# Proving Bishop's Theorem using Krein-Milman Theorem

I am studying the proof of Bishop's theorem (generalization of Stone-Weierstrass) in Rudin's Functional Analysis 2nd edition. He make the following statement on the bottom of page 122,

"Since $\mu \to \int g \, d\mu$ is a weak$^*$-continuous function on K, the Krein-Milman theorem implies that $\int g \, d\mu=0$ for every $\mu$ in $K$."

I do not understand why we know that is a weak$^*$-continuous function nor do I see why the Krein-Milman theorem implies that statement. I think one or both these things might have to do with the fact that K is weak$^*$-compact. Thanks for the help!

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To answer your first question, recall that a sequence $\phi_n$ in $C(S)^*$ (The dual of $C(S)$) converges to $\phi$ in the weak* topology if and only if $\phi_n(g) \to \phi(g)$ for every $f \in C(S)$ (actually we might have to use nets instead of sequences, but the argument is the same). In this specific case, we can identify $C(S)^*$ with a space of measures on $S$ in such a way that the pairing $\phi_n(g)$ is given by $\int_S g\, d\mu_n$ and $\phi(g) = \int_S g\, d\mu$. So to say that a sequence $\mu_n$ converges to $\mu$ in the weak* topology means by definition that $\int_S g\, d\mu_n \to \int_S g\, d\mu$ for every $g$ and hence $\mu \mapsto \int_S g\, d\mu$ is weak* continuous.

To answer your second question, Rudin proved earlier that $\int_S g\, d\mu = 0$ whenever $\mu$ is an extreme point of $K$, from which it immediately follows that $\int_S g\, d\mu = 0$ if $\mu$ is the convex combination of extreme points of $K$. Rudin also showed that $K$ satisfies the hypotheses of the Krein-Millman theorem (convex, balanced, weak*-compact), so the function is $0$ on the whole set ($K$ is the closure of the convex hull of its extreme points).

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(-1) Your explanation is implicitly appeal to the fact that weak-$^*$ continuity can be described in terms of sequential continuity. This is not true in general for non-furst-countable topologies (like weak-$^*$ topology). In the second part you are mistaken Krein-Milman asserrts that convex hull of extreme points is dense (not extreme points themselves). If you find mistakes in this comment I'll remmove my downvote. –  Norbert Apr 6 '13 at 18:59
Regarding your first comment: I included the parenthetical comment that one must use nets instead of sequences to avoid countability considerations in favor of user-friendtly arguments. If you disagree with my claim that the argument is the same for nets, let me know. –  Paul Siegel Apr 7 '13 at 6:05
Regarding your second comment: I suppose an edit is called for. –  Paul Siegel Apr 7 '13 at 6:05
Now it is all ok –  Norbert Apr 7 '13 at 7:57
As for the first part of your question, recall that by Reisz theorem for a compact $S$ the space $C(S)^*$ is isometric to the space of regular complex valued borel measures on $S$ denoted by $M(S)$. Moreover duality between $C(S)$ and $M(S)$ is given by $$\langle\cdot,\cdot\rangle:M(S)\times C(S)\to\mathbb{C}:(\mu,g)\mapsto \int_S gd\mu$$ The map $$\mathrm{ev}_g:M(S)\to\mathbb{C}:\mu\mapsto \int_S gd\mu$$ is weak-$^*$ continuous,because it is continuous with respect to seminorm $$\Vert \mu\Vert_g=\left|\int_S gd\mu\right|$$ Indeed $$|\mathrm{ev}_g(\mu)|=\left|\int_S gd\mu\right|\leq\Vert\mu\Vert_g$$
Now we proceed to the second part. As it was proved earlier on the same page $K$ is weak-$^*$ compact convex subset of $C(S)^*=M(S)$. By Krein Milman theorem $K=\mathrm{cl}_{w^*}(\mathrm{conv}(\mathrm{extr}(K)))$. On the same page it is was proved that for all $\mu\in\mathrm{extr}(K)$ we have $$\int_S gd\mu=0 \tag{1}$$ Obviously this equality holds for all convex combintaions of such points so $(1)$ is valid for all $\mu\in\mathrm{conv}(\mathrm{extr}(K))$. In other words $$\mathrm{conv}(\mathrm{extr}(K))\subset\mathrm{Ker}(\mathrm{ev}_g)$$ Since $\mathrm{ev}_g$ is a weak-$^*$ continuous $\mathrm{Ker}(\mathrm{ev}_g)$ is weak-$^*$ closed. Hence $$K=\mathrm{cl}_{w^*}(\mathrm{conv}(\mathrm{extr}(K)))\subset \mathrm{cl}_{w^*}(\mathrm{Ker}(\mathrm{ev}_g))=\mathrm{Ker}(\mathrm{ev}_g)$$ This means that for all $\mu\in K$ equality $(1)$ holds.