The following is taken from chapter 9.2 from Positive Trigonometric Polynomials and Signal Processing Applications by Bogdan Dumitrescu.
Let $A$ be a symmetric set ($x \in A \implies -x \in A$) such that all atoms $a \in A$ are extreme points of $\text{conv}(A)$, i.e. $a \not\in \text{conv}(A \setminus \{ a \})$ for all $a \in A$. The atomic norm associated with $A$ is $$ \| x \|_A := \inf\{ t \ge 0: x \in t \cdot \text{conv}(A)\}. $$ The dual atomic norm is $$ \| h \|_A^* \overset{\text{Def.}}{=} \sup_{\| x \|_A \le 1} \Re[x^{H} h] = \sup_{a \in A} \Re[a^H h]. $$ Now consider $$ A := \{ a(w, \phi) := e^{i \phi} \psi(w): w, \phi \in [- \pi, \pi]\}, \quad \text{where } \psi(w) := [1, e^{i w}, \ldots, e^{i n w}]. $$ Up until here, everything is fine. But the author now claims $$ \| h \|_{A}^* = \sup_{w, \phi} \Re[h^{H} a(w, \phi)] = \sup_{w, \phi}\left| e^{- i \phi} \sum_{k = 0}^{n} h_k e^{- i k w}\right|. $$ I don't understand how the second equality is derived. I get \begin{align} \Re[h^{H} a(w, \phi)] & = \Re\left[ \sum_{k = 0}^{n} \overline{h}_k e^{i \phi} e^{i k w} \right] = \sum_{k = 0}^{n} \Re\left[\overline{h}_k e^{i \phi} e^{i k w} \right] \\ & = \sum_{k = 0}^{n} \Re(h_k) \cos(\phi + w k) + \Im(h_k) \sin(\phi + w k) \end{align} using the formulae $\Re(a \cdot b) = \Re(a) \Re(b) - \Im(a) \Im(b)$ and $$\Re\left[e^{i \phi} e^{i k w}\right] = \Re\left[e^{i (\phi + k w)}\right] = \cos(\phi + k w).$$
Similarly, writing out the other side of the equality I get \begin{align} \left| e^{-i \phi} \sum_{k = 0}^{n} h_k e^{- i k w} \right|^2 & = \left| \sum_{k = 0}^{n} h_k e^{- i k w} \right|^2 = \Re\left[\sum_{k = 0}^{n} h_k e^{- i k w}\right]^2 + \Im\left[\sum_{k = 0}^{n} h_k e^{- i k w}\right]^2 \\ & = \left( \sum_{k = 0}^{n} \Re(h_k) \cos(w k) + \Im(h_k) \sin(w k) \right)^2 \\ & \qquad + \left( \sum_{k = 0}^{n} \Im(h_k) \cos(w k) + \Re(h_k) \sin(w k) \right)^2, \end{align} which are both very different. What am I missing?
$\Re\left[ \sum_{k = 0}^{n} \overline{h}_k e^{i \phi} e^{i k w} \right]=\Re\left[ \sum_{k = 0}^{n} {h}_k e^{-i \phi} e^{-i k w} \right]$ using ($\Re z = \Re \bar z$)
$\Re\left[ \sum_{k = 0}^{n} {h}_k e^{-i \phi} e^{-i k w} \right]=\Re (e^{-i\phi}S_n)$ where $S_n= \sum_{k = 0}^{n} {h}_k e^{-i k w}$
But now first $\Re (e^{-i\phi}S_n) \le |e^{-i\phi}S_n|=|S_n|$ so the supremum on the real part on $\phi,w$ is at most the supremum on the absolute value of the same and which depends only on $w$ (and on $n$ if that varies too, but not on $\phi$)
However for a fixed $w$ (and a fixed $n$ if needed) hence a fixed $S_n$ there is a (unique modulo $2\pi$ unless $S_n=0$) $\phi_{w,n}$ s.t $\Re (e^{-i\phi_{w,n}}S_n)=|S_n|$ so the supremum on the abolute values on $\phi,w$ is actually attained on the real part, so we get the required equality:
$\| h \|_{A}^* = \sup_{w, \phi} \Re[h^{H} a(w, \phi)] = \sup_{w, \phi}\left| e^{- i \phi} \sum_{k = 0}^{n} h_k e^{- i k w}\right|.$