Parseval's Equality

Theorem. - If $\{e_j\!:\; j \in J \}$ is an orthonormal basis of an Hilbert space $H$ , then for every $x \in H$ the following equality holds:

The above theorem is a more sophisticated form of Bessel's inequality (where the inequality is in fact an equality). The difference is that for Bessel's inequality it is only required that the set $\{e_j : j \in J \}$ is an orthonormal set, not necessarily an orthonormal basis.

Parseval's Theorem

Applying Parseval's equality on the Hilbert space $L^2([-\pi,\pi])$ , of square integrable functions on the interval $[-\pi,\pi]$ , with the orthonormal basis consisting of trigonometric functions, we obtain

Theorem (Parseval's theorem). - Let $f$ be a Riemann square integrable function from $[-\pi,\pi]$ to $\mathbb{R}$ . The following equality holds $$\frac{1}{\pi}\int_{-\pi}^{\pi}f^2(x)dx = \frac{(a_0^f)^2}{2} + \sum_{k=1}^{\infty}[(a_k^f)^2+(b_k^f)^2],$$ where $a_0^f$ , $a_k^f$ , $b_k^f$ are the Fourier coefficients of the function $f$ .

The function $f$ can be a Lebesgue-integrable function, if we use the Lebesgue integral in place of the Riemann integral.