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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,

--
Cesar

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  • 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
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    $\begingroup$ 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
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    $\begingroup$ @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
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+500
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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 $S(\cdot)$.

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