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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 !

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up vote 1 down vote accepted

This is not true.

Let $K$ be, say, the unit ball in a Hilbert space, equipped with its weak topology. Let $\xi$ be a norm-one linear functional, and consider $U := K \cap \{\xi \in [-1, -c) \cup (c, 1]\}$ for $0 < c < 1$. It is an open set, however, its convex hull is $\{x \in K \, | \, \Vert \mathrm{pr}_{\xi^\perp} x \Vert < \sqrt{1 - c^2}\}$, where $\mathrm{pr}_{\xi^\perp}$ is the orthogonal projection on the hyperplane $\xi^\perp$. This set is clearly not open in the weak topology.

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Good example, Thank you ! –  Simon Henry Oct 3 '12 at 7:04
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