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1 Free ebooks ==>

2 Free ebooks ==> Singular Integrals and Related Topics

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4 Free ebooks ==> Singular Integrals and Related Topics Shanzhen Lu Yong Ding Beijing Normal University, China Dunyan Yan Academia Sinica, China World Scientific N E W J E R S E Y L O N D O N S I N G A P O R E B E I J I N G S H A N G H A I H O N G K O N G TA I P E I C H E N N A I

5 Free ebooks ==> Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore USA office: 27 Warren Street, Suite , Hackensack, NJ 0760 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library. SINGULAR INTEGRALS AND RELATED TOPICS Copyright 2007 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Publisher. For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 0923, USA. In this case permission to photocopy is not required from the publisher. ISBN ISBN Printed in Singapore.

6 Free ebooks ==> Preface It is well-known that singular integrals is continuously regarded as a central role in harmonic analysis. There are many nice books related to singular integrals. In this book, there are at least two sides which differ from the other books. One of them is to establish more perfect theory of singular integrals. It includes not only the case of smooth kernels, but also the case of rough kernels. In the same way, we deal with some related operators, such as fractional integral operators and Littlewood-Paley operators. The other is to introduce more new theory on some oscillatory singular integrals with polynomial phases. This book is mainly provided to graduate students in analysis field. However, it is also beneficial to researchers in mathematics. This book consists of five chapters. Let us now illustrate the choice of material in each chapter. Chapter is devoted to the theory of the Hardy-Littlewood maximal operator as the basis of singular integrals and other related operators. It also includes the basic theory of the A p weights. Chapter 2 is related to the theory of singular integrals. Since the theory of singular integrals with Calderón-Zygmund kernel has been introduced in many books, we will pay more attention to the singular integrals with homogeneous kernels. Specially, we will introduce more perfect theory of singular integrals with rough kernels, for instance the L p boundedness of singular integrals with kernels in certain Hardy space on the unit sphere will be fully proved. In addition, the weighted L p boundedness of singular integrals with rough kernels and their commutators will be also established. Chapter 3 is devoted to fractional integrals. In the same way, we will pay more attention to the case of rough kernels. It includes not only the A(p, q) weight theory of fractional integrals with rough kernels, but also the theory of its commutators. Chapter 4 is to introduce a class of oscillatory singular integrals with polynomial phases. Note that this oscillatory singular integral is neither a Calderón-Zygmund operator nor a convolution operator. However there exists certain link between this oscillatory singular integral and the correv

7 Free ebooks ==> vi PREFACE sponding singular integral. Therefore, the criterion on the L p boundedness of oscillatory singular integrals will become a crucial role in this chapter. It will discover an equivalent relation between the L p boundedness of the oscillatory singular integral and that of the corresponding truncated singular integral. Chapter 5 is related to the Littlewood-Paley theory. In this chapter, we will establish two kinds of the weakest conditions on the kernel for the L p boundedness of Marcinkiewicz integral operator with rough kernel. Finally, it is worth pointing out that as space is limited, the theory of singular integrals and related operators in this book is only worked on the Lebesgue spaces although there are many good results on other spaces such as Hardy spaces and BMO space. It should be pointed out that many results in the later three chapters of this book reflect the research accomplishment by the authors of this book and their cooperators. We would like to acknowledge to Jiecheng Chen, Dashan Fan, Yongsheng Han, Yingsheng Jiang, Chin-Cheng Lin, Guozhen Lu, Yibiao Pan, Fernando Soria and Kozo Yabuta for their effective cooperates in the study of singular integrals. On this occasion, the authors deeply cherish the memory of Minde Cheng and Yongsheng Sun for their constant encourage. The first named author of this book, Shanzhen Lu, would like to express his thanks to his former students Wengu Chen, Yong Ding, Zunwei Fu, Yiqing Gui, Guoen Hu, Junfeng Li, Guoquan Li, Xiaochun Li, Yan Lin, Heping Liu, Mingju Liu, Zhixin Liu, Zongguang Liu, Bolin Ma, Huixia Mo, Lin Tang, Shuangping Tao, Huoxiong Wu, Qiang Wu, Xia Xia, Jingshi Xu, Qingying Xue, Dunyan Yan, Dachun Yang, Pu Zhang, and Yan Zhang for their cooperations and contributions to the study of harmonic analysis during the joint working period. Finally, Shanzhen Lu would like to express his deep gratitude to Guido Weiss for his constant encourage and help. Shanzhen Lu Yong Ding Dunyan Yan December, 2006

8 Free ebooks ==> Contents Preface v Hardy-Littlewood Maximal Operator. Hardy-Littlewood maximal operator Calderón-Zygmund decomposition Marcinkiewicz interpolation theorem Weighted norm inequalities Notes and references Singular Integral Operators Calderón-Zygmund singular integral operators Singular integral operators with homogeneous kernels Singular integral operators with rough kernels Commutators of singular integral operators Notes and references Fractional Integral Operators Riesz potential Weighted boundedness of Riesz potential Fractional integral operator with homogeneous kernels Weighted boundedness of T Ω,α Commutators of Riesz potential Commutators of fractional integrals with rough kernels Notes and references Oscillatory Singular Integrals Oscillatory singular integrals with homogeneous smooth kernels Oscillatory singular integrals with rough kernels Oscillatory singular integrals with standard kernels Multilinear oscillatory singular integrals with rough kernels. 202 vii

9 Free ebooks ==> viii CONTENTS 4.5 Multilinear oscillatory singular integrals with standard kernels Notes and references Littlewood-Paley Operator Littlewood-Paley g function Weighted Littlewood-Paley theory Littlewood-Paley g function with rough kernel Notes and references Bibliography 26 Index 27

10 Free ebooks ==> Chapter Hardy-Littlewood Maximal Operator. Hardy-Littlewood maximal operator Let us begin with giving the definition of the Hardy-Littlewood maximal function, which plays a very important role in harmonic analysis. Definition.. (Hardy-Littlewood maximal function) Suppose that f is a locally integrable on R n, i.e., f L loc (Rn ). Then for any x R n, the Hardy-Littlewood maximal function Mf(x) of f is defined by Mf(x) = sup r>0 r n f(x y) dy. (..) y r Moreover, M is also called as the Hardy-Littlewood maximal operator. Sometimes we need to use the following maximal functions. For f L loc (Rn ) and x R n, M f(x) = sup f(y) dy, (..2) r>0 Q(x, r) Q(x,r) where and below, Q(x, r) denotes the cube with the center at x and with side r and its sides parallel to the coordinate axes. Moreover, E denotes

11 Free ebooks ==> 2 Chapter. Hardy-Littlewood Maximal Operator the Lebesgue measure of the set E. More general, M f(x) = sup f(y) dy, (..3) Q x Q where the supremum is taken over all cubes or balls Q containing x. For the Hardy-Littlewood maximal operator M, we would like to give the following some remarks. Remark.. By (..)-(..3), it is easy to see that there exist constants C i (i = 0,, 2, 3) depending only on the dimension n such that Q C 0 Mf(x) C M f(x) C 2 M f(x) C 3 Mf(x) (..4) for any x R n. That is, the Hardy-Littlewood maximal function Mf of f and the maximal functions M f, M f are pointwise equivalent each other. Remark..2 For f L loc (Rn ), the Hardy-Littlewood maximal function Mf(x) is a lower semi-continuous function on R n, and is then a measurable function on R n. By (..4), we only need to show it for M f(x). In fact, it is sufficient to show that for any λ R, the set E = {x R n : M f(x) > λ} is an open set. However, by the definition of M f(x) it suffices to show that E is open for all λ > 0. Equivalently, we only need to show that E c := {x R n : M f(x) λ} is a closed set for all λ > 0. Suppose that {x k } E c satisfying x k x as k. We only need to show that for any r > 0 Q(x, r) Q(x,r) f(y) dy λ. (..5) Denote Q k = Q(x k, r) and f k (y) = f(y)χ Q(x,r) Qk (y) for all k =, 2,, where Q(x, r) Q k = ( Q(x, r)\q k ) ( Qk \Q(x, r) ). Thus, f k (y) f(y) for all k and lim k f k (y) = 0. Applying the Lebesgue dominated convergence theorem, we have lim f k (y) dy = 0. (..6) k Q(x, r) Q(x,r)

