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

Solution 1:

An important piece of information is:

Theorem: $f$ is not continuous.

Proof: Observe that $f$ is invertible, because

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

and so $f \circ f \circ f = f^{-1}$. Any continuous invertible function on $\mathbb{R}$ is either strictly increasing or strictly decreasing.

If $f$ is strictly increasing, then:

• $1 < 2$
• $f(1) < f(2)$
• $f(f(1)) < f(f(2))$
• $-1 < -2$

contradiction! Similarly, if $f$ is strictly decreasing, then:

• $1 < 2$
• $f(1) > f(2)$
• $f(f(1)) < f(f(2))$
• $-1 < -2$

contradiction! Therefore, we conclude $f$ is not continuous. $\square$

For the sake of completeness, the entire solution space for $f$ consists of functions defined as follows:

• Partition the set of all positive real numbers into ordered pairs $(a,b)$
• Define $f$ by, whenever $(a,b)$ is one of our chosen pairs,
  • $f(0) = 0$
  • $f(a) = b$
  • $f(b) = -a$
  • $f(-a) = -b$
  • $f(-b) = a$

To see that every solution is of this form, let $f$ be a solution. Then we must have $f(0) = 0$ because:

• Let $f(0) = a$. Then $f(a) = f(f(0)) = 0$ but $-a = f(f(a)) = f(0) = a$, and so $f(0) = 0$

If $a \neq 0$, then let $f(a) = b$. We have:

• $f(b) = f(f(a)) = -a$
• $f(-a) = f(f(b)) = -b$
• $f(-b) = f(f(-a)) = a$

From here it's easy to see the set $\{ (a,f(a)) \mid a>0, f(a)>0 \}$ partitions the positive real numbers and so is of the form I describe above.

One particular solution is

$$ f(x) = \begin{cases} 0 & x = 0 \\ x+1 & x > 0 \wedge \lceil x \rceil \text{ is odd} \\ 1-x & x > 0 \wedge \lceil x \rceil \text{ is even} \\ x-1 & x < 0 \wedge \lfloor x \rfloor \text{ is odd} \\ -1-x & x < 0 \wedge \lfloor x \rfloor \text{ is even} \end{cases}$$

e.g. $f(1/2) = 3/2$, $f(3/2) = -1/2$, $f(-1/2) = -3/2$, and $f(-3/2) = 1/2$.

(This works out to be Jyrki Lahtonen's example)

Solution 2:

Here is a graph of one such function, piecewise continuous.

Here's another picture to clarify endpoint behavior.

Solution 3:

Hint: If all the real numbers in existence were $1,2,-1,-2$, then the function $f: 1\mapsto 2\mapsto -1\mapsto-2\mapsto 1$ would work.

Spoiler:

Similarly permute $t,1+t,-t,-1-t$ for all $t\in(2n,2n+1]$ and for all natural numbers $n$. Handle $x=0$ separately.

Solution 4:

Another way of phrasing @Haskell Curry's proof:

Assuming the axiom of choice, both $\Bbb{R}$ and $\Bbb{C}$ are $\Bbb{Q}$-vector spaces with bases of the same cardinality. Thus there is some $\Bbb{Q}$-linear isomorphism $\alpha:\Bbb{R} \to \Bbb{C}$.

Then we can let $f$ be $\alpha^{-1} \circ g \circ \alpha$, where $g(z)=iz$.