Smoothness of resolvent of the infinitesimal generator of an Ito diffusion acting on bounded continuous function

Question

Let $dX_t=\sigma(X_t)\,dW_t+\mu(X_t)\,dt$ be an Ito diffusion with Lipschitz coefficients and $\sigma(x)>0$. Let $f(x)$ be a continuous and bounded and non decreasing function. Can we prove that the resolvent $R_rf(x):= \mathbb E_x(\int^\infty_0e^{-rt}f(X_s) \, ds)$ with $r>0$ is at least twice differentiable?

It seems that this is linked to whether the ODE $\frac{1}{2}\sigma(x)u''(x)+\mu(x)u'(x)-ru(x)+f(x)=0$ with boundary condition $u(-\infty)=\frac{\inf f(x)}{r}$ and $u(\infty)=\frac{\sup f(x)}{r}$ has solution in $C^2$ and whether that solution is bounded or at most linear growth. Do we have any results related to that? It seems that most standard ODE literature require the boundary condition to be in finite domain.

Also, we might consider use Feynman Kac equation i.e. Is there any $C^{1,2}$ solution to the PDE $-u_t(t,x)+u(t,x)=\frac{1}{2}\sigma(x)u_{xx}(t,x)+\mu(x)u_x(t,x)-ru(t,x)+f(x)$ with boundary condition of either $u(T,x)=g(x)$ where $g(x)$ is a continuous bounded non-decreasing function, or $u(\infty,x)=0$. And does the solution has polynomian growth condition $\max_{0\leq t\leq T}|u(t,x)|\leq M(1+|x|^{2b})$ for some $M>0$ and $b\geq 1$?

If this is not the case, is there any modification of condition can make this true? for example is $\inf \sigma(x)>0$ enough?

Answer 1

It is sufficient that $\inf_x \sigma(x)>0$ and $\sup_x \sigma(x)<\infty$, and $f$ does not have to be monotone. In this case, denoting $\mathcal L f(x) = \frac{1}{2}\sigma(x)f''(x)+\mu(x)f'(x)$, by Theorem 4.6 and Corollary 4.2 in Chapter 6 of Friedman, Anver, Stochastic differential equations and applications. Vol. 1 Academic Press (1975). ZBL0323.60056, the Cauchy problem \begin{align} -&v_t(t,x)+\mathcal L v(t,x)=0\\ &v(0,x) = f(x) \end{align} has a unique solution satisfying $|v(t,x)|<C(t)\exp\{C(t) x^2\}$ for $t\ge 0, x\in\mathbb{R}$ with some non-decreasing positive $C(t)$, and by Feynman-Kac formula, this solution is given by $$ v(t,x) = \mathbb{E}_x f(X_t) = T_t f(x). $$ Clearly, $v$ is bounded. Therefore, using Fubini and integrating by parts, $$ r\, u(x) = r \,R_r f(x) = r\,\int^\infty_0e^{-rt}T_t f(x) dt = r\int^\infty_0e^{-rt}v(t,x) dt \\ = f(x) - \int^\infty_0e^{-rt}v_t(t,x) dt = f(x)-r\, u(x) + \int^\infty_0e^{-rt}\mathcal L v(t,x) dt. $$ Appealing to boundedness of $v$ again, it is easy to justify that $$\int^\infty_0e^{-rt}\mathcal L v(t,x) dt = \mathcal L\int^\infty_0e^{-rt} v(t,x) dt = \mathcal L u(x).$$ Hence, $$ r\, u(x) - \mathcal L u(x) + f(x) = 0, $$ as required.

The answer to your second question follows immediately from Friedman's results ($u(t,x)$ will be bounded under your assumptions and will have the same polynomial growth as $f$ and $g$ if they grow polynomially; see page 147 ibid).

The results also hold in multidimensional case. Assumptions can be relaxed, thanks to Veretennikov, A. Yu., On strong solutions and explicit formulas for solutions of stochastic integral equations, Mat. Sb., N. Ser. 111(153), 434-452 (1980). ZBL0431.60061. For example, one can assume that $\mu$ is just continuous, and $\sigma$ is just Hölder continuous. (Moreover, $\mu$ can be just measurable or even a generalized function, in which case, naturally, the corresponding PDEs hold in a generalized sense.)