t - s) & + & (e-m)(s, x-t+s) \biggr ] \ ds \\
& & = \int_{x-t}^{x+t} e(0,y) \ dy + 2 \int_0^t \int_{x-t+s}^{x+t-s} \int f(s, y,v) \ dv dy ds.\end{aligned}$$ Using the positivity of the mass and energy, the assumptions on the data, and conservation of mass, the right side satisfies $$\begin{aligned}
\int_{x-t}^{x+t} e(0,y) \ dy + 2\int_0^t \int_{x-t+s}^{x+t-s} \int f(s, y, v) \ dv dy ds & \leq & \int e(0,y) \ dy + 2 \int_0^t \iint f(s,y,v) \ dv dy \ ds \\
& = & \int e(0,y) \ dy + 2 \int_0^t \left ( \iint  f^0(y,v) \ dv \ dy \right ) \ ds \\
& \leq & C(1+t)\end{aligned}$$ and this yields the first result. The other conclusions of the lemma then follow from the first. More specifically, we find $$\label{vineq}
v_0 \pm v_1 = \frac{v_0^2 - v_1^2}{v_0 \mp v_1} = \frac{1+ v_2^2}{v_0 \mp v_1} \geq \frac{2 \vert v_2 \vert}{v_0 \mp v_1} \geq \frac{2 \vert v_2 \vert}{2 v_0} = \vert \hat{v}_2 \vert$$ and by (\[cone\]) $$\begin{aligned}
 \int_0^t \vert j_2(s, x \pm (t-s)) \vert ds & \leq & \int_0^t \int \vert \hat{v}_2 \vert f(s, x \pm (t-s), v) \ dv \ ds\\
&  \leq & \int_0^t \int 