Feller Processes and the resolvent To get you on the same page as I, I am following "Continuous time Markov Processes" by Thomas Liggett, and referring to Chapter 3.  In order to make the question self-contained, I will mention that there is supposedly a correspondence between feller processes (Right continuous, left limit paths with values in a separable locally compact metric space obeying the Markov property and the Feller property), their infinitessimal description (via the theory of unbounded operators called the probability generator) and the semigroup.
I am in the process of understanding the proof of how one obtains a semigroup from a Feller process, and so the generators are not involved here.  Let $C(S)$ denote the vanishing at infinity continuous functions on $S$ which is the metric space mentioned above.  Then we define $T(t)f=E^x(f(X_t))$ where $X_t$ is our Feller process.  I see everything that is required except that in (sup) norm, one has $\lim_{t \downarrow0}T_t(f)=f$.  Toward this, he talks about the Resolvent equation.  I'll remind you what that is:
$U(\alpha)(f)=\int_0^\infty e^{-\alpha t}T_t(f)dt$ where the integral is understood as a that of a continuous Banach-space valued function, and converges in the improper sense. (I don't know anything about Bochner integrals, only the Riemann analog.)  Since $C(S)$ is a complete space, the result is a $C(S)$ element.
Then the resolvent equation is $U(a)U(b)=(U(a)-U(b))/(b-a)$.
Toward the proof of what I am asking for help with, he says some "pointwise" or "nonuniform" version of this can be proven. (The original version "in norm" is proven by applying a linear functional so that all the familiar integral interchanges apply, and then using Hahn Banach to deduce that this was enough.)  I do not know what he means, and I do not see how it can be used to finish the proof. (Toward the latter, he says some words, but I don't get it.  For instance, he says that whatever pointwise version of the Resolvent equation is proven then implies that the range of $U(a)$ is independent of $a>0$.)
So I am asking for help with both the statement and proof of the pointwise resolvent theorem, as well as how to then see that $U(a)$ has range independent of $a$.  I clarify that I don't even know why $U(a)$ would have image inside $C(S)$ if I were to understand the integral as converging for each $x \in S$.  There is no reason a pointwise limit in $C(S)$ stays in $C(S)$, but maybe in this case it does?  This is a pure analysis question, but I can't solve it either.
 A: I was wondering about this same thing myself.  I figured out this much:
As for the range of $U_a$ being independent of $a$: For $f \in C(S)$, it suffices to show that $U_bf \in R_a= \left\{ U_af \mid f \in C(S) \right\}$, where $a,b > 0$, since this shows $R_b \subset R_a$ with $b$ and $a$ arbitrary.
To do so, rearrange the resolvent equation to read 
$\begin{align} U_bf &= U_af - (b-a)U_aU_bf \\
&=U_bf = U_a\left( f -(b-a) U_bf \right)\end{align}$  
Thus $U_bf$ is in the image of $U_a$ provided that the function $f-(b-a)U_bf$ is in $C(S)$.
This is true if $U_b$ has image inside $C(S)$, $R_b \subset C(S)$ (i.e. if $U_b$ is a linear operator on $C(S)$).  
But here I'm a bit stuck too: I can prove continuity of $U_bf$ using continuity of $f \in C(S)$, the Feller property of $\left\{ T_s \right\}_{s \ge 0}$, and Lebesgue's Dominated Convergence Theorem.  Proving that $U_bf$ vanishes at infinity is difficult, and I haven't figure that out yet.  Please let me know if you've found such a proof in the meantime.
