Semigroup derived from abstract Cauchy problem

Question

There is a part in the book 'Stochastic equations in infinite dimensions' by Da Prato and Zabczyk which I can't quite follow. Let's assume that we have the Cauchy problem \begin{equation}\quad \left\{\begin{array}{ll} u'(t)=A_0 u(t), & t\geq 0,\\ u(0)=x, & x\in E,\end{array}\right.\quad(\mathrm{A}.1)\end{equation} where $E$ is a Banach space and $A_0$ is a linear operator defined on a dense linear subspace $D(A_0)$ of $E$. We also assume that the problem is uniformly well posed, meaning:

1. for arbitrary $x\in D(A_0)$ there exists exactly one strongly differentiable function $u(t,x), t\in[0,\infty)$ satisfying $(\mathrm{A}.1)$ for all $t\in[0,\infty)$
2. if $\{x_n\}\in D(A_0)$ and $\lim_{n\to\infty}x_n=0$, then for all $t\in[0,\infty)$ we have $$\lim_{n\to\infty}u(t,x_n)=0\quad (\mathrm{A}.2)$$
3. the limit in $(\mathrm{A}.2)$ is uniform in $t$ on compact subsets of $[0,\infty)$.

We define operators $S(t):D(A_0)\to E$ by the formula $$S(t)x=u(t,x),\quad \forall x\in D(A_0),\forall t\geq 0.$$ For all $t\geq0$ the linear operator $S(t)$ can be uniquely extended to a linear bounded operator on the whole $E$, still denoted by $S(t)$. Now, it's easy to see that $S(0)=I$, but I have a problem proving other two conditions for $C_0$-semigroup. All it says in the book is that $$S(t+s)=S(t)S(s)\quad \forall t,s\geq 0$$ follows by the uniqueness, and that $$S(\cdot)x\ \ \mathrm{is\ continuous\ in}\ \ [0,\infty)\quad \forall x\in E$$ follows from the uniform boundedness theorem. However, I can't see how these implications follow, especially the one with $S(t+s)=S(t)S(s)$. Any help would be appreciated.

Answer 1

The operator $S$ is called the flow $-$ or the evolution operator sometimes $-$ generated by the the linear operator $A_0$. Its effect/role turns out to be the linear operator which "solves" the differential equation by mapping the initial condition to the actual solution, i.e. $u(t) = S(t)u(0) = S(t)x$, since $u(0) = x$ in your case.

More generally, this operator takes the solution at a time $s$ and makes it evolve to a time $t>s$, i.e. $S(t,s)u(s) = u(t)$, hence transitively $S(t+s,t_0) = S(t,s)S(s,t_0)$, which proves to be a semi-group structure. In your book, they dropped the second variable without warning.

It is to be noticed that, depending on the considered space, you could write $S(t,t_0) = S(t-t_0) = e^{(t-t_0)A_0}$ if $A_0$ doesn't depend on $t$, which permits to handle this operator as a transformation in a Lie-theoretical context.

Finally, as for the last implication, it is a well-known result of functional analysis, stating that bounded linear operators are continuous.

Answer 2

I have found the answer to my question (regarding continuity of $S(\cdot)x$ derived from unfiform boundness principle) in the book 'Semigroups of Linear Operators and Applications to Partial Differential Equations' by Amnon Pazy. I'm going to leave it here for posterity.

Let $X$ be a Banach space. A one parameter family $S(t), t\geq0$ of bounded linear operators from $X$ into $X$ is a $C_0$-semigroup if

1. $S(0) = I$,
2. $S(t+s) = S(t)S(s),\quad t,s\geq 0$,
3. $\lim_{t\downarrow 0}S(t)x=x\quad$ for every $\quad x\in X.$

Theorem: Let $S(t)$ be a $C_0$ semigroup. There exist constants $\omega\geq0$ and $M\geq 1$ such that $$\|S(t)\|\leq M e^{\omega t},\quad t\geq 0.$$ Proof: We show first that there is an $\eta>0$ such that $\|S(t)\|$ is bounded on $[0,\eta]$. Let's assume that this statement is false. Then there is a sequence $\{t_n\}$ satisfying $t_n\geq 0, \lim_{n\to\infty}t_n=0$ and $\|S(t_n)\|\geq n$. From the uniform boundness theorem it then follows that for some $x\in X$, $\{\|S(t_n)x\|\}$ is unbounded, which contradicts the third assumption in the definition of $C_0$-semigroup. (The uniform boundness theorem states that if $\{T_n\}$ is a set of bounded linear operators from $X$ to $Y$ and sequence $\{\|T_nx\|\}$ is bounded for all $x\in X$, then $\{\|T_n\|\}$ is bounded. We use the contraposition of this theorem.) Thus, there exists $M\geq 0 $ such that $\|S(t)\|\leq M$ for $0\leq t \leq \eta$. Since $\|T(0)\|=1$, then $M\geq 1$.

Let $\omega:=\eta^{-1}\ln M \geq 0$. Given $t\geq 0$ we have $t = n\eta +\delta$, where $0\leq \delta< \eta$ and $n\in\mathbb{N}$. Therefore by second assumption in the definition of $C_0$-semigroup: $$\|S(t)\| = \|S(n\eta+\delta)\|=\|S(n\eta)S(\delta)\|\leq \|S(\eta)\|^n\|S(\delta\|\leq M^{n+1}\leq MM^{t/\eta}=Me^{\omega t},$$ because $t\geq n\eta\Rightarrow n\leq t/\eta$. This concludes the proof.

Corollary: If $S(t)$ is a $C_0$-semigroup, then for every $x\in X$ function $t\to S(t)x$ is continuous on $[0,\infty)$.

Proof: Given $t,h\geq 0$, we have $$\begin{split}\|S(t+h)x-S(t)x\| &= \|S(t)(S(h)x-x)\|\leq\|S(t)\|\|S(h)x-x\|\leq\\ \\&\leq Me^{\omega t}\|S(h)x-x\|\xrightarrow{h\to 0}0\end{split}$$ and for $t\leq h\leq 0$ $$\begin{split}\|S(t-h)x-S(t)x\|&=\|S(t-h)x-S(t-h+h)x\|=\|S(t-h)(x-S(h)x)\|\leq\\ \\&\leq \|S(t-h)\|\|x-S(h)x\|\leq Me^{\omega (t-h)}\|x-S(h)x\|\xrightarrow{h\to 0} 0\end{split}.$$