Consider for $X = 1,2, \ldots$ the set $\mathcal{S}_X$ of trigonometric polynomials $f(t) := \sum_{|k| \leq X} c_k e^{2\pi i kt}$ on the circle $\mathbb{T} := \mathbb{R}/\mathbb{Z}$ of degree $\leq X$ (Fourier transform supported on $\{-X,\ldots, X\}$), such that $f(0) = 1$ and $c_0 = 0$ (the last assumption is probably inessential). Let $$ M_X(f) := \sup_{\mathbb{T} \setminus [-\frac{1}{X},\frac{1}{X}] }|f|. $$ and $$ B_X := \inf_{f \in \mathcal{S}_X} M_X(f). $$

*Is the $X \to \infty$ limit of $B_X$ strictly positive, or zero?*

The motivation is the same as in this question that I have asked previously. Observe that, for all $n \leq X$, the average of $f(q)$ over $\mu_{n} \setminus \{1\}$, $q := e^{2\pi i t}$, does not exceed $M_X(f)$ in magnitude. As a result, Chebyshev's bound, the orthogonality relation in $\mathbb{Z} / n$ and the Dirichlet convolution identity (definition of the Mangoldt function) $\log = 1 * \Lambda$ yield for any $f \in \mathcal{S}_X$ the estimate $$ \sum_{n \leq X} \frac{\Lambda(n)}{n} = \sum_{k \neq 0} c_k \log{|k|} + O(M_X(f)), $$ where the implied coefficient is absolute and explicit. This would be an asymptotic formula if the $O(\cdot)$ term could be made to approach zero as $X \to \infty$, suggesting that $M_X(f)$ should perhaps be bounded away from zero. If not, then of course the next question would be to ask for the asymptotic computation of the extremal sequence $f_X$ and its decay rate $M_X(f_X) = B_X$.

It occurred to me that the linked question may have possibly been about functions on the circle $\mathbb{T} \leftrightarrow \mathbb{Z}$ rather than on the real line $\mathbb{R} \leftrightarrow \mathbb{R}$. For (I could be wrong about this) it seems to be a rather special situation to have $c_k \sim \frac{1}{X}\varphi(k/X)$ with $\varphi \in \mathcal{S}(\mathbb{R})$ a fixed Schwartz function supported on $[-1,1]$ and with $\varphi(0) = 0$ and $\widehat{\varphi}(0) = 1$. It does follow from the same argument as in Terry Tao's solution of the linked problem that there is an absolute $\epsilon_0 > 0$ such that $\lim_{X \to \infty} M_X(\sum_{k} \frac{1}{X} \varphi(k/X) e^{2\pi i kt}) \geq \epsilon_0$ for all such $\varphi$; for this unpacks to stating that $\sup_{\mathbb{R} \setminus [-1,1]} |\widehat{\varphi}| \geq \epsilon_0 > 0$ whenever $\mathbb{supp}(\varphi) \subset [-1,1]$ and $\varphi(0) = 0, \widehat{\varphi}(0) = 1$: a version of the uncertainty principle on the real line. But it isn't clear to me whether the sequence of solutions to our extremal problem should have such a limiting distribution $\varphi$. Also it would be nice to know of an argument that is directly about trigonometric polynomials.

Is there a version of the uncertainty principle on the circle that would yield the $M_X(f) \geq \epsilon_0 > 0$ answer in the present question too?

[*Note:* The conditions $c_0 = 0$ and $\varphi(0) = 0$ are probably irrelevant to the discussion; but they are convenient, so let me impose them for concreteness' sake. Other natural choices would be to take $c_0 = 1/X$ (corresponding to $\varphi(0) = 1$), or to drop them altogether. ]