Composition Operators on Weighted Orlicz Sequence Spaces

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Composition Operators on weighted Orlicz

sequence spaces

Heera Saini Aditi Sharma and Neetu Singh

Abstract. In this paper, we study the boundedness of composition operators between any two weighted Orlicz sequence spaces

Keywords. Composition operators, Boundedness, weighted Orlicz sequence spaces.

1 Introduction

Let φ : [0, ∞) → [0, ∞) be a Young function, that is, a nondecreasing continuous convex function for which φ(0) = 0 and limx→∞φ(x) = ∞. Let (X,Σ, µ) be a σ-finite and purely

atomic measure space, that is

X =

∞ [

n=1 An,

where An are atoms with the measure µ(An) = an > 0 for all n ∈ N and w = {wn} be a weight in X i.e positive and summable real valued sequence. Then the weighted Orlicz sequence space l_wφ({an}) is defined as the space of all real sequencesf ={fn}∞n=1 such that Iφ,w(λf,{an})<∞for some λ >0, where

Iφ,w(f,{an}) =

∞ X

n=1

φ(|fn|)wnan.

This space is a Banach space with the norm

kfkφ,w,{an} = inf{λ >0 | Iφ,w(f/λ,{an})≤1}.

Throughout the paper, we assume (X,Σ, µ) to be aσ-finite and purely atomic measure space with atoms {An} of measure µ(An) =an>0 for any n∈N, τ :X →X to be a measurable non-singular transformation such that τ(X) = X and bn :=µ(τ−1(An))/µ(An).

Composition operators on Orlicz spaces have also been studied in [3], [4], [5],[9] and [17]. The techniques used in this paper essentially depend on the conditions of embedding of one Orlicz space into another (see, [13, Page 48] and [19] for details).

2010Mathematics Subject Classification. 47B33, 47B37.

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2 Boundedness of Composition Operators

In this section, we study the boundedness of composition operators on weighted Orlicz sequence spaces.

Theorem 2.1. Let w1 = {w1,n} and w2 = {w2,n} be weights in X. Then the composition

operator Cτ : lφw11({an})−→ l

φ2

w2({an}) is bounded if and only if there exist a, b, δ > 0 and a

sequence {cn} of nonnegative integers in l1 such that φ1(u)w1,nan < δ implies

φ2(au)w2,τ−1_(n)a_nb_n ≤bφ₁(u)w_1,na_n+c_n

for all n ∈_N and all u≥0.

We divide the proof in two parts, the sufficient part and the necessary part. The proof of the necessary part is as under:

Proof. Suppose that the given condition holds. Let f = {fn}∞n=1 ∈ lwφ11({an})\ {0}, then

Iφ1,w1

f

kfkφ₁,w₁,{an},{an}

≤ 1. Let M ≥ 1 be a real number satisfying M(b+kck1) ≥ 1,

where kck1 =

∞ P

n=1

cn. Then

Iφ2,w2

Cτf

(M(b+kck1)kfkφ1,w1,{an})/a

,{an}

= ∞ X n=1 φ2

a|Cτfn|

M(b+kck1)kfkφ1,w1,{an}

w2,nan

≤ 1

M(b+kck1)

∞ X

n=1 φ2

a|fτ(n)|

kfkφ1,w1,{an}

w2,nan

= 1

M(b+kck1)

X

n∈τ(X) φ2

a|fn|

kfkφ1,w1,{an}

w2,τ−1_(n)µ(τ−1(A_n))

= 1

M(b+kck1)

∞ X

n=1 φ2

a|fn|

kfkφ1,w1,{an}

w2,τ−1_(n)a_nb_n

≤ 1

M(b+kck1)

∞ X

n=1

b φ1

|fn|

kfkφ1,w1,{an}

w1,nan+cn

≤ 1.

Thus kCτfkφ2,{an} ≤

M

a(b+kck1)kfkφ1,w1,{an}. This shows that Cτ is bounded.

