# The Fourier transform of spherical surface measure and radial functions

Jordan Bell
August 24, 2015

## 1 Notation

For a topological space $X$, we denote by $\mathscr{B}_{X}$ the Borel $\sigma$-algebra of $X$. Let $\rho_{d}$ be the Euclidean metric on $\mathbb{R}^{d}$ and let $m_{d}$ be Lebesgue measure on $\mathbb{R}^{d}$.

## 2 Polar coordinates

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

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

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

 $S(E)=\left\{x\in\mathbb{R}^{d}:\frac{x}{|x|}\in E,0<|x|<1\right\}.$

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

 $S\left(\bigcup E_{\alpha}\right)=\bigcup S(E_{\alpha}),\qquad S\left(\bigcap E% _{\alpha}\right)=\bigcap S(E_{\alpha}),$

and for $E,F\in\mathscr{P}(S^{d-1})$,

 $S(E\setminus F)=S(E)\setminus S(F).$
###### Lemma 1.
 $S(\mathscr{B}_{S^{d-1}})\subset\mathscr{B}_{\mathbb{R}^{d}}.$

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

 $\sigma_{d-1}(E)=d\cdot m_{d}(S(E)),\qquad E\in\mathscr{B}_{S^{d-1}}.$

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

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

The following is proved in Stein and Shakarchi.11 1 Elias M. Stein and Rami Shakarchi, Real Analysis, p. 280, Chapter 6, Theorem 3.4.

###### Theorem 2.

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

 $\gamma\mapsto\int_{0}^{\infty}f^{\gamma}(r)d\mu(r)$

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

 $\int_{\mathbb{R}^{d}}f(x)dm_{d}(x)=\int_{S^{d-1}}\left(\int_{0}^{\infty}f^{% \gamma}(r)d\mu(r)\right)d\sigma(\gamma).$

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

 $f_{r}(\gamma)=f(r\gamma),\qquad\gamma\in S^{d-1}.$
###### Theorem 3.

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

 $r\mapsto\int_{S^{d-1}}f_{r}(\gamma)d\sigma(\sigma)$

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

 $\int_{\mathbb{R}^{d}}f(x)dm_{d}(x)=\int_{0}^{\infty}\left(\int_{S^{d-1}}f_{r}(% \gamma)d\sigma(\gamma)\right)d\mu(r).$

## 3 The Fourier transform of spherical surface measure

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

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

One checks that $J_{\nu}$ satisfies

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

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

 $\displaystyle\gamma_{1}$ $\displaystyle=\cos\phi_{1}$ $\displaystyle\gamma_{2}$ $\displaystyle=\sin\phi_{1}\cos\phi_{2}$ $\displaystyle\gamma_{3}$ $\displaystyle=\sin\phi_{1}\sin\phi_{2}\cos\phi_{3}$ $\displaystyle\cdots$ $\displaystyle\gamma_{d-1}$ $\displaystyle=\sin\phi_{1}\sin\phi_{2}\sin\phi_{3}\cdots\sin\phi_{d-2}\cos\phi% _{d-1}$ $\displaystyle\gamma_{d}$ $\displaystyle=\sin\phi_{1}\sin\phi_{2}\sin\phi_{3}\cdots\sin\phi_{d-2}\sin\phi% _{d-1},$

with

 $0\leq\phi_{1},\ldots,\phi_{d-2}\leq\pi,\qquad 0\leq\phi_{d-1}\leq 2\pi,$

is

 $J=\sin^{d-2}\phi_{1}\sin^{d-3}\phi_{2}\cdots\sin^{2}\phi_{d-3}\sin\phi_{d-2}.$

Then, for $\xi=(\xi_{1},0,\ldots,0)$, $\xi_{1}\neq 0$,

 $\displaystyle\widehat{\sigma}_{d-1}(\xi)$ $\displaystyle=\int_{S^{d-1}}e^{-2\pi i\gamma\cdot\xi}d\sigma(\gamma)$ $\displaystyle=\int_{\phi_{1}=0}^{\pi}\int_{\phi_{2}=0}^{\pi}\cdots\int_{\phi_{% d-2}=0}^{\pi}\int_{\phi_{d-1}=0}^{2\pi}e^{-2\pi i\xi_{1}\cos\phi_{1}}Jd\phi_{d% -1}d\phi_{d-2}\cdots d\phi_{2}d\phi_{1}$ $\displaystyle=2\pi\cdot\int_{\phi_{1}=0}^{\pi}e^{-2\pi i\xi_{1}\cos\phi_{1}}% \sin^{d-2}\phi_{1}d\phi_{1}\cdot\prod_{j=2}^{d-2}\int_{\phi_{j}=0}^{\pi}\sin^{% d-j-1}\phi_{j}d\phi_{j}.$

We work out that

 $\int_{0}^{\pi}\sin^{k}tdt=\frac{\sqrt{\pi}\Gamma\left(\frac{k+1}{2}\right)}{% \Gamma\left(\frac{k+2}{2}\right)}.$

