approach the steady state $f^\infty = M_T$ as $t\to\infty$. This problem is interesting since the collision mechanism drives the local velocity distribution towards $M_T$, but a more complicated mechanism involving the interaction of the streaming term $v \partial_x$ and the collision operator ${\mathbf{Q_{lin}}}$ is responsible for the emergence of spatial uniformity.

To elucidate this key point, let us define the operator ${\mathbf{L}}$ by $${\mathbf{L}}f(x,v) := -v\ \partial_x f (x,v) + {\mathbf{Q_{lin}}}f(x,v)\ .$$ The evolution equation can be written $\partial_t f = {\mathbf{L}}f$. Let $\mathcal{H}$ denote the weighted space $L^2(\tilde{\mathbf{T}}\times{\mathbb{R}}^d;M_T^{-1}(v){\,\operatorname{d}\!\tilde x}{\,\operatorname{d}\!v})$, where in the current discussion $d=1$. Then ${\mathbf{Q_{lin}}}$ is self-adjoint on $\mathcal{H}$, ${\mathbf{L}}f^\infty = 0$, and a simple computation shows that if $f(t)$ is a solution of , $$  \frac{{\rm d}}{{\rm d}t} \| f(t) - f^\infty\|_\mathcal{H}^2 = 2 \langle f(t), {\mathbf{L}}f(t)\rangle_\mathcal{H} = 2 \langle f(t), {\mathbf{Q_{lin}}}f(t)\rangle_\mathcal{H}
    = -2\| f  - M_T \rho\|_{\mathcal{H}}^2\ ,$$ where, as before, $\rho(x,t) := \int_{{\mathbb{R}}}f(x,v,t) {\,\operatorname{d}\!v}$. Thus, while the norm $ \| f(t)- f^\infty\|_\mathcal{H}$ is monotone decreasing, the derivative is zero whenever $f(t)$ has the form $f(t) = M_T \rho$ for [*any*]{} smooth density $\rho$. In particular, the inequality $$\label{coercive}
\langle f - f^\infty , {\mathbf{L}}(f - f^\infty) \rangle_\mathcal{H} \leq - \lambda \|f - f^\infty\|_\mathcal{H}^2$$ is 