How to choose interval for Banach's fixed point theorem?

Question

First of all I don't understand why we need Banach's theorem, as a result I can't make it intuitive for me to understand how it works but I tried to solve an example.

https://en.wikipedia.org/wiki/Banach_fixed-point_theorem#Statement

let $f(x) = x^3$

a) for which fixed points will it converge ?

b) choose an interval $I$ such that all conditions for Banach's theorem are fulfilled (for function $f$)

My try :

a) we have 3 roots $( 0, 1, -1)$ with $0$ being a convergent point. (I think I solved a) but I don't understand what kind of interval are they asking about, and how to compute it)

PS: If some of the text doesn't make sense, please let me know, I translated it from another language.

Answer 1

You need to find an interval that contains $o$ and in wich the conditions of the theorem are satisfied. You are looking for an interval $[a,b]$ such that

1. $0\in[a,b]$
2. $f([a,b])\subset[a,b]$
3. $|f(x)-f(y)|\le L\,|x-y|$ for some constant $L$, $0\le L<1$ and all $x\in[a,b]$.

Since $|x|^3\le|x|$ for $|x|\le1$, it seems reasonable to try an interval $[-a,a]$ with $0<a\le1$. With this choice, 1. and 2. are satisfied. The easiest way to verify 3. is to impose $|f'(x)|\le L$. Since $|f'(x)|=3\,|x|^2$, the maximum of $|f'(x)|$ on the interval $[-a,a]$ is $3\,|a|^2$. This implies that it is enough to take $a<1/\sqrt3$.

Answer 2

Hint: You have to choose a closed interval $I$ such that $f(I) \subseteq I$ (i.e., $f$ maps $I$ into itself) and such that $f$ is a contraction on $I$. If $f$ is differentiable (as is in this case), the last condition is satisfied if there exists a constant $L\in [0,1)$ such that $$ |f'(x)| \leq L \qquad\forall x\in I. $$