Definition 2.2. Let us consider the sequences z={zn},of real numbers. Forα, β, γ, δ >0, define

X_α0 ={z| there exists k∈_N such that

∞ X

n=k

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Y_β0 ={z | there exists k ∈_N such that X n=k

φ2(β|zn|)w2,nan <∞}

and Fn(α, β, γ, δ)

= sup{φ2(β|y|)w2,τ−1_(n)a_nb_n | φ₁(α|y|)w_1,na_n<min(δ, γ−1φ₂(β|y|)w_2,τ−1_(n)a_nb_n)}.

If there is no y for which φ1(α|y|)w1,nan < min(δ, γ−1φ2(β|y|)w2,τ−1_(n)a_nb_n), then we put

Fn(α, β, γ, δ) = 0.

Theorem 2.3. (i) If 0< α0 < α00, then X0 α00 ⊂X

0

α0. Moreover,

T

α X0

α 6=∅.

(ii) Fn(α, β, γ, δ) is nonincreasing with respect to α, γ and nondecreasing with respect to β, δ.

(iii) If φ1(α|y|)w1,nan< δ, then

φ2(β|y|)w2,τ−1_(n)a_nb_n≤γφ₁(α|y|)w_1,na_n+F_n(α, β, γ, δ).

Proof. (i) Let z={zn}n=1∞ ∈X_α000. Then there exists k∈N such that

∞ X

n=k

φ1(α00|zn|)w1,nan<∞.

Since α0 < α00, it follows that for each n ∈_N,

φ1(α0|zn|)w1,nan≤φ1(α00|zn|)w1,nan

which implies

∞ X

n=k

φ1(α0|zn|)w1,nan≤

∞ X

n=k

φ1(α00|zn|)w1,nan<∞.

Thusz={zn}∞n=1 ∈X_α00.

(ii) Follows from the definition of Fn(α, β, γ, δ).

(iii) Let us suppose that φ1(α|y|)w1,nan< δ. If

φ1(α|y|)w1,nan<min(δ, γ−1φ2(β|y|)w2,τ−1_(n)a_nb_n),

then y∈En(α, β, γ, δ). Thus,

φ2(β|y|)w2,τ−1_(n)a_nb_n≤F_n(α, β, γ, δ).

Since γφ1(α|y|)w1,nan≥0,it follows that

If

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then

γ−1φ2(β|y|)w2,τ−1_(n)a_nb_n ≤φ₁(α|y|)w_1,na_n,

which implies that

φ2(β|y|)w2,τ−1_(n)a_nb_n≤γφ₁(α|y|)w_1,na_n.

Again, since Fn(α, β, γ, δ)≥0, we obtain

Theorem 2.4. Let A and B be two nonempty sets of positive numbers. If

Cτ

\

α∈A X_α0

! ⊂ [

β∈B Y_β0,

then there exist α∈A, β ∈B, γ >0, δ >0 and m ∈_N such that

∞ X

n=m

Fn(α, β, γ, δ)<∞.

Proof. Suppose the theorem is not true. Chooseαk ∈Aand β ∈B,k = 1,2, . . . , such that for all α ∈ A there is a k0 with αk ≥ α for all k > k0, and for all β ∈B there is a k00 with βk ≤β for all k > k00. Put

d(n, k) = Fn(αk, βk, k, k−2).

Then

∞ P

n=m

d(n, k) = ∞ for all k and m. Let n1, n2, . . . be an increasing sequence of indices

that partitions the set of natural numbers into segmentsN1 ={1,2, . . ., n1}, Nk={nk−1+ 1, . . . , nk}, k= 2,3, . . . , such that

X

m∈Nk

d(n, k)> 1

k (2.1)

X

m∈Nk\{nk}

d(n, k)≤ 1

k, (2.2)

for k = 1,2, . . . ,where we put P

n∈∅

d(n, k) = 0. Write

N_k+ ={n∈Nk |d(n, k)>0}.

