# $C^{2, \alpha}$ regularity for elliptic equations with Neumann boundary conditons

Say $\Omega\subseteq \mathbb{R}^n$ is a bounded open set and $0<\alpha<1$. I need some $C^{2, \alpha}(\overline\Omega)$ regularity result for elliptic equations with Neumann boundary conditions but that is what I find in Jurgen Jost' book:

If $u$ is a weak solution of $\Delta u=f$ in $\Omega$ and $f\in C^\alpha(\Omega)$, then $u\in C^{2,\alpha}(\Omega)$..........(1)

But it says that we can identify $C^{2, \alpha}(\Omega)$ with $C^{2,\alpha}(\overline\Omega)$ here:
http://www.math.ucsd.edu/~bdriver/231-02-03/Lecture_Notes/Holder-spaces.pdf.

However, Jost used both $C^{2, \alpha}(\Omega)$ and $C^{2,\alpha}(\overline\Omega)$ (when he is doing regularity for Dirichlet boundary) in his book, which indicated these two spaces are not the same?

Or do we have $u\in C^{2,\alpha}(\overline\Omega)$ in (1)? Otherwise where can I find some $C^{2,\alpha}(\overline\Omega)$ regularity result with Neumann boundary conditon?

Let $\Omega\subset\mathbb{R}^N$ be a open bounded domain and assume that $f:\Omega\to\mathbb{R}^N$ satisfies $$\frac{|f(x)-f(y)|}{|x-y|^\alpha}\leq M,\ \forall\ x,y\in\Omega\tag{1}$$

for some fixed $\alpha\in (0,1)$. We can conclude from $(1)$ that $f$ is uniformly continuous in $\Omega$.

Because $f$ is uniformly continuous in $\Omega$, we can extend it in a unique way to the boundary of $\Omega$ and we get a function (still denoted by $f$) $f:\overline{\Omega}\to\mathbb{R}$ which is continuus. The question is: Is this function Holder continuous with exponente $\alpha$?

The answer is yes: take $x,y\in\partial\Omega$ with $x\neq y$ and choose $x_n,y_n\in\Omega$ such that $f(x_n)\to f(x)$ and $f(y_n)\to f(y)$. Note that $$\frac{|f(x)-f(y)|}{|x-y|^\alpha}=\lim_{x_n,y_n\to x,y}\frac{|f(x_n)-f(y_n)|}{|x_n-y_n|^\alpha}\leq M$$

The last equality is true because of the continuity of the function $g(x,y)=\frac{|f(x)-f(y)|}{|x-y|^\alpha}$.

We conclude that (in terms of identification) $C^\alpha(\Omega)=C^\alpha(\overline{\Omega})$.

Now here, come the crucial observation. If $\Omega$ does not have a nice boundary, then it's nonsense to talk about $C^1(\overline{\Omega})$ or $C^2(\overline{\Omega})$, therefore, if we ask some regularity on the boundary, to wit, $\partial\Omega \in C^{2,\alpha}$, hence, the same argument given above we have that $C^{2,\alpha}(\Omega)=C^{2,\alpha}(\overline{\Omega})$

Take a look in Theorem 11.2.2. of the book you mentioned from Jost. Note that the boundary regularity is required.

• Thank you very much for your reply. It helps me a lot. – user100356 Oct 12 '13 at 19:00
• You are welcome @user100356 – Tomás Oct 12 '13 at 19:03