The Fourier transform of spherical surface measure and radial functions

1 Notation

For a topological space $X$, we denote by ${ℬ}_X$ the Borel $\sigma$-algebra of $X$. Let ${\rho }_d$ be the Euclidean metric on ${ℝ}^d$ and let $m_d$ be Lebesgue measure on ${ℝ}^d$.

2 Polar coordinates

Let $X=(0,\mathrm{\infty })$, which is a metric space with the metric inherited from $ℝ$. Define $\mu :{ℬ}_X\to [0,\mathrm{\infty }]$ by

$$d\mu (r)=r^{d-1}d{m}_1(r).$$

Let $S^{d-1}$ be the unit sphere in ${ℝ}^d$. Define $S:𝒫(S^{d-1})\to 𝒫({ℝ}^d)$ by

Namely, $S(E)$ is the sector subtended by the set $E$. $S^{d-1}$ is a metric space with the metric inherited from ${ℝ}^d$, and if $E$ is an open set in $(S^{d-1},{\rho }_d)$, then $S(E)$ is an open set in ${ℝ}^d$. For $E_{\alpha }\in 𝒫(S^{d-1})$,

$$S\left(\bigcup E_{\alpha }\right)=\bigcup S(E_{\alpha }),S\left(\bigcap E_{\alpha }\right)=\bigcap S(E_{\alpha }),$$

and for $E,F\in 𝒫(S^{d-1})$,

$$S(E\setminus F)=S(E)\setminus S(F).$$

Lemma 1.

$$S({ℬ}_{S^{d-1}})\subset {ℬ}_{{ℝ}^d}.$$

We define ${\sigma }_{d-1}:{ℬ}_{S^{d-1}}\to [0,\mathrm{\infty })$ by

$${\sigma }_{d-1}(E)=d\cdot m_d(S(E)),E\in {ℬ}_{S^{d-1}}.$$

For $f:{ℝ}^d\to ℂ$ and $\gamma \in S^{d-1}$, define $f^{\gamma }:(0,\mathrm{\infty })\to ℂ$ by

$$f^{\gamma }(r)=f(r\gamma ),r\in (0,\mathrm{\infty }).$$

The following is proved in Stein and Shakarchi.

Theorem 2.

If $f\mathrm{\in }{L}^{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{,}{m}_d\mathrm{)}$, then (i) for $\sigma$-almost all $\gamma \mathrm{\in }{S}^{d\mathrm{-}\mathrm{1}}$ we have $f^{\gamma }\mathrm{\in }{L}^{\mathrm{1}}\mathit{}\mathrm{(}\mathrm{(}\mathrm{0}\mathrm{,}\mathrm{\infty }\mathrm{)}\mathrm{,}\mu \mathrm{)}$, (ii) the function

$$\gamma \mapsto {\int }_0^{\mathrm{\infty }}{f}^{\gamma }(r)𝑑\mu (r)$$

belongs to $L^{\mathrm{1}}\mathit{}\mathrm{(}{S}^{d\mathrm{-}\mathrm{1}}\mathrm{,}\sigma \mathrm{)}$, and (iii)

$${\int }_{{ℝ}^d}f(x)𝑑m_d(x)={\int }_{S^{d-1}}\left({\int }_0^{\mathrm{\infty }}{f}^{\gamma }(r)𝑑\mu (r)\right)𝑑\sigma (\gamma ).$$

For $r\in (0,\mathrm{\infty })$, define $f_r:S^{d-1}\to ℂ$ by

$$f_r(\gamma )=f(r\gamma ),\gamma \in S^{d-1}.$$

Theorem 3.

If $f\mathrm{\in }{L}^{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{R}}^d\mathrm{,}{m}_d\mathrm{)}$, then (i) for $\mu$-almost all $r\mathrm{\in }\mathrm{(}\mathrm{0}\mathrm{,}\mathrm{\infty }\mathrm{)}$ we have $f_r\mathrm{\in }{L}^{\mathrm{1}}\mathit{}\mathrm{(}{S}^{d\mathrm{-}\mathrm{1}}\mathrm{,}\sigma \mathrm{)}$, (ii) the function

$$r\mapsto {\int }_{S^{d-1}}{f}_r(\gamma )𝑑\sigma (\sigma )$$

belongs to $L^{\mathrm{1}}\mathit{}\mathrm{(}\mathrm{(}\mathrm{0}\mathrm{,}\mathrm{\infty }\mathrm{)}\mathrm{,}\mu \mathrm{)}$, and (iii)

$${\int }_{{ℝ}^d}f(x)𝑑m_d(x)={\int }_0^{\mathrm{\infty }}\left({\int }_{S^{d-1}}{f}_r(\gamma )𝑑\sigma (\gamma )\right)𝑑\mu (r).$$

3 The Fourier transform of spherical surface measure

For real $\nu >-\frac{1}{2}$,

$$J_{\nu }(s)=\frac{{\left(\frac{s}{2}\right)}^{\nu }}{\mathrm{\Gamma }\left(\nu +\frac{1}{2}\right)\sqrt{\pi }}{\int }_{-1}^1{e}^{isx}{(1-x^2)}^{\nu -\frac{1}{2}}𝑑x,s\in ℝ.$$

