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MEASURE OF NON-COMPACTNESS FOR
INTEGRAL OPERATORS IN WEIGHTED
LEBESGUE SPACES
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MEASURE OF NON-COMPACTNESS FOR
INTEGRAL OPERATORS IN WEIGHTED
LEBESGUE SPACES
ALEXANDER MESKHI
Nova Science Publishers, Inc.
New York
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Contents
Preface vii
Basic Notation xi
1 Basic Ingredients 11.1. Homogeneous Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2. Measure of Non–compactness . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3. Hardy–type Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.4. L p( x) Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.5. Schatten–von Neumann Ideals . . . . . . . . . . . . . . . . . . . . . . . . 22
1.6. Singular Integrals in Weighted Lebesgue Spaces . . . . . . . . . . . . . . . 23
1.7. Notes and Comments on Chapter 1 . . . . . . . . . . . . . . . . . . . . . . 25
2 Maximal Operators 272.1. Maximal Functions on Euclidean Spaces . . . . . . . . . . . . . . . . . . . 27
2.2. One–sided Maximal Functions . . . . . . . . . . . . . . . . . . . . . . . . 32
2.3. Maximal Operator on Homogeneous Groups . . . . . . . . . . . . . . . . . 34
2.4. Notes and Comments on Chapter 2 . . . . . . . . . . . . . . . . . . . . . . 35
3 Kernel Operators on Cones 37
3.1. Boundedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
3.2. Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
3.3. Schatten–von Neumann norm Estimates . . . . . . . . . . . . . . . . . . . 45
3.4. Measure of Non–compactness . . . . . . . . . . . . . . . . . . . . . . . . 473.5. Convolution–type Operators with Radial Kernels . . . . . . . . . . . . . . 49
3.6. Notes and Comments on Chapter 3 . . . . . . . . . . . . . . . . . . . . . . 50
4 Potential and Identity Operators 51
4.1. Riesz Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
4.2. Truncated Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
4.3. One–sided Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
4.4. Poisson Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
4.5. Sobolev Embeddings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 634.6. Identity Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
4.7. Partial Sums of Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . 68
v
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vi Contents
4.8. Notes and Comments on Chapter 4 . . . . . . . . . . . . . . . . . . . . . . 69
5 Generalized One-sided Potentials in L p( x) Spaces 71
5.1. Boundedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
5.2. Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
5.3. Measure of Non–compactness . . . . . . . . . . . . . . . . . . . . . . . . 805.4. Notes and Comments on Chapter 5 . . . . . . . . . . . . . . . . . . . . . . 82
6 Singular Integrals 83
6.1. Hilbert Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
6.2. Cauchy Singular Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 86
6.3. Riesz Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
6.4. Calderon–Zygmund Operators . . . . . . . . . . . . . . . . . . . . . . . . 90
6.5. Hilbert Transforms in L p( x) Spaces . . . . . . . . . . . . . . . . . . . . . . 91
6.6. Cauchy Singular Integrals in L p( x) Spaces . . . . . . . . . . . . . . . . . . 98
6.7. Notes and Comments on Chapter 6 . . . . . . . . . . . . . . . . . . . . . . 102
References 103
Index 119
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Preface
One of the important problems of modern harmonic analysis is to establish the bounded-
ness/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 two-
sided 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 theessential norm for maximal functions, fractional integrals, singular and identity operators,
generally speaking, in weighted variable exponent Lebesgue spaces. Such operators nat-
urally 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 stud-
ied in the two-weighted setting and cover the case of non-linear maps, such as, Hardy-
Littlewood and fractional maximal functions. Before, these problems were investigated
only for a restricted class of kernel operators consisting only of Hardy-type and Riemann-
Liouville 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, Calderon–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 eigenvalueestimates 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 includingfractional integrals was developed in the monographs [112], [76], [49]. It should be empha-
sized that the interest in fractional calculus has been stimulated by applications in different
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viii Alexander Meskhi
fields of science, including stochastic analysis of long memory processes, numerical anal-
ysis, 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 exponentLebesgue spaces. Chapter 2 deals with the measure of non–compactness for maximal op-
erators 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 frac-
tional 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 twoweighted estimates of the essential norm for singular integrals (Hilbert transforms, Cauchy
integrals, Riesz transforms, Calderon–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 withboundary value problems for pseudo-convex domains in Cn.
The last two chapters of the monograph is dedicated to the investigation of the com-
pactness 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
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)−1u( x) + f ( x).
In this case one deals with the Dirichlet integral of the form 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 varia-
tional problems with Lagrangians more complicated than is usually assumed in variational
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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. Ruzicka [202] studied the problems in the so called rheological and electrorheolog-ical 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. Kovacik and I. Rakosnık [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 authorand his collaborators regarding the main topics of the monograph, but also contains un-
published 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 theo-
rems, lemmas, formulas etc. follows the natural three-digit system. There are three cat-
egories 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. 06-
1000017-8792, and Georgian National Foundation Grant No. GNSF/ST06/3-010.
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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 ;¯ B( x,r ): closed ball with center x and radius r ;
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;
Bn: volume of the unit ball in Rn, i.e., Bn = 2πn/2
nΓ (n/2) ;
S n−1: area of the unit sphere in Rn, i.e., S n−1 = 2πn/2
Γ (n/2) ;
w( E ) = E w( x)dx, where w is a weight function;| E |: Lebesgue measure of E ;
χ E : 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;
limn→∞
an = liminf n→∞
an, limn→∞
an = limsupn→∞
an for a sequence of real number an;
a ≈ b: there are positive constants c1 and c2 such that c1a ≤ b ≤ c2a;
C m: class of functions whose partial derivatives up to and including those of order m exist
and are continuous;
C m0 : subset of C m of functions with compact support;
C 0: class of continuous functions with compact support;
: end of the proof.
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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 andlinear 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 = ex p( 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 denotedby 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.
In the sequel we denote by B(a,ρ) and ¯ B(a,ρ) open and closed balls respectively with
the center a and radius ρ, i.e.
B(a,ρ) := y ∈ G; r (ay−1) < ρ, ¯ B(a,ρ) := y ∈ G; r (ay−1) ≤ ρ.
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].
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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) ,
G
u( x)dx =
∞ 0
S
u(δt y)t Q−1d σ( y)dt .
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 L pw(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)
f L pw(Ω) = Ω | f ( x)| pw( x)dx1/ p
.
If w ≡ 1, then we denote L p1 (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 Radonmeasure σ on S such that for all u ∈ L1( E ) ,
E
u( x)dx =
∞ 0
A
u(δs ¯ y)sQ−1d σ( ¯ y)ds.
Now we formulate embedding criteria from L pw(Ω) to L
qv (Ω) (q < p), where Ω is a non-
empty open set in G. These results are well-known (see [100]) but we give the proofs for
completeness.
Proposition 1.1.3. Let 0 < q < p <∞ and let v and w be weights on an open set Ω⊆ G.
Then L pw(Ω) is boundedly embedded in L
qv (Ω) if and only if
BΩ :=
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 Holder’s inequality with respect to the exponent pq
we find that
Ω
| f ( x)|qv( x)dx ≤
Ω
| f ( x)| pw( x)dx
q/ p Ω
v( x)
w( x)
p p−q
w( x)dx
q( p−q) pq
.
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Basic Ingredients 3
From this inequality we conclude that
I ≤ BΩ.
To prove the opposite, first we note that from the embedding of L pw(Ω) in L
qv (Ω) it
follows that Ω
v( x)
w( x)
p p−q
w( x)dx < ∞.
Now taking the function f ( x) = v1
p−q ( x)w1
q− p ( x) in the two-weight inequality Ω
| f ( x)|qv( x)dx
1/q
≤ I
Ω
| f ( x)| pw( x)dx
1/ p
we conclude that BΩ < ∞.
The fact that I = BΩ is obvious.
The next statement follows immediately:
Proposition 1.1.4. Let 0 < p <∞. Then L∞(Ω) is boundedly embedded in L pw(Ω) if and
only if
BΩ :=
Ω
w( x)dx
1/ p
< ∞.
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 → L
q
(Ω), where X = L
p
w(Ω). Then the norm of theoperator T is defined as follows:
T = supT f Lq(Ω) : f X ≤ 1.
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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 T κ ( X ,Y ) ( or simply T κ )
and is defined as
T κ ( X ,Y ) := distT ,κ ( X ,Y ) = inf T − K : 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
T K ( X ,Y ) := distT ,K ( X ,Y ) = inf T − K : 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 (U X ) ⊂
2k −1 j=1
(bi +εU Y ) for some b1, . . . ,b2k −1 ∈ Y ,
where U X and U Y 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) T = 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 ) := limn→∞
en(T ). Among other properties we mention that β(T ) = 0 if
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 F = sup x≤1 |F ( x)| < ∞.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):
F L( X , L p(Ω)) := T : T f ( x) =m
∑ j=1
α j( f )u j, m ∈ N, u j ≥ 0, u j ∈ L p(Ω),
u j are linearly independent and α j ∈ X ∗
M ,
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Basic Ingredients 5
F S ( X , L p(Ω)) :=
T : T f ( x) =m
∑ j=1
β j( f )u j, m ∈N, u j ≥ 0, u j ∈ L p(Ω),
u j are linearly independent and β j ∈ S ( X )
M .
If X = L p(Ω), we denote these classes by F L L p(Ω) and F S L p(Ω) respectively. It is
clear that if P ∈ F S X , L p(Ω) resp. P ∈ F L X , L p(Ω) , then P is compact sublinear resp. compact linear
from X to L p(Ω).
We shall use the symbol α(T ) (resp. α(T )) for the distance between the operator T :
X → L p(Ω) and the class F S
X , L p(Ω)
, (resp. F L
X , L p(G)
) i.e.
α(T ) := dist T ,F S
X , L p(Ω)
( resp. α(T ) := dist T ,F L
X , L p(G)
).
For any bounded subset A of L p(Ω) (1 < p < ∞), let
Φ( A) := inf
δ > 0 : A can be covered by finitely many open balls in
L p
(Ω) of radius δand
Ψ( A) := inf P∈F L( L p(Ω))
sup
f − P f L 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, . . . ,g N from L p(Ω) such that for all f ∈ K
and some i ∈ 1,2, . . . , N ,
f − gi L p(Ω) < ε (1.2.2)
Given δ > 0, let Ω be a cube such that for all i ∈ 1,2, . . . , N Ω\Ω |gi( x)| pdx
1/ p
< 1
2δ. (1.2.3)
We assume that all functions from L p(Ω) are equal to zero outside Ω. Let
Ω = ∪m
j=1Q j,
where the Q j are disjoint congruent cubes of diameter h, and define
P f ( x) :=m
∑ j=1
f Q jχ Q j
( x), f Q j:= | Q j|
−1 Q j
f ( y)dy,
where Q j = Ω∩ Q j. Then
gi − Pgi L p(Ω∩Ω) =m
∑ j=1
Q j
1
| Q j|
Q j
[gi( x) − gi( y)]dy
pdx
≤m
∑ j=1 Q j
1
| Q j| Q j
|gi( x) − gi( y)| pdydx
≤ sup| z|<h
Ω
|gi( x) − gi( x + z)| pdx → 0
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6 Alexander Meskhi
as h → 0. From this and (1.2.3) it follows that we may choose h so that
gi − Pgi L p(Ω) < δ, i = 1,2, . . . , N . (1.2.4)
Further,
P f p
L p(Ω) =m
∑ j=1
Q j
| Q j|−1 Q j
f ( y)dy pdx
≤m
∑ j=1
Q j
| Q j|−1
Q j
| f ( y)| pdydx ≤ f p
L p(Ω).
It is obvious that the functionals f → f Q jbelong to ( L p(Ω))∗ ∩M. Consequently, P ∈
F L( L p(Ω)). Finally from (1.2.2)–(1.2.4) and the last inequality we conclude that
f − P f L p(Ω) ≤ f − gi L p(Ω) + gi − Pgi L p(Ω)
+P(gi − f ) L p(Ω) < ε+δ+ gi − f L 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, . . . ,g N ∈
L p(G) such that for all f ∈ K and some i ∈ 1,2, . . . , N
f − gi L 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
G\ ¯ B
|gi( x)| pdx1/ p
< 1
2 δ. (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 sub-
cover B1, B2, . . . , Bn. Further, let us assume that E 1, E 2, . . . , E n is a family of pairwise
disjoint sets of positive measure such that ¯ B =n
i=1
E i and E i ⊂ Bi (we can assume that
E 1 = B1 ∩ ¯ B, E 2 = ( B2\ B1) ∩ ¯ B, . . . , E k = ( Bk \k −1i=1
Bi) ∩ ¯ B, . . .). We define
P f ( x) =n
∑i=1
f E iχ E i ( x), f E i = | E i|−1 E i
f ( x)dx.
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Basic Ingredients 7
Then
gi − Pgi p
L p( ¯ B) =
n
∑ j=1
E j
1
| E j|
E j
[gi( x) − gi( y)]dy
pdx
≤
m
∑ j=1
E j
1| E j|
E j
gi( x) − gi( y) pdy dx
≤ supr ( z)≤2c0h
¯ B
gi( x) − gi( zx) pdx → 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
gi − Pgi L p(G) < δ, i = 1,2,3, . . . , N , (1.2.7)
when h is sufficiently small. Further,
P f p
L p(G) =n
∑ j=1
E j
| E j|−1 ¯ E j
f ( y)dy
pdx
≤n
∑ j=1
E j
| E j|−1 ¯ E j
f ( y) pdy dx ≤ f p
L p( ¯ B) ≤ f p
L p(G).
It is also clear that the functionals f → f E i belong to L p(G)∗
∩ M . Hence P ∈F L( L p(G)). Finally (1.2.5) − (1.2.7) yield
f − P f L p(G) ≤ f − gi L p(G) + gi − Pgi L p(G) + P(gi − f ) L p(G)
< ε+δ+ gi − f L 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 ∈ F L L p(Ω) such that for all f ∈ K,
f − Pε f L 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 ∈ F L
L p(Ω)
such
that
sup
f − Pε f L 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).
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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 U X =
f : f X ≤ 1.
From the compactness of T it follows that T (U X ) 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 ∈ F L
L p(Ω)
such
that for all f ∈ U X ,
T f − PεT f L p(Ω) ≤ ε. (1.2.8)
Let Pε = Pε T . Then Pε ∈ F S
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
Pε f ( x) = Pε(T f )( x) =m
∑ j=1
α j(T f )u j( x) =m
∑ j=1
β j( f )u j( x),
where β j = α j T belongs to S ( X ) ∩ M . By (1.2.8) we have
T f − Pε f L p(Ω) ≤ ε
for all f ∈ U X , 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
T κ = α(T ) (resp. T K = α(T ))
Proof. Let T be bounded, order preserving and sublinear. Suppose that δ > 0. Thenthere exists an operator K ∈ κ ( X , Lq(Ω)), such that T − K ≤ T κ +δ. By Lemma 1.2.2
there is P ∈ F S ( X , Lq(Ω)) for which the inequality K − P < δ holds. This gives
T − P ≤ T − K + K − P ≤ T κ + 2δ.
Hence α(T ) ≤ T κ . The opposite inequality
T κ ≤ α(T )
is obvious.
Lemma 1.2.4. Let G be a homogeneous group 1 ≤ q < ∞ and let P ∈F S ( X , Lq(G)) (resp. P ∈ F L( X , Lq(G))). Then for every a ∈ G and ε > 0 there exist an
operator R ∈ F S ( X , Lq(G)) (resp. R ∈ F L( X , Lq(G))) and positive numbers α , α such that
for all f ∈ X the inequality
(P − R) f Lq(G) ≤ ε f X
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
P f ( x) = N
∑ j=1
β j( f )u j( x), f ∈ X ,
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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
∑ j=1
|β j( f )| ≤ c f X
holds.
Let us choose linearly independent Φ j ∈ Lq(G) and positive real numbers α j, α j such
that
u j −Φ j Lq(G) < ε, j ∈ 1,2, . . . , N ,
and suppΦ j ⊂ B(a, α j)\ B(a,α j). If
R f ( x) = N
∑ j=1
β j( f ) Φ j( x),
then it is obvious that R ∈ F S ( X , Lq(G)) and, moreover,
P f − R f Lq(G) ≤ N
∑ j=1
|β j( f )|u j −Φ j Lq(G) ≤ cε f X
for all f ∈ X . Besides this, supp R f ⊂ B(a, α)\ B(a,α), where α = minα j and α =maxα 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 ∈ F S
X , L p(Ω)
(resp. P ∈
F L( X , Lq(Ω))) , where X = Lr w(Ω) and 1 < r , p < ∞. Then for every a ∈ Ω and ε > 0 there
exists an operator R ∈ F S
X , L p(Ω)
(resp. P ∈ F L( X , Lq(G))) and positive numbers β1 and
β2 , β1 < β2 such that for all f ∈ X the inequality
(P − R) f
L p(Ω)
≤ ε f X
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 = L pv (Ω). Suppose
that P ∈ F L( X ,Y ) and ε > 0. Then there are an operator T ∈ F L( X ,Y ) and a set E α,β := x ∈Ω : 0 < α < r ( x) < β < ∞ such that
P − T < ε
and
supp T f ⊂ E α,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 L pw(G) to
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10 Alexander Meskhi
Lqv (G). Suppose that λ > T κ ( L p
w(G), Lqv (G)) (resp. λ > T K ( L
pw(G), L
qv (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:
T f Lqv ( B(a,τ)) ≤ λ f L
pw(G), (1.2.9)
T f Lqv ( B(a,r )c) ≤ λ f L
pw(G), (1.2.10)
where f ∈ L pw(G).
Proof. Let T be bounded, order preserving and sublinear from L pw(G) to L
qv (G). Let
T (v) be the operator given by
T (v) f = v1q T f .
Then it is easy to see that
T (v)κ ( L pw(G)→ Lq(G)) = T κ ( L p
w(G)→ Lqv (G)).
By Lemma 1.2.3 we have that
λ > α
T (v).
Consequently, there exists P ∈ F S
L
pw(G), Lq(G)
such that
T (v) − P < λ.
Fix a ∈ G. According to Lemma 1.2.4 there are positive constants β1 and β2, β1 < β2, and
R ∈ F S L pw(G), L
qv (G) for which
P − R ≤ λ− T (v) − P
2
and supp R f ⊂ B(a,β2)\ B(a,β1) for all f ∈ L pw(G). Hence,
T (v) − R < 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 ∈ L
p
w(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,
order-preserving and sublinear (resp. bounded linear ) operator from L pw(Ω) to L
qv (Ω).
Assume that λ > T κ ( L pw(Ω), Lq
v (Ω)) (resp. λ > T K ( L pw(Ω), Lq
v (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:
T f Lqv (Ω∩ B(a,τ)) ≤ λ f L pw(Ω); T f Lqv (Ω\ B(a,r )) ≤ λ f L pw(Ω),
where f ∈ L pw(Ω).
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Basic Ingredients 11
1.3. 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 < p ≤ q < ∞ and v and w areweight functions on X . Then the operator
H φ f
( x) =
y∈ X : φ( y)<φ( x)
f ( y)dµ
is bounded from L pw( X ) to L
qv ( X ) if and only if
sup
t >0 x:φ( x)>t
v( x)dµ1/q
x:φ( x)<t
w1− p′( x)dµ
1/ p′
< ∞, p′ = p/( p − 1).
Let E be a cone in G, where G is a homogeneous group (see Section 1.1).
Denote
E t := 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 observ-
ing that
supt >0 E \ E t u( x)dx1/q
t
Q/ p′
≈ sup j∈Z E 2 j+1 \ E
2 j u( x)dx1/q
2
jQ/ p′
,
we have the next statement.
Theorem 1.3.2. Let 1 < p ≤ q < ∞. Then the inequality E
v( x)
E r ( x)
f ( y)dy
qdx
1/q
≤ c
E
| f ( x)| pdx
1/ p
holds if and only if
sup j∈Z
E 2 j+1 \ E
2 j
u( x)dx1/q
2 jQ/ p′< ∞.
We need also the following statement (see [158] for 1 ≤ q < p < ∞ and [222] for 0 <q < 1 < p < ∞).
Theorem 1.3.3. Let 0 < q < p < ∞ and p > 1. Then the inequality ∞0
v( x)
x
0 f (t )dt
q
dx
1/q
≤ c
∞0
( f ( x)) pw( x)dx
1/ p
, f ≥ 0,
holds if and only if ∞0
∞t
v( x)dx
t
0w1− p′
( x)dx
q−1 p/( p−q)
w1− p′(t )dt
( p−q)/( pq)
< ∞.
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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
b
av( x) x
a f (t )dt s
dx1/s
≤ c b
a(w(t ) f (t ))r dt 1/r
, f ≥ 0,
holds if and only if
supa≤t ≤b
b
t vs( x)dx
1/s t
aw−r ′ ( 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 havea( x, y) L
p′ ν (Y )
L
q µ( X )
< ∞.
Then the operator
A f ( x) =
Y
a( x, y) f ( y)d ν( y)
is compact from L p ν(Y ) to L
q µ( 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
A E f ( x) =
E a( x, y) f ( y)dy, x ∈ E ,
is bounded from L pw( E ) to L
qv ( E ) , then A E 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
on Ω belongs to L p(·)ρ (Ω) (or L
p( x)ρ (Ω)) if
S p,ρ( f ) = Ω f ( x)ρ( x) p( x)dx < ∞.
It is known (see, e.g., [138], [111]) that L p( x)ρ (Ω) is a Banach space with the norm
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Basic Ingredients 13
f L
p( x)ρ (Ω)
= inf λ > 0 : S p,ρ
f /λ
≤ 1
.
If ρ ≡ 1, then we use the symbol L p( x)(Ω) (resp. S p) instead of L p( x)ρ (Ω) (resp. S p,ρ). It
is clear that f L
p(·)
ρ (Ω)
= f ρ L p(·)(Ω).
Further, we denote
p−( E ) = inf E
p, p+( E ) = sup E
p
for a measurable set E ⊆ Ω.
Let P (Ω) be the class of all measurable functions p, p : Ω→ Rn, such that the Hardy-
Littlewood maximal operator
M Ω f ( x) = sup
Q∋ x
1
|Q| 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:
f p+( E )
L p(·)( E ) ≤ S p( f χ E ) ≤ f
p−( E )
L p(·)( E ), f L p(·)( E ) ≤ 1;
f p−( E )
L p(·)( E ) ≤ S p( f χ E ) ≤ f
p+( E )
L p(·)( E ), f L p(·)( E ) ≥ 1;
E
f ( x)g( x)dx
≤ 1
p−( E ) +
1
( p+( E ))′
f L p(·)( E ) g
L p′(·)( E ),
where p′( x) = p( x) p( x)−1
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 − Holder 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 M Ω 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)
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14 Alexander Meskhi
Let
H f ( x) = limε→0
| x−t |>ε
f (t )
x − t dt
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 C 0(Ω) of continuous compactly supported functions on Ω is
dense in L p(·)(Ω).
For other properties of L p( x) and L p( x)ρ 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 theinequality
| I | p−( I ∩Ω)− p+( I ∩Ω) ≤ e A
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 ) = e A.
