MHB Laurent series, integral of a holomorphic function

[Samwise1]

We are given $$f = \sum_{n= - \infty} ^{\infty} a_n (z-z_0)^n \in \mathcal{O} (ann (z_0, r, R)), \ \ 0<r<R< \infty $$.

Prove that $$\frac{1}{\pi} \int _{ann (z_0, r, R)} |f(z)|^2 d \lambda(z) = \sum _{n \neq -1} \frac{R^{2n+2} - r^{2n+2}}{n+1}|a_n|^2 + 2 \log \frac{R}{r}|a_{-1}|^2$$.

We know that the series above is convergent, so $$R = \frac{1}{\limsup_ {n \rightarrow \infty} \sqrt[n]{|a_n|}}$$ and $$r = \limsup_ {n \rightarrow \infty} \sqrt[n]{|a_{-n}|}$$.

In the series $$\sum _{n \neq -1} \frac{R^{2n+2}}{n+1}|a_n|^2, \ \ \sum _{n \neq -1} \frac{r^{2n+2}}{n+1}|a_n|^2$$ we have $$b_n = \frac{|a_n|^2}{n+1}$$ and radii of convergence are $R'=\frac{1}{\limsup_ {n \rightarrow \infty} \sqrt[n]{\frac{|a_n|^2}{n+1}}} \ge \frac{1}{\limsup_ {n \rightarrow \infty} \sqrt[n]{|a_n|^2}} \ge \frac{1}{\limsup_ {n \rightarrow \infty} \sqrt[n]{|a_n|}}^2 = R^2$, so $$R' \ge R^2$$.

Similarly, $$r' = \limsup_ {n \rightarrow \infty} \sqrt[n]{\frac{|a_{-n}|^2}{n+1}} \le \limsup_ {n \rightarrow \infty} \sqrt[n]{|a_{-n}|^2} \le r^2$$.

So the two series are convergent - they have the form $$\sum b_n (z-z_0)^n$$ with $z=R^2$ or $-r^2$.

Does that make sense? Could you tell me how to prove the equality of the integral and the series?

 

[Euge]

Hello again Samwise,

Could you please explain the meaning $d\lambda(z)$? Then I can assist you further.

 

[Euge]

If I'm not mistaken, the measure $\lambda$ is a complex measure, viewed as a two-dimensional Lebesgue measure? If so, then by using polar representation you can parametrize the annular region $\text{ann}(z_0, r, R)$ by setting $z = z_0 + \rho e^{it}$, $r \le \rho \le R$ and $0 \le t \le 2\pi$. Then

$$ \frac{1}{\pi} \int_{\text{ann}(z_0, r, R)} |f(z)|^2\, d\lambda(z) = \frac{1}{\pi}\int_r^R \int_0^{2\pi} |f(z_0 + \rho e^{it})|^2 \rho\, dt\, d\rho.$$

Now

$$|f(z)|^2 = f(z) \overline{f(z)} = \sum_{n,m\in \Bbb Z} a_n \overline{a}_m (z - z_0)^n\, \overline{(z - z_0)}^m,$$

which implies

$$|f(z_0 + \rho\, e^{it})|^2 = \sum_{n, m\in \Bbb Z} a_n \overline{a}_m\, \rho^n e^{int} \rho^m e^{-imt} = \sum_{n,m \in \Bbb Z} a_n \overline{a}_m \rho^{n+m} e^{i(n-m)t}.$$

Therefore

$$(*) \frac{1}{\pi} \int_r^R \int_0^{2\pi} |f(z_0 + \rho e^{it})|^2 \rho\, dt\, d\rho = \frac{1}{\pi} \sum_{n,m\in \Bbb Z} a_n \overline{a}_m \int_r^R \rho^{n + m + 1}\, d\rho \int_0^{2\pi} e^{i(n-m)t}\, dt.$$

Since $\int_0^{2\pi} e^{i(n-m)t} dt$ is $2\pi$ when $n = m$ and $0$ when $n \neq m$, the expression on the right hand side of $(*)$ is

$$2\sum_{n\in \Bbb Z} |a_n|^2 \int_r^R \rho^{2n+1}\, d\rho = \sum_{n\neq -1} |a_n|^2 \int_r^R 2\rho^{2n+1}\, d\rho + 2|a_{-1}|^2 \int_r^R \rho^{-1}\, d\rho$$

$$ = \sum_{n \neq -1} |a_n|^2 \frac{R^{2n+2} - r^{2n+2}}{n+1} + 2|a_{-1}|^2 \log \frac{R}{r}.$$