Assume $x_n \to 0$ weakly in a Banach space. Show that for all $\epsilon>0$ and for all $N\in \mathbb{N}$ there exists a $n>N$ s.t. for all $f\in X^\ast, \|f\|\leq 1$ there exists $N<m<n$ s.t. $|f(x_m)|\leq \epsilon$.


Assume the statement doesn't hold, i.e. there is $\epsilon>0, N\in\mathbb{N}$ s.t. $\forall n>N$ there exists a $f_n\in X^\ast, \|f_n\|\leq 1$ with $|f_n(x_m)|\geq \epsilon$ for all $N<m<n$. Set $f$ as the a weak-*-limit point of $(f_n)_n$, which I can do due to Banach-Alaoglu. (I might eventually pass to a subsequence here.)

Fix $m>N$.Then forall $n>m$, I have $\epsilon\leq|f_n(x_m)|$. Then in particular $\epsilon\leq \lim_{n\to\infty}f_n(x_m) = f(x_m)$, contradicting the fact that $x_n\to 0$ weakly.

I am not sure about my proof and think that it is probably wrong, since I didn't use the notion of completeness of $X$.

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    $\begingroup$ The statement in the problem is obviously wrong ($x_n=0$). Is there a typo? $\endgroup$ – Jochen Dec 3 '14 at 7:39
  • $\begingroup$ Sure, I edited it. Thanks for the remark. $\endgroup$ – Peter Dec 3 '14 at 12:01
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    $\begingroup$ I think your proof is okay and that, indeed, the result holds for normed spaces. One has to be a bit careful because Banach-Alaoglu does NOT assert that the dual unit ball is SEQUENTIALLY weak*-compact but, as far as I can see, you do not use this. $\endgroup$ – Jochen Dec 3 '14 at 15:49
  • $\begingroup$ I think, I do use seqeu. compactness, don't I? $\endgroup$ – Peter Dec 3 '14 at 17:17
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    $\begingroup$ All you need is that $f$ is in the weak*-closure of $\lbrace f_n: n\ge m \rbrace$. $\endgroup$ – Jochen Dec 3 '14 at 19:06

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