(v_0 \pm v_1) f(s, x \pm (t-s), v) \ dv \ ds\\
& \leq & C(1 + t).\end{aligned}$$

Next, we represent the fields in terms of the source $j_2$ in the associated transport equations. Either adding or subtracting the equations for $E_2$ and $B$ in (\[RVMFP\]) yields $$\partial_t (E_2 \pm B) \pm \partial_x (E_2 \pm B) = - j_2.$$ Thus, we can write the sum or difference of the fields in terms of initial data and an integral of $j_2$ along one side of the backwards cone, namely $$\label{E2B}
(E_2 \pm B)(t,x) = (E_2 \pm B)(0,x \mp t) - \int_0^t j_2(s,x \mp (t-s)) \ ds.$$ Then, in view of the previous conclusion of the lemma and the assumption on the initial fields, we find $$\Vert (E_2 \pm B)(t) \Vert_\infty \leq C(1+t)$$ and since $$E_2(t,x) = \frac{1}{2} (E_2 + B)(t,x) + \frac{1}{2} (E_2 - B)(t,x),$$ and similarly for $B$, it follows that $\Vert E_2(t) \Vert_\infty$ and $\Vert B(t) \Vert_\infty$ are controlled by this same quantity.

Finally, control of $E_1$ follows from conservation of mass and the assumption on the background density. Integrating the equation for $E_1$ and using the assumption on $E_1(0,x)$ yields $$E_1(t,x) = \int_{-\infty}^x \rho(t,y) \ dy$$ and we 