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This is motivated by a question that was posted last night, but was deleted (I think by the author) before any answers to it appeared. I don't think it has been re-posted since then.

What bijective functions $f$ from $\mathbb{R} \rightarrow \mathbb{R}$ satisfy $$f(x)=\frac{f^{-1}(x)-f^{-1}(-x)}{2}?$$

I first spotted that the RHS was the "odd part" of the inverse function, making $f$ an odd function. But this implies that $f^{-1}$ is odd, making the RHS equal to $f^{-1}$. So I end up with the following information:

  1. $f$ is odd
  2. $f(x)=f^{-1}(x)$.

This is where I got stuck. The only $\mathbb{R} \rightarrow \mathbb{R}$ functions symmetric with respect to both the origin and the line $y=x$ that I can visualize are $f(x)=x$ and $f(x)=-x$, but I am not sure if they are indeed the only ones.

How I can prove/disprove that $f(x)=x$ and $f(x)=-x$ are the only odd, bijective $\mathbb{R} \rightarrow \mathbb{R}$ involutions?

Air Conditioner
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    What about $f(x)=1/x$? Or do you really require domain of definition all real numbers? – Dietrich Burde Apr 07 '18 at 16:59
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    would $x^3$ work? – Arian Apr 07 '18 at 17:00
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    @DietrichBurde Oh oops, $f(x)=1/x$ isn't defined at $0$, so I ignored it. Thanks for pointing that out--I should make the bijectivity requirement more clear. Yes, I'm looking for a function defined on all real numbers. – Air Conditioner Apr 07 '18 at 17:01
  • @Arian $f(x)=x^3$ is not its own inverse, so it does not. – Air Conditioner Apr 07 '18 at 17:03
  • Would $f(x)=1/x$ if $x\neq 0$ and $f(0)=0$ work? – Carl Schildkraut Apr 07 '18 at 17:05
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    How's about $$f(x) = \begin{cases} x & \text{if $x\in\mathbb{Q}$} \ \frac{1}{x} & \text{otherwise?} \end{cases} $$ – Xander Henderson Apr 07 '18 at 17:06
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    If you don't require that $f$ is continuous, then there are a lot of such functions. You can basically construct one freely. $f(0)=0$, but other than that, we can take any real number $x$, and decide freely what we want $f(x)$ to be. This forces $f(-x)$ and $f(f(\pm x))$. Then you can take solve other number, and do the same thing, almost as freely (you can't pick a function value which is already in use). – Arthur Apr 07 '18 at 17:06
  • @Arthur What if I reject the axiom of choice? :P – Xander Henderson Apr 07 '18 at 17:08
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    @XamderHenderson Then you do this a finite number of times, then set $f(x)=x$ for any other number. – Arthur Apr 07 '18 at 17:08
  • Yep, my brain was stuck thinking about continuous functions. Oops. Thanks! You can post these as answers, if you wish. (I'm still curious about the continuous case, but think I'll ask about that in separate question). – Air Conditioner Apr 07 '18 at 17:10
  • Yes, post these as answers, I am awaiting this discussion. – Duchamp Gérard H. E. Apr 07 '18 at 17:46

1 Answers1

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As noted in the comments, there are a lot of possibilities if $f$ is not assumed continuous.

Under the assumption that $f$ is continuous, however, $f(x)=+x$ and $f(x)=-x$ are the only odd involutions of $\mathbb{R}$. Let $f$ be such a function.

As $f$ is a continuous bijection of $\mathbb{R}$, it is strictly monotonic. Since $g(x)=-f(x)$ is also odd, and also an involution since $g(g(x))=-f(-f(x))=x$, we may (and will) assume that $f$ is strictly increasing. We have to show that $f(x)=x$ for all $x$.

Let $E^+=\{x : f(x)>x \}$, $E^-=\{x \, : f(x)<x\}$. These sets are open by continuity of $f$, hence (at most countable) union of disjoint open intervals. Since $f$ is odd, $f(0)=0$ so $0\notin E^+\cup E^-$.

Let $(a,b)$ be such an interval, we first assume this is in $E^+$, let us have a look at the case where $(a,b)\subset E^+$. Since $0\notin E^+$, $a$ or $b$ has to be finite.

Say this is $a$. Since $a\notin E^+$, and $f$ continuous, $f(a)=a$. By continuity, and because $f$ is increasing, $f(a+\epsilon)\in (a,b)\subset E^+$ for small enough $\epsilon>0$. So by definition of $E^+$, $$f(f(a+\epsilon))>f(a+\epsilon)>a+\epsilon.$$ Since $f$ is an involution, we get a contradiction. The four other cases ($b$ is finite but $a$ is not, or $(a,b)\subset E^-$, $a$ or $b$ is finite, are handled similarly.

user120527
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