Talk:Banach-Alaoglu Theorem

I will try to write down a proof this evening, if I find the time. - Giallo (30.1.13, 12:28)


 * Awesome; that'd be greatly appreciated! --Lord_Farin (talk) 11:31, 30 January 2013 (UTC)

I have written a first (much incomplete) outline of the proof. I'll have to work a bit on the proof for the two lemmas, since my source (Royden & Fitzpatrick) is a bit vague about them, but they shouldn't be too hard. Unfortunately I have a couple of exams soon, thus I'll leave it to you until then. By the way, i don't know if this is really the theorem of Banach-Alaoglu or if it should only be named after Alaoglu. This one was proven by Alaoglu, while Banach has shown that if $X$ is separable, then the closed unit ball in $X^*$ is weak* sequentially compact. - Giallo (30.1.13, 21:39)


 * Conway lists it as Alaoglu's Theorem. --Lord_Farin (talk) 21:28, 30 January 2013 (UTC)


 * Ok, I have moved it to the page for Alaoglu's theorem (the discussion, too). - Giallo (30.1.13, 22:45)

I also have written down the true (at least, I hope) Banach-Alaoglu theorem with proof. - Giallo (30.1.13, 23:26)

Retread
As presented in (whether this is accurate or not):
 * in a dual Banach space, the unit ball is weak star compact, and more generally, the polar of a neighborhood of the origin in a topological vector space is also weak star compact.

Whoever takes this on is invited to explore and document all of the redlinks in the above, and to establish whether the above is as fully general as possible, and then if possible to bring it back down to the instance of the real number line (or complex if this is appropriate) and so provide an instance for those for whom the intricacies of Banach spaces are too far over their heads. Like (until I care to do the work to catch up with all this) over mine. --prime mover (talk) 12:54, 20 October 2022 (UTC)


 * Can anyone explain what the polar of a neighborhood of the origin means? --Usagiop (talk) 14:27, 20 October 2022 (UTC)


 * The polar of $U$ is the set $\set {f \in X^\ast : \cmod {\map f x} \le 1 \text { for each } x \in U}$. Caliburn (talk) 20:25, 21 February 2023 (UTC)

I intended to put up a third red-link free proof of this tomorrow, but there is a problem. The general statement (which I will prove for NVSs, and probably come back for TVSs. There obviously are no balls in general TVSs so the statement and proof is very different) states that the unit ball in $X^\ast$ is $w^\ast$-compact. The assumption that $X$ is separable implies that the unit ball in $X^\ast$ with the $w^\ast$ topology is metrizable (this is another theorem that I will be proving - it's also an iff), so sequential compactness is equivalent to compactness. So you get sequential compactness in the separable case, but not in general. How should I deal with this? Caliburn (talk) 20:25, 21 February 2023 (UTC)


 * Sorry but what is the problem? --Usagiop (talk) 21:41, 21 February 2023 (UTC)


 * These proofs assumed that $X$ is separable and make a stronger conclusion. The proof I'm going to put in has weaker conditions and makes a "weaker" conclusion. (well, compactness and sequential compactness don't generally imply each-other but ygm) I could relegate this current page to a corollary, I guess. Separable case and general case? I don't know how heavily these proofs use separability, I haven't really looked at them in detail yet. Caliburn (talk) 22:05, 21 February 2023 (UTC)


 * Sorry, I am very confused. Do you mean the idea of Proof 1 cannot be used in a more general statement where especially $X$ is not separable? But why does this matter?? You can still follow the approach in Proof 2 without separability. --Usagiop (talk) 22:39, 21 February 2023 (UTC)


 * The theorem statement says separable. I want to remove the separability condition and prove weak-* compactness instead, and deduce weak-* sequential compactness in the separable case as a corollary. This would leave Proof 1, if it hinges seriously on separability, homeless, (since it then proves a different statement) and the standard practice on this site is to not delete such things. So I want to know what I should do with it. I just checked and Proof 2 already proves the weaker statement, not using separability and just proving compactness. (without mentioning sequential compactness, seemingly) FWIW I think this page demonstrates quite well the problem with putting up advanced theorems with none of the groundwork, I think I was the first to define even a dual space here... Caliburn (talk) 22:50, 21 February 2023 (UTC)


