# Some further extensions of absolute Cesàro summability for double series

## Abstract

In a recent paper [Savaş and Rhoades in Appl. Math. Lett. 22:1462-1466, 2009], the authors extended the result of Flett [Proc. Lond. Math. Soc. 7:113-141, 1957] to double summability. In this paper, we consider some further extensions of absolute Cesàro summability for double series.

MSC:40F05, 40G05.

Let ${\sum }_{m=0}^{\mathrm{\infty }}{\sum }_{n=0}^{\mathrm{\infty }}{a}_{mn}$ be an infinite double series with real or complex numbers, with partial sums

${s}_{mn}=\sum _{i=0}^{m}\sum _{j=0}^{n}{a}_{ij}.$

For any double sequence $\left({x}_{mn}\right)$ we shall define

${\mathrm{\Delta }}_{11}{x}_{mn}={x}_{mn}-{x}_{m+1,n}-{x}_{m,n+1}+{x}_{m+1,n+1}.$

Denote by ${\mathcal{A}}_{k}^{2}$ the sequence space defined by

${\mathcal{A}}_{k}^{2}=\left\{{\left({s}_{mn}\right)}_{m,n=0}^{\mathrm{\infty }}:\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}{\left(mn\right)}^{k-1}|{a}_{mn}{|}^{k}<\mathrm{\infty };{a}_{mn}={\mathrm{\Delta }}_{11}{s}_{m-1,n-1}\right\}$

for $k\ge 1$.

A four-dimensional matrix $T=\left({t}_{mnij}:m,n,i,j=0,1,\dots \right)$ is said to be absolutely k th power conservative for $k\ge 1$, if $T\in B\left({\mathcal{A}}_{k}^{2}\right)$, i.e., if

$\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}{\left(mn\right)}^{k-1}|{\mathrm{\Delta }}_{11}{s}_{m-1,n-1}{|}^{k}<\mathrm{\infty },$

then

$\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}{\left(mn\right)}^{k-1}|{\mathrm{\Delta }}_{11}{t}_{m-1,n-1}{|}^{k}<\mathrm{\infty },$

where

${t}_{mn}=\sum _{i=0}^{\mathrm{\infty }}\sum _{j=0}^{\mathrm{\infty }}{t}_{mnij}{s}_{ij}\phantom{\rule{1em}{0ex}}\left(m,n=0,1,\dots \right).$

A double infinite Cesáro matrix $C\left(\alpha ,\beta \right)$ is a double infinite Hausdorff matrix with entries

${h}_{mnij}=\frac{{E}_{m-i}^{\alpha -1}{E}_{n-j}^{\beta -1}}{{E}_{m}^{\alpha }{E}_{n}^{\beta }},$

where

${E}_{m}^{\alpha }=\left(\genfrac{}{}{0}{}{m+\alpha }{m}\right)=\frac{\mathrm{\Gamma }\left(\alpha +m+1\right)}{\mathrm{\Gamma }\left(m+1\right)\mathrm{\Gamma }\left(\alpha +1\right)}\approx \frac{{m}^{\alpha }}{\mathrm{\Gamma }\left(\alpha +1\right)}.$

The series $\sum \sum {a}_{mn}$ is said to be summable $|C\left(\alpha ,\beta \right){|}_{k}$, $k\ge 1$, $\alpha ,\beta >-1$, if (see [1])

$\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}{\left(mn\right)}^{k-1}|{\mathrm{\Delta }}_{11}{\sigma }_{m-1,n-1}^{\alpha \beta }{|}^{k}<\mathrm{\infty },$
(1)

where ${\sigma }_{mn}^{\alpha \beta }$ denotes the mn-term of the $C\left(\alpha ,\beta \right)$ transform of a sequence $\left({s}_{mn}\right)$; i.e.,

${\sigma }_{mn}^{\alpha \beta }=\frac{1}{{E}_{m}^{\alpha }{E}_{n}^{\beta }}\sum _{i=0}^{m}\sum _{j=0}^{n}{E}_{m-i}^{\alpha -1}{E}_{n-j}^{\beta -1}{s}_{ij}.$
(2)

Quite recently, Savaş and Rhoades [2] extended the result of Flett [3] to double summability. Their theorem is as follows.

Theorem 1 Let $\alpha \ge \gamma >-1$, $\beta \ge \delta >-1$, and ${\sum }_{m}{\sum }_{n}{a}_{mn}$ be a double series with partial sums ${s}_{mn}$. If ${\sum }_{m}{\sum }_{n}{a}_{mn}$is $|C\left(\gamma ,\delta \right){|}_{k}$-summable, then it is also $|C\left(\alpha ,\beta \right){|}_{k}$-summable, $k\ge 1$.

It then follows that if one sets $\gamma =\delta =0$, then $C\left(\alpha ,\beta \right)\in B\left({\mathcal{A}}_{k}^{2}\right)$ for each $\alpha ,\beta \ge 0$. In this paper, we consider some further extensions of absolute Cesàro summability for double series.

We shall use the following lemmas.

