Operator Semigroup: Resolvent estimates and stabilization, a detail in the paper of Nicoulas Burq and Patrick Gerard

Question

In Appendix A of the paper Stabilization of wave equations on the torus with rough dampings https://msp.org/paa/2020/2-3/p04.xhtml or https://arxiv.org/abs/1801.00983 by Nicoulas Burq and Patrick Gerard.

The author states a equivalent relation:

Supposed that $e^{tA}$ is a strongly continuous semigroups on a Hilbert space $H$ with infinitesimal generator $A$ defined on $D(A)$. The following two properties are equivalent:

(1) There exist $C,\delta>0$ such that the resolvent of $A$, $(A-\lambda)^{-1}$ exists for the real part of $\lambda$, $\rm{Re}\ \lambda\geq-\delta$ and satisfies $\forall\lambda\in\mathbb{C}^{\delta}=\{z\in\mathbb{C}:\rm{Re}\ z\geq-\delta\}$, we have $\|(A-\lambda)^{-1}\|_{\mathscr{L}( H)}\leq C$.

(2)There exist $M,\delta>0$ such that for any $t>0$, $\|e^{tA}\|_{\mathscr{L}( H)}\leq Me^{-\delta t}$.

Confusion of the proof: When proving $(1)\Rightarrow (2)$, the authors say "Since $v(t)$ is supported in $t\in[-1,0]$, the right side hand in $(i\tau+\omega-A)\hat{u}(\tau)=\hat{v}(\tau)$" is holomorphic and bounded in any domain $\mathbb{C}_{\alpha}=\{\tau\in\mathbb{C}:\rm{Im}\ \tau\geq\alpha,\alpha\in\mathbb{R}\}$($\color{red}{why?}$)."

In the next line, the authors say:

"From the assumption on the resolvent, we deduce that $\hat{u}$ admits a holomorphic extension to $\{\tau: \rm{Im} \tau\leq\delta+\omega\}$ which satisfies $\|\hat{u}(\tau)\|_{H}\leq C\|\hat{v}(\tau)\|_{H}$($\color{red}{why?}$)"

Why should we consider the holomorphic extension of $\hat{u}$(the time fourier transform of $u$) thank you very much!!!!

Answer 1

For the first "why": that $\hat{v}(\tau)$ is holomorphic is a version of Paley-Wiener. On the domain $\mathbb{C}_\alpha$ you have $\exp(i\tau t)$ is uniformly bounded (for any $t\in \mathbb{R}$) by $\exp(-\alpha)$), so as long as $v(t)$ is integrable in $t$ the boundedness claim follows.

For the second "why": Setting $\lambda = \omega + i\tau$ then $$ \mathrm{Re}(\lambda) \geq -\delta \iff \mathrm{Re}(i\tau) \geq -\delta-\omega \iff - \mathrm{Im}(\tau) \geq - (\delta + \omega) $$ So the resolvent assumption shows that $(i\tau + \omega - A)$ is invertible with bounded inverse. That the resolvent is holomorphic in $\lambda$ is standard. So you can invert $(i\tau + \omega - A)\hat{u}(\tau) = \hat{v}(\tau)$ to get the claim.

As for the final question, the following is a bit of an imprecise heuristic: using the Fourier inversion formula, you have (up to some constants) $$ u(t) = \int_{-\infty}^\infty \hat{u}(\tau) \exp(it\tau) d\tau$$ If $\hat{u}$ has a holomorphic extension to a strip around the real line, this integral should be equal to (up to terms at infinity) $$ \int_{- \infty}^{\infty} \hat{u}(\tau+i\epsilon) \exp(it(\tau + i\epsilon)) d\tau$$ But now $\exp(it(\tau + i\epsilon))$ has a uniform decay like $\exp(-\epsilon t)$ so you expect the same to hold for $u(t)$. So at least at a formal level you see that having a holomorphic extension to a strip of $\hat{u}$ is connected to the function $u$ having exponential decay.