---
tag: A625F93ANV
kind: definition
name: Fourier transform on locally compact abelian groups
msc: [43A25, 42A38]
defines: VHN5VW5Q92
uses: [ZR09S9R3AW, VY0HXD3GC4, FMW57WNPDB, 6GD25S53ND]
sources:
  - url: https://en.wikipedia.org/wiki/Pontryagin_duality
---

## Definition

Let $G$ be a [[locally compact abelian group|ZR09S9R3AW]] with a [[Haar measure|VY0HXD3GC4]] $\mu$, and let $\widehat{G}$ be its [[Pontryagin dual|FMW57WNPDB]].
Let $f \in L^1(G, \mu)$, the [[group algebra|6GD25S53ND]] of $\mu$-integrable functions.
The *Fourier transform* of $f$ is the function $\widehat{f} \colon \widehat{G} \to \C$ given by
$$\widehat{f}(\chi) = \int_G f(x) \overline{\chi(x)} \, d\mu(x).$$

## Remarks

Example: for $G = \R$ with Lebesgue measure, the continuous characters are $\chi_\xi(x) = e^{2\pi i \xi x}$ for $\xi \in \R$, and the formula gives $\widehat{f}(\chi_\xi) = \int_\R f(x) e^{-2\pi i \xi x} \, dx$, the classical Fourier transform.
For $G = \mathbb{T}$ with normalised arc length and $\chi_n(z) = z^n$, it gives the Fourier coefficients of $f$.
Non-example: for $f(x) = 1$ on $\R$, the integral does not converge because $f \notin L^1(\R)$; the transform of such $f$ only makes sense as a distribution.
The integral converges for $f \in L^1$ because $\abs{\chi(x)} = 1$, and $\abs{\widehat{f}(\chi)} \le \norm{f}_1$.
The function $\widehat{f}$ is continuous on $\widehat{G}$ and vanishes at infinity, and $\widehat{f * g} = \widehat{f}\,\widehat{g}$ for the convolution $f * g$.
Some authors use $\chi(x)$ instead of $\overline{\chi(x)}$; this changes $\widehat{f}(\chi)$ into $\widehat{f}(\overline{\chi})$.