c:/documents and settings/ras o/plocha/diplomka/diplomka bez …rokyta/vyuka/mgr... ·...

Charles University in PragueFaculty of Mathematics and Physics

MASTER THESIS

Rastislav Ol’hava

Optimal pairs of function spaces for weighted Hardy operators

Department of Mathematical Analysis

Supervisor of the master thesis: Prof. RNDr. Lubos Pick, CSc., DSc.

Study programme: Mathematics

Specialization: Mathematical Analysis

Prague 2011

I would like to express my deep gratitude to the supervisor of my master thesisProf. Lubos Pick for concept of my thesis, inspirational consultations, motivation,as well as for a lot of technical advice, that helped me with writing. I would also liketo thank my family for their support.

I declare that I carried out this master thesis independently, and only with the citedsources, literature and other professional sources.

I understand that my work relates to the rights and obligations under the Act No.121/2000 Coll., the Copyright Act, as amended, in particular the fact that the CharlesUniversity in Prague has the right to conclude a license agreement on the use of thiswork as a school work pursuant to Section 60 paragraph 1 of the Copyright Act.

In Prague date August 1, 2011 ..............................

Nazov prace: Optimalne pary priestorov funkciı pre vahove Hardyho operatoryAutor: Rastislav Ol’havaKatedra: Katedra matematicke analyzyVeduci diplomovej prace: Prof. RNDr. Lubos Pick, CSc., DSc., Katedra matem-aticke analyzy, Matematicko-fyzikalnı fakulta, Univerzita Karlova, Sokolovska 83, 18675 Praha 8, Ceska Republika

Abstrakt: Zameriame sa na urcity vahovy Hardyho operator so spojitou kvazikonkav-nou vahou, definovany na Banachovych priestoroch funkciı, v ktorych ma kazda funkciarovnaku normu ako jej prerovnanie. V teorii priestorov funkciı maju tieto operatorysiroke vyuzitie. V predchadzajucom vyskume bolo dokazane, ze platı ekvivalenciamedzi ohranicenost’ou niektorych z tychto operatorov a sobolevovskymi vnoreniami.Nech je nas Hardyho operator ohraniceny z priestoru X do priestoru Y . Tato pracasa venuje hl’adaniu takej dvojice priestorov X a Y , ktora je optimalna. Zmienenaoptimalita by pri d’alsom vyzkume mala viest’ k optimalite v urcitych sobolevovskychvnoreniach. Nasim druhym ciel’om je studium supremalnych operatorov, ktore tiezuzko suvisia s touto tematikou, a odvodenie niektorych ich zakladnych vlastnostı.

Kl’ucove slova: optimalita, vahovy operator Hardyovho typu, supremalny operator

Title: Optimal pairs of function spaces for weighted Hardy operatorsAuthor: Rastislav Ol’havaDepartment: Department of Mathematical AnalysisSupervisor of the master thesis: Prof. RNDr. Lubos Pick, CSc., DSc., Departmentof Mathematical Analysis, Faculty of Mathematics and Physics, Charles University,Sokolovska 83, 186 75 Prague 8, Czech Republic

Abstrakt: We focus on a certain weighted Hardy operator, with a continuous, quasi-concave weight, defined on a rearrangement-invariant Banach function spaces. The op-erators of Hardy type are of great use to the theory of function spaces. The mentionedoperator is a more general version of the Hardy operator, whose boundedness wasshown to be equivalent to a Sobolev-type embedding inequality. This thesis is con-cerned with the proof of existence of domain and range spaces of our Hardy operatorthat are optimal. This optimality should lead to the optimality in the Sobolev-typeembedding equalities. Our another aim is to study supremum operators, which arealso closely related to this issue, and establish some of their basic properties.

Keywords: optimality, weighted Hardy operator, supremum operator

Contents

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 A weighted Hardy operator . . . . . . . . . . . . . . . . . . . . . . . 13

3 Supremum operators . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

Introduction

The first task of this thesis is to study a certain weighted Hardy operator and the op-timality of function spaces on which it acts as a bounded operator. The second task isto study two operators involving supremum and establish some of their properties.

The structure of this thesis is as follows. The first chapter contains all definitionsand results which are required in the next chapters. First, we will define a nonin-creasing rearrangement, which will play a crucial role throughout this paper, and someof its properties. Then, we will introduce the abstract theory of Banach function spaces,which are Banach spaces of measurable functions in which the norm is related to the un-derlying measure in an appropriate way. As the most important terms we can mentiona rearrangement-invariant space (or r.i. space), the associate space and the associatenorm. Next, we will define the dilation operator and also provide the reader with a listof the function spaces which will be mentioned further. At the end of the chapter wewill mention a measure-preserving transformation and its relation to the equimeasur-ability of the functions and introduce some parts of interpolation theory. All proofsexcept that of the last theorem are omitted and the reader can find them togetherwith more detailed information in the monograph [1], which is also the main sourceof information used in this paper.

In the second chapter we will focus on the weighted Hardy operator

Hϕf(t) :=

∫ R

tϕ(s)f(s)

ds

s, t ∈ [0, R],

where R ∈ (0,∞), ϕ is a continuous quasiconcave function and f is measurable andfinite a.e. on [0, R]. In the case when ϕ(t) = t

mn , m ∈ Z+, 1 ≤ m ≤ n − 1, we get

the operator Hn/m, which was introduced in [2]. As we can see from the main resultof [2] the boundedness of the operator Hn/m is equivalent to a certain Sobolev embed-ding, which proves the great importance of the operators of such type. In this specialcase also the problem of the optimality of the particular Sobolev embedding is reducedto the problem of finding an optimal pair of the r.i. spaces XD and XR, such thatthe operator Hn/m is bounded into XR from XD. By the optimality we mean that XD

can not be replaced by a larger r.i. space and XR can not be replaced by any smallerr.i. space. In this chapter we find such optimal pairs of r.i. spaces for the operator Hϕ,that is, in a more general case. In our approach we start with an arbitrary r.i. space Xand then construct the r.i. space XR, which is the smallest r.i. space into which is Hϕ

bounded from X. Having XR we find the r.i. space XD, which is the largest r.i. spacefrom which is Hϕ bounded into XR. Finally, we show that then the pair (XD, XR) isalready optimal for the operator Hϕ.

In the third chapter we will study the operators

(Tϕf)(t) :=1

ϕ(t)supt≤s<R

ϕ(s)f∗(s)

and

(Sϕf)(t) :=ϕ(t)

tsup

0<s≤t

s

ϕ(s)f∗(s), f measurable, t ∈ [0, R]

2

which belong to the group of the so-called Hardy-type operators involving suprema.These operators have proved themselves to be very useful in various research projectsduring last years. In the centre of our interest is the search for optimal pairs of r.i. spacesfor which Sobolev-type embedding holds, in [2], where these supremum operators playedquite an important role. The usage of the supremum operators have two advantages.We can use them in order to make some quantity monotone and then, without any diffi-culty, get rid of them thanks to their boundedness on certain function spaces. The thirdchapter is concerned with the basic mapping properties of the operators Tϕ and Sϕ.We will also study what is the relations between Sϕf , Sϕf∗∗ and (Sϕf)∗∗ and betweenTϕf∗∗ and (Tϕf)∗∗.

Chapter 1

Preliminaries

In this chapter we will introduce all facts which are necessary in the further chapters.Suppose I is the finite closed interval [0, R], where R ∈ (0,∞), and λ is the Lebesgue

measure on R. LetM(I) denote the collection of all extended real-valued, λ-measurablefunctions on I, M0(I) the class of functions in M(I) that are finite λ-a.e. and P(I)the class of nonnegative functions in M0(I). As usual, any two functions coincidingλ-a.e. will be identified. The space M0(I) with the topology of convergence in measureon sets of finite measure is a metrizable topological vector space.

Definition 1.1. Given f ∈ M0(I), we define its nonincreasing rearrangement f∗

on (0, R] by

f∗(t) := inf {s > 0 : λf (s) ≤ t} , t ∈ (0, R],

where the distribution function λf of a function f in M0(I) is defined by

λf (s) := λ {x ∈ I : |f(x)| > s} , s ∈ [0,∞).

Definition 1.2. Functions f and g are said to be equimeasurable (denoted by f ∼ g)if they have the same distribution function.

Proposition 1.1. Suppose f , g and fn, (n = 1, 2, · · · ), belong to P(I) and let C beany real number. The nonincreasing rearrangement f∗ is a nonnegative, nonincreasing,right-continuous function on I. Furthermore,

|g| ≤ |f | a.e. ⇒ g∗ ≤ f∗;

(Cf)∗ = |C|f∗;

(f + g)∗(s+ t) ≤ f∗(s) + g∗(t), 0 < s+ t < R;

|f | ≤ lim infn→∞

|fn| a.e. ⇒ f∗ ≤ lim infn→∞

f∗n;

in particular,|fn| ↑ |f | a.e. ⇒ f∗n ↑ f∗,

where the symbol ↑ denotes monotone pointwise convergence of functions;

f and f∗ are equimeasurable;

(|f |p)∗ = (f∗)p , p ∈ (0,∞).

Proof. See [1, Chapter 2, Proposition 1.7].

While the nonincreasing rearrangement does not necessarily preserve sums or prod-ucts of functions, there are some inequalities that govern these processes. The startingpoint is an elementary inequality due to G.H. Hardy and J.E. Littlewood.

