Let $M \subset \mathbb{R}$ be a compact set and $f: M \rightarrow M$ a continuous strictly decreasing function show $\exists p \in M$ s.t $f(p) = p$, assume $f$ has no periodic orbits of any period, and that it is indeed an attractor fixed point $\forall x_0 \in M$.
I've shown existence and uniqueness of the fixed point, though I'm stuck proving that it is an attractor:
For $x_0 \in M$ split the orbit in $\{f^{2n}(x_0)\}_{n \geq 0}$ and $\{f^{2n+1}(x_0)\}_{n \geq 0}$ if $\{f^{2n}(x_0)\}_{n \geq 0}$ is convergent to $p$ ,which is the only fixed point of both orbits, so is $\{f^{2n+1}(x_0)\}_{n \geq 0}$:
\begin{align} \lim\limits_{n \to \infty} f^{2n+1}(x_0) = \lim\limits_{n \to \infty} f(f^{2n}(x_0)) \stackrel{\text{f continuous}}{=} f(\lim\limits_{n \to \infty}f^{2n}(x_0)) \stackrel{\text{only fixed point}}{=} f(p) = p \end{align}
It remains to show $f^{2n}(x_0)$ is attractor for all $x_0 \in M$:
- $f^2$ is a strictly increasing function.
- $f(x) > x$ if $x < p$ and $f(x) < x$ if $x > p$
- By 2. $f^2(x) < f(x)$ if $x<p$ and $f^2(x) >f(x)$ if $x>p$
I want to show that if $x_0 <p$ then $f^{2n}(x_0)$ is bounded from above by $p$ and that the sequence is increasing so it must converge to $p$, and make a complete analogous statement for $x_0 > p$. I have proven it is bounded from above but I'm not able to prove the sequence is increasing.
Thanks in advance.
Pick $x_0\in M$. By assumption, $f^2(x_0)=x_0$ is forbidden. As $f^2$ is strictly increasing, the sequence $\{f^{2n}(x_0)\}_n$ is strictly monotonic, and by compactness converges to a fixed point $q$ of $f^2$. As $f$ has no orbit of period $2$, we conclude that $q=p$ as desired.