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Let $X$ be any locally compact Hausdorff space and assume that it is not compact. I've heard that the Banach space $(C_0(X),\|\!\cdot\!\|_\infty)$ is not isometrically isomorphic to the (norm) dual of a Banach space. Is there a good book where I can find a proof this result?

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Hint: Show that the unit ball has no extreme points. Use Alaoglu and Krein-Milman to derive a contradiction. – commenter Oct 9 '12 at 21:05
@commenter whu don't you arite this comment as an answer? – Norbert Oct 10 '12 at 9:48
@Norbert: Because I couldn't think of a reference containing that argument. I posted an expanded version as an answer. – commenter Oct 10 '12 at 10:50

Suppose first that $C_0(X) = E^\ast$ for some normed space $E$. By Alaoglu's theorem the closed unit ball $B$ of $C_0(X)$ is compact in the weak*-topology and by the Krein-Milman theorem $B$ has an extreme point (in fact, $B$ is the weak*-closed convex hull of its extreme points).

Thus, in order to prove that $C_0(X)$ is not a dual space, it suffices to show that $B$ has no extreme points:

Let $f \in B$ with $\lVert f\rVert =1$. Since $X$ is not compact and $f$ vanishes at infinity, the set $U = \{x \in X : |f(x)| \lt 1/2\}$ is non-empty open and $C=\{x \in X : \lvert f(x) \rvert \geq 1/2\}$ is non-empty and compact.

Pick $u \in U$ and use Urysohn's lemma to find a function $h\colon X \to [0,1/2]$ such that $h(u) = 1/2$ and $\operatorname{supp}h \subset U$. Then $\left\lVert f \pm h\right\rVert_\infty = 1$ and $f \neq f\pm h$ together with $$ f = \frac{1}{2}\left(f+h\right) + \frac{1}{2}\left(f-h\right) $$ show that $f$ is not an extreme point in $B$.

Remark. It is essential that we assume that $X$ is not compact. Constant functions of norm $1$ are always extremal in the unit ball for compact $X$, so the above argument breaks down. There's a good reason for that: For finite $X$, $C(X) \cong \mathbb{R}^n$ is reflexive, or for $X = \beta\mathbb{N}$ we can show that $C(\beta\mathbb{N}) \cong \ell_\infty = (\ell_1)^\ast$, so these are dual spaces.

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But of course $C[0,1]$ is again not a dual space, so noncompactness is not "if and only if". Real $C[0,1]$ has too few extreme me points. But complex $C[0,1]$ has lots of extreme points so more thought is required for that case. – GEdgar Oct 10 '12 at 13:28

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