Let \(\mathcal S\) be the normalized univalent maps \(f:\Delta\to\mathbb C\) with \(f(0)=0\) and \(f'(0)=1\). Show that the maps \(f\in\mathcal S\) whose images \(f(\Delta)\) are Jordan domains are dense in \(\mathcal S\) for uniform convergence on compact subsets of \(\Delta\).
Solution
Fix \(f\in\mathcal S\). For \(0<r<1\), define \[ f_r(z)=\frac{f(rz)}r. \] Then each \(f_r\) lies in \(\mathcal S\), its image is a Jordan domain, and \[ f_r\longrightarrow f \] uniformly on compact subsets as \(r\to1^-\). Thus the desired maps are dense.
1. Each \(f_r\) is normalized and univalent.
Since \(f\) is univalent and \(z\mapsto rz\) is injective on \(\Delta\), the map \(f_r\) is univalent. Moreover \[ f_r(0)=0, \qquad f_r'(0)=f'(0)=1. \] Hence \(f_r\in\mathcal S\).
2. The image \(f_r(\Delta)\) is a Jordan domain.
Choose \(\rho\) with \[ r<\rho<1. \] The map \(f\) is holomorphic and injective on the open disk \(\rho\Delta\), so it is a homeomorphism there onto its image. In particular, its restriction to the compact circle \[ rS^1 \] is continuous and injective. Therefore \[ \Gamma_r:=f(rS^1) \] is a Jordan curve.
The restriction of \(f\) to the closed disk \(r\overline\Delta\) is a homeomorphism onto its image, so \[ \partial f(r\Delta)=f(rS^1)=\Gamma_r. \] Since \(f(r\Delta)\) is bounded and connected, it is the bounded complementary component of \(\Gamma_r\). Hence \(f(r\Delta)\) is a Jordan domain. Multiplication by \(1/r\) preserves Jordan domains, and therefore \[ f_r(\Delta)=\frac1r f(r\Delta) \] is a Jordan domain as well.
3. One has \(f_r\to f\) uniformly on every compact subset of \(\Delta\) as \(r\to1^-\).
Let \(K\Subset\Delta\). Choose \(R<1\) such that \[ K\subset\{z:|z|\le R\}. \] For \(r\) sufficiently close to \(1\), both \(z\) and \(rz\) lie in a fixed compact disk contained in \(\Delta\). Since \(f\) is uniformly continuous there, \[ \sup_{z\in K}|f(rz)-f(z)|\longrightarrow0. \] Also \(1/r\to1\), so \[ \begin{aligned} \sup_{z\in K}|f_r(z)-f(z)| &\le \frac1r\sup_{z\in K}|f(rz)-f(z)| +\left|\frac1r-1\right|\sup_{z\in K}|f(z)|\\ &\longrightarrow0. \end{aligned} \] Thus \(f_r\to f\) locally uniformly.