Let $f : \mathbb{Z} \to \mathbb{Z}$ be an "almost-homomorphism": The set $\{f(n+m)-f(n)-f(m) : n,m \in \mathbb{Z}\}$ is bounded. We may assume that $f$ is odd, i.e. $f(-n)=-f(n)$ for all $n$. Assume that $\lim_{n \to \infty} \frac{f(n)}{n}=0$. Why is $f$ bounded?
Background: I would like to prove directly that the ring $R$ constructed by Norbert A'Campo arXiv:math/0301015 is isomorphic to the $\mathbb{R}$ (without showing that it is a complete ordered field). The only missing step is the one above. Namely, it shows that $R \to \mathbb{R}$, $[f] \mapsto \lim_{n \to \infty} \frac{f(n)}{n}$ is injective.
Let $c$ denote a bound on $\lvert f(m+n) - f(n) - f(m)\rvert$. As shown in this answer, there is a homomorphism $h\colon \mathbb{Z}\to\mathbb{R}$ with
$$\lvert h(n) - f(n)\rvert \leqslant c\tag{1}$$
for all $n\in\mathbb{Z}$. The condition $\frac{f(n)}{n}\to 0$ together with $(1)$ now implies that $\frac{h(n)}{n}\to 0$, hence $h(n) \equiv 0$, and that means that $f$ is bounded.