If $f_{n+1}(t)\le c\int_0^tf_n(s)\:{\rm d}s$, then $f_n(t)\le\frac{(ct)^n}{n!}\sup_{s\in[0,\:T]}f_0(s)$

Question

Really simple question, but I don't see how to solve it at the moment: Say we know that $$f_{n+1}(t)\le c\int_0^tf_n(s)\:{\rm d}s\tag1$$ for all $t\in[0,T]$. Why can we conclude $$f_n(t)\le\frac{(ct)^n}{n!}\sup_{s\in[0,\:T]}f_0(s)\tag2$$ for all $t\in[0,T]$? I don't get where the factorial comes from.

Answer 1

By induction on $n$.

For $n=0$, the inequality is $f_0(t) \leq \sup_{s \in [0,T]} f_0(s)$ which is true.

Suppose it is true for an integer $n$. Then $$f_{n+1}(t) \leq c \int_0^t f_n(s) \mathrm{ds} \leq c \int_0^t \frac{(cs)^n}{n!} \sup_{s' \in [0,T]} f_0(s') \mathrm{ds} = \frac{c^{n+1}\sup_{s \in [0,T]} f_0(s)}{n!} \int_0^t s^n \mathrm{ds}$$

$$ = \frac{c^{n+1}\sup_{s \in [0,T]} f_0(s)}{n!} \times \frac{t^{n+1}}{n+1} = \frac{(ct)^{n+1}}{(n+1)!}\sup_{s \in [0,T]} f_0(s)$$

and you are done.