Measure of Non-Compactness For Integral Operators in Weighted Lebesgue Spaces

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M EASURE OF N ON - COMPACTNESS FOR I NTEGRAL O PERATORS IN W EIGHTED L EBESGUE S PACES

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M EASURE OF N ON - COMPACTNESS FOR I NTEGRAL O PERATORS IN W EIGHTED L EBESGUE S PACES A LEXANDER M ESKHI

Nova Science Publishers, Inc. New York

c 2009 by Nova Science Publishers, Inc.

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NOTICE TO THE READER The Publisher has taken reasonable care in the preparation of this book, but makes no expressed or implied warranty of any kind and assumes no responsibility for any errors or omissions. No liability is assumed for incidental or consequential damages in connection with or arising out of information contained in this book. The Publisher shall not be liable for any special, consequential, or exemplary damages resulting, in whole or in part, from the readers’ use of, or reliance upon, this material. Independent verification should be sought for any data, advice or recommendations contained in this book. In addition, no responsibility is assumed by the publisher for any injury and/or damage to persons or property arising from any methods, products, instructions, ideas or otherwise contained in this publication. This publication is designed to provide accurate and authoritative information with regard to the subject matter cover herein. It is sold with the clear understanding that the Publisher is not engaged in rendering legal or any other professional services. If legal, medical or any other expert assistance is required, the services of a competent person should be sought. FROM A DECLARATION OF PARTICIPANTS JOINTLY ADOPTED BY A COMMITTEE OF THE AMERICAN BAR ASSOCIATION AND A COMMITTEE OF PUBLISHERS. Library of Congress Cataloging-in-Publication Data Available upon request.

ISBN 978-161728-536-3 (E-Book)

Published by Nova Science Publishers, Inc. ✜ New York

Contents Preface

vii

Basic Notation

xi

1

2

3

4

Basic Ingredients 1.1. Homogeneous Groups . . . . . . . . . . . . . 1.2. Measure of Non–compactness . . . . . . . . . 1.3. Hardy–type Transforms . . . . . . . . . . . . . 1.4. L p(x) Spaces . . . . . . . . . . . . . . . . . . . 1.5. Schatten–von Neumann Ideals . . . . . . . . . 1.6. Singular Integrals in Weighted Lebesgue Spaces 1.7. Notes and Comments on Chapter 1 . . . . . . .

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1 1 3 11 12 22 23 25

Maximal Operators 2.1. Maximal Functions on Euclidean Spaces . . 2.2. One–sided Maximal Functions . . . . . . . 2.3. Maximal Operator on Homogeneous Groups 2.4. Notes and Comments on Chapter 2 . . . . .

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27 27 32 34 35

Kernel Operators on Cones 3.1. Boundedness . . . . . . . . . . . . . . . . . . . 3.2. Compactness . . . . . . . . . . . . . . . . . . . 3.3. Schatten–von Neumann norm Estimates . . . . . 3.4. Measure of Non–compactness . . . . . . . . . . 3.5. Convolution–type Operators with Radial Kernels 3.6. Notes and Comments on Chapter 3 . . . . . . . .

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37 39 43 45 47 49 50

Potential and Identity Operators 4.1. Riesz Potentials . . . . . . . . 4.2. Truncated Potentials . . . . . 4.3. One–sided Potentials . . . . . 4.4. Poisson Integrals . . . . . . . 4.5. Sobolev Embeddings . . . . . 4.6. Identity Operator . . . . . . . 4.7. Partial Sums of Fourier Series

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51 51 55 58 60 63 65 68

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vi

Contents 4.8. Notes and Comments on Chapter 4 . . . . . . . . . . . . . . . . . . . . . .

5 Generalized One-sided Potentials in L p(x) Spaces 5.1. Boundedness . . . . . . . . . . . . . . . . . 5.2. Compactness . . . . . . . . . . . . . . . . . 5.3. Measure of Non–compactness . . . . . . . . 5.4. Notes and Comments on Chapter 5 . . . . . . 6

Singular Integrals 6.1. Hilbert Transforms . . . . . . . . . . . . 6.2. Cauchy Singular Integrals . . . . . . . . . 6.3. Riesz Transforms . . . . . . . . . . . . . 6.4. Calder´on–Zygmund Operators . . . . . . 6.5. Hilbert Transforms in L p(x) Spaces . . . . 6.6. Cauchy Singular Integrals in L p(x) Spaces 6.7. Notes and Comments on Chapter 6 . . . .

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69 71 71 77 80 82

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83 . 83 . 86 . 88 . 90 . 91 . 98 . 102

References

103

Index

119

Preface One of the important problems of modern harmonic analysis is to establish the boundedness/compactness of integral operators in weighted function spaces. When a given integral operator is bounded but non-compact, it is natural and useful for applications to have twosided sharp estimates of the measure of non-compactness (essential norm) for this operator. The book is devoted to the boundedness/compactness and weighted estimates of the essential norm for maximal functions, fractional integrals, singular and identity operators, generally speaking, in weighted variable exponent Lebesgue spaces. Such operators naturally arise in harmonic analysis, boundary value problems for PDE, spectral theory of differential operators, continuum and quantum mechanics, stochastic processes etc. One of the main characteristic features of the monograph is that the problems are studied in the two-weighted setting and cover the case of non-linear maps, such as, HardyLittlewood and fractional maximal functions. Before, these problems were investigated only for a restricted class of kernel operators consisting only of Hardy-type and RiemannLiouville transforms (see, e.g., the monographs [39], [149], [49], [40] and references therein). The book may be considered as a systematic and detailed analysis of a class of specific integral operators from the boundedness/compactness or non-compactness point of view. There is a wide range of problems of mathematical physics whose solutions are closely connected to the subject matter of the book. The main subjects of the monograph (maximal functions, fractional integrals, Hilbert transforms, Riesz transforms, Calder´on–Zygmund singular integrals) are important tools for solving a variety problems in several areas of mathematics and its applications. The problems related to estimates of the measure of the non-compactness for differential and integral operators acting between Banach spaces are closely connected with eigenvalue estimates and other spectral properties for these operators (see, for example, monographs [39], [64], [17], [40]). One of the most challenging problems of the spectral theory of differential operators is the derivation of eigenvalue and singular value estimates for integral operators in terms of their kernels. The works [13], [197], [135] mark an important stage in the development of this theory (see also [39], [64], [149], [49]). Until recently the list of non–trivial cases in which sharp two–sided estimates are available was rather short. Here we present two–sided estimates of the singular numbers for some classes of kernel operators. A weight theory for a wide class of integral transforms with positive kernels including fractional integrals was developed in the monographs [112], [76], [49]. It should be emphasized that the interest in fractional calculus has been stimulated by applications in different

viii

Alexander Meskhi

fields of science, including stochastic analysis of long memory processes, numerical analysis, physics, chemistry, engineering, biology, economics and finance. For the theory of fractional integration and differentiation we refer to the well–known monograph [209]. The book is divided into six chapters and each chapter into sections. The book is started with some background. We have restricted ourself to the concepts of homogeneous groups, measure of non–compactness, Schatten–von Neumann ideals, variable exponent Lebesgue spaces. Chapter 2 deals with the measure of non–compactness for maximal operators defined, generally speaking, on homogeneous groups. Chapter 3 is focused on the boundedness/compactness, Schatten–von Neumann ideal norm estimates and measure of non–compactness for integral operators defined on cones of homogeneous group, while in Chapter 4 we discuss two-weight estimates of the measure of non–compactness for fractional integral and identity operators defined on Euclidean spaces and homogeneous groups. In Chapter 5 we establish boundedness/compactness criteria and two–sided estimates of the measure of non–compactness for the Riemann-Liouville transform in Lebesgue spaces with non–standard growth. In Chapter 6 we present some results regarding one and two weighted estimates of the essential norm for singular integrals (Hilbert transforms, Cauchy integrals, Riesz transforms, Calder´on–Zygmund singular integrals), generally speaking in L p(x) spaces. One of the important examples of homogeneous groups is the Heisenberg group. Lately the theory of function spaces on the nilpotent groups has attracted considerable attention among researchers (see, for example, the monograph by G. Folland and E. Stein [70] ). This attention has mainly been triggered by questions related to solvability of problems for differential equations with variable coefficient occurring on manifolds. For example, the Heisenberg group and function spaces on it have turned out to be closely connected with boundary value problems for pseudo-convex domains in Cn . The last two chapters of the monograph is dedicated to the investigation of the compactness and non–compactness problems for one–sided potentials and singular integrals, generally speaking, in weighted Lebesgue spaces with variable exponent. During the last decade a considerable interest of researchers was attracted to the study of various mathematical problems in the so called spaces with non–standard growth: variable exponent Lebesgue and Sobolev spaces L p(·) ,W n,p(·) . Such spaces naturally arise when one deals with functionals of the form Z

Ω

|∇ f (x)| p(x) dx.

Such a functional appears, for instance, in the study of differential equations of the type div (|∇u(x)| p(x)−2 ∇u) = |u|σ(x)−1 u(x) + f (x). In this case one deals with the Dirichlet integral of the form Z

Ω

(|∇ f (x)| p(x) + |u(x)|σ(x) )dx.

Such mathematical problems and spaces with variable exponent arise in applications to mechanics of the continuum medium. In some problems of mechanics there arise variational problems with Lagrangians more complicated than is usually assumed in variational

Preface

ix

calculus, for example, of the form |ξ|γ(x) when the character of non-linearity varies from point to point (Lagrangians of the plasticity theory, Langrangians of mechanics of the so called rheological fluids and others). Investigation of variational problems with variable exponent started from the papers by V. Zhikov [238], [239] related to the so called Lavrentiev phenomenon. M. Ruˇziˇcka [202] studied the problems in the so called rheological and electrorheological fluids, which lead to the spaces with variable exponent. The variable exponent Lebesgue spaces first appeared in 1931 in the paper by W. Orlicz [189], where the author established some properties of L p(x) spaces on the real line. Further development of these spaces was connected with the theory of modular spaces. The first systematic study of modular spaces is due to H. Nakano [176]. The basis of the variable exponent Lebesgue and Sobolev spaces were developed by J. Musielak (see [173], [174]), H. Hudzik [94], I. I. Sharapudinov [220], O. Kov´acˇ ik and I. R´akosn´ık [138], S. Samko [204], [205], etc (see also the surveys [208], [111] and references therein). The monograph covers not only recent results of the research carried out by the author and his collaborators regarding the main topics of the monograph, but also contains unpublished material. The monograph includes overview of results of other mathematicians working on the topics of the book. The bibliography contains 242 titles. A few words about organization of the book are necessary. The enumeration of theorems, lemmas, formulas etc. follows the natural three-digit system. There are three categories for numbering: theorems, lemmas, propositions and remarks, and the same for formulas. The book is aimed at a rather wide audience, ranging from researchers in functional and harmonic analysis to experts in applied mathematics and graduate students. I express my gratitude to Professor Vakhtang Kokilashvili for his encouragement to prepare this monograph, drawing my attention to the problems studied in Chapter 6 and his remarks and suggestions. The investigation of the measure of non-compactness of maximal operators started in 2001, when I visited the Centre for Mathematical Analysis and its Applications, University of Sussex. I expresses my deep gratitude to Professor David Edmunds and the Centre for support and warm hospitality. Some aspects of Chapters 2 and 6 were discussed with Professor Alberto Fiorenza. I am thankful to him for invitation at the University of Naples. Some results of the monograph were obtained during my stay at the Abdus Salam School of Mathematical Sciences, GC University, Lahore. I am grateful to Professor A. D. Raza Choudary for giving me an opportunity to work with PhD students.

Acknowledgement The monograph was partially supported by INTAS grant No.051000008-8157; No. 061000017-8792, and Georgian National Foundation Grant No. GNSF/ST06/3-010.

Basic Notation Rn : n-dimensional Euclidean space; R = R1 ; R+ = [0, ∞); C: complex plane; B(x, r): open ball with center x and radius r; ¯ r): closed ball with center x and radius r; B(x, I(a, r) = (a − r, a + r); Z: set of all integers; Z+ : set of all non-negative integers; Z− : set of all non-positive integers; N: set of all natural numbers; 2πn/2 Bn : volume of the unit ball in Rn , i.e., Bn = nΓ(n/2) ; n/2

2π Sn−1 : area of the unit sphere in Rn , i.e., Sn−1 = Γ(n/2) ; R w(E) = E w(x)dx, where w is a weight function; |E|: Lebesgue measure of E; χE : characteristic function of a set E; p p′ = p−1 , where p is a constant with 1 < p < ∞;

r′ (x) =

r(x) r(x)−1 ,

where r is a real-valued function;

lim an = lim inf an , lim an = lim sup an for a sequence of real number {an };

n→∞

n→∞

n→∞

n→∞

a ≈ b: there are positive constants c1 and c2 such that c1 a ≤ b ≤ c2 a; Cm : class of functions whose partial derivatives up to and including those of order m exist and are continuous; C0m : subset of Cm of functions with compact support; C0 : class of continuous functions with compact support; : end of the proof.

Chapter 1

Basic Ingredients In this chapter definitions and some auxiliary results are given regarding the main objects of the monograph: homogeneous groups, measure of non–compactness of sublinear and linear operators, Schatten–von Heumann ideals of compact linear operators, Hardy–type inequalities, variable exponent Lebesgue spaces and singular integrals.

1.1.

Homogeneous Groups

A homogeneous group is a simply connected nilpotent Lie group G on a Lie algebra g with an one–parameter group of transformations δt = exp(A log t), t > 0, where A is a diagonalized linear operator in G with positive eigenvalues. In the homogeneous group G the mappings exp o δt o exp−1 , t > 0, are automorphisms in G, which will be again denoted by δt . The number Q = tr A is the homogeneous dimension of G. The symbol e will stand for the neutral element in G. It is possible to equip G with a homogeneous norm r : G → [ 0, ∞) which is continuous on G, smooth on G\{e} and satisfies the conditions: (i) r(x) = r(x−1 ) for every x ∈ G; (ii) r(δt x) = tr(x) for every x ∈ G and t > 0; (iii) r(x) = 0 if and only if x = e ; (iv) There exists a constant c0 > 0 such that r(xy) ≤ c0 (r(x) + r(y)), x, y ∈ G. ¯ ρ) open and closed balls respectively with In the sequel we denote by B(a, ρ) and B(a, the center a and radius ρ, i.e. ¯ ρ) := {y ∈ G; r(ay−1 ) ≤ ρ}. B(a, ρ) := {y ∈ G; r(ay−1 ) < ρ}, B(a, It can be observed that δρ B(e, 1) = B(e, ρ). Let us fix a Haar measure | · | in G such that |B(e, 1)| = 1. Then |δt E| = t Q |E|. In particular, |B(x,t)| = t Q for x ∈ G, t > 0. Examples of homogeneous groups are: the Euclidean n-dimensional space Rn , the Heisenberg group, upper triangular groups, etc. For the definition and basic properties of the homogeneous group we refer to [70].

2

Alexander Meskhi Suppose that S is the unit sphere in G, i.e., S = {x ∈ G : r(x) = 1}.

Proposition 1.1.1. Let G be a homogeneous group. Then There is a (unique) Radon measure σ on S such that for all u ∈ L1 (G), Z

u(x)dx =

Z∞ Z

u(δt y)t Q−1 dσ(y)dt.

0 S

G

For the details see, e.g., [70], p. 14. Let Ω ⊆ G be a set with positive Haar measure. Suppose that w be a locally integrable almost everywhere positive function on Ω (i.e. a weight). Denote by Lwp (Ω) (0 < p < ∞) the weighted Lebesgue space, which is the space of all measurable functions f : Ω → C with the finite norm (quasi-norm if 0 < p < 1) k f kLwp (Ω) =

Z

Ω

| f (x)| p w(x)dx

1/p

.

If w ≡ 1, then we denote L1p (Ω) by L p (Ω). Let A be a measurable subset of S with positive measure. We denote by E a cone in G defined by E := {x ∈ G : x = δs x, 0 < s < ∞, x ∈ A}. It is clear that if A = S, then E = G. The next statement is a consequence of Proposition 1.1.1. Proposition 1.1.2 Let G be a homogeneous group and let A ⊂ S. There is a Radon measure σ on S such that for all u ∈ L1 (E), Z

u(x)dx =

E

Z∞ Z

u(δs y)s ¯ Q−1 dσ(y)ds. ¯

0 A

Now we formulate embedding criteria from Lwp (Ω) to Lv (Ω) (q < p), where Ω is a nonempty open set in G. These results are well-known (see [100]) but we give the proofs for completeness. q

Proposition 1.1.3. Let 0 < q < p < ∞ and let v and w be weights on an open set Ω ⊆ G. q Then Lwp (Ω) is boundedly embedded in Lv (Ω) if and only if BΩ :=

Z Ω

v(x) w(x)

p p−q

w(x)dx

p−q pq

< ∞.

(1.1.1)

When BΩ < ∞, the norm of the embedding I equals BΩ . Proof. Using H¨older’s inequality with respect to the exponent Z

q

Ω

| f (x)| v(x)dx ≤

Z

p

Ω

| f (x)| w(x)dx

q/p Z Ω

v(x) w(x)

p q

p p−q

we find that

w(x)dx

q(p−q) pq

.

Basic Ingredients

3

From this inequality we conclude that kIk ≤ BΩ . q

To prove the opposite, first we note that from the embedding of Lwp (Ω) in Lv (Ω) it follows that p Z v(x) p−q w(x)dx < ∞. Ω w(x) 1

1

Now taking the function f (x) = v p−q (x)w q−p (x) in the two-weight inequality Z

Ω

| f (x)|q v(x)dx

1/q

Z 1/p ≤ kIk | f (x)| p w(x)dx Ω

we conclude that BΩ < ∞. The fact that kIk = BΩ is obvious. The next statement follows immediately: Proposition 1.1.4. Let 0 < p < ∞. Then L∞ (Ω) is boundedly embedded in Lwp (Ω) if and only if Z 1/p BΩ := w(x)dx < ∞. Ω

When BΩ < ∞, the norm of the embedding I equals BΩ .

1.2. Measure of Non–compactness Throughout this section we assume that Ω is either a domain in Rn or a cone in G (of course, Ω might be G itself). Let X be a Banach space which is a certain subclass of all measurable functions on Ω. Denote by X ∗ the space of all bounded linear functionals on X. We say that a real-valued functional F on X is sub-linear if (i) F( f + g) ≤ F( f ) + F(g) for all non-negative f , g ∈ X; (ii) F(α f ) = |α|F( f ) for all f ∈ X and α ∈ C. An operator T : X → L p (Ω) (1 < p < ∞) is said to be sublinear if T ( f + g)(x) ≤ T ( f )(x) + T (g)(x) almost everywhere for arbitrary f , g ∈ X, and T (α f )(x) = |α|T ( f )(x) almost everywhere for all non-negative f ∈ X and α ∈ C. Let T be a sublinear operator T : X → Lq (Ω), where X = Lwp (Ω). Then the norm of the operator T is defined as follows: kT k = sup{kT f kLq (Ω) : k f kX ≤ 1}.

4

Alexander Meskhi

Moreover, T is order-preserving if T f (x) ≥ T g(x) almost everywhere for all non-negative f and g with f (x) ≥ g(x) almost everywhere. Further, if T is sub-linear and order preserving, then obviously it is non-negative, i.e. T f (x) ≥ 0 almost everywhere if f (x) ≥ 0. The measure of non-compactness for an operator T which is bounded, order-preserving and sublinear from X into a Banach space Y will be denoted by kT kκ(X,Y ) ( or simply kT kκ ) and is defined as kT kκ(X,Y ) := dist{T, κ(X,Y )} = inf{kT − Kk : K ∈ κ(X,Y )}, where κ(X,Y ) is the class of all compact sublinear operators from X to Y . For bounded linear operator T : X → Y , where Y is a Banach space, we denote kT kK (X,Y ) := dist{T, K (X,Y )} = inf{kT − Kk : K ∈ K (X,Y )}, where K (X,Y ) is the class of all compact linear operators from X to Y . If X = Y , then we use the symbol κ(X) (resp. K (X)) for κ(X,Y ) (resp. K (X,Y )). Let Y be a Banach spaces and let T be a bounded linear operator from X to Y . The entropy numbers of the operator T are defined as follows:

ek (T ) = inf ε > 0 : T (UX ) ⊂

k−1 2[

j=1

(bi + εUY ) for some b1 , . . . , b2k−1 ∈ Y ,

where UX and UY are the closed unit balls in X and Y respectively. Let us mention some properties of these numbers (see, e.g., [39]). Suppose that S, T : X → Y , R : Y → Z are bounded linear operators, where X,Y, Z are Banach spaces. Then (i) kT k = e1 (T ) ≥ e2 (T ) ≥ · · · ≥ 0; (ii) em+n−1 (S + T ) ≤ em (S) + en (T ) for all m, n ∈ N; (iii) em+n−1 (RS) ≤ em (R)en (S) for all m, n ∈ N. It is known (see, e.g., [64], p. 8) that the measure of non-compactness of T is greater than or equal to β(T ) := lim en (T ). Among other properties we mention that β(T ) = 0 if n→∞

and only if T ∈ K (X,Y ). We denote by S(X) the class of all bounded sublinear functionals defined on X, i.e., S(X) = F : X → R, F is sublinear and kFk = sup |F(x)| < ∞ . kxk≤1

Let M be the set of all bounded functionals F defined on X with the following property: 0 ≤ F f ≤ Fg for any f , g ∈ X with 0 ≤ f (x) ≤ g(x) almost everywhere. We also need the following classes of operators acting from X to L p (Ω): n FL (X, L p (Ω)) := T : T f (x) =

m

∑ α j ( f )u j , m ∈ N, u j ≥ 0, u j ∈ L p (Ω),

j=1

u j are linearly independent and α j ∈ X ∗

\

o M ,

Basic Ingredients n FS (X, L p (Ω)) := T : T f (x) =

5

m

∑ β j ( f )u j , m ∈ N, u j ≥ 0, u j ∈ L p (Ω),

j=1

\ o u j are linearly independent and β j ∈ S(X) M . If X = L p (Ω), we denote these classes by FL L p (Ω) and FS L p (Ω) respectively. It is p clear that if P ∈ FS X, resp. P ∈ FL X, L p (Ω) , then P is compact sublinear L (Ω) resp. compact linear from X to L p (Ω). between the operator T : We shall use the symbol α(T ) (resp. α(T )) for thep distance p p X → L (Ω) and the class FS X, L (Ω) , (resp. FL X, L (G) ) i.e. α(T ) := dist{T, FS X, L p (Ω) } ( resp. α(T ) := dist{T, FL X, L p (G) }).

and

For any bounded subset A of L p (Ω) (1 0 : A can be covered by finitely many open balls in L p (Ω) of radius δ Ψ(A) :=

inf

P∈FL (L p (Ω))

sup k f − P f kL p (Ω) : f ∈ A .

We shall need a statement similar to Theorem V.5.1 of [39].

Theorem 1.2.1. Let Ω be a domain in Rn . For any bounded subset K ⊂ L p (Ω) (1 ≤ p < ∞) the inequality 2Φ(K) ≥ Ψ(K) (1.2.1) holds. Proof. Let ε > Φ(K). Then there exist g1 , g2 , . . . , gN from L p (Ω) such that for all f ∈ K and some i ∈ {1, 2, . . . , N}, k f − gi kL p (Ω) < ε (1.2.2) e be a cube such that for all i ∈ {1, 2, . . . , N} Given δ > 0, let Ω Z 1/p 1 p |gi (x)| dx < δ. e 2 Ω\Ω

(1.2.3)

e = ∪m Q j , We assume that all functions from L p (Ω) are equal to zero outside Ω. Let Ω j=1 where the Q j are disjoint congruent cubes of diameter h, and define m

P f (x) :=

∑

j=1

e j = Ω ∩ Q j . Then where Q

e j |−1 fQe j χQe j (x), fQe j := |Q m

kgi − Pgi kL p (Ω∩Ω) = e m

≤

∑

Z

e j=1 Q j

≤ sup

Z

∑

j=1

1 e j| |Q

|z|
Z

Z

ej Q

f (y)dy,

p 1 Z [gi (x) − gi (y)]dy dx e j |Q e j | Qe j Q

Z

ej Q

|gi (x) − gi (y)| p dydx

|gi (x) − gi (x + z)| p dx → 0

6

Alexander Meskhi

as h → 0. From this and (1.2.3) it follows that we may choose h so that kgi − Pgi kL p (Ω) < δ, i = 1, 2, . . . , N.

(1.2.4)

Further, kP f kLpp (Ω) m

≤

∑

Z

e j=1 Q j

m

=

∑

Z

e j=1 Q j

e j |−1 |Q

Z

ej Q

Z e −1 |Q j |

ej Q

p f (y)dy dx

| f (y)| p dydx ≤ k f kLpp (Ω) .

It is obvious that the functionals f → fQe j belong to (L p (Ω))∗ ∩ M. Consequently, P ∈ FL (L p (Ω)). Finally from (1.2.2)–(1.2.4) and the last inequality we conclude that k f − P f kL p (Ω) ≤ k f − gi kL p (Ω) + kgi − Pgi kL p (Ω) +kP(gi − f )kL p (Ω) < ε + δ + kgi − f kL p (Ω) < 2ε + δ. Thus we have inequality (1.2.1) because δ is arbitrarily small number. For the case of homogeneous groups we have the statement similar to the previous one. Theorem 1.2.2. Let Ω be a cone in G. For any bounded subset K ⊂ L p (Ω) (1 ≤ p < ∞) inequality (1.2.1) holds. Proof. For simplicity assume that E = G. Let ε > Φ(K). Then there are g1 , g2 , . . . , gN ∈ L p (G) such that for all f ∈ K and some i ∈ {1, 2, . . . , N} k f − gi kL p (G) < ε.

(1.2.5)

Further, given δ > 0, let B¯ be the closed ball in G with center e such that for all i ∈ {1, 2, . . . , N} Z

|gi (x)| p dx

G\B¯

1/p

<

1 δ. 2

(1.2.6)

It is known (see [70], p. 8) that every closed ball in G is a compact set. Let us cover B¯ by open balls with radius h. Since B¯ is compact, we can choose a finite subcover {B1 , B2 , . . . , Bn }. Further, let us assume that {E1 , E2 , . . . , En } is a family of pairwise n S disjoint sets of positive measure such that B¯ = Ei and Ei ⊂ Bi (we can assume that ¯ E2 = (B2 \B1 ) ∩ B, ¯ . . . , Ek = E1 = B1 ∩ B,

i=1 k−1 S ¯ . . .). (Bk \ Bi ) ∩ B, i=1

n

P f (x) = ∑ fEi χEi (x), i=1

fEi = |Ei |

−1

Z

Ei

We define

f (x)dx.

Basic Ingredients

7

Then = kgi − Pgi kLpp (B) ¯

n

∑

j=1 Ej m

≤

Z

∑

Z

j=1 Ej

≤

sup

1 |E j |

Z

Ej

p [gi (x) − gi (y)]dy dx

Z 1

|E j |

r(z)≤2c0 h

Ej

Z B¯

p gi (x) − gi (y) dy dx

gi (x) − gi (zx) p dx → 0

as h → 0. The latter fact follows from the continuity of the norm L p (G) (see, e.g., [70], p. 19). From this and (1.2.5) it follows that kgi − Pgi kL p (G) < δ,

i = 1, 2, 3, . . . , N,

(1.2.7)

when h is sufficiently small. Further, kP f kLpp (G) = ≤

n

∑

Z

j=1 Ej n Z

∑

j=1 Ej

|E j |−1 |E j |

Z

E¯ j

Z −1 E¯ j

p f (y)dy dx

f (y) p dy dx ≤ k f k pp

¯ L (B)

≤ k f kLpp (G) .

∗ It is also clear that the functionals f → fEi belong to L p (G) ∩ M. Hence P ∈ FL (L p (G)). Finally (1.2.5) − (1.2.7) yield k f − P f kL p (G) ≤ k f − gi kL p (G) + kgi − Pgi kL p (G) + kP(gi − f )kL p (G) < ε + δ + kgi − f kL p (G) ≤ 2ε + δ. Since δ is arbitrarily small, we have the desired result. Lemma 1.2.1. Let 1 ≤ p < ∞. Assume that K ⊂ L p (Ω) is compact. Then for any given ε > 0 there exists an operator Pε ∈ FL L p (Ω) such that for all f ∈ K, k f − Pε f kL p (Ω) ≤ ε .

Proof. Suppose that Ω is a cone in G and that K is a compact set in L p (Ω). Using Theorem 1.2.2 we see that Ψ(K) = 0. Hence for ε > 0 there exists Pε ∈ FL L p (Ω) such that sup k f − Pε f kL p (Ω) : f ∈ K ≤ ε. Lemma 1.2.2. Let T : X → L p (G) be compact, order-preserving and sublinear (resp. compact linear) operator, where 1 ≤ p < ∞. Then α(T ) = 0 (res. α(T ) = 0).

8

Alexander Meskhi

Proof. We prove the lemma for compact and sublinear operators. The proof is the same in the linear case. For simplicity assume that Ω is a cone in G. Let UX = f : k f kX ≤ 1 . From the compactness of T it follows that T (UX ) is relatively compact in L p (Ω). Using Lemma 1.2.1 we have that for any given ε > 0 there exists an operator Pε ∈ FL L p (Ω) such that for all f ∈ UX , kT f − Pε T f kL p (Ω) ≤ ε. (1.2.8) Let P˜ε = Pε ◦ T. Then P˜ε ∈ FS X, L p (Ω) . Indeed, there exist functionals α j ∈ X ∗ ∩ M, j ∈ {1, 2, . . . , m}, and linearly independent functions u j ∈ L p (Ω), j ∈ {1, 2, . . . , m}, such that P˜ε f (x) = Pε (T f )(x) =

m

m

j=1

j=1

∑ α j (T f )u j (x) = ∑ β j ( f )u j (x),

where β j = α j ◦ T belongs to S(X) ∩ M. By (1.2.8) we have kT f − P˜ε f kL p (Ω) ≤ ε for all f ∈ UX , which on the other hand, implies that α(T ) = 0. We shall also need the following lemma. Lemma 1.2.3. Let T be a bounded, order-preserving and sublinear operator (resp. bounded linear operator) from X to Lq (G), where 1 ≤ q < ∞. Then kT kκ = α(T ) (resp. kT kK = α(T )) Proof. Let T be bounded, order preserving and sublinear. Suppose that δ > 0. Then there exists an operator K ∈ κ(X, Lq (Ω)), such that kT − Kk ≤ kT kκ + δ. By Lemma 1.2.2 there is P ∈ FS (X, Lq (Ω)) for which the inequality kK − Pk < δ holds. This gives kT − Pk ≤ kT − Kk + kK − Pk ≤ kT kκ + 2δ. Hence α(T ) ≤ kT kκ . The opposite inequality kT kκ ≤ α(T ) is obvious. Lemma 1.2.4. Let G be a homogeneous group 1 ≤ q < ∞ and let P ∈ q FS (X, L (G)) (resp. P ∈ FL (X, Lq (G))). Then for every a ∈ G and ε > 0 there exist an operator R ∈ FS (X, Lq (G)) (resp. R ∈ FL (X, Lq (G))) and positive numbers α, α¯ such that for all f ∈ X the inequality k(P − R) f kLq (G) ≤ εk f kX ¯ holds and supp R f ⊂ B(a, α)\B(a, α). Proof. There exist linearly independent non-negative functions u j ∈ Lq (G), j ∈ {1, 2, . . . , N}, such that N

P f (x) =

∑ β j ( f )u j (x),

j=1

f ∈ X,

Basic Ingredients

9

where β j are bounded, order-preserving, sublinear functionals β j : X → R. On the other hand, there is a positive constant c for which the inequality N

∑ |β j ( f )| ≤ ck f kX

j=1

holds. Let us choose linearly independent Φ j ∈ Lq (G) and positive real numbers α j , α¯ j such that ku j − Φ j kLq (G) < ε, j ∈ {1, 2, . . . , N}, and supp Φ j ⊂ B(a, α¯ j )\B(a, α j ). If N

R f (x) =

∑ β j ( f ) Φ j (x),

j=1

then it is obvious that R ∈ FS (X, Lq (G)) and, moreover, N

kP f − R f kLq (G) ≤

∑ |β j ( f )|ku j − Φ j kL (G) ≤ cεk f kX q

j=1

¯ for all f ∈ X. Besides this, supp R f ⊂ B(a, α)\B(a, α), where α = min{α j } and α¯ = max{α¯ j }. Lemmas 1.2.3 and 1.2.4 for Lebesgue spaces defined on Euclidean spaces have been proved in [39] in the linear case. In a similar manner we have the following statements (the proofs are omitted): Lemma 1.2.5. Let Ω be a domain in Rn and let P ∈ FS X, L p (Ω) (resp. P ∈ FL (X, Lq (Ω))), where X = Lwr (Ω) and 1 < r, p < ∞. Then for every a ∈ Ω and ε > 0 there exists an operator R ∈ FS X, L p (Ω) (resp. P ∈ FL (X, Lq (G))) and positive numbers β1 and β2 , β1 < β2 such that for all f ∈ X the inequality

(P − R) f p ≤ εk f kX L (Ω) T

holds and supp R f ⊂ D(a, β2 )\D(a, β1 ), where D(a, s) := Ω B(a, s).