12 Free ebooks ==>. Hardy-Littlewood maximal operator 3 On the other hand, it is clear that f(y) dy = f(y) dy λ. Q(x, r) Q k Q k Q k Hence f(y) dy Q(x, r) Q(x,r) f(y) dy Q(x, r) Q(x,r) Q k + f(y) dy Q(x, r) Q k f k (y) dy + λ. Q(x, r) Q(x,r) Let k, by (..6) we obtain (..5). Remark..3 The Hardy-Littlewood maximal operator M is not a bounded operator from L (R n ) to itself. We only consider the case n =. Take f(x) = χ [0,] (x), then for any x, we have Hence R Mf(x) 2x Mf(x)dx 2x 0 f(y) dy = 2x. Mf(x)dx dx =. 2x Although M is not a bounded operator on L (R n ), however, as its a replacement result we shall see that M is a bounded operator from L (R n ) to L, (R n ), i.e., the weak L (R n ) space (see Definition..2 below). Lemma.. (Vitali type covering lemma) Let E be a measurable subset of R n and let B be a collection of balls B with bounded diameter d(b) covering E in Vitali s sense, i.e. for any x E there exist a ball B x B such that x B x. Then there exist a β > 0 depending only on n, and disjoint countable balls B, B 2,, B k, in B such that B k β E. k

13 Free ebooks ==> 4 Chapter. Hardy-Littlewood Maximal Operator In fact, it will be seen from the proof below that it suffices to take β = 5 n. Proof. Denote l 0 = sup{d(b) : B B} <. Take B B so that d(b ) 2 l 0. Again denote B = {B : B B and B B = } and l = sup{d(b) : B B }, then we choose B 2 B such that d(b 2 ) 2 l. Suppose that B, B 2,, B k have been chosen from B according to the above way, then we denote B k = B : B B with B k B j = and j= l k = sup{d(b) : B B k }. Next we choose B k+ B k such that d(b k+ ) 2 l k. Thus we may choose a sequence B, B 2,, from B such that (i) B, B 2,, B k, are disjoint; (ii) d(b k+ ) 2 sup{d(b) : B B k}, and B k = B : B B and B k for k =, 2,. j= B j = If this process stops at some B k, then it shows that B k =. In this case, for any x E there exists a ball B x B such that x B x and B x B k0 with some k 0 k. Without loss of generality, we may assume that B x B j = for j =, 2,, k 0. So, d(b k0 ) 2 d(b x), and this implies B x 5B k0, where 5B k0 expresses the five times extension of B k0 with the same center. Thus, we have E k j= 5B j, and it leads to k k k E 5B j 5B j 5 n B j. j= j= On the other hand, it is trivial when j= B k =. So, we may assume that j= B k <. Denote Bk = 5B k. We will claim that E j= Bk. (..7) k=

14 Free ebooks ==>. Hardy-Littlewood maximal operator 5 In fact, it suffices to prove that B k= B k for any B B. Since j= B k <, we have d(b k ) 0 as k. Thus there exists k 0 such that d(b k0 ) < 2 d(b). Of course, we may think that the index k 0 is the smallest with the above property. In this case, B x must intersect with some B j for j k 0. Otherwise, d(b k0 ) 2 d(b x). As before, we get B x 5B j = Bj and (..7) follows. Thus E Bk Bk 5n B k. This completes the proof. k= k= k= Definition..2 (Weak L p spaces) Suppose that p < and f is a measurable function on R n. The function f is said to belong to the weak L p spaces on R n, if there is a constant C > 0 such that sup λ {x R n : f(x) > λ} /p C <. λ>0 In other words, the weak L p (R n ) is defined by where L p, (R n ) = { f : f p, < }, f p, := sup λ {x R n : f(x) > λ} /p λ>0 denotes the seminorm of f in the weak L p (R n ). Remark..4 It is easy to verify that for p <, L p (R n ) L p, (R n ). Definition..3 (Operator of type (p, q)) Suppose that T is a sublinear operator and p, q. T is said to be of weak type (p, q) if T is a bounded operator from L p (R n ) to L q, (R n ). That is, there exists a constant C > 0 such that for any λ > 0 and f L p (R n ) ( ) C q {x R n : T f(x) > λ} λ f p ; (..8) T is said to be of type (p, q) if T is a bounded operator from L p (R n ) to L q (R n ). That is, there exists a constant C > 0 such that for any f L p (R n ) T f q C f p, (..9) where and below, f p = f L p (R n ) denotes the L p norm of f(x).

15 Free ebooks ==> 6 Chapter. Hardy-Littlewood Maximal Operator When p = q and the operator T satisfies (..8) or (..9), T is also said to be of weak type (p, p), respectively. Moreover, It is easy to see that an operator of type (p, q) is also of weak type (p, q), but its reverse is not hold generally. Below we shall prove that the Hardy-Littlewood maximal operator M is of weak type (, ) and type (p, p) for < p, respectively. Theorem.. Let f be a measurable function on R n. (a) If f L p (R n ) for p, then Mf(x) < a.e. x R n. (b) There exists a constant C = C(n) > 0 such that for any λ > 0 and f L (R n ) {x R n : Mf(x) > λ} C λ f. (c) There exists a constant C = C(n, p) > 0 such that for any f L p (R n ) < p, Mf p C f p. Proof. Obviously, the conclusion (a) is a direct result of the conclusions (b) and (c). Hence we only give the proof of (b) and (c). Let us first consider (b). For any λ > 0, by Remark.. the set E λ := {x R n : Mf(x) > λ} is an open set, and is then a measurable set. By Definition.., for any x E λ, there exists a ball B x with the center at x such that f(y) dy > λ. B x B x Thus B x < f(y) dy λ B x λ f < for all x E λ. Therefore, if we denote B = {B x : x E λ }, then B covers E λ in Vitali s sense. By Lemma.., we may choose disjoint countable balls B, B 2,, B k, in B such that B k β E λ. Hence β E λ k k B k λ = λ k k B k λ f. B k f(y) dy f(y) dy

16 Free ebooks ==>. Hardy-Littlewood maximal operator 7 Let us now turn to the proof of (c). Clearly, the conclusion (c) holds for p =, we only consider the case < p <. Let f L p (R n ) ( < p < ). For any λ > 0, write f = f + f 2, where { f(x), for f(x) λ/2 f (x) = 0, for f(x) < λ/2. It is easy to see that f L (R n ). Thus we have f(x) f (x) + λ 2 and Mf(x) Mf (x) + λ 2. (..0) Hence, by (..0) and the weak (,) boundedness of M (i.e. the conclusion (b)), we have E λ = {x R n : Mf(x) > λ} {x R n : Mf (x) > λ/2} 2β f (x) dx λ R n = 2β f(x) dx, λ {x R n : f(x) λ/2} where β is the constant in Lemma... Therefore (Mf(x)) p dx R n = p λ p E λ dλ 0 ( ) 2β p λ p f(x) dx dλ 0 λ {x R n : f(x) λ/2} ( 2 f(x) ) 2βp f(x) λ p 2 dλ dx R n 0 = 2βp f(x) p dx. p R n Thus we finish the proof of Theorem... Immediately, by the weak (,) boundedness of the Hardy-Littlewood maximal operator M we may get the Lebesgue differentiation theorem. Theorem..2 (Lebesgue differentiation theorem ) Suppose that f L loc (Rn ). Then lim f(y) dy = f(x) a.e. x R n, r 0 B(x, r) B(x,r)

