# An Extension of Modular Sequence Spaces

Authors: Hemen Dutta; Iqbal H. Jebril
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The main aim of this paper is to present an extension of the modular sequence spaces by means of Cesàro mean of order one, to investigate several relevant algebraic and topological properties, and derive some other spaces in the sequel.


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## Body

## 1. Introduction

Throughout the paperw(X) will represent the spaces of all X valued sequences spaces, where (X,q) is a seminormed space, seminormed by q. For X=C, the space of complex numbers, these represent the corresponding scalar valued sequence spaces. The zero sequence is denoted by θ-=(θ,θ,θ,…), where θ is the zero element of X.An Orlicz function is a functionM:[0,∞)→[0,∞), which is continuous, nondecreasing, and convex with M(0)=0,  M(x)>0, for x>0 and M(x)→∞, as x→∞.If convexity of Orlicz functionM is replaced by M(x+y)≤M(x)+M(y), then this function is called a modulus function introduced by Nakano [1].Lindenstrauss and Tzafriri [2] used the idea of Orlicz function to construct sequence space
(1)ℓM={x∈w:∑k=1∞M(|xk|ρ)<∞,forsomeρ>0∑k=1∞M(|xk|ρ)}.The spaceℓM becomes a Banach space, with the norm
(2)∥x∥=inf{ρ>0:∑k=1∞M(|xk|ρ)≤1}
which is called an Orlicz space. The space ℓM is closely related to the space ℓp which is an Orlicz sequence space with M(x)=xp for 1≤p<∞.Another generalization of Orlicz sequence spaces is due to Woo [3]. Let {Mk} be a sequence of Orlicz functions. Define the vector space ℓ{Mk} by
(3)ℓ{Mk}={x∈w:∑k=1∞Mk(|xk|ρ)<∞,forsomeρ>0}
and equip this space with the norm
(4)∥x∥=inf{ρ>0:∑k=1∞Mk(|xk|ρ)≤1}.Thenℓ{Mk} becomes a Banach space and is called a modular sequence space. The space ℓ{Mk} also generalizes the concept of modular sequence space introduced earlier by Nakano [4], who considered the space ℓ{Mk} when Mk(x)=xαk, where 1≤αk<∞ for k≥1.An Orlicz functionM is said to satisfy the Δ2-condition for all values of u, if there exists a constant K>0, such that M(2u)≤kM(u)(u≥0). The Δ2-condition is equivalent to the satisfaction of inequality M(lu)≤kluM(u) for all values of u and for l>1 (see [5]).The previous Δ2-condition implies M(lu)≤Kllog2KM(u), for all u>0,l>1.Bektaş and Altin [6], Parasar and Choudhary [7], Mursaleen et al. [8], Dutta and Başar [9], Dutta and Bilgin [10], Karakaya and Dutta [11], Tripathy and Dutta [12], Jebril [13], and many others have studied different summable spaces and other sequence spaces using Orlicz functions.ABK-space (introduced by Zeller [14]) (X,∥·∥) is a Banach space of complex sequences x=(xk) in which the coordinate maps are continuous; that is, |xkn-xk|→0, whenever ∥xn-x∥→0 as n→∞, where xn=(xkn), for all n∈N and x=(xk).LetA denote the set of all complex sequences which have only a finite number of nonzero coordinates, and let λ denote a BK-space of sequences x=(xk) which contains A. An element x=(xk) of λ will be called sectionally convergent if
(5)x(n)=∑k=1nxkek⟶x,asn⟶∞,
where ek=(δki), where δkk=1, δki=0 for k≠i.λ will be called AK-space if and only if each of its elements is sectionally convergent.LetM=(Mk) be a sequence of Orlicz functions, let X be a seminormed space with seminorm q, let p=(pk) be sequence of positive real numbers, and let C=(cnk)n,k=0∞ be the Cesàro matrix of order one with cnk=1/(n+1) if 0≤k≤n and cnk=0, otherwise. Then for nonnegative real numbers s we define
(6)ℓ{Mk,p,q,s,C}={x∈w(X):∑k=0∞k-s[Mk(q(∑j=0kxjρ(k+1)))]pk<∞,forsomeρ>0(q(∑j=0kxjρ(k+1)))}.The following inequality will be used throughout the paper. Letp=(pk) be a positive sequence of real numbers with 0<pk≤suppk=G, D=max{1,2G-1}. Then for all ak,bk∈C for all k∈N, we have
(7)|ak+bk|pk≤D{|ak|pk+|bk|pk}.


