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 compact Hausdorff, and $C(X)$ the space of continuous functions over $X$. Denote the closed unit ball in $C(X)$ by $(C(X))_1$, then it can be shown $f$ is an extreme point of $(C(X))_1$ if and only if $|f(x)|=1$ for all $x\in X$.

In Douglas's book, we are asked to show that the convex hull of extreme points of $(C(X))_1$ is dense in $(C(X))_1$. By using a theorem of Fejer we can show this is true when $X=[0,1]$. But I do not know how to do it for general $X$.

Since it is only the 7th problem in the first chapter, I am guessing the solution should be elementary.


share|cite|improve this question
It seems to me wrong. Take $X=[0,1]$ and consider $C(X)$ over $\mathbb{R}$, then you have only two extreme points and their convex hull is a segment which is of course not dense in $(C(X))_1$ – Norbert Nov 14 '12 at 17:10
up vote 2 down vote accepted

Here's a more algebraic approach. The basic idea is this:

If $\lVert f\rVert \leq 1$ and $f$ is real-valued then we can write $g=f+i\sqrt{1-f^2}$. Clearly $|g|^2=g\bar g =1$ so that $g$ is an extreme point of the unit ball. Moreover, $f=\frac12(g+\bar{g})$ so that $f$ is a convex combination of two extreme points. This already shows that every function is a linear (but not necessarily convex) combination of extreme points. However, with some cleverness one can refine this idea to the following statement:

If $\lVert f \rVert \lt 1 - \frac{2}{n}$ then $f$ is the closed convex hull of $n$ extreme points. More precisely, $$ f = \frac{1}{n} (g_1 + \dots + g_n) $$ with $g_k$ extremal for all $1 \leq k \leq n$.

This gives the desired statement by observing that the convergence $(1-\frac{3}{n})f \to f$ exhibits every $f$ with $\lVert f \rVert \leq 1$ as a limit of convex combinations of extreme points.

The statement that every element of norm $\lt 1-\frac{2}{n}$ is an average of $n$ unitary elements holds in an arbitrary unital $C^\ast$-algebra and is not more difficult to prove in general than in the commutative case (given the continuous functional calculus). This result is due to Russo and Dye with a simple proof due to Gardner, see his article. A more detailed proof can be found in Pedersen, Analysis Now, Proposition 3.2.23. Since unitaries are always extreme points (see Pedersen, C*-algebras and their automorphism groups, Proposition 1.4.7), the statement of your question holds in an arbitrary unital $C^\ast$-algebra. Assuming that $A$ has a unit is necessary: the unit ball has an extreme point if and only if $A$ is unital, (see Theorem 1.6.1 in Sakai's C*-algebras and W*-algebras).

share|cite|improve this answer
Great answer! I especially like the many references you give in the end! Thanks! – Hui Yu Nov 17 '12 at 15:58

Norbert pointed out in a comment that this is false for real scalars, so we assume that we consider complex scalars $C(X) = C(X,\mathbb{C})$.

I haven't thought about it deeply, so maybe there are easier approaches. We are going to apply the Stone-Weierstrass theorem.

From the description of extreme points the following are straightforward:

  1. If $f$ is extremal then so are its complex conjugate $\bar{f}$ and $e^{i\alpha} f$ for $\alpha \in \mathbb{R}$.
  2. If $f$ and $g$ are extremal then $f\cdot g$ is extremal.
  3. If $f$ and $g$ are convex combinations of extreme points then so is $\frac{1}{2}(f+g)$.
  4. If $f$ and $g$ are convex combinations of extreme points then so is $f \cdot g$.

    To see the last point, let $f = \sum_{j=1}^m \lambda_j f_j$ and $g = \sum_{k=1}^n \mu_k g_k$ with $\sum \lambda_j = 1 = \sum \mu_k$ and $\lambda_j, \mu_k \geq 0$ and $f_j, g_k$ extremal. Then $f \cdot g = \sum_{j=1}^m \sum_{k=1}^n (\lambda_j \mu_k) \, f_j g_k$ with $\sum_{j=1}^m \sum_{k=1}^n \lambda_j \mu_k = \big( \sum_{j=1}^m \lambda_j \big) \big(\sum_{k=1}^n \mu_k\big)= 1$ and $\mu_j \lambda_k \geq 0$ and $f_j g_k$ is extremal by 2.

It follows from this that the linear span of the extreme points (= the positive multiples of convex combinations of the extreme points by the above) is a self-adjoint subalgebra $A$ of $C(X)$ and since $A$ contains the constant functions, it remains to prove that elements of $A$ separate points of $X$:

If $X$ is empty or has only one point, the statement in the question is very easy to prove. So, let $x_1,x_2 \in K$ be two distinct points. Choose a continuous function $f \colon X \to [0,1]$ such that $f(x_1) = 0$ and $f(x_2) = 1$. Set $g(x) = 1 + i f(x)$. Then $g$ vanishes nowhere, so $h = \frac{g}{\lvert g\rvert}$ is well-defined, continuous and an extreme point of the unit ball. We have $h(x_1) \neq h(x_2)$, so $A$ separates points.

We can now apply Stone-Weierstrass theorem to $A$, showing that $A$ is dense in $C(X)$. From this it follows easily that $A \cap C(X)_1$ is dense in $C(X)_1$.

share|cite|improve this answer
If you had logged in before editing, I think you could have done it without approval. – Ross Millikan Nov 14 '12 at 23:31

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.