Extreme points of the dual ball of C(K)

The measure identification also holds isometrically over $\mathbb{R}$. Indeed, for a bounded real-linear $L:C(K,\mathbb{R})\to \mathbb{R}$, define $L_{\mathbb{C}}(u+iv):=L(u)+iL(v)$. This is complex-linear. Given $h=u+iv$, choose $\theta$ so that $e^{-i\theta }{L}_{\mathbb{C}}(h)=\mathrm{\mid }{L}_{\mathbb{C}}(h)\mathrm{\mid }$; then $\mathrm{\mid }{L}_{\mathbb{C}}(h)\mathrm{\mid }=L(\mathrm{Re}(e^{-i\theta }h))\le \mathrm{\parallel }L\mathrm{\parallel }\mathrm{\parallel }h{\mathrm{\parallel }}_{\mathrm{\infty }}$, while restriction to real-valued functions gives the reverse norm inequality. Thus $\mathrm{\parallel }{L}_{\mathbb{C}}\mathrm{\parallel }=\mathrm{\parallel }L\mathrm{\parallel }$, and [L1] represents $L_{\mathbb{C}}$ by a unique regular complex measure $\mu$. The conjugate measure $\bar{\mu }$ represents the same functional, since for $h=u+iv$, ${\int }_{K}hd\bar{\mu }=\overline{{\int }_K\stackrel{ˉ}{h}d\mu }=\overline{L(u)-iL(v)}=L(u)+iL(v)\mathrm{.}$ Uniqueness in [L1] gives $\mu =\bar{\mu }$, so $\mu$ is real-valued and hence a finite regular signed measure. Conversely, a regular signed measure defines a bounded real functional; applying the same rotation argument to its complex integral shows that its real and complex operator norms agree, so [L1] gives norm $\mathrm{\mid }\mu \mathrm{\mid }(K)$. Together with [L1], this identifies the dual ball with regular signed or complex measures of variation at most one in the respective scalar field.

Conversely every $c{\delta }_x$ with $\mathrm{\mid }c\mathrm{\mid }=1$ is extreme. Suppose $c{\delta }_x=(1-t){\nu }_1+t{\nu }_2$ with $0<t<1$ and $\mathrm{\parallel }{\nu }_j\mathrm{\parallel }\le 1$. Evaluation at the constant function $1$ gives $c=(1-t){\nu }_1(K)+t{\nu }_2(K)$ with $\mathrm{\mid }{\nu }_j(K)\mathrm{\mid }\le 1$, so equality in the triangle inequality forces ${\nu }_1(K)={\nu }_2(K)=c$ and $\mathrm{\mid }{\nu }_j\mathrm{\mid }(K)=1$. For every Borel $E$, the inequalities $1=\mathrm{\mid }{\nu }_j(K)\mathrm{\mid }\le \mathrm{\mid }{\nu }_j(E)\mathrm{\mid }+\mathrm{\mid }{\nu }_j(K\setminus E)\mathrm{\mid }\le \mathrm{\mid }{\nu }_j\mathrm{\mid }(E)+\mathrm{\mid }{\nu }_j\mathrm{\mid }(K\setminus E)=1$ are equalities. Thus ${\nu }_j(E)=c\mathrm{\mid }{\nu }_j\mathrm{\mid }(E)$, so ${\nu }_j=c{\lambda }_j$ for the probability measure ${\lambda }_j:=\mathrm{\mid }{\nu }_j\mathrm{\mid }$. The original equality becomes ${\delta }_x=(1-t){\lambda }_1+t{\lambda }_2$. Positivity gives zero ${\lambda }_j$-mass to every compact subset of $K\setminus \{x\}$; regularity gives ${\lambda }_1={\lambda }_2={\delta }_x$, and hence ${\nu }_1={\nu }_2=c{\delta }_x$.

