When does the cumulative distribution function solve the Kolmogorov backward equation?

Question

For a diffusion $X$ define the cumulative distribution function for $X_T$ started with $X_t=x$:

$$u(t,x):=E^{t,x}(1_{X_T\ge y})$$ Under what conditions does $u$ solve $X$'s Kolmogorov backward equation? Where can I read up on this question?

The Wiki article on Kolmogorov backward equations simply assumes this, but where can I find a proof that it works?

The books listed below give conditions (via the Feynman-Kac Theorem) under which $v(t,x):=E^{t,x}(f(X_T))$ solves the Kolmogorov equation, but only for continuous $f$. So $u$ as defined above does not qualify.

I could find many references for the existence of a transition density that solves the Kolmogorov backward equation, but is this also enough to ensure that $u$ solves it?

References

Oksendal, Stochastic Differential Equations

Karatzas and Shreve, Brownian Motion and Stochastic Calculus

Friedman, Stochastic differential equations and applications

Answer 1

If the function $f : \mathbb{R} \to \mathbb{R}$ is discontinuous, then $u(t,x)= \mathbb{E}_xf(X(t))$ may not satisfy the initial condition, in the sense that the limit statement: $$ \lim_{(t,s) \to (0^+,x)} u(t,s) = f(x) \quad \forall x \in \mathbb{R} \tag{$\star$} $$ may not hold. To be concrete, consider $$ d X(t) = d B(t) \;, \quad X(0) = x $$ where $B(t)$ is a standard Brownian motion on $\mathbb{R}$ and let $$ f(x) = \begin{cases} 1 & \text{if $x\ge 0$} \\ 0 & \text{otherwise} \end{cases} $$ In this case, for any $t>0$: $$ u(t,x) = \mathbb{E} f(x+B(t)) = \frac{1}{2} ( 1 + \operatorname{erf}( \frac{x}{\sqrt{2t}} ) ) \;. $$ However, $ \lim_{(t,s) \to (0^+,0)} u(t,s) $ does not exist, since for any $\alpha>0$ we have that: $$ \lim_{s \to 0^+} u(s^2,s^{\alpha}) = \lim_{s \to 0^+} \frac{1}{2} ( 1 + \operatorname{erf}( \frac{s^{\alpha-1}}{\sqrt{2}} ) ) = \begin{cases} 1/2 & \alpha>1 \\ 1/2 (1 + \operatorname{erf}(1/\sqrt{2}) ) & \alpha=1 \\ 1 & \alpha < 1 \end{cases} \tag{$\star \star$} $$ which depends on $\alpha$ (or in words, the limit depends on the path taken towards the origin). Nevertheless, $u(t,x) \in C^{\infty}((0, \infty)\times \mathbb{R})$, and by dominated convergence, one can show that $u(t,x)$ satisfies the heat equation for any $t>0$. Here is a graphical illustration of the function $u(t,x)$ and three paths from ($\star \star$) corresponding to the choices $\alpha=2$ (red), $\alpha=1$ (black), and $\alpha=1/2$ (blue). Note that each path towards the origin has a different terminus.

To summarize, continuity of $f$ is typically applied to show that the limit statement $(\star)$ holds, which (loosely speaking) is why theorems relating SDEs to parabolic equations often assume that the initial condition $f$ is at least continuous.