= Solution
{c}
Let $z_0\ne0$ be an equality point and set $c=g(z_0)$, so $|c|=1$. For any $|z_0|<r<1$, the same Brownian exit identity gives
$$
1=\operatorname{Re}(\overline c\,g(z_0))
=\mathbb E_{z_0}\operatorname{Re}(\overline c\,g(Z_{T_r})).
$$
Part (i) gives $|g|\leq1$, so the real random variable in the expectation is at most one. Equality of its mean to one forces $g(Z_{T_r})=c$ almost surely. By the scaled version of the exit density from Question 2, every nonempty arc of the exit circle has positive probability. Continuity of $g$ on that circle therefore implies $g=c$ at every boundary point of the radius-$r$ disk.
For any $w$ inside that circle, the Brownian exit identity now gives $g(w)=\mathbb E_wg(Z_{T_r})=c$. Letting $r$ increase to one makes this true throughout $D$. Writing $c=e^{i\theta}$ yields
$$
\boxed{f(w)=we^{i\theta}\quad\text{for all }w\in D.}
$$
This completes the equality case of the <Brownian proof of the Schwarz lemma>.
Back to article page
