# Some New Generalized Difference Spaces of Nonabsolute Type Derived from the Spacesℓp and ℓ∞

Authors: Feyzi Başar; Ali Karaisa
Journal: The Scientific World Journal
 (2013)

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## Abstract

We introduce the sequence spaceℓpλ(B) of none absolute type which is a p-normed space and BK space in the cases 0<p<1 and 1⩽p⩽∞, respectively, and prove that ℓpλ(B) and ℓp are linearly isomorphic for 0<p⩽∞. Furthermore, we give some inclusion relations concerning the space ℓpλ(B) and we construct the basis for the space ℓpλ(B), where 1⩽p<∞. Furthermore, we determine the alpha-, beta- and gamma-duals of the space ℓpλ(B) for 1⩽p⩽∞. Finally, we investigate some geometric properties concerning Banach-Saks type p and give Gurarii's modulus of convexity for the normed space ℓpλ(B).


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## Body

## 1. Introduction

From the summability theory perspective, the role played by the algebraical, geometrical, and topological properties of the new Banach spaces which are defined by the matrix domain of triangle matrices in sequence spaces is well-known.Byw, we denote the space of all real or complex valued sequences. Any vector subspace of w is called a sequence space.A sequence spaceμ with a linear topology is called a K-space provided that each of the maps pi:μ→ℂ defined by pi(x)=xi is continuous for all i∈ℕ, where ℂ denotes the complex field and ℕ={0,1,2,…}. A K-space is called an FK-space provided μ is a complete linear metric space. An FK-space whose topology is normable is called a BK-space (see [1]) which contains ϕ, the set of all finitely nonzero sequences.We writeℓ∞, f, c, and c0 for the spaces of all bounded, almost convergent, convergent, and null sequences, respectively, which are BK-spaces with the usual supnorm defined by(1)∥x∥∞=supk∈ℕ|xk|.
Also, by ℓp and ℓ1, we denote the spaces all p-absolutely and absolutely convergent series, respectively, which are BK-spaces with the usual norm defined by
(2)∥x∥p=(∑k‍|xk|p)1/p,(1⩽p<∞).
Here, and in what follows, the summation without limits runs from 0 to ∞. Further, we write bs and cs for the spaces of all bounded and convergent series, respectively, which are BK-spaces with their natural norm [2].Letμ and γ be two sequence spaces and A=(ank) be an infinite matrix of real or complex numbers ank, where n,k∈ℕ. Then, we say that A defines a matrix transformation from μ into γ and we denote it by writing A:μ→γ, if for every sequence x=(xk)∈μ the sequence Ax={(Ax)n}, the A-transform of x is in γ, where
(3)(Ax)n=∑k‍ankxk∀n∈ℕ.The notation(μ:γ) denotes the class of all matrices A such that A:μ→γ. Thus, A∈(μ:γ) if and only if the series on the right hand side of (3) converges for each n∈ℕ and every x∈μ, and we have Ax={(Ax)n}n∈ℕ∈γ for all x∈μ. The matrix domain μA of an infinite matrix A in a sequence space μ is defined by
(4)μA={x=(xk)∈ω:Ax∈μ}.
An infinite matrix A=(ank) is said to be a triangle if ann≠0 for all n∈ℕ and ank=0 for k>n. The study of matrix domains of triangles has a special importance due to the various properties which they have. For example, if A is a triangle and μ is a BK-space, then μA is also a BK-space with the norm given by ∥x∥μA=∥Ax∥μ for all x∈μA.Throughout the paper, we denote the collection of all finite subsets ofℕ by ℱ. Also, we write e(k) for the sequence whose only nonzero term is a 1 in the kth place for each k∈ℕ.The approach constructing a new sequence space by means of the matrix domain of a particular triangle has recently been employed by several authors in many research papers. For example, they introduced the sequence spaces(ℓ∞)Nq and cNq in [3], (ℓp)C1=Xp and (ℓ∞)C1=X∞ in [4], (c0)C1=c~0 and (c)C1=c~ in [5], (ℓp)Er=epr and (ℓ∞)Er=e∞r in [6], (ℓp)Ar=apr and (ℓ∞)Ar=a∞r in [7], (ℓp)Δ=bvp and (ℓ∞)Δ=bv∞ in [8], μG=Z(u,v;μ) in [9], (c0)Λ=c0λ and cΛ=cλ in [10], and (ℓp)Δ(m)=ℓp(Δ(m)) in [11], where Nq, C1, Rt, Er, Ar, Λ, and Δ(m) denote Nörlund, arithmetic, Riesz, Euler means, Ar matrix, lambda matrix, and generalized difference matrix, respectively, where 1⩽p<∞.Recently, there has been a lot of interest in investigating geometric properties of sequence spaces besides topological and some other usual properties. In the literature, there are many papers concerning the geometric properties of different sequence spaces. For example, in [12], Mursaleen et al. studied some geometric properties of a normed Euler sequence space. Recently, Şimşek and Karakaya [13] have investigated the geometric properties of the sequence space ℓ(u,v,p) equipped with Luxemburg norm. Later, Demiriz and Çakan [14] have studied some geometric properties of the sequence space apr(Δ). For further information on geometric properties of sequence spaces the reader can refer to [15, 16].The main purpose of the present paper is to introduce the difference sequence spacesℓpλ(B) of nonabsolute type and derive some related results. We also establish some inclusion relations, where 0<p⩽∞. Furthermore, we determine the alpha-, beta- and gamma-duals of those spaces and construct their bases. We characterize some classes of infinite matrices concerning the spaces ℓpλ(B) and ℓ∞λ(B) for 1⩽p<∞. Finally, we investigate some geometric properties concerning Banach-Saks type p and give Gurarii's modulus of convexity for the normed space ℓpλ(B).


