# On the New Generalized Hahn Sequence Spacehdp

Authors: Orhan Tuğ; Eberhard Malkowsky; Bipan Hazarika; Taja Yaying
Journal: Abstract and Applied Analysis
 (2022)

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## Abstract

In this article, we define the new generalized Hahn sequence spacehdp, where d=dkk=1∞ is monotonically increasing sequence with dk≠0 for all k∈ℕ, and 1<p<∞. Then, we prove some topological properties and calculate the α−, β−, and γ−duals of hdp. Furthermore, we characterize the new matrix classes hd,λ, where λ=bv,bvp,bv∞,bs,cs,, and μ,hd, where μ=bv,bv0,bs,cs0,cs. In the last section, we prove the necessary and sufficient conditions of the matrix transformations from hdp into λ=ℓ∞,c,c0,ℓ1,hd,bv,bs,cs, and from μ=ℓ1,bv0,bs,cs0 into hdp.


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## Body

## 1. Introduction and Basic Notations

The set of all complex valued sequences is denoted byω and each vector subspace of ω is called a sequence space. The sets ℓ∞, c, and c0 are bounded, convergent and null sequence spaces, respectively. Moreover, ℓp, ℓ1, cs, cs0, bs and bv are the spaces of all absolutely p−summable, absolutely summable, convergent series, null series, bounded series and sequeences of bounded variation, respectively, where 1<p<∞.The alpha-dualλα, beta-dual λβ and gamma-dual λγ of a sequence space λ are defined by
(1)λα≔x=xk∈ω:xy=xkyk∈ℓ1forally=yk∈λ,λβ≔x=xk∈ω:xy=xkyk∈csforally=yk∈λ,λγ≔x=xk∈ω:xy=xkyk∈bsforally=yk∈λ.LetA=ankk,n∈ℕ be an infinite matrix and λ,μ∈ω. We write
(2)yk=Axn=∑kankxkand then we say that A defines a matrix transformation from λ into μ as A:λ⟶μ if Ax=Axn∈μ for every x∈λ. We denote the set of all infinite matrices that map the sequence space λ into the sequence space μ by λ,μ. Thus, A∈λ,μ if and only if the right side of (2) converges for every n∈ℕ, that is, An∈λβ for all n∈ℕ and we have Ax∈μ for all x∈λ. The set λA=x∈ω:Ax∈ω is called the domain of the matrix H in X. It is known that the matrix domain λA is also a sequence space. The readers may refer to these nice papers [1–3] and the textbooks [4–8] concerning domain of special matrices in classical sequence spaces and the theory of summability.If a normed sequence spaceλ contains a sequence bn with the following property that for every x∈λ there is a unique sequence of scalars αn such that
(3)limn⟶∞x−α0b0+α1b1+⋯+αnbn=0then bn is called a Schauder basis for λ. The series ∑kαkbk which has the sum x is then called the expansion of x with respect to bn and written as x=∑kαkbk.Ifλ is an FK-space, ϕ⊂λ and ek is a basis for λ then λ is said to have AK property, where ek is a sequence whose only term in kth place is 1 the others are zero for each k∈ℕ and ϕ=spanek. If ϕ is dense in λ, then λ is called AD-space, thus AK implies AD.The sequence spaceh defined by
(4)h≔x=xk∈ω:∑kkxk−xk+1<∞∩c0is called Hahn sequence space, named after its introducer H. Hahn [9]. The space h is a BK space with the norm
(5)x=∑kkxk−xk+1+supkxkforallx=xk∈h.Rao [10] proved that the space h is a BK space with AK with respect to the norm
(6)xh=∑kkxk−xk+1forallx=xk∈h.Later on Goes [11] introduced a generalised Hahn sequence space hd,d=dk∈ω with dk≠0 for all k, defined by
(7)hd≔x=xk∈ω:∑kdkxk−xk+1<∞∩c0.Quiet recently a scientific study of a more generalized Hahn sequence spacehd, is carried out by Malkowsky et al. [12], defined as follows:
(8)hd≔x=xk∈ω:∑kdkxk−xk+1<∞∩c0.The authors proved that the spacehd is a BK space with AK with respect to the norm
(9)xhd=∑kdkxk−xk+1forallx=xk∈hd,where d=dk is an unbounded and monotonic increasing sequence of positive reals. Besides, the authors stated and proved various significant results concerning characterization of matrix transformations between the space hd and classical BK spaces, and characterization of compact operators on the space hd using Hausdorff measure of non-compactness. We refer to [2, 10, 13–23] and the survey paper [22] for more studies and results related to Hahn sequence space.


