[[fourier transform]]s preserve the [inner product](Inner%20products.md) meaning that, $\int dx\, (f(x))^* g(x) = \frac{1}{2\pi}\int dk\,\mathcal{F}[f(x)]_k^*\,\mathcal{F}[g(x)]_k$ In other words the fourier transform is a [unitary operation](Unitary%20operators.md). This property is referred to as _Parseval's theorem._ %%Note that your statement of Parseval's theorem is probably incomplete. look at wikipedia. break down that statement if need be.%% # Plancherel theorem The application of [Parseval's theorem](Parseval's%20theorem.md) where $f=g$ is referred to as Plancheral's theorem. In this case the [inner product](Inner%20products.md) integral is: $\int dx\, |f(x)|^2 = \frac{1}{2\pi}\int dk\,|\mathcal{F}[f(x)]_k|^2$ --- # Proofs and examples ## Direct proof of Parseval's theorem   ## Proof of Parseval's theorem by treating the Fourier transform as a linear map   %%Note this isn't the same as Parseval's identity%% #MathematicalFoundations/Analysis/FourierAnalysis/Integrals