\frac{1}{4\pi^2(n-1)} \left(\|x\|_L^{-2n+2}\right)^\wedge(\xi).$$ Both sides of the latter equality are even homogeneous functions of degree -2, so these functions are equal as distributions on the whole of $\R^{2n}.$ Since the Fourier transform of even distributions is self-invertible (up to a constant), we get $$\left(\|\cdot\|_{I_c(L)}^{-2}\right)^\wedge = \frac{(2\pi)^{2n}}{4\pi^2(n-1)} \|\cdot\|_L^{-2n+2} >0,$$ so the distribution $\|\cdot\|_{I_c(L)}$ is positive definite. Moreover, if the Fourier transform of $\|\cdot\|_K^{-2}$ is an even strictly positive $R_\theta$-invariant function on the sphere, then one can use the latter equality to construct a complex star body $L$ such that $K=I_c(L).$ This connection holds for arbitrary complex intersection bodies, as shown in the following theorem.

\[posdef\] An origin symmetric complex star body $K$ in $\R^{2n}$ is a complex intersection body if and only if the function $\|x\|_K^{-2}$ represents a positive definite distribution on $\R^{2n}.$

Suppose that $K$ is a complex intersection body with the corresponding measure $\mu.$ To prove that $\|\cdot\|_K^{-2}$ is a positive definite distribution, it is enough to show that $\langle (\|\cdot\|_K^{-2})^\wedge, \phi\rangle \ge 0$ for every even $R_\theta$-invariant non-negative test function $\phi.$ By Definition \[int\] and Proposition \[f-r\], $$\langle (\|\cdot\|_K^{-2})^\wedge, \phi\rangle = \int_{\R^{2n}} \|x\|_K^{-2}\hat{\phi}(x) dx$$ $$= \int_{S^{2n-1}} \|\theta\|_K^{-2} \left(\int_0^\infty r^{2n-3} \hat{\phi}(r\theta) dr \right) d\theta$$ $$= \int_{S^{2n-1}} {\cal{R}}_c\left(\int_0^\infty r^{2n-3} \hat{\phi}(r\cdot) dr\right)(\xi) 