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Let $M\ge 0$ have velocity integral one, and $\rho (g)(x)=\int g(x,v)dv$. On phase-space $L^1$, the normalized relaxation collision operator is $B=P-I$, where $Pg=\rho (g)M$. It obeys $\Vert B\Vert \le 2$ because $\Vert Pg{\Vert }_1\le \Vert g{\Vert }_1$. It preserves mass at each spatial position. The corresponding linear Boltzmann equation combines this bounded collision operator with the free-transport semigroup.

Let $f_1,f_2$ have the same initial data and satisfy the characteristic integral equation. Their difference $g=f_1-f_2$ satisfies $g=\tau g$, hence $g={\tau }^{n}g$ for every $n$.

Fix any $T_0<T$ and let $M={\sup }_{0\le s\le T_0}\Vert g(s){\Vert }_2<\infty$. The factorial bound for a Volterra iterate givesThe scalar factor tends to zero, so $g(t)=0$ throughout this interval. Since $T_0<T$ is arbitrary, the weak solution of the linear Boltzmann equation is unique on $[0,T)$. The same argument gives uniqueness at $T$ whenever solutions are defined there by the integral formula.