# Prove $|T^nx(t)-T^ny(t)|\leq \frac{M^nt^n}{n!}\sup|x(s)-y(s)|$ for $Tx(t)=\int_0^t K(t,s)x(s)ds$

Let $K:[0,1]^2\to \mathbb{R}$ be continuous. Consider $Tx(t)=\int_0^t K(t,s)x(s)ds$ for $0\leq t \leq1$. Suppose $\sup_{0\leq,s\leq 1}|K(t,s)|=M$ and $Tx$ continuous function for $x(t)$ continuous. Let $T^n$ be the $n$-fold composition of $T$ with itself I. want to prove by induction for $n\in \mathbb{N}$, $0\leq t\leq 1$ $$|T^nx(t)-T^ny(t)|\leq \frac{M^nt^n}{n!}\sup_{0\leq t,s\leq 1}|x(s)-y(s)|.$$ For $n=1$, I get $$|Tx(t)-Ty(t)|=|\int_0^t K(t,s)(x(s)-y(s))ds|\leq\int_0^t |K(t,s)(x(s)-y(s))|ds \leq \int_0^t M\sup_{0\leq,s\leq 1}(x(s)-y(s))ds\leq Mt\sup_{0\leq t,s\leq 1}|x(s)-y(s)|$$ so it holds. For the induction step I'm stuck, I get $$|T^{n+1}x(t)-T^{n+1}y(t)|=|\int_0^t K(t,s)(T^nx(s)-T^ny(s))ds|\leq M\cdot t\cdot M^nt^n/n!\sup_{0\leq t,s\leq 1}|x(s)-y(s)|.$$ I don't see how I the $1/(n+1)$ should appear...

When doing estimate in the induction step, you took the $s^n$ outside the integral too early. If you keep it inside, when doing integration, you would have the term $\frac{1}{n+1}$. Note that $$\int_0^t |K(t,s)|\cdot|T^nx(s)-T^ny(s)|ds\leqslant M\cdot \int_0^t \frac{M^n}{n!}\cdot s^n ds\cdot \sup_{s\in[0,1]} |x(s)-y(s)|$$