# Weakly convergent subsequence

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.

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].
• 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) Feb 23 at 0:12
• 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 $x$ Feb 23 at 0:29