Open Access

Some Tauberian theorems for four-dimensional Euler and Borel summability

Advances in Difference Equations20152015:50

https://doi.org/10.1186/s13662-015-0381-2

Received: 29 October 2014

Accepted: 20 January 2015

Published: 24 February 2015

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 \(\ell_{2}\).

Keywords

Tauberian condition Euler-Knopp means Borel method four-dimensional summability method double sequences Pringsheim limit

MSC

40B05 40C05

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 .
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
  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 \(\ell_{2}\).

Declarations

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Authors’ Affiliations

(1)
Department of Mathematics, Afyon Kocatepe University
(2)
Department of Mathematics and Statistics, University of North Florida

References

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Copyright

© Nuray and Patterson; licensee Springer. 2015