Real Harmonic Analysis

Real Harmonic Analysis Published by ANU eView The Australian National University Canberra ACT 0200, Australia Email: [email protected] This title is also available online at http://eview.anu.edu.au

National Library of Australia Cataloguing-in-Publication entry

Author: Auscher, Pascal.

Title: Real harmonic analysis / lectures by Pascal Auscher with the assistance of Lashi Bandara.

ISBN: 9781921934070 (pbk.) 9781921934087 (ebook)

Subjects: Mathematical analysis. Mathematics.

Other Authors/Contributors: Bandara, Lashi.

Dewey Number: 510

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying or otherwise, without the prior permission of the publisher.

Printed by Griffin Press

Lectures by Pascal Auscher with the assistance of Lashi Bandara

Contents

1 Measure Theory 6

2 Coverings and cubes 8

2.1 Vitali and Besicovitch ...... 8

2.2 Dyadic Cubes ...... 12

2.3 Whitney coverings ...... 13

3 Maximal functions 18

n 3.1 Centred Maximal function on R ...... 18

3.2 Maximal functions for doubling measures ...... 21

3.3 The Dyadic Maximal function ...... 23

3.4 Maximal Function on Lp spaces ...... 25

4 Interpolation 30

4.1 Real interpolation ...... 30

4.2 Complex interpolation ...... 33

5 Bounded Mean Oscillation 39

5.1 Construction and properties of BMO ...... 39

5.2 John-Nirenberg inequality ...... 43

5.3 Good λ inequalities and sharp maximal functions ...... 46

i 6 Hardy Spaces 52

6.1 Atoms and H1 ...... 52

6.2 H1 − BMO Duality ...... 55

7 Calder´on-Zygmund Operators 59

7.1 Calder´on-Zygmund Kernels and Operators ...... 59

7.2 The Hilbert Transform, Riesz Transforms, and The Cauchy Operator . . . . 61

7.2.1 The Hilbert Transform ...... 61

7.2.2 Riesz Transforms ...... 63

7.2.3 Cauchy Operator ...... 65

p 7.3 L boundedness of CZOα operators ...... 66

7.4 CZO and H1 ...... 71

7.5 Mikhlin multiplier Theorem ...... 73

7.6 Littlewood-Paley Theory ...... 75

8 Carleson measures and BMO 80

8.1 Geometry of Tents and Cones ...... 81

8.2 BMO and Carleson measures ...... 84

9 Littlewood-Paley Estimates 95

10 T (1) Theorem for Singular Integrals 100

1 Preface

Pascal Auscher recently spent a year at ANU on leave from Universit´eParis-Sud, where he is Professor of Mathematics. During this time he taught this graduate level course covering fundamental topics in modern harmonic analysis.

Auscher has made substantial contributions to harmonic analysis and partial diﬀerential equations, in a wide range of areas including functional calculi of operators, heat kernel estimates, Hardy spaces, weighted norm estimates and boundary value problems.

In particular his contributions were essential to the recent solution of a long-standing conjecture known as the Kato square root problem. This involved substantial new developments concerning the so-called Tb theorems and their application to singular integral operators or more generally to the functional calculus of operators which satisfy Davies- Gaﬀney estimates. These concepts were fundamental in the solution of the Kato problem on square roots of elliptic operators by Auscher, Hofmann, Lacey, Tchamitchian and myself.

Auscher has two published books to his name: Square Root problem for divergence operators and related topics, (with Ph. Tchamitchian) Ast´erisquevol 249, Soc. Math. de France 1998; and On necessary and suﬃcient conditions for Lp estimates of Riesz transforms associated to elliptic operators on Rn and related estimates, Memoirs of the American Math. Society vol. 186, Am. Math. Soc. 2007.

In this lecture course, Auscher provides a basic grounding in the advanced mathematics required for tackling problems in modern harmonic analysis and applying them, for example to partial diﬀerential equations. He does this from the point of view of an expert who fully understands the signiﬁcance of the more basic material. Throughout the manuscript the material is developed in novel ways including some original proofs, derived from the advanced outlook of the author.

Thus the book is a mixture of expository material developed from a contemporary perspective and new work. It is likely to serve two audiences, both graduate students, and established researchers wishing to familiarise themselves with modern techniques of harmonic analysis. The lecture notes were taken and written up by an ANU PhD student, Lashi Bandara, thus providing the ﬁnal polished account of the course material.

Alan McIntosh, Professor of Mathematics, MSI, ANU

2 Introduction

This book presents the material covered in graduate lectures delivered at the Australian National University in 2010. This material had also been partially presented at the Uni- versity Paris-Sud before but never published.

Real Harmonic Analysis originates from the seminal works of Zygmund and Calder´on, pursued by Stein, Weiss, Feﬀerman, Coifman, Meyer and many others.

Moving from the classical periodic setting to the real line, then to higher dimensional Euclidean spaces and ﬁnally to, nowadays, sets with minimal structures, the theory has reached a high level of applicability. This is why it is called real harmonic analysis: the usual exponential functions have disappeared from the picture. Set and function decomposition prevail.

This development has serve to solve famous conjectures in the field. The first one is the boundedness of the Cauchy integral operator on Lipschitz curves by Coifman, McIntosh and Meyer thirthy years ago. In the last ten years, there has been the solution of the Kato conjecture at the intersection of partial differential equations, functional analysis and harmonic analysis, and of the Painlevéconjecture at the border of complex function theory and geometric analysis. We mention also the breakthroughs on weighted norm inequalities with sharp behavior on the weight constants that are occurring at the moment, many of the articles being still in submitted form.

Nowadays all these developments bear on variations of boundedness criteria for singular integral operators and quadratic functionals called Tb theorems after the pionering work of McIntosh and Meyer followed by David, Journéand Semmes. These are powerful and versatile tools. However, they represent more advanced topics than such lectures could cover: these would be a follow-up topic. We have chosen to prepare the grounds to more advanced reading by presenting the basic material that is considered as “well-known” by experts. Nevertheless, anyone who wants to explore this field or apply it to some other problem must know the material we have chosen here. These lecture notes are therefore introductory to the field and accessible to beginners. They do not pretend to cover everything but a selection of important topics. Even if some lecture notes like this one exist, there is a need for updates as they do not cover the same material or address different points.

Let us present it in some more detail. We cover ten chapters: Measure theory, Coverings and cubes, Maximal functions, Interpolation, Bounded Mean Oscillation, Hardy spaces,

3 Calder´on-Zygmund Operators, Carleson measures and BMO, Littlewood-Paley estimates and the T1 theorem for Singular Integrals.

The first chapter is to recall some notation and results from integration theory. The second chapter presents the fundamental tools of modern real harmonic analysis: covering lemmas of various sorts (Vitali, Besicovitch, Whitney). We have chosen to mostly restrict our setting to the Euclidean space to emphasize the ideas rather than the technicalities, Except for the Besicovitch covering lemma, of which we give a proof, they all extend to very general settings and we have tried to give proofs that are not too different from the ones in the extended settings. Maximal functions are treated in Chapter 3. These central objects can be declined in several forms: centred, uncentred, dyadic. They all serve the same goal: decompose the space into sets of controlled sizes according to a function. Chap- ter 4 on interpolation is rather standard and we present a simple as possible treatment. Chapter 5 on bounded mean oscillation restricts its attention to the main properties of BMO functions, the John-Nirenberg inequality and the technique of good lambda inequalities to prove the sharp function inequality of Fefferman-Stein. We also cover the recently developed good lambda inequality with two parameters which is useful for proving inequalities in restricted ranges of the exponents of the Lebesgue spaces. Chapter 6 presents atomic Hardy spaces: Coifman-Weiss’ equivalence of the atomic definitions with p-atoms with 1 < p ≤ ∞ as a consequence of Calderón-Zygmund decompositions on functions, and applications to the Fefferman duality theorem. Chapter 7 is also an introduction to the theory of Calderón-Zygmund operators and some applications to multipliers are given towards Littlewood-Paley theorem. Again more advanced results can be obtained form the references. Chapter 8 covers the notion of a Carleson measure and Carleson’s embedding theorem is proved. Then the connection between Carleson measures and BMO functions is studied in detail with care on the subtle convergence issues that are not often treated in the literature. Applications to the Bony-Coifman-Meyer paraproducts are given. Chapter 9 on Littlewood-Paley estimates could also be called T1 theorem for quadratic functionals and is extracted with details from an important article of Christ-Journé.See also my previous book with Philippe Tchamitchian were not all details are given. Almost orthogonality arguments, in particular the Cotlar-Knapp-Stein lemma, are central here. We finish in Chapter 1 with a proof of the T1 theorem of David-Journé. This proof is not the original one but is the one given by Coifman-Meyer and deserves more publicity. Again the strategy of this proof adapts to very general settings.

Special thanks go to Alan McIntosh and to the Mathematical Science Institute by inviting me as a one year visiting fellow at the Australian National University and also to the CNRS for giving me a “d´el´egation”allowing a leave of absence from my home university. A one year visit that all my family has enjoyed and will remember for years.

Alan McIntosh and I have had and still have a long standing collaboration and it is a pleasure to thank him for sharing so many views on mathematics and life. That he accepted to preface this manuscript is a great privilege.

Thanks also go to Lashi Bandara whose typesetting genius made scratch notes he took from the lectures a readable manuscript. He also corrected some mistakes in the proofs (the remaining ones are my responsibility). My former student Fr´ed´ericBernicot read the proofs entirely and found some more typos. I am grateful for that.

4 Addition to the second version: These lectures have been used again by myself at Universit´e Paris-Sud and also by Thierry Coulhon and Dorothee Frey at ANU. I want to thank them for pointing out a number of remaining typos.

5 Chapter 1

Measure Theory

n While we shall focus our attention primarily on R , we note some facts about measures in an abstract setting and in the absence of proofs.

Let X be a set. The reader will recall that in the literature, a measure µ on X is usually defined on a σ-algebra M ⊂ P(X). This approach is limited in the direction we will take. In the sequel, it will be convenient to forget about measurability and associate a size to arbitrary subsets but still in a meaningful way. We present the Carathéodory’s notion of measure and measurability. In the literature, what we call a measure is sometimes distinguished as an outer measure. Definition 1.0.1 (Measure/Outer measure). Let X be a set. Then, a map µ : P(X) → [0, +∞] is called a measure on X if it satisfies:

(i) µ(∅) = 0, P∞ S∞ (ii) µ(A) ≤ i=1 µ(Ai) whenever A ⊂ i=1 Ai. Remark 1.0.2. Note that (ii) includes the statement: if A ⊂ B, then µ(A) ≤ µ(B).

Certainly, the classical measures are additive on disjoint subsets. This is something we lose in the definition above. However, we can recover both the measure σ-algebra and a notion of measurable set. Definition 1.0.3 (Measurable). Let µ be a measure on X (in the sense of Definition 1.0.1). We say A ⊂ X is µ-measurable if for all Y ⊂ X, µ(A) = µ(A \ Y ) + µ(A ∩ Y ). ∞ Theorem 1.0.4. Let {Ai}i=1 be a countable set of µ-measurable sets. Then,

T∞ S∞ (i) i=1 Ai and i=1 Ai are µ-measurable, ∞ (ii) If the sets {Ai}i=1 are mutually disjoint, then ∞ ! ∞ [ X µ Ai = µ(Ai), i=1 i=1

6 (iii) If Ak ⊂ Ak+1 for all k ≥ 1, then

∞ ! [ lim µ(Ak) = µ Ai , k→∞ i=1

(iv) If Ak ⊃ Ak+1 for all k ≥ 1, then

∞ ! \ lim µ(Ak) = µ Ai , k→∞ i=1

(v) The set M = {A ⊂ X : A is µ − measurable} is a σ-algebra.

A proof of (i) - (iv) can can be found in [GC91, p2]. Then, (v) is an easy consequence and we leave it as an exercise.

The following result illustrates that we can indeed think about classical measures in this framework. Recall that a measure space is a triple (X, M , ν) where M ⊂ P(X) is a σ- algebra and ν : M → [0, +∞] is a measure in the classical sense. Measure spaces are also given as the tuple (X, ν) by which we mean (X, M , ν) where M is the largest σ-algebra containing ν-measurable sets.

Theorem 1.0.5. Let (X, M , ν) be a measure space. Then, there exists a measure µ in the sense of Deﬁnition 1.0.1 on X such that µ = ν on M .

See [G.81, §5.2, Theorem 3].

Of particular importance is the following construction of the Lebesgue measure in our sense.

n Deﬁnition 1.0.6 (Lebesgue Measure). Let A ⊂ R . For a Euclidean ball B, denote the volume of the ball by vol B. Deﬁne the Lebesgue measure L :

( ∞ ∞ ) X [ L (A) = inf vol Bi : A ⊂ Bi and each Bi is an open ball . i=1 i=1

This deﬁnition is justiﬁed by the following proposition.

n 0 0 Proposition 1.0.7. Let (R , M , L ) be the classical Lebesgue measure deﬁned on the largest possible σ-algebra M 0 and let M = {A ⊂ X : A is L − measurable}. Then,

M 0 = M and L (A) = L 0(A) for all A ∈ M 0.

For a more detailed treatment of abstract measure theory, see [GC91, §1] and [G.81, Ch.5].

7 Chapter 2

Coverings and cubes

n We will consider the setting of R with the usual Euclidean norm |· | inducing the standard Euclidean metric dE(x, y) = |x − y|. We note, however, that some of the material that we discuss here can be easily generalised to a more abstract setting.

To introduce some nomenclature, we denote the ball centred at x with radius r > 0 by B(x, r). We are intentionally ambiguous as to whether the ball is open or closed. We will specify when this becomes important.

For a ball B, let rad B denote its radius. For λ > 0, we denote the ball with the same centre but radius λ rad B by λB.

This chapter is motivated by the following two questions.

S (1) Suppose that Ω = B∈B B where B is a family of balls. We wish to extract a subfamily B0 of balls that do not overlap “too much” and still cover Ω.

n (2) Given a set Ω ⊂ R , how can we select a cover of Ω with a given geometric structure.

2.1 Vitali and Besicovitch

n Lemma 2.1.1 (Vitali Covering Lemma). Let {Bα}α∈I be a family of balls in R and suppose that sup rad Bα < ∞. α∈I

Then there exists a subset I0 ⊂ I such that

(i) { } are mutually disjoint. Bα α∈I0 (ii) S ⊂ S 5 . α∈I Bα α∈I0 Bα

Remark 2.1.2. (i) The balls {Bα}α∈I can be open or closed.

8 n (ii) This statement only relies on the metric structure of R with the euclidean metric given by |· |. It can be replaced by a metric space (E, d). As an example, we can set n (E, d) = (R , d∞), where d∞ is the inﬁnity distance given be by the inﬁnity norm. This is equivalent to replacing balls by cubes.

(iii) The condition supα∈I rad Bα < ∞ is necessary. A counterexample is {Bi = B(0, i): i ∈ N}.

Proof. Let M = supα∈I rad Bα. For j ∈ N, deﬁne −j−1 −j I(j) = α ∈ I : 2 M < rad Bα ≤ 2 M .

We inductively extract maximal subsets of each I(j). So, for j = 0, let J(0) be a maximal subset of I(0) such that {Bα}α∈J(0) are mutually disjoint. The existence of such a collection is guaranteed by Zorn’s Lemma. Now, for j = 1, we extract a maximal J(1) ⊂ I(1) such that {Bα}α∈J(1) mutually disjoint and also disjoint from {Bα}α∈J(0). Now, when j = k, we let J(k) ⊂ I(k) be maximal such that {Bα}α∈J(k) mutually disjoint and disjoint from {Bα}α∈∪k−1 J(m). We then let m=0 [ I0 = J(k). k∈N By construction, { } are mutually disjoint. This proves (i). Bα α∈I0

To prove (ii), ﬁx a ball Bα ∈ I. We show that there exists a β ∈ I0 such that Bα ⊂ 5Bβ. We have a k ∈ N such that α ∈ I(k). That is, −k−1 −k 2 M < rad Bα ≤ 2 M.

If i ∈ J(k), then we’re done. So suppose not. By construction, this must mean that Bα must intersect a ball Bβ for β ∈ J(l) where 0 ≤ l ≤ k. But we know that 1 rad B ≥ 2−l−1M ≥ 2−k−1M ≥ rad B β 2 α and by the triangle inequality

d(xα, xβ) ≤ rad Bα + rad Bβ.

Then, rad Bα + d(xα, xβ) ≤ 2 rad Bα + rad Bβ ≤ 5 rad Bβ and so it follows that

Bα = B(xα, rad Bα) ⊂ B(xβ, rad Bα + d(xα, xβ)) ⊂ B(xβ, 5 rad Bβ) = 5Bβ.

Before we introduce our next covering theorem, we require a rigorous notion of a family of balls to not intersect “too much.” First, we note that χX denotes the indicator function of B.

Deﬁnition 2.1.3 (Bounded Overlap). A collection of balls B is said to have bounded overlap if there exists a C ∈ N such that X χB ≤ C. B∈B

9 n Theorem 2.1.4 (Besicovitch Covering Theorem). Let E ⊂ R . For each x ∈ E, let B(x) be a ball centred at x. Assume that E is bounded or that supx∈E rad B(x) < ∞. Then, there exists a countable set E0 ⊂ E and a constant C(n) ∈ N such that

(i) ⊂ S ( ) E x∈E0 B x . (ii) P χ ≤ ( ) x∈E0 B(x) C n . Remark 2.1.5. (i) Here, (ii) means that the set { ( )} forms a bounded covering B x x∈E0 of E with constant C(n). This constant depends only on dimension. The bounded covering property tells us that the balls are “almost disjoint.” In fact, we can organise E = E ∪ ... ∪ E such that each set {B(x)} contain mutually disjoint balls. 0 1 N x∈Ek We will not prove this.

(ii) We can substitute cubes for balls.

(iii) For E unbounded, the condition supx∈E rad B(x) < ∞ is necessary.

n (iv) This theorem is very special to R . A counterexample is the Heisenberg group.

(v) The balls must be centred at each point in E. Otherwise, consider E = [0, 1] and i Bi = [0, 1 − 2 ], i ≥ 1.

Before we proceed to prove the theorem, we require the following Lemma.

n Lemma 2.1.6. For y ∈ R , let Cε(y) be the sector with vertex y with aperture angle ε. Suppose that 0 < ε ≤ π/6. Then, for all R > 0, x, z ∈ Cε(y), if |x| , |z| ≤ R then |x − z| ≤ R.

Proof of Besicovitch Covering Theorem. Let M = supx∈E rad B(x) and suppose that E is bounded and M = ∞. Then, ﬁx an x0 ∈ E and there exists a ball B(x0,R) such that B(x0,R) ⊃ E. Then, we’re done by setting E0 = {x0}.

So, we suppose now that E is bounded and M < ∞. Deﬁne: n o E(k) = x ∈ E : 2−k−1M < rad B(x) ≤ 2−kM .

0 We select points xj inductively from each E(k) to construct a set E (k). So, fix an initial Si−1 x0,0 ∈ E(0) and select x0,i ∈ E(0) by requiring that x0,i 6∈ l=0 B(x0,l). Then, for arbitrary k > 0, assume that E0(0),...,E0(k−1) have already been constructed and construct E0(k) i−1 k−1 0 S 0 by selecting xk,i such that xk,i 6∈ l=0 B(xk,l) ∪m=1 ∪xm,i∈E (m)B(xm,i). Each E (k) must be finite since by the boundedness of E and the definition of E(k), this process must stop after finitely many selections in ( ). Now, let = S 0( ) equipped with the E k E0 k∈N E k natural ordering. That is, E0 = {x1, x2,... } and if i < j then xj 6∈ B(xi). This is the same as saying that xi was selected before xj.

We prove (i). Suppose ∈ but 6∈ S ( ). In particular this means that x E x xi∈E0 B xi x ∈ E0(k) for some k which is a contradiction.

10 n Now, to show (ii), ﬁx y ∈ R , let ε = π/6 and let Cε(y) denote the sector with vertex y and aperture ε. Deﬁne

Ay = {xi ∈ E0 : y ∈ B(xi) and xi ∈ Cε(y)} , and let xi be the ﬁrst element in Ay. Take xj ∈ Ay with j > i. Then,  x , x ∈ C (y)  i j ε |xi − y| ≤ rad B(xi)  |xj − y| ≤ rad B(xj) and so |xi − xj| ≤ max {rad B(xi), rad B(xj)} by application of Lemma 2.1.6. But by the ordering on E0, xj was selected after xi and so xj 6∈ B(xi). Consequently rad B(xi) < |xj − xi| which implies rad B(xj) > rad B(xi).

−k−1 −k Now, suppose that xi was selected at generation k, so 2 M < rad B(xi) ≤ 2 M. Also, −k−1 −k 2 M < rad B(xj) ≤ 2 M for otherwise, xj would be selected at a later generation l > k which implies that rad B(xi) > rad B(xj) which is a contradiction.

−k−2 Also, note that B(xj, 2 M): j ∈ Ay are mutually disjoint and are all contained in −k+1 B(xi, 2 M). It follows that,

X −k−2 −k+1 3 n −k−2 L (B(xj, 2 M)) ≤ L (B(xi, 2 M)) = (2 ) L (B(0, 2 M))

j∈Ay and

−k−2 X X L (B(xj, 2 M)) 3n card Ay = 1 = ≤ 2 L (B(0, 2−k−2M)) j∈Ay j∈Ay and we are done.

We shall not give details for the case that E is unbounded, but it can be obtained from the previous case with some eﬀort. We refer the reader to [GC91, p35].

n n Corollary 2.1.7 (Sard’s Theorem). Let f : R → R and let n L (f(B(x, r))) A = x ∈ R : lim inf = 0 . r→0 L (B(x, r))

Then L (f(A)) = 0. (Recall that by our deﬁnition of L , we can measure every subset of n R ).

Proof. Firstly, we note that for all x ∈ A and ε > 0, there exists an rx ∈ (0, 1] such that

L (f(B(x, rx))) ≤ εL (B(x, rx)).

Let A0 ⊂ A be the set of centres given by Besicovitch applied to the set of balls {Bx = B(x, rx)} for which the above measure condition holds. Then, ( ) ⊂ S ( ( )) and by the f A xi∈A0 f B xi

11 subadditivity of L , X L (f(A)) ≤ L (f(Bxi )) xi∈A0 X ≤ ε L (Bxi ) xi∈A0 X ≤ ε χ (y) d (y) Bxi L ˆA+B(0,1) ≤ εC(n)L (A + B(0, 1)).

Now, if A is bounded, then L (A + B(0, 1)) < ∞ and we can obtain the conclusion by letting ε → 0. Otherwise, we replace A by A ∩ B(0, k) to obtain L (f(A ∩ B(0, k))) = 0 and then by taking k → ∞ we establish L (f(A))) = 0.

It is easy to show that A contains the singular points of f, that is points x at which f is differentiable with vanishing differential df(x) = 0. That is for a differentiable map, almost all points are regular (i.e., non-singular).

2.2 Dyadic Cubes

n We begin with the construction of dyadic cubes on R .

n Deﬁnition 2.2.1 (Dyadic Cubes). Let [0, 1) be the reference cube and let j ∈ Z and n k = (k1, . . . , kn) ∈ Z . Then deﬁne the dyadic cube of generation j with lower left corner 2−jk n j n Qj,k = x ∈ R : 2 x − k ∈ [0, 1) , the set of generation j dyadic cubes

n Qj = {Qj,k : k ∈ Z } , and the set of all cubes

[ n Q = Qj = {Qj,k : j ∈ Z and k ∈ Z } . j∈Z

−j We deﬁne the length of a cube to be its side length `(Qj,k) = 2 .

n Qn Remark 2.2.2. If we were to replace [0, 1) with R = i=1[ai, ai +δ) as a reference cube, n n then the dyadic cubes with respect to R are constructed via the homothety ϕ : R → R where ϕ([0, 1)n) = R.

−j Theorem 2.2.3 (Properties of Dyadic Cubes). (i) For j ∈ Z, `(Qj,k) = 2 , L (Qj,k) = −jn n 2 , and Qj forms a partition of R for each j ∈ Z.

n 0 n (ii) For all j ∈ Z and k ∈ Z , there exists a unique k ∈ Z such that Qj,k ⊂ Qj−1,k0 . We set Qdj,k = Qj−1,k0 and it is called the parent of Qj,k.

12 n (iii) For all j ∈ Z and k ∈ Z , the cubes Qj+1,k0 for which Qj,k is the parent are called the children of Qj,k. n (iv) For all x ∈ R , there exists a unique sequence of dyadic cubes (Qj,kj )j∈Z ⊂ Q such

that Qj+1,kj+1 is a child of Qj,k, Qj,kj is the parent of Qj+1,kj+1 and x ∈ Qj,kj for all j ∈ Z. S (v) Let E ⊂ Q such that Ω = Q∈E Q satisﬁes L (Ω) < ∞. Then, there exists a S 0 collection F ⊂ Q of mutually disjoint dyadic cubes such that Ω = Q0∈F Q .

Proof. We leave it to the reader to verify (i) - (iv). We prove (v).

Deﬁne: n o F = Q0 ∈ Q : Q0 ⊂ Ω but Qc0 6⊂ Ω .

k We prove that F 6= ∅. Let Q ∈ E , and denote the kth parent by Qb . It follows that k n k n k o k0 L (Qb ) = (2 ) L (Q). Now, let k0 = max k : Qb ⊂ Ω and consequently, Qb ∈ F .

S S 0 S 0 Now, note that Q∈E Q ⊂ Q0∈F Q and by the construction of F , Q0∈F Q ⊂ Ω. But S S 0 by hypothesis Ω = Q∈E Q and so it follows that Ω = Q0∈F Q .

To complete the proof, we prove that the cubes in F are mutually disjoint. Let Q0, Q00 ∈ F and assume that Q0 6= Q00. This is exactly that Q0 6⊂ Q00 and Q0 6⊃ Q00 so by (iii), 0 00 Q ∩ Q = ∅. Remark 2.2.4. (i) The way (v) is formulated does not ensure that F ⊂ E . Take for example E = {[0, 1/2), [1/2, 1)} which gives F = {[0, 1)}. (ii) Suppose Ω is an open set with L (Ω) < ∞ and let E = {Q ∈ Q : Q ⊂ Ω}. Then, F ⊂ E and we call F the maximal dyadic cubes in E . Maximality here is with respect to the property Q ⊂ Ω.

k (iii) The assumption L (Ω) < ∞ cannot be dropped. A counterexample is [0, 2 ) ⊂ R for k ∈ N.

2.3 Whitney coverings

The Whitney covering theorems are an important tool in Harmonic Analysis. We initially give a dyadic version of Whitney. But ﬁrst, we recall some terminology from the theory of metric spaces. Recall that the diameter of a set E ⊂ X for a metric space (X, d) is given by diam E = sup d(x, y), x,y∈E and the distance from E to another set F ⊂ X is given by dist(E,F ) = inf d(x, y). x∈E,y∈F Then the distance from E to a point y ∈ X is simply given by dist(x, E) = dist(E,F ) where F = {x}.

13 n Theorem 2.3.1 (Whitney Covering Theorem for Dyadic cubes). Let O $ R be open. Then, there exists a collection of Dyadic cubes F = {Qi}i∈I such that

(i) 1 1 dist(Q , cO) ≤ diam Q ≤ dist(Q , cO), 30 i i 10 i

(ii) [ O = Qi, i∈I

(iii) The dyadic cubes in F are mutually disjoint.

Proof. We deﬁne E as the collection of dyadic cubes Q such that:

(a) Q ⊂ Ω,

1 c (b) diam Q ≤ 10 dist(Q, O), and let F be the maximal subcollection of E as in Remark 2.2.4. This collection F is well deﬁned: let x ∈ O and (Qj,kj )j∈Z the dyadic sequence which contains x. So, there exists a Qj,kj ⊂ Ω and since x is ﬁxed, we can impose the condition (b). This proves (ii), and 1 c (iii) and by construction diam Q ≤ 10 dist(Q, O) for every Q ∈ F . It remains to check the lower bound.

So, take Q ∈ F . Then, by maximality, either Qb 6⊂ Ω or dist(Qb , cO) < 10 diam Qb . Thus, in either case, dist(Q, cO) ≤ 10 diam Qb . Combining this with the fact that diam Qb = 2 diam Q, we ﬁnd

dist(Q, cO) ≤ 10 diam Qb + diam Qb ≤ (20 + 2) diam Q ≤ 30 diam Q, and this completes the proof.

Remark 2.3.2. In (i), the constant 1/10 could be replaced by 1/2 − ε for all ε > 0. However, this would change the constant 1/30 in the lower bound.

