The $\alpha$-Potential-Operator (Definition and resolvent Equation) during my studies I encountered the following Operator ($X_t$ is the standard Browniang Motion, $\alpha>0$ and $f$ is  bounded function )

$U^{\alpha}f(x)=\mathbb{E}^x \int_0^{\infty} e^{-\alpha t}f(X_t) dt$

I have never worked with this type of Operator before and was wondering how
$U^{\alpha}-U^{\beta}$, $U^{\alpha}U^{\beta}$ and $(U^{\alpha})^n,n>1$ are defined.
Furthermore why does the operator Fulfill the fowllowing resolvent equation:

$U^{\alpha}-U^{\beta}=(\beta-\alpha)U^{\alpha}U^{\beta}$

Thanks in advance :)
 A: Let $f$ be a bounded and measurable function. Then
$$|U^\alpha f(x)| \leq \mathbb{E}^x \left(\int_0^\infty e^{-\alpha t} |f(X_t)| \,dt \right) \leq \mathbb{E}^x \left(\int_0^\infty e^{-\alpha \cdot t} \cdot \|f\|_\infty \, dt \right) = \|f\|_\infty \cdot \int_0^\infty e^{-\alpha \cdot t} \, dt < \infty$$
and therefore the function $x \mapsto U^\alpha f(x)$ is still a bounded (and measurable) function. Thus
$$
(U^\beta f)(x)-(U^\alpha f)(x) = \mathbb{E}^x \left(\int_0^{\infty} (e^{-\beta \cdot t}-e^{-\alpha \cdot t}) \cdot f(X_t) \, dt \right) $$
and
$$(U^\beta (U^\alpha g))(x) = \mathbb{E}^x \left( \int_0^{\infty} e^{-\beta \cdot t} \cdot(U^\alpha g)(X_t) \, dt \right)$$
Iterating the last equation you obtain an expression for $(U^\alpha)^n$.
To show that $U^\alpha$ fulfills the resolvent equation you have to know that $U^\alpha = (\alpha \cdot \text{id}- A)^{-1}$ where $A$ is the generator (so you especially have to know some more stuff about generators of semigroups). If you know this, the equation follows like this:
$$U^\beta f-U^\alpha f = U^\alpha((\alpha \cdot \text{id}-A) U^\beta-\underbrace{\text{id}}_{(\beta \cdot \text{id}-A)U^\beta}) f = U^\alpha(\alpha-\beta)U^\beta f = (\alpha-\beta)U^\alpha U^\beta f$$
