= Brownian proof of the Schwarz lemma
{c}
{title2=$f:D\to D,\ f(0)=0\implies |f(z)|\leq|z|$}
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 $|g(z)|\leq1/r$. 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 $\operatorname{Re}(\overline c\,g(Z_{T_r}))=1$ 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.
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