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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?

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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.

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  • $\begingroup$ Thank you for the answer. However, I still can't get to the bottom of it as in general, $A$ is possibly not invertible. (and so is $T(\rho)-I$) So I'm still figuring out how to apply the idea to the more general case. $\endgroup$
    – Aiden Bhe
    Feb 24 at 6:19
  • $\begingroup$ Okay I feel so stupid. I just figure it out. So because $A$ is a bounded linear operator we can let $T(t)=e^{tA}$ now we just have to show that $A(\int_0^\rho T(s)ds)=e^{\rho A}-I$ which is true regardless of $\rho$ since $A(\int_0^\rho T(s)ds)=\int_0^\rho \frac{d}{dt}e^{sA}ds$. $\endgroup$
    – Aiden Bhe
    Feb 24 at 6:59
  • $\begingroup$ In addition to above comment (since I just editted it so I can't re-edit the same comment yet), the rest is to prove that given $A$, the semigroup $T(t)$ is unique. $\endgroup$
    – Aiden Bhe
    Feb 24 at 7:21
  • $\begingroup$ If $T_1(t)$ and $T_2(t)$ have the same generator $A$, fix $x$ in the domain of $A$, $t > 0$ and consider $f(s) = T_1(s) T_2(t-s) x$ for $s \in [0,t]$. This turns out to be continuously differentiable and (as you'd expect with the product/chain rules) $f'(s) = A T_1(s) T_2(t-s)x - T_1(s) A T_2(t-s) x = 0$ (because $A$ commutes with the operators in both semigroups). So $f$ is constant, so $T_2(t)x = f(0) = f(t) = T_1(t)x$. By density of the domain of $A$, $T_2(t)=T_1(t)$. Some details needed to proving the continuity and differentiation rules needed to justify this, but that's the idea. $\endgroup$ Feb 24 at 17:48

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