# Almost absolute weighted summability with index k and matrix transformations

## Abstract

In this paper we generalize the space $$\hat{\ell }_{k}$$ of absolutely almost convergent series (J. Math. Anal. Appl. 161:50–56, 1991) via weighted mean transformations. We study some inclusion relations and their topological properties. Further we characterize certain matrix transformations.

## Introduction

Let w be the set of all sequences of complex numbers. A Banach sequence space X is BK if the map $$p_{n}:X\rightarrow \mathbb{C}$$ defined by $$p_{n}(x)=x_{n}$$ is continuous for all $$n\geq 0$$. A BK- space X is said to have the AK property if $$\phi \subset X$$ and $$( e^{(v)} )$$ is a basis for X, where $$e^{(v)}=\{0,0,0,\ldots,1^{v^{th}\ \mathrm{term}},0,0,\ldots\}$$ and ϕ= span$$\{ e^{(v)} \}$$. If ϕ is dense in X, then it is called an AD-space, so AK implies AD.

Let $$\ell _{\infty }$$ be the space of all bounded sequences. A sequence $$( x_{n} ) \in \ell _{\infty }$$ is said to be almost convergent to γ if all of its Banach limits  coincide to γ. Lorentz  (see  for double sequences) characterized almost convergence by saying that a sequence $$( x_{n} )$$ is almost convergent to γ if and only if

\begin{aligned} \frac{1}{r+1}\sum_{v=0}^{r}x_{n+v} \rightarrow \gamma \quad\text{as }r \rightarrow \infty\ {(\text{uniformly in }n)}. \end{aligned}
(1.1)

This notion plays an important role in summability theory and was investigated by several authors. For example, it was later used to define and study some concepts such as conservative and regular matrices, some sequence spaces, and matrix transformations (see [2, 6, 7, 9, 11, 12, 15]).

Absolute almost convergence emerges naturally as an absolute analogue of almost convergence. To introduce this concept, let $$s_{n}=\Sigma _{v=1}^{n}a_{v}$$ be a partial sum of $$\Sigma a_{v}$$. The series $$\Sigma a_{v}$$ is said to be absolutely almost convergent series if (see )

\begin{aligned} \sum_{m=0}^{\infty } \vert \psi _{m,n} \vert ^{k}< \infty , \quad k>0, \end{aligned}

uniformly in n, where

\begin{aligned} \psi _{m,n}= \textstyle\begin{cases} a_{n},&m=0, \\ \frac{1}{m(m+1)}\sum_{v=1}^{m}va_{n+v},& m\geq 1.\end{cases}\displaystyle \end{aligned}

The space of all absolutely almost convergent series

\begin{aligned} \widehat{\ell }_{k}= \Biggl\{ a=(a_{n})_{n\in \mathbb{N}}: \sum_{m=0}^{\infty } \vert \psi _{m,n} \vert ^{k}< \infty ,\text{ uniformly in }n, k>0 \Biggr\} \end{aligned}

was first defined and studied in . We note an important relation between $$\widehat{\ell }_{k}$$ and absolute Cesaro summability $$\vert C,1 \vert$$ in Flett’s notation , $$\widehat{\ell }_{k}\subset$$ $$\vert C,1 \vert$$ (see ).

The purpose of the present paper is to define an absolute almost weighted summability using some factors and weighted means and to study its topological structures. This new method of summability extends the well-known concept of absolute almost convergence of Das et al. , and space $$\widehat{\ell }_{k}$$ of Das et al.  becomes a special case of our space $$\vert f(\overline{N}_{p}^{u}) \vert _{k}$$. We investigate relations between classical sequence spaces and show that the space $$\vert f(\overline{N}_{p}^{u}) \vert _{k}$$ is not separable for $$k>1$$. Also, we characterize the matrix classes $$(c, \vert f(\overline{N}_{p}) \vert _{k})$$ and $$(c, \vert f(\overline{N}_{p}) \vert _{k}),1\leq k<\infty$$.

## Main results

For any sequence $$( s_{n} )$$, we define $$T_{m,n}$$ by

\begin{aligned} T_{-1,n}(s)=s_{n-1},\qquad T_{m,n}(s)= \frac{1}{P_{m}}\sum_{v=0}^{m}p_{v}s_{n+v}, \quad m \geq 0, \end{aligned}

where $$(p_{n})$$ is a sequence of positive real numbers with

\begin{aligned} P_{n}=p_{0}+p_{1}+\cdots+p_{n} \rightarrow \infty \quad\text{as } n\rightarrow \infty , P_{-1}=p_{-1}=0. \end{aligned}

A straightforward calculation then shows that

\begin{aligned} F_{m,n}(a)=T_{m,n}(s)-T_{m-1,n}(s)=\textstyle\begin{cases} a_{n}, &m=0{,} \\ \frac{p_{m}}{P_{m}P_{m-1}}\sum_{v=1}^{m}P_{v-1}a_{n+v},& m\geq 1{.}\end{cases}\displaystyle \end{aligned}

So, we can give the following definition.

