Recall that Fourier transform \(\hat{f} = \mathcal{F}(f)\) of admissible functions \(f\).

Suppose \(f:\mathbb{R}^{d} \rightarrow \mathbb{R}\) is a real-valued radial function, i.e. \(f(x)\) only depends only on \(\|x\|\). In this case, we have that \(\hat{f}\) is also real-valued radial and moreover \[\begin{align} \hat{f}(t)=2 \pi \| t\|^{-\alpha} \int_{0}^{\infty} f(r) J_{\alpha}(2 \pi r \| t\|)r^{\alpha + 1} dr, \end{align}\]

where \(\alpha=d/2-1\) and \(J_{\alpha}\) is the Bessel function of order \(\alpha\).

Relation to the Hankel transform

In terms of the Hankel tranform \(F_\nu\), we have

\[\begin{align} (2 \pi)^{-1}k^{\alpha}\hat{f}(k) = F_{\alpha}(f_1)(2 \pi k), \text{ where } f_{1}(r) = r^{\alpha} f(r) \end{align}\]

Identity with Laplacian

Consider the Laplacian on \(\mathbb{R}^{d}\) given by \[\begin{align} L = \frac{1}{4 \pi^{2}} \sum_{i=1}^{d} \frac{\partial^{2}}{\partial x_{i}^{2}} \end{align}\]

We then have the following identity for any admissible radial function \(f:\mathbb{R}_{\ge 0} \rightarrow \mathbb{R}\). Suppose \(f_1(x) = - \|x\|^{2}f(\|x\|)\), then \[\begin{align} \mathcal{F}(f_1)(x) = L\hat{f}(t). \end{align}\]

The Laplacian on radial functions acts as \[\begin{align} u \mapsto \frac{1}{4 \pi^{2}} u''(\|t\|) + \frac{d-1}{4 \pi^{2} \|t\|} u'(\|t\|). \end{align}\]

Examples

Gaussians

Fourier transform of \(e^{-\pi C \|x\|^{2}}\) is \({C^{-d/2}}e^{- \tfrac{ \pi}{ C} \|x\|^{2}}\). Here \(d\) is the dimension.

Polynomial times Gaussian

Suppose \(p(r^2)\) is a polynomial in the radius and consider \(f(x) = p(\|x\|^2)e^{- \pi C \|x\|^2}\).

Then, we have that \(\hat{f}(x) ={C^{-d/2}} p(-L) e^{- \tfrac{ \pi}{ C} \|x\|^{2}}\).

From here the action of \(p(-L)\) on \(e^{- \tfrac{\pi}{C} \|x\|^2}\) can be calculated recursively.