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

  1. $S(0)=I$, where $I$ is the identity operator.
  2. $S(t)S(s)=S(t+s)$ for all $t,s \ge 0$.
  3. $t\mapsto S(t)x$ 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,


  • 1
    $\begingroup$ Do you assume some kind of continuity of $S(t)$ with respect to $t$? $\endgroup$
    – timur
    Aug 2 '14 at 3:19
  • 1
    $\begingroup$ @timur Good point. I just assumed that the OP meant a strongly continuous semigroup. $\endgroup$
    – user940
    Aug 2 '14 at 15:11
  • 3
    $\begingroup$ Yes, $t \mapsto S(t)x$ is continuous from $\mathbb{R}^+$ into $H \quad \forall x \in H$. $\endgroup$
    – Luo Kaisa
    Aug 3 '14 at 16:30
  • $\begingroup$ By $\Bbb R_+$ do you mean $\{r \in \Bbb R \mid r \ge 0 \}$ or $\{ r \in \Bbb R \mid r >0 \}$? $\endgroup$ Aug 4 '14 at 4:57
  • 1
    $\begingroup$ @timur Hopefully, it is correct: Let $H$ be the space of bounded continuous functions $f: \mathbb{R} \to \mathbb{R}$ endowed with the norm $$\|f\|_H := \|f\|_{\infty}+ k \cdot |f(0)|$$ for some $k>0$. Moreover, set $S_t f(x) := f(t+x)$. Then $S_t$ is not a contraction semigroup since for any $t>0$ we can choose $f \in H$ such that $f(0)=0$, $f(t)=1$, $\|f\|_{\infty} = 1$. For this $f$, we have $\|f\|_H=1$ and $\|S_t f\|_H = 1+k>1$; hence $\|S_t\| > 1$ for $t>0$. On the other hand, the given inequality holds true since for $f \in H$ with $f(0)=0$ we have $\|f\|_H = \|f\|_{\infty}$. $\endgroup$
    – saz
    Aug 6 '14 at 19:09

Here are some basic facts about strongly continuous semigroups that will be useful:

  • Let $A$ be the infinitesimal generator of $S(\cdot)$. Then $D(A)$ (the domain of $A$) is dense and $A$ is closed.
  • For any $x\in D(A)$ we have $S(t)Ax=\frac{d}{dt}S(t)x$.
  • If two strongly continuous semigroups have the same infinitesimal generator, then in fact they are the same semigroup.

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 $S(\cdot)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.