Orlicz Function Spaces and Composition Operator

ORLICZ FUNCTION SPACES AND COMPOSITION 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 Rourkela

Odisha - 768009

MAY, 2013

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 courseMA592 Projectis 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, 2013

Dr.S Pradhan Assistant Professor Department of Mathematics

NIT Rourkela

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

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 orR . X vector space over the fieldK.

Σ sigma algebra

µ measure defined over Σ.

Ω σ−finite complete measure space.

Φ Young’s function.

∥.∥Φ inf {λ >0 :∫

ΩΦ(|^x_λ|)dµ≤1}. L^Φ(Ω) Orlicz function space.

τ nonsingular measurable transformation from Ω to itself.

C_τ composition operator formL^Φ(Ω) to itself generated by τ.

ν ≪µ ν is absolutely continuous with respect toµ.

∥.∥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 functiont^p in the definitions of the spaces l^p and L^p, 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 u^p(u >0) and v^p(v >0) with ¹_p +¹_q = 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

0 Φ(u)du , Ψ(b) = ∫_b

0 Ψ(v)dv 2

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 Y_Φ and Y_Ψ 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 constantC > 0 such that Φ(2u)≤CΦ(u) hold for allu≥0, Y_Φ becomes a vector space. In the process of norming the spaces Y_Φ, Y_Ψ, Orlicz considered the class L^Φ of all measurable functionsf satisfying

∥f∥Φ= sup{∫

|f g|dx:∫

Ψ(|g|)dx≤1}<∞,

and proved that (L^Φ,∥∥Φ) is a normed linear space. In general, Y_Φ ⊂ L^Φ, however, if Φ satisfies ∆₂−condition defined as above, Y_Φ =L^Φ, cf.[9],[10].

In Mathematics, the composition operatorC_Φwith symbol Φ is a linear operator defined with the help of composition of mapping f◦Φ by the formulaC_Φ(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, L^p spaces, forp≥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, L^Φ(Ω), generated by the Young’s function Φ on an arbitraryσ−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

iii) lim_x→∞Φ(x) = ∞.

Such a function Φis known as a Young function. cf.[9]

Example 2.1.1. i) Φ_p(s) := ^|s_p^|p 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 ifx= 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 0 p(t)dt

Where p(t) is right continuous fort ≥0, positive fort >0and non decreasing which satisfies the conditionp(0) = 0 and p(∞) = lim_t→∞p(t) =∞.

Example 2.1.2. The function M(u) = ^|u_α^|α for α >1 is a N-function for p(t) = t^α−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)>0such 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 L˜^Φ(Ω) be the set of all f : Ω → R, measurable for Σ, such that

∫

ΩΦ(|f|)dµ <∞.

Theorem 2.1.1. 1. The space L˜^Φ(Ω) introduced above is absolutely convex,i.e. if f, g ∈ L˜^Φ(Ω) and α, β are scalars such that |α|+|β| ≤ 1, then αf +βg ∈ L˜^Φ(Ω). Also h∈L˜^Φ(Ω),|f| ≤ |h|, f measurable ⇒f ∈L˜^Φ(Ω).

2. The space L˜^Φ(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. letf, g ∈L˜^Φ(Ω) andα, βare scalars such that|α|+|β| ≤1. letγ =|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_γ^|Φ(|f|) +γ.^|_γ^βΦ(|g|) =

|α|Φ(|f|) +|β|Φ(|g|), but by hypothesis right hand side is integrable. Henceαf+βg ∈ L˜^Φ(Ω). Since Φ is monotonically increasing and |f| ≤ |h| hence Φ(|f|)≤Φ(|h|).

2. To prove ˜L^Φ(Ω) is a vector space it is sufficient to prove, for each f ∈ L˜^Φ(Ω),2f ∈ L˜^Φ(Ω) then nf ∈L˜^Φ(Ω) for any n∈N and then for each α >0, αf ∈L˜^Φ(Ω). Let a, b be any scalars, let γ = |a|+|b|> 0 then we have af +bg =γ(^a_γf + _γ^bg)∈ L˜^Φ(Ω) for any f, g∈L˜^Φ(Ω). Now only remain to prove for each f ∈L˜^Φ(Ω,2f ∈L˜^Φ(Ω).

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

∫

ΩΦ(2|f|)≤K∫

ΩΦ(|f|)<∞hence we have 2f ∈L˜^Φ(Ω).

