 Solution

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.