Fourier Meets Hilbert and Riesz: An Introduction to the Corresponding Transforms 9783110784091, 9783110784053

Table of contents :
Preface
Contents
1 Fundamental concepts
2 Fourier series
3 Schwartz spaces S(ℝn)
4 Distribution functions
5 Three-fold approach to the Hilbert transform
6 Hilbert transform in L2(ℝ)
7 Embedding and strong Lp boundedness for the Hilbert transform
8 Riesz transform
Appendix A
Bibliography
Index

Citation preview

René Erlin Castillo Fourier Meets Hilbert and Riesz

De Gruyter Studies in Mathematics

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Edited by Carsten Carstensen, Berlin, Germany Gavril Farkas, Berlin, Germany Nicola Fusco, Napoli, Italy Fritz Gesztesy, Waco, Texas, USA Niels Jacob, Swansea, United Kingdom Zenghu Li, Beijing, China Karl-Hermann Neeb, Erlangen, Germany

Volume 87

René Erlin Castillo

Fourier Meets Hilbert and Riesz |

An Introduction to the Corresponding Transforms

Mathematics Subject Classification 2020 42-01, 42A10, 42A20, 46-01, 46F12, 47B02 Author Prof. Dr. René Erlin Castillo Universidad Nacional de Colombia Departamento de Matemáticas Carrera 45 No. 26-85 Bogotá 111321 Colombia [email protected]

ISBN 978-3-11-078405-3 e-ISBN (PDF) 978-3-11-078409-1 e-ISBN (EPUB) 978-3-11-078412-1 ISSN 0179-0986 Library of Congress Control Number: 2022934767 Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.dnb.de. © 2022 Walter de Gruyter GmbH, Berlin/Boston Typesetting: VTeX UAB, Lithuania Printing and binding: CPI books GmbH, Leck www.degruyter.com

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A la memoria de Julieta mi mamá-abuela A mi esposa Hilcia del Carmen A mis hijos: René José, Manuel Alejandro, Irene Gabriela y Renzo Rafael

Preface This book is based on lecture notes from a special topics course on advanced analysis offered at Universidad Nacional de Colombia during the spring semester of 2019. Prerequisites are a first-year course on measure and integration theory and functional analysis, as well as some basics of metric spaces and complex variables. Our intention in writing these notes is to prepare students to do research on Fourier analysis. The lecture notes were rather informal with a conversational tone. While emphasizing formality, these notes keep pedagogy in mind as we aim to serve students who might need to teach the material to themselves because they do not have access to an equivalent course at their institutions. We would like to emphasize that this manuscript is meant as a textbook, not as a reference book. The expository paper [1] was an early byproduct of this effort. We have made every effort to keep this book self-contained. To that end, the first chapter includes all fundamental concepts needed for the rest of the book where, indeed, Fourier meets Hilbert and Riesz. The Hilbert transform is essentially the only singular operator in one dimension, in the sense that, without a doubt, it and its generalizations are some of the most important operators in harmonic analysis. We present an elementary introduction to the Hilbert and Riesz transforms in Chapters 5, 6, 7, and 8. An appendix is also given for the sake of completeness. Exercises are included throughout the text to help challenge and motive students as they gauge their progress. While most problems are of a routine nature, a few problems will likely be challenging to beginning students. We have made every effort to keep the manuscript mistake free. However, we apologize in advance for any errors that might have escaped our proof-reading. I especially want to thank Dr. Sergio Lopez-Permouth for his help with the use of English in this introduction. Finally, I would like to express my deepest thanks to Dr. Humbert Rafeiro, for valuable conversations and for suggesting the title of the book. Bogotá, Colombia May 2022

https://doi.org/10.1515/9783110784091-201

René Erlin Castillo

Contents Preface | VII 1 1.1 1.2 1.3 1.4 1.5 1.6

Fundamental concepts | 1 Introduction | 1 Inner products | 1 Strong and weak convergence | 13 The space L2 (ℝ) | 16 A glance at linear operators | 28 The adjoint of an operator | 29

2 2.1 2.2 2.3 2.4 2.5 2.6

Fourier series | 37 Convolution | 60 Fourier transform on ℝ | 67 Fourier transform on L2 (ℝ) | 80 Definition of the Fourier transform in L2 (ℝ) | 83 Fourier transform on L1 (ℝn ) | 86 Bessel functions | 89

3 3.1 3.2

Schwartz spaces S(ℝn ) | 95 Definition of the Fourier transform in L2 (ℝn ) | 114 Some applications of the Fourier transform to partial differential equations | 118

4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11

Distribution functions | 129 Extending the Fourier transform to Lp (ℝn ) | 132 Covering lemma | 145 Maximal function | 146 Weak-Lp space | 150 Lebesgue differentiation theorem | 152 Cube | 159 Properties of dyadic subcubes | 160 Calderón–Zygmund decomposition | 160 Operators of type (p, q) | 164 Marcinkiewicz interpolation theorem | 165 The Cauchy principal value | 167

5 5.1 5.2 5.3

Three-fold approach to the Hilbert transform | 171 Derivation of the Hilbert transform on ℝ | 171 The Hilbert transform | 177 The Hilbert transform on the circle | 181

X | Contents 5.4 5.5 5.5.1 5.5.2 5.5.3 5.6 5.6.1 5.7 5.8

Calderón–Zygmund singular integral operators | 187 Some Hilbert transforms | 197 Hilbert transform of a constant function | 197 Hilbert transform of sin x and cos x | 198 Hilbert transform of eix | 199 Relationship between the Hilbert and Laplace transforms | 200 1 and the Poisson kernel | 202 The Hilbert transform of x+iα Hilbert transform of a periodic function | 205 Properties | 211

6 6.1 6.2 6.3 6.4 6.5

Hilbert transform in L2 (ℝ) | 217 Hilbert transform of a product | 217 Differentiation property of the Hilbert transform in L2 | 224 Hilbert transform as an operator in L2 | 225 The Parseval-type form | 227 A theorem due to E. M. Stein and G. Weiss | 230

7 7.1

Embedding and strong Lp boundedness for the Hilbert transform | 243 Strong Lp boundedness for the Hilbert transform on the circle | 254

8 8.1

Riesz transform | 261 Vectorial Riesz transform | 274

A A.1

Appendix | 277 Convergence in measure | 277

Bibliography | 289 Index | 291

1 Fundamental concepts 1.1 Introduction The Hilbert transform of a sufficiently well-behaved function f (x) is defined to be ∞

1 f (y) Hf (x) = P. V. ∫ dy π x−y 1 = lim+ ϵ→0 π

−∞

∫ |x−y|>ϵ

f (y) dy. x−y

(1.1)

It is not immediately not clear that Hf (x) is well-defined even for nice functions [4]. Though (1.1) “almost” looks like an ordinary convolution, there are, however, certain technical subtleties associated with the definition. The primitive idea behind the definition of the transform is quite simple, namely to transform f (x) by convolving with the 1 kernel πx . It is in doing so rigorously that one encounters technical difficulties as the kernel fails to be absolutely integrable owing to its slow decay and, more importantly, due to the singularity at the origin. The limiting argument in (1.1) is used to avoid the singularity by truncating the kernel around the origin in a systematic fashion. As will be shown in Chapter 5, this indeed works for sufficiently regular functions. The other pathology, namely the slow decay of the kernel, can be circumvented relatively easily, simply by restricting the domain of (1.1) to functions having a sufficiently fast decay. From a historical point of view, the Hilbert transform originated in the work of David Hilbert on integral equations and boundary value problem in 1905 (see [18]). The Hilbert transform was a motivating example for Antoni Zygmund and Alberto Calderón during their study of singular integrals (see [7]). Since then the theory has been expanded widely and found many applications in both mathematics and physics. The topic discussed in this presentation does not follow any specific source. However, we strongly recommend the reader to see the monumental work done by F. W. King in [17] and the references therein.

1.2 Inner products Definition 1.1. Let X be a real vector space. An inner product on X is a function ⟨⋅, ⋅⟩ : X × X → ℝ such that for all x, y, z ∈ X and α, β ∈ ℝ, (a) ⟨x, x⟩ ≥ 0; (b) ⟨x, x⟩ = 0 if and only if x = 0; (c) ⟨αx + βy, z⟩ = α⟨x, z⟩ + β⟨y, z⟩; (d) ⟨x, y⟩ = ⟨y, x⟩. https://doi.org/10.1515/9783110784091-001

2 | 1 Fundamental concepts Example 1. The function ⟨⋅, ⋅⟩ : ℝn × ℝn → ℝ defined by ⟨x, z⟩ = ∑nk=1 xk yk is an inner product on ℝn . Proof. We show that the above formula defines an inner product by verifying that all the properties in Definition 1.1 hold: 1. ⟨x, x⟩ = ∑nk=1 xk2 ≥ 0. 2. ⟨x, x⟩ = 0 implies that xn = 0 for all n, so x = 0. 3. n

⟨αx + βy, z⟩ = ∑ ⟨αxk + βyk ⟩zk k=1 n

n

= α ∑ xk zk + β ∑ yk zk k=1

k=1

= α⟨x, z⟩ + β⟨y, z⟩. 4. ⟨x, y⟩ = ∑k=1 xk yk = ∑nk=1 yk xk = ⟨y, x⟩. By ℂ we denote the set of complex numbers, that is, ℂ = {x + iy : x, y ∈ ℝ}

(i2 = −1).

If z ∈ ℂ with z = x +iy, then x is the real part of z written Re z = x and y is the imaginary part of z written Im z = y. The conjugate of z, denoted z, is defined by z = x + i(−y) = x − iy. We have z + z = 2 Re z and z − z = 2i Im z. If z ∈ ℂ with z = x + iy, we define |z| = √x 2 + y2 . Note 1. One has z ⋅ z = x2 + y2 = |z|2 . For an f : X → ℂ, we define Re and Im as follows: Re f : X → ℝ defined by (Re f )(x) = Re(f (x)), (Im f )(x) = Im(f (x)). Thus f = Re f + i(Im f ). We define f : X → ℂ by f (x) = f (x). Definition 1.2. Let X be a complex vector space. An inner product on X is a function ⟨⋅, ⋅⟩ : X × X → ℂ such that for all x, y, z ∈ X, α, β ∈ ℂ, (a) ⟨x, x⟩ ∈ ℝ and ⟨x, x⟩ ≥ 0; (b) ⟨x, x⟩ = 0 if and only if x = 0; (c) ⟨αx + βy, z⟩ = α⟨x, z⟩ + β⟨y, z⟩; (d) ⟨x, y⟩ = ⟨y, x⟩ (the bar denotes complex conjugate); (e) ⟨x, αy⟩ = α⟨x, y⟩.

1.2 Inner products | 3

The pair (X, ⟨⋅, ⋅⟩) is called an inner product space. We assume familiarity with the concept of a real vector space and related notations. Let X be a vector space. We let 0x denote the zero vector, i. e., x + 0x = x = 0x + x for all x ∈ X. The negative of X is denoted by −x, and then x + (−x) = 0x = (−x) + x, (−1)x = −x, for all x ∈ X, α0x = 0x for all α ∈ ℝ. Example 2. The function ⟨⋅, ⋅⟩ : ℂn × ℂn → ℂ defined by ⟨x, x⟩ = ∑nk=1 xk yk is an inner product on ℂn . This inner product will be called the standard inner product on ℂn . Proof. The proof follows that of Example 1, using the properties of the complex conjugate where appropriate. Remark 1. (1) ⟨0x , y⟩ = 0 for all y ∈ X, where 0x represents the zero vector in X. Proof. ⟨0x , y⟩ = ⟨0x + 0x , y⟩ = ⟨0x , y⟩ + ⟨0x , y⟩. And so ⟨0x , y⟩ = 0. (2) ⟨x, y + z⟩ = ⟨x, y⟩ + ⟨x, z⟩. (3) For all y ∈ X, the mapping φy : X → ℂ defined by φy (x) = ⟨x, y⟩ is a linear functional. This is a consequence of (c) and (e). Indeed, – φx (αy) = αφx (y), – φx (y + z) = ⟨x, y + z⟩ = ⟨x, y⟩ + ⟨x, z⟩ = φx (y) + φx (z). (4) An alternative way to prove the above claim is as follows: φx (αy + βz) = ⟨x, αy + βz⟩

= ⟨x, αy⟩ + ⟨x, βz⟩ = α⟨x, y⟩ + β⟨x, z⟩

= αφx (y) + βφx (z). Theorem 1.1 (Cauchy–Schwarz inequality). Let (X, ⟨⋅, ⋅⟩) be an inner product space. Then 󵄨󵄨 󵄨2 󵄨󵄨⟨x, y⟩󵄨󵄨󵄨 ≤ ⟨x, x⟩⟨y, y⟩ for all x, y ∈ X.

4 | 1 Fundamental concepts Proof. If x = 0x , the conclusion is clear, so assume x ≠ 0x , then ⟨x, y⟩ > 0. For any a ∈ ℂ, 0 ≤ ⟨y − ax, y − ax⟩ = ⟨y, y⟩ − a⟨y, x⟩ − a⟨x, y⟩ + |a|2 ⟨x, x⟩ = ⟨y, y⟩ − [a⟨x, y⟩ + a⟨x, y⟩] + |a|2 ⟨x, x⟩ = ⟨y, y⟩ − 2 Re(a⟨x, y⟩) + |a|2 ⟨x, x⟩. This is true for each a ∈ ℂ. Choosing a = 0 ≤ ⟨y, y⟩ − 2 Re( = ⟨y, y⟩ − 2 = ⟨y, y⟩ −

⟨y,x⟩ , ⟨x,x⟩

we get

|⟨x, y⟩|2 |⟨y, x⟩|2 )+ ⟨x, x⟩ ⟨x, x⟩ ⟨x, x⟩2

|⟨x, y⟩| |⟨x, y⟩|2 + ⟨x, x⟩ ⟨x, x⟩

|⟨x, y⟩|2 . ⟨x, x⟩

Hence |⟨x, y⟩|2 ≤ ⟨y, y⟩, ⟨x, x⟩ and so 󵄨󵄨 󵄨2 󵄨󵄨⟨x, y⟩󵄨󵄨󵄨 ≤ ⟨x, x⟩⟨y, y⟩. Definition 1.3. For ‖ ⋅ ‖ : X → [0, ∞) define ‖x‖ = √⟨x, x⟩, x ∈ X. Remark 2. The Cauchy–Schwarz inequality can be stated as 󵄨󵄨 󵄨 󵄨󵄨⟨x, y⟩󵄨󵄨󵄨 ≤ ‖x‖‖y‖. Theorem 1.2 (Triangle inequality). Let X be a inner product space, then ‖x + y‖ ≤ ‖x‖ + ‖y‖

for all x, y ∈ X.

Proof. Recall that if z = x + iy, then Re z = x ≤ |x| ≤ |z| so Re z ≤ |z|. By the definition of ‖ ⋅ ‖, we have ‖x + y‖2 = ⟨x + y, x + y⟩ = ⟨x, x⟩ + ⟨x, y⟩ + ⟨y, x⟩ + ⟨y, y⟩ = ⟨x, x⟩ + ⟨x, y⟩ + ⟨x, y⟩ + ⟨y, y⟩ = ⟨x, x⟩ + 2 Re⟨x, y⟩ + ⟨y, y⟩

1.2 Inner products | 5

󵄨 󵄨 ≤ ‖x‖2 + 2󵄨󵄨󵄨⟨x, y⟩󵄨󵄨󵄨 + ‖y‖2 ≤ ‖x‖2 + 2‖x‖‖y‖ + ‖y‖2

(by Cauchy–Schwarz inequality)

so 2

‖x + y‖2 ≤ (‖x‖ + ‖y‖) . Hence ‖x + y‖ ≤ ‖x‖ + ‖y‖. Definition 1.4. Let X be a linear vector space and n : X → ℝ+ . Suppose the following statements are true: (1) n(x) ≥ 0 for all x ∈ X. (2) n(x) = 0 if and only if x = 0x . (3) n(αx) = |α|n(x) for all x ∈ X, for all α ∈ ℝ. (4) n(x + y) ≤ n(x) + n(y) for all x, y ∈ X (triangle inequality). Then we call n a norm on X and call the ordered pair (X, n) a normed linear space. We often write n(x) as ‖x‖. A norm is introduced to measure the size or the length of each vector x ∈ X so that properties (1)–(4) should appear as natural requirements. A normed space is a linear vector space endowed with a norm ‖ ⋅ ‖. With a norm, we associate the distance between two vectors, given by d(x, y) = ‖x − y‖, which makes X a metric space and allows defining a topology on X and a notation of convergence in a very simple way. We say that a sequence {xn }n∈ℕ in X converges to x ∈ X, and write xn → x in X, (or limn→∞ xn = x) if d(xn , x) = ‖xn − x‖ → 0

as n → ∞.

An important distinction is between a convergent and Cauchy sequence. A sequence {xn }n∈ℕ in X is a Cauchy sequence if d(xm , xn ) = ‖xm − xn ‖ → 0

as m, n → ∞.

If xn → x in X, from the triangle inequality we get ‖xm − xn ‖ ≤ ‖xm − x‖ + ‖xn − x‖ → 0

6 | 1 Fundamental concepts as m, n → ∞, and therefore {xn }n∈ℕ being convergent implies that {xn }n∈ℕ is a Cauchy sequence. The converse is not true, in general. Take X = ℚ with the usual norm given by |x|. The sequence of rational numbers xn = (1 +

n

1 ) n

is a Cauchy sequence, but lim xn = lim (1 +

n→∞

n→∞

n

1 ) = e ∉ ℚ. n

A normed space in which every Cauchy sequence converges is called complete and deserves a special name. Definition 1.5. A complete normed space is called a Banach space. Proposition 1.1. Let X be an inner product space, then ‖x‖ = √⟨x, x⟩ is a norm on X. Proof. 1. By definition ‖x‖ = √⟨x, x⟩ ∈ [0, ∞), i. e., ‖x‖ ≥ 0 for all x ∈ X. 2. ‖x‖ = 0 means ⟨x, x⟩ = 0, so by an axiom of the inner product, x = 0. 3. By Theorem 1.2, ‖x + y‖ ≤ ‖x‖ + ‖y‖ for all x, y ∈ X. 4. For any α ∈ ℂ, ‖αx‖2 = ⟨αx, αx⟩ = αα⟨x, x⟩ = |α|2 ‖x‖2 . So ‖αx‖ = |α|‖x‖. Hence ‖x‖ = √⟨x, x⟩ is a norm. Definition 1.6. An inner product space (X, ⟨⋅, ⋅⟩) is a Hilbert space if it is a complete metric space relative to the metric induced by the norm. Lemma 1.1 (Polarization identity). Let X be an inner product space. Then for all x, y ∈ X, (a) (Real case) ⟨x, y⟩ = 41 [‖x + y‖2 − ‖x − y‖2 ]. (b) (Complex case) ⟨x, y⟩ = 41 (‖x + y‖2 − ‖x − y‖2 + i‖x + iy‖2 − i‖x − iy‖2 ).

1.2 Inner products | 7

Proof. (a) Since we are in the real case, we have ⟨x, y⟩ = ⟨y, x⟩. Then ‖x + y‖2 − ‖x − y‖2 = ⟨x + y, x + y⟩ − ⟨x − y, x − y⟩ = ⟨x, x⟩ + ⟨x, y⟩ + ⟨y, x⟩ + ⟨y, y⟩ − ⟨x, x⟩ + ⟨x, y⟩ + ⟨y, x⟩ − ⟨y, y⟩ = 4⟨x, y⟩. Hence ⟨x, y⟩ =

1 (‖x + y‖2 − ‖x − y‖2 ). 4

(b) By direct calculation, ‖x + y‖2 − ‖x − y‖2 = 2⟨x, y⟩ + 2⟨y, x⟩. Similarly, i‖x + iy‖2 − i‖x − iy‖2 = 2i⟨x, iy⟩ − 2i⟨y, ix⟩ = −2i2 ⟨x, y⟩ + 2i⟨y, ix⟩ = −2i2 ⟨x, y⟩ + 2i2 ⟨y, x⟩ = 2⟨x, y⟩ − 2⟨y, x⟩. Adding the two expressions gives ‖x + y‖2 − ‖x − y‖2 + i‖x + iy‖2 − i‖x − iy‖2 = 2⟨x, y⟩ + 2⟨y, x⟩ + 2⟨x, y⟩ − 2⟨y, x⟩ = 4⟨x, y⟩. And so ⟨x, y⟩ =

1 [‖x + y‖2 − ‖x − y‖2 + i‖x + iy‖2 − i‖x − iy‖2 ]. 4

Theorem 1.3. A normed vector space X is an inner product space if and only if the parallelogram identity ‖x + y‖2 + ‖x − y‖2 = 2‖x‖2 + 2‖y‖2 holds for all x, y ∈ X.

(1.2)

8 | 1 Fundamental concepts Proof. (󳨐⇒) Let X be a inner product space, then ‖x + y‖2 + ‖x − y‖2 = ⟨x + y, x + y⟩ + ⟨x − y, x − y⟩ = ⟨x, x⟩ + ⟨x, y⟩ + ⟨y, x⟩ + ⟨y, y⟩ + ⟨x, x⟩ − ⟨x, y⟩ − ⟨y, x⟩ + ⟨y, y⟩ = 2⟨x, x⟩ + 2⟨y, y⟩ = 2‖x‖2 + 2‖y‖2 , for all x, y ∈ X. (⇐󳨐) Let X be a vector space in which the parallelogram identity (1.2) holds. We shall only consider the real case. The polarization identity (a) (Lemma 1.1) gives us a hint as to how we should define an inner product. For all x, y ∈ X, define 󵄩󵄩 x + y 󵄩󵄩2 󵄩󵄩 x − y 󵄩󵄩2 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩 ⟨x, y⟩ = 󵄩󵄩󵄩 󵄩 −󵄩 󵄩. 󵄩󵄩 2 󵄩󵄩󵄩 󵄩󵄩󵄩 2 󵄩󵄩󵄩 We claim that ⟨⋅, ⋅⟩ is an inner product on X. Indeed, 1. ⟨x, x⟩ = ‖ x+x ‖2 − ‖ x−x ‖2 = ‖x‖2 ≥ 0. 2 2 2. ⟨x, x⟩ = 0 if and only if ‖x‖2 = 0, if and only if x = 0. 3. 󵄩󵄩 x + y 󵄩󵄩2 󵄩󵄩 x − y 󵄩󵄩2 󵄩󵄩 x + y 󵄩󵄩2 󵄩󵄩 y − x 󵄩󵄩2 󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩 ⟨x, y⟩ = 󵄩󵄩󵄩 󵄩 −󵄩 󵄩 = 󵄩󵄩󵄩 󵄩 −󵄩 󵄩 = ⟨y, x⟩. 󵄩󵄩 2 󵄩󵄩󵄩 󵄩󵄩󵄩 2 󵄩󵄩󵄩 󵄩󵄩 2 󵄩󵄩󵄩 󵄩󵄩󵄩 2 󵄩󵄩󵄩 4. Replacing x by u + v and y by w + v in the parallelogram identity gives ‖u + w + 2v‖2 + ‖u − w‖2 = 2‖u + v‖2 + 2‖w + v‖2 .

(1.3)

Replacing x by u − v and y by w − v in the parallelogram identity yields ‖u + w − 2v‖2 + ‖u − w‖2 = 2‖u − v‖2 + 2‖w − v‖2 .

(1.4)

Subtracting (1.4) from (1.3) then produces ‖u + w + 2v‖2 − ‖u + w − 2v‖2 = 2[‖u + v‖2 − ‖u − v‖2 + ‖w + v‖2 − ‖w − v‖2 ]. Now using the definition of ⟨⋅, ⋅⟩, we get 4⟨u + w, 2v⟩ = 8[⟨u, v⟩ + ⟨w, v⟩] and then 1 ⟨u + w, 2v⟩ = ⟨u, v⟩ + ⟨w, v⟩. 2

(1.5)

1.2 Inner products | 9

Taking w = 0 then yields 1 ⟨u, 2v⟩ = ⟨u, v⟩. 2

(1.6)

Now replace u by x + y and v by z in (1.6) and use (1.5) to get 1 ⟨x + y, z⟩ = ⟨x + y, 2z⟩ = ⟨x, z⟩ + ⟨y, z⟩. 2 5.

We shall show that ⟨λx, y⟩ = λ⟨x, y⟩ for all λ ∈ ℝ and all x, y ∈ X. If λ = n is a nonzero integer, then using (c) we obtain ⟨nx, y⟩ = ⟨x⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ + x + ⋅ ⋅ ⋅ + x , y⟩ n-times

= ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ ⟨x, y⟩ + ⟨x, y⟩ + ⋅ ⋅ ⋅ + ⟨x, y⟩ n-times

= n⟨x, y⟩, thus x nx n⟨ , y⟩ = ⟨ , y⟩ = ⟨x, y⟩, n n that is, x 1 ⟨ , y⟩ = ⟨x, y⟩. n n If λ is a rational number, λ = pq , say, then p x p ⟨ x, y⟩ = p⟨ , y⟩ = ⟨x, y⟩. q q q If λ ∈ ℝ, then there is a sequence {rk }k∈ℕ of rational numbers (since ℚ = ℝ) such that rk → λ as k → ∞. Using the continuity of the norm, we have that 󵄩󵄩2 1 󵄩󵄩󵄩 󵄩󵄩2 1 󵄩󵄩 󵄩󵄩 lim rk x + y󵄩󵄩󵄩 − 󵄩󵄩󵄩 lim rk x − y󵄩󵄩󵄩 󵄩 󵄩 k→∞ 4 󵄩k→∞ 4 󵄩k→∞ 1 1 2 2 = lim ‖rk x + y‖ − lim ‖rk x − y‖ 4 k→∞ 4 k→∞ 󵄩󵄩 r x + y 󵄩󵄩2 󵄩󵄩 r x − y 󵄩󵄩2 󵄩 󵄩󵄩 󵄩󵄩 k 󵄩󵄩 = lim (󵄩󵄩󵄩 k 󵄩󵄩 − 󵄩󵄩 󵄩󵄩 ) 󵄩 󵄩󵄩 󵄩 k→∞ 󵄩 2 2 󵄩 󵄩 󵄩 = lim ⟨rk x, y⟩

⟨λx, y⟩ = ⟨ lim rk x, y⟩ =

k→∞

= lim rk ⟨x, y⟩ k→∞

= λ⟨x, y⟩.

10 | 1 Fundamental concepts Thus ⟨λx, y⟩ = λ⟨x, y⟩ for all λ ∈ ℝ and all x, y ∈ X. Example 3. (a) The normed vector space X = C([a, b]) equipped with the supremum norm ‖ ⋅ ‖∞ is not an inner product space. We shall show that the norm 󵄨 󵄨 ‖x‖∞ = max 󵄨󵄨󵄨x(t)󵄨󵄨󵄨 a≤t≤b does not satisfy the parallelogram identity. To that end, take x(t) = 1

and y(t) =

t−a . b−a

Since x(t) + y(t) = 1 +

t−a b−a

x(t) − y(t) = 1 −

t−a , b−a

and

we have that ‖x‖ = 1 = ‖y‖

and ‖x + y‖ = 2,

‖x − y‖ = 1.

Thus ‖x + y‖2 + ‖x − y‖2 = 5 ≠ 2‖x‖2 + 2‖y‖2 = 4. (b) The normed vector space X = L1 ([0, 2π]) with the norm 2π

󵄨 󵄨 ‖f ‖1 = ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx 0

is not an inner product space. We shall show that the norm 2π

󵄨 󵄨 ‖f ‖1 = ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx 0

does not satisfy the parallelogram identity. To that end, take f (x) = 1 and g(x) = sin x for x ∈ [0, 2π].

1.2 Inner products | 11

Now observe that ‖f +

g‖21

2

2π

= ( ∫ (1 + sin x) dx) = 4π 2 0

and 2

2π

‖f − g‖21 = ( ∫ (1 − sin x) dx) = 4π 2 , 0

while ‖f ‖21

2π

= ( ∫ 1 dx) = 4π 2 0

and 2π

‖g‖21 = ( ∫ sin x dx) = 0. 0

And so ‖f + g‖21 + ‖f − g‖21 = 8π 2 ≠ 4π 2 = 2‖f ‖21 + 2‖g‖21 . Theorem 1.4. Let X be an inner product space. For any fixed y ∈ X, the mappings x → ⟨x, y⟩ and x → ⟨y, x⟩ are continuous on X. Proof. Let us write φy (x) = ⟨x, y⟩ for x ∈ X. Then by the Cauchy–Schwarz inequality, we have 󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨φy (x2 ) − φy (x1 )󵄨󵄨󵄨 = 󵄨󵄨󵄨⟨x2 , y⟩ − ⟨x1 , y⟩󵄨󵄨󵄨 󵄨 󵄨 = 󵄨󵄨󵄨⟨x2 − x1 , y⟩󵄨󵄨󵄨 ≤ ‖x2 − x1 ‖‖y‖.

Given ϵ > 0, choose δ =

ϵ . ‖y‖+1

If ‖x2 − x1 ‖ < δ, then

ϵ‖y‖ 󵄨󵄨 󵄨 < ϵ. 󵄨󵄨φy (x2 ) − φy (x1 )󵄨󵄨󵄨 ≤ ‖x2 − x1 ‖‖y‖ < ‖y‖ + 1 Hence φy is uniformly continuous on X. The continuity of x → ⟨y, x⟩ is established similarly.

12 | 1 Fundamental concepts

Exercises. 1. Let X be an inner product space and let x, y ∈ X . Prove Pythagorean theorem: If ⟨x, y⟩ = 0, then ‖x + y‖2 = ‖x‖2 + ‖y‖2 . 2. 3.

4.

Show that if x, y are vectors in an inner product space and ‖x + y‖ = ‖x‖ + ‖y‖, then one of x, y is a scalar multiple of the other. Prove that in a complex inner product space ℋ the following equalities hold:

(a)

⟨x, y⟩ =

k 󵄩2 k 1 N 󵄩󵄩 ∑ 󵄩󵄩x + ye2πi N 󵄩󵄩󵄩󵄩 e2πi N N k=1󵄩

(b)

⟨x, y⟩ =

1 󵄩 󵄩2 ∫ 󵄩󵄩󵄩x + eiθ y 󵄩󵄩󵄩󵄩 eiθ dθ 2π 󵄩

2π 0

for N ≥ 3. for all x, y ∈ ℋ.

Let X be an inner product space. Prove that ‖x + y‖ ‖x − y‖ ≤ ‖x‖2 + ‖y‖2

5.

for every x, y ∈ X . Let f be a nonnegative function on ℝ having a set of discontinuities which has measure zero. If f ∈ L2 (ℝ), prove that lim ∫ f (x + t)f (x) dx = ‖f ‖2L2 (ℝ) .

t→0

6.

ℝ

Let X be an inner product space. Prove that ‖x‖ = sup‖y‖=1 |⟨x, y⟩|.

Definition 1.7. A real or complex vector space X with an inner product ⟨⋅, ⋅⟩ is called an inner product space. Definition 1.8. Let X be an inner product space. The vectors x, y ∈ X are said to be orthogonal if ⟨x, y⟩ = 0. Definition 1.9. Let X be an inner product space. The set {e1 , . . . , en } ⊂ X is said to be orthonormal if ‖ek ‖ = 1 for 1 ≤ k ≤ n and ⟨ek , ek ⟩ = 0 for all 1 ≤ k, l ≤ n with l ≠ k. Remark 3. An orthonormal set {e1 , . . . , en } in any inner product space X is linearly independent. In particular, if X is n-dimensional then the set {e1 , . . . , en } is a basis for X and any vector x ∈ X can be expressed as n

x = ∑ ⟨x, ek ⟩ek ; k=1

in this case {e1 , . . . , en } is usually called an orthonormal basic and the numbers ⟨x, ek ⟩ are the components of x with respect to this basis.

1.3 Strong and weak convergence

| 13

Theorem 1.5 (Bessel inequality). Let X be an inner product space and let {ek }k∈ℕ be an orthonormal sequence in X. For any x ∈ X, n

󵄨2 󵄨 ∑ 󵄨󵄨󵄨⟨x, ek ⟩󵄨󵄨󵄨 ≤ ‖x‖.

k=1

Proof. For each n ∈ ℕ, let yn = ∑nk=1 ⟨x, ek ⟩ek . Then 0 ≤ ‖x − yn ‖ = ⟨x − yn , x − yn ⟩ n

n

n

k=1 n

k=1 n

k=1 n

󵄨 󵄨2 = ‖x‖2 − ∑ ⟨x, en ⟩⟨x, en ⟩ − ∑ ⟨x, ek ⟩⟨ek , x⟩ + ∑ 󵄨󵄨󵄨⟨x, ek ⟩󵄨󵄨󵄨 󵄨 󵄨2 = ‖x‖2 − ∑ ⟨x, en ⟩⟨x, en ⟩ − ∑ ⟨x, ek ⟩⟨x, en ⟩ + ∑ 󵄨󵄨󵄨⟨x, ek ⟩󵄨󵄨󵄨 k=1 n

n

k=1

n

k=1

󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 = ‖x‖2 − ∑ 󵄨󵄨󵄨⟨x, en ⟩󵄨󵄨󵄨 − ∑ 󵄨󵄨󵄨⟨x, ek ⟩󵄨󵄨󵄨 + ∑ 󵄨󵄨󵄨⟨x, ek ⟩󵄨󵄨󵄨 k=1 n

k=1

k=1

󵄨 󵄨2 = ‖x‖2 − ∑ 󵄨󵄨󵄨⟨x, en ⟩󵄨󵄨󵄨 . k=1

Hence n

󵄨 󵄨2 ∑ 󵄨󵄨󵄨⟨x, ek ⟩󵄨󵄨󵄨 ≤ ‖x‖2 .

k=1

1.3 Strong and weak convergence Since every inner product space is a normed space, it is equipped with a notion of convergence, namely the convergence defined by the norm. This convergence is called the strong convergence. Definition 1.10 (Strong convergence). A sequence {xn }n∈ℕ of vectors in an inner product space X is called strongly convergent to a vector x ∈ X if ‖xn − x‖ → 0 as n → ∞. Definition 1.11 (Weak convergence). A sequence {xn }n∈ℕ of vectors in an inner product X is called weakly convergent to a vector x in X if ⟨xn , y⟩ → ⟨x, y⟩ as n → ∞, for every y ∈ X. The condition in the above definition can be also stated as ⟨xn − x, y⟩ → 0 as n → ∞, for every y ∈ X. It will be convenient to reserve the notation xn → x for the strong convergence, ω

and we will use xn 󳨀→ x to denote the weak convergence.

14 | 1 Fundamental concepts Theorem 1.6. A strongly convergent sequence is weakly convergent (to the same limit), ω i. e., xn → x implies xn 󳨀→ x. Proof. Suppose that a sequence {xn }n∈ℕ converges strongly to x. This means ‖xn −x‖ → 0 as n → ∞. By the Cauchy–Schwarz inequality, we have 󵄨 󵄨󵄨 󵄨󵄨⟨xn − x, y⟩󵄨󵄨󵄨 ≤ ‖xn − x‖‖y‖ → 0

as n → ∞,

and thus ⟨xn − x, y⟩ → 0

as n → ∞ for every y ∈ X.

This proves the theorem. In general, the converse of Theorem 1.6 is not true. A suitable example will be given in Section 1.4. Remark 4. Note that each step function vanishes outside a set of finite measure. Thus, every step function is a member of L1 (ℝ). Moreover, the set of all step functions forms a vector subspace of L1 (ℝ). ω

Theorem 1.7. If xn → x and ‖xn ‖ → ‖x‖, then xn → x. Proof. By the definition of weak convergence, we have for all y ∈ X, ⟨xn , y⟩ → ⟨x, x⟩ as n → ∞. Hence ⟨xn , x⟩ → ⟨x, x⟩ = ‖x‖2 . Now ‖xn − x‖2 = ⟨xn − x, xn − x⟩

= ⟨xn , xn ⟩ − ⟨xn , x⟩ − ⟨x, xn ⟩ + ⟨x, x⟩ = ‖xn ‖2 − [⟨xn , x⟩ + ⟨xn , x⟩] + ‖x‖2

= ‖xn ‖2 − 2 Re⟨xn , x⟩ + ‖x‖2 = ‖xn ‖2 − 2⟨xn , x⟩ + ‖x‖2

= ‖xn ‖2 − 2‖x‖2 + ‖x‖2 = 0

as n → ∞.

The sequence {xn }n∈ℕ is thus strongly convergent to x. The next theorem is often useful in proving weak convergence.

1.3 Strong and weak convergence

| 15

Theorem 1.8. Let E be a subset of an inner product space X such that Span(E) is dense in X. If {xn }n∈ℕ is a bounded sequence in X and ⟨xn , y⟩ → ⟨x, y⟩

for every y ∈ E

ω

then xn 󳨀→ x. Proof. Clearly, if ⟨xn , y⟩ → ⟨x, y⟩ for every y ∈ E, then ⟨xn , x⟩ → ⟨x, y⟩ for every y ∈ Span(E). Let z ∈ E and consider ϵ > 0. Since Span(E) is dense in X, there exists y0 ∈ Span(E) such that ‖z − y0 ‖
n0 . Consequently, for any n > 0, we have 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨󵄨⟨xn , z⟩ − ⟨x, z⟩󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨⟨xn , z⟩ − ⟨xn , y0 ⟩󵄨󵄨󵄨 + 󵄨󵄨󵄨⟨xn , y0 ⟩ − ⟨x, y0 ⟩󵄨󵄨󵄨 + 󵄨󵄨󵄨⟨x, y0 ⟩ − ⟨x, z⟩󵄨󵄨󵄨 ϵ < ‖xn ‖‖z − y0 ‖ + + ‖x‖‖y0 − z‖ 3 ϵ ϵ ϵ |g(x)|}. Then 󵄨󵄨 󵄨 󵄨 󵄨2 󵄨󵄨f (x)g(x)󵄨󵄨󵄨 < 󵄨󵄨󵄨f (x)󵄨󵄨󵄨

for all x ∈ A

and 󵄨󵄨 󵄨 󵄨 󵄨2 󵄨󵄨f (x)g(x)󵄨󵄨󵄨 < 󵄨󵄨󵄨g(x)󵄨󵄨󵄨

for all x ∈ Ac .

Then 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 ∫󵄨󵄨󵄨f (x)g(x)󵄨󵄨󵄨 dm = ∫󵄨󵄨󵄨f (x)g(x)󵄨󵄨󵄨 dm + ∫ 󵄨󵄨󵄨f (x)g(x)󵄨󵄨󵄨 dm ℝ

A

Ac

󵄨 󵄨2 󵄨 󵄨2 < ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dm + ∫󵄨󵄨󵄨g(x)󵄨󵄨󵄨 dm < ∞. ℝ

Hence fg ∈ L1 (ℝ).

ℝ

18 | 1 Fundamental concepts Definition 1.13. The functional ‖ ⋅ ‖ : L2 (ℝ) → ℝ+ is defined by 2

1 2

‖f ‖2 = (∫ |f | dm) . ℝ

Example 4. If f , g ∈ L2 (ℝ) then in general fg ∉ L2 (ℝ). 1 Let f (x) = g(x) = x− 4 for x ∈ (0, 1) and note that 1

dx 󵄨 󵄨2 󵄨 󵄨2 = 2. ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx = ∫󵄨󵄨󵄨g(x)󵄨󵄨󵄨 dx = ∫ √x

ℝ

0

ℝ

But 1

dx 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 = ∞. ∫󵄨󵄨󵄨f (x)g(x)󵄨󵄨󵄨 dx = ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 󵄨󵄨󵄨g(x)󵄨󵄨󵄨 dx = ∫ x

ℝ

0

ℝ

Hence fg ∉ L2 (ℝ). However, fg ∈ L1 (ℝ), as we shall show in Theorem 1.12. Theorem 1.12 (Cauchy–Schwarz inequality). Let f , g ∈ L2 (ℝ). Then ∫ |fg| dm ≤ ‖f ‖2 ‖g‖2 . ℝ

The equality holds if there are constant A and B, not simultaneously zero, such that A|f |2 = B|g|2 m-a. e. Proof. Notice that for any positive numbers a and b, 0 ≤ (b − a)2 = b2 − 2ab + b2 implies that ab ≤ Now let a = have

|f | ‖f ‖2

and b =

|g| ‖g‖2

a2 b2 + . 2 2

(1.7)

for f ≠ 0 and g ≠ 0. Putting back a and b into (1.7), we |fg| |f |2 1 |g| ≤ + . ‖f ‖2 ‖g‖2 2‖f ‖22 2 ‖g‖22

(1.8)

Integrating both sides of (1.8), we get ∫ ℝ

|fg| 1 1 dm ≤ ∫ |f |2 dm + ∫ |g|2 dm ‖f ‖2 ‖g‖2 2‖f ‖2 2‖g‖2 ℝ

1 1 = ( + ) = 1. 2 2

ℝ

1.4 The space L2 (ℝ)

| 19

Thus ∫ |fg| dm ≤ ‖f ‖2 ‖g‖2 . ℝ

Finally, let us check the equality. To this end, choose a = ‖g‖22 and b = ‖f ‖22 such that a|f |2 = b|g|2 m-a. e., hence |f | = ‖f ‖2

|g| . ‖g‖2

(1.9)

Thus |fg| = |f ||g| = ‖f ‖2

|g|2 . ‖g‖2

Integrating both sides of (1.9), we get ∫ |fg| dm =

‖f ‖2 ∫ |g|2 dm = ‖f ‖2 ‖g‖2 , ‖g‖2 ℝ

ℝ

and the proof of the Cauchy–Schwarz inequality is now complete. Theorem 1.13 (Triangle inequality). Let f , g ∈ L2 (ℝ). Then ‖f + g‖2 ≤ ‖f ‖2 + ‖g‖2 . The equality holds if A|f | = B|g| m-a.e. for A and B of the same sign and not simultaneously zero. Proof. Let f , g ∈ L2 (ℝ). Then ∫ |f + g|2 dm = ∫ |f + g||f + g| dm ℝ

ℝ

≤ ∫ |f ||f + g| dm + ∫ |g||f + g| dm. ℝ

(1.10)

ℝ

Notice that f + g ∈ L2 (ℝ), since L2 (ℝ) is a vector space. Coming back to (1.10) and applying the Cauchy–Schwarz inequality, we obtain ∫ |f + g|2 dm ≤ ∫ |f ||f + g| dm + ∫ |g||f + g| dm ℝ

ℝ

ℝ 1 2

≤ [(∫ |f |2 dm) + (∫ |g|2 dm)](∫ |f + g|2 dm), ℝ

ℝ

ℝ

20 | 1 Fundamental concepts and so 2

1 2

1 2

2

2

1 2

(∫ |f + g| dm) ≤ (∫ |f | dm) + (∫ |g| dm) . ℝ

ℝ

ℝ

Hence ‖f + g‖2 ≤ ‖f ‖2 + ‖g‖2 . Now, let us check the equality. Let A and B be numbers of the same sign not simultaneously zero such that A|f | = B|g| m-a. e., then A‖f ‖2 = B‖g‖2 . Hence 󵄩󵄩 B 󵄩󵄩 B + A 󵄩 󵄩 ‖f + g‖2 = 󵄩󵄩󵄩 g + g 󵄩󵄩󵄩 = ‖g‖2 󵄩󵄩 A 󵄩󵄩 A B = ‖g‖2 + ‖g‖2 A = ‖f ‖2 + ‖g‖2 , and the proof of Theorem 1.10 is complete. Lemma 1.2. Let f ∈ L2 (ℝ). Then ‖f ‖2 = 0 if and only if f = 0 m-a. e. Proof. ‖f ‖2 = 0

if and only if ‖f ‖22 = 0, if and only if

∫ |f |2 dm = 0, ℝ

if and only if |f |2 = 0

m-a. e.,

if and only if |f | = 0 if and only if f = 0

m-a. e., m-a. e.

Example 5. Consider the Dirichlet function 1

if x ∈ ℚ,

0

if x ∈ 𝕀.

f (x) = { Then

󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 ‖f ‖22 = ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dm = ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dm + ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dm ℝ

ℚ

𝕀

󵄨 󵄨2 = ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dm = ∫ 1 dm = m(ℚ) = 0. ℚ

ℚ

1.4 The space L2 (ℝ)

| 21

Then f = 0 m-a. e. However, f (x) ≠ 0 for all x ∈ ℝ. Hence Lemma 1.2 shows that ‖ ⋅ ‖2 is not a norm. To remedy this, we will identify functions that are equal almost everywhere. Formally, this is done by introducing an equivalence relation on L2 (ℝ) and then taking the quotient space. Definition 1.14. We say that two functions f , g ∈ L2 (ℝ) are equivalent if f = g m-a. e. We write f ∼g

if and only if f = g

m-a. e.

Remark 5. By Lemma 1.2, f ∼ g if and only if ‖f − g‖2 = 0. Lemma 1.3. The relation of Definition 1.14 is an equivalent relation. Proof. We have to check that ∼ is reflexive, symmetric, and transitive. That it is reflexive and symmetric is clear. To show transitivity, let f , g, h ∈ L2 (ℝ) be such that f ∼ g and g ∼ h, then ‖f − h‖2 = ‖f − g + g − h‖2 ≤ ‖f − g‖2 + ‖g − h‖2 = 0 and so f ∼ h, as we wished to show. We will write [f ] for the equivalence class of functions f ∈ L2 (ℝ) and then Definition 1.14 could be equivalently stated as f ∼g

if and only if f − g ∈ 𝒩 = {f ∈ L2 (ℝ) : f = 0 m-a. e.},

where we defined [f ] = f + 𝒩 ; note also that [0] = 𝒩 . Definition 1.15. Define L2 (ℝ) = L2 (ℝ)/𝒩 = {f + 𝒩 : f ∈ L2 (ℝ)}. Lemma 1.4. With the operations [f ] + [g] = [f + g]

and [αf ] = α[f ]

for f , g ∈ L2 (ℝ), α ∈ ℝ, L2 (ℝ) is a vector space. Proof. First, we must verify that the operations are indeed well defined, that is, if f , g, f ∗ , g ∗ ∈ L2 (ℝ) and f ∼ f ∗ and g ∼ g ∗ , then f + g ∼ f ∗ + g ∗ . We see that 󵄩󵄩 󵄩 󵄩 ∗ ∗ 󵄩 ∗󵄩 ∗󵄩 󵄩󵄩(f + g) − (f + g )󵄩󵄩󵄩2 ≤ 󵄩󵄩󵄩f − f 󵄩󵄩󵄩2 + 󵄩󵄩󵄩g − g 󵄩󵄩󵄩2 = 0, hence f + g ∼ f ∗ + g ∗ .

22 | 1 Fundamental concepts And so the addition is well defined. A similar, but simpler computation shows that scalar multiplication is also well defined. It is tedious, but not hard to check that the operations satisfy all the vector space axioms. Next, we define the norm of g as 󵄩 󵄩 ‖g‖2 = 󵄩󵄩󵄩[f ]󵄩󵄩󵄩2

for g ∈ [f ].

For arbitrary g1 ∈ [f ] and g2 ∈ [f ], we have g1 = g2 m-a. e. Since g1 ∼ f and g2 ∼ f , |g1 |2 = |g2 |2

m-a. e.,

and so 1 2

1 2

(∫ |g1 |2 dm) = (∫ |g2 |2 dm) , ℝ

ℝ

thus ‖g1 ‖2 = ‖g2 ‖2 . This tell us that ‖g‖2 = ‖[f ]‖2 is well defined, being independent of the representative of the class [f ]. Now, it is clear that for f ∈ L2 (ℝ): (1) ‖[f ]‖2 ≥ 0; (2) ‖[αf ]‖2 = ‖αg2 ‖2 = |α|‖g2 ‖2 = |α|‖[f ]‖2 ; (3) 󵄩󵄩 󵄩 󵄩 󵄩 󵄩󵄩[f ] + [g]󵄩󵄩󵄩2 = 󵄩󵄩󵄩[f + g]󵄩󵄩󵄩2 = ‖F + G‖2 ≤ ‖F‖2 + ‖G‖2 󵄩 󵄩 󵄩 󵄩 = 󵄩󵄩󵄩[f ]󵄩󵄩󵄩2 + 󵄩󵄩󵄩[g]󵄩󵄩󵄩2 for F ∈ [f ] and G ∈ [g]. (4) ‖[f ]‖2 = 0 if and only if ‖g‖2 = 0 for any g ∈ [f ], if and only if g = 0

m-a. e.,

if and only if g ∼ 0

and since f ∼ g then also f ∼ 0,

if and only if f = 0

m-a. e.,

if and only if f ∈ 𝒩 , if and only if f ∈ [0]

since [0] ∈ 𝒩 ,

if and only if f = 0 + [0], if and only if f + [0] = 0 + [0] + [0],

1.4 The space L2 (ℝ)

| 23

if and only if f + [0] = [0 + 0], if and only if f + [0] = [0], if and only if [f ] = [0]. Hence ‖[ ⋅ ]‖2 is a norm on L2 (ℝ). When working with L2 (ℝ), the space under consideration is hence not a space whose elements are functions, but rather a space whose elements are equivalence classes of functions. By abusing notation, we simply talk about functions in L2 (ℝ) rather than equivalence classes of functions. Definition 1.16. For f , g ∈ L2 (ℝ), define ⟨f , g⟩L2 = ∫ f (x)g(x) dm. ℝ

Next, we shall check that ⟨⋅, ⋅⟩ is an inner product in L2 (ℝ). Indeed: (a) ⟨f , f ⟩ = ∫ℝ f (x)f (x) dm = ∫ℝ |f (x)|2 dm ≥ 0 since |f (x)|2 ≥ 0 for all x ∈ ℝ. (b) If ⟨f , f ⟩ = 0 then ∫ℝ |f (x)|2 dm = 0 if and only if |f (x)|2 = 0, 󵄨 󵄨 if and only if 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 = 0, if and only if f = 0. (c) ⟨αf + βg, h⟩L2 = ∫ f (x)g(x) dm = ∫ f (x)g(x) dm ℝ

ℝ

= ∫ f (x)g(x) dm = ∫ g(x)f (x) dm ℝ

ℝ

= ⟨g, f ⟩L2 . Remark 6. The L2 -norm is induced by the standard L2 -inner product, since 1

1

2 2 󵄨 󵄨2 ‖f ‖2 = (∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dm) = (∫ f (x)f (x) dm) = √⟨f , f ⟩L2 .

