Does the Fourier transform commute with the spherical average operator? $\newcommand{\Rcal}{\mathcal R}$For $f\colon \mathbb R^d\to \mathbb R$, write $x\in \mathbb R^d$ as $x=r\omega$, with $\omega\in\mathbb S^{d-1}$, and define 
$$
R f(r):=\int_{\mathbb S^{d-1}}f(r\omega)\, d\omega, $$ 
where $d\omega$ is the normalized surface measure on the sphere. Using this, define a function $\Rcal f\colon \mathbb R^d\to \mathbb R$ by 
$$
\Rcal f(x)=Rf(|x|).$$ 

Is it true that $\mathcal F \mathcal R f = \mathcal R \mathcal F f$, where $$\mathcal F g (\xi)=\int_{\mathbb R^d} g(y)\exp(-i y\cdot \xi)\, dy?$$

 A: I think its true: Let us define 
$$ \widehat{d\sigma}(x) := \int\limits_{S^{d}} e^{ix\omega}~d\omega$$
with the normalized surface measure $d\omega$ on $S^{d}$ which denotes the unit sphere in $\mathbb{R}^{d}$(I think there is a little typo at the beginning of your question concerning $S^{d}$ and $S^{d-1}$). Let us note that this defines a radial function on $\mathbb{R}^{d}$ and obviously we have
$ \int\limits_{S^{d}}e^{ix\omega}~d\omega = \int\limits_{S^{d}}e^{-ix\omega}~d\omega. $
Then
\begin{align*}
\mathcal{F}(\mathcal{R}(f))(\xi) &= \int\limits_{\mathbb{R}^{d}}e^{-ix\xi}\mathcal{R}(f)(x)~dx =
\int\limits_{\mathbb{R}^{d}}e^{-ix\xi}\int\limits_{S^{d}}f(|x|\omega)~d\omega~dx \\
&=
\int\limits_{S^{d}}\int\limits_{\mathbb{R}^{d}}e^{-ix\xi}f(|x|\omega)~dx~d\omega 
= \int\limits_{S^{d}}\int\limits^{\infty}_{0}r^{N-1}\int\limits_{S^{d}}e^{-i(r\eta)\xi}f(r\omega)~d\eta~dr~d\omega\\
&=
 \int\limits_{S^{d}}\int\limits^{\infty}_{0}r^{N-1}f(r\omega)\int\limits_{S^{d}}e^{-i(r\eta)\xi}~d\eta~dr~d\omega
=
\int\limits_{S^{d}}\int\limits^{\infty}_{0}r^{N-1}f(r\omega)\widehat{d\sigma}(r\xi)~dr~d\omega \\
&=
\int\limits_{\mathbb{R}^{d}}f(x)\widehat{d\sigma}(|x|\xi)~dx,
\end{align*}
where we have used polar coordinates in the third and fifth line. Also interchanging the order of integration in the second line should be ok. Now note that
\begin{align*}
\mathcal{R}(\mathcal{F}(f))(\xi) &= \int\limits_{S^{d}}\mathcal{F}(f)(|\xi|\omega)~d\omega 
=
\int\limits_{S^{d}}\int\limits_{\mathbb{R}^{d}}e^{-ix(|\xi|\omega)}f(x)~dx~d\omega \\
&=
\int\limits_{\mathbb{R}^{d}}f(x)\int\limits_{S^{d}}e^{-ix(|\xi|\omega)}~dx~d\omega \\
&=
\int\limits_{\mathbb{R}^{d}}f(x)\widehat{d\sigma}(|\xi|x)~dx,
\end{align*}
and the conclusion follows, since $\widehat{d\sigma}$ is radial.
