On Fourier Transforms of Radial Functions and Distributions

Abstract

We find a formula that relates the Fourier transform of a radial function on R ⁿ with the Fourier transform of the same function defined on R ⁿ⁺². This formula enables one to explicitly calculate the Fourier transform of any radial function f(r) in any dimension, provided one knows the Fourier transform of the one-dimensional function t↦f(|t|) and the two-dimensional function (x ₁,x ₂)↦f(|(x ₁,x ₂)|). We prove analogous results for radial tempered distributions.

1 Introduction

The Fourier transform of a function Φ in L ¹(R ⁿ) is defined by the convergent integral

$$F_{n}({\varPhi})(\xi)=\int _{\mathbf{R}^n}{\varPhi}(x) \mathrm {e}^{-2\pi i x\cdot \xi}\, dx\, . $$

If the function Φ is radial, i.e., Φ(x)=φ(|x|) for some function φ on the line, then its Fourier transform is also radial and we use the notation

$$F_{n}({\varPhi})(\xi)= \mathcal{F}_{n}(\varphi )(r), $$

where r=|ξ|. In this article, we will show that there is a relationship between $\mathcal{F}_{n}(\varphi )(r)$ and $\mathcal{F}_{n+2}(\varphi )(r)$ as functions of the positive real variable r.

We have the following result.

Theorem 1.1

Let n≥1. Suppose that f is a function on the real line such that the functions f(|⋅|) are in L ¹(R ⁿ⁺²) and also in L ¹(R ⁿ). Then we have

$$ \mathcal{F}_{n+2}(f)(r) = -\frac{1}{2\pi} \frac{1}{r} \frac{d}{dr} \mathcal{F}_{n}(f)(r),\quad r>0. $$

(1)

Moreover, the following formula is valid for all even Schwartz functions φ on the real line:

$$ \mathcal{F}_{n+2}(\varphi )(r)= \frac {1}{2\pi} \frac {1}{r^2} \mathcal{F}_{n} \biggl(s^{-n+1} \frac {d}{ds}\bigl(\varphi (s)s^n\bigr)\biggr)(r),\quad r>0. $$

(2)

Using the fact that the Fourier transform is a unitary operator on L ²(R ⁿ) we may extend (1) to the case where the functions f(|⋅|) are in L ²(R ⁿ⁺²) and in L ²(R ⁿ). Moreover, in Sect. 4 we extend (1) to tempered distributions. Applications are given in the last section.

Corollary 1.2

Let f(r) be a function on [0,∞) and k some positive integer such the functions x→f(|x|) are absolutely integrable over R ⁿ for all n with 1≤n≤2k+2. Then we have

$$\mathcal{F}_{2k+1}(f)(\rho) = \frac {1}{(2\pi)^k}\sum_{\ell=1}^{k} \frac{(-1)^\ell (2k-\ell-1)!}{2^{k-\ell}(k-\ell)!(\ell-1)!} \frac{1}{\rho^{2k-\ell}} \biggl(\frac{d}{d\rho}\biggr)^\ell \mathcal{F}_{1}(f)(\rho) $$

and

$$\mathcal{F}_{2k+2}(f)(\rho) = \frac {1}{(2\pi)^k}\sum_{\ell=1}^{k} \frac{(-1)^\ell (2k-\ell-1)!}{2^{k-\ell} (k-\ell)!(\ell-1)!} \frac{1}{\rho^{2k-\ell}} \biggl(\frac{d}{d\rho}\biggr)^\ell \mathcal{F}_{2}(f)(\rho). $$

The corollary can be obtained using (1) by induction on k. The simple details are omitted. Again, absolute integrability can be replaced by square integrability.

