If $G$ is a compact abelian group and $f\in C(G)$.
Then $\forall \epsilon >0$,there exists an open neighbourhood $U$ of $0\in G$, such that $\forall g\in G , \forall u_1,u_2\in U$, we have $|f(gu_1)-f(gu_2)|<\epsilon$.
Is this statement right? how to prove it?
I know that if $G$ is a metric group, then I can do it by the lebesgue number.And i try to use the method of proving the existence of lebesgue number to prove the above statement. But I failed.
The statement is true. A sketch of the proof would be like this: for every $g\in G$ we can pick a neighborhood of zero $V_g$ such that if $u\in V_g$ we have $$ |f(ug)-f(g)|<\varepsilon/2 $$ Since the group is abelian, there is a symmetric neighborhood of zero $U_g$ (in that $U_g=U_g^{-1}$) such that $U_g\cdot U_g\subset V_g$. The collection of all the $gU_g$ forms an open cover. Choose a finite subcover indexed by $g_1,\cdots, g_n$. Then, the set $$ U=\bigcap_{i=1}^n U_{g_i} $$ Is the neighborhood you're looking for.