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In the book of Brezis : "Analyse fonctionnelle : Théorie et application", chapter III (i.e. construction of weak topology, weak-* topology reflexives spaces...), why do we need "Banach spaces" ? Isn't normed spaces enough ? The particular example I have in mind if theorem III.16 (named as Kakutani) that says : Let $E$ a Banach spaces. Then $$B_E=\{x\in E\mid \|x\|\leq 1\}$$ is compact for the weak topology $\sigma (E,E')$ $\iff$ $E$ is reflexive.

I read the proof with attention, and I don't see where we use the fact that $E$ is complete for it's norm. So why do we need the assumption to be Banach ? The only reason for me would be that we use Banach-Steinhaus's theorem (BST) (and thus, we need completeness). But in the proof of Kakutani's theorem I don't see anywhere the used of (BST). So maybe the completeness is used somewhere I don't see ?

Bernard
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Peter
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2 Answers2

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Indeed the equivalence still holds if $E$ is an incomplete normed space (over $\mathbb{R}$ or $\mathbb{C}$): both sides are false. This is pretty easy to see directly and really misses the point of the theorem. So the authors probably just decided not to bother to include this relatively uninteresting case.

It's pretty common in functional analysis to write theorems that only cover Banach spaces, even when normed spaces could also be included. This can be for any of several reasons:

  • In most applications, you are working with Banach spaces

  • The theorem may become trivial for incomplete spaces

  • For an incomplete space $X$, the theorem gives you the "right" conclusion if you apply it to the completion of $X$.

Nate Eldredge
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  • Thank you very much for your answer. Could please point me where the author used the fact that $E$ is a Banach space ? (because if both side are false in non Banach space, to justify that at least one side is true in Banach side we must use completeness, don't we ?) – Peter Aug 29 '18 at 20:29
  • As Eduardo point, on $\mathbb R$ as a $\mathbb Q-$vector space, the unit ball is compact. So it hold in incomplete vector space (as I understood, you said that both side are false in incomplete vector space). So could you please give me more explanation ? :) – Peter Aug 29 '18 at 20:36
  • Leaving $\mathbb{Q}$-vector spaces aside, I am not convinced that on a non-complete normed space the unit ball cannot be weakly compact. To me this does not seem 'easy to see' at least :( – Lorenzo Q Aug 29 '18 at 21:32
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    @Peter: No, you misunderstand me. The theorem is true for incomplete spaces as well as complete spaces. (If $E$ is incomplete, then it's true that $E$ has a weakly compact ball iff it is reflexive - because it can never have a weakly compact ball and it is never reflexive). So it's quite possible the author never used the assumption of completeness. (I don't have a copy of the book to see what's actually written.) – Nate Eldredge Aug 29 '18 at 21:40
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    @LorenzoQuarisa Let $E$ be an incomplete normed space (real or complex), $\tilde{E}$ its completion. Then $\tilde{E}' = E'$, and $\sigma(E,E')$ is the subspace topology induced from $\sigma(\tilde{E},E')$. The closed unit ball of $E$ is norm-dense in the closed unit ball of $\tilde{E}$, hence it is a fortiori $\sigma(\tilde{E},E')$-dense in it. In particular, it is not $\sigma(\tilde{E},E')$-closed, which it would be if it were $\sigma(E,E')$-compact. – Daniel Fischer Aug 29 '18 at 21:47
  • @DanielFischer thanks for the explanation. I was trying to construct a weakly continuous functional with no extrema in the unit ball starting from a non-converging Cauchy sequence but with no luck. The approach of considering the completion seems best. – Lorenzo Q Aug 29 '18 at 22:11
  • Alternatively, if you let $x_n$ be a non-converging Cauchy sequence, and $x$ a weak accumulation point (which weak compactness should guarantee), you ought to be able to use the Hahn-Banach separating hyperplane theorem to produce a linear functional $f$ where $f(x)$ is more than $\epsilon$ away from most of the $f(x_n)$. – Nate Eldredge Aug 29 '18 at 22:26
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Consider $\mathbb{R}$ as an vector space over $\mathbb{Q}$ the rationals space endowed with the norm of absolute value, then $\mathbb{R}$ over $\mathbb{Q}$ is an infinite dimensional normed vector space where the ball is compact.

