Weighted almost convergence and related infinite matrices

Abstract

The purpose of this paper is to introduce the notion of weighted almost convergence of a sequence and prove that this sequence endowed with the sup-norm ${\Vert \cdot \Vert } _{\infty}$ is a BK-space. We also define the notions of weighted almost conservative and regular matrices and obtain necessary and sufficient conditions for these matrix classes. Moreover, we define a weighted almost A-summable sequence and prove the related interesting result.

1 Introduction and preliminaries

Let ω denote the space of all complex sequences $s=(s_{j})_{j=0}^{\infty}$ (or simply write $s=(s_{j})$). Any vector subspace of ω is called a sequence space. By $\mathbb {N}$ we denote the set of natural numbers, and by $\mathbb {R}$ the set of real numbers. We use the standard notation $\ell_{\infty}$, c and $c_{0}$ to denote the sets of all bounded, convergent and null sequences of real numbers, respectively, where each of the sets is a Banach space with the sup-norm $\Vert . \Vert _{\infty}$ defined by $\Vert s \Vert _{\infty}=\sup_{j\in\mathbb {N}} \vert s_{j} \vert$. We write the space $\ell_{p}$ of all absolutely p-summable series by

$$\ell_{p}= \Biggl\{ s\in\omega:\sum_{j=0}^{\infty} \vert s_{j} \vert ^{p}< \infty\ (1\leq p< \infty) \Biggr\} . $$

Clearly, $\ell_{p}$ is a Banach space with the following norm:

$$\Vert s \Vert _{p}= \Biggl(\sum_{j=0}^{\infty} \vert s_{j} \vert ^{p} \Biggr)^{1/p}. $$

For $p=1$, we obtain the set $l_{1}$ of all absolutely summable sequences. For any sequence $s=(s_{j})$, let $s^{[n]}=\sum_{j=0}^{n}s_{j}e_{j}$ be its n-section, where $e_{j}$ is the sequence with 1 in place j and 0 elsewhere and $e=(1,1,1,\dots)$.

A sequence space X is called a BK-space if it is a Banach space with continuous coordinates $p_{j}:X\to\mathbb {C}$, the set of complex fields, and $p_{j}(s)=s_{j}$ for all $s=(s_{j})\in X$ and every $j\in \mathbb {N}$. A BK-space $X\supset\psi$, the set of all finite sequences that terminate in zeros, is said to have AK if every sequence $s=(s_{j})\in X$ has a unique representation $s=\sum_{j=0}^{\infty }s_{j}e_{j}$.

Let X and Y be two sequence spaces, and let $A=(a_{n,k})$ be an infinite matrix. If, for each $s=(s_{k})$ in X, the series

$$ A_{n}s=\sum_{k}a_{n,k}s_{k}= \sum_{k=0}^{\infty}a_{n,k}s_{k} $$

(1)

converges for each $n\in\mathbb {N}$ and the sequence $As=(A_{n}s)$ belongs to Y, then we say that matrix A maps X into Y. By the symbol $(X,Y)$ we denote the set of all such matrices which map X into Y. The series in (1) is called A-transform of s whenever the series converges for $n=0,1,\ldots$ . We say that $s=(s_{k})$ is A-summable to the limit λ if $A_{n}s$ converges to λ ($n\to\infty$).

The sequence $s=(s_{k})$ of $\ell_{\infty}$ is said to be almost convergent, denoted by f, if all of its Banach limits [1] are equal. We denote such a class by the symbol f, and one writes $f\mbox{-}\lim s =\lambda$ if λ is the common value of all Banach limits of the sequence $s=(s_{k})$. For a bounded sequence $s=(s_{k})$, Lorentz [2] proved that $f\mbox{-}\lim s =\lambda$ if and only if

