# The Hahn Sequence Space Defined by the Cesáro Mean

Authors: Murat Kirişci
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

TheBK-space of all sequences is given as x=(xk) such that Σk=1∞k|xk−xk+1| converges and xk is a null sequence which is called the Hahn sequence space and is denoted by h. Hahn (1922) defined the h space and gave some general properties. G. Goes and S. Goes (1970) studied the functional analytic properties of this space. The study of Hahn sequence space was initiated by Chandrasekhara Rao (1990) with certain specific purpose in the Banach space theory. In this paper, the matrix domain of the Hahn sequence space determined  by the Cesáro mean first order, denoted by C, is obtained,
and some inclusion relations and some topological properties of this space are investigated. Also dual spaces of this space are computed and, matrix transformations are characterized.


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## Body

## 1. Introduction

By asequence space, we understand a linear subspace of the space ω=ℂℕ of all complex sequences which contains ϕ, the set of all finitely nonzero sequences, where ℂ denotes the complex field and ℕ={0,1,2,…}. We write ℓ∞, c, and c0 for the classical spaces of all bounded, convergent, and null sequences, respectively. Also by bs, cs, ℓ1, and ℓp, we denote the space of all bounded, convergent, absolutely, and p-absolutely convergent series, respectively. Additionally, the spaces bv, bv(C), σ∞, ∫‍λ, and ℓ1(C) are defined by
(1)bv={x=(xk)∈ω:∑k=1∞|xk-xk-1|<∞},bv(C)={x=(xk)∈ω:∑k=1∞‍|xkk+1-1k.(k+1)∑j=0k-1‍xj|<∞},σ∞={x=(xk)∈ω:supnn-1|∑k=1n‍xk|<∞},∫λ={x=(xk)∈ω:(kxk)∈λ},ℓ1(C)={x=(xk)∈ω:∑k=1∞‍|1k+1∑j=0k‍xj|<∞}.A coordinate space (or a K-space) is a vector space of numerical sequences, where addition and scalar multiplication are defined pointwise. That is, a sequence space λ 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. A BK-space is a K-space, which is also a Banach space with continuous coordinate functionals fk(x)=xk, for all k∈ℕ. If a normed sequence space λ contains a sequence (bn) with the property that for every x∈λ there is a unique sequence of scalars (αn) such that
(2)limn→∞∥x-(α0b0+α1b1+⋯+αnbn)∥=0,
then (bn) is called the Schauder basis (or briefly basis) for λ. The series ∑‍αkbk which has the sum x is then called the expansion of x with respect to (bn), and it is written as x=∑‍αkbk. An FK-space  λ is said to have AK property, if ϕ⊂λ and {ek} is a basis for λ, where ek is a sequence whose only nonzero term is 1 in the kth place for each k∈ℕ and ϕ=span{ek}, the set of all finitely nonzero sequences. If ϕ is dense in λ, then λ is called an AD-space; thus, AK implies AD.Letλ and μ be two sequence spaces, and let A=(ank) be an infinite matrix of the complex numbers ank, where k,n∈ℕ. Then, we say that A defines a matrix mapping from λ into μ, and we denote it by writing A:λ→μ if Ax exists and belongs to μ for every sequence x=(xk)∈λ, where Ax={(Ax)n}, the A-transform of x with
(3)(Ax)n=∑k‍ankxkforeachn∈ℕ.For simplicity in notation, here and in what follows, the summation without limits runs from1 to ∞. By (λ:μ), we denote the class of all matrices A such that A:λ→μ. Thus, A∈(λ:μ) if and only if the series on the right side of (3) converges for each n∈ℕ and each x∈λ, and we have Ax={(Ax)n}n∈ℕ∈μ for all x∈λ. A sequence x is said to be A-summable to l if Ax converges to l which is called the A-limit of x.The matrix domainλA of an infinite matrix A in a sequence space λ is defined by
(4)λA={x=(xk)∈ω:Ax∈λ}
which is a sequence space. If A=(ank) is triangle, that is, ann≠0 and ank=0 for all k>n, then one can easily observe that the sequence spaces λA and λ are linearly isomorphic; that is, λA≅λ. There are several examples of the matrix domain λA of an infinite matrix A in a sequence space λ in Chapter  4 in [1]. By ℱ, we will denote the collection of all finite subsets of ℕ.Hahn [2] introduced the space h={x:∑k‍k|xk-xk+1|<∞andlimk→∞xk=0} and proved that the following statements hold:(i)
h is a Banach space with the norm ∥x∥h=∑k‍k|xk-xk+1|+supk∈ℕ|xk|,(ii)
h⊂ℓ1∩∫‍c0,(iii)
hβ=σ∞.


