# Sequence Spaces Defined by Musielak-Orlicz Function overn-Normed Spaces

Authors: M. Mursaleen; Sunil K. Sharma; A. Kılıçman
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

In the present paper we introduce some sequence spaces overn-normed spaces defined by a Musielak-Orlicz function ℳ=(Mk). We also study some topological properties and prove some inclusion relations between these spaces.


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## Body

## 1. Introduction and Preliminaries

An Orlicz functionM is a function, which is continuous, nondecreasing, and convex with M(0)=0, M(x)>0 for x>0 and M(x)→∞ as x→∞.Lindenstrauss and Tzafriri [1] used the idea of Orlicz function to define the following sequence space. Let w be the space of all real or complex sequences x=(xk); then
(1)ℓM={x∈w:∑k=1∞‍M(|xk|ρ)<∞},
which is called as an Orlicz sequence space. The space ℓM is a Banach space with the norm
(2)∥x∥=inf{ρ>0:∑k=1∞‍M(|xk|ρ)≤1}.It is shown in [1] that every Orlicz sequence space ℓM contains a subspace isomorphic to ℓp(p≥1). The Δ2-condition is equivalent to M(Lx)≤kLM(x) for all values of x≥0 and for L>1. A sequence ℳ=(Mk) of Orlicz functions is called a Musielak-Orlicz function (see [2, 3]). A sequence 𝒩=(Nk) defined by
(3)Nk(v)=sup{|v|u-Mk(u):u≥0},k=1,2,…,
is called the complementary function of a Musielak-Orlicz function ℳ. For a given Musielak-Orlicz function ℳ, the Musielak-Orlicz sequence space tℳ and its subspace hℳ are defined as follows:
(4)tℳ={x∈w:Iℳ(cx)<∞forsomec>0},hℳ={x∈w:Iℳ(cx)<∞∀c>0},
where Iℳ is a convex modular defined by
(5)Iℳ(x)=∑k=1∞‍(Mk)(xk),x=(xk)∈tℳ.We considertℳ equipped with the Luxemburg norm
(6)∥x∥=inf{k>0:Iℳ(xk)≤1}
or equipped with the Orlicz norm
(7)∥x∥0=inf{1k(1+Iℳ(kx)):k>0}.LetX be a linear metric space. A function p : X→ℝ is called paranorm if(1)
p(x)≥0 for all x∈X,(2)
p(-x)=p(x) for all x∈X,(3)
p(x+y)≤p(x)+p(y) for all x,y∈X,(4)
(λn) is a sequence of scalars with λn→λasn→∞ and (xn) is a sequence of vectors with p(xn-x)→0asn→∞; then p(λnxn-λx)→0asn→∞.A paranormp for which p(x)=0 implies x=0 is called total paranorm and the pair (X,p) is called a total paranormed space. It is well known that the metric of any linear metric space is given by some total paranorm (see [4], Theorem 10.4.2, pp. 183). For more details about sequence spaces, see [5–12] and references therein.A sequence of positive integersθ=(kr) is called lacunary if k0=0, 0<kr<kr+1 and hr=kr-kr-1→∞ as r→∞. The intervals determined by θ will be denoted by Ir=(kr-1,kr) and qr=kr/kr-1. The space of lacunary strongly convergent sequences Nθ was defined by Freedman et al. [13] as
(8)Nθ={x∈w:limr→∞1hr∑k∈Ir‍|xk-l|=0,forsomel}.Strongly almost convergent sequence was introduced and studied by Maddox [14] and Freedman et al. [13]. Parashar and Choudhary [15] have introduced and examined some properties of four sequence spaces defined by using an Orlicz function M, which generalized the well-known Orlicz sequence spaces [C,1,p], [C,1,p]0, and [C,1,p]∞. It may be noted here that the space of strongly summable sequences was discussed by Maddox [16] and recently in [17].Mursaleen and Noman [18] introduced the notion of λ-convergent and λ-bounded sequences as follows.Letλ=(λk)k=1∞ be a strictly increasing sequence of positive real numbers tending to infinity; that is,
(9)0<λ0<λ1<⋯,λk⟶∞ask⟶∞,
and it is said that a sequence x=(xk)∈w is λ-convergent to the number L, called the λ-limit of x if Λm(x)→L as m→∞, where
(10)λm(x)=1λm∑k=1m‍(λk-λk-1)xk.The sequencex=(xk)∈w is λ-bounded if supm|Λm(x)|<∞. It is well known [18] that if limmxm=a in the ordinary sense of convergence, then
(11)limm(1λm(∑k=1m‍(λk-λk-1)|xk-a|))=0.
