Fourier Inversion
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mathematics Mathematics is an area of knowledge that includes the topics of numbers, formulas and related structures, shapes and the spaces in which they are contained, and quantities and their changes. These topics are represented in modern mathematics ...
, the Fourier inversion theorem says that for many types of functions it is possible to recover a function from its
Fourier transform A Fourier transform (FT) is a mathematical transform that decomposes functions into frequency components, which are represented by the output of the transform as a function of frequency. Most commonly functions of time or space are transformed, ...
. Intuitively it may be viewed as the statement that if we know all
frequency Frequency is the number of occurrences of a repeating event per unit of time. It is also occasionally referred to as ''temporal frequency'' for clarity, and is distinct from ''angular frequency''. Frequency is measured in hertz (Hz) which is eq ...
and
phase Phase or phases may refer to: Science *State of matter, or phase, one of the distinct forms in which matter can exist *Phase (matter), a region of space throughout which all physical properties are essentially uniform * Phase space, a mathematic ...
information about a wave then we may reconstruct the original wave precisely. The theorem says that if we have a function f:\R \to \Complex satisfying certain conditions, and we use the convention for the Fourier transform that :(\mathcalf)(\xi):=\int_ e^ \, f(y)\,dy, then :f(x)=\int_ e^ \, (\mathcalf)(\xi)\,d\xi. In other words, the theorem says that :f(x)=\iint_ e^ \, f(y)\,dy\,d\xi. This last equation is called the Fourier integral theorem. Another way to state the theorem is that if R is the flip operator i.e. (Rf)(x) := f(-x), then :\mathcal^=\mathcalR=R\mathcal. The theorem holds if both f and its Fourier transform are
absolutely integrable In mathematics, an absolutely integrable function is a function whose absolute value is integrable, meaning that the integral of the absolute value over the whole domain is finite. For a real-valued function, since \int , f(x), \, dx = \int f^+ ...
(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 converge in an ordinary sense.


Statement

In this section we assume that f is an integrable continuous function. Use the convention for the Fourier transform that :(\mathcalf)(\xi):=\int_ e^ \, 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 \in \mathbb R^n set :\mathcal^g(x):=\int_ e^ \, g(\xi)\,d\xi. Then for all x \in \mathbb R^n we have :\mathcal^(\mathcalf)(x)=f(x).


Fourier integral theorem

The theorem can be restated as :f(x)=\int_ \int_ e^ \, f(y)\,dy\,d\xi. If is real valued then by taking the real part of each side of the above we obtain :f(x)=\int_ \int_ \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 operatorAn 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. R by :Rg(x):=g(-x). Then we may instead define :\mathcal^f := R\mathcalf = \mathcalRf. It is immediate from the definition of the Fourier transform and the flip operator that both R\mathcalf and \mathcalRf match the integral definition of \mathcal^f, and in particular are equal to each other and satisfy \mathcal^(\mathcalf)(x)=f(x). Since Rf=R\mathcal^\mathcalf =RR \mathcalf we have R=\mathcal^2 and :\mathcal^=\mathcal^3.


Two-sided inverse

The form of the Fourier inversion theorem stated above, as is common, is that :\mathcal^(\mathcalf)(x) = f(x). In other words, \mathcal^ is a left inverse for the Fourier transform. However it is also a right inverse for the Fourier transform i.e. :\mathcal(\mathcal^f)(\xi) = f(\xi). Since \mathcal^ is so similar to \mathcal, this follows very easily from the Fourier inversion theorem (changing variables \zeta := -\zeta): :\begin f & =\mathcal^(\mathcalf)(x)\\ pt & =\int_\int_e^\,e^\, f(y)\, dy\, d\xi\\ pt & =\int_\int_e^\,e^\, f(y)\, dy\, d\zeta\\ pt & =\mathcal(\mathcal^f)(x). \end Alternatively, this can be seen from the relation between \mathcal^f and the flip operator and the
associativity In mathematics, the associative property is a property of some binary operations, which means that rearranging the parentheses in an expression will not change the result. In propositional logic, associativity is a valid rule of replacement f ...
of
function composition In mathematics, function composition is an operation that takes two functions and , and produces a function such that . In this operation, the function is applied to the result of applying the function to . That is, the functions and ...
, since :f = \mathcal^(\mathcalf) = \mathcalR\mathcalf = \mathcal (\mathcal^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 function In mathematics, Schwartz space \mathcal is the function space of all functions whose derivatives are rapidly decreasing. This space has the important property that the Fourier transform is an automorphism on this space. This property enables on ...
s (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 are absolutely integrable (i.e. 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 f be continuous but still require that it and its Fourier transform be absolutely integrable. Then f = g
almost everywhere In measure theory (a branch of mathematical analysis), a property holds almost everywhere if, in a technical sense, the set for which the property holds takes up nearly all possibilities. The notion of "almost everywhere" is a companion notion to ...
where is a continuous function, and \mathcal^(\mathcalf)(x)=g(x) for every x \in \mathbb R^n.


