# estimates of exponential polynomials

Let $p(t) = \Sigma_{k=1}^n c_k e^{i \lambda_k t}$ be an exponential polynomial.

In the paper "Local estimates for exponential polynomials and their applications to inequalities of the uncertainty principle type" http://www.math.msu.edu/~fedja/Published/paper.ps Nazarov proves an estimate on the maximum value attained by the polynomial $p$ in an interval $I$, in terms of the maximum of $p$ in a subset $E \subset I$.

To be precise he obtains the following estimate:-

$$\sup_{t \in I} |p(t)| \leq ( \frac{A \mu(I)}{\mu(E)} )^{n-1} \sup_{t\in E} |p(t)|.$$

At one point he mentions that the result holds true for more general functions of the type $$p(t) = \Sigma_{k=1}^n q_k(t) e^{i \lambda_k t}$$ where $q_k(t)$ are algebraic polynomials of degree d_k$; by an "obvious" approximation argument. It is not clear to me, what exactly is the argument he is suggesting ? One of the method he uses to obtain an estimate as mentioned above is by using Turan's Lemma. Although in that case one gets exponent$2 n^2$instead of$n-1$.$\underline {Turan's Lemma}$Let$ z_1,\dots,z_n$be complex numbers,$|z_j|\geq 1, j=1,\dots,n.$Let$ b_1,\dots, b_n \in \mathbb C $and $$S_j:= \Sigma_{k=1}^n b_k z_k^j$$ Then $$|S_0| \leq \{\frac{4 e (m+n-1)}{n}\}^{n-1} \max_{j=m+1}^{m+n} |S_j|.$$ As a simple consequence of this result when the value of an exponential polynomial (with constant coefficient) is known for$n$consecutive term of an arithmetic progression, then one can get an estimate of the value of the polynomial along that arithmetic progression. i.e., Let$p(t)=\Sigma_{k=1}^n c_k e^{i \lambda_k t}$and assume that the value of the polynomial$p(t)$is known for$t_j=t_0+j \delta$for$ j= m+1,...,m+n$. Then substitute$b_k=c_k e^{i \lambda_j t_0}$and$z_k= e^{i \lambda_k \delta}$and apply Turan's lemma. The result now follows by Lebesgue's density theorem and some averaging argument. We might want to get a similar result like Turan's Lemma for the more general type of exponential polynomial$p(t) = \Sigma_{k=1}^n q_k(t) e^{i \lambda_k t}$where$q_k(t)$are algebraic polynomials of degree d_k$.

But I doubt this is what he is suggesting here, as later in order to get the sharper result (the one with exponent $n-1$) he uses some weak type estimates it seems. (I have not read this part of the proof yet).

So, what exactly is the obvious approximation argument he is trying to suggest here?

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Can't you contact that Nazarov directly? (You even didn't provide a link to the paper.) I suspect that the remark on the more general family would follow from understanding the method of the paper. – Wadim Zudilin Jun 14 '10 at 9:01
I guess your Nazarov is mathoverflow.net/users/1131/fedja on MO. So, you just need to be patient until he comes. – Wadim Zudilin Jun 14 '10 at 12:08
This won't answer the question, but may I suggest writing $$\sup_{a \in A}\text{ and }\max_{a\in A}$$ etc., which \sup and \max, with backslashes. Standard usage and it results in standard formatting conventions being followed: (1) sup and max are not italicized; (2) $a\in A$ appears directly under sup or max; (3) proper spacing precedes and follows sup or max. – Michael Hardy Jun 14 '10 at 14:09
Wadim: can't you sympathise a little with Vagabond here?! fedja and Nazarov are both practically superhuman, whether or not they're the same person (Nazarov's papers are usually fantastically complicated and difficult, unavoidably so because of the subject matter), and I would be far too intimidated or embarrassed ever to ask them for help with one of their papers unless I were VERY sure that I wasn't just being stupid and trivial! The extreme stratification of mathematical ability/knowledge, even amongst serious mathematicians, is very off-putting for me! – Zen Harper Jul 16 '10 at 23:10
@Zen Harper, a few good samaritans did help me out in this case. I asked essentially the same question again here (after guessing what would be the obvious approximation argument most likely be) and got a `very elegant solution' from Andrey Rekalo here mathoverflow.net/questions/29692/… – Vagabond Jul 17 '10 at 2:32