I tried to solve this problem,
Let $K\in L_{loc}^{\infty}$, for $n\in N$, we note $K_{n} = K*K*K...*K$ ($n$ factors), show for every $a > 0$, for every $x \in [0, a]$ and for every $n \in N$ that:
$$ |K_{n}(x)| \le (\sup_{x \in [0, a]}(|K(x)|))^n\frac{x^{n-1}}{(n-1)!} $$
but i don't know where to start. (i tried recurrence by i stuck to prove it for $n+1$)
the problem is long if anyone have any reference to complete solution of this problem this will be a big help.
thanks
Edit:
I think we can solve it like that:
for $n = 1$ it's clear that $|K(x)| \le (\sup_{x \in [0, a]}(|K(x)|))$
for $n = 2$ we have $$ |K_{2}(x)| = \left| \int^{x}_{0}K(x-t)K(t)dt \right| \le (\sup_{x \in [0, a]}(|K(x)|))^2\times x $$ because $ 0 \le x -t \le a$ and $ t \ge 0$
for $n \ge 2$ we have $$ |K_{n}(x)| = \left| \int^{x}_{0}K(x-t_1) \left(\int^{t_1}_{0}K(t_1-t_2) \dots \left(\int^{t_{n-2}}_{0}K(t_{n-2}-t_{n-1})K(t_{n-1}) dt_{n-1}\right) \dots dt_2\right) dt_1 \right| \le (\sup_{x \in [0, a]}(|K(x)|))^n\frac{x^{n-1}}{(n-1)!} $$