\left ( \frac{1}{2}\vert u_-\vert^2 \right ) ,dv\,dx\,dt\\
& = \frac{1}{2} \int_0^T \iint \nabla_v \cdot F(t,x,v)  \vert u_- \vert^2 dv\,dx\,dt\\
& \leq \frac{1}{2} \Vert \nabla_v \cdot F \Vert_\infty  \Vert u_- \Vert_2^2 \end{aligned}$$ We finally have $${\ensuremath{\left\langle  f,   u_- \right\rangle}}  \leq  \left ( \frac{1}{2}\Vert \nabla_v \cdot F \Vert_\infty - \lambda \right ) {\ensuremath{\left\|   u_- \right\|}}_2^2 \leq 0.$$ However, by hypothesis $ f(t,x,v) \geq 0$ and by definition $u_-(t,x,v) \geq 0$, so ${\ensuremath{\left\langle  f,   u_- \right\rangle}} \geq 0$. Therefore, it must be the case that ${\ensuremath{\left\|   u_- \right\|}}_2 = 0$, from which it follows that $  u_- = 0$ and hence $ h(t,x,v) \geq 0$.

Now, we utilize Lemma \[Lnew\] and a very simple argument to finish the proof of Lemma \[L1\]. Let $g \in L^1((0,T); L^\infty(\bfR^{3})), F \in W^{1,\infty}((0,T) \times \bfR^3;\bfR^2)$, and $h_0 \in L^2(\bfR^3) \cap L^\infty(\bfR^3)$ be given. Assume $h \in L^\infty((0,T); L^2(\bfR^3))$ satisfies $\mcL  h =  g(t,x,v)$ in the weak sense and $ h(0,x,v) =  h_0(x,v).$ Define $$w(t,x,v) := {\ensuremath{\left\|  h_0 \right\|}}_\infty + \int_0^t {\ensuremath{\left\|  g(s) \right\|}}_\infty \, ds \ - \  h(t,x,v).$$ Then, we have $$w(0,x,v) = {\ensuremath{\left\|  h_0 \right\|}}_\infty -  h_0(x,v) \geq 0$$ and $$\begin{aligned}
\mcL  w & = & {\ensuremath{\left\| g(t) \right\|}}_\infty - \mcL h\\
& = 