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For the Maxwell molecule collision operator and the Fourier transform with exponent $-i\xi \cdot v$, put ${\xi }^{\pm }=(\xi \pm |\xi |\sigma )/2$. The angular exchange for elastic collisions, rotation invariance of spherical surface area and factorization of the two velocity integrals give the displayed formula. This identity converts the collision integral into a spherical average of products at two related frequencies.

The printed hint with unchanged $\sigma$ is not a valid change of variables: at fixed $\sigma$, the outgoing velocities forget the direction of $v-v_{\ast }$. A correct proof uses the angular exchange for elastic collisions.

Set $C=(v+v_{\ast })/2$ and $v-v_{\ast }=r\omega$, where $r>0$ and $\omega \in {\mathbb{S}}^2$. Then $dvd{v}_{\ast }=dC{r}^{2}drd\omega$, while $v^{\prime }=C+r\sigma /2$ and $v_{\ast }^{\prime }=C-r\sigma /2$. The gain integral becomesExchange the two independently integrated angular variables $\omega$ and $\sigma$. The measure is unchanged, and reverting to $v=C+r\omega /2$, $v_{\ast }=C-r\omega /2$ givesThe measure-zero set $r=0$ causes no difficulty. This proves the requested identity without invoking the false fixed-$\sigma$ Jacobian assertion.

Strict positivity throughout phase space is incompatible with compact support. The following calculation uses positive smooth rapidly decaying solutions with all displayed integrals justified; nonnegative cases use the corresponding entropy limits where justified.

Write $B=|{\mathbb{S}}^2|$ and use integration over $(v,v_{\ast },\sigma )$. The two symmetrized weak collision identities areandThey follow from particle interchange and the angular exchange for elastic collisions. They are the weak identities needed here; no invalid fixed-angle Jacobian is used.

The collision invariants $1$, $v$ and $|v{|}^2$ obey ${\varphi }^{\prime }+{\varphi }_{\ast }^{\prime }=\varphi +{\varphi }_{\ast }$. Momentum follows from $v^{\prime }+v_{\ast }^{\prime }=v+v_{\ast }$, and energy from $|v^{\prime }{|}^2+|v_{\ast }^{\prime }{|}^2=|v{|}^2+|v_{\ast }{|}^2$. HenceFor the Boltzmann H functional $H=\int \int f\log fdxdv$, the spatial transport contribution is a boundary divergence and the term $\int Q$ is zero. Set $a=f^{\prime }{f}_{\ast }^{\prime }$, $b=f{f}_{\ast }>0$. The second weak identity with $\varphi =\log f$ gives the Boltzmann H theoremIndeed this integrand with its negative sign is $b(1-X)\log X$ for $X=a/b>0$, which is nonpositive. The kinetic H functional decreases; the physical entropy with opposite sign increases.