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?
2026-04-12 09:31:38.1775986298
Fourier transform of a function is square integrable
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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 (cf. Sec. 6) by invoking Parseval's theorem.