Composition Operators on Weighted Orlicz
Spaces
Heera Saini
Assistant Professor, P.G. Department of Mathematics,
Govt. M. A. M. College, Jammu, J&K - 180 006, India.
Abstract. In this paper, we study the boundedness of composition operators between any two weighted Orlicz spaces.
Keywords. Composition operators, Boundedness, Orlicz spaces, Musielak-Orlicz spaces, weighted Orlicz spaces.
1
Introduction
Let X = (X,Σ, µ) be a σ- finite complete measure space. Any measurable nonsingular transformation τ induces a composition operator Cτ from L0(X) to itself defined by
Cτf(x) = f(τ(x)), x∈X, f ∈L0(X),
whereL0(X) denotes the linear space of all equivalence classes of all real valued Σ-measurable function on X, where we identify any two functions that are equal µ-almost everywhere on X.
A nondecreasing continuous convex function φ : [0, ∞) → [0, ∞) for which φ(0) = 0 and limx→∞φ(x) =∞is called a Y oung f unction. A function Φ : X×[0,∞)−→[0,∞) is
said to be a generalized Y oung f unction orM usielak−Orlicz f unction if
(i) Φ(x,·) is a Young function for almost every x∈X and
(ii) Φ(·, u) is Σ- measurable for every u≥0.
For any generalized Young function Φ, the M usielak−Orlicz space associated with Φ, denoted by LΦ(X), is defined as the set of all f ∈L0(X) such that
IΦ(λf) =
Z
X
Φ(x, λ|f(x)|)dµ(x)<∞ for some λ=λ(f)>0.
The function IΦ is called the modular. The space LΦ(X) is a Banach space with the Luxemburg−N akano norm
kfkΦ = inf{λ >0| IΦ(f /λ)≤1}.
Letφ be an Orlicz function andw be a weight inX i.e. an a.e. positive and integrable real valued function inX. Then Φ :X×[0,∞)−→[0,∞) defined by
Φ(x, u) = φ(u)w(x), x∈X, u≥0,
is a generalized Young function. The resulting Musielak-Orlicz space is calledweighted Orlicz space and is denoted by Lφ
w(X). In this case, the modular IΦ is given by
IΦ(f) =
Z
X
φ(|f(x)|)w(x)dµ(x).
If Φ is independent ofx, then the resulting Musielak-Orlicz space is simply calledOrlicz space and is denoted by Lφ(X).
Composition operators on Orlicz spaces have also been studied in [3], [4], [5],[8] and [14]. The techniques used in this paper essentially depend on the conditions of embedding of one Orlicz space into another (see, [11, Page 45] for details).
2
Boundedness of Composition Operators
In this section, we study the boundedness of composition operators on weighted Orlicz spaces.
Lemma 2.1. ( [11, Lemma 8.3] ) Let (X,Σ, µ) be a σ-finite nonatomic measure space,
{αn}a sequence of positive numbers and {sn} a sequence of measurable, finite, non-negative
functions on X such that forn = 1,2, . . .
Z
X
sn(x)dµ(x)≥2nαn.
Then there exist an increasing sequence {nk} of integers and a sequence {Ak} of pairwise
disjoint measurable sets such that for k = 1,2, . . .
Z
Ak
snk(x)dµ(x) =αnk.
Theorem 2.2. Let (X,Σ, µ) be a σ-finite nonatomic measure space, w1 and w2 be weights in X and τ : X → X be a measurable non-singular transformation such that τ(X) = X. Denote by gτ the Raydon-Nikodym derivative dµ◦τ−1/dµ. Then the composition operator
Cτ : Lφw11(X) −→ L
φ2
w2(X) is bounded if and only if there exist a, b > 0 and 0 ≤ h ∈ L 1(X)
such that
φ2(au)(w2◦τ−1)(x)gτ(x)≤bφ1(u)w1(x) +h(x)
Proof. Suppose that the given condition holds. Let 0 6=f ∈Lφ1
w1(X). Let M ≥ 1 be a real number satisfying M(b+khk1)≥1.Then
IΦ2
Cτf
(M(b+khk1)kfkΦ1)/a
= Z X φ2
a|Cτf(x)|
M(b+khk1)kfkΦ1
w2(x)dµ(x)
≤ 1
M(b+khk1)
Z
X
φ2
a|f(τ(x))| kfkΦ1
w2(x)dµ(x)
= 1
M(b+khk1)
Z
τ(X) φ2
a|f(y)| kfkΦ1
(w2◦τ−1)(y)d(µ◦τ−1)(y)
= 1
M(b+khk1)
Z
X
φ2
a|f(y)| kfkΦ1
(w2◦τ−1)(y)d(µ◦τ−1)(y)
= 1
M(b+khk1)
Z
X
φ2
a|f(y)| kfkΦ1
(w2◦τ−1)(y)gτ(y)dµ(y)
≤ 1
M(b+khk1)
Z
X
b φ1
|f(y)| kfkΦ1
w1(x) +h(y)
dµ(y)
≤ 1.
