# How to solve the $C^\alpha$ Poisson equation on closed Riemannian manifolds?

To be specific, suppose $M$ is a closed oriented manifold, $g$ is a Riemannian metric of $M$. Let $\Delta_g$ be the Laplace-Beltrami operator w.r.t. $g$.

Prove: Suppose $f\in C^\alpha(M)$ satisfies $\int_M f\, dVol_g=0$, then there exists a function $u\in C^{2,\alpha}(M)$ such that $\Delta_g u=f$ in $M$, and $u$ is unique up to plus a constant, here $0<\alpha<1$.

My attempt is that, firstly use $D(u):=\frac{1}{2}\int_M( |\nabla u|^2+fu)dVol_g$ is a convex functional with a lower bound on $W_0^{1,2}(M)$ to show that there exists a weak solution $u\in W^{1,2}(M)$, next use the $L^2$-regularity theory to show that $u\in W^{2,2}(M)$, but I don't know how to improve the regularity of $u$ further. (Actually, I can use the method to prove that if $f$ is $C^\infty$, then $u$ is also $C^\infty$, but I cannot extend this result to $C^\alpha$ case.)

Another attempt is Schauder estimate. However, in Gilbarg and Trudinger's book they assume that $u\in C^{2,\alpha}(M)$ already to get some interior derivative norm bound of $u$, while I don't know how to establish $u\in C^{2,\alpha}(M)$. They give a continuity method to ensure that, but it seems their discussion works for domains in Euclidean space, not for manifolds. Therefore, I want to split the question into coordinate charts, but I failed, because I don't know how to use the condition $\int_M f\, dVol_g=0$ and how to give boundary conditions in every coordinate charts.

Since I'm a novice in PDE, my presentation of the problem might have some errors. Please correct them by comments or answers. Also, any comments or answers are welcome.

Remark: I've already asked this question on math.stackexchange.com, but nobody replied. Maybe this question is not so suitable for MO, but I really want to get an answer.

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By the way, this is not a Dirichlet problem, since there is no boundary. You can say this is a Poisson equation on a closed manifold. –  timur Jul 20 '12 at 17:35
@timur Thanks for pointing that out~ –  Yuchen Liu Jul 21 '12 at 2:46

Careful. C^\infty is not dense in C^{\alpha}. Nonetheless you can first produce an L^2 solution then use Schauder estimates to show that the solution is C^{2,\alpha}. This last step is localizable -- you can multiply by a smooth cutoff function (and use some preliminary estimates to say that the solution is C^2, for example) and transfer to a coordinate chart, then apply the estimates and continuity method in Gilbarg&Trudinger. There are other cleaner ways to proceed: the solution u = G f is given by a pseudodifferential operator of order -2 and it is not hard to prove (see Taylor, Vol. III) that such a psido is bounded from C^\alpha to C^{2,\alpha}.

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@Rafe Mazzeo: I think my answer is based on your first attempt: math.stackexchange.com/questions/172887/… –  Yuchen Liu Jul 28 '12 at 2:49
what do you mean by "my answer is based on your first attempt"? –  YangMills Jul 28 '12 at 16:03
@YangMills: I mean roughly I use Rafe's method with a little difference: first use $L^2$-theory to find a weak solution $u$, then use Schauder estimate and continuity method to find a $C^{2,\alpha}$ solution $v$ locally, and their difference is a weak solution of $\Delta (u-v)=0$ hence $(u-v)$ is $C^\infty$. Therefore, $u$ is in $C^{2,\alpha}$. –  Yuchen Liu Jul 29 '12 at 3:21
@Rafe: Can you please have a look at my answer? I think I found a way around the density problem. –  timur Aug 23 '12 at 1:19
That is correct, though I guess you need to add a few details that the sequence of solutions $u_j$ which are bounded in $C^{2,\alpha}$ and convergent in $L^2$ (or, say, $C^2$) have limit which lies in $C^{2,\alpha}$. –  Rafe Mazzeo Aug 23 '12 at 2:33

Elliptic regularity in Hölder spaces is precisely what you need here.

If you want to prove it along the lines you described, using $L^2$-theory, then you can first establish that smooth $f$ leads to smooth solution $u$. Then you approximate $f$ in the $L^2$-norm by smooth functions with uniformly bounded $C^{\alpha}$-norms (this is important as smooth functions are not dense in Hölder spaces), and use the Schauder estimate. It suffices to do everything locally.

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