Definition 1: Let $H$ be a Hilbert space. A strongly continuous semigroup is a family
$\{S(t)\}_{t \ge 0}$ of continuous linear operators $S(t): H \rightarrow
H$ such that
- $S(0)=I$, where $I$ is the identity operator.
- $S(t)S(s)=S(t+s)$ for all $t,s \ge 0$.
- $t\mapsto S(t)x$ is continuous on $[0,\infty)$ for all $x\in H$.
A contraction semigroup on a Hilbert space $H$ is a semigroup whose norm
is less or equal than 1; as in $\forall t \in \mathbb{R}^+: \|S(t)\| \le 1.$
Let $S(t)$ be a strongly continuous semigroup defined on a Hilbert space $H$ satisfying:
\begin{align*}
\left\|\int_0^t S(\tau) x\,d\tau\right\|_H \le t\|x\|_H
\end{align*}
How can I show that $S(t)$ is a contraction semigroup, i.e $\|S(t)\| \le
1$ for all $t \ge 0$?
I tried to prove it by contradiction. I supposed that $\exists t_0 \in
\mathbb{R}^+: \|S(t_0)\| \gt 1$ but then I noticed that I couldn't use
the inequality because $\left\|\int_0^{t_0} S(\tau) x\,d\tau\right\|_H$ is
always $\le t_0\|x\|_H$ and not greater than anything else.
Many thanks in advance,
—
Cesar
Best Answer
Here are some basic facts about strongly continuous semigroups that will be useful:
Fix $\lambda>0$, $x\in H$ and define $J(t):=\int_0^t S(\tau)x\,d\tau$ ($J$ depends on $x$ as well, but I will omit it for simplicity).
By hypothesis we have $\|J(t)\|\le t\|x\|$, so the integral $$R(\lambda)x:=\lambda\int_0^\infty e^{-\lambda t}J(t)\,dt$$ makes sense. Let us check that $R(\lambda)=(\lambda I-A)^{-1}$.
$\bullet$ $(S(\epsilon)-I)\int_0^T e^{-\lambda t}J(t)\,dt=\int_0^T e^{-\lambda t}\left(\int_0^t(S(\epsilon)-I)S(\tau)x\,d\tau\right)\,dt$
$=\int_0^T e^{-\lambda t}\left(\int_\epsilon^{t+\epsilon}S(\tau)x\,d\tau-\int_0^t S(\tau)x\,d\tau\right)\,dt$
$=\int_0^T e^{-\lambda t}\left( J(t+\epsilon)-J(\epsilon)-J(t)\right)\,dt$
$=e^{\lambda\epsilon}\int_\epsilon^{T+\epsilon}e^{-\lambda t}J(t)\,dt-\int_0^T e^{-\lambda t}J(t)\,dt-\frac{1-e^{-\lambda T}}{\lambda}J(\epsilon)$
and taking the limit as $T\to\infty$ at the beginning and the end of this chain of equalities and dividing by $\epsilon$ we get $$\frac{S(\epsilon)-I}{\epsilon}\int_0^\infty e^{-\lambda t}J(t)\,dt=\frac{e^{\lambda\epsilon}-1}{\epsilon}\int_0^\infty e^{-\lambda t}J(t)\,dt-\frac{J(\epsilon)}{\lambda\epsilon}-\frac{e^{\lambda\epsilon}}{\epsilon}\int_0^\epsilon e^{-\lambda t}J(t)\,dt$$ But the RHS possesses a limit as $\epsilon\to 0$, namely $R(\lambda)x-\frac{x}{\lambda}$ (the last term tends to $0$ since $e^{-\lambda t}J(t)=o(1)$): thus $\frac{R(\lambda)x}{\lambda}=\int_0^\infty e^{-\lambda t}J(t)\,dt\in D(A)$ and $$A\frac{R(\lambda)x}{\lambda}=R(\lambda)x-\frac{x}{\lambda}$$ i.e. $(\lambda I-A)R(\lambda)x=x$.
$\bullet$ Suppose now $x\in D(A)$. The second fact stated at the beginning gives
$\int_0^T e^{-\lambda t}\left(\int_0^t S(\tau)Ax\,d\tau\right)\,dt =\int_0^T e^{-\lambda t}(S(t)x-x)\,dt$
$=e^{-\lambda T}J(T)+\lambda\int_0^T e^{-\lambda t}J(t)\,dt-\frac{1-e^{-\lambda T}}{\lambda}x$ (in the last equality we integrated by parts).
Sending $T\to\infty$ we obtain $$R(\lambda)Ax=\lim_{T\to\infty}\int_0^T e^{-\lambda t}\left(\int_0^t S(\tau)Ax\,d\tau\right)\,dt=R(\lambda)x-\frac{x}{\lambda}$$ so $(\lambda I-A)R(\lambda)x=x$. Moreover $R(\lambda):H\to D(A)$ is a bounded operator. This proves that $\lambda$ belongs to the resolvent set of $A$ and that $R(\lambda)=(\lambda I-A)^{-1}$.
Finally $\|R(\lambda)x\|\le \lambda\int_0^\infty e^{-\lambda t}\|J(t)\|\,dt \le \lambda\int_0^\infty e^{-\lambda t}t\|x\|\,dt=\frac{\|x\|}{\lambda}$, so $\|(\lambda I-A)^{-1}\|\le\frac{1}{\lambda}$.
So Hille-Yosida theorem for contraction semigroups implies that $A$ generates a contraction semigroup, which coincides with $S(\cdot)$.