Lemma 1.2.6. Let Ω be a cone in G. Suppose that 1 ≤ p < ∞ and Y = Lvp (Ω). Suppose that P ∈ FL (X,Y ) and ε > 0. Then there are an operator T ∈ FL (X,Y ) and a set Eα,β := {x ∈ Ω : 0 < α < r(x) < β < ∞} such that kP − T k < ε and supp T f ⊂ Eα,β for every f ∈ X. Lemma 1.2.7. Let G be a homogeneous group. Suppose that 1 < p, q < ∞ and T a bounded, order-preserving and sublinear (resp. bounded liear) operator from Lwp (G) to

10

Alexander Meskhi

Lv (G). Suppose that λ > kT kκ(Lwp (G),Lvq (G)) (resp. λ > kT kK (Lwp (G),Lvq (G)) ) and a is a point of G. Then there exist constants β1 , β2 , 0 < β1 < β2 < ∞, such that for all τ and r with r > β2 , τ < β1 , the following inequalities hold: q

kT f kLvq (B(a,τ)) ≤ λk f kLwp (G) ,

(1.2.9)

kT f kLvq (B(a,r)c ) ≤ λk f kLwp (G) ,

(1.2.10)

where f ∈ Lwp (G). q

Proof. Let T be bounded, order preserving and sublinear from Lwp (G) to Lv (G). Let T (v) be the operator given by 1 T (v) f = v q T f . Then it is easy to see that kT (v) kκ(Lwp (G)→Lq (G)) = kT kκ(Lwp (G)→Lvq (G)) . By Lemma 1.2.3 we have that

λ > α T (v) .

Consequently, there exists P ∈ FS Lwp (G), Lq (G) such that kT (v) − Pk < λ.

Fix a ∈ G. Accordingto Lemma 1.2.4 there are positive constants β1 and β2 , β1 < β2 , and q R ∈ FS Lwp (G), Lv (G) for which kP − Rk ≤

λ − kT (v) − Pk 2

and supp R f ⊂ B(a, β2 )\B(a, β1 ) for all f ∈ Lwp (G). Hence, kT (v) − Rk < λ. From the last inequality it follows that if 0 < τ < β1 and r > β2 , then (1.2.9) and (1.2.10) hold for f ∈ Lwp (G). The next statement follows in a similar manner as Lemma 1.2.7; therefore the proof is omitted. Lemma 1.2.8. Ω be a domain in Rn . Suppose that 1 < p, q < ∞ and that T is bounded, q order-preserving and sublinear (resp. bounded linear) operator from Lwp (Ω) to Lv (Ω). Assume that λ > kT kκ(Lwp (Ω),Lvq (Ω)) (resp. λ > kT kK (Lwp (Ω),Lvq (Ω)) ) and a ∈ Ω. Then there exist constants β1 , β2 , 0 < β1 < β2 < ∞ such that for all τ and r with r > β2 , τ < β1 , the following inequalities hold: kT f kLvq (Ω∩B(a,τ)) ≤ λk f kLwp (Ω) ; where f ∈ Lwp (Ω).

kT f kLvq (Ω\B(a,r)) ≤ λk f kLwp (Ω) ,

Basic Ingredients

1.3.

11

Hardy–type Transforms

In this section we give some Hardy-type inequalities. Theorem 1.3.1 ([49] (Section 1.1)). Let (X, µ) be a measure space and let φ : X → R be a µ-measureable non-negative function. Suppose that 1 0

Z

{x:φ(x)>t}

Z

1/q v(x)dµ

{x:φ(x)
1/p′ ′ < ∞, p′ = p/(p − 1). w1−p (x)dµ

Let E be a cone in G, where G is a homogeneous group (see Section 1.1). Denote Et := {y ∈ E : r(y) < t}, where t is a positive number. Taking X = E, dµ = dx, v = u, w = 1, φ(x) = r(x) in the previous statement and observing that Z 1/q 1/q Z ′ Q/p′ u(x)dx u(x)dx sup t ≈ sup 2 jQ/p , t>0

E\Et

j∈Z

E2 j+1 \E2 j

we have the next statement.

Theorem 1.3.2. Let 1 < p ≤ q < ∞. Then the inequality Z q 1/q Z Z 1/p p v(x) f (y)dy dx ≤c | f (x)| dx E

Er(x)

holds if and only if

sup j∈Z

E

Z

u(x)dx

E2 j+1 \E2 j

1/q

′

2 jQ/p < ∞.

We need also the following statement (see [158] for 1 ≤ q 1. Then the inequality Z ∞ Z x q 1/q Z ∞ 1/p p v(x) f (t)dt dx ≤c ( f (x)) w(x)dx , f ≥ 0, 0

0

holds if and only if Z ∞ Z ∞ 0

t

v(x)dx

0

Z

t

1−p′

w 0

(x)dx

q−1 p/(p−q)

1−p′

w

(t)dt

(p−q)/(pq)

< ∞.

12

Alexander Meskhi The next statement deals with the Hardy-inequality on an interval.

Theorem 1.3.4. Let r and s be constants such that 1 < r ≤ s < ∞. Suppose that 0 ≤ a < b ≤ ∞. Let v and w be non-negative measurable functions on [a, b). Then the inequality Z b Z v(x) a

x

f (t)dt a

s

dx

1/s

Z b 1/r ≤c (w(t) f (t))r dt , f ≥ 0, a

holds if and only if sup a≤t≤b

Z

b

s

v (x)dx

t

1/s Z

t

−r′

w a

(x)dx

1/r′

< ∞.

The next statements will be useful for us. Theorem 1.3.5 ([101] (Ch. XI)). Let (X, µ) and (Y, ν) be σ-finite measure spaces and let 1 < p, q < ∞. Suppose that for positive function a : X ×Y → R, we have

ka(x, y)k p′ q < ∞. L (Y ) Lµ (X) ν

Then the operator

A f (x) =

Z

a(x, y) f (y)dν(y)

Y q

is compact from Lνp (Y ) to Lµ (X). The next lemma is known as Ando’s theorem (see [4] and [139], Sections 5.3 and 5.4) which in our case is formulated for the set E. Theorem 1.3.6. Let 0 < q < ∞, 1 < p < ∞ and q < p. Suppose that v and w are Haar-measurable almost everywhere positive functions on E. If the operator AE f (x) =

Z

a(x, y) f (y)dy, x ∈ E, E

q

is bounded from Lwp (E) to Lv (E), then AE is compact.

1.4.

L p(x) Spaces

Let Ω be a domain in Rn and let p be a measurable function on Ω. Throughout this section we assume that 1 < p− ≤ p(x) ≤ p+ < ∞, where p− and p+ are respectively the infimum and the supremum of p on Ω. Suppose that ρ is a weight function on Ω, i.e. ρ is an almost everywhere positive locally integrable function on Ω. We say that a measurable function f p(·) p(x) on Ω belongs to Lρ (Ω) (or Lρ (Ω)) if S p,ρ ( f ) =

Z

Ω

f (x)ρ(x) p(x) dx < ∞. p(x)

It is known (see, e.g., [138], [111]) that Lρ (Ω) is a Banach space with the norm

Basic Ingredients

13

k f kL p(x) (Ω) = inf λ > 0 : S p,ρ f /λ ≤ 1 . ρ

p(x)

If ρ ≡ 1, then we use the symbol L p(x) (Ω) (resp. S p ) instead of Lρ (Ω) (resp. S p,ρ ). It is clear that k f kL p(·) (Ω) = k f ρkL p(·) (Ω) . ρ Further, we denote p− (E) = inf p,

p+ (E) = sup p

E

E

for a measurable set E ⊆ Ω. Let P (Ω) be the class of all measurable functions p, p : Ω → Rn , such that the HardyLittlewood maximal operator 1 MΩ f (x) = sup |Q| Q∋x

Z

Q∩Ω

| f (y)|dy, x ∈ Ω,

where the supremum is taken over all cubes Q containing x and satisfying |Q ∩ Ω| > 0, is bounded in L p(·) (Ω). Throughout the paper we will assume that I(a, r) is the interval (a − r, a + r). The following lemma is well-known (see e.g., [138], [204]): Lemma 1.4.1. Let E be a measurable subset of Ω. Then the following inequalities hold: p (E)

p (E)

p (E)

p (E)

k f kL+p(·) (E) ≤ S p ( f χE ) ≤ k f kL−p(·) (E) , k f kL p(·) (E) ≤ 1; k f kL−p(·) (E) ≤ S p ( f χE ) ≤ k f kL+p(·) (E) , k f kL p(·) (E) ≥ 1; Z f (x)g(x)dx ≤ E

where p′ (x) =

p(x) p(x)−1

1 1 + k f kL p(·) (E) kgkL p′ (·) (E) , p− (E) (p+ (E))′

and 1 < p− (E) ≤ p(x) ≤ p+ (E) < ∞.

Definition 1.4.1. Let Ω be a domain in Rn . We say that p satisfies weak Lipschitz (log − H¨older continuity) condition (p ∈ W L(Ω)) if there is a positive constant A such that for all x and y in Ω with 0 < |x − y| < 1/2 the inequality |p(x) − p(y)| ≤ A/(− ln |x − y|)

(1.4.1)

holds. The next result is due to L. Diening [29]. Theorem 1.4.1. Let Ω be bounded domain in Rn . If p satisfies the condition (1.4.1), then the operator MΩ is bounded in L p(·) (Ω). Theorem 1.4.2 ([27], [22]). Let Ω = Rn . Then M is bounded in L p(·) (Rn ) if (1.4.1) holds for all x, y ∈ Rn with 0 < |x − y| < 1/2 and, moreover, there exists a positive constant b such that |p(x) − p(y)| ≤ b/ ln(e + |x|), x, y ∈ Rn , |y| ≥ |x|. (1.4.2)

14

Alexander Meskhi

Let H f (x) = lim

Z

ε→0 |x−t|>ε

f (t) dt x−t

be the Hilbert transform. Theorem 1.4.3 ([26]). Let p ∈ P (R). Then H is bounded in L p(·) (R). The next statement is also valid (see [59], [206], [124], [91]). Theorem 1.4.4. The class C0 (Ω) of continuous compactly supported functions on Ω is dense in L p(·) (Ω). p(x)

For other properties of L p(x) and Lρ spaces see e.g. [138], [220], [204], [111], [208]. We need the following slight modification of Lemma 2.1 from [29] (see also [27]). Proposition 1.4.1. Let Ω be an open set in R. Suppose that p satisfies condition (1.4.1) on Ω with the constant A. Then for all intervals I with |I ∩ Ω| > 0 and |I| < 1/4 the inequality |I| p− (I∩Ω)−p+ (I∩Ω) ≤ eA holds. Proof. Let I := (a − r, a + r) for some a ∈ R and r > 0. It is easy to see that (2r) p− (I∩Ω)−p+ (I∩Ω) ≤ (2r)A/ ln(2r) = eA .

Proposition 1.4.2 ([203, 138]). Let |Ω| < ∞ and 1 ≤ r(x) ≤ p(x) ≤ p < ∞ for x ∈ Ω. Then L p(x) (Ω) ⊆ Lr(x) (Ω) and k f kLr(x) (Ω) ≤ (1 + |Ω|)k f kL p(x) (Ω) . Remark 1.4.1. If p satisfies (1.4.1) with the constant A, then the function 1/p satisfies the same condition with the constant A/(p− )2 . Indeed, we have |1/p(x) − 1/p(y)| = |p(x) − p(y)|/|p(x)p(y)| ≤ −A1 / ln |x − y|, )2

where A1 = A/(p− and |x − y| < 1/2. Analogously we can show that if p satisfies (1.4.1) with the constant A, then p′ satisfies the same condition with the constant A/(p− − 1)2 . Let X be a Banach space and let n FL (X, L p(·) (Ω)) ≡ T : T f (x) =

m

o ,

∑ β j ( f )u j , m ∈ N, u j ≥ 0, u j ∈ L p(·)(Ω)

j=1

where u j are linearly independent and β j are bounded linear functionals on L p(·) (Ω). If X = L p(·) (Ω), then we denote FL (X, L p(·) (Ω)) := FL (L p(·) (Ω)). For any bounded subset A of L p(·) (Ω), let Φ(A) := inf{δ > 0 : A can be covered by finite open balls in L p(·) (Ω) of radius δ}

Basic Ingredients

15

and Ψ(A) :=

inf

P∈FL (L p(·) (Ω))

sup{k f − P f kL p(·) (Ω) : f ∈ A}.

The following lemma is similar to Theorem 1.2.1. Lemma 1.4.2. Let p ∈ P (Ω). Then there exists a positive constant B such that for any bounded subset K ⊂ L p(·) (Ω) the inequality BΦ(K) ≥ Ψ(K) holds. Proof. Let ε > Φ(K). Then there exist g1 , . . . , gN from L p(·) (Ω) such that for all f ∈ K and some j ∈ {1, . . . , N}, k f − g j kL p(·) (Ω) < ε. Let us take δ > 0. Then due to Theorem 1.4.4 there are g¯ j ∈ C0 (Ω), such that for all j ∈ {1, . . . , N}, kg j − g¯ j kL p(·) (Ω) < δ. (1.4.3) Hence for that given δ,

k f − g¯ j kL p(·) (Ω) < ε + δ.

(1.4.4)

By the absolutely continuity of the norm there is a cube Q¯ such that kg¯i kL p(·) (Ω\Q) ¯ < δ.

(1.4.5)

Suppose that σ is a small positive number. Let us divide Q¯ by disjoint cubes Qi so that diam Qi < h and h is sufficiently small for which |g¯ j (x) − g¯ j (y)| < σ when |x − y| < h, x, y ∈ Rn (in fact, we may assume that g¯ j are extended by 0 continuously on Rn ). Further, we take m

Pφ(x) = ∑ φQi χQei (x); φQi = |Qi | i=1

−1

Z

Qi

φ(y)χΩ (y)dy,

ei = Qi ∩ Ω. Taking now into account the fact that g¯ j χΩ = g¯ j , we have where Q ¯ p (Ω∩Q) kg¯ j − Pg¯ j kL+p(·) (Ω∩Q) ¯

≤∑ i

Z ei Q

1 |Q|i

Z

Qi

p(x) |(g¯ j (x) − g¯i (y)|dy dx

m

¯ ei | ≤ σ p− |Q|. ¯ ≤ σ p− (Ω∩Q) ∑ |Q i=1

Consequently,

¯ 1/p+ σ p− /p+ . kg¯ j − Pg¯ j kL p(·) (Ω∩Q) ¯ ≤ |Q|

16

Alexander Meskhi Further, the condition p ∈ P (Ω) yields

N

kPφkL p(·) (Ω) ≤ ∑ φQ j χQe j ≤ kMΩ φkL p(·) (Ω) ≤ kMΩ kkφkL p(·) (Ω) .

i=1

L p(·) (Ω)

Finally, taking into account estimates (1.4.3)–(1.4.5) and the fact that P ∈ FL (L p(·) (Ω)), we find that k f − P f kL p(·) (Ω) ≤ k f − g¯ j kL p(·) (Ω) + kg¯ j − Pg¯ j kL p(·) (Ω) + kPg¯ j − P f kL p(·) (Ω) ¯ 1/p+ σ p− /p+ + kMΩ kkg¯ j − f kL p(·) (Ω) < ε + 2δ + |Q| ¯ 1/p+ σ p− /p+ . ≤ (1 + kMΩ k)ε + (2 + kMΩ k)δ + |Q| ¯ we have the desired result. As δ and σ are arbitrarily small and σ is independent of Q, Lemma 1.4.3. Let p ∈ P (Ω) and let K ⊂ L p(·) (Ω) be compact. Then for a given ε > 0 there exists an operator Pε ∈ FL (L p(·) (Ω)) such that for all f ∈ K, k f − Pε f kL p(·) (Ω) ≤ ε. Proof. Let K be a compact subset of L p(·) (Ω). By Lemma 1.4.2 we have that Ψ(K) = 0. Hence, for ε > 0 there exists Pε ∈ FL (L p(·) (Ω)) such that sup{k f − Pε f kL p(·) (Ω) : f ∈ K} ≤ ε.

Let X and Y be Banach spaces. As before (see Section 1.2 for classical Lebesgue spaces), we denote α(T ) := dist{T, FL (X,Y )}. Lemma 1.4.4. Let p ∈ P (Ω) and let X be a Banach space. Suppose that T : X → L p(·) (Ω) is a compact linear operator. Then α(T ) = 0. Proof. Let UX := { f : k f kX ≤ 1}. From the compactness of T it follows that T (UX ) is relatively compact in L p(·) (Ω). By Lemma 1.4.3 we have that for any ε > 0 there exists an operator Pε ∈ FL (L p(·) (Ω)) such that for all f ∈ UX , kT f − Pε T f kL p(·) (Ω) ≤ ε.

(1.4.6)

Let P¯ε = Pε ◦ T . Then P¯ε ∈ FL (X, L p(·) (Ω)). Indeed, there are bounded linear functionals α j and linearly independent functions u j ∈ L p(·) (Ω), j ∈ {1, . . . , m}, such that m

P¯ε f (x) = Pε (T f )(x) =

∑ α j (T f )u j (x) =

j=1

m

∑ β j ( f )u j (x),

j=1

where β j = α j ◦ T (1 ≤ i ≤ m) are bounded linear functional from X to L p(·) (Ω). Further, inequality (1.4.6) implies kT f − P¯ε f kL p(·) (Ω) ≤ ε for all f ∈ UX , from which it follows that α(T ) = 0.

Basic Ingredients

17

Lemma 1.4.5. Let P ∈ FL (X, L p(·) (Ω)), where X is a Banach space. Then for every ¯ a ∈ Ω and ε > 0 there exist an operator R ∈ FL (X, L p(·) (Ω)) and positive numbers α and α, ¯ such that for all f ∈ X the inequality α < α, k(P − R) f kL p(·) (Ω) ≤ εk f kX ¯ \ Q(a, α), where Q(a, r) is a cube with center a and holds, and moreover, supp R f ⊂ D(a, α) side length r; D(a, r) := Q(a, r) ∩ Ω. Proof. Since P ∈ FL (X, L p(·) (Ω)), there exist linearly independent non-negative functions u j ∈ L p(·) (Ω), j ∈ {1, . . . , m}, such that m

P f (x) =

∑ β j ( f )u j (x),

f ∈ X,

j=1

where β j are bounded linear functionals on X. Further, there is a positive constant c for which N

∑ |β j ( f )| ≤ ck f kX .

j=1

Let us choose linearly independent Φ j ∈ L p(·) (Ω) and positive real numbers α j , α¯ j , such that ku j − Φ j kL p(·) (Ω) < ε, j ∈ {1, . . . , N}, and supp Φ j ⊂ D(a, α¯ j ) \ Q(a, α j ). Let us take

N

R f (x) =

∑ β j ( f )Φ j (x).

j=1

Then it is obvious that R ∈ FL (X, L p(·) (Ω)) and, also N

kP f − R f kL p(·) (Ω) ≤

∑ |β j ( f )|ku j − Φ j kL

p(·) (Ω)

≤ cεk f kX

j=1

for all f ∈ X. Moreover,

¯ \ Q(a, α), supp R f ⊂ D(a, α)

where α = min{α j }, α¯ = max{α¯ j }. The next lemma will be useful. ′

Lemma 1.4.6. If f ∈ / L p(·) (Ω), then there exists a non-negative g ∈ L p (·) (Ω) such that fg ∈ / L1 (Ω). Proof. The easier way to get this result is not a generalization of the arguments from [88]. In our case we use Landau’s resonance theorem (see e.g. Lemma 2.6, p. 10 of [12]) according to which a measurable function f belong to the dual space X ′ of a Banach function space X if and only if f g is integrable for every g in X. In order to use this result ′ in our case it is enough to note that the dual space of L p(·) (Ω) is L p (·) (Ω) (see, e.g., [138], [204]).

18

Alexander Meskhi Lemma 1.4.7([204]). For all f ∈ L p(·) (Ω) the inequality Z k f kL p(·) (Ω) ≤ sup f (y)g(y)dy kgk

≤1 ′ L p (·) (Ω)

Ω

holds.

In the sequel we will assume that Ek := [2k , 2k+1 ); Ik := [2k−1 , 2k+1 ), k ∈ Z. The next statements will be useful in Chapter 5. Lemma 1.4.8. Let 1 < p− (R+ ) ≤ p(x) ≤ q(x) ≤ q+ (R+ ) < ∞ and let p, q ∈ W L(R+ ). Suppose that p(x) ≡ pc = const, q(x) ≡ qc = const when x > a for some positive number a. Then there exists a positive constant c such that

∑ k f χI kL k

p(·) (R

+)

k

kgχIk kLq′ (·) (R+ ) ≤ ck f kL p(·) (R+ ) kgkLq′ (·) (R+ )

′

for all f ∈ L p(·) (R+ ) and g ∈ Lq (·) (R+ ). Proof. For simplicity assume that a = 1. Let us split the sum as follows:

∑ k f χI kL i

i

p(·) (R

+)

kgχIi kLq′ (·) (R+ ) = ∑ + ∑ := J1 + J2 . i≤2

i>2

Taking into account that p(x) = pc = const, q(x) ≡ qc ≡ const on the set (1, ∞), using H¨older’s inequality for series and the fact that pc ≤ qc , we have J2 = ∑ k f χIi kL pc (R+ ) kgχIi kL(qc )′ (R+ ) ≤ ck f kL p(·) (R+ ) kgkLq′ (·) (R+ ) . i>2

Now let us estimate J1 . Suppose that k f kL p(·) (R+ ) ≤ 1 and kgkLq′ (·) (R+ ) ≤ 1. First notice that q, q′ ∈ W L(R+ ). Therefore, by Lemma 1.4.1 and Proposition 1.4.1 we have |Ik |1/q+ (Ik ) ≈ kχIk kLq(·) (R+ ) ≈ |Ik |1/q− (Ik ) ; ′

′

|Ik |1/(q )+ (Ek ) ≈ kχEk kLq′ (·) (R+ ) ≈ |Ik |1/(q )− (Ik ) , where k ≤ 2. Hence H¨older’s inequality (see Lemma 1.4.1) yields J1

≤

c∑

Z 8 k f χ k p(·) Ik L (R+ ) kgχIk kLq′ (·) (R+ )

k≤2 0

Z 8

∑

kχIk kLq(·) (R+ ) kχEk kLq′ (·) (R+ )

k f χIk kL p(·) (R+ ) kgχIk kLq′ (·) (R+ )

χEk (x)dx

χEk (x)dx kχIk kLq(·) (R+ ) kχIk kLq′ (·) (R+ )

k f χIk kL p(·) (R+ )

kgχIk kL p′ (·) (R+ )

≤ c ∑ χEk (·) χEk (·) ∑

q′ (·) Lq(·) ((0,8)) k≤2 kχIk kL p′ (·) (R+ ) L ((0,8)) k≤2 kχIk kLq(·) (R+ ) ≤

c

0 k≤2

:= cS1 · S2 .

Basic Ingredients

19

Now we claim that I(q) ≤ cI(p), where

k f χIk kL p(·) (R+ )

I(q) := ∑ χEk (·) ;

q(·) L ((0,8)) k≤2 kχIk kLq(·) (R+ )

k f χIk kL p(·) (R+ )

I(p) := χEk (·) .

∑ kχI k p(·)

p(·) k L (R+ ) L ((0,8)) k≤2

Indeed, suppose that I(p) ≤ 1. Further, Proposition 1.4.1 and Lemma 1.4.1 yield 1 ∑ |Ik | k≤2

Z

p(x) k f χIk kL p(·) (R ) dx +

≤c

Ek

Z8 0

∑

k f χIk kL p(·) (R+ )

k≤2

kχIk kL p(·) (R+ )

p(x) dx ≤ c. χEk (x)

Consequently, taking into account that q(x) ≥ p(x), Ek ⊂ Ik and k f kL p(·) (R+ ) ≤ 1, we find that Z Z 1 1 q(x) p(x) ∑ |Ik | k f χIk kL p(·) (R+ ) dx ≤ c ∑ |Ik | k f χIk kL p(·) (R+ ) dx ≤ c. k≤2 k≤2 Ek

Ek

This implies that I(q) ≤ c. Let us introduce a function P(t) =

∑ p+ (Ik )χE (t). k

k≤2

It is clear that p(t) ≤ P(t) because Ek ⊂ Ik . Hence, Proposition 1.4.2 for Ω = (0, 8) yields

k f χIk kL p(·) (R+ )

. χEk (·) I(p) ≤ c ∑

P(·) L ((0,8)) k≤2 kχIk kL p(·) (R+ ) p (I )

k Then, using the definition of P and the inequality kχIk kL+p(·) (R

Z 8 0

=

Z 8 0

≤ c

≤ c∑

k≤2

Z

kχIk kL p(·) (R+ )

k≤2

p (I )

∑

k k f χIk kL+p(·) (R

k≤2

Z 8 0

≤ c

∑

k f χIk kL p(·) (R+ )

∑

+)

p (I )

k kχIk kL+p(·) (R

k≤2

≥ c2k , we have

P(x) χEk (x) dx χEk (x) dx

+) p+ (Ik ) k f χIk kL p(·) (R ) + χEk (x) dx 2k

p (Ik ) k f χIk kL+p(·) (R +)

≤c∑

| f (x)| p(x) dx ≤ c. R+

+)

Z

k≤2 Ik

| f (x)| p(x) dx

20

Alexander Meskhi

Consequently, the estimates derived above give us S1 ≤ ck f kL p(·) (R+ ) . Analogously, taking into account the fact that q′ ∈ W L(R+ ) and arguing as above, we find that S2 ≤ ckgkLq′ (·) (R+ ) . Notice that Lemma 1.4.8 is a slight modification of Theorem 2 from [136] (see also [53] for the case p(x) = q(x)). Lemma 1.4.9. Let I = [0, a] be a bounded interval and let p ∈ W L(I). Suppose that 1 < p− (I) ≤ p+ (I) < ∞ and α(x) > 1/p(x) when x ∈ I. Then I(x) := k(x − ·)α(x)−1 χ(x/2,x) (·)kL p′ (·) (R+ ) ≤ cxα(x)−1/p(x) , where the positive constant c does not depend on x. Proof. First notice that the condition p ∈ W L(I) implies p′ ∈ W L(I). Hence we have the following two-sided estimate: ′

′

′

(x − t) p (t) ≤ c1 (x − t) p (x) ≤ c2 (x − t) p (t) , where 0 < t < x < a and the positive constants c1 and c2 depend only on p and a. Consequently, Z x

′

(x − t)(α(x)−1)p (t) dt ≤ c

x/2

Z x

′

(x − t)(α(x)−1)p (x) dt

x/2

= c

Z x/2

= cx

′

′

u(α(x)−1)p (x) du = c(x/2)(α(x)−1)p (x)+1

0 (α(x)−1)p′ (x)+1

:= S(x).

Suppose that I(x) ≤ 1. By the fact that 1/p ∈ W L(I), Proposition 1.4.1 and Lemma 1.4.1 we have (α(x)−1)p′ (x)+1 ′ ′ I(x) ≤ (S(x))1/(p )+ ([x/2,x]) = c (x/2)1/(p )+ ([x/2,x]) (α(x)−1)p′ (x)+1 (α(x)−1)p′ (x)+1 ′ ′ ≤ c (x/2)1/(p )− ([x/2,x]) ≤ c (x/2)1/p (x) ≤ cxα(x)−1/p(x) .

For I(x) > 1, the conclusion is trivial. Now we formulate some Hardy-type inequalities in L p(·) spaces. Let

H f (t) =

Z x 0

be the Hardy transform.

f (t)dt, x > 0,

Basic Ingredients

21

Theorem 1.4.5 ([136]). Let p ∈ W L(I), where I = [0, a], where 0 < a < ∞. Suppose that 1 < p− (I) ≤ p(x) ≤ q(x) ≤ q+ (I) < ∞ and p, q ∈ W L(I). Then the Hardy operator H p(·) q(·) is bounded from Lw (I) to Lv (I) if and only if C := sup kv(·)χ(t,a) (·)kLq(·) (I) kw−1 (·)χ(0,t) (·)kL p′ (·) (I) < ∞. 0
Moreover, there exist positive constants c1 and c2 such that c1C ≤ kH k ≤ c2C. To formulate the next results we need the notation: ( p0 (x), 0 ≤ x ≤ a p0 (x) := inf p(y); pe0 (x) := y∈[0,x] pc ≡ const, x < a

,

where a is a fixed positive number.

Theorem 1.4.6. Let I = [0, a] (0 < a < ∞) and let 1 ≤ p− (I) ≤ p0 (x) ≤ q(x) ≤ q+ (I) < ∞ for almost every x ∈ I. Then the condition sup

Z a

0
′

(v(x))q(x)t q(x)/(p0 ) (x) dx < ∞ q(·)

implies the boundedness of H from L p(·) (I) to Lv (I). Theorem 1.4.7. Let I = R+ and let 1 ≤ p− (I) ≤ p0 (x) ≤ q(x) ≤ q+ (I) < ∞ for almost every x ∈ I. Suppose that q(x) ≡ qc = const, p(x) ≡ pc = const when x > a for some positive number a. Then the condition sup

Z ∞

0
′

p0 ) (x) (v(x))q(x)t q(x)/(f dx < ∞

q(·)

guarantees the boundedness of H from L p(·) (I) to Lv (I). Theorems 1.4.6 and 1.4.7 are special cases of Theorems 3.1 and 3.3 of [52] respectively. Theorem 1.4.8 ([57]). Let p(x) and q(x) be measurable functions on an interval I ⊆ R+ . Suppose that 1 < p− (I) ≤ p+ (I) < ∞ and 1 < q− (I) ≤ q+ (I) < ∞. If

kk(x, y)kL p′ (y) (I) q(x) < ∞, L

(I)

where k is a non-negative kernel, then the operator K f (x) =

Z I

is compact from L p(·) (I) to Lq(·) (I).

k(x, y) f (y)dy

22

1.5.

Alexander Meskhi

Schatten–von Neumann Ideals

Let H be a separable Hilbert space and let σ∞ (H) be the class of all compact linear operators T : H → H, which form an ideal in the normed algebra B of all bounded linear operators on H. To construct a Schatten-von Neumann ideal σ p (H) (0 < p ≤ ∞) in σ∞ (H), the sequence of singular numbers s j (T ) ≡ λ j (|T |) is used, where the eigenvalues λ j (|T |) (|T | ≡ (T ∗ T )1/2 ) are non-negative and are repeated according to their multiplicities and arranged in decreasing order. A Schatten-von Neumann quasinorm (norm if 1 ≤ p ≤ ∞) is defined as follows: kT kσ p (H) ≡

∑ s pj (T )

1/p

, 0 < p < ∞,

j

with the usual modification if p = ∞. Thus we have kT kσ∞ (H) = kT k and kT kσ2 (H) is the Hilbert-Schmidt norm given by the formula Z Z 1/2 kT kσ2 (H) = |a(x, y)|2 dxdy for the integral operator

A f (x) =

Z

a(x, y) f (y)dy.

