How do you show that orbits of $f(x)=x-x^3$ converge to $0$ for$-\sqrt{2} < x < \sqrt{2}$?

Question

It is easy to show when $0\leq|x|\leq1$, but for values of $1<|x|<\sqrt{2}$, I have a hard time showing that $|f(x_{0})| < |x_{0}|$.

I am pretty sure I just have to show that eventually $|f^n(x_{0})| \leq 1$ but I am stuck.

Answer 1

For only the statement in the body, that $|x|\in(1,\sqrt 2)\to |f(x)|\lt |x|,$ note that $|x|\gt 1\to |x^3|\gt |x|\to \frac{|x-x^3|}{x-x^3}=-\frac{|x|}x.$

Since we wish to show that $|f(x)|\lt |x|,$ note the sign swap over the interval of interest from above and use the inequality $x^3-x\lt x$ with positive $x$ and $x-x^3\lt -x$ with negative $x.$ This is true when $x^3\lt 2x$ (positive) and $2x\lt x^3$ (negative) which are both true for all $|x|$ in our interval.

But take caution as this particular inequality does not prove the result that $f^n(x_0)\to 0$ for $x_0\in(-\sqrt 2,\sqrt 2).$