Let
- $b,\sigma\in C_b([0,\infty)\times\mathbb R)$
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$
- $B$ be an $\mathcal F$-Brownian motion on $(\Omega,\mathcal A,\operatorname P)$
- $X$ be a real-valued stochastic process on $(\Omega,\mathcal A,\operatorname P)$ with $${\rm d}X_t=b(t,X_t){\rm d}t+\sigma(t,X_t){\rm d}B_t\;\;\;\text{for all }t\ge0\tag1$$
Now, let $$A_tf:=b(t,\;\cdot\;)f'+\frac12\sigma^2(t,\;\cdot\;)f''\;\;\;\text{for }f\in C^2_b(\mathbb R)$$ By the Itō formula, $$M_t^{[f]}:=f(X_t)-f(X_0)-\int_0^t(A_sf)(X_s)\:{\rm d}s=\int_0^t\sigma(s,X_s)f'(X_s)\:{\rm d}B_s\;\;\;\text{for }t\ge0\tag2$$ for all $f\in C^2(\mathbb R)$ (clearly, the $A_t$ are well-defined for such $f$ too).
I'm new to martingale problems, but if I understood the idea correctly, we're given a family $(\mathcal L_t)_{t\ge0}$ of linear operators from a subspace $\mathcal D\subseteq\left\{g:\mathbb R\to\mathbb R\mid g\text{ is bounded and Borel measurable}\right\}$ to $\left\{g:\mathbb R\to\mathbb R\mid g\text{ is Borel measurable}\right\}$ and say that $X$ is a solution of the $\mathcal F$-martingale problem for $(\mathcal D,(\mathcal L_t)_{t\ge0})$ iff $$N_t^{[f]}:=f(X_t)-f(X_0)-\int_0^t(\mathcal L_sf)(X_s)\;\;\;\text{for }t\ge0$$ is an $\mathcal F$-martingale for all $f\in\mathcal D$.
Now, in the scenario described above, I've seen that people choose the domain for the operators $A_t$ to be $C_0^2(\mathbb R)$ or even $C_0^\infty(\mathbb R)$. What's the reason for that?
Clearly, the important thing is that the right-hand side of $(2)$ is a (strict) $\mathcal F$-martingale. This ensured as long as $(\sigma(t,X_t)f'(X_t))_{t\ge0}$ is square-integrable. Since $\sigma$ is bounded by definition, it should be sufficient to ensure that $f'$ is bounded too. So, isn't $C_b^2(\mathbb R)$ the more natural choice for the domain of the $A_t$?