= Solution
The <Fubini's theorem> and <integral> <triangle inequality> give
$$
\boxed{\|\rho(g)\|_{L^1_x}\leq\int\int|g(x,v)|\,dv\,dx=\|g\|_1.}
$$
Define the <normalized velocity-reset collision operator> using the <normalized velocity-reset projection> $Pg(x,v)=\rho(g)(x)M(v)$ and $B=P-I$. Since $M\geq0$ and $\int M=1$,
$$
\|Pg\|_1=\|\rho(g)\|_{L^1_x}\leq\|g\|_1,
\qquad \|Bg\|_1\leq2\|g\|_1.
$$
Also $P^2=P$; the collision gain replaces the velocity distribution by $M$ while preserving the spatial mass. In <Bochner integral> notation the printed, undamped source operator is
$$
(\tau f)(t)=\int_0^tU_{t-s}Bf(s)\,ds.
$$
Using the transport <isometry> and the <integral> <triangle inequality> proves
$$
\boxed{\|\tau f(t)\|_1\leq2\int_0^t\|f(s)\|_1\,ds
\leq2t\sup_{0\leq s\leq t}\|f(s)\|_1.}
$$
These estimates hold for measurable, locally time-bounded $E$-valued functions. If the displayed supremum is infinite, the numerical bound is interpreted in the extended sense; the construction below works in a space where it is finite.
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