A sequence satisfying $\lvert f_{i+j}-f_i-f_j\rvert \leq M$ bounded?

Question

Let $f:\mathbb N\to\mathbb Z$ be a sequence s.t $$(\exists M\geq 0)(\forall i,j\in\mathbb N)(\lvert f_{i+j}-f_i-f_j\rvert \leq M) $$ Must such a sequence be bounded?
No, it isn't.

Answer 1

A keyword in this question is Hyers-Ulam stability. From the condition in the question one may derive by induction that $\vert f_{n m}-n f_m\vert\leq (n-1)M $. This implies $\vert n f_m-m f_n\vert\leq (n+m-2)M$. Diving by $nm$ shows that the sequence $\left(\frac{f_n}{n}\right)$ is Cauchy. More exactly $\vert\frac{f_n}{n}-\frac{f_m}{m}\vert\leq\left(\frac{1}{n}+\frac{1}{m}\right) M (+)$. Let the real number $f$ its limit. Then $(+)$ for $m\to\infty$ shows that $\vert\frac{f_n}{n}-f\vert\leq \left(\frac{1}{n}+0\right) M$ or $\vert f_n-n f\vert\leq M$. Thus $f_n=n f+r_n$ with some bounded sequence of real numbers $r_n$ such that $n f+r_n\in\mathbb{Z}$.

For example, for any real $c$, the sequence of $f_n$ with $f_n=[c n]$ (the greatest integer not greater than $c n$ satisfies the original inequality with $M=2$ since $x=[x]+ r_x$ with $0\leq r_x<1$.

Answer 2

No such a sequence does not have to be bounded:

Consider $f_n = n$. Then $|f_{i+j} -f_i - f_j| = 0$, but $f_n \to \infty$.