we know that if $g(x)$ is Continuous over $[a,b]$ and
$g(x) \in [a,b], \forall x \in [a,b]$ and
$|g'(x)| < 1 , \forall x \in [a,b]$
then fixed point iteration will converge only into 1 point $p$ , $p \in [a,b] , g(p) = p$.
so my question is, do we have any way to know if the iteration will diverge for any $x0$ ?
i mean this theorem is only 1 sided which means if 1 of these Conditions doesnt occur we cant conclude if the iteration will converge or diverge, so is there any theorem that can tell us if we'll converge or diverge ?
Edit:
im looking for some condition or theorem that if this condition or theorem occurs then for any given $x0$ the iteration will diverge.
Best Answer
In fact, if $g:[a.b]\longrightarrow[a,b]$ is continuous your required divergence for any initial point is impossible because $g$ will have at least fixed point $p$ and $p = g(p) = g(g(p)) = \cdots$
EDIT:
Lat be $F$ the set of fixed points of $g$ and $E =\bigcup_{n=1}^\infty g^{-n}(F)$. If all the fixed points of $g$ are repelling (for all $p\in F$, $|g'(x)| > 1$ for all $x$ in a neighborhood of $p$), then for any $x_0\in[a,b]\setminus E$, the sequence $x_0$, $g(x_0)$, $g(g(x_0))$, $\dots$ diverges.
Proof: Suppose $(x_n) = (g^n(x_0))$ converges. Only can converge to some fixed point $p$. Let be $(p-\epsilon,p+\epsilon)$ a repelling neighborhood. Then, for some $n$, $x_n,x_{n+1}\in(p-\epsilon,p+\epsilon)$ with $p<x_{n+1}<x_n$ or $x_n<x_{n+1}<p$. Then, by the MVT $$|x_{n+1} - p| = |g(x_n) - g(p)| = |g'(\xi)(x_n - p)| > |x_n - p|,$$ contradiction.