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

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

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

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

which may also be written as

$\frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}|f(r{e}^{i\theta }){|}^2\mathrm{d}\theta =\sum _{k=0}^{\mathrm{\infty }}|a_k{r}^k{|}^2.$

Proof

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

$a_n=\frac{1}{2\pi i}\underset{\gamma }{\overset{}{\int }}\frac{f(z)}{z^{n+1}}\mathrm{d}z$

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

$\begin{array}{cc}\underset{0}{\overset{2\pi }{\int }}{\left|f\left(r{e}^{i\theta }\right)\right|}^2\mathrm{d}\theta & =\underset{0}{\overset{2\pi }{\int }}f\left(r{e}^{i\theta }\right)\overline{f\left(r{e}^{i\theta }\right)}\mathrm{d}\theta \\ & =\underset{0}{\overset{2\pi }{\int }}f\left(r{e}^{i\theta }\right)\left(\sum _{k=0}^{\mathrm{\infty }}\overline{a_k{\left(r{e}^{i\theta }\right)}^k}\right)\mathrm{d}\theta & & \text{Using Taylor expansion on the conjugate}\\ & =\underset{0}{\overset{2\pi }{\int }}f\left(r{e}^{i\theta }\right)\left(\sum _{k=0}^{\mathrm{\infty }}\overline{a_k}{\left(r{e}^{-i\theta }\right)}^k\right)\mathrm{d}\theta \\ & =\sum _{k=0}^{\mathrm{\infty }}\underset{0}{\overset{2\pi }{\int }}f\left(r{e}^{i\theta }\right)\overline{a_k}{\left(r{e}^{-i\theta }\right)}^k\mathrm{d}\theta & & \text{Uniform convergence of Taylor series}\\ & =\sum _{k=0}^{\mathrm{\infty }}\left(2\pi \overline{a_k}{r}^{2k}\right)\left(\frac{1}{2\pi i}\underset{0}{\overset{2\pi }{\int }}\frac{f\left(r{e}^{i\theta }\right)}{(r{e}^{i\theta }{)}^{k+1}}ri{e}^{i\theta }\right)\mathrm{d}\theta \\ & =\sum _{k=0}^{\mathrm{\infty }}\left(2\pi \overline{a_k}{r}^{2k}\right)a_k& & \text{Applying Cauchy Integral Formula}\\ & =2\pi \sum _{k=0}^{\mathrm{\infty }}|a_k{|}^2{r}^{2k}\end{array}$

Further Applications

Using this formula, it is possible to show that

$\sum _{k=0}^{\mathrm{\infty }}|a_k{|}^2{r}^{2k}⩽M_r^2$

where

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

This is done by using the integral

$\underset{0}{\overset{2\pi }{\int }}{\left|f\left(r{e}^{i\theta }\right)\right|}^2\mathrm{d}\theta ⩽2\pi {\left|\underset{\theta \in [0,2\pi )}{\max }\left(f\left(r{e}^{i\theta }\right)\right)\right|}^2=2\pi {|\underset{|z|=r}{\max }(f(z))|}^2=2\pi M_r^2$

References

• Template:Cite book

Template:Mathanalysis-stub