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One can associate a strongly continuous contraction semi group (SCCSG) to a Markov process with state space $S$ through its transition function, say $P_t$.

Now one can interpret $P_t$ as a linear operator, acting on the space of bounded functions on S, via

$f(\cdot) \mapsto (P_t f)(\cdot):=\int_S P_t(\cdot,dy)f(dy)$

Or as a map on the space of probability measures via

$\mu(\cdot) \mapsto (\mu P_t)(\cdot):=\int_S \mu(dx)P_t(x,\cdot)$.

Now I encountered the definition of the corresponding potential kernel as

$U = \int_0^{\infty} P_t dt.$

And I am not sure, how to interpret its actions, since I couldn't find an helpful definition.

So, how do I interpret the notation $\mu U$ or $U P_s$ or even $\mu P_s U$?

I think, since $P_t$ is a bounded linear operator, so should U also be a linear operator and thus $U P_s$ seems to be defined as the linear Operator $f\mapsto \int_0^{\infty} (P_t P_s)(f) dt=\int_0^{\infty} (P_{t+s})(f) dt.$

But then, what are $\mu U$ and $\mu P_s U$? Are they, according to the definition of the measure $(\mu P_t)(\cdot)$, also measures, namely $\int_0^{\infty} (\mu P_t)(\cdot) dt$ [Is this finite/ $\sigma$-finite?] and consequently

$\begin{aligned}\int_0^{\infty} [(\mu P_s) P_t](\cdot) dt&=\int_0^{\infty} \int_S \left(\int_S \mu(dy) P_s(y, x)\right) P_t(x,\cdot) dt\\ &= \int_0^{\infty} \int_S \mu(dy)P_{s+t}(y,\cdot) dt=\int_s^{\infty} (\mu P_t)(\cdot) dt \end{aligned},$

is that right?

One further question or rather thoughts: What is the connection between resolvent wrt $P_t$ and the potential kernel defined above? I mean, heuristically the potential kernel is the case $\lambda = 0$ in the $e^{-\lambda t}$-term of the resolvent. But wasn't this case particularly excluded due to convergence issues? Is this connected to the transience of the process? Perhaps, the resolvent is defined only for $\lambda>0$ for existence reasons, since at $\lambda=0$ one has to restrict to transient processes? I don't know, but I would like to know .

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  • $\begingroup$ Your intuitions are right. I suggest to grab a book on the subject. $\endgroup$ – Did Sep 13 '13 at 17:55
  • $\begingroup$ Thank you, I will. Then this "question" is solved. $\endgroup$ – mr.gaussian Sep 13 '13 at 22:31

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