2 The Proof

The Fourier transform of an integrable radial function f(|x|) on R ⁿ is given by

where $\widetilde{J}_{\nu}(x)= x^{-\nu} J_{\nu}(x)$, and J _ν is the classical Bessel function of order ν. This formula can be found in many textbooks, and we refer to, e.g., [3, Sect. B.5] or [10, Sect. IV.1] for a proof. Moreover, this formula makes sense for all integers n≥1, even n=1, in which case

$$J_{-1/2}(t) = \sqrt{\frac{2}{\pi}} \frac{\cos t}{\sqrt{t}}. $$

Let us set

$$\mathcal{H}_{\frac{n}{2} -1}(f)(r) = {(2\pi)^{{{\frac{n}{2}}}}} \int_0^\infty f(s) \widetilde{J}_{\frac{n}{2}-1}(2\pi s r)\,s^{n-1}\,ds. $$

Then we make use of B.2.(1) in [3], i.e., the identity

$$ \frac {d}{dr}\widetilde{J}_\nu (r) = -r \widetilde{J}_{\nu+1}(r), $$

(3)

which is also valid when ν=−1/2, since

$$J_{1/2}(t) = \sqrt{\frac{2}{\pi}} \frac{\sin t}{\sqrt{t}}. $$

In view of (3), it is straightforward to verify that

$$-\frac{1}{r} \frac{d}{dr} \mathcal{H}_{\frac{n}{2}-1}(f)(r) = 2\pi \mathcal{H}_{\frac{n}{2}}(f)(r) = 2\pi \mathcal{H}_{\frac {n+2}{2}-1}(f)(r), $$

provided f is such that interchanging differentiation with the integral defining $\mathcal{H}_{\frac{n}{2} -1}$ is permissible. For this to happen, we need to have that

$$\int_0^\infty \bigl|f(s)\bigr| \biggl|\frac{d}{dr} \bigl(\widetilde{J}_{\frac{n}{2}-1}(rs)\bigr)\biggr|s^{n-1}\,ds < \infty $$

and thus it will be sufficient to have

$$ \int_0^\infty \bigl|f(s)\bigr| rs^{2} \bigl|\widetilde{J}_{\frac{n}{2}}(rs)\bigr| s^{n-1}\,ds \le c \int_0^\infty \bigl|f(s)\bigr| \frac{rs^{2}}{(1+rs)^{\frac {n+1}{2}}} s^{n-1} ds < \infty $$

(4)

since $|\widetilde{J}_{\frac{n}{2}}(s)| \le c (1+s)^{-n/2-1/2}$. But since f(|⋅|) is in L ¹(R ⁿ⁺²) we have

$$ \int_0^{1/r} \bigl|f(s)\bigr| s^{n+1}\,ds + \int_{1/r}^\infty \bigl|f(s)\bigr| s^{\frac {n+1}{2}}\,ds<\infty $$

(5)

and this certainly implies (4) for all r>0. We conclude (1) whenever (5) holds. We note that the appearance of condition (5) is natural as indicated in [8] (Lemma 25.1).

To prove (2) we argue as follows. We have

$$\mathcal{H}_{\frac{n}{2} -1}\biggl(r^{-n+1} \frac {d}{dr}\bigl(\varphi (r)r^n\bigr)\biggr)(r) = {(2\pi)^{{{\frac{n}{2}}}}} \int_0^\infty \frac {d}{ds} \bigl(\varphi (s)s^n\bigr) \widetilde{J}_{\frac{n}{2}-1}(2\pi s r)\,ds $$

and integrating by parts the preceding expression becomes

$$(2\pi)^{\frac{n}{2}+2} \int_0^\infty \varphi (s)s^n sr^2 \widetilde{J}_{\frac {n+2}{2}-1}(2\pi s r)\,ds $$

which is equal to $2\pi r^{2} \mathcal{H}_{\frac {n+2}{2}-1} (\varphi )(r)$. This proves (2).

Remark 2.1

Note that we have

$$\mathcal{H}_\nu(f)(r) = \frac{2\pi}{r^\nu} H_\nu\bigl(f(s)s^\nu\bigr)(2\pi r), $$

where

$$H_{\nu}(f)(r) = \int_0^\infty f(s) J_\nu(r s)s\,ds,\quad \nu\ge -\frac{1}{2}, $$

is the Hankel transform. This of course ties in with the fact that the Hankel transform also arises naturally as the spectral transformation associated with the radial part of the Laplacian −Δ; we refer to [4, Sect. 5] and the references therein for further information. Moreover, note that [6] contains the associated recursion from Theorem 1.1 for the Hankel transform, but only for even Schwartz functions. This recursion was rediscovered in connection with the radial Fourier transform in [9] for the case of Schwartz functions. See also [5] for related results.

