between the Fourier transform of distributions and linear transformations, the Fourier transform of $f\cdot r^{-p}$ is also invariant with respect to all $R_\theta.$ Recall that the two-dimensional space $H_\xi^\bot$ is spanned by vectors $\xi$ and $\xi^\bot$ (see the Introduction). Every vector in $S^{2n-1}\cap H_\xi^\bot$ is the image of $\xi$ under one of the coordinate-wise rotations $R_\theta$, so the Fourier transform of $f\cdot r^{-p}$ is a constant function on $S^{2n-1}\cap H_\xi^\bot.$

The following connection between the Fourier and Radon transforms is well-known; see for example [@K1 Lemma 3.24].

\[f-r\] Let $1\le k <n,$ and let $\phi\in {\cal{S}}(\R^n)$ be an even test function. Then for any $(n-k)$-dimensional subspace $H$ of $\R^n$ $$\int_H \phi(x) dx = \frac1{(2\pi)^k} \int_{H^\bot} \hat{\phi}(x) dx.$$

We also use the spherical version of Proposition \[f-r\]; see [@K1 Lemma 3.25].

\[perp\] Let $\phi$ be an even infinitely smooth function on $S^{n-1},$ let $0<k<n,$ and let $H$ be an arbitrary $(n-k)$-dimensional subspace of $\R^n.$ Then $$\int_{S^{n-1} \cap H} \phi(\theta)d\theta =
 \frac1{(2\pi)^k} \int_{S^{n-1}\cap H^\bot} \left(\phi\cdot r^{-n+k}\right)^\wedge(\theta) d\theta.$$

Let us translate the latter fact to the complex situation.

\[radon-fourier\] Let $\phi\in C_c(S^{2n-1})$ be an even infinitely smooth function. Then for every $\xi\in S^{2n-1}$ $${\cal{R}}_c\phi(\xi)= \frac1{2\pi} \left(\phi\cdot r^{-2n+2}\right)^\wedge (\xi).$$

By Proposition \[perp\], $${\cal{R}}_c\phi(\xi) = 