Kyungpook Mathematical Journal 2018; 58(2): 271-289
On Some Spaces Isomorphic to the Space of Absolutely q-summable Double Sequences
Hüsamettin Çapan and Feyzi Başar*
Graduate School of Natural and Applied Sciences, İstanbul University, Beyazıt Campus, 34134 - Vezneciler/İstanbul, Turkey, e-mail : husamettincapan@gmail.com, İnönü University, 44280 - Malatya, Turkey, e-mail : feyzibasar@gmail.com
*Corresponding Author.
Received: July 15, 2017; Accepted: June 6, 2018; Published online: June 23, 2018.
© Kyungpook Mathematical Journal. All rights reserved.

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Abstract

Let 0 < q < ∞. In this study, we introduce the spaces and of q-bounded variation double sequences and q-summable double series as the domain of four-dimensional backward difference matrix Δ and summation matrix S in the space ℒq of absolutely q-summable double sequences, respectively. Also, we determine their α- and β-duals and give the characterizations of some classes of four-dimensional matrix transformations in the case 0 < q ≤ 1.

Keywords: summability theory, double sequence, difference sequence space, double series, alpha-dual, beta-dual, matrix domain of 4-dimensional matrices, matrix transformations
1. Introduction

We denote the set of all complex valued double sequences by Ω which forms a vector space with coordinatewise addition and scalar multiplication. Any vector subspace of Ω is called as a double sequence space.

By ℳu, we denote the space of all bounded double sequences, that is

$Mu:={x=(xkl)∈Ω:‖x‖∞=supk,l∈ℕ∣xkl∣ <∞},$

which is a Banach space with the norm ‖ · ‖; where ℕ = {0, 1, 2,…}.

If for every ɛ > 0 there exists N = N(ɛ) ∈ ℕ and L ∈ ℂ such that |xklL| < ɛ for all k, l > N, then we call that the double sequence x = (xkl) ∈ Ω is convergent to L in the Pringsheim’s sense (shortly, p-convergent to L) and write p − limk;l→∞xkl = L; where ℂ denotes the complex field (see Pringsheim [16]). We denote the space of all p-convergent double sequences by .

It is well-known that in single sequence spaces a convergent single sequence is bounded. But, in double sequence spaces a p-convergent double sequence may be unbounded. A double sequence is called boundedly convergent to L in the Pringsheim’s sense (shortly, bp-convergent to L), where L is the p-limit of x. We denote the space of such sequences by .

Throughout the text the summation without limits runs from 0 to ∞, for example ∑k;l xkl means that $∑k,l=0∞xkl$, and unless stated otherwise, we assume that ϑ denotes any of the symbols p or bp.

We denote the space of all absolutely q-summable double sequences by ℒq, that is,

$Lq:={x=(xkl)∈Ω:∑k,l∣xkl∣q<∞}, (0

If we take q = 1, we obtain the space ℒu of all absolutely summable double sequences.

Let $ekl=(eijkl)$ be a double sequence defined by

$eijkl:={1,(i,j)=(k,l),0,(i,j)≠(k,l)$

for all i, j, k, l ∈ ℕ and e = ∑k;lekl (coordinatewise sums), is a double sequence that all elements are one. All considered spaces are supposed to contain Φ, the set of all finitely non-zero double sequences; i.e.,

$Φ:={x=(xkl)∈Ω:∃ N∈ℕ ∀ (k,l)∈ℕ2[0,N]2, xkl=0} :=span {ekl:k,l∈ℕ}.$

Let λ be a space of double sequences, converging with respect to some linear convergence rule ϑ − lim : λ → ℂ. The sum of a double series ∑i;j xij with respect to this rule is defined by $ϑ-∑i,jxij=ϑ-limm,n→∞∑i,j=0m,nxij$. Then, the α-dual λα and the β(ϑ)-dual λβ(ϑ) of a double sequence space λ are respectively defined by

$λα:={a=(akl)∈Ω:∑k,l∣aklxkl∣<∞ for all x=(xkl)∈λ},λβ(ϑ):={a=(akl)∈Ω:ϑ-∑k,laklxkl exists for all x=(xkl)∈λ}.$

It is easy to see for any two spaces λ and μ of double sequences that μαλα whenever λμ.

