The sequence space ℓ(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.
1. Introduction
By ω, 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 Space ℓ(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.
Let S(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 space X 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 space X is said to have
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;
the uniform Opial property if for each ϵ>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.
Let X be a real vector space. A functional σ:X→[0,∞) is called a modular if
σ(x)=0 if and only if x=θ;
σ(αx)=σ(x) for all scalars α with |α|=1;
σ(αx+βy)≤σ(x)+σ(y) for all x,y∈X and α,β≥0 with α+β=1;
the modular σ 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 modular σ on X is called
right continuous if limα→1+σ(αx)=σ(x) for all x∈Xσ.
left continuous if limα→1-σ(αx)=σ(x) for all x∈Xσ.
continuous if it is both right and left continuous, where
(7)Xσ={x∈X:limα⟶0+σ(αx)=0}.
We define σp 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 modular σp on ℓ(B~,p) satisfies the following properties with pk≥1 for all k∈ℕ:
if 0<α≤1, then αMσp(x/α)≤σp(x)and σp(αx)≤ασp(x).
If α≥1, then σp(x)≤αMσp(x/α).
If α≥1, then σp(x)≤ασp(x/α).
The modular σpis continuous on the space ℓ(B~,p).
Proof.
Consider the modular σp on ℓ(B~,p).
Let 0<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).
Let α≥1. Then, αM/αpk≥1 for all pk≥1. So, we have
(9)σp(x)≤αMαpkσp(x)=αMσp(xα).
Let α≥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).
By (ii) and (iii), one can immediately see for α>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 any x∈ℓ(B~,p), the following statements hold:
if ∥x∥<1, then σp(x)≤∥x∥.
If ∥x∥>1, then σp(x)≥∥x∥.
∥x∥=1if and only if σp(x)=1.
∥x∥<1if and only if σp(x)<1.
∥x∥>1if and only if σp(x)>1.
Proof.
Let x∈ℓ(B~,p).
Let ϵ>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).
If we choose ϵ>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).
Since σp is continuous, by Theorem 1.4 of [12] we directly have (iii).
This follows from Parts (i) and (iii).
This follows from Parts (ii) and (iii).
Now, we consider the space ℓ(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.
Let Sx={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).
For x=θ, σp(θ)=0 for all α>0. Hence, S0=(0,∞) and ∥θ∥=infS0=inf(0,∞)=0.
Let x≠θ and Y={kx:k∈ℂ and x∈ℓ(B~,p)} be a nonempty subset of ℓ(B~,p). Since YS[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, let k≠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 take 1/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, let x,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 in ℓ(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 space ℓ(B~,p) is rotund if and only if pk>1 for all k∈ℕ.
Proof.
Let ℓ(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, let x∈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.
Let x∈ℓ(B~,p). Then, the following statements hold:
0<α<1 and ∥x∥>α imply σp(x)>αM.
α≥1 and ∥x∥<α imply σp(x)<αM.
Proof.
Let x∈ℓ(B~,p).
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).
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:
limn→∞∥xn∥=1 implies limn→∞σp(xn)=1.
limn→∞σp(xn)=0 implies limn→∞∥xn∥=0.
Proof.
Let (xn) be a sequence in ℓ(B~,p).
Let limn→∞∥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.
We assume that limn→∞∥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.
Let x∈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.
Let ϵ>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 space ℓ(B~,p) has the Kadec-Klee property.
Proof.
Let x∈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→∞.
Since x(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 in ℓ(B~,p) is convergent, the sequence space ℓ(B~,p) has the Kadec-Klee property.
Theorem 13.
For any 1<p<∞, the space (ℓp)B~ has the uniform Opial property.
Proof.
Let ϵ>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 spaces bv(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 space ar(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 space ℓ^(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 spaces ar(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.
Acknowledgments
The main results of this paper were presented in part at the conference First International Conference on Analysis and Applied Mathematics (ICAAM 2012) held on October 18–21, 2012 in Gümüşhane, Turkey, at the University of Gümüşhane.
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