Lemma 1 If $\theta >-1$, $\varphi >-1$, $\theta -\phi >0$ and $\varphi -\psi >0$, then

$\sum _{m=i}^{\mathrm{\infty }}\sum _{n=j}^{\mathrm{\infty }}\frac{{E}_{m-i}^{\phi }{E}_{n-j}^{\psi }}{mn{E}_{m}^{\theta }{E}_{n}^{\varphi }}=\frac{1}{ij{E}_{i}^{\theta -\phi -1}{E}_{j}^{\varphi -\psi -1}}.$
(3)

Proof For $\alpha >-1$, $n\ge 1$ since

$\frac{1}{{E}_{n}^{\alpha }}={\int }_{0}^{1}{\left(1-x\right)}^{\alpha }{x}^{n-1}\phantom{\rule{0.2em}{0ex}}dx$

and

${\left(1-x\right)}^{-\alpha }=\sum _{n=0}^{\mathrm{\infty }}{E}_{n}^{\alpha -1}{x}^{n},$

we obtain

$\begin{array}{rcl}\sum _{m=i}^{\mathrm{\infty }}\sum _{n=j}^{\mathrm{\infty }}\frac{{E}_{m-i}^{\phi }{E}_{n-j}^{\psi }}{mn{E}_{m}^{\theta }{E}_{n}^{\varphi }}& =& \sum _{m=i}^{\mathrm{\infty }}\frac{{E}_{m-i}^{\phi }}{m{E}_{m}^{\theta }}\sum _{n=0}^{\mathrm{\infty }}\frac{{E}_{n}^{\psi }}{\left(n+j\right){E}_{n+j}^{\varphi }}\\ =& \sum _{m=i}^{\mathrm{\infty }}\frac{{E}_{m-i}^{\phi }}{m{E}_{m}^{\theta }}\sum _{n=0}^{\mathrm{\infty }}{E}_{n}^{\psi }{\int }_{0}^{1}{\left(1-x\right)}^{\varphi }{x}^{n+j-1}\phantom{\rule{0.2em}{0ex}}dx\\ =& \sum _{m=i}^{\mathrm{\infty }}\frac{{E}_{m-i}^{\phi }}{m{E}_{m}^{\theta }}{\int }_{0}^{1}{\left(1-x\right)}^{\varphi }{x}^{j-1}\left(\sum _{n=0}^{\mathrm{\infty }}{E}_{n}^{\psi }{x}^{n}\right)\phantom{\rule{0.2em}{0ex}}dx\\ =& {\int }_{0}^{1}{\left(1-x\right)}^{\varphi -\psi -1}{x}^{j-1}dx\sum _{m=0}^{\mathrm{\infty }}\frac{{E}_{m}^{\phi }}{\left(m+i\right){E}_{m+i}^{\theta }}\\ =& \frac{1}{j{E}_{j}^{\varphi -\psi -1}}\sum _{m=0}^{\mathrm{\infty }}{E}_{m}^{\phi }{\int }_{0}^{1}{\left(1-x\right)}^{\theta }{x}^{m+i-1}\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{1}{j{E}_{j}^{\varphi -\psi -1}}{\int }_{0}^{1}{\left(1-x\right)}^{\theta -\phi -1}{x}^{i-1}\phantom{\rule{0.2em}{0ex}}dx\\ =& \frac{1}{ij{E}_{i}^{\theta -\phi -1}{E}_{j}^{\varphi -\psi -1}}.\end{array}$

□

For single series, Lemma 1 due to Chow [4].

Lemma 2 Let $1\le k\le r<\mathrm{\infty }$ and $\alpha ,\beta >-1$. For $i,j\ge 1$, let

${A}_{ij}={A}_{ij}\left(\alpha ,\beta \right)=\sum _{m=i}^{\mathrm{\infty }}\sum _{n=j}^{\mathrm{\infty }}\frac{|{E}_{m-i}^{\alpha -1}{|}^{r/k}|{E}_{n-j}^{\beta -1}{|}^{r/k}}{mn{\left({E}_{m}^{\alpha }\right)}^{r/k}{\left({E}_{n}^{\beta }\right)}^{r/k}}.$
(4)

Then, if $k=r$,

${A}_{ij}=\left\{\begin{array}{ll}O\left({i}^{-\alpha -1}{j}^{-\beta -1}\right),& \mathit{\text{if}}\phantom{\rule{0.1em}{0ex}}\alpha ,\beta <0,\\ O\left({i}^{-\alpha -1}{j}^{-1}\right),& \mathit{\text{if}}\phantom{\rule{0.1em}{0ex}}\alpha <0,\beta \ge 0,\\ O\left({i}^{-1}{j}^{-\beta -1}\right),& \mathit{\text{if}}\phantom{\rule{0.1em}{0ex}}\alpha \ge 0,\beta <0,\\ O\left({i}^{-1}{j}^{-1}\right),& \mathit{\text{if}}\phantom{\rule{0.1em}{0ex}}\alpha ,\beta \ge 0.\end{array}$
(5)

