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Is this statement true, and, if so, does it have a name?

Given a bounded closed convex set, $C\subset \mathbb R^n$, let $C^{int}$ be the interior of $C$ and $C^{bd}=C-C^{int}$ be the boundary of $C$. Given two points $x,y\in C^{bd}$, any "shortest" path between $x,y$ in $\mathbb R^n-C^{int}$ is entirely inside $C^{bd}$.

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It seems true to me, don't know the name though. Proof idea: assume wlog that $0\in C^{int}$, so in particular $C$ is star-convex with centre $0$. Given any $x,y\in C^{bd}$ and any path $\gamma\in\mathbb{R}^n$, construct a homotopy of $\gamma$ which "pushes" in down to $C^{bd}$ along rays coming from $0$. Then try to argue that the new path is shorter, so that any path in $\mathbb{R}^n-C^{int}$ can be projected to a shorter path on $C^{bd}$ –  you Mar 29 '12 at 16:30
    
Yeah, I was thinking it was intuitively obvious, but thought I might be missing something. –  Thomas Andrews Mar 29 '12 at 16:33
    
@you need more than star convexivity to show that the "projected" map to the boundary is continuous, or even well-defined, but yes, this is the basic intuition I had. –  Thomas Andrews Mar 29 '12 at 16:38
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There's a different proof of the result you ask about: one can show that for every closed convex set $C$ in $\mathbb{R}^n$ there is a nearest point projection $\mathbb{R}^n \to C$ which is $1$-Lipschitz, so for any path in $\mathbb{R}^n \smallsetminus C^{\rm int}$ between two points on $C^{\rm bd}$ there is one of at most the same length lying on $C^{\rm bd}$. –  t.b. Mar 29 '12 at 20:33

1 Answer 1

up vote 6 down vote accepted

I am not aware of any name for this property. However, it can be proved rather quickly. I won't deal with the existence of geodesic (some compactness argument may work, but it is not obvious to me which one), and only prove that any geodesic in $X := \mathbb{R}^n-C^{int}$ between points of $C^{bd}$ must lie in $C^{bd}$.

Let $x$, $y$ be in $C^{bd}$. Assume that there exists a geodesic $\gamma : [0,1] \to X$ between $x$ and $y$. If this geodesic does not lie in $C^{bd}$, then there must be some point $z = \gamma (t_0) \in \gamma ([0,1])$ which does not belong to $C^{bd}$.

By the Hahn-Banach theorem, I can find an hyperplane $H$ which strictly separates $C$ and $z$. By the intermediate value theorem, there must be some times $t_1$, $t_2$ with $0 < t_1 < t_0 < t_2 < 1$ and such that $\gamma (t_1)$ and $\gamma (t_2)$ both belong to $H$. The geodesic $\gamma$ restricted to $[t_1, t_2]$ must also be a geodesic.

The shortest path between two points in $H$ is a line. So, $z$ cannot belong to the geodesic between $\gamma (t_1)$ and $\gamma (t_2)$. We get a contradiction. Hence, the geodesic between $x$ and $y$ must lie in $C^{bd}$.

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