# Parseval's identity

In mathematical analysis, Parseval's identity, named after Marc-Antoine Parseval, is a fundamental result on the summability of the Fourier series of a function. Geometrically, it is the Pythagorean theorem for inner-product spaces.

Informally, the identity asserts that the sum of the squares of the Fourier coefficients of a function is equal to the integral of the square of the function,

$\sum _{n=-\infty }^{\infty }|c_{n}|^{2}={\frac {1}{2\pi }}\int _{-\pi }^{\pi }|f(x)|^{2}\,dx,$ where the Fourier coefficients cn of ƒ are given by

$c_{n}={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x){\mathrm {e} }^{-inx}\,dx.$ More formally, the result holds as stated provided ƒ is square-integrable or, more generally, in L2[−π,π]. A similar result is the Plancherel theorem, which asserts that the integral of the square of the Fourier transform of a function is equal to the integral of the square of the function itself. In one-dimension, for ƒL2(R),

$\int _{-\infty }^{\infty }|{\hat {f}}(\xi )|^{2}\,d\xi =\int _{-\infty }^{\infty }|f(x)|^{2}\,dx.$ ## Generalization of the Pythagorean theorem

The identity is related to the Pythagorean theorem in the more general setting of a separable Hilbert space as follows. Suppose that H is a Hilbert space with inner product 〈•,•〉. Let (en) be an orthonormal basis of H; i.e., the linear span of the en is dense in H, and the en are mutually orthonormal:

$\langle e_{m},e_{n}\rangle ={\begin{cases}1&{\mbox{if}}\ m=n\\0&{\mbox{if}}\ m\not =n.\end{cases}}$ Then Parseval's identity asserts that for every x ∈ H,

$\sum _{n}|\langle x,e_{n}\rangle |^{2}=\|x\|^{2}.$ This is directly analogous to the Pythagorean theorem, which asserts that the sum of the squares of the components of a vector in an orthonormal basis is equal to the squared length of the vector. One can recover the Fourier series version of Parseval's identity by letting H be the Hilbert space L2[−π,π], and setting en = e−inx for nZ.

More generally, Parseval's identity holds in any inner-product space, not just separable Hilbert spaces. Thus suppose that H is an inner-product space. Let B be an orthonormal basis of H; i.e., an orthonormal set which is total in the sense that the linear span of B is dense in H. Then

$\|x\|^{2}=\langle x,x\rangle =\sum _{v\in B}\left|\langle x,v\rangle \right|^{2}.$ The assumption that B is total is necessary for the validity of the identity. If B is not total, then the equality in Parseval's identity must be replaced by ≥, yielding Bessel's inequality. This general form of Parseval's identity can be proved using the Riesz–Fischer theorem.