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Is there a result stating that if a function $f:\mathbb{R} \rightarrow \mathbb{R}$ is square integrable and decays at infinity, then its Fourier transform is also square integrable?

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You could leave out the condition that the function decays at infinity which is implied by the fact that it is square integrable. –  Garrett Feb 26 at 21:34
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up vote 2 down vote accepted

More is true: the Fourier transform is an isometry on $L^2(\mathbb{R})$, a fact often noted as a Corollary to the Fourier inversion formula by invoking Parseval's theorem.

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Thanks for reminding me of Parseval's. –  tatterdemalion Feb 12 '13 at 1:57
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