Let \(U\subset\mathbb C\) be a connected open set and let \(p\in U\). Let \(\mathcal F\) be the family of all univalent maps \[ f:U\to\mathbb C \] such that \[ f(p)=0, \qquad f'(p)=1. \] Show that \(\mathcal F\) is compact in the topology of uniform convergence on compact subsets of \(U\).
Solution
We first prove local boundedness uniformly over \(\mathcal F\) by propagating the normalization at \(p\) along finite chains of disks. Montel then gives normality, and Hurwitz gives closedness.
1. Let \(D(q,R)\Subset U\). For \(f\in\mathcal F\), define\[ F_{f,q,R}(\zeta) =\frac{f(q+R\zeta)-f(q)}{R f'(q)}. \]Then \(F_{f,q,R}\in S\), the normalized schlicht class on \(\Delta\).
Because \(f\) is univalent on \(U\), its derivative never vanishes, so the denominator is nonzero. The map \(F_{f,q,R}\) is univalent on \(\Delta\) and satisfies \[ F_{f,q,R}(0)=0, \qquad F_{f,q,R}'(0)=1. \] Thus \(F_{f,q,R}\in S\).
2. Fix \(0<r<1\). There is a constant \(C_r\), independent of\(f,q,R\), such that for \(|\zeta|\le r\),\[ |F_{f,q,R}(\zeta)|\le C_r, \qquad |F_{f,q,R}'(\zeta)|\le C_r. \]
The Koebe distortion theorem for the class \(S\) gives, for \(|\zeta|=s<1\), \[ |F(\zeta)|\le\frac{s}{(1-s)^2} \] and \[ |F'(\zeta)|\le\frac{1+s}{(1-s)^3}. \] Taking the maximum of these expressions for \(0\le s\le r\) gives a constant \(C_r\) depending only on \(r\).
3. Suppose \(D(q,R)\Subset U\) and \(q'\) satisfies\[ |q'-q|\le R/2. \]Uniform upper bounds for \(|f(q)|\) and \(|f'(q)|\) over \(f\in\mathcal F\)imply uniform upper bounds for \(|f(q')|\) and \(|f'(q')|\).
Write \[ q'=q+R\zeta, \qquad |\zeta|\le1/2. \] By 1, \[ f(q')-f(q) =R f'(q)F_{f,q,R}(\zeta), \] so 2 with \(r=1/2\) gives \[ |f(q')| \le |f(q)|+R|f'(q)|C_{1/2}. \] Similarly, \[ f'(q') =f'(q)F_{f,q,R}'(\zeta), \] and hence \[ |f'(q')|\le |f'(q)|C_{1/2}. \] Thus bounds at \(q\) propagate to \(q'\) by constants depending only on the chosen disk.
4. For every compact set \(K\Subset U\), the family \(\mathcal F\) isuniformly bounded on \(K\).
At the basepoint, \[ f(p)=0, \qquad f'(p)=1 \] for every \(f\in\mathcal F\), so uniform bounds are available there.
Because \(U\) is connected and open, it is path connected. For any \(q\in U\), choose a compact path from \(p\) to \(q\). Cover that path by finitely many disks compactly contained in \(U\), chosen successively so that the center of the next disk lies in the half-radius subdisk of the preceding one. Repeated application of 3 propagates uniform bounds for \(f\) and \(f'\) from \(p\) to a neighborhood of \(q\).
Doing this for each point of the compact set \(K\) and then taking a finite subcover gives a constant \(C_K\) such that \[ |f(z)|\le C_K \qquad(z\in K,\ f\in\mathcal F). \]
5. Every sequence in \(\mathcal F\) has a subsequence converging uniformlyon compact subsets of \(U\) to a holomorphic function.
By 4, \(\mathcal F\) is locally uniformly bounded. Montel’s theorem therefore makes \(\mathcal F\) a normal family. Hence every sequence has a subsequence converging uniformly on compact subsets to a holomorphic function \(f:U\to\mathbb C\).
6. Any locally uniform limit of a sequence in \(\mathcal F\) again belongsto \(\mathcal F\).
Let \(f_n\in\mathcal F\) and suppose \(f_n\to f\) locally uniformly. Then derivatives also converge locally uniformly, so \[ f(p)=0, \qquad f'(p)=\lim_{n\to\infty}f_n'(p)=1. \] Thus \(f\) is nonconstant.
Hurwitz’s theorem for univalent functions says that a locally uniform limit of univalent holomorphic functions is either univalent or constant. Since \(f\) is not constant, it is univalent. Therefore \[ f\in\mathcal F. \]
7. Therefore \(\mathcal F\) is compact in the compact-open topology.
By 5 every sequence has a locally uniformly convergent subsequence, and by 6 every such limit remains in \(\mathcal F\). The compact-open topology on holomorphic functions over the second-countable domain \(U\) is metrizable, so sequential compactness is equivalent to compactness. Hence \(\mathcal F\) is compact.