Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 201 / 6

For $m>0$, the Brownian reflection principle givesas $t\to \infty$. Hence Brownian motion hits every positive integer almost surely. Applying the same argument to $-B$ shows that it hits every negative integer almost surely. Taking the countable intersection of these probability-one events provesArbitrarily late positive and negative values occur, and path continuity forces a zero between successive values of opposite sign. Thus recurrence of one-dimensional Brownian motion gives infinitely many visits to zero.

The pair $(B,\tilde{B})$ is a Brownian motion in ${\mathbb{R}}^{2d}$. Assume inductively that $T_{i-1}<\infty$. By the Strong Markov property at $T_{i-1}$, the differenceis a one-dimensional Brownian motion with variance rate two, started from its current value. By recurrence of one-dimensional Brownian motion, it hits zero in finite time almost surely. Hence $T_i<\infty$. Induction through $i=1,\dots ,d$ proves $T_d<\infty$ almost surely.

Start Brownian motions at arbitrary $x,y\in {\mathbb{R}}^d$. Use the successive meeting times from part b, but after coordinate $i$ meets, drive that coordinate of the second process with the first process's increments forever. The Strong Markov property shows that each marginal remains a $d$-dimensional Brownian motion. By part b every coordinate is eventually locked, so the resulting coordinatewise coalescing coupling of Brownian motions has an almost surely finite coalescence time $T$.

Because $f$ is bounded and harmonic, Dynkin formula for Brownian motion shows that $f(B_t^x)$ and $f(B_t^y)$ are bounded martingales. ThereforeSince $T<\infty$ almost surely, the right side tends to zero. Thus $f(x)=f(y)$ for all $x,y$, proving the Brownian coupling proof of the harmonic Liouville theorem.