The Whitney dyadic cubes introduced in the preceding theorem satisfy some important properties. Proposition 2.3.3. The Whitney Dyadic cubes of O satisfy:

(i) For all i ∈ I, 3Qi ⊂ O,

(ii) For all i, j ∈ I, if 3Qi ∩ 3Qj 6= ∅, then

1 diam Qi ≤ ≤ 4, 4 diam Qj

14 (iii) There exists a constant only depending on dimension, C(n), such that

X χ 3Qi ≤ C(n). i∈I

c Proof. (i) Suppose there exists an z ∈ O ∩ 3Qi. So, there exists a y ∈ Qi such that d(y, z) ≤ diam Qi. But,

c c 10 diam Qi ≤ dist(Qi, O) ≤ dist(y, O) ≤ dist(y, z) ≤ diam Qi which is a contradiction.

(ii) By symmetry, it suﬃces to prove that

diam Qi ≤ 4. diam Qj

Let y ∈ 3Qi ∩ 3Qj. We note that dist(y, Qi) ≤ diam Qi since y ∈ 3Qi, and by the triangle inequality,

c c c 10 diam Qi ≤ dist(Qi, O) ≤ dist(y, O) + dist(y, Qi) ≤ dist(y, O) + diam Qi

c which shows that dist(y, O) ≥ 9 diam Qi. c c Also, there exists z ∈ Qj such that d(y, z) ≤ diam Qj and dist(y, O) ≤ dist(z, O) + c diam Qj. We estimate dist(z, O). Fix w ∈ Qj and we ﬁnd

c c c dist(z, O) ≤ d(z, w) + dist(w, O) ≤ diam Qj + dist(w, O).

c By taking an inﬁmum over all w ∈ Qj and using the fact that dist(Qj, O) ≤ c c 30 diam Qj, we ﬁnd that dist(z, O) ≤ 31 diam Qj and consequently dist(y, O) ≤ 32 diam Qj. Putting these estimates together, we get that

diam Qi 32 ≤ < 4. diam Qj 9

(iii) Fix i ∈ I and let Ai = {j ∈ I : 3Qi ∩ 3Qj 6= ∅}. So, for any y ∈ 3Qi ∩ 3Qj,  dist(y, Q ) ≤ diam Q ,  i i dist(y, Qj) ≤ diam Qj,  dist(Qi, Qj) ≤ diam Qi + diam Qj.

k Let K = {−2, −1, 0, 1, 2}. Then, for any j ∈ Ai, diam Qj = 2 diam Qi where k ∈ K. k k Now for such a k ∈ K, deﬁne Ai = j ∈ Ai : diam Qj = 2 diam Qi . k k If j ∈ Ai , then dist(Qi, Qj) ≤ (1 + 2 ) diam Qi ≤ 5 diam Qi, and in particular, this means that Qj ⊂ 10Qi. But all such {Qj} k are mutually disjoint and j∈Ai S n n k so L k Qj ≤ (10) L (Qi) = 40 L (Qj) for any j ∈ A since diam Qi = j∈Ai i −k 2 diam Qj ≤ 4 diam Qj. Then,

X X L (Qj) card Ak = 1 ≤ ≤ 40n i (Q ) k k L i j∈Ai j∈Ai

15 and this completes the proof.

Theorem 2.3.4 (Whitney Covering Theorem for metric spaces). Let (E, d) be a metric space, and let O $ E be open. Then there exists a set of balls E = {Bα}α∈I and a constant c1 < ∞ independent of O such that

(i) The balls in E are mutually disjoint, S (ii) O = α∈I c1Bα,

(iii) 4c1Bα 6⊂ O.

Proof. Let δ(x) = dist(x, cO), and ﬁx 0 < ε < 1/2 to be chosen later. Deﬁne

B = {B(x, εδ(x)) : x ∈ O} and let E = {Bα}α∈I ⊂ B maximal with mutually disjoint balls. The existence of E is guaranteed by Zorn’s Lemma. Now we set rα = εδ(xα) and c1 = 1/2ε. Then,

4c1Bα = B(xα, 4c1rα) = B(xα, 4· (1/2ε)εδ(xi)) = B(xα, 2δ(xi)) 6⊂ O.

This proves (i) and (iii).

S Now, suppose there exists x ∈ O \ α∈I c1Bα. By maximality, there exists a β ∈ I such that ∅ 6= B(x, rx) ∩ B(xβ, rxβ ). In particular, this implies that 1 d(x, x ) ≤ ε(δ(x) + δ(x )) and d(x, x ) ≥ δ(x ) β β β 2 β and so ε δ(xβ) ≤ 1 δ(x). 2 − ε

Now, trivially, B(xβ, 2δ(xα)) ⊂ B(x, 2δ(x) + d(x, xβ)) and by the inequalities above, " # 2ε ε2 2δ(x) + d(x, xβ) ≤ 1 + ε + 1 δ(x). 2 − ε 2 − ε

Let ϕ(ε) denote the quantity within the square brackets and by putting this together, we ﬁnd that B(xβ, 2δ(xβ)) ⊂ B(x, ϕ(ε)δ(x)). Now, note that ϕ(ε) → 0, so we can choose 0 < ε < 1/2 such that ϕ(ε) < 1. For such a choice of ε, we ﬁnd B(xβ, 2δ(xβ)) ⊂ B(x, ϕ(ε)δ(x)) ⊂ O. But this is a contradiction since B(xβ, 2δ(xβ)) 6⊂ O.

n Proposition 2.3.5. Assume that (E, d) = (R , dE), where dE(x, y) = |x − y|, then {c1Bα}α∈I possesses the bounded covering property.

Proof. Fix α ∈ I and let Aα = {β ∈ I : c1Bα ∩ c1Bβ 6= ∅} . Now, take β ∈ Aα and ﬁx z ∈ c1Bα ∩ c1Bβ. Then, 1 1 d(z, x ) ≤ rad c1B = c1 rad B = εδ(x ) = δ(x ). β β β 2ε β 2 β

16 By the triangle inequality, 1 δ(x ) = dist(x , cO) ≤ dist(z, cO) + d(x , z) ≤ dist(z, cO) + δ(x ) β β β 2 β and we conclude 1 ≤ dist(z, cO). 2 Furthermore, 1 3 dist(z, cO) ≤ dist(x , cO) + d(z, x ) ≤ δ(x ) + δ(x ) = δ(x ). α α α 2 α 2 α Combining these two estimates, and by symmetry we conclude that

1 δ(xβ) ≤ ≤ 3. 3 δ(xα)

ε Let E = B(xβ, 3 δ(xα)) : β ∈ Aα . This collection of balls are mutually disjoint by the previous inequality. Now, 1 1 1 3 d(x , x ) ≤ d(x , z) + d(x , z) ≤ δ(x ) + δ(x ) ≤ δ(x ) + δ(x ) ≤ 2δ(x ). α β α β 2 α 2 β 2 α 2 α α Now, set ε 3 + 2 C = ε , 3 ε ε and combining this with the estimate above, B(xβ, 3 δ(xα)) ⊂ B(xα,C 3 δ(xα)). So again, the volumes of the balls can be compared since with constant C since ε is ﬁxed, and using a volume argument as in the proof of Besicovitch (Theorem 2.1.4) we attain a bound on card Aα depending only on dimension. Remark 2.3.6. We note that the preceding proposition can be proved for a metric space having the following structural property: There exists a constant 0 < C < ∞ such that for all R > 0, the number of mutually disjoint balls of radius R contained in a ball of radius 2R is bounded by C.

17 Chapter 3

Maximal functions

3.1 Centred Maximal function on Rn

We begin with the introduction of the classical Hardy-Littlewood maximal function.

Definition 3.1.1 (Hardy-Littlewood Maximal function). Let µ be a reference measure, Borel, positive, and locally finite. Let ν be a second, positive, Borel measure. Define: 1 ν(B(x, r)) Mµ(ν)(x) = sup dν = sup r>0 µ(B(x, r)) ˆB(x,r) r>0 µ(B(x, r))

n 1 for each x ∈ R . If f ∈ Lloc(dµ), then set dν(x) = |f(x)|dµ(x) and deﬁne 1 Mµ(f)(x) = Mµ(ν)(x) = sup |f| dµ. r>0 µ(B(x, r)) ˆB(x,r)

0 Remark 3.1.2. By convention, we take 0 = 0.

The ﬁrst and fundamental question is to ask the size of Mµ(ν) in terms of ν. Theorem 3.1.3 (Maximal Theorem). Suppose that ν is a ﬁnite measure. Then there exists a constant depending only on the dimension C = C(n) > 0 such that for all λ > 0,

n C n µ {x ∈ R : Mµ(ν)(x) > λ} ≤ ν(R ). λ

Remark 3.1.4. (i) The function Mµ(ν) may not be a Borel function but it is a Lebesgue measurable function. This a reason we wanted our measures to be deﬁned on arbitrary n sets. The proof however does illustrate that {x ∈ R : Mµ(ν)(x) > λ} ⊂ B where B C n is a Borel set with µ(B) ≤ λ ν(R ).

(ii) With some regularity assumptions on µ, Mµ(ν) becomes lower semi-continuous. n That is, for all λ > 0, the set {x ∈ R : Mµ(ν)(x) > λ} is open. For example, suppose the map (x, r) 7→ µ(B(x, r)) is continuous (or equivalently n−1 µ(S (x, r)) = 0 (Exercise)). Then, Mµ(ν) is lower semi-continuous (Exercise).

18 n Proof of Maximal Theorem. Let Oλ = {x ∈ R : Mµ(ν)(x) > λ}. So, for all x ∈ Oλ, there exists Bx = B(x, rx) such that ν(Bx) > λµ(Bx). Fix R > 0, and apply Besicovitch to the set Oλ ∩ B(0,R). So, there is an E ⊂ Oλ ∩ B(0,R) at most countable such that

[ X χ Oλ ∩ B(0,R) ⊂ Bx and Bx ≤ C(n). x∈E x∈E Then,

1 1 X X X χ µ(Oλ ∩ B(0,R)) ≤ µ(Bx) ≤ ν(Bx) = Bx dν λ λ ˆ n x∈E x∈E x∈E R 1 ( ) X χ C n n = Bx dν ≤ ν(R ). λ ˆ n λ R x∈E

The sets Oλ ∩ B(0,R) are increasing as R → ∞, so

C(n) n µ(Oλ) = lim µ(Oλ ∩ B(0,R)) ≤ ν(R ) R→∞ λ which completes the proof.

Corollary 3.1.5. For all f ∈ L1(dµ) and for all λ > 0,

n C(n) µ {x ∈ R : Mµ(f)(x) > λ} ≤ |f| dµ. λ ˆRn 1 Theorem 3.1.6 (Lebesgue Diﬀerentiation Theorem). Let f ∈ Lloc(dµ). Then, there exists n c an Lf ⊂ R such that µ( Lf ) = 0 and 1 lim |f(x) − f(y)| dµ(y) = 0 r→0 µ(B(x, r)) ˆB(x,r) for all x ∈ Lf . In particular, 1 f(x) = lim f(y)dµ(y) r→0 µ(B(x, r)) ˆB(x,r) for all x ∈ Lf .

Remark 3.1.7. 1. This is a local statement, and so we can replace f by fχK where K is any compact set. Consequently we can assume that f ∈ L1(dµ).

2. If f is continuous, then we can take Lf = ∅. That is, the Theorem holds everywhere.

Proof. Deﬁne: 1 ωf (x) = lim sup |f(x) − f(y)| dµ(y) r→0 µ(B(x, r)) ˆB(x,r) and we remark that the measurability of ωf is the same as the measurability of Mµ(f). Note that ωf is subadditive, that is, ωf+g ≤ ωf + ωg. Also, ωf ≤ |f| + Mµ(f).

19 0 n 1 Let ε > 0, and since Cc(R ) is dense in L (dµ) (since µ is locally ﬁnite), there exists a 0 n g ∈ Cc(R ) such that |f − g| dµ < ε. ˆRn

Now, observe that from the previous remark (ii), ωg = 0 and so it follows that

ωf ≤ ωf−g + ωg ≤ |f − g| + Mµ(f − g).

Fix λ > 0 and note that λ λ {x ∈ n : ω (x) > λ} ⊂ x ∈ n : |(f − g)| (x) > ∪ x ∈ n : M (f − g)(x) > . R f R 2 R µ 2

From this, it follows that

λ µ {x ∈ n : ω (x) > λ} ≤ µ x ∈ n : |(f − g)| (x) > + R f R 2 λ µ x ∈ n : M (f − g)(x) > R µ 2 1 C(n) ≤ λ |f − g| dµ + λ |f − g| dµ ˆ n ˆ n 2 R 2 R 1 + C(n) ≤ λ ε. 2

n n Now letting ε → 0, we ﬁnd that µ {x ∈ R : ωf (x) > λ} = 0 and µ {x ∈ R : ωf (x) > 0} = n 1 n limn→∞ µ x ∈ R : ωf (x) > n = 0. To complete the proof, we set Lf = {x ∈ R : ωf (x) > 0}.

Remark 3.1.8 (On the measurability of ωf ). Let

1 ωf; 1 (x) = lim sup |f(x) − f(y)| dµ(y) k 1 µ(B(x, r)) ˆB(x,r) r→0, r≤ k and 1 k 1 Mµ (f)(x) = sup |f| dµ. 1 µ(B(x, r)) ˆB(x,r) r>0, r≤ k

Then, note that as n % ∞, ω 1 & ωf . Now, for all ρ ∈ Q + ıQ, f; k

|f(x) − f(y)| ≤ |f(x) − ρ| + |ρ − f(y)| and so it follows that 1 k ω 1 (x) ≤ |f(x) − ρ| + Mµ (ρ − f)(x) f; k and also that 1 k ωf, 1 (x) ≤ inf |f(x) − ρ| + Mµ (ρ − f)(x) . k ρ∈Q+ıQ n By the density of Q + ıQ, equality holds for x ∈ F where F = {x ∈ R : f(x) ∈ C}. Since f ∈ L1(dµ), we have µ(cF ) = 0.

20 Then,

ωf measurable ⇐⇒ ω 1 measurable ∀k f; k 1 k ⇐⇒ Mµ (ρ − f)(x) measurable ∀k ∀ρ

⇐⇒ Mµ(f) measurable. n−1 n Exercise 3.1.9. Prove that if µ(S (x, r)) = 0 for all x ∈ R , r > 0, then ωf is Borel measurable. n Remark 3.1.10. The use of Besicovitch means that this is speciﬁc to R .

3.2 Maximal functions for doubling measures

Definition 3.2.1 (Doubling measure). Let µ be a positive, locally finite, Borel measure. n We say it is doubling if there exists a C > 0 such that for all x ∈ R and all r > 0, µ(B(x, 2r)) ≤ Cµ(B(x, r)). Definition 3.2.2 (Uncentred maximal function). Let µ be positive, locally finite, and Borel and ν positive and Borel. Define

0 ν(B) Mµ(ν)(x) = sup B3x µ(B) n 1 0 0 for all x ∈ R . As before, when f ∈ Lloc(dµ), Mµ(f) = Mµ(ν) where dν = |f| dµ. That is, 0 1 Mµ(f)(x) = sup |f| dµ. B3x µ(B) ˆB Cheap Trick 3.2.3. There exists a constant C = C(n) > 0 such that 0 Mµ(ν) ≤ Mµ(ν) ≤ C(n)Mµ(ν). 0 Lemma 3.2.4. Mµ(ν) is a lower semi-continuous (and hence Borel) function.

n 0 Proof. Fix λ > 0 and ﬁx x ∈ x ∈ R : Mµ(ν)(x) > λ . So, there exists a ball B with x ∈ B such that ν(B) > λ. µ(B) But for any y ∈ B, we have ν(B) M0 (ν)(y) > > λ µ µ(B) n 0 n 0 and so B ⊂ x ∈ R : Mµ(ν)(x) > λ . This exactly means that x ∈ R : Mµ(ν)(x) > λ is open.

Theorem 3.2.5 (Maximal theorem for doubling measures). Let µ be a doubling measure. Then, there exists an constant C(n, µ) > 0 depending on dimension and the constant C in the doubling condition for µ such that for all f ∈ L1(dµ) and all λ > 0,

n 0 C µ x ∈ R : Mµ(f)(x) > λ ≤ |f| dµ. λ ˆRn

21 Cheap proof. We use Cheap Trick 3.2.3 coupled with the Maximal theorem for Mµ(f).

n The preceding proof has the disadvantage that it is inherently tied up with R . The following is a better proof.

0 n 0 Better proof. Deﬁne: Oλ = x ∈ R : Mµ(f)(x) > λ ,

0m ν(B) Mµ (ν)(x) = sup , B3x, rad B≤m µ(B)

0m n 0m 0 S∞ 0m and Oλ = x ∈ R : Mµ (f)(x) > λ . Then, Oλ = m=1 Oλ .

0m Fix m. Then, for all x ∈ Oλ , there exists a ball Bx with x ∈ Bx with rad Bx ≤ m such that 1 |f| dµ > λ. µ(Bx) ˆBx

0m 0 By a repetition of the argument in Lemma 3.2.4, B ⊂ Oλ making Mµ(f) Borel.

Let B = {Bx} 0m , and apply the Vitali Covering Lemma 2.1.1. So, there exists a x∈Oλ 0m countable subset of centres C ⊂ O and mutually disjoint Bx ⊂ B satisfying λ j xj ∈C

0m [ Oλ ⊂ 5Bxj . xj ∈C

Therefore,

0m X 3 X µ(Oλ ) ≤ µ(5Bxj ) ≤ C µ(Bxj ) xj ∈C xj ∈C C3 X C3 C3 C3 ≤ |f| dµ = |f| dµ ≤ |f| dµ ≤ |f| dµ S 0m 0 λ ˆBx λ ˆ Bx λ ˆO λ ˆO xj ∈C j xj ∈C j λ λ

0m Now, Oλ is an increasing sequence of sets and so we obtain the desired conclusion letting m → ∞.

Remark 3.2.6. This proof is not only better because it frees itself from Besicovitch to the 0 more general Vitali, but we get a better estimate since the integral is over Oλ rather than n R .

Corollary 3.2.7 (Lebesgue Diﬀerentiation Theorem for doubling measures). We have

1 f(x) = lim f(y) dµ(y) B3x, rad B→0 µ(B) ˆB

n for µ-almost everywhere x ∈ R .

22 3.3 The Dyadic Maximal function

Let µ be a positive, locally ﬁnite Borel measure.

Deﬁnition 3.3.1 (Dyadic Maximal function). Let Q0 ∈ Q and let D(Q0) be the collection of dyadic subcubes of Q0. Deﬁne:

Q 1 Mµ (f)(x) = sup |f| dµ Q∈D(Q0), Q3x µ(Q) ˆQ

1 for x ∈ Q0 and f ∈ Lloc(Q0; dµ). Theorem 3.3.2 (Maximal theorem for the dyadic maximal function). Suppose f ∈ 1 L (Q0; dµ) and λ > 0. Then,

n Q o 1 µ x ∈ Q0 : Mµ (f)(x) > λ ≤ |f| dµ. λ ˆQ0 Remark 3.3.3. Notice here that the bounding constant here is 1. It is independent of dimension and µ.

We give two proofs.

Q Proof 1. Let Ωλ = x ∈ Q0 : Mµ (f)(x) > λ . If Ωλ 6= ∅ and x ∈ Ωλ, there exists a Q ∈ D(Q0) such that x ∈ Q and 1 |f| dµ > λ. (†) µ(Q) ˆQ

Let C be the collection of all such cubes. Since Q is countable, C is also countable. Then, S for every Q ∈ C , we have that Q ⊂ Ωλ so therefore, Ωλ = Q∈C Q. Let M = {Qi}i∈I ⊂ C be the maximal subcollection. Then, M is a partition of Q0. So, X 1 X 1 µ(Ω ) = µ(Qi) ≤ |f| dµ ≤ |f| dµ. λ λ ˆ λ ˆ i∈I i∈I Qi Ωλ

Remark 3.3.4. (i) The uniqueness of M is a consequence of the disjointness of the dyadic cubes at each generation.

(ii) The proof gives us an even sharper inequality since we are only integrating on the set Ωλ rather than Q0.

Proof 2. If 1 |f| dµ > λ, µ(Q0) ˆQ0 then Ωλ = {Q0} and there’s nothing to do. So assume the converse. We construct a mutually disjoint subset F by the following procedure. Consider a dyadic child of Q0, say, Q. If Q satisﬁes (†), we stop and put Q in F . Otherwise, we apply this procedure to

23 Q in place of Q0. That is, we consider whether a given dyadic child of Q to satisﬁes (†). The collection F is called the stopping cubes for the property (†). 1

We claim that: [ Ωλ = Q. Q∈F

Clearly, Q ∈ F implies that Q ⊂ Ωλ (Proof 1). Suppose that there exists x ∈ Ωλ but S Q 0 x 6∈ Q∈F Q. But we know that Mµ (f)(x) > λ so there exists a dyadic cube Q ∈ D(Q0) such that x ∈ Q0 satisfying the property (†). The stopping time did not stop before Q0 by S 0 0 hypothesis x 6∈ Q∈F Q, but Q satisﬁes (†). By the construction of F , we have Q ∈ F which is a contradiction.

The estimate then follows by the same calculation as in Proof 1.

Remark 3.3.5. The collection F is the maximal collection M from Proof 1. n Deﬁnition 3.3.6 (Dyadic maximal function on R ). Deﬁne:

Q 1 Mµ (f)(x) = sup |f| dµ Q∈Q, Q3x µ(Q) ˆQ n 1 for x ∈ R and f ∈ Lloc(dµ). 1 n Corollary 3.3.7. Suppose f ∈ L (R ; dµ) and λ > 0. Then,

n n Q o 1 µ x ∈ R : Mµ (f)(x) > λ ≤ |f| dµ. λ ˆRn

n Proof. Since we are considering dyadic cubes, we begin by splitting R into quadrants, χ n and let g = f (R+) . We note that it suﬃces to prove the statement for one quadrant since the argument is unchanged for the others.

k k n For each k ∈ N, let Q = [0, 2 ) . Then, note that: Q Mµ (f)(x) = sup gk(x) k∈N where 1 gk(x) = sup |f| dµ. Q∈D(Qk), Q⊂Qk,Q3x µ(Q) ˆQ

k k Now, let Ωλ = x ∈ Q : gk(x) > λ . We compute:

n k 1 1 µ {x ∈ (R+) : gk(x) > λ} = µ(Ωλ) ≤ |f| dµ ≤ |f| dµ. ˆ k ˆ n λ Q λ (R+) The desired conclusion is achieved by letting k → ∞ and summing over all quadrants.

1 Corollary 3.3.8. For f ∈ Lloc(dµ), we have 1 f(x) = lim f(y) dµ(y) Q∈Q, L (Q)→0,Q∈x µ(Q) ˆQ n for µ-almost everywhere x ∈ R . 1This is a stopping time argument, a typical technique in probability.

24 Proof. Exercise.

Remark 3.3.9. For a ﬁxed x, the set {Q ∈ Q : L (Q) → 0, Q 3 x} is really a sequence. Consequently, the limit is really the limit of a sequence.

We have the following important Application.

1 Proposition 3.3.10. For f ∈ Lloc(dµ),  MQ(f)  µ |f| ≤ Mµ(f)  0 Mµ(f)

n for µ-almost everywhere x ∈ R .

n Remark 3.3.11. The centred maximal function uses Besicovitch and is conﬁned to R . The others do not and can be generalised to spaces with appropriate structure. For instance, in the case of the Dyadic maximal function, we must be able to at least perform a dyadic decomposition of the space.

A consequence of the Lebesgue diﬀerentiation is:

0 Mµ(f) ≤ Mµ(f).

Q 0 Exercise 3.3.12. Try to compare Mµ with Mµ (and Mµ). For simplicity, take µ = L (but note that this has nothing to do with the measure but rather the geometry).

3.4 Maximal Function on Lp spaces

0 Q In this section, we let M denote either Mµ, Mµ or Mµ . We ﬁrstly note that for every f ∈ L∞(dµ),

Mf(x) ≤ kfk∞ n ∞ for all x ∈ R . So, Mf is a bounded operator on L (dµ). We investigate the boundedness on Lp(dµ) for p < ∞.

Lemma 3.4.1 (Cavalieri’s principle). Let 0 < p < ∞ and let g be a positive, measurable function. Then,

∞ p n p−1 g dµ = p µ {x ∈ R : g(x) > λ} λ dλ. ˆRn ˆ0

Proof. Assume that µ finite (this is required for the application of Fubini’s theorem in the n following computation). In the case µ is not finite, assuming that µ {x ∈ R : g(x) > λ} < ∞ for all λ > 0, we can restrict µ to a σ-algebra (depending on g) where it is the case. Then this reduces to the assumption µ is finite by approximating g via gN,R = χ χ g {x∈Rn:g(x)≤N} B(0,R) and then applying the Monotone Convergence Theorem.

25 We compute and apply Fubini: ∞ n p−1 p µ {x ∈ R : g(x) > λ}λ dλ ˆ0 ∞ = p 1 dµ(x)λp−1 dλ ˆ0 ˆ{x∈Rn:g(x)>λ} g(x) = pλp−1 dλ dµ(x) ˆRn ˆ0 = gp dµ(x). ˆRn

Theorem 3.4.2 (Boundedness of the Maximal function on Lp). Let 1 < p < ∞. Then, there exists a constant C = C(p, n) > 0 such that whenever f ∈ Lp(dµ) then Mf ∈ Lp(dµ) and

kMfkLp(dµ) ≤ CkfkLp(dµ). 0 For the case of M = Mµ, we assume that µ is doubling and C may also depend on the constants in the doubling condition.

Proof. Assume that f ∈ L1(dµ) ∩ Lp(dµ). Let λ > 0 and deﬁne

( λ f(x) |f(x)| ≥ 2 fλ(x) = λ 0 |f(x)| < 2 and λ |f| ≤ |f | + . λ 2 By the subadditivity of the supremum, λ Mf ≤ Mf + λ 2 and λ {x ∈ n : Mf(x) > λ} ⊂ x ∈ n : Mf (x) > . R R λ 2 Therefore, n n λ C µ {x ∈ R : Mf(x) > λ} ≤ µ x ∈ R : Mfλ(x) > ≤ |fλ| dµ. 2 λ ˆRn Then, ∞ ∞ n p−1 p−2 p µ {x ∈ R : Mf(x) > λ} λ dλ ≤ p C |fλ| λ dµdλ ˆ0 ˆ0 ˆRn 2f(x) p−2 ≤ Cp |fλ| λ dλdµ ˆRn ˆ0 p ≤ C2p−1 |f|p dµ p − 1 ˆRn Now, for a general f ∈ Lp(dµ) by the density of L1(dµ) ∩ Lp(dµ) in Lp(dµ), we can take 1 p a sequence (for example simple functions) fk ∈ L (dµ) ∩ L (dµ) satisfying |fk| % |f|. Then, Mfk %Mf and the proof is complete by invoking the Monotone Convergence Theorem.

26 Remark 3.4.3. (i) The constant C(n, p) satisﬁes:

1 p p C(n, p) ≤ C2p−1 (†) p − 1 So, if we let p → ∞, then we ﬁnd that

1 p p C2p−1 → 1. p − 1 p But as p → 1, C ∼ p−1 and blows up! In fact it is true that if 0 6= f ∈ L1(dL ), then Mf 6∈ L1(dL ) (Exercise). For p = 1, the best result is the Maximal Theorem.

(ii) We remark on the optimality of C(n, p). Consider the inequality (†), and note that the C is the Maximal Theorem constant. This depends on n and the measure µ Q (unless we consider Mµ ). Suppose µ = L . Then the best upper bound for C with respect to n is n log n. Consider the operator norm of M:

Mp,n = sup kMfkLp(dL ). p f∈L (dL ), kfkLp(dL )=1

If M = Mµ (ie, the centred maximal function), then Mp,n can be shown to be bounded in n for any ﬁxed p > 1. This is important in stochastic analysis. Q Now, suppose M = Mµ . Then, p Mp,n = . p − 1 To see this, ﬁrst note that

n 1 L {x ∈ R : Mf(x) > λ} ≤ |f| dL λ ˆ{x∈Rn:Mf(x)>λ} and

kM kp = |M | f Lp(dL ) f dL ˆRn ∞ n p−1 = p L {x ∈ R : Mf(x) > λ} λ dλ ˆ0 ∞ ≤ p |f| dL λp−1 dλ ˆ0 ˆ{x∈Rn:Mf(x)>λ} ∞ Mf = p λp−1 dλ |f| dL ˆ0 ˆ0 p = (Mf)p−1 |f| dL p − 1 ˆRn p ≤ kMfkp−1 kfkp p − 1 Lp(dL ) Lp(dL ) which shows that p kMfk p ≤ kfk p . L (dL ) p − 1 L (dL ) Optimality can be shown via martingale techniques. We will not prove this.