## 2. The New Generalized Hahn Sequence Spacehdp

In this section, we introduce the new generalized Hahn sequence spacehdp as follow
(10)hdp=x∈ω:∑k=1∞dkΔxkp<∞∩c0where dkk=1∞ is an unbounded monotone increasing sequence of positive real numbers with dk≠0 for all k∈ℕ, 1<p<∞, and Δxk=xk−xk+1.Remark 1.
It is clear that if we considerp=1, then hdp becomes hd.
If we consideryk=dkxk−xk+1=Exk where
(11)enk=dn,k=n,−dn,k=n+10,otherwise.Moreover, if we write the terms ofy=yk starting from k to m as follow
(12)ykdk=xk−xk+1yk+1dk+1=xk+1−xk+2yk+2dk+2=xk+2−xk+3⋮ymdm=xm−xm+1and then when we get sum ∑j=kmyj/dj=xk−xm+1. Then we have ∑j=k∞yj/dj=xk whenever xm+1⟶0m⟶∞. Equivalently, the sequence x=xk may also represented by x=Gy where the matrix G=gnk is defined by
(13)gnk=1dk,k≥n,0,0<k<n.ClearlyG=E−1.Theorem 2.
The new Hahn sequence spacehdp is a linear complete normed space space with the norm defined as
(14)xhdp=∑k=1∞dkΔxkp1/p.Proof.
The linearity is clear. Now, we observe that thehdp,xhdp is a Banach space.
(N1) if(15)xhdp=0⇒∑k=1∞dkΔxkp1/p=0⇒dkΔxk=0,since dk≠0 for each k∈ℕ, Δxk=0, that is, xk=xk+1 for all k, and x∈c0 implies xk=0 for all k, that is, x=0.
(N2) For everyx∈hdp and λ∈K,
(16)λxhdp=∑k=1∞dkΔλxkp1/p=λ∑k=1∞dkΔxkp1/p=λxhdp
(N3) For everyx,y∈hdp,
(17)x+yhdp=∑k=1∞dkΔxk+ykp1/p≤∑k=1∞dkΔxkp1/p+∑k=1∞dkΔykp1/p=xhdp+yhdp.
Now we should show that the spacehdp,xhdp is complete. Suppose that x=xnn=1∞ be a Cauchy sequence in the space hdp. Therefore, for every ϵ>0, there exists a natural number N∈ℕ such that
(18)xn−xmhdp<ϵ2,∀m,n≥N.
Since the Cauchy sequencex=xnn=1∞ is in the space hdp, we obtain
(19)∑k=1∞dkΔxknp<∞,andxkn⟶0k⟶∞.
Now we can write the following inequality for fixedℓ≥k, m,n≥N by considering the inequality
(20)a+bp≤2pap+bp
for1<p<∞ that
(21)xkn−xkmp=∑j=kℓΔxkn−Δxkm+xk+1n−xk+1mp≤∑j=kℓdkΔxkn−xkm+xk+1n−xk+1mp≤2p∑j=kℓdkΔxkn−xkmp+xk+1np+xk+1mp.
Then we have the following result sincexn,xm∈c0 for all m,n∈ℕ(22)xkn−xkm≤2xn−xmhdp+xk+1n+xk+1m<ϵ
