Theory and Modern Applications

# Some Tauberian theorems for four-dimensional Euler and Borel summability

## Abstract

The four-dimensional summability methods of Euler and Borel are studied as mappings from absolutely convergent double sequences into themselves. Also the following Tauberian results are proved: if $$x=(x_{m,n})$$ is a double sequence that is mapped into $$\ell_{2}$$ by the four-dimensional Borel method and the double sequence x satisfies $$\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}|\Delta_{10} x_{m,n}|\sqrt {mn}<\infty$$ and $$\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}|\Delta_{01} x_{m,n}|\sqrt {mn}<\infty$$, then x itself is in $$\ell_{2}$$.

## Introduction

The best-known notion of convergence for double sequences is convergence in the sense of Pringsheim. Recall that a double sequence $$x=\{x_{k,l}\}$$ of complex (or real) numbers is called convergent to a scalar L in the sense of Pringsheim (denoted by $$P\mbox{-}\!\lim x=L$$) if for every $$\epsilon> 0$$ there exists an $$N \in\mathbb{N}$$ such that $$\vert x_{k,l} - L\vert < \epsilon$$ whenever $$k,l > N$$. Such an x is described more briefly as ’P-convergent’. It is easy to verify that $$x=\{x_{k,l}\}$$ converges in the sense of Pringsheim if and only if for every $$\epsilon> 0$$ there exists an integer $$N=N(\epsilon)$$ such that $$\vert x_{i,j}-x_{k,l} \vert < \epsilon$$ whenever min$$\{i,j,k,l\}\geq N$$. A double sequence $$x=\{x_{k,l}\}$$ is bounded if there exists a positive number M such that $$|x_{m,n}|\leq M$$ for all m and n, that is, if $$\sup_{m,n}|x_{m,n}|<\infty$$.

A double sequence $$x=\{x_{k,l}\}$$ is said to convergence regularly if it converges in the sense of Pringsheim and, in addition, the following finite limits exist:

\begin{aligned}& \lim_{m\rightarrow\infty}x_{m,n}=\ell_{n} \quad(n=1,2,\ldots),\\& \lim_{n\rightarrow\infty}x_{m,n}=\mathcal{L}_{m} \quad(m=1,2, \ldots). \end{aligned}

Let $$A=(a_{m,n,k,l})$$ denote a four-dimensional summability method that maps the complex double sequence x into the double sequence Ax where the mnth term of Ax is as follows:

$$(Ax)_{m,n}=\sum_{k=0}^{\infty}\sum _{l=0}^{\infty}a_{m,n,k,l}x_{k,l}.$$

In  Robison presented the following notion of regularity for four-dimensional matrix transformation and a Silverman-Toeplitz type characterization of such a notion.

### Definition 1.1

The four-dimensional matrix A is said to be RH-regular if it maps every bounded P-convergent sequence into a P-convergent sequence with the same P-limit.

The assumption of boundedness was added because a double sequence which is P-convergent is not necessarily bounded. Along these same lines, Robison and Hamilton presented a Silverman-Toeplitz type multidimensional characterization of regularity in  and .

### Theorem 1.1

(Hamilton , Robison )

The four-dimensional matrix A is RH-regular if and only if

$$RH_{1}$$::

$$P\mbox{-}\!\lim_{m,n}a_{m,n,k,l} = 0$$ for each k and l;

$$RH_{2}$$::

$$P\mbox{-}\!\lim_{m,n}\sum_{k,l=0,0}^{\infty,\infty }a_{m,n,k,l} = 1$$;

$$RH_{3}$$::

$$P\mbox{-}\!\lim_{m,n}\sum_{k=0}^{\infty} \vert a_{m,n,k,l}\vert = 0$$ for each l;

$$RH_{4}$$::

$$P\mbox{-}\!\lim_{m,n}\sum_{l=0}^{\infty} \vert a_{m,n,k,l}\vert = 0$$ for each k;

$$RH_{5}$$::

$$\sum_{k,l=0,0}^{\infty,\infty} \vert a_{m,n,k,l}\vert$$ is P-convergent;

$$RH_{6}$$::

there exist finite positive integers Δ and Γ such that $$\sum_{k,l>\Gamma} \vert a_{m,n,k,l}\vert <\Delta$$.

