Suppose $X$ is a process given by -


$dX\_t = db\_t$ 
where $b\_t$ is a standard Brownian motion with its filtration $(\mathcal{F}\_t)$.


Suppose an agent earns a payoff given by 


$V(x) = \mathbb{E} [\int\_0^\infty e^{-\int\_0^t r(X\_s)ds} dt|X\_0 =x] $ 


where $r(x) = \begin{cases} 3 & \text{ if } x \ge 0 \\
7 & \text{ otherwise} 
\end{cases}$


I am interested in computing $V(x)$. In particular, I am interested in knowing if $V(x)$ is differentiable at $0$?


I suspect that the answer is no for differentiability. But I don't have a proof. 


Thanks.



Differentiability of value function
I think it's differentiable (everywhere).
Interchanging operations freely, we have
\begin{eqnarray\*}
\newcommand{\D}{\frac{\mathrm d}{{\mathrm d}x}}
\newcommand{\E}{\mathbb E}
V'(x) &=& \D \E\_x \int\_0^\infty \exp\left(-\left(\int\_0^t 3+4[W\_s<0]\mathrm ds\right)\right)\mathrm dt
\\
&=& \int\_0^\infty \E\_0 \D\exp\left(-\left(\int\_0^t 3+4[W\_s<-x]\mathrm ds\right)\right)\mathrm dt
\end{eqnarray\*}
Then we end up needing
$$
L^x(t) := \D \int\_0^t [W\_s<x]\,\mathrm ds.
$$
If we try to interchange the $\D$ with the $\int$ here it seems to not exist at $x=W\_s$. However, we don't need to as, as far as I can tell, this is just the [local time](https://en.wikipedia.org/wiki/Local_time_(mathematics)#Formal_definition) of $W$ at $x$.


