= Solution
For fixed $t$, put $M_t=\sup_{0\leq s\leq t}\|f(s)\|_2$. A <Cauchy-Schwarz inequality> in time strengthens the organization of the estimate in (c):
$$
\|\tau g(t)\|_2^2
\leq C^2t\int_0^t\|g(s)\|_2^2\,ds.
$$
Start with $\|\tau f(t)\|_2^2\leq C^2t^2M_t^2$. If the printed bound holds for $n-1$, then
$$
\|\tau^nf(t)\|_2^2
\leq C^2t\int_0^t
\frac{C^{2n-2}s^{2n-2}}{1\cdot3\cdots(2n-3)}M_t^2\,ds
=\frac{C^{2n}t^{2n}}{1\cdot3\cdots(2n-1)}M_t^2.
$$
Taking square roots proves \b[the <iterated Cauchy-Schwarz bound for a Volterra operator>]:
$$
\boxed{\|\tau^nf(t)\|_2
\leq\frac{C^nt^n}{\sqrt{1\cdot3\cdots(2n-1)}}M_t.}
$$
There is also a <factorial bound for a Volterra iterate>, obtained by iterating the unsquared integral estimate in (c):
$$
\boxed{\|\tau^nf(t)\|_2\leq\frac{C^nt^n}{n!}M_t.}
$$
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 $\prod_{j=1}^n j^2\geq\prod_{j=1}^n(2j-1)$.
Back to article page
