# Showing regularity $(u \in C^2(\overline{\Omega}))$ for the Laplacian Problem.

I am reading Functional Analysis, Sobolev Spaces and Partial Differential Equations, by Haim brezis, and I am a bit confused about the proof. The theorem is stated as:

Let $\Omega \subset \mathbb{R}^n$ be an open set of class $C^2$ with $\partial \Omega$ bounded. Let $f \in L^2(\Omega)$ and $u \in H_0^1(\Omega)$ satisfies the weak formulation (Laplacian): $$\int_\Omega \nabla u \nabla \varphi + \int_{\Omega}u \varphi=\int_{\Omega}f\varphi \quad \forall \varphi \in H_0^1(\Omega)$$ Then $u \in H^2(\Omega)$ and $||u||_{H^2} \leq C ||f||_{L^2}$. Furthermore, if $\Omega$ is of class $C^{m+2}$ and $f \in H^m(\Omega)$, then: $$u \in H^{m+2} \ and \ ||u||_{H^{m+2}} \leq C ||f||_{H^m}$$ In particular, if $f \in H^m(\Omega)$ with $m > n/2$, then: $$u \in C^2(\overline{\Omega})$$

The statement is clear. The problem is that the proof only shows that if $f \in H^m(\Omega)$ then $u \in H^{m+2}(\Omega)$, and it doesn't even mention the last part (Does not show that $u \in C^2(\overline{\Omega})$).

I suppose that this is because it is trivial? Because I haven't been able to show it. Can someone clarify this to me?

• Thanks for the answer. There is a Sobolev Embedding stated in my book as "If $m-\frac{n}{p} \geq 0$ is not an integer (...) $W^{m,p}(\Omega) \subset C^k(\Omega)$, where $k=[m-\frac{n}{p}]$ ([ ] denotes integer part)". The condition: is not an integer ruins the argument, as I cannot prove it. – D1X May 14 '16 at 20:59
• @D1X I don't understand this concern. By assumption you are looking at $u \in H^r$ with $r= m+2$ and $m>n/2$. So $r-n/2>2$ and the theorem can be applied directly. The point about the quantity not being an integer just means that if, e.g., $m-n/p=3$, you cannot conclude $u\in C^3$. For that you would need $> 3$. You can conclude $u\in C^k$ for $k< 3$, however (or, mor precisely $u\in C^{k,\alpha}$ for any $\alpha <1$. I suggest you check the proof of the theorem you are looking at to verify this statent or have a look at the wikipedia page I referred to, where this is detailed, as well. – Thomas May 15 '16 at 7:40