I am studying a paper about Markov semigroup and found an equation about resolvent operator. Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a measurable function, and $P(t):\mathcal{M}(\mathbb{R} )\rightarrow \mathcal{M}(\mathbb{R} )$ be a semigroup act on measurable functions. The resolvent operator $R$ is defined by
$Rf \equiv \int_0^{\infty} e^{-t}P(t)fdt$.
The paper mentioned following identity
$R^n f = \int_0^{\infty} \frac{e^{-t} t^n}{n!} P(t)f dt $
without a proof. Could anyone give me some clue or reference about this? Thanks!
The identity can be proved by induction. Throughout my answer, I will assume that $f$ is bounded and measurable; this ensures that the appearing integrals are well-defined.
By the very definition of the resolvent, the identity holds for $n=1$. Now assume that it holds for some $n \in \mathbb{N}$. Then
$$R^{n+1} f = R(R^n f) = \left( \int_0^{\infty} e^{-t} P_t \left( \int_0^{\infty} e^{-s} \frac{s^n}{n!} P_s f \, ds \right) \, dt \right),$$
and therefore, by Fubini's theorem and the semigroup property,
$$R^{n+1} f = \int_0^{\infty} \int_0^{\infty} e^{-(t+s)} \frac{s^n}{n!} \underbrace{P_t P_s f}_{P_{t+s} f} \, dt \, ds.$$
Performing a change of variables ($u:=s+t$) for the inner integral, we get
$$R^{n+1} f = \frac{1}{n!} \int_0^{\infty} s^n \left( \int_s^{\infty} e^{-u} P_u f\, du \right) \, ds.$$
Now the integration by parts formula yields
\begin{align*} R^{n+1} f &= \frac{1}{n!} \int_0^{\infty} \left( \int_0^s r^n \, dr \right) e^{-s} P_s f \, ds \\ &= \frac{1}{(n+1)!} \int_0^{\infty} s^{n+1} e^{-s} P_s f \, ds \end{align*}
This finishes the inductive step.