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psie
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- 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
Here's the "proof":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.
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?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.