Let λ and μ be two double sequence spaces, and A = (amnkl) be any four-dimensional complex infinite matrix. Then, we say that A defines a matrix mapping from λ into μ and we write A : λμ, if for every sequence x = (xkl) ∈ λ the A-transform Ax = {(Ax)mn}m;n∈ℕ of x exists and belongs to μ; where

$(Ax)mn=ϑ-∑k,lamnklxkl for each m,n∈ℕ.$

We define the ϑ-summability domain$λA(ϑ)$ of A in a space λ of double sequences by

$λA(ϑ):={x=(xkl)∈Ω:Ax=(ϑ-∑k,lamnklxkl)m,n∈ℕexists and is in λ}.$

We say with the notation (1.1) that A maps the space λ into the space μ if $λ⊂μA(ϑ)$ and we denote the set of all four dimensional matrices transforming the space λ into the space μ by (λ : μ). Thus, A = (amnkl) ∈ (λ : μ) if and only if the double series on the right side of (1.1) converges in the sense of ϑ for each m, n ∈ ℕ, i.e, Amnλβ(ϑ) for all m, n ∈ ℕ and every xλ, and we have Axλ for all xλ; where Amn = (amnkl)k;l∈ℕ for all m, n ∈ ℕ. In this paper, we only consider bp-summability domain.

For all k, l, m, n ∈ ℕ, we say that A = (amnkl) is a triangular matrix if amnkl = 0 for k > m or l > n or both, [1]. By following Adams [1], we also say that a triangular matrix A = (amnkl) is called a triangle if amnmn ≠ 0 for all m, n ∈ ℕ. Referring to Cooke [13, Remark (a), p. 22], one can conclude that every triangle matrix has an unique inverse which is also a triangle.

We shall write throughout for simplicity in notation for all k, l, m, n ∈ ℕ that

$Δ10akl=akl-ak+1,l,Δ10klamnkl=amnkl-amn,k+1,l,Δ01akl=akl-ak,l+1,Δ01klamnkl=amnkl-amnk,l+1,Δ11akl=Δ10(Δ01akl),Δ11klamnkl=Δ10kl(Δ01klamnkl), =Δ01(Δ10akl), =Δ01kl(Δ10klamnkl).$

The four dimensional backward difference matrix Δ = (dmnkl) is defined by

$dmnkl:={(-1)m+n-(k+l),m-1≤k≤m and n-1≤l≤n,0,otherwise$

for all k, l, m, n ∈ ℕ. We suppose that the terms of the double sequences x = (xkl) and y = (ykl) are connected with the relation

$ykl=(Δx)kl={x00,k,l=0,x0l-x0,l-1,k=0 and l≥1,xk0-xk-1,0,l=0 and k≥1,xk-1,l-1-xk-1,l-xk,l-1+xkl,k,l≥1$

for all k, l ∈ ℕ. Additionally, a direct calculation gives the inverse Δ−1 = S = (smnkl) of the matrix Δ as follows:

$smnkl:={1,0≤k≤m and 0≤l≤n,0,otherwise$

for all k, l, m, n ∈ ℕ. Here, we can redefine the relation between the double sequences x = (xkl) and y = (ykl) by summation matrix S as follows:

$xkl=(Sy)kl=∑i,j=0k,lyij$

for all k, l ∈ ℕ.

It is worth mentioning here that Altay and Başar [2] have defined the spaces and by using summation matrix S and also Demiriz and Duyar [14] recently defined the spaces ℳu(Δ) and by using backward difference matrix Δ, as folllows:

$BS:={x=(xkl)∈Ω:supk,l∈ℕ∣(Sx)kl∣ <∞},CSϑ:={x=(xkl)∈Ω:Sx={(Sx)kl}k,l∈ℕ∈Cϑ},Mu(Δ):={x=(xkl)∈Ω:supk,l∈ℕ∣(Δx)kl∣ <∞},Cϑ(Δ):={x=(xkl)∈Ω:Δx={(Δx)kl}k,l∈ℕ∈Cϑ}.$

In this study, we introduce the spaces and of all double sequences whose Δ-transforms and S-transforms are absolutely q-summable, that is,

$BVq:={x=(xkl)∈Ω:∑k,l∣(Δx)kl∣q<∞},LSq:={x=(xij)∈Ω:∑k,l∣(Sx)kl∣q<∞}.$

One can easily observe that the sets and are the domain of the backward difference matrix Δ and summation matrix S in the space ℒq which are q-normed spaces with

$‖x‖^BVq=∑k,l∣(Δx)kl∣q and ‖x‖^LSq=∑k,l∣(Sx)kl∣q$

for 0 < q ≤ 1, and normed spaces with

$‖x‖BVq=[∑k,l∣(Δx)kl∣q]1/q and ‖x‖LSq=[∑k,l∣(Sx)kl∣q]1/q$

for 1 < q < ∞, respectively. In the special case q = 1, we obtain the space , defined by Altay and Başar in [2], and the space .

2. New Sequence Spaces

In the present section, we examine some topological properties of the spaces and and also give important inclusion theorems related to them.

### Theorem 2.1

The spacesandare linearly isomorphic to the spaceq, where 0 < q < ∞.

