I was looking at the following lemma from the paper "The weak Harnack inequality for the Boltzmann equation without cut-off" by Cyril Imbert and Luis Silvestre,
I am struggling to prove this lemma.
Thoughts
It seems clear that for each fixed $\sigma$ the integral is over a hyperplane containing zero that is perpendicular to $\sigma$ (this kind of integral transform is known as Radon transform). Thus if we integrate over $\sigma \in \partial B_r$, the domain of integration should be over all of $\mathbb{R}^d$. That said the change of variables and the Jacobian they get is not clear to me.
Any help is appreciated.
Assuming that $r=1$, the left-hand side of (A.5) in the attached image above can be rewritten as $$\tag{1} \iint_{\mathbb R^d\times \mathbb R^d} F(w)\delta(w\cdot \sigma)\delta(\lvert \sigma\rvert -1)\ dwd\sigma,$$ where $\delta$ is the 1-dimensional Dirac generalized function. Since $\delta(\lambda x)=\lambda^{-1}\delta(x)$ for all $\lambda>0$, we have that $$ \delta(w\cdot \sigma) = \frac{1}{\lvert w \rvert}\delta\left(\frac{w}{\lvert w \rvert}\cdot \sigma\right), $$ therefore (1) becomes $$\tag{2}\iint_{\mathbb R^d\times \mathbb R^d} \frac{F(w)}{\lvert w \rvert}\delta(\tfrac{w}{\lvert w \rvert} \cdot \sigma)\delta(\lvert \sigma\rvert -1)\ dwd\sigma.$$ To conclude the proof of (A.5), we observe that, for each fixed $w\ne 0$, $$ \int_{\mathbb R^d} \delta (\tfrac{w}{\lvert w \rvert} \cdot \sigma)\delta(\lvert \sigma\rvert -1)\ d\sigma = \omega_{d-2}.$$ Indeed, this corresponds to integrating along the $d-2$-dimensional sphere obtained by intersecting the hyperplane normal to $w$ with the unit sphere.
And so finally, using Fubini in (2), we obtain $$ \omega_{d-2}\int_{\mathbb R^d} \frac{F(w)}{\lvert w \rvert}\ dw, $$ as we wanted.
The other identities can surely be proven via analogous manipulations, but I must admit that I did not carry out the computations in detail.