27 We give the following important application of Maximal function theory.

n Theorem 3.4.4 (Hardy-Littlewood Sobolev inequality). Let 0 < λ < n, 1 ≤ p < n−λ and q satisfying 1 λ 1 + = 1 + . p n q Then, with ν = 1 , λ |·|λ

(i) When p = 1, there exists a constant C(λ, n) such that for all α > 0, C n { ∈ n : | ∗ ( )| } ≤ k k λ L x R νλ u x > α n u L1(dL ), α λ

(ii) When p > 1, there exists a constant C(p, λ, n) such that

kνλ ∗ ukLp(dL ) ≤ CkukLp(dL ), where 1 (νλ ∗ u)(x) = λ u(y) dL (y) ˆRn |x − y| is the convolution of u with νλ.

Remark 3.4.5 (Motivation). Such a νλ arises when trying to “integrate” functions in n R . Such “potentials” are one way to “anti-derive.” Formally, in the case of λ = n − 2, 1 uˆ(ξ) ∗ u ˆ(ξ) = C(n) |x|n−2 |ξ|2 whereˆdenotes the Fourier Transform. See [Ste71a].

∞ n Proof (Hedberg’s inequality). We prove (i). Let u ∈ Cc (R ), and set w = νλ ∗ u. We take λ > 0 to be chosen later. Then, ∞ 1 X 1 w(x) ≤ λ |u(y)| dL (y) + λ |u(y)| dL (y). ˆ ˆ −k−1 −k |x−y|≥δ |x − y| k=0 δ2 ≤|x−y|≤δ2 |x − y|

First, whenever |x − y| ≥ δ, 1 1 λ |u(y)| dL (y) ≤ λ kukL1(dL ). ˆ|x−y|≥δ |x − y| δ

So, ﬁx k ≥ 0. Then, 1 λ |u(y)| dL (y) ˆδ2−k−1≤|x−y|≤δ2−k |x − y| −k −k−1 −λ L (B(x, δ2 )) ≤ (δ2 ) −k |u(y)| dL (y) L (B(x, δ2 )) ˆ|x−y|≤δ2−k λ −k n−λ ≤ bn2 (δ2 ) ML (u)(x)

28 where bn = L (B(0, 1)), the volume of the n ball. It follows then that

∞ X 1 λ |u(y)| dL (y) ˆ −k−1 −k k=0 δ2 ≤|x−y|≤δ2 |x − y| ∞ X λ −k n−λ ≤ bn2 (δ2 ) ML (u)(x) k=0 λ bn2 ≤ δ−λδn M (u)(x). 1 − 2−(n−λ) L

Now, let λ bn2 A = A(n, λ) = 1 − 2−(n−λ) n and solve for δ such that δ A(n, λ)ML (u)(x) = kukL1(dL ). Since if u 6≡ 0 then ML u(x) 6= 0 almost everywhere (which we leave as an exercise),

1 ! n kukL1(d ) δ = L . A(n, λ)ML (u)(x)

n Then, for almost everywhere x ∈ R ,

−λ −λ n w(x) ≤ δ kukL1(dL ) + δ δ A(n, λ)ML u(x) −λ = δ 2kukL1(dL ) λ !− n kukL1(dL ) = 2 kukL1(dL ) A(n, λ)ML (u)(x) 1 ! q A(n, λ)ML (u)(x) = 2 kukL1(dL ) kukL1(dL ) 1 1 1− 1 = ( ) q M ( )( ) q k k q A n, λ L u x u L1(dL ). From this, it follows that,

( −q) 1− 1 { ∈ n : ( ) } ≤ ∈ n : M ( ) q k k q L x R w x > α L x R L x > α u L1(dL )

1 q C(n) 1− q ≤ kuk kuk 1 αq L1(dL ) L (dL ) C(n) = kukq . αq L1(dL ) ∞ n 1 The density of Cc (R ) in L (dL ) completes the proof.

We leave (ii) as an exercise.

29 Chapter 4

Interpolation

4.1 Real interpolation

Suppose that (M, µ) and (N, ν) are measure spaces and the measures µ and ν are σ-ﬁnite measures.

Recall that when 0 < p < ∞,Lp(M, µ) (or Lp(M, dµ)) is the space of µ-measurable functions f for which |f|p is integrable. When 1 ≤ p < ∞, this space is a Banach space (modulo the almost everywhere equality) and the norm is what we expect:

1/p p kfkp = |f| dµ . ˆM For p = ∞, f ∈ L∞(M, µ) if and only if there exists a λ > 0 such that µ {x ∈ M : |f(x)| > λ} = 0. This is a Banach space and the norm is then given by

kfk∞ = esssup |f| = inf {λ > 0 : µ {x ∈ M : |f(x)| > λ} = 0} .

We define a generalisation of these spaces called the Weak Lp spaces. Definition 4.1.1 (Weak Lp space). Let 0 < p < ∞. Then, define Lp,∞(M, µ) to be the space of µ-measurable functions f satisfying

sup λpµ {x ∈ M : |f(x)| > λ} < ∞. λ>0 For a function f ∈ Lp,∞(M, µ), we deﬁne a quasi-norm:

1 p p kfkp,∞ = sup λ µ {x ∈ M : |f(x)| > λ} . λ>0 When p = ∞, we let Lp,∞(M, µ) = L∞. Remark 4.1.2. (i) The Weak Lp spaces are really a special case of Lorentz spaces Lp,q. They truly generalise the Lp spaces for they are equal to Lp when q = p. See [Ste71b] for a detailed treatment.

30 (ii) We emphasise that for 0 < p < ∞, kfkp,∞ is not a norm, but it is a quasi-norm. That is,

kf + gkp,∞ ≤ 2(kfkp,∞ + kgkp,∞). This is a direct consequence of the following observation: λ λ {x ∈ M : |f + g| > λ} ⊂ x ∈ M : |f| > ∪ x ∈ M : |g| > . 2 2

(iii) When 1 < p < ∞, there exists a norm on Lp,∞(M, µ) which is metric equivalent to k· kp,∞. (iv) When p = 1, there is no such equivalent norm. In fact, L1,∞(M, µ) is not even 1 n 1,∞ n locally convex. For instance when u ∈ L (R , dL ) then ML u ∈ L (R , dL ) by the Maximal theorem.

p Proposition 4.1.3. The Weak L spaces are complete with respect to the metric dp(f, g) = kf − gkp,∞ for 0 < p < ∞. Proposition 4.1.4 (Tchebitchev-Markov inequality). If f is positive, µ-measurable and 0 < p < ∞, then

1 p 1 p µ {x ∈ M : f(x) > λ} ≤ p f dµ(x) ≤ p kfkp. λ ˆ{x∈M:f(x)>λ} λ In particular, Lp(M, µ) ⊂ Lp,∞(M, µ). Remark 4.1.5. The inclusion is strict in general. For instance,

1 n ,∞ n n n ∈ L λ (R , dL ) \ L λ (R , dL ), |x|λ when 0 < λ. Remark 4.1.6 (σ-finiteness of µ). The above definitions do not require σ-finiteness of µ. We will, however, require this in the interpolation theorem to follow.

Deﬁnition 4.1.7 (Sublinear operator). Let K = R or K = C, and let FM denote the space measurable functions f : M → K. Let DM be a subspace of FM and let T : DM → FN . We say that T is sublinear if

|T (f1 + f2)(x)| ≤ |T f1(x)| + |T f2(x)| for ν-almost all x ∈ N. 1 n Example 4.1.8. (i) T : f 7→ ML f where D = Lloc(R , dL ). (ii) Any linear T is sublinear.

Deﬁnition 4.1.9 (Weak/Strong type). Let T : DM → FN be sublinear and let 1 ≤ p, q ≤ ∞. Then,

p q (i) T is of strong type (p, q) if T : DM ∩L (M, µ) → L (N, µ) is a bounded map. That is, p q there exists a C > 0 such that whenever f ∈ DM ∩ L (M, µ) we have T f ∈ L (N, µ) and

kT fkq ≤ Ckfkp.

31 p q,∞ (ii) T is of weak type (p, q) (for q < ∞) if T : DM ∩ L (M, µ) → L (N, µ) is a bounded p map. That is, there exists a C > 0 such that whenever f ∈ DM ∩ L (M, µ) we have T f ∈ Lq,∞(N, µ) and kT fkq,∞ ≤ Ckfkp. (iii) T is of weak type (p, ∞) if it is of strong type (p, ∞). Remark 4.1.10. Note that if T is of strong type (p, q) then it is of weak type (p, q).

Example 4.1.11. (i) The maximal operator T : f → Mµf is of weak type (1, 1) and of strong type (p, p) for 1 < p ≤ ∞.

(ii) The operator u 7→ νλ ∗ u where νλ is the potential in Theorem 3.4.4 is of weak type n n (1, λ ) if 0 < λ < n on (R , L ). Theorem 4.1.12 (Marcinkiewicz Interpolation Theorem). Given (M, µ) and (N, ν) let DM be stable under multiplication by indicator functions. That is, if f ∈ DM then χX f ∈ DM . Let 1 ≤ p1 < p2 ≤ ∞, 1 ≤ q1 < q2 ≤ ∞, with p1 ≤ q1 and p2 ≤ q2. Furthermore, let T : DM → FN be a sublinear map that is of weak type (p1, q1) and (p2, q2). Then, for all p ∈ (p1, p2), T is of strong type (p, q) with q satisfying 1 1 − θ θ 1 1 − θ θ = + and = + . p p1 p2 q q1 q2

Proof. We prove the case when q1 = p1, q2 = p2 < ∞ and leave the general case as an exercise.

First, by the weak type (pi, pi) hypothesis for i = 1, 2 we have constants Ci > 0 such that k k ≤ k k T f pi,∞ Ci f pi p for all f ∈ DM ∩ L i (M, µ). Fix such an f along with p ∈ (p1, p2) and ∞ |T f|p dν = p ν {x ∈ N : |T f(x)| > λ} λp−1 dλ. ˆN ˆ0 Fix λ > 0 and deﬁne ( f(x) |f(x)| > λ gλ(x) = 0 otherwise and hλ(x) = f(x) − gλ(x).

So, gλ = fχ{x∈M:|f(x)|>λ} ∈ DM and hλ = fχ{x∈M:|f(x)|≤λ} ∈ DM by the stability hypothesis on DM .

Now, p p−p1 kfk p1 p1 p1 f p |gλ| dµ = |f| dµ ≤ |f| dµ ≤ p−p ˆM ˆ{x∈M:|f(x)|>λ} ˆM λ λ 1 and by similar calculation,

p2 p2 p2−p p |hλ| = |f| dµ ≤ λ |f| dµ < ∞. ˆM ˆ{x∈M:|f(x)|≤λ} ˆM

32 Since T is sublinear, for ν-almost all x ∈ N,

|T f(x)| ≤ |T gλ(x)| + |T hλ(x)| and so λ λ ν {x ∈ N : |T f(x)| > λ} ≤ ν x ∈ N : |T g (x)| > + ν x ∈ N : |T h (x)| > . λ 2 λ 2

We compute,

∞ λ p−1 ν x ∈ N : |T gλ(x)| > λ dλ ˆ0 2 ∞ p1 C1 p1 p−1 ≤ kgλk λ dλ ˆ λ p1 p1 0 2 ∞ p1 p−p1−1 p1 ≤ (2C1) λ |f| dµdλ ˆ0 ˆ{x∈M:|f(x)|≤λ} |f| ! p1 p−p1−1 = (2C1) λ dλ dµ ˆM ˆ0 p (2C1) 1 = |f|p dµ. p − p1 ˆM

By a similar calculation, but this time integrating from |f| to ∞ with λ−ξ for ξ > 0, we get ∞ p2 λ (2C2) p ν x ∈ N : |T hλ(x)| > dλ ≤ |f| dµ. ˆ0 2 p2 − p ˆM Putting these estimates together,

p1 p2 (2C1) (2C2) |T f|p ≤ p + |f|p dµ ˆN p − p1 p2 − p ˆM which completes the proof.

Remark 4.1.13. If the deﬁnition of gλ was changed to gλ = fχ{x∈M:|f(x)|>aλ} with a ∈ R+, then this leads to better bounds by optimising a. In fact, we can get a log convex combination of C1,C2. That is:

1−θ θ kT fkp ≤ C(p, p1, p2)C1 C2 kfkp.

For a function θ 7→ h(θ), Log convex here means that θ 7→ log h(θ) is convex.

4.2 Complex interpolation

N We begin by considering a “baby” version of complex interpolation. Let a ∈ C . Then, for 1 ≤ p < ∞, 1 N ! p X p kakp = |ai| i=1

33 and for p = ∞

kak∞ = sup |ai| 1≤i≤N are norms. Let M be an N × N matrix, then

kMakp k k N = k k = sup M (L(C ),k·k ) M p,p p N kak a∈C ,kakp6=0 p is the associated operator norm. We leave it as an (easy) exercise to verify that

N X kMk∞,∞ = sup |mij| i j=1 where M = (mij). Then,

N ∗ X kMk1,1 = kM k∞,∞ = sup |mij| j i=1

N where we have used the fact that k· k1 and k· k∞ are dual norms on C . For 1 < p < ∞, there is no such characterisation of kMkp,p but we have the following Schur’s Lemma. Proposition 4.2.1 (Schur’s Lemma).

1 1 p 1− p kMkp,p ≤ kMk1,1kMk∞,∞.

Proof. For matrices, the result follows from the application of H¨older’sinequality.

N Let a ∈ C with kakp = 1. Then,

X |(Ma)i| = mijaj

j X ≤ |mij| |aj| j X 1− 1 1 = |mij| p |mij| p |aj| j 1 1   p0   p X X p ≤  |mij|  |mij| |aj|  j j and so it follows that

p p X p p0 X X p p0 p |(Ma)i| ≤ kMk∞,∞ |mij| |aj| ≤ kMk∞,∞kMk1,1kak i i j which ﬁnishes the proof.

34 This is really a special case of more general interpolation results of the form

1−θ θ kMkq,q ≤ kMkp,p kMkr,r for 1 ≤ p < q < r ≤ ∞ and 1 1 − θ θ = + . q p r A proof of this statement cannot be accessed easily via elementary techniques since there is no explicit characterisation of kMkq,q and kMkr,r. This was the essential ingredient of the preceding proof. This more general statement comes as a consequence of the powerful complex interpolation method.

We prove this by employing the following 3 lines theorem of Hadamard. First note that in the following statement, S˚ denotes the topological interior of S and H(S˚) denotes the holomorphic functions on S˚.

Lemma 4.2.2 (3 lines Theorem of Hadamard). Let S = {ζ ∈ C : 0 ≤ Re ζ ≤ 1}. Let ˚ 0 F ∈ H(S) and F ∈ C (S). Further suppose that kF k∞ < ∞ on S and let

C0 = sup |F (ıt)| and C1 = sup |F (1 + ıt)| . t∈R t∈R Then for x ∈ (0, 1), and t ∈ R, 1−x x |F (x + ıt)| ≤ C0 C1 .

Proof. Assume that C0,C1 > 0. If not, prove the theorem with C0 + δ, C1 + δ in place of 1−Re ζ Re ζ C0,C1 for δ > 0 to conclude that |F (ζ)| ≤ (C0 + δ) (C1 + δ) and letting δ → 0 conclude that 1−Re ζ Re ζ |F (ζ)| = 0 = C0 C1 for ζ ∈ S˚.

ζ−1 −ζ εζ2 ˚ 0 Fix ε > 0 and set Gε(ζ) = F (ζ)C0 C1 e for ζ ∈ S. Then, Gε ∈ H(S),Gε ∈ C (S). Now, ﬁx R > 0 and let QR = S ∩ {ζ ∈ C : Im ζ ≤ R}. By the maximum principle,

sup |Gε(ζ)| = sup |Gε(ζ)| . ζ∈QR ζ∈∂QR

We consider each part of ∂QR. For ζ = ıt with |t| ≤ R,

−1 ε(ıt)2 |Gε(ζ)| = |F (ıt)| C0 e ≤ 1 and for ζ = 1 + ıt with |t| ≤ R,

−1 Re ε(1−t2+2ıt) ε |Gε(ζ)| = |F (1 + ıt)| C1 e ≤ e . Then, for ζ = x + ıR with |t| ≤ R,

x−1 x Re ε(x2−R2+2ıxR) Re ε(1−R2) |Gε(ζ)| ≤ |F (x + ıR)| C0 C1 e ≤ Ce where ! x−1 x C = sup |F (ζ)| sup C0 C1 ζ∈S 0≤x≤1

35 and lastly, when ζ = x − ıR with |t| ≤ R,

Re ε(1−R2) |Gε(ζ)| ≤ Ce by the same calculation since Re ε(x2 − R2 + 2ıxR) = ε(x2 − R2). So, for ε ﬁxed, there exists an R0 such that whenever R ≥ R0,

 2 Ce−εR ≤ 1  ε |Gε(x)| ≤ e  0 ε ∀ζ ∈ S, ∃R ≥ R0 such that ζ ∈ QR0 and |Gε(ζ)| ≤ e and so for all ζ ∈ S and all ε > 0,

2 1−ζ ζ −εζ ε |F (ζ)| ≤ C0 C1 e e .

The proof is then completed by ﬁxing ζ and letting ε → 0.

Remark 4.2.3. This lemma can be proved assuming some growth on F for |Im ζ| → ∞ (rather than F bounded). However, there needs to be some control on the growth.

Lemma 4.2.4 (Phragmen-Lindel¨of). Let Σ ⊂ C be the closed subset between the lines π 1 ˚ 0 R ± ı 2 (obtained from the conformal map ζ 7→ ıπ(ζ − 2 )). Suppose F ∈ H(Σ), C (Σ) and π assume that F is bounded on the lines x ± ı 2 . Suppose there exists an A > 0, β ∈ [0, 1) such that for all ζ ∈ Σ, |F (ζ)| ≤ exp(A exp(β |Re ζ|)). Then, |F (ζ)| ≤ 1 for ζ ∈ Σ.

Proof. The proof is the same as the proof of the preceding lemma, except eε(ıζ)2 = e−εζ2 term needs to be replaced by something decreasing faster to 0 as |Re ζ| → ∞. We leave the details as an exercise (or see [Boa54]).

We are now in a position to state and prove the powerful complex interpolation theorem of Riesz-Thorin.

Theorem 4.2.5 (Riesz-Thorin). Let 1 ≤ p < r ≤ ∞ and let (M, µ), (N, ν) be σ-ﬁnite measure spaces. Let DM denote the space of simple, integrable functions on M, and T : DM → FN be C-linear. Furthermore, assume that T is of strong type (p, p) and (r, r) with bounds Mp and Mr respectively. Then, T is of strong type (q, q) for any q ∈ (p, r) with bound Mq satisfying 1−θ θ Mq ≤ Mp Mr , where 1 1 − θ θ = + q p r with 0 < θ < 1.

Before we prove the theorem, we illustrate the following immediate and important corollary.

36 Corollary 4.2.6. With the above assumptions, T has a continuous extension to a bounded operator Lq(M, dµ) → Lq(N, dν) for each q ∈ (p, r).

q Proof. Fix q ∈ (p, r) and since q < ∞, DM is dense in L (M, dµ). Then, deﬁne the extension by the usual density argument.

q Proof of Riesz-Thorin. Fix p < q < r and note that DM is dense in L (M, dµ), DN is 0 0 dense in Lq (N, dν) and by duality, Lq(N, dν)0 = Lq (N, dν). Thus, it suﬃces to show that

A = sup T f g dν < ∞ ˆ f∈DM , kfkq=1, g∈DN , kgkq0 =1 N as this will imply that T is of strong type (q, q) on DM and kT k ≤ A.

Fix f ∈ DM , g ∈ DN with kfkq = kgkq0 = 1. So,

X χ f = αk Ak k where αk ∈ C, µ(Ak) < ∞ and the sum is ﬁnite. Similarly, X χ g = βl Bl , l and it follows that X X T f g dν = α β T χ χ dν. ˆ k l ˆ Ak Bl N k l N χ p r But µ(Ak) < ∞ which implies that Ak ∈ L (M, dµ) ∩ L (M, dµ) and

χ χ T Ak Bl dν ˆN ˚ 0 is well deﬁned. We will construct an F ∈ H(S), C (S), kF k∞ < ∞ (where S is deﬁned in Lemma 4.2.2) with F (θ) = T f g dν. ˆN

Let ζ ∈ C and write X a(ζ) αk χ fζ = |αk| Ak |αk| k where q q a(ζ) = (1 − ζ) + ζ. p r Similarly, X a(ζ) βl χ gζ = |βl| Bl |βl| l and q0 q0 a(ζ) = (1 − ζ) + ζ. p0 r0

37 Note that a(θ) = 1 if and only if b(θ) = 1, and f = fθ and g = gθ. Set

F (ζ) = T fζ gζ dν ˆN and note that it is well deﬁned for each ζ by the same reasoning as previously for f and g in place of fζ and gζ . Furthermore,

X X a(ζ) b(ζ) αk βl χ χ F (ζ) = |αk| |βk| T Ak Bl dν |αk| |βl| ˆ k l N

0 so F ∈ H(C) ⊂ H(S˚), C (S) and

F (θ) = T f g dν. ˆN

In order to apply Lemma 4.2.2 we estimate supt∈R |F (ıt)| , supt∈R |F (1 + ıt)|. So, by the strong type (p, p) property of T ,

sup |F (ıt)| = sup T fıt gıt dν t∈R t∈R ˆN

≤ sup kT fıtkpkgıtkp0 t∈R

≤ Mp sup kfıtkpkgıtkp0 . t∈R

Using the fact that Ak are mutually disjoint, we can write p p X a(ıt) kfıtkp = |αk| µ(Ak) k

q q q and since a(ıt) = p + ıt r − p ,

p X q q kfıtkp = |αk| µ(Ak) = kfkq. k

p0 q0 0 q0 0 By a similar calculation, kgıtkp0 = kgkq0 if p < ∞ and kgıtk∞ = 1 = kgkq0 if p = ∞.

This shows that supıt |F (ıt)| ≤ Mp. An identical calculation, using the strong type (r, r) property of T gives supıt |F (1 + ıt)| ≤ Mr. Then, set C0 = Mp and C1 = Mr and we invoke Lemma 4.2.2 to ﬁnd 1−Re ζ Re ζ |F (ζ)| ≤ Mp Mr . The proof is then complete by setting ζ = θ.

Exercise 4.2.7. Assume the hypothesis of the previous theorem, but here assume that T is of strong type (p1, p2) and strong type (r1, r2) where 1 ≤ p1, r1 ≤ ∞ and 1 ≤ p2, r2 ≤ ∞. Then, T is of strong type (q1, q2) where 1 1 − θ θ 1 1 − θ θ = + and = + q1 p1 r1 q2 p2 r2 where 0 < θ < 1.

Remark 4.2.8. Notice that in the previous exercise, we do not require p1 < p2 and r1 < r2 as we did in the case of Real interpolation (Theorem 4.1.12).

38 Chapter 5

Bounded Mean Oscillation

n In this chapter, the framework we work within is R with the metric d = d∞ and Lebesgue measure L . By Q(x, r), we always denote a ball with respect to d of radius r centred at n x. That is, Q = Q(x, r) represents an arbitrary cube in R .

n We remark that for the theory, there is nothing special about R and d∞. It is just convenient to work in this setting. The following material could be deﬁned and studied similarly on spaces of homogeneous type.

5.1 Construction and properties of BMO

We introduce some notation. Let mX f denote the mean of the function f on the set X. That is 1 mX f = f dL . L (X) ˆX

1 n Deﬁnition 5.1.1 (Bounded Mean Oscillation (BMO)). Let f ∈ Lloc(R ), real or complex valued. We say that f has bounded mean oscillation if 1 kfk∗ = sup mQ |f − mQ| = sup |f − mQ f| dL < ∞. Q Q L (Q) ˆQ

n Here Q is an arbitrary cube in R . We deﬁne 1 n BMO = f ∈ Lloc(R ): kfk∗ < ∞ .

∞ n Proposition 5.1.2 (Properties of BMO). (i) L (R ) ⊂ BMO and kfk∗ ≤ 2kfk∞.

(ii) BMO is a linear space (over K), and k· k∗ is a semi-norm. That is,

kf + gk∗ ≤ kfk∗ + kgk∗, and

kλfk∗ = |λ| kfk∗

39 for f, g ∈ BMO and λ ∈ K. Furthermore, kfk∗ = 0 if and only if f constant almost n everywhere x ∈ R .

(iii) BMO/K is a normed space with norm

kf + Kk = kfk∗,

making BMO/K a Banach space. BMO convergence is often called convergence modulo constant.

∞ n n (iv) For f ∈ L (R ), x0 ∈ R and t > 0 the function deﬁned by x − x0 ∞ n ft,x0 (x) = f ∈ L (R ). t

Similarly, for f ∈ BMO, ft,x0 ∈ BMO and kft,x0 k∗ = kfk∗.

n Proof. We prove that kfk∗ = 0 if and only if f constant almost everywhere x ∈ R (ii) and leave the rest as an exercise.

Note that the “if” direction is trivial. To prove the “only if” direction, assume that n kfk∗ = 0. Then, for every cube Q ⊂ R , f = mQ f almost everywhere x ∈ Q. Let Qj be n j j n an exhaustion of R by increasing cubes. That is, let Qj = [−2 , 2 ] for j = 0, 1, 2,... .

Then, f = mQj f almost everywhere on Qj and hence mQ0 f = mQj f for all j. Letting n j → ∞ we establish that f is constant almost everywhere on R .

n Exercise 5.1.3. 1. Let A ∈ GLn(K) and x0 ∈ R . Write fA,x0 (x) = f(Ax−x0). Show

that if f ∈ BMO then fA,x0 ∈ BMO and

n −1 kfA,x0 k∗ ≤ 2kAk∞,∞ det A kfk∗.

(Hint: Use the next Lemma). This exercise illustrates that we can indeed change the shape of cubes.

2. Show that 0 kfk∗ = sup mB(|f − mB f|) < ∞ B where B are Euclidean balls deﬁnes an equivalent semi-norm on BMO.

1 n Lemma 5.1.4. Let f ∈ Lloc(R ) and suppose there exists a C > 0 such that for all cubes Q, there exists a cQ ∈ K such that mQ |f − cQ| ≤ C. Then, f ∈ BMO and kfk∗ ≤ 2C.

Proof. We write

f − mQ f = f − cQ + cQ − mQ f = f − cQ + mQ(cQ − f) and it follows that

mQ |f − mQ f| ≤ mQ |f − cQ| + mQ |cQ − f| ≤ 2 mQ |f − cQ| ≤ 2C.

40 ∞ n Example 5.1.5 (ln |x| ∈ BMO). In particular, this implies that L (R ) $ BMO.

We show that this is true for n = 1. Let f(x) = ln |x|. For t > 0, x ft(x) = ln = ln |x| − ln |t| = f(x) + c. t So, for I an interval of length t,

mI |f − mI f| = mI |ft − mI ft| .

x By change of variables, y = t ∈ J where J is of unit length and

mI |ft − mI ft| = mJ |f − mJ f|

1 1 and so we are justiﬁed in assuming that I has unit length. So, let I = [x0 − 2 , x0 + 2 ] and by symmetry, assume x0 ≥ 0.

Consider the case 0 ≤ x0 ≤ 3. Then,

7 2 mI |f| = |f(x)| dx ≤ |ln |x|| dx < ∞. ˆ ˆ 1 I − 2

1 1 Now, suppose x0 ≥ 3. Then, for x ∈ [x0 − 2 , x0 + 2 ],

x x 1 1 1 2 ln(x) − ln x0 − = ln(x) − ln x0 − = dt ≤ dt 2 2 ˆ 1 t ˆ 1 5 x0− 2 x0− 2 2 1 2 1 1 2 ≤ x − x + ≤ x + − x + ≤ 5 0 2 5 0 2 0 2 5

1 2 and by setting CI = ln(x0 − 2 ),C = 5 ,

x + 1 x + 1 0 2 0 2 2 |ln(x) − CI | dL (x) = |ln(x) − CI | dL (x) ≤ = C. 1 ˆ 1 5 x0− 2 x0− 2

Let CI = 0 whenever 0 ≤ x0 < 3 and let

7 ! 2 C0 = max C, |ln |x|| dL (x) . ˆ 1 − 2

0 Then, mI |f − CI | ≤ C and the proof is complete by applying Lemma 5.1.4.

The following proposition highlights an important feature of L∞ which is missing in BMO. Proposition 5.1.6. Let ( ln(x) x > 0 f(x) = . 0 x ≤ 0 Then, f 6∈ BMO and consequently, BMO is not stable under multiplication by indicator functions.

