A question in area theorem

Question

Let $R \in (0 , \infty)$ and let $R_1 \in [0 , R)$ and $R_2 \in (R , \infty)$ We consider $C_R \equiv |z| = R$ and $A(0 , R_1 , R_2) \equiv R_1 < |z| < R_2$. We suppose that $f \in \mathcal{H}(A(0 , R_1 , R_2))$. We know that exists a number sequence ${\{a_n\}}_{n = - \infty}^{\infty}$ such that \begin{equation} f(z) = \sum_{n = - \infty}^{\infty} a_n z^n\mbox{.} \end{equation} We also suppose that $f(C_R)$ is a Jordan curve. Let $J_R = f(C_R)$ and $D_R$ the interior domain whose border is $J_R$. Then $$ \mathcal{V}(D_R) = \pi \left| \sum_{n = - \infty}^{\infty} n {|a_n|}^2 R^{2 n}\right|\mbox{.} $$ It is called area theorem. Using Green's theorem and properties about holomorphic functions in rings (using also that $C_R$ is a compact set in $A(0 , R_1 , R_2)$), I have shown that $$ \mathcal{V}(D_R) = \frac{\pi}{2} \left| \sum_{n = - \infty}^{\infty} (- n a_n a_{- n} - n \overline{a_n a_{- n}}) + 2 \sum_{n = - \infty}^{\infty} n {|a_n|}^2 R^{2 n}\right|\mbox{.} $$ Then I have just to prove that $$ \sum_{n = - \infty}^{\infty} (- n a_n a_{- n} - \overline{n a_n a_{- n}}) $$ converges and it's equal to $0$ but I have no idea. At first I have to prove that $$ \lim_{N \to \infty} \sum_{n = 1}^N (- n a_n a_{- n} - \overline{n a_n a_{- n}}) = - \lim_{M \to \infty} \sum_{n = - M}^{- 1} (- n a_n a_{- n} - \overline{n a_n a_{- n}}) = z_0 \in \mathbb{C} $$ to say that converges. Later, calling $$ S_N = \sum_{n = - N}^N (- n a_n a_{- n} - \overline{n a_n a_{- n}}) $$ for all $N \in \mathbb{Z}$, I can say that $$ \lim_{N \to \infty} S_N = \sum_{n = - \infty}^{\infty} (- n a_n a_{- n} - \overline{n a_n a_{- n}}) $$ and now the conclusion is very easy because obviously $S_N = 0$ for all $N \in \mathbb{Z}$, but I have no idea to prove that $$ \lim_{N \to \infty} \sum_{n = 1}^N (- n a_n a_{- n} - \overline{n a_n a_{- n}}) = - \lim_{M \to \infty} \sum_{n = - M}^{- 1} (- n a_n a_{- n} - \overline{n a_n a_{- n}}) = z_0 \in \mathbb{C}\mbox{.} $$ I only think that I must use ${\{a_n\}}_{n = - \infty}^{\infty}$ is the only sequence such that $f(z) = \sum_{n = - \infty}^{\infty} a_n z^n$, where $f \in \mathcal{H}(A(0 , R_1 , R_2))$. Can you help me?. For example, it isn't true with all the sequences ${\{b_n\}}_{n = - \infty}^{\infty}$ because for $b_n = n$, it's clair that $$ \sum_{n = - N}^N n = 0 $$ but we also know that we can't say that $$ \lim_{N \to \infty} \sum_{n = 0}^N n + \lim_{M \to \infty} \sum_{n = - M}^{- 1} n = \infty - \infty = 0 \mbox{ (it is not correct).} $$

Answer 1

What you need is just the convergence of

$$\sum_{n = 1}^\infty n a_n a_{-n}.\tag{1}$$

Everything else follows easily from the symmetry - for $m = -n$ we have $ma_ma_{-m} = (-n)a_{-n}a_n$ - and conjugation.

To see that $(1)$ is convergent, we note that for $f \in \mathcal{H}(A(0,R_1,R_2))$, the functions $g\colon z \mapsto zf'(z)$ and $h \colon z \mapsto zf'(z)f(z)$ also are holomorphic on the annulus $A(0,R_1,R_2)$, and therefore have convergent Laurent series in that annulus. The Laurent series of $g$ is obtained via termwise differentiation of the Laurent series of $f$ and subsequent multiplication with $z$,

$$g(z) = z\Biggl(\sum_{n = -\infty}^{\infty} a_n z^n\Biggr)' = z\sum_{n = -\infty}^{\infty} n a_n z^{n-1} = \sum_{n = -\infty}^{\infty} n a_n z^n.\tag{2}$$

Then one needs to know that the Laurent series of a product of two holomorphic functions on an annulus is obtained by multiplying the Laurent series of the factors, where the series for the coefficients are absolutely convergent. Knowing that, we see that the Laurent series of $h$ is

$$h(z) = \sum_{k = -\infty}^{\infty} c_k z^k,$$

where the coefficients are given by

$$c_k = \sum_{n = -\infty}^{\infty} n a_n a_{k-n},$$

and the series you are interested in is a part of the absolutely convergent series for $c_0$.

For completeness, we should prove the statement about the absolute convergence of the series for the coefficients of the product. Thus let $g, h \in \mathcal{H}(A(0,R_1,R_2))$ (generic, not [necessarily] the $g$ and $h$ from above) with Laurent series

$$g(z) = \sum_{n = -\infty}^{\infty} \alpha_n z^n\quad\text{and}\quad h(z) = \sum_{k = -\infty}^{\infty} \beta_kz^k$$

in that annulus. By the general theory, these series converge absolutely and locally uniformly on the annulus, in particular on the circle with radius $R$, so

$$\sum_{n = -\infty} \lvert \alpha_n\rvert R^n < +\infty \quad\text{and}\quad \sum_{k = -\infty} \lvert \beta_k\rvert R^k < +\infty.$$

It follows that

$$\sum_{k,n\in \mathbb{Z}} \lvert \alpha_n\rvert\lvert\beta_k\rvert R^{n+k} < +\infty,$$

and hence in particular the subseries over the pairs with $n + k = m$ is absolutely convergent for every $m \in \mathbb{Z}$:

$$\sum_{n = -\infty}^{\infty} \lvert \alpha_n\rvert \lvert \beta_{m-n}\rvert R^m < +\infty.$$

But $R^m$ is just a strictly positive constant, so

$$\sum_{n = -\infty}^{\infty} \lvert \alpha_n \beta_{m-n}\rvert < +\infty$$

for all $m$. The case $m = 0$ and $\alpha_n = na_n,\, \beta_n = a_n$ is the one we're interested in.