Is there a real function that satisfies the relation $f(f(x)) = 1-x$?

Question

As already one user wrote out on Stack Exchange there is a discontinuous function that satisfies the equation $f(f(x)) = -x$. Namely:

Find a real function $f:\mathbb{R}\to\mathbb{R}$ such that $f(f(x)) = -x$?

--> click on the image here to view the function: g(x)

I also know that the function mentioned in that piece is equal to

$$f(x) = g(x-0.5)+0.5$$

Can anyone help me out on this and find the appropriate f(x)?

g(x) is mentioned in the post I linked in my description, name Hurkyl — King, Jul 25 '18 at 20:31
At the end of Hurkyls post there is a function mentioned that satisfies the equation f(f(x)) = -x. I mean that explicit function — King, Jul 25 '18 at 20:35
Please copy the $g$ that you refer to and paste its definition into your post. — md2perpe, Jul 25 '18 at 20:40
Geometric observation $$ 1-f(x) = f(f(f(x))) = f(1-x), $$ so $f(x) + f(1-x) = 1$. — I was suspended for talking, Jul 25 '18 at 20:46
@Sisyphus, you assume $x=f(x)$, your statement final should be $f(1-f^{-1}(x))=1-x$ — Jacob Claassen, Jul 25 '18 at 21:09
Also any solution to $1-f(x) = f(1-x)$ can be written as $f(x) = g(x - 0.5) + 0.5$ where $g$ is odd. But you can't just take any odd function. That's the problem. — md2perpe, Jul 25 '18 at 21:09
No, @JacobClaassen. $f(1-x) = { f(f(x)) = 1-x } = f(f(f(x))) = { f(f(t)) = 1-t, t = f(x) } = 1 - f(x).$ Nothing is assumed. — md2perpe, Jul 25 '18 at 21:11
@christhian-grundmann. That's only a solution to $1-f(x) = f(1-x),$ not to the original problem. — md2perpe, Jul 25 '18 at 21:16
@CristhianGrundmann, that doesn't work, since, for example, we would then have $1=1-0=f(f(0))=f(1/2)=1/2$. — Barry Cipra, Jul 25 '18 at 21:16

score 0 · Answer 1 · answered Jul 25 '18 at 22:04

The same approach as in the accepted answer to the linked question works. We have to have $f(\frac 12)=\frac 12$, which makes $f(f(\frac 12))=\frac 12=1-\frac 12$. Partition the reals greater than $\frac 12$ into disjoint pairs. Given a pair $(a,b)$ define $$f(a)=b\\ f(b)=1-a\\ f(1-a)=1-b\\f(1-b)=a$$ Then $f(f(1-a))=a=1-(1-a)$ and similarly for $f(f(1-b))=b$ as required.

Is there a real function that satisfies the relation $f(f(x)) = 1-x$?

1 Answers1