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


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

then for z = reiθ on the boundary of the disk,


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

which may also be written as


 * $$\frac{1}{2\pi }\int^{2\pi}_0 |f(re^{i\theta}) |^2 \, \mathrm{d}\theta = \sum^\infty_{k = 0} |a_k r^k|^2.$$

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 $$x \in \Complex,$$ we have: $$\overline{x}{x} = |x|^2.$$ Applying both of these facts to the problem starting with the second fact:


 * $$ \begin{align}

\int^{2\pi}_0 \left |f \left (re^{i\theta} \right ) \right |^2 \, \mathrm{d}\theta &= \int^{2\pi}_0 f \left (re^{i\theta} \right ) \overline{f \left (re^{i\theta} \right )} \, \mathrm{d}\theta\\[6pt] &= \int^{2\pi}_0 f \left (re^{i\theta} \right ) \left (\sum^\infty_{k = 0} \overline{a_k \left (re^{i\theta} \right )^k} \right )  \, \mathrm{d}\theta && \text{Using Taylor expansion on the conjugate} \\[6pt] &= \int^{2\pi}_0 f \left (re^{i\theta} \right ) \left (\sum^\infty_{k = 0} \overline{a_k} \left (re^{-i\theta} \right )^k \right )  \, \mathrm{d}\theta \\[6pt] &= \sum^\infty_{k = 0} \int^{2\pi}_0 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^\infty_{k = 0} \left (2\pi \overline{a_k} r^{2k} \right ) \left (\frac{1}{2{\pi}i}\int^{2\pi}_0 \frac{f \left (re^{i\theta} \right )}{(r e^{i\theta})^{k+1}} {rie^{i\theta}} \right ) \mathrm{d}\theta \\ & = \sum^\infty_{k = 0} \left (2\pi \overline{a_k} r^{2k} \right ) a_k && \text{Applying Cauchy Integral Formula} \\ & = {2\pi} \sum^\infty_{k = 0} {|a_k|^2 r^{2k}} \end{align}$$

Further Applications
Using this formula, it is possible to show that


 * $$\sum^\infty_{k = 0} |a_k|^2r^{2k} \leqslant M_r^2$$

where


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

This is done by using the integral


 * $$\int^{2\pi}_0 \left |f \left (re^{i\theta} \right ) \right |^2 \, \mathrm{d}\theta \leqslant 2\pi \left|\max_{\theta \in [0,2\pi)} \left (f \left (re^{i\theta} \right ) \right ) \right |^2 = 2\pi\left |\max_{|z|=r}(f(z)) \right |^2 = 2\pi M_r^2$$