Diffeomorphisms of compact convex subspaces of $\mathbb{R}^n$ with non-empty interior

Question

It is well-known that any convex compact subspace $X \subset \mathbb{R}^n$ with non-empty interior is homeomorphic to the closed unit ball $D^n$. The usual method I know of is to use the Minkowski functional $\rho$ of $X$ and show that the map $f:X \to D^n$ given by $$ f(x) = \frac{\rho(x)}{\|x\|} x$$ is a homeomorphism. The function $f$ need not be a diffeomorphism, as in the case of $X$ being the unit square. In the particular case of the unit square one can show that there is in fact no diffeomorphism to the unit disk, using the diffeomorphism-invariance of corner points of a manifold with corners, but the map $f$ defined above has only a "few" points where it is not differentiable. Inspired by this, I have the following questions:

1. Let $Y \subset X$ be the set of points where the function $f$ above is not differentiable. For example when $X$ is the unit square $$Y = \{(x, y) : x \in [-1, 1], y = x \text{ or } y = -x\}.$$ Does $Y$ always have measure $0$? Is $Y$ a meagre set?
2. Let $\partial Y = Y \cap \partial X$ where $\partial X$ is the topological boundary of $X$. Is $\partial Y$ always finite or countable?
3. Does there always exist a homeomorphism $g$ (not necessarily of the form above) such that the set of points where $g$ is not differentiable is meagre or measure 0?

I would also appreciate partial answers and/or references!

Answer 1

You first have to decide what it means for closed sets $X_1$ and $X_2$ to be diffeomorphic. The usual notion is to find a diffeo from an open neighbourhood of $X_1$ to an open neighbourhood of $X_2$, which is a bijection from $X_1$ to $X_2$.

Then, if $X_1$ is diffeomorphic to a closed disc, necessarily its boundary has to be smooth. If $X_1$ is only convex, then the boundary might be non-differentiable, but only on an measure zero set (Alexandrov Theorem, see for instance http://people.math.sc.edu/howard/Notes/alex.pdf).