Let $S$ be a finite subset of the complex unit circle and $1 \in S$. For each $n \in \mathbb N $, define $f\_n\colon S^{n-1}\to\mathbb R$ by
$$f\_n(x) := \sum\_{w^{n}=1}|x\_1w+ x\_2w^2\cdots+x\_{n-1}w^{n-1}|$$
Denote $z^\*\_n := \min\_{S^{n-1}} f\_n$.


Is it true that for sufficiently large $n$s (which depend on $S$), we have $z\_n^\* = 2(n-1)$?


*Remarks*:


* $f\_n(x)$ is sum of the absolute values of eigenvalues of a circulant matrix generated by $(0,x\_1,x\_2,\ldots,x\_{n-1})$.
* $f\_n(x) = 2(n-1)$, when $x$ is all ones vector.
* Above question stems from [this question](https://mathoverflow.net/questions/302424/is-this-lower-bound-for-a-norm-of-some-complex-matrices-true) and may be related to the Littlewood problem, as [a comment](https://mathoverflow.net/questions/302424/is-this-lower-bound-for-a-norm-of-some-complex-matrices-true#comment754136_302424) of Tao to the last linked question.



An Optimization Problem with Complex Variables, regarding Eigenvalues of Circulant Matrices
The answer is no if there exists $z \in S$ with $z \neq 1$ and $\vert 1 - z\vert \leq \frac{1}{2}$ and $\bar{z} \in S$ :


Let $$p(x) = \sum\_{k=0}^{n-1} b\_k x^k$$ and $x\_k = p(w^k)$ , where $w=e^{\frac{2 \pi i}{n}}$ .


Then the $n b\_k$ are the eigenvalues of the circulant matrix.


Now choose  

$$p(x) = 1 - \frac{1}{n} \sum\_{k=0}^{n-1} x^k + (z-1) \frac{1}{n} \sum\_{k=0}^{n-1} w^k x^k + (\bar{z}-1) \frac{1}{n} \sum\_{k=0}^{n-1} w^{-k} x^k$$


Then $p(1)=0$, $p(w)=\bar{z}$, $p(w^{-1})=z$ and $p(w^k)=1$ otherwise.


Furthermore we have $b\_0 = 1 - \frac{1}{n}(3 - 2 Re (z))$ and $b\_k = \frac{1}{n}(-1 + 2 Re ((z-1)w^k)) \leq 0$ for $k \neq 0$.


Therefore $$\sum\_{k=0}^{n-1} \vert b\_k\vert = 2 b\_0 < 2 (1 - \frac{1}{n})$$


