Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|cite|improve this question
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

2 Answers 2

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 $\nabla u\in L^\infty$.

share|cite|improve this answer
:${W^{2,p}} \in {W^{1,\infty }}$,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

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?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.