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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;


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?


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up vote 5 down vote accepted

Suppose $\varepsilon=|f(t_0)|<1$ for some $t_0$. By continuity there exists $V\ni t_0$, open, with $|f(t)-f (t_0)|<(1-\varepsilon)/2$ for all $t\in V$. Now let $g$ be a continuous function, supported on $V$, such that $g(t_0)=1-\varepsilon$ and $|g|\leq(1-\varepsilon)/2$.

Then $|f\pm g|\leq1$, and $f=\frac12(f+g)+\frac12(f-g)$, so $f$ is not extreme.

Edit: regarding your second question, it is something you already know. $C(X)$ is a unital C$^*$-algebra, so any element is a linear combination of four unitaries, by the usual trick of writing a complex number $a+ib$ with $|a+ib|\leq 1$ as $$ \begin{align} a+ib=&\frac12\left([a+i(1-a^2)^{1/2}]+[a-i(1-a^2)^{1/2}]\right)\\ &+\frac{i}2\left([b+i(1-b^2)^{1/2}]+[b-i(1-b^2)^{1/2}]\right) \end{align} $$

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Thanks very much! – Hui Yu Nov 8 '12 at 16:31
You are welcome :) – Martin Argerami Nov 8 '12 at 18:06
«The usual trick»? I consider myself lucky that I had never seen that identity before! :-) – Mariano Suárez-Alvarez Mar 4 '15 at 19:24

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 $x$.

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thanks for mentioning Urysohn's lemma. – Hui Yu Nov 8 '12 at 16:32

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