Let $V$ be a Banach space. Show that the closed unit ball in $V$ is also closed in the weak topology.
I know this is a consequence of the statement any closed convex subset in $V$ is closed in the weak topology, which the proof used the geometric Hahn-Banach theorem. My question is: does this problem have an elementary proof without using Hahn-Banach? Any help is appreciated.
Best Answer
At first I wanted to follow the reasoning showed in "Brezis - Functional Analysis Sobolev Spaces and Partial Differential Equations" at pages 59-60, riassumed in this identity $$B_V =\bigcap_{f \in V^*,\|f\| \leq 1} \{x \in V \mid |\langle f,x \rangle | \leq 1 \}$$ (two words about it: it relies on the characterization of open/closed set in the weak topology and property of the operatorial norm and -obviously- some corollaries of HB, existence of $f_{x_0}$ such that $f(x_0)=\|x_0\|$ and $\|f_{x_0}\|=1$) and so it would be not suitable for the question here.
But there is a more deep use of HB here. An immediate corollary of HB (adapted to these hypothesis) is
Corollary. Let $V$ a non trivial Banach Space, then $V^*$ is not trivial (it contains other elements than the zero map)
Without HB there is nothing (modulo equivalent forms of it) (as far as I know at least) that grants us the non triviality of $V^*$, in the infinite dimensional case obviously. Without assuming HB the first intersection is non empty but can't be refined more than $$\bigcap_{f \in V^*,\|f\| \leq 1} \{x \in V \mid |\langle f,x \rangle | \leq 1 \} = V$$ because we don't have (we can't write down explicitly) any functional other than the zero map.
In this (rather pathological) case, the weak topology could be the trivial one, the only open set is $V$, and so $B_v$ is not closed, and this observation should show the deep problem in not assuming HB.