Preliminaries 4

Theorem 1.2 (Hardy & Littlewood). Let f and g belong to P(I), then

∫ R

0f(t)g(t) dt ≤

∫ R

0f∗(t)g∗(t) dt,

and in the special case when g belongs to P(I) and E is an arbitrary measurable subsetof I,

∫

Eg(x) dx ≤

∫ λ(E)

0g∗(s) ds, t ∈ I. (1.1)

Proof. See [1, Chapter 2, Lemma 2.1 and Theorem 2.2].

In what follows we will use the maximal function

f∗∗(t) := t−1

∫ t

0f∗(s) ds, t ∈ I,

which has some useful properties. The operator f 7→ f∗∗ is sublinear, namely,

(f + g)∗∗(t) ≤ f∗∗(t) + g∗∗(t), f, g ∈ M0(I), t ∈ I; (1.2)

on the other hand, the operation of the nonincreasing rearrangement is not sublinear.

Example 1.3. The symbol χA denotes the indicator function of a set A ⊂ I, i.e.the function χA : I → {0, 1}, defined by

χA(t) =

{

1 for t ∈ A,0 for t ∈ I\A.

Then, a counterexample to the sublinearity of the nonincreasing rearrangement is ob-tained by

(

χ[0,R2 )+ χ[R2 ,R]

)∗(

3

4R

)

>(

χ[0,R2 )

)∗(

3

4R

)

+(

χ[R2 ,R]

)∗(

3

4R

)

.

Let us introduce the abstract theory of Banach function spaces. The reader canfind more detailed information in [1, Chapter 1].

Definition 1.3. A function on P(I) is called a rearrangement-invariant Banachfunction norm (r.i. norm) if, for all f, g, fn in P(I), and for all constants c ≥ 0,the following properties hold:

(P1) (f) ≥ 0 with (f) = 0 if and only if f = 0 a.e. on P(I);

(P2) (cf) = c(f), c ≥ 0;

(P3) (g + f) ≤ (g) + (f);

(P4) 0 ≤ fn ↑ f a.e. implies (fn) ↑ (f);

Preliminaries 5

(P5) (χI) <∞;

(P6)∫ R0 f(t) dt ≤ C(f) for some constant C ∈ (0,∞), independent of f ;

(P7) (f) = (f∗).

When we omit (P7), we get just Banach function norm or shortly a function norm.We use the symbol . or & in situations similar to the one in (P6). It indicates that

the quantity on the left side is smaller than that on the right side multiplied by certainconstant which does not depend on any function or any other appropriate parametersinvolved.

Definition 1.4. Let be an r.i. norm. The collection X(I) = X()(I) of all functionsin M0(I) for which (|f |) < ∞ is called a rearrangement-invariant space (r.i. space).For each f ∈ X(I), define

‖f‖X(I) = (|f |).

If is just a function norm, the space X(I) is called a Banach function space orshortly a function space. Instead of the notation X(I) we use the abbreviated versionX, so if it is not specified, a Banach function space is over the measure space (I, λ).

The basic example of an r.i. norm is the Lebesgue norm

‖f‖Lp :=

(∫ R

0|f(t)|p dt

)1/p

=

(∫ R

0|f∗(t)|p dt

)1/p

for f ∈ M0(I), p ∈ [1,∞), with

‖f‖L∞ := ess sup0<t<R

|f(t)| = limt→0+

f∗(t).

These norms generate the Lebesgue spaces

Lp := {f ∈ M0(I) : ‖f‖Lp <∞}.

Definition 1.5. Let X,Y be normed linear spaces. Then the space X is said to beembedded in the space Y , denoted by X → Y , if X ⊂ Y and the identity mappingid : X → Y is continuous, i.e. for all x ∈ X:

‖x‖Y & ‖x‖X .

Theorem 1.4. Let be a function norm and let X = X() and ‖·‖X be as in Defini-tion 1.4. Then under the natural vector space operations, (X, ‖·‖X) is a normed linearspace for which the inclusions

S ⊂ X → M0

hold, where S is the set of λ-simple functions on I. In particular, if fn → f in X, thenfn → f in measure on sets of finite measure, and hence some subsequence convergespointwise λ-a.e. to f .

Preliminaries 6

Proof. See [1, Chapter 1, Theorem 1.4].

Definition 1.6. Let be a function norm, then its associate norm ′ is defined by

′(g) = sup

{∫ R

0f(x)g(x) dx; (f) ≤ 1

}

, g ∈ P(I).

Theorem 1.5. Let be a function norm. Then the associate norm ′ is itself afunction norm. Moreover, if is the r.i. norm, then ′ is also the r.i. norm.


Definition 1.7. Let be a function norm and let X = X() be the Banach functionspace determined by . Let ′ be the associate norm of . The Banach functionspace X(′) determined by ′ is called the associate space of X and is denoted by X ′.It follows from the definition of the associate norm that the norm of a function gin the associate space X ′ is given by

‖g‖X′ = sup

{∫ R

0|f(x)g(x)| dx; ‖f‖X ≤ 1

}

.

Remark 1.6. Let g be a positive nonincreasing function in X ′. Notice that accordingto Theorem 1.2 we have

∫ R

0g(t)h(t) dt ≤

∫ R

0g(t)h∗(t) dt,

so we can write

‖g‖X′ = sup

{∫ R

0f∗(x)g(x) dx; ‖f∗‖X ≤ 1

}

.

Theorem 1.7 (Holder’s inequality). Let X be a Banach function space with associatespace X ′. If f belongs to X and g to X ′, then fg is integrable and

∫ R

0|f(x)g(x)| dx ≤ ‖f‖X ‖g‖X′ . (1.3)


Theorem 1.8. Every Banach function space X coincides with its second associatespace X ′′. In other words, a function f belongs to X if and only if it belongs to X ′′,and in that case

‖f‖X = ‖f‖X′′ .


Lemma 1.9. The norm of a function g in the associate space X ′ is given by

‖g‖X′ = sup

{∣

∣

∣

∣

∫ R

0f(x)g(x) dx

∣

∣

∣

∣

; ‖f‖X ≤ 1

}

.

Preliminaries 7

Proof. See [1, Chapter 1, Lemma 2.8].

Proposition 1.10. If X and Y are Banach function spaces and X is embedded in Y ,then Y ′ is embedded in X ′.


Definition 1.8. If f and g belong to P(I), we write f ≺ g if f∗∗ ≤ g∗∗, that is, if

∫ t

0f∗(s) ds ≤

∫ t

0g∗(s) ds, for all t ∈ I.

The relation ≺ is called the Hardy-Littlewood-Polya relation.

Proposition 1.11 (Hardy’s lemma). Let ξ1 and ξ2 be nonnegative measurable func-tions on [0,∞) and suppose

∫ t

0ξ1(s) ds ≤

∫ t

0ξ2(s) ds

for all t ∈ [0,∞). Let η be any nonnegative nonincreasing function on [0,∞). Then

∫ ∞

0ξ1(s)η(s) ds ≤

∫ ∞

0ξ2(s)η(s) ds.


Using these results we get the following relation between the Hardy-Littlewood-Polya relation and the relation between the r.i. norms of two functions.

Theorem 1.12. Let X be a r.i. space. Suppose f1 belongs to M0(I) and f2 belongsto X. If f1 ≺ f2, then f1 belongs to X and ‖f1‖X ≤ ‖f2‖X .

Proof. There is a more general case proved in [1, Chapter 2, Theorem 4.6].

Definition 1.9. The dilation operator Es, s ∈ (0,∞), given at f ∈ P(I), t ∈ I, isdefined by

(Esf)(t) :=

{

f(t/s) if t ∈ [0, s];0 if t ∈ (s,R].

Remark 1.13. The operator Es is bounded on any r.i. space X for every fixed s ∈ (0,∞)as we can see in [1, Chapter 3, Proposition 5.11]. Let us denote by hX(s) the normof Es on X.

Proposition 1.14. The function hX is nondecreasing and submultiplicative on (0,∞),and it satisfies hX(1) = 1 and

hX(t) ≤ max {1, t} , t ∈ (0,∞).


Preliminaries 8

Remark 1.15. Notice that the dilation operator Es preserves equimeasurability. Wehave

λEsf = sλf = sλg = λEsg, for f, g ∈ M0(I) equimeasurable.

We shall now introduce some function spaces which will appear throughout the pa-per.

Definition 1.10. We denote Lp(θ), Γp(θ) and Mθ the spaces generated by the norms

‖f‖Lp(θ) =

(∫ R

0|f(t)|pθ(t) dt

)1/p

;

‖f‖Γp(θ) =

(∫ R

0f∗∗(t)pθ(t) dt

)1/p

;

‖f‖Mθ= sup

0<t<Rθ(t)f∗∗(t),

for f ∈ M0(I) and θ ∈ P(I).

The process of rearranging the finite sequence could be described by the permuta-tion. We are working with measurable functions defined on I, so we need somethingmore general. The notion of permutation is in our case replaced by that of measure-preserving transformation [1, Chapter 2, Section 7] or [4].

