ORLICZ FUNCTION SPACES ANDCOMPOSITION OPERATOR

A Project Report Submitted

in Partial Fulfilment of the Requirements

for the Degree of

MASTER OF SCIENCE

In Mathematics

by

Chinmay kumar Giri

(Roll Number: 411MA2075)

to the

DEPARTMENT OF MATHEMATICS

National Institute Of Technology RourkelaOdisha - 768009

MAY, 2013

DECLARATION

I hereby declare that the project report entitled “ORLICZ FUNCTION SPACES AND

COMPOSITION OPERATOR” submitted for the M.Sc. Degree is a review work car-

ried out by me and the project has not formed the basis for the award of any Degree,

Associateship, Fellowship or any other similar titles.

Place:Date: Chinmay Kumar Giri

Roll No: 411ma2075

ii

iii

CERTIFICATE

This is to certify that the work contained in this report entitled “ORLICZ FUNCTION

SPACES AND COMPOSITION OPERATOR” submitted by Chinmay Kumar

Giri (Roll No: 411MA2075.) to Department of Mathematics, National Institute of Tech-

nology Rourkela for the partial fulfilment of requirements for the degree of master of science

in Mathematics towards the requirement of the course MA592 Project is a bonafide record

of review work carried out by him under my supervision and guidance. The contents of this

project, in full or in parts, have not been submitted to any other institute or university for

the award of any degree or diploma.

May, 2013Dr.S Pradhan

Assistant ProfessorDepartment of Mathematics

NIT Rourkela

Acknowledgement

I would like to thank Dr. S Pradhan for the inspiration, support and guidance he has

given me during the course of this project.

I would like to thank the faculty members of Department of Mathematics for allowing me

to work for this Project in the computer laboratory and for their cooperation. I would like

to thanks to my seniors Ratan Kumar Giri, Karan Kumar Pradhan and Bibekananda Bira,

research scholars, for his timely help during my work.

My heartfelt thanks to all my friends for their invaluable co-operation and constant inspira-

tion during my Project work.

I owe a special debt gratitude to my revered parents, my brother, sister for their blessings

and inspirations.

Place:Date: Chinmay Kumar Giri

Roll No: 411ma2075

iv

Contents

1 Introduction 2

2 Orlicz Function Spaces 4

2.1 Orlicz Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

3 Composition Operators On Orlicz Spaces 15

3.1 Modular and norm continuity of composition operators . . . . . . . . . . . . 20

3.2 Compact Composition Operators in Orlicz space . . . . . . . . . . . . . . . . 23

v

1

English Symbols

N set of natural number .R set of real number .C set of complex number .K field of scalars i.e. either C or R .X vector space over the field K.Σ sigma algebraµ measure defined over Σ.Ω σ−finite complete measure space.Φ Young’s function.∥.∥Φ inf λ > 0 :

∫Ω

Φ(|xλ|)dµ ≤ 1.

LΦ(Ω) Orlicz function space.τ nonsingular measurable transformation from Ω to itself.Cτ composition operator formLΦ(Ω) to itself generated by τ .ν ≪ µ ν is absolutely continuous with respect to µ.∥.∥e essential norm of a bounded linear operator.

Chapter 1

Introduction

Orlicz spaces have their origin in the Banach space researches of 1920. Indeed, after the de-

velopment of Lebesgue theory of integration and inspired by the function tp in the definitions

of the spaces lp and Lp, Orlicz spaces were first proposed by Z.W.Birnbaum and W.Orlicz

in[1] and latter developed by Orlicz himself in[7], [8]. The study and applications of this

theory was picked up again in Poland, USSR and Japan after the war years. Around the

year 1950, H.Nakano [6] studied Orlicz spaces with the name “modulared spaces”. However,

the theory became popular for researches in the western countries after the publication of the

book on “Linear Analysis” by A.C.Zaanen. This possibly resulted in the translation of the

monograph of M.A.Krasnoselskii and Ya.B.Rutickii on Convex Function and Orlicz Spaces

by Leo F. Boron from Russian to English, and after the appearance of English version of

this book in 1961, the theory has been effectively used in many branches of Mathematics

and Statistics e.g, differential and integral equations, harmonic analysis, probability etc.

Prior to the researches of W.Orlicz ,it was W.H.Young [12] who, motivated by the functions

up(u > 0) and vp(v > 0) with 1p

+ 1q

= 1 ,1 < p, q < ∞, introduced a function v = Φ(u) for

u ≥ 0 such that Φ is continuous and strictly increasing with Φ(0) = 0 and Φ(u) → ∞ as

u → ∞. if u = Ψ(v) is the inverse of Φ, he defined

Φ(a) =∫ a

0Φ(u)du , Ψ(b) =

∫ b

0Ψ(v)dv

2

3

for a, b ≥ 0. These functions are known as Young’s function in the literature, and besides

being convex, satisfy the Young’s inequality

ab ≤ Φ(a) + Ψ(b)

for a, b ≥ 0. Young introduced the classes YΦ and YΨ consisting of measurable functions f

for which∫

Φ(|f(x)|)dx < ∞ and∫

Ψ(|f(x)|)dx < ∞, respectively. These spaces failed to

form the vector space. However, if satisfies ∆2−condition in the sense that there exists a

constant C > 0 such that Φ(2u) ≤ CΦ(u) hold for all u ≥ 0, YΦ becomes a vector space. In

the process of norming the spaces YΦ, YΨ, Orlicz considered the class LΦ of all measurable

functions f satisfying

∥f∥Φ = sup∫|fg|dx :

∫Ψ(|g|)dx ≤ 1 < ∞,

and proved that (LΦ, ∥∥Φ) is a normed linear space. In general, YΦ ⊂ LΦ, however, if Φ

satisfies ∆2−condition defined as above, YΦ = LΦ, cf.[9], [10].

In Mathematics, the composition operator CΦ with symbol Φ is a linear operator defined with

the help of composition of mapping f Φ by the formula CΦ(f) = f Φ. Most of the recent

interest in composition operators arises from the study of boundedness, compactness of these

operators (see for example [10],[11]). In analysis this operator has of lost of connection with

the Hardy space, space of analytic functions, Lp spaces, for p ≥ 1.

The material is divided into three chapters. In chapter 2, the basic theory of Orlicz spaces

is presented. Next, chapter 3 contains some results of composition operator on Orlicz spaces

i.e. it’s boundedness and compactness.

Chapter 2

Orlicz Function Spaces

This chapter includes some basic definitions and results which have been used in the next

chapter. We present have the salient features from the theory of Orlicz function spaces,

LΦ(Ω), generated by the Young’s function Φ on an arbitrary σ−finite measurable spaces Ω.

