Show convergence of the Lebesgue integral of some expectations

Question

Let

• $(\mathcal D(A),A)$ be a linear operator on $C_0(\mathbb R)$
• $(\mathcal D(A_n),A_n)$ be a linear operator on the space $\mathcal M_b(\mathbb R)$ of bounded Borel measurable functions from $\mathbb R$ to $\mathbb R$ for $n\in\mathbb N$
• $(Y^n_t)_{t\ge0}$ be a real-valued right-continuous process on a probability space $(\Omega_n,\mathcal A_n,\operatorname P_n)$ for $n\in\mathbb N$

Moreover, let $$\left\|f\right\|_B:=\sup_{x\in B}|f(x)|$$ for $f:\mathbb R\to\mathbb R$ and $B\subseteq\mathbb R$.

Fix $f\in\mathcal D(A)$ and $T>0$. Assume there is a sequence $f_n\in\mathcal D(A_n)$, $n\in\mathbb N$ and a $(B_n)_{n\in\mathbb N}\subseteq\mathcal B(\mathbb R)$ with $$c:=\sup_{n\in\mathbb N}\left\|f_n\right\|_\infty<\infty\tag1,$$ $$\operatorname P_n\left[\forall t\in[0,T]:Y^n_t\in B_n\right]\xrightarrow{n\to\infty}1\tag2$$ and $$\left\|f_n-f\right\|_{B_n}+\left\|A_nf_n-Af\right\|_{B_n}\xrightarrow{n\to\infty}0\tag3.$$ Now let $$\tau_n:=\inf\left\{t>0:\left(\int_0^t\left|(A_nf_n)(Y^n_s)\right|^2\:{\rm d}s\right)^{\frac12}>\sqrt t(\left\|Af\right\|_\infty+1)\right\}$$ for $n\in\mathbb N$. Note that $$\operatorname P_n\left[\tau_n\le T\right]\xrightarrow{n\to\infty}0\tag4.$$ Let $$Z^n_t:=1_{\left\{\:t\:<\:\tau_n\:\right\}}(A_nf_n)(Y^n_t)\;\;\;\text{for }t\ge0.$$

How can we show that $$\int_0^{T-t}\left|\operatorname E_n\left[Z^n_{t+s}-(Af)(Y^n_{t+s})\right]\right|\:{\rm d}s\xrightarrow{n\to\infty}0\tag5$$ for all $t\in[0,T]$?

My idea is to write $$Z^n_s-(Af)(Y^n_s)=1_{\left\{\:s\:<\:\tau_n\:\right\}}(A_nf_n-Af)(Y^n_s)-1_{\left\{\:s\:\ge\:\tau_n\:\right\}}(Af)(Y^n_s)\tag6,$$ but I'm not able to show convergence of the corresponding integral. Note that the left-hand side of $(5)$ is equal to $$\int_t^T\left|\operatorname E_n\left[Z^n_s-(Af)(Y^n_s)\right]\right|\:{\rm d}s\tag7.$$

Answer 1

Clearly,

\begin{align*} |(A_n f_n-Af)(Y_s^n)| &\leq \left( |A_n f_n(Y_s^n)| + |Af(Y_s^n)| \right) 1_{\{Y_s^n \notin B_n\}} \\ &\quad + \sup_{y \in B_n} |(A_n f_n-Af)(y)| 1_{\{Y_s^n \in B_n\}} \\ &\leq \left( |A_n f_n(Y_s^n)| + \|Af\|_{\infty} \right) 1_{\{\exists s \in [0,T]: Y_s^n \notin B_n\}} \\ &\quad + \|A_nf_n-Af\|_{B_n} \end{align*}

for all $s \in [0,T]$. Hence,

\begin{align*}& \int_0^T |E_n[1_{\{s<\tau_n\}} (A_n f_n-Af)(Y_s^n)]| \, ds \\&\leq E_n \left[ 1_{\{\exists s \in [0,T]: Y_s^n \notin B_n\}} \int_{(0,T \wedge \tau_n)} |A_n f_n(Y_s^n)| \, ds \right] + T \|Af\|_{\infty} P_n(\exists s \in [0,T]: Y_s^n \notin B_n) \\ &\quad + T \|A_nf_n-Af\|_{B_n}. \end{align*}

Since, by Jensen's inequality, $$\int_{(0,u)} |A_n f_n(Y_s^n)| \, ds \leq u \sqrt{\int_{(0,u)} |A_n f_n(Y_s^n)|^2 \, ds}$$

it follows from the definition of $\tau_n$ that

$$\int_{(0,T \wedge \tau_n)} |A_n f_n(Y_s^n)| \, ds \leq \sqrt{T} (\|Af\|_{\infty}+1)$$

Thus,

\begin{align*} &\int_t^T |E_n[1_{\{s<\tau_n\}} (A_n f_n-Af)(Y_s^n)]| \, ds \\ &\leq (T+ T^{1/2}) (\|Af\|_{\infty}+1) \left[1 - P_n \left(\forall s \in [0,T]: Y_s^n \in B_n \right) \right] + T \|A_nf_n-Af\|_{B_n} \\ &\xrightarrow[(2),(3)]{n \to \infty} 0 \tag{8}. \end{align*}

On the other hand,

$$1_{\{s>\tau_n\}} |Af(Y_s^n)| \leq \|Af\|_{\infty} 1_{\{s>\tau_n\}} \leq \|Af\|_{\infty} 1_{\{T>\tau_n\}}, \qquad s \leq T,$$

gives

$$\int_0^T |E_n(1_{\{s>\tau_n\}} (Af)(Y_s^n)| \, ds \leq \|Af\|_{\infty} T P_n(\tau_n<T) \xrightarrow[(4)]{n \to \infty} 0. \tag{9}$$

Combining $(8)$ and $(9)$ yields

$$\int_0^T |E_n[Z_s^n-(Af)(Y_s^n)]| \,ds \xrightarrow[]{n \to \infty} 0.$$

In particular,

$$\int_t^T |E_n[Z_s^n-(Af)(Y_s^n)]| \,ds \xrightarrow[]{n \to \infty} 0$$

for any $t \in [0,T]$.