Definition 1.11. Let (R1, µ1) and (R2, µ2) be totally σ-finite measure spaces. A map-ping σ from R1 into R2 is said to be a measure-preserving transformation if, wheneverE is a µ2-measurable subset of R2, the set σ−1E = {x ∈ R1 : σ(x) ∈ E} is a µ1-measurable subset of R1 and

µ1(σ−1E) = µ2(E).

A measure-preserving transformation provides us with a connection between twoequimeasurable functions as we can see in the following result.

Theorem 1.16. Let σ : R1 → R2 be a measure-preserving transformation. If f2is a nonnegative µ2-measurable function on R2, then the function f1 = f2 ◦ σ is anonnegative µ1-measurable function on R1 and f1, f2 are equimeasurable.


The equimeasurability and measure-preserving transformation are connected alsoin an opposite way.

Theorem 1.17. Let (S, µ) be a finite nonatomic measure space and let f be a nonneg-ative µ-measurable function on R. Then there is a measure-preserving transformationσf : S → (0, µ(S)) such that f = f∗ ◦ σf µ-a.e.


Preliminaries 9

Remark 1.18. The measure space (I, λ) is finite and nonatomic. It follows from Theo-rem 1.17 that for two equimeasurable functions f and g there is a measure-preservingtransformation σ such that f = g ◦ σ.

Finally, we introduce some parts of interpolation theory, all from [1, Chapters 3 and5], which will be required in the following sections.

Definition 1.12. A pair (X0, X1) of Banach function spacesX0 andX1 is called a com-patible couple if there is some Hausdorff topological vector space, say H, into whicheach of X0 and X1 is continuously embedded.

According to Theorem 1.4 each of the spaces Lp, Lp(θ), Γp(θ),Mθ is for any weight θand any p ∈ [1,∞] continuously embedded in the topological vector space M0 of λ-measurable functions that are finite λ-a.e. Since metrizable topological vector spaceis also Hausdorff, we obtain that each pair of mentioned spaces forms a compatiblecouple.

Definition 1.13. Let (X0, X1) be a compatible couple of Banach function spaces,with corresponding Hausdorff space H. Let X0 +X1 denote the sum of X0 and X1,that is, the set of elements x in H that are representable in the form x = x0 + x1for some x0 in X0 and x1 in X1. For each x in X0 +X1, set

‖x‖X0+X1= inf

{

‖x0‖X0+ ‖x1‖X1

: x = x0 + x1}

, (1.4)

where the infimum extends over all representations x = x0 + x1 of x where x0 ∈ X0

and x1 ∈ X1.

Notice that X0 +X1 and its norm do not depend on the choice of Hausdorff spaceH associated with the couple (X0, X1).

Theorem 1.19. If (X0, X1) is a compatible couple of Banach function spaces, thenX0 +X1 is Banach function space under the norm (1.4).


Definition 1.14. Let (X0, X1) be a compatible couple of Banach function spaces.The K-functional is defined for each f ∈ X0 +X1 and t > 0 by

K (f, t;X0, X1) = inf{

‖f0‖X0+ t ‖f1‖X1

: f = f0 + f1}

,

where the infimum extends over all representations f = f0+ f1 of f with f0 ∈ X0 andf1 ∈ X1.

Definition 1.15. Let X be an r.i. space. For each t belonging to I, let E be a subsetof I with λ(E) = t and let

ϕX(t) = ‖χE‖X .

The function ϕX so defined is called the fundamental function of X.

Remark 1.20. Observe that the particular choice of set E with λ(E) = t is immaterialsince if F is any subset of I with λ(F ) = t, then χE and χF are equimeasurable and so‖χE‖X = ‖χF ‖X because of rearrangement-invariance of X. Hence, ϕX is well defined.

Preliminaries 10

Proposition 1.21. Let X be an r.i. space. Then the fundamental function ϕX of Xsatisfies:

ϕX is nondecreasing on I; ϕX(t) = 0 if and only if t = 0;

ϕX(t)/t is nonincreasing on (0, R];

ϕX is continuous, except perhaps at the origin.

Proof. See [1, Chapter 2, Corollary 5.3].

The K-functional for a pair which consists of an arbitrary r.i. space X and L∞ canbe equivalently expressed by the norm of X in the following way.

Theorem 1.22. Let X be an r.i. space and suppose f belongs to P(I). Then

K(f, t;X,L∞) ≈∥

∥

∥χ(0,ϕ−1

X (t))f∗∥

∥

∥

X,

for all t ∈ ϕX(I).

Remark 1.23. The fundamental function ϕX do not have to be strictly monotonous,so ϕ−1

X is the generalized inverse of ϕX , defined as

ϕ−1X (t) = sup {x ∈ I;ϕX(x) = t} , t ∈ ϕX(I).

Notice that the following statements hold:

ϕX(

ϕ−1X (t)

)

= t, t ∈ ϕX(I)

and

ϕ−1X (ϕX(t)) ≥ t, t ∈ I.

Hence,

t =∥

∥

∥χ(0,ϕ−1

X (t))

∥

∥

∥

X, t ∈ ϕX(I). (1.5)

Proof of Theorem 1.22. Fix f in P(I) and t > 0 and let αt denote the K-functionalon the left side.

In the first step we show that∥

∥

∥χ(0,ϕ−1

X (t))f∗∥

∥

∥

X≤ αt. (1.6)

We can assume that f belongs to X + L∞ = X. Otherwise the infimum αt would beinfinite and there would be nothing to prove. The spaces X and X + L∞ are indeedthe same because we have L∞ → X, which follows from (P6) in Definition 1.3 andProposition 1.10. Assume that f = g + h with g in X and h in L∞. Then

∥

∥

∥χ(0,ϕ−1

X (t))f∗∥

∥

∥

X≤∥

∥

∥χ(0,ϕ−1

X (t))g∗∥

∥

∥

X+∥

∥

∥χ(0,ϕ−1

X (t))h∗∥

∥

∥

X

≤ ‖g∗‖X + ‖h‖L∞

∥

∥

∥χ(0,ϕ−1

X (t))

∥

∥

∥

X

= ‖g‖X + ϕX(

ϕ−1X (t)

)

‖h‖L∞ = ‖g‖X + t ‖h‖L∞ .

Preliminaries 11

The first inequality holds due to Proposition 1.11. Indeed, we know that for u, v ∈ P(I)it holds

∫ t

0(u+ v)∗(s) ds ≤

∫ t

0u∗(s) ds+

∫ t

0v∗(s) ds, t ∈ I,

so

∫ R

0(u+ v)∗(s)w∗(s) ds ≤

∫ R

0u∗(s)w∗(s) ds+

∫ R

0v∗(s)w∗(s) ds,

where w ∈ M0(I).Hence, using Remark 1.6, we get

∥

∥χ(0,a)(u+ v)∗∥

∥

X= sup

{∫ a

0(u+ v)∗(x)w∗(x) dx; ‖w∗‖X′ ≤ 1

}

≤ sup

{∫ a

0u∗(x)w∗(x) dx+

∫ a

0v∗(x)w∗(x) dx; ‖w∗‖X′ ≤ 1

}

≤ sup

{∫ a

0u∗(x)w∗(x) dx; ‖w∗‖X′ ≤ 1

}

+ sup

{∫ a

0v∗(x)w∗(x) dx; ‖w∗‖X′ ≤ 1

}

=∥

∥χ(0,a)u∗∥

∥

X+∥

∥χ(0,a)v∗∥

∥

X,

for any a ∈ I. Taking the infimum over all possible representations f = g + h, weobtain (1.6).

Next, we will prove

αt ≤ 2∥

∥

∥χ(0,ϕ−1

X (t))f∗∥

∥

∥

X.

It will suffice to find g in X and h in L∞ such that f = g + h and

‖g‖X + t ‖h‖L∞ ≤ 2∥

∥

∥χ(0,ϕ−1

X (t))f∗∥

∥

∥

X. (1.7)

Fix and let∥

∥

∥χ(0,ϕ−1

X (t))f∗

∥

∥

∥

Xbe finite. Then for each subset E of I with λ(E) = ϕ−1

X (t)

we have

‖χEf‖X ≤∥

∥

∥χ(0,ϕ−1

X (t))f∗∥

∥

∥

X<∞,

so the function

g(x) = max{f(x)− f∗(

ϕ−1X (t)

)

, 0}

belongs to X, because λ{x : f(x) > f∗(

ϕ−1X (t)

)

} ≤ ϕ−1X (t), and the function

h(x) = min{f(x), f∗(

ϕ−1X (t)

)

}

Preliminaries 12

belongs to L∞ with the L∞-norm at most f∗(

ϕ−1X (t)

)

. Hence, using (1.5)

‖g‖X + t ‖h‖L∞ = ‖g∗‖X +∥

∥

∥χ(0,ϕ−1

X (t))

∥

∥

∥

X‖h‖L∞

≤∥

∥

∥χ(0,ϕ−1

X (t))f∗∥

∥

∥

X+∥

∥

∥χ(0,ϕ−1

X (t))f∗(

ϕ−1X (t)

)

∥

∥

∥

X

≤ 2∥

∥

∥χ(0,ϕ−1

X (t))f∗∥

∥

∥

X.

Since f = g + h, the proof is complete.

Chapter 2

A weighted Hardy operator

In the present chapter we will introduce a weighted Hardy operator Hϕ and itsdual operator H ′

ϕ for which we will find some mapping properties and also constructan optimal pair of r.i. spaces XD and XR such that

Hϕ : XD −→ XR.

