\int_{S^{2n-1} \cap H_\xi} \phi(\theta)d\theta= \frac1{(2\pi)^2} \int_{S^{2n-1}\cap H_\xi^\bot}
 \left(\phi\cdot r^{-2n+2}\right)^\wedge (\theta) d\theta,$$ and, by Lemma \[const\], the function under the integral in the right-hand side is constant on $S^{2n-1}\cap H_\xi^\bot.$ The value of this constant is the function value at $\xi \in S^{2n-1}\cap H_\xi^\bot.$ Also, recall that $S^{2n-1}\cap H_\xi^\bot$ is the two-dimensional Euclidean unit circle, so $$\int_{S^{2n-1}\cap H_\xi^\bot} \left(\phi\cdot r^{-2n+2}\right)^\wedge (\theta) d\theta =
 2\pi  \left(\phi\cdot r^{-2n+2}\right)^\wedge (\xi).$$

\[self-dual\]The complex spherical Radon transform is self-dual, i.e. for any even functions $f,g\in C_c(S^{2n-1}),$ $$\int_{S^{2n-1}} {\cal{R}}_cf(\xi) g(\xi) d\xi = \int_{S^{2n-1}} f(\theta) {\cal{R}}_c g(\theta) d\theta.$$

By approximation, it is enough to consider the case where $f,g$ are infinitely smooth. For some infinitely smooth even function $h\in C_c(S^{2n-1}),$ we have $g\cdot r^{-2n+2}= \left(h\cdot r^{-2}\right)^\wedge,$ then $ \left(g\cdot r^{-2n+2}\right)^\wedge= (2\pi)^{2n} h\cdot r^{-2}.$ By Lemma \[radon-fourier\] and the spherical Parseval formula (\[parseval\]), $$\int_{S^{2n-1}} {\cal{R}}_cf(\xi) g(\xi) d\xi = \frac1{2\pi} \int_{S^{2n-1}} \left(f\cdot r^{-2n+2}\right)^\wedge(\xi)
(g\cdot r^{-2n+2}) (\xi) d\xi$$ $$= \frac{(2\pi)^{2n}}{2\pi} \int_{S^{2n-1}} \left(f\cdot r^{-2n+2}\right)^\wedge(\xi)
(h\cdot r^{-2})^\wedge (\xi) d\xi$$ $$= \frac1{2\pi} \int_{S^{2n-1}} f(\theta) \left(g\cdot r^{-2n+2}\right)^\wedge(\theta) d\theta = 
\int_{S^{2n-1}} f(\theta) {\cal{R}}_c g(\theta) d\theta.$$

We now prove Lemma \[radon-fourier\] without smoothness assumption. This result is a complex version of [@K1 Lemma 3.7]. We say that a distribution $f$ on $\R^{2n}$ is $R_\theta$-invariant if $\langle f, 