I'm studying the theory of semigroups from Pazy's book. I'm struggling to understand the proof of a theorem stating that
Theorem 1.2. Let $X$ be a banach space. A linear operator $A:X\rightarrow X$ is the infinitesimal generator of a uniformly continuous semigroup iff $A$ is a bounded linear operator.
The proof of $\Leftarrow$ is pretty straight forward by setting
\begin{equation}
T(t)=e^{tA}:=\sum_{n=0}^{\infty}\frac{(tA)^n}{n!}.
\end{equation}
But I don't understand a part in $\Rightarrow$ proof, where we let $T(t)$ be a uniformly continuous semigroup of bounded linear operators on $X$ and some $\rho>0$ be such that $\|I-\rho^{-1}\int_0^\rho T(s)ds\|<1$ so that $\int_0^\rho T(s)ds$ is invertible. We can show that the infinitesimal generator $A:=\lim_{h\searrow 0}h^{-1}(T(h)-I)$ converges in $\mathcal{L}(X)$:
\begin{align}
h^{-1}(T(h)-I)\int_0^\rho T(s)ds = h^{-1}\bigg(\int_\rho^{\rho+h}T(s)ds-\int_0^h T(s)ds\bigg)
\end{align}
and therefore
\begin{equation}
h^{-1}(T(h)-I) = h^{-1}\bigg(\int_\rho^{\rho+h}T(s)ds-\int_0^h T(s)ds\bigg)\bigg(\int_0^\rho T(s)ds\bigg)^{-1}.
\end{equation}
We have $h^{-1}(T(h)-I)\rightarrow (T(\rho)-I)(\int_0^\rho T(s)ds)^{-1}$ in $\mathcal{L}(X)$ as $h\searrow 0$.
By definition of a semigroup, the infinitesimal generator of $T(t)$ is uniquely defined. But it seems to me that there are several ways to construct a different $\int_0^\rho T(s)ds$ by fixing $\rho>0$ at another value.
So my question: how can we show that $(T(\rho)-I)(\int_0^\rho T(s)ds)^{-1}$ is independent of $\rho$ described above?
Best Answer
Realized I was basically writing an answer in comments, so here's the comments as an answer.
Note that $\rho$ appears in two places in your formula (in $T(\rho)$ and in the integral you constructed, to the $−1$ power); so it is not formally inconceivable that the $\rho$- dependence might "cancel out." Running the construction with $T(t)=e^{At}$ for bounded $A$ will show this more explicitly.
For example in the one dimensional case if $T(t) = e^{at}$ is the semigroup, and writing $r$ in place of $\rho$, then $\int_0^r T(s) \, ds = (e^{ar} - 1)/a$, so $(T(r)-1) * (\int_0^r T(s) ds)^{-1} = (e^{ar} - 1) * a/(e^{ar} - 1) = a$.
And of course in general, the very presence of $\rho$ in as a limit of something an operator limit involving only $T(h)$ and $h$ (which do not depend on $\rho$) also shows that the choice of $\rho$ won't matter, subject to the hypotheses. It's just uniqueness of the generator.