Integrable functions in one dimension

; Piecewise smooth; one dimension If the function is absolutely integrable in one dimension (i.e. f \in L^1(\mathbb R)) and is piecewise smooth then a version of the Fourier inversion theorem holds. In this case we define :\mathcal^g(x):=\lim_\int_^R e^\,g(\xi)\,d\xi. Then for all x \in \mathbb R :\mathcal^(\mathcalf)(x) = \frac(f(x_-) + f(x_+)), i.e. \mathcal^(\mathcalf)(x) equals the average of the left and right limits of f at x. 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. 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 :\mathcal^g(x):=\lim_\int_ \varphi(\xi/R)\,e^\,g(\xi)\,d\xi,\qquad\varphi(\xi):=e^. 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 \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. :\mathcal^g(x):=\lim_\int_ \varphi(\xi/R)\,e^\,g(\xi)\,d\xi,\qquad\varphi(\xi):=e^. The conclusion is now simply that for all x \in \mathbb R^n :\mathcal^(\mathcalf)(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 defined with the smooth cut off, but with the conclusion that :\mathcal^(\mathcalf)(x) = f(x) for
almost every In measure theory (a branch of mathematical analysis), a property holds almost everywhere if, in a technical sense, the set for which the property holds takes up nearly all possibilities. The notion of "almost everywhere" is a companion notion t ...
x \in \mathbb R^n.


Square integrable functions

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 :g_k(\xi):=\int_ e^ \, f(y)\,dy,\qquad k\in\mathbb, we can set \textstyle\mathcalf := \lim_g_k where the limit is taken in the 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 :f(x)=\mathcal(\mathcal^f)(x)=\mathcal^(\mathcalf)(x) in the mean squared norm. In one dimension (and one dimension only), it can also be shown that it converges for
almost every In measure theory (a branch of mathematical analysis), a property holds almost everywhere if, in a technical sense, the set for which the property holds takes up nearly all possibilities. The notion of "almost everywhere" is a companion notion t ...
- this is
Carleson's theorem Carleson's theorem is a fundamental result in mathematical analysis establishing the pointwise (Lebesgue) almost everywhere convergence of Fourier series of functions, proved by . The name is also often used to refer to the extension of the res ...
, but is much harder to prove than convergence in the mean squared norm.


Tempered distributions

The Fourier transform may be defined on the space of tempered distributions \mathcal'(\mathbb^n) by duality of the Fourier transform on the space of Schwartz functions. Specifically for f\in\mathcal'(\mathbb^n) and for all test functions \varphi\in\mathcal S(\mathbb^n) we set :\langle \mathcalf,\varphi\rangle := \langle f,\mathcal\varphi\rangle, where \mathcal\varphi is defined using the integral formula. If 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 \mathcal^\colon\mathcal'(\mathbb^n)\to\mathcal'(\mathbb^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\mathcal^ = \mathcal^\mathcal = \operatorname_.


Relation to Fourier series

The Fourier inversion theorem is analogous to the
convergence of Fourier series In mathematics, the question of whether the Fourier series of a periodic function converges to a given function is researched by a field known as classical harmonic analysis, a branch of pure mathematics. Convergence is not necessarily given in th ...
. In the Fourier transform case we have :f\colon\mathbb^n\to\mathbb,\quad\hat f\colon\mathbb^n\to\mathbb, :\hat f(\xi):=\int_ e^ \, f(y)\,dy, :f(x)=\int_ e^ \, \hat f(\xi)\,d\xi. In the Fourier series case we instead have :f\colon ,1n\to\mathbb,\quad\hat f\colon\mathbb^n\to\mathbb, :\hat f(k):=\int_ e^ \, f(y)\,dy, :f(x)=\sum_ e^ \, \hat f(k). In particular, in one dimension k \in \mathbb Z and the sum runs from - \infty to \infty.