 * Yes that was me, I was experimenting with putting up everything from a dictionary, starting at A and working through to Z. This one uses terms deeper than I immediately understood even by tracing the links through the rest of the dictionary, at which point I decided that approach was not a good idea. --prime mover (talk) 23:24, 21 February 2023 (UTC)


 * OK, I got it. I think there are two options:
 * -Create a separate page Banach-Alaoglu Theorem (Separable NVS) and move Proof 1 there
 * -Remove the separability assumption and rename Proof 1 to e.g. Proof 1 (separable case, only)
 * But I do not know what is the 's way. Anyway it is worth to keep Proof 1, which is explicit and seemingly does not depend on Axiom of Choice. --Usagiop (talk) 23:40, 21 February 2023 (UTC)


 * Actually I just learnt of the Eberlein–Šmulian theorem - which seems to imply that weak sequential compactness and weak compactness are equivalent in Banach spaces. It's only mentioned in the appendix of Rudin and did not appear in my functional analysis course, apparently it may appear in Conway. But completeness is not in the hypotheses currently. Caliburn (talk) 23:21, 21 February 2023 (UTC)
 * I now realized that Eberlein–Šmulian theorem does only say compact iff relatively sequentially compact. In fact, the non-separable space $X=\map {\ell^\infty} \N$ is a counterexample. --Usagiop (talk) 19:02, 16 June 2023 (UTC)
 * I think I misspoke because this is the $w^\ast$ topology not the weak topology. But it's true that for a Banach space, weak compactness is equivalent to weak sequential compactness. Caliburn (talk) 19:08, 16 June 2023 (UTC)
 * I now also realized that Eberlein–Šmulian theorem does not say anything for $w^\ast$-topology. And the citation above was not correct, either. That theorem only asserts with respect to the weak topology that relatively compact iff relatively sequentially compact. In particular, even for the weak topology, unless we know that the unit ball is closed, the theorem does not help. --Usagiop (talk) 21:12, 16 June 2023 (UTC)
 * I believe my statement is correct. (or Wikipedia and a few other sources are wrong) The unit ball is always norm and weak closed, this follows from Mazur's Theorem. What I am unsure about is how weakly sequentially compact implies weakly closed, which would be needed for the equivalence of our two statements. All the stuff I can find just says "and in particular, a subset is weakly compact iff weakly sequentially compact" without explanation. Caliburn (talk) 22:29, 16 June 2023 (UTC)
 * OK, then, for the unit ball, in weak topology, compact iff sequentially compact. By the way, I think the sequential compactness always implies the closedness. If $A$ is not closed, then there is a $\sequence {a_n} \subseteq A$ with $\lim_{n\to\infty}a_n \not \in A$. Then $\sequence {a_n}$ cannot have a subseuquence which converges in $A$. So, $A$ is not sequentially compact. --Usagiop (talk) 22:58, 16 June 2023 (UTC)
 * Of course, as far as the limit is unique. --Usagiop (talk) 23:00, 16 June 2023 (UTC)
 * Well, $A$ not closed means there is a net (here called a "Moore-Smith sequence") with terms in $A$ that converges to a value outside $A$. Sufficient condition for a sequence to work would be metrizability but the closed unit ball is metrizable the dual of $X$ is separable. A proof that weakly sequentially compact plus weakly relatively compact implies weakly closed can be found in page 130, "Functional Analysis and Infinite Dimensional Geometry" which I have cited a few times. The proof does not seem very deep once everything is defined, but I haven't read it yet. Caliburn (talk) 23:09, 16 June 2023 (UTC)
 * OK, I think I got your point. --Usagiop (talk) 23:25, 16 June 2023 (UTC)