ℝ

ℝ

Theorem 1.14. L2 (ℝ) is a complete space. Proof. Let {fn }n∈ℕ be a Cauchy sequence in L2 (ℝ). Then for all ϵ > 0, there exists an n0 ∈ ℕ such that ‖fm − fn ‖22 < ϵ2

24 | 1 Fundamental concepts if m, n ≥ n0 . Then by Chebyschev inequality (see Appendix A, Theorem A.6) 󵄨 󵄨 ϵ2 μ({x ∈ ℝ : 󵄨󵄨󵄨fm (x) − fn (x)󵄨󵄨󵄨 ≥ ϵ}) ≤ ‖fm − fn ‖22

if m, n ≥ n0 .

The latter tells us that {fn }n∈ℕ is a Cauchy sequence in measure, therefore, by Riesz theorem (see Appendix A, Theorem A.5), there exists a subsequence {fnk }k∈ℕ of {fn }n∈ℕ that converges m-a. e. to a measurable function f . By Fatou lemma, we have ‖f ‖22 = ∫ |f |2 dm ≤ lim inf ∫ |fnk |2 dm < ∞. n→∞

ℝ

ℝ

So f ∈ L2 (ℝ). Invoking again Fatou lemma, we have ‖fn − f ‖22 = ∫ |fn − f |2 dm ≤ lim inf ∫ |fn − fnk |2 dm < ϵ2 n→∞

ℝ

ℝ

whenever n ≥ n0 . This means that {fn }n∈ℕ converges to f in L2 (ℝ). Theorem 1.15. Let (ℝ × ℝ, L × L , m × m) be a Lebesgue measure space. Suppose that f is L × L -measurable and f (⋅, y) ∈ L2 (m(x)) for all y ∈ ℝ. Then 1

1

󵄨󵄨 󵄨󵄨2 2 2 󵄨 󵄨 󵄨 󵄨2 (∫󵄨󵄨󵄨∫ f (x, y) dm(y)󵄨󵄨󵄨 dm(x)) ≤ ∫(∫󵄨󵄨󵄨f (x, y)󵄨󵄨󵄨 dm(x)) dm(y). 󵄨󵄨 󵄨󵄨 ℝ ℝ

ℝ

ℝ

Proof. By the Cauchy–Schwarz inequality and Fubini theorem, we obtain 󵄨󵄨 󵄨󵄨󵄨2 󵄨 ∫󵄨󵄨󵄨∫ f (x, y) dm(y)󵄨󵄨󵄨 dm(x) 󵄨󵄨 󵄨󵄨

ℝ ℝ

󵄨󵄨 󵄨󵄨󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 󵄨 = ∫󵄨󵄨󵄨∫ f (x, y) dm(y)󵄨󵄨󵄨󵄨󵄨󵄨∫ f (x, y) dm(y)󵄨󵄨󵄨 dm(x) 󵄨󵄨 󵄨󵄨󵄨󵄨 󵄨󵄨 ℝ ℝ

ℝ

󵄨 󵄨󵄨 󵄨 󵄨󵄨󵄨 󵄨 ≤ ∫ ∫󵄨󵄨󵄨f (x, y)󵄨󵄨󵄨󵄨󵄨󵄨∫ f (x, y) dm(y)󵄨󵄨󵄨 dm(y) dm(x) 󵄨󵄨 󵄨󵄨 ℝℝ

ℝ

󵄨 󵄨󵄨 󵄨 󵄨󵄨󵄨 󵄨 = ∫ ∫󵄨󵄨󵄨f (x, y)󵄨󵄨󵄨󵄨󵄨󵄨∫ f (x, y) dm(y)󵄨󵄨󵄨 dm(x) dm(y) 󵄨󵄨 󵄨󵄨 ℝℝ

ℝ

1

1

󵄨󵄨 󵄨󵄨2 2 2 󵄨 󵄨 󵄨 󵄨2 ≤ ∫[(∫󵄨󵄨󵄨f (x, y)󵄨󵄨󵄨 dm) (∫󵄨󵄨󵄨∫ f (x, y) dm(y)󵄨󵄨󵄨 dm(x)) ] dm(y). 󵄨󵄨 󵄨󵄨 ℝ

ℝ

ℝ ℝ

Hence 1

1

󵄨󵄨 󵄨󵄨2 2 2 󵄨 󵄨2 󵄨 󵄨 (∫󵄨󵄨󵄨∫ f (x, y) dm(y)󵄨󵄨󵄨 dm(x)) ≤ ∫(∫󵄨󵄨󵄨f (x, y)󵄨󵄨󵄨 dm(x)) dm(y), 󵄨󵄨 󵄨󵄨 ℝ ℝ

which ends the proof.

ℝ

ℝ

1.4 The space L2 (ℝ)

| 25

Definition 1.17 (Hilbert space). A vector space with an inner product is called a Hilbert space if it is complete with respect to the norm generated by the inner product. Theorem 1.16. The space (L2 (ℝ), ‖ ⋅ ‖L2 (R) ) is a Hilbert space. Definition 1.18. A step function is a function of the form ∑nk=1 αk χIk , where αk ≠ 0, 1 ≤ k ≤ n, and each Ik is a bounded interval. Theorem 1.17. The set of all step functions is dense in L1 (ℝ). Proof. See [8]. Theorem 1.18 (Riemann–Lebesgue). If f : ℝ → ℝ is Lebesgue integrable, then lim ∫ f (x) cos(nx) dx = lim ∫ f (x) sin(nx) dx = 0.

n→∞

n→∞

ℝ

ℝ

Proof. Note first that 󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨f (x) cos(nx)󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 ∈ L1 (ℝ) thus f (x) cos(nx) ∈ L1 (ℝ) for all n ∈ ℕ. Case 1. If f (x) = χ[a,b) (x) then 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨󵄨 b 󵄨󵄨 1 󵄨 󵄨 󵄨󵄨 󵄨 󵄨 󵄨󵄨∫ f (x) cos(nx) dx󵄨󵄨󵄨 = 󵄨󵄨󵄨∫ cos(nx) dx󵄨󵄨󵄨 = 󵄨󵄨󵄨sin(nb) − sin(na)󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨󵄨 a 󵄨󵄨󵄨 n ℝ 2 ≤ → 0 as n → ∞. n Hence limn→∞ ∫ℝ f (x) cos(nx) = 0. Case 2. Let f be a step function such that f = ∑k=1 αk χIk . Hence m

∫ f (x) cos(nx) = ∑ αk ∫ χIk (x) cos(nx) dx. k=1

ℝ

ℝ

By case 1, each term on the right-hand side has limit zero as n → ∞. This proves case 2. Case 3. Let f ∈ L1 (ℝ). By Theorem 1.17, there is a step function φ such that 󵄨 󵄨 ϵ ∫󵄨󵄨󵄨f (x) − φ(x)󵄨󵄨󵄨 < . 2

ℝ

By case 2, lim ∫ φ(x) cos(nx) dx = 0.

n→∞

ℝ

26 | 1 Fundamental concepts Hence, given ϵ > 0, there exists n0 ∈ ℕ such that 󵄨󵄨 󵄨󵄨󵄨 ϵ 󵄨󵄨 󵄨󵄨∫ φ(x) cos(nx)󵄨󵄨󵄨 < 󵄨󵄨 2 󵄨󵄨

whenever n ≥ n0 .

ℝ

If n ≥ n0 then 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨 󵄨󵄨∫ f (x) cos(nx) dx 󵄨󵄨󵄨 = 󵄨󵄨󵄨∫(f (x) − φ(x)) cos(nx) dx + ∫ φ(x) cos(nx) dx󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ℝ

ℝ

ℝ

󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 ≤ ∫󵄨󵄨󵄨f (x) − φ(x)󵄨󵄨󵄨 dx + 󵄨󵄨󵄨∫ φ(x) cos(nx) dx󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 ℝ

ℝ

ϵ ϵ < + = ϵ. 2 2 This means

lim ∫ f (x) cos(nx) dx = 0.

n→∞

ℝ

Similarly, limn→∞ ∫ℝ f (x) sin(nx) dx = 0. In the following example we shall show that the converse of Theorem 1.6 is false. Example 6. Let {xn }n∈ℕ be a sequence in L2 ([0, 2π]) defined by xn =

sin(nt) n

for n = 1, 2, 3, . . . .

Then by the Riemann–Lebesgue theorem 2π

1 lim ∫ y(t) sin(nt)dt = 0 lim ⟨x , y⟩ = n→∞ n π n→∞ 0

ω

for all y ∈ L2 ([0, 2π]). Thus ⟨xn , y⟩ → ⟨0, y⟩, which means that xn → 0. But ‖xn −

0‖2L2 ([0,2π])

for every n ∈ ℕ. Hence xn ↛ 0.

2π

1 1 = 2 lim ∫ sin2 (nt)dt = π π n→∞ 0

1.4 The space L2 (ℝ)

| 27

Exercises. 1. Prove that, for any f ∈ C([0, 1]), 1

2 󵄨󵄨 1 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 1 󵄨󵄨 󵄨󵄨2 󵄨󵄨 󵄨󵄨 󵄨󵄨󵄨∫ f (x) sin(πx) dx 󵄨󵄨󵄨 ≤ √ (∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx) , 󵄨󵄨 󵄨󵄨 2 󵄨0 󵄨 0

2.

and describe the functions f for which equality holds. Find an inner product on the space of complex polynomials such that the corresponding norm is given by 1 2

1

󵄨 󵄨2 󵄨 󵄨2 ‖f ‖ = ( ∫ |x|󵄨󵄨󵄨󵄨f (x)󵄨󵄨󵄨󵄨 + 3󵄨󵄨󵄨󵄨f ′ (x)󵄨󵄨󵄨󵄨 dx) . −1

3.

Prove that, for any continuously differentiable function f on [−π, π], 1

2 󵄨󵄨 π 󵄨󵄨 π 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨2 󵄨󵄨 ′ 󵄨󵄨2 󵄨󵄨 󵄨󵄨 √ ′ 󵄨 󵄨 󵄨 󵄨 󵄨󵄨󵄨 ∫ (f (t) cos t − f (t) sin t) dt 󵄨󵄨󵄨 ≤ 2π( ∫ (󵄨󵄨f (t)󵄨󵄨 + 󵄨󵄨f (t)󵄨󵄨 ) dt) . 󵄨󵄨−π 󵄨󵄨 −π 󵄨 󵄨

4.

Prove that, for any polynomial f , 1

2 󵄨󵄨 1 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 5 󵄨󵄨 󵄨󵄨2 󵄨󵄨 ′ 󵄨󵄨2 󵄨󵄨 󵄨󵄨 3 ′ 󵄨 󵄨 󵄨 󵄨 󵄨󵄨󵄨 ∫ (|x| f (x) + 6xf (x)) dx 󵄨󵄨󵄨 ≤ √ ( ∫ (|x|󵄨󵄨f (x)󵄨󵄨 + 3󵄨󵄨f (x)󵄨󵄨 ) dx) . 3 󵄨󵄨 󵄨󵄨 󵄨−1 󵄨 −1

5.

For f , g ∈ L2 (ℝ), show that ‖f + g‖22 + ‖f − g‖22 = 2(‖f ‖22 + ‖g‖22 ).

6.

Let f , fn ∈ L2 ([0, 1]). Assume (1)

lim ‖fn ‖2 = ‖f ‖2

n→∞

and (2)

7.

1

1

0

0

lim ∫ fn g dm = ∫ fg dm

n→∞

for all g ∈ L2 ([0, 1]).

Show limn→∞ ‖fn − f ‖2 = 0. Let fn , gn ∈ L2 (ℝ) with n ∈ ℕ. If lim ∫(fn − f )2 dm = lim ∫(gn − g)2 dm = 0,

n→∞

n→∞

ℝ

ℝ

prove that lim ∫ fn gn dm = ∫ fg dm.

n→∞

ℝ

ℝ

28 | 1 Fundamental concepts

8.

1

Let f ∈ L2 ([0, 1]) be such that ‖f ‖2 = 1 and ∫0 f dm ≥ α > 0. Also for β ∈ ℝ, let Eβ = {x ∈ [0, 1] : f (x) ≥ β}. If 0 < β < α, prove that m(Eβ ) ≥ (β − α)2 .

9.

This inequality is known in the literature as the Paley–Zygmund inequality. Let us consider a set E ⊂ ℝ with m(E) = 1 and let f , g ∈ L2 (E). If ∫E f dm = 0, show that 2

2

(∫ fg) ≤ [∫ g2 dm − (∫ g dm) ] ∫ f 2 dm. E

E

E

E

10. Let f be a function defined on ℝ such that f (x) and xf (x) belong to L2 (ℝ). Prove that ∞

2

∞

1 2

∞

1 2

󵄨 󵄨 󵄨 󵄨 󵄨2 2󵄨 ( ∫ 󵄨󵄨󵄨󵄨f (x)󵄨󵄨󵄨󵄨) ≤ 8( ∫ 󵄨󵄨󵄨󵄨f (x)󵄨󵄨󵄨󵄨 dx) ( ∫ |x| 󵄨󵄨󵄨󵄨f (x)󵄨󵄨󵄨󵄨 dx) . −∞

−∞

−∞

1.5 A glance at linear operators Let X be a space with an inner product. If A is a nonempty subset of X then the orthogonal complement A⊥ of A is the set of all vectors which are orthogonal to any vector of A, i. e., A⊥ = {x ∈ X : x ⊥ y for all y ∈ A}, where x ⊥ y means that ⟨x, y⟩ = 0. From the linearity and continuity of the inner product (check this!), it is clear that A⊥ is a closed subspace of X such that: (a) 0 ∈ A⊥ . (b) A⊥ = (A)⊥ and A ∩ A⊥ = {0}. (c) X = {0}⊥ and X ⊥ = {0}. Now we gather some more properties of the orthogonal complement Y ⊥ of a subspace Y, and we will write Y ⊥⊥ = (Y ⊥ )⊥ . Theorem 1.19. If Y is a closed vector subspace of a Hilbert space ℋ, then Y ⊥⊥ = Y. Corollary 1.1. If Y is any vector subspace of a Hilbert space ℋ, then Y ⊥⊥ = Y. A mapping T : X → Y between two vector spaces (over the same field) is said to be a linear operator if T(αx + βy) = αT(x) + βT(y) holds for all x, y ∈ X and all scalars α, β.

1.6 The adjoint of an operator

| 29

If X and Y are normed spaces and T : X → Y is an operator, then its (operator) norm is defined by 󵄩 󵄩 󵄩 󵄩 ‖T‖ = sup 󵄩󵄩󵄩T(x)󵄩󵄩󵄩Y = sup 󵄩󵄩󵄩T(x)󵄩󵄩󵄩Y . ‖x‖X =1

‖x‖X ≤1

An operator T : X → Y between two normed spaces is called an isometry (or, more precisely, a linear isometry) if ‖T(x)‖ = ‖x‖ for each x ∈ X. A linear operator T : X → Y between two normed spaces is said to be bounded if there exists M > 0 such that 󵄩󵄩 󵄩 󵄩󵄩T(x)󵄩󵄩󵄩Y ≤ M‖x‖X . The collection of all bounded operators from a normed space X into another normed space Y will be denoted by ℬ(X, Y). We shall write ℬ(X) instead of ℬ(X, X). It turns out that ℬ(X, Y), with the usual algebraic operations and with the operator norm, is a normed space. If X and Y are normed spaces and T : X → Y, the null space, or kernel, and the range of T will be denoted by 𝒩 = Ker(T) and T(X) = ℛ(T) = rang(T), respectively: (a) Ker(T) = 𝒩 (T) = {x ∈ X : T(x) = 0}; (b) T(X) = rang(T) = ℛ(T) = {y ∈ Y : T(x) = y for some x ∈ X}. A linear operator T : X → Y is injective, or one-to-one, if and only if Ker(T) = {0}.

1.6 The adjoint of an operator Definition 1.19. Let X be a normed space over ℝ. The space ℬ(X, ℝ) is called the dual space of X and denoted by X ∗ . Theorem 1.20 (F. Riesz theorem). If ℋ is a Hilbert space and f : ℋ → ℂ is a continuous and linear function, then there exists a unique vector y ∈ ℋ such that f (x) = ⟨x, y⟩ for all x ∈ ℋ. Moreover, ‖f ‖ = ‖y‖. Proof. (Existence). If f (x) = 0 for all x ∈ ℋ then y = 0 will be a suitable choice. Otherwise Ker(f ) = {x ∈ ℋ : f (x) = 0} is a proper closed subspace of ℋ so that [Ker(f )]⊥ ≠ 0. Therefore there exists z ∈ [Ker(f )]⊥ such that f (z) = 1. In particular, z ≠ 0, so we may z define y = ‖z‖ 2 . Since f is a linear transformation, f (x − f (x)z) = f (x) − f (x)f (z) = 0,

30 | 1 Fundamental concepts and hence x − f (x)z ∈ Ker(f ). However, z ∈ (Ker(f ))⊥ so ⟨x − f (x)z, z⟩ = 0. Therefore ⟨x, z⟩ − f (x)⟨z, z⟩ = 0, and so ⟨x, z⟩ = f (x)‖z‖2 . Hence f (x) = ⟨x,

z ⟩ = ⟨x, y⟩. ‖z‖2

Moreover, if ‖x‖ ≤ 1 then by the Cauchy–Schwarz inequality, 󵄨 󵄨 󵄨 󵄨󵄨 󵄨󵄨f (x)󵄨󵄨󵄨 = 󵄨󵄨󵄨⟨x, y⟩󵄨󵄨󵄨 ≤ ‖x‖‖y‖ ≤ ‖y‖. Hence ‖f ‖ ≤ ‖y‖. On the other hand, if x =

y ‖y‖

y then ‖x‖ = ‖ ‖y‖ ‖ = 1, and so

󵄨 󵄨 |f (y)| ⟨y, y⟩ = = ‖y‖. ‖f ‖ ≥ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 = ‖y‖ ‖y‖ Therefore ‖f ‖ ≥ ‖y‖. (Uniqueness). If y and w are such that f (x) = ⟨x, y⟩ = ⟨x, w⟩ for all x ∈ ℋ, then ⟨x, y − w⟩ = 0 for all x ∈ ℋ. If x = y − w, then ⟨y − w, y − w⟩ = 0, and hence y − w = 0 by part (e) in the definition of the inner product. Thus y = w, as required. Theorem 1.21. Let ℋ and 𝒦 be complex Hilbert spaces and let T ∈ ℬ(ℋ, 𝒦). There exists a unique T ∗ ∈ ℬ(ℋ, 𝒦) such that ⟨T(x), y⟩ = ⟨x, T ∗ (y)⟩ for all x ∈ ℋ and y ∈ 𝒦. Proof. Let y ∈ 𝒦 and let f : ℋ → ℂ be defined by f (x) = ⟨Tx, y⟩. Then f is a linear transformation and, by the Cauchy–Schwarz inequality and boundedness of T, 󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨f (x)󵄨󵄨󵄨 = 󵄨󵄨󵄨⟨Tx, y⟩󵄨󵄨󵄨 ≤ ‖Tx‖‖y‖ ≤ ‖T‖‖x‖‖y‖. Hence f is bounded, and so, by the F. Riesz theorem, there exists a unique z ∈ ℋ such that f (x) = ⟨x, z⟩ for all x ∈ ℋ. We define T ∗ (y) = z, so that T ∗ is a function from 𝒦 to ℋ such that ⟨T(x), y⟩ = ⟨x, T ∗ (y)⟩

(1.11)

for all x ∈ ℋ and all y ∈ 𝒦. Thus T ∗ is a function which satisfies the equation in the statement of the theorem, but we have yet to show that it is in ℬ(𝒦, ℋ). It is perhaps not even clear yet that T ∗ is a linear transformation, so the first step is to show this. Let y1 , y2 ∈ 𝒦, α, β ∈ ℂ, and consider x ∈ ℋ. Then by (1.11), ⟨x, T ∗ (αy1 + βy2 )⟩ = ⟨T(x), αy1 + βy2 ⟩ = α⟨T(x), y1 ⟩ + β⟨T(x), y2 ⟩

1.6 The adjoint of an operator

| 31

= α⟨x, T ∗ (y1 )⟩ + β⟨x, T ∗ (y2 )⟩

= ⟨x, αT ∗ (y1 )⟩ + ⟨x, βT ∗ (y2 )⟩. Hence T ∗ (αy1 , βy2 ) = αT ∗ (y1 ) + βT ∗ (y2 ), and so T ∗ is a linear transformation. Next, we shall show that T ∗ is bounded. Using the Cauchy–Schwarz inequality, 󵄩󵄩 ∗ 󵄩󵄩2 ∗ ∗ 󵄩󵄩T (y)󵄩󵄩 = ⟨T (y), T (y)⟩

= ⟨TT ∗ (y), y⟩ 󵄩 󵄩 ≤ 󵄩󵄩󵄩TT ∗ (y)󵄩󵄩󵄩‖y‖ 󵄩 󵄩 ≤ ‖T‖󵄩󵄩󵄩T ∗ (y)󵄩󵄩󵄩‖y‖.

If ‖T ∗ (y)‖ > 0, then ‖T ∗ (y)‖ ≤ ‖T‖‖y‖. While if T ∗ (y) = 0 then trivially ‖T ∗ (y)‖ ≤ ‖T‖‖y‖. Hence for all y ∈ 𝒦, 󵄩󵄩 ∗ 󵄩󵄩 󵄩󵄩T 󵄩󵄩 ≤ ‖T‖‖y‖, and so T ∗ is bounded and 󵄩󵄩 ∗ 󵄩󵄩 󵄩󵄩T 󵄩󵄩 ≤ ‖T‖.

(1.12)

Finally, we have to show that T ∗ is unique. Suppose that B1 an B2 are in ℬ(𝒦, ℋ) and that for all x ∈ 𝒦 and y ∈ 𝒦, ⟨T(x), y⟩ = ⟨x, B1 (y)⟩ = ⟨x, B2 (y)⟩. Hence for x = B1 (y) − B2 (y) we have ⟨B1 (y) − B2 (y), B1 (y) − B2 (y)⟩ = 0, thus B1 (y) = B2 (y) for all y ∈ 𝒦. Therefore B1 = B2 . Hence T ∗ is unique. Definition 1.20. If ℋ and 𝒦 are complex Hilbert spaces and T ∈ ℬ(ℋ, 𝒦) is the operator T ∗ constructed in Theorem 1.16, then it is called the adjoint of T. Example 7. Let ℋ be a complex Hilbert space. If I is the identity operator on ℋ, then I ∗ = I. Proof. If x, y ∈ ℋ. Then ⟨I(x), y⟩ = ⟨x, y⟩ = ⟨x, I(y)⟩. By the uniqueness of adjoint, I ∗ = I. Further properties of the adjoint are given in Theorem 1.22.

32 | 1 Fundamental concepts Theorem 1.22. Let ℋ and 𝒦 be a complex Hilbert spaces and let T ∈ ℬ(ℋ, 𝒦). (a) (T ∗ )∗ = T. (b) ‖T ∗ ‖ = ‖T‖. Proof. (a) We shall show that the adjoint of the adjoint is the original operator ⟨y, (T ∗ ) (x)⟩ = ⟨T ∗ y, x⟩ ∗

(by definition of (T ∗ )∗ )

= ⟨x, T ∗ y⟩

(by definition of an inner product)

= ⟨Tx, y⟩

(by definition of T ∗ )

= ⟨y, Tx⟩

(by definition of an inner product).

Hence (T ∗ )∗ (x) = T(x) for all x ∈ ℋ. Thus (T ∗ )∗ = T. (b) By (1.12), we have ‖T ∗ ‖ ≤ ‖T‖. Applying this result to (T ∗ )∗ and using (a), we have ‖T‖ = ‖(T ∗ )∗ ‖ ≤ ‖T ∗ ‖ ≤ ‖T‖. And so ‖T‖ = ‖T ∗ ‖. Theorem 1.23. Let ℋ and 𝒦 be a complex Hilbert spaces and let T ∈ ℬ(ℋ, 𝒦). Then: (a) Ker(T ∗ ) = [rang(T)]⊥ . (b) Ker(T) = [rang(T ∗ )]⊥ . Proof. (a) y ∈ Ker(T ∗ ) if and only if T ∗ (y) = 0, if and only if ⟨x, T ∗ (y)⟩ = 0

for all x,

if and only if ⟨T(x), y⟩ = 0

for all x,

if and only if y ∈ [rang(T)] . ⊥

(b)

Hence Ker(T ∗ ) = [rang(T)]⊥ . x ∈ Ker(T)

if and only if T(x) = 0, if and only if ⟨T(x), y⟩ = 0

for all y,

if and only if ⟨x, T (y)⟩ = 0

for all y,

∗

if and only if x ∈ [rang(T )] , ∗

hence Ker(T) = [rang(T ∗ )]⊥ .

⊥

1.6 The adjoint of an operator

| 33

Definition 1.21. Let ℋ be a complex Hilbert space. A linear operator T : ℋ → ℋ is an isometry if 󵄩 󵄩󵄩 󵄩󵄩T(x)󵄩󵄩󵄩 = ‖x‖ for all x ∈ ℋ. Definition 1.22. Let ℋ be a complex Hilbert space and T : ℋ 󳨀→ ℋ be a linear operator. The operator T is said to be a unitary operator if T ∗ T = TT ∗ = I. Theorem 1.24. Let ℋ be a complex Hilbert space and T : ℋ → ℋ be a linear operator. The following are equivalent: (a) T is unitary. (b) T ∗ is unitary. (c) T and T ∗ are isometries. (d) T and T ∗ are one-to-one. Proof. (a) ⇒ (b) Since T = (T ∗ )∗ TT ∗ = T ∗ T = I, one gets (T ∗ )∗ T ∗ = T ∗ (T ∗ )∗ = I. (b) ⇒ (c) 󵄩󵄩 󵄩2 ∗ 2 󵄩󵄩T(x)󵄩󵄩󵄩 = ⟨T(x), T(x)⟩ = ⟨x, T T(x)⟩ = ⟨x, I(x)⟩ = ⟨x, x⟩ = ‖x‖ for all x ∈ ℋ. Thus ‖T(x)‖ = ‖x‖ for all x ∈ ℋ. Also 󵄩󵄩 ∗ 󵄩󵄩2 ∗ ∗ ∗ ∗ ∗ 󵄩󵄩T (x)󵄩󵄩 = ⟨T (x), T (x)⟩ = ⟨x, (T ) T (x)⟩ = ⟨x, I(x)⟩ = ⟨x, x⟩ for all x ∈ ℋ = ‖x‖2 . And so ‖T ∗ (x)‖ = ‖x‖ for all x ∈ ℋ. (c) 󳨐⇒ (d) If T(x) = T(y) then 󵄩 󵄩 󵄩 󵄩 0 = 󵄩󵄩󵄩T(x) − T(y)󵄩󵄩󵄩 = 󵄩󵄩󵄩T(x − y)󵄩󵄩󵄩 = ‖x − y‖ and 󵄩 󵄩 󵄩 󵄩 0 = 󵄩󵄩󵄩T ∗ (x) − T ∗ (y)󵄩󵄩󵄩 = 󵄩󵄩󵄩T ∗ (x − y)󵄩󵄩󵄩 = ‖x − y‖, and so in both cases we have x = y. Remark 7. It is clear that T and T ∗ are uniformly continuous. Theorem 1.25. Let T : ℋ → ℋ be unitary, then rang(T) is a closed subset of ℋ. Proof. Since T is linear, rang(T) is a subspace of ℋ. Letting y ∈ rang(T), we shall prove that y ∈ rang(T). Indeed, if y ∈ rang(T) then there exists a sequence {yn }n∈ℕ in ℋ such that yn → y in ℋ where yn = T(xn ) for xn ∈ ℋ (n ∈ ℕ). Since 󵄩 󵄩 ‖xn − xm ‖ = 󵄩󵄩󵄩T(xn ) − T(ym )󵄩󵄩󵄩 = ‖yn − ym ‖ → 0

as n, m → ∞,

34 | 1 Fundamental concepts {xn }n∈ℕ is a Cauchy sequence in ℋ. The completeness of ℋ implies that xn → x ∈ ℋ. By Remark 7 and the uniqueness of the limit, we conclude y = lim yn = lim T(xn ) n→∞

n→∞

= T( lim xn ) n→∞

= T(x), hence y ∈ rang(T). Exercises. 1. Define ϕ : L2 ([a, b]) 󳨀→ L2 ([a, b]) by ϕ(f )(x) = xf (x)

2.

for all x ∈ [a, b].

Prove that ϕ is a linear and bounded operator with ‖ϕ‖ = max{|a|, |b|}. Let T : L2 ((a, b)) → ℝ be defined by b

T (f ) = ∫ f (x) dx a

3.

for f ∈ L2 ((a, b)). Show that T is linear and bounded, and compute ‖T ‖. Show that 󵄨󵄨 ∞ ∞ 󵄨󵄨 󵄨󵄨 󵄨󵄨 f (x)g(y) 󵄨󵄨 󵄨 dxdy 󵄨󵄨󵄨 ≤ π‖f ‖2 ‖g‖2 󵄨󵄨 ∫ ∫ 󵄨󵄨 󵄨󵄨 x + y 󵄨󵄨 0 0 󵄨󵄨

4.

for f , g ∈ L2 ((0, ∞)). Let K be a nonnegative measurable function on (0, ∞) such that ∞

∫ K(x)x s−1 dx = φ(s) 0

for 0 < s < 1. Let f be a nonnegative measurable function in (0, ∞). Let us define ∞

Tf (x) = ∫ k(xy)f (y) dy. 0

Prove that 1 ‖Tf ‖2 ≤ φ( )‖f ‖2 . 2 What can be said about Tf and φ(s) if K(x) = e−x .

1.6 The adjoint of an operator

5.

| 35

Let K : [a, b] × [a, b] → ℝ be defined by 0

K(s, t) = {

1

if a ≤ t ≤ s ≤ b, if a ≤ s ≤ t ≤ b,

and V : L2 ([a, b]) → L2 ([a, b]) be the operator defined by b

t

a

a

Vx(t) = ∫ K(s, t)x(s) ds = ∫ x(s) ds for x ∈ L2 ([a, b]). This operator is known as the Volterra operator. (a) Show that ‖V ‖ ≤ b − a. (b) Show that the adjoint operator of the Volterra operator is given by b

V ∗ (y(s)) = ∫ y(t) dt. s

6.

(Hardy inequality) Let f be a positive and measurable function in (0, ∞). The Hardy operator is defined by Hf (x) =

x

1 ∫ f (y) dy. x 0

The Hardy operator is an average operator. Let f ∈ L2 ((0, ∞)). Prove that ‖Hf ‖2 ≤ 2‖f ‖2 . 7. 8.

Let f be a differentiable function such that f (−∞) = 0 and f (∞) = 0. Let D be the differentiation d operator D = dx . Show that the adjoint operator of D is −D. Show that the adjoint operator of the Hardy operator H : L2 (0, ∞) → L2 (0, ∞) is given by ∞

H∗ f (y) = ∫ f (x) y

9.

dx x

for f ≥ 0. Let T : L2 (ℝ) → L2 (ℝ) be defined by T (f )(x) = |x|−1 f ( x1 ). Prove that 󵄩󵄩 󵄩 󵄩󵄩T (f )󵄩󵄩󵄩 = ‖f ‖2 . 󵄩 󵄩2

2 Fourier series We will consider Fourier series on 𝕋 := ℝ \ 2πℤ, that is, a function f : 𝕋 → ℝ is a 2πperiodic function on ℝ. Conveniently, we shall abuse notation at various points and consider the domain of such functions to be [0, 2π] or [−π, π]. We also define C(𝕋) to be the set of continuous functions on [−π, π] which are 2π-periodic. Also L1 (T) will be the set of L1 -functions on [−π, π] which are 2π-periodic, and so on. A function f : 𝕋 → ℝ is 2π-periodic if it satisfies the equation f (x + 2π) = f (x)

for all x ∈ ℝ.

Let us consider the following set of trigonometric functions: {1, cos(nx), sin(nx) : n ∈ ℤ+ }. Notice that for n ∈ ℤ+ , π

π

∫ cos(nx) cos(mx) dx = {

0

−π

if n = m, if n ≠ m,

and π

π

∫ sin(nx) sin(mx) dx = { −π

0

if n = m,

if n ≠ m.

Also π

∫ cos(nx) sin(mx) dx = 0, −π

π

∫ cos(nx) dx = 0, −π π

∫ sin(mx) dx = 0. −π

This shows that {1, cos(nx), sin(nx) : n ∈ ℤ+ } is an orthogonal system. Moreover, as a consequence of Weierstrass approximation theorem (see [9, Theorem 3.23]), we have that the system is actually an orthogonal basis for the space L2 ([−π, π]). Hence, we have that the set {

1 cos(nx) sin(nx) , , : n ∈ ℤ+ } √2π √π √π

https://doi.org/10.1515/9783110784091-002

38 | 2 Fourier series is an orthogonal basis for L2 (−π, π). Thus, given f ∈ L2 ([−π, π]), we have, by Remark 3 and Theorem 1.5, that f (x) =

a0 ∞ + ∑ a cos(nx) + bn sin(nx). 2 k=1 n

(2.1)

The coefficients a0 , a1 , . . . , b1 , b2 , . . . can be determined as follows: integrating both sides of (2.1) from −π to π gives π

∫ f (x) dx = −π

π

π

π

−π

−π

−π

∞ a0 ∫ dx + ∑ (an ∫ cos(nx) dx + bn ∫ sin(nx) dx). 2 n=1

(2.2)

Since π

π

∫ cos(nx) dx = 0

and

∫ sin(nx) dx = 0,

−π

−π

we get that π

π

−π

−π

a ∫ f (x) dx = 0 = ∫ dx = πa0 . 2

Solving for a0 yields π

1 a0 = ∫ f (x) dx. π −π

Now we multiply (2.1) by cos(mx) and integrate: π

∫ f (x) cos(mx) dx −π

π

π

π

−π

−π

−π

∞ a = 0 ∫ cos(mx)dx + ∑ (an ∫ cos(nx) cos(mx)dx + bn ∫ sin(nx) cos(mx)dx). 2 n=1

By orthogonality, we have π

∫ cos(mx) dx = 0, −π

π

m > 0,

∫ cos(mx) sin(nx) dx = 0, −π

(2.3)

2 Fourier series | 39

and π

0

∫ cos(nx) cos(mx) dx = { −π

π

if n ≠ m,

if n = m.

Thus (2.3) reduces to π

∫ f (x) cos(nx) dx = an π, −π

and so π

1 an = ∫ f (x) cos(nx) dx. π −π

Finally, if we multiply (2.2) by sin(nx), integrate, and make use of the results π

π

∫ sin(mx) dx = 0,

m > 0,

∫ sin(mx) sin(nx) dx = 0, −π

−π

and π

0 ∫ sin(nx) sin(mx) dx = { π −π

if n ≠ m,

if n = m,

we find that π

1 bn = ∫ f (x) sin(nx) dx. π −π

Definition 2.1 (Fourier coefficients). Given f : T → ℝ, define the nth Fourier coefficient by π

̂f (n) = 1 ∫ f (y)e−iny dy. 2π −π

Note that, since |e−inx | ≤ 1, we have 1 󵄨󵄨̂ 󵄨󵄨 ‖f ‖ 󵄨󵄨f (n)󵄨󵄨 ≤ 2π L1 (T) if f ∈ L1 (T).

40 | 2 Fourier series Theorem 2.1 (Riemann–Lebesgue lemma). If f ∈ L1 (T), then 󵄨 󵄨 lim 󵄨󵄨̂f (n)󵄨󵄨󵄨 = 0.

n→∞󵄨

Proof. For a ∈ ℝ, we define fa (x) = f (x − a). Its nth Fourier coefficient is given by 2π

̂f (n) = 1 ∫ f (y)e−iny dy a a 2π 0

2π

1 = ∫ f (y − a)einy dy. 2π 0

We now let z = y − a. Then the limits of integration do not change because f is 2πperiodic. So we obtain 2π

2π

0

0

̂f (n) = 1 ∫ f (z)e−in(z+a) dz = e−ina 1 ∫ f (z)e−inz dz = e−ina ̂f (n). a 2π 2π Now, choose a = πn . Then e−ina = e−iπ = 1, and for such a, we have 󵄨 󵄨 󵄨 󵄨 2󵄨󵄨󵄨̂f (n)󵄨󵄨󵄨 = 󵄨󵄨󵄨̂f (n) − (−̂f (n))󵄨󵄨󵄨 󵄨 󵄨 = 󵄨󵄨󵄨̂f − e−ina ̂f (n)󵄨󵄨󵄨 󵄨 󵄨 = 󵄨󵄨󵄨̂f (n) − ̂fa (n)󵄨󵄨󵄨

2π 2π 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 1 1 −iny 󵄨 dy − = 󵄨󵄨 ∫ f (y)e ∫ f (y − a)e−iny dy󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 2π 2π 󵄨 󵄨 0 0

=

󵄨 2π 󵄨󵄨 󵄨󵄨 1 󵄨󵄨󵄨󵄨 −iny 󵄨󵄨 (f (y) − f (y − a))e dy 󵄨∫ 󵄨󵄨 2π 󵄨󵄨󵄨󵄨 󵄨󵄨 0

2π

≤

1 󵄨󵄨 󵄨 ∫ 󵄨f (y) − f (y − a)󵄨󵄨󵄨 dy. 2π 󵄨 0

If f ∈ C(𝕋), then |f (y) − f (y − πn )| → 0 as n → ∞ for every y ∈ [0, 2π]. Hence by the Lebesgue dominated convergence theorem, 2π

󵄨󵄨 π 󵄨󵄨󵄨 󵄨 lim ∫ 󵄨󵄨󵄨f (y) − (y − )󵄨󵄨󵄨 dx = 0. n→∞ 󵄨󵄨 n 󵄨󵄨 0

2 Fourier series | 41

In general, if f ∈ L1 (T), there exists g ∈ C(T) ∩ L1 (T) such that ‖f − g‖L1 (T) < ϵ2 . Take N0 large enough so that |ĝ (n)| < ϵ2 whenever n ≥ N0 , then 󵄨 󵄨 󵄨 󵄨󵄨̂ 󵄨󵄨 󵄨󵄨̂ 󵄨󵄨f (n)󵄨󵄨 = 󵄨󵄨f (n) − ĝ (n)󵄨󵄨󵄨 + 󵄨󵄨󵄨ĝ (n)󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 − g)(n)󵄨󵄨󵄨 + 󵄨󵄨󵄨ĝ (n)󵄨󵄨󵄨 = 󵄨󵄨󵄨(f? 󵄨 󵄨 = ‖f − g‖L1 (T) + 󵄨󵄨󵄨ĝ (n)󵄨󵄨󵄨 ϵ ϵ < + = ϵ whenever n ≥ N. 2 2 Next, the fundamental question is whether we can recover f as a Fourier series. Define N

SN f (x) = ∑ ̂f (k)eikx . k=−N

In order to study the convergence of Fourier series, it will be helpful to write SN f (x) as follows: π

N N 1 SN f (x) = ∑ ̂f (k)eikx = ∑ ( ∫ f (y)e−iky dy)eikx 2π k=−N k=−N −π

π

=

π

N 1 1 ∫ f (y) ∑ eik(x−y) dy = ∫ f (y)DN (x − y) dy, 2π 2π k=−N −π

−π

where DN (t) = ∑Nk=−N eikt is the Dirichlet kernel. The Dirichlet kernel is a somewhat awkward sum. In the following result, we can see that it can be reduced to a single quotient of sines. Lemma 2.1. N

DN (x) = ∑ einx =

sin[(N + 21 )x] sin( x2 )

k=−N

.

Proof. We compute N

2N

2N

k=−N

k=0

k=0

DN (x) = ∑ eikx = ∑ ei(k−N)x = eiNx ∑ eikx = eiNx [1 + eix + ⋅ ⋅ ⋅ + ei(2N)x ] = e−iNx [

1 − ei(2N+1)x ]. 1 − eix

Notice that ei(2N+1)x = ei(

2N+1 2N+1 + 2 )x 2 1

1

1

= ei(N+ 2 +N+ 2 )x 1

= ei(N+ 2 x) ei(N+ 2 x) =

1

ei(N+ 2 x) 1

e−i(N+ 2 x)

.

42 | 2 Fourier series Hence

e−iNx [

i(2N+1)x

1−e 1 − eix

] = e−iNx [

1

ei(N+ 2 x)

1−

1

e−i(N+ 2 x) ix e2 ix e− 2

1− 1

= eiNx [

] 1

e−i[(N+ 2 )x] −ei(N+ 2 )x 1 e−i(N+ 2 )x ix

]

ix

e− 2 −e 2 ix e− 2

1

1

1

e−i(N+ 2 )x e−(N+ 2 )x − ei(N+ 2 )x = [ ] 1 ix ix e−i(N+ 2 )x e− 2 − e 2 1

= =

1

ei(N+ 2 )x − ei(N+ 2 )x ix

ix

e2 −e2 2i sin[(N + 21 )x] 2i sin( x2 )

=

sin[(N + 21 )x] sin( x2 )

where we use 1 cos x = (eix + e−ix ) 2

and

sin x =

1 ix (e − eix ). 2i

Hence DN (x) =

sin[(N + 21 )x] sin( x2 )

.

Note that π

π

N

∫ DN (x) dx = ∫ ∑ eiky dy = 2π. −π

−π k=−N

Lemma 2.2. The sequence {DN }n∈ℤ is unbounded in L1 ([−π, π]). Proof. π

π 󵄨󵄨 sin[(N + 1 )x] 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 2 2π‖DN ‖1 = ∫ 󵄨󵄨DN (x)󵄨󵄨 dx = ∫ 󵄨󵄨󵄨 󵄨󵄨 dx 󵄨󵄨 󵄨󵄨 sin( x2 ) −π −π π

π

2 󵄨󵄨 sin(2N + 1) x 󵄨󵄨 󵄨󵄨 sin(2N + 1)y 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 2 󵄨󵄨 = ∫ ∫ 󵄨󵄨 󵄨󵄨 dx ≥ 4 ∫󵄨󵄨󵄨 󵄨󵄨 dy x 󵄨󵄨 sin( ) 󵄨󵄨 󵄨󵄨 󵄨󵄨 y 2 −π

0

,

2 Fourier series | 43

2N

(2N+1) π2

=4∑

k=0

=

∫ k π2

2N 󵄨󵄨 sin z 󵄨󵄨 2 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 dz ≥ 4 ∑ (k + 1)π 󵄨󵄨 z 󵄨󵄨 k=0

(2N+1) π2

∫

| sin z| dz

k π2

8 2N 1 . ∑ π k=0 k + 1

Hence supN ‖DN ‖1 = ∞. Theorem 2.2. There exists an f ∈ C([−π, π]) whose Fourier series diverges at the origin. Proof. For all N ∈ ℕ, consider UN : C([−π, π]) → ℝ defined by π

1 UN (f ) = SN f (0) = ∫ f (x)DN (x) dx. 2π −π

Hence π

π

−π

−π

1 1 󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨 󵄨 󵄨 ∫ 󵄨󵄨f (x)󵄨󵄨󵄨󵄨󵄨󵄨DN (x)󵄨󵄨󵄨 dx ≤ ‖f ‖∞ ( ∫ 󵄨󵄨D (x)󵄨󵄨 dx) 󵄨󵄨UN (f )󵄨󵄨󵄨 ≤ 2π 󵄨 2π 󵄨 N 󵄨 implies that UN is continuous with π

󵄨 󵄨 ‖UN ‖ ≤ ∫ 󵄨󵄨󵄨DN (x)󵄨󵄨󵄨 dx = ‖DN ‖1 −π

for all N ∈ ℕ. We define fN (x) = sign (DN (x)), which is a discontinuous function at the zeros of DN (x), which has a finite number of zeros. Indeed, DN (x) = 0 if and only if sin[(N + 1 )x] = 0, x ≠ 0, if and only if (N + 21 )x = ±kπ with k = 1, 2, . . . , N, if and only if x = ± πk1 2 (N+ 2 )

with k = 1, 2, . . . , N. Hence DN (x) has 2N zeros. Given ϵ > 0 small enough, let fNϵ : [−π, π] → ℝ denote the continuous piecewise affine function that is equal to fN on [−π, π] \ INϵ where INϵ denotes the intersection of [−π, π] with the union of the open intervals of length ϵ centered at the 2N zeros of the Dirichlet kernel DN that belong to the interval [−π, π]. In Figure 2.1 we can see how the function fNϵ behaves for n = 3. Also note that ‖fNϵ ‖ = 1. Then, by the dominated convergence theorem, 󵄨󵄨 ϵ 󵄨 󵄨 󵄨 󵄨󵄨(fN (x) − f (x))DN (x)󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨DN (x)󵄨󵄨󵄨,

44 | 2 Fourier series

Figure 2.1: Graph of the continuous function fNϵ for N = 3.

hence π 󵄨󵄨 󵄨󵄨 π 󵄨󵄨 󵄨 󵄨󵄨UN (f ϵ ) − ∫ 󵄨󵄨󵄨DN (x)󵄨󵄨󵄨 dx 󵄨󵄨󵄨 ≤ ∫ 󵄨󵄨󵄨(f ϵ (x) − fN (x))DN (x)󵄨󵄨󵄨 dx → 0, N 󵄨󵄨 󵄨 󵄨 󵄨󵄨 󵄨 N 󵄨 󵄨󵄨 󵄨󵄨 −π −π

ϵ → 0,

since ‖fNϵ ‖ = 1. And so 󵄨 󵄨 ‖UN ‖ ≥ 󵄨󵄨󵄨UN (fNϵ )󵄨󵄨󵄨 → ‖DN ‖1

as ϵ → 0.

By Lemma 2.1, sup ‖UN ‖ = sup ‖DN ‖1 = ∞. N

N

Finally, by Banach–Steinhaus theorem (see Theorem A.8), there exists an f C([−π, π]) such that

∈

󵄨 󵄨 sup󵄨󵄨󵄨UN (f )󵄨󵄨󵄨 = ∞, N

which proves Theorem 2.2. Theorem 2.3. Let f ∈ L1 (T). Suppose f (y) = 0 for all y ∈ (x − δ, x + δ). Then SN f (x) → 0 as N → ∞. Proof. Let A = [−π, π] \ [−δ, δ] and consider SN f (x) =

=

π π

π

−π −π π

−π

1 ∫ ∫ f (y)DN (x − y) dy = ∫ f (y)DN (x − y) dy 2π 1 1 ∫ f (x − y)DN (y) dy = ∫ f (x − y)DN (y) dy 2π 2π −π

A

1 f (x − y) 1 = ∫ y sin[(N + )y] dy. 2π 2 sin( 2 ) A

2 Fourier series | 45

Note that y 󳨃→

f (x−y) sin( y2 )

is an L1 -function on A, so π

1

1

1 f (x − y) ei(N+ 2 )y − e−i(N+ 2 )y SN f (x) = dy ∫ χA 2π 2i sin( y2 ) −π

=

π

π

−π

−π

1 1 f (x − y) iy2 iNy 1 1 f (x − y) −iy2 −iNy dy − e e dy. ∫ χA ∫ χA y e e 2π 2i 2π 2i sin( 2 ) sin( y2 ) iy

iy

(x−y) 2 (x−y) − 2 Setting g(y) = χA (y) fsin( and h(y) = χA (y) fsin( , we have y e y e ) ) 2

2

SN f (x) =

π

π

−π

−π

1 1 1 1 ∫ g(y)eiNy dy − ∫ h(y)e−iNy dy. 2π 2i 2π 2i

1 1̂ = ĝ (−N) − h(N) → 0, 2i 2i

N → ∞,

by the Riemann–Lebesgue lemma. The following result give us the minimal hypothesis to ensure that SN f (x) converge to f (x). Theorem 2.4 (Dini’s convergence theorem). Let f ∈ L1 (T). Suppose that there exists δ > 0 such that 󵄨󵄨 f (x − t) − f (x) 󵄨󵄨 󵄨 󵄨󵄨 ∫ 󵄨󵄨󵄨 󵄨󵄨 dt < +∞. 󵄨󵄨 󵄨󵄨 t

|t|a

󵄨 󵄨󵄨 󵄨 󵄨 󵄨 ≤ ∫ 󵄨󵄨󵄨x(ξ0 − ξ )󵄨󵄨󵄨󵄨󵄨󵄨f (x)󵄨󵄨󵄨dx + 2 ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨dx |x|≤a

|x|>a

70 | 2 Fourier series 󵄨 󵄨 󵄨 󵄨 = |ξ − ξ0 | ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨|x|dx + 2 ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨dx |x|>a

|x|≤a

󵄨 󵄨 󵄨 󵄨 ≤ a|ξ − ξ0 | ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨dx + 2 ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨dx. |x|>a

|x|≤a

If ϵ > 0, we can select a > 0 such that we have the estimate ϵ 󵄨 󵄨 ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨dx < . 4

|x|>a

For this fixed number a, it follows that if δ=

ϵ , 2a[∫|x|≤a |f (x)|dx + 1]

then |ξ − ξ0 | < δ implies |̂f (ξ ) − ̂f (ξ0 )| < ϵ, and the proof of continuity is complete. Theorem 2.20. Suppose f ∈ L1 (ℝ), and α and λ are real numbers. (a) If g(x) = f (x)eiαx , then ĝ (ξ ) = ̂f (ξ − α). (b) If g(x) = f (x − α), then ĝ (ξ ) = ̂f (ξ )e−iαx . (c) If g ∈ L1 (ℝ), then (f? ∗ g)(ξ ) = ̂f (ξ )ĝ (ξ ). Thus the Fourier transform converts multiplication by a character into translation, and vice versa, and it converts convolution to pointwise product, and the product into a pointwise convolution. (d) If g(x) = f (−x), then ĝ (ξ ) = ̂f (ξ ). (e) If g(x) = f (x/λ) and λ > 0, then ĝ (ξ ) = λ̂f (λξ ). d̂f (ξ ) (f) If xf (x) ∈ L (ℝ), then ̂f ∈ C 1 and = (−2ixf )̂(ξ ). 1

dξ

(g) If f (x) → 0 as |x| → ∞, then (f̂′ )(ξ ) = 2iξ ̂f (ξ ).