## 2. The New Hahn Sequence Space

Following Hahn [2], we introduce the sequence space h(C) as follows:
(5)h(C)={x=(xk)∈w:∑k‍k|1(k+1)(k+2)∑j=0k‍xj-xk+1k+2|xk+k+<∞,1k+1∑j=0k‍xj⟶0(k⟶∞)}.
With the notation of (4), we may redefine the space h(C) as follows:
(6)h(C)=(h)C.
We define a sequence y=(yk) as the C-transform of a sequence x=(xk); that is,
(7)yk=1n+1∑k=0n‍xk∀n∈ℕ.Hahn [2] proved that h⊂ℓ1. Now, we give some inclusion relations.Theorem 1.
The following inclusions are strict:(a)
h(C)⊂ℓ1(C),(b)
h⊂bv,(c)
h(C)⊂bv(C).Proof.
(a) It is clear thath(C)⊂ℓ1(C) from h⊂ℓ1 [2]. Now, we show that this inclusion is strict. Let us consider the sequence x=(xk), x0=1, and xk=(-1)k(2k+1)/k(k+1)(k≥1). Then, y=Cx={(-1)k/(k+1)2}. Since the sequence y is in ℓ1 but not in h, then x∈ℓ1(C)∖h(C).
(b) Sinceh⊂ℓ1 [2] and ℓ1⊂bv, then h⊂bv.
(c) We choose the sequence(xk)=e=(1,1,…). Since limk→∞(1/(k+1))∑j=0k‍xj=1, then x=e is not in h(C), but it is in bv(C). Thus, we see that h(C)⊂bv(C) is strict.Theorem 2.
The sequence spaceh(C) is a BK-space with the norm
(8)∥x∥h(C)=∑kk|1k+1∑j=0kxj-1k+2∑j=0k+1xj|+supk∈ℕ|1k+1∑j=0k‍xj|.Proof.
Since (6) holds, h is a BK-space with the norm ∥·∥h [2, 3], and the matrix C is triangle matrix, then Theorem 4.3.2 of Wilansky [4] gives the fact that the space h(C) is a BK-space.Lemma 3 (see [5]).
TheBK-space h has an AK property.Since{ek:k∈ℕ}⊄h(C), then one has the following.Theorem 4.
TheBK-space h(C) does not have an AK property.Theorem 5.
One has the following:(9)h(C)=ℓ1(C)∩∫‍bv(C)=ℓ1(C)∩∫‍bv0(C).Proof.
It is similar to the proof of [3, Theorem 3.2].Theorem 6.
The sequence spaceh(C) is norm isomorphic to the space h; that is, h(C)≅h.Proof.
To prove this, we will show the existence of a linear bijection between the spacesh(C) and h. Consider the transformation T defined, with the notation of (7), from h(C) to h by x↦y=Tx. The linearity of T is clear. Furthermore, it is trivial that x=θ=(0,0,0,…) whenever Tx=θ, and, hence, T is injective.
Lety∈h, and define the sequence x=(xk) by xk=(k+1)yk-kyk-1(k∈ℕ). Then, we have
(10)∥x∥h(C)=∑kk|1k+1∑j=0k‍[(j+1)yj-jyj-1]ddddddi-1k+2∑j=0k+1‍[(j+1)yj-jyj-1]|=∑k‍k|yk-yk+1|=∥y∥h<∞.
Consequently, we see from here that T is surjective and is norm preserving. Hence, T is a linear bijection which, therefore, shows that the spaces h(C) and h are norm isomorphic, as desired.Theorem 7.
Define a sequenceb(k)={bn(k)}n∈ℕ of elements of the space h(C) for every fixed k∈ℕ by
(11)bn(k)={(-1)n-k(k+1),k≤n≤k+1,0,0≤n<korn>k+1.
Then, the sequence {bn(k)}n∈ℕ is a basis for the space h(C), and any x∈h(C) has a unique representation of the form
(12)x=∑k‍λkb(k),
where λk=(C1x)k for all k∈ℕ.Proof.
It is clear that{b(k)}⊂h(C), since
(13)C1b(k)=ek∈ℓ(k=0,1,2,…).
Letx∈h(C) be given. For every nonnegative integer m, we put
(14)x[m]=∑k=0m‍λkb(k).
Then, we obtain by applying C1 to (14) with (13) that
(15)C1x[m]=∑k=0m‍λkC1b(k)=∑k=0m‍(C1x)kek,{C1(x-x[m])}i={0,(0≤i≤m),(C1x)i,(i>m),(i,m∈ℕ).
Given that ε>0, then there is an integer m0 such that
(16)|(C1x)j|<ε2
for all m≥m0. Hence,
(17)∥x-x[m]∥h(C)=supn≥m|(C1)n|≤supn≥m0|(C1)n|≤ε2<ε
for all m≥m0, which proves that x∈h(C) is represented as in (12).
To show the uniqueness of this representation, we assume thatx=∑k‍μkb(k). Since the linear transformation T, from h(C) to h, used in Theorem 6 is continuous, then we have at this stage that
(18)(C1x)n=∑k‍μk{C1b(k)}n=∑k‍μkenk=μn(n∈ℕ),
which contradicts the fact that (C1x)n=λn for all n∈ℕ. Hence, the representation (12) of x∈h(C) is unique.


