**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](http://nbviewer.ipython.org/github/asanjuanc/wavenotebooks/blob/master/conjeture.ipynb) is an IPython notebook I made with the corresponding experimentation 


**Related Previous Works**


* [This paper](http://www.maths.ed.ac.uk/~carbery/analysis/papers/cw-vc.pdf) is about Van Der Corput's Lemma in higher dimension. We don't know how to apply It.
* [This paper](http://eccc.hpi-web.de/report/2010/182/download/) 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?



A conjecture about the measure estimates of a trigonometric polynomial
This solution is based on the suggestion of [Ian](https://mathoverflow.net/users/1840/ian-morris). First we need an extension of the Nazarov-Turán Lema in infinite dimentions. It can be found [here](http://www.sciencedirect.com/science/article/pii/S0021904505002340). 


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 $\psi$