## 2. The Sequence Spacesℓpλ(B) and ℓ∞λ(B) of Nonabsolute Type

This section is devoted to the examination of the basic topological properties of the setsℓpλ(B) and ℓ∞λ(B). Let throughout that (λk) be strictly increasing sequence of positive reals tending to ∞; that is
(5)0<λ1<λ2<⋯,limk→∞λk=∞.
Let us define the lambda matrix Λ=(λnk) by
(6)λnk={λk-λk-1λn,0⩽k⩽n,0,k>n.
Recently, Mursaleen and Noman [17, 18] have studied the sequence spaces ℓ∞λ and ℓpλ of nonabsolute type as follows:
(7)ℓpλ:={x=(xk)∈ω:∑n‍|∑k=0n‍λk-λk-1λnxk|p<∞};x=(xk)∈ω1:∑n‍|∑k=0n‍λk-λk-1λn|p∞(0<p<∞),ℓ∞λ:={x=(xk)∈ω:supn∈ℕ|∑k=0n‍λk-λk-1λnxk|<∞}.
With the notation of (4), we can redefine the spaces ℓ∞λ and ℓpλ by ℓpλ=(ℓp)Λ and ℓ∞λ=(ℓ∞)Λ, where 0<p<∞.Letr and s be non-zero real numbers, and define the generalized difference matrix B(r,s)={bnk(r,s)} by
(8)bnk(r,s)={r,k=n,s,k=n-1,0,otherwise,
for all n,k∈ℕ. The B(r,s)-transform of a sequence x=(xk) is {B(r,s)x}k=rxk+sxk-1 for all k∈ℕ. We note that the matrix B(r,s) can be reduced to the difference matrix Δ in the case r=1 and s=-1. So, the results related to the domain of the matrix B(r,s) are more general and comprehensive than those of the matrix domain of Δ and include them.Now, we introduce the new sequence spacesℓpλ(B) and ℓ∞λ(B) as follows:
(9)ℓpλ(B):={|∑k=0n‍λk-λk-1λn(rxk+sxk-1)|px=(xk)∈ω:x=(xk)(xk)∑n‍|∑k=0n‍λk-λk-1λn(rxk+sxk-1)|p<∞∑k=0n‍λk-λk-1λn};∑n‍|∑k=0n‍λk-λk-1λn1(rxk+sxk-1)|p1212(0<p<∞),ℓ∞λ(B):={|∑k=0n‍λk-λk-1λn(rxk+sxk-1)|x=(xk)∈ω:x=(xk)(xk)supn∈ℕ|∑k=0n‍λk-λk-1λn(rxk+sxk-1)|<∞|∑k=0n‍λk-λk-1λn(rxk+sxk-1)|}.
By the notation of (4), we can redefine the spaces ℓ∞λ(B) and ℓpλ(B) as follows:
(10)ℓpλ(B)=(ℓpλ)B(0<p<∞),ℓ∞λ(B)=(ℓ∞λ)B,
where B denotes the generalized difference matrix B(r,s)={bnk(r,s)} defined by (8). Now, we may define the triangle matrix Λ^=(λ^nk) by
(11)λ^nk={r(λk-λk-1)+s(λk+1-λk)λn,k<n,r(λn-λn-1)λn,k=n,0,k>n.
Define the sequence y=(yk) as the Λ^-transform of a sequence x=(xk); that is,
(12)yk=(Λ^x)k=∑i=0k-1‍r(λi-λi-1)+s(λi+1-λi)λkxi+r(λk-λk-1)λkxk∀k∈ℕ.
Now, we can redefine the spaces ℓ∞λ(B) and ℓpλ(B) with the notation of (4) as
(13)ℓpλ(B)=(ℓp)Λ^(0<p<∞),ℓ∞λ(B)=(ℓ∞)Λ^.
Also, we derive from the equality (12) that
(14)yk=∑i=0k‍(λi-λi-1)(rxi+sxi-1)λk∀k∈ℕ.
Then, since the sequence spaces ℓpλ(B) and ℓp are linearly isomorphic; that is, ℓpλ(B)≅ℓp; it is trivial that the two-sided implication “x∈ℓpλ(B) if and only if y∈ℓp” holds, where 0<p⩽∞.We have the following result which is essential in the text.Theorem 1.
The following statements hold.(a)
If0<p<1, then ℓpλ(B) is a complete p-normed space with the p-norm ∥x∥ℓpλ(B)=∥Λ^(x)∥p; that is,
(15)∥x∥ℓpλ(B)=∑n‍|(Λ^x)n|p;(0<p<1).(b)
If1⩽p⩽∞, then ℓpλ(B) is BK-space with the norm ∥x∥ℓpλ(B)=∥Λ^(x)∥p; that is,
(16)∥x∥ℓpλ(B)=[∑n‍|(Λ^x)n|p]1/p;(1⩽p<∞),(17)∥x∥ℓ∞λ(B)=supn∈ℕ|(Λ^x)n|.Proof.
(a) Let0<p<1. It is immediate by the fact ℓpλ(B)≅ℓp that ℓpλ(B) is a complete p-normed linear space with the p-norm ∥x∥ℓpλ(B)=∑n‍|(Λ^x)n|p.
(b) Since the setsℓp and ℓ∞ endowed with the norms ∥·∥p and ∥·∥∞ are BK-spaces (see [2, Example 7.3.2 (b), (c)]) and the matrix Λ^ is triangle, Theorem 4.3.2 of Wilansky [19, page 61] gives the fact that the spaces ℓpλ(B) and ℓ∞λ(B) are BK-spaces with the norms in (16) and (17), respectively.One can easily check that the absolute property does not hold on the spaceℓpλ(B); that is, ∥x∥ℓ∞λ(B)≠∥|x|∥ℓ∞λ(B) for at least one sequence in the space ℓpλ(B), and this tells us that ℓpλ(B) is nonabsolute type, where |x|=(|xk|) and 0<p⩽∞.Now, one may expect the similar result for the spaceℓpλ(B) as was observed for the space ℓp and ask the natural question: Is not the space ℓpλ(B) a Hilbert space with p≠2? The answer is positive and is given by the following theorem.Theorem 2.
Except the casep=2, the space ℓpλ(B) is not an inner product space and hence it is not a Hilbert space, where 1⩽p<∞.Proof.
We have to prove that the spaceℓ2λ(B) is the only Hilbert space among the ℓpλ(B) spaces for 1⩽p<∞. Since the space ℓpλ(B) is a BK-space with the norm ∥x∥ℓ2λ(B)=∥Λ^x∥2 by Part (b) of Theorem 1 and its norm can be obtained from an inner product; that is,
(18)∥x∥ℓ2λ(B)=〈x,x〉1/2=〈Λ^x,Λ^x〉1/2
holds for every x∈ℓ2λ(B), the space ℓ2λ(B) is a Hilbert space. Let us define the sequences z=(zk) and t=(tk) by
(19)zk:={1r,(-sr)+(-sr)2-12(-sr)(-sr)2k=0,r-sr2,(-sr)+(-sr)21(-sr)(-sr)2k=1,(-sr)k-21r[(-sr)+(-sr)2-(λ1λ2-λ1)],(-sr)+(-sr)2(-sr)2*(λ1λ2-λ1)k⩾2,tk:={1r,(-sr)2(-sr)2(-sr)2(-sr)121(-sr)2k=0,-1r(λ1+λ0λ1-λ0+sr),(-sr)1(-sr)(-sr)2k=1,(-sr)k-21r[λ1λ2-λ1+(sr)2+sr(λ1+λ0λ1-λ0)],(-sr)+(-sr)2(-sr)2(-sr)2(λ1λ2-λ1)k⩾2.
Then, we have
(20)Λ^z=(1,1,0,0,…),Λ^t=(1,-1,0,0,…).
Thus, it can easily be seen that
(21)∥z+t∥ℓpλ(B)2+∥z-t∥ℓpλ(B)2=8≠4(22/p)=2(∥z∥ℓpλ(B)2+∥t∥ℓpλ(B)2);(p≠2),
that is, the norm of the space ℓpλ(B) with p≠2 does not satisfy the parallelogram equality which means that this norm cannot be obtained from an inner product. Hence, the space ℓpλ(B) with p≠2 is a Banach space which is not a Hilbert space. This completes the proof.