This gives

 $\displaystyle\prod_{j=2}^{d-2}\int_{\phi_{j}=0}^{\pi}\sin^{d-j-1}\phi_{j}d\phi% _{j}$ $\displaystyle=\prod_{j=2}^{d-2}\frac{\sqrt{\pi}\Gamma\left(\frac{d-j}{2}\right% )}{\Gamma\left(\frac{d-j+1}{2}\right)}=\pi^{\frac{d-3}{2}}\frac{\Gamma\left(% \frac{2}{2}\right)}{\Gamma\left(\frac{d-1}{2}\right)}=\frac{\pi^{\frac{d-3}{2}% }}{\Gamma\left(\frac{d-1}{2}\right)}.$

With this we have, for $\xi=(\xi_{1},0,\ldots,0)$, $\xi_{1}\neq 0$,

 $\widehat{\sigma}_{d-1}(\xi)=2\pi\frac{\pi^{\frac{d-3}{2}}}{\Gamma\left(\frac{d% -1}{2}\right)}\int_{0}^{\pi}e^{-2\pi i\xi_{1}\cos t}\sin^{d-2}tdt.$

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

 $\displaystyle\int_{0}^{\pi}e^{is\cos t}\sin^{d-2}tdt$ $\displaystyle=\int_{0}^{\pi}e^{is\cos t}(1-\cos^{2}t)^{\frac{d-2}{2}}dt$ $\displaystyle=\int_{1}^{-1}e^{isx}(1-x^{2})^{\frac{d-2}{2}}\frac{-dx}{\sqrt{1-% x^{2}}}$ $\displaystyle=\int_{-1}^{1}e^{isx}(1-x^{2})^{\frac{d}{2}-1-\frac{1}{2}}dx$ $\displaystyle=\frac{\Gamma\left(\frac{d}{2}-\frac{1}{2}\right)\sqrt{\pi}}{% \left(\frac{s}{2}\right)^{\frac{d}{2}-1}}J_{\frac{d}{2}-1}(s).$

Thus, taking $s=-2\pi\xi_{1}$,

 $\displaystyle\widehat{\sigma}_{d-1}(\xi)$ $\displaystyle=2\pi\frac{\pi^{\frac{d-3}{2}}}{\Gamma\left(\frac{d-1}{2}\right)}% \frac{\Gamma\left(\frac{d}{2}-\frac{1}{2}\right)\sqrt{\pi}}{\left(\frac{-2\pi% \xi_{1}}{2}\right)^{\frac{d}{2}-1}}J_{\frac{d}{2}-1}(-2\pi\xi_{1})$ $\displaystyle=2\pi\cdot(-\xi_{1})^{-\frac{d}{2}+1}J_{\frac{d}{2}-1}(-2\pi\xi_{% 1}).$

For $\xi_{1}<0$ 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 $\xi\in\mathbb{R}^{d}$. Let $T:\mathbb{R}^{d}\to\mathbb{R}^{d}$ be the rotation that sends $\xi$ ti $(0,\ldots,0,-|\xi|)$. Since $\sigma_{d-1}\circ T=\sigma_{d-1}$ (namely, surface measure $\sigma_{d-1}$ is invariant under rotations),

 $\widehat{\sigma}_{d-1}(\xi)=\widehat{\sigma}_{d-1}((0,\ldots,0,-|\xi|))=2\pi|% \xi|^{-\frac{d}{2}+1}J_{\frac{d}{2}-1}(2\pi|\xi|).$

For real $\nu>-\frac{1}{2}$, we use the following asymptotic formula for $J_{\nu}(s)$:22 2 Elias M. Stein and Rami Shakarchi, Complex Analysis, p. 319, Appendix A.1.

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

We get from this that

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

## 4 The Fourier transform of radial functions

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

 $f(x)=f_{0}(|x|),\qquad x\in\mathbb{R}^{d}.$

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

 $\displaystyle\widehat{f}(\xi)$ $\displaystyle=\int_{\mathbb{R}^{d}}e^{-2\pi ix\cdot\xi}f(x)dx$ $\displaystyle=\int_{0}^{\infty}\left(\int_{S^{d-1}}e^{-2\pi ir\sigma\cdot\xi}f% (r\sigma)d\sigma(\gamma)\right)d\mu(r)$ $\displaystyle=\int_{0}^{\infty}\left(\int_{S^{d-1}}e^{-2\pi ir\gamma\cdot\xi}d% \sigma(\gamma)\right)f_{0}(r)d\mu(r)$ $\displaystyle=\int_{0}^{\infty}\widehat{\sigma}_{d-1}(r\xi)f_{0}(r)d\mu(r)$ $\displaystyle=\int_{0}^{\infty}2\pi(r|\xi|)^{-\frac{d}{2}+1}J_{\frac{d}{2}-1}(% 2\pi r|\xi|)f_{0}(r)d\mu(r)$ $\displaystyle=2\pi|\xi|^{-\frac{d}{2}+1}\int_{0}^{\infty}r^{-\frac{d}{2}+1}J_{% \frac{d}{2}-1}(2\pi r|\xi|)f_{0}(r)d\mu(r).$