World Library  
Flag as Inappropriate
Email this Article

Fourier inversion formula

Article Id: WHEBN0007193615
Reproduction Date:

Title: Fourier inversion formula  
Author: World Heritage Encyclopedia
Language: English
Subject: Fourier transform, Radon transform, Stone–von Neumann theorem
Collection:
Publisher: World Heritage Encyclopedia
Publication
Date:
 

Fourier inversion formula

In mathematics, the Fourier inversion theorem says that for many types of function it is possible to recover a function from its Fourier transform. Intuitively it may be viewed as the statement that if we know all frequency and phase information about a wave then we may reconstruct the original wave precisely.

The theorem says that if we have a function f: ℝn→ℂ satisfying certain conditions, and we use the convention for the Fourier transform that

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

then

f(x)=\int_{\mathbb{R}^n} e^{2\pi ix\cdot\xi} \, (\mathcal{F}f)(\xi)\,d\xi.

In other words, the theorem says that

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

This last equation is called the Fourier integral theorem.

Another way to state the theorem is note that, if R is the flip operator i.e. Rf(x):=f(−x), then

\mathcal{F}^{-1}=\mathcal{F}R=R\mathcal{F}.

The theorem holds if both f and its Fourier transform are absolutely integrable (in the Lebesgue sense) and f is continuous at the point x. However, even under more general conditions versions of the Fourier inversion theorem hold. In these cases the integrals above may not make sense, or the theorem may hold for almost all x rather than for all x.

Statement

In this section we assume that f is an integrable continuous function. Use the convention for the Fourier transform that

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

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 g and all x∈ℝn set

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

Then for all x∈ℝn we have

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

Fourier integral theorem

The theorem can be restated as

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

If f is real valued then by taking the real part of each side of the above we obtain

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

Inverse transform in terms of flip operator

For any function g define the flip operator[note 1] R by

Rg(x):=g(-x).

Then we may instead define

\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 R\mathcal{F}f and \mathcal{F}Rf match the integral definition of \mathcal{F}^{-1}f, and in particular are equal to each other and satisfy \mathcal{F}^{-1}(\mathcal{F}f)(x)=f(x).

Two sided inverse

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

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

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

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

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

\begin{align}

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

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

\end{align}

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

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

Conditions on the function

When used in physics and engineering, the Fourier inversion theorem is often used under the assumption that everything "behaves nicely". In mathematics such heuristic arguments are not permitted, and the Fourier inversion theorem includes an explicit specification of what class of functions is being allowed. However, there is no "best" class of functions to consider so several variants of the Fourier inversion theorem exist, albeit with compatible conclusions.

Schwartz functions

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).

Integrable functions with integrable Fourier transform

The Fourier inversion theorem holds for all continuous functions that absolutely integrable (i.e. L1(ℝn)) with absolutely integrable Fourier transform. This includes all Schwartz functions, so is a strictly stronger form of the theorem than the previous one mentioned. These conditions have the benefit that the integrals that define the Fourier transform and its inverse are absolutely integrable. This condition is the one used above in the statement section.

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

Integrable functions in one dimension

Piecewise smooth; one dimension

If the function is absolutely integrable in one dimension (i.e. fL1(ℝ)) and is piecewise smooth then a version of the Fourier inversion theorem holds. In this case we define

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

Then for all x∈ℝ

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

i.e. \mathcal{F}^{-1}(\mathcal{F}f)(x) equals the average of the left and right limits of f at x. Note that at points where f is continuous this simply equals 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. fL1(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

\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 f is continuous and absolutely integrable on 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.

\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 x∈ℝn

\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 f and assume merely that it is absolutely integrable, then a version of the theorem still holds. The inverse transform is again be define the smooth cut off, but with the conclusion that

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

for almost every x∈ℝ.

Square integrable functions

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

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 \textstyle\mathcal{F}f := \lim_{k\to\infty}g_k where the limit is taken in the L2 norm. The inverse transform may be defined by density in the same way or by defining it terms of the Fourier transform and the flip operator. We then have

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

for almost every x∈ℝ.

Tempered distributions

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

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

where \mathcal{F}\varphi is defined using the integral formula. If f is a function in L1+L2 then this agrees with the usual definition. We may define the inverse transform \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

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

Relation to Fourier series

When considering the Fourier series of a function it is conventional to rescale it so that it acts on [0,2π] (or is 2π periodic). In this section we instead use the somewhat unusual convention taking f to act on [0,1], since that matches the convention of the Fourier transform used here.

The Fourier inversion theorem is analogous to the convergence of Fourier series. In the Fourier transform case we have

f\colon\mathbb{R}^n\to\mathbb{C},\quad\hat f\colon\mathbb{R}^n\to\mathbb{C},
\hat f(\xi):=\int_{\mathbb{R}^n} e^{-2\pi iy\cdot\xi} \, f(y)\,dy,
f(x)=\int_{\mathbb{R}^n} e^{2\pi ix\cdot\xi} \, \hat f(\xi)\,d\xi.

