= Solution
For a <harmonic function> $u$ on $\mathbb C\cong\mathbb R^2$, the <Itô formula> gives
$$
u(B_t)=u(B_0)+\int_0^t\nabla u(B_s)\cdot dB_s+\frac12\int_0^t\Delta u(B_s)ds=u(B_0)+\int_0^t\nabla u(B_s)\cdot dB_s.
$$
Stop on leaving increasing discs and at increasing deterministic times. On each such interval the <gradient> is bounded, so the <stochastic integral> is a <martingale>. This proves that \b[$u(B)$ is a <continuous local martingale>].
If $|u|\leq K$, this is a bounded <local martingale> and hence a true <martingale>. Start <Brownian motion> at an arbitrary $z$. Its <Gaussian heat kernel> gives
$$
u(z)=\mathbb E_z u(B_t)=\int_{\mathbb R^2}u(v)p_t(v-z)dv,\qquad p_t(v)=\frac1{2\pi t}e^{-|v|^2/(2t)}.
$$
For a unit vector $e$, $\partial_e p_t(v)=-(e\cdot v)p_t(v)/t$, so $\int|\partial_ep_t(v)|dv=\sqrt{2/\pi}/\sqrt t$. Integrating the directional derivative along the segment from $z$ to $w$ yields
$$
|u(z)-u(w)|\leq K\int|p_t(v-z)-p_t(v-w)|dv\leq K\sqrt{\frac2\pi}\frac{|z-w|}{\sqrt t}.
$$
Letting $t\to\infty$ gives \b[the <harmonic Liouville theorem>]:
$$
\boxed{u(z)=u(w)\text{ for all }z,w\in\mathbb C;\quad u\text{ is constant}.}
$$
This <Gaussian heat-kernel proof of the harmonic Liouville theorem> uses the bounded <martingale> identity established above.
Back to article page