### Definition 2.1

Let $$\Sigma a_{v}$$ be an infinite series with partial summations $$(s_{n})$$. Let $$(p_{n})$$ and $$( u_{n} )$$ be sequences of positive real numbers. The series $$\Sigma a_{v}$$ is said to be absolute almost weighted summable $$\vert f(\overline{N}_{p}),u_{m} \vert _{k}$$, $$k\geq 1$$, if

\begin{aligned} \sum_{m=0}^{\infty }u_{m}^{k-1} \bigl\vert F_{m,n}(a) \bigr\vert ^{k}< \infty \end{aligned}

uniformly in n.

For $$\vert f(\overline{N}_{p}^{u}) \vert _{k},k\geq 1$$, we write the set of all series summable by the method $$\vert f(\overline{N}_{p}),u_{m} \vert _{k}$$. Then $$\Sigma a_{v}$$ is summable $$\vert f(\overline{N}_{p}),u_{m} \vert _{k}$$ iff the series $$\Sigma a_{v}\in$$ $$\vert f(\overline{N}_{p}^{u}) \vert _{k}$$. Note that, in the case $$u_{m}=p_{m}=1$$ for $$m\geq 0$$, it reduces to the set of absolutely almost convergent series $$\widehat{\ell }_{k}$$ given by Das, Kuttner, and Nanda . Further, it is clear that the space $$\vert \overline{N}_{p}^{u} \vert _{k}$$ is derived from $$\vert f(\overline{N}_{p}^{u}) \vert _{k}$$ by putting $$n=0$$ [9, 13, 14], and also $$\vert f(\overline{N}_{p}^{u}) \vert _{k}\subset \vert \overline{N}_{p}^{u} \vert _{k}$$, but the converse is not true.

First we give some relations between the new method and classical sequence spaces such as bs and $$\ell _{\infty }$$, which are the sets of all bounded series and bounded sequences, respectively.

### Theorem 2.2

Let $$(p_{m})$$ and $$(u_{m})$$ be sequences of positive numbers.

1. (i)

If

\begin{aligned} \biggl( \frac{1}{u_{m}} \biggr) \in \ell _{\infty }, \end{aligned}
(2.1)

then $$\vert f(\overline{N}_{p}^{u}) \vert _{k}\subset \ell _{ \infty }$$, $$k\geq 1$$.

2. (ii)

If

\begin{aligned} \sum_{m=0}^{\infty }u_{m}^{k-1} \biggl( \frac{p_{m}}{P_{m}} \biggr) ^{k}< \infty , \end{aligned}
(2.2)

then $$bs\subset \vert f(\overline{N}_{p}^{u}) \vert _{k}$$, $$k>1$$.

### Proof

(i) Let $$a=(a_{v})\in$$ $$\vert f(\overline{N}_{p}^{u}) \vert _{k}$$. Then, by the definition, there exists an integer M such that

\begin{aligned} \sum_{m=M}^{\infty }u_{m}^{k-1} \bigl\vert F_{m,n}(a) \bigr\vert ^{k} \leq 1{} \end{aligned}
(2.3)

holds for all n. So, it is sufficient to show that the sequence $$( \vert F_{m,n}(a)) \vert )$$ is bounded for a fixed number m. By (2.1) and (2.3), we have $$\vert F_{m,n}(a) \vert \leq u_{m}^{-1+1/k}$$ for $$m\geq M$$ and all n. On the other hand, for $$m\geq 1$$,

\begin{aligned} a_{m+n}=\frac{P_{m}}{p_{m}}F_{m,n}(a)-\frac{P_{m-2}}{p_{m-1}}F_{m-1,n}(a). \end{aligned}
(2.4)

It follows by applying (2.4) for any $$m\geq M+1$$ that $$a=(a_{v})\in \ell _{\infty }$$, which completes the proof.