## 3. The Inclusion Relations

In this section, we give some inclusion relations between the spacesℓp and ℓ∞ and the spaces ℓpλ(B) and ℓ∞λ(B), where 0<p<∞. We essentially prove that ℓ∞⊂ℓ∞λ(B) holds and characterize the case in which the inclusion ℓp⊂ℓpλ(B) holds for 1⩽p<∞.Theorem 3.
Let0<p<s<∞. Then, the inclusions ℓpλ(B)⊂ℓsλ(B) strictly holds.Proof.
Let0<p<s<∞ and x=(xk)∈ℓpλ(B). This implies that Λ^x∈ℓp. Since ℓp⊂ℓs, Λ^x∈ℓs. So, we have ℓpλ(B)⊂ℓsλ(B).
Let us consider the sequencez=(zk) with the aid of the sequence x=(xk)={(k+1)-(s+1-p)}k=0∞∈ℓs∖ℓp defined by
(22)zk:=1r∑j=0k‍(-sr)k-j∑i=j-1j‍(-1)j-iλi(λj-λj-1)(i+1)-(s+1-p),∑i=j-1j‍(-1)()j-i121λi(λj-λj-1)(i+1)∑j=0k‍(-sr)k-j∀k∈ℕ.
Then, since Λ^z=x∈ℓs∖ℓp,z∈ℓsλ(B)∖ℓpλ(B) which shows that the inclusion ℓpλ(B)⊂ℓsλ(B) is strict.Theorem 4.
The inclusionsℓpλ(B)⊂c0λ(B)⊂cλ(B)⊂ℓ∞λ(B) strictly hold.Proof.
The inclusionc0λ(B)⊂cλ(B) strictly holds by Theorem 3.1 of [20]. So, it is enough to show that the inclusions ℓpλ(B)⊂c0λ(B) and cλ(B)⊂ℓ∞λ(B) are strict, where 0<p<∞. Assume that x=(xk)∈ℓpλ(B). This means that Λ^x∈ℓp. Since ℓp⊂c0, Λ^x∈c0 which implies that x∈c0λ(B). Hence, the inclusion ℓpλ(B)⊂c0λ(B) holds for 0<p<∞. Now, we show that this inclusion is strict. Let 0<p<∞ and define the sequence x=(xk) as follows:
(23)xk:=1r∑j=0k‍(-sr)k-j∑i=j-1j‍(-1)j-iλi(λj-λj-1)(i+1)1/p1r∑j=0k‍(-sr)k-j∑i=j-1j‍(-1)j-i1r1r1r1r1r∑j=0k‍foreachk∈ℕ.
Then, we have for every n∈ℕ that
(24)(Λ^x)n:=1λn∑k=0n‍(λk-λk-1)(rxk+sxk-1)=1(n+1)1/p,
which shows that Λ^x∉ℓp but Λ^x∈c0. Thus, the sequence x is in c0λ(B) but not in ℓpλ(B). Hence, ℓpλ(B)⊂c0λ(B) is strict.
Sincec⊂ℓ∞ holds, we have the inclusion cλ(B)⊂ℓ∞λ(B). Let us define the sequence y=(yk) by
(25)yk:=1r∑j=0k‍(-sr)k-j∑i=j-1j‍(-1)j-iλiλj-λj-1(-1)i1r∑j=0k‍(-sr)k-j122∑i=j-1j‍λiλj-λj-1foreachk∈ℕ.
Then, one can easily see for every n∈ℕ that
(26)(Λ^y)n:=1λn∑k=0n‍(λk-λk-1)(rxk+sxk-1)=(-1)n,
which shows that Λ^y∈ℓ∞∖c. Thus, y is in ℓ∞λ(B) but not in cλ(B); that is, y∈ℓ∞λ(B)∖cλ(B). That is to say that the inclusion cλ(B)⊂ℓ∞λ(B) is strict. This completes the proof.Theorem 5.
The inclusionℓ∞⊂ℓ∞λ(B) strictly holds.Proof.
Letx=(xk)∈ℓ∞. Then, we have
(27)∥x∥ℓ∞λ(B)=supn∈ℕ|(Λ^x)n|=supn∈ℕ|1λn∑k=0n‍(λk-λk-1)(rxk+sxk-1)|⩽supn∈ℕ1λn∑k=0n‍(λk-λk-1)|sxk-1+rxk|⩽supn∈ℕ1λn∑k=0n‍(λk-λk-1)(|s||xk-1|+|r||xk|)⩽(|r|+|s|)∥x∥∞supn∈ℕ1λn∑k=0n‍(λk-λk-1)=(|r|+|s|)∥x∥∞,
which means that x=(xk)∈ℓ∞λ(B). So, the inclusion ℓ∞⊂ℓ∞λ(B) holds. Furthermore, consider the sequence v=(vk) defined by
(28)vk:=1r∑i=0k‍(-sr)k-i∀k∈ℕwith|-sr|>1.
Clearly, v∉ℓ∞. Then, we obtain by (12) that
(29)(Λ^v)n:=1λn∑k=0n‍(λk-λk-1)=1,∀n∈ℕ,
which shows that Λ^v=e∈ℓ∞. This implies that v∈ℓ∞λ(B)∖ℓ∞. Hence, the inclusion ℓ∞⊂ℓ∞λ(B) is strict and this completes the proof.Theorem 6.
If the inclusionℓp⊂ℓpλ(B) holds, then (1/λn)∈ℓp, where 0<p<∞.Proof.
Assume that the inclusionℓp⊂ℓpλ(B) holds and consider sequence e(0)=(1,0,0,…)∈ℓp. So, by this hypothesis, e(0)=(1,0,0,…)∈ℓpλ(B). Hence, Λ^e(0)∈ℓp. Therefore,
(30)1λn∑k=0n‍(λk-λk-1)(rxk+sxk-1)=1λnλ0r,
and we obtain that
(31)∑n‍|(Λ^e(0))n|p=|λ0r|p∑n‍(1λn)p<∞,
which shows that (1/λn)∈ℓp. This completes the proof.Lemma 7 (see [17, Lemma 4.11, page, 43]).
If(1/λn)∈ℓ1, then
(32)M=supn∈ℕ∑n=k∞‍λk-λk-1λn<∞.Theorem 8.
If(1/λn)∈ℓ1, the inclusion ℓp⊂ℓpλ(B) strictly holds for 1⩽p<∞.Proof.
Letx=(xk)∈ℓp for 1<p<∞. Then, by applying Hölder's inequality, we derive from (12) that
(33)|(Λ^x)n|=|1λn∑k=0n‍(λk-λk-1)(rxk+sxk-1)|⩽∑k=0n‍λk-λk-1λn|sxk-1+rxk|=∑k=0n‍(λk-λk-1λn)1/p×(λk-λk-1λn)p-1/p|sxk-1+rxk|⩽{∑k=0n‍[(λk-λk-1λn)1/p|sxk-1+rxk|]p}1/p×{∑k=0n‍[(λk-λk-1λn)p-1/p]q}1/q=[∑k=0n‍(λk-λk-1λn)|sxk-1+rxk|p]1/p×[∑k=0n‍(λk-λk-1λn)]p-1/p,
which gives that
(34)|(Λ^x)n|p⩽∑k=0n‍(λk-λk-1λn)|sxk-1+rxk|p[∑k=0n‍(λk-λk-1λn)]p-1.
By (34) and Lemma 7, we have
(35)∑n‍|(Λ^x)n|p⩽∑n‍1λn∑k=0n‍(λk-λk-1)|sxk-1+rxk|p=∑k‍|sxk-1+rxk|p∑n=k∞‍λk-λk-1λn⩽M∑k‍|sxk-1+rxk|p.
Therefore, combining the inequality (35) with Minkowski's inequality, we derive that
(36)∥x∥ℓpλ(B)=(∑n‍|(Λ^x)n|p)1/p⩽M1/p|r|(∑k‍|xk|p)1/p+M1/p|s|(∑k‍|xk-1|p)1/p⩽M1/p(|r|+|s|)∥x∥p<∞.
This shows that x∈ℓpλ(B). So, the inclusion ℓp⊂ℓpλ(B) holds. Now, let us consider the sequence v=(vk) defined by
(37)vk:={1r,k=0,-1r(λ0λ1-λ0+sr)(-sr)k-1,k⩾1,
with |-s|>|r|. Then, since Λ^v=e(0)∈ℓp, one can immediately observe that v is in ℓpλ(B) but not in ℓp. That is, v∈ℓpλ(B)∖ℓp. Thus, we have showed that the inclusion ℓp⊂ℓpλ(B) is strict. Similarly, the inclusion ℓp⊂ℓpλ(B) also strictly holds in the case p=1, so we omit the details. This completes the proof.Theorem 9.
The sequence spacesℓ∞ and ℓpλ(B) do not include each other.Proof.
It is clear by Theorem8 that the sequence spaces ℓ∞ and ℓpλ(B) are not disjointed. Let us consider the sequence v=(vk) defined by (37). Then, v is in ℓpλ(B) but not in ℓ∞. Now, let us define the sequence x=(xk):={(1/r)∑i=0k‍(-s/r)k-i}k∈ℕ with |-s/r|<1. Then, since Λ^x=e∉ℓp, x is in ℓ∞ but not in ℓpλ(B). This completes the proof.


