Basin of attraction of the fixed map $f(x) = x-x^3$ Prove that the interval $(-\sqrt 2 ,\sqrt 2 )$ is the basin of attraction of the fixed point $0$ of the map $f(x)=x-x^3$, for $x \in \mathbb{R}$.
How one would prove this? In the examples I've seen so far they usually prove that a fixed point has a certain basin of attraction by proving that the function is decreasing or increasing for certain values within the basin of attraction. In this case however the values 'jump' from positive to negative making it impossible to use that method. I've been trying some options using the absolute value, but I can't figure it out. Could you please show me a plausible proof for this situation?
 A: Let $F(x)=\left|f(x)\right|$.  I claim that $F(x)<\left|x\right|$ for all $x\in\left(-\sqrt{2},\sqrt{2}\right)$. This is quite easy to see from a graph:

It can be proved using the factorization
$$F(x) = \left|x-x^3\right| = \left|x\right|\,\left|1-x^2\right|.$$
Now, the claim is trivial for $0 < x \leq 1$ since then $0<1-x^2<1$ so
$$F(x) =  \left|x\right|\,\left|1-x^2\right| < |x|.$$
If $1<x<\sqrt{2}$, then $1<x^2<2$ so that $0<x^2-1<1$ and, again, $\left|1-x^2\right|<1$.
With this lemma out of the way, your problem is easy.  Any seed $x_1\in\left(0,\sqrt{2}\right)$ leads to a decreasing sequence, that is bounded below by zero, under iteration of $F$.  Thus, there is a limit; that limit must be zero, since zero is the only fixed point of $F$ in $[0,\sqrt{2})$.  Any seed in $\left(-\sqrt{2},0\right)$ leads to a positive first iterate to which the previous analysis applies.  These results extend to $f$ since the absolute value of an orbit of $f$ is exactly an orbit of $F$.  Finally, the basin is no larger than $\left(-\sqrt{2},\sqrt{2}\right)$, since those endpoints form an orbit of period 2 for $f$.
A: Here is a roadmap:


*

*If $B$ is the basin of attraction of the fixed point $0$, then $f(B)\subseteq B$.

*If $I=(-\sqrt 2 ,\sqrt 2 )$, then $f(I)\subseteq I$.

*If $J$ is open and contains any point not in $I$, then $J$ contains points that diverge under iterates of $f$.
A: Here's a half-assed attempt, doing it by parts for two domains. Feel free to downvote for any strange handwaving.
$\mathbf{First\text{ }part:}$
$f$ is a contraction on the interval $(-\sqrt{6}/3,0)\cup (0,\sqrt{6}/3)$.
Proof:
It suffices to show that for $x,y\in (-\sqrt{6}/3,0)\cup (0,\sqrt{6}/3)$:
$$|x^2+y^2+xy-1|\lt 1$$
This is standard Calculus, left as an exercise.
Verification:
g := proc (x, y) options operator, arrow; x^2+x*y-1+y^2 end proc
with(plots)
plot3d(abs(g(x, y)), x = -(1/3)*sqrt(6) .. (1/3)*sqrt(6),
y = -(1/3)*sqrt(6) .. (1/3)*sqrt(6), axes = box, scaling = constrained)


This covers the domain $(-\sqrt{6}/3,0)\cup (0,\sqrt{6}/3)$.
$\mathbf{Second\text{ }part:}$
For the rest of the domain, $(-\sqrt{2},-\sqrt{6}/3]\cup[\sqrt{6}/3,\sqrt{2})$, it suffices to prove that for each $x$ in this domain:
$$|f(x)|\lt |x|\text{  (1)}$$
and this also standard Calculus:
Verification:
p1 := plot(abs(f(x)), x = (1/3)*sqrt(6) .. sqrt(2),
scaling = constrained, color = red)
p2 := plot(x, x = (1/3)*sqrt(6) .. sqrt(2),
scaling = constrained, color = blue)
display(p1, p2)


The end points, $\pm\sqrt{2}$ give a two cycle, so they are excluded.
The above covers the entire domain.
Note: (1) is sufficient to ensure this domain is covered in its entirety, because there are no fixed points of $f$ in there, which prevents the iterates of $f$ from converging to a rogue point and staying on this domain. Therefore, any point $x$ in there, will eventually end up in the domain of the first part, where $f$ is a contraction and we are done.
