# $Conv(Ex((C(X))_1))$ is dense in $(C(X))_1$?

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.

Thanks!

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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

## 2 Answers

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).

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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 $C(X)_1$.

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