see [Leo22c]). By linearity ∆m h uε = (∆h u) ∗ ϕε . Since
11.2. Some Equivalent Seminorms
421
kv ∗ ϕε kLp (RN ) ≤ kvkLp (RN ) for any v ∈ Lp (RN ) (see [Leo22c]), by (11.10) we get, Z Z dh dh p p m k∆h uε kLp k∆m ≤ < ∞. h ukLp N +sp khk khkN +sp RN RN We claim that uε ∈ W s,p (RN ). Since ∂ α uε = ∂ α ϕε ∗ u for every multiindex α (see [Leo22c]), it follows that ∂ α uε belongs to Lp (RN ), so uε ∈ W n,p (RN ) for every n. In turn, uε ∈ W s,p (RN ) by Theorem 11.15. We can now apply Theorems 6.52 and 11.23 to obtain Z Z Z ∞ dt dh bsc+1 bsc+1 p ∆h u(x)p dx 1+sp k∆h uε kLp (RN ) sup N +sp khk t RN 0 khk≤t RN ∇bsc+1 uε pW s−bsc,p (RN ) < ∞. Since condition (6.37) holds with n = bsc + 1, by Theorem 6.55, Z Z dh dh bsc+1 p p k∆h uε kLp (RN ) k∆m h uε kLp N +sp khk khkN +sp N RN ZR dh p . ≤ k∆m h ukLp khkN +sp RN Since uε → u pointwise LN a.e. in RN (see [Leo22c]), letting ε → 0+ in the previous inequality and using Fatou’s lemma on the left gives Z Z dh dh bsc+1 p p k∆m < ∞. k∆h ukLp (RN ) h ukLp N +sp khk khkN +sp RN RN We can now use the first part of the proof (the case m = bsc + 1) to obtain that u ∈ W s,p (RN ) with Z dh bsc+1 bsc p ∇ uW s−bsc,p (RN ) k∆h ukpLp khkN +sp N ZR Z dh dh bsc+1 p k∆m uk k∆h ukpLp h Lp N +sp khk khkN +sp RN RN ∇bsc upW s−bsc,p (RN ) , where the third inequality follows from Theorem 6.55.
The previous theorem allow us to use slicing techniques. Theorem 11.28. Let 1 ≤ p < ∞, s > 0, s ∈ / N, and let m ∈ N be such that m ≥ bsc + 1. A function u ∈ Lp (RN ) belongs to W s,p (RN ) if and only if (11.13)
N Z Z X i=1
R
p ∆m t,i u(x) dxdt < ∞. t1+sp RN
422
11. Higher Order Fractional Sobolev Spaces
Moreover, for all u ∈ W s,p (RN ), ∇
bsc
upW s−bsc,p (RN )
≈
N Z Z X i=1
R
p ∆m t,i u(x) dxdt. t1+sp RN
Proof. Let u ∈ Lp (RN ) be such that (11.13) holds. By Theorem 6.61, Z Z N Z Z p p X ∆m ∆mN t,i u(x) h u(x) dxdh dxdt < ∞. N +sp t1+sp RN RN khk R RN i=1
Hence, we can apply Theorem 11.27 to conclude that u ∈ W s,p (RN ). Moreover, by Theorems 6.61 and 11.27, Z dh p ∇bsc upW s−bsc,p (RN ) k∆mN h ukLp khkN +sp RN Z Z N Z Z p p X ∆m ∆m t,i u(x) h u(x) dxdt dxdh N +sp t1+sp RN RN khk R RN i=1 bsc
∇
upW s−bsc,p (RN ) .
11.3. Slicing As mentioned in Section 6.2, slicing techniques are very useful, since they allow us to reduce proofs for functions of several variables to the case of functions of one variable. We recall that given x = (x1 , . . . , xN ) ∈ RN and i = 1, . . . , N , we denote by x0i ∈ RN −1 the vector obtained from x by removing the ith component xi . With an abuse of notation, we write x = (x0i , xi ) ∈ RN −1 × R.
(11.14)
When i = N , we also use the notation x = (x0 , xN ) ∈ RN −1 × R.
(11.15)
Theorem 11.29. Let 1 ≤ p < ∞ and s > 1, s ∈ / N. A function u ∈ Lp (RN ) s,p N belongs to W (R ) if and only if it admits a representative u ¯ such that u ¯(x0i , ·) ∈ W s,p (R) for LN −1 a.e. x0i ∈ RN −1 and for all i = 1, . . . , N , with (11.16)
N Z X i=1
RN −1
(k¯ u(x0i , ·)kpW m,p (R) + ∂im u ¯(x0i , ·)pW s−m,p (R) ) dx0i < ∞,
where m = bsc. Moreover, for all u ∈ W s,p (RN ), N Z X m p ∇ uW s−m,p (RN ) ≈ ∂im u ¯(x0i , ·)pW s−m,p (R) dx0i . i=1
RN −1
11.3. Slicing
423
Proof. Step 1: Assume that u ∈ W s,p (RN ) and let u ¯ be a representative of u. By Theorem 11.28, (11.17)
N Z Z X R
i=1
∆m ¯(x)p t,i u dxdt ∇m upW s−m,p (RN ) < ∞. 1+sp t N R
Writing x = (x0i , xi ) ∈ RN −1 × R, it follows by Fubini’s theorem that there exists a set Ei ⊂ RN −1 such that LN −1 (Ei ) = 0 and for all x0i ∈ RN −1 \ Ei , Z Z Z ∆m ¯(x0i , xi )p t,i u ¯ u(x0i , xi )p dxi + dxi dt < ∞. t1+sp R R R By applying Theorem 11.27 to the function u ¯(x0i , ·), we obtain that u ¯(x0i , ·) ∈ W s,p (R) with Z Z ∆m ¯(x0i , xi )p t,i u p m 0 ∂i u ¯(xi , ·)W s−m,p (R) dxi dt. t1+sp R R By the Gagliardo–Nirenberg interpolation theorem (Exercise 5.9), Z Z Z Z ∆m ¯(x0i , xi )p t,i u n 0 p 0 p dxi dt < ∞ ∂i u ¯(xi , xi ) dxi ¯ u(xi , xi ) dxi + t1+sp R R R R for all n = 1, . . . , m. Integrating in x0i over RN −1 and using (11.17) and Tonelli’s theorem gives Z Z Z ∆m ¯(x)p t,i u p 0 m 0 ∂i u ¯(xi , ·)W s−m,p (R) dxi ≤ dxdt t1+sp R RN RN −1 Z RN
∇bsc upW s−m,p (RN ) , Z n p ∂i u ¯(x) dx ¯ u(x)p dx + ∇bsc upW s−m,p (RN ) RN
for all i = 1, . . . , N and n = 1, . . . , m. Hence, (11.16) holds. Step 2: Assume that u ∈ Lp (RN ) admits a representative u such that u ¯(x0i , ·) ∈ W s,p (R) for LN −1 a.e. x0i ∈ RN −1 and for all i = 1, . . . , N , and (11.16) holds. By applying Theorem 11.27 to the function u ¯(x0i , ·), we obtain that Z Z ∆m ¯(x0i , xi )p t,i u dxi dt ∂im u ¯(x0i , ·)pW s−m,p (R) . 1+sp t R R Summing over all i and using (11.16) allows us to use Theorem 11.28 to conclude that u ∈ W s,p (RN ), with ∇
m
upW s−m,p (RN )
N Z Z X i=1
R
p ∆m t,i u(x) dxdt. t1+sp RN
424
11. Higher Order Fractional Sobolev Spaces
Moreover, by Corollary 11.26, kukpW m,p (RN ) kukpLp (RN ) + ∇m upW s−m,p (RN ) N Z Z p X ∆m t,i u(x) kukpLp (RN ) + dxdt. t1+sp R RN
i=1
Next, we extend Theorem 6.49 to the case 1 ≤ s < 2. We recall that, given ν ∈ SN −1 , we define ν ⊥ := {y ∈ RN : y · ν = 0}. Given y ∈ ν ⊥ and E ⊆ RN , we set E(x, ν) := {t ∈ R : x + tν ∈ E}. In what follows, for brevity, we will write Z Z f dΣ := f dHN −1 , where we recall that HN −1 is the (N − 1)dimensional Hausdorff measure. Theorem 11.30 (Slicing). Let 1 ≤ p < ∞ and 1 < s < 2. Then u ∈ W 1,p (RN ) belongs to W s,p (RN ) if and only if u admits a representative u ¯ such that for HN −1 a.e. ν ∈ SN −1 and HN −1 a.e. y ∈ ν ⊥ , the function u ¯(y + ·ν) belongs to W s,p (R), with Z Z (11.18) (u(y + tν)p + ∂ν u(y + tν)p ) dtdΣ(y) < ∞, ν⊥
and (11.19) Z SN −1
Z ν⊥
Z Z R
R
R
∂ν u ¯(x + tν) − ∂ν u ¯(x + τ ν)p dtdτ dΣ(y)dΣ(ν) < ∞. t − τ 1+(s−1)p
Moreover, ∇upW s−1,p (RN ) Z Z Z Z ∂ν u ¯(x + tν) − ∂ν u ¯(x + τ ν)p ≈ dtdτ dΣ(y)dΣ(ν). t − τ 1+(s−1)p SN −1 ν ⊥ R R Proof. Step 1: Assume that u ∈ W s,p (RN ). Then in particular, u ∈ W 1,p (RN ), so by Theorem 6.49, there exists a representative u ¯ of u such that, for every ν ∈ SN −1 and for HN −1 a.e. y ∈ ν ⊥ , the function u(y + ·ν) belongs to W 1,p (R), with Z Z (u(y + tν)p + ∂ν u(y + tν)p ) dtdΣ(y) ≤ kukpW 1,p (RN ) < ∞. ν⊥
R
11.3. Slicing
425
By Theorem 6.47 applied to ∇¯ u, Z Z Z Z k∇¯ u(y + tν) − ∇¯ u(y + τ ν)kp dtdτ dΣ(y)dΣ(ν) t − τ 1+(s−1)p SN −1 ν ⊥ R R = 2∇upW s−1,p (RN ) < ∞. Hence, by Tonelli’s theorem, there exists a set M0 ⊂ SN −1 , with HN −1 (M0 ) = 0, such that for every ν ∈ SN −1 \ M0 , Z Z Z k∇¯ u(y + tν) − ∇¯ u(y + τ ν)kp dtdτ dΣ(y) < ∞. t − τ 1+(s−1)p ν⊥ R R Fix ν ∈ SN −1 \ M0 . Then, again by Tonelli’s theorem, there exists a set F0 ⊂ ν ⊥ , with HN −1 (F0 ) = 0, such that for all y ∈ ν ⊥ \ F0 , Z (u(y + tν)p + ∂ν u(y + tν)p ) dt < ∞, Z Z R ∂ν u ¯(y + tν) − ∂ν u ¯(y + τ ν)p dtdτ < ∞. t − τ 1+(s−1)p R R This implies that u(y + ·ν) ∈ W s,p (R). Step 2: Let u ∈ W 1,p (RN ) and assume that u admits a representative u ¯ such that for HN −1 a.e. ν ∈ SN −1 and HN −1 a.e. y ∈ ν ⊥ , the function u ¯(y + ·ν) belongs to W s,p (R), and that (11.18) and (11.19) hold. Then, by Theorem 5.28 and a change of variables, Z Z Z ∞Z ∆2τ,ν u ¯(y + tν)p ∆2r,ν u ¯(y + tν)p 1 dtdτ = dtdr 2 R R τ 1+sp r1+sp 0 R Z Z ∂ν u ¯(y + tν) − ∂ν u ¯(y + τ ν)p dtdτ. t − τ 1+(s−1)p R R Integrating both sides in y over ν ⊥ and in ν over SN −1 gives Z Z Z Z ¯(y + tν)p ∆2τ,ν u 1 A := dtdτ dΣ(y)dΣ(ν) 2 SN −1 ν ⊥ R R τ 1+sp Z Z Z Z ∂ν u ¯(y + tν) − ∂ν u ¯(y + τ ν)p dtdτ dΣ(y)dΣ(ν). t − τ 1+(s−1)p SN −1 ν ⊥ R R Using spherical coordinates, we write h = rν, where r > 0 and ν ∈ SN −1 . Then Z Z Z Z ∞Z ∆2h u(x)p ∆2rν u(x)p N −1 dxdh = r dxdrdΣ(ν) N +sp rN +sp RN RN khk SN −1 0 RN Z Z Z 1 ∆2τ ν u(x)p = dτ dxdΣ(ν), 2 SN −1 RN R τ 1+sp where we use Tonelli’s theorem and the change of variables τ = −r. For every fixed ν ∈ SN −1 and for every x ∈ RN , we write x = y + tν, where
426
11. Higher Order Fractional Sobolev Spaces
y ∈ ν ⊥ , and t ∈ R. Since the Jacobian matrix of this change of variables has determinant 1, the righthand side of the previous equality equals Z Z Z Z 1 ∆2τ ν u(y + tν)p dtdτ dΣ(y)dΣ(ν), 2 SN −1 ν ⊥ R R τ 1+sp which equals A. Thus, we have shown that Z Z ∆2h u(x)p dxdh = A N +sp RN RN khk Z Z Z Z ∂ν u ¯(x + tν) − ∂ν u ¯(x + τ ν)p dtdτ dΣ(y)dΣ(ν) < ∞. t − τ 1+(s−1)p SN −1 ν ⊥ R R We now use Theorem 11.27 to conclude that u ∈ W s,p (RN ).
Exercise 11.31. Let 1 ≤ p < ∞, 1 < s < ∞, s ∈ / N, and m = bsc. Prove that u ∈ W m,p (RN ) belongs to W s,p (RN ) if and only if u admits a representative u ¯ such that for HN −1 a.e. ν ∈ SN −1 and HN −1 a.e. y ∈ ν ⊥ , the function u ¯(y + ·ν) belongs to W s,p (R), with Z Z m X (u(y + tν)p + ∂νk u(y + tν)p ) dtdΣ(y) < ∞ ν⊥
and Z SN −1
Z ν⊥
R
k=1
Z Z R
R
∂νm u ¯(x + tν) − ∂νm u ¯(x + τ ν)p dtdτ dΣ(y)dΣ(ν) < ∞. t − τ 1+(s−m)p
11.4. Embeddings In this section we extend the embedding results obtained in Chapter 7 to W s,p (RN ), where s > 1, s ∈ / N. In this section, we use a different notation for Lr norms to give a unified treatment. To be precise, given r ∈ [−∞, ∞], r 6= 0, and a function u : RN → R, we define if r > 0, kukLr (RN ) k∇n ukL∞ (RN ) if r < 0 and a = 0, (11.20) ur := ∇n uC 0,a (RN ) if r < 0 and 0 < a < 1, where if r < 0 we set n := b−N/rc and a := −n − N/r ∈ [0, 1), provided the righthand sides are well defined. We recall that if 0 < σ < 1 and σp < N , the Sobolev critical exponent p∗σ is defined by (11.21)
p∗σ :=
Np . N − σp
11.4. Embeddings
427
Given 1 ≤ p < ∞ and s > 1 such that s ∈ / N and (s − k)p < N for some k = 0, . . . , bsc, we define the Sobolev critical exponent. p∗s,k :=
(11.22)
Np . N − (s − k)p
Note that p∗s,bsc = p∗s−bsc . Theorem 11.32 (Sobolev–Gagliardo–Nirenberg embedding in W s,p ). Let 1 ≤ p < ∞ and let s > 1 be such that s ∈ / N and sp < N . Then for every function u ∈ W s,p (RN ) and every k = 0, . . . , bsc, k∇k uk
p∗ s,k (RN )
L
uW s,p (RN ) .
We begin with a preliminary result, which is of interest in itself. Here we use the notation in (11.20). Lemma 11.33. Let 1 ≤ p < ∞ and s > 1 be such that s ∈ / N and let s,p N (s − bsc)p < N . Then for every u ∈ W (R ), (11.23)
k∇bsc uk
p∗ s−bsc (RN )
L
uW s,p (RN ) .
Moreover, for every k = 0, . . . , bsc − 1 and 0 ≤ θ ≤ 1 − k/bsc, , ∇k ur kukθLp (RN ) u1−θ W s,p (RN ) where (11.24)
(1 − θ)
1 p
−
1 s − k k 1 +θ + = ∈ (−∞, 1], N p N r
except when s − k − N/p is a nonnegative integer, in which case (11.23) only holds for 0 < θ ≤ 1 − k/bsc. Proof. Set m := bsc and consider a multiindex α ∈ NN 0 , with α = m. Then ∂ α u ∈ W s−m,p (RN ), with s − m ∈ (0, 1) and (s − m)p < N . Hence, we may apply the Sobolev–Gagliardo–Nirenberg embedding theorem (Theorem 7.6) to ∂ α u to conclude that k∂ α uk p∗s−m N ∂ α uW s−m,p (RN ) . Therefore, L (R ) m,p∗s−m p N N ˙ u ∈ L (R ) ∩ W (R ). We can now apply the Gagliardo–Nirenberg interpolation result (see [Leo22d]) to estimate the intermediate derivatives: ∇k ur kukθLp (RN ) k∇m uk1−θ p∗ L
s−m (RN )
kukθLp (RN ) u1−θ , W s,p (RN )
1 k where 0 ≤ k < m, (1 − θ) p∗1 − m−k + θ + = 1r , and 0 ≤ θ ≤ N p N s−m 1 − k/m, except that when m − k − N/p∗s−m is a nonnegative integer, we 1 s−k take 0 < θ ≤ 1 − k/m. To conclude, observe that p∗1 − m−k N = p − N s−m and that m − k − N/p∗s−m = s − k − N/p, so, if s − k − N/p is a nonnegative integer, then we take 0 < θ ≤ 1 − k/m.
428
11. Higher Order Fractional Sobolev Spaces
We turn to the proof of Theorem 11.32. Proof of Theorem 11.32. Given k = 0, . . . , bsc − 1, taking θ = 0 in Np Lemma 11.33, we get r = N −(s−k)p = p∗s−k . Observe that we are not in the exceptional case of Lemma 11.33, that is, s − k − N/p = l ∈ N0 . Indeed, this would imply that p = N/(s − k − l). But then sp = N s/(s − k − l) > N , which contradicts the hypothesis that sp < N . Next we consider the critical case sp = N . Theorem 11.34. Let 1 ≤ p < N . Then for every u ∈ W N/p,p (RN ) and p ≤ r < ∞, kukLr (RN ) uW N/p,p (RN ) , while for every k = 1, . . . , bN/pc, k∇k ukLN/k (RN ) uW N/p,p (RN ) . Proof. We apply Lemma 11.33 with s = N/p, observing that (s − bsc)p = N − bN/pcp < N . If k = 0, then s − k − N/p = 0, so (11.23) holds for every 0 < θ ≤ 1 with r = p/θ (but not for θ = 0). On the other hand, for every k = 1, . . . , bN/pc, taking θ = 0 in (11.24), we get r = N/k. Finally, we study the super critical case sp > N . Theorem 11.35 (Morrey’s embedding in W s,p ). Let 1 ≤ p < ∞ and s > 1 be such that s ∈ / N and sp > N . Then every u ∈ W s,p (RN ) admits a representative u ¯ such that u ¯ ∈ C n,a (RN ), with (11.25)
k∇k u ¯k∞ kukW s,p (RN ) ,
∇n u ¯C 0,a (RN ) uW s,p (RN )
for k = 0, . . . , n, where (i) if (s − bsc)p < N and s − N/p ∈ / N, then n = bs − N/pc and a = s − N/p − n; (ii) if (s − bsc)p < N and s − N/p ∈ N, then n = s − N/p − 1 and 0 < a < 1; (iii) if (s − bsc)p = N , then n = bsc − 1 and 0 < a < 1 − N/p; (iv) if (s − bsc)p > N , then n = bsc and a = s − bsc − N/p. We begin with some preliminary results. Lemma 11.36. Let 1 ≤ p < ∞ and s > 1 be such that s ∈ / N and let (s − bsc)p = N . Then for every p ≤ q < ∞ and u ∈ W s,p (RN ), k∇bsc ukLq (RN ) kukW s,p (RN ) . Moreover, u admits a precise representative u ¯ such that u ¯ ∈ C bsc−1,γ (RN ) for every 0 < γ < 1 − N/p, with k∇k u ¯k∞ kukW s,p (RN ) ,
∇bsc−1 uC 0,γ (RN ) uW s,p (RN )
11.4. Embeddings
429
for k = 0, . . . , bsc − 1. Proof. Set m := bsc and consider a multiindex α ∈ NN 0 with α = m. Then ∂ α u ∈ W s−m,p (RN ), with (s − m)p = N . Hence, we may apply Theorem 7.12 to ∂ α u to conclude that for every p ≤ q < ∞, k∂ α ukLq (RN ) k∂ α ukW s−m,p (RN ) kukW s,p (RN ) .
(11.26)
˙ m,q (RN ). We can now Taking p < q < ∞, we have that u ∈ Lp (RN ) ∩ W apply the Gagliardo–Nirenberg interpolation theorem (see [Leo22d]) with k = 0 to obtain , ur kukθLp (RN ) k∇m uk1−θ Lq (RN )
(11.27) where 0 ≤ θ ≤ 1 and
(1 − θ)
1 q
−
m θ 1 + = ∈ (−∞, 1]. N p r
Note that since N < p < q < ∞, we are not in one of the exceptional cases. If θ = 0, then n = b−N/rc = bm − N/qc = m − 1 and a = 1 − N/q ∈ (0, 1) (since q > p > N ). Hence, by (11.20), (11.26), and (11.27), u admits a representative u ¯ such that ∇m−1 u ¯C 0,1−N/q (RN ) = ur k∇m ukLq (RN ) kukW s,p (RN ) , while if θ = 1, then r = p. Hence, by the continuity in θ we can find θ ∈ (0, 1) such that 1r = 0, that is, by (11.20), (11.26), and (11.27), kukW s,p (RN ) . k¯ uk∞ = ur kukθLp (RN ) k∇m uk1−θ Lq (RN ) By interpolating in spaces of H¨ older continuous functions or by applying Gagliardo–Nirenberg interpolation results (see [Leo22d]) for other values of θ, we obtain that all derivatives of u ¯ of order less than or equal to m − 1 are bounded, with k¯ ukC m−1,1−N/q (RN ) kukW s,p (RN ) . Finally, by taking q sufficiently large, we can make sure that 1 − N/q ≥ γ for every 0 < γ < 1 − N/p. Lemma 11.37. Let 1 ≤ p < ∞ and s > 1 be such that s ∈ / N and (s − bsc)p > N . Then u admits a precise representative u ¯ such that u ¯ ∈ bsc,γ N C (R ), with k¯ ukC bsc (RN ) kukW s,p (RN ) , b
where γ = s − bsc − N/p.
∇bsc uC 0,γ (RN ) uW s,p (RN ) ,
430
11. Higher Order Fractional Sobolev Spaces
Proof. Set m := bsc and consider a multiindex α ∈ NN 0 with α = m. Then ∂ α u ∈ W s−m,p (RN ), with (s − m)p > N . Hence, we may apply Morrey’s embedding theorem (Theorem 7.23) to ∂ α u to conclude that ∂ α u admits a representative vα that is H¨older continuous with exponent γ = s − m − N/p, with (11.28)
kvα k∞ kukW s,p (RN ) ,
vα C 0,γ (RN ) uW s,p (RN ) .
˙ m,∞ (RN ). Let u It follows that u ∈ Lp (RN ) ∩ W ¯ be the continuous representative of u. We can now apply the Gagliardo–Nirenberg interpolation result (see [Leo22d]) to obtain that for k = 0 and 0 ≤ θ ≤ 1, (11.29)
ur kukθLp (RN ) k∇m uk1−θ , L∞ (RN )
where −(1 − θ)
1 m θ + = ∈ (−∞, 1]. N p r
m When θ = 0, − N = 1r , and so n := b−N/rc = bmc = m and a = 0. On the other hand, if θ = 1, then r = p. Hence, by the continuity in θ we can find θ ∈ (0, 1) such that 1r = 0, that is, by (11.20), (11.28), and (11.29),
kukW s,p (RN ) . k¯ uk∞ = ur kukθLp (RN ) k∇m uk1−θ L∞ (RN ) As in Lemma 11.36, we conclude that u ¯ and all its derivatives of order up to m are bounded, with k∇k u ¯k∞ kukW s,p (RN ) for k = 1, . . . , m. Together with (11.28), this shows that u ¯ ∈ C m,γ (RN ). We turn to the proof of Morrey’s embedding theorem. Proof of Theorem 11.35. Let m = bsc. We consider three cases. Step 1: Assume that (s − m)p < N and s − N/p ∈ / N. Then by Lemma 11.33, with k = 0 and 0 ≤ θ ≤ 1, (11.30)
uLr kukθLp (RN ) u1−θ , W s,p (RN )
where (1−θ) p1 − Ns + pθ = 1r ∈ (−∞, 1]. Taking θ = 0 gives p1 − Ns = 1r < 0, so n = b−N/rc = b−N/p + sc and a = −N/p + s − b−N/p + sc ∈ [0, 1). Hence, by (11.20) and (11.30), u admits a representative u ¯ such that (11.31)
∇n u ¯C 0,a (RN ) = ur uW s,p (RN ) ,
while if θ = 1, then r = p. By the continuity of θ we can find θ ∈ (0, 1) such that such that 1r = 0, that is, by (11.20) and (11.30), (11.32)
k¯ ukL∞ (RN ) = ur kukθLp (RN ) u1−θ kukW s,p (RN ) . W s,p (RN )
11.4. Embeddings
431
As in the proof of Lemmas 11.36, we conclude that u ¯ and all its derivatives of order up to n are bounded with (11.33)
k∇k u ¯kL∞ (RN ) kukW s,p (RN )
for k = 1, . . . , n. Together with (11.31), this shows that u ¯ ∈ C n,a (RN ). Step 2: Assume that (s − m)p < N and s − N/p = l ∈ N. Then 1/r = −(1 − θ)l/N + θ/p → −l/N as θ → 0+ and so n := b−N/rc = l − 1 for all θ sufficiently small. In turn, a = 1 − θl − θN/p, which can be made arbitrarily close to 1 by taking θ small. The inequalities in (11.25) follow from (11.31), (11.32), and (11.33). Step 3: Assume that (s − m)p = N . Then by Lemma 11.36, u admits a precise representative u ¯ such that u ¯ ∈ C m−1,γ (RN ) for every 0 < γ < 1 − N/p, with k¯ ukC m−1 (RN ) kukW s,p (RN ) ,
∇m−1 uC 0,γ (RN ) uW s,p (RN ) .
Step 4: Assume that (s − m)p > N . Then by Lemma 11.37, u admits a precise representative u ¯ such that u ¯ ∈ C bsc,γ (RN ), with k¯ ukC bsc (RN ) kukW s,p (RN ) ,
∇bsc uC 0,γ (RN ) uW s,p (RN ) ,
where γ = s − bsc − N/p.
We recall that F denotes the Fourier transform. Exercise 11.38. Let s = N/2 + α, where 0 < α < 1, and u ∈ H s (RN ). (i) Prove that F(u) ∈ L1 (RN ), with kF(u)kL1 (RN ) kukH s (RN ) . Deduce that u has admits a representative u ¯ ∈ C0 (RN ), with k¯ uk∞ kukH s (RN ) . Hint: See Exercise 11.6. (ii) Let x, h ∈ RN , with 0 < khk ≤ 1/2. Prove that Z e2πih·y − 12 dy khk2α . 2 s RN (1 + kyk ) Hint: Split the integral over B(0, 1/khk) and RN \ B(0, 1/khk). (iii) Let x, h ∈ RN , with 0 < khk ≤ 1/2. Prove that Z 1/2 α 2 s 2 ¯ u(x + h) − u ¯(x) khk . (1 + kyk ) F(u) dy RN
Deduce that k¯ ukC 0,α (RN ) kukH s (RN ) . Next, we extend Theorems 7.28 and 7.35. We recall that if s ∈ N, we set uW s,p (Ω) := k∇s ukLp (Ω) .
432
11. Higher Order Fractional Sobolev Spaces
Theorem 11.39. Let 1 ≤ p1 , p2 < ∞, 0 < s1 < s2 be such that s2 −
(11.34)
N N = s1 − , p2 p1
˙ s2 ,p2 (RN ) be such that ∂ α u vanishes at infinity for all multiand let u ∈ W indexes α with 1 ≤ α ≤ m, where m = bs2 c if s2 ∈ / N and m = s2 − 1 if s2 ∈ N. If s2 ∈ N and p2 = 1, we further assume that N ≥ 2. Then (11.35)
uW s1 ,p1 (RN ) uW s2 ,p2 (RN ) .
Moreover, W s2 ,p2 (RN ) ,→ W s1 ,p1 (RN ). Proof. For simplicity, we study first the case 0 < s2 < 2, although this step is not needed in Step 2. Step 1: Assume 0 < s2 < 2. The inequality in (11.35) follows from Theorem 7.35 if s2 < 1 and from Theorems 7.28 and 7.31 if s2 = 1. Thus, assume that 1 < s2 < 2. If s1 = 1, then by (11.34), (s2 − 1)p2 < N and p1 = N p2 ∗ N −(s2 −1)p2 = (p2 )s2 −1 , and so, (11.35) follows from the Sobolev–Gagliardo– Nirenberg embedding theorem ([Leo22d]) applied to ∇u, which vanishes at infinity by hypothesis. If 1 < s1 < s2 < 2, then by (11.34), s1 − 1 + pN2 = s2 − 1 + pN1 , and so we can apply Theorem 7.35 to ∇u. It remains to consider the case 0 < s1 < 1 < s2 < 2. We need to prove uW s1 ,p1 (R) ∇uW s2 −1,p2 (RN ) . Observe that by (11.34), pN2 = s2 − s1 + pN1 > s2 − 1, which implies that p2 (s2 − 1) < N . Hence, we are in the subcritical case. Since ∇u vanishes at infinity, by applying the Sobolev–Gagliardo–Nirenberg embedding theorem (Theorem 7.6) to ∇u, we obtain (11.36)
k∇uk
(p2 )∗ s2 −1
L
∇uW s2 −1,p2 (RN ) .
∗
˙ 1,(p2 )s2 −1 (RN ). Since Thus, u belongs to W N N N N − s1 − +1= − s1 − + s2 = 0 ∗ p1 (p2 )s2 −1 p1 p2 by (11.34), we can apply Theorem 7.28 to conclude that (11.37)
uW s1 ,p1 (RN ) k∇uk
L
(p2 )∗ s2 −1
.
Inequality (11.35) follows by combining (11.36) and (11.37). Step 2: We now consider the general case. Let σ2 = s2 − bs2 c and σ1 = s1 − bs1 c. Since s1 < s2 , we have that bs2 c ≥ bs1 c =: m, and so, k = bs2 c − bs1 c ∈ N0 . Using this notation, (11.35) becomes (11.38)
∇m uW σ1 ,p1 (R) ∇m+k uW σ2 ,p2 (RN ) ,
11.4. Embeddings
433
where, by (11.34), σ1 + pN2 = σ2 + k + pN1 . The proof is by induction on k. If k = 0, then σ1 + pN2 = σ2 + pN1 and σ1 < σ2 . Hence, we are in a position to apply Theorems 7.28, 7.31, and 7.35 to ∇m u to conclude that ∇m uW σ1 ,p1 (RN ) ∇m uW σ2 ,p2 (RN ) , which gives (11.38) in this case. Next, given k0 ∈ N assume that (11.38) holds for all 0 ≤ k ≤ k0 , 1 ≤ p1 , p2 < ∞, and 0 ≤ σ1 , σ2 < 1 satisfying σ1 + pN2 = σ2 + k + pN1 , with the caveat that if σ2 = 0 and p2 = 1, we assume that N ≥ 2. Consider 0 ≤ σ1 , σ2 < 1 and 1 ≤ p1 , p2 < ∞ satisfying σ1 + pN2 = σ2 + k0 + 1 + pN1 . 1 )p1 +N If (k0 −σN ∈ / N, we define σ0 := 0 and p0 := (k0 −σN1p)p1 1 +N . Observe that p0 > 1 since N (k0 − σ1 )p1 + N = − 1 − σ2 p1 < N p1 . p2 1 )p1 +N On the other hand, if (k0 −σN ∈ N, then we take 0 < σ0 < 1 so small (k0 +σ0 −σ1 )p1 +N p1 that ∈ / N and p0 := (k0 +σ0N N −σ1 )p1 +N . In both cases, we have
σ1 +
N p0
= σ0 + k0 +
N p1 .
Hence, by the induction hypothesis,
∇m uW σ1 ,p1 (R) ∇m+k0 uW σ0 ,p0 (RN ) .
(11.39) Since
N N − σ2 = −σ1 + k0 + 1 + > 0, p2 p1 0 we are in the subcritical case σ2 p2 < N . Let q11 = p10 + 1−σ N . Observe that q1 > 1. To see this, we use the identities σ1 + pN2 = σ2 + k0 + 1 + pN1 and σ1 + pN0 = σ0 + k0 + pN1 to get
N N N = + 1 − σ0 = −σ1 + σ0 + k0 + + 1 − σ0 q1 p0 p1 N N = −σ1 + k0 + +1= − σ2 , p1 p2 so q1 =
N p2 N −σ2 p2
= p∗σ2 > 1. If σ0 = 0, then ∇m+k0 uW σ0 ,p0 (RN ) = k∇m+k0 ukLp0 (RN )
and, since q11 − N1 = p10 , we have that p0 = q1∗ . Hence, by the Sobolev– Gagliardo–Nirenberg embedding theorem applied to ∇m+k0 u ([Leo22d]), (11.40)
k∇m+k0 ukLp0 (RN ) = k∇m+k0 ukLq1∗ (RN ) k∇m+k0 +1 ukLq1 (RN ) .
434
11. Higher Order Fractional Sobolev Spaces
On the other hand, if 0 < σ0 < 1, then since Theorem 7.28 applied to ∇m+k0 u to obtain (11.41)
N q1
+ σ0 =
N p0
+ 1 we can use
∇m+k0 uW σ0 ,p0 (RN ) k∇m+k0 +1 ukLq1 (RN ) .
Thus, we have obtained the same upper bound in both cases. Since q1 = p∗σ2 , by the Sobolev–Gagliardo–Nirenberg embedding theorem (Theorem 7.6) applied to ∇m+k0 +1 u, (11.42)
k∇m+k0 +1 ukLq1 (RN ) ∇m+k0 +1 uW σ2 ,p2 (RN ) .
Combining (11.39), (11.40), (11.41), and (11.42) gives (11.38).
Remark 11.40. If in Step 1 we do not assume that ∇u vanishes at infinity, then in the case 0 < s1 < 1 < s2 < 2 in place of (11.36) we obtain k∇u − Au k
(p2 )∗ s2 −1
L
∇uW s2 −1,p2 (RN )
for some Au ∈ RN depending on u. Hence, the function v(x) = u(x) − Au x ∗ ˙ 1,(p2 )s2 −1 (RN ). Since belongs to W N N N N − s1 − +1= − s1 − + s2 = 0, ∗ p1 (p2 )s2 −1 p1 p2 we can apply Theorem 7.28 to conclude that vW s1 ,p1 (RN ) k∇u − Au k
(p2 )∗ s2 −1
L
.
Hence, (11.35) should be replaced by u − pu W s1 ,p1 (RN ) uW s2 ,p2 (RN ) , where pu is a polynomial of degree one. More generally, in Step 2, (11.35) should be replaced by u − pu W s1 ,p1 (RN ) uW s2 ,p2 (RN ) , where pu is a polynomial of degree bs2 c (respectively, s2 − 1) if s2 ∈ / N (respectively, if s2 ∈ N). Using the previous theorems, we can show that W s,p (RN ) is an algebra when sp > N . Theorem 11.41 (Product). Let 1 ≤ p < ∞ and s > 1 be such that s ∈ /N and sp > N . If u, v ∈ W s,p (RN ), then uv ∈ W s,p (RN ) with kuvkW s,p (RN ) kukW s,p (RN ) kvkW s,p (RN ) . Proof. Let m = bsc. Given α ∈ NN 0 with α = m, by Exercise 11.10, X uvW s−m,p (RN ) ∂ α−β u∂ β vW s−m,p (RN ) . β≤α
11.4. Embeddings
435
Since s − m ∈ (0, 1), we can apply the slicing theorem (Theorem 6.35) to estimate ∂ α−β u∂ β vW s−m,p (RN ) N Z ∞ Z X ∆h,i (∂ α−β u∂ β v)(x)p dx RN
0
i=1
1/p
dh h1+(s−m)p
,
where ∆h,i u(x) = u(x + hei ) − u(x). Write ∆h,i (∂ α−β u∂ β v)(x) = ∂ β v(x + hei )∆h,i (∂ α−β u)(x) + ∂ α−β u(x)∆h,i ∂ β v(x). Then Z
∞Z
∆h,i (∂ α−β u∂ β v)(x)p dx
RN
0
∞Z
Z
∂ β v(x + hei )∆h,i (∂ α−β u)(x)p dx
0
Z
RN ∞Z
∂ α−β u(x)∆h,i (∂ β v)(x)p dx
+ 0
dh h1+(s−m)p
RN
dh h1+(s−m)p dh
h1+(s−m)p
=: A + B.
