PlanetPhysics/Generalized Fourier and Measured Groupoid Transforms

\newcommand{\sqdiagram}[9]{Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikiversity.org/v1/":): {\displaystyle \diagram #1 \rto^{#2} \dto_{#4}& \eqno{\mbox{#9}}} }

Generalized Fourier transforms

Fourier-Stieltjes transforms and measured groupoid transforms are useful generalizations of the (much simpler) Fourier transform, as concisely shown in the following table (see also [TableOfFourierTransforms Fourier transforms] ) - for the purpose of direct comparison with the latter transform. Unlike the more general Fourier-Stieltjes transform, the Fourier transform exists if and only if the function to be transformed is Lebesgue integrable over the whole real axis for $t\in {\mathbb {R} }$, or over the entire ${\mathbb {C} }$ domain when ${\check {m}}(t)$ is a complex function.

Fourier-Stieltjes transform .

Given a positive definite, measurable function $f(x)$ on the interval $(-\infty ,\infty )$ there exists a monotone increasing, real-valued bounded function $\alpha (t)$ such that:

$f(x)=\int _{\mathbb {R} }e^{itx}d(\alpha (t),$

for all $x\in {\mathbb {R} }$ except a small set. When $f(x)$ is defined as above and if $\alpha (t)$ is nondecreasing and bounded then the measurable function defined by the above integral is called the Fourier-Stieltjes transform of $\alpha (t)$, and it is continuous in addition to being positive definite.

\subsubsection*{FT and FT-Generalizations}

$f(t)$	${\mathcal {F}}{f(t)}={\hat {f}}(x)=(2\pi )^{-1}\int {e{(-itx)}dx}$	Conditions*	Explanation	Description
$e^{-t}\theta (t)$	${\mathcal {F}}{[f(t)]}(x)=(2\pi )^{-1}\int {\theta (t)e{(it^{2}x)}dx}$	from $-\infty$ to +$\infty$	From $Mathematica^{TM**}$
$c$	$({\sqrt {2\pi }})^{-1}c$
		Notice on the next line the overline	bar (${\overline {}}$) placed above $t(x)$
$f(t)$	$\int {\hat {f}}(x){\overline {t(x)}}dx$	$f(t)\in {L^{1}(G_{l})}$, with $G_{l}$ a	Fourier-Stieltjes transform	${\hat {f}}(x)\in {C_{0}({\hat {G_{l}}})}$
		locally compact groupoid ;
		$\int$ is defined via
		a left Haar measure on $G_{l}$
${\hat {m}}(x)$	${\check {m}}(t)=\int e^{itx}d{\hat {m}}(x)$	as above	Inverse Fourier-Stieltjes	${\check {m}}(t)\in {L^{1}(G_{l})}$,
			transform	(, ).
${\hat {m}}(x)$	${\check {m}}(t)=\int e^{itx}d{\hat {m}}(x)$	When $G_{l}=\mathbb {R}$, and it exists	This is the usual	${\check {m}}(t)\in {\mathbb {R} }$
		only when ${\hat {m}}(x)$ is	Inverse Fourier transform
		Lebesgue integrable on
		the entire real axis

• Note the 'slash hat' on ${\hat {f}}(x)$ and ${\hat {G_{l}}}$;
  • Calculated numerically using this link to $Mathematica^{TM}$