Therefore,x=xnn=1∞ is a Cauchy sequence of Complex numbers for each k∈ℕ. Since ℂ is complete, then x=xnn=1∞ converges to an arbitrary sequence x=xk∈ℂ as n⟶∞, that is, for every ϵ>0 there exists N0∈ℕ such that
(23)xkn−xk<ϵ
for eachk∈ℕ. Therefore, we have the following with an arbitrary ℓ that
(24)∑k=1ℓdkΔxkn−xkmp1/p≤xn−xmhdp<ϵhence, by letting limit as m⟶∞ over (24) we obtain
(25)∑k=1ℓdkΔxkn−xkp1/p≤xn−xhdp<ϵ
Moreover, we obtain by (23) for any n≥N0 and 1<p<∞(26)0≤xkp≤2pxk−xknp+xknp⟶0k⟶∞
which says thatx=xk∈c0. Furthermore,
(27)xhdp=∑k=1∞dkΔxkp1/p=∑k=1∞dkΔxk−xkn+xknp1/p≤∑k=1∞dkΔxk−xknp1/p+∑k=1∞dkΔxknp1/p<∞.
So, it shows thatx∈hdp.Corollary 3.
The Hahn sequence spacehdp is a BK space.Theorem 4.
The spacehdp does not have AK property.Proof.
If the sequence spacehdp has AK property, then every sequence x=xkk=1∞∈hdp is the limit of its m−section xm=∑k=1mxkek, i.e.,
(28)x=limm⟶∞xm=limm⟶∞∑k=1mxkek=∑k=1∞xkek.
Let us considerdk=k and xk=1/k for all k∈ℕ. Then obviously x=xk∈c0 and, since p>1,
(29)∑k=1∞dkxk−xk+1p=∑k=1∞k1k−1k+1p=∑k=1∞1k+1p<∞,that is, x∈hdp. But, if we write xm=∑k=1mxkek for every m∈ℕ as the m−section of the sequence x, then we obtain for all m∈ℕ(30)x−xmhdpp=mxm+1p+∑k=m+1∞dkxk−xk+1p≥mm+1p.
Hence,xm½x as m⟶∞. It completes the proof.Theorem 5.
The spacehdp is linearly isomorphic to the space ℓp.Proof.
Let us define the mappingT:hdp⟶ℓp by y=Tx=Ex. Linearity of T is obvious. Since the triangle E is the matrix representation of T, therefore T is invertible too. Now, let y∈ℓp and consider the following equality:
(31)xhdp=∑k=1∞dkΔxkp1/p=∑k=1∞ykp1/p=yℓp<∞.
This implies thatT is onto and preserves the norm. Consequently T defines an isomorphism from hdp to ℓp. This completes the proof.Theorem 6.
Let us define a sequenceqkd=qnkdn∈ℕ of elements of the space hdp for every fixed k∈ℕ by
(32)qnkd=1dk,k≥n,0,1≤k≤n−1.
Then the sequenceqkdk∈ℕ is a basis for the space hdp and any x∈hdp has unique representation of the form
(33)x=∑kλkqkd,where λk=Exk for all k∈ℕ and 1<p<∞.