If $k, then

${A}_{ij}=\left\{\begin{array}{ll}O\left({i}^{-\frac{\alpha r}{k}-1}{j}^{-\frac{\beta r}{k}-1}\right),& \mathit{\text{if}}\phantom{\rule{0.1em}{0ex}}\alpha <1-k/r,\beta <1-k/r,\\ O\left({i}^{-\frac{\alpha r}{k}-1}{j}^{-\frac{r}{k}}logj\right),& \mathit{\text{if}}\phantom{\rule{0.1em}{0ex}}\alpha <1-k/r,\beta =1-k/r,\\ O\left({i}^{-\frac{\alpha r}{k}-1}{j}^{-\frac{r}{k}}\right),& \mathit{\text{if}}\phantom{\rule{0.1em}{0ex}}\alpha <1-k/r,\beta >1-k/r,\\ O\left({i}^{-\frac{\alpha r}{k}-1}{j}^{-\frac{\beta r}{k}-1}logi\right),& \mathit{\text{if}}\phantom{\rule{0.1em}{0ex}}\alpha =1-k/r,\beta <1-k/r,\\ O\left({i}^{-\frac{r}{k}}{j}^{-\frac{r}{k}}logilogj\right),& \mathit{\text{if}}\phantom{\rule{0.1em}{0ex}}\alpha =1-k/r,\beta =1-k/r,\\ O\left({i}^{-\frac{\alpha r}{k}-1}{j}^{-\frac{r}{k}}logi\right),& \mathit{\text{if}}\phantom{\rule{0.1em}{0ex}}\alpha =1-k/r,\beta >1-k/r,\\ O\left({i}^{-\frac{r}{k}}{j}^{-\frac{\beta r}{k}-1}\right),& \mathit{\text{if}}\phantom{\rule{0.1em}{0ex}}\alpha >1-k/r,\beta <1-k/r,\\ O\left({i}^{-\frac{r}{k}}{j}^{-\frac{\beta r}{k}-1}logj\right),& \mathit{\text{if}}\phantom{\rule{0.1em}{0ex}}\alpha >1-k/r,\beta =1-k/r,\\ O\left({i}^{-\frac{r}{k}}{j}^{-\frac{r}{k}}\right),& \mathit{\text{if}}\phantom{\rule{0.1em}{0ex}}\alpha >1-k/r,\beta >1-k/r.\end{array}$
(6)

Proof Since $|{E}_{n}^{\alpha }|\le K\left(\alpha \right){n}^{\alpha }$ for all α, and ${E}_{n}^{\alpha }\ge M\left(\alpha \right){n}^{\alpha }$ for $\alpha >-1$, where $K\left(\alpha \right)$ and $M\left(\alpha \right)$ are positive constants depending only on α, if $k, then

$\begin{array}{rl}{A}_{ij}=& O\left(1\right)\sum _{m=i}^{\mathrm{\infty }}\sum _{n=j}^{\mathrm{\infty }}{m}^{-\frac{\alpha r}{k}-1}{n}^{-\frac{\beta r}{k}-1}{\left(m-i+1\right)}^{\frac{\left(\alpha -1\right)r}{k}}{\left(n-j+1\right)}^{\frac{\left(\beta -1\right)r}{k}}\\ =& O\left(1\right)\left(\sum _{m=i}^{2i-1}{m}^{-\frac{\alpha r}{k}-1}{\left(m-i+1\right)}^{\frac{\left(\alpha -1\right)r}{k}}+\sum _{m=2i}^{\mathrm{\infty }}{m}^{-\frac{\alpha r}{k}-1}{\left(m-i+1\right)}^{\frac{\left(\alpha -1\right)r}{k}}\right)\\ ×\left(\sum _{n=j}^{2j-1}{n}^{-\frac{\beta r}{k}-1}{\left(n-j+1\right)}^{\frac{\left(\beta -1\right)r}{k}}+\sum _{n=2j}^{\mathrm{\infty }}{n}^{-\frac{\beta r}{k}-1}{\left(n-j+1\right)}^{\frac{\left(\beta -1\right)r}{k}}\right)\\ =& O\left(1\right)\left({i}^{-\frac{\alpha r}{k}-1}\sum _{m=i}^{2i-1}{\left(m-i+1\right)}^{\frac{\left(\alpha -1\right)r}{k}}+\sum _{m=2i}^{\mathrm{\infty }}{m}^{-\frac{r}{k}-1}\right)\\ ×\left({j}^{-\frac{\beta r}{k}-1}\sum _{n=j}^{2j-1}{\left(n-j+1\right)}^{\frac{\left(\beta -1\right)r}{k}}+\sum _{n=2j}^{\mathrm{\infty }}{n}^{-\frac{r}{k}-1}\right)\\ =& O\left(1\right){i}^{-\frac{\alpha r}{k}-1}{j}^{-\frac{\beta r}{k}-1}\sum _{m=1}^{i}\sum _{n=1}^{j}{m}^{\frac{\left(\alpha -1\right)r}{k}}{n}^{\frac{\left(\beta -1\right)r}{k}}+O\left(1\right){i}^{-\frac{\alpha r}{k}-1}{j}^{-r/k}\sum _{m=1}^{i}{m}^{\frac{\left(\alpha -1\right)r}{k}}\\ +O\left(1\right){i}^{-r/k}{j}^{-\frac{\beta r}{k}-1}\sum _{n=1}^{j}{n}^{\frac{\left(\beta -1\right)r}{k}}+O\left(1\right){\left(ij\right)}^{-r/k}.\end{array}$

According as $\frac{\left(\alpha -1\right)r}{k}$ and $\frac{\left(\beta -1\right)r}{k}=-1$, $<-1$ or $>-1$, we have (6). The case $k=r$ is proved similarly. □

For single series, Lemma 2 due to Mehdi [5].

We now prove the following theorem.

Theorem 2 Let $r\ge k\ge 1$.

(i) It holds that $C\left(\alpha ,\beta \right)\in \left({\mathcal{A}}_{k}^{2},{\mathcal{A}}_{r}^{2}\right)$ for each $\alpha ,\beta >1-k/r$.

