Fourier transform of a radial function

Question

Consider a function $f \in L^2(\mathbb{R}^n)$ such that $f$ is radial. My question is, is the Fourier transform $\hat{f}(\xi)$ automatically radial (I can see it is even in each variable $x_i$), or we need some conditions on $f$? Thanks for your help.

Answer 1

It is: you can decompose the integral as follows: write $x= r n$, where $n$ is a unit vector, and then $$ \int_{\mathbb{R}^n} e^{i\xi \cdot x} f(\lvert x \rvert) d^n x = \int_{r=0}^{\infty} f(r) \left( \int_{|n|=1} e^{i(r \xi) \cdot n} \, dn \right) r^{n-1} \, dr, $$ by considering the volume as composed of spherical shells.

Now, the inner integral is in fact only a function of $r\xi$. You can take my word for it, or evaluate it: setting $\rho=\lvert \xi \rvert$, then we can write $\xi \cdot x = r\rho \cos{\theta} $, where $0 \leqslant \theta \leqslant \pi$, and then integrate over the remaining angles on the unit sphere. This gives $$ S_{n-2} \int_0^{\pi} e^{i r \rho \cos{\theta}} \sin^{n-2}{\theta} d\theta, $$ where $S_{n-2}$ is the area of the $(n-2)$-dimensional sphere. The remaining integral can be done in terms of Bessel functions; I have a note on it here (last section), although it's not very detailed.

Answer 2

It indeed will be radial. It is called a Hankel transform:

Where $J_{\nu}$ is a bessel function of the first kind such that $\nu>-1/2$. For it to match fourier transform you use $J_0$ in 2 dimensional radial functions. For three dimensional radial functions, you use spherical bessel functions instead. https://en.wikipedia.org/wiki/Hankel_transform

There is a nice table on wiki. It is quite difficult to do this transform.

If you are interested in how fourier transform connects to Hankel transform, the 4th part derives it (quite short): http://math.arizona.edu/~faris/methodsweb/hankel.pdf

Answer 3

Preliminaries: i) $f$ is radial iff $f\circ T = f$ for every orthogonal transformation $T$ on ${R}^n.$ ii) An orthogonal transformation $T$ preserves the inner product: $\langle Tx,Ty \rangle = \langle x,y \rangle$ for all $x,y \in \mathbb {R}^n.$ iii) If $T$ is orthogonal, then $|\det J_T|=1,$ where $J_T$ is the Jacobian matrix of $T.$

So suppose $f\in L^1(\mathbb {R}^n)$ and $f$ is radial. Fix an orthogonal transformation $T.$ Then

$$\hat {f} (Tx) = \int_{\mathbb {R}^n} f(t) e^{-i\langle Tx,t \rangle}\,dt = \int_{\mathbb {R}^n} f(Ts) e^{-i\langle Tx,Ts \rangle}\,ds= \int_{\mathbb {R}^n} f(s) e^{-i\langle x,s \rangle}\,ds = \hat f (x).$$

Thus $\hat f$ is radial as desired.

That was for $L^1,$ but the question was about $L^2$ as @AdamHughes reminded me. See the comments below to get the result for $L^2.$