A transference theorem for radial multipliers which exploits the connection between the Fourier transform of radial functions on R ⁿ and R ⁿ⁺² was obtained in [1]. This multiplier theorem is based on an identity dual to (3).

3 Radial Distributions

We denote by $\mathcal{S}(\mathbf{R}^{n})$ the space of Schwartz functions on R ⁿ and by $\mathcal{S}' (\mathbf{R}^{n})$ the space of tempered distributions on R ⁿ. A Schwartz function is called radial if for all orthogonal transformations A∈O(n) (that is, for all rotations on R ⁿ) we have

$$\varphi = \varphi \circ A. $$

We denote the set of all radial Schwartz functions by $\mathcal{S}_{\mathit{rad}}(\mathbf{R}^{n})$. For further background on radial distributions we refer to Treves [13, Lect. 5]. Observe that in the one-dimensional case the radial Schwartz functions are precisely the even Schwartz functions, that is:

$$\mathcal{S}_{\mathit{rad}}(\mathbf{R}) = \mathcal{S}_{\mathit{even}}(\mathbf{R}) = \bigl\{\varphi \in \mathcal{S}(\mathbf{R}):\varphi (x)=\varphi (-x)\bigr\}. $$

Similarly, a distribution $u\in \mathcal{S}'(\mathbf{R}^{n})$ is called radial if for all orthogonal transformations A∈O(n) we have

$$u = u\circ A. $$

This means that

$$\langle u,\varphi \rangle = \langle u,\varphi \circ A\rangle $$

for all Schwartz functions φ on R ⁿ. We denote by $\mathcal{S}'_{\mathit{rad}}(\mathbf{R}^{n})$ the space of all radial tempered distributions on R ⁿ. We also denote by S ⁿ⁻¹ the (n−1)-dimensional unit sphere on R ⁿ and by ω _n−1 its surface area.

Given a general, non necessarily radial, Schwartz function there is a natural homomorphism

$$\mathcal{S}\bigl(\mathbf{R}^n\bigr)\to \mathcal{S}_{\mathit{rad}}(\mathbf{R}),\qquad \varphi (x)\mapsto \varphi ^o(r) = \frac {1}{\omega _{n-1}}\int_{\mathbf{S}^{n-1}} \varphi (r \theta)\,d\theta $$

with the understanding that when n=1, then $\varphi ^{o}(x)=\frac{1}{2}(\varphi (x)+\varphi (-x))$. Conversely, given an even Schwartz function on R we can define a corresponding radial Schwartz function via

$$\mathcal{S}_{\mathit{rad}}(\mathbf{R})\to \mathcal{S}_{\mathit{rad}}\bigl(\mathbf{R}^n\bigr),\qquad \varphi (r) \mapsto \varphi ^O(x)=\varphi \bigl(|x|\bigr). $$

The map φ↦φ ^O is a homomorphism; the proof of this fact is omitted since a stronger statement is proved at the end of this section. Both facts require the following lemma:

Lemma 3.1

Suppose that f is a smooth even function on R. Then there is a smooth function g on the real line such that

$$f(x) = g\bigl(x^2\bigr) $$

for all x∈R. Moreover, one has for t≥0

$$ \bigl|g^{(k)}(t)\bigr|\le C(k) \sup_{0\le s\le \sqrt{t}} \bigl|f^{(2k)}(s)\bigr|. $$

(6)

Proof

By Whitney’s theorem [14], there is a smooth function g on the real line such that

$$f(t) = g\bigl(t^2\bigr) $$

for all real t.

To see the last assertion we use the following representation of the remainder in Taylor’s theorem:

from which one easily derives (6). This yields in particular that

$$\frac {g^{(k)}(0)}{k!} = \frac {f^{(2k)}(0)}{(2k)!} $$

since

$$2^{-2k} k {2k\choose k} \int_0^1 \bigl(1-s^2\bigr)^{k-1}\,ds = 2^{-2k} k {2k\choose k} \frac{{\varGamma}(k){\varGamma}(1/2)}{{\varGamma}(k+1/2)} = 1. $$

 □

The composition φ↦(φ ^o)^O=φ ^rad gives rise to a homomorphism from $\mathcal{S}(\mathbf{R}^{n}) \to \mathcal{S}_{\mathit{rad}}(\mathbf{R}^{n})$ which reduces to the identity map on radial Schwarz functions. In particular, the map φ↦φ ^o defines a one-to-one correspondence between radial Schwartz functions on R ⁿ and even Schwartz functions on the real line. Moreover, φ is radial if and only if φ=φ ^rad.

