Fourier transform of a radial function

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.

Solution 1:

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 $L^2.$

Solution 2:

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.