Let $K$ be a compact convex subset of locally convex topological vector space $E$. Let $U$ be an open subset of $K$.
Is $conv(U)$ (the convex hull of $U$) an open subset of $K$ ?
You see, it is well know that if $U$ is an open subset of $E$, then $conv(U)$ is an open subset of E. the argument for this is based on the fact that the addition is an open map for $E \times E$ to $E$, hence if you take $(a_1,...,a_n)$ such that $\sum a_i =1$, then the set of $\sum a_i u_i$ for $u_i \in U$ is an open set, and hence $conv(U)$ is an union of open set. But when we restrict ourselves to $K$ the former argument no longer hold : for exemple if $U = K$, $n=2$ and $a_1=a_2= \frac{1}{2}$, then the set of $\sum a_i u_i$ is the set of non-extremal point of $K$, which may be non open even in a finite dimensional case... but still i can't find any counterexample to my question.
Note : Using local convexity we can show that it is enough to proove that if $U$ and $V$ are two open convex subset of $K$ then the set of $a u +(1-a)v$ with a in $[0,1]$ is open in $K$.
Thanks !