# Fourier transform

The printable version is no longer supported and may have rendering errors. Please update your browser bookmarks and please use the default browser print function instead.

One of the integral transforms (cf. Integral transform). It is a linear operator $F$ acting on a space whose elements are functions $f$ of $n$ real variables. The smallest domain of definition of $F$ is the set $D=C_0^\infty$ of all infinitely-differentiable functions $\phi$ of compact support. For such functions

$$(F\phi)(x) = \frac{1}{(2\pi)^{\frac{n}{2}}} \cdot \int_{\mathbf R^n} \phi(\xi) e^{-i x \xi} \, \mathrm d\xi.$$

In a certain sense the most natural domain of definition of $F$ is the set $S$ of all infinitely-differentiable functions $\phi$ that, together with their derivatives, vanish at infinity faster than any power of $\frac{1}{|x|}$. Formula (1) still holds for $\phi\in S$, and $(F \phi)(x) \equiv \psi(x)\in S$. Moreover, $F$ is an isomorphism of $S$ onto itself, the inverse mapping $F^{-1}$ (the inverse Fourier transform) is the inverse of the Fourier transform and is given by the formula:

$$\phi(x) = (F^{-1} \psi)(x) = \frac{1}{(2\pi)^{\frac{n}{2}}} \cdot \int_{\mathbf R^n} \psi(\xi) e^{i x \xi} \, \mathrm d\xi.$$

Formula (1) also acts on the space $L_{1}\left(\mathbf{R}^{n}\right)$ of integrable functions. In order to enlarge the domain of definition of the operator $F$ generalization of (1) is necessary. In classical analysis such a generalization has been constructed for locally integrable functions with some restriction on their behaviour as $|x|\to\infty$ (see Fourier integral). In the theory of generalized functions the definition of the operator $F$ is free of many requirements of classical analysis.

The basic problems connected with the study of the Fourier transform $F$ are: the investigation of the domain of definition $\Phi$ and the range of values $F \Phi = \Psi$ of $F$; as well as studying properties of the mapping $F: \ \Phi \rightarrow \Psi$( in particular, conditions for the existence of the inverse operator $F ^ {\ -1}$ and its expression). The inversion formula for the Fourier transform is very simple:

$$F ^ {\ -1} [g (x)] \ = \ F [g (-x)].$$

Under the action of the Fourier transform linear operators on the original space, which are invariant with respect to a shift, become (under certain conditions) multiplication operators in the image space. In particular, the convolution of two functions $f$ and $g$ goes over into the product of the functions $Ff$ and $Fg$:

$$F (f * g) \ = \ Ff \cdot Fg;$$

and differentiation induces multiplication by the independent variable:

$$F (D^ \alpha f \ ) \ = \ (ix)^ \alpha Ff.$$

In the spaces $L _{p} ( \mathbf R^{n} )$, $1 \leq p \leq 2$, the operator $F$ is defined by the formula (1) on the set $D _{F} = (L _{1} \cap L _{p} ) ( \mathbf R^{n} )$ and is a bounded operator from $L _{p} ( \mathbf R^{n} )$ into $L _{q} ( \mathbf R^{n} )$, $p^{-1} + q^{-1} = 1$:

$$\left\{\frac{1}{(2 \pi)^{n / 2}} \int_{\mathbf{R}^{n}}|(F f)(x)|^{q} d x\right\}^{1 / q} \leq\left\{\frac{1}{(2 \pi)^{n / 2}} \int_{\mathbf{R}^{n}}|f(x)|^{p} d x\right\}^{1 / p}$$

(the Hausdorff–Young inequality). $F$ admits a continuous extension onto the whole space $L _{p} ( \mathbf R^{n} )$ which (for $1 < p \leq 2$) is given by

$$\tag{3} (Ff \ ) (x) \ = \ \lim\limits _ {R \rightarrow \infty} {}^{q} \ { \frac{1}{(2 \pi ) ^ n/2} } \int\limits _ {| \xi | < R} f ( \xi ) e ^ {-i \xi x} \ d \xi \ = \ \widetilde{f} (x).$$

Convergence is understood to be in the norm of $L _{q} ( \mathbf R^{n} )$. If $p \neq 2$, the image of $L _{p}$ under the action of $F$ does not coincide with $L _{q}$, that is, the imbedding $FL _{p} \subset L _{q}$ is strict when $1 \leq p < 2$( for the case $p = 2$ see Plancherel theorem). The inverse operator $F ^ {\ -1}$ is defined on $FL _{p}$ by

$$(F ^ {\ -1} \widetilde{f} \ ) \ = \ \lim\limits _ {R \rightarrow \infty} {}^{p} \ { \frac{1}{(2 \pi ) ^ n/2} } \int\limits _ {| \xi | < R} \widetilde{f} ( \xi ) e ^ {i \xi x} \ d \xi ,\ \ 1 < p \leq 2.$$

The problem of extending the Fourier transform to a larger class of functions arises constantly in analysis and its applications. See, for example, Fourier transform of a generalized function.

#### References

 [1] E.C. Titchmarsh, "Introduction to the theory of Fourier integrals" , Oxford Univ. Press (1948) [2] A. Zygmund, "Trigonometric series" , 2 , Cambridge Univ. Press (1988) [3] E.M. Stein, G. Weiss, "Fourier analysis on Euclidean spaces" , Princeton Univ. Press (1971)

Instead of "generalized function" the term "distributiondistribution" is often used.

If $x = (x _{1} \dots x _{n} )$ and $\xi = ( \xi _{1} \dots \xi _{n} )$ then $x \cdot \xi$ denotes the scalar product $\sum _{ {i = 1}^{n}} x _{i} \xi _{i}$.

If in (1) the "normalizing factor" $(1/ {2 \pi} )^{n/2}$ is replaced by some constant $\alpha$, then in (2) it must be replaced by $\beta$ with $\alpha \beta = (1/ {2 \pi} )^{n}$.

At least two other conventions for the "normalization factor" are in common use:

$$\tag{a1} (F \phi ) (x) \ = \ \int\limits _ {\mathbf R ^ n} \phi ( \xi ) e ^ {- ix \cdot \xi} \ d \xi ,$$

$$(F ^ {\ -1} \phi ) (x) \ = \ \frac{1}{(2 \pi ) ^ n} \int\limits _ {\mathbf R ^ n} \phi ( \xi ) e ^ {ix \cdot \xi} \ d \xi ,$$

$$\tag{a2} (F \phi ) (x) \ = \ \int\limits _ {\mathbf R ^ n} \phi ( \xi ) e ^ {- 2 \pi ix \cdot \xi} \ d \xi ,$$

$$(F ^ {\ -1} \phi ) (x) \ = \ \int\limits _ {\mathbf R^{n} } \phi ( \xi ) e ^ {2 \pi ix \cdot \xi} \ d \xi .$$

The convention of the article leads to the Fourier transform as a unitary operator from $L _{2} ( \mathbf R^{n} )$ into itself, and so does the convention (a2). Convention (a1) is more in line with harmonic analysis.

#### References

 [a1] W. Rudin, "Functional analysis" , McGraw-Hill (1973)
How to Cite This Entry:
Fourier transform. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Fourier_transform&oldid=44378
This article was adapted from an original article by P.I. Lizorkin (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article