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
f Lr ( x)(Ω) ≤ (1 + |Ω|) f L 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|,
where A1 = A/( p−)2 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
F L( X , L p(·)(Ω)) ≡
T : T f ( x) =m
∑ j=1
β j( f )u j, m ∈N, u j ≥ 0, u j ∈ L p(·)(Ω),
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 F L( X , L p(·)(Ω)) := F L( 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
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Basic Ingredients 15
and
Ψ( A) := inf P∈F L( L p(·)(Ω))
sup f − P f L 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 anybounded subset K ⊂ L p(·)(Ω) the inequality
BΦ(K ) ≥Ψ(K )
holds.
Proof. Let ε > Φ(K ). Then there exist g1, . . . ,g N from L p(·)(Ω) such that for all f ∈ K
and some j ∈ 1, . . . , N ,
f − g j L p(·)(Ω) < ε.
Let us take δ > 0. Then due to Theorem 1.4.4 there are g j ∈ C 0(Ω), such that for all
j ∈ 1, . . . , N ,
g j − g j L p(·)(Ω) < δ. (1.4.3)
Hence for that given δ, f − g j L p(·)(Ω) < ε+δ. (1.4.4)
By the absolutely continuity of the norm there is a cube Q such that
gi L 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
Pφ( x) =m
∑i=1
φQiχ Qi( x); φQi = |Qi|−1
Qi
φ( y)χΩ( y)dy,
where Qi = Qi ∩Ω. Taking now into account the fact that g jχΩ = g j, we have
g j − Pg j p+(Ω∩ Q)
L p(·)(Ω∩ Q) ≤∑
i
Qi
1
|Q|i
Qi
|(g j( x) − gi( y)|dy
p( x)
dx
≤ σ p−(Ω∩ Q)m
∑i=1
| Qi| ≤ σ p− | Q|.
Consequently,
g j − Pg j L p(·)(Ω∩ Q) ≤ | Q|1/ p+σ p−/ p+.
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16 Alexander Meskhi
Further, the condition p ∈ P (Ω) yields
Pφ L p(·)(Ω) ≤
N
∑i=1
φQ jχ Q j
L p(·)(Ω)
≤ M Ωφ L p(·)(Ω) ≤ M Ωφ L p(·)(Ω).
Finally, taking into account estimates (1.4.3)–(1.4.5) and the fact that P ∈F L( L p(·)(Ω)), we find that
f − P f L p(·)(Ω) ≤ f − g j L p(·)(Ω) + g j − Pg j L p(·)(Ω) + Pg j − P f L p(·)(Ω)
< ε+ 2δ+ | Q|1/ p+σ p−/ p+ + M Ωg j − f L p(·)(Ω)
≤ (1 + M Ω)ε+ (2 + M Ω)δ+ | Q|1/ p+σ p−/ p+ .
As δ and σ are arbitrarily small and σ is independent of Q, we have the desired result.
Lemma 1.4.3. Let p ∈ P (Ω) and let K ⊂ L p(·)
(Ω) be compact. Then for a given ε > 0there exists an operator Pε ∈ F L( L p(·)(Ω)) such that for all f ∈ K,
f − Pε f L 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 ∈ F L( L p(·)(Ω)) such that
sup f − Pε f L 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 ) := distT ,F L( 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 U X := f : f X ≤ 1. From the compactness of T it follows that T (U X ) is
relatively compact in L p(·)(Ω). By Lemma 1.4.3 we have that for any ε > 0 there exists an
operator Pε ∈ F L( L p(·)(Ω)) such that for all f ∈ U X ,
T f − PεT f L p(·)(Ω) ≤ ε. (1.4.6)
Let Pε = Pε T . Then Pε ∈ F L( 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
Pε f ( x) = Pε(T f )( x) =m
∑ j=1
α j(T f )u j( x) =m
∑ j=1
β j( f )u j( x),
where β j = α j T (1 ≤ i ≤ m) are bounded linear functional from X to L p(·)(Ω). Further,
inequality (1.4.6) impliesT f − Pε f L p(·)(Ω) ≤ ε
for all f ∈ U X , from which it follows that α(T ) = 0.
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Basic Ingredients 17
Lemma 1.4.5. Let P ∈ F L( X , L p(·)(Ω)) , where X is a Banach space. Then for every
a ∈Ω and ε > 0 there exist an operator R ∈ F L( X , L p(·)(Ω)) and positive numbers α and α ,
α < α , such that for all f ∈ X the inequality
(P − R) f L p(·)(Ω) ≤ ε f X
holds, and moreover, supp R f ⊂ D(a, α) \ Q(a,α) , where Q(a,r ) is a cube with center a and
side length r; D(a,r ) := Q(a,r ) ∩Ω.
Proof. Since P ∈ F L( X , L p(·)(Ω)), there exist linearly independent non-negative func-
tions u j ∈ L p(·)(Ω), j ∈ 1, . . . ,m, such that
P f ( x) =m
∑ j=1
β j( f )u j( x), f ∈ X ,
where β j are bounded linear functionals on X . Further, there is a positive constant c forwhich N
∑ j=1
|β j( f )| ≤ c f X .
Let us choose linearly independent Φ j ∈ L p(·)(Ω) and positive real numbers α j, α j,
such that
u j −Φ j L p(·)(Ω) < ε, j ∈ 1, . . . , N ,
and suppΦ j ⊂ D(a, α j) \ Q(a,α j).
Let us take
R f ( x) = N
∑ j=1
β j( f )Φ j( x).
Then it is obvious that R ∈ F L( X , L p(·)(Ω)) and, also
P f − R f L p(·)(Ω) ≤ N
∑ j=1
|β j( f )|u j −Φ j L p(·)(Ω) ≤ cε f X
for all f ∈ X . Moreover,
supp R f ⊂
D(
a, α
) \Q
(a,α
),where α = minα j, α = maxα 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
f g /∈ 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]).
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18 Alexander Meskhi
Lemma 1.4.7([204]). For all f ∈ L p(·)(Ω) the inequality
f L p(·)(Ω) ≤ supg
L p′(·)(Ω)≤1
Ω
f ( y)g( y)dy
holds.
In the sequel we will assume that
E k := [2k ,2k +1); I k := [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 χ I k L p(·)(R+)gχ I k
Lq′(·)(R+) ≤ c f L p(·)(R+)g Lq′(·)(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:
∑i
f χ I i L p(·)(R+)gχ I i Lq′(·)(R+) =∑i≤2
+∑i>2
:= J 1 + J 2.
Taking into account that p( x) = pc = const, q( x) ≡ qc ≡ const on the set (1,∞), using
Holder’s inequality for series and the fact that pc ≤ qc, we have
J 2 =∑i>2
f χ I i L pc (R+)gχ I i L(qc)′(R+) ≤ c f L p(·)(R+)g
Lq′(·)(R+).
Now let us estimate J 1. Suppose that f L p(·)(R+) ≤ 1 and g Lq′(·)(R+) ≤ 1. First notice that
q,q′ ∈ W L(R+). Therefore, by Lemma 1.4.1 and Proposition 1.4.1 we have
| I k |1/q+( I k ) ≈ χ I k
Lq(·)(R+) ≈ | I k |1/q−( I k );
| I k |1/(q′)+( E k ) ≈ χ E k
Lq′(·)(R+) ≈ | I k |
1/(q′)−( I k ),
where k ≤ 2. Hence Holder’s inequality (see Lemma 1.4.1) yields
J 1 ≤ c∑k ≤2
8
0
f χ I k L p(·)(R+)gχ I k
Lq′(·)(R+)
χ I k Lq(·)(R+)χ E k
Lq′(·)(R+)
χ E k ( x)dx
≤ c
8
0∑k ≤2
f χ I k L p(·)(R+)gχ I k
Lq′(·)(R+)
χ I k Lq(·)(R+)χ I k
Lq′(·)(R+)
χ E k ( x)dx
≤ c∑k ≤2
f χ I k L p(·)(R+)
χ I k Lq(·)(R+)
χ E k (·)
Lq(·)((0,8))
∑k ≤2
gχ I k L p′(·)(R+)
χ I k
L p′(·)(R+)
χ E k (·)
Lq′(·)((0,8))
:= cS 1 · S 2.
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Basic Ingredients 19
Now we claim that
I (q) ≤ cI ( p),
where
I (q) := ∑k ≤2
f χ I k L p(·)(R+)
χ I k Lq(·)(R+) χ
E k (·) Lq(·)((0,8))
;
I ( p) :=
∑k ≤2
f χ I k L p(·)(R+)
χ I k L p(·)(R+)
χ E k (·)
L p(·)((0,8))
.
Indeed, suppose that I ( p) ≤ 1. Further, Proposition 1.4.1 and Lemma 1.4.1 yield
∑k ≤2
1
| I k |
E k
f χ I k
p( x)
L p(·)(R+)dx ≤ c
8 0 ∑k ≤2
f χ I k L p(·)(R+)
χ I k L p(·)(R+)
χ E k ( x)
p( x)
dx ≤ c.
Consequently, taking into account that q( x) ≥ p( x), E k ⊂ I k and f L p(·)(R+) ≤ 1, we find
that
∑k ≤2
1
| I k |
E k
f χ I k
q( x)
L p(·)(R+)dx ≤ c∑
k ≤2
1
| I k |
E k
f χ I k
p( x)
L p(·)(R+)dx ≤ c.
This implies that I (q) ≤ c.
Let us introduce a function
P(t ) = ∑k ≤2
p+( I k )χ E k (t ).
It is clear that p(t ) ≤ P(t ) because E k ⊂ I k . Hence, Proposition 1.4.2 for Ω = (0,8) yields
I ( p) ≤ c
∑k ≤2
f χ I k L p(·)(R+)
χ I k L p(·)(R+)
χ E k (·)
LP(·)((0,8))
.
Then, using the definition of P and the inequality χ I k
p+( I k )
L p(·)(R+) ≥ c2k , we have
8
0∑
k ≤2
f χ I k L p(·)(R+)χ I k
L p(·)(R+)
χ E k ( x)P( x)
dx
=
8
0
∑k ≤2
f χ I k
p+( I k )
L p(·)(R+)
χ I k
p+( I k )
L p(·)(R+)
χ E k ( x)
dx
≤ c
8
0
∑k ≤2
f χ I k
p+( I k )
L p(·)(R+)
2k χ E k
( x)
dx
≤ c∑k ≤2 f χ I k
p+( I k )
L p(·)(R+) ≤ c∑k ≤2 I k | f ( x)|
p( x)
dx
≤ c
R+
| f ( x)| p( x)dx ≤ c.
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20 Alexander Meskhi
Consequently, the estimates derived above give us
S 1 ≤ c f L p(·)(R+).
Analogously, taking into account the fact that q′ ∈ W L(R+) and arguing as above, we
find thatS 2 ≤ cg
Lq′(·)(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) := ( x − ·)α( x)−1χ( x/2, x)(·) L 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. Conse-
quently,
x
x/2( x − t )(α( x)−1) p′(t )dt ≤ c x
x/2( x − t )(α( x)−1) p′( x)dt
= c
x/2
0u(α( x)−1) p′( x)du = c( x/2)(α( x)−1) p′( x)+1
= cx(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
I ( x) ≤ (S ( x))1/( p′)
+([ x/2, x])
= c( x/2)1/( p′)
+([ x/2, x])(α( x)−1) p′( x)+1
≤ c
( x/2)1/( p′)−([ x/2, x])(α( x)−1) p′( x)+1
≤ c
( x/2)1/ p′( x)(α( x)−1) p′( x)+1
≤ 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 ) = x
0 f (t )dt , x > 0,
be the Hardy transform.
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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
is bounded from L p(·)w ( I ) to L
q(·)v ( I ) if and only if
C := sup
0<t <a
v(·)χ(t ,a)(·) Lq(·)( I )w−1(·)χ(0,t )(·) L p′(·)( I ) < ∞.
Moreover, there exist positive constants c1 and c2 such that c1C ≤ H ≤ c2C.
To formulate the next results we need the notation:
p0( x) := inf y∈[0, x]
p( y); p0( x) :=
p0( x), 0 ≤ x ≤ a
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
sup0<t <a
a
t (v( x))q( x)t q( x)/( p0)′( x)dx < ∞
implies the boundedness of H from L p(·)( I ) to Lq(·)v ( 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
sup0<t <∞
∞t
(v( x))q( x)t q( x)/( p0)′( x)dx < ∞
guarantees the boundedness of H from L p(·)( I ) to Lq(·)v ( 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 k ( x, y) L p′( y)( I )
Lq( x)( I )
< ∞,
where k is a non-negative kernel, then the operator
K f ( x) = I
k ( x, y) f ( y)dy
is compact from L p(·)( I ) to Lq(·)( I ).
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22 Alexander Meskhi
1.5. 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 eigenvaluesλ 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:
T σ p( H ) ≡∑
j
s p j (T )
1/ p
, 0 < p < ∞,
with the usual modification if p = ∞. Thus we have T σ∞( H ) = T and T σ2( H ) is the
Hilbert-Schmidt norm given by the formula
T σ2( H ) = |a( x, y)|2
dxdy1/2
for the integral operator
A f ( x) =
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
E t := y ∈ E : r ( y) < t ,where t is a positive number.
We denote by l p L2ρ( E )
the set of all measurable functions g : E → R for which
gl p
L2ρ( E ) =
∑
n∈Z
E
2n+1 \ E 2n
|g( x)|2ρ( x)dx p/21/ p
< ∞.
The next statement deals with interpolation result which is a consequence of more gen-
eral statements from [229]( p.127, p. 147) (see also [11]).
Proposition 1.5.1. Let 1 ≤ p0, p1 ≤ ∞, 0 ≤ θ ≤ 1, 1 p = 1−θ p0 + θ p1 . Suppose that ρ is aweight function on E. If A is a bounded operator from l pi
L2ρ( E )
into σ pi
L2ρ( E )
, where
i = 0,1 , then it is also bounded from l p L2ρ( E )
to σ p
L2ρ( E )
. Moreover,
Al p( L2ρ( E ))→σ p( L2( E )) ≤ T 1−θ
l p0 ( L2ρ( E ))→σ p0
( L2( E ))T θ
l p1 ( L2ρ( E ))→σ p1
( 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 f k , gk be orthonormal systems in a
Hilbert space H . If T ∈ σ p( H ) , then
T σ p( H ) ≥∑
n
|T f n,gn| p1/ p
.
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Basic Ingredients 23
1.6. Singular Integrals in Weighted Lebesgue Spaces
In this section we give some well-known statements regarding singular integrals in weighted
Lebesgue spaces. Let
H f ( x) = p.v. R f ( y)
x − ydy
and
S Γ f (t ) = p.v. 1
πi
Γ
f (τ)
τ− t d τ
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 < p < ∞ , if
supa∈Rr >0
A(r ,a) p (R) := sup
a∈Rr >0
1
2r
a+r
a−r w(s)ds1/ p 1
2r
a+r
a−r w1− p′
(s)ds1/ p′
< ∞.
Theorem 1.6.1 ([95]). Let 1 < p < ∞. Then the Hilbert transform H is bounded in
L pw(R) if and only if w ∈ A p(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 l
0|S Γ f (t (s))| pw(s)ds ≤ c
l
0| f (t (s))| pw(s)ds
holds if and only if w ∈ A p(0, l) , that is,
Bl := sup
1
| I |
I
w(s)ds
1/ p 1
| I |
I
w1− p′(s)ds
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 < p < ∞ and let the Hilbert transform H be bounded from L
pw(R) to L
pv (R). Then
supt >0
| x|>t
v( x)
| x| p dx
1/ p | x|<t
w1− p′( x)dx
1/ p′
< ∞
and
supt >0
| x|<t
v( x)dx
1/ p | x|>t
w1− p′( x)
| x| p′ dx
1/ p′
< ∞.
Suppose that for the operator
K f ( x) = p.v.
Rn
k ( x − y) f ( y),dy (1.6.1)
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24 Alexander Meskhi
the inequality K f
L2(Rn) ≤ c
f
L2(Rn), f ∈ C ∞0 (Rn) (1.6.2)
holds and the kernel k satisfies the following two conditions:
(i) there exists a positive constant A such that the inequality
∂∂ x
αk ( x)
≤ A| x|−n−α (1.6.3)
holds for all x ∈Rn, x = 0, and |α| ≤ 1;
(ii) there exists a positive constant b and an unit vector u0 such thatk ( 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) = limr →0γ n
Rn\ B( x,r )
x j − y j
| x − y|n+1 f ( y) dy, j = 1, . . . ,n, (1.6.5)
where x = ( x1, . . . , xn) ∈ Rn, γ n = 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
B := sup 1
| B| B w( x) dx1/ p
1
| B| B w1− p′( x) dx
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 L pw(Rn). Further, if (1.6.2) – (1.6.4)
hold and K is bounded in L pw(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 inR
n
. Then by choosing u = tu0 , with t fixed sufficiently large we can guaranteethat
|k (r (u + v)) − k (ru)| ≤ 1
2|k (ru)|
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 L pw(Rn) to L
pv (Rn) it follows
that w1− p′is locally integrable.
Proof. Let B := B(0,r ). Suppose that I (r ) := B w1− p′( x)dx = ∞ for some positive
number r . Then there exists g ∈ L p( B), g ≥ 0, such that B w−1/ pg = ∞. Let us assume that
f r ( y) = g( y)w−1/ p( y)χ B( 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′
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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 f r ( x)| ≥ 12
( f r ) B|k (ru)| for all x ∈ B′. Hence by (1.6.4) the following
estimates hold:
K f r L pv (Rn) ≥ χ B′ ( x)K f r ( x) L
pv (Rn) ≥
b
2rt
(v B′ )1/ p f B = ∞.
On the other hand, f r L pw(Rn) = g L 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 ) = I , where I is the identity operator from
W k , p
0 (Ω) to L p∗(Ω), k ∈N, 1 ≤ p <∞, k p < n and p∗ = np
n−kp. From this fact it follows that
all entropy numbers of I are equal to I .
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∗],
where p∗ = npn−kp , then the entropy numbers ek ( I ) of I : W m, p0 (Ω) → Lq(Ω) satisfy ek ( I ) ≈
k −mn .
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 opera-
tors 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 asymp-
totic 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 vari-
able 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 forclassical 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
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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 S Γ is bounded in L p( x)(Γ ) if and only if Γ is a Carleson curve, i.e.,
supt ∈Γ , r >0
ν(Γ ∩ B(t ,Γ ))
r < ∞.
The one-weight problem for the Hilbert transform in classical Lebesgue spaces was
solved in [95]. In [24] it was proved that the Calderon- Zygmund singular operator is
bounded in L pw(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 S T K ( L p(T )), where where T is the unit
circle, has been calculated in [79],[80] for p = 2n and p = 2n
2n−1, where Γ = T is the unit
circle (see also [81]). In these papers a lower estimate for S T K ( L p(T )) has been also
derived for all p ∈ (1,∞). The value of the norm of S T acting in L
p
(T ) (1 < p < ∞) wasfound 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].
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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 par-ticular, 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
M Ω as follows:
M Ω f ( x) = sup B∩Ω∋ x
1| B|
B∩Ω
| f ( y)|dy,
where the supremum is taken over all balls B in Rn with x ∈ B ∩Ω.
If Ω = Rn, then we use the notation
M Ω := M .
The next result is due to B. Muckenhoupt ([169]).
Theorem 2.1.1. Let 1 < p < ∞. Then the operator M is bounded in L pw(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 M Ω is bounded in L pw(Ω) if and only
if w ∈ A p(Ω) , i.e.
sup B
1
| B ∩Ω|
B∩Ω
w( x)dx
1/ p 1
| B ∩Ω|
B∩Ω
w1− p′( x)dx
1/ p′
< ∞. (2.1.1)
Now we formulate and prove our main results concerning the maximal operators defined
on Ω.
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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 L pw(Ω) to L
pv (Ω).
Moreover, if M Ω is bounded from L pw(Ω) to L
pv (Ω) , then the following estimate holds
M Ωκ ≥ supa∈Ω limr →0
1
| B(a,r )| B(a,r ) v( x)dx1/ p
B(a,r ) w
1− p′
( x)dx1/ p′
.
Proof. It is known (see e.g. [169], [212], [214]) that if M Ω is bounded from L pw(Ω) to
L pv (Ω), then v,w1− p′
∈ Lloc(Ω). Further, let λ > M Ωκ ( L pw(Ω), L
pv (Ω)) and a ∈Ω. By Lemma
1.2.8 we have that there exists a constant β such that if 0 < τ < β, then B(a,τ)∩Ω
v( x) M Ω f
p( x)dx ≤ λ p
Ω
| f ( x)| pw( x)dx. (2.1.2)
Consequently, putting f ( x) = χ B(a,τ)∩Ω( x)w1− p′( x) in (2.1.2) we find that
| B(a,τ)|− p
B(a,τ)
v( x)dx
B(a,τ)
w1− p′( x)dx
p−1
≤ λ p
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
M Ωκ ≥ 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
Rn such that M is compact from L
pw(Rn) to L
pv (Rn). Moreover, if M is bounded from L
pw(Rn)
to L pv (Rn) , then the estimate
M κ ≥ max supa∈Rn
limτ→0
I 1(a,τ); supa∈Rn
limτ→∞
I 2(a,τ)
holds, where
I 1(a,τ) = 1
| B(a,τ)|
B(a,τ)
v( x)dx
1/ p B(a,τ)
w1− p′( x)dx
1/ p′
;
I 2(a,τ) = 1
| B(a,2τ)|
B(a,2τ)\ B(a,τ)
v( x)dx
1/ p B(a,τ)
w1− p′( x)dx
1/ p′
.
Proof. Let λ > M κ
. By Lemma 1.2.7 there are constants β1
and β2
such that B(a,τ)
v( x)( M f ( x)) pdx ≤ λ p Rn
| f ( x)| pw( x)dx (2.1.3)
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Maximal Operators 29
and Rn\ B(a,t )
v( x)( M f ( x)) pdx ≤ λ p Rn
| f ( x)| pw( 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
supa∈Rn
limτ→0
I 1(a,τ) ≤ λ,
while (2.1.4) implies that B(a,2t )\ B(a,t )
v( x)( M f ( x)) pdx ≤ λ p Rn
| f ( x)| pw( x)dx. (2.1.5)
Hence, due to the definition of M and putting f ( x) = χ B(a,t )( x)w1− p′( x) in (2.1.5), we find
that
B(a,2t )\( B(a,t ) v( x)dx B(a,t ) w1− p′
( x)dx p−1
| 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,η1r ) ≥ 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 I 2(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
M K ( L pw(Rn)) ≥ max sup
a∈Rn
limτ→∞
I (a,τ);C B,n supa∈Rn
limτ→0
I (a,τ),
where
I (a,τ) := | B(a,τ)|−1
B(a,τ)
w( x)dx
1/ p B(a,τ)
w1− p′( x)dx
1/ p′
and C B,n is a constant depending only on ¯ B and n.
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30 Alexander Meskhi
Proof. By Theorem 2.1.4 it suffices to show that
w( B(a,η1τ) \ B(a,τ)) ≥ cw( B(a,τ)),
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,η1τ)) \ 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
M Ωκ ( L pw(Ω))
≥ supa∈Rn
limτ→0
| B(a,τ)|−1
B(a,τ)
w( x)dx
1/ p B(a,τ)
w1− p′( x)dx
1/ p′
.