Proof

We will only show with 0 < q < ∞.

Let 0 < q < ∞. With the notation of (1.2), consider the transformation T from to ℒq defined by xTx = Δx. Then, clearly T is linear and injective. Let y ∈ ℒq and define the sequence x = Sy as in (1.3). Then, we have Δx = Δ(Sy) = y which gives with 0 < q ≤ 1 and with 1 < q < ∞, i.e., . Hence, T is surjective and is norm preserving.

This completes the proof.

Since and , we can give following theorem without proof.

### Theorem 2.2

The setsandare linear spaces with the coordinatewise addition and scalar multiplication, and the following statements hold:

(i) Let 0 < q < 1. Then,andare complete q-normed spaces withand, respectively.

(ii) Let 1 ≤ q <. Then,andare Banach spaces withand, respectively.

Now, we define the double sequences $bkl=(bijkl)$ and $dkl=(dijkl)$ by

$bijkl:={1,i≥k and j≥l,0,otherwise,dijkl:={1,(i,j)=(k,l),(k+1,l+1),-1,(i,j)=(k+1,l),(k,l+1),0,otherwise$

for all i, j, k, l ∈ ℕ. Then it is obvious that the sets {e, ekl, bkl, dkl; k, l ∈ ℕ} ⊂ and {dkl; k, l ∈ ℕ} ⊂ . These double sequences will be used in the rest of the study.

### Definition 2.3

([18, p. 225]) A double sequence space λ is said to be monotone if xu = (xklukl) ∈ λ (coordinatwise product) for every x = (xkl) ∈ λ and u = (ukl) ∈ {0, 1}ℕ×ℕ, where {0, 1}ℕ×ℕ denotes the set of all double sequences consisting of 0’s and 1’s.

If λ is monotone, then λα = λβ(ϑ), but the converse is not true in general.

### Theorem 2.4

The spacesandare not monotone, where 0 < q <.

Proof

Let λ be a double sequence space. To show λ is not monotone, we must find a sequence u = (ukl) ∈ {0, 1}ℕ×ℕ such that xu = (xklukl) =λ for a sequence x = (xkl) ∈ λ.

Let us define the double sequence u = (ukl) by

$ukl:={0,k or l odd,1,otherwise$

for all k, l ∈ ℕ. Then , but . Hence, the space is not monotone.

To show is not monotone take u = ekl. Then, , but .

### Theorem 2.5

Let 0 < q <. Then, the inclusionis strict.

Proof

Let x = (xkl) ∈ ℒq. Then, by neglecting negative indexed terms of x, we obtain

$‖x‖^BVq=∑k,l∣xk-1,l-1-xk-1,l-xk,l-1+xkl∣q ≤4∑k,l∣xkl∣q=4‖x‖^Lq$

for 0 < q ≤ 1 and by using Minkowski’s inequality

$‖x‖BVq=(∑k,l∣xk-1,l-1-xk-1,l-xk,l-1+xkl∣q)1/q ≤4(∑k,l∣xkl∣q)1/q=4‖x‖Lq$

for 1 < q < ∞, that is, for 0 < q < ∞. Also, by , the inclusion is strict.

Since backward difference matrix Δ and summation matrix S are opposite working matrices we can give the following inclusion theorem without proof.

### Theorem 2.6

Let 0 < q <. Then, the inclusionis strict.

### Theorem 2.7

Let 1 < q <. Then, the setsq and do not contain each other.

Proof

It is immediate that and . Consider the sequence x = (xkl) defined by

$xkl:=(-1)k+l(k+1)(l+1)$

for all k, l ∈ ℕ. Since q > 1, the series

$∑k,l∣xkl∣q=∑k,l1[(k+1)(l+1)]q$

is convergent, that is, x ∈ ℒq. Nevertheless, we get from

$(Δx)kl={1,k,l=0,(-1)l2l+1l(l+1),k=0 and l≥1,(-1)k2k+1k(k+1),l=0 and k≥1,(-1)k+l(k+1)(l+1)+(k+1)l+k(l+1)+klkl(k+1)(l+1),k,l≥1$

that the series

$∑k,l∣(Δx)kl∣=1+∑l=1∞2l+1l(l+1)+∑k=1∞2k+1k(k+1)+∑k,l=1∞(k+1)(l+1)+(k+1)l+k(l+1)+klkl(k+1)(l+1) ≥1+∑l=1∞l+1l(l+1)+∑k=1∞k+1k(k+1)+∑k,l=1∞(k+1)(l+1)kl(k+1)(l+1) =1+∑l=1∞1l+∑k=1∞1k+∑k,l=1∞1kl$

diverges which gives the fact that . Therefore, .

