# Some Seminormed Difference Sequence Spaces overn-Normed Spaces Defined by a Musielak-Orlicz Function of Order (α,β)

Authors: S. A. Mohiuddine; Sunil K. Sharma; Dina A. Abuzaid
Journal: Journal of Function Spaces
 (2018)

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## Abstract

We first define the notion of lacunary statistical convergence of order(α,β), and taking this notion into consideration, we introduce some seminormed difference sequence spaces over n-normed spaces with the help of Musielak-Orlicz function M=(Mk) of order (α,β). We also examine some topological properties and prove inclusion relations between the resulting sequence spaces.


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## Body

## 1. Introduction and Preliminaries

Gähler [1] extended the usual notion of normed spaces into 2-normed spaces, while the notion was again extended to n-normed spaces by Misiak [2]. Assume that X is a linear space over the field K of real or complex numbers of dimension d≥n≥2, n∈N (N denotes the set of natural numbers). A real valued function ·,…,· on Xn satisfying the conditions,(N1)
x1,x2,…,xn=0 if and only if x1,…,xn are linearly dependent in X,(N2)
x1,x2,…,xn is invariant under permutation,(N3)
αx1,x2,…,xn=|α|x1,x2,…,xn for any α∈K,(N4)
x1+x1′,x2,…,xn≤x1,x2,…,xn+x1′,x2,…,xn,is called an-norm on X, and the pair (X,·,…,·) is called a n-normed space over K.A sequence(xk)k∈N in a n-normed space (X,·,…,·) is said to be(i)
convergent toξ∈X if limk→∞xk-ξ,z1,…,zn-1=0 holds for every z1,…,zn-1∈X,(ii)
Cauchy iflimk,i→∞xk-xi,z1,…,zn-1=0 holds for every z1,…,zn-1∈X.Ann-Banach space (X,·,…,·) is a complete n-normed space, where completeness means that, for every (xk) in X with xk-xi,z1,…,zn-1→0(k,i→∞), there exists ξ∈X such that xk-ξ,z1,…,zn-1=0(k→∞).Kızmaz [3] was the first who introduced the idea of difference sequence spaces and studied Z(Δ)={x=(xk)∈w:Δx∈Z}(Z=l∞,c,c0), where Δx=xk-xk+1 for all k∈N, w is the space of all complex or real sequences, and the standard notations l∞, c, and c0 denote bounded, convergent, and null sequences, respectively. Et and Çolak [4] presented a generalization of these difference sequence spaces and introduced the space Z(Δn); in this case, Δnx is given by Δnx=Δ(Δn-1x)=Δn-1xk-Δn-1xk+1 for n≥2. Another generation is given by Tripathy and Esi [5] by considering Δm instead as Δ in Kızmaz spaces and is defined by Δmx=xk-xk+m. In view of Δn and Δm, Tripathy et al. [6] introduced the difference sequence space as follows:(1)ZΔmn=x=xk∈w:Δmnxk∈Z,where Δmnx=(Δmnxk)=(Δmn-1xk-Δmn-1xk) and Δ0xk=xk, which is equivalent to the following binomial representation:(2)Δmnxk=∑v=0n-1vnvxk+mv.We remark that if we take m=1, then difference sequence space Z(Δmn) is reduced to Z(Δn) and for the choice of n=m=1, we obtain the difference sequence space Z(Δ). For recent work related to various kinds of difference sequence spaces, we refer to [7–13] and references therein.IfX is a linear space and g:X→R such that (i) x=0⇒g(x)=0, (ii) g(x+y)≤g(x)+g(y), (iii) g(-x)=g(x), and (iv) if tk→t(k→∞) and xk→x(k→∞) in the sense that g(xk-x)→0(k→∞) for scalars tk,t and the vectors xk,x∈X, then tkxk→tx(k→∞) in the sense that g(tkxk-tx)→0(k→∞); then g is said to be a paranorm on X and the pair (X,g) is called a paranormed space.An Orlicz function is a functionM:[0,∞)→[0,∞) which is continuous, nondecreasing, and convex with M(0)=0, M(x)>0 as x>0 and M(x)→∞(x→∞). Clearly, if M is a convex function and M(0)=0, then M(λx)≤λM(x) for all λ∈(0,1). An Orlicz function M is said to satisfy Δ2-condition for all values of u, if there exists