The set of all absolutely convergent double sequences will be denoted $$\ell_{2}$$, that is,

$$\ell_{2}=\Biggl\{ x=\{x_{k,l}\}: \sum _{m=0}^{\infty}\sum_{n=0}^{\infty}|x_{k,l}|< \infty\Biggr\} .$$

For single sequences, in  Fridy and Roberts proved the following Tauberian theorem.

### Theorem 1.2

If B is a Borel matrix and $$x=(x_{k})$$ is a sequence such that Bx in $$\ell=\{x=(x_{k}): \sum_{k=1}^{\infty }|x_{k}|<\infty\}$$ and

$$\sum_{r=1}^{\infty}|\Delta x_{r}| \sqrt{r}< \infty,$$

then x is in .

Our aim is to extend the results in  from single absolutely convergent sequences to double absolutely convergent sequences. In , Patterson proved that the matrix $$A=(a_{m,n,k,l})$$ determines an $$\ell_{2}\mbox{-}\ell_{2}$$ method if and only if

$$\sup_{k,l}\sum_{m=0}^{\infty} \sum_{n=0}^{\infty }|a_{m,n,k,l}|< \infty.$$
(1.1)

## Euler-Knopp and Borel $$\ell_{2}\mbox{-}\ell_{2}$$ methods

The four-dimensional Euler-Knopp method, for any complex numbers $$r_{1}$$ and $$r_{2}$$, is defined by

\begin{aligned} &E_{r_{1},r_{2}}[m,n,k,l]\\ &\quad=\left \{ \begin{array}{@{}l@{\quad}l} \binom{m}{k}\binom{n}{l} r_{1}^{k+1}(1-r_{1})^{m-k}r_{2}^{l+1}(1-r_{2})^{n-l}, & \mbox{if }k\leq m, l\leq n,\\ 0, & \mbox{otherwise}. \end{array} \right . \end{aligned}

An application of the Maclaurin series expansion of $$(1-z_{1})^{k+1}(1-z_{2})^{l+1}$$ shows that each column sum of $$E_{r_{1},r_{2}}$$ converges absolutely to $$\frac{1}{r_{1}r_{2}}$$ provided that $$0< r_{1}\leq1$$ and $$0< r_{2}\leq1$$. If $$0< r_{1}< 1$$ and $$0< r_{2}< 1$$, then $$P\mbox{-}\!\lim_{m,n} E_{r_{1},r_{2}}[m,n,m,n]=0$$, so $$E_{r_{1},r_{2}}^{-1}$$ is not an $$\ell_{2}\mbox{-}\ell_{2}$$ matrix. We summarize this as follows.

### Theorem 2.1

The four-dimensional Euler-Knopp method $$E_{r_{1},r_{2}}$$ is a sum-preserving $$\ell_{2}\mbox{-}\ell_{2}$$ matrix for which $$\ell_{2_{E_{r_{1},r_{2}}}}\neq\ell_{2}$$ if and only if $$0< r_{1}< 1$$ and $$0< r_{2}< 1$$, where $$\ell_{2_{E_{r_{1},r_{2}}}}$$ is the summability field of $$E_{r_{1},r_{2}}$$.

The four-dimensional Borel method B is given by the matrix

$$b_{m,n,k,l}=\frac{e^{-(m+n)}m^{k}n^{l}}{k!l!},\quad m,n,k,l=0,1,2,3,\ldots.$$

By a direct application of Theorem 3.1 in , one can show that B is an $$\ell_{2}\mbox{-}\ell_{2}$$ matrix.

### Theorem 2.2

If $$r_{1}>0$$ and $$r_{2}>0$$ and $$x=(x_{k,l})$$ is a double sequence such that $$E_{r_{1},r_{2}}x$$ is in  $$\ell_{2}$$, then Bx is in $$\ell_{2}$$.

