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
Part 1: use the reflection principle to define \[ F(z) \da \begin{cases} f(z) & \abs{z} \leq 1 \\ {1\over \bar{f\qty{1/\bar{z}}} } & \abs{z} \geq 1 \end{cases} .\]
Now \(F:\CP^1\to \CP^1\) is holomorphic and all such functions are rational. As a consequence, \(f\) is rational.
Part 2: As in the proof of Schwarz, define \(g(z) \da {f(z)\over z^n}\) where \(n = \ord_{f}(0)\). Then \(g\) is holomorphic on \(\DD\) since the singularity at \(z=0\) is removable. On \(\abs{z} = r<1\), \[ \abs{g(z)} = { \abs{f(z)} \over \abs{z} } = {\abs{f(z)} \over r} \leq {1\over r} \convergesto{r\to 1^-} 1 ,\] using that \(\abs{f} \leq 1\) on \(\DD\). By the MMP, \(\abs{g} \leq 1\) on all of \(\DD\). Note that \(\abs{g} = 1\) when \(\abs{z}=1\), so \(\abs{1/g}\leq 1\) in \(\DD\) by the MMP, forcing \(\abs{g} = 1\). Unwinding this, \(\abs{f} = \abs{z}^n\), go \(f(z) = \lambda z^n\) for some \(\abs{\lambda} = 1\).
Part 3: Define \(\Psi(z) \da \prod_{k\leq n} \psi_{a_k}(z)\) where \(\psi_a(z) \da {a-z\over 1-\bar a z}\). Set \(g(z) \da {f(z) \over \Psi(z)}\), then by the same argument as above, \(\abs{g} \leq 1\) and \(\abs{g} = 1\) on \(\abs{z} = 1\). Then \(g\) has no zeros, since they’ve all been divided out, and no poles since \(f\) is holomorphic on \(\DD\), so \(1/g\) is holomorphic on \(\DD\). Since \(\abs{1/g} = 1\) on \(S^1\), this forces \(g\) to be constant. Equality in the Schwarz lemma implies \(g(z) = \lambda z\) is a rotation, and unwinding this yields \(f(z) = \lambda \Psi(z)\).