Improving Semigroup Bounds with Resolvent Estimates

Abstract

The purpose of this paper is to revisit the proof of the Gearhart–Prüss–Huang–Greiner theorem for a semigroup S(t), following the general idea of the proofs that we have seen in the literature and to get an explicit estimate on $\Vert S(t) \Vert$ in terms of bounds on the resolvent of the generator. A first version of this paper was presented by the two authors in ArXiv (2010) together with applications in semi-classical analysis and some of these results has been subsequently published in two books written by the authors. Our aim is to present new improvements, partially motivated by a paper of D. Wei. On the way we discuss optimization problems confirming the optimality of our results.

Appendix A. Optimization with \(\epsilon _1=\epsilon _2=+\)

In this section we let $\epsilon _1=\epsilon _2=+$ in Theorem 1.6 and assume that $\mathrm {supp\,}(r(\omega )^2-\phi '^2)_-\subset [0,a]$, $\mathrm {supp\,}(r(\omega )^2-\psi '^2)_-\subset [0,b]$ for some $a,b>0$, where $\Phi =e^\phi$, $\Psi =e^\psi$. The results we get in this case seem less conclusive, but are perhaps still of some interest. Assuming, to start with, that $\phi$ and $\psi$ are given on [0, a] and [0, b] respectively, we shall discuss how to choose $\phi$ on $]a,+\infty [$ and $\psi$ on $]b,+\infty [$, for every given $t>a+b$, in order to optimize the estimate on $\Vert S(t)\Vert$. A later problem will be to choose a, b with $a+b<t$ and the restrictions of $\phi$, $\psi$ to [0, a] and [0, b] respectively.

From (1.8) we get with $r=r(\omega )$,

$$\begin{aligned} \Vert S(t)\Vert \le e^{\omega t} \frac{\Vert (r^2-\phi '^2)_-^{1/2}m\Vert _{\phi -\omega \cdot } \Vert (r^2-\psi '^2)_-^{1/2}m\Vert _{\psi -\omega \cdot }}{I(\phi ,\psi )}, \end{aligned}$$

(5.1)

where

$$\begin{aligned} I(\phi ,\psi )=I_{a,b,t}(\phi ,\psi )=\int _a^{t-b} e^{\phi +\iota \psi }(r^2-\phi '^2)_+^{1/2}(r^2-\iota \psi '^2)_+^{1/2} ds. \end{aligned}$$

(5.2)

Here we put $\iota _t \psi (s)=\psi (t-s)$, and write simply $\iota$ when the choice of t is clear. We try to choose $\phi (s)$ for $s\ge a$ and $\psi (s)$ for $s\ge b$ so that $I(\phi ,\psi )$ is as large as possible. Write

$$\begin{aligned} \phi (s)=\phi (a)+{\widetilde{\phi }}(s-a),\ \psi (s)=\psi (b)+{\widetilde{\psi }}(s-b). \end{aligned}$$

For $s\in [a,t-b]$, set $s=a+{\widetilde{s}}$, $0\le {\widetilde{s}}\le t-a-b$. Then with ${\widetilde{t}}=t-a-b$,

$$\begin{aligned} \phi (s)+\iota \psi (s)= & {} \phi (s)+\psi (t-s)=\phi (a+{\widetilde{s}})+\psi (t-a-{\widetilde{s}})\\= & {} \phi (a){+}\psi (b){+}(\phi (a{+}{\widetilde{s}}){-}\phi (a)){+}(\psi (b+(t-a-b)-{\widetilde{s}})-\psi (b)\\= & {} \phi (a)+\psi (b)+{\widetilde{\phi }}({\widetilde{s}})+{\widetilde{\psi }} (t-a-b-{\widetilde{s}})\\= & {} \phi (a)+\psi (b)+{\widetilde{\phi }}({\widetilde{s}})+\iota _{{\widetilde{t}}}{\widetilde{\psi }}({\widetilde{s}}) \end{aligned}$$

and we get with ${\widetilde{\iota }} =\iota _{{\widetilde{t}}}$,

$$\begin{aligned} \begin{aligned} I(\phi ,\psi )&=e^{\phi (a)+\psi (b)}\int _0^{{\widetilde{t}}} e^{{\widetilde{\phi }}({\widetilde{s}})+{\widetilde{\iota }} {\widetilde{\psi }}({\widetilde{s}})} (r^2-{\widetilde{\phi }}'^2)_+^{1/2} (r^2-{\widetilde{\iota }} \widetilde{\psi } '^2)_+^{1/2}d{\widetilde{s}}\\&= e^{\phi (a)+\psi (b)}I_{0,0,{\widetilde{t}}}(\widetilde{\phi },{\widetilde{\psi }}). \end{aligned} \end{aligned}$$

(5.3)

We wish to choose ${\widetilde{\phi }}$, ${\widetilde{\psi }}$ with ${\widetilde{\phi }}(0)={\widetilde{\psi }}(0)=0$ such that ${\widetilde{I}}({\widetilde{\phi }} ,\widetilde{\psi })=I_{0,0,{\widetilde{t}}}({\widetilde{\phi }},{\widetilde{\psi }})$ is as large as possible.