By the definition of Fn and d(n, k) and the inequality (2.1), for every n ∈N_k+, there exists

zn with the property that

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such that

X

n∈N_k+

φ2(β|zn|)w2,τ−1_(n)a_nb_n >

1

k. (2.3)

By the definitions of Fn and d(n, k) and the inequality (2.2), we have

X

n∈N_k+

φ1(αk|zn|)w1,nan <

X

n∈N_k+\{nk}

1

kφ2(β|zn|)w2,τ−1(n)anbn+ 1 k2

≤ 1

k

X

n∈N_k+\{nk}

d(n, k) + 1 k2

≤ 2

k2. (2.4)

Thus, for eachn ∈N_k+,there exists zn ∈Rsuch that (2.3) and (2.4) hold. For n∈Nk\Nk+, choose zn ∈ R such that Pn∈Nk\Nk+φ1(α|zn|)w1,nan < ∞. Let z = {zn}. We now take an

arbitrary α ∈ A and choose k0 so large that for all k ≥ k0, we have αk ≥k. Then by using (2.4), we obtain

∞ X

n=n_k₀₋₁+1

φ1(α|zn|)w1,nan =

∞ X

k=k0

X

n∈Nk

φ1(α|zn|)w1,nan

≤ ∞ X k=k0   X

n∈Nk

φ1(αk|zn|)w1,nan+

X

n∈Nk\Nk+

φ1(α|zn|)w1,nan





< ∞.

This shows that z∈ T

α∈A X0

α. On the other hand, for any arbitraryβ ∈B and m∈N, choose k00large enough that for allk ≥k00, we haveβ ≥βk andnk−1+ 1≥m. By change of variable in the second step below and then readjusting the variable, and using (2.3) in the last step, we obtain

∞ X

n=m

φ2(β|Cτ(zn)|)w2,nan =

∞ X

n=m

φ2(β|zτ(n)|)w2,nan

=

∞ X

n=m

φ2(β|zn|)w2,τ−1_(n)µ(τ−1(A_n))

=

∞ X

n=m

φ2(β|zn|)w2,τ−1_(n)a_nb_n

≥ ∞ X

k=k0

X

n∈N_k+

φ2(βk|zn|)w2,τ−1_(n)a_nb_n

= ∞.

Hence Cτ(z)6∈ S

β∈B

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Corollary 2.5. (i) If Cτ(T α∈A

X0 α) ⊂

S

β∈B Y0

β, then there exist α ∈ A, β ∈ B, γ > 0, δ > 0

and a sequence {cn} of nonnegative integers such that

∞ P

n=m

cn < ∞ for some m ∈ N

and that φ1(α|y|)w1,nan< δ implies

φ2(β|y|)w2,τ−1_(n)anbn≤γφ1(α|y|)w1,nan+cn,

for all n∈_N.

(ii) Suppose there exist α ∈ A, β ∈ B, γ > 0, δ > 0 and a sequence {cn} of nonnegative

integers such that

∞ P

n=m

cn <∞ for somem ∈_N and that φ1(α|y|)w1,nan< δ implies

φ2(β|y|)w2,τ−1_(n)anbn≤γφ1(α|y|)w1,nan+cn,

for all n∈_N, then Cτ(Xα0)⊂Yβ0.

(iii) If Cτ(T

α∈A

X_α0)⊂ S

β∈B

Y_β0, then there exist α∈A and β ∈B such that Cτ(X_α0)⊂Y_β0.

(iv) Cτ(Xα0)⊂Yβ0 if and only if there exist γ >0, δ >0and a sequence {cn}of nonnegative

integers such that

∞ P

n=m

cn <∞ for somem ∈N and that φ1(α|y|)w1,nan< δ implies

φ2(β|y|)w2,τ−1_(n)a_nb_n≤γφ₁(α|y|)w_1,na_n+c_n,

for all n∈_N.

(v) Let Xα = {z | P

∞

n=1φ1(α|zn|)w1,nan < ∞}. If Cτ(Xα) ⊂

S

β∈B Y0

β, then there exists

β ∈B such that Cτ(Xα0)⊂Yβ0. In particular, if Cτ(Xα)⊂Yβ0, then Cτ(Xα0)⊂Yβ0.

