Can I show that this function is surjective?

Question

Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that

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

for all $x \in \mathbb{R}$. Can I conclude that $f$ is surjective? If so, how can I prove it?

Answer 1

Note that for all $x \in \Bbb R$, there exist real numbers $a = x + f(x)$ and $b = x - f(x)$ such that $$ x = f(a) + f(b) $$ Let $S$ denote $f(\Bbb R)$, i.e. the image of $f$. It's clear that $\{c + d : c,d \in S\} = \Bbb R$ (more concisely, $S + S = \Bbb R$).

If we know that $f$ is continuous, then the above allows us to conclude that $f$ is surjective. In particular, it suffices to state that because $S + S$ is not bounded above and not bounded below, $S$ is also not bounded above and not bounded below. Because $S= f(\Bbb R)$ is also connected, we can conclude that $f(\Bbb R) = \Bbb R$.

If we are to answer this without the assumption of continuity, things get tricky. In order for $f$ to fail to be surjective, $S$ would need to be such that $S + S = \Bbb R$ with $S \neq \Bbb R$. It is not obvious to me that this should be possible without invoking the axiom of choice. Even if we could find such an $S$, it is unclear how one could get $f$ to actually satisfy the functional equation.