$$\lim_{k\to\infty}\frac{s_{m}+s_{m+1}+\cdots+s_{m+k}}{k+1}=\lambda $$

uniformly in m. This notion was later used to (i) define and study conservative and regular matrices [3]; (ii) introduce related sequence spaces derived by the domain of matrices [4–6]; (iii) study some related matrix transformations [7–9]; (iv) define related sequence spaces derived as the domain of the generalized weighted mean and determine duals of these spaces [10, 11]. As an extension of the notion of almost convergence, Kayaduman and Şengönül [12, 13] defined Cesàro and Riesz almost convergence and established related core theorems. The almost strongly regular matrices for single sequences were introduced and characterized [14], and for double sequences they were studied by Mursaleen [15] (also refer to [16–19]). As an application of almost convergence, Mohiuddine [20] proved a Korovkin-type approximation theorem for a sequence of linear positive operators and also obtained some of its generalizations. Başar and Kirişçi [21] determined the duals of the sequence space f and other related spaces/series and investigated some useful characterizations.

We now recall the following result.

Lemma 1.1

([22])

Let X and Y be BK-spaces. (i) Then $(X,Y)\subset B(X,Y)$, that is, every $A\in(X,Y)$ defines an operator ${\mathcal{L}}_{A}\in B(X,Y)$ by ${\mathcal{L}}_{A}(x)=Ax$ for all $x\in X$, where $B(X,Y)$ denotes the set of all bounded linear operators from X into Y. (ii) Then $A\in(X,\ell_{\infty})$ if and only if $\Vert A \Vert _{(X,\ell_{\infty})}=\sup_{n} \Vert A_{n} \Vert _{X}<\infty$. Moreover, if $A\in(X,\ell _{\infty})$, then $\Vert {\mathcal{L}}_{A} \Vert = \Vert A \Vert _{(X,\ell_{\infty})}$.

2 Weighted almost convergence

Definition 2.1

Let $t=(t_{k})$ be a given sequence of nonnegative numbers such that $\liminf_{k} t_{k}>0$ and $T_{m}=\sum_{k=0}^{m-1}t_{k}\neq0$ for all $m\geq1$. Then the bounded sequence $s=(s_{k})$ of real or complex numbers is said to be weighted almost convergent, shortly $f(\bar{N})$-convergent, to λ if and only if

$$\lim_{m\to\infty}\frac{1}{T_{m}}\sum_{k=r}^{r+m-1}t_{k}s_{k}= \lambda\quad\mbox{uniformly in }r. $$

We shall use the notation $f(\bar{N})$ for the space of all sequences which are $f(\bar{N})$-convergent, that is,

$$ f(\bar{N})= \Biggl\{ s\in l_{\infty}:\exists\lambda\in\mathbb {C} \ni \lim _{m\to\infty}\frac{1}{T_{m}}\sum_{k=r}^{r+m-1}t_{k}s_{k}= \lambda\mbox{ uniformly in }r; \lambda=f(\bar{N})\mbox{-}\lim s \Biggr\} . $$

(2)

We remark that if we take $t_{k}=1$ for all k, then (2) is reduced to the notion of almost convergence introduced by Lorentz [2]. Clearly, a convergent sequence is $f(\bar{N})$-convergent to the same limit, but its converse is not always true.

Example 2.2

Consider a sequence $s=(s_{k})$ defined by $s_{k}=1$ if k is odd and 0 for even k. Also, let $t_{k}=1$ for all k. Then we see that $s=(s_{k})$ is $f(\bar{N})$-convergent to $1/2$ but not convergent.

Definition 2.3

The matrix A (or a matrix map A) is said to be weighted almost conservative if $As\in f({\bar {N}})$ for all $s=(s_{k})\in c$. One denotes this by $A\in(c,f({\bar {N}}))$. If $A\in(c,f({\bar{N}}))$ with $f({\bar{N}})\mbox{-}\lim As=\lim s$, then we say that A is weighted almost regular matrix; one denotes such matrices by $A\in(c,f({\bar{N}}))_{R}$.

Theorem 2.4

The space $f(\bar{N})$ of weighted almost convergence endowed with the norm ${\Vert \cdot \Vert } _{\infty}$ is a BK-space.