We refer, for example, to [13], [14], [16], [135], for more information concerning Schatten-von Neumann ideals. Let ρ be a weight function on E, where E is a cone in a homogeneous group G (see the previous section). Suppose that Et := {y ∈ E : r(y) < t}, where t is a positive number. We denote by l p Lρ2 (E) the set of all measurable functions g : E → R for which Z p/2 1/p kgkl p L2 (E) = ∑ |g(x)|2 ρ(x)dx < ∞. ρ

n∈Z

E2n+1 \E2n

The next statement deals with interpolation result which is a consequence of more general statements from [229]( p.127, p. 147) (see also [11]). θ Proposition 1.5.1. Let 1 ≤ p0 , p1 ≤ ∞, 0 ≤ θ ≤ 1, 1p = 1−θ ρ is a p0 + p1 . Suppose that weight function on E. If A is a bounded operator from l pi Lρ2 (E) into σ pi Lρ2 (E) , where i = 0, 1, then it is also bounded from l p Lρ2 (E) to σ p Lρ2 (E) . Moreover,

kAkl p (Lρ2 (E))→σ p (L2 (E)) ≤ kT k1−θ l p0 (L2 (E))→σ p ρ

0

(L2 (E))

kT kθl p1 (L2 (E))→σ p ρ

1

(L2 (E)) .

The following statement is obvious for p = ∞; when 1 ≤ p < ∞ it follows from Lemma 2.11.12 of [197]. Proposition 1.5.2. Let 1 ≤ p ≤ ∞ and let { fk }, {gk } be orthonormal systems in a Hilbert space H. If T ∈ σ p (H), then 1/p . kT kσ p (H) ≥ ∑ |hT fn , gn i| p n

Basic Ingredients

1.6.

23

Singular Integrals in Weighted Lebesgue Spaces

In this section we give some well-known statements regarding singular integrals in weighted Lebesgue spaces. Let Z f (y) H f (x) = p.v. dy R x−y and Z f (τ) 1 dτ SΓ f (t) = p.v. πi Γ τ − t be the Hilbert transform and Cauchy singular integral operator respectively, where Γ = {t ∈ C : t = t(s), 0 ≤ s ≤ l ≤ ∞} is a smooth Jordan Curve on which the arc–length is chosen as a parameter and l is the length of Γ. Definition 1.6.1. We say that the weight w ∈ A p (R), 1 0

1 := sup a∈R 2r r>0

Z a+r

w(s)ds

a−r

1/p

1 2r

Z a+r

1−p′

w

(s)ds

a−r

1/p′

< ∞.

Theorem 1.6.1 ([95]). Let 1 < p < ∞. Then the Hilbert transform H is bounded in if and only if w ∈ A p (R).

Lwp (R)

The next statement is from [109]. Theorem 1.6.2. Let 1 < p < ∞, l ≤ ∞. Suppose that Γ is a smooth curve. Then the inequality Z l 0

|SΓ f (t(s))| p w(s)ds ≤ c

holds if and only if w ∈ A p (0, l), that is, Bl := sup

1 |I|

Z

w(s)ds I

1/p

Z l

| f (t(s))| p w(s)ds

0

1 |I|

Z

′

w1−p (s)ds I

1/p′

< ∞,

where the supremum is taken over all intervals in (0, l). For the next statement see, e.g., [172], [46], [49], p. 553. Proposition 1.6.1. Let 1 0

and sup t>0

Z

Z

|x|>t

v(x) dx |x| p

v(x)dx |x|
1/p Z

1/p Z

1−p′

w |x|
′

|x|>t

1/p′

<∞

1/p′

< ∞.

(x)dx

w1−p (x) dx |x| p′

Suppose that for the operator K f (x) = p.v.

Z

Rn

k(x − y) f (y), dy

(1.6.1)

24

Alexander Meskhi

the inequality

K f 2 n ≤ c f 2 n , f ∈ C0∞ (Rn ) L (R ) L (R )

(1.6.2)

holds and the kernel k satisfies the following two conditions: (i) there exists a positive constant A such that the inequality ∂ α k(x) ≤ A|x|−n−α ∂x

(1.6.3)

holds for all x ∈ Rn , x 6= 0, and |α| ≤ 1; (ii) there exists a positive constant b and an unit vector u0 such that k(x) ≥ b|x|−n

(1.6.4)

when x = λ · u0 with −∞ < λ < +∞. It is easy to see that the Riesz transforms R j f (x) = lim γn r→0

Z

Rn \B(x,r)

xj −yj f (y) dy, j = 1, . . . , n, |x − y|n+1

(1.6.5)

where x = (x1 , . . . , xn ) ∈ Rn , γn = Γ[(n + 1)/2]/π(n+1)/2 , satisfy conditions (1.6.2)–(1.6.4). If n = 1, then R1 f (x) is the Hilbert transform H. Definition 1.6.2. Let 1 < p < ∞. We say that the weight w belongs to A p (Rn ) if

1 B := sup |B|

Z

w(x) dx B

1/p

1 |B|

Z

1−p′

w

(x) dx

B

1/p′

< ∞,

(1.6.6)

where the supremum is taken over all balls B in Rn and |B| is a volume of B. Theorem 1.6.3 ([224], p. 205, p. 210). If conditions (1.6.2) and (1.6.3) are satisfied and w ∈ A p (Rn ), then the operator K is bounded in Lwp (Rn ). Further, if (1.6.2)–(1.6.4) hold and K is bounded in Lwp (Rn ), then w ∈ A p (Rn ). The following statements will be useful in Chapter 6. Lemma 1.6.1 ([224], Section 4.6). Let condition (1.6.3) be satisfied and let u0 be the unit vector in Rn . Then by choosing u = tu0 , with t fixed sufficiently large we can guarantee that 1 |k(r(u + v)) − k(ru)| ≤ |k(ru)| 2 whenever r ∈ R \ {0} and |v| ≤ 2. Lemma 1.6.2. Let 1 < p < ∞ and let conditions (1.6.3) and (1.6.4) be satisfied. Then from the boundedness of the operator K given by (1.6.1) from Lwp (Rn ) to Lvp (Rn ) it follows ′ that w1−p is locally integrable. R

′

Proof. Let B := B(0, r). Suppose that I(r) := B wR1−p (x)dx = ∞ for some positive number r. Then there exists g ∈ L p (B), g ≥ 0, such that B w−1/p g = ∞. Let us assume that fr (y) = g(y)w−1/p (y)χB (y) and B′ = B(ru, r), where u = tu0 (t is from Lemma 1.6.1 and u0 is the unit vector taken so that (1.6.4) holds). Obviously, x = ru + ux′ for x ∈ B′ and y = ry′

Basic Ingredients

25

for y ∈ B, where |x′ | < 1 and |y′ | < 1. Thus x − y = r(u + v) with |v| < 2 and consequently, Lemma 1.6.1 yields |K fr (x)| ≥ 12 ( fr )B |k(ru)| for all x ∈ B′ . Hence by (1.6.4) the following estimates hold: kK fr kLvp (Rn ) ≥ kχB′ (x)K fr (x)kLvp (Rn ) ≥

b (vB′ )1/p fB = ∞. 2rt

On the other hand, k fr kLwp (Rn ) = kgkL p (B) < ∞. Finally we conclude that I(r) < ∞ for all r > 0.

1.7.

Notes and Comments on Chapter 1

Necessary and sufficient conditions guaranteeing the two-weight inequality for the Hardy operator defined on the semi–axis were derived in [168], [18], [108], [158] (see also the survey paper [140], monographs [188], [145], [144], [49], [40] and references therein). For the estimates of the measure of non-compactness for various operators in Lebesgue spaces we refer to [39], [98], [99], [228], [41], [187], [62], [161], [163], [150] (see also [40], [49] and references therein). In the paper [92] it was proved that β(I) = kIk, where I is the identity operator from ∗ k,p np . From this fact it follows that W0 (Ω) to L p (Ω), k ∈ N, 1 ≤ p < ∞, kp < n and p∗ = n−kp all entropy numbers of I are equal to kIk. Two-sided estimates of entropy and approximation numbers of the embedding operators between l p and between Sobolev spaces were established in [15], [64], [77], [104], [151], etc. In the monograph [64] it was shown that if Ω is a bounded domain in Rn and q ∈ [p, p∗ ], np where p∗ = n−kp , then the entropy numbers ek (I) of I : W0m,p (Ω) → Lq (Ω) satisfy ek (I) ≈ m k− n . Lower and upper estimates of a Schatten-von Neumann ideal norms for the Hardy-type operators were derived in [182], [63], [152] (see also the survey paper [226] and references therein). This problem for one–sided potentials was studied in [181] and for kernel operators involving one–sided potentials was investigated in [162] (see also the monograph [49] and references therein). For two-sided estimates of singular, entropy and approximation numbers for the Hardy and potential-type operators we refer to the papers [13], [14], [16], [41], [42], [45], [55], [65], [175], [149], [67], [181], [34], [82], [83], [164], [49], etc. For estimates of the entropy numbers for the weighted discrete Hardy operators see [25]. In this paper some probabilistic applications of these estimates are also given. The asymptotic behaviour of the entropy numbers of diagonal operators generated by logarithmically decreasing sequences were presented in [20], [146]. The space L p(·) is the special case of the Musielak-Orlicz space. The basis of the variable exponent Lebesgue spaces were developed by W. Orlicz and J. Musielak, H. Hudzik (see [189], [174], [94]). The boundedness of the Hardy-Littlewood maximal operator in unweighted L p(x) spaces was established in the papers [29], [180], [27], [22], [89], [137]. The same problem for classical integral operators has been investigated in the papers [26], [30], [32], [196], [204], [205], [120]–[123], [60], [61], [167], [54] (see also [111], [208] and references therein). In

26

Alexander Meskhi

[121] it was shown that if p satisfies the weak Lipschitz condition on Γ, where Γ is a finite rectifiable curve, then SΓ is bounded in L p(x) (Γ) if and only if Γ is a Carleson curve, i.e., ν(Γ ∩ B(t, Γ)) < ∞. r t∈Γ, r>0 sup

The one-weight problem for the Hilbert transform in classical Lebesgue spaces was solved in [95]. In [24] it was proved that the Calder´on- Zygmund singular operator is bounded in Lwp (Rn ), 1 < p < ∞, if w ∈ A p (Rn ). The necessity of the condition w ∈ A p (Rn ) for the boundedness of R j was established in [73], p. 417 (see also [224] for the related topics). The essential norm of the Cauchy integral kST kK (L p (T )) , where where T is the unit n circle, has been calculated in [79],[80] for p = 2n and p = 2n2−1 , where Γ = T is the unit circle (see also [81]). In these papers a lower estimate for kST kK (L p (T )) has been also derived for all p ∈ (1, ∞). The value of the norm of ST acting in L p (T ) (1 < p < ∞) was found in [195]. Estimates of the norm for the Ahlfors-Beurling operators and Riesz transforms were studied in [36]–[38], [96], [10], [177], [192], [193], [194], [35].

Chapter 2

Maximal Operators This chapter deals with lower estimates of the the measure of non-compactness for the Hardy–Littlewood and fractional maximal operators in weighted Lebesgue spaces. In particular, we conclude that there is no weight pair for which these operators are compact from one weighted Lebesgue space into another one. Examples of weight pairs for which appropriate estimates hold are also given.

2.1.

Maximal Functions on Euclidean Spaces

Given any measurable function f on a domain Ω ⊆ Rn we define the maximal operators MΩ as follows: Z 1 | f (y)|dy, MΩ f (x) = sup B∩Ω∋x |B| B∩Ω

where the supremum is taken over all balls B in Rn with x ∈ B ∩ Ω. If Ω = Rn , then we use the notation MΩ := M. The next result is due to B. Muckenhoupt ([169]).

Theorem 2.1.1. Let 1 < p < ∞. Then the operator M is bounded in Lwp (Rn ) if and only if w ∈ A p (Rn ) (see Definition 1.6.2). The following statement can be found, e.g., in [73]: Theorem 2.1.2. Let 1 < p < ∞. Then the operator MΩ is bounded in Lwp (Ω) if and only if w ∈ A p (Ω) , i.e.

1 sup |B ∩ Ω| B

Z

w(x)dx B∩Ω

1/p

1 |B ∩ Ω|

Z

1−p′

w B∩Ω

1/p′ (x)dx < ∞.

(2.1.1)

Now we formulate and prove our main results concerning the maximal operators defined on Ω.

28

Alexander Meskhi

Theorem 2.1.3 Let 1 < p < ∞. Suppose that Ω be a bounded domain in Rn . Then there is no weight pair (v, w) on Ω for which the operator M is compact from Lwp (Ω) to Lvp (Ω). Moreover, if MΩ is bounded from Lwp (Ω) to Lvp (Ω), then the following estimate holds 1 kMΩ kκ ≥ sup lim a∈Ω r→0 |B(a, r)|

Z

v(x)dx B(a,r)

1/p Z

1−p′

w

(x)dx

B(a,r)

1/p′

.

Proof. It is known (see e.g. [169], [212], [214]) that if MΩ is bounded from Lwp (Ω) to ′ then v, w1−p ∈ Lloc (Ω). Further, let λ > kMΩ kκ(Lwp (Ω),Lvp (Ω)) and a ∈ Ω. By Lemma 1.2.8 we have that there exists a constant β such that if 0 < τ < β, then Lvp (Ω),

Z

B(a,τ)∩Ω

v(x) MΩ f

p

(x)dx ≤ λ

p

Z

Ω

| f (x)| p w(x)dx.

(2.1.2)

′

Consequently, putting f (x) = χB(a,τ)∩Ω (x)w1−p (x) in (2.1.2) we find that |B(a, τ)|

−p

Z

v(x)dx B(a,τ)

Z

1−p′

w

(x)dx

B(a,τ)

p−1

≤ λp

for all a ∈ Ω and sufficiently small τ. The Lebesgue differentiation theorem completes the proof. Corollary 2.1.1. Let p = 2, Ω = (−1, 1), v(x) = w(x) = |x|α , where −1 < α < 1. Then kMΩ kκ ≥

1 . (1 − α2 )1/2

For Ω = Rn we have Theorem 2.1.4. Let 1 < p < ∞ and let Ω = Rn . Then there is no weight pair (v, w) on such that M is compact from Lwp (Rn ) to Lvp (Rn ). Moreover, if M is bounded from Lwp (Rn ) to Lvp (Rn ), then the estimate Rn

kMkκ ≥ max{ sup lim I1 (a, τ); sup lim I2 (a, τ)} a∈Rn τ→∞

a∈Rn τ→0

holds, where 1 I1 (a, τ) = |B(a, τ)| 1 I2 (a, τ) = |B(a, 2τ)|

Z

Z

v(x)dx B(a,τ)

1/p Z

v(x)dx B(a,2τ)\B(a,τ)

1−p′

(x)dx

w B(a,τ)

1/p Z

1−p′

w B(a,τ)

1/p′

(x)dx

;

1/p′

.

Proof. Let λ > kMkκ . By Lemma 1.2.7 there are constants β1 and β2 such that Z

B(a,τ)

v(x)(M f (x)) p dx ≤ λ p

Z

Rn

| f (x)| p w(x)dx

(2.1.3)

Maximal Operators and

Z

v(x)(M f (x)) dx ≤ λ p

Rn \B(a,t)

p

Z

Rn

29

| f (x)| p w(x)dx

(2.1.4)

for all τ ≤ β1 and t > β2 . Now observe that by (2.1.3) and the arguments used in the proof of Theorem 1.2.3 we find that sup lim I1 (a, τ) ≤ λ, a∈Rn τ→0

while (2.1.4) implies that Z

B(a,2t)\B(a,t)

v(x)(M f (x)) p dx ≤ λ p

Z

Rn

| f (x)| p w(x)dx.

(2.1.5)

′

Hence, due to the definition of M and putting f (x) = χB(a,t) (x)w1−p (x) in (2.1.5), we find that Z

Z

v(x)dx

B(a,2t)\(B(a,t)

′

p−1

w1−p (x)dx

B(a,t)

|B(a, 2t)|−p ≤ λ p .

Passing t → ∞ and taking the supremum over all a ∈ Rn we derive the desired result.

We say that the measure µ on Rn satisfies a doubling condition (µ ∈ DC(Rn )) if there exists a positive constant b such that µB(a, 2r) ≤ bµ(a, r) for all a ∈ Rn and r > 0. Remark 2.1.1. It is known (see e.g. [227], p.21, [236]) that if µ ∈ DC(Rn ), then µ ∈ RD(Rn ) (reverse doubling condition), i.e. there exist constants η1 , η2 > 1 such that µB(a, η1 r) ≥ η2 µB(a, r) for all a ∈ Rn and r > 0. Remark 2.1.2. Analyzing the proof of Theorem 2.1.4 we notice that the constant 2 in |B(a, 2τ)| of the expression I2 (a, τ) might be replaced by some constant σ > 1. Theorems 2.1.3, 2.1.4 and Remarks 2.1.1 and 2.1.2 yield the next statement. Corollary 2.1.2. Let 1 < p < ∞ and let B be the constant defined by (1.6.6). Suppose that Ω = Rn and (1.6.6) holds. Then kMkK (Lwp (Rn )) ≥ max{ sup lim I(a, τ);CB,n ¯ sup lim I(a, τ)}, a∈Rn τ→∞

where I(a, τ) := |B(a, τ)|

−1

Z

w(x)dx B(a,τ)

a∈Rn τ→0

1/p Z

¯ and n. and CB,n ¯ is a constant depending only on B

1−p′

w B(a,τ)

(x)dx

1/p′

30

Alexander Meskhi Proof. By Theorem 2.1.4 it suffices to show that w(B(a, η1 τ) \ B(a, τ)) ≥ cw(B(a, τ)), R

where w(E) := E w and the positive constant c depends only on B¯ and n. But it is easy to verify that if w ∈ A p (Rn ), then ¯ w(B(a, 2τ)) ≤ cn Bw(B(a, τ)) for all τ > 0 and a ∈ Rn . By Remark 2.1.1 we have w(B(a, η1 τ)) \ B(a, τ)) ≥ (η2 − 1)w(B(a, τ)), ¯ where η2 depends only on n and B. Corollary 2.1.3. Let 1 < p < ∞ and let Ω be a bounded domain. Suppose that w ∈ A p (Ω) (i.e. (2.1.1) holds ). Then kMΩ kκ(Lwp (Ω)) Z 1/p Z 1/p′ −1 1−p′ ≥ sup lim |B(a, τ)| w(x)dx w (x)dx . a∈Rn τ→0

B(a,τ)

B(a,τ)

Let us now estimate the measure of non-compactness for the fractional maximal function Z 1 | f (y)|dy, Mα,Ω f (x) = sup 1−α/n B∩Ω B∩Ω∋x |B| where Ω ⊆ Rn is a domain.

The next statement is well-known (see [171]): Theorem 2.1.5. Let 1 < p < ∞ and let 0 < α < n/p. Suppose that p∗ = Ω=

Rn .

Then Mα is bounded from

1 B˜ := sup |B| B

Z

B

Lρpp/p∗ (Rn )

ρ(x)dx

1/p∗

to

∗ Lρp (Rn )

1 |B|

Z

and

if and only if

−p′ /p∗

B

np n−αp

ρ

(x)dx

1/p′

< ∞,

(2.1.6)

where the supremum is taken over all balls B ⊂ Rn . The next statement can be derived in the same way as in the case of the HardyLittlewood maximal operator; therefore we omit the proofs. np Theorem 2.1.6. Let 1 < p < ∞, 0 < α < n/p, p∗ = n−αp . Suppose that Ω is a bounded n domain in R . Then there is no weight pair (v, w) on Ω such that Mα,Ω is compact from ∗ ∗ Lwp (Ω) to Lvp (Ω). Further, suppose that Mα,Ω is bounded from Lwp (Ω) to Lvp (Ω). Then

kMα,Ω kκ(L p (Ω)→L p∗ (Ω)) w

≥ sup lim |B(a, τ)|α/n−1 a∈Ω τ→0

Z

v(x)dx B(a,τ)

v

1/p∗ Z

′

w1−p (x)dx B(a,τ)

1/p′

.

Maximal Operators

31

Corollary 2.1.4. Let 1 < p < ∞, 0 < α < n/p, p∗ =

that Ω is a bounded

domain in

Rn

and Mα,Ω is bounded from

Lρpp/p∗ (Ω)

to

np n−αp . Suppose ∗ Lρp (Ω). Then

kMα,Ω kκ(L p

p∗ ∗ (Ω)→Lρ (Ω)) ρ p/p

≥ sup lim |B(a, τ)|α/n−1 a∈Ω τ→0

Z

B(a,τ)

ρ(x)dx

1/p∗ Z

′

B(a,τ)

∗

ρ−p /p (x)dx

1/p′

.

In the case of Ω = Rn we have the following statement. np Theorem 2.1.7.Let 1 < p < ∞, 0 < α < n/p, p∗ = n−αp . Suppose that Ω = Rn and that (2.1.6) holds. Then there is no weight pair (v, w) such that the operator Mα,Ω is compact ∗ Lwp (Rn ) to Lvp (Rn ). Moreover, the following estimate holds: (α)

(α)

kMα,Ω kκ(L p (Ω),L p∗ (Ω)) ≥ max{ sup lim I1 (a, τ); sup lim I2 (a,t)}, w

a∈Rn τ→0

v

a∈Rn t→∞

where (α) I1 (a, τ)

α/n−1

:= |B(a, τ)|

Z

v(x)dx B(a,τ)

1/p∗ Z

1−p′

(x)dx

w B(a,τ)

1/p′

,

and (α) I2 (a, τ) :=

α/n−1

|B(a, 2τ)|

Z

v(x)dx B(a,2τ)\B(a,τ)

1/p∗ Z

1−p′

w

(x)dx

B(a,τ)

1/p′

.

np Corollary 2.1.5. Let 1 < p < ∞, 0 < α < n/p and p∗ = n−αp . Suppose that B˜ < ∞, where B˜ is the constant defined by (2.1.6). Then the inequality

kMα kκ(L p

∗ (R ρ p/p

n ),L p (Rn )) ρ

≥ max{ sup lim J (α) ,CB,n,α,p sup lim J (α) (a, τ)} ˜ a∈Rn τ→∞

a∈Rn τ→0

holds, where J

(α)

α/n−1

(a, τ) = |B(a, τ)|

Z

B(a,τ)

ρ(x)dx

1/p∗ Z

−p′ /p∗

B(a,τ)

e n, α and p. and the positive constant CB,n,α,p depends on B, ˜

ρ

(x)dx

1/p′

Proof. First observe that the condition (2.1.6) implies ρ ∈ A1+p∗ /p′ (Rn ). Hence the R measure ρ(E) = E ρ(x)dx satisfy the doubling condition. Applying Remark 2.1.1, Theorems 2.1.5, 2.1.7 and the arguments from the proof of Corollary 2.1.2 we have the desired result.

32

Alexander Meskhi

2.2.

One–sided Maximal Functions

In this section we deal with the one-sided maximal functions: Nα+ f (x) = sup h>0

Nα− f (x)

1 h1−α

= sup h>0

1 h1−α

Z x+h

| f (y)|dy, x ∈ R,

x

Z x

| f (y)|dy, x ∈ R,

x−h

where 0 ≤ α < 1. If α = 0, then we denote N + := N0+ and N − := N0− . Definition 2.2.1. Let 1 0 such that Zx 1/p Zx+h 1/p′ 1 1 1−p′ w(t)dt w (t)dt ≤ c; x ∈ R, h > 0. h h x

x−h

Further, w ∈

A− p (R)

if

Zx+h 1/p Zx 1/p′ 1 1 1−p′ ≤ C; x ∈ R, h > 0, w(t)dt w (t)dt h h x

x−h

for some positive constant C. Theorem 2.2.1 ([216], [3]). Let 1 < p < ∞. Then (i) N + is bounded in Lwp (R) if and only if w ∈ A+ p (R); (ii) N − is bounded in Lwp (R) if and only if w ∈ A− p (R). Definition 2.2.2. Let p and q be constants such that 1 < p < ∞, 1 < q < ∞. We say that a weight ρ ∈ A+ pq (R) if Zx 1q Zx+h 1′ p 1 1 q −p′ < ∞. ρ (t)dt ρ (t)dt sup h 0
x

Further, ρ ∈ A− pq (R+ ) if Zx+h 1q Zx 1′ p 1 1 q −p′ sup ρ (t)dt ρ (t)dt < ∞. h 0
x−h

p Theorem 2.2.2 ([3]). Suppose that 0 < α < 1, 1 < p < α1 and q = 1−αp . Then Nα+ is q − bounded from Lρpp (R) to Lρq (R) if and only if ρ ∈ A+ pq (R). Further, Nα is bounded from q p Lρ p (R) to Lρq (R) if and only if ρ ∈ A− pq (R). In the next statement we assume that the symbol Nα denotes one of the operators Nα+ , Nα− .

Maximal Operators

33

Theorem 2.2.3. Let 1 < p ≤ q < ∞,and α = 1/p − 1/q. Then there is no weight pair q (v, w) defined on R for which the operator Nα is compact from Lwp (R) to Lv (R). Moreover, q p if Nα is bounded from Lw (R) to Lv (R) for some weight functions v and w, then Z a 1/q Z a+h 1/p′ 1 1 + α−1 1−p′ kNα kκ ≥ 2 sup lim v(x)dx w (x)dx ; h a a∈R h→0 h a−h Z a+h 1/q Z a 1/p′ ′ 1 1 v(x)dx w1−p (x)dx . h a−h a∈R h→0 h a

kNα− kκ ≥ 2α−1 sup lim

Proof. We prove the theorem for Nα ≡ Nα+ . Let λ > kNα+ kκ . First observe that the q boundedness of Nα+ from Lwp (R) to Lv (R) implies J(t) :=

Z t

−t

′

w1−p (τ)dτ < ∞

for every t > 0. Indeed, if J(t) = ∞ for some t > 0, then there is a non-negative function g on (−t,t) belonging to L p ([−t,t]) such that Z t

−t

g(τ)w−1/p (τ)dτ = ∞.

Assuming now that ft (y) = g(y)w−1/p (y)χ(−t,t) (y) we find that kNα+ ft kLvq (R) ≥ kχ(−2t,−t) Nα+ ft kLvq (R) ≥ ct On the other hand,

α−1

Z

−t

v(x)dx −2t

k ft k

p Lw (R)

1/q Z

=

Z t

−t

t −t

g(y)w−1/p (y)dy = ∞

g p (x)dx < ∞ q

which contradicts the boundedness of Nα+ from Lwp (R) to Lv (R). Consequently, we conclude that J(t) < ∞ for every t > 0. Repeating the arguments of the proof of Theorem 2.1.4 we have that Z q/p Z a+τ + q q p v(x)(Nα f (x)) dx ≤ λ | f (x)| w(x)dx . a−τ

R

for all a ∈ R and small τ, where λ > kNα+ kκ . Hence Z a Z a+τ q Z q/p 1 q p v(x)dx f (t)dt dx ≤ λ f (x)w(x)dx , (2τ)1−α a−τ a R

(2.2.1)

where f ≥ 0. ′ Assuming f (t) = χ(a,a+τ) (t)w1−p (t) in (2.2.1), passing now to the limit when τ → 0 and taking the supremum with respect to all a we have the desired estimate for Nα+ . The Lebesgue differentiation theorem completes the proof. The proof for Nα− is similar to that for Nα+ .

34

Alexander Meskhi

2.3.

Maximal Operator on Homogeneous Groups

Let G be a homogeneous group and let Mα f (x) = sup B∋x

1 α

|B|1− Q

Z

| f (y)|dy, x ∈ G, 0 ≤ α < Q,

B

where the supremum is taken over all balls B ⊂ G containing x. If α = 0 then Mα becomes the Hardy-Littlewood maximal function which will be denoted by M. The following statements hold (see, e.g., [70], [76]). Theorem 2.3.1. Let 1 < p < ∞. Then the Hardy-Littlewood maximal function M is bounded in Lρp (G) if and only if ρ ∈ A p (G) i.e. 1 Z 1p 1 Z 1′ ′ ρ(x)dx ρ1−p (x)dx p < ∞, sup |B| |B| B B

(2.3.1)

B

where the supremum is taken over all balls B ⊂ G. Theorem 2.3.2. Let 1 < p < ∞ , 0 < α < Qp . Then the fractional maximal function Mα q

is bounded from Lρpp (G) to Lρq (G), where q = sup B

Qp Q−αp ,

if and only if ρ ∈ A pq (G) i.e.

1q 1 Z 1′ 1 Z ′ ρq (x)dx ρ−p (x)dx p < ∞. |B| |B| B

(2.3.2)

B

Now we formulate and prove the main results of this subsection. Theorem 2.3.3. Let 1 < p < ∞. Suppose that M is bounded from Lwp (G) to Lvp (G). Then there is no weight pair (v, w) such that M is compact from Lwp (G) to Lvp (G). Moreover, the inequality 1 a∈G τ→0 |B(a, τ)|

kMkκ(Lwp (G),Lvp (G)) ≥ sup lim

Z

v(x)dx

B(a,τ)

1p Z

′

w1−p (x)dx

B(a,τ)

1′ p

holds. Proof. Suppose that λ > kMkκ(Lwp (G)→Lvp (G)) and a ∈ G. By Lemma 1.2.7 we have that Z

v(x)(M f (x)) dx ≤ λ p

B(a,τ)

p

Z

| f (x)| p w(x)dx

B(a,τ)

for all τ (τ ≤ β) and all f supported in B(a, τ). First observe that Z

′

w1−p (x)dx < ∞

B(a,τ)

for all τ > 0 (see also, for example, [168], [227], [76], Ch. 4).

(2.3.3)

Maximal Operators

35

′

Further, substituting f (y) = χB(a,r) (y) w1−p (y) in inequality (2.3.3)we find that 1 |B(a, τ)| p

Z

v(x)dx

Z

′

w1−p (x)dx

B(a,τ)

B(a,τ)

p−1

≤ λp.

This inequality and Lebesgue differentiation theorem for homogeneous groups (see [70], p. 67) yield the desired result. Qp . Suppose that Mα is Theorem 2.3.4. Let 1 < p < ∞, 0 < α < Q/p and let q = Q−αp q p bounded from Lw (G) to Lv (G). Then there is no weight pair (v, w) such that Mα is compact q from Lwp (G) to Lv (G). Moreover, the inequality

1

kMα kκ ≥ sup lim

α

−1 a∈G τ→0 |B(a, τ)| Q

Z

v(x)dx

1q Z

′

w1−p (x)dx

B(a,τ)

B(a,τ)

1′ p

holds. The proof of this statement is similar to that of Theorem 2.3.3; therefore it is omitted. Example 2.3.1. Let 1 < p < ∞, v(x) = w(x) = r(x)γ , where −Q < γ < (p − 1)Q. Then 1 1 ′ ′ −1 kMkκ(Lwp (G)) ≥ Q (γ + Q) p (γ(1 − p ) + Q) p .

Indeed, first recall that the fact that |B(e, 1)| = 1 and Proposition 1.1.1 implies σ(S) = Q, where S is the unit sphere in G and σ(S) is its measure. Taking into account Theorems 2.3.1 and 2.3.3 we have Z 1/p Z 1/p′ 1 1−p′ p w(x)dx w (x)dx kMkκ(Lw (G)) ≥ lim τ→0 |B(e, τ)| B(e,τ) B(e,τ) = σ(S) lim τ

−Q

τ→0

Z

τ

t 0

γ+Q−1

dt

1/p Z

τ

t

γ(1−p′ )+Q−1

0

1 1 ′ ′ −1 = Q (γ + Q) p (γ(1 − p ) + Q) p .

dt

1/p′

2.4. Notes and Comments on Chapter 2 This chapter is based on the results of the papers [58], [43], [5]. A result analogous to that of [58] has been obtained in [184], [185] for the HardyLittlewood maximal operators with more general differentiation bases on symmetric spaces. The one-weight problem for the Hardy-Littlewood maximal functions was solved by B. Muckenhoupt [169] (for maximal functions defined on quasimetric measure spaces with doubling condition see, e.g., [227]) and for fractional maximal functions and Riesz potentials by B. Muckenhoupt and R. L. Wheeden [171]. Theorem 2.3.2 for Euclidean spaces was derived in [171] and for homogeneous groups and quasimetric measure spaces with doubling condition, for instance, in [70], Ch. 6; [76], Ch. 4.

