## Differential operator estimate

Suppose I have an a priori estimate of the form

$\|u\|\substack{H^1(\mathbb{R}^n)} \le C \| Pu \|_{L^2(\mathbb{R}^n)},$

where $H^1=W^{1,2}$ is the first order $L^2$ Sobolev space (which I suppose is irrelevant for my question). Moreover $P = \sum a_{(\alpha )}(x) \partial ^{(\alpha )}$ is a partial differential operator (say of order 1). Now, let $\Omega \subset \mathbb{R}^n$. Can I claim that

$\|u \| \substack{H^1(\Omega )}\leq C \| Pu \|_{L^2(\tilde{\Omega })}$

for any $\tilde{\Omega }$ such that $\overline {\Omega } \subset \tilde{\Omega }$ (the bar means closure)?

-
not in general, because $P$ will have a finite dimensional but non trivial kernel in general. This is because you did not impose boundary conditions. – Denis Serre Nov 29 2011 at 12:19
Yes, you are right. Actually my $P$ is of the form $P-z$ and I happen to know that $P-z$ is invertible with a bounded inverse. Especially the kernel of $P-z$ is trivial. – Alex A Nov 29 2011 at 12:36
The kernel of which operator is trivial? The one acting on functions on $R^n$ or the the one acting on functions on $\overline{\Omega}$? – Deane Yang Nov 29 2011 at 15:53
Sorry, I meant the original one acting on $L^2(\mathbb{R}^n)$ has trivial kernel. Moreover I do not need that $\Omega$ is compactly contained in $\tilde{\Omega }$. If $\tilde{\Omega }$ is an $\varepsilon$-neighborhood of $\Omega$ that would suffice. – Alex A Nov 30 2011 at 8:47