# Integration in Orlicz-Bochner Spaces

Authors: Marian Nowak
Journal: Journal of Function Spaces
 (2018)

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## Abstract

Let(Ω,Σ,μ) be a complete σ-finite measure space, φ be a Young function, and X and Y be Banach spaces. Let Lφ(X) denote the Orlicz-Bochner space, and Tφ∧ denote the finest Lebesgue topology on Lφ(X). We study the problem of integral representation of (Tφ∧,·Y)-continuous linear operators T:Lφ(X)→Y with respect to the representing operator-valued measures. The relationships between (Tφ∧,·Y)-continuous linear operators T:Lφ(X)→Y and the topological properties of their representing operator measures are established.


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## Body

## 1. Introduction and Preliminaries

Throughout the paper,(X,·X) and (Y,·Y) denote real Banach spaces and X∗ and Y∗ denote their Banach duals, respectively. By BX and BY∗ we denote the closed unit ball in X and in Y∗. Let L(X,Y) stand for the space of all bounded operators from X and Y, equipped with the uniform operator norm ·.We assume that(Ω,Σ,μ) is a complete σ-finite measure space. Denote by Σf(μ) the δ-ring of sets A∈Σ with μ(A)<∞. By L0(X) we denote the linear space of μ-equivalence classes of all strongly Σ-measurable functions f:Ω→X, equipped with the topology T0 of convergence in measure on sets of finite measure.Now we recall the basic concepts and properties of Orlicz-Bochner spaces (see [1–6] for more details).By aYoung function we mean here a continuous convex mapping φ:[0,∞)→[0,∞) that vanishes only at 0 and φ(t)/t→0 as t→0 and φ(t)/t→∞ as t→∞. Let φ∗ stand for the complementary Young function of φ in the sense of Young.LetLφ(X) (resp., Lφ) denote theOrlicz-Bochner space (resp.,Orlicz space) defined by a Young function φ; that is,(1)LφX≔f∈L0X:∫ΩφλfωXdμ<∞for  someλ>0=f∈L0X:f·X∈Lφ.Then Lφ(X), equipped with the topology Tφ of the norm(2)fφ≔infλ>0:∫ΩφfωXλdμ≤1,is a Banach space. For a sequence (fn) in Lφ(X), fnφ→0 if and only if ∫Ωφ(λfnωX)dμ→0 for all λ>0. Let(3)BLφX≔f∈LφX:fφ≤1.Let(4)EφX=f∈L0X:∫ΩφλfωXdμ<∞∀λ>0.Then Eφ(X) is a ·φ-closed subspace of Lφ(X).Recall that a subsetH of Lφ(X) is said to besolid whenever f1(ω)X≤f2(ω)Xμ-a.e. and f1∈Lφ(X), f2∈H imply f1∈H. A linear topology ξ on Lφ(X) is said to belocally solid if it has a local basis at 0 consisting of solid sets (see [4]).According to [7, Definition 2.2] and [6] we have the following definition.Definition 1.
A locally solid topologyξ on Lφ(X) is said to be aLebesgue topology if for a net (fα) in Lφ(X), fα(·)X→(o)0 in the Banach lattice Lφ implies fα→0 in ξ.In view of the super Dedekind completeness ofLφ one can restrict in the above definition to usual sequences (fn) in Lφ(X) (see [7, Definition 2.2, p. 173]).Note that, for a sequence(fn) in Lφ(X), fn(·)X→(o)0 in Lφ if and only if fn(ω)X→0  μ-a.e. and fnωX≤uω  μ-a.e. for some 0≤u∈Lφ.Forε>0 let Uφ(ε)={f∈Lφ(X):∫Ωφ(fωX)dμ≤ε}. Then the family of all sets of the form:∗⋃n=1∞∑i=1nUφεi,where (εn) is a sequence of positive numbers and is a local basis at 0 for a linear topology Tφ∧ on Lφ(X) (see [4, 6] for more details). Using [4, Lemma 1.1] one can show that the sets of the form ∗ are convex and solid, so Tφ∧ is a locally convex-solid topology.We now recall terminology and basic facts concerning the spaces of weak∗-measurable functions g:Ω→X∗ (see [8, 9]). Given a function g:Ω→X∗ and x∈X, let gx(ω)=g(ω)(x) for ω∈Ω. By L0(X∗,X) we denote the linear space of the weak∗-equivalence classes of all weak∗-measurable functions g:Ω→X∗. In view of the super Dedekind completeness of L0 the set {|gx|:x∈BX} is order bounded in L0 for each g∈L0(X∗,X). Thus one can define the so-calledabstract normϑ:L0(X∗,X)→L0 by(5)ϑg≔supgx:x∈BXinL0.One can easy check that the following properties of ϑ hold:ϑ(g)=0 if and only if g=0 and g∈L0(X∗,X),ϑ(λg)=|λ|φ(g) for λ∈R and g∈L0(X∗,X),ϑ(g1+g2)≤ϑ(g1)+ϑ(g2) if g1,g2∈L0(X∗,X),ϑ(1Ag)=1Aϑ(g) for A∈Σ and g∈L0(X∗,X).It is known that, forf∈L0(X), g∈L0(X∗,X), the function 〈f,g〉:Ω→R defined by 〈f,g〉(ω)=〈f(ω),g(ω)〉 is measurable and(6)fω,gω≤fωXϑgωμ-a.e.Moreover, ϑ(g)=g(·)X∗ for g∈L0(X∗). Let(7)Lφ∗X∗,X≔g∈L0X∗,X:ϑg∈Lφ∗.Clearly Lφ∗(X∗)⊂Lφ∗(X∗,X). If, in particular, X∗ has the Radon-Nikodym property (i.e., X is anAsplund space; see [10, p. 213]), then Lφ∗(X∗,X)=Lφ∗(X∗).LetLφ(X)∗ stand for the Banach dual of Lφ(X), equipped with the conjugate norm ·φ∗.Recall that a Young functionφ satisfies the Δ2-condition if φ(2t)≤dφ(t) for some d>1 and all t≥0. We shall say that a Young function ψ iscompletely weaker than another φ (in symbols, ψ⊲φ) if for an arbitrary c>1 there exists d>1 such that ψ(ct)≤dφ(t) for all t≥0. Note that a Young function φ satisfies the Δ2-condition if and only if φ⊲φ. If ψ⊲φ, then Lφ⊂Eψ and it follows that Lφ(X)⊂Eψ(X).Now we present basic properties of the topologyTφ∧ on Lφ(X).Theorem 2.