Proof

To prove our results, first we have to prove that $f(\bar{N})$ is a Banach space normed by

$$ \Vert s \Vert _{f(\bar{N})}=\sup_{m,r} \bigl\vert \Psi_{m,r}(s) \bigr\vert , $$

(3)

where

$$\Psi_{m,r}(s)=\frac{1}{T_{m}}\sum_{k=r}^{r+m-1}t_{k}s_{k}. $$

It is easy to verify that (3) defines a norm on $f(\bar{N})$. We have to show that $f(\bar{N})$ is complete. For this, we need to show that every Cauchy sequence in $f(\bar{N})$ converges to some number in $f(\bar{N})$. Let $(s^{k})$ be a Cauchy sequence in $f(\bar{N})$. Then $(s_{j}^{k})$ is a Cauchy sequence in $\mathbb {R}$ (for each $j=1,2,\dots$). By using the notion of the norm of $f(\bar{N})$, it is easy to see that $(s^{k})\to s$. We have only to show that $s\in f(\bar{N})$.

Let $\epsilon>0$ be given. Since $(s^{k})$ is a Cauchy sequence in $f(\bar{N})$, there exists $M\in\mathbb {N}$ (depending on ϵ) such that

$$\bigl\Vert s^{k}-s^{i} \bigr\Vert < \epsilon/3\quad \mbox{for all }k,i>M, $$

which yields

$$\sup_{m,r} \bigl\vert \Psi\bigl(s^{k}-s^{i} \bigr) \bigr\vert < \epsilon/3. $$

Therefore we have $\vert \Psi(s^{k}-s^{i}) \vert <\epsilon/3$. Taking the limit as $m\to\infty$ gives that $\vert \lambda ^{k}-\lambda^{i} \vert <\epsilon/3$ for each m, r and $k,i>M$, where $\lambda^{k}=f(\bar{N})\mbox{-}\lim_{m} s^{k}$ and $\lambda ^{i}=f(\bar{N})\mbox{-}\lim_{m} s^{i}$. Let $\lambda=\lim_{r\to \infty}\lambda^{i}$. Letting $i\to\infty$, one obtains

$$ \bigl\vert \Psi_{mr} \bigl(s^{k}-s^{i} \bigr) \bigr\vert < \epsilon/3\quad\mbox{and}\quad \vert\lambda^{k}- \lambda \vert< \epsilon/3 $$

(4)

for each m, r and $k>M$. Now, for fixed k, the above inequality holds. Since $s^{k}\in f(\bar{N})$, for fixed k, we get

$$\lim_{m\to\infty}\Psi_{mr}\bigl(s^{k}\bigr)= \lambda^{k} \quad\mbox{uniformly in }r. $$

For given $\epsilon>0$, there exists positive integers $M_{0}$ (independent of r, but dependent upon ϵ) such that

$$ \bigl\vert \Psi_{mr} \bigl(s^{k} \bigr)- \lambda^{k} \bigr\vert < \epsilon/3 $$

(5)

for $m>M_{0}$ and for all r. It follows from (4) and (5) that

$$\begin{aligned} \bigl\vert \Psi_{mr}(s)-\lambda \bigr\vert =& \bigl\vert \Psi_{mr}(s)- \Psi_{mr} \bigl(s^{k} \bigr)+ \Psi_{mr} \bigl(s^{k} \bigr)-\lambda^{k}+ \lambda^{k}-L \bigr\vert \\ \leq& \bigl\vert \Psi_{mr}(s)-\Psi_{mr} \bigl(s^{k} \bigr) \bigr\vert + \bigl\vert \Psi_{mr} \bigl(s^{k} \bigr)- \lambda^{k} \bigr\vert + \bigl\vert \lambda^{k}-L \bigr\vert < \epsilon. \end{aligned}$$

This proves that $f(\bar{N})$ is a Banach space normed by (3).

