Let $1 \leq p < \infty$ and let $(\alpha_{ij})_{i,j=1}^{\infty}$ be an 'infinite matrix' such that for all $\xi = (x_n) \in \ell^p$ $(A\xi)_n := \sum_{k=1}^\infty \alpha_{nk}x_k$ converges and defines a sequence $A\xi \in \ell^p$.
First I had to show that the function $F_N^n : \ell^p \rightarrow \mathbb{F}$ defined by $F_N^n\xi := \sum_{k=1}^N \alpha_{nk}x_k$ is continuous (for any N and n).
Next I have to use this result to prove that the map $\xi \mapsto (A\xi)_n$ is continuous. My idea was to take a sequence $(x_m)$ with elements in $\ell^p$ such that $x_m \rightarrow x$. Then we have $(Ax)_n = \lim_N F_N^n(x) = \lim_N \lim_m F_N^n(x_m)$ and $\lim_m (A(x_m))_n = \lim_m \lim_N F_N^n(x_m)$.
My first question here: can we interchange the limits here to complete the proof?
After this I have to use the result to show that $A$ is continuous. I was wondering whether I could use a similar idea. Define the function $G_N: \ell^p \rightarrow \ell^p$ that works like $A$ on the first $N$ coordinates and sends the remaining coordinates to zero. Then we would have $A(x) = \lim_N G_N(x) = \lim_N \lim_m G_N(x_m)$ which we would like to be equal to $\lim_m \lim_N (x_m) = A(x_m)$.
Again my question is whether it is possible to interchange the limits in this way. And if it does not work, what would be a good way to prove the statement?