Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

According to Gamelin's $\textit{Complex Analysis}$, a finite Blaschke product is a rational function of the form $B(z)= e^{i \varphi} (\frac{z-a_1}{1-\bar{a_1} z} \cdots \frac{z-a_n}{1-\bar{a_n} z})$ where $a_1, ..., a_n \in \mathbb{D}$ and $0 \leq \varphi \leq 2\pi$. Similarly, I would guess that an infinite Blaschke product would be of the form $e^{i \varphi} \prod_{n=1}^\infty\frac{z-a_n}{1-\bar{a_n} z}$. I believe this is supposed to satisfy what is known as the Blaschke condition, i.e. $\sum_{n=1}^\infty (1-|a_n|) < \infty$, but how is that so? Can this be verified using the log function on the infinite product?

share|improve this question
    
There are problems defining what one means by an infinite product of complex numbers unless some condition is imposed: otherwise the partial products won't converge. I think the particular issue of how you define infinite BPs is discussed in Rudin's RCA book but I don't have my copy at hand right now –  user16299 Feb 2 '12 at 2:15
1  
Yes, this is shown in Chapter 15 of Rudin's Real and complex analysis, starting on page 310 under the subheading "Blaschke products". The Blaschke condition applies to the zero sets of more general classes of analytic functions on the disk, including all bounded analytic functions. For the particular case of Blaschke product a more general theorem on convergence of infinite products (Theorem 15.6 on pp 300-301) is applied. –  Jonas Meyer Feb 2 '12 at 2:30
add comment

1 Answer

up vote 3 down vote accepted

Actually, the infinite Blaschke product, for $|a_n|\le1$ and $|z|<1$, is defined as $$ e^{i\varphi}\prod_{n=1}^\infty\frac{|a_n|}{a_n}\frac{z-a_n}{\overline{a}_n z-1}\tag{1} $$ The factor of $\;{-}\dfrac{|a_n|}{a_n}$ simply rotates $\dfrac{z-a_n}{1-\overline{a}_n z}$, which, for finite products, is incorporated into $e^{i\varphi}$. However, for infinite products, it is needed for convergence.

First, note that $$ \begin{align} \frac{|a_n|}{a_n}\frac{z-a_n}{\overline{a}_n z-1} &=|a_n|\frac{z-a_n}{|a_n|^2 z-a_n}\\ &=(1-(1-|a_n|))\left(1+\frac{z(1-|a_n|^2)}{|a_n|^2 z-a_n}\right)\\ &=(1-(1-|a_n|))\left(1+\frac{z(1+|a_n|)}{|a_n|^2\left(z-\frac{1}{\overline{a}_n}\right)}(1-|a_n|)\right)\tag{2} \end{align} $$ where $$ \begin{align} \left|\frac{z(1+|a_n|)}{|a_n|^2\left(z-\frac{1}{\overline{a}_n}\right)}\right| &\le\frac{1+|a_n|}{|a_n|^2}\frac{|z|}{1-|z|}\\ &\le6\frac{|z|}{1-|z|}\tag{3} \end{align} $$ when $|a_n|\ge\frac12$.

Equations $(2)$ and $(3)$ say that the infinite product in $(1)$ converges absolutely when $|z|<1$ and $$ \sum_{n=1}^\infty(1-|a_n|)\tag{4} $$ converges. That is, the infinite product $\prod\limits_{n=1}^\infty(1+z_n)$ converges absolutely when $\sum\limits_{n=1}^\infty|z_n|$ converges.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.