Hausdorff measure of noncompactness of matrix operators on some new difference sequence spaces

Abstract

The new sequence spaces $X(r,s,t;\Delta)$ for $X\in\{l_{\infty}, c, c_{0}\}$ have been defined by using generalized means and difference operator. In this work, we establish identities or estimates for the operator norms and the Hausdorff measure of noncompactness of certain matrix operators on some new difference sequence spaces $X(r,s,t;\Delta )$ where $X\in\{l_{\infty}, c, c_{0},l_{p}\}$ ($1\leq{p}<\infty$), as derived by using generalized means. Further, we find the necessary and sufficient conditions for such operators to be compact by applying the Hausdorff measure of noncompactness. Finally, as applications we characterize some classes of compact operators between these new difference sequence spaces and some other BK-spaces.

1 Preliminaries and background

The study of sequence spaces has been very useful in many branches of analysis. Recently, some new sequence spaces have been defined by using matrix domain of a suitable matrix. Beside this, the Hausdorff measure of noncompactness is very useful in the classification of compact operators between Banach spaces.

The difference sequence spaces were introduced for the first time by Kizmaz in [1]. Afterwards, many authors have introduced and studied some new sequence spaces defined by using the difference operator. For example, in [2] Ahmad and Mursaleen, in [3] Colak and Et, in [4] Basar and Altay, in [5] Orhan, in [6] Polat and Altay, in [7] Aydin and Basar, and in [8] Basar and Altay have introduced and studied some new sequence spaces defined by using a difference operator. Some authors like Malkowsky and Savas [9], Altay and Basar [10], Mursaleen and Noman [11] and Basarir and Kara [12] have defined the sequence spaces by using the generalized weighted means. For the first time, in 2011 Polat et al. [13] have introduced the new sequence spaces by combining both the weighted means and the difference operator. Then in 2015 Manna et al. [14] have introduced new sequence spaces defined by using both the generalized means and the difference operator.

In this paper we obtain some identities or estimates for the operator norms and the Hausdorff measure of noncompactness of certain matrix operators on new difference sequence spaces defined by Manna et al. Further, we find the necessary and sufficient conditions for such operators to be compact and we consider some matrix mappings between the spaces of the generalized means (these have been defined in [15]) and difference sequence spaces derived by using the generalized means. Finally, as applications we characterize some classes of compact operators between these new difference sequence spaces and some other BK-spaces.

In this section, we give some related definitions and preliminary results.

We shall write w for the set of all complex sequences $x=(x_{k})_{k=0}^{\infty}$. Any vector subspace of w is called a sequence space. Let $l_{\infty}$, c, and $c_{0}$ denote the sets of all bounded, convergent and null sequences, respectively, and $l_{p}=\{ x\in{w}:\sum_{k=0}^{\infty}|x_{k}|^{p}<{\infty}, 1\leq{p}<{\infty}\}$. For an infinite matrix A and a sequence space X, the matrix domain of A is defined by $X_{A}=\{x\in{w}:Ax\in{X}\}$. By e and $e^{n}$ ($n\in{\mathbb{N}}$), we denote the sequences such that $e_{k}=1$ for $k=0,1,\ldots$ and $e_{n}^{(n)}=1$ and $e_{k}^{n}=0$ ($k\neq{n}$). Let bs and cs be the sequence spaces of all bounded and convergent series, respectively. A sequence $(b_{n})$ in a normed linear space $(X,\|{\cdot}\|)$ is called a Schauder basis for X if for every $x\in{X}$, there is a unique sequence $(\delta_{n})_{n=0}^{\infty }$ of scalars such that $x=\sum_{n=0}^{\infty}\delta_{n}b_{n}$. For definitions of K-space, FK-space, BK-space and AK-property we refer the reader to [16].

Throughout, the matrices are infinite matrices of complex number. Let $A=(a_{nk})_{n,k}$ be an infinite matrix with real or complex entries $a_{nk}$. We write $A_{n}$ as the sequence of the nth row of A, that is, $A_{n}=(a_{nk})_{k}$ for every n. In addition, if $x=(x_{k})\in{w}$ then we define the A-transform of x as the sequence $Ax=(A_{n}(x))_{n=0}^{\infty}$, that is,

$$ A_{n}(x)=\sum_{k=0}^{\infty}a_{nk}x_{k} \quad (n\in{\mathbb{N}}), $$

(1)

provided the series on the right converges for each $n\in{\mathbb{N}}$. For any two sequence spaces X and Y, we denote by $(X,Y)$ the class of all infinite matrices A that map X into Y. As is well known the β-dual a subset X of w is defined by

$$ {X^{\beta}}=\bigl\{ a=(a_{k})\in{w} :ax=(a_{k}x_{k}) \in{\mathit{cs}}\text{ for all } x=(x_{k})\in{X}\bigr\} . $$

Thus $A\in{(X,Y)}$ if and only if $A_{n}\in{X^{\beta}}$ for all $n\in {\mathbb{N}}$ and $Ax\in{Y}$ for all $x\in{X}$. An infinite matrix $T=(t_{nk})$ is said to be triangle if $t_{nk}=0$ for $k>n$ and $t_{nn}\neq0$, $n\in{\mathbb{N}_{0}}$ (where $\mathbb{N}_{0}=\{0, 1, 2,\ldots\}$).

If $X\supset{\phi}$ is a BK-space and $a=(a_{nk})\in {w}$, then we write

$$ \|{a}\|_{X}^{*}=\sup_{x\in{S_{X}}} \Biggl\vert \sum_{k=0}^{\infty}a_{k}x_{k} \Biggr\vert , $$

(2)

provided the expression on the right is defined and finite which is the case whenever $a\in{X^{\beta}}$ [17].

2 The Hausdorff measure of noncompactness

As is well known, one of the best methods in the characterization of compact operators between the Banach spaces depends on applying the Hausdorff measure of noncompactness. Therefore, many authors like Kara and Basarir in [18, 19], Basarir and Kara in [20, 21], Kara et al. in [22] and Alotaibi et al. in [23] characterized some classes of compact operators on the spaces in their papers by using the Hausdorff measure of noncompactness.

Let S and M be subsets of a metric space $(X,d)$ and $\epsilon>0$. We say that S is an ϵ-net of M if for every $x\in{M}$ there exists $s\in{S}$ such that $d(x,s)<\epsilon$. The ϵ-net S of M is a finite ϵ-net if S is a finite set.

We denote by $M_{X}$ the collection of all bounded subsets of a metric space $(X,d)$. If $Q\in{M_{X}}$, we denote by $\chi(Q)$ the Hausdorff measure of noncompactness of the set Q and define it by

$$ \chi(Q)=\inf\{\epsilon>0: Q\text{ has a finite }\epsilon\text{-net in }X\}. $$

The function $\chi:M_{X}\rightarrow[0,\infty )$ is called the Hausdorff measure of noncompactness [24]. We can find the basic properties of the Hausdorff measure of noncompactness in [25].

We give an estimate for the Hausdorff measure of noncompactness in Banach spaces with a Schauder basis with the following theorem.

