Find all bijections $\,\,f:[0,1]\rightarrow[0,1],\,$ which satisfy $\,\,f\big(2x-f(x)\big)=x$.

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.

Solution 1:

In the formulation of the question the function $f$ is one-to-one and onto. Below two proofs are presented. The first one requires the one-to-one and assumption, while the second one does not.

First answer requiring that $f$ is one-to-one.

Clearly, $f$ is also onto, as an arbitrary $x\in[0,1]$, is the image of $2x-f(x)$, and since $f$ is one-to-one and onto, $f$ possesses an inverse $f^{-1}:[0,1]\to [0,1]$.

The functional relation $f\big(2x-f(x)\big)=x$, implies $$ f(x)-x=x-f^{-1}(x). \tag{1} $$ We shall show that $f(x)=x$. Assume not. Then $$ f(x_0)-x_0\ne 0, $$ for some $x_0\in(0,1)$. Let's assume that $f(x_0)-x_0=a>0$. Then $(1)$ implies that $$ f(x_0)-x_0=x_0-f^{-1}(x_0)=f^{-1}(x_0)-f^{-2}(x_0)=\cdots=f^{-k}(x_0)-f^{-(k+1)}(x_0), $$ for every $k\in\mathbb N$, where $f^{-k}$ is $f^{-1}\circ\cdots\circ f^{-1}$ $k$ times. But this means that $$ f^{-1}(x_0)=x_0-a,\,\,f^{-2}(x_0)=x_0-2a,\ldots,f^{-k}(x_0)=x_0-ka, $$ which means that $\lim_{k\to\infty}f^{-k}(x_0)=-\infty$. A contradiction, $f^{-k}(x_0)\in [0,1]$.

We would reach to a contradiction even if we had assumed that $a<0$.

Second answer not requiring that $f$ is one-to-one.

As $f\big(2x - f(x)\big) = x,\,$ for all $x \in \left[ {0,1} \right]$, then $$ 0 \le 2x - f(x) \le 1 \quad \Longrightarrow\quad 2x - 1 \le f(x) \le 2x,\,\,\, \text{for all $x \in \left[ {0,1} \right]$.} $$ Replacing in the above $x$ with $2x - f(x)$ we obtain \begin{align} 2\big( {2x - f(x)} \big) - 1 \le f\big( {2x - f(x)} \big) \le 2\big( {2x - f(x)} \big), \end{align} or \begin{align} 2\big( {2x - f(x)} \big) - 1 \le x \le 2\big( {2x - f(x)} \big) \end{align} which implies that $$ \frac{3x-1}{2}\le f(x) \le \frac{3x}{2}. $$ Repetition of this process produces the following inequalities $$ x+\frac{x-1}{n}\le f(x)\le x+\frac{x}{n},\,\,\, \text{for all $x \in \left[ {0,1} \right]\,\,$ and $\,\,n\in\mathbb N$,} $$ and therefore $$ f(x)=x ,\,\,\, \text{for all $x \in \left[ {0,1} \right]$.} $$

Solution 2:

OK guys, thanks to @woso's suggestion I found an answer. Let me share it with you.

First Answer

By $(2)$ we have $$f(x)-x=x-f^{-1}(x)\tag{3}$$ Now let $x_0\in[0,1]$ be arbitrary. We define the sequence $(x_n)_{n=1}^{\infty}$ as follows $$x_n=f(x_{n-1})\tag{4}$$ for all $n\ge1$. From $(3)$ and $(4)$ we have $$\begin{align*}x_n-x_{n-1}&=x_{n-1}-x_{n-2}\\ &= x_{n-2}-x_{n-3}\\ &= x_{n-3}-x_{n-4}\\&\vdots\\ &= x_{1}-x_{0}\tag{5}\end{align*}$$ Now from $(5)$ we have $$x_n-x_0=\sum_{k=1}^{n}{(x_k-x_{k-1})}=\sum_{k=1}^{n}{(x_1-x_0)}=n(x_1-x_0)\tag{6}$$ Now since $x_n,x_0\in[0,1]$ so by $(6)$ we have $$n|x_1-x_0|=|x_n-x_0|\le 1\tag{7}$$ so by $(7)$ for alll $n\in\Bbb{N}$ we have $|x_1-x_0|\le \frac{1}{n}$, but it's true only if $|x_1-x_0|=0$, so finaly $$f(x_0)=x_1=x_0$$ since $x_0\in[0,1]$ was arbitrary so $f(x)=x$ for all $x\in[0,1]$ and it's the only solution.

Second Answer

We can show that $f(x)=x$ is the only solution even without considering $1$-$1$ and onto conditions.

For all $x\in[0,1]$ consider $g(x)=x-f(x)$. Let $x_0\in[0,1]$ be arbitrary, then $g(x_0)=x_0-f(x_0)$ and by assumption $x_1=x_0+g(x_0)=2x_0-f(x_0)\in[0,1]$, also we can see that $g(x_1)=g(x_0)$, the same process shows that $x_2=x_1+g(x_1)\in[0,1]$ and $g(x_2)=g(x_1)=g(x_0)$,so the sequence $(x_n)$ defined by $x_n=x_{n-1}+g(x_{n-1})$ has these properties:

1. for all $n\ge0$, $x_n\in[0,1]$;

2. for all $n\ge1$, $g(x_n)=g(x_0)$.

so $x_n-x_{n-1}=g(x_{n-1})=g(x_0)$, and $$x_n-x_0=\sum_{k=1}^{n}{(x_k-x_{k-1})}=\sum_{k=1}^{n}{g(x_0)}=ng(x_0) \tag{*}$$

Now since $x_n,x_0\in[0,1]$ so by $(*)$ $$n|g(x_0)|=|x_n-x_0|\le 1\tag{**}$$ so by $(**)$ for all $n\in\Bbb{N}$ we have $|g(x_0)|\le\frac{1}{n}$, but it's true only if $g(x_0)=0$, so finally $$f(x_0)=x_0-g(x_0)=x_0$$ since $x_0\in[0,1]$ was arbitrary so $f(x)=x$ for all $x\in[0,1]$and it's the only solution.

If you find something wrong with it, please add a comment.

Solution 3:

I think the answers so far are great, but I thought I'd record a geometrically motivated idea I had and see if people could make it work. The given equality is equivalent to (2) which is equivalent to $$ \frac{f(x) + f^{-1}(x)}2 = x. $$ In other words, the graph of $y=f(x)$, when reflected across $y=x$, has the property that vertical line segments from $(t,f(t))$ to $(t,f^{-1}(t))$ have their midpoints at $(t,t)$ for all $t\in[0,1]$. But that shouldn't be compatible with the fact that all northwest-southeast line segments $(t,f(t))$ and $(f(t),t)$ are also bisected by $y=x$.

If $f(t)>t$ ever, maybe we should choose $t$ such that $f(t)-t$ is maximized, in fact letting $t$ be the smallest such $t$ that maximizes $f(t)-t$. Then we should be able to show that $t-f^{-1}(t)$ is strictly smaller than $f(t)-t$. And since the problem is symmetric in $f$ and $f^{-1}$, the case $f(t)>t$ suffices. (My intuition here has $f$ continuous, but it might be possible to make it work without assuming continuity a priori, by replacing max/min by sup/inf.)