Let Φ∈∆₂ locally, then we haveµ(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 thatf =f₁+f₂ and Φ(2|f|) = Φ(2|f₁|) + Φ(2|f₂|)≤Φ(2|f₁|) +KΦ(|f₂|).

Hence

∫

Ω

(2|f|)dµ≤Φ(2x0)µ(Ω) +K

∫

Ω

Φ(|f|)dµ <∞ ,

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

Conversely, letE ∈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 ∈ L˜^Φ(Ω) such that 2f /∈ L˜^Φ(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 x_n ≥ n such that Φ(2x_n)≥nΦ(x_n), n ≥1. Letn₀ ∈N such that

∑

n≥n0

1

n² < α and Φ(x_n)≥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 D_n ∈Σ such that µ(D_n) = 1/(n₀ +n−1)², n ≥1.

LetF_k ⊂D_k, F_k ∈Σ such that µ(F_k) = ^µ(D_Φ(x^K)

n). Letf =

∑∞ n=1

x_nχ_Fn then clearly f is measurable. Now

∫

Ω

Φ(|f|)dµ=

∑∞ n=1

Φ(x_n)µ(F_n)

= ∑

n≥n0

1/n² <∞

sof ∈L˜^Φ(Ω).

8 Now

∫

Ω

Φ(2f)dµ=

∑∞ n=1

Φ(2x_n)µ(F_n)

≥ ∑

n≥n0

nΦ(x_n)µ(F_n)

= ∑

n≥n0

1 n =∞ .

so 2f /∈L˜^Φ(Ω) So ∆₂ condition is necessary to prove ˜L^Φ(Ω) is a vector space.

Example 2.1.3. Let Ω ={1,2, ...},Σ = the power set of Ω, andµbe the counting measure, i.e., µ({i}) = 1, i≥1. Let Φ(x) =e^x2 −1. Then Φ is a N-function and Φ ∈∆₂. We assume that ˜L^Φ(Ω) is a linear space. In fact, if f∈L˜^Φ(Ω) then

∫

ΩΦ(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 ∈L˜^Φ(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.

Definition 2.1.10. LetL^Φ(Ω)be the set of all measurable functions such that∫

ΩΦ(α|f|)dµ <

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

Thus L^Φ(Ω) ={f : Ω→[0,∞], measurable: ∫

ΩΦ(α|f|)dµ <∞ for someα >0}. Theorem 2.1.2. The set L^Φ(Ω) is a vector space.

Proof. Letf₁, f₂ ∈L^Φ(Ω).

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

ΩΦ(α₁|f|)dµ < ∞and ∫

ΩΦ(α₂|f|)dµ <∞. Letα=min{α₁, α₂}, then α >0.

Now by using convexcity of Φ we get

∫

ΩΦ(^α₂(f1+f2))dµ≤ ¹₂∫

ΩΦ(α1f1)dµ+¹₂∫

ΩΦ(α2f2)dµ.

Hence ∫

ΩΦ(^α₂(f₁+f₂))dµ <∞, Where ^α₂ >0.

Hence f₁ +f₂ ∈L^Φ(Ω). ThusL^Φ(Ω) is closed under addition.

Now we have to prove L^Φ(Ω) is closed under scalar multiplication.

Let f ∈ L^Φ(Ω) ⇒ f +f = 2f ∈ L^Φ(Ω), hence nf ∈L^Φ(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 ∈L^Φ(Ω). Hence L^Φ(Ω) is a vector space.

Definition 2.1.11. If f, g ∈ L˜^Φ(Ω) and α, β are scalars such that |α| +|β| ≤ 1 then αf+βg ∈L˜^Φ(Ω). Also h∈L˜^Φ(Ω), |f| ≤ |h|, f is measurable then f ∈L˜^Φ(Ω). Any space of function with the above property is called circled and solid space.

Lemma 2.1.1. Let

B_Φ={g ∈L˜^Φ(Ω) :∫

ΩΦ(g)dµ≤1},

and B_Φ is a circled and solid subset of L˜^Φ(Ω) and f ∈ L^Φ(Ω) if and only if αf ∈ B_Φ for someα >0.