(ii) If $\alpha ,\beta =1-k/r$ and the condition ${\sum }_{m=1}^{\mathrm{\infty }}{\sum }_{n=1}^{\mathrm{\infty }}{\left(mn\right)}^{k-1}logmlogn|{a}_{mn}{|}^{k}=O\left(1\right)$ is satisfied then $C\left(\alpha ,\beta \right)\in \left({\mathcal{A}}_{k}^{2},{\mathcal{A}}_{r}^{2}\right)$.

(iii) If the condition ${\sum }_{m=1}^{\mathrm{\infty }}{\sum }_{n=1}^{\mathrm{\infty }}{m}^{k+\left(1-\alpha \right)\frac{r}{k}-2}{n}^{k+\left(1-\beta \right)\frac{r}{k}-2}|{a}_{mn}{|}^{k}=O\left(1\right)$ is satisfied then $C\left(\alpha ,\beta \right)\in \left({\mathcal{A}}_{k}^{2},{\mathcal{A}}_{r}^{2}\right)$ for each $-k/r<\alpha ,\beta <1-k/r$.

(iv) If $\beta =1-k/r$ and the condition ${\sum }_{m=1}^{\mathrm{\infty }}{\sum }_{n=1}^{\mathrm{\infty }}{m}^{k+\left(1-\alpha \right)\frac{r}{k}-2}{n}^{k-1}logn|{a}_{mn}{|}^{k}=O\left(1\right)$ is satisfied then $C\left(\alpha ,\beta \right)\in \left({\mathcal{A}}_{k}^{2},{\mathcal{A}}_{r}^{2}\right)$ for each $-k/r<\alpha <1-k/r$.

(v) If the condition ${\sum }_{m=1}^{\mathrm{\infty }}{\sum }_{n=1}^{\mathrm{\infty }}{m}^{k+\left(1-\alpha \right)\frac{r}{k}-2}{n}^{k-1}|{a}_{mn}{|}^{k}=O\left(1\right)$ is satisfied then $C\left(\alpha ,\beta \right)\in \left({\mathcal{A}}_{k}^{2},{\mathcal{A}}_{r}^{2}\right)$ for each $-k/r<\alpha <1-k/r$ and $\beta >1-k/r$.

(vi) If $\alpha =1-k/r$ and the condition ${\sum }_{m=1}^{\mathrm{\infty }}{\sum }_{n=1}^{\mathrm{\infty }}{m}^{k-1}{n}^{k+\left(1-\beta \right)\frac{r}{k}-2}logm|{a}_{mn}{|}^{k}=O\left(1\right)$ is satisfied then $C\left(\alpha ,\beta \right)\in \left({\mathcal{A}}_{k}^{2},{\mathcal{A}}_{r}^{2}\right)$ for each $-k/r<\beta <1-k/r$.

(vii) If the condition ${\sum }_{m=1}^{\mathrm{\infty }}{\sum }_{n=1}^{\mathrm{\infty }}{m}^{k-1}{n}^{k-1}logm|{a}_{mn}{|}^{k}=O\left(1\right)$ is satisfied then $C\left(\alpha ,\beta \right)\in \left({\mathcal{A}}_{k}^{2},{\mathcal{A}}_{r}^{2}\right)$ for each $\alpha >1-k/r$, $\beta <1-k/r$.

(viii) If the condition ${\sum }_{m=1}^{\mathrm{\infty }}{\sum }_{n=1}^{\mathrm{\infty }}{m}^{k-1}{n}^{k+\left(1-\beta \right)\frac{r}{k}-2}|{a}_{mn}{|}^{k}=O\left(1\right)$ is satisfied then $C\left(\alpha ,\beta \right)\in \left({\mathcal{A}}_{k}^{2},{\mathcal{A}}_{r}^{2}\right)$ for each $\alpha >1-k/r$ and $-k/r<\beta <1-k/r$.

(ix) If $\beta =1-k/r$ and the condition ${\sum }_{m=1}^{\mathrm{\infty }}{\sum }_{n=1}^{\mathrm{\infty }}{m}^{k-1}{n}^{k-1}logn|{a}_{mn}{|}^{k}=O\left(1\right)$ is satisfied then $C\left(\alpha ,\beta \right)\in \left({\mathcal{A}}_{k}^{2},{\mathcal{A}}_{r}^{2}\right)$ for each $\alpha >1-k/r$.

Proof We shall show that $\left({\sigma }_{mn}^{\alpha \beta }\right)\in {\mathcal{A}}_{r}^{2}$, i.e.,

$\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}{\left(mn\right)}^{r-1}|{\mathrm{\Delta }}_{11}{\sigma }_{m-1,n-1}^{\alpha \beta }{|}^{r}<\mathrm{\infty }.$

Let ${\tau }_{mn}^{\alpha \beta }$ denote the mn-term of the $C\left(\alpha ,\beta \right)$-transform in terms of $\left(mn{a}_{mn}\right)$, i.e.,

${\tau }_{mn}^{\alpha \beta }=\frac{1}{{E}_{m}^{\alpha }{E}_{n}^{\beta }}\sum _{i=1}^{m}\sum _{j=1}^{n}{E}_{m-i}^{\alpha -1}{E}_{n-j}^{\beta -1}ij{a}_{ij}.$

For $\alpha ,\beta >-1$, since

${\tau }_{mn}^{\alpha \beta }=mn\left({\sigma }_{mn}^{\alpha \beta }-{\sigma }_{m,n-1}^{\alpha \beta }-{\sigma }_{m-1,n}^{\alpha \beta }+{\sigma }_{m-1,n-1}^{\alpha \beta }\right),$

to prove the theorem, it will be sufficient to show that

$\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}{\left(mn\right)}^{-1}|{\tau }_{mn}^{\alpha \beta }{|}^{r}<\mathrm{\infty }.$
(7)

