The proof of the Area Theorem for Conformal Maps

Question

The Area Theorem:
Suppose $f(z)$ is one-to-one and analytic on the punctured unit disk, and is given by $f(z) = 1/z + \sum_0^\infty a_nz^n$

Then $\sum_0^\infty n|a_n|^2 \le 1$

I'm reading the proof on wikipedia https://en.wikipedia.org/wiki/Area_theorem_(complex_analysis) which seems to work except for the very last step in which they commute a limit and a sum.

They had proved up until this point that $f(z)$ takes a circle of radius $r$ to a bounded region of area $-\pi \sum_{-1}^\infty n r^{2n} |a_n|^2$ where they defined $a_{-1} = 1$.

The result would be proven if only we could take the limit $r\rightarrow 1$ inside the summation. But how can this be justified? It suffices to show that $\sum n|a_n|^2$ converges in which case we'd have the result by Abel's theorem.

Answer 1

The last step (before the limit $r\to1$ is taken) of the proof shows $$ \sum_{n=0}^\infty nr^{2n}|a_n|^2\leq\frac{1}{r^2} $$ for each $r\in(0,1)$. Now for the left hand side we have $$ \sum_{n=0}^Nnr^{2n}|a_n|^2\leq\sum_{n=0}^\infty nr^{2n}|a_n|^2\leq\frac{1}{r^2} $$ for each $N\in\mathbb N$, since the coefficients of the sum are non-negative. For fixed $N\in\mathbb N$ we can pass to $r\to1$ and obtain $$ \sum_{n=0}^Nn|a_n|^2\leq1 $$ for this and therefore for each $N\in\mathbb N$. Now let $N\to\infty$ to conclude the proof.

Answer 2

I want to show $\lim_{r \rightarrow 1^-}-\pi \sum_{-1}^\infty n r^{2n} |a_n|^2 = -\pi \sum_{-1}^\infty n |a_n|^2$

Either $-\pi \sum_{-1}^\infty n |a_n|^2$ is finite, or it is equal to $-\infty$. In the first case, we are done, by Abel's theorem.

In the second case, there is a more general version of Abel's theorem that says $\lim_{r \rightarrow 1^-}-\pi \sum_{-1}^\infty n r^{2n} |a_n|^2 = -\infty$.

Since $-\pi \sum_{-1}^\infty n r^{2n} |a_n|^2$ is the area of a region, it is positive. Also as $r \rightarrow 1^-$, the sequence is decreasing because the negative terms get larger. So the limit must be finite as $r \rightarrow 1^-$. So the first case is the only possible one.