 Solution

Let $z_0\ne 0$ be an equality point and set $c=g(z_0)$, so $|c|=1$. For any $|z_0|<r<1$, the same Brownian exit identity givesPart (i) gives $|g|\le 1$, 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 $g(w)={\mathbb{E}}_{w}g(Z_{T_r})=c$. Letting $r$ increase to one makes this true throughout $D$. Writing $c=e^{i\theta }$ yieldsThis completes the equality case of the Brownian proof of the Schwarz lemma.