contraction mapping and convergence proof I am having difficulties arranging and concluding the proof...
Suppose $f$ maps the open interval $E$ into itself, $0 < b < 1$, $f$ has property $X(b)$ (that property is Lipschitz continuous), and
$x_0 \in E$ Prove that the sequence $\{ x_k\}$ defined recursively by $x_k = f(x_{k−1})$ for $k \geq 1$
converges.
what I started with:
Given: $x_0 \in E$ $f(x_{k-1}) = x_k$ $0 < b <1$, $f$ has the property $X(b) \in E$, $|f'(x)| \leq b$, $x_0 \in E$
from $|f'(x)| \leq b$
$f(x_0) - f(x) \leq y+x_1$ becomes $f(x_0) = x_1$
$x_k = f(x_{k-1})$. for $ K\geq 1$
I am stuck and not sure if whatI am starting with is the right track :(

I have no idea about what you mean by $y$
I think you should notice that Lipchitz condition doesn't mean the existence of derivative, so you are not allowed to use $f'(x)$
Here's my proof
By the condition, given $x_0\in E$, we have $x_n\in E,\forall n\in \mathbb{N}$
By Lipschitz condition, we have
$$|f(x)-f(y)|\leq b|x-y|,\forall x,y\in E$$
specially,
$$|x_{n+2}-x_{n+1}|=|f(x_{n+1})-f(x_n)|\leq b|x_{n+1}-x_n|,\forall n\in \mathbb{N}$$
let $n=0,1,...,k-1$, and multiply the inequalities together, we have
$$|x_{k+1}-x_{k}|\leq b^k|x_1-x_0|$$
note that given $x_0$, $x_1-x_0$ is a constant, so when $m>n$
$$|x_m-x_n|\leq \sum_{k=n}^{m-1}|x_{k+1}-x_k|\leq \sum_{k=n}^{m-1}b^k|x_1-x_0|\leq \sum_{k=n}^\infty b^k|x_1-x_0|=\frac{b^n}{1-b}|x_1-x_0|$$
$\forall\epsilon>0,\exists N,\frac{b^N}{1-b}|x_1-x_0|<\epsilon$, then we have $|x_m-x_n|<\epsilon,\forall m>n>N$
by Cauchy's criterion, it is equivalent to the convergence of $\lbrace x_n\rbrace$
