Minimal set of condition for the Feynmann-Kac formula to hold

Question

The Feynmann Kac formula tell us that the solution of the PDE $$ \frac{\partial u}{\partial t}(x,t) + \mu(x,t) \frac{\partial u}{\partial x}(x,t) + \tfrac{1}{2} \sigma^2(x,t) \frac{\partial^2 u}{\partial x^2}(x,t) -V(x,t) u(x,t) + f(x,t) = 0$$ with terminal condition $u(x,T)=\psi(x)$

Can be represented as $$ u(x,t) = E^Q\left[ \int_t^T e^{- \int_t^\tau V(X_s,s)\, ds}f(X_\tau,\tau)d\tau + e^{-\int_t^T V(X_\tau,\tau)\, d\tau}\psi(X_T) \,\Bigg|\, X_t=x \right] $$ where $X_\tau$ satisfies the SDE $$ dX_t = \mu(X,t)\,dt + \sigma(X,t)\,dW^Q_t, $$ The article does not specify the regularity of the various coefficients of the PDE

In the wikipedia article is quoted that:

"A proof that the above formula is a solution of the differential equation is long, difficult and not presented here"

Where I can find this "long difficult proof" quoted in the article?

What are the minimal set of conditions on the coefficients for which the FK formula still holds (and so we have unicity of the PDE)

---

In my notes about SDE sufficient conditions are that:

1. $\mu,\sigma \in L^\infty_{\text{loc}} (\Omega)$ where $\Omega$ is a bounded domain such that the exit time of the Brownian motion is finite a.s.
2. There exists a strong solution of the SDE for any initial condition $X_0=x \in \Omega$
3. $V(x,t) \ge 0$
4. There exists $u \in C^2(\Omega) \cap C (\bar{\Omega})$ that is a solution of the PDE with boundary condition $u_{\partial \Omega}=\psi \in C(\partial \Omega)$

If the statement above would be true this means that condition $4$ is always satisfied. Moreover also the boundness of the domain would not be necessary

Answer 1

In Shreve-Karatzas pg.366, they study the Feynman Kac formula for continuous and linear growth coefficients.

In the work "Feynman-Kac Formulas for Solutions to Degenerate Elliptic and Parabolic Boundary-Value and Obstacle Problems with Dirichlet Boundary Conditions" the authors develop a Feynman-Kac formulation in the case of weak regularity for the coefficients (including Hölder)

$$Au:=\frac{1}{2}\operatorname{tr}(a(x)D^{2}u(x))-\langle b(x),Du(x)\rangle+c(x)u(x)$$

and boundary data $g$, they obtain for the elliptic case $f=Au$

and the parabolic case $u_{t}=Au-f$