This is not a homework problem, so I am not sure whether this has a "good" answer or not--I came up with this question when I am now learning functional analysis and wonder whether my "freshman's intuition" for exponential works.

If $Q$ is some bounded linear operator on some Banach space(and maps to the same space), then since it is bounded, we can define the exponential operator $$P(t)=\exp(tQ)$$ for every $t>0$. It is again a bounded linear operator for every fixed $t$.

My question is that what is a sufficient condition on $Q$(if necessary, better!) for $\{P(t)\}$ to converge in operator norm to some operator as $t\rightarrow\infty$?

My intuition tells me there should be something of $Q$ being negative--at least non-positive, resulting in $P(t)\rightarrow 0$ or some other operator. If the spectrum of $Q$ is contained in $\{z\in\mathbb{C}:Re(z)<0\}$, will this convergence happen? However I failed to link this convergence to $Q$'s spectrum...

This is not a homework problem, so I am not sure whether this has a "good" answer or not. I came up with this question when I am now learning functional analysis and wonder whether my "freshman's intuition" for exponential works.

If $Q$ is some bounded linear operator on some Banach space(and maps to the same space), then since it is bounded, we can define the exponential operator $$P(t)=\exp(tQ)$$ for every $t>0$. It is again a bounded linear operator for every fixed $t$.

Question. What is a sufficient condition on $Q$ (if necessary, better!) for $\{P(t)\}$ to converge in operator norm to some operator as $t\rightarrow\infty$?

My intuition tells me there should be something of $Q$ being negative--at least non-positive, resulting in $P(t)\rightarrow 0$ or some other operator. If the spectrum of $Q$ is contained in $\{z\in\mathbb{C}\, :\, \operatorname{Re}(z)<0\}$, will this convergence happen? However, I failed to link this convergence to $Q$'s spectrum.