On calculating the generator of Brownian motion

Define $$C_0^2(\mathbb{R}^d)=\{f\in C^2(\mathbb{R}^d):f,\Delta f\in C_0(\mathbb{R}^d)\}$$.

I wanted to calculate the infinitesimal generator of d-dimensional Brownian motion (restricted to $$C_0^2(\mathbb{R}^d)$$).

Let $$\{B_t\}$$ be a d-dimensional Brownian motion starting at $$x\in \mathbb{R}^d$$.

According to Le-Gall (2016), I need to prove that $$h(B_t)-\frac{1}{2}\int_0^t\Delta h(B_s)ds$$ is a martingale for any function $$h\in C_0^2(\mathbb{R}^d)$$.

$$\textbf{My attempt:}$$ use Ito's formula to obtain$$h(B_t)-\frac{1}{2}\int_0^t\Delta h(B_s)ds=h(x) +\int_0^t \sum_{i=1}^d \frac{\partial h}{\partial x_i}(B_s)dB_s^{(i)}$$ if the gradient $$\nabla h$$ is bounded, then the last term of the above equation is a martingale and the proof is complete.

Please help by proving that $$h\in C_0^2(\mathbb{R}^d)$$ guarantees a bounded $$\nabla h$$ or giving an alternative approach to conclude that $$h(B_t)-\frac{1}{2}\int_0^t\Delta h(B_s)ds$$ is a martingale for functions $$h\in C_0^2(\mathbb{R}^d)$$.

• The function $\sin(x^3)/x \in C_0^2(x \in \mathbb R)$ but its derivative is unbounded. Jul 20, 2023 at 13:28
• But the second derivative of the function is also unbounded, which contradicts the definition of $C_0^2(\mathbb{R})$ in this question. Jul 20, 2023 at 13:36
• That's what I get for not reading the definition well. It seems that sending only the Laplacian to zero and not the Hessian is not enough, but I haven't found any counterexamples. Jul 20, 2023 at 16:48

Since $$f,\Delta f$$ are bounded, the
$$M_{t}:=f(B_t)-\frac{1}{2}\int_0^t\Delta f(B_s)ds$$
$$E[M_{t}|\mathcal{F}_{s}]=M_{s}+E[f(B_t)-f(B_s)-\frac{1}{2}\int_{s}^{t}\Delta f(B_r)dr|\mathcal{F}_{s}]$$ and by Markov property and using the semigroup $$P_{t}$$ we get $$=M_{s}+E_{B_{s}}[f(B_{t-s})-f(B_0)-\frac{1}{2}\int_{0}^{t-s}\Delta f(B_r)dr|\mathcal{F}_{s}]$$ $$=M_{s}+P_{t-s}f(B_{s})-f(B_{s})-\frac{1}{2}\int_{0}^{t-s}P_{r}\Delta f(B_s)dr$$ $$=M_{s}+0,$$
where in the last step we used $$\frac{d}{dt}P_{t}f(y)=P_{t}\Delta f(y)$$.