I'm trying to understand the proof of the implication "(g) $\Rightarrow$ (f)" in Corollary 8.7 of Chapter 4 in the book Markov Processes: Characterization and Convergence by Stewart N. Ethier, Thomas G. Kurtz. Here is the theorem and its proof:

Corollary 8.7

And here are the relevant parts they refer to:

Corollary 8.6

Theorem 8.2

For simplicity, I will assume that $E$ is a locally compact separable metric space, $(T(t))_{t\ge0}$ is a strongly continuous contraction semigroup on $C_0(E)$, $(\mathcal D(A),A)$ is the generator of $(T(t))_{t\ge0}$, $(T_n(t))_{t\ge0}$ is a strongly continuous contraction semigroup on the space of bounded measurable functions $E_n\to\mathbb R$ and $(\mathcal D(A_n),A_n)$ is the generator of $(T_n(t))_{t\ge0}$.

The approach described in the proof of Ethier and Kurtz is analogous to the usual proof of the Trotter-Kato approximaiton theorem. As it is done there, I would like to show that (g) implies the following about the resolvents as an intermediate step: If $f\in C_0(E)$ and $T\ge0$, there is a $G_n\in\mathcal B(E_n)$ with (8.35) and $$\left\|R_\lambda(A_n)\pi_n-\pi_nR_\lambda(A)f\right\|_{G_n}\xrightarrow{n\to\infty}0\tag1,$$ where $R_\lambda(A)$ denotes the resolvent of $(\mathcal D(A),A)$, $R_\lambda(A_n)$ denotes the resolvent of $(\mathcal D(A_n),A_n)$ and for simplicity I write $\left\|g\right\|_G:=\sup_{x\in G}|g(x)|$. How can we show this?

We may note that $(0,\infty)$ is contained in the resolvent set of $(\mathcal D(A_n),A_n)$ and $(\mathcal D(A),A)$ and $$R_\lambda(A_n)f=\int_0^\infty e^{-\lambda t}T_n(t)f\:{\rm d}t\tag2$$ for all bounded measurable $f:E_d\to\mathbb R$ and $\lambda>0$ and $$R_\lambda(A)=\int_0^\infty e^{-\lambda t}T(t)f\:{\rm d}t\tag3$$ for all $f\in C_0(E)$ and $\lambda>0$.

So, given $f\in C_0(E)$, we obtain \begin{equation}\begin{split}|R_\lambda(A_n)\pi_nf-\pi_nR_\lambda(A)|&=\left|\int_0^\infty e^{-\lambda t}(T_n(t)\pi_nf-\pi_nT(t)f)\:{\rm d}t\right|\\&\le\int_0^Te^{-\lambda t}|T_n(t)\pi_nf-\pi_nT(t)f|\:{\rm d}t+\frac{2\left\|f\right\|_\infty}\lambda e^{-\lambda T}\end{split}\tag4\end{equation} for all $T\ge0$ and $n\in\mathbb N$.

By assumption, for all $T\ge0$, there is a $G_T^{(n)}\in\mathcal B(E_n)$ for all $n\in\mathbb N$ with $$\operatorname P\left[Y_n(t)\in G_T^{(n)}\text{ for all }t\in[0,T]\right]\xrightarrow{n\to\infty}1\tag5$$ and $$\left\|T_n(t)\pi_nf-\pi_nT(t)f\right\|_{G_T^{(n)}}\xrightarrow{n\to\infty}0\;\;\;\text{for all }t\in[0,T]\tag6.$$ We may clearly assume $G^{(n)}_T\subseteq G^{(n)}_{T'}$ for all $n\in\mathbb N$ and $T\le T'$.

Question 1: How can we conclude from that? Is it sufficient to consider a single $T\ge0$ and use that the integral converges by dominated convergence and $e^{-\lambda T}\xrightarrow{T\to\infty}0$? Ethier and Kurtz write this as we would need to alter the definition of $G_T^{(n)}$, but by the argumentation above we should be able to stay with $G_T^{(n)}$ ...

Question 2: While Ethier/Kurtz prove that (g) implies (f), I guess that the other direction fails to hold (otherwise they would surely mention that they are equivalent). Are we at least able to show that if (f) holds, then for all $f\in L$ and $t>s\ge0$, there is a $G_n\subseteq E_n$ with $\lim_{n\to\infty}\operatorname P\left[Y_n(s)\in G_n\right]=1$ and $\lim_{n\to\infty}\sup_{y\in G_n}|T_n(t-s)\pi_nf(y)-\pi_nT(t)f(y)|=0$?

  • $\begingroup$ The screenshots of Ethier and Kurtz you include appear to cut out the statement $(g)$. You might want to fix this. $\endgroup$ – Rhys Steele May 19 at 18:10
  • $\begingroup$ @RhysSteele Thanks for nothing that. Fixed it. $\endgroup$ – 0xbadf00d May 20 at 4:58
  • $\begingroup$ @RhysSteele Do you have some thoughts on question 2? $\endgroup$ – 0xbadf00d Jun 7 at 13:54
  • $\begingroup$ @RhysSteele or on this question: math.stackexchange.com/q/3243051/47771, which is strongly related? $\endgroup$ – 0xbadf00d Jun 7 at 14:03
  • $\begingroup$ @RhysSteele I've tried to simplify everything and I've got the feeling that the result should hold, but I'm unable to finish things up: math.stackexchange.com/q/3254532/47771 $\endgroup$ – 0xbadf00d Jun 9 at 4:19

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