a constant K>0 such that M(Lu)≤KLM(u) for all values of L>1 (see Krasnoselskii and Rutitsky [14] and, more recent, Giannetti et al. [15]). Using the idea of Orlicz function, Lindenstrauss and Tzafriri [16] constructed the sequence space(3)lM=xk∈w:∑k=1∞Mxkρ<∞,forsomeρ>0,which is called Orlicz sequence space and showed that lM is a Banach space with the following norm:(4)x=infρ>0:∑k=1∞Mxkρ≤1.The space lM is closely related to the space lp which is an Orlicz sequence space with M(t)=|t|p for 1≤p<∞.A sequenceM=(Mk) of Orlicz functions is said to be a Musielak-Orlicz function [17]. A sequence N=(Nk) is defined by Nk(v)=sup{|v|u-(Mk):u≥0}(k=1,2,…) which is said to be complementary function of M. For a given M, the Musielak-Orlicz sequence space tM and its subspace hM are defined by(5)tM=x∈w:IMcx<∞forsomec>0,hM=x∈w:IMcx<∞∀c>0,where IM denotes the convex modular and is defined by(6)IMx=∑k=1∞Mkxkx=xk∈tM.It is noted that tM can be considered equipped with the Luxemburg norm or equipped with the Orlicz norm, where Luxemburg and Orlicz norms are given by(7)x=infk>0:IMxk≤1,x0=inf1k1+IMkx:k>0,respectively.A sequencex=(xk)∈l∞ is said to be almost convergent if all of its Banach limits coincide. The space c^ of almost convergent sequences was defined by Lorentz [18] (also see [19]) as follows:(8)c^=x=xk∈l∞:limn→∞1n∑k=1nxk+sexists,uniformlyins.Recall that x is strongly almost convergent [20] to some number ξ if(9)limn→∞1n∑k=1nxk+s-ξ=0uniformlyins.The idea of statistical convergence first appeared, under the name of almost convergence, in the first edition of Zygmund [21]. Later, this idea was introduced by Fast [22] and Steinhaus [23], independently, and some of its basic properties were studied by Schoenberg [24], Šalát [25], and Fridy [26]. Some useful results, applications, and various developments on this topic have been presented by many authors; we refer to [27–35]. Çolak [36] extended this notion with the help of α-density (for α=1, α-density reduced to natural density) and called it statistical convergence of order α. In the recent past, Şenül [37] presented an interesting generalization of this notion by taking into account (α,β) instead of α and defined statistical convergence of order (α,β) as follows.The sequencex=(xk) is said to be statistically convergent of order (α,β), briefly Sαβ-convergence, to L if for each ϵ>0, we have(10)limn→∞1nαk≤n:xk-L≥ϵβ=0,where |{k≤n:k∈E}|β denotes the βth power of number of elements of E not exceeding n, and we write Sαβ-limxk=L.


## 2. Construction of Difference Sequence Spaces

Before defining some difference sequence spaces, let us first introduce the notion of lacunary statistical convergence of order(α,β). Recall that the standard notation θ=(kr) denotes the lacunary sequence, where (kr) is a sequence of positive integers such that k0=0, 0<kr<kr+1, and hr≔kr-kr-1→∞(r→∞), throughout the article, the intervals determined by the lacunary sequence θ will be denoted by Ir=(kr-1,kr] and the ratio qr=kr/kr-1(r≠1) by qr (see [38]).Suppose thatE⊂N and 0<α≤β≤1. We define the θ(α,β)-density of E by(11)δθα,βE=limn1hrαk∈kr-1,kr:k∈Eβ,in case this limit exists, where hrα denotes the αth power of hr and |{k∈Ir:k∈E}|β denotes the βth power of number of elements of E in Ir.Suppose that0<α≤β≤1 and let θ=(kr) be a lacunary sequence. Then, we say that the sequence x=(xk) is lacunary statistically convergent of order (α,β) (shortly, Sαβ(θ)-convergence) to the number L and we write Sαβ(θ)-limxk=L if for each ϵ>0, the set A(ϵ)=k∈N:xk-L≥ϵ has θ(α,β)-density zero; that is,(12)limr→∞1hrαk∈kr-1,kr:xk-L≥ϵβ=0.Example 1.
