Function such that $f(f(f(x))) = f(f(x)) \neq f(x)$

Question

Give an example of a function $f\colon\Bbb R\rightarrow \Bbb R$ such that $f\circ f\circ f=f\circ f\neq f$, that is, $f(f(f(x))) = f(f(x))$ on $\Bbb R$ but $f(f(x))\ne f(x)$ for some $x\in\mathbb R$.

Answer 1

You may try the continuous function $f(x)=|x|-x$. Then $f(f(x))=0$.

Answer 2

Good old rational indicator function

$$f(x) =\mathbf 1_{\mathbb Q}(x)= \begin{cases} 1 & \text{if $x\in\mathbb{Q}$} \\ 0 & \text{otherwise} \\ \end{cases} $$

does the trick.

Answer 3

Given any integer $n\ge2$, we can define $f:\mathbb R\to\mathbb R$ such that $f,f^2,\ldots,f^n$ are distinct functions but $f^n=f^{n+1}$. Your question is just the case where $n=2$.

We first define $f$ on a smaller domain and then extend its definition to the whole real line. Let $f:\{0,1,\ldots,n+1\}\to\mathbb R$ be defined by $$f(x)=\begin{cases} x,&x=0,1,\\ x-1,&x=2,3,\ldots,n+1. \end{cases}$$ Then $f^k(n+1)=n+1-k$ for $k=1,2,\ldots,n$. Hence $f,f^2,\ldots,f^n$ are distinct functions. However, as $f^n(0)=0$ and $f^n(x)=1$ for all $x>0$, we have $f^{n+1}=f^n$.

Next, we extend the definition of $f$ to the nonnegative real line. When $x\ge0$ but $x\not\in\{0,1,\ldots,n+1\}$, we define $f$ such that \begin{cases} f(x)=x,&x\in(0,1),\\ f(x)\le k-1,&x\in(k-1,k),\,k=2,3,\ldots,n+1,\\ f(x)\le n,&x>n+1. \end{cases} (Note that a $f$ can be defined as a continuous function, although this is not required by the OP.) Then $f(\mathbb R_{\ge0})\subset[0,n],\,f^2(\mathbb R_{\ge0})\subset[0,n-1],\,\ldots$ and so on, so that $f^n(\mathbb R_{\ge0})\subset[0,1]$. Since $f$ is the identity function on $[0,1]$, we have $f^{n+1}=f^n$.

Finally, when $x<0$, we define $f(x)=x$ or $f(x)=-f(-x)$.