Using Hölder’s inequality, we have

$\begin{array}{rl}\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{mn}|{\tau }_{mn}^{\alpha \beta }{|}^{r}=& \sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{mn}|\frac{1}{{E}_{m}^{\alpha }{E}_{n}^{\beta }}\sum _{i=1}^{m}\sum _{j=1}^{n}{E}_{m-i}^{\alpha -1}{E}_{n-j}^{\beta -1}ij{a}_{ij}{|}^{r}\\ \le & \sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{mn{\left({E}_{m}^{\alpha }\right)}^{r}{\left({E}_{n}^{\beta }\right)}^{r}}{\left\{\sum _{i=1}^{m}\sum _{j=1}^{n}|{E}_{m-i}^{\alpha -1}||{E}_{n-j}^{\beta -1}|{i}^{k}{j}^{k}|{a}_{ij}{|}^{k}\right\}}^{r/k}\\ ×{\left\{\sum _{i=1}^{m}\sum _{j=1}^{n}|{E}_{m-i}^{\alpha -1}||{E}_{n-j}^{\beta -1}|\right\}}^{\left(k-1\right)r/k}.\end{array}$

Since

$\begin{array}{rl}\sum _{i=1}^{m}\sum _{j=1}^{n}|{E}_{m-i}^{\alpha -1}||{E}_{n-j}^{\beta -1}|=& \left(|{E}_{0}^{\alpha -1}|+\sum _{i=1}^{m-1}|{E}_{m-i}^{\alpha -1}|\right)\left(|{E}_{0}^{\beta -1}|+\sum _{j=1}^{n-1}|{E}_{n-j}^{\beta -1}|\right)\\ =& \left(|{E}_{0}^{\alpha -1}|+|\sum _{i=1}^{m-1}{E}_{m-i}^{\alpha -1}|\right)\left(|{E}_{0}^{\beta -1}|+|\sum _{j=1}^{n-1}{E}_{n-j}^{\beta -1}|\right)\\ =& \left(|{E}_{0}^{\alpha -1}|+|\sum _{i=0}^{m}{E}_{m-i}^{\alpha -1}-{E}_{m}^{\alpha -1}-{E}_{0}^{\alpha -1}|\right)\\ ×\left(|{E}_{0}^{\beta -1}|+|\sum _{j=0}^{n}{E}_{n-j}^{\beta -1}-{E}_{n}^{\beta -1}-{E}_{0}^{\beta -1}|\right)\\ =& \left(|{E}_{0}^{\alpha -1}|+|{E}_{m-1}^{\alpha }-{E}_{0}^{\alpha -1}|\right)\left(|{E}_{0}^{\beta -1}|+|{E}_{n-1}^{\beta }-{E}_{0}^{\beta -1}|\right),\end{array}$

and using the fact that

$|\frac{{E}_{m-1}^{\alpha }}{{E}_{m}^{\alpha }}|=O\left(1\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}|\frac{{E}_{n-1}^{\beta }}{{E}_{n}^{\beta }}|=O\left(1\right),$

we obtain

$\begin{array}{r}\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{mn}|{\tau }_{mn}^{\alpha \beta }{|}^{r}\\ \phantom{\rule{1em}{0ex}}\le \sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{{\left({E}_{m}^{\alpha }\right)}^{\left(k-1\right)r/k}{\left({E}_{n}^{\beta }\right)}^{\left(k-1\right)r/k}}{mn{\left({E}_{m}^{\alpha }\right)}^{r}{\left({E}_{n}^{\beta }\right)}^{r}}\\ \phantom{\rule{2em}{0ex}}×{\left\{\sum _{i=1}^{m}\sum _{j=1}^{n}|{E}_{m-i}^{\alpha -1}||{E}_{n-j}^{\beta -1}|{i}^{k}{j}^{k}|{a}_{ij}{|}^{k}\right\}}^{r/k}\\ \phantom{\rule{1em}{0ex}}\le \sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{{\left({E}_{m}^{\alpha }\right)}^{-r/k}{\left({E}_{n}^{\beta }\right)}^{-r/k}}{mn}\\ \phantom{\rule{2em}{0ex}}×{\left\{\sum _{i=1}^{m}\sum _{j=1}^{n}|{E}_{m-i}^{\alpha -1}||{E}_{n-j}^{\beta -1}|{\left(ij\right)}^{1-k/r+{k}^{2}/r}|{a}_{ij}{|}^{{k}^{2}/r}{\left(ij\right)}^{-\left(r-k\right)+k\left(r-k\right)/r}|{a}_{ij}{|}^{k\left(r-k\right)/r}\right\}}^{r/k}.\end{array}$

Applying Hölder’s inequality with indices $r/k$, $r/\left(r-k\right)$, we deduce that

$\begin{array}{rl}\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{mn}|{\tau }_{mn}^{\alpha \beta }{|}^{r}\le & \sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{{\left({E}_{m}^{\alpha }\right)}^{-r/k}{\left({E}_{n}^{\beta }\right)}^{-r/k}}{mn}\sum _{i=1}^{m}\sum _{j=1}^{n}|{E}_{m-i}^{\alpha -1}{|}^{r/k}|{E}_{n-j}^{\beta -1}{|}^{r/k}{\left(ij\right)}^{k-1+r/k}|{a}_{ij}{|}^{k}\\ ×{\left\{\sum _{i=1}^{m}\sum _{j=1}^{n}{\left(ij\right)}^{k-1}|{a}_{ij}{|}^{k}\right\}}^{\left(r-k\right)/k}.\end{array}$