(ii) Let $$a=(a_{v})\in bs$$. Denote $$M=\sup_{v} \vert \sum_{j=0}^{v}a_{j} \vert$$. Then we have

\begin{aligned} \Biggl\vert \sum_{v=1}^{m}P_{v-1}a_{n+v} \Biggr\vert &= \Biggl\vert \sum_{v=1}^{m-1}(-p_{v}) \sum_{j=1}^{v}a_{n+j}+P_{m-1} \sum_{j=1}^{m}a_{n+j} \Biggr\vert \\ &\leq \sum_{v=1}^{m-1}p_{v} \Biggl\vert \sum_{j=n+1}^{n+v}a_{j} \Biggr\vert +P_{m-1} \Biggl\vert \sum_{j=n+1}^{n+m}a_{n+j} \Biggr\vert \\ &\leq \sum_{v=1}^{m-1}p_{v} \Biggl\vert \sum_{j=0}^{n+v}a_{j}- \sum_{j=0}^{n}a_{j} \Biggr\vert +P_{m-1} \Biggl\vert \sum_{j=0}^{n+m}a_{n+j}- \sum_{j=0}^{n}a_{j} \Biggr\vert \\ &\leq 4MP_{m-1}, \end{aligned}

which implies that

\begin{aligned} \sum_{m=0}^{\infty }u_{m}^{k-1} \bigl\vert F_{m,n}(a) \bigr\vert ^{k} &=u_{0}^{k-1} \vert a_{n} \vert ^{k}+\sum_{m=1}^{\infty }u_{m}^{k-1} \Biggl\vert \frac{p_{m}}{P_{m}P_{m-1}}\sum_{v=1}^{m}P_{v-1}a_{n+v} \Biggr\vert ^{k} \\ &\leq (4M)^{k}\sum_{m=0}^{\infty }u_{m}^{k-1} \biggl( \frac{p_{m}}{P_{m}} \biggr) ^{k}< \infty . \end{aligned}

This completes the proof. □

For the special case $$u_{m}=p_{m}=1$$ for all $$m\geq 0$$, $$\vert f(\overline{N}_{p}^{u}) \vert _{k}=\widehat{\ell }_{k}$$ and (2.2) reduces to $$\sum_{m=0}^{\infty } ( m+1 ) ^{-k}<\infty$$. So we have the following result in .

### Corollary 2.3

For $$k>1$$, $$bs\subset \widehat{\ell }_{k}$$.

### Theorem 2.4

Let $$(u_{m})$$ be a sequence of positive numbers such that (2.4) holds. Then $$\vert f(\overline{N}_{p}^{u}) \vert _{k}$$, $$k\geq 1$$, is a BK space with respect to the norm

\begin{aligned} \Vert a \Vert _{ \vert f(\overline{N}_{p}^{u}) \vert _{k}}{:}=\sup \Biggl\{ \sum _{m=0}^{\infty }u_{m}^{k-1} \bigl\vert F_{m,n}(a) \bigr\vert ^{k}:n\in \mathbb{N} \Biggr\} ^{1/k}. \end{aligned}
(2.5)

### Proof

It is routine to prove that the norm conditions are satisfied by (2.5). We only note that (2.5) is well defined. In fact, if $$a\in \vert f(\overline{N}_{p}^{u})\vert _{k}$$, then, as in the proof of part (ii) of Theorem 2.2, there exists an integer M such that, for all n,

\begin{aligned} \sum_{m=M}^{\infty }u_{m}^{k-1} \bigl\vert F_{m,n}(a) \bigr\vert ^{k} \leq 1, \end{aligned}

and $$( u_{m}^{1-1/k} \vert F_{m,n}(a) \vert )$$ is bounded for all $$m,n\geq 0$$. This gives

\begin{aligned} \sup \Biggl\{ \sum_{m=0}^{\infty }u_{m}^{k-1} \bigl\vert F_{m,n}(a) \bigr\vert ^{k}:n\in \mathbb{N} \Biggr\} \leq 1+\sup \Biggl\{ \sum_{m=0}^{M}u_{m}^{k-1} \bigl\vert F_{m,n}(a) \bigr\vert ^{k}:n,m\in \mathbb{N} \Biggr\} < \infty . \end{aligned}