We study the term A since the second term is similar. Let ` = β. We distinguish three cases. If (s − `)p > N , then by Morrey’s embedding theorem (Theorem 7.23) applied to ∂ β v, k∂ β vkL∞ (RN ) k∂ β vkW s−`,p (RN ) , so, also by Theorem 6.35, A ≤ k∂ β vkpL∞ (RN ) ∂ α−β upW s−m,p (RN ) kvkpW s,p (RN ) ∂ α−β upW s−m,p (RN ) . Now, if β = 0, then ∂ α−β uW s−m,p (RN ) = ∂ α uW s−m,p (RN ) ≤ uW s,p (RN ) . On the other hand, if ` > 0, then α − β = m − ` < m, and so, by Theorem 6.21, ∂ α−β uW s−m,p (RN ) k∂ α−β ukW 1,p (RN ) kukW m,p (RN ) . If (s − `)p < N , we use H¨ older’s inequality to bound (11.43) 1/q Z ∞ Z ∆h,i (∂ α−β u)(x)pq dx A ≤ k∂ β vkpLpq0 (RN ) 0
Taking q ≥ ∗
N (s−`)p ,
RN
we get pq 0 ≤ p∗s−` =
Np N −(s−`)p .
dh . h1+(s−m)p
Since ∂ β v ∈ Lp (RN ) ∩
Lps−` (RN ) by the Sobolev–Gagliardo–Nirenberg embedding theorem (Theorem 7.6) applied to ∂ β v, it follows that k∂ β vkLpq0 (RN ) kvkW s,p (RN ) . Let n ∈ N with n ≥ `+1. By Corollary 7.37, with p2 = p, s2 = s−m+N/p−N/(pq), p1 = pq, s1 = s − m, r = p, we have 1/q Z ∞ Z dh n α−β pq ∆h,i (∂ u)(x) dx 1+(s−m)p h 0 RN Z Z N X ∞ dh ∆nh,j (∂ α−β u)(x)p dx 1+(s−m)p+N −N/q . N h R j=1 0
436
11. Higher Order Fractional Sobolev Spaces
Taking q =
N N −`p
≥
N (s−`)p
∞ Z
Z 0
since sp > N , the previous inequality reduces to
∆nh,i (∂ α−β u)(x)pq dx
RN N Z ∞Z X j=1
0
RN
1/q
dh h1+(s−m)p
∆nh,j (∂ α−β u)(x)p dx
dh h1+[s−(m−`)]p
∇` (∂ α−β u)pW s−m,p (RN ) , where the last inequality follows from a slicing argument (Theorem 11.28) and the facts that bs − (m − `)c = `. Since the righthand side is finite, we can now apply Exercise 6.59 to obtain 1/q Z ∞ Z dh α−β pq ∆h,i (∂ u)(x) dx 1+(s−m)p h 0 RN 1/q Z ∞ Z dh ∆nh,i (∂ α−β u)(x)pq dx ∇m upW s−m,p (RN ) . 1+(s−m)p N h 0 R It follows from (11.43) that A kvkpW s,p (RN ) kukpW s,p (RN ) . Finally, If (s − `)p = N , we proceed as in the case (s − `)p < N , using Theorem 7.12 applied to ∂ β v to obtain k∂ β vkLr (RN ) kvkW s,p (RN ) for every p ≤ r < ∞.
11.5. Interpolation Inequalities In this section we prove the Gagliardo–Nirenberg inequality (11.44)
uW s,p (RN ) uθW s1 ,p1 (RN ) u1−θ . W s2 ,p2 (RN )
Here, 0 < θ < 1, 1 ≤ p, p1 , p2 ≤ ∞, and s, s1 , s2 ≥ 0. When s = 0, we set ˙ 0,p (RN ) := Lp (RN ) and, for m ∈ N, W (11.45)
uW 0,p (RN ) := kukLp (RN ) ,
uW m,p (RN ) := k∇m ukLp (RN ) .
We leave it as an exercise to check (see Section 7.7) that a necessary condition ˙ s1 ,p1 (RN ) ∩ W ˙ s2 ,p2 (RN ) is for (11.44) to hold for all u ∈ W N N N = θ s1 − + (1 − θ) s2 − . (11.46) s− p p1 p2 The main result of this section is the following interpolation theorem. Theorem 11.42 (GagliardoNirenberg). Let 1 < p1 , p2 < ∞, 0 ≤ s1 < s2 , and 0 < θ < 1. Then (11.47)
uW s,p (RN ) uθW s1 ,p1 (RN ) u1−θ W s2 ,p2 (RN )
11.5. Interpolation Inequalities
437
˙ s1 ,p1 (RN ) ∩ W ˙ s2 ,p2 (RN ), where for all u ∈ W (11.48)
1−θ 1 θ + , = p p1 p2
s = θs1 + (1 − θ)s2 ,
and we are using the notation in (11.45). We begin with the case s1 , s ∈ N0 and s2 ∈ / N. Theorem 11.43. Let 1 < p1 < ∞, 1 ≤ p2 < ∞, m1 , m ∈ N0 , and s2 > 1, s2 ∈ / N, with m1 < m < s2 . Then k∇m ukLp (RN ) k∇m1 ukθLp1 (RN ) u1−θ W s2 ,p2 (RN ) ˙ m1 ,p1 (RN ) ∩ W ˙ s2 ,p2 (RN ), where for all u ∈ W (11.49)
1 θ 1−θ = + , p p1 p2
m = θm1 + (1 − θ)s2 .
The proof relies on two preliminary results. Lemma 11.44. Let u ∈ Lqloc (RN ), 1 ≤ q < ∞, 0 < σ < 1, and 0 ≤ a < N + σ. Then for every x ∈ RN and h > 0, !1/q Z Z u(y) − u(x) u(y) − u(x)q N +σ−a dy h dy . N +σq ky − xka B(x,h) B(x,h) ky − xk Proof. Write Z Z u(y) − u(x) ky − xkN/q+σ−a (11.50) dy = u(y) − u(x) dy. N/q+σ ky − xka B(x,h) B(x,h) ky − xk If q > 1, we use H¨ older’s inequality to bound the righthand side from above by !1/q0 Z !1/q Z u(y) − u(x)q (N/q+σ−a)q 0 ky − xk dy dy N +σq B(x,h) B(x,h) ky − xk !1/q Z u(y) − u(x)q N +σ−a h dy , N +σq B(x,h) ky − xk where we use the fact that Z Z 0 ky − xk(N/q+σ−a)q dy B(x,h)
h
0
0
rN −1+(N/q+σ−a)q dr h(N +σ−a)q .
0
On the other hand, if q = 1, the righthand side of (11.50) is bounded from above by Z u(y) − u(x) dy. hN +σ−a N +σ B(x,h) ky − xk
438
11. Higher Order Fractional Sobolev Spaces
Lemma 11.45. Let u ∈ C ∞ (RN ), ψ ∈ Cc∞ (RN ), with supp ψ ⊆ B(0, 1), N and h > 0, and α ∈ NN 0 be a multiindex. Then for every x ∈ R Z M(u)(x) α N ψh (x − y)∂ u(y) dy ψ hα , R 1 where ψh (x) := hN ψ hx and M(u) is the maximal function of u. Moreover, if α > 0, 1 ≤ q < ∞, and 0 < σ < 1, then for every x ∈ RN and h > 0, !1/q Z Z q 1 u(y) − u(x) α dy . N ψh (x − y)∂ u(y) dy ψ hα−σ N +σq R B(x,h) ky − xk Proof. Let m := α. If m ≥ 1, we integrate by parts and use the fact that supp ψα ⊆ B(0, 1) to write Z Z (−1)m ∂α y−x α ψh (x − y)∂ u(y) dy = ψ u(y) dy α hN h RN RN ∂y Z ∂αψ y − x (−1)m (11.51) = N +m u(y) dy. α h h B(x,h) ∂y Hence, Z Z k∂ α ψk∞ k∂ α ψk∞ α u(y) dy ≤ ψ (x − y)∂ u(y) dy ≤ M(u)(x). N h hN +m B(x,h) hm R To prove the second inequality, observe that if m ≥ 1, then Z ∂αψ y − x dy = 0 α h B(x,h) ∂y by the divergence theorem and the fact that supp ψ ⊆ B(0, 1). Hence, we can write (11.51) as Z Z ∂αψ y − x (−1)m α [u(y) − u(x)] dy. ψh (x − y)∂ u(y) dy = N +m α h h B(x,h) ∂y RN In turn, Z
RN
Z k∂ α ψk∞ ψh (x − y)∂ u(y) dy ≤ N +m u(y) − u(x) dy. h B(x,h) α
Using Lemma 11.44 with a = 0, we can bound the righthand side from above by !1/q Z hN +σ u(y) − u(x)q C N +m dy . N +σq h B(x,h) ky − xk We turn to the proof of Theorem 11.43.
11.5. Interpolation Inequalities
439
Proof of Theorem 11.43. Step 1: Let m2 := bs2 c. Assume that u ∈ C ∞ (RN ), m1 = 0, m = m2 . Let β ∈ NN with β = m2 . For h > 0, let ϕh be a standard mollifier and write Z Z ∂ β u(x) = ϕh (x − y)∂ β u(y) dy + ϕh (x − y)(∂ β u(y) − ∂ β u(x)) dy RN
RN
=: A + B. 1 hm2
By Lemma 11.45, A B ≤
kϕk∞ hN
M(u)(x), while by Lemma 11.44 with a = 0,
Z
∂ β u(y) − ∂ β u(x) dy
B(x,h)
hs2 −m2
Z B(x,h)
∂ β u(y) − ∂ β u(x)p2 dy ky − xkN +(s2 −m2 )p2
!1/p2 .
Hence, (11.52) Z
M(u)(x) + hs2 −m2 ∂ u(x) hm2 β
B(x,h)
∂ β u(y) − ∂ β u(x)p2 dy ky − xkN +(s2 −m2 )p2
!1/p2 .
By Exercise 1.26, β
∂ u(x) (M(u)(x))
1−m2 /s2
Z RN
∂ β u(y) − ∂ β u(x)p2 dy ky − xkN +(s2 −m2 )p2
m2 /(s2 p2 ) .
Using the fact that m = m2 = (1 − θ)s2 (see (11.49)), raising both sides to power p, and integrating in x over RN gives Z (1−θ)p/p2 Z Z ∂ β u(y) − ∂ β u(x)p2 β p θp ∂ u(x) dx (M(u)(x)) dy dx. N +(s2 −m2 )p2 RN RN RN ky − xk (1−θ)p By (11.49), 1 = θp p1 + p2 , which implies that θp < p1 . Hence, we can use H¨older’s inequality with exponents p1 /(θp) and p2 /[(1 − θ)p] to bound from above the righthand side of the previous inequality with Z θp/p1 Z Z (1−θ)p/p2 ∂ β u(y) − ∂ β u(x)p2 p1 C (M(u)(x)) dx dydx . N +(s2 −m2 )p2 RN RN RN ky − xk
To conclude in this case, we use the continuity of the maximal operator M in Lp1 (RN ) for p1 > 1 (see [Leo22c]). Step 2: Assume that u ∈ C ∞ (RN ), m1 = 0 < m < m2 := bs2 c. Let ¯ 2 and let p¯ be given by 1 = θ¯ + 1−θ¯ . θ¯ ∈ (0, 1) be such that m2 = (1 − θ)s p¯ p1 p2 By the previous step, ¯
¯
θ k∇m2 ukLp¯(RN ) kukθLp1 (RN ) u1− . W s2 ,p2 (RN )
440
11. Higher Order Fractional Sobolev Spaces
˙ m2 ,¯p (RN ). Hence, by the Gagliardo–NirenIt follows that u ∈ Lp1 (RN ) ∩ W berg interpolation theorem (see [Leo22d]), 1−m/m
where 1 p¯
=
1 r
θ¯ p1
m/m
2 k∇m ukLr (RN ) kukLp1 (RN 2) k∇m2 ukLp¯(RN , ) ¯ 2 , m = (1 − θ)s2 , and = mm2 p1¯ + 1 − mm2 p11 . Since m2 = (1 − θ)s
+
1−θ¯ p2 ,
by (11.49) we have ¯ 1 1 m 1 − θ¯ m 1 m θ = + + 1− = + 1− r m2 p¯ m2 p1 m2 p1 p2 m 1 m 1 − θ¯ θ 1−θ ¯ = − (1 − θ) + 1 + = + = m2 p1 m2 p2 p1 p2
m m2 1 , p
1 p1
so r = p. Combining the last two inequalities yields ¯ 1−m/m +θm/m 2
k∇m ukLp (RN ) kukLp1 (RN 2)
¯ (1−θ)m/m
uW s2 ,p2 (RN2)
. = kukθLp1 (RN ) u1−θ W s2 ,p2 (RN ) Step 3. Assume that u ∈ C ∞ (RN ). If m1 ≥ 1, we write (11.49)2 as m − m1 = (1 − θ)(s2 − m1 ) and we apply the previous steps to ∂ γ u ∈ ˙ s2 −m1 ,p1 (RN ), where γ = m1 , to obtain Lp1 (RN ) ∩ W k∇m−m1 (∂ γ u)kLp (RN ) k∂ γ ukθLp1 (RN ) ∂ γ u1−θ W s2 −m1 ,p2 (RN ) . k∇m1 ukθLp1 (RN ) u1−θ W s2 ,p2 (RN ) Given the arbitrariness of γ, we obtain . k∇m ukLp (RN ) k∇m1 ukθLp1 (RN ) u1−θ W s2 ,p2 (RN ) Finally, to remove the additional hypothesis that u ∈ C ∞ (RN ), we apply the previous inequality to u ∗ ϕε , where ϕε is a standard mollifier. We leave the details as an exercise. Next we consider the case s, s2 ∈ N0 and s1 ∈ / N. Theorem 11.46. Let 1 ≤ p1 < ∞, 1 < p2 < ∞, m, m2 ∈ N0 and s1 > 0, s1 ∈ / N, with s1 < m < m2 . Then k∇m ukLp (RN ) uθW s1 ,p1 (RN ) k∇m2 uk1−θ Lp2 (RN ) ˙ s1 ,p1 (RN ) ∩ W ˙ m2 ,p2 (RN ), where for all u ∈ W (11.53)
1 θ 1−θ = + , p p1 p2
m = θs1 + (1 − θ)m2 .
Lemma 11.47 plays the same role as Lemma 5.7.
11.5. Interpolation Inequalities
441
Lemma 11.47. Let u ∈ C ∞ (RN ) and let ϕh be a standard mollifier, h > 0. Then for every m ∈ N and x ∈ RN , (11.54) m−1 X X 1 Z y−x u(x) = u(y) dy ψα hN R N h k=0 α=k Z h Z X 1 y−x α α ∂ u(y)(y − x) dα + ϕ dtdy, N +1 t kx−yk t B(x,h) α=m
where ψα ∈ Cc∞ (RN ), with supp ψα ⊆ B(0, 1), and dα ∈ R. Proof. Fix x ∈ RN and define Z Z (11.55) f (r) := ϕr (y − x)u(y) dy = RN
ϕ(z)u(x + rz) dz,
r > 0,
RN = y−x r .
Since supp ϕ ⊆ B(0, 1), where we make the change of variables z by differentiating under the integral sign (see [Leo22c]), we have that f ∈ C ∞ (R+ ), with (exercise) Z dk f (k) (r) = ϕ(z) k (u(x + rz)) dy (11.56) dr RN Z X = cα ϕ(z)z α ∂ α u(x + rz) dz α=k
RN
for every r > 0 and k ∈ N. Define v(z) := u(x + rz), z ∈ RN . Then for α α every multiindex α, ∂∂z αv (z) = ∂∂xαu (x + rz)rα , so (11.56) becomes X cα Z (k) f (r) = ϕ(z)z α ∂ α v(z) dz. r k RN α=k
By integrating by parts and the change of variables y = x + rz, so that dy = rN dz, we obtain Z X cα α (k) (11.57) f (r) = (−1) ∂ α (ϕ(z)z α )v(z) dz rk N R α=k X 1 Z =: ψ˜α (z)u(x + rz) dz r k RN α=k X 1 Z y−x ˜ = ψα u(y) dy. r rN +k RN α=k
By Taylor’s formula with integral remainder, for 0 < r < h, Z h m−1 X f (k) (h) 1 k f (r) = (r − h) − f (m) (t)(r − t)m−1 dt. k! (m − 1)! r k=0
442
11. Higher Order Fractional Sobolev Spaces
Using (11.55), (11.56), and (11.57), we can rewrite the previous identity as m−1 X
X 1 (r − h)k Z y−x ˜ (11.58) (ϕr ∗ u)(x) = u(y) dy ψα k! hN +k RN h k=0 α=k Z Z h X 1 ϕ(z)z α ∂ α u(x + tz) dzdt. (r − t)m−1 − cα (m − 1)! N R r α=m
By the change of variables y = x + tz, the fact that supp ϕ ⊆ B(0, 1), and Fubini’s theorem, for 0 < r < h, (11.59) Z h
(r − t)m−1
Z
ϕ(z)z α ∂ α u(x + tz) dzdt r Z h Z (r − t)m−1 y−x y−x α α = ϕ ∂ u(y) dydt tN t t r B(x,t) Z Z h (r − t)m−1 y−x α α = ∂ u(y)(y − x) ϕ dtdy. tN +m t B(x,h) max{kx−yk,r} RN
Observe that Z (11.60)
∞
tm−1 tN +m
y − x ∂ u(y)kx − yk ϕ dtdy t kx−yk B(x,h) Z Z ∞ 1 α m ≤ kϕk∞ ∂ u(y)kx − yk dtdy N +1 B(x,h) kx−yk t Z ∂ α u(y) dy. =C N −m B(x,h) kx − yk α
m
Z
By Tonelli’s theorem, for every x0 ∈ RN , Z Z ∂ α u(y) dydx N −m B(x0 ,h) B(x,h) kx − yk Z Z 1 α ≤ ∂ u(y) dxdy N −m B(x0 ,2h) B(y,h) kx − yk Z βN hm < ∞, = ∂ α u(y) dy m B(x0 ,2h) and so,
R
∂ α u(y) B(x,h) kx−ykN −m dy
< ∞ for LN a.e. x ∈ B(x0 , h). Since x0 is
arbitrary, we have that the previous integral is finite for LN a.e. x ∈ RN . Hence, in view of (11.59) and (11.60), we can use the Lebesgue dominated
11.5. Interpolation Inequalities
443
convergence theorem to obtain Z Z r ϕ(z)z α ∂ α u(x + tz) dzdt (r − t)m−1 lim r→0+
RN
h m−1
Z
α
∂ u(y)(y − x)
= (−1)
α
Z
h
1
kx−yk
B(x,h)
tN +1
ϕ
y−x t
dtdy.
In turn, since (ϕr ∗ u)(x) → u(x) as r → 0+ at every x (see [Leo22c]), k ˜ letting r → 0+ in (11.58) gives (11.54) with ψα := (−1) k! ψα . We turn to the proof of Theorem 11.46. Proof of Theorem 11.46. Step 1: Let m1 := bs1 c. Assume that u ∈ C ∞ (RN ) and that m1 = 0. Let β ∈ NN with β = m. We apply Lemma 11.47 to ∂ β u to obtain ∂ β u(x) =
m2X −m−1
X 1 Z y−x ψ ∂ β u(y) dy α hN RN h k=0 α=k Z X + dα ∂ α+β u(y)(y − x)α B(x,h)
α=m2 −m
Z
h
1
×
tN +1
kx−yk
ϕ
y−x t
dtdy =: A + B.
By Lemma 11.45, A
Z
1 hm−s1
RN
u(y) − u(x)p1 dy ky − xkN +s1 p1
1/p1 .
On the other hand, Z
h
1
(11.61) kx−yk
tN +1
ϕ
y−x t
dt ϕ
1 . kx − ykN
Hence, by (11.61) and the fact that m2 − m ≥ 1, ∂ α+β u(y) kx − ykm2 −m dy kx − ykN B(x,h) α=m2 −m Z X ∂ α+β u(y) m2 −m−1 h dy. N −1 B(x,h) kx − yk
B
X
α=m2 −m
Z
444
11. Higher Order Fractional Sobolev Spaces
We can now apply Lemma 7.30 to obtain Z 2h Z X 1 hm2 −m−1 B ∂ α+β u(y) dydr N r 0 B(x,r) α=m2 −m X hm2 −m M(∂ α+β u)(x) hm2 −m M(∇m2 u)(x). α=m2 −m
Hence, β
∂ u(x)
Z
1 hm−s1
RN
u(y) − u(x)p1 dy ky − xkN +s1 p1
1/p1
+ hm2 −m M(∇m2 u)(x).
By Exercise 1.26, ∂ β u(x)
Z RN
u(y) − u(x)p1 dy ky − xkN +s1 p1
(m2 −m)/[(m2 −s1 )p1 ]
× (M(∇m2 u)(x))(m−s1 )/(m2 −s1 ) . Using the fact that m2 − m = θ(m2 − s1 ), raising both sides to power p, and integrating in x over RN gives θp/p1 Z Z Z u(y) − u(x)p1 β p dy ∂ u(x) dx N +s1 p1 RN RN RN ky − xk × (M(∇m2 u)(x))(1−θ)p/p2 dx. (1−θ)p By (11.49), 1 = θp p1 + p2 , which implies that θp < p1 . Hence, we can use H¨older’s inequality with exponents p1 /(θp) and p2 /[(1 − θ)p] to bound from above the righthand side of the previous inequality with Z Z θp/p1 Z (1−θ)p/p2 u(y) − u(x)p1 m2 p2 C dy (M(∇ u)(x)) dx . N +s1 p1 RN RN ky − xk Rn
To conclude in this case, we use the continuity of the maximal operator M in Lp2 (RN ) for p2 > 1 (see [Leo22c]). Step 2: The case m1 ≥ 1 is left as an exercise (see Step 3 of the proof of Theorem 11.43). Finally, we consider the case s ∈ N0 and s1 , s2 ∈ / N. Theorem 11.48. Let 1 < p1 < ∞, 1 ≤ p2 < ∞, m ∈ N0 , s1 > 0, s2 > 1, s1 , s2 ∈ / N, with s1 < m < s2 . Then k∇m ukLp (RN ) uθW s1 ,p1 (RN ) u1−θ W s2 ,p2 (RN ) ˙ m1 ,p1 (RN ) ∩ W ˙ s2 ,p2 (RN ), where for all u ∈ W (11.62)
θ 1−θ 1 = + , p p1 p2
m = θs1 + (1 − θ)s2 .
11.5. Interpolation Inequalities
445
Proof. Let m1 := bs1 c and m2 := bs2 c. Step 1: Assume that u ∈ C ∞ (RN ), m1 = 0, m = m2 . Let β ∈ NN with β = m2 . For h > 0, write Z Z ∂ β u(x) = ϕh (x − y)∂ β u(y) dy + ϕh (x − y)(∂ β u(y) − ∂ β u(x)) dy RN
RN
=: A + B, where ϕh is a standard mollifier. By Lemma 11.45, A
Z
1 hm2 −s1
B(x,h)
u(y) − u(x)p1 dy ky − xkN +s1 p1
!1/p1 ,
while by Lemma 11.44 with a = 0, B ≤ h
s2 −m2
∂ β u(y) − ∂ β u(x)p2 dy ky − xkN +(s2 −m2 )p2
Z B(x,h)
!1/p2 .
Hence, 1
∂ β u(x)
(11.63)
hm2 −s1 + hs2 −m2
Z B(x,h)
u(y) − u(x)p1 dy ky − xkN +s1 p1
Z B(x,h)
!1/p1
∂ β u(y) − ∂ β u(x)p2 dy ky − xkN +(s2 −m2 )p2
!1/p2 .
By Exercise 1.26, β
Z
∂ u(x) RN
u(y) − u(x)p1 dy ky − xkN +s1 p1
Z × RN
(s2 −m2 )/[(s2 −s1 )p1 ]
∂ β u(y) − ∂ β u(x)p2 dy ky − xkN +(s2 −m2 )p2
(m2 −s1 )/[(s2 −s1 )p2 ] .
Using the fact that m = m2 = θs1 + (1 − θ)s2 , raising both sides to power p, and integrating in x over RN gives θp/p1 Z Z Z u(y) − u(x)p1 β p ∂ u(x) dx dy N +s1 p1 RN RN RN ky − xk Z (1−θ)p/p2 ∂ β u(y) − ∂ β u(x)p2 dy × dx. N +(s2 −m2 )p2 RN ky − xk (1−θ)p By (11.62), 1 = θp p1 + p2 , which implies that θp < p1 . Hence, we can use H¨older’s inequality with exponents p1 /(θp) and p2 /[(1 − θ)p] to bound from
446
11. Higher Order Fractional Sobolev Spaces
above the righthand side of the previous inequality with Z Z θp/p1 u(y) − u(x)p1 C dydx N +s1 p1 RN RN ky − xk (1−θ)p/p2 Z Z ∂ β u(y) − ∂ β u(x)p2 dydx . × N +(s2 −m2 )p2 Rn RN ky − xk Step 2: Assume that u ∈ C ∞ (RN ), m1 = 0 < m < m2 . Let θ¯ ∈ (0, 1) be ¯ 1 + (1 − θ)s ¯ 2 and let p¯ be given by 1 = θ¯ + 1−θ¯ . By the such that m2 = θs p¯
p1
previous step, ¯
p2
¯
θ k∇m2 ukLp¯(RN ) uθW s1 ,p1 (RN ) u1− . W s2 ,p2 (RN )
(11.64)
˙ s1 ,p1 (RN ) ∩ W ˙ m2 ,¯p (RN ). By Theorem 11.46, It follows that u ∈ W ˜
where
1 r
˜
θ k∇m ukLr (RN ) uθW s1 ,p1 (RN ) k∇m2 uk1− , Lp¯(RN )
(11.65) =
θ˜ p1
+
1−θ˜ p¯
˜ 1 + (1 − θ)m ˜ 2 . Observe that and m = θs
˜ 1 + (1 − θ)m ˜ 2 = θs ˜ 1 + (1 − θ)( ˜ θs ¯ 1 + (1 − θ)s ¯ 2) m = θs ˜ θ]s ¯ 1 + (1 − θ)(1 ˜ − θ)s ¯ 2, = [θ˜ + (1 − θ) ˜ θ. ¯ In turn, so θ = θ˜ + (1 − θ) ¯ 1 θ˜ 1 − θ˜ θ˜ θ 1 − θ¯ ˜ = + = + (1 − θ) + r p1 p¯ p1 p1 p2 ˜ ¯ ˜ ˜ ¯ θ 1−θ 1 θ + (1 − θ)θ (1 − θ)(1 − θ) + = + = . = p1 p2 p1 p2 p Therefore, if we combine inequalities (11.64) and (11.65), we get ˜ ˜ θ¯ θ+(1− θ)
˜ ¯ (1−θ)(1− θ)
1−θ k∇m ukLp (RN ) uW s1 ,p1 (RN ) uW s2 ,p2 (RN ) = uθ. W s1 ,p1 (RN ) uW s2 ,p2 (RN ) .
Step 3. The case m1 ≥ 1 is left as an exercise (see Step 3 of the proof of Theorem 11.43). We turn to the proof of Theorem 11.42. Proof of Theorem 11.42. Let m2 ∈ N0 be such that s2 = m2 + σ2 , where 0 ≤ σ2 < 1. The proof of (11.47) is by induction on m2 . If m2 = 0 or m2 = 1 and σ2 = 0, then (11.47) follows from Theorem 7.41. Let m2 ∈ N, with σ2 > 0 if m2 = 1, and assume that (11.47) holds for all 1 < p1 , p2 < ∞, 0 ≤ s1 < s2 , and 0 < θ < 1 satisfying (11.48) and such that s2 = n + σ, where 0 ≤ σ < 1 and n = 0, . . . , m2 − 1. Let s1 = m1 + σ1 , where m1 ∈ N0 and 0 ≤ σ1 < 1. If m1 ≥ 1, we write (11.48)2 as s − m1 = θ(s1 − m1 ) + (1 − θ)(s2 − m1 ). Since s2 − m1 = m2 − m1 + σ2 , with m2 − m1 ≤ m2 − 1, we can apply
11.5. Interpolation Inequalities
447
˙ s1 −m1 ,p1 (RN ) ∩ W ˙ s2 −m1 ,p2 (RN ), where the induction hypothesis to ∂ γ u ∈ W N γ ∈ N is a multiindex with γ = m1 , to obtain ∂ γ uW s−m1 ,p (RN ) ∂ γ uθW s1 −m1 ,p1 (RN ) ∂ γ u1−θ W s2 −m1 ,p2 (RN ) . uθW s1 ,p1 (RN ) u1−θ W s2 ,p2 (RN ) Given the arbitrariness of γ, we obtain uW s,p (RN ) = ∇m1 uW s−m1 ,p (RN ) uθW s1 ,p1 (RN ) u1−θ . W s2 ,p2 (RN ) Thus, it remains to consider the case m1 = 0. Since s1 < 1, we can write ¯ 1 + (1 − θ)s ¯ 2 . Let 1 = θ¯ + 1−θ¯ . By Theorems 11.43, 11.48, 11.46 and 1 = θs q p1 p2 the Gagliardo–Nirenberg interpolation theorem (see [Leo22d]), (11.66)
¯
¯
θ . k∇ukLq (RN ) uθW s1 ,p1 (RN ) u1− W s2 ,p2 (RN )
If s = 1, then q = p and θ¯ = θ, and there is nothing else to prove. If s > 1 ˜ + (1 − θ)s ˜ 2 . Then and s2 ∈ / N, write s = θ1 ˜ + (1 − θ)s ˜ 2 = θ[ ˜ θs ¯ 1 + (1 − θ)s ¯ 2 ] + (1 − θ)s ˜ 2 = θ˜θs ¯ 1 + (1 − θ˜θ)s ¯ 2, s = θ1 ¯ In turn, which implies that θ = θ˜θ. 1 θ 1−θ θ˜θ¯ 1 − θ˜θ¯ = + = + p p1 p2 p1 p2 ¯ ˜ 1 − θθ¯ θ˜ 1 − θ˜ 1 1−θ − = + . = θ˜ + q p2 p2 q p2 ˜ 2 − 1), we can apply the induction hypothesis to Since s − 1 = (1 − θ)(s q N s −1,p 2 (RN ) to obtain ˙ 2 ∂i u ∈ L (R ) ∩ W ˜
˜
˜
˜
θ θ ∂i uW s−1,p (RN ) k∂i ukθLq (RN ) ∂i u1− ≤ k∇ukθLq (RN ) u1− W s2 ,p2 (RN ) W s2 −1,p2 (RN ) ¯˜
¯ θ˜ (1−θ)
˜
θ uθWθ s1 ,p1 (RN ) uW s2 ,p2 (RN ) u1− W s2 ,p2 (RN )
= uθW s1 ,p1 (RN ) u1−θ , W s2 ,p2 (RN ) where we use (11.66) in the third inequality and the fact that θ = θ˜θ¯ in the last equality. ˜ 1 + (1 − θ)1. ˜ If 0 < s < 1, write s = θs Then ˜ 1 + (1 − θ)1 ˜ = θs ˜ 1 + (1 − θ)[ ˜ θs ¯ 1 + (1 − θ)s ¯ 2] s = θs ˜ θ]s ¯ 1 + (1 − θ)(1 ˜ − θ)s ¯ 2, = [θ˜ + (1 − θ)
448
11. Higher Order Fractional Sobolev Spaces
˜ θ. ¯ In turn, which implies that θ = θ˜ + (1 − θ) ˜ θ¯ (1 − θ)(1 ˜ − θ) ¯ 1−θ θ˜ + (1 − θ) 1 θ + = + = p p1 p2 p1 p2 ˜ ˜ ¯ ¯ ˜ ˜ θ + (1 − θ)θ ˜ 1 − θ = θ + 1 − θ. = + (1 − θ) p1 q p1 p1 q Hence, we apply Theorem 7.41 to obtain ˜
˜
θ uW s,p (RN ) uθW s1 ,p1 (RN ) k∇uk1− Lq (RN ) ˜
¯ ˜ θ(1− θ)
¯ ˜ (1−θ)(1− θ)
uθW s1 ,p1 (RN ) uW s1 ,p1 (RN ) uW s2 ,p2 (RN ) = uθW s1 ,p1 (RN ) u1−θ , W s2 ,p2 (RN ) ˜ ˜ θ¯ where we use (11.66) in the second inequality and the fact that θ = θ+(1− θ) in the last equality. Remark 11.49. For the cases p1 = 1, or p2 = 1, or θ ∈ {0, 1}, we refer to the paper of Brezis and Mironescu [BM18].
11.6. Superposition In this section we study superposition properties for fractional Sobolev spaces. When 0 < s < 1, we have seen in Theorem 6.27 that if f : R → R is Lipschitz continuous, with f (0) = 0 when Ω has infinite measure, then f ◦ u ∈ W s,p (Ω) for every u ∈ W s,p (Ω). This is no longer the case when s > 1. Indeed, we have seen in Exercise 5.40 that the function u(x) = x, x ∈ (0, 1) does not belong to W 1+1/p,p ((0, 1)). Theorem 11.50 (Truncation). Let 1 ≤ p < ∞ and 1 < s < 1 + 1/p. If u ∈ W s,p (RN ), then u ∈ W s,p (RN ), with kukW s,p (RN ) kukW s,p (RN ) . Proof. Given u ∈ W s,p (RN ), by Theorem 11.29, u admits a representative u ¯ such that u ¯(x0i , ·) ∈ W s,p (R) for LN −1 a.e. x0i ∈ RN −1 and for all i = 1, . . . , N , with N Z X (k¯ u(x0i , ·)kpW 1,p (R) + ∂i u ¯(x0i , ·)pW s−1,p (R) )dx0i < ∞. i=1
RN −1
Let i = 1, . . . , N and x0i ∈ RN −1 be such that u ¯(x0i , ·) ∈ W s,p (R). Then by 0 Theorem 5.37 and Remark 5.38, ¯ u(xi , ·) belongs to W s,p (R) and ∂i ¯ u(x0i , ·)W s−1,p (R) ∂i u ¯(x0i , ·)W s−1,p (R) .
11.6. Superposition
449
Raising both sides to power p and integrating in x0i over RN −1 gives Z Z p 0 0 ∂i u ¯(x0i , ·)pW s−1,p (R) dx0i . ∂i ¯ u(xi , ·)W s−1,p (R) dxi RN −1 in W 1,p (RN )
RN −1
On the other hand, by the chain rule (see [Leo22d]), ¯ u ∈ 1,p N W (R ), with kukW 1,p (RN ) ≤ kukW 1,p (RN ) . Hence, we can apply Theorem 11.29 to conclude that ¯ u ∈ W s,p (RN ). Next we prove that by assuming more regularity on f , we have that f ◦ u ∈ W s,p (RN ) whenever u ∈ W s,p (RN ) ∩ L∞ (RN ). Theorem 11.51 (Superposition). Let 1 < p < ∞, 1 < s < 2, and u ∈ W s,p (RN ) ∩ L∞ (RN ). If f ∈ Cb2 (R) is such that f (0) = 0, then f ◦ u ∈ W s,p (RN ). We begin with a preliminary result. Lemma 11.52. Let 1 < p < ∞, 1 < s < 2, 0 < σ < 1, and u ∈ W s,p (RN ) ∩ W σ,sp/σ (RN ). If f ∈ Cb2 (R) is such that f (0) = 0, then f ◦ u ∈ W s,p (RN ). Proof. By Taylor’s formula of orders one and two, for every t, t0 ∈ R, Z t f (t) − f (t0 ) − f 0 (t0 )(t − t0 ) = (t − r)f 00 (r) dr, f (t) − f (t0 ) − f 0 (t0 )(t − t0 ) =
t0 t
Z
(f 0 (r) − f 0 (t0 )) dr,
t0
and so f (t) − f (t0 ) − f 0 (t0 )(t − t0 ) ≤ kf 00 k∞ t − t0 2 , f (t) − f (t0 ) − f 0 (t0 )(t − t0 ) ≤ kf 0 k∞ t − t0 . It follows that for every 1 ≤ a ≤ 2, (11.67)
f (t) − f (t0 ) − f 0 (t0 )(t − t0 ) ≤ (kf 0 k∞ + kf 00 k∞ )t − t0 a ,
or equivalently, f (t) − f (t0 ) = f 0 (t0 )(t − t0 ) + R2 (t, t0 ), where R2 (t, t0 ) f t − t0 a . By replacing t with t1 and adding the two equalities, we get f (t) − 2f (t0 ) + f (t1 ) = f 0 (t0 )(t − 2t0 + t1 ) + R2 (t, t0 ) + R2 (t1 , t0 ). Taking t = u(x + 2h), t0 = u(x + h), and t1 = u(x), we obtain ∆2h (f ◦ u)(x) = f (u(x + 2h)) − 2f (u(x + h)) + f (u(x)) = f 0 (u(x + h))∆2h u(x) + R2 (u(x + 2h), u(x + h)) + R2 (u(x), u(x + h)).
