**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 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, 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 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*t** ^{p}* in the deﬁnitions
of the spaces

*l*

*and*

^{p}*L*

*, 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.*

^{p}Prior to the researches of W.Orlicz ,it was W.H.Young [12] who, motivated by the functions
*u** ^{p}*(u >0) and

*v*

*(v >0) with*

^{p}^{1}

*+*

_{p}^{1}

*= 1 ,1*

_{q}*< 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
constant*C >* 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
functions*f* 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 operator*C*_{Φ}with symbol Φ is a linear operator deﬁned with
the help of composition of mapping *f◦*Φ by the formula*C*_{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, 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,
*L*^{Φ}(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

*iii) lim*_{x}* _{→∞}*Φ(x) =

*∞.*

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

**Example 2.1.1.** i) Φ* _{p}*(s) :=

^{|}

^{s}

_{p}

^{|}*with p*

^{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*
0 *p(t)dt*

*Where p(t) is right continuous fort* *≥*0, positive for*t >*0*and 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, *u*_{0} *≥*0 *such that*

*M*(2u)*≤kM*(u), (u*≥u*_{0})

**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*

*(A) = 0*

^{−1}*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* *D*1 *⊂D, D*1 *∈*Σ *such that* *µ(D*1) = *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. let*f, g* *∈L*˜^{Φ}(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}_{γ}* ^{|}*Φ(

*|f|*) +

*γ.*

^{|}

_{γ}*Φ(*

^{β}*|g|*) =

*|α|*Φ(*|f|*) +*|β|*Φ(*|g|*), but by hypothesis right hand side is integrable. Hence*αf*+*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}_{γ}*f* + _{γ}^{b}*g)∈* *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|*) *≤* *K*Φ(*|f|*), K > 0 *⇒*

∫

ΩΦ(2*|f|*)*≤K*∫

ΩΦ(*|f|*)*<∞*hence we have 2f *∈L*˜^{Φ}(Ω).

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

Now let *f*_{1} =*f* if *|f| ≤x*_{0} and 0 otherwise. Let *f*_{2} =*f−f*_{1} so that*f* =*f*_{1}+*f*_{2} and
Φ(2*|f|*) = Φ(2*|f*_{1}*|*) + Φ(2*|f*_{2}*|*)*≤*Φ(2*|f*_{1}*|*) +*K*Φ(*|f*_{2}*|*).

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, 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* *∈* *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. Let

*n*

_{0}

*∈*N such that

∑

*n**≥**n*0

1

*n*^{2} *< α* and Φ(x* _{n}*)

*≥*1 for all

*n≥n*

_{0}.

Since *µ* is diﬀuse on F, there is a measurable *F*_{0} *⊂* *F* such that *µ(F*_{0}) = ∑ 1
*n*^{2} *< α.*

We can ﬁnd a set *D*_{1} *∈* Σ, D_{1} *⊂* *F*_{0} such that *µ(D*_{1}) = 1/n^{2}_{0} . Similarly again we can
ﬁnd set *D*_{2} *∈* Σ, D_{2} *⊂* *F*_{0} *−D*_{1} such that *µ(D*_{2}) = 1/(n_{0} + 1)^{2}. On repeating this
process, we can ﬁnd disjoint sets *D*_{n}*∈*Σ such that *µ(D** _{n}*) = 1/(n

_{0}+

*n−*1)

^{2}

*, n*

*≥*1.

Let*F*_{k}*⊂D*_{k}*, F*_{k}*∈*Σ such that *µ(F** _{k}*) =

^{µ(D}_{Φ(x}

^{K}^{)}

*n*).
Let*f* =

∑*∞*
*n=1*

*x*_{n}*χ*_{F}* _{n}* then clearly f is measurable. Now

∫

Ω

Φ(*|f|*)dµ=

∑*∞*
*n=1*

Φ(x* _{n}*)µ(F

*)*

_{n}= ∑

*n**≥**n*0

1/n^{2} *<∞*

so*f* *∈L*˜^{Φ}(Ω).

8 Now

∫

Ω

Φ(2f)dµ=

∑*∞*
*n=1*

Φ(2x* _{n}*)µ(F

*)*

_{n}*≥* ∑

*n**≥**n*0

*nΦ(x** _{n}*)µ(F

*)*

_{n}= ∑

*n**≥**n*0

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, and*µ*be the counting measure,
i.e., *µ({i}*) = 1, *i≥*1. Let Φ(x) =*e*^{x}^{2} *−*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.* Let*f*_{1}*, f*_{2} *∈L*^{Φ}(Ω).

