Suppose \(f:\DD\to \DD\) is analytic. Prove that \[ \forall a\in \DD, \qquad {\abs{f'(a)} \over 1 - \abs{f(a)}^2 } \leq {1 \over 1 - \abs{a}^2} .\]
Solution
For \(a\in\DD\), let \(\psi_a(z)\coloneqq{a-z\over1-\bar az}\), an automorphism of \(\DD\) with \(\psi_a(0)=a\), \(\psi_a(a)=0\) and \(\inverseof{\psi_a}=\psi_a\).
For \(z,w\in\DD\), \[ \abs{ \psi_w(z) } \geq \abs{\psi_{f(w)}(f(z)) } ,\] i.e. \[ \abs{f(w) - f(z) \over 1 - \bar{f(w)}f(z)} \leq \abs{w-z \over 1-\bar{w} z} .\]
Make a change of variables \(a\da \psi_w(z)\) so \(z=\inverseof{\psi_w}(a) = \psi_w(a)\), then the desired inequality follows if we can show \[ \abs{ \psi_{f(w)}(f(\psi_w(a))) } \leq \abs{a} .\]
So define \(F \da \psi_{f(w)} \circ f \circ \psi_w\), then since \(\psi_w(0) = w\), \[ F(0) = \psi_{f(w)}(f(w)) = 0 .\] Moreover \(\abs{F(z)}\leq 1\) since each constituent is a map \(\DD\to \DD\). So \(F\) satisfies Schwarz and the claim follows.
Dividing the claim by \(\abs{z-w}\) for \(z\neq w\): \[ \abs{f(w) - f(z) \over 1 - \bar{f(w)}f(z)} &\leq \abs{w-z \over 1-\bar{w} z} \\ \implies \abs{ 1\over 1-\bar{f(w)}f(z) } \cdot \abs{f(z) - f(w) \over z-w} &\leq \abs{1\over 1-\bar{w}z} \\ ,\] and taking \(z\to w\) on both sides yields \[ {\abs{f'(w)}\over 1 - \abs{f(w)}^2 } \leq {1\over 1-\abs{w}^2} \implies \abs{f'(w)} \leq {1-\abs{f(w)}^2\over 1-\abs{w}^2 } .\]