We will discussed these are the Banach spaces equipped with the equivalent orlicz and gauge

norms.

2.1 Orlicz Spaces

Before going to the main results of this chapter, let us begin with the following definitions.

Definition 2.1.1. A real function Φ defined on an interval (a, b), where −∞ ≤ a < b ≤ ∞

is called convex if the following inequality hold

Φ((1 − λ)x + λy) ≤ (1 − λ)Φ(x) + λΦ(y)

whenever a < x < b, a < y < b and 0 ≤ λ ≤ 1.

Definition 2.1.2. Let Φ : R → R+ be a convex function such that

i) Φ(−x) = Φ(x)

ii) Φ(x) = 0 iff x=0

4

5

iii) limx→∞Φ(x) = ∞.

Such a function Φ is known as a Young function. cf.[9]

Example 2.1.1. i) Φp(s) := |s|pp

with p≥ 1;

ii) Let Φ(x) = |x|p, p ≥ 1. Then Φ is a continuous Young function such that Φ(x) = 0 if and

only if x = 0, and Φ(x) → ∞ as x → ∞ while Φ(x) < ∞ for all x ∈ R.

Definition 2.1.3. A function is called N-function if it admits the representation

M(u) =∫ u

0p(t)dt

Where p(t) is right continuous for t ≥ 0, positive for t > 0 and non decreasing which satisfies

the condition p(0) = 0 and p(∞) = limt→∞ p(t) = ∞.

Example 2.1.2. The function M(u) = |u|αα

for α > 1 is a N-function for p(t) = tα−1.

Definition 2.1.4. We say that N-function M(u) satisfies the ∆2 condition for the large

values of u if there exists constant k > 0, u0 ≥ 0 such that

M(2u) ≤ kM(u), (u ≥ u0)

Definition 2.1.5. Let Ω = (Ω,Σ, µ) be a σ-finite measurable space and let τ : Ω → Ω be a

measurable transformation, that is τ−1(A) ∈ Σ for any A ∈ Σ. If µ(τ−1(A) = 0 for all

Aϵ Σ with µ(A)=0, then τ is said to be nonsingular.

Definition 2.1.6. An atom of the measure µ is an element A∈ Σ with µ(A) > 0 such that

for each F ∈ Σ,if F ⊂ A then either µ(F ) = 0 or µ(F ) = µ(A).

6

Definition 2.1.7. A set A ∈ Σ is an atom for µ if µ(A) > 0 and for each B ⊂ A,B ∈ Σ

either µ(B) = 0 or µ(A − B) = 0. A set D ∈ Σ is diffuse for µ if it does not contain

anyµ−atom.i.e. for 0 ≤ λ ≤ µ(D) we can find a set D1 ⊂ D,D1 ∈ Σ such that µ(D1) = λ.

Definition 2.1.8. Let LΦ(Ω) be the set of all f : Ω → R, measurable for Σ, such that∫Ω

Φ(|f |)dµ < ∞.

Theorem 2.1.1. 1. The space LΦ(Ω) introduced above is absolutely convex,i.e. if f, g ∈

LΦ(Ω) and α, β are scalars such that |α| + |β| ≤ 1, then αf + βg ∈ LΦ(Ω). Also

h ∈ LΦ(Ω), |f | ≤ |h|, f measurable ⇒ f ∈ LΦ(Ω).

2. The space LΦ(Ω) is linear space if Φ ∈ ∆2 globally when µ(Ω) = ∞, and locally if

µ(Ω) < ∞ and ∆2−condition is necessary if µ is diffuse on a set of positive measure.

Proof. 1. let f, g ∈ LΦ(Ω) and α, β are scalars such that |α|+|β| ≤ 1. let γ = |α|+|β| ≤ 1.

Then by using the monotonicity and convexcity of Φ we get

Φ(|αf + βg|) ≤ Φ(|α||f | + |β||g|) ≤ γΦ( |α|γ|f | + |β|

γ|f |) = γ. |α|

γΦ(|f |) + γ. |β

γΦ(|g|) =

|α|Φ(|f |) + |β|Φ(|g|), but by hypothesis right hand side is integrable. Hence αf +βg ∈

LΦ(Ω). Since Φ is monotonically increasing and |f | ≤ |h| hence Φ(|f |) ≤ Φ(|h|).

2. To prove LΦ(Ω) is a vector space it is sufficient to prove, for each f ∈ LΦ(Ω), 2f ∈

LΦ(Ω) then nf ∈ LΦ(Ω) for any n ∈ N and then for each α > 0, αf ∈ LΦ(Ω). Let a, b

be any scalars, let γ = |a| + |b| > 0 then we have af + bg = γ( aγf + b

γg) ∈ LΦ(Ω) for

any f, g ∈ LΦ(Ω). Now only remain to prove for each f ∈ LΦ(Ω, 2f ∈ LΦ(Ω).

Since Φ ∈ ∆2 globally, then we have µ(Ω) = ∞,Φ(2|f |) ≤ KΦ(|f |), K > 0 ⇒∫Ω

Φ(2|f |) ≤ K∫Ω

Φ(|f |) < ∞ hence we have 2f ∈ LΦ(Ω).

Let Φ ∈ ∆2 locally, then we have µ(Ω) < ∞, then Φ(2x) ≤ KΦ(x) for each 0 ≤ x0 ≤ x.

Now let f1 = f if |f | ≤ x0 and 0 otherwise. Let f2 = f − f1 so that f = f1 + f2 and

Φ(2|f |) = Φ(2|f1|) + Φ(2|f2|) ≤ Φ(2|f1|) + KΦ(|f2|).

7

Hence ∫Ω

(2|f |)dµ ≤ Φ(2x0)µ(Ω) + K

∫Ω

Φ(|f |)dµ < ∞

,

Thus 2f ∈ LΦ(Ω). Hence LΦ(Ω) is a vector space when Φ ∈ ∆2.

Conversely, let E ∈ Σ be a set of positive measure and let µ diffuse on E and Φ ∈ ∆2

is not regular.

To prove necessity of ∆2 condition we have to construct a f ∈ LΦ(Ω) such that 2f /∈

LΦ(Ω). Let assume that 0 < α < µ(E) ≤ ∞. Then by given statement on µ,there

is an F ⊂ E,F ∈ Σ with µ(F ) = α. Since Φ /∈ ∆2, ∃ a sequence xn ≥ n such that

Φ(2xn) ≥ nΦ(xn), n ≥ 1. Let n0 ∈ N such that

∑n≥n0

1

n2< α and Φ(xn) ≥ 1 for all n ≥ n0.

