= Solution
The <Riesz-Markov-Kakutani representation theorem> says that a positive <linear functional> on $C(K)$ has the form $f\mapsto\int_Kf\,d\nu$ for a unique finite positive <regular Borel measure> $\nu$, with <norm> $\nu(K)$. Its complex form says that every bounded complex <linear functional> is represented by a unique finite regular <complex measure> $\mu$, and
$$
\boxed{C(K)^*\cong M(K),\qquad L_\mu(f)=\int_Kf\,d\mu,\qquad\|L_\mu\|=|\mu|(K).}
$$
The extension from positive to arbitrary <linear functionals> follows by positive/negative decomposition of real <linear functionals> and then real/imaginary decomposition. The last quantity is the <total variation norm of a measure>, so this identifies the dual isometrically.
An <extreme point> $e$ of a <convex set> $C$ cannot be written as $e=ta+(1-t)b$ with $0<t<1$ and distinct $a,b\in C$. The <Krein-Milman theorem>, applied with the underlying real locally convex <weak-star topology>, states
$$
\boxed{C=\overline{\operatorname{conv}(\operatorname{ext}C)}^{\,w^*},\qquad\operatorname{ext}C\neq\varnothing.}
$$
We next prove <Milman's converse to the Krein-Milman theorem>. Suppose an <extreme point> $e$ were outside the weak-star closure of $S$. A basic weak-star neighborhood of $e$ disjoint from $S$ uses finitely many real coordinates: real and imaginary parts of evaluations at elements of $X$. Its complement is the union of finitely many closed half-spaces
$$
\ell_j(c)-\ell_j(e)\geq\varepsilon_j\quad\text{or}\quad\ell_j(c)-\ell_j(e)\leq-\varepsilon_j.
$$
Intersect these with $C$, discard empty intersections, and call the resulting compact <convex sets> $C_1,\ldots,C_m$. They cover $S$, and none contains $e$.
The <convex hull> of their union is compact. Every point in it can be written $\sum_{i=1}^mt_ic_i$ with $(t_i)$ in the finite simplex and $c_i\in C_i$, by combining terms from the same <convex set>. The map from the simplex times $\prod_iC_i$ to that sum is weak-star continuous, so its image is compact and closed. It therefore contains $\overline{\operatorname{conv}S}^{\,w^*}=C$, in particular $e$. But extremality forces every $c_i$ having a positive coefficient in a representation of $e$ to equal $e$, contradicting $e\notin C_i$. Hence
$$
\boxed{\operatorname{ext}C\subseteq\overline S^{\,w^*}.}
$$
Assume $K\neq\varnothing$. We claim that the <extreme points of the dual unit ball of C(K)> are
$$
\boxed{\operatorname{ext}B_{C(K)^*}=\{\alpha\delta_t:t\in K,\ |\alpha|=1\}.}
$$
A measure of <norm> less than one is not extreme, since it admits a small nonzero perturbation $\pm\eta\delta_t$ within the ball. For a measure $\mu$ of <norm> one, suppose $|\mu|$ is not a point mass. There is a Borel set $E$ with $0<a=|\mu|(E)<1$: if the <measure support> has two points, choose disjoint neighborhoods of positive mass; a regular probability measure supported at just one point is the corresponding point mass. Then
$$
\mu=a\frac{\mu|_E}{a}+(1-a)\frac{\mu|_{K\setminus E}}{1-a}
$$
is a convex combination of distinct norm-one measures. Thus an extreme measure must have variation concentrated at one point, and must be $\alpha\delta_t$ with $|\alpha|=1$.
Conversely, if $\alpha\delta_t=(\nu+\eta)/2$ with $\|\nu\|,\|\eta\|\leq1$, then
$$
1=\left|\frac{\nu(\{t\})+\eta(\{t\})}{2}\right|\leq\frac{|\nu(\{t\})|+|\eta(\{t\})|}{2}\leq1.
$$
Equality throughout forces both measures to have all their variation at $t$, and forces their phases to be $\alpha$. Hence $\nu=\eta=\alpha\delta_t$, proving extremality. If $K=\varnothing$, the dual ball is $\{0\}$ and its sole <extreme point> is $0$.
Every finite <Borel measure> on $[0,1]$ is regular, so the given $\mu$ belongs to the dual <unit ball> of $C[0,1]$. Apply <Banach-Alaoglu theorem> and <Krein-Milman theorem> to that ball. It is the weak-star closed <convex hull> of these phased point masses. The <unit ball> of the finite-dimensional <vector subspace> $F$ is <norm> compact. Choose a finite $\varepsilon/4$-net $f_1,\ldots,f_m$ in it. There is a convex combination
$$
\nu=\sum_{i=1}^Na_i\alpha_i\delta_{w_i},\qquad a_i\geq0,\quad\sum_i a_i=1,\quad|\alpha_i|=1,
$$
whose integrals differ from those of $\mu$ by less than $\varepsilon/2$ on every $f_j$. Put $t_i=a_i\alpha_i$. Then \b[$\sum_i|t_i|=1$] and $\|\nu\|\leq1$. For any $f$ in the <unit ball> of $F$, choose $f_j$ with $\|f-f_j\|<\varepsilon/4$. The <linear functional> $L_\mu-L_\nu$ has <norm> at most two, so
$$
\left|\int f\,d\mu-\sum_it_if(w_i)\right|<\frac\varepsilon2+2\frac\varepsilon4=\varepsilon.
$$
Scaling yields the requested estimate for every $f\in F$. This is <atomic approximation on finite-dimensional spaces of continuous functions>. Repeated nodes are allowed: keeping their individual terms preserves the exact sum of coefficient magnitudes even if their phases cancel. If $F=\{0\}$, one point with coefficient one suffices.
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