Proposition 3.2

For $u\in \mathcal{S}'_{\mathit{rad}}(\mathbf{R}^{n})$ and $\varphi \in \mathcal{S}(\mathbf{R}^{n})$ we have

$$\langle u,\varphi \rangle = \bigl\langle u,\varphi ^{\mathit{rad}}\bigr\rangle. $$

Proof

By a simple change of variables the formula holds for any u which is a polynomially bounded locally integrable function. Next we fix a tempered distribution u on R ⁿ and we consider a radial Schwartz function ψ with integral 1 and we set ψ _ε (x)=ε ⁻ⁿ ψ(x/ε). Then we notice that the convolution of ψ _ε ∗u converges to u in $\mathcal{S}'(\mathbf{R}^{n})$ as ε→0. Hence, since the claim holds if u is replaced by ψ _ε ∗u by the first observation, it remains true in the limit ε→0. □

In particular, note that a radial distribution is uniquely determined by its action on radial Schwartz functions. Furthermore, given a distribution $u\in \mathcal{S}'(\mathbf{R}^{n})$ we can define a radial distribution $u^{\mathit{rad}}\in \mathcal{S}'_{\mathit{rad}}(\mathbf{R}^{n})$ via

$$\bigl\langle u^{\mathit{rad}},\varphi \bigr\rangle := \bigl\langle u,\varphi ^{\mathit{rad}} \bigr\rangle. $$

Moreover, u is radial if and only if u=u ^rad.

For n∈Z ⁺ we denote by $\mathcal{R}_{n} = r^{n-1}\mathcal{S}_{\mathit{even}}(\mathbf{R})$ the space of functions of the form ψ(r)r ⁿ⁻¹, where ψ is an even Schwartz function on the line. This space inherits the topology of S(R) and its dual space is denoted by $\mathcal{R}_{n}'$. Two distributions $w_{1}, w_{2}\in \mathcal{S}'(\mathbf{R})$ are equal in the space $\mathcal{R}_{n}'$ if for all even Schwartz functions ψ on the line we have:

$$\bigl\langle w_1, r^{n-1} \psi(r)\bigr\rangle = \bigl\langle w_2, r^{n-1} \psi(r)\bigr\rangle. $$

Note that in dimension n≥2 we have that all distributions of order n−2 supported at the origin equal the zero distribution in the space $\mathcal{R}_{n}'$. Thus two radial distributions w ₁ and w ₂ are equal in $\mathcal{R}_{n}'$ whenever w ₁−w ₂ is a sum of derivatives of the Dirac mass at the origin of order at most n−2.

One may build radial distributions on R ⁿ starting from distributions in $\mathcal{R}_{n}'$. Indeed, given u _⋄ in $\mathcal{R}_{n}'$ and φ in $\mathcal{S}(\mathbf{R}^{n})$ we define a radial distribution u by setting

$$\langle u,\varphi \rangle := \frac{\omega _{n-1}}{2} \bigl\langle u_\diamond, \varphi ^o(r) r^{n-1}\bigr\rangle. $$

The converse is the content of the following proposition.

Proposition 3.3

The map $\mathcal{R}_{n}\to \mathcal{S}_{\mathit{rad}}(\mathbf{R}^{n})$, ψ(r)r ⁿ⁻¹↦ψ ^O(x) is a homeomorphism and hence for every radial distribution u we can define u _⋄ in $\mathcal{R}_{n}'$ via

$$\bigl\langle u_\diamond, \psi(r)r^{n-1}\bigr\rangle := \frac {2}{\omega _{n-1}} \bigl\langle u,\psi^O\bigr\rangle. $$

Proof

It suffices to show the first claim. To this end we will show that for all multiindices α and β we have

$$\sup_{x\in \mathbf{R}^n} \bigl|x^\alpha \partial ^\beta _x\bigl(\psi\bigl(|x|\bigr)\bigr)\bigr|\le \sum_{0\le \ell,m\le 4(|\beta |+|\alpha |+n)} \sup_{r>0} \biggl|r^m \biggl(\frac {d}{dr}\biggr)^\ell \bigl(r^{n-1}\psi(r)\bigr)\biggr|. $$

First we consider the case |x|≤1. Setting r=|x|≤1 we have

using Lemma 3.1 with ψ(t)=g(t ²).

