= Solution
Fix $T$ and set $A_T=\sup_{0\leq s\leq T}\|f(s)\|_1$. The preceding <integral> estimate gives the result for one iterate. If for some $n\geq0$ the bound $\|\tau^nf(s)\|_1\leq A_T2^ns^n/n!$ holds for $s\leq T$, then
$$
\|\tau^{n+1}f(t)\|_1\leq2\int_0^t\|\tau^nf(s)\|_1\,ds
\leq\frac{2^{n+1}A_T}{n!}\int_0^t s^n ds
=\frac{2^{n+1}t^{n+1}}{(n+1)!}A_T.
$$
Taking $T=t$ yields exactly
$$
\boxed{\|\tau^nf(t)\|_1\leq\frac{(2t)^n}{n!}
\sup_{0\leq s\leq t}\|f(s)\|_1.}
$$
The 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.
Back to article page
