TITLE: Is each closed convex set a manifold with corners?
QUESTION [10 upvotes]: Assume that $C$ is a convex set in $\mathbb{R}^{n}$ with non empty interior.
Then consider its closure, is it a smooth manifold with corners?
Edit: 
1) The closure of $C$ should be a smooth manifold with corners equipped with the induced smooth structure from $\mathbb{R}^{n}$.
2) With non empty interior I mean that $C$ contains a non empty open set.

REPLY [14 votes]: See de Rham's "cutting corners" curve.  



Picture from an MO answer by Bill Thurston; also see a description there.
The limiting curve is $C^1$ but not $C^2$.  It has a tangent everywhere, but curvature zero almost everywhere.  When you say "manifold with boundary" what do you require?
Plug: English translation of de Rham's paper is in my book Classics on Fractals

REPLY [14 votes]: Let $C \subset S^1 \subset \mathbb{R}^2$ be a Cantor set. Let $H_C$ be its convex hull, the smallest closed subset of $\mathbb{R}^2$ containing $C$. Then $H_C$ is not a manifold with corners. Its boundary $\partial H_C$ is $C^1$ at every point except for the countably many points which are endpoints of components of $S^1-C$, these points forming a dense subset of $C$. Also, $\partial H_C$ fails to be $C^2$ at every point of $C$. But $H_C$ is homeomorphic to the closed 2-disc, indeed there is an ambient homeomorphism $\mathbb{R}^2 \mapsto \mathbb{R}^2$ taking $H_C$ to the unit closed disc.