# Bounded almost-homomorphisms on the integers

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.
• Thank you. But this is kind of cheating. If there is a homomorphism equivalent to $f$, there is really nothing to prove (and actually this is exactly what I would like to prove here!). Is there any more direct argument? – Martin Brandenburg Jun 20 '14 at 15:37
• Well, for all $n$, we have $\lvert f(2n) - 2f(n)\rvert \leqslant c$. Iterating gives $$\left\lvert \frac{f(2^k n)}{2^k} - \frac{f(2^{k-1}n)}{2^{k-1}}\right\rvert \leqslant \frac{c}{2^k},$$ and hence $$\left\lvert \frac{f(2^kn)}{2^k} - f(n)\right\rvert \leqslant c.$$ Since the first term tends to $0$ for $k\to\infty$ if $n > 0$, it follows that $\lvert f(n)\rvert \leqslant c$ for $n > 0$. Then $f(-n) = - f(n)$ shows the same for $n < 0$. Would you consider that more direct? – Daniel Fischer Jun 20 '14 at 15:50