41 Proof. Let I = [−ε, ε] for ε > 0 and small. We compute, 1 ε 1 0 mI |f − mI f| = |ln |x| − mI f| dL (x) + |mI f| dL (x). 2ε ˆ0 2ε ˆ−ε Then, 0 0 ε 1 1 1 |mI f| dL (x) = ln |x| dL x dL (y) 2ε ˆ−ε 2ε ˆ−ε 2ε ˆ0 1 1 = |ε ln ε + ε| 2 2ε 1 = |ln ε + 1| . 4 But this is not bounded as ε → 0. Proposition 5.1.7 (Further properties of BMO). (i) Whenever f ∈ BMO, then |f| ∈ BMO and k |f| k∗ ≤ 2kfk∗.

(ii) Suppose f, g ∈ BMO are real valued. Then, f+, f−, max(f, g), min(f, g) ∈ BMO. Furthermore, 3 k max(f, g)k , k min(f, g)k ≤ (kfk + kgk ) . ∗ ∗ 2 ∗ ∗ (iii) Let f ∈ BMO real valued. Then we have the following Approximation by truncation. Let  N f(x) > N  fN (x) = f(x) −N ≤ f(x) ≤ N  −N f(x) < N ∞ n for N ∈ R+. Then, fN ∈ L (R ), kfN k∗ ≤ 2kfk∗ and fN → f almost everywhere n in R . (iv) Assume f is complex valued. Then f ∈ BMO if and only if Im f, Re f ∈ BMO and

kIm fk∗, kRe fk∗ ≤ kfk∗ ≤ kIm fk∗ + kRe fk∗.

Proof. (i) Let CQ = |mQ f|. Then, ||f| − CQ| ≤ |f − mQ f| and so mQ ||f| − CQ| ≤ mQ |f − mQ f| ≤ kfk∗. Then, apply Lemma 5.1.4. (ii) Apply (i). Exercise.

(iii) Pick Q a cube, and let x, y ∈ Q. Then, |fN (x) − fN (y)| ≤ |f(x) − f(y)| and

fN (x) − mQ fN = (fN (x) − fN (y)) dL (y). Q So,

|fN (x) − mQ fN | ≤ |fN (x) − fN (y)| dL (x)dL (y) Q Q Q

≤ |f(x) − mQ f + mQ f − f(y)| dL (x)dL (y) Q Q

≤ |f(x) − mQ f| + |mQ f − f(y)| dL (x)dL (y) Q Q

≤ 2kfk∗.

42 (iv) Exercise.

1 n Exercise 5.1.8. Find a function f ∈ Lloc(R ) with |f| ∈ BMO, but f∈ / BMO. Proposition 5.1.9. Let Q, R be cubes with Q ⊂ R. Let f ∈ BMO. Then,

L (R) |mQ f − mR f| ≤ kfk . L (Q) ∗

Proof. We compute,

Corollary 5.1.10. (i) Suppose that L (R) ≤ 2L (Q) with Q ⊂ R. Then |mR f − mQ f| ≤ 2kfk∗.

(ii) Suppose that Q, R are arbitrary cubes (not necessarily Q ⊂ R). Then, |mR f − mQ f| ≤ Cnkfk∗ ρ(Q, R) where ! L (R) L (Q) dist(Q, R) ρ(Q, R) = ln 2 + + + 1 . L (Q) L (R) (L (Q) ∧ L (R)) n

Proof. We leave the proof as an exercise but note that the proof of (i) is easy. The proof of (ii) requires a “telescoping argument.”

5.2 John-Nirenberg inequality

BMO was invented by John-Nirenberg for use in partial diﬀerential equations.

Theorem 5.2.1 (John-Nirenberg inequality). There exist constants C = C(n) ≥ 0 and α = α(n) > 0 depending on dimension such that for all f ∈ BMO with kfk∗ 6= 0 and for all cubes Q and λ > 0,

λ −α kfk L {x ∈ Q : |f(x) − mQ f| > λ} ≤ Ce ∗ L (Q)

43 Remark 5.2.2 (On the exponential decay). Note that by the deﬁnition of the BMO norm 1 combined with Tchebitchev-Markov inequality (Proposition 4.1.4), we get decay in λ since

L {x ∈ Q : |f(x) − mQ f| > λ} 1 1 ≤ |f − mQ f| dL ≤ kfk L (Q). ˆ ∗ λ {x∈Q:|f(x)−mQ f|>λ} λ

It is natural to ask the question why we get extra gain into exponential decay. The reason is as follows. The expression above is for a single cube. But the exponential decay comes from the fact that we have scale and translation invariant estimates. This is typical in harmonic analysis.

Proof of the John-Nirenbeg inequality. First, note that it is enough to assume that Q = n Q0 = [0, 1) by the scale and translation invariance of BMO. Furthermore, we can assume that mQ0 f = 0 since kfk∗ = kf − mQ fk∗. By multiplying by a constant coupled with the fact that k· k∗ is a semi-norm, we need to only consider kfk∗ = 1.

−αλ Let Fλ = {x ∈ Q : |f(x)| > λ} . We show that L (Fλ) ≤ Ce . We prove this for fN (the truncation of f) and let N → ∞ to establish the claim via the monotone convergence ∞ n theorem. So, without loss of generality, assume that f ∈ L (R ).

Consider the case when λ > 1. As before, let D(Q) denote the dyadic subcubes of Q Q. Let Eλ = x ∈ Q : M f(x) > λ and we have that Fλ ⊂ Eλ up to a set of null Q measure (since |f| ≤ M f almost everywhere). Hence, L (Fλ) ≤ L (Eλ) and we show −αλ that L (Eλ) ≤ Ce . Then, by deﬁnition

Q sup mQ |f| = M f(x) Q∈D(Q), Q3x and

mQ |f| = mQ |f − mQ f| ≤ kfk∗ = 1.

Coupled with this and the assumption that λ > 1, we have that Eλ $ Q. Let C = {Qi,λ} be the maximal collection of subcubes Q ∈ D(Q) such that Eλ = tQi,λ. So,

mQ |f| > λ and m |f| ≤ λ i,λ Qdi,λ with Qdi,λ ⊂ Q.

n For each i, we estimate mQ f − m f . Let {Rk} be a set of cubes such that i,λ Qdi,λ k=0

Qi,λ = R0 ⊂ R1 ⊂ · · · ⊂ Rn = Qdi,λ and L (R ) k+1 ≤ 2. L (Rk) It is trivial that such a collection exists. By this we have that for all k,

mRk+1 f − mRk f ≤ 2kfk∗ = 2

44 and summing over k,

mQ f − m f ≤ 2n. i,λ Qdi,λ Therefore,

mQ f ≤ 2n + m f ≤ 2n + λ. i,λ Qdi,λ

0 Now, pick a δ > 2n + 1 to be chosen later. Let C = {Qj,λ+δ} be the maximal disjoint covering for Eλ+δ. Then, Eλ+δ ⊂ Eλ and for each j, there exists a unique i such that Qj,λ+δ ⊂ Qi,λ. Fix i, and estimate: L (Eλ+δ ∩ Qi,λ) = L t Qj,λ+δ {j:Qj,λ+δ⊂Qi,λ} X = L (Qj,λ+δ)

{j:Qj,λ+δ⊂Qi,λ} 1 X ≤ |f| dL ˆ λ + δ Qj,λ+δ {j:Qj,λ+δ⊂Qi,λ} 1 ≤ |f| dL ˆ λ + δ Eλ+δ∩Qi,λ 1 1 ≤ f − mQ f dL + mQ f L (Eλ+δ ∩ Qi,λ) ˆ i,λ i,λ λ + δ Qi,λ λ + δ 1 2n + λ ≤ |Q | kfk + L (E ∩ Q ). λ + δ i,λ ∗ λ + δ λ+δ i,λ

Therefore, 1 L (E ∩ Q ) ≤ L (Q ) λ+δ i,λ δ − 2n i,λ and summing over i, 1 L (E ) ≤ L (E ). λ+δ δ − 2n λ

Now, ﬁx δ such that 1 < δ − 2n and observe that

1 k L (E ) ≤ L (E ) λ+kδ δ − 2n λ for k ∈ N. If λ ≥ 2, then there exists a unique k ∈ N such that 2 + kδ ≤ λ < 2 + (k + 1)δ and so it follows that

−k ln(δ−2n) −k ln(δ−2n) −k ln(δ−2n) −αλ L (Eλ) ≤ L (E2+2kδ) ≤ e L (E2) ≤ e L (Q) ≤ e ≤ Ce .

For the case 0 < λ ≤ 2,

α2 −α2 −αλ L (Fλ) ≤ L (Q) ≤ e · e ≤ Ce and the proof is complete.

45 p n Deﬁnition 5.2.3 (BMOp). For f ∈ Lloc(R ), deﬁne

1 p p kfk∗,p = sup(mQ |f − mQ f| ) Q and deﬁne n p n o BMOp = f ∈ Lloc(R ): kfk∗,p < ∞ .

Remark 5.2.4. Note that BMOp ⊂ BMO and kfk∗ . kfk∗,p.

Corollary 5.2.5. For all 1 < p < ∞, BMOp = BMO and kfk∗,p ' kfk∗.

Proof. Fix a cube Q and f ∈ BMO. Then,

∞ p p−1 |f − mQ f| dL = pλ L {x ∈ Q : |f(x) − mQ f| > λ} dλ ˆQ ˆ0 ∞ λ p−1 −α ≤ p λ Ce kfk∗ dλ L (Q) ˆ0 and noting that ∞ λ p−1 −α kfk p p λ Ce ∗ dλ ≤ Ckfk∗ ˆ0 completes the proof.

Exercise 5.2.6. For f ∈ BMO, there exists a β > 0 such that

sup exp(β |f − mQ f|) dL < ∞. Q Q

5.3 Good λ inequalities and sharp maximal functions

We introduce the following variants on centred and uncentred maximal function. They are constructed using arbitrary cubes rather than balls.

1 n Deﬁnition 5.3.1 (Cubic maximal functions). For f ∈ Lloc(R ), deﬁne the centred cubic maximal function: Mf(x) = sup |f(y)| dL (y) r>0 Q(x,r) and the uncentred cubic maximal function:

0 M f(x) = sup |f(y)| dL (y). Q3x Q

Proposition 5.3.2. There exist C1,C2 > 0 such that

C1ML f ≤ M f ≤ C2ML f and 0 0 0 C1ML f ≤ M f ≤ C2ML f

46 Proof. The proof follows easily noting that there exist constants A and B such that for every cube Q, there exist balls B1,B2 with the same centre such that B1 ⊂ Q ⊂ B2 and AL (B1) = L (Q) = BL (B2). 0 Remark 5.3.3. In particular, this means that we can simply substitute M and M in 0 place of ML and ML in Theorems and obtain the same conclusions.

We now introduce a new type of maximal function which will be the primary tool of this section.

1 n n Deﬁnition 5.3.4 (Sharp maximal function). For f ∈ Lloc(R ) and x ∈ R , deﬁne ] M f(x) = sup mQ |f − mQ f| . Q3x

] ∞ n Remark 5.3.5. (i) f ∈ BMO if and only if M f ∈ L (R ). Furthermore, kfk∗ = ] kM fk∞. 0 (ii) M]f ≤ 2M f.

p n ] p n In particular (ii) means that if f ∈ L (R ) with 1 < p < ∞, then M f ∈ L (R ).

It is natural to ask whether there is a converse to (ii) in the previous remark. There is no pointwise inequality - consider f constant. This is also true for Lp functions (see [Ste93]). 0 ] p n The only hope is to prove kM fkp . kM fkp for f ∈ L (R ).

Good λ inequalities (originally from probability theory) help us to establish such a bound. These are distributional inequalities of the following type: Deﬁnition 5.3.6 (Good λ inequality). A good λ inequality is of the form

n n L {x ∈ R : |f(x)| > κλ, |g(x)| ≤ γλ} ≤ ε(κ, γ) L {x ∈ R : |f(x)| > λ} (Iλ) where f, g are measurable, λ > 0, κ > 1, γ ∈ (0, 1) and ε(κ, γ) > 0. Proposition 5.3.7. Suppose there exists a ∈ (0 ∞) such that k k ∞ and assume p0 , f p0 < (Iλ) holds for all λ > 0. Then, for all p ∈ [p0, ∞) satisfying 1 = sup (1 − κpε(κ, γ)) > 0 Cp κ>1, γ∈(0,1) we have 1 κ kfk ≤ (Cp) p kgk p γ p for some κ > 1 and γ < 1.

χ Proof. Let fN = f {x∈Rn:|f(x)|

N p p−1 n |fN | = p λ L {x ∈ R : |f(x)| > λ} dλ ˆ ˆ0 N p κ p−1 n = pκ λ L {x ∈ R : |f(x)| > κλ} dλ. ˆ0

47 Also,

n n n {x ∈ R : |f(x)| > κλ} ⊂ {x ∈ R : |f(x)| > κλ, |g(x)| ≤ γλ} ∪ {x ∈ R : |g(x)| > γλ} and so it follows that

n n L {x ∈ R : |f(x)| > κλ} ≤ L {x ∈ R : |f(x)| > κλ, |g(x)| ≤ γλ} n + L {x ∈ R : |g(x)| > γλ} n ≤ ε(κ, γ)L {x ∈ R : |f(x)| > λ} n + L {x ∈ R : |g(x)| > γλ} by invoking (Iλ). Therefore, it follows that

N p κ p−1 n pκ λ L {x ∈ R : |f(x)| > κλ} dλ ˆ0 N p k p−1 n ≤ pκ ε(κ, γ) λ L {x ∈ R : |f(x)| > λ} dλ ˆ0 N p k p−1 n + pκ λ L {x ∈ R : |g(x)| > γλ} dλ ˆ0 N p p−1 n ≤ pκ ε(κ, γ) λ L {x ∈ R : |f(x)| > λ} dλ ˆ0 p ∞ κ p−1 n + p λ L {x ∈ R : |g(x)| > λ} dλ. γ ˆ0 By the assumption that k k ∞, we have that k k ∞ for all ∈ [ ∞) and so f p0 < fN p < p p0, N p p−1 n (1 − κ ε(κ, γ))p λ L {x ∈ R : |f(x)| > λ} dλ ˆ0 p ∞ κ p−1 n ≤ p λ L {x ∈ R : |g(x)| > λ} dλ. γ ˆ0 Then, apply the monotone convergence theorem to obtain the conclusion.

The goal is to prove the following important inequality. 1 n Theorem 5.3.8 (Feﬀerman-Stein inequality). Let p0 ∈ (0, ∞) and f ∈ Lloc(R ) such that 0 kM k ∞. Then, for all ∈ [ ∞) there exists a 0 (independent of ) such f p0 < p p0, Cp > f 0 ] that kM fkp ≤ CpkM fkp. Corollary 5.3.9. Let p ∈ (1, ∞). Then, there exists a C(p, n) > 0 such that for all p n f ∈ L (R ), 0 ] kM fkp ≤ C(p, n)kM fkp. 0 ] p n In particular, kfkp ' kM fkp ' kM fkp on L (R ).

p n 0 p n Proof. Apply the theorem with p0 = p since f ∈ L (R ) if and only if M f ∈ L (R ).

To prove the Feﬀerman-Stein inequality, by Proposition 5.3.7, it suﬃces to prove (Iλ) with 0 f replaced with M f and g replaced with M]f. First, we need two key Lemmas.

48 Lemma 5.3.10 (Localisation for maximal functions). There exists κ0 = κ0(n) > 1 such 1 n that for all f ∈ Lloc(R ), for all cubes Q, and all λ > 0 if there exists C > 1 and x˜ ∈ CQ 0 with M f(˜x) ≤ λ, then for all κ ≥ κ0

n 0 o 0 n n κ Q ∩ x ∈ R : M f(x) > κλ ⊂ x ∈ R : M (fχMQ)(x) > λ κ0 with M = C + 2.

0 Proof. We know that M f ≤ κ0Mf for some κ0 = κ0(n) > 1. Let

n n 0 o x ∈ Q ∩ x ∈ R : M f(x) > κλ and so κ Mf(x) > λ > λ κ0 since κ > κ0. So, there exists an r > 0 such that 1 κ |f(y)| dL (y) > λ. |Q(x, r)| ˆQ(x,r) κ0

0 First,x ˜ 6∈ Q(x, r) since M f(x) ≤ λ. This implies that kx − x˜k∞ ≥ r. Secondly, by hypothesis,x ˜ ∈ CQ and letting xQ be the centre of Q,

kx − x˜k∞ ≤ kx − xQk∞ + kxQ − x˜k∞ ≤ rad Q + C rad Q ≤ (C + 1) rad Q.

So, for any y ∈ Q(x, r)

ky − xQk∞ ≤ ky − xk∞ + kx − xQk∞ ≤ r + rad Q ≤ (C + 1) rad Q + rad Q ≤ M rad Q Thus, κ M(fχMQ)(x) ≥ |fχMQ| dL = |f| dL > λ Q(x,r) Q(x,r) κ0 and completes the proof.

1 n Lemma 5.3.11 (Proving (Iλ)). Fix q ∈ (1, ∞] and α ≥ 1. Let F ∈ Lloc(R ), F ≥ 0 such n that for all cubes Q there exists GQ,HQ : R → R+ measurable with

(i) F ≤ GQ + HQ almost everywhere x ∈ Q,

n (ii) For all x ∈ R ,

 1 q q 0 sup H q < ∞ αM F (x) ≥ Q3x Q Q . ﬄ kHQkL∞(Q) q = ∞

Set

G(x) = sup GQ. Q3x Q

49 0 0 0 Then, there exists C = C(q, n) > 0 and κ0 = κ0(α, n) ≥ 1 such that for all λ > 0, κ > κ0, 0 and γ ∈ (0, 1],(Iλ) holds for M F and G in place of f and g respectively. That is,

n 0 o n 0 o n n L x ∈ R : M F (x) > κλ, |G(x)| ≤ γλ ≤ ε(κ, γ) L x ∈ R : M F (x) > λ where αq γ ε(κ, γ) = C + . κ κ α Remark 5.3.12. When q = ∞, κ is replaced by 0.

n n 0 o Proof. Let λ > 0 and Eλ = x ∈ R : M F (x) > λ is open by the lower semi-continuity 0 of M F .

n n If Eλ = R , there’s nothing to do. So, suppose that Eλ 6= R and use a Whitney decomposition with dyadic cubes (Theorem 2.3.1). So, Eλ = tQi, mutually disjoint with diam Qi c comparable with dist(Qi, Eλ). In particular, there exists a constant C = C(n) > 1 such c that for all i, CQi ∩ Eλ 6= ∅. That is, for each i, there exists ax ˜i ∈ CQi such that 0 n n 0 o M f(˜xi) ≤ λ. Set Di = Qi ∩ x ∈ R : M f(x) > κλ, G ≤ γλ , and so

n 0 o X n L (Di) = L x ∈ R : M f(x) > κλ, G ≤ γλ i since κ ≥ 1.

We estimate each Di for each i. Assume Di 6= ∅. So, there exists a yi ∈ Qi such that G(yi) ≤ γλ. So, by the Localisation Lemma 5.3.10

n 0 o n L (Di) ≤ L (Qi ∩ x ∈ R : M f(x) > κλ ) n χ κ ≤ L x ∈ R : M(f MQi )(x) > λ κ0 ≤ A + B where 0 κ n χ A = L x ∈ R : M (GMQi MQi )(x) > λ 2κ0 and 0 κ n χ B = L x ∈ R : M (HMQi MQi )(x) > λ . 2κ0

We estimate A by invoking the Maximal theorem (weak type (1, 1)):

κ0 χ κ0 A ≤ 2C GMQi MQi dL ≤ 2C GMQi dL ˆ n ˆ κλ R κλ MQi κ0 G(yi) κ0 ≤ 2C L (MQi) ≤ 2C L (MQi)γ. κ λ κ

50 By the Maximal theorem (weak type (q, q)) for q < ∞,

2 q 2 q κ0 χ q κ0 q B ≤ C (HMQi MQi ) dL = C (HMQi ) dL ˆ n ˆ κλ R κλ MQi q q q 2 0 q 2 2 κ0 κ0 q κ0 q ≤ C L (MQi M f(˜xi) ≤ C L (MQi)(αλ) = L (MQi)C L (MQi)α . κλ κλ κ

If q = ∞, then 0 kH k ∞ ≤ αM F (˜x ) ≤ αλ. MQi L (MQi) i

Thus, if κ > α2κ0, then

0 κ n χ x ∈ R : M (HMQi MQi )(x) > λ = ∅. κ0

We are now in a position to prove the Feﬀerman-Stein inequality.

(i) F ≤ GQ + HQ, and 0 0 (ii) kHQkL∞(Q) = |mQ f| ≤ M f(x) = M F (x) for all x ∈ Q.

] γ Apply Lemma 5.3.11 to get the inequality (Iλ) with G = M f and ε(κ, γ) = C κ .

p Then, apply Proposition 5.3.7 with p ≥ p0 since 1 − κ ε(κ, γ) > 0 for ﬁxed κ and small γ. Thus, we conclude that for some Cp > 0,

0 0 ] kM fkp = kM F kp ≤ CpkGkp = CpkM fkp and the proof is complete.

We have the following Corollary to the Feﬀerman-Stein inequality.

Corollary 5.3.13 (Stampacchia). Suppose that T is sublinear on DRn , a subspace of the space of measurable functions stable under multiplication by indicator functions. Suppose p n p n ∞ n further that T :L (R ) → L (R ) for some p ∈ [1, ∞) and T : DRn ∩ L (R ) → BMO are bounded. Then for all q ∈ (p, ∞), T is strong type (q, q) with log convex control of operators “norms.” f

51 Chapter 6

Hardy Spaces

6.1 Atoms and H1

Hardy spaces are function spaces designed to be better suited to some applications than L1. We consider atomic Hardy spaces. n Deﬁnition 6.1.1 (∞-atom). Let Q be a cube in R . A measurable function a : Q → C is called an ∞-atom on Q if

(i) spt a ⊂ Q, (ii) k k ≤ 1 , a ∞ L (Q)

(iii) Q a dL = 0. ´

∞ ∞ ∞ We denote the collection of ∞-atoms on Q by AQ and A = ∪QAQ .

Remark 6.1.2. Note that (i) along with (ii) implies that kak1 ≤ 1. n Deﬁnition 6.1.3 (p-atom). Let Q be a cube in R . A measurable function a : Q → C is called an p-atom on Q if

(i) spt a ⊂ Q, (ii) k k ≤ 1 , a p 1− 1 L (Q) p p p p (iii) Q a dL = 0. We denote the collection of p-atoms on Q by AQ and A = ∪QAQ. ´ 1,p 1 n 1,p Deﬁnition 6.1.4 (H ). Let 1 < p ≤ ∞ and f ∈ L (R ). We say that f ∈ H if there exist -atoms { } and ( ) ∈ 1( ) such that p ai i∈N λi i∈N ` N ∞ X f = λiai i=0 almost everywhere.

52 P P Remark 6.1.5. The fact that |λi| < ∞ combined with kaik1 ≤ 1 implies that λiai 1 n 1 n P 1 n converges in L (R ). For f ∈ L (R ), then the equality f = λiai is in L (R ). We 0 n could certainly give a similar definition in the more general setting of f ∈ S (R ). Then, 0 n we would ask the convergence of the series in the sense of S (R ) in the definition. Then 1 n 0 n 1 P as L (R ) embeds in S (R ) it coincides with the L function i λiai after identification.

1,p On noting that H is a K vector space, we define a norm. Definition 6.1.6 (H1,p norm). We define

nX X o kfkH1,p = inf |λi| : f = λiai , where the inﬁmum is taken over all possible representations of f.

1,p Proposition 6.1.7. (i) (H , k· kH1,p ) is a Banach space, (ii) Whenever 1 < p < r < ∞, we have

1,∞ 1,r 1,p 1 n H ⊂ H ⊂ H ⊂ L (R ),

Proof. Exercise.

Theorem 6.1.8 (Equivalence of H1,p spaces). For 1 < p < ∞, H1,∞ = H1,p with equivalence of norms.

Proof. (i) We establish what is called the Calder´on-Zygmunddecomposition of functions. More precisely, we show given a p-atom a that there exists a decomposition a = b+g 1,p 1 1,∞ with b ∈ H with kbkH1,p ≤ 2 and g ∈ H with kgkH1,∞ ≤ C(n, p) (Note that 1 kakH1,p ≤ 1 so kbkH1,p ≤ 2 is better). n p Let Q be a cube in R such that a ∈ AQ. As before, let D(Q) denote the dyadic subcubes of Q. We have

p 1 p 1 mQ |a| = |a| ≤ p . L (Q) ˆQ L (Q)

p p Fix α > 0 with α > mQ |a| to be chosen later and let

1 n Q p o Eα = x ∈ Q :(M |a| )(x) p > α .

∞ Note then that Q 6= Eα and Eα = ti=1Qi where Qi ∈ D(Q) maximal for the property p p that mQi |a| > α . Set χ bi = (a − mQi a) Qi P∞ and b = i=1 bi. Let g = a − b.

We note the properties of bi. First, spt bi ⊂ Qi and

bi dL = 0. ˆQi

53 Also, 1 p p 1 1 kbik ≤ |a| + |mQ a| L (Qi) p = λi p ˆ i 1− 1 Qi L (Qi) p

1 p p where λi = 2 |mQi |a| | L (Qi). We note that λi 6= 0. For otherwise, if λi = 0, 1 p p then a = 0 on Qi but we assume that (mQi |a| ) > α > 0 which is a contradiction. Therefore, a = 1 b is a p-atom. i λi i Now, by H¨older’sinequality and the Maximal Theorem,

1 1 1 1− ∞ ∞ ∞ ! p ∞ ! p X X p p 1− 1 X p X λi = 2 |a| L (Qi) p ≤ 2 |a| L (Qi) ˆ ˆ i=1 i=1 Qi i=1 Qi i=1 1 p−1 p 1 C 1 p 1− p p ≤ 2 |a| L (Eα) ≤ 2 |a| p−1 ≤ 2 . ˆQ ˆQ α αL (Q)

Now, we choose α such that

1 p−1 1 2 = αL (Q) 2 to ﬁnd cp α = , L (Q)

1 p−1 1 with cp = 4 . It then follows that kbkH1,p ≤ 2 . Now, consider g, ( a on Q \Eα g = . mQi a on Qi for each i

p Q p p Certainly, on Q \Eα, |a| ≤ M (|a| ) ≤ α almost everywhere and so |g| ≤ α almost everywhere. On Qi, by maximality and H¨older’s inequality,

n n |mQ a| ≤ mQ |a| ≤ 2 m |a| ≤ 2 α. i i Qci Hence, |g| ≤ 2nα. It then follows that

cp kgk ≤ 2nα = 2n . ∞ L (Q)

1 ∞ 1,∞ We also have g = a = 0, so n g ∈ which implies that g ∈ H with Q Q 2 cp AQ n ´ ´ kgkH1,∞ ≤ 2 cp.

1,p 0 (ii) Fix f0 ∈ H with f0 6= 0. We show there exists a decomposition f0 = f1 + g with

2 0 4 n kf k 1,p ≤ kf k 1,p and kg k 1,∞ ≤ 2 c kf k 1,p . 1 H 3 0 H H 3 p 0 H

P∞ For every ε > 0 there exists an atomic decomposition f0 = i=1 λjaj with

∞ X |λj| ≤ kf0kH1,p + ε. i=1

54 1 We apply (i) to each aj to ﬁnd a decomposition aj = bj + gj with kbjkH1,p ≤ 2 . P∞ 1,p Certainly, f1 = j=1 λjbj exists since H is a Banach space (Proposition 6.1.7), and ∞ 1 X 1 kf k 1,p ≤ |λ | ≤ kf k 1,p + ε . 1 H 2 j 2 0 H j=1

1 0 P∞ 1,∞ So, choose ε = 3 kf0kH1,p . Setting g = j=0 λjgj (the sum converging in H n because it is a Banach space and kgjkH1,∞ ≤ 2 cp), we ﬁnd

0 n 4 n kg k 1,∞ ≤ 2 c kf k 1,p + ε = 2 c kf k 1,p . H p 0 H 3 p 0 H

(iii) We iterate

0 f0 = f1 + g 1 f1 = f2 + g 2 f2 = f3 + g . . . = .

and so for each k, 0 1 k−1 f0 = fk + g + g + ··· + g .