Proof. (a) ĝ (ξ ) = ∫ g(x)e−ixξ dx ℝ

= ∫ f (x)eiαx e−ixξ dx ℝ

= ∫ f (x)e−ix(ξ −α) dx ℝ

= ̂f (ξ − α).

2.2 Fourier transform on ℝ

(b) ĝ (ξ ) = ∫ g(x)e−ixξ dx ℝ

= ∫ f (x − α)e−ixξ dx ℝ

= ∫ f (y)e−i(y+α)ξ dy ℝ

= e−iαx ∫ f (y)e−iyξ dy =e

ℝ −iαx ̂

f (ξ ).

(c) By Fubini theorem, we have (f? ∗ g)(ξ ) = ∫(f? ∗ g)(x)e−ixξ dx ℝ

= ∫[∫ f (x − y)g(y)dy]e−ixξ dx ℝ ℝ

= ∫[∫ f (x − y)e−ixξ dx]g(y)dy ℝ ℝ

= ∫[∫ f (w)e−i(w+y)ξ dw]g(y)dy ℝ ℝ

= ∫[∫ f (w)e−iwξ dw]g(y)e−iyξ dy ℝ ℝ

= (∫ f (w)e−iwξ dw)(∫ g(y)e−iyξ dy) ℝ

= ̂f (ξ )ĝ (ξ ).

ℝ

(d) ĝ (ξ ) = ∫ g(x)e−ixξ dx ℝ

= ∫ f (−x)e−ixξ dx ℝ

= ∫ f (x)eixξ dx ℝ

= ∫ f (x)e−ixξ dx = ̂f (ξ ). ℝ

| 71

72 | 2 Fourier series (e) ĝ (ξ ) = ∫ g(x)e−ixξ dx ℝ

= ∫ f (x/λ)e−ixξ dx ℝ

= ∫ f (u)e−iλuξ du ℝ

= λ ∫ f (u)e−iuλξ du ℝ

= λ̂f (λξ ). (f) By the Leibniz integral rule, we have d̂ d f (ξ ) = ∫ f (x)e−ixξ dx dξ dξ ℝ

= ∫ f (x) ℝ

d −ixξ e dx dξ

= ∫ f (x)(−ix)e−ixξ dx ℝ

= (−ixf )̂(ξ ). (g) Integrating by parts, we get f̂′ (ξ ) = ∫ f ′ (x)e−ixξ dx ℝ

󵄨∞ = f (x)e−ixξ 󵄨󵄨󵄨−∞ − ∫ f (x)(−iξ )e−ixξ dx ℝ

= iξ ∫ f (x)e

−ixξ

dx

ℝ

= iξ ̂f (ξ ). And the proof of Theorem 2.20 is complete. Theorem 2.21. Let a ≠ 0 a real number. If T : ℝ → ℝ is given by T(x) = ax then (f ∘ T)̂(ξ ) = provided the Fourier transform of f exists.

1 ̂ −1 (f ∘ T )(ξ ), |a|

2.2 Fourier transform on ℝ

| 73

Proof. Suppose ̂f exists. Then (f ∘ T)̂(ξ ) = ∫(f ∘ T)(x)e−ixξ dx ℝ

r

= lim ∫ f (ax)e−ixξ dx r→0

−r ar

= lim ∫ f (y)e−yξ /a r→0

=

−ar ∞

dy a

1 ∫ f (y)e−iyξ /a dy |a| −∞

1 ̂ ξ f( ) |a| a 1 ̂ −1 = f (T (ξ )) |a| 1 ̂ −1 = (f ∘ T )(ξ ). |a| =

And so the proof is complete. Theorem 2.22 (Riemman–Lebesgue theorem). Let f ∈ L1 (ℝ). Then lim ̂f (ξ ) = 0.

|ξ |→∞

Proof. By Theorem 2.20, we have eiyξ ̂f (ξ ) = ∫ f (x + y)e−ixξ dx. ℝ

For a fixed ξ , choose y =

πξ |ξ |2

so that eiyξ = eiπ = −1. Therefore, ̂f (ξ ) = − ∫ f (x + πξ )e−ixξ dx. |ξ |2 ℝ

Adding ̂f (ξ ) = ∫ℝ f (x)e−ixξ dx to (2.19), we obtain πξ 2̂f (ξ ) = ∫(f (x) − f (x + 2 ))e−ixξ dx. |ξ | ℝ

(2.19)

74 | 2 Fourier series Hence 󵄨 πξ 󵄨󵄨󵄨 󵄨󵄨̂ 󵄨󵄨 1 󵄨󵄨󵄨 󵄨󵄨f (ξ )󵄨󵄨 ≤ ∫󵄨󵄨f (x) − f (x + 2 )󵄨󵄨󵄨dx 2 󵄨󵄨 |ξ | 󵄨󵄨 ℝ

1 󵄨 󵄨 = ∫󵄨󵄨󵄨f (x) − f (x + y)󵄨󵄨󵄨dx. 2 ℝ

Then 󵄨 󵄨 󵄨 󵄨 lim 󵄨󵄨󵄨̂f (ξ )󵄨󵄨󵄨 ≤ lim ∫󵄨󵄨󵄨f (x) − f (x + y)󵄨󵄨󵄨dx = 0. y→0

|ξ |→∞

ℝ

The equality follows from the continuity of transform in the L1 -norm. 2 2 Theorem 2.23. Let f (x) = e−|x| , then ̂f (ξ ) = √πe−ξ .

Proof. By Theorem 2.20(f), we can write 2 d̂f (ξ ) = (−2ixe−x )̂(ξ ) dξ ? 2 ′ = i[(e−x ) ](ξ )

= i(f̂′ )(ξ ) = i(2iξ )̂f (ξ ) = −2ξ ̂f (ξ ). 2 Hence ln ̂f (ξ ) = −ξ 2 + C and so ̂f (ξ ) = Ce−ξ . Now, we shall evaluate the constant C. 2 Having ̂f (0) = C implies ∫ℝ e−x dx = C, thus C = √π. 2 Finally, ̂f (ξ ) = √πe−ξ , and the proof is complete.

Now, let H(x) = have

1 −|x| e 2π

and define hλ (ξ ) = ∫ℝ H(λx)e−ixξ dx. By Example 8(b), we hλ (ξ ) =

λ 1 , 2 π λ + ξ2

and hence ∫ℝ hλ (ξ )dξ = 1. Note also that 0 < H(x) ≤ 1 and limλ→0 H(λx) = 1. Theorem 2.24. Let f ∈ L1 (ℝ), then (f ∗ hλ )(x) = ∫ℝ H(λξ )̂f (ξ )eixξ dξ .

2.2 Fourier transform on ℝ

Proof. Applying Fubini theorem, we have (f ∗ hλ )(x) = ∫ f (x − y)hλ (y)dy ℝ

= ∫ f (x − y)(∫ H(λξ )e−iξy dξ )dy ℝ

ℝ

= ∫ H(λξ )[∫ f (x − y)e−iξy dy]dξ ℝ

ℝ

= ∫ H(λξ )[∫ f (w)e−i(x−w)ξ dw]dξ ℝ

ℝ

= ∫ H(λξ )e

ixξ

[∫ f (w)e−iwξ dw]dξ

ℝ

ℝ

= ∫ H(λξ )̂f (ξ )eixξ dξ . ℝ

Theorem 2.25. If g ∈ L∞ (ℝ) and g is continuous at a point x, then lim (g ∗ hλ )(x) = g(x).

λ→0

Proof. Since ∫ℝ hλ (y)dy = 1, we get (g ∗ hλ )(x) − g(x) = ∫ g(x − y)hλ (y)dy − ∫ g(x)hλ (y)dy ℝ

ℝ

= ∫(g(x − y) − g(x))hλ (y)dy. ℝ

On the one hand, note that ξ ξ h1 (x) = ∫ H(λ )ei λ x dξ = λ ∫ H(λw)eiwx dw = λhλ (x), λ

ℝ

ℝ

hence λ−1 h1 ( xλ ) = hλ (x). On the other hand, y ∫(g(x − y) − g(x))hλ (y)dy = ∫(g(x − y) − g(x))λ−1 h1 ( )dy. λ

ℝ

ℝ

Hence (g ∗ hλ )(x) − g(x) = ∫(g(x − λw) − g(x))h1 (w)dw. ℝ

| 75

76 | 2 Fourier series Moreover, 󵄨 󵄨 󵄨 󵄨󵄨 󵄨󵄨(g(x − λw) − g(x))h1 (w)󵄨󵄨󵄨 ≤ 2‖g‖∞ 󵄨󵄨󵄨h1 (w)󵄨󵄨󵄨 and 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 lim 󵄨󵄨󵄨(g ∗ hλ )(x) − g(x)󵄨󵄨󵄨 = lim 󵄨󵄨󵄨∫(g(x − λw) − g(x))h1 (w)dw󵄨󵄨󵄨 󵄨󵄨 λ→0󵄨󵄨 λ→0 ℝ

󵄨 󵄨󵄨 󵄨 ≤ lim ∫󵄨󵄨󵄨g(x − λw) − g(x)󵄨󵄨󵄨󵄨󵄨󵄨h1 (w)󵄨󵄨󵄨dw λ→0

= 0.

ℝ

And so limλ→0 (g ∗ hλ )(x) = g(x), which ends the proof. Theorem 2.26. If f ∈ L1 (ℝ) and ̂f ∈ L1 (ℝ), then f (x) = ∫ ̂f (ξ )eixξ dξ

a. e.

ℝ

Proof. If f ∈ L1 (ℝ), by Theorem 2.24, we know that (f ∗ hλ )(x) = ∫ H(λξ )̂f (ξ )eixξ dξ , ℝ 2

where H(λξ ) = e−|λξ | . Moreover, since 0 ≤ H(λξ ) ≤ 1 and limλ→0 H(λξ ) = 1, we have: (a) |H(λξ )̂f (ξ )eixξ | ≤ |̂f (ξ )|. (b) limλ→0 H(λξ )̂f (ξ )eixξ = ̂f (ξ )eixξ . So by the Lebesgue dominated convergence theorem, we get that lim (f ∗ hλ )(x) = ∫ ̂f (ξ )eixξ .

λ→0

(2.20)

ℝ

Since ∫ℝ hλ (y) dy = 1, we have 󵄨 󵄨 󵄨 󵄨 ∫󵄨󵄨󵄨f ∗ hλ (x) − f (x)󵄨󵄨󵄨 dx ≤ ∫ ∫󵄨󵄨󵄨f (x − y) − f (x)󵄨󵄨󵄨hλ (y) dydx. ℝ

ℝℝ

Since the Lebesgue measure is invariant with respect to translations, for fy (x) = f (x−y) we have ‖fy − f ‖L1 (ℝ) ≤ 2‖f ‖L1 (ℝ) ,

2.2 Fourier transform on ℝ

| 77

and so 󵄨 󵄨 󵄨 󵄨 ‖fy − f ‖L1 (ℝ) 󵄨󵄨󵄨hλ (y)󵄨󵄨󵄨 ≤ 2‖f ‖L1 (ℝ) 󵄨󵄨󵄨hλ (y)󵄨󵄨󵄨. Since 󵄨 󵄨 lim ‖fy − f ‖L1 (ℝ) 󵄨󵄨󵄨hλ (y)󵄨󵄨󵄨 = 0,

λ→0

we may invoke one more time the Lebesgue dominated convergence theorem and hence we have 󵄨 󵄨 󵄨 󵄨 lim ∫󵄨󵄨󵄨(f ∗ hλ )(x) − f (x)󵄨󵄨󵄨dx ≤ lim ∫ ‖fy − f ‖L1 (ℝ) 󵄨󵄨󵄨hλ (y)󵄨󵄨󵄨dy

λ→0

λ→0

ℝ

ℝ

󵄨 󵄨 = ∫ lim ‖fy − f ‖L1 (ℝ) 󵄨󵄨󵄨hλ (y)󵄨󵄨󵄨dy ℝ

λ→0

= 0. Hence lim ‖f ∗ hλ − f ‖L1 (ℝ) = 0,

λ→0

then lim (f ∗ hλ )(x) = f (x)

a. e.

f (x) = ∫ ̂f (ξ )eixξ dξ

a. e.

λ→0

Finally, by (2.20), we get

ℝ

Theorem 2.27 (Inversion theorem). If f ∫ℝ ̂f (ξ )eixξ dξ (x ∈ ℝ), then f = g a. e.

∈ L1 (ℝ) and ̂f ∈ L1 (ℝ), and if g(x) =

Proof. By Theorem 2.24, (f ∗ hλ )(x) = ∫ℝ H(λξ )̂f (ξ )eixξ dx. Then by the Lebesgue dominated convergence theorem, we have lim ∫(f ∗ hλ )(x)dx = ∫ ̂f (ξ )eixξ dx = g(x).

λ→0

ℝ

ℝ

Again by Theorem 2.24, we obtain ‖f − g‖L1 (ℝ) ≤ ‖f ∗ hλ − f ‖ + ‖f ∗ hλ − g‖ → 0 Thus f = g a. e. And the proof is complete.

as λ → 0.

78 | 2 Fourier series Theorem 2.28 (Uniqueness). If f ∈ L1 (ℝ) and ̂f (ξ ) = 0 for all ξ ∈ ℝ, then f (x) = 0 a. e. Proof. Since ̂f (ξ ) = 0, then ̂f ∈ L1 (ℝ). Applying Theorem 2.27, we have f = 0 a. e., as we wished. ̂ ) = (̂f ∗ Theorem 2.29. If f and g belong to L1 (ℝ) and ̂f ∈ L1 (ℝ), ĝ ∈ L1 (ℝ), then fg(ξ ĝ )(ξ ). Proof. By Theorem 2.26 and Fubini theorem, we have ̂ ) = ∫ f (x)g(x)e−ixξ dx = ∫(∫ ̂f (t)e−ixt dt)g(x)e−ixξ dx fg(ξ ℝ

ℝ

ℝ

= ∫ ̂f (t) ∫ g(x)e−i(ξ −t)x dxdt = ∫ ̂f (t)ĝ (ξ − t)dt = (̂f ∗ ĝ )(ξ ). ℝ

ℝ

ℝ

̂ ) = (̂f ∗ ĝ )(ξ ), and the proof is complete. Hence fg(ξ Exercises. 1. Let e−x

f (x) = {

2.

if 0 < x < ∞,

0

if −∞ < x < 0.

Find ̂f (ξ ). Let xe−x

if 0 < x < ∞,

f (x) = {

0

3. 4.

Find ̂f (ξ ). ∞ If f is odd, prove that ̂f (ξ ) = −2i ∫0 f (x) sin(xξ )dx. If f ∈ L1 (ℝ) and f is odd, prove that b

lim ∫

b→∞ a→0+ a

5. 6. 7.

if −∞ < x < 0.

∞

̂f (ξ ) dξ = −πi ∫ f (x)dx. ξ 0

x Let f ∈ L1 (ℝ). Prove that (∫−∞ f (t)dt)̂(ξ ) = − iξ1 ̂f (ξ ). Let ̂f (ξ ) = max(1 − |ξ |, 0). Prove that f (x) = ( sin(2x) )2 . x

Let

1 − |x| if |x| ≤ 1,

f (x) = { Find ̂f (ξ ).

0

if |x| > 1.

2.2 Fourier transform on ℝ

8.

| 79

7.1. Is ̂f (ξ ) continuous? 7.2. It is Lebesgue integrable?

Show that |eiθ − 1| ≤ |θ| for 0 ≤ θ ≤ 1.

Theorem 2.30 (Shifting hats). If f , g ∈ L1 (ℝ). Then ∫ ̂f (x)g(x)dx = ∫ f (x)ĝ (x)dx. ℝ

(2.21)

ℝ

Proof. Note that both integrals in (2.21) are finite, since having f , g ∈ L1 (ℝ) implies ̂f , ĝ ∈ L (ℝ) (see Theorem 2.19) and ̂f g, f ĝ ∈ L (ℝ). Since F(x, y) = e−2πix⋅y f (x)g(x) is a ∞ 1 measurable function on ℝ × ℝ and 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 ∬ 󵄨󵄨󵄨F(x, y)󵄨󵄨󵄨 dxdy = ∫󵄨󵄨󵄨f (y)󵄨󵄨󵄨 dy ∫󵄨󵄨󵄨g(x)󵄨󵄨󵄨 dx = ‖f ‖L1 (ℝ) ‖g‖L1 (ℝ) < ∞, ℝ×ℝ

ℝ

ℝ

the function F(x, y) is integrable on ℝ × ℝ, and so we can apply Fubini theorem to obtain ∫ ̂f (x)g(x)dx = ∫(∫ f (y)e−ixy dy)g(x)dx ℝ

ℝ

ℝ

= ∫ f (y)(∫ g(x)e−ixy dx)dy ℝ

ℝ

= ∫ f (y)ĝ (y)dy. ℝ

Exercises. 1.1. Prove that if f is continuous at 0, then lim

ϵ→0

1.2. 2.1.

2 1 ̂ ∫ f (x)eϵ|x| dx = f (0). π

ℝ

Prove that if f ∈ L1 (ℝ), ̂f ≥ 0, and f is continuous at zero, then ̂f ∈ L1 (ℝ). Prove that there exists a constant B > 0 such that for t > 0 we have 󵄨󵄨 t 󵄨 󵄨󵄨 sin(ξ ) 󵄨󵄨󵄨 󵄨󵄨 󵄨 dξ 󵄨󵄨󵄨 ≤ B. 󵄨󵄨∫ 󵄨󵄨 󵄨󵄨 ξ 󵄨󵄨0 󵄨󵄨

2.2.

If f is an odd L1 (ℝ) function, conclude that for t > 0, 󵄨 󵄨󵄨 t 󵄨󵄨 ̂f (ξ ) 󵄨󵄨󵄨 󵄨 󵄨󵄨 dξ 󵄨󵄨󵄨 ≤ B‖f ‖L1 (ℝ) . 󵄨󵄨∫ 󵄨󵄨 󵄨󵄨 ξ 󵄨󵄨0 󵄨󵄨

80 | 2 Fourier series

3.

Let f ∈ L1 (ℝ). Prove that 1 󵄨󵄨̂ 󵄨󵄨 󵄨󵄨f (ξ )󵄨󵄨 = O( ). 󵄨 󵄨 |ξ |

4.

a sin x x

Prove that lima→∞ ∫0

dx =

π . 2

2.3 Fourier transform on L2 (ℝ) The Fourier transform does not apply directly to every function of L2 (ℝ) since not every function which belongs to L2 (ℝ) is in L1 (ℝ), and vice versa. Example 9. If x−2/3

f (x) = {

0

if x > 1,

if x ≤ 1,

note that ∫ℝ |f (x)|2 dx = ∫1 x−4/3 dx = 3, thus f ∈ L2 (ℝ). But ∞

∞ 󵄨󵄨∞ 󵄨 󵄨 󵄨 ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨dx = ∫ x−2/3 dx = 3x 1/3 󵄨󵄨󵄨 = ∞, 󵄨󵄨1

ℝ

1

hence f ∉ L1 (ℝ). On the other hand, if x−2/3

g(x) = {

0

if x ∈ (0, 1),

if x ∉ (0, 1),

then 1

󵄨󵄨1 󵄨 󵄨 󵄨 ∫󵄨󵄨󵄨g(x)󵄨󵄨󵄨dx = ∫ x−2/3 dx = 3x 1/3 󵄨󵄨󵄨 = 3, 󵄨󵄨0 0

ℝ

thus g ∈ L1 (ℝ). But 1

󵄨󵄨1 󵄨 󵄨2 󵄨 ∫󵄨󵄨󵄨g(x)󵄨󵄨󵄨 dx = ∫ x−4/3 dx = −3x −1/3 󵄨󵄨󵄨 = ∞, 󵄨󵄨0

ℝ

hence g ∉ L2 (ℝ).

0

2.3 Fourier transform on L2 (ℝ)

| 81

Now observe that, for the given f , its Fourier transform in L1 (ℝ) for ξ = 0 does not exists since ̂f (0) = ∫ f (x)e−ix(0) dx = ∫ f (x)dx ℝ

ℝ

it is not defined, because f ∉ L1 (ℝ). Theorem 2.31. If f ∈ L1 (ℝ) ∩ L2 (ℝ) and ̂f ∈ L1 (ℝ), then ̂f ∈ L2 (ℝ) and ‖̂f ‖L2 (ℝ) = ‖f ‖L2 (ℝ) . Proof. Let f ∈ L1 (ℝ) ∩ L2 (ℝ) and ̂f ∈ L1 (ℝ). By the inversion theorem and Fubini theorem, we have 󵄨 󵄨2 ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx = ∫ f (x)f (x)dx = ∫ f (x)[∫ ̂f (ξ )e−ixξ dξ ]dx ℝ

ℝ

ℝ

ℝ

= ∫ ̂f (ξ )[∫ f (x)eixξ dx]dξ = ∫ ̂f (ξ )[∫ f (x)e−ixξ dx]dξ ℝ

ℝ

ℝ

󵄨2

ℝ

󵄨 = ∫ ̂f (ξ )̂f (ξ )dξ = ∫󵄨󵄨󵄨̂f (ξ )󵄨󵄨󵄨 dξ . ℝ

ℝ

Hence 󵄨 󵄨2 󵄨 󵄨2 ∫󵄨󵄨󵄨̂f (ξ )󵄨󵄨󵄨 dξ = ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx < ∞, ℝ

ℝ

which proves that ̂f ∈ L2 (ℝ), and so ‖̂f ‖L2 (ℝ) = ‖f ‖L2 (ℝ) . Theorem 2.32. L1 (ℝ) ∩ L2 (ℝ) is dense in L2 (ℝ). Proof. For any f ∈ L2 (ℝ), we only need to find a sequence of functions {fn }n∈ℕ ⊂ L1 (ℝ)∩ L2 (ℝ) such that ‖fn − f ‖L2 (ℝ) → 0 as n → ∞. If for every n ∈ ℕ, we define fn (x) = f (x)χ[−n,n] (x), then for each n ∈ ℕ we have 󵄨 󵄨 󵄨 󵄨 ∫󵄨󵄨󵄨fn (x)󵄨󵄨󵄨dx = ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨χ[−n,n] (x)dx ℝ

ℝ

1/2

1/2

󵄨 󵄨2 󵄨 󵄨2 ≤ (∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx) (∫󵄨󵄨󵄨χ[−n,n] (x)󵄨󵄨󵄨 dx) ℝ

ℝ

= √2n‖f ‖L2 (ℝ) < ∞,

82 | 2 Fourier series and 󵄨2 󵄨 󵄨2 󵄨 ∫󵄨󵄨󵄨fn (x)󵄨󵄨󵄨 dx ≤ ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx < ∞. ℝ

ℝ

Hence {fn }n∈ℕ ⊂ L1 (ℝ) ∩ L2 (ℝ). Also for all x ∈ ℝ, lim 󵄨f (x) n→∞󵄨 n 󵄨󵄨

󵄨2 − f (x)󵄨󵄨󵄨 = 0

and 󵄨󵄨 󵄨2 󵄨 󵄨 󵄨 󵄨2 󵄨 󵄨2 󵄨󵄨fn (x) − f (x)󵄨󵄨󵄨 ≤ 2(󵄨󵄨󵄨fn (x)󵄨󵄨󵄨 + 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 ) ≤ 4󵄨󵄨󵄨f (x)󵄨󵄨󵄨 . Note that 4|f |2 ∈ L1 (ℝ), and so, by the Lebesgue dominated convergence theorem, we have 󵄨 󵄨2 lim ‖fn − f ‖2L2 (ℝ) = lim ∫󵄨󵄨󵄨fn (x) − f (x)󵄨󵄨󵄨 dx n→∞

n→∞

ℝ

󵄨 󵄨2 = ∫ lim 󵄨󵄨󵄨fn (x) − f (x)󵄨󵄨󵄨 dx = 0, n→∞ ℝ

and the proof is complete. Theorem 2.33 (Plancherel). The mapping ℱ : L2 (ℝ) → L2 (ℝ) defined by ℱ (f ) = ̂f is a one-to-one and onto linear transformation, and ‖̂f ‖L2 (ℝ) = ‖f ‖L2 (ℝ) ,

f ∈ L2 (ℝ).

Proof. Let f ∈ L2 (ℝ). By Theorem 2.32, we can choose as sequence {fn }n∈ℕ ⊂ L1 (ℝ) ∩ L2 (ℝ) such that ‖fn − f ‖L2 (ℝ) → 0 as n → ∞, and then ‖̂fn − ̂f ‖L2 (ℝ) → 0 as n → ∞. Since 󵄨󵄨 󵄨 󵄨󵄨‖fn ‖L2 (ℝ) − ‖f ‖L2 (ℝ) 󵄨󵄨󵄨 ≤ ‖fn − f ‖L2 (ℝ) → 0,

as n → ∞,

󵄨󵄨 ̂ 󵄨 󵄨󵄨‖fn ‖L2 (ℝ) − ‖̂f ‖L2 (ℝ) 󵄨󵄨󵄨 ≤ ‖̂fn − ̂f ‖L2 (ℝ) → 0,

as n → ∞,

and

we get ‖fn ‖L2 (ℝ) → ‖f ‖L2 (ℝ)

and ‖̂fn ‖L2 (ℝ) → ‖̂f ‖L2 (ℝ) ,

as n → ∞.

Hence ‖̂f ‖L2 (ℝ) = lim ‖̂fn ‖L2 (ℝ) = lim ‖fn ‖L2 (ℝ) = ‖f ‖L2 (ℝ) . n→∞

n→∞

2.4 Definition of the Fourier transform in L2 (ℝ)

| 83

Now, we shall show the linearity of ℱ on L2 (ℝ). To this end, let f , g ∈ L2 (ℝ) and choose {fn }n∈ℕ , {gn }n∈ℕ ⊂ L1 (ℝ) ∩ L2 (ℝ) such that fn → f and gn → g in L2 (ℝ). For any constants α, β, we have (αfn + βgn )̂ = α̂fn + βĝn . Since αfn + βgn → αf + βg in L2 (ℝ), we have (αfn + βgn )̂ → (αf + βg)̂ in L2 (ℝ), and since ̂fn → ̂f , ĝn → ĝ in L2 (ℝ), α̂fn + βĝn → α̂f + βĝ

in L2 (ℝ).

Hence (αf + βg)̂ = α̂f + βĝ , as desired. Now 󵄩󵄩 󵄩 󵄩 󵄩 󵄩󵄩ℱ (f − g)󵄩󵄩󵄩L2 (ℝ) = 󵄩󵄩󵄩(f − g)̂󵄩󵄩󵄩L2 (ℝ) = ‖f − g‖L2 (ℝ) shows that ℱ is an isometry, so, by Theorem 1.23, ℱ and ℱ ∗ are unitary and one-toone, then L2 (ℝ) = {0}⊥ = [ker(ℱ ∗ )] = rang(ℱ ) = ℱ (L2 (ℝ)), ⊥

which shows that ℱ is onto. And so the proof is complete.

2.4 Definition of the Fourier transform in L2 (ℝ) Let f ∈ L2 (ℝ). Since L1 (ℝ) ∩ L2 (ℝ) is dense in L2 (ℝ), there exists a sequence {fn }n∈ℕ in L1 (ℝ) ∩ L2 (ℝ) such that fn → f in L2 (ℝ). By Plancherel Theorem 2.33, ‖̂fm − ̂fn ‖L2 (ℝ) = ‖f? m − fn ‖L2 (ℝ)

= ‖fm − fn ‖L2 (ℝ) → 0

as m, n → ∞, which means that {̂fn }n∈ℕ is a Cauchy sequence in L2 (ℝ). Theorem 1.14 guarantees the existence of F ∈ L2 (ℝ) such that lim f n→∞ n

̂ = F.

84 | 2 Fourier series Again by Theorem 1.14, ‖f ‖L2 (ℝ) = lim ‖fn ‖L2 (ℝ) = lim ‖̂fn ‖L2 (ℝ) = ‖F‖L2 (ℝ) . n→∞

n→∞

This function F is what we call the Fourier transform of f . But first we have to demonstrate that F is independent of the choice of the approximating sequence fn . To prove this, suppose that fn′ is another sequence in L1 (ℝ)∩L2 (ℝ) which converges to f in L2 (ℝ). This sequence also gives rise to a function F ′ such that ̂f → F ′ in L (ℝ). n 2 But then the sequence fn − fn′ → 0 in L2 (ℝ) and ̂fn − f̂′ n → F − F ′ in L2 (ℝ). By Theorem 2.33, 󵄩 󵄩 󵄩 󵄩 ‖0‖L2 (ℝ) = lim 󵄩󵄩󵄩fn − fn′ 󵄩󵄩󵄩L (ℝ) = lim 󵄩󵄩󵄩f? − fn′ 󵄩󵄩󵄩L (ℝ) 2 2 n→∞ n→∞ n 󵄩 󵄩 󵄩 󵄩 = lim 󵄩󵄩󵄩̂fn − f̂′ n 󵄩󵄩󵄩L (ℝ) = 󵄩󵄩󵄩F − F ′ 󵄩󵄩󵄩L (ℝ) . 2 2 n→∞ Thus F = F ′ m-a. e. So F and F ′ represent the same element of L2 (ℝ). Summarizing, for any f ∈ L2 (ℝ) there exists a unique ̂f ∈ L2 (ℝ) such that if fn is any sequence satisfying fn ∈ L1 (ℝ) ∩ L2 (ℝ) and fn → f in L2 (ℝ) then ̂fn → ̂f in L2 (ℝ). This function ̂f is called the Fourier transform of f . Theorem 2.34 (Heisenberg inequality). If f , xf (x), and ξ f ̂(ξ ) belong to L2 (ℝ), then 1 4 󵄨 󵄨2 󵄨 󵄨2 ‖f ‖ ≤ (∫ x2 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx)(∫ ξ 2 󵄨󵄨󵄨f ̂(ξ )󵄨󵄨󵄨 dξ ). 4 L2 (ℝ) ℝ

ℝ

Proof. Since ξ f ̂ ∈ L2 (ℝ), note that 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 ∫󵄨󵄨󵄨f ′ (x)󵄨󵄨󵄨 dx = ∫󵄨󵄨󵄨f ′̂ (ξ )󵄨󵄨󵄨 dξ = ∫󵄨󵄨󵄨iξ f ̂(ξ )󵄨󵄨󵄨 dξ = ∫󵄨󵄨󵄨ξ f ̂(ξ )󵄨󵄨󵄨 dξ < ∞. ℝ

ℝ

ℝ

ℝ

Then f is absolutely continuous. Hence f ′ exists a. e. Before continuing with the proof, we claim that limx→∞ x|f (x)|2 = 0. Proof of claim. Assume that limx→∞ x|f (x)|2 = ∞, then for all M > 0 there exists N > 0 such that if x > N, then x|f (x)|2 > M. Hence dx 󵄨 󵄨2 󵄨 󵄨2 → ∞, ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx ≥ ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx > M ∫ x

ℝ

x>N

x>N

which lead to a contradiction, since ∫ℝ |f (x)|2 dx < ∞. Therefore, limx→∞ x|f (x)|2 = 0.

2.4 Definition of the Fourier transform in L2 (ℝ)

|

Integrating by parts, we have 󵄨a 󵄨2 󵄨2 󵄨󵄨 󵄨 󵄨 ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx = ∫ f (x)f (x)dx = lim x󵄨󵄨󵄨f (x)󵄨󵄨󵄨 󵄨󵄨󵄨 − ∫ x(f (x)(f )′ (x) + f (x)f ′ (x)) dx a→∞ 󵄨󵄨−a ℝ

ℝ

ℝ

= − ∫ x(f (f ) (x) + f (x)f (x)) dx = − ∫ x(f f ′ (x) + f (x)f ′ (x)) dx ′

′

ℝ

ℝ

= −x ∫ x(f f ′ (x) + f (x)f ′ (x)) dx = − ∫ 2Re xf f ′ (x) dx. ℝ

ℝ

Hence, by the Cauchy–Schwarz inequality and Plancherel theorem, we have 󵄨󵄨 󵄨󵄨 1 󵄨󵄨 󵄨2 󵄨 󵄨 ∫󵄨󵄨f (x)󵄨󵄨󵄨 dx = − ∫ Re xf (x)f ′ (x) dx ≤ 󵄨󵄨󵄨∫ Re xf (x)f ′ (x) dx󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 2 ℝ

ℝ

ℝ

󵄨 󵄨 ≤ ∫󵄨󵄨󵄨Re xf (x)f ′ (x)󵄨󵄨󵄨 dx ℝ

󵄨 󵄨 󵄨󵄨 󵄨 󵄨 ≤ ∫󵄨󵄨󵄨xf (x)f ′ (x)󵄨󵄨󵄨 dx = ∫󵄨󵄨󵄨xf (x)󵄨󵄨󵄨󵄨󵄨󵄨f ′ (x)󵄨󵄨󵄨 dx ℝ

1 2

ℝ

1

2 󵄨 󵄨2 󵄨2 ≤ (∫ x 󵄨󵄨f (x)󵄨󵄨󵄨 dx) (∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx)

2 󵄨󵄨

ℝ

1 2

ℝ

1

2 󵄨 󵄨2 󵄨2 = (∫ x 󵄨󵄨f (x)󵄨󵄨󵄨 dx) (∫󵄨󵄨󵄨f ′ (x)󵄨󵄨󵄨 dx)

2 󵄨󵄨

ℝ

1 2

ℝ

1

2 󵄨2 󵄨 󵄨2 = (∫ x 󵄨󵄨f (x)󵄨󵄨󵄨 dx) (∫󵄨󵄨󵄨f ′̂ (ξ )󵄨󵄨󵄨 dξ )

2 󵄨󵄨

ℝ

1 2

ℝ

1

2 󵄨 󵄨2 󵄨 󵄨2 = (∫ x2 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx) (∫󵄨󵄨󵄨ξ f ̂(ξ )󵄨󵄨󵄨 dξ )

ℝ

1 2

ℝ

1

2 󵄨 󵄨2 󵄨2 = (∫ x 󵄨󵄨f (x)󵄨󵄨󵄨 dx) (∫ ξ 2 󵄨󵄨󵄨f ̂(ξ )󵄨󵄨󵄨 dξ ) .

2 󵄨󵄨

ℝ

ℝ

Finally, 1 4 󵄨 󵄨2 󵄨 󵄨2 ‖f ‖ ≤ (∫ x2 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx)(∫ ξ 2 󵄨󵄨󵄨f ̂(ξ )󵄨󵄨󵄨 dξ ), 4 L2 (ℝ) ℝ

which ends the proof.

ℝ

85

86 | 2 Fourier series Remark 9. Recall that the Cauchy–Schwarz inequality ∫ℝ ‖fg‖ dx ≤ ‖f ‖L2 ‖g‖L2 is an equality if and only if f and g are scalar multiples of one another. Hence Heisenberg 2 ′ − αx2 inequality is an equality if f (x) = −αxf (x). That is, f (x) = Ce where C > 0. Exercises. 2 1. For f (x) = e−α|x| with α > 0, prove that 2

1 󵄨 󵄨2 󵄨 󵄨2 󵄨 󵄨2 (∫󵄨󵄨󵄨󵄨f (x)󵄨󵄨󵄨󵄨 dx) = (∫ x 2 󵄨󵄨󵄨󵄨f (x)󵄨󵄨󵄨󵄨 dx)(∫ ξ 2 󵄨󵄨󵄨󵄨̂f (ξ )󵄨󵄨󵄨󵄨 dξ ). 4 ℝ

2.

ℝ

ℝ

Under the hypotheses of Theorem 2.34, prove that 󵄩󵄩 󵄩2 󵄩 󵄩 󵄩󵄩 ̂ 󵄩󵄩󵄩 󵄩󵄩f (x)󵄩󵄩󵄩 ≤ 4π 󵄩󵄩󵄩(x − x0 )f (x)󵄩󵄩󵄩 󵄩 󵄩 󵄩L2 󵄩 󵄩L2 (ℝ) 󵄩󵄩(ξ − ξ0 )f (ξ )󵄩󵄩L2 (ℝ) .

2.5 Fourier transform on L1 (ℝn ) In the sequel, for technical reasons only, we will present the definition of the Fourier transform with a slight modification. Definition 2.7. Let f ∈ L1 (ℝn ). We define ̂f : ℝn → ℂ, the Fourier transform of f , by ̂f (ξ ) = ∫ f (x)e−2πix⋅ξ dx, ℝn

where x ⋅ ξ = ∑nk=1 xk ξk for x and ξ ∈ ℝn ; x ⋅ ξ is known as the inner product. In the following theorem we gather some properties of the Fourier transform. Theorem 2.35. The following properties of the Fourier transform hold: (a) (Linearity) Let f , g ∈ L1 (ℝn ) and α, β ∈ ℂ. Then (αf? + βg)(ξ ) = α̂f (ξ ) + βĝ (ξ ). (b) (Continuity) Let f ∈ L1 (ℝn ). Then ̂f is continuous and satisfies ‖̂f ‖L∞ (ℝn ) ≤ ‖f ‖L1 (ℝn ) . (c) (Riemann–Lebesgue) Let f ∈ L1 (ℝn ). Then lim|ξ |→∞ |̂f (ξ )| = 0. (d) (Convolution) Let f , g ∈ L (ℝn ). Then f? ∗ g(ξ ) = ̂f (ξ )ĝ (ξ ). 1

2πi h⋅ξ ̂ (e) (Shift) Let f ∈ L1 (ℝn ) and h ∈ ℝn . For τh f (x) = f (x + h), we have τ? f (ξ ), h f (ξ ) = e 2πi x⋅h and for σh f (x) = f (x)e , we have

̂f (ξ ) = ̂f (ξ − h). σ h

2.5 Fourier transform on L1 (ℝn )

| 87

(f) (Rotation) Let Θ be an orthogonal matrix and set (Θf )(x) = f (Θx), x ∈ ℝn . If f ∈ L1 (ℝn ), then f? (Θ⋅)(ξ ) = ̂f (Θξ ). (g) (Scaling) Let f ∈ L1 (ℝn ), λ ∈ ℝn , and define g(x) = λ1n f (x/λ). Then g ∈ L1 (ℝn ) and ĝ (ξ ) = ̂f (λξ ). 𝜕f (h) (Differentiation) Let f ∈ L1 (ℝn ) be such that 𝜕x ∈ L1 (ℝn ). Then j

(

̂ 𝜕f )(ξ ) = (2πiξj )̂f (ξ ). 𝜕xj

(i) (Multiplication) Let f ∈ L1 (ℝn ) be such that gj (x) = −2πixj f (x) is in L1 (ℝn ). If ̂f is differentiable in the ξj direction, then ĝ (ξ ) =

𝜕̂f (ξ ) . 𝜕ξj

Proof. We shall prove property (f) because the other properties are proved in the same way as in the one-dimensional case. Proof of property (f). Let Θ be an orthogonal matrix. Then ΘΘT = I where ΘT denote the transpose of Θ. Then we have: (a) Θv ⋅ w = v ⋅ ΘT w for any v, w ∈ ℝn . (b) | det Θ| = 1. (c) ΘT = Θ−1 . Now, with all of this in hand, we make the change of variable y = Θx, which leads us to the following: f̂ (Θ)(ξ ) = ∫ f (Θx)e−2iπ x⋅ξ dx = ∫ f (Θx)e−2iπΘ

−1

ℝn

Θx⋅ξ

dx

ℝn −1

= ∫ f (y)e−2iπΘ

y⋅ξ

T

| det Θ| dy = ∫ f (y)e−2iπΘ

ℝn

ℝn

= ∫ f (y)e−2iπ y⋅Θξ dy = ̂f (Θξ ). ℝn

Therefore, f̂ (Θ)(ξ ) = ̂f (Θξ ) = (̂f ∘ Θ)(ξ ), and the proof of property (f) is complete. n 2 Theorem 2.36. If a > 0 and f (x) = e−πa|x| , then ̂f (ξ ) = a− 2 e−π

|ξ |2 a

.

y⋅ξ

dy

88 | 2 Fourier series Proof. First, consider the case n = 1. By Theorem 1.24(f), we have 2 ̂ ′ )(ξ ) = (−2πixe−πax )̂(ξ ) (f

2 ′ ̂ i = ( )((e−πax ) ) (ξ ) a i = ( )(2πiξ )̂f (ξ ) a 2π = −( )ξ ̂f (ξ ). a

Now, we shall solve the following differential equation using standard methods: d̂f (ξ ) 2π = (− )ξ ̂f (ξ ). dξ a In doing so, we have 󵄨󵄨 ̂f (ξ ) 󵄨󵄨 π 2 󵄨 󵄨󵄨 ln󵄨󵄨󵄨 󵄨=− ξ , 󵄨󵄨 ̂f (0) 󵄨󵄨󵄨 a therefore ξ

(2.22)

̂f (ξ ) = ̂f (0)e−π a .

1 2 π 21 Now, note that ̂f (0) = ∫ℝ e−πax dx = ( πa ) = ( a1 ) 2 . Hence, for the case n = 1, (2.22) can be written as 1 2

̂f (ξ ) = ( 1 ) e−π a . a ξ2

The n-dimensional case follows from Fubini theorem, since if |x| = ∑nk=1 xk2 , then 2

̂f (ξ ) = ∫ e−πa|x| e−2πix⋅ξ dx ℝn

n

2

n

= ∫ e−πa ∑k=1 xk −2πi ∑k=1 xk ξk dx1 dx2 ⋅ ⋅ ⋅ dxn ℝn 2

2

= ( ∫ e−πax1 e−2πix1 ξ1 dx1 )( ∫ e−πax2 e−2πix2 ξ2 dx2 ) ℝn

ℝn

× ⋅ ⋅ ⋅ × (∫ e n

−πaxn2

− e2πixn ξn dxn )

ℝn 2

= ∏ ∫ e−πaxk − e2πixk ξk dxk . k=1 ℝn

2.6 Bessel functions |

89

Using the case n = 1, we get 1

n

2

ξk 1 2 = ∏( ) e−π a a k=1 n

ξk 1 2 n = ( ) ∏ e−π a a k=1

2

n

n 1 2 = ( ) e−π ∑k=1 a

ξk2 a

n

ξ2 1 2 = ( ) e−π a , a

and the proof is complete.

2.6 Bessel functions This section will be entirely based in Section G from [15]. Among other interesting results, we will see in this section that the Fourier transform of χB , were B ⊂ ℝn is a ball, is a Bessel function – what a surprise! Definition 2.8. The Bessel function of first kind of order m is ∞

Jm (z) = ∑ (−1)k k=0

( z2 )m+2k

k!Γ(k + m + 1)

,

where z ∈ ℂ and −1 < m < ∞. By the ratio test, we can check that the radius of convergence of the above power series is ∞. Lemma 2.3. For α > −1, 1

α

∫ (1 − t 2 ) e−ist dt = Γ(α + 1)√π −1

Tα+ 1 (S) 2

1

( 2s )α+ 2

.

Proof. We use the power series for e−ist and integrate term by term (this will be justified at the end of the proof): 1

α

∞ 1

−1

k=1 −1

k

α (−ist)

∫ (1 − t 2 ) e−ist dt = ∑ ∫ (1 − t 2 )

k!

dt

1

(−is)2k α =∑ ∫ (1 − t 2 ) t 2k dt (2k)! k=1 ∞

−1

(odd terms = 0)

90 | 2 Fourier series 1

k s2k α u −1 2 ∫(1 − u2 ) u 2 dt = ∑ (−1) (2k)! 2 k=1 ∞

k

∞

= ∑ (−1)k k=0

2k

(u = t 2 )

0

s 1 B(α + 1, k + ). (2k)! 2

By the properties of the beta function, 1

1 Γ(α + 1) Γ(k + 2 ) B(α + 1, k + ) = 2 Γ(α + k + 32 ) (2k)! Γ(α + 1)

((k − 21 ) ⋅ ⋅ ⋅ 32 ⋅ 21 √π)

=

Γ(α + k + 32 )

=

Γ(α + k + 32 )

=

Γ(α + k + 32 ) 22k k!

Γ(α + 1)

Γ(α + 1)

(2k)! (2k − 1) ⋅ ⋅ ⋅ 3 ⋅ 1√π 2k(2k − 2) ⋅ ⋅ ⋅ 4 ⋅ 2 2k (2k)! 2k k! 1

.

Thus 1 2k ∞ Jα+ 1 (s) s Γ(α + 1)√π = Γ(α + 1) 2 1 . ∫ (1 − t 2 )e−ist dt = ∑ ( ) 3 2 Γ(α + k + 2 ) ( s )α+ 2 k=0 2

−1

The partial sums of the first series at the beginning of the proof can be dominated as follows: 󵄨󵄨 N ∞ k 󵄨󵄨󵄨 k 󵄨󵄨 󵄨󵄨(1 − t 2 )α ∑ (−ist) 󵄨󵄨󵄨 ≤ (1 − t 2 )α ∑ |s| = e|s| (1 − t 2 )α . 󵄨󵄨 󵄨 k! 󵄨󵄨󵄨 k! 󵄨󵄨 k=0 k=0 Since e|s| (1−t 2 )α ∈ L1 (ℝ) for α > −1, we can apply the Lebesgue dominated convergence theorem to justify the interchange of summation and integration, which we employed in the calculation. And the proof is complete. Next, let us define the following function: (1 − |x|2 )δ

δ

(1 − |x|2 )+ = {

if |x| < 1,

0

if |x| > 1,

for δ > −1. Theorem 2.37. For δ > −1, the Fourier transform in ℝn satisfies: δ

(1? − |x|2 )+ (ξ ) = Γ(α + 1)

J n +δ (|ξ |) 2

(

|ξ | n +δ )2 2

.

(2.23)

2.6 Bessel functions |

91

Proof. For the case n = 1, the result follows immediately from Lemma 2.3. Hence for n ≥ 2, we write ℝn = ℝn−1 × ℝ and use the notation x = (y, t) where y ∈ ℝn−1 and t ∈ ℝ. Now, we assume that ξ = (0, 0, . . . , ρ) where ρ = |ξ |. Thus δ

δ (1? − |x|2 )+ (ξ ) = ∫ (1 − |x|2 ) e−2πiρxn dx |x| 0. 2 a (2) f (x) = e−a(1+x ) ∈ S(ℝn ) for any a > 0. (3) C0∞ (ℝn ) ⊂ S(ℝn ) where C0∞ (ℝn ) = {f ∈ C ∞ (ℝn ) : supp f is compact in ℝn } and supp f = {x ∈ ℝn : f (x) ≠ 0}. The space S(ℝn ) is not a normed space because |ϕ|α,β is only a seminorm for α ≥ 0 and β > 0 as the condition |ϕ|α,β = 0 if only if ϕ = 0 fails to hold for, e. g., ϕ(x) = c ∈ ℝ. But the space (S, ρ) is a metric space if the metric is defined by ρ(f , g) = ∑ 2−|α|−|β| α,β≥0

|f − g|α,β

1 + |f − g|α,β

.

Exercise. Prove that ρ is a metric, that is, (1) ρ(f , g) ≥ 0 and ρ(f , g) = 0 if and only if f = g; (2) ρ(f , g) = ρ(g, f ); (3) ρ(g, h) ≤ ρ(g, f ) + ρ(f , h); (4) (In addition) |ρ(f , h) − ρ(g, h)| ≤ ρ(f , g).

Theorem 3.2 (Completeness). The space (S, ρ) is a complete metric space, i. e., every Cauchy sequence converges.