We will make use of the inequality

$$ \bigl|\psi(s)\bigr|\le \sup_{0<t<s} \biggl|\biggl(\frac {d}{dt}\biggr)^{M} \bigl(t^{M}\psi (t)\bigr)(s)\biggr| $$

(7)

which follows by applying the fundamental theorem of calculus M times and of the identity:

$$ s^M \frac {d^m\psi}{ds^m}(s) = \sum_{\ell=0}^m (-1)^\ell \ell! {m\choose \ell} {M\choose \ell} \biggl(\frac {d}{ds}\biggr)^{m-\ell} \bigl(s^{M-\ell}\psi(s)\bigr) $$

(8)

which is valid for M≥m and is easily proved by induction.

Applying (7) to ψ ^(2k)(s) we obtain

$$ \bigl|\psi^{(2k)}(s)\bigr|\le \sup_{0<t<s} \biggl|\biggl(\frac {d}{dt}\biggr)^{M}\bigl(t^{M} \psi^{(2k)}(t) \bigr)(s)\biggr| $$

(9)

and using (8) for s ^M ψ ^(2k)(s) with M=2|β|+n−1 and m=2k we deduce that |ψ ^(2k)(s)| is pointwise bounded by a sum of derivatives of terms s ⁿ⁻¹ ψ(s) multiplied by powers of s. It follows that sup _s>0|ψ ^(2k)(s)| is controlled by a finite sum of Schwartz seminorms of the function s ⁿ⁻¹ ψ(s).

The case |x|≥1 is easier since when |β|≠0

$$\bigl|\partial ^\beta _x\bigl(\psi\bigl(|x|\bigr)\bigr)\bigr|\le \sum_{j=1}^{|\beta |} \bigl|\psi{(j)}\bigl(|x|\bigr)\bigr| \frac {C_{j,\beta }}{|x|^{|\beta |-j}}, $$

and taking M=max(|α|,|β|+n−1) we have

$$ \sup_{|x|\ge 1} \bigl|x^\alpha \partial ^\beta _x\bigl(\psi\bigl(|x|\bigr)\bigr)\bigr|\le C_{\beta } \sum_{j=1}^{|\beta |}\sup_{s\ge 1} \bigl\{ s^{M} \bigl|\psi^{(j)}(s)\bigr|\bigr\}, $$

(10)

which is certainly controlled by a finite sum of Schwartz seminorms of s ⁿ⁻¹ ψ(s) in view of (8). □

Note that if u is given by a function f(x), then u _⋄ is given by the function f ^o(x). We also remark that the map $\frac{1}{r} \frac{d}{dr}$ is a homomorphism from $\mathcal{R}_{n}'$ to $\mathcal{R}_{n+1}'$ defined as the dual map of $- \frac{d}{dr}\frac{1}{r}$.

A related approach defining u _⋄ for a given distribution u supported in R ⁿ∖{0} can be found in [11]. Our approach does not impose restrictions on the support of the distribution.

4 The Extension to Tempered Distributions

Let u be a radial distribution on R ^k and let F _k (u) be the k-dimensional Fourier transform of u.

Theorem 4.1

Given an even tempered distribution v ₀ on the real line, define radial distributions v _n on R ⁿ and v _n+2 on R ⁿ⁺² via the identities

$$ \langle v_n, \varphi \rangle = \biggl\langle v_0, \frac{1}{2}\omega_{n-1} r^{n-1} \varphi ^o\biggr\rangle $$

(11)

for all radial Schwartz functions φ(x) = φ ^o(|x|) on R ⁿ and

$$\langle v_{n+2}, \varphi \rangle = \biggl\langle v_0, \frac{1}{2} \omega_{n+1} r^{n+1} \varphi ^o \biggr\rangle $$

for all radial Schwartz functions φ(x) = φ ^o(|x|) on R ⁿ⁺².

