Theorem 2.9.2603F670D5FB5A

Proof. On $H$ we use the Fourier transform defined by the Euclidean inner product $\operatorname{Re}\langle \cdot ,\cdot \rangle $ and the measure $m_ H$; in an orthonormal frame it is the transform of subsection 2.9.1 on $\mathbb R^{2n-2}$. For $y\in \mathbb C^ n$ and $\xi \in H$ we have $\operatorname{Re}\langle P_ Hy,\xi \rangle =\operatorname{Re}\langle y,\xi \rangle $, because $y-P_ Hy$ is orthogonal to $H$. Hence the Fourier transform of $(P_ H)_*\mu $ at $\xi \in H$ is $\int \mathrm e^{-2\pi \mathrm i\operatorname{Re}\langle P_ Hy,\xi \rangle }\, \mathrm d\mu (y)=\widehat\mu (\xi )$; that is, it is the restriction of $\widehat\mu $ to $H$. Apply lemma 2.9.24 to the continuous function $|\widehat\mu |^2$, and then theorem 2.9.6 with $d=2n$ and $s=2n-2$, for which $|\xi |^{s-d}=|\xi |^{-2}$. We obtain

The inner integral is a Borel function of $H$, by the frame construction in subsection 2.9.17; hence it is finite for $\nu $-almost every $H$. For each such $H$, we apply proposition 2.9.5 on $\mathbb C^{n-1}=\mathbb R^{2n-2}$ to the image of $\mu $ under $T_ H^{-1}P_ H$. Since $T_ H$ preserves $\operatorname{Re}\langle \cdot ,\cdot \rangle $, the Fourier transform of this image is $w\mapsto \widehat\mu (T_ Hw)$, whose squared $L^2$ norm is $\int _ H|\widehat\mu |^2\, \mathrm d m_ H<\infty $. Hence the image has a density in $L^1\cap L^2$, and composing it with $T_ H^{-1}$ we obtain a density $g_ H$ of $(P_ H)_*\mu $ with respect to $m_ H$, with $\| g_ H\| _2^2=\int _ H|\widehat\mu |^2\, \mathrm d m_ H$. In particular, $H\mapsto \| g_ H\| _2^2$ is Borel, being equal to this Fourier integral, so no measurable choice of the densities themselves is needed. Integrating over $H$ proves the identity.

To compute $\beta _ n$, recall from (2.9.8) that $s_{k-1}=2\pi ^{k/2}/\Gamma (k/2)$ for every $k\geq 1$. Hence $a_ n=\Gamma (n)/(\pi \Gamma (n-1))$, while $b_{2n,2n-2}=\pi ^{n-2}/\Gamma (n-1)$, and therefore $\beta _ n=\Gamma (n)/\pi ^{n-1}$. $\square$