# Find all real and continuous functions that are a 3-involution.

Find all continuous functions $f: \mathbb R \rightarrow \mathbb R$such that $f(f(f(x)))=x$.

Obviously one solution to this functional equation is $f(x)=x$.

If the function is NOT continuous, there are also other solutions such as $f(x)=\frac{1}{1-x}$, but I'm not sure how to find all solutions that are continuous.

• $x\longmapsto\dfrac1{1-x}$ is continuous, but its domain is not $\mathbb{R}$. – gniourf_gniourf Feb 20 '16 at 13:18
• You could let f(1)=1 – John Smith Feb 21 '16 at 10:12
• No, because with $f:\mathbb{R}\longrightarrow\mathbb{R}:x\longmapsto\begin{cases}1&\text{if$x=1$}\\\dfrac1{1-x}&\text{if$x\neq1$}\end{cases}$ the function is not injective. The only way to have $f$ injective is $$f:\mathbb{R}\longrightarrow\mathbb{R}:x\longmapsto\begin{cases}0&\text{if x=1}\\\dfrac1{1-x}&\text{if x\neq1}\end{cases}$$ but then $f(f(1))=f(0)=1$ and hence $f(f(f(1)))=f(1)=0\neq1$. – gniourf_gniourf Feb 21 '16 at 10:21
• – Martin Sleziak Dec 4 '16 at 14:08

Observe that $f$ must be injective. Now, since $f$ is continuous and injective on the interval $\mathbb{R}$, we conclude that $f$ is monotone. Now the variations of $f\circ f\circ f$ are that of $f$, and we conclude that $f$ must be increasing.

Now, let $x\in\mathbb{R}$. There are three cases:

1. $f(x)<x$,
2. $f(x)>x$,
3. $f(x)=x$.

We'll show that cases 1. and 2. are impossible:

1. Assume that $f(x)<x$. Then, since $f$ is increasing, $f(f(x))<f(x)$, and applying $f$ again yields $x=f(f(f(x)))<f(f(x))$ from which we conclude that $x<x$, which is impossible.
2. Similarly, assume that $f(x)>x$. Then, $f(f(x))>f(x)$ and $x=f(f(f(x)))>f(f(x))$ from which we conclude that $x>x$, which is impossible.
3. Hence we must have $f(x)=x$.

This proof easily generalizes to the case of a continuous function $f:\mathbb{R}\longrightarrow\mathbb{R}$ such that $$\forall x\in\mathbb{R},\ f^{[2p+1]}(x)=x,$$ where $p\in\mathbb{N}$ and $f^{[2p+1]}$ stands for the $(2p+1)$-th iterate of $f$.