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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.

Separate the transported initial datum from the gain term. Defineand the Boltzmann Volterra operatorThe Duhamel principle makes the linear Boltzmann equation equivalent, whenever the integrals and solution are legitimate, toThe gain samples velocities $v_{\ast }$ at the backward spatial position of the outgoing velocity $v$; replacing that spatial shift by one depending on $v_{\ast }$ would give a different equation.

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.