## 4. The Basis for the Spaceℓpλ(B)

In this section, we begin with defining the concept of the Schauder basis for a normed sequence space and then give the basis of the sequence spaceℓpλ(B), where 1⩽p<∞. Now, we define the Schauder basis of a normed space. If a normed sequence space μ contains a sequence (bn) with the property that for every x∈μ there is a unique sequence of scalars (αn) such that
(38)limn→∞∥x-(α0b0+α1b1+⋯+αnbn)∥=0,
then (bn) is called a Schauder basis (or briefly basis) for μ. The series ∑k‍αkbk which has the sum x is called the expansion of x with respect to (bn), and written as x=∑k‍αkbk.Theorem 10.
The following statements hold.(i)
The spaceℓ∞λ(B) has no Schauder basis.(ii)
Define the sequenceb(k)={bn(k)}n∈ℕ of elements of the space ℓpλ(B) by
(39)bn(k)={(-sr)n-k[λkr(λk-λk-1)+λks(λk+1-λk)],k<n,λnr(λn-λn-1),k=n,0,k>n,for all n,k∈ℕ. Then, the sequence {bn(k)} is a basis for the space ℓpλ(B) and every x∈ℓpλ(B) has a unique representation of the form
(40)x=∑k‍(Λ^x)kb(k).Proof.
(i) It is known that the matrix domainμA of a normed sequence space μ has a basis if and only if μ has a basis whenever A=(ank) is a triangle [21, Remark 2.4]. Since the space ℓ∞ has no Schauder basis, ℓ∞λ(B) has no Schauder basis.
(ii) Let1⩽p<∞. It is clear that b(k)={bn(k)}⊂ℓpλ(B), since Λ^b(k)=e(k)∈ℓp for all k∈ℕ. Furthermore, let x∈ℓpλ(B) be given. For every nonnegative integer m, we put
(41)x[m]=∑k=0m‍(Λ^x)kb(k).
Then, by applying Λ^ to (41), we get that
(42)Λ^x[m]=∑k=0m‍(Λ^x)kΛ^b(k)=∑k=0m‍(Λ^x)ke(k),
and therefore, we have
(43){Λ^(x-x[m])}n={0,0⩽n⩽m,(Λ^x)n,n>m,
for all n,m∈ℕ. Now, for any given ϵ>0, there is a nonnegative integer m0 such that
(44)∑n=m0+1∞‍|(Λ^x)n|p<(ϵ2)p.
Thus, we have for every m⩾m0 that
(45)∥x-x[m]∥ℓpλ(B)=(∑n=m+1∞‍|(Λ^x)n|p)1/p⩽(∑n=m0+1∞‍|(Λ^x)n|p)1/p⩽ϵ2<ϵ,
for all m⩾m0 which proves that x∈ℓpλ(B) is represented as in (40).
Let us show that the uniqueness of representation forx∈ℓpλ(B) is given by (40). Suppose, on the contrary, that there exists a representation x=∑k‍μk(x)b(k). Since the transformation T defined from ℓpλ(B) to ℓp by Tx=Λ^x=y is continuous, we have
(46)(Λ^x)k=∑n‍μn{Λ^b(k)}n=∑n‍μn(x)δnk=αk(x);k∈ℕ,
which contradicts the assumption that (Λ^x)k≠αk(x) for all k∈ℕ. That is to say that the representation (40) of x∈ℓpλ(B) is unique.


