= Solution
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 <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>.
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