Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
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
2  
@commenter whu don't you arite this comment as an answer? –  userNaN 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

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.