# Someℓ(p)-Type New Sequence Spaces and Their Geometric Properties

Authors: Ekrem Savaş; Vatan Karakaya; Necip Şimşek
Journal: Abstract and Applied Analysis
 (2009)

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## Abstract

We introduce anℓ(p)-type new sequence space and investigate its some topological properties including AK and AD properties. Besides, we examine some geometric properties of this space concerning
Banach-Saks type p and Gurarii's modulus of convexity.


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## Body

## 1. Introduction

In general, theℓ(p)-type spaces have many useful applications because of the properties of the spaces ℓ(p) and ℓp. In [1], it was shown that the subspaces of Orlicz spaces, which have rich geometric properties, are isomorphic to the space ℓp. Also since the space ℓp is reflexive and convex, it is natural to consider the geometric structure of these spaces.Recently there has been a lot of interest in investigating geometric properties of sequence spaces besides topological and some other usual properties. In literature, there are many papers concerning the geometric properties of different sequence spaces. For example; in [2], Mursaleen et al. studied some geometric properties of normed Euler sequence space. Sanhan and Suantai [3] investigated the geometric properties of Cesáro sequence space ces(p) equipped with Luxemburg norm. Further information on geometric properties of sequence spaces can be found in [4–7].The main purpose of our work is to introduce anℓp-type new sequence space together with matrix domain and its summability methods. Also we investigate some topological properties of this new space as the paranorm, AK and AD properties, and furthermore characterize geometric properties concerning Banach-Saks type p and Gurarii's modulus of convexity.


