When the boundary of convex set is convex?

Question

I think the boundary of a convex set cannot be convex, since the segment between tow points on the boundary lies inside the convex set, so the boundary does not contain the segment?

Answer 1

Take $C = [0,\infty) \subset \mathbb R$. This is a convex set whose boundary is $\{0\}$ which is convex.

In $\mathbb R^n$ any affine subspace $A$ of dimension $< n$ is convex and we have $A = bd(A)$. Moreover any affine halfspace $H$ is convex and $bd(H)$ is an affine hyperplane which is convex.

You see that the boundary of a convex set may very well be convex. Analyzing the above examples we see that that the case that $C$ is convex such that $C = bd(C)$ is a trivial case. So let us restrict to the case that $C$ has interior points. The halfspace example (including $C = [0,\infty) \subset \mathbb R$) shows that even then $bd(C)$ may be convex.

You do not say anything about $C$, but I guess you consider the case that $C$ is closed (in which case $bd(C) \subset C$). If not, observe that the closure of $C$ is also convex and has the same boundary as $C$. See Is closure of convex subset of $X$ is again a convex subset of $X$?

Here is a positive result:

If $C$ is compact with non-empty interior, then $bd(C)$ is not convex.

To see this, let $x \in C$ an interior point. Consider any line $L$ through $x$. Since $L$ is convex and closed, the set $D = L \cap C$ is convex and compact. Thus $D$ is a line segment. Its two boundary points are contained in $bd(C)$, but the "open interval" between the boundary points is not contained in $bd(C)$. Thus $bd(C)$ is not convex.