Let \(U\subset\mathbb C\) be open and \(P\in U\). Let \(\mathcal F\) be the family of all holomorphic maps \(f:U\to D=\{z:|z|<1\}\) such that \(f(P)=0\).
Prove that \(S=\sup_{f\in\mathcal F}|f'(P)|\) is finite.
Prove that there are \(f_n\in\mathcal F\) converging uniformly on compact subsets of \(U\) to a holomorphic \(f_0:U\to D\) with \(f_0'(P)=S\).
Part (b) fails for an arbitrary subfamily. For \(U=D\), \(P=0\) and \(\mathcal F=\{z\mapsto iz\}\), the supremum is one but every sequence has derivative \(i\), not one. Disconnected domains cause another obstruction: on \(U=D(0,1)\cup D(3,1)\) take \(P=0\) and, for \(n\geq2\), let \(h_n(z)=(1-1/n)z\) on the first component and \(h_n=1-1/n\) on the second. The family \(\{h_n:n\geq2\}\) has derivative supremum one. A sequence approaching that supremum must have its indices tend to infinity, hence converges to one on the second component, outside the open target disk. The full family in the statement allows both needed normalizations.
Solution
1. The derivative supremum is a finite nonnegative real number.
The zero function belongs to \(\mathcal F\). Choose \(r>0\) with \(\overline{D(P,r)}\subset U\). Cauchy’s derivative formula and \(|f|\leq1\) on that circle give \[ |f'(P)|\leq\frac1r\qquad(f\in\mathcal F) \] [@SS03]. Thus \(0\leq S\leq1/r<\infty\). This bound also applies to every nonempty subfamily, proving (a).
2. A maximizing sequence can have real nonnegative derivatives and vanish off the component of \(P\).
Let \(V\) be the connected component of \(U\) containing \(P\). It is open, and \(U\setminus V\) is open as a union of the other components. For each \(n\geq1\), choose \(g_n\in\mathcal F\) with \(|g_n'(P)|>S-1/n\), using the definition of supremum. Multiply \(g_n\) on \(V\) by \(\lambda_n=\overline{g_n'(P)}/|g_n'(P)|\) when its derivative is nonzero, and use \(\lambda_n=1\) otherwise. Define \(h_n=\lambda_ng_n\) on \(V\) and \(h_n=0\) on \(U\setminus V\). These are holomorphic maps into \(D\) with \(h_n(P)=0\), so belong to the full family, and \[ h_n'(P)=|g_n'(P)|\longrightarrow S. \] This construction also covers \(S=0\).
3. A subsequence has the required limit on all of \(U\).
The restrictions \(h_n|_V\) are uniformly bounded by one. Montel’s theorem gives a subsequence \(h_{n_j}\) converging uniformly on compact subsets of \(V\) to a holomorphic \(h\) with \(|h|\leq1\) and \(h(P)=0\) [@SS03]. If \(|h|=1\) at any point of \(V\), the maximum modulus principle would make \(h\) a constant of modulus one, contradicting \(h(P)=0\). Therefore \(h(V)\subset D\).
Set \(f_0=h\) on \(V\) and \(f_0=0\) on \(U\setminus V\). This is holomorphic and disk-valued. For any compact \(K\subset U\), the set \(K\cap V\) is compact because \(V\) is closed relative to \(U\). On that set the subsequence converges uniformly, and on \(K\setminus V\) it is identically zero. Thus convergence to \(f_0\) is uniform on every such \(K\). Finally, on a fixed small circle centered at \(P\) inside \(V\), Cauchy’s derivative formula gives \[ |h_{n_j}'(P)-f_0'(P)| \leq\frac1r\sup_{|z-P|=r}|h_{n_j}(z)-f_0(z)|\longrightarrow0. \] Step 2 now yields \(f_0'(P)=S\). Relabeling this subsequence as \((f_j)\) proves (b).