1,p 0 1 k 1,∞ Certainly fk → 0 in H , and g + g + ··· + g converges to g in H with

∞ j 4 n X 2 kgk 1,∞ ≤ 2 c kf k 1,p . H 3 p 3 0 H j=0

1,∞ 1,p By Proposition 6.1.7, convergence in H implies convergence in H and so f0 = g 1,∞ and f0 ∈ H .

This motivates the following deﬁnition.

Deﬁnition 6.1.9 (Hardy space H1). We deﬁne H1 to be any H1,p for 1 < p ≤ ∞ with the corresponding norm.

6.2 H1 − BMO Duality

We show that the dual space of H1 and BMO are isomorphic with equivalent norms. This relationship was first established by C. Fefferman but using a different characterisation of H1.

Theorem 6.2.1 (H1 − BMO duality). The dual of H1 is isomorphic to BMO with equivalent norms.

55 1 1,2 Not quite a correct “proof”. We work with H = H and BMO = BMO2 with corre- 2 sponding norms k· kH1,2 and k· k∗. Let b ∈ BMO2 and a ∈ A . Set

Lb(a) = b(x)a(x) dL (x). ˆRn

Then,L b(a) is well deﬁned and since for all cubes Q we have spt a ⊂ Q and Q a = 0, ´ 1 2 |Lb(a)| ≤ |b − mQ b| kak2 ≤ kbk∗. ˆQ

2 2 Pm By linearity, Lb is deﬁned on Vect A . Let f ∈ Vect A . Then, f = i=1 λiai and

m X Lb(f) ≤ kbk∗ λiai. i=1

2 1,2 1,2 By density of Vect A in H ,L b can be deﬁned on all of H .

2 Remark 6.2.2. This proof is correct if we know that for all f ∈ Vect A , kfkH1,2 ' P P inf finite |λi|. Note that we automatically have kfkH1,2 ≤ inf finite |λi|. This subtlety P went unnoticed for some time, and it has only been recently that kfkH1,2 & inf finite |λi| was proved by an intricate argument. Also this equivalence is not true for all atoms - there exists a counter example for ∞-atoms.

We take a diﬀerent approach in proving the theorem.

1 1 1,2 Proof of the H − BMO duality. We work with H = H and BMO = BMO2 with corresponding norms k· kH1,2 and k· k∗.

∞ n 1,2 (i) Take b ∈ L (R ) and f ∈ H . Deﬁne,

Lb(a) = b(x)a(x) dL (x) ˆRn 1,2 1 n P∞ and note that it is well deﬁned since H ⊂ L (R ). If f = i=1 λiai, we can apply Dominated Convergence since

∞ ∞ X X |b(x)λiai(x)| dL (x) ≤ kbk∞ |λi| . ˆ n R i=1 i=1 Also,

|Lb(a)| = (b(x) − mQ b) ai(x) dL (x) ≤ kbk∞ ˆRn P∞ if spt ai ⊂ Q. This implies that |Lb(f)| ≤ kbk∗ i=1 |λi| and taking an inﬁmum over all possible λi, |Lb(f)| ≤ kbk∗kfkH1,2 .

56 2 ∞ (ii) Now take b ∈ BMO2 and f ∈ Vect A . Let (bk)k=1 be the truncation of b. So, 1 |bk| ≤ |b| almost everywhere, bk % b almost everywhere, and kbkk∗ ≤ 2kbk∗. Now, Pm Pm suppose f = i=1 λiai and Lbk (f) = i=1 λiLbk (ai). Since b ∈ BMO2, we have 2 n 2 1 n that b ∈ Lloc(R ) which implies b ∈ L (spt ai). So, certainly |bkai| ≤ |bai| ∈ L (R ) almost everywhere. Thus,

bkai dL → bai dL , ˆRn ˆRn

ie., Lbk (ai) → Lb(a). This implies that

|Lb(f)| ≤ sup |Lbk (f)| ≤ 2kbk∗kfkH1,2 . k

1,2 (iii) We now apply a density argument and extend Lb to the whole of H . Let L˜b denote this extension. 1,2 0 So we have shown that whenever b ∈ BMO2 we have L˜b ∈ (H ) . Let T : BMO2 → 1,2 0 (H ) denote the map b 7→ L˜b. (iv) We leave it as an exercise to show that T is linear.

(v) We show that T is injective. Let b ∈ BMO2 such that L˜b = 0. We show that b is constant. 2 Fix a cube Q and let f ∈ L (Q) with Q f dL = 0. Then, there exists a λ ∈ K such f 2 ´ 2 that λ ∈ AQ which implies that f ∈ Vect A . So,

0 = L˜b(f) = Lb(f) = bf dL ˆQ

n and since we assume Q f = 0, it follows that b|Q is constant. By exhaustion of R by increasing Q, we deduce´ that b is constant.

1,2 0 2 (vi) Lastly, we show that T is surjective. Let L ∈ (H ) and ﬁx a cube Q. Let L0(Q) = n 2 o 2 1,2 2 f ∈ L (Q): Q f dL = 0 and note that L0(Q) ⊂ H . Take f ∈ L0(Q) and ´ 1 λ = kfk2L (Q) 2 . Then, f 1 k k ≤ 1 λ 2 L (Q) 2 f so λ is a 2-atom. Thus,

f 1 2 kfkH1,2 = λk k ≤ kfk2L (Q) . λ H1,2

2 0 Furthermore, L| 2 ∈ (L (Q)) and so by the Riesz Representation Theorem, there L0(Q) 0 2 2 exists a bQ ∈ L0(Q) such that for all f ∈ L0(Q),

L(f) = bQf dL ˆQ

1 2 and kbQk2 ≤ kLk(H1,2)0 L (Q) .

1Here we assume b real valued. For complex valued b, separate real and imaginary parts.

57 0 0 2 Let Q, Q denote two cubes with Q ⊂ Q . Then, whenever f ∈ L0(Q),

L(f) = bQf dL = bQ0 f dL ˆQ ˆQ0

and so bQ − bQ0 is constant almost everywhere in Q. Deﬁne b as follows:

( n b n (x) x ∈ [−1, 1] ( ) = [−1,1] b x j j n j−1 j−1 n b[−2j ,2j ]n (x) + cj x ∈ [−2 , 2 ] \ [−2 , 2 ] , j ≥ 1

n where cj is the constant such that b[−2j ,2j ]n − b[−1,1]n = −cj on [−1, 1] . ˜ We show that b ∈ BMO2, kbk∗ ≤ kLk(H1,2)0 and L = Lb. Fix Q, and let j ∈ N j j n such that Q ⊂ [−2 , 2 ] . Let k be such that 2 ≤ k ≤ j. Then, ck − ck−1 = k−1 k−1 b[−2k,2k]n − b[−2k−1,2k−1]n which is constant on [−2 , 2 ] and in particular on k−1 k−1 k−2 k−2 j j n [−2 , 2 ] \ [−2 , 2 ]. Therefore, b(x) = b[−2j ,2j ]n (x) + cj on all of [−2 , 2 ] and in particular on Q. Also, there exists a constant c such that b[−2j ,2j ]n − bQ = c on the cube Q and so b = bQ + c + cj on Q. Then,

b − mQ b = bQ + c + cj − mQ bQ − c − cj = bQ

since mQ bQ = 0. Therefore,

2 2 2 |b − mQ b| dL = |bQ| dL ≤ kLk(H1,2)0 L (Q). ˆQ ˆQ

2 The fact that L = L˜b follows from the fact that L(a) = L˜b(a) for all a ∈ A .

Remark 6.2.3. BMO and “atomic H1” can be deﬁned on any space of homogeneous type. The results of this section go through in such generality. See [CR77].

58 Chapter 7

Calder´on-Zygmund Operators

7.1 Calder´on-Zygmund Kernels and Operators

n n n We denote the diagonal of R × R by ∆ = {(x, x): x ∈ R }. Definition 7.1.1 (Calderón-Zygmund Kernel). Let 0 < α ≤ 1.A Calderón-Zygmund c Kernel of order α is a continuous function K : ∆ → K such that there exist a C > 0 and satisfies:

(i) For all (x, y) ∈ c∆, C |K(x, y)| ≤ , |x − y|n

0 n 0 1 (ii) For all x, y, y ∈ R satisfying |y − y | ≤ 2 |x − y| when x 6= y, 0 α 0 |y − y | 1 K(x, y) − K(x, y ) ≤ C , |x − y| |x − y|n

0 0 n 0 1 (iii) For all x, x , y ∈ R satisfying |x − x | ≤ 2 |x − y| when x 6= y, 0 α 0 |x − x | 1 K(x, y) − K(x , y) ≤ C , |x − y| |x − y|n

We write K ∈ CZKα and norm it via kKkα = inf {C : (i) to (iii) hold}. 1 Remark 7.1.2. (i) The constant 2 can be replaced by any θ ∈ (0, 1). Then, the constant C changes. (ii) The Euclidean norm |· | can be changed to any other norm. Again, C changes.

(iii) When α = 1, ∇yK(x, y) exists almost everywhere and satisﬁes: C0 |∇yK(x, y)| ≤ |x − y|n+1 for all (x, y) ∈ c∆.

59 (iv) When α = 1, ∇xK(x, y) exists almost everywhere and satisﬁes:

C0 |∇xK(x, y)| ≤ |x − y|n+1

for all (x, y) ∈ c∆.

∗ ∗ (v) Deﬁne K (x, y) = K(y, x). Then, K ∈ CZKα implies K ∈ CZKα.

2 n Deﬁnition 7.1.3 (Kernel associated to an operator). Let T ∈ L(L (R )). We say that a c 2 n kernel K : ∆ → K is associated to T if for all f ∈ L (R ), with spt f compact,

T f(x) = K(x, y)f(y) dL (y) ˆRn for almost every x ∈ c(spt f).

Remark 7.1.4. This integral is a Lebesgue integral for all x ∈ c(spt f). Moreover, this says that T f can be represented by this integral away from the support of f.

Definition 7.1.5 (Calderón-Zygmund Operator). A Calderón-ZygmundOperator of order 2 n α is an operator T ∈ L(L (R )) that is associated to a K ∈ CZKα. We define CZOα to be the collection of all Calderón-Zygmundoperators of order . Also, k k = α T CZOα k k 2 n + k k . T L(L (R )) K α ∗ Remark 7.1.6. (i) T ∈ CZOα if and only if T ∈ CZOα. 2 n Also, let f, g ∈ L (R ) with spt f, spt g compact and spt f ∩ spt g = ∅. Then,

hT ∗g, fi = hg, T fi = g(x) K(x, y)f(y) dL (y) dL (x) ˆRn ˆRn = f(y) K(x, y)g(x) dL (x) dL (y) ˆRn ˆRn

∗ and since f was arbitrary, T g(y) = n K(x, y)g(x) d (x) for almost every y ∈ R L c(spt g). That is, T ∗ has associated kernel´ K∗.

tr tr (ii) T ∈ CZOα if and only if T ∈ CZOα, where T is the real transpose of T . The associated kernel to T tr is Ktr(x, y) = K(y, x).

(iii) The map T 7→ K, where T ∈ CZOα and K ∈ CZKα the associated kernel, is not injective. Consequently, one cannot deﬁne a CZOα uniquely given a kernel K ∈ CZOα. The following is an important illustration. ∞ n Let m ∈ L (R ) and let Tm be the map f 7→ mf. It is easy to check that this is 2 n c 2 n a bounded operator on L (R ). Let K = 0 on ∆ and let f ∈ L (R ) with spt f compact. Then, whenever x 6∈ spt f, Tmf(x) = m(x)f(x) = 0. Therefore,

Tmf(x) = K(x, y)f(y) dL (y) ˆRn

c whenever x ∈ (spt f), which shows the associated kernel to Tm is 0.

60 7.2 The Hilbert Transform, Riesz Transforms, and The Cauchy Operator

We discuss three important examples that have motivated the theory.

7.2.1 The Hilbert Transform

n 0 n Deﬁnition 7.2.1 (Hilbert Transform). We deﬁne a map H : S (R ) → S (R ) by 1 H(ϕ) = p. v. ∗ ϕ. πx That is, 1 1 p. v. ∗ ϕ (y) = lim ϕ(y − x) dL (x). πx ε→0 ˆ{x:|x|>ε} πx

2 Proposition 7.2.2. H extends to a bounded operator on L (R).

Proof. We can analyse this convolution via the Fourier Transform. For a function ϕ ∈ n S (R ), the Fourier transform is given by

ϕˆ(ξ) = e−ıx·ξ dL (x). ˆRn 0 n We can extend this naturally to T ∈ S (R ) by defining Tˆ via hTˆ , ϕi = hT, ϕî for every n ϕ ∈ S (R ). So, when ϕ ∈ S (R), 1 1 hp. v. ˆ, ϕi = hp. v. , ϕî πx πx 1 = lim ϕˆ(x) dL (x) ε→0 ˆ{x:|x|>ε} πx 1 = lim ϕˆ(x) dL (x) ε→0 ˆ{x:ε−1>|x|>ε} πx ! 1 = lim ϕ(ξ) eıx·ξ dL (x) dL (ξ). ε→0 ˆRn ˆ{x:ε−1>|x|>ε} πx n Now, fix ξ ∈ R . Then, 1 1 eıx·ξ dL (x) = −ı sin(x· ξ) dL (x) ˆ{x:ε−1>|x|>ε} πx ˆ{x:ε−1>|x|>ε} πx

= −2ı sin(x· ξ) dL (x) ˆ{x:ε−1>x>ε}

= −2ı sin(x |ξ|) sgn(ξ) dL (x) ˆ{x:ε−1>x>ε} 1 2ı ε|ξ| sin u = − sgn(ξ) dL (u). π ˆ ε u |ξ|

61 The integral appearing on the right hand side is uniformly bounded on ε and ξ. Thus, by Dominated Convergence,

1 hp. v. ˆ, ϕi = −ı sgn(ξ)ϕ(ξ) dL (ξ) πx ˆR and so for all ϕ ∈ S (R), Hϕd(ξ) = −ı sgn(ξ)ϕ ˆ(ξ). Since the Fourier transform is bounded 2 2 on L (R), we extend H to the whole of L (R) by deﬁning Hfd(ξ) = −ı sgn(ξ)fˆ(ξ) almost everywhere in R. Then, this extension agrees on S (R) and by Plancherel Theorem, kHfk2 = kfk2.

Proposition 7.2.3. H ∈ CZO1.

Proof. Let K ∈ CZK1 be deﬁned by 1 K(x, y) = , π(x − y)

2 c when x 6= y. Fix f ∈ L (R) with spt f compact. Then, ﬁx x ∈ (spt f) and choose ∞ r such that B(x, r) ∩ spt f = ∅. So, there exists a sequence ϕn ∈ Cc (R) such that 2 n spt ϕn ∩ B(x, r) = ∅ and ϕn → f in L (R ). Then, for every z ∈ B(x, r),

Hϕn(z) = K(z, y)ϕn(y) dL (y) → K(z, y)f(y) dL (y) ˆR ˆR 2 n c and Hϕn → Hf in L (R ). Covering (spt f) with countably many such balls, we conclude that Hf(x) = K(x, y)f(y) dL (y) ˆRn c almost everywhere x ∈ (spt f). Therefore H ∈ CZO1.

∞ The Hilbert Transform comes from Complex Analysis. Let f ∈ Cc (R) and take the Cauchy extension f to C \ R. That is, 1 f(t) F (z) = dL (t) 2πı ˆR z − t where z = x + ıy, y 6= 0. It is an easy fact that F is holomorphic on C \ R. But C \ R is not connected. So, 1 lim F (x ± ıy) = (f(x) ± ıHf(x)) y→0+ 2 and consequently F (x + ıy) − F (x − ıy) Hf(x) = lim , y→0+ ı and f(x) = lim F (x + ıy) + F (x − ıy). y→0+

We have the following Theorem of M.Riesz:

62 Theorem 7.2.4 (Boundedness of the Hilbert Transform). H has a bounded extension to p L (R) for 1 < p < ∞.

Corollary 7.2.5. Let F±(x) = limy→0+ F (x ± ıy). Then the decomposition f = F+ + F− p n is topological in L (R ). That is kfkp ' kF+kp + kF−kp. 1 Remark 7.2.6. When f is real valued, 2 Hf is the imaginary part of F+.

7.2.2 Riesz Transforms

Motivated by the symbol side of the Hilbert Transform, we define operators Rj for j = n 1, . . . , n on R . 2 n 2 n Definition 7.2.7 (Riesz Transform). Define Rj :L (R ) → L (R ) by

ξj (Rjf)ˆ(ξ) = −ı fˆ(ξ) |ξ| for j = 1, . . . , n.

We note that by Plancherel’s Theorem, Rj is well deﬁned and in particular kRjfk2 ≤ kfk2.

Proposition 7.2.8. Rj ∈ CZO1.

Proof. Consider xj Kj(x) = p. v. cn |x|n+1 0 n for some cn > 0. Then Kj ∈ S (R ). If we can show that for appropriate cn,

ξj Kcj(ξ) = −ı |ξ|

0 n in S (R ), by the same argument as for the Hilbert Transform,

xj − yj Rjf = cn n+1 f(y) dL (y) ˆRn |x − y| 2 n c for all f ∈ L (R ) with spt f compact and for almost every x ∈ (spt f).

n We compute the Fourier Transform of Kj. Fix ϕ ∈ S (R ). Then,

xj −ıx·ξ hKcj, ϕi = hKj, ϕˆi = lim cn e ϕ(ξ) dL (ξ) dL (x). ε→0 n+1 ˆ{x:ε<|x|<ε−1} |x| ˆRn For ξ 6= 0, let xj −ıx·ξ Iε = cn n+1 e dL (x). ˆ{x:ε<|x|<ε−1} |x|

As before, we show that |Iε| is uniformly bounded in ξ and ε and that

ξj Iε → −ı |ξ|

63 as ε → 0.

As previously, write e−ıx·ξ = cos(x· ξ) − ı sin(x· ξ). We only need to regard the imaginary ξ part. By a change of variables, let w = |ξ| and x = |ξ| y. Then,

yj Iε = −ıcn n+1 sin(y· w) dL (y) ˆn ε 1 o | | x: |ξ| <|x|< |ξ|ε y since the Jacobian factor of the change of variables is cancelled by the homogeneity of xj . |x|n+1

We change variables again, this time to polar coordinates. Let y = rθ, for r > 0 and θ ∈ Sn−1. Then,

1 ! ε|ξ| sin(rθ· w) Iε = −ıcn θj dr dσ(θ) ˆ n−1 ˆ ε r S |ξ|

n−1 where dσ is the surface measure on S . So, |Iε| is uniformly bounded since

1 ε|ξ| sin(rθ· w) dr ˆ ε r |ξ| is uniformly bounded in ε, |ξ| and θ. Furthermore,

1 ε|ξ| sin(rθ· w) dr → sgn(θ· w) ˆ ε r |ξ| as ε → 0 and so π Iε → −ıcn θj sgn(ω· w) dσ(θ). 2 ˆSn−1 Write

aj = θj sgn(θ· w) dσ(θ) ˆSn−1 and let π a = (a1, . . . , an) = −ıcn ((θ − (θ· w)w) + (θ· w)w) sgn(θ· w) dσ(θ) 2 ˆSn−1 π = −ıcn |θ· w| dσ(θ) w. 2 ˆSn−1 because (θ − (θ· w)w) sgn(θ· w) is odd in the symmetry with respect to the hyperplane {w}⊥ and Sn−1 is invariant under this symmetry. By rotational invariance,

|θ· w| dσ(θ) = |θ1| dσ(θ) ˆSn−1 ˆSn−1 and so we deﬁne cn by π cn |θ1| dσ(θ) = 1. 2 ˆSn−1 Then, it follows that ξj aj = −ıwj = −ı |ξ| and the proof is complete.

64 p n Theorem 7.2.9. Rj is bounded on L (R ) whenever 1 < p < ∞. n Corollary 7.2.10 (Application to PDEs). Let ϕ ∈ S (R ). Then, ∂i∂jϕ = −RiRj∆ϕ and

k∂i∂jϕkp ≤ C(n, p)k∆ϕkp whenever 1 < p < ∞.

n Proof. We note that for all ξ ∈ R ,

(∂i∂jϕ)ˆ(ξ) = (−ıξi)(−ıξj)ϕ ˆ(ξ) ξi ξj = −ı −ı |ξ|2 ϕˆ(ξ) |ξ| |ξ|   n ξi ξj X 2 = −ı −ı  ξj ϕˆ(ξ) |ξ| |ξ| j=1 ξi ξj = −ı −ı (−∆ϕ)ˆ(ξ) |ξ| |ξ| and by application of the theorem, the proof is complete.

7.2.3 Cauchy Operator

The Cauchy Operator is an example of an operator that is not of convolution type.

2 Identify R ' C. Let ϕ : R → R be a Lipschitz map. That is, there exists an M > 0 such that |ϕ(t) − ϕ(s)| ≤ M |t − s|. By Rademacher’s Theorem [Fed96, Theorem 3.1.6], 0 ∞ 0 ϕ is diﬀerentiable almost everywhere and ϕ ∈ L (R) with kϕ k∞ ≤ M. Now, let Γ = {t + ıϕ(t): t ∈ R} ⊂ C.

If f is smooth in a neighbourhood of Γ and has compact support, then whenever z 6∈ Γ, deﬁne 1 f(w) f(s + ıϕ(s)) F (z) = dw = (1 + ıϕ0(s)) ds 2πı ˆ z − w ˆR z − (s + ıϕ(s)) where z = Z(t) + ıα and Z(t) = t + ıϕ(t). Fix t. Then, 1 lim = F (Z(t) + ıα) = f(z(t)) + Cf(z(t)) α→0± 2 (which are the Plemelj formula’s - details in [Mey92, Volume 2]) where the Cauchy operator is given by

1 f(z(s)) 1 f(w) Cf(z(t)) = lim z0(s) ds = lim dw. ε→0 2πı ˆ{s:|z(t)−z(s)|>ε} z(t) − z(s) ε→0 2πı ˆ{w∈Γ:|z−w|>ε} z − w

Let f˜(s) = f(z(s))z0(s). Then, 1 Cf(z(t)) = p. v. f˜(s) ds = T f˜(t). ˆ{s:|z(t)−z(s)|>ε} z(t) − z(s)

65 Theorem 7.2.11 (Coifman, McIntosh, Meyer (1982)). T ∈ CZO1 with kernel 1 p. v. ∈ CZK1. z(t) − z(s)

˜ ˜ The hard part of the theorem is to show kT fk2 ≤ Ckfk2. As a consequence, Corollary 7.2.12. (i) C is bounded on L2(Γ, |dw|) where |dw| is the arclength measure, (ii) The decomposition

f(z) = lim F (z(t) + ıα) + lim F (z(t) + ıα) α→0+ α→0−

is topological in Lp(Γ, |dw|).

These results have important applications in boundary value problems, geometric measure theory and partial diﬀerential equations.

Remark 7.2.13. We emphasise that this operator C is not of convolution type. Unlike in the previous two examples, we cannot employ the Fourier transform or simple techniques.

p 7.3 L boundedness of CZOα operators

2 The L boundedness of CZOα operators comes for free by deﬁnition. It is an interesting q n p n question to ask when T ∈ CZOα is a bounded map from L (R ) to L (R ). But ﬁrst, we have the following proposition which shows that at least for the Hilbert transform, p = q.

q p Proposition 7.3.1. Suppose the Hilbert transform H :L (R) → L (R) for some p, q > 1 is bounded. Then, p = q.

q n Proof. Let f ∈ L (R ) and consider the function g(x) = f(λx) for λ > 0. Then,

α β λ kHfkq ≤ Cλ kfkp

1 1 and so α = − q , β = − p and α = β which implies p = q.

As a heuristic, we cannot hope to prove Lq to Lp boundedness unless p = q.

1 c Deﬁnition 7.3.2 (H¨ormanderkernel). Let K ∈ Lloc( ∆) and suppose there exists CH > 0 such that

0 esssup(y,y0)∈R2n K(x, y) − K(x, y ) dL (x) ≤ CH . ˆ{x:|x−y|≥2|y−y0|}

Then, K is called a H¨ormanderkernel. Remark 7.3.3. The number 2 appearing in the set of integration is irrelevant. This can be replaced by any A > 1 at the cost of changing CH .

66 Lemma 7.3.4. (i) Every CZKα kernel is H¨ormander.

(ii) The adjoint of a CZKα kernel is H¨ormander.

∗ Proof. The proof of (ii) follows easily from (i) observing that K ∈ CZKα implies K ∈ CZKα.

We prove (i). Let K ∈ CZKα and so we have that

0 α 0 |y − y | 1 K(x, y) − K(x, y ) ≤ Cα |x − y| |x − y|n

0 1 whenever |y − y | ≤ 2 |x − y| and x 6= y. So,

|y − y0|α 1 n dL (x) ˆ{x:|x−y|≥2|y−y0|} |x − y| |x − y| ∞ α X |y − y0| 1 = n dL (x) ˆ j 0 j+1 0 |x − y| |x − y| j=0 {x:2 2|y−y |≤|x−y|≤2 2|y−y |} 0 α ∞ |y − y | X j+1 0 ≤ (B(y, 2 y − y )) 0 α+n L 2 |y − y | j=0 ≤ A(α, n).

We now present the following important and main lemma.

1 n Lemma 7.3.5 (Calder´on-Zygmund decomposition). Let f ∈ L (R ) and λ > 0. Then n there exists a C(n) > 0 and a decomposition of f = g + b almost everywhere on R where ∞ n P∞ g ∈ L (R ) with kgk∞ ≤ C(n)λ, and b = i=1 bi where

(i) spt bi ⊂ Bi with Bi a ball,

(ii) |bi| d ≤ C(n)λ (Bi), Bi L L ´ (iii) n bi = 0, R ´ (iv) {Bi} have the bounded overlap property

∞ X χ Bi ≤ C(n), i=1

P∞ 1 (v) i=1 L (Bi) ≤ C(n) λ kfk1. Remark 7.3.6. (i) The constant C(n) depends only on the dimension n.

P∞ P∞ 2 P∞ (ii) Note that i=1 kbik1 ≤ C(n)λ i=1 L (Bi) ≤ C(n) kfk1 which shows that i=1 bi 1 1 n 2 converges in L . Hence, b ∈ L (R ) with kbk1 ≤ C(n) kfk1.

67 ∞ n 1 n p n (iii) The fact that g ∈ L (R ) ∩ L (R ) implies g ∈ L (R ) for all p ∈ [1, ∞]. In the case of p = 2, q q p 2 kgk2 ≤ kgk1kgk∞ ≤ (1 + C(n)) C(n) λkfk1.

Proof of the Calder´on-Zygmunddecomposition. Recall that M0f is the uncentred maxi- n n 0 mal function of f on balls of R . We know that the set Ωλ = {x ∈ R : M f(x) > λ} is open and of ﬁnite measure by the Maximal Theorem: C L (Ω ) ≤ kfk . λ λ 1 n o n ˜ ˜ Also, Ωλ 6= R . Let E be a Whitney covering of Ωλ. Set Bi = c1Bi : Bi ∈ E where c1 is the constant in the Whitney Covering Lemma 2.3.4. Then, (iv) is proved and ∞ ∞ X X C(n) L (Bi) = χB dL ≤ C(n)χΩ dL ≤ C kfk ˆ i ˆ λ λ 1 i=1 i=1 which proves (v).

−1 We can now take c ∈ (0, 1) (say, c = c1 ) and so {cBi} are mutually disjoint. Then, P construct a partition of unity ϕi so that i ϕi = 1 on Ωλ and ϕi = 1 on cBi. Explicitly, χ = Bi ϕi P χ . j Bj Now, set ( fϕi − B fϕi dL on Bi bi = i . 0ﬄ otherwise

Since we allow Bi to be closed we (i) is proved and (iii) is apparent from the construction of bi.

Now, to prove (ii), we note that |bi| d ≤ 2 |f| d and Bi L Bi L ´ ´ c ˜ c 4Bi ∩ Ωλ = 4c1Bi ∩ Ω 6= ∅. 0 c Then, |f| d ≤ M f(z) (4Bi) for all z ∈ 4Bi. Choosing z ∈ Ω we observe that 4Bi L L λ M0f(z´) ≤ λ and so n |bi| dL ≤ 2λL (4Bi) ≤ 2 4 λL (Bi) ˆBi which establishes (ii).

Deﬁne: ( c f on Ωλ g = P . ( fϕi d )χ on Ω i Bi L Bi λ c 0 ﬄ Then, on Ωλ, f ≤ M f ≤ λ almost everywhere. On Ωλ, by invoking the bounded overlap property, X n fϕi dL χB ≤ C(n) sup fϕi dL ≤ C(n) sup |f| dL ≤ C(n)4 λ. i i Bi i Bi i Bi This completes the proof.

