Let $G$ be a locally compact Abelian group and $\hat{G}$ be its dual group, that is the group of all complex functions $\gamma:G\to\mathbb C$ such that $$\gamma(x+y)=\gamma(x)\gamma(y)\\|\gamma(x)|=1$$ for all $x,y\in G$.
I'm trying to understand the proof that shows F-S transform $$\hat{\mu}(\gamma)=\int\gamma(-x)d\mu(x)$$ is uniformly continuous.
We have $$|\hat{\mu}(\gamma_1)-\hat{\mu}(\gamma_2)|= \big |\int\big (\gamma_1(-x)-\gamma_2(-x)\big )d\mu(x)\big |\\ \le \int|1-(\gamma_1+\gamma_2)(x)|d|\mu|$$
and then every thing will follow, am having a hard time seeing why the inequality used here is correct.
Appreciate your help, thank you.