Analysis Problem of the Day 29

Today’s problem is Problem 10 on the UCLA Spring 2023 Analysis Qual:

Problem 10. A function of the form

$B(z) = \lambda \prod_{i=1}^n \frac{z-a_j}{1-\overline{a_j} z}, \quad a_j \in \mathbb{D}, |\lambda|=1$

is called a Blaschke product of degree n. Show that for $\alpha \in \mathbb{D},$

$z \to \frac{B(z)+\alpha}{1+\overline{\alpha} B(z)}$

is a Blaschke product of degree $n.$

Solution: Notice that the map

$g: z \to \frac{B(z)+\alpha}{1+\overline{\alpha} B(z)}$

is given by $g = \phi_{-\alpha} \circ B,$ where $\phi_{-\alpha}:= \frac{z+\alpha}{1+\overline{\alpha}z}$ is the automorphism of the unit disc sending $-\alpha \to 0.$ In particular, the images of the zeros $a_1,..,a_n$ of $B$ under $\phi_{-\alpha}$ are $b_1,...,b_n \in \mathbb{D}$ (up to multiplicity). Moreover, each factor of the Blaschke product satisfies

$\left|\frac{z-a_j}{1-\overline{a_j} z}\right| = 1$

when $|z|=1,$ so therefore this also holds for the product $B$ and correspondingly also the composition $g = \phi_{-\alpha} \circ B.$ Now, define $\widetilde{g}(z) = \frac{g(z)}{\prod_{j=1}^n \phi_{b_j}(z)}.$ Then, $\widetilde{g}(z)$ is a holomorphic function on the unit disc with no zeros inside $\mathbb{D}$ and $|\widetilde{g}(z)|=1$ when $|z|=1.$ By the maximum modulus principle applied to $\widetilde{g}$ and $\frac{1}{\widetilde{g}},$ we conclude that $|\widetilde{g}|=1$ on $\mathbb{D},$ i.e. $\widetilde{g}(z) = \lambda$ is constant with $|\lambda|=1.$ This immediately implies that $g(z) = \widetilde{g}(z) \prod_{i=1}^n \phi_{b_j}(z)$ is a Blaschke product of degree $n.$

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