68 Theorem 7.3.7. Every T ∈ CZOα is of weak type (1, 1).

We have the following immediate consequence.

Corollary 7.3.8. Let T ∈ CZOα. Then, for all p ∈ (1, ∞), T is strong type (p, p).

Proof. Since T is weak type (1, 1) by the Theorem and strong type (2, 2) by deﬁnition, we have that T is strong type (p, p) for p ∈ (1, 2).

∗ ∗ Now, note that T ∈ CZOα implies that T ∈ CZOα and so T has a bounded extension p n 0 p0 n 0 p n to L (R ) = L (R ) for 1 < p < 2. Therefore, T has a bounded extension to L (R ) for 2 < p < ∞.

1 n 2 n Proof of Theorem 7.3.7. Let f ∈ L (R ) ∩ L (R ) and ﬁx λ > 0. We show that

n C L ({x ∈ R : |T f(x)| > λ}) ≤ kfk λ 1 with C independent of f and λ. Since we only know T f(x) when x 6∈ spt f, we use the Calder´on-Zygmund decomposition to localise. Let f = g + b this decomposition at level 2 n λ with the properties of g and b from Lemma 7.3.5. Since f, g ∈ L (R ), we also have 2 n P∞ χ b ∈ L (R ). Since b = i=1 bi, with bi = (fϕi − mBi fϕi) Bi we have that this series 2 n converges in L (R ). So, T f = T g + T b and we estimate by Markov’s inequality n λ 4 2 4 2 = ∈ : | ( )| ≤ | | ≤ k k 2 n k k A L x R T g x > 2 T g dL 2 T L(L (R )) g 2 2 λ ˆRn λ

2 and use kgk2 ≤ C(n)λkfk1.

P∞ P∞ 2 n Now, T (b) = T ( i=1 bi) = i=1 T (bi) with the series on the right converging in L (R ) P and |T (b)| ≤ i |T (bi)| almost everywhere. So, with c > 1 to be chosen later,

Consequently, it is enough to prove that

sup |T bi| dL ≤ C(T )kbik1 ˆ n i R \cBi

69 since kbik1 ≤ C(n)λL (Bi) which gives

∞ ∞ X 1 X C(n)2 kbik ≤ C(n) L (Bi) ≤ kfk . λ 1 λ 1 i=1 i=1

n We note that for almost everywhere x ∈ \ cBi, T bi(x) = K(x, y)bi(y) d (y). Let R Bi L n yi be the centre of the ball Bi. Since n bi d = 0, for almost´ all x ∈ \ cBi, R L R ´

T bi(x) = (K(x, y) − K(x, yi))bi(y) dL (y). ˆBi

We choose c = 2 since 2 |y − yi| ≤ 2 rad Bi ≤ |x − y| and

! |T bi| dL ≤ |bi(y)| |K(x, y) − K(x, yi)| dL (x) dL (y) ˆ n ˆ ˆ R \2Bi y∈Bi {x:|x−y|≥2|y−yi|}

≤ |bi(y)| CH (K) dL (y) ≤ CH (K)kbik1 ˆRn where CH (K) is the H¨ormanderconstant associated with K. Taking an inﬁmum on the right hand side, we have

| | ≤ k k k k T bi dL K CZKα bi 1. ˆ n R \2Bi

The sum + gives us the desired conclusion with constant ≤ ( )(k k 2 n + A B C C n T L(L (R )) k k = ( )k k . K CZKα C n T CZOα

1 n 1 n For a general f ∈ L (R ) let fk → f be a sequence which converges in L (R ) with each 2 n fk ∈ L (R ). Without loss of generality, assume that fk → f almost everywhere (since we can pass to a subsequence). The weak type (1, 1) condition gives that T fk is Cauchy 1,∞ n in measure and call T˜ f the limit. This exists almost everywhere and T˜ f ∈ L (R ). Furthermore, T˜(f)(x) = K(x, y)f(y) dL (y) ˆRn for almost every x ∈ c(spt f) with spt f compact.

Remark 7.3.9. It would also suﬃce to prove for general f in the previous Theorem by 1 n 2 n 1 n 1 n 2 n 1,∞ n noting L (R ) ∩ L (R ) is dense in L (R ) and that T :L (R ) ∩ L (R ) → L (R ) is 1,∞ n 1 n 1,∞ n bounded. Since L (R ) is complete, T extends to a bounded map T˜ :L (R ) → L (R ). Example 7.3.10. We note that

1 x − 1 H(χ[0,1])(x) = − ln π x whenever x 6∈ [0, 1].

1 1 This example is of importance because H is a CZO, χ[0,1] ∈ L (R) but H(χ[0,1]) 6∈ L (R).

70 7.4 CZO and H1

1 A natural question to ask is: what subspace of L should we choose so that a CZOα maps that space back into L1?

1 1 n Theorem 7.4.1. Let T ∈ CZOα. Then, T induces a bounded operator H → L (R ). ∞ n Corollary 7.4.2. Let T ∈ CZOα. Then, T extends to a bounded operator from L (R ) to BMO.

∞ n 1 tr 1 Proof. Let f ∈ L (R ) and g ∈ H . Then, Lg = hf, T gi is a linear functional on H satisfying

tr tr tr h i = ≤ k k k k 1 1 n k k 1 f, T g fT g dL f ∞ T L(H ,L (R )) g H . ˆRn By duality, there exists a β ∈ BMO such that L = Lβ. Deﬁne T f = β, with β identiﬁed with Lβ. Remark 7.4.3. (i) This was originally proved directly, without alluding to duality. See [Ste93].

(ii) We apply T f to p-atoms. Let a ∈ A p. Then,

hT f, ai = Lβ(a) = βa dL . ˆRn

Let B = B(yB, rB) = spt a. Then,

hT f, ai = fT tr(a) dL ˆRn = fT tr(a) dL + fT tr(a) dL ˆ2B ˆRn\2B = T (fχ2B)a dL + f(y) K(x, y)a(x) dL (x) dL (y) ˆRn ˆRn\2B ˆRn

= T (fχ2B)a dL + ˆRn ! f(y)(K(x, y) − K(yB, y)) dL (y) a(x) dL (x) dL (y) ˆRn ˆRn\2B

by the application of Fubini. So, on B there exists a constant CB such that

β(y) = T (fχ2B)(y) + f(y)(K(x, y) − K(yB, y)) dL (y) + CB. ˆRn\2B

∞ 1 n Proof of Theorem 7.4.1. We show that whenever a ∈ A , then T a ∈ L (R ) with kT ak1 ≤ 2 n ∞ n C(n, T ). We automatically have T a ∈ L (R ) since a ∈ L (R ) and spt a ⊂ B a ball. 1 Then, since kak2 ≤ 1 , L (B) 2

1 1 n/2 | | ≤ (2 ) 2 k k 2 ≤ (2 ) 2 k k 2 n k k ≤ 2 k k 2 n T a dL L B T a L (2B) L B T L(L (R )) a 2 T L(L (R )). ˆ2B

71 As in the proof of Theorem 7.3.7,

| | ≤ ( )k k k k T a dL C n K CZOα a 1. ˆRn\2B

1 1 n 1,∞ n 1 1 Since H ⊂ L (R ), T f ∈ L (R ) for every f ∈ H . So, ﬁx f ∈ H and pick a P∞ P∞ ∞ representation: f = j=1 λjaj where j=1 |λj| ≤ 2kfkH1 with aj ∈ A . This series 1 P∞ converges almost everywhere in L to f. Thus, T j=1 λjaj = T f almost everywhere. Also, ∞ ∞ X X kλjT ajk1 ≤ |λj| C(n, T ) ≤ 2kfkH1 C(n, T ). j=1 j=1 P∞ 1 Thus, j=1 λjT aj converges in L and hence,

∞  ∞  X X λjT aj = T  λjaj j=1 j=1 holds almost everywhere (remark that the equality for ﬁnite sums is trivial). Hence T f ∈ 1 n L (R ) and kT fk1 ≤ 2C(n, T )kfkH1 .

Proposition 7.4.4. Let T ∈ CZOα. Then, T 1 is deﬁned as a BMO function.

∞ n ∞ n Proof. Follows easily from the fact that 1 ∈ L (R ), and T :L (R ) → BMO is bounded.

Remark 7.4.5. To compute T 1, use the formula for T f on each ball B for f = 1.

1 1 tr Corollary 7.4.6. Let T ∈ CZOα. Then T maps H to H if and only if T 1 = 0 (in BMO).

Before we prove this corollary, we need the following lemmas.

∞ Lemma 7.4.7. Let T ∈ CZOα with associated kernel K ∈ CZKα and a ∈ A with j+1 j spt a ⊂ B = B(yB, rB). For each j ∈ N with j ≥ 1, let Cj(B) = 2 B \ 2 B. Then, for all x ∈ Cj(B), | ( )| ≤ k k 2−j(n+α) −n T a x K CZKα rB .

Proof. We compute and use the α regularity of K,

|y − y | α 1 | ( )| ≤ k k B | ( )| ( ) T a x K CZKα n a y dL y ˆy∈B |x − yB| |x − yB| 1 ≤ k k α | ( )| ( ) K CZKα rB j n+α a y dL y (2 rB) ˆy∈B and the result follows since |a(y)| dL (y) ≤ 1. ˆy∈B

72 n Lemma 7.4.8. Let m : R → C and B = B(yB, rB) a ball such that

1. | |2 ≤ C 2B m dL L (B) , ´ j+1 j 2. For every j ∈ N, j ≥ 1, and x ∈ Cj(B) = 2 B \ 2 B we have

| ( )| ≤ k k 2−j(n+α) −n m x K CZKα rB .

1 Then, ∈ H and k k 1 does not exceed a constant depending on , k k and 0. m m H n K CZKα α >

The proof is left as an exercise.

Proof of Corollary 7.4.6. Let a be an ∞-atom with spt a ⊂ B. By Lemma 7.4.7 and previously, we have shown that a satisﬁes the hypothesis of Lemma 7.4.8. We conclude 1 that T a ∈ H with a uniform norm (with respect to a) if and only if n T a dL = 0. But, ∞ tr R n T a d = 0 for all atoms a ∈ if and only if T 1 = 0. ´ R L A ´

7.5 Mikhlin multiplier Theorem

∞ n Deﬁnition 7.5.1 (Fourier multiplier operator). Let m ∈ L (R ). Deﬁne the Fourier 2 n 2 n multiplier operator operator Tm :L (R ) → L (R ) associated to m by

(Tmf)ˆ= mf.ˆ

2 n Remark 7.5.2. By Plancherel theorem, ∈ L(L ( )) and k k 2 n = k k . Tm R Tm L(L (R )) m ∞

p n We want to consider when such an operator is bounded on L (R ) for p 6= 2. Such operators arise naturally - for instance when studying PDE with constant or smooth coeﬃcients. Also, we have the following important example. Example 7.5.3 (The Riesz Transform). Consider the Fourier multiplier

ξj m(ξ) = −ı . |ξ|

th Then, Tm = Rj, the j Riesz transform.

∞ n Theorem 7.5.4 (Mikhlin Multipler Theorem). Let m ∈ L (R ). Assume that m ∈ ∞ n n C (R \{0}) and for all α ∈ N there exists a Cα < ∞ such that

∂|α| C ( ) ≤ α α m ξ |α| ∂ξ |ξ| when ξ 6= 0, and where |α| = α1 + ··· + αn. Then, Tm ∈ CZO1, hence Tm has a bounded p n extension to L (R ) for all p ∈ (1, ∞). tr 1 Remark 7.5.5. Note that Tm (and Tm ) is bounded on H .

73 Proof. It is enough to show that Tm has an associated K ∈ CZK1. We use the Littlewood- n 1 Paley decomposition. The idea is to split up R such that |ξ|α is essentially constant on each part.

∞ 1 Take w ∈ C (R+, [0, 1]) with spt w ⊂ [ 2 , 4] and w = 1 on [1, 2]. Deﬁne: w(t) W (t) = P (2−j ) j∈Z w t for 0. Observe that 1 ≤ P (2−j ) ≤ 4 for all 0 and 7→ P (2−j ) ∈ t > j∈Z w t t > t j∈Z w t ∞ ∞ ∞ n C (0, ∞). In fact W ∈ C (0, ∞) ∩ L (R ),

k d W (t) ≤ Ck dtk

n ∞ n for all t ∈ (0, ∞). Deﬁne ϕ : R → [0, ∞] by ϕ(ξ) = W (|ξ|). Then, ϕ ∈ C (R \{0}), radial bounded and 1 spt ϕ ⊂ C = ξ ∈ n : ≤ |ξ| ≤ 4 0 R 2 n for all α ∈ N , and ∂|α| k α ϕk ≤ C(n, α). ∂ξ ∞ The ( ) = (2−j ) form a partition of unity: P ( ) = 1, ∈ n \{0}. ϕj ξ ϕ ξ j∈Z ϕj ξ ξ R

0 n c C We want to compute k =m ˇ in ( ) and show that |k(x)| ≤ n and |∇k(x)| ≤ S R |x| |x|n+1 0 2n (x 6= 0). This implies the Schwartz kernel of Tm will be given by k(x − y) (in S (R )) c and its restriction to ∆ is CZK1.

n Let ψ ∈ S (R ) then

hk, ψi = hm,ˇ ψi = hm, ψˇi

= m(ξ)ψˇ(ξ) dL (ξ) ˆRn X ˇ = m(ξ)ϕj(ξ)ψ(ξ) dL (ξ) ˆ n j∈Z R X 1 ıx·ξ = m(ξ)ϕj(ξ) n e ψ(x) dL (x) dL (ξ) ˆ n (2π) ˆ n j∈Z R R X = hkj, ψi j∈Z where 1 ıx·ξ hkj, ψi = n e m(ξ)ϕj(ξ) ψ(x) dL (ξ) dL (x). (2π) ˆRn ˆRn ∞ n ∞ In fact, mϕj ∈ L (R ) with compact support, so kj identiﬁes with a bounded C function and we write 1 ıx·ξ kj(x) = n m(ξ)ϕj(ξ)e dL (ξ) (2π) ˆRn

74 for ∈ n. Also, = P in 0( n). x R k j∈Z kj S R

We estimate kj. First,

jn |kj(x)| ≤ kmk∞ |ϕj(ξ)| dL (ξ) = kmk∞kϕjk1 = kmk∞kϕ1k2 . ˆRn Then, for x large, (−1)|α| ∂|α| eıx·ξ kj(x) = n α (mϕj)(ξ) α dL (ξ). (2π) ˆRn ∂ξ (ıx) n We cover R by conical sectors. Then, in the ﬁrst section, |x| ' |x1|. So, choose α such that αj = 0 for j > 1 and |α| −jα jn ∂ |kj(x)| ≤ C2 2 k α ϕk ∂ξ 1 n where C depends on the sector. Now, we cover R by a ﬁnite number of sections to obtain 2jn |kj(x)| ≤ CM (1 + 2j |x|)M for all M ∈ N. Then, for x 6= 0, X C |k(x)| ≤ |kj(x)| ≤ . |x|n j∈Z

A repetition of this argument with ∇k in place of k yields the desired estimate for ∇k.

7.6 Littlewood-Paley Theory

We extend Calder´on-Zygmund Operators to a Hilbert space valued functions. In this section, we let H be a separable complex Hilbert space and we denote its norm by |· | = p H h· , · iH .

We require the following notions. n Definition 7.6.1 (Strongly measurable). Let f : R → H . Then, we say that f is strongly measurable if given an orthonormal Schauder basis {ei} of H , the Fourier coefficients n hf, eii : R → C are measurable. p n p Definition 7.6.2 (L (R ; H )). For 1 ≤ p < ∞, we define the Hilbert valued L space p n n denoted by L (R ; H ) to be the set of strongly measurable functions f : R → H such p p that n |f| d < ∞. Then, the L norm is defined as kfk = k |f| k . R H L p H p ´ ∞ n Similarly, for p = ∞, we say that f ∈ L (R ; H ) if f is strongly measurable and |f|H ∈ ∞ n L (R ). Then, kfk∞ = k |f|H k∞.

A deeper discussion of these ideas can be found in [Yos95].

We also extend the notion of a CZK kernel and CZO operator to a Hilbert valued setting.

75 Definition 7.6.3 (Hilbert CZKα). Let H1, H2 be separable Hilbert spaces. Then, we say a c K is a Hilbert valued Calderón-ZygmundKernel if K(x, y) ∈ L(H1, H2) for all (x, y) ∈ ∆ and satisfies (i) to (iii) in Definition 7.1.1 with the absolute value replaced by L(H1, H2) norm. We denote the set of all such K by CZKα(H1, H2).

Deﬁnition 7.6.4 (Hilbert CZOα). Let H1, H2 be separable Hilbert spaces. We say that 2 n 2 n a T ∈ L(L (R ; H1), L (R ; H2)) is a Calder´on-Zygmundoperator of order α if it is associated to a K ∈ CZOα(H1, H2) by:

T f(x) = K(x, y)f(y) dL (y) ˆRn c 2 n almost everywhere x ∈ (spt f) with f ∈ L (R ; H1) with spt f compact. We denote the set of all such operators by CZOα(H1, H2).

Theorem 7.6.5. If T ∈ CZOα(H1, H2), then T induces a continuous extension to a p n p n bounded operator from L (R ; H1) to L (R ; H2) for 1 < p < ∞.

Proof. Same as the “scalar” case.

n n 1 Theorem 7.6.6. Let ϕ ∈ S (R ) such that sptϕ ˆ ⊂ ξ ∈ R : 2 ≤ |ξ| ≤ 4 and ϕˆ(ξ) = 1 if 1 ≤ |ξ| ≤ 2. Let 1 < p < ∞. Then, there exist constants C1,C2 < ∞ depending only on n, ϕ and p such that

1   2 X 2 C1kfkp ≤ ||  |∆jf|  ||p ≤ C2kfkp (LP) j∈Z p n jn j for all f ∈ L (R ) where ∆jf = ϕj ∗ f with ϕj(x) = 2 ϕ(2 x). n o Remark 7.6.7. Note that (∆ )ˆ= ˆ, ( ) = ˆ( ξ ) and spt ⊂ : 1 ≤ |ξ| ≤ 4 . jf ϕcjf ϕcj ξ ϕ 2j ϕcj ξ 2 2j

The pieces ∆j are “almost-orthogonal” for “good” f, g: 1 ∆jf(x)∆lg(x) dL = n (∆jf)ˆ(ξ)(∆lg)ˆ(ξ) dL (ξ) = 0 ˆRn (2π) ˆRn if |j − l| ≥ 3.

p This theorem says how to “pack” the ∆jf pieces to recover the L norm.

2 n 2 n 2 Proof. Deﬁne T :L (R ; C) → L (R ; ` (Z)) as the map f 7→ (∆jf)j∈Z.

2 n 2 n 2 (i) T ∈ L(L (R ; C), L (R ; ` (Z))) because:

2 X 2 |T f(x)|`2( ) dL (x) = |∆jf(x)| dL (x) ˆ n Z ˆ n R R j∈Z 1 2 X 2 ˆ = n |ϕj(ξ)| f(ξ) dL (ξ) (2π) ˆ n c j∈Z R 2 ˆ = m(ξ) f(ξ) dL (ξ) ˆRn

76 2 where m(ξ) = P 1 ϕˆ( ξ ) . Also, j∈Z (2π)n 2j 4 m(ξ) ≤ kϕˆk2 (2π)n ∞ for all ξ 6= 0. Thus,

2 2 2 | | 2 ≤ 4k ˆk | | T f ` (Z) dL ϕ ∞ f dL . ˆRn ˆRn

1 (ii) We have m(ξ) ≥ (2π)n for all ξ 6= 0 sinceϕ ˆ(ξ) = 1 for 1 ≤ |ξ| ≤ 2. However,

2 2 | | 2 ≥ k k T f ` (Z) dL f 2 ˆRn so (LP) holds for p = 2.

2 (iii) We apply Calder´on-Zygmund theory to show that T ∈ CZOα(C, ` (Z)). n Let Kj(x, y) = ϕj(x − y) for j ∈ Z and x, y ∈ R and K(x, y) = (Kj(x, y))j∈Z for c 2 P 2 ∈ ∆. We note that k ( )k 2 = | ( )| . We use the fact that x, y K x, y L(C,` (Z)) j∈Z Kj x, y jn j −M |ϕj(x)| = Cϕ2 (1 + 2 |x|) whenever M > n. We split the above sum according to 2j |x − y| ≥ 1 or < 1 and this gives us

X 2 1 |Kj(x, y)| ≤ Cϕ,n . |x − y|2n j∈Z

Also, ∇xK(x, y) = (∇xKj(x, y))j∈Z and

j(n+1) j j(n+1) j −M |∇xK(x, y)| ≤ C2 |∇ϕ| 2 (x − y) ≤ Cϕ2 (1 + 2 |x|)

where M > n + 1. This implies that

Cϕ k∇xK(x, y)k 2 ≤ L(C,` (Z)) |x − y|n+1

and ∇yK(x, y) = −∇xK(x, y).

∞ n ∞ n 2 (iv) Let f ∈ Cc (R ) and g ∈ Cc (R , ` (Z)) with spt f ∩ spt g = ∅. First, note that this implies spt f ∩ spt |g| = ∅. Then, ∞ X T f(x)· g(x) dL (x) = ∆jf(x)· gj(x) dL (x) ˆ n ˆ n R R j=−∞ ∞ X = kj(x, y)f(y) dL (y) gj(x) dL (x) ˆ n ˆ n j=−∞ R R ∞ X = kj(x, y)gj(x) dL (y) f(y) dL (x) ˆ n ˆ n j=−∞ R R ∞ X = kj(x, y)gj(x)f(y) dL (y) dL (x) ˆ n ˆ n R j=−∞ R

77 by applying Fubini and by the disjoint supports of f and g. Since T f(x) ∈ L2(`2) 2 and n K(x, y)f(y) is an ` valued integral, we have that R ´ T f(x) = K(x, y)f(y) dL (y) ˆRn 2 c 2 c in Lloc( (spt f); ` ) and hence almost everywhere x ∈ (spt f). p n p n 2 (v) We conclude that T extends to an operator from L (R ; C) to L (R ; ` (Z)) for P 2 1 p n 1 < p < ∞. Hence, ( |∆jf| ) 2 = |T f| 2 ∈ L ( ; ) with k |T f| 2 k ≤ j∈Z ` (Z) R C ` (Z) p C(p, n, CZO1)kfkp. (vi) We prove the left hand side of the (LP) estimate. We show that there exists a n n ϕ˜ ∈ S (R ) such that for all ξ ∈ R \{0}, X ξ ξ ϕˆ( )ϕ˜ˆ( ) = 1. 2j 2j j∈Z Let  ϕˆ(ξ) n 2 ξ ∈ \{0}  ξ R P ϕˆ( ) w(ξ) = j∈Z 2j 0 ξ = 0 ∞ n n and note that w ∈ C (R ), with spt w ⊂ sptϕ ˆ. Deﬁneϕ ˜ =w ˇ ∈ S (R ). By construction, P ˆ( ξ ) ˜ˆ( ξ ) = 1. j∈Z ϕ 2j ϕ 2j p n 2 n ∞ n Let f ∈ L (R ; C) ∩ L (R ; C) and g ∈ Cc (R ). Then,

1 ˆ f(x)g(x) dL (x) = n f(ξ)gˆ(ξ) dL (ξ) ˆRn (2π) ˆRn X ˆ = ϕˆj(ξ)f(ξ)· ϕ˜ˆ(ξ)g(ξ) dL (ξ) ˆ n j∈Z R X = ∆jf(x)∆˜ jg(x) dL (x) ˆ n j∈Z R X = ∆jf(x)∆˜ jg(x) dL (x) ˆ n R j∈Z

where ∆˜jg =ϕ ˜j ∗ g and this implies

˜ fg dL ≤ k |T f| 2 k k|T g| 2 k . ` (Z) p ` (Z) p0 ˆRn

A repetition of previous steps applied to ˜ yields k| ˜ | 2 k ≤ k k 0 . By duality T T g ` (Z) p C g p p p0 ∞ n p n of L and L and density of Cc (R ) in L (R ), 0 kfk ≤ C(p , n, ϕ) ≤ k |T f| 2 k . p ` (Z) p

2 n Then, we can remove the f ∈ L (R ; C) again by density and boundedness of T .

78 Remark 7.6.8. One can prove a “continuous time” Littlewood Paley estimates with same ϕ. Deﬁne for t > 0 1 x ϕt(x) = ϕ( ) and Qt = ϕt ∗ f. tn t p n Then, for all p ∈ (1, ∞), there exists constants C1,C2 < ∞ such that for all f ∈ L (R ), ∞ 2 dt C1kfkp ≤ k |Qtf| k ≤ C2kfkp. ˆ0 t p

j 1 2 2 dt That is, we sent 2 to tn and ` (Z) to L (R+, t ). This is because

−(j+1) ∞ X 2 dt h(t) dt = h(t) . ˆ0 ˆ2−j t j∈Z

2 2 dt This perspective tells us that ` (Z) is really the discrete version of L (R+, t ).

79 Chapter 8

Carleson measures and BMO

Deﬁnition 8.0.1 (Carleson measure). A Carleson measure is a positive locally ﬁnite Borel n+1 measure µ on R+ such that there exists a constant C < ∞ for which

µ(B × (0, r]) ≤ CL (B) for all B = B(x, r). We call B × (0, r] the Carleson window and deﬁne the Carleson norm

µ(B × (0, rad B] kµkC = sup . B L (B))

The following measures are not Carleson measures.

Example 8.0.2. (i) dµ(x, t) = dL (x)dt since no such constant C is possible for large balls.

dt r dt (ii) dµ(x, t) = dL (x) t since µ(B × (0, r]) = L (B) 0 t = ∞. ´ dL (x)dt (iii) dµ(x, t) = tα for α ∈ R. Note that

r ( r1−α dt L (B) 1−α 1 − α > 0 µ(B × (0, r]) = L (B) α = . ˆ0 t ∞ otherwise

So, we only need to consider the situation 1 − α > 0 but in this case, we cannot get uniform control via a constant C.

n Deﬁnition 8.0.3 (Cone). Let x ∈ R . We deﬁne the cone over x:

n+1 Γ(x) = (y, t) ∈ R+ : |x − y| < t .

The following are examples of Carleson measures.

χ dt Example 8.0.4. (i) dµ(x, t) = [a,b](t)dL (x) t where 0 < a < b < ∞. Then, the b constant C = ln a .

80 χ dt (ii) dµ(y, t) = Γ(x)(y)dL (y) t . Then, r dt r dt L (B) µ(B × (0, r]) ≤ L (B(x, t)) = tnL (B(0, 1)) = . ˆ0 t ˆ0 t n n Deﬁnition 8.0.5 (Tent). Let B = B(xB, rB) ⊂ R be an open ball. We deﬁne the Tent over the ball:

n+1 c n+1 T (B) == (y, t) ∈ R+ : 0 < t ≤ d(y, B) = (y, t) ∈ R+ : B(y, t) ⊂ B .

(Balls are open in this deﬁnition) Similarly, using d∞ instead of d, and `(Q) instead of r, we can deﬁne a Tent over a cube Q.

Remark 8.0.6. 1. B × (0, r] can be changed to B × (0, cr] for any ﬁxed c > 0.

2. B × (0, r] can be changed to T (B).

8.1 Geometry of Tents and Cones

We begin with the following observation.

c Proposition 8.1.1. Let B be an open ball. Then, T (B) = ∪z6∈BΓ(z).

This lead to the following deﬁnition.

n Deﬁnition 8.1.2 (Tent over open set). Let Ω ⊂ R be open. Then, deﬁne n+1 c c T (Ω) = (y, t) ∈ R+ : 0 < t ≤ d(y, Ω) = (∪z6∈ΩΓ(z)) .

Remark 8.1.3. Observe that (y, t) ∈ T (Ω) implies that y ∈ Ω.

n+1 Deﬁnition 8.1.4 (Non-tangential maximal function). Let f : R+ → C and deﬁne the non-tangential maximal function of f:

M∗f(x) = sup |f(y, t)| ∈ [0, ∞]. (y,t)∈Γ(x)

n+1 Remark 8.1.5. Given a Borel measure µ on R+ , we can deﬁne the non-tangential ∗ maximal function Mµ with respect to µ by replacing the sup with an esssup. Note then ∗ that Mµ is deﬁned µ almost everywhere. Proposition 8.1.6. M∗f is lower semicontinous and hence Borel.

n ∗ Proof. Let α ≥ 0 and x ∈ R such that M f(x) > α. Now, there exists a (y, t) ∈ Γ(x) such that |f(y, t)| > α. Therefore, for all z ∈ B(y, t) we have (y, t) ∈ Γ(z) and hence ∗ n ∗ M (z) ≥ |f(y, t)| > α. That is, x ∈ B(y, t) ⊂ {x ∈ R : M f(x) > α}. Proposition 8.1.7. Fix α ≥ 0. Then,

n+1 n ∗ (y, t) ∈ R+ : |f(y, t)| > α ⊂ T ({x ∈ R : M f(x) > α}).

