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](https://mathoverflow.net/questions/285575/the-supremum-value-of-int-ft-log-frac1t-dt-for-normalized-fouri) 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](https://mathoverflow.net/questions/285575/the-supremum-value-of-int-ft-log-frac1t-dt-for-normalized-fouri/285929#285929) 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. ]



An extremal problem related either to an uncertainty principle on the circle, or else to the prime number theorem
A compactness argument shows that for sufficiently large $X$ one has the bound
$$ \sup\_{x \in {\mathbb T} \backslash [-1/X,1/X]} |f(x)| \gg \sup\_{x \in [-1/X,1/X]} |f(x)|$$
whenever $f$ is a trigonometric polynomial of degree at most $X$; this would imply that $B\_X \gg X$. (Perhaps there is a normalising factor of $1/X$ missing in your question?)


Proof: Suppose the claim failed, then one could (after normalising) find a sequence $X\_n \to \infty$ and a sequence $f\_n$ of trigonometric polynomials of degree at most $X\_n$ such that
$$ \sup\_{x \in [-1/X\_n,1/X\_n]} |f\_n(x)| = 1$$
and
$$ \sup\_{x \in {\mathbb T} \backslash [-1/X\_n,1/X\_n]} |f\_n(x)| = o(1).$$
We can find $x\_n \in [-1/X\_n,1/X\_n]$ such that $|f\_n(x\_n)|=1$. Writing $F\_n: {\mathbf R} \to {\mathbf C}$ for the $X\_n$-periodic function 
$$ F\_n(x) := f\_n( \frac{x}{X\_n} - x\_n \hbox{ mod } 1)$$
we see that $|F\_n(0)|=\|F\_n\|\_{L^\infty}=1$, that $\sup\_{2 \leq |x| \leq X\_n/2} |F\_n(x)| = o(1)$, and that $F\_n$ is band-limited to $[-1,1]$ (i.e., its Fourier transform is supported in $[-1,1]$). In particular, if $\varphi$ is a Schwartz function whose Fourier transform equals $1$ on $[-1,1]$, then $|\langle F\_n, \varphi \rangle| = |F\_n(0)| = 1$.


By passing to a subsequence one can assume that $F\_n$ converges weakly to another function $F$ in the unit ball of $L^\infty$, which is then non-zero by testing against $\varphi$. On the other hand, $F\_n$ vanishes outside of $[-2,2]$ and its distributional Fourier transform is supported on $[-1,1]$, which is a contradiction as the Fourier transform is also analytic.


The compactness argument does not give an effective bound for the implied constant, but presumably one can do so if one uses a suitable effective version of the uncertainty principle, such as the one in [this blog post of mine](https://terrytao.wordpress.com/2009/02/18/hardys-uncertainty-principle/). 


