TITLE: Bounded weak derivative
QUESTION [5 upvotes]: Let $f \in L^{\infty}$ be a function such that $f$ and the weak derivatives $D^{\alpha}f\in L^{\infty}$ exist for all $\vert \alpha\vert\ge 2$. Does this imply that also $D^{\alpha}f$ with $\vert \alpha\vert=1$ exists in $L^{\infty}?$
It sounds somehow natural and I could not find a counterexample, but I don't quite know.

REPLY [4 votes]: It depends on the geometry of the domain. Consider e.g. the function $f(x,\,y) = xy$ on $\{|xy| < 1\}$, which is bounded with constant Hessian on this domain but has linearly growing gradient. Or take $f$ defined on a union of disjoint intervals in $\mathbb{R}$ of length $2^{-k}$ to be linear with slope $2^k$ on each and vanishing at the midpoints.
The result will hold locally by mollification and interpolation, and globally on domains where interpolation estimates between $C^0$ and $C^2$ hold, e.g. convex domains (like $\mathbb{R}^n$, as Iosif shows) or bounded $C^2$ domains.