 Solution

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 ${\prod }_{j=1}^n{j}^2\ge {\prod }_{j=1}^n(2j-1)$.