### Theorem 2.8

Let 1 < q <. Then, the setsu and do not contain each other.

Proof

One can easily see that and . Consider the sequence Δx = {(Δx)kl} as in (2.1) for all k, l ∈ ℕ. Then, we obtain

$∑k,l∣{S(Δx)}kl∣q=∑k,l∣xkl∣q=∑k,l1[(k+1)(l+1)]q<∞,$

i.e., , but Δx =∈ ℒu by Theorem 2.7.

### Theorem 2.9

Let 0 < q < 1. Then, the setsu and do not contain each other.

Proof

It is clear that and . Define x = (xkl) by

$xkl:=(-1)k+l[(k+1)(l+1)]1/q$

for all k, l ∈ ℕ. Since 1=q > 1, the series

$∑k,l∣xkl∣=∑k,l1[(k,l)(l+1)]1/q$

is convergent. On the other hand, we see from

$(Δx)kl={1,k,l=0,(-1)ll1/q+(l+1)1/q[l(l+1)]1/q,k=0 and l≥1,(-1)kk1/q+(k+1)1/q[k(k+1)]1/q,l=0 and k≥1,(-1)k+l[(k+1)(l+1)]1/q+[(k+1)l]1/q[kl(k+1)(l+1)]1/q+(-1)k+l[k(l+1)]1/q+(kl)1/q[kl(k+1)(l+1)]1/q,k,l≥1$

that the series

$∑k,l∣(Δx)kl∣q=1+∑l=1∞|l1/q+(l+1)1/q[l(l+1)]1/q|q+∑k=1∞|k1/q+(k+1)1/q[k(k+1)]1/q|q+∑k,l=1∞|[(k+1)(l+1)]1/q+[(k+1)l]1/q+[k(l+1)]1/q+(kl)1/q[kl(k+1)(l+1)]1/q|q≥1+∑l=1∞|(l+1)1/q[l(l+1)]1/q|q+∑k=1∞|(k+1)1/q[k(k+1)]1/q|q+∑k,l=1∞|[(k+1)(l+1)]1/q[kl(k+1)(l+1)]1/q|q=1+∑l=1∞1l+∑k=1∞1k+∑k,l=1∞1kl$

is divergent. Hence, .

### Theorem 2.10

Let 0 < q < 1. Then, the setsq and do not contain each other.

Proof

It is easy to see that and . If we consider the sequence x in (2.2), then it is immediate that .

Let 0 < q < s < ∞. It is known that the inclusions ℒq ⊂ ℒs ⊂ ℳu strictly hold. By combining this fact with Theorem 2.1, we can give the following theorem without proof.

### Theorem 2.11

Let 0 < q < s <. Then, the inclusionsandstrictly hold.

### Theorem 2.12

Let λ denotes any of the spacesu or and 1 < q <. Then, neither of the spacesand λ includes the other one.

Proof

It is clear that λ. Define x = (xkl) and y = (ykl) by

$xkl:=∑i,j=0k,l1(i+1)(j+1) and ykl:={1,k=0 and l even,0,otherwise$

for all k, l ∈ ℕ. Then, since

$(Δx)kl:=1(k+1)(l+1) and (Δy)kl:={(-1)k+l,k=0,1 and l∈ℕ,0,otherwise$

one can conclude that and . Hence, the spaces and λ are overlap but neither contains the other.

3. Dual Spaces

In this section, we give the α- and β(bp)-duals of the spaces and in the case 0 < q ≤ 1. It is worth mentioning that although the alpha dual of a double sequence space is unique its beta dual may be more than one with respect to ϑ-convergence rule. By λ, we mean that {λ(n−1)ζ}ζ for a double sequence space λ and n ∈ ℕ1, the set of positive integers. It is well-known that $Luα=Mu$ and $Muα=Lu$.

### Theorem 3.1

Let 0 < q ≤ 1. Then, the followings hold for all k ∈ ℕ1:

$BVqnα:={Lu,n=2k-1,Mu,n=2k,LSqnα:={Mu,n=2k-1,Lu,n=2k.$
Proof

Let 0 < q ≤ 1.

(i) $BVqα=Lu$.

$Lu⊂BVqα$: Let us consider a = (akl) ∈ ℒu and . Then, we have by relation (1.3) that y ∈ ℒq ⊂ ℒu which gives

$∑k,l∣aklxkl∣≤∑k,l∣akl∣∑i,j=0k,l∣yij∣≤ ‖a‖Lu‖y‖Lu,$

that is, ax ∈ ℒu. Hence, $a∈BVqα$.

$BVqα⊂Lu$: Suppose that $a∈BVqαLu$. Then, we have ax ∈ ℒu for but a =∈ ℒu. If we consider , then we obtain ae = a =∈ ℒu, that is, $a∉BVqα$, a contradiction. Hence, a must be in ℒu.

