Problem: Suppose that the function $f:bR^d obR$ is radial, that is, for any $x,yinbR^d$ with $|x|=|y|$, we have $f(x)=f(y)$. Show that the Fourier transform $calF(xi)$ is also radial.
Proof: For any $xi,etainbR^d$ with $|xi|=|eta|$, there exists an orthogonal matrix $A$ such that $eta=Axi
a eta^T=xi^T A^T$. Thus, $$eexea calF f(eta) &=int_{bR^d} f(x) e^{-i eta^Tx}
d x\ &=int_{bR^d} f(x) e^{-i xi^T A^T x}
d x\ &=int_{bR^d} f(Ay) e^{-i xi^T y}
d y\ & qx{A^TA=E
a A^T=A^{-1}, y=A^T x=A^{-1}x
a x=Ay}\ &=int_{bR^d} f(y) e^{-i xi^Ty}
d yqwz{$f$ is radial}\ &=calF f(xi). eeaeeex$$