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
Define \(F:\CC\to \CC\) by \(F(z)=f(z)\) for \(\abs z\le1\) and \(F(z) = 1/\bar{f(1/\bar{z})}\) for \(\abs z>1\). This is defined because \(f\) is nonvanishing on \(\bar\DD\).
- \(F\) is holomorphic on \(\abs z>1\), as the composite of the holomorphic map \(w\mapsto 1/f(w)\) on \(\DD\) with the two antiholomorphic maps \(z\mapsto1/\bar z\) and \(w\mapsto\bar w\).
- \(F\) is continuous on \(\CC\): for \(\abs{z_0}=1\) we have \(1/\bar{z_0}=z_0\) and \(\abs{f(z_0)}=1\), so \(1/\bar{f(z_0)}=f(z_0)\), and both formulas tend to \(f(z_0)\) as \(z\to z_0\).
- A continuous function on \(\CC\) that is holomorphic off the circle \(S^1\) is entire, by Morera’s theorem as in the proof of the Schwarz reflection principle.
- \(F\) is bounded: with \(M=\max_{\bar\DD}\abs f\) and \(m=\min_{\bar\DD}\abs f>0\), we have \(\abs F\le\max(M,1/m)\).
By Liouville’s theorem \(F\) is constant, so \(f\) is constant.