Suppose that $K$ is a compact convex set in a topological vectorial space locally convex such that $K$ is the closed convex hull of its exposed points $(x_\mu)$. I would like to know if it is true that there exists a set of functionals $(f_\mu)$, such that $f_\mu$ exposes $x_\mu$, that separes the points of $K$, i.e. for all distinct points x and y of K, it exists $\mu$ such that $f_\mu(x)\neq f_\mu(y)$?
Someone has a tip to show this or un conterexample?
Thank you for your help
This answer works with extreme points instead of exposed points and is therefore not an answer to the original question! (I'll leave it here for reference)
The statement is
Let $I$ be an uncountably infinite index set. Consider the cube $C=\prod_{n\in I} [0,1]\subset\mathbb R^I$ in the product topology. By Tychonov $C$ is compact, because all $[0,1]$ are compact. Because $\mathbb R$ is locally convex, the product space is also locally convex.
Assume there is some continuous functional $f:\mathbb R^I\to\mathbb R$ that "exposes" the extreme point $0$. Then we can assume $0\equiv f(0) < f(e_i)$ for all $i\in I$, where $e_i$ is the $i$-th unit vector and also an extreme point of $C$. For all $J\subset I$ let $e_J$ be the vector with elements $$ (e_J)_j = \begin{cases} 1,& j\in J\\ 0,& j\notin J\end{cases}$$ One can see that $$ \mathbb R\ni f(e_I) = \sum_{i\in I} f(e_i) \overset{\text{$I$ uncountable, $f(e_i)>0$}} = \infty $$ which is a contradiction.
Proof: Let $\Lambda$ be the set of all finite subsets of $I$ ordered by inclusion. Then $(e_J)_{J\in\Lambda}$ is a net that converges to $e_I$ (because it converges pointwise). Therefore $\lim_{J\in\Lambda} f(e_J)\to f(e_I)$. We get the contradiction $$ \infty = \sum_{i\in I} f(e_i) \equiv \lim_{J\in\Lambda} \sum_{j\in J} f(e_j) \overset{\text{$J$ finite}}= \lim_{J\in\Lambda} f(\sum_{j\in J} e_j) = \lim_{J\in\Lambda} f(e_J) = f(e_I) \in \mathbb R $$
Alternative proof without nets: Let $\varepsilon>0$ be arbitrary and let $U$ be an open neighbourhood of $e_I$ such that $|f(v-e_I)|<\varepsilon$ for all $v\in U$. By definition of the product topology we can choose $U$ such that $U=\prod_{i\in I_0} U_i\times\prod_{i\in I\setminus I_0}\mathbb R$ for a finite subset $I_0\subset I$ and open neighbourhoods $U_i$ of $1\in\mathbb R$.
Let $N\in\mathbb N$ be arbitrary and let $J_N\supset I_0$ be finite with $\sum_{j\in J_N} f(e_j)>N$ (exists because $I$ is uncountable and all $f(e_i)>0$). Then $J_N\supset I_0$ implies $e_{J_N}\in U$, therefore $$f(e_I)=f(e_{J_N})+f(e_I-e_{J_N}) > N - \varepsilon$$ Since $N$ was arbitrary we get a contradiction to $f(e_I)\in\mathbb R$ being finite.
