Let \(D = \{z \in \mathbb{C} : |z| < 1\}\) and \(f : D \to D\) be a holomorphic function. Prove
\[\frac{|f(0)| - |z|}{1 + |f(0)| \cdot |z|} \leq |f(z)| \leq \frac{|f(0)| + |z|}{1 - |f(0)| \cdot |z|}, \quad \forall z \in D.\]
Solution
1. Normalization at the origin gives a Schwarz-lemma estimate.
Put \(c=f(0)\) and define \[ H(z)=\frac{f(z)-c}{1-\overline c f(z)}. \] The denominator is nonzero because \(|c|,|f(z)|<1\). For every \(w\in D\), \[ 1-\left|\frac{w-c}{1-\overline c w}\right|^2 =\frac{(1-|c|^2)(1-|w|^2)}{|1-\overline c w|^2}>0. \] Thus \(H:D\to D\) is holomorphic and \(H(0)=0\). Schwarz’s lemma gives \(|H(z)|\leq|z|\) [@SS03], or \[ |f(z)-c|\leq |z|\,|1-\overline c f(z)|. \]
2. The triangle inequalities give both requested bounds.
For fixed \(z\in D\), write \(a=|c|\), \(b=|f(z)|\) and \(r=|z|\). Step 1 implies \[ |b-a|\leq |f(z)-c|\leq r(1+ab). \] Using \(b-a\leq r(1+ab)\) gives \(b(1-ar)\leq a+r\). Since \(ar<1\), division yields \(b\leq(a+r)/(1-ar)\). Using \(a-b\leq r(1+ab)\) instead gives \(a-r\leq b(1+ar)\), hence \(b\geq(a-r)/(1+ar)\). The latter denominator is positive even when \(a-r<0\), so no additional case assumption is needed. Substituting back gives the two stated inequalities at every point, including \(z=0\).