0
$\begingroup$

I am studying Pazy's book "Semigroups of Linear Operators and Applications to Partial Differential Equations" and when considering an operator like:

A linear differential operator, $$A : W^{2m,p}(\Omega)\cap W^{m,p}_0(\Omega) \subset L^p(\Omega) \to L^p(\Omega),$$ where $\Omega$ is a bounded open set on $\mathbb{R}^n$ with smooth boundary, $1 < p < \infty$. If one assumes that $A$ is strongly elliptic, in order to show (via his argument) that $-A$ generates an analytic semigroup one must assume that $0\in \rho(A)$ (the resolvent set) (i.e, $A$ is invertible). So, my question is: is this operator necessarily invertible? He does not mention it anywhere.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Well, $0\in\rho(A)$ means $A$ is invertible, by definition. But this condition is certainly not necessary for $A$ to generate an analytic semigroup. $\endgroup$
    – Michael Renardy
    Commented Jun 17, 2020 at 19:42
  • $\begingroup$ @MichaelRenardy, yeah, I am sorry, I think I wrote it badly, the point is, is it true that $0\in\rho(A)$? I didn't want to write it as a claim, this is indeed my question. I totally agree that it is not necessary to show that $A$ generates an analytic semigroup, but it is definitely necessary on his argument. $\endgroup$
    – L.F. Cavenaghi
    Commented Jun 17, 2020 at 19:52

0

Reset to default

You must log in to answer this question.