Show that if \(f: D(0, R) \rightarrow \mathbb{C}\) is holomorphic, with \(|f(z)| \leq M\) for some \(M>0\), then \[ \left|\frac{f(z)-f(0)}{M^{2}-\overline{f(0)} f(z)}\right| \leq \frac{|z|}{M R} . \]
The strategy:
Write the RHS as \(a\). Note that we need to get rid of the \(M^2\) on the LHS, so keep the \(M\) around and write \(a \da z/R\) so \(z = aR\).
Make the substitution to get \[ \abs{f(aR) - f(0) \over M^2 - \bar{f(0)} f(aR) } \leq \inverseof{M} \abs{a} \\ \implies \abs{M\qty{ f(aR) - f(0)} \over M^2 - \bar{f(0)} f(aR) } \leq \abs{a} \\ \abs{ f(aR)/M - f(0)/M \over 1 - \bar{f(0)} f(aR)/M^2 } \leq \abs{a} .\]
- Recognize the LHS as \(\psi_w(g(a))\) for \(w\da f(0)/M\) and \(g(a) \da f(aR)/M\).
Solution
Proof due to Swaroop Hegde!
Fix \(R, M\) and make a clever choice: define \[ F: \DD &\to \CC \\ z &\mapsto {f(Rz) \over M} .\] Write \(a\da F(0)\) and consider the Blaschke factor \[ \psi_a(z) \da {a-z \over 1-\bar{a} z} \in \Aut(\DD) ,\] and define \[ g: \DD &\to \DD \\ z &\mapsto (\psi_a \circ F)(z) .\] Then \(g(0) = 0\) and \(\abs{g(z)} \leq 1\) for all \(z\in \DD\), so by Schwarz we have \(\abs{g(z)} \leq \abs{z}\) for all \(z\in \DD\). Thus for all \(z\in \DD\), \[ &\abs{g(z)} \leq z \\ \\ \iff & \abs{\psi_a(F(z)) } \leq \abs{z} \\ \\ \iff & \abs{ {f(Rz) \over M} - a \over 1 - {\bar a f(Rz) \over M} } \leq \abs{z} \\ \\ \iff & \abs{f(Rz) - f(0) \over 1 - {\bar{f(0)} f(Rz) \over M^2 } } \leq \abs{z} \\ \\ \iff & \abs{f(Rz) - f(0) \over M^2 - \bar{f(0)} f(Rz) } \leq {\abs{z} \over M} \\ \\ \iff & \abs{f(w) - f(0) \over M^2 - \bar{f(0)} f(w) } \leq {\abs{w} \over MR} ,\] which holds for all \(w\in \DD\) by replacing \(Rz\) with \(w\) (i.e. to show this equality for arbitrary \(w\in \DD\), write \(w = Rz\) for some \(z\in \DD\) and run this chain of inequalities backward).