10 Proof. Letf, g ∈B_Φ 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 ∈ L^Φ(Ω) so that αf ∈ L˜^Φ(Ω) for some α > 0. Let a_n ↘ 0 be arbitrary and let α_n = min{α, a_n}. Then α_n ↘ 0 and Φ(α_nf) ≤ Φ(αf) and Φ(α_nf) → 0 when Φ is a continuous Young function. Hence by dominated convergence then ∫

Ω(α_nf)dµ→ 0 so that there is somen₀ ∈Nsuch that ∫

ΩΦ(α_n0f)dµ≤1. Thus α_n0f ∈B_Φ.

Definition 2.1.12. For f ∈L^Φ(Ω) define ∥f∥Φ =inf{λ >0 : I_Φ(^f_λ)≤ 1}, where I_Φ(^f_λ) =

∫

ΩΦ(|^f_λ|)dµ is called the modular of Φ.

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

Proof. Clearly L^Φ(Ω) is a linear space.

Next we have to verify∥.∥Φ is a norm onL^Φ(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

µ(A)>0⇒ ^f_k ∈B_Φ for all k >0⇒nf ∈B_Φ 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 α̸= 0.

∥αf∥Φ =

∫

Ω

Φ(|αx

λ |)dµ≤1

=|α|inf{ λ

|α| >0 :

∫

Ω

Φ(| f

|λα|

|)dµ≤1}

=|α|inf{β >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+₂^ϵ and ∥f₂∥Φ < α₂+₂^ϵ.

Letβ =α₁+α₂

Sincef1+f2 ∈L^Φ(Ω),∥f1+∥f2 <∞. consider∫

ΩΦ((f₁+f₂)/β)dµ=∫

ΩΦ(_α^f1

1.^α_β¹ + _α^f2

2.^α_β²)dµ

≤ ^α_β¹ ∫

Ω Φ(_α^f1

1)dµ+^α_β² ∫

ΩΦ(_α^f2

2)( by using convexity of Φ)

≤ ^α_β¹ +^α_β² = 1

⇒ _β¹(f₁+f₂)∈L^Φ(Ω)

12 Hence ∥¹_β(f₁+f₂)∥Φ = _β¹∥(f₁+f₂)∥Φ ≤1

⇒ ∥(f1+f2)∥Φ ≤β

Butβ =α1+α2,∥f1∥< α1+₂^ϵand∥f2∥< α2 +₂^ϵ

⇒ ∥f₁∥Φ+∥f₂∥Φ < α₁+α₂+ϵ

⇒ ∥f₁∥Φ+∥f₂∥Φ < β+ϵ

⇒ ∥f₁∥Φ+∥f₂∥Φ <|(f₁ +f₂)∥Φ+ϵ Sinceϵ >0 be arbitrary

⇒ ∥f₁∥Φ+∥f₂∥Φ ≤ ∥(f₁+f₂)∥Φ

Hence (iii) follows.

Thus∥x∥Φ =inf{λ > 0 : ∫

ΩΦ(|^x_λ|)dµ≤ 1} is the norm defined on L^Φ(Ω). Hence L^Φ(Ω) is a normed linear space.

Remark 2.1.1. The above norm∥.∥Φ defined on the spaceL^Φ(Ω) is known asLuxemburg- Nakano Norm.

Lemma 2.1.2. ∥f∥Φ ≤1 if and only if ∫

ΩΦ(f)dµ≤1.

Proof. Letα=∥f∥Φ, f ∈L^Φ(Ω). Ifα = 0 then it is trivial so letα >0. Then by definition,

1

α ∈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 ∈ B_Φ then by definition∥f∥Φ≤1 holds.

Remark 2.1.2. If α > 1 then ∫

ΩΦ(_α^f)dµ ≤ 1 but ∫

ΩΦ(f)dµ = ∞ is possible. Thus only 0≤α ≤1 is possible here.

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

Proof. Let{f_n, n >1}be a cauchy sequence inL^Φ(Ω) such that∥f_n−f_m∥Φ →0 asm, n→ ∞ we have to construct af ∈L^Φ(Ω) which satisfy∥fn−f∥Φ →0 asn → ∞. Since we considered Φ is a Young function there are two cases.

