Proof of a property of the uniformly continuous semigroups.

Question

Let $X$ be a Banach space and $T:=\{T(t)\}_{t\geq 0}$ an uniformly continuous semigroup of bounded linear operators on $X$.

So, we know that

(i) $T(t)$ is linear and bounded for all $t\geq 0$;
(ii) $T(0)=I$;
(iii) $T(t + s)= T(t)T(s)$ for every $t,s\geq 0$;
(iv) $\lim_{t\downarrow 0}\|T(t)-I\|=0.$

I need help to prove that the following equality holds.

$$\lim_{h\to0^+}\left\|\frac{1}{h}\int_t^{t+h}T(s)ds-T(t)\right\|=0.$$

Thanks.

Answer 1

Obviously, we have by the semigroup property (iii)

$$\begin{align*} \frac{1}{h} \int_t^{t+h} T(s) \, ds - T_t &=\frac{1}{h} \int_t^{t+h} (T(s)-T(t)) \, ds \\ &=\frac{1}{h} \int_0^h T_t(T(s)-I) \, ds. \end{align*}$$

Since $T(t)$ is a bounded operator, this implies

$$\begin{align*} \left\| \frac{1}{h} \int_t^{t+h} T(s) \,ds - T(t) \right\| &\leq \frac{1}{h} \int_0^h \|T(t)\| \cdot \|T(s)-I\| \, ds \\ &\leq \|T(t)\| \cdot \sup_{s \leq h} \|T(s)-I\|.\end{align*}$$

As $\lim_{h \downarrow 0+} \|T(h)-I\|=0$, this finishes the proof.