Since µ is diffuse on F, there is a measurable F0 ⊂ F such that µ(F0) =∑ 1

n2< α.

We can find a set D1 ∈ Σ, D1 ⊂ F0 such that µ(D1) = 1/n20 . Similarly again we can

find set D2 ∈ Σ, D2 ⊂ F0 − D1 such that µ(D2) = 1/(n0 + 1)2. On repeating this

process, we can find disjoint sets Dn ∈ Σ such that µ(Dn) = 1/(n0 + n− 1)2, n ≥ 1.

Let Fk ⊂ Dk, Fk ∈ Σ such that µ(Fk) = µ(DK)Φ(xn)

.

Let f =∞∑n=1

xnχFn then clearly f is measurable. Now

∫Ω

Φ(|f |)dµ =∞∑n=1

Φ(xn)µ(Fn)

=∑n≥n0

1/n2 < ∞

so f ∈ LΦ(Ω).

8

Now ∫Ω

Φ(2f)dµ =∞∑n=1

Φ(2xn)µ(Fn)

≥∑n≥n0

nΦ(xn)µ(Fn)

=∑n≥n0

1

n= ∞

.

so 2f /∈ LΦ(Ω) So ∆2 condition is necessary to prove LΦ(Ω) is a vector space.

Example 2.1.3. Let Ω = 1, 2, ...,Σ = the power set of Ω, and µ be the counting measure,

i.e., µ(i) = 1, i ≥ 1. Let Φ(x) = ex2 − 1. Then Φ is a N-function and Φ ∈ ∆2. We assume

that LΦ(Ω) is a linear space. In fact, if f∈ LΦ(Ω) then∫Ω

Φ(f)dµ =∞∑n=1

(e|f(n)|2 − 1) < ∞

So terms of right hand side are bounded. Let K > 0 be the bound so that the e(|f(n)|2) ≤

K + 1, n ≥ 1. Then∫Ω

Φ(2f)dµ =∞∑n=1

(e(4|f(n)|2) − 1)

≤∞∑n=1

(e|f(n)|2 − 1)(K + 2)((K + 1)2 + 1)

= (K + 2)((K + 1)2 + 1)

∫Ω

Φ(f)dµ < ∞.

Hence 2f ∈ LΦ(Ω), and the space is linear.

Definition 2.1.9. Let Φ : [0,∞) → [0,∞) be a continuous, non-decreasing and convex

function with Φ(0) = 0, Φ(x) > 0 for x > 0 and Φ(x) → ∞ as x → ∞. Such function is

known as an Orlicz function.

9

Definition 2.1.10. Let LΦ(Ω) be the set of all measurable functions such that∫Ω

Φ(α|f |)dµ <

∞ for some α >0. The space LΦ(Ω) is called Orlicz Space.

Thus LΦ(Ω) = f : Ω → [0,∞], measurable:∫Ω

Φ(α|f |)dµ < ∞ for some α > 0.

Theorem 2.1.2. The set LΦ(Ω) is a vector space.

Proof. Let f1, f2 ∈ LΦ(Ω).

Then there exist α1 > 0 and α2 > 0 such that∫Ω

Φ(α1|f |)dµ < ∞ and∫Ω

Φ(α2|f |)dµ < ∞.

Let α = minα1, α2, then α > 0.

Now by using convexcity of Φ we get∫Ω

Φ(α2(f1 + f2))dµ ≤ 1

2

∫Ω

Φ(α1f1)dµ+12

∫Ω

Φ(α2f2)dµ.

Hence∫Ω

Φ(α2(f1 + f2))dµ < ∞, Where α

2> 0.

Hence f1 + f2 ∈ LΦ(Ω). Thus LΦ(Ω) is closed under addition.

Now we have to prove LΦ(Ω) is closed under scalar multiplication.

Let f ∈ LΦ(Ω) ⇒ f + f = 2f ∈ LΦ(Ω), hence nf ∈ LΦ(Ω) for all integers n > 1. Now for

any β ∈ R there exists n0 ∈ N such that |β| ≤ n0 ⇒ |βf | ≤ |n0f |. So by theorem 2.1.1

βf ∈ LΦ(Ω). Hence LΦ(Ω) is a vector space.

Definition 2.1.11. If f, g ∈ LΦ(Ω) and α, β are scalars such that |α| + |β| ≤ 1 then

αf + βg ∈ LΦ(Ω). Also h ∈ LΦ(Ω), |f | ≤ |h|, f is measurable then f ∈ LΦ(Ω). Any space of

function with the above property is called circled and solid space.

Lemma 2.1.1. Let

BΦ = g ∈ LΦ(Ω) :∫Ω

Φ(g)dµ ≤ 1,

and BΦ is a circled and solid subset of LΦ(Ω) and f ∈ LΦ(Ω) if and only if αf ∈ BΦ for

some α > 0.

10

Proof. Let f, g ∈ BΦ and α, β are scalars such that |α| + |β| ≤ 1. Then∫Ω

Φ(|αf + βg|)dµ ≤ |α|∫Ω

Φ(|f |)dµ + |β|∫Ω

Φ(|g|)dµ

≤ |α| + |β|

≤ 1

Let f ∈ LΦ(Ω) so that αf ∈ LΦ(Ω) for some α > 0. Let an 0 be arbitrary and let

αn = minα, an. Then αn 0 and Φ(αnf) ≤ Φ(αf) and Φ(αnf) → 0 when Φ is a

continuous Young function. Hence by dominated convergence then∫Ω

(αnf)dµ → 0 so that

there is some n0 ∈ N such that∫Ω

Φ(αn0f)dµ ≤ 1. Thus αn0f ∈ BΦ.

Definition 2.1.12. For f ∈ LΦ(Ω) define ∥f∥Φ = infλ > 0 : IΦ(fλ) ≤ 1, where IΦ(f

λ) =∫

ΩΦ(|f

λ|)dµ is called the modular of Φ.

Theorem 2.1.3. (LΦ(Ω), ∥.∥Φ), is a normed linear space.

Proof. Clearly LΦ(Ω) is a linear space.

Next we have to verify ∥.∥Φ is a norm on LΦ(Ω), i.e. to verify the following three conditions.

(i) ∥f∥Φ = 0 iff f = 0 a.e.

(ii) ∥αf∥Φ = |α∥f∥Φ for all α ∈ K

(iii) ∥f + g∥Φ ≤ ∥f∥Φ + ∥g∥Φ

Clearly if f = 0 a.e. then ∥f∥Φ = 0.