Definition 2.1. Let ϕ be a nonnegative, strictly increasing function defined on the in-terval I. If

(i) ϕ(0) = 0;

(ii) ϕ(t)/t is nonincreasing on (0, R],

then ϕ is said to be quasiconcave.

Definition 2.2. Let ϕ be a continuous, quasiconcave function on I. Then we definethe weighted Hardy operators Hϕ and H ′

ϕ by

(Hϕg)(t) :=

∫ R

tϕ(s)g(s)

ds

s,

(H ′ϕg)(t) :=

ϕ(t)

t

∫ t

0g(s) ds,

where g ∈ P(I), t ∈ I.

Remark 2.1. There is a certain duality between H ′ϕ and Hϕ. Indeed, using the Fubini

theorem, we get for f, g ∈ P(I)

∫ R

0f(t)H ′

ϕ g(t) dt =

∫ R

0f(t)

ϕ(t)

t

∫ t

0g(s) ds dt =

∫ R

0

∫ t

0f(t)

ϕ(t)

tg(s) ds dt

=

∫ R

0

∫ R

sf(t)

ϕ(t)

tg(s) dt ds =

∫ R

0g(s)

∫ R

sf(t)

ϕ(t)

tdt ds

=

∫ R

0g(s)Hϕ f(s) ds.

Applying this, we obtain the following simple result.

Lemma 2.2. The operator H ′ϕ is dual to the operator Hϕ in a sense that, for X,Y

two r.i. spaces and for f, g ∈ P(I)

||Hϕ f ||Y . ||f ||X if and only if ||H ′ϕ g||X′ . ||g||Y ′ .

A weighted Hardy operator 14

Proof. Let f, g ∈ P(I). We know that the norm in the associate space is given by

||g||X′ = sup

{∣

∣

∣

∣

∫ R

0f(t)g(t) dt

∣

∣

∣

∣

: f ∈ X, ||f ||X ≤ 1

}

.

We use the duality from Remark 2.1 and the Holder inequality (1.3) to prove the secondstatement from the first one:

||H ′ϕ g||X′ = sup

{∣

∣

∣

∣

∫ R

0f(t)H ′

ϕ g(t) dt

∣

∣

∣

∣

: f ∈ X, ||f ||X ≤ 1

}

= sup

{∣

∣

∣

∣

∫ R

0Hϕ f(t)g(t) dt

∣

∣

∣

∣

: f ∈ X, ||f ||X ≤ 1

}

≤ sup{||g||Y ′ ||Hϕ f ||Y : f ∈ X, ||f ||X ≤ 1}

. sup{||g||Y ′ ||f ||X : f ∈ X, ||f ||X ≤ 1} = ||g||Y ′ .

Proof of the reverse implication is similar.

Lemma 2.3. Let ϕ be defined as above and, in addition, suppose t/ϕ(t) is absolutelycontinuous on I. Then,

Hϕ : L1 → L1

(

d

dt

t

ϕ(t)

)

and Hϕ : L1

(

ϕ(t)

t

)

→ L∞

and

H ′ϕ : L1 →M 1

(ϕ(t)/t)∗∗and H ′

ϕ :Mϕ → L∞.

Also

H ′ϕ : L1 →M b

1(ϕ(t)/t)∗∗

,

where Xb represents the closure of the bounded functions in an r.i. space X of functionsin P(I).

Remark 2.4. By the symbol ddt

tϕ(t) we mean the derivative of t

ϕ(t) , hence the classof functions which are defined a.e. on I such that

∫ s

0

d

dt

t

ϕ(t)dt =

s

ϕ(s).

Let us remind from the basic course of calculus that the absolute continuity of thefunction t/ϕ(t) is equivalent with the existence of an integrable representative of thisclass, so the absolute continuity of t/ϕ(t) is exactly what we need to guarantee the ex-

istence of the space L1(

ddt

tϕ(t)

)

.


Proof of Lemma 2.3. All assertions except the last one are easy to prove, using justthe Fubini theorem. Assuming f belongs to L1 we get

‖Hϕf‖L1(

ddt

tϕ(t)

) ≤

∫ R

0

∫ R

tϕ(s)|f(s)|

ds

s

(

d

dt

t

ϕ(t)

)

dt

=

∫ R

0

∫ s

0

(

d

dt

t

ϕ(t)

)

dtϕ(s)

s|f(s)| ds =

∫ R

0|f(s)| ds <∞

and

‖Hϕf‖L∞ = sup0<t<R

∣

∣

∣

∣

∫ R

tϕ(s)f(s)

ds

s

∣

∣

∣

∣

≤

∫ R

0ϕ(s)

|f(s)|

sds = ‖f‖

L1(

ϕ(t)t

) .

As for the operator H ′ϕ, we have

∥

∥H ′ϕf∥

∥

M 1(ϕ(t)/t)∗∗

= sup0<t<R

1

(ϕ(s)/s)∗∗(t)

(

ϕ(s)

s

∫ s

0f(z) dz

)∗∗

(t)

≤ sup0<t<R

1

(ϕ(s)/s)∗∗(t)

(

ϕ(s)

s

∫ R

0f(z) dz

)∗∗

(t)

=

∣

∣

∣

∣

∫ R

0f(z) dz

∣

∣

∣

∣

≤ ‖f‖L1 <∞,

and

∥

∥H ′ϕf∥

∥

L∞= sup

0<t<R

∣

∣

∣

∣

ϕ(t)

t

∫ t

0f(s) ds

∣

∣

∣

∣

≤ sup0<t<R

ϕ(t)

t

∫ t

0|f(s)| ds

≤ sup0<t<R

ϕ(t)

t

∫ t

0f∗(s) ds = ‖f‖Mϕ

<∞.

To prove the last assertion, we suppose {fn}n∈N is a sequence of bounded functionsconverging to f in L1. Then the functions H ′

ϕfn are also bounded and

|(H ′ϕfn)(t)| ≤

ϕ(t)

t

∫ t

0|fn(s)| ds ≤ ϕ(R) ‖fn‖L∞ , t ∈ I,

while

|(H ′ϕf)(t)− (H ′

ϕfn)(t)| ≤ϕ(t)

t

∫ t

0|f(s)− fn(s)| ds

≤ϕ(t)

t‖f − fn‖L1 , t ∈ I,

so

(H ′ϕf −H ′

ϕfn)∗∗(t) ≤

(

ϕ(t)

t

)∗∗

‖f − fn‖L1 , t ∈ I,

which means∥

∥H ′ϕf −H ′

ϕfn∥

∥

M 1(ϕ(s)/s)∗∗(t)

≤ ‖f − fn‖L1 → 0, as n→ ∞.


In what follows we will try to determine the optimal pair of the r.i. spaces XD andXR, such that the operator Hϕ is bounded into XR from XD.

Definition 2.3. Let XD and XR be r.i. spaces and suppose T is an operator definedon XD such that

T : XD → XR.

We say that XD and XR form an optimal pair for T if

T : Y → XR implies Y → XD,

and

T : XD → Z implies XR → Z,

when Y and Z are r.i. spaces.

Remark 2.5. In other words, r.i. spaces XD and XR form an optimal pair for the oper-ator T if XD can not be replaced in T : XD → XR by any essentially larger r.i. spaceand XR can not be replaced in T : XD → XR by any essentially smaller r.i. spacepreserving the boundedness of T .

First, we start with the domain X and construct the optimal range XR. Second,we construct the optimal domain XD for XR. Finally, we obtain the optimality of thispair for the operator Hϕ.

In view of Proposition 1.10, we define the r.i. space determined by the norm as-sociated with the norm ‖·‖X using the operator H ′

ϕ. It will be convenient to usethe nonincreasing rearrangement of a function. So the next simple statement will beof use.

Lemma 2.6. Let A,B be Banach function spaces. If B is an r.i. space, then the fol-lowing two statements are equivalent:

(i) ||H ′ϕ g||A . ||g||B, g ∈ P(I),

(ii) ||H ′ϕ g

∗||A . ||g||B, g ∈ P(I).

Proof. In an r.i. space B, one always has ||g∗||B = ||g||B. Thus, the second statementis the reduction of the first one, considering just decreasing functions, so we have (ii)from (i). Conversely, it follows from (1.1) that

∥

∥

∥

∥

ϕ(t)

t

∫ t

0g(s) ds

∥

∥

∥

∥

A

≤

∥

∥

∥

∥

ϕ(t)

t

∫ t

0g∗(s) ds

∥

∥

∥

∥

A

. ||g||B, g ∈ P(I).

Theorem 2.7. Let X be an r.i. space. Defining the functional ‖·‖X′[H′

ϕ]by

‖g‖X′[H′

ϕ]:=∥

∥H ′ϕg

∗∥

∥

X′=

∥

∥

∥

∥

ϕ(t)

t

∫ t

0g∗(s) ds

∥

∥

∥

∥

X′

= ‖ϕg∗∗‖X′


for all g ∈ P(I), we get an r.i. norm and(

X ′[H ′ϕ])′, the associate space of X ′[H ′

ϕ], isthe optimal range space for X with respect to Hϕ, in the sense that every r.i. space Zsatisfying

‖Hϕf‖Z . ‖f‖X , f ∈ P(I); (2.1)

is essentially larger than(

X ′[H ′ϕ])′, which means

‖g‖Z . ‖g‖(X′[H′

ϕ])′ , g ∈ P(I).