Under the measure identification of [step 1.1] and with the selection licensed by [L3], a measure $\mu$ with $\mathrm{\mid }\mu \mathrm{\mid }(K)<1$ is not extreme: choose $x\in K$ and $0<\epsilon <1-\mathrm{\mid }\mu \mathrm{\mid }(K)$; then $\mu =\frac{1}{2}(\mu +\epsilon {\delta }_x)+\frac{1}{2}(\mu -\epsilon {\delta }_x)$ with both summands of variation at most $\mathrm{\mid }\mu \mathrm{\mid }(K)+\epsilon <1$, and the summands differ from $\mu$. Hence every extreme point has norm and total variation one.

If $\mathrm{\mid }\mu \mathrm{\mid }(K)=1$ and there is a Borel set $E$ with $0<\mathrm{\mid }\mu \mathrm{\mid }(E)<1$, then $\mu$ is not extreme. Write ${\mu }_1:=\mu {\mathrm{\mid }}_E$ and ${\mu }_2:=\mu {\mathrm{\mid }}_{K\setminus E}$. Both are nonzero with $\mathrm{\parallel }{\mu }_j\mathrm{\parallel }=\mathrm{\mid }{\mu }_j\mathrm{\mid }(K)\in (0,1)$, and $\mu =\mathrm{\parallel }{\mu }_1\mathrm{\parallel }{\nu }_1+\mathrm{\parallel }{\mu }_2\mathrm{\parallel }{\nu }_2$, where ${\nu }_j:={\mu }_j\mathrm{/}\mathrm{\parallel }{\mu }_j\mathrm{\parallel }$ have norm one. They are distinct because $\mathrm{\mid }{\nu }_1\mathrm{\mid }(E)=1$ whereas $\mathrm{\mid }{\nu }_2\mathrm{\mid }(E)=0$. The positive coefficients sum to one, so this is a proper convex combination inside the ball.

Suppose $\mathrm{\mid }\mu \mathrm{\mid }(K)=1$ and $\mathrm{\mid }\mu \mathrm{\mid }(E)\in \{0,1\}$ for every Borel $E$. Then $\mu =c{\delta }_y$ for a unique $y\in K$ and some $\mathrm{\mid }c\mathrm{\mid }=1$. If $\mathrm{\mid }\mu \mathrm{\mid }(\{x\})=0$, outer regularity gives an open neighbourhood $U$ of $x$ with $\mathrm{\mid }\mu \mathrm{\mid }(U)<1$, and the two-valued hypothesis forces $\mathrm{\mid }\mu \mathrm{\mid }(U)=0$. If every singleton had measure zero, these open zero-measure sets would cover $K$, so compactness would give a finite such cover and contradict $\mathrm{\mid }\mu \mathrm{\mid }(K)=1$. Thus $\mathrm{\mid }\mu \mathrm{\mid }(\{y\})=1$ for some $y$. Additivity gives $\mathrm{\mid }\mu \mathrm{\mid }(K\setminus \{y\})=0$, so $\mathrm{\mid }\mu \mathrm{\mid }={\delta }_y$ and $y$ is unique. Since $\mu$ is concentrated on $\{y\}$, putting $c:=\mu (\{y\})$ gives $\mu =c{\delta }_y$ and $\mathrm{\mid }c\mathrm{\mid }=1$.

Let $\mu$ be an extreme point. By [step 2.2] $\mathrm{\mid }\mu \mathrm{\mid }(K)=1$. If $\mathrm{\mid }\mu \mathrm{\mid }$ were not $\{0,1\}$-valued then [step 2.3] would give a proper convex combination, contradicting extremality; hence [step 2.4] gives $\mu =c{\delta }_y$ with $\mathrm{\mid }c\mathrm{\mid }=1$.

By [step 3.1] every extreme point is $c{\delta }_y$ with $\mathrm{\mid }c\mathrm{\mid }=1$, and by [step 2.1] every such point is extreme; hence $\mathrm{ext}{B}_{C(K{)}^{\ast }}=\{c{\delta }_x:x\in K,\mathrm{\mid }c\mathrm{\mid }=1\}$.