## 2. Preliminaries and Notations

Letw be the space of all real-valued sequences. Each linear subspace of w is called a sequence space denoted by λ. We denote by ℓ1 and ℓp absolutely and p-absolutely convergent series, respectively.A sequence spaceλ with a linear topology is called a K-space provided that each of the maps pi:λ→ℂ defined by pi(x)=xi is continuous for all i∈ℕ, where ℂ denotes the complex field, and ℕ={0,1,2,3,…}. A K-space λ is called FK-space provided that λ is a complete linear metric space. An FK-space whom topology is normable is called BK-space. An FK-space λ is said to have AK property, if ϕ⊂λ and {e(k)} is a basis for λ, where e(k) is a sequence whose only nonzero term is 1, kth place for each k∈ℕ, and ϕ=span{e(k)}, the set of all AD-space, thus AK implies AD.A linear topological spaceX over the real field ℝ is said to be a paranormed space if there is a subadditivity function g:X→ℝ such that g(θ)=0, g(-x)=g(x) and scalar multiplication is continuous. It is well known that the space ℓp is AK-space where 1≤p<∞.Throughout this work, we suppose that(pk) is a bounded sequence of strictly positive real numbers with suppk=H and M=max{1,H}. Also the summation without limits runs from 0 to ∞. In [8], the linear space ℓ(p) was defined by Maddox (see also Simons [9] and Nakano [10]) as follows:
(2.1)ℓ(p)={x=(xk)∈w:∑n|xn|pn<∞}
which is a complete space paranormed by
(2.2)g(x)=(∑n|xn|pn)1/M.Letλ, μ be any two sequence spaces, and let  A=(ank) be an infinite matrix of real numbers ank, where n,k∈ℕ. Then we write Ax=((Ax)n), the A-transform of x, if (Ax)n=∑kankxk converges for each n∈ℕ.A matrixA=(ank) is called a triangle if ank=0 for k>n and ann≠0 for all n∈ℕ. It is trivial that A(Bx)=(AB)x holds for the triangle matrices A, B and a sequence x. Further, a triangle matrix P uniquely has an invert P-1=Q which is also a triangle matrix. Then if Px=y,(2.3)x=P(Qx)=Q(Px),x=Qy
hold for all x∈w.By(λ,μ), we denote the class of all infinite matrices A such that A:λ→μ. The matrix domain λA of an infinite matrix A in a sequence space λ is defined by λA={x=(xk)∈w:Ax∈λ} which is a sequence space. It is well known that the new sequence space λA generated by the limitation matrix A from a sequence space λ is the expansion or the contraction of original space λ.IfA is triangle, then one can easily observe that the sequence spaces λA and λ are linearly isomorphic, that is, λA=̃λ. Let λ be a sequence space. Then the continuous dual λA′ of the space λA is defined by λA′={f:f=g∘A;g∈λ′}. Let X be a seminormed space. A set Y⊂X is called fundamental set if the span of Y is dense in X. An application of Hahn-Banach theorem on fundamental set is as follows: if Y is the subset of a seminormed space X and f(Y)=0 implies f=0 for f∈X′, then Y is a fundamental set (see [11]).By the idea mentioned above, let us give the definitions of some matrices to construct a new sequence space in sequel to this work. We denoteΔ=(δnk) and S=(snk) by
(2.4)δnk={(-1)n-k,ifn-1≤k≤n,0,otherwise,snk={1,if0≤k≤n,0,otherwise.