To show that it is a Banach space, let us take an arbitrary Cauchy sequence $$( a^{m} ) = ( a_{v}^{m} ) =(a_{v}^{1},a_{v}^{2},\ldots)= ( (a_{1}^{1},a_{2}^{1},\ldots),\ldots,(a_{1}^{m},a_{2}^{m},a_{3}^{m},\ldots),\ldots )$$, where $$a^{m}= ( a_{v}^{m} ) \in \vert f(\overline{N}_{p}^{u}) \vert _{k}$$ for $$m\geq 0$$. Given $$\varepsilon >0$$. Then there exists an integer $$m_{0}$$ such that $$\Vert a^{m_{1}}-a^{m_{2}} \Vert _{ \vert f( \overline{N}_{p}^{u}) \vert _{k}}<\varepsilon$$ for $$m_{1},m_{2}>m_{0}$$ and all v, or, equivalently,

\begin{aligned} \Biggl\{ u_{0}^{k-1} \bigl\vert a_{n}^{m_{1}}-a_{n}^{m_{2}} \bigr\vert ^{k}+\sum_{m=1}^{\infty }u_{m}^{k-1} \Biggl\vert \frac{p_{m}}{P_{m}P_{m-1}}\sum_{v=1}^{m}P_{v-1} \bigl( a_{n+v}^{m_{1}}-a_{n+v}^{m_{2}} \bigr) \Biggr\vert ^{k} \Biggr\} ^{1/k}< \varepsilon . \end{aligned}
(2.6)

This also gives that $$\vert a_{v}^{m_{1}}-a_{v}^{m_{2}} \vert ^{k}< \varepsilon /u_{0}^{1-1/k}$$ holds for all $$m_{1},m_{2}>m_{0}$$, i.e., the sequence $$( a_{v}^{m} )$$ is a Cauchy sequence in the set of complex numbers $$\mathbb{C}$$. So, it converges to a number $$a_{v}$$ $$(v=0,1,\ldots)$$, i.e., $$\lim_{m\rightarrow \infty }a_{v}^{m}=a_{v}$$. Now, letting $$m_{2}\rightarrow \infty$$, by (2.6) we have for $$\Vert a^{m_{1}}-a \Vert _{ \vert f(\overline{N}_{p}^{u}) \vert _{k}}<\varepsilon$$ for $$m_{1}>m_{0}$$. This means $$\lim_{m\rightarrow \infty }a^{m}=a$$. Further, since

\begin{aligned} \Vert a \Vert _{ \vert f(\overline{N}_{p}^{u}) \vert _{k}}\leq \bigl\Vert a^{m_{1}}-a \bigr\Vert _{ \vert f(\overline{N}_{p}^{u}) \vert _{k}}+ \bigl\Vert a^{m_{1}} \bigr\Vert _{ \vert f(\overline{N}_{p}^{u}) \vert _{k}}< \infty , \end{aligned}

then $$a\in \vert f(\overline{N}_{p}^{u}) \vert _{k}$$. So, $$\vert f(\overline{N}_{p}^{u}) \vert _{k}$$ is a Banach space. This completes the proof. □

We note that if E is a BK-space such that $$bs\subset E\subset \ell _{\infty }$$, then E is not separable (and hence not reflexive) (see ). Hence the following result at once follows from Theorem 2.2.

### Corollary 2.5

If $$(p_{m})$$ and $$(u_{m})$$ are sequences of positive numbers satisfying (2.1) and (2.2), then $$\vert f(\overline{N}_{p}^{u}) \vert _{k}$$ is not separable for $$k>1$$.

## Matrix transformations on space $$\vert f(\overline{N}_{p}) \vert _{k}$$

In this section we characterize certain matrix transformations on the space $$\vert f(\overline{N}_{p}) \vert _{k}$$. First we recall some notations. Let X, Y be any subsets of ω and $$A=(a_{nv})$$ be an infinite matrix of complex numbers. By $$A(x){:}=(A_{n}(x))$$, we indicate the A-transform of a sequence $$x= ( x_{v} )$$ if the series

\begin{aligned} A_{n} ( x ) {:}=\sum_{v=0}^{\infty }a_{nv}x_{v} \end{aligned}

are convergent for $$n\geq 0$$. If $$Ax\in Y$$, whenever $$x\in X$$, then we say that A defines a matrix mapping from X into Y and denotes the class of all infinite matrices A such that $$A:X\rightarrow Y$$ by $$(X,Y)$$. Also, we denote the set of all k-absolutely convergent series by $$\ell _{k},1\leq k<\infty$$, i.e.,

\begin{aligned} \ell _{k}= \Biggl\{ x= ( x_{v} ) \in w:\sum _{v=0}^{ \infty } \vert x_{v} \vert ^{k}< \infty \Biggr\} , \end{aligned}

which is a BK-space with respect to the norm

\begin{aligned} \Vert x \Vert _{\ell _{k}}= \Biggl( \sum_{v=0}^{ \infty } \vert x_{v} \vert ^{k} \Biggr) ^{1/k}. \end{aligned}

Also we make use of the following lemma in .

