Let $\phi: \mathbb{R} \to \mathbb{R}$ a homeomorphism of finite order, it is $\phi^p(x)=x$ for $p>0$.Prove that $\phi$ has a fixed point.
Attempt
Let $g: \mathbb{R} \to \mathbb{R}^p$ given by
$$g(x)=(\phi(x)-x,\phi^2(x)-\phi(x), \cdots, \phi^p(x)-\phi^{p-1}(x))$$
$(i)$ Notice that $g$ is continuous since is linear combination of continuous functions.
$(ii)$ WLOG assume that $0< \phi(0)$ in otherwise ($\phi(0)<0$) and notice that if $g$ don't have a fixed point in $[0,\phi(0)]$ then for each $i\in 0,1,\cdots,p-1$ we will have $\phi^{i+1}(x)-\phi(x)>0$
(in otherwise in $[ \phi(0),0]$we should have $\phi^{i+1}(x)-\phi(x)<0)$
But by the continuity from $(i)$ it implies that $\phi$ maintain the sign in each enter of $g$.
$(iii)$ Finally notice that if $g$ maintain the sign in all the entries then it would implies that $\phi$ is not periodic which is not true, then $g$ don't maintain the sign in all their entries and therefore by $(ii)$ we get that $\phi$ have a fixed point.
I have two questions about my proof.
In $(ii)$ is valid the arguments to show that if $\phi$ have a fixed point then each of the entries maintain the same sign (Wich depends that if $0< \phi(0)$)?.
The part $(iii)$ I felt that is the more frail part of the proof.
There are other more strong argue to show that is impossible that the signs of all the entries of $g$ maintain constant ?
Best Answer
In (ii), I don't see why $g$ not having a fixed point would imply that each component is strictly positive, it could be that one of them is strictly positive and another is strictly negative. In (iii), it seems like you are not arguing the point.
Here's a neat proof of the claim: Assume that $\phi$ does not have a fixed point, then WLOG we can assume that $ \phi(x)-x > 0$ for all $x \in \mathbb{R}.$(By the mean value theroem) This implies that
$$ \phi(\phi^{k-1}(x)) -\phi^{k-1}(x) >0$$ for $1 \le k\le p.$ But now $$0= \phi^p(x) -x= \sum_{k=1}^p (\phi^{i}(x) -\phi^{i-1}(x)) >0,$$ a contradiction.