Suppose \(f\) is continuous and nonvanishing on \(\bar \DD\), and holomorphic in \(\DD\). Prove that if \(\abs{z} = 1 \implies \abs{f(z)} = 1\), then \(f\) is constant.
Hint: Extend \(f\) to all of \(\CC\) by \(f(z) = 1/ \bar{f(1/\bar z)}\) for any \(\abs{z} > 1\), and argue as in the Schwarz reflection principle.
Solution
Goal: If \(f\) is continuous and nonvanishing on \(\bar\DD\), holomorphic in \(\DD\), and \(\abs{f(z)} = 1\) for \(\abs{z} = 1\), prove \(f\) is constant.
1. \(\abs{f(z)} \leq 1\) for \(z \in \DD\).
By the maximum modulus principle applied to \(f\) on \(\DD\), since \(\abs f = 1\) on the boundary \(\abs{z} = 1\).
2. \(\abs{f(z)} \geq 1\) for \(z \in \DD\).
\(f\) is nonvanishing on \(\bar\DD\), so \(1/f\) is holomorphic on a neighborhood of \(\bar\DD\); the maximum modulus principle applied to \(1/f\) gives \(\abs{1/f(z)} \leq \max_{\abs{\zeta}=1}\abs{1/f(\zeta)} = 1\), i.e. \(\abs{f(z)} \geq 1\).
3. \(\abs{f(z)} = 1\) for all \(z \in \DD\).
Combine 1 and 2.
4. Define the reflection \(F\) on \(\CC\) by \(F(z) = f(z)\) for \(\abs{z} \leq 1\) and \(F(z) = \frac{1}{\overline{f(1/\bar z)}}\) for \(\abs{z} > 1\); then \(F\) is well defined and continuous.
For \(\abs{z} > 1\), \(1/\bar z\) lies in \(\DD \setminus \theset{0}\) (indeed \(\abs{1/\bar z} = 1/\abs z < 1\)), where \(f \neq 0\) by assumption, so the formula makes sense.
As \(\abs{z} \to 1^+\), \(1/\bar z \to 1/z = \bar z\) … precisely, \(1/\bar z \to \bar\zeta\) with \(\abs\zeta = 1\); using \(\abs{f} = 1\) on the circle, \(F(z) = 1/\overline{f(1/\bar z)} \to 1/\overline{f(\bar\zeta)} = f(\zeta)\), so \(F\) matches \(f\) continuously across \(\abs{z} = 1\).
5. \(F\) is holomorphic on \(\CC \setminus \partial\DD\) and satisfies \(F(z) = \overline{F(1/\bar z)}^{-1}\) (the reflection functional equation); by the Schwarz reflection principle \(F\) is entire.
On \(\abs{z} < 1\), \(F = f\) is holomorphic.
On \(\abs{z} > 1\), \(z \mapsto 1/\bar z\) is antiholomorphic, so \(f(1/\bar z)\) is antiholomorphic and its conjugate \(\overline{f(1/\bar z)}\) is holomorphic; since \(f \neq 0\) there, \(F = 1/\overline{f(1/\bar z)}\) is holomorphic. The identity \(F(z) = 1/\overline{F(1/\bar z)}\) holds by definition (and extends across the circle), and \(\abs F = 1\) on \(\partial\DD\), so the Schwarz reflection principle applies: the two holomorphic pieces glue into a single entire function.
6. \(F\) is bounded on \(\CC\).
On \(\abs z \leq 1\), \(\abs F = \abs f \leq 1\) by 1 (or by compactness, \(f\) bounded).
On \(\abs z > 1\), \(\abs{F(z)} = \frac{1}{\abs{f(1/\bar z)}} \leq 1\) by 2 applied to \(1/\bar z \in \DD\). Hence \(\abs F \leq 1\) everywhere.
7. \(F\) is constant, hence \(f\) is constant.
By Liouville’s theorem, the bounded entire function \(F\) from 5 and 6 is constant; restricting to \(\DD\) shows \(f\) is constant.
8. Q.E.D.
7 proves the claim.