TITLE: Find all bijections $\,\,f:[0,1]\rightarrow[0,1],\,$ which satisfy $\,\,f\big(2x-f(x)\big)=x$.
QUESTION [19 upvotes]: A friend of mine gave me the following problem: 
Find all functions $f:[0,1]\to[0,1]$, which are one-to-one and onto and satisfy the following functional relation:
$$
f\big(2x-f(x)\big)=x, \tag{1}
$$
for all $x\in [0,1]$.
Clearly, the identity function $f(x)=x$ is one such function.
Also, as $f$ is a bijection $f^{-1}$ exists, and by $(1)$ we have
$$
f^{-1}(x)=2x-f(x), \tag{2}
$$
but I have no idea that how should I continue. It will be great if someone can give me some hints. 
Thanks in advance.

REPLY [2 votes]: Suppose that, for some $x_0$,
$$
f(x_0)-x_0=\Delta\tag{1}
$$
where $\Delta\ne0$. By the hypothesis,
$$
f^{-1}(x_0)=2x_0-f(x_0)=x_0-\Delta\tag{2}
$$
Thus, subtracting $x_0$ from $(2)$ gives
$$
f^{-1}(x_0)-x_0=-\Delta\tag{3}
$$
Considering $(1)$ and $(3)$, we may assume $\Delta\gt0$ by switching the roles of $f$ and $f^{-1}$.

Applying $f$ to $(2)$ yields
$$
f(x_0-\Delta)=x_0\tag{4}
$$
which gives
$$
f(x_0-\Delta)-(x_0-\Delta)=\Delta\tag{5}
$$
which is $(1)$ shifted by $\Delta$. Therefore, inductively, we get
$$
f(x_0-k\Delta)-(x_0-k\Delta)=\Delta\tag{6}
$$
and
$$
f^{-1}(x_0-k\Delta)-(x_0-k\Delta)=-\Delta\tag{7}
$$
$\hspace{3cm}$

Since $\Delta\gt0$, we can find a $k$ so that $x_0-k\Delta\ge0$ and $x_0-(k+1)\Delta\lt0$. For this $k$, $(7)$ says
$$
f^{-1}(x_0-k\Delta)=x_0-(k+1)\Delta\lt0\tag{8}
$$
but $f^{-1}:[0,1]\mapsto[0,1]$.
Thus, $\Delta=0$, and therefore, $f(x)=x$.