# 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.

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The only one as I think is $y=x$ – Sameh Shenawy Jan 28 '14 at 18:03
@Semsem, so you suggest to prove that $f(x)=x?$ – Xabat Jan 28 '14 at 18:07
let us do it together I have to prove or you have to give one different example – Sameh Shenawy Jan 28 '14 at 18:11
How do I interpret (1) when $2x-f(x)\notin[0,1]$, or are we assuming that $f$ has the property that $2x-f(x)\in [0,1]$ for all $x$? – Jason DeVito Jan 28 '14 at 19:03
@JasonDeVito, The assumption says $f$ is a function on $[0,1]$, so "Yes", $2x-f(x)\in [0,1]$ for all $x\in [0,1]$. – Xabat Jan 28 '14 at 19:09

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 $$f(x)=x ,\,\,\, \text{for all x \in \left[ {0,1} \right].}$$

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Can you explain a little more that why $f$ is increasing? – Xabat Jan 28 '14 at 21:22
@Xabat: If a function $f:[a,b]\to[a,b]$ is continuous and 1-1, then it is necessarily strictly increasing or strictly decreasing. – Yiorgos S. Smyrlis Jan 28 '14 at 21:25
@Xabat: Actually, I am not using this in the proof, after all! – Yiorgos S. Smyrlis Jan 28 '14 at 21:30
@Xabat: See update. – Yiorgos S. Smyrlis Jan 28 '14 at 21:31
I see, that's nice and perfect. Thanks and (+1). What do you think about @woso's claim, that 1-1 condition is not necessary? – Xabat Jan 28 '14 at 21:41

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

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.

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.

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(+1) for me too - We need to make sure that the 1-1 condition is necessary! – Yiorgos S. Smyrlis Jan 28 '14 at 23:08

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.)

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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, $f(x)=x$.

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