Letφ be a Young function. Then the following statements hold:(i)
Tφ∧⊂Tφ and Tφ∧=Tφ if φ satisfies the Δ2-condition.(ii)
Tφ∧ is the finest Lebesgue topology on Lφ(X).(iii)
Tφ∧ is generated by the family of norms {·ψ|Lφ(X):ψ⊲φ}.(iv)
(Lφ(X),Tφ∧)∗={Fg:g∈Lφ∗(X∗,X)}, where for g∈Lφ∗(X∗,X),(8)Fgf=∫Ωfω,gωdμforf∈LφX,Fgφ∗=sup∫ΩfωXϑgωdμ:f∈BLφX=ϑgφ∗.(v)
(Lφ(X),Tφ∧) is a closed subset of the Banach space (Lφ(X),·φ∗).(vi)
IfX∗ has the Radon-Nikodym property, then the space (Lφ(X),Tφ∧) is strongly Mackey; hence Tφ∧ coincides with the Mackey topology τ(Lφ(X),Lφ∗(X∗)).Proof.
(i)–(iii) See [4, Theorems 6.1, 6.3 and 6.5].
(iv) In view of [6, Corollary 4.4 and Theorem 1.2], we get (Lφ(X),Tφ∧)∗=Lφ(X)n~, where Lφ(X)n~ stands for the order continuous dual of Lφ(X) (see [7, 8, 11] for more details). According to [8, Theorem 4.1] Lφ(X)n~={Fg:g∈Lφ∗(X∗,X)}.
Using [11, Theorem 1.3] for g∈Lφ∗(X∗,X) we have(9)Fgφ∗≔sup∫Ωfω,gωdμ:f∈BLφX=sup∫ΩfωXϑgωdμ:f∈BLφX=ϑgφ∗.
(v) See [12, § 3, Theorem 2].
(vi) See [6, Theorem 4.5].Letγφ[Tφ,T0] (briefly γφ) denote the naturalmixed topology on Lφ(X); that is, γφ is the finest linear topology that agrees with T0 on ·φ-bounded sets in Lφ(X) (see [5, 13, 14] for more details). Then γφ is a locally convex-solid Hausdorff topology (see [14, Theorem 3.2]) and γφ and Tφ have the same bounded sets. This means that (Lφ(X),γφ) is a generalized DF-space (see [15]) and its follows that (Lφ(X),γφ) is quasinormable (see [15, p. 422]). Moreover, for a sequence (fn) in Lφ(X), fn→0 in γφ if and only if fn→0 in T0 and supnfnφ<∞ (see [14, Theorem 3.1]).We say that a Young functionφincreases essentially more rapidly than another ψ (in symbols, ψ≪φ) if for arbitrary c>0, ψ(ct)/φ(t)→0 as t→0 and t→∞.Theorem 3.
Letφ be a Young function. Then the mixed topology γφ on Lφ(X) is generated by the family of norms {·ψ|Lφ(X):ψ≪φ}.Proof.
It is known that the mixed topologyγφ on Lφ is generated by the family of norms {·ψLφ:ψ≪φ} (see [16, Theorem 2.1]). Since fψ=f(·)Xψ for f∈Lφ(X), by [14, (54), p. 97], the mixed topology γφ on Lφ(X) is generated by the family of norms {·ψLφ(X):ψ≪φ}.Sinceψ≪φ implies ψ⊲φ, in view of Theorems 2 and 3, we get(10)γφ⊂Tφ∧.The problem of integral representation of bounded linear operators on Banach function spaces of vector-valued functions to Banach spaces in terms of the corresponding operator-valued measures has been the object of much study (see [5, 17–24]). In particular, Dinculeanu (see [19, § 13, Sect. 3], [20], [21, § 8, Sect. B]) studied the problem of integral representation of bounded linear operators from Lp(X) to a Banach space Y. It is known that if 1≤p<∞, μ(Ω)<∞ and an operator measure m:Σ→L(X,Y) vanishes on μ-null sets and has the finite q-semivariation m~q(Ω)(1<q≤∞,1/p+1/q=1), then one can define the integral ∫Ωfdm for all f∈Lp(X). Moreover, if T:Lp(X)→Y is a bounded linear operator, then the associated operator measure m:Σ→L(X,Y) has the finite q-semivariation m~q(Ω) and T(f)=∫Ωfdm for all f∈Lp(X) (see [19, § 13, Theorem 1 p. 259], [20, Theorem 4]). The relationships of the q-semivariation m~q to the properties of operators from Lp(X) to Y were studied in [22]. Diestel [23] found the integral representation of bounded linear operators from an Orlicz-Bochner space Lφ(X) to a Banach spaces if μ(Ω)<∞ and a Young φ satisfies the Δ2-condition.The present paper is a continuation of [5], where we establish integral representation of (γφ,·Y)-continuous linear operators T:Lφ(X)→Y. We study the problem of integration of functions in Lφ(X) with respect to the representing operator measures of (Tφ∧,·Y)-continuous linear operators T:Lφ(X)→Y. An integral representation theorem for (Tφ∧,·Y)-continuous linear operators T:Lφ(X)→Y is established (see Theorem 9 below). We study the relationships between (Tφ∧,·Y)-continuous operators T:Lφ(X)→Y and the properties of their representing measures m:Σf(μ)→L(X,Y).


