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Let $\{T_{t}\}_{t \geq0}$ be a $C_{0}$-semigroup and we will denote

$$D(A)=\{x\in X : \exists \lim_{h\rightarrow 0_{+}} \frac{T(h)x-x}{h}\}$$ and define $$A:D(A) \rightarrow X; Ax=\lim_{h\rightarrow 0_{+}} \frac{T(h)x-x}{h}.$$

The operator $A$ is named the infinitezimal generator of $C_{0}$ semigroup.

How can I prove that:

$$\int^{t}_{0}{T(\tau)xd\tau} \in D(A)$$ for every $x \in X$ and $\forall$ $t \geq0$ and more $$A\int^{t}_{0}{T(\tau)xd\tau}=T(t)-x.$$

I found a proof which is not so clear for me:

$$\frac{T(h)\int^{t}_{0}{T(\tau)xd\tau}-\int^{t}_{0}{T(\tau)xd\tau}}{h}=\frac{1}{h}\int^{t}_{0}{T(h+\tau)xd\tau}-\frac{1}{h}\int^{t}_{0}{T(\tau)xd\tau}=$$ $$=\frac{1}{h}\int^{t+h}_{h}{T(s)xds}-\frac{1}{h}\int^{t}_{0}{T(\tau)xd\tau}\color{red}=$$ $$\color{red}{=\frac{1}{h}\int^{t+h}_{t}{T(s)xds}-\frac{1}{h}\int^{h}_{0}{T(s)xds}\rightarrow T(t)x-x}, $$ so $\displaystyle \int^{t}_{0}{T(\tau)xd\tau} \in D(A). $

What is written with $\color{red}{red}$ is what I cannot understand. How can I obtain that integral with $\color{red}{red}$ and why if $h \to 0$ then I obtain $T(t)x-x$ ?

thanks :)

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1 Answer 1

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To arrive at the integrals in red note that $$ \int_h^{h+t}-\int_0^t = \int_0^{h+t}-\int_0^h-\int_0^t=\int_t^{t+h}-\int_0^h $$

After that the idea is that $\frac{1}{h}\int_t^{t+h} T(s)ds \to T(t)$. It is just like differentiating an integral, but the function under the integral is not real valued, but with values in a Banach Space. That is a Bochner integral.

If you have a $C_0$-semigroup then the mapping $t \mapsto T(t)x$ is continuous, and a similar theorem holds, namely that the Bochner integral $\frac{1}{h}\int_t^{t+h} T(s)ds$ is differentiable and its derivative is $T(t)$.

In the end you use that $T(0)=I$ to get the final result.

A proof of the result about the differentiability of the integral function can be found in this link: Corollary B.5.2

(I guess you have acces to P. Preda books on this subject at Timisoara. They do have an introductory chapter on Bochner integration. Maybe you can find the proof of what I said above.)

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Thanks again Beni :) I will read what you recommend me. It's not so clear for me this notions, one of them seems really hard. If you have time, can you tell me, plese why : $\displaystyle \frac{e^{th}}{h} \int^{h}_{0}{e^{-\lambda s}T(s)xds}\to x$ when $h \to 0_{+}$? I asked you this, because I want to prove that $R_{\lambda}x\in D(A)$. Thanks :-) –  Iuli Jan 27 '13 at 18:41
    
It's exactly the same proof as above. $e^{th} \to 1$ and the rest is again just an integral differentiation of the same form. –  Beni Bogosel Jan 27 '13 at 21:35
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