Fourier transform preserves the $L^2$ norm.Suggest a slogan
If \(f\in L^1(\RR^d)\cap L^2(\RR^d)\), then its Fourier transform \(\widehat f\) lies in \(L^2(\RR^d)\) and \[ \norm{f}_{L^2}^2 = \int_{\RR^d} \abs{f}^2 = \int_{\RR^d} \abs{\widehat f}^2 = \norm{\widehat{f}}_{L^2}^2 . \]
The analogue for Fourier series is Parseval’s identity: for \(f\in L^2([0,1])\) with Fourier coefficients \(c_k\coloneqq\int_0^1 f(x)e^{-2\pi i kx}\,dx\), one has \(\int_0^1\abs{f}^2=\sum_{k\in\ZZ}\abs{c_k}^2\).