## 2.φ∗-Semivariation of Operator Measures

Assume thatm:Σf(μ)→L(X,Y) is an additive measure such that m≪μ; that is, m(A)=0 if μ(A)=0.LetS(Σf(μ),X) denote the space of all X-valued Σf(μ)-simple functions on Ω. Then s∈S(Σf(μ),X) if s=∑(1Ai⊗xi), where (Ai) is a finite pairwise disjoint sequence in Σf(μ) and xi∈X. For s=∑i=1n(1Ai⊗xi)∈S(Σf(μ),X) and A∈Σ, we can define theintegral∫Asdm by(11)∫Asdm≔∑i=1nmAi∩Axi.Note that(12)∫Asdm=∫Ω1Asdm.For y∗∈Y∗, we define a measure my∗:Σf(μ)→X∗ by the equality(13)my∗Ax≔y∗mAxforA∈Σfμ,x∈X.Fors=∑i=1n(1Ai⊗xi)∈S(Σf(μ),X) and A∈Σ, we define the integral ∫Asdmy∗ by the equality:(14)∫Asdmy∗≔∑i=1nmy∗Ai∩Axi.Then(15)y∗∫Asdm=∫Asdmy∗.Following [23], [19, § 13] one can define the φ∗-semivariationm~φ∗(A) of m on A∈Σ by(16)m~φ∗A≔sup∑i=1nmA∩AixiY,where the supremum is taken over all finite pairwise disjoint sets {A1,…,An} in Σf(μ) and xi∈X for i=1,…,n such that ∑i=1n1Ai⊗xiφ≤1.One can observe that(17)m~φ∗A=sup∫AsdmY:s∈SΣfμ,X,sφ≤1.Note that(18)m~φ∗A≤m~φ∗BifA,B∈ΣwithA⊂B,m~φ∗A∪B≤m~φ∗A+m~φ∗BforA,B∈Σ.Let (my∗~)φ∗(A) stand for the φ∗-semivariation of my∗ on A∈Σ; that is,(19)my∗~φ∗A=sup⁡∫Asdmy∗:s∈SΣfμ,X,sφ≤1.The following lemma will be useful.Lemma 4.
Letφ be a Young function and m:Σf(μ)→L(X,Y) be a measure with m≪μ and m~φ∗(Ω)<∞. Then the following statements hold:(i)
Iff∈Eφ(X), then there exists a ·φ-Cauchy sequence (sn) in S(Σf(μ),X) such that sn(ω)-f(ω)X→0μ-a.e.(ii)
If(sn) is a ·φ-Cauchy sequence in S(Σf(μ),X), then for A∈Σ, (∫Asndm) is a Cauchy sequence in a Banach space Y and for every y∗∈Y∗, (∫Asndmy∗) is a Cauchy sequence in R.(iii)
Iff∈Eφ(X) and (sn′) and (sn′′) are ·φ-Cauchy sequence in S(Σf(μ),X) such that sn′(ω)-f(ω)X→0μ-a.e. and sn′′(ω)-f(ω)X→0μ-a.e., then for A∈Σ, one has(20)lim∫Asn′dm=lim∫Asn′′dm,and for everyy∗∈Y∗, one has(21)lim∫Asn′dmy∗=lim∫Asn′′dmy∗.Proof.
(i) Letf∈Eφ(X). Then there exists a sequence (sn) in S(Σf(μ),X) such that sn(ω)-f(ω)X→0μ-a.e. and snωX≤fωXμ-a.e. for all n∈N (see [21, Theorem 6, p. 4]). Using the Lebesgue dominated convergence theorem, we obtain that ∫Ωφ(λ(sn(ω)-f(ω)X)dμ→0 for all λ>0, so sn-fφ→0. Hence (sn) is a ·φ-Cauchy sequence.
(ii) Assume that(sn) is a ·φ-Cauchy sequence in S(Σf(μ),X). Hence for n,k∈N, we have(22)∫Asndm-∫AskdmY=∫Asn-skdmY≤sn-skφm~φ∗A≤sn-skφm~φ∗Ω.It follows that (∫Asndm) is a Cauchy sequence in Y. Hence in view of (15), for y∗∈Y∗, (∫Asndmy∗) is a Cauchy sequence in R.
(iii) Note that(sn′-sn′′) is a ·φ-Cauchy sequence and sn′ω-sn′′ωX→0μ-a.e. Hence there exists h∈Eφ(X) such that (sn′-sn′′)-hφ→0. Note that T0Eφ(X)⊂TφEφ(X). Hence (sn′-sn′′)-h→0 in T0 and it follows that there exists a subsequence (skn′-skn′′) of (sn′-sn′′) such that skn′ω-skn′′ω-hωX→0μ-a.e. Then h(ω)=0μ-a.e., so sn′-sn′′φ→0 and for A∈Σ, we get(23)∫Asn′dm-∫Asn′′dmY=∫Asn′-sn′′dmY≤sn′-sn′′φm~φ∗A.It follows that(24)lim∫Asn′dm=lim∫Asn′′dmand hence, in view of (15) for every y∗∈Y∗, we have(25)lim∫Asn′dmy∗=lim∫Asn′′dmy∗.Following [21, § 13, Definition 1, p. 254], in view of Lemma 4 we have the following.Definition 5.
Letφ be a Young function and m:Σf(μ)→L(X,Y) be an additive measure such that m≪μ and m~φ∗(Ω)<∞. Then for every f∈Eφ(X) and A∈Σ, we can define theintegral∫Afdm by the equality(26)∫Afdm≔lim∫Asndmand for y∗∈Y∗, we can define theintegral∫Afdmy∗ by the equality(27)∫Afdmy∗≔lim∫Asndmy∗,where (sn) is an arbitrary ·φ-Cauchy sequence in S(Σf(μ),X) such that snω-fωX→0μ-a.e.


