- Suppose \(f\) is a non-vanishing continuous function on \(\overline{\mathbb{D}}\) that is holomorphic in \(\mathbb{D}\). Prove that if
\[ | f (z) | = 1 \quad \text { whenever } | z | = 1, \]
then \(f\) is constant.
[Hint: Extend \(f\) to all of \(\mathbb{C}\) by \(f ( z ) = 1 / \overline { f ( 1 / \overline { { z } } ) }\) whenever \(|z| > 1\), and argue as in the Schwarz reflection principle.]
Solution
Because \(f\) is continuous on \(\overline{\mathbb D}\), holomorphic on \(\mathbb D\), and satisfies \(|f|=1\) on \(\partial\mathbb D\), the maximum-modulus principle gives \[ |f(z)|\le1\qquad(z\in\mathbb D). \tag{1} \]
Since \(f\) is non-vanishing on \(\overline{\mathbb D}\), the function \(1/f\) is continuous on \(\overline{\mathbb D}\) and holomorphic on \(\mathbb D\). On the boundary, \[ \left|\frac1{f(z)}\right|=1. \] Applying the maximum-modulus principle to \(1/f\) yields \[ \frac1{|f(z)|}\le1, \] so \[ |f(z)|\ge1\qquad(z\in\mathbb D). \tag{2} \] Combining (1) and (2) gives \(|f(z)|=1\) throughout \(\mathbb D\).
If \(f\) were nonconstant, the open mapping theorem would imply that \(f(\mathbb D)\) is open in \(\mathbb C\), but it is contained in the unit circle, which has empty interior. This is impossible. Hence \(f\) is constant.