Let us now estimate the measure of non-compactness for the fractional maximal func-
tion
M α,Ω f ( x) = sup
B∩Ω∋ x
1
| B|1−α/n B∩Ω| f ( y)|dy,
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∗ = npn−α p and
Ω = Rn. Then M α is bounded from L
p
ρ p/ p∗ (Rn) to L p∗
ρ (Rn) if and only if
˜ B := sup B
1
| B|
Bρ( x)dx
1/ p∗ 1
| B|
Bρ− p′/ p∗
( 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 Hardy-
Littlewood maximal operator; therefore we omit the proofs.
Theorem 2.1.6. Let 1 < p <∞ , 0 < α< n/ p, p∗ = npn−α p . Suppose that Ω is a bounded
domain in Rn. Then there is no weight pair (v,w) on Ω such that M α,Ω is compact from
L pw(Ω) to L
p∗
v (Ω). Further, suppose that M α,Ω is bounded from L pw(Ω) to L
p∗
v (Ω). Then
M α,Ωκ ( L pw(Ω)→ L
p∗v (Ω))
≥ supa∈Ω
limτ→0
| B(a,τ)|α/n−1
B(a,τ)
v( x)dx
1/ p∗ B(a,τ)
w1− p′( x)dx
1/ p′
.
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Maximal Operators 31
Corollary 2.1.4. Let 1 < p <∞ , 0 < α< n/ p, p∗ = npn−α p . Suppose that Ω is a bounded
domain in Rn and M α,Ω is bounded from L p
ρ p/ p∗ (Ω) to L p∗
ρ (Ω). Then
M α,Ωκ ( L p
ρ p/ p∗ (Ω)→ L p∗ρ (Ω))
≥ supa∈Ω
limτ→0
| B(a,τ)|α/n−1
B(a,τ)ρ( x)dx
1/ p∗ B(a,τ)ρ− p′/ p∗
( x)dx
1/ p′
.
In the case of Ω = Rn we have the following statement.
Theorem 2.1.7. Let 1 < p <∞ , 0 < α< n/ p, p∗ = npn−α p . Suppose that Ω=R
n and that
(2.1.6) holds. Then there is no weight pair (v,w) such that the operator M α,Ω is compact
L pw(Rn) to L
p∗
v (Rn). Moreover, the following estimate holds:
M α,Ωκ ( L pw(Ω), L p∗
v (Ω)) ≥ max supa∈Rn
limτ→0
I (α)1 (a,τ); supa∈Rn
limt →∞
I (α)2 (a,t ),
where
I (α)1 (a,τ) := | B(a,τ)|α/n−1
B(a,τ)
v( x)dx
1/ p∗ B(a,τ)
w1− p′( x)dx
1/ p′
,
and
I (α)2 (a,τ) := | B(a,2τ)|α/n−1 B(a,2τ)\ B(a,τ)
v( x)dx1/ p∗
B(a,τ)w1− p
′
( x)dx1/ p′
.
Corollary 2.1.5. Let 1 < p < ∞ , 0 < α < n/ p and p∗ = npn−α p . Suppose that ˜ B < ∞ ,
where ˜ B is the constant defined by (2.1.6). Then the inequality
M ακ ( L p
ρ p/ p∗ (Rn), L pρ(Rn)) ≥ max sup
a∈Rn
limτ→0
J (α),C B,n,α, p supa∈Rn
limτ→∞
J (α)(a,τ)
holds, where
J (α)(a,τ) = | B(a,τ)|α/n−1
B(a,τ)ρ( x)dx
1/ p∗ B(a,τ)ρ− p′/ p∗
( x)dx1/ p′
and the positive constant C ˜ B,n,α, p depends on B, n, α and p.
Proof. First observe that the condition (2.1.6) implies ρ ∈ A1+ p∗/ p′ (Rn). Hence the
measure ρ( E ) =
E ρ( x)dx satisfy the doubling condition. Applying Remark 2.1.1, Theo-
rems 2.1.5, 2.1.7 and the arguments from the proof of Corollary 2.1.2 we have the desired
result.
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32 Alexander Meskhi
2.2. One–sided Maximal Functions
In this section we deal with the one-sided maximal functions:
N +α f ( x) = sup
h>0
1
h1−α
x+h
x
| f ( y)|dy, x ∈R,
N −α f ( x) = suph>0
1
h1−α
x
x−h| f ( y)|dy, x ∈R,
where 0 ≤ α < 1. If α = 0, then we denote N + := N +0 and N − := N −0 .
Definition 2.2.1. Let 1 < p <∞. We say that w ∈ A+ p (R) if there exists a constant c > 0
such that
1
h
x
x−h
w(t )dt
1/ p
1
h
x+h
x
w1− p′(t )dt
1/ p′
≤ c; x ∈ R, h > 0.
Further, w ∈ A− p (R) if
1
h
x+h x
w(t )dt
1/ p1
h
x x−h
w1− p′(t )dt
1/ p′
≤ C ; x ∈ R, h > 0,
for some positive constant C .
Theorem 2.2.1 ([216], [3]). Let 1 < p < ∞. Then
(i) N + is bounded in L p
w(R) if and only if w
∈ A+
p (R);
(ii) N − is bounded in L pw(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
sup0<h≤ x
1
h
x x−h
ρq(t )dt
1q
1
h
x+h x
ρ− p′(t )dt
1 p′
< ∞.
Further, ρ ∈ A− pq(R+) if
sup0<h≤ x
1
h
x+h x
ρq(t )dt
1q
1
h
x x−h
ρ− p′(t )dt
1 p′
< ∞.
Theorem 2.2.2 ([3]). Suppose that 0 < α < 1 , 1 < p < 1α and q = p
1−α p . Then N +α is
bounded from L pρ p (R) t o L
qρq (R) if and only if ρ ∈ A+
pq(R). Further, N −α is bounded from
L pρ p (R) to L
qρq (R) if and only if ρ ∈ A−
pq(R).In the next statement we assume that the symbol N α denotes one of the operators N +α ,
N −α .
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Maximal Operators 33
Theorem 2.2.3. Let 1 < p ≤ q < ∞ ,and α = 1/ p − 1/q. Then there is no weight pair
(v,w) defined on R for which the operator N α is compact from L pw(R) to L
qv (R). Moreover,
if N α is bounded from L pw(R) to L
qv (R) for some weight functions v and w, then
N +
ακ
≥ 2α−1 supa∈R
limh→01
h a
a−hv( x)dx
1/q
1
h a+h
aw1− p′
( x)dx1/ p′
;
N −α κ ≥ 2α−1 supa∈R
limh→0
1
h
a+h
av( x)dx
1/q1
h
a
a−hw1− p′
( x)dx
1/ p′
.
Proof. We prove the theorem for N α ≡ N +α . Let λ > N +α κ . First observe that the
boundedness of N +α from L pw(R) to L
qv (R) implies
J (t ) := t
−t w1− p′
(τ)d τ < ∞
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 t
−t g(τ)w−1/ p(τ)d τ = ∞.
Assuming now that f t ( y) = g( y)w−1/ p( y)χ(−t ,t )( y) we find that
N +α f t Lqv (R) ≥ χ(−2t ,−t ) N +α f t L
qv (R)
≥ ct α−1 −t
−2t v( x)dx1/q t
−t g( y)w−1/ p( y)dy = ∞
On the other hand,
f t L pw(R) =
t
−t g p( x)dx < ∞
which contradicts the boundedness of N +α from L pw(R) to L
qv (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
a+τ
a−τv( x)( N +α f ( x))qdx ≤ λq
R| f ( x)| pw( x)dxq/ p
.
for all a ∈ R and small τ, where λ > N +α κ .Hence
1
(2τ)1−α
a
a−τv( x)dx
a+τ
a f (t )dt
q
dx ≤ λq
R
f p( x)w( x)dx
q/ p
, (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 N +α . TheLebesgue differentiation theorem completes the proof.
The proof for N −α is similar to that for N +α .
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34 Alexander Meskhi
2.3. Maximal Operator on Homogeneous Groups
Let G be a homogeneous group and let
M α f ( x) = sup
B∋ x
1
| B|1− α
Q B
| f ( y)|dy, x ∈ G, 0 ≤ α < Q,
where the supremum is taken over all balls B ⊂ G containing x. If α = 0 then M α 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 L pρ (G) if and only if ρ ∈ A p(G) i.e.
sup B 1
| B| B
ρ( x)dx1 p 1
| B| B
ρ1− p′
( x)dx 1
p′< ∞, (2.3.1)
where the supremum is taken over all balls B ⊂ G.
Theorem 2.3.2. Let 1 < p < ∞ , 0 < α < Q p
. Then the fractional maximal function M α
is bounded from L pρ p (G) to L
qρq (G) , where q = Qp
Q−α p , if and only if ρ ∈ A pq(G) i.e.
sup B
1
| B|
B
ρq( x)dx 1
q 1
| B|
B
ρ− p′
( x)dx 1
p′< ∞. (2.3.2)
Now we formulate and prove the main results of this subsection.
Theorem 2.3.3. Let 1 < p <∞. Suppose that M is bounded from L pw(G) to L
pv (G). Then
there is no weight pair (v,w) such that M is compact from L pw(G) to L
pv (G). Moreover, the
inequality
M κ ( L pw(G), L p
v (G)) ≥ supa∈G
limτ→0
1
| B(a,τ)|
B(a,τ)
v( x)dx 1
p
B(a,τ)
w1− p′
( x)dx 1
p′
holds.
Proof. Suppose that λ > M κ ( L pw(G)→ L
pv (G)) and a ∈ G. By Lemma 1.2.7 we have that
B(a,τ)
v( x)( M f ( x)) pdx ≤ λ p
B(a,τ)
| f ( x)| pw( x)dx (2.3.3)
for all τ (τ≤ β) and all f supported in B(a,τ). First observe that
B(a,τ)
w
1− p′
( x)dx < ∞
for all τ > 0 (see also, for example, [168], [227], [76], Ch. 4).
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Maximal Operators 35
Further, substituting f ( y) = χ B(a,r )( y) w1− p′
( y) in inequality (2.3.3)we find that
1
| B(a,τ)| p
B(a,τ)
v( x)dx
B(a,τ)
w1− p′
( x)dx p−1
≤ λ p.
This inequality and Lebesgue differentiation theorem for homogeneous groups (see
[70], p. 67) yield the desired result.
Theorem 2.3.4. Let 1 < p < ∞ , 0 < α < Q/ p and let q = QpQ−α p . Suppose that M α is
bounded from L pw(G) to L
qv (G). Then there is no weight pair (v,w) such that M α is compact
from L pw(G) to L
qv (G). Moreover, the inequality
M ακ ≥ supa∈G
limτ→0
1
| B(a,τ)|αQ
−1
B(a,τ)
v( x)dx 1
q
B(a,τ)
w1− p′
( x)dx 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
M κ ( L pw(G)) ≥ Q
(γ + Q)
1 p (γ (1 − p
′) + Q)
1
p′ −1
.
Indeed, first recall that the fact that | B(e,1)| = 1 and Proposition 1.1.1 impliesσ(S ) = Q,
where S is the unit sphere in G and σ(S ) is its measure. Taking into account Theorems 2.3.1and 2.3.3 we have
M κ ( L pw(G)) ≥ lim
τ→0
1
| B(e,τ)|
B(e,τ)
w( x)dx
1/ p B(e,τ)
w1− p′( x)dx
1/ p′
= σ(S ) limτ→0τ−Q
τ0
t γ +Q−1dt
1/ p τ0
t γ (1− p′)+Q−1dt
1/ p′
= Q(γ + Q)1 p (γ (1 − p
′) + Q)
1
p′
−1
.
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 Hardy-
Littlewood 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 poten-
tials 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.
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36 Alexander Meskhi
Two-weight criteria involving the operator itself for the Hardy-Littlewood maximal op-
erator have been established in [212] (see [172], [191] for another type of sufficient condi-
tions).
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 M L
pw(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 M L
pw(Rn) ≤
C p,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.
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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
E t := y ∈ E : r ( y) < t .
In the sequel we also use the notation:
S x := E r ( x)/2c0, F x := E r ( x)\S x,
where the constant c0 comes from the triangle inequality for r (see Section 1.1).
In this chapter boundedness/compactness criteria from L p( E ) to Lqv ( E ) are established
for the operator
K f ( x) = E r ( x)
k ( x, y) f ( y)dy, x ∈ E , ( A)
with positive kernel k , where 1 < p,q < ∞ or 0 < q ≤ 1 < p < ∞, E r ( 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) E r ( x)
k ( x, y) f ( y)dy, x ∈ E , ( B)
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 ∈ V λ , if there exist positive constants c1 , c2 and c3 such that
(i) k ( x, y) ≤ c1k ( x,δ1/(2c0) x) (C )
for all x, y ∈ E with r ( y) < r ( x)/(2c0);
k ( x, y) ≥ c2k ( x,δ1/(2c0) x) (C ′)
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38 Alexander Meskhi
for all x, y ∈ E with r ( x)/(2c0) < r ( y) < r ( x);
(ii) F x
k λ′( x, y)dy ≤ c3r Q( x)k λ
′( x,δ1/(2c0) x), λ′ = λ/(λ− 1), ( D)
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 ∈ V λ, where n/λ < α ≤ n. Indeed, it is easy to see that if y ∈ S x, 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
F x
| x − y|(α−n)λ′ dy = ∞
0
y ∈ F x : | x − y|(α−n)λ′ > εd ε
≤ | x|(α−n)λ′
0
(· · · ) +
∞
| x|(α−n)λ′
(· · · ) := I1 + I2.
For I1 we have
I1 ≤
| x|(α−n)λ′ 0
B(0, | x|)d ε = c| x|(α−n)λ′+n,
while for I2 we observe that
I2 ≤
∞
| x|(α−n)λ′ y : | y| < | x|, | x − y| ≤ ε1/(α−n)λ′
d ε
≤ c
∞ | x|(α−n)λ′
εn/(α−n)λ′ d ε = cα,n
| x|(α−n)λ′
1+n/(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 :
(i) k (δt ¯ x,δτ ¯ y) ≤ c1k (δt ¯ x,δs ¯ z)
for all t , τ, s, ¯ x, ¯ y, ¯ z with 0 < τ < s < t ; ¯ x, ¯ y, ¯ z ∈ A;
(ii)
t
t /(2c0)k λ
′(δt ¯ x,δτ ¯ y)τQ−1d τ≤ c2t Q · k λ
′(δt ¯ x,δt /(2c0) ¯ x), t > 0, ¯ x ∈ A,
then k ∈ V λ.
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
(i) k (s, l) ≤ c1k (s,t ), 0 < l < t < s,
(ii) t
t /(2c0)k λ
′(t ,s)sQ−1ds ≤ c2t Qk λ
′(t ,t /(2c0)), 1 < λ < ∞,
then k ∈ V λ.
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Kernel Operators on Cones 39
3.1. Boundedness
In this section we establish boundedness/compactness criteria for K acting from L p( E ) to
Lqv ( E ), 1 < p ≤ q < ∞.
Theorem 3.1.1. Let 1 < p ≤ q < ∞ and let v be a weight on E. Suppose that k ∈ V p.
Then K is bounded from L p( E ) to Lqv ( E ) if and only if
B := sup j∈Z
B( j) := sup j∈Z
E
2 j+1 \ E 2 j
v( x)k q( x,δ1/(2c0) x)dx1/q
2 jQ/ p′
< ∞. (3.1.1)
Proof. Sufficiency. Let f ≥ 0 on E . We have
K f q
Lqv ( E )
≤ c
E
v( x)
S x
k ( x, y) f ( y)dyq
dx +
E
v( x)
F x
k ( x, y) f ( y)dyq
dx
:= c(I1 + I2).
By condition (C) and Theorem 1.3.2 we find that
I1 ≤ c
E
v( x)k q( x,δ1/(2c0) x)
E r ( x)
f ( y)dyq
dx ≤ c
E
f p ( x)dxq/ p
,
while Holder’s inequality and condition (D) yield
I2 ≤
E v( x)
F x
k p′( x, y)dy
q/ p′ F x
f p ( y)dyq/ p
dx
≤ c E
v( x)k q( x,δ1/(2c0) x)r Qq/ p′ ( x) F x
f p ( y)dyq/ p
dx
≤ c∑ j∈Z
E
2 j+1 \ E 2 j
v( x)k q( x,δ1/(2c0) x)r Qq/ p′( x)
F x
f p ( y)dyq/ p
dx
≤ c∑ j∈Z
E
2 j+1 \ E 2 j
v( x)k q( x,δ1/(2c0) x)r Qq/ p′( x)dx
× E
2 j+1 \ E 2 j−1/c0
f
p
( y)dyq/ p
≤ cB
q
f
q
L p( E ).
Sufficiency has been proved.
Necessity. To prove necessity we take the functions f j( x) = χ E 2 j+1
( x). Then simple
calculations show that f j L p( E ) = c2 jQ/ p. Further, condition (C) yields
K f j( x)q
Lqv ( E )
≥
E 2 j+1 \ E
2 j
v( x)
F x
f j( y)k ( x, y)dyq
dx
≥ c E
2 j+1 \ E 2 j
v( x)k q
( x,δ1/(2c0) x)dx2 jQq
.
Finally, due to the boundedness of K we have the desired result.
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40 Alexander Meskhi
Now we consider the case q < p.
Theorem 3.1.2. Let 0 < q < p < ∞ and let p > 1. Suppose that k ∈ V p. Then the
following conditions are equivalent:
(i) K is bounded from L p( E ) to Lqv ( E );
(ii)
D :=
E
E \ E r ( x)
k q( y,δ1/(2c0) y)v( y)dy
p/( p−q)
r ( x)Qp(q−1)/( p−q)dx
( p−q)/ pq
< ∞.
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
I1 ≤ c E v( x)k q( x,δ1/(2c0) x) S
x
f ( y)dyq
dx
= c
∞0
V (t )
t /(2c0)
0F (τ)d τ
q
dt ,
where
V (t ) := t Q−1
Av(δt ¯ x)k q(δt ¯ x,δt /(2c0) ¯ x)d σ( ¯ x),
F (t ) := t Q−1
A
F (δt ¯ x)d σ( ¯ x).
Notice that
D = ∞
0
∞t
V (τ)d τ p/( p−q)
t Qp(q−1)/( p−q)+Q−1dt ( p−q)/( pq)
= c
∞0
∞t
V (τ)d τ
p/( p−q)
×
t
0τ(Q−1)(1− p)(1− p′)d τ
p(q−1)/( p−q)
t (Q−1)(1− p)(1− p′)dt
( p−q)/( pq)
.
Consequently, due to Theorem 1.3.3, Holder’s inequality and Proposition 1.1.2 we find
that
I1 ≤ c
∞0
t (Q−1)(1− p)F p(t )dt
q/ p
= c
∞0
t (Q−1)(1− p)
A
f (δt ¯ x)d σ( ¯ x)
p
t (Q−1) pdt
q/ p
≤ c
∞0
t Q−1
A
f p(δt ¯ x)d σ( ¯ x)
dt
q/ p
= c
E
f p( x)dx
q/ p
.
Further, applying Holder’s inequality twice and condition (D) we find that
I2 ≤
E v( x)
F x
f p ( y)dy
q/ p F x
k p′( x, y)dy
q/ p′
dx
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Kernel Operators on Cones 41
≤ c
E
v( x)
F x
f p( y)dy
q/ p
r Qq/ p′( x)k q( x,δ1/(2c0) x)dx
= c∑ j∈Z
E
2 j+1 \ E 2 j
v( x)
F x
f p( y)dy
q/ p
r ( x)Qq/ p′k q( x,δ1/(2c0) x)dx
≤ c∑ j∈Z
E
2 j+1 \ E 2 j−1/c0
f p ( y)dy
q/ p E
2 j+1 \ E 2 j
v( x)r Qq/ p′( x)k q( x,δ1/(2c0) x)dx
≤ c
∑
j∈Z
E
2 j+1 \ E 2 j−1/c0
f p( y)dy
q/ p
×
∑
j∈Z
E
2 j+1 \ E 2 j
v( x)r Qq/ p′( x)k q( x,δ1/(2c0) x)dx
p/( p−q)( p−q)/ p
≤ c f q
L p( E )
∑
j∈Z
E
2 j+1 \ E 2 j
v( x)r Qq/ p′( x)k q( x,δ1/(2c0) x)dx
p/( p−q)( p−q)/ p
=: c f q
L p( E )( ¯ D)q.
Besides this,
( ¯ D) pq/( p−q)≤c∑ j∈Z
2Qq( p−1) j/( p−q)
E
2 j+1 \ E 2 j
v( x)k q( x,δ1/(2c0) x)dx
p/( p−q)
≤ c∑ j∈Z
E
2 j \ E 2 j−1
r ( y)Qp(q−1)/( p−q)
×
E \ E r ( y)
v( x)k q( x,δ1/(2c0) x)dx
p/( p−q)
dy
≤ cD pq/( p−q) < ∞.
Now we show that (i)⇒(iii). Let vn( x) = v( x)χ E n\ E 1/n( x), where n is an integer with
n ≥ 2. Let
f n( x) =
E \ E r ( x)
vn( z)k q( z,δ1/(2c0) z)dz
1/( p−q)
r ( x)Q(q−1)/( p−q).
It is easy to see that
f L p( E ) =
E
E \ E r ( x)
k q( y,δ1/(2c0) y)vn( y)dy
p/( p−q)
×r ( x)Qp(q−1)/( p−q)dx1/ p
< ∞.
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42 Alexander Meskhi
On the other hand, by condition (C ′) and integration by parts we find that
K f Lqv ( E ) ≥
E
vn( x)
F x
f n( y)k ( x, y)dy
q
dx
1/q
≥ c E
vn( x)k q( x,δ1/(2c0) x) F x
f n( y)dyq
dx1/q
= c
E
vn( x)k q( x,δ1/(2c0) x)
F x
E \ E r ( y)
vn( z)k q( z,δ1/(2c0) z)dz
1/( p−q)
×r ( y)Q(q−1)/( p−q)dy
q
dx
1/q
≥ c E vn( x)k q( x,δ1/(2c0) x) E \ E
r ( x)
vn( z)k q( z,δ1/(2c0) z)dzq/( p−q)
×
F x
r ( y)Q(q−1)/( p−q)dy
q
dx
1/q
≥ c
E
vn( x)k q( x,δ1/(2c0) x)
E \ E r ( x)
vn( z)k q( z,δ1/(2c0) z)dz
q/( p−q)
×r ( x)Qq( p−1)/( p−q)dx
1/q
c ∞0
t Q−1 A
vn(δt ¯ x)k q(δt ¯ x,δt /(2c0) ¯ x)d σ( ¯ x) ∞t τQ−1
×
A
vn(δτ ¯ z)k q(δτ ¯ z,δτ/(2c0) ¯ z)d σ(¯ z)
d τ
q/( p−q)
t Qq( p−1)/( p−q)dt
1/q
= c
∞0
d
∞t τQ−1
A
vn(δτ ¯ x)k q(δτ ¯ x,δτ/(2c0) ¯ x)d σ( ¯ x)
d τ
p/( p−q)
×t Qq( p−1)/( p−q)dt
1/q
= c
∞
0 ∞
t τQ−1
Avn(δτ ¯ x)k q(δτ ¯ x,δτ/(2c0))d σ( ¯ x)
d τ
p/( p−q)
×t Qq( p−1)/( p−q)dt 1/q
= c
∞0
t Q−1
∞t τQ−1
A
vn(δτ ¯ x)k q(δτ ¯ x,δτ/(2c0) ¯ x)d σ( ¯ x)
d τ
p/( p−q)
×t Qq( p−1)/( p−q)−Qdt
1/q
= c E E \ E r ( x)
vn( x)k q( x,δ1/(2c0) x)dx p/( p−q)
r ( x)Qp(q−1)/( p−q)dx1/q
.
