When are spectral values a pole of the resolvent?

Question

Let $(T_t)_{\geq 0}$ be a $C_0$-semigroup on a Banach space $X$ with the generator $A.$

What are some conditions sufficient for $\lambda \in \sigma(A)$ to be a pole of resolvent?

I'm looking for conditions on the semigroup or $A.$

Background: I'm trying to prove some stability results for $(T_t)_{t \geq 0}$ and it seems to me that not knowing beforehand that the spectral bound $s(A)$ is a pole of $R(\cdot, A)$ is a roadblock in most cases. Hence I'm looking for conditions that ensure it is a pole, so I can somehow try to prove those conditions.

Answer 1

One sufficient condition for this is that the semigroup is eventually compact. This is proved in Corollary V.3.2 in the book by Engel & Nagel.

Section 5.8 of the book by Taylor is a good source for other conditions.

I'm aware of $2$ sufficient conditions in the case $X$ is a Banach lattice and the resolvent set of $A$ is non-empty:

$1.)\, u \in E_+$ and $D(A^n)\subseteq E_u$ for some $n \in \mathbb{N}.$

$2.)\, u \in E_+$ and $T(t_0)E\subseteq E_u$ for some $t_0>0.$

The proof of these two can be found in Corollary 2.4 and 2.5 of a paper by D. Daners and J. Glück.

Answer 2

A $C_0$ semigroup $S(t)$ on a Banach space is uniformly bounded in operator norm near $0$. That's enough to be able to prove there there is a $\rho > 0$ such that $e^{-\rho t}S(t)$ is a bounded $C_0$ semigroup. A bounded $C_0$ semigroup $S(t)$ has a generator $A$ with resolvent $$ -(A-\lambda I)^{-1}= \int_0^{\infty}e^{t(A-\lambda I)}dt=\int_0^{\infty}e^{-\lambda t}S(t)dt. $$ That gives you a bound on the resolvent in terms of the uniform bound on $e^{-\rho t}S(t)$, and the bound is applicable for $\Re\lambda > \rho$. If there is a pole for the resolvent on the line $\Re\lambda=\rho$, then it is a pole of order $1$ because of the bound that is obtained by the above integral.