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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?

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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.

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  • $\begingroup$ Many thanks for your answer. This is extremely useful. A quick question before I fully digest your answer. What exactly is uniform degeneracy? $\endgroup$
    – avk255
    Sep 8, 2015 at 12:57
  • $\begingroup$ @Aditiya, uniforn non-degeneracy, sorry. This is $v(x) > c > 0$. $\endgroup$
    – zhoraster
    Sep 8, 2015 at 13:41
  • $\begingroup$ Brilliant. This is the intuition I too had. Your reply is extremely useful. Sorry to bug you more but since I didn't know that $E(e^{-r \tau^y_x}) \rightarrow 1$ I tried to prove it myself. My proof is very complicated as it involves time change of the stochastic integral. The reason why I have to do it is because $v(x)$ is a function of $x$ and not a constant. So I was wondering if you had an easy proof of this, seemingly obvious, claim. $\endgroup$
    – avk255
    Sep 8, 2015 at 14:02
  • $\begingroup$ @Aditiya, this is a standard fact about one-dimensional diffusions, see Section 4.7 in Ikeda-Watanabe or Ito-McKean (here can't say where exactly as I don't have the book at hand). But yes, you can change the time to get a Wiener process + bounded drift, and here this is just the consequence of ILL. And don't hesitate to ask, I will be glad to help (although I am not a great specialist in diffusions). $\endgroup$
    – zhoraster
    Sep 8, 2015 at 14:09
  • $\begingroup$ Thanks. The "problem" with Ito-McKean, from what I see, is that most of the work on diffusion assumes continuous drift and volatility. Part of the reason, I guess, they do that is because having a strong solution could be an issue otherwise for the associated SDE. In my case, my starting point is that I know that I have a strong solution to the SDE despite the coefficients not being continuous. But I also know they're bounded the way I said. So I am not able to find the exact result I can readily use (which I would like) that'll tell me that $E(e^{-r\tau^y_x}) \rightarrow 1$. $\endgroup$
    – avk255
    Sep 8, 2015 at 16:18

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