Since $c\subset f(\bar{N})\subset l_{\infty}$, there exist positive real numbers α and β with $\alpha<\beta$ such that $\alpha \Vert s \Vert _{\infty}\leq \Vert s \Vert _{f(\bar{N})}\leq\beta \Vert s \Vert _{\infty}$. That is to say, two norms ${\Vert \cdot \Vert } _{\infty}$ and ${\Vert \cdot \Vert } _{f(\bar{N})}$ are equivalent. It is well known that the spaces c and $l_{\infty}$ endowed with the norm ${\Vert \cdot \Vert } _{\infty}$ are BK-spaces, and hence the space $f(\bar{N})$ endowed with the norm ${\Vert \cdot \Vert } _{\infty}$ is also a BK-space. □

We prove the following characterization of weighted almost conservative matrices.

Theorem 2.5

The matrix $A=(a_{n,k})$ is weighted almost conservative, that is, $A\in(c,f({\bar{N}}))$ if and only if

$$\begin{aligned}& \sup \Biggl\{ \sum_{k=0}^{\infty} \frac{1}{T_{m}} \Biggl\vert \sum_{n=r}^{r+m-1}t_{n}a_{n,k} \Biggr\vert :m\in\mathbb {Z}^{+} \Biggr\} < \infty; \end{aligned}$$

(6)

$$\begin{aligned}& \lim_{m\to\infty}\frac{1}{T_{m}}\sum_{n=r}^{r+m-1}t_{n}a_{n,k}= \lambda_{k}\quad\textit{exists }(k=0,1,2,\dots) \textit{ uniformly in }r; \end{aligned}$$

(7)

$$\begin{aligned}& \lim_{m\to\infty}\frac{1}{T_{m}}\sum_{n=r}^{r+m-1} \sum_{k=0}^{\infty}t_{n}a_{n,k}= \lambda\quad\textit{exists uniformly in }r. \end{aligned}$$

(8)

Proof

Necessity. Let $A\in(c,f({\bar{N}}))$. Since the sequences e and $e_{k}$ both are convergent, so A-transforms of the sequences $e_{k}$ and e belong to $f({\bar{N}})$ and exist uniformly in r. It follows that (7) and (8) are valid. Let r be any nonnegative integer. One writes

$$\Phi_{mr}(s)=\frac{1}{T_{m}}\sum_{n=r}^{r+m-1}t_{n} \alpha_{n}(s), $$

where

$$\alpha_{n}(s)=\sum_{k=0}^{\infty}a_{n,k}s_{k}. $$

It follows that $\alpha_{n}\in c'$ for all $n\in\mathbb {N}_{0}:=\mathbb {N}\cup\{0\}$ and this yields $\Phi_{mr}\in c'$ ($m\geq 1$). Since $A\in(c,f({\bar{N}}))$,

$$\lim_{m\to\infty}\Phi_{mr}(s)=\Phi(s)\quad \mbox{exists uniformly in }r. $$

It is clear that $(\Phi_{mr}(s))$ is bounded for $s=(s_{k})\in c$ and fixed r. Hence, by the uniform boundedness principle, $( \Vert \Phi_{mr} \Vert )$ is bounded. For each $p\in\mathbb {Z}^{+}$ (the positive integers), the sequence $x=(x_{k})$ is defined by

$$ x_{k}= \textstyle\begin{cases} \operatorname{sgn}\sum_{n=r}^{r+m-1}t_{n}a_{n,k}&\mbox{if }0\leq k\leq p,\\ 0&\mbox{if }k>p. \end{cases} $$

Then a sequence $x\in c$, $\Vert x \Vert =1$ and

$$ \bigl\vert \Phi_{mr}(x) \bigr\vert =\frac{1}{T_{m}}\sum _{k=0}^{p} \Biggl\vert \sum_{n=r}^{r+m-1}t_{n}a_{n,k} \Biggr\vert . $$

(9)

Therefore, we obtain

$$ \bigl\vert \Phi_{mr}(x) \bigr\vert \leq \Vert \Phi_{mr} \Vert \Vert x \Vert = \Vert \Phi_{mr} \Vert . $$

(10)

Equations (9) and (10) give that

$$ \frac{1}{T_{m}}\sum_{k=0}^{p} \Biggl\vert \sum_{n=r}^{r+m-1}t_{n}a_{n,k} \Biggr\vert \leq \Vert \Phi_{mr} \Vert < \infty, $$

it follows that (6) is valid.