36

Alexander Meskhi

Two-weight criteria involving the operator itself for the Hardy-Littlewood maximal operator have been established in [212] (see [172], [191] for another type of sufficient conditions). The two-weight problem for the fractional maximal functions has been solved in [212], [214], [217], [236] (see also [78], [76], Ch. 4, for quasimetric measure spaces). Sharp estimates for the Hardy-Littlewood maximal functions were obtained in [84], [85]. In [19] the author found the sharp dependence of kMkLwp (Rn ) on B (see (1.6.6) for the definition of B ) in Theorem 2.1.2 for Ω = Rn . In particular, if 1 < p < ∞, then kMkLwp (Rn ) ≤ ′ Cp,n (B) p and the exponent p′ is the best possible. This result was used in [35] to establish sharp one-weighted estimates for the Hilbert, Beurling and martingale transforms.

Chapter 3

Kernel Operators on Cones Let E be a cone in a homogeneous group G (see Section 1.1 for the definition). We denote Et := {y ∈ E : r(y) < t}. In the sequel we also use the notation: Sx := Er(x)/2c0 , Fx := Er(x) \Sx , where the constant c0 comes from the triangle inequality for r (see Section 1.1). q In this chapter boundedness/compactness criteria from L p (E) to Lv (E) are established for the operator Z K f (x) =

k(x, y) f (y)dy, x ∈ E,

(A)

Er(x)

with positive kernel k, where 1 < p, q < ∞ or 0 < q ≤ 1 < p < ∞, Er(x) and E are certain cones in homogeneous groups and k satisfies conditions which in the one-dimensional case are similar to those of [163]. The measure of non-compactness for K is also estimated from the both sides. We present also two-sided estimates of Schatten-von Neumann norms for the operator with positive kernel Ku f (x) = u(x)

Z

k(x, y) f (y)dy, x ∈ E,

(B)

Er(x)

where u is a measurable function on E. We need some definitions regarding the kernel k. Definition A. Let k be a positive function on {(x, y) ∈ E × E : r(y) < r(x)} and let 1 < λ < ∞. We say that k ∈ Vλ , if there exist positive constants c1 , c2 and c3 such that (i)

k(x, y) ≤ c1 k(x, δ1/(2c0 ) x)

(C)

for all x, y ∈ E with r(y) < r(x)/(2c0 ); k(x, y) ≥ c2 k(x, δ1/(2c0 ) x)

(C′ )

38

Alexander Meskhi

for all x, y ∈ E with r(x)/(2c0 ) < r(y) < r(x); Z

(ii)

kλ (x, y)dy ≤ c3 rQ (x)kλ (x, δ1/(2c0 ) x), λ′ = λ/(λ − 1), ′

′

(D)

Fx

for all x ∈ E. Example A. Let G = Rn , r(xy−1 ) = |x − y|, δt x = tx, x, y ∈ Rn . If k(x, y) = |x − y|α−n , then k ∈ Vλ , where n/λ < α ≤ n. Indeed, it is easy to see that if y ∈ Sx , then |x| ≤ |x − y| + |y| ≤ |x − y| + |x|/2. Hence |x|/2 = k(x, x/2) ≤ k(x, y). Consequently, (C) holds. Further, it ′ is easy to see that (C ) is also satisfied. Moreover, we have Z

Z ∞

′

|x − y|(α−n)λ dy = Fx

0

{y ∈ Fx : |x − y|(α−n)λ′ > ε} dε

Z |x|(α−n)λ′

≤

(· · · ) +

0

For I1 we have I1 ≤

(α−n)λ |x|Z

0

while for I2 we observe that Z ∞

I2 ≤

(α−n)λ′

|x|

≤ c

Z∞

|x|(α−n)λ

′

Z∞

(· · · ) := I1 + I2 .

(α−n)λ′

|x|

′

B(0, |x|) dε = c|x|(α−n)λ′ +n ,

y : |y| < |x|, |x − y| ≤ ε1/(α−n)λ′ } dε

′ ′ 1+n/(α−n)λ ′ εn/(α−n)λ dε = cα,n |x|(α−n)λ ′

= cα,n |x|n+(α−n)λ . Example B. It is easy to see that if the following two conditions are satisfied for k: k(δt x, ¯ δτ y) ¯ ≤ c1 k(δt x, ¯ δs z¯)

(i)

for all t, τ, s, x, ¯ y, ¯ z¯ with 0 < τ < s < t; x, ¯ y, ¯ z¯ ∈ A; (ii)

Z t

kλ (δt x, ¯ δτ y)τ ¯ Q−1 dτ ≤ c2t Q · kλ (δt x, ¯ δt/(2c0 ) x), ¯ t > 0, x¯ ∈ A, ′

t/(2c0 )

′

then k ∈ Vλ . Example C. Let k(x, y) = k(r(x), r(y)) be a radial kernel. It is easy to check that if there exist positive constants c1 and c2 such that ¯ l) ≤ c1 k(s,t), ¯ k(s, 0 < l < t < s,

(i) (ii) then k ∈ Vλ .

Z t

t/(2c0 )

′ ′ k¯ λ (t, s)sQ−1 ds ≤ c2t Q k¯ λ (t,t/(2c0 )), 1 < λ < ∞,

Kernel Operators on Cones

3.1.

39

Boundedness

In this section we establish boundedness/compactness criteria for K acting from L p (E) to q Lv (E), 1 < p ≤ q < ∞. Theorem 3.1.1. Let 1 < p ≤ q < ∞ and let v be a weight on E. Suppose that k ∈ Vp . q Then K is bounded from L p (E) to Lv (E) if and only if Z 1/q ′ Q/p 2j B := sup B( j) := sup v(x)kq (x, δ1/(2c0 ) x)dx < ∞. (3.1.1) j∈Z

j∈Z

E2 j+1 \E2 j

Proof. Sufficiency. Let f ≥ 0 on E. We have Z hZ Z q Z q i q kK f kLq (E) ≤ c v(x) k(x, y) f (y)dy dx + v(x) k(x, y) f (y)dy dx v

E

E

Sx

Fx

:= c(I1 + I2 ). By condition (C) and Theorem 1.3.2 we find that Z Z q Z q/p q , I1 ≤ c v(x)k (x, δ1/(2c0 ) x) f (y)dy dx ≤ c f p (x)dx E

Er(x)

E

while H¨older’s inequality and condition (D) yield Z Z ′ q/p′ Z q/p p I2 ≤ v(x) k (x, y)dy f p (y)dy dx E

≤ c

Z

Fx

Fx

v(x)k (x, δ1/(2c0 ) x)r q

E

j∈Z E2 j+1 \E2 j

≤ c∑

j∈Z

×

Z

Z q/p ′ v(x)kq (x, δ1/(2c0 ) x)rQq/p (x) f p (y)dy dx Fx

′

v(x)kq (x, δ1/(2c0 ) x)rQq/p (x)dx

E2 j+1 \E2 j

Z q/p (x) f p (y)dy dx Fx

Z

≤ c∑

Qq/p′

Z

q/p q f (y)dy ≤ cBq k f kL p (E) . p

E2 j+1 \E2 j−1 /c

0

Sufficiency has been proved. Necessity. To prove necessity we take the functions f j (x) = χE2 j+1 (x). Then simple calculations show that k f j kL p (E) = c2 jQ/p . Further, condition (C) yields Z Z q q kK f j (x)kLq (E) ≥ v(x) f j (y)k(x, y)dy dx v

E2 j+1 \E2 j

≥ c

Z

E2 j+1 \E2 j

Fx

v(x)kq (x, δ1/(2c0 ) x)dx 2 jQq .

Finally, due to the boundedness of K we have the desired result.

40

Alexander Meskhi Now we consider the case q < p.

Theorem 3.1.2. Let 0 < q 1. Suppose that k ∈ Vp . Then the following conditions are equivalent: q (i) K is bounded from L p (E) to Lv (E); (ii) D :=

Z Z E

p/(p−q) (p−q)/pq Qp(q−1)/(p−q) k (y, δ1/(2c0 ) y)v(y)dy r(x) dx < ∞. q

E\Er(x)

Proof. First we show that (ii) ⇒ (i). Suppose that f ≥ 0. Keeping the notation of sufficiency of the proof of Theorem 3.1.1, using Proposition 1.1.2 and condition (C) we have Z q Z q I1 ≤ c v(x)k (x, δ1/(2c0 ) x) f (y)dy dx E

=c

Z ∞ 0

where V (t) := t Q−1

Z V (t)

Z

A

Sx

q F(τ)dτ dt,

t/(2c0 ) 0

v(δt x)k ¯ q (δt x, ¯ δt/(2c0 ) x)dσ( ¯ x), ¯

F(t) := t

Q−1

Z

A

F(δt x)dσ( ¯ x). ¯

Notice that D=

Z ∞ Z 0

∞

t

p/(p−q) (p−q)/(pq) Qp(q−1)/(p−q)+Q−1 V (τ)dτ t dt

Z =c

∞ 0

Z

t

∞

p/(p−q) V (τ)dτ

Z t p(q−1)/(p−q) (p−q)/(pq) (Q−1)(1−p)(1−p′ ) (Q−1)(1−p)(1−p′ ) τ dτ × t dt . 0

Consequently, due to Theorem 1.3.3, H¨older’s inequality and Proposition 1.1.2 we find that

Z =c

Z I1 ≤ c ∞

t

∞

t

p

F (t)dt

0

(Q−1)(1−p)

0

(Q−1)(1−p)

Z

A

q/p

p q/p (Q−1)p f (δt x)dσ( ¯ x) ¯ t dt

Z ∞ Z q/p Z q/p Q−1 p p ≤c t f (δt x)dσ( ¯ x) ¯ dt =c f (x)dx . 0

A

E

Further, applying H¨older’s inequality twice and condition (D) we find that I2 ≤

Z

E

Z q/p Z q/p′ p p′ v(x) f (y)dy k (x, y)dy dx Fx

Fx

Kernel Operators on Cones ≤c =c∑

Z

E

Z

Z q/p ′ p v(x) f (y)dy rQq/p (x)kq (x, δ1/(2c0 ) x)dx Fx

j∈Z E2 j+1 \E2 j

≤ c∑

Z

j∈Z

Z q/p ′ p v(x) f (y)dy r(x)Qq/p kq (x, δ1/(2c0 ) x)dx Fx

q/pZ f (y)dy

′

v(x)rQq/p (x)kq (x, δ1/(2c0 ) x)dx

p

E2 j+1 \E2 j−1 /c

0

Z ≤c ∑

E2 j+1 \E2 j

j∈Z

≤

q ck f kL p (E)

v(x)r

Qq/p′

E2 j+1 \E2 j

∑

Z

j∈Z

q/p f (y)dy p

j∈Z E2 j+1 \E2 j−1 /c0

Z × ∑

41

p/(p−q) (p−q)/p (x)k (x, δ1/(2c0 ) x)dx

v(x)r

q

p/(p−q) (p−q)/p (x)k (x, δ1/(2c0 ) x)dx

Qq/p′

q

E2 j+1 \E2 j

q ¯ q. =: ck f kL p (E) (D)

Besides this, ¯ pq/(p−q) ≤c ∑ 2Qq(p−1) j/(p−q) (D) j∈Z

≤c∑

Z

Z

v(x)k (x, δ1/(2c0 ) x)dx q

E2 j+1 \E2 j

p/(p−q)

r(y)Qp(q−1)/(p−q)

j∈Z E2 j \E2 j−1

Z ×

v(x)k (x, δ1/(2c0 ) x)dx q

E\Er(y)

p/(p−q)

dy

≤ cD pq/(p−q) < ∞. Now we show that (i)⇒(iii). Let vn (x) = v(x)χEn \E1/n (x), where n is an integer with n ≥ 2. Let fn (x) =

Z

vn (z)k (z, δ1/(2c0 ) z)dz q

E\Er(x)

1/(p−q)

r(x)Q(q−1)/(p−q) .

It is easy to see that k f kL p (E) =

Z Z E

p/(p−q) k (y, δ1/(2c0 ) y)vn (y)dy q

E\Er(x)

×r(x)Qp(q−1)/(p−q) dx

1/p

< ∞.

42

Alexander Meskhi ′

On the other hand, by condition (C ) and integration by parts we find that Z Z q 1/q q kK f kLv (E) ≥ vn (x) fn (y)k(x, y)dy dx E

Fx

Z Z q 1/q q ≥c vn (x)k (x, δ1/(2c0 ) x) fn (y)dy dx E

Fx

Z Z Z q =c vn (x)k (x, δ1/(2c0 ) x) E

Fx

1/(p−q) vn (z)k (z, δ1/(2c0 ) z)dz q

E\Er(y)

Q(q−1)/(p−q)

×r(y)

Z Z q ≥c vn (x)k (x, δ1/(2c0 ) x) E

q 1/q dy dx

vn (z)k (z, δ1/(2c0 ) z)dz q

E\Er(x)

Z q 1/q Q(q−1)/(p−q) × r(y) dy dx

q/(p−q)

Fx

Z Z q ≥c vn (x)k (x, δ1/(2c0 ) x) E

q/(p−q) vn (z)k (z, δ1/(2c0 ) z)dz q

E\Er(x)

Qq(p−1)/(p−q)

×r(x) Z c

∞

t

Q−1

0

Z

dx

1/q

Z ∞ vn (δt x)k ¯ (δt x, ¯ δt/(2c0 ) x)dσ( ¯ x) ¯ τQ−1 q

A

t

Z q/(p−q) 1/q q Qq(p−1)/(p−q) × vn (δτ z¯)k (δτ z¯, δτ/(2c0 ) z¯)dσ(¯z) dτ t dt A

Z ∞ Z ∞ Z p/(p−q) Q−1 q d τ vn (δτ x)k ¯ (δτ x, ¯ δτ/(2c0 ) x)dσ( ¯ x) ¯ dτ =c 0

×t Qq(p−1)/(p−q) dt

t

1/q

A

Z p/(p−q) Z ∞Z ∞ vn (δτ x)k ¯ q (δτ x, ¯ δτ/(2c0 ) )dσ(x) ¯ dτ =c τQ−1 0

A

t

×t

Qq(p−1)/(p−q)

dt

1/q

Z ∞ Z ∞ Z Q−1 Q−1 q =c t τ vn (δτ x)k ¯ (δτ x, ¯ δτ/(2c0 ) x)dσ( ¯ x) ¯ dτ p/(p−q) 0

t

A

×t Z Z =c E

Qq(p−1)/(p−q)−Q

dt

1/q

p/(p−q) 1/q Qp(q−1)/(p−q) vn (x)k (x, δ1/(2c0 ) x)dx r(x) dx . q

E\Er(x)

Now the boundedness of K and Fatou’s lemma completes the proof of the implication (i)⇒ (ii).

Kernel Operators on Cones

3.2.

43

Compactness

This section deals with the compactness of the operator K. Theorem 3.2.1. Let 1 < p ≤ q < ∞. Suppose that k ∈ Vp . Then K is compact from q L p (E) to Lv (E) if and only if (i) (3.1.1) holds; (ii) lim B( j) = lim B( j) = 0 (3.2.1) j→−∞

j→+∞

(see (3.1.1) for B( j)). Proof. Sufficiency. Let 0 < a < b < ∞ and represent K f as follows: K f = χEa K ( f χEa ) + χEb \Ea K( f χEb ) +χE\Eb K( f χEb/2c0 ) + χE\Eb K( f χE\Eb/2c ) 0

:= P1 f + P2 f + P3 f + P4 f . For P2 we have P2 f (x) =

Z

k∗ (x, y) f (y)dy,

E

where

k∗ (x, y) =

χEb \Ea (x)χEr(x) (y)k(x, y). Further, observe that

Z Z

S :=

E

≤c

E

Z

Eb \Ea

Z p′ q/p′ k∗ (x, y) dy v(x)dx =

Z

Z q/p′ ′ k p (x, y)dy v(x)dx + c

Z

Eb \Ea

Z

Sx

Eb \Ea

Er(x)

Fx

q/p′ ′ k p (x, y)dy v(x)dx

q/p′ ′ k p (x, y)dy v(x)dx

:= S1 + S2 . Taking into account the condition k ∈ Vp we have Si ≤ c

Z

′

kq (x, δ1/(2c0 ) x)rqQ/p (x)v(x)dx

Eb \Ea

≤ cbq/p

′

Z

Eb \Ea

kq (x, δ1/(2c0 ) x)v(x)dx < ∞, i = 1, 2.

Finally we have S < ∞ and consequently, by Theorem 1.3.5 we conclude that P2 is compact for every a and b. Analogously, we obtain the compactness of P3 . Further, by the arguments used in the proof of Theorem 3.1.1 we have kP1 k ≤ cB

(a)

:= c sup t≤a

Z

Ea \Et

v(x)kq (x, δ1/(2c0 ) x)dx

1/q

′

t Q/p ;

44

Alexander Meskhi kP4 k ≤ cB(b) := c sup t≥b

Therefore

Z

v(x)kq (x, δ1/(2c0 ) x)dx

E\Et

1/q

t Q − bQ

1/p′

.

kK − P2 − P3 k ≤ kP1 k + kP4 k ≤ c B(a) + B(b) .

It remains to show that condition (3.2.1) implies

lim B(a) = lim B(b) = 0.

a→0

b→∞

Let a > 0. Then a ∈ [2m , 2m+1 ) for some integer m. Consequently, B(a) ≤ B(2 for 0 < t < 2m+1 , there exists j ∈ Z, j ≤ m, such that t ∈ [2 j , 2 j+1 ). Hence t qQ/p

′

Z

m+1 )

. Further,

v (x)kq (x, δ1/(2c0 ) x))dx

E2m+1 \Et

≤ 2( j+1)qQ/p

′

Z

m

∑

k= j

≤ cBq (k)2 jqQ/p

′

v(x)kq (x, δ1/(2c0 ) x)dx

E2k+1 \E2k ∞

∑ 2−kqQ/p

′

= cBq (k).

k= j

Taking into account this estimate we shall find that m

B(2 ) ≤ sup(B(k))q =: S(m). k≤m m

If a → 0, then m → −∞. Consequently, S(m) → 0, which implies that B(2 ) → 0 as a → 0. Finally we have that B(a) → 0 as a → 0. Now we take arbitrary b > 0. Then b ∈ [2m , 2m+1 ) for some m ∈ Z. Hence B(b) ≤ B(2m ) . Further, for t > 2m , there exists k ≥ m, k ∈ Z such that t ∈ [2k , 2k+1 ). For such a t we have Z q/p′ v(x) kq (x, δ1/(2c0 ) x)dx t Q − 2mQ E\Et

≤

Z

v(x)kq (x, δ1/(2c0 ) x)dx

E\E2k

≤ c2kqQ/p

′

∞

∑

j=k

Z

2(k+1)Q − 2mQ

q/p′

v(x)k(x, δ1/(2c0 ) x)q (x)dx

E2 j+1 \E2 j

≤ cBq (k)2kqQ/p

′

∞

∑ 2− jqQ/p

′

≤ cBq (k).

j=k

Consequently, B(2m ) ≤ sup B(k). From the last inequality we conclude that the condition k≥m

lim B(k) = 0 implies lim B(b) = 0.

k→+∞

b→∞

Kernel Operators on Cones

45

Necessity. Let K be compact. As (3.1.1) is obvious we have to prove (3.2.1). Let j ∈ Z ′ and put f j (y) = χE2 j+1 \E2 j−1 /c (y)2− jQ/p . Then for φ ∈ L p (E), we have 0

Z f j (y)φ(y)dy ≤ E

Z

E2 j+1 \E2 j−1 /c

=

Z

0

E2 j+1 \E2 j−1 /c

0

Z

1/p | f j (y)| p dy

1/p′ ′ |φ(y)| p dy

E2 j+1 \E2 j−1 /c

0

1′ ′ p →0 |φ(y)| p dy

as j → −∞ or j → +∞. On the other hand, kK f j kLvq (E) ≥

Z

E2 j+1 \E2 j

Z

h ≥ c

E2 j+1 \E2 j

h ≥ c

Z

E2 j+1 \E2 j

≥ c

Z

1/q q K f j (x) v(x)dx

Z

Fx

q i1/q k(x, y) f j (y)dy v(x)dx

Z q i1/q kq (x, δ1/(2c0 ) x) f j (y)dy v(x)dx Fx

kq (x, δ1/(2c0 ) x)v(x)dx

E2 j+1 \E2 j

1/q

′

2 jQ/p = cB( j).

As a compact operator maps a weakly convergent sequence into a strongly convergent one, we conclude that (3.2.1) holds. For q 1. Suppose that k ∈ Vp . Then the following conditions are equivalent: q (i) K is compact from L p (E) to Lv (E); (ii) D :=

Z Z E

p/(p−q) (p−q)/pq Qp(q−1)/(p−q) k (y, δ1/(2c0 ) y)v(y)dy r(x) dx < ∞. q

E\Er(x)

Proof follows immediately from Theorems 3.1.2 and 1.3.6.

3.3.

Schatten–von Neumann norm Estimates

In this section we give necessary and sufficient conditions guaranteeing two-sided estimates of the Schatten-von Neumann norms for the operator given by (B). We denote by k0 the function r(x)Q k2 (x, δ1/(2c0 ) x).

46

Alexander Meskhi

Theorem 3.3.1. Let 2 ≤ p < ∞ and let k ∈ V2 . Then Ku ∈ σ p (L2 (E)) if and only if u ∈ l p (Lk20 (E)). Moreover, there exist positive constants b1 and b2 such that b1 kukl p (L2

≤ kKu kσ p (L2 (E)) ≤ b2 kukl p (L2

k0 (E))

k0 (E))

.

Proof. Sufficiency. First observe that J(x) :=

Z

k2 (x, y)dy ≤ ck0 (x),

Er(x)

where c is a positive constant. Indeed, splitting the integral over Er(x) into two parts and taking into account the condition k ∈ V2 we have J(x) =

Z

k2 (x, y)dy +

Sx

Z

Fx

k2 (x, y)dy ≤ c1 k0 (x) + c2 k0 (x) ≤ c3 k0 (x).

Hence, the Hilbert-Schmidt formula yields kKu kσ2 (L2 (E)) =

Z

E×Er(x)

=

Z E

1/2 2 k(x, y)u(x) dxdy

Z 1/2 Z 1/2 2 u (x) k (x, y)dy dx ≤c u2 (x)k0 (x)dx 2

Er(x)

= c ∑

Z

E

u2 (x)k0 (x)dx

n∈Z E2n+1 \E2n

1/2

= ckukl 2 (L2

k0 (E))

.

On the other hand, by Theorem 3.1.1 and Proposition 1.5.1 we have that there is a positive constants c such that kKu kσ p (L2 (E)) ≤ ckukl p (L2 (E)) , k0

where 2 ≤ p < ∞. Necessity. Let Ku ∈ σ p L2 (E) . We set

1

fn (x) = χE2n+1 \E2n (x)|E2n+1 \ E2n |− 2 ; 1/2

−1/2

gn (x) = u(x)χE2n+1 \E3·2n−1 (x)k0 (x)αn where αn =

Z

,

u2 (x)k0 (x)dx.

E2n+1 \E3·2n−1

Then it is easy to verify that { fn } and {gn } are orthonormal systems in L2 (E). By Proposi-

Kernel Operators on Cones

47

tion 1.5.2 we find that ∞ > kKu kσ p (L2 (E)) ≥ =

h

∑

n∈Z

Z

h Z ≥ c ∑ n∈Z

h Z ≥ c ∑ n∈Z

h ≥ c ∑ n∈Z

p 1/p < K f , g > u n n ∑

k∈Z

−1/2

1/2

E2n+1 \E3·2n−1

k0 (x)(K fn )(x)u2 (x)αn

1/2 −1/2 k0 (x)u2 (x)αn

E2n+1 \E3·2n−1

Z

dx

i p 1/p k(x, y) fn (y)dy dx

Er(x) \E2n

−1/2 −nQ/2

k(x, δ1/(2c0 ) x)u2 (x)αn

E2n+1 \E3·2n−1

Z

−1/2

u2 (x)k0 (x)αn

dx

E2n+1 \E3·2n−1

p i1/p

2

p i1/p

(r(x)Q − 2nQ )dx

h ∞ i p/2 1/p = c ∑ αn .

i1/p p

n=0

Let us now take 1

fn′ (x) = χE3·2n−1 \E3·2n−2 (x)|E3·2n−1 \E3·2n−2 |− 2 ; −1/2

1/2

g′n (x) = u(x)χE3·2n−1 \E2n (x)k0 (x)βn where βn =

Z

,

u2 (x)k0 (x)dx

E3·2n−1 \E2n

and argue as above. Then we conclude that ∞ > kKu kσ

p

h ≥c

L2 (E)

∞

i p/2 1/p

∑ βn

n=0

.

Summarizing the estimates obtained above we have Z p/2 1/p 2 p/2 1/p u (x)k0 (x)dx ≤ ∑ (αn + βn ) ∑ n∈Z

E2n+1 \E2n

n∈Z

≤ c(kKu kσ p (L2 (0,∞)) + kKu kσ p (L2 (0,∞)) )

≤ 2ckKu kσ p (L2 (0,∞)) .

3.4.

Measure of Non–compactness

In this section we present two-sided estimates for the measure of non-compactness kKkK (L p (E),Lvq (E)) of the operator K. Theorem 3.4.1. Let 1 < p ≤ q < ∞ and let k ∈ Vp . Assume that K is bounded from X to q Y , where X = L p (E) and Y = Lv (E). Then there exist positive constants b1 (depending on c1 , c3 , p and q) and b2 (depending on c2 p and q) such that the inequality b1 J ≤ kKkK (X,Y ) ≤ b2 J

48

Alexander Meskhi

holds, where J = lim B( j) + lim B( j), j→+∞

j→−∞

B( j) is defined by (3.1.1) and the constants c1 , c2 and c3 are defined in Definition A. Proof. From the proof of Theorem 3.2.1 we see that (n,m)

kK − P1 (n,m)

where P1

(n,m)

and P2

(n,m)

− P2

k ≤ b2 sup B( j) + sup B( j) , j≥m

j≤n

are compact operators for every n, m ∈ Z, n < m. Consequently, kKkK (X,Y ) ≤ b2 J,

where b2 depends only on p, q, c1 and c3 . To obtain the lower estimate kKkK (X,Y ) ≥ b1 J, we take λ > kKkK (X,Y ) . Then by Lemma 1.2.3 there exists P ∈ FL (X,Y ) such that kK − Pk < λ. On the other hand, using Lemma 1.2.6, for ε = (λ − kK − Pk)/2, there exist T ∈ FL (X,Y ) and Eα,β := {x ∈ E : 0 < α < r(x) < β < ∞} such that kP − T k < ε

(3.4.1)

and supp T f ⊂ Eα,β . From (3.4.1) we obtain kK f − T f kY ≤ λk f kX for every f ∈ X. Thus, Z

|K f (x)|q v(x)dx +

Eα

Z

|K f (x)|q v(x)dx ≤ λq k f kX q

(3.4.2)

E\Eβ

for every f ∈ X. Let us choose n ∈ Z such that 2n < α. Assume that j ∈ Z, j ≤ n and f j (y) = χE2 j+1 . ′ Then using condition (C ) we find that Z q Z Z q |K f j (x)| v(x)dx ≥ k(x, y) f j (y)dy v(x)dx E2 j+1 \E2 j

E2 j+1 \E2 j

≥c

Z

Fx

kq (x, δ1/(2c0 ) x)v(x)r(x)q dx.

E2 j+1 \E2 j q

On the other hand, k f j kX = c2 jQq/p and consequently, (3.4.2) yields cB( j) ≤ λ

Kernel Operators on Cones

49

for every integer j, j ≤ n. Hence sup j≤n B( j) ≤ cλ for all integers n with the condition 2n < α. Therefore lim sup j≤n B( j) ≤ cλ. n→−∞

Now we take m ∈ Z such that 2m > β. Then for f j (y) = χE2 j+1 ) (y) ( j ≥ m), we obtain Z

|K f j (x)|q v(x)dx ≥ c

E2 j+1 \E2 j

Z

kq (x, δ1/(2c0 ) x)v(x)r(x)q dx.

E2 j+1 \E2 j q

On the other hand, k f j kX = c2Q jq/p . Hence sup j≥m B( j) ≤ cλ, where c depends only on p, q and c1 . Consequently, lim sup j≥m B( j) ≤ cλ from which it follows the desired estimate. m→+∞

3.5.

Convolution–type Operators with Radial Kernels

Let ϕ be a positive function on [0, ∞) and let ¯ f (x) = K

Z

Er(x)

ϕ(r(xy−1 )) f (y)dy, x ∈ E.

We say that ϕ belongs to Uλ , 1 < λ < ∞, if (a) there exists a positive constants c1 and c2 such that ϕ(r(xy−1 )) ≤ c1 ϕ(r(x)), y ∈ Sx , ϕ(r(xy−1 )) ≥ c2 ϕ(r(x)), y ∈ Er(x) ; (b) there is a positive constant c3 for which the inequality Z

ϕλ (r(xy−1 ))dy ≤ c3 (r(x))Q ϕλ (r(x)). ′

Fx

′

Example 3.5.1. Let ϕ(t) = t α−Q , where Q/λ < α < Q. Then it is easy to see that ϕ ∈ Uλ . Indeed, (a) is obvious due to the properties of the quasi-norm r. Let us show (b). We have

I := =

Z ∞

Z

ϕλ (r(xy−1 ))dy = ′

Fx

Z

′

(r(xy−1 ))(α−Q)λ dy Fx ′

|B(0, r(x))| ∩ {y ∈ E : (r(xy−1 ))(α−Q)λ > s}|ds

0

=

Z r(x)(α−Q)λ′ 0

(· · · ) +

Z ∞

r(x)(α−Q)λ

′

(· · · ) := I1 + I2 .

It is clear that I1 ≤ (r(x))Q ϕλ (r(x)), while for I2 we find that ′

I2 ≤

Z ∞

Q

(α−Q)λ′

(r(x))

s (α−Q)λ′ ds = c(r(x))(α−Q)λ +Q = c(r(x))Q ϕλ (r(x)). ′

′

50

Alexander Meskhi

The statements of this section are proved just in the same way as for the operator K (see the previous sections); therefore the proofs are omitted. Theorem 3.5.1. Let 1 < p ≤ q < ∞ and let k ∈ Up . Then ¯ is bounded from L p (E) to Lvq (E) if and only if (i) K ¯ j) := sup B¯ := sup B( j∈Z

j∈Z

Z

v(x)ϕq (r(x))dx

E2 j+1 \E2 j

1/q

2j

Q/p′

< ∞;

¯ is compact from L p (E) to Lvq (E) if and only if B¯ < ∞ and lim j→−∞ B¯ j = K lim j→+∞ B¯ j = 0. (ii)

Theorem 3.5.2. Let 0 < q 1. Suppose that ϕ ∈ Up . Then the following conditions are equivalent: ¯ is bounded from L p (E) to Lvq (E); (i) K ¯ is compact from L p (E) to Lvq (E); (ii) K (p−q)/(pq) p/(p−q) R R q Qp(q−1)/(p−q) r(x) dx < ∞. (iii) E E\Er(x)ϕ (r(y))v(y)dy ¯ ¯ = t Q ϕ2 (t) and let k(x) ¯ Let ϕ(t) = ϕ(r(x)). Suppose that ¯ u f (x) = u(x) K

Z

Er(x)

ϕ(r(xy−1 )) f (y)dy, x ∈ E,

where u is a measurable function on E. ¯ u ∈ σ p (L2 (E)) if and only if Theorem 3.5.3. Let 2 ≤ p < ∞ and let ϕ ∈ Up . Then K u ∈ l p (Lk2¯ (E)). Moreover, there exist positive constants b1 and b2 such that ¯ u kσ (L2 (E)) ≤ b2 kuk p 2 . b1 kukl p (L2¯ (E)) ≤ kK l (L ¯ (E)) p k

3.6.

k

Notes and Comments on Chapter 3

In this chapter we use the material from [7] and [8]. Section 3.4 is published first time. The two-weight problem for higher-dimensional Hardy-type operators defined on cones in Rn involving Oinarov [183] kernels was studied in [234], [97] (see also [221], for Hardytype transforms on star-shaped regions). A full characterization of a class of weight pairs (v, w) governing the boundedness of q integral operators with positive kernels from Lwp to Lv , 1 0,

0

from L p (R+ ) to Lv (R+ ), 1 < p, q < ∞, 1/p < α < 1 have been obtained in [160] (see also [198]). This result was generalized in [163] (see also [49], Ch. 2) for integral operators with positive kernels involving fractional integrals. q

Chapter 4

Potential and Identity Operators This chapter is devoted to estimates of the measure of non-compactness for potential operators in weighted Lebesgue spaces defined on Euclidean spaces and homogeneous groups, partial sums of Fourier series, Poisson integrals. The same problem for the identity operator is also investigated. In some cases we conclude that there is no weight pair for which a potential operator is compact from one weighted Lebesgue space into another one. Here keep the notation of Section 1.2.