100 | 3 Schwartz spaces S(ℝn ) Proof. Let {ϕk }k∈ℕ ⊂ S be a Cauchy sequence. For any σ > 0 and any γ ∈ ℕn0 , let

ϵ=

2−|γ| σ . 1+2σ

Then there exists an n0 (ϵ) ∈ ℕ such that ρ(φk , φm ) < ϵ,

k, m ≥ n0 (ϵ), since {φk }k∈ℕ is a Cauchy sequence. Thus, we have |φk − φm |σ,γ

1 + |φk − φm |σ,γ


0 there exists δ > 0 such that η 󵄨󵄨 󵄨 󵄨󵄨f (x)󵄨󵄨󵄨 < ‖φ‖L1 (ℝn ) + 1 whenever |x| < δ. Also we have 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨∫ φ(x) dx󵄨󵄨󵄨 ≤ ‖φ‖L1 (ℝn ) , 󵄨󵄨 󵄨󵄨 ℝ

which is why we may conclude that 󵄨󵄨 󵄨󵄨 x 󵄨󵄨 󵄨󵄨 η x 󵄨 󵄨 󵄨 󵄨󵄨 −n ξ −n ∫ 󵄨󵄨󵄨φ( )󵄨󵄨󵄨 dx 󵄨󵄨 ∫ ξ φ( )f (x) dx󵄨󵄨󵄨 ≤ 󵄨󵄨 ‖φ‖L (ℝn ) + 1 󵄨󵄨 ξ 󵄨󵄨 󵄨󵄨 ξ 1 n ℝ

|x|δ

108 | 3 Schwartz spaces S(ℝn )

≤

η

‖φ‖L1 (ℝn ) + 1

η−n ‖φ‖L1 (ℝn )

󵄨 󵄨 󵄨 󵄨 + sup 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 ∫ 󵄨󵄨󵄨φ(y)󵄨󵄨󵄨 dy n x∈ℝ

|y|> ξδ

󵄨 󵄨 = η + (sup 󵄨󵄨󵄨f (x)󵄨󵄨󵄨)Iξ , x∈ℝn

where Iξ = ∫|y|> δ |φ(y)| dy. But Iξ →0 as ξ → 0, so this proves the theorem. ξ

Remark 11. Now, for f and g ∈ S(ℝn ), we have F(x, ξ ) = f (x)g(x)e−2iπx⋅ξ ∈ L1 (ℝn ) × L1 (ℝn ). Thus by the Fubini theorem, we then have ⟨ℱ (f ), g⟩ = ∫ f ̂(ξ )g(ξ ) dξ ℝn

= ∫ g(ξ )( ∫ f (x)e−2πix⋅ξ dx) dξ ℝn

ℝn

= ∫ f (x)( ∫ e−2πix⋅ξ g(ξ ) dξ ) dx ℝn

ℝn

= ⟨f , ℱ g⟩L (ℝn ) , ∗

2

where ℱ g(x) = ℱ g(−x). Here ℱ ∗ is the adjoint operator (in the sense of L2 ) which maps S(ℝn ) into S(ℝn ) since ℱ : S(ℝn ) → S(ℝn ). The inverse Fourier transform ℱ −1 is defined as ℱ −1 := ℱ ∗ . In order to justify this definition, we shall prove the following theorem. Theorem 3.6 (Fourier inverse formula). Let f ∈ S(ℝn ). Then ℱ ℱ (f ) = f . ∗

Proof. Let us consider v(x) = e−

|x|2 2

n

. We already know that ∫ℝn v(x) dx = (2π) 2 and v̂ = v. n

We apply Theorem 1.7 with φ(x) = (2π)− 2 v(x) and f ∈ S(Rn ). Then n x (2π) 2 f (0) = lim+ ∫ ξ −n v( )f (x) dx ξ ξ →0

ℝn

= lim+ ⟨f , ξ −n σ 1 v ⟩L (ℝn ) ξ →0

= lim+ ⟨f , ξ ξ →0

n

−n

2

σ 1 ℱ (v)⟩L (ℝn ) n

2

3 Schwartz spaces S(ℝn )

|

109

= lim+ ⟨f , ℱ (σξ v)⟩L (ℝn ) ξ →0

2

= lim+ ⟨ℱ (f ), σξ v⟩L (ℝn ) ξ →0

2

= ⟨ℱ (f ), 1⟩L (ℝn ) 2

= ∫ ℱ (f )(ξ )e−iξ ⋅(0) dξ , ℝn

after using the Lebesgue dominated convergence theorem in the last step. This proves that n

f (0) = (2π)− 2 ∫ ℱ (f )(ξ )e−iξ ⋅0 dx = ℱ ∗ ℱ (0). ℝn

It remains to observe that f (x) = (τ−x f )(0) = (ℱ ∗ ℱ (τ−x )(0) n

= (2π)− 2 ∫ ℱ (τ−x f )(ξ )eiξ ⋅0 dξ − n2

= (2π)

ℝn

∫ (ℱ f )(ξ )eix⋅ξ dξ ℝn

= ℱ ℱ f (x), ∗

thus the proof is complete. Corollary 3.2. The Fourier transform is an isometry (in the sense of L2 ). Proof. Let f ∈ S(ℝn ). Then ‖ℱ f ‖2L2 (ℝn ) = ⟨ℱ f , ℱ f ⟩L2 (ℝn )

= ⟨f , ℱ ∗ ℱ f ⟩L (ℝn ) = ⟨f , f ⟩L2 (ℝn )

2

= ‖f ‖2L2 (ℝn ) . Remark 12. Note that

⟨ℱ (f ), g⟩L (ℝn ) = ⟨f , ℱ ∗ g⟩L (ℝn ) 2

2

means that ∫ f ̂(ξ )g(ξ ) dξ = ∫ f (x)ℱ ∗ g(x) dx = ∫ f (x)ℱ g dx, ℝn

ℝn

ℝn

110 | 3 Schwartz spaces S(ℝn ) which implies that ̂ ) dξ , ∫ f ̂(ξ )g(ξ ) dξ = ∫ f (ξ )g(ξ ℝn

ℝn

or ⟨ℱ f , g⟩L2 (ℝn ) = ⟨f , ℱ g⟩L2 (ℝn ) . Lemma 3.2. Let A be a real symmetric, positive definite n × n matrix. Then 1

n

∫ e−xAx e−2πix⋅ξ dx = π 2 (det(A)) 2 eπ

2

ξ ⋅A−1 ξ

.

ℝn

Proof. Since A is real symmetric and positive definite, it is diagonalizable. Explicitely, there is an orthogonal matrix O and a diagonal matrix D = diag(λ1 , . . . , λn ),

λj > 0,

so that A = OT DO. Let y = Ox and η = Oξ . Then x ⋅ Ax = x ⋅ OT DOx = Ox ⋅ DOx = y ⋅ Dy n

= ∑ λj yj2 j=1

and x ⋅ ξ = OT y ⋅ OT η = y ⋅ η. Thus n

2

∫ e−x⋅Ax e−2πix⋅ξ dx = ∫ e− ∑j=1 λj yj −2πiy⋅η dy ℝn

ℝn n

2

2

= ∏ ∫ e−λj yj e−2πiyj ηj dyj . j=1 ℝ

Computing each of these integrals, we have ∫ e−λy ℝ

2

−2πiyη

dy = ∫ e−λ(y− ℝ

πiη 2 πη2 )− λ λ

dy = e−

πη2 λ

2

∫ e−λy dy. ℝ

3 Schwartz spaces S(ℝn )

| 111

So n

∫ e−x⋅Ax e−2πix⋅ξ dx = ∏ e j=1

ℝn

−

π 2 ηj λj

1

− 21

π 2 λj

n

= π 2 (λ1 , . . . , λn )e n

− ∑nj=1

−1

= π 2 (det(A)) 2 eπη⋅D

−1

π 2 ηj λj

η

.

The result follows from the fact that η ⋅ D−1 η = Oξ ⋅ D−1 Oξ = ξ ⋅ A−1 ξ . Exercise. Let f ∈ Cc∞ (ℝ) and suppose f (x) = 0 whenever |x| > R. Prove that ̂f can be extended to a holomorphic function on all ℂ and we have the decay estimate −n 󵄨󵄨̂ 󵄨󵄨 󵄨󵄨f (z)󵄨󵄨 ≤ Cn (1 + |z|) e2π|Im (z)|R . 󵄨 󵄨

Theorem 3.7 (Plancherel’s theorem). Suppose f ∈ L1 (ℝn ) ∩ L2 (ℝn ). Then ̂f ∈ L2 (ℝn ) and ‖̂f ‖L2 (ℝn ) = ‖f ‖L2 (ℝn ) . ̂ ) = ̂f (ξ )ĝ (ξ ). Now h ∈ ℝn and Proof. Let g(x) = f (−x). Define h = f ∗ g, so we have h(ξ h is continuous. Next, ĝ (ξ ) = ∫ f (−x)e−2πix⋅ξ dx. ℝn

Let y = −x, so ĝ (ξ ) = ∫ f (−x)e−2πix⋅ξ dx = ∫ f (−y)e−2πiy⋅ξ dy = ̂f (y). ℝn

ℝn

Now, we claim that for any g, h ∈ L1 (ℝn ), ̂ dx = ∫ ĝ (ξ )h(ξ ) dξ . ∫ g(x)h ℝn

ℝn

112 | 3 Schwartz spaces S(ℝn ) Proof of claim. ̂ dx = ∫ g(x)[ ∫ h(ξ )e−2πix⋅ξ dξ ] dx ∫ g(x)h(x) ℝn

ℝn

ℝn

= ∫ [ ∫ g(x)e−2πix⋅ξ dx]h(ξ ) dξ ℝn ℝn

= ∫ ĝ (ξ )h(ξ ) dξ . ℝ n

2

|x|2

Now, set g(ξ ) = e−ϵ|ξ | and then ĝ (x) = (2ϵ)− 2 e 4ϵ . Thus |x|2

n

2

̂ )e−ϵ|ξ | dϵ = (2ϵ)− 2 ∫ h(x)e− 4ϵ dx ∫ h(ξ ℝn

ℝn

x 2 | − n2 −| 2ϵ

e

= ∫ (2ϵ)

h(x) dx.

ℝn

Invoking Theorem 3.5, as ϵ → 0, we get ̂ ) dξ = h(0). ∫ h(ξ ℝ

But, we know h = f ∗ g and ̂ ) = ̂f (ξ )ĝ (ξ ) = ̂f (ξ )̂f (ξ ) = 󵄨󵄨󵄨̂f (ξ )󵄨󵄨󵄨2 . h(ξ 󵄨 󵄨 So ̂ ) dξ = ∫ |̂f |2 dξ . h(0) = ∫ h(ξ ℝn

ℝ

We also know that h(ξ ) = (f ∗ g)(ξ )

implies h(0) = ∫ f (y)g(−y) dy ℝn

= ∫ f (y)f (y) dy ℝn

󵄨 󵄨2 = ∫ 󵄨󵄨󵄨f (y)󵄨󵄨󵄨 dy. ℝn

3 Schwartz spaces S(ℝn )

| 113

Thus 󵄨2 󵄨 󵄨2 󵄨 ∫ 󵄨󵄨󵄨̂f (ξ )󵄨󵄨󵄨 dξ = h(0) = ∫ 󵄨󵄨󵄨f (y)󵄨󵄨󵄨 dy. ℝn

ℝn

Hence ‖̂f ‖L2 (ℝn ) = ‖f ‖L2 (ℝn ) . Now, by virtue of Theorem 3.7 (Plancherel’s theorem), we shall define the Fourier transform in L2 (ℝn ) in the same way as in Definition 2.7. Theorem 3.8. The inverse map ℱ −1 of ℱ = ̂f is a one-to-one linear mapping of L2 (ℝn ) onto itself, and 󵄩󵄩 −1 󵄩󵄩 󵄩󵄩ℱ (f )󵄩󵄩2 = ‖f ‖2 . Proof. By definition (see Remark 12), ℱ ∗ = ℱ −1 , where ℱ ∗ is the adjoint operator of ℱ in the sense of L2 . Since L2 is a Hilbert space, we may apply Theorem 1.24(c) to obtain 󵄩󵄩 −1 󵄩󵄩 󵄩 ∗ 󵄩 󵄩󵄩ℱ (f )󵄩󵄩2 = 󵄩󵄩󵄩ℱ (f )󵄩󵄩󵄩2 = ‖f ‖2 . Part (d) of Theorem 1.24 shows that ℱ −1 is one-to-one. Lemma 3.3. The space C0∞ (ℝn ) is dense in S(ℝn ). Proof. Choose φ ∈ C0∞ (ℝn ) with φ(x) = 1 whenever |x| ≤ 1. Then, given ψ ∈ S(ℝn ), consider the sequence ψm (x) = φ(

x )ψ(x). m

Clearly, ψm = ψ as long as |x| ≤ m, so if it converges in S(ℝn ) it must converge to ψ. Suppose m ≥ n. Then by Leibniz formula, x α x Dα (ψm (x) − ψn (x)) = ∑ ( )Dα (φ( ) − φ( )) ⋅ Dα−β ψ(x). m m β β≤α All the derivatives of φ( mx ) are bounded, independent of n, and φm = φn whenever |x| ≤ m. So for any p, 0 󵄨󵄨 α 󵄨 󵄨󵄨D (φm (x) − ψn (x))󵄨󵄨󵄨 ≤ { Cα,p ⟨x⟩−2 1

if |x| ≤ m, if |x| > m,

where ⟨x⟩ = (1 + |x|2 ) 2 . Hence ψm is Cauchy sequence in S(ℝn ).

114 | 3 Schwartz spaces S(ℝn ) Theorem 3.9. The following holds C0∞ (ℝn ) = L2 (ℝn ), that is, C0∞ (ℝn ) is dense in L2 (ℝn ) in the sense of L2 -norm. Proof. We will use the fact that the set of finite linear combinations of characteristic functions of bounded measurable sets in ℝn is dense in L2 (ℝn ). This is a well-know fact from measure and integration theory. Let Ω ⊂ ℝn be a bounded measurable set and let ϵ > 0. Then for any ϵ > 0 there exist a closed set F and an open set O such that F⊂Ω⊂O

and

m(O \ F) < ϵ2

(or only m(O) < ϵ2 if there is no closed set F ⊂ Ω). Here m is the Lebesgue measure in ℝn . Let now φ be a function from C0∞ (ℝn ) such that sup(φ) ⊂ O, φ|F = 1, and 0 ≤ φ ≤ 1. Then 󵄨 󵄨2 ‖φ − χΩ ‖2L2 (ℝn ) = ∫ 󵄨󵄨󵄨φ(x) − χΩ (x)󵄨󵄨󵄨 dm ℝn

≤ ∫ 1 dm = m(O \ F) < ϵ2 , O\F

where χΩ denotes the characteristic function of Ω, i. e., 1

χΩ (x) = {

0

if x ∈ Ω, if x ∉ Ω.

Thus we may conclude that C0∞ (ℝn ) = L2 (ℝn ). Note 2. Also we have that C0∞ (ℝn ) = L1 (ℝn ).

3.1 Definition of the Fourier transform in L2 (ℝn ) If f ∈ L2 (ℝ) then there exists {φk }k∈ℕ in C0∞ (ℝn ) such that φk → f in L1 (ℝn ) and in L2 (ℝn ), thus ̂k − ̂f ‖∞ ≤ ‖φk − f ‖L1 (ℝn ) → 0 ‖φ as k → ∞. So that ̂k (ξ ) = ̂f (ξ ) lim φ

k→∞

(uniformly)

3.1 Definition of the Fourier transform in L2 (ℝn )

| 115

for all ξ ∈ ℝn . Since C0∞ (ℝn ) ⊂ S(ℝn ), we then get ̂k − φ ̂j ‖L2 (ℝn ) = ‖φk − φj ‖L2 (ℝn ) ‖φ as j, k → ∞. Thus for any ϵ > 0 there exists N0 such that 2 󵄨̂ ̂j (ξ )󵄨󵄨󵄨󵄨 dξ < ϵ ∫ 󵄨󵄨󵄨φ k (ξ ) − φ

ℝn

if j, k ≥ N0 . Now, let j → ∞ and apply Fatou’s lemma to obtain 2 2 󵄨̂ ̂j (ξ )󵄨󵄨󵄨󵄨 dξ = ∫ lim inf󵄨󵄨󵄨󵄨φ ̂k (ξ ) − φ ̂j (ξ )󵄨󵄨󵄨󵄨 dξ ∫ 󵄨󵄨󵄨φ k (ξ ) − φ

ℝn

j→∞

ℝn

2 󵄨̂ ̂j (ξ )󵄨󵄨󵄨󵄨 dξ ≤ lim inf ∫ 󵄨󵄨󵄨φ k (ξ ) − φ j→∞

0,

ut (x, 0) = ψ(x).

We apply the Fourier transform with respect to the spatial variable to the equation and both initial conditions. Thus we obtain the transformed problem ̂ (ξ , t) = 0, utt (ξ , t) − c2 ξ 2 u

{

̂(x), u(ξ , 0) = φ

x ∈ ℝn , t > 0, ̂ ̂ t (ξ , 0) = ψ(x). and u

Its solution is the function ̂(ξ ) cos(cξt) + ̂ (ξ , t) = φ u

1 ̂ ψ sin(cξt). cξ

122 | 3 Schwartz spaces S(ℝn ) The solution of original problem is then found by the inverse Fourier transform: ∞

1 ̂ 1 ̂(ξ ) cos(cξt) + ψ(ξ u(x, t) = ) sin(cξt))eixξ dξ . ∫ (φ 2π cξ

(3.15)

−∞

Substituting the complex representation of the sine and cosine functions into (3.15), we obtain ∞

1 1 ̂(ξ )(eicξt + e−icξt )eiξ ⋅x dξ u(x, t) = ∫ φ 2π 2 −∞

∞

1 ̂ 1 ψ(ξ )(eicξt − e−icξt )eiξ ⋅x dξ + ∫ 2π 2icξ −∞ ∞

=

1 1 ̂(ξ )[ei(x−ct)ξ + e−i(x−ct)ξ ] dξ ∫ φ 4π 2 −∞

∞

+

1 1 ̂(ξ )[ei(x−ct)ξ − e−i(x−ct)ξ ] dξ ∫ φ 4πc iξ −∞ ∞

=

∞

1 ̂(ξ )ei(x−ct)ξ dξ ] ̂(ξ )ei(x+ct)ξ dξ + ∫ φ [∫ φ 4π −∞

x+ct

∞

+

−∞

1 ̂ ) ∫ eiyξ dydξ ∫ ψ(ξ 4πc x−ct

−∞ ∞

∞

1 ̂(ξ )ei(x−ct)ξ dξ ] ̂(ξ )ei(x+ct)ξ dξ + ∫ φ = [∫ φ 4π −∞

−∞

x+ct ∞

+

1 ̂ )eiy⋅ξ dξdy. ∫ ∫ ψ(ξ 4πc x−ct −∞

Applying again the inverse Fourier transform, we obtain x+ct

1 1 u(x, t) = (φ(x + ct) + φ(x − ct)) + ∫ ψ(y) dy. 2 4c x−ct

Lemma 3.4. For β ∈ ℝ, we have e

β2 1 eu − 4u = ∫ e du. π √u

∞

−β

0

Cauchy problem for the wave equation

| 123

Proof. First, we claim that e−β =

∞

2 cos(βx) dx. ∫ π 1 + x2 0

Proof of claim. Considering the right-hand side, we obtain ∞

∞

0

−∞ ∞

cos(βx) 2 cos(βx) 1 dx = dx ∫ ∫ π π 1 + x2 1 + x2 eiβx 1 dx, ∫ π 1 + x2

=

−∞

as sine is an odd function. Consider the contour C in the complex plane given by [−R, R] followed by semi1 circle of radius R in the upper half plane. Note that 1+x 2 has poles at x = ±i and so eiβx 1 eiβx dx = 2i Resx=i ( ) ∫ 2 π 1+x 1 + x2 C

= 2i Resx=i ( 2

ei β 2i = e−β .

eiβx 1 ⋅ ) x+i x−i

= 2i

Since

eiβx 1+x2

→ 0 as |x| → ∞, we obtain that 1 eiβx dx = e−β , ∫ π 1 + x2 ∞

−∞

as claimed. Now observe that ∞

∫ e−(1+x

2

)u

du =

0

∞

1 1 , ∫ es ds = 1 + x2 1 + x2 0

using the substitution s = (1 + x2 )u. Hence we have that e−β =

=

∞

∞

∞

0

0

2 2 cos(βx) 2 dx = ∫ cos(βx) ∫ e−(1+x )u dudx ∫ π π 1 + x2

0 ∞

∞

0

0

2 2 ∫ e−u ∫ e−ux cos(βx) dxdu π

(by Fubini’s theorem)

124 | 3 Schwartz spaces S(ℝn )

=

=

∞

∞

0 ∞

−∞ ∞

0 ∞

−∞

2 1 ∫ e−u ∫ e−ux cos(βx) dxdu (by symmetry) π 2 1 ∫ e−u ∫ e−ux eiβx dxdu (as sine is odd) π

∞

β 2 1 π )y 2πi( = ∫ e−u √ ∫ e−πy e √4πu dydu π u

0

−∞

∞

=

β 1 e −π( )2 e √4πu du ∫ √π √u

u (putting y = x √ , and by Theorem 2.23) π

−u

0 ∞

=

β2 e−u − 4u 1 e du, ∫ √π √u

0

as required. Note 3. The above result is known as the principle of subordination. Remark 13. In the sequel we will look for a function P(x, t) such that ? P(ξ , t) = e−2πt|ξ | . In order to find such a function, we shall apply Lemma 3.4 with β = 2π|ξ |. First, we shall calculate P(x, 1) = ∫ e−2π|ξ | e2πix⋅ξ dξ ℝ

eu −4π 2 |ξ4u|2 1 du)e2πix⋅ξ dξ e ∫ √π √u

= ∫(

ℝn

ℝ

2 2 |ξ | 1 eu ∫ ∫ e−π u e2πix⋅ξ dξdu (by Fubini’s theorem) √π √u

∞

=

0 ∞

ℝn

n

2 1 eu u 2 2πi(x√ πu ⋅η) = dηdu ( ) ∫ e−π|η| e ∫ √π √u π

0 ∞

=

n 2

ℝn

2 1 eu u ( ) e−u|x| du (by Theorem 2.23) ∫ √π √u π

0

=

=

π (putting y = √ ξ ) n

1 π

(n+1) 2

(1 +

Γ( n+1 ) 2 π

n+1 2

∞

1 |x|2 )

n+1 2

1 (1 + |x|2 )

n+1 2

∫s 0

.

(n−1) 2

e−s ds

Laplace and Poisson equations | 125

Finally, for t ≠ 1, we have P(x, t) = =

) Γ( n+1 2 π

n+1 2

Γ( n+1 ) 2 π

n+1 2

1

(1 +

n+1 | xt |2 ) 2

t (t 2

+ |x|2 )

n+1 2

.

Note 4. The function P above is known as the Poison kernel.

Laplace and Poisson equations The Laplace and Poisson equations can also be solved in some cases, by the Fourier transform. As an example, let us consider the problem uxx + uyy = 0, x ∈ ℝ, y > 0, { { { u(x, 0) = f (x), { { { {u(x, y) bounded as y → +∞. We shall apply the Fourier transform with respect to x, which in our problem leads to the equation ̂ yy − ξ 2 u ̂ = 0, u whose general solution is ̂ (ξ , y) = a(ξ )e−ξy + b(ξ )eξy , u for arbitrary functions a and b. The boundedness assumption implies b(ξ ) = 0

a(ξ ) = 0

for ξ > 0, for ξ < 0.

Hence ̂ (ξ , y) = c(ξ )e−|ξ |y , u with c being an arbitrary function. If we take into account the boundary condition, we derive c(ξ ) = ̂f (ξ ), and thus ̂ (ξ , y) = e−|ξ |y ̂f (ξ ). u

126 | 3 Schwartz spaces S(ℝn ) By Remark 13, ̂ , y)̂f (ξ ). ̂ (ξ , y) = P(ξ u The inverse Fourier transform for n = 1 leads to the solution of the original problem in the form of convolution u(x, y) = P(⋅, y) ∗ f (x) ∞

=

y f (t) dt. ∫ π (x − t)2 + y2 −∞

Remark 14. In some cases, the Fourier transform is applicable also to equations with nonconstant coefficients. Let us consider, for example, the Cauchy problem for the transport equation tux + ut = 0,

{

u(x, 0) = f (x).

x ∈ ℝn , t > 0,

Since the varying coefficient in this case is the time variable t, we use the Fourier transform with time playing the role of a parameter. We have ̂. ℱ (tux ) = t ℱ (ux ) = iξt u Transforming the equation and the initial condition, we obtain ̂+u ̂ t = 0, iξt u { ̂ (ξ , 0) = ̂f (ξ ). u Hence its solution is given by it

̂ (ξ , t) = ̂f (ξ )e− 2 ξ . u By the inverse Fourier transform, e. g., using Theorem 2.35(e), we obtain the solution of the original equation in the form u(x, t) = f (x −

t2 ). 2

Exercises. 1. Given g ∈ L2 (ℝ), consider the solution of the Schrödinger equation iut + uxx = 0 x ∈ ℝ, t > 0, { u(x, 0) = g(x).

Laplace and Poisson equations | 127

Show that 󵄨 󵄨2 󵄨 󵄨2 ̂ (ξ )󵄨󵄨󵄨 dξ . lim ∫ 󵄨󵄨󵄨󵄨u(x, t)󵄨󵄨󵄨󵄨 dx = ∫ 󵄨󵄨󵄨󵄨g 󵄨

t→∞

|x|< 21

|x| 0,

x ∈ ℝn ,

for unknown function u(x, t). (a) Derive an explicit representation for a solution u(x, t) of the above problem in terms of the initial data (you should prove that your u is well defined, but you need not prove that it represents a smooth function). (b) Assume that g is the Fourier transform of a function which vanishes in the ball of radius R > 0 centered at the origin. Show then that the solution u satisfies 󵄩󵄩 󵄩 (1−R 2 )t 󵄩󵄩u(⋅, t)󵄩󵄩󵄩 ‖g‖L2 (ℝn ) . 󵄩 󵄩L2 (ℝn ) ≤ e 4.

Assume 0 < s < ∞ and u ∈ L2 (ℝn ). Then u ∈ Hs (ℝn )

̂ ∈ L2 (ℝn ). if (1 + |ξ |2 )u

For a noninteger s, we set 󵄩 󵄩 ̂ 󵄩󵄩󵄩 ‖u‖Hs (ℝn ) = 󵄩󵄩󵄩󵄩(1 + |ξ |s )u 󵄩L2 (ℝn ) .

128 | 3 Schwartz spaces S(ℝn ) Use the Fourier transform to prove that if u ∈ Hs (ℝn ) for s > n2 , then u ∈ L∞ (ℝn ) with the bound ‖u‖L∞ (ℝn ) ≤ C‖u‖Hs (ℝn ) , 5.

the constant C depending only on s and n. Let s > n2 and 0 < α < min(s − n2 , 1). Prove that there is a constant C such that 󵄨󵄨 󵄨 󵄨󵄨u(x) − u(y)󵄨󵄨󵄨 ≤ C‖u‖Hs (ℝn ) |x − y|α . 󵄨 󵄨

6.

Consider the space 󵄨󵄨 󵄨2 n { } 󵄨󵄨 𝜕f (x) 󵄨󵄨󵄨 󵄨󵄨 dx < ∞ . W 1,2 (ℝn ) = {f ∈ L2 (ℝn ) : ∑ ∫ 󵄨󵄨󵄨 } 󵄨󵄨 𝜕xj 󵄨󵄨󵄨 󵄨 j=1 n 󵄨 { } ℝ If f ∈ W 1,2 (ℝn ), show that 󵄨󵄨2 n 󵄨󵄨󵄨 |f (x + t) + f (x − t) − 2f (x)|2 󵄨󵄨 𝜕f (x) 󵄨󵄨󵄨 dxdt = an ∫ ∑󵄨󵄨 ∫ ∫ 󵄨 dx. 󵄨 𝜕xj 󵄨󵄨󵄨 |t|n+2 󵄨 j=1󵄨󵄨 ℝn ℝn ℝn

7.

Find the solution to the two-dimensional Laplace equation 𝜕2 ϕ 𝜕2 ϕ + 2 = 0, 𝜕x 2 𝜕y with

𝜕ϕ 𝜕x

y > 0,

→ 0 and ϕ → 0 as √x 2 + y 2 → ∞, and

Use the Fourier transform in x.

ϕ(x, 0) = 1

if |x| ≤ 1,

ϕ(x, 0) = 1

if |x| > 0.

4 Distribution functions Suppose that (X, A , μ) is a measure space and let ℱ (X, A ) denote the set of all A -measurable functions on X. Definition 4.1. The distribution function Df of a function f in ℱ (X, A ) is given by 󵄨 󵄨 Df (λ) = μ({x ∈ X : 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 > λ})

(4.1)

for λ ≥ 0. In what follows, we gather some useful properties of distribution functions. Theorem 4.1. Let f and g be two functions in ℱ (X, A ). Then for all λ, λ1 , λ2 ≥ 0, we have: (a) Df is decreasing and continuous from the right; (b) |g| ≤ |f | μ-a. e. implies Dg (λ) ≤ Df (λ); λ (c) Dcf (λ) = Df ( |c| ) for all c ∈ ℂ \ {0}; (d) Df +g (λ1 + λ2 ) ≤ Df (λ1 ) + Dg (λ2 ); (e) If |f | ≤ lim infn→∞ |fn | μ-a. e. then Df (λ) ≤ lim inf Dfn (λ); n→∞

(f) If |fn | ↑ |f |, then limn→∞ Dfn (λ) = Df (λ). Proof. (a) Let λ1 ≤ λ2 be arbitrary. Then 󵄨 󵄨 󵄨 󵄨 {x ∈ X : 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 > λ2 } ⊂ {x ∈ X : 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 > λ1 }. Hence, by the monotonicity of a measure, we have that Df (λ2 ) ≤ Df (λ1 ), that is, Df is decreasing. To prove that Df is continuous from the right, let λ0 ≥ 0 and define 󵄨 󵄨 Ef (λ) = {x ∈ X : 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 > λ}, then, by the monotone convergence theorem for a measure, we have lim Df (λ0 +

n→∞

1 1 ) = lim μ(Ef (λ0 + )) n→∞ n n ∞

= μ( ⋃ Ef (λ0 + n=1

= μ(Ef (λ0 )) = Df (λ0 ).

https://doi.org/10.1515/9783110784091-004

1 )) n

130 | 4 Distribution functions Since Ef (λ1 ) ⊆ Ef (λ3 ) ⊆ Ef (λ3 ) ⊆ ⋅ ⋅ ⋅ when λ1 ≥ λ2 ≥ λ3 ≥ ⋅ ⋅ ⋅ , this establishes the right-continuity. (b) Let f and g be two functions in ℱ (X, A ) such that |g| ≤ |f | μ-a. e. Then 󵄨 󵄨 󵄨 󵄨 {x ∈ X : 󵄨󵄨󵄨g(x)󵄨󵄨󵄨 > λ} ⊆ {x ∈ X : 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 > λ}, and thus Dg (λ) ≤ Df (λ). (c) Let f ∈ ℱ (X, A ) and c ∈ ℂ \ {0}. Then 󵄨 󵄨 󵄨 󵄨 {x ∈ X : 󵄨󵄨󵄨cf (x)󵄨󵄨󵄨 > λ} = {x ∈ X : |c|󵄨󵄨󵄨f (x)󵄨󵄨󵄨 > λ} λ 󵄨 󵄨 = {x ∈ X : 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 > } |c| and thus λ 󵄨 󵄨 󵄨 󵄨 μ({x ∈ X : 󵄨󵄨󵄨cf (x)󵄨󵄨󵄨 > λ}) = μ({x ∈ X : 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 > }), |c| which is simply Dcf (λ) = Df (

λ ). |c|

(d) Let f , g ∈ ℱ (X, A ) and λ1 , λ2 ≥ 0. Then 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 {x ∈ X : 󵄨󵄨󵄨f (x) + g(x)󵄨󵄨󵄨 > λ1 + λ2 } ⊆ {x ∈ X : 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 + 󵄨󵄨󵄨g(x)󵄨󵄨󵄨 > λ1 + λ2 } 󵄨 󵄨 󵄨 󵄨 ⊆ {x ∈ X : 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 > λ} ∪ {x ∈ X : 󵄨󵄨󵄨g(x)󵄨󵄨󵄨 > λ2 }. Thus 󵄨 󵄨 μ({x ∈ X : 󵄨󵄨󵄨f (x) + g(x)󵄨󵄨󵄨 > λ1 + λ2 }) 󵄨 󵄨 󵄨 󵄨 ≤ μ({x ∈ X : 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 > λ}) + μ({x ∈ X : 󵄨󵄨󵄨g(x)󵄨󵄨󵄨 > λ2 }), which gives Df +g ≤ Df (λ1 ) + Dg (λ2 ). (e) Fix λ ≥ 0 and let E = {x ∈ X : |f (x)| > λ} and En = {x ∈ X : |fn (x)| > λ}, where n = 1, 2, . . . . Clearly, since E ⊂ ⋃∞ m=1 ⋂n>m En , we get ∞

μ(E) ≤ μ( ⋃ ⋂ En ) = lim μ( ⋂ En ) m=1 n>m

≤ lim inf μ(En ), n→∞

m→∞

n>m

4 Distribution functions | 131

which gives Df (λ) ≤ lim inf Dfn (λ). n→∞

(f) If |fn | ↑ |f |, then Ef1 (λ) ⊆ Ef2 (λ) ⊆ Ef3 (λ) ⊆ ⋅ ⋅ ⋅ . Hence ∞

Ef (λ) = ⋃ Efn (λ), n=1

and thus ∞

Df (λ) = μ(Ef (λ)) = μ( ⋃ Efn (λ)) = lim μ(Efn (λ)) = lim Dfn (λ). n=1

n→∞

n→∞

Example 11. Let ([0, ∞), L , m) be a Lebesgue measure space. Let f : [0, ∞) → [0, ∞) be given by 0 if x = 0, { { { { 1 { {ln( 1−x ) if 0 < x < 1, f (x) = { 1 { {ln( 2−x ) if 2 < x < 3, { { { if x ≥ 3. {0 Find Df . Solution. After some routine calculations, we get, for λ > 0, 󵄨 󵄨 Df (λ) = m({x ∈ [0, ∞) : 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 > λ}) = m({x = 0 : 0 > λ}) + m({x ∈ (0, 1) : ln( ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ =0

1 )λ}) 1−x

+ m({x ∈ [1, 2] ∞ > λ} ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ ⏟⏟⏟:⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ ⏟⏟) + m({x ∈ (2, 3) : ln( = [1,2]

1 ) > λ}) x−2

+ m({x ∈ [3, ∞) : 0 > λ}) ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ =0

1 ) > λ}) 1−x 1 + m({x ∈ (2, 3) : ln( ) > λ}) 2−x

= 1 + m({x ∈ (0, 1) : ln(

= 1 + m({x ∈ (0, 1) : 1 − e−λ < x < 1}) + m({x ∈ (2, 3) : 2 < x < 2 + e−λ }) = 1 + eλ + eλ = 1 + 2e−λ .

132 | 4 Distribution functions

Exercises. 1. Let ([0, a], L , m) be a Lebesgue space with a ∈ ℝ+ and f : [0, a] → [0, an ] be the function given by f (x) = x n (n ∈ ℕ). Find Df . 2. Let f (x) = ex for x ∈ [0, 1]. Find Df . 3. Let f : [1, +∞) → ℝ+ be defined by f (x) = x12 . Find Df . 4. Let ([0, ∞), L , m) be a measure space. Define f : [0, ∞) → [0, ∞) as 1 − (x − 1)2 f (x) = { 0

if 0 ≤ x ≤ 2,

if x > 2.

Find Df .

4.1 Extending the Fourier transform to Lp (ℝn ) Having defined the Fourier transform as a continuous map ⋅ ̂ : S(ℝn ) → S(ℝn ) and shown that it is invertible there, we now seek to extend it to Lp (ℝn ) for suitable values of p. To begin with, we shall recall some facts about Lp spaces. For 1 ≤ p < ∞ and for any open subset Ω ⊂ ℝn , we define 󵄨

󵄨p

ℒp (Ω) = {f : Ω → ℝ : f is Lebesgue measurable and ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx < +∞}. Ω

The quantity 1

‖f ‖ℒp (Ω)

p 󵄨 󵄨p = (∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx)

Ω

is then a seminorm on ℒp (Ω). Indeed, let Ω = [0, 1] and consider the Dirichlet function 1

if x ∈ ℚ ∩ [0, 1],

0

if x ∈ (ℝ \ ℚ) ∩ [0, 1].

f (x) = { For p = 1, we have

‖f ‖ℒ1 (Ω) = ∫ f (x) dm =

∫

[0,1]

ℚ∩[0,1]

1 dm

= m(ℚ ∩ [0, 1]) = 0. But f (x) ≠ 0 for all x ∈ [0, 1]. To correct this nuisance, we resort to the notation of quotient space, i. e., we will split all the elements of ℒp (Ω) into equivalence classes. In other words, two functions f and g in ℒp (Ω) are said to belong to the same equivalent class if and only if f = g m-almost everywhere; in symbols, f ∼ g if and only if f = g

4.1 Extending the Fourier transform to Lp (ℝn )

| 133

m-almost everywhere. It is just a matter of routine calculations to verify that ∼ defines an equivalence relation. Once this is verified, we denote the class generated by f as [f ] = {g ∈ ℒp (Ω) : g ∼ f }

(4.2)

and define the norm of g as ‖g‖p = ‖[f ]‖p for g ∈ [f ]. For arbitrary g1 ∈ [f ] and g2 ∈ [f ], we have that g1 = g2 m-a. e. Since g1 ∼ f and g2 ∼ f , this tells us that ‖[f ]‖p = ‖g‖p is well defined, being independent of the representative of the class [f ]. And so we define the Lebesgue space Lp (Ω) as the set of the equivalence classes Lp (Ω) = {[f ] : f ∈ ℒp (Ω)}, where [ ⋅ ] is defined in (4.2). Note 5. The developments in the Lp space it is going to be mutatis mutandi the same as those in Section 2.2, which takes a lot of work just to have ‖f ‖p = 0 if and only if f = [0], but in practice we never think of the Lp space as the space of equivalence classes. We do the usual abuse of notation f = [f ]. Similarly, one can also define the space of measurable, essentially bounded functions 󵄨

󵄨

ℒ∞ (Ω) = {f : Ω → ℝ : ∃ M > 0 such that 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 < M a. e.},

and, as previously, L∞ (Ω) = {[f ] : f ∈ ℒ∞ (Ω)}. Theorem 4.2. If 1 ≤ p < ∞, then (Lp (Ω), ‖ ⋅ ‖p ) is a Banach space. For a proof, see [8]. The next theorem is quite important since it permits calculating an integral in a general space via one-dimensional integral. Formula (4.3) is sometimes called Cavalieri’s principle, or even has the fancy designation of a layered cake representation. Theorem 4.3. Let f ∈ Lp (Ω), 0 < p < ∞. Then ∞

󵄨 󵄨p 󵄨 󵄨 ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dm = p ∫ λp−1 m({x ∈ Ω : 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 > λ} dλ) Ω

0 ∞

= p ∫ λp−1 Df (λ) dλ. 0

For proof see [8]. Now we state without proof the following well-know results.

(4.3)

134 | 4 Distribution functions Theorem 4.4 (Hölder inequality). Let p and q be nonnegative real numbers such that 1 + q1 = 1 and f ∈ Lp (Ω), g ∈ Lq (Ω). Then fg ∈ L1 (Ω) and p ∫ |fg|dm ≤ ‖f ‖p ‖g‖q . Ω

The equality holds if there are constants A and B, not simultaneously zero, such that A|f |p = B|g|q m-a. e. For a proof, see [8]. Theorem 4.5 (Minkowski inequality). Let 1 ≤ p ≤ ∞ and f , g ∈ Lp (Ω). Then f + g ∈ Lp (Ω) and ‖f + g‖p ≤ ‖f ‖p + ‖g‖p . The equality holds if A|f | = B|g| m-a. e. for A and B of the same sign and not simultaneously zero. For a proof, see [8]. Theorem 4.6. C0∞ (ℝn ) = Lp (ℝn ) for 1 ≤ p < ∞. Proof. After minor modifications, the proof goes line by line the same as that of Theorem 3.9. Remark 15. Theorem 4.6 does not hold for p = ∞. Indeed, for a function f = C ≠ 0 and for any φ ∈ C0∞ (ℝn ), we have ‖f − φ‖L∞ (ℝn ) ≥ |C| > 0. Hence, we cannot approximate any function from L∞ (ℝn ) by functions from C0∞ (ℝn ). It means that C0∞ (ℝn ) ≠ L∞ (ℝn ). The Fourier transform of L1 and L2 functions will be used to define the Fourier transform ̂f when f belongs to Lp for some p with 1 < p < 2 and also to show that the map ℱ : f → ̂f is bounded from Lp into Lq with p1 + q1 = 1. Let us first show that the set L1 (ℝn ) + L2 (ℝn ) defined by L1 (ℝn ) + L2 (ℝn ) = {f : f = f1 + f2 , f1 ∈ L1 (ℝn ), f2 ∈ L2 (ℝn )} contains Lp (ℝn ) for 1 ≤ p ≤ 2. This is obvious if p is either 1 or 2. If 1 < p < 2, and f ∈ Lp (ℝn ), the functions f1 and f2 in the simple decomposition f = fχ{|f |≥1} + fχ{|f | 0 and define Fϵ (z) =

f (z) . 1 + ϵz

Further, let Ω = {z ∈ ℂ : 0 < Re(z) ≤ 1} and R = {z = x + iy : x ∈ [0, 1], y ∈ [

−K K , ]}. ϵ ϵ

136 | 4 Distribution functions Let z = x + iy ∈ 𝜕Ω. Hence, if x = 0 then |f (z)| |f (z)| |f (z)| 󵄨 󵄨 󵄨󵄨 󵄨 = = ≤ 󵄨󵄨f (z)󵄨󵄨󵄨 ≤ M0 = 1. 󵄨󵄨Fϵ (z)󵄨󵄨󵄨 = |1 + ϵz| |1 + ϵiy| √1 + (ϵy)2 󵄨 Now suppose x = 1. Then |f (z)| |f (z)| 󵄨 󵄨󵄨 = 󵄨󵄨Fϵ (z)󵄨󵄨󵄨 = |1 + ϵz| |1 + ϵ + iϵy| |f (z)| 󵄨󵄨 󵄨 ≤ 󵄨f (z)󵄨󵄨󵄨 ≤ M1 = 1. = 1+ϵ 󵄨 Also, |f (z)| |f (z)| 󵄨󵄨 󵄨 = 󵄨󵄨Fϵ (z)󵄨󵄨󵄨 = |1 + ϵz| |1 + ϵx + iϵy| |f (z)| |f (z)| K = ≤ ≤ . |Im (1 + ϵx + iϵy)| ϵ|y| ϵ|y| So, for z ∈ 𝜕Ω \ Ω, we have 󵄨󵄨 K 󵄨󵄨󵄨 |f (x ± 󵄨󵄨 󵄨󵄨Fϵ (x ± i )󵄨󵄨󵄨 ≤ 󵄨󵄨 ϵ 󵄨󵄨 K

iK )| ϵ

≤

K = 1. K

Then for a fixed ϵ > 0, we have that 󵄨󵄨 󵄨 󵄨󵄨Fϵ (z)󵄨󵄨󵄨 ≤ 1

on 𝜕R.

So, by the maximum modulus principle, 󵄨 󵄨 max󵄨󵄨󵄨Fϵ (z)󵄨󵄨󵄨 ≤ 1. z∈R

Also, for z = x + iy ∈ Ω \ R, we have |y| >

K , ϵ

hence

K K 󵄨󵄨 󵄨 ≤ = 1. 󵄨󵄨Fϵ (z)󵄨󵄨󵄨 ≤ ϵ|y| ϵ( K ) ϵ Thus |Fϵ (z)| ≤ 1 for all z ∈ Ω. Then we have that f (z)| ≤ |1 + ϵz| ≤ 1 + ϵ|z|. Since this is true for all ϵ > 0, we have that |f (z)| ≤ 1, and so Mθ ≤ 1 = M01−θ M1θ . Case 2. M0 > 0 and M1 > 0.

4.1 Extending the Fourier transform to Lp (ℝn )

| 137

Let G(z) =

f (z)

M01−Re(z) M1Re(z)

.

Since M01−Re(z) and M1Re(z) are both entire functions, G(z) is continuous on 0 ≤ Re(z) ≤ 1 and analytic in the interior. Further, for 0 ≤ Re(z) ≤ 1, 󵄨󵄨 󵄨 󵄨󵄨G(z)󵄨󵄨󵄨 = ≤

|f (z)|

M01−Re(z) M1Re(z)

K , min(1, M0 ) min(1, M1 )

so G(z) is bounded on 0 ≤ Re(z) ≤ 1, Also |f (iy)| M0 󵄨 󵄨 sup󵄨󵄨󵄨G(iy)󵄨󵄨󵄨 = sup = =1 M0 y∈ℝ y∈ℝ M0 and |f (1 + iy)| M1 󵄨 󵄨 sup󵄨󵄨󵄨G(1 + iy)󵄨󵄨󵄨 = sup = = 1. M1 M1 y∈ℝ y∈ℝ Hence, by Case 1, for 0 ≤ θ ≤ 1, we have sup y∈ℝ

|f (θ + iy)| 󵄨 󵄨 = sup󵄨󵄨󵄨G(θ + iy)󵄨󵄨󵄨 ≤ 1. M01−θ M1θ y∈ℝ

Therefore Mθ = sup |f (θ + iy)| ≤ M01−θ M1θ . Lemma 4.2. If f ∈ Lp (X, μ) and 1 ≤ p < ∞, then 󵄨󵄨 󵄨󵄨 󵄨 󵄨 ‖f ‖p = sup 󵄨󵄨󵄨∫ fg dμ󵄨󵄨󵄨, 󵄨 󵄨󵄨 ‖g‖q =1󵄨 X

where

1 p

+

1 q

= 1.

Proof. Let g ∈ Lq with ‖g‖q = 1. Then by Hölder inequality, 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨∫ fg dμ󵄨󵄨󵄨 ≤ ∫ |fg| dμ ≤ ‖f ‖p ‖g‖q = ‖f ‖p , 󵄨󵄨 󵄨󵄨 X

X

so 󵄨󵄨 󵄨󵄨 󵄨 󵄨 sup 󵄨󵄨󵄨∫ fg dμ󵄨󵄨󵄨 ≤ ‖f ‖p . 󵄨 󵄨󵄨 ‖g‖q =1󵄨 X

138 | 4 Distribution functions Now, letting g =

|f (x)|p−1 sgn f ‖f ‖p−1 p

, we have 󵄨󵄨 |f |p−1 sgn f 󵄨󵄨q 󵄨󵄨 󵄨 ‖g‖qq = ∫󵄨󵄨󵄨 󵄨 dμ 󵄨󵄨 ‖f ‖p−1 󵄨󵄨󵄨 p X

= =

1

∫ |f |q(p−1) dμ

‖f ‖q(p−1) p X

1 ∫ |f |p dμ = 1. ‖f ‖pp X

Also, 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 |f (x)|p−1 sgn f 󵄨󵄨 󵄨󵄨 󵄨 󵄨 dμ 󵄨󵄨 󵄨󵄨∫ fg dμ󵄨󵄨󵄨 = 󵄨󵄨󵄨∫ f 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ‖f ‖p−1 p X

X

󵄨󵄨 |f ||f |p−1 󵄨󵄨 󵄨󵄨 󵄨 = 󵄨󵄨󵄨∫ 󵄨 󵄨󵄨 ‖f ‖p−1 󵄨󵄨󵄨 p X

=

1

‖f ‖p−1 p

∫ |f |p dμ X

= ‖f ‖p . Then, we have 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 ‖f ‖p = 󵄨󵄨󵄨∫ fg dμ󵄨󵄨󵄨 ≤ sup 󵄨󵄨󵄨∫ fg dμ󵄨󵄨󵄨 ≤ ‖f ‖p , 󵄨󵄨 󵄨󵄨 ‖g‖q =1󵄨󵄨 󵄨󵄨 X

X

and so 󵄨󵄨 󵄨󵄨 󵄨 󵄨 ‖f ‖p = sup 󵄨󵄨󵄨∫ fg dμ󵄨󵄨󵄨. 󵄨 󵄨󵄨 ‖g‖q =1󵄨 X

Exercise. Suppose that f ∈ Lp (ℝn ) and g ∈ Lp (ℝn ). Prove that f ∗ g vanishes at infinity, that is, prove that for each ϵ > 0, there exist R > 0 such that (f ∗ g)(x) < ϵ

for all |x| > R.

We are now ready to prove the Riesz–Thorin interpolation theorem, which allows us to establish boundedness of a linear operator on certain Lp spaces. Theorem 4.7 (Riesz–Thorin). Let T be a linear operator and let 1 ≤ p0 , p1 , q0 , q1 ≤ ∞, where p0 ≠ p1 and q0 ≠ q1 . Suppose T : Lp0 (μ) → Lq0 (ν) is bounded with norm M1 . Let

4.1 Extending the Fourier transform to Lp (ℝn )

1 p

=

1−θ p0

+

norm M ≤

1 = 1−θ q q0 M01−θ M1θ .