Since $\left({s}_{mn}\right)\in {\mathcal{A}}_{k}^{2}$, we have

$\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{mn}|{\tau }_{mn}^{\alpha \beta }{|}^{r}=O\left(1\right)\sum _{i=1}^{\mathrm{\infty }}\sum _{j=1}^{\mathrm{\infty }}{\left(ij\right)}^{k-1}|{a}_{ij}{|}^{k}{\left(ij\right)}^{r/k}\sum _{m=i}^{\mathrm{\infty }}\sum _{n=j}^{\mathrm{\infty }}\frac{|{E}_{m-i}^{\alpha -1}{|}^{r/k}|{E}_{n-j}^{\beta -1}{|}^{r/k}}{mn{\left({E}_{m}^{\alpha }\right)}^{r/k}{\left({E}_{n}^{\beta }\right)}^{r/k}}.$

(i) From Lemma 1, if $\alpha ,\beta >1-k/r$, then

$\begin{array}{rl}\sum _{m=i}^{\mathrm{\infty }}\sum _{n=j}^{\mathrm{\infty }}\frac{{\left({E}_{m-i}^{\alpha -1}\right)}^{r/k}{\left({E}_{n-j}^{\beta -1}\right)}^{r/k}}{mn{\left({E}_{m}^{\alpha }\right)}^{r/k}{\left({E}_{n}^{\beta }\right)}^{r/k}}& ={\left(ij\right)}^{-1}\frac{1}{{E}_{i}^{\frac{r}{k}-1}{E}_{j}^{\frac{r}{k}-1}}\\ =O\left({\left(ij\right)}^{-r/k}\right).\end{array}$

Therefore, for the case $\alpha ,\beta >1-k/r$, we have

$\begin{array}{rl}\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{mn}|{\tau }_{mn}^{\alpha \beta }{|}^{r}& =O\left(1\right)\sum _{i=1}^{\mathrm{\infty }}\sum _{j=1}^{\mathrm{\infty }}{\left(ij\right)}^{k-1}|{a}_{ij}{|}^{k}{\left(ij\right)}^{r/k}{\left(ij\right)}^{-r/k}\\ =O\left(1\right)\sum _{i=1}^{\mathrm{\infty }}\sum _{j=1}^{\mathrm{\infty }}{\left(ij\right)}^{k-1}|{a}_{ij}{|}^{k}=O\left(1\right),\end{array}$

since $\left({s}_{mn}\right)\in {\mathcal{A}}_{k}^{2}$.

(ii) If $\alpha ,\beta =1-k/r$, from Lemma 2, then

$\sum _{m=i}^{\mathrm{\infty }}\sum _{n=j}^{\mathrm{\infty }}\frac{|{E}_{m-i}^{\alpha -1}{|}^{r/k}|{E}_{n-j}^{\beta -1}{|}^{r/k}}{mn{\left({E}_{m}^{\alpha }\right)}^{r/k}{\left({E}_{n}^{\beta }\right)}^{r/k}}=O\left({i}^{-\frac{r}{k}}{j}^{-\frac{r}{k}}logilogj\right).$

Hence,

$\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{mn}|{\tau }_{mn}^{\alpha \beta }{|}^{r}=O\left(1\right)\sum _{i=1}^{\mathrm{\infty }}\sum _{j=1}^{\mathrm{\infty }}{\left(ij\right)}^{k-1}logilogj|{a}_{ij}{|}^{k}=O\left(1\right).$

(iii) If $-k/r<\alpha$, $\beta <1-k/r$, from Lemma 2, then

$\sum _{m=i}^{\mathrm{\infty }}\sum _{n=j}^{\mathrm{\infty }}\frac{|{E}_{m-i}^{\alpha -1}{|}^{r/k}|{E}_{n-j}^{\beta -1}{|}^{r/k}}{mn{\left({E}_{m}^{\alpha }\right)}^{r/k}{\left({E}_{n}^{\beta }\right)}^{r/k}}=O\left({i}^{-\frac{\alpha r}{k}-1}{j}^{-\frac{\beta r}{k}-1}\right),$

and then

$\begin{array}{rcl}\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{mn}|{\tau }_{mn}^{\alpha \beta }{|}^{r}& =& O\left(1\right)\sum _{i=1}^{\mathrm{\infty }}\sum _{j=1}^{\mathrm{\infty }}{\left(ij\right)}^{k+r/k-1}{i}^{-\frac{\alpha r}{k}-1}{j}^{-\frac{\beta r}{k}-1}|{a}_{ij}{|}^{k}\\ =& O\left(1\right)\sum _{i=1}^{\mathrm{\infty }}\sum _{j=1}^{\mathrm{\infty }}{i}^{k+\left(1-\alpha \right)\frac{r}{k}-2}{j}^{k+\left(1-\beta \right)\frac{r}{k}-2}|{a}_{ij}{|}^{k}=O\left(1\right).\end{array}$

