# A uniformly continuous semigroup has a unique infinitesimal generator

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 $$$$T(t)=e^{tA}:=\sum_{n=0}^{\infty}\frac{(tA)^n}{n!}.$$$$

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 $$$$h^{-1}(T(h)-I) = h^{-1}\bigg(\int_\rho^{\rho+h}T(s)ds-\int_0^h T(s)ds\bigg)\bigg(\int_0^\rho T(s)ds\bigg)^{-1}.$$$$ We have $$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?

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.
• 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. Feb 24 at 6:19
• 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 $A(\int_0^\rho T(s)ds)=\int_0^\rho \frac{d}{dt}e^{sA}ds$. Feb 24 at 6:59
• 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. Feb 24 at 7:21
• 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. Feb 24 at 17:48