Sufficiency. Let conditions (6)-(8) hold. Let r be any nonnegative integer, and let $s_{k}\in c$. Then

$$\begin{aligned} \Phi_{mr}(s) =&\frac{1}{T_{m}}\sum_{n=r}^{r+m-1} \sum_{k=0}^{\infty}t_{n}a_{n,k}s_{k} \\ =&\frac{1}{T_{m}}\sum_{k=0}^{\infty}\sum _{n=r}^{r+m-1}t_{n}a_{n,k}s_{k}, \end{aligned}$$

which gives

$$\bigl\vert \Phi_{mr}(s) \bigr\vert \leq\frac{1}{T_{m}}\sum _{k=0}^{\infty} \Biggl\vert \sum_{n=r}^{r+m-1}t_{n}a_{n,k} \Biggr\vert \Vert s \Vert . $$

It follows from hypothesis (6) that $\vert \Phi_{mr}(s) \vert \leq B_{r} \Vert s \Vert$, where $B_{r}$ is a constant independent of r. Thus we have $\Phi_{mr}\in c^{\prime}$ for each $m\geq1$, which gives that a sequence $( \Vert \Phi_{mr} \Vert )$ is bounded for each nonnegative integer r. Hypotheses (7) and (8) imply that the limit of $\Phi_{mr}(e_{k})$ and $\Phi_{mr}(e)$ must exist for all nonnegative integers k and r. Since $\{e,e_{0},e_{1},\dots\}$ is a fundamental set in c, it follows from [23, p. 252] that $\lim_{m}\Phi_{mr}(s)=\Phi _{r}(s)$ exists and $\Phi_{r}\in c^{\prime}$. Therefore $\Phi_{r}$ has the following form (see [23, p. 205]):

$$\Phi_{r}(s)=\xi\Biggl(\Phi_{r}(e)-\sum _{k=0}^{\infty}\Phi _{r}(e_{k}) \Biggr)+\sum_{k=0}^{\infty}s_{k} \Phi_{r}(e_{k}), $$

where $\xi=\lim s_{k}$. From (7) and (8), we see that $\Phi_{r}(e_{k})=\lambda_{k}$ for a nonnegative integer k and $\Phi_{r}(e)=\lambda$. Therefore, for each $s\in c$ and a nonnegative integer r, we have

$$\lim_{m\to\infty}\Phi_{mr}(s)=\Phi(s) $$

with the following expression:

$$ \Phi(s)=\xi \Biggl(\lambda-\sum_{k=0}^{\infty} \lambda_{k} \Biggr)+\sum_{k=0}^{\infty}s_{k} \lambda_{k}. $$

(11)

Since $\Phi_{mr}\in c^{\prime}$, so it has the representation

$$ \Phi_{mr}(s)=\xi \Biggl(\Phi_{mr}(e)-\sum _{k=0}^{\infty}\Phi_{mr}(e_{k}) \Biggr)+ \sum_{k=0}^{\infty}s_{k} \Phi_{mr}(e_{k}). $$

(12)

We observe from (11) and (12) that the convergence of $\Phi_{mr}(s)$ to $\Phi(s)$ is uniform since $\lim_{m\to\infty }\Phi_{mr}(e_{k})=\lambda_{k}$ and $\lim_{m\to\infty}\Phi _{mr}(e)=\lambda$ uniformly in r. Hence, A is a weighted almost conservative matrix. □

In the following theorem, we obtain the characterization of weighted almost regular matrices.

