# Some Geometric Properties of the Domain of the Double Sequential Band MatrixB(r~,s~) in the Sequence Space ℓ(p)

Authors: Havva Nergiz; Feyzi Başar
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The sequence spaceℓ(p) was introduced by Maddox (1967). Quite recently, the sequence space ℓ(B~,p) of nonabsolute type has been introduced and studied which is the domain of the double sequential band matrix B(r~,s~) in the sequence space ℓ(p) by Nergiz and Başar (2012). The main purpose of this paper is to investigate the geometric properties of the space ℓ(B~,p), like rotundity and Kadec-Klee and the uniform Opial properties. The last section of the paper is devoted to the conclusion.


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## Body

## 1. Introduction

Byω, we denote the space of all real-valued sequences. Any vector subspace of ω is called a sequence space. We write ℓ∞, c, and c0 for the spaces of all bounded, convergent, and null sequences, respectively. Also by bs, cs, ℓ1, and ℓp; we denote the spaces of all bounded, convergent, absolutely convergent, and p-absolutely convergent series, respectively, where 1<p<∞.Assume here and after that(pk) is a bounded sequence of strictly positive real numbers with suppk=H and M=max{1,H}. Then, the linear space ℓ(p) was defined by Maddox [1] (see also Simons [2] and Nakano [3]) as follows:
(1)ℓ(p)={x=(xk)∈w:∑k|xk|pk<∞}(0<pk≤H<∞)
which is complete paranormed space paranormed by
(2)g(x)=(∑k|xk|pk)1/M.
For simplicity in notation, here and in what follows, the summation without limits runs from 1 to ∞.Quite recently, Nergiz and Başar [4] have introduced the space ℓ(B~,p) of nonabsolute type which consists of all sequences whose B(r~,s~)-transforms are in the space ℓ(p), where B(r~,s~)={bnk(rk,sk)} is defined by
(3)bnk(rk,sk)={rk,k=n,sk,k=n-1,0,otherwise
for all k,n∈ℕ, where r~=(rk) and s~=(sk) are the convergent sequences. We should record that the double sequential band matrices were used for determining its fine spectrum over some sequence spaces by Kumar and Srivastava in [5, 6], Panigrahi and Srivastava in [7], and Akhmedov and El-Shabrawy in [8]. The reader may refer to Nergiz and Başar [4, 9] for relevant terminology and additional references on the space ℓ(B~,p), since the present paper is a natural continuation of them. Here and after, for short we write B~ instead of B(r~,s~). In the special case pk=p for all k∈ℕ, the space ℓ(B~,p) is reduced to the space (ℓp)B~; that is,(4)(ℓp)B~:={(xk)∈ω:∑k|sk-1xk-1+rkxk|p<∞},(0<p<∞).


## 2. The Rotundity of the Spaceℓ(B~,p)

The rotundity of Banach spaces is one of the most important geometric property in functional analysis. For details, the reader may refer to [10–12]. In this section, we characterize the rotundity of the space ℓ(B~,p) and give some results related to this concept.Definition 1.
LetS(X) be the unit sphere of a Banach space X. Then, a point x∈S(X) is called an extreme point if 2x=y+z implies y=z for every y,z∈S(X). A Banach space X is said to be rotund (strictly convex) if every point of S(X) is an extreme point.Definition 2.
A Banach spaceX is said to have Kadec-Klee property (or property (H)) if every weakly convergent sequence on the unit sphere is convergent in norm.Definition 3.
A Banach spaceX is said to have  (i)
the Opial property if every sequence(xn) weakly convergent to x0∈X satisfies
(5)liminfn→∞∥xn-x0∥<liminfn→∞∥xn+x∥
for every x∈X with x≠x0;(ii)
the uniform Opial property if for eachϵ>0, there exists an r>0 such that
(6)1+r≤liminfn→∞∥xn+x∥
for each x∈X with ∥x∥≥ϵ and each sequence (xn) in X such that xn→0 and liminfn→∞∥xn∥≥1.Definition 4.