## 2. Main Results

In this section we give the theorems that characterize the structure of the class of sequencesℓ{Mk,p,q,s,C} and some other spaces which can be derived from this space.Theorem 1.
Letp=(pk) be bounded sequence of positive reals; then ℓ{Mk,p,q,s,C} is a linear space over the field of complex numbers.Proof.
Letx,y∈ℓ{Mk,p,q,s,C} and α,β∈C. Then there exist some ρ1>0 and ρ2>0 such that
(8)∑k=0∞k-s[Mk(q(∑j=0kxjρ1(k+1)))]pk<∞,∑k=0∞k-s[Mk(q(∑j=0kyjρ2(k+1)))]pk<∞.
We consider ρ3=max(2|α|ρ1,2|β|ρ2). Since each Mk is non-decreasing and convex, and since q is a seminorm,
(9)∑k=0∞k-s[Mk(q(∑j=0k(αxk+βyk)ρ3(k+1)))]pk≤∑k=0∞k-s[Mk(q(∑j=0k(αxj)ρ3(k+1))+q(∑j=0k(βyj)ρ3(k+1)))]pk≤∑k=0∞k-s[Mk(q(∑j=0kxj2ρ1(k+1))+q(∑j=0kyj2ρ2(k+1)))]pk≤D{∑k=0∞k-s[Mk(q(∑j=0kxjρ1(k+1)))]pk+∑k=0∞k-s[Mk(q(∑j=0kyjρ2(k+1)))]pk}<∞.
This completes the proof.Theorem 2.
ℓ
{
M
k
,
p
,
q
,
s
,
C
} is a paranormed space (need not total paranorm) space with paranorm g, defined as follows
(10)g(x)=inf{ρpn/H:∑k=0∞k-s[Mk(q(∑j=0kxjρ(k+1)))]≤1,n=1,2,…[Mk(q(∑j=0kxjρ(k+1)))]},
where H=max{1,supkpk}.Proof.
Clearlyg(x)=g(-x). Since Mk(0)=0, for all k∈N we get inf{ρpn/H}=0 for x=θ.
Now letx,y∈ℓ{Mk,p,q,s,C}, and let us choose ρ1>0 and ρ2>0 such that
(11)∑k=0∞k-s[Mk(q(∑j=0kxjρ1(k+1)))]≤1,∑k=0∞k-s[Mk(q(∑j=0kyjρ2(k+1)))]≤1.
Let ρ=ρ1+ρ2. Then we have
(12)∑k=0∞k-s[Mk(q(∑j=0k(xj+yj)ρ(k+1)))]≤(ρ1ρ1+ρ2)∑k=0∞k-s[Mk(q(∑j=0kxjρ1(k+1)))]+(ρ2ρ1+ρ2)∑k=0∞k-s[Mk(q(∑j=0kyjρ2(k+1)))]≤1.
Hence g(x+y)≤g(x)+g(y).
Finally letλ be a given non-zero scalar; then the continuity of the scalar multiplication follows from the following equality
(13)g(λx)=inf{ρpn/H:∑k=0∞k-s[Mk(q(∑j=0k(λxj)ρ(k+1)))]≤1,n=1,2,…(q(∑j=0k(λxj)ρ(k+1)))}=inf{(|λ|s)pn/H:∑k=0∞k-s[Mk(q(∑j=0kxjs(k+1)))]≤1,n=1,2,…(q(∑j=0k(λxj)ρ(k+1)))},wheres=ρ|s|.
This completes the proof.The proof of the following theorem is easy, so omitted.Theorem 3.
LetM=(Mk) and T=(Tk) be sequences of Orlicz functions. For any two sequences p=(pk) and t=(tk) of bounded positive real numbers and for any two seminorms q1 and q2 one has(i)
ifq1 is stronger than q2, then ℓ{Mk,p,q1,s,C}⊂ℓ{Mk,p,q2,s,C},(ii)
ℓ{Mk,p,q1,s,C}∩ℓ{Mk,p,q2,s,C}⊂ℓ{Mk,p,q1+q2,s,C},(iii)
ℓ{Mk,p,q,s,C}∩ℓ{Tk,p,q,s,C}⊂ℓ{Mk+Tk,p,q,s,C},(iv)
ℓ{Mk,p,q1,s,C}∩ℓ{Mk,t,q2,s,C}≠φ,(v)
Ifs1≤s2, then ℓ{Mk,p,q,s1,C}⊂ℓ{Mk,p,q,s2,C}.Theorem 4.
