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I am seeking a proof for the following statement of the Parseval identity:

If $f$ is a holomorphic function defined on the ball $B(0,r)$, with power series $f(z) = \sum_n a_n z^n$, then $$2 \pi \left( \sum_n |a_n|^2 r^{2n} \right) = \int_0^{2\pi}|f(re^{it})|^2dt.$$

I have seen some proofs of other statements of Parseval identity (specifically using Fourier series), but not of this. I guess I'm looking for one of two possible answers:

1) A proof of the quoted result above (or an outline thereof), or

2) A reference to a standard proof of some statement of the Parseval identity, along with some hints as to how to obtain "my" statement of it from the other (having trouble seeing how to transition).

Thanks in advance for any help!

Bach
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Bachmaninoff
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1 Answers1

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$$\begin{align}\int_0^{2\pi} |f(re^{it})|^2dt &= \int_0^{2\pi}\overline{f(re^{it})}f(re^{it})dt\\ &=\int_0^{2\pi}\sum_m \overline{a_m}r^me^{-imt}\sum_n {a_n}r^ne^{int}\ dt\\ &=\sum_n\sum_m\overline{a_m}a_nr^{m+n}\int_0^{2\pi}e^{it(n-m)}dt\\ &=\sum_n\sum_m\overline{a_m}a_nr^{m+n}2\pi\delta_{mn}\\ &=2\pi\sum_n\overline{a_n}a_nr^{n+n}\\ &=2\pi\sum_n |a_n|^2r^{2n}\\ \end{align}$$

Some justification is needed for exchanging the integration and summations. But the summation is just a power series in $(re^{it})$, and the acceptability of integrating power series term-by-term is a basic analysis result.

Paul Sinclair
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  • The interchanging of operation is always valid when $r$ is less than the radius of convergence. – TravorLZH Aug 06 '21 at 14:50
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    @TravorLZH - that only applies to power series, as I said. (The concept of "radius of convergence" isn't even defined more generally.) In general, the exchange of any two limiting operations must be justified. But in many cases, as here, it isn't hard to do. – Paul Sinclair Aug 06 '21 at 15:12
  • I agree. It becomes a bit annoying to study the uniform convergence of certain series. – TravorLZH Aug 07 '21 at 08:59