Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $p = \infty$ ?

share|improve this question
    
In the $(\Leftarrow)$ implication, where do you use the hypothesis that $x_k\to x$ in $X$? –  wisefool Dec 28 '12 at 13:53
1  
The problem is that $N$ depends on $\epsilon$. –  Davide Giraudo Dec 28 '12 at 14:53
add comment

1 Answer 1

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 $y=0$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.