Theorem 2.1

[25]

Let X be a Banach space with a Schauder basis $(b_{k})_{k=0}^{\infty }$, $P_{n}:X\rightarrow{X}$ ($n\in{\mathbb{N}}$) be the projector onto the linear span of $\{b_{0}, b_{1}, \ldots, b_{n}\}$ and ${Q\in {M_{X}}}$. Then we have

$$ \frac{1}{a}\cdot\limsup\Bigl(\sup_{x\in{Q}}\bigl\Vert {(I-P_{n}) (x)}\bigr\Vert \Bigr)\leq \chi{(Q)}\leq{\limsup{\Bigl(\sup _{x\in{Q}}\bigl\Vert {(I-P_{n}) (x)}\bigr\Vert \Bigr)}}, $$

where $a=\limsup_{n\to{\infty}}\|{I-P_{n}}\|$ and I denotes the identity operator on X.

Now, we show how to compute the Hausdorff measure of noncompactness in the spaces $c_{0}$ and $l_{p}$ ($1\leq{p}<\infty$).

Theorem 2.2

[26]

Let Q be a bounded subset of the normed space X and X is $c_{0}$ or $l_{p}$ ($1\leq{p}<\infty$). If $P_{n}:X\rightarrow{X}$ ($n\in{\mathbb {N}}$) is the operator defined by $P_{n}(x)=x^{[n]}=(x_{0},x_{1},\ldots ,x_{n},0,0,0,\ldots)$ for all $x=(x_{k})_{k=0}^{\infty}\in{X}$, then

$$ \chi(Q)=\lim_{n\to{\infty}}\Bigl(\sup_{x\in{Q}}\bigl\Vert {(I-P_{n}) (x)}\bigr\Vert \Bigr). $$

We have the following result concerning with the Hausdorff measure of noncompactness in the matrix domain of triangles in normed sequence spaces.

Theorem 2.3

[27]

Let X be a normed sequence space, T a triangle, and ${\chi}_{T}$ and χ denote the Hausdorff measure of noncompactness on $M_{{X}_{T}}$ and $M_{X}$. Then ${{\chi}_{T}(Q)={\chi}({T}(Q))}$ for all $Q\in{M_{{X}_{T}}}$.

Let X and Y be Banach spaces and $L\in{B(X,Y)}$. Then we denote by $\|{L}\|_{\chi}$ the Hausdorff measure of noncompactness of L and define it by

$$ \|{L}\|_{\chi}={\chi} {\bigl(L(S_{X})\bigr)}={ \chi} {\bigl(L({\overline{B}}_{X})\bigr)}, $$

(3)

where $S_{X}=\{x\in{X}:\|{x}\|=1\}$ is the unit sphere and $\overline{B}_{X}=\{x\in{X}:\|{x}\|\leq{1}\}$ is the unit ball in X, and

$$ L\text{ is compact if and only if }\|{L}\|_{\chi}=0\quad \mbox{[27]}. $$

(4)

3 The sequence spaces $X(r,s,t;\Delta)$ for $X\in\{ l_{\infty}, c, c_{0}, l_{p}\}$ ($1\leq{p}<\infty$)

In this subsection we first introduce the generalized means that given by Mursaleen and Noman in [15].

We first define the sets ℧ and $\mho_{0}$ as follows:

$$\begin{aligned}& \mho=\bigl\{ u=(u_{n})_{n=0}^{\infty} \in{w}:u_{n}\neq{0}\ \forall{n}\bigr\} , \\& \mho _{0}=\bigl\{ u=(u_{n})_{n=0}^{\infty}\in{w}:u_{0}\neq{0} \bigr\} . \end{aligned}$$

Let $r=(r_{n})\in{\mho}$, $t=(t_{n})\in{\mho}$, and $s=(s_{n})\in{\mho _{0}}$. For any sequence $x=(x_{n})\in{w}$, we define the sequence $y=(y_{n})$ of generalized means of x by:

$$ y_{n}=\frac{1}{r_{n}}\sum _{k=0}^{n}{s_{n-k}t_{k}x_{k}} \quad (n\in{\mathbb{N}_{0}}). $$

(5)

Further, we define the infinite matrix $A(r,s,t)$ of generalized means by

$$ \bigl(A(r,s,t)_{nk}\bigr)=\left \{ \textstyle\begin{array}{l@{\quad}l} s_{n-k}t_{k}/r_{n} ,& 0\leq{k}\leq{n}, \\ 0 ,& k>n \end{array}\displaystyle \right . $$

for all $n,k\in{\mathbb{N}}$. By using the notation (1) and (3), we see that y is the $A(r,s,t)$-transform of x, that is, $y=(A(r,s,t))x$ for all $x\in{w}$. Let $D_{0}^{(s)}=\frac{1}{s_{0}}$ and

$$ D_{n}^{(s)}=\frac{1}{s_{0}^{n+1}}\left \vert \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} s_{1} & s_{0} & 0 & 0 & \cdots& 0\\ s_{2} & s_{1} & s_{0} & 0 & \cdots& 0\\ \vdots& \vdots& \vdots& \vdots& \cdots& \vdots\\ s_{n-1} & s_{n-2} & s_{n-3} & s_{n-4} & \cdots& s_{0}\\ s_{n} & s_{n-1} & s_{n-2} & s_{n-3} & \cdots& s_{1} \end{array}\displaystyle \right \vert \quad (n=1, 2, 3, \ldots), $$

so the inverse of $A(r,s,t)$ is the triangle $B=(b_{nk})_{nk}$, which is defined by

$$ b_{nk}=\left \{ \textstyle\begin{array}{l@{\quad}l} (-1)^{n-k}\frac{D_{n-k}^{(s)}}{t_{n}}r_{k} ,& 0\leq{k}\leq{n}, \\ 0 ,& k>n. \end{array}\displaystyle \right . $$

We now ready to introduce the sequence space $X(r,s,t;\Delta)$ where $X\in\{l_{\infty}, c, c_{0}\}$ as the matrix domain of triangle $A(r,s,t;\Delta)$ as follows:

$$ X(r,s,t;\Delta)=\Biggl\{ x=(x_{n})\in{w}: \Biggl(\frac{1}{r_{n}}\sum _{k=0}^{n}{s_{n-k}t_{k}} {\Delta}x_{k}\Biggr)_{n}\in{X}\Biggr\} , $$

which is a combination of the generalized means and the difference operator $\Delta({\Delta}x_{k}=x_{k}-x_{k-1}, x_{-1}=0)$ and by using notation (1) we have $X(r,s,t;\Delta)=X_{A(r,s,t;\Delta)}$ where $A(r,s,t;\Delta)=A(r,s,t)\cdot{\Delta}$. It is clear that if X is a sequence space, then $X(r,s,t;\Delta)$ is a sequence space too and we call that the difference sequence space derived by using generalized means.

