Does $f:\mathbb{R} \rightarrow \mathbb{R}$ satisfies $f(f(x)) = x$ have a fixed point?

Question

Given continuous mapping $f:\mathbb{R} \rightarrow \mathbb{R}$ satisfies $f(f(x)) = x$. Does this function has a fixed point?

I know it has fixed point on domain [0,1].

Answer 1

It is not possible that we always have $f(x) > x,$ for then $f(f(x)) > f(x) > x$

It is not possible that we always have $f(x) < x,$ for then $f(f(x)) < f(x) < x$

Thus $g(x) = f(x) - x$ is not always negative or always positive.

If $g$ is always zero we are done. If not, there is some $a$ with $g(a) \neq 0.$ If $g(a) > 0,$ there is some $b$ with $g(b) \leq 0.$ If $g(b)=0$ we are done. If $g(b) < 0,$ there is some $c$ between $a,b$ where $g(c) =0;$ this is the intermediate value theorem for continuous functions.

Same thing if the first one we found had $g(a) < 0.$

Answer 2

Interpreting the condition to be existential too is sufficient for the existence of a fixed point: say $x_0$ is such that $f\circ f(x_0) = x_0$. Put $x_1=f(x_0)$, so that the points $p=(x_0,x_1)$ and $q=(x_1,x_0)$ lie on the graph of $f$. If $x_1=x_0$, then $x_0$ is already a fixed point (and $p=q$). If $x_1 \neq x_0$, then by the continuity of $f$ the part of the graph of $f$ connecting $p$ to $q$ must intersect the line $y=x$.

As a final remark this is a special case of the Sharkovskii's Theorem.