I have seen that, if $f \in L^2 (\Omega)$ with $\Omega$ bounded and regular then if $u$ solves $- \Delta u =f$ in $H^1_0 (\Omega)$, then $u \in H^2$. This is standard regularity theory. I've also seen that these arguments work, in general, if $f \in L^2 _{loc} $ and $u \in H^1 _{loc}$, and the thesis would be that $u \in H^2 _{loc}$, where this time the domain can be unbounded.
My question is: does global regularity also hold for unbounded domains? For instance, $f \in L^2 (\mathbb{R}^n)$ and $u \in H^1 (\mathbb{R}^n)$ solves the previous equation, then do I have $u \in H^2 (\mathbb{R}^n)$? Or do I only have $ u \in H^2 _{loc} (\mathbb{R}^n)$?
I have a feeling that this is true, and it could be proved by contradiction, but I wanted to ask before attempting to prove something false.
Also, if the thing above is true, does it also hold for global $L^p$ regularity? Do you have any suggested reference where thsese facts are proved? (if true, of course)
Best Answer
Depending on your background, I would try one of the two:
Recover second order derivatives of $u$ by applying Riesz transforms to $f$. One uses Fourier transform to justify that Riesz transforms are bounded operators on $L^2$.
Apply integration by parts twice: $\int \partial_{ii} u \partial_{jj} u = \int \partial_{ij} u \partial_{ij} u$. This shows that $\int |\Delta u|^2 = \int |D^2 u|^2$.
In both cases, one needs to check some details related to regularity of $u$ (or rather lack thereof) and the sense in which derivatives are taken.