If $f:\mathbb{R}\to\mathbb{R}$ is continuous and $f(x)\neq x$ for all $x$, must it be true that $f(f(x))\neq x$ for all $x$?

Question

Let $f: \Bbb R → \Bbb R$ be a continuous function such that $f(x)=x$ has no real solution . Then is it true that $f(f(x))=x$ also has no real solution ?

Answer 1

Hint: We know that $f(x)\neq x$ for all $x\in\mathbb{R}$. Without loss of generality, let's say that at $x_0\in\mathbb{R}$, we have that $f(x_0)<x_0$. Can it ever be the case that $f(x_1)>x_1$ for any other $x_1\in\mathbb{R}$?

Answer 2

Since $f$ is continuous, saying that $f(x) = x$ has no solution means that either $f(x) < x$ for all $x$ or $f(x) > x$ for all $x$. Let's assume wlog that $f(x) > x$, then $f(f(x)) > f(x) > x$ for all $x$ so $f(f(x))=x$ has no solution either.

Answer 3

Just a variation of the proof. Suppose $f(f(x))=x$. Define $g:x\mapsto f(x)-x$, which is continuous, and has $g(f(x))=f(f(x))-f(x)=-(f(x)-x)=-g(x)$. Then by the intermediate value theorem $g(y)=0$ for some $y$ in the closed interval bounded by $x$ and $f(x)$, and then of course $f(y)=y$.

Answer 4

If $f(f(c))=c$ and $f(c)=r$ then $f(r)=c$.
Since $c\neq r$ it follows that 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 $f(c)-c=r-c<0$...) $\Rightarrow\Leftarrow$.