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Unit-mass solutions of the isotropic Maxwell molecule collision operator with matching first moments obey this comparison. The Bobylev identity, the Fourier bound by one, and $|{\xi }^{+}{|}^2+|{\xi }^{-}{|}^2=|\xi {|}^2$ give a scalar damped differential inequality. Its integral form and the Gronwall inequality yield nonexpansion in the Fourier distance of order two. This does not alone establish strict contraction or equilibrium convergence.

Apply the spherical identity of part (b) with $x=v-v_{\ast }$, $y=\xi$ and $\varphi (s)=e^{-is/2}$. The angular exponential in part (c) can be replaced, after angular integration, by $e^{-i|\xi |(v-v_{\ast })\cdot \sigma /2}$. The full phase is thenThe Fubini theorem factors the two velocity integrals into Fourier transforms. With ${\xi }^{\pm }=(\xi \pm |\xi |\sigma )/2$, the gain integral is exactly ${\int }_{{\mathbb{S}}^2}\hat{f}({\xi }^{+})\hat{f}({\xi }^{-})d\sigma$.

The loss Fourier transform is $\hat{f}(\xi )\hat{f}(0)$. Dividing the gain by the sphere area gives the Bobylev identity for the Maxwell molecule collision operator:The initial Fourier datum is ${\hat{f}}_0$. The surface measure on a sphere here is two-dimensional surface area, not the restriction of ambient three-dimensional Lebesgue measure, which would give the sphere measure zero.

The reference to part (f) within this part is a printed self-reference; the needed product estimate is part (e). Since both masses are one, subtracting the Bobylev identities gives, for $\xi \ne 0$,The normalized angular average and part (e) giveFor completeness, the Duhamel principle for this scalar equation yields $d(t)\le e^{-t}d(0)+{\int }_0^t{e}^{-(t-s)}d(s)ds$. Apply the Gronwall inequality to $e^{t}d(t)$ to obtain the Fourier nonexpansion for Maxwell molecules, $\boxed{d(f_t,g_t)\le d(f_0,g_0)}$. This estimate is nonexpansion; by itself it does not prove strict decay or convergence to a specified equilibrium.