## 5. The Alpha-, Beta- and Gamma-Duals of the Spaceℓpλ(B) and ℓ∞λ(B)

In this section, we give some theorems determining the alpha-, beta- and gamma-duals of the spacesℓpλ(B) and ℓ∞λ(B). We start with the definition of the alpha-, beta- and gamma-duals of a sequence space.Ifx and y are sequences and X and Y are subsets of ω, then we write x·y=(xkyk)k=0∞,x-1*Y={a∈ω:a·x∈Y}, and
(47)x·y=(xkyk)k=0∞,x-1*Y={a∈ω:a·x∈Y},M(X,Y)=⋂x∈X‍x-1*Y={a:a·x∈Y∀x∈X},
for the multiplier space of X and Y. One can easily observe for a sequence space Z with Y⊂Z and Z⊂X that the inclusions M(X,Y)⊂M(X,Z) and M(X,Y)⊂M(Z,Y) hold. The alpha-, beta- and gamma-duals of a sequence space, which are, respectively, denoted by Xα, Xβ, and Xγ are defined by
(48)Xα=M(X,ℓ1),Xβ=M(X,cs),Xγ=M(X,bs).
It is obvious that Xα⊂Xβ⊂Xγ. Also, it can be easily seen that the inclusions Xα⊂Yα, Xβ⊂Yβ, and Xγ⊂Yγ hold, whenever Y⊂X. Now, we may begin with quoting the following lemmas [22] which are needed in proving Theorems 14–16.Lemma 11.
A
=
(
a
n
k
)
∈
(
ℓ
p
:
ℓ
1
) if and only if (i)
For1<p⩽∞(49)supF∈ℱ∑k‍|∑n∈F‍ank|q<∞.(ii)
Forp=1(50)supk∈ℕ∑n‍|ank|<∞.Lemma 12.
LetA=(ank) be an infinite matrix. Then, the following statements hold.  (i)
Let1<p<∞. Then, A∈(ℓp:c) if and only if
(51)limn→∞ankexistsforeachfixedk∈ℕ,(52)supn∈ℕ∑k‍|ank|q<∞.(ii)
A∈(ℓ1:c) if and only if (51) holds and
(53)supn,k∈ℕ|ank|<∞.(iii)
A∈(ℓ∞:c) if and only if (51) holds and
(54)supn∈ℕ∑k‍|ank|<∞,limn→∞∑k‍|ank-limn→∞ank|=0.Lemma 13.
LetA=(ank) be an infinite matrix. Then, the following statements hold.  (i)
Let1<p⩽∞. Then, A∈(ℓp:ℓ∞) if and only if (52) holds(ii)
A∈(ℓ1:ℓ∞) if and only if (53) holds.Theorem 14.
Define the setstqλ and t∞λ by
(55)tqλ={a=(ak)∈ω:supF∈ℱ∑k‍|∑n∈F‍tnkλ|q<∞},t∞λ={a=(ak)∈ω:supk∈ℕ∑n‍|tnkλ|<∞}.
Then, [ℓ1λ(B)]α=t∞λ and [ℓpλ(B)]α=tqλ for 1<p⩽∞, where the matrix T=(tnkλ) is defined via the sequence a=(an)∈w by
(56)tnkλ={(-sr)n-k[λkr(λk-λk-1)+λks(λk+1-λk)]an,(-sr)n-k(-sr)n-k1212(-sr)n-k0⩽k⩽n-1,λnr(λn-λn-1)an,(-sr11111)n-kk=n,0,(-sr)n-k(-sr)n-k1111111121k>n,
for all n,k∈ℕ.Proof.
Leta=(an)∈w and 1<p<∞. Then, by using (12), we immediately derive for every n∈ℕ that
(57)anxn=1r∑k=0n‍(-sr)n-k∑i=k-1k‍(-1)k-iλiλk-λk-1yian=(Ty)n.
Thus, we observe by (57) that ax=(anxn)∈ℓ1 whenever x=(xk)∈ℓpλ(B) if and only if Ty∈ℓ1 whenever y=(yk)∈ℓp. This means that a=(ak)∈[ℓpλ(B)]α if and only if T∈(ℓp:ℓ1). Therefore, we get by Lemma 11 with T instead of A that a=(ak)∈[ℓpλ(B)]α if and only if
(58)supF∈ℱ∑k‍|∑n∈F‍tnkλ|q<∞,
which leads us to the consequence that [ℓpλ(B)]α=tqλ, for 1<p⩽∞. Similarly, we get from (57) that a=(ak)∈[ℓ1λ(B)]α if and only if T∈(ℓ1:ℓ1) which is equivalent to (50) of Lemma 11 that
(59)supk∈ℕ∑n‍|tnkλ|<∞.Theorem 15.
Define the setsd1λ, d2λ, d3λ, d4λ, and eqλ as follows:
(60)d1λ={∑j=k+1∞‍(-sr)k-jaja=(ak)∈ω:a∈∑j=k+1∞‍(-sr)k-jajexistsforeachk∈ℕ(-sr)k-j},eqλ={a=(ak)∈ω:supn∈ℕ∑k=0n-1‍|a~k(n)|q<∞},d2λ={a=(ak)∈ω:supk,n∈ℕ|a~k(n)|<∞},d3λ={a=(ak)∈ω:limn→∞∑k‍|a~k(n)|=∑k‍|a~k|},d4λ={a=(ak)∈ω:supk∈ℕ|λkr(λk-λk-1)ak|<∞},
where
(61)a~k(n)=λk{∑j=k+1n‍(-sr)k-jajakr(λk-λk-1)a~k(n)λka~ka~ka~k+[1r(λk-λk-1)+1s(λk+1-λk)]∑j=k+1n‍ajajaj×∑j=k+1n‍(-sr)k-jaj}fork<n,a~k=limn→∞a~k(n).
Then,
(62)[ℓpλ(B)]β={e1λ∩d4λ∩d3λ,p=∞,eqλ∩d1λ∩d4λ,1<p<∞,d1λ∩d2λ∩d4λ,p=1.Proof.
Let us consider the equality(63)∑k=0n‍akxk=∑k=0n‍{1r∑j=0k‍(-sr)k-j[∑i=j-1j‍(-1)j-iλiλj-λj-1yi]}ak=∑k=0n-1‍λk[∑j=k+1n‍(-sr)k-jaj(1r(λk-λk-1)+1s(λk+1-λk))∑k=0n-1‍∑k=0n-1‍×∑j=k+1n‍(-sr)k-jaj]yk+anr(λn-λn-1)yn=∑k=0n-1‍a~k(n)yk+anr(λn-λn-1)yn=(Dy)n∀n∈ℕ,
where the matrix D=(dnkλ) is defined for all n,k∈ℕ by
(64)dnkλ={a~k(n),0⩽k⩽n-1,λnr(λn-λn-1)an,k=n,0,k>n.
Then, we deduce from (63) with Lemma 12 that ax=(anxn)∈cs whenever x=(xk)∈ℓpλ(B) if and only if Dy∈c whenever y=(yk)∈ℓp. This means that a=(ak)∈[ℓpλ(B)]β if and only if D∈(ℓp:c), where 1⩽p⩽∞. Therefore, we derive from (51) and (52) that
(65)∑j=k+1∞‍(-sr)k-jajexistsforeachk∈ℕ,supn∈ℕ∑kn-1‍|a~k(n)|q<∞,supk∈ℕ|λkr(λk-λk-1)ak|<∞,
which shows that [ℓpλ(B)]β=eqλ∩d1λ∩d4λ for 1<p<∞. Since beta-dual of the space ℓpλ(B) for the cases p=1 and p=∞ can be similarly computed, we omit the details. This completes the proof.Theorem 16.
Let1<p⩽∞. Then, [ℓpλ(B)]γ={d2λ∩d4λ,p=1,e1λ∩d4λ,p>1.Proof.
This may be obtained in the similar way used in the proof of Theorem15 with Lemma 13 instead of Lemma 12. So, we omit the details.


