Computation of domain of infinitesimal generator of Brownian motion

Question

Let $L:C_{0}(\mathbb{R}) \supseteq D(L) \rightarrow C_{0}(\mathbb{R})$ be the generator of the standard one dimensional Brownian motion.

I calculated that $\{f \in C_{0}(\mathbb{R}):f'' \in C_{0}(\mathbb{R})\} \subseteq D(L)$ holds and that $Lf=\frac{1}{2}f''$ on this subset. But how can I show that $\{f \in C_{0}(\mathbb{R}):f'' \in C_{0}(\mathbb{R})\} = D(L)$ holds? I would be really grateful for some hints or a solution.

Answer 1

A mantra: "The domain of the generator is the range of the resolvent".

For standard Brownian motion the ($\alpha$-)resolvent is the integral operator $$ U^\alpha g(x):=\int_{\Bbb R} {1\over \sqrt{2\alpha}}e^{-\sqrt{2\alpha}|y-x|} g(y) dy \quad x\in\Bbb R. $$ (Here $\alpha>0$.)

It's not too hard to check that $D(L)=U^\alpha(C_0(\Bbb R))\subset \{f\in C_0(\Bbb R): f''\in C_0(\Bbb R)\}$.