Characteristics of a convex set if its boundary is convex

Question

If $A$ is a convex set in $\mathbb R^n$, when is its boundary convex as well?

I think $\partial A$ must be either contained in a hypersurface or must equal $\mathbb R^n$.

Answer 1

Here is a proof for $n=2$. If $\partial A = \emptyset$ or $\partial A = \{x_1\}$, you are done (although I don't think that a singleton can be the boundary of a convex set in two dimensions). Let $x_1,x_2\in\partial A$ and assume that there is some $x_3\in\partial A$ such that $x_1,x_2,x_3$ form a non-degenerated triangle $\Delta$. Then the whole closed triangle is a subset of $\partial A$. That means:

$(*)$ Each neighborhood of each point in $\Delta$ contains points in $A$ and $A^c$.

In particular, there are infinitely many point of both $A$ and $A^c$ in $\Delta$. Now, choose a point $z$ in the interior of $\Delta$ which belongs to $A^c$. Draw any line through $z$. Then it is not possible that there are points from $A$ on this line on both sides from $z$ (since then $z\in A$, which is a contradiction). Choose points $a,b\in A\cap\Delta$ that form another non-degenerated triangle $\Delta'$ with $z$. Then any line from $z$ to a point on $[a,b]$ contains a point from $A$ on one side, hence on the other side there are only points from $A^c$. Therefore, a whole sector between the lines $[a,z]$ and $[b,z]$ (the one opposite to $\Delta'$) contains only points in $A^c$, which contradicts $(*)$.

EDIT: This proof can easily be raised to higher dimensions.

Answer 2

Yes my guess is correct. In fact according to my book one can prove that a convex set in Euclidean space contains a ball if it is not contained in an (n-1)-dimensional affine subspace.