I'm reading Fourier Analysis in groups by Walter Rudin, and A Course on Abstract Harmonic Analysis by Folland. In particular, I'm reading the Bochner Theorem's proof. However, my question doesn't need about that. First, some considerations.
- Let $G$ be a locally compact (Hausdorff) group.
- $M(G)$ is the Banach Algebra of complex Radon Measures on $G$, with the norm $||\mu||$ (the total variation, I can include the definition here if you need it).
In the final part of Bochner Theorem's proof, it says something like (not exactly, I want to simplify in order to be understood):
($\mu \in M(G)$) Since $\mu(G) = ||\mu||$, then $\mu \geq 0$.
Why?
Thank you.
In the proof p. 21 from Rudin's book, we have a complex-valued Radon measure $\mu$ on $G$ such that $\mu(G) = 1 = \| \mu \| := |\mu|(G)$. We prove that this implies $\mu \geq 0$. Consider $\phi \in C^0(G, \Bbb R_{ \geq 0})$. We have
\begin{align*} \| \varphi \|_{\infty} - \mathrm{Im}\left( \int_{G} \varphi\,d\mu \right) &\stackrel{\mu(G) = 1}{=} \text{Im}\left( \int_{G} [\| \varphi \|_{\infty} \mathbb 1_{G} - \varphi]\,d\mu \right) \leq \Big\| \| \varphi\|_{\infty} \mathbb 1_{G} - \varphi \Big\|_{\infty} \|\mu\| \stackrel{\phi \geq 0}{\leq} \| \varphi \|_{\infty} \end{align*}
Thus $$\mathrm{Im}\left( \int_{G} \varphi\,d\mu \right) \geq 0$$ and hence $\mathrm{Im}(\mu(E)) \geq 0$ for any Borel set $E \subset G$. Moreover, $\mathrm{Im}(\mu(G)) = 0$, which implies that $\mu$ is real-valued. Using the same argument as above, one shows that $\mathrm{Re}(\mu(E)) \geq 0$ for any Borel set $E \subset G$. Therefore, $\mu$ is a positive measure.