## 3. Integral Representation of Continuous Operators on Orlicz-Bochner Spaces

For a bounded linear operatorT:Lφ(X)→Y let(28)Tφ≔supTfY:f∈BLφX.Proposition 6.
LetT:Lφ(X)→Y be a bounded linear operator and(29)mAx≔T1A⊗xforA∈Σfμ,x∈X.Then the following statements hold:(i)
ForA∈Σf(μ)m(A)∈L(X,Y) and m(A)≤Tφ·1Aφ.(ii)
m≪μ.(iii)
m(An)→0 if An↓∅ with An∈Σf(μ).(iv)
m:Σf(μ)→L(X,Y) is countably additive; that is, m(⋃n=1∞Bn)=∑n=1∞m(Bn) if (Bn) is a pairwise disjoint sequence in Σf(μ) with ⋃n=1∞Bn∈Σf(μ).(v)
m~φ∗(Ω)≤Tφ.Proof.
(i) LetA∈Σf(μ). Then for x∈BX, we have 1A⊗xφ≤1Aφ and hence(30)mAxY=T1A⊗xY≤Tφ·1A⊗xφ≤Tφ1Aφ,so m(A)≤Tφ·1Aφ.
(ii) This follows from (i) because1Aφ=0 if μ(A)=0.
(iii) Assume thatAn↓∅ with An∈Σf(μ). Then 1A1(ω)≥1An(ω)↓0 for ω∈Ω. By the Lebesgue dominated convergence theorem, we obtain that ∫Ωφ(λ1An(ω))dμ→0 for every λ>0. This means that 1Anφ→0 and by (i), m(An)→0.
(iv) Assume that(Bn) is a pairwise disjoint sequence in Σf(μ) with B=⋃n=1∞Bn∈Σf(μ). Let An=B∖⋃i=1nBi for n∈N. Then An∈Σf(μ) and An↓∅. Hence by (iii) m(B)-∑i=1nm(Bi)=m(B)-m(⋃i=1nBi)=m(An)→0.
Statement (v) is obvious.Definition 7.
LetT:Lφ(X)→Y be a bounded linear operator and(31)mAx≔T1A⊗xforA∈Σfμ,x∈X.Then the measure m:Σf(μ)→L(X,Y) will be called arepresenting measure of T.Proposition 8.
LetT:Lφ(X)→Y be a (Tφ∧,·Y)-continuous linear operator and m:Σf(μ)→L(X,Y) be its representing measure. Then there exists a Young function ψ such that ψ⊲φ and m~ψ∗(Ω)<∞.Proof.
According to Theorem2 there exist a finite set {ψi:i=1,…,n} of Young functions with ψi⊲φ for i=1,…,n and a>0 such that(32)TfY≤amax1≤i≤n⁡fψi∀f∈LφX.Let ψ(t)=max1≤i≤nψi(t) for t≥0. Then ψ is a Young function with ψ⊲φ and(33)TfY≤afψ∀f∈LφX.Hence(34)m~ψ∗Ω=supTsY:s∈SΣfμ,X,sψ≤1≤a<∞.For a linear operatorT:Lφ(X)→Y and A∈Σ, let(35)TAf≔T1Afforf∈LφX.Now we can state our main result that extends the classical results concerning the integral representation of operators on Lebesgue-Bochner spacesLp(X)(1≤p<∞) (see [19, § 13, Theorem 1, pp. 259–261]) to operators on Orlicz-Bochner spaces Lφ(X).Theorem 9.
LetT:Lφ(X)→Y be a (Tφ∧,·Y)-continuous linear operator and m:Σf(μ)→L(X,Y) be its representing measure. Then for A∈Σ the following statements hold:(i)
TA:Lφ(X)→Y is a (Tφ∧,·Y)-continuous linear operator.(ii)
Forf∈Lφ(X), one has(36)TAf=∫Afdmand fory∗∈Y∗, one has(37)y∗TAf=∫Afdmy∗.(iii)
Forf∈Lφ(X), the measure mf:Σ→Y defined by the equality(38)mfA≔∫AfdmforA∈Σis countably additive.(iv)
TAφ=m~φ∗(A)and fory∗∈Y∗, y∗∘TAφ∗=(y∗∘T)Aφ∗=(my∗~)φ∗(A).(v)
m~φ∗(A)=sup{my∗~)φ∗A:y∗∈BY∗.(vi)
Forf∈Lφ(X), one has(39)∫AfdmY≤m~φ∗Afφand fory∗∈Y∗, one has(40)∫Afdmy∗≤my∗~φ∗Afφ.Proof.
