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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.

Expansion of the two squares givesLet $D=d(f,g)$ be the Fourier distance of order two, with the supremum taken over $\xi \ne 0$. Both Fourier transforms have modulus at most one. Add and subtract $\hat{g}({\xi }^{+})\hat{f}({\xi }^{-})$, then use the triangle inequality:Dividing by $|\xi {|}^2$ proves the bound by $D$, and splitting the last expression into its two weighted terms gives the requested intermediate inequality. If one of ${\xi }^{\pm }$ vanishes, its unweighted difference is zero by equal mass; its weighted term is interpreted as zero, avoiding a $0/0$ quotient.

Equal mass and first moment also explain finiteness of this distance: subtract the constant and linear Taylor terms in the Fourier integral and use $|e^{-ia}-1+ia|\le a^2/2$. This bounds the difference by $|\xi {|}^2\int |v{|}^2(f+g)/2$. No direction-independent extension of the quotient at zero is required.