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)
[cross-posted] https://mathoverflow.net/q/394839/172526