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.

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 $L^2.$
