Let $D$ be a subset of $\mathbb Z/n \mathbb Z$ containing $0$. For $m$ an integer, set $$\alpha(m,D)=\sum\_{d \in D} e\left (\frac{m d }{n}\right ),$$
where as usual $e(x) = e^{2 i \pi x}$ This is an exponential sum (or if you like:
character sum). Obviously $|\alpha(m,D)| \leq |D|$.


Now consider
$$\sigma(D) = \frac{1}{n} \sum\_{m=0}^{n-1} |\alpha(m,D)|.$$


A simple upper bound using Cauchy-Schwarz for $\sigma(D)$ is $\sigma(D) \leq \sqrt{ |D|}$,
which is better than the trivial bound $\sigma(D) \leq |D|$. But here I am interested in a lower bound:



> 
> is it true that $\sigma(D) \geq 1$, with equality only if $D$ is a subgroup of $\mathbb Z/n \mathbb Z$ ?
> 
> 
> 


This seems true on some numerical tests I have done, and this seems a very elementary question , but I don't see how to prove it at this time (I might be missing something trivial).
More generally, I am interested in any information or reference on this number $\sigma(D)$.
It appears in the error term in the formula for the number of primes $p \leq x$ which residues modulo $n$ is in $D$.



Lower bound for exponential sums
This seems easier than you might have expected. Up to normalization, your
quantities $\alpha(m,D)$ are Fourier coefficients of the indicator function
of $D$ (for which reason many people would rather use the notation $\hat
1\_D(m)$). As such, they satisfy the Parseval identity
 $$ \sum\_m |\alpha(m,D)|^2 = n|D|. $$
In view of $|\alpha(m,D)|\le|D|$, this yields
 $$ n|D| \le \sum\_m |D| |\alpha(m,D)|, $$
whence
 $$ \sigma = \frac1n \sum\_m |\alpha(m,D)| \ge 1. $$
Moreover, for equality to hold, one needs all $|\alpha(m,D)|$ to
be equal to either $0$ or $|D|$, which is only possible if $D$ is a coset of
a subgroup of ${\mathbb Z}/n{\mathbb Z}$.


