A question in the proof of $C(X)$ is not a dual space of a Banach space. Let $X$ be a non-singleton compact connected space.I want to show that $C_{\mathbb{R}}(X)$ is not the  dual space of a Banach space,$\mathbb{R}$ is real number field.
I already know that, extreme points of $Ball(C_{\mathbb{R}}(X))$ are $\{f\equiv 1, f\equiv -1\}$. Then by Krein-Milman theorem, 
$Ball(C_{\mathbb{R}}(X))$ = weak*-closure of convex conbination set of $\{f\equiv 1, f\equiv -1\}$ = weak* closure of $\{f\equiv a : a\in[-1,1]\}$.
How can I conclude that $X$ is a singleton from above?
Any help would be appreciated!
 A: The set of constant functions $\{f\equiv a: a\in[-1,1]\}$ is already weak*-closed (for instance, it is compact, since the map $[-1,1]\to C_\mathbb{R}(X))$ sending $a$ to the constant function taking value $a$ is continuous and $[-1,1]$ is compact), so you've shown that in fact every element of $Ball(C_\mathbb{R}(X))$ is a constant function.  This means every continuous function $X\to\mathbb{R}$ is constant.  But by Urysohn's lemma, if $x,y\in X$ are distinct points, there is a continuous function $f:X\to\mathbb{R}$ such that $f(x)=0$ and $f(y)=1$.  Such a function is not constant, so it must be impossible to have distinct points $x,y\in X$.  That is, $X$ must have only one point.
A: Even more is true. This space is not isomorphic to a dual Banach space. Indeed, this space is separable and separable dual Banach spaces have the Radon-Nikodym property (which is an isomorphic invariant). This space clearly lacks this property as it contains a copy of $c_0$ (the Radon-Nikodym property passes to closed subspaces).
