Let $\sigma \in C(\mathbb R)$, and $X$ be a solution of \begin{equation}\label{eq:1} X_{t} = x + t + \int_{0}^{t} \sigma(X_{s}) dB_{s} \end{equation} where $B$ is 1-d Brownian motion under filtered probability space $(\Omega, \mathcal F, \mathbb P, \{\mathcal F_{t}\}_{t\ge 0})$. We denote by $\mathbb P^{x}$ the probability on $C([0, \infty), \mathbb R)$ induced by the process $X$ starting from $x$. Consider a stopping time $\tau = \inf\{t>0: X(t) <0\}$.
[Q.] What is the sufficient condition on $\sigma$ to have the continuity of $u(x) = \mathbb E^{x}[e^{-\tau}]$?
[A.] $\sigma \in C(0,1)$ and $\sigma(0) \neq 0$.
[Sketch of Proof.] If $\sigma \in C^{0,1}$, then the following two conditions are standard:
(C1) The above SDE has unique strong solution;
(C2) $\mathbb P^{x_{n}} \Rightarrow \mathbb P^{x}$ whenever $x_{n} \to x$;
Moreover, according to the [Discussion 2] of the post (regularity of zero point), we also have
(C3) $\mathbb P^{0} (\tau = 0) = 1$.
In fact, (C3) implies that the mapping $\omega \mapsto \tau(\omega)$ is $\mathbb P^{0}$ almost surely continuous function on $C([0, \infty), \mathbb R)$ w.r.t. Skorohod topology. (In fact, it's equivalent to the a.s. continuity w.r.t. max norm on $C([0, T], \mathbb R)$ for each $T<\infty$ in this case.) Together with (C2), we have $\tau^{x_{n}}$ converge to $\tau^{x}$ in distribution by mapping theorem. This shows the continuity of $u$. END.
My question is then, can we have weaker sufficient condition on $\sigma$ to have continuity of $u$?
