1
$\begingroup$

Let $f : \mathbb{D} → \mathbb{D},$ $f(z) = \sum_{n = 0}^{\infty} a_nz^n$ be a bounded analytic function

a.) Prove that for any $r < 1, \sum_{n = 0}^\infty |a_n|^2r^{2n} = \frac{1}{2π}\int_0^{2\pi} |f(re^{it})|^2 dt.$

b.) Show that the series $\sum_{n=0}^\infty |a_n|^2$ converges.

I am having trouble with the following complex qual problem. Any suggestions? Thanks

$\endgroup$

closed as off-topic by Andrés E. Caicedo, Cookie, Kirill, Moishe Kohan, user91500 Jun 26 '14 at 5:33

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Andrés E. Caicedo, Cookie, Kirill, Moishe Kohan, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.

2
$\begingroup$

For (a), begin with $$|f(z)|^2 = f(z)\overline{f(z)} = \sum_{n=0}^\infty a_n z^n \sum_{m=0}^\infty \bar a_m \bar z^m = \sum_{n=0}^\infty\sum_{m=0}^\infty a_n \bar a_m z^n z^m$$ (Convergence is absolute and uniform on compact subsets, no problem with rearrangement and integration term-by-term.) When plugging in $z=r e^{it}$, you get a constant multiple of $r^{2n} e^{i(n-m)t}$. Over $[0,2\pi]$ this integrates to zero unless $n=m$. The reuslt follows.

(b) If $|f|\le M$, then the integral from (a) is at most $M$, for any $r$. For every $N$ we have $$\sum_{n=0}^N |a_n|^2 = \lim_{r\to 1} \sum_{n=0}^N |a_n|^2 r^{2n} \le M$$ The partial sums are bounded, hence the series converges.

$\endgroup$

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