Lemma 3.1

[14]

The sequence spaces $X(r,s,t;\Delta)$ where $X\in{\{l_{\infty}, c, c_{0}\}}$ are BK-spaces with the norm given by

$$\begin{aligned} \|{x}\|_{X(r,s,t;\Delta)} =&\sup_{n}\Biggl\vert \frac{1}{r_{n}}\sum_{k=0}^{n}{s_{n-k}t_{k}} {\Delta}x_{k}\Biggr\vert \\ =&\sup_{n}\bigl\vert \bigl(A(r,s,t;\Delta)x \bigr)_{n}\bigr\vert \end{aligned}$$

and so they are linearly isomorphic to the spaces $X\in{\{l_{\infty}, c, c_{0}\}}$, respectively.

We mention the following basic lemmas, which will be used in proving our results.

Lemma 3.2

Let X be a BK-space with AK or $X=l_{\infty}$ and $\overline {\Delta{X}}=X(r,s,t;\Delta)$. If $a=(a_{k})\in{(\overline{\Delta {X}})^{\beta}}$, then $\tilde{a}=(\tilde{a}_{k})\in{X^{\beta}}$ and we have

$$ \sum_{k=0}^{\infty}a_{k}x_{k}= \sum_{k=0}^{\infty}\tilde{a}_{k}y_{k}, $$

(6)

which holds for every $x=(x_{k})\in{X(r,s,t;\Delta)}$ with $y=A(r,s,t)\Delta{x}$ where

$$ \tilde{a}_{k}=\sum_{j=0}^{n} \sum_{k=0}^{n-j}(-1)^{k} \frac {D_{k}^{(s)}}{t_{k+j}}a_{j}r_{j}\quad (k\in{\mathbb{N}}). $$

(7)

Proof

It is immediate by the same technique as ([11], Theorem 4.5). □

Lemma 3.3

Let X be a BK-space with AK property or $X=l_{\infty}$ and $\overline{{\Delta}{X}}=X(r,s,t;\Delta)$. Then we have

$$ \|{a}\|_{\overline{{\Delta}{X}}}^{*}{}={}\|{\tilde {a}}\|_{X^{\beta}} $$

for all $a=(a_{k})\in{(\overline{{\Delta}{X}})^{\beta}}$, where $\tilde {a}=(\tilde{a}_{k})$ is the sequence defined by (7).

Proof

Let $a=(a_{k})\in{(\overline{{\Delta}{X}})^{\beta}}$. By applying Lemma 3.2 we have $\tilde{a}=(\tilde{a}_{k})\in{X^{\beta}}$ and equality (6) holds for all sequences $x=(x_{k})\in{\overline{{\Delta }{X}}}$ and $y=(y_{k})\in{X}$. Further, it follows by the equality in Lemma 3.1 that $x\in{S_{\overline{{\Delta}{X}}}}$ if and only if $y\in {S_{X}}$. So, we derive from (2) and (6) that

$$\begin{aligned} \|{a}\|_{\overline{{\Delta}{X}}}^{*} = &\sup_{x\in {S_{\overline{{\Delta}{X}}}}}\Biggl\vert \sum_{k=0}^{\infty}a_{k} \Delta{x_{k}}\Biggr\vert \\ = &\sup_{y\in{S_{X}}}\Biggl\vert \sum _{k=0}^{\infty}\tilde{a}_{k}y_{k} \Biggr\vert \\ = &\|{\tilde{a}}\|_{X^{\beta}}. \end{aligned}$$

This completes the proof. □

Throughout this paper we assume $A=(a_{nk})$ is an infinite matrix and $\tilde{A}=(\tilde{a}_{nk})$ is the associated matrix which is defined by

$$ \tilde{a}_{nk}=\sum_{j=0}^{n} \sum_{k=0}^{n-j}(-1)^{k} \frac {D_{k}^{(s)}}{t_{k+j}}a_{nj}r_{j} \quad (n,k\in{\mathbb{N}}), $$

(8)

and provided the series on the right converge for all $n,k\in{\mathbb {N}}$. Then we have the following.

Lemma 3.4

Let X be a BK-space with AK property or $X=l_{\infty}$ and ${\overline{{\Delta}{X}}}=X(r,s,t;\Delta)$, Y be any sequence space and $A=(a_{nk})$ be an infinite matrix. If ${A\in{(\overline{\Delta {X}},Y)}}$, then $\tilde{A}\in{(X,Y)}$ such that $Ax=\tilde{A}y$ for all $x\in{X}$ and connected sequence $y=A(r,s,t)\cdot{\Delta}x$, where $\tilde{A}=(\tilde{a}_{nk})$ is the associated matrix defined by (8).

Proof

Let $x\in{\overline{{\Delta}{X}}}$ and $y\in{Y}$ be connected by equation (5) and $A\in(\overline{{\Delta}{X}},Y)$. By applying Lemma 3.3, we have $\tilde{A}_{n}\in{X^{\beta}}$ for all $n \in{ \mathbb{N}}$ and the equality $Ax=\tilde{A}y$ holds, hence $\tilde{A}y\in{Y}$. Because every $y \in{Y}$ is the associated sequence of some $x \in{\overline{{\Delta}{X}}}$, we conclude that $\tilde{A}\in{(X,Y)}$. This completes the proof. □

Theorem 3.1

Let $\overline{\Delta{X}}=X(r,s,t;\Delta)$ where X be a BK-space with AK or ${X=l_{\infty}}$, $A=(a_{nk})$ is an infinite matrix and $\tilde{A}=(\tilde{a}_{nk})$ is the associated matrix. If A is in any of the classes $(\overline{\Delta{X}},l_{\infty})$, $(\overline{\Delta {X}},c)$, or $(\overline{\Delta{X}},c_{0})$, then we have

$$ \|{L_{A}}\|=\|{A}\|_{(\overline{\Delta {X}},l_{\infty})}=\sup_{n}\|{ \tilde{A}_{n}}\| _{X}^{*}< {\infty}. $$

Proof

By using Lemma 3.1 we know that the spaces $\overline {\Delta{X}}$ and X are linearly isomorphic. Now by combining ([16], Remark 1.2(a)), and Lemma 3.3 the proof is complete. □

Theorem 3.2

Let X be a BK-space with AK or $X=l_{\infty}$ and $\overline {\Delta{X}}=X(r,s,t;\Delta)$. If A is in $(\overline{\Delta {X}},l_{1})$, then

$$ \|{A}\|_{(\overline{\Delta{X}},l_{1})}\leq\| {L_{A}}\|\leq4\cdot\|{A} \|_{(\overline{\Delta{X}},l_{1})}, $$

where

$$ \Vert {A}\Vert _{(\overline{\Delta{X}},l_{1})}=\sup_{N\in {F}}\biggl\Vert \sum_{n\in{N}}\tilde{A}_{n}\biggr\Vert _{X}^{*}< {\infty}. $$

Proof

By using Lemma 3.1 we know the spaces $\overline{\Delta {X}}$ and X are linearly isomorphic. Now by combining ([16], Remark 1.2(b)) and Lemma 3.3 the proof is complete. □