17 Free ebooks ==> 8 Chapter. Hardy-Littlewood Maximal Operator where B(x, r) denotes the ball with the center at x and radius r. Proof. Since for any R > 0, fχ B(0,R) L (R n ), we may assume that f L (R n ). Denote L r (f)(x) = f(y) dy, B(x, r) B(x,r) and let Then Let us first show that for any λ > 0 Λ(f)(x) = liml r (f)(x) lim L r (f)(x). r 0 r 0 Λ(f)(x) 2 sup L r (f)(x) = 2Mf(x). r>0 E λ (Λf) := {x R n : Λ(f)(x) > λ} = 0. (..) In fact, for any ε > 0 we may decompose f = g + h, where g is a continuous function with compact support set and h < ε. Thus and it leads to By Theorem.. (b), we have Λ(f)(x) Λ(g)(x) + Λ(h)(x) = Λ(h)(x), E λ (Λf) E λ (Λh) E λ/2 (Mh). E λ (Λf) 2C λ h < 2Cε λ. Thus, by the arbitrariness of ε we know (..) holds, and (..) shows that the limit lim r 0 L r (f)(x) exists for a.e. x R n. On the other hand, by the integral continuity, we have lim L r(f) f r 0 = lim r 0 R n B(x, r) = lim r 0 R n B(0, r) lim r 0 B(0, r) B(0,r) B(x,r) B(0,r) f(y) dy f(x) dx [f(x y) f(x)]dy dx R n f(x y) f(x) dx dy = 0.

18 Free ebooks ==>.2 Calderón-Zygmund decomposition 9 Hence there exists a subsequence {r k } satisfying r k 0 as k, such that lim k L r k (f)(x) = f(x) for a.e. x R n. Because lim L r (f)(x) exists for a.e. x R n, thus r 0 lim f(y) dy = f(x) a.e. x R n, r 0 B(x, r) B(x,r) which is the desired conclusion. Remark..5 Clearly, by the equivalence of (..) - (..3), it is easy to see that the conclusion of Theorem..2 still holds if we replace the ball B(x, r) by cube Q(x, r), even more generally, by a cube Q containing x..2 Calderón-Zygmund decomposition Applying Lebesgue differentiation theorem, we may give a decomposition of R n, called as Calderón-Zygmund decomposition, which is extremely useful in harmonic analysis. Theorem.2. (Calderón-Zygmund decomposition of R n ) Suppose that f is nonnegative integrable on R n. Then for any fixed λ > 0, there exists a sequence {Q j } of disjoint dyadic cubes (here by disjoint we mean that their interiors are disjoint) such that () f(x) λ for a.e. x / j Q j; (2) Q j λ f ; j (3) λ < Q j Q j f(x) dx 2 n λ. Proof. By f L (R n ), we may decompose R n into a net of equal cubes with whose interiors are disjoint such that for every Q in the net f(x) dx λ. Q Q

19 Free ebooks ==> 0 Chapter. Hardy-Littlewood Maximal Operator Let Q be any fixed cube in the net. We divide it into 2 n equal cubes, and denote Q is one of these cubes. Then there are the follwing two cases. Case (i) Q f(x) dx > λ. Q Case (ii) In the case (i) we have λ < Q f(x) dx Q Q f(x) dx λ. Q 2 n Q f(x) dx 2 n λ. Q Hence, we do not sub-divide Q any further, and Q is chosen as one of the sequence {Q j }. For the case (ii) we continuously sub-divide Q into 2 n equal subcubes, and repeat this process until we are forced into the case (i). Thus we get a sequence {Q j } of cubes obtained from the case (i). By Theorem..2, f(x) = lim f(x) dx λ for a.e. x / Q j. Q x Q Q Q 0 j This proves the theorem. Remark.2. In place of R n by a fixed cube Q 0, we may similarly discuss the Calderón-Zygmund decomposition on Q 0 for f L (Q 0 ) and λ > 0. Moreover, we also may obtain the similar decomposition for f L p (R n ) (p > ). An application of the Calderón-Zygmund decomposition on R n is that it may be used to give the L boundedness of the Hardy-Littlewood maximal operator M in some sense. More precisely, we have the following conclusion. Theorem.2.2 Suppose that f L (R n ). (i) If then Mf L loc (Rn ). R n f(x) log + f(x) dx <,

20 Free ebooks ==>.2 Calderón-Zygmund decomposition (ii) If f is supported in a ball B and Mf L (B), then f(x) log + f(x) dx <, B where { log f(x), for f(x) > log + f(x) = 0, for f(x). Proof. We first give two estimates on the maximal operator. {x R n : Mf(x) > 2λ} C λ {x: f(x) >λ} f(x) dx, (.2.) and {x R n : M f(x) > λ} C λ {x: f(x) >λ} f(x) dx. (.2.2) The constant C appearing in (.2.) and (.2.2) depends only on the dimension n. In fact, (.2.) has appeared in the proof of Theorem... As for (.2.2), using the Calderón-Zygmund decomposition on R n (Theorem.2.) for λ > 0, we get the sequence {Q j } of cubes to satisfy λ < f(x) dx 2 n λ, Q j Q j and f(x) λ for a.e. x / j Q j. Thus for x Q j, M f(x) > λ and {x R n : M f(x) > λ} j 2 n λ 2 n λ Q j j Q j f(x) dx {x: f(x) >λ} f(x) dx.

21 Free ebooks ==> 2 Chapter. Hardy-Littlewood Maximal Operator Let us return to the proof of Theorem.2.2. First we consider (i). For any compact set K R n, it follows from (.2.) that Mf(x)dx = {x K : Mf(x) > λ} dλ K 0 = 2 {x K : Mf(x) > 2λ} dλ 0 { } 2 K dλ + {x R n : Mf(x) > 2λ} dλ 0 { } C 2 K + f(x) dxdλ λ {x: f(x) >λ} f(x) { } dλ = 2 K + C f(x) R n λ dx { } = 2 K + C f(x) log + f(x) dx R n <. Thus we obtain the conclusion (i). As for (ii), without loss of generality, we may assume that the radius of B is R. we denote by B the ball with center as one of B and the radius 2R. For any x B \ B, we take x B such that x is the point symmetric to x with respect to the boundary of B. Then it is easy to check that for any r > 0, B(x, r) B B(x, 0r). Thus, B(x, r) B(x,r) f(y) dy = B(x, r) C n B(x, 0r) C n Mf(x). B(x,r) B B(x,0r) f(y) dy f(y) dy Hence Mf(x ) C n Mf(x) for any x B \ B. By the estimates above, it is easy to see that Mf L (B ). On the other hand, for any x R n \ B, when r < R we have f(y) dy = 0 B(x, r) B(x,r) by the support set condition. Thus Mf is bounded on R n \ B, since Mf(x) B(x, R) f, for any x R n \ B.