LetX be a real vector space. A functional σ:X→[0,∞) is called a modular if  (i)
σ(x)=0 if and only if x=θ;(ii)
σ(αx)=σ(x) for all scalars α with |α|=1;(iii)
σ(αx+βy)≤σ(x)+σ(y) for all x,y∈X and α,β≥0 with α+β=1;(iv)
the modularσ is called convex if σ(αx+βy)≤ασ(x)+βσ(y) for all x,y∈X and α,β>0 with α+β=1.A modularσ on X is called (a)
right continuous iflimα→1+σ(αx)=σ(x) for all x∈Xσ.(b)
left continuous iflimα→1-σ(αx)=σ(x) for all x∈Xσ.(c)
continuous if it is both right and left continuous, where(7)Xσ={x∈X:limα⟶0+σ(αx)=0}.We defineσp on ℓ(B~,p) by σp(x)=∑k‍|sk-1xk-1+rkxk|pk. If pk≥1 for all k∈ℕ={1,2,3,…}, by the convexity of the function t↦|t|pk for each k∈ℕ, σp is a convex modular on ℓ(B~,p).Proposition 5.
The modularσp on ℓ(B~,p) satisfies the following properties with pk≥1 for all k∈ℕ:  (i)
if0<α≤1, thenαMσp(x/α)≤σp(x)    andσp(αx)≤ασp(x).(ii)
Ifα≥1, thenσp(x)≤αMσp(x/α).(iii)
Ifα≥1, thenσp(x)≤ασp(x/α).(iv)
The modularσp   is continuous on the spaceℓ(B~,p).Proof.
Consider the modularσp on ℓ(B~,p).  (i)
Let0<α≤1, then αM/αpk≤1. So, we have
(8)αMσp(xα)=αM∑k‍1αpk|sk-1xk-1+rkxk|pk=∑k‍αMαpk|sk-1xk-1+rkxk|pk≤∑k‍|sk-1xk-1+rkxk|pk=σp(x),σp(αx)=∑k‍αpk|sk-1xk-1+rkxk|pk≤α∑k‍|sk-1xk-1+rkxk|pk=ασp(x).(ii)
Letα≥1. Then, αM/αpk≥1 for all pk≥1. So, we have
(9)σp(x)≤αMαpkσp(x)=αMσp(xα).(iii)
Letα≥1. Then, α/αpk≥1 for all pk≥1. So, we have
(10)σp(x)=∑k‍|sk-1xk-1+rkxk|pk≤∑k‍ααpk|sk-1xk-1+rkxk|pk=ασp(xα).(iv)
By (ii) and (iii), one can immediately see forα>1 that
(11)σp(x)≤ασp(x)≤σp(αx)≤αMσp(x).
By passing to limit as α→1+ in (11), we have limα→1+σp(αx)=σp(x). Hence, σp is right continuous. If 0<α<1, by (i) we have
(12)αMσp(x)≤σp(αx)≤ασp(x).
By letting α→1- in (12), we observe that limα→1-σp(αx)=σp(x). Hence, σp is also left continuous, and so, it is continuous.Proposition 6.
For anyx∈ℓ(B~,p), the following statements hold:  (i)
if∥x∥<1, thenσp(x)≤∥x∥.(ii)
If∥x∥>1, thenσp(x)≥∥x∥.(iii)
∥x∥=1   if and only ifσp(x)=1.(iv)
∥x∥<1   if and only ifσp(x)<1.(v)
∥x∥>1   if and only ifσp(x)>1.Proof.
Letx∈ℓ(B~,p).  (i)
Letϵ>0 be such that 0<ϵ<1-∥x∥. By the definition of ∥·∥, there exists an α>0 such that ∥x∥+ϵ>α and σp(x)≤1. From Parts (i) and (ii) of Proposition 5, we obtain
(13)σp(x)≤σp[(∥x∥+ϵ)xα]≤(∥x∥+ϵ)σp(xα)≤∥x∥+ϵ.
Since ϵ is arbitrary, we have (i).(ii)
If we chooseϵ>0 such that 0<ϵ<1-(1/∥x∥), then 1<(1-ϵ)∥x∥<∥x∥. By the definition of ∥·∥ and Part (i) of Proposition 5, we have
(14)1<σp[x(1-ϵ)∥x∥]≤1(1-ϵ)∥x∥σp(x).
