Does the operator need to be closed in the Lumer-Phillips Theorem?

48 Views Asked by At

enter image description here

enter image description here

Query. The Lumer-Phillips Theorem (Theorem 3.4.5) requires as a hypothesis that the operator $A$ is closed? I ask this because in Theorem 3.15, it is not required to be closed, but the operator $\overline{A}$ is closed by definition. So I suppose Theorem 3.4.5 needs to require $A$ to be closed? (Theorem 3.4.5 in Vector-valued Laplace transforms and Cauchy problems, and Theorem 3.15 in One-parameters semigroups for linear evolution equations) Thank you very much.

1

There are 1 best solutions below

0
On

The closedness of the operator in Theorem 3.4.5 is included in the two conditions a) and b). Indeed, a dissipative operator $A$ is closed iff $(\lambda-A)D(A)$ is closed in $X$ for one (and hence all) $\lambda>0$. See the book of Pazy for more details.