I asked this question on [Math SE](https://math.stackexchange.com/q/3683600/385720) but didn't receive any response.


Let $(T\_t)$ be a $C\_0$-semigroup on a Banach space $X$ with generator $A.$ If $\lambda\_0\in \mathbb{C}$ is such that $e^{\lambda\_0 t}$ is a pole of $R(\cdot,T(t)),$ then $\lambda\_0$ is a pole of $R(\cdot,A)$ and $$k(e^{\lambda\_0 t},T(t))\geq k(\lambda\_0,A)\tag{1}$$ where $k(\cdot,\cdot)$ denotes the order of the corresponding pole. This is proved for example in Theorem IV.3.6 in the [book by Engel and Nagel](https://link.springer.com/book/10.1007/b97696).


**My question:** Suppose we a priori know that $\lambda\_0$ is a pole of $R(\cdot,A)$ and $e^{\lambda\_0 t}$ is a pole of $R(\cdot,T(t))$ for all $t\geq 0.$ Are there any *known* conditions which guarantee an equality in $(1)$ for at least some $t>0?$



Equality in spectral inclusion theorem
Here is a positive solution if the semigroup is eventually compact:


Consider, say, the open right halfplane
$$
 H := \{\lambda \in \mathbb{C}: \, \operatorname{Re}\lambda > \operatorname{Re}\lambda\_0 - 1\}.
$$
Then $H$ contains only finitely many spectral values of $A$, and all these spectral values are poles of the resolvent with finite-rank spectral projections. So if we denote the sum of these finitely many spectral projections by $P$, then the range $PX$ is a finite dimensional space, and the restriction of the semigroup to the complementary subspace $\ker P$ only contains spectral values with real part $\le \operatorname{Re}\lambda\_0 - 1$.


All these properties follow from Corollary V.3.2 in the semigroup book of Engel and Nagel (2000).


The restriction of the semigroup to $\ker P$ is also eventually compact, so the spectral mapping theorem holds for it [op. cit., Corollary IV.3.12(i)]. Hence, $e^{t\lambda\_0}$ is not in the spectrum of $T\_t|\_{\ker P}$ for any $t$.


So we only need to deal with the restriction of the semigroup to the finite dimensional space $PX$. Hence, we may asume from now on that $X$ itself is finite dimensional and that $A$ is a matrix. After a coordinate transformation, $A$ is in Jordan normal form. The matrix $A$ has distinct eigenvalues $\lambda\_0, \dots, \lambda\_m$, and for each $k \in \{0, \dots, m\}$ there are Jordan blocks $J\_{k,1}, \dots, J\_{k,\ell\_k}$.


Since $A$ is the direct sum of all these Jordan blocks, we can compute $e^{tA}$ by computing $e^{tJ\_{k,i}}$ for all Jordan blocks separately.


Now, if $t \in [0,\infty)$ is such that the numbers $e^{t\lambda\_0}, \dots, e^{t\lambda\_m}$ are all distinct, then out of all direct summands
$$
 e^{tJ\_{0,1}} , \dots, e^{tJ\_{0,\ell\_0}} , \quad \dots \quad , e^{tJ\_{m,1}}, \dots, e^{tJ\_{m,\ell\_m}}
$$
of $e^{tA}$, only the matrices $e^{tJ\_{0,1}} , \dots, e^{tJ\_{0,\ell\_0}}$ have the eigenvalue $e^{t\lambda\_0}$. Hence, the dimension of largest Jordan block in $e^{tA}$ that belongs to the eigenvalue $e^{t\lambda\_0}$ (i.e., the order of the pole $e^{t\lambda\_0}$ of $R(\cdot,e^{tA})$) cannot be larger than the largest dimension of the matrices $e^{tJ\_{0,1}} , \dots, e^{tJ\_{0,\ell\_0}}$ - which is the same as the largest dimension of the matrices $J\_{0,1}, \dots, J\_{0,\ell\_0}$ (and this is, in turn, the order the pole $\lambda\_0$ of $R(\cdot,A)$).


This solves the question for all those times $t$ for which the numbers $e^{t\lambda\_0}, \dots, e^{t\lambda\_m}$ are distinct.


