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.
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Davide Giraudo
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Falang
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2Shorter than what? Do you have a long proof? – Jonas Meyer May 20 '13 at 05:46
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1I 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 05:59
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3There is a short proof using Banach Alaoglu... – copper.hat May 20 '13 at 06:06
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4@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 06:12
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Thanks for you hints and explanation. – Falang May 20 '13 at 08:56
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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 $H^{\ast \ast}$.)
Zach L.
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4@N.U. yes, and in fact reflexive Banach spaces are characterized by the property in the OP. This follows from the Eberlein-Smulyan theorem combined with Kakutani's theorem that a Banach space is reflexive iff its unit ball is weakly compact. – Martin May 20 '13 at 07:02