Conversly, Let ∥f∥Φ = 0, to show that f = 0 a.e. If possible let |f | > 0 on a set of positive

measure. Then there exists a number δ > 0 such that A = x : |f(x)| ≥ δ satisfies

11

µ(A) > 0 ⇒ fk∈ BΦ for all k > 0 ⇒ nf ∈ BΦ for all n ≥ 1. Hence, for n ≥ 1

Φ(nδ)µ(A) =

∫A

Φ(nδ)dµ

≤∫A

Φ(nf)dµ

≤∫Ω

Φ(nf)dµ

≤ 1

Since µ(A) > 0, we have Φ(nδ) → ∞ as n → ∞ which is a contradiction. Hence f = 0 a.e.

for (ii) consider the non-trivial case α = 0.

∥αf∥Φ =

∫Ω

Φ(|αxλ|)dµ ≤ 1

= |α|inf λ

|α|> 0 :

∫Ω

Φ(| fλ|α|

|)dµ ≤ 1

= |α|infβ > 0 :

∫Ω

Φ(β|f |)dµ ≤ 1

= |α|. ∥ f ∥Φ

finally for triangle inquality (iii),by definition of infimum there exists α1 > 0 and α2 > 0

such that ∥f1∥Φ < α1 + ϵ2

and ∥f2∥Φ < α2 + ϵ2.

Let β = α1 + α2

Since f1 + f2 ∈ LΦ(Ω), ∥f1 + ∥f2 < ∞.

consider∫Ω

Φ((f1 + f2)/β)dµ =∫Ω

Φ( f1α1.α1

β+ f2

α2.α2

β)dµ

≤ α1

β

∫Ω

Φ( f1α1

)dµ + α2

β

∫Ω

Φ( f2α2

)( by using convexity of Φ)

≤ α1

β+ α2

β= 1

⇒ 1β(f1 + f2) ∈ LΦ(Ω)

12

Hence ∥ 1β(f1 + f2)∥Φ = 1

β∥(f1 + f2)∥Φ ≤ 1

⇒ ∥(f1 + f2)∥Φ ≤ β

But β = α1 + α2, ∥f1∥ < α1 + ϵ2and∥f2∥ < α2 + ϵ

2

⇒ ∥f1∥Φ + ∥f2∥Φ < α1 + α2 + ϵ

⇒ ∥f1∥Φ + ∥f2∥Φ < β + ϵ

⇒ ∥f1∥Φ + ∥f2∥Φ < |(f1 + f2)∥Φ + ϵ

Since ϵ > 0 be arbitrary

⇒ ∥f1∥Φ + ∥f2∥Φ ≤ ∥(f1 + f2)∥Φ

Hence (iii) follows.

Thus ∥x∥Φ = infλ > 0 :∫Ω

Φ(|xλ|)dµ ≤ 1 is the norm defined on LΦ(Ω). Hence LΦ(Ω) is

a normed linear space.

Remark 2.1.1. The above norm ∥.∥Φ defined on the space LΦ(Ω) is known as Luxemburg-

Nakano Norm.

Lemma 2.1.2. ∥f∥Φ ≤ 1 if and only if∫Ω

Φ(f)dµ ≤ 1.

Proof. Let α = ∥f∥Φ, f ∈ LΦ(Ω). If α = 0 then it is trivial so let α > 0. Then by definition,

1α∈ BΦ. If α ≤ 1,then ∫

Ω

Φ(f)dµ ≤∫Ω

Φ(f

α)dµ ≤ 1

so that ∥f∥Φ ≤ 1 implies that left hand side is bounded by 1 on other hand, f ∈ BΦ then

by definition ∥f∥Φ ≤ 1 holds.

Remark 2.1.2. If α > 1 then∫Ω

Φ( fα

)dµ ≤ 1 but∫Ω

Φ(f)dµ = ∞ is possible. Thus only

0 ≤ α ≤ 1 is possible here.

Theorem 2.1.4. The normed linear space (LΦ(Ω), ∥.∥Φ) is a Banach Space.

13

Proof. Let fn, n > 1 be a cauchy sequence in LΦ(Ω) such that ∥fn−fm∥Φ → 0 as m,n → ∞

we have to construct a f ∈ LΦ(Ω) which satisfy ∥fn−f∥Φ → 0 as n → ∞. Since we considered

Φ is a Young function there are two cases.

Let x0 = supx ∈ R+ : Φ(x) = 0. Then by definition of Φ the above set define in the

braces is compact so 0 ≤ x0 < ∞. Then by hypothesis there exist numbers km,n ≥ 0

(k−1m,n ≤ ∥fn − fm∥Φ) such that∫

Ω

Φ(km,n|Fn − fm|)dµ ≤ 1 (1)

Let define Amn = ω : km,n|Fn − fm|(ω) > x0 ∈ Σ is at most σ−finite for µ. Let Bk =

Bmnk = ω : km,n|Fn − fm|(ω) > x0 + k−1, then clearly Amn = ∪∞

k=1Bk and for each k

µ(Bk) < ∞. Since by condition(1)

µ(Bk) ≤ 1

Φ(x0 + k−1)

∫Bk

Φ(km,n|Fn − fm|)dµ ≤ 1 (2)

Hence each Amn is σ−finite and let A = ∪m,n≥1Amn. Thus on Ac, km,n|Fn − fm|(ω) ≤ x0, so

that ω ∈ Ac, |fn(ω) − fm(ω)| → 0 uniformly. Thus there is a measurable function g0 on Ac

such that fn(ω) → g0(ω), and |g0(ω)| ≤ x0, ω ∈ Ac.

Let us take Ω for A then fn is a cauchy sequence on LΦ(Ω) and hence for each B ∈

Σ, µ(B) < ∞ by condition (2), we have

µ(B ∩ |fn − fm| ≥ ϵ) = µ(B ∩ Φ(kmn|fn − fm|) ≥ Φ(kmnϵ))

≤ 1

Φ(kmnϵ)

∫B

Φ(km,n|Fn − fm|)dµ

≤ [Φ(kmnϵ)]−1

Since ϵ > 0 be fixed and and kmn → ∞, from we get that fn is a cauchy sequence in

µ−measure on each B by using the σ−finiteness property we have fn is a cauchy sequence

14

in measure. If fn → f in measure. Then there is subsequence fni such that then fni

→ f

a.e. But fni is a cauchy sequence in ∥f∥Φ, we get ∥fn∥Φ → ρ and hence ∥fni

∥Φ → ρ. By

using Fatou’s lemma we get∫Φ

(|f |ρ

) ≤ limt→∞inf

∫Ω

Φ(fni

∥fni∥Φ

) ≤ 1

.