Theorem 2.6

The matrix $A\in(c,f({\bar{N}}))_{R}$ if and only if

$$\begin{aligned}& \sup \Biggl\{ \sum_{k=0}^{\infty} \frac{1}{T_{m}} \Biggl\vert \sum_{n=r}^{r+m-1}t_{n}a_{n,k} \Biggr\vert :m\in\mathbb {Z}^{+} \Biggr\} < \infty; \end{aligned}$$

(13)

$$\begin{aligned}& \lim_{m\to\infty}\frac{1}{T_{m}}\sum_{n=r}^{r+m-1}t_{n}a_{n,k}=0 \quad\textit{uniformly in }r~(k\in\mathbb {N}_{0}); \end{aligned}$$

(14)

$$\begin{aligned}& \lim_{m\to\infty}\frac{1}{T_{m}}\sum_{n=r}^{r+m-1} \sum_{k=0}^{\infty}t_{n}a_{n,k}=1 \quad\textit{uniformly in }r. \end{aligned}$$

(15)

Proof

Necessity. Let $A\in(c,f({\bar{N}}))_{R}$. We see that condition (13) holds by using the fact that A is also weighted almost conservative. Take $e_{k},e\in c$. Then A-transforms of the sequences $e_{k}$ and e are weighted almost convergent to 0 and 1, respectively, since $e_{k}\to0$ and $e\to1$. Hence $e_{k}\in c$ gives condition (14) and $e\in c$ proves the validity of (15).

Sufficiency. Let conditions (13)-(15) hold. It is easy to see that A is weighted almost conservative. So, for each $(s_{k})\in c$, $\lim_{m\to\infty}\Phi_{mr}(s)=\Phi(s)$ uniformly in r. Thus we obtain from (11) and our hypotheses (13)-(15) that $\Phi(s)=\xi=\lim s_{k}$. This yields A is weighted almost regular. □

We now obtain necessary and sufficient conditions for the matrix A which transform the absolutely convergent series into the space of weighted almost convergence.

Theorem 2.7

The matrix $A\in(l_{1},f({\bar{N}}))$ if and only if

$$\begin{aligned}& \sup_{k,m,r} \Biggl\vert \frac{1}{T_{m}}\sum _{n=r}^{r+m-1}t_{n}a_{n,k} \Biggr\vert < \infty, \end{aligned}$$

(16)

$$\begin{aligned}& \lim_{m\to\infty}\frac{1}{T_{m}}\sum_{n=r}^{r+m-1}t_{n}a_{n,k}= \lambda_{k}\quad\textit{exists for each }k\in\mathbb {N}_{0} \textit{ uniformly in }r. \end{aligned}$$

(17)

Proof

Necessity. Let $A\in(l_{1},f({\bar{N}}))$. Condition (17) follows since $e_{k}\in l_{1}$. Let $\Phi_{mr}$ be a continuous linear functional on $l_{1}$ defined by

$$ \Phi_{mr}(s)=\frac{1}{T_{m}}\sum _{k=0}^{\infty}\sum_{n=r}^{r+m-1}t_{n}a_{n,k}s_{k}. $$

Then we have

$$ \bigl\vert \Phi_{mr}(s) \bigr\vert \leq\sup_{k} \Biggl\vert \frac{1}{T_{m}}\sum_{n=r}^{r+m-1}t_{n}a_{n,k} \Biggr\vert \Vert s \Vert _{1}, $$

which yields

$$ \Vert \Phi_{mr} \Vert \leq\sup_{k} \Biggl\vert \frac{1}{T_{m}}\sum_{n=r}^{r+m-1}t_{n}a_{n,k} \Biggr\vert . $$

(18)

For any fixed $k\in\mathbb {N}_{0}$, we define a sequence $s=(s_{j})$ by

$$ s_{j}= \textstyle\begin{cases} \operatorname{sgn}\sum_{n=r}^{r+m-1}t_{n}a_{n,k}&\mbox{if }j=k,\\ 0&\mbox{if }j\neq k. \end{cases} $$

Then we have $\Vert s \Vert _{1}=1$ and

$$\bigl\vert \Phi_{mr}(s) \bigr\vert = \Biggl\vert \frac{1}{T_{m}}\sum _{n=r}^{r+m-1}t_{n}a_{n,k}s_{k} \Biggr\vert = \Biggl\vert \frac{1}{T_{m}}\sum_{n=r}^{r+m-1}t_{n}a_{n,k} \Biggr\vert \Vert s \Vert _{1}, $$

so

$$ \Vert \Phi_{mr} \Vert \geq\sup_{k} \Biggl\vert \frac{1}{T_{m}}\sum_{n=r}^{r+m-1}t_{n}a_{n,k} \Biggr\vert . $$