θ p1

and

+

θ q1

| 139

where 0 < θ < 1. Then T : Lp (μ) → Lq (ν) is bounded with

Proof. Our general strategy will be to construct a function F that satisfies the assumptions of the three-lines lemma. Then we will use F to bound Tf in Lq (ν) for an f , which is simple. Next we will extend the result from a simple f to all of Lp . Case 1. 1 ≤ p0 , p1 , q0 , q1 < ∞. Let f ∈ Lp where ‖f ‖p = 1. We want to prove that ‖Tf ‖q ≤ M01−θ M1θ . By Lemma 4.2. if suffices to show, for any simple function g, that 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 1−θ θ 󵄨󵄨∫ T(f )g dμ󵄨󵄨󵄨 ≤ M0 M1 . 󵄨󵄨 󵄨󵄨 X

Case (a). f is simple. Let f = ∑nj=1 aj χAj and g = ∑m k=1 bk χBk , where aj , bk ∈ ℂ and {An }n∈ℕ and {Bn }n∈ℕ are both disjoint collections of subset of X (not necessarily disjoint from each other) where μ(A) < ∞ and ν(B) < ∞ for all j, k. + pz and β(z) = 1−z + pz . For any z = x + iy, where 0 ≤ x ≤ 1, define Let α(z) = 1−z p q 0

1

0

1

󵄨 󵄨 α(z) fz (t) = 󵄨󵄨󵄨f (t)󵄨󵄨󵄨 α(θ) ei arg f (t) and 󵄨 󵄨 1−β(z) gz (t) = 󵄨󵄨󵄨g(t)󵄨󵄨󵄨 1−β(θ) ei arg g(t) . Also, let F(z) = ∫ T(fz )gz dν. Y

Claim. The function F(z) is analytic on 0 < Re(z) < 1. Indeed, 󵄨 󵄨 󵄨 α(z) 󵄨 1−β(z) F(z) = ∫ T(fz )gz dν = ∫ T(󵄨󵄨󵄨f (t)󵄨󵄨󵄨 α(θ) ei arg f (t) )󵄨󵄨󵄨g(t)󵄨󵄨󵄨 1−β(θ) ei arg g(t) dν Y

Y n

m

α(z)

1−β(z)

= ∫ T(∑ |aj | α(θ) ei arg aj χAj )( ∑ |bk | 1−β(θ) ei arg bk χBk ) dν j=1

Y

n

k=1

m

α(z)

1−β(z)

= ∫(∑ |aj | α(θ) ei arg aj T(χAj ))( ∑ |bk | 1−β(θ) ei arg bk χBk ) dν Y j=1 n m

k=1

α(z)

1−β(z)

= ∑ ∑ |aj | α(θ) |bk | 1−β(θ) ei(arg aj +arg bk ) ∫ T(χAj )χBk dν. j=1 k=1

Y

140 | 4 Distribution functions If we recall how α and β are defined, then we see that the above is just a linear combination of exponential functions, so we actually have that F(z) is entire. From this, we also see that F(z) is bounded on 0 ≤ Re(z) ≤ 1. Claim. We have |F(iy)| ≤ M0 and |F(1 + iy)| ≤ M1 for all y ∈ ℝ. Then by Hölder inequality, 󵄨󵄨 󵄨 󵄨 󵄨󵄨 󵄨 󵄨󵄨 󵄨󵄨F(iy)󵄨󵄨󵄨 = 󵄨󵄨󵄨∫ T(fiy )giy dν󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 Y

1 ′

1

q0 q ′ 0 󵄨 󵄨q ≤ (∫󵄨󵄨󵄨T(fiy )󵄨󵄨󵄨 0 dμ) (∫ |giy |q0 dμ)

Y

󵄩 󵄩 = 󵄩󵄩󵄩T(fiy )󵄩󵄩󵄩q ‖giy ‖q0′

Y

0

≤ ‖T‖q0 ‖fiy ‖p0 ‖giy ‖q0′

= M0 ‖fiy ‖p0 ‖giy ‖q0′ ,

since T is bounded from Lp0 with norm M0 . Further, 󵄨 󵄨p ‖fiy ‖pp00 = ∫󵄨󵄨󵄨|f | α(θ) ei arg f (t) 󵄨󵄨󵄨 0 dμ α(iy)

X

󵄨p 󵄨 = ∫󵄨󵄨󵄨|f | α(θ) 󵄨󵄨󵄨 0 dμ α(iy)

X

= ∫ |f |p dμ X

= 1,

(4.4)

where we get (4.4) because 󵄨󵄨󵄨󵄨 󵄨 α(iy) 󵄨p0 󵄨󵄨 󵄨Re( α(iy) ) 󵄨 󵄨Im ( α(iy) ) 󵄨p0 󵄨󵄨󵄨󵄨f (z)󵄨󵄨󵄨 α(θ) 󵄨󵄨󵄨 = 󵄨󵄨󵄨󵄨󵄨󵄨f (z)󵄨󵄨󵄨 α(θ) 󵄨󵄨󵄨f (z)󵄨󵄨󵄨 α(θ) 󵄨󵄨󵄨 󵄨 󵄨Re( α(iy) )p = 󵄨󵄨󵄨f (z)󵄨󵄨󵄨 α(θ) 0 and Re(

α(iy) 1 − iy iy 1−θ θ ) = Re[( + )( + ) ] α(θ) p0 p1 p0 p1 −1

p1 − ip1 y + ip0 y )p] p0 p1 pp p = 1 = , p0 p1 p0

= Re[(

4.1 Extending the Fourier transform to Lp (ℝn )

|

141

so Re(

α(iy) p )p0 = ( )p0 = p. α(θ) p0

Similarly, ‖giy ‖q0′ = 1. So we have that |F(iy)| ≤ M0 . Also, again by Hölder inequality, 󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨F(1 + iy)󵄨󵄨󵄨 = 󵄨󵄨󵄨∫ T(f1+iy )g1+iy dν󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 Y

󵄩 󵄩 ≤ 󵄩󵄩󵄩T(f1+iy )󵄩󵄩󵄩q ‖g1+iy ‖q1′ 1 󵄩 󵄩 ≤ ‖T‖󵄩󵄩󵄩(f1+iy )󵄩󵄩󵄩q ‖g1+iy ‖q1′ 1 󵄩󵄩 󵄩󵄩 = M1 󵄩󵄩(f1+iy )󵄩󵄩q ‖g1+iy ‖q1′ . 1

(4.5)

Further, q′ 󵄨󵄨 󵄨 1−β(1+iy) 󵄨q′ ‖g1+iy ‖q1′ = ∫󵄨󵄨󵄨󵄨󵄨󵄨g(t)󵄨󵄨󵄨 1−β(θ) 󵄨󵄨󵄨 1 dν 1 Y

)q1′ 󵄨 󵄨Re( 1−β(1+iy) 1−β(θ) = ∫󵄨󵄨󵄨g(t)󵄨󵄨󵄨 dν Y

󵄨 󵄨q = ∫󵄨󵄨󵄨g(t)󵄨󵄨󵄨 1 dν ′

Y

= 1,

(4.6)

where we get (4.6) because Re(

1 − β(1 + iy) 1 − (1 + iy) (1 + iy) 1−θ θ ) = Re[(1 − − )(1 − ( + )) ] 1 − β(θ) q0 q1 70 q1 −1

1 1 )( ′ ) q1 q

−1

= (1 −

=

q′ . q1′

Similarly, we have that ‖f1+iy ‖p1 = 1, so from (4.5) we obtain |F(1 + iy)| ≤ M1 . Then by Lemma 4.1, for 0 ≤ θ ≤ 1, we have 󵄨 󵄨 sup󵄨󵄨󵄨F(θ + iy)󵄨󵄨󵄨 ≤ M01−θ M1θ . y∈ℝ

Note that fθ (t) = |f (t)|ei arg f (t) = f (t) and similarly gθ (t) = g(t). Then 󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 󵄨 󵄨󵄨F(θ)󵄨󵄨󵄨 = 󵄨󵄨󵄨∫ T(f )g dν󵄨󵄨󵄨. 󵄨󵄨 󵄨󵄨 Y

142 | 4 Distribution functions Thus 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨 1−θ θ 󵄨󵄨∫ T(f )g dν󵄨󵄨󵄨 ≤ sup󵄨󵄨󵄨F(θ + iy)󵄨󵄨󵄨 ≤ M0 M1 . 󵄨󵄨 y∈ℝ 󵄨󵄨 Y

Since this is true for any simple function g, we have then that M = ‖T‖ ≤ M01−θ M1θ . Case (b). f ∈ Lp . Since f ∈ Lp , there exists a sequence of simple functions {fn }n∈ℕ so that |fn | ≤ |f | and fn → f pointwise. Let E = {x : |f (x)| > 1}, g = fχE , and gn = fn χE . Further, let h = f − g and hn = fn − gn . Without loss of generality, we might assume p0 ≤ p1 . Then ‖g‖pp00 = ∫ |g|p0 dμ = ∫ |f |p0 χE dμ X

X

p

≤ ∫ |f | χE dμ ≤ ∫ |f |p dμ < ∞, X

X

so g ∈ Lp0 . Also 󵄨 󵄨p ‖h‖pp11 = ∫ |f − fχE |p1 dμ = ∫󵄨󵄨󵄨f (1 − χE )󵄨󵄨󵄨 1 dμ X

X

p1

= ∫ |f | χE c dμ ≤ ∫ |f |p χE c dμ X

X p

≤ ∫ |f | dμ < ∞, X

so h ∈ Lp1 . Since fn → f a. e., |fn | ≤ |f | and |f | ∈ Lp , by the dominated convergence theorem, we have ‖fn − f ‖p → 0, and so ‖fn ‖p → ‖f ‖p . Similarly, by the dominated convergence theorem, we also have that ‖gn − g‖p0 → 0. And, lastly, since 󵄨 󵄨 |hn | = |fn − gn | = 󵄨󵄨󵄨fn (1 − χE )󵄨󵄨󵄨 ≤ |f |(1 − χE ) = |f − fχE |

= |h|, hn → h a. e., and h ∈ Lp1 , we have that ‖hn − h‖p1 → 0. Since T is bounded from Lp0 to Lq0 and from Lp1 to Lq1 , we then have that ‖T(gn )−T(g)‖q0 → 0 and ‖T(hn )−T(h)‖q1 → 0. Then by Chebyshev inequality, q

‖T(gn ) − T(g)‖q00 󵄨 󵄨 → 0, μ({x : 󵄨󵄨󵄨T(gn ) − T(g)󵄨󵄨󵄨 > ϵ}) ≤ ϵ

4.1 Extending the Fourier transform to Lp (ℝn ) μ

|

143

μ

which means that T(gn ) → T(g), and T(hn ) → T(h), too. So there exists a subsequence {T(hnk )} of {T(hn )}, so that T(hnk ) → T(h) a. e., and then T(fn ) = T(gn ) + T(hn ) → T(g) + T(h) = T(f ). So by Fatou’s lemma, we have 󵄩 󵄩 󵄩 󵄩󵄩 inf󵄩󵄩T(fn )󵄩󵄩󵄩q 󵄩󵄩T(f )󵄩󵄩󵄩q ≤ lim n→∞ 󵄩

≤ lim inf M01−θ M1θ ‖fn ‖q ≤

n→∞ M01−θ M1θ ‖f ‖q .

Finally if p = ∞, then p0 = p1 = ∞ and q = 1, then q0 = q1 = 1. Hence there is nothing to prove. The following result is the main in this section, and we shall give two proof of it. Theorem 4.8 (Haussdorff–Young). If 1 ≤ p ≤ 2, then there is a constant C depending only on p such that ‖̂f ‖Lq (ℝn ) ≤ C‖f ‖Lq (ℝn ) ,

1 1 + = 1, p q

(4.7)

for all f ∈ Lp (ℝn ). Proof. It suffices to prove the theorem for 1 < p < 2 since the endpoint cases p = 1 and p = 2 have already been established. Fix f and p with f ∈ Lp (ℝn ) and 1 < p < 2. Since the Fourier transform is a linear operator, we may assume without loss of generality that ‖f ‖p = 1. Let us consider the distribution function of |̂f |, namely, 󵄨 󵄨 D(λ) = m({x ∈ ℝn : 󵄨󵄨󵄨̂f (x)󵄨󵄨󵄨 > λ}),

λ > 0.

For each λ > 0, define g and h by g = gλ = fχ{|f |p−1 ≥ 1 } λ

and h = hλ = fχ{|f |p−1 < 1 } . λ

Note that f = g + h. Also, consider g ∈ L1 (ℝn ) and h ∈ L2 (ℝn ) with norms satisfying ‖g‖[L1 (ℝn ) =

∫ {|f |p−1 ≥λ−1 }

|f |dx ≤ λ

∫ {|f |p−1 ≥λ−1 }

|f |p dx ≤ λ‖f ‖pL

n)

p (ℝ

=λ

(4.8)

144 | 4 Distribution functions and ‖h‖2[L2 (ℝn ) =

|f |2 dx ≤ λ

∫ − 1 {0 λ} ⊂ {|h| 2 ̂ > D(λ) = m({|̂f | < λ}) ≤ m({|h| ≤

1

λ }) 2

̂ 2 dx ∫ |h|

( λ2 )2 n ℝ

4 1 ‖h‖22 λ2 (2π)n 1 |f |2 dx, ≤ 2 ∫ λ =

{0 0 be arbitrary and considering |x| > a results in ℳf (x) ≥

1 m(B(x, 2|x|))

󵄨 󵄨 ∫ 󵄨󵄨󵄨f (y)󵄨󵄨󵄨 dy B(x,2|x|)

1 󵄨 󵄨 ≥ ∫ 󵄨󵄨󵄨f (y)󵄨󵄨󵄨 dy m(B(0, 2|x|)) B(0,a)

const 󵄨 󵄨 = ∫ 󵄨󵄨󵄨f (y)󵄨󵄨󵄨 dy. |x|n B(0,a)

Since |x|−n is not integrable for |x| > a, this yields that ∫B(0,a) |f (y)| dy = 0. From the arbitrariness of a, we conclude that f = 0. Let us give a particular example for the one-dimensional case.

4.3 Maximal function

| 149

Example 12. Take 1

f (x) =

x log2 x

χ(0, 1 ) , 2

and let us use radius r = x. Note that f ∈ L1 (ℝ) and 2x

1 󵄨󵄨 󵄨 ∫󵄨f (y)󵄨󵄨󵄨 dy 2x 󵄨

ℳf (x) ≥

0 x

1 dy ∫ 2x y log2 y

≥

0

−1 = . 2x log x Since 2x −1 is not integrable in the neighborhood of x = 0, we obtain that ℳf ∉ L1 (ℝ). log x Theorem 4.10. If f ∈ L1 (ℝn ), then m({x ∈ ℝn : ℳf (x) > λ}) ≤

3n 󵄨󵄨 󵄨 ∫ 󵄨f (y)󵄨󵄨󵄨 dy λ 󵄨 ℝn

i. e., ℳf ∈ Weak-L1 (see Definition 4.4). Proof. Fix λ > 0 and let Aλ = {x ∈ ℝn : ℳf (x) > λ}. By the inner regularity of Lebesgue measure m(Aλ ) = sup{m(K) : K ⊂ Aλ is compact}. So it is enough to prove that m(K) ≤

3n 󵄨󵄨 󵄨 ∫ 󵄨f (y)󵄨󵄨󵄨 dy λ 󵄨 ℝn

for every compact subset K of Aλ . If x ∈ K, then there is an open ball Bx centered at x such that 1 󵄨 󵄨 ∫ 󵄨󵄨f (y)󵄨󵄨󵄨 dy > λ. m(Bx ) 󵄨 Bx

Since K is compact, we may extract a finite subcover {B1 , B2 , . . . , BN } from the open cover {Bx : x ∈ K}. By Lemma 4.3, there is a finite subfamily of disjoint balls

150 | 4 Distribution functions {B′1 , B′2 , . . . , B′M } such that N

m(K) ≤ ∑ m(Bi ) i=1

M

≤ 3n ∑ m(B′j ) j=1

n M

≤

3 󵄨 󵄨 ∑ ∫󵄨󵄨f (y)󵄨󵄨󵄨 dy λ j=1 󵄨 B′j

≤

3n 󵄨󵄨 󵄨 ∫ 󵄨f (y)󵄨󵄨󵄨 dy, λ 󵄨 ℝn

which proves the result.

4.4 Weak-Lp space Definition 4.4. Let (ℝn , L , m) be a Lebesgue measure space. Let f ∈ ℱ (ℝn , L ). Suppose 1 ≤ p < ∞. The function f is said to belong to the weak-Lp space on ℝn if 1 p

‖f ‖(p,∞) = (sup λp Df (λ)) < ∞. λ>0

In other words, the weak-Lp (ℝn ) space is defined by L(p,∞) (ℝn ) = {f : ‖f ‖(p,∞) < ∞}. Two functions will be considered equal if they are equal m-a. e. as in the case of Lp (ℝn ) spaces. The weak Lebesgue spaces are larger than the Lebesgue spaces, which is just a restatement of the Chebyshev inequality. Theorem 4.11. For any 1 ≤ p < ∞ and any f ∈ Lp (ℝn ), we have Lp (ℝn ) ⊂ L(p,∞) (ℝn ),

(4.11)

‖f ‖(p,∞) ≤ ‖f ‖p .

(4.12)

and hence

Proof. If f ∈ Lp (ℝn ), then 󵄨 󵄨 λp m({x ∈ ℝn : 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 > λ}) ≤ ∫ |f |p dm ≤ ∫ |f |p dm, {|f |>λ}

ℝn

4.4 Weak-Lp space

| 151

therefore 󵄨 󵄨 λp m({x ∈ ℝn : 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 > λ}) ≤ ‖f ‖pp ,

(4.13)

and from (4.13) we get 1 p

(sup{λp Df (λ)}) ≤ ‖f ‖p .

(4.14)

λ>0

Hence f ∈ L(p,∞) (ℝn ), which means that (4.11) is true. From (4.14) we get (4.12). The inclusion (4.11) is strict. Indeed, let f (x) = x Lebesgue measure on ℝ. Then m({x ∈ (0, ∞) :

1 |x|

1 p

− p1

on (0, ∞) and m be the

> λ}) = m({x ∈ (0, ∞) : |x|
λ} ⊂ {x : ℳfλ (x) >

λ }. 2

From the latter relation, we obtain m({x : ℳf (x) > λ}) ≤

3n 3n 󵄨󵄨 󵄨 ∫ 󵄨󵄨fλ (x)󵄨󵄨󵄨 dm = λ λ ℝn

󵄨 󵄨󵄨 󵄨󵄨f (x)󵄨󵄨󵄨 dm.

∫ {x:|f (x)|≥ λ2 }

Now, using Theorem 4.3 and (4.15), we obtain ∞

󵄨 󵄨p ∫ 󵄨󵄨󵄨ℳf (x)󵄨󵄨󵄨 dm = p ∫ λp−1 m({x : ℳf (x) > λ}) dλ 0

ℝn

∞

≤ 3n p ∫ λp−2 (

∫

0

{x:f (x)≥ λ2 }

∞

2|f (x)|

0

0

n

󵄨󵄨 󵄨 󵄨󵄨f (x)󵄨󵄨󵄨 dm) dλ

󵄨 󵄨 ≤ 3 p ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨( ∫ λp−2 dλ) dm 3n p 󵄨󵄨 3 p 󵄨󵄨 󵄨󵄨 󵄨p 󵄨p−1 = ∫ 󵄨󵄨f (x)󵄨󵄨󵄨󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dm = ∫ 󵄨󵄨f (x)󵄨󵄨󵄨 dm, p−1 p−1 n

∞

∞

0

0

which ends the proof.

Exercise. Let f , g ∈ L(p,∞) . Prove that: (a) ‖cf ‖(p,∞) = |c|‖f ‖(p,∞) for any constant c; p p (b) ‖f + g‖(p,∞) ≤ 2(‖f ‖(p,∞) + ‖g‖(p,∞) ).

4.5 Lebesgue differentiation theorem The maximal function provides a crucial estimate in the following proof. Theorem 4.13. If f ∈ L1,loc (ℝn ), then for a. e. x ∈ ℝn , lim[

r→0

1 ∫ f (y) dy] = f (x). m(B(x, r)) B(x,r)

(4.15)

4.5 Lebesgue differentiation theorem

| 153

Moreover, for a. e., x ∈ ℝn , lim[

r→0

1 󵄨 󵄨 ∫ 󵄨󵄨󵄨f (y) − f (x)󵄨󵄨󵄨 dy] = 0. m(B(x, r)) B(x,r)

Proof. Since 󵄨󵄨 󵄨󵄨 1 󵄨󵄨 󵄨 ∫ (f (y) − f (x)) dy]󵄨󵄨󵄨 󵄨󵄨lim[ 󵄨󵄨r→0 m(B(x, r)) 󵄨󵄨 B(x,r)

≤ lim[ r→0

1 󵄨 󵄨 ∫ 󵄨󵄨󵄨f (y) − f (x)󵄨󵄨󵄨 dy], m(B(x, r)) B(x,r)

we just need to prove the second result. We define f ∗ : ℝn → [0, ∞] by f ∗ (x) = lim sup[ r→0+

1 󵄨 󵄨 ∫ 󵄨󵄨󵄨f (y) − f (x)󵄨󵄨󵄨 dy] m(B(x, r)) B(x,r)

and shall show that f ∗ = 0 pointwise a. e. If g ∈ Cc (ℝn ) then, given any ϵ > 0, there exists a δ > 0 such that |g(x) − g(y)| < ϵ whenever |x − y| < δ. Hence, if r < δ, 1 󵄨 󵄨 ∫ 󵄨󵄨󵄨g(y) − g(x)󵄨󵄨󵄨 dy < ϵ, m(B(x, r)) B(x,r)

which implies that g ∗ = 0. We prove the result for general f by approximation with a continuous function. First, note that we can assume that f ∈ L1 (ℝn ) is integrable without loss of generality; for example, if the result holds for fχBk (0) ∈ L1 (ℝn ) for each k ∈ ℕ except on a set Ek of measure zero, then it holds for f ∈ L1,loc (ℝn ) except on ⋃∞ k=1 Ek which has measure zero. Next, observe that since 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨󵄨f (y) + g(y) − (f (x) + g(x))󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨f (y) − f (x)󵄨󵄨󵄨 + 󵄨󵄨󵄨g(y) − g(x)󵄨󵄨󵄨 and lim sup(A + B) ≤ lim sup A + lim sup B, we have (f + g)∗ ≤ f ∗ + g ∗ . Thus, if f ∈ L1 (ℝn ) and g ∈ Cc (ℝn ), we have (f − g)∗ ≤ f ∗ + g ∗ = f ∗ = (f − g + g)∗ ≤ (f − g)∗ + g ∗ = (f − g)∗ , which shows that (f − g)∗ = f ∗ .

154 | 4 Distribution functions If f ∈ L1 (ℝn ) then we claim that there is a constant C, depending only on n, such that for every 0 < λ < ∞, m({x ∈ ℝn : f ∗ > λ}) ≤

C ‖f ‖ n . λ L1 (ℝ )

(4.16)

To show this, we estimate f ∗ (x) ≤ sup[ r>0

1 󵄨 󵄨 ∫ 󵄨󵄨󵄨f (y) − f (x)󵄨󵄨󵄨 dy] m(B(x, r)) B(x,r)

1 󵄨 󵄨 󵄨 󵄨 ≤ sup[ ∫ 󵄨󵄨󵄨f (y)󵄨󵄨󵄨 dy] + 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 r>0 m(B(x, r)) 󵄨 󵄨 ≤ ℳf (x) + 󵄨󵄨󵄨f (x)󵄨󵄨󵄨.

B(x,r)

It follows that {f ∗ > λ} ⊂ {ℳf + |f |} ⊂ {ℳf >

λ λ } ∪ {|f | > }. 2 2

By Theorem 4.10, m({x ∈ ℝ : ℳf (x) >

λ 2 ⋅ 3n }) ≤ ‖f ‖L1 (ℝn ) . 2 λ

Combining these estimates, we conclude that (4.16) holds with C = 2(3n + 1). Finally, suppose that f ∈ L1 (ℝn ) and 0 < λ < ∞. It is well known (see [8]) that for any ϵ > 0 there exists g ∈ Cc (ℝn ) such that ‖f − g‖L1 (ℝn ) < ϵ. Then m({x ∈ ℝn : f ∗ (x) > λ}) = m({x ∈ ℝn : (f − g)∗ > λ}) C ≤ ‖f − g‖L1 (ℝn ) λ C < ϵ. λ Since ϵ > 0 is arbitrary, it follows that m({x ∈ ℝn : f ∗ (x) > λ}) = 0, and hence, since ∞

{x ∈ ℝn : f ∗ (x) > 0} = ⋃ {x ∈ ℝn : f ∗ (x) > k=1

we have m({x ∈ ℝn : f ∗ (x) > 0}) = 0. This proves the result.

1 }, k

4.5 Lebesgue differentiation theorem

| 155

Definition 4.5. For a function f ∈ L1,loc (ℝn ), the Lebesgue set of f is defined by Lf = {x ∈ ℝn :

1 󵄨 󵄨 ∫ 󵄨󵄨󵄨f (y) − f (x)󵄨󵄨󵄨 dy = 0}. m(B(x, r)) B(x,r)

Theorem 4.14. For any f ∈ L1,loc (ℝn ), m((Lf )c ) = 0. Proof. For any z ∈ ℂ, we can apply Theorem 4.13 to gz (x) = |f (x) − z| to conclude that, except on a Lebesgue null set Ez , we have lim

r→0

1 󵄨 󵄨 󵄨 󵄨 ∫ 󵄨󵄨󵄨f (y) − z 󵄨󵄨󵄨 dy = 󵄨󵄨󵄨f (x) − z 󵄨󵄨󵄨. m(B(x, r)) B(x,r)

Let E = ⋃z∈ℚ+iℚ Ez . Then m(E) = 0, and if x ∈ E c , for any ϵ > 0 we can choose z ∈ ℚ+iℚ with |f (x) − z| < ϵ2 , so that 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 ϵ 󵄨󵄨f (y) − f (x)󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨f (y) − z 󵄨󵄨󵄨 + 󵄨󵄨󵄨f (x) − z 󵄨󵄨󵄨 < 󵄨󵄨󵄨f (y) − z 󵄨󵄨󵄨 + . 2 It follows then that lim sup r→0

1 󵄨 󵄨 󵄨 󵄨 ∫ 󵄨󵄨󵄨f (y) − f (x)󵄨󵄨󵄨 dy ≤ lim ∫ 󵄨󵄨󵄨f (y) − z 󵄨󵄨󵄨 dy + ϵ r→0 m(B(x, r)) B(x,r)

B(x,r)

󵄨 󵄨 ϵ = 󵄨󵄨󵄨f (x) − z 󵄨󵄨󵄨 + < ϵ. 2

Since ϵ is arbitrary, x ∈ Lf . And so E c ⊂ Lf which implies (Lf )c ⊂ E, hence m(Lcf ) = 0 we are done. Remark 16. Note that if K ∈ L1 (ℝn ) with K ≥ 0 and ∫ℝn K(x) dx = 1, then we can obtain an approximate identity {Kϵ }ϵ>0 for L1 (ℝn ) by putting Kϵ (x) =

1 x K( ). ϵn ϵ

We have x ∫ Kϵ (x) dx = ∫ ϵ−n K( ) dx = ∫ K(y) dy = 1 ϵ

ℝn

ℝn

ℝn

by the change of variable y = xϵ , where we used the fact that ϵ−n is the determinant of the Jacobian of the map x 󳨃→

x ϵ

156 | 4 Distribution functions from ℝn to ℝn . And 󵄨 󵄨 ∫ |Kϵ(x) | dx = ∫ 󵄨󵄨󵄨K(y)󵄨󵄨󵄨 dy → 0

as ϵ → 0.

|y|> δϵ

|x|>δ

Next, for ϕ ∈ L1 (ℝn ) with ∫ℝn ϕ(x) dx = 1, it is assumed that if 󵄨 󵄨 ψ(x) = sup 󵄨󵄨󵄨ϕ(y)󵄨󵄨󵄨, |y|≥|x|

then ψ ∈ L1 (ℝn ).

It is clear that: (I) ψ is radial, i. e., |x| = |y| implies ψ(x) = ψ(y). (II) For |x| = r, the function ψ0 (r) = ψ(x) is decreasing in r > 0. (III) |x|n ψ(x) → 0 when |x| tends to zero or to infinity and, in particular, there exists a constant M > 0 such that |x|n ψ(x) for all 0 < |x| < ∞. (IV) For every 0 < η < ∞ and 1 ≤ p ≤ ∞, ‖χη ψϵ ‖q → 0

as ϵ → 0,

where χη is the characteristic function of the set {x : |x| > η} and x ψϵ (x) = ϵ−n ψ( ). ϵ In fact, to show (III), note that, by (II), r

∫

ψ(x) dx = ∫ dx ′ ∫ ψ0 (ρ)ρn−1 dρ ≥ wn ψ0 (r)n (1 − 2−n ). Σ

r η

q

η p ≤ (ϵ ψ0 ( )) ∫ ψϵ (x) dx, ϵ −n

|x|>η

η

η

and, by (III), η−n ( ϵ )n ψ0 ( ϵ ) → 0 as ϵ → 0 for any fixed η.

4.5 Lebesgue differentiation theorem

| 157

Theorem 4.15. Let x ∈ Lf be such that ∫ℝn f (x) dx = a, where a is a positive real number. Suppose ϕ ∈ L1 (ℝn ). Let ψ = sup|y|≥|x| |ϕ(y)| and, for ϵ > 0, let ϕϵ (x) = ϵ−n ϕ( xϵ ). If ψ ∈ L1 (ℝn ) and f ∈ Lp (ℝn ) with 1 ≤ p ≤ ∞, then lim (f ∗ ϕϵ )(x) = f (x) ∫ ϕ(y) dy.

ϵ→0

ℝn

Proof. Let us fix x ∈ Lf and δ > 0. We can find η > 0 such that for all 0 < r < η, 󵄨 󵄨 r −n ∫ 󵄨󵄨󵄨f (x − y) − f (x)󵄨󵄨󵄨 dy > δ.

(4.18)

|y|≤r

Since ∫ℝn ϕϵ (y) dy = ∫ℝn ϕ(y) dy = a for all ϵ > 0, we have 󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 󵄨 󵄨󵄨(f ∗ ϕϵ )(x) − af (x)󵄨󵄨󵄨 = 󵄨󵄨󵄨 ∫ (f (x − y) − f (x))ϕϵ (y) dy󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 n ℝ

󵄨󵄨 󵄨󵄨 󵄨 󵄨 ≤ 󵄨󵄨󵄨 ∫ (f (x − y) − f (x))ϕϵ (y) dy󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 |y|≤η

󵄨󵄨 󵄨󵄨 󵄨 󵄨 + 󵄨󵄨󵄨 ∫ (f (x − y) − f (x))ϕϵ (y) dy󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 |y|>η

= I1 + I2 . Changing variables to polar coordinates y → (y′ , s), (4.18) becomes r

r

−n

󵄨 󵄨 ∫ ∫󵄨󵄨󵄨f (x − sy′ ) − f (x)󵄨󵄨󵄨sn−1 dsdy′ < δ, 0 Σ

and denoting 󵄨 󵄨 g(s) = ∫󵄨󵄨󵄨f (x − sy′ ) − f (x)󵄨󵄨󵄨 dy′ ,

(4.19)

Σ r

G(r) = ∫ g(s)sn−1 ds, 0 −n

Δx (r) = r G(r),

(4.20) (4.21)

158 | 4 Distribution functions we get Δx (r) < δ for 0 < r < η. Thus y 󵄨 󵄨 |I1 | ≤ ∫ 󵄨󵄨󵄨f (x − y) − f (x)󵄨󵄨󵄨ϵ−n ψ( ) dy ϵ |y|≤η η

r = ∫ r n−1 g(r)ϵψ0 ( ) dr ϵ 0

η

η r 󵄨󵄨󵄨 r = G(r)ϵ ψ0 ( )󵄨󵄨󵄨 − ∫ G(r)d(ϵ−k ψ0 ( )) (integrating by parts) ϵ 󵄨󵄨0 ϵ −n

0

η ϵ

n η r r 󵄨󵄨󵄨 = Δx (r)( ) ψ0 ( )󵄨󵄨󵄨 − ∫ G(ϵs)ϵ dψ0 (s) ϵ ϵ 󵄨󵄨0 0

η ϵ

≤ Δx (η)M − ∫ Δx (ϵs)sn dψ0 (s) 0 ∞

< δ(M − ∫ sn dψ0 (s)) 0

= δ(M + B), where M is the constant in (III) and ∞

∞

0

0

B = − ∫ sn dψ0 (s) = n ∫ sn−1 ψ0 (s) ds =

n ∫ ψ(x) dx > 0. wn ℝn

Since ψ ∈ L1 (ℝn ), B < ∞. To estimate I2 , we let ψϵ (y) = ϵ−n ψ( yϵ ) and then 󵄨 󵄨 󵄨 󵄨 |I2 | ≤ ∫ 󵄨󵄨󵄨f (x − y)󵄨󵄨󵄨ψϵ (y) dy + 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 ∫ ψϵ (y) dy. |y|>η

(4.22)

|y|>η

Since ∫ ψϵ (y) dy = ∫ ψ(y) dy → 0

as ϵ → 0,

η |y|> 2

|y|>η

due to ψ ∈ L1 (ℝn ), the second term on the right-hand side of (4.22) tends to zero with ϵ. The same is true for the first term since applying Hölder inequality to the functions |f (x − y)| and χη ψϵ , we get 󵄨 󵄨 ∫ 󵄨󵄨󵄨f (x − y)󵄨󵄨󵄨ψϵ (y) dy ≤ ‖f ‖p ‖χη ψϵ ‖q |y|>η

4.6 Cube

| 159

for p1 + q1 = 1, where the last factor tends to zero by property (IV). Collecting the estimates for I1 and I2 , we see that 󵄨 󵄨󵄨 󵄨󵄨(f ∗ ϕϵ )(x) − af (x)󵄨󵄨󵄨 → 0

as ϵ → 0,

as desired.

4.6 Cube A closed cube is a bounded interval in ℝn , whose sides are parallel to the coordinate axes and equally long, that is, Q = [a1 , b1 ] × [a2 , b2 ] × ⋅ ⋅ ⋅ × [an , bn ] with b1 − a1 = b2 − a2 = ⋅ ⋅ ⋅ = bn − an . The side length of a cube Q is denoted by l(Q). In case we want to specify the center, we write l Q(x, l) = {y ∈ ℝn : |yk − xk | ≤ , k = 1, 2, . . . , n} 2 for a cube with center at x ∈ ℝn and side length l > 0. If Q = Q(x, l), we denote αQ = Q(x, αl) for α > 0. Let Q = [a1 , b1 ] × [a2 , b2 ] × ⋅ ⋅ ⋅ × [an , bn ] be a closed cube in ℝn with side length l. We decompose Q into subcubes recursively. Denote D0 = {Q}. Bisect each interval [ai , bi ], i = 0, 1, 2, . . . and obtain 2n congruent subcubes of Q. Denote this collection of cubes by D1 . Bisect every cube in D1 and obtain 2n subcubes. Continuing this way, we obtain generations of dyadic cubes Dk , k = 0, 1, 2, . . . . The dyadic subcubes in Dk are of the form m1 l (m + 1)l ml (m + 1)l , b1 + 1 k ] × [a2 + k2 , b2 + 2 k ] 2k 2 2 2 m l (m + 1)l × ⋅ ⋅ ⋅ × [an + kn , bn + n k ], 2 2

[a1 +

where k = 0, 1, 2, . . . and mj = 0, 1, . . . , 2k , j = 1, 2, . . . , n. The collection of all dyadic subcubes of Q is ∞

D = ⋃ Dk . k=0

A cube Q′ ∈ D is called a dyadic subcube of Q. Figure 4.2 bellow illustrates some dyadic subcubes

160 | 4 Distribution functions

Figure 4.2: Collections of dyadic subcubes.

4.7 Properties of dyadic subcubes Dyadic subcubes of Q have the following properties: 1. Every Q′ ∈ D is a subcube of Q. 2. Cubes in Dk cover Q and the interiors of the cubes in Dk are pairwise disjoint for every k = 0, 1, 2, . . . . 3. If Q′ , Q′′ ∈ D , either one is contained in the other or the interiors of the cubes are disjoint. This is called the nesting property; see Figure 4.3. 4. If Q′ ∈ D and j < k, there is exactly one parent cube in Dj which contains Q′ . 5. Every cube Q′ ∈ Dk is a union of exactly 2n children cubes Q′′ ∈ Dk+1 with m(Q′ ) = 2n m(Q′′ ). 6. If Q′ ∈ Dk , then l(Q′ ) = 2−k l(Q) and m(Q′ ) = 2−nk m(Q).

Figure 4.3: Nestedness property.

4.8 Calderón–Zygmund decomposition Applying Lebesgue differentiation theorem, we may give a decomposition of ℝn , called Calderón–Zygmund decomposition, which is extremely useful in harmonic analysis.

4.8 Calderón–Zygmund decomposition

| 161

Theorem 4.16 (Calderón–Zygmund decomposition of ℝn ). Suppose that f is a nonnegative integrable function on ℝn . Then for any fixed λ > 0, there exists a sequence {Qj }j∈ℕ of disjoint dyadic cubes such that (1) f (x) ≤ λ for a. e. x ∉ ⋃∞ i=1 Qj ; 1 ‖f ‖ ; (2) m(⋃∞ Q ) ≤ 1 j i=1 λ 1 (3) λ < m(Q ∫ f (x) dx ≤ 2n λ. ) Q j

j

Proof. For f ∈ L1 (ℝn ), we may decompose ℝn into a net of equal cubes whose interiors are disjoint and such that for every Q in the net 1 ∫ f (x) dx ≤ λ. m(Q) Q

Let Q′ be any fixed cube in the net. We divide it into 2n equal cubes, and denote by Q′′ one of these cubes. Then there are the following two cases: Case 1. 1 ∫ f (x) dx > λ. m(Q′′ ) Q′′

Case 2. 1 ∫ f (x) dx ≤ λ. m(Q′′ ) Q′′

In Case 1, we have λ
1, if |f (x)| ≤ 1.

Proof. Let f ∈ L1 (ℝn ). By Theorem 4.10, m({x ∈ ℝn : ℳf (x) > 2λ}) ≤

C λ

∫

󵄨󵄨 󵄨 󵄨󵄨f (x)󵄨󵄨󵄨 dx.

(4.24)

{x:|f (x)|>λ}

For any compact set K ⊂ ℝn , it follows from (4.24) that ∞

∫ ℳf (x) dx = ∫ m({x ∈ K : ℳf (x) > λ}) dλ K

0 ∞

= 2 ∫ m({x ∈ K : ℳf (x) > 2λ}) dλ 0

1

∞

≤ 2{∫ m(K) + ∫ m({x ∈ K : ℳf (x) > λ}) dλ} 1

0 ∞

≤ 2{m(K) + ∫ 0

C λ

󵄨󵄨 󵄨 󵄨󵄨f (x)󵄨󵄨󵄨 dxdλ}

∫ {x:|f (x)|>λ}

|f (x)|

dλ 󵄨 󵄨 dx} = 2{m(K) + C ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 ∫ λ ℝn

1

󵄨 󵄨 󵄨 󵄨 = 2{m(K) + C ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 log+ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx} < ∞. ℝn

Thus we obtain the conclusion (4.23). Another important application of the Calderón–Zygmund decomposition on ℝn is that may be used to deduce the Calderón–Zygmund decomposition of a function. The later is a very important tool in harmonic analysis.

4.8 Calderón–Zygmund decomposition

| 163

Theorem 4.18 (Calderón–Zygmund decomposition for a function). Let f be a nonnegative integrable function on ℝn . Then for any fixed λ > 0, there exist a sequence {Qj }j∈ℕ of disjoint dyadic cubes and functions g, b such that: (1) f (x) = g(x) + b(x).

(The function g is called the “good” function since it will belong to the Lp (ℝn ) space. The

function b is called the “bad” function. It will turn out that it is not a bad function since it will have mean value zero by construction.) (2) g(x) ≤ 2λ for a. e. x ∈ ℝ;

(3) ‖g‖pp ≤ Cλp−1 ‖f ‖1 , 1 < p < ∞;

(4) b(x) = 0 a. e. x ∈ ℝn \ ⋃∞ j=1 Qj ;

(5) ∫Q b(x) dx = 0, j = 1, 2, . . . . j

Proof. Applying the Calderón–Zygmund decomposition on ℝn for f and λ > 0, we obtain a sequence {Qj }j∈ℕ of disjoint dyadic cubes such that f (x) ≤ λ

∞

for a. e. x ∈ ℝn \ ⋃ Qj , j=1

1 ∑ m(Qj ) ≤ ‖f ‖1 , λ j=1 ∞

and λ
0 and f ∈ Lp (ℝn ), q

C 󵄨 󵄨 m({x ∈ ℝn : 󵄨󵄨󵄨Tf (x)󵄨󵄨󵄨 > λ}) ≤ ( ‖f ‖p ) . λ

(4.25)

Operator T is said of type (p, q) if T is bounded from Lp (ℝn ) to Lq (ℝn ). That is, there exist a constant C > 0 such that for any f ∈ Lp (ℝn ), ‖Tf ‖q ≤ C‖f ‖p .

(4.26)

Although the maximal function of an integrable function is not integrable, it is much worse than an integrable function. As we shall show in the next theorem, it belongs to the weak-L1 (ℝn ) space.

4.10 Marcinkiewicz interpolation theorem

| 165

4.10 Marcinkiewicz interpolation theorem Notice that in the proof of Lp -boundedness of the Hardy–Littlewood maximal operator ℳ, we use the weak-L1 and L∞ -boundedness of ℳ. The following result considers a similar problem for a more general sublinear operator. Theorem 4.19 (Marcinkiewicz interpolation theorem). Let 1 ≤ p0 ≤ p1 < ∞. Suppose that T is a sublinear operator defined on Lp0 (ℝn ) + Lp1 (ℝn ), which is simultaneously of weak type (p0 , p0 ) and weak type (p1 , p1 ). Then T is of strong type (p, p) with p0 < p < p1 , that is, there exists a constant C > 0 such that ‖Tf ‖p ≤ C‖f ‖p . Proof. Let f ∈ Lp (ℝn ). For each λ > 0, we can write f (x) = fλ (x) + f λ (x), where fλ (x) = f (x)χ{x:|f (x)|≤λ} (x) and f λ (x) = f (x)χ{x:|f (x)|>λ} (x). Let us consider only the case p1 < ∞. Now, we want to show that f λ (x) ∈ Lp0 (ℝn ) and fλ (x) ∈ Lp1 (ℝn ). For that, let us estimate 󵄨 󵄨p ∫ 󵄨󵄨󵄨f λ (x)󵄨󵄨󵄨 0 dm = λp0 ℝn

∫ {x:|f (x)|>λ}

< λp0

∫ {x:|f (x)|>λ}

󵄨󵄨 f (x) 󵄨󵄨p0 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 dm 󵄨󵄨 λ 󵄨󵄨󵄨 󵄨󵄨 f (x) 󵄨󵄨p 󵄨󵄨 󵄨󵄨 󵄨 dm 󵄨󵄨 󵄨󵄨 λ 󵄨󵄨󵄨

󵄨 󵄨p < λp0 −p ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dm λ} ⊂ {x : 󵄨󵄨󵄨Tf λ (x)󵄨󵄨󵄨 > } ∪ {x : 󵄨󵄨󵄨Tf 2 (x)󵄨󵄨󵄨 > }. 2 2 2

166 | 4 Distribution functions On the other hand, due to Cavalieri’s principle and the weak type estimate for the operator T, we obtain 󵄨p 󵄨 ‖Tf ‖pp = ∫ 󵄨󵄨󵄨Tf (x)󵄨󵄨󵄨 dm ℝn

∞

󵄨 󵄨 = p ∫ λp−1 m({x : 󵄨󵄨󵄨Tf (x)󵄨󵄨󵄨 > λ}) dλ 0 ∞

󵄨 󵄨 λ ≤ p ∫ λp−1 m({x : 󵄨󵄨󵄨Tf λ (x)󵄨󵄨󵄨 > })dλ 2 2 0

∞

󵄨 λ 󵄨 λ + p ∫ λp−1 m({x : 󵄨󵄨󵄨Tf 2 (x)󵄨󵄨󵄨 > })dλ 2 0

∞

≤ p ∫ λp−1 0

C p0 󵄨p 󵄨 ( ∫ 󵄨󵄨󵄨f λ (x)󵄨󵄨󵄨 0 dm) dλ 2 λp0 ℝn

∞

+ p ∫ λp−1 0

C p1 󵄨 λ 󵄨p ( ∫ 󵄨󵄨󵄨f 2 (x)󵄨󵄨󵄨 1 dm) dλ p 1 λ ℝn

∞

= p ∫ λp−p0 −1 C p0 ( 0

󵄨󵄨 󵄨p0 󵄨󵄨f (x)󵄨󵄨󵄨 dm) dλ

∫ {x:|f (x)|≤ λ2 }

∞

+ p ∫ λp−p0 −1 C p0 ( 0

∫

󵄨󵄨 󵄨p0 󵄨󵄨f (x)󵄨󵄨󵄨 dm) dλ.

{x:|f (x)|> λ2 }

By the Fubini theorem, we get ‖Tf ‖p ≤ pC

2|f (x)|

󵄨 󵄨p ∫ (󵄨󵄨󵄨f (x)󵄨󵄨󵄨 0 ∫ λp−p0 −1 dλ) dm

p0

0

ℝn

+ pC =

p1

p−p0

2

∞

󵄨 󵄨p ∫ (󵄨󵄨󵄨f (x)󵄨󵄨󵄨 1 ∫ λp−p1 −1 dλ) dm ℝn p0

pC p − p0

2|f (x)|

󵄨 󵄨p 󵄨 󵄨p−p ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 0 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 0 dm ℝn

2p−p1 pC p1 󵄨󵄨 󵄨p 󵄨 󵄨p−p + ∫ 󵄨󵄨f (x)󵄨󵄨󵄨 1 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 1 dm, p1 − p ℝn

i. e., ‖Tf ‖pp ≤ p(

2p−p0 pC p0 2p−p1 pC p1 󵄨 󵄨p + ) ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dm. p − p0 p1 − p ℝn

4.11 The Cauchy principal value | 167

Then ‖Tf ‖p ≤ C(p0 , p, p1 )‖f ‖p , where 1

p 2p−p0 pC p0 2p−p1 pC p1 C(p0 , p, p1 ) = [p( + )] , p − p0 p1 − p

which ends the proof.

Exercises. 1. Show that if T is an operator of strong type (p, q) then T is of weak type (p, q). 2. (Kolmogorov inequality) If E ⊂ ℝn with m(E) < ∞, if T is of weak type (p, q) and 0 < r < q, then 1− r 󵄨 󵄨r ∫󵄨󵄨󵄨󵄨Tf (x)󵄨󵄨󵄨󵄨 dm ≤ C(q, r)[m(E)] q ‖f ‖rp . E

3.

Let E ⊂ ℝn with m(E) < ∞ be such that 1− r 󵄨 󵄨r ∫󵄨󵄨󵄨󵄨Tf (x)󵄨󵄨󵄨󵄨 dm ≤ C(p, r)[m(E)] q ‖f ‖rp , E

with 0 < r < p. Prove that q

C 󵄨 󵄨 m({x : 󵄨󵄨󵄨󵄨Tf (x)󵄨󵄨󵄨󵄨 > λ}) ≤ ( ‖f ‖p ) . λ 4.

Let T be a sublinear operator defined on L1 + L∞ such that T is of weak type (1, 1) and satisfies ‖Tf ‖∞ ≤ A‖f ‖∞ , where A is positive constant. Prove that if f ∈ L1 + L∞ and λ > 0 then C 󵄨 󵄨 m({x : 󵄨󵄨󵄨󵄨T (x)󵄨󵄨󵄨󵄨 > λ}) ≤ λ

󵄨 󵄨 ∫ m({x : 󵄨󵄨󵄨󵄨f (x)󵄨󵄨󵄨󵄨 > t}) dt.

[ Aλ ,∞)

4.11 The Cauchy principal value In what follows, we shall review the notation of the Cauchy principal value needed in the definition of the Hilbert transform. Let us consider a real-valued function f and its

168 | 4 Distribution functions integral over an interval [a, b], b

∫ f (x) dx. a

Suppose that for some x0 ∈ [a, b], f is unbounded, i. e., 󵄨 󵄨 lim sup󵄨󵄨󵄨f (x)󵄨󵄨󵄨 = ∞.

x→x0

Then one usually considers the integral as x0 −ϵ1

b

b

∫ f (x) dx = lim ∫ f (x) dx + lim ∫ f (x) dx, ϵ1 →0

a

ϵ2 →0

a

x0 +ϵ2

where the limits are taken independently of each other. These limits may, however, not exist. Another possibility is to look at a symmetric limit, called the Cauchy principal value of the integral, x0 −ϵ

b

a

x0 +ϵ

b

lim [ ∫ f (x) dx + ∫ f (x) dx] = P. V. ∫ f (x) dx,

ϵ→0

a

which may exist even if the individual limits do not. This is due to the fact that using a single parameter in the limit allows cancellations between the integrals. The Cauchy principal value of an improper integral ∞

∫ f (x) dx −∞

is defined in a similar fashion as a

∞

P. V. ∫ f (x) dx = lim ∫ f (x) dx. a→∞

−∞

−a

It should be noted that if even both the Cauchy principal value and the asymmetric limit of an integral exist they do not need to yield the same result (see [1]). The Cauchy principal value is a useful tool enabling one to extract finite and meaningful quantities of an otherwise ill-defined expression. Now, let us consider, for example, the integral 6

∫ 0

dx , 5−x

4.11 The Cauchy principal value | 169

where the integrand has a singularity at x = 5. The standard asymmetric limiting procedure results in 6

∫ 0

5−ϵ1

6

dx dx dx = lim ∫ + lim ∫ 5 − x ϵ1 →0 5 − x ϵ2 → 5−x 5+ϵ2

0

ϵ 1 = − lim [ln( 1 )] − lim [ln( )], ϵ1 →0 ϵ2 →0 5 ϵ2 an expression formally of the form ∞ − ∞, and hence not defined. The Cauchy principal value of the integral gives, on the other hand, the neat result 6

5−ϵ

6

0

0

5+ϵ

dx dx dx P. V. ∫ = lim ( ∫ + ∫ ) 5 − x ϵ→0 5−x 5−x ϵ 1 = − lim [ln( ) + ln( )] ϵ→0 5 ϵ 1 = − lim ln( ) ϵ→0 5 = ln 5.

The symmetry of the limit is indeed important for the consistency of the results: if the upper limit of the first integral in the above calculation had been 5 − 2ϵ while the other integral still had 5 + ϵ as its lower limit, the result would have been ln(2/5), which has really no meaning at all. For further reading on the Cauchy integral, see [3, 10, 14, 19, 24].

5 Three-fold approach to the Hilbert transform 5.1 Derivation of the Hilbert transform on ℝ Let f ∈ Lp (ℝ), 1 ≤ p < ∞ and consider the integral F(z) =

1 f (t) dt, ∫ 2πi t − z

(5.1)

ℝ

where z = x + i is a complex number whose imaginary part is 1. Decomposing into its real and imaginary parts, we can write F(z) =

1 i(t−z)

∞

1 f (t) dt ∫ 2π 1 − i(x − t) −∞ ∞

=

1 f (t)(1 + i(x − t)) dt ∫ 2π (x − t)2 + 1 −∞ ∞

=

∞

(x − t) 1 f (t) i dt + ∫ f (t) dt ∫ 2 2 π((x − t)2 + 1) π((x − t)2 + 1) −∞

1 i = (f ∗ P)(x) + (f ∗ Q)(x), 2 2

−∞

where P(x) = π(x12 +1) is the Poisson kernel and Q(x) = Poisson conjugate kernel. In

x π(x 2 +1)

ℝ2+ = {x + iy : x, y ∈ ℝ, y > 0}, for P and Q, we define Py (x) =

1 x y P( ) = y y π(x2 + y2 )

Qy (x) =

1 x x Q( ) = . y y π(x2 + y2 )

and

Applying these expressions to (5.1), we get ∞

1 f (t) F(z) = dt ∫ 2πi t−z −∞

1 i = (f ∗ Py )(x) + (f ∗ Qy )(x). 2 2 https://doi.org/10.1515/9783110784091-005

is its conjugate, known as

172 | 5 Three-fold approach to the Hilbert transform Next, observe that for N > 0, the function N

f (t) 1 FN (z) = dt ∫ 2πi t−z −N

is an analytic function in ℝ2+ and, for z belonging to the upper halfplane, it goes uniformly to ∞

1 f (t) F(z) = dt ∫ 2πi t−z

as N → ∞.

