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In connection with reading a proof of a theorem in a real-analysis textbook, I've stumbled upon the following theorem.

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 have not studied much convex-analysis, and I've been trying to find a beginner's friendly proof of this theorem. I've looked up Boyd's and Vandenberghe's textbook, and I've seen their proof of this theorem, but I think some parts of it are unclear, as highlighted in this question. From my search, in particular inspired from this answer, I have collected the following "proof", which I would love to see verified (for the definition of a supporting hyperplane, see Wikipedia).

Proof. Suppose the assumptions hold and it is not convex. Then there exists points $p,q$ in the interior of $C$ such that a line segment joining $p,q$ leaves the set somewhere in-between. In particular, the line segment must pass 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 $C$ must be convex.

Consulting chat, 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 to what extent this proof makes use of the fact that $C$ has non-empty interior. Moreover, I'm doubting its legitimacy because it's not guaranteed that the points $p,q$ will lie in the interior. If this proof does not work, can it be improved to work?

psie
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  • The theorem in my real-analysis textbook that uses the above theorem is the one proved here. At the end, they conclude that every boundary point has a supporting hyperplane, from which it must follow that it is convex. – psie Jan 30 '24 at 13:49
  • You just rephrased the answer you linked which has a slick proof and a picture. What exactly do you want us to verify? Finally: I hope you have not consulted chat*GPT*. – Kurt G. Jan 30 '24 at 13:56
  • Yes, it is inspired from the answer you linked. What I'm asking is, how can we know that there are points in the interior of $C$ such that there is a line segment between them that leaves the set somewhere along the line? This proof claims that such points exist in the interior, but this is not justified. And no, I consulted the chat on this site. – psie Jan 30 '24 at 14:08
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    Good. Strictly speaking that short proof shows that that the interior of $C$ is convex. From that it follows that the closure of that interior is convex. This does not necessarily show that $C$ is convex. Example: take a closed disk to which you attach a line segment. This is not convex. For sets who are not the closure of their interior I think another proof is needed. – Kurt G. Jan 30 '24 at 14:50

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