Hence f ∈ LΦ(Ω). Let m be fixed and k ≥ 0 be given, then Φ(|fni− fnj

|k) → Φ(|f − fnj|k)

as i → ∞, a.e. let ni, nj ≥ n0 and kni,nj≥ k then∫

Ω

Φ(k|Fni− fnj

|)dµ ≤∫Ω

Φ(ki,nj

|Fni− fnj

|)dµ ≤ 1

let ni → ∞ then by using Fatou’s lemma we get ∥f − fnj∥Φ ≤ 1

k. Since k > 0 is arbitrary

∥f − fnj∥Φ → 0 If fnj

is any other subsequence with limit f ′, then fn′j, fni

, i ≥ 1, j ≥ 1 ⊂

fn so that f = f ′ a.e. because fn → f in measure. So for every convergent subsequence

and hence for the whole sequence,∥fn − f∥Φ → 0. This shows that every cauchy sequence of

(LΦ(Ω), ∥.∥Φ) converges to an element in the space.

Chapter 3

Composition Operators On OrliczSpaces

This chapter is devoted to the study of Composition operators Cτ between Orlicz spaces

LΦ(Ω) generated by measurable and non-singular transformations τ from Ω into itself. We

also investigate the Boundedness and compactness of the composition operators on the Orlicz

spaces by using different types of ∆2 conditions of the orlicz function Φ.

Definition 3.0.13. A measure µ on Ω is complete if whenever E ∈ Ω, F ⊆ E and µ(E) = 0,

then F ∈ Ω.

Definition 3.0.14. A measure µ on Ω is σ−finite if for every set E ∈ Ω,we have E = ∪En

for some sequence En such that En ∈ Ω and µ(En) < ∞ for each n.

Example 3.0.4. The Lebesgue measure m defined on R,the class of mesurable sets of R, is

σ−finite and complete.

Definition 3.0.15. If µ and ν are measures on the measure space (Ω,Σ) and ν(E) = 0

whenever µ(E) = 0, then we say that ν is absolutely continuous with respect to µ and we

write ν ≪ µ.

Theorem 3.0.5. If (Ω,Σ, µ) is a σ−finite measure space and ν is a σ−finite measure on Ω

such that ν ≪ µ, then there exists a finite-valued non-negative measurable function f on Ω

15

16

such that for each E∈ Σ, ν(E) =∫Efdµ. Also f is unique in the sense that if ν(E) =

∫Egdµ

for each E ∈ Ω, then f=g a.e.(µ).

Definition 3.0.16. Let µ and ν be σ−finite measure on (Ω,Σ) and suppose that ν ≪

µ. Then the Radon-Nikodym derivative dνdµ, of ν with respect to µ, is any measurable

function f such that ν(E) =∫Efdµ for each E ∈ Ω.

Definition 3.0.17. Let X and Y be two non-empty sets and let F(X) and F(Y) be two

topological vector spaces of complex valued functions on X and Y respectively. Suppose T:

X → Y is a mapping such that f T ∈ F (Y ) whenever F ∈ F (X). Then we define a

composition transformation CT : F (X) → F (Y ) by CT = f T for every f ∈ F (X).If CT is

continuous, we call it a composition operators induced by T.

Definition 3.0.18. Let B be a Banach space and K be the set of all compact operators on

B. For T ∈ L(B), the Banach algebra of all bounded linear operators on B into itself, the

essential norm of T means the distance from T to K in the operator norm,namely

∥T∥e = inf∥T − S∥ : S ∈ K.

Clearly, T is compact iff ∥T∥e = 0.

We need the following result for proving continuity of the composition operator.

Theorem 3.0.6. (Closed Graph Theorem) Let X and Y be Banach spaces and F : X →

Y be a closed linear map, then F is continuous.

Theorem 3.0.7. The composition map Cτ : LΦ(Ω) → LΦ(Ω) is continuous.

Proof. Let fn and Cτfn be sequence in LΦ(Ω) such that fn → f and Cτfn → g for some

f, g ∈ LΦ(Ω). Then we can find a subsequence fnk of fn such that

Φ(|fnk− f |)(x) → 0 for µ−almost all x ∈ Ω.

17

from non-singularity of τ ,

Φ(|fnk− f |τ)(x) → 0 for µ− almost all x ∈ Ω.

From the above two relation, we conclude that Cτf = g. This proved that the graph is

closed and by using closed graph theorem Cτ is continuous.

Theorem 3.0.8. Let Ω2 consists of infinitely many atoms, Φ be an Orlicz function and τ

be a non-singular measurable transformation from Ω into itself. Put

α = infϵ > 0 : N(h, ϵ) consists offinitely many atoms

where N(h, ϵ) = x ∈ Ω : h(x) > ϵ.If Cτ : LΦ(Ω) → LΦ(Ω) is a composition operator, then

1. ∥Cτ∥e = 0 if and only if α = 0.

2. ∥Cτ∥e ≥ α if 0 < α ≤ 1 and Φ(x) ≻≻ x.

3. ∥Cτ∥e ≤ α if α > 1

Proof. 1. From the above theorem we can conclude that Cτ is compact iff α = 0.

2. Let 0 < α ≤ 1 and Φ(x) ≻≻ x.Let 0 < ϵ < 2α be arbitrary. Let F = N(h, α− ϵ2), then

by definition of α either F contains a non-atomic subset or has infinitely many atoms.

If F contains a non-atomic subset then there are measurable subsets En, n ∈ N , such

that En+1 ⊆ En ⊆ F, 0 < µ(En) < 1n. Let us define fn = Φ−1(1/µ(En))χEn . Then

∥fn∥Φ = 1 for all n ∈ N . We have to prove fn → 0 weakly.To prove this we have to

show that∫Ωfng → 0 for all g ∈ LΨ(Ω), where Ψ is the complementary function to Φ.

Let A ⊆ F with 0 < µ(A) < ∞ and g = χA since Φ(x) ≻≻ x, then we have

|∫ΩfnχAdµ| = Φ−1( 1

µ(En))µ(A ∩ En) ≤ Φ−1( 1

µ(En))µ(En) = Φ−1(1/µ(En))

1/µ(En)→ 0, n → ∞

Since simple functions are dense in LΨ(Ω), thus fn is converge to 0 weakly. Now assume

that F consists of infinitely many atoms. Let (En)∞n=0 be disjoint atoms in F. Again on

18

putting fn as above. If µ(En) → 0, then by using the similar argument we had above,∫ΩfnχAdµ → 0. Now we have to prove that ∥Cτfn∥Φ ≥ α− ϵ

2. Since 0 < α− ϵ

2< 1 we

have

∥Cτfn∥Φ = infδ > 0 :

∫Ω

Φ(|fn τ |

δ)dµ ≤ 1

= infδ > 0 :

∫Ω

hΦ(|fn|δ

)dµ ≤ 1

≥ infδ > 0 :

∫Ω

(α− ϵ

2)Φ(

|fn|δ

)dµ ≤ 1

≥ infδ > 0 :

∫Ω

Φ((α− ϵ/2)|fn|

δ)dµ ≤ 1

= (α− ϵ/2) infδ > 0 :

∫Ω

Φ(|fn|δ

)dµ ≤ 1

= α− ϵ

2

.