450
11. Higher Order Fractional Sobolev Spaces
Using the bounds on f (see (11.67)) and R2 gives ∆2h (f ◦ u)(x) f ∆2h u(x) + ∆h u(x + h)a + ∆h u(x)a . In turn, by the change of variables y = x + h, Z Z Z Z ∆2h (f ◦ u)(x)p ∆2h u(x)p dxdh dxdh f N +sp khkN +sp RN RN RN RN khk Z Z ∆h u(x)ap + dxdh =: A + B. N +sp RN RN khk Note that A < ∞ in view of Theorems 6.52 and 11.23 and the fact that u ∈ W s,p (RN ). It remains to estimate B. Since u ∈ W σ,sp/σ (RN ) and N + σsp/σ = N + sp, by Theorem 6.55, Z Z Z Z ∆2h u(x)sp/σ ∆h u(x)sp/σ dxdh dxdh < ∞. khkN +sp khkN +sp RN RN RN RN Let p ≤ r ≤ sp/σ and write
1 r
=
θ p
+
1−θ sp/σ .
By interpolation,
kukLr (RN ) ≤ kukθLp (RN ) kuk1−θ < ∞. Lsp/σ (RN ) Similarly, Z RN
×
1/r Z Z θ/p ∆2h u(x)r ∆2h u(x)p dxdh ≤ dxdh N +sp N +sp RN khk RN RN khk !(1−θ)/(sp/σ) Z Z ∆2h u(x)sp/σ < ∞. dxdh khkN +sp R N RN
Z
It follows from Theorem 11.27 that u ∈ W sp/r,r (RN ). Taking a = min{2, s/σ} ∈ [1, 2] and r = ap ∈ [p, sp/σ] proves that B < ∞. This shows that Z Z ∆2h (f ◦ u)(x)p dxdh < ∞. khkN +sp R N RN On the other hand, by the chain rule in W 1,p (RN ) (see [Leo22d]), f ◦ u ∈ W 1,p (RN ). Hence, in view of Theorems 6.52 and 11.23, f ◦ u ∈ W s,p (RN ). We turn to the proof of Theorem 11.51. Proof of Theorem 11.51. In view of Lemma 11.52, it suffices to prove that if u ∈ W s,p (RN ) ∩ L∞ (RN ) and 0 < σ < 1, then u ∈ W σ,sp/σ (RN ). By Theorems 6.52 and 11.23, Z Z Z Z ∆2h u(x)p ∆h ∇u(x)p dxdh dxdh < ∞. N +sp N +(s−1)p RN RN khk RN RN khk
11.7. Extension
451
Then for p ≤ q < ∞, u(x)q ≤ kukq−p u(x)p , L∞ (RN )
∆2h u(x)q kukq−p ∆2h u(x)p . L∞ (RN )
Taking q = sp/σ and observing that N + σq = N + sp, we have Z Z sp/σ−p sp/σ u(x)p dx u(x) dx ≤ kukL∞ (RN ) RN
RN
and Z RN
Z RN
∆2h u(x)sp/σ sp/σ−p dxdh kukL∞ (RN ) khkN +sp
Z RN
Z RN
∆2h u(x)p dxdh. khkN +sp
Using Theorem 6.55, we conclude that u ∈ W σ,sp/σ (RN ).
Remark 11.53. More generally, one can prove that if 1 < p < ∞, s > 1, bsc+1 s∈ / N, u ∈ W s,p (RN ) ∩ L∞ (RN ), and f ∈ Cb (R) is such that f (0) = 0, s,p N then f ◦ u ∈ W (R ). We refer to [BM01b] for a proof. In the critical and supercritical cases, sp = N and sp > N , we can remove the additional condition that u ∈ L∞ (RN ). Theorem 11.54. Let 1 < p < ∞, 1 < s < 2, with sp ≥ N , and u ∈ W s,p (RN ). If f ∈ Cb2 (R) is such that f (0) = 0, then f ◦ u ∈ W s,p (RN ). Proof. Step 1: Assume sp > N . Then by Morrey’s embedding theorem (Theorem 11.35), u ∈ L∞ (RN ). Hence, we can apply Theorem 11.51 to obtain that f ◦ u ∈ W s,p (RN ). Step 2: Assume sp = N . By Theorem 11.39, taking 0 < σ < 1, we have that W N/p,p (RN ) ,→ W σ,N/σ (RN ). Since sp/σ = N/σ, we can apply Lemma 11.52 with s = N/p to conclude that f ◦ u ∈ W s,p (RN ).
11.7. Extension In this section we prove some extension results. We begin by showing that s,p extension domain for 1 < s < 2 (see Definition 11.14). RN + is a W Theorem 11.55. Let 1 ≤ p < ∞, 1 < s < 2, and u ∈ W s,p (RN + ). Then the N function v : R → R, given by 3u(x0 , −xN ) − 2u(x0 , −2xN ) if xN < 0, v(x) := u(x) if xN > 0, belongs to W s,p (RN ), with kvkLp (RN ) kukLp (RN ) , +
k∇vkLp (RN ) k∇ukLp (RN ) , +
∇vW s−1,p (RN ) ∇uW s−1,p (RN ) . +
452
11. Higher Order Fractional Sobolev Spaces
Proof. We leave it as an exercise to show that v ∈ W 1,p (RN ), with 3∂i u(x0 , −xN ) − 2∂i u(x0 , −2xN ) if xN < 0, ∂i v(x) = ∂i u(x) if xN > 0, for i = 1, . . . , N − 1 and LN a.e. x ∈ RN , −3∂N u(x0 , −xN ) + 4∂N u(x0 , −2xN ) if xN < 0, ∂N v(x) = u(x) if xN > 0, and that kvkLp (RN ) kukLp (RN ) and k∇vkLp (RN ) k∇ukLp (RN ) . On the + + other hand, by Tonelli’s theorem, for every i = 1, . . . , N we have Z Z Z Z ∂i v(x) − ∂i v(y)p ∂i u(x) − ∂i u(y)p dxdy = dxdy N +(s−1)p N kx − ykN +(s−1)p RN RN kx − yk RN R + + Z Z Z Z ∂i u(x) − ∂i v(y)p ∂i v(x) − ∂i v(y)p +2 dxdy + dxdy kx − ykN +(s−1)p kx − ykN +(s−1)p RN RN RN RN + − − − =: A + B + C. Writing ∂i u = 3∂i u − 2∂i u for i = 1, . . . , N − 1 and ∂N u = −3∂N u + 2∂N u, N for x ∈ RN + and y ∈ R− and i = 1, . . . , N , we have, p p ∂i u(x) − ∂i v(y)p ∂i u(x) − ∂i u(y 0 , −yN ) + ∂i u(x) − ∂i u(y 0 , −2yN ) . In turn, Z
Z
B RN −
RN +
Z
Z
∂i u(x) − ∂i u(y 0 , −yN )p dxdy kx − ykN +(s−1)p
+ RN −
RN +
∂i u(x) − ∂i u(y 0 , −2yN )p dxdy. kx − ykN +(s−1)p
We now make the change of variables (y 0 , −yN ) = w in the first integral on the right and (y 0 , −2yN ) = z in the second, to obtain Z Z ∂i u(x) − ∂i u(w)p B dxdw k(x0 − w0 , xN + wN )kN +(s−1)p RN RN + + Z Z ∂i u(x) − ∂i u(z)p + dxdz, k(x0 − z 0 , xN + zN /2)kN +(s−1)p RN RN + + and observe that xN + wN  = xN + wN ≥ xN − wN  and xN + zN /2 = xN + zN /2 ≥ (xN + zN )/2 ≥ xN − zN /2, so that B ∂i upW s−1,p (RN ) . +
For x, y ∈ RN −, p ∂i v(x) − ∂i v(y)p ∂i u(x0 , −xN ) − ∂i u(y 0 , −yN ) p + ∂i u(x0 , −2xN ) − ∂i u(y 0 , −2yN ) .
11.7. Extension
453
Hence, Z
Z
C RN −
RN −
Z
Z
∂i u(x0 , −xN ) − ∂i u(y 0 , −yN )p dxdy kx − ykN +(s−1)p
+ RN −
RN −
∂i u(x0 , −2xN ) − ∂i u(y 0 , −2yN )p dxdy. kx − ykN +(s−1)p
We now make the change of variables (x0 , −xN ) = z and (y 0 , −yN ) = w in the first integral on the right and (x0 , −2xN ) = z and (y 0 , −2yN ) = w in the second to obtain C ∂i upW s−1,p (RN ) . +
Exercise 11.56. Given m ∈ N, 1 ≤ p < ∞, 1 < s < ∞, with s ∈ / N, s,p N u ∈ W (R+ ), and m = bsc, prove that there exist c1 , . . . , cm+1 ∈ R such that the function v : RN → R, given by Pm+1 0 n=1 cn u(x , −nxN ) if xN < 0, v(x) := u(x) if xN > 0, belongs to W s,p (RN ), that k∇k vkLp (RN ) k∇k ukLp (RN ) for every 0 ≤ k ≤ + m, and that ∇m vW s−m,p (RN ) ∇m uW s−m,p (RN ) . +
Next we treat the case of supergraphs. Theorem 11.57. Let 1 ≤ p < ∞, 1 < s < ∞, s ∈ / N, m = bsc, f ∈ m,1 N −1 C (R ), and Ω := {(x0 , xN ) ∈ RN −1 × R : xN > f (x0 )}. Then there exists a continuous linear operator E : W s,p (Ω) → W s,p (RN ) such that for all u ∈ W s,p (Ω), E(u)(x) = u(x) for LN a.e. x ∈ Ω, and kE(u)kW s,p (RN ) f kukW s,p (Ω) . Proof. Step 1: Assume that 1 < s < 2. By Corollary 11.19, the function s,p (RN ), with w(y) := u(y 0 , yN + f (y 0 )), y ∈ RN + , belongs to W + kwkW s,p (RN ) f kukW s,p (Ω) . +
RN
Define the function v : → R as follows: 3w(y 0 , −yN ) − 2w(y 0 , −2yN ) if yN < 0, v(y) := w(y) if yN > 0. Then, by Theorem 11.55, v ∈ W s,p (RN ), with kvkW s,p (RN ) kwkW s,p (RN ) f kukW s,p (Ω) . +
In turn, by Theorem 11.18, the function E(u)(x) := v(x0 , xN − f (x0 )), x ∈ RN , belongs to W s,p (RN ), with kE(u)kW s,p (RN ) f kvkW s,p (RN ) f kukW s,p (Ω) .
454
11. Higher Order Fractional Sobolev Spaces
Step 2: The general case in similar and makes use of Exercise 11.56. The details are left to the reader. Exercise 11.58. Let 1 ≤ p < ∞, 1 < s < ∞, s ∈ / N, m = bsc, and let Ω ⊂ RN be an open bounded set with boundary of class C m,1 . Prove that there exists a continuous linear operator E : W s,p (Ω) → W s,p (RN ) such that for all u ∈ W s,p (Ω), E(u)(x) = u(x) for LN a.e. x ∈ Ω, and kE(u)kW s,p (RN ) Ω kukW s,p (Ω) . Hint: Use a partition of unity. Remark 11.59. Theorem 11.57 continues to hold if f : RN −1 → R is only assumed to be Lipschitz continuous. More generally, the theorem holds for uniformly Lipschitz domains. The proof is more involved, so I refer the interested reader to the paper of De Vore and Sharpley [DS93, Section 6].
11.8. Trace Theory In this section we extend the trace theory developed in Chapter 9 to fractional Sobolev spaces W s,p (Ω), where 1 < s < 2. Let Ω ⊆ RN be open with Lipschitz continuous boundary. In view of Rademacher’s theorem [Leo17, Theorem 9.14], for HN −1 a.e. x ∈ ∂Ω, there exists a tangent space T∂Ω (x) to ∂Ω at x. We say that a vector ν ∈ RN \ {0} is a normal vector to ∂Ω at x if it is orthogonal to all vectors in T∂Ω (x), that is, ν · t = 0 for all t ∈ T∂Ω (x). A unit normal vector ν to ∂Ω at x is called a unit outward normal to Ω at x if there exists δ > 0 such that x0 − tν ∈ Ω and x0 + tν ∈ RN \ Ω for all 0 < t < δ. Given a function u : Ω → R and a point x ∈ ∂Ω for which the unit outward normal ν is defined, we define the normal derivative ∂ν u(x) of u at x as the limit u(x + tν) − u(x) ∂ν u(x) = lim . t→0 t In particular, if u ∈ C 1 (Ω), then we can extend u to a C 1 function on an open set U that contains Ω. Since the extended function is differentiable at x ∈ ∂Ω, it follows that (11.68)
∂ν u(x) =
N X
∂i u(x)νi (x).
i=1
Let Ω ⊆ RN , N ≥ 2, be an open set whose boundary ∂Ω is Lipschitz continuous, and let u ∈ W s,p (Ω), 1 < s < 2. Since u ∈ W 1,p (Ω), the trace of u is well defined (see [Leo22d]). We now show that if (s − 1)p > 1, the trace of the normal derivative is also well defined.
11.8. Trace Theory
455
Theorem 11.60. Let Ω ⊆ RN , N ≥ 2, be an open set whose boundary ∂Ω is Lipschitz continuous, 1 < p < ∞ and 1 < s < 2 with (s − 1)p > 1. There exists a unique linear operator Tr1 : W s,p (Ω) → Lploc (∂Ω) such that (i) Tr1 (∂ν u) = ∂ν u on ∂Ω for all u ∈ W s,p (Ω) ∩ C 1 (Ω), (ii) for every R ≥ 1, 0 < ε < 1, and u ∈ W s,p (Ω), Z Z p N −1 −1 (11.69)  Tr1 (∂ν u) dH Ω,R ε B(0,R)∩∂Ω
+
k∇ukp dx
B(0,2R)∩Ω (s−1)p−1 ε k∇ukpW s−1,p (Ω) .
The function Tr1 (∂ν u) is called the trace of the normal derivative of u on ∂Ω. Proof. If u ∈ W s,p (Ω), then ∂i u ∈ W s−1,p (Ω) for every i = 1, . . . , N , and so, by Theorem 9.14, Tr(∂i u) is well defined with Z Z  Tr(∂i u)p dHN −1 Ω,R ε−1 ∂i up dx B(0,R)∩∂Ω
+ε
B(0,2R)∩Ω k∂i ukpW s−1,p (Ω) .
(s−1)p−1
Define (11.70)
Tr1 (∂ν u) :=
N X
Tr(∂i u)νi .
i=1
Then Z B(0,R)∩∂Ω
p
 Tr1 (∂ν u) dH Z Ω,R ε−1
N −1
Z =
B(0,2R)∩Ω
 Tr(∇u) · νp dHN −1
B(0,R)∩∂Ω
k∇ukp dx + ε(s−1)p−1 k∇ukpW s−1,p (Ω) .
Note that if u ∈ W s,p (Ω) ∩ C 1 (Ω), then by (11.68), (11.70), and the fact that Tr(∂i u) = ∂i u by Theorem 9.14, we have that Tr1 (∂ν u) = ∂ν u. Finally, the uniqueness of Tr1 follows from the uniqueness of the trace operator (see Theorem 9.14) and (11.70). To characterize Tr1 , as usual, we begin with the case in which Ω is a halfspace. As in Section 9.3, with an abuse of notation, we identify Tr(u) and Tr1 (∂N u) with the functions Tr(u)(·, 0) and Tr1 (∂N u)(·, 0). In view of this identification, we can write Tr(u)(x0 ), Tr1 (∂N u)(x0 ) and Tr(u), Tr1 (∂N u) ∈ Lp (RN −1 ) rather than Tr(u)(x0 , 0), Tr1 (∂N u)(x0 , 0) and Tr(u), Tr1 (∂N u) ∈ Lp (RN −1 × {0}).
456
11. Higher Order Fractional Sobolev Spaces
Theorem 11.61. Let N ≥ 2, 1 < p < ∞, and 1 < s < 2, with (s − 1)p > 1. Then for all u ∈ W s,p (RN + ), k Tr(u)kW s−1/p,p (RN −1 ) kukW s,p (RN ) , +
k Tr1 (∂N u)kLp (RN −1 ) kukW s,p (RN ) ,
(11.71)
+
 Tr1 (∂N u)W s−1−1/p,p (RN −1 ) uW s,p (RN ) . +
Moreover, for all i = 1, . . . , N − 1, (11.72)
∂i Tr(u) = Tr(∂i u).
1 N Proof. Step 1: Assume that u ∈ W s,p (RN + ) ∩ C (R+ ). Since ∂i u ∈ N W s−1,p (RN + ) ∩ C(R+ ), by Theorem 9.19, for all i = 1, . . . , N ,
k∂i u(·, 0)kLp (RN −1 ) k∂i ukW s−1,p (RN ) , +
∂i u(·, 0)W s−1−1/p,p (RN −1 ) ∂i uW s−1,p (RN ) . +
˙ s−1/p,p (RN −1 ). W
In particular, this implies that u(·, 0) ∈ On the other hand, 1,p N since u ∈ W (R+ ), by the trace theory in Sobolev spaces (see [Leo22d]), u(·, 0) ∈ Lp (RN −1 ), with ku(·, 0)kLp (RN −1 ) kukLp (RN ) + k∂N ukLp (RN ) . +
+
Thus, u(·, 0) ∈ W s−1/p,p (RN −1 ). This implies that for every ϕ ∈ Cc∞ (RN −1 ) and i = 1, . . . , N − 1, Z Z 0 0 0 (11.73) ∂i u(x , 0)ϕ(x ) dx = − u(x0 , 0)∂i ϕ(x0 ) dx0 . RN −1
RN −1
Step 2: Using a reflection argument (see Theorem 11.55) and mollifying, 1 N we can construct a sequence of functions un ∈ W s,p (RN + ) ∩ C (R+ ) such s,p N that un → u in W (R+ ). Then kun (·, 0) − uk (·, 0)kW s−1/p,p (RN −1 ) kun − uk kW s,p (RN ) , +
k∂N un (·, 0) − ∂N uk (·, 0)kW s−1−1/p,p (RN −1 ) kun − uk kW s,p (RN ) , +
which implies that {un (·, 0)}n and {∂N un (·, 0)}n are Cauchy sequences in W s−1/p,p (RN −1 ) and W s−1−1/p,p (RN −1 ), respectively. By the uniqueness of the trace operator (Theorems 9.14 and 11.60), it follows that Tr(u) ∈ W s−1/p,p (RN −1 ) and Tr1 (∂N u) ∈ W s−1−1/p,p (RN −1 ), and that the estimates in (11.71) hold. Finally, replacing u with un in (11.73) and letting n → ∞ gives Z Z 0 0 0 Tr(∂i u)(x )ϕ(x ) dx = − Tr(u)(x0 )∂i ϕ(x0 ) dx0 RN −1
for every ϕ ∈
Cc∞ (RN −1 ),
RN −1
which implies (11.72).
11.8. Trace Theory
457
In what follows, given u ∈ W s,p (RN ), by its trace on RN −1 or, more precisely, RN −1 × {0}, we mean the trace of the restriction of u to RN +. Theorem 11.62. Let N ≥ 2, 1 < p < ∞, 1 < s < 2 with (s − 1)p > 1, and g ∈ W s−1/p,p (RN −1 ). Then there exists a function u ∈ W s,p (RN ) such that Tr(u) = g in RN −1 and kukW s,p (RN ) kgkW s−1/p,p (RN −1 ) . Proof. Step 1:RLet ϕ ∈ Cc∞ (RN −1 ) be nonnegative and such that supp ϕ ⊆ BN −1 (0, 1) and RN −1 ϕ(x0 ) dx0 = 1. For x0 ∈ RN −1 and xN > 0 define 0 Z 1 x − y0 (11.74) v(x) := N −1 ϕ g(y 0 ) dy 0 . x N −1 xN N R Since g ∈ W 1,p (RN −1 ), by the Lebesgue dominated convergence theorem and integration by parts, for all i = 1, . . . , N − 1 and x ∈ RN + , we have 0 Z 0 x −y ∂g 0 ∂v 1 ϕ (x) = N −1 (y ) dy 0 . ∂xi x ∂x N −1 xN i N R Since (s−1)p > 1, we can reason as in Steps 1 and 2 of the proof of Theorem ∂g 9.21, with g replaced by ∂x , to obtain i ∂g ∂v ∂xi s−1,p N ∂xi s−1−1/p,p N −1 . W (R ) W (R ) +
Similarly, N −1
X 1 ∂u (x) = − ∂xN xN −1 i=1 N
Z ϕ RN −1
x0 − y 0 xN
xi − yi ∂g 0 (y ) dy 0 . xN ∂xi
We can now reason as in Steps 1 and 2 of the proof of Theorem 9.21, with ∂g and ϕ by ϕ(x0 )xi , to get g replaced by ∂x i ∂v ∂xN
W s−1,p (RN +)
N −1 X i=1
∂g ∂xi
. W s−1−1/p,p (RN −1 )
˙ s,p (RN ). To construct a Step 2: The function v in (11.74) belongs to W + s,p N ∞ function in W (R+ ), let ψ ∈ Cc ([0, ∞)) with ψ(t) = 1 for t ∈ [0, 1), ψ(t) = 0 for t ≥ 2, and 0 ≤ ψ ≤ 1. Define 0 Z ψ(xN ) x − y0 (11.75) u(x) := N −1 ϕ g(y 0 ) dy 0 , x ∈ RN +. xN xN RN −1 Then u(x) = v(x) for 0 < xN < 1, and so u and v have the same trace. Moreover, for i = 1, . . . , N − 1, ∂u ∂v (x) = ψ(xN ) (x), ∂xi ∂xi
∂u ∂v (x) = ψ(xN ) (x) + ψ 0 (xN )v(x). ∂xN ∂xN
458
11. Higher Order Fractional Sobolev Spaces
Reasoning as in Step 3 of the proof of Theorem 9.21, we get
∂u
∂g
kukLp (RN ) ≤ kgkLp (RN −1 ) , , + ∂xi Lp (RN ) ∂xi Lp (RN −1 ) +
N −1 X
∂u
∂g
∂xN p N kgkLp (RN −1 ) +
∂xi p N −1 L (R ) L (R ) i=1
+
and
∂u
∂g
∂xi s−1,p N ∂xi s−1−1/p,p N −1 , W (R+ ) W (R )
N −1 X ∂u
∂g
∂xN s−1,p N
∂xi s−1−1/p,p N −1 + kgkW s−1−1/p,p (RN −1 ) . W (R ) W (R ) +
i=1
Note that by Theorem 11.16, kgkW s−1−1/p,p (RN −1 ) kgkW s−1/p,p (RN −1 ) . Reasoning as in Step 4 of the proof of Theorem 9.21, we conclude that the trace of u is g. Finally, by Theorem 11.55 we can use a reflection argument to extend u to W s,p (RN ). Next we study the traces of normal derivatives on ∂Ω. In what follows, given v ∈ W s,p (RN ), by the trace of its normal derivative on RN −1 or, more precisely, RN −1 × {0}, we mean the trace of the normal derivative of the restriction of v to RN +. Theorem 11.63. Let N ≥ 2, 1 < p < ∞, 1 < s < 2, with (s − 1)p > 1, and h ∈ W s−1−1/p,p (RN−1 ). Then there exists a function v ∈ W s,p (RN ) such ∂v that Tr(v) = 0, Tr1 ∂x = h, and N kvkW s,p (RN ) khkW s−1/p,p (RN −1 ) .
(11.76)
+
Proof. Step 1: RLet ϕ ∈ Cc∞ (RN −1 ) be nonnegative and such that supp ϕ ⊆ BN −1 (0, 1) and RN −1 ϕ(x0 ) dx0 = 1, and ψ ∈ Cc∞ ([0, ∞)) with ψ(t) = 1 for t ∈ [0, 1), ψ(t) = 0 for t ≥ 2, and 0 ≤ ψ ≤ 1. For x0 ∈ RN −1 and xN > 0 define 0 Z ψ(xN ) x − y0 ϕ (11.77) v1 (x) := N −2 h(y 0 ) dy 0 . xN xN RN −1 Writing
0 Z xN ψ(xN ) x − y0 ϕ h(y 0 ) dy 0 , v1 (x) = −1 x N −1 xN N R N and reasoning as in Step 3 of the proof of Theorem 9.21 with ψ replaced by xN ψ(xN ), we have that Z Z ∞ Z p p v1 (x) dx ≤ xN ψ(xN ) dxN h(y 0 )p dy 0 . RN +
0
RN −1
11.8. Trace Theory
459
By the properties of mollifiers (see [Leo22c]) where xN plays the role of ε), for any i = 1, . . . , N − 1 we have that Z ∂v1 ψ(xN ) ∂ϕ x0 − y 0 (x) = N −1 h(y 0 ) dy 0 . ∂xi xN xN RN −1 ∂xi Note that the righthand side of the previous identity has the same form of the function u defined in Step 3 of the proof of Theorem 9.21, with the only ∂ϕ difference that ϕ is replaced by ∂x . Hence, following that proof, we get i ∂v1 ∂xi s−1,p N khkW s−1−1/p,p (RN −1 ) . W (R ) +
On the other hand, by differentiating with respect to xN (see [Leo22c]), we obtain that 0 Z x − y0 xN ψ 0 (xN ) ∂v1 ϕ (x) = h(y 0 ) dy 0 N −1 ∂xN x N −1 xN N R 0 Z x − y0 ψ(xN ) ϕ − (N − 2) N −1 h(y 0 ) dy 0 x N −1 x N R 0 0 Z N 0 x −y x − y0 ψ(xN ) ∇ x0 ϕ · h(y 0 ) dy 0 . − N −1 xN xN xN RN −1 Each of the three integrals on the righthand side has the same form of the function u defined in Step 3 of the proof of Theorem 9.21. Note that in the first term we have xN ψ 0 (xN ) in place of ψ(xN ) and in the last term we have φ(x0 ) = ∇x0 ϕ(x0 ) · x0 in place of ϕ(x0 ). Hence, following that proof, we get ∂v1 ∂xN s−1,p N khkW s−1−1/p,p (RN −1 ) . W (R+ ) ∂v1 = h. Assume that It remains to show that Tr(v1 ) = 0 and Tr1 ∂x N ∞ N −1 ). Since ψ(t) = 1 for t ∈ [0, 1), supp ϕ ⊆ B N −1 (0, 1) and Rh ∈ Cc (R 0 0 = 1, by the change of variables (x0 − y 0 )/x ϕ(x ) dx = z 0 for 0 < N −1 N R xN < 1, we have 0 Z x − y0 1 (11.78) xN = N −2 ϕ dy 0 . xN xN RN −1
In turn, ! x0 − y 0 ϕ dy 0 −2 x N −1 xN N R N 0 0 0 Z 1 x − y0 x − y0 x − y0 = − N −1 ϕ + (N − 2)∇x0 ϕ · dy 0 . x x x N −1 xN N N N R
∂ 1= ∂xN
1
Z
460
11. Higher Order Fractional Sobolev Spaces
Thus, we can proceed as in Step 4 of Theorem 9.21, with ϕ replaced by the function φ(x0 ) = −ϕ(x0 ) − (N − 2)∇x0 ϕ(x0 ) · x0 to show that h is the trace of ∂v1 + ∂xN . On the other hand, by (11.77), v1 (x) ≤ khk∞ xN → 0 as xN → 0 . Thus, the trace of v1 is zero in this case. The case h ∈ W s−1−1/p,p (RN −1 ) follows by a density argument and the continuity of the trace operators. We leave the details as an exercise. Finally, by Theorem 11.55 we can use a reflection argument to extend v to W s,p (RN ). Corollary 11.64. Let N ≥ 2, 1 < p < ∞, 1 < s < 2 with (s − 1)p > 1, g ∈ W s−1/p,p (RN −1 ), and h ∈ W s−1−1/p,p (RN −1 ). Then there exists a function ∂u u ∈ W s,p (RN ) such that Tr(u) = g, Tr1 ∂x = h in RN −1 , and N kukW s,p (RN ) kgkW s−1/p,p (RN −1 ) + khkW s−1−1/p,p (RN −1 ) . Proof. By Theorem 11.62 there exists a function w ∈ W s,p (RN ) such that Tr(w) = g and kwkW s,p (RN ) kgkW s−1/p,p (RN −1 ) . In particular, ∂N w ∈ W s−1,p (RN ), so by Theorem 11.61, Tr1 (∂N w) ∈ W s−1−1/p,p (RN −1 ), with k Tr1 (∂N w)kW s−1−1/p,p (RN −1 ) k∂N wkW s−1,p (RN ) . We now apply Theorem 11.63, with h replaced by h − Tr1 (∂N w), to find a function v ∈ W s,p (RN ) such that Tr(v) = 0, Tr1 (∂N v) = h − Tr1 (∂N w), and kvkW s,p (RN ) khkW s−1−1/p,p (RN −1 ) + kTr1 (∂N w)kW s−1−1/p,p (RN −1 ) . It suffices to take u = w + v.
Theorem 11.65 (Traces and W0s,p ). Let N ≥ 2, 1 < p < ∞, 1 < s < 2, with (s − 1)p > 1, and u ∈ W s,p (RN + ). Then Tr(u) = 0 and Tr1 (∂N u) = 0 in RN −1 if and only if u ∈ W0s,p (RN + ). Proof. If u ∈ Cc∞ (RN + ), then Tr(u) = 0 and Tr1 (∂N u) = 0, and so, since s,p N W0 (R+ ) is the closure of Cc∞ (RN , + ) with respect to the norm k · kW s,p (RN +) it follows from the continuity of the trace operators that Tr(u) = 0 and Tr1 (∂N u) = 0 for all u ∈ W0s,p (RN + ). To prove the converse implication, let u ∈ W s,p (RN + ) be such that Tr(u) = 0 and Tr1 (∂N u) = 0. By Exercise 11.20, we can also assume that u = 0 for all x ∈ RN + with kxk > R, where R > 0. Consider the function u(x) if xN > 0, u ˜(x) := 0 otherwise. Since u ∈ W 1,p (RN ˜ ∈ W 1,p (RN ). In view + ) and Tr(u) = 0, it follows that u of (11.72) and the fact that Tr1 (∂N u) = 0, we have Tr(∂i u) = 0 for all
11.8. Trace Theory
461
i = 1, . . . , N . By Remark 6.80 and the fact that Tr(∂i u) = 0, we have that Z Z Z ∂i u(x) − ∂i u(y)p ∂i u(x)p dx dxdy. (s−1)p kx − ykN +(s−1)p RN RN RN xN + + + s−1,p It follows from Theorem 6.99 that ∂i u ∈ W00 (RN + ). This shows that the s,p N function u ˜ ∈ W (R ). Since translation is continuous in Lp spaces, we have that the function vδ (x) := u ˜(x0 , xN − δ), x ∈ RN , δ > 0, belongs to s,p N W (R ) and kvδ − u ˜kW s,p (RN ) → 0 as δ → 0+ . Note that vδ (x) = 0 for all x ∈ RN with xN < δ and vδ has bounded support (since u ˜ vanishes outside a B(0, R)). By mollifying vδ and using a diagonal argument, we can construct a sequence of functions vn ∈ Cc∞ (RN + ) such that vn → u in s,p N ). ). Hence, u ∈ W (R W s,p (RN + + 0
In the case 1 < s < 2 and (s − 1)p < 1, we can no longer define the trace of the normal derivative, but we can still prove that Tr(W s,p (RN + )) = W s−1/p,p (RN −1 ). Theorem 11.66. Let N ≥ 2, 1 < p < ∞, and 1 < s < 2, with (s − 1)p < 1. Then for all u ∈ W s,p (RN + ), k Tr(u)kLp (RN −1 ) kukW s,p (RN ) , +
 Tr(u)W s−1/p,p (RN −1 ) uW s,p (RN ) . +
0 N −1 N Proof. Step 1: Assume that u ∈ W s,p (RN + ) ∩ C(R+ ). Given x ∈ R and t > 0, write
u(x0 , 0) = −u(x0 , 2t) + 2u(x0 , t) + ∆2t,N u(x0 , 0). Averaging in t over (0, τ ) yields Z τ Z τ Z τ u x0 , 2t dt + 2 u x0 , t dt + 1 u(x0 , 0) ≤ 1 ∆2t,N u(x0 , 0) dt. τ 0 τ 0 τ 0 Raising both sides to power p and using H¨ older’s inequality, we get Z τ p−1 u(x0 , 0) p τ u x0 , 2t p dt p τ 0 Z Z p τ p−1 τ τ p−1 τ 2 0 + p u x , t dt + p ∆t,N u(x0 , 0)p dt. τ τ 0 0 We now replace u with ∆2τ,n u, where n = 1, . . . , N − 1, to obtain Z τ Z τ 2 2 2 ∆τ,n u(x0 , 0) p 1 ∆τ,n u x0 , 2t p dt + 1 ∆τ,n u x0 , t p dt τ 0 τ 0 Z τ 1 + ∆2τ,n ∆2t,N u(x0 , 0)p dt. τ 0
462
11. Higher Order Fractional Sobolev Spaces
Dividing by τ sp and integrating in τ over R+ and in x0 over RN −1 , it follows Z Z ∞ Z Z ∞Z τ 2 2 0 p dτ 0 ∆τ,n u x0 , 2t p dt dτ dx0 ∆τ,n u(x , 0) sp dx τ τ 1+sp RN −1 0 RN −1 0 0 Z Z ∞Z τ 2 ∆τ,n u x0 , t p dt dτ dx0 + τ 1+sp N −1 0 ZR Z ∞ Z0 τ dτ + ∆2τ,n ∆2t,N u(x0 , 0)p dt 1+sp dx0 =: A + B + C. τ N −1 R 0 0 Since k∆τ,n vkLp (RN −1 ) ≤ 2kvkLp (RN −1 ) , using Tonelli’s theorem, we can estimate Z Z ∞Z τ dτ C ∆2t,N u(x0 , 0)p dt 1+sp dx0 τ N −1 0 ZR Z ∞ 0 Z ∞ 1 ∆2t,N u(x0 , 0)p dτ dtdx0 1+sp τ N −1 0 t ZR Z ∞ dt ∆2t,N u(x0 , 0)p sp dx0 . t RN −1 0 We now apply Hardy’s inequality (see Theorem 5.34) to the function u(x0 , ·) to obtain that Z Z ∞Z ∞ 2 ∆τ,N u(x0 , t)p C dtdτ dx0 . 1+sp τ N −1 R 0 0 In conclusion we have shown that (11.79) Z Z RN −1
0
∞
∆2τ,n u(x0 , 0)p dτ dx0 τ sp
Z
∞Z ∞
Z
RN −1
0
Z
Z
0 ∞Z ∞
+ RN −1
0
2 ∆τ,n u (x0 , t) p
0
dtdτ dx0 τ 1+sp ∆2τ,N u(x0 , t)p dtdτ dx0 . τ 1+sp
By reflecting u and applying Theorem 11.28, we obtain that the righthand side is bounded from above by CupW s,p (RN ) . Summing these inequalities +
over n = 1, . . . , N − 1, by Theorem 6.61, we obtain that the lefthand side is bounded from below by Z Z 0 p ∆2N h0 u(x , 0) dh0 dx0 . N −1+sp 0 N −1 N −1 kh k R R N −1 Since u(·, 0) ∈ Lp (RN −1 ), by Corollary 6.60 we conclude that u(·, 0) ∈ W s−1/p,p (RN −1 ). Step 2: The case u ∈ W s,p (RN + ) is left as an exercise.
