Dirac measures are extreme points of unit ball of $C(K)^*$.

$\newcommand{\norm}[1]{\left\lVert#1\right\rVert}$ Let $(X, \mathcal{A})$ by a measurable space and denote by $\mathcal{M}$ the space of (finite) complex measures on $(X, \mathcal{A})$, endowed with the variation norm. Let $B$ be the closed unit ball in $\mathcal{M}$. I will show the following:

For any $x \in X$ and $a \in \mathbb{C}$ with $|a| =1$, the measure $a \delta_x$ is an extreme point of $B$ (where $\delta_x$ is the Dirac measure centered at $x$).

Proof:

Let $\mu := a \delta_x$ for some $|a| =1$. The idea of the proof is that the variation measure $| \mu |$ is equal to $\delta_x$, so that $| \mu |$ is an extreme point of the convex set $\mathcal{P}$ of probability measures on $(X, \mathcal{A})$.

Suppose that $\mu = s \nu_1 + (1-s) \nu_2$, where $\nu_1, \nu_2 \in B$ and $0 <s <1$. We want to prove that $\nu_1 = \nu_2 = \mu$.

First, we can assume that $\norm{ \nu_1 } = \norm{ \nu_2 } = 1$ (so that $| \nu_1 |$ and $| \nu_2 |$ are probabilities). Otherwise, we would have $\norm{ \mu } = \norm{ s \nu_1 + (1-s) \nu_2 } \leq s \norm{\nu_1} +(1-s) \norm{ \nu_2} < s + (1-s) = 1$, a contradiction.

We have by definition of $\mu$ that: $$ \delta_x = | \mu | = | s \nu_1 +(1-s) \nu_2 | \leq s |\nu_1| +(1-s) |\nu_2|. $$

The measure $\nu := s | \nu_1 | + (1-s) |\nu_2|$ is a probability measure, so the inequality $\delta_x \leq \nu$ is in fact an equality. Indeed, if $A$ is measurable and contains $x$ we must have $$ 1 = \delta_x (A) \leq \nu(A) \leq 1, $$ so $\nu(A)=1$. On the other hand, if $x$ is not in $A$, then $\nu(A) = \nu(X) - \nu(X \setminus A) = 1 - 1 = 0$. This means that $\nu = \delta_x$.

Therefore we have $\delta_x = s | \nu_1 | + (1-s) | \nu_2 |$. Since $| \nu_1 |$ and $| \nu_2 |$ are probabilites and $\delta_x$ is an extreme point of $\mathcal{P}$, we must have $| \nu_1 | = | \nu_2 | = \delta_x$.

The equality $| \nu_1 | = | \nu_2 | = \delta_x$ implies that $\nu_1$ and $\nu_2$ are multiples of $\delta_x$. Indeed, if $A$ is a measurable set that does not contain $x$, then $$ | \nu_i (A) | \leq |\nu_i|(A) = \delta_x (A) = 0, $$ so that $\nu_i(A) =0$. On the other hand, if $A$ contains $x$ then $\nu_i(A) = \nu_i(X) - \nu_i(X \setminus A ) = \nu_i(X)$. Therefore we have $\nu_i = b_i \delta_x$, where $b_i = \nu_i(X)$. Furthermore, we have $| \nu_i | = |b_i \delta_x | = |b_i| \delta_x$, which implies that $|b_i|=1$.

Our assumption that $\mu = s \nu_1 + (1-s) \nu_2$ thus gives $$ a \delta_x = (s b_1 + (1-s) b_2) \delta_x, $$ and therefore $a = s b_1 +(1-s) b_2$, where $0< s <1$ and $a$, $b_1$ and $b_2$ are complex numbers of modulus $1$. This is only possible if $a = b_1 = b_2$, and we conclude that $\nu_1 = \nu_2 = a \delta _x$. $\square$

Here is part of the proof. Can you finish it?

Notation. $K$ is a compact Hausdorff space. $C(K)$ is the Banach space of continuous, complex-valued, functions on $K$ with norm $$ \|f\| = \sup \{|f(x)\;:\; x \in K\} $$ An element $f \in C(K)$ is called nonnegative if $f(x) \ge 0$ for all $x \in K$. We write $f \ge 0$ if that happens.

Write $\mathbb{1}$ for the constant function with value $1$. Then of course $\mathbb{1} \in C(K)$ and $\|\mathbb{1}\| = 1$ and $\mathbb{1} \ge 0$.

$C(K)^*$ is the Banach space of bounded linear functionals on $C(K)$ with norm $$ \|\alpha\| = \sup \{|\alpha(f)|\;:\; f \in C(K), \|f\| \le 1\} $$ A functional $\alpha \in C(K)^*$ is called nonnegative iff $\alpha(f) \ge 0$ for all nonnegative $f \in C(K)$. We write $\alpha \ge 0$ if that happens.

For $a \in K$ let $\delta_a$ be the "Dirac functional" defined by $$ \delta_a (f) = f(a)\qquad\text{for all } f \in C(K). $$ Then $\delta_a \in C(K)^*$ and $\|\delta_a\| = 1$ and $\delta_a \ge 0$.

Lemma. Let $\alpha \in C(K)^*$. Suppose $\|\alpha\| \le 1$ and $\alpha(\mathbf{1}) = 1$. Then $\alpha \ge 0$.

Proof goes here

Proposition. Let $B = \{\alpha \in C(K)^*\;:\;\|\alpha\|\le 1\}$ be the unit ball of $C(K)^*$. Let $a \in K$. Then $\delta_a$ is an extreme point of $B$.

Proof. Suppose $\alpha, \beta \in B$ and $\frac{1}{2}(\alpha+\beta) = \delta_a$. We must show that $\alpha = \beta= \delta_a$. First we apply the Lemma to $\alpha, \beta$. Compute $$ 1 = |1| = \left|\delta_a(\mathbb{1})\right| =\left|\frac{1}{2}\big(\alpha(\mathbb{1})+\beta(\mathbb{1})\big)\right| \le \frac{1}{2}\left(\big|\alpha(\mathbb{1})\big|+\big|\beta(\mathbb{1})\big|\right) \le \frac{1}{2}(1+1) = 1 . $$ Equality holds in the triangle inequality, so in fact $\alpha(\mathbb{1}) = 1 = \beta(\mathbb{1})$. Also, $\alpha, \beta \in B$ by hypothesis. So by the Lemma, we conclude that $\alpha \ge 0, \beta \ge 0$.

To be continued...