Let u ⁿ=F _n (v _n ) and u ⁿ⁺²=F _n+2(v _n+2). Then the identity

$$ -\frac {1}{2\pi r}\frac {d}{dr} u_\diamond^n = u_\diamond^{n+2} $$

(12)

holds on $\mathcal{R}_{n+2}'$.

Proof

We denote by 〈⋅,⋅〉 _n the action of the distribution on a function in dimension n. Let ψ(r) be an even Schwartz function on the real line. Then we need to show that

$$ \biggl\langle -\frac {1}{2\pi r}\frac {d}{dr} u_\diamond^n, \omega _{n+1} r^{n+1} \psi(r)\biggr\rangle_1 = \bigl\langle u_\diamond^{n+2}, \omega _{n+1} r^{n+1} \psi(r) \bigr\rangle_1. $$

(13)

This is equivalent to showing that

$$ \frac {1}{2\pi} \bigl\langle u_\diamond^n, \omega _{n+1} \bigl(r^{n } \psi(r)\bigr)'\bigr\rangle_1 = \bigl\langle u_\diamond^{n+2}, \omega _{n+1} r^{n+1} \psi(r) \bigr\rangle_1. $$

(14)

We introduce the even Schwartz function η(r)=r ⁻ⁿ⁺¹(r ⁿ ψ(r))′=nψ(r)+rψ′(r) on the real line and functions η ^O on R ⁿ and ψ ^O on R ⁿ⁺² by setting

$$\psi^O(x) = \psi\bigl(|x|\bigr),\qquad \eta^O(y)=\eta\bigl(|y|\bigr) $$

for y∈R ⁿ and x∈R ⁿ⁺². Then (14) is equivalent to

$$ \frac {1}{2\pi} \frac {\omega _{n+1}}{\omega _{n-1}} \bigl\langle u_\diamond^n, \omega _{n-1} r^{n-1} \eta(r) \bigr\rangle_1 = \bigl\langle u_\diamond^{n+2}, \omega _{n+1} r^{n+1} \psi(r) \bigr\rangle_1 $$

(15)

which is in turn equivalent to

$$ \frac {1}{2\pi} \frac {\omega _{n+1}}{\omega _{n-1}} \bigl\langle F_n(v_n), \eta^O \bigr\rangle_n = \bigl\langle F_{n+2}(v_{n+2}), \psi^O \bigr\rangle_{n+2} $$

(16)

and also to

$$ \frac {1}{2\pi} \frac {\omega _{n+1}}{\omega _{n-1}} \bigl\langle v_n , F_n\bigl(\eta^O\bigr) \bigr\rangle_n = \bigl\langle v_{n+2}, F_{n+2}\bigl(\psi^O\bigr)\bigr\rangle_{n+2}. $$

(17)

We now switch to dimension one by writing (17) equivalently as

$$ \frac {1}{2\pi} \frac {\omega _{n+1}}{\omega _{n-1}} \bigl\langle v_0, \omega _{n-1}r^{n-1} \mathcal{F}_n(\eta)(r)\bigr\rangle_1 = \bigl\langle v_{0}, \omega _{n+1} r^{n+1}\mathcal{F}_{n+2}(\psi)(r)\bigr\rangle_{1}. $$

(18)

But this identity holds if

$$\frac {1}{2\pi} \mathcal{F}_n(\eta)(r) = r^2 \mathcal{F}_{n+2}(\psi)(r), $$

which is valid as a restatement of (2); recall that $\eta(r)= r^{-n+1} \frac {d}{dr}(r^{n}\psi(r))$. This proves (13). □

It is straightforward to check that for polynomially bounded smooth functions all operations coincide with the usual ones. We end this section with a few more illustrative examples. Let δ _n be the Dirac mass on R ⁿ.

Examples

• (a) Let v _n =δ _n . One can see that

  $$v_0 = \frac {2(-1)^{n-1}}{\omega_{n-1}(n-1)!} \biggl(\frac {d}{dr}\biggr)^{(n-1)}(\delta_1) $$

  satisfies (11). Acting v ₀ on r ⁿ⁺¹ φ ^o(r) yields that v _n+2=0 and thus $u_{\diamond}^{n+2}=0$. Also $u_{\diamond}^{n}=1$; so both sides of (12) are equal to zero.