This implies that(12)limm|Λm(x)-a|=limm|1λm∑k=1m‍(λk-λk-1)(xk-a)|=0,
which yields that limmΛm(x)=a and hence x=(xk)∈w is λ-convergent to a.The concept of 2-normed spaces was initially developed by Gähler [19] in the mid 1960s, while for that of n-normed spaces one can see Misiak [20]. Since then, many others have studied this concept and obtained various results; see Gunawan ([21, 22]) and Gunawan and Mashadi [23]. Let n∈ℕ and let X be a linear space over the field 𝕂, where 𝕂 is the field of real or complex numbers of dimension d, where d≥n≥2. A real valued function ∥·,…,·∥ on Xn satisfying the following four conditions (1)
∥x1,x2,…,xn∥=0 if and only if x1,x2,…,xn are linearly dependent in X;(2)
∥x1,x2,…,xn∥ is invariant under permutation;(3)
∥αx1,x2,…,xn∥=|α|∥x1,x2,…,xn∥ for any α∈𝕂;(4)
∥x+x′,x2,…,xn∥≤∥x,x2,…,xn∥+∥x′,x2,…,xn∥is called an n-norm on X, and the pair (X,∥·,…,·∥) is called an n-normed space over the field 𝕂.For example, if we may takeX=ℝn being equipped with the n-norm ∥x1,x2,…,xn∥E = the volume of the n-dimensional parallelepiped spanned by the vectors x1,x2,…,xn which may be given explicitly by the formula
(13)∥x1,x2,…,xn∥E=|det(xij)|,
where xi=(xi1,xi2,…,xin)∈ℝn for each i=1,2,…,n, leting (X,∥·,…,·∥) be an n-normed space of dimension d≥n≥2 and {a1,a2,…,an} be linearly independent set in X, then the following function ∥·,…,·∥∞ on Xn-1 defined by
(14)∥x1,x2,…,xn-1∥∞=max{∥x1,x2,…,xn-1,ai∥:i=1,2,…,n}
defines an (n-1)-norm on X with respect to {a1,a2,…,an}.A sequence(xk) in an n-normed space (X,∥·,…,·∥) is said to converge to some L∈X if
(15)limk→∞∥xk-L,z1,…,zn-1∥=0limk→∞foreveryz1,…,zn-1∈X.A sequence(xk) in an n-normed space (X,∥·,…,·∥) is said to be Cauchy if
(16)limp→∞k→∞∥xk-xp,z1,…,zn-1∥=0limxxp→∞k→∞foreveryz1,…,zn-1∈X.If every Cauchy sequence inX converges to some L∈X, then X is said to be complete with respect to the n-norm. Any complete n-normed space is said to be n-Banach space.Letℳ=(Mk) be a Musielak-Orlicz function, and let p=(pk) be a bounded sequence of positive real numbers. We define the following sequence spaces in the present paper:
(17)w0θ(ℳ,Λ,p,s,∥·,…,·∥)={Λk(x)ρx=(xk)∈w:limr→∞1hr=×∑k∈Ir‍k-s[Mk(∥Λk(x)ρ,z1,z2,…,zn-1∥)]pk=0,(∥Λk(x)ρ,z1,z2z2z2z2z2z2z2z2,∥)ρ>0,s≥0Λk(x)ρ},wθ(ℳ,Λ,p,s,∥·,…,·∥)={Λk(x)-Lρx=(xk)∈w:limr→∞1hr=×∑k∈Ir‍k-s[Mk(∥Λk(x)-Lρ,z1,z2,…,zn-1∥)]pk=0,forsomeL,ρ>0,s≥0Λk(x)-Lρ},w∞θ(ℳ,Λ,p,s,∥·,…,·∥)={Λk(x)ρx=(xk)∈w:supr1hr=×∑k∈Ir‍k-s[Mk(∥Λk(x)ρ,z1,z2,…,zn-1∥)]pk<∞,(∥Λk(x)ρ,z1,z2z2z2⋯,zn-1∥)ρ>0,s≥0Λk(x)ρ}.If we takeℳ(x)=x, we get
(18)w0θ(Λ,p,s,∥·,…,·∥)={Λk(x)ρx=(xk)∈w:limr→∞1hr=×∑k∈Ir‍k-s[∥Λk(x)ρ,z1,z2,…,zn-1∥]pk=0,∥Λk(x)ρ,z1,z2z2,⋯,zn-1∥ρ>0,s≥0Λk(x)ρ},wθ(Λ,p,s,∥·,…,·∥)={Λk(x)-Lρx=(xk)∈w:limr→∞1hr=×∑k∈Ir‍k-s[∥Λk(x)-Lρ,z1,z2,…,zn-1∥]pk=0,∥,z1,z2z2…,zn-1∥forsomeL,ρ>0,s≥0Λk(x)-Lρ},w∞θ(Λ,p,s,∥·,…,·∥)={Λk(x)ρx=(xk)∈w:supr1hr=×∑k∈Ir‍k-s[∥Λk(x)ρ,z1,z2,…,zn-1∥]pk<∞,∥Λk(x)ρ,z1,z2z2z2,…,zn-1∥ρ>0,s≥0Λk(x)ρ}.If we takep=(pk)=1 for all k∈ℕ, we have
(19)w0θ(ℳ,Λ,s,∥·,…,·∥)={Λk(x)ρx=(xk)∈w:limr→∞1hr=×∑k∈Ir‍k-s[Mk(∥Λk(x)ρ,z1,z2,…,zn-1∥)]=0,∥Λk(x)ρ,z1,z2,z1,z2,z1,…,zn-1∥ρ>0,s≥0Λk(x)ρ},wθ(ℳ,Λ,s,∥·,…,·∥)={x=(xk)∈w:limr→∞1hr=×∑k∈Ir‍k-s[Mk(∥Λk(x)-Lρ,z1,z2,…,zn-1∥)]=0,forsomeL,ρ>0,s≥0Λk(x)-Lρ},w∞θ(ℳ,Λ,s,∥·,…,·∥)={Λk(x)ρx=(xk)∈w:supr1hr=×∑k∈Ir‍k-s[Mk(∥Λk(x)ρ,z1,z2,…,zn-1∥)]<∞,∥Λk(x)ρ,z1,z2,z1,,z1z1…,zn∥ρ>0,s≥0Λk(x)ρ}.The following inequality will be used throughout the paper. If0≤pk≤suppk=H, K=max(1,2H-1), then
(20)|ak+bk|pk≤K{|ak|pk+|bk|pk}
for all k and ak,bk∈ℂ. Also |a|pk≤max(1,|a|H) for all a∈ℂ.In this paper, we introduce sequence spaces defined by a Musielak-Orlicz function overn-normed spaces. We study some topological properties and prove some inclusion relations between these spaces.


