# Intuition for the definition of a probability generator of a Feller process

I am working with the definition of a probability generator of a Feller process as stated in Liggett's book, "Continuous time Markov processes":

Let $S$ be a compact state space and denote by $C(S)$ the continuous functions on $S$. A probability generator is a linear operator $L$ satisfying:

i) $\mathcal{D}(L)$ is dense in $C(S)$

ii) If $f\in \mathcal{D}(L), \lambda \geq 0$, and $f-\lambda L f =g$, then $$\inf_{x\in S} f(x) \geq \inf_{x\in S} g(x)$$

iii) $Range(1-\lambda L) = C(S)$ for all sufficiently small $\lambda > 0$

iv) $1\in \mathcal{D}(L)$ and $L1 = 0$

Now, properties i) and iv) make sense to me as the domain should cover the space in question and no (probability) mass should be generated out of nowhere.

Property iii) is rather abstract, but I think this acts an extension for boundedness in theoretical statements. As generators are often differential operators, they are often not bounded. Liggett uses this property to show that a generator can be approximated by a bounded operator. On the other hand, every bounded operator fulfills this property.

Unfortunately, I don't get an intuition for property ii).

Is there an intuitive reason why this should hold for a Feller process or is it purely technical as for example just described for property iii)?

• I do not have the book handy, but I am guessing he will go on to show that if $L$ satisfies these conditions, then it generates a semigroup $P_t$ ((iii) seems to be saying something about the spectrum of $L$ and will probably let you invoke Hille-Yosida) and $P_t$ will have the necessary properties to be a (Feller) transition function, so that $P_t f(x) = E_x f(X_t)$ where $X_t$ is the associated Markov process. (iv) will give you $P_t 1 = 1$ and I suspect (ii) will be related to showing that $P_t$ is positivity preserving. – Nate Eldredge Jul 14 '14 at 15:09

Property ii) is natural to ask if you keep in mind the martingale characterization of the generator. Indeed, let $x$ such that $f(x)=\inf f$. Let now $X_t$ be a Markov process with generator $L$ started at $x$. We have

$\mathbb{E}(f(X_t) )\ge f(x)$.

But the process $f(X_t) -\int_0^t Lf(X_s) ds$ is a martingale so

$\mathbb{E}(f(X_t)) =f(x) +\int_0^t \mathbb{E}(Lf(X_s)) ds$

We deduce

$\int_0^t \mathbb{E}(Lf(X_s)) ds \ge 0$

If you divide by $t$ and let $t$ go to 0 you get

$Lf(x) \ge 0$

This implies $f(x) -g(x) \ge 0$ and thus $\inf f \ge \inf g$. In other words the generator of a Markov semigroup satisfies a maximum principle.