Show that a continuous mapping $f:[0,1]\to [0,1]$ which satisfies $f(f(x))=x$ for each $x \in [0,1]$, and for which $f(x)$ does not equal x for at least one x in [0,1], must have exactly one fixed point.
I was thinking to go by contradiction: if there are two fixed points $x$ and $x'$ then plug into definition, but I came out with nothing. The fact that $f(x)$ doesn't equal $x$ for at least one $x$ tells me that we have not the identity function; however this doesn't assure me that the function is increasing/decreasing everywhere i.e $[0,1]$. So, I have nothing.
Thanks a lot!
You know that $f(f(x))=x$ so $f$ is bijective. A bijective continuous function is strictly monotone.
If $f$ is strictly decreasing you have a unique fixed point. If $f$ is increasing, then if eventually $f(x) \neq x$ it follows that $f(f(x))\neq x$. For example if $f(x)>x$ we have $f(f(x))>f(x)>x$ (contradiction).