In "Analyse fonctionnelle" of Brezis, in chapter III why do we need Banach spaces ? (especially for Kakutani's theorem)

Question

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 ?

Answer 1

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 $X$.

Answer 2

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.