Silverman–Toeplitz theorem

In mathematics, the Silverman–Toeplitz theorem, first proved by Otto Toeplitz, is a result in series summability theory characterizing matrix summability methods that are regular. A regular matrix summability method is a linear sequence transformation that preserves the limits of convergent sequences.^[1] The linear sequence transformation can be applied to the divergent sequences of partial sums of divergent series to give those series generalized sums.

An infinite matrix ${\displaystyle (a_{i,j})_{i,j\in \mathbb {N} }}$ with complex-valued entries defines a regular matrix summability method if and only if it satisfies all of the following properties:

${\displaystyle {\begin{aligned}&\lim _{i\to \infty }a_{i,j}=0\quad j\in \mathbb {N} &&{\text{(Every column sequence converges to 0.)}}\\[3pt]&\lim _{i\to \infty }\sum _{j=0}^{\infty }a_{i,j}=1&&{\text{(The row sums converge to 1.)}}\\[3pt]&\sup _{i}\sum _{j=0}^{\infty }\vert a_{i,j}\vert <\infty &&{\text{(The absolute row sums are bounded.)}}\end{aligned}}}$

An example is Cesàro summation, a matrix summability method with

${\displaystyle a_{mn}={\begin{cases}{\frac {1}{m}}&n\leq m\\0&n>m\end{cases}}={\begin{pmatrix}1&0&0&0&0&\cdots \\{\frac {1}{2}}&{\frac {1}{2}}&0&0&0&\cdots \\{\frac {1}{3}}&{\frac {1}{3}}&{\frac {1}{3}}&0&0&\cdots \\{\frac {1}{4}}&{\frac {1}{4}}&{\frac {1}{4}}&{\frac {1}{4}}&0&\cdots \\{\frac {1}{5}}&{\frac {1}{5}}&{\frac {1}{5}}&{\frac {1}{5}}&{\frac {1}{5}}&\cdots \\\vdots &\vdots &\vdots &\vdots &\vdots &\ddots \\\end{pmatrix}}.}$

Formal statement

Let the aforementioned inifinite matrix ${\displaystyle (a_{i,j})_{i,j\in \mathbb {N} }}$ of complex elements satisfy the following conditions:

1. ${\displaystyle \lim _{i\to \infty }a_{i,j}=0}$ for every fixed ${\displaystyle j\in \mathbb {N} }$.
2. ${\displaystyle \sup _{i\in \mathbb {N} }\sum _{j=1}^{i}\vert a_{i,j}\vert <\infty }$;

and ${\displaystyle z_{n}}$ be a sequence of complex numbers that converges to ${\displaystyle \lim _{n\to \infty }z_{n}=z_{\infty }}$. We denote ${\displaystyle S_{n}}$ as the weighted sum sequence: ${\displaystyle S_{n}=\sum _{m=1}^{n}{\left(a_{n,m}z_{n}\right)}}$.

Then the following results hold:

1. If ${\displaystyle \lim _{n\to \infty }z_{n}=z_{\infty }=0}$, then ${\displaystyle \lim _{n\to \infty }{S_{n}}=0}$.
2. If ${\displaystyle \lim _{n\to \infty }z_{n}=z_{\infty }\neq 0}$ and ${\displaystyle \lim _{i\to \infty }\sum _{j=1}^{i}a_{i,j}=1}$, then ${\displaystyle \lim _{n\to \infty }{S_{n}}=z_{\infty }}$.^[2]

Proof

Proving 1.

For the fixed ${\displaystyle j\in \mathbb {N} }$ the complex sequences ${\displaystyle z_{n}}$, ${\displaystyle S_{n}}$ and ${\displaystyle a_{i,j}}$ approach zero if and only if the real-values sequences ${\displaystyle \left|z_{n}\right|}$, ${\displaystyle \left|S_{n}\right|}$ and ${\displaystyle \left|a_{i,j}\right|}$ approach zero respectively. We also introduce ${\displaystyle M=\sup _{i\in \mathbb {N} }\sum _{j=1}^{i}\vert a_{i,j}\vert >0}$.

Since ${\displaystyle \left|z_{n}\right|\to 0}$, for prematurely chosen ${\displaystyle \varepsilon >0}$ there exists ${\displaystyle N_{\varepsilon }=N_{\varepsilon }\left(\varepsilon \right)}$, so for every ${\displaystyle n>N_{\varepsilon }\left(\varepsilon \right)}$ we have ${\displaystyle \left|z_{n}\right|<{\frac {\varepsilon }{2M}}}$. Next, for some ${\displaystyle N_{a}=N_{a}\left(\varepsilon \right)>N_{\varepsilon }\left(\varepsilon \right)}$ it's true, that ${\displaystyle \left|a_{n,m}\right|<{\frac {M}{N_{\varepsilon }}}}$ for every ${\displaystyle n>N_{a}\left(\varepsilon \right)}$ and ${\displaystyle 1\leqslant m\leqslant n}$. Therefore, for every ${\displaystyle n>N_{a}\left(\varepsilon \right)}$

${\displaystyle {\begin{aligned}&\left|S_{n}\right|=\left|\sum _{m=1}^{n}\left(a_{n,m}z_{n}\right)\right|\leqslant \sum _{m=1}^{n}\left(\left|a_{n,m}\right|\cdot \left|z_{n}\right|\right)=\sum _{m=1}^{N_{\varepsilon }}\left(\left|a_{n,m}\right|\cdot \left|z_{n}\right|\right)+\sum _{m=N_{\varepsilon }}^{n}\left(\left|a_{n,m}\right|\cdot \left|z_{n}\right|\right)<\\&<N_{\varepsilon }\cdot {\frac {M}{N_{\varepsilon }}}\cdot {\frac {\varepsilon }{2M}}+{\frac {\varepsilon }{2M}}\sum _{m=N_{\varepsilon }}^{n}\left|a_{n,m}\right|\leqslant {\frac {\varepsilon }{2}}+{\frac {\varepsilon }{2M}}\sum _{m=1}^{n}\left|a_{n,m}\right|\leqslant {\frac {\varepsilon }{2}}+{\frac {\varepsilon }{2M}}\cdot M=\varepsilon \end{aligned}}}$

which means, that both sequences ${\displaystyle \left|S_{n}\right|}$ and ${\displaystyle S_{n}}$ converge zero.^[3]

Proving 2.

${\displaystyle \lim _{n\to \infty }\left(z_{n}-z_{\infty }\right)=0}$. Applying the already proven statement yields ${\displaystyle \lim _{n\to \infty }\sum _{m=1}^{n}{\big (}a_{n,m}\left(z_{n}-z_{\infty }\right){\big )}=0}$. Finally,

${\displaystyle \lim _{n\to \infty }S_{n}=\lim _{n\to \infty }\sum _{m=1}^{n}{\big (}a_{n,m}z_{n}{\big )}=\lim _{n\to \infty }\sum _{m=1}^{n}{\big (}a_{n,m}\left(z_{n}-z_{\infty }\right){\big )}+z_{\infty }\lim _{n\to \infty }\sum _{m=1}^{n}{\big (}a_{n,m}{\big )}=0+z_{\infty }\cdot 1=z_{\infty }}$, which completes the proof.