(ii) $LSqα=Mu$.

$Mu⊂LSqα$: Take a ∈ ℳu and . Then, we get by

$‖ax‖Lu≤ ‖a‖∞‖x‖Lu$

that $a∈LSqα$.

$LSqα⊂Mu$: Consider $a∈LSqαMu$. Since a =∈ ℳu, there exist the subsequences of natural numbers {k(i)} and {l(i)}, at least one of them is strictly increasing, such that

$ak(i),l(i)>(i+1)2/q$

for all i ∈ ℕ. If we define the sequence x by using the double sequence dkl as

$x=∑i(i+1)-2/qdk(i),l(i),$

then we obtain that

$∑i,j=0k,lxij:={(i+1)2/q,k=k(i) and l=l(i),0,k≠k(i) and l≠l(i)$

which leads us to the fact that

$∑k,l|∑i,j=0k,lxij|q=∑i1(i+1)2<∞,$

i.e., . Nevertheless, by choosing k(i) + 1 < k(i + 1) or l(i) + 1 < l(i + 1) we get

$∑k,l∣aklxkl∣>∑i∣ak(i),l(i)xk(i),l(i)∣ >∑i(i+1)2/q∣xk(i),l(i)∣ =∑i1=∞,$

i.e., $a∉LSqα$, a contradiction. Hence, a ∈ ℳu.

Now, by using the facts $Luα=Mu$ and $Muα=Lu$, one can easily show that (3.1) holds with mathematical induction.

We give the following lemma which is needed in proving the β(ϑ)-dual of the spaces and .

### Lemma 3.2

([17]) Let 0 < q ≤ 1. Then, a four-dimensional matrix A = (amnkl) ∈ ( ) if and only if the following conditions hold:

$supm,n,k,l∈ℕ∣amnkl∣ <∞,$$∃αkl∈Cϑ-limm,n→∞ amnkl=αkl for each k,l∈ℕ.$

Now, we may give the beta-duals of the new spaces with respect to the ϑ-convergence rule using the technique in [4] and [5] for the spaces of single sequences.

Let us define the sets and via the double sequence u, as follows:

$BS(u):={a=(aij)∈Ω:au=(aijuij)i,j∈ℕ∈BS},CSϑ(u):={a=(aij)∈Ω:au=(aijuij)i,j∈ℕ∈CSϑ}.$

### Theorem 3.3

Let 0 < q ≤ 1. Then, the β(ϑ)-dual of the spaceis.

Proof

We will determine the necessary and sufficient conditions in order to the sequence t = (tmn) defined by

$tmn:=∑i,j=0m,naijxij; x=(xij)∈BVq$

for all m, n ∈ ℕ to be ϑ-convergent for a sequence a = (aij) ∈ Ω.

Let us define the sequence by the relation (1.3) which gives y = (ykl) ∈ ℒq. Then, we can write t = (tmn) in the matrix form, as follows:

$tmn=∑i,j=0m,nxijaij=∑i,j=0m,n(∑k,l=0i,jykl) aij =∑k,l=0m,n(∑i,j=k,lm,naij) ykl =∑k,l=0m,nbmnklykl=(By)mn,$

where the four-dimensional matrix B = (bmnkl) is defined by

$bmnkl:={∑i,j=k,lm,naij,0≤k≤m and 0≤l≤n,0,otherwise$

for all k, l, m, n ∈ ℕ. Now, it is easy to see that whenever if and only if whenever y = (ykl) ∈ ℒq which leads us to the fact that . Therefore, by using the conditions (3.2) and (3.3) of Lemma 3.2, we obtain the conditions

$supm,n,k,l∈ℕ∣bmnkl∣ =supm,n,k,l∈ℕ|∑i,j=k,lm,naij|=supm,n,k,l∈ℕ|∑i,j=0m,naijbijkl|<∞,$$ϑ-limm,n→∞ bmnkl=ϑ-limm,n→∞∑i,j=0m,naijbijkl exists$

for all k, l ∈ ℕ. By means of (3.5) and (3.6), we can say that $abkl=(aijbijkl)∈{BS,CSϑ}$, in other words, .

This completes the proof.

### Theorem 3.4

Let 0 < q ≤ 1. Then, the β(ϑ)-dual of the spaceis the setu.

Proof

We prove the theorem it by the similar way used in the proof of Theorem 3.3.

