Prove that $f(f(x))=x$ has no roots .... $f$ having a general form

Question

This problem gave me some headache, especially because $f$ have its own general form :

let $f(x) = ax^2 + bx + c$. Suppose that $f(x) = x$ has no real roots.

Show that equation $f(f(x))=x$ has also no real roots.

Answer 1

Consider $g(x) = f(x)-x$. If $f$ is continuous, then $g$ will be continuous as well. A continuous function without zeroes is either strictly positive or strictly negative. This means that either $f(x)>x$ for all $x$ or $f(x)<x$ for all $x$.

Hence either $f(f(x))<f(x)<x$ for all $x$ or $f(f(x))>f(x)>x$ for all $x$. Either way, $f(f(x))\neq x$ for all $x$.

Answer 2

Let $\mbox{Id}_\mathbb{R}:\mathbb{R}\to \mathbb{R}$ the function $\mbox{Id}_\mathbb{R}(x)=x$. Note that :

• $\mbox{Im}(\mbox{id}_{\mathbb{R}})\supset \mbox{Im}(f)$,

• $\mbox{Im}(f)=\mbox{Im}(f|_{\mbox{id}_{\mathbb{R}}})\supset \mbox{Im}(f|_{\mbox{Im}(f)})=\mbox{Im}(f\circ f)$

• $\mbox{Im}(f\circ f-\mbox{id}_\mathbb{R})\supset\mbox{Im}(f|_{\mbox{id}_\mathbb{R}} -\mbox{id}_\mathbb{R})$.

Then if $f|_{\mbox{id}_\mathbb{R}} -\mbox{id}_\mathbb{R}$ no have real roots then $f\circ f-\mbox{id}_\mathbb{R}$ too no have real roots.

Answer 3

let g(x)=f(x)-x, because f(x)=x has no real roots, then g(x)>0 or g(x)<0 for all x belongs to R.

suppose g(x)>0 for all x. then g(f(x))>0 for all x.

f(f(x))-x=f(f(x))-f(x)+f(x)-x=g(f(x))+g(x)>0 for all x

then f(f(x))-x>0 for all x, so f(f(x))=x also has no real roots.

the same proof for g(x)<0 for all x.

Answer 4

This can be generalized to: If $f:\mathbb R\rightarrow \mathbb R$ is a continuous function and $f(f(x))$ has a real fixed point then $f(x)$ has a real fixed point:

If $f(f(c))=c$ and $f(c)=r$ then $f(r)=c$.
If $c\neq r$ then there is an $s$ between $r$ and $c$ such that $f(s)=s$ (In the case $c<r\Rightarrow f(r)-r=c-r>0$ and $f(c)-c=r-c<0$...).