# 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 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,

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

More formally, 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 ${\displaystyle f\in L^{2}(\mathbb {R} ),}$

${\displaystyle \int _{-\infty }^{\infty }|{\hat {f}}(\xi )|^{2}\,d\xi =\int _{-\infty }^{\infty }|f(x)|^{2}\,dx.}$

## 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 ${\displaystyle e_{n}=e^{-inx}}$ for ${\displaystyle n\in \mathbb {Z} .}$

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.