Parseval 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 a generalized Pythagorean theorem for inner-product spaces (which can have an uncountable infinity of basis vectors).

Informally, 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,
$\Vert f \Vert^2_= \int_^\pi , f(x), ^2 \, dx=2\pi\sum_^\infty , c_n, ^2$
where the Fourier coefficients $c_n$ of $f$ are given by
$c_n = \frac \int_^ f(x) e^ \, dx.$

More formally, the result holds as stated provided $f$ is a square-integrable function or, more generally, in Lp space 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 $f \in L^2(\R),$
$\int_^\infty , \hat(\xi), ^2\,d\xi = \int_^\infty , f(x), ^2\, dx.$

Another similar result is the Hesham identity which gives the integral of the fourth power of the function f \in L^4 \pi, \pi/math> in terms of its Fourier coefficients given $f$ has a finite-length discrete Fourier transform with $M$ number of coefficients $c \in \C$.
\Vert f \Vert^4_= \int_^\pi , f(x), ^4 \, dx=2\pi\sum_^ c_k \sum_^ c_l^* \Bigg \underset c_m^* c_ + \underset c_^* c_m \Bigg/math>
if $c \in \R$ the identity is simplified to
$\Vert f \Vert^4_= \int_^\pi , f(x), ^4 \, dx=2\pi\sum_^ c_k \sum_^ c_l \sum_^ c_m c_$

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 $\langle \,\cdot\,, \,\cdot\, \rangle.$ Let $\left(e_n\right)$ 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:
:$\langle e_m, e_n\rangle = \begin 1 & \mbox~ m = n \\ 0 & \mbox~ m \neq n. \end$

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

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 L^2 \pi, \pi and setting $e_n = e^$ for $n \in \Z.$

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 $H$; that is, an orthonormal set which is in the sense that the linear span of $B$ is dense in $H.$ Then
$\, x\, ^2 = \langle x,x\rangle = \sum_\left, \langle x, v\rangle\^2.$

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

See also

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References

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{{Hilbert space

Fourier series
Theorems in functional analysis