11.8. Trace Theory
463
Remark 11.67. When 1 < p < ∞ and s = 1 + 1/p, the previous proof shows that Z Z ∞ 2 ∆τ,n u(x0 , 0)p dτ dx0 upW 1+1/p,p (RN ) , 1+p + τ N −1 R 0 1,p N −1 ), where B 1,p (RN −1 ) is the that is, that Tr(W 1+1/p,p (RN + )) ⊆ B (R Besov space defined as the space of all functions v ∈ Lp (RN −1 ) such that Z Z ∆2h0 v(x0 )p 0 0 dx dh < ∞ N −2+p RN −1 RN −1 kh0 kN −1
(see [Leo17]). Note that Hardy’s inequality continues to hold for B 1,p (R+ ) (see Remark 5.35). The proof of the inclusion W s−1/p,p (RN −1 ) ⊆ Tr(W s,p (RN + )) is more s,p N N involved. As before, given u ∈ W (R ), by its trace on R −1 or, more precisely, RN −1 × {0}, we mean the trace of the restriction of u to RN +. Theorem 11.68. Let N ≥ 2, 1 < p < ∞, 1 < s < 2, with (s − 1)p < 1, and g ∈ W s−1/p,p (RN −1 ). Then there exists a function u ∈ W s,p (RN ) such that Tr(u) = g in RN −1 and kukW s,p (RN ) kgkW s−1/p,p (RN −1 ) . Proof. Step 1: Assume g ∈ Cc∞ (RN −1 ). Let ϕR ∈ Cc∞ (RN −1 ) be nonnegative and such that supp ϕ ⊆ BN −1 (0, 1) and RN −1 ϕ(x0 ) dx0 = 1. For x0 ∈ RN −1 and xN > 0 define 0 Z 1 x − y0 (11.80) v(x) := N −1 ϕ g(y 0 ) dy 0 . x N −1 xN N R By standard trace theory (see [Leo17, Proof of Theorem 18.28]), v ∈ ˙ 1,p (RN W + ), with Z Z (11.81) v(x)p dx0 ≤ g(x0 )p dx0 for every xN > 0, N −1 N −1 R ZR (11.82) ∂i v(x)p dx gpW 1−1/p,p (RN −1 ) RN +
for all i = 1, . . . , N . Observe that (11.83)
gW 1−1/p,p (RN −1 ) kgkW s−1/p,p (RN −1 )
by Lemma 6.17. We claim that Z Z ∞ Z ∞Z dt 2 p dt (11.84) ∆t,i v(x) 1+sp dx ∆t,i g(x0 )p dx0 sp t t RN 0 0 RN −1 +
464
11. Higher Order Fractional Sobolev Spaces
for every i = 1, . . . , N − 1. To see this, we use Tonelli’s theorem and write Z Z ∞ Z ∞Z tZ dt dt ∆2t,i v(x)p 1+sp dx = ∆2t,i v(x)p dx0 dxN 1+sp t t RN 0 0 0 RN −1 + Z ∞Z ∞Z dt + ∆2t,i v(x)p dx0 dxN 1+sp =: A + B. t N −1 0 t R By the change of variables (x0 − y 0 )/xN = w0 we can write (11.80) as Z (11.85) v(x) = ϕ(w0 )g(x0 − xN w0 ) dw0 . RN −1
Hence, Z
ϕ(w0 )∆2t,i g(x0 − xN w0 ) dw0 . R It follows from H¨older’s inequality and the fact that RN −1 ϕ(x0 ) dx0 = 1 that Z ∆2t,i v(x)p ≤ ϕ(w0 )∆2t,i g(x0 − xN w0 )p dw0 . ∆2t,i v(x) =
RN −1
RN −1
Integrating in x0 over RN −1 , using Tonelli’s theorem and the change of variables x0 − xN w0 = z 0 , so that dx0 = dz 0 , gives Z Z Z 2 p 0 ∆t,i v(x) dx ≤ ϕ(w0 )∆2t,i g(x0 − xN w0 )p dw0 dx0 RN −1 RN −1 RN −1 Z Z Z = ϕ(w0 ) dw0 ∆2t,i g(z 0 )p dz 0 = ∆2t,i g(z 0 )p dz 0 . RN −1
RN −1
RN −1
Since k∆t,i vkLp (RN −1 ) ≤ 2kvkLp (RN −1 ) , we have Z ∞Z tZ Z ∞Z dt dt 2 0 p 0 A≤ ∆t,i g(z ) dz dxN 1+sp = ∆2t,i g(z 0 )p dz 0 sp t t N −1 N −1 0 R Z0 ∞ Z0 R dt ∆t,i g(z 0 )p dz 0 sp . t 0 RN −1 To estimate B, we use the fundamental theorem of calculus, Fubini’s theorem, and the change of variables (x0 − y 0 + rei )/xN = w0 , so that N1−1 dy 0 = xN
dw0 , to write ∆2t,i v(x)
Z
1
x0 − y 0 xN
∆ ϕ g(y 0 ) dy 0 −1 t,i xN N Z t Z 1 ∂ϕ x0 − y 0 + rei = ∆t,i g(y 0 ) drdy 0 N ∂x x N −1 x i N R 0 N Z Z t 1 ∂ϕ 0 = ∆t,i (w )g(x0 + rei − xN w0 ) drdw0 xN ∂x N −1 i R 0 Z Z t ∂ϕ 0 1 = (w )∆t,i g(x0 + rei − xN w0 ) drdw0 . xN RN −1 0 ∂xi = ∆t,i
RN −1
11.8. Trace Theory
465
1/p0 1/p ∂ϕ ∂ϕ ∂ϕ Writing ∂x = ∂xi ∂xi , it follows from H¨older’s inequality that i ∆2t,i v(x)p
tp−1 p xN
Z t ∂ϕ 0 0 0 p 0 ∂xi (w ) ∆t,i g(x + rei − xN w ) drdw . N −1 R 0
Z
Integrating in x0 over RN −1 , using Tonelli’s theorem and the change of variables x0 + rei − xN w0 = z 0 , so that dx0 = dz 0 , gives Z ∆2t,i v(x)p dx0 N −1 R Z Z Z t ∂ϕ 0 tp−1 p (w ) ∆t,i g(x0 + rei − xN w0 )p drdw0 dx0 xN RN −1 RN −1 0 ∂xi Z tp p ∆t,i g(z 0 )p dz 0 . xN RN −1 Hence, Z B
∞
tp
Z
t1+sp 0 Z ∞Z
0
RN −1
t
∞
1 xpN
Z RN −1
∆t,i g(z 0 )p dz 0
∆t,i g(z 0 )p dz 0 dxN dt dt , tsp
R∞ N 1 where we use the fact that t dx tp−1 . Combining the estimates for A xpN and B gives (11.84) when i = 1, . . . , N − 1. Step 2: We claim that (11.86) Z Z ∞
RN +
∆2t,N v(x)p
0
dt
dx t1+sp
Z RN −1
Z RN −1
g(y 0 ) − g(x0 )p kx0
−
−2+sp y 0 kN N −1
dx0 dy 0 .
Fix x0 ∈ RN −1 . By Theorem 5.32 applied to v(x0 , ·), we get Z ∞Z ∞ Z ∞ dt (2−s)p 2 ∆2t,N v(x0 , xN )p 1+sp dxN xN ∂N v(x0 , xN )p dxN . t 0 0 0 Integrating both sides in x0 over RN −1 and using Tonelli’s theorem gives Z Z ∞ Z dt (2−s)p 2 ∆2t,N v(x)p 1+sp dx xN ∂N v(x0 , xN )p dx. N N t R+ 0 R+ Hence, to prove (11.86), it remains to show that Z Z Z g(y 0 ) − g(x0 )p (2−s)p 2 p (11.87) xN ∂N v(x) dx dx0 dy 0 . N −2+sp N −1 RN −1 kx0 − y 0 k RN R + N −1
466
11. Higher Order Fractional Sobolev Spaces
By making the change of variables x0 − y 0 = z 0 and then differentiating v under the integral sign (see [Leo22c]), we obtain that 0 ! Z 2 ∂ 1 z 2 ∂N v(x) = g(x0 − z 0 ) dz 0 ϕ 2 N −1 xN xN RN −1 ∂xN 0 ! Z 1 ∂2 z = ϕ [g(x0 − z 0 ) − g(x0 )] dz 0 , 2 N −1 x N −1 ∂x x N R N N where in the second equality we use the fact that ! 0 Z ∂2 ∂2 1 z 0= dz 0 (1) = ϕ N −1 xN ∂x2N ∂x2N RN −1 xN 0 ! Z ∂2 1 z = dz 0 . ϕ 2 N −1 xN xN RN −1 ∂xN Since supp ϕ ⊆ BN −1 (0, 1), we have that 2 ∂N v(x)
Z = BN −1 (0,xN )
∂2 ∂x2N
1 −1 xN N
ϕ
z0 xN
!
[g(x0 − z 0 ) − g(x0 )] dz 0 .
We leave as an exercise to check that 0 ! 0 1 ∂2 1 z z = N +1 Φ ϕ 2 N −1 xN xN ∂xN xN xN for some function Φ ∈ Cc∞ (RN −1 ), with supp Φ ⊆ BN −1 (0, 1). Therefore, Z 1 2 g(x0 − z 0 ) − g(x0 ) dz 0 . (11.88) ∂N v(x) N +1 xN BN −1 (0,xN ) (2−s)p
Raising both sides to the power p, multiplying by xN over RN older’s inequality, we get + , and using H¨ Z (2−s)p 2 xN ∂N v(x)p dx
, integrating in x
RN +
Z RN +
Z RN +
(2−s)p
xN
(N +1)p
xN
Z
g(x0 − z 0 ) − g(x0 ) dz 0
p
dx
BN −1 (0,xN )
(2−s)p+(N −1)(p−1)
xN
(N +1)p xN
Z BN −1 (0,xN )
g(x0 − z 0 ) − g(x0 )p dz 0 dx =: A.
11.8. Trace Theory
467
By Tonelli’s theorem, we obtain the estimate Z ∞Z Z 1 A= g(x0 − z 0 ) − g(x0 )p dz 0 dx0 dxN N −1+sp 0 RN −1 BN −1 (0,xN ) xN Z ∞ Z Z 1 0 0 0 p g(x − z ) − g(x ) = dxN dz 0 dx0 N −1+sp kz 0 kN −1 xN RN −1 RN −1 Z Z g(x0 − z 0 ) − g(x0 )p 0 0 dx dz . −2+sp kz 0 kN RN −1 RN −1 N −1 Hence, we have shown (11.87). ˙ s,p (RN ). Step 3: The function v constructed in the previous steps belongs to W + ) with the same trace. Let We now modify it to obtain a function in W s,p (RN + ψ ∈ Cc∞ ([0, ∞)) be such that ψ(t) = 1 for t ∈ [0, 1), ψ(t) = 0 for t ≥ 2, and 0 ≤ ψ ≤ 1. Define u(x) := ψ(xN )v(x), where v is the function constructed in Step 1. By standard trace theory (see [Leo17, Proof of Corollary 18.29]), u ∈ W 1,p (RN + ), with kukW 1,p (RN ) kgkW 1−1/p,p (RN −1 ) kgkW s−1/p,p (RN −1 ) . +
If i = 1, . . . , N − 1, then ∆2t,i u(x) = ψ(xN )∆2t,i v(x), so, since 0 ≤ ψ ≤ 1, by the previous step, Z Z ∞ Z Z ∞ dt 2 p dt ∆t,i u(x) 1+sp dx ≤ ∆2t,i v(x)p 1+sp dx N N t t R+ 0 R+ 0 Z Z 0 g(y ) − g(x0 )p dx0 dy 0 . N −2+sp 0 0 N −1 N −1 kx − y k R R N −1 On the other hand, if i = N , by Exercise 1.10, ∆2t,N u(x)p ∆2t,N v(x)p + ∆t ψ(xN + t)p ∆t,N v(x)p + ∆2t ψ(xN + 2t)p v(x)p . Therefore, Z Z RN +
∞
0
Z
dt ∆2t,N u(x)p 1+sp dx t
Z
Z + RN +
0
∆2t,N v(x)p
RN +
0
∆t ψ(xN + t)p ∆t,N v(x)p
0
Z
∞
Z
∞
+ RN +
Z
∞
∆2t,N ψ(xN )p v(x)p
dt t1+sp
dt t1+sp
dt t1+sp
dx
dx
dx =: A + B + C.
By the previous step, A gpW s−1/p,p (RN −1 ) , while by (11.81) and Tonelli’s theorem, Z ∞Z ∞ Z Z p dt 2 0 p 0 C ∆t,N ψ(xN ) 1+sp dxN g(x ) dx g(x0 )p dx0 , t RN −1 0 0 RN −1
468
11. Higher Order Fractional Sobolev Spaces
where the last inequality follows from Theorem 5.32 applied to the function ψ ∈ Cc∞ (R). Since ψ(t) = 0 for t ≥ 2, we have that ∆t ψ(xN + t) = 0 if xN ≥ 2 or t ≥ 2. Hence, Z Z 2Z 2 dt ∆t ψ(xN + t)p ∆t,N v(x)p 1+sp dxN dx0 . B≤ t N −1 R 0 0 By the fundamental theorem of calculus and H¨older’s inequality, Z 1 p p ∂N v(x0 , xN + tr)p dr, ∆t,N v(x) ≤ t 0
while, by the mean value theorem, ∆t ψ(xN + t) ≤ kψ 0 k∞ t. Therefore, by Tonelli’s theorem and the change of variables yN = xN + tr, so that dyN = dxN , Z 1 Z 2 Z 2 2p Z t ∂N v(x0 , xN + tr)p drdtdxN dx0 B 1+sp RN −1 0 0 0 t Z 2 Z Z ∞ tp(2−s)−1 dt ∂N v(x0 , yN )p dyN dx0 gpW 1−1/p,p (RN −1 ) , 0
RN −1
0
where in the last inequality we use (11.82) and the fact that 0 < s < 2. Combining the estimates for A, B, and C and using (11.83), we obtain Z Z ∞ dt ∆2t,i u(x)p 1+sp dx ≤ kgkpW s−1/p,p (RN −1 ) . t RN 0 + Step 4: The general case g ∈ W s−1/p,p (RN −1 ) is left as an exercise.
Remark 11.69. In the case 1 < p < ∞ and s = 1 + 1/p, one can adapt the previous proof to show that B 1,p (RN −1 ) = Tr(W 1+1/p,p (RN + )). In Step 1, inequality (11.84) should be replaced by Z Z ∞ Z ∞Z dt dt ∆3t,i v(x)p 2+p dx ∆2t,i g(x0 )p dx0 1+p . t t 0 RN 0 RN −1 + In Step 3, we take ϕ to be even. Then we can write 0 0 Z 1 x − y0 y − x0 1 v(x) = N −1 ϕ +ϕ g(y 0 ) dy 0 xN xN xN RN −1 2 0 Z 1 1 z = N −1 ϕ [g(x0 − z 0 ) + g(x0 + z 0 )] dz 0 , 2 x N −1 xN N R where we made the changes of variables z 0 = x0 − y 0 and z 0 = y 0 − x0 . By differentiating twice with respect to xN , we get 0 ! Z 2 ∂ 1 z 2 ∂N v(x) = ϕ [g(x0 − z 0 ) + g(x0 + z 0 )] dz 0 . 2 N −1 xN 2xN RN −1 ∂xN
11.8. Trace Theory
Since
R
RN −1
∂2 ∂x2N
469
1
N −1 ϕ
2xN
z0 xN
dz 0 = 0 (if we take g = 1 in (11.80), we
obtain that v = 1), we have 0 ! Z 2 ∂ 1 z 2 ∂N v(x) = [g(x0 − z 0 ) − 2g(x0 ) + g(x0 + z 0 )] dz 0 . ϕ 2 N −1 xN 2xN RN −1 ∂xN In turn, the estimate (11.88) becomes Z 1 2 ∂N v(x) N +1 g(x0 − z 0 ) − 2g(x0 ) + g(x0 + z 0 ) dz 0 . xN BN −1 (0,xN ) We can now continue as before to conclude that Z Z Z ∆2z 0 g(x0 )p 0 0 p−1 2 p xN ∂N v(x) dx dx dz . N −1+p RN RN −1 RN −1 kz 0 kN −1 + Finally, the proof in Step 3 continues to hold in view of the embedding B 1,p (RN −1 ) ⊆ W 1,p (RN −1 ) (see [Leo17, Corollary 17.68]). Exercise 11.70. Let N ≥ 2, 1 < p < ∞, 1 < s < 2, with (s − 1)p < 1, and s,p N u ∈ W s,p (RN + ). Prove that Tr(u) = 0 if and only if u ∈ W0 (R+ ). In the following exercises, we denote by F the Fourier transform and by F 0 the Fourier transform in RN −1 , that is, Z 0 0 0 0 F (v)(x ) := e−2πix ·y v(y 0 ) dy 0 , x ∈ RN −1 . RN −1
Exercise 11.71. Let x0 ∈ RN −1 .
1 2
< s < 1. Given u ∈ S(RN ), let v(x0 ) := u(x0 , 0),
(i) Prove that F 0 (v)(x0 ) = (ii) Prove that for all x0 ∈ (1 +
R
R F(u)(x RN −1 ,
kx0 k2N −1 )s−1/2 F 0 (v)(x0 )2
Z
0 , x ) dx , N N
x0 ∈ RN −1 .
(1 + kxk2 )s F(u)(x0 , xN )2 dxN .
R
Deduce that vH s−1/2 (RN −1 ) uH s (RN ) . Hint: Use Lemma 6.36 and Exercise 11.6. Exercise 11.72. Let 12 < s < 1. Given v ∈ H s−1/2 (RN −1 ), let u ∈ L2 (RN ) be the function given by x 1 N q F(u)(x) := F 0 (v)(x0 )ϕ q , x ∈ RN , 2 2 0 0 1 + kx kN −1 1 + kx kN −1 where ϕ ∈ Cc∞ (RN −1 ) is a standard mollifier.
470
11. Higher Order Fractional Sobolev Spaces
Prove that for all x0 ∈ RN −1 , Z 1 (1 + kxk2 )s F(u)(x0 , xN )2 dxN (1 + kx0 k2N −1 )s− 2 F 0 (v)(x0 )2 . R
Deduce that uH s (RN ) vH s−1/2 (RN −1 ) . Hint: Use Lemma 6.36 and Exercise 11.6. To characterize the trace operators for more general domains Ω ⊂ RN with smooth boundary, we need to define higher order fractional Sobolev spaces W σ,p on the manifold ∂Ω. The situation is more complicated than in Chapter 9 since when σ > 1, we need to talk about derivatives of functions defined only on ∂Ω. One possibility would be to introduce the tangential gradient of a function. Another avenue is to use charts to parametrize ∂Ω. We will follow the second approach. As in Chapter 9, we begin with the case in which Ω is the supergraph of a smooth function f : RN −1 → R. A chart for the manifold M = {(x0 , f (x0 )) : x0 ∈ RN −1 } is given by ϕ(x0 ) = (x0 , f (x0 )), x0 ∈ RN −1 . Note that ϕ is unbounded. For this reason, we introduce the homogeneous spaces C˙ m,α . Definition 11.73. Let Ω ⊆ RN be an open set, let 0 < α ≤ 1 and m ∈ N. A function u : Ω → R belongs to the homogeneous space C˙ m,α (Ω) if it admits continuous partial derivatives of every order up to m, with X ∂ α u(x) − ∂ α u(x) < ∞. uC m,α (Ω) := sup kx − ykα x,y∈Ω, x6=y N α∈N0 , α=m
Definition 11.74. Given an integer 1 ≤ k < N , m ∈ N ∪ {∞}, and 0 < a ≤ 1, a nonempty set M ⊆ RN is called a kdimensional parametrized surface or parametrized manifold of class C˙ m,α if there exist an open set V ⊆ Rk and a function ϕ : V → RN of class C˙ m,α such that (i) ϕ(V ) = M ; (ii) ϕ : V → M is biLipschitz continuous, that is, it is invertible and Lipschitz continuous together with its inverse ϕ−1 : M → V ; (iii) the Jacobian matrix Jϕ (y) has rank k for all y ∈ V . The function ϕ is called a chart or a parametrization or a system of coordinates of M . Remark 11.75. If M is as in Definition 11.74, given y0 ∈ V , since Jϕ (y0 ) has rank k, by relabeling the coordinates, we can assume that the k × k 1 ,...,ϕk ) minor ∂(ϕ ∂(y1 ,...,yk ) (y0 ) has a determinant different from zero. Hence, if we define Ψ : V × RN −k → RN by Ψ(y, z) := ϕ(y) + (0, z), we have that det JΨ (y0 , 0) = det
∂(ϕ1 , . . . , ϕk ) (y0 ) 6= 0. ∂(y1 , . . . , yk )
11.8. Trace Theory
471
Since Ψ is of class C˙ m,α , it follows by the inverse function theorem that Ψ−1 is of class C˙ m,α in a neighborhood of ϕ(y0 ), with JΨ−1 (x) = (JΨ (Ψ−1 (x)))−1 . Hence, if we assume that there exist λ, Λ > 0 such that kJϕ k∞ ≤ Λ, and for every y ∈ V there is a k × k minor A(y) of Jϕ (y) such that  det A(y) ≥ λ, then ϕ−1 is Lipschitz continuous with Lip ϕ−1 λ,Λ 1. Remark 11.76. If M is as in Definition 11.74, reasoning as in the previous remark and using the inverse function theorem, it can be shown that if ψ : W → M is another chart, then the function ψ −1 ◦ ϕ : V → W is a diffeomorphism of class C˙ m,α , that is, ψ −1 ◦ ϕ and ϕ−1 ◦ ψ are biLipschitz continuous and of class C˙ m,α (exercise). The function ψ −1 ◦ ϕ is called a change of parameters or a change of coordinates. Example 11.77. Given a Lipschitz continuous function f : RN −1 → R of class C˙ 1,α , the graph of f , M = {(x0 , f (x0 )) : x0 ∈ RN −1 }, is an (N − 1)dimensional parametrized surface of class C˙ 1,α . To see this, take ϕ : RN −1 → RN to be ϕ(x0 ) = (x0 , f (x0 )). As in Example 9.24, we have that ϕ : RN −1 → M is biLipschitz continuous, and ϕ is of class C˙ 1,α . Definition 11.78. Given 1 ≤ p < ∞, σ > 1, σ ∈ / N, and a kdimensional N m,1 ˙ parametrized surface M ⊂ R of class C , where 1 ≤ k < N , m = bσc, we say that a function v : M → R belongs to the fractional Sobolev space W σ,p (M ) if the function v ◦ ϕ belongs to W σ,p (V ) and we set kvkLp (M ) := kv ◦ ϕkLp (V ) ,
vW σ,p (M ) := v ◦ ϕW σ,p (V ) ,
kvkW σ,p (M ) := kv ◦ ϕkW σ,p (V ) . Here ϕ : V → RN is a chart of M . Remark 11.79. If ψ : W → M is another chart, since the function ψ −1 ◦ϕ : V → W is biLipschitz continuous and of class C˙ m,1 , by Remark 11.76 we can apply Theorem 11.18 to obtain that v ◦ ψ ∈ W σ,p (W ), with kv ◦ ψkW σ,p (W ) ψ,ϕ kv ◦ ϕkW σ,p (V ) ψ,ϕ kv ◦ ψkW σ,p (W ) . Hence, if we change charts, we obtain an equivalent norm, but the space W σ,p (M ) is independent of the particular chart. Exercise 11.80. Let Ω := {(x0 , xN ) ∈ RN −1 × R : xN > f (x0 )}, where f : RN −1 → R is a Lipschitz continuous function of class C˙ 1,1 . Given a function u ∈ C 1 (Ω), the tangential gradient or surface gradient ∇T u : ∂Ω → RN is defined as ∇T u = (IN − ν ⊗ ν)∇u, where IN is the N × N identity matrix and ν : ∂Ω → RN is the unit outward normal to ∂Ω. (i) Prove that ∇u = ∇T u + ∂ν u. (ii) Prove that 1 ∇T u = 2 J
J 2 ∇x0 u + (∇x0 u · ∇x0 f )∇x0 f − ∂N u∇x0 f J 2 ∂ N u + ∂ N u − ∇ x0 u · ∇ x0 f
,
472
11. Higher Order Fractional Sobolev Spaces
where J(x0 ) :=
q
1 + k∇x0 f (x0 )k2N −1 , u is evaluated at (x0 , f (x0 ))
and f and J at x0 . (iii) Let v(x0 ) := u(x0 , f (x0 )), x0 ∈ RN −1 . Prove that ∇T u = A∇x0 v, where A is a matrix depending on f . In what follows, given u ∈ W s,p (RN ), by its trace and the trace of its normal derivative on ∂Ω, we mean the traces of the restriction of u to Ω. Theorem 11.81. Let 1 < p < ∞, 1 < s < 2, with (s − 1)p > 1, and Ω := {(x0 , xN ) ∈ RN −1 × R : xN > f (x0 )}, where f : RN −1 → R is a Lipschitz continuous function of class C˙ 1,1 . Then for every u ∈ W s,p (Ω), (11.89)
k Tr(u)kW s−1/p,p (∂Ω) + k Tr1 (∂ν u)kW s−1−1/p,p (∂Ω) f kukW s,p (Ω) .
Conversely, given g ∈ W s−1/p,p (∂Ω) and h ∈ W s−1−1/p,p (∂Ω), there exists u ∈ W s,p (RN ) such that Tr(u) = g and Tr1 (∂ν u) = h on ∂Ω, and kukW s,p (RN ) f kgkW s−1/p,p (∂Ω) + khkW s−1−1/p,p (∂Ω) . Proof. Step 1: Assume that u ∈ W s,p (Ω)∩C 1 (Ω). Since ∂i u ∈ W s−1,p (Ω)∩ C(Ω), by Theorem 9.29, for all i = 1, . . . , N , (11.90)
k∂i ukW s−1−1/p,p (∂Ω) f k∂i ukW s−1,p (Ω) .
Since f : RN −1 → R is of class C˙ 1,1 , we have that ν ∈ C 0,1 (∂Ω). Hence, the function ∂ν u(x) = ∇u(x) · ν(x), x ∈ ∂Ω, belongs to W s−1−1/p,p (∂Ω) (exercise), with (11.91)
k∂ν ukW s−1−1/p,p (∂Ω) f k∇ukW s−1,p (Ω) .
On the other hand, since u ∈ W 1,p (Ω), by the trace theory in Sobolev spaces (see [Leo22d]), u∂Ω ∈ Lp (∂Ω), with (11.92)
kukLp (∂Ω) f kukW 1,p (Ω) .
Let w(x0 ) = u(x0 , f (x0 )), x0 ∈ RN −1 . Then ∂i w(x0 ) = ∂i u(x0 , f (x0 )) + ∂N u(x0 , f (x0 ))∂i f (x0 ). Since ∂i f is of class C 0,1 , by (11.90), ∂i w ∈ W s−1−1/p,p (RN −1 ) (exercise), with k∂i wkW s−1−1/p,p (RN −1 ) f k∂i ukW s−1−1/p,p (∂Ω) + k∂N ukW s−1−1/p,p (∂Ω) f k∂i ukW s−1,p (Ω) + k∂N ukW s−1,p (Ω) .
11.8. Trace Theory
473
Hence, also by (11.92), w ∈ W s−1/p,p (RN −1 ). It follows from Definition 11.78 that u∂Ω ∈ W s−1/p,p (∂Ω), with (11.93)
kukW s−1/p,p (∂Ω) f kukW s,p (Ω) .
Step 2: Assume that u ∈ W s,p (Ω). By Corollary 11.19 the function 0 0 N v : RN + → R, given by v(y) := u(y , yN + f (y )), y ∈ R+ , belongs to s,p N W (R+ ) and kvkW s,p (RN ) f kukW s,p (Ω) . Using a reflection argument + (see Theorem 11.55) and mollifying, we can construct a sequence of funcs,p (RN ). Define 1 N tions vn ∈ W s,p (RN + ) ∩ C (R+ ) such that vn → v in W + un (x) := vn (x0 , xN −f (x0 )), x ∈ Ω. By Theorem 11.18, un ∈ W s,p (Ω)∩C 1 (Ω) and un → u in W s,p (Ω). Then kun − uk kW s−1/p,p (∂Ω) f kun − uk kW s,p (Ω) , k∂ν un − ∂ν uk kW s−1−1/p,p (∂Ω) f kun − uk kW s,p (Ω) , which implies that { uk ∂Ω }n and { ∂ν uk ∂Ω }n are Cauchy sequences in the spaces W s−1/p,p (∂Ω) and W s−1−1/p,p (∂Ω), respectively. By the uniqueness of the trace operator (Theorem 9.14), it follows that Tr(u) ∈ W s−1/p,p (∂Ω) and Tr1 (∂ν u) ∈ W s−1−1/p,p (∂Ω). Moreover, by (11.91) and (11.93), the estimate (11.89) holds. Step 3: Conversely, given g ∈ W s−1/p,p (∂Ω) and h ∈ W s−1−1/p,p (∂Ω), set ˜ 0 ) := h(x0 , f (x0 )), x0 ∈ RN −1 . By Definition g˜(x0 ) := g(x0 , f (x0 )) and h(x s−1/p,p N −1 ˜ ∈ W s−1−1/p,p (RN −1 ). Since s − 1/p > 1, 11.78, g˜ ∈ W (R ) and h we have that ∂i g˜ ∈ W s−1−1/p,p (RN −1 ), while ∂i f ∈ C 0,1 . Hence, we can apply Theorem 6.23 to obtain that the function (11.94) ˜ 0) ∇x0 g˜(x0 ) · ∇x0 f (x0 ) h(x h1 (x0 ) := − , x0 ∈ RN −1 , 1 + k∇x0 f (x0 )k2N −1 (1 + k∇x0 f (x0 )k2N −1 )1/2 belongs to W s−1−1/p,p (RN −1 ). Now we use Corollary 11.64 to find a function w ∈ W s,p (RN ) such that Tr(w)(0, ·) = g˜, Tr1 (∂N w)(0, ·) = h1 in RN −1 , and kwkW s,p (RN ) k˜ g kW s−1/p,p (RN −1 ) + kh1 kW s−1−1/p,p (RN −1 ) f kgkW s−1/p,p (∂Ω) + khkW s−1−1/p,p (∂Ω) . Setting u(x) := w(x0 , xN − f (x0 )), x ∈ Ω, by Theorem 11.18 (see also the proof of Corollary 11.19), we have that u ∈ W s,p (RN ), with kukW s,p (RN ) f kwkW s,p (RN ) f kgkW s−1/p,p (∂Ω) + khkW s−1−1/p,p (∂Ω) . To conclude, observe that for x ∈ RN and i = 1, . . . , N − 1, ∂i u(x) = ∂i w(x0 , xN − f (x0 )) − ∂N w(x0 , xN − f (x0 ))∂i f (x0 ), ∂N u(x) = ∂N w(x0 , xN − f (x0 )),
474
11. Higher Order Fractional Sobolev Spaces
so, using the fact that ν(x0 , f (x0 )) =
(∇x0 f (x0 ),−1) (1+k∇x0 f (x0 )k2N −1 )1/2
and (11.94), we
have Tr1 (∂ν u)(x0 , f (x0 )) =
∇x0 Tr(w)(x0 ) · ∇x0 f (x0 ) − Tr1 (∂N w)(x0 )(1 + k∇x0 f (x0 )k2N −1 )1/2 2 1/2 0 0 (1 + k∇x f (x )kN −1 )
∇x0 g˜(x0 ) · ∇x0 f (x0 ) − h1 (x0 )(1 + k∇x0 f (x0 )k2N −1 )1/2 2 1/2 0 0 (1 + k∇x f (x )kN −1 ) ˜ 0 ) = h(x0 , f (x0 )). = h(x
=
Exercise 11.82. Let 1 < p < ∞, 1 < s < 2, with (s − 1)p < 1, and let Ω be as Theorem 11.81. Prove that k Tr(u)kW s−1/p,p (∂Ω) f kukW s,p (Ω) for every u ∈ W s,p (Ω). Conversely, given g ∈ W s−1/p,p (∂Ω), prove that there exists u ∈ W s,p (RN ) such that Tr(u) = g on ∂Ω and kukW s,p (RN ) f kgkW s−1/p,p (∂Ω) . Theorem 11.83 (Traces and W0s,p ). Let 1 < p < ∞, 1 < s < 2, with (s − 1)p > 1, and Ω := {(x0 , xN ) ∈ RN −1 × R : xN > f (x0 )}, where f : RN −1 → R is a Lipschitz continuous function of class C˙ 1,1 . Then u ∈ W s,p (Ω) belongs to W0s,p (Ω) if and only if Tr(u) = 0 and Tr1 (∂N u) = 0 on ∂Ω. Proof. If u ∈ Cc∞ (Ω), then Tr(u) = 0 and Tr1 (∂N u) = 0 on ∂Ω, and since W0s,p (Ω) is the closure of Cc∞ (Ω) with respect to the norm k · kW s,p (Ω) , it follows from the continuity of the trace operators that Tr(u) = 0 and Tr1 (∂N u) = 0 for all u ∈ W0s,p (Ω). Conversely, let u ∈ W s,p (Ω) be such that Tr(u) = 0 and Tr1 (∂N u) = 0. Then ∂i u ∈ W s−1,p (Ω) and since (s − 1)p > 1, by Theorem 9.29, there exists Tr(∂i u) on Ω. On the other hand, since Tr(u) = 0 and Tr1 (∂N u) = 0, it follows from items (i) and (iii) in Exercise 11.80 (which continues to hold for u ∈ W s,p (Ω) by a density argument) that Tr(∂i u) = 0 on ∂Ω. Consider 0 0 N the function v : RN + → R, given by v(y) := u(y , yN + f (y )), y ∈ R+ . By s,p N Corollary 11.19, v ∈ W (R+ ), with ∂i v(y) = ∂i u(y 0 , yN + f (y 0 )) + ∂N u(y 0 , yN + f (y 0 ))∂i f (y 0 ), ∂N u(y) = ∂N u(y 0 , yN + f (y 0 )) for i = 1, . . . , N − 1. Since Tr(u) = 0 and Tr(∂i u) = 0 on ∂Ω, we have that Tr(v) = 0 and Tr(∂i v) = 0 on {yN = 0}. Hence, by Theorem 11.65, v ∈ W0s,p (RN ). Hence, we can find a sequence of functions vn ∈ Cc∞ (RN +) 0 , x − f (x0 )), x ∈ Ω. such that vn → v in W s,p (RN ). Define u (x) := v (x n n N +
11.8. Trace Theory
475
Then vn ∈ Cc1 (Ω) and vn → u in W s,p (Ω). By mollifying each un and using a diagonal argument (exercise), we conclude that u ∈ W0s,p (Ω). Exercise 11.84. Let 1 < p < ∞, 1 < s < 2, with (s − 1)p < 1, and let Ω be as in Theorem 11.83. Prove that u ∈ W s,p (Ω) belongs to W0s,p (Ω) if and only if Tr(u) = 0 on ∂Ω. Definition 11.85. Given an integer 1 ≤ k < N , m ∈ N ∪ {∞}, and 0 < a ≤ 1, a nonempty set M ⊂ RN is called a kdimensional surface or manifold of class C˙ m,α if for every x0 ∈ M there exist an open set U containing x0 , an open set V ⊆ Rk , and a function ϕ : V → RN of class C˙ m,α such that (i) ϕ(V ) = M ∩ U ; (ii) ϕ is injective; (iii) ϕ : V → M ∩ U is biLipschitz continuous, that is, ϕ : V → M ∩ U and ϕ−1 : M ∩ U → V are Lipschitz continuous. The function ϕ is called a local chart or a local parametrization or a local system of coordinates of M . An atlas for S M is a family of local charts N ϕα : Vα → R , α ∈ Λ, with the property that α∈Λ ϕα (Vα ) = M . Exercise 11.86. Let Ω ⊂ RN be an open set with boundary of class C m,1 , m ∈ N. Prove that ∂Ω is an (N − 1)dimensional manifold of class C˙ m,1 . Hint: See the proof of Proposition 9.35 and Remark 11.75. Definition 11.87. Given 1 ≤ p < ∞, σ > 1, σ ∈ / N, and a bounded kdimensional surface M ⊂ RN of class C m,1 , where 1 ≤ k < N , m = bσc, let {ϕn }`n=1 be an atlas, with ϕn : Vn → RN , let Un ⊆ RN be open sets such that ϕn (Vn ) = M ∩ Un , and let {ψn }`n=1 be a smooth partition of unity subordinated to {Un }`n=1 , with supp ψn ⊂ Un . We say that a function v : M → R belongs to the fractional Sobolev space W σ,p (M ) if the function (ψn v) ◦ ϕn belongs to W σ,p (Vn ) for every n = 1, . . . , `, and we set kvkLp (M ) :=
` X
!1/p k(ψn v) ◦ ϕn kpLp (Vn )
,
n=1
vW σ,p (M ) :=
` X
!1/p (ψn v) ◦
ϕn pW σ,p (Vn )
,
n=1
kvkW σ,p (M ) := kvkLp (M ) + vW σ,p (M ) . Exercise 11.88. Prove that the space W σ,p (M ) in the previous definition does not depend on the particular atlas and partition of unity. Hint: Use Theorem 11.18.
476
11. Higher Order Fractional Sobolev Spaces
Next we consider domains with a bounded boundary of class C 1,1 . In what follows, given u ∈ W s,p (RN ), by its trace and the trace of its normal derivative on ∂Ω we mean the traces of the restriction of u to Ω. Theorem 11.89. Let Ω ⊂ RN be an open set with bounded boundary of class C 1,1 , N ≥ 2, 1 < p < ∞, and 1 < s < 2, with (s − 1)p > 1. Then for all u ∈ W s,p (Ω), k Tr(u)kW s−1/p,p (∂Ω) + k Tr1 (∂ν u)kW s−1−1/p,p (∂Ω) Ω kukW s,p (Ω) . Conversely, for every g ∈ W s−1/p,p (∂Ω) and h ∈ W s−1−1/p,p (∂Ω), there exists a function u ∈ W s,p (RN ) such that Tr(u) = g and Tr1 (∂ν u) = h on ∂Ω, and kukW s,p (RN ) Ω kgkW s−1/p,p (∂Ω) + khkW s−1−1/p,p (∂Ω) . Proof. Step 1: Let u ∈ W s,p (Ω). For every x ∈ ∂Ω there exist a rigid motion Tx : RN → RN , with Tx (x) = 0, a function fx : RN −1 → R of class C 1,1 , with fx (0) = 0, and rx > 0 such that Tx (∂Ω ∩ B(x, 2rx )) = {(y 0 , yN ) ∈ B(0, 2rx ) : yN = fx (y 0 )} =: Wx . We consider ϕx (y 0 ) = Tx−1 ((y 0 , fx (y 0 ))), y 0 ∈ RN −1 , and we define Vx := ϕ−1 x (B(x, 2rx )). Since ∂Ω is compact, there exist x1 , . . . , x` ∈ ∂Ω such that ∂Ω ⊆
` [
B(xn , rxn ).
n=1
Write fn := fxn , ϕn := ϕxn , rn := rxn , Tn := Txn , Vn := Vxn , and Wn := Wxn . Consider a smooth partition of unity {ψn }`n=1 subordinated to {B(xn , rn )}`n=1 , with supp ψn ⊂ B(xn , rn ). Since ∂i u ∈ W s−1,p (∂Ω) for every i = 1, . . . , N , by Theorem 9.39, k Tr(∂i u)kW s−1−1/p,p (∂Ω) Ω k∂i ukW s−1,p (Ω) Ω kukW s,p (Ω) . By Definition 9.36, k Tr(∂i u)kW s−1−1/p,p (∂Ω) =
` X
k(ψn Tr(∂i u)) ◦ ϕn kpW s−1−1/p,p (V
n)
1/p
.
n=1
Since ∂Ω is of class C 1,1 , the unit outward normal ν : ∂Ω → RN is Lipschitz continuous. In turn, (ψn νi ) ◦ ϕn : Vn → RN is Lipschitz continuous. Hence, by Theorem 6.23, k(ψn Tr(∂i u)νi ) ◦ ϕn kW s−1−1/p,p (Vn ) Ω k(ψn Tr(∂i u)) ◦ ϕn kW s−1−1/p,p (Vn ) Ω kukW s,p (Ω) . It follows that k(ψn Tr(∇u) · ν) ◦ ϕn kW s−1−1/p,p (Vn ) Ω kukW s,p (Ω) .