−∞

And so F is an analytic function in ℝ2+ , which implies that u(x, y) = f ∗ Py (x) and v(x, y) = f ∗ Qy (x) are conjugate harmonic; u(x, y) and v(x, y) are known as the Poisson integral of f and the conjugate Poisson integral of f , respectively. By the previous result, we already known that the integral u(x, y) it is well defined. However, {Qy } is not an approximation of the identity. Indeed, ∞

∞

−∞

−∞ ∞

|x| 1 󵄨 󵄨 dx ∫ 2 ∫ 󵄨󵄨󵄨Qy (x)󵄨󵄨󵄨 dx = π x + y2 =

2 x dx ∫ π x2 + y2 0

b

=

2 x lim ∫ dx π b→∞ x 2 + y2 0

b 2 ln |x 2 + y2 | 󵄨󵄨󵄨󵄨 lim 󵄨󵄨 󵄨󵄨0 π b→∞ 2 2 󵄨󵄨 󵄨 󵄨 1 󵄨 b + y 󵄨󵄨 = lim ln󵄨󵄨 󵄨 → ∞. π b→∞ 󵄨󵄨󵄨 y2 󵄨󵄨󵄨

=

Definition 5.1. Given x0 ∈ ℝ and α > 0, the cone Γα (x0 ) in ℝ2+ with vertex (x0 , 0) and angle α is the region Γα (x0 ) = {(x, y) ∈ ℝ2+ : |x − x0 | < αy}, see Figure 5.1.

(5.2)

5.1 Derivation of the Hilbert transform on ℝ

| 173

Figure 5.1: Cone with vertex x0 and angle α.

Definition 5.2. A function U : ℝ2 → ℝ has a nontangential limit l at x0 ∈ ℝ if for α > 0, lim

(x,y)→(x0 ,0)

U(x, y) = l

inside the cone Γα (x0 ). Next, the following proposition is proved. Proposition 5.1. If f ∈ Lp (ℝ), 1 ≤ p < ∞, then the Poisson integral ∞

∫ f (t)Py (x − t) dx = U(x, y) −∞

has a nontangential limit equal to f for almost all x ∈ ℝ. Proof. Let x0 ∈ ℝ be fixed and for α > 0 take (x, y) ∈ Γα (x0 ). Then ∞

U(x, y) − f (x0 ) = ∫ (f (t) − f (x0 ))Py (x − t) dt, −∞

since ∫−∞ Py (x − t) dt = 1. For any x such that |x − x0 | < αy, we get ∞

|x0 | < αt + |x|. Hence x02 < α2 y2 + x2 + 2αt|x|, and so 0 ≤ (αy − x)2 = α2 y2 − 2αt|x| + x 2 , where 2αt|x| ≤ α2 y2 + x2 and hence x02 + y2 ≤ (1 + 2α2 )y2 + 2x2 .

174 | 5 Three-fold approach to the Hilbert transform Therefore x02 + y2 ≤ max{1 + 2α2 , 2}(y2 + 2x2 ). Also we obtain Py (x) =

y y 1 ≤ Cα = Cα Py (x0 ), 2 2 2 π x +y π(x0 + y2 )

where Cα = max{1 + 2α2 , 2}. Thus, if (x, y) ∈ Γα (x0 ), we have ∞

󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨U(x, y) − f (x0 )󵄨󵄨󵄨 ≤ ∫ 󵄨󵄨󵄨f (t) − f (x0 )󵄨󵄨󵄨Py (x − t) dt −∞

∞

󵄨 󵄨 ≤ Cα ∫ 󵄨󵄨󵄨f (t) − f (x0 )󵄨󵄨󵄨Py (x − t) dt −∞ ∞

󵄨 󵄨 ≤ Cα ∫ 󵄨󵄨󵄨f (t) − f (x0 )󵄨󵄨󵄨Py (x0 − t) dt. −∞

By the previous result, we known that ∞

󵄨 󵄨 lim ∫ 󵄨󵄨󵄨f (t) − f (x0 )󵄨󵄨󵄨Py (x0 − t) dt = 0,

y→0

−∞

which implies U(x, y) → f

a. e. as (x, y) → (x0 , 0)

inside Γα (x0 ). Theorem 5.1. For f ∈ Lp (ℝ), 1 ≤ p < ∞, the conjugate Poisson integral of f , that is, ∞

1 x−t v(x, y) = (f ∗ Qy )(x) = dt, ∫ f (t) π (x − t)2 + y2 −∞

goes to a finite limit as z → x nontangentially. Proof. The technique used to prove this result rests on analytic function theory. So we omit it. Imporant. The next theorem guarantees the existence of the Hilbert transform for a function f ∈ Lp (ℝ) with 1 ≤ p < ∞.

5.1 Derivation of the Hilbert transform on ℝ

| 175

Theorem 5.2. For f ∈ Lp (ℝ), 1 ≤ p < ∞, the Hilbert transform Hf exists and is finite a. e. Moreover, it is equal to the limit of the conjugate Poisson integral of f , that is, ∞

lim ∫ f (x − t)

y→0

−∞

f (x − t) t dt = lim ∫ dt. y→0 t t 2 + y2

(5.3)

{|t|>y}

Proof. For y > 0, we can write ∞

∫ f (x − t) −∞

t f (x − t) dt dt − ∫ t t 2 + y2 |t|>y

f (x − t) t f (x − t) = ∫ 2 dt + ∫ f (x − t) 2 dt − ∫ dt t t + y2 t + y2 |t|>y

|t|≤y

= ∫ f (x − t)( |t|>y

|t|>y

1 t t − ) dt + ∫ f (x − t) 2 dt t 2 + y2 t t + y2 |t|≤y

t y

t

1 1 1 y = ∫ f (x − t) ( t − ) dt + ∫ f (x − t) ( t ) dt y ( )2 + 1 t y ( )2 + 1 y

| yt |>y

y

| yt |≤y

∞

= ∫ f (x − t)ϕy (t) dt, −∞

where ϕy (t) = y1 ϕ( yt ) and 2

t

ϕ(t) = { t t+1

−

1 t

t 2 +1

if |t| > 1, if |t| ≤ 1.

Furthermore, note that, if |t| > 1, the function 1 |t| 1 − = t 2 + 1 |t| |t|(t 2 + 1) is decreasing and so 󵄨 󵄨 sup 󵄨󵄨󵄨ϕ(t)󵄨󵄨󵄨 =

|t|≥|x|

1 |x|(x2 + 1)

if |x| > 1.

On the other hand, if |t| ≤ 1, we have 󵄨󵄨 󵄨 󵄨󵄨ϕ(t)󵄨󵄨󵄨 =

1 |t| ≤ , x2 + 1 2

y

176 | 5 Three-fold approach to the Hilbert transform so sup|t|≥|x| |ϕ(t)| =

1 2

if |x| ≤ 1. In summary, 1 2 󵄨 { 󵄨 ψ(x) = sup 󵄨󵄨󵄨ϕ(t)󵄨󵄨󵄨 = { |x|(1+x ) 1 |t|≥|x| {2

if |x| > 1, if |x| ≤ 1.

Observe that ϕ is odd, thus ∞

∫ ϕ(t) dt = 0

and

ϕ ∈ L1 (ℝ).

−∞

We shall show that ψ ∈ L1 (ℝ). In fact, ∞

∫ ψ(x) dx = ∫ ψ(x) dx + ∫ ψ(x) dx −∞

|x|>1 −1

−∞

=∫ 1

∞

1

1

−1

dx dx 1 +∫ + ∫ dx x(x2 + 1) x(x2 + 1) 2

= ∫ ∞

|x|≤1

1

dx 1 + ∫ dx. 2 x(x + 1) 2 −1

Let us calculate the first integral: ∞

∞

1

1

∞

dx x dx −∫ 2 =∫ dx ∫ 2 x x(x + 1) x +1 1

b

b

1

1

x dx −∫ 2 = lim (∫ dx) b→∞ x x +1 ln(b2 + 1) + ln 2 ) b→∞ 2 b ln 2 = lim (ln[ + ]) √b2 + 1 b→∞ 2 ln 2 = . 2 = lim (ln(b) −

Finally, ∞

∞

1

−∞

−1

−1

dx 1 + ∫ dx = ln(2) + 1. ∫ ψ(x) dx = 2 ∫ x(x2 + 1) 2

5.2 The Hilbert transform

| 177

Note that ∫ℝ ϕ(y) dy = 0 (since ϕ is odd), so by Theorem 4.15, ∞

lim ∫ f (x − t)ϕy (t) dt = f (x) ∫ ϕ(y) dy = 0,

y→0

−∞

ℝ

and thus the proof of the theorem is now complete. Definition 5.3. The Hilbert transform of a measurable function f on ℝ is defined by the Cauchy principal value integral Hf (x) =

f (y) 1 P. V. ∫ dy π x−y ℝ

f (x − y) 1 dy. = P. V. ∫ π y ℝ

5.2 The Hilbert transform The Hilbert transform of a sufficiently well-behaved function f (x) is defined to be ∞

1 f (y) 1 Hf (x) = P. V. ∫ dy = lim+ π x−y ϵ→0 π −∞

= − lim+ ϵ→0

1 f (x − t) dt. ∫ π t

∫ |x−y|>ϵ

f (y) dy x−y (5.4)

|t|>ϵ

Now, we will check that Hf in (5.4) is well defined. In order to do that, let us take f belonging to the Schwartz class S = S(ℝ). Then Hf (x) = =

1 lim π ϵ→0+ 1 lim π ϵ→0+

1 = lim π ϵ→0+ =

1 lim π ϵ→0+

∫ |x−t|>ϵ

f (t) dt x−t

∫ ϵϵ

= Hϵ f (−x). Thus Hf (x) = lim Hϵ f (x) = lim Hϵ f (−x) = Hf (−x). ϵ→0

ϵ→0

Now, we study the convolution property of the Hilbert transform. Theorem 5.13. H(f ∗ g)(x) = (Hf ∗ g)(x) = (f ∗ Hg)(x). Proof. By Fubini theorem, on the one hand, we have H(f ∗ g)(x) =

∞

(f ∗ g)(y) 1 P. V. ∫ dy π x−y −∞ ∞

∞

1 1 = P. V. ∫ ( ∫ f (s)g(y − s) ds) dy π x−y −∞

∞

= ∫ f (s)( −∞ ∞

= ∫ f (s)( −∞ ∞

−∞ ∞

1 g(y − s) P. V. ∫ dy) ds π x−y −∞ ∞

1 g(y) P. V. ∫ dy) ds π x−s−y −∞

= ∫ f (s)Hg(x − s) ds −∞

= (f ∗ Hg)(x).

(5.22)

On the other hand, ∞

1 (f ∗ g)(y) H(f ∗ g)(x) = P. V. ∫ dy π x−y −∞ ∞

∞

1 1 = P. V. ∫ ( ∫ f (y − s)g(s) ds) dy π x−y −∞

∞

= ∫( −∞

−∞

∞

1 f (y − s) P. V. ∫ dy)g(s) ds π x−y −∞

5.8 Properties | 215

∞

= ∫( −∞ ∞

∞

1 f (y) P. V. ∫ dy)g(s) ds π x−s−y −∞

= ∫ Hf (x − s)g(s) ds = (Hf ∗ g)(x). −∞

Finally, combining (5.22) and (5.23), we have H(f ∗ g)(x) = (Hf ∗ g)(x) = (f ∗ Hg)(x).

(5.23)

6 Hilbert transform in L2 (ℝ) We treat the Hilbert transform as an operator to perform functional analysis technique, among other things, and prove that this operator is a one-to-one, bounded, and unitary operator in L2 (ℝ); moreover, we prove that the operator H is onto and so has an inverse, namely H −1 = −H, and precisely its adjoint H ∗ is given by H ∗ = H −1 = −H. We end this section by calculating the exact Lp -norm of H(χE ).

6.1 Hilbert transform of a product The Hilbert transform of a product of two functions is now considered. The following result is due to Bedrosian (see [2]) and is referred to as the product theorem for Hilbert transforms, or Bedrosian’s theorem. Let f and g be two functions belonging to L2 (ℝ). Suppose that the Fourier transform of f (x), denoted by ℱ (ξ ), vanishes for |ξ | > a with a > 0, and the Fourier transform of g(x), denoted by 𝒢 (ξ ), vanishes for |ξ | < a, then H{f (x)g(x)} = f (x)Hg(x). The Hilbert transform of the product is given in terms of the Fourier transforms ℱ and 𝒢 . The Fubini theorem allows us to write ∞

∞

H{f (x)g(x)} = H[ ∫ ℱ (ξ )eiξx dξ ∫ 𝒢 (t)eitx dt] −∞

−∞ ∞

∞

−∞ ∞

−∞ ∞

−∞

−∞

= ∫ ℱ (ξ ) dξ ∫ 𝒢 (t)H{ei(t+ξ )x } dt = ∫ ℱ (ξ ) dξ ∫ 𝒢 (t){−i sgn(ξ + t)ei(t+ξ )x } dt. The last integral simplifies upon noting that the support of ℱ (ξ ) is [−a, a] and the support of 𝒢 (ξ ) is (−∞, −a] ∪ [a, ∞), hence H{f (x)g(x)} a

−a

∞

= −i ∫ ℱ (x)eiξx dξ [ ∫ 𝒢 (t)eixt sgn(t + ξ ) dt + ∫ 𝒢 (t)eixt sgn(t + ξ ) dt] −a

a

−∞

a

−a+ξ

−a

−∞

∞

= −i ∫ ℱ dξ [ ∫ 𝒢 (ν − ξ )eixν sgn(ν) dν + ∫ 𝒢 (ν − ξ )eixν sgn(ν) dν] https://doi.org/10.1515/9783110784091-006

a+ξ

218 | 6 Hilbert transform in L2 (ℝ) a

−a

∞

= −i ∫ F(ξ )eixξ dξ [− ∫ 𝒢 (y)eixy dy + ∫ 𝒢 (y)eixy dy] −a

a

−∞ −a

∞

= f (x)[ ∫ 𝒢 (y)(−i sgn y)eixy dy + ∫ 𝒢 (y)(−i sgn y)eixy dy] a

−∞ −a

∞

= f (x)[ ∫ 𝒢 (y)H{ixy} dy + ∫ 𝒢 (y)H{ixy} dy] a ∞

−∞ −a

= f (x)[H[ ∫ 𝒢 (y)eixy dy + ∫ 𝒢 (y)eixy dy]] a

−∞ ∞

= f (x)H( ∫ 𝒢 (y)eixy ) = f (x)Hg(x). −∞

And so H{f (x)g(x)} = f (x)Hg(x).

(6.1)

Remark 21. The restriction on the class of functions for which equation (6.1) holds is rather severe, but there is an important practical application. The following examples will illustrate the approach. Let us consider the function sinc(x) = sin(πx) . The Fourier transform of sinc(ax), where a is a real number, is given πx by ℱ {sinc(ax)} =

1 {sgn(πa + x) + sgn(πa − x)}. 2a

The sinc function therefore satisfies the condition that its Fourier transform has a support of a finite interval around the origin, specifically (−πa, πa), from which it follows that H(sin(ax) sinc(bx)) = H(sin(ax)) sinc(bx) = − cos(ax) sinc(bx) for 0 < bπ < a. In a similar fashion, H(cos(ax) sinc(bx)) = H(cos(ax)) sinc(bx) = sin(ax) sinc(bx) for 0 < bπ < a. Theorem 6.1 (Hilbert inversion theorem). Given f ∈ L2 (ℝ), ∞

1 Hf (y) f (x) = − P. V. ∫ dy. π x−y −∞

6.1 Hilbert transform of a product | 219

Proof. Using the Fourier inversion theorem, we can rewrite f as follows: ∞

1 f (x) = ∫ ̂f (ξ )e−ixξ dξ √2π −∞ ∞

∞

1 1 = ∫( ∫ f (μ)eiξμ dμ)e−ixξ dξ √2π √2π =

−∞ ∞

∞

−∞

−∞

−∞

1 ∫ ( ∫ {f (μ) cos (ξμ) + if (μ) sin (ξμ)}dμ)(cos (xξ ) − i sin (xξ ))dξ . √2π

If we consider the case of real-valued functions, we have ∞

∞

1 f (x) = ∫ ( ∫ f (μ) cos (ξμ)dμ) cos (xξ )dξ 2π −∞ −∞ ∞ ∞

+

1 ∫ ( ∫ f (μ) sin (ξμ)dμ) sin (xξ )dξ . 2π −∞ −∞

Thus, we can finally write ∞

f (x) = ∫ {a(ξ ) cos (xξ ) + b(ξ ) sin (xξ )} dξ , 0

where ∞

1 a(ξ ) = ∫ f (μ) cos (ξμ)dμ π −∞

and ∞

1 b(ξ ) = ∫ f (μ) sin (ξμ)dμ. π −∞

Now, we define ∞

u(x, y) = ∫ {a(ξ ) cos (xξ ) + b(ξ ) sin (xξ )}e−yξ dξ 0

and observe that, for y ≥ 0, u(x, y) is well defined and its the real part of ∞

ϕ(z) = ∫ {a(ξ ) − ib(ξ )}eizξ dξ , 0

220 | 6 Hilbert transform in L2 (ℝ) where z = x + iy. The imaginary part of ϕ(z) is then ∞

v(x, y) = − ∫ {b(ξ ) cos(xξ ) − a(ξ ) sin(xξ )} dξ . 0

Writing g(x) = −v(x, 0), we have ∞

g(x) = ∫ {b(ξ ) cos(xξ ) − a(ξ ) sin(xξ )} dξ 0 ∞ ∞

=

1 ∫ ∫ {f (μ)(sin(μξ ) cos(μξ ) − cos(μξ ) sin(μξ ))} dμdξ π 0 −∞ ∞ ∞

=

1 ∫ ∫ {f (μ) sin[(μ − x)ξ ]} dμdξ . π 0 −∞

This can be written as λ ∞

1 ∫ ∫ {f (μ) sin[(μ − x)ξ ]} dμdξ . λ→∞ π

g(x) = lim

0 −∞

Applying Fubini theorem and solving for ξ gives λ

∞

− cos(μ − x)ξ 1 )] dμ ∫ [f (μ)( λ→∞ π μ−x 0

g(x) = lim = lim

λ→∞

−∞ ∞

1 − cos(μ − x)λ 1 ] dμ ∫ f (μ)[ π μ−x −∞ ∞

∞

0

0

1 − cos(−μ0 − x)λ 1 − cos(μ − x)λ 1 (∫ f (μ) dμ + ∫ f (μ) dμ0 ). λ→∞ π μ−x −μ0 − x

= lim

Now using the change of coordinates μ − x = t and μ0 + x = t0 , we obtain g(x) =

∞

∞

1 − cos(λt0 ) 1 − cos(λt) 1 (∫ f (x + t) dt − ∫ f (x − t0 ) dt0 ) π t t0 0

= lim

λ→∞

0

∞

1 1 − cos(λt) (f (x + t) − f (x − t))dt ∫ π t 0

∞

∞

0

0

1 (f (x + t) − f (x − t) 1 cos(λt) dt − lim ∫ (f (x + t) − f (x − t))dt. = ∫ λ→∞ π π t t

6.1 Hilbert transform of a product | 221

Now, let us take care of ∞

1 cos(λt) (f (x + t) − f (x − t))dt. lim ∫ λ→∞ π t 0

In order to compute this limit, setting F(t, x) = f (x + t) − f (x − t) and applying the Riemann–Lebesgue lemma, we have ∞

∞

0

0

cos(λt) cos(u) u lim ∫ F(t, x) dt = lim ∫ F( , x)dt = 0, λ→∞ λ→∞ t u λ since F(0, x) = 0. Thus ∞

1 f (x + t) − f (x − t) dt. g(x) = ∫ π t 0

In a similar manner, f (x) = −

∞

1 g(x + t) − g(x − t) dt. ∫ π t 0

Now, note that ∞

1 g(x + t) − g(x − t) f (x) = − ∫ dt π t 0

∞

=−

1 g(x + t) − g(x − t) lim ∫ dt π ϵ→0 t

=−

1 g(x + t) g(x − t) lim [ ∫ dt − ∫ dt] π ϵ→0 t t

=−

1 g(y) g(y) lim [ ∫ dy + ∫ dy] π ϵ→0 y−x y−x

ϵ

∞

= P. V. ∫

−∞

∞

∞

ϵ ∞

ϵ x−ϵ

x+ϵ

−∞

g(y) dy. y−x

In a similar way, we have ∞

g(x) = P. V. ∫

−∞

f (z) dz. z−y

222 | 6 Hilbert transform in L2 (ℝ) Finally, ∞

g(y) 1 f (x) = − P. V. ∫ dy π y−x −∞ ∞

∞

1 f (z) 1 (P.V. ∫ dz) dy = − P. V. ∫ π y−x z−y −∞ ∞

=−

−∞

1 Hf (y) P. V. ∫ dy. π y−x −∞

The following result will have an important role in the development of the next result. Theorem 6.2. The Fourier transform of

1 x

is given by

1 x

ℱ ( ) = −iπ sgn(ξ ),

where sgn is the signum function −1, x < 0, { { { sgn(x) = {0, x = 0, { { x > 0. {1, Proof. Taking the Fourier transform of

1 x

for x ≠ 0, we have

∞

1 −2πixξ 1 dx ℱ ( ) = P. V. ∫ e x x −∞ ∞

= P. V. ∫

−∞ ∞

= P. V. ∫

−∞

cos (2πxξ ) − i sin (2πxξ ) dx x ∞

cos (2πxξ ) sin (2πxξ ) dx − i P. V. ∫ dx. x x −∞

The first integral vanishes as a principal value because the integrand is an odd function. The latter one has no issue with divergence since the integrand approaches the value 1 near the origin and so the principal value plays no role. Instead, it has to be considered separately for different values of ξ . For ξ = 0, the integral vanishes trivially. Assuming that ξ < 0 and making a change of variable t = 2πxξ , the integral

6.1 Hilbert transform of a product | 223

becomes ∞

∫ −∞

∞

sin (2πxξ ) sin t dx = − ∫ dt = −π. x t ∞

For ξ > 0, the same calculation yields ∞

∫ −∞

sin (2πxξ ) = π. x

Combining all these results, we can conclude that 1 x

ℱ ( ) = −iπ sgn(ξ ).

Sometimes the Hilbert transform is defined by the following result. Theorem 6.3. Let f ∈ L2 (ℝ) and consider Hf (x) = (f ∗ g)(x), where g(x) =

1 . πx

Then ̂ (ξ ) = −i sgn(ξ )̂f (ξ ). Hf

Proof. Taking the Fourier transform of (6.2), we have ̂ (ξ ) = f? Hf ∗ g(ξ ) = ̂f (ξ )ĝ (ξ ) 1 = (−i π sgn(ξ ))̂f (ξ ) π = −i sgn(ξ )̂f (ξ ). Theorem 6.4. Let f ∈ L2 (ℝ). Then f (x) and Hf (x) are orthogonal functions, that is, ⟨f , Hf ⟩ = 0. Proof. Using the Parseval identity, we get ∞

⟨f , Hf ⟩ = ∫ f (x)Hf (x) dx −∞ ∞

? = ∫ ̂f (ξ )Hf (ξ ) dξ −∞

(6.2)

224 | 6 Hilbert transform in L2 (ℝ) ∞

= ∫ ̂f (ξ )[−i sgn ξ ̂f (ξ )] dξ −∞ ∞

= i ∫ sgn ξ ̂f (ξ )̂f (ξ ) dξ −∞ ∞

󵄨2 󵄨 = i ∫ sgn ξ 󵄨󵄨󵄨̂f (ξ )󵄨󵄨󵄨 dξ . −∞

Since |̂f (ξ )|2 = |̂f (−ξ )|2 , the function sgn |̂f (ξ )|2 is odd as sgn ξ is odd and |̂f (ξ )|2 is even. Further, with a symmetric interval of integration, the integral is zero and we have the desired result, ⟨f , Hf ⟩ = 0, which means that f (x) and Hf (x) are orthogonal functions, completing the proof. The Hilbert transform can be studied by replacing the usual convolution with (πx) by convolution with (πx)−1 |x|−iγ , where γ is a constant. It might be expected that this would be an easier kernel to study, by virtue of the less singular nature of this choice and with the expectation that the properties of the Hilbert transform could be recovered in the limit γ → 0. This approach turns out not to be so simple, see [16]. −1

6.2 Differentiation property of the Hilbert transform in L2 Theorem 6.5. Let f ∈ L2 (ℝ) such that f is differentiable with f ′ ∈ L2 (ℝ) and Hf be differentiable with (Hf )′ ∈ L2 (ℝ). Then H(

d df (x) )= H(f (x)). dx dx

Proof. By the differentiation property of the Fourier transform and from Theorem 6.3, we get by some simple algebra ′ (ξ ) = (−i sign ξ )f? ′ (ξ ) ? Hf

= (−i sign ξ )(2πiξ )̂f (ξ ) = 2πiξ (−i sign ξ )̂f (ξ ) ̂ )(ξ ) = 2πiξ H(f ′ (f )(ξ ). = H?

Taking the inverse Fourier transform, we obtain H( dfdx(x) ) = complete.

d H(f (x)), and the proof is dx

6.3 Hilbert transform as an operator in L2

Example 15. Let f (x) =

x , 1+x 2

| 225

then

H(

1 d x 1 ) = H(− )) ( 2 dx 1 + x 2 1 + x2 1 d 1 =− (H( )) 2 dx 1 + x2 1 d x ) =− ( 2 dx 1 + x 2 x2 − 1 = . 2(1 + x2 )2

6.3 Hilbert transform as an operator in L2 Now it is time to do some functional analysis. Definition 6.1. If ℋ is a Hilbert space and T ∈ B(ℋ), then T is unitary if TT ∗ = T ∗ T = I. Here T ∗ stands for the adjoint of T. – –

In what follows, we consider the Hilbert transform H as an operator. Note that ̂‖ ‖Hf ‖L2 = ‖Hf L2 󵄩󵄩 󵄩 = 󵄩󵄩−i sign(⋅)̂f (⋅)󵄩󵄩󵄩L 2 ̂ = ‖f ‖ = ‖f ‖ . L2

L2

Thus H ∈ B(ℋ). Also observe that if Hf1 = Hf2 , then 󵄩 󵄩 0 = ‖Hf1 − Hf2 ‖L2 = 󵄩󵄩󵄩H(f1 − f2 )󵄩󵄩󵄩L = ‖f2 − f1 ‖L2 . 2 Hence f1 = f2 , and so H is one-to-one. Now, we like to prove that (H ∗ )∗ = H. Indeed, ⟨y, (H ∗ ) x⟩ = ⟨H ∗ y, x⟩ ∗

= ⟨x, H ∗ y⟩ = ⟨Hx, y⟩

= ⟨y, Hx⟩. Hence (H ∗ )∗ x = Hx for all x ∈ L2 , so (H ∗ ) = H. ∗

(6.3)

226 | 6 Hilbert transform in L2 (ℝ) Now, let us consider ⟨H ∗ (Hf ), f ⟩ = ⟨Hf , Hf ⟩

= ‖Hf ‖2L2 = ‖f ‖2L2

= ⟨f , f ⟩

= ⟨If , f ⟩. Therefore H ∗ (H)f = I, ∀f ∈ L2 , and hence H ∗ (H) = I.

(6.4)

Next, by (6.3) and (6.4), we have (H ∗ ) (H ∗ ) = I, ∗

and then H(H ∗ ) = I.

(6.5)

By (6.3) and (6.5), we obtain H(H ∗ ) = H ∗ (H) = I.

(6.6)

In this way we have shown that H is a unitary operator. By (6.6), H ∗ ((H ∗ ) ) = (H ∗ ) (H ∗ ) = I. ∗

∗

(6.7)

Thus H ∗ is unitary, and ‖f ‖2L2 = ⟨f , f ⟩ = ⟨(H ∗ )H ∗ f , f ⟩ = ⟨H ∗ f , H ∗ f ⟩ 󵄩 󵄩2 = 󵄩󵄩󵄩H ∗ f 󵄩󵄩󵄩L , 2

so ‖H ∗ f ‖L2 = ‖f ‖L2 , which means that H ∗ is unitary and so H ∗ is one-to-one. From this fact observe that L2 = {0}⊥ = ker(H ∗ ) = rang(H) ⊥

= rang(H) = H(L2 ),

hence H(L2 ) = L2 , which means that H is onto. Therefore its inverse exists, name it S = H −1 , then SH = HS = I, also H ∗ = H ∗ I = H ∗ (HS) = (H ∗ H)S

6.4 The Parseval-type form

| 227

= IS = S. Hence H ∗ = H −1 . Now, H(Hf ) (̂ ξ ) = −(i sgn ξ )(Hf ) (̂ ξ ) = (i sgn ξ )2 ̂f (ξ )

= −̂f (ξ ). Taking the inverse Fourier transform in this way, we have proved that H(H) = −I.

(6.8)

Next, H(H) = −I implies H −1 (H(H)) = −H −1 (I). Hence H = −H −1 , i. e., H −1 = −H. And so H ∗ = −H.

(6.9)

Now that we know the fact that H −1 = −H, we will provide an easy, but no less interesting, proof of Theorem 6.1. Theorem 6.6. Let f ∈ L2 (ℝ). Then f (x) = −

∞

Hf (y) 1 P. V. ∫ dy. π x−y −∞

Proof. Let f ∈ L2 (ℝ). Then f (x) can be written as f (x) = H −1 (Hf )(x) = −H(Hf )(x) ∞

1 Hf (y) = − P. V. ∫ dy. π x−y −∞

6.4 The Parseval-type form Theorem 6.7. Let f , g ∈ L2 (ℝ), then ∞

∞

−∞

−∞

∫ Hf (x)g(x) dx = − ∫ f (x)H(g(x)) dx.

(6.10)

228 | 6 Hilbert transform in L2 (ℝ) Proof. Let f , g ∈ L2 (ℝ). Then ∞

∫ Hf (x)g(x) dx = ⟨Hf , g⟩L2 −∞

= ⟨Hf , −H(H(g))⟩L

2

= ⟨Hf , H(H(g))⟩L

2

= −⟨H ∗ (H(f )), H(g)⟩L

2

= −⟨H −1 (H(f )), H(g)⟩L

2

= −⟨f , H(g)⟩L

2

∞

= − ∫ f (x)Hg(x) dx. −∞

And hence ∞

∞

−∞

−∞

∫ Hf (x)g(x) dx = − ∫ f (x)Hg(x) dx

for f , g ∈ L2 (ℝ).

Theorem 6.8. If f , g ∈ L2 (ℂ), then ∞

∞

−∞

−∞

∫ Hf (x)g(x) dx = − ∫ f (x)Hg(x) dx. Proof. Let us begin by considering f and g belonging to L2 (ℂ). Then ∞

∫ Hf (x)g(x) dx = ⟨Hf (x), g(x)⟩L

2

−∞

̂ (ξ ), ĝ (ξ )⟩ = ⟨Hf L

2

∞

̂ (ξ )ĝ (ξ ) dξ = ∫ Hf −∞ ∞

= ∫ (−i sign(ξ )̂f (ξ ))ĝ (ξ ) dξ −∞ ∞

= ∫ ̂f (ξ )[−i sign(ξ )ĝ (ξ )] dξ −∞ ∞

= − ∫ ̂f (ξ )[i sign(ξ )ĝ (ξ )] dξ −∞

(6.11)

6.4 The Parseval-type form

∞

= − ∫ ̂f (ξ )[−i sign(ξ )ĝ (ξ )] dξ

| 229

(i sign ξ = −i sign ξ )

−∞ ∞

̂ ) dξ = − ∫ ̂f (ξ )Hg(ξ −∞

̂ )⟩ = −⟨̂f (ξ ), Hg(ξ L = −⟨f (x), Hg(x)⟩L

2

2

∞

= − ∫ f (x)Hg(x) dx, −∞

i. e., ∞

∞

−∞

−∞

∫ Hf (x)g(x) dx = − ∫ f (x)Hg(x) dx. Note that, in what follows, we will exhibit another way to prove results (6.10) and (6.11). In order to do that, observe that ̂ , HG⟩ ̂ L ⟨Hf , HG⟩L2 = ⟨Hf 2 ̂ )⟩ = ⟨−i sign(ξ )̂f (ξ ), −i sign(ξ )G(ξ L

2

̂ )⟩ = −i sign(ξ )(−i sign(ξ ))⟨̂f (ξ ), G(ξ L

2

̂ )⟩ = −i sign(ξ )(i sign(ξ ))⟨̂f (ξ ), G(ξ L

2

2 ̂ )⟩ = −i2 (sign(ξ )) ⟨̂f (ξ ), G(ξ L

2

̂ )⟩ = ⟨̂f (ξ ), G(ξ L

2

= ⟨f , G⟩L2 .

Now, taking G = Hg for f and g real, then gives ⟨Hf , H(Hg)⟩L = ⟨f , Hg⟩L2 , 2

⟨Hf , −g⟩L2 = ⟨f , Hg⟩L2 , ∞

∞

−∞

−∞

− ∫ Hf g = ∫ f Hg, i. e., ∞

∞

−∞

−∞

∫ Hf (x) g(x) dx = − ∫ f (x) Hg(x) dx.

230 | 6 Hilbert transform in L2 (ℝ) And for complex functions f and g, we get ∞

∞

−∞

−∞

∫ Hf (x) g(x) dx = − ∫ f (x) Hg(x) dx, thus ending the proof.

6.5 A theorem due to E. M. Stein and G. Weiss Let us start by stating and demonstrating the following useful lemma. Lemma 6.1. Let P(x) = xn + an xn−1 + ⋅ ⋅ ⋅ + a2 x + a1 be polynomial of degree n. Let r1 , r2 , . . . , rn be the roots of P(x) = 0, then n

∑ rk = −an .

k=1

Proof. By induction for n = 2, if the polynomial x 2 + a2 x + a1 = 0 has two roots r1 and r2 such that (x − r1 )(x − r2 ) = 0 then x2 − r1 x − r2 x + r1 r2 = 0, and so x2 − (r1 + r2 )x + r1 r2 = 0. Hence r1 + r2 = −a2 . Now, for n = 3, the polynomial x3 + a3 x 2 + a2 x + a1 = 0 has three roots, named r1 , r2 , and r3 , such that 2

(x − r1 )(x − r2 )(x − r3 ) = 0,

(x − (r1 + r2 )x + r1 r2 )(x − r3 ) = 0,

x3 − (r+ r2 )x2 + (r1 r2 )x − r3 x 2 + (r1 + r2 )r3 x − r1 r2 r3 = 0,

x3 − (r1 + r2 + r3 )x2 + (r1 r2 + r1 r3 + r2 r3 )x − r1 r2 r3 = 0,

so from the latter equality we have r1 + r2 + r3 = −a3 .

6.5 A theorem due to E. M. Stein and G. Weiss |

231

Next, suppose that for xn + an xn−1 + ⋅ ⋅ ⋅ + a2 x + a1 = 0, n

∑ rk = −an

k=1

holds.

Now, the polynomial xn+1 + an+1 xn + ⋅ ⋅ ⋅ + a2 x + a1 = 0 has n + 1 roots named r1 , r2 , . . . , rn+1 such that (x − r1 )(x − r2 ) ⋅ ⋅ ⋅ (x − rn )(x − rn+1 ) = 0, n

[xn − ( ∑ rk )x n−1 + ⋅ ⋅ ⋅](x − rn+1 ) = 0, k=1

n

n

k=1

k=1

[xn+1 − ( ∑ rk )x n + ⋅ ⋅ ⋅] + [−rn+1 xn + ( ∑ rk )rn+1 xn−1 + ⋅ ⋅ ⋅] = 0, n+1

xn+1 − ( ∑ rk )x n + ⋅ ⋅ ⋅ = 0. k=1

Therefore n+1

∑ rk = −an+1 .

k=1

Lemma 6.2. Suppose that ai , bi (i = 1, 2, 3, . . . , n) are real numbers satisfying a1 < b1 < a2 < b2 < ⋅ ⋅ ⋅ < an < bn and let g be the rational function n

g(x) = ∏ k=1

x − ak x − bk

(6.12)

(x ∈ ℝ).

If Δ is a real number such that Δ ≠ 1, then the equation g(x) = |Δ| has n distinct roots r1 , r2 , . . . , rn which satisfy n

n

n

k=1

k=1

k=1

∑ bk = ∑ rk + (1 − Δ)−1 ∑ (bk − ak ).

(6.13)

Furthermore, if Δ > 1, then n

(Δ − 1)m({g > Δ}) = (Δ + 1)m({g < −Δ}) = ∑ (bk − ak ). k=1

(6.14)

Proof. Since g has a simple pole at each bk (k = 1, 2, 3, . . . , n) and n

lim g(x) = lim ∏

|x|→∞

|x|→∞

k=1

n x − ak x − ak = ∏ lim = 1, x − bk k=1 |x|→∞ x − bk

(6.15)

232 | 6 Hilbert transform in L2 (ℝ) there are exactly n distinct solutions, say, r1 , r2 , . . . , rn , to the equation g(x) = |Δ| (Δ ≠ 1). Then n

∏ k=1

For ∏nk=1

x−ak x−bk

x − ak =Δ x − bk

n

and

∏ k=1

x − ak = −Δ. x − bk

= Δ, then n

n

k=1

k=1

∏(x − ak ) = Δ ∏(x − bk ) and so n

n

k=1

k=1

∏(x − ak ) − Δ ∏(x − bk ) = 0, where n

n

n

k=0

k=1

k=1

P(x) = ∑ pk xk = ∏(x − ak ) − Δ ∏(x − bk ) = 0.

(6.16)

Then n

n

k=1

k=1

∏(x − ak ) − Δ ∏(x − bk ) = 0 implies n

n

[xn − ( ∑ ak )x n+1 + ⋅ ⋅ ⋅] − Δ[xn − ( ∑ bk )xn+1 + ⋅ ⋅ ⋅] = 0, k=1

k=1

n

n

(1 − Δ)xn + (− ∑ ak + Δ ∑ bk )xn−1 + ⋅ ⋅ ⋅ = 0, xn +

k=1 n − ∑k=1 ak

k=1 Δ ∑nk=1 bk n−1

+ (1 − Δ)

x

+ ⋅ ⋅ ⋅ = 0.

Since r1 , r2 , . . . , rn are the roots of the polynomial P(x) = 0, then, by Lemma 6.1, we have n

∑ rk = −

k=1

− ∑nk=1 ak + Δ ∑nk=1 bk . (1 − Δ)

6.5 A theorem due to E. M. Stein and G. Weiss |

And then n

n

k=1

k=1 n

∑ rk = (1 − Δ)−1 ∑ ak −

n Δ ∑ bk (1 − Δ) k=1

= (1 − Δ)−1 ∑ ak + ( k=1 n

1−Δ−1 n )∑b (1 − Δ) k=1 k

= (1 − Δ)−1 ∑ ak + (1 − k=1 n

n 1 ) ∑ bk (1 − Δ) k=1 n

= (1 − Δ)−1 ∑ ak + (1 − (1 − Δ)−1 ) ∑ bk k=1 n

n

k=1

k=1

k=1

k=1

n

= (1 − Δ)−1 ∑ ak + ∑ bk − (1 − Δ)−1 ∑ bk . Hence n

n

n

n

k=1

k=1 n

k=1 n

k=1

k=1

k=1

∑ bk = ∑ rk + (1 − Δ)−1 ∑ bk − (1 − Δ)−1 ∑ ak = ∑ rk + (1 − Δ)−1 ∑ (bk − ak ).

If Δ > 1, then {g > Δ} = ⋃nk=1 (bk , rk ) (see Figure 6.1), and so n

n

k=1

k=1

m({g > Δ}) = m( ⋃ (bk , rk )) = ∑ (rk − bk ). Since n

n

n

k=1

k=1

k=1

∑ bk = ∑ rk + (1 − Δ)−1 ∑ (bk − ak ),

we then obtain n

n

k=1

k=1

∑ (bk − rk ) = (1 − Δ)−1 ∑ (bk − ak ),

and so n

n

k=1

k=1 n

− ∑ (rk − bk ) = (1 − Δ)−1 ∑ (bk − ak ), −m({g > Δ}) = (1 − Δ)−1 ∑ (bk − ak ), k=1

233

234 | 6 Hilbert transform in L2 (ℝ)

Figure 6.1: Graph of a rational function ∏nk=1

x−ak x−bk

= g(x). n

(Δ − 1)m({g > Δ}) = ∑ (bk − ak ). k=1

Moreover, if −Δ < −1, then {g < −Δ} = ⋃nk=1 (rk , bk ) (see Figure 6.1). Hence n

n

k=1

k=1

m({g < −Δ}) = m( ⋃ (rk , bk )) = ∑ (bk − rk ). Now, we have n

n

k=1

k=1

n

∑ bk = ∑ rk + (Δ + 1)−1 ∑ (bk − ak ),

n

n

k=1

∑ (bk − rk ) = (Δ + 1)−1 ∑ (bk − ak ),

k=1

k=1 n

m({g < −Δ}) = (Δ + 1)−1 ∑ (bk − ak ), n

k=1

(Δ + 1)m({g < −Δ}) = ∑ (bk − ak ). k=1

And thus we conclude that n

(Δ − 1)m({g > Δ}) = (Δ + 1)m({g < −Δ}) = ∑ (bk − ak ). k=1

(6.17)

235

6.5 A theorem due to E. M. Stein and G. Weiss |

In the following result we can observe that the distribution function of HχE depends only on the measure of E and not on the way in which E happens to be distributed over the real line. Theorem 6.9 (Stein–Weiss, [27]). Let E be the union of finitely many disjoint intervals, each of finite length. Then DHχE (λ) =

2m(E) , sinh (πλ)

λ > 0,

(6.18)

where DHχE (λ) = m({|HχE | > λ}) which is known as the distribution function of HχE . Proof. We may express E in the form n

E = ⋃(aj , bj ) j=1

(6.19)

such that a1 < b1 < a2 < b2 < ⋅ ⋅ ⋅ < an < bn . We already known that bi

1 n dy HχE (x) = [∑ ∫ ] π i=1 x − y ai

󵄨󵄨 x − a 󵄨󵄨 1 n 󵄨 i 󵄨󵄨 [∑ log󵄨󵄨󵄨 󵄨] 󵄨󵄨 x − bi 󵄨󵄨󵄨 π i=1 󵄨 󵄨󵄨 n 󵄨󵄨 x − ai 󵄨󵄨󵄨󵄨 1 󵄨 = log󵄨󵄨∏ 󵄨󵄨󵄨. 󵄨󵄨 π 󵄨 i=1 x − bi 󵄨󵄨 =

Fix λ > 0 and F = {|HχE | > λ}, then m(F) = DHχE (λ). Since 󵄨󵄨 n 󵄨 x − ai 󵄨󵄨󵄨󵄨 1 󵄨󵄨󵄨 HχE (x) = [log󵄨󵄨∏ 󵄨󵄨], 󵄨󵄨 󵄨 π 󵄨 i=1 x − bi 󵄨󵄨 we have e If we set g(x) = ∏ni=1

x−ai , x−bi

πHχE (x)

󵄨 󵄨󵄨 n 󵄨󵄨 x − ai 󵄨󵄨󵄨󵄨 󵄨 = 󵄨󵄨∏ 󵄨󵄨. 󵄨 󵄨󵄨 󵄨 i=1 x − bi 󵄨󵄨

then F may be decomposed into the disjoint union

F = {|g| > eπλ } ∪ {|g| < e−πλ } = F1 ∪ F2 .

(6.20)

236 | 6 Hilbert transform in L2 (ℝ) Now, applying Lemma 6.2 to g, we obtain m(F1 ) = m({|g| > eπλ })

= m({g > eπλ }) + m({g < −eπλ })

=

∑ni=1 (bi − ai )

+

∑ni=1 (bi − ai )

eπλ − 1 eπλ + 1 m(E) m(E) + = πλ e − 1 eπλ + 1 2eπλ m(E) m(E) . = 2πλ = sinh (πλ) e −1 Next, for F2 , m(F2 ) = m({|g| < e−πλ })

= m({g > −e−πλ }) + m({g < e−πλ }) =

∑ni=1 (bi − ai )

+

∑ni=1 (bi − ai )

−e−πλ − 1 −e−πλ + 1 m(E) m(E) = −πλ + −e − 1 −e−πλ + 1 −2eπλ m(E) = −2πλ e −1 −m(E) m(E) = = . sinh (−πλ) sinh (πλ) Finally, m({|HχE | > λ}) = m(F1 ) + m(F2 ) =

2m(E) , sinh (πλ)

(6.21)

i. e., m({|HχE | > λ}) =

2m(E) . sinh (πλ)

A short proof of the Stein–Weiss theorem using complex-variable methods can be found Calderon [6] and Garnett [12]. Also, an additional discussion can be found in Sagher and Xiang [25]. The Stein–Weiss theorem has been discussed for the ergodic Hilbert transform by Ephremidze [11]. Let us recall the Gamma and Riemann zeta functions defined by ∞

Γ(p) = ∫ up−1 e−u du 0

and

1 p n n=1 ∞

ζ (p) = ∑

for p > 1.

6.5 A theorem due to E. M. Stein and G. Weiss |

237

Theorem 6.10. Let E ⊂ ℝ be the union of finitely many disjoint intervals, each of finite length. Then 4p m(E) 󵄨p 󵄨 (1 − 2−p )ζ (p)Γ(p) ∫󵄨󵄨󵄨HχE (x)󵄨󵄨󵄨 dx = πp

for p > 1.

ℝ

Proof. For p > 1, by the Cavalieri principle (Theorem 4.3), we have ∞

󵄨 󵄨p ∫󵄨󵄨󵄨HχE (x)󵄨󵄨󵄨 dx = p ∫ λp−1 DHχE (λ) dλ 0 ∞

ℝ

= p ∫ λp−1 0

2m(E) dλ sinh (πλ) ∞

= 2p m(E) ∫ w = eπλ ⇒ λ =

dw ln w ⇒ dλ = π πw

0 ∞

= 2p m(E) ∫

0 ∞

= 4p m(E) ∫ w=

dr 1 ⇒ dw = − 2 r r

λp−1 dλ sinh (πλ) λp−1 eπλ dλ e2πλ − 1 ( lnπw )p−1 w

(w2 − 1)πw

dw

1 ∞

=

4p m(E) (ln w)p−1 dw ∫ πp w2 − 1

=

(ln 1r )p−1 dr 4p m(E) ] [− ∫ 1 πp r2 2 − 1

1

1

=

0

r

1

4p m(E) (− ln r)p−1 dr ∫ πp 1 − r2 0

1

∞ 4p m(E) = ∫(− ln r)p−1 ∑ r 2n dr p π n=0 0

− ln r = u ⇒ dr = −e−u du =

1

4p m(E) ∞ ∑ ∫(− ln r)p−1 r 2n dr πp n=0 0

∞ 0

=

4p m(E) ∑ ∫ −(u)p−1 e−2nu e−u du πp n=0 ∞ ∞ ∞

=

4p m(E) ∑ ∫ up−1 e−(2n+1)u du πp n=0 0

238 | 6 Hilbert transform in L2 (ℝ) p−1

∞

=

4p m(E) ∞ s ) ∑ ∫( πp 2n +1 n=0

e−s

0

ds 2n + 1

∞

4p m(E) 1 = ∑ ∫ sp−1 e−s ds p πp (2n + 1) n=0 ∞

0

∞ 1 4p m(E) = Γ(p) ∑ p p π n=1 (2n + 1)

= = =

∞ ∞ 1 1 4p m(E) Γ(p)[ − ] ∑ ∑ p p πp n (2n) n=1 n=1 ∞ 4p m(E) 1 −p Γ(p)(1 − 2 ) ∑ p πp n n=1

4p m(E) (1 − 2−p )ζ (p)Γ(p), πp

as we wished to prove. Exercises. 1. For f ∈ S(ℝ), show that: ∞ (a) Hf (0) = lima→∞ π1 ∫−∞ f (y) cos(ay)−1 dy; y 1 ∞ ∫ (f (y) π 0 f (y) 1 limξ →0 π ∫|y|>ξ −y

(b) Hf (0) = lima→∞

2.

(c)

Hf (0) =

− f (−y)) cos(ay)−1 dy; y dy.

Suppose f has a well-defined Hilbert transform. Then: (I) If f is even, show that ∞

Hf (x) =

f (y) 2x P. V. ∫ 2 dy π x − y2

H(f ) =

2 yf (y) P. V. ∫ 2 dy. π x − y2

0

(II) If f is odd, show that

1

∞

0

3.

Let f (y) =

4.

Let ϕ ∈ S(ℝ). Show that Hϕ ∈ L1 (ℝ) if and only if ∫ℝ ϕ(x) dx = 0.

5. 6.

√1−y 2

χ(−1,1) (y). Show that Hf (0) = 0 for |x| < 1 and Hf (x) =

Let f (y) = √1 − y 2 χ(−1,1) (y). Prove that Hf (x) = x for |x| < 1. Let f , g ∈ S(ℝ) and show that

∫ Hf (x)g(x) dx = − ∫ f (x)Hg(x) dx. ℝ

ℝ

1 √x 2 −1

for |x| > 1.

6.5 A theorem due to E. M. Stein and G. Weiss |

7.

Let g(t) = χ(0,x) (t) and show that x

∫ H(t) dt = 0

8.

239

󵄨󵄨 󵄨 󵄨󵄨 1 x 󵄨󵄨󵄨 ∫ f (t) ln󵄨󵄨󵄨1 − 󵄨󵄨󵄨 dt. 󵄨󵄨 π t 󵄨󵄨󵄨 󵄨 ℝ

Suppose f is nonnegative, even, and supported on |x| ≥ δ, 0 < δ < 1, for y ∈ [2δ, ∞). Prove that ∫ |x−y|≥δ

f (x) dx ≤ C(Mf (√⋅))y 2 , (x − y)2

where C is a constant depending on δ, but not on f , and M denotes the Hardy–Littlewood maximal operator b

} { 1 ∫ |g| : y ∈ (a, b)}. (Mg)(y) = sup{ b−a a } { 9.