Proof. We first show that ‖·‖X′[H′

ϕ]is indeed an r.i. norm.

Properties (P1) and (P2) are satisfied due to the linearity of the integral, the mono-tonicity of ϕ and the properties of the norm ‖·‖X .

The triangle inequality (P3) holds because of the subadditivity of the maximaloperator f 7→ f∗∗, since

‖f + g‖X′[H′

ϕ]= ‖ϕ(f + g)∗∗‖X′ ≤ ‖ϕf∗∗ + ϕg∗∗‖X′ ≤ ‖ϕf∗∗‖X′ + ‖ϕg∗∗‖X′ .

If fn ↑ f a.e., then f∗n ↑ f∗. Using the monotone convergence theorem, we obtainthe pointwise convergence H ′

ϕf∗n ↑ H ′

ϕf∗, which implies ‖fn‖X′[H′

ϕ]↑ ‖f‖X′[H′

ϕ]and we

have (P4).The function ϕ is bounded because it is a continuous function defined on the closed

interval I. In the sequel, we use the boundedness of ϕ:

‖χI‖X′[H′

ϕ]= ‖ϕ(χI)

∗∗‖X′ ≤ ϕ(R) ‖χI‖X′ <∞.

Property (P7) is obvious, because our functional is defined using nonincreasingrearrangement.

To prove (P6), we use property (P6) for the norm ‖·‖X′ and the Fubini theorem.For some positive C which does not depend on the choice of f , one has

∫ R

0ϕ(t)f∗∗(t) dt ≤ C ‖ϕf∗∗‖X′ .

We can express the left side in another way as

∫ R

0ϕ(t)f∗∗(t) dt =

∫ R

0

ϕ(t)

t

∫ t

0f∗(s) ds dt =

∫ R

0f∗(s)

∫ R

s

ϕ(t)

tdt ds.

The function 1tϕ(t) is positive on interval (0, R], so the function φ(s) =

∫ Rs

ϕ(t)t dt

is decreasing, continuous, positive on [0, R] and with φ(R) = 0. It remains to finda constant K such that

∫ R

0f(t) dt =

∫ R

0f∗(t) dt ≤ K

∫ R

0f∗(t)φ(t) dt.


We can assume that φ(0) > 1, otherwise we would just multiply the right side of previ-ous inequality with a suitable constant. Due to the intermediate value property of con-tinuous functions, there is t′ ∈ (0, R) such that φ(t′) = 1. With K = R

t′ = 1+ R−t′

t′ andwith the monotonicity of φ, we are quickly done:

∫ R

0f∗(t) dt =

∫ t′

0f∗(t) dt+

∫ R

t′f∗(t) dt ≤

∫ t′

0f∗(t)φ(t) dt+ f∗(t′)(R− t′)

≤

∫ R

0f∗(t)φ(t) dt+

R− t′

t′

∫ t′

0f∗(t′) dt

≤

∫ R

0f∗(t)φ(t) dt+

R− t′

t′

∫ R

0f∗(t)φ(t) dt

=R

t′

∫ R

0f∗(t)φ(t) dt.

Indeed, we have the third inequality because φ(t) > 1 and f∗(t) > f∗(t′) on (0, t′).The optimality of

(

X ′[H ′ϕ])′

follows from its definition. If Z is an r.i. space suchthat

‖Hϕf‖Z . ‖f‖X , f ∈ P(I);

we have also∥

∥H ′ϕg∥

∥

X′. ‖g‖Z′ ,

using Lemma 2.2. Hence,

‖g‖X′[H′

ϕ]= ‖ϕg∗∗‖X′ . ‖g∗‖Z′ = ‖g‖Z′

and using duality one more time, we are done.

Example 2.8. Let us find the optimal range of operator Hϕ for the Lebesgue spaceLp([0, 1]), where p ∈ (1,∞). Then we have (Lp)′([0, 1]) = Lp

′

([0, 1]), where 1p +

1p′ = 1,

so the norm, which generates the space associated to the desired optimal range spaceis the following

‖g‖Y ′ = ‖ϕg∗∗‖Lp′ ([0,1]) .

In other words, we can express our norm as ‖g‖Y ′ = ‖g‖Γp′(ϕp′ )([0,1]) . We see that

the desired optimal range is the space(

Γp′

(ϕp′

))′

([0, 1]), which was described for ex-

ample in [3] by

‖g‖(Γp′ (ϕp′ ))′

([0,1])=

∫ 1

0

g∗∗(t)p(

∫∞

t

(

ϕ(s)s

)p′

ds

)p−1

dt

t.

It remains to find the optimal domain ZD for a given r.i. space Z with respect

to Hϕ. Since Hϕ : L1 → L1(

ddt

tϕ(t)

)

, we should assume the space L1(

ddt

tϕ(t)

)

is

embedded in the space Z.


Theorem 2.9. Suppose t/ϕ(t) is an absolutely continuous function on (0, R] and ϕ(t)/tis integrable function on (0, R]. Let Z be an r.i. space, such that

Z → L1

(

d

dt

t

ϕ(t)

)

.

Defining the functional ‖‖sZ[Hϕ] by

‖g‖sZ[Hϕ] := suph∼g

‖Hϕh‖Z = suph∼g

∥

∥

∥

∥

∫ R

t

ϕ(s)

sh(s) ds

∥

∥

∥

∥

Z

, h, g ∈ P(I),

we get an r.i. norm on P(I), which is essentially the smallest r.i. norm satisfying

‖Hϕf‖Z . ‖f‖sZ[Hϕ] , f ∈ P(I). (2.2)

Proof. We get the optimality of sZ[Hϕ] and (2.2) immediately from its definition.Indeed, supposing the norm generating an r.i. space Y satisfies

‖Hϕh‖Z . ‖h‖Y , h ∈ P(I),

we get

‖f‖sZ[Hϕ] = suph∼f

‖Hϕh‖Z . suph∼f

‖h‖Y = ‖f‖Y , f ∈ P(I).

As in the previous case properties (P1) and (P2) are obvious.We use Remark 1.18 which asserts the existence of a measure-preserving trans-

formation for every pair of equimeasurable functions such that the first function isa composition of the second one and this transformation. If h ∼ f + g, there ex-ists a measure-preserving transformation σ1 such that h = (f + g) ◦ σ1. We definethe functions hf = f ◦ σ1 and hg = g ◦ σ1, which are equimeasurable with f and g dueto Theorem 1.16 and h = hf + hg.

Now, (P3) follows from

‖f + g‖sZ[Hϕ] = suphf+hg∼f+g

‖Hϕ(hf + hg)‖Z

≤ suphf+hg∼f+g

[

‖Hϕhf‖Z + ‖Hϕhg‖Z]

≤ ‖f‖sZ[Hϕ] + ‖g‖sZ[Hϕ] .

To prove (P4), let us suppose fn ↑ f . Again, h ∼ fn implies the existenceof a measure-preserving transformation σ2 such that h = fn ◦ σ2 ≤ fn+1 ◦ σ2 = k,where k ∼ fn+1. Hence

‖fn‖sZ[Hϕ] = suph∼fn

‖Hϕh‖Z ≤ supk∼fn+1

‖Hϕk‖Z = ‖fn+1‖sZ[Hϕ] .

Next, if σ3 is a measure-preserving transformation such that h = f ◦ σ3, thenhn := fn ◦ σ3 ↑ f ◦ σ3 = h. So (Hϕhn) ↑ (Hϕh), which shows τ(fn) ↑ τ(f).


For (P5) and (P6) we use also the integrability of ϕ(t)/t:

‖χI‖sZ[Hϕ] = ‖HϕχI‖Z =

∥

∥

∥

∥

∫ R

t

ϕ(s)

sds

∥

∥

∥

∥

Z

≤ ‖χI‖Z

∫ R

0

ϕ(s)

sds <∞,

and

‖f‖sZ[Hϕ] ≥

∥

∥

∥

∥

∫ R

tf(s)

ϕ(s)

sds

∥

∥

∥

∥

Z

&

∥

∥

∥

∥

∫ R

tf(s)

ϕ(s)

sds

∥

∥

∥

∥

L1(

ddt

tϕ(t)

)

=

∫ R

0f(t) dt, f ∈ P(I).

Finally, for every pair of equimeasurable functions f, g ∈ P(I) we have the equality‖f‖sZ[Hϕ] = ‖g‖sZ[Hϕ], so (P7) holds.

Combining Theorems 2.7 and 2.9, we get an optimal pair of r.i. function normswith desired properties. Let us summarize these results in the following theorem.

Theorem 2.10. Suppose t/ϕ(t) is an absolutely continuous function on (0, R] andϕ(t)/t is integrable function on (0, R] and let X be an r.i. space. Then r.i. spaces XR

and XD generated by

‖f‖XR := ‖f‖(X′[H′

ϕ])′ , (2.3)

and

‖f‖XD := ‖f‖sXR[Hϕ] , f ∈ P(I), (2.4)

form an optimal pair for the operator Hϕ.