4.1.

Riesz Potentials

Let G be a homogeneous group and let Iα f (x) =

Z

G

f (y) dy, r(xy−1 )Q−α

0 < α < Q,

be the Riesz potential operator. It is well known (see [70], Ch. 6) that Iα is bounded from L p (G) to Lq (G), 1 < p, q < ∞, if and only if Qp q= . (4.1.1) Q − αp q

Moreover, if (4.1.1) holds, then Iα is bounded from Lρpp (G) to Lρq (G) if and only if sup B

1 |B|

Z

B

ρ(x)q dx

1/q

1 |B|

Z

′

B

ρ(x)−p dx

1/p′

< ∞,

where the supremum is taken over all balls B in G (see [171] for Euclidean spaces and [76] for quasimetric measure spaces with doubling condition). Our first result in this section is the following statement: Theorem 4.1.1. Let 1 < p ≤ q < ∞, 0 < α < Q. Let Iα be bounded from Lwp (G) to Then the following inequality holds

q Lv (G).

kIα kK ≥ Cα,Q max{A1 , A2 , A3 },

52

Alexander Meskhi

where Cα,Q = A1 = sup lim rα−Q a∈G r→0

Z

v(x)dx

B(a,r)

A2 = sup lim

a∈G r→0

Z

1 , (2c0 )Q−α

1/q Z

′

w1−p (x)dx

B(a,r)

v(x)dx

1/q

Z

(B(a,r))c

B(a,r)

1/p′

;

1/p′ ′ ′ r(ay−1 )(α−Q)p w1−p (y)dy

and A3 = sup lim

a∈G r→0

Z

1−p

w

′

(x)dx

Z

1/p′

(B(a,r))c

B(a,r)

1/q r(ay−1 )(α−Q)q v(y)dy

(c0 is the constant from the triangle inequality for the homogeneous norms ). The next statement is formulated for the Riesz potentials JΩ,α f (x) =

Z

f (y)|x − y|α−n dy,

x ∈ Ω,

Ω

where Ω is a domain in

Rn .

Theorem 4.1.2. Let Ω ⊆ Rn be a domain in Rn . Let 1 < p ≤ q < ∞. If JΩ,α is bounded q from Lwp (Ω) to Lv (Ω), then we have kJΩ,α kK ≥ 2α−n B1 , where B1 = sup lim r a∈Ω r→0

Further, if Ω =

Rn ,

α−n

Z

v(x)dx

B(a,r)

1q Z

′

w1−p (x)dx B(a,r)

1′ p

.

then kJΩ,α kK ≥ 2α−n max{B2 , B3 },

where B2 = sup lim

a∈Rn r→0

B3 = sup lim

Z

v(x)dx

Z

w1−p (x)dx

a∈Rn r→0

Z

(α−n)p′

|a − y|

Rn \B(a,r)

B(a,r)

B(a,r)

1/q

′

1/p′

Z

Rn \B(a,r)

1−p

w

′

1/p′ (y)dy ,

1/q |a − y|(α−n)q v(y)dy .

pQ Corollary 4.1.1. Let 1 < p < ∞, 1 < p < Qα , q = Q−αp , then there is no weight pair q (v, w) for which Iα is compact from Lwp (G) to Lv (G). Moreover, if Iα is bounded from Lwp (G) q to Lv (G), then kIα kK ≥ Cα,Q A1 ,

Potential and Identity Operators

53

where Cα,Q and A1 are defined in Theorem 4.1.1. Proof of Theorem 4.1.1. By Lemma 1.2.7 we have that for a ∈ G and λ > kIα kK (Lwp (G),Lvq (G)) there are positive constants β1 and β2 , β1 < β2 , such that for all τ, s (τ < β1 , s > β2 ), Z q Z q/p q f (x) p w(x)dx v(x) Iα f (x) dx ≤ λ , (4.1.2) G

B(a,τ)

for f ∈ Lwp (G), and

Z

q Z q v(x) Iα f (x) dx ≤ λ

B(a,s)c

B(a,s)

q/p f (x) p w(x)dx ,

(4.1.3)

for f with supp f ⊂ B(a, s). ′ Now assuming f (x) = χB(a,r) (x)w1−p (x) in (4.1.2) and observing that Z

′

w1−p (x)dx < ∞

B(a,r)

for all r > 0 (see also [76], Ch. 3 for this fact), we find that Z

B(a,r)

Z v(x)

B(a,r)

′ q Z w1−p (y) q dy dx ≤ λ r(xy−1 )Q−α

1−p

′

w

(x)dx

B(a,r)

q/p

< ∞.

Further, if x, y ∈ B(a, τ), then r(xy−1 ) ≤ c0 r(xa−1 ) + r(ay−1 ) ≤ 2c0 τ.

Hence

kIα kK ≥ Cα,Q A1 . ′ w1−p (x)

If f (x) = χB(a,τ)c (x) r(ay−1 )(Q−α)(p′ −1) , then Z

B(a,τ)

Z v(x)

′

w1−p (y)dy ′

B(a,τ)c

≤λ

q

r(xy−1 )Q−α r(ay−1 )(Q−α)(p −1)

Z

B(a,τ)c

′ w1−p (x)dx q/p

r(ay−1 )(Q−α)p

′

q

dx

< ∞.

Let r(xa−1 ) < τ and r(ya−1 ) > τ. Then r(xy−1 ) ≤ c0 r(xa−1 ) + r(ay−1 ) ≤ c0 τ + r(ay−1 ) ≤ 2c0 r(ay−1 ).

Hence, by (4.1.2) we have

1 (2c0 )q(Q−α)

Z

B(a,τ)

v(x)dx

Z

B(a,τ)c

′ w1−p (y)dy q

r(ay−1 )(Q−α)p

′

54

Alexander Meskhi ≤λ

q

Z

′ w1−p (x)dx q/p

r(ay−1 )(Q−α)p

B(a,τ)c

′

.

The latter inequality implies 1 A2 . (2c0 )Q−α

kIα kK ≥

Further, observe that (4.1.3) means that the norm of the operator I¯α f (x) =

Z

f (y)dy r(y−1 a)Q−α

B(a,s)

can be estimated as follows: kI¯α kLwp (B(a,s))→Lvq (B(a,s)c ) ≤ λ. By the duality arguments we find that kI¯α kLwp (B(a,s))→Lvq (B(a,s)c ) = kI˜α k

L

′ q

v1−q

where I˜α g(y) =

Z

B(a,s)c

′

′ p ′ w1−p

(B(a,s)c )→L

(B(a,s))

,

g(x)dx . r(xy−1 )Q−α

Indeed, by Fubini’s theorem and H¨older’s inequality we have Z g(x)(I¯α f (x)) dx ¯ kIα f kLvq (B(a,s)c ) ≤ sup kgk

≤

sup kgk

≤1 ′ q c L ′ (B(a,s) ) 1−q v

Z

sup kgk

≤

≤1 ′ q B(a,s)c Lv (B(a,s)c )

| f (y)| I˜α (|g|)(y) dy

≤1 ′ q c B(a,s) L ′ (B(a,s) ) v1−q

Z

1 Z 1 ′ p′ | f (y)| p w(y)dy p I˜α (|g|) (y)w1−p (y)dy p′

B(a,s)

B(a,s)

Z ≤ kI˜α k

B(a,s)

1p | f (y)| p w(y)dy .

Hence kI¯α k ≤ kI˜α k. Analogously, kI˜α k ≤ kI¯α k. Further, (4.1.3) implies ′ Z Z ′ ′ g(y)dy p p 1−p dx ≤ λ w (x) r(xy−1 )Q−α B(a,s)

(B(a,s))c

Z

(B(a,s)c )

′

q 1−q

|g(x)| v

′

(x)dx

p′ /q′

.

Now taking g(x) = χB(a,s)c (x)r(xa−1 )(Q−α)(1−q) v(x) in the last inequality we conclude that kIα kK ≥ (2c01)Q−α A3 . Theorem 4.1.2 follows in the same manner as Theorem 4.1.1 was obtained. We only need to use Lemma 1.2.8 instead of Lemma 1.2.7.

Potential and Identity Operators

4.2.

55

Truncated Potentials

This subsection is devoted to two-sided estimates of the essential norm for the operator Z

Tα f (x) =

f (y) , x ∈ G. r(xy−1 )Q−α

B(e,2r(x))

A necessary and sufficient condition guaranteeing the trace inequality for Tα defined on Rn was established in [215]. This result was generalized in [117], [49] (Ch. 6) for the spaces of homogeneous type. From the latter result (it is also a consequence of Theorem 3.5.1 for E = G) we have Proposition 4.2.1. Let 1 Q/p. Then q (i) Tα is bounded from L p (G) to Lv (G) if and only if B := sup B(t) := sup t>0

t>0

Z

v(x)r(x)(α−Q)q dx

r(x)>t

1/q

′

t Q/p < ∞;

(4.2.1)

q

(i) Tα is compact from L p (G) to Lv (G) if and only if lim B(t) = lim B(t) = 0. t→∞

t→0

Theorem 4.2.1. Let 1 b

Z

CQ,α = (2c0 )α−Q ; 1/q Z 1/p′ ′ v(x)r(x)(α−Q)q dx w1−p (x)dx ;

B(e,a)\B(e,t)

Z

b→∞

B(e,t)

(α−Q)q

v(x)r(x)

B(e,t)c

dx

1/q

Z

B(e,t)\B(e,b)

1−p′

w

(x)dx

1/p′

.

To prove Theorem 4.2.1 we need the following lemma. Lemma 4.2.1. Let p, q and α satisfy the conditions of Theorem (4.2.1). Then from the ′ q boundedness of Tα from Lwp (G) to Lv (G) it follows that w1−p is locally integrable on G. Proof. Let I(t) =

Z

B(e,t)

′

w1−p (x)dx = ∞

56

Alexander Meskhi R for some t > 0. Then there exists g ∈ L p B(e,t) such that gw−1/p = ∞. Let us assume B(e,t)

that ft (y) = g(y)w−1/p (y)χB(e,t) (y). Then we have

kTα ft kLvq (G) ≥ kχB(e,t)c Tα ft kLvq (G) Z 1/q Z ′ (α−Q)q ≥c v(x)r(x) dx g(y)w−1/p (y)dy = ∞. B(e,t)c

B(e,t)

On the other hand, k ft kLwp (G) =

Z

g p (x)dx < ∞.

B(e,t)

Finally we conclude that I(t) < ∞ for all t, t > 0. Proof of Theorem 4.2.1. Let λ > kTα kK (Lwp (G),Lvq (G)) . Then by Lemma 1.2.7 there exists a positive constant β such that for all τ1 , τ2 satisfying 0 < τ1 < τ2 < β1 , and all f with supp f ⊂ B(e, τ1 ), the inequality kTα f kLvq (B(e,τ2 )\B(e,τ1 )) ≤ λk f kLwp (B(e,τ1 )) . holds. Observe that if r(x) > τ1 and r(y) < τ1 , then r(xy−1 ) ≤ 2c0 r(x). Consequently, taking ′ f = w1−p χB(e,τ1 ) and using Lemma 4.2.1 we find that Z

1 (2c0 )Q−α

1′ (α−Q)q q1 Z 1−p′ p (x)dx w ≤λ v(x) r(x) dx B(e,τ1 )

B(e,τ2 )\B(e.τ1 )

for all τ1 , τ2 , 0 < τ1 < τ2 < β1 . Hence 1 lim A(a) ≤ λ. (2c0 )(Q−α)q a→0 Further, by virtue of Lemma 1.2.7 (see (1.2.10)) there exists β2 such that for all s1 , s2 with β2 < s1 < s2 the inequality kTα f kLvq (B(e,s2 )c ) ≤ λk f kLwp (B(e,s2 )\B(e,s1 )) ′

holds, where supp f ⊂ B(e, s2 )\B(e, s1 ). Hence taking f = w1−p χB(e,s2 )\B(e,s1 ) in the previous inequality and using Lemma 4.2.1 we find that 1 (2c0 )Q−α

Z

B(e,s2 )c

(α−Q)q 1q v(x) r(x) dx

Z

B(e,s2 )\B(e,s1 )

′

w1−p (x)dx

1′ p

≤λ

which leads us to the estimate 1 lim A ≤ λ. (2c0 )Q−α b→∞ (b) Thus we have the desired result.

Potential and Identity Operators Theorem 4.2.2. Let 1 < p ≤ q < ∞ and let Then there is a positive constant C such that

57

< α < Q. Suppose that (4.2.1) holds.

Q p

kTα kK (L p (G)→Lvq (G)) ≤ C lim B(a) + lim B(b) , a→0

where B

(a)

= sup t≤a

Z

b→∞

v(x)r(x)(α−Q)q dx

B(e,a)\B(e,r)

B(b) = sup t≥b

Z

v(x)r(x)(α−Q)q dx

B(e,t)c

1/q

1/q

′

rQ/p ;

rQ − bQ

1/p′

.

Proof. Let 0 < a < b < ∞ and represent Tα f as follows: Tα f = χB(e,a) Tα ( f χB(e,a) ) + χB(e,b)\B(e,a) Tα ( f χB(e,b) ) +χG\B((e,b) Tα ( f χB(e,b/2c0 ) ) + χG\B(e,b) Tα ( f χG\B(e,b/2c0 ) ) ≡:= P1 f + P2 f + P3 f + P4 f , where B(e,t) (t > 0) denotes the closed ball in G with center e and radius t. For P2 , we have Z P2 f (x) = k(x, y) f (y)dy, G

where k(x, y) = χB(e,b)\B(e,a) (x)χB(e,2r(x)) (y)r(xy−1 )α−Q . Further observe that Z Z q ′

(k(x, y)) p dy

G

=

Z

B(e,b)\B(e,a)

≤c

Z

B(e,b)\B(e,a)

≤c

p′

v(x)dx

G

Z

−1

(α−Q)p′

(r(xy ))

B(e,2r(x))

Z

B(e,r(x)/2c0 )

Z

q′ p dy v(x)dx

q′ ′ p (r(xy−1 ))(α−Q)p dy v(x)dx. ′

r(x)(α−Q)q+q/p v(x)dx < ∞.

B(e,b)\B(e,a)

Hence by Lemma 1.3.5 we conclude that P2 is compact for every a and b. Now we observe that if r(x) > b and r(y) < b/2c0 , then r(x) ≤ 2c0 r(xy−1 ). Further, Lemma 1.3.5 implies that P3 is compact. Further, repeating the arguments of sufficiency of the proof of Theorem 3.1.1 (see also proof of Theorem 3.4.1 or [49], Ch. 6) we find that kP1 k ≤ C1 B(a) ;

kP4 k ≤ C2 B(b/2c0 ) ,

58

Alexander Meskhi

where the constants C1 and C2 depend only on p, q, Q and α. Therefore kTα − P2 − P3 k ≤ kP1 k + kP4 k ≤ c B(a) + B(b/2c0 ) . The last inequality completes the proof.

Theorem 4.2.3. Let p and q satisfy the conditions of Theorem 4.2.2. Suppose that (4.2.1) holds. Then we have the following two-sided estimate: c2 lim B(a) + lim B(b) ≤ kTα kK (L p (G),Lvq (G)) ≤ c1 lim B(a) + lim B(b) a→0

a→0

b→∞

b→∞

for some positive constants c1 and c2 depending only on Q, α, p, and q.

Theorem 4.2.3 follows immediately from Theorems 4.2.1 and 4.2.2.

4.3.

One–sided Potentials

Let Rα f (x) =

Zx 0

Z∞

f (t) dt, Wα f (x) = (x − t)1−α

x

f (t) dt, (t − x)1−α

where x ∈ R+ and α is a constant satisfying the condition 0 < α < 1. Theorem 4.3.1. Let 1 kRα kK (Lwp (R+ ),Lvp (R+ )) and a ∈ R+ . By Lemma 1.2.8 we have that a+r Z

v(x)(Rα f (x)) dx ≤ λ q

q

a

Za

qp p f (x) w(x)dx

a−r

′

for small r and non-negative f with supp f ⊂ (a − r, r). Hence assuming f (x) = w1−p (x) in the latter inequality we find that Zx v(x)

a+r Z a

0

′

w1−p dt (x − t)1−α

q

dx ≤ λ

q

Za

1−p′

w

(x)dx

a−r

qp

For x ∈ (a, a + r) and t ∈ (a − r, r), we have that x − t < 2r. Hence (α−1)q

(2r)

a+r Z

v(x)dx

a

Za

a−r

1−p′

w

(t)dt

q

≤λ

q

Za

1−p′

w

a−r q

(x)dx

qp

. ′

Taking into account the boundedness of Rα from Lwp (R+ ) to Lv (R+ ) we have w1−p ∈ Lloc (R+ ) (see, e.g., [3]). Consequently,

Potential and Identity Operators

α−1

λ ≥ (2r)

59

Za+r 1q Za 1′ p 1−p′ v(x)dx w (x)dx . a

a−r

Taking the limit when r → 0 and the supremum over all a ∈ R+ in the latter expression, we have the desired result. In an analogous manner we can obtain the next statement. Theorem 4.3.2. Let 1 < p ≤ q < ∞. Suppose that Wα is bounded from Lwp (R+ ) to Lvp (R+ ). Then α−1

kWα kK ≥ 2

α−1

sup lim τ

a∈R+ τ→0

Za

a−τ

1q Za+τ 1′ p 1−p′ v(x)dx w (x)dx . a

Theorem 4.3.3. Let 1 kWα kK (Lwp (R+ ),Lvq (R+ )) . Then by Lemma 1.2.8 and the estimate t − x ≤ t − a which holds for x ∈ (a, a + r) and t > a + r, the inequality a+r Z∞ q Z∞ qp Z f (t) q p v(x)dx dt ≤ λ ( f (x)) w(x)dx (t − a)1−α a

a+r

a+r

holds, where f ≥ 0, f ∈ Lwp (R+ ), supp f ⊂ (a + r, ∞), a ∈ R+ and r is a small positive ′ ′ number. Assuming that f (t) = w1−p (t)(t − a)(p −1)(α−1) χ(a+r,∞) (t) in the last inequality and observing that the integral on the right-hand side is finite for this f (see, e.g., [49], Section 2.2), we have Za+r 1q Z∞ 1′ ′ p w1−p (x) λ≥ v(x)dx . ′ (1−α) dx p (x − a) a

a+r

Taking the supremum over all a ∈ R+ and passing to the limit when r → 0 in the righthand side of the latter inequality, we obtain the desired estimate. Analogously can be established the following statement for Rα . Theorem 4.3.4. Let 1 < p < ∞, q < ∞. Suppose that Rα is bounded from Lwp (R+ ) to Lvp (R+ ). Then kRα kK ≥ sup lim

Za

a∈R+ τ→0

a−r

v(x)dx

1q Za−r −∞

′

w1−p (x) ′ (a − x) p (1−α)

1′ p

.

60

Alexander Meskhi

4.4.

Poisson Integrals

Here we discuss the essential norm of the Poisson integral P f (x,t) =

Z

f (y)P(x − y,t)dy, x ∈ Rn , t > 0,

Rn

where, P(x,t) = t(t 2 + |x|2 )−

n+1 2

.

In this section we use the following notation: n

2π 2 T B(a, r) := {(x,t) ∈ Rn+1 , + ; x ∈ B(a, r),t < r}; Bn := |B(0, 1)| = nΓ( n2 )

where B(a, r) ⊂ Rn is a ball with center a and radius r and |B(0, 1)| is the volume of the ball B(0, 1). Lemma 4.4.1 If (x,t) ∈ T (B) and f (y) = χB (y), then P f (x,t) ≥

Bn 5

n+1 2

.

Proof. Let B = B(a, r). Then we have, P f (x,t) ≥ t

Z

B(a,t)

≥t

Z

dy

B(a,t)

(t 2 + |2t|2 )

n+1 2

≥

dy (t 2 + |x − y|2 ) Bnt n+1 5

n+1 2

t n+1

n+1 2

= Bn 5−

n+1 2

.

In the following statements we keep the notation of Section 1.2. p n+1 n q Lemma 4.4.2. Let w be a weight function on Rn+1 + and let S ∈ FL (Lw (R ), L (R+ )), p n n q where 1 ≤ p, q < ∞. Then for every a ∈ R and ε > 0, there exist R ∈ FL (Lw (R ), L (Rn+1 + )) p n and positive numbers α, α, 0 < α < α < ∞, such that for all f ∈ Lw (R ) the inequality

k(S − R) f kLq (Rn+1 ≤ εk f kLwp (Rn ) + ) holds and suppR f ⊂ T (B(a, α))\T (B(a, α)). Proof. It is clear that there exists linearly independent non-negative functions U j ∈ Lq (Rn+1 + ), j = 1, . . . , N, such that N

S f (x,t) =

∑ β j ( f )U j (x,t),

j=1

where β j are bounded linear functionals on Lwp (Rn ). Further there is a positive constant C such that N

∑ |β j ( f )| ≤ Ck f kL (R ) . p w

j=1

n

Potential and Identity Operators

61

Simple geometric observation shows that we can choose linearly independent Φ j ∈ Lq (Rn+1 + ) and numbers α j , α j so that kU j − Φ j kLq (Rn+1 < ε/C, + )

j = 1, 2, . . . , N,

and suppΦ j ⊂ T (B(a, α))\T (B(a, α)). Let N

∑ β j ( f )Φ j (x,t).

R f (x,t) =

j=1

Then we have N

∑ |β j ( f )|kU j − Φ j kL (R

kS f − R f kLq (Rn+1 ≤ + )

q

j=1

n+1 + )

≤ εk f kLwq (Rn )

for all f ∈ Lwp (Rn ). Moreover, it is clear that supp R f ⊂ T (B(a, α))\T (B(a, α)), where α = max {α j }, α = max {α j }. The statement below is a slight modification of Lemma 1.2.3; therefore we omit the proof. Proposition 4.4.1. Let T be a sublinear and bounded operator from Lwp (Rn ) to where 1 < p, q < ∞. Then kT kK = α(T ).

Lq (Rn+1 + ),

Theorem 4.4.1.Let 1 < p ≤ q < ∞ and let the operator P be bounded from Lwp (Rn ) to Then the following inequality holds:

q Lv (Rn+1 + ).

q kPkK (Lwp (Rn+1 ≥ max{D1 , D2 , D3 }, n+1 + ),Lv (R+ ))

where D1 =

Bn 5

n+1 2

sup lim

a∈Rn r→0

Z

v(x,t)dxdt

a∈Rn r→0

D3 = sup lim

a∈Rn r→0

Z

Z

t q v(x,t)dxdt

T (B(a,r))

v(x,t)dxdt

1q Z

− 1p

;

′

w(x)1−p dx

B(a,r)

1q

Z

(B(a,r))c

T (B(a,r))

w(x)dx

B(a,r)

T B(a,r)

D2 = 5−(n+1)/2 sup lim r−n−1

1q Z

′

w1−p (y)dy (r2 + 4|y − a|2 )

1

Proof. Denote Pv f (x,t) = v p (x,t)P f (x,t). Then

n+1 2

1′

1′ p

p′

p

;

.

kPv kK (Lwp (Rn ),Lq (Rn+1 = kPkK (Lwp (Rn ),Lvq (Rn+1 )) . + )) Let λ > kPkK (Lwp (Rn ),Lvq (Rn+1 . Then we see that λ > kPv kK (Lwp (Rn ),Lq (Rn+1 . Hence there + )) + ))

exists S ∈ FL (Lwp (Rn ), Lq (Rn+1 + )) for which

kPv − Sk < λ.

62

Alexander Meskhi

Let a ∈ Rn . Then by Lemma 4.4.2 there exist positive numbers α and α and an operator R ∈ FL (Lwp (Rn ), Lq (Rn+1 + ))) such that supp R f ⊂ T (B(a, α))\T (B(a, α)) and kS − Rk ≤

λ − kPv − Sk . 2

Hence, kPv − Rk < λ. Therefore, k(Pv − R) f kLq (Rn+1 ≤ λk f kLwp (Rn ) + ) for all f ∈ Lwp (Rn ). Now the latter inequality implies that if r < α, then Z

v(x,t)(Pv f ) (x,t)dxdt ≤ λ q

q

Z

p

( f (x)) w(x)dx

Rn

T (B(a,r))

qp

(4.4.1)

for all f ≥ 0, f ∈ Lwp (Rn ). If we take f (x) = χB(a,r) (x) in (4.4.1) and use Lemma 4.4.1 we find that B q Z Z qp n q v(x,t)dxdt ≤ λ w(x)dx n+1 5 2 T (B(a,r))

B(a,r)

which gives the estimate kPkK ≥ D1 . ′ Assuming that f (x) = χB(a,r) (x)w1−p (x) in (4.4.1) we have Z

Z t v(x,t)

′

w1−p (y)dy

q

T (B(a,r))

B(a,r)

≤ λq

Z

(t 2 + |x − y|2 ) ′

w1−p (x)dx

B(a,r)

qp

n+1 2

q

dxdt

< ∞.

Further, it is easy to see that for y ∈ B(a, r) and (x,t) ∈ T (B(a, r)) we have t 2 + |x − y|2 ≤ r2 + (2r)2 = 5r2 , which implies 5−(n+1)/2 rn+1

Z

T (B(a,r))

t q v(x,t)dxdt

1q Z

B(a,t)

′

w1−p (x)dx

1′ p

≤λ

for all r < α. The latter inequality yields the inequality kPkK ≥ D2 . To get the estimate kPkK ≥ D3 , we observe that if (x,t) ∈ T (B(a, r)) and |y − a| > r, then t 2 + |x − y|2 ≤ r2 + 4|a − y|2 . n+1 (1−p′ ) ′ Taking f (x) = χ{x:|x−a|>r} (x)w1−p (x) r2 + 4|a − y|2 2 in (4.4.1), we find that kPv kK ≥ D3 .

Potential and Identity Operators

63

Remark 4.4.1. If w ∈ A p (Rn ) (see (1.6.6) ) then it is easy to see that the inequality kPkK ≥ D1 of Theorem 4.3.1 can be replaced by kPkK ≥

1 5

n+1 2

sup lim r−n

B

a∈Rn r→0

Z

v(x,t)dxdt

T (B(a,r))

1q Z

′

w1−p (x)dx

B(a,r)

1′ p

,

where B is defined in (1.6.6).

4.5.

Sobolev Embeddings

In this section we deal with the identity operator from a weighted Sobolev space into a Lebesgue space. Let Ω ⊆ Rn be a domain and let 1 ≤ p < ∞. Suppose that m is non-negative integer. Assume that a weight w on R satisfies the condition w ∈ A p (Rn ) (see (1.6.6)). We define the weighted Sobolev space Wwm,p (Ω) as the set of functions u ∈ Lwp (Ω) with weak derivatives Dα u ∈ Lwp (Ω) for |α| ≤ m. Then norm of u in Wwm,p (Ω) is given by kuk

m,p Ww (Ω)

=

∑

Z

|Dα u(x)| p w(x)dx

|α|≤m Ω

1p

.

ewm,p as the closure of C∞ (Ω) in Wwm,p (Ω). Together with It is also defined the space W 0 we consider the space Vwm,p (Ω) with the norm

Wwm,p (Ω)

kuk

m,p Vw (Ω)

=

∑

Z

α

p

|D u(x)| w(x)dx

|α|=m Ω

1p

.

For weighted Sobolev inequalities we refer, e.g., to [2], [158],[230], [241]. ewm,p (Ω), It is well-known (see, e.g., [230], p. 16) that if w ∈ A p (Rn ) then Wwm,p (Ω), W m,p Vw (Ω) are Banach spaces. Now we formulate and prove the main statements of this section. In the sequel we keep the notation of Section 1.2. Theorem 4.5.1. Let 1 ≤ p ≤ q < ∞ and let m be any integer such that 0 ≤ m < n. Suppose that w is a weight function on Rn satisfying the condition w ∈ A p (Rn ). If Wwm,p (Ω) q q is embedded in Lv (Ω), i.e., I : Wwm,p (Ω) → Lv (Ω) is bounded, then R

−1 B(a,r) kIkK ≥ sup lim Sm,p (r, ψ) a∈Ω r→0

R

B(a,r)

where Sm,p (r, ψ) =

v(x)dx

1q

w(x)dx

1p ,

1 |α|p sup Dα ψ(x) p p , ψ is a function from C ∞ (Rn ) whose r ∑ 0

|α|≤m

1≤|x|≤2

support is in B(0, 2) and equal to 1 in B(0, 1).

64

Alexander Meskhi Proof. Let Iv u = vu, then kIv kK (Wwm,p (Ω),Lq (Ω)) = kIkK (Wwm,p (Ω),Lvq (Ω)) .

Let λ > α(Iv ), then there is P ∈ FL (Wwm,p (Ω), Lq (Ω)) such that kIv − Pk < λ. For a ∈ Ω there exist a positive number α and an operator R ∈ FL (Wwm,p (Ω), Lq (Ω)) such that λ − kIv − Pk 2

kP − Rk ≤

and suppRu ⊂ Ω\B(a, α). Hence kIv − Rk ≤ λ. Consequently, k(Iv − R)ukLq (Ω) ≤ λkukWwm,p (Ω) . If r < α, then the latter inequality implies Z

v(x)|u(x)|q dx

B(a,r)

1q

≤ λkukWwm,p (Ω) .

Let us now take ψ ∈ C0∞ (Rn ) which is equal 1 in B(0, 1) , supp ψ ⊂ B(0, 2) and set φ = ψ( x−a r ). Then taking u = ψ in the last inequality we find that Z

v(x)dx

B(a,r)

1q

≤ λ

∑

Z 1 Dα ψ(x) p w(x)dx p

|α|≤m Ω

Z ≤ λSm,p (r, ψ)

w(x)dx

B(a,r)

1p

.

Hence, R

1 B(a,r) kIkK ≥ R Sm,p (r, ψ) B(a,r)

v(x)dx

1q

w(x)dx

1p

for all a ∈ Ω and small r. The next statement follows similarly: Theorem 4.5.2. Let 1 ≤ p ≤ q < ∞, m ≤ n. Suppose that w ∈ A p (Rn ). If Vwm,p (Ω) is q q embedded in Lv (Ω), i.e., I : Vwm,p (Ω) → Lv (Ω) is bounded, then R

−1 B(a,r) kIkK (Vwm,p (Ω),Lwq (Ω)) ≥ sup lim Sm,p (ψ) a∈Rn r→0

R

B(a,r)

v(x)dx

1q

w(x)dx

1p

,

Potential and Identity Operators where Sm,p (ψ) = 2m/p

∑

65

p 1p sup Dα ψ(x) and ψ is a function from C0∞ (Rn ) with

|α|=m 1≤|x|≤2

supp in B(0, 2) and value 1 on B(0, 1).

Corollary 4.5.1. Let 1 ≤ p < ∞, 0 ≤ m < np . Suppose that q = βq

np n−mp ,

w(x) = |x|β and

v(x) = |x| p , where β > − np q . Then − 1p −1 q1 − 1p βq 1 kIkK (Vwm,p (Ω),Lwq (Ω)) ≥ Sm,p (ψ) Sn−1 +n (β + n) q p where Sm,p (ψ) is defined as in the previous statement and Sn−1 =

2πn/2 Γ(n/2) .

