Proving differentiability of a payoff

Question

Let $F = \Phi (X^x_T)$. $X^x_T$ behaves as a risky asset such that $dX^x_t=rX^x_tdt+\sigma X^x_tdW_t$ with $W_t$ standard Brownian Motion; $X_0^x=x$. $\Phi$ is continuous with linear growth and $P(x) = e^{−rT}\mathbb{E}[F]$. Prove that $P \in C^2(\mathbb{R}_+;\mathbb{R}) $.

I've already proved that $P \in C^1(\mathbb{R}_+;\mathbb{R}$). Also, solving the SDE, we obtain $X^x_T=x e^{(r-\frac{\sigma^2}{2})T+\sigma \sqrt{T}Y}$ with $Y \sim \mathcal{N}(0,1)$ but I have no idea on how to continue. Could some one help me? Thanks in advance.

Answer 1

This is shown in Le Gall Brownian Motion, Martingales, and Stochastic Calculus theorem 7 (for continuous and bounded but the proof is similar). Another source is here lecture 12: stochastic differential equations, diffusion:

Define the infinitesimal generator of the process $X_{t}$ to be the differential generator

$$G=\frac{1}{2}\sigma^2(x)\partial_{xx}+\mu(x)\partial_{x}.$$

Theorem 5 Assume that $\mu(x),\sigma(x)$ are Lipschitz with linear growth. Let $f(x),K(x)$ be continuous such that $K\geq 0$ and $f(x)=O(|x|)$ as $|x|\to +\infty$. Then the function defined by $$u(t,x)=E_{x}\left[exp(-\int_{0}^{t}K(X_{s})ds)f(X_{t})\right]$$ satisfies the diffusion equation $u_{t}=Gu-Ku$ with $u(0,x)=f(x)$.

As mentioned there one needs to make a modification for unbounded. In particular, we need integrability but thanks to the linear growth we have

$$E[\Phi(X_{t}^{x})]\leq cE[|X_{t}^{x}|]=cE[X_{t}^{x}]=cxe^{rT}<\infty.$$

The rest of the proof is the same as in the case of $f$ being bounded and continuous.