Uniform convergence of resolvents of Markov processes

Question

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. 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.

Answer 1

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 $p_t f(x) = f(x + t)$.