***[cross-posting] <https://math.stackexchange.com/q/4153692/522463>***


Let $G$ be an infinite group and $\varphi\colon G\longrightarrow\mathbb{C}$ such that
\begin{eqnarray\*}
\exists\,\delta>0, \forall\,a,b\in G,\, |\varphi(ab)-\varphi(a)\varphi(b)|\leq\delta\quad(\mathcal{P})
\end{eqnarray\*}
What can be said about $\varphi$? Precisely, is it true that   

$(1)\quad\varphi$ is bounded, or   

$(2)\quad\varphi(ab)=\varphi(a)\varphi(b)$ for all $a,b\in G$?   

Let $e$ be the neutral element of $G$. Clearly, if $\varphi(e)\neq 1$, then $|\varphi(a)|\leq\delta|1-\varphi(e)|^{-1}$ for all $a\in G$; hence $\varphi$ is bounded. However, if $\varphi(e)=1$, it’s not clear to me if $\varphi$ remains bounded. I also think that when it’s not bounded, $\varphi$ may end up satisfying $\varphi(ab)=\varphi(a)\varphi(b)$ for all $a,b\in G$ (my attempts to build counterexamples failed, which makes me feel that the claim might be true)


Note: $\mathcal{P}$ does not define a *quasimorphism* (the landing set is the complex numbers and the property is multiplicative instead of the usual additive one for *quasimorphisms*)



On almost-(multiplicative)-morphisms between infinite groups and the complex numbers
If $\varphi$ is not bounded, there is a sequence $(x\_n)\_{n\in\mathbb{N}}\in G^{\mathbb{N}}$ such that $|\varphi(x\_n)|>2^n$ for all $n\in\mathbb{N}$. Let $x,y\in G$ and $n\in\mathbb{N}$.
\begin{eqnarray\*}
|\varphi(x)\varphi(y)-\varphi(xy)||\varphi(x\_n)|&=&|\varphi(x)\varphi(y)\varphi(x\_n)-\varphi(xy)\varphi(x\_n)|\\
&\leq&|\varphi(x)\varphi(y)\varphi(x\_n)-\varphi(x)\varphi(yx\_n)|+\\
&&|\varphi(x)\varphi(yx\_n)-\varphi(xyx\_n)|+\\
&&|\varphi(xyx\_n)-\varphi(xy)\varphi(x\_n)|\\
&\leq&\delta(|\varphi(x)|+2)
\end{eqnarray\*}
Hence, $|\varphi(x)\varphi(y)-\varphi(xy)|\leq\delta(|\varphi(x)|+2)2^{-n}$. Take the limit as $n\longrightarrow\infty$, and you're done.


