What can be $f$ so that $f(f(x)) = -x$?

Question

What can be $f$ so that $f^2(x) = -x$ for all $x\in R$?

I know that if $f^2(x) = -x$ then $f(x)$ is injective and $f$ can not be continuous.

But I can not find an example of discontinuous function so that $f^2(x) = -x$ for all $x\in R$.

Can anyone help me?

See https://math.stackexchange.com/questions/312385/continuous-function-f-mathbbr-to-mathbbr-such-that-ffx-x — Robert Z, Oct 04 '17 at 15:45
First of all, clarify your notation. Presumably, $f^2(x) = f(f(x))$. Second, how do you know that $f$ cannot be continuous? Finally, what have you tried? — Ben Grossmann, Oct 04 '17 at 15:46
@amWhy it's clear that the asker is looking for a potentially discontinuous solution. — Ben Grossmann, Oct 04 '17 at 15:47
@Omnomnomnom Yes, but the answer there provides a discontinuous solution as well — Mike Earnest, Oct 04 '17 at 15:48
@MikeEarnest just took a closer look as you commented, thanks — Ben Grossmann, Oct 04 '17 at 15:48
@sani don't just look at titles of duplicates, look at the answers!. — amWhy, Oct 04 '17 at 15:52

score 0 · Accepted Answer · answered Oct 04 '17 at 15:49

0

Just look at the cycle representation of $f\circ f=-x$, there is one fixed element and a bunch of $2$-cycles, So to build a suitable permutation just pair up the $2$-cycles and for each $(a,b)$ and $(c,d)$ add the $4$-cycle $(a,c,b,d)$.

answered Oct 04 '17 at 15:49

Asinomás

105,651

In summary, if $A,B$ is a partition of the positive reals and $g:A\rightarrow B$ is a bijection then the permutation with cycles $(a,g(a),-a,-g(a))$ works. – Asinomás Oct 04 '17 at 15:51

What can be $f$ so that $f(f(x)) = -x$?

1 Answers1