Convex hull of the extreme points of the unit ball in $C(K)$ where $K$ is the Cantor set. Let $K$ denote the usual $1/3$ Cantor set and let $B=B_{C(K)}$ (here $B_{C(K)}$ = {$v \in C(K) : \|v\| \leq 1 $} denotes the closed unit ball of $C(K))$. Then how to prove that $B$ coincides with the norm closure of the convex hull of its set of extreme points?
Do I need to explicitly identify the entire set of extreme points of $B$?
Please help me out. Thank you.
 A: Hints:


*

*The Cantor set has a basis consisting of clopen sets.

*The characteristic function $[F]$ of a clopen set $F$ is continuous and $f_{F} = [F] - [K \smallsetminus F]$ is an extremal point of the unit ball. In fact, all extremal points are of this form, see below.

*Note that $f_{F} \cdot f_{G} = - f_{F \mathop{\Delta} G}$, where $F \mathop{\Delta} G = (F \smallsetminus G) \cup (G \smallsetminus F)$.

*Observe that for $x \neq y$ there exists a clopen set $F$ such that $x \in F$ and $y \in K \smallsetminus F$.

*By 3. and 4. the subspace $\mathscr{A}$ generated by $\mathscr{E} = \{f_{F}\,:\,F\text{ clopen}\}$ is a unital subalgebra of $C(K)$ separating the points of $K$, hence it is norm dense in $C(K)$ by the Stone-Weierstrass theorem. 

*Therefore the convex hull $\operatorname{conv}{\mathscr{E}} = B \cap \mathscr{A}$ is  norm dense in $B$.
Of course, it is a bit of an overkill to appeal to Stone-Weierstrass and it is a good exercise to show the density of $\mathcal{A}$ by bare hands.
Added:
There is the following general fact:

Let $K$ be a compact Hausdorff space and let $B$ be the closed unit ball of $C(K)$ then $f \in B$ is an extremal point if and only if $f(K) \subset \{\pm 1\}$. In particular $f$ is an extremal extremal if and only if $f = [F]-[K\smallsetminus F]$ for some clopen set.

It is not hard to see that $f(K) = \{\pm 1\}$ implies that $f$ is extremal. Conversely, suppose there is $x \in K$ such that $|f(x)| \lt 1$. Then there is $\varepsilon \gt 0$ and a neighborhood $U$ of $x$ such that $|f(y)| \lt 1-\varepsilon$ for all $y \in U$. By Urysohn's lemma, we can find find a continuous $g: K \to [0,1]$ vanishing outside $U$ and $g(x)=1$. Put $h_\pm = f \pm \varepsilon g$. Then $f_{\pm} \in B$ and $f = \frac{1}{2} f_+ + \frac{1}{2} h_-$ so that $f$ is not extremal.
If $f = f_{F}$ then $f(K) \subset \{\pm 1\}$, hence $f$ is extremal. If $f$ is extremal then $f(K) \subset \{\pm 1\}$, hence $F = f^{-1}(0,\infty) = \{x\,:\,f(x) = 1\}$ is a clopen set and thus $f = f_{F}$ for a clopen set.
This also shows that $K$ is connected if and only if $B$ has precisely two extremal points: the constant functions $\pm 1$.
A: why $conv\{f_F: F \;is\; clopen\}=B∩span\{f_F: F \;is\; clopen\}$ may be explained as solving the following linear equation :
Since each $f\in B∩span\{f_F: F \;is\; clopen\}$ can be written as a finite combination $\sum_{i=1}^{m}\alpha_if_{F_i}$ , the equation:$\sum_{i=1}^{2^m}\lambda_if_{K_i}=\sum_{i=1}^{m}\alpha_if_{F_i}$ has a solution $\{\lambda_i\}$ such that $\sum_{i=1}^{2^m}|\lambda_i |\leq1$, where $\{K_i\}_{i=1}^{2^m}$ is just a mutually disjoint partition of $K$.