(i) Assume that(fα) is a net in Lφ(X) such that fα→0 in Tφ∧. Since Tφ∧ is a locally solid topology on Lφ(X), we get 1Afα→0 in Tφ∧. Hence(41)TAfαY=T1AfαY⟶0.(ii) In view of Proposition 8 there exists a Young function ψ such that ψ⊲φ and m~ψ∗(Ω)<∞. Then Lφ(X)⊂Eψ(X). Let f∈Lφ(X). Then there exists a sequence (sn) in S(Σf(μ),X) such that sn(ω)-f(ω)X→0μ-a.e. and sn(ω)X≤fωXμ-a.e. for all n∈N (see [21, Theorem 6, p. 4]). Then sn→f in Tφ∧ because Tφ∧ is a Lebesgue topology. Hence sn-fψ→0. In view of Lemma 4 we can define the integral ∫Afdm by the equality(42)∫Afdm≔lim∫Asndm.Since TA(sn)=∫Asndm and by (i), TA is (Tφ∧,·Y)-continuous, we get(43)TAf=lim∫Asndm.Hence(44)TAf=∫Afdmand for y∗∈Y∗, we have(45)y∗TAf=limy∗∫Asndm=lim∫Asndmy∗=∫Afdmy∗. (iii) Let f∈Lφ(X) and (An) be a sequence in Σ such that An↓∅. Then 1An(ω)↓0 for ω∈Ω, and hence 1An(ω)f(ω)X→0μ-a.e. and 1AnωfωX≤fωXμ-a.e. Hence 1Anf→0 in Tφ∧ because Tφ∧ is a Lebesgue topology, and by (i) we get(46)mfAnY=∫AnfdmY=T1AnfY⟶0. (iv) Note that m~φ∗(A)≤TAφ. To show that TAφ≤m~φ∗(A), assume that f∈BLφ(X). Choose a sequence (sn) in S(Σf(μ),X) such that snω-fωX→0μ-a.e. and sn(ω)X≤fωXμ-a.e. for all n∈N. Since Tφ∧ is a Lebesgue topology, we have sn→f in Tφ∧ and hence TAsn-TAfY→0. Note that TA(sn)=∫Asndm.
Letε>0 be given. Choose n0∈N such that TA(f)-∫Asn0dmY≤ε. Then(47)TAfY≤TAf-∫Asn0dmY+∫Asn0dmY≤ε+m~φ∗A.It follows that TAφ≤m~φ∗(A), so m~φ∗(A)=TAφ. Hence for y∗∈Y∗, we easily get(48)y∗∘TAφ∗=y∗∘TAφ∗=my∗~φ∗A. (v) Using (iv) we have(49)m~φ∗A=TAφ=supTAfY:f∈LφX,fφ≤1=supy∗∈BY∗y∗∘TAf:f∈LφX,fφ≤1=supy∗∈BY∗y∗∘TAφ∗=supy∗∈BY∗my∗~φ∗A. (vi) This follows from (ii) and (iv).For a sequence(An) in Σ, we will write An↘μ∅ if An↓ and μ(An∩A)→0 for every A∈Σf(μ).Definition 10.
A measurem:Σf(μ)→L(X,Y) with m≪μ and m~φ∗(Ω)<∞ is said to be φ∗-semivariationallyμ-continuous if m~φ∗(An)→0 whenever An↘μ∅, (An)⊂Σ.Using a standard argument we can show the following.Proposition 11.
Letm:Σ→L(X,Y) be an additive measure such that m≪μ and m~φ(Ω)<∞. Then the following statements are equivalent:(i)
m is φ∗-semivariationally μ-continuous.(ii)
The following two conditions hold simultaneously:
(a)
For everyε>0 there exists δ>0 such that m~φ∗(A)≤ε whenever μ(A)≤δ, A∈Σ.(b)
For everyε>0 there exists A0∈Σf(μ) such that m~φ∗(Ω∖A0)≤ε.The following theorem characterizesφ∗-semivariationally μ-continuous representing measures.Theorem 12.
LetT:Lφ(X)→Y be a (Tφ∧,·Y)-continuous linear operator and m:Σf(μ)→L(X,Y) be its representing measure. Then the following statements are equivalent:(i)
m is φ∗-semivariationally μ-continuous.(ii)
T is (γφ,·Y)-continuous.(iii)
T(fn)Y→0 if fn→0 in T0 and supnfnφ<∞.(iv)
TAnφ→0 if An↘μ∅, (An)⊂Σ.Proof.
(i)⇔ (ii)⇔ (iii) See [5, Corollary 2.8 and Proposition 1.1].
(i)⇔ (iv) This follows from Theorem 9.Now assume thatΩ is a completely regular Hausdorff space. Let Ba denote the σ-algebra of Baire sets in Ω, which is the σ-algebra generated by the class Z of all zero sets of bounded continuous positive functions on ω. By P we denote the family of all cozero (=positive) in Ω (see [25, p. 108]).Letμ:Ba→[0,∞) be a countably additive measure. Then μ is zero-set regular; that is, for every A∈Ba and ε>0 there exists Z∈Z with Z⊂A such that μ(A∖Z)≤ε (see [25, p. 118]). It follows that for every A∈Ba and ε>0 there exist U∈P, U⊃A such that μ(U∖A)≤ε.We can assume thatμ to be complete (if necessary we can take the completion (Ω,Ba¯,μ¯) of the measure space (Ω,Ba,μ)).Proposition 13.
