Orlicz Function Spaces and Composition Operator

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A Project Report Submitted in Partial Fulfilment of the Requirements

for the Degree of


In Mathematics


Chinmay kumar Giri (Roll Number: 411MA2075)

to the


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.


Date: Chinmay Kumar Giri

Roll No: 411ma2075





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.


Date: Chinmay Kumar Giri

Roll No: 411ma2075




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



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.

σ−finite complete measure space.

Φ Young’s function.

∥.∥Φ inf {λ >0 :∫

Φ(|xλ|)dµ1}. 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 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 functiontp 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

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 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 ∆2condition in the sense that there exists a constantC > 0 such that Φ(2u)≤CΦ(u) hold for allu≥0, YΦ becomes a vector space. In the process of norming the spaces YΦ, YΨ, Orlicz considered the class LΦ of all measurable functionsf satisfying

∥f∥Φ= sup{

|f g|dx:∫


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

In Mathematics, the composition operatorCΦwith symbol Φ is a linear operator defined with the help of composition of mapping f◦Φ by the formulaCΦ(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, 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, LΦ(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 :RR+ be a convex function such that i)Φ(−x) = Φ(x)

ii) Φ(x) = 0 iff x=0



iii) limx→∞Φ(x) = ∞.

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

Example 2.1.1. i) Φp(s) := |sp|p with p1;

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(∞) = limt→∞p(t) =∞.

Example 2.1.2. The function M(u) = |uα|α for α >1 is a N-function for p(t) = tα−1.

Definition 2.1.4. We say that N-function M(u) satisfies the2 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 be a measurable transformation, that is τ−1(A)Σ for any A∈Σ. If µ(τ−1(A) = 0 for all 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

Φ(|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)<∞ and2−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(|αγ||f|+ |βγ||f|) = γ.|αγ|Φ(|f|) +γ.|γβΦ(|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 ˜LΦ(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(aγf + γbg)∈ 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


Φ(|f|)<∞hence we have 2f ∈L˜Φ(Ω).

Let Φ2 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|).



(2|f|)dµΦ(2x0)µ(Ω) +K

Φ(|f|)dµ < ,

Thus 2f ∈L˜Φ(Ω). Hence ˜LΦ(Ω) is a vector space when Φ2.

Conversely, letE Σ be a set of positive measure and let µ diffuse on E and Φ2

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 xn n such that Φ(2xn)≥nΦ(xn), n 1. Letn0 N such that



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.

LetFk ⊂Dk, Fk Σ such that µ(Fk) = µ(DΦ(xK)

n). Letf =


xnχFn then clearly f is measurable. Now




= ∑


1/n2 <∞

sof ∈L˜Φ(Ω).


8 Now






= ∑


1 n = .

so 2f /∈L˜Φ(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, andµbe 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∈L˜Φ(Ω) then



(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





(e|f(n)|2 1)(K+ 2)((K+ 1)2+ 1)

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

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. LetLΦ(Ω)be the set of all measurable functions such that

Φ(α|f|)dµ <

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

Thus LΦ(Ω) ={f : Ω[0,], measurable: ∫

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

Proof. Letf1, f2 ∈LΦ(Ω).

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

Φ(α1|f|)dµ < and ∫

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

Now by using convexcity of Φ we get

Φ(α2(f1+f2))dµ 12



Hence ∫

Φ(α2(f1+f2))dµ <, Where α2 >0.

Hence f1 +f2 ∈LΦ(Ω). ThusLΦ(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 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

BΦ={g ∈L˜Φ(Ω) :∫


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


10 Proof. Letf, g BΦ and α, β are scalars such that|α|+|β| ≤1. Then

Φ(|αf +βg|)dµ≤ |α|



≤ |α|+|β|


Let f LΦ(Ω) so that αf L˜Φ(Ω) for some α > 0. Let an 0 be arbitrary and let αn = min{α, 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 ∫

nf)dµ 0 so that there is somen0 Nsuch that ∫

Φ(αn0f)dµ≤1. Thus αn0f BΦ.

Definition 2.1.12. For f ∈LΦ(Ω) define ∥f∥Φ =inf{λ >0 : IΦ(fλ) 1}, where IΦ(fλ) =

Φ(|fλ|)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 onLΦ(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 fk BΦ for all k >0⇒nf BΦ for all n≥1. Hence, for n≥1 Φ(nδ)µ(A) =







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 =


λ |)dµ1

=|α|inf{ λ

|α| >0 :

Φ(| f



=|α|inf{β >0 :


=|α|.∥f Φ

finally for triangle inquality (iii),by definition of infimum there exists α1 > 0 and α2 > 0 such that∥f1Φ < α1+2ϵ and ∥f2Φ < α2+2ϵ.

