& {\ensuremath{\left\| g(t) \right\|}}_\infty -  g(t,x,v)\\
& \geq & 0.\end{aligned}$$ Thus, by Lemma \[Lnew\], we find $ w(t,x,v) \geq 0$, by which it follows that $$h(t,x,v) \leq {\ensuremath{\left\|  h_0 \right\|}}_\infty + \int_0^t {\ensuremath{\left\|  g(s) \right\|}}_\infty\,ds$$ for all $t,x,v$. Finally, taking the supremum in $(x,v)$, the conclusion follows.

To prove the cone estimate, we begin by using conservation of mass. Integrating the Vlasov equation over all $(x,v)$ we find $$\frac{d}{dt} \iint f(t,x,v) \ dv \ dx = 0.$$ Thus, using the decay of $f^0$ we find for every $t \in [0,T]$ $$\label{mass}
\iint f(t,x,v) \ dv \ dx = \iint  f^0(x,v) \ dv \ dx < \infty.$$

To derive the necessary energy identities, we first rewrite the Fokker-Planck term in the Vlasov equation as $$\nabla_v \cdot ( D \nabla_v f) = v_0^{-1} \biggl ( \partial_{v_1} ( v_1^2 \partial_{v_1} f) + \partial_{v_2} ( v_2^2 \partial_{v_2} f) + 2v_1v_2\partial_{v_1v_2} f + \Delta_v f \biggr ).$$ Then, multiplying the Vlasov equation by $v_0$ and integrating in $v$, the Fokker-Planck term becomes $$\begin{aligned}
\int v_0 \nabla_v \cdot ( D \nabla_v f) \ dv & = & \int \left ( \partial_{v_1} ( v_1^2 \partial_{v_1} f) + \partial_{v_2} ( v_2^2 \partial_{v_2} f) + 2v_1v_2\partial_{v_1v_2} f + \Delta_v 