Malkowsky and Savas [12], Choudhary and Mishra [13], and Altay and Basar [14] have defined the sequence spaces Z(u,v;X), ℓ(p),¯ and ℓ(u,v;p), respectively. By using the matrix domain, the spaces Z(u,v;X),ℓ(p)¯, and ℓ(u,v;p) may be redefined by Z(u,v;X)=XG(u,v),ℓ(p)¯=(ℓp)S, and ℓ(u,v;p)=(ℓ(p))G(u,v), respectively.Ifλ⊂w is a sequence space and x=(xk)∈λ, (Sx)-transform with (2.4) corresponds to nth partial sum of the series ∑nxn and it is denoted by s=(sn).By using (2.4) and any infinite lower triangular matrix A, we can define two infinite lower triangular matrices A¯ and Â as follows: A¯=AS and Â=ΔA¯. Let x=(xk) be a sequence in λ. By considering the multiplication of infinite lower triangular matrices, we have A(Sx)=A¯x, that is,
(2.5)tn=∑v=0nanvsv=∑v=0na¯nvxv.
Also since Â=ΔA¯, we have Âx=(ΔA¯)x, that is,
(2.6)tn-tn-1=∑v=0nânvxv.
Now let us write the following equality:
(2.7)zn=(Âx)n=∑v=0nânvxv.
It can be seen that for any sequences x,  y and scalar α∈ℝ, (Â(x+y))n=(Âx)n+(Ây)n and (Â(αx))n=α(Âx)n. We now define new sequence space as follows:
(2.8)ℓ(Â;p)={x=(xk)∈w:∑n|(Âx)n|pn<∞}.
For some special cases of the infinite lower triangular matrix A and the sequence (pk), we obtain the following spaces.(i) Ifpk=p for all k∈ℕ, the space ℓ(Â;p) reduces to the normed space ℓp(Â) denoted by(2.9)ℓp(Â)={x=(xk)∈w:∑n|∑v=0nânvxv|p<∞}.(ii) IfA=(C,1), which is Cesáro matrix order 1, then the space ℓ(Â;p) corresponds to the space ℓ(Ĉ;p) denoted by
(2.10)ℓ(Ĉ;p)={x=(xk)∈w:∑n|(Ĉx)n|pn<∞},
where (Ĉx)n=(1/n(n+1))∑k=1nkxk for n≥1 and (Ĉx)0=x0.(iii) IfA=(N,pn), which is Nörlund type matrix, then the space ℓ(Â;p) reduces to the space ℓ(N̂;p)=|N¯,pn|(r) (see [15, 16]) denoted by
(2.11)ℓ(N̂;p)={x=(xk)∈w:∑n|(N̂x)n|pn<∞},
where (N̂x)n=(pn/PnPn-1)∑k=1nPk-1xk for n≥1 and (N̂x)0=x0.Also ifpk=p for all k∈ℕ, then the spaces ℓ(Ĉ;p) and ℓ(N̂;p)=|N¯,pn|(r) reduce to the spaces ℓp(Ĉ) and ℓp(N̂)=|N¯p| (see [17]), respectively.Now let us introduce some definitions of geometric properties of sequence spaces.Let(X,∥·∥) be a normed linear space, and let S(X) and B(X) be the unit sphere and unit ball of X (for the brevity X=(X,∥·∥)), respectively. Consider Clarkson's modulus of convexity (Clarkson [18] and Day [19]) defined by
(2.12)δX(ε)=inf{1-∥x+y∥2;x,y∈S(X),∥x-y∥=ε},
where 0≤ε≤2. The inequality δX(ε)>0 for all ε∈(0,2] characterizes the uniformly convex spaces.In [20], Gurariĭ's modulus of convexity is defined by
(2.13)βX(ε)=inf{1-infα∈[0,1]∥αx+(1-α)y∥;x,y∈S(X),∥x-y∥=ε},
where 0≤ε≤2. It is easily shown that δX(ε)≤βX(ε)≤2δX(ε) for any 0≤ε≤2. Also if 0<βX(ε)<1, then X is uniformly convex, and if βX(ε)<1, then X is strictly convex.A Banach spaceX is said to have the Banach-Saks property if every bounded sequence (xn) in X admits a subsequence (zn) such that the sequence {tk(z)} is convergent in the norm in X (see [21]), where
(2.14)tk(z)=1k+1(z0+z1+z2+⋯+zk)(k∈ℕ).Let1<p<∞. A Banach space is said to have the Banach-Saks typep or property (BSp), if every weakly null sequence (xk) has a subsequence (xkl) such that for some C>0(2.15)∥∑l=0nxkl∥<C(n+1)1/p
for all n∈ℕ (see [22]).