Theorem 3.3

Let ${\overline{\Delta{l_{1}}}}=l_{1}(r,s,t;\Delta)$ and $1\leq {p}<{\infty}$. If $A\in(\overline{\Delta{l_{1}}},l_{p})$, then

$$ \|{L_{A}}\|=\|{A}\|_{(\overline{\Delta {l_{1}}},l_{p})}=\sup_{k} \Biggl(\sum_{n=0}^{\infty}|\tilde {a}_{nk}|^{p}\Biggr)^{\frac{1}{p}}< {\infty}. $$

Proof

By using [14] and [11] we deduce $A\in {(\overline{\Delta{l_{1}}},l_{p})}$ if and only if the expression on the right of (8) converges and

$$ \sup_{k}\Biggl(\sum_{n=0}^{\infty}| \tilde{a}_{nk}|^{p}\Biggr)< {\infty} $$

and

$$ \sup_{k}\Bigl(\max_{0\leq{k}\leq{m}}\bigl\vert a_{mk}^{(n)}\bigr\vert \Bigr)< {\infty} \quad (\forall n\in { \mathbb{N}}), $$

where

$$ a_{mk}^{(n)}=\sum_{j=0}^{n} \sum_{k=0}^{n-j}(-1)^{k} \frac {D_{k}^{(s)}}{t_{k+j}}a_{nj}r_{j} \quad ({0\leq{k}\leq{m}}, m\in{ \mathbb{N}}). $$

 □

By using Theorem 3.2, we immediate obtain the following result.

Corollary 3.1

Let ${\overline{\Delta{l_{p}}}}=l_{p}(r,s,t;\Delta)$ ($1\leq{p}<{\infty }$) and $A\in{(\overline{\Delta{l_{p}}},l_{1})}$. Then we have

$$ \|{A}\|_{(\overline{\Delta{l_{p}}},l_{1})}^{(r)}\leq \|{L_{A}}\|\leq4\cdot\|{A} \|_{(\overline{\Delta {l_{p}}},l_{1})}^{(r)}, $$

where

$$ \|{A}\|_{(\overline{\Delta{l_{p}}},l_{1})}^{(r)}=\sup_{N\in{F_{r}}}\Biggl(\sum _{k=0}^{\infty}\biggl|\sum_{n\in{N}} \tilde {a}_{nk}\biggr|^{q}\Biggr)^{\frac{1}{q}}\quad (r\in{ \mathbb{N}}). $$

Proof

Let $X=l_{p}$ in Theorem 3.2. □

Now, as a consequence of Corollary 3.1 we have the following lemma.

Lemma 3.5

Let ${\overline{\Delta{l_{1}}}}=l_{1}(r,s,t;\Delta)$ and $1\leq {p}<{\infty}$. If $A\in{(l_{p},\overline{\Delta{l_{1}}})}$, then

$$ \|{A}\|_{(l_{p},\overline{\Delta{l_{1}}})}^{(r)}\leq \|{L_{A}}\|\leq4\cdot\|{A} \|_{(l_{p},\overline {\Delta{l_{1}}})}^{(r)}, $$

where

$$ \|{A}\|_{(l_{p},\overline{\Delta{l_{1}}})}^{(r)}=\sup_{N\in{F_{r}}}\Biggl(\sum _{k=0}^{\infty}\biggl\vert \sum _{n\in{N}}\tilde {a}_{nk}\biggr\vert ^{q} \Biggr)^{\frac{1}{q}}\quad (r\in{\mathbb{N}}), $$

and $\tilde{a}_{nk}$ is defined by (8).

Proof

Because the sequence spaces $\overline{\Delta{l_{1}}}$ and $l_{1}$ are norm-isomorphic,

$$ A\in{(l_{p},\overline{\Delta{l_{1}}})}\cong{A \in{(l_{p},l_{1})}}. $$

Now this is a special case of ([16], Remark 1.2(b)) when $X=l_{p}$.

This completes the proof. □

4 Compact operators on the spaces $X(r,s,t;\Delta)$ for $X\in \{l_{\infty}, c, c_{0}, l_{p}\}$ ($1\leq{p}<\infty$)

In this section, we apply our results to obtaining some identities or estimates for the Hausdorff measure of noncompactness of certain matrix operators on the spaces $X(r,s,t;\Delta)$ (where $X\in \{l_{\infty}, c, c_{0}, l_{p}\}$ ($1\leq{p}<\infty$)). Also, we consider the necessary and sufficient conditions for such operators to be compact.

Lemma 4.1

[28]

Let $X\supset{\phi}$ be a BK-space. Then we have:

• (a) If $A\in{(X,l_{\infty})}$, then

  $$ 0\leq{\|{L_{A}}\|_{\chi}}\leq{\limsup_{n\to{\infty }}} \|{A_{n}}\|_{X}^{*}. $$

• (b) If $A\in{(X,c_{0})}$, then

  $$ \|{L_{A}}\|_{\chi}={\limsup_{n\to{\infty}}}\| {A_{n}}\|_{X}^{*}. $$

• (c) If X has AK or $X=l_{\infty}$ and $A\in {(X,c)}$, then

  $$ \frac{1}{2}\cdot\limsup_{n\to{\infty}}\|{A_{n}-\alpha} \| _{X}^{*}\leq\|{L_{A}}\|_{\chi}\leq{ \limsup_{n\to{\infty }}}\|{A_{n}-\alpha}\|_{X}^{*}, $$

  where $\alpha=(\alpha_{k})$ with $\alpha_{k}=\lim_{n\to{\infty}} a_{nk}$ for all $k\in{\mathbb{N}}$.

Now, by combining Lemmas 3.3, 3.4, and 4.1 we have the next result.

Theorem 4.1

Let X be a BK-space with AK or $X=l_{\infty}$ and $\overline {\Delta{X}}=X(r,s,t;\Delta)$. Then we have:

• (a) If $A\in{(\overline{\Delta{X}},l_{\infty})}$, then

  $$ 0\leq{\|{L_{A}}\|_{\chi}}\leq{\limsup _{n\to{\infty}}\biggl(\sum_{n}| \tilde{A}_{n}|\biggr)}, $$

  (9)

  and

  $$ L_{A}\textit{ is compact if }\lim_{n\to{\infty}} \biggl(\sum_{n}|\tilde{A}_{n}|\biggr)=0. $$

  (10)

• (b) If $A\in{(\overline{\Delta{X}},c_{0})}$, then

  $$ {\|{L_{A}}\|_{\chi}}={\limsup _{n\to{\infty}}}\biggl(\sum_{n}| \tilde{A}_{n}|\biggr), $$

  (11)

  and

  $$ L_{A}\textit{ is compact if and only if } \lim _{n\to{\infty}}\biggl(\sum_{n}| \tilde{A}_{n}|\biggr)=0. $$

  (12)

• (c) If $A\in{(\overline{\Delta{X}},c)}$, then

  $$ \frac{1}{2}\cdot{\limsup_{n\to{\infty}}\biggl(\sum _{n}|\tilde{A}_{n}-\tilde{\alpha }|\biggr) \leq\|{L_{A}}\|_{\chi}}\leq{\limsup_{n\to{\infty }}} \biggl(\sum_{n}|\tilde{A}_{n}-\tilde{ \alpha}|\biggr) $$

  (13)

  and

  $$ L_{A}\textit{ is compact if and only if }\lim _{n\to{\infty}}\biggl(\sum_{n}| \tilde{A}_{n}-\tilde{\alpha}|\biggr)=0 , $$

  (14)

  where $\tilde{\alpha}=(\tilde{\alpha}_{k})$ with $\tilde{\alpha }_{k}=\lim_{n\to{\infty}}{ \tilde{a}_{nk}}$ for all $k\in{\mathbb{N}}$.