So, (1-ϵ)∥x∥<σp(x) for all ϵ∈(0,1-(1/∥x∥)). This implies that ∥x∥<σp(x).(iii)
Sinceσp is continuous, by Theorem 1.4 of [12] we directly have (iii).(iv)
This follows from Parts (i) and (iii).(v)
This follows from Parts (ii) and (iii).Now, we consider the spaceℓ(B~,p) equipped with the Luxemburg norm given by
(15)∥x∥=inf{α>0:σp(xα)≤1}.Theorem 7.
ℓ
(
B
~
,
p
) is a Banach space with Luxemburg norm.Proof.
LetSx={α>0:σp(x/α)≤1} and ∥x∥=infSx for all x∈ℓ(B~,p). Then, Sx⊂(0,∞). Therefore, ∥x∥≥0 for all x∈ℓ(B~,p).
Forx=θ, σp(θ)=0 for all α>0. Hence, S0=(0,∞) and ∥θ∥=infS0=inf(0,∞)=0.
Letx≠θ and Y={kx:k∈ℂ and x∈ℓ(B~,p)} be a nonempty subset of ℓ(B~,p). Since YS[ℓ(B~,p)], there exists k1∈ℂ such that k1x∉S[ℓ(B~,p)]. Obviously, k1≠0. We assume that 0<α<1/k1 and α∈Sx. Then, (x/α)∈S[ℓ(B~,p)]. Since |k1α|<1, we get
(16)k1x=k1αxα∈S[ℓ(B~,p)]
which contradicts the assumption. Hence, we obtain that if α∈Sx, then α>1/|k1|. This means that ∥x∥≥1/|k1|>0. Thus, we conclude that ∥x∥=0 if and only if x=θ.
Now, letk≠0 and α∈Skx. Then, we have
(17)σp(kxα)≤1,kxα∈S[ℓ(B~,p)].
Therefore, we obtain
(18)|k|xα=|k|k×kxα∈S[ℓ(B~,p)],α|k|∈Sx.
That is, ∥x∥≤α/|k| and |k|∥x∥≤α for all α∈Skx. So, |k|∥x∥≤∥kx∥.
If we take1/k and kx instead of k and x, respectively, then we obtain that
(19)|1kx|∥kx∥≤∥1kkx∥=∥x∥,∥kx∥≤|k|∥x∥.
Hence, we get ∥kx∥=|k|∥x∥. This also holds when k=0.
To prove the triangle inequality, letx,y∈ℓ(B~,p) and ϵ>0 be given. Then, there exist α∈Sx and β∈Sy such that α<∥x∥+ϵ and β<∥y∥+ϵ. Since S[ℓ(B~,p)] is convex,
(20)xα∈S[ℓ(B~,p)],yβ∈S[ℓ(B~,p)],(x+y)α+β=αα+β(xα)+βα+β(yβ)∈S[ℓ(B~,p)].
Therefore, α+β∈Sx+y. Then, we have ∥x+y∥≤α+β<∥x∥+∥y∥+2ϵ. Since ϵ>0 was arbitrary, we obtain ∥x+y∥≤∥x∥+∥y∥. Hence, ∥x∥=inf{α>0:σp(x/α)≤1} is a norm on ℓ(B~,p).
Now, we need to show that every Cauchy sequence inℓ(B~,p) is convergent according to the Luxemburg norm. Let {xk(n)} be a Cauchy sequence in ℓ(B~,p) and ϵ∈(0,1). Thus, there exists n0 such that ∥x(n)-x(m)∥<ϵ for all n,m≥n0. By Part (i) of Proposition 6, we have
(21)σp(x(n)-x(m))≤∥x(n)-x(m)∥<ϵ
for all n,m≥n0. This implies that
(22)∑k|[B~(x(n)-x(m))]k|pk<ϵ.
Then, for each fixed k and for all n,m≥n0,
(23)|[B~(x(n)-x(m))]k|=|(B~x(n))k-(B~x(m))k|<ϵ.