In the Fourier series case we instead have

f\colon[0,1]^n\to\mathbb{C},\quad\hat f\colon\mathbb{Z}^n\to\mathbb{C},
\hat f(k):=\int_{[0,1]^n} e^{-2\pi iy\cdot k} \, f(y)\,dy,
f(x)=\sum_{k\in\mathbb{Z}^n} e^{2\pi ix\cdot k} \, \hat f(k).

In particular, in one dimension k is simply an integer and the sum runs from −∞ to .

Applications


In applications of the Fourier transforrm the Fourier inversion theorem often plays a critical role. In many situations the basic strategy is to apply the Fourier transform, perform some operation or simplification, and then apply the inverse Fourier transform.

More abstractly, the Fourier inversion theorem is a statement about the Fourier transform as an operator (see Fourier transform on function spaces). For example, the Fourier inversion theorem on fL2(ℝn) shows that the Fourier transform is a unitary operator on fL2(ℝn).

Properties of inverse transform

The inverse Fourier transform is extremely similar to the original Fourier transform: as discussed above, it differs only in the application of a flip operator. For this reason the properties of the Fourier transform hold for the inverse Fourier transform, such as the Convolution theorem and the Riemann–Lebesgue lemma.

Tables of Fourier transforms may easily be used for the inverse Fourier transform by composing the looked-up function with the flip operator. For example, looking up the Fourier transform of the rect function we see that

f(x)=\operatorname{rect}(a x) \quad \Rightarrow \quad (\mathcal{F}f)(\xi)=\frac{1}{|a|} \operatorname{sinc}\left(\frac{\xi}{a}\right)\!,

so the corresponding fact for the inverse transform is

g(\xi)=\operatorname{rect}(a \xi) \quad \Rightarrow \quad (\mathcal{F}^{-1}g)(x)=\frac{1}{|a|} \operatorname{sinc}\left(-\frac{x}{a}\right)\!.

Proof

The proof uses a few facts.

  1. If η∈ℝn and g(x) = e2πixηf(x), then (\mathcal{F}g)(\xi) = (\mathcal{F}f)(\xi - \eta).
  2. If a∈ℝ and g(x) = f(ax), then (\mathcal{F}g)(\xi) = (\mathcal{F}f)(\xi/a)/a^n.
  3. For f and g in L1(ℝn), Fubini's theorem implies that \textstyle\int f (\mathcal{F}g) = \int(\mathcal{F}f) g.
  4. Define φ(x):=eπ|x|2; then \mathcal{F}\varphi = \varphi.
  5. Define φε(x):=φ(x/ε)/εn. Then with denoting convolution, φε is an approximation to the identity: for any continuous fL1(ℝn) and point x∈ℝn, limε→0φεf(x)=f(x) (where the convergence is pointwise).

First note that, since, by assumption, \mathcal{F}f\in L^1(\mathbb{R}^n), then it follows by the dominated convergence theorem that

\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.

Define g(ξ)=eπε2|ξ|2+2πixξ. Applying facts 1, 2 and 4 from we obtain

(\mathcal{F}g)(y) = \frac{1}{\varepsilon^n}e^{-\frac{\pi}{\varepsilon^2}|x - y|^2}.

Using fact 3 from above on f and g we thus have

\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 = (\phi_{\varepsilon} * f)(x),

the convolution of f with an approximate identity. But since fL1(ℝn) fact 5 says that

\lim_{\varepsilon\to 0}\phi_{\varepsilon} * f (x) = f(x).

Putting together the above we have shown that

\int_{\mathbb{R}^n} e^{2\pi i x\cdot\xi}(\mathcal{F}f)(\xi)\,d\xi = f(x). \qquad\square

Notes

References

This article was sourced from Creative Commons Attribution-ShareAlike License; additional terms may apply. World Heritage Encyclopedia content is assembled from numerous content providers, Open Access Publishing, and in compliance with The Fair Access to Science and Technology Research Act (FASTR), Wikimedia Foundation, Inc., Public Library of Science, The Encyclopedia of Life, Open Book Publishers (OBP), PubMed, U.S. National Library of Medicine, National Center for Biotechnology Information, U.S. National Library of Medicine, National Institutes of Health (NIH), U.S. Department of Health & Human Services, and USA.gov, which sources content from all federal, state, local, tribal, and territorial government publication portals (.gov, .mil, .edu). Funding for USA.gov and content contributors is made possible from the U.S. Congress, E-Government Act of 2002.
 
Crowd sourced content that is contributed to World Heritage Encyclopedia is peer reviewed and edited by our editorial staff to ensure quality scholarly research articles.
 
By using this site, you agree to the Terms of Use and Privacy Policy. World Heritage Encyclopedia™ is a registered trademark of the World Public Library Association, a non-profit organization.
 


Copyright © World Library Foundation. All rights reserved. eBooks from Project Gutenberg are sponsored by the World Library Foundation,
a 501c(4) Member's Support Non-Profit Organization, and is NOT affiliated with any governmental agency or department.