Bounded weak derivative 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.
 A: 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.
A: $\newcommand{\R}{\mathbb{R}}
\newcommand{\al}{\alpha}$
This is a partial answer: Assuming additionally that $D^\al f$ exists in $C(\R^n)$ for $|\al|=1$, let us show that $D^\al f$ is bounded. 
Indeed, without loss of generality $\al=e:=(1,0,\dots,0)\in\R^n$. 
We have $|f|\le M$ and $|D^{2e}f|\le M_2$ for some real $M,M_2\ge0$. 
By fixing arbitrary values of the last $n-1$ arguments of the functions, we reduce the consideration to the case $n=1$, so that $D^e f=f'$, the derivative of $f$. Then, by an appropriate version of Taylor's theorem, for any real $x$ and some $c_\pm=c_\pm(x)\in[-1,1]$ we have 
\begin{equation}
 f(x\pm1)=f(x)\pm f'(x)+c_\pm M_2/2, 
\end{equation}
whence, by subtraction,
\begin{equation}
 |f'(x)|\le\tfrac12|f(x+1)-f(x-1)|+M_2/2\le M+M_2/2,
\end{equation}
so that $|f'|\le M+M_2/2$.