(19)

We obtain from (18) and (19) that

$$ \Vert \Phi_{mr} \Vert =\sup_{k} \Biggl\vert \frac{1}{T_{m}}\sum_{n=r}^{r+m-1}t_{n}a_{n,k} \Biggr\vert . $$

Since $A\in(l_{1},f({\bar{N}}))$, for any $s\in l_{1}$, we have

$$ \sup_{m,r} \bigl\vert \Phi_{mr}(s) \bigr\vert = \sup _{m,r} \Biggl\vert \frac{1}{T_{m}}\sum_{k=0}^{\infty} \sum_{n=r}^{r+m-1}t_{n}a_{n,k}s_{k} \Biggr\vert < \infty. $$

(20)

By using the uniform boundedness theorem, Equation (20) becomes

$$ \sup_{m,r} \Vert \Phi_{mr} \Vert =\sup _{k,m,r} \Biggl\vert \frac{1}{T_{m}}\sum_{n=r}^{r+m-1}t_{n}a_{n,k} \Biggr\vert < \infty. $$

This proves the validity of (16).

Sufficiency. Let conditions (16) and (17) hold, and let $s=(s_{k})\in l_{1}$. In virtue of these conditions, we see that

$$ \lim_{m\to\infty}\frac{1}{T_{m}}\sum_{k=0}^{\infty} \sum_{n=r}^{r+m-1}t_{n}a_{n,k}s_{k}= \sum_{k=0}^{\infty}\lambda_{k}s_{k} \quad\mbox{uniformly in }r, $$

(21)

it also converges absolutely. Furthermore, $\frac{1}{T_{m}}\sum_{k=0}^{\infty}\sum_{n=r}^{r+m-1}t_{n}a_{n,k}s_{k}$ converges absolutely for each m and r.

Let $\epsilon>0$ be given. Then there exists $k_{0}\in\mathbb {N}$ such that

$$ \sum_{k>k_{0}} \vert s_{k} \vert< \epsilon. $$

(22)

By condition (17), we can find some $m_{0}\in\mathbb {N}$ such that

$$ \Biggl\vert \sum_{k\leq k_{0}} \Biggl[\frac{1}{T_{m}}\sum _{n=r}^{r+m-1}t_{n}a_{n,k}- \lambda_{k} \Biggr]s_{k} \Biggr\vert < \epsilon $$

(23)

for all $m>m_{0}$ uniformly in r. Now

$$\begin{aligned} \Biggl\vert \sum_{k=0}^{\infty} \Biggl[ \frac{1}{T_{m}}\sum_{n=r}^{r+m-1}t_{n}a_{n,k}- \lambda_{k} \Biggr]s_{k} \Biggr\vert \leq& \Biggl\vert \sum _{k\leq k_{0}} \Biggl[\frac{1}{T_{m}}\sum _{n=r}^{r+m-1}t_{n}a_{n,k}- \lambda_{k} \Biggr]s_{k} \Biggr\vert \\ &{}+\sum_{k>k_{0}} \Biggl\vert \frac{1}{T_{m}}\sum _{n=r}^{r+m-1}t_{n}a_{n,k}- \lambda_{k} \Biggr\vert \vert s_{k} \vert \end{aligned}$$

(24)

for all $m>m_{0}$ uniformly in r. By using Equations (22) and (23) and our hypotheses in the above inequality, we see that (21) holds, and hence the sufficiency part. □

Theorem 2.8

If the matrix A in $(l_{1},f({\bar {N}}))$, then $\Vert {\mathcal{L}}_{A} \Vert = \Vert A \Vert$.

