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Let $\{t_k\}_{k=-\infty}^\infty$ be a sequence of real numbers. I'm interested in finding the largest number A such that \begin{equation*} \int_{-\Omega}^\Omega|\sum_{k=-\infty}^{+\infty}c_ke^{2\pi i t_kf}|df\ge A \sum_{k=-\infty}^{+\infty}|c_k| \end{equation*} holds for every sequence $\{c_k\}_{k=-\infty}^\infty$. Clearly, $A$ depends on the sequence $t_k$. In particular I'm interested in the case where the sequence $t_k$ are separated. That is, \begin{equation*} \min_{k\neq\ell}|t_k-t_\ell|\ge \delta. \end{equation*} So I would like to calculate $A$ as a function of $\delta$.

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I don't see where the $t_k$ appear in your inequality... – Yemon Choi Jul 15 '13 at 0:17
thanks, I corrected my question it appears in the power of exponentials. – mohi Jul 15 '13 at 0:19
Since continuous functions on the circle can have Fourier series which are not absolutely convergent, my first guess is that you will need some stronger assumptions on your sequence $(t_k)$ to even guarantee the existence of such an $A$ – Yemon Choi Jul 15 '13 at 0:29
My conjecture is that if $\delta>\frac{1}{\Omega}$ then one has $A\ge 1$. I believe that some work of Arne Beurling implies that if $\delta<\frac{1}{2\Omega}$ then $A=0$. – mohi Jul 15 '13 at 0:37
A trivial remark which shows at least where the threshold $1/(2\Omega)$ you mention might somehow come from: for $\Omega=\pi$ and $t_k=k/(2\pi)$, the Dirichlet kernel gives $\simeq \log n$ on the lhs and $2n+1$ on the rhs. So there does not exist $A>0$ in this case. – 1015 Jul 15 '13 at 2:19

Here's something in the vein of a counterexample, at least to what you're conjecturing. Let $r > 1$, and consider the sequence $t_{k} = r k / \Omega$, which has $\min_{k \neq \ell} | t_{k} - t_{l} | = r / \Omega > 1 / \Omega$. By changing variable in the integral, the inequality becomes $$ \frac{\Omega}{r} \int_{-r}^{r} \biggl| \sum_{k} c_{k} e^{2 \pi i k x} \biggr| \, dx \geq A \sum_{k} |c_{k}|. $$ This inequality in turn implies $$ \frac{2 \Omega \lceil r \rceil}{r} \int_{0}^{1} \biggl| \sum_{k} c_{k} e^{2 \pi i k x} \biggr| \, dx \geq A \sum_{k} |c_{k}|. $$

I'm not familiar with the intricacies of convergence of Fourier series, but this latter inequality is essentially of the form $$ \| \hat{g} \|_{1} \leq C \| g \|_{1} $$ for functions $g$ on the unit circle $\mathbb{T}$, and no such inequality holds.

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As Julien (in the comments) and Jason (in the previous answer) point out, the inequality you want is impossible, which is easily seen by considering the case of a Dirichlet kernel. Also I am perplexed by comment about a result of Arne Beurling. Clearly, the left hand side of your proposed inequality is non-negative, so the result is trivial with $A=0$.

However, there is an much weaker inequality of the form you want. The inequality states (for $\Omega >1/2$) that:

$$\int_{-\Omega}^\Omega|\sum_{k=1}^{\infty}c_ke^{2\pi i t_k x}|dx \ge A \sum_{k=1}^{\infty}\frac{|c_k|}{k} $$

where $t_{k+1}-t_{k} \geq 1$ and $A$. Note that you can rescale to consider the case of smaller $\delta$ (after adjusting $\Omega$ appropriately).

This was proven by Fedja Nazarov in On a proof of the Littlewood conjecture by McGehee, Pigno and Smith. Algebra i Analiz 7 (1995), no.2, pp. 106-120.

As the title suggests, the proof is a generalization of the McGehee-Pigno-Smith proof of Littlewood's conjecture on exponential sum which is precisely the case with $\Omega =1/2$ and the $t_i$'s are taken to be distinct integers.

One can see that the left side of Nazarov's inequality is of the order $\log(n)$ in the case of the Dirichlet kernel of size $n$. Thus, apart from the implicit constant, this inequality is best possible.

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