Now the boundedness of K and Fatou’s lemma completes the proof of the implication
(i)⇒ (ii).
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Kernel Operators on Cones 43
3.2. Compactness
This section deals with the compactness of the operator K.
Theorem 3.2.1. Let 1 < p ≤ q < ∞. Suppose that k ∈ V p. Then K is compact from
L p( E ) to Lqv ( E ) if and only if
(i) (3.1.1) holds;(ii)
lim j→−∞
B( j) = lim j→+∞
B( j) = 0 (3.2.1)
(see (3.1.1) for B( j)).
Proof. Sufficiency. Let 0 < a < b < ∞ and represent K f as follows:
K f = χ E aK ( f χ E a ) +χ E b\ E aK( f χ E b )
+χ E \ E bK( f χ E b/2c0 ) +χ E \ E bK( f χ E \ E b/2c0 )
:= P1 f + P2 f + P3 f + P4 f .
For P2 we have
P2 f ( x) = E
k ∗( x, y) f ( y)dy,
where k ∗( x, y) = χ E b\ E a ( x)χ E r ( x)( y)k ( x, y). Further, observe that
S := E
E k ∗( x, y) p′
dyq/ p′
v( x)dx = E b\ E a
E r ( x)
k p′( x, y)dy
q/ p′
v( x)dx
≤ c
E b\ E a
S x
k p′( x, y)dy
q/ p′
v( x)dx + c
E b\ E a
F x
k p′( x, y)dy
q/ p′
v( x)dx
:= S 1 + S 2.
Taking into account the condition k ∈ V p we have
S i ≤ c E b\ E a
k q
( x,δ1/(2c0) x)r qQ/ p′
( x)v( x)dx
≤ cbq/ p′
E b\ E a
k q( 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
P1 ≤ cB(a)
:= c supt ≤a E a\ E t
v( x)k q
( x,δ1/(2c0) x)dx1/q
t Q/ p′
;
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44 Alexander Meskhi
P4 ≤ cB(b) := c supt ≥b
E \ E t
v( x)k q( x,δ1/(2c0) x)dx1/q
t Q − bQ1/ p′
.
Therefore
K− P2 − P3 ≤ P1 + P4 ≤ c B(a) + B(b).It remains to show that condition (3.2.1) implies
lima→0
B(a) = limb→∞
B(b) = 0.
Let a > 0. Then a ∈ [2m,2m+1) for some integer m. Consequently, B(a) ≤ B(2m+1). Further,
for 0 < t < 2m+1, there exists j ∈ Z , j ≤ m, such that t ∈ [2 j,2 j+1). Hence
t qQ/ p′
E 2m+1 \
E t
v ( x)k q( x,δ1/(2c0) x))dx
≤ 2( j+1)qQ/ p′ m
∑k = j
E
2k +1 \ E 2k
v( x)k q( x,δ1/(2c0) x)dx
≤ cBq(k )2 jqQ/ p′ ∞
∑k = j
2−kqQ/ p′= cBq(k ).
Taking into account this estimate we shall find that
B(2m) ≤ supk ≤m
( B(k ))q =: S (m).
If a → 0, then m → −∞. Consequently, S (m) → 0, which implies that B(2m) → 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 E \ E t
v( x) k q ( x,δ1/(2c0) x)dx
t Q − 2mQq/ p′
≤ E \ E
2k
v( x)k q( x,δ1/(2c0) x)dx2(k +1)Q − 2mQq/ p′
≤ c2kqQ/ p′ ∞
∑ j=k
E
2 j+1 \ E 2 j
v( x)k ( x,δ1/(2c0) x)q( x)dx
≤ cBq(k )2kqQ/ p′ ∞
∑ j=k
2− jqQ/ p′≤ cBq(k ).
Consequently, B(2
m
) ≤ supk ≥m
B(
k )
. From the last inequality we conclude that the condition
limk →+∞
B(k ) = 0 implies limb→∞
B(b) = 0.
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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) = χ E 2 j+1 \ E
2 j−1/c0
( y)2− jQ/ p. Then for φ ∈ L p′( E ), we have
E
f j( y)φ( y)dy
≤
E 2 j+1 \ E 2 j−1/c0
| f j( y)| pdy
1/ p
E 2 j+1 \ E 2 j−1/c0
|φ( y)| p′dy
1/ p′
=
E 2 j+1 \ E
2 j−1/c0
|φ( y)| p′dy 1
p′
→ 0
as j → −∞ or j → +∞. On the other hand,
K f j Lqv ( E ) ≥
E
2 j+1 \ E 2 j
K f j( x)
qv( x)dx
1/q
≥ c E
2 j+1 \ E 2 j
F x
k ( x, y) f j( y)dyq
v( x)dx1/q
≥ c E
2 j+1 \ E 2 j
k q( x,δ1/(2c0) x)
F x
f j( y)dyq
v( x)dx1/q
≥ c
E 2 j+1 \ E
2 j
k q( x,δ1/(2c0) x)v( x)dx1/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 < p we have the next statement:
Theorem 3.2.2. Let 0 < q < p < ∞ and let p > 1. Suppose that k ∈ V p. Then the
following conditions are equivalent:
(i) K is compact from L p( E ) to Lqv ( E );
(ii)
D :=
E
E \ E r ( x)
k q( y,δ1/(2c0) y)v( y)dy p/( p−q)
r ( x)Qp(q−1)/( p−q)dx( p−q)/ pq
< ∞.
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 k 0 the function r ( x)Qk 2( x,δ1/(2c0) x).
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46 Alexander Meskhi
Theorem 3.3.1. Let 2 ≤ p < ∞ and let k ∈ V 2. Then Ku ∈ σ p( L2( E )) if and only if
u ∈ l p( L2k 0
( E )). Moreover, there exist positive constants b1 and b2 such that
b1ul p( L2k 0
( E )) ≤ Kuσ p( L2( E )) ≤ b2ul p( L2k 0
( E )).
Proof. Sufficiency. First observe that
J ( x) :=
E r ( x)
k 2( x, y)dy ≤ ck 0( x),
where c is a positive constant. Indeed, splitting the integral over E r ( x) into two parts and
taking into account the condition k ∈ V 2 we have
J ( x) =
S x
k 2( x, y)dy +
F x
k 2( x, y)dy ≤ c1k 0( x) + c2k 0( x) ≤ c3k 0( x).
Hence, the Hilbert-Schmidt formula yields
Kuσ2( L2( E )) =
E × E r ( x)
k ( x, y)u( x)
2dxdy
1/2
=
E
u2( x)
E r ( x)
k 2( x, y)dy
dx1/2
≤ c
E
u2( x)k 0( x)dx1/2
= c∑n∈Z
E
2n+1 \ E 2n
u2( x)k 0( x)dx1/2
= cul2( L2k 0
( 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
Kuσ p( L2( E )) ≤ cul p( L2k 0
( E )),
where 2 ≤ p < ∞.
Necessity. Let Ku ∈ σ p
L2( E )
. We set
f n( x) = χ E 2n+1 \ E 2n ( x)| E 2n+1 \ E 2n |− 12 ;
gn( x) = u( x)χ E 2n+1 \ E
3·2n−1( x)k
1/20 ( x)α
−1/2n ,
where
αn =
E
2n+1 \ E 3·2n−1
u2( x)k 0( x)dx.
Then it is easy to verify that f n and gn are orthonormal systems in L2( E ). By Proposi-
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Kernel Operators on Cones 47
tion 1.5.2 we find that
∞ > Kuσ p( L2( E )) ≥∑
k ∈Z
<Ku f n, gn > p1/ p
= ∑n∈Z E 2n+1 \ E 3·2n−1
k 1/2
0 ( x)(K f n)( x)u2( x)α
−1/2n dx
p
1/ p
≥ c∑
n∈Z
E
2n+1 \ E 3·2n−1
k 1/20 ( x)u2( x)α
−1/2n
E r ( x)\ E 2n
k ( x, y) f n( y)dy
dx p1/ p
≥ c∑
n∈Z
E
2n+1 \ E 3·2n−1
k ( x,δ1/(2c0) x)u2( x)α−1/2n 2−nQ/2(r ( x)Q− 2nQ)dx
p1/ p
≥ c∑
n∈Z
E
2n+1 \ E 3·2n−1
u2( x)k 0( x)α−1/2n dx
p1/ p
= c ∞∑
n=0
α p/2n
1/ p
.
Let us now take
f ′n( x) = χ E 3·2n−1 \ E
3·2n−2( x)| E 3·2n−1 \ E 3·2n−2 |− 1
2 ;
g′n( x) = u( x)χ E
3·2n−1 \ E 2n ( x)k 1/20 ( x)β
−1/2n ,
where
βn =
E
3·2n−1 \ E 2n
u2( x)k 0( x)dx
and argue as above. Then we conclude that
∞ > Kuσ p
L2( E )
≥ c ∞∑
n=0
β p/2n
1/ p
.
Summarizing the estimates obtained above we have∑
n∈Z
E
2n+1 \ E 2n
u2( x)k 0( x)dx
p/21/ p
≤
∑
n∈Z
(αn +βn) p/2
1/ p
≤ c(Kuσ p( L2(0,∞)) + Kuσ p( L2(0,∞)))
≤ 2cKuσ p( L2(0,∞)).
3.4. Measure of Non–compactness
In this section we present two-sided estimates for the measure of non-compactness
KK ( L p( E ), Lqv ( E )) of the operator K.
Theorem 3.4.1. Let 1 < p ≤ q <∞ and let k ∈ V p. Assume that K is bounded from X to
Y, where X = L p
( E ) and Y = L
q
v ( E ). Then there exist positive constants b1 (depending onc1 , c3 , p and q) and b2 (depending on c2 p and q) such that the inequality
b1 J ≤ KK ( X ,Y ) ≤ b2 J
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48 Alexander Meskhi
holds, where
J = lim j→+∞
B( j) + lim j→−∞
B( 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
K− P(n,m)1 − P
(n,m)2 ≤ b2
sup j≥m
B( j) + sup j≤n
B( j),
where P(n,m)1 and P
(n,m)2 are compact operators for every n,m ∈ Z, n < m. Consequently,
KK ( X ,Y ) ≤ b2 J ,
where b2 depends only on p, q, c1 and c3.
To obtain the lower estimate
KK ( X ,Y ) ≥ b1 J ,
we take λ > KK ( X ,Y ). Then by Lemma 1.2.3 there exists P ∈ F L( X ,Y ) such that K−P < λ. On the other hand, using Lemma 1.2.6, for ε = (λ− K− P)/2, there exist
T ∈ F L( X ,Y ) and E α,β := x ∈ E : 0 < α < r ( x) < β < ∞ such that
P − T < ε (3.4.1)
and
supp T f ⊂ E α,β.
From (3.4.1) we obtain
K f − T f Y ≤ λ f X
for every f ∈ X . Thus, E α
|K f ( x)|qv( x)dx +
E \ E β
|K f ( x)|qv( x)dx ≤ λq f q
X (3.4.2)
for every f ∈ X .
Let us choose n ∈ Z such that 2n < α. Assume that j ∈ Z , j ≤ n and f j( y) = χ E 2 j+1
.
Then using condition (C ′) we find that
E 2 j+1 \ E
2 j
|K f j( x)|qv( x)dx ≥
E 2 j+1 \ E
2 j
F x
k ( x, y) f j( y)dy
q
v( x)dx
≥ c
E
2 j+1 \ E 2 j
k q( x,δ1/(2c0) x)v( x)r ( x)qdx.
On the other hand, f jq
X = c2 jQq/ p and consequently, (3.4.2) yields
cB( j) ≤ λ
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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 limn→−∞
sup j≤n B( j) ≤ cλ.
Now we take m ∈ Z such that 2m > β. Then for f j( y) = χ E 2 j+1)
( y) ( j ≥ m), we obtain
E
2 j+1 \ E 2 j
|K f j( x)|q
v( x)dx ≥ c E
2 j+1 \ E 2 j
k q
( x,δ1/(2c0) x)v( x)r ( x)q
dx.
On the other hand, f jq
X = c2Q jq/ p. Hence sup j≥m B( j) ≤ cλ, where c depends only on p,
q and c1. Consequently, limm→+∞
sup j≥m B( j) ≤ cλ from which it follows the desired estimate.
3.5. Convolution–type Operators with Radial Kernels
Let ϕ be a positive function on [0,∞) and let
K f ( x) =
E r ( x)
ϕ(r ( xy−1)) f ( y)dy, x ∈ E .
We say that ϕ belongs to U λ, 1 < λ < ∞, if
(a) there exists a positive constants c1 and c2 such that
ϕ(r ( xy−1)) ≤ c1ϕ(r ( x)), y ∈ S x,
ϕ(r ( xy−1
)) ≥ c2ϕ(r ( x)), y ∈ E r ( x);
(b) there is a positive constant c3 for which the inequality F x
ϕλ′(r ( xy−1))dy ≤ c3(r ( x))Qϕλ
′(r ( x)).
Example 3.5.1. Let ϕ(t ) = t α−Q, where Q/λ < α < Q. Then it is easy to see that
ϕ ∈ U λ. Indeed, (a) is obvious due to the properties of the quasi-norm r . Let us show (b).
We have
I :=
F x
ϕλ′(r ( xy−1))dy =
F x
(r ( xy−1))(α−Q)λ′ dy
=
∞0
| B(0,r ( x))| ∩ y ∈ E : (r ( xy−1))(α−Q)λ′ > s|ds
=
r ( x)(α−Q)λ′
0(· · · ) +
∞r ( x)(α−Q)λ′
(· · · ) := I 1 + I 2.
It is clear that I 1 ≤ (r ( x))Qϕλ′(r ( x)), while for I 2 we find that
I 2 ≤ ∞
(r ( x))(α−Q)λ′ s
Q
(α−Q)λ′ ds = c(r ( x))(α−Q)λ′+Q = c(r ( x))Qϕλ′(r ( x)).
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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 ∈ U p. Then
(i) K is bounded from L p( E ) to Lqv ( E ) if and only if
¯ B := sup j∈Z
¯ B( j) := sup j∈Z
E
2 j+1 \ E 2 j
v( x)ϕq(r ( x))dx1/q
2 jQ/ p′
< ∞;
(ii) K is compact from L p( E ) to Lqv ( E ) if and only if ¯ B < ∞ and lim j→−∞ ¯ B j =
lim j→+∞ ¯ B j = 0.
Theorem 3.5.2. Let 0 < q < p < ∞ and let p > 1. Suppose that ϕ ∈ U p. Then the
following conditions are equivalent:
(i) K is bounded from L p( E ) to Lqv ( E );
(ii) K is compact from L p( E ) to Lqv ( E );
(iii)
E
E \ E r ( x)ϕq(r ( y))v( y)dy
p/( p−q)
r ( x)Qp(q−1)/( p−q)dx
( p−q)/( pq)
< ∞.
Let ϕ(t ) = t Qϕ2(t ) and let k ( x) = ϕ(r ( x)). Suppose that
Ku f ( x) = u( x)
E r ( x)
ϕ(r ( xy−1)) f ( y)dy, x ∈ E ,
where u is a measurable function on E .
Theorem 3.5.3. Let 2 ≤ p < ∞ and let ϕ ∈ U p. Then K
u ∈ σ p( L2
( E )) if and only if u ∈ l p( L2
k ( E )). Moreover, there exist positive constants b1 and b2 such that
b1ul p( L2k
( E )) ≤ Kuσ p( L2( E )) ≤ b2ul p( L2k
( E )).
3.6. 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 Hardy-type transforms on star-shaped regions).
A full characterization of a class of weight pairs (v,w) governing the boundedness of
integral operators with positive kernels from L pw to L
qv , 1 < p < q <∞, have been established
in [75] (see also [76], Ch. 3). Criteria guaranteeing the boundedness/compactness of the
operator
R α f ( x) =
x
0( x − t )α−1 f (t )dt , x > 0,
from L p(R+) to Lqv (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 operatorswith positive kernels involving fractional integrals.
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Chapter 4
Potential and Identity Operators
This chapter is devoted to estimates of the measure of non-compactness for potential oper-
ators in weighted Lebesgue spaces defined on Euclidean spaces and homogeneous groups,partial sums of Fourier series, Poisson integrals. The same problem for the identity opera-
tor 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
I α f ( x) = G
f ( y)
r ( xy−1)Q−αdy, 0 < α < Q,
be the Riesz potential operator.
It is well known (see [70], Ch. 6) that I α is bounded from L p(G) to Lq(G), 1 < p,q <∞,
if and only if
q = Qp
Q −α p. (4.1.1)
Moreover, if (4.1.1) holds, then I α is bounded from L pρ p (G) to L
qρq (G) if and only if
sup B
1
| B|
Bρ( x)qdx
1/q 1
| B|
Bρ( 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 I α be bounded from L pw(G) to
Lqv (G). Then the following inequality holds
I αK ≥ C α,Q max A1, A2, A3,
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52 Alexander Meskhi
where
C α,Q = 1
(2c0)Q−α,
A1 = supa∈G
limr →0
r α−Q
B(a,r )
v( x)dx
1/q
B(a,r )
w1− p′
( x)dx
1/ p′
;
A2 = supa∈G
limr →0
B(a,r )
v( x)dx1/q
( B(a,r ))c
r (ay−1)(α−Q) p′w1− p
′
( y)dy1/ p′
and
A3 = supa∈G
limr →0
B(a,r )
w1− p′
( x)dx1/ p
′ ( B(a,r ))c
r (ay−1)(α−Q)qv( y)dy1/q
(c0 is the constant from the triangle inequality for the homogeneous norms ).The next statement is formulated for the Riesz potentials
J Ω,α f ( x) = Ω
f ( y)| x − y|α−ndy, x ∈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 J Ω,α is bounded
from L pw(Ω) to L
qv (Ω), then we have
J Ω,αK ≥ 2α−n B1,
where
B1 = supa∈Ω
limr →0
r α−n
B(a,r )
v( x)dx 1
q
B(a,r )w1− p
′
( x)dx 1
p′.
Further, if Ω = Rn , then
J Ω,αK ≥ 2α−n max B2, B3,
where
B2 = supa∈Rn
limr →0
B(a,r )
v( x)dx1/q
Rn\ B(a,r )
|a − y|(α−n) p′w1− p
′
( y)dy1/ p′
,
B3 = supa∈Rn
limr →0
B(a,r )
w1− p′( x)dx
1/ p′ Rn\ B(a,r )
|a − y|(α−n)qv( y)dy1/q
.
Corollary 4.1.1. Let 1 < p < ∞ , 1 < p < Qα , q = pQ
Q−α p , then there is no weight pair
(v,w) for which I α is compact from L pw(G) to Lq
v (G). Moreover, if I α is bounded from L pw(G)
to Lqv (G) , then
I αK ≥ C α,Q A1,
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Potential and Identity Operators 53
where C α,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 > I αK ( L
pw(G), L
qv (G)) there are positive constants β1 and β2, β1 < β2, such that for all τ,s
(τ < β1, s > β2),
B(a,τ)
v( x) I α f ( x)qdx ≤ λq G
f ( x) pw( x)dxq/ p, (4.1.2)
for f ∈ L pw(G), and
B(a,s)c
v( x) I α f ( x)
qdx ≤ λq
B(a,s)
f ( x) pw( x)dx
q/ p
, (4.1.3)
for f with supp f ⊂ B(a,s).
Now assuming f ( x) = χ B(a,r )( x)w1− p
′
( x) in (4.1.2) and observing that B(a,r )
w1− p′
( x)dx < ∞
for all r > 0 (see also [76], Ch. 3 for this fact), we find that B(a,r )
v( x)
B(a,r )
w1− p′( y)
r ( xy−1)Q−αdyq
dx ≤ λq
B(a,r )
w1− p′
( x)dxq/ p
< ∞.
Further, if x, y ∈ B(a,τ), then
r ( xy−1) ≤ c0
r ( xa−1) + r (ay−1)
≤ 2c0τ.
Hence
I αK ≥ C α,Q A1.
If f ( x) = χ B(a,τ)c ( x) w1− p′( x)
r (ay−1)(Q−α)( p′−1), then
B(a,τ)
v( x) B(a,τ)c
w1− p′
( y)dy
r ( xy−1
)Q−α
r (ay−1
)(Q−α)( p
′−1)
q
dx
≤ λq
B(a,τ)c
w1− p′
( x)dx
r (ay−1)(Q−α) p′
q/ p
< ∞.
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)
≤ 2c0r (ay−1).
Hence, by (4.1.2) we have
1
(2c0)q(Q−α)
B(a,τ)
v( x)dx
B(a,τ)c
w1− p′
( y)dy
r (ay−1)(Q−α) p′
q
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54 Alexander Meskhi
≤ λq
B(a,τ)c
w1− p′
( x)dx
r (ay−1)(Q−α) p′
q/ p
.
The latter inequality implies
I αK ≥ 1
(2c0)Q−α A2.
Further, observe that (4.1.3) means that the norm of the operator
¯ I α f ( x) =
B(a,s)
f ( y)dy
r ( y−1a)Q−α
can be estimated as follows:
¯ I α L pw( B(a,s))→ L
qv ( B(a,s)c) ≤ λ.
By the duality arguments we find that
¯ I α L pw( B(a,s))→ L
qv ( B(a,s)c) = ˜ I α
Lq
′
v1−q′ ( B(a,s)c)→ L
p′
w1− p′ ( B(a,s))
,
where
˜ I αg( y) =
B(a,s)c
g( x)dx
r ( xy−1)Q−α.
Indeed, by Fubini’s theorem and Holder’s inequality we have
¯ I α f Lqv ( B(a,s)c) ≤ sup
g L
q′
v ( B(a,s)c)≤1 B(a,s)c g( x)( ¯ I α f ( x))dx
≤ supg
Lq
′
v1−q′ ( B(a,s)c)
≤1
B(a,s)
| f ( y)| ˜ I α(|g|)( y) dy
≤ supg
Lq
′
v1−q′ ( B(a,s)c)
≤1
B(a,s)
| f ( y)| p w( y)dy
1 p
B(a,s)
˜ I α(|g|)
p′
( y)w1− p′
( y)dy
1 p′
≤ ˜ I α B(a,s)
| f ( y)| p w( y)dy 1 p
.
Hence ¯ I α ≤ ˜ I α. Analogously, ˜ I α ≤ ¯ I α.Further, (4.1.3) implies
B(a,s)
w1− p′
( x) ( B(a,s))c
g( y)dy
r ( xy−1)Q−α
p′
dx ≤ λ p′
( B(a,s)c)
|g( x)|q′
v1−q′
( x)dx p
′/q
′
.