Example 4.5.2. Let 1 ≤ p ≤ q < ∞ and let 0 ≤ m < np . Suppose that V 1,p (Ω) is continq uously embedded in Lv (Ω). Then n

kIkK ≥ sup limCn,p rm− p a∈Ω r→0

Z

v(x)dx

B(a,r)

1q

,

31/2

− 1p 6e1+ 1−31/2 Sn−1 (1−31/2 )2

where Cn,p = . This follows from Theorem 4.5.2 taking w ≡ 1 and  for |x| < 1,   1 1 1+ 2 −1 (|x|−1) ψ(x) = for 1 ≤ |x| ≤ 2, e   0 for |x| > 2.

4.6.

Identity Operator

This section is devoted to lower estimates of the measure of non–compactness for the idenq tity operator I acting from Lwp (Ω) to Lv (Ω), where Ω = [0, π] and q < p. To prove the main statement we need some lemmas. Lemma 4.6.1. Let fn (x) = sin 2n x, 0 ≤ x ≤ π. Assume that v ∈ C1 ([0, π[) and v, v′ ∈ L1 ([0, π[). Then Z π Z 1 π lim v(x)| fn (x) − fm (x)|dx ≥ v(x)dx. (4.6.1) 2 0 n,m→∞ 0 n6=m

Proof. Let us denote In,m :=

Rπ

In,m ≥ =

1 2

Z π 0

0

v(x)| fn (x) − fm (x)|dx. We have

1 2

v(x) fn2 (x)dx −

Z π 0

Z π 0

v(x)( fn (x) − fm (x))2 dx v(x) fn (x) fm (x)dx +

1 1 := I1 − I2 + I3 . 2 2

1 2

Z π 0

v(x) fm2 (x)dx

66

Alexander Meskhi For I1 we have I1 = Z π

1 2

Z π

v(x)(1 − cos 2n+1 x)dx

0

Z

1 π v(x) cos 2n+1 xdx 2 0 0 Z Z π 1 π 1 = v(x)dx − n+2 v′ (x) sin 2n+1 xdx, 2 0 2 0 while for I2 we find that =

1 I2 = 2 =

1 2

Z π

1 2

Z π

v(x)dx −

v(x)[cos(2n − 2m )x − cos(2n + 2m )x]dx

0

Z

1 π v(x) cos(2n + 2m )xdx 2 0 1 1 := I21 − I22 . 2 2

v(x) cos(2n − 2m )xdx −

0

It is easy to see that I21 = −

1 n 2 − 2m

1 I22 = − n 2 + 2m Hence

v′ (x) sin(2n − 2m )xdx;

0

Z π

v′ (x) sin(2n − 2m )xdx.

0

1 I2 = − n 2(2 − 2m ) 1 + n 2(2 + 2m )

and

Z π

Z π

Z π

v′ (x) sin(2n − 2m )xdx

0

v′ (x) sin(2n + 2m )xdx

0

Z

Z

π 1 1 π v(x)dx − m+2 v′ (x) sin 2n xdx. 2 0 2 0 Finally, passing n and m to infinity we conclude that

I3 =

lim In,m ≥

n,m→∞ n6=m

1 4

Z π 0

v(x)dx +

1 4

Z π 0

v(x)dx =

1 2

Z π 0

v(x)dx.

Note that in the unweighted case the statement that { fn } contains no subsequence convergent in L1 is made in [143], p. 90. Lemma 4.6.2. Let fn (x) = sin 2n x, 0 ≤ x ≤ π. Suppose that v is a weight on [0, π]. Then the inequality (4.6.1) holds.

Potential and Identity Operators

67

Proof. Since v is a weight function, we have that v is almost everywhere positive on [0, π], and v ∈R L1 ([0, π]). Then there exists a sequence {vk }, where 0 ≤ vk (x) ≤ v(x), vk ∈ C∞ (0, π) and 0π v′k (x)dx < ∞ such that lim

Z π

k→∞ 0

vk (x)dx =

Z π

v(x)dx,

0

Let us take k so large that 1 2

Z π 0

1 vk (x)dx > 2

Z π 0

v(x)dx − ε.

By Lemma 4.6.1 we can choose n and m so that Z π 0

vk (x)| fn (x) − fm (x)|dx ≥

1 2

Z π 0

vk (x)dx − ε >

1 2

Z π

v(x)dx − 2ε.

0

From this we conclude that (4.6.1) holds. Theorem 4.6.1. Let 1 < q < p < ∞ and let Ω = [0, π]. Then there is no pair of weights q (v, w) for which I is compactly embedded from Lwp (Ω) to Lv (Ω). Moreover, if (1.1.1) holds for some weights v and w on Ω, then Z π 1/q . Z π 1/p 1 v(x)dx kIkK (Lwp (Ω),Lvq (Ω)) ≥ w(x)dx . (4.6.2) 4 0 0 Proof. By Lemma 4.6.2 there exists a sequence { fn } ⊂ BL∞ (BL∞ is the closed unit ball in L∞ ) such that k(I : L∞ → Lv1 )( fn ) − (I : L∞ → Lv1 )( fm )kLv1 = k fn − fm kLv1 > λ − ε, R

where λ := 21 0π v(x)dx and ε is a small positive number. Hence (I : L∞ → Lv1 )(BL∞ ) cannot be covered by a finite number of balls of radius λ/2 − ε/2. Thus for entropy numbers of I, we have en (I : (L∞ → Lv1 )) ≥ λ/2 − ε/2 for any n ∈ N. Consequently, using the inequality kIkK (L∞ ,Lv1 ) ≥ β(I) (see Section 1.2 for some properties of entropy numbers of bounded linear operators) we find that kIkK (L∞ ,Lv1 ) ≥ λ/2 − ε/2. Further, by Propositions 1.1.3 and 1.1.4 we have kI :

kI :

Lwp

Lw∞

→

Lwp k

=

π

w(x)dx 0

1/p

;

p 1/q−1/p v(x) p−q → = w(x)dx ; w(x) 0 Z π 1/q′ q 1 kI : Lv → Lv k = v(x)dx . Lvq k

Z π

Z

0

68

Alexander Meskhi Hence (I : L∞ → Lv1 ) = (I : Lvq → Lv1 ) ◦ (I : Lwp → Lvq ) ◦ (I : L∞ → Lwp ).

From this it follows that λ−ε ≤ kIkK (L∞ ,Lv1 ) ≤ kIkLvq →L1 kIkL∞ →Lwp kIkK (Lwp ,Lvq ) . 2 Therefore

λ−ε 2kIkLvq →Lv1 kIkL∞ →Lwp Z π −1/q′ Z π −1/p 1 = (λ − ε) v(x)dx w(x)dx . 2 0 0 kIkK (Lwp ,Lvq ) ≥

But ε can be taken arbitrarily small. Hence we have (4.6.2). Since v(x) > 0 almost everywhere on Ω, we have the desired result.

4.7.

Partial Sums of Fourier Series

Here we investigate lower estimates of the essential norm for the partial sums 1 Sn f (x) = π

Zπ

f (t)Dn (x − t)dt, n ∈ N,

−π

of the Fourier series of f ∞ 1 f ∼ a0 + ∑ (ak cos kx + bk sin kx), 2 k=1 n

where Dn = 21 + ∑ cos kt. k=1

For basic properties of Sn see, for instance, [242]. Theorem 4.7.1. Let 1 < p < ∞. Then there is no n ∈ N and weight pair (w, v) on T := (−π, π) such that Sn is compact from Lwp (T ) to Lvp (T ). Moreover, if Sn is bounded from Lwp (T ) to Lvp (T ), then kSn k ≥

1 (2 + 21/2 )1/2 sup lim 2π a∈T r→0 2r

Z a+r

v(x)dx

a−r

1p 1 Z a+r 1′ ′ p w1−p (x)dx . 2r a−r

(4.7.1)

Proof. Taking λ > kSn kκ(Lwp (T ),Lvp (T )) , by Lemma 1.2.8 we find that Z I

v(x)|Sn f (x)| p dx ≤ λ p

Z

| f (x)| p w(x)dx

(4.7.2)

I

for the intervals I := (a − r, a + r), where r is a small positive number and supp f ⊂ I.

Potential and Identity Operators Let J1 =

Z

p

|Sn f (x)| v(x)dx

J2 =

and

I

Suppose that |I| ≤ (see [95]),

| f (x)| p w(x)d(x).

I

π 4

and n is the greatest integer less than or equal to |Sn f (x)| ≥ 1−p′

Using this estimate and taking f := w J1 ≥

Z

69

1

π

sin

1 π

Z

| f (θ)| sin 3π 8 π 4n

I

π 4|I| .

Then for x ∈ I (4.7.3)

dθ.

(x)χI (x) we find that

3π p −p |I| 8

Z

v(x)dx

I

Z

′

w1−p (x)dx

I

R

p

.

′

On the other hand, due to (4.7.3) it is easy to see that J2 = w1−p (x)dx < ∞. I

Hence, by (4.7.2) we conclude that λ≥

1 3π 1 sin π 8 |I|

Z I

v(x)dx

1p 1 Z 1′ ′ p w1−p (x)dx . |I| I

Now passing r to 0, taking the supremum over all a ∈ T and using the fact that sin 3π 8 =

(2+21/2 )1/2 2

we find that (4.7.1) holds.

Corollary 4.7.1. Let 1 < p < ∞ and let n ∈ N. Then kSn kκ(L p (T )) ≥

(2 + 21/2 )1/2 . 2π

Corollary 4.7.2. Let 1 < p < ∞ and let n ∈ N. Suppose that w(x) = v(x) = |x|α . Then we have kSn kκ(Lwp (T )) ≥

4.8.

1′ (2 + 21/2 )1/2 1 1p 1 p . 2π α+1 α(1 − p′ ) + 1

Notes and Comments on Chapter 4

Sections 4.1, 4.2 and 4.7 are based on the paper [5]. The results of Section 4.6 are were derived in the paper [43]. q Criteria for the trace inequality (L p → Lv boundedness) for the Riesz potentials were established in [1], [159] (see also the monographs [2], [158], [76] and references therein). The two-weight problem for the Riesz potentials was solved by E. Sawyer [213], M. Gabidzashvili and V. Kokilashvili [71], [72] (see also [112]), however, the conditions established by M. Gabidzashvili and V. Kokilashvili are more transparent than those of E. Sawyer.

70

Alexander Meskhi

Necessary and sufficient conditions guaranteeing the compactness of the Reisz potentials from one weighted Lebesgue space into another one have been derived in [47] (see also [49], Section 5.2). The two-weight problem for integral transforms with positive kernels defined on quasimetric doubling measure spaces were found in [75], [76]. The same problem was solved in [217], [219] for spaces having a group structure (see also the survey paper [110]). For two–weight inequalities for Poisson integrals we refer to [170], [218] (see also [76], Ch. 3 for integral operators with more general positive kernel). A full characterization of a class of weight pairs (v, w) governing the boundedness of q one–sided potentials from Lwp to Lv (1 < p < q < ∞) was established in [76], [50] (see also [49], Ch. 2). We refer also to [153], [154] for the Sawyer–type two-weight criteria for one–sided potentials. The one-weight problem for the partial sums of the Fourier series was solved by R. A. Hunt, Muckenhoupt and R. L. Wheeden [95] (see also the monograph [76]). Finally we point out that the non–compactness for the majorants of partial sums of the Fourier series T f (x) = supn |Sn f (x)| was investigated in [186]. We are indebted to Professor Peter Bushell for a key idea which led to the proof of Lemma 4.6.1.

Chapter 5

Generalized One-sided Potentials in L p(x) Spaces This chapter deals with boundedness/compactness criteria and measure of non– compactness for the generalized Riemann-Liouville operator Rα(x) f (x) =

Z x

f (t)(x − t)α(x)−1 dt, x > 0,

0

in the L p(x) spaces, where 0 < inf α ≤ sup α < 1. In particular, necessary and sufficient conditions on a weight v guaranteeing the boundedness/ compactness of Rα(x) from L p(x) q(·)

to Lv are established provided that p satisfies weak Lipschitz condition. When p is an arbitrary measurable function, we derive sufficient conditions (which are also necessary for constant exponents) governing the trace inequality for the operator Rα(x) . Two-sided weighted estimates of the measure of non–compactness for Rα(x) are also established. Throughout this chapter we assume that I is either a bounded interval [0, a] or R+ . We use the notation: Ek := [2k , 2k+1 ); Ik := [2k−1 , 2k+1 ), k ∈ Z.

5.1.

Boundedness

In this section we establish necessary and sufficient conditions for the boundedness of the q(x) operator Rα(x) from L p(x) to Lv . Theorem 5.1.1. Let I = [0, a] be a bounded interval and let 1 < p− (I) ≤ p(x) ≤ q(x) ≤ q+ (I) < ∞. Suppose that (α − 1/p)− (I) > 0. Further, assume that p, q ∈ W L(I). Then the inequality kvRα(x) f kLq(x) (I) ≤ ck f kL p(x) (I) , f ∈ L p(·) (I) (5.1.1) holds if and only if

′ v(x)

Aa := sup Aa (t) := sup χ(t,a) (x) 1−α(x) q(x) t 1/p (0) < ∞. L (I) x 0
72

Alexander Meskhi

Moreover, there exist positive constants c1 and c2 such that c1 Aa ≤ kRα(x) kL p(x) (I)→Lq(x) (I) ≤ c2 Aa . v

Proof. For simplicity assume that a = 1. Sufficiency. Suppose that f ≥ 0. We represent Rα(x) f as follows: (Rα(x) f )(x)

Z x/2

=

f (t)(x − t)α(x)−1 dt +

0

:=

f (t)(x − t)α(x)−1 dt

x/2

(1) (2) (Rα(x) f )(x) + (Rα(x) f )(x).

Hence

Z x

(1)

(2)

kvRα(x) f kLq(x) (I) ≤ kvRα(x) f kLq(x) (I) + kvRα(x) f kLq(x) (I) := S(1) + S(2) . It is easy to see that if 0 < t < x/2, then x/2 ≤ x −t. Consequently, (x −t)α(x)−1 ≤ cxα(x)−1 , where the positive constant c does not depend on x. Hence, taking into account Theorem 1.4.5 we have

v(x)

(1) S ≤ c 1−α(x) H f (x) q(x) ≤ cAa k f kL p(·) (I) . L (I) x Suppose now that kgkLq′ (x) (I) ≤ 1. Using Lemmas 1.4.8 and 1.4.9 we find that Z 1 0

≤c

Z v(x) Z

∑

x

α(x)−1

f (t)(x − t)

dt g(x)dx

x/2

k∈Z− Ek−1

v(x)kχ(x/2,x) (·) f (·)kL p(·) (I)

× kχ(x/2,x) (·)(x − ·)α(x)−1 kL p′ (·) (I) g(x)dx ≤c

∑

kχIk−1 (·) f (·)kL p(·) (I)

k∈Z−

≤c

∑

k∈Z−

Z

v(x)xα(x)−1/p(x) g(x)dx Ek−1

kχIk−1 (·) f (·)kL p(·) (I) χEk−1 (x)v(x)xα(x)−1/p(x)

Lq(x) (I)

× kχEk−1 (·)g(·)kLq′ (·) (I) ′

≤ c2k/p (0)

∑

kv(x)xα(x)−1 χEk−1 (x)kLq(x) (I) kχIk−1 (·) f (·)kL p(·) (I)

k∈Z−

× kχEk−1 (·)g(·)kLq′ (·) (I) ≤ cAa k f kL p(·) (I) kgkLq′ (·) (I) ≤ cAa k f kL p(·) (I) . Taking the supremum with respect to g and applying Lemma 1.4.7, we have the desired result. Necessity. Let us take fk (x) = χ[0,2k−2 ] (x), where k ∈ Z− . Then by the condition p ∈ W L(I), Lemma 1.4.1 and Proposition 1.4.1 we have ′

k fk kL p(·) (I) ≤ c2k/(p )+ ([0,2

k+2 ])

′

≤ c2k/p (0) .

On the other hand, kRα(x) f kLq(x)

(I) vq(x)

≥ ckχEk−1 (x)v(x)xα(x)−1 kLq(x) (I) .

Generalized One-sided Potentials

73

Here we used the estimate (x − t)α(x)−1 ≥ cxα(x)−1 when x ∈ [2k−1 , 2k ], t < 2k−2 . Hence ′ A¯ := sup A¯ k := sup kχEk−1 (x)v(x)xα(x)−1 kLq(x) (I) 2k/p (0)

k∈Z−

k∈Z−

≤ ckRα(x) kL p(x) (I)→Lq(x) (I) . v

Let us now take t ∈ I. Then t ∈ [2m−1 , 2m ) for some m ∈ Z− . Consequently, 0

A1 (t) ≤

∑ kχE

k−1

′ ¯ m/p′ (0) (x)v(x)xα(x)−1 kLq(x) (I) 2m/p (0) ≤ A2

k=m

0

∑ 2−k/p (0) ≤ cA¯ ′

k=m

≤ ckRα(x) kL p(x) (I)→Lq(x) (I) . v

Hence A1 ≤ ckRα(x) kL p(x) (I)→Lq(x) (I) . v

Theorem 5.1.2. Let I = R+ and let 1 < p− (I) ≤ p(x) ≤ q(x) ≤ q+ (I) < ∞. Suppose that (α − 1/p)− (I) > 0. Further, assume that p, q ∈ W L(I) and that there is a positive number a such that q(x) ≡ qc = const, p(x) ≡ pc = const outside [0, a]. Then inequality (5.1.1) holds if and only if

′ v(x)

A∞ := sup A∞ (t) := sup χ(t,∞) (x) 1−α(x) q(x) t 1/P (t) < ∞, L (I) x t>0 t>0 where

P(t) =

(

p(0), 0 ≤ t ≤ a, pc , t > a.

Moreover, there are positive constants c1 and c2 such that c1 A∞ ≤ kRα(x) kL p(x) (I)→Lq(x) (I) ≤ c2 A∞ . v

Proof. For simplicity we assume that a = 1. First we prove sufficiency. Suppose that f ≥ 0. We have kvRα(x) f kLq(x) (I) ≤ kvRα(x) f kLq(x) ([0,2]) + kvRα(x) f kLq(x) ((2,∞)) := I1 + I2 . Taking into account Theorem 5.1.1 we find that the condition A∞ < ∞ implies I1 ≤ cA∞ k f kL p(x) ([0,2]) ≤ cA∞ k f kL p(x) (I) . For I2 , we have I2

≤

Z 1

v(x) (x − t)α(x)−1 f (t)dt

0

Z

+ v(x)

Lq(x) ((2,∞))

(x − t) f (t)dt

q(x) 1 L ((2,∞))

Z x

α(x)−1 + f (t)dt

v(x) x/2 (x − t)

q(x) x/2

α(x)−1

L

:= I2,1 + I2,2 + I2,3 .

((2,∞))

74

Alexander Meskhi

Notice that when t ≤ 1 and x ≥ 2, then (x − t)α(x)−1 ≤ cxα(x)−1 . Consequently, using H¨older’s inequality (see Lemma 1.4.1) we find that

I2,1 ≤ c v(x)xα(x)−1 q(x) k f χ[0,1] kL p(·) (I) kχ[0,1] kL p(·) (I) L ((2,∞))

≤ c v(x)xα(x)−1 q(x) k f kL p(·) (I) ≤ cA∞ k f kL p(·) (I) . ([1,∞))

L

It is easy to see that the estimate (x − t)α(x)−1 ≤ cxα(x)−1 and Theorem 1.3.4 implies Z x

α(x)−1 I2,2 ≤ c v(x)x f (t)dt q(x) L

1

([1,∞))

≤ cA∞ k f kLq(x) ([1,∞)) ≤ cA∞ k f kLq(x) (I) ,

while H¨older’s inequality for the classical Lebesgues spaces yields qc

(I2,3 )

+∞ Z Z qc /pc ′ qc (α(x)−1)qc pc ≤ c ∑ v(x) x dx f (t)dt 2k/(pc ) ≤

k=1 Ek q cAq∞c k f kLcp(·) (I) .

Ik

Necessity follows in the same way as in the case of Theorem 5.1.1. In this case we take the test functions ft (x) = χ(t/2,t) (x), t > 0. The details are omitted. To formulate the next statements we recall that the functions p0 (x) and pe0 (x) are defined as follows: ( p0 (x), 0 ≤ x ≤ a p0 (x) := inf p(y); pe0 (x) := , y∈[0,x] pc ≡ const, x < a where a is a fixed positive number.

Theorem 5.1.3. Let I = [0, a], where a < ∞. Suppose that p and q are measurable functions on I and 1 < p− (I) ≤ p0 (x) ≤ q(x) ≤ q+ (I) < ∞. Suppose also that α− (I) > 1/p− (I). If Z a

Ba := sup Ba (t) := sup

0
0
(v(x)xα(x)−1 )q(x)t q(x)/(p0 ) (x) dx < ∞, ′

(5.1.2)

then inequality (5.1.1) holds. Proof. For simplicity assume that a = 1. Suppose that S p ( f ) ≤ 1, where f ≥ 0. We have

Sq,v (Rα(x) )

≤ +

2q− (I)−1 Z 1

v(x)

0

:=

Z 1 Z v(x) 0

Z x

x/2 0

α(x)−1

(x − y)

x/2

2q− (I)−1 (I1 + I2 ).

q(x) (x − y)α(x)−1 f (y)dy dx q(x) f (y)dy dx

Generalized One-sided Potentials

75

If 0 < y < x/2, then (x − y)α(x)−1 ≤ cxα(x)−1 , where the positive constant c does not depend on x. Consequently, using Theorem 1.4.6 we find that q(x) Z 1 Z x I1 ≤ c v(x)(x/2)α(x)−1 f (y)dy dx ≤ C. 0

0

By H¨older’s inequality with respect to the exponent p0 (x) we have Z x q(x)/p0 (x) Z 1 q(x) p0 (x) I2 ≤ (v(x)) ( f (y)) dy 0

× Now observe that Z x

f (y)

p0 (x)

Z

Z

dy ≤

x/2

x/2

x

(α(x)−1)(p0 )′ (x)

(x − y) x/2

[x/2,x]∩{ f ≤1} Z x

( f (y))

p0 (x)

q(x)/(p0 )′ (x) dy dx.

dy +

Z

[x/2,x]∩{ f >1}

( f (y)) p0 (x) dy

( f (y)) p(y) dy;

≤ cx +

x/2

Z x

x/2

′

′

(x − y)(α(x)−1)(p0 ) (x) dy = c p,α x(α(x)−1)(p0 ) (x)+1 .

The latter equality holds because the condition α− (I) > 1/p− (I) guarantees α(x) > 1/p0 (x). Therefore Z 1 (v(x))q(x) xq(x)α(x) dx I2 ≤ c p,q 0

Z 1

+

q(x)

v(x)

0

Z

x

f (y)

p(y)

x/2

c p,q [I2,1 + I2,2 ].

:=

q(x)/p0 (x) (α(x)−1)q(x)+q(x)/(p0 )′ (x) dy x dx

For I2,1 , we find that I2,1 =

∑

k∈Z−

≤

∑

Z

v(x)q(x) x(α(x)−1)q(x) xq(x) dx Ek−1

kq− (I)/p+ (I)

2

k∈Z−

≤ cB1

∑

Z

′

v(x)q(x) x(α(x)−1)q(x) 2(k−1)q(x)/(p0 ) (x) dx Ek−1

2kq− (I)/p+ (I) ≤ cB1 < ∞,

k∈Z−

while taking into account the fact that q(x)/p0 (x) ≥ 1 we have Z x Z ′ α(x)−1 q(x) p(y) I2,2 ≤ c ∑ (v(x)x ) ( f (y)) dy xq(x)/(p0 ) (x) dx k∈Z− Ek−1

≤ c

∑

k∈Z−

Z

x/2

v(x)x Ek−1

α(x)−1 q(x) (k−1)q(x)/(p0 )′ (x)

≤ cB1 S p ( f ) ≤ cB1 < ∞.

2

dx

Z

( f (y))

p(y)

Ik−1

Combining the above-derived estimates we find that I2 < ∞. The proof follows.

dy

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Alexander Meskhi

Theorem 5.1.4. Let I = R+ . Suppose that p(x) and q(x) are measurable functions on I and 1 < p− (I) ≤ p0 (x) ≤ q(x) ≤ q+ (I) < ∞. Suppose also that α− (I) > 1/p− (I) and there exists a positive number a such that q(x) ≡ qc = const, p(x) ≡ pc = const outside [0, a]. If B∞ := sup B∞ (t) := sup

Z ∞

0
0
(v(x)xα(x)−1 )q(x)t q(x)/( pe0 ) (x) dx < ∞, ′

(5.1.3)

q(x)

then Rα(x) is bounded from L p(x) (I) to Lvq(x) (I). Remark 5.1.1. Notice that (5.1.3) is also necessary for the boundedness of Rα(x) from q to Lvq (R+ ), where p and q are constants (see Theorem 5.1.1).

L p (R+ )

Proof of Theorem 5.1.4. Suppose that f ≥ 0 and S p ( f ) ≤ 1. For simplicity assume that a = 1. We have Sq,v (Rα(x) f ) =

Z 2 0

Z q(x) q(x) v(x) (Rα f ) (x)dx +

:= I1 + I2 .

∞ 2

q(x) v(x) (Rα(x) f )qc (x)dx

Observe that the condition B∞ < ∞ implies Ba < ∞. Consequently, by Theorem 5.1.3 we conclude that I1 ≤ c < ∞, while for I2 , we find that I2

Z ∞ Z 1 qc qc α(x)−1 ≤ c (v(x)) (x − y) f (y)dy dx 2 0 Z x/2 qc Z ∞ qc α(x)−1 + (v(x)) (x − y) f (y)dy dx 2 1 Z x qc Z ∞ qc α(x)−1 + (v(x)) (x − y) f (y)dy dx 2

x/2

c[I2,1 + I2,2 + I2,3 ].

:=

Using H¨older’s inequality for Lebesgue spaces with variable exponent (see Lemma 1.4.1) we have Z 1 qc Z ∞ qc α(x)−1 I2,1 ≤ c (v(x)) (x − y) f (y)dy dx 2 0 Z ∞ α(x)−1 qc ≤ c (v(x)(x/2) ) dx k f kL p(·) (I) kχ[0,1] kL p′ (·) (I) ≤ cB∞ , 2

while Theorem 1.3.4 yields I2,2 ≤ c

Z ∞ 1

v(x)x

α(x)−1 qc

Z

x 1

qc f (y)dy dx ≤ cB∞ .

Generalized One-sided Potentials

77

Now applying the condition α(x) > 1/p− (I), we find that Z x qc ∞ Z qc α(x)−1 I2,3 = f (y)(x − y) dy dx ∑ (v(x)) k=1 Ek

≤

∞ Z

∑

x/2

qc

(v(x))

k=1 Ek

× ≤

Z

x

x x/2

qc /pc ( f (y)) dy pc

(α(x)−1)(pc )′

(x − y)

x/2 ∞ Z

c∑

k=1

≤

Z

∞

cB∞ ∑

qc

qc /(pc )′ dy

(α(x)−1)qc +qc /(pc )′

(v(x)) (x/2) Ek

Z

k=1 Ek

dx

Z

( f (y))

p(y)

Ek

dy

( f (y)) p(y) dy ≤ cB∞ < ∞.

Summarizing the estimates for I1 and I2 we have the desired result.

5.2.

Compactness

In this section we give the criteria for which the operator Rα(x) is compact from L p(·) (I) to q(·)

Lv (I). Theorem 5.2.1. Let I = [0, a], 0 < a < ∞, and let 1 < p− (I) ≤ p+ (I) ≤ q− (I) ≤ q+ (I) < ∞. Suppose that (α − 1/p)− (I) > 0. Further, assume that p, q ∈ W L(I). Then Rα(x) is q(·)

compact from L p(·) (I) to Lv (I) if and only if Aa < ∞;

(i)

lim Aa (t) = 0,

(ii)

t→0

where Aa and Aa (t) are defined in Theorem 5.1.1. Proof. Sufficiency. For simplicity assume that a = 1 (In this case Aa = A1 ). We represent Rα(x) as follows: (1)

(2)

Rα(x) f (x) = Rα(x) f (x) + Rα(x) f (x), where

(2)

(1)

Rα(x) f (x) = χ[0,β] (x)Rα(x) f (x), Rα(x) f (x) = χ(β,1] (x)Rα(x) f (x), and 0 < β < 1. Observe that by Lemma 1.4.9 we have the following estimates:

χ(β,1] (x)v(x) χ[0,x] (y)(x − y)α(x)−1 L p′ (y) (I) q(x) L (I)

α(x)−1

p′ (y) ≤ χ(β,1] (x)v(x) χ[0,x/2] (y)(x − y) L (I) Lq(x) (I)

+ χ(β,1] (x)v(x) χ(x/2,x] (y)(x − y)α(x)−1 L p′ (y) (I) q(x) L (I)

α(x)−1/p(x)

≤ c χ(β,1] (x)v(x)x Lq(x) (I)

+c χ(β,1] (x)v(x)xα(x)−1/p(x) Lq(x) (I) < ∞,

78

Alexander Meskhi (1)

because A1 < ∞. Consequently, by Theorem 1.4.8, Rα(x) is compact. Further, observe that the arguments of the proof of Theorem 5.1.1 enable us to conclude that (1)

(2)

kRα(x) − Rα(x) kL p(·) (I)→Lq(·) (I) ≤ kRα(x) kL p(·) (I)→Lq(·) (I) ≤ c sup A1 (t), v

v

0
where the positive constant c depends only on p, q and α. Passing β to 0 we have that Rα(x) is compact as a limit of compact operators. Necessity. Suppose that ft (x) = t −1/p(0) χ[0,t/2) (x). Hence, using H¨older’s inequality for spaces (see Lemma 1.4.1), Proposition 1.4.1 and the absolutely continuity of the norm k · kL p′ (·) , we have Z 1 ≤ k(p)k ft (·)k p(·) kϕ(·)χ[0,t/2) (·)k p′ (·) f (x)ϕ(x)dx t L (I) L (I) L p(·)

0

≤ ct −1/p(0)t 1/p+ ([0,t/2]) kϕ(·)χ[0,t/2) (·)kL p′ (·) (I)

≤ ckϕ(·)χ[0,t/2) (·)kL p′ (·) (I) → 0

′

as t → 0 for all ϕ ∈ L p (x) (I). Hence, ft converges weakly to 0 as t → 0. Further, it is obvious that

Z t/2

kRα(x) ft kLq(·) (I) ≥ χ[t,1) (x)v(x) (x − t)α(x)−1 dt q(·) t −1/p(0) v

L

0

′

≥ ct −1/p (0) χ[t,1) (x)v(x)xα(x)−1

Lq(x) (I)

(I)

.