Let x₀ = sup{x ∈ 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 k_m,n ≥ 0 (k⁻_m,n¹ ≤ ∥f_n−f_m∥Φ) such that

∫

Ω

Φ(k_m,n|F_n−f_m|)dµ≤1 (1)

Let define A_mn = {ω : k_m,n|F_n−f_m|(ω) > x₀} ∈ Σ is at most σ−finite for µ. Let B_k = B_k^mn = {ω : km,n|Fn −fm|(ω) > x0 +k⁻¹}, then clearly Amn = ∪^∞k=1Bk and for each k µ(B_k)<∞. Since by condition(1)

µ(B_k)≤ 1 Φ(x₀+k⁻¹)

∫

B_k

Φ(k_m,n|F_n−f_m|)dµ≤1 (2)

Hence eachA_mn isσ−finite and let A=∪m,n≥1A_mn. Thus on A^c, k_m,n|F_n−f_m|(ω)≤x₀, so that ω ∈ A^c,|f_n(ω)−f_m(ω)| →0 uniformly. Thus there is a measurable function g₀ on A^c such thatf_n(ω)→g₀(ω), and |g₀(ω)| ≤x₀, ω∈A^c.

Let us take Ω for A then {f_n} is a cauchy sequence on L^Φ(Ω) and hence for each B ∈ Σ, µ(B)<∞ by condition (2), we have

µ(B∩ {|f_n−f_m| ≥ϵ}) =µ(B∩ {Φ(k_mn|f_n−f_m|)≥Φ(k_mnϵ)})

≤ 1

Φ(k_mnϵ)

∫

B

Φ(km,n|Fn−fm|)dµ

≤[Φ(k_mnϵ)]⁻¹

Since ϵ > 0 be fixed and and k_mn → ∞, from we get that {f_n} is a cauchy sequence in µ−measure on each B by using theσ−finiteness property we have{f_n} is a cauchy sequence

14 in measure. If f_n→f˜in measure. Then there is subsequence {f_ni} such that thenf_ni →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 )≤lim_t→∞inf

∫

Ω

Φ( f_ni

∥fni∥Φ

)≤1 .

Hence f ∈L^Φ(Ω). Let m be fixed and k ≥0 be given, then Φ(|f_ni−f_nj|k)→Φ(|f−f_nj|k) asi→ ∞, a.e. let n_i, n_j ≥n₀ and k_ni,nj ≥k then

∫

Ω

Φ(k|F_ni −f_nj|)dµ≤

∫

Ω

Φ(k_i,nj|F_ni −f_nj|)dµ≤1

let n_i → ∞ then by using Fatou’s lemma we get ∥f −f_nj∥Φ ≤ ¹_k. Since k > 0 is arbitrary

∥f−f_nj∥Φ →0 If f_nj is any other subsequence with limit f^′, then {f_n′

j, f_ni, i≥1, j ≥1} ⊂ {f_n} so that f =f^′ a.e. because f_n →f in measure. So for every convergent subsequence and hence for the whole sequence,∥f_n−f∥Φ →0. This shows that every cauchy sequence of (L^Φ(Ω),∥.∥Φ) converges to an element in the space.

Composition Operators On Orlicz Spaces

This chapter is devoted to the study of Composition operators Cτ between Orlicz spaces L^Φ(Ω) generated by measurable and non-singular transformationsτ 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 measureµonΩiscompleteif whenever E∈Ω, F⊆E andµ(E) = 0, then F ∈Ω.

Definition 3.0.14. A measure µon Ωisσ−finiteif for every set E ∈Ω,we have E =∪E_n for some sequence {E_n} such that E_n ∈Ω and µ(E_n)<∞ 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 wheneverµ(E) = 0, then we say thatν isabsolutely continuous with respect toµ 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) = ∫

Ef dµ. 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 ^dν_dµ, of ν with respect to µ, is any measurable function f such that ν(E) =∫

Ef dµ 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).IfCT 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 mapC_τ :L^Φ(Ω)→L^Φ(Ω) is continuous.

Proof. Let{f_n} and{C_τf_n} be sequence in L^Φ(Ω) such thatf_n →f and C_τf_n→g for some f, g∈L^Φ(Ω). Then we can find a subsequence {fn_k} of {fn}such that

Φ(|f_nk −f|)(x)→0 forµ−almost all x∈Ω.

from non-singularity ofτ,

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

From the above two relation, we conclude that C_τf = g. This proved that the graph is closed and by using closed graph theorem C_τ 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=inf{ϵ >0 :N(h, ϵ) consists of f initely many atoms}

where N(h, ϵ) ={x∈Ω :h(x)> ϵ}.If C_τ :L^Φ(Ω)→L^Φ(Ω) is a composition operator, then 1. ∥C_τ∥e= 0 if and only if α = 0.

2. ∥C_τ∥e≥α if 0< α≤1 and Φ(x)≻≻x.

