# A continuous involutive function

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 $$f(b)=a$$.

• How about $(1-x^n)^{1/n}$ for $n>0$? Mar 21, 2022 at 19:44
• Thanks Raad, this is a nice example. Honestly, I concentrated on consctructing a function as mentioned above, but I was afraid "it may ruin the continuity". Mar 21, 2022 at 19:52
• You can start with an arbitrary decreasing function from $[a, 1]$ onto $[0, a]$ and “mirror” it along the $y=x$ line. Mar 21, 2022 at 19:54

$$x \mapsto 1-x$$ is certainly not the only example. Another one is $$x \mapsto 1 - \sqrt{2x-x^2}$$
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 $$f_k(x)=\frac{1-x}{1+kx}$$
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 $$0\le f_a(x) for $$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 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 $$f_{0.5}(x)=1-x$$.