Feynman-Kac theorem: probabilistic proof of existence of solution to parabolic PDE

Question

Friedman (in his book: PDEs of Parabolic Type) shows how to construct a solution to the Cauchy problem $$ \partial_t u(t,x) = b(x) \partial_x u(t,x) + \frac{1}{2} \sigma(x)^2 \partial_{x,x} u(t,x) $$ with initial condition $u(0,x) = f(x)$ using the parametrix method (no probability involved) under the conditions: $b, \sigma, f$ are $\alpha$-Holder continuous. In particular, $u \in C^{1,2}$.

It is also well-know by the Feynman-Kac theorem that if $u$ has polynomial growth, then it has a stochastic representation as $u(t,x) = \mathbb{E}[f(X_t^x)]$, where $X$ solves the SDE $$ X_t^x = x + \int_0^t b(X_s^x) \, ds + \int_0^t \sigma(X_s^x) \, dB_s . $$

As Friedman shows (in another book: SDEs and Applications), one can define $v(t,x) := \mathbb{E}[f(X_t^x)]$ and prove directly that $v \in C^{1,2}$ and it solves the above PDE when the following conditions are satisfied: $b, \sigma, f$ are all twice continuously differentiable with bounded derivatives.

My questions:

(1): Why do we make different assumptions about the regularity of $b, \sigma, f$ in each case?

(2): Can one define $v(t,x) := \mathbb{E}[f(X_t^x)]$ and, under the assumptions $b, \sigma, f$ are $\alpha$-Holder continuous (possibly also need linear growth), prove that $v$ solves the PDE?

Answer 1

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 $u_{t}=Au-f$