Let \(f(z)\) be holomorphic in an open set containing the closed unit disc \(\overline{D}\). Suppose that \(|f(z)| < 1\) for all \(z\) on the unit circle. Show that there is exactly one point \(z \in D\) (the interior of \(\overline{D}\)) for which \(f(z) = z\).
Solution
1. The fixed-point equation has exactly one zero in the unit disk.
Set \[ F(z)=f(z)-z. \] Both \(F\) and \(-z\) are holomorphic on a neighborhood of the closed unit disk. On \(|z|=1\) the hypothesis gives \[ |F(z)-(-z)|=|f(z)|<1=|-z|. \] Hence Rouché’s theorem implies that \(F\) and \(-z\) have the same number of zeros in \(D\), counted with multiplicity. The function \(-z\) has exactly one zero there, namely the simple zero at \(0\). Therefore \(F\) has exactly one zero in \(D\), counted with multiplicity.
Moreover, \(F\) has no zero on the unit circle: if \(F(z)=0\) and \(|z|=1\), then \(f(z)=z\) would give \(|f(z)|=1\), contrary to the strict boundary hypothesis. Thus the single zero counted by Rouché is an interior point, and because its total multiplicity is one it is a unique point. Equivalently, there exists exactly one \(z\in D\) with \(f(z)=z\).