valid in general for $\lambda = 0$, but for no positive value of $\lambda$. If (\[coercive\]) were valid for some $\lambda>0$, we would have had $\|f(t) - f^\infty\|_\mathcal{H}^2 \leq e^{-t\lambda}\|f^I - f^\infty\|_\mathcal{H}^2$ for all solutions of our equation, and we would say that the evolution equation is [*coercive*]{}. However, while this is not the case, it does turn out that one still has constants $1< c < \infty$ and $\lambda> 0$ such that $$\|f(t) - f^\infty\|_\mathcal{H}^2 \leq ce^{-t\lambda}\|f^I - f^\infty\|_\mathcal{H}^2\ .$$ (The fact that there exist initial data $f(0) \neq f^\infty $ for which the derivative of the norm is zero shows that necessarily $c>1$.) In Villani’s terminology (see §3.2 of [@ViH06]), this means that our evolution equation is [*hypocoercive*]{}.

Since $f(t)$ and $f^\infty$ are probability densities, a natural norm in which to measure the distance between them is the $L^1$ distance, or, what is the same up to a factor of $2$, the total variation distance between the corresponding probability measures. However, as is well known, the norm $\|\cdot \|_\mathcal{H}$ controls the $L^1$ norms. Specifically, by the Cauchy-Schwarz inequality, $$\begin{aligned}
\label{L1dom}
\|f(t) - f^\infty\|_\mathcal{H}^2 &=&  \int_{\tilde{\mathbf{T}}\times{\mathbb{R}}^d}  | f(x,v,t)M_T^{-1}(v)   -1|^2 M_T(v){\,\operatorname{d}\!\tilde x} {\,\operatorname{d}\!v}\nonumber\\
&\geq& \left(\int_{\tilde{\mathbf{T}}\times{\mathbb{R}}^d}  | f(x,v,t)M_T^{-1}(v)   -1| M_T(v){\,\operatorname{d}\!\tilde x} 