proof of Silverman-Toeplitz theorem

First, we shall show that the series $\\sum_{n=0}^{\\infty}a_{mn}z_{n}$ converges. Since the sequence $\\{z_{n}\\}$ converges, it must be bounded in absolute value — there must exist a constant $K>0$ such that $|z_{n}|\\leq K$ for all $n$ . Hence, $|a_{mn}z_{n}|\\leq K|a_{mn}|.$ Summing this gives

\\sum_{n=0}^{\\infty}|a_{mn}z_{n}|\\leq K\\sum_{n=0}^{\\infty}|a_{mn}|\\leq KB.

Hence, the series $\\sum_{n=0}^{\\infty}a_{mn}z_{n}$ is absolutely convergent which, in turn, implies that it converges.

Let $z$ denote the limit of the sequence $\\{z_{n}\\}$ as $n\\to\\infty$ . Then $|z|\\leq K$ .We need to show that, for every $\\epsilon>0$ , there exists an integer $M$ such that

\\left|\\sum_{n=0}^{\\infty}a_{mn}z_{n}-z\\right|<\\epsilon

whenever $m>M$ .

Since the sequence $\\{z_{n}\\}$ converges, there must exist an integer $n_{1}$ such that $|z_{n}-z|<\\frac{\\epsilon}{4B}$ whenever $n>n_{1}$ .

By condition 3, there must exist constants $m_{o},m_{1},\\ldots,m_{n_{1}}$ such that

|a_{mn}|<{\\epsilon\\over 4(n_{1}+1)(K+1)}\\quad\\textrm{for}\\quad 0\\leq n\\leq n_{%1}\\quad\\textrm{and}\\quad m>m_{n}.

(1)

Choose $m^{\\prime}=\\max\\{m_{0},m_{1},\\ldots,m_{n_{1}}\\}.$ Then

\\left|\\sum_{n=0}^{n_{1}}a_{mn}z_{n}\\right|\\leq\\sum_{n=0}^{n_{1}}|a_{mn}z_{n}|<%\\sum_{n=0}^{n_{1}}{|z_{n}|\\epsilon\\over 4(n_{1}+1)(K+1)}<{\\epsilon\\over 4}

(2)

when $m>m^{\\prime}.$

By condition 2, there exists a constant $m^{\\prime\\prime}$ such that

\\left|\\sum_{n=0}^{\\infty}a_{mn}-1\\right|<{\\epsilon\\over 6(|z|+1)}

whenever $m>m^{\\prime\\prime}$ . By (1),

\\left|\\sum_{n=0}^{n_{1}}a_{mn}\\right|<{\\epsilon\\over 4(K+1)}\\leq{\\epsilon\\over4%(|z|+1)}

when $m>m^{\\prime}$ . Hence, if $m>m^{\\prime}$ and $m>m^{\\prime\\prime}$ , we have

\\left|\\sum_{n=n_{1}+1}^{\\infty}a_{mn}-1\\right|<{\\epsilon\\over 2(|z|+1)}.

(3)

and

\\left|\\sum_{n=n_{1}+1}^{\\infty}a_{mn}(z_{n}-z)\\right|\\leq\\sum_{n=n_{1}+1}^{%\\infty}|a_{mn}|~{}|z_{n}-z|<{\\epsilon\\over 4B}\\sum_{n=0}^{\\infty}|a_{mn}|\\leq{%\\epsilon\\over 4}.

(4)

By the triangle inequality and (2), (3), (4) it follows that

\\left|\\sum_{n=0}^{\\infty}a_{mn}z_{n}-z\\right|\\leq\\left|\\sum_{n=0}^{n_{1}}a_{mn%}z_{n}\\right|+\\left|\\sum_{n=n_{1}+1}^{\\infty}a_{mn}(z_{n}-z)\\right|+|z|\\left|%\\sum_{n=n_{1}+1}^{\\infty}a_{mn}-1\\right|

<\\frac{\\epsilon}{4}+\\frac{\\epsilon}{4}+\\frac{\\epsilon}{2}=\\epsilon.