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 $\ell_{2}$.

1 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 [1] 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 [2] and [1].

Theorem 1.1

(Hamilton [2], Robison [1])

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 [3] 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 ℓ.

Our aim is to extend the results in [3] from single absolutely convergent sequences to double absolutely convergent sequences. In [4], 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)

2 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 [4], 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 $-1< s_{2}<-1+\frac{2}{r_{2}}$.

3 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

• (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

• (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 $\ell_{2}$.