Parseval's identity

In mathematical analysis, Parseval's identity 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 square integral of the function. To wit,:$sum\_\{n=-infty\}^infty\; |a\_n|^2\; =\; frac\{1\}\{2pi\}int\_\{-pi\}^pi\; |f(x)|^2,dx$where the Fourier coefficients "a"_"n" of ƒ are given by:$a\_n\; =\; frac\{1\}\{2pi\}int\_\{-pi\}^\{pi\}\; f(x)e^\{-inx\},dx.$More formally, the result holds as stated provided ƒ is square-integrable or, more generally, in [Lp space|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 ƒ ∈ L²(R),:$int\_\{-infty\}^infty\; |hat\{f\}(xi)|^2,dxi\; =\; int\_\{-infty\}^infty\; |f(x)|^2,\; dx.$

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",

:$sum\_n\; |langle\; x,\; e\_n\; angle|^2\; =\; |x|^2.$

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 "e"_n = "e"^-i"n"πx 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

:$|x|^2=langle\; x,x\; angle=sum\_\{vin\; B\}left|langle\; x,v\; angle\; ight|^2.$

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 ≥, thus yielding Bessel's inequality. This general form of Parseval's identity can be proved using the Riesz–Fischer theorem.

References

*citation|last1=Johnson|first1=Lee W.|first2=R. Dean|last2=Riess|title=Numerical Analysis|year=1982|edition=2nd|publisher=Addison-Wesley|location=Reading, Mass.|id=ISBN 0-201-10392-3.
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