### Lemma 3.1

Suppose that $$A= ( a_{nv} )$$ is an infinite matrix with complex numbers and $$p= ( p_{v} )$$ is a bounded sequence of positive numbers such that $$H=\sup_{v}p_{v}$$ and $$C=\max \{ 1,2^{H-1} \}$$. Then

\begin{aligned} \bigl( 4C^{2} \bigr) ^{-1}U_{p} ( A ) \leq L_{p}(A)\leq U_{p} ( A ) \end{aligned}

provided that

\begin{aligned} U_{p} ( A ) =\sum_{v=0}^{\infty } \Biggl( \sum_{n=0}^{\infty } \vert a_{nv} \vert \Biggr) ^{p_{v}}< \infty {} \end{aligned}

and

\begin{aligned} L_{p}(A)=\sup \Biggl\{ \sum_{v=0}^{\infty } \biggl\vert \sum_{n\in N}a_{nv} \biggr\vert ^{p_{v}}:N\subset N_{0} \textit{ is finite} \Biggr\} < \infty . \end{aligned}

Now we begin with the first theorem given the characterization of the class $$(\ell _{1}, \vert f(\overline{N}_{p}) \vert _{k})$$.

### Theorem 3.2

Let $$u=(u_{n})$$ be a sequence of positive numbers, and let $$A=(a_{vj})$$ be an infinite matrix. Then $$A\in (\ell _{1}, \vert f(\overline{N}_{p}) \vert _{k}),1\leq k<\infty$$, if and only if

\begin{aligned} ( a_{vj} ) \in \bigl\vert f(\overline{N}_{p}) \bigr\vert _{k}\quad\textit{for each }j \end{aligned}
(3.1)

and

\begin{aligned} L{:}=\sup_{n,j}\sum_{m=0}^{\infty }u_{m}^{k-1} \bigl\vert b(m,n,j) \bigr\vert ^{k}< \infty {,} \end{aligned}
(3.2)

where

\begin{aligned} b(m,n,j)=\textstyle\begin{cases} a_{nj}, & m=0{,} \\ \frac{p_{m}}{P_{m}P_{m-1}}\sum_{v=1}^{m}P_{v-1}a_{n+v,,j},& m\geq 1.\end{cases}\displaystyle \end{aligned}

### Proof

Necessity. Suppose $$A\in (\ell _{1}, \vert f(\overline{N}_{p}) \vert _{k})$$. Then $$A(x)\in \vert f(\overline{N}_{p}) \vert _{k}$$ for all $$x\in \ell _{1}$$, i.e.,

\begin{aligned} &\sum_{m=0}^{\infty }u_{m}^{k-1} \bigl\vert F_{m,n} \bigl(A(x) \bigr) \bigr\vert ^{k} \\ &\quad=u_{0}^{k-1} \bigl\vert A_{n}(x) \bigr\vert ^{k}+\sum_{m=1}^{ \infty }u_{m}^{k-1} \Biggl\vert \sum_{v=1}^{m} \frac{p_{m}P_{v-1}}{P_{m}P_{m-1}}A_{n+v}(x) \Biggr\vert ^{k} \\ &\quad=u_{0}^{k-1} \Biggl\vert \sum _{j=0}^{\infty }a_{n,j}x_{j} \Biggr\vert ^{k}+\sum_{m=1}^{\infty }u_{m}^{k-1} \Biggl\vert \sum_{j=0}^{ \infty } \Biggl( \sum _{v=1}^{m}\frac{p_{m}P_{v-1}}{P_{m}P_{m-1}}a_{n+v,j} \Biggr) x_{j} \Biggr\vert ^{k} \\ &\quad=\sum_{m=0}^{\infty }u_{m}^{k-1} \Biggl\vert \sum_{j=0}^{\infty }b(m,n,j)x_{j} \Biggr\vert ^{k} \end{aligned}

converges uniformly in n.

