Editing Fourier inversion theorem (section)

===Integrable functions in one dimension===

; Piecewise smooth; one dimension
If the function is absolutely integrable in one dimension (i.e. <math> f \in L^1(\mathbb R)</math>) and is piecewise smooth then a version of the Fourier inversion theorem holds. In this case we define

:<math>\mathcal{F}^{-1}g(x):=\lim_{R\to\infty}\int_{-R}^R e^{2\pi ix\xi}\,g(\xi)\,d\xi.</math>

Then for all <math> x \in \mathbb R</math>

:<math>\mathcal{F}^{-1}(\mathcal{F}f)(x) = \frac{1}{2}(f(x_-) + f(x_+)),</math>

i.e. <math>\mathcal{F}^{-1}(\mathcal{F}f)(x)</math> equals the average of the left and right limits of <math> f</math> at <math> x</math>. At points where <math> f</math> is continuous this simply equals <math> f(x)</math>.

A higher-dimensional analogue of this form of the theorem also holds, but according to Folland (1992) is "rather delicate and not terribly useful".

; Piecewise continuous; one dimension

If the function is absolutely integrable in one dimension (i.e. <math> f \in L^1(\mathbb R)</math>) but merely piecewise continuous then a version of the Fourier inversion theorem still holds. In this case the integral in the inverse Fourier transform is defined with the aid of a smooth rather than a sharp cut off function; specifically we define

:<math>\mathcal{F}^{-1}g(x):=\lim_{R\to\infty}\int_{\mathbb{R}} \varphi(\xi/R)\,e^{2\pi ix\xi}\,g(\xi)\,d\xi,\qquad\varphi(\xi):=e^{-\xi^2}.</math>

The conclusion of the theorem is then the same as for the piecewise smooth case discussed above.

; Continuous; any number of dimensions

If <math> f</math> is continuous and absolutely integrable on <math>\mathbb R^n</math> then the Fourier inversion theorem still holds so long as we again define the inverse transform with a smooth cut off function i.e.  

:<math>\mathcal{F}^{-1}g(x):=\lim_{R\to\infty}\int_{\mathbb{R}^n} \varphi(\xi/R)\,e^{2\pi ix\cdot\xi}\,g(\xi)\,d\xi,\qquad\varphi(\xi):=e^{-\vert\xi\vert^2}.</math>

The conclusion is now simply that for all <math>x \in \mathbb R^n</math>

:<math>\mathcal{F}^{-1}(\mathcal{F}f)(x)=f(x).</math>

; No regularity condition; any number of dimensions

If we drop all assumptions about the (piecewise) continuity of <math>f</math> and assume merely that it is absolutely integrable, then a version of the theorem still holds. The inverse transform is again defined with the smooth cut off, but with the conclusion that 

:<math>\mathcal{F}^{-1}(\mathcal{F}f)(x) = f(x)</math> 

for [[almost every]] <math>x \in \mathbb R^n.</math>