# Non-trivial faces of the closed convex hull of a non-convex closed set with connected complement

I'm trying to prove or disprove a problem, but I'm struggling to make headway. Any help is appreciated.

Suppose $X$ is a Hilbert Space, and $C \subseteq X$ is closed, bounded, non-convex, and $X \setminus C$ is connected. Does the boundary of $\operatorname{conv} C$ necessarily contain a line segment?

The following are my thoughts on the problem:

The connectedness and openness of the complement gives us an open subset of the boundary of $\operatorname{conv} C$ that doesn't intersect $C$. The Bishop-Phelps theorem tells us that support points are dense in the boundary of convex sets, so there must exist support points of $\overline{\operatorname{conv}} C$ that aren't contained in $C$.

If we choose one such point, by translation, we can say it is $0$ without loss of generality. If this point is not extreme in $\overline{\operatorname{conv}} C$, then we are done. However, such points can be extreme (e.g. take $C$ to be the standard orthonormal basis in $l^2$, in which case $0$ is such a point).

I decided to look at the tangent cone of $\overline{\operatorname{conv}} C$ from the point $0$. I then picked a support point $x$ of this cone, other than $0$, but close enough that we could guarantee that it would not be in $C$. The supporting hyperplane must support the set $\overline{\operatorname{conv}} C$ at $0$. I'm hoping to show that $x \in \overline{\operatorname{conv}} C$.

This is where I get stuck. If the functional that supports at $x$ achieves its maximum on $C$, then I get to the result, but this may not be the case.

Again, any help is appreciated. Thanks in advance.

• @Desire: Ah yes, that's right. Boundedness is necessary. I'll edit to include that. – Theo Bendit Oct 29 '17 at 15:56
• You said : "The connectedness and openness of the complement gives us an open subset of the boundary of $\bar{\text{conv}(C)}$ that doesn't intersect $C.$ " ?! what if $C$ just contains two points.? – Red shoes Oct 31 '17 at 0:34
• @Redshoes Then $\overline{\operatorname{conv}}C$ is a line segment. The open line segment is open in the boundary (in the subspace topology), and doesn't intersect $C$. – Theo Bendit Oct 31 '17 at 0:46
• OK .. I misunderstood it. – Red shoes Oct 31 '17 at 1:03
• @TheoBendit I thought from the beginning you meant non empty and not a singleton; that is, prove that the boundary contains a segment. This is an interesting and I think non trivial question (and also natural to me). The example by Red shoes satisfies this, so it's ok – Del Jan 4 '18 at 18:01