ThuskCτfkΦ2 ≤ Ma(b+khk1)kfkΦ1. This shows that Cτ is bounded.
Consider the function
hn(x) = sup u≥0
φ2(2−nu)(w2◦τ−1)(x)gτ(x)−2nφ1(u)w1(x)
.
Write X =
∞
S
i=1
Xi,where {Xi}∞i=1 is a pairwise disjoint sequence of measurable subsets ofX
with µ(Xi)<∞ for every i= 1,2, . . .
For every q ∈ Q+,we put f
q,i(x) = qχXi(x), where χXi is the characteristic function of Xi.
Then it can be shown that
hn(x) = sup
q∈Q+
i∈N
φ2(2−nfq,i(x))(w2 ◦τ−1)(x)gτ(x)−2nφ1(fq,i(x))w1(x)
.
Taking (fk) to be a rearrangement of (fq,i) withf1 =f0,i, the above equation can be rewritten
as
hn(x) = sup k∈N
φ2(2−nfk(x))(w2◦τ−1)(x)gτ(x)−2nφ1(fk(x))w1(x)
.
Then hn are measurable and hn(x)≥ 0 for eachx ∈X. To complete the proof, we have to
show that RXhn(x)dµ(x)<∞ for some n. Suppose this is not true.
Denote
rm,n(x) = max
1≤k≤m φ2(2 −n
fk(x))(w2 ◦τ−1)(x)gτ(x)−2nφ1(fk(x))w1(x)
Then rm,n are measurable, rm,n(x) ≥ 0 and rm,n(x) is a nondecreasing sequence
tend-ing to hn(x) as m → ∞ for every x ∈ X. Thus for any n, there exists mn such that
R
X rmn,n(x)dµ(x)≥2
n. Taking r
n=rmn,n, we have
R
Xrn(x)dµ(x)≥2
n for n = 1,2, . . .
Let
En,k =
x∈X | φ2(2−nfk(x))(w2◦τ−1)(x)gτ(x)−2nφ1(fk(x))w1(x) = rn(x)
and
En =X\(En,1∪En,2∪ . . .∪En,mn).
Then µ(En) = 0.
Let
˜ fn(x) =
0 if x∈En,1∪En
fk(x) if x∈En,k\ k−1
S
j=1
En,j, k= 2,3, . . . , mn.
Then
rn(x) = φ2(2−nf˜n(x))(w2 ◦τ−1)(x)gτ(x)−2nφ1( ˜fn(x))w1(x)
≥ 0.
Therefore,
Z
X
φ2(2−nf˜n(x))(w2◦τ−1)(x)gτ(x)dµ(x) = 2n
Z
X
φ1( ˜fn(x))w1(x)dµ(x)
+
Z
X
rn(x)dµ(x)
≥
Z
X
rn(x)dµ(x)
≥ 2n.
By Lemma 2.1, we obtain an increasing sequence {nk} and a sequence {Ak} of pairwise
disjoint measurable sets such that
Z
Ak
φ2(2−nkf˜nk(x))(w2◦τ
−1)(x)g
τ(x)dµ(x) = 1, k= 1,2, . . .
Put
f(x) =
˜
fnk(x) if x∈Ak
Letλ >0. Choose p large enough that 2−np ≤λ. Then
Z
X
φ2(λCτf(x))w2(x)dµ(x) =
Z
X
φ2(λf(τ(x)))w2(x)dµ(x)
=
Z
τ(X)
φ2(λf(y)) (w2◦τ−1)(y)d(µ◦τ−1)(y)
=
Z
X
φ2(λf(y))(w2◦τ−1)(y)gτ(y)dµ(y)
=
∞
X
k=1
Z
Ak
φ2(λf˜nk(y))(w2◦τ
−1)(y)g
τ(y)dµ(y)
≥
∞
X
k=p
Z
Ak
φ2(2−nkf˜nk(y))(w2◦τ
−1)(y)g
τ(y)dµ(y)
= ∞.
And
Z
X
φ1(f(x))w1(x)dµ(x) =
∞
X
k=1 2−nk
Z
Ak
φ2(2−nkf˜nk(x))(w2◦τ
−1)(x)g
τ(x)dµ(x)
−
∞
X
k=1 2−nk
Z
Ak
rnk(x)dµ(x)
≤
∞
X
k=1 2−nk
Z
Ak
φ2(2−nkf˜nk(x))(w2◦τ
−1)(x)g
τ(x)dµ(x)
=
∞
X
k=1 2−nk
≤ 1.
Thus,f ∈Lφ1
w1(X) but Cτ(f)∈/ L
φ2
w2(X),which is a contradiction. Hence,
Z
X
hn(x)dµ(x)<∞for somen.
This completes the proof.
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