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. ]

  • $\begingroup$ @reuns: It doesn't exist, according to the answer. All this amounts is to observe and prove that a degree $X$ trigonometric polynomial having $f(0) = 1$ is not too small (i.e., is bounded below by a positive absolute constant) in the complement of the $1/X$-neighborhood of $0$. $\endgroup$ – Vesselin Dimitrov Dec 8 '17 at 20:06
  • $\begingroup$ So there is no concrete number-theoretic $f$ you really want to study ? If I ask it is because that's the kind of questions I'm interested in about $\sum_{n \le N}\frac{ \mu(n)}{n} \{ nx\}, \sum_n e^{2i \pi n x} 1_{\gcd(n,N!)=1}$, $\min_a |1-\zeta(s)\sum_{n=1}^N a_n(N) n^{-s}|^2$ $\endgroup$ – reuns Dec 8 '17 at 20:10

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.

  • 1
    $\begingroup$ So it is the same as in $\mathbb{R}$: extending functions by periodicity, and applying the same kind of compactness argument. Hence I was wrong to think it could be a different problem. Yes, I had neglected the $1/X$ factor: it should be simply about the bound $M_X(f) \gg 1$ among the degree-$X$ trigonometric polynomials with $f(0) = 1$. I edited my post to reflect that. Thanks for this very quick answer! $\endgroup$ – Vesselin Dimitrov Dec 8 '17 at 6:47

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