Consider by (1.2). Then, x = (xkl) ∈ ℒq. Therefore, we obtain by applying Abel’s generalized transformation for double sequences that

$tmn=∑k,l=0m,naklykl =∑k,l=0m-1,n-1(Δ11akl)xkl+∑k=0m-1(Δ10akn)xkn+∑l=0n-1(Δ01aml)xml+amnxmn =∑k,l=0m,ncmnklxkl=(Cx)mn,$

where the four-dimensional matrix C = (cmnkl) is defined by

$cmnkl:={Δ11akl,0≤k≤m-1 and 0≤l≤n-1,Δ10akn,0≤k≤m-1 and l=n,Δ01aml,0≤l≤n-1 and k=m,amn,k=m and l=n,0,otherwise$

for all k, l, m, n ∈ ℕ. By using similar approach in Theorem 3.3, . Therefore, we get by Lemma 3.2 that

$ϑ-limm,n→∞ cmnkl=Δ11akl,$

i.e., ϑ − limm;n→∞cmnkl always exists for each k, l ∈ ℕ. Also, from the condition

$supm,n,k,l∈ℕ∣ cmnkl∣ <∞$

we have (amn) ∈ ℳu, (Δ01aml) ∈ ℳu, (Δ10akn) ∈ ℳu and (Δ11akl) ∈ ℳu for all k, l, m, n ∈ ℕ. It is easy to show that the condition a = (amn) ∈ ℳu is sufficient for the matrix C = (cmnkl) to be bounded for all k, l, m, n ∈ ℕ.

This completes the proof.

4. Matrix Transformations

In the present section, we characterize the classes (ℒq : ℒq1), ( ), ( ), ( ) and ( ) together with a corollary characterizing some classes of four-dimensional matrices without proof; where 0 < q ≤ 1 and 0 < q1< ∞.

### Theorem 4.1

Let 0 < q ≤ 1 and 0 < q1<. Then, A = (amnkl) ∈ (ℒq : ℒq1) if and only if the following condition holds:

$supk,l∈ℕ∑m,n∣amnkl∣q1 <∞.$
Proof

Let us consider A = (amnkl) ∈ (ℒq : ℒq1) with 0 < q ≤ 1 and 0 < q1< ∞. Then, Ax exists and belongs to ℒq1 for all x ∈ ℒq. Since ekl ∈ ℒq, we obtain

$∑m,n|∑i,jamnijeijkl|q1=∑m,n∣amnkl∣q1<∞$

for all k, l ∈ ℕ. Hence, (4.1) is necessary.

Conversely, suppose that the condition (4.1) holds and x = (xkl) ∈ ℒq. Since ℒq ⊂ ℒu for 0 < q ≤ 1, x also belongs to ℒu. Thus, we have

$∑m,n|∑k,lamnijxkl|q1≤∑m,n(∑k,l∣amnkl∣∣xkl∣)q1 ≤∑m,n(∣amnk0l0∣∑k,l∣xkl∣)q1 ≤‖x‖Luq1∑m,n∣amnk0l0∣q1<∞$

for any fixed k0, l0 ∈ ℕ. Therefore, A ∈ (ℒq : ℒq1).

This completes the proof.

### Theorem 4.2

Let 0 < q ≤ 1 and 0 < q1<. Then, A = (amnkl) ∈ ( ) if and only if the following condition holds:

$supk,l∈ℕ∑m,n|∑i,j=k,l∞amnij|q1<∞.$
Proof

We obtain the necessity of the condition (4.2) by choosing the double sequence .

Let us define by (1.3) which gives y = (ykl) ∈ ℒq. Then, we derive by the s, t-th rectangular partial sum of the series ∑i;j amnijxij that

$(Ax)mn[s,t]=∑i,j=0s,t(∑k,l=0i,jykl) amnij =∑k,l=0s,t(∑i,j=k,ls,tamnij) ykl$

for all m, n, s, t ∈ ℕ. Therefore, we see by letting s, t → ∞ that

$(Ax)mn=∑k,l(∑i,j=k,l∞amnij) ykl$

for all m, n ∈ ℕ. Thus, we get the desired result by the same way used in proving Theorem 4.1.

### Theorem 4.3

Let 0 < q ≤ 1 and 0 < q1<. Then, A = (amnkl) ∈ ( ) if and only if the following conditions hold:

$supk,l∈ℕ∑m,n∣Δ11klamnkl∣q1<∞,$$supk,l∈ℕ∑m,n∣Δ10klamnkl∣q1<∞,$$supk,l∈ℕ∑m,n∣Δ01klamnkl∣q1<∞,$$supk,l∈ℕ∑m,n∣amnkl∣q1<∞.$
Proof

Take by the relation x = Δy which gives y = (ykl) ∈ ℒq. Let us define the four-dimensional matrix $Ast=(amnklst)$ by

$amnklst:={amnkl,0≤k≤s and 0≤l≤t,0,k>s or l>t$

for each s, t ∈ ℕ and all m, n, k, l ∈ ℕ. By using generalized Abel transformation for double series, we obtain the equalities

