Proof Silverman-Toeplitz theorem Proof Silverman-Toeplitz theorem: Let $A$ be an infinite matrix with entries $(a_{ij})$. Two sequences $\sigma $ and $s$ are related by this matrix as follows
$$\sigma_i =\sum_{j=0}^{\infty} a_{ij}s_i$$
Prove that for a convergent sequence $s$, $\sigma$ converges to the same value iff 
$$\lim_{i\to \infty} a_{ij} =0 \quad \text{for each } j$$
$$\lim_{i\to \infty} \sum_{j=0}^{\infty} a_{ij}=1$$
$$\sup_i \sum_{j=0}^{\infty} |a_{ij}|<\infty$$
I have no problem proving $\leftarrow$, and the first two conditions for $\rightarrow$ are easy: for the first one put $s_k=\delta_{kj}$ then $0=\lim_{i\to \infty }\sigma_i =\lim_{i\to \infty }\sum_{j=0}^{\infty} a_{ij}s_i=\lim_{i\to \infty } a_{ij}$. For the second one put $s_k=1$ then $1=\lim_{i\to \infty }\sigma_i =\lim_{i\to \infty }\sum_{j=0}^{\infty} a_{ij}$.

The third one is more problematic. I think the Banach-Steinhaus theorem could be applied here. I already used it to prove that $\sum_{j=0}^{\infty} |a_{ij}| <\infty$, so that the hypothesis is satisfied if the linear operators are taken to be the rows of $A$. That leaves me with the two possibilities: either they are all bounded or they diverge to infinity on a dense $G_{\delta}$. How can I eliminate the latter?

Although this answer is very late, I still think it might be useful to others. Let $A_n$ be the $n$th row of $A$ and let $a_k$ denote its $k$th element. If $\sum_k |a_k|$ does not converge we can choose an index sequence $k_r$ such that $k_0 = 0$, and
$$ \sum_{k = k_{r-1}}^{k_r-1} |a_k| > r \quad \text{for $r \geq 1.$}$$ 
Let $$s_k = \frac{\operatorname{sgn}(a_k)}{r}, \quad k_{r-1} \leq k < k_r.$$
Then $$\sum_{k=0}^\infty a_ks_k = \sum_{r=1}^\infty \sum_{k=k_{r-1}}^{k_r-1} \frac{|a_{nk}|}{r} > \sum_{r=1}^\infty 1 = \infty.$$ This is impossible since $(A_ns)_{n=0}^\infty$ is a convergent sequence and hence $A_ns$ must be finite. So in fact $A_n$ is absolutely summable, and hence defines a bounded linear functional on the space $l^\infty$ of bounded sequences, and hence on the space $c$ of convergent sequences with the inherited supremum norm. The family $(A_n)$ is pointwise bounded on each convergent sequence $s$, and by the Uniform Boundedness Principle we get $\sup_n \|A_n\|_c <\infty$, where $\|\cdot\|_c$ is the operator norm $c \to \mathbb{C}$. This operator norm is easily shown to be equal to the absolute sum, but we only need to show that it is at least as large as the absolute sum. For $\epsilon > 0$ choose $r$ such that $$\sum_{k = r+1}^\infty |a_{nk}| < \epsilon$$ and set $$x = \begin{cases}
\operatorname{sgn}(a_{nk}) & k \leq r\\
0 & \text{else}
\end{cases}
$$
Then $$|A_nx| = \sum_{k=0}^r |a_{nk}| \geq \sum_{k=0}^\infty |a_{nk}| - \epsilon.$$
This shows that $\|A_n\|_c \geq \sum_{k=0}^\infty |a_{nk}|$, which completes the proof. 