• (b) Let v _n+2=δ _n+2. Then

  $$v_0 = \frac {2(-1)^{n+1}}{\omega_{n+1}(n+1)} \biggl(\frac {d}{dr}\biggr)^{(n+1)}(\delta_1). $$

Let $\Delta =\partial _{1}^{2}+\cdots + \partial _{n}^{2}$ be the Laplacian. We claim that the distribution

$$ v_n = \frac {\omega_{n-1}}{\omega_{n+1}} \frac {1}{n}{\varDelta}(\delta_n) $$

(19)

satisfies (11). Then $u_{\diamond}^{n+2}=1$ and also $u_{\diamond}^{n}=-r^{2}(2\pi)^{2} \omega_{n-1}/(2n\omega_{n+1})$. Thus (12) is valid since 2πω _n−1=nω _n+1.

It remains to prove that the distribution v _n in (19) satisfies (11). For φ(x)=φ ^o(|x|) in $\mathcal{S}(\mathbf{R}^{n})$ we have

$$ \langle v_n, \varphi \rangle = \bigl\langle v_0, \omega _{n-1} r^{n-1} \varphi ^o(r)\bigr\rangle = \frac {\omega _{n-1}}{\omega _{n+1}} \frac {2}{(n+1)!} \bigl\langle \delta _1, \bigl(r^{n-1} \varphi ^o(r)\bigr)^{(n-1)}\bigr\rangle $$

(20)

and one notices that the (n−1)st derivative of r ⁿ⁻¹ φ ^o(r) evaluated at zero is equal to $\frac{1}{2} (n+1)!(\varphi ^{o})''(0)$. To compute the value of this derivative we use Lemma 3.1 to write φ(x)=φ ^o(|x|)=g(|x|²) where $g'(0)=\frac{1}{2}(\varphi ^{o})''(0)$. It follows that $g'(0)=\frac{1}{2n} \Delta (\varphi )(0)$. Combining these observations yields that the expression in (20) is equal to

$$\frac {\omega _{n-1}}{\omega _{n+1}} \frac {1}{n} \Delta (\varphi )(0) = \biggl\langle \frac {\omega _{n-1}}{\omega _{n+1}} \frac {1}{n} \Delta (\delta _n), \varphi \biggr\rangle, $$

which proves the claim.

Remark 4.2

As pointed out in Remark 2.1, the action of the Fourier transform on the associated function on the reals φ ^o is given by the Hankel transform. In particular, the results in this section also give a natural extension of the Hankel transform (for half-integer order) to distributions. Of course this coincides with the usual approach, see [6, 15, 16] and the references therein. To this end observe that the space F used in [6] is precisely the set of functions on [0,∞) which extend to an even Schwartz function on R.

5 Applications

We begin with a simple example. In dimension one we have that the Fourier transform of sech(π|x|) is sech(π|ξ|). It follows from (1) that in dimension three we have

$$F_3\bigl(\mathrm {sech}\bigl(\pi|x|\bigr)\bigr)(\xi) = \frac{1}{2|\xi|} \mathrm {sech}\bigl(\pi|\xi|\bigr) \tanh\bigl(\pi|\xi|\bigr). $$

since

$$\frac{d}{dr} \frac{2}{\mathrm {e}^{\pi r}+\mathrm {e}^{-\pi r}} = -2\pi \frac{\mathrm {e}^{\pi r}-\mathrm {e}^{-\pi r}}{(\mathrm {e}^{\pi r}+\mathrm {e}^{-\pi r})^2} = -2\pi \frac{1}{2} \mathrm {sech}(\pi r) \tanh(\pi r). $$

Continuing this process, one can explicitly calculate the Fourier transform of sech(π|x|) in all odd dimensions.

More sophisticated applications of our formulas appear in computations of functions of the Laplacian −Δ, which arise in numerous applications. For example, in quantum mechanics the Laplacian −Δ arises as the free Schrödinger operator (cf., e.g., [7, 12]) and functions f(−Δ) are defined via the spectral theorem by

$$f(-{\varDelta}) \varphi = K * \varphi ,\quad \varphi \in \mathcal{S}\bigl(\mathbf{R}^n\bigr), $$

where K is the tempered distribution given by the inverse Fourier transform of the radial function f(4π ²|ξ|²), which is assumed polynomially bounded. Knowledge of the inverse Fourier transform of f(4π ²|ξ|²), for ξ∈R and ξ∈R ², yields explicit formulas for the kernel K of f(−Δ) in all dimensions.

