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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:


Thanks in advance :)

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up vote 1 down vote accepted

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

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


$$(U^\beta (U^\alpha g))(x) = \mathbb{E}^x \left( \int_0^{\infty} e^{-\beta \cdot t} \cdot(U^\alpha g)(X_t) \, dt \right)$$

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:

$$U^\beta f-U^\alpha f = U^\alpha((\alpha \cdot \text{id}-A) U^\beta-\underbrace{\text{id}}_{(\beta \cdot \text{id}-A)U^\beta}) f = U^\alpha(\alpha-\beta)U^\beta f = (\alpha-\beta)U^\alpha U^\beta f$$

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Do you by any chance have a reference where it is shown that $U^{\alpha}=(\alpha\cdot id - a)^{-1}$ holds ?? – Probabilitnator Jul 26 '12 at 20:06
It's contained in "Brownian Motion: An Introduction to Stochastic Processes" - René L. Schilling/Lothar Partzsch, chapter 7. – saz Jul 27 '12 at 14:51

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