 Brownian proof of the Schwarz lemma

For a holomorphic function on the unit disk fixing zero, $g(z)=f(z)/z$ has a removable singularity at zero and a bounded holomorphic extension. A bounded holomorphic function evaluated along planar Brownian motion is a bounded stopped martingale before domain exit. Stop at the first exit from a concentric radius-$r$ disk with $|z|<r<1$. Optional stopping gives $g(z)={\mathbb{E}}_{z}g(Z_{T_r})$, hence $|g(z)|\le 1/r$. Letting $r$ increase to one proves the displayed bound.

If $|g(z_0)|=1$ at a nonzero point, put $c=g(z_0)$. The same expectation identity forces $\mathrm{Re}(\overline{c}g(Z_{T_r}))=1$ almost surely. The exit density is strictly positive on every boundary arc by the Poisson kernel for the unit disk, and $g$ is continuous there, so $g=c$ on that circle. The expectation identity for every starting point inside it then gives $g=c$ throughout that smaller disk. Letting $r$ increase to one shows $f(z)=cz$ on the full disk, with $|c|=1$. This derives both the inequality and its rigidity using Brownian harmonic measure.