On almost-(multiplicative)-morphisms between infinite groups and the complex numbers [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)
 A: It seems like it's actually true that $\phi$ must either be bounded or multiplicative (I'm a little surprised!). For each $g,h\in G$, set $\varepsilon_{g,h}:=\phi(gh)-\phi(g)\phi(h)$, so the assumption is that $|\varepsilon_{g,h}|\le \delta$ for all $g,h\in G$.
Suppose $\phi$ is not multiplicative, so we can choose $a,b\in G$ with $\varepsilon_{a,b}\ne 0$. Now let $c\in G$ be arbitrary, and we claim there is a uniform bound on $\phi(c)$. On the one hand we have $\phi(abc)=\phi(a)\phi(bc)+\varepsilon_{a,bc} = \phi(a)\phi(b)\phi(c)+\phi(a)\varepsilon_{b,c}+\varepsilon_{a,bc}$. On the other hand we have $\phi(abc)=\phi(ab)\phi(c)+\varepsilon_{ab,c} = \phi(a)\phi(b)\phi(c)+\varepsilon_{a,b}\phi(c)+\varepsilon_{ab,c}$. These are equal, so we have $\phi(a)\varepsilon_{b,c}+\varepsilon_{a,bc} = \varepsilon_{a,b}\phi(c)+\varepsilon_{ab,c}$. Solving for $\phi(c)$ we get $\phi(c)=(\phi(a)\varepsilon_{b,c}+\varepsilon_{a,bc}-\varepsilon_{ab,c})/\varepsilon_{a,b}$. Since $|\varepsilon_{g,h}|\le\delta$ for all $g,h\in G$, we get $|\phi(c)|\le (|\phi(a)|+2)\delta/|\varepsilon_{a,b}|$, a uniform bound.
A: 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.