Finally we conclude that limt→0 A1 (t) = 0 because the compact operator maps a weakly convergent sequence into strongly convergent one. Theorem 5.2.2. Let I = R+ and let 1 < p− (I) ≤ p(x) ≤ q(x) ≤ q+ (I) < ∞. Suppose that p(x) ≡ pc = const and q(x) ≡ qc = const when x > a for some positive constant a. Let (α − 1/p)− (I) > 0. Further, assume that p, q ∈ W L(I). Then Rα(x) is compact from L p(·) (I) q(·)

to Lv (I) if and only if A∞ < ∞;

(i)

lim A∞ (t) = lim A∞ (t) = 0,

(ii)

t→0

t→∞

where A∞ and A∞ (t) are defined in Theorem 5.1.2. Proof. For simplicity assume that a = 1. To prove sufficiency we use the representation (n) Rα(x) f = ∑5n=1 Rα(x) f , where (1)

Rα(x) f (x) = χ[0,β) (x)(Rα(x) f )(x), (2)

Rα(x) f (x) = χ[β,γ) (x)Rα(x) (χ[0,β/2] f )(x), (3)

Rα(x) f (x) = χ[β,γ) (x)Rα(x) (χ[β/2,∞) f )(x), (4)

Rα(x) f (x) = χ[γ,∞) Rα(x) (χ[0,γ/2) f )(x), (5)

Rα(x) f (x) = χ[γ,∞) (x)Rα(x) (χ[γ/2,∞) f )(x),

Generalized One-sided Potentials where 0 < β < 1/2 < 2 < γ < ∞. Now observe that

α(x)−1

χ[β,γ) (x)v(x)

χ[0,β/2) (y)(x − y)

≤ c χ[β,γ) (x)v(x)xα(x)−1

′

L p (x) (I)

79

′ L p (y) (I) Lq(x) (I)

χ[0,β/2) p′ (·) < ∞ (I)

L

because A∞ < ∞. Further,

α(x)−1

χ[β,γ) (x)v(x) χ[β/2,∞) (y)(x − y)

p′ (y)

L (I) Lq(x) (I)

α(x)−1

≤ χ[β,γ) (x)v(x) χ[β/2,x/2) (y)(x − y)

p′ (y)

L

(I) Lq(x) (I)

α(x)−1 + χ (x)v(x) χ (y)(x − y)

[β,γ) [x/2,x)

′ L p (y) (I)

:= I1 + I2 .

It is easy to see that Lemma 1.4.1 (for f = χ[β/2,γ/2] ) implies

I1 ≤ c χ[β,γ) (x)v(x)xα(x)−1 q(x) χ[β/2,γ/2) (·) L

Lq(x) (I)

′

L p (·) (I)

(I)

< ∞.

Analogously, by Lemma 1.4.9 we can see that I2 < ∞ because A∞ < ∞. Applying Theorem 5.1.1 we find that (1)

kRα(x) kL p(·) (I)→Lq(·) (I) = kRα(x) kL p(·) ([0,β))→Lq(·) ([0,β)) ≤ c sup A∞ (t) → 0 v

v

0
as β → 0. Arguing as in the proof of sufficiency of Theorem 5.1.2 (see also [160] for constant α), we conclude that the inequality

(5)

Rα(x) f (x)

q(x)

L p(x) ([γ,∞))→Lv

≤ c sup

([γ,∞))

t>γ/2

Z

∞

qc (α(x)−1)qc

(v(x)) x

t

1/qc ′ dx (t − γ)1/(pc )

holds. The latter term tends to 0 when γ → ∞ because limt→∞ A∞ (t) = 0. Finally we conclude that Rα(x) is compact. Necessity. The condition A∞ < ∞ is a consequence of Theorem 5.1.2. The condition lim A∞ (t) = 0 follows in the same manner as in the proof of necessity of Theorem 5.2.1. To

t→0

show that lim A∞ (t) = 0, we can (as above) use the facts that p and q are constants outside t→∞

q(·)

[0, a] and Rα(x) is compact from L p(·) (I) to Lv (I) if and only if the operator Wα,v f (x) = ′

′

is compact from Lq (x) (I) to L p (x) (I).

Z ∞ x

v(y) f (y)(y − x)α(y)−1 dy

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Alexander Meskhi

Measure of Non–compactness

This section deals with two-sided estimates of the distance between the operator Rα(x) and q(·)

the class of compact linear operators from L p(·) (I) to Lv (I) when I is either finite interval [0, a] or R+ provided that p and q satisfy the weak Lipschitz condition on I (see Definition 1.4.1). Let X and Y be Banach spaces. Recall that (see Section 1.4) K (X,Y ) (resp. FL (X,Y )) denotes the class of compact linear operators (resp. finite rank operators) acting from X to Y . Let kT kK (X,Y ) := dist{T, K (X,Y )}; α(T ) := dist{T, FL (X,Y )}, where T is a bounded linear operator from X to Y . Recall that the symbol P (Ω) denotes the class of those p for which the HardyLittlewood maximal operator MΩ is bounded in L p(x) (Ω) (see section 1.4). Theorem 5.3.1. Let Ω ⊆ Rn be a domain. Assume that X is a Banach space. Suppose that 1 < q− (Ω) ≤ q+ (Ω) < ∞ and q ∈ P (Ω). Then kT kK (X,Lq(·) (Ω)) = α(T ), where T is a bounded linear operator from X to Y . Proof. Let δ > 0. Then there exists an operator K ∈ K (X, L p(·) (Ω)) such that kT − Kk < kT kK + δ. By Lemma 1.4.4 there is P ∈ FL (X, L p(·) (Ω)) for which the inequality kK − Pk < δ holds. This gives kT − Pk ≤ kT − Kk + kK − Pk ≤ kT kK + 2δ. Hence α(T ) ≤ kT kK . The reverse inequality is obvious. Theorem 5.3.2. Let I = [0, a], where 0 < a < ∞. Suppose that 1 < p− (I) ≤ p(x) ≤ q(x) ≤ q+ (I) < ∞. Assume that (α − 1/p)− (I) > 0. Let p ∈ W L(I) and let Aa < ∞ (see Theorem 5.1.1). Then there exists two positive constants b1 and b2 such that b1 A ≤ kT kK (L p(·) (I),Lq(·) (I)) ≤ b2 A , v

where A := lim Aβ , Aβ := sup Aa (t) and Aa (t) is defined in Theorem 5.1.1. β→0

0
Theorem 5.3.3. Let I := R+ and let 1 < p− (I) ≤ p(x) ≤ q(x) ≤ q+ (I) < ∞. Suppose that (α − 1/p)− (I) > 0. Further, assume that p ∈ W L(I). Suppose also that A∞ < ∞ (see Theorem 5.1.2). Then there exists two positive constants b1 and b2 such that b1 A ∞ ≤ kT kK (L p(·) (I),Lq(·) (I)) ≤ b2 A ∞ , v

where A ∞ := lim Aβ + lim A(γ) , Aβ := sup A∞ (t) and A(γ) := sup A∞ (t) and A∞ (t) is defined β→0

in Theorem 5.1.2.

γ→∞

0
t>γ

Generalized One-sided Potentials

81

Proof of Theorem 5.3.2. For simplicity we assume that a = 1. The upper estimate follows immediately from the estimate (2)

(1)

kRα(x) − Rα(x) kL p(·) (I)→Lq(·) (I) ≤ kRα(x) kL p(·) (I)→Lq(·) (I) ≤ cAβ , v

(1)

v

(2)

where Rα(x) = χ[0,β] (x)Rα(x) f (x), Rα(x) = χ[β,1] (x)Rα(x) f (x), 0 < β < 1 (see the proof of (2)

Theorem 5.2.1 for the details) and the fact that Rα(x) is compact (see Theorem 1.4.8). To get the lower estimate we take a positive number λ so that λ > kT kK (L p(·) (I),Lq(·) (I)) . Notice v that kRα(x) kK L p(·) (I),Lq(·) (I) = kRα(x),v kK (L p(·) (I),Lq(·) (I)) , v

where

Rα(x),v f (x) = v(x) Consequently, by Theorem 5.3.1

Z x

f (t)(x − t)α(x)−1 dt.

0

λ > α(Rα(x),v ). Hence, there exist g1 , . . . , gN ∈ Lq(·) (I) such that α(Rα(x),v ) ≤ kRα(x),v − Fk < λ, where

N

F f (x) =

∑ α j ( f )g j (x),

j=1

α j are linear bounded functionals in and gi are linearly independent. Further, there exist g¯1 , . . . , g¯N such that support of g¯i is in [σi , a], 0 < σi < a, and L p(·) (I)

kRα(x),v − F0 k < λ, where F0 f (x) = ∑Nj=1 α j ( f )g¯ j (x). Suppose that σ = min{σ j }. Then suppF0 f ⊂ [σ, a]. Let 0 < t < β < σ and let f be a non-negative function with support in [0,t/2] such that k f kL p(·) (I) ≤ 1. Consequently, for such an f we have λ ≥ λk f kL p(·) (I) ≥ kχ[0,β] (x)(Rα(x),v f (x) − F0 f (x))kL p(x) (I) ≥ kχ[t,β] (x)(Rα(x),v f )(x)kLq(x) (I)

Z t/2

α(x)−1

≥ χ[t,β] (x)v(x) (x − y) f (y)dy

q(x) 0 L (I)

Z t/2

α(x)−1 ≥ c f (y)dy .

χ[t,β] (x)v(x)x

Lq(x) (I)

0

Taking the supremum with respect to f we find that

χ[t,β] (x)v(x)xα(x)−1 kχ[0,t/2] (·)kL p′ (·) (I) ≥ cAβ .

Lq(x) (I)

Taking into account the condition p ∈ W L(I) and Lemma 1.4.1 we have the desired result. In a similar way can be proved Theorem 5.3.3; therefore we omit it.

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Alexander Meskhi

Notes and Comments on Chapter 5

The statements of Sections 5.1 and 5.2 were presented in [9]. Necessary and sufficient conditions on a weight function v guaranteeing the boundedq ness of the Riemann-Liouville operator Rα from L p (R+ ) to Lv (R+ ), where α, p and q are constants were established in [160] for α > 1/p and in [118] for 0 < α < 1/p (see also [163], [49], Ch. 2). Later the same problems were investigated independently in [198], [199]. For weighted inequalities for the classical integral operators in variable exponent function spaces we refer to the papers [122]–[133], [57], [52], [136], [31], [53], [33], [51], [119], [105], [106], [210], [211], [90], etc (see also the surveys [111], [208] and references therein). Integral–type necessary conditions and sufficient conditions governing the compactness q(·) of the Hardy operator H from L p(·) (I) to Lv (I) were established in [52]. We refer also to [57] for the compactness of the potential-type operators in weighted L p(·) spaces with p(·) special weights. A dominated compactness theorem in Lρ (Ω, µ), where µ(Ω) < ∞ and ρ is a power-type weight was established in [200]. This result was applied to fractional integral operators over bounded sets. In [163] (see also [49], Ch. 2) two-sided estimates of the measure of non–compactness for one-sided potentials acting from the classical Lebesgue space into the classical weighted Lebesgue space were obtained. Lower and upper estimates of the measure of non– compactness for the Hardy operator in variable exponent Lebesgue spaces were studied in [52].

Chapter 6

Singular Integrals In this chapter the essential norm for singular integrals (Hilbert transforms, Cauchy integrals, Riesz transforms, Calde´on-Zygmund operators), generally speaking, in weighted Lebesgue spaces with non–standard growth is estimated from below. We keep the notation of Chapter 1.

6.1.

Hilbert Transforms

Suppose that H is the Hilbert transform (see Section 1.6 for the definition). The following statements give the lower estimate of the essential norm for H in classical weighted Lebesgue spaces. Theorem 6.1.1. Let 1 < p < ∞. Suppose that H is bounded from Lwp (R) to Lvp (R). Then kHkK

q p (Lw (R)→Lv (R))

≥ sup lim

a∈R r→0

Z

a+r

v(x)dx a

1/p Z

∞

′

w1−p (x) dx p′ a+r (x − a)

1/p′

.

Theorem 6.1.2. Let 1 < p < ∞ and let w ∈ A p (R). Then kHkK (Lwp (R)) ≥ where

1 max{A¯ 1 , A¯ 2 }, 2

Z a+r 1/p Z a 1/p′ 1 1 1−p′ ¯ A1 = sup lim w(x)dx w (x)dx ; r a−r a∈R r→0 r a Z a 1/p Z a+r 1/p′ ′ 1 1 1−p A¯2 = sup lim w(x)dx w (x)dx . r a a∈R r→0 r a−r

Proof of Theorem 6.1.1. Let a ∈ R and λ > kHkK (Lwp (R)→Lvp (R)) . By Lemma 1.2.8 there exists a positive number β such that for all r < β and all f with the support in ⊂ (a + r, ∞)we have Z a+r Z ∞ v(x)|H f (x)| p dx ≤ λ p w(x)| f (x)| p dx. (6.1.1) a

a+r

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Alexander Meskhi

It is obvious that if y ∈ (a + r, ∞) and x ∈ (a, a + r), then y − x ≤ y − a. If we assume that ′

f (y) = w1−p (y)χ(a+r,∞) (y)(y − a)1−p

′

in (6.1.1), then we find that λp

Z ∞

′

′

w1−p (x)(x − a)−p dx ≥

a+r

≥ Now if we show that

Z a+r

v(x)|H f (x)| p dx

a

Z

a+r

v(x)dx a

J(a, r) :=

Z ∞

a+r

Z

′

∞

w1−p (x) dx p′ a+r (x − a)

′

p

.

′

w1−p (x)(x − a)−p dx < ∞

for all a ∈ R and r < β, then we are done because (6.1.1) implies the inequality 1/p′ Z a+r 1/p Z ∞ 1−p′ w (x) ≤λ v(x)dx dx p′ a a+r (x − a)

for all a ∈ R and r < α. Suppose the opposite: there exists a ∈ R and r > 0 such that J(a, r) = ∞. By duality arguments there exists a function g ∈ L p (a + r, ∞) such that g ≥ 0 and Z ∞ g(x)w−1/p (x) a+r

x−a

dx = ∞.

Further, we take the function φ(x) = g(x)χ(a+r,∞) (x)w−1/p (x) in the two-weight inequality kHφkLvp ≤ ckφkLwp and, consequently, we conclude that Z a+r Z ∞ p Z ∞ g(x)w−1/p (x) ∞= v(x)dx dx ≤ c (g(x)) p dx < ∞ x−a a a+r a+r which is impossible unless v(x) = 0 almost everywhere on (a, a + r).

Proof of Theorem 6.1.2. First notice that by Theorem 1.6.1 the condition w ∈ A p (R) implies the boundedness of H in Lwp (R). Let λ > kHkK (Lwp (R) and let a ∈ R. Using again Lemma 1.2.8 we have that there is β > 0 such that if 0 < r < β and supp f ⊂ (a − r, a), then Z a+r a

w(x)|H f (x)| p dx ≤ λ p

Z a

w(x)| f (x)| p dx.

(6.1.2)

a−r

It is clear that x − y < 2r when y ∈ (a − r, a), x ∈ (a, a + r). Let us put f (y) = in (6.1.2). Then we observe that Z a+r Z a p Z a Z a+r 1 p 1−p′ p 1−p′ λ w ≥ w(x)|H f (x)| dx ≥ w(x)dx w (x)dx (2r) p a−r a a a−r

′ w1−p (y)χ(a−r,a) (y)

Singular Integrals

85

′

from which, taking into account the fact that w1−p ∈ Lloc (R) (see e.g. [95]), it follows 1 ¯ 1 ¯ 2 A1 ≤ λ. Analogously we have 2 A2 ≤ λ. Corollary 6.1.1. Let p = 2, w(x) = |x|α , −1 < α < 1. Then kHkK (Lw2 (R)) ≥

1 . 1+α

This follows immediately from Theorem 6.1.2 if we assume that v(x) ≡ w(x) ≡ |x|α and a = 0. Corollary 6.1.2. Let 1 < p < ∞ and let w ∈ A p (R). Then kHkK (Lwp (R)→Lwp (R))) ≥

1 (r,a) sup lim A p (R), 2(1 + 2kHkLwp (R) ) a∈R r→0

(r,a)

where A p (R) is defined in Definition 1.6.1, and kHkLwp (R) := kHkLwp (R)→Lwp (R) .

Proof. First note that the condition w ∈ A p (R) and Theorem 1.6.1 imply the inequalities: Z a+r Z a p p w(x)dx ≤ 2 kHkL p (R) w(x)dx, (6.1.3) w

a

Z a

a−r

w(x)dx ≤ 2 p kHkLpp (R) w

a−r

Z a+r

(6.1.4)

w(x)dx.

a

Indeed, if we put f (y) = χ(a−r,a) (y) in the one-weight inequality Z

p

w(x)|H f (x)| dx ≤ R

kHkLpp (R) w

Z

w(x)| f (x)| p dx,

(6.1.5)

R

then we find that Z

w(x)|H f (x)| p dx ≥ R

Z a+r a

R

Z w(x)

a

dy x a−r − y

p

dx ≥

1 2p

Z a+r

w(x)dx.

a

a On the other hand, k f kLpp (R) = a−r w < ∞. Hence (6.1.3) holds. w Analogously we can show that (6.1.4) holds. Let us introduce the notation:

W (b, c) :=

Z

c

w(x)dx b

1/p

; V (b, c) :=

Z

c

1−p′

w b

(x)dx

1/p′

Further, due to Theorem 6.1.2 and (6.1.3) − (6.1.4) we conclude that 1 1 W (a − r, a + r)V (a − r, a + r) ≤ W (a − r, a)V (a − r, a) 2r 2r +

1 1 W (a − r, a)V (a, a + r) + W (a, a + r)V (a − r, a) 2r 2r

.

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Alexander Meskhi

2kHkLwp (R) 1 W (a, a + r)V (a, a + r) ≤ W (a, a + r)V (a − r, a) 2r 2r 1 1 + W (a − r, a)V (a, a + r) + W (a, a + r)V (a − r, a) 2r 2r p 2kHkLw (R) + W (a − r, a)V (a, a + r) ≤ 2(1 + 2kHkLwp (R) )kHkK (Lwp (R)) 2r when r is small. +

6.2.

Cauchy Singular Integrals

Let Γ be a smooth Jordan curve and let SΓ be the Cauchy singular integral operator along Γ (see Section 1.6 for the definition). We begin with the following Lemma: Lemma 6.2.1. Let 1 < p < ∞. Suppose that SΓ is bounded from Lwp (0, l) to Lvp (0, l). Then Z ′ SI := w1−p (s)ds < ∞, for all subintervals I of (0, l). I

Proof. Let S = ∞ for some I. Consequently, it follows that there exists g ∈ L p (I), g ≥ R I −1/p 0, such that g(t)w (t)dt = ∞. Now let φ(s) = f (t(s)) = g(s)w−1/p (s)χI (s). Let I = I

(a − r, a) ⊂ (0, l). Without loss of generality we can assume that I ′ = (a, a + r) ⊂ (0, l). We have (see [109], [107], p. 56) |SΓ f (t(σ))| ≥ for σ ∈

I′.

1 2π

Z I

φ(s) 1 ds ≥ s−σ 4πr

Z

φ(s)ds

(6.2.2)

I

Consequently, |SΓ f (t(σ))| ≥

1 Z φ(s)ds χI ′ (σ) 4πr I

for any σ ∈ (0, l). Hence, using inequality (6.2.3), we find that 1 kSΓ f kLvp (0,l) ≥ 4πr =

On the other hand,

1 4πr

Z |I

Z I

φ(s)ds kχI ′ kLvp (0,l)

g(s)w−1/p (s)ds kχI ′ kLvp (0,l) = ∞. {z

=∞

}

kφkLwp (0,l) = kgkL p (I) < ∞.

This contradicts the boundedness of SΓ from Lwp (0, l) to Lvp (0, l).

(6.2.3)

Singular Integrals

87

Theorem 6.2.1. Let 1 < p < ∞. Suppose that Γ is a Jordan smooth curve. Then there exists no weight pair (v, w) such that the operator SΓ is compact from Lwp (0, l) to Lvp (0, l). Moreover, if SΓ is bounded from Lwp (0, l) to Lvp (0, l), then the inequality kSΓ kK (Lwp (0,l)) ≥ holds, where

1 e1 , A e2 } max{A 4π

Z a+r 1/p Z a 1/p′ 1 1 1−p′ e v(s)ds w (s)ds A1 := sup lim ; r a−r a∈(0,l) r→0 r a

1/p Z a 1/p′ Z a+r ′ 1 1 1−p e2 := sup lim w(s)ds w (s)ds . A r a−r a∈(0,l) r→0 r a

Proof. Let SΓ be bounded from Lwp (0, l) to Lvp (0, l), λ > kSΓ kK (Lwp (0,l),Lvp (0,l)) and a ∈ (0, l). Then, using Lemma 1.2.8 there exists a positive number β such that for all r < β we have kSΓ f kLvq (I(a,r)) ≤ λk f kLwp (0,l) , f ∈ Lwp (0, l), (6.2.4) where I(a, r) = (a − r, a + r). Let I1 := (a − r, a), I2 := (a, a + r) ϕ(s) = f (t(s)) ≥ 0 ) and supp ϕ ⊂ I2 . Then we have the estimate similar to (6.2.2): Z Z 1 ϕ(s) 1 |SΓ f (t(σ))| ≥ ds χI1 (σ) ≥ ϕ(s) ds χI1 (σ), 2π I2 s − σ 4rπ I2 Analogously,

|SΓ f (t(σ))| ≥ 1−p′

1 4rπ

Z

I1

ϕ(s)ds χI2 (σ).

′

By Lemma 6.2.1 we have that w is locally integrable. Let ϕ(s) = w1−p (s)χI1 (s). Then by (6.2.4) we have Z a+r p−1 Z a 1 1 1 1−p′ w (s)ds v(s)ds ≤ λ p , a ∈ (0, l). (4π) p r a r a−r The latter inequality implies

Analogously, it follows that

This completes the proof.

1 e A2 ≤ λ. 4π 1 e A1 ≤ λ. 4π

Theorems 6.2.1 and 1.6.2 imply the next statement: Theorem 6.2.2. Let 1 < p < ∞. Suppose that Bl < ∞, where Bl is defined in Theorem 1.6.2. Then the inequality kSΓ kK (Lwp (0,l)) ≥

1 e1 , A e2 } max{A 4π

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Alexander Meskhi

holds, where Z a+r 1/p Z a 1/p′ 1 1 1−p′ e A1 := sup lim w(s)ds w (s)ds ; r a−r a∈(0,l) r→0 r a Z a+r 1/p Z a 1/p′ 1 1 1−p′ e A2 := sup lim w(s)ds w (s)ds . r a−r a∈(0,l) r→0 r a

The next two corollaries follow in the same manner as in the case of the Hilbert transform. Corollary 6.2.1. Let p = 2, w(x) = xα , where −1 < α < 1. Then kSΓ kK (Lwp (0,l)) ≥

1 . 4π(1 − α2 )1/2

Corollary 6.2.2. Let 1 < p < ∞ and let w ∈ A p (0, l) (see Theorem 1.6.2). Then kSΓ kK (Lwp (0,l)) ≥ where (r,a) Ap

=

1 2r

Z a+r a−r

1 (r,a) sup lim A p , 4π(4πkSΓ kLwp (0,l) + 1) a∈(0,l) r→0

w(s)ds

1/p

1 2r

Z a+r

1−p′

w

(s)ds

a−r

and kSΓ kLwp (0,l) is the norm of SΓ in Lwp (0, l).

6.3.

1/p′

Riesz Transforms

Let R j f , 1 ≤ j ≤ n, be the Riesz transforms of f defined by (1.6.5). Theorem 6.3.1. Let 1 < p < ∞. Then there are no pair of weights (v, w) and integer j, 1 ≤ j ≤ n, such that the operator R j is compact from Lwp (Rn ) to Lvp (Rn ). Moreover, if R j is bounded from Lwp (Rn ) to Lvp (Rn ) for some j, then the following inequality holds

v(a) kR j kK ≥ An ess sup n w(a) a∈R where An =

γn Bn , 2n+1 n3/2

Bn =

1/p

,

πn/2 Γ(1+n/2) .

Proof. Let R j be bounded from Lwp (Rn ) to Lvp (Rn ) for some 1 ≤ j ≤ n. By Lemma ′ 1.6.2 we have that w1−p ∈ Lloc (R). Using Lemma 1.2.8, for λ > kR j kK L p (Rn ),L p (Rn ) and w

w

a∈ there exists a positive number β such that for all 0 < τ < β and f ∈ Lwp (Rn ) the inequality kR j,v f kL p (B(a,r)) ≤ λk f kLwp (Rn ) (6.3.1) Rn ,

holds, where R j,v f = vR j f .

Singular Integrals

89

Let a = (a1 , . . . , an ) and let n o E j,a := x = (x1 , . . . , xn ) ∈ Rn : max{|xi − ai |, 1 ≤ i ≤ n} = x j − a ∩ B(a, τ). It is obvious that

B(a, τ) = ∪nj=1 E j,a ∩ (−E j,a ) .

Let f¯ = f χ−E j,a and let x ∈ E j,a . Then |R j f¯(x)| = γn ≥ γn

Z

−E j,a

f (t)

Z

−E j,a

f (t)

xj −tj dt |x − t|n+1

(x j − t j ) γn dt = 1/2 1/2 n |x − t| n (x j − t j ) n Z γn ≥ 1/2 f (t)dt, n (2τ)n −E j,a

Z

−E j,a

f (t) dt |x − t|n

where f ≥ 0. ′ Further, using the latter estimates and assuming f = w1−p in (6.3.1) we have Z Z p Z ′ 1−p′ cn v(x)dx w (x)dx ≤ λ p w1−p (x)dx, −E j,a

E j,a

−E j,a

where cn =

γnp . (n1/2 2n ) p τnp

On the other hand, notice that 1 |E j,a | Indeed, we have

Z

v(x)dx → v(a) a.e.. Ej

|B(a, τ)| Bn τn = . 2n 2n Z 1 Z ≤ 1 v(x) − v(a) |E j,a | E |E j,a | E |v(x) − v(a)|dx j,a j,a |E j,a | =

Therefore

2n ≤ |B(a, τ)|

as τ → 0. Analogously,

Z

(6.3.2)

|v(x) − v(a)|dx −→ 0 a.e.

B(a,τ)

Z 1 1−p′ 1−p′ (x) − w (a)|dx → 0 | − E j,a | −E |w j,a

when τ → 0. Hence

cn |E j,a | p

v(a) ≤ λp w(a)

for almost every a ∈ Rn , which on the other hand, together with (6.3.2) implies the desired estimate.

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Alexander Meskhi

6.4.

Calder´on–Zygmund Operators

In this section we discuss the essential norm of the Calder´on-Zygmund singular integral operator K (see (1.6.1) for the definition of K). Our aim in this section is to prove the following statement: Theorem 6.4.1. Let 1 < p < ∞. Suppose that conditions (1.6.2)-(1.6.4) are satisfied. Then there exists no weight pair (v, w) such that the singular integral operator K is compact from Lwp (Rn ) to Lvp (Rn ). Moreover, if K is bounded from Lwp (Rn ) to Lvp (Rn ), then the inequality v(a) 1/p kKkK ≥ c ess sup (6.4.1) w(a) a∈Rn holds, where the positive constant c depends only on n, b and t (see (1.6.4) and Lemma 1.6.1 for b and t). ′

Proof. Let K be bounded from Lwp (Rn ) to Lvp (Rn ). Lemma 1.6.2 implies that w1−p is locally integrable. Further, repeating the arguments of Theorem 6.3.1 we see that by Lemma 1.2.7 for λ > kKkK (Lwp (Rn ),Lwp (Rn ) ) and a ∈ Rn , there exists β > 0 and R ∈ FL Lwp (Rn ), Lvp (Rn ) with supp R ⊂ Rn \ B(a, β) for all f ∈ Lwp (Rn ) such that for all f ∈ Lwp (Rn ) the inequality kK f kL p (Rn ) ≤ λk f kLwp (Rn ) (6.4.2) holds. Let B := B(a, r), where r < β. Suppose that B′ is the translation of B in the direction of u, i.e. B′ = B(a + ru, r), where u = tu0 , t is taken so that the conditions of Lemma 1.6.1 are satisfied and u0 is the unit vector chosen so that (1.6.4) holds. Let f be any non-negative function supported in B. Consider T f (x) for x ∈ B′ . We have K f (x) =

Z

k(x − y) f (y)dy B

with x = a + ru + rx′ , |x′ | < 1. Since y ∈ B, we find that y = a + ry′ for |y′ | < 1. Thus x − y = r(u + r(y′ − x′ )) = r(u + v) with |v| < 2. Further, Lemma 1.6.1 and condition (1.6.4) yield 1 1 (6.4.3) |K f (x)| ≥ fB |k(ru)| ≥ c fB , 2 |B| for all x ∈ B′ , where |B| denotes a measure of B and c is the positive constant depending only on n, b and t. Due to inequality (6.4.2) we obtain Z p Z Z p v(x) k(x − y) f (y)dy dx ≤ λ ( f (y)) p w(y)dy B′

B

B ′

for all non-negative f with supp f ⊂ B. Let f (x) = w1−p (x)χB (x). Then using (6.4.3), we find that Z Z cp ′ p p v(x)dx f ≤ λ w1−p (y)dy. B ′ |B| p B B

Singular Integrals

91

Consequently by Lemma 1.6.2 we have ′

c p vB′ ((w1−p )B ) p−1 ≤ λ p .

(6.4.4)

Further, observe that the equality lim vB′ = v(a)

(6.4.5)

r→0

holds for almost all a. This follows from the obvious fact |vB′ − v(a)| ≤ c

1 e |B|

Z

Be

|v(x) − v(a)|dx → 0

as r → 0, where Be = B(a, r(t + 1)) and c is a positive constant. Inequalities (6.4.4) and (6.4.5) yield v(a) 1/p c ≤λ w(a)

for almost every a (here the positive constant c depends only on a, n and t). As λ is an arbitrary number greater than kKkK , we conclude that (6.4.1) holds.

6.5.

Hilbert Transforms in L p(x) Spaces

Here we estimate from below the essential norm of the Hilbert transform acting between two weighted Lebesgue spaces with variable exponent. In particular, we show that there is p(·) no weight pair (v, w) and a function p ∈ W L(R) for which H is compact from Lw (R) to p(·) Lv (R). First we formulate the main results of this section p(·)

p(·)

Theorem 6.5.1. Let p ∈ P (R) and let H be bounded from Lw (R) to Lv (R). Then the following estimate holds kHkK ≥ (1/2) max{A1 , A2 }, where

(6.5.1)

1 A1 = sup lim kχ(a−r,a) vkL p(·) (R) kχ(a,a+r) w−1 kL p′ (·) (R) , a∈R r→0 r 1 A2 = sup lim kχ(a,a+r) vkL p(·) (R) kχ(a−r,a) w−1 kL p′ (·) (R) . r→0 r a∈R p(·)

p(·)

Theorem 6.5.2. Let p ∈ P (R). Suppose that H is bounded from Lw (R) to Lv (R). Then kHkK ≥ max{B1 , B2 }, (6.5.2) where B1 = sup lim kχ(a,a+r) (·)v(·)kL p(·) (R) kχ(a+r,+∞) (·)w−1 (·)(· − a)−1 kL p′ (·) (R) ; a∈R r→0

92

Alexander Meskhi B2 = sup lim kχ(a−r,a) (·)v(·)kL p(·) (R) kχ(−∞,a−r) (·)w−1 (·)(a − ·)−1 kL p′ (·) (R) . a∈R r→0

The next statement also holds. p(·)

p(·)

Theorem 6.5.3. Let p ∈ P (R). Suppose that H is bounded from Lw (R) to Lv (R). Then kHkK ≥ (1/4) max{C1 ,C2 }, where C1 = sup lim kχI(a,r) (·)v(·)kL p(·) (R) kχR\I(a,r) (·)w−1 (·)| · −a|−1 kL p′ (·) (R) ; a∈R r→0

C2 = sup lim kχR\I(a,r) (·)v(·)| · −a|−1 kL p(·) (R) kχI(a,r) (·)w−1 (·)kL p′ (·) (R) . a∈R r→∞

Now we give another estimate of the essential norm of H. p(·)

p(·)

Theorem 6.5.4. Let p ∈ P (R). Assume that H is bounded from Lw (R) to Lv (R). Then kHkK ≥ (1/4) sup lim kχ(a−r,a+r) vkL p(·) (R) kw−1 (·)(r + |a − ·|)−1 kL p′ (·) (R) . a∈R r→0

Corollary 6.5.1. Let p satisfy (1.4.1) and (1.4.2). Then there is no weight pair (v, w) p(·) p(·) such that H is compact from Lw (R) to Lv (R). Moreover, if H is bounded from L p(·) (R) to L p(·) (R), then the inequality Z a+r 1/p− (I(a,r)) 2 e−A/(p− ) 1 p(t) kHkK L p(·) (R),L p(·) (R) ≥ sup lim (v(t)) dt w v 4 a>0 r→0 2r a−r Z a+r 1/(p′ )− (I(a,r)) 1 −p′ (t) × (w(t)) dt >0 2r a−r

holds. Corollary 6.5.2. Let p satisfy conditions (1.4.1) and (1.4.2). Then kHkK

≥ (1/4)e−A/(p− )2 ,

L p(·) (R) )

where the positive constant A is from (1.4.1).

