Parseval–Gutzmer formula

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

$f(z) = \sum^\infty_{k = 0} a_k z^k,$

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

$\int^{2\pi}_0 |f(re^{i\vartheta}) |^2 \, \mathrm{d}\vartheta = 2\pi \sum^\infty_{k = 0} |a_k|^2r^{2k}.$

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 0 of radius r. We also have that, for x in the complex plane C,

$\overline{x}{x} = |x|^2$

We can apply both of these facts to the problem. Using the second fact,

$\int^{2\pi}_0 |f(re^{i\vartheta}) |^2 \, \mathrm{d}\vartheta = \int^{2\pi}_0 {f(re^{i\vartheta})}\overline{f(re^{i\vartheta})} \, \mathrm{d}\vartheta$

Now, using our Taylor Expansion on the conjugate,

$= \int^{2\pi}_0 {f(re^{i\vartheta})}{\sum^\infty_{k = 0} \overline{a_k (re^{i\vartheta})^k}} \, \mathrm{d}\vartheta$

Using the uniform convergence of the Taylor Series and the properties of integrals, we can rearrange this to be

$= \sum^\infty_{k = 0} \int^{2\pi}_0 \frac{{f(re^{i\vartheta})}\overline{a_k} (r^k)}{(e^{i\vartheta})^k} , \mathrm{d}\vartheta$

With further rearrangement, we can set it up ready to use the Cauchy Integral Formula statement

$= \sum^\infty_{k = 0} ({2\pi}{\overline{a_k} r^{2k}})(\frac{1}{2{\pi}i}\int^{2\pi}_0 \frac{{f(re^{i\vartheta})}}{(r e^{i\vartheta})^{k+1}} {rie^{i\vartheta}}) \mathrm{d}\vartheta$

Now, applying the Cauchy Integral Formula, we get

$= \sum^\infty_{k = 0} ({2\pi}{\overline{a_k} r^{2k}}){a_k} = {2\pi} \sum^\infty_{k = 0} {|a_k|^2 r^{2k}}$

Further Applications

Using this formula, it is possible to show that

$\sum^\infty_{k = 0} |a_k|^2r^{2k} \le {M_r}^2$ where $M_r = \sup\{|f(z)| : |z| = r\}$

This is done by using the integral

$\int^{2\pi}_0 |f(re^{i\vartheta}) |^2 \, \mathrm{d}\vartheta \le 2\pi |max_{\vartheta \in [0,2\pi)}(f(re^{i\vartheta}))|^2 = 2\pi |max_{|z|=r}(f(z))|^2 = 2\pi(M_r)^2$