Let
- $(\Omega,\mathcal A,\operatorname P)$ be a complete probability space
- $W$ be a Brownian motion on $(\Omega,\mathcal A,\operatorname P)$
- $b,\sigma:\mathbb R\to\mathbb R$ be Borel measurable with $$|b(t,x)|^2+|\sigma(t,x)|^2\le C_1(1+|x|^2)\;\;\;\text{for all }t\ge0\text{ and }x\in\mathbb R\tag1$$ for some $C_1\ge0$ and $$|b(t,x)-b(t,y)|^2+|\sigma(t,x)-\sigma(t,y)|^2\le C_2|x-y|^2\;\;\;\text{for all }t\ge0\text{ and }x,y\in\mathbb R\tag2$$ for some $C_2\ge0$
- $(X_t)_{t\ge0}$ be a strong solution of $${\rm d}X_t=b(t,X_t){\rm d}t+\sigma(t,X_t){\rm d}W_t\tag3$$
Moreover, let $$(L_tf)(x):=b(t,x)f'(x)+\frac12\sigma^2(t,x)f''(x)\;\;\;\text{for }x\in\mathbb R\text{ and }f\in C^2(\mathbb R)$$ for $t\ge0$. Now, let $f\in C^2(\mathbb R)$. By the Itō formula, $$f(t,X_t)=f(0,X_0)+\int_0^t(L_sf)(X_s)\:{\rm d}s+\underbrace{\int_0^t\sigma f'(X_s)\:{\rm d}W_s}_{=:\:M_t}\;\;\;\text{for all }t\ge0\tag4.$$ Assuming $f'$ is bounded, we obtain that $M$ is a martingale.
Are we able to show that $$\operatorname E\left[\int_0^t\left|(L_sf)(X_s)\right|\:{\rm d}s\right]<\infty\tag5$$ for all $t\ge0$?
I've read here (in Corollary 6.5) that this would be the case. However, I don't see how we can bound $\left|f''(X_t)\right|$.