<Home>Qualifying Exams><Analysis>Spring 2019 - Problem 9

Spring 2019 - Problem 9

Blaschke products

Let $\D = \set{z \in \C \mid \abs{z} < 1}$ and let $\mathcal{A}\p{\D}$ be the space of functions holomorphic in $\D$ and continuous $\cl{\D}$ . Let

\mathcal{U} = \set{f \in \mathcal{A}\p{\D} \st \abs{f\p{z}} = 1 \text{ for all } z \in \partial\D}.

Show that $f \in \mathcal{U}$ if and only if $f$ is a finite Blaschke product,

f\p{z} = \lambda \prod_{j=1}^N \frac{z - a_j}{1 - \conj{a_j}z},

for some $a_j \in \D$ , $1 \leq j \leq N < \infty$ and $\abs{\lambda} = 1$ .

Solution.

If $f$ is a Blaschke product, then it's clear that $f \in \mathcal{U}$ . Conversely, suppose $\abs{f\p{z}} = 1$ on $\partial\D$ and continuous on $\cl{\D}$ . By uniform continuity, we see that $f$ does not vanish on a neighborhood of $\partial\D$ , so because $f$ only has finitely many zeroes $a_1, \ldots, a_N$ . Let

B\p{z} = \prod_{j=1}^N \frac{z - a_j}{1 - \conj{a_j}z}

so that $\frac{f}{B}$ is non-zero. Thus, $\frac{B}{f}$ is holomorphic, so by the maximum principle applied to both $\frac{f}{B}$ and $\frac{B}{f}$ , we see that $\abs{\frac{f}{B}} = 1$ on $\D$ . Thus, $\frac{f}{B} \equiv \lambda$ with $\abs{\lambda} = 1$ , so

f\p{z} = \lambda \prod_{j=1}^N \frac{z - a_j}{1 - \conj{a_j}z},

which was what we wanted to show.