Proof. (i) Follows from Theorem 2.4 and Corollary 2.3(iii) with Fn(α, β, γ, δ) = cn.

(ii) Let z={zn}n=1∞ ∈Xα0. Then there exists k∈N such that

∞ X

n=k

φ1(α|zn|)w1,nan<∞

where we suppose k≥m and φ1(α|zn|)w1,nanδ for n≥k. Thus

∞ X

n=k

φ2(β|Cτ(zn)|)w2,nan =

∞ X

n=k

φ2(β|zτ(n)|)w2,nan

=

∞ X

n=k

φ2(β|zn|)w2,τ−1_(n)µ(τ−1(An))

=

∞ X

n=k

φ2(β|zn|)w2,τ−1_(n)a_nb_n

≤ γ

∞ X

n=k

φ1(α|zn|)w1,nan+

∞ X

n=k cn

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Thus,Cτ(z)∈Yβ0 and hence Cτ(Xα0)⊂Yβ0.

(iii) Follows from (i) and Theorem 2.3(iii) with Fn(α, β, γ, δ) =cn.

(iv) The direct part follows from (i) and the converse part from (ii).

(v) Suppose Cτ(Xα)⊂ S β∈B

Y0

β.Let z∈Xα0. Then there exists k∈N such that

∞ X

n=k

φ1(α|zn|)w1,nan <∞.

We put zn = yn for n < k and zn = zn for n ≥ k, where yn is the sequence given by Definition 2.2. Then z = {zn}∞n=1 ∈ Xα. Thus, by our supposition Cτ(z) ∈ S

β∈B Y0

β,

so there exists β ∈ B such that Cτ(z) ∈ Y_β0. It then follows that Cτ(z) ∈ Y_β0 which implies that Cτ(z) ∈ S

β∈B Y0

β. Therefore Cτ(Xα0) ⊂

S

β∈B Y0

β. Finally, using (iii), there

existsβ ∈B such that Cτ(Xα0)⊂Yβ0.

We now prove the sufficient part of Theorem 2.1.

Theorem 2.6. If a composition operatorCτ :lwφ11({an})−→l

φ2

w2({an})is bounded, then there

exista, b, δ >0 and a sequence{cn}∞n=1 of nonnegative integers inl1 such thatφ1(u)w1,nan< δ implies

φ2(au)w2,τ−1_(n)a_nb_n≤bφ₁(u)w_1,na_n+c_n,

for all n ∈_N and all u≥0.

Proof. Observe that S

α>0 X0

α =lφ1({an}) and S β>0

Y0

β =lφ2({an}). Suppose Cτ :lφ1({an})−→

lφ2₍_{_an_}_{) is bounded. Then, by Corollary 2.5, we see that for each} _{α >} _{0, there exist}

β, γ, δ > 0 and a sequence {cn} of nonnegative integers such that

∞ P

n=m

cn < ∞ for some

m∈_N and that φ1(αu)w1,nan< δ implies

φ2(βu)w2,τ−1_(n)a_nb_n≤γφ₁(αu)w_1,na_n+c_n,

for all n ∈_Nand all u≥0.

Puttingαu=v, β/α=aandγ =b, the above condition can be rewritten asφ1(v)w1,nan < δ implies

φ2(av)w2,τ−1_(n)anbn≤bφ1(v)w1,nan+cn,

for all n ∈_Nand all v ≥0.

Forn < m, we takecn= max[supv∈S(φ2(av)w2,τ−1_(n)a_nb_n−bφ₁(v)w_1,na_n), 0], whereS is the

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Heera Saini,Department of Mathematics, Govt. M. A. M. College, Jammu, J&K - 180 006, India.

E-mail : [email protected]

Aditi Sharma, Department of Mathematics,University of Jammu, Jammu, J&K - 180 006, India.

E-mail :[email protected]

Neetu Singh, Department of Mathematics,University of Jammu, Jammu, J&K - 180 006, India.