Letβ =α1+α2

Sincef1+f2 ∈LΦ(Ω),∥f1+∥f2 <∞. consider∫



1.αβ1 + αf2






2)( by using convexity of Φ)

αβ1 +αβ2 = 1



12 Hence 1β(f1+f2)Φ = β1(f1+f2)Φ 1

⇒ ∥(f1+f2)Φ ≤β

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

⇒ ∥f1Φ+∥f2Φ < α1+α2+ϵ

⇒ ∥f1Φ+∥f2Φ < β+ϵ

⇒ ∥f1Φ+∥f2Φ <|(f1 +f2)Φ+ϵ Sinceϵ >0 be arbitrary

⇒ ∥f1Φ+∥f2Φ ≤ ∥(f1+f2)Φ

Hence (iii) follows.

Thus∥x∥Φ =inf{λ > 0 : ∫

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 spaceLΦ(Ω) is known asLuxemburg- Nakano Norm.

Lemma 2.1.2. ∥f∥Φ 1 if and only if


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


α BΦ. If α≤1,then




so that ∥f∥Φ 1 implies that left hand side is bounded by 1 on other hand, f BΦ then by definition∥f∥Φ1 holds.

Remark 2.1.2. If α > 1 then

Φ(αf)dµ 1 but

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.


Proof. Let{fn, n >1}be a cauchy sequence inLΦ(Ω) such that∥fn−fmΦ 0 asm, n→ ∞ we have to construct af ∈LΦ(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 km,n 0 (km,n1 ≤ ∥fn−fmΦ) such that

Φ(km,n|Fn−fm|)dµ1 (1)

Let define Amn = : km,n|Fn−fm|(ω) > x0} ∈ Σ is at most σ−finite for µ. Let Bk = Bkmn = : km,n|Fn −fm|(ω) > x0 +k1}, then clearly Amn = k=1Bk and for each k µ(Bk)<∞. Since by condition(1)

µ(Bk) 1 Φ(x0+k1)


Φ(km,n|Fn−fm|)dµ1 (2)

Hence eachAmn isσ−finite and let A=m,n1Amn. 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 thatfn(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)})






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 theσ−finiteness property we have{fn} is a cauchy sequence


14 in measure. If fn→f˜in measure. Then there is subsequence {fni} such that thenfni →f a.e. But {fni} is a cauchy sequence in ∥f∥Φ, we get ∥fnΦ →ρ and hence ∥fniΦ →ρ. By using Fatou’s lemma we get



ρ )≤limt→∞inf

Φ( fni


)1 .

Hence f ∈LΦ(Ω). Let m be fixed and k 0 be given, then Φ(|fni−fnj|k)→Φ(|f−fnj|k) asi→ ∞, a.e. let ni, nj ≥n0 and kni,nj ≥k then

Φ(k|Fni −fnj|)dµ

Φ(ki,nj|Fni −fnj|)dµ1

let ni → ∞ then by using Fatou’s lemma we get ∥f −fnjΦ 1k. Since k > 0 is arbitrary

∥f−fnjΦ 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.


Composition Operators On Orlicz Spaces

This chapter is devoted to the study of Composition operators Cτ between Orlicz spaces LΦ(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 measureµoniscompleteif whenever E∈Ω, F⊆E andµ(E) = 0, then F Ω.

Definition 3.0.14. A measure µonisσ−finiteif for every set E Ω,we have E =∪En for some sequence {En} such that En 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 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 onsuch that ν µ, then there exists a finite-valued non-negative measurable function f on



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 , 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 mapCτ :LΦ(Ω)→LΦ(Ω) is continuous.

Proof. Let{fn} and{Cτfn} be sequence in LΦ(Ω) such thatfn →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 forµ−almost all x∈Ω.


from non-singularity ofτ,

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. Let2 consists of infinitely many atoms, Φ be an Orlicz function and τ be a non-singular measurable transformation frominto 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 Cτ :LΦ(Ω)→LΦ(Ω) is a composition operator, then 1. ∥Cτe= 0 if and only if α = 0.

2. ∥Cτe≥α if 0< α≤1 and Φ(x)≻≻x.

3. ∥Cτe≤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. LetF =N(h, α2ϵ), 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 En, n N, such that En+1 En F,0 < µ(En) < n1. Let us define fn = Φ1(1/µ(En))χEn. Then

∥fnΦ = 1 for all n N. We have to prove fn 0 weakly.To prove this we have to show that∫

fng 0 for all g ∈LΨ(Ω), where Ψ is the complementary function to Φ.

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


fnχAdµ|= Φ1(µ(E1


n))µ(En) = Φ1/µ(E1(1/µ(En))

n) 0, n→ ∞ Since simple functions are dense inLΨ(Ω), thusfnis 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 puttingfn as above. Ifµ(En)0, then by using the similar argument we had above,

fnχAdµ→0. Now we have to prove that∥CτfnΦ ≥α−ϵ2.Since 0 < α−ϵ2 <1 we have

∥CτfnΦ =inf{δ >0 :


δ )dµ1}

=inf{δ >0 :


δ )dµ1}

≥inf{δ >0 :



δ )dµ1}

≥inf{δ >0 :


δ )dµ1}

= (α−ϵ/2)inf{δ >0 :


δ )dµ1}

=α− ϵ 2 .