Assume thatΩ is a completely regular Hausdorff space and (Ω,Ba,μ) is a complete finite measure space. Let T:Lφ(X)→Y be a (Tφ∧,·Y)-continuous linear operator and m:Ba→L(X,Y) be its representing measure. Then the following statements are equivalent:(i)
m is φ∗-semivariationally μ-continuous.(ii)
For every sequence(An) in Ba such that An↓ and μ(An)→0 there exists a sequence (Un) in P with An⊂Un↓ such that m~φ∗(Un)→0.(iii)
For every sequence(An) in Ba such that An↓ and μ(An)→0 there exists a sequence (Un) in P with An⊂Un↓ such that(50)supTfY:f∈BLφX,suppf⊂Un⟶0.Proof.
(i) ⇒ (ii) Assume that (i) holds and(An) is a sequence in Ba such that An↓ and μ(An)→0. Then there exists a sequence (Un) in P such that An⊂Un↓ and μ(Un∖An)≤1/n for n∈N.
Letε>0 be given. Then in view of Proposition 11 there exists δ>0 such that m~φ∗(A)≤ε/2 if μ(A)≤δ with A∈Ba. Choose n1∈N such that μ(Un∖An)≤δ for n≥n2. Then m~φ∗(Un∖An)≤ε/2 for n≥n1. Since m~φ∗(An)→0, we can choose n2∈N such that m~φ∗(An)≤ε/2 for n≥n2. Then for n≥n0=max(n1,n2), we get(51)m~φ∗Un≤m~φ∗Un∖An+m~φ∗An≤ε;that is, (ii) holds.
(ii) ⇒ (iii) Assume that (ii) holds and(An) is a sequence in Bo such that An↓ and μ(An)→0. Then there exists a sequence (Un) in P with An⊂Un↓ such that m~φ∗(Un)→0. Note that, for f∈BLφ(X) with suppf⊂Un for n∈N, by Theorem 9 we have(52)TfY=∫ΩfdmY=∫UnfdmY≤m~φ∗Un.It follows that (iii) holds.
(iii) ⇒ (i) Assume that (iii) holds andAn↓ with μ(An)→0. Then there exists a sequence (Un) in P with An⊂Un↓ such that(53)supTfY:f∈BLφX,suppf⊂Un⟶0.Assume on the contrary that (i) fails to hold. Then without loss of generality we can assume that(54)m~φ∗An>ε0for  someε0>0,alln∈N.Choose n0∈N such that(55)supTfY:f∈BLφX,suppf⊂Un0<ε02.In view of (54) there exists a pairwise disjoint set {B1,…,Bk} in Ba, xi∈X for i=1,…,k and y∗∈BY∗ such that ∑i=1k1Bi⊗xiφ≤1 and(56)y∗∑i=1kmAn0∩Bixi≥ε0.Let s0=∑i=1k(1An0∩Bi⊗xi). Then s0φ≤1 and supps0⊂An0⊂Un0. Then by (55) we get T(s0)Y<ε0/2.
On the other hand, in view of (56) we have Ts0Y≥ε0. This contradiction establishes that (i) holds.Corollary 14.
Assume thatΩ is a completely regular Hausdorff space and (Ω,Ba,μ) is complete finite measure space. Let T:Lφ(X)→Y be a (γφ,·Y)-continuous linear operator and m:Ba→L(X,Y) be its representing measure. Then m~φ∗ is regular; that is, for every A∈Ba and ε>0 there exist Z∈Z and U∈P with Z⊂A⊂U such that m~φ∗(U∖Z)≤ε.Proof.
In view of Theorem12  m is φ∗-semivariationally μ-continuous. Let A∈Ba and ε>0 be given. Then by Proposition 11 there exists δ>0 such that m~φ∗(B)≤ε whenever B∈Ba and μ(B)≤δ. By the regularity of μ one can choose Z∈Z and U∈P with Z⊂A⊂U such that μ(U∖Z)≤δ. Hence m~φ∗(U∖Z)≤ε, as desired.


