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I am trying to understand the following computation for the Fourier Transform of a Radial Function on $\mathbb R^n$. I shall ask questions in-line.

Suppose $n\ge 2$, $f\in L^1(\Bbb R^n)$, and $f(x) = \psi(|x|)$ for all $x\in \Bbb R^n$, for some $\psi:[0,\infty)\to \Bbb C$. We know that if $f\in L^1(\Bbb R^n)$, then $$\int_{\Bbb R^n} f \, d\mathcal L^n = \int_{S^{n-1}} \left(\int_0^\infty f(rx) r^{n-1}\, dr \right)\, d\sigma^{n-1}(x) \tag{1}$$ Fix $e\in S^{n-1}$ and let $S_\theta = \{x\in S^{n-1}: e\cdot x = \cos\theta\}$ for $0\le \theta\le \pi$. The set $S_\theta$ is a $(n-2)$ dimensional sphere of radius $\sin\theta$, so $$\sigma_{\sin\theta}^{n-2}(S_\theta) = b(n) (\sin\theta)^{n-2} \tag{1.1}$$

  1. What does $(1.1)$ mean, and where does it come from?

where $b(n) = \sigma^{n-2}(S^{n-2})$. Then for $g\in L^1(S^{n-1})$, $$\int_{S^{n-1}} g\, d\sigma^{n-1} = \int_0^{\pi}\left( \int_{S_\theta} g(x)\, d\sigma_{\sin\theta}^{n-2}(x) \right)\, d\theta \tag{2}$$

  1. Where does $(2)$ come from?

Applying $(1)$ and Fubini's theorem, $$\hat f(re) = \int f(y) e^{-2\pi ir e\cdot y}\, dy =\int_0^{\infty} \psi(s) s^{n-1}\left(\int_{S^{n-1}} e^{-2\pi irse\cdot x}\, d\sigma^{n-1}(x) \right) \, ds$$

  1. I can see that $$\int f(y) e^{-2\pi ir e\cdot y}\, dy = \int_{S^{n-1}} \left( \int_0^{\infty} \psi(s)e^{-2\pi irse\cdot x} s^{n-1} \, ds\right) \,\, d\sigma^{n-1}(x)$$ Why is Fubini's theorem applicable? We need it to show that $$\int_{S^{n-1}} \left( \int_0^{\infty} \psi(s)e^{-2\pi irse\cdot x} s^{n-1} \, ds\right) \,\, d\sigma^{n-1}(x)=\int_0^{\infty} \psi(s) s^{n-1}\left(\int_{S^{n-1}} e^{-2\pi irse\cdot x}\, d\sigma^{n-1}(x) \right) \, ds$$

The inside integral can be computed with the help of $(2)$, since $e^{-2\pi irse\cdot x}$ is constant in $S_\theta$: $$\int_{S^{n-1}} e^{-2\pi irse\cdot x}\, d\sigma^{n-1}(x) = \int_0^\pi e^{-2\pi irs \cos\theta} \sigma^{n-2}_{\sin\theta} (S_\theta)\, d\theta = b(n)\int_0^\pi e^{-2\pi irs \cos\theta} (\sin\theta)^{n-2}\, d\theta$$

  1. I believe understanding the above computation is related to my first two questions in the post, which is why I'm stuck here.

Changing variable $\cos\theta\mapsto -t$ and introducing for $m > -1/2$ the Bessel functions $J_m:[0,\infty)\to \Bbb R$, we obtain $$\int_{S^{n-1}} e^{-2\pi irse\cdot x}\, d\sigma^{n-1}(x) = b(n)\int_{-1}^1 e^{2\pi irst} (1-t^2)^{(n-3)/2}\, dt = c(n)(rs)^{-(n-2)/2} J_{(n-2)/2} (2\pi rs)$$ leading to the formula for the Fourier transform of the radial function $f$: $$\hat f(x) = c(n) |x|^{-(n-2)/2} \int_0^\infty \psi(s) J_{(n-2)/2} (2\pi |x|s) s^{n/2}\, ds$$

Voilà! The computation finally ends.

Reference: Fourier Analysis and Hausdorff Dimension by Pertti Mattila.

stoic-santiago
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1 Answers1

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Throughout this posing, we denote by $\mathbf{e}_k$, $1\leq k\leq n$ the canonical basic vectors in $\mathbb{R}^n$.

