Let $G$ a group, $\delta>0$ and $f:G\longrightarrow \mathbb{C}$ bounded such that $$ \forall (x,y)\in G^2,\,|f(xy)-f(x)f(y)|\leqslant\delta $$ How can I find the lowest possible value of $C$ such that $|f(x)|\leqslant C$ for all $x\in G$ ?
What I did : if $C$ works, then $C+C^2$ works as a $\delta$ since $$(1)\ \ \ |f(xy)-f(x)f(y)|\leqslant|f(xy)|+|f(x)|\cdot|f(y)|\leqslant C+C^2$$ Moreover, if $f(1)\neq 1$, $\lambda=|1-f(1)|>0$ and if $\delta$ works, then $\frac{\delta}{\lambda}$ works as a $C$ : $$(2)\ \ \ \forall x\in G,\,\lambda|f(x)|=|f(x)-f(x)f(1)|\leqslant\delta$$ so that $|f(x)|\leqslant\frac{\delta}{\lambda}$. Let $$\Omega=\{ \delta >0\ |\ \forall (x,y)\in G^2,\,|f(xy)-f(x)f(y)|\leqslant\delta \}$$ then $\Omega$ is a closed set and if $\delta_0=\inf\Omega$, then $\Omega=[\delta_0,+\infty[$. Since $\forall x\in G,\,|f(x)|\leqslant\frac{\delta_0}{\lambda}$, we have $\delta_0=0$ iff $f=0$, we will now suppose that $f\neq 0$. Furthermore, since $\delta_0\in\Omega$, then $\frac{\delta_0}{\lambda}+\frac{\delta_0^2}{\lambda^2}\in\Omega$ by using $(1)$ and $(2)$. Thus $$\frac{\delta_0}{\lambda}+\frac{\delta_0^2}{\lambda^2}\geqslant\delta_0 $$ and $\delta_0\geqslant\lambda^2-\lambda$. I think that $\delta_0=\lambda^2-\lambda$ but I don't know how to prove it, and I don't know what to do in the case $f(1)=1$.