# Find a real function f:R→Rf:\mathbb{R}\to\mathbb{R} such that f(f(x))=−xf(f(x)) = -x?

I’ve been perusing the internet looking for interesting problems to solve. I found the following problem and have been going at it for the past 30 minutes with no success:

Find a function $f: \mathbb{R} \to \mathbb{R}$ satisfying $f(f(x)) = -x$ for all $x \in \mathbb{R}$.

I am also wondering, can we find $f$ so that is continuous?

I was thinking of letting $f$ be a periodic function, and adding half the period to x each time. I had no success with this, and am now thinking that such a function does not exist.

An important piece of information is:

Theorem: $f$ is not continuous.

Proof: Observe that $f$ is invertible, because

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

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)