orlicz function spaces and composition …ethesis.nitrkl.ac.in/5436/1/411ma2075.pdforlicz function...
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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).
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