$(Astx)mn=∑k,l=0s,tamnklxkl =∑k,l=0s-1,t-1(Δ11klamnkl) ykl+∑k=0s-1(Δ10klamnkl) ykt+∑l=0t-1(Δ01klamnsl) ysl+amnstyst =(Dsty)mn$

which gives that if and only if Dst ∈ (ℒq : ℒq1), where the four-dimensional matrix $Dst=(dmnklst)$ defined by

$dmnkllst:={Δ11klamnkl,0≤k≤s-1 and 0≤l≤t-1,Δ11klamnkl,0≤k≤s-1 and l=t,Δ01klamnkl,0≤l≤t-1 and k=s,amnkl,k=s and l=t,0,k>s or l>t$

for each s, t ∈ ℕ and all m, n, k, l ∈ ℕ. Now, one can easily derive the conditions (4.3)–(4.6) for all s, t ∈ ℕ.

By using Theorem 4.2 and Theorem 4.3, we can give the following two theorems without proof.

### Theorem 4.4

Let 0 < q ≤ 1. Then, A = (amnkl) ∈ ( ) if and only if the following conditions hold:

$supm,n,k,l∈ℕ|∑i,j=k,l∞amnij|<∞,$$ϑ-limm,n→∞∑i,j=k,l∞amnij exists for all k,l∈ℕ.$

### Theorem 4.5

Let 0 < q ≤ 1. Then, A = (amnkl) ∈ ( ) if and only if the conditions (3.2) and (3.3) hold.

### Theorem 4.6

Suppose that the elements of the four-dimensional infinite matrices E = (emnkl) and F = (fmnkl) are connected with the relation

$eklij=∑m,n=0k,lfmnij$

for all i, j, k, l, m, n ∈ ℕ and λ, μ be any given two double sequence spaces. Then, E ∈ (λ : μΔ) if and only if F ∈ (λ : μ) and also F ∈ (λ : μS) if and only if E ∈ (λ : μ).

Proof

Let x = (xij) ∈ λ. By using (4.9), we derive that

$∑i,j=0s,teklijxij=∑m,n=0k,l∑i,j=0s,tfmnijxij$

for all k, l, m, n, s, t ∈ ℕ and by letting s, t → ∞ that

$∑i,jeklijxij=∑m,n=0k,l∑i,jfmnijxij$

which lead us to the fact

$(Ex)kl=∑m,n=0k,l(Fx)mn$

for all k, l ∈ ℕ. Then, it is easy to see by (4.10) that ExμΔ whenever xλ if and only if Fxμ whenever xλ and similarly, FxμS whenever xλ if and only if Exμ whenever xλ.

This completes the proof.

As a consequence of Theorem 4.6, we can give the following corollary.

### Corollary 4.7

Let 0 < q ≤ 1, 0 < q1<and the elements of the fourdimensional matrices E = (emnkl) and F = (fmnkl) are connected with the relation (4.9). Then, the following statements hold:

(i) E = (emnkl) ∈ (ℒq : (Δ)) if and only if the conditions (3.2) and (3.3) hold with fmnkl instead of amnkl.

(ii) F = (emnkl) ∈ (ℒq : ) if and only if the conditions (3.2) and (3.3) hold with emnkl instead of amnkl.

(iii) E = (emnkl) ∈ (ℒq : ) if and only if the condition (4.1) holds with fmnkl instead of amnkl.

(iv) F = (emnkl) ∈ (ℒq : ) if and only if the condition (4.1) holds with emnkl instead of amnkl.

(v) E = (emnkl) ∈ ( ) if and only if the condition (4.2) holds with fmnkl instead of amnkl.

(vi) F = (emnkl) ∈ ( ) if and only if the condition (4.2) holds with emnkl instead of amnkl.

(vii) E = (emnkl) ∈ ( ) if and only if the conditions (4.3)–(4.6) hold with fmnkl instead of amnkl.

(viii) F = (emnkl) ∈ ( ) if and only if the conditions (4.3)–(4.6) hold with emnkl instead of amnkl.

(ix) E = (emnkl) ∈ ( ) if and only if the conditions (4.7) and (4.8) hold with fmnkl instead of amnkl.

(x) F = (emnkl) ∈ ( ) if and only if the conditions (4.7) and (4.8) hold with emnkl instead of amnkl.

(xi) E = (emnkl) ∈ ( : (Δ)) if and only if the conditions (3.2) and (3.3) hold with fmnkl instead of amnkl.

(xii) F = (emnkl) ∈ ( ) if and only if the conditions (3.2) and (3.3) hold with emnkl instead of amnkl.