Applications

In applications of the Fourier transform 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 f \in L^2(\mathbb R^n) shows that the Fourier transform is a unitary operator on L^2(\mathbb R^n).


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 In mathematics, the convolution theorem states that under suitable conditions the Fourier transform of a convolution of two functions (or signals) is the pointwise product of their Fourier transforms. More generally, convolution in one domain (e.g. ...
and the
Riemann–Lebesgue lemma In mathematics, the Riemann–Lebesgue lemma, named after Bernhard Riemann and Henri Lebesgue, states that the Fourier transform or Laplace transform of an ''L''1 function vanishes at infinity. It is of importance in harmonic analysis and asymptot ...
. 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(a x) \quad \Rightarrow \quad (\mathcalf)(\xi)=\frac \operatorname\left(\frac\right), so the corresponding fact for the inverse transform is g(\xi)=\operatorname(a \xi) \quad \Rightarrow \quad (\mathcal^g)(x)=\frac \operatorname\left(-\frac\right) .


Proof

The proof uses a few facts, given f(y) and \mathcalf (\xi) = \int_ e^ f(y)\,dy. # If x \in \mathbb R^n and g(\xi) = e^ \psi(\xi), then (\mathcalg)(y) = (\mathcal\psi)(y - x). # If \varepsilon \in \mathbb R and \psi(\xi) = \varphi(\varepsilon\xi), then (\mathcal\psi)(y) = (\mathcal\varphi)(y/\varepsilon)/, \varepsilon, . # For f, g \in L^1(\mathbb R^n),
Fubini's theorem In mathematical analysis Fubini's theorem is a result that gives conditions under which it is possible to compute a double integral by using an iterated integral, introduced by Guido Fubini in 1907. One may switch the order of integration if th ...
implies that \textstyle\int g(\xi) \cdot (\mathcalf)(\xi)\,d\xi = \int(\mathcalg)(y) \cdot f(y)\,dy. # Define \varphi(\xi) = e^; then (\mathcal\varphi)(y) = \varphi(y). # Define \varphi_\varepsilon(y) = \varphi(y/\varepsilon)/\varepsilon^n. Then with \ast denoting
convolution In mathematics (in particular, functional analysis), convolution is a operation (mathematics), mathematical operation on two function (mathematics), functions ( and ) that produces a third function (f*g) that expresses how the shape of one is ...
, \varphi_\varepsilon is an approximation to the identity: for any continuous f \in L^1(\mathbb R^n) and point x \in \mathbb R^n, \lim_ (\varphi_\varepsilon \ast f)(x) = f(x) (where the convergence is pointwise). Since, by assumption, \mathcalf\in L^1(\mathbb^n), then it follows by the
dominated convergence theorem In measure theory, Lebesgue's dominated convergence theorem provides sufficient conditions under which almost everywhere convergence of a sequence of functions implies convergence in the ''L''1 norm. Its power and utility are two of the primary ...
that :\int_ e^(\mathcalf)(\xi)\,d\xi = \lim_\int_ e^(\mathcalf)(\xi)\,d\xi. Define g_x(\xi) = e^. Applying facts 1, 2 and 4, repeatedly for multiple integrals if necessary, we obtain :(\mathcalg_x)(y) = \frace^=\varphi_\varepsilon(x-y). Using fact 3 on f and g_x, for each x\in\mathbb R^n, we have :\int_ e^(\mathcalf)(\xi)\,d\xi = \int_ \frace^ f(y)\,dy = (\varphi_\varepsilon * f)(x), the convolution of f with an approximate identity. But since f \in L^1(\mathbb R^n), fact 5 says that :\lim_(\varphi_ * f) (x) = f(x). Putting together the above we have shown that : \int_ e^(\mathcalf)(\xi)\,d\xi = f(x). \qquad\square


Notes


References

* * {{cite book, last=Folland, first=G. B., authorlink=Gerald Folland, year=1995, title=Introduction to Partial Differential Equations, edition=2nd, publisher=Princeton Univ. Press, location=Princeton, USA, isbn=978-0-691-04361-6 Generalized functions Theorems in Fourier analysis