Proof of a theorem about generators of uniformly continuous semigroup I´m reading Vrabie and I'm have some problems understanding this proof.



We have that $\{S(t); t\geq 0\}$ is uniformly continuous, so $\displaystyle\lim_{t\downarrow 0} S(t)=I$.

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*Question 1: This implies that $\exists\rho >0$ such that $||\displaystyle\frac{1}{\rho}\displaystyle\int_0^\rho S(t) dt -I||_{\mathcal{L}(X)} <1$.

Because of the limit we have that $\exists\rho >0$ such that $|| S(t)  -I||_{\mathcal{L}(X)} <1$. Now $$||\displaystyle\int_0^\rho S(t) dt -\displaystyle\int_0^\rho I dt||_{\mathcal{L}(X)}\leq \displaystyle\int_0^\rho ||S(t)  - I|| dt<\displaystyle\int_0^\rho  1 dt$$
this is equivalent to: $$||\displaystyle\int_0^\rho S(t) dt -\rho I ||_{\mathcal{L}(X)}<\rho$$
and just dividing by $\rho$ we have the inequality. I assume that $\displaystyle\int_0^\rho I dt=\rho I$, but I don't know if it is true. Is it?

We notice that the integral here is a Riemann integral of a continuous function $S:[0,\rho]\to\mathcal{L}(X)$, which is defined by a simple analogy with its scalar counterpart.

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*Question 2: About the sentence in bold, how do we define it? We define it as a operator given by $$x\to \displaystyle\frac{1}{x}\displaystyle\int_0^xS(t) dt$$ A bit lost here.


Consecuently, the operator $\displaystyle\frac{1}{\rho}\displaystyle\int_0^\rho S(t) dt$ is invertible and accordingly $\displaystyle\int_0^\rho S(t) dt$ has the same property.

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*Question 3: The phrase in bold just stunned me. We know that $S(t)$ is invertible as it belongs to a uniformly continuous semigroup by how can we deduce that the operator $\displaystyle\frac{1}{\rho}\displaystyle\int_0^\rho S(t) dt$ is invertible?


For the following part:



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*Question 4: we can change the letter $\rho$ highlighted in blue by the letter $h$ highlighted in blue, just because of question 2. Is jut notation.



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*Question 5: If we take limit when $h\to 0$ in $$\displaystyle\frac{1}{h}\displaystyle\int_h^{\rho+h} S(t) dt-\displaystyle\frac{1}{h}\displaystyle\int_0^h S(t) dt$$ why do we get $S(\rho)-I$?. Lets do a little of work, $$\displaystyle\frac{1}{h}\displaystyle\int_h^{\rho+h} S(t) dt-\displaystyle\frac{1}{h}\displaystyle\int_0^h S(t) dt=\displaystyle\frac{1}{h}\displaystyle\int_0^{\rho+h} S(t) dt-\displaystyle\frac{1}{h}\displaystyle\int_0^h S(t) dt-\displaystyle\frac{1}{h}\displaystyle\int_0^h S(t) dt$$, so it's clear that
$$\displaystyle\frac{1}{h}\displaystyle\int_0^{\rho+h} S(t) dt-\displaystyle\frac{1}{h}\displaystyle\int_0^h S(t) dt=S(\rho)\;\text{when}\; h\to 0$$.

But where where does the $I$ come from? It comes from $\displaystyle\frac{1}{h}\displaystyle\int_0^h S(t) dt$?
 A: For a continuous mapping $S:[0,\infty )\to \mathcal{L}(X)$ and $0\le a<b$ we define the integral sums $S_n(a,b)$ by
$$S_n(a,b)={b-a\over n}\sum_{k=1}^nS\left (a+k{b-a\over n}  \right ).$$
It can be shown that the sequence $S_n(a,b)$ satisfies the Cauchy condition hence it is convergent and its limit is called the integral
$$\int\limits_a^b S(x)\,dx=\lim_{n\to \infty}S_n(a,b).$$
The integral shares all the properties of the standard Riemann integral: it is linear with respect to $S(x)$ and is additive with respect to the domain of integration, i.e.
$$\int\limits_a^bS(x)\,dx + \int\limits_b^cS(x)\,dx=\int\limits_a^cS(x)\,dx, \quad a<b<c$$
Defintion of the integral implies
$$\left \|\int\limits_a^bS(x)\,dx \right \|\le (b-a)\max_{a\le x\le b}\|S(x)\|$$
For $S(x)\equiv I$ we have $S_n(a,b)=(b-a)I,$ hence $\int\limits_a^b\,I\,dx=(b-a)\,I.$
Let $S(x)$ be continuous and $S(0)=I$ (every continuous semigroup satisfies that). Then
for $\varrho>0$ sufficiently small the operator $U_\varrho:=\varrho^{-1}\int\limits_0^\varrho S(x)\,dx $ is invertible. To this end it suffices to show that $\|U_\varrho -I\|<1.$ Observe that
$$U_\varrho -I={1\over \varrho}\int\limits_0^\varrho [S(x)-I]\,dx.$$
Therefore
$$\|U_\varrho -I\|\le {1\over \varrho}\int\limits_0^\varrho \|S(x)-I\|\,dx\le \max_{0\le x\le \varrho}\|S(x)-I\|$$
By continuity of $S(x)$ at $0,$ there exists $\delta>0$ such that for any $0\le x\le \delta$ we have $\|S(x)-I\|<1.$ Thus for $0<\varrho \le \delta$ we get $\|U_\varrho -I\|<1,$ and $U_\varrho$ is invertible.
Other questions can be answered basing on the above. At places semigroup property should be applied if necessary.
For example
$$S(h)\int\limits_0^\varrho S(t)\,dt=\int\limits_0^\varrho S(t+h)\,dt=\int\limits_h^{h+\varrho} S(t)\,dt$$
I hope it clarifies some issues.
