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For $E=L_{x,v}^1$, let $(\tau f)(t)={\int }_0^t{U}_{t-s}(P-I)f(s)ds$. The factorial bound for a Volterra iterate is $\Vert {\tau }^{n}f(t){\Vert }_1\le (2t{)}^n{\sup }_{s\le t}\Vert f(s){\Vert }_1/n!$. Thus the Neumann series converges in $C([0,T];E)$ for every finite $T$ and solves $f=U_{\cdot }{f}_0+\tau f$. The same estimate applied to a difference proves uniqueness among locally time-bounded integral solutions. The whole collision term remains inside this undamped integral, rather than absorbing loss into the free propagator.

Starting from $\Vert \tau g(t){\Vert }^2\le C^{2}t{\int }_0^t\Vert g(s){\Vert }^{2}ds$, induction integrates $s^{2n-2}$ at the $n$th step. Its denominator accumulates the odd integers $1,3,\dots ,2n-1$. The factorial bound for a Volterra iterate is stronger because $(n!{)}^2\ge (2n-1)!!$.

Fix $T$ and set $A_T={\sup }_{0\le s\le T}\Vert f(s){\Vert }_1$. The preceding integral estimate gives the result for one iterate. If for some $n\ge 0$ the bound $\Vert {\tau }^{n}f(s){\Vert }_1\le A_T{2}^n{s}^n/n!$ holds for $s\le T$, thenTaking $T=t$ yields exactlyThe case $n=0$ uses the identity operator. This factorial bound for a Volterra iterate comes from time ordering, so no commutation between free transport and the collision projection is assumed.

On each finite interval use the Banach space $X_T=C([0,T];E)$ with the supremum norm. The free term $g(t)=U_t{f}_0$ belongs to $X_T$, and the time-integral operator $\tau$ maps $X_T$ to itself with norm at most $2T$. The strongly continuous semigroup property and boundedness of $B$ justify continuity of the Bochner integral.

DefineThe factorial bound for a Volterra iterate gives $\Vert {\tau }^{j}g{\Vert }_{X_T}\le (2T{)}^j\Vert f_0{\Vert }_1/j!$. The series therefore converges absolutely in $X_T$. Since $\tau$ is a bounded linear operator on $X_T$, it can be passed through the convergent sum, givingThus $f=g+\tau f$, the required integral formulation, and $f(0)=f_0$. Each term on a larger interval restricts to the identical term on a smaller interval, so these constructions define a single global solution without having to restart at successive times. This is an integrable Volterra solution for normalized velocity relaxation. It is a mild solution of an abstract Cauchy problem in $L^1$ and hence an $L^1$ weak solution in the paper's integral-formulation sense. No smallness condition such as $2T<1$ is needed.

If $f$ and $\tilde{f}$ are two weak solutions with the same initial value and the required local time bound, their difference $w$ satisfies $w=\tau w$. By linearity this implies $w={\tau }^{n}w$ for every $n$. On $[0,T]$ put $A_T={\sup }_{s\le T}\Vert w(s){\Vert }_1<\infty$. The factorial bound for a Volterra iterate now givesFor fixed $T$, the factor tends to zero; its successive-term ratio is $2T/(n+1)$. Therefore $w(t)=0$ as an $L^1$ element at every time on this interval. Since $T$ is arbitrary, the locally time-bounded $L^1$ weak solution is unique globally. This proof uses precisely the additional condition requested, rather than assuming arbitrary pointwise-in-time integrability alone supplies a finite uniform bound.

For fixed $t$, put $M_t={\sup }_{0\le s\le t}\Vert f(s){\Vert }_2$. A Cauchy-Schwarz inequality in time strengthens the organization of the estimate in (c):Start with $\Vert \tau f(t){\Vert }_2^2\le C^2{t}^2{M}_t^2$. If the printed bound holds for $n-1$, thenTaking square roots proves the iterated Cauchy-Schwarz bound for a Volterra operator:There is also a factorial bound for a Volterra iterate, obtained by iterating the unsquared integral estimate in (c):Its factor $t^n/n!$ is the volume of the time-ordered simplex $0<s_n<\cdots <s_1<t$. It is stronger than the printed estimate because ${\prod }_{j=1}^n{j}^2\ge {\prod }_{j=1}^n(2j-1)$.

Work in the Banach space ${\mathcal{B}}_T$ of bounded strongly measurable maps $[0,T]\to L_{x,v}^2$, with norm $\Vert g{\Vert }_{{\mathcal{B}}_T}={\sup }_{0\le t\le T}\Vert g(t){\Vert }_2$. Keeping actual representatives at every time matches the pointwise-in-time mild formulation. The Boltzmann Volterra operator is bounded on this space, and the factorial bound for a Volterra iterate givesThus the Volterra series for the linear Boltzmann equation converges in operator norm for every finite $T$, even when $CT\ge 1$. SetFor its partial sums, $(I-\tau ){\sum }_{n=0}^N{\tau }^{n}F=F-{\tau }^{N+1}F$. The remainder tends to zero by the factorial estimate. Hence $(I-\tau )f=F$, precisely the required characteristic integral equation.

The norm bound in (b) gives an explicit choice of the existence constant:Also $f(0)=f_0$, since every term with $n\ge 1$ vanishes at zero and the damping interval has length zero. This proves existence in the paper's weak, characteristic-integral sense. With merely measurable nonnegative $a$, that sense does not itself require a continuous initial trace; that trace follows under the additional local characteristic-integrability condition described in (b).

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.

The factorial bound for a Volterra iterate gives convergence in the space of bounded strongly measurable $L^2$-valued functions on $[0,T]$. The sum solves $(I-\tau )f=F$, where $F=U_a(t,0)f_0$, and satisfies ${\sup }_{t\le T}\Vert f(t){\Vert }_2\le e^{CT}\Vert f_0{\Vert }_2$. Applying the same iterate bound to a difference $g=\tau g$ proves uniqueness on every finite interval.