3. ∥C_τ∥e≤α if α >1

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

2. Let 0< α≤1 and Φ(x)≻≻x.Let 0< ϵ <2α be arbitrary. LetF =N(h, α−₂^ϵ), then by definition ofα either F contains a non-atomic subset or has infinitely many atoms.

If F contains a non-atomic subset then there are measurable subsets E_n, n ∈ N, such that E_n+1 ⊆ E_n ⊆ F,0 < µ(E_n) < _n¹. Let us define f_n = Φ⁻¹(1/µ(E_n))χ_En. Then

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

Ωf_ng →0 for all g ∈L^Ψ(Ω), where Ψ is the complementary function to Φ.

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

|∫

Ωf_nχ_Adµ|= Φ⁻¹(_µ(E¹

n))µ(A∩E_n)≤Φ⁻¹(_µ(E¹

n))µ(E_n) = ^Φ−_1/µ(E^1(1/µ(En))

n) →0, n→ ∞ Since simple functions are dense inL^Ψ(Ω), thusf_nis converge to 0 weakly. Now assume that F consists of infinitely many atoms. Let (E_n)^∞_n=0 be disjoint atoms in F. Again on

18 puttingf_n as above. Ifµ(E_n)→0, then by using the similar argument we had above,

∫

ΩfnχAdµ→0. Now we have to prove that∥Cτfn∥Φ ≥α−^ϵ₂.Since 0 < α−^ϵ₂ <1 we have

∥Cτfn∥Φ =inf{δ >0 :

∫

Ω

Φ(|f_n◦τ|

δ )dµ≤1}

=inf{δ >0 :

∫

Ω

hΦ(|f_n|

δ )dµ≤1}

≥inf{δ >0 :

∫

Ω

(α− ϵ

2)Φ(|f_n|

δ )dµ≤1}

≥inf{δ >0 :

∫

Ω

Φ((α−ϵ/2)|f_n|

δ )dµ≤1}

= (α−ϵ/2)inf{δ >0 :

∫

Ω

Φ(|f_n|

δ )dµ≤1}

=α− ϵ 2 .

Finally let a compact Operator T on L^Φ(Ω) such that ∥C_τ −T∥ < ∥C_τ∥e+ ₂^ϵ. Then we have

∥C_τ∥e>∥C_τ−T∥ − ϵ 2

≥ ∥C_τf_n−T f_n∥Φ− ϵ 2

≥ ∥C_τf_n∥Φ− ∥T f_n∥Φ− ϵ 2

≥(α− ϵ

2)− ∥T f_n∥Φ− ϵ 2.

for alln ∈N. Since a compact operator maps weakly convergent sequences into norm convergent ones, it follows that∥T f_n∥Φ →0. Hence∥C_τ∥e≥α−ϵ. Sinceϵis arbitrary, we obtain∥C_τ∥e ≥α.

3. Let α >1 and take ϵ > 0 be arbitrary and put K =N(h, α+ϵ). The definition ofα implies that K consist of finitely many atoms. Hence we can writeK ={E₁, E₂, ..., E_m}

where E₁, E₂, ..., E_m are distinct. Since (M_χkC_τf)(x) = ∑m

i=1χ_k(E_i)f(τ(E_i)), for all f ∈L^Φ(Ω), henceMχ_kCτ has finite rank. Now, letF ⊆X K such that 0 < µ(F)<∞, then we have

µ◦τ⁻¹(F) =∫

Fhdµ≤(α+ϵ)µ(F).

Sinceα+ϵ >1 and Φ⁻¹ is a concave function, we obtain that Φ⁻¹(_µ◦τ−¹1(F))≥ _α+ϵ¹ Φ⁻¹(_µ(F¹ ₎)

That is

{Φ⁻¹(_µ◦τ−¹1(F))}⁻¹ ≤(α+ϵ){Φ⁻¹(_µ(F¹ ₎)}⁻¹.

It follows that∥χ_F ◦τ∥Φ ≤(α+ϵ)∥χ_F∥Φ. Since simple functions are dense in L^Φ(Ω), we obtain

sup_{∥f∥Φ≤1}∥χ_X/Kf◦τ∥Φ ≤ sup_{∥f∥Φ≤1}∥χ_x/kf∥Φ ≤α+ϵ.

finally, sinceMχ_KCτ is a compact operator, we get

∥C_τ −M_χKC_τ = sup_{∥f∥Φ≤1}∥(1−χ_k)C_τf∥Φ = sup_{∥f∥Φ≤1}∥χ_x/kC_τf∥Φ ≤α+ϵ.