## 3. Some Topological Properties of the Spaceℓ(Â;p)

In this section, we investigate some topological properties of the sequence spaceℓ(Â;p) as the paranorm AK property and AD property. Let us begin the following theorem.Theorem 3.1.
(i) The spaceℓ(Â;p) is complete linear metric space with respect to the paranorm defined by
(3.1)h(x)=(∑n|(Ax)n|pn)1/M.
(ii) If the sequence(pn) is constant sequence and p≥1, then ℓp(Â) is a Banach space normed by
(3.2)∥z∥ℓp=∥x∥ℓp(Â)=(∑n|(Ax)n|p)1/p.Proof.
The proof of (ii) is routine verification by using standard techniques and hence it is omitted.
The proof of (i) is that the linearity ofℓ(Â;p) with respect to the coordinatewise addition and scalar multiplication follows from the following inequalities which are satisfied for x,y∈ℓ(Â;p):
(3.3)(∑n|(Â(x+y))n|pn)1/M≤(∑n|(Âx)n|pn)1/M+(∑n|(Ây)n|pn)1/M
and |α|pn≤max{1,|α|M} for any α∈ℝ (see [23]). After this step, we must show that the space ℓ(Â;p) holds the paranorm property and the completeness with respect to given paranorm. It is easy to show that h(θ)=0, and h(x)=h(-x) for all x∈ℓ(Â;p). Besides, from (3.3) we obtain h(x+y)≤h(x)+h(y) for all x,y∈ℓ(Â;p). To complete the paranorm conditions for the space ℓ(Â;p), it remains to show the continuity of the scalar multiplication. Let (xm) be any sequence in ℓ(Â;p) such that h(xm-x)→0, and let (αm) be also any sequence of scalars such that |αm-α|→0(m→∞). From subadditivity of h, we give the inequality h(xm)≤h(x)+h(xm-x). Hence {h(xm)} is bounded and we have
(3.4)h(αmxm-αx)=(∑n|(αm-α)∑v=0nânvxvm+α∑v=0nânv(xvm-xv)|pn)1/M
which tends to zero as m→∞. Consequently we obtain that h is a paranorm over the space ℓ(Â;p). To prove the completeness of the space ℓ(Â;p), let us take any Cauchy sequence (xi) in the space ℓ(Â;p). Then for a given ε>0, there exists a positive integer n0(ε) such that h(xi-xj)<ε for all i,j≥n0(ε). By using the definitions of the Cauchy sequence and the paranorm, we have, for each fixed n,
(3.5)|(Âxi)n-(Âxj)n|≤(∑n|(Â(xi)n-(Â(xj)n|pn)1/M<ε
for every i,j≥n0(ε). Hence we obtain that the sequence {Â(x0)n,Â(x1)n,Â(x2)n,…} is a Cauchy sequence of real numbers for every fixed n∈ℕ. Since ℝ is complete, it converges, that is, (Â(xj)n→(Âx)n as j→∞, where {(Âx)n}={(Âx)0,(Âx)1,(Âx)2,…}. Now let us choose m∈ℕ such that ∑n=0m|(Âxi)n-(Âxj)n|pn<εM for each m∈ℕ and i,j≥n0(ε). By taking j→∞ and for every i≥n0(ε), we get
(3.6)∑n=0m|(Âxi)n-(Âx)n|pn<εM.
Again taking m→∞ and for every i≥n0(ε), it is obtained that h(xi-x)<ε. We write the following equality:
(3.7)|(Âx)n|=|(Âx)n+(Âxi)n-(Âxi)n|.
By using (3.7) and Minkowski's inequality, we get
(3.8)(∑n|(Âx)n|pn)1/M≤h(xi)+h(xi-x)
which implies x∈ℓ(Â;p). It follows xi→x as i→∞. Consequently, since (xi) is any Cauchy sequence, we obtain that the space ℓ(Â;p) is complete. This completes the proof.Theorem 3.2.
The spaceℓ(Â;p) is linearly isomorphic to the space ℓ(p).Proof.
Let us defineÂ-transform between the spaces ℓ(Â;p) and ℓ(p) such that x→z=Âx. We have to show that the transformation Â is linear, injective and surjective. The linearity of Â is obvious. Moreover it is injective because of x=θ whenever Âx=θ. For the surjective property, let y∈ℓ(p). From (2.3) and (2.7), there exists a matrix B̂ such that xn=(B̂y)n. We have
(3.9)h(x)=(∑n|(Â(B̂y))n|pn)1/M=(∑n|yn|pn)1/M=g(y)<∞.
Hence we obtain that the transformation Â is surjective. Consequently, the spaces ℓ(Â;p) and ℓ(p) are linearly isomorphic spaces.Theorem 3.3.
The spaceℓp(Â) has AD property.Proof.
Letf∈(ℓp(Â))′. Then f(x)=g(Âx) for some g∈ℓp′. Since ℓp has AK property and ℓp′≅ℓq where 1/p+1/q=1,
(3.10)f(x)=∑nan(Âx)n
for some a=(an)∈ℓq. Also since ℓp(Â)≅ℓp and the inclusion ϕ⊂ℓp holds, we have ϕ⊂ℓp(Â). For any f∈(ℓp(Â))′ and e(k)∈ϕ, we have
(3.11)f(e(k))=∑nan(Âe(k))n=(Ĥa)k,
where Ĥ is transpose of the matrix Â. Hence from Hahn-Banach theorem, ϕ⊂ℓp(Â) is dense in ℓp(Â) if and only if Ĥa=θ for a∈ℓq implies a=θ. Besides, since the null space of the operator Ĥ on w is {θ}, ℓp(Â) has AD property. Hence the proof is completed.


