Let $f(x)$ be $x^2-x-2$. I want to find the root using FPI in an interval where it will converge. I have chosen $g(x)=x^2-2$ and so $g'(x)=2x$. The convergence condition, $|g'(x)|<1$ is obviously satisfied in $-0.5<x<0.5$.
Problem: I have failed to find a consistent interval outside of this convergence interval up to $\pm 1$ for which the iteration consistently diverges.
Question: Does this convergence condition only guarantee convergence (in the bounds) but not divergence (outside the bounds)?
Let us consider the fixed point iterations associated to the function $g: x \mapsto x^2-2$, defined by the quadratic map $$ x_{n+1} = {x_n}^2 - 2, \qquad x_0 \in \Bbb R . $$ This map has many periodic points, even with large period. The period-one fixed points $-1$, $2$ are both repelling fixed points (indices $2>1$ and $4>1$, respectively). Thus, fixed-point iterations will not converge towards these values unless the starting value $x_0$ is exactly equal to $-1$ or $2$. Setting $y_n = -\frac{1}{4} x_n + \frac{1}{2}$, the logistic map $y_{n+1} = r y_n (1-y_n)$ with parameter $r=4$ is obtained, which exact solution is bounded and exhibits chaotic behavior (see also this article). Therefore, $$ x_n = 2\cos\left(2^n\cos^{-1}(x_0/2)\right) $$ is bounded and exhibits chaotic behavior too (the sequence does not diverge to infinity).