# If $B$ is a Brownian motion and $f \in \mathcal C^2(\mathbb R^n)$ harmonic, then $f(B)$ is a continuous martingale

I'm reading about Itô's formula for a multi-dimensional Brownian motion at page $$39$$ of these notes, i.e.,

Let $$\underline{B}$$ be a standard $$n$$-dimensional Brownian motion and $$f \in \mathcal{C}^2\left(\mathbb{R}^n\right)$$. Then $$f\left(\underline{B}_t\right)-f\left(\underline{B}_0\right)=\sum_{i=1}^n \int_0^t f_{x_i}^{\prime}\left(\underline{B}_s\right) d B_s^{(i)}+\frac{1}{2} \int_0^t \Delta f\left(\underline{B}_s\right) d s \quad \text { a.s., } \quad \forall t \in \mathbb{R}_{+} .$$

Again, the process $$\left(f\left(\underline{B}_t\right), t \in \mathbb{R}_{+}\right)$$is a continuous semi-martingale. In particular:

• if $$\Delta f(\underline{x})=0, \forall \underline{x} \in \mathbb{R}^n$$, then $$f(\underline{B})$$ is a continuous local martingale.
• if $$\Delta f(\underline{x}) \geq 0, \forall \underline{x} \in \mathbb{R}^n$$, then $$f(\underline{B})$$ is a continuous local submartingale.
• if $$\Delta f(\underline{x}) \leq 0, \forall \underline{x} \in \mathbb{R}^n$$, then $$f(\underline{B})$$ is a continuous local supermartingale.

Whether the word "local" can be removed or not in the above sentences depends now on technical conditions. From what we have already seen, we know that if $$\Delta f(\underline{x})=0, \forall \underline{x} \in \mathbb{R}^n$$ and $$\mathbb{E}\left(\int_0^t\left(f_{x_i}^{\prime}\left(\underline{B}_s\right)\right)^2 d s\right)<\infty, \quad \forall t \in \mathbb{R}_{+}, \quad \forall 1 \leq i \leq n, \qquad (18)$$ then $$f(\underline{B})$$ is a continuous square-integrable martingale. Since $$\langle f(\underline{B})\rangle_t=\sum_{i=1}^n \int_0^t\left(f_{x_i}^{\prime}\left(\underline{B}_s\right)\right)^2 d s$$ (notice that this process is always well defined, even in the case where $$f(\underline{B})$$ is not a martingale), we see that condition $$(18)$$ is equivalent to $$\mathbb{E}\left(\langle f(\underline{B})\rangle_t\right)<\infty, \quad \forall t \in \mathbb{R}_{+}.$$

My understanding Assume $$\Delta f(\underline{x})=0, \forall \underline{x} \in \mathbb{R}^n$$. Then $$\mathbb E [f\left(\underline{B}_t\right)] - f\left(0\right) = \sum_{i=1}^n \mathbb E \bigg [ \int_0^t f_{x_i}^{\prime}\left(\underline{B}_s\right) d B_s^{(i)} \bigg ] =0 \quad \forall t \in \mathbb{R}_{+} .$$

It follows that $$f(\underline{B})$$ is not just a continuous local martingale but a continuous martingale.

Could you confirm if my above understanding is correct?

• You probably need that $\int_0^t \left|\nabla f\right|^2 ds$ is finite for any $t > 0$ to conclude that $\Delta f (x) = 0$ implies that $f(B)$ is a martingale. It is not clear from what you wrote that you assume this condition in your final statement. Commented Mar 18, 2023 at 13:03

$$\mathbb E \bigg [ \int_0^t f_{x_i}^{\prime}\left(\underline{B}_s\right) d B_s^{(i)} \bigg ]$$ has no reason to be $$0$$.
And even if it was, having $$\mathbb E [f\left(\underline{B}_t\right)] - f\left(0\right)=0$$ for all $$t\in\mathbb R_+$$ is not enough to deduce that $$f(\underline{B})$$ is a martingale. It would be true for instance if you showed that for any bounded stopping time $$T$$, $$f\left(\underline{B}_T\right)$$ is integrable and $$\mathbb E [f\left(\underline{B}_T\right)] - f\left(0\right)=0$$. But you would not be able to show it with the only assumption that $$\Delta f(\underline{x})=0, \forall \underline{x} \in \mathbb{R}^n$$. As you wrote above, this implies that $$f(\underline{B})$$ is a local martingale, not necessarily a martingale.