0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

This is a consequence of the Sobolev Embedding Theorems, which is explained, e.g., here

(and it's not trivial, but well known. If this is a textbook what you are reading it should be mentioned somewhere).

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – D1X May 14 '16 at 20:59
  • 1
    $\begingroup$ @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. $\endgroup$ – Thomas May 15 '16 at 7:40
  • $\begingroup$ You are right, thank you, Thomas. $\endgroup$ – D1X May 16 '16 at 12:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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