Proof

It is clear that (10), (12), and (14) are, respectively, obtained from (9), (11), and (13) by using (4). So, to prove (9), (11), and (13) we have the following.

Because $\overline{\Delta{X}}$ is a BK-space, hence by combining Lemma 3.3 and parts (a) and (b) of Lemma 4.1 we obtain (9) and (11).

To prove (13), if $A\in{(\overline{\Delta{X}},c)}$, then by Lemma 3.4 we have ${\tilde {A}\in{(X,c)}}$. So, by part (c) of Lemma 4.1 we have

$$ \frac{1}{2}\cdot{\limsup_{n\to{\infty}}\biggl(\sum _{n}|\tilde{A}_{n}-\tilde{\alpha }|\biggr) \leq\|{L_{\tilde{A}}}\|_{\chi}}\leq{\limsup_{n\to {\infty}}} \biggl(\sum_{n}|\tilde{A}_{n}-\tilde{ \alpha}|\biggr), $$

(15)

where $\tilde{\alpha}=(\tilde{\alpha}_{k})$ and $\tilde{\alpha}_{k}=\lim_{n\to{\infty}}{ \tilde{a}_{nk}}$ for all $k\in{\mathbb{N}}$. Now, we write $S=S_{X}$ and $\overline{\Delta{S}}=S_{\overline{\Delta {X}}}$ for short. Then we obtain by (3) and ([16], Remark 1.2(c))

$$ \|{L_{A}}\|_{\chi}=\chi\bigl(L_{A}( \overline{\Delta{S}})\bigr)=\chi (A{\overline{\Delta{S}}}) $$

(16)

and

$$ \|{L_{\tilde{A}}}\|_{\chi}=\chi\bigl(L_{\tilde{A}}(S) \bigr)=\chi (\tilde{A}S). $$

(17)

Also, by using the equality in Lemma 3.1 we have $x\in{\overline{\Delta {S}}}$ if and only if $y\in{S}$, and since $Ax=\tilde{A}y$ by Lemma 3.4, we derive that $A\overline{\Delta{S}}=\tilde{A}S$. This result together (16) and (17) leads to $\|{L_{A}}\|_{\chi }=\|{L_{\tilde{A}}}\|_{\chi}$. So, we get (13) from (15). This completes the proof. □

The following example will show that it is possible for $L_{A}$ in (10) to be compact but $\lim_{n\to{\infty}}(\sum_{n}|\tilde{A}_{n}|)\neq {0}$. Hence, in general, we have just ‘if’ in (10) of Theorem 4.1(a).

Example 4.1

Let $\overline{\Delta{X}}=X(r,s,t;\Delta)$ where $X\in\{l_{\infty}, c, c_{0}\}$. Then let for $n=1$ the sequence $s=(s_{n})=0$ and for every $n\neq{1}$, $s=(s_{n})=1$. Also, let the sequences $r=(r_{n})=1$, $t=(t_{n})=1$ are constant. Define the matrix $A=(a_{nk})$ by

$$ a_{nk}=\left \{ \textstyle\begin{array}{l@{\quad}l} \frac{s_{n-k}t_{k}}{r_{n}}& (0\leq{k}\leq{2}), \\ 0& (k\geq{3}). \end{array}\displaystyle \right . $$

Then, for every $x=(x_{n})\in{\overline{\Delta{X}}}$, we have $Ax=[\frac{s_{2}t_{0}}{r_{2}}x_{0}e^{(0)}+\frac {s_{1}t_{1}}{r_{2}}x_{1}e^{(1)}+\frac{s_{0}t_{2}}{r_{2}}x_{2}e^{(2)}]$. So $A\in{(\overline{\Delta{X}},l_{\infty})}$. It is clear that $L_{A}$ is of finite rank so is compact. By using (8) we obtain $\tilde{A}_{n}=1$, for all $n\in{\mathbb{N}}$. Thus, we have that $\sum_{n}{|\tilde{A}_{n}|}=1$ for all $n\in{\mathbb{N}}$, which implies that $\lim_{n\to{\infty}}\sum_{n}{|\tilde{A}_{n}|}=1\neq{0}$.

Let $F_{r}$ ($r\in{\mathbb{N}}$) be the subcollection of F consisting of all nonempty and finite subsets of $\mathbb{N}$ with elements that are greater than r, that is,

$$ F_{r}=\{N\in{F}:n>r \text{ for all }n\in{N}\} \quad (r\in{ \mathbb{N}}). $$

Lemma 4.2

[11]

Let $X\supset{\phi}$ be a BK-space. If $A\in{(X,l_{1})}$, then

$$ \lim_{r\to{\infty}}\biggl(\sup_{N\in{F_{r}}}\biggl\Vert { \sum_{n\in {N}}A_{n}}\biggr\Vert _{X}^{*}\biggr)\leq \Vert {L_{A}}\Vert _{\chi}\leq 4\cdot\lim_{r\to{\infty}}\biggl(\sup _{N\in{F_{r}}}\biggl\Vert {\sum_{n\in {N}}A_{n}} \biggr\Vert _{X}^{*}\biggr). $$

By using Lemma 4.2, we have the next result.

Theorem 4.2

Let X be a BK-space with AK and $\overline{\Delta {X}}=X(r,s,t;\Delta)$. If ${A\in(\overline{\Delta{X}},l_{1})}$, then

$$ \lim_{r\to{\infty}}\|{A}\|_{(\overline{\Delta {X}},l_{1})}^{(r)} \leq\|{L_{A}}\|_{\chi}\leq4\cdot\lim_{r\to {\infty}} \|{A}\|_{(\overline{\Delta{X}},l_{1})}^{(r)}, $$

(18)

where

$$ \Vert {A}\Vert _{(\overline{\Delta{X}},l_{1})}^{(r)}=\sup_{N\in {F_{r}}} \biggl\Vert {\sum_{n\in{N}}\tilde{A}_{n}} \biggr\Vert _{X}^{*}\quad ({r\in {\mathbb{N}}}) $$

and

$$ L_{A}\textit{ is compact if and only if }\lim _{r\to{\infty}}\| {A}\|_{(\overline{\Delta{X}},l_{1})}^{(r)}=0. $$

(19)

Proof

By combining Lemma 3.3 and Lemma 4.2 we obtain (18). By using (4), we get (19) from (18). □

By using the previous results we have the next result.

