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I'm learning about the weak and weak* topologies on a normed vector space following the book of Brezis. He limits his discussion to case where $E$ is a Banach space, and my question is most simply stated as, "Why?". I can't find an example of a theorem where the completeness of $E$ is a necessary hypothesis. Most of the basic results on weak and weak* topologies proceed by applications of Hahn-Banach, which holds for a much wider class of spaces than just Banach spaces.

Are there any examples of reasonably elementary (i.e. of relevance to a first-year graduate student who does not anticipate having heavy contact with functional analysis in the future) facts about weak or weak* topologies that are true for Banach spaces but not all normed vector spaces?

EDIT: I should add that, as Yemon Choi points out, dual spaces of normed vector spaces are complete, so the weak-star topology will never be defined on a space that isn't Banach. With regards to the weak-star topology, then, my question should refer to aspects of the weak-star topology on a space $E^*$ where the original $E$ is not Banach.

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    Did you actually learn any interesting theorems about the weak topology, other than perhaps the principle of uniform boundedness? Most of the useful theorems in a first course on functional analysis that I know about are about the weak-* topology (e.g. Banach-Alaoglu, Krein-Millman), which only exists on Banach spaces as Yemon Choi explained. My guess is that almost any interesting theorem you know that actually is about the weak topology will use completeness.2012-01-19

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I don't know whether the property I discuss is advanced for the first year student. This example shows that weak$^*$ boundness in $X^*$ imply uniform boundness provided $X$ is complete (see uniform boundness principle). But this is not true in general

Consider the space $c_{00}(\mathbb{N})$ of finitely supported sequences with uniform norm. This is not a complete space, and its completion is $c_0(\mathbb{N})$. Consider family of functionals $\{f_n:n\in\mathbb{N}\}$ defined by equality $ f_n :c_{00}(\mathbb{N})\to\mathbb{C}:x\mapsto\sum\limits_{k=0}^nkx(k) $ It is easy to check that

Each functional $f_n$ is bounded and $\Vert f_n\Vert=\frac{1}{2}n(n+1)$. Hence $ \sup\{f_n:n\in\mathbb{N}\}=+\infty $

For each $x\in c_{00}(\mathbb{N})$ the sum $\sum\limits_{k=0}^\infty k|x(k)|$ is finite. Hence we can obtain $ \sup\{|f_n(x)|:n\in\mathbb{N}\}\leq\sum\limits_{k=0}^\infty k|x(k)|<\infty $ Thus we constructed a family of bounded linear operators (in particular - functionals) which are pointwise bounded but not uniformly bounded. The raison d'etre of such a family is that our space $c_{00}(\mathbb{N})$ is not complete.

On the other hand uniform boundness principle guarantees boundness of any family of functionals in $X^*$ provided the space $X$ is complete.

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    Do we need the summation here? Can't we just define $f_n(x) = nx(n)$?2017-04-11