### Proof

We use the familiar technique of showing that $$BE_{r_{1},r_{2}}$$ is an $$\ell_{2}\mbox{-}\ell_{2}$$ matrix. Since $$Bx=BE_{r_{1},r_{2}}^{-1}E_{r_{1},r_{2}}x$$, this will ensure that Bx is in $$\ell_{2}$$ whenever $$E_{r_{1},r_{2}}x$$ in $$\ell_{2}$$. Since $$E_{r_{1},r_{2}}^{-1}=E_{\frac{1}{r_{1}},\frac{1}{r_{2}}}$$ we replace $$s_{1}=\frac{1}{r_{1}}$$ and $$s_{2}=\frac{1}{r_{2}}$$ and show that $$BE_{s_{1},s_{2}}$$ is an $$\ell_{2}\mbox{-}\ell_{2}$$ matrix for all positive $$s_{1}$$ and $$s_{2}$$. The $$mnkl$$th term of $$BE_{s_{1},s_{2}}$$ is given by

\begin{aligned} &BE_{s_{1},s_{2}}[m,n,k,l]\\ &\quad=\sum_{i=k}^{\infty}\sum _{j=l}^{\infty }\frac{e^{-(m+n)}m^{i}n^{j}}{i!j!} \binom{i}{k}\binom{j}{l} (1-s_{1})^{i-k}s_{1}^{k}(1-s_{2})^{j-l}s_{2}^{l} \\ &\quad= \frac{e^{-(m+n)}m^{k}n^{l}s_{1}^{k}s_{2}^{l}}{k!l!}\sum_{i=k}^{\infty}\sum _{j=l}^{\infty}\frac {m^{i-k}n^{j-k}}{(i-k)!(j-l)!}(1-s_{1})^{i-k} (1-s_{2})^{j-l}\\ &\quad=\frac{(ms_{1})^{k}(ns_{2})^{l}e^{-(ms_{1}+ns_{2})}}{k!l!}. \end{aligned}

Summing the $$(k,l)$$th column of $$BE_{s_{1},s_{2}}$$, we get

\begin{aligned} \sum_{m=0}^{\infty}\sum _{n=0}^{\infty}\bigl|BE_{s_{1},s_{2}}[m,n,k,l]\bigr| &= \frac{1}{k!l!}\sum_{m=0}^{\infty}\sum _{n=0}^{\infty}(ms_{1})^{k}(n s_{2})^{l}e^{-(ms_{1}+ns_{2})} \\ &= O\biggl(\frac{1}{k!l!}\int_{0}^{\infty}\int _{0}^{\infty }(t_{1}s_{1})^{k}(t_{2}s_{2})^{l}e^{-(t_{1}s_{1}+t_{2}s_{2})}\,dt_{1}\,dt_{2} \biggr) \\ &= O\biggl(\frac{1}{s_{1}s_{2}}\biggr). \end{aligned}

Hence,

$$\sup_{k,l}\sum_{m=0}^{\infty} \sum_{n=0}^{\infty }\bigl|BE_{s_{1},s_{2}}[m,n,k,l]\bigr|< \infty,$$

so $$BE_{s_{1},s_{2}}$$ is an $$\ell_{2}\mbox{-}\ell_{2}$$ matrix. □

Theorem 2.2 and the $$\ell_{2}\mbox{-}\ell_{2}$$ property of $$E_{r_{1},r_{2}}$$ lead to the following result.

### Theorem 2.3

The four-dimensional Borel matrix determines an $$\ell_{2}\mbox{-}\ell_{2}$$ method.

In addition to the inclusion relation given in Theorem 2.2, we can also show that the $$\ell_{2}\mbox{-}\ell_{2}$$ method B is strictly stronger than all $$E_{r_{1},r_{2}}$$ methods by the following example.

### Example 2.1

Suppose $$r_{1}>0$$ and $$r_{2}>0$$ and $$x_{k,l}=(-s_{1})^{k}(-s_{2})^{l}$$ where $$s_{1}\geq-1+\frac{2}{r_{1}}$$ and $$s_{2}\geq-1+\frac {2}{r_{2}}$$; then Bx is in $$\ell_{2}$$ but $$E_{r_{1},r_{2}}$$ is not in $$\ell_{2}$$. Let us consider the following methods:

\begin{aligned} (Bx)_{m,n} =&\sum_{k=0}^{\infty}\sum _{l=0}^{\infty }e^{-(m+n)}\frac{m^{k}}{k!} \frac{n^{l}}{l!}(-s_{1})^{k}(-s_{2})^{l} \\ =&e^{-(m+n)}e^{-(s_{1}m+s_{2}n)}=e^{-[(s_{1}+1)m+(s_{2}+1)n]} \end{aligned}

and

\begin{aligned} (E_{r_{1},r_{2}}x)_{m,n} =&\sum_{k=0}^{m} \sum_{l=0}^{n}\binom{m}{k}\binom{n}{l} (1-r_{1})^{m-k}(-r_{1}s_{1})^{k}(1-r_{2})^{n-l}(-r_{2}s_{2})^{l}\\ =&(1-r_{1}-r_{1}s_{1})^{m} (1-r_{2}-r_{2}s_{2})^{n}. \end{aligned}