22 Free ebooks ==>.2 Calderón-Zygmund decomposition 3 It is also clear that Mf(x) 0, as x. Thus, for any λ 0 > 0 the set {x R n : Mf(x) > λ 0 } must be contained in some ball. So I = Mf(x)dx <. {x R n :Mf(x)>λ 0 } Now, if we take λ 0 = C n, then we have I C n {x R n : Mf(x) > λ} dλ. On the other hand, by (..4) Mf(x) C n M f(x) for any x R n. If we denote λ = λ/c n, then by (.2.2) {x R n : Mf(x) > λ} {x R n : M f(x) > λ } C λ f(x) dx = C nc λ {x: f(x) >λ } {x: C n f(x) >λ} By two inequalities above we have c n C I f(x) dx dλ C n λ {x: c n f(x) >λ} cn f(x) dλ = c n C f(x) R n c n λ dx = c n C f(x) log + f(x) dx. R n The proof of Theorem.2.2 is finished. f(x) dx. Another important application of the Calderón-Zygmund decomposition on R n is that it may be used to deduce the Calderón-Zygmund decomposition on function. The later is a very important tool in harmonic analysis. Theorem.2.3 (Calderón-Zygmund decomposition for function) Let f be nonnegative integrable on R n. Then for any fixed λ > 0, there exists a sequence {Q j } of disjoint dyadic cubes and the functions g, b such that

23 Free ebooks ==> 4 Chapter. Hardy-Littlewood Maximal Operator (i) f(x) = g(x) + b(x); (ii) g(x) 2 n λ for a.e. x R n ; (iii) g p p Cλ p f for < p < ; (iv) b(x) = 0, a.e. x R n \ j Q j ; (v) Q j b(x) dx = 0, j =, 2,. Proof. Applying the Calderón-Zygmund decomposition on R n for f and λ > 0, we obtain a sequence {Q j } of disjoint dyadic cubes such that f(x) λ for a.e. x R n \ j Q j ; Q j λ f ; j and λ < f(x) dx 2 n λ, for j =, 2,. Q j Q j Now we define g(x) and b(x) as follows. g(x) = f(x) x R n \ j Q j Q j f(x) dx, x Q j, j =, 2, ; Q j and 0 x R n \ Q j j b(x) = f(x) f(x) dx, x Q j, j =, 2,. Q j Q j Thus f(x) = g(x) + b(x), and by the definition of b, both (iv) and (v) hold. Moreover, by the definition of g, (ii) holds. Finally, if denote F = R n \ j Q j,

24 Free ebooks ==>.3 Marcinkiewicz interpolation theorem 5 then g p p = g(x) p g(x) dx + g(x) p g(x) dx j Q j F (2 n λ) p g(x) dx + λ p g(x) dx j Q j F = (2 n λ) p f(x) dx + λ p f(x) dx j Q j F Cλ p f. This is just (iii), and the proof of Theorem.2.3 is finished..3 Marcinkiewicz interpolation theorem Notice that in the proof of L p boundedness of the Hardy-Littlewood maximal operator M, we only use the weak L and L boundedness of M. This inspires us to consider the similar problem for more general sublinear operator. Theorem.3. (Marcinkiewicz interpolation theorem) Let (X, X, µ) and (X, Y, ν) be two measure spaces, and let the sublinear operator T be both of weak type (p 0, p 0 ) and weak type (p, p ) for p 0 < p. That is, there exist the constants C 0, C > 0 such that for any λ > 0 and f ( ) p0 C0 (i) ν({x X : T f(x) > λ}) λ f p 0,µ ; (ii) ν({x X : T f(x) > λ}) ( ) p C λ f p,µ for p <. If p =, then the weak type and strong type coincide by definition: T f,ν C f,µ. Then T is also of type (p, p) for all p 0 < p < p, i.e. there exist a constants C > 0 such that for any f L p (X, µ) ( X /p ( /p T f(x) dν) p C f(x) dµ) p. X

25 Free ebooks ==> 6 Chapter. Hardy-Littlewood Maximal Operator Proof. For f L p (X, µ) and λ > 0, write f(x) = f λ (x) + f λ (x), where f λ (x) = { f(x), for f(x) > λ, 0, for f(x) λ. Thus, f λ L p 0 (X, µ) and f λ L p (X, µ). Moreover, T f(x) T f λ (x) + T f λ (x). Case I: p <. By the weak type (p i, p i ) of T (i =, 2), we have ν({x X : T f(x) > λ}) By (.3.), we have that X ( ) ν {x X : T f λ (x) > λ/2} ( ) +ν {x X : T f λ (x) > λ/2} ( ) p0 ( ) p 2C0 λ f λ 2C p0,µ + λ f λ p,µ. T f(x) p dν = p λ p ν({x X : T f(x) > λ})dλ 0 2 p 0 C 0 p λ p p 0 f(x) p 0 dµ(x)dλ + 2 p C p 2 p 0 C 0 p X 0 {x X: f(x) >λ} λ p p 0 {x X: f(x) <λ} f(x) f(x) p p C p f(x) p = 2p 0 C 0 p p p 0 = C p X X X λ p p 0 dλdµ(x) f(x) f(x) p dµ + 2p C p p p f(x) p dµ. (.3.) f(x) p dµ(x)dλ λ p p dλdµ(x) X f(x) p dµ Case II: p =. In this case, if we denote α = /(2C ), then ν({x X : T f(x) > λ}) ν({x X : T f αλ (x) > λ/2})

26 Free ebooks ==>.3 Marcinkiewicz interpolation theorem 7 and X T f(x) p dν 2 p 0 C 0 p 0 λ p p 0 = 2 p 0 C 0 p f(x) p 0 X = C p f(x) p dµ. This completes the proof of Theorem.3.. X {x X: f(x) >αλ} f(x) /α 0 f(x) p 0 dµ(x)dλ λ p p 0 dλdµ(x) As an application of the Marcinkiewicz interpolation theorem, below we will prove the Fefferman-Stein inequality on the Hardy-Littlewood maximal operator M. Theorem.3.2 (Fefferman-Stein inequality) Let < p <. Then there exists a constant C = C n,p such that for any nonnegative measurable function ϕ(x) on R n and f, (Mf(x)) p ϕ(x)dx C f(x) p Mϕ(x)dx. (.3.2) R n R n Proof. Without loss of generality, we may assume that Mϕ(x) < a.e. x R n and Mϕ(x) > 0. If denote dµ(x) = M ϕ(x)dx and dν(x) = ϕ(x)dx, then by the Marcinkiewicz interpolation theorem in order to get (.3.2), it suffices to prove that M is both of type (L (µ), L (ν)) and of weak type (L (µ), L (ν)). Let us first show that M is of type (L (µ), L (ν)). In fact, if f,µ a, then {x R n : f(x) >a} Mϕ(x)dx = µ({x R n : f(x) > a}) = 0. Since Mϕ(x) > 0 for any x R n, we have {x R n : f(x) > a} = 0, equivalently, f(x) a a.e. on R n. Thus Mf(x) a a.e. on R n and this follows Mf,ν a. Therefore, Mf,ν f,µ. Before proving that M is also of weak type (L (µ), L (ν)), we give the following lemma.

