Are there good bounds for the norm of a uniformly continuous semigorup $e^{tA}$ in terms of its generator $A$?

Question

Are there good bounds for the norm of a uniformly continuous semigorup $e^{tA}$ in terms of its generator $A$? I'm only looking for references. In the standard literature, I wasn't able to find something useful.

Answer 1

As long as $A$ is normal (which is implied by self-adjointness), we have the continuous (or, more generally, Borel) functional calculus and spectral mapping theorem, which guarantee for $f \in C(\sigma(A))$: $$\|f(A)\| = \sup_{x \in \sigma(A)} |f(x)|$$

Applying this to $f(x) = e^{tx}$ gives you your answer.

$$\|e^{tA}\| = \sup_{x \in \sigma(A)} |e^{tx}|$$ That’s the exact answer for an arbitrary $t$ and normal $A$.

Note that the right-hand side is bounded above by $\sup_{x \in \sigma(A)} e^{|tx|} = e^{|t|\|A\|}$. So you have the general inequality $$\|e^{tA}\| \le e^{|t|\|A\|}$$ This is an equality if $t$ and $A$ are positive, but there could be a drastic difference between the two sides in other cases (say, if $t$ is a big negative number).

Edit: I assumed here that you were working with operators on a Hilbert space. If you’re in the more general Banach space setting, then you want the holomorphic functional calculus, but the above discussion should nearly work in that setting. The main issue is that the norm and spectral radius might not be the same there.