Prove that T is a contraction in M Hello I have problems with this exercise
Let $M,d$ a complete metric space

*

*Let $T:M  \longrightarrow M $ an application such that for a certain $K \in \mathbb{N} $, the aplication $T^k = T \circ T \circ ... \circ T (K \; factors)$ is a contraction in $M$. Prove that T has a unique fixed point


*In the metric space $M= (C[0,1],d_\infty]$ Let the Volterra ecuation:
$$x(t)- \mu \displaystyle\int_{0}^{t} k(t,s) x(s)ds=v(t) , t \in[0,b] $$
where $c\in C[0,b] , k \in C ([o,b]x [o,b])$ and $\mu \in \mathbb{C}$ define the operator:
$$Tx(t)=v(t)+ \mu\displaystyle\int_{0}^{t} k(t,s) x(s)ds=v(t) , x \in[0,b]$$
Prove that $T^m$ satisfy :
$$ |T^mx(t)-T^my(t) | \leq{} |\mu|^m c^m \displaystyle\frac{t^m}{m!} d_\infty (x,y)$$
$\forall m \in \mathbb{N}$ where $c=max \{ k(t,s) : (t,s) \in [0,b]^2 \}$
My attempt:
(1) I don't know how to prove it
(2) We have $c^m \displaystyle\frac{t^m}{m!} = \displaystyle\frac{ (|n |ct)^n}{n!} $
Thanks
 A: Let me assume $T$ to be continuous for the first point. Consider the sequences $x_{n+1}:=T^k(x_n)$ and $y_{n+1}=T^k(y_n), y_0=T(x_0)$. By the Banach fixed point theorem you get (for $x_\infty$ being the unique fixed point of $T^k$) $$x_\infty=\lim_{n \rightarrow \infty} x_n =\lim_{n\rightarrow\infty} y_n = \lim_{n\rightarrow\infty} T(x_n)=T(x_\infty).$$ If $T$ had another fixed point, then it was also a fixed point of $T^k$, which is not possible.
For the second point we compute for $t\geq 0$
$$ \vert Tx(t)-Ty(t)\vert =\vert \mu \int_0^t k(s,t)(x(s)-y(s))ds \vert \leq \vert \mu\vert \cdot \int_0^t \vert k(s,t) \vert \cdot \vert x(s)-y(s)\vert ds \leq \vert \mu\vert \int_0^t c \vert x(s) -y(s)\vert \leq \vert \mu\vert c d(x,y) \int_0^t 1 ds.$$
This allows you to do induction. I leave that to you.
Added: In fact we do not need continuity for first part. We simply note that $$ T^k(T(x_\infty))=T(T^k(x_\infty))=T(x_\infty)$$
and by uniqueness of the fixed point of $T^k$ we get $T(x_\infty)=x_\infty$.
It is quite fun to construct maps such that $T^2$ is a contraction, but $T$ is nowhere continuous. We can for example take $X=[-1,1]$ with the euclidean metric and
$$ T(x)=\begin{cases} 0,& x=\pm 1, \\ 1,& x\in (-1,1)\cap \mathbb{Q}, \\ -1,& x\in (-1,1)\setminus \mathbb{Q}. \end{cases}$$
Then $T^2(x)=0$, so is a contraction, but $T$ is nowhere continuous.
