I have a question on an inequality written on the bottom of this page. Let $G$ be a locally compact group and let $\gamma : G \to S^1 \subset \Bbb C$ be a character of $G$.
If $\mu$ is a complex Borel measure on $G$, then we define $$\widehat \mu (\gamma) = \int_G \gamma(-x)d\mu(x)$$
The inequality that I don't understand is the following :
$$|\widehat \mu(\gamma_1) - \widehat \mu(\gamma_2)| \leq \int_G |1-(\gamma_1-\gamma_2)(x)| \;d|\mu|(x) $$
For me, it should be $$|\widehat \mu(\gamma_1) - \widehat \mu(\gamma_2)| \\\stackrel{1}{=} \left| \int_G \gamma_1(-x) - \gamma_2(-x) \;d\mu(x) \right| \\\stackrel{2}{\leq} \int_G |\gamma_1(-x) - \gamma_2(-x)| \;d|\mu|(x) \\ \\\stackrel{3}{=} \int_G |(\gamma_1 - \gamma_2)(-x)| \;d|\mu|(x) \\\stackrel{4}{=} \int_G |(\gamma_1 - \gamma_2)(y)| \;d|\mu'|(y)$$
where $\mu'(E) = \mu(-E)$.
Where am I wrong? Thank you!
If $z,w\in\Bbb C$ and $|z|=|w|=1$ then $|z-w|=|1-z/w|$.
Now, the group operation on the group of characters is pointwise multiplication. But it's an abelian group, so the operation is written with a plus sign. Hence the curious-looking fact that $$(\gamma_1-\gamma_2)(x)=\gamma_1(x)/\gamma_2(x).$$
And it gets worse; when he writes $\gamma_1(x)-\gamma_2(x)$ he's talking about actual subtraction. So $$\gamma_1(x)-\gamma_2(x)\ne(\gamma_1-\gamma_2)(x);$$in fact $$|\gamma_1(x)-\gamma_2(x)|=|1-\gamma_1(x)/\gamma_2(x)|=1-(\gamma_1-\gamma_2)(x)|. $$
This is hideous - when I'm in charge the group operation is written as multiplication even though it's an abelian group....