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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 $h\in C_0^2(\mathbb{R}^d)$.

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  • $\begingroup$ The function $\sin(x^3)/x \in C_0^2(x \in \mathbb R)$ but its derivative is unbounded. $\endgroup$
    – Olius
    Jul 20, 2023 at 13:28
  • $\begingroup$ But the second derivative of the function is also unbounded, which contradicts the definition of $C_0^2(\mathbb{R})$ in this question. $\endgroup$
    – tfatree
    Jul 20, 2023 at 13:36
  • $\begingroup$ 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. $\endgroup$
    – Olius
    Jul 20, 2023 at 16:48

1 Answer 1

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This is covered in Revuz-Yor (1.6) Proposition chapter VII.

Since $f,\Delta f$ are bounded, the

$$M_{t}:=f(B_t)-\frac{1}{2}\int_0^t\Delta f(B_s)ds$$

is integrable. We write the difference

$$E[M_{t}|\mathcal{F}_{s}]=M_{s}+E[f(B_t)-f(B_s)-\frac{1}{2}\int_{s}^{t}\Delta f(B_r)dr|\mathcal{F}_{s}]$$ 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 $\frac{d}{dt}P_{t}f(y)=P_{t}\Delta f(y)$.

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