d\mu(\xi)$$ $$= \int_{S^{2n-1}} \left( \int_{H_\xi} \hat{\phi}(x) dx \right) d\mu(\xi)$$ $$= (2\pi)^{2n-2} \int_{S^{2n-1}} \left( \int_{H_\xi^\bot} \phi(x) dx \right) d\mu(\xi) \ge 0.$$

Now suppose that $\|\cdot\|_K^{-2}$ is a positive definite distribution. By Proposition \[posdef\], there exists a finite Borel measure $\mu$ on $S^{2n-1}$ such that for any even test function $\phi$ $$\label{eq2}
\int_{\R^{2n}} \|x\|_K^{-2} \phi(x)\ dx = \int_{S^{2n-1}} \left(\int_0^\infty t \hat\phi(t\xi) dt \right) d\mu(\xi).$$ Recall that $K$ is $R_\theta$-invariant, so we can assume that $\mu$ is $R_\theta$-invariant and the latter equality holds only for even $R_\theta$-invariant test functions $\phi.$ For each such test function, we have by (\[eq1\]) that $$\int_0^\infty t \hat{\phi}(t\xi) dt  = \frac1{2\pi}\int_{H_\xi^\bot} \hat{\phi}(x) dx.$$ Using this and Proposition \[f-r\] and then writing the interior integral in polar coordinates, we get that the right-hand side of (\[eq2\]) is equal to $$\frac1{2\pi} \int_{S^{2n-1}} \left( \int_{H_\xi^\bot} \hat{\phi}(x) dx \right) d\mu(\xi)=
2\pi \int_{S^{2n-1}} \left( \int_{H_\xi} \phi(x) dx \right) d\mu(\xi)$$ $$= 2\pi  \int_{S^{2n-1}} {\cal{R}}_c\left(\int_0^\infty r^{2n-3} \phi(r\cdot) dr \right)(\xi)\ d\mu(\xi).$$ Writing the left-hand side of (\[eq2\]) in polar coordinates we get $$\int_{S^{2n-1}} \|\theta\|_K^{-2} \left(\int_0^\infty r^{2n-3} \phi(r\theta) dr \right) d\theta$$ $$\label{eq3}
=2\pi  \int_{S^{2n-1}} {\cal{R}}_c\left(\int_0^\infty r^{2n-3} \phi(r\cdot) dr \right)(\xi)\ d\mu(\xi)$$ for any even $R_\theta$-invariant test function $\phi.$ Now let $\phi(x)=u(r)v(\theta)$ for every $x\in \R^{2n},$ where $x=r\theta,\ r\in 