Well-defined elastic reflection gain and collision identities

Put $c=(v+w)\mathrm{/}2$ and $r=(v-w)\mathrm{/}2$. The formulas in [F1] give $v^{\mathrm{\prime }}=c+r-2(r\cdot n)n$ and $w^{\mathrm{\prime }}=c-r+2(r\cdot n)n$. The map $r\mapsto r-2(r\cdot n)n$ is reflection across the plane perpendicular to $n$: applying it twice returns $r$, and its squared norm is $\mathrm{\mid }r{\mathrm{\mid }}^2$. The center $c$ is fixed. Hence the pair map is an orthogonal involution on ${\mathbb{R}}^6$, its determinant has absolute value one, $v^{\mathrm{\prime }}+w^{\mathrm{\prime }}=v+w$, and $\mathrm{\mid }{v}^{\mathrm{\prime }}{\mathrm{\mid }}^2+\mathrm{\mid }{w}^{\mathrm{\prime }}{\mathrm{\mid }}^2=\mathrm{\mid }v{\mathrm{\mid }}^2+\mathrm{\mid }w{\mathrm{\mid }}^2$.

The defining integral of $\sigma$ has total mass $(4\pi {)}^{-1}(2)(2\pi )=1$. Every nonempty relative-open subset of $S^2$ contains a patch away from a coordinate pole and seam; its preimage contains an open rectangle on which $\sin \theta >0$, so it has positive $\sigma$-measure. Every nonempty open subset of the full parameter space contains a product of two Euclidean balls and such a sphere patch; each factor has positive measure, so $\mu$ has full support. For an elastic pair, set $c=(v+w)\mathrm{/}2$, $r=(v-w)\mathrm{/}2$, and $\tilde{r}=(\tilde{v}-\tilde{w})\mathrm{/}2$. Momentum and squared-speed conservation give the same center and $\mathrm{\mid }\tilde{r}\mathrm{\mid }=\mathrm{\mid }r\mathrm{\mid }$. If $\tilde{r}\mathrm{\ne }r$, choose $n=(r-\tilde{r})\mathrm{/}\mathrm{\mid }r-\tilde{r}\mathrm{\mid }$; the equal-norm identity gives $2(r\cdot n)n=r-\tilde{r}$, hence the reflection sends $r$ to $\tilde{r}$. If $\tilde{r}=r$, choose any unit $n$ perpendicular to $r$ when $r\mathrm{\ne }0$, and any unit $n$ when $r=0$. Thus every elastic pair is attained.

For each fixed $n$, let $T_n(v,w)=(v^{\mathrm{\prime }},w^{\mathrm{\prime }})$. If $h\in L^1({\mathbb{R}}^6)$ is real-valued, write $h=h^{+}-h^{-}$, where $h^{+}=\max (h,0)$ and $h^{-}=\max (-h,0)$. Applying [F4] to $\mathrm{\mid }h\mathrm{\mid }$ shows that $h\circ T_n$ is absolutely integrable because $T_n$ has unit absolute Jacobian. Applying [F4] separately to $h^{+}$ and $h^{-}$, whose integrals are finite, and subtracting gives ${\int }_{{\mathbb{R}}^6}h(T_n(v,w))dvdw={\int }_{{\mathbb{R}}^6}h(v,w)dvdw\mathrm{.}$ By [F2], choose a finite Borel representative $h_0$ equal to $h$ almost everywhere. For every fixed $n$, [F4] gives the same change-of-variables equality for $h_0$, hence for $h$. The map $(v,w,n)\mapsto h_0(T_n(v,w))$ is Borel, and [F4] gives ${\int }_{{\mathbb{R}}^6}\mathrm{\mid }{h}_0(T_n(v,w))\mathrm{\mid }dvdw=\mathrm{\parallel }{h}_0{\mathrm{\parallel }}_1$ for every $n$. Since $\sigma (S^2)=1$, [F3] permits integration of this absolutely integrable identity in $n$. Thus the cited nonnegative change-of-variables result has the signed $L^1$ consequence used below.

Replace the representatives of $f$ and $g$ by Borel representatives using [F2]. Tonelli [F3] makes the sections measurable and permits integration in any order. For each fixed $n$, the pair reflection in step 1.1 is a $C^1$ diffeomorphism with unit absolute Jacobian, so [F4] gives ${\int }_{{\mathbb{R}}^6}h(v^{\mathrm{\prime }},w^{\mathrm{\prime }})dvdw={\int }_{{\mathbb{R}}^6}h(v,w)dvdw\mathrm{.}$ It follows that $\int \mathrm{\mid }\mathcal{G}(f,g)\mathrm{\mid }dv\le \mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_1$. The same change of variables shows that altering either representative on a Lebesgue-null set changes the integrand only on a $\mu$-null set; therefore the output class is independent of representatives. Bilinearity follows from linearity of the integral. Since $\mathrm{\mid }v{\mathrm{\mid }}^2\le \mathrm{\mid }{v}^{\mathrm{\prime }}{\mathrm{\mid }}^2+\mathrm{\mid }{w}^{\mathrm{\prime }}{\mathrm{\mid }}^2$, one has $w_0(v)\le w_0(v^{\mathrm{\prime }})+w_0(w^{\mathrm{\prime }})$. Applying Tonelli and [F4] again gives $\mathrm{\parallel }\mathcal{G}(f,g){\mathrm{\parallel }}_{1,w_0}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{1,w_0}\mathrm{\parallel }g{\mathrm{\parallel }}_1+\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }g{\mathrm{\parallel }}_{1,w_0}\le 2\mathrm{\parallel }f{\mathrm{\parallel }}_{1,w_0}\mathrm{\parallel }g{\mathrm{\parallel }}_{1,w_0}\mathrm{.}$ The same bounds applied to $\mathrm{\mid }f\mathrm{\mid },\mathrm{\mid }g\mathrm{\mid }$ define the signed bilinear extension.