Now taking g( x) = χ B(a,s)c ( x)r ( xa−1)(Q−α)(1−q)v( x) in the last inequality we conclude
that I αK ≥ 1
(2c0)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.
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Potential and Identity Operators 55
4.2. Truncated Potentials
This subsection is devoted to two-sided estimates of the essential norm for the operator
T α f ( x) = B(e,2r ( x))
f ( y)
r ( xy−1
)Q−α
, x ∈ G.
A necessary and sufficient condition guaranteeing the trace inequality for T α 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 < p ≤ q < ∞ and let α > Q/ p. Then
(i) T α is bounded from L p(G) to Lqv (G) if and only if
B := supt >0
B(t ) := supt >0
r ( x)>t
v( x)r ( x)(α−Q)qdx1/qt Q/ p′ < ∞; (4.2.1)
(i) T α is compact from L p(G) to Lqv (G) if and only if
limt →0
B(t ) = limt →∞
B(t ) = 0.
Theorem 4.2.1. Let 1 < p ≤ q <∞ and let 0 < α< Q. Suppose that T α is bounded from
L pw(G) to L
qv (G). Then the inequality
T αK ( L pw(G)→ L
qv (G)) ≥ C Q,α
lima→0
A(a) + limb→∞
A(b)
holds, where
C Q,α = (2c0)α−Q;
A(a) = sup0<t <a
B(e,a)\ B(e,t )
v( x)r ( x)(α−Q)qdx1/q
B(e,t )
w1− p′( x)dx
1/ p′
;
A(b) = supt >b
B(e,t )c
v( x)r ( x)(α−Q)qdx1/q B(e,t )\ B(e,b)
w1− p
′
( x)dx1/ 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
boundedness of T α from L pw(G) to L
qv (G) it follows that w1− p′
is locally integrable on G.
Proof. Let
I (t ) = B(e,t )
w1− p′( x)dx = ∞
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56 Alexander Meskhi
for some t > 0. Then there exists g ∈ L p B(e,t )
such that
B(e,t )
gw−1/ p =∞. Let us assume
that f t ( y) = g( y)w−1/ p( y)χ B(e,t )( y). Then we have
T α f t Lqv (G) ≥ χ B(e,t )c T α f t L
qv (G)
≥ c B(e,t )c
v( x)r ( x)(α−Q)qdx1/q
B(e,t )
g( y)w−1/ p′( y)dy = ∞.
On the other hand,
f t L pw(G) =
B(e,t )
g p( x)dx < ∞.
Finally we conclude that I (t ) < ∞ for all t , t > 0.
Proof of Theorem 4.2.1. Let λ > T αK ( L p
w(G), Lq
v (G)). Then by Lemma 1.2.7 thereexists 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
T α f Lqv ( B(e,τ2)\ B(e,τ1)) ≤ λ f L
pw( B(e,τ1)).
holds. Observe that if r ( x) > τ1 and r ( y) < τ1, then r ( xy−1) ≤ 2c0r ( x). Consequently, taking
f = w1− p′χ B(e,τ1) and using Lemma 4.2.1 we find that
1
(2c0)Q−α B(e,τ2)\ B(e.τ1)
v( x)r ( x)(α−Q)q
dx1q
B(e,τ1)
w1− p′( x)dx
1 p′
≤ λ
for all τ1,τ2,0 < τ1 < τ2 < β1. Hence
1
(2c0)(Q−α)q lim
a→0 A(a) ≤ λ.
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
T α f Lqv ( B(e,s2)c) ≤ λ f L pw( B(e,s2)\ B(e,s1))
holds, where supp f ⊂ B(e,s2)\ B(e,s1). Hence taking f = w1− p′χ B(e,s2)\ B(e,s1) in the previ-
ous inequality and using Lemma 4.2.1 we find that
1
(2c0)Q−α
B(e,s2)c
v( x)
r ( x)(α−Q)q
dx 1
q
B(e,s2)\ B(e,s1)
w1− p′( x)dx
1 p′
≤ λ
which leads us to the estimate
1(2c0)Q−α
limb→∞
A(b) ≤ λ.
Thus we have the desired result.
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Potential and Identity Operators 57
Theorem 4.2.2. Let 1 < p ≤ q < ∞ and let Q p < α < Q. Suppose that (4.2.1) holds.
Then there is a positive constant C such that
T αK ( L p(G)→ Lqv (G)) ≤ C
lima→0
B(a) + limb→∞
B(b)
,
where
B(a) = supt ≤a
B(e,a)\ B(e,r )
v( x)r ( x)(α−Q)qdx1/q
r Q/ p′;
B(b) = supt ≥b
B(e,t )c
v( x)r ( x)(α−Q)qdx1/q
r Q − bQ1/ p′
.
Proof. Let 0 < a < b < ∞ and represent T α f as follows:
T α f = χ B(e,a)T α( f χ B(e,a)) +χ B(e,b)\ B(e,a)T α( f χ B(e,b))
+χG\ B((e,b)T α( f χ B(e,b/2c0)) +χG\ B(e,b)T α( 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
P2 f ( x) = G
k ( x, y) f ( y)dy,
where k ( x, y) = χ B(e,b)\ B(e,a)( x)χ B(e,2r ( x))( y)r ( xy−1)α−Q.
Further observe that G
G
(k ( x, y)) p′dy q
p′
v( x)dx
=
B(e,b)\ B(e,a)
B(e,2r ( x))
(r ( xy−1))(α−Q) p′dy q
p′
v( x)dx
≤ c B(e,b)\ B(e,a)
B(e,r ( x)/2c0)
(r ( xy−1))(α−Q) p′dy
q
p′
v( x)dx.
≤ c
B(e,b)\ B(e,a)
r ( x)(α−Q)q+q/ p′v( x)dx < ∞.
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) ≤ 2c0r ( 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
P1 ≤ C 1 B(a); P4 ≤ C 2 B(b/2c0),
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58 Alexander Meskhi
where the constants C 1 and C 2 depend only on p, q, Q and α.Therefore
T α− P2 − P3 ≤ P1 + P4 ≤ 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
lima→0
B(a) + limb→∞
B(b)
≤ T αK ( L p(G), L
qv (G)) ≤ c1
lima→0
B(a) + limb→∞
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 PotentialsLet
Rα f ( x) =
x 0
f (t )
( x − t )1−αdt , W α f ( x) =
∞ x
f (t )
(t − x)1−αdt ,
where x ∈ R+ and α is a constant satisfying the condition 0 < α < 1.
Theorem 4.3.1. Let 1 < p ≤ q < ∞. Suppose that Rα is bounded from L pw(R+) to
L pv (R+). Then
RαK ≥ 2α−1 supa∈R+
limτ→0τα−1 a+τ
a
v( x)dx 1
q a a−τ
w1− p′( x)dx 1
p′
.
Proof. Let λ > RαK ( L pw(R+), L p
v (R+)) and a ∈R+. By Lemma 1.2.8 we have that
a+r a
v( x)( Rα f ( x))qdx ≤ λq a
a−r
f ( x)
pw( x)dx
q p
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
a+r a
v( x)
x 0
w1− p′
( x − t )1−αdt
q
dx ≤ λq a
a−r
w1− p′( x)dx
q p
For x ∈ (a,a + r ) and t ∈ (a − r ,r ), we have that x − t < 2r . Hence
(2r )(α−1)q
a+r
a
v( x)dx
a
a−r
w1− p′(t )dt
q
≤ λq
a
a−r
w1− p′( x)dx
q p
.
Taking into account the boundedness of Rα from L pw(R+) to L
qv (R+) we have w1− p′
∈ Lloc(R+) (see, e.g., [3]). Consequently,
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Potential and Identity Operators 59
λ≥ (2r )α−1
a+r a
v( x)dx
1q a
a−r
w1− p′( x)dx
1 p′
.
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 W α is bounded from L pw(R+) to
L pv (R+). Then
W αK ≥ 2α−1 supa∈R+
limτ→0τα−1
a a−τ
v( x)dx
1q a+τ
a
w1− p′( x)dx
1 p′
.
Theorem 4.3.3. Let 1 < p ≤ q < ∞. Suppose that W α is bounded from L pw(R+) to L
pv (R+). Then
W αK ≥ supa∈R+
limr →0
a+r a
v( x)dx 1
q ∞
a+r
w1− p′( x)
( x − a) p′(1−α)
1 p′.
Proof. Let
λ > W αK ( L pw(R+), L
qv (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 inequalitya+r a
v( x)dx
∞ a+r
f (t )
(t − a)1−αdt
q
≤ λq ∞
a+r
( f ( x)) pw( x)dx
q p
holds, where f ≥ 0, f ∈ L pw(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
λ≥ a+r a
v( x)dx 1
q ∞ a+r
w1− p′ ( x)( x − a) p′(1−α)
dx 1
p′
.
Taking the supremum over all a ∈ R+ and passing to the limit when r → 0 in the right-
hand 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 L pw(R+) to
L pv (R+). Then
RαK ≥ supa∈R+
limτ→0
a a−r
v( x)dx 1
q a−r −∞
w1− p′( x)
(a − x) p′(1−α)
1 p′
.
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60 Alexander Meskhi
4.4. Poisson Integrals
Here we discuss the essential norm of the Poisson integral
P f ( x,t ) =
Rn
f ( y)P( x − y,t )dy, x ∈Rn, t > 0,
where,
P( x,t ) = t (t 2 + | x|2)− n+12 .
In this section we use the following notation:
T B(a,r )
:= ( x, t ) ∈ R
n+1+ ; x ∈ B(a,r ),t < r ; Bn := | B(0,1)| =
2πn2
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) = χ B( y) , then
P f ( x,t ) ≥ Bn
5n+1
2
.
Proof. Let B = B(a,r ). Then we have,
P f ( x, t ) ≥ t
B(a,t )
dy
(t 2 + | x − y|2)n+1
2
≥ t
B(a,t )
dy
(t 2 + |2t |2)n+1
2
≥ Bnt n+1
5n+1
2 t n+1= Bn5− n+1
2 .
In the following statements we keep the notation of Section 1.2.
Lemma 4.4.2. Let w be a weight function on Rn+1+ and let S ∈ F L( L
pw(Rn), Lq(Rn+1
+ )) ,where 1 ≤ p,q <∞. Then for every a ∈Rn and ε> 0, there exist R ∈ F L( L
pw(Rn), Lq(Rn+1
+ ))and positive numbers α,α, 0 < α < α < ∞ , such that for all f ∈ L
pw(Rn) the inequality
(S − R) f Lq(Rn+1+ ) ≤ ε f L pw(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
S f ( x, t ) = N
∑ j=1
β j( f )U j( x,t ),
where β j are bounded linear functionals on L pw(Rn). Further there is a positive constant C
such that N
∑ j=1
|β j( f )| ≤ C f L pw(Rn).
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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
U j −Φ j Lq(Rn+1+ ) < ε/C , j = 1,2, . . . , N ,
and suppΦ j ⊂ T ( B(a,α))\T ( B(a,α)). Let
R f ( x, t ) = N
∑ j=1
β j( f )Φ j( x, t ).
Then we have
S f − R f Lq(Rn+1+ ) ≤
N
∑ j=1
|β j( f )|U j −Φ j Lq(Rn+1+ ) ≤ ε f L
qw(Rn)
for all f ∈ L pw(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 L pw(Rn) to
Lq(Rn+1+ ) , where 1 < p,q < ∞. Then T K = α(T ).
Theorem 4.4.1. Let 1 < p ≤ q < ∞ and let the operator P be bounded from L pw(Rn) to
Lq
v (R
n+1
+ ). Then the following inequality holds:
PK ( L
pw(Rn+1
+ ), Lqv (Rn+1
+ )) ≥ max D1, D2, D3,
where
D1 = Bn
5n+1
2
supa∈Rn
limr →0
T
B(a,r )
v( x, t )dxdt 1
q
B(a,r )
w( x)dx− 1
p
;
D2 = 5−(n+1)/2 supa∈Rn
limr →0
r −n−1
T ( B(a,r ))
t qv( x, t )dxdt
1q
B(a,r )
w( x)1− p′dx
1
p′
;
D3 = supa∈Rn
limr →0
T ( B(a,r ))
v( x, t )dxdt 1
q
( B(a,r ))c
w1− p′( y)dy
(r 2 + 4| y − a|2)
n+1
2
p′
1 p′
.
Proof. Denote Pv f ( x, t ) = v1 p ( x,t )P f ( x, t ). Then
PvK ( L
pw(Rn), Lq(Rn+1
+ )) = PK ( L pw(Rn), Lq
v (Rn+1)).
Let λ > PK ( L
pw(Rn), Lq
v (Rn+1
+ )). Then we see that λ > Pv
K ( L pw(Rn), Lq(Rn+1
+ )). Hence there
exists S ∈ F L( L pw(Rn), Lq(Rn+1
+ )) for which
Pv − S < λ.
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62 Alexander Meskhi
Let a ∈ Rn. Then by Lemma 4.4.2 there exist positive numbers α and α and an operator
R ∈ F L( L pw(Rn), Lq(Rn+1
+ ))) such that supp R f ⊂ T ( B(a,α))\T ( B(a,α)) and
S − R ≤ λ− Pv − S
2 .
Hence,Pv − R < λ.
Therefore,
(Pv − R) f Lq(Rn+1+ ) ≤ λ f L
pw(Rn)
for all f ∈ L pw(Rn). Now the latter inequality implies that if r < α, then
T ( B(a,r ))
v( x, t )(Pv f )q( x,t )dxdt ≤ λq Rn
( f ( x)) pw( x)dx q
p
(4.4.1)
for all f ≥ 0, f ∈ L pw(Rn). If we take f ( x) = χ B(a,r )( x) in (4.4.1) and use Lemma 4.4.1 we
find that Bn
5n+1
2
q
T ( B(a,r ))
v( x,t )dxdt ≤ λq
B(a,r )
w( x)dx q
p
which gives the estimate PK ≥ D1.
Assuming that f ( x) = χ B(a,r )( x)w1− p′( x) in (4.4.1) we have
T ( B(a,r ))
t
q
v( x,t ) B(a,r )
w1− p′( y)dy
(t 2 + | x − y|2) n+12 q
dxdt
≤ λq
B(a,r )
w1− p′( x)dx
q p
< ∞.
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 ≤ r 2 + (2r )2 = 5r 2,
which implies
5−(n+1)/2
r n+1
T ( B(a,r ))
t qv( x, t )dxdt 1
q
B(a,t )
w1− p′( x)dx
1 p′
≤ λ
for all r < α.The latter inequality yields the inequality PK ≥ D2.
To get the estimate PK ≥ D3, we observe that if ( x,t ) ∈ T ( B(a,r )) and | y − a| > r ,
then
t 2 + | x − y|2 ≤ r 2 + 4|a − y|2.
Taking f ( x) = χ x:| x−a|>r ( x)w1− p′( x)r 2 + 4|a − y|2 n+1
2 (1− p′)in (4.4.1), we find that
PvK ≥ D3.
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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
PK ≥ D1 of Theorem 4.3.1 can be replaced by
PK ≥ 1
5n+1
2 Bsup
a∈Rn
limr →0
r −n
T ( B(a,r ))
v( x,t )dxdt
1q
B(a,r )
w1− p′( x)dx
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 theweighted Sobolev space W
m, pw (Ω) as the set of functions u ∈ L
pw(Ω) with weak derivatives
Dαu ∈ L pw(Ω) for |α| ≤ m. Then norm of u in W
m, pw (Ω) is given by
uW m, p
w (Ω) = ∑|α|≤m
Ω
| Dαu( x)| pw( x)dx 1
p
.
It is also defined the space
W
m, pw as the closure of C ∞0 (Ω) in W
m, pw (Ω). Together with
W m, p
w (Ω) we consider the space V m, p
w (Ω) with the norm
uV m, p
w (Ω) = ∑|α|=m
Ω
| Dαu( x)| pw( x)dx 1
p.
For weighted Sobolev inequalities we refer, e.g., to [2], [158],[230], [241].
It is well-known (see, e.g., [230], p. 16) that if w ∈ A p(Rn) then W m, p
w (Ω), W m, p
w (Ω),
V m, p
w (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 W
m, pw (Ω)
is embedded in Lqv (Ω) , i.e., I : W
m, pw (Ω) → L
qv (Ω) is bounded, then
I K ≥ supa∈Ω
limr →0
S m, p(r ,ψ )
−1
B(a,r )
v( x)dx 1
q
B(a,r )
w( x)dx 1
p
,
where S m, p(r ,ψ ) = ∑|α|≤m
r |α| p sup1≤| x|≤2
Dαψ ( x) p 1
p , ψ is a function from C ∞0 (Rn) whose
support is in B(0,2) and equal to 1 in B(0,1).
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64 Alexander Meskhi
Proof. Let I vu = vu, then
I vK (W m, p
w (Ω), Lq(Ω)) = I K (W m, p
w (Ω), Lqv (Ω)).
Let λ > α( I v), then there is P ∈ F L(W m, p
w (Ω), Lq(Ω)) such that I v − P < λ. For a ∈ Ω
there exist a positive number α and an operator R ∈ F L(W m, p
w (Ω), Lq(Ω)) such that
P − R ≤ λ− I v − P
2
and supp Ru ⊂Ω\ B(a,α). Hence I v − R ≤ λ. Consequently,
( I v − R)u Lq(Ω) ≤ λuW m, p
w (Ω).
If r < α, then the latter inequality implies
B(a,r )
v( x)|u( x)|qdx 1
q
≤ λuW m, p
w (Ω).
Let us now take ψ ∈ C ∞0 (Rn) which is equal 1 in B(0,1) , supp ψ ⊂ B(0,2) and set
φ = ψ ( x−ar
). Then taking u = ψ in the last inequality we find that B(a,r )
v( x)dx 1
q
≤ λ ∑|α|≤m
Ω
Dαψ ( x)
p
w( x)dx 1
p
≤ λS m, p(r ,ψ ) B(a,r )
w( x)dx 1 p
.
Hence,
I K ≥ 1
S m, p(r ,ψ )
B(a,r )
v( x)dx 1
q
B(a,r )
w( x)dx1 p
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 V m, p
w (Ω) is
embedded in Lqv (Ω) , i.e., I : V
m, pw (Ω) → L
qv (Ω) is bounded, then
I K (V m, pw (Ω), Lqw(Ω)) ≥ supa∈Rn limr →0S m, p(ψ )−1 B(a,r )
v( x)dx
1q
B(a,r )
w( x)dx 1
p ,
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Potential and Identity Operators 65
where S m, p(ψ ) = 2m/ p ∑
|α|=m
sup1≤| x|≤2
Dαψ ( x) p 1
p
and ψ is a function from C ∞0 (Rn) with
supp in B(0,2) and value 1 on B(0,1).
Corollary 4.5.1. Let 1 ≤ p < ∞, 0 ≤ m < n p. Suppose that q = np
n−mp, w( x) = | x|β and
v( x) = | x|
βq
p , where β > −npq . Then
I K (V m, p
w (Ω), Lqw(Ω)) ≥
S m, p(ψ )
−1S
1q
− 1 p
n−1
βq p
+ n− 1
p
(β+ n)1q
where S m, p(ψ ) is defined as in the previous statement and S n−1 = 2πn/2
Γ (n/2) .
Example 4.5.2. Let 1 ≤ p ≤ q < ∞ and let 0 ≤ m < n p
. Suppose that V 1, p(Ω) is contin-
uously embedded in Lqv (Ω). Then
I K ≥ supa∈Ω
limr →0
C n, p r m− n
p B(a,r )
v( x)dx 1q,
where C n, p = S − 1
p
n−16e
1+ 31/2
1−31/2
(1−31/2)2 .
This follows from Theorem 4.5.2 taking w ≡ 1 and
ψ ( x) =
1 for | x| < 1,
e1+ 1
(| x|−1)2−1 for 1 ≤ | x| ≤ 2,
0 for | x| > 2.
4.6. Identity Operator
This section is devoted to lower estimates of the measure of non–compactness for the iden-
tity operator I acting from L pw(Ω) to L
qv (Ω), where Ω = [0,π] and q < p.
To prove the main statement we need some lemmas.
Lemma 4.6.1. Let f n( x) = sin 2n x, 0 ≤ x ≤ π. Assume that v ∈ C 1([0,π[) and v,v′ ∈
L1
([0,π[). Thenlim
n,m→∞n=m
π0
v( x)| f n( x) − f m( x)|dx ≥ 1
2
π0
v( x)dx. (4.6.1)
Proof. Let us denote I n,m := π
0 v( x)| f n( x) − f m( x)|dx. We have
I n,m ≥ 1
2
π0
v( x)( f n( x) − f m( x))2dx
= 1
2 π
0
v( x) f 2
n
( x)dx − π
0
v( x) f n( x) f m( x)dx + 1
2 π
0
v( x) f 2
m
( x)dx
:= 1
2 I 1 − I 2 +
1
2 I 3.
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66 Alexander Meskhi
For I 1 we have
I 1 = 1
2
π0
v( x)(1 − cos2n+1 x)dx
= 1
2 π
0v( x)dx −
1
2 π
0v( x) cos2n+1 xdx
= 1
2
π0
v( x)dx − 1
2n+2
π0
v′( x) sin2n+1 xdx,
while for I 2 we find that
I 2 = 1
2
π0
v( x)[cos(2n − 2m) x − cos(2n + 2m) x]dx
= 1
2
π0
v( x) cos(2n − 2m) xdx −1
2
π0
v( x) cos(2n + 2m) xdx
:= 12
I 21 − 12
I 22.
It is easy to see that
I 21 = − 1
2n − 2m
π0
v′( x) sin(2n − 2m) xdx;
I 22 = − 1
2n + 2m
π0
v′( x) sin(2n − 2m) xdx.
Hence
I 2 = − 12(2n − 2m)
π0
v′( x) sin(2n − 2m) xdx
+ 1
2(2n + 2m)
π0
v′( x) sin(2n + 2m) xdx
and
I 3 = 1
2
π0
v( x)dx − 1
2m+2
π0
v′( x) sin2n xdx.
Finally, passing n and m to infinity we conclude that
limn,m→∞
n=m
I n,m ≥ 14 π
0v( x)dx + 1
4 π
0v( x)dx = 1
2 π
0v( x)dx.
Note that in the unweighted case the statement that f n contains no subsequence con-
vergent in L1 is made in [143], p. 90.
Lemma 4.6.2. Let f n( x) = sin 2n x, 0 ≤ x ≤ π. Suppose that v is a weight on [0,π]. Then
the inequality (4.6.1) holds.
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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 ∈ L1([0,π]). Then there exists a sequence vk , where 0 ≤ vk ( x) ≤ v( x), vk ∈C ∞(0,π) and
π0 v′
k ( x)dx < ∞ such that
limk →∞ π
0
vk ( x)dx = π
0
v( x)dx,
Let us take k so large that
1
2
π0
vk ( x)dx > 1
2
π0
v( x)dx − ε.
By Lemma 4.6.1 we can choose n and m so that π0
vk ( x)| f n( x) − f m( x)|dx ≥ 1
2
π0
vk ( x)dx −ε > 1
2
π0
v( x)dx − 2ε.