LetM=(Mk) and T=(Tk) be sequences of Orlicz functions which satisfy Δ2-condition and s>1, then
(14)ℓ{Mk,p,q,s,C}⊆ℓ{Tk∘Mk,p,q,s,C}.Proof.
Letx∈ℓ{Mk,p,q,s,C} and ε>0. We choose 0<δ<1 such that each Mk(u)<ε for 0≤u≤δ. We write yk=Mk(q((∑j=0kxj)/ρ(k+1))) and consider
(15)∑k=0∞k-s[Tk(yk)]pk=∑1k-s[Tk(yk)]pk+∑2k-s[Tk(yk)]pk,
where the first summation is over yk≤δ and the second is over yk>δ. Now we have
(16)∑1k-s[Tk(yk)]pk<max(1,εH)∑k=1∞k-s<∞,sinces>1.
For yk>δ, we use the fact that
(17)yk<ykδ≤1+(ykδ).
Since each Tk is non-decreasing and convex, it follows that
(18)Tk(yk)<Tk(1+ykδ)<12Tk(2)+12Tk(2ykδ),foreachkinN.
Since each Tk satisfies Δ2-condition, we have
(19)Tk(yk)<12KykδTk(2)+12KykδTk(2)=Kykδ-1Tk(2).
Hence
(20)∑2k-s[Tk(yk)]pk≤max(1,(Kδ-1M(2))H)×∑k=1∞k-s(yk)pk<∞.
Thus(21)∑k=0∞k-s[Tk(yk)]pk=∑1k-s[Tk(yk)]pk+∑2k-s[Tk(yk)]pk≤max(1,εH)×∑k=1∞k-s+max(1,(Kδ-1M(2))H)×∑k=1∞k-s(yk)pk<∞.
Hence x∈ℓ{Tk∘Mk,p,q,s,C}.
This completes the proof.TakingMk(x)=x,forall  x∈[0,∞) and k in N, in Theorem 4, we get the next corollary.Corollary 5.
LetM=(Mk) be any sequence of Orlicz functions which satisfy Δ2-condition and s>1, then
(22)ℓ{p,q,s,C}⊆ℓ{Mk,p,q,s,C}.We will writef≈g for non-negative functions fand g whenever C1f≤g≤C2f for some Cj>0,j=1,2.Theorem 6.
LetM=(Mk) and T=(Tk) be sequences of Orlicz functions. If Mk≈Tk for each k∈N, then ℓ{Mk,p,q,s,C}=ℓ{Tk,p,q,s,C}.Proof.
The proof is obvious.Theorem 7.
LetM=(Mk) be a sequence of Orlicz functions. If limt→0(Mk(t)/t)>0 and limt→0(Mk(t)/t)<∞, for each k∈N, then ℓ{Mk,p,q,s,C}=ℓ{p,q,s,C}.Proof.
If the given conditions are satisfied, we haveMk(t)≈t for each k, and the proof follows from Theorem 6.If we takes=0, the sequence space ℓ{Mk,p,q,s,C} reduces to the following sequence space:
(23)ℓ{Mk,p,q,C}={x∈w(X):∑k=0∞[Mk(q(∑j=0kxjρ(k+1)))]pk<∞,forsomeρ>0[Mk(q(∑j=0kxjρ(k+1)))]}.Theorem 8.
Letp=(pk) be bounded sequence of positive reals, and let (X,q) be a complete seminormed space, then ℓ{Mk,p,q,C} is a complete paranormed space paranormed by h, defined by
(24)h(x)=inf{ρpn/H:∑k=0∞[Mk(q(∑j=0kxjρ(k+1)))]≤1,n=1,2,…[Mk(q(∑j=0kxjρ(k+1)))]},
where H=max{1,supkpk}.Proof.
Let(xi) be a Cauchy sequence in ℓ{Mk,p,q,C}. Let δ>0 be fixed, and let r>0 be such that for a given 0<ε<1,ε/rδ>0, and rδ≥1. Then there exists a positive integer n0 such that
(25)h(xs-xt)<εrδ,∀s,t≥n0.⟹inf{ρpn/H:∑k=0∞[Mk(q(∑j=0k(xjs-xjt)ρ(k+1)))]≤1,n=1,2,…Mk(q(∑j=0k(xjs-xjt)ρ(k+1)))}<εrδ,∀s,t≥n0.
Hence we have
(26)∑k=0∞[Mk(q(∑j=0k(xjs-xjt)h(xs-xt)(k+1)))]≤1,∀s,t≥n0.