11.8. Trace Theory
477
Hence, k Tr1 (∂ν u)kW s−1−1/p,p (∂Ω) =
` X
k(ψn Tr(∇u) · ν) ◦ ϕn kpW s−1−1/p,p (V
1/p
n)
n=1
Ω kukW s,p (Ω) . Fix n ∈ {1, . . . , `}. By Exercise 11.13, ψn u ∈ W s,p (Ω), and so, by Theorem 11.18, (ψn u) ◦ Tn−1 ∈ W s,p (Tn (Ω)), with k(ψn u) ◦ Tn−1 kW s,p (Tn (Ω)) n kψn ukW s,p (Ω) . Define (11.95)
Ωn := {(y 0 , yN ) ∈ RN −1 × R : yN > fn (y 0 )}.
Since supp ψn ⊂ B(xn , rn ), we leave it as an exercise to check that if we extend (ψn u) ◦ Tn−1 to be zero in Ωn \ Wn , the extended function (ψn u) ◦ Tn−1 belongs to W s,p (Ωn ), with k(ψn u) ◦ Tn−1 kW s,p (Ωn ) n k(ψn u) ◦ Tn−1 kW s,p (Tn (Ω)) n kψn ukW s,p (Ω) . Hence, by Theorem 11.81, k(ψn u) ◦ Tn−1 kW s−1/p,p (∂Ωn ) n k(ψn u) ◦ Tn−1 kW s,p (Ωn ) n kψn ukW s,p (Ω) . Note that by Definition 11.78, k(ψn u) ◦ Tn−1 kW s−1/p,p (∂Ωn ) = k((ψn u) ◦ Tn−1 ) ◦ φn kW s−1/p,p (RN −1 ) , where φn (y 0 ) = (y 0 , fn (y 0 )). Since ϕn (y 0 ) = Tn−1 ((y 0 , fn (y 0 ))), it follows that Tn−1 ◦ φn = ϕn , and so, by Definition 11.87, kukpW s−1/p,p (∂Ω) =
` X
k(ψn u) ◦ ϕn kpW s−1/p,p (V
n)
n=1
Ω
` X
kψn ukpW s,p (Ω) Ω kukpW s,p (Ω) ,
n=1
where the last inequality follows from Exercise 11.13. Step 2: Let g ∈ W s−1/p,p (∂Ω) and h ∈ W s−1−1/p,p (∂Ω). Then !1/p ` X p kgkW s−1/p,p (∂Ω) = k(ψn g) ◦ ϕn kW s−1/p,p (V ) < ∞, n
n=1
khkW s−1−1/p,p (∂Ω) =
` X
!1/p k(ψn h) ◦ ϕn kpW s−1−1/p,p (V
n)
n=1
< ∞.
478
11. Higher Order Fractional Sobolev Spaces
Since supp ψn ⊂ B(xn , rn ) and Vn = ϕ−1 n (B(xn , 2rn )), where we recall that ϕn is biLipschitz continuous and of class C˙ 1,1 , if we extend (ψn g) ◦ ϕn and (ψn h) ◦ ϕn to be zero outside of Vn , we have that (exercise) k(ψn g) ◦ ϕn kW s−1/p,p (RN −1 ) n k(ψn g) ◦ ϕn kpW s−1/p,p (V ) , n
k(ψn h) ◦ ϕn kW s−1−1/p,p (RN −1 ) n k(ψn h) ◦
ϕn kpW s−1−1/p,p (V ) . n
Recalling that ϕn (y 0 ) = Tn−1 ((y 0 , fn (y 0 ))) and that ωn (y 0 ) := (y 0 , fn (y 0 )) is a chart for ∂Ωn (see (11.95) and Example 11.77), we have that (ψn g) ◦ Tn−1 ∈ W s−1/p,p (∂Ωn ) and (ψn h) ◦ Tn−1 ∈ W s−1−1/p,p (∂Ωn ), with k(ψn g) ◦ Tn−1 kW s−1/p,p (∂Ωn ) = k(ψn g) ◦ ϕn kW s−1/p,p (RN −1 ) n k(ψn g) ◦ ϕn kW s−1/p,p (Vn ) and, similarly, k(ψn h) ◦ Tn−1 kW s−1−1/p,p (∂Ωn ) n k(ψn h) ◦ ϕn kW s−1−1/p,p (Vn ) . Hence, by Theorem 11.81 there exists a function wn ∈ W s,p (RN ) such that Tr(wn ) = (ψn g) ◦ Tn−1 and Tr1 (∂ν wn ) = (ψn h) ◦ Tn−1 on ∂Ωn and kwn kW s,p (RN ) n k(ψn g) ◦ ϕn kW s−1/p,p (Vn ) + k(ψn h) ◦ ϕn kW s−1−1/p,p (Vn ) . Setting vn = wn ◦ Tn , by Theorem 11.18, we have that vn ∈ W s,p (RN ), with (11.96)
kvn kW s,p (RN ) kwn kW s,p (RN ) n k(ψn g) ◦ ϕn kLp (Vn ) + k(ψn )h ◦ ϕn kW s−1−1/p,p (Vn ) .
Moreover, Tr(vn ) = ψn g and Tr1 (∂ν vn ) = ψn h on Tn (∂Ωn ) = ∂Tn (Ωn ), where we used the fact that Tn is a rigid motion, so the outward normal to ∂Tn (Ωn ) is obtained by rotating the outward normal to ∂Ωn . Since supp ψn ⊂ B(xn , rn ), we can find a function φn ∈ Cc∞ (RN ) such that supp φn ⊂ B(xn , rn ) and φn ψn = ψn on ∂Ω ∩ B(xn , rn ). Define un := φn vn . By Theorem 11.11, kun kW s,p (RN ) n kvn kW s,p (RN ) .
(11.97)
Note that if x ∈ ∂Ω ∩ B(xn , rn ), (φn ψn )(x) = ψn (x), and so Tr(un )(x) = Tr(vn )(x) = (ψn g)(x) and Tr1 (∂ν un )(x) = Tr1 (∂ν vn )(x) = (ψn h)(x). Hence, P if we define u := `n=1 un , then by (11.96) and (11.97), kukW s,p (RN ) Ω
` X n=1
k(ψn g) ◦ ϕn kW s−1/p,p (Vn ) +
` X
k(ψn h) ◦ ϕn kW s−1−1/p,p (Vn )
n=1
Ω kgkW s−1/p,p (∂Ω) + khkW s−1−1/p,p (∂Ω) .
11.9. Notes
479
Since Tr and Tr1 are linear, Tr(u) =
` X n=1
Tr(un ) =
` X
ψn g = g
on ∂Ω,
n=1
and, similarly Tr1 (∂ν u) = h, where we use the fact that ∂Ω.
P`
n=1 ψn
= 1 in
Exercise 11.90. Let Ω ⊂ RN be an open set with bounded boundary of class C 1,1 , N ≥ 2, 1 < p < ∞, and 1 < s < 2, with (s − 1)p < 1. Prove that for all u ∈ W s,p (Ω), k Tr(u)kW s−1/p,p (∂Ω) Ω kukW s,p (Ω) . Conversely, prove that for every g ∈ W s−1/p,p (∂Ω), there exists a function u ∈ W s,p (RN ) such that Tr(u) = g on ∂Ω and kukW s,p (RN ) f kgkW s−1/p,p (∂Ω) . Next we show that if the domain Ω is sufficiently regular, we may characterize W0s,p (Ω) as the subspace of functions in W s,p (Ω) with trace zero. Theorem 11.91 (Traces and W01,p ). Let Ω ⊂ RN , N ≥ 2, be an open set whose boundary ∂Ω is bounded and of class C 1,1 , 1 < p < ∞, 1 < s < 2, with (s − 1)p > 1, and u ∈ W s,p (Ω). Then Tr(u) = 0 and Tr1 (∂ν u) = 0 on ∂Ω if and only if u ∈ W0s,p (Ω). Proof. This follows from Theorem 11.83 and using a partition of unity. We leave the details as an exercise. Exercise 11.92. Let Ω ⊂ RN be an open set with bounded boundary of class C 1,1 , N ≥ 2, 1 < p < ∞, and 1 < s < 2, with (s − 1)p < 1. Prove that u ∈ W s,p (Ω) belongs to W0s,p (Ω) if and only if Tr(u) = 0 on ∂Ω.
11.9. Notes Exercise 11.7 is adapted from the paper of del Teso, G´ omezCastro, and V´azquez [dTGCV20]. Step 1 of the proof of Theorem 11.23 and Exercise 11.80 are due to Ian Tice. Exercise 11.38 is adapted from the book of Taylor [Tay96]. The proof of Theorem 11.39 is adapted from the paper of A.D. Nguyen, J.I. D´ıaz, and Q.H. Nguyen [NDN20]. Theorem 11.41 draws upon the paper of Zolesio [Zol77]; see also the papers of Brezis and Mironescu [BM01b] and of Behzadan and Holst [BH21] for recent results. The representation in Lemma 11.47 is adapted from the book of Besov, Il’in, and Nikol’ski˘ı [BIN78, Section 7.2]. For more information on Gagliardo–Nirenberg interpolation inequalities we refer to the papers the Brezis and Mironescu [BM18], [BM19] and the references therein. Theorem 11.50 has been adapted from the review paper of Mironescu [Mir18]. The proofs of Theorem 11.51 and Lemma 11.52 are drawn from the paper of Brezis
480
11. Higher Order Fractional Sobolev Spaces
and Mironescu [BM01a]. We refer to the paper of De Vore and Sharpley [DS93, Section 6] for an extension theorem in unbounded domains. The proof of Theorem 11.66 is adapted from the paper of Solonnikov [Sol67], while that of Theorem 11.68 is from the paper of Il’in and Solonnikov [IS69]. Exercises 11.71 and 11.72 are drawn from Tartar’s book [Tar07].
Chapter 12
Some Equivalent Seminorms Some people have got a mental horizon of radius zero and call it their point of view. — David Hilbert
In this chapter we briefly describe two alternative approaches to the theory of fractional Sobolev spaces. The first approach is based on the abstract theory of interpolation. It relies on the fact that the spaces W s,p (Ω) can be regarded as “intermediate spaces” between W m,p (Ω) and W m+1,p (Ω) provided the open set Ω is sufficiently regular. Here m = bsc and W 0,p (Ω) := Lp (Ω). We refer to the books of Adams and Fournier [AF03], Bahouri, Chemin, and Danchin [BCD11], Bennett and Sharpley [BS88], Bergh and L¨ ofstr¨om [BL76], Leoni [Leo17], Lions and Magenes (mostly for the case p = 2) [LM72b], [LM72a], Lunardi [Lun18], Peetre [Pee76], Sawano [Saw18], Tartar [Tar07], and Triebel [Tri95] for more information on abstract interpolation and the study of Besov and fractional Sobolev spaces in this setting. The second approach relies on Fourier analysis, or, more specifically, on Littlewood–Paley theory. This approach allows us to give elegant proofs that work for all s ∈ R \ N. We refer to the books of Grafakos [Gra14], Runst and Sickel [RS96], and Triebel [Tri10], [Tri06], [Tri01].
12.1. The Interpolation Seminorm In this section, using the real method of interpolation, we will show that the fractional Sobolev space W s,p (RN ) can be regarded as the interpolation space between Lp (RN ) and W 1,p (RN ). 481
482
12. Some Equivalent Seminorms
Definition 12.1. Two normed spaces (X0 , k·kX0 ), (X1 , k·kX1 ) are called an admissible pair if they are embedded into a common Hausdorff topological vector space X. Given an admissible pair (X0 , k · kX0 ), (X1 , k · kX1 ), for t > 0 we define the norm (12.1)
x ∈ X0 + X1 7→ K(x, t) := inf{kx0 kX0 + tkx1 kX1 },
where the infimum is taken over all possible decompositions x = x0 + x1 , x0 ∈ X0 , x1 ∈ X1 . For 1 ≤ p ≤ ∞ and 0 < s < 1, we define the real interpolation space (X0 , X1 )s,p := {x ∈ X0 + X1 : kxks,p < ∞}, where if 1 ≤ p < ∞, (12.2)
kxks,p :=
Z
∞ p
(K(x, t))
0
dt 1/p t1+sp
,
while if p = ∞, (12.3)
kxks,∞ := sup t−s K(x, t). t>0
Note that these definitions make sense because the function K(x, ·) is increasing. We refer to [Leo17, Chapter 16] for the proofs of the following theorems and for more information about interpolation spaces. Theorem 12.2. Let (X0 , k · kX0 ), (X1 , k · kX1 ) be an admissible pair, and let 1 ≤ p ≤ ∞ and 0 < s < 1. Then (X0 , X1 )s,p is a normed space and the following embeddings hold X0 ∩ X1 ,→ (X0 , X1 )s,p ,→ X0 + X1 . Moreover, if X0 and X1 are Banach spaces, then so is (X0 , X1 )s,p . Proposition 12.3. Let (X0 , k·kX0 ), (X1 , k·kX1 ) be an admissible pair, with X1 ,→ X0 , 1 ≤ p ≤ ∞, and 0 < s < 1. Then k · kX0 is an equivalent norm in X0 + X1 and for every T > 0, Z T dt 1/p x 7→ kxkX0 + (K(x, t))p 1+sp t 0 is an equivalent norm in (X0 , X1 )s,p for 1 ≤ p < ∞, while x 7→ kxkX0 + sup t−s K(x, t) 0 1, then 1,p s,p s,p N N N s,p (Lp (RN (RN + ), W0 (R+ ))s,p = W00 (R+ ) = W0 (R+ ) ⊂ W + ),
where the inclusion is strict.
488
12. Some Equivalent Seminorms
1,p N Proof. Let X := (Lp (RN + ), W0 (R+ ))s,p . In view of Corollary 6.101, it s,p suffices to prove that X = W00 (RN + ). s,p Step 1: We prove that X ⊆ W00 (RN + ). Consider the linear map T : p N p N L (R+ ) → L (R ) given by
T (u)(x) :=
u(x) if x ∈ RN +, 0 otherwise.
∞ N Note that if u ∈ Cc∞ (RN + ), then T (u) ∈ Cc (R ), and so by the definition of 1,p N 1,p (RN ) is linear and continuous. It follows W01,p (RN + ), T : W0 (R+ ) → W from by Theorem 12.4, that
T : X → (Lp (RN ), W 1,p (RN ))s,p = W s,p (RN ), where in the last equality we use Theorem 12.5. This shows that if u ∈ X, s,p (RN then T (u) ∈ W s,p (RN ), which implies that u ∈ W00 + ), by the definition s,p N of W00 (R+ ). s,p (RN ¯ be a representative of u and define Step 2: Given u ∈ W00 + ), let u
u ˜(x) :=
¯(x) if x ∈ RN u +,
if x ∈ RN \ RN +.
0
s,p ˜ belongs to u ∈ W s,p (RN ). For By definition of W00 (RN + ), we have that u N t > 0 and x ∈ R write Z Z 1 1 (12.8) ∆y u ˜(x) dy + N u ˜(x + y) dy u ˜(x) = − N t t (−t,0)N (−t,0)N
=: vt (x) + wt (x). By H¨older’s inequality we have that vt (x)p ≤
tN (p−1) tN p
Z
∆y u ˜(x)p dy.
(−t,0)N
Integrating in x over RN + gives Z RN +
vt (x)p dx ≤ ≤
1 tN
Z
1 tN
Z
RN +
RN
Z
∆y u ˜(x)p dydx
(−t,0)N
Z √
B(0, N t)
∆y u ˜(x)p dydx.
12.1. The Interpolation Seminorm
489
In turn, by Tonelli’s theorem, Z ∞ Z ∞ Z Z dt 1 kvt kpLp (RN ) 1+sp ≤ ∆y u ˜(x)p dydxdt √ N +1+sp + t t N 0 0 R B(0, N t) Z ∞ Z Z 1 ≤ ∆y u ˜(x)p dtdxdy √ N +1+sp kyk/ N t RN RN Z Z ∆y u ˜(x)p dxdy. N +sp RN RN kyk To estimate wt , given i = 1, . . . , N , we use Fubini’s theorem to find a set Ei ⊆ RN −1 such that LN −1 (RN −1 \ Ei ) = 0 and for every x0i ∈ Ei , the function u ˜(x0i , ·) ∈ Lp (R). Here, we are R x using0 notation (6.1). It follows that ˜(xi , t) dt is absolutely continuous, for every x0i ∈ Ei , the function xi 7→ 0 i u with Z xi ∂ u ˜(x0i , t) dt = u ˜(x0i , xi ) for L1 a.e. xi ∈ R ∂xi 0 (see [Leo22d]). In turn, if x0i + yi0 ∈ Ei , then by the change of variables xi + yi = τi , we have Z 0 Z xi 0 0 u ˜(x0i + yi0 , τi ) dτi . u ˜(xi + yi , xi + yi ) dyi = −t
xi −t
Therefore Z 0 ∂ u ˜(x0i + yi0 , xi + yi ) dyi = ∆−t,i u ˜(x0i + yi0 , xi ) ∂xi −t
for L1 a.e. xi ∈ R.
1,p We leave it as an exercise to check that this implies that wt ∈ Wloc (RN ), with Z 1 ∂i wt (x) = N ∆−t,i u ˜(x0i + yi0 , xi ) dyi0 for LN a.e. x ∈ RN . t N −1 (−t,0)
By Minkowski’s inequality for integrals (see [Leo22c]), Z 1 1 k∂i wt kLp (RN ) ≤ N k∆−t,i u ˜kLp (RN ) dyi0 ≤ sup k∆h u ˜kLp (RN ) . t t khk≤√N t (−t,0)N −1 older’s inequality and On the other hand, kwt kLp (RN ) ≤ kukLp (RN ) by H¨ +
Tonelli’s theorem. Hence, wt ∈ W 1,p (RN ). Moreover, if xN < 0, then by (12.8), we have that wt = 0. In particular, wt has trace zero on xN = 0 and so, wt ∈ W01,p (RN + ). In turn, by Theorem 6.52 and the previous inequality, Z ∞ ∞ dt dt p p t k∂i wt kLp (RN ) 1+sp ≤ sup k∆h u ˜kpLp (RN ) 1+sp ˜ upW s,p (RN ) . √ + t t 0 0 khk≤ N t
Z
490
12. Some Equivalent Seminorms
On the other hand, since kwt kLp (RN ) ≤ kukLp (RN ) , + + Z 1 Z 1 dt t(1−s)p−1 dt kukpLp (RN ) . tp kwt kpLp (RN ) 1+sp ≤ kukpLp (RN ) + + + t 0 0 1,p (RN ), and In conclusion, we have shown that vt ∈ Lp (RN + ), wt ∈ W + Z 1 Z 1 dt dt (kvt kpLp (RN ) + tp kwt kpW 1,p (RN ) ) 1+sp (K(u, t))p 1+sp ≤ + + t t 0 0 p p uW s,p (RN ) . kukLp (RN ) + ˜ +
Since
W01,p (RN +)
,→
Lp (RN + ),
it follows from Proposition 12.3 that
kukX kukLp (RN ) + ˜ uW s,p (RN ) , +
where X :=
1,p N (Lp (RN + ), W0 (R+ ))s,p .
Exercise 12.11. For 1 < p < ∞ and α > 0, let ( ) Z dx dx p Lp R N u(x)p α < ∞ , := u ∈ Lloc (RN +; α +) : xN xN RN + be endowed with the norm Z u 7→ RN +
dx u(x)p α xN
!1/p .
p p N For 0 < s < 1, characterize the space (Lp (RN + ), L (R+ ; dx/xN ))s,p .
Exercise 12.12. Let u ∈ W01,p (RN + ). Prove that Z Z p dx u(x) p ∂N u(x)p dx. N x R RN N + + Hint: Use Corollary 1.5. Exercise 12.13. Use Exercises 12.11 and 12.12 to give an alternative proof of Step 1 of Theorem 12.10.
12.2. The Littlewood–Paley Seminorm In this section we give a characterization of fractional Sobolev spaces based on Fourier analysis and Littlewood–Paley theory. We refer to [Leo22c] for the definition and basic properties of the Fourier transforms, the space of realvalued rapidly decreasing functions S(RN ), and its dual space S 0 (RN ) (the space of tempered distributions). We remark that a function u in Lp (RN ) can be identified with the tempered distribution Z Tu (φ) := u(x)φ(x) dx, φ ∈ S(RN ). RN
12.2. The Littlewood–Paley Seminorm
491
In particular, this implies that W s,p (RN ) ⊆ S 0 (RN ). Next we show that ˙ s,p (RN ) ⊆ S 0 (RN ). Given n, m ∈ N0 and φ ∈ S(RN ), we define W (12.9)
X
kφkm,n :=
X
sup xα ∂ β φ(x).
0≤α≤n 0≤β≤n x∈R
N
˙ s,p (RN ). Then the Theorem 12.14. Let 1 ≤ p < ∞, 0 < s < 1, and u ∈ W N N linear function Tu : S(R ) → R , defined by Z Tu (φ) :=
u(x)φ(x) dx,
φ ∈ S(RN ),
RN
is well defined and continuous, with Tu (φ) u kφkm,0 for all φ ∈ S(RN ) and for some m ∈ N. Proof. Let u ¯ be a representative of u. Since Z
Z
RN
RN
¯ u(x) − u ¯(y)p dxdy < ∞, kx − ykN +sp
by Fubini’s theorem, for LN a.e. y ∈ RN , Z (12.10) RN
¯ u(x) − u ¯(y)p dx < ∞. kx − ykN +sp
Fix y ∈ RN for which (12.10) holds. Then, Z
Z u(x)φ(x) dx =
¯ u(x)φ(x) dx Z Z ≤ ¯ u(y) φ(x) dx +
RN
RN
RN
¯ u(x) − u ¯(y)φ(x) dx
RN
=: A + B. Using (12.9), we have (1 + kxk)N +1 φ(x) dx N +1 RN (1 + kxk) Z 1 ¯ u(y)kφkN +1,0 dx. N +1 RN (1 + kxk) Z
A ≤ ¯ u(y)
492
12. Some Equivalent Seminorms
On the other hand, by H¨older’s inequality Z ¯ u(x) − u ¯(y) kx − ykN/p+s φ(x) B≤ dx N/p+s kx − yk RN Z 1/p Z 1/p0 ¯ u(x) − u ¯(y)p N p0 /p+sp0 p0 ≤ dx kx − yk φ(x) dx N +sp RN kx − yk RN 1/p Z ¯ u(x) − u ¯(y)p dx ≤ N +sp RN kx − yk !1/p0 Z 0 0 kx − ykN p /p+sp 0 0 × (1 + kxk)np φ(x)p dx (1 + kxk)np0 RN !1/p0 1/p Z Z 0 0 kx − ykN p /p+sp ¯ u(x) − u ¯(y)p dx dx , kφkn,0 N +sp (1 + kxk)np0 RN RN kx − yk where n ∈ N is taken so large that the last integral on the righthand side converges. Combining the estimates for A and B completes the proof. Next we introduce the LittlewoodPaley decomposition. Proposition 12.15. There is a nonnegative radial function ϕ0 ∈ Cc∞ (RN ) such that (12.11) (12.12)
supp ϕ0 = B(0, 2) \ B(0, 1/2), ϕ0 > 0 in B(0, 2) \ B(0, 1/2), X ϕ0 (2−k x) = 1 for every x ∈ RN \ {0}. k∈Z
Moreover, (12.13)
supp ϕ0 (2−k ·) = B(0, 2k+1 ) \ B(0, 2k−1 )
for every k ∈ Z, and for every x ∈ RN \ {0} there is a unique kx ∈ Z such that X (12.14) 1= ϕ0 (2−k x) = ϕ0 (2−kx −1 x) + ϕ0 (2−kx x) + ϕ0 (2−kx +1 x). k∈Z
Proof. Construct a C ∞ function φ : [0, ∞) → [0, 1] such that φ(r) = 1 for r ∈ [0, 1], φ is strictly decreasing in [1, 2], and φ(r) = 0 for r > 2. Define ϕ(x) := φ(kxk) − φ(2kxk), x ∈ RN . Then ϕ(x) = 0 for kxk ≤ 1/2 and kxk ≥ 2 and 0 < ϕ(x) < 1 for 1/2 < kxk < 2. Since ϕ = 0 in B(0, 1/2) and outside B(0, 2), the function ϕ belongs to Cc∞ (RN ). Moreover, for every k ∈ Z, (12.15)
supp ϕ(2−k ·) = B(0, 2k+1 ) \ B(0, 2k−1 ).
12.2. The Littlewood–Paley Seminorm
493
For every x ∈ RN \ {0}, we can find a unique kx ∈ Z such that 2kx −1 < kxk ≤ 2kx . In view of (12.15), it follows that ϕ(2−kx x) > 0 and ϕ(2−k x) = 0 for all k 6= kx − 1, kx , kx + 1. Hence, X ϕ(2−k x) = ϕ(2−kx −1 x) + ϕ(2−kx x) + ϕ(2−kx +1 x) > 0, k∈Z
and so, we can define ϕ(x) , −k k∈Z ϕ(2 x)
x ∈ RN \ {0},
ϕ0 (x) := P
ϕ0 (0) = 0.
In view of (12.15), that in a small neighborhood of each x ∈ RN \ P we have −k {0} the series k∈Z ϕ(2 x) is given by a finite sum, so ϕ0 is C ∞ in that neighborhood. Using the fact that ϕ = 0 in B(0, 1/2), we have that ϕ0 is C ∞ also near zero. Hence, ϕ0 ∈ Cc∞ (RN ), and (12.11) and (12.12) hold. Property (12.14) now follows from (12.12) and (12.13). Corollary 12.16. Let ϕ0 be the function constructed in Proposition 12.15 and define ω0 (x) := 1 −
(12.16)
∞ X
ϕ0 (2−k x),
x ∈ RN .
k=0
Then ω0 is radial, supp ω0 = B(0, 1), ω0 ∈ Cc∞ (RN ), and for all x ∈ RN and j ∈ Z, ∞ X ω0 (2−j x) + ϕ0 (2−k x) = 1. k=j
Proof. By (12.13), ω0 = 1 in B(0, 1/2), while by (12.11), (12.13), and (12.14), supp ω0 = B(0, 1). Moreover, by the definition of ω0 , ω0 (x/2) − ω0 (x) =
∞ X
−k
ϕ0 (2
x) −
k=0
∞ X
ϕ0 (2−k−1 x) = ϕ0 (x).
k=0
It follows that for every j, n ∈ Z, with j < n, we have ω0 (2−j x) +
n X
ϕ0 (2−k x) = ω0 (2−j x) +
k=j
n X
ω0 (2−k−1 x) −
k=j −n−1
= ω0 (2
n X
ω0 (2−k x)
k=j
x).
In particular, for all x ∈ RN and j ∈ Z, by the definition of ω0 and the previous identity, 1 = ω0 (x) +
∞ X k=0
ϕ0 (2−k x) = ω0 (2−j x) +
∞ X k=j
ϕ0 (2−k x).
494
12. Some Equivalent Seminorms
Let ϕ0 be as in Proposition 12.15. Then the inverse Fourier transform of the function ϕ0 (2−k ·) belongs to the space of rapidly decreasing functions S(RN ; C) (see [Leo22c]) and is given by (12.17) −k
(ϕ0 (2
∨
Z
e2πix·y ϕ0 (2−k y) dy Z k k e2πix2 ·z ϕ0 (z) dz = 2kN ϕ∨ = 2kN 0 (2 x) =: ψk (x),
·)) (x) =
RN
RN
where we use the change of variables z = 2−k y. Since ϕ0 is radial, it follows that ψk is realvalued and radial (see [Leo22c]). Note that by the change of variables y = 2k x, Z Z k ϕ∨ (2 x) dx = ϕ∨ (12.18) kψk kL1 (RN ) = 2kN 0 0 (y) dy. RN
RN
Hence, by Young’s inequality (see [Leo22c]) for every u ∈ Lp (RN ), 1 ≤ p ≤ ∞, we have that u ∗ ψk ∈ Lp (RN ), and so, if 1 ≤ p < ∞ and s ∈ R \ N, we may define 1/p X p ksp 2 ku ∗ ψ k , := (12.19) u∨ k Lp (RN ) W s,p (RN ) k∈Z
while if p = ∞, (12.20)
ks u∨ W s,∞ (RN ) := sup 2 ku ∗ ψk kL∞ (RN ) . k∈Z
Observe that the seminorms (12.19) and (12.20) have been defined also for s ≤ 0. More generally, given a realvalued tempered distribution T ∈ S 0 (RN ) (see [Leo22c]), we have that Z (12.21) (T ∗ ψk )(φ) = T (ψk (x − ·))φ(x) dx, φ ∈ S(RN ). RN
Hence, the tempered distribution T ∗ ψk can be identified with the C ∞ function x 7→ T (ψk (x − ·)). Thus, with a slight abuse of notation, given 1 ≤ p ≤ ∞ and s ∈ R \ N, for T ∈ S 0 (RN ) and 1 ≤ p < ∞, we can define X 1/p p ∨ ksp (12.22) T W s,p (RN ) := 2 kT ∗ ψk kLp (RN ) , k∈Z
while for p = ∞, (12.23)
ks T ∨ W s,∞ (RN ) := sup 2 kT ∗ ψk kL∞ (RN ) . k∈Z
Exercise 12.17. Let 1 ≤ p ≤ ∞, s ∈ R \ N, and ϕ0 ∈ Cc∞ (RN ) satisfy (12.11)–(12.12). Given T ∈ S 0 (RN ), prove that T ∨ = 0 if and only W s,p (RN ) if T can be identified with a polynomial.
12.2. The Littlewood–Paley Seminorm
495
Next we prove that if T ∈ S 0 (RN ) is such that T ∨ < ∞ for W s,p (RN ) s > 0, then T can be identified with a function in L1loc (RN ). Theorem 12.18. Let 1 ≤ p < ∞ and s > 0, s ∈ / N. If T ∈ S 0 (RN ) is such ∨ that T W s,p (RN ) < ∞, then T = T ∗ ω0∨ +
(12.24)
∞ X
T ∗ ψk
in S 0 (RN ),
k=0
and there exists a function u ∈ L1loc (RN ) such that T (φ) = every φ ∈ Cc∞ (RN ).
R
RN
φu dx for
Lemma 12.19. Let ω0 be the function in Corollary 12.16 and T ∈ S 0 (RN ). Then for every j ∈ Z, ∞ X −j ∨ T ∗ ψk in S 0 (RN ). (12.25) T = T ∗ (ω0 (2 ·)) + k=j
Proof. By Corollary 12.16, for every j, n ∈ Z, with j < n, and every x ∈ RN , n X −j ϕ0 (2−k x) = ω0 (2−n−1 x). ω0 (2 x) + k=j
Since T and the convolution are linear, it follows by the fact that ω0 and ϕ0 are radial (and so are their inverse Fourier transforms, which is left as an exercise) that for every φ ∈ S(RN ; C), n X T ∗ (ω0 (2−j ·))∨ + T ∗ ψk (φ) k=j
= T φ ∗ (ω0 (2−j ·))∨ +
n X
(ϕ0 (2−k ·))∨
k=j −n−1
= T (φ ∗ (ω0 (2
∨
·)) ) = (T ∗ (ω0 (2−n−1 ·))∨ )(φ).
This shows that (12.26)
−j
T ∗ (ω0 (2
∨
·)) +
n X
T ∗ ψk = T ∗ (ω0 (2−n−1 ·))∨ .
k=j
We leave it as an exercise to check that since ω0 (2−n−1 x) = 1 for x ∈ b 0 (2−n−1 ·) → φb in S(RN ; C) as n → ∞. B(0, 2n−1 ) by (12.11), we have that φω In turn, since F is an isomorphism from S(RN ; C) to S(RN ; C) ([Leo22c]), it follows that φ ∗ (ω0 (2−n−1 ·))∨ → φ in S(RN ; C), and so, (T ∗ (ω0 (2−n−1 ·))∨ )(φ) = T (φ ∗ (ω0 (2−n−1 ·))∨ ) → T (φ) as n → ∞, which, together with (12.26), proves (12.25).
496
12. Some Equivalent Seminorms
We turn to the proof of Theorem 12.18. Proof of Theorem 12.18. By Lemma 12.19 we can write ∞ X T = T ∗ ω0∨ + T ∗ ψk . k=0
If p > 1, by H¨older’s and Minkowski’s inequalities (12.27) X 1/p0 X 1/p ∞ ∞ ∞ X p −ksp0 ksp kT ∗ ψk kLp (RN ) ≤ 2 2 kT ∗ ψk kLp (RN ) k=0
k=0 T ∨ W s,p (RN ) .
k=0
The same inequality holds if p = 1. On the other hand, since ω0 ∈ S(RN ), T ∗ ω0∨ can be identified with a C ∞ function with polynomial growth (see [Leo22c]). 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. In the following theorem, we will prove that  · ∨ is an equivalent W s,p (RN ) seminorm in W s,p (RN ). Theorem 12.20. Let 1 ≤ p < ∞, 0 < s < 1, and ϕ0 ∈ Cc∞ (RN ) be as in Proposition 12.15. Then for every u ∈ W s,p (RN ), uW s,p (RN ) u∨ W s,p (RN ) uW s,p (RN ) . We begin with a preliminary result. Lemma 12.21. Let u ∈ Lp (RN ), 1 ≤ p < ∞. Then for every x ∈ RN \ {0} and every k ∈ Z, k+1 X
(u ∗ ψk )(x) =
(12.28)
(u ∗ (ψk ∗ ψj ))(x).
j=k−1
In addition, kukLp (RN ) ≤
X
ku ∗ ψk kLp (RN ) .
k∈Z
Proof. Step 1: Assume first that u ∈ S(RN ). By standard properties of the Fourier transform ([Leo22c]), and by (12.12), (12.17), for x ∈ RN \ {0}, X ck (x) = u u\ ∗ ψk (x) = u b(x)ψ b(x)ϕ0 (2−k x) = u b(x) ϕ0 (2−j x)ϕ0 (2−k x) j∈Z
=u b(x)
k+1 X j=k−1
ϕ0 (2−j x)ϕ0 (2−k x) = u b(x)
k+1 X j=k−1
ϕ0 (2−k x)ϕ0 (2−j x),
12.2. The Littlewood–Paley Seminorm
497
where we use (12.13). Taking the inverse Fourier transform on both sides, we obtain (see [Leo22c]) (u ∗ ψk )(x) =
k+1 X
(b uϕ0 (2−k ·)ϕ0 (2−j ·))∨
j=k−1
=
k+1 X
(u ∗ ((ϕ0 (2−k ·))∨ ∗ (ϕ0 (2−j ·))∨ ))(x)
j=k−1
=
k+1 X
(u ∗ (ψk ∗ ψj ))(x),
j=k−1
which proves (12.28) under the additional hypothesis that u ∈ S(RN ). Next, if u ∈ Lp (RN ), consider a sequence of functions un ∈ Cc∞ (RN ) such that un → u in Lp (RN ), un → u pointwise LN a.e. in RN and un (x) ≤ g(x) for LN a.e. x ∈ RN , for all n ∈ N, and for some function g ∈ Lp (RN ). Since Z (un ∗ ψk )(x) = un (y)ψk (x − y) dy RN
and the function gψk (x − ·) belongs to Lp (RN ) by H¨ older’s inequality, we can use the Lebesgue dominated convergence theorem to show that the sequence {un ∗ ψk }n converges pointwise to u ∗ ψk in RN \ {0} as n → ∞. Similarly, {un ∗ (ψk ∗ ψj )}n converges pointwise to u ∗ (ψk ∗ ψj ) in RN \ {0} as n → ∞. Since (12.28) holds for each un , letting n → ∞, it follows that (12.28) holds also for u. Step 2: Assume that X
ku ∗ ψk kLp (RN ) < ∞
k∈Z
and define Sn (x) :=
n X
(u ∗ ψk )(x),
x ∈ RN .
k=−n
Then for n > l ≥ 1, kSn − Sl kLp (RN ) ≤
X
ku ∗ ψk kLp (RN ) → 0
l≤k≤n
as l → ∞, which implies that {Sn }n is a Cauchy sequence in Lp (RN ), so Sn → v in Lp (RN ) as n → ∞ for some function v ∈ Lp (RN ). This shows that X (12.29) v= (u ∗ ψk ) in Lp (RN ), k∈Z
498
12. Some Equivalent Seminorms
with kvkLp (RN )
X
= (u ∗ ψk )
k∈Z
≤
X
ku ∗ ψk kLp (RN ) .
k∈Z
Lp (RN )
It remains to show that v = u. We begin by proving that ku ∗ (ω0 (2−l ·))∨ kL∞ (RN ) → 0
(12.30)
as l → −∞.