Finally let a compact Operator T on LΦ(Ω) such that ∥Cτ − T∥ < ∥Cτ∥e + ϵ2. Then

we have

∥Cτ∥e > ∥Cτ − T∥ − ϵ

2

≥ ∥Cτfn − Tfn∥Φ − ϵ

2

≥ ∥Cτfn∥Φ − ∥Tfn∥Φ − ϵ

2

≥ (α− ϵ

2) − ∥Tfn∥Φ − ϵ

2.

for all n ∈ N . Since a compact operator maps weakly convergent sequences into norm

convergent ones, it follows that ∥Tfn∥Φ → 0. Hence ∥Cτ∥e ≥ α−ϵ. Since ϵ is arbitrary,

we obtain ∥Cτ∥e ≥ α.

3. Let α > 1 and take ϵ > 0 be arbitrary and put K = N(h, α + ϵ). The definition of α

implies that K consist of finitely many atoms. Hence we can write K = E1, E2, ..., Em

19

where E1, E2, ..., Em are distinct. Since (MχkCτf)(x) =

∑mi=1 χk(Ei)f(τ(Ei)), for all

f ∈ LΦ(Ω), hence MχkCτ has finite rank. Now, let F ⊆ X K such that 0 < µ(F ) < ∞,

then we have

µ τ−1(F ) =∫Fhdµ ≤ (α + ϵ)µ(F ).

Since α + ϵ > 1 and Φ−1 is a concave function, we obtain that

Φ−1( 1µτ−1(F )

) ≥ 1α+ϵ

Φ−1( 1µ(F )

)

That is

Φ−1( 1µτ−1(F )

)−1 ≤ (α + ϵ)Φ−1( 1µ(F )

)−1.

It follows that ∥χF τ∥Φ ≤ (α + ϵ)∥χF∥Φ. Since simple functions are dense in LΦ(Ω),

we obtain

sup∥f∥Φ≤1∥χX/Kf τ∥Φ ≤ sup∥f∥Φ≤1∥χx/kf∥Φ ≤ α + ϵ.

finally, since MχKCτ is a compact operator, we get

∥Cτ −MχKCτ = sup∥f∥Φ≤1∥(1 − χk)Cτf∥Φ = sup∥f∥Φ≤1∥χx/kCτf∥Φ ≤ α + ϵ.

Example 3.0.5. Let Φ be an Orlicz function such that Φ−1(2n)2n

→ 0 as n → ∞. Put Ω and

µ as above. Define τ(1) = τ(2) = τ(3) = 1, τ(4) = 2, τ(5) = τ(6) = 3, τ(2n + 1) = 5 , for

n ≥ 3, τ(2n) = 2n−2 for n ≥ 4, and τ(x) = 5x for all x ∈ (−∞, 0]. Then a simple function

gives h = 7/4χ1 + 1/4χ2 + 3/8χ3 + 1/3χ2n+1:n≥3 + 1/4χ2n:n≥4 + 1/5χ(−∞,0], and α = 13. Thus

∥Cτ∥e ≥ 13on LΦ(Ω).

20

3.1 Modular and norm continuity of composition op-

erators

For the modular and norm continuity of composition operators Cτ in an Orlicz spaces LΦ(Ω),

we represent necessary and sufficient conditions for any Orlicz function Φ and any σ−finite

measure space (Ω,Σ, µ). For any Orlicz function Φ which satisfies ∆2 condition for all u,

the same is done for norm continuity of the composition operator Cτ in LΦ(Ω). If Φ satisfies

∆2 condition for large u, then the problem of continuity of the composition operator Cτ in

LΦ(Ω) is completely solved if the measure space is nonatomic of finite or infinite measure.

Without any regularity condition on Φ, the conditions for continuity of Cτ from LΦ(Ω) into

itself are explained in terms of the Radon-Nikodym derivative dµτ−1

dµ.

Theorem 3.1.1. Assume that τ : Ω → Ω is a measurable nonsingular transformation.

1. if 0 < aΦ = bΦ < ∞ then IΦ(Cτx) = IΦ(x) whenever IΦ(x) < ∞.

2. if 0 ≤ aΦ < bΦ ≤ ∞ then the inequality

IΦ(Cτx) ≤ KIΦ(x) (1)

holds for all x such that IΦ(x) < ∞ with some K > 0 independent of x if and only if

µ(τ−1(A)) ≤ Kµ(A) (2)

for all A ∈ Σ with µ(A) < ∞.

Proof. 1. In this case the function Φ is 0 in the interval [0,aΦ) and ∞ on (aΦ,∞).

Therefore, IΦ < ∞ iff ∥x∥∞ ≤ aΦ ⇒ ∥Cτx∥∞ ≤ aΦ ⇒ IΦ(Cτx) = 0 = IΦ.

2. Let assume that 0 ≤ aΦ < bΦ ≤ ∞.

Necessary condition:

21

Let assume that the condition IΦ(Cτx) ≤ KIΦ(x) holds. If A ∈ Σ and µ(A) = 0,then

non singularity of τ gives µ(τ−1(A)) = 0 and we have µ(τ−1(A)) = Kµ(A). Thus

suppose that A ∈ Σ and 0 < µ(A) < ∞. Let a ∈ (aΦ, bΦ) and taking x = aχA. Then

IΦ =∫A

Φ(a)dµ(s) = Φ(a)µ(A) < ∞.

Since CτχA = χτ−1(A) then by (1) we have

Φ(a)µ(τ−1(A)) = IΦ(Cτx) ≤ KIΦ(x) = KΦ(a)µ(A).

Since 0 < Φ(a) < ∞, then we have µ(τ−1(A)) ≤ Kµ(A).

Sufficient condition:

Let assume that 0 ≤ aΦ < bΦ ≤ ∞. and condition (2) satisfied. From this we have

µτ−1 ≪ µ by Radon-Nikodym theorem we have, µτ−1(A) =∫Afτ (t)dµ(t) for A ∈ Σ

and for some function fτ locally integrable on Ω and fτ ∈ L∞(Ω) and ∥fτ∥∞ ≤ K.