Consider a sequencex=(xk) defined by xk=k if kr-[hrα]+1≤k≤kr(r∈N) and xk=0 otherwise. Also, let β=1. Then for any ϵ>0,(13)limr→∞1hrαk∈kr-1,kr:xk-0≥ϵ≤hrαhrα⟶0r⟶∞.We see that x=(xk) is Sαβ(θ)-convergent to zero but not convergent.Remark 2.
Forθ=(2r), the notion of Sαβ(θ)-convergence coincides with Sαβ-convergence in [37]. Also, if we take θ=(2r) and β=1, then Sαβ(θ)-convergence reduces to the notion of statistical convergence of order α due to Çolak [36]. Further, the choice of θ=(2r) and α=β=1 in the definition of lacunary statistically convergent of order (α,β) gives us the notion of statistical convergence due to Fast [22]. Moreover, if α=β=1 in the definition of Sαβ(θ)-convergence, then we obtain lacunary statistical convergence introduced by Fridy and Orhan [39].We denote byw(n-X) the space of all sequences defined over (X,·,…,·). Let X be a seminormed space, seminormed by q=(qk), and let p=(pk) be any bounded sequence of positive real numbers. We are now ready to define the following sequence spaces:(14)cθM,p,Δnm,q,·,…,·αβ=x=xk∈wn-X:limr→∞1hrα∑k∈IrMkqkΔnmxk+s-Lρ,z1,…,zn-1pkβ=0uniformly  ins,for  someL,ρ>0,c0θM,p,Δnm,q,·,…,·αβ=x=xk∈wn-X:limr→∞1hrα∑k∈IrMkqkΔnmxk+sρ,z1,…,zn-1pkβ=0uniformly  ins,for  someρ>0,c∞θM,p,Δnm,q,·,…,·αβ=x=xk∈wn-X:supr,s1hrα∑k=1nMkqkΔnmxk+sρ,z1,…,zn-1pkβ<∞uniformlyins,forsomeρ>0.Note that if we considerM(x)=x; then the above difference sequence spaces reduce to the following spaces:(15)cθp,Δnm,q,·,…,·αβ=x=xk∈wn-X:limr→∞1hrα∑k∈IrqkΔnmxk+s-Lρ,z1,…,zn-1pkβ=0uniformlyins,forsomeL,ρ>0,c0θp,Δnm,q,·,…,·αβ=x=xk∈wn-X:limr→∞1hrα∑k∈IrqkΔnmxk+sρ,z1,…,zn-1pkβ=0uniformlyins,forsomeρ>0,c∞θp,Δnm,q,·,…,·αβ=x=xk∈wn-X:supr,s1hrα∑k=1nqkΔnmxk+sρ,z1,…,zn-1pkβ<∞,uniformlyins,forsomeρ>0.It is to further notice that if we takepk=1 for all k∈N, then the spaces cθ[M,p,Δnm,q,·,…,·]αβ, c0θ[M,p,Δnm,q,·,…,·]αβ, and c∞θ[M,p,Δnm,q,·,…,·]αβ reduce to the following spaces:(16)cθM,Δnm,q,·,…,·αβ=x=xk∈wn-X:limr→∞1hrα∑k∈IrMkqkΔnmxk+s-Lρ,z1,…,zn-1β=0,uniformlyins,forsomeL,ρ>0,c0θM,Δnm,q,·,…,·αβ=x=xk∈wn-X:limr→∞1hrα∑k∈IrMkqkΔnmxk+sρ,z1,…,zn-1β=0,uniformlyins,forsomeρ>0,c∞θM,Δnm,q,·,…,·αβ=x=xk∈wn-X:supr,s1hrα∑k=1nMkqkΔnmxk+sρ,z1,…,zn-1β<∞,uniformlyins,forsomeρ>0.Remark 3.