We temporarily drop the tildes. The problem is then to choose $\phi$, $\psi$ with $\phi (0)=\psi (0)=0$ such that

$$\begin{aligned} I(\phi ,\psi )=\int _0^t e^{\phi +\iota \psi }(r^2-\phi '^2)_+^{1/2} (r^2-\iota \psi '^2)_+^{1/2} ds \end{aligned}$$

(5.4)

is as large as possible.

At this moment we do not know how to solve this general problem, so we restrict the class of functions (satisfying $\phi (0)=\psi (0)=0$) by requiring that

$$\begin{aligned} \phi +\iota \psi =\mathrm {Const.}\hbox { on }[0,t]. \end{aligned}$$

(5.5)

In other words, we require that $(\iota \psi )'=-\phi '$. The constant in (5.5) is then equal to $\phi (t)=\int _0^t \phi '(s)ds$. With $\Vert \cdot \Vert$ denoting the $L^2([0,t])$-norm, we get

$$\begin{aligned} I(\phi ,\psi )=\exp \left( \int _0^t \phi '(s)ds\right) \int _0^t (r^2-\phi '(s)^2)ds\le \exp (\sqrt{t}\Vert \phi '\Vert )(tr^2-\Vert \phi '\Vert ^2),\nonumber \\ \end{aligned}$$

(5.6)

requiring also that $|\phi '|\le r$. Here we have equality precisely when $\phi '$ is equal to some constant $\alpha \in [0,r]$, so for any given value $\beta \in [0,r\sqrt{t}]$ of $\Vert \phi '\Vert$, we should choose

$$\begin{aligned} \phi '=\alpha , \ \ \phi (s)=\psi (s)=\alpha s,\hbox { with } \sqrt{t}\alpha =\beta \,. \end{aligned}$$

(5.7)

The corresponding maximal value of $I(\phi ,\psi )$ is given by

$$\begin{aligned} J(\alpha , t)=te^{\alpha t}(r^2-\alpha ^2). \end{aligned}$$

(5.8)

We look for the maximum of this function of $\alpha$:

$$\begin{aligned} \partial _\alpha J(\alpha ,t)=-t^2e^{\alpha t}\left( \alpha ^2+\frac{2}{t}\alpha -r^2 \right) . \end{aligned}$$

The two critical points are given by a local maximum at

$$\begin{aligned} \alpha _+=\alpha _+(t)=\frac{1}{t}(\sqrt{1+(rt)^2}-1)\in \, ]0,r[ \end{aligned}$$

(5.9)

and a local minimum at

$$\begin{aligned} \alpha _-=\alpha _-(t)=-\frac{1}{t}(\sqrt{1+(rt)^2}+1)<0. \end{aligned}$$

We see that $\alpha _+$ is a global maximum. The corresponding maximal value is given by

$$\begin{aligned} J_{\mathrm {max}}(t)=J(\alpha _+,t)=e^{\sqrt{1+(rt)^2}-1}\frac{2}{t}(\sqrt{1+(rt)^2}-1). \end{aligned}$$

(5.10)

Let us compute the asymptotic behaviour of $J_{\mathrm {max}}(t)$ when $t\rightarrow +\infty$. We get

$$\begin{aligned} \sqrt{1+(rt)^2}= & {} rt+{{\mathcal {O}}}\left( \frac{1}{rt} \right) ,\ \alpha _+=r-\frac{1}{t}+{{\mathcal {O}}}\left( \frac{1}{trt} \right) ,\nonumber \\ J_{\mathrm {max}}(t)= & {} \left( 1+{{\mathcal {O}}}\left( \frac{1}{rt} \right) \right) \frac{2}{e}e^{rt}r,\ rt\rightarrow +\infty . \end{aligned}$$

(5.11)

Here we recall that the new $t={\widetilde{t}}$ is equal to $t-a-b$ for the original t. Returning to the original $I(\phi ,\psi )=I_{a,b,t}(\phi ,\psi )$ (cf. (5.3)) with ${{\phi }_\vert }_{[0,a]}$, ${{\psi }_\vert }_{[0,b]}$ prescribed, we get with the choice

$$\begin{aligned}&{\left\{ \begin{array}{ll} \phi (s)-\phi (a)=\alpha _+(s-a),\ s\ge a,\\ \psi (s)-\psi (b)=\alpha _+(s-b),\ s\ge b, \end{array}\right. }\end{aligned}$$