By solving $$-1< 1-r_{1}-r_{1}s_{1}<1$$ and $$-1< 1-r_{2}-r_{2}s_{2}<1$$, we see that $$E_{r_{1},r_{2}}x$$ is in $$\ell_{2}$$ if and only if $$-1< s_{1}<-1+\frac{2}{r_{1}}$$ and $$-1< s_{2}<-1+\frac{2}{r_{2}}$$.

## Tauberian theorems

To prove Theorem 3.1 we need the following lemma.

### Lemma 3.1

If

$$b_{m,n,k,l}=\frac{e^{-(m+n)}m^{k}n^{l}}{k!l!}$$

and r and s are positive integers, then

1. (i)
$$\sum_{m=r+1}^{\infty}\sum _{n=s+1}^{\infty}\sum_{k=0}^{r} \sum_{l=0}^{s}b_{m,n,k,l}=O(\sqrt{rs})$$

and

1. (ii)
$$\sum_{m=0}^{r}\sum _{n=0}^{s}\sum_{k=r+1}^{\infty} \sum_{l=s+1}^{\infty}b_{m,n,k,l}=O(\sqrt{rs}).$$

### Proof

Let $$p=[\sqrt{r}]$$ and $$q=[\sqrt{s}]$$, and let us write the sum in (i) as

$$\sum_{m=r+1}^{\infty}\sum _{n=s+1}^{\infty}\sum_{k=0}^{r-p} \sum_{l=0}^{s-q}b_{m,n,k,l}+ \sum _{m=r+1}^{\infty}\sum_{n=s+1}^{\infty} \sum_{k=r-p+1}^{r}\sum _{l=s-q+1}^{s}b_{m,n,k,l}=\phi_{r,s}+ \varphi_{r,s}.$$

If $$s_{1}< m$$ and $$s_{2}< n$$, then

\begin{aligned} \sum_{k=0}^{s_{1}}\sum _{l=0}^{s_{2}}\frac{m^{m}n^{l}}{k!l!} =&\frac {m^{s_{1}}n^{s_{2}}}{s_{1}!s_{2}!} \biggl(1+\frac{s_{1}}{m}+\frac {s_{1}}{m}\frac{s_{1}-1}{m}+\cdots\biggr) \biggl(1+\frac{s_{2}}{n}+\frac{s_{2}}{n}\frac{s_{2}-1}{n}+\cdots\biggr) \\ \leq& \frac{m^{s_{1}}n^{s_{2}}}{s_{1}!s_{2}!}\biggl(1+\frac{s_{1}}{m}+\biggl(\frac {s_{1}}{m} \biggr)^{2}+\cdots\biggr) \biggl(1+\frac{s_{2}}{m}+\biggl( \frac{s_{2}}{m}\biggr)^{2}+\cdots\biggr) \\ =&\frac{m^{s_{1}}n^{s_{2}}}{s_{1}!s_{2}!}\frac{m}{m-s_{1}}\frac{n}{n-s_{2}}. \end{aligned}

In $$\phi_{r,s}$$, let $$s_{1}=r-p$$, $$s_{2}=s-q$$, and

$$\max_{m\geq r+1;n\geq s+1}\frac{mn}{(m-r+p)(n-s+q)}=\frac {(r+1)(s+1)}{(p+1)(q+1)}\leq( \sqrt{r}+1) (\sqrt{s}+1),$$

thus

\begin{aligned} \phi_{r,s}< (\sqrt{r}+1) (\sqrt{s}+1)\frac {1}{(r-p)!(s-q)!}\sum _{m=r+1}^{\infty}\sum_{n=s+1}^{\infty }e^{-(m+n)}m^{r-p}n^{s-q} \leq (\sqrt{r}+1) (\sqrt{s}+1). \end{aligned}

