Suppose $p$ is an attracting fixed point under a continuous map $f$ and that the basin of attraction of $p$ is the interval $(a,b)$. How do I show that $f(a,b) \subset (a,b)$?
I said that since $f$ is continuous, we can involve derivatives. (Edit: apparently that's false, and I don't know what I am talking about...) Since $p$ is an attracting fixed point, $|f'(p)| < 1$. So $\frac{|f(x)-f(y)|}{|x-y|}<1$ for every $x,y \in (a,b)$. This implies $f(a,b) \subset (a,b)$.
Is this correct?
I'm going to assume the basin of attraction $B$ of $p$ is defined to be the set of points $x$ for which $f^n(x)\to p$ as $n\to \infty$. Then, if $x\in B$, we want to show that $f(x)\in B$ as well, since this shows $f(B)\subseteq B$. So why is $f(x)$ in the basin of attraction? Well, let's look at its orbit: $$f^n(f(x)) = f^{n+1}(x).$$ That is, the orbit of $f(x)$ is really just a subsequence of the orbit $f^n(x)$ of $x$. Since the orbit $f^n(x)$ of $x$ converges to $p$ by assumption, it follows that the orbit $f^n(f(x))$ of $f(x)$ also converges to $p$, and thus $f(x)$ is in the basin $B$.