## 6. Certain Matrix Mapping Related to the Spacesℓpλ(B) and ℓ∞λ(B)

In this section, we characterize the matrix classes(ℓpλ(B):ℓ∞), (ℓpλ(B):c0), (ℓpλ(B):c), (ℓpλ(B):ℓ1), (ℓ1λ(B):ℓp), (ℓ∞λ(B):ℓp), where 1⩽p⩽∞. Also, by means of a given basic lemma, we derive the characterizations of certain other classes. Since the characterization of matrix mapping on the space ℓpλ(B) can be proved in a similar way, we omit the proof for the cases p=1 and p=∞ and consider only the case 1<p<∞ in the proofs of theorems given in this section.For an infinite matrixA=(ank), we write for brevity that
(66)a~nk(m)=λk{∑j=k+1m‍(-sr)k-janjankr(λk-λk-1)+[1r(λk-λk-1)+1s(λk+1-λk)]ankr(λk-λk-1)+×∑j=k+1m‍(-sr)k-janj}ifk<m,a~nk=λk{∑j=k+1∞‍(-sr)k-janjankr(λk-λk-1)+[1r(λk-λk-1)+1s(λk+1-λk)]ankr(λk-λk-1)+×∑j=k+1∞‍(-sr)k-janj},a(n,k,m)=1m+1∑j=0m‍an+j,k∀n,k,m∈ℕ.
Now, we may begin with quoting the following lemmas (see [22]) which are needed for proving our main results.Lemma 17.
LetA=(ank) be an infinite matrix. Then, the following statements hold.  (i)
A∈(ℓ1:c0) if and only if supn,k∈ℕ|ank|<∞ and
(67)limn→∞ank=0foreachfixedk∈ℕ.(ii)
Let1<p<∞. Then, A∈(ℓp:c0) if and only if (67) holds and supk∈ℕ∑n‍|ank|q<∞.(iii)
A∈(ℓ∞:c0) if and only if (67) holds and ∑k‍|ank| is uniformly convergent.Lemma 18.
Let1⩽p<∞. Then, A=(ank)∈(ℓ1:ℓp) if and only if supn∈ℕ∑k‍|ank|q<∞.Lemma 19.
Let1<p<∞. Then, A=(ank)∈(ℓ∞:ℓp) if and only if supK∈ℱ∑n‍|∑k∈K‍ank|p<∞.Theorem 20.
LetA=(ank) be an infinite matrix. Then, the following statements hold.(i)
Let1<p<∞. Then, A∈(ℓpλ(B):ℓ∞) if and only if
(68)∑j=k+1∞‍(-sr)k-jajexistsforeachfixedk∈ℕ,(69){λkr(λk-λk-1)ank}∈ℓ∞foreveryn∈ℕ,(70)supk∈ℕ∑n‍|a~nk|q<∞.(71)(ank)k∈ℕ∈eqλ∩d1λ∩d4λforeachn∈ℕ.(ii)
A∈(ℓ1λ(B):ℓ∞) if and only if (68) and (69) hold and
(72)supn,k∈ℕ|a~nk|<∞.(iii)
A∈(ℓ∞λ(B):ℓ∞) if and only if (68) and (69) hold and
(73)supk∈ℕ∑n‍|a~nk|<∞,(74)limm→∞∑n‍|a~nk(m)|=∑n‍|a~nk|.Proof.
(i) Assume that the conditions (68)–(71) hold and take any x∈ℓpλ(B), where 1<p<∞. Then, we have by Theorem 16 that (ank)k∈ℕ∈[ℓpλ(B)]β for all n∈ℕ and this implies the existence of Ax. Also, it is clear that the associated sequence y=(yk) is in the space ℓp⊂c0.
Let us now consider the following equality derived by using relation (12) from mth partial sum of the series ∑k‍ankxk as follows:
(75)∑k=0m‍ankxk=∑k=0m-1‍a~nk(m)yk+λmr(λm-λm-1)anmym∑k=0m‍ankxk=∑km‍a~nk(m)yk(m)ykxk∀n,m∈ℕ.
Therefore, by using (68)–(70), we obtain from (75) as m→∞ that
(76)∑k‍ankxk=∑k‍a~nkyk∀n∈ℕ.
Furthermore, since the matrix A~=(a~nk) is in the class (ℓp:ℓ∞) by Lemma 13, we have A~y∈ℓ∞. Now, by passing to supremum over n in (76), we derive by applying Hölder's inequality that
(77)∥Ax∥∞=supn∈ℕ|∑k‍ankxk|⩽supn∈ℕ(∑k‍|a~nk|q)1/q(∑k‍|yk|p)1/p<∞,
which shows that Ax∈ℓ∞ and hence A∈(ℓpλ(B):ℓ∞).
Conversely, assume thatA=(ank)∈(ℓpλ(B):ℓ∞), where 1<p<∞. Then, since (ank)k∈ℕ∈[ℓpλ(B)]β for all n∈ℕ by the hypothesis, the necessity of (71) is obvious. Since (ank)k∈ℕ∈[ℓpλ(B)]β, (76) holds for all sequences x∈ℓpλ(B) and y∈ℓp which are connected by relation (12). Let us now consider the continuous linear functionals fn on ℓpλ(B) by
(78)fn(x)=∑k‍ankxk∀n∈ℕ.
Then, since ℓpλ(B) and ℓp are norm isomorphic, it should follow with (76) that
(79)∥fn∥=∥A~n∥q=(∑k‍|a~nk|q)1/q,
for all n∈ℕ. This shows that the functionals defined by the rows of A on ℓpλ(B) are pointwise bounded. Thus, we deduce by Banach-Steinhaus theorem that these functionals are uniformly bounded, which yields that there exists a constant K>0 such that ∥fn∥⩽K for all n∈ℕ. This shows the necessity of the condition (70) which completes the proof of part (i).Theorem 21.
LetA=(ank) be an infinite matrix. Then, the following statements hold.  (i)
A∈(ℓ1λ(B):c) if and only if (68) and (69) hold and
(80)limn→∞a~nk=αkforeachk∈ℕ.(ii)
Let1<p<∞. Then, A∈(ℓpλ(B):c) if and only if (68)–(71) hold and (80) also holds.(iii)
A∈(ℓ∞λ(B):c) if and only if (68), (69), and (74) hold, and
(81)limn→∞∑k‍|a~nk-αk|=0.Proof.
We consider only part (ii). Assume thatA satisfies the conditions (68)–(71) and (80), and x∈ℓpλ(B), where 1<p<∞. Then, Ax exists and by using (80), we have for every k∈ℕ that |a~nk|q→|αk|q as n→∞ which leads us with (70) to the following inequality:
(82)∑j=0k‍|αj|q⩽∑j=0k‍|a~nj|q=M<∞,
which holds for every k∈ℕ. This shows that (αk)∈ℓq. Since x∈ℓpλ(B), we have y∈ℓp. Therefore, we derive by applying Hölder's inequality that (αkyk)∈ℓ1 for each y∈ℓp.
Now, for any givenϵ>0, choose a fixed k0∈ℕ such that
(83)(∑k=k0+1k‍|yk|p)1/p⩽ϵ4M1/q.
Then, it follows (80) that there is m0∈ℕ such that
(84)|∑k=0k0‍(a~nk-αk)yk|⩽ϵ2,
for every m⩾m0. Thus, by using (76), we get that
(85)|∑k‍ankxk-∑k‍αkyk|=|∑k‍(a~nk-αk)yk|⩽|∑k=0k0‍(a~nk-αk)yk|+|∑k=k0+1∞‍(a~nk-αk)yk|<ϵ2+[∑k=k0∞‍(|a~nk|+|αk|)q]1/q[∑k=k0∞‍|yk|p]1/p<ϵ2+ϵ4M1/q[(∑k=k0∞‍(|a~nk|)q)1/q+(∑k=k0∞‍|αk|p)1/p]<ϵ2+ϵ4M1/q2M1/q=ϵ,
for all sufficiently large m⩾m0. Hence, (Ax)n→∑k‍αkyk as n→∞ which means that Ax∈c; that is, A=(ank)∈(ℓpλ(B):c).
Conversely, suppose thatA∈(ℓpλ(B):c), where 1<p<∞. Then, since c⊂ℓ∞, A∈(ℓpλ(B):ℓ∞). Thus, the necessity of (68)–(71) is immediately obtained from Theorem 20 which together imply that (76) holds for all sequences x∈ℓpλ(B). Since Ax∈c by our assumption, we derive by (76) that A~y∈c which means that A~=(a~nk)∈(ℓp:c). Thus the necessity of (80) is immediate by (51) of Lemma 12. This completes the proof of part (ii).Now, we can mention the sequence spacef of almost convergent sequences. The shift operator P is defined on ω by (Px)n=xn+1 for all n∈ℕ. A Banach limit L is defined on ℓ∞ such that  (i)
L(x)⩾0 for x=(xk), where xk⩾0 for all k∈ℕ,(ii)
L(Px)=L(x),(iii)
L(e)=1, where e=(1,1,1,etc.).A sequence x=(xk)∈ℓ∞ is said to be almost convergent to the generalized limit α if all Banach limits of x is α [23], and denoted by f-limxk=α. Let Pj be the composition of P with itself for j times and define tmn(x) for a sequence x=(xk) by
(86)tmn(x):=1m+1∑j=0m‍(Pjx)n∀m,n∈ℕ.