## 2. Main Results

Theorem 1.
Letℳ=(Mk) be a Musielak-Orlicz function, and let p=(pk) be a bounded sequence of positive real numbers, then the spaces w0θ(ℳ,Λ,p,s,∥·,…,·∥),wθ(ℳ,Λ,p,s,∥·,…,·∥), and w∞θ(ℳ,Λ,p,s,∥·,…,·∥) are linear spaces over the field of complex number ℂ.Proof.
Letx=(xk),lety=(yk)∈w0θ(ℳ,Λ,p,s,∥·,…,·∥), and let α,β∈ℂ. In order to prove the result, we need to find some ρ3 such that
(21)limr→∞1hr∑k∈Ir‍k-s[Mk(∥Λk(αx+βy)ρ3,z1,z2,…,zn-1∥)]pk=0.
Sincex=(xk),y=(yk)∈w0θ(ℳ,Λ,p,s,∥·,…,·∥), there exist positive numbers ρ1,ρ2>0 such that
(22)limr→∞1hr∑k∈Ir‍k-s[Mk(∥Λk(x)ρ1,z1,z2,…,zn-1∥)]pk=0,limr→∞1hr∑k∈Ir‍k-s[Mk(∥Λk(y)ρ2,z1,z2,…,zn-1∥)]pk=0.
Define ρ3=max(2|α|ρ1,2|β|ρ2). Since (Mk) is nondecreasing, convex function and by using inequality (20), we have
(23)1hr∑k∈Ir‍k-s[Mk(∥Λk(αx+βy)ρ3,z1,z2,…,zn-1∥)]pk≤1hr∑k∈Ir‍k-s[Mk(∥αΛk(x)ρ3,z1,z2,…,zn-1∥≤1hr1hr1hr=-∑k∈Ir‍k-s+∥βΛk(y)ρ3,z1,z2,…,zn-1∥)]pk≤K1hr∑k∈Ir‍12pkk-s[Mk(∥Λk(x)ρ1,z1,z2,…,zn-1∥)]pk≤+K1hr∑k∈Ir‍12pkk-s[Mk(∥Λk(y)ρ2,z1,z2,…,zn-1∥)]pk≤K1hr∑k∈Ir‍k-s[Mk(∥Λk(x)ρ1,z1,z2,…,zn-1∥)]pk≤+K1hr∑k∈Ir‍k-s[Mk(∥Λk(y)ρ2,z1,z2,…,zn-1∥)]pk⟶0asr⟶∞.
Thus, we have αx+βy∈w0θ(ℳ,Λ,p,s,∥·,…,·∥). Hence, w0θ(ℳ,Λ,p,s,∥·,…,·∥) is a linear space. Similarly, we can prove that wθ(ℳ,Λ,p,s,∥·,…,·∥) and w∞θ(ℳ,Λ,p,s,∥·,…,·∥) are linear spaces.Theorem 2.
Letℳ=(Mk) be a Musielak-Orlicz function, and let p=(pk) be a bounded sequence of positive real numbers. Then w0θ(ℳ,Λ,p,s,∥·,…,·∥) is a topological linear space paranormed by
(24)g(x)=inf{∑k∈Ir‍k-sρpr/H:=infinf(1hr∑k∈Ir‍k-s[Mk(∥∑k∈Ir‍k-sΛk(x)ρ,(1hr∑k∈Ir‍(1hr1hr1hr=∑k∈Ir‍z1,z2,…,zn-1∑k∈Ir‍k-s∥)]pk)1/H≤1ef22(efsdf1hr∑k∈Ir‍k-s},
where H=max(1,supkpk)<∞.Proof.
