# Dirichlet problem for manifold, how to prove $W^{1,2}_0(\Omega)$ solution is $C^{2,\alpha}(\bar{\Omega})$?

Let $$M$$ be an n dimensional Riemannian manifold without boundary. Let $$\Omega \subset M$$ be a bounded domain with smooth boundary. Let $$f \in C^{\alpha}(\bar{\Omega})$$, Consider the Dirichlet problem. $$\Delta u=f \ \text{ in }\Omega, u|_{\partial \Omega}=0.$$ Do we have a solution $$u\in C^{2,\alpha}(\bar{\Omega})$$?

The following argument is inspired by How to solve the $C^\alpha$ Poisson equation on closed Riemannian manifolds?.

Since $$f$$ is bounded, $$f\in L^2(\Omega)$$, by the existence of minizer of the energy functional (just like the Euclidean case), there exists a weak solution $$u\in W_0^{1,2}(U)$$. For $$x\in \Omega$$, let $$U$$ be a small neighborhood contained in a coordinate chart, consider $$\Delta v=f \text{ in }U, v|_{\partial U}=0.$$ By the theory of Dirichlet problem for Euclidean space, we have a solution $$v\in C^{2,\alpha}(\bar{U})$$. Since $$\Delta (u-v)=0 \text{ in } U,$$ then $$u-v\in C^{\infty}(U)$$, so $$u\in C^{2,\alpha}(U)$$. By the arbitrariness of x, we know $$u\in C^{2,\alpha}(\Omega)$$.

But how to prove that $$u\in C^{2,\alpha}(\bar{\Omega})$$?

• Going via the $L^2$/Sobolev theory seems a bit odd to me (temporarily 'forgetting' that $f$is $C^\alpha$ seems silly); I'd have thought one could stick to Holder space estimates here. Maybe I am wrong though. – DCM Jun 23 at 12:43
• @DCM If I recall correctly, this is what is done in Gilbarg-Trudinger. The existence theorem in Schauder theory is by continuity, so one needs to start with an existence theorem for the Laplacian. The best way to get this seems to be to pass through the $L^2$ theory. – Ryan Unger 21 hours ago

Your link contains a comment that $$C^\infty(\overline\Omega)$$ is not dense in $$C^{2,\alpha}(\overline\Omega)$$, which is true. But it is true that given $$f\in C^{2,\alpha}(\overline\Omega)$$ and $$\beta\in (0,\alpha)$$, there is a sequence $$f_i\in C^{\infty}(\overline\Omega)$$ converging to $$f$$ in the $$C^{2,\beta}(\Omega)$$ norm.
So let $$u_i\in C^\infty(\overline\Omega)$$ be the Dirichlet solution of $$\Delta u_i=f_i$$. We can apply Schauder estimates $$\|u_i\|_{2,\alpha;\Omega}\le C(\|u_i\|_{0;\Omega}+\|f_i\|_{0,\alpha;\Omega}),\tag{*}$$ where $$C$$ is a constant independent of $$i$$. By the maximum principle, it is possible to bound $$\|u_i\|_{0;\Omega}\le C\|f_i\|_{0;\Omega}\le C\|f_i\|_{0,\alpha;\Omega},$$ see Theorem 3.7 in Gilbarg-Trudinger. So $$(*)$$ becomes $$\|u_i\|_{2,\alpha;\Omega}\le C\|f_i\|_{0,\alpha;\Omega}.\tag{**}$$ Now we can actually choose the $$f_i$$'s so that $$\|f_i\|_{0,\alpha;\Omega}\le 2\|f\|_{0,\alpha;\Omega}$$ for any $$i$$, so $$(**)$$ just says that $$(u_i)$$ is bounded in $$C^{2,\alpha}(\overline\Omega)$$. By the Arzela-Ascoli theorem, we obtain a convergent subsequence to a function $$u$$ in $$C^{2,\beta}(\overline\Omega)$$ and since $$f_i\to f$$ in $$C^{0,\beta}(\overline\Omega)$$, $$\Delta u=f$$.
The key observation now is that even though the convergence of the $$u_i$$'s is in $$C^{2,\beta}(\overline\Omega),$$ the function $$u$$ itself is in $$C^{2,\alpha}(\overline\Omega)$$. This follows just from the uniform convergence of $$u_i$$, $$\nabla u_i$$, and $$\nabla^2u_i$$, and the fact that these all satisfy the $$\alpha$$-Hölder condition with the same constants.