## 3.α−, β− And γ− Duals of hdp

In this section, we first calculate theγ−dual of the space hd and then we give some basic lemmas to prove the α−, β− and γ− duals of the space hdp. The following calculations on the β−dual of the space hd has been stated by Goes ([11], 4.1 Theorem, p. 485) and it has recently been proved by Malkowsky et al. [12], Proposition 2.3., p.5) as
(34)hdβ=bsd=a∈ω:supn1dn∑k=1nak<∞.Moreover, Goes ([11], 4.2 Corolary (b)) showed the α− dual of the space hd as follows:
(35)hdα=a∈ω:supn1dn∑k=1nak<∞.Now we can calculate theγ−dual fo the space hd.Lemma 7.
hdγ=a∈ω:supn1/dn∑k=1nak<∞.Proof.
Suppose thata=ak∈ω and x=xk∈hd be given. Then we have the nth−partial sum of the series ∑kakxk as
(36)zn=∑k=1nakxk=∑k=1nak∑j=knyjdj=∑k=1n−1∑j=1kajdkyk+yndn−1∑j=1naj=Byn,where the infinite matrix B=bnk is defined as
(37)bnk=1dk∑j=1kaj,k≥n,0,k<n.
It gives thatax∈bs, that is, z∈ℓ∞ whenever x∈hd if and only if By∈ℓ∞ whenever y∈ℓ1. Therefore, a∈hdγ if and only if B∈ℓ1,ℓ∞. By Lemma 10(d) and since xhd=yℓ1=∑kyk<∞ we have that
(38)supk1dk∑j=1kaj<∞.
This completes the proof.Now we have the following useful Lemmas for the further computations.Lemma 8 (see [12], Theorem 11., p.8).
We have(a)
A∈hd,ℓ∞ if and only if(39)Ahd,ℓ∞=supnAnbsd=supn,m1dm∑k=1mank<∞.(b)
A∈hd,c if and only if (39) holds, and the limits(40)αk=limn⟶∞ankexistforallk.Lemma 9 (see [12], Theorem 13., p.9).
We haveA∈hd,ℓ1 if and only if
(41)Ahd,ℓ1=supm1dm∑n=1∞∑k=1mank<∞.Lemma 10 (see [4], Theorem 9.7.3, p.356).
Let1<p<∞, q=p/p−1. Then we have
(a)
A∈ℓp,ℓ1 if and only if(42)supN∈ℕFinite∑k=0∞∑n∈Nankq<∞.(b)
A∈ℓp,c if and only if (40) holds and(43)supn∑kankq<∞.(c)
A∈ℓp,ℓ∞ if and only if (43) holds(d)
A∈ℓ1,ℓ∞ if and only if(44)supn,kank<∞.Theorem 11.
Let1<p<∞, q=p/p−1. Then
(45)hdpα=a=ak∈ω:supN∈ℕfinite∑k∑n∈Nandkq<∞.Proof.
Suppose thata=an∈ω be given and consider the following equation that
(46)anxn=ak∑k=n∞ykdk=∑k=n∞andkyk=Dyn,n∈ℕ,where the matrix D=dnk is defined as
(47)dnk=andk,k≥n,0,k<nfor all k,n∈ℕ. It gives us that ax=anxn∈ℓ1 whenever x∈hdp if and only if Dy∈ℓ1 whenever y=yn∈ℓp. Therefore, a=an∈hdpα whenever x∈hdp if and only if D∈ℓp,ℓ1. Therefore, the condition in (42) of Lemma 10(a) holds with dnk instead of ank, that is,
(48)supN∈ℕFinite∑k=0∞∑n∈Nandkq<∞.
It gives theα−dual of the space hdp.Theorem 12.
hdpβ=a=ak∈ω:supn1/dnq∑k∑j=knajq<∞.Proof.
Suppose thata=ak∈ω and x=xk∈hdp are given. Let us consider the following equality.
(49)sn=∑k=1nakxk=∑k=1nak∑j=knyjdj=∑k=1n∑j=1kajdkyk=Byn,where the matrix B=bnk is defined as in (37) for all k,n∈ℕ. Then we can say that ax=akxk∈cs whenever x=xk∈hdp if and only if By∈c whenever y=yk∈ℓp. Therefore, a=ak∈hdpβ if and only if B∈ℓp,c. Then the condition in Lemma 10(b) holds with bnk instead of ank, that is
(50)limn⟶∞bnkexistforallk;supn∑kbnkq<∞.
Thus, we can obtain thathdpβ=a=ak∈ω:supn1/dnq∑k∑j=knajq<∞∩cs. It is clearly seen that
(51)a=ak∈ω:supn1dnq∑k∑j=knajq<∞⊂csand then
(52)hdpβ=a=ak∈ω:supn1dnq∑k∑j=knajq<∞as we desired.Theorem 13.
hdpγ=a=ak∈ω:supn1/dnq∑k∑j=knajq<∞.Proof.
Since we have similar pattern with Theorem12, we can conclude the proof with the following computation only.
Let us say thatax=akxk∈bs whenever x=xk∈hdp if and only if By∈ℓ∞ whenever y=yk∈ℓp. Therefore, a=ak∈hdpβ if and only if B∈ℓp;ℓ∞. Then the condition in Lemma 10(c) holds with bnk instead of ank, that is
(53)supn∑kbnkq<∞.
Thus, we can obtain thathdpγ=a=ak∈ω:supn1/dnq∑k∑j=knajq<∞ as we desired.