Clearlyg(x)≥0 for x=(xk)∈w0θ(ℳ,Λ,p,s,∥·,…,·∥). Since Mk(0)=0, we get g(0)=0. Again if g(x)=0, then
(25)inf{(1hr∑k∈Ir‍k-s[Mk(∥Λk(x)ρ,z1,z2,…,zn-1∥)]pk)ρpr/H:infinf(1hr∑k∈Ir‍k-s[Mk(∥1hr∑k∈Ir‍k-sΛk(x)ρ,(1hr∑k∈Ir‍(1hrz21hrz21hrz1,z2,…,zn-11hr∑k∈Ir‍k-sΛk(x)ρ∥)]pk)1/H≤1(1hr∑k∈Ir‍k-s[Mk(∥Λk(x)ρ,z1,z2,…,zn-1∥)]pk)}=0.
This implies that for a given ϵ>0, there exist some ρϵ(0<ρϵ<ϵ) such that
(26)(1hr∑k∈Ir‍k-s[Mk(∥Λk(x)ρϵ,z1,z2,…,zn-1∥)]pk)1/H≤1.
Thus,
(27)(1hr∑k∈Ir‍k-s[Mk(∥Λk(x)ϵ,z1,z2,…,zn-1∥)]pk)1/H≤(1hr∑k∈Ir‍k-s[Mk(∥Λk(x)ρϵ,z1,z2,…,zn-1∥)]pk)1/H.
Suppose that (xk)≠0 for each k∈ℕ. This implies that Λk(x)≠0 for each k∈ℕ. Let ϵ→0, then
(28)∥Λk(x)ϵ,z1,z2,…,zn-1∥⟶∞.
It follows that
(29)(1hr∑k∈Ir‍k-s[Mk(∥Λk(x)ϵ,z1,z2,…,zn-1∥)]pk)1/H⟶∞,
which is a contradiction. Therefore, Λk(x)=0 for each k, and thus (xk)=0 for each k∈ℕ. Let ρ1>0 and ρ2>0 be the case such that
(30)(1hr∑k∈Ir‍k-s[Mk(∥Λk(x)ρ1,z1,z2,…,zn-1∥)]pk)1/H≤1,(1hr∑k∈Ir‍k-s[Mk(∥Λk(y)ρ2,z1,z2,…,zn-1∥)]pk)1/H≤1.
Let ρ=ρ1+ρ2; then, by using Minkowski's inequality, we have
(31)(1hr∑k∈Ir‍k-s[Mk(∥Λk(x+y)ρ,z1,z2,…,zn-1∥)]pk)1/H≤(1hr∑k∈Ir‍k-s[Mk({z1,z2,…,zn-11hr∑k∈Ir‍k-sΛk(x)+Λk(y)ρ1+ρ2∥}∥{z1,z2,…,zn-11hr∑k∈Ir‍k-sΛk(x)+Λk(y)ρ1+ρ2∥}Λk(x)+Λk(y)ρ1+ρ2,1hr1hr1hr1hr1hr1hr1hr∑k∈Ir‍∑k∈Ir‍k-sz1,z2,…,zn-11hr∑k∈Ir‍k-sΛk(x)+Λk(y)ρ1+ρ2∥)]pk)1/H≤({[∥Λk(y)ρ2,z1,z2,…,zn-11hr∑k∈Ir‍k-s∥]}1hr∑k∈Ir‍k-s[Λk(y)ρ2{[∥Λk(y)ρ2,z1,z2,…,zn-11hr∑k∈Ir‍k-s∥]}Mk(ρ1ρ1+ρ2)1hr∑k∈Ir‍k-s1hr∑k∈Ir‍×[∥Λk(x)ρ1,z1,z2,…,zn-1∥]+(ρ2ρ1+ρ2)1hr∑k∈Ir‍k-s1hr-∑k∈Ir‍×[∥Λk(y)ρ2,z1,z2,…,zn-11hr∑k∈Ir‍k-s∥]]pk)1/H≤(ρ1ρ1+ρ2)1hr1hr×(1hr∑k∈Ir‍k-s[Mk(∥Λk(x)ρ1,z1,z2,…,zn-1∥)]pk)1/H+(ρ2ρ1+ρ2)1hr1hr×(1hr∑k∈Ir‍k-s[Mk(∥Λk(y)ρ2,z1,z2,…,zn-1∥)]pk)1/H≤1.
Since ρ,ρ1, and ρ2 are nonnegative, we have
(32)g(x+y)=inf{1hr∑k∈Ir‍k-s{z1,z2,…,zn-11hr∑k∈Ir‍k-sΛk(x+y)ρ∥)]pk)1/H}ρpr/H:infinf(1hr∑k∈Ir‍k-s[Mk(∥1hr∑k∈Ir‍k-sΛk(x+y)ρ,infinfinfinfiniinfinnfinz1,z2,…,zn-11hr∑k∈Ir‍k-sΛk(x+y)ρ∥)]pk)1/H≤1}≤inf{1hr∑k∈Ir‍k-s{z1,z2,…,zn-11hr∑k∈Ir‍k-sΛk(x+y)ρ∥)]pk)1/H}Λk(x)ρ1(ρ1)pr/H:infinfinf(1hr∑k∈Ir‍k-s[Mk(∥1hr∑k∈Ir‍k-sΛk(x)ρ1,infinfinfinfinfinfinfinfinfiz1,z2,…,zn-11hr∑k∈Ir‍k-sΛk(x)ρ1∥)]pk)1/H≤1}≤f+inf{{z1,z2,…,zn-11hr∑k∈Ir‍k-sΛk(x+y)ρ∥)]pk)1/H}Λk(y)ρ21hr∑k∈Ir‍k-s(ρ2)pr/H:=infinf(1hr∑k∈Ir‍k-s[Mk(∥1hr∑k∈Ir‍k-sΛk(y)ρ2,infinfinfininifinfinfinfinfz1,z2,…,zn-11hr∑k∈Ir‍k-sΛk(y)ρ2∥)]pk)1/H≤1}.