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
(v,w) for which I is compactly embedded from L pw(Ω) to L
qv (Ω). Moreover, if (1.1.1) holds
for some weights v and w on Ω , then
I K ( L pw(Ω), Lq
v (Ω)) ≥ 1
4
π0
v( x)dx
1/q π0
w( x)dx
1/ p
. (4.6.2)
Proof. By Lemma 4.6.2 there exists a sequence f n ⊂ B L∞ ( B L∞ is the closed unit ball
in L∞) such that
( I : L∞→ L1v )( f n) − ( I : L∞→ L1
v )( f m) L1v
= f n − f m L1v> λ−ε,
where λ := 12
π0 v( x)dx and ε is a small positive number. Hence ( I : L∞→ L1
v )( B L∞) 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∞→ L1v )) ≥ λ/2 − ε/2
for any n ∈ N. Consequently, using the inequality I K ( L∞, L1v ) ≥ β( I ) (see Section 1.2 for
some properties of entropy numbers of bounded linear operators) we find that
I K ( L∞, L1v ) ≥ λ/2 −ε/2.
Further, by Propositions 1.1.3 and 1.1.4 we have
I : L∞w → L pw =
π0
w( x)dx
1/ p
;
I : L pw → Lq
v = π
0 v( x)
w( x) p
p−q
w( x)dx1/q−1/ p
;
I : Lqv → L1
v =
π0
v( x)dx
1/q′
.
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68 Alexander Meskhi
Hence
( I : L∞→ L1v ) = ( I : Lq
v → L1v ) ( I : L p
w → Lqv ) ( I : L∞→ L p
w).
From this it follows that
λ−ε
2 ≤ I K ( L∞, L1
v ) ≤ I Lqv → L1 I L∞→ L
pw
I K ( L pw, L
qv ).
Therefore
I K ( L pw, L
qv ) ≥ λ−ε
2 I Lqv → L1
v I L∞→ L
pw
= 1
2(λ− ε)
π0
v( x)dx
−1/q′ π0
w( x)dx
−1/ p
.
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
S n f ( x) = 1
π
π −π
f (t ) Dn( x − t )dt , n ∈ N,
of the Fourier series of f
f ∼ 1
2a0 +∞
∑k =1
(ak cos kx + bk sin kx),
where Dn = 12 +
n
∑k =1
cos kt .
For basic properties of S n 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 S n is compact from L pw(T ) to L p
v (T ). Moreover, if S n is bounded from
L pw(T ) to L
pv (T ) , then
S n ≥ (2 + 21/2)1/2
2π sup
a∈T
limr →0
1
2r
a+r
a−r v( x)dx
1 p 1
2r
a+r
a−r w1− p′
( x)dx 1
p′. (4.7.1)
Proof. Taking λ > S nκ ( L pw(T ), L p
v (T )), by Lemma 1.2.8 we find that
I
v( x)|S n f ( x)| pdx ≤ λ p
I
| f ( x)| pw( x)dx (4.7.2)
for the intervals I := (a − r ,a + r ), where r is a small positive number and supp f ⊂ I .
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Potential and Identity Operators 69
Let
J 1 =
I
|S n f ( x)| pv( x)dx and J 2 =
I
| f ( x)| pw( x)d ( x).
Suppose that | I | ≤ π4 and n is the greatest integer less than or equal to π
4| I | . Then for x ∈ I
(see [95]),|S n f ( x)| ≥
1
π
I
| f (θ)| sin 3π8
π4n
d θ. (4.7.3)
Using this estimate and taking f := w1− p′( x)χ I ( x) we find that
J 1 ≥1
π sin
3π
8
p
| I |− p
I
v( x)dx
I
w1− p′( x)dx
p
.
On the other hand, due to (4.7.3) it is easy to see that J 2 = I
w1− p′( x)dx < ∞.
Hence, by (4.7.2) we conclude that
λ≥ 1
π sin
3π
8
1
| I |
I
v( x)dx 1
p 1
| I |
I
w1− p′( x)dx
1 p′.
Now passing r to 0, taking the supremum over all a ∈ T and using the fact that sin 3π8
=(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
S nκ ( 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
S nκ ( L pw(T )) ≥
(2 + 21/2)1/2
2π 1
α+ 11 p
1
α(1 − p′) + 11
p′.
4.8. 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].
Criteria for the trace inequality ( L p → Lqv 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 es-
tablished by M. Gabidzashvili and V. Kokilashvili are more transparent than those of E.
Sawyer.
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70 Alexander Meskhi
Necessary and sufficient conditions guaranteeing the compactness of the Reisz poten-
tials 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 quasi-
metric 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]). Fortwo–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
one–sided potentials from L pw to L
qv (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 theFourier series T f ( x) = supn |S n 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.
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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) =
x
0 f (t )( x − t )α( x)−1dt , x > 0,
in the L p( x) spaces, where 0 < inf α ≤ supα < 1. In particular, necessary and sufficient
conditions on a weight v guaranteeing the boundedness/ compactness of Rα( x) from L p( x)
to Lq(·)
v 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:
E k := [2k ,2k +1); I k := [2k −1,2k +1), k ∈ Z.
5.1. Boundedness
In this section we establish necessary and sufficient conditions for the boundedness of the
operator Rα( x) from L p( x) to Lq( x)v .
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
vRα( x) f Lq( x)( I ) ≤ c f L p( x)( I ), f ∈ L p(·)( I ) (5.1.1)
holds if and only if
Aa := sup0<t <a
Aa(t ) := sup0<t <a
χ(t ,a)( x) v( x) x1−α( x)
Lq( x)( I )
t 1/ p′(0) < ∞.
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72 Alexander Meskhi
Moreover, there exist positive constants c1 and c2 such that
c1 Aa ≤ Rα( x) L p( x)( I )→ L
q( x)v ( I )
≤ c2 Aa.
Proof. For simplicity assume that a = 1.
Sufficiency. Suppose that f ≥ 0. We represent Rα( x) f as follows:
( Rα( x) f )( x) =
x/2
0 f (t )( x − t )α( x)−1dt +
x
x/2 f (t )( x − t )α( x)−1dt
:= ( R(1)α( x) f )( x) + ( R
(2)α( x) f )( x).
Hence
vRα( x) f Lq( x)( I ) ≤ vR(1)α( x) f Lq( x)( I ) + vR
(2)α( x) f Lq( 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 Theorem1.4.5 we have
S (1) ≤ c
v( x)
x1−α( x) H f ( x)
Lq( x)( I )
≤ cAa f L p(·)( I ).
Suppose now that g Lq′( x)( I ) ≤ 1. Using Lemmas 1.4.8 and 1.4.9 we find that
1
0v( x)
x
x/2 f (t )( x − t )α( x)−1dt
g( x)dx
≤ c
∑k ∈ Z − E k −1
v( x)χ( x/2, x)
(·) f (·) L
p(·)
( I )
× χ( x/2, x)(·)( x − ·)α( x)−1 L p′(·)( I )g( x)dx
≤ c ∑k ∈ Z −
χ I k −1(·) f (·) L p(·)( I )
E k −1
v( x) xα( x)−1/ p( x)g( x)dx
≤ c ∑k ∈ Z −
χ I k −1(·) f (·) L p(·)( I )
χ E k −1( x)v( x) xα( x)−1/ p( x)
Lq( x)( I )
× χ E k −1(·)g(·)
Lq′(·)( I )
≤ c2
k / p′(0)
∑k ∈ Z −v( x) xα( x)−1
χ E k −1 ( x) Lq( x)( I )χ I k −1 (·) f (·) L p(·)( I )
× χ E k −1(·)g(·)
Lq′(·)( I ) ≤ cAa f L p(·)( I )g Lq′(·)( I ) ≤ cAa f L p(·)( I ).
Taking the supremum with respect to g and applying Lemma 1.4.7, we have the desired
result.
Necessity. Let us take f k ( 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
f k L p(·)( I ) ≤ c2k /( p′)+([0,2k +2]) ≤ c2k / p′(0).
On the other hand,
Rα( x) f L
q( x)
vq( x)( I )
≥ cχ E k −1( x)v( x) xα( x)−1 Lq( x)( I ).
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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 := supk ∈ Z −
¯ Ak := supk ∈ Z −
χ E k −1( x)v( x) xα( x)−1 Lq( x)( I )2k / p′(0)
≤ c Rα( x) L p( x)( I )→ L
q( x)v ( I )
.
Let us now take t ∈ I . Then t ∈ [2m−1,2m) for some m ∈ Z −. Consequently,
A1(t ) ≤0
∑k =m
χ E k −1( x)v( x) xα( x)−1 Lq( x)( I )2m/ p′(0) ≤ ¯ A2m/ p′(0)
0
∑k =m
2−k / p′(0) ≤ c ¯ A
≤ c Rα( x) L p( x)( I )→ L
q( x)v ( I )
.
Hence A1 ≤ c Rα( x) L p( x)( I )→ L
q( x)v ( I )
.
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
A∞ := supt >0
A∞(t ) := supt >0
χ(t ,∞)( x) v( x)
x1−α( x)
Lq( x)( I )
t 1/P′(t ) < ∞,
where
P(t ) = p(0), 0 ≤ t ≤ a,
pc, t > a. Moreover, there are positive constants c1 and c2 such that
c1 A∞ ≤ Rα( x) L p( x)( I )→ L
q( x)v ( I )
≤ c2 A∞.
Proof. For simplicity we assume that a = 1. First we prove sufficiency. Suppose that
f ≥ 0. We have
vRα( x) f Lq( x)( I ) ≤ vRα( x) f Lq( x)([0,2]) + vRα( x) f Lq( x)((2,∞)) := I 1 + I 2.
Taking into account Theorem 5.1.1 we find that the condition A∞ < ∞ implies
I 1 ≤ cA∞ f L p( x)([0,2]) ≤ cA∞ f L p( x)( I ).
For I 2, we have
I 2 ≤
v( x)
1
0( x − t )α( x)−1 f (t )dt
Lq( x)((2,∞))
+
v( x) x/2
1( x − t )α( x)−1 f (t )dt
Lq( x)((2,∞))
+v( x) x
x/2( x − t )α( x)−1 f (t )dt
Lq( x)((2,∞))
:= I 2,1 + I 2,2 + I 2,3.
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74 Alexander Meskhi
Notice that when t ≤ 1 and x ≥ 2, then ( x − t )α( x)−1 ≤ cxα( x)−1. Consequently, using
Holder’s inequality (see Lemma 1.4.1) we find that
I 2,1 ≤ c
v( x) xα( x)−1
Lq( x)((2,∞)) f χ[0,1] L p(·)( I )χ[0,1] L p(·)( I )
≤ cv( x) xα( x)−1 Lq( x)([1,∞))
f L p(·)( I ) ≤ cA∞ f L p(·)( I ).
It is easy to see that the estimate ( x − t )α( x)−1 ≤ cxα( x)−1 and Theorem 1.3.4 implies
I 2,2 ≤ c
v( x) xα( x)−1
x
1 f (t )dt
Lq( x)([1,∞))
≤ cA∞ f Lq( x)([1,∞)) ≤ cA∞ f Lq( x)( I ),
while Holder’s inequality for the classical Lebesgues spaces yields
( I 2,3)qc ≤ c +∞
∑k =1
E k
v( x)qc x(α( x)−1)qc dx
I k
f pc (t )dt qc/ pc
2k /( pc)′
≤ cAqc∞ f qc
L p(·)( I ).
Necessity follows in the same way as in the case of Theorem 5.1.1. In this case we take the
test functions f t ( 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
p0( x) are defined
as follows:
p0( x) := inf y∈[0, x]
p( y); p0( x) := p0( x), 0 ≤ x ≤ a
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
Ba := sup
0<t <a
Ba(t ) := sup
0<t <a a
t
(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
S q,v( Rα( x)) ≤ 2q−( I )−1
1
0
v( x)
x/2
0( x − y)α( x)−1 f ( y)dy
q( x)
dx
+ 1
0
v( x) x
x/2( x − y)α( x)−1 f ( y)dyq( x)
dx:= 2q−( I )−1( I 1 + I 2).
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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
I 1 ≤ c
1
0
v( x)( x/2)α( x)−1
x
0 f ( y)dy
q( x)
dx ≤ C .
By Holder’s inequality with respect to the exponent p0( x) we have
I 2 ≤ 1
0(v( x))q( x)
x
x/2( f ( y)) p0( x)dy
q( x)/ p0( x)
×
x
x/2( x − y)(α( x)−1)( p0)′( x)dy
q( x)/( p0)′( x)
dx.
Now observe that
x
x/2
f ( y) p0( x)dy ≤ [ x/2, x]∩ f ≤1
( f ( y)) p0( x)dy + [ x/2, x]∩ f >1
( f ( y)) p0( x)dy
≤ cx + x
x/2( f ( y)) p( y)dy;
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
I 2 ≤ c p,q 1
0
(v( x))q( x) xq( x)α( x)dx
+ 1
0v( x)q( x)
x
x/2 f ( y) p( y)dy
q( x)/ p0( x)
x(α( x)−1)q( x)+q( x)/( p0)′( x)dx
:= c p,q[ I 2,1 + I 2,2].
For I 2,1, we find that
I 2,1 = ∑k ∈ Z −
E k −1
v( x)q( x) x(α( x)−1)q( x) xq( x)dx
≤
∑k ∈ Z −
2kq−( I )/ p+( I ) E k −1
v( x)q( x) x(α( x)−1)q( x)2(k −1)q( x)/( p0)′( x)dx
≤ cB1 ∑k ∈ Z −
2kq−( I )/ p+( I ) ≤ cB1 < ∞,
while taking into account the fact that q( x)/ p0( x) ≥ 1 we have
I 2,2 ≤ c ∑k ∈ Z −
E k −1
(v( x) xα( x)−1)q( x)
x
x/2( f ( y)) p( y)dy
xq( x)/( p0)′( x)dx
≤ c ∑k ∈ Z − E
k −1 v( x) xα( x)−1
q( x)
2(k −1)q( x)/( p0)′( x)dx I k −1
( f ( y)) p( y)dy≤ cB1S p( f ) ≤ cB1 < ∞.
Combining the above-derived estimates we find that I 2 < ∞. The proof follows.
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76 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∞ := sup0<t <∞
B∞(t ) := sup0<t <∞
∞
t
(v( x) xα( x)−1)q( x)t q( x)/(
p0)′( x)dx < ∞, (5.1.3)
then Rα( x) is bounded from L p( x)( I ) to Lq( x)
vq( x) ( I ).
Remark 5.1.1. Notice that (5.1.3) is also necessary for the boundedness of Rα( x) from
L p(R+) to Lqvq (R+), where p and q are constants (see Theorem 5.1.1).
Proof of Theorem 5.1.4. Suppose that f ≥ 0 and S p( f ) ≤ 1. For simplicity assume that
a = 1. We have
S q,v( Rα( x)
f ) = 2
0 v( x)q( x)( Rα f )q( x)( x)dx +
∞
2 v( x)q( x)( Rα( x)
f )qc ( x)dx
:= I 1 + I 2.
Observe that the condition B∞ < ∞ implies Ba < ∞. Consequently, by Theorem 5.1.3 we
conclude that I 1 ≤ c < ∞, while for I 2, we find that
I 2 ≤ c
∞2
(v( x))qc
1
0( x − y)α( x)−1 f ( y)dy
qc
dx
+ ∞
2
(v( x))qc x/2
1
( x − y)α( x)−1 f ( y)dyqc
dx
+
∞2
(v( x))qc
x
x/2( x − y)α( x)−1 f ( y)dy
qc
dx
:= c[ I 2,1 + I 2,2 + I 2,3].
Using Holder’s inequality for Lebesgue spaces with variable exponent (see Lemma
1.4.1) we have
I 2,1 ≤ c
∞
2(v( x))qc
1
0( x − y)α( x)−1 f ( y)dy
qc
dx
≤ c ∞
2(v( x)( x/2)α( x)−1)qc dx
f L p(·)( I )χ[0,1] L p′(·)( I ) ≤ cB∞,
while Theorem 1.3.4 yields
I 2,2 ≤ c
∞1
v( x) xα( x)−1
qc
x
1 f ( y)dy
qc
dx ≤ cB∞.
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Generalized One-sided Potentials 77
Now applying the condition α( x) > 1/ p−( I ), we find that
I 2,3 =∞
∑k =1
E k
(v( x))qc
x
x/2 f ( y)( x − y)α( x)−1dy
qc
dx
≤
∞
∑k =1 E k
(v( x))qc x
x/2( f ( y)) pc
dyqc/ pc
×
x
x/2( x − y)(α( x)−1)( pc)′
dy
qc/( pc)′
≤ c∞
∑k =1
E k
(v( x))qc ( x/2)(α( x)−1)qc+qc/( pc)′dx
E k
( f ( y)) p( y)dy
≤ cB∞
∞
∑k =1
E k
( f ( y)) p( y)dy ≤ cB∞ < ∞.
Summarizing the estimates for I 1 and I 2 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
Lq(·)v ( 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
compact from L
p(·)
( I ) to L
q(·)
v ( I ) if and only if (i) Aa < ∞;
(ii) limt →0
Aa(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 repre-
sent Rα( x) as follows:
Rα( x) f ( x) = R(1)α( x) f ( x) + R
(2)α( x) f ( x),
where
R(2)α( x) f ( x) = χ[0,β]( x) Rα( x) f ( x), R
(1)α( 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 )
Lq( x)( I )
≤χ(β,1]( x)v( x)
χ[0, x/2]( y)( x − y)α( x)−1
L p′( y)( I )
Lq( x)( I )
+χ(β,1]( x)v( x)χ( x/2, x]( y)( x − y)α( x)−1
L p′( y)( I ) Lq( x)
( I
)≤ cχ(β,1]( x)v( x) xα( x)−1/ p( x)
Lq( x)( I )
+cχ(β,1]( x)v( x) xα( x)−1/ p( x)
Lq( x)( I )
< ∞,
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78 Alexander Meskhi
because A1 < ∞. Consequently, by Theorem 1.4.8, R(1)α( x) is compact. Further, observe that
the arguments of the proof of Theorem 5.1.1 enable us to conclude that
Rα( x) − R(1)α( x)
L p(·)( I )→ Lq(·)v ( I )
≤ R(2)α( x)
L p(·)( I )→ Lq(·)v ( I )
≤ c sup0<t <β
A1(t ),
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 f t ( x) = t −1/ p(0)χ[0,t /2)( x). Hence, using Holder’s inequality for
L p(·) spaces (see Lemma 1.4.1), Proposition 1.4.1 and the absolutely continuity of the norm
· L p′(·) , we have 1
0 f t ( x)ϕ( x)dx
≤ k ( p) f t (·) L p(·)( I )ϕ(·)χ[0,t /2)(·) L p′(·)( I )
≤ ct −1/ p(0)t 1/ p+([0,t /2])ϕ(·)χ[0,t /2)(·) L p′(·)( I )
≤ cϕ(·)χ[0,t /2)(·) L p′(·)( I ) → 0
as t → 0 for all ϕ ∈ L p′( x)( I ). Hence, f t converges weakly to 0 as t → 0. Further, it is
obvious that
Rα( x) f t Lq(·)v ( I )
≥χ[t ,1)( x)v( x)
t /2
0( x − t )α( x)−1dt
Lq(·)( I )
t −1/ p(0)
≥ ct −1/ p′(0)χ[t ,1)( x)v( x) xα( x)−1
Lq( x)( 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 )
to Lq(·)v ( I ) if and only if
(i) A∞ < ∞;
(ii) limt →0
A∞(t ) = limt →∞
A∞(t ) = 0,
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
Rα( x) f = ∑5n=1 R
(n)α( x)
f , where
R(1)α( x) f ( x) = χ[0,β)( x)( Rα( x) f )( x),
R(2)α( x) f ( x) = χ[β,γ )( x) Rα( x)(χ[0,β/2] f )( x),
R(3)α( x) f ( x) = χ[β,γ )( x) Rα( x)(χ[β/2,∞) f )( x),
R(4)α( x) f ( x) = χ[γ ,∞) Rα( x)(χ[0,γ /2) f )( x),
R(5)α( x) f ( x) = χ[γ ,∞)( x) Rα( x)(χ[γ /2,∞) f )( x),
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Generalized One-sided Potentials 79
where 0 < β < 1/2 < 2 < γ < ∞. Now observe thatχ[β,γ )( x)v( x)
χ[0,β/2)( y)( x − y)α( x)−1
L p′( y)( I )
Lq( x)( I )
≤ cχ[β,γ )( x)v( x) xα( x)−1 L p′( x)( I )χ[0,β/2) L p′(·)( I ) < ∞
because A∞ < ∞. Further,χ[β,γ )( x)v( x)χ[β/2,∞)( y)( x − y)α( x)−1
L p′( y)( I )
Lq( x)( I )
≤
χ[β,γ )( x)v( x)χ[β/2, x/2)( y)( x − y)α( x)−1
L p′( y)( I )
Lq( x)( I )
+χ[β,γ )( x)v( x)χ[ x/2, x
)
( y)( x − y)α( x)−1 L p′( y)( I ) Lq( x)( I )
:= I 1 + I 2.
It is easy to see that Lemma 1.4.1 (for f = χ[β/2,γ /2]) implies
I 1 ≤ c
χ[β,γ )( x)v( x) xα( x)−1
Lq( x)( I )
χ[β/2,γ /2)(·)
L p′(·)( I )< ∞.
Analogously, by Lemma 1.4.9 we can see that I 2 < ∞ because A∞ < ∞. Applying Theorem
5.1.1 we find that
R(1)α( x)
L p(·)( I )→ Lq(·)v ( I )
= Rα( x) L p(·)([0,β))→ L
q(·)v ([0,β))
≤ c sup0<t <β
A∞(t ) → 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
R(5)α( x) f ( x)
L p( x)([γ ,∞))→ Lq( x)v ([γ ,∞))
≤ c supt >γ /2
∞t
(v( x))qc x(α( x)−1)qc dx
1/qc
(t − γ )1/( pc)′
holds. The latter term tends to 0 when γ → ∞ because limt →∞ A∞(t ) = 0. Finally we con-clude that Rα( x) is compact.
Necessity. The condition A∞ < ∞ is a consequence of Theorem 5.1.2. The condition
limt →0
A∞(t ) = 0 follows in the same manner as in the proof of necessity of Theorem 5.2.1. To
show that limt →∞
A∞(t ) = 0, we can (as above) use the facts that p and q are constants outside
[0,a] and Rα( x) is compact from L p(·)( I ) to Lq(·)v ( I ) if and only if the operator
W α,v f ( x) =
∞ x
v( y) f ( y)( y − x)α( y)−1dy
is compact from Lq′( x)( I ) to L p′( x)( I ).
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80 Alexander Meskhi
5.3. Measure of Non–compactness
This section deals with two-sided estimates of the distance between the operator Rα( x) and
the class of compact linear operators from L p(·)( I ) to Lq(·)v ( 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. F L( X ,Y ))
denotes the class of compact linear operators (resp. finite rank operators) acting from X to
Y . Let
T K ( X ,Y ) := distT ,K ( X ,Y ); α(T ) := distT ,F L( 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 Hardy-
Littlewood maximal operator M Ω is bounded in L p( x)(Ω) (see section 1.4).
