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Let $\Omega\subset\mathbb{R}^2$ a bounded set with Lipshitz continuous boundary.

If $$z\in L_0^2(\Omega)=\{v\in L^2(\Omega):\int_\Omega v\,dx=0\},$$

it is true that exists $\phi\in H^{1+\delta}(\Omega)$, $\delta\in (\frac{1}{2},1]$, such that $\Delta \phi=\nabla\cdot (\nabla \phi)=z$ and $||\nabla \phi||_{\delta,\Omega}\leq c||z||_{0,\Omega}$?

I known that when $\Omega$ is a convex set, what is written above holds with $\delta=1$, but i'm not sure if it holds in a more general case. Thanks.

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No, I believe pure homogeneous Neumann problem does not have higher than $H^{1+\frac{1}{2}}$ regularity, assuming the worst case, in nonconvex domain. – Shuhao Cao Apr 28 '13 at 20:19

As stated, the problem can be reduced to the convex case by letting $B$ be a ball containing $\Omega$, and extending $z$ by $0$ to $B\setminus \Omega$. Or, just convolve $z$ with the fundamental solution of the Laplacian on $\mathbb R^2$; the convolution will be in $H^2$ and the shape of $\Omega$ does not matter.

The comment by Shuhao Cao mentions the Neumann problem. If $\partial v/\partial n$ is required to vanish on $\partial\Omega$ then the story is different, but I don't see any such condition in the OP. (Maybe it's missing.)

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