## 4. Compact Operators on Orlicz-Bochner Spaces

The following theorem presents necessary conditions for a(Tφ∧,·Y)-continuous operator T:Lφ(X)→Y to be compact.Theorem 15.
Assume that a Young functionφ such that φ∗ satisfies the Δ2-condition. Let T:Lφ(X)→Y be a (Tφ∧,·Y)-continuous linear operator and m:Σf(μ)→L(X,Y) be its representing measure. If T is compact, then m is φ∗-semivariationally μ-continuous.Proof.
Assume thatT is compact and m fails to be φ∗-semivariationally μ-continuous. Then there exist ε>0 and a sequence (An) in Σ with An↘μ∅ such that TAn=mφ∗~(An)>ε for n∈N (see Theorem 9). Hence one can choose a sequence (yn∗) in BY∗ such that(57)yn∗∘TAnφ∗≥ε∀n∈N.
By Schauder’s theorem the conjugate mappingT∗:Y∗→Lφ(X)∗ is compact. Note that T∗(yn∗)=yn∗∘T∈Lφ(X)n~ for all n∈N, where Lφ(X)n~ is a closed subspace of the Banach space (Lφ(X)∗,·φ∗) (see Theorem 2). Then for every n∈N there exists gn∈Lφ∗(X∗,X) such that(58)yn∗∘Tf=∫Ωfω,gnωdμforf∈LφX,yn∗∘Tφ∗=sup∫ΩfωXϑgnωdμ:f∈BLφX=ϑgnφ∗.Hence we obtain that, for each n∈N,(59)yn∗∘TAnφ∗=1Anϑgnφ∗=ϑ1Angnφ∗.Since T∗(BY∗) is a relatively sequentially compact subset of ((Lφ(X)n~,·φ∗), there exist a subsequence (gkn) of (gn) and g∈Lφ∗(X∗,X) such that(60)Fgn-Fgφ∗=ϑgkn-gφ∗⟶0.Choose nε∈N such that ϑ(gkn-g)φ∗≤ε/2 for n≥nε. Hence for n≥nε,(61)ϑ1Akngφ∗-ϑ1Akngknφ∗≤ϑ1Akngkn-gφ∗=1Aknϑgkn-gφ∗≤ϑgkn-gφ∗≤ε2.Using (57) and (61), for n≥nε, we get(62)ε≤y∗∘TAknφ∗=ϑ1Akngknφ∗≤ε2+ϑ1Akngφ∗and hence(63)1Aknϑgφ∗=ϑ1Akngφ∗≥ε2.
On the other hand, sinceφ∗ is supposed to satisfy the Δ2-condition, we have that 1Aknϑ(g)φ∗→0 (see [26, Theorem 3, pp. 58-59]). This contradiction establishes that m is φ∗-semivariationally μ-continuous.Corollary 16.
Assume thatφ is a Young function such that φ∗ satisfies the Δ2-condition. Let T:Lφ(X)→Y be a (Tφ∧,·Y)-continuous linear operator. Then the following statements are equivalent:(i)
T is compact.(ii)
T is (γφ,·Y)-compact; that is, there exists a γφ-neighborhood V of 0 in Lφ(X) such that T(V) is a relatively norm compact set in Y.(iii)
There exists a Young functionψ with ψ≪φ such that {∫Ωfdm:f∈Lφ(X), fψ≤1} is a relatively norm compact set in Y.Proof.
(i) ⇒ (ii) Assume that (i) holds. Then by Theorems12 and 15  T is (γφ,·Y)-continuous. Since the space (Lφ(X),γφ) is quasinormable, by Grothendieck's classical result (see [15, p. 429]), we obtain that T is (γφ,·Y)-compact.
(ii) ⇒ (i) The implication is obvious.
(ii)⇔ (iii) This follows from Theorem 3.