It follows that
(27)Mk(q(∑j=0k(xjs-xjt)h(xs-xt)(k+1)))≤1,∀s,t≥n0,k∈N.
For r>0 with Mk(rδ/2)≥1, we have
(28)Mk(q(∑j=0k(xjs-xjt)h(xs-xt)(k+1)))≤Mk(rδ2),∀s,t≥n0,k∈N.
SinceMk is non-decreasing for each k∈N, we have
(29)q(∑j=0k(xjs-xjt)k+1)≤rδ2·εrδ=ε2.
Hence it follows that (xks) is a Cauchy sequence in (X,q) for each k∈N. But (X,q) is complete, and so (xks) is convergent in (X,q) for each k∈N.
Letlims→∞xks=xk exists for each k∈N.
Now we have for alls,t≥n0,
(30)inf{ρpn/H:∑k=0∞[Mk(q(∑j=0k(xjs-xjt)ρ(k+1)))]≤1,n=1,2,…(q(∑j=0k(xjs-xjt)ρ(k+1)))}<ε.
Then we have
(31)limt→∞{inf{ρpn/H:∑k=0∞[Mk(q(∑j=0k(xjs-xjt)ρ(k+1)))]≤1,n=1,2,…[Mk(q(∑j=0k(xjs-xjt)ρ(k+1)))]}}<ε,∀s≥n0.
Using the continuity of Orlicz functions, we have
(32)inf{ρpn/H:∑k=0∞[Mk(q(∑j=0k(xjs-limt→∞xjt)ρ(k+1)))]≤1,n=1,2,…[Mk(q(∑j=0k(xjs-limt→∞xjt)ρ(k+1)))]}<ε,∀s≥n0.
This implies
(33)inf{ρpn/H:∑k=0∞[Mk(q(∑j=0k(xjs-xj)ρ))]≤1,n=1,2,…[Mk(q(∑j=0k(xjs-xj)ρ))]}<ε,∀s≥n0.
It follows that (xs-x)∈ℓ{Mk,p,q,C}.
Since(xs)∈ℓ{Mk,p,q,C} and ℓ{Mk,p,q,C} is a linear space, so we have x=xs-(xs-x)∈ℓ{Mk,p,q,C}.
This completes the proof.If we takes=0 and pk=l, a constant the sequence space ℓ{Mk,p,q,s,C} reduces to the following sequence space:
(34)ℓ{Mk,q,C}={x∈w(X):∑k=0∞[Mk(q(∑j=0kxjρ(k+1)))]<∞,forsomeρ>0[Mk(q(∑j=0kxjρ(k+1)))]}.Theorem 9.
Let(X,q) be a complete normed space; then ℓ{Mk,q,C} is a Banach space normed by ∥·∥, defined by
(35)∥x∥=inf{ρ:∑k=0∞[Mk(q(∑j=0kxjρ(k+1)))]≤1}.Proof.
We prove that∥·∥ is a norm on ℓ{Mk,q,C}. The completeness part can be proved using similar arguments as applied to prove the previous theorem.
Ifx=θ, then it is obvious that ∥x∥=0. Conversely assume ∥x∥=0. Then using the definition of norm, we have
(36)inf{ρ:∑k=0∞[Mk(q(∑j=0kxjρ(k+1)))]≤1}=0.
This implies that for a given ε>0, there exists some ρε(0<ρε<ε) such that
(37)∑k=0∞[Mk(q(∑j=0kxjρε(k+1)))]≤1.
It follows that
(38)Mk(q(∑j=0kxjρε(k+1)))≤1,∀k∈N.
Thus
(39)Mk(q(∑j=0kxjε(k+1)))≤Mk(q(∑j=0kxjρε(k+1)))≤1,∀k∈N.
Suppose (∑j=0nixj)/(ni+1)≠0, for some i. Let ε→0 then (∑j=0nixj)/(ε(ni+1))→∞.
It follows thatMk(q((∑j=0kxj)/(ε(k+1))))→∞ as ε→0 for some ni∈N. This is a contradiction.
Therefore(∑j=0kxj)/(k+1)=0,forallk∈N.
It follows thatxk=0 for all k≥1. Hence x=θ.
Again proof of the properties∥x+y∥≤∥x∥+∥y∥ and for any scalar α,∥αx∥=|α|∥x∥ are similar to that of Theorem 2.It is easy to see that∥xi∥→0 implies that xki→0 for each i≥1. Hence we have the following proposition.Proposition 10.