Example 3.0.5. Let Φ be an Orlicz function such that ^Φ−₂¹n⁽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−2for n≥4, andτ(x) = 5xfor all x∈(−∞,0]. Then a simple function gives h = 7/4_χ1+ 1/4_χ2 + 3/8_χ3+ 1/3_χ2n+1:n≥3 + 1/4_χ2n:n≥4+ 1/5_χ(−∞,0], and α = ¹₃. Thus

∥C_τ∥e ≥ ¹₃ on L^Φ(Ω).

20

3.1 Modular and norm continuity of composition op- erators

For the modular and norm continuity of composition operatorsC_τ in an Orlicz spacesL^Φ(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 operatorC_τ inL^Φ(Ω). If Φ satisfies

∆₂ condition for large u, then the problem of continuity of the composition operatorC_τ in L^Φ(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 C_τ fromL^Φ(Ω) into itself are explained in terms of the Radon-Nikodym derivative ^dµ◦_dµ^τ−1.

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

1. if 0< a_Φ =b_Φ <∞ then I_Φ(C_τx) =I_Φ(x) whenever I_Φ(x)<∞. 2. if 0≤a_Φ < b_Φ ≤ ∞ then the inequality

I_Φ(C_τx)≤KI_Φ(x) (1)

holds for all x such that I_Φ(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,a_Φ) and ∞ on (a_Φ,∞).

Therefore,IΦ <∞ iff ∥x∥∞≤aΦ ⇒ ∥Cτx∥∞≤aΦ ⇒IΦ(Cτx) = 0 =IΦ. 2. Let assume that 0≤a_Φ < b_Φ ≤ ∞.

Necessary condition:

Let assume that the conditionI_Φ(C_τx)≤KI_Φ(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

I_Φ =∫

AΦ(a)dµ(s) = Φ(a)µ(A)<∞. SinceC_τχ_A=χ_τ−1(A) then by (1) we have

Φ(a)µ(τ⁻¹(A)) = I_Φ(C_τx)≤KI_Φ(x) =KΦ(a)µ(A).

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

Sufficient condition:

Let assume that 0 ≤ a_Φ < b_Φ ≤ ∞. and condition (2) satisfied. From this we have µ◦τ⁻¹ ≪µby Radon-Nikodym theorem we have,µ◦τ⁻¹(A) = ∫

Af_τ(t)dµ(t) forA∈Σ and for some function f_τ locally integrable on Ω and f_τ ∈ L^∞(Ω) and ∥f_τ∥_∞ ≤ K.

Otherwise,there isA ∈Σ with 0< µ(A)<∞ such that f_τ(t)> K for any t∈A.This implies thatµ◦τ⁻¹(A) = ∫

Afτ(t)dµ(t)> Kµ(A), which is a contradiction to (2).

Therefore,

I_Φ(C_τx) =

∫

Ω

Φ(|C_τx(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)|)f_τ(t)dµ(t)

≤K

∫

Ω

Φ(|x(t)|)dµ(t)

=KI_Φ(x).

22

Theorem 3.1.2. Assume thatτ : Ω→Ωis a measurable nonsingular transformation. Then the composition operatorC_τ is bounded from an Orlicz space L^Φ(Ω) into itself, that is, there exists M >0 such that

∥C_τx∥Φ ≤M∥x∥Φ for all x∈L^Φ(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 puttingx=χ_A whereA∈Σ and 0< µ(A)<∞ we get

1

Φ⁻¹(1/µ(τ⁻¹(A))) ≤ _Φ−1(1/µ(A))^M

⇒ Φ⁻¹(_µ(A)¹ )≤MΦ⁻¹(_µ(τ−¹1(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(M u)

Φ(u) <∞,

and Φ(M u)≤LΦ(u) for all u >0, which gives foru= Φ⁻¹(v) that Φ(MΦ⁻¹(v))≤LΦ(Φ⁻¹(v))≤Lv and so

MΦ⁻¹(v)≤Φ⁻¹{Φ(MΦ⁻¹(v))} ≤Φ⁻¹(Lv) for all v >0

From the condition (4) we get

Φ⁻¹(_µ(A)¹ )≤MΦ⁻¹(_µ(τ−¹_1(A)))≤Φ⁻¹(_µ(τ−^L1(A)))