Therefore, g(x+y)≤g(x)+g(y). Finally we prove that the scalar multiplication is continuous. Let μ be any complex number. By definition,
(33)g(μx)=inf{1hr∑k∈Ir‍k-sΛk(μx)ρρpr/H:=innii(1hr∑k∈Ir‍k-s[Mk(∥1hr∑k∈Ir‍k-sΛk(μx)ρ,(1hr∑k∈Ir‍k-s(1hr=∑k∈Ir‍iik-sz1,z2,…,zn-11hr∑k∈Ir‍k-sΛk(μx)ρ∥)]pk)1/Hinin&fin≤11hr∑k∈Ir‍k-s}.
Thus,
(34)g(μx)=inf{∑k∈Ir‍1hrΛk(x)k-st(|μ|t)pr/H:=inin(1hr∑k∈Ir‍k-s[Mk(∥1hr∑k∈Ir‍k-s1hr∑k∈Ir‍k-sΛk(x)t,(1hr∑k∈Ir‍(1hr1hr1hr∑k∈Ir‍iik-sz1,z2,…,zn-1Λ1hr∑k∈Ir‍∥)]pk)1/H=in&finfin≤11hr∑k∈Ir‍k-s1hr∑k∈Ir‍k-s},
where 1/t=ρ/|μ|. Since |μ|pr≤max(1,|μ|suppr), we have
(35)g(μx)≤max(1,|μ|suppr)×inf{Λk(x)t1hr∑k∈Ir‍k-stpr/H:=inf==(1hr∑k∈Ir‍k-s[Mk(∥1hr∑k∈Ir‍k-sΛk(x)t,(1hr∑k∈Ir‍=k-s(1hr=∑k∈Ir‍iik-sz1,z2,…,zn-11hr∑k∈Ir‍k-sΛk(μx)ρ∥)]pk)1/H=n&fn&=≤11hr∑k∈Ir‍k-s}.
So the fact that scalar multiplication is continuous follows from the above inequality. This completes the proof of the theorem.Theorem 3.
Letℳ=(Mk) be a Musielak-Orlicz function. If supk[Mk(x)]pk<∞ for all fixed x>0, then w0θ(ℳ,Λ,p,s,∥·,…,·∥)⊆w∞θ(ℳ,Λ,p,s,∥·,…,·∥).Proof.
Letx=(xk)∈w0θ(ℳ,Λ,p,s,∥·,…,·∥); then there exists positive number ρ1 such that
(36)limr→∞1hr∑k∈Ir‍k-s[Mk(∥Λk(x)ρ1,z1,z2,…,zn-1∥)]pk=0.
Define ρ=2ρ1. Since (Mk) is nondecreasing and convex and by using inequality (20), we have
(37)supr1hr∑k∈Ir‍k-s[Mk(∥Λk(x)ρ,z1,z2,…,zn-1∥)]pk=supr1hr∑k∈Ir‍k-s[Mk(∥Λk(x)+L-Lρ,=supr1hr∑k∈Ir‍k-s=supr∑k∈Ir‍k-sz1,z2,…,zn-1Λk(x)+L-Lρ∥)]pk≤Ksupr1hr∑k∈Ir‍k-s12pk[Mk(∥Λk(x)-Lρ1,z1,z2,…,zn-1{[Mk(∥Λk(x)-Lρ1,}∥{[Mk(∥Λk(x)-Lρ1,})]pkKsupr+Ksupr1hr∑k∈Ir‍k-s12pk[Mk(∥Lρ1,z1,z2,…,zn-1∥)]pk≤Ksupr1hr∑k∈Ir‍k-s[Mk(∥Λk(x)-Lρ1,z1,z2,…,zn-1∥)]pkKsupr+Ksupr1hr∑k∈Ir‍k-s[Mk(∥Lρ1,z1,z2,…,zn-1∥)]pk<∞.
Hence, x=(xk)∈w∞θ(ℳ,Λ,p,s,∥·,…,·∥).Theorem 4.
Let0<infpk=h≤pk≤suppk=H<∞ and let ℳ=(Mk),ℳ'=(Mk′) be Musielak-Orlicz functions satisfying Δ2-condition, then one has(i)
w0θ(ℳ',Λ,p,s,∥·,…,·∥)⊂w0θ(ℳ∘ℳ',Λ,p,s,∥·,…,·∥);(ii)
wθ(ℳ',Λ,p,s,∥·,…,·∥)⊂wθ(ℳ∘ℳ',Λ,p,s,∥·,…,·∥);(iii)
w∞θ(ℳ',Λ,p,s,∥·,…,·∥)⊂w∞θ(ℳ∘ℳ',Λ,p,s,∥·,…,·∥).Proof.
