Assume $A$ is a closed linear operator on a Banach space $X$ and is densely defined. Assume the spectral bound $s(A) = \sup\{Re\lambda: \lambda\in \sigma(A)\}$ is finite. For example, if $A$ is the generator of a semigroup $T(t)$ with growth bound $\omega$, i.e., $\T(t)\\le Ce^{\omega t}$, we have $s(A) \le \omega$. Let $\alpha > s(A)$ be a real numer. Let $R(\lambda, A) = (\lambda  A)^{1}$ be the resolvent. Do we have the following spectral representation of the semigroup $e^{tA}$? $$e^{tA} = \frac{1}{2\pi i}\int_{\Gamma}e^{tz}R(z,A)dz,$$ where $\Gamma \subset \mathbb{C}$ is the vertical line $\{z\in\mathbb{C}, Re(z) = \alpha\}$.

1$\begingroup$ In which sense would you expect the integral to converge? (Absolute convergence does certainly not hold.) $\endgroup$– Jochen GlueckMay 12, 2020 at 9:02

$\begingroup$ Yes, you are right, the integral cannot be absolutely convergent. My question is whether the integral is welldefined (convergent in the usual functional calculus sense) and equals the semigroup generated by $A$. $\endgroup$– Jacob LuMay 13, 2020 at 1:07

$\begingroup$ I guess there is no way since the resolvent cannot decay faster than $1/z$. The above integral does not converge even for multiplication operators. $\endgroup$– Giorgio MetafuneMay 13, 2020 at 8:10

1$\begingroup$ @GiorgioMetafune: Well, since the exponential function oscillates along the imaginary axis, there might be a chance for this to converge as an improper integral. I wouldn't be too surprised if it holds, say, in finite dimensions (or even for analytic semigroups). But there are a lot of details to check... $\endgroup$– Jochen GlueckMay 13, 2020 at 11:46

1$\begingroup$ Alright, I've done the calculation in the onedimensional case. First note that there's the factor $\frac{1}{2\pi i}$ missing in front of the integral. But this little issue aside, we can use integration by parts and Cauchy's integral formula to see that the formula does indeed hold in the onedimensional case (and thus also for diagonalizable matrices). I'd suspect that the same works for bounded generators $A$, but I have not checked this in detail. For analytic semigroups, things seem to be more involved since we cannot simply add a large circle on the left of the imaginary axis. $\endgroup$– Jochen GlueckMay 13, 2020 at 12:10
1 Answer
I've looked it up now. The formula in question does indeed hold in the following sense:
Theorem. Let $(e^{tA})_{t \in [0,\infty)}$ be a $C_0$semigroup on a complex Banach space $X$. Let $\omega \in \mathbb{R}$ be a real number that is strictly larger than the growth bound of our semigroup and let, for each $k > 0$, $\Gamma_k$ denote the complex line from $\omega  ik$ to $\omega +ik$. Then:
(i) The formula $$ (*) \qquad e^{tA}x = \lim_{k \to \infty} \frac{1}{2\pi i} \int_{\Gamma_k} e^{zt} R(z,A)x \, dz $$ holds for each $x$ in the domain of $A$.
(ii) If $X$ is an UMD space, then $(*)$ does even hold for each $x \in X$.
Reference: Proposition 3.12.1 and Theorem 3.12.2 in "Arendt, Batty, Hieber, Neubrander: VectorValued Laplace Transforms and Cauchy Problems (2011)".
The same reference contains a counterexample that shows that one cannot drop the assumption in (ii) that $X$ be a UMD space, in general (this is Example 3.12.3 there; hardly surprising, the counterexample is a shift semigroup...).

$\begingroup$ You are right, nice to have this. $\endgroup$ May 13, 2020 at 13:02

$\begingroup$ Many thanks for the reference, this is very helpful！ $\endgroup$– Jacob LuMay 13, 2020 at 20:13