For $f\ge 0$, use Tonelli and the unit-Jacobian involution to obtain $\int \mathcal{G}(f,f)dv=\int f(v)f(w)dvdw=M^2$. For the first moment, the same change of variables gives $\int v\mathcal{G}(f,f)dv=\int v^{\mathrm{\prime }}f(v)f(w)d\mu$. The simultaneous exchange $(v,w,n)\mapsto (w,v,n)$ preserves $\mu$ and swaps $v^{\mathrm{\prime }}$ with $w^{\mathrm{\prime }}$, while $f(v)f(w)$ is unchanged. Averaging the two expressions replaces $v^{\mathrm{\prime }}$ by $(v^{\mathrm{\prime }}+w^{\mathrm{\prime }})\mathrm{/}2=(v+w)\mathrm{/}2$, yielding $M\int vfdv$. Applying the same averaging to the second moment uses $(\mathrm{\mid }{v}^{\mathrm{\prime }}{\mathrm{\mid }}^2+\mathrm{\mid }{w}^{\mathrm{\prime }}{\mathrm{\mid }}^2)\mathrm{/}2=(\mathrm{\mid }v{\mathrm{\mid }}^2+\mathrm{\mid }w{\mathrm{\mid }}^2)\mathrm{/}2$ and gives $M\int \mathrm{\mid }v{\mathrm{\mid }}^{2}fdv$. The first-moment change of variables is applied coordinatewise to signed integrable functions; finiteness follows from the finite mass and second moment.

Let $M=1$ and let $\phi$ satisfy the four absolute-integrability conditions in the Statement. By [F2], choose a finite Borel representative ${\phi }_0$ equal to $\phi$ almost everywhere. The Lebesgue integrals weighted by $f$ or $\mathcal{G}(f,f)$ are unchanged, and the products with $A$ or $B$ agree $\mu$-almost everywhere: the unprimed exceptional sets are null by Tonelli, while for each fixed $n$ the primed exceptional sets are null by the measure-preserving involution in step 1.1, so Tonelli makes their union $\mu$-null. Thus it suffices to prove the identity for ${\phi }_0$. Tonelli and the gain definition give $\int {\phi }_0(v)\mathcal{G}(f,f)(v)dv=\int {\phi }_0(v)Bd\mu$. By Tonelli, $h_n(v,w)={\phi }_0(v)f(v^{\mathrm{\prime }})f(w^{\mathrm{\prime }})$ belongs to $L^1({\mathbb{R}}^6)$ for $\sigma$-almost every $n$, since its absolute integral over $n,v,w$ is $\int \mathrm{\mid }{\phi }_0\mathrm{\mid }\mathcal{G}(f,f)dv<\mathrm{\infty }$. For those $n$, step 2.1 applied to $h_n$ gives $\int {\phi }_0(v)f(v^{\mathrm{\prime }})f(w^{\mathrm{\prime }})dvdw=\int {\phi }_0(v^{\mathrm{\prime }})f(v)f(w)dvdw$; the fixed-$n$ change of variables also shows the right side is absolutely integrable. Fubini then gives $\int {\phi }_0(v)Bd\mu =\int {\phi }_0(v^{\mathrm{\prime }})Ad\mu$. The loss term is $\int {\phi }_0(v)Ad\mu$ because $\int f=1$ and $\sigma (S^2)=1$. The stated integrability conditions make these terms absolutely integrable; particle exchange gives the same for the $w^{\mathrm{\prime }}$ and $w$ terms, and the conditions on $A\mathrm{\Delta }{\phi }_0$ and $B\mathrm{\Delta }{\phi }_0$ make the incoming/outgoing change of variables integrable. Averaging the difference with its image under simultaneous particle exchange yields $\frac{1}{2}\int A[{\phi }_0(v^{\mathrm{\prime }})+{\phi }_0(w^{\mathrm{\prime }})-{\phi }_0(v)-{\phi }_0(w)]d\mu =-\frac{1}{2}\int A\mathrm{\Delta }{\phi }_{0}d\mu$. The incoming/outgoing involution sends $A$ to $B$ and $\mathrm{\Delta }{\phi }_0$ to $-\mathrm{\Delta }{\phi }_0$, so $\int (B-A)\mathrm{\Delta }{\phi }_{0}d\mu =-2\int A\mathrm{\Delta }{\phi }_{0}d\mu$. Combining these equalities gives the claimed factor $1\mathrm{/}4$ identity.