## 4. Some Matrix Transformations from and into the Spacehd

In this section, we characterize the classeshd,λ, where λ=bv,bvp,bv∞,bs,cs,, and μ,hd, where μ=bv,bv0,bs,cs0,cs.Remark 14.
Ifdk=k for every k∈ℕ, then characterization of Theorem 17-Corollary 23 yields the characterisation of h,λ, where λ=bv,bvp,bv∞,bs,cs,Definition 15 (see [8], Definition 7.4.2).
LetX be a BK space. A subset E of the set ϕ called a determining set for X if DX=B¯X∩ϕ is the absolutely convex hull of E.Proposition 16 (see [12], Proposition 3.2, p.8).
Let(54)sd,k=1dk·ekforeachk∈ℕ,andE=sd,k:k∈ℕ.
ThenE is a determining set for hd.Theorem 17.
The infinite matrixA=ank∈hd,bv if and only if
(55)supm∈ℕ1dm∑n∑k=1mank−an−1,k<∞.Proof.
Suppose thatE=1/dmem:m∈ℕ is a determining set for the sequence space hd. The sequence space bv is BK space with .bv. Let ym=1/dmem∈E. Then
(56)Anym=∑k=1∞ankykm=1dm∑k=1mankforalln=1,2,⋯
Hence,(57)Anymbv=∑nAnym−An−1ym=∑n1dm∑k=1mank−1dm∑k=1man−1,k=1dm∑n∑k=1mank−an−1,k<∞which gives the condition in (55).Proposition 18 (see [1], Lemma 5.3).
LetX,Y be any two sequence spaces, A be an infinite matrix and U a triangle matrix. Then A∈X,YU if and only if UA∈X,Y.Corollary 19.
The infinite matrixA=ank∈hd,bvp=hd,ℓpΔ if and only if
(58)supm∈ℕ1dmp∑n∑k=1mank−an−1,kp<∞,1≤p<∞.Corollary 20.
The infinite matrixA=ank∈hd,bv∞=hd,ℓ∞Δ if and only if
(59)supm,n∈ℕ1dm∑n∑k=1mank−an−1,k<∞.Theorem 21.
The infinite matrixA=ank∈hd,bs if and only if
(60)supm,n∈ℕ1dm∑k=1n∑j=1makj<∞.Proof.
The proof is similar to that of Theorem17 with the norm .bs replaced by the norm .bv. Therefore, we have only the following norm
(61)Anymbs=supm∈ℕ∑k=1nAkym=supm∈ℕ∑k=1n1dm∑j=1makj=supm∈ℕ1dm∑k=1n∑j=1makj.If we take supremumn∈ℕ from both sides of above equality, then we can easily see the condition in (60).Theorem 22.
The infinite matrixA=ank∈hd,cs if and only if the condition in (60) holds and
(62)limn⟶∞∑k=1nank=αkexistsforeachk∈ℕ.Proof.
Suppose thatA=ank∈hd,cs. Then Ax exists and is in cs for each x=xk∈hd. Thus, the partial sum of the series Ax converges, that is the condition in (62) is immediate. The rest of the proof is similar to that of Theorem 17 by replacing the norm .bv with the norm .bs=.cs. Hence, we obtain the necessity of the condition in (4.4).Corollary 23.
The infinite matrixA=ank∈hd,cs0 if and only if the conditions in (60) and (62) hold with αk=0 for every k∈ℕ.Theorem 24.
The infinite matrixA=ank∈bv0,hd if and only if
(63)limn⟶∞∑k=1mank=0(64)supk∈ℕ∑n=1∞dn∑k=1mank−an+1,k<∞.Proof.
Suppose thatE=em:m∈ℕ be a determining set of bv0 and ym=em∈E for each m∈ℕ. Then the kth column of Aym where
(65)Akym=∑k=1∞ankykm=∑k=1mank,forn=1,2,⋯
Then,(66)Akymhd=∑n=1∞dn∑k=1mank−an+1,k,forallk∈ℕ.Hence, the sequenceAkymk=1∞ is bounded in hd if and only if supkAkymhd<∞ which shows the necessity of the condition in (64). Moreover, Akym∈c0 for all k∈ℕ, then the condition in (63) is also immediate.Corollary 25.
The infinite matrixA=ank∈bv,hd if and only if the conditions in (63) and (64) hold, and
(67)Ae∈hdTheorem 26.
The infinite matrixA=ank∈bs,hd if and only if
(68)supK∈ℕfinite∑k=1∞∑n∈Kdnbnk−bn+1,k,(69)limn⟶∞∑k=1∞bnk=0where the matrix B=bnk is defined as bnk=ank−an,k+1.Proof.
Suppose thatA=ank∈bs,hd and Ax exists and is in hd for every x∈bs. Thus, An∈bsβ=bv0⊂c0. Let x be a sequence such that vk=∑i=1kxi. If v∈ℓ∞ and xn=vn−vn−1, then x∈bs. By ([8], Lemma 8.5.3(i)), we can say that Ax=Bv. It says that B∈ℓ∞,hd. So the necessity of the conditions in (68) and (69) are immediate.Theorem 27.
The infinite matrixA=ank∈cs0,hd if and only if the condition in (68) holds and
(70)limn⟶∞bnk=0where the matrix B=bnk is defined as bnk=ank−an,k+1.Proof.
The proof is similar to that of Theorem26 with An∈cs0β=csβ=bv⊂c and by ([8], Lemma 8.5.3(i)), we can say that B∈c0,hd. So the necessity of the conditions in (68) and (70) are immediate.Corollary 28.
The infinite matrixA=ank∈cs,hd if and only if the conditions in (67), (68), (70) hold.Remark 29.
If we considerdk=k for every k∈ℕ, then the characterization of Theorem 17, Corollary 19-Corollary 23 yields the characterisation of h,λ, where λ=bv,bvp,bv∞,bs,cs,, and the characterization of Theorem 24-Corollary 28 yields the characterisation ofμ,h, where μ=bv,bv0,bs,cs0,cs (see [18], p.13-16).


