I want to show that for $X$ compact and Hausdorff, any $f\in C(X)$ satisfying $|f(x)|=1$ for all $x\in X$ is an extreme point of the unit ball of $C(X)$. Here, $C(X)$ is the space of continuous complex-valued functions on $X$.
I tried to prove it by contradiction, writing $f = \frac12(f_1+f_2)$ for distinct $f_1,f_2$ in the unit ball of $C(X)$. Then by the triangle inequality, $|f_1(x)|=|f_2(x)|=1$ for all $x\in X$. But I cannot see how this leads anywhere. Am I missing something obvious?
Anyway, I am quite stuck on this and would appreciate a hint.
Best Answer
Hints:
$1).\ $ This seems to be more or less your idea: first prove the more general fact that a point $x$ is an extreme point of the unit ball in a Banach space if and only if for all $t\in X$ we have $\|x\pm t\|\le 1\Rightarrow t=0.$ For one direction, set $y=x+t$ and $z=x-t$ and apply the definition. For the other, write $x=\alpha z+(1-\alpha)w$ and suppose $\alpha\neq 0,1.$
$2).\ $ Suppose $|f(x)|=1.$ Take a $g\neq 0$ so there is an $x_0$ such that $g(x_0)\neq 0$. Now consider what happens if $\|f\pm g\|\le 1.$