# Parseval's identity

Short description: The energy of a periodic function is the same in the time and frequency domain.

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. The identity asserts the equality of the energy of a periodic signal (given as the integral of the squared amplitude of the signal) and the energy of its frequency domain representation (given as the sum of squares of the amplitudes). Geometrically, it is a generalized Pythagorean theorem for inner-product spaces (which can have an uncountable infinity of basis vectors).

The identity asserts that the sum of squares of the Fourier coefficients of a function is equal to the integral of the square of the function, $\displaystyle{ \Vert f \Vert^2_{L^2(-\pi,\pi)}= \int_{-\pi}^\pi |f(x)|^2 \, dx=2\pi\sum_{n=-\infty}^\infty |c_n|^2 }$ where the Fourier coefficients $\displaystyle{ c_n }$ of $\displaystyle{ f }$ are given by $\displaystyle{ c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) e^{-inx} \, dx. }$

The result holds as stated provided $\displaystyle{ f }$ is a square-integrable function or, more generally, in Lp space $\displaystyle{ L^2[-\pi, \pi]. }$ 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 $\displaystyle{ f \in L^2(\R), }$ $\displaystyle{ \int_{-\infty}^\infty |\hat{f}(\xi)|^2\,d\xi = \int_{-\infty}^\infty |f(x)|^2\, dx. }$

## 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 $\displaystyle{ H }$ is a Hilbert space with inner product $\displaystyle{ \langle \,\cdot\,, \,\cdot\, \rangle. }$ Let $\displaystyle{ \left(e_n\right) }$ be an orthonormal basis of $\displaystyle{ H }$; i.e., the linear span of the $\displaystyle{ e_n }$ is dense in $\displaystyle{ H, }$ and the $\displaystyle{ e_n }$ are mutually orthonormal:

$\displaystyle{ \langle e_m, e_n\rangle = \begin{cases} 1 & \mbox{if}~ m = n \\ 0 & \mbox{if}~ m \neq n. \end{cases} }$

Then Parseval's identity asserts that for every $\displaystyle{ x \in H, }$ $\displaystyle{ \sum_n \left|\left\langle x, e_n \right\rangle\right|^2 = \|x\|^2. }$

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 $\displaystyle{ H }$ be the Hilbert space $\displaystyle{ L^2[-\pi, \pi], }$ and setting $\displaystyle{ e_n = \frac{e^{-i n x}}{\sqrt{2 \pi}} }$ for $\displaystyle{ n \in \Z. }$

More generally, Parseval's identity holds in any inner product space, not just separable Hilbert spaces. Thus suppose that $\displaystyle{ H }$ is an inner-product space. Let $\displaystyle{ B }$ be an orthonormal basis of $\displaystyle{ H }$; that is, an orthonormal set which is total in the sense that the linear span of $\displaystyle{ B }$ is dense in $\displaystyle{ H. }$ Then $\displaystyle{ \|x\|^2 = \langle x,x\rangle = \sum_{v\in B}\left|\langle x, v\rangle\right|^2. }$

The assumption that $\displaystyle{ B }$ is total is necessary for the validity of the identity. If $\displaystyle{ B }$ is not total, then the equality in Parseval's identity must be replaced by $\displaystyle{ \, \geq, }$ yielding Bessel's inequality. This general form of Parseval's identity can be proved using the Riesz–Fischer theorem.