Let $X = \{X_t : t \geq 0\}$ be a reflected Brownian motion at level $a > 0$, i.e., we can write \begin{equation} X_t = W_t - A_t, \end{equation} where $W = \{W_t : t \geq 0\}$ is a standard Brownian Motion and where $ A = \{A_t : t \geq 0\}$ is a (nonnegative and nondecreasing) process satisfying $A_t = \max\left(\sup_{0 \leq s \leq t}[W_s - a]; 0\right)$. In words, the process $X$ acts like a Brownian motion on $(-\infty, a)$, and when $X$ tries to travel above $a$, it is reflected at $a$, and the surplus is compensated by the process $A$.
Fix $q > 0$ and $r > 0$. Given $X_0 = x \in [0, a]$ and given an nondecreasing, nonnegative function with bounded derivative $f \in C^2(-\infty, a]$ such that $f(0) > 0$. Furthermore, $f$ is a function with at most linear growth. I am trying to study the following expression when $t \to \infty$: \begin{equation*} F(t,x) := \mathbf{E}_x[e^{-q(t \wedge \tau^r)}f(X_{t \wedge \tau^r})] - \mathbf{E}_x\left[ \int_0^{t \wedge \tau^r} e^{-qs} \left( \frac{1}{2}f''(X_s) - qf(X_s) \right) \mathrm{d} s \right], \end{equation*}
where $\tau^r := \inf\{t > 0 : X_t < 0 \text{ for a longer amount of time } r\}$ and where $\mathbf{E}_x$ denotes the expectation with the starting point. In other words, excursions below $0$ for $X$ cannot last more than $r$ time units. Note that $\tau^r$ can be defined rigorously.
I believe that given the additional assumption that $\frac12 f''(y) - qf(y) = 0$ for all $y \in [0, a]$, we have $\lim_{t \to \infty} F(t,x) = 0$. Does somebody have a clue about how to proceed? I suspect it can be related to a martingale.
My attempt:
The function $g(t,x) := e^{-qt}f(x)$ is also twice continuously differentiable, and Itô's formula yields \begin{equation*} F(t,x) = f(x) - \mathbf{E}_x\left[ \int_0^{t \wedge \tau^r} e^{-qs} f'(X_s) \mathrm{d} A_s\right] \end{equation*}
Note that the (martingale) term $\mathbf{E}_x \left[\int_0^{t \wedge \tau^r} f'(X_s) \mathrm{d} W_s \right]$ vanishes, as $f$ has bounded derivative by assumption. Hence, $\lim_{t \to \infty}F(t,x)$ should be independent of the behaviour of $f$ on the negative, as $x \geq 0$ and we recall that $\mathrm{supp}(\mathrm{d} A_s) \subseteq \{s : X_s = a\}$. Can we find one particular example of $f$ (satisfying the requirements above) for which the limit gives $0$?