One checks that $J_{\nu }$ satisfies

$$J_{\nu }(-s)=e^{i\pi \nu }{J}_{\nu }(s),s\in ℝ.$$

We remind ourselves of spherical coordinates for $S^{d-1}$. The Jacobian of the transformation

${\gamma }_1$ $=\cos {\varphi }_1$

${\gamma }_2$ $=\sin {\varphi }_1\cos {\varphi }_2$

${\gamma }_3$ $=\sin {\varphi }_1\sin {\varphi }_2\cos {\varphi }_3$

$\mathrm{\cdots }$

${\gamma }_{d-1}$ $=\sin {\varphi }_1\sin {\varphi }_2\sin {\varphi }_3\mathrm{\cdots }\sin {\varphi }_{d-2}\cos {\varphi }_{d-1}$

${\gamma }_d$ $=\sin {\varphi }_1\sin {\varphi }_2\sin {\varphi }_3\mathrm{\cdots }\sin {\varphi }_{d-2}\sin {\varphi }_{d-1},$

with

$$0\le {\varphi }_1,\mathrm{\dots },{\varphi }_{d-2}\le \pi ,0\le {\varphi }_{d-1}\le 2\pi ,$$

is

$$J={\sin }^{d-2}{\varphi }_1{\sin }^{d-3}{\varphi }_2\mathrm{\cdots }{\sin }^2{\varphi }_{d-3}\sin {\varphi }_{d-2}.$$

Then, for $\xi =({\xi }_1,0,\mathrm{\dots },0)$, ${\xi }_1\ne 0$,

${\widehat{\sigma }}_{d-1}(\xi )$ $={\displaystyle {\int }_{S^{d-1}}}{e}^{-2\pi i\gamma \cdot \xi }𝑑\sigma (\gamma )$

$={\displaystyle {\int }_{{\varphi }_1=0}^{\pi }}{\displaystyle {\int }_{{\varphi }_2=0}^{\pi }}\mathrm{\cdots }{\displaystyle {\int }_{{\varphi }_{d-2}=0}^{\pi }}{\displaystyle {\int }_{{\varphi }_{d-1}=0}^{2\pi }}{e}^{-2\pi i{\xi }_1\cos {\varphi }_1}J𝑑{\varphi }_{d-1}𝑑{\varphi }_{d-2}\mathrm{\cdots }𝑑{\varphi }_2𝑑{\varphi }_1$

$=2\pi \cdot {\displaystyle {\int }_{{\varphi }_1=0}^{\pi }}{e}^{-2\pi i{\xi }_1\cos {\varphi }_1}{\sin }^{d-2}{\varphi }_{1}d{\varphi }_1\cdot {\displaystyle \prod _{j=2}^{d-2}}{\displaystyle {\int }_{{\varphi }_j=0}^{\pi }}{\sin }^{d-j-1}{\varphi }_{j}d{\varphi }_j.$

We work out that

$${\int }_0^{\pi }{\sin }^{k}tdt=\frac{\sqrt{\pi }\mathrm{\Gamma }\left(\frac{k+1}{2}\right)}{\mathrm{\Gamma }\left(\frac{k+2}{2}\right)}.$$

This gives

${\displaystyle \prod _{j=2}^{d-2}}{\displaystyle {\int }_{{\varphi }_j=0}^{\pi }}{\sin }^{d-j-1}{\varphi }_{j}d{\varphi }_j$ $={\displaystyle \prod _{j=2}^{d-2}}{\displaystyle \frac{\sqrt{\pi }\mathrm{\Gamma }\left(\frac{d-j}{2}\right)}{\mathrm{\Gamma }\left(\frac{d-j+1}{2}\right)}}={\pi }^{\frac{d-3}{2}}{\displaystyle \frac{\mathrm{\Gamma }\left(\frac{2}{2}\right)}{\mathrm{\Gamma }\left(\frac{d-1}{2}\right)}}={\displaystyle \frac{{\pi }^{\frac{d-3}{2}}}{\mathrm{\Gamma }\left(\frac{d-1}{2}\right)}}.$

With this we have, for $\xi =({\xi }_1,0,\mathrm{\dots },0)$, ${\xi }_1\ne 0$,

$${\widehat{\sigma }}_{d-1}(\xi )=2\pi \frac{{\pi }^{\frac{d-3}{2}}}{\mathrm{\Gamma }\left(\frac{d-1}{2}\right)}{\int }_0^{\pi }{e}^{-2\pi i{\xi }_1\cos t}{\sin }^{d-2}tdt.$$

But doing the change of variable $x=\cos t$, for nonzero real $s$ we have

${\displaystyle {\int }_0^{\pi }}{e}^{is\cos t}{\sin }^{d-2}tdt$ $={\displaystyle {\int }_0^{\pi }}{e}^{is\cos t}{(1-{\cos }^{2}t)}^{\frac{d-2}{2}}𝑑t$

