Boundary of convex set is piecewise $C^1$ 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 $\gamma :I\to\mathbb{R}^2$?

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 $(1,0)$.