5. Conclusion

As the domain of backward difference matrix in the space p of absolutely p-summable sequences, the space bvp of p-bounded variation single sequences were studied in the case 1 < p ≤ ∞ by Başar and Altay [5], and in the case 0 < p ≤ 1 by Altay and Başar [3]. Later, by introducing the space ℓ̂p as the domain of double band matrix B(r, s) in the space p Kirişçi and Başar [15] generalized the space bvp. Besides this, the space bvp was extended to the paranormed space bv(u, p) of single sequences by Başar et al. [6].

Recently, the space ℒq of absolutely q-summable double sequences with q > 1 was introduced by Başar and Sever [9], and some complementary results related to the space ℒq have been recently given by Yeşilkayagil and Başar [17]. We introduce the space as the domain of four dimensional summation matrix S in the space ℒq with 0 < q < ∞. It is natural to expect the extension of the space to the paranormed space as a generalization of the space $ℓ(p)¯$ derived by Choudhary and Misra [12] as the domain of the two dimensional summation matrix in the paranormed space (p) of single sequences. Our main goal is to investigate the space of q-bounded variation double sequences and is to extend to the results obtained for the space bvq. Of course, it is worth mentioning here that the domain of the backward difference matrix Δ in the paranormed space (p) and also the investigation of the results for double sequences corresponding to Başar et al. [6] remains open.

Additionally, one can generalize the main results of the present paper related to the space by using the four dimensional triangle matrix B(r, s, t, u) = {bmnkl(r, s, t, u)} instead of the four dimensional backward difference matrix Δ, where r, s, t, u ∈ ℝ with r, t ≠ 0 and

$bmnkl(r,s,t,u):={rt,(k,l)=(m,n),st,(k,l)=(m-1,n),ru,(k,l)=(m,n-1),su,(k,l)=(m-1,n-1),0,otherwise$

for all m, n, k, l ∈ ℕ. Furthermore, following Başar and Çapan [7, 8] and Çapan and Başar [10, 11], one can also extend the main results of this paper to the paranormed case.

References
1. Adams, CR (1933). On non-factorable transformations of double sequences. Proc Natl Acad Sci USA. 19, 564-567.
2. Altay, B, and Başar, F (2005). Some new spaces of double sequences. J Math Anal Appl. 309, 70-90.
3. Altay, B, and Başar, F (2007). The fine spectrum and the matrix domain of the difference operator Δ on the sequence space ℓp, (0 < p < 1). Commun Math Anal. 2, 1-11.
4. Başar, F, and Altay, B (2002). Matrix mappings on the space bs(p) and its β-, γ-, and -duals. Aligarh Bull Math. 21, 79-91.
5. Başar, F, and Altay, B (2003). On the space of sequences of p-bounded variation and related matrix mappings. Ukrainian Math J. 55, 136-147.
6. Başar, F, Altay, B, and Mursaleen, M (2008). Some generalizations of the space bvp of p-bounded variation sequences. Nonlinear Anal. 68, 273-287.
7. Başar, F, and Çapan, H (2017). On the paranormed spaces of regularly convergent double sequences. Results Math. 72, 893-906.
8. Başar, F, and Çapan, H (2019). On the paranormed space Mu(t) of double sequences. Bol Soc Parana Mat. 37, 99-111.
9. Başar, F, and Sever, Y (2009). The space ℒq of double sequences. Math J Okayama Univ. 51, 149-157.
10. Çapan, H, and Başar, F (2016). Some paranormed difference spaces of double sequences. Indian J Math. 58, 405-427.
11. Çapan, H, and Başar, F (2018). On the paranormed space ℒ(t) of double sequences. Filomat. 32, 1043-1053.
12. Choudhary, B, and Mishra, SK (1993). On Köthe-Toeplitz duals of certain sequence spaces and their matrix transformations. Indian J Pure Appl Math. 24, 291-301.
13. Cooke, RC (1950). Infinite matrices and sequence spaces. London: Macmillan and Co. Limited
14. Demiriz, S, and Duyar, O (2015). Domain of difference matrix of order one in some spaces of double sequences. Gulf J Math. 3, 85-100.
15. Kirişçi, M, and Başar, F (2010). Some new sequence spaces derived by the domain of generalized difference matrix. Comput Math Appl. 60, 1299-1309.
16. Pringsheim, A (1900). Zur Theorie der zweifach unendlichen Zahlenfolgen. Math Ann. 53, 289-321.
17. Yeşilkayagil, M, and Başar, F (2017). Domain of Riesz mean in the space . Filomat. 31, 925-940.
18. Zeltser, M (2001). Weak sequential completeness of β– duals of double sequence spaces. Anal Math. 27, 223-238.

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