(5.12)

$$\begin{aligned}&\alpha _+=\alpha _+({\widetilde{t}}),\ \ {\widetilde{t}}=t-a-b, \end{aligned}$$

(5.13)

that

$$\begin{aligned} \begin{aligned} I(\phi ,\psi )&=J_{\mathrm {max}}(t-a-b)e^{\phi (a)+\psi (b)}\\&=\left( 1+{{\mathcal {O}}}\left( \frac{1}{t-a-b} \right) \right) \frac{2}{e} e^{\phi (a)+\psi (b)} \, r \, e^{r(t-a-b)},\ \ t-a-b\rightarrow +\infty , \end{aligned} \end{aligned}$$

(5.14)

when $r=r(\omega )>0$ is fixed.

Summing up the discussion so far, we get from (5.1), (5.14):

Proposition 5.1

Let $a,b>0$ and let $\Phi \in C([0,a]; {\mathbb {R}})$, $\Psi \in C([0,b])$ be increasing, piecewise $C^1$ with $\Phi (0)=\Psi (0)=0$,

$$\begin{aligned} {\left\{ \begin{array}{ll} r(\omega )^2\Phi ^2-\Phi '^2\le 0,\hbox { on }[0,a],\\ r(\omega )^2\Psi ^2-\Psi '^2\le 0,\hbox { on }[0,b]. \end{array}\right. } \end{aligned}$$

Write $\Phi = e^\phi$, $\Psi =e^\psi$, with $\phi$, $\psi$ real. Then for $t>a+b$, with $r=r(\omega )$,

$$\begin{aligned} e^{-\omega t}\Vert S(t)\Vert \le \frac{\Vert (r^2\Phi ^2-\Phi '^2)^{\frac{1}{2}}_-m\Vert _{e^{\omega \cdot }L^2([0,a])}\Vert (r^2\Psi ^2-\Psi '^2) ^{\frac{1}{2}}_-m\Vert _{e^{\omega \cdot }L^2([0,b])}}{J_{\mathrm {max}}(t-a-b)\Phi (a)\Psi (b)},\nonumber \\ \end{aligned}$$

(5.15)

where $J_{\mathrm {max}}({\widetilde{t}})$ is given in (5.10) and has the asymptotics (5.11). In particular for large values of $t-a-b$,

$$\begin{aligned} \begin{aligned} e^{-\omega t}\Vert S(t)\Vert&\le \left( 1+{\mathcal O}\left( \frac{1}{r(t-a-b)} \right) \right) \frac{e}{2r}e^{-r(t-a-b)}\\&\quad \times \frac{\Vert (r^2\Phi ^2-\Phi '^2) ^{\frac{1}{2}}_-m\Vert _{e^{\omega \cdot }L^2([0,a])}\Vert (r^2\Psi ^2-\Psi '^2) ^{\frac{1}{2}}_-m\Vert _{e^{\omega \cdot }L^2([0,b])}}{\Phi (a)\Psi (b)}. \end{aligned} \end{aligned}$$

(5.16)

Here we meet the same quantities as in the previous section. Hence we obtain (cf Theorem 1.9), if $a^{*} (m)$ is bounded, $r=1$, $\omega =0$ and $a, b \le a^{*} (m)$, $t> a+b$,

$$\begin{aligned} || e^t S(t)|| \le \left( 1+{\mathcal O}\left( \frac{1}{(t-a-b)} \right) \right) \frac{e}{2} m(a) m(b) e^{a+b} \psi _0(a)^\frac{1}{2} \psi _0 (b)^\frac{1}{2}. \end{aligned}$$

(5.17)

As $t\rightarrow +\infty$, we have lost a factor (e/2) in comparison with the statement of Theorem 1.9. Nevertheless it is conceivable that for some t the estimate obtained by this approach is better.

Non optimality. Possible improvements? We have solved the optimization problem for $I(\phi ,\psi )$ in (5.4) for $(\phi ,\psi )$ varying in a restricted class. The purpose of this remark is to show that the solution $(\phi ,\psi )$ in (5.7) with $\alpha =\alpha _+$ is not a critical point for $I(\phi ,\psi )$ when $(\phi ,\psi )$ varies more freely and hence we can perturb our special solution slightly (leaving the restricted class) and find an even larger value of $I(\phi ,\psi )$.