Ifα=β=1, n=1, and qk(x)=x∀k in our difference sequence spaces defined above, then these spaces reduce to the difference sequence spaces defined by Raj et al. [40]. Moreover, if we take α=β=1 in our difference sequence spaces and take uk=1 for all k in the difference sequence spaces defined by Raj and Sharma [41], then we observe that difference sequence spaces of both the manuscripts coincide.The following inequality will be used to prove some of our results in the next section. If0≤pk≤suppk=H, D=max(1,2H-1), then(17)ak+bkpk≤Dakpk+bkpk,for all k and ak,bk∈C. Also |a|pk≤max(1,|a|H) for all a∈C.


## 3. Main Results

In this section, we are going to study some topological properties and to obtain some inclusion relations between the difference sequence spaces defined in the previous section.Theorem 4.
LetM = (Mk) be a Musielak-Orlicz function and let p=(pk) be a bounded sequence of positive real numbers. Then the spaces cθ[M,p,Δnm,q,·,…,·]αβ, c0θ[M,p,Δnm,q,·,…,·]αβ and c∞θ[M,p,Δnm,q,·,…,·]αβ are linear over the field of complex numbers C.Proof.
Letx=(xk),y=(yk)∈c0θ[M,p,Δnm,q,·,…,·]αβ and α,β∈C. Then there exist positive numbers ρ1 and ρ2 such that(18)limr→∞1hrα∑k∈IrMkqkΔnmxk+sρ1,z1,…,zn-1pkβ=0uniformlyins,limr→∞1hrα∑k∈IrMkqkΔnmxk+sρ2,z1,…,zn-1pkβ=0uniformlyins.Let ρ3=max(2|α|ρ1,2|β|ρ2). Since M=(Mk) is nondecreasing convex function, by (17), we obtain(19)1hrα∑k∈IrMkqkΔnmαxk+s+βyk+sρ3,z1,…,zn-1pkβ≤D1hrα∑k∈Ir12pkMkqkΔnmxk+sρ1,z1,…,zn-1pkβ+D1hrα∑k∈Ir12pkMkqkΔnmyk+sρ2,z1,…,zn-1pkβ≤D1hrα∑k∈IrMkqkΔnmxk+sρ1,z1,…,zn-1pkβ+D1hrα∑k∈IrMkqkΔnmyk+sρ2,z1,…,zn-1pkβ⟶0asr⟶∞,uniformlyins.Thus, we have (αx+βy)∈c0θ[M,p,Δnm,q,·,…,·]αβ. Hence, c0θ[M,p,Δnm,q,·,…,·]αβ is a linear space. In a similar way, one can prove that cθ[M,p,Δnm,q,·,…,·]αβ and c∞θ[M,p,Δnm,q,·,…,·]αβ are linear spaces.Theorem 5.
For any Musielak-Orlicz functionM=(Mk) and a bounded sequence of positive real numbers p=(pk), the space c0θ[M,p,Δnm,q,·,…,·]αβ is a topological linear space paranormed by(20)gx=infρpr/K:1hrα∑k∈IrMkqkΔnmxk+sρ,z1,…,zn-1pkβ1/K≤1,r,s∈N,where K=max(1,supkpk<∞).Proof.
