Explanation of an answer in “Fourier transform of a radial function”.

Question

I was working on showing that if $f \in L^1(\mathbb{R}^d)$ is a radial function, then so if its Fourier transform. I had a difficult time proving that, and sought help on MSE. Something had already asked a very similar question over here: Fourier transform of a radial function.

Now, one of the proofs is this one:

So suppose $f \in L^1(\mathbb{R}^d)$ and $f$ is radial. Fix an orthogonal transformation $T$. Then: $$\hat {f} (Tx) = \int_{\mathbb {R}^n} f(t) e^{-i\langle Tx,t \rangle}\,dt \overset{(1)}{=} \int_{\mathbb {R}^n} f(Ts) e^{-i\langle Tx,Ts \rangle}\,ds\overset{(2)}{=} \int_{\mathbb {R}^n} f(s) e^{-i\langle x,s \rangle}\,ds = \hat f (x).$$

I don't understand how the $(1)$ and $(2)$ appear. Can someone explicit better the line of thought given here?

Answer 1

I'd put it like so: Say $T$ is orthogonal. This means that $T^*=T^{-1}$ and it also implies that $x\mapsto Tx$ is a measure-preserving transformation on $\Bbb R^n$, which is to say that $\int\phi\circ T=\int\phi$. So $$\begin{aligned}\hat f(Tx)&=\int f(t)e^{-it<t,Tx>}dt\\&= \int f(t)e^{-i<T^{-1}t,x>}dt\\&=\int f(Tt)e^{-i<t,x>}dt\\&=\int f(t)e^{-i<t,x>}dt\\&=\hat f(x).\end{aligned}$$