$H^2$ does not contain any rational functions with poles on the unit circle

Question

The Hardy-hilbert space, $H^2$, consists of all analytic functions having power series representations with square-summable complex coefficients. That is,

$$H^2=\{f : f(z)=\sum_{n=0}^{\infty} a_n z^ns.t \sum_n |a_n|^2 < \infty \}$$

I would like to show that $H^2$ does not contain any rational functions with poles on the unit circle. May I have some ideas to show it ?

Thanks !

Answer 1

The following two observations do the job.

1. If $f\in H^2$, then it follows from Cauchy-Schwartz inequality that $$|f(z)|\leq\sum_{n\geq 0}|a_n| |z|^n \leq \big(\sum_{n\geq 0}|a_n|^2\big)^{1/2} \big(\sum_{n\geq 0} |z|^{2n}\big)^{1/2}=\frac{\|f\|_{H^2}}{\sqrt{1-|z|^2}}=O\big(\frac{1}{\sqrt {1-|z|}}\big).$$
2. If $f$ is a rational function and $w$ is a pole of $f$ with order $k\geq 1$, then $$|f(z)| \sim\frac{1}{|z-w|^k} \quad z\to w.$$