Let \(f,g:\Delta\to\mathbb C\) be univalent maps satisfying \[ f(0)=g(0)=0, \qquad f'(0)=g'(0)=1. \] If \[ f(\Delta)\supseteq g(\Delta), \] prove that \(f=g\).
Solution
Since \(g(\Delta)\subseteq f(\Delta)\) and \(f\) is univalent, the map \[ h=f^{-1}\circ g \] is a holomorphic self-map of \(\Delta\). The normalizations force equality in the Schwarz lemma.
1. The map\[ h=f^{-1}\circ g:\Delta\to\Delta \]is well-defined and holomorphic.
Because \(f\) is univalent, it is a biholomorphism from \(\Delta\) onto the domain \(f(\Delta)\), so its inverse \[ f^{-1}:f(\Delta)\to\Delta \] is holomorphic. The inclusion \[ g(\Delta)\subseteq f(\Delta) \] therefore makes \(h=f^{-1}\circ g\) a holomorphic map from \(\Delta\) to \(\Delta\).
2. One has\[ h(0)=0 \qquad\text{and}\qquad h'(0)=1. \]
Since \(f(0)=g(0)=0\), \[ h(0)=f^{-1}(g(0))=f^{-1}(0)=0. \] Differentiating \(g=f\circ h\) at \(0\) gives \[ g'(0)=f'(h(0))h'(0)=f'(0)h'(0). \] The normalization \(f'(0)=g'(0)=1\) therefore yields \[ h'(0)=1. \]
3. The Schwarz lemma forces \(h\) to be the identity.
The Schwarz lemma says that a holomorphic self-map of \(\Delta\) fixing \(0\) satisfies \[ |h'(0)|\le1, \] with equality only for a rotation \[ h(z)=e^{i\theta}z. \] By 2, \(h'(0)=1\), so the rotation factor is \(e^{i\theta}=1\). Hence \[ h(z)=z \] for all \(z\in\Delta\).
4. Therefore \(f=g\).
Since \(g=f\circ h\) and \(h=\operatorname{id}_\Delta\) by 3, \[ g=f. \]