Proof

Let $A\in(l_{1},f({\bar{N}}))$. Then we have

$$\bigl\Vert {\mathcal{L}}_{A}(s) \bigr\Vert =\sup _{m,r} \Biggl\vert \frac {1}{T_{m}}\sum_{k=0}^{\infty} \sum_{n=r}^{r+m-1}t_{n}a_{n,k}s_{k} \Biggr\vert \leq\sup_{m,r}\sum_{k=0}^{\infty} \Biggl\vert \frac {1}{T_{m}}\sum_{n=r}^{r+m-1}t_{n}a_{n,k} \Biggr\vert \vert s_{k} \vert , $$

which gives $\Vert {\mathcal{L}}_{A}(s) \Vert \leq \Vert A \Vert \Vert s \Vert _{1}$. This implies that $\Vert {\mathcal{L}}_{A} \Vert \leq \Vert A \Vert$. Also, ${\mathcal{L}}_{A}\in B(l_{1},f({\bar{N}}))$ gives

$$\bigl\Vert {\mathcal{L}}_{A}(s) \bigr\Vert = \Vert As \Vert \leq \Vert {\mathcal{L}}_{A} \Vert \Vert s \Vert _{1}. $$

Taking $s=(e_{k})$ and using the fact that $\Vert e_{k} \Vert _{1}=1$ ∀k, one obtains $\Vert A \Vert \leq \Vert {\mathcal{L}}_{A} \Vert$. Hence we conclude that $\Vert {\mathcal{L}}_{A} \Vert = \Vert A \Vert$. □

Definition 2.9

Let $t=(t_{k})_{k\in\mathbb {N}}$ be a given sequence of nonnegative numbers such that $\liminf_{k} t_{k}>0$ and $T_{m}=\sum_{k=0}^{m-1}t_{k}\neq0$ for all $m\geq1$. A sequence $s=(s_{k})$ is said to be weighted almost A-summable to $\lambda\in\mathbb {C}$ if the A-transform of sequence $s=(s_{k})$ is weighted almost convergent to λ; equivalently, we can write

$$\lim_{m}\sigma_{mr}(s)=\lambda\quad\mbox{uniformly in }r, $$

where

$$\sigma_{mr}(s)=\frac{1}{T_{m}}\sum_{n=r}^{r+m-1} \sum_{k=0}^{\infty }t_{n}a_{n,k}s_{k}. $$

In the applications of summability theory to function theory, it is important to know the region in which $S=(S_{k}(z))$, the sequence of partial sums of the geometric series is A-summable to $\frac {1}{1-z}$ for a given matrix A. In the following theorem, we find the region in which S is weighted almost A-summable to $\frac{1}{1-z}$.

Theorem 2.10

Let $A=(a_{n,k})$ be a matrix such that (15) holds. The sequence $(S_{k}(z))$ is weighted almost A-summable to $\frac{1}{1-z}$ if and only if $z\in R$, where

$$ R= \Bigl\{ z= \bigl(z^{k} \bigr):\lim_{m} \sigma_{mr}(z)=0 \textit{ uniformly in }r \Bigr\} . $$

Proof

One writes

$$\begin{aligned} \sigma_{mr} =&\frac{1}{T_{m}}\sum_{n=r}^{r+m-1} \sum_{k=0}^{\infty}t_{n}a_{n,k}S_{k}(z) \\ =&\frac{1}{T_{m}}\sum_{n=r}^{r+m-1}\sum _{k=0}^{\infty}t_{n}a_{n,k} \frac{1-z^{k+1}}{1-z} \\ =&\frac{1}{(1-z)T_{m}}\sum_{n=r}^{r+m-1}\sum _{k=0}^{\infty}t_{n}a_{n,k}- \frac{z}{(1-z)T_{m}}\sum_{n=r}^{r+m-1}\sum _{k=0}^{\infty}t_{n}a_{n,k}z^{k}. \end{aligned}$$

Taking the limit as $m\to\infty$ in the above equality and using condition (15), one obtains

$$\lim_{m\to\infty}\sigma_{mr}=\frac{1}{1-z} \quad \mbox{uniformly in }r $$

if and only if $z\in R$. This completes the proof. □