# Bounded sequence in Hilbert space contains weak convergent subsequence

In Hilbert space $H$, $\{x_n\}$ is a bounded sequence then it has a weak convergent subsequence.

Is there any short proof? Thanks a lot.

• Shorter than what? Do you have a long proof? – Jonas Meyer May 20 '13 at 5:46
• I think you'll want to assume a separable Hilbert space with an infinite basis $\{e_n\}_{n\in\mathbb N}$. Assuming that, work towards a diagonal argument. – Ian Coley May 20 '13 at 5:59
• There is a short proof using Banach Alaoglu... – copper.hat May 20 '13 at 6:06
• @FrankMcGovern: Separability isn't needed, but if desired one could reduce to the separable case by considering the Hilbert subspace generated by $\{x_n\}$. – Jonas Meyer May 20 '13 at 6:12
• Thanks for you hints and explanation. – Falang May 20 '13 at 8:56

Suppose $M$ bounds the sequence. Then, if we think of $H$ as sitting inside $H^{**}$, then for any $T \in H^\ast$ with $\|T\| \leq 1$, we have $\|x_n(T)\| = \|Tx_n\| \leq \|x_n\| \leq M$, so the operator norms of the $x_n$ thought of as operators on $H^\ast$ are bounded by $M$. Apply Banach-Alaoglu. (To the unit ball of $H^{\ast \ast}$.)