Let $X$ be a compact Hausdorff space and $C(X)$ be the space of continuos functions in sup-norm.
I read in Douglas' Banach algebra techniques in operator theory that the followings are equivalent:
1)$f\in C(X)$ is an extreme point of the unit ball;
and
2)$|f(x)|=1$ for all $x\in X$.
It is easy to show that 2) implies 1). However, I am unable to show the converse.
I could show that $f$ is extreme implies $\|f\|=1$ but this is far from what we need.
My guess: if 2) is false. We try to construct a nonnegative function $r$ on $X$ which is strictly positive when $|f(x)|\neq 1$ and meanwhile \begin{equation} |(1+r)f|\le 1 \end{equation}on $X$. Then we can decompose \begin{equation} f=\frac{1}{2}(1+r)f+\frac{1}{2}(1-r)f. \end{equation} If $|f|$ is bounded away from $0$, then $r=-1+1/|f|$ would be a good choice, but I cannot rule out the case when $f$ vanishes at certain points.
Can somebody help? Thanks!
Also a related problem, Douglas then says these extreme points of the unit ball spans the entire $C(X)$. Can somebody also give a hint on this?
Thanks!