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If $A\subseteq\mathbb{R}^n$ is an non empty set and $H$ is the convex hull of $A$, how can I prove that the boundary of $H$ consists only of points that lie in the boundary of $A$ and segments that join points from the boundary of $A$?

Student
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  • Is this the exact statement to be proved? If $A$ is a stellated dodecahedron, for example, the convex hull $H$ is an icosahedron; what points on the boundary of $A$ are endpoints of a segment through the centroid of one of the faces of $H$? – David K Jun 13 '15 at 18:11
  • @DavidK I don't quite understand. Could you explain what could be a correct "version" of this statement. E.g. Is it true that $\partial H \setminus \partial A$ consists of hyperplanes? I'm more interested in figuring out the infinite dimensional version of this where the convex hull is replaced with the closed convex hull. I'll be happy to ask a separate question if you think that is required. – Canine360 Aug 02 '21 at 05:33
  • @Canine360 I don't know what the correct statement should have been. Your idea might be correct; at least I have not thought of a counterexample yet. – David K Aug 02 '21 at 11:50

1 Answers1

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Prove this by induction on the dimension $n$. When $n=1$ this is clear.

Now for $n > 1$, by considering closest point projection, you can prove that every point on the boundary of a convex body has a supporting hyperplane $L$. So let $x \in \partial H $ and let $L$ be a supporting hyperplane at $x$ (we know the supporting hyperplane at a point on the boundary of a convex set need not be unique).

Then $L\cap H$ is the convex hull of $L\cap A$. By the induction hypothesis, we have either $x \in L\cap A$ or $x = \lambda y_1 + (1-\lambda)y_2$ for $y_1, y_2 \in \partial (L \cap A)$. Now notice that $\partial (L \cap A) \subset L \cap \partial A$.

Canine360
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TYS
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  • Why is $n=1$ clear? I don't see how... – Student Jun 13 '15 at 19:56
  • In 1D, your convex hull will be a line segment, with the end points belonging to $A$. – TYS Jun 13 '15 at 20:02
  • That's the point. For $n=1$ why we have that $\partial A\supset\partial H$? – Student Jun 13 '15 at 20:12
  • Any point in $H$ must be a convex combination of points in $A$, this implies in particular that it must be less than or equal to some point in $A$ and greater than or equal to some point in $A$. – TYS Jun 13 '15 at 20:44
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    I don't think this works. $x$ need not be on the boundary of $L \cap H$ viewed as a subspace of $\mathbb{R}^{n-1}$, e.g. triangle with basis supported by the hyperplane. – B. Freitas Apr 20 '17 at 12:24