Cesaro average semigroup Let $(P_t)_{t\geq 0}$ be a symmetric strongly continuous semigroup on $L^2(X,\mu)$ with generator $(L,\mathcal{D}(L))$. Given $f\in\mathcal{D}(L^*)$, define the Cesaro average
$$
A_tf=\frac{1}{t}\int_0^tP_sf\,ds,\qquad t>0,
$$
where the integral is understood in the Bochner sense.
Why is $A_t f\in\mathcal{D}(L)$ for all $t>0$ and 
$$
LA_tf=\frac1t\int_0^tLP_sf\,ds?
$$
 A: We have $y\in D(L)$ if the limit
$$\lim_{h\to 0^+} \frac{P_hy-y}{h}$$
exists. In this case, $Ly$ equals to the limit. Therefore, you have to show that the limit
$$\lim_{h\to 0^+} \frac{P_hA_tf-A_tf}{h}$$
exists and is equal to
$$\frac1t\int_0^tLP_sf\,ds.$$
For this, note that
\begin{align*}
\frac{P_hA_tf-A_tf}{h}&=\frac{P_h\frac{1}{t}\int_0^tP_sf\,ds-\frac{1}{t}\int_0^tP_sf\,ds}{h}\\
&=\frac{1}{th}\left(\int_0^tP_{s+h}f\,ds-\int_0^tP_s f\,ds\right)\\
&=\frac{1}{th}\left(\int_h^{t+h}P_{\tau}f\,d\tau-\int_0^tP_s f\,ds\right)\\
&=\frac{1}{th}\left(\int_0^{t+h}P_{\tau}f\,d\tau-\int_0^{h}P_{\tau}f\,d\tau-\int_0^tP_s f\,ds\right)\\
&=\frac{1}{th}\left(\int_t^{t+h}P_{\tau}f\,d\tau-\int_0^hP_\tau f\,d\tau\right)\\
&=\frac{1}{t}\left(\frac{1}{h}\int_t^{t+h}P_{\tau}f\,d\tau-\frac{1}{h}\int_0^hP_\tau f\,d\tau\right)
\end{align*}
Then, taking the limit as $h\to0^+$, we conclude that
$$LA_t f=\frac{1}{t}\left(P_tf-P_0f\right)=\frac{1}{t}\int_0^t\frac{d}{ds}(P_s f)\,ds=\frac{1}{t}\int_0^tLP_s f\,ds.$$
because
$$\frac{1}{h}\int_t^{t+h} P_sf\,ds\overset{h\to 0^+}{\longrightarrow}P_t f\quad\text{and}\quad \frac{d}{ds}(P_sf)=LP_sf.$$
