Let $A=(a_{ij})$ be an infinite complex matrix. $A$ associates each sequence $(s_j)$ with a sequence $(\sigma _i)$ defined by $$\sigma _i=\sum_{j=0}^{\infty}a_{ij}s_j \ \ \ \ \ \ (i=1,2,3,...)$$ provided that these series converges.
Suppose that $A$ transforms every convergent sequence $(s_j)$ to a sequence $(\sigma_i)$ which converges to the same limit. (Edit: So $A$ maps a convergent sequence to a convergent sequence. I guess it is OK to let some finite coordinates of the image to be infinity; otherwise the problem becomes trivial.)
I want to show that
$$\sup_{i}\sum_{j=0}^{\infty}|a_{ij}|<\infty$$
I think I should proceed by constructing a convergent sequence $(s_j)$ assuming $\sup_{i}\sum_{j=0}^{\infty}|a_{ij}|=\infty$ but $(\sigma_i)$ does not converge to the same limit. However, I cannot come up with any good sequence that lead to the contradiction... Could you please give me some hint? I am totally stuck with this problem now and would really appreciate your help.
For all $i\in\mathbb{N}$, define the following family $$ A_i:(c_0,\|\cdot\|_\infty)\to \mathbb{R},(s_j)_j\mapsto \sum_j a_{i,j}s_j. $$ Since the summation $\sum_j a_{i,j}s_j$ is finite whenever $s_j$ admits a limit, by picking $s_j=1$ we find that $\sum_j a_{i,j}<\infty$, that is $(a_{i,j})_j\in l^1$ for all $i\in\mathbb{N}$ (I am assuming that the summation is the Lebesgue integral with respect to the counting measure, so that absolute convergence implies convergence and viceversa). $A_i$ is clearly linear and continuous since $$ |A_i (s_j)_j| = \left| \sum_j a_{i,j}s_j\right| \leq \sum_j |a_{i,j}||s_j|\leq \|(s_j)_j\| \sum_j |a_{i,j}| \leq C_i \|(s_j)_j\|. $$ To recover the opposite inequality let $(s_j^i)_j = (sign(a_{i,j}))_j$. Truncate the sequence up to the $n$-th term, so that $$(s^{n,i}_j)_j = (sign(a_{i,j}) \chi_{j\leq n})_j \in c_0, $$ as it is identically equal to $0$ for $j\geq n$. We can compute $$ \|A_i\|\geq \lim_n |A_i(s_j^{i,n})_j| \overset{\text{monotone convergence}}{=} \lim_n \sum_{j=0}^n |a_{i,j}| = C_i, \quad \|A_i\| = C_i = \sum_j |a_{i,j}|. $$
Notice that $c_0 = \{(s_j)_j\in l^\infty: \lim_j s_j=0\}$ is a Banach space (easy to prove) w.r.t $\|\cdot\|_\infty$ norm. Fix now any $(s_j)_j\in c_0$, since $s_j\to 0$ then $A_i(s_j)_j\to 0$ as $i\to\infty$ by hypothesis on the matrix, so that $$ \sup_i |A_i(s_j)_j| <\infty. $$ Apply Banach–Steinhaus to conclude that $$ \sup_i \|A_i\| = \sup_i \sum_j |a_{i,j}| <\infty. $$