Theorem: $$ f(f(x)) = f^{-1}(x) \implies f(x) = x, \ \ f : \Bbb R \to \Bbb R $$


$f(x)$ is a real, continuous function that satisfies $f(f(x)) = f^{-1}(x)$.

That means that either $f : [x,f(x)] \to [f(x), f^{-1}(x)]$ or $f : [f(x),x] \to [ f^{-1}(x), f(x)]$. This is because the function is strictly monotonic, since it has an inverse.

This interval mapping shows us that $f(x) - x = f^{-1}(x) - f(x)$, which means $2f(x) -x = f^{-1}(x)$, yielding the equality:

$$f(f(x)) = 2f(x) -x$$

From that, we gather some info about the function itself:

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

This leads to the fact of $f(f(x) -2x) = x$, with the consequence of:

$$f(f(f(f(f(x) -2x)))) = f(f(x) -2x) = x$$

This can only be true if $f(x) = x$.


The above proof has been edited a lot. In the first revision, I added a vital component I had forgotten. However, the revised proof was still incorrect, although the conclusion was correct. I have now reworked the proof quite drastically, and I believe it is correct.

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    $\begingroup$ Doesn't the identity function satisfy this? $\endgroup$
    – Randall
    Apr 27 at 16:12
  • 1
    $\begingroup$ It's not clear what constraints on $f$ beside $f^2=f^{-1}$ you're assuming that rule out options such as $f(x)=zx$ with $z\in\Bbb C,\,z^3=1$. $\endgroup$
    – J.G.
    Apr 27 at 16:18
  • $\begingroup$ I also don't see the "which finally means $-x=f^{-1}(x)$" line. $\endgroup$
    – Randall
    Apr 27 at 16:19
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    $\begingroup$ Here he means that $f(x) \neq \ x \forall x$ , or the proof wouldn't make sense. $\endgroup$
    – RicardoMM
    Apr 27 at 16:52
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    $\begingroup$ what about $ f(x) = \frac{-1}{x+1} \; ? \; \;$ If you want to rule this out based on the asymptote at $x = -1,$ you need to discuss the domain, somehow $\endgroup$
    – Will Jagy
    Apr 27 at 18:01

2 Answers 2


As other people have mentioned in the comments, there are several issues with the clarity of your proof. It can be assumed that by $f(x)$ is a function, you mean a continuous real valued function $f:\mathbb{R}\to S\subseteq\mathbb{R}$. I will continue under this assumption.

At the beginning of the second paragraph you are making an unjustified step. Why would $[x,f(x)]$ map to $[f(x),f^{2}(x)]=[f(x),f^{-1}(x)]$? There is a property of $f$ you are trying to use implicitly, but confused slightly:

that property is that continuous functions with an inverse are monotone

From this you should be able to directly derive some inequalities that show that $f(x)=x.$

Hope this helps.

P.S. It might also help to think of your conditions as saying that $f^3(x)=x$ AND $f$ is invertible.


well, no. $$ f(x) = \frac{-1}{x+1} $$

function composition for Moebius transformations is nothing more complicated then matrix multiplication.

Take $$ P = \left( \begin{array}{rr} 0 & -1 \\ 1 & 1 \\ \end{array} \right) $$ $$ P^3 = \left( \begin{array}{rr} -1 & 0 \\ 0 & -1 \\ \end{array} \right) $$ which gives the identity function $ \frac{-x}{-1} $


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