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. □