If we put $$x={(e^{j})}= ( e_{r}^{j} ) \in \ell _{1}$$ such that $$e_{r}^{j}=1$$ for $$r=j$$ and zero otherwise, $$A(e^{j})\in \vert f(\overline{N}_{p}) \vert _{k}$$, which gives that (3.1) holds. Further, since $$\ell _{1}$$ is a Banach space, by the Banach–Steinhaus theorem, $$A:\ell _{1}\rightarrow \mathbb{C}$$ is a continuous linear map. So, for fixed n and s,

\begin{aligned} q_{sn}(x)= \Biggl( \sum_{m=0}^{s}u_{m}^{k-1} \bigl\vert F_{m,n} \bigl(A(x) \bigr) \bigr\vert ^{k} \Biggr) ^{1/k} \end{aligned}

is a continuous seminorm on $$\ell _{1}$$, which implies that $$\lim_{s\rightarrow \infty }q_{sn}(x)=q_{n}(x)$$ is a continuous seminorm, or, equivalently, there exists a constant K such that

\begin{aligned} q_{n}(x)= \Biggl( \sum_{m=0}^{\infty }u_{m}^{k-1} \bigl\vert F_{m,n} \bigl(A(x) \bigr) \bigr\vert ^{k} \Biggr) ^{1/k}\leq K \Vert x \Vert _{ \ell _{1}} \end{aligned}
(3.3)

for every $$x\in \ell _{1}$$. Applying (3.3) with $$x= ( e_{v}^{j} ) \in \ell _{1}$$ we have, for all $$j,n\geq 0$$,

\begin{aligned} \Biggl( \sum_{m=0}^{\infty }u_{m}^{k-1} \bigl\vert b (m,n,j)\bigr\vert ^{k} \Biggr) ^{1/k}\leq K, \end{aligned}

which gives (3.2).

Sufficiency. Suppose that (3.1) and (3.2) hold. Take $$x\in \ell _{1}$$. Then we should show $$A(x)\in \vert f(\overline{N}_{p}) \vert _{k}$$. For this, it is enough to prove that

\begin{aligned} \sum_{m=l}^{\infty }u_{m}^{k-1} \bigl\vert F_{m,n} \bigl(A(x) \bigr))\bigr\vert ^{k}\rightarrow 0\quad \text{as }l\rightarrow \infty , \end{aligned}

uniformly in n. By applying generalized Minkowski’s inequality, we get

\begin{aligned} \Biggl( \sum_{m=l}^{\infty }u_{m}^{k-1} \bigl\vert F_{m,n} \bigl(A(x) \bigr) \bigr\vert ^{k} \Biggr) ^{1/k}&= \Biggl( \sum_{m=l}^{\infty }u_{m}^{k-1} \Biggl\vert \sum_{j=0}^{\infty }b(m,n,j)x_{j} \Biggr\vert ^{k} \Biggr) ^{1/k} \\ &\leq \sum_{j=0}^{\infty } \vert x_{j} \vert \Biggl( \sum_{m=l}^{ \infty }u_{m}^{k-1} \bigl\vert b(m,n,j) \bigr\vert ^{k} \Biggr) ^{1/k} \\ &=\sum_{j=0}^{\infty } \vert x_{j} \vert R(l,n,j), \end{aligned}
(3.4)

where

\begin{aligned} R(l,n,j)= \Biggl( \sum_{m=l}^{\infty }u_{m}^{k-1} \bigl\vert b(m,n,j) \bigr\vert ^{k} \Biggr) ^{1/k}. \end{aligned}

On the other hand, it follows from (3.2) that, for all $$l,n,j\geq 0$$,

\begin{aligned} R(l,n,j) \vert x_{j} \vert \leq L^{1/k} \vert x_{j} \vert , \end{aligned}

which gives

\begin{aligned} \sum_{j=0}^{\infty } \vert x_{j} \vert R(l,n,j)\leq L^{1/k} \sum_{j=0}^{\infty } \vert x_{j} \vert < \infty . \end{aligned}

Now, let $$\varepsilon >0$$. Then there exists an integer $$j_{0}$$ such that, for all l and n,

\begin{aligned} \sum_{j=j_{0}}^{\infty } \vert x_{j} \vert R(l,n,j)< \frac{\varepsilon }{2}. \end{aligned}

Also, by (3.1), for each j,

\begin{aligned} R(l,n,j)\rightarrow 0\quad\text{as }l\rightarrow \infty \text{ uniformly in }n, \end{aligned}

there exists an integer $$l_{0}$$ so that, for $$l\geq l_{0}$$ and all n,

\begin{aligned} \sum_{j=0}^{j_{0}-1} \vert x_{j} \vert R(l,n,j)< \frac{\varepsilon }{2}. \end{aligned}

So, we have, for $$l\geq l_{0}$$ and all n,

\begin{aligned} \sum_{j=0}^{\infty } \vert x_{j} \vert R(l,n,j)< \varepsilon , \end{aligned}

which implies, by (3.4),

\begin{aligned} \Biggl( \sum_{m=l}^{\infty }u_{m}^{k-1} \bigl\vert F_{m,n} \bigl(A(x) \bigr) \bigr\vert ^{k} \Biggr) ^{1/k}< \varepsilon . \end{aligned}

This states that

\begin{aligned} \sum_{m=l}^{\infty }u_{m}^{k-1} \bigl\vert F_{m,n} \bigl(A(x) \bigr) \bigr\vert ^{k} \rightarrow 0\quad\text{as }l\rightarrow \infty \text{ uniformly in }n. \end{aligned}

□

This completes the proof.