(iv) If $\beta =1-k/r$ and $-k/r<\alpha <1-k/r$, from Lemma 2, then

$\sum _{m=i}^{\mathrm{\infty }}\sum _{n=j}^{\mathrm{\infty }}\frac{|{E}_{m-i}^{\alpha -1}{|}^{r/k}|{E}_{n-j}^{\beta -1}{|}^{r/k}}{mn{\left({E}_{m}^{\alpha }\right)}^{r/k}{\left({E}_{n}^{\beta }\right)}^{r/k}}=O\left({i}^{-\frac{\alpha r}{k}-1}{j}^{-\frac{r}{k}}logj\right),$

therefore, we have

$\begin{array}{rl}\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{mn}|{\tau }_{mn}^{\alpha \beta }{|}^{r}& =O\left(1\right)\sum _{i=1}^{\mathrm{\infty }}\sum _{j=1}^{\mathrm{\infty }}{\left(ij\right)}^{k-1}|{a}_{ij}{|}^{k}{\left(ij\right)}^{r/k}{i}^{-\frac{\alpha r}{k}-1}{j}^{-\frac{r}{k}}logj\\ =O\left(1\right)\sum _{i=1}^{\mathrm{\infty }}\sum _{j=1}^{\mathrm{\infty }}{i}^{k+\left(1-\alpha \right)\frac{r}{k}-2}{j}^{k-1}logj|{a}_{ij}{|}^{k}=O\left(1\right).\end{array}$

(v) If $-k/r<\alpha <1-k/r$ and $\beta >1-k/r$, then

$\begin{array}{rcl}\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{mn}|{\tau }_{mn}^{\alpha \beta }{|}^{r}& =& O\left(1\right)\sum _{i=1}^{\mathrm{\infty }}\sum _{j=1}^{\mathrm{\infty }}{\left(ij\right)}^{k-1}|{a}_{ij}{|}^{k}{\left(ij\right)}^{r/k}{i}^{-\frac{\alpha r}{k}-1}{j}^{-\frac{r}{k}}\\ =& O\left(1\right)\sum _{i=1}^{\mathrm{\infty }}\sum _{j=1}^{\mathrm{\infty }}{i}^{k+\left(1-\alpha \right)\frac{r}{k}-2}{j}^{k-1}|{a}_{ij}{|}^{k}=O\left(1\right),\end{array}$

by using Lemma 2.

(vi) If $\alpha =1-k/r$ and $-k/r<\beta <1-k/r$, then

$\begin{array}{rcl}\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{mn}|{\tau }_{mn}^{\alpha \beta }{|}^{r}& =& O\left(1\right)\sum _{i=1}^{\mathrm{\infty }}\sum _{j=1}^{\mathrm{\infty }}{\left(ij\right)}^{k-1}|{a}_{ij}{|}^{k}{\left(ij\right)}^{r/k}{i}^{-\frac{r}{k}}{j}^{-\frac{\beta r}{k}-1}logi\\ =& O\left(1\right)\sum _{i=1}^{\mathrm{\infty }}\sum _{j=1}^{\mathrm{\infty }}{i}^{k-1}{j}^{k+\left(1-\beta \right)\frac{r}{k}-2}logi|{a}_{ij}{|}^{k}=O\left(1\right),\end{array}$

by using Lemma 2.

(vii) If $\alpha =1-k/r$ and $\beta >1-k/r$, then

$\begin{array}{rcl}\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{mn}|{\tau }_{mn}^{\alpha \beta }{|}^{r}& =& O\left(1\right)\sum _{i=1}^{\mathrm{\infty }}\sum _{j=1}^{\mathrm{\infty }}{\left(ij\right)}^{k-1}|{a}_{ij}{|}^{k}{\left(ij\right)}^{r/k}{i}^{-\frac{r}{k}}{j}^{-\frac{r}{k}}logi\\ =& O\left(1\right)\sum _{i=1}^{\mathrm{\infty }}\sum _{j=1}^{\mathrm{\infty }}{i}^{k-1}{j}^{k-1}logi|{a}_{ij}{|}^{k}=O\left(1\right),\end{array}$

by using Lemma 2.

(viii) If $\alpha >1-k/r$ and $-k/r<\beta <1-k/r$, then

$\begin{array}{rcl}\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{mn}|{\tau }_{mn}^{\alpha \beta }{|}^{r}& =& O\left(1\right)\sum _{i=1}^{\mathrm{\infty }}\sum _{j=1}^{\mathrm{\infty }}{\left(ij\right)}^{k-1}|{a}_{ij}{|}^{k}{\left(ij\right)}^{r/k}{i}^{-\frac{r}{k}}{j}^{-\frac{\beta r}{k}-1}\\ =& O\left(1\right)\sum _{i=1}^{\mathrm{\infty }}\sum _{j=1}^{\mathrm{\infty }}{i}^{k-1}{j}^{k+\left(1-\beta \right)\frac{r}{k}-2}|{a}_{ij}{|}^{k}=O\left(1\right),\end{array}$

by using Lemma 2.

(ix) If $\alpha >1-k/r$ and $\beta =1-k/r$, then

$\begin{array}{rcl}\sum _{m=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{mn}|{\tau }_{mn}^{\alpha \beta }{|}^{r}& =& O\left(1\right)\sum _{i=1}^{\mathrm{\infty }}\sum _{j=1}^{\mathrm{\infty }}{\left(ij\right)}^{k-1}|{a}_{ij}{|}^{k}{\left(ij\right)}^{r/k}{i}^{-\frac{r}{k}}{j}^{-\frac{\beta r}{k}-1}logj\\ =& O\left(1\right)\sum _{i=1}^{\mathrm{\infty }}\sum _{j=1}^{\mathrm{\infty }}{i}^{k-1}{j}^{k-1}logj|{a}_{ij}{|}^{k}=O\left(1\right),\end{array}$

by using Lemma 2. □

The one-dimensional version of Theorem 2 appears in [6]. By (5), Theorem 2 includes the following theorem with the special case $r=k$.