In $$\varphi_{r,s}$$, observe that

$$\sum_{k=r-p+1}^{r}\sum _{l=s-q+1}^{s}b_{m,n,k,l}\leq\sqrt{rs}\max _{k\geq r;l\geq s}b_{m,n,k,l}=\sqrt{rs}e^{-(m+n)} \frac{m^{r}n^{s}}{r!s!},$$

thus

$$\varphi_{r,s}\leq\sqrt{rs}\frac{1}{r!s!}\sum _{m=r+1}^{\infty}\sum_{n=s+1}^{\infty}e^{-(m+n)}m^{r}n^{s} \leq\sqrt{rs}.$$

Hence, (i) is proved. Next write the sum in (ii) as

$$\sum_{m=0}^{r}\sum _{n=0}^{s}\sum_{k=r+1}^{r+p-1} \sum_{l=s+1}^{s+q-1}b_{m,n,k,l}+ \sum _{m=0}^{r}\sum_{n=0}^{s} \sum_{k=r+p}^{\infty}\sum _{l=s+q}^{\infty}b_{m,n,k,l}=\lambda_{r,s}+ \mu_{r,s}.$$

Assume that $$\lambda_{r,s}=0$$ if $$p=1$$, $$q=1$$. Then

\begin{aligned} \lambda_{r,s} \leq&(p-1) (q-1)\sum_{m=0}^{r} \sum_{n=0}^{s}e^{-(m+n)}\max _{k>r;l>s}\frac{m^{k}n^{l}}{k!l!} \\ \leq&(\sqrt{r}-1) (\sqrt{s}-1)\frac{1}{(r+1)!(s+1)!}\sum _{m=0}^{r}\sum_{n=0}^{s}e^{-(m+n)}m^{r+1}n^{s+1} \\ \leq&(\sqrt{r}-1) (\sqrt{s}-1). \end{aligned}

If $$s_{1}\geq m$$ and $$s_{2}\geq n$$, then

\begin{aligned} &\sum_{k=s_{1}}^{\infty}\sum _{l=s_{2}}^{\infty}\frac {m^{k}n^{l}}{k!l!}\\ &\quad=\frac{m^{s_{1}}n^{s_{2}}}{s_{1}!s_{2}!}\biggl(1+\frac {m}{s_{1}+1}+ \frac{m}{s_{1}+1} \frac{m}{s_{1}+2}+\cdots\biggr) \biggl(1+\frac{n}{s_{2}+1}+\frac {n}{s_{2}+1} \frac{n}{s_{2}+2}+\cdots\biggr)\\ &\quad\leq \frac{m^{s_{1}}n^{s_{2}}}{s_{1}!s_{2}!}\biggl(1+\frac {m}{s_{1}+1}+\biggl(\frac{m}{s_{1}+1} \biggr)^{2}+\cdots\biggr) \biggl(1+\frac{n}{s_{2}+1}+\biggl( \frac {n}{s_{2}+1}\biggr)^{2}+\cdots\biggr)\\ &\quad=\frac{m^{s_{1}}n^{s_{2}}}{s_{1}!s_{2}!}\frac {(s_{1}+1)(s_{2}+1)}{(s_{1}+1-m)(s_{2}+1-n)}. \end{aligned}

Let $$s_{1}=r+p$$ and $$s_{2}=s+q$$, we have

\begin{aligned} \mu_{r,s} \leq&\frac{1}{(r+p)!(s+q)!}\sum_{m=0}^{r} \sum_{n=0}^{s} e^{-(m+n)}m^{r+p}n^{s+q} \biggl(\frac{r+p+1}{r+p+1-m}\biggr) \biggl(\frac{s+q+1}{s+q+1-n}\biggr) \\ \leq&\frac{r+p+1}{p+1}\frac{s+q+1}{q+1}\frac{1}{(r+p)!(s+q)!}\sum _{m=0}^{r}\sum_{n=0}^{s} e^{-(m+n)}m^{r+p}n^{s+q} \\ \leq&(\sqrt{r}+1) (\sqrt{s}+1). \end{aligned}

Thus the lemma is proved. □

We are now ready to prove the following result.

### Theorem 3.1

If x is a double sequence such that Bx is in $$\ell_{2}$$,

$$\sum_{r=0}^{\infty}\sum _{s=0}^{\infty}|\Delta_{10}x_{s,r}|\sqrt {rs}< \infty$$
(3.1)

and

$$\sum_{r=0}^{\infty}\sum _{s=0}^{\infty}|\Delta_{01}x_{s,r}|\sqrt {rs}< \infty,$$
(3.2)

then x in $$\ell_{2}$$ where $$\Delta_{10}x_{r,s}=x_{r,s}-x_{r+1,s}$$ and $$\Delta_{01}x_{r,s}=x_{r,s}-x_{r,s+1}$$.

