If $f(f(x))=-f(x)$,does it necessarily follow that $f(x)=-x$?

Question

Is there any counterexample other than $f(x)=0$? The actual problem states: Find a function $f: \mathbb{R} \to \mathbb{R}$, such that $f(f(x+y))=x-f(y)$. By substituting $x=0$, we get $f(f(y))=-f(y)$. Is it safe to conclude that $f(y)= -y$ for all $y$?

Answer 1

If $f$ is surjective, absolutely. Then, for any $y \in \mathbb{R}$, we have $f(x) = y$ for some $x \in \mathbb{R}$, so $$f(y) = f(f(x)) = -f(x) = -y.$$ If you look back at the original equation, $f(f(x + y)) = x - f(y)$, it's not difficult to show that this is surjective; fix $y$ and let $x$ vary. So, indeed, $f(x) = -x$ for all $x$.

Here's an example of a function satisfying $f(f(x)) = -f(x)$ that isn't surjective:

$$f(x) = \begin{cases}-x & \text{if $-1 \le x \le 1$} \\ -\frac{2x}{x^2+1} & \text{otherwise}\end{cases}.$$

The key is that such a function must be $-x$ on its range, but free to be anything else outside the range. I opted for a continuous odd function!