Hence, the sequence {(B~x(n))k} is a Cauchy sequence in ℝ. Since ℝ is complete, there is a (B~x)k∈ℝ such that (B~x(m))k→(B~x)k as m→∞. Therefore, as m→∞ by (22), we have
(24)∑k‍|[B~(x(n)-x)]k|pk<ϵ
for all n≥n0.
Now, we have to show that(xk) is an element of ℓ(B~,p). Since (B~x(m))k→(B~x)k as m→∞, we have
(25)limm⟶∞σp(x(n)-x(m))=σp(x(n)-x).
Then, we see by (21) that σp(x(n)-x)≤∥x(n)-x∥<ϵ for all n≥n0. This implies that xn→x as n→∞. So, we have x=x(n)-(x(n)-x)∈ℓ(B~,p). Therefore, the sequence space ℓ(B~,p) is complete with respect to Luxemburg norm. This completes the proof.Theorem 8.
The spaceℓ(B~,p) is rotund if and only if pk>1 for all k∈ℕ.Proof.
Letℓ(B~,p) be rotund and choose k∈ℕ such that pk=1 for k<3. Consider the following sequences given by
(26)x=(0,1r1,-s1r1r2,s1s2r1r2r3,…),y=(0,0,1r2,-s2r2r3,s2s3r2r3r4,…).
Then, obviously x≠y and
(27)σp(x)=σp(y)=σp(x+y2)=1.
By Part (iii) of Proposition 6, x,y,(x+y)/2∈S[ℓ(B~,p)] which leads us to the contradiction that the sequence space ℓ(B~,p) is not rotund. Hence, pk>1 for all k∈ℕ.
Conversely, letx∈S[ℓ(B~,p)] and v,z∈S[ℓ(B~,p)] with x=(v+z)/2. By convexity of σp and Part (iii) of Proposition 6, we have
(28)1=σp(x)≤σp(v)+σp(z)2≤12+12=1,
which gives that σp(v)=σp(z)=1, and
(29)σp(x)=σp(v)+σp(z)2.
Also, we obtain from (29) that
(30)∑k|sk-1xk-1+rkxk|pk=12(∑k|sk-1vk-1+rkvk|pk+∑k|sk-1zk-1+rkzk|pk).
Since x=(v+z)/2, we have
(31)∑k|sk-1(vk-1+zk-1)+rk(vk+zk)|pk=12(∑k|sk-1vk-1+rkvk|pk+∑k|sk-1zk-1+rkzk|pk).
This implies that
(32)|sk-1(vk-1+zk-1)+rk(vk+zk)|pk=12|sk-1vk-1+rkvk|pk+12|sk-1zk-1+rkzk|pk
for all k∈ℕ. Since the function t↦|t|pk is strictly convex for all k∈ℕ, it follows by (32) that vk=zk for all k∈ℕ. Hence, v=z. That is, the sequence space ℓ(B~,p) is rotund.Theorem 9.
Letx∈ℓ(B~,p). Then, the following statements hold:  (i)
0<α<1 and ∥x∥>α imply σp(x)>αM.(ii)
α≥1 and ∥x∥<α imply σp(x)<αM.Proof.
Letx∈ℓ(B~,p).  (i)
Suppose that∥x∥>α with 0<α<1. Then, ∥x/α∥>1. By Part (ii) of Proposition 6, ∥x/α∥>1 implies σp(x/α)≥∥x/α∥>1. That is, σp(x/α)>1. Since 0<α<1, by Part (i) of Proposition 5, we get αMσp(x/α)≤σp(x). Thus, we have αM<σp(x).(ii)
Let∥x∥<α and α≥1. Then, ∥x/α∥<1. By Part (i) of Proposition 6, ∥x/α∥<1 implies σp(x/α)≤∥x/α∥<1. That is, σp(x/α)<1. If α=1, then σp(x/α)=σp(x)<1=αM. If α>1, then by Part (ii) of Proposition 5, we have σp(x)≤αMσp(x/α). This means that σp(x)<αM.Theorem 10.
Let(xn) be a sequence in ℓ(B~,p). Then, the following statements hold:  (i)
limn→∞∥xn∥=1 implies limn→∞σp(xn)=1.(ii)
limn→∞σp(xn)=0 implies limn→∞∥xn∥=0.Proof.
