I need to identify an interval $[a, b]$ for which the Contraction Mapping Theorem guarantees convergence to the positive fixed point (or verify that there is no such interval) for the function $g(x) = x^3+3x^2-3$.
Clearly, the only positive fixed point is $1$. However, according to the mean value theorem, the positive interval will be $[0,0.464]$. Does that mean that there is no such interval guaranteed by the CMT for the function $g(x)$? Am I thinking about this right?
Recall that $g(x) : I \to \mathbb{R}$ is Lipschitz iff there exists a $k>0$ such that:
$$k = \sup_{x \in I} |g'(x)|.$$
Let $a = 1 - \delta$, $b = 1+ \delta$ and $I = [a, b]$. The derivative of $g$ is:
$$g'(x) = 3x^2 + 6x.$$
$g'(x)$ as a stationary point at $x=-1$ (absolute minimum). In addition, $g''(x) > 0$ for all $x > -1.$ Then, for $1- \delta > -1 \Rightarrow 0 < \delta < 2$, we have that:
$$g'(1-\delta) < g'(1+\delta)$$
and hence
$$k = \sup_{x \in I} |g'(x)| = |g'(1+\delta)| = |3(1+\delta)^2 + 6(1 + \delta)| = (1+\delta)|9 + 3\delta|.$$
As a consequence:
$$k \in [9, 30].$$
$k$ is also the contraction factor, but it is always bigger than $1$, and hence, there is no basin of attraction for $x^* = 1.$
Alternatively, notice that, for $x^* = 1$: $$g'(x^*) = 3\cdot 1^2 + 6 \cdot 1 = 9 > 1.$$
This demonstrates that there is no neighborhood of $x^*$, let's say $I$, such that:
$$\lim_{n \to +\infty} x_n = x^* ~\text{provided that}~ x_0 \in I \setminus \{x^*\},$$ where
$$x_{n+1} = g(x_{n}).$$
Sources for the latter argument are here and here, among the thousands found on Google.