Lorentz [23] proved that f-limxk=α if and only if limm→∞tmn(x)=α, uniformly in n. It is well-known that a convergent sequence is almost convergent such that its ordinary and generalized limits are equal. By f and fs, we denote the space of all almost convergent sequences and series, respectively, that is,
(87)f={∃α∈ℂ∋limm→∞∑k=0m‍xn+km+1x=(xk)∈ω:(xk)∃α∈ℂ∋limm→∞∑k=0m‍xn+km+1=αuniformlyinn∑k=0m‍xn+km+1},fs={∃α∈ℂ∋limm→∞∑k=0m‍∑j=0n+k‍xjm+1x=(xk)∈ω:(xk)∃α∈ℂ∋limm→∞∑k=0m‍∑j=0n+k‍xjm+1=αuniformlyinn∑k=0m‍∑j=0n+k‍xjm+1}.Theorem 22.
LetA=(ank) be an infinite matrix. Then, the following statements hold.(i)
A∈(ℓ1λ(B):f) if and only if (68) and (69) hold, and
(88)f-limn→∞a~nk=αkforeveryk∈ℕ.(ii)
Let1<p<∞. Then, A∈(ℓpλ(B):f) if and only if (68)–(71) and (88) hold.(iii)
A∈(ℓ∞λ(B):f) if and only if (68) and (69) hold, and
(89)limm→∞∑n‍|a~(n,k,m)-αk|=0uniformlyn.Proof.
Theorem22 can be similarly proved by the same technique used in the proof of Theorem 21.Theorem 23.
LetA=(ank) be an infinite matrix. Then, the following statements hold.  (i)
A=(ank)∈(ℓ1λ(B):c0) if and only if (68) and (69) hold, and
(90)limn→∞a~nk=0foreveryk∈ℕ.(ii)
Let1<p<∞. Then, A=(ank)∈(ℓpλ(B):c0) if and only if (68)–(71) and (90) hold.(iii)
A=(ank)∈(ℓ∞λ(B):c0) if and only if (68), (69), and (74) hold and
(91)limn→∞∑k‍|a~nk|=0.Proof.
It is natural that Theorem23 can be proved by the same technique used in the proof of Theorem 21 with Lemma 12 instead of Lemma 17 and so we omit the proof.Theorem 24.
LetA=(ank) be an infinite matrix. Then, the following statements hold.  (i)
A∈(ℓ1λ(B):ℓ1) if and only if (68), (69), and (72) hold and
(92)supk∈ℕ∑n‍|a~nk|<∞.(ii)
Let1<p<∞. Then, A∈(ℓpλ(B):ℓ1) if and only if (68)–(71) hold and
(93)supF∈ℱ∑k‍|∑n∈F‍a~nk|q<∞.(iii)
A∈(ℓ∞λ(B):ℓ1) if and only if (68), (69), and (74) hold and
(94)supF∈ℱ∑k‍|∑n∈F‍a~nk|<∞.Proof.
Since Parts (i) and (iii) can be proved in a similar way, to avoid the repetition of the similar statements, we consider only part (ii).
Suppose thatA satisfies the conditions (68)–(71), (93) and take any x∈ℓpλ(B), where 1<p<∞, then y∈ℓp. We have by Theorem 16 that (ank)k∈ℕ∈[ℓpλ(B)]β for all n∈ℕ and this implies that Ax exists. Besides, it follows by combining (93) and Lemma 11 that the matrix A~∈(ℓp:ℓ1) and so we have A~y∈ℓ1. Additionally, we derive from (68)–(71) that the relation (76) holds which yields that Ax∈ℓ1 and so we have A∈(ℓpλ(B):ℓ1).
Conversely, assume thatA∈(ℓpλ(B):ℓ1), where 1<p<∞. Since ℓ1⊂ℓ∞, A∈(ℓpλ(B):ℓ∞). Thus, Theorem 20 implies the necessity of (68)–(71) which imply the relation (76). Since Ax∈ℓ1 by the hypothesis, we deduce by (76) that A~y∈ℓ1 which means that A~∈(ℓp:ℓ1). Now, the necessity of (93) is immediate by the condition (49) of Lemma 11. This completes the proof of part (ii).Theorem 25.
Let1⩽p<∞. Then, A=(ank)∈(ℓ1λ(B):ℓp) if and only if (68) and (69) hold, and
(95)supk∈ℕ∑n‍|a~nk|p<∞.Proof.
Suppose that the conditions (68), (69), and (95) hold and take x∈ℓ1λ(B). Then, y∈ℓ1. We have by Theorem 16 that (ank)k∈ℕ∈[ℓ1λ(B)]β for each n∈ℕ and this implies that Ax exists. Furthermore, by (95), one can obtain that
(96)supk∈ℕ|a~nk|⩽supk∈ℕ(∑n‍|a~nk|p)1/p<∞foreveryn∈ℕ.
Hence, the series ∑n‍|a~nk| absolutely converges for each fixed n∈ℕ. Therefore; since (68) and (69) hold, if we let to limit in (75) as m→∞, the relation (76) holds. Thus, by applying Minkowski's inequality and using (76) and (95), we obtain
(97)(∑n‍|∑k‍ankxk|p)1/p=(∑n‍|∑k‍a~nkyk|p)1/p⩽∑k‍yk(∑n‍|a~nk|p)1/p<∞,
which means that Ax∈ℓp and so A∈(ℓ1λ(B):ℓp).
Conversely, assume thatA∈(ℓ1λ(B):ℓp), where 1⩽p<∞. Since ℓp⊂ℓ∞, then A∈(ℓ1λ(B):ℓ∞). Thus, Theorem 20 implies that the necessity of (68) and (69) is clear by the relation (76). Since Ax∈ℓp by our assumption, we deduce by (76) that A~y∈ℓp which means that A~∈(ℓ1:ℓp). Now, the necessity of (95) is immediate by Lemma 18. This completes the proof.Theorem 26.
Let1<p<∞. Then, A=(ank)∈(ℓ∞λ(B):ℓp) if and only if (68) and (69) hold and
(98)∑k‍|a~nk|convergesforeveryn∈ℕ,supK∈ℱ∑n‍|∑k∈K‍a~nk|p<∞.Proof.
Theorem26 can be proved by the same technique used in the proof of Theorem 25 with Lemma 19 instead of Lemma 18 and so we omit the details.Lemma 27 (see [8, Lemma 5.3]).
Letλ and μ be any two sequence spaces, let A be an infinite matrix and B a triangle matrix. Then, A∈(λ:μB) if and only if BA∈(λ:μ).It is trivial that Lemma27 has several consequences. Indeed, combining Lemma 27 with Theorems 20–26, one can derive the following results.Corollary 28.
LetA=(ank) be an infinite matrix and u=(un) and v=(vn) be sequences of non-zero numbers, and define the matrix C=(cnk) by cnk=un∑j=0n‍vjajk for all n,k∈ℕ. Then, the necessary and sufficient conditions in order A belongs to any of the classes (ℓpλ(B):ℓ∞(u,v)), (ℓpλ(B):ℓp(u,v)), and (ℓpλ(B):c(u,v)) are obtained from respective ones in Theorems 20–26 by replacing the entries of the matrix A by those of the matrix C. The spaces ℓ∞(u,v), ℓp(u,v), and c(u,v) are defined in [9] as the spaces of all sequences whose generalized weighted means are in the spaces ℓ∞, c, and ℓp. Since the spaces ℓ∞(u,v), c(u,v), and ℓp(u,v) can be reduce in the cases vk=rk, un=1/Rn and vk=1, un=1/n to the Riesz sequence spaces r∞t, rct, and rpt and to the Cesàro sequence spaces X∞, c~, and Xp, respectively, Corollary 28 also includes the characterizations of classes (ℓpλ(B):r∞t), (ℓpλ(B):rpt), (ℓpλ(B):rct) and (ℓpλ(B):X∞), (ℓpλ(B):Xp) and (ℓpλ(B):c~), where 1⩽p⩽∞.Corollary 29.
LetA=(ank) be an infinite matrix and define the matrix C=(cnk) by
(99)cnk=∑j=0n‍(nj)(1-r)n-jr-jajk∀n,k∈ℕ.
Then, the necessary and sufficient conditions in order A which belongs to any of the classes (ℓpλ(B):e∞r), (ℓpλ(B):e0r), (ℓpλ(B):epr), and (ℓpλ(B):ecr) are obtained from respective ones in Theorems 20–26 by replacing the entries of the matrix A by those of the matrix C; where e∞r, epr and ecr, and e0r denote the Euler spaces of all sequences whose Er-transforms are in the spaces ℓ∞, ℓp and c, and c0 which were introduced in [6, 12], where 1⩽p<∞.


