Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 29 / 5

Let $B$ be a standard planar Brownian motion started at $z$. Neighbourhood recurrence means that, with probability one, every nonempty open subset of ${\mathbb{R}}^2$ is visited at arbitrarily large times. Not hitting points means that for each fixed $w\ne z$, ${\mathbb{P}}_z(\exists t\ge 0:B_t=w)=0$. If $w=z$, time zero must be excluded, and there is still no return at any positive time. This assertion concerns each prescribed point; it does not say that the path contains no points.

We prove both claims from the planar Brownian annulus hitting probability. Fix a center $w$ and radii $0<r<\rho =|z-w|<R$, and define the stopping times ${\tau }_r=\inf \{t:|B_t-w|=r\}$ and ${\tau }_R=\inf \{t:|B_t-w|=R\}$. Their minimum $\sigma$ is finite almost surely: a coordinate of Brownian motion cannot stay forever in a bounded interval, by the Brownian reflection principle, so exit from the outer disk is finite. On the annulus,The Itô formula therefore makes $\log |B_{t\wedge \sigma }-w|$ a local martingale. It is bounded between $\log r$ and $\log R$, hence is a genuine martingale. The optional stopping theorem at $t\wedge \sigma$, followed by the bounded convergence theorem, givesThusThe exit can only be at one of the two distinct boundaries by path continuity.

For a fixed $r$, let $R\to \infty$. These events increase to $\{{\tau }_r<\infty \}$: any path up to a finite hitting time is bounded and so avoids some sufficiently large outer circle. The ratio tends to one, proving that any closed disk of positive radius is hit almost surely from any starting point outside it. An initial point inside it already counts as a hit.

To prove the stronger recurrence statement, take a closed disk contained in any prescribed nonempty open set. At each integer time $n$, the Markov property and the preceding disk-hitting result imply that some time $t\ge n$ visits this disk with probability one. Intersecting these events over all $n$ proves visits at arbitrarily large times. A countable basis of disks with rational centers and rational radii makes this simultaneous for all nonempty open sets. This is recurrence of planar Brownian motion, or equivalently neighbourhood recurrence.

For point avoidance, keep $R>\rho$ fixed and let $r\downarrow 0$. Hitting $w$ before ${\tau }_R$ requires hitting every inner circle first. The displayed ratio tends to zero, so this point-hitting event has probability zero. Any finite hit of $w$ would occur before exit from some outer circle with integer radius, again because the preceding path is bounded. A countable union now gives $\boxed{{\mathbb{P}}_z(T_w<\infty )=0(z\ne w)}$, the polar point for planar Brownian motion assertion.

Finally, even if the process starts at $w$, for each deterministic $s>0$ its normal distribution has a density, so $\mathbb{P}(B_s=w)=0$. The Markov property at $s$ and the result from other starting points exclude any visit to $w$ at or after $s$. Taking $s=1/n$ excludes all positive-time returns. Thus planar Brownian motion repeatedly approaches every point, while almost surely never hitting any prescribed point at positive time.