7
$\begingroup$

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 !

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Good example, Thank you ! $\endgroup$ – Simon Henry Oct 3 '12 at 7:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.