Extreme points of $B_{C(K)^{*}}$ I am currently working on the following problem.
Let $K$ be any compact Hausdorff space. Show that any extreme point of $B_{C(K)^*}$ is of the form $\pm \delta_s$ where $\delta_s$ is the probability measure defined on the Borel sets of $K$ by $\delta_s(B)=1$ if $s \in B$ and $0$ otherwise. 
I know those sets form an extreme point, but don't know how to do the reverse point. Any help would be appreciated.
Thanks in advance!
 A: While a correct approach is to invoke the Riesz representation theorem, here is an argument tailored for the specific question. It is an operator algebra approach.
For any $L\in C(K)^*$ and $f\in C(K)$ define $L_f\in C(K)^*$ by $L_f(g)=L(fg).$ Note $L_{fg}=(L_f)_g.$ A nice property is that given any non-negative $f_1,\dots,f_k\in C(K)$ with $\sum f_i=1,$ we have $\sum\|L_{f_i}\|=\|L\|.$ The direction $\sum\|L_{f_i}\|\geq \|L\|$ is the triangle inequality. For the other direction take a small $\epsilon>0.$ Pick $g_i$ with $\|g_i\|_{\infty}\leq 1$ and $L_{f_i}(g_i)\geq (1-\epsilon)\|L_{f_i}\|.$ Since $|\sum f_i(x)g_i(x)|\leq \sum_i f_i(x)\cdot\max_i |g_i(x)|\leq 1$ for each $x\in K,$ we have $\|\sum f_ig_i\|_\infty\leq 1.$ So $(1-\epsilon)\sum\|L_{f_i}\|\leq \sum L_{f_i}(g_i)=\sum L(f_ig_i)\leq \|L\|.$ Then take $\epsilon\to 0.$
Let $L$ be an extreme point of $B_{C(K)^*}.$ Let $\mathcal A$ be the set of open subsets of $K$ of the form $f^{-1}((0,\infty))$ where $f\geq 0$ satisfies $L_f=0.$
I first claim that $\bigcup \mathcal A\neq K.$ For any finite list $f_1,\dots,f_k\geq 0$ with $\bigcup f^{-1}((0,\infty))=K$ we have $f_1+\dots+f_k>0.$ Let $g=1/(f_1+\dots+f_k).$ Then $L=(L_{f_1}+\dots+L_{f_k})_g$ is non-zero and hence some $L_{f_i}$ is non-zero. By compactness, $\bigcup \mathcal A$ does not cover $K.$
Now consider any two distinct points $x_1,x_2\in K.$ I claim that at least one of these points is in $\bigcup\mathcal A.$ By regularity of $K$ there is an $f:K\to[-1,1]$ with $f(x_1)=-1$ and $f(x_2)=1.$ Set $f_1=\max(-f,0)$ and $f_2=\max(f,0).$ So $f_1,f_2$ are $[0,1]$-valued functions satisfying $f_1f_2=0$ and $f_1(x_1),f_2(x_2)\neq 0.$ We can decompose $L$ as $L=L_{f_1}+L_{f_2}+L_{1-f_1-f_2}.$ As mentioned earlier, we have the nice property $\|L_{f_1}\|+\|L_{f_2}\|+\|L_{1-f_1-f_2}\|=\|L\|=1.$ 
Since $L$ is an extreme point this implies $L_{f_i}=\|L_{f_i}\|L.$ But then $0=L_{f_1f_2}=(L_{f_1})_{f_2}=\|L_{f_1}\|L_{f_2}=\|L_{f_1}\|\cdot\|L_{f_2}\|L$ so $L_{f_i}$ is zero for some $i.$ But $L_{f_i}=0$ implies $x_i\in \bigcup\mathcal A,$ which is what we wanted to show.
We have shown that $\bigcup \mathcal A$ consists of a single point. Let's call it $x.$
I claim $L(f)=0$ for any $f$ with $f(x)=0.$ Take a small $\epsilon>0.$ Define $f_\epsilon(x)=\max(-\epsilon,\min(\epsilon,f(x))$ - clamping to $[-\epsilon,\epsilon].$ The set $K_\epsilon=\{y\in K\mid |f(y)|\geq\epsilon\}$ is a compact subset of $\bigcup\mathcal A,$ so there exist $g_1,\dots,g_k\geq 0$ with $L_{g_i}=0$ and such that the function $g=g_1+\dots+g_k$ is strictly positive on $K_\epsilon.$ Let $h$ be a continuous extension of $1/g$ from $K_\epsilon$ to $K,$ which exists by the Tietze extension theorem. Then $L(f-f_\epsilon)=L((1-gh)(f-f_\epsilon))=0$ because $f-f_\epsilon$ is supported on $K_\epsilon$ where $1-gh=0.$ This gives $|L(f)|=|L(f_\epsilon)|\leq\epsilon\to 0.$
For any $f\in C(K)$ we get $L(f-f(x)1)=0,$ giving $L=L(1)\delta_x.$ Since $1=\|L\|=|L(1)|\|\delta_x\|$ the coefficient $L(1)$ must be $\pm 1.$
