The following is an exercise I am trying to solve:
"Let $X$ be a locally compact Hausdorff topological space, and $CB(X)$ denote the space of all bounded continuous functions on $X$, with the supremum norm $||\cdot||_\infty$. For each $x\in X$, let $E_x\in CB(X)^*$ denote the evaluation map $f\mapsto f(x)$. Prove that the weak$^*$-closure of the convex circled hull $cco(\{E_x:x\in X\})$ equals the closed unit ball of $CB(X)^*$."
Note: Here by 'circled hull' of a set $S$, it is meant the balanced hull of $S$, i.e, the set $\{\alpha s: \alpha\in\mathbb{C}, |\alpha|\leq 1, s\in S\}$.
My progress so far:
I have been trying to prove it using contradiction via separation theorems. For example, take $V:=(CB(X)^*, \mbox{weak}^*\mbox{-topology})$; $A:= \overline{cco}^{w^*} (\{E_x: x\in X\})$. Now if possible, let there exists some $\phi$ in the closed unit ball of $CB(X)^*$ such that $\phi \notin A $. Let $B:= cco(\{\phi\})$. Then $A\cap B =\varnothing$ where $A$ is compact and $B$ is closed. Hence using separation theorems on locally convex linear toplogical spaces (here $V$), we see that there exists a linear functional $\omega \in V^*$ such that: $$\sup\{|\omega(f)|: f \in B\} < \inf \{|\omega(f)|: f\in A\}. $$
I was hoping using the fact that $||\phi||\leq 1$ and the properties of evaluation maps, I could arrive at some contradiction. But I have been unable to do it yet.
Any comment/suggestion is welcome. Thanks in advance !
By the bipolar theorem, the weak$^*$-closure of the absolutely (or as you say, circled) convex hull of $M=\{E_x:x\in \Omega\}$ is the (absolute) bipolar $M^{\bullet\circ}$ where $M^\bullet=\{f\in CB(X): |E_x(f)|\le 1\text{ for all }x\in X\}$ is the unit ball of $CB(X)$. That's it.
(Your attempt is quite similar to the proof of the bipolar theorem, you should apply the Hahn-Banach separation theorem to $A$ and $B=\{\phi\}$ without the absultely convex hull.)