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

Now, using our Taylor Expansion on the conjugate,

=\int _{0}^{2\pi }{f(re^{i\vartheta })}{\sum _{k=0}^{\infty }{\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 _{k=0}^{\infty }\int _{0}^{2\pi }{\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 _{k=0}^{\infty }({2\pi }{{\overline {a_{k}}}r^{2k}})({\frac {1}{2{\pi }i}}\int _{0}^{2\pi }{\frac {f(re^{i\vartheta })}{(re^{i\vartheta })^{k+1}}}{rie^{i\vartheta }})\mathrm {d} \vartheta

Now, applying the Cauchy Integral Formula, we get

=\sum _{k=0}^{\infty }({2\pi }{{\overline {a_{k}}}r^{2k}}){a_{k}}={2\pi }\sum _{k=0}^{\infty }{|a_{k}|^{2}r^{2k}}

Using this formula, it is possible to show that

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

where

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

This is done by using the integral

\int _{0}^{2\pi }|f(re^{i\vartheta })|^{2}\,\mathrm {d} \vartheta \leq 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}

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