Talk:Banach–Alaoglu theorem

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Latest comment: 10 years ago by Noix07

This showed up in Wikipedia:Deletion_log_archive/June_2003 where it seems someone had unsuccessfully started to put in a proof. I hope this is more acceptable, even if it is a bare sketch. -- AndrewKepert 08:36, 5 Apr 2004 (UTC)

Why is it so complicated!!!!!!!!! I'm copying the first lines (that are clear) from the intro and put it as a section — Preceding unsigned comment added by Noix07 (talkcontribs) 15:05, 12 December 2013 (UTC)Reply

Sequencially compact

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All the applications of the Banach-Alaoglu theorem I know make use of sequencially compactness. Is this a consequence of compactness in this case (or for certain spaces)?--Trigamma 18:41, 17 January 2007 (UTC)Reply

Consequences Section

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It says there

 The corresponding result for p=1 is not true, as L1(μ) is not reflexive. (*)

however, this does not follow from

 If X is a reflexive Banach space, then every bounded sequence in X has a weakly convergent subsequence.

(the two are actually equivalent, so (*) holds true, but that's not clear from the article). —Preceding unsigned comment added by 84.191.204.239 (talk) 15:45, 17 March 2009 (UTC)Reply