# 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,

where the Fourier coefficients *c*_{n} of *ƒ* are given by

More formally, the result holds as stated provided *ƒ* is square-integrable or, more generally, in *L*^{2}[−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 *ƒ* ∈ *L*^{2}(**R**),

## 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 (*e*_{n}) be an orthonormal basis of *H*; i.e., the linear span of the *e*_{n} is dense in *H*, and the *e*_{n} are mutually orthonormal:

Then Parseval's identity asserts that for every *x* ∈ *H*,

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 *L*^{2}[−π,π], and setting *e*_{n} = e^{−inx} for *n* ∈ **Z**.

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

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.

## See also

## References

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