- Show that if \(f : D ( 0 , R ) \to \mathbb { C }\) is holomorphic, with \(| f ( z ) | < M\) for some \(M > 0 ,\) then
\[ \left| \frac {f (z) - f (0)}{M ^ {2} - \overline {{f (0)}} f (z)} \right| \leq \frac {| z |}{M R}. \]
[Hint: Use the Schwarz lemma.]
Solution
Define \[ g(\zeta)=\frac{f(R\zeta)}{M},\qquad \zeta\in\mathbb D. \] Then \(g:\mathbb D\to\mathbb D\). Put \(a=g(0)=f(0)/M\) and let \[ \phi_a(w)=\frac{w-a}{1-\overline a\,w}, \] the automorphism of \(\mathbb D\) carrying \(a\) to \(0\). The function \[ h=\phi_a\circ g \] is holomorphic from \(\mathbb D\) to itself and satisfies \(h(0)=0\). Schwarz’s lemma yields \[ |h(\zeta)|\le|\zeta|. \] For \(z=R\zeta\), this becomes \[ \left|\frac{f(z)/M-f(0)/M}{1-\overline{f(0)}f(z)/M^2}\right| \le \frac{|z|}{R}. \] Simplifying, \[ M\left|\frac{f(z)-f(0)}{M^2-\overline{f(0)}f(z)}\right| \le\frac{|z|}{R}, \] and hence \[ \boxed{ \left|\frac{f(z)-f(0)}{M^2-\overline{f(0)}f(z)}\right| \le\frac{|z|}{MR}}. \]