(1) and (2): The key is to give an explicit representation for $\boldsymbol{x} \in\mathbb{R}^n\setminus\{{\bf 0}\}$ in polar coordinates in terms of the angle between ${\bf x}$ and ${\bf e}_n$, and the orthoganal projection of $\boldsymbol{x}$ onto the subspace $\mathbb{R}^{n-1}\times\{0\}$. Let $\varphi_{n-1}\in[0,\pi]$ be the angle between $\boldsymbol{x}$ and ${\bf e}_n$,, $\rho=|\boldsymbol{x}|$, and let $P$ be the orthogonal projection from $\mathbb{R}^n$ onto $\mathbb{R}^{n-1}\times\{0\}$. Then, $\boldsymbol{x}\cdot{\bf e}_n=\rho \cos\varphi_{n-1}$ and ${\bf x}\cdot P{\bf x}=\rho \sin\varphi_{n-1}|P{\bf x}|$; hence, \begin{align} {\bf x}=P{\bf x} + \rho\cos\varphi_{n-1}{\bf e}_n= \rho\sin\varphi_{n-1} \tfrac{1}{|P{\bf x}|}P{\bf x} + \rho\cos\varphi_{n-1}{\bf e}_n. \end{align} Starting with $n=2$, and proceeding by induction we obtain the following parameterization $\Phi$ of points in $\mathbb{R}^n$ \begin{align} x_n&=\rho\cos\varphi_{n-1}, \quad x_k=\rho\Big(\prod^{n-1}_{j=k}\sin\varphi_j\Big) \, \cos\varphi_{k-1},\quad 2<k\leq n-1,\\ x_2&=\rho\prod^{n-1}_{j=1}\sin\varphi_j, \quad x_1=\rho\Big(\prod^{n-1}_{j=2}\sin\varphi_j\Big)\,\cos\varphi_1 \end{align} where $\rho\geq0$ and $(\varphi_1,\ldots,\varphi_{n-1})\in[0,2\pi]\times[0,\pi]^{n-2}$. It is easy to check that the parameterization $\Phi:(0,\infty)\times (0,2\pi)\times(0,\pi)^{n-2}\rightarrow\mathbb{R}^n\setminus(\{0\}\times\mathbb{R}^{n-2}_+\times\mathbb{R})$ defined above is a diffeomorphism, and that \begin{align} |\det(\Phi')|=\rho^{n-1}\,\prod^{n-1}_{j=2} \sin^{j-1}\varphi_j \end{align} If $\rho=1$, we obtain a representation of the surface area $d\sigma_{n-1}$ on $\mathbb{S}^{n-1}\setminus (\{0\}\times\mathbb{R}^{n-2}_+\times\mathbb{R})$ in terms of the parameters $(\varphi_1,\ldots,\varphi_{n-1})\in(0,\pi)^{n-2}\times(0,2\pi)$: \begin{align} \sigma_{n-1}(d\,\varphi_1,\ldots,d\,\varphi_{n-1})&= \sin^{n-2}\varphi_{n-1}\cdot\ldots\cdot\sin\varphi_2\, \,d\varphi_1\cdots d\varphi_{n-1}\nonumber\\ &=\sin^{n-2}\varphi_{n-1}\cdot \sigma_{n-2}(d\,\varphi_1,\ldots,d\,\varphi_{n-2}).\label{polar-volume} \end{align}

(3) and (4): Now, if $f$ is an integrable radial function on $\mathbb{R}^n$, then $\widehat{f}(\boldsymbol{t})=\widehat{f}(|\boldsymbol{t}|\boldsymbol{e}_n)$. Hence \begin{align} \widehat{f}(\boldsymbol{t})&=\int e^{-2\pi i \boldsymbol{x}\cdot\boldsymbol{t}} f(|\boldsymbol{x}|)\,d\boldsymbol{x}=\int e^{-2\pi i |\boldsymbol{t}|x_n} f(|\boldsymbol{x}|)\,d\boldsymbol{x}=\\ &= \int^\infty_0f(\rho)\rho^{n-1} \Big(\int_{S_{n-1}}e^{-2\pi i|\boldsymbol{t}|\rho\cos\phi_{n-1}} \prod^{n-1}_{j=2}\sin^{j-1} \phi_j \,d\phi_{n-1}\ldots d\phi_1\Big) \,d\rho\\ &=\sigma_{n-2}\int^\infty_0 f(\rho)\rho^{n-1}\Big(\int^\pi_0 e^{-2\pi i|\boldsymbol{t}|\rho\cos\theta}\sin^{n-2}\theta\,d\theta\Big)\,d\rho\tag{*}\label{bessel} \end{align} The integral inside \eqref{bessel} is related to Bessel functions, and has been discussed in MSE before (see for example the Lemma in this posting) and we summarized this as

Lemma For $p>-1$ \begin{align} \int^\pi_0e^{-it\cos\theta}\sin^{2p+1}\theta\,d\theta = \frac{\Gamma(p+1)\sqrt{\pi}}{(t/2)^{p+\tfrac12}}J_{p+\tfrac12}(t)=\int^1_{-1}(1-x^2)^p e^{-xti}\,dx\tag{0}\label{zero} \end{align} where $J_m$ is the Bessel function of order $m$.

Finally, the Lemma above and the identity $\sigma_{n-2}=\frac{2\pi^{(n-1)/2}}{\Gamma\big(\frac{n-1}{2}\big)}$ yields

Theorem (Hankel) If $f$ is a radial function in $\mathcal{L}_1(\mathbb{R}^n,\lambda_n)$, then \begin{align*} \widehat{f}(\boldsymbol{t}) =\frac{2\pi}{|\boldsymbol{t}|^{(n/2) -1}} \int^\infty_0 f(\rho)\rho^{n/2} J_{(n/2)-1}(2\pi|\boldsymbol{t}|\rho)\,d\rho \end{align*} where $J_m$ is the Bessel function of order $m$

Mittens
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