Theorem 4.3

Let $\overline{\Delta{l_{1}}}=l_{1}(r,s,t;\Delta)$ and $1\leq{p}\leq {\infty}$. If $A\in{\overline{(\Delta{l_{1}}},l_{p})}$, then

$$ \|{L_{A}}\|_{\chi}=\lim_{r\to{\infty}} \Biggl(\sup_{k}\Biggl(\sum_{n=r}^{\infty}| \tilde{a}_{nk}|^{p}\Biggr)^{\frac{1}{p}}\Biggr) $$

(20)

and

$$ L_{A}\textit{ is compact if and only if }\lim _{r\to{\infty}}\Biggl(\sup_{k}\Biggl(\sum _{n=r}^{\infty}|\tilde{a}_{nk}|^{p} \Biggr)^{\frac{1}{p}}\Biggr)=0. $$

(21)

Proof

It can be similarly proved by the same technique as Theorem 3.9 in [15]. □

In the following example we show that there exist matrix operators in the class $B(\overline{\Delta{l_{1}}},l_{p})$ which are not compact ($1\leq{p}<\infty$), that is, the limit in (20) may not be zero.

Example 4.2

Let $A=(a_{nk})$ be the infinite matrix defined by

$$ a_{nk}=\left \{ \textstyle\begin{array}{l@{\quad}l} s_{n-k}t_{k}/r_{n} ,& 0\leq{k}\leq{1}, \\ 0 ,& k>2, \end{array}\displaystyle \right . $$

where $s=(s_{n})=1$ if $n=0$ and $s=(s_{n})=0$ for every $n\neq0$ and $r=(r_{n})=1=t=(t_{n})$. Then $A\in(\overline{\Delta{l_{1}}},l_{1})$ because $\overline{\Delta{l_{1}}}$ is the matrix domain of A in $l_{1}$ and hence $A\in{(\overline{\Delta{l_{1}}},l_{p})}$ ($1\leq {p}<\infty$). Further, it is easy to show that the associated matrix à is the identity matrix, that is, $\tilde{a}_{nn}=1$ and $\tilde{a}_{nk}=0$ for $k\neq{n}$ ($n\in{\mathbb{N}}$). Now, let $r\in {\mathbb{N}}$ be given. Then we have, for every $k\in{\mathbb{N}}$,

$$ \sum_{n=r}^{\infty}|\tilde{a}_{nk}|^{p}= \left \{ \textstyle\begin{array}{l@{\quad}l} 1 ,& k\geq{r}, \\ 0 ,& k< r. \end{array}\displaystyle \right . $$

This implies that

$$ \sup_{k}\Biggl(\sum_{n=r}^{\infty}| \tilde{a}_{nk}|^{p}\Biggr)^{\frac{1}{p}}=1\quad (r\in { \mathbb{N}}), $$

which shows that by (20) that $\|{L_{A}}\|_{\chi}=1\neq {0}$ and hence $L_{A}$ is not compact.

By using Lemmas 3.5 and 4.2 we have the following result on the Hausdorff measure of noncompactness.

Corollary 4.1

Let $\overline{\Delta{l_{1}}}=l_{1}(r,s,t;\Delta)$ and $1\leq{p}\leq {\infty}$. If $A\in{(l_{p},{\overline{{\Delta}l_{1}})}}$, then

$$ \|{A}\|_{(l_{p},\overline{{\Delta}l_{1}})}^{(r)}\leq \|{L_{A}} \|_{\chi}\leq4\cdot\|{A}\| _{(l_{p},\overline{{\Delta}l_{1}})}^{(r)}, $$

(22)

where

$$ \|{A}\|_{(l_{p},\overline{{\Delta}l_{1}})}^{(r)}=\sup_{k}\Biggl(\sum _{n=r}^{\infty}|\tilde{a}_{nk}|^{q} \Biggr)^{\frac{1}{q}} \quad ({r\in {\mathbb{N}}}) $$

and

$$ L_{A}\textit{ is compact if and only if } \lim _{r\to{\infty}}\| {A}\|_{(l_{p},\overline{{\Delta}l_{1}})}^{(r)}=0. $$

(23)

Proof

We know the spaces $l_{1}$ and $\overline{\Delta {l_{1}}}$ are norm-isomorphic. So, the proof is complete by Lemmas 3.5 and 4.2. □

5 Some applications

In this section, by applying our results in the previous sections we consider some relations between the spaces of generalized means and difference sequence spaces derived by using the generalized means and we characterize some classes of compact operators between these new difference sequence spaces and some other BK-spaces. Also, we derive some identities and estimates for the operator norms and the Hausdorff measure of noncompactness on these spaces.

Throughout, $X(r,s,t)$ is the sequence space of the generalized means and for more details we refer the reader to [11]; further, for an arbitrary BK-space X, we put ${{\overline{X}=X(r,s,t)}}$, and $\overline{\Delta{X}}=X(r,s,t;\Delta)$.

The first result is a consequence of Theorem 3.1 and Theorem 4.1.

Corollary 5.1

Let X be a BK-space with AK or ${X=l_{\infty}}$. Then:

If A is in any of the classes $(\overline{\Delta{X}},\overline {l_{\infty}})$, $(\overline{\Delta{X}},\overline{c})$, or $(\overline {\Delta{X}},\overline{c_{0}})$ then

$$ \|{L_{A}}\|=\|{A}\|_{(\overline{\Delta {X}},\overline{l_{\infty}})}=\sup_{n}\|{ \tilde{A}_{n}}\| _{X}^{*}< {\infty}, $$

where $\tilde{A}_{n}=(\tilde{a}_{nk})$ is the associated matrix defined by

$$ \tilde{a}_{nk}=\sum_{j=k}^{\infty }{(-1)}^{j-k}D_{j-k}^{(s)}r_{k}a_{nj}/t_{j} \quad (n,k\in{\mathbb{N}}). $$

(24)

Further:

• (a) If $A\in{(\overline{\Delta{X}},\overline{l_{\infty }})}$, then

  $$ 0\leq\|{L_{A}}\|_{\chi}\leq\limsup_{n\to{\infty}}\| {\tilde{A}_{n}}\|_{X}^{*} $$

  and

  $$ L_{A}\textit{ is compact if }\lim_{n\to{\infty}}\|{\tilde {A}_{n}}\|_{X}^{*}=0. $$

• (b) If $A\in{(\overline{\Delta{X}},\overline{c})}$, then

  $$ \frac{1}{2}\cdot\limsup_{n\to{\infty}}\|{\tilde{A}_{n}- \tilde{\alpha }}\|_{X}^{*}\leq\|{L_{A}} \|_{\chi}\leq\limsup_{n\to {\infty}}\|{\tilde{A}_{n}- \tilde{\alpha}}\|_{X}^{*} $$

  and

  $$ L_{A}\textit{ is compact if and only if }\lim_{n\to{\infty}} \| {\tilde{A}_{n}-\tilde{\alpha}}\|_{X}^{*}=0, $$

  where $\tilde{\alpha}=(\tilde{\alpha}_{k})$ with $\tilde{\alpha }_{k}=\lim_{n\to{\infty}}{ \tilde{a}_{nk}}$ for all $k\in{\mathbb {N}}$.

