Quasi homomorphism from integers to reals. Let $f:\mathbb{Z}\rightarrow \mathbb{R}$ be a quasi-homomorphism, i.e $|f(a+b)-f(a)-f(b)|\leq D$ $\forall$ $a$ and $b$ in $\mathbb{Z}$ ($\mathbb{R}$ and $\mathbb{Z}$ are here considered as additive groups and so you see the plus sign). I have to prove that there exists a unique number $\tau$ $\epsilon $ $\mathbb{R}$ such that $f(n)-n\tau$ is bounded.
I have separately shown the uniqueness part and have found bounds that work in following restrictive cases : 
(i) If $n$ is positive, I have bounded it above by $f(0)$ using $\tau=f(1)+D$.
(ii) If $n$ is positive, I have bounded it below by $f(0)$ using $\tau=f(1)-D$.
(iii) If $n$ is negative, I have bounded it above by $f(0)$ using $\tau=f(-1)+D$.
(iv) If $n$ is negative, I have bounded it below by $f(0)$ using $\tau=f(-1)-D$. I understand that I have always used $f(0)$ to bound them, but that is only what I can see since I am splitting $n$ as $n$ times the generator $1$ $\epsilon$ $\mathbb{Z}$. Please help me bound this universally using just one unique $\tau$.
 A: Define the function $f^*:\mathbb Z\to\mathbb R$ by:
$$f^*(a)=\lim_{n\to +\infty}\frac{f(na)}{n}.$$
One has to show that the limit exists, and let us leave that. We can notice that $f^*(0)=0$ and $f^*(ka)=kf^*(a)$ for every $k>0$:
$$f^*(ka)= \lim_{n\to+\infty}\frac{f(nka)}{n}= k\lim_{n\to+\infty}\frac{f(nka)}{nk}= k\lim_{n\to+\infty}\frac{f(na)}{n}= kf^*(a).$$
Note that by induction on $n>0$ we have:
$$|f(na)-nf(a)|\leqslant (n-1)D.$$
For $n=1$, this is trivial, so assume that the inequality holds for $n$, we have:
$$|f((n+1)a)-(n+1)f(a)|\leqslant |f((n+1)a)-f(na)-f(a)|+ |f(na)-nf(a)|\leqslant D+(n-1)D=nD.$$
Therefore:
$$\frac{|f(na)-nf(a)|}{n}\leqslant \frac{n-1}{n}D<D,$$
and so: 
$$|f^*(a)-f(a)|= |\lim_{n\to+\infty}\frac{f(na)}{n}-f(a)|= \lim_{n\to+\infty}|\frac{f(na)}{n}-f(a)|= \lim_{n\to+\infty}\frac{|f(na)-nf(a)|}{n}\leqslant D.$$
Put $\tau=f^*(1)$. For positive $a$, we have:
$$|f(a)-a\tau|= |f(a)-af^*(1)|= |f(a)-f^*(a)|\leqslant D.$$
It remains to prove that $|f(-a)+a\tau|$ is bounded for positive $a$. By induction on $n>0$ we see that:
$$|f(0)-f(-na)-nf(a)|\leqslant nD.$$
For $n=1$ this is trivial. Assume that the inequality holds for $n$, we have:
$$|f(0)-f(-(n+1)a)-(n+1)f(a)|\leqslant |f(0)-f(-na)-nf(a)|+ |f(-na)- f(-(n+1)a)-f(a)|\leqslant nD+D=(n+1)D.$$
Therefore:
$$-|f(0)|-nD\leqslant f(0)-nD\leqslant f(-na)+nf(a)\leqslant f(0)+nD\leqslant |f(0)|+nD,$$
so:
$$\frac{|f(-na)+nf(a)|}{n}\leqslant \frac{|f(0)|}{n}+D,$$
and we have:
$$|f^*(-a)+f(a)|= |\lim_{n\to\infty}\frac{f(-na)}{n}+f(a)|= \lim_{n\to+\infty}\frac{|f(-na)+nf(a)|}{n}\leqslant \lim_{n\to+\infty}\frac{|f(0)|}{n}+D=D.$$
Finally:
$$|f(-a)+a\tau|= |f(-a)+af^*(1)|=|f(-a)+f^*(a)|\leqslant |f(-a)-f^*(-a)|+|f^*(-a)+f(a)|+|f^*(a)-f(a)|\leqslant D+D+D=3D.$$
So $|f(a)-a\tau|$ is bounded by $3D$ for every $a$.
