Let $k:[0,1] \times [0,1] \to \mathbb{R}$ be continous, and $x(t) = \int_0^t k(t,s)x(s)ds$ for $0 \leq t \leq 1$. Not let $Tx(t) = \int_0^t k(t,s)x(s)ds$ and suppose $sup_{0 \leq t, s \leq 1}|k(t,s)|= M$. Assuming $Tx$ is a continous function for $x$ continous,and letting $T^{(n)}$ be the n-fold composition of $T$ with itself, show by induction for $n \in \mathbb{N}$ and $0 \leq t \leq 1$ we have $$|T^{(n)}x(t)-T^{(n)}y(t)|\leq \frac{M^n t^n}{n!} sup_{0 \leq s \leq 1}|x(s)-y(s)$$
Now I have got stuck even on the simplest case, for $n=1$, I have:
$$|Tx(t) - Ty(t)| = |\int_0^t k(t,s)x(s)ds - \int_0^t k(t,s)y(s)ds|$$ $$= |\int_0^t k(t,s)(x(s)-y(s))ds|$$ $$\leq |M\int_0^t(x(s) - y(s)ds|$$
But I can't find out where the $t$ which should be in the equation comes from.
I'm also struggling with the induction part:
Assuming the statement is true for $n = k$, i.e. $$|T^{(k)}x(t)-T^{(k)}y(t)|\leq \frac{M^k t^k}{k!} sup_{0 \leq s \leq 1}|x(s)-y(s)|$$
Then for $n = k+1$,
$$|T{(k+1)}x(t) - T^{(k+1)}y(t)| = |T \circ T^k x(s) - T \circ T^k y(s)|$$ $$ \leq T(\frac{M^k t^k}{k!} sup_{0 \leq s \leq 1}|x(s)-y(s)|$$
Then I am not sure how to take the $T$ of the above function.
All hints and tips much appreciated thanks