Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
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 $$ \begin{align} a+ib=&\frac12\left([a+i(1-a^2)^{1/2}]+[a-i(1-a^2)^{1/2}]\right)\\ &+\frac{i}2\left([b+i(1-b^2)^{1/2}]+[b-i(1-b^2)^{1/2}]\right) \end{align} $$

share|cite|improve this answer
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 $x$.

share|cite|improve this answer
thanks for mentioning Urysohn's lemma. – Hui Yu Nov 8 '12 at 16:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.