Then there exist *α*_{1} *>*0 and *α*_{2} *>*0 such that ∫

ΩΦ(α_{1}*|f|*)dµ < *∞*and ∫

ΩΦ(α_{2}*|f|*)dµ <*∞*.
Let*α*=*min{α*_{1}*, α*_{2}*}*, then *α >*0.

Now by using convexcity of Φ we get

∫

ΩΦ(^{α}_{2}(f1+*f*2))dµ*≤* ^{1}_{2}∫

ΩΦ(α1*f*1)dµ+^{1}_{2}∫

ΩΦ(α2*f*2)dµ.

Hence ∫

ΩΦ(^{α}_{2}(f_{1}+*f*_{2}))dµ <*∞*, Where ^{α}_{2} *>*0.

Hence *f*_{1} +*f*_{2} *∈L*^{Φ}(Ω). Thus*L*^{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 *n*_{0} *∈* N such that *|β| ≤* *n*_{0} *⇒ |βf| ≤ |n*_{0}*f|*. 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.* Let*f, 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

_{n}*f)*

*≤*Φ(αf) and Φ(α

_{n}*f)*

*→*0 when Φ is a continuous Young function. Hence by dominated convergence then ∫

Ω(α_{n}*f*)dµ*→* 0 so that
there is some*n*_{0} *∈*Nsuch that ∫

ΩΦ(α_{n}_{0}*f)dµ≤*1. Thus *α*_{n}_{0}*f* *∈***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 on*L*^{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*

*|**λ**α**|*

*|*)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 *α*_{1} *>* 0 and *α*_{2} *>* 0
such that*∥f*_{1}*∥*Φ *< α*_{1}+_{2}* ^{ϵ}* and

*∥f*

_{2}

*∥*Φ

*< α*

_{2}+

_{2}

*.*

^{ϵ}Let*β* =*α*_{1}+*α*_{2}

Since*f*1+*f*2 *∈L*^{Φ}(Ω),*∥f*1+*∥f*2 *<∞.*
consider∫

ΩΦ((f_{1}+*f*_{2})/β)dµ=∫

ΩΦ(_{α}^{f}^{1}

1*.*^{α}_{β}^{1} + _{α}^{f}^{2}

2*.*^{α}_{β}^{2})dµ

*≤* ^{α}_{β}^{1} ∫

Ω Φ(_{α}^{f}^{1}

1)dµ+^{α}_{β}^{2} ∫

ΩΦ(_{α}^{f}^{2}

2)( by using convexity of Φ)

*≤* ^{α}_{β}^{1} +^{α}_{β}^{2} = 1

*⇒* _{β}^{1}(f_{1}+*f*_{2})*∈L*^{Φ}(Ω)

12
Hence *∥*^{1}* _{β}*(f

_{1}+

*f*

_{2})

*∥*Φ =

_{β}^{1}

*∥*(f

_{1}+

*f*

_{2})

*∥*Φ

*≤*1

*⇒ ∥*(f1+*f*2)*∥*Φ *≤β*

But*β* =*α*1+*α*2*,∥f*1*∥< α*1+_{2}^{ϵ}*and∥f*2*∥< α*2 +_{2}^{ϵ}

*⇒ ∥f*_{1}*∥*Φ+*∥f*_{2}*∥*Φ *< α*_{1}+*α*_{2}+*ϵ*

*⇒ ∥f*_{1}*∥*Φ+*∥f*_{2}*∥*Φ *< β*+*ϵ*

*⇒ ∥f*_{1}*∥*Φ+*∥f*_{2}*∥*Φ *<|*(f_{1} +*f*_{2})*∥*Φ+*ϵ*
Since*ϵ >*0 be arbitrary

*⇒ ∥f*_{1}*∥*Φ+*∥f*_{2}*∥*Φ *≤ ∥*(f_{1}+*f*_{2})*∥*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 as Luxemburg-*

**Nakano Norm.****Lemma 2.1.2.** *∥f∥*Φ *≤*1 *if and only if* ∫

ΩΦ(f)dµ*≤*1.

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

1

*α* *∈***B**Φ. 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 in*L*^{Φ}(Ω) such that*∥f*_{n}*−f*_{m}*∥*Φ *→*0 as*m, n→ ∞*
we have to construct a*f* *∈L*^{Φ}(Ω) which satisfy*∥f**n**−f∥*Φ *→*0 as*n* *→ ∞*. Since we considered
Φ is a Young function there are two cases.