Theorem 5.3.1. Let Ω ⊆ R
n
be a domain. Assume that X is a Banach space. Supposethat 1 < q−(Ω) ≤ q+(Ω) < ∞ and q ∈ P (Ω). Then
T K ( 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 T −K < T K + δ. By Lemma 1.4.4 there is P ∈ F L( X , L p(·)(Ω)) for which the inequality
K − P < δ holds. This gives
T − P ≤ T − K + K − P ≤ T K + 2δ.
Hence
α(T ) ≤ T K .
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
b1A ≤ T K ( L p(·)( I ), L
q(·)v ( I ))
≤ b2A ,
where A := limβ→0
Aβ , Aβ := sup0<t <β
Aa(t ) and Aa(t ) is defined in Theorem 5.1.1.
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
b1A ∞ ≤ T K ( L p(·)( I ), L
q(·)v ( I ))
≤ b2A ∞,
where A ∞ := limβ→0
Aβ+ limγ →∞
A(γ ) , Aβ := sup0<t <β
A∞(t ) and A(γ ) := supt >γ
A∞(t ) and A∞(t ) is defined
in Theorem 5.1.2.
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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
Rα( x) − R(2)α( x)
L p(·)( I )→ L
q(·)v ( I )
≤ R(1)α( x)
L p(·)( I )→ L
q(·)v ( I )
≤ cAβ,
where R(1)α( x) = χ[0,β]( x) Rα( x) f ( x), R
(2)α( x) = χ[β,1]( x) Rα( x) f ( x), 0 < β < 1 (see the proof of
Theorem 5.2.1 for the details) and the fact that R(2)α( x) is compact (see Theorem 1.4.8). To
get the lower estimate we take a positive number λ so that λ > T K ( L p(·)( I ), L
q(·)v ( I ))
. Notice
that
Rα( x)K
L p(·)( I ), L
q(·)v ( I )
= Rα( x),vK ( L p(·)( I ), Lq(·)( I )),
where
Rα( x),v f ( x) = v( x) x
0 f (t )( x − t )α( x)−1dt .
Consequently, by Theorem 5.3.1 λ > α( Rα( x),v).
Hence, there exist g1, . . . ,g N ∈ Lq(·)( I ) such that
α( Rα( x),v) ≤ Rα( x),v − F < λ,
where
F f ( x) = N
∑ j=1
α j( f )g j( x),
α j are linear bounded functionals in L p(·)( I ) and gi are linearly independent. Further, there
exist g1, . . . , g N such that support of gi is in [σi,a], 0 < σi < a, and
Rα( x),v − F 0 < λ,
where F 0 f ( x) = ∑ N j=1α j( f )g j( x). Suppose that σ = minσ j. Then suppF 0 f ⊂ [σ,a]. Let
0 < t < β < σ and let f be a non-negative function with support in [0, t /2] such that
f L p(·)( I ) ≤ 1. Consequently, for such an f we have
λ ≥ λ f L p(·)( I ) ≥ χ[0,β]( x)( Rα( x),v f ( x) − F 0 f ( x)) L p( x)( I )
≥ χ[t ,β]( x)( Rα( x),v f )( x) Lq( x)( I )
≥χ[t ,β]( x)v( x)
t /2
0( x − y)α( x)−1 f ( y)dy
Lq( x)( I )
≥ c
χ[t ,β]( x)v( x) xα( x)−1
Lq( x)( I )
t /2
0 f ( y)dy
.
Taking the supremum with respect to f we find thatχ[t ,β]( x)v( x) xα( x)−1
Lq( x)( I )
χ[0,t /2](·) L p′(·)( I ) ≥ cAβ.
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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82 Alexander Meskhi
5.4. 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 bounded-
ness of the Riemann-Liouville operator Rα from L p(R+) to Lqv (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 func-
tion 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
of the Hardy operator H from L p(·)( I ) to Lq(·)v ( I ) were established in [52]. We refer also
to [57] for the compactness of the potential-type operators in weighted L p(·)
spaces withspecial weights. A dominated compactness theorem in L
p(·)ρ (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].
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Chapter 6
Singular Integrals
In this chapter the essential norm for singular integrals (Hilbert transforms, Cauchy in-
tegrals, Riesz transforms, Caldeon-Zygmund operators), generally speaking, in weightedLebesgue 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 L pw(R) to L
pv (R). Then
H K ( L pw(R)→ L
qv (R)) ≥ sup
a∈Rlimr →0
a+r
av( x)dx
1/ p ∞a+r
w1− p′( x)
( x − a) p′ dx
1/ p′
.
Theorem 6.1.2. Let 1 < p < ∞ and let w ∈ A p(R). Then
H K ( L pw(R)) ≥
1
2 max ¯ A1, ¯ A2,
where
¯ A1 = supa∈R
limr →0
1
r
a+r
aw( x)dx
1/ p1
r
a
a−r w1− p′
( x)dx
1/ p′
;
¯ A2 = supa∈R
limr →0
1
r
a
a−r w( x)dx
1/ p1
r
a+r
aw1− p′
( x)dx
1/ p′
.
Proof of Theorem 6.1.1. Let a ∈R and λ> H K ( L pw(R)→ L
pv (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 a+r
av( x)| H f ( x)| pdx ≤ λ p
∞a+r
w( x)| f ( x)| pdx. (6.1.1)
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84 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
λ p ∞a+r
w1− p′ ( x)( x − a)− p′ dx ≥ a+
r
av( x)| H f ( x)| pdx
≥
a+r
av( x)dx
∞a+r
w1− p′( x)
( x − a) p′ dx
p
.
Now if we show that
J (a,r ) := ∞
a+r 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 a+r
av( x)dx
1/ p ∞a+r
w1− p′( x)
( x − a) p′ dx
1/ p′
≤ λ
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 ∞a+r
g( x)w−1/ p( x)
x − adx = ∞.
Further, we take the function
φ( x) = g( x)χ(a+r ,∞)( x)w−1/ p( x)
in the two-weight inequality
H φ L pv ≤ cφ L
pw
and, consequently, we conclude that
∞ =
a+r
av( x)dx
∞
a+r
g( x)w−1/ p( x)
x − adx
p
≤ c
∞
a+r (g( x)) pdx < ∞
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 L
pw(R). Let λ > H K ( L
pw(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 a+r
aw( x)| H f ( x)| pdx ≤ λ p
a
a−r w( x)| f ( x)| pdx. (6.1.2)
It is clear that x − y < 2r when y ∈ (a − r ,a), x ∈ (a,a + r ). Let us put f ( y) =
w1− p′
( y)χ(a−r ,a)( y) in (6.1.2). Then we observe that
λ p a
a−r w1− p′
≥ a+r
aw( x)| H f ( x)| pdx ≥
1
(2r ) p
a+r
aw( x)dx
a
a−r w1− p′
( x)dx
p
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Singular Integrals 85
from which, taking into account the fact that w1− p′∈ Lloc(R) (see e.g. [95]), it follows
12
¯ A1 ≤ λ. Analogously we have 12
¯ A2 ≤ λ.
Corollary 6.1.1. Let p = 2 , w( x) = | x|α, −1 < α < 1. Then
H K ( L2w(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
H K ( L pw(R)→ L
pw(R))) ≥
1
2(1 + 2 H L pw(R))
supa∈R
limr →0
A(r ,a) p (R),
where A(r ,a
) p (R) is defined in Definition 1.6.1 , and
H L pw(R) := H L
pw(R)→ L
pw(R).
Proof. First note that the condition w ∈ A p(R) and Theorem 1.6.1 imply the inequali-
ties: a+r
aw( x)dx ≤ 2 p H p
L pw(R)
a
a−r w( x)dx, (6.1.3)
a
a−r w( x)dx ≤ 2 p H p
L pw(R)
a+r
aw( x)dx. (6.1.4)
Indeed, if we put f ( y) = χ(a−r ,a)( y) in the one-weight inequality R
w( x)| H f ( x)| pdx ≤ H p
L pw(R)
R
w( x)| f ( x)| pdx, (6.1.5)
then we find that
Rw( x)| H f ( x)| pdx ≥ a+r
a
w( x) a
a−r
dy
x − y p
dx ≥ 1
2 p
a+r
a
w( x)dx.
On the other hand, f p
L pw(R)
= a
a−r w < ∞. Hence (6.1.3) holds.
Analogously we can show that (6.1.4) holds. Let us introduce the notation:
W (b,c) :=
c
bw( x)dx
1/ p
; V (b,c) :=
c
bw1− p′
( x)dx
1/ p′
.
Further, due to Theorem 6.1.2 and (6.1.3) − (6.1.4) we conclude that
1
2r W (a − r ,a + r )V (a − r ,a + r ) ≤
1
2r W (a − r ,a)V (a − r ,a)
+ 1
2r W (a − r ,a)V (a,a + r ) +
1
2r W (a,a + r )V (a − r ,a)
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86 Alexander Meskhi
+ 1
2r W (a,a + r )V (a,a + r ) ≤
2 H L pw(R)
2r W (a,a + r )V (a − r ,a)
+ 1
2r W (a − r ,a)V (a,a + r ) +
1
2r W (a,a + r )V (a − r ,a)
+2
H
L
p
w(R)2r
W (a − r ,a)V (a,a + r ) ≤ 2(1 + 2 H L pw(R)) H K ( L
pw(R))
when r is small.
6.2. Cauchy Singular Integrals
Let Γ be a smooth Jordan curve and let S Γ 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 S Γ is bounded from L pw(0, l) to L p
v (0, l).
Then
S I :=
I w1− p′
(s)ds < ∞, for all subintervals I of (0, l).
Proof. Let S I = ∞ for some I . Consequently, it follows that there exists g ∈ L p( I ), g ≥0, such that
I
g(t )w−1/ p(t )dt = ∞. Now let φ(s) = f (t (s)) = g(s)w−1/ p(s)χ I (s). Let 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)
|S Γ f (t (σ))| ≥ 1
2π
I
φ(s)
s −σds ≥
1
4πr
I
φ(s)ds (6.2.2)
for σ ∈ I ′. Consequently,
|S Γ f (t (σ))| ≥ 1
4πr
I
φ(s)dsχ I ′ (σ) (6.2.3)
for any σ ∈ (0, l). Hence, using inequality (6.2.3), we find that
S Γ f L pv (0,l) ≥
1
4πr
I
φ(s)ds
χ I ′ L pv (0,l)
= 1
4πr
I
g(s)w−1/ p(s)ds
=∞
χ I ′ L
pv (0,l) = ∞.
On the other hand,
φ L pw(0,l) = g L p( I ) < ∞.
This contradicts the boundedness of S Γ from L pw(0, l) to L
pv (0, l).
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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 S Γ is compact from L pw(0, l) to L
pv (0, l).
Moreover, if S Γ is bounded from L pw(0, l) to L
pv (0, l) , then the inequality
S Γ K ( L pw(0,l)) ≥
1
4π
max A1, A2
holds, where
A1 := supa∈(0,l)
limr →0
1
r
a+r
av(s)ds
1/ p1
r
a
a−r w1− p′
(s)ds
1/ p′
;
A2 := supa∈(0,l)
limr →0
1
r
a+r
aw(s)ds
1/ p1
r
a
a−r w1− p′
(s)ds
1/ p′
.
Proof. Let S Γ be bounded from L pw(0, l) to L
pv (0, l), λ > S Γ K ( L
pw(0,l), L p
v (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
S Γ f Lqv ( I (a,r )) ≤ λ f L
pw(0,l), f ∈ L p
w(0, l), (6.2.4)
where I (a,r ) = (a − r ,a + r ).
Let I 1 := (a − r ,a), I 2 := (a,a + r )ϕ(s) = f (t (s)) ≥ 0
) and suppϕ⊂ I 2. Then we have
the estimate similar to (6.2.2):
|S Γ f (t (σ))| ≥
1
2π I 2
ϕ(s)
s −σds
χ I 1 (σ) ≥
1
4r π I 2
ϕ(s) ds
χ I 1 (σ),
Analogously,
|S Γ f (t (σ))| ≥
1
4r π
I 1
ϕ(s)ds
χ I 2 (σ).
By Lemma 6.2.1 we have that w1− p′is locally integrable. Let ϕ(s) = w1− p′
(s)χ I 1 (s). Then
by (6.2.4) we have
1
(4π) p
1
r
a+r
aw1− p′
(s)ds
p−11
r
a
a−r v(s)ds
≤ λ p, a ∈ (0, l).
The latter inequality implies 1
4π A2 ≤ λ.
Analogously, it follows that1
4π A1 ≤ λ.
This completes the proof.
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
S Γ K ( L pw(0,l)) ≥
1
4π max A1, A2
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88 Alexander Meskhi
holds, where
A1 := sup
a∈(0,l)
limr →0
1
r
a+r
aw(s)ds
1/ p1
r
a
a−r w1− p′
(s)ds
1/ p′
;
A2 := supa∈(0,l)
limr →0
1
r
a+r
aw(s)ds
1/ p1
r
a
a−r w1− p′
(s)ds1/ p′
.
The next two corollaries follow in the same manner as in the case of the Hilbert trans-
form.
Corollary 6.2.1. Let p = 2 , w( x) = xα , where −1 < α < 1. Then
S Γ K ( L pw(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
S Γ K ( L pw(0,l)) ≥
1
4π(4πS Γ L pw(0,l) + 1)
supa∈(0,l)
limr →0
A(r ,a) p ,
where
A(r ,a) p =
1
2r
a+r
a−r w(s)ds
1/ p 1
2r
a+r
a−r w1− p′
(s)ds
1/ p′
and S Γ L pw(0,l) is the norm of S Γ in L pw(0, l).
6.3. 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 L pw(Rn) to L
pv (Rn). Moreover, if R j is
bounded from L pw(Rn) to L
pv (Rn) for some j, then the following inequality holds
R jK ≥ An esssupa∈Rn
v(a)
w(a)
1/ p
,
where An = γ n Bn
2n+1n3/2 , Bn = πn/2
Γ (1+n/2).
Proof. Let R j be bounded from L pw(Rn) to L
pv (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 > R j
K
L
pw(Rn), L p
w(Rn) and
a ∈ Rn, there exists a positive number β such that for all 0 < τ < β and f ∈ L pw(Rn) the
inequality
R j,v f L p( B(a,r )) ≤ λ f L pw(Rn) (6.3.1)
holds, where R j,v f = vR j f .
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Singular Integrals 89
Let a = (a1, . . . ,an) and let
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) = ∪n
j=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)| = γ n
− E j,a
f (t ) x j − t j
| x − t |n+1dt
≥ γ n
− E j,a
f (t ) ( x j − t j)
| x − t |nn1/2( x j − t j)dt = γ n
n1/2
− E j,a
f (t )
| x − t |ndt
≥ γ n
n1/2(2τ)n − E j,a
f (t )dt ,
where f ≥ 0.
Further, using the latter estimates and assuming f = w1− p′in (6.3.1) we have
cn
E j,a
v( x)dx
− E j,a
w1− p′( x)dx
p
≤ λ p
− E j,a
w1− p′( x)dx,
where
cn = γ
pn
(n1/22n) pτnp.
On the other hand, notice that
1
| E j,a|
E j
v( x)dx → v(a) a.e..
Indeed, we have
| E j,a| = | B(a,τ)|
2n=
Bnτn
2n. (6.3.2)
Therefore
1
| E j,a|
E j,a
v( x) − v(a)
≤ 1
| E j,a|
E j,a
|v( x) − v(a)|dx
≤ 2n
| B(a,τ)|
B(a,τ)|v( x) − v(a)|dx −→ 0 a.e.
as τ→ 0.
Analogously, 1
| − E j,a|
− E j,a
|w1− p′( x) − w1− p′
(a)|dx
→ 0
when τ→ 0.
Hence
cn| E j,a|
p v(a)
w(a) ≤ λ
p
for almost every a ∈ Rn, which on the other hand, together with (6.3.2) implies the desired
estimate.
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6.4. Calderon–Zygmund Operators
In this section we discuss the essential norm of the Calderon-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 com-
pact from L pw(Rn) to L
pv (Rn). Moreover, if K is bounded from L
pw(Rn) to L
pv (Rn) , then the
inequality
K K ≥ c esssupa∈Rn
v(a)
w(a)
1/ p
(6.4.1)
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 L pw(Rn) to L
pv (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 > K K ( L pw(Rn), L p
w(Rn)) and a ∈ Rn, there exists β > 0 and R ∈
F L
L
pw(Rn), L p
v (Rn)
with supp R ⊂ Rn \ B(a,β) for all f ∈ L
pw(Rn) such that for all f ∈
L pw(Rn) the inequality
K f L p(Rn) ≤ λ f L pw(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 aresatisfied 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) =
B
k ( x − y) f ( y)dy
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
|K f ( x)| ≥ 12
f B|k (ru)| ≥ c f B 1| B| , (6.4.3)
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 B′
v( x)
B
k ( x − y) f ( y)dy
pdx ≤ λ p
B( f ( y)) pw( y)dy
for all non-negative f with supp f ⊂ B. Let f ( x) = w1− p′( x)χ B( x). Then using (6.4.3), we
find thatc p
| B| p
B′v( x)dx
f
p B ≤ λ p
B
w1− p′( y)dy.
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Singular Integrals 91
Consequently by Lemma 1.6.2 we have
c pv B′ ((w1− p′) B) p−1 ≤ λ p. (6.4.4)
Further, observe that the equality
limr →0 v B′ = v(a) (6.4.5)
holds for almost all a. This follows from the obvious fact
|v B′ − v(a)| ≤ c 1
| B|
B |v( x) − v(a)|dx → 0
as r → 0, where B = B(a,r (t + 1)) and c is a positive constant.
Inequalities (6.4.4) and (6.4.5) yield
c v(a)w(a)1/ p
≤ λ
for almost every a (here the positive constant c depends only on a, n and t ). As λ is an
arbitrary number greater than K K , 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
no weight pair (v,w) and a function p ∈ W L(R) for which H is compact from L p(·)w (R) to
L p(·)v (R).
First we formulate the main results of this section
Theorem 6.5.1. Let p ∈ P (R) and let H be bounded from L p(·)w (R) to L
p(·)v (R). Then
the following estimate holds
H K ≥ (1/2) max A1, A2, (6.5.1)
where
A1 = supa∈R
limr →0
1
r χ(a−r ,a)v L p(·)(R)χ(a,a+r )w−1
L p′(·)(R),
A2 = supa∈R
limr →0
1
r χ(a,a+r )v L p(·)(R)χ(a−r ,a)w−1 L p′(·)(R).
Theorem 6.5.2. Let p ∈ P (R). Suppose that H is bounded from L p(·)w (R) to L
p(·)v (R).
Then
H K ≥ max B1, B2, (6.5.2)
where
B1 = supa∈R
limr →0
χ(a,a+r )(·)v(·) L p(·)(R)χ(a+r ,+∞)(·)w−1(·)(· − a)−1 L p′(·)(R);
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92 Alexander Meskhi
B2 = supa∈R
limr →0
χ(a−r ,a)(·)v(·) L p(·)(R)χ(−∞,a−r )(·)w−1(·)(a − ·)−1 L p′(·)(R).
The next statement also holds.
Theorem 6.5.3. Let p ∈ P (R). Suppose that H is bounded from L p(·)w (R) to L
p(·)v (R).
Then H K ≥ (1/4) maxC 1,C 2,
where
C 1 = supa∈R
limr →0
χ I (a,r )(·)v(·) L p(·)(R)χR\ I (a,r )(·)w−1(·)| · −a|−1 L p′(·)(R);
C 2 = supa∈R
limr →∞
χR\ I (a,r )(·)v(·)| · −a|−1 L p(·)(R)χ I (a,r )(·)w−1(·) L p′(·)(R).
Now we give another estimate of the essential norm of H .
Theorem 6.5.4. Let p ∈ P (R). Assume that H is bounded from L p(·)w (R) to L
p(·)v (R).
Then
H K ≥ (1/4) supa∈R
limr →0
χ(a−r ,a+r )v L p(·)(R)w−1(·)(r + |a − ·|)−1 L p′(·)(R).
Corollary 6.5.1. Let p satisfy (1.4.1) and (1.4.2). Then there is no weight pair (v,w)
such that H is compact from L p(·)w (R) to L
p(·)v (R). Moreover, if H is bounded from L p(·)(R)
to L p(·)(R) , then the inequality
H K
L
p(·)w (R), L
p(·)v (R)
≥ e− A/( p−)2
4 sup
a>0
limr →0
1
2r
a+r
a−r (v(t )) p(t )dt
1/ p−( I (a,r ))
×
1
2r
a+r
a−r (w(t ))− p′(t )dt
1/( p′)−( I (a,r ))
> 0
holds.
Corollary 6.5.2. Let p satisfy conditions (
1.4.1
) and
(1.4.2
). Then
H K
L p(·)(R)
) ≥ (1/4)e− A/( p−)2
,
where the positive constant A is from (1.4.1).
Remark 6.5.1. It is known that if
v( x) =
| x|−1/ p ln−1 e
| x| , if 0 < x ≤ 1,
1, if x > 1;
w( x) = | x|−1/ p, if 0 < x ≤ 1,1, if x > 1
,
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Singular Integrals 93
where p is a constant with 1 < p < ∞, then H is bounded from L pw(R) to L
pv (R) (see [46]).
Based on this fact and Theorem 6.5.2 we have the following estimate:
H K
L
pw(R), L p
v (R)
≥ ( p − 1)−1/ p.
To prove the main results of this section, we need some lemmas.
Lemma 6.5.1. Let T be a linear map from L p(·)w to L
p(·)v . Then T is bounded (resp.
compact ) if and only if T v,w is bounded (resp. compact ) in L p(·) , where T v,w f := vT ( f w−1).
Moreover, T L
p(·)w (R)→ L
p(·)v (R)
= T v,w L p(·)(R)→ L p(·)(R). Further, if T is bounded, then
T K
L
p(·)w (R), L
p(·)v (R)
= T v,wK ( L p(·)(R)).
Proof. The first part of the lemma can be checked immediately. For the second part
observe that
T v,w − P L p(·)(R)→ L p(·)(R) = T − Pv,w L
p(·)w (R)→ L
p(·)v (R)
,
where Pv,w f = 1/vP( f w).
Lemma 6.5.2. Let H be bounded from L p(·)w (R) to L
p(·)v (R). Then
G I (·) L p′(·)(R) < ∞,
for all bounded intervals I, where G I ( x) = w−1( x)(| I |/2 + | x − a I |)−1 and a I is the center
of I.
Proof. Suppose that G I /∈ L p′(·)(R) for some interval I := (a I − τ,a I + τ). By Lemma
1.4.6 we have that there exists g ∈ L p′(·)(R) such that g ≥ 0 and R
G I ( x)g( x)dx = ∞.
Hence either S I := +∞
a I G I g =∞ or
a I
−∞G I g =∞. Suppose that S I =∞. Then we take f ( x) =g( x)χ(a I ,+∞)( x). Then using Lemma 6.5.1 we find that
∞ > χ(a I ,+∞)(·)g(·) L p(·)(R) ≥ H
−1
L p(·)w → L p(·)v H v,w f L p(·)(R)
≥ H −1
L p(·)w → L
p(·)v
χ(a I −τ,a I )(·)v(·) L p(·)(R)
+∞ a I
g(t )G I (t )dt = ∞.
