Let $X$ be a compact Hausdorff space and $C(X)$ the Banach space of continuous $\mathbb K$-valued functions equipped with the supremum norm. We denote the dual space of $C(X)$ by $C(X)^*$.

A well-known theorem, due to Arens & Kelley (1947), states that the extreme points of the unit ball of $C(X)^*$ are precisely the functionals of the form$$\alpha= \theta\cdot\delta_x,$$ where $\theta$ is an unitary scalar and $\delta_x$ is the functional "evaluation at $x$", that is, $ \delta_x(f)=f(x), $ for every $f\in C(X)$.

I wonder if there is a version of this theorem for $C_0(X)$ spaces, that is, the Banach space of continuous $\mathbb K$-valued functions which vanish at infinity for some locally compact (not necessarily compact) Hausdorff space $X$.

Actually, I'm pretty sure I have seen people using this theorem for the $C_0(X)$ spaces, however I still fail to find any proof of this. I was able to prove, by myself, a version of this theorem for $C_0(X)$ spaces in the case $\mathbb K=\mathbb R$. I still have no answer for the case $\mathbb K=\mathbb C$.

  • $\begingroup$ Well, your hint seems fine to me. This way is so much better. The onliest proof I knew (the one at Dunford's book) didn't use the representation theorem. Instead it strongly depends on the compactness of $X$. $\endgroup$ – André Porto Dec 6 '17 at 18:59

Using a form of Riesz representation theorem, one can argue as follows. A unit-norm functional on $C_0(X)$ is represented by a complex measure $\mu$ of total variation $1$. If the support of $\mu$ consists of one point, it is a unimodular multiple of Dirac mass. Otherwise, $\operatorname{supp}\mu$ contains two distinct points $a\ne b$. Let $U,V$ be disjoint neighborhoods of $a$ and $b$. By the definition of support, $|\mu|(U)>0$ and $|\mu|(V)>0$. Let $t = |\mu|(U)$; by the above $0<t<1$. Define $$ \lambda = t^{-1} \mu_{|U}\quad \text{and} \quad \nu = (1-t)^{-1} \mu_{|U^c} $$ Then both $\lambda$ and $\nu$ represent unit-norm functionals, and $$ \mu = t \lambda + (1-t)\nu $$ proving that $\mu$ is not extreme.


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