Letx=(xk)∈w0θ(ℳ′,Λ,p,s,∥·,…,·∥), then we have
(38)limr→∞1hr∑k∈Ir‍k-s[Mk′(∥Λk(x)ρ,z1,z2,…,zn-1∥)]pk=0.
Letϵ>0 and choose δ with 0<δ<1 such that Mk(t)<ϵ for 0≤t≤δ. Let (yk)=Mk′[∥Λk(x)/ρ,z1,z2,…,zn-1∥] for all k∈ℕ. We can write
(39)1hr∑k∈Ir‍k-s(Mk[yk])pk=1hr∑yk≤δk∈Ir‍k-s(Mk[yk])pk+1hr∑yk≥δk∈Ir‍k-s(Mk[yk])pk.
So, we have
(40)1hr∑yk≤δk∈Ir‍k-s(Mk[yk])pk≤[Mk(1)]H1hr∑yk≤δk∈Ir‍k-s(Mk[yk])pk≤[Mk(2)]H1hr∑yk≤δk∈Ir‍k-s(Mk[yk])pk.
For yk>δ,yk<yk/δ<1+yk/δ. Since (Mk)′s are nondecreasing and convex, it follows that
(41)Mk(yk)<Mk(1+ykδ)<12Mk(2)+12Mk(2ykδ).
Since ℳ=(Mk) satisfies Δ2-condition, we can write
(42)Mk(yk)<12TykδMk(2)+12TykδMk(2)=TykδMk(2).
Hence,
(43)1hr∑yk≥δk∈Ir‍k-sMk[yk]pk≤max(1,(TMk(2)δ)H)1hr∑yk≤δk∈Ir‍k-s[yk]pk.
From (40) and (43), we have x=(xk)∈w0θ(ℳ∘ℳ',Λ,p,s,∥·,…,·∥). This completes the proof of (i). Similarly we can prove that
(44)wθ(ℳ′,Λ,p,s,∥·,…,·∥)⊂wθ(ℳ∘ℳ′,Λ,p,s,∥·,…,·∥),w∞θ(ℳ′,Λ,p,s,∥·,…,·∥)⊂w∞θ(ℳ∘ℳ′,Λ,p,s,∥·,…,·∥).Theorem 5.
Let0<h=infpk=pk<suppk=H<∞. Then for a Musielak-Orlicz function ℳ=(Mk) which satisfies Δ2-condition, one has(i)
w0θ(Λ,p,s,∥·,…,·∥)⊂w0θ(ℳ,Λ,p,s,∥·,…,·∥);(ii)
wθ(Λ,p,s,∥·,…,·∥)⊂wθ(ℳ,Λ,p,s,∥·,…,·∥);(iii)
w∞θ(Λ,p,s,∥·,…,·∥)⊂w∞θ(ℳ,Λ,p,s,∥·,…,·∥).Proof.
It is easy to prove, so we omit the details.Theorem 6.
Letℳ=(Mk) be a Musielak-Orlicz function and let 0<h=infpk. Then w∞θ(ℳ,Λ,p,s,∥·,…,·∥)⊂w0θ(Λ,p,s,∥·,…,·∥) if and only if
(45)limr→∞1hr∑k∈Ir‍k-s(Mk(t))pk=∞
for some t>0.Proof.
Letw∞θ(ℳ,Λ,p,s,∥·,…,·∥)⊂w0θ(Λ,p,s,∥·,…,·∥). Suppose that (45) does not hold. Therefore, there are subinterval Ir(j) of the set of interval Ir and a number t0>0, where
(46)t0=∥Λk(x)ρ,z1,z2,…,zn-1∥∀k,
such that
(47)1hr(j)=∑k∈Ir(j)‍k-s(Mk(t0))pk≤K<∞,m=1,2,3,….
Let us define x=(xk) as follows:
(48)Λk(x)={ρt0,k∈Ir(j)0,k∉Ir(j).
Thus, by (47),x∈w∞θ(ℳ,Λ,p,s,∥·,…,·∥). But x∉w∞0(Λ,p,s,∥·,…,·∥). Hence, (45) must hold.
Conversely, suppose that (45) holds and let x∈w∞θ(ℳ,Λ,p,s,∥·,…,·∥). Then for each r,
(49)1hr∑k∈Ir‍k-s[Mk(∥Λk(x)ρ,z1,z2,…,zn-1∥)]pk≤K<∞.
Suppose that x∉w0θ(Λ,p,s,∥·,…,·∥). Then for some number ϵ>0, there is a number k0 such that for a subinterval Ir(j), of the set of interval Ir,
(50)∥Λk(x)ρ,z1,z2,…,zn-1∥>ϵfork≥k0.