• (c) If $A\in{(\overline{\Delta{X}},\overline{c_{0}})}$, then

  $$ \|{L_{A}}\|_{\chi}=\limsup_{n\to{\infty}}\|{\tilde {A}_{n}}\|_{X}^{*} $$

  and

  $$ L_{A}\textit{ is compact if and only if }\lim_{n\to{\infty}} \| {\tilde{A}_{n}}\|_{X}^{*}=0. $$

Proof

We know that the spaces $\overline{l_{\infty}}$, c̅, $\overline{c_{0}}$ and $l_{\infty}$, c, $c_{0}$ are norm-isomorphic, respectively. Now by Theorems 3.1 and 4.1 the proof is completed. □

The next result is a consequence of Theorem 3.2 and Theorem 4.2.

Corollary 5.2

Let X be a BK-space with AK or ${X=l_{\infty}}$. If $A\in {(\overline{\Delta{X}},\overline{l_{1}})}$, then

$$ \|{A}\|_{(\overline{\Delta{X}},\overline{l_{1}})}\leq \|{L_{A}}\|\leq4\cdot\|{A} \|_{(\overline{\Delta {X}},\overline{l_{1}})}, $$

where

$$ \Vert {A}\Vert _{(\overline{\Delta{X}},\overline{l_{1}})}=\sup_{N\in{F}}\biggl\Vert { \sum_{n\in{N}}\tilde{A}_{n}}\biggr\Vert _{X}^{*}< {\infty}, $$

and $\tilde{A}_{n}=(\tilde{a}_{nk})$ is similarly defined by (24).

Further:

If $A\in{(\overline{\Delta{X}},\overline{l_{1}})}$, then

$$ \|{A}\|_{(\overline{\Delta{X}},\overline {l_{1}})}^{(r)}\leq\|{L_{A}}\|_{\chi} \leq4\cdot\| {A}\|_{(\overline{\Delta{X}},\overline{l_{1}})}^{(r)}, $$

where

$$ \Vert {A}\Vert _{(\overline{\Delta{X}},\overline {l_{1}})}^{(r)}=\sup_{N\in{F_{r}}} \Bigl\Vert {\sum\tilde{A}_{n}}\Bigr\Vert _{X}^{*}< {\infty} $$

and

$$ L_{A}\textit{ is compact if and only if }\lim_{r\to{\infty}} \| {A}\|_{(\overline{\Delta{X}},\overline{l_{1}})}^{(r)}=0. $$

Proof

We know that the spaces $\overline{l_{1}}$ and $l_{1}$ are norm-isomorphic. Now by Theorems 3.2 and 4.2 the proof is complete. □

Now we have a special case of Theorem 3.3 in [28] in the new spaces $\overline{\Delta{X}}$ where $X\in\{l_{\infty}, c, c_{0}\}$.

Corollary 5.3

Let X be a BK-space with AK or ${X=l_{\infty}}$ and Y̅ denote any of the spaces $\overline{l_{\infty}}$, c̅, or $\overline{c_{0}}$ we and let A be an infinite matrix. Then:

If $A\in{(\overline{Y},\overline{\Delta{X}})}$ then we have

$$ \|{L_{A}}\|=\|{A}\|_{(\overline{Y},\overline {\Delta{l_{\infty}}})}=\sup_{n}\Biggl( \sum_{k=0}^{\infty}|\tilde {a}_{nk}| \Biggr)< {\infty}, $$

where $\tilde{A}_{n}=(\tilde{a}_{nk})$ is the associated matrix defined by (8).

Further:

• (a) If $A\in{(\overline{l_{\infty}},\overline{\Delta {l_{\infty}}})}$, then

  $$ 0\leq\|{L_{A}}\|_{\chi}\leq\limsup_{n\to{\infty}} \Biggl(\sum_{k=0}^{\infty}|\tilde{a}_{nk}| \Biggr) $$

  and

  $$ L_{A}\textit{ is compact if }\lim_{n\to{\infty}}\Biggl(\sum _{k=0}^{\infty }|\tilde{a}_{nk}| \Biggr)=0. $$

• (b) If $A\in{(\overline{c},\overline{\Delta{l_{\infty }}})}$, then

  $$ \frac{1}{2}\cdot\limsup_{n\to{\infty}}\Biggl(\sum _{k=0}^{\infty}|\tilde {a}_{nk}-\tilde{ \alpha}_{k}|\Biggr)\leq\|{L_{A}}\|_{\chi}\leq \limsup_{n\to{\infty}}\Biggl(\sum_{k=0}^{\infty}| \tilde{a}_{nk}-\tilde{\alpha}_{k}|\Biggr) $$

  and

  $$ L_{A}\textit{ is compact if and only if }\lim_{n\to{\infty}} \Biggl(\sum_{k=0}^{\infty}|\tilde{a}_{nk}- \tilde{\alpha}_{k}|\Biggr)=0, $$

  where $\tilde{\alpha}=(\tilde{\alpha}_{k})$ with $\tilde{\alpha }_{k}=\lim_{n\to{\infty}}{ \tilde{a}_{nk}}$ for all $k\in{\mathbb {N}}$.

• (c) If $A\in{(\overline{c},\overline{\Delta{l_{\infty }}})}$, then

  $$ \|{L_{A}}\|_{\chi}=\limsup_{n\to{\infty}}\Biggl(\sum _{k=0}^{\infty}|\tilde{a}_{nk}|\Biggr) $$

  and

  $$ L_{A}\textit{ is compact if and only if }\lim_{n\to{\infty}} \Biggl(\sum_{k=0}^{\infty}|\tilde{a}_{nk}| \Biggr)=0. $$

The spaces bs, cs, and $\mathit{cs}_{0}$ are BK-spaces with the same norm given by $\|{x}\|_{\mathit{bs}}=\sup_{n}|\sum_{k=0}^{n}x_{k}|$. Now, by using Theorems 3.1, 4.1 and 3.2, 4.2 we derive the next consequence.

Corollary 5.4

Let $X\supset{\phi}$ be a BK-space and A an infinite matrix. If A is in any of the classes $(\overline{\Delta{X}},\mathit{cs}_{0})$, $(\overline {\Delta{X}},\mathit{cs})$, or $(\overline{\Delta{X}},\mathit{bs})$, then

$$ \|{L_{A}}\|=\sup_{n}\|{\tilde{A}}\| _{(\overline{\Delta{X}},\mathit{bs})}^{(n)}< {\infty}, $$

(25)

where

$$ \|{\tilde{A}}\|_{(\overline{\Delta{X}},\mathit{bs})}^{(n)}=\sup_{n}\Biggl( \sum_{k=0}^{\infty}\Biggl|\sum _{m=0}^{n}\tilde{a}_{mk}\Biggr|\Biggr) \quad (n \in{\mathbb{N}}). $$

Furthermore, we have:

• (a) If $A\in{(\overline{\Delta{X}},\mathit{cs}_{0})}$, then

  $$ \|{L_{A}}\|_{\chi}=\limsup_{n\to{\infty}} \|{\tilde {A}}\|_{(\overline{\Delta{X}},\mathit{bs})}^{(n)} $$