## 3. Duals of the Sequence Spaceh(C)

In this section, we state and prove the theorems determining theα-, β-, and γ-duals of the sequence space h(C).The setS(λ,μ) defined by
(19)S(λ,μ)={z=(zk)∈ω:xz=(xkzk)∈μdid∀x=(xk)∈λ}
is called the multiplier space of the sequence spaces λ and μ. One can easily observe for a sequence space υ with λ⊃υ⊃μ that the inclusions
(20)S(λ,μ)⊂S(υ,μ),S(λ,μ)⊂S(λ,υ)
hold. With the notation of (19), the alpha-, beta-, and gamma-duals of a sequence space λ, which are, respectively, denoted by λα, λβ, and λγ are defined by
(21)λα=S(λ,ℓ1),λβ=S(λ,cs),λγ=S(λ,bs).The alpha-, beta-, and gamma-duals of a sequence space are also referred to as theKöthe-Toeplitz dual, the generalized Köthe-Toeplitz dual, and the Garling dual of a sequence space, respectively.Given anFK-space X containing Φ, its conjugate is denoted by X′, and its f-dual or sequential dual is denoted by Xf and is given by Xf={all sequences (f(ek)):f∈X′}.We need the following lemmas.Lemma 8 (see [6]).
LetBU=(bnk) be defined via a sequence a=(ak)∈ω, and let the inverse v=(vnk) of the triangle matrix U=(unk) be defined by bnk=∑j=kn‍ajvjk for all k,n∈ℕ. Then,
(22)λUβ={a=(ak)∈ω:BU∈(λ:c)},λUγ={a=(ak)∈ω:BU∈(λ:ℓ∞)}.Lemma 9 (see [5]).
(i)A∈(h:ℓ) if and only if
(23)∑n=1∞‍|ank|converges(k=1,2,…),supk1k∑n=1∞‍|∑υ=1k‍anυ|<∞.
(ii)A∈(h:c) if and only if
(24)supn,k1k|∑υ=1k‍anυ|<∞,(25)limn→∞ankexists(k=1,2,…).
(iii)A∈(h:c0) if and only if
(26)limn→∞ank=0
and (24) hold.
(iv)A∈(h:ℓ∞) if and only if (24) holds.
(v)A∈(h:h) if and only if (26) holds and
(27)∑n=1∞‍n|ank-an+1,k|converges(k=1,2,…),supk1k∑n=1∞‍n|∑v=1k‍(anv-an+1,v)|<∞.Theorem 10.
Theα-dual of the space h(C) is the set
(28)d1={|∑k∈K(-1)n-k(k+1)an|a=(ak)∈ω:supN,K∈ℱ1kh×∑n∈N|∑k∈K(-1)n-k(k+1)an|<∞}.Proof.
Leta=(an)∈ω. We define the matrix B=(bnk) via the sequence a=(an) by
(29)bnk={(-1)n-k(k+1)an(n-1≤k≤n+1),0,(0≤k<n-1ork>n),(n,k∈ℕ).
Bearing in mind the relation (7), we immediately derive that
(30)anxn=∑k=n-1n‍(-1)n-k(k+1)anyk=(By)n(n∈ℕ).
We, therefore, observe by (30) that ax=(anxn)∈ℓ1 whenever x∈h(C) if and only if By∈ℓ1 whenever y∈h. Then, we derive by Lemma 9 (i) that
(31)supN,K∈ℱ1k∑n∈N‍|∑k∈K‍(-1)n-k(k+1)an|<∞,
which yields the result that [h(C)]α=d1.Hahn [2] proved that hβ=σ∞.Theorem 11.
Consider the following:(32)[h(C)]β=σ∞∩ℓ∞.Proof.
Consider the equation(33)∑k=0n‍akxk=∑k=0n‍ak(∑j=k∞‍yjj)=∑k=0n‍(∑j=k∞‍aj)yk=(Ey)n(n∈ℕ),
where E=(enk) is defined by
(34)en(k)={∑j=k∞‍ajj(0≤k≤n),0(k>n),(n,k∈ℕ).
Thus, we deduce from (33) that ax=(akxk)∈cs whenever x=(xk)∈h(C) if and only if Ey∈c whenever y=(yk)∈h. Therefore, we derive the consequence from Lemma 9 (ii) that [h(C)]β=σ∞∩ℓ∞.Theorem 12.
One has the following:(35)[h(C)]γ=σ∞∩ℓ∞.Proof.
This is obtained in the similar way used in the proof of Theorem11.