27 Free ebooks ==> 8 Chapter. Hardy-Littlewood Maximal Operator Lemma.3. Let f L (R n ) and λ > 0. If the sequence {Q k } of cubes is chosen from the Calderón-Zygmund decomposition on R n for f and λ > 0, then {x R n : M f(x) > 7 n λ} Q k, k where Q k = 2Q k. Then we have {x R n : M f(x) > 7 n λ} 2 n k Q k. Proof. Suppose that x / k Q k. Then there are two cases for any cube Q with the center x. If Q R n \ k Q k, then f(x) dx λ. Q Q If Q Q k for some k, then it is easy to check that Q k 3Q, and {Q k : Q k Q } 3Q. Hence, for F = R n \ k Q k, we have f(x) dx f(x) dx + Q k Q F λ Q + Q k Q λ Q + 2 n λ 3Q 7 n λ Q. Q k Q 2 n λ Q k Q k f(x) dx Thus we know that M f(x) 7 n λ for any x / k Q k, and it yields that {x R n : M f(x) > 7 n λ} Q k = 2n Q k. k k Let us return to the proof of weak type (L (µ), L (ν)). We need to prove that there exists a constant C such that for any λ > 0 and f L (µ) ϕ(x)dx = ν({x R n : Mf(x) > λ}) {x R n : Mf(x)>λ} C (.3.3) f(x) Mϕ(x)dx. λ R n

28 Free ebooks ==>.3 Marcinkiewicz interpolation theorem 9 We may assume that f L (R n ). In fact, if take f k = f χ B(0,k), then f k L (R n ), 0 f k (x) f k+ (x) for x R n and k =, 2,. Moreover, lim f k(x) = f(x). k By (..4), there exists c n > 0 such that Mf(x) c n M f(x) for all x R n. Applying the Calderón-Zygmund decomposition on R n for f and λ = λ c n7, we get a sequence {Q n k } of cubes satisfying By Lemma.3., we have that {x R n : Mf(x)>λ} λ < f(x) dx 2 n λ. Q k Q k ϕ(x)dx {x R n : M f(x)>7 n λ } ϕ(x)dx ϕ(x)dx k Q k k ( )( ) ϕ(x)dx Q k k Q λ f(y) dy k Q k = c n7 n ( 2 n ) f(y) λ k Q k Q k ϕ(x)dx dy Q k c n4 n f(y) M ϕ(y)dy λ k Q k = C f(y) Mϕ(y)dy. λ R n Q k ϕ(x)dx Thus, M is of weak type (L (µ), L (ν)), and the Fefferman-Stein inequality can be obtained by applying Theorem.3. with p 0 = and p =. The Marcinkiewicz interpolation theorem shows that if a sublinear operator T is the weak boundedness of at two ends of indexes, then T is also the strong boundedness for all index point between the two ends. The following conclusion shows that the weak boundedness of a sublinear operator T at an index point is equivalent to its strong boundedness in some sense for all index which are less than this point. Theorem.3.3 (Kolmogorov inequality) Suppose that T is a sublinear operator from L p (R n ) to measurable function spaces and p <. Then

29 Free ebooks ==> 20 Chapter. Hardy-Littlewood Maximal Operator (a) If T is of weak type (p, p), then for all 0 < r < p and all set E with finite measure, there exists a constant C > 0 such that T f(x) r dx C E r/p f r p. (.3.4) E (b) If there exists 0 < r < p and constant C > 0, such that (.3.4) holds for all set E with finite measure and f L p (R n ), then T is of weak type (p, p). Proof of (a). Since T is of weak type (p, p), for any λ > 0 Thus for any set E with E < {x R n : T f(x) > λ} C λ p f p p. {x E : T f(x) > λ} {x R n : T f(x) > λ} C λ p f p p. Hence for any 0 < r < p, we have T f(x) r dx E = r r r 0 λ r {x E : T f(x) > λ} dλ λ r min{ E, 0 C f p E /p 0 C E r/p f r p. C λ p f p p}dλ λ r E dλ + Cr C f p E /p λ r p f p pdλ Proof of (b). For any λ > 0, take E = {x R n : T f(x) > λ}, then E is a measurable set and E <. Otherwise, there is a sequence {E k } of measurable sets such that E k E and E k = k for k =, 2,. Thus for every k, we have λ r k = λ r E k T f(x) r dx C E k r/p f r p = Ck r/p f r p. E k However, it is not true. Thus by (.3.4), we obtain λ r E T f(x) r dx C E r/p f r p. {x R n : T f(x) >λ}

30 Free ebooks ==>.4 Weighted norm inequalities 2 From this it follows Hence T is of weak type (p, p). {x R n : T f(x) > λ} = E C λ p f p p..4 Weighted norm inequalities In this section we shall extend the conclusions of Theorem.. to a general measure spaces (R n, ω(x)dx), where ω is called a weight function, which is a nonnegative locally integrable function on R n. For the sake of convenience, in this section we shall use (..3) as the definition of the Hardy-Littlewood maximal operator M. Indeed, by (..4) the maximal operator M is essentially the same as the operator M. Now we give the definition of A p weight. Definition.4. (A p weights ( p < )) Let ω(x) 0 and ω(x) L loc (Rn ). We say that ω A p for < p < if there is a constant C > 0 such that sup Q ( ) ( p ω(x) dx ω(x) dx) p C, (.4.) Q Q Q Q where and below, /p + /p =. We say that ω A if there is a constant C > 0 such that Mω(x) Cω(x). (.4.2) The smallest constant appearing in (.4.) or (.4.2) is called the A p constant of ω. Remark.4. Clearly, ω A if and only if there is a constant C > 0 such that for any cube Q ω(x)dx C inf ω(x), (.4.3) Q x Q Q where and below, inf is the essential infimum. Moreover, it is easy to see that for p < and any ω A p, its A p constant C. In fact, for any

31 Free ebooks ==> 22 Chapter. Hardy-Littlewood Maximal Operator cube Q, we have that = ω(x) /p ω(x) /p dx Q Q {( )( ) p } /p ω(x) dx ω(x) p dx Q Q Q Q C /p. Let us now give some elementary properties of A p weights. Proposition.4. (Properties (I) of A p weights) (i) A p A q, if p < q <. (ii) For < p <, ω A p if and only if ω(x) p A p. (iii) If ω 0, ω A, then ω 0 ω p A p for < p <. (iv) If ω A p ( p < ), then for any 0 < ε <, ω ε A p. (v) If ω A p ( p < ), then for any f L loc (Rn ), ( p f(x) dx) ω(q) C f(x) p ω(x)dx. (.4.4) Q Q (vi) If ω A p ( p < ), then for any δ > there exists a constant C(n, p, δ) such that for any cube Q, ω(δq) C(n, p, δ)ω(q). In particular, the case taking δ = 2 shows that A p weights satisfy double condition. (vii) If ω A p ( p < ), then for any 0 < α < there exists 0 < β < such that for any measurable subset E Q, E α Q and ω(e) βω(q). Proof. (i) For p >, this is a direct consequence of Definition.4. and Hölder s inequality. If p =, then by (.4.3) ( q ω(x) dx) q sup ω(x) Q x Q Q = ( inf C Q x Q ω(x)) ( Q Q ω(x)dx).