Proof. According to (P6) in Definition 1.3, each r.i. space is a subset of L1, so using

Hϕ : L1 → L1(

ddt

tϕ(t)

)

from Lemma 2.3 together with Hϕ : X → XR, we obtain

that XR is embedded in L1(

ddt

tϕ(t)

)

. Hence, we can use Theorem 2.9 and we get

the optimality of the domain XD in Hϕ : XD → XR.Assume that an r.i. space Z satisfies Hϕ : XD → Z. Since the domain XD is optimal

in Hϕ : XD → XR and Hϕ : X → XR we obtain X → XD. Hence, we get Hϕ : X → Zwhich implies XR → Z, because the r.i. space XR is optimal in Hϕ : X → XR whichfollows from Theorem 2.7.

Chapter 3

Supremum operators

Definition 3.1. Let ϕ be a continuous, quasiconcave function. Then, we definethe supremum operators Tϕ and Sϕ by

(Tϕf)(t) :=1

ϕ(t)supt≤s<R

ϕ(s)f∗(s),

and

(Sϕf)(t) :=ϕ(t)

tsup

0<s≤t

s

ϕ(s)f∗(s) f ∈ P(I), t ∈ I.

Definition 3.2. We say the function ϕ satisfies the B1-condition if

∫ t

0

1

ϕ(s)ds .

t

ϕ(t), t ∈ I,

and the B2-condition if

∫ t

0

ϕ(s)

sds . ϕ(t), t ∈ I.

We can easily show that the supremum operators Tϕ and Sϕ have a property similarto the sublinearity.

Proposition 3.1. The supremum operators Tϕ and Sϕ satisfy

(Tϕ[f + g]) (t) ≤ 2 (Tϕf + Tϕg)

(

t

2

)

, (3.1)

and

(Sϕ[f + g]) (t) ≤ 2 (Sϕf + Sϕg)

(

t

2

)

, f, g ∈ P(I), t ∈ I. (3.2)

Proof. Using the inequality (f+g)∗(s+t) ≤ f∗(s)+g∗(t) from Proposition 1.1 togetherwith the sublinearity of the supremum, we get

(Tϕ[f + g]) (t) =1

ϕ(t)supt≤s<R

ϕ(s)(f + g)∗(s)

≤1

ϕ(t)supt≤s<R

ϕ(s)[

f∗(s

2

)

+ g∗(s

2

)]

≤ supt≤s<R

ϕ(s)

ϕ(t)f∗(s

2

)

+ supt≤s<R

ϕ(s)

ϕ(t)g∗(s

2

)

.

Supremum operators 22

The function ϕ is quasiconcave, so we haveϕ( s2)s2

≥ ϕ(s)s , which implies ϕ(s) ≤ 2ϕ

(

s2

)

and ϕ(s)ϕ(t) ≤ 2

ϕ( s2)ϕ( t2)

, because ϕ is strictly increasing. Hence,

supt≤s<R

ϕ(s)

ϕ(t)f∗(s

2

)

+ supt≤s<R

ϕ(s)

ϕ(t)g∗(s

2

)

≤ 21

ϕ(

t2

)

(

supt≤s<R

ϕ(s

2

)

f∗(s

2

)

+ supt≤s<R

ϕ(s

2

)

g∗(s

2

)

)

≤ 21

ϕ(

t2

)

(

supt2≤s<R

ϕ (s) f∗ (s) + supt2≤s<R

ϕ (s) g∗ (s)

)

= 2 (Tϕf + Tϕg)

(

t

2

)

.

The second inequality can be proved by the same method, namely,

(Sϕ[f + g]) (t) =ϕ(t)

tsup0<s≤t

s

ϕ(s)(f + g)∗(s)

≤ sup0<s≤t

sϕ(t)

tϕ(s)f∗(s

2

)

+ sup0<s≤t

sϕ(t)

tϕ(s)g∗(s

2

)

≤ 2ϕ(

t2

)

t2

(

sup0<s≤t

s2

ϕ(

s2

)f∗(s

2

)

+ sup0<s≤t

s2

ϕ(

s2

)g∗(s

2

)

)

≤ 2 (Sϕf + Sϕg)

(

t

2

)

.

Lemma 3.2. Let ϕ satisfy the B1-condition, then the operator Tϕ satisfies

Tϕ : L1 → L1 and Tϕ :Mϕ →Mϕ. (3.3)

Next, let ϕ satisfy the B2-condition, then the operator Sϕ satisfies

Sϕ :M tϕ(t)

→M tϕ(t)

and Sϕ : L∞ → L∞. (3.4)

Proof. The set of all continuous functions with support in I, C0(I), is dense in L1(I).Hence, it suffices to verify

‖Tϕf‖L1 . ‖f‖L1

for f ∈ C0(I). In this case, we can extend the domain of f∗ on I by

f∗(0) := limt→0+

f∗(t).

Assuming f 6≡ 0, define

(Qf)(t) := supt≤s<R

ϕ(s)f∗(s), t ∈ I,


and set A :={

k ∈ Z+ : (Qf)(

2−kR)

> (Qf)(

21−kR)}

. Then, A is nonempty, becauseif max0<t<R ϕ(t)f

∗(t) is attained in[

2−kR, 21−kR)

, then k ∈ A. Given k ∈ A we define

tk =

{

0 if (Qf)(t) = (Qf)(

2−kR)

, t ∈(

0, 2−kR]

;min

{

2−jR; (Qf)(

2−jR)

= (Qf)(

2−kR)}

otherwise.

Hence,

(Qf)(t) = (Qf)(

2−kR)

, k ∈ A, t ∈[

tk, 2−kR

]

.

By the definition of A, sup2−kR≤s<1 ϕ(s)f∗(s) is attained in

[

2−kR, 21−kR)

when k ∈ A.Thus, for every k ∈ A and t ∈

[

tk, 21−kR

)

, together with the monotocity of (Qf)(t) weget

(Qf)(t) ≤ (Qf)(

2−kR)

= sup2−kR≤s<21−kR

ϕ(s)f∗(s)

≤ ϕ(

21−kR)

f∗(

2−kR)

.

(3.5)

It is obvious that⋃

k∈A

[

tk, 21−kR

)

⊇ (0, R). So, using all this, the monotocity of f∗

and the B1-condition for ϕ we have

‖Tϕf‖L1 =

∫ R

0

(Qf)(t)

ϕ(t)dt ≤

∑

k∈A

∫ 21−kR

tk

(Qf)(t)

ϕ(t)dt

≤∑

k∈A

ϕ(

21−kR)

f∗(

2−kR)

∫ 21−kR

0

1

ϕ(t)dt

.∑

k∈A

21−kRf∗(

2−kR)

= 4∑

k∈A

∫ 2−kR

2−1−kRdtf∗

(

2−kR)

≤ 4∑

k∈A

∫ 2−kR

2−1−kRf∗(t) dt ≤ 4 ‖f‖L1 .

Next, Tϕf is nonincreasing for all f ∈ P(I), so (Tϕf)∗∗(t) = 1

t

∫ t0 (Tϕf)(s) ds. Then,

for every f ∈Mϕ(I) we have

‖Tϕf‖Mϕ= sup

0<t<R{ϕ(t)(Tϕf)

∗∗(t)}

= sup0<t<R

{

ϕ(t)

t

∫ t

0

1

ϕ(s)sup

s≤r<Rϕ(r)f∗(r) ds

}

≤ sup0<t<R

{

ϕ(t)

t

∫ t

0

1

ϕ(s)sup

0<r<Rϕ(r)f∗∗(r) ds

}

= ‖f‖Mϕsup

0<t<R

{

ϕ(t)

t

∫ t

0

1

ϕ(s)ds

}

,

but the last supremum is finite because of the B1-condition for ϕ.


Next, suppose f belongs to L∞. We have

‖Sϕf‖L∞ = ess sup0<t<R

ϕ(t)

tsup0<s≤t

s

ϕ(s)f∗(s)

≤ ‖f‖L∞ ess sup0<t<R

sup0<s≤t

ϕ(t)

t

s

ϕ(s)= ‖f‖L∞ ,

because sϕ(s) is nondecreasing, so the supremum is attained in t.

Finally, we will observe that, for every g ∈ P(I) and t ∈ I,

sup0<s≤t

s

ϕ(s)g∗(s) ≈ sup

0<s≤t

s

ϕ(s)g∗∗(s). (3.6)

Indeed, setting

M = sup0<s≤t

s

ϕ(s)g∗(s),

we get

g∗(s) ≤Mϕ(s)

s, s ∈ (0, t].

Using B2-condition for ϕ, we obtain

g∗∗(y) =1

y

∫ y

0g∗(s) ds ≤

M

y

∫ y

0

ϕ(s)

sds .

M

yϕ(y), y ∈ (0, t],

and, therefore,

sup0<y≤t

y

ϕ(y)g∗∗(y) .M, t ∈ I.

The second inequality is obvious since g∗ ≤ g∗∗.Applying (3.6), we get

‖Sϕf‖M tϕ(t)

= sup0<t≤R

t

ϕ(t)

(

ϕ(y)

ysup

0<s<y

s

ϕ(s)f∗(s)

)∗∗

(t)

. sup0<t≤R

t

ϕ(t)

(

ϕ(y)

ysup

0<s<y

s

ϕ(s)f∗(s)

)∗

(t)

≤ ‖f‖M tϕ(t)

sup0<t<R

t

ϕ(t)

(

ϕ(y)

y

)∗

(t) = ‖f‖M tϕ(t)

,

using again that ϕ(t)t is nonincreasing.