Otherwise,there is A ∈ Σ with 0 < µ(A) < ∞ such that fτ (t) > K for any t ∈ A.This

implies that µ τ−1(A) =∫Afτ (t)dµ(t) > Kµ(A), which is a contradiction to (2).

Therefore,

IΦ(Cτx) =

∫Ω

Φ(|Cτx(s)|)dµ(s)

=

∫Ω

Φ(|x(τ(s))|)dµ(s)

=

∫τ(Ω)

Φ(|x(t)|)d(µ τ−1)(t)

≤∫Ω

Φ(|x(t)|)d(µ τ−1)(t)

=

∫ω

Φ(|x(t)|)fτ (t)dµ(t)

≤ K

∫Ω

Φ(|x(t)|)dµ(t)

= KIΦ(x).

22

Theorem 3.1.2. Assume that τ : Ω → Ω is a measurable nonsingular transformation. Then

the composition operator Cτ is bounded from an Orlicz space LΦ(Ω) into itself, that is, there

exists M > 0 such that

∥Cτx∥Φ ≤ M∥x∥Φ for all x ∈ LΦ(Ω) (3)

If condition (2) holds. If, in addition, Φ satisfies the condition ∆2 for all u > 0,then (3) and

(2) are equivalent.

Proof. Necessary condition:

Let assume that condition (3) hold and by putting x = χA where A ∈ Σ and 0 < µ(A) < ∞

we get

1Φ−1(1/µ(τ−1(A)))

≤ MΦ−1(1/µ(A))

⇒ Φ−1( 1µ(A)

) ≤ MΦ−1( 1µ(τ−1(A))

) (4)

for all A ∈ Σ with 0 < µ(A) < ∞

Since Φ satisfies ∆2 condition for all u > 0,it follows that

L := supu>0Φ(Mu)Φ(u)

< ∞,

and Φ(Mu) ≤ LΦ(u) for all u > 0, which gives for u = Φ−1(v) that

Φ(MΦ−1(v)) ≤ LΦ(Φ−1(v)) ≤ Lv

and so

MΦ−1(v) ≤ Φ−1Φ(MΦ−1(v)) ≤ Φ−1(Lv)

for all v > 0

From the condition (4) we get

Φ−1( 1µ(A)

) ≤ MΦ−1( 1µ(τ−1(A))

) ≤ Φ−1( Lµ(τ−1(A))

)

23

or equivalently µ(τ−1(A)) ≤ Lµ(A) for all A ∈ Σ with 0 < µ(A) < ∞, which finish the proof

of the necessary condition with K=L.

Sufficient condition:

From Theorem 3.1.1 we know that if the condition (2) satisfied with K ≥ 1 then condition

(1) holds and

IΦ( CτxK∥x∥Φ

) ≤ 1KIΦ( Cτx

∥x∥Φ) ≤ IΦ( x

∥x∥Φ) ≤ 1.

hence ∥Cτx∥Φ ≤ K∥x∥Φ for all 0 = x ∈ LΦ(Ω) that is condition (3) hold with M=K.

Remark 3.1.1. The sufficient condition of theorem 3.1.2 can be proved in two different ways,

namely by using simple functions and by the Orlicz interpolation theorem which is saying that

any Orlicz space LΦ(Ω) is an exact interpolation space between L1(Ω) and L∞(Ω).

Remark 3.1.2. Condition (2) is sufficient for the continuity of any composition operator

from any symmetric space X into it self if X has either the Fatou property or an absolutely

continuous norm, because X is then an interpolation space between L1 and L∞.

Remark 3.1.3. If 0 < aΦ ≤ bΦ < ∞, then the Orlicz spaces LΦ is equal to L∞ with an

equivalent norm. Hence the composition operator Cτ is norm-continuous on LΦ(Ω) for every

nonsingular transformation τ . However, in order to obtain modular continuity of Cτ which

is stronger than norm continuity, we need the additional assumption (2) on τ as shown in

the above theorem.

3.2 Compact Composition Operators in Orlicz space

Theorem 3.2.1. Let Φ be an Orlicz function and τ be a non singular measurable transfor-

mation from Ω to itself. Then the operator Cτ : LΦ(Ω) → LΦ(Ω) is compact if and only if

for any ϵ > 0, the set N(h, ϵ)=x ∈ Ω : h(x) > ϵ consist of finitely many atoms.

24

Proof. We shall prove it by method of contradiction. Let assume that ϵ > 0 be given, the set

N(h, ϵ) either contains a non-atomic subset or has infinitely many atoms. In both cases we

can find a sequence of pairwise disjoint measurable subsets An with 0 < An < ∞ for every

n. Let define fn = Φ−1( 1µ(An)

)χAn . Then IΦ(fn) =∫Ω

Φ(|fn|)dµ =∫Ω

Φ(Φ−1( 1µ(An)

)χAn)dµ =

1,whence fn ∈ LΦ(Ω) and ∥fn∥Φ = 1. So we have

IΦ(fn τ) =

∫Ω

Φ(|fn τ |)dµ

=

∫Ω

hΦ(|fn|)dµ

=

∫An

Φ(Φ−1(1

µ(An))hdµ

≥ ϵ

∫An

1

µ(An))dµ

= ϵ

whence Cτ ∈ L(LΦ(Ω) and ∥Cτfn∥ ≥ ϵ. Consiquently we have for m = n:

IΦ(Cτfm − Cτfn) =

∫Ω

Φ(|fm − fn|)hdµ

=

∫Am

Φ(|fm|)hdµ +

∫An

Φ(|fn|)hdµ

= IΦ(fm τ) + IΦ(fn τ)

≥ 2ϵ

Therefore, ∥Cτfm −Cτfn∥Φ ≥ ϵ for m,n ∈ N with m = n. This means that Cτfn contains

no Cauchy subsequence, that is Cτ (U(LΦ(Ω))) is not relatively compact, where U(LΦ(Ω))

is the unit ball of LΦ(Ω). Consiquently, the operator Cτ is not compact, and so this is a

contradiction.

Conversely, let ϵ > 0 be given and the set N(h,ϵ) consists of finitely many atoms, Mχτ−1ACτ

25

is a finite rank operator. Since h < ϵ on Ω A, for each f∈ LΦ(Ω) with ∥f∥Φ ≤ 1 we have

IΦ(f τ − χτ−1Af τ) = IΦ((1 − χτ−1A)f τ)

=

∫Ω

|(χΩ−A τ)f τ |dµ

=

∫Ω−A

hΦ(|f |)dµ

≤∫Ω−A

ϵΦ(|f |)dµ

≤ ϵ∥f∥Φ ≤ ϵ.

it follows that

∥Cτ −Mχτ−1ACτ∥ = sup∥Cτf − χτ−1ACτf∥Φ ≤ ϵ whenever ∥f∥Φ ≤ 1.

