Showing that $||K^nf||_p \leq \frac{1}{n!}||k||_\infty^n ||f||_p$ where $K$ is the integral operator coming from the kernel $k(x,y) = \max\{0,x-y\}$

Question

Here is what is written in my notes:

$\bf{Remark.}$ There are quite a few ways to manufacture operators that contain only $0$ in the spectrum: Let $I = [0,1]$ and let $k \in \mathcal{C}(I^2)$ such that $k(x,y) = \max\{0,x-y \}$. Define $K$ to be the integral operator coming from this kernel, i.e. $Kf(x) = \int_0^1k(x,y)f(y)dy$. We can write down a formula for what happens when we iterate this operator. Check that $$ ||K^nf||_p \leq \frac{1}{n!}||k||_\infty^n ||f||_p $$ Hence, $r\{K \} = 0$

Now I understand everything here but how do I get the inequality above? I've trying a simple $n=2$, but I still can't see how to obtain it.

Answer 1

$$\|Kf\|_p^p = \int_0^1 \left|\int_0^1 k(x,y) f(y) \, dy\right|^p \, dx \le \int_0^1 \int_0^1 |k(x,y)|^p |f(y)|^p \, dy \, dx \le \|k\|_\infty^p \|f\|_p^p$$