## 5. Matrix Transformations on the Hahn Sequence Spacehdp

In this section, we characterize matrixhdp;λ and μ;hdp where λ=ℓ∞,c,c0,ℓ1,hd,bv,bs,cs and μ=ℓ1,bv0,bs,cs0, respectively. The following lemma is significant for our investigation:Lemma 30 (see [1]).
Matrix transformation betweenBK-spaces are continuous.Theorem 31.
A=ank∈hdp,ℓ∞ if and only if
(71)supn,m1dmq∑k=1∞∑j=kmanjq<∞,(72)supn∑k=1∞1dkq∑j=1kanjq<∞.Proof.
Let1<p<∞ and A∈hdp,ℓ∞. This means that Ax exists and is contained in the space ℓ∞ for each x∈hdp. Apparently ankk=1∞∈hdpβ. This establishes the necessity of condition in (71).
We recall thatyk=dkΔxk or equivalently xk=∑j=k∞yj/dj for each k∈ℕ. This leads us to the following equality
(73)∑k=1∞ankxk=∑k=1∞ank∑j=k∞yjdj=∑k=1ank∑j=k∞Δxj=∑j=1∞∑k=1jankyjdjfor each n∈ℕ. Since hdp and ℓ∞ are BK spaces, Lemma 30 ascertains that there exists a positive number M<∞ such that
(74)Axℓ∞≤Mxhdpfor all x∈hdp. Therefore, by the aid of Hölder’s inequality together with (73) and Theorem 5, we get
(75)Axℓ∞yℓp=supn∑k=1∞∑j=1kanjyk/dkyℓp≤supn∑k=1∞1dkq∑j=1kanjq1/q<∞for all x∈hdp. This establishes the necessity of condition in (72).
Conversely, assume that conditions in (71) and (72) hold and take any x∈hdp. Then ankk=1∞∈hdpβ which implies that Ax exists. Again applying Hölder’s inequality and the fact that y=yk∈ℓp, we deduce that
(76)Axℓ∞=supn∑k=1∞∑j=1kanjykdk≤supn∑k=1∞1dkq∑j=1kanjq1/q∑k=1ykp1/p<∞.
Consequently,A∈hdp,ℓ∞. This completes the proof.Theorem 32.
A∈hdp,c if and only if (71) and (72) hold, and for each k∈ℕ, there exists αk∈ℝ such that
(77)limn⟶∞1dk∑j=1kanj=αk.Proof.
LetA∈hdp,c. Then Ax exists and is contained in the space c for all x∈hdp. Since c⊂ℓ∞,Ax∈ℓ∞. Thus the necessity of conditions in (71) and (72) follows from Theorem 31. Let us consider x=Gek and define the matrix D=dnk by
(78)dnk=1dk∑j=1kanjfor all n,k=1,2,3,⋯. Then by virtue of equality (73), we get that
(79)Ax=AGek=DEGek=Dek=dnkn∈ℕ.
SinceAx∈c, so dnkn∈ℕ∈c. This proves the necessity of the condition in (77).
Conversely, assume that the conditions in (71), (72) and (77) hold. Then ankk∈ℕ∈hdpβ for each n∈ℕ which implies that Ax exists for all x∈hdp. Therefore we again obtained equality (73). Then, one can notice that the conditions in (71) and (77) correspond to the conditions in (43) and (40), respectively, with dnk instead of ank. This concludes that Dy=Ax∈c. Hence A∈hdp,c.Replacingc by c0 in Theorem 32, we obtain the following corollary:Corollary 33.
A∈hdp,c0 if and only if (71) and (72) hold, and (77) also holds with αk=0 for all k.Theorem 34.
A=ank∈hdp,ℓ1 if and only if (71) holds, and
(80)supN∈ℕFinite∑k=1∞∑n∈N1dk∑j=1kanjq<∞,1<p<∞.Proof.
This is similar to the proof of Theorem31. Hence details are omitted.Theorem 35.
A=ank∈hdp,hd if and only if the condition in (71) holds and
(81)limn⟶∞1dk∑j=1kanj=0,forallk∈ℕ(82)supk∈ℕ1dk∑n=1∞dn∑j=1kanj−an+1,jq1/q<∞,1<p<∞.Proof.
