Let \(U\subseteq\mathbb C\) be an open set, let \(D=\{z\in\mathbb C:|z|<1\}\), and let \(\mathcal F\) be the set of all holomorphic functions \(f\colon U\to D\). Given \(z_0\in U\), show that there exists \(f_0\in\mathcal F\) such that \[ |f_0''(z_0)|=\sup_{f\in\mathcal F}|f''(z_0)|. \]
Solution
1. \(M=\sup_{f\in\mathcal F}|f''(z_0)|\) is finite.
Choose \(r>0\) with \(\overline{D(z_0,r)}\subset U\). Every \(f\in\mathcal F\) satisfies \(|f|<1\), so Cauchy’s estimate gives \(|f''(z_0)|\leq2/r^2\) [@SS03].
2. Some holomorphic \(f_0:U\to\overline D\) satisfies \(|f_0''(z_0)|=M\).
Choose \(f_n\in\mathcal F\) with \(|f_n''(z_0)|\to M\). The family \(\mathcal F\) is uniformly bounded, so Montel’s theorem gives a subsequence \((f_{n_k})\) converging uniformly on compact subsets of \(U\) to a holomorphic \(f_0\) [@SS03]. Derivatives of all orders then converge locally uniformly, so \(f_{n_k}''(z_0)\to f_0''(z_0)\) and \(|f_0''(z_0)|=M\). Pointwise limits give \(|f_0|\leq1\) on \(U\).
3. On each connected component \(W\) of \(U\), either \(|f_0|<1\) on \(W\) or \(f_0\) is a constant of modulus one on \(W\).
If \(|f_0(w)|=1\) for some \(w\in W\), then \(|f_0|\) attains its maximum over \(W\) at an interior point. The maximum modulus principle makes \(f_0\) constant on \(W\) [@SS03].
4. Q.E.D.
Define \(g\) to equal \(f_0\) on every component of \(U\) on which \(|f_0|<1\), and to equal \(0\) on every other component. Components are open, so \(g\) is holomorphic on \(U\), and step 3 gives \(g(U)\subset D\). Thus \(g\in\mathcal F\). If \(|f_0|<1\) on the component of \(z_0\), then \(g=f_0\) near \(z_0\) and \(|g''(z_0)|=M\). Otherwise \(f_0\) is constant near \(z_0\), so \(M=|f_0''(z_0)|=0=|g''(z_0)|\). In both cases \(g\) attains the supremum.