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Assume $M$ is a Riemannian manifold, and $\Omega $ is a bounded domain. Consider the Poisson equation: $$\Delta u = f \qquad \text{with }u \in {W^{1,2}}$$ What is the worst regularity of $f$ which ensures $u$ is locally Lipschitz?

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Along the lines of what Ray Yang suggested, for $f \in L^p(\Omega)$ with $\Omega \subset \mathbb{R}^n$ and $p > n$ the gradient in fact has a Holder $1-n/p$ modulus of continuity. One obtains this either by estimating the Newtonian potential or by applying $W^{2,p}$ estimates and Morrey's embedding. –  Connor Mooney Jul 10 '13 at 12:40
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Of course the regularity theory is local, then is is the same on a smooth manifold than in an open set of $\mathbb{R}^n$. Then the answer of your question is found in Sobolev embedding and regularity theory. If $f\in L^p$ with $p>1$ then $u\in W^{2,p}$ , see chapter 9 of Gilbarg&Trudinger. Then $W^{2,p}\subset W^{1,\infty}$ if $p> n$. But using Lorentz space, see Grafakos's book , you can improved a bit this result. $f\in L^{n,1}$ implies $u \in W^{2,(n,1)}$, then with the improved Sobolev embedding: $W^{1,(n,1)}\subset L^{\infty}$, we get $\nabla u\in L^\infty$.

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:${W^{2,p}} \in {W^{1,\infty }}$,if p>n?I just know it's Holder continuous,how can it be Lipshitz?Please show the book where I can find this proposition. –  jiangsaiyin Jul 12 '13 at 9:25
    
$\nabla u\in W^{1,p}\subset C^{0,\eta}\subset L^\infty$ it is the classical Sobolev embedding, see Adams on Sobolev spaces. –  Paul Jul 12 '13 at 13:29
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On $\mathbb{R}^n$, to be locally $C^1$ you only need $f$ to be bounded (this is done by estimates on the Newtonian potential). I don't see why things should be different on a nice manifold?

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