Clearlyg(x)≥0 for x=(xk)∈c0θ[M,p,Δnm,q,·,…,·]αβ. Since Mk(0)=0, we obtain g(0)=0. Again, if g(x)=0, then we have(21)infρpr/K:1hrα∑k∈IrMkqkΔnmxk+sρ,z1,…,zn-1pkβ1/K≤1,r,s∈N=0.This implies that, for a given ϵ>0, there exists some ρϵ(0<ρϵ<ϵ) such that(22)1hrα∑k∈IrMkqkΔnmxk+sρϵ,z1,…,zn-1pkβ1/K≤1.Thus,(23)1hrα∑k∈IrMkqkΔnmxk+sϵ,z1,…,zn-1pkβ1/K≤1hrα∑k∈IrMkqkΔnmxk+sρϵ,z1,…,zn-1pkβ1/K≤1,for each r and s. Assume that xk≠0 for each k∈N. This implies that Δnmxk+s≠0 for each k,s∈N. Let ϵ→0; then qk(Δnmxk+s/ϵ,z1,…,zn-1)→∞. It follows that(24)1hrα∑k∈IrMkqkΔnmxk+sϵ,z1,…,zn-1pkβ1/K⟶∞,which is a contradiction. Consequently, we have Δnmxk+s=0 for each k,s and so xk=0 for each k∈N. Suppose that ρ1>0 and ρ2>0 be such that(25)1hrα∑k∈IrMkqkΔnmxk+sρ1,z1,…,zn-1pkβ1/K≤1,1hrα∑k∈IrMkqkΔnmxk+sρ2,z1,…,zn-1pkβ1/K≤1,for each r and s. Assume that ρ=ρ1+ρ2. Then, by Minkowski’s inequality, we have(26)1hrα∑k∈IrMkqkΔnmxk+s+yk+sρ,z1,…,zn-1pkβ1/K≤∑k∈Irρ1ρ1+ρ2MkqkΔnmxk+sρ1,z1,…,zn-1+ρ2ρ1+ρ2MkqkΔnmyk+sρ2,z1,…,zn-1pkβ1/K≤ρ1ρ1+ρ21hrα∑k∈IrMkqkΔnmxk+sρ1,z1,…,zn-1pkβ1/K+ρ2ρ1+ρ21hrα∑k∈IrMkqkΔnmyk+sρ2,z1,…,zn-1pkβ1/K≤1.Since ρ’s are nonnegative, one obtains(27)gx+y=infρpr/K:1hrα∑k∈IrMkqkΔnmxk+s+yk+sρ,z1,…,zn-1pkβ1/K≤1,r,s∈N≤infρ1pr/K:1hrα∑k∈IrMkqkΔnmxk+sρ1,z1,…,zn-1pkβ1/K≤1,r,s∈N+infρ2pr/K:1hrα∑k∈IrMkqkΔmyk+sρ2,z1,…,zn-1pkβ1/K≤1,r,s∈N.Consequently, g(x+y)≤g(x)+g(y). It is only left to show that the scalar multiplication is continuous. Let us consider a complex number λ. Then, by definition, we obtain(28)gλx=infρpr/K:1hrα∑k∈IrMkqkΔnmλxk+sρ,z1,…,zn-1pkβ1/K≤1,r,s∈N.Then(29)gλx=infλtpr/K:1hrα∑k∈IrMkqkΔmxk+st,z1,…,zn-1pkβ1/K≤1,r,s∈N,where t=ρ/|λ|. Since |λ|pr≤max(1,|λ|suppr), we have(30)gλx≤max1,λsupprinftpr/K:1hrα∑k∈IrMkqkΔnmxk+st,z1,…,zn-1pkβ1/K≤1,r,s∈N.So, the fact that scalar multiplication is continuous follows from the above inequality.Theorem 6.
LetM=(Mk) be a Musielak-Orlicz function. If supk[Mk(x)]pk<∞ for all fixed x>0, then c0θ[M,p,Δnm,q,·,…,·]αβ⊂c∞θ[M,p,Δnm,q,·,…,·]αβ.Proof.
