How to show that $f(x)\equiv x$ for an injective and continuous $ f $ satisfying $f\big(2x-f(x)\big)\equiv x$?

Question

Let $f:\mathbb R\to \mathbb R$ be a bijective function with a fixed point $a$. i.e. $f(a)=a$. Suppose $f$ satisfies the following relation $$f\big(2x-f(x)\big)=x \text,\qquad \forall x\in \mathbb R \text.$$ Then how can I show that $f(x)\equiv x$?

Additional question: what about if $f$ is injective and continuous?

I saw a bit long solution of this before somewhere (I don't know where). I am looking for a beautiful solution.

Edit: I think one can work on the following way for solution:

Suppose there exist a $b$ such that $f(b)\ne b$. Then the set $$K:=\{b,f(b),f\circ f(b), \dots \}$$ is a infinite set of distinct points. I have no Idea how to use this!

Answer 1

This is false. To construct a counterexample, let us first see what the functional equation says about a single orbit of $f$. Applying $f^{-1}$ to both sides and substituting $y=f^{-1}(x)$, the functional equation is equivalent to $2f(y)-f(f(y))=y$ or $$f(f(y))=2f(y)-y.$$ Now fix some $x\in\mathbb{R}$ and write $x_n=f^n(x)$ for $n\in\mathbb{Z}$. The functional equation is then equivalent to saying that every such orbit $(x_n)$ satisfies the linear recurrence $$x_{n+2}=2x_{n+1}-x_n.$$ The solutions to this recurrence are just $x_n=cn+d$ for constants $c$ and $d$. Or, since $d$ must be $x_0=x$, given our initial point $x$ we can take any sequence of the form $x_n=cn+x$.

So to construct a counterexample, we just have to construct a bijection whose orbits all have this form. That is easy by transfinite recursion: if we've already defined $f$ on a set of orbits of size less than $\mathfrak{c}$, we can extend $f$ to one more point $x$ by picking $c$ so that the orbit $\{cn+x:n\in\mathbb{Z}\}$ of $x$ will not intersect any of the orbits we have chosen so far. So, building $f$ one orbit at a time like this by transfinite recursion (starting from a single orbit that gives $f$ a fixed point), we can build $2^{\mathfrak{c}}$ different solutions (since we have more than one choice of $c$ that works at each step).

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On the other hand, it is true if you assume $f$ is continuous and injective instead of bijective. First, note that injectivity implies bijectivity, since the functional equation immediately implies $f$ is surjective. Now suppose $f$ is additionally continuous and $f(x)\neq x$ for some $x$. By the analysis above, $f^n(x)$ must have the form $cn+x$ for some constant $c$, and $c\neq 0$ since $f(x)\neq x$. Now there exists $n\in\mathbb{Z}$ such that the fixed point $a$ is between $cn+x$ and $c(n+1)+x$. Let us assume for convenience of notation that $c>0$; the case $c<0$ is similar. We then have $$cn+x<a$$ but $$f(a)=a<c(n+1)+x=f(cn+x),$$ which means $f$ must be decreasing (it must be monotone since it is a continuous bijection). But $$x<c+x$$ and $$f(x)=c+x<2c+x=f(c+x)$$ so $f$ must be increasing. This is a contradiction.

Answer 2

As Eric Wofsey wrote in his comment, the statement as it stands is false. Here is a counterexample. Let $$f(x)=\begin{cases}x,&x\in\mathbb Z;\\ x-1,&x\not\in\mathbb Z.\end{cases}$$ Then $f$ is bijective, it has a fixed point (any integer will do) and $$f(2x-f(x))=\begin{cases}f(2x-x)=f(x)=x,&x\in\mathbb Z;\\ f(2x-(x-1))=f(x+1)=x,&x\not\in\mathbb Z.\end{cases}$$ However, the statement is true if both the domain and codomain of $f$ are equal to some bounded set $S$. See the question mentioned by Tsemo Aristide in another comment.