On a proof of the converse of the supporting hyperplane theorem

[psie]

TL;DR Summary

I'm reading a proof of the supporting hyperplane theorem, but I'm getting stuck at a small detail and I'm doubting its legitimacy.

The converse of the supporting hyperplane theorem states

Theorem. Suppose $C$ is closed with non-empty interior and has a supporting hyperplane at each point of it's boundary. Then it is convex.

Here's the "proof":

Suppose the assumptions hold and it is not convex. Then there exists a line segment joining $p,q\in C$ such that some of the line segment is outside of $C$. In particular, the line segment passes through a boundary point $b$ where's there's a hyperplane separating $p$ and $q$. But that is a contradiction that $C$ has a supporting hyperplane at $b$, namely that $C$ is contained entirely in one side of the two closed half-spaces bounded by the hyperplane. So $C$ is convex indeed.

I've been told that any proof that does not use the fact that $C$ has non-empty interior will not work, because it easy to construct counterexamples of sets that will fail if they have empty interior. I'm not sure about this proof; it's not guaranteed that the points $p,q$ lie in the interior. On the other hand, it's short and simple, maybe too good to be true? Anyway, in case this does not work, what is the correct way of proving this?

 

[fresh_42]

What is a supporting hyperplane? I imagine a closed ring or a closed circle. Both must be excluded somehow since the ring isn't convex and the circle has no interior points.

 

[psie]

What is a supporting hyperplane?

I should have included a short description. Anyway, I'm quoting from Wikipedia:

A supporting hyperplane of a set $S$ in $\mathbb R^n$ is a hyperplane that has both of the following two properties:

• $S$ is entirely contained in one of the two closed half-spaces bounded by the hyperplane.
• $S$ has at least one boundary point on the hyperplane.

 

[fresh_42]

I should have included a short description. Anyway, I'm quoting from Wikipedia:

A supporting hyperplane of a set $S$ in $\mathbb R^n$ is a hyperplane that has both of the following two properties:

• $S$ is entirely contained in one of the two closed half-spaces bounded by the hyperplane.
• $S$ has at least one boundary point on the hyperplane.

I don't understand this definition. It does not refer to single points of $S$ whereas the condition in the theorem does.

Definitely not convex.

 

[fresh_42]

I assume the condition reads: For every point $p\in \partial C$ on the boundary of $C$ there is a hyperplane $H_p$ such that a) $C$ is entirely contained in one closed halfspaces created by $H_p$ and b) $p\in H_p.$ The second condition is crucial, as well as the dependency of the hyperplane of the point $p$ is. Different point, different hyperplane. The hyperplanes form the complex hull of $C$, like in this example (but only on one half):

This rules out the ring, but not the circle. However, if there is a point $p$ within the circle, and the circle isn't a disc, then there is a boundary point of the circle within the disc that has no supporting hyperplane since condition a) is violated.

 

[fresh_42]

The proof basically follows my example: If $C$ is not convex, then it is a "ring". Then we take a boundary point of the inner ring and its hyperplane doesn't partition the space anymore in such a way that the entire $C$ is in one half.

 

[psie]

The proof basically follows my example: If $C$ is not convex, then it is a "ring". Then we take a boundary point of the inner ring and its hyperplane doesn't partition the space anymore in such a way that the entire $C$ is in one half.

Hmm, ok. I feel like what needs to be justified about the proof is the following; how can we know that the closed, non-convex set $C$, with non-empty interior, has two points such that the line segment between those two points leaves the set?

 

[fresh_42]

Hmm, ok. I feel like what needs to be justified about the proof is the following; how can we know that the closed, non-convex set $C$, with non-empty interior, has two points such that the line segment between those two points leaves the set?

Isn't it an indirect proof? We assume that $C$ is not convex. Since $C$ has a non-empty interior, it cannot be a singleton or a line segment. This means that there are at least two points $p\neq q$ on the boundary of $C$ such that the line segment $g$ between them is completely outside of $C$ except for the endpoints $p$ and $q$.

Since $C$ is on the same side of both hyperplanes, it must be in the area $\mathrm{I}$ or the area $\mathrm{II}.$ In any case, the center between $p$ and $q$ is not in $C$. If we take the perpendicular of that center on $C$ then we get another boundary point $r \in \partial \,C$ whose hyperplane does not fulfill the requirement that $C$ is entirely on one side.

 

[fresh_42]

P.S.: I think we only need one point of $b\in g$ that is outside of $C$, not the entire line segment. Suits better for "not convex".

 

[Office_Shredder]

Just to say it out loud since I don't think it has been stated yet, a circle (not a filled disk) is an example of a set which is obviously not convex but appears to pass the proof in the original post - any two points on the circle that you connect, do not pass through a boundary point anywhere that would allow you to draw a contradictory hyperplane. You need a non empty interior to force a line segment between some pair of points on the circle to pass through a boundary point somewhere inside the circle.

I'm not convinced fresh has really posted a complete proof.

This means that there are at least two points p≠q on the boundary of C such that the line segment g between them is completely outside of C except for the endpoints p and q.

It probably does, but as a matter of rigor this feels like the kind of thing that might be surprisingly tricky.

I think the first thing to observe is if p is in the interior and q is a boundary point, the whole line between p and q must be in C. Otherwise there is a boundary point of C in the interior of the line, call it r. We know r has a supporting hyperplane of C by hypothesis. Interior points cannot be on the supporting hyperplane, which means the line segment [pq] cannot be contained in it. Hence the hyperplane must separate p and q which is a contradiction (this statement also probably deserves a short proof which I will leave to psie to attempt, you have to actually think about definitions here).

This also means that if q is a boundary point and there exists a sequence of interior points $q_n\to q$, and p is any boundary point, then the line segment [pq] is also contained in C, as we know $tp+(1-t)q_n\in C$ for all n and any $t\in [0,1]$. Since C is closed taking the limit as n goes to infinity gives us $tp+(1-t)q\in C$ for all t which is what we needed.

So the only possible failure mode is that two points on the boundary which are both not the limit of interior points may fail the convexity test, which is the example that fresh covered already (though you may need to truncate your line at the first boundary point it meets)