Suppose \(f:\DD\to \DD\) with \(f(0) = 0\) and that there exists an \(r\in (0, 1)\) with \(f(r) = f(-r) = 0\). Show that \[ \abs{f(z)} \leq \abs{z^2-r^2 \over 1-r^2 z^2} .\]
Solution
The key observation is that this factors: \[ {z^2 - r^2 \over 1-r^2 z^2} = {r-z\over 1+rz}{-r-z\over 1-rz} = \psi_{r}(z) \psi_{-r}(z) ,\] so this inequality will follow from Schwarz on \(g(z) \da f(z)/\psi_a(z)\psi_{-a}(z)\). Schwarz does apply since \(\abs{f}\leq 1\) in \(\DD\) and \(\abs{\psi_a(z)} = 1\) on \(S^1\), so \(\abs{g(z)} \leq 1\) on \(S^1\) and by the MMP this inequality holds in all of \(\DD\). So \(\abs{g(z)}\leq \abs{z}\) and unwinding gives the desired inequality.