# 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  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  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 ; (ii) introduce related sequence spaces derived by the domain of matrices ; (iii) study some related matrix transformations ; (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 , and for double sequences they were studied by Mursaleen  (also refer to ). As an application of almost convergence, Mohiuddine  proved a Korovkin-type approximation theorem for a sequence of linear positive operators and also obtained some of its generalizations. Başar and Kirişçi  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

()

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 . 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. □

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## Acknowledgements

This work was supported by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, under grant No. (130-694-D1435). The authors, therefore, gratefully acknowledge the DSR technical and financial support.

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Correspondence to Syed Abdul Mohiuddine.

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