# Is my proof of $\neg (f(f(x)) = f^{-1}(x)) \land (f(x) = x)$ correct?

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

Proof:

$$f(x)$$ 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 $$f(x) = x$$.

EDIT:

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.

• Doesn't the identity function satisfy this? Apr 27 at 16:12
• 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 $z\in\Bbb C,\,z^3=1$.
– J.G.
Apr 27 at 16:18
• I also don't see the "which finally means $-x=f^{-1}(x)$" line. Apr 27 at 16:19
• Here he means that $f(x) \neq \ x \forall x$ , or the proof wouldn't make sense. Apr 27 at 16:52
• 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 Apr 27 at 18:01

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 $$\frac{-x}{-1}$$