In the following part of this section we introduce some further properties of the op-erators Sϕ and Tϕ.


Lemma 3.3. Let (X0, X1) and (Y0, Y1) be a compatible couples of Banach functionspaces. Then, for every operator T satisfying T : X0 → Y0, T : X1 → Y1 and

(T [f + g]) (t) ≤ 2 (Tf + Tg)

(

t

2

)

, f, g ∈ P(I), t ∈ I,

we have

K (Tf, t;Y0, Y1) . K (f, t;X0, X1) , f ∈ X0 +X1, t > 0.

Proof. It follows from the embeddings that we have two positive constants Ci, i = 0, 1,such that

‖Tf‖Yi ≤ Ci ‖f‖Xi , f ∈ P(I), i = 0, 1.

Fix f ∈ X0 + X1 and assume that f = f0 + f1 where fi ∈ Xi, i = 0, 1. Then,by decomposing Tf in an appropriate way, we get

Tf = T [f0 + f1] ≤ 2E2Tf0 + 2E2Tf1,

so Tf − 2E2Tf0 ≤ 2E2Tf1. Note that, according to Proposition 1.14, we have‖E2f‖X ≤ 2 ‖f‖X , because

1

‖f‖X‖E2f‖X =

∥

∥

∥

∥

1

‖f‖XE2f

∥

∥

∥

∥

X

=

∥

∥

∥

∥

E2f

‖f‖X

∥

∥

∥

∥

X

≤ 2, f ∈ P(I). (3.7)

Applying this and the sublinearity of T , we get

‖2E2Tf0‖Y0 ≤ 4 ‖Tf0‖Y0 ≤ 4C0 ‖f0‖X0,

and similarly

‖Tf − 2E2Tf0‖Y1 ≤ ‖2E2Tf1‖Y1 ≤ 4C1 ‖f1‖X1.

Finally,

K (Tf, t;Y0, Y1) ≤ ‖2E2Tf0‖Y0 + t ‖Tf − 2E2Tf0‖Y1≤ 4max{C0, C1}

(

‖f0‖X0+ t ‖f1‖X1

)

.

Taking the infimum over all possible representations f = f0 + f1, we obtain the re-quired inequality.

Theorem 3.4. Suppose that T is an operator defined on M tϕ(t)

satisfying

(T [f + g]) (t) ≤ 2 (Tf + Tg)

(

t

2

)

, f, g ∈ P(I), t ∈ I,

as well as

T :M tϕ(t)

→M tϕ(t)

and T : L∞ → L∞. (3.8)


Then,

(Tf)∗∗(t) . (Sϕf∗∗)(t) . (Sϕ)(t), f ∈ P(I), t ∈ I. (3.9)

Moreover, if ϕ satisfies B2-condition, we have

(Sϕf)∗∗(t) ≈ (Sϕf

∗∗)(t) ≈ (Sϕf)(t), f ∈ P(I), t ∈ I. (3.10)

Remark 3.5. We denote tϕ(t) by ψ(t). Notice that every nonincreasing g ∈ P(I) satisfies

sup0<s≤t

ψ(s)g∗∗(s) = sup0<s<R

(

χ(0,t)g)∗∗

(s)ψ(s), t ∈ (0, R]. (3.11)

Indeed, since

(

χ(0,t)g)∗∗

(s)ψ(s) =1

s

∫ s

0

(

χ(0,t)g)∗

(y) dys

ϕ(s)=

∫ t

0

(

χ(0,t)g)∗

(y) dy1

ϕ(s)

when s ∈ [t, R) and 1ϕ(s) is strictly decreasing, we have that

sup0<s<R

(

χ(0,t)g)∗∗

(s)ψ(s) = sup0<s≤t

(

χ(0,t)g)∗∗

(s)ψ(s).

This yields (3.11) because

(

χ(0,t)g)∗∗

(s) = g∗∗(s), s ∈ (0, t].

Proof of Theorem 3.4. Using (3.8) and Lemma 3.3 we get the following K-functionalinequality

K (Tf, t;Mψ, L∞) . K (f, t;Mψ, L

∞) , t ∈ I.

We may assume C > 1 and replace t with ψ(t) in this inequality, and we get

K (Tf, ψ(t);Mψ, L∞) . K (f, ψ(t);Mψ, L

∞) , t ∈ I.

Then, together with Theorem 1.22, the equality (3.11) in the case when g = (Tf)∗ andthe fact that the fundamental function of Mψ is

ϕMψ(t) = ψ(t),

we get

sup0<s≤t

ψ(s)(Tf)∗∗(s) =∥

∥χ(0,t)(Tf)∗∥

∥

Mψ≤

∥

∥

∥

∥

∥

χ(

0,ϕ−1Mψ

(ψ(t))

)(Tf)∗

∥

∥

∥

∥

∥

Mψ

≈ K (Tf, ψ(t);Mψ, L∞) . K (f, ψ(t);Mψ, L

∞)

≈

∥

∥

∥

∥

∥

χ(

0,ϕ−1Mψ

(ψ(t))

)f∗

∥

∥

∥

∥

∥

Mψ

= sup0<s≤ϕ−1

Mψ(ψ(t))

ψ(s)f∗∗(s), t ∈ I.


Hence,

ψ(t)(Tf)∗∗(t) . sup0<s≤ϕ−1

Mψ(ψ(t))

ψ(s)f∗∗(s),

or

(Tf)∗∗(t) .1

ψ(t)

ϕ−1Mψ

(ψ(t))

ϕ(

ϕ−1Mψ

(ψ(t))) (Sϕf

∗∗)(

ϕ−1Mψ

(ψ(t)))

= (Sϕf∗∗)(

ϕ−1Mψ

(ψ(t)))

≤ (Sϕf∗∗) (t),

since Sϕf∗∗ is nonincreasing and we have Remark 1.23. The first part of (3.9) is proved

and the second part follows readily from (3.6).Now, in view of (3.2) and (3.4) the operator Sϕ satisfies all the assumptions neces-

sary to replace T by Sϕ, so (3.9) also holds when T = Sϕ.Finally, it follows from (3.6) and the monotocity of Sϕf

∗∗ that

(Sϕf∗∗) (t) ≤ (Sϕf

∗∗)∗∗ (t) ≈ (Sϕf)∗∗ (t), t ∈ I,

so we have (3.10).

Theorem 3.6. Let Tϕ be the supremum operator defined in Definition 3.1 and let ϕsatisfy the B1-condition. Then, for all f ∈ P(I) and t ∈ I,

(Tϕf)∗∗(t) . (Tϕf

∗∗)(t). (3.12)

Proof. For an arbitrary pair of f ∈ P(I) and t ∈ I, set

ft(s) = min[f(s), f∗(t)]

and

f t(s) = max[f(s)− f∗(t), 0], s ∈ I.

Then

f(s) = ft(s) + f t(s),

f∗(s) = f∗t (s) + (f t)∗(s), s ∈ I,

where

f∗t (s) = min[f∗(s), f∗(t)],

(f t)∗(s) = max[f∗(s)− f∗(t), 0] = (f∗(s)− f∗(t))χ(0,t), s ∈ I.

Since (Tϕf)(t) is nonincreasing in t

(Tϕf)∗∗(t) =

1

t

∫ t

0

1

ϕ(s)sup

s≤r<Rϕ(r)f∗(r) ds

≤1

t

∫ t

0

1

ϕ(s)sup

s≤r<Rϕ(r)f∗t (r) ds

+1

t

∫ t

0

1

ϕ(s)sup

s≤r<Rϕ(r)(f t)∗(r) ds

= I + II.


Next, using the B1-condition for ϕ, we get

I =1

t

∫ t

0

1

ϕ(s)sup

s≤r<Rϕ(r)min[f∗(r), f∗(t)] ds

=1

t

∫ t

0

1

ϕ(s)max

[

supt≤r<R

ϕ(r)f∗(r), sups≤r<t

ϕ(r)f∗(t)

]

ds

=1

t

∫ t

0

1

ϕ(s)supt≤r<R

ϕ(r)f∗(r) ds =1

t

∫ t

0

1

ϕ(s)ds sup

t≤r<Rϕ(r)f∗(r)

.1

ϕ(t)supt≤r<R

ϕ(r)f∗(r) = (Tϕ)(t)

and

II =1

t

∫ t

0

1

ϕ(s)sup

s≤r<Rϕ(r)(f t)∗(r) ds =

1

t

∫ t

0(Tϕf

t)(s) ds

≤1

t

∫ R

0(Tϕf

t)(s) ds.

We use Tϕ : L1 → L1,

1

t

∫ R

0(Tϕf

t)(s) ds .1

t

∫ R

0(f t)∗(s) ds =

1

t

∫ t

0[f∗(s)− f∗(t)] ds . f∗∗(t).

Combining these two results and

f∗∗(t) = supt≤s<R

f∗∗(s) =1

ϕ(t)ϕ(t) sup

t≤s<Rf∗∗(s) ≤

1

ϕ(t)supt≤s<R

ϕ(s)f∗∗(s),

we obtain

(Tϕf)∗∗(t) . [f∗∗(t) + (Tϕ)(t)] . (Tϕf

∗∗)(t), f ∈ P(I), t ∈ I. (3.13)

Now we are in a position to state and prove the main result of this section.