Letx=(xk)∈c0θ[M,p,Δnm,q,·,…,·]αβ. There exists some positive ρ1 such that(31)limr→∞1hrα∑k∈IrMkqkΔnmxk+sρ1,z1,…,zn-1pkβ=0uniformlyins.Define ρ=2ρ1. Since M=(Mk) is nondecreasing and convex, by using (17), we obtain(32)supr,s1hrα∑k∈IrMkqkΔnmxk+sρ,z1,…,zn-1pkβ≤Dsupr,s1hrα∑k∈Ir12pkMkqkΔnmxk+s-Lρ1,z1,…,zn-1pkβ+Dsupr,s1hrα∑k∈Ir12pkMkqkLρ1,z1,…,zn-1pkβ≤Dsupr,s1hrα∑k∈IrMkqkΔnmxk+s-Lρ1,z1,…,zn-1pkβ+Dsupr,s1hrα∑k∈IrMkqkLρ1,z1,…,zn-1pkβ,which gives(33)supr,s1hrα∑k∈IrMkqkΔnmxk+sρ,z1,…,zn-1pkβ<∞.Hence, x=(xk)∈c∞θ[M,p,Δnm,q,·,…,·]αβ.Theorem 7.
Let0<infpk=h≤pk≤suppk=H<∞ and let M=(Mk) and M′=(Mk′) be two Musielak-Orlicz functions satisfying Δ2-condition. Then(i)
c0θ[M′,p,Δnm,q,·,…,·]αβ⊂c0θ[M∘M′,p,Δnm,q,·,…,·]αβ,(ii)
cθ[M′,p,Δnm,q,·,…,·]αβ⊂cθ[M∘M′,p,Δnm,q,·,…,·]αβ,(iii)
c∞θ[M′,p,Δnm,q,·,…,·]αβ⊂c∞θ[M∘M′,p,Δnm,q,·,…,·]αβ.Proof.
Letx=(xk)∈cθ[M,p,Δnm,q,·,…,·]αβ. Then, we have(34)limr→∞1hrα∑k∈IrMk′qkΔnmxk+s-Lρ,z1,…,zn-1pkβ=0,uniformly in s and for some L∈C. Let ϵ>0 and choose δ with 0<δ<1 such that Mk(t)<ϵ for 0≤t≤δ. Let(35)yk,s=Mk′qkΔnmxk+s-Lρ,z1,…,zn-1∀k,s∈N.We can write(36)1hrα∑k∈IrMkyk,spkβ=1hrα∑k∈Ir,yk,s≤δMkyk,spkβ+1hrα∑k∈Ir,yk,s>δMkyk,spkβ.Since M=(Mk) satisfies Δ2-condition, we have(37)1hrα∑k∈Ir,yk,s≤δMkyk,spkβ≤Mk1H1hrα∑k∈Ir,yk,s≤δMkyk,spkβ. For yk,s>δ, we obtain(38)yk,s<yk,sδ<1+yk,sδ.It follows from the fact that M=(Mk) is nondecreasing and convex that(39)Mkyk,s<Mk1+yk,sδ<12Mk2+12Mk2yk,sδ.Since (Mk) satisfies Δ2-condition, we can write(40)Mkyk,s<12Tyk,sδMk2+12Tyk,sδMk2=Tyk,sδMk2.Hence,(41)1hrα∑k∈Ir,yk,s>δMkyk,spkβ≤max1,TMk2δH1hrα∑k∈Ir,yk,s>δyk,spkβ. Equation (37) together with (41) gives(42)x=xk∈c0θM∘M′,p,Δnm,q,·,…,·αβ.This completes the proof of (i). In a similar way, using Δ2-condition for (Mk), we can prove that(43)cθM′,p,Δnm,q,·,…,·αβ⊂cθM∘M′,p,Δnm,q,·,…,·αβ,c∞θM′,p,Δnm,q,·,…,·αβ⊂c∞θM∘M′,p,Δnm,q,·,…,·αβ.Corollary 8.
Let0<infpk=h≤pk≤suppk=H<∞ and let M=(Mk) be Musielak-Orlicz function satisfying Δ2-condition. Then(44)c0θp,Δnm,q,·,…,·αβ⊂c0θM,p,Δnm,q,·,…,·αβ,c∞θp,Δnm,q,·,…,·αβ⊂c∞θM,p,Δnm,q,·,…,·αβ.Proof.
The proof follows by takingM′(x)=x in the above Theorem 7.Theorem 9.
