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Is it true that any bounded sequence of a (separable) Hilbert space contains a weakly convergent subsequence? I would say yes because, by Banach-Alaoglu theorem I can say that any bounded sequence contains a subsequence weakly* convergent, and since an Hilbert space is reflexive by Milman-Pettis theorem, the weak* convergence implies weak convergence.

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Your proof is incomplete. Compact topological spaces need not be sequentially compact. So Banach Alaoglu Theorem only gives you a convergent subnet, not necessarily a convergent subsequence.

The answer is YES. By Banach Alaoglu Theorem the closed unit ball of $H$ is weakly compact. By separability this implies that the ball is also metrizable. Hence, the closed unit ball is a compact metric space in weak topology and any sequence has a convergent subsequence. Same is true of close ball of any radius.

[The closed unit ball of $X^{*}$ is a compact metric space in the weak* topology whenever $X$ is separable Banach space].

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  • $\begingroup$ Is also my solution correct or not? $\endgroup$
    – hobal51825
    Feb 22 at 23:53
  • $\begingroup$ @hobal51825 I have edited my answer. $\endgroup$ Feb 23 at 0:02
  • $\begingroup$ In my notes I found this proposition: If X is separable and reflexive then any bounded sequence $\{x_n\}_{n \in \mathbb N} \subset X$ has a weakly convergent subsequence in $X$. Can I say that: Hilbert space is reflexive (for the same reason explained in the main question) and separable by hypothesis, then for the proposition any bounded seq contains a weakly convergent subseq.? (Banach Alaoglu theorem is used in the proof of the proposition) $\endgroup$
    – hobal51825
    Feb 23 at 0:12
  • $\begingroup$ Yes, that is correct but you should have quoted that result explcitly in the proof. Separability is essential for this result and yo had to say exactly how it is used in the proof. @hobal51825 $\endgroup$ Feb 23 at 0:16
  • $\begingroup$ Since $X$ is separable and reflexive, then $X^*$ (dual) is separable and I can apply Banach Alaoglu theorem to a sequence $\{\tau(x_n)\}_{n \subset \mathbb N} \subset X^{**}$ (bidual) finding $\tau(x_{n_h})$ weak* convergent to $\Lambda$. Finally, since X is reflexive I have that $x=\tau^{-1}(\Lambda)$ and $x_{n_h}$ weakly converges to $x$ $\endgroup$
    – hobal51825
    Feb 23 at 0:29

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