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Some Tauberian theorems for four-dimensional Euler and Borel summability

Advances in Difference Equations20152015:50

  • Received: 29 October 2014
  • Accepted: 20 January 2015
  • Published:


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 \(\ell_{2}\).


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


  • 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

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


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


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


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


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


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

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 \(\ell_{2}\).


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

$$b_{m,n,k,l}=\frac{e^{-(m+n)}m^{k}n^{l}}{k!l!} $$
and r and s are positive integers, then
  1. (i)
    $$\sum_{m=r+1}^{\infty}\sum _{n=s+1}^{\infty}\sum_{k=0}^{r} \sum_{l=0}^{s}b_{m,n,k,l}=O(\sqrt{rs}) $$
  1. (ii)
    $$\sum_{m=0}^{r}\sum _{n=0}^{s}\sum_{k=r+1}^{\infty} \sum_{l=s+1}^{\infty}b_{m,n,k,l}=O(\sqrt{rs}). $$


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
$$\max_{m\geq r+1;n\geq s+1}\frac{mn}{(m-r+p)(n-s+q)}=\frac {(r+1)(s+1)}{(p+1)(q+1)}\leq( \sqrt{r}+1) (\sqrt{s}+1), $$
$$\begin{aligned} \phi_{r,s}< (\sqrt{r}+1) (\sqrt{s}+1)\frac {1}{(r-p)!(s-q)!}\sum _{m=r+1}^{\infty}\sum_{n=s+1}^{\infty }e^{-(m+n)}m^{r-p}n^{s-q} \leq (\sqrt{r}+1) (\sqrt{s}+1). \end{aligned}$$
In \(\varphi_{r,s}\), observe that
$$\sum_{k=r-p+1}^{r}\sum _{l=s-q+1}^{s}b_{m,n,k,l}\leq\sqrt{rs}\max _{k\geq r;l\geq s}b_{m,n,k,l}=\sqrt{rs}e^{-(m+n)} \frac{m^{r}n^{s}}{r!s!}, $$
$$\varphi_{r,s}\leq\sqrt{rs}\frac{1}{r!s!}\sum _{m=r+1}^{\infty}\sum_{n=s+1}^{\infty}e^{-(m+n)}m^{r}n^{s} \leq\sqrt{rs}. $$
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 \(\ell_{2}\),
$$ \sum_{r=0}^{\infty}\sum _{s=0}^{\infty}|\Delta_{10}x_{s,r}|\sqrt {rs}< \infty $$
$$ \sum_{r=0}^{\infty}\sum _{s=0}^{\infty}|\Delta_{01}x_{s,r}|\sqrt {rs}< \infty, $$
then x in \(\ell_{2}\) where \(\Delta_{10}x_{r,s}=x_{r,s}-x_{r+1,s}\) and \(\Delta_{01}x_{r,s}=x_{r,s}-x_{r,s+1}\).


It is suffices to show that \(Bx-x\) is in \(\ell_{2}\); that is,
$$\sum_{m=0}^{\infty}\sum _{n=0}^{\infty}\Biggl\vert \sum _{k=0}^{\infty}\sum_{l=0}^{\infty}b_{m,n,k,l}x_{k,l}-x_{m,n} \Biggr\vert < \infty. $$
$$\sum_{k=0}^{\infty}\sum _{l=0}^{\infty}b_{m,n,k,l}=1 $$
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
$$S_{1}=\sum_{m=0}^{\infty}\sum _{n=0}^{\infty}\sum_{k=0}^{m} \sum_{l=0}^{n}b_{m,n,k,l}|x_{k,l}-x_{m,n}| $$
$$S_{2}=\sum_{m=0}^{\infty}\sum _{n=0}^{\infty}\sum_{k=m+1}^{\infty} \sum_{l=n+1}^{\infty}b_{m,n,k,l}|x_{k,l}-x_{m,n}|. $$
$$\begin{aligned}& |x_{k,l}-x_{m,n}|=|x_{m,n}-x_{k,l}|=\Biggl\vert \sum_{s=m}^{k-1}\Delta _{10}x_{s,n}+\sum_{r=n}^{l-1} \Delta_{01}x_{k,r}\Biggr\vert ,\\& \begin{aligned}[b] S_{1}\leq{}&\sum_{m=0}^{\infty}\sum _{n=0}^{\infty}\sum_{k=0}^{m-1} \sum_{l=0}^{n-1}b_{m,n,k,l} \Biggl(\sum _{s=m}^{k-1}|\Delta_{10}x_{s,n}|+ \sum_{r=n}^{l-1}|\Delta _{01}x_{k,r}| \Biggr)\\ \leq{}&\sum_{r=0}^{\infty}\sum _{s=0}^{\infty}|\Delta_{10}x_{r,s}|\sum _{m=r+1}^{\infty}\sum_{n=s+1}^{\infty} \sum_{k=0}^{r}\sum _{l=0}^{s}b_{m,n,k,l}\\ &{}+\sum_{r=0}^{\infty}\sum _{s=0}^{\infty}|\Delta_{01}x_{r,s}|\sum _{m=r+1}^{\infty}\sum_{n=s+1}^{\infty} \sum_{k=0}^{r}\sum _{l=0}^{s}b_{m,n,k,l}\\ ={}& \Biggl(\sum_{r=0}^{\infty}\sum _{s=0}^{\infty}|\Delta_{10}x_{r,s}|+\sum _{r=0}^{\infty}\sum_{s=0}^{\infty}| \Delta_{01}x_{r,s}| \Biggr)\zeta _{r,s}, \quad\mbox{say}. \end{aligned} \end{aligned}$$
$$\begin{aligned} S_{2} \leq&\sum_{m=0}^{\infty}\sum _{n=0}^{\infty}\sum_{k=m+1}^{\infty} \sum_{l=n+1}^{\infty}b_{m,n,k,l} \Biggl(\sum _{s=m}^{k-1}| \Delta_{10}x_{s,n}|+ \sum_{r=n}^{l-1}|\Delta_{01}x_{k,r}| \Biggr) \\ \leq&\sum_{r=0}^{\infty}\sum _{s=0}^{\infty}|\Delta_{10}x_{r,s}|\sum _{m=0}^{r}\sum_{n=0}^{s} \sum_{k=r+1}^{\infty}\sum _{l=s+1}^{\infty }b_{m,n,k,l} \\ &{}+\sum_{r=0}^{\infty}\sum _{s=0}^{\infty}|\Delta_{01}x_{r,s}|\sum _{m=0}^{r}\sum_{n=0}^{s} \sum_{k=r+1}^{\infty}\sum _{l=s+1}^{\infty }b_{m,n,k,l} \\ =& \Biggl(\sum_{r=0}^{\infty}\sum _{s=0}^{\infty}|\Delta_{10}x_{r,s}|+\sum _{r=0}^{\infty}\sum_{s=0}^{\infty}| \Delta_{01}x_{r,s}| \Biggr)\varsigma _{r,s}, \quad\mbox{say}. \end{aligned}$$
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 \(\ell_{2}\).


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Authors’ Affiliations

Department of Mathematics, Afyon Kocatepe University, Afyonkarahisar, Turkey
Department of Mathematics and Statistics, University of North Florida, Jacksonville, FL, USA


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© Nuray and Patterson; licensee Springer. 2015