Let(xn) be a sequence in ℓ(B~,p).  (i)
Letlimn→∞∥xn∥=1 and ϵ∈(0,1). Then, there exists n0∈ℕ such that 1-ϵ<∥xn∥<ϵ+1 for all n≥n0. By Parts (i) and (ii) of Theorem 9, 1-ϵ<∥xn∥ implies σp(xn)>(1-ϵ)M and ∥xn∥<ϵ+1 implies σp(xn)<(1+ϵ)M for all n≥n0. This means ϵ∈(0,1) and for all n≥n0 there exists n0∈ℕ such that (1-ϵ)M<σp(xn)<(1+ϵ)M. That is, limn→∞σp(xn)=1.(ii)
We assume thatlimn→∞∥xn∥≠0 and ϵ∈(0,1). Then, there exists a subsequence (xnk) of (xn) such that ∥xnk∥>ϵ for all k∈ℕ. By Part (i) of Theorem 9, 0<ϵ<1 and ∥xnk∥>ϵ imply σp(xnk)>ϵM. Thus, limn→∞σp(xn)≠0 for all k∈ℕ. Hence, we obtain that limn→∞σp(xn)=0 implies limn→∞∥xn∥=0.Theorem 11.
Letx∈ℓ(B~,p) and (x(n))⊂ℓ(B~,p). If σp(x(n))→σp(x) as n→∞ and xk(n)→xk as n→∞ for all k∈ℕ, then x(n)→x as n→∞.Proof.
Letϵ>0 be given. Since σp(x)=∑k‍|(B~x)k|pk<∞, there exists k0∈ℕ such that
(33)∑k=k0+1∞‍|(B~x)k|pk<ϵ3(2M+1).
It follows from the fact
(34)limn⟶∞[σp(x(n))-∑k=1k0|(B~x(n))k|pk]=σp(x)-∑k=1k0|(B~x)k|pk
that there exists n0∈ℕ such that for all n≥n0 and for all k∈ℕ,
(35)σp(xnk)-∑k=1k0|(B~x(n))k|pk<σp(x)-∑k=1k0|(B~x)k|pk+ϵ3(2M),
and for all n≥n0,
(36)∑k=1k0|{B~(x(n)-x)}k|pk<ϵ3.
Therefore, we obtain from (33), (35), and (36) that
(37)σp(xn-x)=∑k=1∞|{B~(x(n)-x)}k|pk<∑k=1k0|{B~(x(n)-x)}k|pk+∑k=k0+1∞|{B~(x(n)-x)}k|pk<ϵ3+2M[∑k=k0+1∞|(B~x(n))k|pk+∑k=k0+1∞|(B~x)k|pk]<ϵ3+2M[∑k0+1∞|(B~x)k|pkσp(xn)-∑k=1k0|(B~x(n))k|pk+∑k0+1∞|(B~x)k|pk]<ϵ3+2M[∑k=k0+1∞|(B~x)k|pkσp(x)-∑k=1k0|(B~x)k|pk+ϵ3(2M)+∑k=k0+1∞|(B~x)k|pk]<ϵ3+2M[2∑k=k0+1∞|(B~x)k|pk+ϵ3(2M)]<ϵ3+2M[2ϵ3(2M+1)+ϵ3(2M)]=ϵ.
This means that σp(x(n)-x)→0 as n→∞. By Part (ii) of Theorem 10, σp(x(n)-x)→0 as n→∞ implies ∥xn-x∥→0 as n→∞. Hence, xn→x as n→∞.Theorem 12.
The sequence spaceℓ(B~,p) has the Kadec-Klee property.Proof.
Letx∈S[ℓ(B~,p)] and (x(n))⊂ℓ(B~,p) such that ∥x(n)∥→1 and x(n)→wx are given. By Part (ii) of Theorem 10, we have σp(x(n))→1 as n→∞. Also x∈S[ℓ(B~,p)] implies ∥x∥=1. By Part (iii) of Proposition 6, we obtain σp(x)=1. Therefore, we have σp(x(n))→σp(x) as n→∞.
Sincex(n)→wx and qk:ℓ(B~,p)→ℝ defined by qk(x)=xk is continuous, xk(n)→xk as n→∞ for all k∈ℕ. Therefore, x(n)→x as n→∞.
