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.

  • 2
    $\begingroup$ Shorter than what? Do you have a long proof? $\endgroup$ – Jonas Meyer May 20 '13 at 5:46
  • 1
    $\begingroup$ 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. $\endgroup$ – Ian Coley May 20 '13 at 5:59
  • 3
    $\begingroup$ There is a short proof using Banach Alaoglu... $\endgroup$ – copper.hat May 20 '13 at 6:06
  • 4
    $\begingroup$ @FrankMcGovern: Separability isn't needed, but if desired one could reduce to the separable case by considering the Hilbert subspace generated by $\{x_n\}$. $\endgroup$ – Jonas Meyer May 20 '13 at 6:12
  • $\begingroup$ Thanks for you hints and explanation. $\endgroup$ – 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 $H^{\ast \ast}$.)

  • $\begingroup$ So you only needed a reflexive Banach space? That's very nice! $\endgroup$ – N.U. May 20 '13 at 6:56
  • 3
    $\begingroup$ @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. $\endgroup$ – Martin May 20 '13 at 7:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.