I am reading Ewing and Schober's paper on analytically computing the area of the Mandelbrot set and I hope, in a shred of such, that someone might have an idea why swapping an integral and sum is justified. The paper can be foud on https://link.springer.com/article/10.1007%2FBF01385497, p.69.
A diffeomorphism $$\phi(z)=\sum_{m=-1}^{\infty} b_mz^{-m}$$ maps the exterior of the unit disk to the exterior of the Mandelbrot set. The image $\phi(C_r)$ of a circle of radius $r>1$ forms a region $M( R)$ and we wish to find the area of $M(1)$.
Using Green's Theorem, they give the following expression for the area $$ -\frac12 \int \sum_{n,m\geq -1} m\overline{b_n}b_m(1/r)^{n+m}e^{(m-n)i\theta} d\theta. $$ I am trying to understand:
Why does the sum $$\sum_{n,m\geq -1} m\overline{b_n}b_m(1/r)^{n+m}e^{(m-n)i\theta}$$ converges uniformly so that swapping the integral and the sum is justified?
I tried to bound it by a geometric series but the trouble is that the coefficients are not constant and also $r>1$. The first coeffients are $b_0=-1/2,b_2=-1/4,b_3=15/128,b4=0,b_5=-47/1024$ but computing these coefficients seems to be a delicate matter. Does the sum converge because these coefficients converge very quickly to zero?
Idea: the (three) terms with $m = -1$ or $n = -1$ are irrelevant, ignore it. Now, group the terms with $m + n = k$: $$ \sum_{k\ge 0}\left(\sum_{n+m = k} m\overline{b_n}b_m e^{(m-n)i\theta}\right)r^{-k} $$ If $|b_n|\le 1$, we have $$\left|\sum_{n+m = k} m\overline{b_n}b_m e^{(m-n)i\theta}\right|\le\sum_{m=0}^k m = k(k+1)/2.$$ Finally, take a fixed $ R > 1$ and apply the M-test in the interval $r\ge R$.