= Solution
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 $(v,v_*,\sigma)$. The two symmetrized weak collision identities are
$$
\int Q(f,f)\phi(v)\,dv
=\frac1{2B}\int ff_*\bigl(\phi'+\phi_*'-\phi-\phi_*\bigr)\,dv\,dv_*\,d\sigma
$$
and
$$
\int Q(f,f)\phi(v)\,dv
=\frac1{4B}\int(f'f_*'-ff_*)\bigl(\phi+\phi_*-\phi'-\phi_*'\bigr)\,dv\,dv_*\,d\sigma.
$$
They 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 $\phi'+\phi_*'=\phi+\phi_*$. <Momentum> follows from $v'+v_*'=v+v_*$, and energy from $|v'|^2+|v_*'|^2=|v|^2+|v_*|^2$. Hence
$$
\boxed{\int Q(f,f)\,dv=0,\qquad\int vQ(f,f)\,dv=0,\qquad\int|v|^2Q(f,f)\,dv=0}.
$$
For the <Boltzmann H functional> $H=\int\!\!\int f\log f\,dx\,dv$, the spatial transport contribution is a boundary <divergence> and the term $\int Q$ is zero. Set $a=f'f_*'$, $b=ff_*>0$. The second weak identity with $\phi=\log f$ gives the <Boltzmann H theorem>
$$
\boxed{H'(t)=-\frac1{4B}\int\!\!\int\!\!\int(a-b)\log(a/b)\,dx\,dv\,dv_*\,d\sigma\leq0}.
$$
Indeed 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.
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