32 Free ebooks ==>.4 Weighted norm inequalities 23 On the other hand, since x α A p if and only if n < α < n(p ) (see Proposition.4.4 in this chapter), we have A p A q. (ii) Since (p )(p ) =, for any cube Q we have ( )( ω(x) p dx Q Q Q = Q {( ω(x) dx Q Q C p. ) p [ω(x) p ] p dx )( Q Q ) p } p ω(x) p dx (iii) For any Q, by ω i A (i =, 2) and (.4.3) we have ω i (x) sup ω i (x) ( inf ω i(x) ) ( ) ωi (Q) C. (.4.5) x Q x Q Q From (.4.5), it immediately follows that ( )( p ω 0 (x)ω (x) p dx ω 0 (x) p ω (x) dx) C. Q Q Q Q (iv) For p =, by Hölder s inequality and (.4.3) ( ε ω(x) ε dx ω(x)dx) ( C inf Q Q Q ω(x)) ε = C ε inf Q x Q x Q ω(x)ε. Similarly, we can get the conclusion (iv) for < p < by Definition.4. and Hölder s inequality. (v) For p =, by (.4.3), we have that f(x) dx ω(q) = Q Q f(x) dx Q C Q C f(x) ω(x)dx. Q ω(x)dx Q Q f(x) dx inf x Q ω(x)

33 Free ebooks ==> 24 Chapter. Hardy-Littlewood Maximal Operator When < p <, by Hölder s inequality Q = f(x) dx Q f(x) ω(x) /p ω(x) /p dx Q Q ( ) /p ( f(x) p ω(x)dx Q Q Q Q ( ) /p ( C /p f(x) p ω(x)dx Q Q Q ) (p )/p ω(x) p dx Q ω(x)dx) /p. Thus, (.4.4) follows from this. (vi) If we replace Q by δq and f(x) by χ Q (x) in (.4.4), respectively, then this is just the conclusion of (vi). (vii) Let S = Q \ E and f(x) = χ S (x) in (.4.3). Then Thus ( α) p ω(q) ( ) S p ω(q) C ω(x)dx. Q S Notice that C (see Remark.4.), ( E ) p ) ω(q) C( ω(x)dx ω(x)dx. Q Q E ω(e) C ( α)p ω(q). C So, we get the conclusion of (vii) with β = (C ( α) p )/C. The following theorem gives a very important and useful property of A p weights. Theorem.4. (Reverse Hölder inequality) Let ω A p, p <. Then there exist a constant C and ε > 0 depending only on p and the A p constant of ω, such that for any cube Q ( ) /(+ε) ω(x) +ε dx C ω(x)dx. (.4.6) Q Q Q Q

34 Free ebooks ==>.4 Weighted norm inequalities 25 Proof. Fix a cube Q, by Remark.2. we apply the Calderón-Zygmund decomposition with respect to Q for ω and the increasing sequence {ω(q)/ Q = λ 0 < λ < < λ k < } For each λ k we may get a sequence {Q k,i } of disjoint cubes such that ω(x) λ k for x / Λ k = i Q k,i, and λ k < ω(x)dx 2 n λ k. Q k,i Q k,i Since λ k+ > λ k, for every Q k+,j it is either to equal some Q k,i or a subcube of Q k,i for some i. Thus Q k+,j < ω(x)dx λ k+ Q k+,j = Q k,i ω(x)dx λ k+ Q k,i Q k+,j Q k,i ω(x)dx λ k+ Q k,i Q k,i From this, it follows λ k 2 n Q k,i. λ k+ Q k,i Λ k+ 2 n λ k λ k+ Q k,i. For fixed α <, we choose {λ k } such that 2 n λ k /λ k+ = α. Equivalently, λ k = (2 n /α) k λ 0. Thus Q k,i Λ k+ α Q k,i. By Proposition.4. (vii), there exists 0 < β < such that ω(q k,i Λ k+ ) βω(q k,i ). Summing up with respect to the index i, we obtain ω(λ k+ ) βω(λ k ), and it yields ω(λ k+ ) β k ω(λ 0 ).

35 Free ebooks ==> 26 Chapter. Hardy-Littlewood Maximal Operator Similarly, we also have Λ k+ α Λ k and Λ k+ α k Λ 0. Hence Λ k = lim Λ k = 0. k Thus Q k=0 ω(x) +ε dx = ω(x) +ε dx + ω(x) +ε dx Q\Λ 0 Λ k \Λ k+ λ ε 0ω(Q \ Λ 0 ) + k=0 λ ε k+ ω(λ k \ Λ k+ ) k=0 ( ) λ ε 0 ω(q \ Λ 0 ) + (2 n /α) (k+)ε β k ω(λ 0 ) k=0 ( λ ε 0 ω(q \ Λ 0 ) + (2 n /α) ε k=0 ) [(2 n /α) ε β] k ω(λ 0 ). Let ε > 0 be small enough such that (2 n /α) ε β <. Then the series converges. Therefore we have ω(x) +ε dx Cλ ε 0( ω(q \ Λ0 ) + ω(λ 0 ) ) ) Q ( ε ( ) = C ω(x)dx) ω(x)dx Q. Q Q Q Q Thus we get (.4.6) and the proof of Theorem.4. is finished. As a corollary of Theorem.4., we get some further properties of A p weights. Proposition.4.2 (Properties (II) of A p weights) (viii) If ω A p ( < p < ), then there is an ε > 0 such that p ε > and ω(x) A p ε. (ix) A p = q<p A q, if < p <. (x) If ω A p ( p < ), then there is an ε > 0 such that ω(x) +ε A p. (xi) If ω A p ( p < ),then there is δ > 0 and C > 0 such that for any cube Q and a measurable subset E Q ( ) ω(e) E δ ω(q) C. (.4.7) Q

36 Free ebooks ==>.4 Weighted norm inequalities 27 Proof. (viii) Let ω A p, then ω p A p by the property (ii). Applying the Reverse Hölder inequality (Theorem.4.) for ω p, then ( ) (p )/(+θ) ( p ω(x) ( p )(+θ) dx C p ω(x) dx) p, Q Q Q Q where θ > 0. Now multiplying the factor ω(x)dx Q on two sides of the above inequality, we have Q ( )( (p )/(+θ) ω(x)dx ω(x) dx) ( p )(+θ) C. Q Q Q Q Denote ( p )( + θ) = q, then < q < p and ω A q. Thus we get the property (viii) with ε = p q. (ix) The property (ix) is a direct corollary of the properties (i) and (viii). Indeed, we know that A p q<p A q by the property (i). On the other hand, by (viii), A p q<p A q. (x) Suppose that ω A, then by (.4.6) ( C +ε ω(x) +ε dx ω(x)dx) C ω(x) +ε a.e. x R n. Q Q Q Q Hence ω(x) +ε A. If ω A p, p >, then ω(x) p A p by (ii). Using Theorem.3.3, it is clear that there exists an ε > 0 such that and Q Q ω(x) +ε dx C ( Q Q ) +ε ω(x)dx ( +ε ω(x) ( p )(+ε) dx C 2 ω(x) dx) p. Q Q Q Q hold at the same time. Thus, ( ) ( ) p ω(x) +ε dx [ω(x) +ε ] p dx Q Q Q Q {( ) ( } p +ε C ω(x)dx ω(x) dx) p C. Q Q Q Q

37 Free ebooks ==> 28 Chapter. Hardy-Littlewood Maximal Operator This shows that ω +ε A p. (xi) Since ω(x) A p ( p < ), using Hölder s inequality for + ε and ( + ε)/ε, where ε is fixed by (.4.6), we have ( ) /(+ε) ω(x)dx ω(x) +ε dx E ε/(+ε) E E ( ) /(+ε) = ω(x) +ε dx Q /(+ε) E ε/(+ε) Q E C ω(x)dx Q /(+ε) E ε/(+ε) Q Q ( ) E ε/(+ε) = Cω(Q). Q Thus we have showed that ω satisfies (.4.7) if taking δ = ε/( + ε). Remark.4.2 Let ω be a nonnegative locally integrable function on R n. We say that ω A if ω satisfies (.4.7). The property (xi) shows that p< A p A. However, it can be proved that the above containing relationship may be reversed. So, we have indeed A = p< A p. In this section, we shall see that A p weights give a characterization of weighted weak L p and strong L p boundedness for the Hardy-Littlewood maximal operator M. Theorem.4.2 (Characterization of the weighted weak type (p, p)) Suppose that p <. Then the Hardy-Littlewood maximal operator M is of weak type (L p (ωdx), L p (ωdx)) if and only if ω A p. That is, there exists a constant C > 0 such that for any λ > 0 and f(x) L p (ωdx) ( p < ) ω(x)dx C λ R p f(x) p ω(x)dx (.4.8) n if and only if ω A p. {x R n : Mf(x)>λ} Proof. Let us first prove that ω A p is the necessary condition of (.4.8). For p =, let Q be any cube and Q Q. If denote f = χ Q, then for any 0 < λ < Q / Q, we have Q {x R n : Mf(x) > λ}. By (.4.8) λ ω(x)dx λ ω(x)dx C ω(x)dx. Q {x R n : Mf(x)>λ} Q