Write $f=\iota \psi$ for simplicity. We then want to find $\phi ,f\in C^2([0,1])$ with

$$\begin{aligned} \phi (0)=f(t)=0, \end{aligned}$$

(5.18)

$\phi$ increasing, f decreasing (i.e. $\phi '\ge 0$, $f'\le 0$) with

$$\begin{aligned} r^2-\phi '^2>0,\ \ r^2-f'^2>0, \end{aligned}$$

(5.19)

such that $I(\phi ,\iota f)$ is as large as possible and in particular such that $(\phi ,f)$ is a critical point for I. We make a variational calculation considering infinitessimal variations $(\phi +\delta \phi ,f+\delta f)$ with $\delta \phi (0)=\delta f(t)=0$. Then

$$\begin{aligned} \delta I(\phi ,\iota f)=\mathrm {I}+\mathrm {II}+\mathrm {III}, \end{aligned}$$

where with $K(\phi ,f)(s):=e^{\phi +f}(r^2-\phi '^2)^{\frac{1}{2}}(r^2-f '^2)^{\frac{1}{2}}:$

$$\begin{aligned} \mathrm {I}= & {} \int _0^t K(\phi ,f)(s)(\delta \phi (s)+\delta f(s))ds,\\ \mathrm {II}= & {} \int _0^t K(\phi ,f)(s) \frac{\delta \left( (r^2-\phi '^2)^{1/2} \right) }{(r^2-\phi '^2)^{1/2}} ds,\\ \mathrm {III}= & {} \int _0^t K(\phi ,f)(s) \frac{\delta \left( (r^2-f '^2)^{1/2} \right) }{(r^2-f'^2)^{1/2}} ds. \end{aligned}$$

Here,

$$\begin{aligned} \delta \left( (r^2-\phi '^2)^{1/2} \right) =-(r^2-\phi '^2)^{-1/2}\phi '\delta \phi ' \end{aligned}$$

and similarly for f, so

$$\begin{aligned} \mathrm {II}= & {} -\int _0^t K(\phi ,f)(s)(r^2-\phi '^2)^{-1}\phi '\delta \phi ' ds,\\ \mathrm {III}= & {} -\int _0^t K(\phi ,f)(s)(r^2-f '^2)^{-1}f '\delta f ' ds. \end{aligned}$$

Here we integrate by parts, using that $\delta \phi (0)=\delta f(t)=0$:

$$\begin{aligned} \mathrm {II}= & {} \int _0^t \left( \partial _s\circ K(\phi ,f)(r^2-\phi '^2)^{-1}\circ \partial _s\phi \right) \delta \phi ds -K(\phi ,f)(r^2-\phi '^2)\phi '\delta \phi (t),\\ \mathrm {III}= & {} \int _0^t \left( \partial _s\circ K(\phi ,f)(r^2-f '^2)^{-1}\circ \partial _sf \right) \delta f ds +K(\phi ,f)(r^2-f '^2)f '\delta f (0). \end{aligned}$$

This gives,

$$\begin{aligned} \begin{aligned} \delta (\phi ,\iota f)&=\int _0^t\left( K(\phi ,f)+\partial _s\circ K(\phi ,f)(r^2-\phi '^2)^{-1}\circ \partial _s\phi \right) \delta \phi ds\\&\quad -K(\phi ,f)(r^2-\phi '^2)\phi '\delta \phi (t)\\&\quad +\int _0^t\left( K(\phi ,f)+\partial _s\circ K(\phi ,f)(r^2-f '^2)^{-1}\circ \partial _sf \right) \delta f ds\\&\quad +K(\phi ,f)(r^2-f '^2)f'\delta f (0). \end{aligned} \end{aligned}$$

Assumption (5.19) implies that $K(\phi ,f)(r^2-\phi '^2)>0$, $K(\phi ,f)(r^2-f '^2)>0$ and we see that $(\phi ,f)$ is a critical point precisely when

$$\begin{aligned} \phi '(t)=0,\ \ f'(0)=0, \end{aligned}$$

(5.20)

(in addition to (5.18)) and

$$\begin{aligned} {\left\{ \begin{array}{ll} K(\phi ,f)+\partial _s\circ K(\phi ,f)(r^2-\phi '^2)^{-1}\circ \partial _s\phi =0\\ K(\phi ,f)+\partial _s\circ K(\phi ,f)(r^2-f '^2)^{-1}\circ \partial _sf=0 \end{array}\right. } \hbox { on }[0,t]. \end{aligned}$$

(5.21)

We conclude that $(\phi ,\psi )$ in (5.7) with $\alpha =\alpha _+$, is not a critical point for $I(\cdot , \cdot )$, since it does not satisfy (5.19). Hence by modifying $\phi$ slightly near $s=t$, we can increase $I(\phi ,\psi )$ further.