# Show that if a convex $C$ has a supporting hyperplane at every point of its boundary, then it's convex

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?

Thanks!

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 $t p$.
• @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. Oct 7 '16 at 11:09