I am currently studying iteration sequences and I am a bit hung up on one specific bit which involves determining intervals of attraction of fixed points.
I've been given a graphical method to determine the intervals of attraction, and a more formal method called the "gradient criterion" which states that if $f$ is a smooth function with an attracting fixed point $a$, then if an interval $I$ with midpoint $a$, exists for which $|f'(x)| < 1$, for $x \in I$ then $I$ is a interval of attraction.
Right, so far so good. Now, what I have a harder time understanding are is that if I solve the inequality $-1 < f'(x) < 1$ and then use the result to identify a interval $I$ with midpoint $a$ (where a is an attracting fixed point), then isn't the complete interval, its simply a interval where I know that a start value from this interval will tend to $a$. However, larger intervals exits, which is mentioned in my text book, where its simply pointed out that other start values might end up producing terms which are in $I$, and they will then tend to $a$ from there. I get that, but its rather informal, and I feel like I'm missing how I can formally attack the problem of finding the full interval.
By looking various graphs of functions being iterated, it seems to me like I need to look at the boundries of the open interval $I$, if for example, the gradient is increasing though the lower boundry, I need to extend the interval with the largest possible interval where f is increasing and no other fixed points are present.
Am I going in the right direction here?
If a maximal interval of attraction exists, then by continuity it has to be open. If that interval is $(b,c)$, and if the function is still defined in the closed interval $[b,c]$, then $b$ and $c$ can not be attracted to $a$, and since $f$ maps $(b,c)$ into itself, it has to map the endpoints into the closure $[b,c]$. This shows that $f(\{b,c\}) \subseteq \{b,c\}$, so that either $b$ and $c$ are both fixed points, or one of them is fixed and the other one its preimage, or they form a two-cycle.
These possibilities are realized with $f_1(x) = x^3$, $f_2(x) = x^2$, $f_3(x) = -x^3$, where in each case $[b,c]=[-1,1]$. So you can find the maximal interval of attraction by finding all fixed points, their preimages, and all two-cycles.