In the last inequality we used the inequality t − x ≤ (t − a I ) +τ which is true for all x, t with
x ∈ (a I − τ,a I ) and t > a I .
Proposition 6.5.1. Let H be bounded from L p(·)w (R) to L
p(·)v (R). Then
sup I
χ I v L p(·)(R)w−1(·)(| I |/2 + |a I − ·|)−1 L p′(·)(R)
≤ 4 H L
p(·)w (R)→ L
p(·)v (R)
, (6.5.3)
where I is a bounded interval and a I is the center of I .
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94 Alexander Meskhi
Proof. Due to Lemma 6.5.2 we have that G I ∈ L p′(·)(R), where
G I ( x) = w−1( x)(| I |/2 + | x − a I |)−1.
Let g ≥ 0 and let g ∈ L p(·)(R). Then by Holder’s inequality for L p(·) spaces (see Lemma
1.4.1) we see that J I :=
R
g(t )G I (t )dt < ∞. (6.5.4)
Let us choose r ∈R so that
+∞ r
g(t )G I (t )dt = (1/2) J I . (6.5.5)
Now observe that if x ∈ I (−∞,r ) and t ∈ (r ,+∞), then 0 < t − x ≤ |t − a I | + |a I − x| <
|t − a I | + | I |/2. Hence for such an x we have (recall that H v,w f = vH ( f /w))
H v,wg( x) ≥ v( x)
+∞ r
g(t )G I (t )dt = ( J I /2)v( x). (6.5.6)
Due to Lemma 6.5.1 we have
g L p(·)(R) ≥ H −1
L p(·)w → L
p(·)v
H v,wg L p(·)(R) ≥ ( J I /2) H −1χ I
(−∞,r )(·)v(·) L p(·)(R).
In a similar manner we can find that
g L p(·)(R) ≥ ( J I /2) H −1χ I
(r ,+∞)(·)v(·) L p(·)(R).
Now taking the supremum with respect to g with g L 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
H v,wK ( L p(·)) = H K ( L
p(·)w , L
p(·)v )
.
Let λ> H K ( L
p(·)w , L
p(·)v )
. Then by the previous equality and Theorem 5.3.1 we have that
λ > α( H v,w). Hence there exists P ∈ F L( L p(·)) such that
H v,w − P < λ.
Let us take an arbitrary a ∈ R. By Lemma 1.4.5 there exist a positive number β and R ∈F L( L p(·)) such that
R − P < λ− H v,w − P
2
and
supp R f ⊂R\ I (a,β)
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Singular Integrals 95
for all f ∈ L p(·), where I (a,β) = (a −β,a +β). Consequently,
( H v,w − R) f L p(·) ≤ λ f L p(·), f ∈ L p(·)(R),
where supp R f ⊂R\ I (a,β). From the latter inequality it follows that if 0 < τ < β, then
χ I (a,τ) H v,w f L p(·)(R) ≤ λ f L 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
w−1(·)χ I (a,τ)(·) L p(·)(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
∞ > λgχ(a,a+τ) L p(·)(R) ≥ χ(a−τ,a)(·) H v,w f (·) L p(·)(R)
≥ 1
2τχ(a−τ,a)(·)v(·) L p(·)(R) a+τ
agw−1.
Taking the supremum with respect to all g with g L p(·)(R) ≤ 1, applying Lemma 1.4.7
and passing to the limit as τ→ 0, we have that
H K ≥ (1/2) A1.
In a similar manner we can show that
H K ≥ (1/2) A2.
Proof of Theorem 6.5.2. Using the arguments from the proof of Theorem 6.5.1, for
λ > H K 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 g L p(·)(R) ≤ 1. Due to Lemma 6.5.2 we have
χ(a+τ,+∞)(·)w−1(·)(· − a)−1 L p′(·)(R) < ∞.
This implies
+∞
a+τ
g(t )w−1(t )(t − a)−1dt < ∞.
Further,
∞ > λgχ(a,a+τ) L p(·)(R) ≥ χ(a,a+τ)(·) H v,w f (·) L p(·)(R)
≥ χ(a,a+τ)(·)v(·) L p(·)(R)
+∞
a+τg(t )(t − a)−1w−1(t )dt
.
Taking the supremum with respect to all such a g we conclude that
H K ≥ B1.
Analogously, H K ≥ B2.
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Proof of Theorem 6.5.3. Repeating the arguments of Theorem 6.5.1 we arrive at in-
equality (6.5.7). Further, assume that supp f ⊂ (a + τ,+∞) in (6.5.7), where f ≥ 0 and
f L p(·)(R) ≤ 1. Then we observe that
∞ > λ f χ(a+τ,a) L p(·)(R) ≥ χ I (a,r )(·) H v,w f (·) L p(·)(R)
≥ (1/2)χ I (a,τ)(·)v(·) L p(·)(R)
+∞
a+τ f (t )(t − a)−1w−1(t )dt
.
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)χ I (a,τ)(·)v(·) L p(·)(R)χ(a+τ,+∞)(·)w−1(·)(· − a)−1 L p′(·)(R).
Arguing in the same manner as above we shall see that
λ≥ (1/2)χ I (a,τ)(·)v(·) L p(·)(R)χ(−∞,a−τ)(·)w−1(·)(a − ·)−1 L p′(·)(R).
Summarazing the estimates derived above, we conclude that
χ I (a,τ)(·)v(·) L p(·)(R)χR\ I (a,r )(·)w−1(·)| · −a|−1 L p′(·)(R)
≤ χ I (a,τ)(·)v(·) L p(·)(R)χ(−∞,a−r )(·)w−1(·)| · −a|−1 L p′(·)(R)
+χ I (a,τ)(·)v(·) L p(·)(R)χ(a+r ,+∞)(·)w−1(·)| · −a|−1 L p′(·)(R) ≤ 4λ.
These estimates lead us to the conclusion H K ≥ (1/4)C 1.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 ∈ F L( L pw) such that
H v,w f − R f L p(·)(R) ≤ λ f L p(·)(R), f ∈ L p(·)(R),
where λ > H K
L
p(·)w , L
p(·)v
, and supp R f ⊂ I (a,γ ). Consequently,
χR\ I (a,s) H v,w f L p(·)v (R) ≤ λ f L p(·)(R), f ∈ L p(·)(R),
for all s, s > γ . Let f be a non-negative function and let supp f ⊂ I (a,s). Then
∞ > λ f χ I (a,s) L p(·)(R) ≥ χ(a+s,+∞)(·) H v,w f (·) L p(·)(R)
≥ (1/2)χ I (a+s,+∞)(·)v(·) L p(·)(R)
a+s
a−s f (t )w−1(t )dt
.
Taking the supremum with respect to f with f L p(·)(R) ≤ 1 we find that
λ≥ (1/2)χ(a+s,+∞)(·)(· − a)−1v(·) L p(·)(R)χ I (a,s)(·)w−1(·) L p′(·)(R).
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Singular Integrals 97
Analogously,
λ≥ (1/2)χ(−∞,a−s)(·)(a − ·)−1v(·) L p(·)(R)χ I (a,s)(·)w−1(·) L p′(·)(R).
Consequently, H K ≥ (1/4)C 2.
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 Proposi-
tion 6.5.1. Let g be a non-negative function such that g L 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
I ∩ (−∞,r ) = /0. Observe that 0 < (t − x) ≤ (t − a) + τ whenever x ∈ I ∩ (−∞,r ) and t > r .
Using the arguments similar to those of Theorem 6.5.1 we have that (6.5.7) holds. Substi-
tuting f = gχ(r ,+∞) in (6.5.7) we find that
∞ > λχ(r ,+∞)g L p(·)(R) ≥ ( J I /2)χ I ∩(−∞,r )(·)v(·) L p(·)(R),
where J I is defined by (6.5.4).Analogously, if I ∩ (r ,+∞) = /0, then
∞ > λχ(−∞,r )g L p(·)(R) ≥ ( J I /2)χ I ∩(r ,+∞)(·)v(·) L 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
H K L p(·)w (R), L p(·)
v (R)≥ (1/4)lim
r →0(1/2r )χ I (a,r )(·)v(·) L p(·)(R)w−1(·)χ I (a,r )(·)
L p′(·)(R)
≥ (1/4)limr →0
(2r )−1/ p+( I (a,r ))−1/( p+( I (a,r )))′
a+r
a−r (v(t )) p(t )dt
1/ p−( I (a,r ))
×
a+r
a−r (w(t ))− p′(t )dt
1/( p′)−( I (a,r ))
≥ e− A/( p
−)2
4 lim
r →0 1
2r a+r
a−r (v(t )) p(t )dt 1/ p−( I (a,r ))
×
1
2r
a+r
a−r (w(t ))− p′(t )dt
1/( p′)−( I (a,r ))
> 0.
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
H K ( L p(·)) ≥ (1/4) sup
a∈Rlimr →0
1
2r χ I (·) L p(·)(R)(·)χ I (·)
L p′(·)(R)
≥ e− A/( p−)2
4 sup
a∈Rlimr →0
1
2r (2r )1/ p+( I )(2r )1/( p+( I ))′
= e− A/( p−)2
4 .
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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 S Γ along a smooth Jordan curve Γ on which the arc length is chosen as a parameter.
We begin with the following Lemma:Lemma 6.6.1. Let 1 < p− ≤ p( x) ≤ p+ < ∞. If S Γ is bounded from L
p(·)w (0, l) to
L p(·)v (0, l) , then
S I := w−1χ I L p′(·)(0,l) < ∞,
for all subintervals I of (0, l).
Proof. Let S I = ∞ for some I . This implies that there exists some g ∈ L p(·)( I ), g ≥ 0,such that
I
g(t )w−1(t )dt = ∞. Let
φ(s) = f (t (s)) = g(s)w−1(s)χ I (s)
and let I = (a − r ,a) ⊂ (0, l). We can assume that I ′ = (a,a + r ) ⊂ (0, l). We have (see
[107], [109])
|S Γ f (t (σ))| ≥ 1
2π
I
φ(s)
s −σds ≥
1
4πr
I
φ(s)ds
for σ ∈ I ′ and sufficiently small r . Thus
|S Γ f (t (σ))| ≥ 1
4πr I
φ(s)dsχ I ′ (σ) (6.6.1)
for any σ.Hence using inequality (6.6.1) we find that
v(σ)(S Γ f )(σ) L p(σ)(0,l) ≥χ I ′ (σ)v(σ)
1
2π
I
φ(s)
s −σds
L p(σ)(0,l)
≥ 1
4πr I
φ(s)dsχ I
′ (σ)v(σ) L
p(σ)
(0,l)
= 1
4πr
I
g(s)w−1(s)ds
=∞
χ I ′ (σ)v(σ) L p(σ)(0,l) = ∞.
On the other hand,
w(·)φ(·) L p(·)(0,l) = χ I (·)g(·) L p(·)(0,l) < ∞
Now the inequalityv(S Γ f ) L p(·)(0,l) ≤ cw(·)φ(·) L p(·)(0,l)
implies the desired result.
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Singular Integrals 99
Theorem 6.6.1. Let 1 < p− ≤ p( x) ≤ p+ < ∞ and let S Γ be bounded from L p(·)w (0, l) to
L p(·)v (0, l). Then
S Γ K ( L
p(·)w (0,l), L
p(·)v (0,l))
≥ 1
4π max ˜ A1, ˜ A2,
where
˜ A1 = supa∈(0,l)
limr →0
1r
χ(a−r ,a)v L p(·)(0,l)χ(a,a+r )w−1 L p′(·)(0,l)
and
˜ A2 = supa∈(0,l)
limr →0
1
r χ(a,a+r )v L p(·)(0,l)χ(a−r ,a)w−1
L p′(·)(0,l).
Proof. Let
λ > S Γ K ( L
p(·)w (0,l), L
p(·)v (0,l))
.
Then using the fact that
S Γ K ( L
p(·)w (0,l), L
p(·)v (0,l))
= S Γ ,v,wK ( L p(·)(0,l)),
where
S Γ ,v,w f
(t (s)) = v(s)
S Γ f w−1
(t (s)) and the equality (see Lemma 5.3.1)
S Γ ,v,wK ( L p(·)(0,l)) = α(S Γ ,v,w),
we have
λ > α(S Γ ,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 ∈ F L( L p(·)(0, l)) such that
R − P < λ− S Γ ,v,w − P
2
and supp R f ⊂ (0, l)\ I (a,β). Consequently,
(S Γ ,v,w − R) f (t (·)) L p(·)(0,l) ≤ λφ L 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
χ I (a,r )S Γ ,v,w f (t (·)) L p(·)(0,l) ≤ λφ L p(·)(0,l) (6.6.2)
which holds for all φ ∈ L p(·)(0, l). According to Lemma 6.6.1, w−1χ I L p′(·)(0,l) < ∞ for all
subintervals I ⊂ (0, l). Let g(s) ≥ 0, g ∈ L p(·)(0, l). Then by Holder’s inequality for L p(·)
spaces we find that I (a,r )
gw−1 ≤ cgχ I L p(·)(0,l)w−1χ I L p′(·)(0,l) < ∞.
Further, if s ∈ I 2 = (a,a + r ) and σ ∈ I 1 = (a − r ,a), then using (6.6.1) we haveS Γ ,v,w f (t (σ))= v(σ)
S Γ ( f w−1)(t (σ))≥
v(σ)
4πr
I 2
w−1(s)φ(s)ds. (6.6.3)
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100 Alexander Meskhi
Taking φ(s) = g(s)χ I 2 (s) in (6.6.2) and taking into account (6.6.3) we see that
∞ > λg(s)χ I 2 (s) L p(·)(0,l) ≥ χ I 1 (·)S Γ ,v,w f (·) L p(·)(0,l)
≥ χ I 1 v L p(·)(0,l) I 2
w−1(s)φ(s)ds.
Taking the supremum over all such a g we get
λ≥ 1
4πr v(·)χ(a−r ,a)(·) L p(·)(0,l)w−1(·)χ(a,a+r )(·)
L p′(·)(0,l).
This inequality implies
S Γ K ( L
p(·)w (0,l), L
p(·)v (0,l))
≥ 1
4π˜ A1.
Let us now take σ ∈ I 2 and let φ(s) = f (t (s)) be nonnegative function with suppφ ⊂ I 1.
Then S Γ f (t (σ))≥
1
2π
I 1
φ(s)
s −σds ≥
1
4πr
I 1
φ(s)ds.
Thus, for σ ∈ I 2 and sufficiently small r ,S Γ f (t (σ))≥
1
4πr
I 1
φ(s)dsχ I 2 (σ).
Taking φ(s) = g(s)χ I 1 (s) in (6.6.2) we get
∞ > g(s)χ I 1 (s) L p(·)(0,l) ≥ 14πr
v(·)χ I 2 (·) L p(·)(0,l) I 1
w−1(s)g(s)ds.
If we take the supremum with respect to g and use the fact that
w−1 L p′(·)( I 1) ≤ supg
L p(·)(0,l)≤1
l
0χ I 1 (t )g(t )w−1(t )dt
,we obtain
λ≥ 1
4πr χ(a,a+r )v L p(·)(0,l)χ(a−r ,a)w−1
L p′(·)(0,l).
Taking the supremum over a ∈ (0, l) and passing to the limit when r → 0, we conclude that
S Γ K ( L
p(·)w (0,l), L
p(·)v (0,l))
≥ 1
4π˜ A2.
Theorem 6.6.2. Let 1 < p− ≤ p( x) ≤ p+ < ∞ and let S Γ be bounded in L p(·)w (Γ ). Then
S Γ K ( L
p(·)w (0,l))
≥ 1
4π(4πS Γ + 1) supa∈(0,l)
limr →0
A(r ,a) p(·)
,
where A
(r ,a) p(·) =
1
2r χ(a−r ,a+r )(·)w(·) L p(·)(0,l)χ(a−r ,a+r )(·)w−1(·)
L p′(·)(0,l)
and S Γ is the operator norm.
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Singular Integrals 101
Proof. Let f (t (s)) = χ I 1 (s), where I 1 = (a − r ,a). Suppose that I 2 := (a,a + r ). We
have
J := S Γ f L
p(·)w (0,l)
≥ 1
2π
χ I 2 (·)
I 1
f (t (s))
· − sds
L p(·)w (0,l)
≥ 1
4πχ I 2 (·)
L p(·)w (0,l)
.
Also,
J ≤ S Γ χ I 1 (·) L
p(·)w (0,l)
.
Combining these inequalities we obtain
χ I 2 (·)w(·) L p(·)(0,l) ≤ 4πS Γ χ I 1 (·)w(·) L p(·)(0,l). (6.6.4)
Analogously,
χ I 1 (·)w(·) L p(·)(0,l) ≤ 4πS Γ χ I 2 (·)w(·) L p(·)(0,l). (6.6.5)
Now applying (6.6.4) and (6.6.5) we find that
A(r ,a) p(·) =
1
2r χ(a−r ,a+r )(·)w(·) L p(·)(0,l)χ(a−r ,a+r )(·)w−1(·)
L p′(·)(0,l)
≤ 1
2r
χ I 1 (·)w(·) L p(·)(0,l) + χ I 2 (·)w(·) L p(·)(0,l)
×
χ I 1 (·)w−1(·) L p′(·)(0,l) + χ I 2 (·)w−1(·)
L p′(·)(0,l)
=
1
2r χ I 1
(·)w(·) L
p(·)
(0,l)χ
I 1(·)w−1(·)
L p′(·)
(0,l)
+χ I 1 (·)w(·) L p(·)(0,l)χ I 2 (·)w−1(·) L p′(·)(0,l)
+χ I 2 (·)w(·) L p(·)(0,l)χ I 1 (·)w−1(·) L p′(·)(0,l)
+χ I 2 (·)w(·) L p(·)(0,l)χ I 2 (·)w−1(·) L p′(·)(0,l)
≤
1
2
4πS Γ
1
r χ I 2 (·)w(·) L p(·)(0,l)χ I 1 (·)w−1(·)
L p′(·)(0,l)
+
1
r χ I 1 (·)w(·) L p(·)(0,l)χ I 2 (·)w−1
(·) L p′(·)(0,l)
+1
r χ I 2 (·)w(·) L p(·)(0,l)χ I 1 (·)w−1(·)
L p′(·)(0,l)
+4πS Γ 1
r χ I 1 (·)w(·) L p(·)(0,l)χ I 2 (·)w−1(·)
L p′(·)(0,l)
.
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
supa∈(0,l)
limr →0
A(r ,a)
p(·) ≤
1
216π2
S Γ
S Γ K ( L
p(·)w (0,l))
+ 4π
S Γ K ( L
p(·)w (0,l))
4πS Γ K ( L
p(·)w (0,l))
+ 16π2S Γ S Γ K ( L
p(·)w (0,l))
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102 Alexander Meskhi
= 4πS Γ K ( L
p(·)w (0,l))
+ 16π2S Γ S Γ K ( L
p(·)w (0,l)))
.
Therefore
supa∈(0,l)
limr →0
A(r ,a) p(·) ≤ 4πS Γ
K ( L p(·)w (0,l))
4πS Γ + 1
.
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 S Γ K ( L pw(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
S T K ( L pw(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 S Γ 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 S Γ K ( L p
w), 1 < p < ∞,
where w is the Muckenhoupt weight.
The one-weight problem for the Hilbert transform and Caldeon-Zygmund singular in-
tegrals 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 sameproblems 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 S Γ in classical Lebesgue spaces we refer to
[109], [107], [76], [49].
Weighted estimates for S Γ in L p(·) spaces were obtained in [120], [121], [127]–[130],
[132], [134].
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Index
A
Adams, 103
AMS, 104Amsterdam, 108, 118
application, 112, 113applied mathematics, ix
asymptotic, 25, 105, 113asymptotics, 105, 108
averaging, 105
B
Banach spaces, vii, 4, 16, 63, 80
behavior, 113, 115Boston, 103, 108, 112
boundary value problem, vii, viii bounded linear operators, 4, 22, 67
Brownian motion, 113
C
calculus, vii, ix, 118Cauchy 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, 113conjecture, 105
continuity, 7, 13, 15, 78corona, 115
Czech Republic, 110
D
decomposition, 115
definition, 1, 19, 29, 36, 37, 83, 86, 90density, 109, 111
derivatives, xi, 63differential equations, viiidifferentiation, viii, 28, 33, 35
distribution, 112
duality, 54, 84
E
economics, viiielasticity, 118
encouragement, ixentropy, 4, 25, 67, 107, 113, 114
equality, 75, 91, 94, 99Euclidean space, viii, xi, 9, 35, 51
F
family, 6finance, viii
fluid, 105Fourier, 51, 68, 70, 106, 110
fractional integrals, vii, 50, 113, 114, 117, 118
G
gene, 104generalization, 17
generalizations, 104graduate students, ix
groups, viii, 1, 6, 35, 37, 51, 102, 103, 107, 108, 110
growth, viii, 83, 111
H
Harmonic analysis, 117Heisenberg, 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, 118Hong Kong, 110
House, 110, 113, 115
I
identity, vii, viii, 25, 51, 63, 65
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Index120
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,
117integration, viii, 42, 108, 116
interval, 12, 13, 20, 21, 71, 80, 93
J
Jordan, 23, 86, 87, 98
K
kernel, vii, 21, 24, 25, 37, 38, 70, 103, 107Kolmogorov, 116
L
lead, ix, 96Lebesgue measure, xi
Lie algebra, 1
Lie group, 1linear, 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, 60London, 104, 105, 106, 107, 108, 110, 115, 116, 117
Lyapunov, 102
M
manifold, viii
manifolds, viiimartingale, 36, 105
mathematicians, ixmathematics, vii
measures, 115
memory, viiimemory processes, viii
metric, 70, 109
Mexico, 116Mexico City, 116
modeling, 116monograph, vii, viii, ix, 1, 25, 70
Moscow, 112, 114motion, 113
N
natural, vii, ix, xi
New Jersey, 107, 110, 117
New York, 105, 110, 117, 118
nonlinear, vii, 110non-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, 115organization, ix
orthogonality, 117
P
paper, ix, 13, 25, 69, 70
parameter, 23, 98Paris, 105
partial differential equations, 113 physics, vii, viii, 111
plasticity, ixPoisson, 51, 60, 70, 114, 117
Poland, 110
power, 117 property, 4
Q
quantum, vii
quantum mechanics, vii
R
radius, xi, 1, 5, 6, 14, 57, 60, 67
random, 108random matrices, 108
range, vii
real numbers, 9, 17recall, 35, 74, 94
research, ixresearchers, viii, ix
Russian, 104, 107, 108, 109, 110, 112, 114, 115,117, 118
S
series, 18, 51, 68, 70, 106
Singapore, 110
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Index 121
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,
118solutions, vii
spatial, 117
spectrum, 104, 108, 114St. Petersburg, 114
stochastic, vii, viiistochastic processes, vii
students, ixsymbols, 104
systems, 22, 46
T
theory, vii, viii, ix, 103, 104, 105, 107, 108, 110,
116, 118time, 50Tokyo, 114
topological, 114
topology, 117
transformations, 1, 109
translation, 90transparent, 69
trees, 107
U
unification, 107
USSR, 108, 118
V
values, 107, 108, 115, 116variable, 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
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