38 Free ebooks ==>.4 Weighted norm inequalities 29 By the arbitrariness of λ < Q / Q, we get ω(x)dx C ω(x)dx. Q Q Q Q Applying Lebesgue differentiation theorem (Theorem..2), we have that ω(y)dy Cω(x) a.e. x Q. Q By the arbitrariness of cube Q again, we have Q Mω(x) Cω(x) a.e. x R n. Thus ω A. For p >, let Q be any cube. If we take f = ω p χ Q, then for any 0 < λ < ω p (Q)/ Q, Q {x R n : Mf(x) > λ}. By (.4.8) λ p Q Q ω(x)dx λp Q Q C Q = C Q Hence by arbitrariness of λ, we get ( ω(x)dx C Q Q {x R n : Mf(x)>λ} ω(x)dx [ω(x) p χ Q (x)] p ω(x)dx R n ω(x) p dx. Q Q ω(x) p dx) p. So ω A p. Below we will prove that ω A p is also a sufficient condition of (.4.8). When p =, (.4.8) is just a direct corollary of (.3.3). Now let us consider the case p >. By Hölder s inequality and the condition ω A p, we have ( ) ( ) p ω(x)dx f(x) dx Q Q ( = Q ( Q C Q Q Q Q Q Q ) ( ω(x)dx Q ) ( ω(x)dx Q f(x) p ω(x)dx. Q Q ) p f(x) ω(x) /p ω(x) /p dx ) ( ) p f(x) p ω(x)dx ω(x) p dx Q Q

39 Free ebooks ==> 30 Chapter. Hardy-Littlewood Maximal Operator Thus, this shows that if f L p (ωdx) then f L loc (Rn ), and Q ( ) p ( ) ω(x)dx C f(x) dx f(x) p ω(x)dx. (.4.9) Q Q Q So without loss of generality, we may assume that f 0 and f L (R n ). Applying the Calderón-Zygmund decomposition (Theorem.2.) for f at height 7 n λ, we get a cube sequence {Q k } such that By Lemma.3., we have that 7 n λ < f(x)dx for all Q k. (.4.0) Q k Q k {x R n : Mf(x) > λ} k Q k, where Q k = 2Q k. Thus by the property (vi) of A p weights and (.4.9) and (.4.0), we conclude that {x R n : Mf(x)>λ} ω(x)dx ω(x)dx C2 np ω(x)dx k 2Q k k Q k C2 ( ) p ( ) np f(x) dx f(x) p ω(x)dx Q k k Q k Q k C 4np λ p f(x) p ω(x)dx. R n This shows that the Hardy-Littlewood maximal operator M is of weak type (L p (ωdx), L p (ωdx)). Hence we finish the proof of Theorem.4.2. Theorem.4.3 (Characterization of the weighted strong type (p, p)) Suppose that < p <. Then the Hardy-Littlewood maximal operator M is of strong type (L p (ωdx), L p (ωdx)) if and only if ω A p. Proof. The necessity is a corollary of Theorem.4.2. In fact, since M is of strong type (L p (ωdx), L p (ωdx)), and is also weak type (L p (ωdx), L p (ωdx)). Then by Theorem.4.2, ω A p.

40 Free ebooks ==>.4 Weighted norm inequalities 3 On the other hand, if ω A p for < p <, then by the property (viii) there exists < q < p such that ω A q. By Theorem.4.2, M is of weak type (L p (ωdx), L p (ωdx)). That is, {x R n : Mf(x)>λ} ω(x)dx C λ q R n f(x) q ω(x)dx. (.4.) Notice that if ω A p, then for any measurable set E R n ω(e) = 0 if and only if E = 0. So, L (ωdx) = L in the sense of equality of norms. Hence by the L (R n )-boundedness of M (Theorem..), we have Mf,ω f,ω. Using the Marcinkiewicz interpolation theorem (Theorem.3.) between this and (.4.) we obtain Mf(x) p ω(x)dx C f(x) p ω(x)dx. R n R n In other words, M is of strong type (L p (ωdx), L p (ωdx)). In this section, we shall discuss further some important properties of A weights, such as the construction of A weights, the relationship between A weights and A p weights and its some application, etc. Theorem.4.4 (Constructive characterization of A weights) (a) If f(x) L loc (Rn ) and Mf(x) <, a.e. x R n, where M is the Hardy-Littlewood maximal operator. Then for any 0 < ε <, ω(x) = ( Mf(x)) ε A and whose A constant depends only on ε. (b) If ω(x) A, then there exists f(x) L loc (Rn ), 0 < ε < and a function b(x) such that (i) 0 < C b(x) C 2, a.e. x R n ; ( ε (ii) Mf(x)) <, a.e. x R n ; ( ε. (iii) ω(x) = b(x) Mf(x))

41 Free ebooks ==> 32 Chapter. Hardy-Littlewood Maximal Operator Proof of (a). By Remark.4., we need to show that for any 0 < ε <, there exists a constant C such that for any cube Q in R n ( ) εdy ( ε Mf(y) C Mf(x)) for a.e. x Q. Q Q Fix Q and 0 ε <, we denote f = fχ 2Q + fχ R n \2Q := f + f 2. Then ( ) ε ( ) ε ( ε. Mf(x) Mf (x) + Mf 2 (x)) By the weak (, ) boundedness of M and Kolmogorov s inequality (Theorem.3.3) Mf (x) ε dx C Q Q Q Q ε f ε ( ) ε C f (y) dy Q for any x Q. Now we give the estimate of 2Q C2 nε( Mf(x) ) ε ( Mf 2 (x)) ε. Clearly, if y Q and a cube Q y such that Q (R n \ 2Q), then 4Q Q. Hence if x Q Q Q f 2 (y) dy 4n 4Q f 2 (y) dy 4 n Mf(x). 4Q This shows that Mf 2 (y) 4 n Mf(x) for any y Q, and so is ( εdy Mf 2 (y)) ( 4 nε ε. Mf(x)) Q Q Proof of (b). Since ω(x) A, then by the Reverse Hölder inequality (Theorem.4.) there exists η > 0 such that for any cube Q This implies that ( ) /(+η) ω(x) +η dx ω(x)dx Cω(x). Q Q Q Q M(ω +η )(x) /(+η) Cω(x) for a.e. x R n. On the other hand, by the Lebesgue differentiation theorem (Theorem..2) ω(x) +η M(ω +η )(x).