81 n ∗ Proof. Note that |f(y, t)| > α implies B(y, t) ⊂ {x ∈ R : M f(x) > α} and hence t < d(y, cΩ).

Deﬁnition 8.1.8. Deﬁne:

µ(T (B)) C(µ)(x) = sup ∈ [0, ∞]. B3x L (B)

n+1 Theorem 8.1.9 (Carleson’s Lemma). Let µ be a Carleson measure and f : R+ → C n ∗ be a µ-measurable function. Let α > 0 such that L {x ∈ R : M f(x) > α} < ∞. Then there exists a C(n) such that

n+1 µ (y, t) ∈ R+ : |f(y, t)| > α ≤ C(n) C(µ)(x) dL (x). ˆ{x∈Rn:M∗f(x)>α}

n Remark 8.1.10. (i) For all x ∈ R C(µ)(x) ≤ kµkC (with k· kC deﬁned using tents or Carleson windows).

n (ii) For all x ∈ R , µ(T (B)) ≤ inf C(µ)(x). L (B) x∈B

(iii) C(µ) is lower semicontinous and non negative.

n ∗ Proof of Carleson’s Lemma. Set Ω = {x ∈ R : M f(x) > α} and note that Ω is open n with L (Ω) < ∞. Hence, Ω $ R and so we can apply the Whitney Covering Lemma (with balls) to obtain the existence of c = c(n),C = C(n) with c(n) < 1 < C(n) < ∞ and {Bi = Bi(yi, ri)}i∈I where each Bi is a ball, Ω = ∪iBi and cBi are mutually disjoint and c CBi ∩ Ω 6= ∅.

We note that it is enough to estimate µ(T (Ω)). Let (y, t) ∈ T (Ω). Then, y ∈ Ω and there c exists an i ∈ I such that y ∈ Bi. Let z ∈ CBi ∩ Ω. Then,

c c t ≤ d(y, Ω) ≤ |y − z| ≤ |y − yi| + |yi − z| ≤ (1 + C)ri ≤ d(y, (2 + C)Bi).

Thus, (y, t) ∈ T ((2 + C)Bi). Then,

µ(T (Ω)) ≤ µ(∪iT ((2 + C)Bi)) X ≤ µ(T ((2 + C)Bi)) i X ≤ inf C(µ)(x)L ((2 + C)Bi) z∈(2+C)B i i n X 2 + C ≤ inf C(µ)(z) L (cBi) z∈cBi c i n 2 + C X ≤ C(µ)(z) dL (z) c ˆ i cBi 2 + C n ≤ C(µ)(z) dL (z) c ˆΩ since the balls cBi are mutually disjoint in Ω.

82 Corollary 8.1.11. For all open Ω, µ(T (Ω)) ≤ CkµkC L (Ω).

n ∗ Proof. Note that in the previous proof, we only used the fact that Ω = {x ∈ R : M f(x) > α} n n to obtain that Ω $ R . So, the argument works when Ω $ R . Certainly, the claim is n trivially true when Ω = R . Corollary 8.1.12. With the assumptions of Carleson’s Lemma,

n+1 n ∗ µ (y, t) ∈ R+ : |f(y, t)| > α ≤ ckµkC L ({x ∈ R : M f(x) > α}). n+1 Corollary 8.1.13. Assume f : R+ → C is µ-measurable and µ is a Carleson measure. Then, ∗ |f(y, t)| dµ(y, t) ≤ CkµkC M f(x) dL (x). ¨ n+1 ˆ n R+ R

∗ ∗ Proof. If kM fk1 = ∞, there’s nothing to prove. So, assume not. Then, kM fk1 < ∞ n ∗ implies L {x ∈ R : M f(x) > α} < ∞ for all α > 0. Thus, n+1 n ∗ µ (y, t) ∈ R+ : |f(y, t)| > α ≤ C(n)L {x ∈ R : M f(x) > α}kµkC and integrating both sides from 0 to ∞ in α ﬁnishes the proof.

n Deﬁnition 8.1.14. Fix t > 0 and let ϕ be a function on R . We deﬁne 1 x ϕt(x) = ϕ . tn t

We state the following technical Lemma. −n−ε n Lemma 8.1.15. Let ϕ(x) = (1 + |x|) , ε > 0, x ∈ R . Then, there exists a c = n c(n, ε) < ∞ such that for all x ∈ R and for all (y, t) ∈ Γ(x)

|ϕt ∗ f(y)| ≤ cMf(x)

1 n for all f ∈ L ( ) such that n ϕ(y) |f(y)| d (y) < ∞. loc R R L ´ We deﬁne the following operator family which will be of interest to us.

Deﬁnition 8.1.16 ((Rt)t>0). For t > 0, let

Rtf(x) = Kt(x, y)f(y) dL (y) ˆRn n 1 n almost everywhere x ∈ whenever f ∈ L ( ) such that n ϕ(y) |f(y)| d (y) < ∞ R loc R R L and with ´ |Kt(x, y)| ≤ Cϕt(x − y) n for all t > 0 and almost everywhere x, y ∈ R . p n Corollary 8.1.17 (Carleson Embedding). For all 1 < p < ∞ and f ∈ L (R ),

p p |Rtf(y)| dµ(y, t) ≤ C(n, ε, C)kµk kfkp ¨ n+1 C R+ whenever µ is a Carleson measure.

83 p Proof. Let g(x, t) = |Rtf(x)| . By application of Corollary 8.1.13,

p ∗ |Rtf(x)| dµ(x, t) ≤ C(n)kµkC M g(x) dL (x). ¨ n+1 ˆ n R+ R

Then, by Lemma 8.1.15, |Rtf(y)| ≤ C(n, ε)Mf(x) if |x − y| < t. Thus,

∗ p p p C(n)kµkC M g(x) dL (x) ≤ C(n)C(n, ε) (Mf) dL ≤ C(p, n)kfkp. ˆRn ˆRn

8.2 BMO and Carleson measures

Theorem 8.2.1. Suppose that

n (i) Rt(1)(x) = 0 almost everywhere x ∈ R and for all t > 0 (note that this is saying n Kt(x, y) d (y) = 0 almost everywhere), R L ´ 2 n (ii) For all f ∈ L (R ),

2 dL (x)dt 2 |Rtf(x)| ≤ C1 |f(x)| dL (x). ¨ n+1 t ˆ n R+ R

n Let b ∈ BMO. Then Rt(b)(x) is deﬁned almost everywhere x ∈ R and for all t > 0 and 2 2 there exists a C2 < ∞ such that |Rt(b)(x)| ≤ C2kbk∗ and

2 dL (x)dt |Rt(b)(x)| t is a Carleson measure.

Remark 8.2.2. The estimate in condition (ii) of the Theorem is one of Littlewood-Paley type. It does not follow from the kernel bounds on Rt. That is, kRtfk2 ≤ C(n, ε)kfk2 is dt not enough to integrate from 0 to ∞ in t .

n Proof of Theorem 8.2.1. Fix x ∈ R . Pick a ball B centred at x. Write b = b1 + b2 + mB b where b1 = (b − mB b)χB and b2 = (b − mB b)χcB. Let

Ij = |Kt(x, y)| |bj(y)| dL (y) ˆRn for j = 1, 2. We estimate these two quantities.

First, assume that rad B = t. Then,

1 x − y I1 ≤ n ϕ |b(y) − mB b| dL (y) ≤ C(n)kbk∗ ˆB t t

84 and using the fact that when |x − y| ∼ 2nt,

1 x − y 1 x − y −n−ε 1 ϕ ≤ 1 + ≤ 2−j(n+ε) tn t tn t tn we have

Therefore, Rtb1,Rtb2 are well deﬁned with uniform estimates and Rt(mB b) = 0 since mB b is constant by (i). Hence, Rtb is well deﬁned and |Rtb| ≤ C2kbk∗. In fact, Rtb(x) does not depend on the choices of B 3 x. That is, it is independent of decomposition.

Now, we do the Carleson measure estimate. Fix B0 = B0(x0, r) and B = B(x0, 2r). We use (ii),

n and L (B) = 2 L (B0).

For Rt(b2), we do the same as before for I2, but this time with x ∈ B0, y 6∈ B and j+1 rad B = 2r0. So, noting that when |x − y| > 2 r0 we have

1 x − y 1 |x − y|−n−ε ϕ ≤ 1 + tn t tn t n+ε ε n 1 t t 1 −(j+1)ε ≤ n j+1 n+ε = j+1 2 t (2 r0) r0 2 r0

85 we calculate

Thus,

2ε 2 dL (x)dt 2 t dL (x)dt |Rt(b2)(x)| ≤ C(n, ε)kbk∗ ¨B0×(0,r] t ¨B0×(0,r] r0 t 2 1 ≤ C(n, ε)kbk L (B0) ∗ 2ε the proof is completed by adding up the two estimates.

We deﬁne the following family of operators.

n Deﬁnition 8.2.3 ((Qt)t>0). Let ϕ ∈ S (R ) with n ϕ dL = 0. Deﬁne Qtf = ϕt ∗ f 1 n R when f ∈ Lloc(R ). ´ Remark 8.2.4. We note that

1 x − y Qtf(x) = n ϕ f(y) dL (y). ˆRn t t We have (i):

Qt(1) = ϕt(x − y) dL (y) = 0. ˆRn And for (ii):

( ) 1 ∞ 2 2 dL x dt 2 dt ˆ |Qtf(x)| = n |ϕˆ(tξ)| f(ξ) dL (ξ) ¨ n+1 t (2π) ˆ n ˆ t R+ R 0 2 ≤ Akfk2. since ϕˆ(0) = 0 and ( c |ξ| |ξ| ≤ 1 |ϕˆ(ξ)| ≤ c |ξ| |ξ| ≥ 1 with ∞ dt A = sup |ϕˆ(tξ)| < ∞. ξ∈Rn ˆ0 t

So, we can apply the Theorem with Rt = Qt.

86 ∞ n+1 dL (x)dt Theorem 8.2.5. Let β ∈ L (R+ , t ) and assume that

dL (x)dt dµ(x, t) = |β(x, t)|2 t

1 n is a Carleson measure. Let βt(x) = β(x, t). Then for all f ∈ H (R ),

(i) For all ε > 0 and R < ∞,

R dL (x)dt Iε,R(f) = Qt(βt)(x)f(x) ˆRn ˆε t

1 2 is well deﬁned and |Iε,R(f)| ≤ C(n, ϕ)kfkH1 kµkC .

(ii) We have that

lim Iε,R(f) ε→0,R→∞

1 2 exists and deﬁnes an element b ∈ BMO with kbk∗ ≤ C(n, ϕ)kµkC . We write,

∞ dt b = Qtβt . ˆ0 t

Proof. To prove (i), we note that |Qt(βt)(x)| ≤ kβk∞kϕk1 almost everywhere (x, t) ∈ n+1 1 n 1 n R+ . Now, f ∈ H (R ) implies f ∈ L (R ) and

R dL (x)dt R dt Iε,R(f) = Qt(βt)(x)f(x) ≤ kβk∞kϕk1kfk1 < ∞. ˆRn ˆε t ˆε t

Now for (ii), let f = a ∈ A ∞ with spt a ⊂ B with B a ball. Then,

R tr dt Iε,R(a) = βt(x)Qt (a)(x) dL (x) ˆε ˆRn t

tr where Qt = ψt ∗ where ψ(y) = ϕ(−y). Let

tr dL (x)dt I = |βt(x)| Qt (a)(x) . ¨ n+1 t R+

Then clearly, |Iε,R| ≤ I.

n+1 j j We compute this integral by covering R+ with square annuli. Let A2j B = 2 B × (0, 2 r], 2 j j+1 C0 = A2B and Cj = A2j+1B \ A2j B when j > 0 Also, deﬁne Cj = A2j+1B ∩ 2 B × (0, 2 r] 1 2 when j > 0 and Cj = Cj \ Cj .

87 2 Cj

1 1 Cj Cj ...

......

For j = 0, by application of Cauchy-Schwarz,

1 1 2 2 tr dL (x)dt 2 dL (x)dt tr 2 dL (x)dt |βt(x)| Qt (a)(x) ≤ |βt(x)| Qt (a)(x) ¨C0 t ¨C0 t ¨C0 t 1 2 ≤ kµkC L (2B) C(n, ϕ)kak2.

− 1 But kak2 ≤ C(n, ϕ)L (B) 2 .

Now, we consider j > 0. Let 1 y − x yB − x A(x, t) = sup n ϕ − ϕ y∈B t t t

tr 1 j so that we have Qt (a)(x) ≤ A(x, t)kak1. Consider (x, t) ∈ Cj . Then, 2 ≤ |x − yB| ≤ 2j+1r and 0 < t ≤ 2j+1r. If y ∈ B then |x − y| ∼ 2jr and we have the mean value inequality −M r 1 2j |A(x, t)| ≤ C(ϕ, n) 1 + t tn t for all M > 0. Therefore,

2 2j+1r j −M ! 2 dL (x)dt j+1 j r 1 2 dt |A(x, t)| ≤ L (2 B \ 2 B) C n 1 + ¨ 1 t ˆ t t t t Cj 0 ∞ j+1 n 1 1 n+1 −M 2 du ≤ C(2 r) 2j j 2n (u (1 + u) ) 2 (2 r) ˆ1 u 1 1 ≤ C 22j L (2j+1B)

∞ n+1 −M 2 du since 1 (u (1 + u) ) u < ∞ when M > n + 1. ´ 88 2 j j j+1 Now, if (x, t) ∈ Cj , we have |x − yB| ≤ 2 r, and 2 r ≤ t < 2 r and the mean value inequality takes the form r 1 z − x A(x, t) ≤ sup n ∇ϕ z∈B t t t and when x ∈ 2B, r 1 z − x r ∇ϕ ≤ k∇ϕk t tn t (2jr)n+1 ∞ and when x 6∈ 2B,

r |y − x|−M A(x, t) ≤ C(n) 1 + B . (2jr)n+1 2jr Therefore,

dL (x)dt 2j+1r dL (x)dt |A(x, t)|2 ≤ |A(x, t)|2 ¨ 2 t ˆ ˆ j t Cj 2j B 2 r

≤ ln 2 |A(x, t)|2 dL (x) ˆ2j B ! ≤ ln 2 |A(x, t)|2 dL (x) + |A(x, t)|2 dL (x) ˆ2B ˆ2j B\2B ! 1 r2 |y − x|−2M ≤ + 1 + B ( ) C j(n+2) j 2(n+1) j dL x 2 L (B) (2 r) ˆ2j B 2 r 1 1 ≤ C + 22jL (B) 22jL (2jB) Therefore, 2 dL (x)dt C 1 |A(x, t)| ≤ 2j j+1 ¨ 2 t 2 (2 B) Cj L and

tr dL (x)dt I = |βt(x)| Qt (a)(x) ¨ n+1 t R+

X tr dL (x)dt = |βt(x)| Q (a)(x) ¨ t t j Cj 1 1 ! 2 ! 2 X 2 dL (x)dt tr 2 dL (x)dt ≤ |βt(x)| Q (a)(x) ¨ t ¨ t t j Cj Cj 1 1 2 j+1 2 C 1 ≤ kµk L (2 B) C 22j L (2j+1B) 1 2 ≤ CkµkC .

This tells us that limε→0,R→∞ Iε,R exists and

tr dL (x)dt lim Iε,R = βt(x)Qt (a)(x) . ε→0,R→∞ ¨ n+1 t R+

89 1 ∞ P Now, take a general f ∈ H . Then, we can ﬁnd λj and aj ∈ A such that f = j λjaj 1 1 n P in H and in particular L (R ) and also such that j |λj| ≤ 2kfkH1 . Using the fact that Iε,R ≤ Cε,Rkfk1, we can write X Iε,R(f) = λjIε,R(aj). j

Therefore,

1 1 X 2 X 2 |Iε,R(f)| ≤ |λj| |Iε,R(aj)| ≤ CkµkC |λj| ≤ CkµkC 2kfkH1 j j with C independent of ε and R.

Let R dt bε,R = Qtβt ˆε t and note that it is a measurable function. Also, hbε,R, fi = Iε,R(f). Thus, bε,R ∈ BMO 1 2 10 with kbε,Rk∗ ≤ CkµkC with C independent of ε, R. Now since BMO = H there exists a convergent subsequence (bεj ,Rj ) and hbεj ,Rj , fi → hβ, fi for some β ∈ BMO. The ∞ uniqueness of β then follows from the uniqueness of limε→0,R→∞ Iε,R(a) for atoms a ∈ A ∞ 1 and then by the density of Vect A in H so limε→0,R→∞hbε,R, fi = hβ, fi.

tr Exercise 8.2.6. Replace Qt by Rt with kernel Kt satisfying:

C |x − y|−n−ε Kt(x, y) ≤ 1 + tn t when ε > 0 and

|y − z|δ 1 |x − y|−n−ε−δ |Kt(x, y) − Kt(x, z)| ≤ C 1 + t tn t

1 for δ > 0 when |y − z| ≤ 2 (t + |x − y|). ˜ n Theorem 8.2.7 (Paraproducts of J.M. Bony). Let ψ, ψ, ϕ ∈ ( ) such that n ψ d = S R R L ˜ n ψ d = 0. Deﬁne, ´ R L ´ Qt = ψt ∗ , Q˜t = ψ˜t ∗ ,Pt = ϕt ∗ and let b ∈ BMO. Let ∞ dt πb(f) = Q˜t(Qt(b)Pt(f)) . ˆ0 t Then,

2 n (i) We have πb ∈ L(L (R )) and ˜ kπb(f)k2 ≤ C(n, ϕ, ψ, ψ)kbk∗kfk2

2 n whenever f ∈ L (R ),

(ii) πb ∈ CZO1,

90 tr (iii) πb (1) = 0 in BMO, ˜ (iv) If n ϕ d = 1 and ψ, ψ are radial functions satisfying R L ´ ∞ dt ψ˜(tξ)ψ(tξ) = 1 ˆ0 t

then πb(1) = b in BMO.

Proof. (i) Let R dt Iε,R = h Q˜t(Qt(b)Pt(f)) , gi ˆε t 2 n ˜ ˜tr whenever g ∈ L (R ). Then, by noting that hQt(Qt(b)Pt(f)), gi = hQt(b)Pt(f), Qt (g)i and applying Fubini, R ˜tr dL (x)dt |Iε,R| = Qt(b)(x)Pt(f)(x)Qt (g)(x) ˆε ˆRn t R ( ) ˜tr dL x dt ≤ Qt(b)(x)Pt(f)(x)Qt (g)(x) ˆε ˆRn t ≤ I1 I2 where 2 2 dL (x)dt I1 = |Pt(f)(x)Qt(b)(x)| ¨ n+1 t R+ and 2 2 ˜tr dL (x)dt I2 = Qt (g)(x) . ¨ n+1 t R+ ˜ ˜ n We have the Littlewood-Paley estimate I2 ≤ C(ψ, n)kgk2 since ψ ∈ S (R ) and n ψ d = 0. Then, note that R L ´ 2 dL (x)dt |Qt(b)(x)| t is a Carleson measure and so by application of Carleson Embedding Theorem with Rt = Pt, I1 ≤ C(n, ϕ, ψ)kfk2kbk∗. Hence, ˜tr dL (x)dt lim Iε,R = Qt(b)(x)Pt(f)(x)Qt (g)(x) ε→0,R→∞ ¨ n+1 t R+ 2 n is a bounded bilinear form on L (R ) and we conclude (i) by invoking the Riesz Representation Theorem.

∞ n (ii) Let f, g ∈ Cc (R ). Then,

˜tr dL (x)dt hπb(f), gi = Qt(b)(x)Pt(f)(x)Qt (g)(x) = hK, g ⊗ fi ¨ n+1 t R+ where g ⊗ f(y, z) = g(y)f(z) and ∞ dL (x)dt K(y, z) = ϕt(z − x)Qt(b)(x)ψ˜t(y − x) ˆ0 ˆRn t

91 n 0 where this equality is in S (R ) . ˜ Now by using the fact that |Qt(b)(x)| ≤ Ckbk∗ and by the decay of ϕ, ψ using that −n−ε ηε(x) ≤ (1 + |x|) and ηε ∗ ηε ≤ C(n, ε)ηε, we have

∞ 1 y − z dt C(n, ε) n ηε ≤ n . ˆ0 t t t |y − z|

˜ 1 The estimate for ∇yK follows since under diﬀerentiation in y, ψt gives an extra t factor and so we get C(n, ε) |∇yK| ≤ . |y − z|n+1

So, K ∈ CZK1. (iv) We already have that

˜tr dL (x)dt hπb(f), gi = Qt(b)(x)Pt(f)(x)Qt (g)(x) ¨ n+1 t R+

2 n ∞ n whenever f, g ∈ L (R ). First, we show this same equality when f ∈ L (R ) and when g = a ∈ A ∞. Recall that when b ∈ BMO and a ∈ A ∞, we have ( ) ˜tr dL x dt |Qt(b)(x)| Qt (a)(x) ≤ Ckbk∗. ¨ n+1 t R+

n+1 Also, |Pt(f)(x)| ≤ kϕk1kfk∞ for all (x, t) ∈ R+ . Therefore,

˜tr dL (x)dt Qt(b)(x)Pt(f)(x)Qt (a)(x) ¨ n+1 t R+ exists. tr 1 1 n tr 1 n tr Note that πb :H → L (R ) and so πb (a) ∈ L (R ) and hence hf, πb (a)i well tr χ deﬁned and by deﬁnition hf, πb (a)i = hπb(f), ai. Let fk = f [−2k,2k]n when k ∈ N 2 ∞ n tr tr and so fk ∈ L ∩ L (R ). So by Dominated convergence hfk, πb (a)i → hf, πb (a)i n+1 and Pt(fk)(x) → Pt(f)(x) for all (x, t) ∈ R+ .

Now, set f = 1, and since we assumed that n ϕ d = 1, we have Pt(1) = 1 for all R L n+1 (x, t) ∈ R+ . Then, ´

˜tr dL (x)dt hπb(1), ai = Qt(b)(x)Qt (a) ¨ n+1 t R+ R ˜tr dL (x)dt = lim Qt(b)(x)Qt (a) = lim hb, Uε,Ri ε→0,R→∞ n ε→0,R→∞ ˆε ˆR t where R tr ˜tr dt Uε,R = Qt Qt (a) . ˆε t 1 It suﬃces to prove that Uε,R → a in H since this gives hπb(1), ai = hb, ai for all ∞ a ∈ A and hence πb(1) = b.

92 2 n −1 We note that a ∈ L (R ) since spt a ⊂ B for some ball B and kak∞ ≤ L (B) . Using the assumption that ψ, ψ˜ are radial, we have R ˆ ˜ˆ dt U\ε,R(a)(ξ) = ψ(tξ)ψ(tξ) aˆ(ξ) ˆε t 2 n and so Uε,R(a) → a in L (R ). n Deﬁne a function ϕ on R by ∞ ˆ dt 1 ˆ dt ϕ(ξ) = ψˆ(tξ)ψ˜(tξ) = 1 − ψˆ(tξ)ψ˜(tξ) ˆ1 t ˆ0 t n ∞ n whenever ξ ∈ R \{0}. It is an easy fact that ϕ ∈ C (R \{0}), with decay at ∞. ∞ n n Then, ϕ extends to a C (R ) function with ϕ(0) = 1. This implies that ϕ ∈ S (R ) n and soϕ ˇ ∈ S (R ). Then, U\ε,R(a)(ξ) = (−ϕ(Rξ) + ϕ(εξ))ˆa(ξ) so that

Uε,R =ϕ ˇR ∗ a +ϕ ˇε ∗ a.

1 Then, it is enough to prove that ϕR ∗ a → a in H as R → ∞ and ϕε ∗ a → a in H1 as ε → 0.

Now, using n a d = 0, spt a ⊂ B = B(x , r ) and kak ≤ 1, we leave it as an R L B B 1 exercise to show´ that −n−1 !! C(n, ϕ) 1 |x − xB| |ϕR ∗ a(x)| ≤ 1 + R Rn R when R is large,

ϕR ∗ a dL = ϕR dL a dL = 0 ˆRn ˆRn ˆRn and R 1 ϕR ∗ a ∈ H C(n, ϕ) 1 with uniform norm with respect to R (ie., kϕR ∗ akH1 = O( R )). 2 n Set hε = ϕε ∗ a − a. Then, we know that hε → 0 in L (R ). Deﬁne functions 1 2 1 2 1 hε = (hε − m2B hε)χ2B and hε by hε = hε + hε. Then, spt hε ⊂ 2B,

1 2 2 hε dL ≤ |hε| dL → 0 ˆ2B ˆ2B 1 1 1 and 2B hε dL = 0. These facts then imply that hε → 0 in H using the fact that 2 n 1 f ∈´L (R ) : spt f ⊂ 2B continuously embeds into H . Also,

2 1 hε dL = hε dL − hε dL = 0 − 0 = 0. ˆRn ˆRn ˆRn We leave it as an exercise to show that

2 −n−1 hε(x) ≤ C(n, ϕ)C(ε)(1 + |x − xB|) n for all x ∈ R and ε < rB where

C(ε) = sup(m2B hε, ε).

With this fact in hand, khεkH1 ≤ C(n, ϕ)C(ε) → 0.

93 (iii) We note that as before,

tr ˜tr dL (x)dt hπb (f), gi = Qt(b)(x)Pt(g)(x)Qt (f)(x) ¨ n+1 t R+

2 n ∞ n for all f, g ∈ L (R ). We want to show that this holds when we have f ∈ L (R ) ∞ and g ∈ A . As before, let fk = f[−2k,2k]n , and so

tr hπb (f), gi = hf, πb(g)i = lim hfk, πb(g)i k→∞ ˜tr dL (x)dt = Qt(b)(x)Pt(g)(x)Qt (fk)(x) ¨ n+1 t R+

˜tr ˜tr Now, although Qt (fk)(x) → Qt (f)(x) for all (x, t) we cannot apply Dominated n+1 convergence theorem since n g d 6= 0. Instead, cover by set Cj as in the R L R+ proof of Theorem ??. ´ For j > 0, we have the same estimates of P tr(g) as for for Q˜tr(g) in (iv). So, t t ˜tr Qt (g) ≤ C uniformly with respect to t, x, k. Then, take the limit. For the situation when j = 0,

1 ( ) 2 ( ) 2 ˜tr dL x dt ˜tr dL x dt Qt(b)(x)Pt(g)(x)Qt (f − fk)(x) ≤ Qt (f − fk)(x) ¨C0 t ¨C0 t 1 2 dL (x)dt 2 |Qt(b)(x)Pt(g)(x)| ¨C0 t

If k is large, f − fk = 0 on a neighbourhood of 2B. The region C0 is above 2B, so we can apply decay estimates to prove

2 ( ) ˜tr dL x dt Qt (f − fk)(x) → 0. ¨C0 t

2 dL (x)dt The other term is dealt with by noting that |Qt(b)(x)| t is a Carleson measure. ˜tr n+1 ˜ Now, put f = 1, Qt (1)(x) = 0 for all (x, t) ∈ R+ since n ψ dL = 0.Thus, tr ∞ tr R hπb (1), gi = 0 for all g ∈ A and so πb (1) = 0 in BMO. ´

94 Chapter 9

Littlewood-Paley Estimates

Deﬁnition 9.0.1 (Littlewood-Paley Estimate). Let (Rt)t>0 be a family of operators on a Hilbert space. Suppose there exists a C > 0 such that

∞ 2 dt 2 kRtfk2 ≤ Ckfk2. (LPE) ˆ0 t Such an estimate is called a Littlewood-Paley Estimate.

n Example 9.0.2. Let Rt = ψt ∗ where ψ ∈ ( ) with n ψ d = 0. Then, S R R L ´ ∞ ∞ 2 2 2 dt ˆ dt ˆ 2 kRtfk2 = C ψ(tξ) f(ξ) dL (ξ) ≤ C(ψ)kfk2. ˆ0 t ˆRn ˆ0 t Remark 9.0.3. Unlike in the example, we will not have the luxury of the Fourier Trans- form in the general theory. This is our goal.