## 7. Some Geometric Properties of the Spaceℓpλ(B)

In the present section, we investigate some geometric properties of the spaceℓpλ(B). First, we define some geometric properties of the spaces. Let (X,∥·∥) be a normed space and let S(x) and B(x) be the unit sphere and unit ball of X, respectively. Consider Clarkson's modulus of convexity (see [24, 25]) defined by
(100)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 [26], Gurarii's modulus of convexity is defined by
(101)βX(ε)=inf{1-infα∈[0,1]∥αx+(1-α)y∥;=inf11infx,y∈S(x),∥x-y∥=εinfα∈[0,1]∥αx+(1-α)y∥},
where 0⩽ε⩽2. It is easily shown that δX(ε)⩽βX(ε)⩽2δX(ε) for any 0⩽ε⩽2. Further, 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 sequence (zn) such that the sequence {tk(z)} is convergent in the norm in X [27], where
(102)tk(z)=1k+1(z0+z1+⋯+zk)∀k∈ℕ.A Banach spaceX is said to have the weak Banach-Saks property whenever given any weakly null sequence (xn) in X and there exists a subsequence (zn) of (xn) such that the sequence {tk(z)} is strongly convergent to zero.In [28], García-Falset introduced the following coefficient:
(103)R(X)=sup{liminfn→∞∥xn+x∥;(xn)⊂B(x),xn⟶w0}.Remark 30 (see [29]).
A Banach spaceX with R(X)<2has a weak fixed point property.Theorem 31.
The spaceℓpλ(B) has Banach-Saks 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(ℓpλ(B)). Set u0=x0 and u1=xn1=x1. Then, there exists t1∈ℕ such that
(104)∥∑i=t1+1∞‍u1(i)e(i)∥ℓpλ(B)<ε1.
The assumption “(xn) is a weakly null sequence” implies that xn→0 with respect to the coordinatewise, there exists n2∈ℕ such that
(105)∥∑i=0t1‍xn(i)e(i)∥ℓpλ(B)<ε1,
where n⩾n2. Set u2=xn2. Then, there exists t2>t1 such that
(106)∥∑i=t2+1∞‍u2(i)e(i)∥ℓpλ(B)<ε2.
By using the fact that xn→0 with respect to the coordinatewise, there exists n3>n2 such that
(107)∥∑i=0t2‍xn(i)e(i)∥ℓpλ(B)<ε2,
where n⩾n3. If we continue this process, we can find two increasing sequences (ti) and (ni) of natural numbers such that
(108)∥∑i=0tj‍xn(i)e(i)∥ℓpλ(B)<εj,
for each n⩾nj+1 and
(109)∥∑i=tj+1∞‍uj(i)e(i)∥ℓpλ(B)<εj,
where uj=xnj. Hence,
(110)∥∑j=0n‍uj∥ℓpλ(B)=∥∑j=0n‍(∑i=0tj-1‍uj(i)e(i)∥∑j=0n‍uj∥+∑i=tj-1tjuj(i)e(i)+∑i=tj+1∞uj(i)e(i))∥ℓpλ(B)⩽∥∑j=0n‍∑i=0tj-1‍uj(i)e(i)∥ℓpλ(B)+∥∑j=0n‍∑i=tj-1tj‍uj(i)e(i)∥ℓpλ(B)+∥∑j=0n‍∑i=tj+1∞‍uj(i)e(i)∥ℓpλ(B)⩽∥∑j=0n‍(∑i=tj-1+1tj‍uj(i)e(i))∥ℓpλ(B)+2∑j=0n‍εj.
On the other hand, one can see that ∥x∥ℓpλ(B)<1. Thus, ∥x∥ℓpλ(B)p<1, and we have
(111)∥∑j=0n‍∑i=tj-1+1tj‍uj(i)e(i)∥ℓpλ(B)p=∑j=0n‍∑i=tj-1tj‍|∑k=0i-1‍a~ikxj(k)+λir(λi-λi-1)xi(k)|p+∑j=0n‍∑i=0∞‍|∑k=0i‍a~ikxj(k)+λir(λi-λi-1)xi(k)|p⩽(n+1).
Therefore, we obtain
(112)∥∑j=0n‍∑i=tj-1+1tj‍uj(i)e(i)∥ℓpλ(B)⩽(n+1)1/p.
By using that fact that 1⩽(n+1)1/p for all n∈ℕ and 1⩽p<∞, we have
(113)∥∑j=0n‍uj∥ℓpλ(B)⩽(n+1)1/p+1⩽2(n+1)1/p.
Therefore, the space ℓpλ(B) has Banach-Saks type p.Remark 32.
Note thatR(ℓpλ(B))=R(ℓp)=21/p, since ℓpλ(B) is linearly isomorphic to ℓp.
Thus, by Remarks30 and 32, we have the following.Corollary 33.
Let1<p<∞. Then, the sequence space ℓpλ(B) has the weak fixed point property.Theorem 34.
Gurarii's modulus of convexity for the normed spaceℓpλ(B) is
(114)βℓpλ(B)(ε)⩽1-[1-(ε2)p]1/p,
where 0⩽ε⩽2.Proof.
Letx∈ℓpλ(B). Then, we have
(115)∥x∥ℓpλ(B)=∥Λ^x∥p=[∑n‍|(Λ^x)n|p]1/p.
Let 0⩽ε⩽2 and consider the following sequences:
(116)z=(zn)={Λ^-1[[1-(ε2)p]1/p],Λ^-1(ε2),0,0,…},t=(tn)={Λ^-1[[1-(ε2)p]1/p],Λ^-1(-ε2),0,0,…}.
Since un=(Λ^z)n and vn=(Λ^t)n, one can see that
(117)u=(un)={[1-(ε2)p]1/p,ε2,0,0,…},v=(vn)={[1-(ε2)p]1/p,-ε2,0,0,…}.
By using the sequences z=(zn) and t=(tn), we obtain the following equalities:
(118)∥z∥ℓpλ(B)p=∥Λ^z∥p=|[(1-ε2)p]p|1/p+|ε2|p=1-(ε2)p+(ε2)p=1,∥t∥ℓpλ(B)p=∥Λ^t∥p=|[(1-ε2)p]p|1/p+|-ε2|p=1-(ε2)p+(ε2)p=1,∥z-t∥ℓpλ(B)p=∥Λ^z-Λ^t∥p={|[1-(ε2)p]1/p-[1-(ε2)p]1/p|p+|ε2-(-ε2)|p[1-(ε2)p]1/p}1/p=ε.
For 0⩽α⩽1(119)infα∈[0,1]∥αz+(1-α)t∥ℓpλ(B)=infα∈[0,1]∥αΛ^z+(1-α)Λ^t∥p=infα∈[0,1]{|α[1-(ε2)p]1/p+(1-α)[1-(ε2)p]1/p|pinfα∈[0,1]12infα∈[0,1]+|αε2+(1-α)(-ε2)|p[1-(ε2)p]1/p}1/p=infα∈[0,1][1-(ε2)p+(2α-1)p(ε2)p]1/p=[1-(ε2)p]1/p.
Therefore, for1⩽p<∞, we have
(120)βℓpλ(B)(ε)⩽1-[1-(ε2)p]1/p.
This step concludes the proof.Corollary 35.
The following statements hold.(i)
Forε>2, βℓpλ(B)(ε)=1. Thus, ℓpλ(B) is strictly convex.(ii)
For0<ε⩽2, βℓpλ(B)(ε)⩽1. Thus, ℓpλ(B) is uniformly convex.Corollary 36.
Forα=1/2, βℓpλ(B)(ε)=δℓpλ(B)(ε).


## 8. Conclusion

The domain of Euler meansEr of order r, the method Ar, and the generalized difference matrix B(r,s) in the sequence spaces ℓp and ℓ∞ investigated by Altay et al. [6], Aydın and Başar [7], and Kirişçi and Başar [30], respectively. Since Λ^ is the composition of Λ and B(r,s), our corresponding results are much more general than the results given by Kirişçi and Başar [30]. Additionally, we emphasize on some geometric properties of the new space ℓpλ(B). It is obvious that the matrix Λ^ is not comparable with the matrices Er, Ar, or B(r,s). So, the present results are new. As a natural continuation of this paper, one can study the domain of the matrix Λ^ in Maddox's spaces ℓ∞(p), c(p), c0(p), and ℓ(p).



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