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

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.