Let *x*_{0} = *sup{x* *∈* R^{+} : Φ(x) = 0*}*. Then by deﬁnition of Φ the above set deﬁne in the
braces is compact so 0 *≤* *x*_{0} *<* *∞*. Then by hypothesis there exist numbers *k*_{m,n}*≥* 0
(k^{−}_{m,n}^{1} *≤ ∥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)

*> x*

_{0}

*} ∈*Σ is at most

*σ−*ﬁnite for

*µ. Let*

*B*

*=*

_{k}*B*

_{k}*=*

^{mn}*{ω*:

*k*

*m,n*

*|F*

*n*

*−f*

*m*

*|*(ω)

*> x*0 +

*k*

^{−}^{1}

*}*, then clearly

*A*

*mn*=

*∪*

^{∞}*k=1*

*B*

*k*and for each k

*µ(B*

*)*

_{k}*<∞*. Since by condition(1)

*µ(B** _{k}*)

*≤*1 Φ(x

_{0}+

*k*

^{−}^{1})

∫

*B*_{k}

Φ(k_{m,n}*|F*_{n}*−f*_{m}*|*)dµ*≤*1 (2)

Hence each*A** _{mn}* is

*σ−*ﬁnite and let

*A*=

*∪*

*m,n*

*≥*1

*A*

*. Thus on*

_{mn}*A*

^{c}*, k*

_{m,n}*|F*

_{n}*−f*

_{m}*|*(ω)

*≤x*

_{0}, so that

*ω*

*∈*

*A*

^{c}*,|f*

*(ω)*

_{n}*−f*

*(ω)*

_{m}*| →*0 uniformly. Thus there is a measurable function

*g*

_{0}on

*A*

*such that*

^{c}*f*

*(ω)*

_{n}*→g*

_{0}(ω), and

*|g*

_{0}(ω)

*| ≤x*

_{0}

*, 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_{mn}*ϵ)}*)

*≤* 1

Φ(k_{mn}*ϵ)*

∫

*B*

Φ(k*m,n**|F**n**−f**m**|*)dµ

*≤*[Φ(k_{mn}*ϵ)]*^{−}^{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 the*σ−*ﬁniteness property we have*{f*_{n}*}* is a cauchy sequence

14
in measure. If *f*_{n}*→f*˜in measure. Then there is subsequence *{f*_{n}_{i}*}* such that then*f*_{n}_{i}*→f*
a.e. But *{f**n**i**}* is a cauchy sequence in *∥f∥*Φ, we get *∥f**n**∥*Φ *→ρ* and hence *∥f**n**i**∥*Φ *→ρ. By*
using Fatou’s lemma we get

∫

Φ

(*|f|*

*ρ* )*≤lim*_{t}_{→∞}*inf*

∫

Ω

Φ( *f*_{n}_{i}

*∥f**n**i**∥*Φ

)*≤*1
*.*

Hence *f* *∈L*^{Φ}(Ω). Let m be ﬁxed and *k* *≥*0 be given, then Φ(*|f*_{n}_{i}*−f*_{n}_{j}*|k)→*Φ(*|f−f*_{n}_{j}*|k)*
as*i→ ∞*, a.e. let *n*_{i}*, n*_{j}*≥n*_{0} and *k*_{n}_{i}_{,n}_{j}*≥k* then

∫

Ω

Φ(k*|F*_{n}_{i}*−f*_{n}_{j}*|*)dµ*≤*

∫

Ω

Φ(k_{i}_{,n}_{j}*|F*_{n}_{i}*−f*_{n}_{j}*|*)dµ*≤*1

let *n*_{i}*→ ∞* then by using Fatou’s lemma we get *∥f* *−f*_{n}_{j}*∥*Φ *≤* ^{1}* _{k}*. Since

*k >*0 is arbitrary

*∥f−f*_{n}_{j}*∥*Φ *→*0 If *f*_{n}* _{j}* is any other subsequence with limit

*f*

*, then*

^{′}*{f*

_{n}*′*

*j**, f*_{n}_{i}*, 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 *C**τ* between Orlicz spaces
L^{Φ}(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 measureµon*Ω*is completeif whenever E∈*Ω, F

*⊆E andµ(E) = 0,*

*then F*

*∈*Ω.

**Definition 3.0.14.** *A measure* *µon* Ω*isσ− 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* *is absolutely 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) =* ∫

*E**f dµ. Also f is unique in the sense that if* *ν(E) =* ∫

*E**gdµ*
*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) =*∫

*E**f 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* *C**T* :*F*(X)*→F*(Y) *by* *C**T* =*f◦T* *for every* *f* *∈F*(X).If*C**T* *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*_{τ}*f*_{n}*}* be sequence in L^{Φ}(Ω) such that*f*_{n}*→f* and *C*_{τ}*f*_{n}*→g* for some
*f, g∈L*^{Φ}(Ω). Then we can ﬁnd a subsequence *{f**n*_{k}*}* of *{f**n**}*such that

Φ(*|f*_{n}_{k}*−f|*)(x)*→*0 for*µ−*almost all *x∈*Ω.

from non-singularity of*τ*,

Φ(*|f**n*_{k}*−f|τ*)(x)*→*0 for *µ−* almost all *x∈*Ω.

