0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $$\begin{aligned}\hat f(Tx)&=\int f(t)e^{-it<t,Tx>}dt\\&= \int f(t)e^{-i<T^{-1}t,x>}dt\\&=\int f(Tt)e^{-i<t,x>}dt\\&=\int f(t)e^{-i<t,x>}dt\\&=\hat f(x).\end{aligned}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.