Since any weakly convergent sequence inℓ(B~,p) is convergent, the sequence space ℓ(B~,p) has the Kadec-Klee property.Theorem 13.
For any1<p<∞, the space (ℓp)B~ has the uniform Opial property.Proof.
Letϵ>0 and ϵ0∈(0,ϵ) be given such that 1+(ϵp/2)>(1+ϵ0)p. Also let x∈(ℓp)B~ and ∥x∥≥ϵ. There exists k1∈ℕ such that
(38)∑k=k1+1∞|(B~x)k|p<(ϵ04)p.
Hence, we have
(39)∥∑k=k1+1∞xkek∥<ϵ04.
Furthermore, we have
(40)ϵp≤∑k=1k1|(B~x)k|p+∑k=k1+1∞|(B~x)k|p<∑k=1k1|(B~x)k|p+(ϵ04)p<∑k=1k1|(B~x)k|p+ϵp4,
which yields that
(41)3ϵp4<∑k=1k1|(B~x)k|p.
For any weakly null sequence (x(m))⊂S[(ℓp)B~], since xk(m)→0 as m→∞ for each k∈ℕ, there exists m0∈ℕ such that for all m>m0,
(42)∥∑k=1k1xk(m)ek∥<ϵp4.
Therefore, for all m>m0,
(43)∥x(m)+x∥=∥∑k=1k1(xk(m)+xk)ek+∑k=k1+1∞(xk(m)+xk)ek‍∥≥∥∑k=1k1xkek+∑k=k1+1∞xk(m)ek∥-∥∑k=1k1xk(m)ek∥-∥∑k=k1+1∞xkek∥≥∥∑k=1k1xkek+∑k=k1+1∞xk(m)ek∥-ϵp4-ϵp4.
Moreover,
(44)∥∑k=1k1xkek+∑k=k1+1∞xk(m)ek∥p=∑k=1k1|(B~x)kek|p+∑k=k1+1∞|(B~x(m))kek|p≥3ϵp4+(1-ϵp4)=1+ϵp2>(1+ϵ0)p.
Then, we have
(45)∥x(m)+x∥≥∥∑k=1k1xkek+∑k=k1+1∞xk(m)ek∥-ϵp2≥1+ϵ0-ϵp2>1+ϵ0p2.
This means that (ℓp)B~ has the uniform Opial property.


## 3. Conclusion

The sequence spacesbv(u,p) and bv∞(u,p) of nonabsolute type consisting of all sequences x=(xk) such that {uk(xk-xk-1)} is in the Maddox' spaces ℓ(p) and ℓ∞(p) were introduced by Başar et al. [13], where u=(uk) is a sequence such that uk≠0 for all k∈ℕ and the rotundity of the space bv(u,p) was examined.The sequence spacear(u,p) of nonabsolute type consisting of all sequences x=(xk) such that Arx={∑k=0n‍(1+rk)xk/(n+1)}∈ℓ(p) was studied by Aydın and Başar [14], and some results related to the rotundity of the space ar(u,p) were given.Quite recently, the sequence spaceℓ^(p) of nonabsolute type consisting of all sequences x=(xk) such that B(r,s)x=(sxk-1+rxk)∈ℓ(p) was defined by Aydın and Başar [15], and emphasized the rotundity of the space ℓ^(p) together with some related results.Although the sequence spacesar(u,p) and ℓ(B~,p) are not comparable, since the double sequential band matrix B(r~,s~) reduces to the generalized difference matrix B(r,s) in the special case r~=re and s~=se, the new space ℓ(B~,p) is more general than the space ℓ^(p). Similarly, the sequence space ℓ(B~,p) is also reduced to the space bv(u,p) in the case r~=(uk) and s~=(-uk). So, the results on the space ℓ(B~,p) are much more comprehensive than the results on the space bv(u,p). Additionally, the corresponding theorems on the Kadec-Klee property of the space ℓ(B~,p) and the uniform Opial property of the space (ℓp)B~ were not given by Başar et al. [13] and Aydın and Başar [15] which make the present paper significant.



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