Let1<p<∞ and suppose that A∈hdp,hd such that Ax exists and is in the space hd for each x∈hdp. Thus, we clerly see that ankk=1∞∈hdpβ which proves the necessity of condition in (71).
By using the relationyk=dkΔxk or equivalently xk=∑j=k∞yj/dj for each k∈ℕ between the terms of the sequences x=xk and y=yk, we reach the equality in (73) which can be written as in the following
(83)∑k=1∞ankxk=∑k=1∞ank∑j=k∞yjdj=∑k=1∞∑j=1kanjykdk=∑k=1∞dnkykwhere the infinite matrix D=dnk is defined as dnk=1/dk∑j=1kanj for each n,k∈ℕ. Since hdp and hd are BK spaces, we revisit Lemma 30 that there exists a positive real number N such that
(84)Axhd≤Nxhdpfor all x∈hdp. Therefore, by using Hölder’s inequality together with (73) and Theorem 5, we get
(85)Axhdxhdp=supk∑n=1∞dnΔAnxq1/qyℓp≤supk1dk∑n=1∞dn∑j=1kanj−an+1,jq1/q<∞for all x∈hdp. This shows the necessity of condition in (82). Moreover, since Ax=Dy∈hd for every x∈hdp the necessity of condition in (81) is also seen.
Conversely, suppose that the conditions in (71), (81), and (82) hold. Let us take any x∈hdp. Then An∈hdpβ which implies that Ax exists. The rest of proof follows the similar path to that of Theorem 31 with the following inequality
(86)Axhd=supk∑n=1∞dnΔAnx≤supk1dk∑n=1∞dn∑j=1kanj−an+1,jq1/q∑k=1ykp1/p<∞.
Thus,A∈hdp,hd. This completes the proof.Since the proof of the following results can be done similary to that of Theorem35, we give them without their proofs.Corollary 36.
The followings hold.(i)
The infinite matrixA=ank∈hdp,bv if and only if the condition in (71) holds and(87)supm∈ℕ1dmq∑n∑k=1mank−an−1,kq<∞(ii)
The infinite matrixA=ank∈hdp,bs if and only if the condition in (71) holds and(88)supm,n∈ℕ1dmq∑k=1n∑j=1makjq<∞.(iii)
The infinite matrixA=ank∈hdp,cs if and only if the conditions in (71) and (88) hold, and(89)limn⟶∞∑k=1m1dk∑j=1kanj=αkexistsforeachk∈ℕ.Theorem 37.
A=ank∈ℓ1,hdp if and only if
(90)limn⟶∞ank=0,forallk∈ℕ,(91)supn∑n=1∞dnank−an+1,kp1/p<∞,1<p<∞.Proof.
Sincehdp is a BK space by Corollary 3, we apply ([8], Example 8.4.1) to obtain that A∈ℓ1,hdp if and only if the columns of A form a bounded set in hdp.
We have(92)Akhdp=∑n=1∞dnank−an+1,kp1/pforallk.
Hence the setAk:k∈ℕ is bounded in hdp if and only if supkAkhdp<∞, which gives the condition in (91). Moreover Ak∈c0 for all k, which gives the condition in (87). This completes the proof.Now we give the following corollaries without their proofs since the proofs are followed similarly to that of Theorem37.Corollary 38.
The followings hold.(i)
The infinite matrixA=ank∈bv0,hdp if and only if the condition in (63) holds, and(93)supk∈ℕ∑n=1∞dn∑k=1mank−an+1,kp<∞.(ii)
The infinite matrixA=ank∈bs,hdp if and only if(94)supK∈ℕfinite∑k=1∞∑n∈Kdnbnk−bn+1,kp,(95)limn⟶∞∑k=1∞bnk=0where the matrix B=bnk is defined as bnk=ank−an,k+1.
(iii)
The infinite matrixA=ank∈cs0,hdp if and only if the condition in (94) holds and(96)limn⟶∞bnk=0where the matrix B=bnk is defined as bnk=ank−an,k+1.