### Proof

It is suffices to show that $$Bx-x$$ is in $$\ell_{2}$$; that is,

$$\sum_{m=0}^{\infty}\sum _{n=0}^{\infty}\Biggl\vert \sum _{k=0}^{\infty}\sum_{l=0}^{\infty}b_{m,n,k,l}x_{k,l}-x_{m,n} \Biggr\vert < \infty.$$

Since

$$\sum_{k=0}^{\infty}\sum _{l=0}^{\infty}b_{m,n,k,l}=1$$

for each m, n, the above sum can be written as

$$\sum_{m=0}^{\infty}\sum _{n=0}^{\infty}\Biggl\vert \sum _{k=0}^{\infty}\sum_{l=0}^{\infty}b_{m,n,k,l}(x_{k,l}-x_{m,n}) \Biggr\vert$$

and we need only show the following:

$$S=\sum_{m=0}^{\infty}\sum _{n=0}^{\infty}\sum_{k=0}^{\infty} \sum_{l=0}^{\infty}b_{m,n,k,l}|x_{k,l}-x_{m,n}|< \infty.$$

Let $$S=S_{1}+S_{2}$$, where

$$S_{1}=\sum_{m=0}^{\infty}\sum _{n=0}^{\infty}\sum_{k=0}^{m} \sum_{l=0}^{n}b_{m,n,k,l}|x_{k,l}-x_{m,n}|$$

and

$$S_{2}=\sum_{m=0}^{\infty}\sum _{n=0}^{\infty}\sum_{k=m+1}^{\infty} \sum_{l=n+1}^{\infty}b_{m,n,k,l}|x_{k,l}-x_{m,n}|.$$

Since

\begin{aligned}& |x_{k,l}-x_{m,n}|=|x_{m,n}-x_{k,l}|=\Biggl\vert \sum_{s=m}^{k-1}\Delta _{10}x_{s,n}+\sum_{r=n}^{l-1} \Delta_{01}x_{k,r}\Biggr\vert ,\\& \begin{aligned}[b] S_{1}\leq{}&\sum_{m=0}^{\infty}\sum _{n=0}^{\infty}\sum_{k=0}^{m-1} \sum_{l=0}^{n-1}b_{m,n,k,l} \Biggl(\sum _{s=m}^{k-1}|\Delta_{10}x_{s,n}|+ \sum_{r=n}^{l-1}|\Delta _{01}x_{k,r}| \Biggr)\\ \leq{}&\sum_{r=0}^{\infty}\sum _{s=0}^{\infty}|\Delta_{10}x_{r,s}|\sum _{m=r+1}^{\infty}\sum_{n=s+1}^{\infty} \sum_{k=0}^{r}\sum _{l=0}^{s}b_{m,n,k,l}\\ &{}+\sum_{r=0}^{\infty}\sum _{s=0}^{\infty}|\Delta_{01}x_{r,s}|\sum _{m=r+1}^{\infty}\sum_{n=s+1}^{\infty} \sum_{k=0}^{r}\sum _{l=0}^{s}b_{m,n,k,l}\\ ={}& \Biggl(\sum_{r=0}^{\infty}\sum _{s=0}^{\infty}|\Delta_{10}x_{r,s}|+\sum _{r=0}^{\infty}\sum_{s=0}^{\infty}| \Delta_{01}x_{r,s}| \Biggr)\zeta _{r,s}, \quad\mbox{say}. \end{aligned} \end{aligned}