An important application is the explicit calculation of the n-dimensional kernel G _n (x) for the resolvent associated with the function f(r)=(r−z)⁻¹, z∈C∖[0,∞). In the one-dimensional case, an easy computation shows that

$$G_1(x) = \frac{1}{2\sqrt{-z}} \mathrm {e}^{-\sqrt{-z}|x|}. $$

Hence, by the L ² version of Theorem 1.1 (cf. the discussion right after Theorem 1.1) the three-dimensional kernel is given by

$$G_3(x) = -\frac{1}{2\pi r} \frac{d}{dr} G_1(r)\bigg|_{r=|x|} = \frac{1}{4\pi|x|} \mathrm {e}^{-\sqrt{-z} |x|}. $$

The computation of G ₅(x),G ₇(x),… requires Theorem 4.1 since the assumptions of Theorem 1.1 are no longer satisfied. For instance, Theorem 4.1 gives

$$G_5(x) = \frac{1+|x| \sqrt{-z}}{8\pi^2|x|^3} \mathrm {e}^{-\sqrt{-z}|x|}. $$

Another interesting situation where our theorem is useful are the spectral projections associated with the function f(r)=χ _[0,E](r), E>0. Again in the one-dimensional case the kernel for the resolvent can be easily computed and found to be

$$P_1(x) = \frac{\sin(x\sqrt{E})}{\pi x}. $$

Thus by Theorem 1.1 the three-dimensional kernel is given by

$$P_3(x) = -\frac{1}{2\pi r} \frac{d}{dr} P_1(r)\bigg|_{r=|x|} = \frac{\sin(|x|\sqrt{E})-|x|\sqrt{E}\cos(|x|\sqrt{E})}{2\pi^2|x|^3}. $$

Finally, the Fourier transform is a crucial tool in solving constant coefficient linear partial differential equations (cf., e.g., [2]). Using the above trick one can of course derive the fundamental solution for the heat (or Schrödinger) equation in three dimensions from the one-dimensional one. However, since the three-dimensional case is no more difficult than the one-dimensional case we rather turn to the Cauchy problem for the wave equation

$$u_{tt}-{\varDelta} u = 0,\qquad u(0,x)=\psi(x),\qquad u_t(0,x)=\varphi (x), $$

in R ⁿ, whose solution is given by

$$u(t,x) = \cos(t\sqrt{-{\varDelta}}) \psi(x) + \frac{\sin(t \sqrt{-{\varDelta}})}{\sqrt{-{\varDelta}}} \varphi (x). $$

Since the first term can be obtained by differentiating the second (with respect to t) it suffices to look only at the second and assume ψ=0. Moreover, since the Fourier transform of $f(x)=\frac{\sin(a \pi x)}{a \pi x}$ is F ₁(f)(ξ)=|a|⁻¹ χ _[−1/2,1/2](ξ/a), we obtain

$$u(t,x) = \int_\mathbf{R}\frac{1}{2} \chi_{[-t,t]}(x-y) \varphi (y)\,dy, $$

which is of course just d’Alembert’s formula. In order to apply Theorem 4.1 we use $v_{0}(r)= \frac{\sin(t r)}{r}$ such that $u^{1}=F_{1}^{-1}(v_{1})$ as well as $u^{1}_{\diamond}$ are associated with the function $\frac{1}{2} \chi_{[-t,t]}(x)$. Hence by Theorem 4.1

$$\bigl\langle F_3^{-1}(v_3), \varphi \bigr\rangle = \frac{\omega _2}{2} \biggl\langle -\frac{1}{2\pi r} \frac{d}{dr} \frac{1}{2} \chi_{[-t,t]}(r), r^2 \varphi ^o(r)\biggr\rangle = \frac{\omega _2}{4\pi} t \varphi ^o(t) $$

and we obtain Kirchhoff’s formula

$$u(t,x) = \frac{t}{4\pi} \int_{\mathbf{S}^2} \varphi (x-t\theta)\,d\theta. $$