IfM=(Mk) be the Musielak-Orlicz function, then the following statements are equivalent:(i)
c∞θ[p,Δnm,q,·,…,·]αβ⊂c∞θ[M,p,Δnm,q,·,…,·]αβ,(ii)
c0θ[p,Δnm,q,·,…,·]αβ⊂c∞θ[M,p,Δnm,q,·,…,·]αβ,(iii)
supr1/hrα[∑k∈Ir[Mk(t/ρ)]pk]β<∞(t,ρ>0).Proof.
(i) ⇒ (ii). It is obvious in view of the following fact that(45)c0θp,Δnm,q,·,…,·αβ⊂c∞θp,Δnm,q,·,…,·αβ.
(ii) ⇒ (iii). Let c0θ[p,Δnm,q,·,…,·]αβ⊂c∞θ[M,p,Δnm,q,·,…,·]αβ. Assume that (iii) does not hold. Then, for some t,ρ>0, we obtain(46)supr1hrα∑k∈IrMktρpkβ=∞,and so one can find a subinterval Ir(j) of the set of interval Ir such that(47)1hrjα∑k∈IrjMkj-1ρpkβ>j,j=1,2.Let us define the sequence x=(xk) by(48)Δnmxk+s=j-1,ifk∈Irj,0,ifk∉Irj,∀s∈N.Then x=(xk)∈c0θ[p,Δnm,q,·,…,·]αβ but, by (47), x=(xk)∉c∞θ[M,p,Δnm,q,·,…,·]αβ which contradicts (ii). Hence, the condition (iii) must hold.
(iii) ⇒ (i). Suppose that (iii) holds and x=(xk)∈c0θ[p,Δnm,q,·,…,·]αβ. Also assume that x=(xk)∉c∞θ[M,p,Δnm,q,·,…,·]αβ. Then(49)supr,s1hrα∑k∈IrMkΔmxk+sρ,z1,…,zn-1pkβ=∞.Let t=qk(Δnmxk+s,z1,…,zn-1) for each k and fixed s. It follows from (49) that(50)supr1hrα∑k∈IrMktρβ=∞,which contradicts (iii). Hence, the condition (i) holds.Theorem 10.
Let1≤pk≤suppk<∞ and let M=(Mk) be a Musielak-Orlicz function. Then the following statements are equivalent:(i)
c0θ[M,p,Δnm,q,·,…,·]αβ⊂c0θ[p,Δnm,q,·,…,·]αβ,(ii)
c0θ[M,p,Δnm,q,·,…,·]αβ⊂c∞θ[p,Δnm,q,·,…,·]αβ,(iii)
infr1/hrα[∑k∈Ir[Mk(t/ρ)]pk]β>0(t,ρ>0).Proof.
It is trivial to obtain that (i) implies (ii). We are now proving that (ii) implies (iii). Let the condition (ii) hold. Assume that (iii) does not hold. Therefore,(51)infr1hrα∑k∈IrMktρpkβ=0t,ρ>0and so we can find a subinterval Ir(j)(j=1,2) of Ir such that(52)1hrjα∑k∈IrjMkjρpkβ<j-1. For all s∈N, define Δnmxk+s=j if k∈Ir(j) and Δnmxk+s=0 if k∉Ir(j). If follows from (52) that x=(xk)∈c0θ[p,Δnm,q,·,…,·]αβ but x=(xk)∉c∞θ[p,Δnm,q,·,…,·]αβ which contradicts (ii). Thus, the condition (iii) holds. It is left to prove that (iii) implies (i). Suppose that (iii) holds and also assume that x=(xk)∈c0θ[M,p,Δnm,q,·,…,·]αβ. Therefore,(53)limr→∞1hrα∑k∈IrMkqkΔnmxk+sρ,z1,…,zn-1pkβ=0uniformlyins,forsomeρ>0.Again, assume that x=(xk)∉c0θ[p,Δnm,q,·,…,·]αβ. Then, for some number ϵ>0 and a subinterval Ir(j) of Ir, one obtains Δnmxk+s,z1,…,zn-1≥ϵ for all k∈N and for some s≥s0. Therefore, in view of the properties of the Orlicz function, one writes(54)MkqkΔnmxk+sρ,z1,…,zn-1pkβ≥Mkϵρpkβ.By taking into account (53), we obtain(55)limr→∞1hrα∑k∈IrMkϵρpkβ=0,which contradicts (iii). Hence, (i) must hold.Theorem 11.
