Let \(\bar{B}(a, r)\) denote the closed disk of radius \(r > 0\) centered at \(a \in \mathbb{C}\). Let \(f\) be holomorphic on an open neighborhood containing \(\bar{B}(a, r)\), and define \[ M = \sup_{z \in \bar{B}(a, r)} |f(z)|. \] Prove that for all \(z \in \bar{B}(a, r/2)\) with \(z \ne a\): \[ \frac{|f(z) - f(a)|}{|z - a|} \le \frac{2M}{r}. \]
Solution
Goal: Let \(\bar B(a, r)\) be the closed disk of radius \(r\) about \(a\), and let \(f\) be holomorphic on an open set containing it, with \(M := \sup_{z \in \bar B(a,r)} \abs{f(z)}\). Prove that for \(z \in \bar B\qty{a, \frac{r}{2}}\), \(z \neq a\): \[\frac{\abs{f(z) - f(a)}}{\abs{z - a}} \leq \frac{2M}{r}.\]
1. Reduce to the unit disk by scaling: define \(g(w) := f(a + rw)\) for \(\abs w \leq 1\).
\(g\) is holomorphic on a neighborhood of the closed unit disk, and \(\abs{g(w)} \leq M\) there.
2. Define \(h(w) := g(w) - g(0)\); then \(h(0) = 0\) and \(\abs{h(w)} \leq 2M\) on \(\abs w \leq 1\).
\(\abs{h(w)} = \abs{g(w) - g(0)} \leq \abs{g(w)} + \abs{g(0)} \leq 2M\).
3. Apply Schwarz’s lemma to \(\frac{h}{2M}\).
3.1. \(\frac{h}{2M}\) is holomorphic on \(\DD\) and maps \(\DD\) into \(\overline{\DD}\), with value \(0\) at \(w = 0\).
2; if \(M = 0\) then \(f \equiv 0\) and \(\abs{h(w)} \leq 0\) holds, so assume \(M > 0\).
3.2. \(\abs{h(w)} \leq 2M \abs w\) for \(\abs w < 1\).
Schwarz’s lemma applied to \(h/2M\).
4. Translate back to \(z\).
4.1. For \(z \in \bar B(a, r/2)\), set \(w := (z - a)/r\); then \(\abs w \leq 1/2 < 1\).
\(\abs{z - a} \leq r/2\).
4.2. \(\abs{f(z) - f(a)} = \abs{h(w)} \leq 2M\abs w = \frac{2M}{r}\abs{z - a}\).
1, 2 and 3.2, with \(f(z) - f(a) = g(w) - g(0) = h(w)\).
5. Q.E.D.
4.2 is exactly the claimed inequality (both sides well-defined since \(z \neq a\)).