# 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 $$LA_tf=\frac1t\int_0^tLP_sf\,ds?$$

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 $$\frac{1}{h}\int_t^{t+h} P_sf\,ds\overset{h\to 0^+}{\longrightarrow}P_t f\quad\text{and}\quad \frac{d}{ds}(P_sf)=LP_sf.$$