Suppose \(f:\DD\to\DD\) is analytic and admits a continuous extension \(\tilde f: \bar \DD \to \bar \DD\) such that \(\abs{z} = 1 \implies \abs{f(z)} = 1\).
Prove that \(f\) is a rational function.
Suppose that \(z=0\) is the unique zero of \(f\). Show that \[ \exists n\in \NN, \lambda \in S^1 \qtext{ such that }f(z) = \lambda z^n .\]
Suppose that \(a_1, \cdots, a_n \in \DD\) are the zeros of \(f\) and prove that \[ \exists \lambda \in S^1 \qtext{such that} f(z) = \lambda \prod_{j=1}^n {z - a_j \over 1 - \bar{a_j} z} .\]
Solution
Because \(|f|=1\) on \(\partial\mathbb D\) and the extension is continuous, there is an annular neighborhood of the boundary on which \(f\) has no zeros. Hence all zeros of \(f\) lie in a compact subset of \(\mathbb D\). Since zeros of a nonzero holomorphic function are isolated, there are only finitely many of them, counted with multiplicity; write them as \(a_1,\dots,a_n\).
For each zero define the Blaschke factor \[ B_{a_j}(z)=\frac{z-a_j}{1-\overline{a_j}z}. \] It has the same zero multiplicity contribution as \(f\) and satisfies \(|B_{a_j}(z)|=1\) on \(|z|=1\). Therefore \[ g(z)=\frac{f(z)}{\prod_{j=1}^n B_{a_j}(z)} \] extends holomorphically and without zeros to \(\mathbb D\), continuously to the closed disk, and satisfies \(|g|=1\) on the boundary.
The maximum modulus principle gives \(|g|\le1\) in \(\mathbb D\). Since \(g\) has no zeros, \(1/g\) is holomorphic and has boundary modulus \(1\), so the same principle gives \(|1/g|\le1\), i.e. \(|g|\ge1\). Thus \(|g|\equiv1\), and the open mapping theorem forces \(g\) to be constant. Hence for some \(\lambda\in S^1\), \[ \boxed{f(z)=\lambda\prod_{j=1}^n \frac{z-a_j}{1-\overline{a_j}z}.} \] This is rational, proving the first assertion as well.
If \(0\) is the unique zero and has multiplicity \(n\), every \(a_j=0\), so the formula reduces to \[ \boxed{f(z)=\lambda z^n.} \]