[Fall 2012, Problem \(\# 7 ]\) Write \(\mathbb { D } : = \{ z \in \mathbb { C } : | z | < 1 \}\) for the open unit disk. Suppose \(f : \mathbb { D } \to \mathbb { D }\) is analytic, and admits a continuous extension \(\widetilde { f } : \overline { { \mathbb { D } } } \to \overline { { \mathbb { D } } }\) such that \(| f ( z ) | = 1\) whenever \(| z | = 1\) .
Prove that \(f\) is a rational function.
Suppose that \(z = 0\) is the unique zero of \(f\). Prove that \(f(z) = \lambda z^n\) for some \(\lambda \in \mathbb { C }\) of absolute value 1 and some natural number \(n\) .
More generally, suppose that \(a _ { 1 } , \dots , a _ { n } \in \mathbb { D }\) are the zeros of \(f\), listed with multiplicity. Prove that
\[ f ( z ) = \lambda \prod _ { j = 1 } ^ { n } { \frac { z - a _ { j } } { 1 - { \overline { { a } } } _ { j } z } } , \quad | \lambda | = 1 . \]
Solution
Let \[ \DD=\{z\in\CC:\abs{z}<1\}. \]
1. The function \(f\) has only finitely many zeros in \(\DD\).
The continuous extension \(\widetilde f\) satisfies \[ \abs{\widetilde f(\zeta)}=1 \] for every \(\zeta\) on the unit circle. By continuity, for every such \(\zeta\) there is a neighborhood \(U_\zeta\) in \(\overline\DD\) on which \[ \abs{\widetilde f}>\frac12. \] The unit circle is compact, so finitely many of the \(U_\zeta\) cover it. Consequently there is an \(r<1\) such that \[ \abs{f(z)}>\frac12 \] whenever \(r<\abs{z}<1\).
Thus every zero of \(f\) lies in the compact disk \[ \{z:\abs{z}\leq r\}\subset\DD. \] Since \(f\) is not identically zero, its zeros are isolated. An infinite set of zeros in this compact disk would have an accumulation point in \(\DD\), contradicting the identity theorem. Hence the zero set is finite.
2. List the zeros of \(f\), with multiplicity, as\[ a_1,\ldots,a_n \]and define\[ B(z)=\prod_{j=1}^n \frac{z-a_j}{1-\overline{a_j}z}. \]Then \(B\) is analytic on a neighborhood of \(\overline\DD\), has exactly thesame zeros in \(\DD\) as \(f\) with the same multiplicities, and\[ \abs{B(\zeta)}=1 \]for every \(\abs{\zeta}=1\).
For each \(j\), one has \(\abs{a_j}<1\), so the denominator \[ 1-\overline{a_j}z \] does not vanish on \(\overline\DD\). Hence every factor, and therefore \(B\), is analytic on a neighborhood of \(\overline\DD\).
Its numerator shows that the factor indexed by \(j\) has a simple zero at \(a_j\); listing the \(a_j\) with multiplicity therefore makes the zero multiplicities of \(B\) agree with those of \(f\).
If \(\abs{\zeta}=1\), then \[ \abs{\zeta-a_j} = \abs{\zeta}\, \abs{1-\overline{\zeta}a_j} = \abs{1-\overline{a_j}\zeta}. \] Thus every factor has modulus one on the unit circle, and so does \(B\).
3. The quotient\[ h(z)=\frac{f(z)}{B(z)} \]extends to a zero-free analytic function on \(\DD\), continuous on\(\overline\DD\), and satisfies\[ \abs{h(\zeta)}=1 \]for every \(\abs{\zeta}=1\). Consequently\[ h\equiv\lambda \]for some \(\lambda\in\CC\) with \(\abs{\lambda}=1\).
By step 2, \(f\) and \(B\) have the same zeros with the same multiplicities, so every apparent singularity of \(f/B\) at an \(a_j\) is removable. After removing them, \(h\) is analytic and zero-free on \(\DD\). It is continuous on \(\overline\DD\) because both \(f\) and \(B\) are continuous there and \(B\) is nonzero on the unit circle.
For \(\abs{\zeta}=1\), the boundary hypotheses on \(f\) and step 2 give \[ \abs{h(\zeta)} = \frac{\abs{f(\zeta)}}{\abs{B(\zeta)}} = 1. \] The maximum modulus principle applied on the disk gives \[ \abs{h(z)}\leq1 \] for \(z\in\DD\). Since \(h\) has no zeros, \(1/h\) is analytic on \(\DD\) and continuous on \(\overline\DD\), with boundary modulus one as well. Applying the same argument to \(1/h\) gives \[ \abs{h(z)}\geq1. \] Therefore \(\abs{h(z)}=1\) throughout \(\DD\). By the open mapping theorem, \(h\) is constant. Write \(h\equiv\lambda\); its boundary modulus gives \(\abs{\lambda}=1\).
4. Part (a): \(f\) is a rational function.
By step 3, \[ f(z) = \lambda B(z) = \lambda \prod_{j=1}^n \frac{z-a_j}{1-\overline{a_j}z}. \] The right-hand side is rational.
5. Part (b): if \(0\) is the unique zero of \(f\), then\[ \boxed{ f(z)=\lambda z^n } \]for some \(n\in\NN\) and some \(\lambda\in\CC\) with\(\abs{\lambda}=1\).
Let \(n\) be the multiplicity of the zero at \(0\). Then the zero list in step 2 is \[ a_1=\cdots=a_n=0. \] Hence \[ B(z) = \prod_{j=1}^n\frac{z}{1} = z^n. \] Step 3 therefore gives \[ f(z)=\lambda z^n, \qquad \abs{\lambda}=1. \]
6. Part (c): in general,\[ \boxed{ f(z)=\lambda \prod_{j=1}^n \frac{z-a_j}{1-\overline{a_j}z}, \qquad \abs{\lambda}=1. } \]
This is exactly the identity \(f=\lambda B\) established in step 3, with \(B\) defined in step 2.
7. Q.E.D.
Steps <1>4–<1>6 prove parts (a)–(c), respectively.