Proof Silverman-Toeplitz theorem

Question

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?