42 Free ebooks ==>.4 Weighted norm inequalities 33 ( ε, Therefore, if let ε = /( + η), f(x) = ω(x) +η and b(x) = ω(x)/ Mf(x)) then we obtain the conclusions of (b). Thus we finish the proof of Theorem.4.4. Below we shall turn to discuss the relationship between A weights and A p weights. By Remark.4., if ω A, then we have the following equivalent form of (.4.2) Q Q ω(x)dx sup(ω(x) ) C, x Q where and below, sup is the essential supremun. On the other hand, it is easy to see that when p ( p ω(x) dx) p ω,q = sup(ω(x) ). Q x Q Q Therefore, the A weights can be seen as the limit of A p weights when p. By the property (iii) of A p weights, we have known that if ω, ω 2 A then ω ω p 2 A p. A very deep result is that the converse of above conclusion is also true. Theorem.4.5 (Jones decomposition of A p weights) Let ω be nonnegative locally integrable function. Then for < p <, ω A p if and only if there are ω, ω 2 A such that ω(x) = ω (x) ω 2 (x) p. Proof. we only consider the necessity. Since ω A p, by the property (ii) ω(x) p A p. First let < p 2, then s = p >. Take a nonnegative function u 0 (x) L p (R n, ωdx), let us construct a function sequence {u j (x)} by the Hardy-Littlewood maximal operator M as follows. u j+ (x) = [M(u s j )(x)]/s + ω(x) M(u j ω)(x), for j = 0,,. Then we have the following two conclusions. (a) There exists a constant C such that u j+ p, ωdx C u j p, ωdx, for j = 0,,. (b) For the constant C appearing in (a), take δ > C and denote U(x) = δ j u j (x), then Uω A and U s A. j=0

43 Free ebooks ==> 34 Chapter. Hardy-Littlewood Maximal Operator In fact, by Theorem.4.3, we have that u j+ p p, ωdx ( ) C [M(u s j /s )(x)]p ω(x)dx + [ω(x) M(u j ω)(x)] p ω(x)dx R n R ( n ) = C [M(u s j)(x)] p ω(x)dx + [M(u j ω)(x)] p ω(x) p dx R n R ( n ) C u p j (x)ω(x)dx + [u j (x)ω(x)] p ω(x) p dx R n R n C u p j (x)ω(x)dx. R n So the conclusion (a) holds and it is easy to check that the constant C is independent of u j for all j 0. Now let us verify (b). By the definition of u j+, we have that M(Uω)(x) = δ δ j M(u j ω)(x) j=0 δ j u j+ (x)ω(x) j=0 δ j u j (x)ω(x) j= δ(u(x)ω(x)). Hence Uω A. On the other hand, we have that [M(U s )(x)] /s δ j [M(u s j )(x)]/s δ j u j+ (x) δu(x). j=0 So U s A too. Thus, if denote ω = Uω and ω 2 = U s, then ω, ω 2 A and ω = (Uω) (U s ) /s = (Uω) (U s ) p = ω ω p 2. Hence the necessity of theorem holds for < p 2. If 2 < p < and ω A p, then ω p A p and < p 2. Thus by the above proof process, there are ν, ν 2 A such that ω p = ν ν p 2. Equivalently, ω = ν 2 ν p. Thus we complete the proof of Theorem.4.5. j=0 Below we give a sharp result about A weights.

44 Free ebooks ==>.5 Notes and references 35 Proposition.4.3 Let x R n. Then x α A if and only if n < α 0. Proof. If x α A. Then x α L loc (Rn ), hence the condition α > n is necessary. On the other hand, if α > 0 and x α A, we take < p + α/n, then x α A p by the property (i) of A p weights. However, x α( p ) / L loc (Rn ) for the choice of p. Conversely, suppose that n < α 0, we will show that x α A. Indeed, for any fixed cube Q, we denote Q 0 is the translation of Q with the center at origin. The case (i): 2Q 0 Q. In this case, we have 4Q 0 Q, and x α dx x α dx C Q α/n C inf Q Q Q 4Q 0 x Q x α. Thus The case (ii): 2Q 0 Q =. Notice that if x, y Q, then x x y + y C Q /n + y (C + ) y, x α dx C inf Q Q x Q x α. It is easy to see that the constants C in the cases both (i) and (ii) depend only on n. Therefore, we have proved that x α A. By the property (iii) of A p weights, we get immediately the following sharp result for A p weights. Proposition.4.4 Let x R n. Then for < p <, x α A p if and only if n < α < n(p )..5 Notes and references Theorem.. was first proved by Hardy and Littlewood [HaL] for n = and then by Wiener [Wi] for n >. The idea of proof given here was taken from Stein [St4] which is one of the most important monograph in harmonic analysis. The Calderón-Zygmund decomposition (Theorems.2. and.2.3) first appeared in Calderón and Zygmund [CaZ] which is regarded as the foundation of several variables singular integrals theory.

45 Free ebooks ==> 36 Chapter. Hardy-Littlewood Maximal Operator Theorem.2.2 is due to Stein [St2]. The idea of proof given here was taken from Garcia-Cuerva and Rubia de Francia [GaR], which is a nice monograph on the topic of weighted norm inequalities. The Marcinkiewicz interpolation theorem was first announced by Marcinkiewicz [Ma]. Its complete proof can be found in Zygmund [Zy]. Theorem.3. in this chapter is general form of the Marcinkiewicz interpolation theorem, whose proof was taken from Duoandikoetxea [Du2]. Theorem.3.2 is due to Fefferman and Stein [FeS]. The Reverse Hölder inequality (Theorem.4.) first was proved by Coifman and Fefferman [CoiF]. Theorem.4.2 and Theorem.4.3 were first proved by Muckenhoupt [Mu] for n =. So, the A p weights are also called as Muckenhoupt A p weights. Precisely, a function ω is called to satisfy Muckenhoupt A p condition if ω A p. The idea of proving Theorems.4.2 and.4.3 given here was taken from Journé [Jou]. Theorem.4.4 is due to Coifman and Rochberg [CoiR]. Theorem.4.5 was first proved by Jones [Jon]. The proof of Theorems.4.5 given here was taken from Coifman, Jones and Rubia de Francia [CoiJR].

46 Free ebooks ==> Chapter 2 Singular Integral Operators Calderón-Zygmund singular integral operator is a direct generalization of the Hilbert transform and the Riesz transform. The former is originated from researches of boundary value of conjugate harmonic functions on the upper half-plane, and the latter is tightly associated to the regularity of solution of second order elliptic equation. Now we will introduce their backgrounds briefly. Suppose f L p (R) ( p < ). Consider the Cauchy integral on R: F (z) = f(t) 2πi R t z dt, where z = x + iy, y > 0. It is clear to see that F (z) is analytic on R 2 +. Note that F (z) = y 2π R (x t) 2 + y 2 f(t)dt + i x t 2π R (x t) 2 + y 2 f(t)dt := [ ] (P y f)(x) + i(q y f)(x), 2 where is called the Poisson kernel, and P y (t) = π Q y (t) = π 37 y t 2 + y 2 t t 2 + y 2

47 Free ebooks ==> 38 Chapter 2. Singular Integral Operators is called the conjugate Poisson kernel. Correspondingly, P y f is called the Poisson integral of f, and Q y f is called the conjugate Poisson integral of f. From the property of boundary values of harmonic functions, it follows that P y f f, a.e. as y 0 and Q y f Hf, a.e. as y 0. Hf is called the Hilbert transform of f. Hf(x) = p.v. π It can be proved that if f L 2 (R), then R f(t) dt. (2.0.) x t Ĥf(ξ) = isgnξ ˆf(ξ). (2.0.2) Let K(x) = p.v., then Hf = (K f). x Now suppose that f L 2 (R n ) and give the Poisson equation u = f, where n 2 = x 2 j= j is the Laplacian operator on R n. By taking the Fourier transform on both sides of the equation we have that ˆf(ξ) = 4π 2 ξ 2 û(ξ). This is equivalent to Thus, for j, k n, û(ξ) = 4π 2 ξ 2 ˆf(ξ). If we define the operator then it implies that ( 2 u x j x k ) (ξ) = 4π 2 ξ j ξ k û(ξ) = ξ jξ k ξ 2 ˆf(ξ). R j f(ξ) = i ξ j ξ ˆf(ξ), j =, 2,, n, (2.0.3) ( 2 u x j x k ) (ξ) = R j R k f(ξ).