Let f be a function in C 1 (ℝ) with compact support and b > 0. Prove that Hb f (x) = lim

ϵ→0+

∫ ℝ\[−ϵ,bϵ]

f (x − y) 1 dy = f (x) ln( ). y b

10. Prove that, for any f ∈ L2 ([−π, π]), ? Hf (n) = −i sgn(n)̂f (n) for all n ∈ ℤ. 11. If f (x) =

a 2

+ ∑∞ n=1 (an cos(nx) + bn sin(nx)). Prove that ∞

Hf (x) = ∑ (an sin(nx) − bn cos(nx)). n=1

π

12. If f ∈ L2 ([−π, π]) is such that ∫−π f (x) dx = 0, prove that ‖Hf ‖L2 (−π,π) = ‖f ‖L2 (−π,π) . π

13. If ∫−π f (x) dx = 0, prove that H(Hf ) = −f . 14. Let 0 < r < 1. Prove that ix ix (a) H( e ix ) = −i e ix ; 1−re 1−re −ix e−ix ) = i e −ix ; −ix 1−re 1−re 2 sin x H( 1−2r1−r ) = 1−2r2rcos ; cos x+r 2 x+r 2

(b) H(

(c) (d) HDN (x) = sin(Nx) + (1 − cos(Nx)) cot( 2x ), where DN (x) is the Dirichlet kernel. (e) Prove that x N+1 HKN (x) = cot( ) − cot[( )x]KN (x), 2 2 where KN (x) is the kernel defined in (1.8).

240 | 6 Hilbert transform in L2 (ℝ)

15. In what follows, we will write ∫ f (x) dx = T

Consider f (x) = (a) Prove that

a0 2

2π

1 ∫ f (x) dx. 2π 0

= ∑∞ n=1 (an cos(nx) + bn sin(nx)).

an = 2 ∫ f (x) cos(nx) dx T

and bn = 2 ∫ f (x) sin(nx) dx. T

(b) Prove that ∞

Hf (0) = −2 ∑ ∫ f (x) sin(nx) dx. n=1

(c)

T

Prove that Hf (0) = lim ∫ N→∞

cos[(N + 21 )x] − cos( 2x ) sin( 2x )

T

f (x) dx.

(d) Prove that Hf (0) = lim ∫ N→∞

T

cos[(N + 21 )x] − cos( 2x ) f (x) − f (−x) [ ] dx. 2 sin( 2x )

(e) Prove that Hf (0) = − ∫ T

(f)

Verify Hf (0) = limϵ→0+

1 2π

∫ϵ 0, if ξ = η > 0, if ξ > η > 0, if ξ > η = 0, if ξ = 0,

if ξ < η = 0, if η < ξ < 0, if ξ = η < 0, if η < ξ < 0

η { 1 if ξ > 1 or ξ = 0, { { { = {0 if ξη = 1 or η = 0, { { { η −1 if ξ < 1 { = m(η)m(ξ − η), https://doi.org/10.1515/9783110784091-007

244 | 7 Embedding and strong Lp boundedness for the Hilbert transform it follows from the convolution theorem that 2

̂(η)φ ̂(ξ − η)m(η)m(ξ − η) dη ℱ [φ + 2H(φH(φ))](ξ ) = ∫ φ ̂ Hφ(ξ ̂ − η) dη = ∫ Hφ(η) ̂ ∗ Hφ)(ξ ̂ ) = (Hφ ?2 (ξ ). = (Hφ) Applying the inverse Fourier transform on both sides, we obtain that (Hφ)2 = φ2 + 2H(φH(φ)). We are now ready to establish the Lp -boundedness of the Hilbert transform. Theorem 7.1 (M. Riesz). If 1 < p < ∞ and f ∈ Lp , then ‖Hf ‖p ≤ Cp ‖f ‖p for some constant Cp > 0. Proof. We will prove by induction that ‖Hf ‖2k ≤ (2k − 1)‖f ‖2k for f ∈ L2k . Indeed, 󵄨 󵄨2 󵄨 󵄨2 ‖f ⋅ Hf ‖22 = ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 󵄨󵄨󵄨Hf (x)󵄨󵄨󵄨 dx ℝ

1

1

2 2 󵄨 󵄨2 2 󵄨 󵄨2 2 ≤ (∫(󵄨󵄨󵄨f (x)󵄨󵄨󵄨 ) dx) (∫(󵄨󵄨󵄨Hf (x)󵄨󵄨󵄨 ) dx)

ℝ

ℝ

2 22

2 2

2 󵄨 󵄨22 󵄨 󵄨22 = (∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx) (∫󵄨󵄨󵄨Hf (x)󵄨󵄨󵄨 dx)

ℝ

= ‖f ‖222 ‖Hf ‖222 ,

ℝ

where ‖Hf ‖222 ≤ ‖f ‖222 + 2‖f ‖22 ‖Hf ‖22 . Next, if ‖Hf ‖22 ≤ ‖f ‖22 , we have the inequality sought, that is, ‖Hf ‖222 ≤ ‖f ‖222 + 2‖f ‖222 = (22 − 1)‖f ‖222 . Conversely, if ‖f ‖22 ≤ ‖Hf ‖22 , we have ‖Hf ‖222 ≤ ‖f ‖22 ‖Hf ‖22 + 2‖f ‖22 ‖Hf ‖22 = 3‖f ‖22 ‖Hf ‖22 , and so ‖Hf ‖22 ≤ (22 − 1)‖f ‖22 .

7 Embedding and strong Lp boundedness for the Hilbert transform

| 245

Suppose that the inequality is satisfied for k = n and let us see that is valid for k = n+1. For this, we do a procedure analogous to the case k = 2. That is, ‖Hf ‖22n+1

2 n+1

2 󵄨2n+1 󵄨 = (∫󵄨󵄨󵄨Hf (x)󵄨󵄨󵄨 dx)

ℝ

1 n

2 󵄨2 2n 󵄨 = (∫(󵄨󵄨󵄨Hf (x)󵄨󵄨󵄨 ) dx)

ℝ

󵄩 󵄩 = 󵄩󵄩󵄩(Hf )2 󵄩󵄩󵄩2n 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩f 2 󵄩󵄩󵄩2n + 2󵄩󵄩󵄩H(f ⋅ Hf )󵄩󵄩󵄩2n 󵄩 󵄩 ≤ 󵄩󵄩󵄩f 2 󵄩󵄩󵄩2n+1 + 2(22 − 1)‖f ⋅ Hf ‖2n 󵄩 󵄩 ≤ 󵄩󵄩󵄩f 2 󵄩󵄩󵄩2n+1 + 2(22 − 1)‖f ‖2n+1 ‖Hf ‖2n+1 . And hence ‖Hf ‖2n+1 ≤ (2n+1 − 1)‖f ‖2n+1 . Now for k ≥ 0, the latter inequality implies that ‖Hf ‖2k ≤ (2k − 1)‖f ‖2k

for all f ∈ L2k

and ‖Hf ‖2k+1 ≤ (2k+1 − 1)‖f ‖2k+1

for all f ∈ L2k+1 .

Then by the Riez–Thorin theorem, we have 1−t

‖Hf ‖p ≤ (2k − 1) where k

1 p

=

1−t t + 2k+1 2k k+1

t

(2k − 1) ‖f ‖p

and 0 < t < 1, that is, p =

2k+1 2−t

and 0 < t < 1. As for these values of

t, 2 ≤ p ≤ 2 , H has to be bounded in Lp with 2 ≤ p ≤ 2k+1 . As the result is valid for all k ≥ 1, we conclude that it is bounded in Lp for p ≥ 2. We might check all of this, in fact, if 0 < t < 1 󳨐⇒ 2 − t < 2 󳨐⇒ 2 − t < 2 k

and and

󳨐⇒ 2 (2 − t) < 2

k+1

k

for all f ∈ Lp ,

1 λ}) ≤ ‖f ‖L1 . λ Proof. From the Calderón–Zygmund decomposition at height λ, we obtain a sequence of disjoint intervals {Ij } such that (I) f (x) ≤ λ for almost all x ∉ ⋃j Ij ; (II) m(⋃j Ij ) ≤ λ1 ‖f ‖L1 ; (III) λ < m(I1 ) ∫I f (x) dx ≤ 2λ. j

j

Now, we decompose f into a sum f = g + b, where {f (x) g(x) = { 1 ∫ f (x) dx { m(Ij ) Ij

if x ∉ ⋃j Ij ,

if x ∈ Ij

and b(x) = ∑ bj (x) = ∑(f (x) − j

j

1 ∫ f (x) dx)χIj (x). m(Ij ) Ij

We then observe that 0 ≤ g(x) ≤ λ if x ∉ ⋃j Ij , while 0 ≤ g(x) =

1 ∫ f (x) dx ≤ 2λ m(Ij )

if x ∈ Ij .

Ij

Hence g(x) ≤ 2λ, and so g ∈ L1 . Since it is obviously integrable on (⋃j Ij )c and on each Ij (that are disjoint), ∫ g(x) dx = ∫ f (x) dx < ∞. Ij

Ij

Notice that ∫ g(x) dx = ∫ f (x) dx < ∞. ℝ

ℝ

7 Embedding and strong Lp boundedness for the Hilbert transform

| 247

On the other hand, 1 1 ∫ b(x) dx = ∑ ∫(f (x) − m(Ij ) ∫ f (x) dx)χIj (x) dx m(Ij ) m(Ij ) j Ij

Ij

=∑ j

Ij

1 1 ∫ f (x) dx − ∫ f (x) dx m(Ij ) m(Ij ) Ij

Ij

= 0. We now proceed as follows. Since 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨󵄨Hf (x)󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨Hg(x)󵄨󵄨󵄨 + 󵄨󵄨󵄨Hb(x)󵄨󵄨󵄨, we have 󵄨 󵄨 󵄨 󵄨 λ m({x ∈ ℝ : 󵄨󵄨󵄨Hf (x)󵄨󵄨󵄨 > λ}) ≤ m({x ∈ ℝ : 󵄨󵄨󵄨Hg(x)󵄨󵄨󵄨 > }) 2 󵄨 󵄨 λ + m({x ∈ ℝ : 󵄨󵄨󵄨Hb(x)󵄨󵄨󵄨 > }). 2 Now, for the “good” function g, we use the L2 -boundedness of the Hilbert transform to obtain 󵄨 󵄨 λ m({x ∈ ℝ : 󵄨󵄨󵄨Hg(x)󵄨󵄨󵄨 > }) = 2

dx

∫ {x:|Hg(x)|> λ2 }

≤

∫ {x:|Hg(x)|> λ2 }

≤

|Hg(x)|2 dx (λ/2)2

4 󵄨󵄨 4 󵄨 󵄨2 󵄨2 ∫󵄨Hg(x)󵄨󵄨󵄨 dx = 2 ∫󵄨󵄨󵄨g(x)󵄨󵄨󵄨 dx λ2 󵄨 λ ℝ

ℝ

8 8 ≤ ∫ g(x) dx = ∫ f (x) dx. λ λ ℝ

(7.1)

ℝ

Next, let 2Ij be the interval with the same center cj as Ij but twice the length, and let Ω = ⋃j Ij and Ω∗ = ⋃j 2Ij . Then m(Ω∗ ) ≤ 2m(Ω) ≤

2 ‖f ‖ . λ L1

248 | 7 Embedding and strong Lp boundedness for the Hilbert transform Then we estimate 󵄨 λ 󵄨 λ 󵄨 󵄨 m({x ∈ ℝ : 󵄨󵄨󵄨Hb(x)󵄨󵄨󵄨 > }) = m({x ∈ Ω∗ : 󵄨󵄨󵄨Hb(x)󵄨󵄨󵄨 > }) 2 2

c 󵄨 󵄨 λ + m({x ∈ (Ω∗ ) : 󵄨󵄨󵄨Hb(x)󵄨󵄨󵄨 > }) 2

c 󵄨 󵄨 λ ≤ m(Ω∗ ) + m({x ∈ (Ω∗ ) : 󵄨󵄨󵄨Hb(x)󵄨󵄨󵄨 > }) 2 2 2 󵄨 󵄨 ≤ ‖f ‖L1 + ∫ 󵄨󵄨󵄨Hb(x)󵄨󵄨󵄨 dx. λ λ ∗

(7.2)

ℝ\Ω

Since |Hb(x)| ≤ ∑j |bj (x)| a. e., it will suffice to prove that 󵄨 󵄨 ∑ ∫ 󵄨󵄨󵄨Hbj (x)󵄨󵄨󵄨 dx ≤ C‖f ‖L1 . j ℝ\2I k

Now, bj ∈ S(ℝ), nonetheless, for x ∉ Ω∗ , we have the formula Hbj (x) = ∫ Ij

bj (y)

x−y

dy.

We recall that bj has mean zero and denote by Cj the center of the interval Ij . Notice that y ∈ Ij implies |y − Cj | ≤ m(Ij )/2 and that x ∉ Ij and y ∈ Ij imply |x − y| > |x − Cj |/2. Then 󵄨󵄨 bj (y) 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 dy󵄨󵄨 dx ∫ 󵄨󵄨󵄨Hbj (x)󵄨󵄨󵄨 dx = ∫ 󵄨󵄨󵄨∫ 󵄨󵄨 x − y 󵄨󵄨󵄨 ℝ\2Ij Ij

ℝ\2Ij

󵄨󵄨 󵄨󵄨 1 1 󵄨 󵄨 − )dy󵄨󵄨󵄨 dx = ∫ 󵄨󵄨󵄨∫ bj (y)( 󵄨󵄨 󵄨󵄨 x − y x − Cj ℝ\2Ij Ij

󵄨 󵄨 ≤ ∫󵄨󵄨󵄨bj (x)󵄨󵄨󵄨 ∫ Ij

ℝ\2Ij

󵄨 󵄨 ≤ ∫󵄨󵄨󵄨bj (x)󵄨󵄨󵄨 ∫ Ij

ℝ\2Ij

|y − Cj |

|x − y||x − Cj | m(Ij )

|x − Cj |2

dx dy

󵄨 󵄨 = 2 ∫󵄨󵄨󵄨bj (y)󵄨󵄨󵄨 dy, Ij

since the inner integral equals ∞

2m(Ij ) ∫

m(Ij )

dx dy

1 dr = 2m(Ij ) = 2. 2 m(Ij ) r

7 Embedding and strong Lp boundedness for the Hilbert transform

| 249

Therefore 󵄨 󵄨 󵄨 󵄨 ∑ ∫ 󵄨󵄨󵄨Hbj (x)󵄨󵄨󵄨 dx ≤ 2 ∑ ∫󵄨󵄨󵄨bj (x)󵄨󵄨󵄨 dy ≤ 4‖f ‖L1 . j I j

j ℝ\2I j

(7.3)

Thus by (7.1), (7.2), and (7.3), 20 󵄨 󵄨 m({x ∈ ℝ : 󵄨󵄨󵄨Hf (x)󵄨󵄨󵄨 > λ}) ≤ ‖f ‖L1 . λ This concludes the proof. Theorem 7.3. Let f ∈ Lp with 1 < p < 2. Then ‖Hf ‖pL ≤ C‖f ‖pL . p

p

Proof. The first step is to break up a function f ∈ Lp for 1 < p < 2 as f = g + b, where g and b are in L1 and L2 , respectively. One way of achieving this decomposition is by truncating |f | on its range, that is, by setting g = fχ{|f |>λ} and b = fχ{|f |≤λ} for some λ > 0 (typically, b is the tail part of f and g is the singular part of f ). Since 1 − p < 0 and 2 − p > 0, we see that 󵄨 󵄨p 󵄨 󵄨1−p 󵄨 󵄨 ∫󵄨󵄨󵄨g(x)󵄨󵄨󵄨 dx = ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx ℝ

|f |>λ

󵄨 󵄨p ≤ λ1−p ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx ≤λ

1−p

|f |>λ

‖f ‖Lp

and 󵄨 󵄨2 󵄨 󵄨p 󵄨 󵄨2−p ∫󵄨󵄨󵄨b(x)󵄨󵄨󵄨 dx = ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx ℝ

|f |≤λ

󵄨 󵄨p ≤ λ2−p ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx λ

∞

|f |

∞

0

∞

0

0

0

|f |

∞

|f |

∞

∞

0

0

|f |

|f |≤λ

󵄨 󵄨 󵄨 󵄨2 = B1 ∫ λp−2 (∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx)dλ + B2 ∫ λp−3 ( ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx)dλ = B1 ∫ λ 0

p−2

󵄨 󵄨2 󵄨 󵄨 (∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx)dλ + B2 ∫ λp−3 ( ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx)dλ

∞

|f |

0

0

∞

∞

0

|f |

󵄨 󵄨 󵄨 󵄨2 = B1 ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨(∫ λp−2 dλ)dx + B2 ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 ( ∫ λp−3 dλ)dx ∞

=(

B B1 󵄨 󵄨p + 2 ) ∫ 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx. p−1 p−2 0

Finally, ‖Hf ‖Lp ≤ C‖f ‖Lp , where C =

B1 p−1

+

B2 , p−2

and the proof is complete.

Observation. If f ∈ Lp with 1 < p < ∞ and {fn }n∈ℕ is a sequence in C0∞ (ℝ) such that limn→∞ ‖fn − f ‖ = 0, then, by the M. Riesz theorem, we have ‖Hfn − Hf ‖p ≤ Cp ‖fn − f ‖p → 0 as n → ∞, for some constant Cp > 0. Using the M. Riesz theorem and Minkowski inequality, one has ‖Hf − Hϵ f ‖p ≤ ‖Hf − Hfn ‖p + ‖Hfn − Hϵ fn ‖p + ‖Hϵ fn − Hϵ f ‖p ≤ Cp ‖fn − f ‖p + ‖Hϵ fn − Hfn ‖p + C‖fn − f ‖p = (Cp + C)‖fn − f ‖p + ‖Hϵ fn − Hfn ‖p .

7 Embedding and strong Lp boundedness for the Hilbert transform

| 251

If {fn }n∈ℕ is a sequence with compact support, then there exists a bounded A ⊂ ℝ such that if x ∈ Ac then Hϵ fn (x) = 0 and Hfn (x) = 0, and so 󵄨p 󵄨p 󵄨 󵄨 ‖Hϵ fn − Hfn ‖pp = ∫󵄨󵄨󵄨Hϵ fn (x) − Hfn (x)󵄨󵄨󵄨 dx = ∫󵄨󵄨󵄨Hϵ fn (x) − Hfn (x)󵄨󵄨󵄨 dx A

ℝ

󵄨p 󵄨 = ∫󵄨󵄨󵄨Hϵ fn (x) − Hfn (x)󵄨󵄨󵄨 χA (x) dx. ℝ

Moreover, |Hϵ fn (x)| and |Hfn (x)| are bounded for x ∈ ℝ. Thus 󵄨󵄨 󵄨p p 󵄨󵄨Hϵ fn (x) − Hfn (x)󵄨󵄨󵄨 χA (x) ≤ C χA (x), where C p χA ∈ L1 (ℝ) for some constant C p > 0. By the dominated convergence theorem, one has 󵄨 󵄨p lim ‖Hϵ fn − Hfn ‖p = lim ∫󵄨󵄨󵄨Hϵ fn (x) − Hf (x)󵄨󵄨󵄨 χA (x) dx = 0.

ϵ→0

ϵ→0

ℝ

Therefore ‖Hf − Hϵ f ‖p → 0 if n → ∞ and ϵ → 0. This means that H is continuous. Theorem 7.4 (Inverse formula). If f ∈ Lp (ℝ), 1 < p < ∞, then H(Hf ) = −f

a. e.

Proof. Since H(Hf ) = −f a. e. in L2 (ℝ), it is in particular valid in C0∞ (ℝ). It is well known that C0∞ (ℝ) is dense in Lp (ℝ), so for f ∈ Lp (ℝ) there exists a sequence {fn }n∈ℕ in C0∞ (ℝ) such that limn→∞ fn = f . Then H(Hfn ) = −fn a. e. Hence H(Hf ) = H(H( lim fn )) = H( lim H(fn )) n→∞

n→∞

= lim H(Hfn ) = − lim fn = −f . n→∞

n→∞

Theorem 7.5. Let f , g ∈ Lp (ℝ), then ∫ℝ Hf (x)g(x) dx = − ∫ℝ f (x)Hg(x) dx. Proof. Let f ∈ Lp (ℝ) and g ∈ Lp (ℝ). Then there exist sequences {fn }n∈ℕ and {gn }n∈ℕ in C0∞ (ℝ) such that limn→∞ fn = f and limn→∞ gn = g. Thus lim H(fn ) = H( lim fn ) = H(f )

n→∞

n→∞

and lim H(gn ) = H( lim gn ) = H(g).

n→∞

n→∞

252 | 7 Embedding and strong Lp boundedness for the Hilbert transform Since C0∞ (ℝ) ⊂ S(ℝ) ⊂ L2 (ℝ), we have that {fn }n∈ℕ and {gn }n∈ℕ belong to L2 (ℝ). Then, by Theorem 6.7, we have ∫ H(f (x))g(x) dx = ∫ H( lim fn (x)) lim gn (x) dx n→∞

ℝ

n→∞

ℝ

= ∫ lim H(fn (x))gn (x) dx ℝ

= − ∫ lim fn (x)H(gn (x)) dx ℝ

= − ∫ lim fn (x)H( lim gn (x))dx n→∞

n→∞

ℝ

= − ∫ f (x)Hg(x) dx. ℝ

Theorem 7.6. Let f ∈ Lp (ℝ) with 2 < p < ∞. Then 󵄩󵄩 󵄩 󵄩󵄩H(f )󵄩󵄩󵄩L ≤ C‖f ‖Lp . p

Proof. Let f ∈ Lp (ℝ) with Theorem 7.3,

1 p

+

1 q

= 1. Since 2 < p < ∞, we get that 1 < q < 2. Hence by 󵄩󵄩 󵄩 󵄩󵄩H(g)󵄩󵄩󵄩Lq ≤ C‖g‖Lq

(7.4)

for g ∈ Lq (ℝ). By Theorems 7.5 and 4.2, as well as Hölder inequality and (7.4), we have 󵄨󵄨 󵄨󵄨 󵄩󵄩 󵄩 󵄨 󵄨 󵄩󵄩H(f )󵄩󵄩󵄩Lp = sup 󵄨󵄨󵄨∫ Hf (x)g(x) dx󵄨󵄨󵄨 󵄨󵄨 ‖g‖Lq =1󵄨󵄨 ℝ

󵄨󵄨 󵄨󵄨 󵄨 󵄨 = sup 󵄨󵄨󵄨− ∫ f (x)Hg(x) dx󵄨󵄨󵄨 󵄨 󵄨󵄨 ‖g‖Lq =1󵄨 ℝ

󵄩 󵄩 ≤ sup ‖f ‖Lp 󵄩󵄩󵄩H(g)󵄩󵄩󵄩L q ‖g‖Lq =1

≤ C‖f ‖Lp , which proof that H is bounded on Lp for 2 < p < ∞. Theorem 7.7. If f ∈ Lp (ℝ) for ϵ > 0, let us consider the function g(x, y) = If: (I)

f is derivable, in which case

f (x − y) χ{|y|>ϵ} (y). y

𝜕g (x, y) 𝜕x

=

d f (x−y) χ{|y|>ϵ} (y) dx y

exists;

7 Embedding and strong Lp boundedness for the Hilbert transform

(x, y)| ≤ h(y) for some h ∈ L1 (ℝ) for all x ∈ ℝ and all ϵ > 0; (II) | 𝜕g 𝜕x

| 253

(III) The integrals ∫−∞ g(x, y) dy and ∫ϵ g(x, y) dy are uniformly convergent for all ϵ > 0, ∞

−ϵ

then d df Hf (x) = H( )(x). dx dx Proof. By definition of the derivative, d Hf (x + h) − Hf (x) Hf (x) = lim h→0 dx h 1 1 f (x − y + h) − f (x − y) = lim lim dy ∫ h→0 ϵ→0 π y h |y|>ϵ

= lim

ϵ→0

1 g(x + h, y) − g(x, y) lim ∫ dy. π h→0 h ℝ

By the mean value theorem, there exists ξ ∈ (x, x + h) such that g(x + h, y) − g(x, y) 𝜕g = (ξ , y). h 𝜕x By condition (II), 󵄨󵄨 g(x + h, y) − g(x, y) 󵄨󵄨 󵄨󵄨 𝜕g 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 󵄨󵄨 = 󵄨󵄨 (ξ , y)󵄨󵄨󵄨 ≤ h(y), 󵄨󵄨 h 󵄨󵄨 󵄨󵄨 󵄨󵄨 𝜕x with h ∈ L1 (ℝ). Finally, by the Lebesgue dominated convergence theorem, 1 g(x + h, y) − g(x, y) d Hf (x) = lim ∫ lim dy ϵ→0 π h→0 dx h ℝ

1 𝜕g = lim ∫ (x, y) dy ϵ→0 π 𝜕x ℝ

1 1 df (x − y) dy = lim ∫ ϵ→0 π y dx |y|>ϵ

= H(

df )(x). dx

254 | 7 Embedding and strong Lp boundedness for the Hilbert transform

7.1 Strong Lp boundedness for the Hilbert transform on the circle Lemma 7.2. If − π2 ≤ θ ≤ π2 , 0 < γ < π2 , 1 < p < 2, Then | sin θ|p ≤ Ap (γ) cosp (θ) − Bp (γ) cos(pθ), with Ap (γ) =

tanp−1 tan((p−1)γ)

and Bp (γ) =

(7.5)

sinp−1 (γ) . sin((p−1)γ)

Proof. First note that for p = 2, (7.5) reduces to the well-known trigonometric equality sin2 θ = cos θ − cos(2θ), with A2 (γ) =

tan γ =1 tan γ

and B2 (γ) =

sin(γ) = 1. sin(γ)

Hence it suffices to restrict consideration to the case 1 < p < 2. The functions appearing in (7.5) are even, and hence the discussion may be restricted to the interval 0 < θ ≤ π2 . Consider the function h(θ) = with − π2 < p ≤ 2, 0 < γ < to θ, we have h′ (θ) =

π , 2

sinp (θ) − a cos(pθ) cosp (θ)

0 < p ≤ 1 (1 < p ≤ 2). Next, differentiating h with respect

[p sinp−1 (θ)(cos(θ) + ap sin(pθ)] cosp (θ) cos2p (θ) −

p(cosp−1 (θ))(− sin(θ))[sinp −a cos(pθ)] cos2p (θ)

taking common factor in the numerator cosp−1 (θ) then gives = = = =

[p sinp−1 (θ) cos(θ) + ap sin(pθ)] cosp (θ) + p sin(θ)[sinp (θ) − a cos(pθ)] cosp+1 (θ) p sinp−1 (θ)[cos2 (θ) + sin2 (θ)] + ap sin((p − 1)θ) cosp+1 (θ)

p sinp−1 (θ) + ap sin((p − 1)θ) cosp+1 (θ)

a sin((p − 1)θ) p sinp−1 (θ) [1 + ]. p+1 cos (θ) sinp−1 (θ)

Let g(θ) = 1 +

a sin((p − 1)θ) sinp−1 (θ)

.

7.1 Strong Lp boundedness for the Hilbert transform on the circle

| 255

If a ≠ 0 then g(θ) is strictly monotone on the interval 0 < θ ≤ π2 , since g ′ (θ) =

a(p − 1) sin((2 − p)θ) sinp (θ)

is of constant sign in the specified interval. Select a = −Bp (γ), then the only solution of g(θ) = 0 is θ = γ and hence h′ (θ) vanishes only at θ = γ. Now, p(p − 1) sinp−2 (γ) sin((2 − p)γ) 󵄨 h′′ (θ)󵄨󵄨󵄨θ=γ = , sin((p − 1)γ) cosp−1 (γ) which is negative for 0 < γ < occurs for

π 2

and 0 < p < 2, p ≠ 1, and hence the maximum of h(θ)

󵄨 h(θ)󵄨󵄨󵄨θ=γ =

tanp−1 (γ) = Ap (γ). tan((p − 1)γ)

It follows that h(θ) ≤ Ap (γ), and hence sinp (θ) + Bp (γ) cos(pθ) cosp (θ)

≤ Ap (γ),

and so (7.5) is established. Theorem 7.8 (M. Riesz [23]). Let R > 0 and let F(z) = U(z) + iV(z) be holomorphic in |z| < R with V(0) = 0. Then for z = reiθ , |z| = r < R, and 1 < p < ∞, 2π

2π

1 󵄨󵄨 1 󵄨󵄨 󵄨p 󵄨p ∫ 󵄨V(reiθ )󵄨󵄨󵄨 dθ ≤ Mpp ∫ 󵄨U(reiθ )󵄨󵄨󵄨 dθ, 2π 󵄨 2π 󵄨 0

(7.6)

0

where Mp = max{tan( π4 ), cot( π4 )} is the best possible. The proof we give is due to Pichorides [22] who also obtained the best constant. Proof. Let F(z) be analytic inside the unit disk |z| < 1 and suppose F(z) = U(z) + iV(z) = Reiz = R[cos θ + i sin θ], with U(z) > 0 and V(0) = 0 on the boundary of the unit disc. Because of the restriction U(z) > 0, − π2 < θ < π2 , consider the contour integral ∫ C

[F(z)]p dz, z

256 | 7 Embedding and strong Lp boundedness for the Hilbert transform where C is the circular contour |z| = r, r < 1, centered at the origin. The Cauchy integral formula gives ∫ C

[F(z)]p p dz = 2πi[F(0)] , z

which on using z = reiθ gives 2π

∫ [F(reiϕ )]i dϕ = 2πi[F(0)]

p

0

and hence 2π

p

∫ Rp eipθ dϕ = 2π[F(0)] . 0

Thus 2π

p

∫ Rp cos(pθ) dϕ = 2π[U(0)] . 0

Now multiply (7.5) by Rp and integrate over ϕ to get 2π

2π

2π

0

0

0

∫ |R sin θ|p dϕ ≤ Ap (γ) ∫ Rp cosp (θ) dϕ − Bp (γ) ∫ Rp cos(pθ) dϕ. Thus 2π

2π

2π

0

0

0

󵄨 󵄨p 󵄨 󵄨p ∫ 󵄨󵄨󵄨V(reiϕ )󵄨󵄨󵄨 dϕ ≤ Ap (γ) ∫ 󵄨󵄨󵄨U(reiϕ )󵄨󵄨󵄨 dϕ − Bp (γ) ∫ Rp cos(pθ) dθ. And so 2π

2π

0

0

p 󵄨 󵄨p 󵄨 󵄨p ∫ 󵄨󵄨󵄨V(reiϕ )󵄨󵄨󵄨 dϕ ≤ Ap (γ) ∫ 󵄨󵄨󵄨U(reiϕ )󵄨󵄨󵄨 dϕ − Bp (γ)[U(0)] .

Since Bp (γ) > 0 and [U(0)]p > 0, we obtain 2π

2π

0

0

󵄨 󵄨p 󵄨 󵄨p ∫ 󵄨󵄨󵄨V(reiϕ )󵄨󵄨󵄨 dϕ ≤ Ap (γ) ∫ 󵄨󵄨󵄨U(reiϕ )󵄨󵄨󵄨 dϕ.

(7.7)

7.1 Strong Lp boundedness for the Hilbert transform on the circle

| 257

The minimum value for Ap (γ) with 1 < p < 2 is determined in the following manner. Since Ap (γ) =

tanp−1 (γ) , tan((p − 1)γ)

we then get 𝜕Ap (γ) 𝜕γ which equals zero for γ = 𝜕2 Ap (γ) 𝜕γ 2

=

𝜕Ap (γ) 𝜕γ

=

π 2p

(p − 1) tanp (γ) sin((p − 2)γ) cos(pγ) sin2 (γ) sin2 ((p − 1)γ)

,

and

[2p csc(2γ) − 2 cot(γ) + (p − 2) cot((p − 2)γ)

− 2(p − 1) cot((p − 1)γ)] +

p(p − 1) tanp (γ) sin((2 − p)γ) sin(pγ) sin2 (γ) sin2 ((p − 1)γ)

,

so that π π p(p − 1) tanp ( 2p ) cos( 2p ) 𝜕2 Ap (γ) 󵄨󵄨󵄨 󵄨󵄨 = − > 0. 󵄨 π π 2 2 𝜕γ 󵄨󵄨γ= π sin ( ) cos2 ( ) 2p

2p

2p

Hence, the optimal constant Ap (γ) is given by Ap (

π π ) = tanp ( ) 2p 2p

and then (7.7) can be written as 2π

2π

π 󵄨 󵄨p 󵄨 󵄨p ∫ 󵄨󵄨󵄨V(reiϕ )󵄨󵄨󵄨 dϕ ≤ tanp ( ) ∫ 󵄨󵄨󵄨U(reiϕ )󵄨󵄨󵄨 dϕ. 2p

(7.8)

0

0

π ). Note that the case p > 2 is similar, and the This establishes (7.6) with Mp = tan( 2p proof for general U can be found in Pichorides [22].

Theorem 7.9 (M. Riesz [23]). The Hilbert transform is a bounded linear operator on Lp (T) (1 < p < ∞), that is, for f ∈ Lp (T), π

π

−π

−π

1 π 󵄨 󵄨p 󵄨 󵄨p ∫ 󵄨󵄨󵄨Hf (eiθ )󵄨󵄨󵄨 dθ ≤ tanp ( ) ∫ 󵄨󵄨󵄨f (eiθ )󵄨󵄨󵄨 dθ. 2π 2p

(7.9)

258 | 7 Embedding and strong Lp boundedness for the Hilbert transform Proof. For f ∈ Lp (T), construct π

eiθ + z 1 F(z) = dθ. ∫ f (eiθ ) iθ 2π e −z −π

By Theorem 5.3, F(z) is analytic in |z| < 1. Also, writing F(z) = U(z) + iV(z), we have U(reiθ ) = (f ∗ Pr )(eiθ ) and V(reiθ ) = (f ∗ Qr )(eiθ ). When integrating over the unit circle T, we shall write θ instead of eiθ to simplify notation. Now by Hölder inequality, we have 󵄨󵄨p 󵄨󵄨 π 1 1 󵄨󵄨 󵄨 󵄨p 󵄨󵄨 (2π)p 󵄨󵄨󵄨U(reiθ )󵄨󵄨󵄨 = 󵄨󵄨󵄨 ∫ f (θ + t)Prp (t)Prq (t) dt 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨−π

p q

π

π

≤ ( ∫ f (θ + t)Pr (t))( ∫ Pr (t) dt) . −π

−π π

Since ∫−π Pr (t) dt = 2π, we obtain p 󵄨󵄨

iθ 󵄨󵄨p

p−1

(2π) 󵄨󵄨U(re )󵄨󵄨 ≤ (2π)

π

∫ Pr (θ − t) dt. −π

Hence π

π π

−π

−π −π

1 󵄨 󵄨 󵄨p 󵄨p ∫ 󵄨󵄨󵄨U(reiθ )󵄨󵄨󵄨 dt ≤ ∫ ∫ 󵄨󵄨󵄨f (θ)󵄨󵄨󵄨 Pr (θ − t) dtdθ. 2π

By Fubini theorem, π

π

π

−π

−π

−π

p

1 󵄨 󵄨p 󵄨 󵄨 ∫ 󵄨󵄨󵄨U(reiθ )󵄨󵄨󵄨 dθ ≤ ( ∫ Pr (θ − t) dt)( ∫ 󵄨󵄨󵄨f (θ)󵄨󵄨󵄨) dθ, 2π

and so π

π

−π

−π

󵄨 󵄨p 󵄨 󵄨p ∫ 󵄨󵄨󵄨U(reiθ )󵄨󵄨󵄨 dθ ≤ ∫ 󵄨󵄨󵄨f (θ)󵄨󵄨󵄨 dθ.

Combining (7.8) and (7.10), we have π

π

−π

−π

π 󵄨 󵄨p 󵄨 󵄨p ∫ 󵄨󵄨󵄨V(reiθ )󵄨󵄨󵄨 dθ ≤ tanp ( ) ∫ 󵄨󵄨󵄨f (reiθ )󵄨󵄨󵄨 dθ, 4

(7.10)

7.1 Strong Lp boundedness for the Hilbert transform on the circle

| 259

for 0 < r < 1. But by Remark 18, limr→0 V(reiθ ) = Hf (eiθ ) a. e., so by Fatou’s lemma, π

π

−π

−π

π 󵄨p 󵄨p 󵄨 󵄨 ∫ 󵄨󵄨󵄨Hf (eiθ )󵄨󵄨󵄨 dθ ≤ tanp ( ) ∫ 󵄨󵄨󵄨f (eiθ )󵄨󵄨󵄨 dθ. 2p

The proof is now complete. Exercises. 1. Use Theorem 5.6 to prove that for f ∈ Lp (ℝ), ‖Hf ‖Lp (ℝ) ≤ tanp ( 2.

π )‖f ‖Lp (ℝ) . 2p

For f ∈ Lp (T ), prove that π tanp ( 2p ) 󵄨 󵄨 ‖f ‖Lp (T ) . m({θ ∈ [−π, π] : 󵄨󵄨󵄨󵄨Hf (eiθ )󵄨󵄨󵄨󵄨 > λ}) ≤ p λ

3.

Let f ∈ Lp (ℝ) and g ∈ Lq (ℝn ) (1 < p < ∞,

1 p

+

1 q

= 1). Prove that

󵄨󵄨 󵄨󵄨 ∞ ∞ 󵄨󵄨 󵄨󵄨 π f (x)g(x) 󵄨 󵄨󵄨 dydx 󵄨󵄨󵄨 ≤ tanp ( )‖f ‖Lp (ℝ) ‖g‖Lq (ℝ) . 󵄨󵄨 ∫ ∫ 󵄨󵄨 󵄨󵄨 x − y 2p 󵄨󵄨 󵄨󵄨 −∞ −∞ 4.

Let f ∈ Lp (ℝ) and g ∈ Lq (ℝ). Prove that ∞

∞

−∞

−∞

∫ f (x)Hg(x) dx = − ∫ Hf (x)g(x) dx.

8 Riesz transform Theorem 8.1. Let 0 < α < n. Then (π −(

n−α ) 2

Γ(

α n−α 1 α 1 ) n−α ) (ξ ) = π − 2 Γ( ) α . 2 |x| 2 |x|

̂

(8.1)

Proof. First, observe that ∞

∫e

−πt|x|2

t

n−α 2

0

∞

u dt = ∫ e−u ( ) t π(x)2 0

= π −(

n−α ) 2

Γ(

n−α 2

du u

n−α 1 ) n−α . 2 |x|

Let f ∈ S′ (Rn ), then (π −(

n−α ) 2

Γ(

n−α n−α 1 1 )(f ) ) n−α ) (f ) = (π − 2 2 |x| |x|n−α n−α n−α 1 = ∫ π −( 2 ) Γ( ) n−α ̂f (x) dx. 2 |x|

̂

(8.2)

ℝn

Our first observation, together with the definition of the Fourier transform, gives ∞

2

= ∫ ∫ e−πt|x| t 0 ℝn ∞

n−α 2

dt ∫ e2πix⋅y f (y) dydx t ℝn

2

= ∫ ∫ ( ∫ e−πt|x| e−2πix⋅y dx)t −( ℝn 0

n−α ) 2

ℝn

dt f (y) dy, t 2

where the quantity in the parenthesis is precisely the Fourier transform of e−π|x| . Recall that (see Lemma 3.2) −1

n

(e−x⋅Ax ) (ξ ) = π 2 (det(A)) 2 e−π ̂

2

ξ ⋅A−1 ξ

.

Thus 2

n

2

y

n

∫ e−πt|x| −2πix⋅y dx = π 2 (πt)e−π e−πy πt = t 2 e−π ℝn

https://doi.org/10.1515/9783110784091-008

|y|2 t

.

262 | 8 Riesz transform Continuing the computation, we get ∞

= ∫ ∫ e−π ℝn 0 ∞

|y|2 t

α

t − 2 f (y)dy

n

= ∫ ∫ t − 2 e−π ℝn 0 ∞

|y|2

t −(

− α2

π|y|2 ) = ∫ ∫( u

dt f (y) dy t

n−α ) 2

e−u

ℝn 0

du f (y) dy u

α = ∫ π −α Γ( )|y|−α f (y) dy 2 ℝn

α α = (π − 2 Γ( |y|−α ))(f ). 2

This proves (8.2) in the sense of distributions. We now define the main generalization of the Hilbert transform, the Riesz transform Rj , j = 1, 2, . . . , n. Given f ∈ S(ℝn ), we set yj

Rj f (x) = Cn P. V. ∫ where Cn =

Γ( n+1 ) 2 π(

n+1 ) 2

f (x − y) dy,

|y|n+1

ℝn

.

Notice that for n = 1, we have C1 = transform.

1 π

and recover the definition of the Hilbert

Lemma 8.1. For f ∈ L2 (ℝn ) and j = 1, . . . , k, we have ̂f )(ξ ) = −i (R j

ξj |ξ |

̂f (ξ ).

Proof. We claim that for φ ∈ S(ℝn ), (𝜕xj |x|−(1+n) )(φ) = (1 − n) P. V.

xj

|x|n+1

(φ).

Proof of claim. (1 − n) P. V.

(8.3)

xj

|x|n+1

(φ) = lim+ (1 − n) ∫ ϵ→0

ℝn

= lim+ ∫ 𝜕xj ( ϵ→0

ℝn

xj (ϵ2 + |x|2 ) 1

(ϵ2

+ |x|2 )

n−1 2

n−1 2

φ(x) dx

)φ(x) dx

8 Riesz transform

= − lim+ ∫ ϵ→0

=−∫ ℝn

ℝn

1 (ϵ2 + |x|2 )

n−1 2

𝜕xj φ(x) dx

(by the Green formula)

1 𝜕x φ(x) dx |x|n−1 j

= −(𝜕xj |x|−(n−1) )(φ). Now, we shall prove that (Cn P. V.

xj |x|

ξj

̂

) (ξ ) = −i n+1

|ξ |

.

Indeed, from the above computation, applying Fourier transform, (P. V.

xj

1 ̂ ) (ξ ) = (𝜕x |x|−(1+n) ) (ξ ) 1−n j |x|n+1 2πi ̂ = (|x|−(1+n) ) (ξ ) 1−n n −1 1 2πi ξj π 2 Γ( 2 ) −1 = |ξ | 1 − n Γ( n−1 ) 2 ̂

n

2πiξj π 2 −1 Γ( 21 ) 1 =− (n − 1)Γ( n−1 ) |ξ | =−

2ππ

= −i

n −1 2

√2 2

) Γ( n+1 2

π

n+1 2

) Γ( n+1 2

ξj

|ξ |

2

ξj |ξ | ,

as we wished to prove. Remark 22. For f ∈ L2 (ℝ), we may write yj

Rj (f )(x) = lim+ Cn ∫ ϵ→0

|y|n+1

|y|>ϵ

= Cn P. V. ∫ ℝn

|y|n+1

= (f ∗ Kj )(x), with Cn =

) Γ( n+1 2 π

n+1 2

yj

f (x − y) dy

f (x − y) dy

j = 1, 2, . . . , n,

x

j , where Ki (x) = Cn P. V. |x|n+1 .

Theorem 8.2. Let f ∈ L2 (ℝn ). Then ‖Rj ∗ f ‖L2 (ℝn ) ≤ Cn ‖f ‖L2 (ℝn ) .

| 263

264 | 8 Riesz transform Proof. Since ̂f (ξ ) = (R ? R j j ∗ f )(ξ ) ̂j (ξ )̂f (ξ ) =R = −iCn

ξj

|ξ |

̂f (ξ ),

by Plancharel’s identity, we have ? ‖Rj ∗ f ‖L2 (ℝ) = ‖R ∗ f ‖L2 (ℝn ) 󵄩󵄩 ξj 󵄩󵄩󵄩 󵄩 = 󵄩󵄩󵄩−iCn ̂f 󵄩󵄩󵄩 󵄩󵄩 |ξ | 󵄩󵄩L2 (ℝn ) ξj = Cn ‖̂f ‖L2 (ℝn ) |ξ | ξ ≤ Cn ‖̂f ‖L2 (ℝn ) |ξ | = Cn ‖f ‖L2 (ℝn ) . Hence ‖Rj ∗ f ‖L2 ℝn ≤ Cn ‖f ‖L2 (ℝn ) . The latter inequality tells us that Rj is bounded, but it is not unitary. Theorem 8.3. Let f ∈ L2 (ℝn ). Then n

∑ ‖Rj f ‖L2 (ℝn ) = ‖f ‖L2 (ℝn ) . j=1

ξ

̂f (ξ ) = −i j ̂f (ξ ), we get Proof. Since R j |ξ | ̂f (ξ ) = −i R j

ξj

̂f (ξ ) R |ξ | j ξj ξj = (−i )(−i )̂f (ξ ) |ξ | |ξ |

= i2 =−

|ξj |2 |ξ |2

|ξj |2 |ξ |2

̂f (ξ ) ̂f (ξ ).

8 Riesz transform

Hence 󵄩󵄩 󵄩 󵄩 󵄩󵄩 󵄩󵄩Rj (Rj f )󵄩󵄩󵄩L2 (ℝn ) = 󵄩󵄩󵄩R? j (Rj f )󵄩 󵄩L2 (ℝn ) |ξj |2

=

|ξ | |ξj |2

=

|ξ |

‖̂f ‖L2 (ℝn ) ‖f ‖L2 (ℝn ) ,

and so n

󵄩 󵄩 󵄩 󵄩 ∑󵄩󵄩󵄩R2j f 󵄩󵄩󵄩L (ℝn ) = ∑󵄩󵄩󵄩Rj (Rj f )󵄩󵄩󵄩L (ℝn ) 2

j=1

2

j=1 n

=∑ j=1

|ξj |2 |ξ |2

‖f ‖L2 (ℝn )

= ‖f ‖L2 (ℝ) , completing the proof. Corollary 8.1. We have that ∑nj=1 R2j = −I. 2

̂ 2 f )(ξ ) = − |ξj | ̂ Proof. Since (R f (ξ ), we obtain j |ξ |2 n

n

̂ 2 f (ξ ) = − ∑ ∑R j j=1

j=1

|ξj |2 |ξ |2

̂f (ξ ),

thus n

̂f (ξ ) = −̂f (ξ ). ∑R j j=1

Taking the inverse Fourier transform, we get n

∑ R2j f = −f j=1

or, equivalently,

n

∑ R2j = −I. j=1

n

Theorem 8.4. Let f , g ∈ S(ℝ ). Then ∫ Rj f (x)g(x) dx = − ∫ f (x)Rj g(x) dx. ℝn

Moreover, R∗j = −Rj .

ℝn

| 265

266 | 8 Riesz transform Proof. Let f , g ∈ S(ℝn ). Then ̂̂ ̂̌ dx = ∫ R ̂f (x)g(ξ ̌ ) dξ ∫ Rj f (x)g(x) dx = ∫ Rj f (x)g(x) j ℝn

ℝn

ℝn

̂f (x)g(ξ ̂j (ξ )g(ξ ̌ ) dξ = ∫ ̂f (ξ )K ̌ ) dξ = ∫R j ℝn

ℝn

? ̃ (ξ ) dξ = ∫ ̂f (ξ )K j∗g

(g̃ (x) = g(−x))

ℝn

̂ ? ? = ∫ ̂f (ξ )K j ∗ g(−ξ ) dξ = − ∫ f (ξ )Kj ∗ g(ξ ) dξ ℝn

ℝn

̂ = − ∫ ̂f (ξ )R j g(ξ ) dξ = − ∫ f (x)Rj g(x) dx. ℝn

ℝn

Hence ∫ Rj f (x)g(x) dx = − ∫ f (x)Rj g(x) dx. ℝn

(8.4)

ℝn

By (8.4), we have ⟨Rj f , g⟩ = ⟨f , −Rj g⟩, which implies that R∗j g = −Rj g, thus R∗j = −Rj , as we wished to prove. Theorem 8.5. For Kj (x) =

xj , |x|n+1

j = 1, 2, . . . , n, there exists B such that

󵄨 󵄨 ∫ 󵄨󵄨󵄨Kj (x − y) − Kj (x)󵄨󵄨󵄨 dx ≤ B. |x|>2|y|

Proof. Let Kj (x) =

xj |x|n+1

= xj |x|−(n+1) for j = 1, 2, . . . , n. Then 󵄨 x 󵄨󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 −(n+1) − (n + 1)xj |x|−(n+2) 󵄨󵄨󵄨 󵄨󵄨∇Kj (x)󵄨󵄨󵄨 = 󵄨󵄨󵄨|x| 󵄨󵄨 |x| 󵄨󵄨 󵄨󵄨 −(n+1) −(n+3) 󵄨󵄨 = 󵄨󵄨|x| − (n + 1)xj x|x| 󵄨󵄨 ≤ |x|−(n+1) + (n + 1)|xj ||x||x|−(n+3) ≤ |x|−(n+1) + (n + 1)|x|2 |x|−(n+3) = |x|−(n+1) + (n + 1)|x|−(n+1) = 2|x|−(n+1) + n|x|−(n+1) =

(2 + n) . |x|n+1

Hence 󵄨󵄨 󵄨 (2 + n) . 󵄨󵄨∇Kj (x)󵄨󵄨󵄨 ≤ |x|n+1

8 Riesz transform

| 267

Let σ(x, y) denote the segment with end points in x − y and x. By the mean value theorem, there exists z ∈ σ(x, y) such that Kj (x − y) − Kj (x) = [∇Kj (z)]y. Hence if |x| > 2|y| then |x − y| ≥ |x| −

|x| 2

=

|x| , 2

and so

|z| ≥ min{|x − y|, |x|} ≥

|x| . 2

Therefore 󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨Kj (x − y) − Kj (x)󵄨󵄨󵄨 ≤ |y| sup 󵄨󵄨󵄨∇Kj (z)󵄨󵄨󵄨 z∈σ(x,y)

1 n+1 |z| z∈σ(x,y)

≤ |y| sup ≤ 2n+1

|y| , |x|n+1

and hence 󵄨 󵄨 ∫ 󵄨󵄨󵄨Kj (x − y) − Kj (x)󵄨󵄨󵄨 dx ≤ 2n+1 |y| ∫

dx |x|

|x|>2|y|

|x|>2|y|

+∞

= 2n+1 ωn |y| ∫ r −2 dr = 2n ωn . 2|y|

Thus 󵄨 󵄨 ∫ 󵄨󵄨󵄨Kj (x − y) − Kj (x)󵄨󵄨󵄨 dx ≤ B, |x|>2|y|

where B = 2n ωn , and the proof is complete. Theorem 8.6 (Kolmogorov). As an operator, the Riesz transform Rn (n = 1, 2, . . . ) is of weak type (1, 1) on S(ℝn ), that is, c 󵄨 󵄨 m({x ∈ ℝn : 󵄨󵄨󵄨Rn f (x)󵄨󵄨󵄨 > λ}) ≤ ‖f ‖L1 (ℝn ) . λ

(8.5)

Proof. From the Calderón–Zygmund decomposition at height λ, we obtain a sequence of disjoint cubes {Qj }j∈ℕ such that: (I) f (x) ≤ λ for almost all x ∉ ⋃j Qj ; (II) m(⋃j Qj ) ≤ λ1 ‖f ‖L1 (ℝn ) ; 1 (III) λ < m(Q ∫ f (x) dx ≤ 2λ. ) Q j

j

268 | 8 Riesz transform Now, we decompose f as a sum f = g + b, where {f (x) g(x) = { 1 ∫ f (x) dx { m(Qj ) Qj

if x ∉ ⋃j Qj , if x ∈ Qj

and b(x) = ∑ bj (x) = ∑(f (x) − j

j

1 ∫ f (x))χQj (x). m(Qj ) Qj

We observe that 0 ≤ g(x) ≤ λ if x ∉ ⋃j Qj , while 0 ≤ g(x) =

1 ∫ f (x) dx ≤ 2λ m(Qj ) Qj

if x ∈ Qj . Since it is obviously integrable on (⋃j Qj )c and on each Qj (that are disjoint), ∫ g(x) dx = ∫ f (x) dx < ∞. Qj

Qj

Notice that ∫ g(x) dx = ∫ f (x) dx < ∞ ℝn

ℝn

and 󵄨 󵄨 󵄨 󵄨 ∫󵄨󵄨󵄨b(x)󵄨󵄨󵄨 dx ≤ ∑ ∫󵄨󵄨󵄨bj (x)󵄨󵄨󵄨 dx j Q j

Qj

󵄨 󵄨 ≤ ∑ ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx + j Q j

1 󵄨 󵄨 ∫󵄨󵄨f (x)󵄨󵄨󵄨 dx ∫ dx m(Qj ) 󵄨 Qj

Qj

󵄨 󵄨 ≤ 2 ∑ ∫󵄨󵄨󵄨f (x)󵄨󵄨󵄨 dx j Q j

≤ 2‖f ‖L1 (ℝn ) . Also note that 1 1 1 ∫ g(x) dx = ∑ ∫(f (x) dx − ∫ f (x) dx)χQj (x) dx m(Qj ) m(Qj ) m(Qj ) j Qj

Qj

= ∑( j

Qj

1 1 ∫ f (x) dx − ∫ f (x) dx) = 0. m(Qj ) m(Qj ) Qj

Qj

8 Riesz transform

| 269

Since Rm f (x) = Rm g(x) + Rm b(x) and |Rm f (x)| ≤ |Rm g(x)| + |Rm b(x)|, we get 󵄨 λ 󵄨 󵄨 󵄨 m({x ∈ ℝn : 󵄨󵄨󵄨Rm f (x)󵄨󵄨󵄨 > λ}) ≤ m({x ∈ ℝn : 󵄨󵄨󵄨Rm g(x)󵄨󵄨󵄨 > }) 2 󵄨 λ 󵄨 + m({x ∈ ℝn : 󵄨󵄨󵄨Rm b(x)󵄨󵄨󵄨 > }). 2 Now, on the good function g we use the L2 boundedness of the Riesz transform to obtain 󵄨 󵄨 λ m({x ∈ ℝn : 󵄨󵄨󵄨Rm g(x)󵄨󵄨󵄨 > }) 2 =

∫

dx

{x∈ℝn :|Rm b(x)|> λ2 }

≤

∫ {x∈ℝn :|Rm b(x)|> λ2 }

≤

(

Rm g(x) λ 2

2

) dx

4 󵄨󵄨 4 󵄨 󵄨2 󵄨2 ∫ 󵄨R g(x)󵄨󵄨󵄨 dx ≤ 2 ∫ 󵄨󵄨󵄨g(x)󵄨󵄨󵄨 dx λ2 n 󵄨 m λ n ℝ

8 ≤ ∫ g(x) dx λ

ℝ

󵄨 󵄨2 (g(x) ≤ λ a. e. 󳨐⇒ 󵄨󵄨󵄨g(x)󵄨󵄨󵄨 ≤ 2λg(x) a. e.)

