Let \(D=\{z\in\mathbb C:|z|<1\}\), let \(U\) be an open set containing \(\overline D\), and let \(f:U\to\mathbb C\) be holomorphic and nonconstant on the connected component of \(U\) containing \(\overline D\). If \(|f(z)|=1\) for every \(z\in\partial D\), prove that \(D\subset f(\overline D)\).
Nonconstancy on all of a disconnected \(U\) is insufficient. On \(U=D(0,2)\cup D(4,1)\), set \(f=1\) on the first component and \(f(z)=z\) on the second. This function is holomorphic and nonconstant on \(U\), but \(f(\overline D)=\{1\}\). The component qualification excludes precisely this obstruction.
Solution
In fact \(f(D)=D\).
1. The function has at least one zero in \(D\) and satisfies \(|f|<1\) there.
The maximum modulus principle applied on the closed disk gives \(|f|\leq1\) in \(D\) [@SS03]. If equality held at an interior point, \(f\) would be constant on \(D\) and hence, by the identity theorem, on its component in \(U\). This contradicts the stated nonconstancy. Thus \(|f|<1\) in \(D\).
Suppose \(f\) had no zero in \(D\). It has none on the boundary either, since the boundary modulus is one. Consequently \(1/f\) is holomorphic on a neighborhood of \(\overline D\). Applying the maximum modulus principle to \(1/f\) would give \(|1/f|\leq1\) in \(D\), contradicting \(|f|<1\). Thus \(f\) has an interior zero. Its number \(N\) of zeros in \(D\), counting multiplicities, is finite and positive: otherwise compactness of \(\overline D\) would give an accumulation point in \(U\), forcing \(f\) to vanish identically on its component by the identity theorem [@SS03].
2. Every point of \(D\) is attained inside \(D\).
Fix \(w\) with \(|w|<1\). On \(\partial D\), \[ |(f-w)-f|=|w|<1=|f|. \] Rouche’s theorem says that \(f-w\) and \(f\) have the same number \(N\geq1\) of zeros in \(D\), counting multiplicity [@SS03]. Hence \(f(z)=w\) for some \(z\in D\). This proves \(D\subset f(D)\), while step 1 gives \(f(D)\subset D\). Therefore \(f(D)=D\) and, in particular, \(D\subset f(\overline D)\) as required.