Fourier inversion theorem

The most common statement of the Fourier inversion theorem is to state the inverse transform as an integral. For any integrable function ${\displaystyle g}$ and all ${\displaystyle x\in \mathbb {R} ^{n}}$ set

${\displaystyle {\mathcal {F}}^{-1}g(x):=\int _{\mathbb {R} ^{n}}e^{2\pi ix\cdot \xi }\,g(\xi )\,d\xi .}$

Then for all ${\displaystyle x\in \mathbb {R} ^{n}}$ we have

${\displaystyle {\mathcal {F}}^{-1}({\mathcal {F}}f)(x)=f(x).}$

The theorem can be restated as

${\displaystyle f(x)=\int _{\mathbb {R} ^{n}}\int _{\mathbb {R} ^{n}}e^{2\pi i(x-y)\cdot \xi }\,f(y)\,dy\,d\xi .}$

By taking the real part[1] of each side of the above we obtain

${\displaystyle f(x)=\int _{\mathbb {R} ^{n}}\int _{\mathbb {R} ^{n}}\cos(2\pi (x-y)\cdot \xi )\,f(y)\,dy\,d\xi .}$

For any function ${\displaystyle g}$ define the flip operator[note 1] ${\displaystyle R}$ by

${\displaystyle Rg(x):=g(-x).}$

Then we may instead define

${\displaystyle {\mathcal {F}}^{-1}f:=R{\mathcal {F}}f={\mathcal {F}}Rf.}$

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

Since ${\displaystyle Rf=R{\mathcal {F}}^{-1}{\mathcal {F}}f=RR{\mathcal {FF}}f}$ we have ${\displaystyle R={\mathcal {F}}^{2}}$ and

${\displaystyle {\mathcal {F}}^{-1}={\mathcal {F}}^{3}.}$

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

${\displaystyle {\mathcal {F}}^{-1}({\mathcal {F}}f)(x)=f(x).}$

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

${\displaystyle {\mathcal {F}}({\mathcal {F}}^{-1}f)(\xi )=f(\xi ).}$

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

${\displaystyle {\begin{aligned}f&={\mathcal {F}}^{-1}({\mathcal {F}}f)(x)\\[6pt]&=\int _{\mathbb {R} ^{n}}\int _{\mathbb {R} ^{n}}e^{2\pi ix\cdot \xi }\,e^{-2\pi iy\cdot \xi }\,f(y)\,dy\,d\xi \\[6pt]&=\int _{\mathbb {R} ^{n}}\int _{\mathbb {R} ^{n}}e^{-2\pi ix\cdot \zeta }\,e^{2\pi iy\cdot \zeta }\,f(y)\,dy\,d\zeta \\[6pt]&={\mathcal {F}}({\mathcal {F}}^{-1}f)(x).\end{aligned}}}$

Alternatively, this can be seen from the relation between ${\displaystyle {\mathcal {F}}^{-1}f}$ and the flip operator and the associativity of function composition, since

${\displaystyle f={\mathcal {F}}^{-1}({\mathcal {F}}f)={\mathcal {F}}R{\mathcal {F}}f={\mathcal {F}}({\mathcal {F}}^{-1}f).}$

The Fourier inversion theorem holds for all Schwartz functions (roughly speaking, smooth functions that decay quickly and whose derivatives all decay quickly). This condition has the benefit that it is an elementary direct statement about the function (as opposed to imposing a condition on its Fourier transform), and the integral that defines the Fourier transform and its inverse are absolutely integrable. This version of the theorem is used in the proof of the Fourier inversion theorem for tempered distributions (see below).

The Fourier inversion theorem holds for all continuous functions that are absolutely integrable (i.e. ${\displaystyle L^{1}(\mathbb {R} ^{n})}$) with absolutely integrable Fourier transform. This includes all Schwartz functions, so is a strictly stronger form of the theorem than the previous one mentioned. This condition is the one used above in the statement section.

A slight variant is to drop the condition that the function ${\displaystyle f}$ be continuous but still require that it and its Fourier transform be absolutely integrable. Then ${\displaystyle f=g}$ almost everywhere where g is a continuous function, and ${\displaystyle {\mathcal {F}}^{-1}({\mathcal {F}}f)(x)=g(x)}$ for every ${\displaystyle x\in \mathbb {R} ^{n}}$.