Using (12.17), with ω0 in place of ϕ0 , we have (ω0 (2−l ·))∨ (x) = 2lN ω0∨ (2l x), and thus, Z
0
(ω0 (2−l ·))∨ (x)p dx = 2lN p
0
Z
0
ω0∨ (2l x)p dx N R Z 0 lN (p0 −1) =2 ω0∨ (y)p dy,
RN
RN
where we make the change of variables y = 2l x. By Young’s inequality (see [Leo22c]), ku ∗ (ω0 (2−l ·))∨ kL∞ (RN ) ≤ kukLp (RN ) k(ω0 (2−l ·))∨ kLp0 (RN ) 0
= kukLp (RN ) kω0∨ kLp0 (RN ) 2lN (1−1/p ) → 0 as l → −∞ since p0 > 1. Thus, (12.30) holds. Define Z (12.31)
Tu (φ) :=
φ ∈ S(RN ).
uφ dx, RN
Since Tu is a tempered distribution (Why?), we can apply (12.25) to obtain (12.32)
−l
∨
Tu = Tu ∗ (ω0 (2 ·)) +
∞ X
Tu ∗ ψk
in S 0 (RN ).
k=l
By the definition of convolution of a tempered distribution (see [Leo22c]), Fubini’s theorem, and the fact that ω0 is radial, it follows that for every φ ∈ S(RN ), (12.33)
(Tu ∗ (ω0 (2−l ·))∨ )(φ) = Tu (φ ∗ (ω0 (2−l ·))∨ ) Z = u(x)(φ ∗ (ω0 (2−l ·))∨ )(x) dx N ZR = φ(x)(u ∗ (ω0 (2−l ·))∨ )(x) dx. RN
Hence, −l
∨
−l
∨
(Tu ∗ (ω0 (2 ·)) )(φ) ≤ ku ∗ (ω0 (2 ·)) kL∞ (RN )
Z φ dx → 0 RN
12.2. The Littlewood–Paley Seminorm
499
as l → −∞ by (12.30). It follows from (12.32) that X Tu = Tu ∗ ψk in S 0 (RN ). k∈Z
On the other hand, (12.29) implies that Tv = that is, u = v.
P
k∈Z Tu
∗ ψk , so Tu = Tv ,
Remark 12.22. Note that the inequality fails if u ∈ / Lp (RN ). Indeed, if u is a constant different from zero, then kukLp (RN ) = ∞, while u ∗ ψk = 0 (see (12.34)) for every k. We turn to the proof of Theorem 12.20. Proof. Step 1: We prove that u∨ uW s,p (RN ) for every u ∈ W s,p (RN ) W s,p (RN ). By the definition of a Fourier transform and the theorem on the inverse Fourier transform (see [Leo22c]), it follows from (12.11) and (12.17) that Z (12.34) 0 = ϕ0 (2−k 0) = ψbk (0) = ψk (y) dy. RN
Hence, for every k ∈ Z and x ∈ RN , Z Z (u ∗ ψk )(x) = ψk (y)(u(x − y) − u(x)) dy = RN
ψk (y)∆−y u(x) dy.
RN
By (12.19) and Minkowski’s inequality for integrals (see [Leo22c]), we can write Z p p X ksp ∨ uW s,p (RN ) ≤ 2 ψk (y)k∆−y ukLp (RN ) dy (12.35) RN
k∈Z
A + B, where A :=
X
2
ksp
B(0,2−k )
k∈Z
B :=
X
Z
2ksp
k∈Z
ψk (y)k∆−y ukLp (RN ) dy
Z RN \B(0,2−k )
p
,
ψk (y)k∆−y ukLp (RN ) dy
p
.
By H¨older’s inequality, (12.17), and the change of variables x = 2k y, Z p ψk (y)k∆−y ukLp (RN ) dy B(0,2−k ) Z p−1 Z kN ∨ k p0 (2 ϕ0 (2 y)) dy k∆−y ukpLp (RN ) dy ≤ B(0,2−k ) B(0,2−k ) Z Z p−1 p0 ≤ 2kN (ϕ∨ k∆−y ukpLp (RN ) dy. 0 (x)) dx B(0,1)
B(0,2−k )
500
12. Some Equivalent Seminorms
It follows from the Lebesgue monotone convergence theorem and the fact that k0 X 2k0 (N +sp) 2k(N +sp) = , −(N +sp) 1 − 2 k=−∞ where k0 = −blog kyk/ log 2c, that Z X k(N +sp) (12.36) A 2 B(0,2−k )
k∈Z
Z
2k(N +sp) k∆−y ukpLp (RN ) dy
X
RN
Z RN
k∆−y ukpLp (RN ) dy
k: kyk N + sp, Z p kykr ψk (y) k∆ uk p (RN ) dy −y L kykr RN \B(0,2−k ) Z p−1Z dy k p0 dy ≤ 2kN p (kykr ϕ∨ (2 y)) k∆−y ukpLp (RN ) 0 r r kyk kyk N −k N −k R \B(0,2 ) R \B(0,2 ) Z Z p−1 dy p0 dx ≤ 2kN −kr (kxkr ϕ∨ k∆−y ukpLp (RN ) . 0 (x)) r kxk kykr RN \B(0,1) RN \B(0,2−k ) Again by the Lebesgue monotone convergence theorem and the fact that for r > N + sp, ∞ X 2k0 (N +sp−r) , 2k(N +sp−r) = 1 − 2N +sp−r k=k0
where k0 = −blog kyk/ log 2c, we obtain Z X k(N +sp−r) B 2 RN \B(0,2−k )
k∈Z
Z RN
Z RN
X
k∆−y ukpLp (RN )
2k(N +sp−r) k∆−y ukpLp (RN )
k: kyk≥2−k
k∆−y ukpLp (RN )
dy kykr
dy kykr
dy . kykN +sp
By combining this estimate with (12.35) and (12.36), we obtain u∨ W s,p (RN ) uW s,p (RN ) . Step 2: We will show that uW s,p (RN ) ≤ Cu∨ . By (12.17), (12.28), W s,p (RN ) the fundamental theorem of calculus, and the fact that u ∗ (ψk ∗ ψj ) =
12.2. The Littlewood–Paley Seminorm
501
ψk ∗ (u ∗ ψj ), (u ∗ ψk )(x + h) − (u ∗ ψk )(x) =
k+1 X
[(u ∗ (ψk ∗ ψj ))(x + h) − (u ∗ (ψk ∗ ψj ))(x)]
j=k−1
=
k+1 X
2
kN
RN
j=k−1
=
k+1 X
2
kN
k+1 X
2kN
j=k−1
Z RN
j=k−1
=
Z
N X
k ∨ k [ϕ∨ 0 (2 (x + h − y)) − ϕ0 (2 (x − y))](u ∗ ψj )(y) dy
Z
1
k k [∇ϕ∨ 0 (2 (x + th − y)) · (2 h)] dt(u ∗ ψj )(y) dy
0
2k hn [gn (2k ·) ∗ (u ∗ ψj )](x),
n=1
where 1
Z gn (z) :=
k ∂n ϕ∨ 0 (z + 2 th) dt.
0
By Minkowski’s inequality and Young’s inequality (see [Leo22c]),
k∆h (u ∗ ψk )kLp (RN ) 2kN
k+1 X N X
2k khkkgn (2k ·)kL1 (RN ) ku ∗ ψj kLp (RN )
j=k−1 n=1 k+1 X
2k khk
(12.37)
ku ∗ ψj kLp (RN ) ,
j=k−1
where we use Fubini’s theorem and the change of variables 2k z + 2k th = y, so that 2kN dz = dy, to estimate Z
1Z
Z
k
gn (2 z) dz ≤ RN
=
0
RN
1
Z
2kN
k k ∂n ϕ∨ 0 (2 z + 2 th) dzdt
RN
∂n ϕ∨ 0 (y) dy
1 2kN
.
On the other hand, by Minkowski’s inequality and the change of variables x + h = y, (12.38)
k∆h (u ∗ ψk )kLp (RN ) ≤ 2ku ∗ ψk kLp (RN ) .
502
12. Some Equivalent Seminorms
Fix k0 ∈ Z. Since ∆h (u ∗ ψk ) = ψk ∗ (∆h u), we can apply Lemma 12.21 to ∆h u to obtain, together with (12.37) and (12.38), X k∆h (u ∗ ψk )kLp (RN ) k∆h ukLp (RN ) k∈Z
(12.39)
X
=
k∆h (u ∗ ψk )kLp (RN ) +
k∆h (u ∗ ψk )kLp (RN )
k>k0
k≤k0
khk
X
X
k
2
k≤k0
X
ku ∗ ψj kLp (RN ) +
j=k−1
X
khk
k+1 X
ku ∗ ψk kLp (RN )
k>k0
2k ku ∗ ψk kLp (RN ) +
k≤k0 +1
X
ku ∗ ψk kLp (RN ) ,
k>k0
where in the last inequality we use the fact that X k≤k0
2
k
k+1 X j=k−1
+
X
ku ∗ ψj kLp (RN ) = 2
X
k≤k0
2k ku ∗ ψk kLp (RN ) +
k≤k0
≤4
2k−1 ku ∗ ψk−1 kLp (RN )
X
1 X k+1 2 ku ∗ ψk+1 kLp (RN ) 2 k≤k0
k
2 ku ∗ ψk kLp (RN ) .
k≤k0 +1
If u∨ > 0, define W s,p (RN ) 2ks ku ∗ ψk kLp (RN ) ck := . u∨ W s,p (RN ) Then by (12.19), X p (12.40) ck = k∈Z
1
X
2ksp ku p (u∨ ) s,p N W (R ) k∈Z
∗ ψk kpLp (RN ) = 1.
Moreover, ku ∗ ψk kLp (RN ) ≤ 2−ks ck u∨ W s,p (RN )
for all k ∈ Z.
Using this inequality in (12.39) gives (12.41) X X k(1−s) ∨ k∆h ukLp (RN ) u∨ khk 2 c + u 2−ks ck . s,p N s,p N k W (R ) W (R ) k≤k0 +1
Let k0 = kh ∈ Z be such that (12.42)
1 2 ≤ 2kh < . khk khk
k>k0
12.2. The Littlewood–Paley Seminorm
503
Raising both sides of (12.41) to power p, dividing by khkN +sp , and integrating in h over RN yields Z dh k∆h ukpLp (RN ) khkN +sp RN p Z X dh p khkp 2k(1−s) ck (u∨ W s,p (RN ) ) N +sp khk N R k≤kh +1 p Z X dh p (12.43) 2−ks ck + (u∨ W s,p (RN ) ) N +sp khk N R k>kh
=:
p (u∨ W s,p (RN ) ) A
+
p (u∨ W s,p (RN ) ) B.
By H¨older’s inequality for series (see [Leo22c]) we have p p X X 0 2k(1−s) ck = 2k(1−s)/p 2k(1−s)/p ck k≤kh +1
k≤kh +1
p/p0
≤
X k≤kh +1
2k(1−s) cpk
X
2k(1−s)
k≤kh +1
1
X
khk(1−s)(p−1)
2k(1−s) cpk ,
k≤kh +1
where in the last inequality we use geometric series and (12.42). Using (12.42) once more, it follows from the Lebesgue monotone convergence theorem and (12.40) that Z X A khkp−N −sp−(1−s)(p−1) 2k(1−s) cpk dh RN
Z
k≤kh +1
X
khk1−s−N
RN
=
X
k∈Z
2k(1−s) cpk
k∈Z
X
χB(0,1/2k−1 ) (h)2k(1−s) cpk dh
2k(1−s) cpk
k∈Z
Z
khk1−s−N dh
B(0,1/2k−1 )
1 2(1−s)(k−1)
1.
Similarly, again by H¨older’s inequality for series (see [Leo22c]) and (12.42), p p X X 0 2−ks ck = 2−ks/p 2−ks/p ck k>kh
k>kh
p/p0
≤
X k>kh
2−ks
X k>kh
2−ks cpk khks(p−1)
X k>kh
2−ks cpk .
504
12. Some Equivalent Seminorms
In turn, by (12.42) once more, the Lebesgue monotone convergence theorem, and (12.40), Z X 2−ks cpk dh khk−N −sp+s(p−1) B RN
Z
k>kh
khk−N −s
RN
=
X
X
χRN \B(0,1/2k ) (h)2−ks cpk dh
k∈Z
2−ks cpk
k∈Z
Z
khk−N −s dh
RN \B(0,1/2k )
X
2−ks cpk 2ks 1.
k∈Z
Combining the estimates for A and B with (12.43) shows Z dh p k∆h ukpLp (RN ) (u∨ W s,p (RN ) ) khkN +sp RN provided u∨ > 0. If u∨ = 0, then ku ∗ ψk kLp (RN ) = 0 for W s,p (RN ) W s,p (RN ) every k, and so u = 0 by Lemma 12.21. Remark 12.23. Theorem 12.20 continues to hold for higher order fractional Sobolev spaces, namely, one can show that if 1 < p < ∞ and s > 0, s ∈ / N, then for every u ∈ W s,p (RN ), uW s,p (RN ) u∨ W s,p (RN ) uW s,p (RN ) . We refer to [Leo17, Proposition 17.21 and Theorem 17.77] for a proof. Exercise 12.24. Let ϕ0 ∈ Cc∞ (RN ) be as in Proposition 12.15 and let ψk be defined as in (12.17). (i) Prove that (u ∗ ψk )(x) =
1 2
Z ψk (y)[u(x + y) − 2u(x) + u(x − y)] dy RN
for every u ∈ B˙ 1,1 (RN ) and x ∈ RN . Hint: ψk is even. (ii) Prove that u∨ ≤ CuB 1,1 (RN ) . B 1,1 (RN ) Given 1 ≤ p ≤ ∞, and s ∈ R\N, in view of (12.25), for every u ∈ Lp (RN ), we may define for 1 ≤ p < ∞, X 1/p p g ∨ ksp (12.44) kukW s,p (RN ) := ku ∗ ω0 kLp (RN ) + 2 ku ∗ ψk kLp (RN ) , k∈N0
and for p = ∞, ∨ ks kukg W s,∞ (RN ) := ku ∗ ω0 kL∞ (RN ) + sup 2 ku ∗ ψk kL∞ (RN ) . k∈N0
12.2. The Littlewood–Paley Seminorm
505
Here ω0 is the function given in (12.16). In view of (12.21), these norms make sense also for tempered distributions T ∈ S 0 (RN ), so we define 1/p X p g ∨ ksp l ≥ 1, X kSn − Sl kLp (RN ) ≤ 2−ks 2ks ku ∗ ψk kLp (RN ) l≤k≤n
≤
1/p0
1/p
X
2ksp ku ∗ ψk kpLp (RN )
l≤k≤n
X
0
2−ksp
→0
l≤k≤n
as l → ∞, where we use the facts that s > 0 and kukg < ∞. This W s,p (RN ) implies that {Sn }n is a Cauchy sequence in Lp (RN ), so Sn → v as n → ∞ for some function v ∈ Lp (RN ). This shows that X (12.51) v = u ∗ ω0∨ + (u ∗ ψk ) in Lp (RN ), k∈N0
with kvkLp (RN ) = lim kSn kLp (RN ) n→∞ X ∨ ≤ ku ∗ ω0 kLp (RN ) + ku ∗ ψk kLp (RN ) k∈N0
≤ ku ∗ ω0∨ kLp (RN ) +
1/p0
1/p
X
2ksp ku ∗ ψk kpLp (RN )
k∈N0
X
2
−ksp0
.
k∈N0
The identities (12.50) and (12.51) imply that u = v. Hence, kukLp (RN ) kukg by the previous inequality. W s,p (RN ) Remark 12.28. Note that in the proof we never used the fact that s < 1. Hence, in view of Remark 12.23, Theorem 12.26 continues to hold for s > 1, s∈ / N. Using the norm k · kg , we are able to prove a density results for W s,p (RN ) all non integers s. Theorem 12.29 (Density). Let s ∈ R, s ∈ / N, and 1 ≤ p < ∞. Then for every T ∈ S 0 (RN ) such that kT kg < ∞, there exists a sequence W s,p (RN ) {un }n in S(RN ) such that kT − Tun kg → 0 as n → ∞. W s,p (RN ) Proof. Step 1: Let T ∈ S 0 (RN ) be such that kT kg < ∞. For n ∈ N W s,p (RN ) define n X ∨ Tn := T ∗ ω0 + T ∗ ψk ∈ S 0 (RN ). k=0
12.2. The Littlewood–Paley Seminorm
509
In view of (12.25), (12.52)
∞ X
T − Tn =
in S 0 (RN ).
T ∗ ψk
k=n+1
By the definition of convolution of a tempered distributions (see [Leo22c]), for f, g ∈ S(RN ), (T ∗ f )(g) = T (f˜∗ g), where f˜(x) = f (−x), x ∈ RN . Since ψj and ω0∨ are radial, by Lemma 12.27, (T ∗ ψk ) ∗ ψj = 0 for k − j ≥ 2 and (T ∗ ψk ) ∗ ω0∨ = 0 for k ≥ 2. It follows from (12.52) that for every j ∈ N0 , if n ≤ j, j+1 X (T − Tn ) ∗ ψj = (T ∗ ψk ) ∗ ψj k=max{j−1,n+1}
and (T − Tn ) ∗ ψj = 0 if j < n, while (T − Tn ) ∗ ω0∨ = 0 for n ≥ 1. By the Minkowski and Young inequalities (see [Leo22c]), j+1 X
k(T − Tn ) ∗ ψj kLp (RN ) ≤
k(T ∗ ψk ) ∗ ψj kLp (RN )
k=max{j−1,n+1} j+1 X
≤ kT ∗ ψj kLp (RN )
kψk kL1 (RN )
k=max{j−1,n+1}
kT ∗ ψj kLp (RN ) , where we use the fact that (T ∗ ψk ) ∗ ψj = (T ∗ ψj ) ∗ ψk and (12.18). In turn, by (12.45), for n ≥ 1, X 1/p p jsp kT − Tn kg 2 kT ∗ ψ k → 0 as n → ∞. j Lp (RN ) W s,p (RN ) j≥n
Step 2: To conclude the proof, it remains to approximate each Tn with a function in Cc∞ (RN ). Fix n ∈ N. The tempered distributions T ∗ ω0∨ and T ∗ ψk can be identified with the C ∞ functions x 7→ T (ψk (x − ·)), and x 7→ T (ω0∨ (x − ·)), respectively (see [Leo22c]). It follows from (12.45), that Tn can be identified with a function in Lp (RN ) ∩ C ∞ (RN ). By the density of Cc∞ (RN ) in Lp (RN ), given ε > 0, we can find vn ∈ Cc∞ (RN ) such that (12.53)
kTn − vn kLp (RN ) ≤ ε.
Define un := vn ∗ ω0∨ +
n X
vn ∗ ψ k .
k=0
Fix j ∈ N0 . If j > n + 1, by Lemma 12.27, un ∗ ψj = 0, while by the previous step Tn ∗ ψj = 0. Again by Lemma 12.27, if 1 ≤ j ≤ n + 1, (vn ∗ ω0∨ ) ∗ ψj = 0
510
12. Some Equivalent Seminorms
and (vn ∗ ψk ) ∗ ψj = 0 for k − j ≥ 2. Hence, for 1 ≤ j ≤ n + 1, we can write un ∗ ψj =
n X
j+1 X
(vn ∗ ψk ) ∗ ψj =
k=0
(vn ∗ ψk ) ∗ ψj = vn ∗ ψj ,
k=j−1
where in the last equality we use Lemma 12.21. Hence, by Young’s inequality (see [Leo22c]), (12.18), and (12.53), (12.54)
k(Tn − un ) ∗ ψj kLp (RN ) = k(Tn − vn ) ∗ ψj kLp (RN ) ≤ kTn − vn kLp (RN ) kψj kL1 (RN ) ε.
On the other hand, if j = 0, un ∗ ψ0 = (vn ∗ ω0∨ ) ∗ ψ0 +
n X
(vn ∗ ψk ) ∗ ψ0
k=0 1 X
= (vn ∗ (ω0∨ − ψ−1 )) ∗ ψ0 +
(vn ∗ ψk ) ∗ ψ0 = vn ∗ ψ0 ,
k=−1
where (vn ∗(ω0∨ −ψ−1 ))∗ψ0 = 0 by (12.16) and we use Lemma 12.21. Hence, as before, (12.55)
k(Tn − un ) ∗ ψ0 kLp (RN ) = k(Tn − vn ) ∗ ψ0 kLp (RN ) ≤ kTn − vn kLp (RN ) kψ0 kL1 (RN ) ε.
By collecting the inequalities (12.54) and (12.55), it follows that X
2jsp k(Tn − un ) ∗ ψj kpLp (RN ) =
n+1 X
2jsp k(Tn − un ) ∗ ψj kpLp (RN ) n εp
j=0
j∈N0
(recall that n is fixed). By (12.49) and the fact that (vn ∗ ω0∨ ) ∗ ψj = 0 for j ≥ 1, we can write un ∗
ω0∨
= (vn ∗
ω0∨ )
= (vn ∗
ω0∨ )
∗
ω0∨
∗
ω0∨
+
n X
(vn ∗ ψk ) ∗ ω0∨
k=0
+ (vn ∗ ψ0 ) ∗ ω0∨ = vn ∗ ω0∨ .
Using Young’s inequality (see [Leo22c]), k(Tn − un ) ∗ ω0∨ kLp (RN ) = k(Tn − vn ) ∗ ω0∨ kLp (RN ) ≤ kTn − vn kLp (RN ) kω0∨ kL1 (RN ) n ε. This shows that kTn − un kg n ε. W s,p (RN )
0
We recall that the dual space of W0s,p (Ω) is denoted by W −s,p (Ω).
12.2. The Littlewood–Paley Seminorm
511
Theorem 12.30 (Dual spaces). Let s > 0, s ∈ / N, and 1 < p < ∞. Then 0 for every T ∈ W −s,p (RN ), kT kg kT kW −s,p0 (RN ) kT kg . W −s,p0 (RN ) W −s,p0 (RN ) Conversely, every T ∈ S 0 (RN ) such that kT kg < ∞ can be uniquely W −s,p0 (RN ) 0
extended to an element in the dual of W s,p (RN ). In particular, W −s,p (RN ) may be identified with the space {T ∈ S 0 (RN ) : kT kg < ∞}. MoreW s,p (RN ) over, W s,p (RN ) is reflexive. Given a Lebesgue measurable set E ⊆ RN and 1 ≤ p < ∞, we define the space X p of all sequences {un }n∈N0 of functions in Lp (E) such that (12.56)
k{un }n∈N0 kX p :=
∞ X
!1/p kun kpLp (E)
< ∞.
n=0
We leave it as an exercise to check that X p is a normed space. Lemma 12.31. Let E ⊆ RN be a Lebesgue measurable set, 1 ≤ p < ∞, and 0 s ∈ R. Then for every L ∈ (X p )0 there exists {vn }n∈N0 in X p such that L({un }n∈N0 ) =
∞ Z X
un vn dx,
{un }n∈N0 ∈ X p ,
n=0 E
and k{vn }n∈N0 kX p0 = kLk(X p )0 . Moreover, X p is reflexive for 1 < p < ∞. Proof. Given k ∈ N0 , define Tk : Lp (E) → R by Tk (u) := L({uδk,n }n ),
u ∈ Lp (E),
where δk,n = 0 if n 6= k and δk,k = 1. Since {uδk,n }n∈N0 ∈ X p , by (12.56) we have Tk (u) = L({uδk,n }n ) ≤ kLk(X p )0 k{uδk,n }n kX p = kLk(X p )0 kukLp (E) . This shows that the linear function Tk : Lp (E) → R is continuous, so it belongs to the dual of Lp (E). By the Riesz representation theorem in Lp (E) 0 (see [Leo22c]), there exists vk ∈ Lp (E) such that Z (12.57) Tk (u) = L({uδk,n }n ) = uvk dx, E
with kvk kLp0 (E) = kTk k(Lp (E))0 ≤ kLk(X p )0 .
512
12. Some Equivalent Seminorms
By the linearity of L, it follows that for every ` ∈ N0 and {un }n∈N0 ∈ X p , ! ` Z ` ` X X X uk vk dx = Tk (uk ) = L {uk δk,n }n k=0 E k=0 k=0
`
X
≤ kLk(X p )0 {uk δk,n }n .
k=0
Xp
p0 −1
Fix ` and take uk := vk  sgn vk , k = 0, . . . , `. We have that R p0 dx, so by the previous inequality, v  E k
` Z `
X X
p0 vk  dx ≤ kLk(X p )0 {uk δk,n }n
E k=0
k=0
= kLk(X p )0
` X
0
R E
uk p dx =
Xp !1/p
kvn kpLp0 (E)
< ∞.
n=0
Hence, after dividing (if the lefthand side is different from zero), we obtain !1−1/p ` X p0 kvn kLp0 (E) ≤ kLk(X p )0 . n=0
Letting ` → ∞ gives ∞ X
(12.58)
p0
!1−1/p
kvn kLp0 (E)
≤ kLk(X p )0 .
n=0 0
This shows that {vn }n∈N0 ∈ X p . Given {un }n∈N0 ∈ X p , by the Lebesgue monotone convergence theorem and H¨older’s inequality (see [Leo22c]), Z X ∞ ∞ Z ∞ X X un vn  dx = un vn  dx ≤ kun kLp (E) kvn kLp0 (E) E n=0
(12.59)
n=0 E
n=0
≤ k{un }n kX p k{vn }n kX p0 < ∞.
It follows that w(x) :=
∞ X
un (x)vn (x) < ∞
n=0
for LN a.e. x ∈ E and that w is Lebesgue integrable. Thus, we can apply the Lebesgue dominated convergence theorem to obtain ` Z ∞ Z X X lim uk vk dx = uk vk dx. `→∞
k=0
E
k=0
E
12.2. The Littlewood–Paley Seminorm
513
On the other hand, by the continuity and linearity of L, ` X L ({uk δk,n }n ) = L ({wn,` }n ) L({un }n ) − k=0
∞ X
≤ kLk(X p )0
!1/p kun kpLp (E)
→0
n=`+1
as ` → ∞, where wn,` := un if n ≥ ` + 1 and wn,` := 0 if n < ` + 1. These two limits and (12.57) imply that ∞ Z X un vn dx (12.60) L({un }n ) = n=0 E
for every {un }n∈N0 ∈ X p . In view of (12.58), it remains to show that kLk(X p )0 ≤ k{vn }n kX p0 .
(12.61) By (12.59) and (12.60), L({un }n ) ≤
Z X ∞ E n=0
un vn  dx ≤ k{un }n kX p k{vn }n kX p0 .
Dividing by k{un }n∈N0 kX p when {un }n 6= 0 and taking the supremum over all {un }n∈N0 ∈ Xp proves (12.61). We turn to the proof of Theorem 12.30. < ∞ and Proof. Step 1: Let T ∈ S 0 (RN ) be such that kT kg W −s,p0 (RN ) P ∞ φ ∈ S(RN ). By Lemma 12.19, T = T ∗ ω0∨ + k=0 T ∗ ψk in S 0 (RN ). By the definition of convolution of tempered distributions and the fact that ω0∨ and ψk are radial (see [Leo22c]), we can write (12.62)
T (φ) = T (ω0∨ ∗ φ) +
∞ X
T (ψk ∗ φ).
k=0 P∞ j=0 φ
By Corollary 12.16, φ = φ ∗ ω0∨ + ∗ ψj , while by Lemma 12.27, ∨ ψk ∗ ψj = 0 for k − j ≥ 2 and ω0 ∗ ψj = 0 for j ≥ 1. Hence, ω0∨ ∗ φ = ω0∨ ∗ φ ∗ ω0∨ + ω0∨ ∗ φ ∗ ψ0 , ψk ∗ φ = ψk ∗ φ ∗ ω0∨ +
k+1 X
ψk ∗ φ ∗ ψj .
j=(k−1)+
Plugging these expressions into (12.62), we get T (φ) = (T ∗ ω0∨ )(φ ∗ ω0∨ ) + 2(T ∗ ω0∨ )(φ ∗ ψ0 ) +
∞ X
k+1 X
(T ∗ ψk )(φ ∗ ψj ).
k=0 j=(k−1)+
514
12. Some Equivalent Seminorms
In turn, T (φ) (T ∗ ω0∨ )(φ ∗ ω0∨ ) + (T ∗ ω0∨ )(φ ∗ ψ0 ) +
∞ X
k+1 X
(T ∗ ψk )(φ ∗ ψj ) =: A + B + C.
k=0 j=(k−1)+
By (12.21), Z T (ψk (x − ·))(φ ∗ ψj )(x) dx,
(T ∗ ψk )(φ ∗ ψj ) = RN
and so, by H¨ older’s inequality, (T ∗ ψk )(φ ∗ ψj ) ≤ kT ∗ ψk kLp0 (RN ) kφ ∗ ψj kLp (RN ) . In turn, using Holder’s inequality for series (see [Leo22c]), C≤
∞ X
2
−ks
ks
kT ∗ ψk kLp0 (RN ) 2
k=0
≤
k+1 X
kφ ∗ ψj kLp (RN )
j=(k−1)+
∞ X
−ksp0
2
kT ∗
0 ψk kpLp0 (RN )
!1/p0
k=0
∞ X 2ksp ×
p 1/p
k+1 X
kφ ∗ ψj kLp (RN )
k=0 j=(k−1)+ g kT kW −s,p0 (RN ) kφkg W s,p (RN ) ,
where we use the fact that, by Young’s inequality, p ∞ k+1 X X 2ksp kφ ∗ ψj kLp (RN ) k=0
j=(k−1)+
≤ 3p−1
∞ X
k+1 X
2ksp kφ ∗ ψj kpLp (RN )
k=0 j=(k−1)+ p−1
=3
∞ X j=0
∞ X
kφ ∗
ψj kpLp (RN )
j+1 X
2ksp
k=(j−1)+
2jsp kφ ∗ ψj kpLp (RN ) kφkg W s,p (RN ) .
j=0
Simpler estimates show that A + B kT kg kφkg W s,p (RN ) . W −s,p0 (RN )
12.2. The Littlewood–Paley Seminorm
515
In conclusion, we have shown that T (φ) kT kg kφkg 0 W s,p (RN ) W −s,p (RN ) for all φ ∈ S(RN ). By the density of S(RN ) in W s,p (RN ) (Theorem 12.29), it follows that the linear function T : S(RN ) → R can be uniquely extended to a continuous linear function T : W s,p (RN ) → R, with T (φ) kT kg kφkg W s,p (RN ) W −s,p0 (RN ) 0
for all φ ∈ W s,p (RN ). Hence, T ∈ (W s,p (RN ))0 = W −s,p (RN ) and kT kW −s,p0 (RN ) kT kg . W −s,p0 (RN ) 0
Step 2. Let T ∈ W −s,p (RN ). Let X p be defined as in Lemma 12.31. For every u ∈ W s,p (RN ), consider the sequence {un }n∈N0 given by u0 := u ∗ ω0∨ and un := 2ns (u ∗ ψn ). In view of (12.44), {un }n∈N0 ∈ X p , with g p kukg W s,p (RN ) k{un }n kX kukW s,p (RN ) .
Let Y p be the subspace of X p consisting of all sequences {un }n∈N0 constructed as above, as u varies over all W s,p (RN ). Then the function L : Y p → R, defined by L({un }n ) := T (u),
{un }n∈N0 ∈ Y p ,
is linear and L({un }n ) = T (u) ≤ kT kW −s,p0 (RN ) kuk∨ W s,p (RN ) kT kW −s,p0 (RN ) k{un }n kX p . By the Hahn–Banach extension theorem (see [Leo22b]), we can extend L to a linear continuous function L : X p → R with L({un }n ) kT kW −s,p0 (RN ) k{un }n kX p for every {un }n∈N0 ∈ X p . By Lemma 12.31, there exists {vn }n∈N0 ∈ X p such that ∞ Z X L({un }n ) = un vn dx n=0 E
for every {un }n∈N0 ∈
Xp
and
k{vn }n kX p0 = kLk(X p )0 kT kW −s,p0 (RN ) . Define wn := 2ns vn . Then (12.63)
L({un }n ) =
∞ Z X
2−ns un wn dx
n=0 E
and (12.64)
k{2−ns wn }n kX p0 = kLk(X p )0 kT kW −s,p0 (RN ) .
0
516
12. Some Equivalent Seminorms
In particular, if u ∈ W s,p (RN ), then by (12.63), Z ∞ Z X ∨ w0 (u ∗ ω0 ) dx + wn (u ∗ ψn ) dx. T (u) = RN
N n=1 R
By the definition of the convolution of tempered distribution (see [Leo22c]) and the fact that ψj is radial, for φ ∈ S(RN ), Z w0 ((ψj ∗ φ) ∗ ω0∨ ) dx (T ∗ ψj )(φ) = T (ψj ∗ φ) = RN
+
∞ Z X
wn ((ψj ∗ φ) ∗ ψn ) dx.
N n=1 R
By Lemma 12.27, we can write Z (T ∗ ψ0 )(φ) = RN
w0 ((ψ0 ∗ φ) ∗ ω0∨ ) dx
Z w1 ((ψ0 ∗ φ) ∗ ψ1 ) dx
+ RN
and (12.65)
j+1 X
(T ∗ ψj )(φ) =
Z wn ((ψj ∗ φ) ∗ ψn ) dx
N n=max{1,(j−1)+ } R
for j ≥ 1. Similarly, (T ∗ ω0∨ )(φ) =
Z RN
w0 ((ω0∨ ∗ φ) ∗ ω0∨ ) dx
Z + RN
w0 ((ω0∨ ∗ φ) ∗ ψ0 ) dx.
By Young’s inequality (see [Leo22c]) and the fact that (ψj ∗ φ) ∗ ψn = (ψj ∗ ψn ) ∗ φ by Fubini’s theorem, k(ψj ∗ ψn ) ∗ φkLp (RN ) ≤ kψj kL1 (RN ) kψn kL1 (RN ) kφkLp (RN ) 2 = kϕ∨ 0 kL1 (RN ) kφkLp (RN ) ,
where we use (12.18). Assume that j ≥ 1. By (12.65), H¨older’s inequality, and the previous inequality (T ∗ ψj )(φ) ≤
j+1 X
kwn kLp0 (RN ) k(ψj ∗ ψn ) ∗ φkLp (RN )
n=(j−1)+
j+1 X n=(j−1)+
kwn kLp0 (RN ) kφkLp (RN ) .
12.2. The Littlewood–Paley Seminorm
517
This implies that T ∗ ψj can be uniquely extended to a continuous linear function T ∗ ψj : Lp (RN ) → R, with j+1 X
kT ∗ ψj k(Lp (RN ))0
kwn kLp0 (RN ) .
n=(j−1)+
By the Riesz representation theorem in Lp (RN ) (see [Leo22c]), we can 0 identify T ∗ ψj with a function in Lp (RN ), with j+1 X
kT ∗ ψj kLp0 (RN )
kwn kLp0 (RN ) .
n=(j−1)+
Similarly, for j = 0, kT ∗ ψ0 kLp0 (RN ) kw0 kLp0 (RN ) and kT ∗ ω0∨ kLp0 (RN ) kw0 kLp0 (RN ) . In view of (12.45), it follows that kT kg W −s,p0 (RN )
= kT ∗
ω0∨ kLp0 (RN )
+
X
2
−jsp0
kT ∗
0 ψj kpLp0 (RN )
1/p0
j∈N
kw0 kLp0 (RN ) +
X
2
−jsp0
j∈N
∞ X
0
j+1 X
p0 1/p0 kwn kLp0 (RN )
n=(j−1)+
!1/p0
0
2−nsp kwn kpLp0 (RN )
,
n=0
where we use the fact that p0 j+1 ∞ ∞ X X X −jsp0 p0 −1 0 2 kwn kLp (RN ) ≤3 j=0 ∞ X n=0
0
kwn kpLp0 (RN )
0
0
2−jsp kwn kpLp0 (RN )
j=0 n=(j−1)+
n=(j−1)+
=
j+1 X
j+1 X
0
2−jsp
n=(j−1)+
∞ X
0
0
2−nsp kwn kpLp0 (RN ) .
n=0
In turn, by (12.64), kT kg kT kW −s,p0 (RN ) . Together with Step 1, 0 W −s,p (RN ) 0
this shows that the dual W −s,p (RN ) of W s,p (RN ) may be identified with the space of tempered distributions T ∈ S 0 (RN ) such that kT kg < ∞. W −s,p0 (RN ) Step 3: Since the subspace Y p defined in Step 2 is closed and X p is reflexive, the reflexivity of W s,p (RN ) follows (see [Leo22b]). Remark 12.32. For an alternative approach on duality, which relies on abstract interpolation theory, we refer to [Leo17, Theorem 17.41].
518
12. Some Equivalent Seminorms
12.3. Notes The proof of Theorem 12.5 is adapted from the lecture notes of G¨ otze [G¨ ot10]. Exercise 12.8 was suggested by Ian Tice.
