Past exam of the Brownian motion in complex analysis course of the University of Oxford / 2011 / 3 / b

Since $f(0)=0$, the function $g(z)=f(z)/z$ extends analytically to zero by $g(0)=f^{\prime }(0)$. It is bounded on the disk: outside $|z|<1/2$, $|g(z)|\le 2$, while on the closed half-radius disk its analytic extension is continuous and hence bounded.

Fix $z\ne 0$ and choose $|z|<r<1$. Let $T_r$ be the first exit time of Brownian motion started at $z$ from the radius-$r$ disk. The bounded analytic martingale identity in part (a) givesOn this exit circle, $|g(Z_{T_r})|=|f(Z_{T_r})|/r\le 1/r$. The triangle inequality gives $|g(z)|\le 1/r$. Letting $r\uparrow 1$ proves $|g(z)|\le 1$, and thereforeAt zero the claim holds by the assumed zero value. This is the Schwarz lemma, obtained from Brownian optional stopping rather than an analytic maximum principle.

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.