In the special case $$p_{m}=u_{m}=1$$ for all $$m\geq 0$$, we have $$\vert f(\overline{N}_{p}) \vert =\widehat{\ell }_{k}$$, and so the following result follows from Theorem 3.2.

### Corollary 3.3

$$A\in (\ell _{1},\widehat{\ell }_{k}),1\leq k< \infty$$, if and only if

\begin{aligned} ( a_{vj} ) \in \widehat{\ell }_{k}\quad\textit{for each }j \end{aligned}

and

\begin{aligned} \sup_{n,j}\sum_{m=0}^{\infty } \bigl\vert b^{\prime }(m,n,j) \bigr\vert ^{k}< \infty {,} \end{aligned}

where

\begin{aligned} b^{\prime }(m,n,j)=\textstyle\begin{cases} a_{nj}, & m=0, \\ \frac{1}{m(m+1)}\sum_{v=1}^{m}va_{n+v,,j},& m\geq 1.\end{cases}\displaystyle \end{aligned}
(3.5)

### Theorem 3.4

Let $$u=(u_{n})$$ be a sequence of positive numbers, and let $$A=(a_{vj})$$ be an infinite matrix. Then $$A\in (c, \vert f(\overline{N}_{p}) \vert _{k}),1\leq k<\infty$$, if and only if conditions (3.1),

\begin{aligned} B=\sup_{n}\sum_{m=0}^{\infty }u_{m}^{k-1} \Biggl( \sum_{j=0}^{ \infty } \bigl\vert b(m,n,j) \bigr\vert \Biggr) ^{k}< \infty , \end{aligned}
(3.6)

and

\begin{aligned} \sum_{m=0}^{\infty }u_{m}^{k-1} \Biggl\vert \sum_{j=0}^{\infty }b(m,n,j) \Biggr\vert ^{k}\quad\textit{converges uniformly in }n \end{aligned}
(3.7)

hold.

### Proof

Necessity. Let $$A\in (c, \vert f(\overline{N}_{p}) \vert _{k})$$. Then $$A(x)\in \vert f(\overline{N}_{p}) \vert _{k}$$ for every $$x\in c$$. Now, take $$x=e^{ ( j ) }$$ and $$x=e=(1,1,\ldots)$$. Then (3.1) and (3.7) hold, respectively. Also, it follows as in the proof of Theorem 3.2 that

\begin{aligned} \Biggl( \sum_{m=0}^{\infty }u_{m}^{k-1} \bigl\vert F_{m,n} \bigl(A(x) \bigr) \bigr\vert ^{k} \Biggr) ^{1/k}\leq K \Vert x \Vert _{ \infty }. \end{aligned}
(3.8)

Let N be an arbitrary finite set of natural numbers, and define a sequence x by

\begin{aligned} x_{j}=\textstyle\begin{cases} 1,&j\in N, \\ 0,& j\notin N,\end{cases}\displaystyle \end{aligned}
(3.9)

then $$x\in c$$ and $$\Vert x \Vert =1$$. Applying (3.8) with this sequence (3.9), we have

\begin{aligned} \Biggl\{ \sum_{m=0}^{\infty }u_{m}^{k-1} \biggl( \biggl\vert \sum_{j \in N}b(m,n,j) \biggr\vert \biggr) ^{k} \Biggr\} ^{1/k}\leq K. \end{aligned}
(3.10)

Hence, it is seen from Lemma 3.1 together with $$p_{v}=1$$ for all v that (3.10) is equivalent to (3.6).

