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


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

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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.


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