Exercise 2.27 in Boyd and Vanderberghe:

Suppose the set C is closed, has nonempty interior, and has a supporting hyperplane at every point in its boundary. Show that C is convex.

Seems to me one approach is to prove that the intersection of all the supporting hyperplanes is exactly C. Clearly this intersection contains C. Geometrically the other direction seems obvious, but any hint how to argue it rigorously?



2 Answers 2


Hint: Suppose wlog 0 is in C. Suppose $p$ is not in C. Take $0 < t < 1$ such that $t p$ is on the boundary of C, and consider a supporting hyperplane at $t p$.

  • $\begingroup$ Sure, I agree, but it seems one needs to argue that p is on the other side of that supporting hyperplane, which is the formal part that I'm missing :-) How might one argue that? $\endgroup$
    – BoB
    Apr 4, 2011 at 19:31
  • 5
    $\begingroup$ @BoB It's 5 years late, but I was puzzled by the same question and resolved it. So I'll post a comment here, in case other new comers also have similar confusion. Suppose $a^T(x-tp)=0$ is the supporting hyperplane at $tp$. WLOG, assume the origin is an interior point of $C$. Since the origin is in $C$, it follows that $a^T(-tp)<0$, which implies that $a^Tp>0$. Now note that $a^T(p-tp)=(1-t)a^Tp>0$. So $p$ must be on the other side of the hyperplane, which is the desired contradiction. $\endgroup$
    – syeh_106
    Oct 7, 2016 at 11:09
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    $\begingroup$ I have plagiarised your answer here math.stackexchange.com/a/2784052/27978. $\endgroup$
    – copper.hat
    May 16, 2018 at 18:21

Consider a hypothetical line segment that starts and ends at points p, q inside C, but temporarily passes outside C somewhere in the middle. To exit the set, the line segment must passes through the boundary at some point b, and the boundary hyperplane there separates p from q , thus a contradiction.


edit: some changes for clarity edit2: add image


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