From the above two relation, we conclude that C_{τ}*f* = *g. This proved that the graph is*
closed and by using closed graph theorem C* _{τ}* is continuous.

**Theorem 3.0.8.** *Let* Ω_{2} *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. Let*F* =*N*(h, α*−*_{2}* ^{ϵ}*), 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 *E*_{n}*, n* *∈* *N*, such
that *E*_{n+1}*⊆* *E*_{n}*⊆* *F,*0 *< µ(E** _{n}*)

*<*

_{n}^{1}. Let us deﬁne

*f*

*= Φ*

_{n}

^{−}^{1}(1/µ(E

*))χ*

_{n}

_{E}*. Then*

_{n}*∥f*_{n}*∥*Φ = 1 for all *n* *∈* *N*. We have to prove *f*_{n}*→* 0 weakly.To prove this we have to
show that∫

Ω*f*_{n}*g* *→*0 for all *g* *∈L*^{Ψ}(Ω), where Ψ is the complementary function to Φ.

Let*A⊆F* with 0*< µ(A)<∞* and *g* =*χ** _{A}* since Φ(x)

*≻≻x, then we have*

*|*∫

Ω*f*_{n}*χ*_{A}*dµ|*= Φ^{−}^{1}(_{µ(E}^{1}

*n*))µ(A*∩E** _{n}*)

*≤*Φ

^{−}^{1}(

_{µ(E}^{1}

*n*))µ(E* _{n}*) =

^{Φ}

^{−}_{1/µ(E}

^{1}

^{(1/µ(E}

^{n}^{))}

*n*) *→*0, n*→ ∞*
Since simple functions are dense in*L*^{Ψ}(Ω), thus*f** _{n}*is converge to 0 weakly. Now assume
that F consists of inﬁnitely many atoms. Let (E

*)*

_{n}

^{∞}*be disjoint atoms in F. Again on*

_{n=0}18
putting*f** _{n}* as above. If

*µ(E*

*)*

_{n}*→*0, then by using the similar argument we had above,

∫

Ω*f**n**χ**A**dµ→*0. Now we have to prove that*∥C**τ**f**n**∥*Φ *≥α−*^{ϵ}_{2}*.*Since 0 *< α−*^{ϵ}_{2} *<*1 we
have

*∥C**τ**f**n**∥*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*_{τ}*f*_{n}*−T f*_{n}*∥*Φ*−* *ϵ*
2

*≥ ∥C*_{τ}*f*_{n}*∥*Φ*− ∥T f*_{n}*∥*Φ*−* *ϵ*
2

*≥*(α*−* *ϵ*

2)*− ∥T f*_{n}*∥*Φ*−* *ϵ*
2*.*

for all*n* *∈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. Sinceϵ*is 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 write*K* =*{E*_{1}*, E*_{2}*, ..., E*_{m}*}*

where *E*_{1}*, E*_{2}*, ..., E** _{m}* are distinct. Since (M

_{χ}

_{k}*C*

_{τ}*f)(x) =*∑

*m*

*i=1**χ** _{k}*(E

*)f(τ(E*

_{i}*)), for all*

_{i}*f*

*∈L*

^{Φ}(Ω), hence

*M*

*χ*

_{k}*C*

*τ*has ﬁnite rank. Now, let

*F*

*⊆X K*such that 0

*< µ(F*)

*<∞*, then we have

*µ◦τ*^{−}^{1}(F) =∫

*F**hdµ≤*(α+*ϵ)µ(F*).

Since*α*+*ϵ >*1 and Φ^{−}^{1} is a concave function, we obtain that
Φ^{−}^{1}(_{µ}_{◦}_{τ}_{−}^{1}1(F))*≥* _{α+ϵ}^{1} Φ^{−}^{1}(_{µ(F}^{1} _{)})

That is

*{*Φ* ^{−1}*(

_{µ}

_{◦}

_{τ}

_{−}^{1}1(F))

*}*

^{−1}*≤*(α+

*ϵ){*Φ

*(*

^{−1}

_{µ(F}^{1}

_{)})

*}*

*.*

^{−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/K}*f◦τ∥*Φ *≤* sup_{∥}_{f}_{∥}_{Φ}_{≤}_{1}*∥χ*_{x/k}*f∥*Φ *≤α*+*ϵ.*

