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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

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It is true that $f\in W^{k,2}$ implies that $u\in W^{k+2,2}$ (see theorem 4 here), but this is not true if you replace the power 2 with $\infty$ (see remark 9 here).

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. 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.

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  • $\begingroup$ I recently saw a new citation stating that it is true for $\partial \Omega \in C^{2,\alpha}$ for $\alpha > 0$, but again no primary source. Still your answer and sources are very valuable. $\endgroup$ Aug 18, 2014 at 20:07
  • $\begingroup$ @StefanReiterer, unfortunately I don't have primary sources for the counterexample either. The remark 9 I cited states: "Details can be found in any textbook in real harmonic analysis." Maybe regularity is better under some extra assumptions like positivity of $f$; this would explain the discrepancy. $\endgroup$ Aug 18, 2014 at 20:33

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