## 4. Matrix Transformations

Let us suppose throughout that the sequencesx=(xk) and y=(yk) are connected with (7), and let the A-transform of the sequence x=(xk) be r=(rn), and let the B-transform of the sequence y=(yk) be s=(sn); that is,
(36)rn=(Ax)n=∑k‍ankxk(n∈ℕ),sn=(By)n=∑k‍bnkyk(n∈ℕ).
It is clear here that the method B is applied to the C-transform of the sequence x=(xk), while the method A is directly applied to the terms of the sequence x=(xk). So the methods A and B are essentially different.Following Şengönül and Başar [7], we give some knowledge about the dual summability methods of the new type. Let us assume the existence of the matrix product BC. We will say in this situation that the methods A and B in (36) are the dual of the new type if r=(rn) is reduced to s=(sn) (or s=(sn) becomes r=(rn)) under the application of the formal summation by parts. This leads us to the fact that BC exists and is equal to A and (BC)x=B(Cx) formally holds, if one side exists. This statement is equivalent to the relation
(37)ank=∑j=k∞‍1j+1bnj(orbnk=(k+1)Δank,dwhereΔank=ank-an,k+1)(n∈ℕ).Now, we may give the following theorem.Theorem 13.
LetA=(ank) and B=(bnk) be the dual matrices of the new type, and μ be any given sequence space. Then, let A∈(h(C):μ) if and only if B∈(h:μ) and
(38){(n+1)ank}n∈ℕ∈c0
for every fixed k∈ℕ.Proof.
Suppose thatA=(ank) and B=(bnk) are dual matrices of the new type; that is to say that (37) holds; let μ be any given sequence space, and take account that the spaces h(C) and h are linearly isomorphic.
LetA∈(h(C):μ), and take any y∈h. Then, BC exists, and (ank)k∈ℕ∈d2∩d3∩cs, which yields that (ank)k∈ℕ∈ℓ1 for each n∈ℕ. Hence, By exists, and, thus, letting m→∞ in the equality
(39)∑k=0m‍bnkyk=∑k=0m‍∑j=km‍1j+1bnjxk(n,m∈ℕ),
we have by (37) that By=Ax, which leads us to the consequence that B∈(h:μ).
Conversely, letB∈(h:μ), and (38) hold, and take any x∈h(C). Then, we have (bnk)k∈ℕ∈ℓ1, which gives together with (38) that (ank)k∈ℕ∈[h(C)]β for each n∈ℕ. Hence, Ax exists. Therefore, we obtain from the equality
(40)∑k=0m‍ankxk=∑k=0m-1‍(k+1)Δnkyk+(m+1)anmym=∑k=0m‍bnkyk(n,m∈ℕ)
as m→∞ that Ax=By, and this shows that A∈(h(C):μ). This completes the proof.By the changing roles of the spacesh(c) and μ in Theorem 13, we have the following.Theorem 14.
Suppose that the elements of the infinite matricesF=(fnk) and G=(gnk) are connected with the relation
(41)gnk=∑j=0n‍fjkn+1
for all k,m∈ℕ, and let μ be any given sequence space. Then, F∈(μ:h(C)) if and only if G∈(μ:h).Proof.
Letx=(xk)∈μ, and consider the following equality with (41):