## 6. Conclusion

Most recently, Malkowsky at al [12] defined the generalized Hahn sequence space hd, where d is an unbounded monotone increasing sequence of positive real numbers, and characterized several classes of bounded linear operators or matrix transformations from hd into λ=ℓ∞,c,c0,ℓ1,hd, and also from λ=ℓ∞,c,c0,ℓ1 into hd. Moreover, the norms of the corresponding bounded linear operators and the Hausdorff measure of non-compactness for the operators in the above classes, and one application given by a tridiagonal matrix to present a Fredholm operator from hd into itself were studied in [12].Malkowsky [14] established the characterisations of the classes of bounded linear operators from the generalised Hahn sequence space hd into the spaces c0, c and c∞. Moreover, he proved estimates for the Hausdorff measure of noncompactness of bounded linear operators from hd into c, and identities for the Hausdorff measure of noncompactness of bounded linear operators from hd to c0, and then he used these results to characterise the classes of compact operators from hd to c and c0.Dolic′anin-Dekic′ and Gilic′ [20] established the characterisations of the classes of bounded linear operators from the generalised Hahn sequence space hd into the spaces w0, w and w∞. Then they proved estimates for the Hausdorff measure of noncompactness of bounded linear operators from hd into w, and identities for the Hausdorff measure of noncompactness of bounded linear operators from hd into w0, and then they used these results to characterise the classes of compact operators from hd to w0 and w.In this article, we defined new generalized Hahn sequence spacehdp, where d=dkk=1∞ is an unbounded monotone increasing sequence of positive real numbers with dk≠0 for all k∈ℕ, and 1<p<∞. Then, we proved some topological properties and showed some inclusion relations. Moreover, we calculate the α−, β−, and γ−duals of hdp. Furthermore, we characterize the new matrix classes hd,λ, where λ=bv,bvp,bv∞,bs,cs,, and μ,hd, where μ=bv,bv0,bs,cs0,cs. In the last section, we prove the necessary and sufficient conditions of the matrix transformations from hdp into λ=ℓ∞,c,c0,ℓ1,hd,bv,bs,cs, and from μ=ℓ1,bv0,bs,cs0 into hdp.



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