Plancherel theorem: Difference between revisions

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{{Short description|Theorem in harmonic analysis}}

In [[mathematics]], the '''Plancherel theorem''' (sometimes called the [[Marc-Antoine Parseval|Parseval]]–Plancherel identity<ref>{{cite book |author1=Cohen-Tannoudji, Claude |author2=Dupont-Roc, Jacques |author3=Grynberg, Gilbert |title=Photons and Atoms : Introduction to Quantum Electrodynamics |year=1997 |url=https://archive.org/details/photonsatomsintr00cohe_398 |url-access=limited |publisher=Wiley |isbn=0-471-18433-0 |page=[https://archive.org/details/photonsatomsintr00cohe_398/page/n39 11]}}</ref>) is a result in [[harmonic analysis]], proven by [[Michel Plancherel]] in 1910. It states that the integral of a function's [[squared modulus]] is equal to the integral of the squared modulus of its [[frequency spectrum]]. That is, if <math>f(x) </math> is a function on the real line, and <math>\widehat{f}(\xi)</math> is its frequency spectrum, then

== Statement of the theorem ==

In [[mathematics]], the '''Plancherel theorem''' (sometimes called the [[Marc-Antoine Parseval|Parseval]]–Plancherel identity<ref>{{cite book |author1=Cohen-Tannoudji, Claude |author2=Dupont-Roc, Jacques |author3=Grynberg, Gilbert |title=Photons and Atoms : Introduction to Quantum Electrodynamics |year=1997 |url=https://archive.org/details/photonsatomsintr00cohe_398 |url-access=limited |publisher=Wiley |isbn=0-471-18433-0 |page=[https://archive.org/details/photonsatomsintr00cohe_398/page/n39 11]}}</ref>) is a result in [[harmonic analysis]], proven by [[Michel Plancherel]] in 1910. It states that the integral of a function's [[squared modulus]] is equal to the integral of the squared modulus of its [[frequency spectrum]]. That is, if <math>f(x) </math> is aan integrable function on the real line, and <math>\widehat{f}(\xi)</math> is its frequency spectrum, theni.e.

<math>\int_{-\infty}^{\infty}|f(x)|dx<\infty</math>

and <math>\widehat{f}(\xi)</math> is its frequency spectrum, i.e.

<math>\hat{f}(\xi)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)\exp(-i\xi x)dx</math>

then

{{Equation box 1

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