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 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 <martingale>[martingales]. Therefore
$$
\begin{aligned}
|f(x)-f(y)|
&=\left|\mathbb E[f(B_t^x)-f(B_t^y)]\right|\\
&\leq2\|f\|_\infty\mathbb P(T>t).
\end{aligned}
$$
Since $T<\infty$ 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>.

Solution

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

{c}

For a bounded harmonic function $f$ on $\mathbb R^d$, the processes $f(B_t^x)$ and $f(B_t^y)$ are bounded martingales. Couple the Brownian motions from $x$ and $y$ so that they coalesce in finite time almost surely. Then
$$
|f(x)-f(y)|
\leq2\lVert f\rVert_\infty\mathbb P(T>t)
\longrightarrow0,
$$
which proves that $f$ is constant in every dimension.


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 Incoming links: Brownian coupling proof of the harmonic Liouville theorem