From properties of sequence of Orlicz functions, we obtain
(51)[Mk(∥Λk(x)ρ,z1,z2,…,zn-1∥)]pk≥Mk(ϵ)pk,
which contradicts (45), by using (49). Hence, we get
(52)w∞θ(ℳ,Λ,p,s,∥·,…,·∥)⊂w0θ(Λ,p,s,∥·,…,·∥).
This completes the proof.Theorem 7.
Letℳ=(Mk) be a Musielak-Orlicz function. Then the following statements are equivalent:(i)
w∞θ(Λ,p,s,∥·,…,·∥)⊂w∞θ(ℳ,Λ,p,s,∥·,…,·∥);(ii)
w0θ(Λ,p,s,∥·,…,·∥)⊂w∞θ(ℳ,Λ,p,s,∥·,…,·∥);(iii)
supr1/hr∑k∈Ir‍k-s(Mk(t))pk<∞forallt>0.Proof.
(i)⇒ (ii). Let (i) hold. To verify (ii), it is enough to prove
(53)w0θ(Λ,p,s,∥·,…,·∥)⊂w∞θ(ℳ,Λ,p,s,∥·,…,·∥).
Let x=(xk)∈w0θ(Λ,p,s,∥·,…,·∥). Then for ϵ>0, there exists r≥0, such that
(54)1hr∑k∈Ir‍k-s[∥Λk(x)ρ,z1,z2,…,zn-1∥]pk<ϵ.
Hence, there exists K>0 such that
(55)supr1hr∑k∈Ir‍k-s[∥Λk(x)ρ,z1,z2,…,zn-1∥]pk<K.
So, we get x=(xk)∈w∞θ(ℳ,Λ,p,s,∥·,…,·∥).
(ii)⇒ (iii). Let (ii) hold. Suppose (iii) does not hold. Then for some t>0(56)supr1hr∑k∈Ir‍k-s(Mk(t))pk=∞,
and therefore we can find a subinterval Ir(j), of the set of interval Ir, such that
(57)1hr(j)∑k∈Ir(j)‍k-s(Mk(1j))pk>j,j=1,2,3,…
Let us define x=(xk) as follows:
(58)Λk(x)={ρj,k∈Ir(j)0,k∉Ir(j).
Thenx=(xk)∈w0θ(Λ,p,s,∥·,…,·∥). But by (57), x∉w∞θ(ℳ,Λ,p,s,∥·,…,·∥), which contradicts (ii). Hence, (iii) must holds.
(iii)⇒ (i). Let (iii) hold and suppose that x=(xk)∈w∞θ(Λ,p,s,∥·,…,·∥). Suppose that x=(xk)∉w∞θ(ℳ,Λ,p,s,∥·,…,·∥); then
(59)supr1hr∑k∈Ir‍k-s[Mk(∥Λk(x)ρ,z1,z2,…,zn-1∥)]pk=∞.
Let t=∥Λk(x)/ρ,z1,z2,…,zn-1∥ for each k; then by (59),
(60)supr1hr∑k∈Ir‍k-s(Mk(t))pk=∞,
which contradicts (iii). Hence, (i) must hold.Theorem 8.
Letℳ=(Mk) be a Musielak-Orlicz function. Then the following statements are equivalent:(i)
w0θ(ℳ,Λ,p,s,∥·,…,·∥)⊂w0θ(Λ,p,s,∥·,…,·∥);(ii)
w0θ(ℳ,Λ,p,s,∥·,…,·∥)⊂w∞θ(Λ,p,s,∥·,…,·∥);(iii)
infr1/hr∑k∈Ir‍k-s(Mk(t))pk>0forallt>0.Proof.
(i)⇒ (ii). It is obvious.
(ii)⇒ (iii). Let (ii) hold. Suppose that (iii) does not hold. Then
(61)infr1hr∑k∈Ir‍k-s(Mk(t))pk=0forsomet>0,
and we can find a subinterval Ir(j), of the set of interval Ir, such that
(62)1hr(j)∑k∈Ir(j)‍k-s(Mk(j))pk<1j,j=1,2,3,…
Let us define x=(xk) as follows:
(63)Λk(x)={ρj,k∈Ir(j)0,k∉Ir(j).
Thus, by (62), x=(xk)∈w0θ(ℳ,Λ,p,s,∥·,…,·∥), but x=(xk)∉w∞θ(Λ,p,s,∥·,…,·∥), which contradicts (ii). Hence, (iii) must hold.
(iii)⇒ (i). Let (iii) hold. Suppose that x=(xk)∈w0θ(ℳ,Λ,p,s,∥·,…,·∥). Then
(64)1hr∑k∈Ir‍k-s[Mk(∥Λk(x)ρ,z1,z2,…,zn-1∥)]pk⟶0(∥Λk(x)ρ,z1,z2,⋯,∥)(∥Λk(x)ρ,∥)asr⟶∞.
Again suppose that x=(xk)∉w0θ(Λ,p,s,∥·,…,·∥); for some number ϵ>0 and a subinterval Ir(j), of the set of interval Ir, we have
(65)∥Λk(x)ρ,z1,z2,…,zn-1∥≥ϵ∀k.