## 5. Topology Associated with theφ∗-Semivariation of a Representing Measure

Assume thatT:Lφ(X)→Y be a (Tφ∧,·Y)-continuous linear operator. Let m:Σf(μ)→L(X,Y) be its representing measure. Let us put(64)pmy∗≔my∗~φ∗Ωfory∗∈Y∗.Note that pm is a seminorm on Y∗. Following [22, 27] let δm,φ∗ stand for the topology on BY∗ defined by the seminorm pm restricted to BY∗.The following theorem characterizes(Tφ∧,·Y)-continuous compact operators T:Lφ(X)→Y in terms of the topological properties of the space (BY∗,δm,φ∗) (see [22, Theorem 3]).Theorem 17.
LetT:Lφ(X)→Y be a (Tφ∧,·Y)-continuous linear operator and m:Σf(μ)→L(X,Y) be its representing measure. Then the following statements are equivalent:(i)
The space(BY∗,δm,φ∗) is compact.(ii)
T is compact.Proof.
(i) ⇒ (ii) Assume that(BY∗,δm,φ∗) is compact. Let (yn∗) be a sequence in BY∗. Without loss of generality we can assume that yn∗→y0∗ in δm,φ∗ for some y∗∈BY∗. Then using Theorem 9 for f∈Lφ(X), we have(65)T∗yn∗-T∗y0∗f=yn∗-y0∗Tf=∫Ωfdmyn∗-y0∗≤myn∗-y0∗~φ∗Ωfφ.It follows that T∗(yn∗)-T∗y0∗φ∗≤(myn∗-y0∗~)φ∗(Ω), where pm(yn∗-y0∗)=(myn∗-y0∗~)φ∗(Ω)→n0. This means that T∗ is compact and hence T is compact.
(ii) ⇒ (i) Assume thatT is compact and (yα∗) is a net in BY∗. Since BY∗ is σ(Y∗,Y)-compact, without loss of generality we can assume that yα∗→αy0∗ in σ(Y∗,Y) for some y0∗∈BY∗. In view of the compactness of the conjugate operator T∗:Y∗→Lφ(X)∗, there exists a subset (yβ∗) of (yα∗) and Φ0∈Lφ(X)∗ such that T∗(yβ∗)-Φ0φ∗→β0. On the other hand, since T∗ is (σ(Y∗,Y),σ(Lφ(X)∗,Lφ(X)))-continuous, we get T∗(yβ∗)→βT∗(y0∗) in σ(Lφ(X)∗,Lφ(X)). Hence Φ0=T∗(y0∗); that is, T∗(yβ∗)-T∗(y0∗)φ∗→β0.
Letε>0 be given. Then there exist a pairwise disjoint set {A1,…,An} in Σf(μ) and xi∈X for i=1,…,n such that ∑i=1n1Ai⊗xiφ≤1 and(66)myβ∗-y0∗~φ∗Ω≤∑i=1nyβ∗-y0∗mAixi+ε.Hence(67)myβ∗-y0∗~φ∗Ω≤∑i=1nyβ∗-y0∗T1Ai⊗xi+ε≤yβ∗-y0∗T∑i=1n1Ai⊗xi+ε=T∗yβ∗-y0∗∑i=1n1Ai⊗xi+ε≤T∗yβ∗-y0∗φ∗∑i=1n1Ai⊗xiφ+ε≤T∗yβ∗-Ty0∗φ∗+ε.Hence pm(yβ∗-y0∗)=(myβ∗-y0∗~)φ∗(Ω)→β0, and this means that the space (BY∗,δm,φ∗) is compact.As a consequence of Theorems17 and 15, we have the following.Corollary 18.
Assume thatφ is a Young function such that φ∗ satisfies the Δ2-condition. Let T:Lφ(X)→Y be a (Tφ∧,·Y)-continuous linear operator and m:Σfμ→LX,Y be its representing measure. If the space (BY∗,δm,φ∗) is compact, then m is φ∗-semivariationally μ-continuous.



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