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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?

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Think about what it means for $x_m\rightarrow x$. That means $\|x_m-x\|_p$ is getting arbitrarily small. What can you say about the relative differences between $x_m(i)$ and $x(i)$, where $x(i)$ is the i'th term of your $\ell_p$ sequence – Alex R. Nov 28 '12 at 5:23
up vote 1 down vote accepted

I don't think your "exchange limits" plan is a good one. What you have to do is use the continuity of $F_N^n$ and the fact that $A\xi$ is in $\ell^p$ (so somehow tails of sequences don't matter much).

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