Part 3
Applications
Chapter 13
Interior Regularity for the Poisson Problem It is selfevident that any and all paths must be open to a researcher during the actual course of his [or her] investigations. — Karl Weierstrass
In this chapter, we will study the use of fractional Sobolev spaces in the regularity theory of elliptic equations. For simplicity, we will consider interior regularity (away from the boundary) and will only study the Poisson problem −∆u(x) = f (x)
in Ω.
The goal is to show that if f ∈ W s,p (Ω), for 1 < p < ∞ and 0 < s < 1, then s+2,p (Ω). a weak solution u to the Poisson problem belongs to Wloc
13.1. Cacciopoli’s Inequality We recall some important properties of harmonic functions. Given PN an2 open N 2 set Ω ⊆ R , a function u ∈ C (Ω) is harmonic in Ω if ∆u := n=1 ∂n u = 0 in Ω. We recall that SN −1 = ∂B(0, 1),
αN = LN (B(0, 1)),
βN = HN −1 (SN −1 ),
with βN = N αN . Theorem 13.1 (Mean value formulas). Let Ω ⊆ RN be an open set and let u ∈ C 2 (Ω) be a harmonic function. Then for every ball B(x, R) b Ω, Z Z 1 1 N −1 u(y) dH (y), u(x) = u(y) dy. u(x) = βN RN −1 ∂B(x,R) αN RN B(x,R) 521
522
13. Interior Regularity for the Poisson Problem
Proof. Define the function Z 1 (13.1) f (r) = u(y) dHN −1 (y), βN rN −1 ∂B(x,r)
0 < r ≤ R.
Using spherical coordinates and writing y = x + rz, we get Z 1 f (r) = u (x + rz) dHN −1 (z). βN SN −1 Define v(z) := u(x + rz), z ∈ B(0, 1), and observe that v is harmonic. By differentiating under the integral sign (see [Leo22c]) and the divergence theorem, we have Z 1 0 f (r) = ∇u (x + rz) · z dHN −1 (z) βN SN −1 Z Z 1 1 N −1 (13.2) ∂ν v (z) dH (z) = ∆v(z) dz = 0. = βN r SN −1 βN B(0,1) This implies that f is constant in (0, R]. On the other hand, by the continuity of u, given ε > 0 there exists δ > 0 such that u(y) − u(x) ≤ ε for all y ∈ B (x, δ). Hence, for 0 < r < δ, Z 1 N −1 f (r) − u(x) = (u(y) − u(x)) dH (y) ≤ ε. N −1 ∂B(x,r) βN r Since f (R) = f (r), it follows that f (R) − u(x) ≤ ε for all ε > 0, which shows that f (R) = u(x). Using spherical coordinates and what we just proved, we have Z Z RZ u(y) dy = u(y) dHN −1 (y)dr B(x,R)
0
∂B(x,r) Z R
= u(x)βN
rN −1 dr = u(x)αN RN .
0
Using mean value formulas, we can give a useful characterization of harmonic functions. Theorem 13.2. Let Ω ⊆ RN be an open set and let u ∈ C(Ω) be such that for every ball B(x, R) b Ω, Z 1 u(y) dHN −1 (y). (13.3) u(x) = βN RN −1 ∂B(x,R) Then u ∈ C ∞ (Ω) and is harmonic in Ω. Proof. For ε > 0, let uε = ϕε ∗ u be a mollification of u, where ϕ is a standard mollifier.
13.1. Cacciopoli’s Inequality
523
Since the function ϕ is radial, we can write ϕε (x) = gε (kxk), x ∈ Ω. Fix x ∈ Ω and 0 < ε < dist(x, ∂Ω). Then by the change of variables z = x − y and the fact that supp ϕε ⊂ B(0, ε), Z Z ϕε (z)u (x − z) dz ϕε (x − y)u(y) dy = uε (x) = Ω ε
Z =
B(0,ε)
rN −1
Z ∂B(0,1)
0
Z
ϕε (rw) u (x − rw) dHN −1 (w) dr
ε
=
rN −1 gε (r)
0
Z
u (x − rw) dHN −1 (w) dr
∂B(0,1) ε
r
=
Z
N −1
gε (r)
0
Z
ε
= u(x)βN
1 rN −1
Z
u(y) dHN −1 (y)dr
∂B(x,r)
rN −1 gε (r) dr = u(x),
0
where we use the fact that u ∈ C ∞ (Ω).
R
RN
ϕε dx = 1 and (13.3). This proves that
Define the function f as in (13.1). Since f is constant by (13.3), it follows that f 0 = 0. Hence, by (13.2) and the change of variables y = x + rz, Z 1 0 ∆u(y) dy, 0 = f (r) = βN rN −1 B(x,r) R and so B(x,r) ∆u(y) dy = 0 for every r > 0. Since ∆u is continuous, it follows that ∆u = 0. This shows that u is harmonic. Remark 13.3. With a similar proof, one can show that if u : Ω → R is measurable, integrable on compact sets, and satisfies Z u(y)∆ϕ(y) dy = 0 Ω
for all ϕ ∈ Cc∞ (Ω), then u is harmonic. We leave the details as an exercise. Corollary 13.4. Let Ω ⊆ RN be an open set and let u : Ω → R be a harmonic function. Then for every ball B(x0 , 2R) b Ω and 1 < p < ∞, !1/p Z 1 p sup u(x) ≤ u(y) dy . (αN RN )1/p B(x0 ,2R) B(x0 ,R) Proof. By the mean value formula for x ∈ B(x0 , R), Z 1 u(x) = u(y) dy, αN RN B(x,R)
524
13. Interior Regularity for the Poisson Problem
and so, by H¨ older’s inequality and the fact that B(x, R) ⊂ B(x0 , 2R), !1/p Z Z 0 1 (αN RN )1/p u(x) ≤ u(y) dy ≤ u(y)p dy αN RN B(x,R) αN R N B(x,R) !1/p Z 1 u(y)p dy . ≤ (αN RN )1/p B(x0 ,2R) Taking the supremum over x ∈ B(x0 , R) concludes the proof.
Using the mean value formulas, we can obtain interior estimates of the derivatives of a harmonic function. Since a harmonic function u in Ω is of class C ∞ (Ω), for every multiindex α, we have that α ∂α ∂ u 0= (∆u) = ∆ , ∂xα ∂xα α
and thus ∂∂xαu is also harmonic in Ω. This simple observation has important applications. Theorem 13.5. Let Ω ⊆ RN be an open set and let u : Ω → R be a harmonic function. Then for every ball B(x, R) b Ω, k∇u(x)k ≤
N max u . R ∂B(x,R)
Proof. Since ∇u is harmonic, by the mean value formula applied to ∇u, Z 1 ∇u(x) = ∇u(y) dy αN RN B(x,R) Z 1 = u(y)ν (y) dHN −1 (y), αN RN ∂B(x,R) where we use the divergence theorem. Hence, Z 1 N k∇u(x)k ≤ max u max u . 1 dHN −1 (y) = αN RN ∂B(x,R) R ∂B(x,R) ∂B(x,R)
Remark 13.6. Note that Theorem 13.5 uses only the fact that ∇u is harmonic, and thus, it continues to apply if u ∈ C 3 (Ω) satisfies ∆u(x) = c for all x ∈ Ω and for some constant c ∈ R. We recallR that Q(x, r) is a cube centered at x and sidelength r and that uE = LN1(E) E u(y) dy. Theorem 13.7 (Cacciopoli’s inequality). Let Ω ⊆ RN be an open set and let u : Ω → R be a harmonic function. Then for every Q(x0 , R) b Ω and
13.2. H 2 Regularity
525
0 < r < R, Z
2
Q(x0 ,r)
(u − uQ(x0 ,r) ) dx
r N +2 Z R
Q(x0 ,R)
(u − uQ(x0 ,R) )2 dx.
Proof. Set Qr := Q(x0 , r). By a rescaling argument, it suffices to take R = 1 (exercise). Also, since u − uQ1 is harmonic, without loss of generality we may assume that uQ1 = 0. Suppose first that r < 12 . Since u ∈ C ∞ (Ω), there exists x1 ∈ Qr such that uQr = u(x1 ), and so, by the mean value theorem for every x1 ∈ Qr , u(x) − uQr  = u(x) − u(x1 ) rk∇u(z)k for some z in the segment joining x to x1 . It follows from Theorem 13.5, that u(x) − uQr  k∇u(z)kr r max u = ru(w) ∂B(z,1/4) Z Z 1/2 Z 2 u dy r u dy r u dy , r B (w, 1 ) Q1 Q1 4
where we use the mean value formula and H¨ older’s inequality. Squaring both sides and integrating over Qr , we get Z Z Z 2 N +2 2 N +2 (u − uQ1 )2 dy. u dy = r (u − uQr ) dx r Q1
Q1
Qr 1 2
< r ≤ 1, then using the fact that (exercise) Z 2 min (u − c) dx = (u − uQr )2 dx,
On the other hand, if
Z
(13.4)
c∈R
we have that Z
Qr
2
Qr
Z
2
(u − uQ1 ) dx ≤ (u − uQ1 )2 dx Q1 Z n+2 r (u − uQ1 )2 dx,
(u − uQr ) dx ≤ Qr
Z
Qr
Q1
where we use the fact that 1 ≤ 2r.
13.2. H 2 Regularity In this section we will prove that if f ∈ L2 (RN ), then there exists a unique solution (up to affine functions) u ∈ H˙ 2 (RN ) to the Poisson problem (13.5)
−∆u(x) = f (x)
with k∇2 ukL2 (RN ) kf kL2 (RN ) .
in RN ,
526
13. Interior Regularity for the Poisson Problem
Definition 13.8. Given an open set Ω ⊆ RN , 1 ≤ p ≤ ∞, µ ∈ R, and 1,p f ∈ Lploc (Ω), we say that u ∈ Wloc (Ω) is a weak solution to −∆u + µu = f in Ω if Z (13.6) (∇u · ∇w + µuw − f w) dx = 0 Ω
for all w ∈
0 W 1,p (Ω)
with compact support in Ω.
We begin by proving the existence of weak solutions when f ∈ L2 (RN ). Theorem 13.9. Let f ∈ L2 (RN ). Then (13.5) admits a weak solution u ∈ H˙ 2 (RN ) such that −∆u(x) = f (x) for LN a.e. x ∈ RN and k∇2 ukL2 (RN ) kf kL2 (RN ) . Moreover, any other weak solution to (13.5) that belongs to H˙ 2 (RN ) differs from u by an affine function. The main difficulty in proving the existence of solutions is that Poincar´e’s inequality fails in unbounded domains, so we cannot control lower order terms with ∇2 u. To circumvent this problem we consider the perturbed problem −∆u + µu = f
(13.7)
in RN ,
where µ > 0. Lemma 13.10. Let f ∈ L2 (RN ) and µ > 0. Then (13.7) admits a unique weak solution u ∈ H 2 (RN ) such that −∆u(x) + µu(x) = f (x) for LN a.e. x ∈ RN and Z Z 2 2 2 2 2 f 2 dx. µ u + µk∇uk + k∇ uk dx RN
RN
Proof. Step 1: Existence and uniqueness of solutions. Equation (13.7) is the Euler–Lagrange equation of the convex functional Z µ 2 1 2 k∇v(x)k + v (x) − f (x)v(x) dx, v ∈ H 1 (RN ). F(v) = 2 2 RN By H¨older’s inequality, 1 µ (13.8) F(v) ≥ k∇vk2L2 + kvk2L2 − kvkL2 kf kL2 , 2 2 and thus, F(v) → ∞ as kvkH 1 → ∞. Since F(0) = 0, this implies that inf
v∈H 1 (RN )
F(v) = ` ∈ R.
Consider a minimizing sequence {vk }k , that is, a sequence of functions in H 1 (RN ) such that lim F(vk ) = `. k→∞
13.2. H 2 Regularity
527
In view of (13.8), {vk }k is bounded in H 1 (RN ). Thus by reflexivity, there exist a subsequence {vkj }j and u ∈ H 1 (RN ) such that vkj * u in H 1 (RN ) as j → ∞. In particular, ∇vkj * ∇u in L2 (RN ; RN ) and vkj * u in L2 (RN ). By the lower semicontinuity of the norm with respect to weak convergence, we have Z Z lim inf j→∞
RN
v 2 dx.
vk2j dx ≥
RN
By extracting a further subsequence, not relabeled, we can assume that there exists the limit Z Z v 2 dx. vk2j dx ≥ lim j→∞ RN
RN
Hence, Z lim
µ 2
j→∞ RN
vk2j
− f vkj
Z dx ≥
µ
v 2 − f v dx,
RN 2 2 L (RN ) and
where we use the facts that vkj * u in that f ∈ L2 (RN ). In turn, Z Z 1 µ 2 ` = lim F(vkj ) = lim k∇vkj k2 dx + lim v − f v dx j→∞ j→∞ RN 2 j→∞ RN 2 Z µ 1 k∇uk2 + u2 − f u dx = F(u) ≥ `, ≥ 2 2 RN again by lower semicontinuity of the norm with respect to weak convergence. This implies that F(u) = `. Note that since F is given by the sum of quadratic terms, which are strictly convex, and a linear term, F is strictly convex, that is, F(θu1 + (1 − θ)u2 ) < θF(u1 ) + (1 − θ)F(u2 ), whenever u1 6= u2 and θ ∈ (0, 1). Hence, minimizers are unique. Step 2: Weak Euler–Lagrange equations. Let u ∈ H 1 (RN ) be the unique minimizer of F constructed in Step 1. Given w ∈ H 1 (RN ) and t ∈ R, we have that g(t) := F(u + tw) ≥ F(u) = g(0). Hence, the function g has a minimum at t = 0. If g is differentiable at t = 0, then g 0 (0) = 0. But Z g(t) − g(0) t µt 2 2 = k∇wk + ∇u · ∇w + w + µuw − f w dx t−0 2 2 N ZR → (∇u · ∇w + µuw − f w) dx, RN
and so, (13.9)
0
Z (∇u · ∇w + µuw − f w) dx,
0 = g (0) = RN
528
13. Interior Regularity for the Poisson Problem
which shows that u is the weak solution to (13.7). In particular, taking w = µu in (13.9) gives Z Z Z Z 1 1 2 2 2 2 2 µk∇uk + µ u dx = µf u dx ≤ µ u dx + f 2 dx. 2 RN 2 RN RN RN In turn,
Z (13.10) RN
µ2 µk∇uk + u2 2 2
1 dx ≤ 2
Z
f 2 dx.
RN
Step 3: Regularity of the solution. To prove the regularity of u, we will use Nirenberg’s difference quotient method. We recall that ∆h v(x) = v(x + h) − v(x),
x ∈ RN ,
where h ∈ RN . Note that by the change of variables x + h = y for v, g ∈ L2 (RN ), Z Z (13.11) ∆h v(x)g(x) dx = (v(x + h) − v(x))g(x) dx RN RN Z Z = v(y)g(y − h) dy − v(x)g(x) dx N RN ZR = v(y)∆−h g(y) dy. RN
Taking w = ∆−h ∆h u in (13.9) and using (13.11), we have Z 0= (∇u · ∇(∆−h ∆h u) + µu∆−h ∆h u − f ∆−h ∆h u) dx N ZR = k∇∆h uk2 + µ(∆h u)2 − f ∆−h ∆h u dx. RN
Hence, by H¨older’s inequality Z (k∇∆h uk2 + µ(∆h u)2 ) dx ≤ kf kL2 k∆−h ∆h ukL2 . RN
By the standard properties of Sobolev functions (see Step 1 of the proof of Theorem 12.5), Z Z 2 2 (∆−h ∆h u) dx ≤ khk k∇∆h uk2 dx. RN
RN
Dividing by khk2 and combining the previous two inequalities, we get Z k∇∆h uk2 (∆h u)2 1 + µ dx ≤ kf kL2 k∇∆h ukL2 2 2 khk khk khk N R Z Z 1 1 2 f dx + k∇∆h uk2 dx. ≤ 2 RN 2khk2 RN
13.2. H 2 Regularity
529
In turn, Z 1 k∇∆h uk2 (∆h u)2 1 +µ dx ≤ f 2 dx. 2 2 2 khk khk 2 N N R R Again by standard properties of Sobolev functions (see [Leo22d]), we obtain Z Z κ 2 2 1 2 (13.12) f 2 dx, k∇ uk + µκk∇uk dx ≤ 2 RN RN 2 Z
where κ = κ(N ) > 0. In particular, we can integrate by parts in (13.9) to get Z (13.13) (−∆u + µu − f ) w dx = 0 RN
for all w ∈ H 1 (RN ). This implies that −∆u(x) + µu(x) − f (x) = 0
for LN a.e. x ∈ RN .
We turn to the proof of Theorem 13.9. Proof of Theorem 13.9. Step 1: Existence and regularity of weak solutions. For each µ > 0 let uµ ∈ H 2 (RN ) be the function given in Lemma 13.10. By Poincar´e’s inequality (see [Leo22d]) for R ≥ 1, Z Z 2 uµ (x) − (uµ )B(0,1) − (∇uµ )B(0,1) · x dx R k∇2 uµ (x)k2 dx, B(0,R) B(0,R) Z Z k∇uµ (x) − (∇uµ )B(0,1) k2 dx R k∇2 uµ (x)k2 dx, B(0,R)
where we recall that wE =
B(0,R) 1 LN (E)
R E
w(x) dx. Hence, if we define
vµ (x) := uµ (x) − (uµ )B(0,1) − (∇uµ )B(0,1) · x, we have that the {vµ }µ is bounded in H 2 (B(0, R)) for every R ≥ 1. In view of (13.12) and the previous estimates, taking µ = n1 and using a diagonal argument, we can find a function u and a subsequence of {vµ }µ , not relabeled, 2 (RN ) and ∂ 2 v * ∂ 2 u in L2 (RN ) as µ → 0+ . such that vµ * u in Hloc i,j µ i,j On the other hand, if we define vµ := µuµ , by (13.10) and (13.12), we have that Z Z 1 2 2 dx vµ + k∇vµ k f 2 dx, µ N N R R and so, up to a further subsequence, we can assume that vµ * v in H 1 (RN ) as µ → 0+ , where ∇v = 0. Since v ∈ H 1 (RN ), necessarily, v = 0. Hence, if we let µ → 0+ in (13.13), we obtain Z (−∆u − f ) w dx = 0 RN
530
13. Interior Regularity for the Poisson Problem
for all w ∈ H 1 (RN ). Integrating by parts, we get that u is a weak solution to (13.5). Moreover, letting µ → 0+ in (13.12) and using the lower semicontinuity of the L2 norm with respect to weak convergence, we obtain Z Z f 2 dx. k∇2 uk2 dx κ RN
RN
Step 2: Uniqueness of weak solutions. Let u1 , u2 ∈ H˙ 2 (RN ) be two weak solutions to (13.5). Then Z ∇(u1 − u2 ) · ∇w dx = 0 RN
H 1 (RN )
for all w ∈ with compact support. Hence, by Remark 13.3, the 2 (u − u ). In turn, by Corollary function u1 − u2 is harmonic and so is ∂i,j 1 2 13.4, !1/2 Z 1 2 2 2 sup ∂i,j (u1 − u2 )(x) ≤ ∂i,j (u1 − u2 )(y) dy (αN R)N/2 B(x0 ,2R) B(x0 ,R) for every x0 ∈ RN and R > 0. Letting R → ∞ and using the fact that 2 (u −u ) ∈ L2 (RN ), we obtain that ∂ 2 (u −u ) = 0 for all i, j = 1, . . . , N . ∂i,j 1 2 2 i,j 1 This implies that u1 − u2 is an affine function. Definition 13.11. Given an open set Ω ⊆ RN , 1 ≤ p ≤ ∞, µ ∈ R, and 1,p (Ω) is a weak solution to −∆u+µu = F ∈ Lploc (Ω; RN ), we say that u ∈ Wloc div F in Ω if Z (∇u · ∇w + µuw + F · ∇w) dx = 0 Ω
for all w ∈
0 W 1,p (Ω)
with compact support in Ω.
Exercise 13.12. Let F ∈ L2 (RN ; RN ). (i) Let µ > 0 and consider the functional Z 1 µ G(v) = k∇v(x)k2 + v 2 (x) + F (x) · ∇v(x) dx, 2 2 RN
v ∈ H 1 (RN ).
Prove that the minimization problem inf v∈H 1 (RN ) G(v) admits a unique solution uµ ∈ H 1 (RN ). (ii) Prove that uµ is the unique weak solution to −∆u + µu = div F . (iii) Prove that Z
2
µu + k∇uk RN
2
Z dx
kF k2 dx.
RN
Exercise 13.13. Let F ∈ L2 (RN ; RN ) and for every µ > 0 let uµ ∈ H 1 (RN ) be the function given in Exercise 13.12. Define vµ (x) := uµ (x) − (uµ )B(0,1) , x ∈ RN .
13.2. H 2 Regularity
531
1 (RN ) and ∇v * (i) Prove that up to a subsequence, vµ * u in Hloc µ 2 N N + 1 N ˙ ∇u in L (R ; R ) as µ → 0 for some function u ∈ H (R ).
(ii) Prove that u is a weak solution to −∆u = div F . Prove that any other weak solution in H˙ 1 (RN ) differs from u by a constant. As a corollary of Theorem 13.9, we can prove interior regularity to the Poisson problem in an open set Ω ⊆ RN . Theorem 13.14 (Interior regularity). Let Ω ⊆ RN be an open set, let 1 (Ω) be a weak solution to f ∈ L2loc (Ω), and let u ∈ Hloc −∆u = f
(13.14)
in Ω.
2 (Ω) and −∆u(x) = f (x) for LN a.e. x ∈ Ω. Moreover, for Then u ∈ Hloc all U b V b Ω, kukL2 (V \U ) k∇ukL2 (V \U ) (13.15) k∇2 ukL2 (U ) + + kf kL2 (V ) . 2 dist(U, ∂V ) dist (U, ∂V )
Proof. Construct a function η ∈ Cc∞ (Ω; [0, 1]) such that η = 1 in U , η = 0 outside V , and 1 1 (13.16) k∇η(x)k , k∇2 η(x)k 2 dist(U, ∂V ) dist (U, ∂V ) for all x ∈ Ω. Take w(x) := η(x)v(x), where v ∈ H 1 (RN ) has bounded support. Then ∇w(x) = v(x)∇η(x) + η(x)∇v(x), and so, using (13.6), we get Z 0 = (η∇u · ∇v + v∇u · ∇η + f ηv) dx ZΩ = (∇(ηu) · ∇v − u∇η · ∇v + v∇u · ∇η + f ηv) dx ZΩ = (∇(ηu) · ∇v + uv∆η + 2v∇u · ∇η + f ηv) dx, Ω
where in the last equality we integrate by parts. Since η has compact support in Ω, by extending ηu to be zero outside Ω, we have that ηu is a weak solution to ∆(ηu) = u∆η + 2∇u · ∇η + f η =: f1 ∈ L2 (RN ).
(13.17)
Hence, by Theorem 13.9 and (13.16), Z Z 2 2 k∇ (ηu)k dx f12 dx RN
Z 1 u2 dx dist4 (U, ∂V ) V \U RN Z Z 1 2 + k∇uk dx + f 2 dx. 2 dist (U, ∂V ) V \U V
532
13. Interior Regularity for the Poisson Problem
Using the fact that η = 1 in U , we have that Z Z 2 2 k∇2 (ηu)k2 dx. k∇ uk dx ≤ RN
U
Next we prove that if f ∈ H k (RN ), then u ∈ H˙ k+2 (RN ). The idea here is to differentiate the equation −∆u = f with respect to xi to get −∆(∂i u) = ∂i f ∈ L2 (RN ) and then to apply Theorem 13.9 with ∂i u in place of u. Theorem 13.15. Let f ∈ H m (RN ), m ∈ N, and u ∈ H˙ 2 (RN ) be a weak solution to (13.5). Then u ∈ H˙ m+2 (RN ) and k∇2+i ukL2 (RN ) k∇i f kL2 (RN ) for every i = 1, . . . , m. Proof. We only give the proof for m = 1. Given w ∈ Cc∞ (RN ) take v := ∂i w ∈ Cc∞ (RN ) in (13.6) to find Z Z ∇(∂i w) · ∇u dx + ∂i wf dx = 0. RN
RN
Since u ∈ H˙ 2 (RN ), we can integrate by parts the previous equation to get Z Z ∇w · ∇(∂i u) dx + w∂i f dx = 0. RN
RN
By density, this equation holds for all w ∈ H 1 (RN ) with bounded support. Hence, ∂i u ∈ H˙ 1 (RN ) is a weak solution to −∆(∂i u) = ∂i f , and so, we are in a position to apply Theorem 13.9 to the function ∂i u to conclude that ∂i u ∈ H˙ 2 (RN ), with k∇2 (∂i u)kL2 (RN ) k∂i f kL2 (RN ) . m (Ω) for m ∈ N, Exercise 13.16. Let Ω ⊆ RN be an open set, let f ∈ Hloc 1 (Ω) be a weak solution to (13.14). Prove that u ∈ H m+2 (Ω) and let u ∈ Hloc loc and that for all U b V b Ω and j = 2, . . . , m + 2, j
k∇ ukL2 (U )
j−2 X i=0
+
1 dist
j−2−i
(U, ∂V )
k∇i f kL2 (V )
1 1 kukL2 (V \U ) + k∇ukL2 (V ) . j−1 dist (U, ∂V ) dist (U, ∂V ) j
13.3. W 2,p Regularity In this section we study the Poisson equation −∆u = f for f ∈ Lp (RN ). If u ∈ L1loc (RN ) we define the sharp maximal function of u as Z 1 u(y) − uQ  dy. (13.18) M# (u)(x) := sup N Q: x∈Q L (Q) Q
13.3. W 2,p Regularity
533
We also define the dyadic maximal function of u as Z 1 d (13.19) M (u)(x) := sup u(y) dy, N Q dyadic: x∈Q L (Q) Q A dyadic cube Q has the form Q = 2k (j + [0, 1)N ), where k ∈ Z and j ∈ ZN . Theorem 13.17 (Fefferman–Stein). If 1 < p ≤ q < ∞ and u ∈ Lp (RN ), then k Md (u)kLq (RN ) k M# (u)kLq (RN ) . We begin with an auxiliary result. Lemma 13.18. Let u ∈ L1loc (RN ) and let u ¯ be a representative of u. Then d N N ¯ u(x) ≤ M (u)(x) for L a.e. x ∈ R . Proof. For every Lebesgue point x of u ¯, we have Z 1 lim N ¯ u(y) − u ¯(x) dy = 0. + r→0 r Q(x,r) Given 0 < ε < 1, by the density of dyadic numbers, for every r > 0 we can find a dyadic cube Qr containing x such that Q(x, r) ⊆ Qr and LN (Qr ) ≤ (1 + ε)rN , and so, Z Z 1 1 ¯ u(x) ≤ N ¯ u(y) − u ¯(x) dy + N ¯ u(y) dy r r Q(x,r) Q(x,r) Z Z 1 (1 + ε) ¯ u(y) dy ≤ N ¯ u(y) − u ¯(x) dy + N r L (Qr ) Qr Q(x,r) Z 1 ≤ N ¯ u(y) − u ¯(x) dy + (1 + ε) Md (u)(x). r Q(x,r) Letting r → 0+ gives ¯ u(x) ≤ (1 + ε) Md (u)(x). Finally, we let ε → 0+ to obtain that ¯ u(x) ≤ Md (u)(x) for every Lebesgue point of u ¯. Exercise 13.19. Let u ∈ L1 (RN ). Prove that for every t > 0, Z 1 LN ({x ∈ RN : Md (u)(x) > t}) u(x) dx. t RN We turn to the proof of the Fefferman–Stein theorem (Theorem 13.17). Proof of Theorem 13.17. Step 1: Let t, δ > 0. We claim that (13.20)
LN ({x ∈ RN : Md (u)(x) > 2t, M# (u)(x) ≤ δt}) ≤ 2N δLN ({x ∈ RN : Md (u)(x) > t}).
534
13. Interior Regularity for the Poisson Problem
Let Et := {x ∈ RN : Md (u)(x) > t}. If x ∈ Et , then there exists a dyadic cube Q such that x ∈ Q and uQ > t. Let Q be the largest such dyadic cube containing x such that uQ > t. Note that such a cube exists, since by H¨older’s inequality if Qr is a dyadic cube containing x and with sidelength r, then
uQr
1 = N r
0
rN/p 1 u dy ≤ N kukLp (Qr ) ≤ N/p kukLp (RN ) → 0 r r Qr
Z
as r → ∞. Hence, uQr ≤ t for all r sufficiently large. Note that if y ∈ Q, then Md (u)(y) > t and Q is the largest such dyadic cube containing y such that uQ > t (a larger cube would contain also x). This shows that Q ⊆ Et and that Et can be written as a countable disjoint union of maximal dyadic cubes. Fix one of these cubes, say Q0 . To prove (13.20), it is enough to show that
(13.21)
LN ({x ∈ Q0 : Md (u)(x) > 2t, M# (u)(x) ≤ δt}) ≤ δ2N LN (Q0 ).
If the lefthand side of the previous inequality is zero, then there is nothing to prove. Thus, assume that there exists x1 ∈ Q0 such that Md (u)(x1 ) > 2t and M# (u)(x1 ) ≤ δt. Consider the dyadic cube 2Q0 containing Q0 and of twice the sidelength of Q0 . By the maximality of Q0 , we have that u2Q0 ≤ t. We claim that if x ∈ Q0 and Md (u)(x) > 2t, then Md (uχQ0 )(x) > 2t. To see this, note that by the maximality of Q0 , for every dyadic cube Q containing Q0 , we must have uQ0 ≥ uQ . Hence, in the definition of Md (u)(x) it suffices to consider dyadic cubes contained in Q0 . But if Q ⊆ Q0 , then in Q we have that uχQ0 = u, and so, uQ = (χQ0 u)Q , which shows that Md (uχQ0 )(x) = Md (u)(x). This proves the claim. In turn, Md ((u − u2Q0 )χQ0 )(x) ≥ Md (uχQ0 )(x) − u2Q0 > 2t − t = t, where we used the fact the sublinearity of Md and the fact that Md (u2Q0 χQ0 ) ≤ Md (u2Q0 ) ≤ u2Q0 ≤ t.
13.3. W 2,p Regularity
535
This shows that if Md (u)(x) > 2t, then Md ((u − u2Q0 )χQ0 )(x) > t. By the weak L1 inequality for Md (see Exercise 13.19), LN ({x ∈ Q0 : Md (u)(x) > 2t, M# (u)(x) ≤ δt}) ≤ LN ({x ∈ Q0 : Md ((u − u2Q0 )χQ0 )(x) > t}) Z Z 1 LN (2Q0 ) 1 u − u2Q0  dx = u − u2Q0  dx ≤ t Q0 t LN (2Q0 ) 2Q0 ≤
LN (2Q0 ) # M (u)(x1 ) ≤ 2N δLN (Q0 ), t
where we use the fact that M# (u)(x1 ) ≤ δt. Hence, (13.21) holds. Step 2: By the continuity of Md in Lp (RN ) for p > 1 (see [Leo22c]), we have that Md (u) ∈ Lp (RN ). By the layercake representation, k Md (u)kpLp (RN ) Z ∞ =p λp−1 LN ({x ∈ RN : Md (u)(x) > λ}) dλ < ∞. 0
In particular, for n ∈ N and q ≥ p, we have Z n In := q λq−1 LN ({x ∈ RN : Md (u)(x) > λ}) dλ 0 Z n q−p λp−1 LN ({x ∈ RN : Md (u)(x) > λ}) dλ < ∞. ≤ qn 0
On the other hand, by the change of variables λ = 2t, we can estimate Z n/2 In = 2q q tq−1 LN ({x ∈ RN : Md (u)(x) > 2t}) dt 0
≤ 2q q
n/2
Z
tq−1 LN ({x ∈ RN : Md (u)(x) > 2t, M# (u)(x) ≤ δt}) dt
0
+ 2q q
n/2
Z
tq−1 LN ({x ∈ RN : M# (u)(x) > δt}) dt.
0
By (13.20) and the change of variables s = δt, we see that the righthand side of the previous inequality is bounded from above by Z n/2 q+N δ2 q tq−1 LN ({x ∈ RN : Md (u)(x) > t}) dt 0
2q + qq δ
Z
δn/2
sq−1 LN ({x ∈ RN : M# (u)(x) > s}) ds 0 Z ∞ 2q q+N ≤ δ2 In + q q sq−1 LN ({x ∈ RN : M# (u)(x) > s}) ds. δ 0
536
13. Interior Regularity for the Poisson Problem
Let δ be such that δ2q+N = 12 . Since In < ∞, it follows that Z ∞ 1 2q 2q sq−1 LN ({x ∈ RN : M# (u)(x) > s}) ds = q k M# (u)kqLq (RN ) . In ≤ q q 2 δ δ 0 Letting n → ∞ and using the Lebesgue monotone convergence theorem and the layercake representation, we get k Md (u)kqLq (RN ) k M# (u)kqLq (RN ) . We are ready to study the Poisson problem in the case f ∈ Lp (RN ), p > 2. Theorem 13.20 (Lp regularity, p > 2). Let f ∈ Lp (RN ), p > 2. Then ˙ 2,p (RN ) such that −∆u(x) = f (x) (13.5) admits a weak solution u ∈ W for LN a.e. x ∈ RN and k∇2 ukLp (RN ) kf kLp (RN ) . Moreover, any other ˙ 2,p (RN ) differs from u by an affine weak solution to (13.5) that belongs to W function. Proof. Step 1: Let f ∈ Cc∞ (RN ). By Theorem 13.9 there exists a weak solution u ∈ H˙ 2 (RN ) to (13.5). Moreover, by Theorem 13.15, u ∈ H˙ m+2 (RN ) for every m ∈ N, with k∇2+i ukL2 (RN ) k∇i f kL2 (RN ) for every i = 1, . . . , m. It follows from Morrey’s embedding theorem applied to ∇2 u (see [Leo22d]) that ∇2 u belongs to L∞ (RN ; RN ×N ). Since ∇2 u ∈ L2 (RN ; RN ×N ), by interpolation, ∇2 u ∈ Lq (RN ; RN ×N ) for every 2 ≤ q ≤ ∞. Fix x0 ∈ RN and let Qr be a cube containing x0 and sidelength r. We denote by Qs the cube with the same center of Qr and sidelength s > 0. For R ≥ r, write u = v + w, where v solves Dirichlet’s problem −∆v = 0 in Q2R , v=u on ∂Q2R . To prove the existence of v ∈ H 1 (Q2R ), it is enough to minimize the functional Z 1 k∇ψk2 dx F (ψ) := 2 Q2R over all ψ ∈ H 1 (Q2R ) such that Tr(ψ) = Tr(u) on ∂QR . We leave this as an 2v exercise. By Cacciopoli’s inequality applied to the harmonic function ∂x∂i ∂x j (see Theorem 13.7), Z r N +2 Z 2 2 2 (13.22) k∇ v − (∇ v)Qr k dx k∇2 v − (∇2 v)QR k2 dx. R Qr QR On the other hand, since w is a solution of the problem −∆w = f in Q2R , (13.23) w=0 on ∂Q2R ,
13.3. W 2,p Regularity
537
by interior regularity (Theorem 13.14 with Ω = Q2R , U = QR , and V = Q3R/2 ), k∇2 wk2L2 (QR ) kf k2L2 (Q2R ) +
(13.24)
1 k∇wk2L2 (Q2R ) , R2
where we use Poincar´e’s inequality in H01 (Q2R ) (see [Leo22d]), that is, kwkL2 (Q2R ) Rk∇wkL2 (Q2R ) .
(13.25)
Multiplying the equation by w and integrating by parts, we get Z Z 2 wf dx ≤ kf kL2 (Q2R ) kwkL2 (Q2R ) k∇wk dx = Q2R
Q2R
Rkf kL2 (Q2R ) k∇wkL2 (Q2R ) , again by Poincar´e’s inequality (13.25). Rkf kL2 (Q2R ) , and, in turn,
It follows that k∇wkL2 (Q2R )
k∇2 wk2L2 (QR ) kf k2L2 (Q2R ) .
(13.26)
By (13.22) and the fact that u = v + w, Z k∇2 u − (∇2 u)Qr k2 dx Qr Z Z 2 2 2 ≤2 k∇ v − (∇ v)Qr k dx + 2 k∇2 w − (∇2 w)Qr k2 dx Qr Qr Z r N +2 Z 2 2 2 k∇ v − (∇ v)QR k dx + k∇2 wk2 dx, R QR QR where we use the fact that, by H¨older’s inequality, Z Z 2 2 N 2 2 k(∇ w)Qr k dx = r k(∇ w)Qr k Qr
k∇2 wk2 dx.