  (26)

  and

  $$ L_{A}\textit{ is compact if and only if }\lim _{n\to{\infty}}\| {\tilde{A}}\|_{(\overline{\Delta{X}},\mathit{bs})}^{(n)}=0. $$

  (27)

• (b) If $A\in{(\overline{\Delta{X}},\mathit{bs})}$, then

  $$ 0\leq\|{L_{A}}\|_{\chi}\leq\limsup _{n\to{\infty}}\| {\tilde{A}}\|_{(\overline{\Delta{X}},\mathit{bs})}^{(n)} $$

  (28)

  and

  $$ L_{A}\textit{ is compact if }\lim_{n\to{\infty}} \|{\tilde {A}}\|_{(\overline{\Delta{X}},\mathit{bs})}^{(n)}=0. $$

  (29)

• (c) If X has AK and $A\in{(\overline{\Delta {X}},\mathit{cs})}$, then

  $$ \frac{1}{2}\cdot\limsup_{n\to{\infty}}\|{\tilde{A}- \tilde{\alpha }}\|_{(\overline{\Delta{X}},\mathit{bs})}^{(n)}\leq\| {L_{A}} \|_{\chi}\leq\limsup_{n\to{\infty}}\|{\tilde {A}-\tilde{\alpha}} \|_{(\overline{\Delta{X}},\mathit{bs})}^{(n)}, $$

  (30)

  where $\tilde{\alpha}=(\tilde{\alpha}_{k})$ with $\tilde{\alpha }_{k}=\lim_{m\to{\infty}}\tilde{a}_{mk}$ for all $k\in{\mathbb{N}}$ and

  $$ L_{A}\textit{ is compact if and only if }\lim _{n\to{\infty}}\limsup_{n\to{\infty}}\|{\tilde{A}-\tilde{ \alpha}}\|_{(\overline {\Delta{X}},\mathit{bs})}^{(n)}=0. $$

  (31)

Proof

The proof is quite similar to Corollary 5.1 in [16] for the new spaces $\overline{\Delta{X}}$ where $X\in{\{l_{\infty }, c, c_{0}\}}$. □

We denote the space of all sequences of bounded variation by bν and define

$$ b\nu=\bigl\{ x=(x_{k})\in{w}: (x_{k}-x_{k-1}) \in{l_{1}}\bigr\} . $$

So, the space bν is a BK-space with

$$ \|{x}\|_{b\nu}=\sum_{k=0}^{\infty}|x_{k}-x_{k-1}|. $$

Now, we have the following result.

Corollary 5.5

Let $X\supset{\phi}$ be a BK-space. If $A\in{(\overline{\Delta {X}},b\nu)}$, then

$$ \|{A}\|_{(\overline{\Delta{X}},b\nu)}\leq\| {L_{A}}\|\leq4\cdot\|{A} \|_{(\overline{\Delta{X}},b\nu)}, $$

where

$$ \|{A}\|_{(\overline{\Delta{X}},b\nu)}=\sup_{N\in{F}}\Biggl(\sum _{n\in{N}}\Biggl\vert \sum_{k=0}^{n}( \tilde{A}_{k}-\tilde{A}_{k-1})\Biggr\vert \Biggr)< \infty. $$

Furthermore:

If $A\in{(\overline{\Delta{X}},b\nu)}$, then

$$ \|{A}\|_{(\overline{\Delta{X}},b\nu)}^{(r)}\leq\| {L_{A}}\|_{\chi} \leq4\cdot\|{A}\|_{(\overline{\Delta {X}},b\nu)}^{(r)}, $$

where

$$ \|{A}\|_{(\overline{\Delta{X}},b\nu)}^{(r)}=\sup_{N\in {F_{r}}}\Biggl(\sum _{n\in{N}}\Biggl|\sum_{k=0}^{n}( \tilde{A}_{k}-\tilde {A}_{k-1})\Biggr|\Biggr)< \infty\quad (r\in{ \mathbb{N}}) $$

and

$$ L_{A}\textit{ is compact if and only if }\lim_{r\to{\infty}} \| {A}\|_{(\overline{\Delta{X}},b\nu)}^{(r)}=0. $$

Proof

We proceed by Theorem 4.4 in [28] and Theorem 4.2. □

We denote $b\upsilon^{p}$ for the space of all sequences of p-bounded variation and define it by

$$ b\upsilon^{p}=\bigl\{ x=(x_{k})\in{w}:(x_{k}-x_{k-1}) \in{l_{p}}\bigr\} \quad (1< p< \infty). $$

$b\upsilon^{p}$ is a BK-space with its natural norm (cf. [17]). For every $a=(a_{k})\in{(b\upsilon^{p})}^{\beta}$, we have

$$ \|{a}\|_{b\upsilon^{p}}^{*}=\Biggl(\sum _{k=0}^{\infty}\Biggl\vert \sum _{j=k}^{\infty}a_{j}\Biggr\vert ^{q}\Biggr)^{\frac{1}{q}}. $$

(32)

Corollary 5.6

Let X denote any of the spaces $l_{\infty}$ or $c_{0}$, and let A be an infinite matrix, $1< p<\infty$, and $q=p/(p-1)$. If $A\in {(b\upsilon^{p},\overline{\Delta{X}})}$, then

$$ \|{L_{A}}\|=\|{A}\|_{(b\upsilon ^{p},\overline{\Delta{l_{\infty}}})}^{(r)} =\sup_{n>r}\Biggl(\sum_{k=0}^{\infty} \Biggl\vert \sum_{j=k}^{\infty}\tilde {a}_{nj}\Biggr\vert ^{q}\Biggr)^{\frac{1}{q}}\quad (r\in{ \mathbb{N}}). $$

Further:

• (a) If $A\in{(b\upsilon^{p},\overline{\Delta{l_{\infty }}})}$, then

  $$ 0\leq\|{L_{A}}\|_{\chi}\leq\lim_{r\to{\infty}}\| {A} \|_{(b\upsilon^{p},\overline{\Delta{l_{\infty}}})}^{(r)} $$

  and

  $$ L_{A}\textit{ is compact if }\lim_{r\to{\infty}}\|{A}\| _{(b\upsilon^{p},\overline{\Delta{l_{\infty}}})}^{(r)}=0. $$

• (b) If $A\in{(b\upsilon^{p},\overline{\Delta {c_{0}}})}$, then

  $$ \|{L_{A}}\|_{\chi}=\lim_{r\to{\infty}}\| {A} \|_{(b\upsilon^{p},\overline{\Delta{l_{\infty}}})}^{(r)} $$

  and

  $$ L_{A}\textit{ is compact if and only if }\lim_{r\to{\infty}} \| {A}\|_{(b\upsilon^{p},\overline{\Delta{l_{\infty}}})}^{(r)}=0. $$

Proof

The proof is a special case of ([28], Theorem 3.3) when $X=b\upsilon^{p}$. Then this completes the proof Lemma 3.1. □