Also,

\begin{aligned} S_{2} \leq&\sum_{m=0}^{\infty}\sum _{n=0}^{\infty}\sum_{k=m+1}^{\infty} \sum_{l=n+1}^{\infty}b_{m,n,k,l} \Biggl(\sum _{s=m}^{k-1}| \Delta_{10}x_{s,n}|+ \sum_{r=n}^{l-1}|\Delta_{01}x_{k,r}| \Biggr) \\ \leq&\sum_{r=0}^{\infty}\sum _{s=0}^{\infty}|\Delta_{10}x_{r,s}|\sum _{m=0}^{r}\sum_{n=0}^{s} \sum_{k=r+1}^{\infty}\sum _{l=s+1}^{\infty }b_{m,n,k,l} \\ &{}+\sum_{r=0}^{\infty}\sum _{s=0}^{\infty}|\Delta_{01}x_{r,s}|\sum _{m=0}^{r}\sum_{n=0}^{s} \sum_{k=r+1}^{\infty}\sum _{l=s+1}^{\infty }b_{m,n,k,l} \\ =& \Biggl(\sum_{r=0}^{\infty}\sum _{s=0}^{\infty}|\Delta_{10}x_{r,s}|+\sum _{r=0}^{\infty}\sum_{s=0}^{\infty}| \Delta_{01}x_{r,s}| \Biggr)\varsigma _{r,s}, \quad\mbox{say}. \end{aligned}

By Lemma 3.1, $$\zeta_{r,s}=O(\sqrt{rs})$$ and $$\varsigma_{r,s}=O(\sqrt{rs})$$, we have

$$S_{1}+S_{2}\leq\lambda \Biggl(\sum _{r=0}^{\infty}\sum_{s=0}^{\infty }| \Delta_{10}x_{s,r}|\sqrt{rs}+ \sum _{r=0}^{\infty}\sum_{s=0}^{\infty }| \Delta_{01}x_{s,r}|\sqrt{rs} \Biggr)< \infty,$$

which proves the theorem. □

Combining Theorem 3.1 with Theorem 2.2, we are lead to the following $$\ell_{2}\mbox{-}\ell_{2}$$ Tauberian theorem for the four-dimensional Euler-Knopp means.

### Theorem 3.2

If $$r_{1}>0$$, $$r_{2}>0$$, and x is a double sequence satisfying (3.1) such that $$E_{r_{1},r_{2}}$$ is in $$\ell_{2}$$, then x is in $$\ell_{2}$$.

### Example 3.1

The following double sequence is not mapped into $$\ell_{2}$$ by B or by $$E_{r_{1},r_{2}}$$, with $$r_{1}>0$$, $$r_{2}>0$$. Define $$x=\{x_{k,l}\}$$ by

$$x_{0,0}=\frac{\pi^{2}}{3}\quad\mbox{and}\quad \Delta_{01}x_{k,j}= \frac {1}{(j+1)^{2}},\qquad \Delta_{10}x_{i,0}=\frac{1}{(i+1)^{2}},\quad i,j=1,2,3,\ldots.$$

Then x satisfies (3.1) and (3.2), but x is not in $$\ell_{2}$$ because if $$k\geq1$$ and $$l\geq1$$,

\begin{aligned} x_{k,l} =& x_{0,0}-\sum_{i=0}^{k-1} \Delta _{10}x_{i,0}-\sum_{j=0}^{l-1} \Delta_{01}x_{k,j} \\ =& \frac{\pi^{2}}{3}-\sum_{i=0}^{k-1} \frac{1}{(i+1)^{2}}-\sum_{j=0}^{l-1} \frac{1}{(j+1)^{2}} \\ =& \frac{\pi^{2}}{6}-\sum_{i=0}^{k-1} \frac{1}{(i+1)^{2}}+\frac{\pi ^{2}}{6}-\sum_{j=0}^{l-1} \frac{1}{(j+1)^{2}} \\ =& \sum_{i=k}^{\infty}\frac{1}{(i+1)^{2}}+\sum _{j=l}^{\infty}\frac {1}{(j+1)^{2}}\sim \frac{1}{k}-\frac{1}{l}. \end{aligned}

Hence, by Theorem 3.1, Bx is not in $$\ell_{2}$$.

## References

1. Robison, GM: Divergent double sequences and series. Trans. Am. Math. Soc. 28, 50-73 (1926)

2. Hamilton, HJ: Transformations of multiple sequences. Duke Math. J. 2, 29-60 (1936)

3. Fridy, JA, Roberts, KL: Some Tauberian theorems for Euler and Borel summability. Int. J. Math. Math. Sci. 3(4), 731-738 (1980)

4. Patterson, RF: Four dimensional matrix characterization of absolute summability. Soochow J. Math. 30(1), 21-26 (2004)

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Correspondence to Fatih Nuray.

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The authors declare that they have no competing interests.

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All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

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