A continuous involutive function

Question

Suppose that $f:[0,1]\rightarrow [0,1]$ is a continuous function such that $f(f(x))=x$ for all $x\in [0,1]$.

We know f is one to one and onto. Morover, it has a fixed point.

If we assume further that $f$ is strictly decreasing, we conclude that $f(0)=1, f(1)=0$, do these conditions imply that $f(x)=1-x$ for all $x$?

I think the above information is not sufficient to claim that f is of this form. However, I was not able to construct another function satisfying the given properties, for example, by partitioning the domain into pairs $\{a,b\}$ where $f(a)=b$ and $f(b)=a$.

Please help. Thanks alot.

Answer 1

$x \mapsto 1-x$ is certainly not the only example. Another one is $$x \mapsto 1 - \sqrt{2x-x^2}$$

Answer 2

I believe every function that passes through $(0,1)$ and $(1,0)$ and is symmetric along the bisectrix $y=x$ is a suitable example, for instance $$f_k(x)=\frac{1-x}{1+kx}$$

Answer 3

For $a \in (0.1)$, define $f_a: [0,1] \to [0,1]$ by

$$ f_a(x)= \begin{cases} -mx+1 & \mbox{ for } 0 \le x\le a\cr -\frac{x-1}{m} &\mbox{ for } a< x \le 1 \end{cases}, $$ where $\displaystyle m=\frac{1-a}{a}>0$.

It is clear that $f_a $ is decreasing, $f_a(0)=1, f_a(1)=0$ and $f_a(a)=a$.

Notice that $a \le f_a(x) \le 1$ for $0 \le x \le a$, and $0\le f_a(x)<a $ for $a< x \le 1$. It follows for $0\le x \le a$, we have $$ f_a(f_a(x))=f_a(-mx+1)=-\frac{-mx+1-1}{m}=x, $$ and for $a<x\le 1$ we have $$ f_a(f_a(x))=f_a\left(-\frac{x-1}{m}\right)=-m\left(-\frac{x-1}{m}\right)+1=x. $$ Hence $f_a(f_a(x))=x$ for all $ x \in [0,1]$. In particular $f_{0.5}(x)=1-x$.