The spaceℓ{Mk,q,C} is a BK-space.Now we study theAK-characteristic of the space ℓ{Mk,q,s,C}. Before that we give a new definition and prove some results which  will be required.Definition 11.
For any sequence of Orlicz functionsM=(Mk), we define
(40)h{Mk,q,C}={x∈w(X):∑k=0∞[Mk(q(∑j=0kxjρ(k+1)))]<∞,foreveryρ>0∑k=0∞[Mk(q(∑j=0kxjρ(k+1)))]}.
Clearly h{Mk,q,C} is a subspace of ℓ{Mk,q,C}. The topology of h{Mk,q,C} is the one it inherits from ∥·∥.Proposition 12.
LetM=(Mk) be a sequence of Orlicz functions which satisfy Δ2-condition. Then
(41)ℓ{Mk,q,C}=h{Mk,q,C}.Proof.
It is enough to prove thatℓ{Mk,q,C}⊆h{Mk,q,C}.
Letx∈ℓ{Mk,q,C}, then for some ρ>0,
(42)∑k=0∞[Mk(q(∑j=0kxjρ(k+1)))]<∞.
Choose an arbitrary η>0. If ρ≤η then
(43)∑k=0∞[Mk(q(∑j=0kxjη(k+1)))]<∑k=0∞[Mk(q(∑j=0kxjρ(k+1)))]<∞.
Let now η<ρ and put l=ρ/η>1.
Since eachMk satisfies the Δ2-condition, there exist constants Kk such that
(44)∑k=0∞[Mk(q(∑j=0kxjη(k+1)))]≤∑k=0∞Kk(ρη)log2Kk[Mk(q(∑j=0kxjρ(k+1)))].
Let S=supkKk(ρ/η)log2Kk. Then for every η>0(45)∑k=0∞[Mk(q(∑j=0kxjη(k+1)))]≤S∑k=0∞[Mk(q(∑j=0kxjρ(k+1)))]<∞.
This completes the proof.Proposition 13.
Let(X,q) be a complete normed space, then h{Mk,q,C} is an AK-space.Proof.
Letx∈h{Mk,q,C}. Then for each ε,0<ε<1, we can find an s0 such that
(46)∑k≥s0[Mk(q(∑j=0kxjε(k+1)))]≤1.
Hence for s≥s0,
(47)∥x-x(s)∥=inf{ρ>0:∑k≥s+1[Mk(q(∑j=0kxjρ(k+1)))]≤1}≤inf{ρ>0:∑k≥s[Mk(q(∑j=0kxjρ(k+1)))]≤1}<ε.
Thus we can conclude that h{Mk,q,C} is an AK space.Combining Proposition10 and Proposition 12, we have the following theorem.Theorem 14.
LetM=(Mk) be a sequence of Orlicz functions which satisfy Δ2-condition, then ℓ{Mk,q,C} is an AK-space.Proposition 15.
The spaceh{Mk,q,C} is a closed subspace of ℓ{Mk,q,C}.Proof.
Let{xs} be a sequence in h{Mk,q,C} such that ∥xs-x∥→0, where x∈ℓ{Mk,q,C}.
To complete the proof we need to show thatx∈h{Mk,q,C}; that is,
(48)∑k≥0[Mk(q(∑j=0kxjρ(k+1)))]<∞foreveryρ>0.
To ρ>0 there corresponds an l such that ∥xl-x∥≤ρ/2. Then using convexity of each Mk,
(49)∑k≥0[Mk(q(∑j=0kxjρ(k+1)))]=∑k≥0[Mk(q(2∑j=0kxjl-2(∑j=0kxjl-∑j=0kxj)2ρ(k+1)))]≤12∑k≥0[Mk(q(2∑j=0kxjlρ(k+1)))]+12∑k≥0[Mk(q(2∑j=0k(xjl-xj)ρ(k+1)))]≤12∑k≥0[Mk(q(2∑j=0kxjlρ(k+1)))]+12∑k≥0[Mk(q(2∑j=0k(xjl-xj)∥xl-x∥(k+1)))].
Now from Theorem 9, using the definition of norm ∥·∥, we have
(50)∑k≥0[Mk(q(2∑j=0k(xjl-xj)∥xl-x∥(k+1)))]≤1.
It follows that
(51)∑k≥0[Mk(q(∑j=0kxjρ(k+1)))]<∞foreveryρ>0.
Thus x∈h{Mk,q,C}.Hence we have the following corollary.Corollary 16.
The spaceh{Mk,q,C} is a BK-space.



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