Parseval–Gutzmer formula
# Parseval–Gutzmer formula

In mathematics, the Parseval–Gutzmer formula states that, if $f$ is an analytic function on a closed disk of radius r with Taylor series

    $f(z)=\sum _{k=0}^{\infty }a_{k}z^{k},$

then for z = $re^{iθ}$ on the boundary of the disk,

    $\int _{0}^{2\pi }|f(re^{i\theta })|^{2}\,\mathrm {d} \theta =2\pi \sum _{k=0}^{\infty }|a_{k}|^{2}r^{2k},$

which may also be written as

    ${\frac {1}{2\pi }}\int _{0}^{2\pi }|f(re^{i\theta })|^{2}\,\mathrm {d} \theta =\sum _{k=0}^{\infty }|a_{k}r^{k}|^{2}.$

## Proof

The Cauchy Integral Formula for coefficients states that for the above conditions:

    $a_{n}={\frac {1}{2\pi i}}\int _{\gamma }^{\frac {f(z)}{z^{n+1}}}\,\mathrm {d} z$

where γ is defined to be the circular path around origin of radius r. Also for $x\in \mathbb {C} ,$ we have: ${\overline {x}}{x}=|x|^{2}.$ Applying both of these facts to the problem starting with the second fact:

    ${\begin{aligned}\int _{0}^{2\pi }\left|f\left(re^{i\theta }\right)\right|^{2}\,\mathrm {d} \theta &=\int _{0}^{2\pi }f\left(re^{i\theta }\right){\overline {f\left(re^{i\theta }\right)}}\,\mathrm {d} \theta \\[6pt]&=\int _{0}^{2\pi }f\left(re^{i\theta }\right)\left(\sum _{k=0}^{\infty }{\overline {a_{k}\left(re^{i\theta }\right)^{k}}}\right)\,\mathrm {d} \theta &&{\text{Using Taylor expansion on the conjugate}}\\[6pt]&=\int _{0}^{2\pi }f\left(re^{i\theta }\right)\left(\sum _{k=0}^{\infty }{\overline {a_{k}}}\left(re^{-i\theta }\right)^{k}\right)\,\mathrm {d} \theta \\[6pt]&=\sum _{k=0}^{\infty }\int _{0}^{2\pi }f\left(re^{i\theta }\right){\overline {a_{k}}}\left(re^{-i\theta }\right)^{k}\,\mathrm {d} \theta &&{\text{Uniform convergence of Taylor series}}\\[6pt]&=\sum _{k=0}^{\infty }\left(2\pi {\overline {a_{k}}}r^{2k}\right)\left({\frac {1}{2{\pi }i}}\int _{0}^{2\pi }{\frac {f\left(re^{i\theta }\right)}{(re^{i\theta })^{k+1}}}{rie^{i\theta }}\right)\mathrm {d} \theta \\&=\sum _{k=0}^{\infty }\left(2\pi {\overline {a_{k}}}r^{2k}\right)a_{k}&&{\text{Applying Cauchy Integral Formula}}\\&={2\pi }\sum _{k=0}^{\infty }{|a_{k}|^{2}r^{2k}}\end{aligned}}$

## Further Applications

Using this formula, it is possible to show that

    $\sum _{k=0}^{\infty }|a_{k}|^{2}r^{2k}\leqslant M_{r}^{2}$

where

    $M_{r}=\sup\{|f(z)|:|z|=r\}.$

This is done by using the integral

    $\int _{0}^{2\pi }\left|f\left(re^{i\theta }\right)\right|^{2}\,\mathrm {d} \theta \leqslant 2\pi \left|\max _{\theta \in [0,2\pi )}\left(f\left(re^{i\theta }\right)\right)\right|^{2}=2\pi \left|\max _{|z|=r}(f(z))\right|^{2}=2\pi M_{r}^{2}$
