Parseval equality PersevalEquality 2003-09-10 10:05:39 2008-09-07 16:10:03 Theorem Parseval theorem % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \subsection{Parseval's Equality} {\bf 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: \begin{displaymath} \|x\|^2 = \sum_{j \in J} | \langle x , e_j \rangle |^2 . \end{displaymath} 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. \subsection{Parseval's Theorem} Applying Parseval's equality on the Hilbert space \PMlinkname{$L^2([-\pi,\pi])$}{LpSpace}, of square integrable functions on the interval $[-\pi,\pi]$, with the orthonormal basis consisting of trigonometric functions, we obtain {\bf 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.