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Let $X$ be a Banach space. Prove that a linear map $M\colon X\mapsto \ell^p, \; p\geqslant 1$ is continuous iff for every sequence $(x_k)$ that converges in $X$ to $x \in X$, we have that the $n$-th term of the sequence $Mx_k$ converges to the $n$-th term of $Mx$ for all $n$.

My try: $(\Rightarrow)$ If $M$ is continuous $\|Mx_n - Mx\| \leq \epsilon$ hence $|(Mx_n)_i - (Mx)_i| \leq \epsilon$

$(\Leftarrow)$ $$y\in \ell^p \Rightarrow \|y\|_p < \infty \Rightarrow \exists N : \|y\|_p^p = \sum_{i = 1}^N |y_i|^p + \underbrace{\sum_{i = N+1}^\infty |y_i|^p}_{\leq \epsilon}$$

So we can find an $N$ such that $$\|Mx-y\|_p^p = \underbrace{\sum_{i = 1}^N |(Mx)_i - y_i|^p}_{\leq N\epsilon} + \underbrace{\sum_{i = N+1}^\infty |(Mx)_i - y_i|^p}_{\leq 2\epsilon} \leq (N+2)\epsilon.$$ Hence by closed graph we are done. Is this correct? What happens when $p = \infty$ ?

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In the $(\Leftarrow)$ implication, where do you use the hypothesis that $x_k\to x$ in $X$? – wisefool Dec 28 '12 at 13:53
The problem is that $N$ depends on $\epsilon$. – Davide Giraudo Dec 28 '12 at 14:53
up vote 2 down vote accepted

Closed graph theorem is indeed the idea. Let $\{x_k\}$ a sequence of elements of $X$ which converges to $0$ and such that $Mx_k\to y\in\ell^p$. As the $n$-th terms of $Mx_k$ converges to the $n$-th one of the null sequence, we have $(Mx_k)^{(n)}\to 0$. As for $1\leqslant p\leqslant+\infty$, we have for all $k$ and for all $n$, $$|y^{(n)}|\leqslant |y^{(n)}-(Mx_k)^{(n)}|+|(Mx_k)^{(n)}|\leqslant \lVert y-Mx_k\rVert_{\ell^p}+|(Mx_k)^{(n)}|$$ so $y=0$.

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Thanks, So we can always change convergent sequences $x_k \rightarrow x$ to some that converges to 0 i suppose. Can you please tell me what I did wrong? – Johan Dec 28 '12 at 14:38

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