Let0<pk≤rk for all k∈N and let (rk/pk) be bounded. Then(56)cθM,r,Δnm,q,·,…,·αβ⊂cθM,p,Δnm,q,·,…,·αβ.Proof.
Suppose thatx∈cθ[M,r,Δnm,q,·,…,·]αβ. Write(57)tk=MkqkΔnmxk+s-Lρ,z1,…,zn-1rkβ,and μk=pk/rk for all k∈N. Then 0<μk≤1 for all k∈N. Take 0<μ<μk for k∈N. Consider two sequences (uk) and (vk) defined by uk=tk and vk=0 if tk≥1 and uk=0 and vk=tk for tk<1. Clearly(58)tk=uk+vk,tkμk=ukμk+vkμk,∀k∈N.It follows that ukμk≤uk≤tk and vkμk≤vkμ. Therefore,(59)1hrα∑k∈Irtkμk=1hrα∑k∈Irukμk+vkμk≤1hrα∑k∈Irtk+1hrα∑k∈Irvkμ.Now, for each k, we obtain(60)1hrα∑k∈Irvkμ=∑k∈Ir1hrαvkμ1hrα1-μ≤∑k∈Ir1hrαvkμ1/μμ∑k∈Ir1hrα1-μ1/1-μ1-μ=1hrα∑k∈Irvkμ.By taking the above two inequalities into account, we get(61)1hrα∑k∈Irtkμk≤1hrα∑k∈Irtk+1hrα∑k∈Irvkμ.Hence, we conclude that x∈cθ[M,p,Δnm,q,·,…,·]αβ.Theorem 12.
(i)
If0<infpk≤pk≤1forallk∈N, then(62)cθM,p,Δnm,q,·,…,·αβ⊂cθM,Δnm,q,·,…,·αβ. (ii)
If1≤pk≤suppk<∞forallk∈N, then(63)cθM,Δnm,q,·,…,·αβ⊂cθM,p,Δnm,q,·,…,·αβ.Proof.
(i) Assume thatx∈cθ[M,p,Δnm,q,·,…,·]αβ. Then(64)limr→∞1hrα∑k∈IrMkqkΔnmxk+s-Lρ,z1,…,zn-1pkβ=0.Since 0<infpk≤pk≤1, we have(65)limr→∞1hrα∑k∈IrMkqkΔnmxk+s-Lρ,z1,…,zn-1β≤limr→∞1hrα∑k∈IrMkqkΔnmxk+s-Lρ,z1,…,zn-1pkβ.Thus, we have(66)limr→∞1hrα∑k∈IrMkqkΔnmxk+s-Lρ,z1,…,zn-1β=0,which gives x∈cθ[M,Δnm,q,·,…,·]αβ. Therefore, we conclude that(67)cθM,p,Δnm,q,·,…,·αβ⊂cθM,Δnm,q,·,…,·αβ.
(ii) Suppose thatpk≥1 for each k and suppk<∞. Let x∈cθ[M,Δnm,q,·,…,·]αβ. Then, for each ϵ>0 there exists a positive integer N such that(68)limr→∞1hrα∑k∈IrMkqkΔnmxk+s-Lρ,z1,…,zn-1pkβ=0<1.Since 1≤pk≤suppk<∞, we have(69)limr→∞1hrα∑k∈IrMkqkΔnmxk+s-Lρ,z1,…,zn-1pkβclimr→∞1hrα≤limr→∞1hrα∑k∈IrMkqkΔmxk+s-Lρ,z1,…,zn-1β=0<1.Hence, x∈cθ[M,p,Δnm,q,·,…,·]αβ.



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