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!
Best Answer
You are quite close: don't look at $(1\pm r)f$, look at $f\pm r$ instead.
There is $\varepsilon \gt 0$ such that the set $U = \{x : \lvert f(x)\rvert \lt 1-\varepsilon\}$ is non-empty. Let $u \in U$ be arbitrary. Urysohn's lemma gives a function $g$ such that $g(u) = 1$ and $g|_{U^c} \equiv 0$. Take $r = \varepsilon g$. Then $f = \frac{1}{2}(f+r) + \frac{1}{2}(f-r)$ shows that $f$ is not extremal because $\lvert f(x) \pm r(x)\rvert \leq 1$ for all $x$.