Boundary of convex set is piecewise $C^1$

Question

Let $K$ be a convex and compact subset of $\mathbb{R}^2$. Is it true that the boundary of $K$ can be parameterized by a piecewise $C^{1}$ application $\gamma :I\to\mathbb{R}^2$?

Answer 1

No, this is not true. In fact, the set of corners can be dense on the boundary.

Here is a simple example that the boundary is not piecewise $C^1$: Define $C$ to be the closed convex hull of the points $$ x_n = (\cos(1/n), \sin(1/n), \ n\in \mathbb N, $$ $(-1,0)$, $(-1,0)$, and $(0,1)$. Then $C$ is compact and convex. But does not have piecewise smooth boundary near $(1,0)$.

Answer 2

Another example is to look at Julia sets, and take convex hulls of these. You can easily end up with a situation where the boundary has countably many corners, and thus accumulation points of corners.