# A conjecture about the measure estimates of a trigonometric polynomial

Formulation of the Conjecture

Let $\Omega =(0,\pi)\times (0,2\pi)\subset\mathbb R^2$ and let $\psi:\Omega\to \mathbb{R}$ defined by $$\psi(x,t)=\sum_{k\in S \,j\in S'} \sin(kx)\left( a_{kj}\sin(jt)+b_{kj}\cos(jt)\right),$$ where $\int_\Omega \psi^2 = 1$ and $S,S'\subset \mathbb N$ are finite subsets of $\mathbb N$.

Note: $\psi$ is a given and fixed trigonometric polynomial with norm in $L^2(\Omega)$ equal to 1. For the conjecture $\Omega$ could be other region, for example $(0,2\pi)^2$, and $\psi$ could be a more general trigonometric polynomial. But this simpler case is the case we are interested in.

There are $\epsilon_0,\alpha>0$ such that if $\epsilon\in(0,\epsilon_0)$, then $$\mu(\{(x,t)\in\Omega\,:\,|\psi(x,t)|<\epsilon\})<\epsilon^\alpha$$

where $\mu$ is the Lebesgue Measure in $\mathbb R^2$.

Experimental Approach

Here is an IPython notebook I made with the corresponding experimentation

Related Previous Works

• This paper is about Van Der Corput's Lemma in higher dimension. We don't know how to apply It.
• This paper is about polynomials in random variables. We don't know how to use It eather in this case.

Ideas given by experts

Experts in orthogonal polynomials recommended the use of Chebyshev Polynomials. But it seems to be a not very obvious problem.

Question

¿How to prove the conjecture exposed above?

• Editing: I think that you need encompassing parentheses after $\Sigma$. And a factorization would make the formula a bit easier to grasp visually. Sep 2 '14 at 4:22
• Can you please review the new formulation of the question to see if It is more understandable? Sep 2 '14 at 23:12
• Can't you simply take $\beta=1$ and then assert that the measure is bounded by $M\epsilon^{\alpha}$ for $\alpha <1$ and $\epsilon$ small enough? (All I've done is call $\epsilon^{\beta}$ as $\epsilon$.) This might be a little bit clearer. In any case, the question does seem reasonable to me, and I'll add my vote to reopen. Sep 3 '14 at 0:39
• Yes, you can take $\beta=1$. Moreover, you can take $M=1$ in this case we have that there is $\alpha\in(0,1)$ such that $\mu(E_1)<\epsilon^\alpha$. Sep 3 '14 at 0:57
• It seems to me that you are asking for a two-dimensional version of the Nazarov-Turan inequality (see Wikipedia: take $J$ to be the whole set and $E$ the set on which $|\psi|<\varepsilon$, then rearrange and take the power $1/(n-1)$). I suggest asking Fedya Narazov. He's a regular contributor here, so he might just turn up spontaneously. Sep 3 '14 at 8:57

This solution is based on the suggestion of Ian. First we need an extension of the Nazarov-Turán Lema in infinite dimentions. It can be found here.

The formulation of the Nazarov-Turán Lema in higher dimensions is the following:

(Nazarov-Turán Lemma) Let $p:\mathbb T^n\to\mathbb C$ a trigonometric polynomial definded by $$p(\boldsymbol z)=\sum_k c_k\boldsymbol z^{r_k}.$$

The index $k$ of the polynomial is given by $k=(k_1,\dots,k_n)$ for $0\leq k_i\leq m_i$ donde $m_i\in \mathbb N$. The exponents $r_k$ are defined by $(r_{1,k_1},\dots,r_{n,k_n})$ where $r_{i,0}<\dots<r_{i,m_i}\in \mathbb Z$. We take $z^{r_k}:=z_1^{r_{1,k_1}}\dots z_{n}^{r_{n,k_n}}$. Suppose that $c_k\neq 0$ for every $k$.

Let $E\subset \mathbb T^n$ a measurable set with $\tilde\mu(E)>0$, where $\tilde\mu$ is the Haar measure on $\mathbb T^n$ ($\tilde\mu(\mathbb T^n)=1$). Then $$\sup_{\boldsymbol z\in\mathbb T^n}{|p(\boldsymbol z)|}\leq \Big(\frac{14n}{\tilde\mu(E)}\Big)^{m_1+\cdots+m_n}\sup_{\boldsymbol z \in E}{|p(\boldsymbol z)|} .$$

The conjecture is an inmediate consecuence of this Lemma. Let's prove it:

Proof: Let $\tilde\psi:\mathbb T^2\to\mathbb R$ the natural periodic extension of $\psi:\Omega\to \mathbb R$. Because of the equivalence of norms in finite dimensional spaces, there is a $c>0$ such that $\|\psi\|_\infty = c$. Define $$E:=\{(x,t)\in\mathbb T^2\,:\,|\tilde \psi(x,t)|<\epsilon\}.$$

Applying the Nazarov-Turán Lemma with $n=2$ we have $$\mu(E)\leq 112\pi^2\epsilon^{\frac1{(m_1+m_2)}}c^{\frac{-1}{m_1+m_2}},$$ where $\mu$ is the Lebesgue measure.

It's clear that $\mu(\{(x,t)\in\Omega\,:\,|\psi(x,t)|<\epsilon\})\leq \mu(E)$. Taking $\alpha>0$ and $$\alpha<\frac{\log\Big(112\pi^2\epsilon^{\frac1{(m_1+m_2)}}c^{\frac{-1}{m_1+m_2}}\Big)}{\log(\epsilon)}$$ we obtain the desired estimation. Moreover $\alpha$ don't deppend on $\psi$