Editing Fourier inversion theorem (section)

==Statement==

In this section we assume that <math>f</math> is an integrable continuous function. Use the [[Fourier transform#convention|convention for the Fourier transform]] that 

:<math>(\mathcal{F}f)(\xi):=\int_{\mathbb{R}} e^{-2\pi iy\cdot\xi} \, f(y)\,dy.</math>

Furthermore, we assume that the Fourier transform is also integrable.

===Inverse Fourier transform as an integral===

The most common statement of the Fourier inversion theorem is to state the inverse transform as an integral. For any integrable function <math>g</math> and all <math>x \in \mathbb R</math> set

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

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

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

{{collapse top|title=Proof}}
Given <math>f(y)</math> and <math>\mathcal{F}f (\xi) = \int_{\mathbb{R}^n} e^{-2\pi i y\cdot\xi} f(y)\,dy</math>, the proof uses the following facts:

# If <math>x \in \mathbb R^n</math> and <math>g(\xi) = e^{2 \pi \mathrm{i}x \cdot \xi} \psi(\xi)</math>, then <math display="block">(\mathcal{F}g)(y) = (\mathcal{F}\psi)(y - x).</math>
# If <math>\varepsilon \in \mathbb R</math> and <math>\psi(\xi) = \varphi(\varepsilon\xi)</math>, then <math display="block">(\mathcal{F}\psi)(y) = (\mathcal{F}\varphi)(y/\varepsilon)/|\varepsilon|^n.</math>
# For <math>f, g \in L^1(\mathbb R^n)</math>, [[Fubini's theorem]] implies <math display="block">\textstyle\int g(\xi) \cdot (\mathcal{F}f)(\xi)\,d\xi = \int(\mathcal{F}g)(y) \cdot f(y)\,dy.</math>
# Define <math>\varphi(\xi) = e^{-\pi \vert \xi \vert^2}</math> such that <math display="block">(\mathcal{F}\varphi)(y) = \varphi(y).</math>
# Define <math>\varphi_\varepsilon(y) = \varphi(y/\varepsilon)/\varepsilon^n</math>; an [[nascent delta function|approximation to the identity]]. That is, <math display="block">\lim_{\varepsilon \to 0} (\varphi_\varepsilon \ast f)(x) = f(x),</math> converges pointwise for any continuous <math>f \in L^1(\mathbb R^n)</math> and point <math>x \in \mathbb R^n</math>.
Since, by assumption, <math>\mathcal{F}f\in L^1(\mathbb{R}^n)</math>, it follows by the [[dominated convergence theorem]] that
<math display="block">\int_{\mathbb{R}^n} e^{2\pi i x\cdot\xi}(\mathcal{F}f)(\xi)\,d\xi = \lim_{\varepsilon \to 0}\int_{\mathbb{R}^n} e^{-\pi\varepsilon^2|\xi|^2 + 2\pi i x\cdot\xi}(\mathcal{F}f)(\xi)\,d\xi.</math>
Define
<math display="block">g_x(\xi) = e^{-\pi\varepsilon^2\vert \xi \vert^2 + 2 \pi \mathrm{i} x \cdot \xi}.</math>
Applying facts 1, 2 and 4, repeatedly for multiple integrals if necessary, we obtain
<math display="block">(\mathcal{F}g_x)(y) = \frac{1}{\varepsilon^n}e^{-\frac{\pi}{\varepsilon^2}|x - y|^2}=\varphi_\varepsilon(x-y).</math>
Using fact 3 on <math>f</math> and <math>g_x</math>, for each <math>x\in\mathbb R^n</math>, we have
<math display="block">\int_{\mathbb{R}^n} e^{-\pi\varepsilon^2|\xi|^2 + 2\pi i x\cdot\xi}(\mathcal{F}f)(\xi)\,d\xi = \int_{\mathbb{R}^n} \frac{1}{\varepsilon^n}e^{-\frac{\pi}{\varepsilon^2}|x - y|^2} f(y)\,dy = (\varphi_\varepsilon * f)(x),</math>
the [[convolution]] of <math>f</math> with an approximate identity. But since <math>f \in L^1(\mathbb R^n)</math>, fact 5 says that
<math display="block">\lim_{\varepsilon\to 0}(\varphi_{\varepsilon} * f) (x) = f(x).</math>
Putting together the above we have shown that
<math display="block">\int_{\mathbb{R}^n} e^{2\pi i x\cdot\xi}(\mathcal{F}f)(\xi)\,d\xi = f(x). \qquad\square</math>
{{collapse bottom}}
===Fourier integral theorem===

The theorem can be restated as 

:<math>f(x)=\int_{\mathbb{R}} \int_{\mathbb{R}} e^{2\pi i(x-y)\cdot\xi} \, f(y)\,dy\,d\xi.</math>

By taking the real part<ref>[[without loss of generality|w.l.o.g]] {{math|''f''}} is real valued, as any complex-valued function can be split into its real and imaginary parts and every operator appearing here is linear in {{math|''f''}}.</ref> of each side of the above we obtain 

:<math>f(x)=\int_{\mathbb{R}} \int_{\mathbb{R}} \cos (2\pi (x-y)\cdot\xi) \, f(y)\,dy\,d\xi.</math>

===Inverse transform in terms of flip operator===

For any function <math>g</math> define the flip operator<ref>An [[operator (mathematics)|operator]] is a transformation that maps functions to functions. The flip operator, the Fourier transform, the inverse Fourier transform and the identity transform are all examples of operators.</ref> <math>R</math> by

:<math>Rg(x):=g(-x).</math>

Then we may instead define 

:<math>\mathcal{F}^{-1}f := R\mathcal{F}f = \mathcal{F}Rf.</math>

It is immediate from the definition of the Fourier transform and the flip operator that both <math>R\mathcal{F}f</math> and <math>\mathcal{F}Rf</math> match the integral definition of <math>\mathcal{F}^{-1}f</math>, and in particular are equal to each other and satisfy <math>\mathcal{F}^{-1}(\mathcal{F}f)(x)=f(x)</math>.

Since <math>Rf=R\mathcal{F}^{-1}\mathcal{F}f =RR \mathcal{FF}f</math> we have <math>R=\mathcal{F}^2</math> and

:<math>\mathcal{F}^{-1}=\mathcal{F}^3.</math>

===Two-sided inverse===

The form of the Fourier inversion theorem stated above, as is common, is that

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

In other words, <math>\mathcal{F}^{-1}</math> is a left inverse for the Fourier transform. However it is also a right inverse for the Fourier transform i.e.

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

Since <math>\mathcal{F}^{-1}</math> is so similar to <math>\mathcal{F}</math>, this follows very easily from the Fourier inversion theorem (changing variables <math>\zeta := -\xi</math>):

:<math>\begin{align}
f & =\mathcal{F}^{-1}(\mathcal{F}f)(x)\\[6pt]
 & =\int_{\mathbb{R}}\int_{\mathbb{R}}e^{2\pi ix\cdot\xi}\,e^{-2\pi iy\cdot\xi}\, f(y)\, dy\, d\xi\\[6pt]
 & =\int_{\mathbb{R}}\int_{\mathbb{R}}e^{-2\pi ix\cdot\zeta}\,e^{2\pi iy\cdot\zeta}\, f(y)\, dy\, d\zeta\\[6pt]
 & =\mathcal{F}(\mathcal{F}^{-1}f)(x).
\end{align}</math>

Alternatively, this can be seen from the relation between <math>\mathcal{F}^{-1}f</math> and the flip operator and the [[associativity]] of [[function composition]], since

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