Piecewise smooth; one dimension

If the function is absolutely integrable in one dimension (i.e. ${\displaystyle f\in L^{1}(\mathbb {R} )}$) and is piecewise smooth then a version of the Fourier inversion theorem holds. In this case we define

${\displaystyle {\mathcal {F}}^{-1}g(x):=\lim _{R\to \infty }\int _{-R}^{R}e^{2\pi ix\xi }\,g(\xi )\,d\xi .}$

Then for all ${\displaystyle x\in \mathbb {R} }$

${\displaystyle {\mathcal {F}}^{-1}({\mathcal {F}}f)(x)={\frac {1}{2}}(f(x_{-})+f(x_{+})),}$

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

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. ${\displaystyle f\in L^{1}(\mathbb {R} )}$) 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

${\displaystyle {\mathcal {F}}^{-1}g(x):=\lim _{R\to \infty }\int _{\mathbb {R} }\varphi (\xi /R)\,e^{2\pi ix\xi }\,g(\xi )\,d\xi ,\qquad \varphi (\xi ):=e^{-\xi ^{2}}.}$

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

Continuous; any number of dimensions

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

${\displaystyle {\mathcal {F}}^{-1}g(x):=\lim _{R\to \infty }\int _{\mathbb {R} ^{n}}\varphi (\xi /R)\,e^{2\pi ix\cdot \xi }\,g(\xi )\,d\xi ,\qquad \varphi (\xi ):=e^{-\vert \xi \vert ^{2}}.}$

The conclusion is now simply that for all ${\displaystyle x\in \mathbb {R} ^{n}}$

${\displaystyle {\mathcal {F}}^{-1}({\mathcal {F}}f)(x)=f(x).}$

No regularity condition; any number of dimensions

If we drop all assumptions about the (piecewise) continuity of ${\displaystyle f}$ 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

${\displaystyle {\mathcal {F}}^{-1}({\mathcal {F}}f)(x)=f(x)}$

for almost every ${\displaystyle x\in \mathbb {R} ^{n}.}$ [2]

In this case the Fourier transform cannot be defined directly as an integral since it may not be absolutely convergent, so it is instead defined by a density argument (see the Fourier transform article). For example, putting

${\displaystyle g_{k}(\xi ):=\int _{\{y\in \mathbb {R} ^{n}:\left\vert y\right\vert \leq k\}}e^{-2\pi iy\cdot \xi }\,f(y)\,dy,\qquad k\in \mathbb {N} ,}$

we can set ${\displaystyle \textstyle {\mathcal {F}}f:=\lim _{k\to \infty }g_{k}}$ where the limit is taken in the ${\displaystyle L^{2}}$-norm. The inverse transform may be defined by density in the same way or by defining it in terms of the Fourier transform and the flip operator. We then have

${\displaystyle f(x)={\mathcal {F}}({\mathcal {F}}^{-1}f)(x)={\mathcal {F}}^{-1}({\mathcal {F}}f)(x)}$

in the mean squared norm. In one dimension (and one dimension only), it can also be shown that it converges for almost every x∈ℝ- this is Carleson's theorem, but is much harder to prove than convergence in the mean squared norm.

The Fourier transform may be defined on the space of tempered distributions ${\displaystyle {\mathcal {S}}'(\mathbb {R} ^{n})}$ by duality of the Fourier transform on the space of Schwartz functions. Specifically for ${\displaystyle f\in {\mathcal {S}}'(\mathbb {R} ^{n})}$ and for all test functions ${\displaystyle \varphi \in {\mathcal {S}}(\mathbb {R} ^{n})}$ we set

${\displaystyle \langle {\mathcal {F}}f,\varphi \rangle$ :=\langle f,{\mathcal {F}}\varphi \rangle ,}

where ${\displaystyle {\mathcal {F}}\varphi }$ is defined using the integral formula. If ${\displaystyle f\in L^{1}(\mathbb {R} ^{n})\cap L^{2}(\mathbb {R} ^{n})}$ then this agrees with the usual definition. We may define the inverse transform ${\displaystyle {\mathcal {F}}^{-1}\colon {\mathcal {S}}'(\mathbb {R} ^{n})\to {\mathcal {S}}'(\mathbb {R} ^{n})}$, either by duality from the inverse transform on Schwartz functions in the same way, or by defining it in terms of the flip operator (where the flip operator is defined by duality). We then have

${\displaystyle {\mathcal {F}}{\mathcal {F}}^{-1}={\mathcal {F}}^{-1}{\mathcal {F}}=\operatorname {Id} _{{\mathcal {S}}'(\mathbb {R} ^{n})}.}$