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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 an arbitrary finite $T$ and use the Banach space of bounded continuous functions $C_b([0,T]\times {\mathbb{R}}^{2d})$ with the supremum norm. Set $A=\Vert k_1{\Vert }_{\infty }\Vert k_2{\Vert }_1$. For $f$ in this space, the velocity integralis bounded by $\Vert k_2{\Vert }_1\Vert f{\Vert }_{\infty }$. If $(t_n,x_n)\to (t,x)$, the integrands converge pointwise and are bounded by the integrable function $\Vert f{\Vert }_{\infty }{k}_2$. The dominated convergence theorem proves continuity. Thus the rank-one gain on bounded continuous functions, $Kf=k_1{H}_f$, is continuous and bounded. The same theorem applied to the time integral shows that $\tau$ maps this space into itself, including at $t=0$.

Time ordering gives the factorial bound for a Volterra iterate:Consequently the Volterra series for the linear Boltzmann equationconverges uniformly on the whole finite time slab, so its sum is continuous and bounded and satisfies the fixed point equation. Tracking the damping factors gives the sharper pointwise bound $\Vert f(t){\Vert }_{\infty }\le e^{(A-1)t}\Vert f_0{\Vert }_{\infty }$. This proves existence for every finite $T$, without requiring $AT<1$ or uniform continuity of $f_0$. Boundedness on finite slabs does not assert a uniform bound over infinite time.

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 space $C([a,b])$ with the supremum norm is a Banach space. Continuity of the kernel makes $K$ a bounded linear operator. Induction on $n$ givesso $A^{n}y-A^{n}z=K^n(y-z)$. The ordered integration region gives the factorial bound for a Volterra iterate:For completeness, if this holds at $n$, then integrating $M{M}^n(s-a{)}^n/n!$ from $a$ to $t$ gives the bound at $n+1$; the initial case is the defining integral. Thusand ${\sum }_{n\ge 1}{\alpha }_n=e^{M(b-a)}-1<\infty$. Part (b) now gives uniform convergence of every iteration sequence to the unique fixed point. Since $y=Ay$ is precisely the Volterra integral equation, this is its unique continuous solution. No restriction $M(b-a)<1$ is needed: the factorial controls all iterates on the whole finite interval.

A bounded continuous factor $k_1$ and an integrable factor $k_2$ give a bounded velocity-integral gain on bounded continuous functions, with bound $\Vert Kf{\Vert }_{\infty }\le \Vert k_1{\Vert }_{\infty }\Vert k_2{\Vert }_1\Vert f{\Vert }_{\infty }$. The dominated convergence theorem proves joint continuity of the velocity integral. Combined with damped transport, the factorial bound for a Volterra iterate proves existence on arbitrary finite time slabs.

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.