if it is the limit (in the radial metric) of complex radial sums of complex ellipsoids.

To prove this result we need a few lemmas. For fixed $\xi\in S^{2n-1}$, $a>0$, $b>0$, let $E_{a,b}(\xi)$ be an ellipsoid in $\R^{2n}$ with the norm $$\|x\|_{E_{a,b}(\xi)}=\left(\frac{( x, \xi )^2+ (x, \xi^\bot)^2}{a^2}+ \frac{|x|_2-( x, \xi )^2- (x,\xi^\bot)^2}{b^2}\right)^{\frac12}, \ x\in \R^{2n}.$$

Clearly, $E_{a,b}(\xi)$ is a complex ellipsoid. In fact, $(x, \xi )^2+ (x, \xi^\bot)^2 = |(x,\xi)_c|^2$ is the modulus squared of the complex scalar product of $x$ and $\xi$ considered as vectors from $\C^n.$ The latter does not change when $x$ is multiplied by any complex number of modulus 1, which means that the norm of $E_{a,b}(\xi)$ is invariant with respect to all rotations $R_\theta.$

Using the formula for the Fourier transform of powers of the Euclidean norm in $\R^{2n}$ (see [@GS p.192]), we get $$\left(|x|_2^{-2}\right)^\wedge(\theta)=C(n) |\theta|_2^{ -2n+2},$$ where $C(n) = 2^{2n-3}\pi^n \Gamma(n-1).$ By the connection between linear transformations and the Fourier transform, $$\label{linear transform}
\left( \|Tx\|^{-1}\right)^\wedge(y)=|\det T|^{-1}\left( \|x\|^{-1} \right)^\wedge((T^{*})^{-1} y)$$ one can easily compute the following:

\[Lem:FTellipsoid\] For all $\theta\in S^{2n-1}$, $$\left(   \|x\|^{-2}_{E_{a,b}(\xi)}\right)_x^\wedge(\theta) = \frac{C(n)}{a^{2n-4}}
\|\theta\|_{E_{b,a}(\xi)}^{-2n+2}.$$

By (\[linear transform\]) with $T$ being the composition of a rotation and a diagonal operator, $$\begin{aligned}
\left(   \|x\|^{-2}_{E_{a,b} (\xi)}\right)_x^\wedge(\theta)  