I have a question about convergence of resolvents of Markov processes. 


Let $X$, $X^n$ be Markov processes on a locally compact separable metric space $E$.
We denote $\{ R\_{\alpha}\}\_{\alpha>0}$ and $\{ R\_{\alpha}^n\}\_{\alpha>0}$ by the resolvents of $X$ and $X^n$, respectively. 


We assume the following:


* for any bounded measurable function $f$ and $\alpha>0$, $R\_{\alpha}f$ and $R\_{\alpha}^nf$ are continuous functions on $E$.
* for any bounded measurable function $f$ and $\alpha>0$, $\lim\_{n \to \infty}\sup\_{x \in E}|R\_{\alpha}f(x)-R\_{\alpha}^nf(x)|=0$.


**My question** 


Let $\{p\_t\}\_{t>0}$ and $\{p\_t^n\}$ be the semigroups of $X$ and $X^n$, respectively. Then, can we have the following?


* for any bounded measurable function $f$ and $t>0$, $\lim\_{n \to \infty}\sup\_{x \in E}|p\_tf(x)-p\_t^nf(x)|=0$.


There exists a related result in Theorem 3.4.2 of [Pagy's book](https://link.springer.com/book/10.1007%2F978-1-4612-5561-1). However we cannot apply this results to my question since the space of continuous functions on $E$ is not a Banach space. If you know any other related results, please let me know. 



Uniform convergence of resolvents of Markov processes
In general: *no*.


Let $X$ be the uniform motion to the right with speed $1$ (so $X\_t = X\_0 + t$), and let $X^n$ be the uniform motion to the right with speed $1$ plus an independent Brownian motion with variance $t/(2 n^2)$. Then $R\_\alpha$ is the convolution operator with kernel $e^{\alpha x} \mathbf{1}\_{(-\infty, 0)}(x)$, while $R\_\alpha^n$ is the convolution operator with kernel $$\frac{n}{\sqrt{\alpha + n^2}} \left( e^{2 n ((\alpha + n^2)^{1/2} - n) x} \mathbf{1}\_{(-\infty, 0)}(x) + e^{-2 n ((\alpha + n^2)^{1/2} + n) x} \mathbf{1}\_{(0, \infty)}(x) \right) .$$
Your first condition is satisfied: convolution of a bounded function with an integrable kernel is always continuous. It is also easy to see that the kernels of $R\_\alpha^n$ converge in $L^1$ to the kernel of $R\_\alpha$, which implies your second condition.


On the other hand it is clear that the distribution of $X\_t^n$ is absolutely continuous, so $p\_t^n f$ is continuous as a convolution of a bounded measurable function with an integrable kernel. It follows that if $f$ is discontinuous, $p\_t^n f(x)$ cannot converge uniformly to $p\_t f(x) = f(x + t)$.