ﬁnally, since*M**χ*_{K}*C**τ* is a compact operator, we get

*∥C*_{τ}*−M*_{χ}_{K}*C** _{τ}* = sup

_{∥}

_{f}

_{∥}_{Φ}

_{≤}_{1}

*∥*(1

*−χ*

*)C*

_{k}

_{τ}*f∥*Φ = sup

_{∥}

_{f}

_{∥}_{Φ}

_{≤}_{1}

*∥χ*

_{x/k}*C*

_{τ}*f∥*Φ

*≤α*+

*ϵ.*

**Example 3.0.5.** *Let* Φ *be an Orlicz function such that* ^{Φ}^{−}_{2}^{1}*n*^{(2}^{n}^{)} *→*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* = 7/4* _{χ1}*+ 1/4

*+ 3/8*

_{χ2}*+ 1/3*

_{χ3}

_{χ2n+1:n}

_{≥}_{3}+ 1/4

_{χ2n:n}

_{≥}_{4}+ 1/5

_{χ(}

_{−∞}

_{,0]}*, and*

*α*=

^{1}

_{3}

*. Thus*

*∥C*_{τ}*∥**e* *≥* ^{1}_{3} *on* *L*^{Φ}(Ω).

20

**3.1** **Modular and norm continuity of composition op-** **erators**

For the modular and norm continuity of composition operators*C** _{τ}* in an Orlicz spaces

*L*

^{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 operator

*C*

*in*

_{τ}*L*

^{Φ}(Ω). If Φ satisﬁes

∆_{2} condition for large u, then the problem of continuity of the composition operator*C** _{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}* from

*L*

^{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_{τ}*x) =I*_{Φ}(x) *whenever* *I*_{Φ}(x)*<∞.*
*2. if* 0*≤a*_{Φ} *< b*_{Φ} *≤ ∞* *then the inequality*

*I*_{Φ}(C_{τ}*x)≤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,*I*Φ *<∞* iﬀ *∥x∥**∞**≤a*Φ *⇒ ∥C**τ**x∥**∞**≤a*Φ *⇒I*Φ(C*τ**x) = 0 =I*Φ*.*
2. Let assume that 0*≤a*_{Φ} *< b*_{Φ} *≤ ∞*.

*Necessary condition:*

Let assume that the condition*I*_{Φ}(C_{τ}*x)≤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*, b*Φ) and taking *x*=*aχ**A*. Then

*I*_{Φ} =∫

*A*Φ(a)dµ(s) = Φ(a)µ(A)*<∞.*
Since*C*_{τ}*χ** _{A}*=

*χ*

_{τ}*−*1(A) then by (1) we have

Φ(a)µ(τ^{−}^{1}(A)) = *I*_{Φ}(C_{τ}*x)≤KI*_{Φ}(x) =*K*Φ(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} *≪µ*by Radon-Nikodym theorem we have,*µ◦τ*^{−}^{1}(A) = ∫

*A**f** _{τ}*(t)dµ(t) for

*A∈*Σ and for some function

*f*

*locally integrable on Ω and*

_{τ}*f*

_{τ}*∈*

*L*

*(Ω) and*

^{∞}*∥f*

_{τ}*∥*

_{∞}*≤*

*K.*

Otherwise,there is*A* *∈*Σ with 0*< µ(A)<∞* such that *f** _{τ}*(t)

*> K*for any

*t∈A.This*implies that

*µ◦τ*

^{−}^{1}(A) = ∫

*A**f**τ*(t)dµ(t)*> Kµ(A), which is a contradiction to (2).*

Therefore,

*I*_{Φ}(C_{τ}*x) =*

∫

Ω

Φ(*|C*_{τ}*x(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*→*Ω*is 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*_{τ}*x∥*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 putting*x*=*χ** _{A}* where

*A∈*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}_{−}^{1}1(A))) (4)
for all *A∈*Σ with 0*< µ(A)<∞*

Since Φ satisﬁes ∆_{2} condition for all *u >*0,it follows that
*L*:=*sup**u>0*Φ(M u)

Φ(u) *<∞*,

and Φ(M u)*≤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

*M*Φ* ^{−1}*(v)

*≤*Φ

^{−1}*{*Φ(MΦ

*(v))*

^{−1}*} ≤*Φ

*(Lv) for all*

^{−1}*v >*0

From the condition (4) we get

Φ^{−}^{1}(_{µ(A)}^{1} )*≤M*Φ^{−}^{1}(_{µ(τ}_{−}^{1}_{1(A))})*≤*Φ^{−}^{1}(_{µ(τ}_{−}* ^{L}*1(A)))