3-dim Brownian motion, harmonic function and its expectation

Question

Given $f(x)=\frac{1}{|x+z|}$, a function from $\mathbb{R}^3\backslash \{z\}$ to $\mathbb{R}$, $z \in \mathbb{R}^3\backslash \{0\}$ and $B$ a 3-dim Brownian motion. I had succes showing that this function is a harmonic function and with help of Ito formula that $f(B_t)$ is a local martingale. But what is needed is to show that $f(B_t)$ isn't a martingale. There is a hint to show $\lim\limits_{t \to \infty} \mathbb{E}(f(B_t))=0$.

Can someone help me with an upper bound of this? Because i tried to calculate the distribution of $f(B_t)$, but this is something like 1 divided by a big sum of some normally distributed variables and some modificated Chi-squared distributed. No idea how i could do it.

Answer 1

Consider the functions $g=f\mathbf 1_{f\geqslant1}$ and $h=f\mathbf 1_{f\lt1}$, hence $f=g+h$.

• The function $g$ is nonzero only on the ball centered at $-z$ with radius $1$ and the density of $B_t$ is uniformly bounded by $c(t)=(2\pi t)^{-3/2}$ hence $$E(g(B_t))\leqslant c(t)\int_{\|x\|\leqslant1}\frac{\mathrm dx}{\|x\|}=c(t)4\pi\int_0^1\frac{r^2\mathrm dr}{r}=2\pi c(t),$$ which implies that $E(g(B_t))\to0$ when $t\to\infty$.
• In dimension $3$, $\|B_t\|\to\infty$ almost surely hence $h(B_t)\to0$ almost surely. Since $h\lt1$, this implies that $E(h(B_t))\to0$ by dominated convergence.

This shows that $E(f(B_t))\to0$ when $t\to\infty$. But if $(f(B_t))$ is a martingale then $E(f(B_t))=f(0)$ for every $t$. Since $f(0)\ne0$, this is impossible, QED.