Thus Cτ is the limit of some finite rank operators and is therefore compact.

Example 3.2.1. Let Φ be an Orlicz function and let Ω = (−∞, 0] ∪ N , where N is the

set of natural numbers. Let µ be the Lebesque measure on (−∞, 0] and µ(n) = 12n

if n ∈

N . Define τ : N → N as τ(1) = 2, τ(2) = τ(3) = 3,τ(4) = τ(5) = τ(6) = 4 and

τ(n) = n for n ≥ 7, and τ(x) = 23x, for all x ∈ (−∞, 0]. Direct computation shows that

h = 2χ2 + 3χ3 + 74χ4

+ 1χn:n≥7 + 32χ(−∞,0]

,and α = 32.So ∥Cτ∥e ≤ 3

2on LΦ(Ω).

Theorem 3.2.2. Let Cτ ∈ B(LΦ(Ω)). Then Cτ is compact if and only if LΦ(χϵµτ−1) is

finite dimensional for each ϵ > 0, where χϵ = x : dµτ−1

dµ(x) ≥ ϵ.

Proof. Let f ∈ LΦ(Ω), then we have

∥Cτf∥Φ = infϵ > 0 :

∫Φ(

|f τ |ϵ

)dµ ≤ 1

= infϵ > 0 :

∫Φ(

f

ϵ)dµτ−1

= ∥f∥Φ,µτ−1

26

Hence Cτ is a compact operator if and only if I : LΦ(χϵµτ−1) → LΦ(χϵµτ

−1) is compact

operator if and only if LΦ(χϵµτ−1) is finite dimensional, where I is the identity operator.

Lemma 3.2.1. If (Ω,Σ, µ) is a non-atomic measure space, then no non zero composition

operator on LΦ(Ω) is compact.

Lemma 3.2.2. If (Ω,Σ, µ) is a σ−finite atomic measure space, then Cτ on LΦ(Ω) is compact

if and only if the set n :∑

m∈τ−1(an)

µ(am) ≥ ϵµ(an) is a finite, where a1, a2, ... are the atoms

of the space.

Theorem 3.2.3. Let Φ be an Orlicz function vanishing only at zero with finite values, that

is aΦ = 0 and bΦ = ∞. Let τ be a measurable nonsingular transformation from Ω into itself

such that τ(Ω) = Ω. If Cτ is a compact operator from LΦ into itself, then the measure µ is

purely atomic.

Proof. Let Ω = Ω1 ∪ Ω2 where Ω1 and Ω2 are disjoint, let µΩ1 is non atomic and µΩ2 is

purely atomic. Since µ τ−1 ≪ µ , then by Radon-Nikodym theorem there exists a function

h locally integrable on Ω1 such that µ τ−1(A) =∫Ah(t)dµ for any A ∈ Σ ∩ Ω1. Let define

A0 = t ∈ Ω1 : h(t) > 0. We have to prove that µ τ−1(A0) = 0. We shall prove it by

method of contradiction let assume that the statement is not true i.e. µ τ−1(A0) > 0, then

there exist a ϵ > 0 such that the set A1 = t ∈ A0 : h(t) ≥ ϵ has positive measure. Let us

take a sequence Bn of pairwise disjoint subsets of Σ∩A1 where 0 < µ(Bn) < 1/2n for n ∈ N.

Let us define

xn = Φ−1(1/µ(Bn))χBn for n > n0

27

Thus IΦ = 1 when xn ∈ LΦ(Ω) and ∥xn∥ = 1 for n > n0. For m,n > n0 and m = n

IΦ(Cτxm − Cτxn) =

∫Ω

Φ(|Cτxm(s) − Cτxn(s)|)dµ(s)

=

∫Ω

Φ(|xm(τ(s)) − xn(τ(s))|)dµ(s)

=

∫τ(Ω)

Φ(|xm(t) − xn(t)|)dµ τ−1(t)

=

∫Ω

Φ(|xm(t) − xn(t)|)dµ τ−1(t)

=

∫Ω

Φ(|xm(t) − xn(t)|)h(t)dµ(t)

=

∫Bm

Φ(|xm(t)|h(t)dµ(t)) +

∫Bn

Φ(|xn(t)|h(t)dµ(t))

≥ 1

µ(Bm)ϵµ(Bm) +

1

µ(Bn)ϵµ(Bn) = 2ϵ

Hence,∥Cτxm − Cτxn∥Φ ≥ 2ϵ for m,n > n0 and m = n. That means Cτxn contains no

cauchy subsequence, that means Cτ (B(LΦ(Ω))), where B(LΦ(Ω)) is the unit ball of LΦ(Ω)

is not compact.Hence Cτ is not compact which is a contradiction. Hence our assumption is

wrong and µ τ−1(A0) = 0. This complete the proof.

Bibliography

[1] Birnbaum, Z. and Orlicz, W., Uber die Verallgemeinerung des Begriffes der zueinader

konjugierten potenzen, Studia math 3(1931), 1-67.

[2] Cui, Y., Hudzik, H., Kumar, R. and Maligranda, L., Composition Operators in Orlicz

Spaces. J. Aust. Math. Soc 76 (2004), 189-206.

[3] Jabbarzadeh, M. R. The Essential norm Of a composition Operator on Orlicz spaces ;

TUBITAK (2010), 537-542.

[4] Komal, B. S. and Gupta, S., Composition Operators On Orlicz Spaces. Indian J.Pure

appl. Math 32(7)(2001), 1117-1122.

[5] Leo F. Boron, Convex function and Orlicz Spaces (1961).

[6] Nakano, H., Modulared Sequence Spaces , Proc. Jap. Acad.27 (1951), 508-512.

[7] Orlicz , W., Uber eine gewisse Klasse von Raume vom Typus b , Bull. int’l. de I’Acad.

Pol. Serie 8 (1932), 207-220.

[8] Orlicz, W., Uber Raume (LM) vom Typus b , Bull. int’l. de I’Acad. Pol. Serie 10 (1936),

93-107.

[9] Rao, M.M. and Ren, Z.D., Theory Of Orlicz Spaces. Marcel Dekker, New York (1991).

[10] Shapiro, J. H., Composition Operators and Classical Theory, Springer- verlog (1993).

28

29

[11] Shapiro, J. H., Smith, W. and stegerga, D.A., Geometric models and Compactness of

Composition Operator, J. Functional analysis (1995), 127.

[12] Youn, W. H., On classes of summable functions and their Fourier series , Proc. Royal

Soc. London 87 (1912), 225-229.