We have a strongly continuous semigroup $\{T(t)\}_{t\ge 0}$ on a Banach space X with a generator $ A:D(A)\subset X\to X$.

I need to show that $\displaystyle T(t)x = \sum\limits_{n=0}^{\infty} \frac{t^n}{n!}A^nx $ holds for all $t\geq0 $ and all $x\in X$ that are entire.

I know that an element $x\in \bigcap^\infty_{k=1} D(A^k)$ is called entire if the inequality
$$ \sum\limits_{n=0}^{\infty}\frac{1}{n!} |t|^n ||A^nx||_X < \infty $$

holds for $t\in\mathbb{R}$

How do I use this inequality and how should I proceed?

I have started with

$\displaystyle \|T(t)x\|=\|\sum\limits_{n=0}^{\infty} \frac{t^n}{n!}A^nx\|=...$

But I don't really know where to go from here.


If $x$ is analytic, \begin{align} T(t)x & = x+\int_{0}^{t}T(s)Ax ds \\ & = x + (s-t) T(s)Ax|_{s=0}^{t}-\int_{0}^{t}(s-t)T(s)A^{2}xds \\ & = x + t Ax -\int_{0}^{t}(s-t)T(s)A^{2}xds \\ & = x + t Ax +\frac{t^{2}}{2!}A^{2}x+\int_{0}^{t}\frac{(s-t)^{2}}{2!}T(s)A^{3}xds \end{align} The remainder tends to $0$ as the process of integration-by-parts is continued, assuming $x$ is analytic. Therefore $T(t)x=S(t)x$ for all $t \ge 0$ if $x$ is analytic.

  • $\begingroup$ Thanks for the answer. I understand the first line, but not the second when you start integrating by parts. I get why its $A^2$, but not how you got $(s-t)$ in front when integrating $T(s)$ ? (I have written it out too see that I get $tAx$) $\endgroup$ May 25 '15 at 15:45
  • $\begingroup$ @Leibniz1337 : An antiderivative is unique only up to a constant. I chose the constant so that one of the evaluation terms would vanish; I wanted the evaluation term at $s=t$ to vanish. It's a standard trick for deriving a truncated Taylor series. $\frac{d}{ds}(s-t) = \frac{d}{ds}s$. $\endgroup$ May 25 '15 at 16:23
  • $\begingroup$ @user3482534 : No. You're evaluating in $s$. $(s-t)T(s)|_{s=0}^{t}=(t-t)T(t)Ax-(0-t)T(0)Ax=tAx$. $\endgroup$ May 30 '15 at 16:13
  • $\begingroup$ In the last step of your prove, it should read $\left.\frac{(s-t)^{2}}{2!}A^{2}x\right]_{s=0}^{t}$ $\endgroup$
    – nippon
    Jun 8 '15 at 17:50
  • $\begingroup$ @nippon : Actually, I had meant to leave off the evaluation, now that the pattern had been established. I fixed that now. $\endgroup$ Jun 8 '15 at 17:55

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