Sufficiency. Suppose that (3.1), (3.6), and (3.7) hold. Given $$x\in c$$ and say $$\lim_{j}x_{j}=\beta$$. Then, by (3.7), as in Theorem 3.2,

\begin{aligned} \sum_{m=0}^{\infty }u_{m}^{k-1} \bigl\vert F_{mn} \bigl(A(x) \bigr) \bigr\vert ^{k}= \sum _{m=0}^{\infty }u_{m}^{k-1} \Biggl\vert \sum_{j=0}^{\infty }b(m,n,j)x_{j} \Biggr\vert ^{k}< \infty . \end{aligned}

Now it is enough to show that the tail of this series tends to zero uniformly in n. To see that, we write

\begin{aligned} &\sum_{m=M}^{\infty }u_{m}^{k-1} \bigl\vert F_{mn} \bigl(A(x) \bigr) \bigr\vert ^{k} \\ &\quad= \sum_{m=M}^{\infty }u_{m}^{k-1} \Biggl\vert \sum_{j=0}^{\infty } \beta b(m,n,j)+ \sum_{j=0}^{\infty }b(m,n,j) ( x_{j}- \beta ) \Biggr\vert ^{k} \\ &\quad\leq 2^{k} \Biggl\{ \sum_{m=M}^{\infty }u_{m}^{k-1} \Biggl\vert \sum_{j=0}^{\infty }\beta b(m,n,j) \Biggr\vert ^{k}+\sum_{m=M}^{ \infty }u_{m}^{k-1} \Biggl\vert \sum_{j=0}^{\infty }b(m,n,j) ( x_{j}- \beta ) \Biggr\vert ^{k} \Biggr\} \\ &\quad=2^{k} \bigl\{ F_{M,n}^{1}+F_{M,n}^{2} \bigl(x^{\prime } \bigr) \bigr\} , \quad\text{say}. \end{aligned}

It is clear from (3.7) that $$F_{M,n}^{1}\rightarrow 0$$ as $$M\rightarrow \infty$$ uniformly in n. On the other hand, since $$x_{j}\rightarrow \beta$$, for any $$\varepsilon >0$$, there exists an integer $$j_{0}$$ such that

\begin{aligned} \vert x_{j}-\beta \vert < \frac{1}{2} \biggl( \frac{\varepsilon }{B} \biggr) ^{1/k}\quad\text{for }j\geq j_{0}, \end{aligned}

which gives us, by (3.6), for all $$n\geq 0$$,

\begin{aligned} F_{M,n}^{2} \bigl(x^{\prime } \bigr) &=\sum _{m=M}^{\infty }u_{m}^{k-1} \Biggl\vert \sum_{j=0}^{\infty }b(m,n,j) ( x_{j}- \beta ) \Biggr\vert ^{k} \\ &\leq 2^{k}\sum_{m=M}^{\infty }u_{m}^{k-1} \Biggl\{ \Biggl\vert \sum_{j=0}^{j_{0}-1}b(m,n,j) ( x_{j}-\beta ) \Biggr\vert ^{k}+ \Biggl\vert \sum _{j=j_{0}}^{\infty }b(m,n,j) ( x_{j}- \beta ) \Biggr\vert ^{k} \Biggr\} \\ &=2^{k}\sup_{j} \vert x_{j}-\beta \vert \sum_{m=M}^{ \infty }u_{m}^{k-1} \Biggl( \sum_{j=0}^{j_{0}-1} \bigl\vert b(m,n,j) \bigr\vert \Biggr) ^{k}+\frac{\varepsilon }{2}. \end{aligned}

By (3.1), the first term of the equality is smaller than $$\varepsilon /2$$ for sufficiently large M and all n. This means $$F_{M,n}^{2}\rightarrow 0$$ as $$M\rightarrow \infty$$ uniformly in n. Hence, the theorem is established.

For $$p_{m}=u_{m}=1$$, Theorem 3.4 also gives the characterization of the class $$(c,\widehat{\ell }_{k})$$ as follows. □

### Corollary 3.5

Let $$A=(a_{vj})$$ be an infinite matrix and $$(b^{\prime }(m,n,j))$$ be as in (3.5). Then $$A\in (c,\widehat{\ell }_{k}),1\leq k<\infty$$, if and only if conditions (3.1),

\begin{aligned} \sup_{n}\sum_{m=0}^{\infty } \Biggl( \sum_{j=0}^{\infty } \bigl\vert b^{\prime }(m,n,j) \bigr\vert \Biggr) ^{k}< \infty , \end{aligned}

and

\begin{aligned} \sum_{m=0}^{\infty } \Biggl\vert \sum _{j=0}^{\infty }b^{\prime }(m,n,j) \Biggr\vert ^{k}{}\quad\textit{converges uniformly in }n \end{aligned}

hold.

Not applicable.

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## Acknowledgements

The authors are grateful to the learned reviewers for their valuable comments which improved the paper.

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