I am reading Katznelson's book on harmonic analysis, and I reached the following statement regarding a characterization of the Fourier-Stieltjes series of positive measures:
Taken from page 38: (7.5) A series $S \sim \sum a_ne^{int}$ is the Fourier-Stieltjes series of a positive measure if and only if, for all $n\in\mathbb{N}$:
$$\sigma_n(S,t) := \sum_{j=-n}^n(1-\dfrac{|j|}{n+1})a_je^{ijt}\ge 0$$
The implication $\sigma_n(S,t) \ge 0$ for all $n,t$ implies the existence of a positive measure is unclear to me. What Katznelson is trying to employ is a theorem he proves earlier saying that if $S$ is a series satisfying $\Vert \sigma_n(S)\Vert_{B^*}\le C$ for all $n\in\mathbb{N}$, then there exists a norm whose Fourier-Stieltjes series coincide with $S$'s. However, it is unclear to me how to bound $\Vert \sigma_n(S)\Vert_{B^*}$. Katznelson says that the norm of the measure $\sigma_n(S)$ is $\dfrac{1}{2\pi}\int_{\mathbb{T}}\sigma_n(S,t)dt = a_0$. The computation is clearly trivial, but why is that the $B^*$-norm of $\sigma_n(S)$? This is like saying that the trigonometric polynomial stretched the most by $\sigma_n(S)$ is 1. Why should that be true? Clearly I am missing something.
Thanks in advance.
TLDR: Why is the norm of a non-negative measure $\mu$ over a Banach space $B$ is $\hat{\mu}(0)$?
Thank you David C. Ullrich. However, in fact, what one needs here is far simpler than the Riesz Representation Theorem. It so happens that the $B^*$ norm of $\sigma_n(S)$ is defined as $$\sup_{f:\Vert f\Vert_\infty = 1}\{|\int_\mathbb{T}f(t)\bar{\sigma_n(S,t)}dt|\}$$
However, since by the triangle inequality, the integral $|\int_\mathbb{T}f(t)\bar{\sigma_n(S,t)}dt|$ is always upper bounded by $\int_\mathbb{T}|f(t)\sigma_n(S,t)|dt$, and $\sigma_n(S,t)\ge 0$, we have that the integral is in fact upper bounded by $\int_\mathbb{T}|f(t)|\sigma_n(S,t)dt$. Since $\Vert f \Vert_{\infty} = 1$, the integrand is upper bounded by $\sigma_n(S,t)$, which shows that $\Vert \sigma_n(S)\Vert \le a_0$.
Thanks!