## 4. Some Geometric Properties of the spaceℓp(Â)

In this section, we give some geometric properties for the spaceℓp(Â).Theorem 4.1.
The spaceℓp(Â) has the Banach-Saks of type p.Proof.
Let(εn) be a sequence of positive numbers for which ∑n=1∞εn≤1/2. Let (xn) be a weakly null sequence in B(ℓp(Â)). Set z0=x0=0 and z1=xn1=x1. Then there exists s1∈ℕ such that
(4.1)∥∑i=s1+1∞z1(i)e(i)∥ℓp(Â)<ε1.
Since (xn) is a weakly null sequence implies that xn→0 with respect to the coordinatwise, there is an n2∈ℕ such that
(4.2)∥∑i=0s1xn(i)e(i)∥ℓp(Â)<ε1,
where n≥n2. Set z2=xn2. Then there exists an s2>s1 such that
(4.3)∥∑i=s2+1∞z2(i)e(i)∥ℓp(Â)<ε2.
By using the fact that xn→0 (coordinatwise), there exists an n3>n2 such that
(4.4)∥∑i=0s2xn(i)e(i)∥ℓp(Â)<ε2,
where n≥n3.
If we continue this process, we can find two increasing subsequences(si) and (ni) such that
(4.5)∥∑i=0sjxn(i)e(i)∥ℓp(Â)<εj
for each n≥nj+1 and
(4.6)∥∑i=sj+1∞zj(i)e(i)∥ℓp(Â)<εj,
where zj=xnj. Hence,
(4.7)∥∑j=0nzj∥ℓp(Â)=∥∑j=0n(∑i=0sj-1zj(i)e(i)+∑i=sj-1+1sjzj(i)e(i)+∑i=sj+1∞zj(i)e(i))∥ℓp(Â)≤∥∑j=0n(∑i=sj-1+1sjzj(i)e(i))∥ℓp(Â)+∥∑j=0n(∑i=0sj-1zj(i)e(i))∥ℓp(Â)+∥∑j=0n(∑i=sj+1∞zj(i)e(i))∥ℓp(Â)≤∥∑j=0n(∑i=sj-1+1sjzj(i)e(i))∥ℓp(Â)+2∑j=0nεj.
On the other hand, since xn∈B(ℓp(Â)) and ∥x∥ℓp(Â)=(∑i=0∞|∑v=0iâivxv|p)1/p, it can be seen that ∥x∥ℓp(Â)<1. Therefore ∥x∥ℓp(Â)p<1. We have
(4.8)∥∑j=0n(∑i=sj-1+1sjzj(i)e(i))∥ℓp(Â)p=∑j=0n∑i=sj-1+1sj|∑v=0iâivzj(v)|p≤∑j=0n∑i=0∞|∑v=0iâivzj(v)|p≤(n+1).
Hence we obtain
(4.9)∥∑j=0n(∑i=sj-1+1sjzj(i)e(i))∥ℓp(Â)≤(n+1)1/p.
By using the fact 1≤(n+1)1/p for all n∈ℕ and 1≤p<∞, we have
(4.10)∥∑j=0nzj∥ℓp(Â)≤(n+1)1/p+1≤2(n+1)1/p.
Hence ℓp(Â) has the Banach-Saks type p. This completes the proof.Theorem 4.2.
Gurarii's modulus of convexity for the normed spaceℓp(Â) is
(4.11)βℓp(Â)(ε)≤1-(1-(ε2)p)1/p,
where 0≤ε≤2.Proof.
We havex∈ℓp(Â). Then we have
(4.12)∥x∥ℓp(Â)=∥Âx∥lp=(∑n|(Âx)n|p)1/p.
Let 0≤ε≤2 and consider the following sequences:
(4.13)x=(xn)=(B̂((1-(ε2)p)1/p),B̂(ε2),0,0,…),y=(yn)=(B̂((1-(ε2)p)1/p),B̂(-ε2),0,0,…),
where B̂ is the inverse of the matrix Â. Since zn=(Âx)n and tn=(Ây)n, we have
(4.14)z=(zn)=((1-(ε2)p)1/p,(ε2),0,0,…),t=(tn)=((1-(ε2)p)1/p,(-ε2),0,0,…).
By using sequences given above, we obtain the following equalities:
(4.15)∥x∥ℓp(Â)p=∥Âx∥lpp=|(1-(ε2)p)1/p|p+|ε2|p=1-(ε2)p+(ε2)p=1,∥y∥ℓp(Â)p=∥Ây∥lpp=|(1-(ε2)p)1/p|p+|-ε2|p=1-(ε2)p+(ε2)p=1,∥x-y∥ℓp(Â)=∥Âx-Ây∥lp=(|(1-(ε2)p)1/p-(1-(ε2)p)1/p|p+|ε2-(-ε2)|p)1/p=ε.
To complete the conditions of βℓp(Â)(ε) for Gurarii's modulus of convexity, it remains to show the infimum of ∥αx+(1-α)t∥ℓp(Â) for 0≤α≤1. We have
(4.16)inf0≤α≤1∥αx+(1-α)y∥ℓp(Â)=inf0≤α≤1∥αÂx+(1-α)Ây∥lp=inf0≤α≤1[|α(1-(ε2)p)1/p+(1-α)(1-(ε2)p)1/p|p+|α(ε2)+(1-α)(-ε2)|p]1/p=inf0≤α≤1[1-(ε2)p+|2α-1|p(ε2)p]1/p=(1-(ε2)p)1/p.
Consequently we get for p≥1(4.17)βℓp(Â)(ε)≤1-(1-(ε2)p)1/p.
This is the desired result. Hence the proof is completed.Corollary 4.3.
(i) Ifε=2, then βℓp(Â)(ε)≤1 and hence ℓp(Â) is strictly convex.
(ii) If0<ε<2, then 0<βℓp(Â)(ε)<1 and hence ℓp(Â) is uniformly convex.Corollary 4.4.
Ifα=1/2, then δℓp(Â)(ε)=βℓp(Â)(ε).



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