Editing Orthogonal functions (section)

{{Short description|Type of function}}
In [[mathematics]], '''orthogonal functions''' belong to a [[function space]] that is a [[vector space]] equipped with a [[bilinear form]]. When the function space has an [[interval (mathematics)|interval]] as the [[domain of a function|domain]], the bilinear form may be the [[integral]] of the product of functions over the interval:
:<math> \langle f,g\rangle = \int \overline{f(x)}g(x)\,dx  .</math>

The functions <math>f</math> and <math>g</math> are [[Orthogonality_(mathematics)|orthogonal]] when this integral is zero, i.e. <math>\langle f, \, g \rangle = 0</math> whenever <math>f \neq g</math>. As with a [[basis (linear algebra)|basis]] of vectors in a finite-dimensional space, orthogonal functions can form an infinite basis for a function space.  Conceptually, the above integral is the equivalent of a vector [[dot product]]; two vectors are mutually independent (orthogonal) if their dot-product is zero.

Suppose <math> \{ f_0, f_1, \ldots\}</math> is a sequence of orthogonal functions of nonzero [[L2-norm|''L''<sup>2</sup>-norm]]s <math display="inline"> \left\| f_n \right\| _2 = \sqrt{\langle f_n, f_n \rangle} = \left(\int f_n ^2 \ dx \right) ^\frac{1}{2} </math>.  It follows that the sequence <math>\left\{ f_n / \left\| f_n \right\| _2 \right\}</math> is of functions of ''L''<sup>2</sup>-norm one, forming an [[orthonormal sequence]].  To have a defined ''L''<sup>2</sup>-norm, the integral must be bounded, which restricts the functions to being [[square-integrable function|square-integrable]].