I am struggling with this problem:
Given a stochastic differential equation $$ dX_t = b(X_t) dt + \sigma (X_t) \,dW_t $$ where $W$ is a Brownian motion and the functions $b$ and $\sigma$ are bounded and smooth. Assume that for every square-integrable $\xi$ independent of $W$, there exists a unique strong solution $X$ such that $X_0 = \xi$ and $\sup_{t \geq 0} \mathbb{E} ( X^2_t) < + \infty$. Assume that for some constant $k>0$, \begin{equation} \mathbb{E} [(X_t - Y_t)^2] \leq \mathbb{E} [ (X_0 - Y_0)^2] e^{-kt}, \end{equation} for any two strong solutions $X$ and $Y$. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a smooth function and let $u: [0, \infty) \times \mathbb{R}$ be a bounded and smooth solution of the (Feynman-Kac) PDE $$ \frac{\partial u }{\partial t} = b(x) \frac{\partial u}{\partial x} + \frac{1}{2} \sigma(x)^2 \frac{\partial^2 u}{\partial x^2}, $$ with boundary condition $$u(0,x) = f(x), \quad \forall x \in \mathbb{R}.$$ Now let $\frac{b}{\sigma^2}$ be locally integrable and let $$ p(x) = \frac{C}{\sigma(x)^2} \exp \bigg\{ \int_0^x \frac{2 b(s)}{\sigma(s)^2} \,ds \bigg\},$$ where $C>0$ is chosen so that $p$ is a density function. Using the fact that $$ \int_{\mathbb{R}} u(t,x) p(x) \,dx = \int_{\mathbb{R}} f(x) p(x) \,dx \quad \forall t \geq 0,$$ show that $$u(t,x) \rightarrow \int_{\mathbb{R}} f(y) p(y) \,dy \quad \quad \text{ as } \, \, t \rightarrow \infty,$$ for all $x \in \mathbb{R}.$
I know from standard results that $u(t,x) = \mathbb{E} [ f(X_t) | X_0=x]$. However, how does this relate to the integral equality involving $p(x)$ and the bound involving $\mathbb{E}[ (X_t -Y_t)^2 ]$? Any ideas to this complicated problem?