Then from properties of the Orlicz function, we can write
(66)[Mk(∥Λk(x)ρ,z1,z2,…,zn-1∥)]pk≥(Mk(ϵ))pk.
Consequently, by (64), we have
(67)limr→∞1hr∑k∈Ir‍k-s(Mk(ϵ))pk=0,
which contradicts (iii). Hence, (i) must hold.Theorem 9.
Let0≤pk≤qk for all k and let (qk/pk) be bounded. Then
(68)wθ(ℳ,Λ,q,s,∥·,…,·∥)⊆wθ(ℳ,Λ,p,s,∥·,…,·∥).Proof.
Letx=(xk)∈wθ(ℳ,Λ,q,s,∥·,…,·∥); write
(69)tk=[Mk(∥Λk(x)ρ,z1,z2,…,zn-1∥)]qk
and μk=pk/qk for all k∈ℕ. Then 0<μk≤1 for all k∈ℕ. Take 0<μ≤μk for k∈ℕ. Define sequences (uk) and (vk) as follows.
Fortk≥1, let uk=tk and vk=0, and for tk<1, let uk=0 and vk=tk. Then clearly for all k∈ℕ, we have
(70)tk=uk+vk,tkμk=ukμk+vkμk.
Now it follows that ukμk≤uk≤tk and vkμk≤vkμ. Therefore,
(71)1hr∑k∈Ir‍tkμk=1hr∑k∈Ir‍(ukμk+vkμk)≤1hr∑k∈Ir‍tk+1hr∑k∈Ir‍vkμ.
Now for each k,
(72)1hr∑k∈Ir‍vkμ=∑k∈Ir‍(1hrvk)μ(1hr)1-μ≤(∑k∈Ir‍[(1hrvk)μ]1/μ)μ×(∑k∈Ir‍[(1hr)1-μ]1/(1-μ))1-μ=(1hr∑k∈Ir‍vk)μ,
and so
(73)1hr∑k∈Ir‍vkμ≤1hr∑k∈Ir‍tk+(1hr∑k∈Ir‍vk)μ.
Hence, x=(xk)∈wθ(ℳ,Λ,p,s,∥·,…,·∥). This completes the proof of the theorem.Theorem 10.
(
i) If 0<infpk≤pk≤1 for all k∈ℕ, then
(74)wθ(ℳ,Λ,p,s,∥·,…,·∥)⊆wθ(ℳ,Λ,s,∥·,…,·∥).
(
ii) If 1≤pk≤suppk=H<∞, for all k∈ℕ, then
(75)wθ(ℳ,Λ,s,∥·,…,·∥)⊆wθ(ℳ,Λ,p,s,∥·,…,·∥).Proof.
(i) Letx=(xk)∈wθ(ℳ,Λ,p,s,∥·,…,·∥); then
(76)limr→∞1hr∑k∈Ir‍k-s[Mk(∥Λk(x)-Lρ,z1,z2,…,zn-1∥)]pk=0.
Since 0<infpk≤pk≤1, this implies that
(77)limr→∞1hr∑k∈Ir‍k-s[Mk(∥Λk(x)-Lρ,z1,z2,…,zn-1∥)]≤limr→∞1hr∑k∈Ir‍k-s[Mk(∥Λk(x)-Lρ,z1,z2,…,zn-1∥)]pk;
therefore,
(78)limr→∞1hr∑k∈Ir‍k-s[Mk(∥Λk(x)-Lρ,z1,z2,…,zn-1∥)]=0.
Hence,
(79)wθ(ℳ,Λ,p,s,∥·,…,·∥)⊆wθ(ℳ,Λ,s,∥·,…,·∥).
(ii) Let pk≥1 for each k and suppk<∞. Let x=(xk)∈wθ(ℳ,Λ,s,∥·,…,·∥); then for each ρ>0, we have
(80)limr→∞1hr∑k∈Ir‍k-s[Mk(∥Λk(x)-Lρ,limr→∞1hr∑k∈Ir‍k-slimr→∞1hr=z1,z2,…,zn-1Λk(x)-Lρ∥)]pk=0<1.
Since 1≤pk≤suppk<∞, we have
(81)limr→∞1hr∑k∈Ir‍k-s[Mk(∥Λk(x)-Lρ,z1,z2,…,zn-1∥)]pk≤limr→∞1hr∑k∈Ir‍k-s[Mk(∥Λk(x)-Lρ,z1,z2,…,zn-1∥)]=0<1.
Therefore, x=(xk)∈wθ(ℳ,Λ,p,s,∥·,…,·∥), for each ρ>0. Hence,
(82)wθ(ℳ,Λ,s,∥·,…,·∥)⊆wθ(ℳ,Λ,p,s,∥·,…,·∥).
This completes the proof of the theorem.Theorem 11.
If0<infpk≤pk≤suppk=H<∞, for all k∈ℕ, then
(83)wθ(ℳ,Λ,p,s,∥·,…,·∥)=wθ(ℳ,Λ,s,∥·,…,·∥).Proof.
It is easy to prove so we omit the details.



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*Source: 364743-2013-11-06.xml*