Fix a discount rate $r>0$, and let $m,v,f:\mathbb{R} \rightarrow \mathbb{R}$ be bounded measurable functions of locally bounded variation, with $v$ globally bounded below by some strictly positive constant i.e. $v(x) > c > 0$ 


Given a standard Brownian motion B and a starting value x, I know the SDE
$dX\_t = m(X\_t) dt + v(X\_t) dB\_t$
has a unique strong solution with $X\_0 = x$. 


Let 
$U(x):= E [ \int\_0^\infty re^{-rt} f(X\_t) dt | X\_0=x]$, a well-defined number for each $x.$


Question: 
How nice is U guaranteed to be, without imposing further conditions on m,v,f? [In particular, I don't want to assume any of them to be continuous. If further boundedness-type conditions help, I'm more okay imposing those.]
Is U guaranteed to be continuous? Differentiable?



Continuity of expected payoff from a diffusion
This is just a partial answer addressing the continuity: $U$ is continuous, even uniformly continuous. Below is the idea.


Denote $\tau\_x^y = \inf\{ t\ge 0: X\_t = y\,|\, X\_0 = x\}$, then $\tau\_x^y<\infty$ a.s. in view of the uniform non-degeneracy of $v$ and boundedness of $m$, so, using the strong Markov property, $$U(x) = rE\left[\int\_0^{\tau\_x^y} e^{-rs}f(X\_t)dt \,\Big|\, X\_0=x\right] + E\big[e^{-r\tau\_x^y}\big]U(y).$$
The first term converges to $0$ uniformly as $|y-x|\to 0$ in view of the uniform non-degeneracy of $v$ and boundedness of $m$. For the same reason, $E\big[e^{-r\tau\_x^y}\big]\to 1$ uniformly.


Update: $U'$ is in general discontinuous. Start by noting that we expect $U$ to solve equation $$rU - \mathcal A U - f = 0,\tag{1}$$ where $\mathcal A$ is the generator of $X$, at least in some weak sense. Then it is already clear that $U'$ or $U''$ must be discontinuous in the point where $f$, $m$ or $v$ are discontinuous.