Eduardo
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    Interesting, but I think that in the definition of normed space we require either a real or complex vector space, so this is not a normed space. – Lorenzo Q Aug 29 '18 at 20:21
  • I think OP means "normed linear space," which at least Wikipedia takes to mean a vector space over $\mathbb R$ or $\mathbb C.$ https://en.wikipedia.org/wiki/Normed_vector_space – Thomas Andrews Aug 29 '18 at 20:22
  • Can you provide me a reference, I meant if I'm looking for this definition in the Bourbaki or other classical book I will find this warning? I have this doubt, but I have never found a reference that said this with all words... – Eduardo Aug 29 '18 at 20:22
  • Normed spaces are generally required to have $\Bbb{R}$ or $\Bbb{C}$ as the field of scalars. You can generalise this in various ways, e.g., to normed spaces over an ordered field $\Bbb{F}$ with the norm taking values in $\Bbb{F}$ or (as you are doing) to normed spaces over an ordered field $\Bbb{F}$ with the norm taking values in a superfield of $\Bbb{F}$. This probably isn't what the OP is interested in, but it would be interesting if you commented on how your example relates to the original question. – Rob Arthan Aug 29 '18 at 20:25
  • As for references, I think it is down to you to provide a reference that supports your example - the restriction to $\Bbb{R}$ or $\Bbb{C}$ as the field of scalars of a normed space is quite standard. – Rob Arthan Aug 29 '18 at 20:32
  • In fact this example is very interesting. So indeed, the theorem should be true in non Banach space... – Peter Aug 29 '18 at 20:33
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    @RobArthan I just trying to learn with you guys, If is told to me that I'm wrong It is my pleasure to kown why... I'm not pretend know everything, I'm free to be failure... – Eduardo Aug 29 '18 at 20:35
  • Even assuming that considering a vector space over $\mathbb{Q}$ makes sense in this context, are we even sure that the ball is weakly compact? Ok, if $\mathbb{R}$ is taken as a real vector space then its weak and strong topology are the same, but is this true when $\mathbb{R}$ is taken as a vector space on $\mathbb{Q}$? – Lorenzo Q Aug 29 '18 at 20:36
  • @LorenzoQuarisa: If the unit ball is compact in $\mathbb R$ over $\mathbb Q$ then of course it's weakly compact, no ? – Peter Aug 29 '18 at 20:37
  • @Peter Ah, yes, I agree, because the weak topology would be weaker regardless. But isn't $\mathbb{R}$ over $\mathbb{Q}$ still a Banach space? The metric is the same as in $\mathbb{R}$ over $\mathbb{R}$ and completeness only depends on the metric. – Lorenzo Q Aug 29 '18 at 20:41
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    How? The definition of the unit ball and its compactness only depend on the topology of $\mathbb{R}$, and not on which field it is considered as a vector space. – Lorenzo Q Aug 29 '18 at 20:55
  • @Eduardo: apologies if I sounded confrontational. I didn't mean to. You'd asked for a reference, but I think you will find most standard texts do restrict the field of scalars to $\Bbb{R}$ or $\Bbb{C}$. So I was interested to know where you had learnt about possible generalisations. – Rob Arthan Aug 29 '18 at 21:24
  • @RobArthan French wikipedia is more general. You need a non-discrete field with an absolute value. But things like duality theory crash hard without further requirements on the scalar field. The (topological) dual of $\mathbb{R}$ as a $\mathbb{Q}$-vector space is ${0}$, for example. Some things carry over to $p$-adics, maybe local fields in general. But I don't know much about non-Archimedian local fields. – Daniel Fischer Aug 29 '18 at 22:03
  • @DanielFischer: sure - as I said in my first comment, the usual definition of a normed space can be generalised in several ways. – Rob Arthan Aug 29 '18 at 22:50