ℝn

=

8 ∫ f (x) dx λ ℝn

≤

8 󵄨󵄨 8 󵄨 ∫ 󵄨f (x)󵄨󵄨󵄨 dx = ‖f ‖L1 (Rn ) . λ 󵄨 λ ℝn

Hence 8 󵄨 󵄨 λ m({x ∈ ℝn : 󵄨󵄨󵄨Rm g(x)󵄨󵄨󵄨 > }) ≤ ‖f ‖L1 (Rn ) . 2 λ

(8.6)

Next, let 2Qj be the cube with the same center cj as Qj and twice the side length, let Ω = ⋃j Qj and Ω∗ = ⋃j 2√nQj . Then m(Ω∗ ) ≤ 2√n ∑ m(Qj ) ≤ j

2 ‖f ‖ n . λ L1 (ℝ )

Then we estimate 󵄨 󵄨 λ 󵄨 󵄨 λ m({x ∈ ℝn : 󵄨󵄨󵄨Rm b(x)󵄨󵄨󵄨 > }) = m({x ∈ Ω∗ : 󵄨󵄨󵄨Rm b(x)󵄨󵄨󵄨 > }) 2 2

c 󵄨 󵄨 λ + m({x ∈ (Ω∗ ) : 󵄨󵄨󵄨Rm b(x)󵄨󵄨󵄨 > }) 2

270 | 8 Riesz transform c 󵄨 󵄨 λ ≤ m(Ω∗ ) + m({x ∈ (Ω∗ ) : 󵄨󵄨󵄨Rm b(x)󵄨󵄨󵄨 > }) 2 2 2 󵄨 󵄨 ≤ ‖f ‖L1 (ℝn ) + ∫ 󵄨󵄨󵄨Rm b(x)󵄨󵄨󵄨 dx. λ λ ℝn \Ω∗

Since 󵄨 󵄨 󵄨 󵄨󵄨 󵄨󵄨Rm b(x)󵄨󵄨󵄨 ≤ ∑󵄨󵄨󵄨Rm bj (x)󵄨󵄨󵄨 j

a. e.,

it will suffice to prove that ∑ j

󵄨󵄨 󵄨 󵄨󵄨Rm bj (x)󵄨󵄨󵄨 dx ≤ C‖f ‖L1 (ℝn ) .

∫ ℝn \2√nQj

Now, bj ∈ S(ℝn ) and, for x ∉ Ω∗ , we have Rm bj (x) = ∫ bj (y)Km (x − y) dy, ℝn

since 1 ∫ bj (x) dx = 0 m(Qj ) Qj

for x ∈ Ω∗ . Then Rm bj (x) = ∫ Km (x − y)bj (y) dy = ∫[Km (x − y) − Km (x − cj )]bj (y) dy Qj

Qj

and hence 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 ∫ 󵄨󵄨󵄨Rm bj (x)󵄨󵄨󵄨 dx ≤ ∫󵄨󵄨󵄨bj (y)󵄨󵄨󵄨 ∫ 󵄨󵄨󵄨Km (x − y) − Km (x − cj )󵄨󵄨󵄨 dxdy. ℝn \Ω∗

Qj

ℝn \Ω∗j

Since ℝn \ Ω∗j ⊆ {x ∈ ℝn : |x − cj | > 2|y − cj |}, we obtain 󵄨 󵄨 ∫ 󵄨󵄨󵄨Km (x − y) − Km (x − cj )󵄨󵄨󵄨 dx ≤ ℝn \Ω∗j

󵄨 󵄨󵄨 󵄨󵄨Km (x − y) − Km (x − cj )󵄨󵄨󵄨 dx

∫ |x−cj |>2|y−cj |

=

∫ |ω|>2|y−cj |

󵄨󵄨 󵄨 󵄨󵄨Km (ω − (y − cj )) − Km (ω)󵄨󵄨󵄨 dx.

8 Riesz transform | 271

By Theorem 5.5, 󵄨󵄨 󵄨 󵄨󵄨Km (ω − (y − cj )) − Km (ω)󵄨󵄨󵄨 dω ≤ B,

∫ |ω|>2|y−cj |

and so ∑ j

󵄨 󵄨 󵄨 󵄨󵄨 󵄨󵄨Rm bj (x)󵄨󵄨󵄨 ≤ ∑ ∫󵄨󵄨󵄨bj (y)󵄨󵄨󵄨 dy ≤ B ∑ ‖f ‖L1 (ℝn ) .

∫

j

j Q j

ℝn \2√nQj

Hence 󵄨 󵄨 λ m({x ∈ ℝn : 󵄨󵄨󵄨Rm b(x)󵄨󵄨󵄨 > }) ≤ B‖f ‖L1 (ℝn ) . 2

(8.7)

Combining (8.6) and (8.7), we get (8.5). Theorem 8.7. Let f ∈ Lp (ℝn ), 1 < p < ∞. Then ‖Rj f ‖Lp (ℝn ) ≤ C‖f ‖Lp (ℝn ) . Proof. For 1 < p < 2, we proceed as in the proof of Theorem 4.7. Now if 2 < p < ∞ and 1 + q1 = 1, then 1 < q < 2. Thus for f ∈ Lp (ℝn ), by Theorem 5.4 and applying Hölder p inequality, we have 󵄨󵄨 󵄨󵄨 󵄨 󵄨 ‖Rj f ‖Lp (Rn ) = sup 󵄨󵄨󵄨 ∫ Rj f (x)g(x) dx󵄨󵄨󵄨 󵄨󵄨 ‖g‖q ≤1󵄨󵄨 n ℝ

󵄨󵄨 󵄨󵄨 󵄨 󵄨 = sup 󵄨󵄨󵄨− ∫ f (x)Rj g(x) dx󵄨󵄨󵄨 󵄨󵄨 ‖g‖q ≤1󵄨󵄨 n ℝ

≤ ‖f ‖p sup ‖Rj g‖q ‖g‖q ≤1

≤ Ap ‖f ‖p . Here Ap is independent of j and f . This finishes the proof. Theorem 8.8. Let f ∈ C02 (ℝn ) and consider Δf = ∑nj=1 bound

𝜕2 f . 𝜕xj2

󵄩󵄩 𝜕2 f 󵄩󵄩 󵄩󵄩 󵄩󵄩 ≤ Ap ‖Δf ‖Lp (ℝn ) 󵄩󵄩 󵄩󵄩 󵄩󵄩 𝜕xj 𝜕xk 󵄩󵄩Lp (ℝn ) for 1 < p < ∞.

Then we have an a priori

272 | 8 Riesz transform Proof. If f ∈ C02 (ℝn ) then, for j = 1, 2, . . . , n, we have ̂f )(x) = −i (R j Since

̂ 𝜕f 𝜕xj

xj |x|

̂f .

x

(x) = −i |x|j ̂f (x), we get ? 𝜕? 𝜕f 𝜕2 f (x) = ( )(x) 𝜕xk 𝜕xj 𝜕xk 𝜕xj ̂ 𝜕f (x) 𝜕xj = 4π 2 xk xj ̂f (x) = −2πixk

xj xk )(i )(−4π|x|2 )̂f (x) |x| |x| xj xk ̂ (x) = (−i )(i )Δf |x| |x| ix = −( k )R̂ Δf (x) |x| j = −R? R Δf (x). = (−i

k j

Taking the inverse Fourier transform of this, we obtain 𝜕2 f (x) = −Rk Rj Δf . 𝜕xk 𝜕xj Hence by Theorem 5.7, we have 󵄩󵄩 𝜕2 f 󵄩󵄩 󵄩󵄩 󵄩󵄩 = ‖ − Rk Rj Δf ‖Lp (ℝn ) 󵄩󵄩 󵄩󵄩 󵄩󵄩 𝜕xj 𝜕xk 󵄩󵄩Lp (ℝn ) ≤ Bp ‖Rj Δf ‖Lp (ℝn ) ≤ Bp Cp ‖Δf ‖Lp (ℝn ) = Ap ‖Δf ‖Lp (ℝn ) , where Ap = Bp Cp . Theorem 8.9. Let f ∈ C01 (ℝn ). Then 󵄩󵄩 𝜕f 󵄩󵄩 󵄩󵄩 𝜕f 󵄩󵄩 󵄩󵄩 𝜕f 𝜕f 󵄩󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩 󵄩󵄩 󵄩 + 󵄩󵄩󵄩 ≤ Ap 󵄩󵄩󵄩 +i , 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩 󵄩󵄩 𝜕x1 󵄩󵄩Lp (ℝn ) 󵄩󵄩 𝜕x2 󵄩󵄩Lp (ℝn ) 󵄩󵄩 𝜕x1 𝜕x2 󵄩󵄩󵄩Lp (ℝn )

1 < p < ∞.

8 Riesz transform | 273

Proof. The proof is similar to the previous one. Indeed, ̂ 𝜕f (ξ ) = 2iξj ̂f (ξ ) 𝜕xj = 2iξj

|ξ |2 ̂ f (ξ ) |ξ |2

ξ12 + ξ22 ̂ )f (ξ ) |ξ | |ξ | ξj (ξ − iξ2 )(ξ1 − iξ2 ) = 2i ( 1 )̂f (ξ ) |ξ | |ξ | ξj (ξ1 − iξ2 ) = (2iξ1 ̂f (ξ ) + i((2iξ2 )̂f (ξ ))) |ξ | |ξ | ̂ ̂ ξj (ξ1 − iξ2 ) 𝜕f 𝜕f = ( (ξ ) + i (ξ )) |ξ | |ξ | 𝜕x1 𝜕x2 = 2i

ξj

(

= −ℱ (Rj (R1 − iR2 ))(

𝜕f 𝜕f +i )(ξ ). 𝜕x1 𝜕x2

Hence taking the inverse Fourier transform, we obtain 𝜕f 𝜕f 𝜕f = −Rj (R1 − iR2 )( +i ). 𝜕xj 𝜕x1 𝜕x2 Thus 󵄩󵄩 𝜕f 󵄩󵄩 󵄩󵄩 𝜕f 󵄩󵄩󵄩󵄩 𝜕f 󵄩󵄩 󵄩󵄩 󵄩󵄩 +i )󵄩 󵄩󵄩 󵄩󵄩 = 󵄩󵄩−R1 (R1 − iR2 )( 󵄩󵄩 𝜕x1 󵄩󵄩 󵄩󵄩 𝜕x1 𝜕x2 󵄩󵄩󵄩Lp (ℝn) 󵄩󵄩 𝜕f 󵄩󵄩󵄩󵄩 𝜕f 󵄩 ≤ Dp 󵄩󵄩󵄩(R1 − iR2 )( +i )󵄩 𝜕x1 𝜕x2 󵄩󵄩󵄩Lp (ℝn ) 󵄩󵄩 󵄩󵄩 󵄩󵄩 𝜕f 󵄩󵄩󵄩󵄩 𝜕f 󵄩󵄩󵄩󵄩 𝜕f 𝜕f 󵄩 󵄩 +i )󵄩󵄩 +i )󵄩 ≤ DP [󵄩󵄩󵄩R1 ( + 󵄩󵄩󵄩R2 ( ] 󵄩󵄩 𝜕x1 𝜕x2 󵄩󵄩Lp (ℝn ) 󵄩󵄩 𝜕x1 𝜕x2 󵄩󵄩󵄩Lp (ℝn ) 󵄩󵄩 𝜕f 󵄩󵄩 𝜕f 𝜕f 󵄩󵄩󵄩󵄩 𝜕f 󵄩󵄩󵄩󵄩 󵄩 󵄩 ≤ DP Cp 󵄩󵄩󵄩 +i +i + Dp Bp 󵄩󵄩󵄩 󵄩󵄩 󵄩 󵄩󵄩 𝜕x1 󵄩󵄩 𝜕x1 𝜕x2 󵄩󵄩Lp (ℝn ) 𝜕x2 󵄩󵄩󵄩Lp (ℝn ) 󵄩󵄩󵄩 𝜕f 𝜕f 󵄩󵄩󵄩󵄩 +i . = (Dp Cp + Dp Bp )󵄩󵄩󵄩 󵄩 𝜕x2 󵄩󵄩󵄩Lp (ℝn ) 󵄩󵄩 𝜕x1 Similarly, we have 󵄩󵄩 𝜕f 󵄩󵄩 𝜕f 󵄩󵄩 𝜕f 󵄩󵄩󵄩󵄩 󵄩󵄩 󵄩 󵄩󵄩 ≤ (Ep Fp + Ep Gp )󵄩󵄩󵄩 +i . 󵄩󵄩 󵄩 󵄩󵄩 󵄩󵄩 𝜕x1 󵄩󵄩 𝜕x2 󵄩󵄩Lp (ℝn ) 𝜕x2 󵄩󵄩󵄩Lp (ℝn )

274 | 8 Riesz transform Finally, 󵄩󵄩 𝜕f 󵄩󵄩 󵄩󵄩 𝜕f 󵄩󵄩 𝜕f 󵄩󵄩 𝜕f 󵄩󵄩󵄩󵄩 󵄩 󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 + 󵄩󵄩󵄩 ≤ Ap 󵄩󵄩󵄩 , +i 󵄩󵄩 󵄩󵄩 󵄩 󵄩󵄩 󵄩󵄩 𝜕x1 𝜕x2 󵄩󵄩󵄩Lp (ℝn ) 󵄩󵄩 𝜕x1 󵄩󵄩Lp (ℝn ) 󵄩󵄩 𝜕x2 󵄩󵄩Lp (ℝn ) where Ap = (Dp Cp + Dp Bp ) + (Ep Fp + Ep Gp ).

8.1 Vectorial Riesz transform Definition 8.1. The Riesz transform n n R : L2 (ℝn ) → L ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ 2 (ℝ ) × ⋅ ⋅ ⋅ × L2 (ℝ ) n−times

is defined as (Rf )(x) = (R1 f (x), . . . , Rn f (x)), through its single components Rk f (x) =

) Γ( n+1 2 π

n+1 2

∫ ℝn

yk f (x − y) dy |y|n+1

and ̂f (ξ ) = −i ξk ̂f (ξ ). R k |ξ | Proposition 8.1. Let f , g ∈ L2 (ℝn ). Then ⟨Rf , Rg⟩Yn [L2 (ℝn )]n = ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ L2 (ℝn ) × ⋅ ⋅ ⋅ × L2 (ℝn )

= ⟨f , g⟩L2 (ℝn ) , where Yn =

n-times

Proof. Let f , g ∈ L2 (ℝn ). Note that n

̂, Rg⟩ ̂f , R ̂ Y = ∑ ⟨R ̂ ⟨Rf , Rg⟩Yn = ⟨Rf k k g⟩L2 (ℝn ) n n

k=0

̂f (ξ )R? = ∑ ∫R k g(ξ )(ξ ) dξ k k=0 ℝn n

= ∑ ∫ (−i k=0 ℝn n

= ∑ ∫ (−i k=0 ℝn

ξ ξk ̂ f (ξ ))(−i k ĝ (ξ )) dξ |ξ | |ξ | ξk ̂ ξ f (ξ ))(i k ĝ (ξ )) dξ |ξ | |ξ |

8.1 Vectorial Riesz transform n

= ∑ ∫ −i2

=

| 275

ξk2 ̂ f (ξ )ĝ (ξ ) dξ |ξ |2

k=0 ℝn n ξ2 ∑ k2 ∫ |ξ | n k=0 ℝ

̂f (ξ )ĝ (ξ ) dξ

= ∫ ̂f (ξ )ĝ (ξ ) dξ = ⟨̂f , ĝ ⟩L2 (Rn ) = ⟨f , g⟩L2 (Rn ) , ℝn

and the proof is complete. Corollary 8.2. Let f ∈ L2 (ℝn ). Then n

‖Rf ‖2Yn = ∑ ‖Rk f ‖2L2 (ℝn ) = ‖f ‖2L2 (ℝn ) . k=0

Proof. Let f ∈ L2 (ℝn ). Then ̂, Rf ̂⟩ ‖Rf ‖2Yn = ⟨Rf , Rf ⟩Yn = ⟨Rf Yn n

̂f , R ̂f ⟩ n = ∑ ⟨R k k L2 (ℝ ) k=0 n

̂f ‖ n = ∑ ‖R k L2 (ℝ ) k=0 n

= ∑ ‖Rk f ‖L2 (ℝn ) k=0

= ‖f ‖2L2 (ℝn ) . Theorem 8.10. Let f ∈ Yn . Then the adjoint operator R∗ : Yn → L2 (ℝn ) is given by n

(R∗ f )(x) = ∑ (R∗ fk )(x). k=1

Proof. Let f and φ be in Yn . Then n

⟨Rf , φ⟩L2 (ℝn ) = ∑ ⟨Rk f , φk ⟩Yk k=1 n

= ∑ ⟨f , R∗k φk ⟩Y

n

k=1

n

= ⟨f , ∑ R∗ φk ⟩ . k=1

Thus R∗ φ = ∑nk=1 R∗k φk .

Yn

276 | 8 Riesz transform Corollary 8.3. We have that ‖R∗ f ‖L2 (ℝn ) = ‖f ‖Yn . Theorem 8.11. We have that R∗ = R−1 . Proof. Let us consider ⟨R∗ (Rf ), f ⟩L (ℝn ) = ⟨Rf , Rf ⟩L2 (Rn ) 2

= ‖Rf ‖2L2 (ℝn )

= ⟨f , f ⟩

= ⟨If , f ⟩. Thus R∗ (Rf ) = If . Hence R∗ (R) = I for all f ∈ L2 (ℝn ), and so R∗ (R)R−1 = IR−1 , thus R∗ = R−1 . Exercises. 1. Let Δ = ∑nk=1

𝜕2 𝜕xk

be the Laplacian operator. If f ∈ L2 (ℝn ), prove that 2

̂ (ξ ) = (2π‖ξ ‖) ̂f (ξ ). −Δf 2.

3.

1

Let f ∈ L2 (ℝn ). Prove that (Rk f )(x) = −(−Δ)− 2 n

𝜕f (x) . 𝜕xk

Let f ∈ Lp (ℝ ), 1 < p < ∞. For any Borel subset A of ℝn , prove that 1

󵄨 󵄨 ∫󵄨󵄨󵄨󵄨Rj f (x)󵄨󵄨󵄨󵄨 dx ≤ Cp ‖f ‖Lp (ℝn ) (m(A)) q . A

Here

1 p

+

1 q

= 1.

Appendix A A.1 Convergence in measure Definition A.1. Let (X, A , μ) be a measure space. Let f , fn , n ≥ 1 be measurable functions in X. The sequence {fn }n∈ℕ is said to converge in measure to f if, for all ϵ > 0, 󵄨 󵄨 lim μ({x ∈ X : 󵄨󵄨󵄨fn (x) − f (x)󵄨󵄨󵄨 ≥ ϵ}) = 0.

n→∞ μ

We write this as fn → f . Remark 23. The above definition is equivalent to each of the following statements: (a) For all ϵ > 0 there exists N(ϵ) ∈ ℕ such that 󵄨 󵄨 μ({x ∈ X : 󵄨󵄨󵄨fn (x) − f (x)󵄨󵄨󵄨 ≥ ϵ}) < ϵ,

for n ≥ N(ϵ).

(b) For all m ∈ ℕ there exists N(m) ∈ ℕ such that 1 󵄨 󵄨 1 μ({x ∈ X : 󵄨󵄨󵄨fn (x) − f (x)󵄨󵄨󵄨 ≥ }) < , m m

for m ≥ N(m).

Example 16. (a) Let X = [0, 1] and A = L ∩ [0, 1] = {E ∩ [0, 1] : E ∈ L } and let μ = m|L . Consider f1 = χ[0,1] ,

f5 = χ[ 3 ,1] ,

f2 = χ[0, 1 ] ,

and so on;

f3 = χ[ 1 ,1] ,

let fn = χ[

f4 = χ[0, 1 ] ,

0 ≤ j ≤ 2k .

2

2

4

4

j

2k

, j+1 k ] 2

if n = 2k + j,

Each fn is A -measurable. μ

Claim. We have that fn → 0. Proof of claim. Let ϵ > 0. Then 1 󵄨 󵄨 μ({x ∈ X : 󵄨󵄨󵄨fn (x) − 0󵄨󵄨󵄨 ≥ ϵ}) = k 2 if n = 2k + j, 0 ≤ j ≤ 2k ; thus n < 2k + 2k = 2k+1 , and hence 1 1 < 2k+1 n https://doi.org/10.1515/9783110784091-009

or

1 2 < . 2k n

278 | Appendix A And so 2 󵄨 󵄨 μ({x ∈ X : 󵄨󵄨󵄨fn (x) − 0󵄨󵄨󵄨 ≥ ϵ}) < , n which means 󵄨 󵄨 lim μ({x ∈ X : 󵄨󵄨󵄨fn (x) − 0󵄨󵄨󵄨 ≥ ϵ}) = 0.

n→∞ μ

So fn → 0. Next, we show that {fn (x)}n∈ℕ does not converge to 0 for any x ∈ [0, 1]. Observe that 2k −1

[0, 1] = ⋃ [ j=0

j j+1 , ]. 2k 2k

Hence, if x ∈ [0, 1] then x ∈ [ 2jk , j+1 ] for some j; 0 ≤ j ≤ 2k − 1 < 2k , this is true for 2k all k ∈ ℕ. Hence fn (x) = 1 for infinitely many n. This means lim sup fn (x) = 1. n→∞

So limn→∞ fn (x) = 0 is not true for each x ∈ X. (b) Let fn , n ≥ 1 be measurable functions on the Lebesgue measure space (ℝ, L , m), defined by fn = χ[n,n+1] . Then limn→∞ fn (x) = f (x) = 0 for all x ∈ ℝ. But 󵄨 󵄨 m({x ∈ ℝ : 󵄨󵄨󵄨fn (x)󵄨󵄨󵄨 ≥ 1}) = m([n, n + 1]) = 1 for all n ∈ ℕ and hence {fn (x)}n∈ℕ does not converge to f in measure. Thus pointwise convergence or convergence a. e. does not imply convergence in measure. Note that the above examples show that convergence in measure neither implies nor is implied by convergence pointwise or convergence a. e. Theorem A.1 (Uniqueness). Let (X, A , μ) be a measurable space and let {fn }n∈ℕ be a sequence of real-valued measurable functions defined on X. Let f and g be a real-valued μ

μ

measurable functions defined on X such that fn → f and fn → g. Then f = g μ-a. e.

A.1 Convergence in measure

| 279

Proof. Let ϵ > 0. Then 󵄨 ϵ 󵄨 󵄨 󵄨 {x ∈ X : 󵄨󵄨󵄨f (x) − g(x)󵄨󵄨󵄨 ≥ ϵ} ⊂ {x ∈ X : 󵄨󵄨󵄨fn (x) − f (x)󵄨󵄨󵄨 ≥ } 2 󵄨 ϵ 󵄨 ∪ {x ∈ X : 󵄨󵄨󵄨fn (x) − g(x)󵄨󵄨󵄨 ≥ }. 2 And we deduce that 󵄨 󵄨 μ({x ∈ X : 󵄨󵄨󵄨f (x) − g(x)󵄨󵄨󵄨 ≥ ϵ}) = 0 μ

μ

since fn → f and fn → g, respectively. The result follows from the relation ∞ 󵄨 󵄨 󵄨 󵄨 1 {x ∈ X : 󵄨󵄨󵄨f (x) − g(x)󵄨󵄨󵄨 > 0} = ⋃ {x ∈ X : 󵄨󵄨󵄨f (x) − g(x)󵄨󵄨󵄨 ≥ }. n n=1

Theorem A.2. Let {fn }n∈ℕ converge almost uniformly to f . Then {fn }n∈ℕ converges in measure to f . Proof. Let {fn }n∈ℕ be a sequence of measurable functions such that fn → f almost uniformly on X. Hence given ϵ > 0 there exists A ∈ A such that μ(X \ A) < ϵ and fn → f uniformly on A. Hence there exists N ∈ ℕ such that |fn (x) − f (x)| < ϵ for all x ∈ A whenever n ≥ N. Thus for n ≥ N, 󵄨 󵄨 {x ∈ X : 󵄨󵄨󵄨fn (x) − f (x)󵄨󵄨󵄨 ≥ ϵ} ⊂ (X \ A). Hence for n ≥ N, 󵄨 󵄨 μ({x ∈ X : 󵄨󵄨󵄨fn (x) − f (x)󵄨󵄨󵄨 ≥ ϵ}) ≤ μ(X \ A) < ϵ, μ

which means that fn → f . Theorem A.3. Let (X, A , μ) be a measure space with μ(X) < ∞ and let {fn }n∈ℕ converge a. e. to f . Then {fn }n∈ℕ converges in measure to f . Proof. Let En (ϵ) = {x ∈ X : |fn (x) − f (x)| ≥ ϵ}. Then ∞

En (ϵ) ⊆ ⋃ Em (ϵ). m≥n

Note that {⋃∞ m≥n Em (ϵ)}n∈ℕ is a decreasing sequence of sets in A . Since we have ∞

∞ ∞

m≥n

n=1 m=n

lim μ( ⋃ Em (ϵ)) = μ( ⋂ ⋃ Em (ϵ))

n→∞

280 | Appendix A due to fn → f a. e. and ∞ ∞

⋂ ⋃ Em (ϵ) ⊆ {x ∈ X : fn (x) does not converge to f (x)},

n=1 m=n

we have ∞ ∞

μ( ⋂ ⋃ Em (ϵ)) = 0. n=1 m=n

Hence ∞

∞ ∞

m=n

n=1 m=n

lim μ(En (ϵ)) ≤ lim μ( ⋃ Em (ϵ)) = μ( ⋂ ⋃ Em (ϵ)) = 0,

n→∞

n→∞

μ

which implies limn→∞ μ(En (ϵ)) = 0, thus also fn → f . Definition A.2. Let (X, A , μ) be a measure space. Let {fn }n∈ℕ be a sequence of measurable functions on X. We say that {fn }n∈ℕ is Cauchy in measure if for every ϵ > 0, 󵄨 󵄨 lim μ({x ∈ X : 󵄨󵄨󵄨fn (x) − fm (x)󵄨󵄨󵄨 ≥ ϵ}) = 0.

n,m→∞

Theorem A.4. If fn → f in measure, then {fn }n∈ℕ is Cauchy in measure. Proof. Let ϵ > 0 be given. If x ∈ X is such that 󵄨󵄨 󵄨 ϵ 󵄨󵄨fn (x) − f (x)󵄨󵄨󵄨 < 2

󵄨 󵄨 ϵ and 󵄨󵄨󵄨fm (x) − f (x)󵄨󵄨󵄨 < , 2

then 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨󵄨fn (x) − fm (x)󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨fn (x) − f (x)󵄨󵄨󵄨 + 󵄨󵄨󵄨fm (x) − f (x)󵄨󵄨󵄨 < ϵ, which implies that 󵄨 󵄨 󵄨 󵄨 ϵ {x ∈ X : 󵄨󵄨󵄨fn (x) − fm (x)󵄨󵄨󵄨 < ϵ} ⊃ {x ∈ X : 󵄨󵄨󵄨fn (x) − f (x)󵄨󵄨󵄨 < } 2 󵄨 󵄨 ϵ ∩ {x ∈ X : 󵄨󵄨󵄨fm (x) − f (x)󵄨󵄨󵄨 < }. 2 And so 󵄨 󵄨 󵄨 󵄨 ϵ {x ∈ X : 󵄨󵄨󵄨fn (x) − fm (x)󵄨󵄨󵄨 ≥ ϵ} ⊂ {x ∈ X : 󵄨󵄨󵄨fn (x) − f (x)󵄨󵄨󵄨 ≥ } 2 󵄨 󵄨 ϵ ∪ {x ∈ X : 󵄨󵄨󵄨fm (x) − f (x)󵄨󵄨󵄨 ≥ }. 2

A.1 Convergence in measure

| 281

Thus we also have 󵄨 ϵ 󵄨 󵄨 󵄨 μ({x ∈ X : 󵄨󵄨󵄨fn (x) − fm (x)󵄨󵄨󵄨 ≥ ϵ}) ≤ μ({x ∈ X : 󵄨󵄨󵄨fn (x) − f (x)󵄨󵄨󵄨 ≥ }) 2 󵄨 ϵ 󵄨 + μ({x ∈ X : 󵄨󵄨󵄨fm (x) − f (x)󵄨󵄨󵄨 ≥ }), 2 which converges to 0 as n, m → ∞, so that {fn }n∈ℕ is Cauchy in measure. Theorem A.5 (Riesz). Let fn → f in measure. Then there exists a subsequence {fnk }k∈ℕ of {fn }n∈ℕ such that fnk → f a. e. [μ] in X. Proof. We shall construct by induction a strictly increasing sequence of positive integers {nk }k∈ℕ such that 1 󵄨 󵄨 1 μ({x ∈ X : 󵄨󵄨󵄨fnk − f (x)󵄨󵄨󵄨 ≥ }) < k . k 2 Since fn → f in measure, we apply the definition and find n1 ∈ ℕ such that 1 󵄨 󵄨 μ({x ∈ X : 󵄨󵄨󵄨fn1 (x) − f (x)󵄨󵄨󵄨 ≥ 1}) < . 2 Suppose n1 < n2 < ⋅ ⋅ ⋅ < nk have been chosen such that 1 󵄨 󵄨 1 μ({x ∈ X : 󵄨󵄨󵄨fni (x) − f (x)󵄨󵄨󵄨 ≥ }) < i , i 2 with 1 ≤ i ≤ k. Since fn → f in measure, there exists N ∈ ℕ such that 1 1 󵄨 󵄨 }) < k+1 μ({x ∈ X : 󵄨󵄨󵄨fnk (x) − f (x)󵄨󵄨󵄨 ≥ k+1 2 whenever n ≥ N. Choose nk+1 > max{N, nk }. Then nk+1 > nk and 1 1 󵄨 󵄨 }) < k+1 . μ({x ∈ X : 󵄨󵄨󵄨fnk+1 (x) − f (x)󵄨󵄨󵄨 ≥ k+1 2 Thus the construction of {nk }k∈ℕ is complete by induction. Now, we show that fnk → f a. e. [μ]. For each k ∈ ℕ, let 󵄨 󵄨 1 Ek = {x ∈ X : 󵄨󵄨󵄨fnk − f (x)󵄨󵄨󵄨 ≥ }. k

(A.1)

282 | Appendix A Then each Ek ∈ A and μ(Ek ) < Ak+1 ⊂ Ak and

1 . Let Ak 2k

= ⋃n≥k En (k ∈ ℕ) and B = ⋂∞ k=1 Ak . Note that

∞

∞

n=1

n=1

1 = 1. n 2 n=1 ∞

μ(A1 ) = μ( ⋃ En ) ≤ ∑ μ(En ) ≤ ∑ Hence μ(A1 ) < ∞. We have μ(B) = lim μ(Ak ). k→∞

But ∞

μ(Ak ) = μ(⋃ En ) ≤ ∑ n≥k

n=k

1 1 = . 2n 2k−1

Hence μ(B) = limk→∞ μ(Ak ) = 0. Claim. The following holds limk→∞ fnk (x) = f (x) for all x ∈ X \ B. Proof of claim. Let x ∈ X \ B. Then there exists Nx ∈ ℕ such that x ∈ ANx , and thus x ∉ Ek for k ≥ Nx . Letting ϵ > 0, choose M > 0 such that M1 < ϵ. Let N = max{Nx , M}. For k ≥ N, we have x ∉ Ek , which means 1 1 󵄨󵄨 󵄨 1 ≤ t}) ≤ ∫ |f |dμ t

(A.2)

X

for every t > 0. Proof. Since tχ{x∈X:|f (x)|>t} ≤ |f |, we have 󵄨 󵄨 tμ({x ∈ X : 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 > t}) = ∫ tχ{x∈X:|f (x)|>t} dμ ≤ ∫ |f |dμ. X

X

A.1 Convergence in measure

| 283

Hence 1 󵄨 󵄨 μ({x ∈ X : 󵄨󵄨󵄨f (x)󵄨󵄨󵄨 > t}) ≤ ∫ |f |dμ. t X

Remark 24. The significance of inequality (A.2) is that it estimates the size of f in terms of the integral of f . Recall that A ⊂ X is nowhere dense in X if int(A) = 0. Note that A is nowhere dense in X, then X \ A is dense in X. Definition A.3. A subset E of X is a set of the first category in Y if E can be expressed as a countable union of nowhere dense sets. Example 17. ℚ is of the first category in ℝ. A set is said to be of the second category it is not of the first category. Theorem A.7 (Baire). Let (X, d) be a complete metric space. The intersection of a countable number of dense open sets in X is dense in X. Corollary A.1 (Baire category theorem). A complete metric space is a set of the second category. (As a subset of itself.) Proof. Assume the conclusion of the theorem is not true. Let X be a complete metric space which is of the first category, then X = ⋃∞ n=1 En where each En is nowhere dense in X. We have ∞

∞

∞

n=1

n=1

n=1

⋃ E ⊂ X = ⋃ En ⊂ ⋃ E n ,

i. e., ∞

X = ⋃ E. n=1

Hence ∞

0 = ⋂ (E)c . n=1

Write Fn = (E)c = X \ E n , then each Fn is dense in X. Thus 0 = ⋂∞ n=1 Fn , a contradiction to Baire’s theorem. This proves Baire category theorem. Lemma A.1. Let {fi : i ∈ I} be a family of real-valued continuous functions defined on a complete metric space X. Suppose that for all x ∈ X, there exists Mx ∈ ℝ such that

284 | Appendix A |fi (x)| ≤ Mx for i ∈ I. Then there exist a nonempty open set O ⊂ X and M > 0 such that 󵄨 󵄨󵄨 󵄨󵄨fi (x)󵄨󵄨󵄨 ≤ M

for all x ∈ O,

for i ∈ I. Proof. For each k ∈ ℕ and all i ∈ I, define Eki = {x ∈ X : |fi (x)| ≤ k} and let Ek = ⋂ Eki

(k ∈ ℕ).

i∈I

Since fi is continuous, each Eki is a closed set in X. This means each Ek is closed, being an intersection of closed sets. Claim. What follows it is true X = ⋃∞ k=1 Ek . Proof of claim. Clearly, ∞

⋃ Ek ⊂ X.

n=1

(A.3)

Now, let x ∈ X and choose k ≥ Mx , k ∈ ℕ. Then |fi (x)| ≤ Mx ≤ k, so ∞

X ⊂ ⋃ Ek . n=1

(A.4)

By (A.3) and (A.4), we have ∞

X = ⋃ Ek . n=1

Since X is a complete metric space, by the category theorem, there exists M ∈ ℕ such that EM is not nowhere dense, i. e., int(E M ) ≠ 0 thus int(EM ) ≠ 0. Let O = int(EM ). Then O is nonempty open and x ∈ O implies x ∈ EM , so 󵄨󵄨 󵄨 󵄨󵄨fi (x)󵄨󵄨󵄨 ≤ M

for all i ∈ I,

proving Lemma A.3. Theorem A.8 (Uniform boundedness principle or Banach–Steinhaus theorem). Let X be a Banach space and {Ti : i ∈ I} a family of bounded linear transformations from X into a normed linear vector space Y. Suppose that for all x ∈ X there exists Mx ∈ ℝ such that ‖Ti x‖ ≤ Mx for all i ∈ I. Then {Ti : i ∈ I} is uniformly bounded, i. e., there exists M such that ‖Ti ‖ ≤ M for i ∈ I. Proof. For each i ∈ I, define fi : X 󳨀→ ℝ by fi (x) = ‖Ti (x)‖ (x ∈ X). Each fi is continuous on X (fi = ‖⋅‖∘Ti ). By hypothesis, for all x ∈ X there exists Mx ∈ ℝ such that ‖Ti (x)‖ ≤ Mx for all i ∈ I, i. e., |fi (x)| ≤ Mx for all x ∈ X and all i ∈ I.

A.1 Convergence in measure

| 285

Hence by Lemma A.1, there exist a nonempty open set O ⊂ X and C such that |fi (x)| ≤ C for all x ∈ O and all i ∈ I. Since O ≠ 0, there exist y ∈ O and r > 0 such that B(y, r) ⊂ O. Hence 󵄩 󵄩󵄩 󵄩󵄩Ti (z)󵄩󵄩󵄩 ≤ C

for all z ∈ B(y, r) and all i ∈ I.

(A.5)

Let x ∈ X be arbitrary with ‖x‖ ≤ 1, then y + 2r x ∈ B(y, r) and thus, by (A.5), 󵄩󵄩 r 󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩Ti (y + x)󵄩󵄩󵄩 ≤ C 󵄩󵄩 2 󵄩󵄩

for all i ∈ I.

Hence 󵄩󵄩 󵄩󵄩 r 󵄩󵄩󵄩 󵄩󵄩󵄩 r 󵄩 󵄩󵄩 󵄩󵄩Ti ( x)󵄩󵄩󵄩 = 󵄩󵄩󵄩Ti (y + x − Ti (y))󵄩󵄩󵄩 󵄩󵄩 2 󵄩󵄩 󵄩󵄩 2 󵄩󵄩 󵄩󵄩 󵄩󵄩 r 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩Ti (y + x)󵄩󵄩󵄩 + 󵄩󵄩󵄩Ti (y)󵄩󵄩󵄩 󵄩󵄩 2 󵄩󵄩 ≤ 2C for all i ∈ I, i. e., r 󵄩󵄩 󵄩 󵄩T (x)󵄩󵄩 ≤ 2C 2󵄩 i 󵄩

for all i ∈ I,

󵄩󵄩 󵄩 4C 󵄩󵄩Ti (x)󵄩󵄩󵄩 ≤ r

for all i ∈ I.

or

But x ∈ X was arbitrary with ‖x‖ ≤ 1. Hence by the definition of norm, 4C r

for all i ∈ I.

‖Ti ‖ ≤ M

for all i ∈ I.

‖Ti ‖ ≤ Choosing M =

4C , r

we have

This proves the Banach–Steinhaus theorem. Theorem A.9. Suppose f : X × [a, b] → ℂ with −∞ < a < b < ∞ and that f (⋅, t) : X → ℂ is integrable for each t ∈ [a, b]. Define F(t) = ∫ f (x, t)dμ(x) X

for each t ∈ [a, b]. Then:

286 | Appendix A (I)

If there exists g ∈ L1 (μ) such that |f (x, t)| ≤ g(x) for all x, t and limt→t0 f (x, t) = f (x, t0 ) for every x, then lim F(t) = F(t0 ).

t→t0

In particular, if f (x, ⋅) is continuous for every x, then F is continuous. exists and there exists h ∈ L1 (μ) such that | 𝜕f (x,t) | ≤ h(x) for all x, t, then F is (II) If 𝜕f 𝜕t 𝜕t differentiable and 𝜕F(t) 𝜕f (x, t) =∫ dμ(x) 𝜕t 𝜕t X

for all t ∈ [a, b]. We may write the above equality as d 𝜕f (x, t) dμ(x). ∫ f (x, t)dμ(x) = ∫ dt 𝜕t X

X

Proof. (I) Let {tn }n∈ℕ be any sequence in [a, b] such that tn → t0 and define fn (x) = f (x, tn ) and f0 (x) = f (x, t0 ) for every x ∈ X. Hence fn → f0 pointwise and |fn | ≤ g for all n ∈ ℕ. Thus applying Lebesgue dominated convergence theorem, we have ∫ f0 dμ = lim ∫ fn dμ, n→∞

X

X

that is, F(t0 ) = ∫ f (x, t0 )dμ(x) = ∫ f0 (x)dμ(x) X

X

= lim ∫ fn (x)dμ(x) n→∞

X

= lim ∫ f (x, tn )dμ(x) n→∞

X

= lim F(tn ). n→∞

Since this is true for every sequence {tn }n∈ℕ in [a, b] converging to t0 , the result follows. (II) Take again any sequence {tn }n∈ℕ in [a, b] with tn → t0 such that tn ≠ t0 for all n ∈ ℕ and define hn (x) =

f (x, tn ) − f (x, t0 ) t − t0

A.1 Convergence in measure

| 287

for all n ∈ ℕ and x ∈ X. Define also h0 (x) =

𝜕f (x, t0 ) 𝜕t

for all x ∈ X. Hence hn → h0 and, since each hn is measurable, it follows that h0 is measurable. By the mean value theorem, we have 󵄨󵄨 󵄨 |f (x, tn ) − f (x, t0 )| 󵄨󵄨hn (x)󵄨󵄨󵄨 = |t − t | n

0

󵄨󵄨 𝜕f (x, t) 󵄨󵄨 󵄨 󵄨󵄨 ≤ sup 󵄨󵄨󵄨 󵄨 𝜕t 󵄨󵄨󵄨 t∈[a,b]󵄨󵄨

≤ h(x),

for all x ∈ X, and we can again apply Lebesgue dominated convergence theorem to obtain ∫ h0 dμ = lim ∫ hn dμ, n→∞

X

X

that is, F(tn ) − F(t0 ) dF(t) = lim n→∞ dt tn − t = lim ∫ hn (x)dμ(x) n→∞

X

= ∫ h0 (x)dμ(x) X

=∫ X

𝜕f (x, t0 ) dμ(x). 𝜕t

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https://doi.org/10.1515/9783110784091-010

Index L2 (ℝ) 21 ⟨f , g⟩L2 23 σ(E) 60

Dirichlet function 132 Distribution function 129 Dual space 29 Dyadic subcubes properties 160

Adjoint 32 – properties 32

Equivalence – class of functions 21

Banach space 6 Bessel functions 89 – definition 89 Bounded – sequence 15

Fejer kernel 47 Fourier coefficients 39 Fourier inverse formula 108 Fourier series 37 – diverges 43 Fourier transform – basic properties 68 – example 67 – L2 (ℝ) 83 – on L1 (ℝn ) 86 – properties 86 – on L2 (ℝ) 80 – on ℝ 67 – properties 104 Fourier transform in L2 (ℝn ) 114 Function – Dirichlet 20 – distribution 129 – equivalent 21 – good 163, 247 – Hardy–Littlewood maximal 146 – locally integrable 146 – maximal 146 – periodic 37 – square integrable 17 – steep 25 Functional – ‖ ⋅ ‖ 17

Calderón–Zygmund – kernel 187 – singular integral operators 187 Calderón–Zygmund decomposition 160 Calderón–Zygmund decomposition for a function 163 Calderón–Zygmund decomposition of ℝn 160 Cauchy in measure 280 Cauchy sequence 5 Cauchy–Schwarz – inequality 30 Cavalieri principle 237 Cavalieri’s principle 133, 144 Cesáro means 46 Complete 25 Complete normed space 6 Completeness 99 Cone 172 Convergence – in measure 277 – strong 13 – weak 13, 16 Convergent sequence 5 Convolution 60, 64 Covering lemma 145 Cube 159 – closed 159 – dyadic 159 – side length 159 Dense 15 Derivation of the Hilbert transform on ℝ 171

https://doi.org/10.1515/9783110784091-011

Gamma function 236 Hilbert, David 1 Hilbert transform 177 – a constant function 197 – as an operator in L2 225 – basic properties 211 – convolution property 214 – differentiation property 224

292 | Index 1 – of x+iα 202 – of sin x and cos x 198 – of periodic function 205, 206 – on the circle 181 – on the real line 209 – product 217 – truncated 203 Hölder – condition 203 – inequality 134 Hormander condition 187

Identity – Parseval I 58 – Parseval II 59 – polarization 6 Inequality – Bessel 13, 54 – Cauchy–Schwarz 18, 19, 24, 30, 31 – Chebyschev 24 – Hardy 35 – Heisenberg 84 – Kolmogorov 167 – Peley–Zygmund 28 – Tchebyshev 282 – triangle 19 – Wirtinger 60 Inner product 2, 28 – L2 (ℝ) 23 Inner product space 12 – orthonormal 12 Inner products 1 Inverse formula 251 Isometry 29, 33 Kernel 29 – Poisson 202 Laplace equation 125 Laplace transform 200 Lebesgue set 155 Lemma – Fatou 24 – Riemann–Lebesgue 40 Limit – nontangential 173 Linear isometry 29 Linear operator 28 – bounded 29

– injective 29 Linear transformation 30 Marcinkiewicz interpolation theorem 165 Minkowski’s inequality 134 Norm 21, 22 – operator 29 Norm on X 5 Normed linear space 5 Operator – adjoint 29, 31 – Hardy 35 – Hardy–Littlewood maximal 165 – heat 127 – identity 31 – strong type (p, q) 167 – unitary 33 – Volterra 35 – weak type (p, q) 164 Operator of type (p, q) 164 Orthogonal basis 37 Orthogonal complement 28 Orthogonal system 37 Parseval-type form 227 Poisson – conjugate Poisson kernel 181 – kernel 182, 186 Poisson conjugate kernel 204 Poisson equation 125 Properties – orthogonal complement 28 Reflexive 21 Relation – equivalence 21 – equivalent 21 Riemann zeta function 236 Riesz transform – vectorial 274 Schwartz spaces 95 Seminorm 99 Sequence – Cauchy 23 – weakly convergent 15 Space – L2 (ℝ) 17

Index | 293

– complete 23 – Hilbert 6, 25, 29 – inner product 7 – Lebesgue measure 24 – normed 29 – null 29 – quotient 21 – vector 21 Subcube 159 – dyadic 159 Subset – first category 283 – second category 283 Subspace – closed 28 – closed vector of a Hilbert space 28 Symmetric 21 The Cauchy principal value 167 Theorem – Alaouglu’s 185 – Baire 283 – Baire category 283 – Banach–Steinhaus 284 – Bedrosian 217 – Cauchy–Schwarz’s inequality 3

– Dini’s convergence 45 – E. M. Stein and G. Weiss 230 – Fejér 51 – Fubini 24 – Hadamard’s three-lines 135 – Haussdorff–Young 143 – Hilbert inversion 218 – inversion 77 – Kolmogorov 267 – Lebesgue differentiation 152 – M. Riesz 244 – Marcinkiewicz interpolation 165 – Plancherel 82 – Riemann–Lebesgue 25 – Riemman–Lebesgue 73 – Riesz 24, 29, 30, 281 – Riesz–Thorin 138 – shifting hats 79 – Stein–Weiss 235 Transitive 21 Triangle inequality 4 Uniform boundedness principle 284 Weak-Lp space 150

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