Suppose $A \subseteq \mathbb{Z}_N := \mathbb{Z}/N\mathbb{Z}$ with $|A| \geq N/2$. Let $$ \hat{A}(h) := \sum_{a \in A} e_N(ha), $$ where $e_N(x) := e^{2\pi i x/N}$. Clearly $|\hat{A}(h)| \leq |A|$ for each $h$.
Question: Is it true that $$ S := \sum_{h=1}^{N-1} \left| \hat{A}(h) \right| $$ is smaller than the trivial bound $N|A|$? I feel the large size of $A$ should imply greater cancellation in the sum. Perhaps one can show that $|S| \leq N \log N$ or so?
Yes. By Cauchy's inequality, \begin{align*} (S+\#A)^2 = \bigg( \sum_{h=0}^{N-1} |\hat A(h)| \bigg)^2 &\le \bigg( \sum_{h=0}^{N-1} |\hat A(h)|^2 \bigg) \bigg( \sum_{h=0}^{N-1} 1^2 \bigg) \\ &= N \sum_{h=0}^{N-1} |\hat A(h)|^2 \\ &= N^2 \sum_{a\in A} 1^2 = N^2 \#A, \end{align*} where the last line follows by Plancherel's theorem applied to $x_a = 1$ if $a\in A$ and $x_a=0$ otherwise. Therefore $$ S \le N\sqrt{\#A} - \#A. $$