(42)1i+1∑j=0i‍∑k=0n‍fjkxk=∑k=0n‍gikxk(i,n∈ℕ)
which yields as n→∞ that
(43)1i+1∑j=0i‍(Fx)j=(Gx)i(i∈ℕ).
Therefore, one can easily see by (43) that Fx∈h(C) whenever x∈μ if and only if Gx∈c whenever x∈μ.Corollary 15.
(i)A=(ank)∈(h(C):ℓ) if and only if (23) hold with ∑j=k∞‍(1/(k+1))bnj instead of ank.
(ii)A=(ank)∈(h(C):c) if and only if (24) and (25) hold with ∑j=k∞‍(1/(k+1))bnj instead of ank.
(iii)A=(ank)∈(h(C):c0) if and only if  limn→∞ank=0 and (24) hold with ∑j=k∞‍(1/(k+1))bnj instead of ank.
(iv)A=(ank)∈(h(C):ℓ∞) if and only if (24) hold with ∑j=k∞‍(1/(k+1))bnj instead of ank.
(v)A=(ank)∈(h(C):h) if and only if (26) and (27) hold with ∑j=k∞‍(1/(k+1))bnj instead of ank.


## 5. Conclusion

Hahn [2] defined the space h and gave some of its general properties. G. Goes and S. Goes [3] studied the functional analytic properties of the space h. The study of the Hahn sequence space was initiated by Chandrasekhara Rao [5] with a certain specific purpose in the Banach space theory. Also Chandrasekhara Rao [5] computed some matrix transformations. Chandrasekhara Rao and Subramanian [8] introduced a new class of sequence spaces called semi-replete spaces. Chandrasekhara Rao and Subramanian [8] defined the semi-Hahn space and proved that the intersection of all the semi-Hahn spaces is the Hahn space. Balasubramanian and Pandiarani [9] defined the new sequence space h(F) called the Hahn sequence space of fuzzy numbers and proved that β- and γ-duals of h(F) is the Cesàro space of the set of all fuzzy bounded sequences.The sequence spaceh was defined by Hahn [2], and G. Goes and S. Goes [3] and Chandrasekhara Rao et al. [5, 8, 10] investigated some properties of the space h. In exception of these works, there has not been any work related to the Hahn sequence space. In this paper, the Hahn sequence space h defined by the Cesáro mean worked as follows. In Section 2, the new Hahn sequence space is determined by the Cesáro mean, and some properties of this space are investigated. In Section 3, α-, β-, and γ-duals of the new Hahn sequence space are computed. In Section 4, the matrix classes (h(C):μ) and (μ:h(C)) are characterized, where μ is an arbitrary sequence space, and some results of these characterizations are given.We can define the matrix domainhA of an arbitrary triangle A, compute its α-, β-, and γ-duals, and characterize the matrix transformations on them into the classical sequence spaces, and almost the convergent sequence space is a new result.



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