Remark 6.5.1. It is known that if ( e |x|−1/p ln−1 |x| , if 0 < x ≤ 1, v(x) = 1, if x > 1 ( |x|−1/p , if 0 < x ≤ 1, w(x) = , 1, if x > 1

;

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93

where p is a constant with 1 < p < ∞, then H is bounded from Lwp (R) to Lvp (R) (see [46]). Based on this fact and Theorem 6.5.2 we have the following estimate: kHkK

p

≥ (p − 1)−1/p .

p

Lw (R),Lv (R)

To prove the main results of this section, we need some lemmas. p(·)

p(·)

Lemma 6.5.1. Let T be a linear map from Lw to Lv . Then T is bounded (resp. compact) if and only if Tv,w is bounded (resp. compact) in L p(·) , where Tv,w f := vT ( f w−1 ). Moreover, kT kL p(·) (R)→L p(·) (R) = kTv,w kL p(·) (R)→L p(·) (R) . Further, if T is bounded, then w

v

kT kK

p(·)

p(·)

= kTv,w k

K (L p(·) (R)) .

Lw (R), Lv (R)

Proof. The first part of the lemma can be checked immediately. For the second part observe that kTv,w − PkL p(·) (R)→L p(·) (R) = kT − P˜v,w kL p(·) (R)→L p(·) (R) , w

v

where P˜v,w f = 1/vP( f w). p(·)

p(·)

Lemma 6.5.2. Let H be bounded from Lw (R) to Lv (R). Then kGI (·)kL p′ (·) (R) < ∞, for all bounded intervals I, where GI (x) = w−1 (x)(|I|/2 + |x − aI |)−1 and aI is the center of I. ′

Proof. Suppose that GI ∈ / L p (·) (R) for some interval I := (aI − τ, aI + τ). By Lemma ′ 1.4.6 we have that there exists g ∈ L p (·) (R) such that g ≥ 0 and Z

GI (x)g(x)dx = ∞.

R

R +∞

R aI

Hence either SI := aI GI g = ∞ or −∞ GI g = ∞. Suppose that SI = ∞. Then we take f (x) = g(x)χ(aI ,+∞) (x). Then using Lemma 6.5.1 we find that ∞ > kχ(aI ,+∞) (·)g(·)kL p(·) (R) ≥ kHk−1p(·)

p(·)

Lw →Lv

≥

kHk−1p(·) p(·) kχ(aI −τ,aI ) (·)v(·)kL p(·) (R) Lw →Lv

kHv,w f kL p(·) (R)

Z+∞

g(t)GI (t)dt = ∞.

aI

In the last inequality we used the inequality t − x ≤ (t − aI ) + τ which is true for all x, t with x ∈ (aI − τ, aI ) and t > aI . p(·)

p(·)

Proposition 6.5.1. Let H be bounded from Lw (R) to Lv (R). Then

sup kχI vkL p(·) (R) w−1 (·)(|I|/2 + |aI − ·|)−1 L p′ (·) (R) ≤ 4kHkL p(·) (R)→L p(·) (R) , I

where I is a bounded interval and aI is the center of I.

w

v

(6.5.3)

94

Alexander Meskhi ′

Proof. Due to Lemma 6.5.2 we have that GI ∈ L p (·) (R), where GI (x) = w−1 (x)(|I|/2 + |x − aI |)−1 . Let g ≥ 0 and let g ∈ L p(·) (R). Then by H¨older’s inequality for L p(·) spaces (see Lemma 1.4.1) we see that Z g(t)GI (t)dt < ∞.

(6.5.4)

g(t)GI (t)dt = (1/2)JI .

(6.5.5)

JI :=

R

Let us choose r ∈ R so that Z+∞ r

T

Now observe that if x ∈ I (−∞, r) and t ∈ (r, +∞), then 0 < t − x ≤ |t − aI | + |aI − x| < |t − aI | + |I|/2. Hence for such an x we have (recall that Hv,w f = vH( f /w)) Hv,w g(x) ≥ v(x)

Z+∞

g(t)GI (t)dt = (JI /2)v(x).

(6.5.6)

r

Due to Lemma 6.5.1 we have kgkL p(·) (R) ≥ kHk−1p(·)

p(·)

Lw →Lv

kHv,w gkL p(·) (R) ≥ (JI /2)kHk−1 kχI T(−∞,r) (·)v(·)kL p(·) (R) .

In a similar manner we can find that kgkL p(·) (R) ≥ (JI /2)kHk−1 kχI T(r,+∞) (·)v(·)kL p(·) (R) .

Now taking the supremum with respect to g with kgkL p(·) (R) ≤ 1 and using Lemma 1.4.7 we conclude that (6.5.3) holds. Proof of Theorem 6.5.1. By Lemma 6.5.1 we have kHv,w kK (L p(·) ) = kHkK (L p(·) ,L p(·) ) . w

v

Let λ > kHkK (L p(·) ,L p(·) ) . Then by the previous equality and Theorem 5.3.1 we have that w

v

λ > α(Hv,w ). Hence there exists P ∈ FL (L p(·) ) such that kHv,w − Pk < λ.

Let us take an arbitrary a ∈ R. By Lemma 1.4.5 there exist a positive number β and R ∈ FL (L p(·) ) such that λ − kHv,w − Pk kR − Pk < 2 and suppR f ⊂ R \ I(a, β)

Singular Integrals

95

for all f ∈ L p(·) , where I(a, β) = (a − β, a + β). Consequently, k(Hv,w − R) f kL p(·) ≤ λk f kL p(·) , f ∈ L p(·) (R), where suppR f ⊂ R \ I(a, β). From the latter inequality it follows that if 0 < τ < β, then kχI(a,τ) Hv,w f kL p(·) (R) ≤ λk f kL p(·) (R) , f ∈ L p(·) (R).

(6.5.7)

Let g be a non-negative function such that g ∈ L p(·) (R). By Lemma 6.5.2 we have that R < ∞. Hence I(a,τ) gw−1 < ∞. Now observe that for t ∈ (a, a + τ) and x ∈ (a − τ, a), 0 < t − x < 2τ. Consequently, assuming f = gχ(a,a+τ) in (6.5.7) we find that ∞ > λkgχ(a,a+τ) kL p(·) (R) ≥ kχ(a−τ,a) (·)Hv,w f (·)kL p(·) (R) kw−1 (·)χI(a,τ) (·)kL p(·) (R)

1 ≥ kχ(a−τ,a) (·)v(·)kL p(·) (R) 2τ

Z

a+τ a

gw . −1

Taking the supremum with respect to all g with kgkL p(·) (R) ≤ 1, applying Lemma 1.4.7 and passing to the limit as τ → 0, we have that kHkK ≥ (1/2)A1 . In a similar manner we can show that kHkK ≥ (1/2)A2 .

Proof of Theorem 6.5.2. Using the arguments from the proof of Theorem 6.5.1, for λ > kHkK and a ∈ R we have that inequality (6.5.7) holds. Let us take f = gχ(a+τ,+∞) in (6.5.7), where g is non-negative and kgkL p(·) (R) ≤ 1. Due to Lemma 6.5.2 we have kχ(a+τ,+∞) (·)w−1 (·)(· − a)−1 kL p′ (·) (R) < ∞. This implies

Z +∞ a+τ

g(t)w−1 (t)(t − a)−1 dt < ∞.

Further, ∞ > λkgχ(a,a+τ) kL p(·) (R) ≥ kχ(a,a+τ) (·)Hv,w f (·)kL p(·) (R) Z +∞ −1 −1 ≥ kχ(a,a+τ) (·)v(·)kL p(·) (R) g(t)(t − a) w (t)dt . a+τ

Taking the supremum with respect to all such a g we conclude that kHkK ≥ B1 . Analogously, kHkK ≥ B2 .

96

Alexander Meskhi

Proof of Theorem 6.5.3. Repeating the arguments of Theorem 6.5.1 we arrive at inequality (6.5.7). Further, assume that supp f ⊂ (a + τ, +∞) in (6.5.7), where f ≥ 0 and k f kL p(·) (R) ≤ 1. Then we observe that ∞ > λk f χ(a+τ,a) kL p(·) (R) ≥ kχI(a,r) (·)Hv,w f (·)kL p(·) (R)

≥ (1/2)kχI(a,τ) (·)v(·)kL p(·) (R)

Z

+∞ a+τ

f (t)(t − a) w (t)dt . −1 −1

In the latter inequality we used the estimate t − x ≤ 2(t − a) which holds for all t and x with t > a + τ, |x − a| < τ. Consequently, taking the supremum over all such an f we conclude that λ ≥ (1/2)kχI(a,τ) (·)v(·)kL p(·) (R) kχ(a+τ,+∞) (·)w−1 (·)(· − a)−1 kL p′ (·) (R) . Arguing in the same manner as above we shall see that λ ≥ (1/2)kχI(a,τ) (·)v(·)kL p(·) (R) kχ(−∞,a−τ) (·)w−1 (·)(a − ·)−1 kL p′ (·) (R) . Summarazing the estimates derived above, we conclude that kχI(a,τ) (·)v(·)kL p(·) (R) kχR\I(a,r) (·)w−1 (·)| · −a|−1 kL p′ (·) (R) ≤ kχI(a,τ) (·)v(·)kL p(·) (R) kχ(−∞,a−r) (·)w−1 (·)| · −a|−1 kL p′ (·) (R) +kχI(a,τ) (·)v(·)kL p(·) (R) kχ(a+r,+∞) (·)w−1 (·)| · −a|−1 kL p′ (·) (R) ≤ 4λ. These estimates lead us to the conclusion kHkK ≥ (1/4)C1 . Further, notice that due to Theorem 5.3.1 and Lemma 1.4.5 we have that there exists a sufficiently large positive number γ and R ∈ FL (Lwp ) such that kHv,w f − R f kL p(·) (R) ≤ λk f kL p(·) (R) , f ∈ L p(·) (R), where λ > kHkK

p(·)

p(·)

Lw ,Lv

, and suppR f ⊂ I(a, γ). Consequently,

kχR\I(a,s) Hv,w f kL p(·) (R) ≤ λk f kL p(·) (R) , f ∈ L p(·) (R), v

for all s, s > γ. Let f be a non-negative function and let supp f ⊂ I(a, s). Then ∞ > λk f χI(a,s) kL p(·) (R) ≥ kχ(a+s,+∞) (·)Hv,w f (·)kL p(·) (R) ≥ (1/2)kχI(a+s,+∞) (·)v(·)kL p(·) (R)

Z

a+s a−s

f (t)w−1 (t)dt .

Taking the supremum with respect to f with k f kL p(·) (R) ≤ 1 we find that λ ≥ (1/2)kχ(a+s,+∞) (·)(· − a)−1 v(·)kL p(·) (R) kχI(a,s) (·)w−1 (·)kL p′ (·) (R) .

Singular Integrals

97

Analogously, λ ≥ (1/2)kχ(−∞,a−s) (·)(a − ·)−1 v(·)kL p(·) (R) kχI(a,s) (·)w−1 (·)kL p′ (·) (R) . Consequently, kHkK ≥ (1/4)C2 .

Proof of Theorem 6.5.4. Let a ∈ R and let τ be so small positive number that (6.5.7) holds. Let us denote I := (a − τ, a + τ). We repeat the arguments from the proof of Proposition 6.5.1. Let g be a non-negative function such that kgkL p(·) (R) ≤ 1. According to Lemma 6.5.2 we have that (6.5.4) holds. Now we choose r, r ∈ R, so that (6.5.5) is fulfilled. Let / Observe that 0 < (t − x) ≤ (t − a) + τ whenever x ∈ I ∩ (−∞, r) and t > r. I ∩ (−∞, r) 6= 0. Using the arguments similar to those of Theorem 6.5.1 we have that (6.5.7) holds. Substituting f = gχ(r,+∞) in (6.5.7) we find that ∞ > λkχ(r,+∞) gkL p(·) (R) ≥ (JI /2)kχI∩(−∞,r) (·)v(·)kL p(·) (R) , where JI is defined by (6.5.4). / then Analogously, if I ∩ (r, +∞) 6= 0, ∞ > λkχ(−∞,r) gkL p(·) (R) ≥ (JI /2)kχI∩(r,+∞) (·)v(·)kL p(·) (R) . Summarazing these inequalities and taking the supremum with respect to g and a, and passing to the limit as τ → 0 we have the desired result. Proof of Corollary 6.5.1. Let a ∈ R. Suppose that v(a) > 0 and w(a) < ∞. Due to the condition p ∈ W L(R), Proposition 1.4.1, Theorems 6.5.4 and Remark 1.4.1 we have kHkK

p(·)

p(·)

Lw (R),Lv (R)

≥ (1/4)lim (1/2r)kχI(a,r) (·)v(·)kL p(·) (R) kw−1 (·)χI(a,r) (·)kL p′ (·) (R) r→0

−1/p+ (I(a,r))−1/(p+ (I(a,r)))′

≥ (1/4)lim (2r) r→0

Z ×

a+r

−p′ (t)

(w(t)) a−r

dt

Z

a+r

(v(t)) a−r

1/(p′ )− (I(a,r))

p(t)

dt

1/p− (I(a,r))

Z a+r 1/p− (I(a,r)) 2 e−A/(p− ) 1 p(t) ≥ lim (v(t)) dt r→0 2r a−r 4 Z a+r 1/(p′ )− (I(a,r)) 1 −p′ (t) × (w(t)) dt > 0. 2r a−r

Proof of Corollary 6.5.2. Let I := (a − r, a + r). Applying the condition p ∈ W L(R), Theorems 6.5.4, Proposition 1.4.1 and Remark 1.4.1 we have kHkK (L p(·) ) ≥ (1/4) sup lim

1

a∈R r→0 2r

2

≥

kχI (·)kL p(·) (R) k(·)χI (·)kL p′ (·) (R) 2

′ e−A/(p− ) 1 e−A/(p− ) sup lim (2r)1/p+ (I) (2r)1/(p+ (I)) = . 4 4 a∈R r→0 2r

98

Alexander Meskhi

6.6.

Cauchy Singular Integrals in L p(x) Spaces

Here we discuss lower estimates of the essential norm for the Cauchy singular integral operator SΓ along a smooth Jordan curve Γ on which the arc length is chosen as a parameter. We begin with the following Lemma: p(·)

Lemma 6.6.1. Let 1 < p− ≤ p(x) ≤ p+ < ∞. If SΓ is bounded from Lw (0, l) to p(·) Lv (0, l), then SI := kw−1 χI kL p′ (·) (0,l) < ∞, for all subintervals I of (0, l). Proof.R Let SI = ∞ for some I. This implies that there exists some g ∈ L p(·) (I), g ≥ 0, such that g(t)w−1 (t)dt = ∞. Let I

φ(s) = f (t(s)) = g(s)w−1 (s)χI (s) and let I = (a − r, a) ⊂ (0, l). We can assume that I ′ = (a, a + r) ⊂ (0, l). We have (see [107], [109]) Z Z φ(s) 1 1 ds ≥ φ(s)ds |SΓ f (t(σ))| ≥ 2π s − σ 4πr I

I

for σ ∈ I ′ and sufficiently small r. Thus 1 Z φ(s)ds χI ′ (σ) |SΓ f (t(σ))| ≥ 4πr I

for any σ. Hence using inequality (6.6.1) we find that

1 Z φ(s)

kv(σ)(SΓ f )(σ)kL p(σ) (0,l) ≥ χI ′ (σ)v(σ) ds p(σ) 2π s − σ L (0,l) I

≥

1 4πr

Z I

φ(s)ds kχI ′ (σ)v(σ)kL p(σ) (0,l)

Z 1 = g(s)w−1 (s)ds kχI ′ (σ)v(σ)kL p(σ) (0,l) = ∞. 4πr |I {z } =∞

On the other hand,

kw(·)φ(·)kL p(·) (0,l) = kχI (·)g(·)kL p(·) (0,l) < ∞ Now the inequality kv(SΓ f )kL p(·) (0,l) ≤ ckw(·)φ(·)kL p(·) (0,l) implies the desired result.

(6.6.1)

Singular Integrals

99 p(·)

Theorem 6.6.1. Let 1 < p− ≤ p(x) ≤ p+ < ∞ and let SΓ be bounded from Lw (0, l) to p(·) Lv (0, l). Then 1 kSΓ kK (L p(·) (0,l),L p(·) (0,l)) ≥ max{A˜ 1 , A˜ 2 }, w v 4π where 1 A˜ 1 = sup lim kχ(a−r,a) vkL p(·) (0,l) kχ(a,a+r) w−1 kL p′ (·) (0,l) a∈(0,l) r→0 r and

1 A˜ 2 = sup lim kχ(a,a+r) vkL p(·) (0,l) kχ(a−r,a) w−1 kL p′ (·) (0,l) . r→0 r a∈(0,l) Proof. Let λ > kSΓ kK (L p(·) (0,l),L p(·) (0,l)) . w

v

Then using the fact that kSΓ kK (L p(·) (0,l),L p(·) (0,l)) = kSΓ,v,w kK (L p(·) (0,l)) , w

v

where SΓ,v,w f (t(s)) = v(s) SΓ f w−1 (t(s)) and the equality (see Lemma 5.3.1) kSΓ,v,w kK (L p(·) (0,l)) = α(SΓ,v,w ),

we have λ > α(SΓ,v,w ). Let us take an arbitrary a ∈ (0, l). By Lemma 1.4.5 there exists a positive number β and an operator R ∈ FL (L p(·) (0, l)) such that kR − Pk <

λ − kSΓ,v,w − Pk 2

and supp R f ⊂ (0, l)\I(a, β). Consequently, k(SΓ,v,w − R) f (t(·))kL p(·) (0,l) ≤ λkφkL p(·) (0,l) for all f ∈ L p(·) (0, l). If we choose r so small that 0 < r < β then the inequality above leads us to the estimate kχI(a,r) SΓ,v,w f (t(·))kL p(·) (0,l) ≤ λkφkL p(·) (0,l) (6.6.2) which holds for all φ ∈ L p(·) (0, l). According to Lemma 6.6.1, kw−1 χI kL p′ (·) (0,l) < ∞ for all subintervals I ⊂ (0, l). Let g(s) ≥ 0, g ∈ L p(·) (0, l). Then by H¨older’s inequality for L p(·) spaces we find that Z

gw−1 ≤ ckgχI kL p(·) (0,l) kw−1 χI kL p′ (·) (0,l) < ∞.

I(a,r)

Further, if s ∈ I2 = (a, a + r) and σ ∈ I1 = (a − r, a), then using (6.6.1) we have Z SΓ,v,w f (t(σ)) = v(σ) SΓ ( f w−1 )(t(σ)) ≥ v(σ) w−1 (s)φ(s)ds. 4πr I2

(6.6.3)

100

Alexander Meskhi

Taking φ(s) = g(s)χI2 (s) in (6.6.2) and taking into account (6.6.3) we see that ∞ > λkg(s)χI2 (s)kL p(·) (0,l) ≥ kχI1 (·)SΓ,v,w f (·)kL p(·) (0,l) ≥ kχI1 vkL p(·) (0,l)

Z

w−1 (s)φ(s)ds.

I2

Taking the supremum over all such a g we get 1 kv(·)χ(a−r,a) (·)kL p(·) (0,l) kw−1 (·)χ(a,a+r) (·)kL p′ (·) (0,l) . 4πr This inequality implies λ≥

kSΓ kK (L p(·) (0,l),L p(·) (0,l)) ≥ w

v

1 ˜ A1 . 4π

Let us now take σ ∈ I2 and let φ(s) = f (t(s)) be nonnegative function with supp φ ⊂ I1 . Then Z Z 1 φ(s) SΓ f (t(σ)) ≥ 1 ds ≥ φ(s)ds. 2π s − σ 4πr I1

I1

Thus, for σ ∈ I2 and sufficiently small r, Z SΓ f (t(σ)) ≥ 1 φ(s)ds χI2 (σ). 4πr I1

Taking φ(s) = g(s)χI1 (s) in (6.6.2) we get ∞ > kg(s)χI1 (s)kL p(·) (0,l) ≥

1 kv(·)χI2 (·)kL p(·) (0,l) 4πr

Z

w−1 (s)g(s)ds.

I1

If we take the supremum with respect to g and use the fact that Z l −1 −1 kw kL p′ (·) (I1 ) ≤ sup χI1 (t)g(t)w (t)dt , kgkL p(·) (0,l) ≤1

0

we obtain

1 kχ vk p(·) kχ w−1 kL p′ (·) (0,l) . 4πr (a,a+r) L (0,l) (a−r,a) Taking the supremum over a ∈ (0, l) and passing to the limit when r → 0, we conclude that λ≥

kSΓ kK (L p(·) (0,l),L p(·) (0,l)) ≥ w

v

1 ˜ A2 . 4π

p(·)

Theorem 6.6.2. Let 1 < p− ≤ p(x) ≤ p+ < ∞ and let SΓ be bounded in Lw (Γ). Then kSΓ kK (L p(·) (0,l)) ≥ w

where

1 (r,a) sup lim A p(·) , r→0 4π(4πkSΓ k + 1) a∈(0,l)

1 kχ (·)w(·)kL p(·) (0,l) kχ(a−r,a+r) (·)w−1 (·)kL p′ (·) (0,l) 2r (a−r,a+r) and kSΓ k is the operator norm. (r,a)

A p(·) =

Singular Integrals

101

Proof. Let f (t(s)) = χI1 (s), where I1 = (a − r, a). Suppose that I2 := (a, a + r). We have Z f (t(s)) 1 1

J := kSΓ f kL p(·) (0,l) ≥ χI2 (·) ds p(·) ≥ kχI2 (·)kL p(·) (0,l) . w w 2π ·−s 4π Lw (0,l) I1

Also,

J ≤ kSΓ kkχI1 (·)kL p(·) (0,l) . w

Combining these inequalities we obtain kχI2 (·)w(·)kL p(·) (0,l) ≤ 4πkSΓ kkχI1 (·)w(·)kL p(·) (0,l) .

(6.6.4)

kχI1 (·)w(·)kL p(·) (0,l) ≤ 4πkSΓ kkχI2 (·)w(·)kL p(·) (0,l) .

(6.6.5)

Analogously, Now applying (6.6.4) and (6.6.5) we find that (r,a)

1 kχ (·)w(·)kL p(·) (0,l) kχ(a−r,a+r) (·)w−1 (·)kL p′ (·) (0,l) 2r (a−r,a+r) i 1h ≤ kχI1 (·)w(·)kL p(·) (0,l) + kχI2 (·)w(·)kL p(·) (0,l) 2r i h × kχI1 (·)w−1 (·)kL p′ (·) (0,l) + kχI2 (·)w−1 (·)kL p′ (·) (0,l)

A p(·) =

=

1h kχI1 (·)w(·)kL p(·) (0,l) kχI1 (·)w−1 (·)kL p′ (·) (0,l) 2r

+kχI1 (·)w(·)kL p(·) (0,l) kχI2 (·)w−1 (·)kL p′ (·) (0,l)

+kχI2 (·)w(·)kL p(·) (0,l) kχI1 (·)w−1 (·)kL p′ (·) (0,l) i +kχI2 (·)w(·)kL p(·) (0,l) kχI2 (·)w−1 (·)kL p′ (·) (0,l)

1h 1 4πkSΓ k kχI2 (·)w(·)kL p(·) (0,l) kχI1 (·)w−1 (·)kL p′ (·) (0,l) 2 r 1 + kχI1 (·)w(·)kL p(·) (0,l) kχI2 (·)w−1 (·)kL p′ (·) (0,l) r 1 + kχI2 (·)w(·)kL p(·) (0,l) kχI1 (·)w−1 (·)kL p′ (·) (0,l) r i 1 +4πkSΓ k kχI1 (·)w(·)kL p(·) (0,l) kχI2 (·)w−1 (·)kL p′ (·) (0,l) . r Using Theorem 6.6.1 for v ≡ w, taking the supremum over all a ∈ (0, l) and passing to the limit as r → 0 and we find that 1h (r,a) sup lim A p(·) ≤ 16π2 kSΓ kkSΓ kK (L p(·) (0,l)) + 4πkSΓ kK (L p(·) (0,l)) w w 2 a∈(0,l) r→0 ≤

4πkSΓ kK (L p(·) (0,l)) + 16π2 kSΓ kkSΓ kK (L p(·) (0,l)) w

w

i

102

Alexander Meskhi = 4πkSΓ kK (L p(·) (0,l)) + 16π2 kSΓ kkSΓ kK (L p(·) (0,l))) . w

Therefore

w

h i (r,a) sup lim A p(·) ≤ 4πkSΓ kK (L p(·) (0,l)) 4πkSΓ k + 1 .

a∈(0,l) r→0

w

Finally we have the desired result.

6.7.

Notes and Comments on Chapter 6

This chapter is based on the papers [165], [166], [43], [6]. For the estimates of the essential norm kSΓ kK (Lwp (Γ)) , where Γ is a Lyapunov curve and w is a power-type weight, see [141], [142]. In [68] it was shown that when w ∈ A2 (Γ), then kST kK (Lwp (T )) = 1 if and only if log w ∈ V MO(T ), where T is the unit circle. It should be pointed out that in the one-weight case the lower estimates of the essential norm of SΓ in Banach function spaces, where Γ is a Carlesson curve, have been derived in [102], [103]. In particular, these results give the lower estimates of kSΓ kK (Lwp ) , 1 < p < ∞, where w is the Muckenhoupt weight. The one-weight problem for the Hilbert transform and Calde´on-Zygmund singular integrals was solved in [95], [24] (see also the monographs [73], [224], [76] and references therein). For two-weight inequalities for the Hilbert transform and singular integrals on Rn in Lebesgue spaces we refer to the papers [172], [46], [190], [23], [178], [179], [201], [28], [147] (see also the monographs [76], [49], [233] and references therein). We notice that the conditions of [178] and [147] on weight pairs involve the operator itself. The same problems for singular integrals defined on nilpotent groups were studied in [113], [114], [86] (see also [49], [76], [87] and references therein). It should be emphasized that the two-weight problem for the Hilbert transform remains still open. For weighted inequalities for the operator SΓ in classical Lebesgue spaces we refer to [109], [107], [76], [49]. Weighted estimates for SΓ in L p(·) spaces were obtained in [120], [121], [127]–[130], [132], [134].

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Index duality, 54, 84

A E Adams, 103 AMS, 104 Amsterdam, 108, 118 application, 112, 113 applied mathematics, ix asymptotic, 25, 105, 113 asymptotics, 105, 108 averaging, 105

economics, viii elasticity, 118 encouragement, ix entropy, 4, 25, 67, 107, 113, 114 equality, 75, 91, 94, 99 Euclidean space, viii, xi, 9, 35, 51

F B Banach spaces, vii, 4, 16, 63, 80 behavior, 113, 115 Boston, 103, 108, 112 boundary value problem, vii, viii bounded linear operators, 4, 22, 67 Brownian motion, 113

C calculus, vii, ix, 118 Cauchy integral, 26, 83 classes, vii, 4, 5, 109, 115 classical, 16, 25, 26, 74, 82, 83, 102, 104, 105, 106, 109, 110, 111, 115 closure, 63 composition, 113 conjecture, 105 continuity, 7, 13, 15, 78 corona, 115 Czech Republic, 110

D decomposition, 115 definition, 1, 19, 29, 36, 37, 83, 86, 90 density, 109, 111 derivatives, xi, 63 differential equations, viii differentiation, viii, 28, 33, 35 distribution, 112

family, 6 finance, viii fluid, 105 Fourier, 51, 68, 70, 106, 110 fractional integrals, vii, 50, 113, 114, 117, 118

G gene, 104 generalization, 17 generalizations, 104 graduate students, ix groups, viii, 1, 6, 35, 37, 51, 102, 103, 107, 108, 110 growth, viii, 83, 111

H Harmonic analysis, 117 Heisenberg, viii, 1, 108 Heisenberg group, viii, 1, 108 Hilbert, vii, viii, 14, 22, 23, 24, 26, 36, 83, 88, 91, 102, 103, 105, 108, 109, 110, 114, 115, 116 Hilbert space, 22 Holland, 118 Hong Kong, 110 House, 110, 113, 115

I identity, vii, viii, 25, 51, 63, 65

120

Index

Indiana, 105, 109, 117 inequality, 2, 3, 5, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 23, 24, 25, 31, 34, 35, 37, 39, 40, 44, 47, 49, 51, 52, 54, 55, 56, 58, 59, 60, 61, 62, 63, 64, 66, 67, 69, 71, 73, 74, 75, 76, 78, 79, 80, 84, 85, 86, 87, 88, 90, 92, 93, 94, 95, 96, 98, 99, 100, 103, 104, 105, 106, 108, 109, 110, 112, 114, 116, 117 integration, viii, 42, 108, 116 interval, 12, 13, 20, 21, 71, 80, 93

J Jordan, 23, 86, 87, 98

K

New York, 105, 110, 117, 118 nonlinear, vii, 110 non-linearity, ix norms, 25, 37, 45, 52, 104, 106, 113 numerical analysis, viii

O operator, vii, 1, 3, 4, 5, 7, 8, 9, 10, 11, 12, 13, 16, 17, 21, 22, 23, 24, 25, 26, 27, 28, 30, 31, 33, 36, 37, 43, 45, 47, 50, 51, 54, 55, 61, 62, 63, 64, 65, 71, 77, 78, 79, 80, 82, 86, 87, 88, 90, 98, 99, 100, 102, 104, 105, 108, 109, 111, 112, 114, 115, 116, 117 Operators, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 90, 105, 110, 111, 112, 115 organization, ix orthogonality, 117

kernel, vii, 21, 24, 25, 37, 38, 70, 103, 107 Kolmogorov, 116

P L lead, ix, 96 Lebesgue measure, xi Lie algebra, 1 Lie group, 1 linear, 1, 3, 4, 5, 7, 8, 9, 10, 14, 16, 17, 22, 60, 67, 80, 81, 93, 108, 114, 118 linear function, 3, 14, 16, 17, 60 London, 104, 105, 106, 107, 108, 110, 115, 116, 117 Lyapunov, 102

paper, ix, 13, 25, 69, 70 parameter, 23, 98 Paris, 105 partial differential equations, 113 physics, vii, viii, 111 plasticity, ix Poisson, 51, 60, 70, 114, 117 Poland, 110 power, 117 property, 4

Q

M manifold, viii manifolds, viii martingale, 36, 105 mathematicians, ix mathematics, vii measures, 115 memory, viii memory processes, viii metric, 70, 109 Mexico, 116 Mexico City, 116 modeling, 116 monograph, vii, viii, ix, 1, 25, 70 Moscow, 112, 114 motion, 113

quantum, vii quantum mechanics, vii

R radius, xi, 1, 5, 6, 14, 57, 60, 67 random, 108 random matrices, 108 range, vii real numbers, 9, 17 recall, 35, 74, 94 research, ix researchers, viii, ix Russian, 104, 107, 108, 109, 110, 112, 114, 115, 117, 118

N natural, vii, ix, xi New Jersey, 107, 110, 117

S series, 18, 51, 68, 70, 106 Singapore, 110

Index singular, vii, viii, 1, 22, 23, 25, 26, 83, 86, 90, 98, 102, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116 Sobolev space, viii, ix, 25, 63, 105, 109, 113, 116, 118 solutions, vii spatial, 117 spectrum, 104, 108, 114 St. Petersburg, 114 stochastic, vii, viii stochastic processes, vii students, ix symbols, 104 systems, 22, 46

T theory, vii, viii, ix, 103, 104, 105, 107, 108, 110, 116, 118 time, 50 Tokyo, 114 topological, 114 topology, 117

121

transformations, 1, 109 translation, 90 transparent, 69 trees, 107

U unification, 107 USSR, 108, 118

V values, 107, 108, 115, 116 variable, vii, viii, ix, 1, 25, 76, 82, 91, 104, 105, 106, 107, 109, 111, 112, 114, 116, 117 vector, 24, 90

Y yield, 7, 19, 29, 35, 39, 90, 91