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.

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.
