If $\|(z-A_n)^{-1}-(z-A)^{-1}\|\to 0$ and $e^{-tA}$ decays, will $e^{-tA_n}$ decay for large $n$? Let $A,A_n$ be closed operators on a Hilbert space $\mathcal H$ and assume that


*

*$-(A-{\rm Id}_{\cal H})$ generates a strongly continuous contraction semigroup 

*$-A_n$ generates a strongly continuous contraction semigroup for every $n$.

*$\|(z-A_n)^{-1}-(z-A)^{-1}\|\to 0$ as $n\to\infty$ for every $z$ with $\Re(z)<0$.


Can one show that there exists a $\mu\in (0,1)$ such that $-(A_n-\mu)$ generates a bounded semigroup for $n$ large enough?
If not, are there simple counterexamples?
 A: I think, I've found a counterexample. Let $\mathcal H:=L^2(\mathbb R)$ and consider the multiplication operators
$$(Au)(t) = \left(1+it\right)u(t)$$
$$(A_nu)(t) = \left(1+it-\tanh(\tfrac{t}{n})^2\right)u(t)$$
on the common domain $\mathcal D=\{u\in L^2(\mathbb R) : tu\in L^2(\mathbb R)\}$. 
Then $-A+1$ generates a contraction semigroup by the Lumer-Phillips theorem and its spectrum is the essential range of $(1+it)$ which is a vertical line through the point 1.
Similarly, $A_n$ generates a contraction semgroup by Lumer-Phillips, however, here the tanh-term prevents us from subtracting an identity operator. For every $n$ the spectrum of $A_n$ is a curve passing through 1 and approaching the imaginary axis for $\text{Im}(z)\to\pm\infty$. 
The resolvents of $A$, $A_n$ are simply given by the multiplication operators
$$(z-A)^{-1}f = \frac{f}{z-1-it}$$
$$(z-A_n)^{-1}f = \frac{f}{z-1-it+\tanh(\frac{t}{n})^2}$$
for $\text{Re}(z)<0.$
It is easy to check that $\|(z-A)^{-1}-(z-A_n)^{-1}\|_{\mathcal L(L^2(\mathbb R))}\to 0$ as $n\to\infty$.
However, since $\tanh(t/n)^2\to 1$ as $t\to\pm\infty$ for every $n$, one has $\sigma(A_n-\mu)\cap\{\text{Re}(z)<0\}\neq\emptyset$ for every $\mu>0$.