Finally let a compact Operator T on LΦ(Ω) such that ∥Cτ −T∥ < ∥Cτe+ 2ϵ. Then we have

∥Cτe>∥Cτ−T∥ − ϵ 2

≥ ∥Cτfn−T fnΦ ϵ 2

≥ ∥CτfnΦ− ∥T fnΦ ϵ 2


2)− ∥T fnΦ ϵ 2.

for alln ∈N. Since a compact operator maps weakly convergent sequences into norm convergent ones, it follows that∥T fnΦ 0. Hence∥Cτe≥α−ϵ. Sinceϵis arbitrary, we obtain∥Cτe ≥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, ..., Em}


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

i=1χk(Ei)f(τ(Ei)), for all f ∈LΦ(Ω), henceMχkCτ has finite rank. Now, letF ⊆X K such that 0 < µ(F)<∞, then we have

µ◦τ1(F) =∫


Sinceα+ϵ >1 and Φ1 is a concave function, we obtain that Φ1(µτ11(F)) α+ϵ1 Φ1(µ(F1 ))

That is

{Φ−1(µτ11(F))}−1 (α+ϵ){Φ−1(µ(F1 ))}−1.

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

supfΦ1∥χX/Kf◦τ∥Φ supfΦ1∥χx/kf∥Φ ≤α+ϵ.

finally, sinceMχKCτ is a compact operator, we get

∥Cτ −MχKCτ = supfΦ1(1−χk)Cτf∥Φ = supfΦ1∥χx/kCτf∥Φ ≤α+ϵ.

Example 3.0.5. Let Φ be an Orlicz function such that Φ21n(2n) 0 as n → ∞. Putand 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) = 2n2for n≥4, andτ(x) = 5xfor all x∈(−∞,0]. Then a simple function gives h = 7/4χ1+ 1/4χ2 + 3/8χ3+ 1/3χ2n+1:n3 + 1/4χ2n:n4+ 1/5χ(−∞,0], and α = 13. Thus

∥Cτe 13 on LΦ(Ω).



3.1 Modular and norm continuity of composition op- erators

For the modular and norm continuity of composition operatorsCτ in an Orlicz spacesLΦ(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 operatorCτ inLΦ(Ω). If Φ satisfies

2 condition for large u, then the problem of continuity of the composition operatorCτ 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 fromLΦ(Ω) into itself are explained in terms of the Radon-Nikodym derivative τ1.

Theorem 3.1.1. Assume that τ : Ωis 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(τ1(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:


Let assume that the conditionIΦ(Cτx)≤KIΦ(x) holds. If A Σ and µ(A) = 0,then non singularity of τ gives µ(τ1(A)) = 0 and we have µ(τ1(A)) = Kµ(A). Thus suppose that A∈Σ and 0< µ(A)<∞. Let a∈(aΦ, bΦ) and taking x=A. Then

IΦ =∫

AΦ(a)dµ(s) = Φ(a)µ(A)<∞. SinceCτχA=χτ1(A) then by (1) we have

Φ(a)µ(τ1(A)) = IΦ(Cτx)≤KIΦ(x) =KΦ(a)µ(A).

Since 0<Φ(a)<∞, then we have µ(τ1(A))≤Kµ(A).

Sufficient condition:

Let assume that 0 aΦ < bΦ ≤ ∞. and condition (2) satisfied. From this we have µ◦τ1 ≪µby Radon-Nikodym theorem we have,µ◦τ1(A) = ∫

Afτ(t)dµ(t) forA∈Σ and for some function fτ locally integrable on Ω and fτ L(Ω) and ∥fτ K.

Otherwise,there isA Σ with 0< µ(A)<∞ such that fτ(t)> K for any t∈A.This implies thatµ◦τ1(A) = ∫

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


IΦ(Cτx) =
















Theorem 3.1.2. Assume thatτ : Ωis a measurable nonsingular transformation. Then the composition operatorCτ 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 condition2 for all u >0,then (3) and (2) are equivalent.

Proof. Necessary condition:

Let assume that condition (3) hold and by puttingx=χA whereA∈Σ and 0< µ(A)<∞ we get


Φ1(1/µ(τ1(A))) Φ1(1/µ(A))M

Φ1(µ(A)1 )≤MΦ1(µ(τ11(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= Φ1(v) that Φ(MΦ1(v))≤LΦ(Φ1(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

Φ1(µ(A)1 )≤MΦ1(µ(τ11(A)))Φ1(µ(τL1(A)))




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