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.