Qr
Using the fact that v = u − w and (13.26), it follows that Z k∇2 u − (∇2 u)Qr k2 dx Qr Z r N +2 Z k∇2 u − (∇2 u)QR k2 dx + k∇2 wk2 dx R QR QR Z r N +2 Z k∇2 uk2 dx + f 2 dx. R QR Q2R Taking R = kr, where k ∈ N, we get Z Z Z 1 2 2 2 2 2 k∇ u − (∇ u)Qr k dx N +2 k∇ uk dx + f 2 dx, k Qr Qkr Q2kr
538
13. Interior Regularity for the Poisson Problem
and so, 1 rN
Z
k∇2 u − (∇2 u)Qr k2 dx
Qr
Z
1
2
2
k∇ uk dx +
k 2 (kr)N
kN
Z N
(kr) Qkr Q2kr 1 2 M(k∇2 uk2 )(x0 ) + k N M(f 2 )(x0 ), k
f 2 dx
where M is the maximal function. Finally, by H¨older’s inequality, Z 1/2 Z 1 rN/2 2 2 2 2 2 k∇ u − (∇ u)Qr k dx ≤ N k∇ u − (∇ u)Qr k dx r N Qr r Qr 1 [M(k∇2 uk2 )(x0 )]1/2 + k N/2 [M f 2 (x0 )]1/2 . k Since the previous inequality holds for every cube Qr of sidelength r containing x0 , by (13.18) we get M# (∇2 u)(x0 )
1 [M(k∇2 uk2 )(x0 )]1/2 + k N/2 [M(f 2 )(x0 )]1/2 . k
By the Fefferman–Stein theorem (Theorem 13.17) and Lemma 13.18, k∇2 ukLp (RN ) ≤ k Md (∇2 u)kLp (RN ) ≤ k M# (∇2 u)kLp (RN ) 1 k[M(k∇2 uk2 )]1/2 kLp (RN ) + k N/2 k[M(f 2 )]1/2 kLp (RN ) . k Since
p 2
> 1, by the continuity of the maximal function (see [Leo22c]) "Z 2/p #1/2 k[M(k∇2 uk2 )]1/2 kLp (RN ) = [M(k∇2 uk2 )]p/2 dx RN
"Z
k∇2 ukp dx
2/p #1/2
RN
= k∇2 ukLp (RN ) .
Similarly, k[M(f 2 )]1/2 kLp (RN ) kf kLp (RN ) . Hence, k∇2 ukLp (RN ) ≤
C k∇2 ukLp (RN ) + Ck N/2 kf kLp (RN ) . k
Taking k so large that Ck < 12 , we conclude that k∇2 ukLp (RN ) kf kLp (RN ) . Note that here we used the fact that ∇2 u ∈ Lp (RN ; RN ×N ). Step 2: Take fn ∈ Cc∞ (RN ) such that fn → f in Lp (RN ) and let un be the solutions constructed in the previous step corresponding to fn . Then k∇2 (un − um )kLp (RN ) kfn − fm kLp (RN ) → 0
13.3. W 2,p Regularity
539
as n, m → ∞. Thus, ∇2 un → v in Lp (RN ; RN ×N ). By considering un − Tn , where Tn is an affine function with Z Z Z Z un dx = Tn dx, ∇un dx = ∇Tn dx, B(0,1)
B(0,1)
B(0,1)
B(0,1)
it follows from Poincare’s inequality in bounded domains ([Leo22d]) and a diagonal argument that a subsequence, not relabeled, of {un −Tn }n converges 2,p 2,p (RN ) with ∇2 u = v. Using the weak (RN ) to a function u ∈ Wloc in Wloc form of the equation −∆un = −∆(un − Tn ) = fn , we get that −∆u = f . Since k∇2 un kLp (RN ) kfn kLp (RN ) , letting n → ∞ we get k∇2 ukLp (RN ) kf kLp (RN ) . ˙ 2,p (RN ) be two weak solutions to Step 3: Uniqueness. Let u1 , u2 ∈ W (13.5). Then Z ∇(u1 − u2 ) · ∇w dx = 0 RN 0
for all w ∈ W 1,p (RN ) with compact support. Hence, by Remark 13.3 the 2 (u − u ). In turn, by Corollary function u1 − u2 is harmonic and so is ∂i,j 1 2 13.4, !1/p Z 1 2 2 p sup ∂i,j (u1 − u2 )(x) ≤ ∂i,j (u1 − u2 )(y) dy (αN R)N/p B(x0 ,2R) B(x0 ,R) for every x0 ∈ RN and R > 0. Letting R → ∞ and using the fact that 2 (u −u ) ∈ Lp (RN ), we obtain that ∂ 2 (u −u ) = 0 for all i, j = 1, . . . , N . ∂i,j 1 2 2 i,j 1 This implies that u1 − u2 is an affine function. Exercise 13.21. Let 1 < p < ∞ and f : RN → R be a Lebesgue measurable 0 function. Assume that f g ∈ L1 (RN ) for every g ∈ Lp (RN ) ∩ L2 (RN ) and that there exists a constant C > 0 such that Z f g dx ≤ CkgkLp0 (RN ) . RN
Prove that f ∈ Lp (RN ) with kf kLp (RN ) ≤ C. Exercise 13.22. Let F ∈ Lp (RN ), p > 2. (i) Assume that F ∈ L2 (RN ; RN ) ∩ Lp (RN ; RN ) and let u be the solution to −∆u = div F in RN given in Exercise 13.13. Fix x0 ∈ RN and let Qr be a cube containing x0 . For R ≥ r write u = v + w, where v solves Dirichlet’s problem −∆v = 0 in Q2R , v=u on ∂Q2R .
540
13. Interior Regularity for the Poisson Problem
Prove that Z r N +2 Z k∇v − (∇v)Qr k2 dx k∇v − (∇v)QR k2 dx R Qr QR and that k∇wkL2 (Q2R ) kF kL2 (Q2R ) . (ii) Let F and u be as in item (i). Prove that Z Z r N +2 Z 2 2 k∇u − (∇u)Qr k dx k∇uk dx + kF k2 dx. R Qr QR Q2R Deduce that 1 [M(k∇uk2 )(x0 )]1/2 + k N/2 [M(kF k2 )(x0 )]1/2 . k (iii) Let F and u be as in item (i). Prove that k∇ukLp (RN ) kF kLp (RN ) . M# (∇u)(x0 )
(iv) Let F ∈ Lp (RN ), p > 2. Prove that there exists a weak solution u to −∆u = div F in RN and that k∇ukLp (RN ) kF kLp (RN ) . Prove that ˙ 1,p (RN ) differs from u by a constant. any other weak solution in W Theorem 13.23 (Lp regularity, 1 < p < 2). Let f ∈ Lp (RN ), 1 < p < 2. ˙ 2,p N Then (13.5) admits a weak solution
u ∈ W (R ) such that −∆u(x) = f (x) 2 N N
for L a.e. x ∈ R and ∇ u Lp (RN ) kf kLp (RN ) . Moreover, any other ˙ 2,p (RN ) differs from u by an affine weak solution to (13.5) that belongs to W function. Proof. Step 1: Let f ∈ Lp (RN ) ∩ L2 (RN ). By Theorem 13.9, there exists a weak solution u ∈ H˙ 2 (RN ) to (13.5). Given w ∈ Cc∞ (RN ), integrating by parts, we have Z Z Z Z 2 2 2 2 ∂i,j u∆w dx = u∆∂i,j w dx = ∆u∂i,j w dx = f ∂i,j w dx. RN
RN
RN
RN
Given w ∈ H˙ 2 (RN ) we can construct a sequence wn ∈ Cc∞ (RN ) such that 2 w → ∂ 2 w in L2 (RN ) for all i, j = 1, . . . , N (see [Leo22d]). Then by ∂i,j n i,j what we just proved Z Z 2 2 ∂i,j u∆wn dx = f ∂i,j wn dx RN
RN
2 u, f ∈ L2 (RN ), we can let n → ∞ to obtain for every n. Since ∂i,j Z Z 2 2 ∂i,j u∆w dx = f ∂i,j w dx (13.27) RN
RN
for all w ∈ H˙ 2 (RN ). 0
Given ϕ ∈ Lp (RN ) ∩ L2 (RN ), by Theorems 13.9 and Theorem 13.20, ˙ 2,p (RN ) to Poisson’s probthere exists a weak solution w ∈ H˙ 2 (RN ) ∩ W N 2 lem −∆w = ϕ in R , with k∇ wkL2 (RN ) kϕkL2 (RN ) , k∇2 wkLp0 (RN )
13.4. W s,p Regularity
541
kϕkLp0 (RN ) . Hence, by (13.27) and H¨older’s inequality Z Z Z 2 2 2 f ∂i,j w dx ∂i,j u∆w dx = − ∂i,j uϕ dx = − N N N R R R
≤ kf kLp (RN ) ∇2 w Lp0 (RN ) kf kLp (RN ) kϕkLp0 (RN ) . It follows from Exercise 13.21 that 2 k∂i,j ukLp (RN ) Z 2 p0 N 2 N = sup ∂i,j uϕ dx : ϕ ∈ L (R ) ∩ L (R ), kϕkLp0 (RN ) ≤ 1 RN
kf kLp (RN ) . Step 2: If f ∈ Lp (RN ), 1 < p < 2, we construct a sequence {fn }n of functions in L2 (RN ) ∩ Lp (RN ) such that fn → f in Lp (RN ) and proceed as in Step 2 of the proof of Theorem 13.20. The proof of the uniqueness of solutions is identical to that of Theorem 13.20. Exercise 13.24. Let F ∈ Lp (RN ), 1 < p < 2. Prove that there exists a ˙ 1,p (RN ) to −∆u = div F in RN , that k∇ukLp (RN ) weak solution u ∈ W ˙ 1,p (RN ) differs from u by kF kLp (RN ) , and that any other weak solution in W a constant. Theorem 13.25 (Interior Lp regularity). Let Ω ⊆ RN be an open set, 1,p (Ω) be a weak solution to let f ∈ Lp (Ω), 1 < p < ∞, and let u ∈ Wloc 2 Poisson’s equation −∆u = f in Ω. Then ∇ u ∈ Lploc (Ω). Moreover, for all U b V b Ω, 1 k∇2 ukLp (U ) kukLp (V \U ) 2 dist (U, ∂V ) 1 + k∇ukLp (V \U ) + kf kLp (V ) . dist(U, ∂V ) The proof is similar to that of Theorem 13.14 and is left as an exercise. Reasoning as in the proof of Theorem 13.15, we have the following result. Theorem 13.26. Let f ∈ W m,p (RN ), m ∈ N, 1 < p < ∞, and let u ∈ ˙ 2,p (RN ) be a weak solution to (13.5). Then u ∈ W ˙ m+2,p (RN ) and W k∇2+i ukLp (RN ) k∇i f kLp (RN )
for every i = 1, . . . , m.
13.4. W s,p Regularity In this section we will prove that if f ∈ W s,p (RN ), 1 < p < ∞, 0 < s < 1, ˙ s+2,p (RN ) then there exists a unique solution (up to affine functions) u ∈ W 2 to the Poisson problem (13.5), with k∇ ukW s,p (RN ) kf kW s,p (RN ) .
542
13. Interior Regularity for the Poisson Problem
Theorem 13.27 (W s,p regularity). Let f ∈ W s,p (RN ), 1 < p < ∞, 0 < ˙ s+2,p (RN ) such that s < 1. Then (13.5) admits a weak solution u ∈ W −∆u(x) = f (x) for LN a.e. x ∈ RN and k∇2 ukLp (RN ) kf kLp (RN ) ,
∇2 uW s,p (RN ) f W s,p (RN ) .
˙ s+2,p (RN ) ∩ Moreover, every other weak solution to (13.5) that belongs to W 2,p N ˙ (R ) differs from u by an affine function. W Proof. Since f ∈ Lp (RN ) by Theorems 13.9, 13.20, and 13.23 there exists ˙ 2,p (RN ) to (13.5) with a weak solution u ∈ W Z Z 2 p (13.28) k∇ u(x)k dx f (x)p dx. RN
RN
RN
Fix h ∈ \ {0}. Then the function u(· + h) is a weak solution to (13.5) with f replaced by f (· + h). Hence, by linearity, u(· + h) − u(·) is a weak solution to ∆[u(x + h) − u(x)] = f (x + h) − f (x). Again, by Theorems 13.9, 13.20, and 13.23, we have Z Z 2 2 p k∇ u(x + h) − ∇ u(x)k dx f (x + h) − f (x)p dx. RN
RN
khkN +sp
Dividing both sides by and integrating in h gives Z Z Z Z f (x + h) − f (x)p k∇2 u(x + h) − ∇2 u(x)kp dxdh dxdh. khkN +sp khkN +sp RN RN RN RN 2 u ∈ W s,p (RN ). This shows that ∂i,j
˙ s+2,p (RN ) ∩ The uniqueness of u up to affine functions in the class W follows from uniqueness in Theorems 13.9, 13.20, and 13.23.
˙ 2,p (RN ) W
Exercise 13.28. Let F ∈ W s,p (RN ; RN ), 1 < p < ∞, 0 < s < 1. Prove that ˙ s+1,p (RN ) ∩ W ˙ 1,p (RN ) to −∆u = div F in there exists a weak solution u ∈ W N R , with k∇ukLp (RN ) kF kLp (RN ) and ∇uW s,p (RN ) F W s,p (RN ) . Prove ˙ s+1,p (RN ) ∩ W ˙ 1,p (RN ) that every other weak solution that belongs to W differs from u by a constant. Theorem 13.29 (Interior W s,p regularity). Let Ω ⊆ RN be an open set, let 1,p f ∈ W s,p (Ω), 1 < p < ∞, 0 < s < 1, and let u ∈ Wloc (Ω) be a weak solution to Poisson’s equation −∆u = f in Ω, s,p where f ∈ Lp (Ω). Then ∇2 u ∈ Wloc (Ω). Moreover, for all U b V b Ω,
∇2 uW s,p (U ) δ kukW s,p (V ) + k∇ukW s,p (V ) + kf kW s,p (V ) , where δ := dist(U, ∂V ).
13.4. W s,p Regularity
543
Proof. As in the proof of Theorem 13.14, the function ηu, extended to be zero outside Ω, is a weak solution to Poisson’s equation ∆(ηu) = u∆η + 2∇u · ∇η + f η =: f1
in RN .
Moreover, k∇n ηk∞
1 dist (U, ∂V ) n
for n = 1, 2, 3.
1,p s,p Since u ∈ Wloc (Ω), by Theorem 6.21, we have that u ∈ Wloc (Ω). In turn, s,p N by Theorem 6.23, f1 ∈ W (R ), with 3 s 2 f1 W s,p (RN ) k∇2 ηk1−s ∞ k∇ ηk∞ kukLp (V ) + k∇ ηk∞ uW s,p (V ) 2 s + k∇ηk1−s ∞ k∇ ηk∞ k∇ukLp (V ) + k∇ηk∞ ∇uW s,p (V )
+ k∇ηks∞ kf kLp (V ) + f W s,p (V ) 1 1 1 2+s kukLp (V ) + 2 uW s,p (V ) + 1+s k∇ukLp (V ) δ δ δ 1 1 + ∇uW s,p (V ) + s kf kLp (V ) + f W s,p (V ) . δ δ We can now apply Theorem 13.27 to obtain ∇2 (ηu)W s,p (RN ) f1 W s,p (RN ) .
Reasoning as in the proof of Theorem 13.15, we have the following result. Theorem 13.30. Let f ∈ W s,p (RN ), let s > 0, s ∈ / N, 1 < p < ∞, and ˙ s−bsc+2,p (RN ) ∩ W ˙ 2,p (RN ) be a weak solution to (13.5). Then let u ∈ W ˙ s+2,p (RN ), with u∈W k∇2+i ukLp (RN ) k∇i f kLp (RN ) , ∇2+bsc uW s−bsc,p (RN ) ∇bsc f W s−bsc,p (RN ) for every i = 1, . . . , bsc. ˙ s,p (RN ). Next we consider the case in which f ∈ W ˙ s,p regularity). Let f ∈ W ˙ s,p (RN ), 1 < p < ∞, 0 < Theorem 13.31 (W ˙ s+2,p (RN ) such that s < 1. Then (13.5) admits a weak solution u ∈ W N N 2 −∆u(x) = f (x) for L a.e. x ∈ R and ∇ uW s,p (RN ) f W s,p (RN ) . ˙ s,p (RN ), by Theorem 6.107 we can find a sequence Proof. Given f ∈ W {fn }n of functions in Cc∞ (RN ) such that fn − f W s,p (RN ) → 0. Let un ∈ ˙ s+2,p (RN ) be a weak solution to −∆un = fn in RN with W k∇2 un kLp (RN ) kfn kLp (RN ) ,
∇2 un W s,p (RN ) fn W s,p (RN ) .
˙ s,p (Theorem 6.33) for every R ≥ 2, By Poincar´e’s inequality in W Z (13.29) k∇2 un (x) − (∇2 un )B(0,1) kp dx RN +sp ∇2 un pW s,p (RN ) . B(0,R)
544
13. Interior Regularity for the Poisson Problem
Set An := (∇2 un )B(0,1) and define 1 vn (x) := un (x) − An xT · x. 2 Then Z B(0,R)
k∇2 vn (x)kp dx RN +sp ∇2 un pW s,p (RN )
for every R ≥ 2. Moreover, since (∇2 vn )B(0,1) = 0, Z B(0,R)
k∇2 (vn − vk )(x)kp dx RN +sp ∇2 (vn − vk )pW s,p (RN ) = RN +sp ∇2 (un − uk )pW s,p (RN ) RN +sp fn − fk pW s,p (RN ) → 0
2 v } is a Cauchy sequence in Lp (B(0, R)) for every as n, k → ∞. Thus, {∂i,j n n R ≥ 2. 2,p By Poincar´e’s inequality in Wloc (RN ) (see [Leo22d]), we can find a polynomial pn of degree one such that for every R ≥ 2, Z Z p vn (x) − pn (x) dx R k∇2 vn (x)kp dx, B(0,R) B(0,R) Z Z k∇vn (x) − ∇pn (x)kp dx R k∇2 vn (x)kp dx. B(0,R)
B(0,R)
Also, (vn −vk )(x)−(pn −pk )(x)p dx R
Z
k∇(vn −vk )(x)−∇(pn −pk )(x)kp dx R
Z
Z B(0,R)
Z B(0,R)
k∇2 (vn −vk )(x)kp dx,
B(0,R)
k∇2 (vn −vk )(x)kp dx.
B(0,R)
Hence, if we define the function wn := vn − pn , we have that {wn }n is a Cauchy sequence in W 2,p (B(0, R)) for every R ≥ 2 with ∇2 wn W s,p (RN ) = ˙ s+2,p (RN ) such that ∇2 un W s,p (RN ) . Thus, we can find a function w ∈ W 2,p wn → w in Wloc (RN ). By extracting a subsequence, not relabeled, we may also assume that ∇2 wn → ∇2 w pointwise LN a.e. in RN . ˙ s,p , for every R ≥ 2, Again by Poincar´e’s inequality in W Z B(0,R)
(fn − f )(x) − (fn − f )B(0,1) kp dx RN +sp fn − f pW s,p (RN ) → 0.
13.5. Notes
545
R Hence, if ϕ ∈ Cc∞ (RN ) is such that RN ϕ dx = 0, by also using the fact that 2,p wn → w in Wloc (RN ), we have Z Z Z ϕ(∆un − trace An ) dx ϕ∆wn dx = lim ϕ∆w dx = lim n→∞ RN n→∞ RN RN Z Z = − lim ϕfn dx = − lim ϕ(fn − (fn )B(0,1) ) dx n→∞ RN n→∞ RN Z Z ϕf dx. ϕ(f − fB(0,1) ) dx = − =− RN RN R R Since RN ϕ(∆w + f ) dx = 0 for all ϕ ∈ Cc∞ (RN ) such that RN ϕ dx = 0, there exists a constant c0 such that ∆w(x) + f (x) = c0 for LN a.e. x ∈ RN . Consider a symmetric matrix A0 such that trace A0 = c0 and define u(x) := w(x) − 12 A0 xT · x. Then −∆u(x) = f (x) for LN a.e. x ∈ RN . Moreover, since ∇2 wn → ∇2 w pointwise LN a.e. in RN , it follows by Fatou’s lemma that ∇2 uW s,p (RN ) = ∇2 wW s,p (RN ) ≤ lim inf ∇2 wn W s,p (RN ) n→∞
2
= lim inf ∇ un W s,p (RN ) n→∞
lim inf fn W s,p (RN ) = f W s,p (RN ) . n→∞
˙ s,p (RN ; RN ), 1 < p < ∞, 0 < s < 1. Prove Exercise 13.32. Let F ∈ W ˙ s+1,p (RN ) to −∆u = div F in RN , that there exists a weak solution u ∈ W with ∇uW s,p (RN ) F W s,p (RN ) .
13.5. Notes The proof of Theorem 13.9 is adapted from the lecture notes of S. M¨ uller [M¨ ul18]. The general approach in this chapter is inspired by Krylov’s book [Kry08]. It avoids the theory of singular integrals to obtain Lp regularity.
Chapter 14
The Fractional Laplacian Geometry is the art of correct reasoning from incorrectly drawn figures. — Henri Poincare
In this chapter we introduce the fractional Laplacian and study some of its properties. We are only going to scratch the surface here. As I explained in the introduction, there is a massive amount of literature on the fractional Laplacian. We refer to the survey papers of Abatangelo and Valdinoci [AV19], Garofalo [Gar19], RosOton [RO16a], and the book of Bucur and Valdinoci [BV16] for more information on the subject.
14.1. Definition and Main Properties In this section we will use Fourier transforms. We refer to [Leo22c] for relevant definitions and properties. Given φ ∈ S(RN ; C), for x ∈ RN , c −∆φ(x) =−
N d N X X ∂2φ b b (2πxj )2 φ(x) = (2π)2 kxk2 φ(x). (x) = 2 ∂x j j=1 j=1
The fractional Laplacian (−∆)s is obtained by replacing (2π)2 kxk2 with b (2π)2s kxk2s . In order for x 7→ kxk2s φ(x) to be locally integrable, one needs to assume that either −N < 2s or −N ≥ 2s and that φb and a sufficient number of its derivatives vanish at x = 0. For simplicity, we will focus here on the first case. Given φ ∈ S(RN ; C) and −N/2 < s < N/2, consider the function b g(x) := (2π)2s kxk2s φ(x), x ∈ RN . In general, the function g does not belong to S(RN ; C) since kxk2s or its derivatives eventually become singular at x = 0. However, g belongs to 547
548
14. The Fractional Laplacian
L1 (RN ; C) (exercise), so the inverse Fourier transform of g is well defined. Hence, we can define the fractional Laplacian (−∆)s of φ as Z s ∨ 2s b dy. e2πix·y kyk2s φ(y) (14.1) (−∆) φ(x) := g (x) = (2π) RN
Since g does not belong to S(RN ; C), neither does (−∆)s φ. However, by differentiating under the integral sign, we have that (−∆)s φ is a C ∞ function. More precisely, let Ss (RN ; C) be the space of all functions ψ : RN → C of class C ∞ such that for all multiindexes α ∈ NN 0 , (14.2)
kψkα,s := sup (1 + kxkN +2s )∂ α ψ(x) < ∞. x∈RN
N The family of seminorms k·kα,s , α ∈ NN 0 , generates a topology for Ss (R ; C). 0 N We denote by Ss (R ; C) the corresponding dual space.
Exercise 14.1. Let −N/2 < s < N/2. Prove that a linear functional T : Ss (RN ; C) → C is continuous if and only if there exists m ∈ N0 such that (14.3)
kT (φ)k kφkα,s
for every φ ∈ S(RN ; C)
for every multiindex α ∈ NN 0 , with 0 ≤ α ≤ m. Theorem 14.2. Let 0 < s < 1 and φ ∈ S(RN ; C). Then for x ∈ RN , Z cN,s φ(x + h) − 2φ(x) + φ(x − h) (14.4) (−∆)s φ(x) = − dh, 2 RN khkN +2s where (14.5)
cN,s := − R 2 RN
(2π)s 1−cos(2πy1 ) kykN +2s
. dy
Moreover, (−∆)s φ belongs to Ss (RN ; C) and for every α ∈ NN 0 , k(−∆)s φkα,s sup (1 + kykN +2 )k∇2 ∂ α φ(y)k y∈RN
+ sup (1 + kykN +1 )∂ α φ(y). y∈RN
Proof. Step 1: We claim that the function Z φ(x + h) − 2φ(x) + φ(x − h) ψ(x) := dh, khkN +2s RN
x ∈ RN ,
14.1. Definition and Main Properties
549
is well defined with kψk0,s < ∞, where k · k0,s is defined in (14.2). For r > 0, write Z φ(x + h) − 2φ(x) + φ(x − h) (14.6) ψ(x) = dh khkN +2s B(0,r) Z φ(x + h) − 2φ(x) + φ(x − h) dh =: A + B. + khkN +2s N R \B(0,r) To estimate A, we use Taylor’s formula of order 2 with remainder term centered at x to write φ(x + h) = φ(x) + ∇φ(x) · h +
N 1 X 2 ∂i,j φ(x)hi hj + R2 (x, h), 2 i,j=1
N 1 X 2 ∂i,j φ(x)hi hj + R2 (x, −h), φ(x − h) = φ(x) − ∇φ(x) · h + 2 i,j=1
where R2 (x, h) :=
N Z X
1 2 2 (1 − t)[∂i,j φ(x + th) − ∂i,j φ(x)]hi hj dt.
i,j=1 0
Adding these two identities gives φ(x + h) + φ(x − h) − 2φ(x) =
N X
2 ∂i,j φ(x)hi hj + R2 (x, h) + R2 (x, −h).
i,j=1
Hence, for h ∈ B(0, r), φ(x + h) + φ(x − h) − 2φ(x) khkN +2s 1 khk2 sup (1 + kykN +2 )k∇2 φ(y)k. 1 + rN +2 khkN +2s y∈B(x,r) In turn, 1 A 1 + rN +2
r−2s 1 + rN
N +2
sup (1 + kyk
2
Z
)k∇ φ(y)k
y∈B(x,r)
B(0,r)
sup (1 + kykN +2 )k∇2 φ(y)k. y∈B(x,r)
On the other hand, by a change of variables, Z φ(x + h) − φ(x) B≤2 dh. khkN +2s N R \B(0,r)
khk2 dh khkN +2s
550
14. The Fractional Laplacian
For h ∈ RN \ B(0, r), φ(x + h) − φ(x) 1 1 ≤ sup (1 + kykN )φ(y) khkN +2s khkN +2s 1 + kxkN y∈RN +
φ(x + h) . khkN +2s
Using spherical coordinates, Z RN \B(0,r)
1 1 dh 2s . khkN +2s r
Hence, B
1 1 sup (1 + kykN )φ(y) r2s 1 + kxkN y∈RN Z 1 + N +2s φ(x + h) dh. r RN \B(0,r)
If x 6= 0, taking r = 12 kxk and combining the estimates for A and B, we get 0 ≤ ψ(x) +
1 kxkN +2s
1 kxkN +2s
sup (1 + kykN +2 )k∇2 φ(y)k y∈RN
sup (1 + kykN )φ(y) + y∈RN
kxkN +2s
while if x = 0, take r = 1. Since Z Z 1 + kykN +1 φ(y) dy = φ(y) dy 1 + kykN +1 RN RN Z N +1 ≤ sup (1 + kξk )φ(ξ) ξ∈RN
Z
1
RN
φ(y) dy, RN
1 dy, 1 + kykN +1
this shows that kψk0,s sup (1 + kykN +2 )k∇2 φ(y)k + sup (1 + kykN +1 )φ(y). y∈RN
y∈RN
Step 2: By the previous step Z Z ψ(y) dy =
1 + kykN +2s ψ(y) dy N +2s RN 1 + kyk Z 1 ≤ kψk0,s dy < ∞. N +2s RN 1 + kyk
RN
Hence, if we set Z ω(x) := RN
φ(x + h) − 2φ(x) + φ(x − h) dh, khkN +2s
x ∈ RN ,
14.1. Definition and Main Properties
551
we have that ω(x) ≤ ψ(x), and so we can apply Fubini’s theorem to obtain Z ω b (x) = e−2πix·y ω(y) dy N Z ZR φ(y + h) − 2φ(y) + φ(y − h) −2πix·y = e dhdy khkN +2s N N R ZR Z 1 e−2πix·y [φ(y + h) − 2φ(y) + φ(y − h)] dydh = N +2s RN khk RN Z Z 1 [e2πix·h − 2 + e−2πix·h ]e−2πix·z φ(z) dzdh = N +2s khk N N R R Z 2πix·h − 2 + e−2πix·h e b = φ(x) dh, khkN +2s RN where we make the change of variables y + h = z and y − h = z. Since for t ∈ R, eit = cos t + i sin t, Z Z e2πix·h − 2 + e−2πix·h 1 − cos(2πx · h) dh = −2 dh N +2s khk khkN +2s RN RN Z 1 − cos(2π x · y) kxkN +2s kxk = −2 dy, N +2s kxkN kyk N R where in the last equality, when x 6= 0, we make the change of variables y h = kxk . We leave it as an exercise to check that Z −2 RN
x · y) 1 − cos(2π kxk
kykN +2s
Z dy = −2 RN
1 − cos(2πy1 ) dy =: DN,s . kykN +2s
b In conclusion, we have shown that ω b (x) = DN,s kxk2s φ(x). Multiplying s both sides by (2π) , taking the inverse Fourier transform (see [Leo22c]), and using (14.1), we obtain Z s s b dy = DN,s (−∆)s φ(x). (2π) ω(x) = (2π) DN,s e2πix·y kyk2s φ(y) RN
This proves (14.4). Step 3: It remains to show that (−∆)s φ ∈ Ss (RN ; C). Since ∂j φ ∈ S(RN ; C), by (14.1) and the formula for the Fourier transform of partial derivatives (see [Leo22c]), we have Z d ∂φ s ∂φ 2s (−∆) (x) = (2π) (y) dy e2πix·y kyk2s ∂xj ∂xj RN Z b dy = (2π)2s e2πix·y (2πiyj )kyk2s φ(y) N R Z 2s ∂ b dy = ∂ ((−∆)s φ) (x), = (2π) e2πix·y kyk2s φ(y) ∂xj RN ∂xj
552
14. The Fractional Laplacian
where we differentiate under the integral sign (see [Leo22c]). Hence, by the ∂φ previous two steps with φ replaced by ∂x , we get that j k(−∆)s ∂j φk0,s sup (1 + kykN +2 )k∇2 ∂j φ(y)k y∈RN
+ sup (1 + kykN +1 )∂j φ(y). y∈RN
We now proceed by induction. We omit the details.
Exercise 14.3. Let 0 < s < 1 and u : RN → R be a bounded function of class C 2 . Prove that for every x ∈ RN and r > 0, Z u(x + h) − 2u(x) + u(x − h) 1 dh 2s kukL∞ (RN ) N +2s khk r RN + r2(1−s) k∇2 ukL∞ (B(x,r)) . Exercise 14.4. Prove that the constant cN,s given in (14.5) is equal to s22s Γ N +2s 2 , π N/2 Γ(1 − s) where Γ is the Gamma function Z ∞ (14.7) Γ(y) := τ y−1 e−τ dτ,
y > 0.
0
Note that the classical Laplacian −∆ is a local operator in the sense that the definition of −∆φ at the point x is determined only by the values of φ in a small neighborhood of x. In contrast, the fractional Laplacian is nonlocal, in the sense that, by (14.4), the value of (−∆)s φ at the point x uses the values of φ far away from x. At the same time, the presence of the term khkN +2s creates a singularity since in order for (−∆)s φ to be well defined, we will need to impose some regularity on the function φ. For this reason, operators of the type (−∆)s are called integraldifferential operators. Exercise 14.5. Let α, s ∈ (0, 1) be such that α + 2(1 − s) < 1. Prove that if u : RN → R is a bounded function of class C 2 with ∇2 u H¨older continuous with exponent α, then (−∆u)s is H¨older continuous with exponent β = α + 2(1 − s). Using Theorem 14.2 we can extend (−∆)s to Ss0 (RN ; C). Theorem 14.6. Let 0 < s < 1 and T ∈ Ss0 (RN ; C). Define (−∆)s T : S(RN ; C) → C by (−∆)s T (φ) := T ((−∆)s φ),
φ ∈ S(RN ; C).
14.1. Definition and Main Properties
553
Then there exists m ∈ NN 0 such that (−∆)s T (φ) sup (1 + kxkN +2 )k∇2 ∂ α φ(x)k x∈RN
+ sup (1 + kxkN +1 )∂ α φ(x) x∈RN s for all α ∈ NN 0 with 0 ≤ α ≤ m. In particular, (−∆) T is a tempered distribution.
Proof. By Exercise 14.1, there exists m ∈ N0 such that T (ψ) kψkα,s for all ψ ∈ Ss (RN ; C) and for all multiindexes α ∈ NN 0 with 0 ≤ α ≤ m. Hence, by Theorem 14.2, (−∆)s T (φ) = T ((−∆)s φ) k(−∆)s φkα,s sup (1 + kykN +2 )k∇2 ∂ α φ(y)k + sup (1 + kykN +1 )∂ α φ(y). y∈RN
y∈RN
This implies that (−∆)s T : S(RN ; C) → C is continuous.
Remark 14.7. In particular, if u ∈ L1loc (RN ) is such that Z u(x) dx < ∞, N +2s RN 1 + kxk then we define Z Tu (φ) := u(x)φ(x) dx, φ ∈ Ss (RN ; C). RN
We have Z Tu (φ) ≤
Z u(x)φ(x) dx =
RN
Z ≤ RN
RN
u(x) (1 + kxkN +2s )φ(x) dx 1 + kxkN +2s
u(x) dxkφk0,s . 1 + kxkN +2s
This shows that Tu ∈ Ss0 (RN ; C). We can now use Theorem 14.6 to show that (−∆)s Tu is a tempered distribution. Next we show that we can write (−∆)s using first order differences instead of second order. Theorem 14.8. Let s ∈ (0, 1) and φ ∈ S(RN ; C). (i) If s ∈ (0, 1/2), then the integral Z φ(x) − φ(y) dy kx − ykN +2s N R
(14.8)
is well defined for every x ∈ RN , with Z φ(x) − φ(y) 1 1−2s k∇φkL∞ (B(x,r)) N kx − ykN +2s dy r2s kφkL∞ (RN ) + r R
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14. The Fractional Laplacian
for every r > 0. Moreover, (14.9)
Z
s
(−∆φ) (x) = cN,s RN
φ(x) − φ(y) dy. kx − ykN +2s
(ii) If s ∈ [1/2, 1), then the ( principal value) limit Z Z φ(x) − φ(y) φ(x) − φ(y) PV dy := lim dy N +2s N +2s + ε→0 RN kx − yk RN \B(x,ε) kx − yk exists for every x ∈ RN and (14.10) Z φ(x) − φ(y) 1 dy kφkL∞ (RN ) + r2(1−s) k∇2 φkL∞ (B(x,r)) RN \B(x,ε) kx − ykN +2s r2s for every r > 0 and every 0 < ε < r. Moreover, Z φ(x) − φ(y) s dy. (−∆φ) (x) = cN,s PV kx − ykN +2s RN Proof. (i) For r > 0, write Z Z Z φ(x) − φ(y) φ(x) − φ(y) φ(x) − φ(y) dy = dy + dy N +2s N +2s kx − yk kx − yk kx − ykN +2s N N R B(x,r) R \B(x,r) =: A + B. To estimate A, we use Taylor’s formula of order 1 with remainder term at x to write φ(y) = φ(x) + ∇φ(x) · (y − x) + R1 (x, y), where Z
1
[∇φ((1 − t)x + ty) − ∇φ(x)] · (x − y) dt.
R1 (x, y) := 0
Hence, for y ∈ B(x, r), φ(x) − φ(y) k∇φkL∞ (B(x,r)) kx − yk, and so, Z A k∇φkL∞ (B(x,r)) B(x,r)
r
1−2s
kx − yk dh kx − ykN +2s
k∇φkL∞ (B(x,r)) ,
where we used the fact that 0 < s < 1/2. On the other hand, Z 1 1 B ≤ 2kφkL∞ (RN ) dy 2s kφkL∞ (RN ) . N +2s kx − yk r N R \B(x,r)
14.1. Definition and Main Properties
555
Combining the estimates for A and B shows that the integral is well defined. In turn, by the changes of variables y = x + h and y = x − h, we can write Z Z Z φ(x) − φ(y) φ(x) − φ(y) φ(x) − φ(y) 1 1 dy = dy + dy N +2s N +2s 2 RN kx − yk 2 RN kx − ykN +2s RN kx − yk Z Z 1 φ(x) − φ(x + h) φ(x) − φ(x − h) 1 = dh + dh N +2s 2 RN khk 2 RN khkN +2s Z φ(x + h) + φ(x − h) − 2φ(x) 1 =− dh, 2 RN khkN +2s which shows (14.9). (ii) For 0 < ε < r, write Z Z φ(x) − φ(y) φ(x) − φ(y) dy = dy N +2s kx − yk kx − ykN +2s B(x,r)\B(x,ε) RN \B(x,ε) Z φ(x) − φ(y) + dy =: D + E. kx − ykN +2s N R \B(x,r) To estimate D, we use Taylor’s formula of order 1 with remainder term at x to write φ(y) = φ(x) + ∇φ(x) · (y − x) + T1 (x, y), where N Z 1 X 2 T1 (x, y) := (1 − t)[∂i,j φ((1 − t)x + ty)](yi − xi )(yj − xj ) dt. i,j=1 0
By the change of variables y