If $f:\mathbb Q\to\mathbb Q$ is "almost-linear" so $\{f(x+y)-f(x)-f(y)\}$ is bounded, does $f$ map bounded sets to bounded sets? If a function $f:\mathbb Q\to\mathbb Q$ is "almost-linear", in the sense that
$$\exists M>0:\forall x\in\mathbb Q:\forall y\in\mathbb Q:|f(x+y)-f(x)-f(y)|\leq M,$$
does it follow that
$$\forall L>0:\exists N>0:\forall x\in\mathbb Q:|x|\leq L\implies|f(x)|\leq N\;?$$
Two functions being boundedly close to each other is an equivalence relation. Almost-linearity says that $f(x+y)$ is boundedly close to $f(x)+f(y)$. Taking $y=nx$ and using induction, we can see that, for any fixed $n\in\mathbb Z$, $f(nx)$ is boundedly close to $nf(x)$. Rescaling $x$ to $\tfrac mnx$, we can further see that $f(\tfrac mnx)$ is boundedly close to $\tfrac mnf(x)$; but the "distance" between these functions may depend on $\tfrac mn$....
This is a variation on my previous question.
 A: Paying closer attention to the exact values of the bounds, using induction and triangle inequalities and rescaling, we obtain, for positive integers $n,m$,
$$|f(nx)-nf(x)|\leq(n-1)M,$$
$$|f(-nx)-(-n)f(x)|\leq(3n-1)M,$$
$$|f(\tfrac mnx)-\tfrac mnf(x)|\leq\frac{n+m-2}{n}M,$$
$$|f(-\tfrac mnx)-(-\tfrac mn)f(x)|\leq\frac{3n+m-2}{n}M,$$
$$|f(-\tfrac mnx)-(-\tfrac mn)f(x)|\leq\frac{n+3m-2}{n}M.$$
Set $x=1$. If we interpret $\pm\tfrac mn\in\mathbb Q$ as the variable, bounded by $\tfrac mn\leq L$, then
$$|f(+\tfrac mn)|=|f(\tfrac mn1)-\tfrac mnf(1)+\tfrac mnf(1)|$$
$$\leq\frac{n+m-2}{n}M+\tfrac mn|f(1)|$$
$$\leq M+(M+|f(1)|)L$$
and
$$|f(-\tfrac mn)|=|f(-\tfrac mn1)+\tfrac mnf(1)-\tfrac mnf(1)|$$
$$\leq\frac{3n+m-2}{n}M+\tfrac mn|f(1)|$$
$$\leq3M+(M+|f(1)|)L.$$
Now we can take $N=3M+(M+|f(1)|)L$, so that $|f(\pm\tfrac mn)|\leq N$ is bounded.

It looks like this generalizes to finite-dimensional normed vector spaces over $\mathbb Q$. If $\vec r=\sum_{i=1}^dr_i\hat e_i$ is bounded by $\lVert\vec r\rVert\leq L$, then all of the coefficients $r_i$ are bounded. (The exact values will depend on the norm.) Thus
$$\lVert f(\vec r)\rVert=\lVert f(\sum_{i=1}^dr_i\hat e_i)\rVert$$
$$=\lVert f(r_1\hat e_1+\sum_{i=2}^dr_i\hat e_i)-f(r_1\hat e_1)-f(\sum_{i=2}^dr_i\hat e_i)+f(r_1\hat e_1)+f(\sum_{i=2}^dr_i\hat e_i)\rVert$$
$$\leq M+\lVert f(r_1\hat e_1)\rVert+\lVert f(\sum_{i=2}^dr_i\hat e_i)\rVert$$
$$\leq M+\lVert f(r_1\hat e_1)\rVert+M+\lVert f(r_2\hat e_2)\rVert+\lVert f(\sum_{i=3}^dr_i\hat e_i)\rVert$$
$$\leq\cdots\leq(d-1)M+\sum_{i=1}^d\lVert f(r_i\hat e_i)\rVert,$$
and by the 1-dimensional proof, with "$x=\hat e_i$" instead of "$x=1$", all of the terms $\lVert f(r_i\hat e_i)\rVert$ are bounded.
