Let $u$ be solution of $-\Delta u = f$ in $\Omega$ and $\frac{\partial u}{\partial n} = 0$ on $\partial \Omega$.


Is it true that if $f \in L^{\infty}(\Omega)$ then $u \in W^{2,\infty}(\Omega)$?
(Assuming a 'nice' boundary of course.)


I think that I already found such a result in quite an old book, but I currently don't have proper literature at hand, and would need the following:


1. Which are the assumptions.
2. Where could I find a proper citation (I've forgotten it ...)


Help would really appreciated



W^{2,∞} regularity of solutions of Poisson's equation if the right hand side is in L^{∞}
It is true that $f\in W^{k,2}$ implies that $u\in W^{k+2,2}$ (see theorem 4 [here](https://www.math.ualberta.ca/~xinweiyu/527.1.11f/lec14.pdf)), but this is not true if you replace the power 2 with $\infty$ (see remark 9 [here](https://www.math.ualberta.ca/~xinweiyu/527.1.11f/lec11.pdf)).


I think the boundary conditions are not very important for interior regularity.
Suppose you had a counterexample $v\in W^{1,\infty}\setminus W^{2,\infty}$ in the ball $B(0,2)$ solving $-\Delta v=g$ for $g\in L^\infty$.
Let $\phi\in C^\infty(B(0,2))$ be a compactly supported function with $\phi|\_{B(0,1)}\equiv1$, and define $u=v\phi\in W^{1,\infty}\setminus W^{2,\infty}$.
Now $u$ vanishes in a neighborhood of the boundary and if we define $f=\phi g-v\Delta\phi-2\nabla v\cdot\nabla\phi$ (which is in $L^\infty$), then $-\Delta u=f$.


I assumed here that there is a counterexample with local Lipschitz regularity, but such regularity holds in [great generality](http://arxiv.org/abs/1004.1101).
I also implicitly assumed that the non-$W^{2,\infty}$ singularity of $v$ occurs in $B(0,1)$, but this is just a matter of scaling and translating any counterexample.