Deﬁnition 9.0.4 (Operator ε-family). Let ε > 0 and (Rt)t>0 be a family of Operators with kernels Kt(x, y) where the map (t, x, y) 7→ Kt(x, y) is measurable satisfying:

3n (i) For all (t, x, y, z) ∈ R+ × R ,

C |x − y|−n−ε |Kt(x, y)| ≤ 1 + , tn t

(ii) For δ > 0,

|y − z|δ 1 |x − y|−n−ε−δ |Kt(x, y) − Kt(x, z)| ≤ C 1 + , t tn t where C < ∞. Such a family of operators is called an ε-family.

Theorem 9.0.5 (Christ, Journ´e,Coifman, Meyer). Let (Rt)t>0 be an ε-family. Then,

(i) (LPE) holds,

95 (ii) The measure 2 dL (x)dt |Rt(1)(x)| t is a Carleson measure, are equivalent statements.

Remark 9.0.6. The condition in (ii) only involves one test function, namely the constant function 1.

The following proof of 9.0.5 is due to Feﬀerman-Stein.

Proof of 9.0.5 (i) =⇒ (ii). Fix a Carleson box R = B × (0, r] where r = rad B and let ∞ n f ∈ L (R ). Let f1 = fχ2B and f2 = fχc2B so that f = f1 + f2. Then,

2 dL (x)dt 2 |Rt(f1)(x)| ≤ Ckf1k2 ≤ Ckfk∞L (2B). ¨R t

By noting that whenever (x, t) ∈ R implies x ∈ B, we have

|Rt(f2)(x)| = Kt(x, y)f(y) dL (y) ˆy∈c2B C |x − y|−n−ε t ε ≤ kfk∞ n 1 + dL (y) ≤ Ckfk∞ ˆy∈c2B t t r

Therefore,

2ε 2 dL (x)dt 2 t dL (x)dt 2 1 |Rt(f2)(x)| ≤ Ckfk∞ ≤ Ckfk∞L (B) . ¨R t ¨R r t 2ε

Before we proceed to prove the converse, we require the following important Lemma.

Lemma 9.0.7. Let (Vt)t>0 be an ε-family. Suppose that Vt(1) = 0. Then, there exist an ε0 ∈ (0, ε) and a C > 0 such that for all (t, s) ∈ (0, ∞)2,

∗ s kVtV k 2 n ≤ Ch( ) s L(L (R ) t

0 1 ε where h(x) = inf x, x .

∞ n Proof of 9.0.5 (ii) =⇒ (i). First, deﬁne Vt = Rt−Rt(1)Pt where Pt = ϕt ∗ , ϕ ∈ Cc (R ) with n ϕ d = 1. Then, R L ´

(Rt(1)Pt)f(x) = Rt(1)(x)Pt(f)(x) = Rt(1)(x)ϕt(x − y)f(y) dL (y) ˆRn

96 2 n which shows that (Vt)t>0 is an ε-family with the same ε as Rt. Now, take f ∈ L (R ). Then, 2 dL (x)dt 2 |Rt(1)(x)Pt(f)(x)| ≤ Ckµk kfk2 ¨ n+1 t C R+ where 2 dL (x)dt µ = |Rt(1)(x)| . t 2 dL (x)dt Hence, (ii) implies that |Rt(1)(x)Pt(1)(x)| t satisﬁes (i) and so proving (i) for Rt reduces to proving (i) for Vt.

n+1 Firstly, we note that Vt(1)(x) = 0 for all (x, t) ∈ R+ . We use a technique developed for a diﬀerent purpose called the Cotlar-Knapp-Stein inequality or T ∗T -argument. We have

∞ ∞ 2 dL (x)dt dt ∗ dt |Vt(f)(x)| = hVtf, Vtfi = hVt Vtf, fi ¨ n+1 t ˆ t ˆ t R+ 0 0

∗ and for each t, Vt Vt is a bounded (from ε-family hypothesis) self-adjoint operator on 2 n L (R ). We want the right hand side to be equal to hSf, fi for some bounded operator S.

Fix r, R such that 0 < r ≤ R < ∞, and let

R ∗ dt Sr,R = Vt Vt ˆr t

2 n ∗ and it is a bounded operator on L ( ) since by the -family hypothesis k k 2 n R ε Vt Vt L(L (R )) is uniformly bounded in t and so

R ∗ dt R kS k 2 n ≤ kV V k 2 n ≤ C ln . r,R L(L (R )) t t L(L (R )) ˆr t r

Also, Sr,R is self adjoint and therefore, operator theory tells us that

m m kSr,Rk 2 n = kS k . L(L (R )) r,R L(L2(Rn)) We write out Sm:

R R m ∗ ∗ ∗ dt1 dtm Sr,R = ... Vt1 (Vt1 Vt2 )Vt2 ... (Vtn−1 Vtm )Vtm ... . ˆr ˆr t1 tm and on application of Lemma 9.0.7, we get

kSm k r,R L(L2(Rn)) R R ∗ t1 tn−1 dt1 dtm ≤ ... kV k Ch . . . Ch kV k 2 n ... t1 L(L2( n)) tm L(L (R )) ˆr ˆr R t2 tm t1 tm R R m−1 ∗ t1 tn−1 dt1 dtm ≤ C ... kV k h . . . h kV k 2 n ... . t1 L(L2( n)) tm L(L (R )) ˆr ˆr R t2 tm t1 tm Since the integral is a convolution, we have the technical estimate that

∞ tj−1 tj dtj tj−1 h h = Cε0 h . ˆ0 tj tj+1 tj tj+1

97 ∗ ∞ t1 1 dt Combining this with the facts that kVt k 2 n ∈ L (R+) and h( ) ∈ L (R+, ), we j L(L (R )) tm t have R R m m−1 m−2 ∗ t1 dt1 dtm kS k ≤ C C 0 kV k h kV k 2 n r,R L(L2( n)) ε t1 L(L2( n)) tm L(L (R )) R ˆr ˆr R tm t1 tm R m−1 m−2 dtm ≤ k k 2 n C Cε0 C Vtm L(L (R )) ˆr tm 00 m R ≤ C (CCε) ln r and 1 m 00 m R kSr,Rk 2 n ≤ lim inf C (CCε) ln ≤ CCε0 . L(L (R )) m→∞ r Then, we ﬁnd that R ∗ dt lim Vt Vt ≤ CCε0 r→0,R→∞ ˆ t 2 n r L(L (R )) which proves

∞ R 2 dt ∗ dt 2 kVtfk2 = lim hVt Vtf, fi ≤ CCε0 kfk2 ˆ0 t r→0,R→∞ ˆr t

2 n for all f ∈ L (R ).

Now, we return to the proof of the Lemma.

∗ ∗ Proof of Lemma 9.0.7. We study VtVs and its kernel. Let Vt(x, y) and Vs (x, y) denote ∗ the kernels of Vt and Vs respectively and

∗ ∗ VtVs (f)(x) = Vt(x, z) Vs (z, y)f(y) dL (y)dL (z) = Ks,t(x, y)f(y) dL (y) ˆRn ˆRn ˆRn where

Ks,t(x, y) = Vt(x, z)Vs(y, z) dL (z). ˆRn p n We note that by Fubini, this expression holds for all f ∈ ∪1≤p≤∞L (R ).

Now, assume that 0 < s < t. Then we have the estimates

C |x − z|−n−ε C |x − z|−n−ε |Vt(x, z)| ≤ 1 + and |Vs(x, z)| ≤ 1 + . tn t sn s We leave it as an exercise to prove

C |x − z|−n−ε C |x − z|−n−ε n 1 + n 1 + dL (z) ˆRn t t s s C |x − y|−n−ε ≤ C(n, ε) 1 + (†) tn t for 0 < s ≤ t.

98 Now, using oscillation of y 7→ Vs(y, z), we can write

Ks,t(x, y) = (Vt(x, z) − Vt(x, y)) Vs(y, z) dL (z) ˆRn and hence

0 C0 |y − z|δ |Vt(x, z) − Vt(x, y)| ≤ . tn t

Therefore, by choosing 0 < δ0 < ε,

C |x − y|−n−ε |Ks,t(x, y)| ≤ 1 + . tn t

Now, if X ≤ A and X ≤ B, then for all θ ∈ [0, 1] the estimate X ≤ AθB1−θ holds. We apply this to |Ks,t(x, y)| with θ such that (1 − θ)(n + ε) = n + ν > n to obtain

C sδ0θ |x − y|−n−ν |Ks,t(x, y)| ≤ 1 + tn t t for all 0 < s ≤ t. By symmetry, when t ≤ s,

0 C t δ θ |x − y|−n−ν |Ks,t(x, y)| ≤ 1 + . sn s s

Using Young’s inequality that L2 ∗ L1 ⊂ L2, we obtain

δ0θ ∗ s t kVtV k 2 n ≤ C inf , . s L(L (R )) t s

99 Chapter 10

T (1) Theorem for Singular Integrals

2 n This chapter concerns itself with the question of proving the boundedness of L (R ) operators that are associated to Calderón-Zygmund kernels. ∞ n ∞ n 0 Definition 10.0.1 (Schwartz kernel). Let T :Cc (R ) → Cc (R ) be linear and contin- ∞ 2n 0 uous. Then, the uniquely given K ∈ Cc (R ) defined by hK, g ⊗ fi = hT f, gi is called the Schwartz kernel of T .

We begin with the following deﬁnition. ∞ n ∞ n 0 Deﬁnition 10.0.2 (Singular integral operator). Let T :Cc (R ) → Cc (R ) be linear and continuous. Then T is said to be a Singular integral operator if its Schwartz kernel c when restricted to ∆ is a CZKα for some α > 0. We write T ∈ SIO.

We emphasise that while a Singular integral operator has CZK kernel, it is not a Calder´on- 2 n Zygmund operator. Recall that a Calder´on-Zygmund operator is bounded on L (R ). The following examples emphasise that a Singular integral operator need not be bounded on 2 n L (R ). 2 n Example 10.0.3. (i) T f(x) = |x| f(x) when x ∈ R is not bounded. But T is an SIO 2 and the Schwartz kernel is given by K(x, y) = |x| δ0(x − y) and K|c∆ = 0 ∈ CZKα. 0 ∞ n 2 n (ii) T f = f whenever f ∈ Cc (R ) is not bounded on L (R ). The Schwartz kernel here 0 is given by K(x, y) = −δ0(x − y). Again, K|c∆ = 0 ∈ CZK

The previous examples illustrated that one of the problem for the boundedness of a T ∈ 2 n SIO on L (R ) is the behaviour of the Schwartz kernel on the diagonal ∆. We will prove

100 2 n tr a theorem that T ∈ SIO is bounded in L (R ) if and only if both T (1),T (1) ∈ BMO and when T has the “weak boundedness property.” In the sequel, we will give a rigorous formulation of this property. This is the key property that will give control on the diagonal of the Schwartz kernel of T .

n ∞ n Deﬁnition 10.0.4 (NB,q). For q ∈ N and B a ball in R and any ϕ ∈ Cc (R ) with spt ϕ ⊂ B, deﬁne α |α| NB,q(ϕ) = sup k∂ ϕk∞(rad B) . α∈Nn, |α|≤q Remark 10.0.5. This is a scale invariant quantity. Let spt ϕ ⊂ B(0, 1) and x − x0 ψ(x) = ϕ r

n for some x0 ∈ R and r > 0. Then, spt ψ ⊂ B(x0, r) and NB(x0,r),q(ψ) = NB(0,1),q(ϕ). n Proposition 10.0.6. Let T ∈ SIO and ﬁx x0 ∈ R , r > 0 and deﬁne Tx0,r by · −x0 · −x0 hTx ,rϕ, ψi = hT ϕ , ψ i. 0 r r

n Then, Tx0,r ∈ SIO and its kernel is given by Kx0,r(x, y) = r K(rx + x0, ry + x0).

Proof. Exercise.

2 n 2 n Corollary 10.0.7. T ∈ L(L ( )) if and only if T ∈ L(L ( )) and kT k 2 n ≤ R x0,r R x0,r L(L (R )) n k k 2 n r T L(L (R )). Deﬁnition 10.0.8 (Weak boundedness property). Let T ∈ SIO. We say that T has the weak boundedness property (or WBP) if there exists a q ∈ N, q > 0, and a C > 0 such ∞ that for all balls B and all ϕ, ψ ∈ Cc (B),

|hT ϕ, ψi| ≤ CL (B)NB,q(ϕ)NB,q(ψ).

1 1 Remark 10.0.9. We note that kϕk2 ≤ L (B) 2 kϕk∞ ≤ L (B) 2 NB,q(ϕ). Therefore, if 2 n T ∈ L(L (R )), then

|h i| ≤ k k 2 n k k k k ≤ k k 2 n ( )N ( )N ( ) T ϕ, ψ T L(L (R )) ϕ 2 ψ 2 T L(L (R ))L B B,q ϕ B,q ψ .

We now deal with the issue of giving meaning to T acting on the constant function f ≡ 1. Note that this is non trivial since T is an operator defined on compactly supported functions. Definition 10.0.10. Define n ∞ n D0(R ) = ψ ∈ Cc (R ): ψ dL = 0 . ˆRn n Lemma 10.0.11. Let T ∈ SIO and ψ ∈ D0(R ). Let B be a ball such that spt ψ ⊂ B. Then, T ψ ∈ L1(c2B) and

k k 1 c ≤ k k ( )k k 1 T ψ L ( 2B) K CZKα C n, B ψ L (B).

101 1 c 1 c Remark 10.0.12. In writing T ψ ∈ L ( 2B), we mean the distribution T ψ|c2B ∈ L ( 2B).

∞ n c Proof. Fix ϕ ∈ Cc (R ) and spt ϕ ⊂ 2B. Then,

hT ψ, ϕi = hK, ϕ ⊗ ψi = hK|c∆ , ϕ ⊗ ψi since spt ϕ ⊗ ψ ⊂ c∆. Certainly, hK|c∆ , ϕ⊗ψi = K(x, y)ϕ(x)ψ(y) dL (y)dL (x) = K(x, y)ψ(y) dL (y) ϕ(x) dL (x) ¨c∆ ˆRn ˆRn and so T ψ|c2B agrees with K(x, y)ψ(y) dL (y). ˆRn

Now, since n ψ d = 0, R L ´

K(x, y)ψ(y) dL (y) = (K(x, y) − K(x, yB)) ψ(y) dL (y) ˆRn ˆRn where yB is the centre of B and

(rad B)α ( ) ( ) ( ) ≤ k k | ( )| ( ) K x, y ψ y dL y K CZKα n+α ψ y dL y ˆRn ˆRn |x − yB| (rad B)α ≤ k k k k 1 K CZKα n+α ψ L (B). |x − yB|

The conclusion is achieved by integrating over x ∈ c2B.

∞ ∞ n n Proposition 10.0.13. Let f ∈ C ∩ L (R ) and ψ ∈ D0(R ) and let B be a ball such ∞ n that spt ψ ⊂ B. Then, whenever η ∈ Cc (R ) such that spt η ⊂ 4B and η ≡ 1 on 3B,

hT (fη), ψi + (1 − η)(x)f(x)T tr(ψ)(x) dL (x) ˆRn is well deﬁned and does not depend on the choice of η.

∞ n Proof. We note that ψ and fη ∈ Cc (R ) and this implies that hT (fη), ψi exists. By the tr 1 c ∞ n application of Lemma 10.0.11, we have T (ψ)|c2B ∈ L ( 2B). Also, (1 − η)f ∈ L (R ) and spt (1 − η)f ⊂ c2B.

To show the independence of η, choose η1 and η2 with the desired properties. Then, we note that

tr hT (fη1), ψi + (1 − η1)(x)f(x)T (ψ)(x) dL (x) ˆRn tr = hT (fη2), ψi + (1 − η2)(x)f(x)T (ψ)(x) dL (x) ˆRn is equivalent to

tr hT (f(η1 − η2)), ψi = (η1 − η2)(x)f(x)T (ψ)(x) dL (x). ˆRn

102 Noting that spt ψ ⊂ B and spt (η1 − η2)f ⊂ 4B \ 3B, we compute

hT (f(η1 − η2)), ψi = K(x, y)f(y)(η1 − η2)(x)ψ(x) dL (x)dL (y) ¨Rn = (η1 − η2)(x)f(x) K(y, x)ψ(y) dL (y) dL (x) ˆRn ˆRn by the application of Fubini.

As a consequence, we are able to make the following deﬁnition.

∞ ∞ n Deﬁnition 10.0.14. Whenever f ∈ C ∩ L (R ), deﬁne T f by

hT f, ψi = hT (fη), ψi + (1 − η)(x)f(x)T tr(ψ)(x) dL (x) ˆRn n ∞ n whenever ψ ∈ D0(R ) and η ∈ Cc (R ) such that spt ψ ⊂ B, spt η ⊂ 4B and η ≡ 1 on 3B.

∞ n Remark 10.0.15. If f ∈ Cc (R ) and η = 1 on spt f, the right hand side agrees with hT (fη), ψi which is hT f, ψi. Hence, this deﬁnition is an extension of the original one.

∞ ∞ n n 0 Proposition 10.0.16. Whenever f ∈ C ∩ L (R ), we have T f ∈ D0(R ) .

n 0 Proof. We want to show that T f ∈ D0(R ) with bounds depending only on f. Let B be n a ball and ψ ∈ D0(R ) a such that spt ψ ⊂ B. Fix an η as in the deﬁnition of T f.

∞ n ∞ n 0 As T ∈ L(Cc (R ), Cc (R ) ), there exists an integer M > 0 such that

α α |hT (fη), ψi| ≤ C(M, 4B) sup k∂ (fη)k∞ sup k∂ ψk∞ |α|≤M |α|≤M 0 α α ≤ C (M, 4B) sup k∂ fkL∞(B) sup k∂ ψk∞ |α|≤M |α|≤M

Now,

tr tr (1 − η)(x)f(x)T (ψ)(x) dL (x) ≤ k1 − ηk∞kfk∞ T (ψ)(x) dL (x) ˆRn ˆc2B ≤ k1 − k k k ( )k k k k η ∞ f ∞C n, α K CZKα ψ 1 ≤ ( )(1 + k k )k k k k k k ( ) C n, α η ∞ f ∞ K CZKα ψ ∞L B . Altogether, we have shown

α |hT f, ψi| ≤ C(M,B) sup k∂ ψk∞. |α|≤M

n This shows that ψ 7→ hT f, ψi is continuous on D0(R ).

We now present the crucial theorem of this chapter.

103 Theorem 10.0.17 (T (1) Theorem of David-Journ´e(’84)). Let T ∈ SIO. Then, T ∈ 2 n tr L(L (R )) if and only if T possesses the weak boundedness property, T (1),T (1) ∈ BMO.

2 n Remark 10.0.18. When we write T ∈ L(L (R )), we mean T extends to such an operator since what we prove is that

|hT f, gi| ≤ Ckfk2kgk2 ∞ n whenever f, g ∈ Cc (R ).

The following Lemma is of use in the proof of the T (1) Theorem. We leave its proof as an exercise.

∞ n Lemma 10.0.19. Let ϕ ∈ Cc (R ) be a radial function such that n ϕ dL = 1, and 1 ∞ n R spt ϕ ⊂ B(0, 2 ). Let Pt = ϕt ∗ . Then for all f ∈ Cc (R ), ´

2 2 lim Pt (f) = f and lim Pt (f) = 0 t→0 t→∞

∞ n in Cc (R ).

2 n Proof of the T (1) Theorem. We already have one direction: if T ∈ L(L (R )) we have already seen that T has WBP and T (1),T tr(1) ∈ BMO. We prove the converse.

tr Let T0 = T − πT (1) − πT tr(1) where πb is the paraproduct with b ∈ BMO. We note tr we can choose πb such that πb ∈ CZO1 ⊂ SIO, πb(1) = b in BMO and πb = 0 in BMO. Therefore, T0 ∈ SIO and possesses WBP. Also, T0(1) = T (1) − T (1) − 0 = 0 tr tr tr 2 n and T0 (1) = T (1) − 0 − T (1) = 0 in BMO. Since T ∈ L(L (R )) if and only if 2 n tr T0 ∈ L(L (R )), we can assume that T (1) = 0 and T (1) = 0 in BMO.

∞ n Take a ϕ and Pt as in the hypothesis of Lemma 10.0.19. Fix f, g ∈ Cc (R ) and by the same Lemma,

2 2 2 2 2 2 2 2 hT f, gi = limhTPε f, Pε gi − lim hTPRf, PRgi = lim hTPε f, Pε gi − hTP 1 f, P 1 gi . ε→0 R→∞ ε→0 ε ε

2 ∞ n 1 Also, the map t ∈ (0, ∞) 7→ Pt f ∈ Cc (R ) is C and therefore,

1 ε d 2 2 hT f, gi = lim hTPt ,Pt gi dt. ε→0 ˆε dt Here, d d d hTP 2,P 2gi dt = hT P 2f ,P 2gi + hTP 2f, P 2gi dt t t dt t t t dt t

2 d 2 tr d 2 2 d 2 2 and since hTPt f, dt Pt gi = hT dt Pt g ,Pt fi, it is enough to just treat hT dt Pt f ,Pt gi.

2 ∞ n n Now, we note that Pt f ∈ Cc (R ) ⊂ S (R ) and we consider the spatial Fourier Trans- form. On noting that ϕ is radial if and only ifϕ ˆ is radial, we have

n \d d d X P 2f(ξ) = Pd2f(ξ) = ϕˆ(tξ)2fˆ(ξ) = 2 ξ (∂ ϕˆ(tξ))ϕ ˆ(tξ). dt t dt t dt k k k=1

104 ˜ ˜ ∞ 1 So deﬁne ψk(x) = −xkϕ(x) and ψk(x) = ∂kϕ(x) so that ψk, ψk ∈ Cc (B(0, 2 )) with ˜ ˆ ˜ˆ n ψ d = n ψ d = 0. Also,ψ (ξ) = ı∂ ϕˆ(ξ) and ψ (ξ) = −ıξ ϕˆ(ξ) and therefore, R k L R k L k k k k ´ ´ n n \d 2 X 2 X ˆ P 2f(ξ) = t ξ ∂ ϕ(ˆtξ) ϕˆ(tξ)fˆ(ξ) = ψˆ (tξ)ψ˜ (tξ)fˆ(ξ). dt t t k k t k k k=1 k=1 ˜ Letting Qk,t = ψk,t ∗ and Q˜k,t = ψk,t ∗ we have

n d 2 X P 2f = Q Q˜ f. dt t t k,t k,t k=1

We consider Qk,tQ˜k,tf for each k. So ﬁx k and ε > 0. We prove that there exists a C > 0 (independent of ε) such that

1 ε ˜ 2 dt hTQk,tQk,tf, Pt gi ≤ Ckfk2kgk2. ˆε t Let 1 1 ε ˜ 2 dt ε ˜ tr tr 2 dt Iε = hTQk,tQk,tf, Pt gi = hQk,tf, Qk,tT Pt gi , ˆε t ˆε t tr tr 2 and Rk,t = Qk,tT Pt . Then, by Cauchy-Schwarz

1 1 1 ! 2 1 ! 2 ε 2 dt ε ˜ 2 dt |Iε| ≤ kRk,tgk2 kQk,tfk2 . ˆε t ˆε t

By an LPE estimate, we get that

1 1 ! 2 ε ˜ 2 dt kQk,tfk2 ≤ C(ψk)kfk2 ˆε t and so it is enough to show that

∞ 2 dt 2 kRk,tgk2 ≤ C(ϕ, ψk,T )kgk2. ˆ0 t

∞ n We examine the kernel of Rk,t. Pick a h ∈ Cc (R ). First, we note that hRk,tg, hi = 2 n hTQk,th, Pt gi. For u, v ∈ R , set

u v ψk,t(x) = ψk,t(x − u) and ϕk,t(x) = ϕt(x − v) so that u 2 v Qk,th = ψk,th(u) dL (u) and Pt g = ϕt g(v) dL (v). ˆRn ˆRn ∞ n ∞ n 0 Using the continuity of T :Cc (R ) → Cc (R ) , we can write

2 u v hTQk,th, Pt gi = h(u)hT ψk,t, ϕt ig(v) dL (u)dL (v) ˆRn ˆRn

105 u v and so the kernel of Rk,t is Kk,t(u, v) = hT ψk,t, ϕt i.

u t v We estimate Kk,t(u, v). First, suppose |u − v| ≤ 3t. then, spt ψk,t ⊂ B(u, 2 ) and spt ϕt ⊂ u v B(v, t) and by setting B = B(v, 4t) we have that spt ψk,t, spt ϕt ⊂ B. By the fact that T has WBP, u v |Kk,t(u, v)| ≤ CL (B)NB,q(ψk,t)NB,q(ϕt ) and so C(n, ψ , ϕ) |K (u, v)| ≤ C k . k,t tn u t u Now, suppose |u − v| ≥ 3t. As before, we have spt ψ ⊂ B(u, ) and also n ψ d = 0 k,t 2 R k,t L which gives ´ α u t u T ψk,t(x) ≤ CkKk kψk,tk CZKα |x − u|n+α 1 v c where K is the kernel of T and when x 6∈ B(u, t). Now, spt ϕt ⊂ B(v, t) ⊂ B(u, t) and so tα h u vi = u ( ) v( ) ( ) ≤ k k k k k k T ψk,t, ϕt T ψk,t x ϕt x dL x C K CZKα n+α ψk 1 ϕ 1 ˆRn |v − u| since kψu k = kψ k = kψ k and kϕ k = kϕk . k,t 1 k,t 1 k 1 t 1 1

v Now, we consider the regularity of Kk,t(u, v) with respect to v. We note that v 7→ ϕt ∈ ∞ n 1 Cc (R ) is a C function and

v 1 1 u v ∂v ϕ = − (∂ ϕ) and ∂v K (u, v) = − hT ψ , (∂ ϕ) )i. l t t l t l k,t t k,t l t This gives C |u − v|−n−ε |∂v K (u, v)| ≤ 1 + . l k,t tn+1 t 0 0 So (Rt)t>0 is an ε -family for some ε < ε.

∞ n 3t We compute Rk,t(1). Choose η ∈ Cc (R ) with η ≡ 1 on B(u, 2 ) and spt η ⊂ B(u, 2t). Then,

Rk,t(1)(u) = Kk,t(u, v) dL (v) ˆRn u v = hT ψk,t, ϕt i dL (v) ˆRn u v u v = hT ψk,t, ηϕt i dL (v) + hT ψk,t, (1 − η)ϕt i dL (v) ˆRn ˆRn u v u v = hT ψk,t, ηϕt dL (v)i + T ψk,t(x)(1 − η)(x)ϕt (x) dL (x) dL (v). ˆRn ˆRn ˆRn First, we have v ηϕt dL (v) = η ˆRn ∞ n u v u u 1 in C ( ) so hT ψ , n ηϕ d (v)i = hT ψ , ηi. Also, T ψ ∈ L (spt (1 − η)) in x, c R k,t R t L k,t k,t ∞ n v 1 n v ∞ n (1 − η) ∈ L (R ) in x´, ϕt ∈ L (R ) in v and ϕt ∈ L (R ) in x. By Fubini coupled with v n ψ d (v) = 1 R k,t L ´ u v u T ψk,t(x)(1 − η)(x)ϕt (x) dL (x) dL (v) = T ψk,t(x)(1 − η)(x) dL (x). ˆRn ˆRn ˆRn

106 By deﬁnition of T tr(1) and using T tr(1) = 0 in BMO, u u tr u Rk,t(1)(u) = T ψk,t(x)(1 − η)(x)dL (x) + hT ψk,t, ηi = hT (1), ψk,ti = 0. ˆRn ˆRn

0 Combining these facts, we have shown that Rk,t is an ε -family with Rk,t(1) = 0 for all t and k. In particular dL (x)dt |R (1)(x)|2 k,t t is Carleson. Therefore, by Theorem 9.0.5,

∞ 2 dt 2 kRk,tgk2 ≤ Ckgk2 ˆ0 t and the proof is complete.

Remark 10.0.20 (Avoiding the reduction to T (1) = T tr(1) = 0). We note that for all n tr u tr t > 0 and all u ∈ R the equality Rk,t(1)(u) = hT (1), ψk,ti holds when T (1) ∈ BMO. tr u Going back again to the deﬁnition of hT (1), ψk,ti and comparing with the deﬁnition of ˇ tr tr ˇ ˆˇu Qk,t(T (1))(u) when T (1) ∈ BMO (since Qk,t = ψk,t ∗ ), we can see that

tr u ˇ tr hT (1), ψk,ti = Qk,t(T (1))(u).

tr Thus, Rk,t(1) = Qˇk,t(T (1)). As the Littlewood-Paley estimates combined with the decay tr of Qˇk,t and T (1) ∈ BMO imply that

ˇ tr 2 dL (x)dt Qk,t(T (1))(x) t is a Carleson measure we can conclude by applying Theorem 9.0.5.

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