Theorem 3.7. Let ϕ satisfy the B2-condition. Then there exists C > 0 such that, forany r.i. space X and f in P(I),

∥

∥

∥

∥

∥

supt≤s<R

ϕ(s)f∗∗(s)

∥

∥

∥

∥

∥

X

≤ C ‖ϕ(t)f∗∗(t)‖X . (3.14)

Proof. According to Theorem 1.12 and (3.7) we have g ≺ h implies ‖g‖X ≤ C ‖h‖Xand ‖E2f‖X ≤ 2 ‖f‖X . Thus, it suffices to find C > 0, independent of f ∈ P(I) and ton some right neighbourhood of 0, say, t ∈

(

0, R3)

, for which

∫ t

0sup

s≤z<Rϕ(z)f∗∗(z) ds ≤ C

∫ t

0(ϕ(z)f∗∗(z))∗

(s

2

)

ds. (3.15)


Fix f ∈ P(I) and t ∈(

0, R3)

and take ft and ft from the proof of previous Theorem.

We use the decomposition of f∗∗ on f∗∗t and (f t)∗∗, so we have

sups≤z<R

ϕ(z)f∗∗(z) = sups≤z<R

ϕ(z)[f∗∗t (z) + (f t)∗∗(z)]

≤ sups≤z<R

ϕ(z)f∗∗t (z) + sups≤z<R

ϕ(z)(f t)∗∗(z).

Now, we estimate each of these summands. Firstly, we rewrite the functions f∗∗tand (f t)∗∗ to see what are their forms on (0, t) and (t, R):

f∗∗t (z) =

{

f∗(t), 0 < z < t;tzf

∗(t) + 1z

∫ zt f

∗(y) dy, t ≤ z < R;

(f t)∗∗(z) =

{

f∗∗(z)− f∗(t), 0 < z < t;tz [f

∗∗(t)− f∗(t)], t ≤ z < R.

Hence, for 0 < s ≤ t, using the monotonicity of ϕ(t)/t we get

sups≤z<R

ϕ(z)(f t)∗∗(z) = max

[

sups≤z<t

ϕ(z)[f∗∗(z)− f∗(t)],

supt≤z<R

ϕ(z)t

z[f∗∗(t)− f∗(t)]

]

≤ max

[

sups≤z<t

ϕ(z)f∗∗(z), supt≤z<R

ϕ(z)

ztf∗∗(t)

]

≤ sups≤z≤t

ϕ(z)f∗∗(z),

so using once more time the monotonicity of ϕ(t)/t and the B2-condition for ϕ, wehave

∫ t

0sup

s≤z<Rϕ(z)(f t)∗∗(z) ds ≤

∫ t

0sups≤z≤t

ϕ(z)

z

∫ z

0f∗(y) dy ds

≤

(∫ t

0f∗(y) dy

)(∫ t

0sups≤z≤t

ϕ(z)

zds

)

=

(∫ t

0f∗(y) dy

)(∫ t

0

ϕ(s)

sds

)

≤ Cϕ(t)

∫ t

0f∗(s) ds.

(3.16)

The monotonicity of g∗ implies∫ t0 g

∗ ≥∫ 2tt g and clearly g

(

·2

)∗(s) = g∗

(

s2

)

. Sinceϕ(t)/t is nonincreasing,

ϕ(2t)

ϕ(t)≤ 2. (3.17)


All this together with the monotonicity of f∗∗ asserts

ϕ(t)

∫ t

0f∗(s) ds ≤ 2ϕ

(

t

2

)

tf∗∗(t) = 2

∫ 2t

tϕ

(

t

2

)

f∗∗(

2t

2

)

ds

≤ 2

∫ 2t

tϕ(s

2

)

f∗∗(s

2

)

ds ≤ 2

∫ t

0(ϕ(z)f∗∗(z))∗

(s

2

)

ds,

which together with (3.16) yields

∫ t

0sup

s≤z<Rϕ(z)(f t)∗∗(z) ds ≤ 2C

∫ t

0(ϕ(z)f∗∗(z))∗

(s

2

)

ds.

Now, we estimate the second summand, that is,

∫ t

0sup

s≤z<Rϕ(z)f∗∗t (z) ds ≤ C

∫ t

0(ϕ(z)f∗∗(z))∗

(s

2

)

ds.

For an arbitrary s ∈ (0, t) we have

sups≤z<R

ϕ(z)f∗∗t (z) = max

[

sups≤z<t

ϕ(z)f∗∗t (z), supt≤z<R

ϕ(z)f∗∗t (z)

]

≤ max

[

ϕ(t)f∗(t), supt≤z<R

[

ϕ(z)

ztf∗(t) +

ϕ(z)

z

∫ z

tf∗(y) dy

]

]

≤ max

[

ϕ(t)f∗(t), ϕ(t)f∗(t) + supt≤z<R

ϕ(z)

z

∫ z

tf∗(y) dy

]

= ϕ(t)f∗(t) + supt≤z<R

ϕ(z)

z

∫ z

tf∗(y) dy

≤ 3 supt≤z<R

ϕ(z)

z

∫ z

tf∗(y) dy,

since ϕ is nondecreasing, ϕ(t)t is nonincreasing and

ϕ(t)f∗(t) ≤ 2ϕ(2t)

2ttf∗(t) = 2

ϕ(2t)

2t

∫ 2t

tf∗(t) dy

≤ 2ϕ(2t)

2t

∫ 2t

tf∗(y

2

)

dy ≤ 2 supt≤z<R

ϕ(z)

z

∫ z

tf∗(y) dy.

Thus,

∫ t

0sup

s≤z<Rϕ(z)f∗∗t (z) ds ≤ 3t sup

t≤z<R

ϕ(z)

z

∫ z

tf∗(y) dy.

It will suffice to prove that

supt≤z<R

ϕ(z)

z

∫ z

tf∗(y

2

)

dy ≤ Cϕ(s)

s

∫ s

0f∗(y

2

)

dy


on a set E of measure at least t, (for all s ∈ E), since using this inequality, the mono-tonicity of f∗, (3.17) and (1.1), we get

3t supt≤z<R

ϕ(z)

z

∫ z

tf∗(y) dy ≤ 3

∫

Esupt≤z<R

ϕ(z)

z

∫ z

tf∗(y) dy ds

≤ 3

∫

Esupt≤z<R

ϕ(z)

z

∫ z

tf∗(y

2

)

dy ds

≤ 3C

∫

E

ϕ(s)

s

∫ s

0f∗(y

2

)

dy ds

≤ 6C

∫

Eϕ(s

2

)

f∗∗(s

2

)

ds ≤ C ′

∫ t

0(ϕ(z)f∗∗(z))∗

(s

2

)

ds.

Next suppose z0 ∈ (t, R] is such that

supt≤z<R

ϕ(z)

z

∫ z

tf∗(y

2

)

dy =ϕ(z0)

z0

∫ z0

tf∗(y

2

)

dy.

This supremum is attained in (t, R] because the function inside is positive, continuouson [t, R] and vanishing for z = t. We consider two cases.

Let z0 ∈ (t, 2t). Since f∗ is nonincreasing,

∫ z0

0f∗(y

2

)

dy ≥ 2

∫ z0

tf∗(y

2

)

dy.

Thus, for s ∈ (z0, z0 + t), we have

ϕ(s)

s

∫ s

0f∗(y

2

)

dy ≥ϕ(z0 + t)

z0 + t

∫ z0

0f∗(y

2

)

dy

≥ϕ(2z0)

2z02

∫ z0

tf∗(y

2

)

dy ≥ϕ(z0)

z0

∫ z0

tf∗(y

2

)

dy

= supt≤z<R

ϕ(z)

z

∫ z

tf∗(y

2

)

dy,

since ϕ(t)/t is nonincreasing and ϕ(t) is increasing. In this case, we set E = (z0, z0+t).Let z0 ∈ [2t, R]. Similarly, for s ∈ (z0 − t, z0), we have

ϕ(s)

s

∫ s

0f∗(y

2

)

dy ≥ϕ(z0)

z0

∫ z0−t

0f∗(y

2

)

dy

≥1

2

ϕ(z0)

z0

∫ z0

0f∗(y

2

)

dy ≥1

2

ϕ(z0)

z0

∫ z0

tf∗(y

2

)

dy

=1

2supt≤z<R

ϕ(z)

z

∫ z

tf∗(y

2

)

dy.

So, in this case it suffices to set E = (z0− t, z0). Choosing maximal constant among allthose which we have used, each inequality holds and the proof is complete.

Bibliography

[1] Bennett C., Sharpley R.: Interpolation of Operators. Pure and Applied Mathe-matics 129. Academic Press, Boston 1988.

[2] Kerman R., Pick L.: Optimal Sobolev imbeddings. Forum Math., 2006, vol. 18,no. 4, p. 535–570.

[3] Gogatishvili A., Pick L.: Discretization and antidiscretization of rearrangement-invariant norms. Publ. Mat., 2003, vol. 47, no. 2, p. 311–358.

[4] Ryff J.V.: Measure preserving transformations and rearrangements. J. Math.Anal. Appl., 1970, vol. 31, no. 3, p. 449–458.

c:/documents and settings/ras o/plocha/diplomka/diplomka bez …rokyta/vyuka/mgr... ·...

Documents