$={\displaystyle {\int }_1^{-1}}{e}^{isx}{(1-x^2)}^{\frac{d-2}{2}}{\displaystyle \frac{-dx}{\sqrt{1-x^2}}}$

$={\displaystyle {\int }_{-1}^1}{e}^{isx}{(1-x^2)}^{\frac{d}{2}-1-\frac{1}{2}}𝑑x$

$={\displaystyle \frac{\mathrm{\Gamma }\left(\frac{d}{2}-\frac{1}{2}\right)\sqrt{\pi }}{{\left(\frac{s}{2}\right)}^{\frac{d}{2}-1}}}{J}_{\frac{d}{2}-1}(s).$

Thus, taking $s=-2\pi {\xi }_1$,

${\widehat{\sigma }}_{d-1}(\xi )$ $=2\pi {\displaystyle \frac{{\pi }^{\frac{d-3}{2}}}{\mathrm{\Gamma }\left(\frac{d-1}{2}\right)}}{\displaystyle \frac{\mathrm{\Gamma }\left(\frac{d}{2}-\frac{1}{2}\right)\sqrt{\pi }}{{\left(\frac{-2\pi {\xi }_1}{2}\right)}^{\frac{d}{2}-1}}}{J}_{\frac{d}{2}-1}(-2\pi {\xi }_1)$

$=2\pi \cdot {(-{\xi }_1)}^{-\frac{d}{2}+1}{J}_{\frac{d}{2}-1}(-2\pi {\xi }_1).$

For this is

$${\widehat{\sigma }}_{d-1}(\xi )=2\pi {|\xi |}^{-\frac{d}{2}+1}{J}_{\frac{d}{2}-1}(2\pi |\xi |).$$

In general, take nonzero $\xi \in {ℝ}^d$. Let $T:{ℝ}^d\to {ℝ}^d$ be the rotation that sends $\xi$ ti $(0,\mathrm{\dots },0,-|\xi |)$. Since ${\sigma }_{d-1}\circ T={\sigma }_{d-1}$ (namely, surface measure ${\sigma }_{d-1}$ is invariant under rotations),

$${\widehat{\sigma }}_{d-1}(\xi )={\widehat{\sigma }}_{d-1}((0,\mathrm{\dots },0,-|\xi |))=2\pi {|\xi |}^{-\frac{d}{2}+1}{J}_{\frac{d}{2}-1}(2\pi |\xi |).$$

For real $\nu >-\frac{1}{2}$, we use the following asymptotic formula for $J_{\nu }(s)$:

$$J_{\nu }(s)=\sqrt{\frac{2}{\pi s}}\cos \left(s-\frac{\pi \nu }{2}-\frac{\pi }{4}\right)+O(s^{-3/2}),s\to +\mathrm{\infty }.$$

We get from this that

$$|{\widehat{\sigma }}_{d-1}(\xi )|=O({|\xi |}^{-\frac{d}{2}+\frac{1}{2}}),|\xi |\to \mathrm{\infty }.$$

4 The Fourier transform of radial functions

A function $f:{ℝ}^d\to ℂ$ is said to be radial if there is a function $f_0:[0,\mathrm{\infty })\to ℂ$ such that

$$f(x)=f_0(|x|),x\in {ℝ}^d.$$

For $f\in L^1({ℝ}^d)$, Using polar coordinates we determine the Fourier transform of a radial function. For $\xi \in {ℝ}^d$,

$\widehat{f}(\xi )$ $={\displaystyle {\int }_{{ℝ}^d}}{e}^{-2\pi ix\cdot \xi }f(x)𝑑x$

$={\displaystyle {\int }_0^{\mathrm{\infty }}}\left({\displaystyle {\int }_{S^{d-1}}}{e}^{-2\pi ir\sigma \cdot \xi }f(r\sigma )𝑑\sigma (\gamma )\right)𝑑\mu (r)$

$={\displaystyle {\int }_0^{\mathrm{\infty }}}\left({\displaystyle {\int }_{S^{d-1}}}{e}^{-2\pi ir\gamma \cdot \xi }𝑑\sigma (\gamma )\right)f_0(r)𝑑\mu (r)$

$={\displaystyle {\int }_0^{\mathrm{\infty }}}{\widehat{\sigma }}_{d-1}(r\xi )f_0(r)𝑑\mu (r)$

$={\displaystyle {\int }_0^{\mathrm{\infty }}}2\pi {(r|\xi |)}^{-\frac{d}{2}+1}{J}_{\frac{d}{2}-1}(2\pi r|\xi |)f_0(r)𝑑\mu (r)$

$=2\pi {|\xi |}^{-\frac{d}{2}+1}{\displaystyle {\int }_0^{\mathrm{\infty }}}{r}^{-\frac{d}{2}+1}{J}_{\frac{d}{2}-1}(2\pi r|\xi |)f_0(r)𝑑\mu (r).$