Theorem 3 Let $k\ge 1$.

(i) It holds that $C\left(\alpha ,\beta \right)\in B\left({\mathcal{A}}_{k}^{2}\right)$ for each $\alpha ,\beta \ge 0$.

(ii) If the condition ${\sum }_{m=1}^{\mathrm{\infty }}{\sum }_{n=1}^{\mathrm{\infty }}{m}^{k-\alpha -1}{n}^{k-\beta -1}|{a}_{mn}{|}^{k}=O\left(1\right)$ is satisfied then $C\left(\alpha ,\beta \right)\in B\left({\mathcal{A}}_{k}^{2}\right)$ for each $-1<\alpha <0$ and $-1<\beta <0$.

(iii) If the condition ${\sum }_{m=1}^{\mathrm{\infty }}{\sum }_{n=1}^{\mathrm{\infty }}{m}^{k-\alpha -1}{n}^{k-1}|{a}_{mn}{|}^{k}=O\left(1\right)$ is satisfied then $C\left(\alpha ,\beta \right)\in B\left({\mathcal{A}}_{k}^{2}\right)$ for each $-1<\alpha <0$ and $\beta \ge 0$.

(iv) If the condition ${\sum }_{m=1}^{\mathrm{\infty }}{\sum }_{n=1}^{\mathrm{\infty }}{m}^{k-1}{n}^{k-\beta -1}|{a}_{mn}{|}^{k}=O\left(1\right)$ is satisfied then $C\left(\alpha ,\beta \right)\in B\left({\mathcal{A}}_{k}^{2}\right)$ for each $\alpha \ge 0$ and $-1<\beta <0$.

Remark 1 Theorem 3 moderates Theorem 1 of [7]. Since Holder’s inequality is valid if each of the terms is nonnegative, it should be added the absolute values of the binomial coefficients in the proof of Theorem 1 of [7], when $-1<\alpha <0$ and/or $-1<\beta <0$. Therefore, if we replace the binomial coefficients with their absolute values, then the inequality (15) of [7] will be true. So, we should add the conditions, given above in (ii), (iii) and (iv) of Theorem 3 in the statement of Theorem 1 of [7], for the cases $-1<\alpha <0$ and/or $-1<\beta <0$.

Corollary 1 Let ${\theta }_{mn}^{\alpha }=\frac{1}{{E}_{m}^{\alpha }}{\sum }_{i=0}^{m}{E}_{m-i}^{\alpha -1}{s}_{in}=C\left(\alpha ,0\right)\left({s}_{mn}\right)$.

(i) It holds that $C\left(\alpha ,0\right)\in B\left({\mathcal{A}}_{k}^{2}\right)$ for each $\alpha \ge 0$.

(ii) If the condition ${\sum }_{m=1}^{\mathrm{\infty }}{\sum }_{n=1}^{\mathrm{\infty }}{m}^{k-\alpha -1}{n}^{k-1}|{a}_{mn}{|}^{k}=O\left(1\right)$ is satisfied then $C\left(\alpha ,0\right)\in B\left({\mathcal{A}}_{k}^{2}\right)$ for each $-1<\alpha <0$.

Corollary 2 Let ${\theta }_{mn}^{\beta }=\frac{1}{{E}_{n}^{\beta }}{\sum }_{j=0}^{n}{E}_{n-j}^{\beta -1}{s}_{mj}=\left(C,0,\beta \right)\left({s}_{mn}\right)$.

(i) It holds that $C\left(0,\beta \right)\in B\left({\mathcal{A}}_{k}^{2}\right)$ for each $\beta \ge 0$.

(ii) If the condition ${\sum }_{m=1}^{\mathrm{\infty }}{\sum }_{n=1}^{\mathrm{\infty }}{m}^{k-1}{n}^{k-\beta -1}|{a}_{mn}{|}^{k}=O\left(1\right)$ is satisfied then $C\left(0,\beta \right)\in B\left({\mathcal{A}}_{k}^{2}\right)$ for each $-1<\beta <0$.

Corollary 3 Let ${\sigma }_{mn}=\frac{1}{\left(m+1\right)\left(n+1\right)}{\sum }_{i=0}^{m}{\sum }_{j=0}^{n}{s}_{ij}=\left(C,1,1\right)\left({s}_{mn}\right)$. Then $C\left(1,1\right)\in B\left({\mathcal{A}}_{k}^{2}\right)$.

## References

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6. Şevli H, Savaş E: On absolute Cesàro summability. J. Inequal. Appl. 2009., 2009: Article ID 279421

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## Acknowledgements

Dedicated to Professor Hari M. Srivastava.

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Correspondence to Hamdullah Şevli.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

The authors contributed equally and significantly in writing this paper. Both the authors read and approved the final manuscript.

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Şevli, H., Savaş, E. Some further extensions of absolute Cesàro summability for double series. J Inequal Appl 2013, 144 (2013). https://doi.org/10.1186/1029-242X-2013-144

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• DOI: https://doi.org/10.1186/1029-242X-2013-144

### Keywords

• absolute summability
• bounded operator
• Cesáro matrix
• double series
• Hausdorff matrix