\phi(R_\theta(\cdot))\rangle = \langle f, \phi \rangle$ for every test function $\phi \in {\cal{S}}(\R^{2n})$ and every $\theta\in [0,2\pi].$ If $f$ and $g$ are $R_\theta$-invariant distributions, and $\langle f, \phi \rangle=\langle g, \phi \rangle$ for any test function $\phi$ that is invariant with respect to all $R_\theta,$ then $f=g.$ This follows from the observation that the value of an $R_\theta$-invariant distribution on a test function $\phi$ does not change if $\phi$ is replaced by the function $\frac1{2\pi}\int_0^{2\pi} \phi(R_\theta(\cdot))d\theta.$

\[r-s\] Let $f\in C_c(S^{2n-1})$ be an even function. Then the Fourier transform of $f\cdot r^{-2n+2}$ is a continuous function on the sphere extended to a homogeneous function of degree -2 on the whole $\R^{2n}.$ Moreover, on the sphere this function is equal (up to a constant) to the complex spherical Radon transform of $f$: for any $\xi\in S^{2n-1},$ $${\cal{R}}_cf(\xi) = \frac1{2\pi} \left(f\cdot r^{-2n+2}\right)^\wedge(\xi).$$

Let $\phi \in {\cal{S}}(\R^{2n})$ be any even $R_\theta$-invariant test function. Then $\hat{\phi}$ is also an even $R_\theta$-invariant test function. By Lemma \[const\], for any $\xi\in S^{2n-1},$ $$\label{eq1}
 \int_{H_\xi^\bot} \hat{\phi}(x) dx = 
\int_{S^{2n-1}\cap H_\xi^\bot} \left(\int_0^\infty r\hat{\phi}(r\theta)dr \right) d\theta =
2\pi \int_0^\infty r\hat{\phi}(r\xi)dr.$$ By Proposition \[f-r\] and Lemma \[self-dual\], $$\langle \left(f\cdot r^{-2n+2}\right)^\wedge, \phi \rangle = \int_{\R^{2n}} |x|_2^{-2n+2} f(x/|x|_2) \hat{\phi}(x) dx$$ $$= \int_{S^{2n-1}} f(\xi) 