For sake of notation, let $Q = [0,1]\times [0,1]$. Additionally, define $$ \rho_{x_0,\xi_0}(x,\xi) = \sqrt{(x-x_0)^2+(\xi-\xi_0)^2}. $$
Definition. Let $g\in L^2(Q)$. We say that the point $(x_0, \xi_0) \in Q$ is a weak $L^2$-zero of $g$ if the following holds:
- $g(x_0,\xi_0) = 0$.
- There exists constants $\delta > 0$, $k\in \mathbf{N}$ such that $$ \frac{\rho_{x_0,\xi_0}^k}{g}\chi_{B_{\delta}} \in L^2(Q) $$ where $B_{\delta}$ is an open ball centered at $(x_0, \xi_0)$ of radius $\delta$.
Furthermore, define the double trigonometric polynomial as follows. Let $F \subseteq \mathbf{Z}^2$ be a finite subset. Then, $$ p_m(x,\xi) = e^{2\pi i m_1 x}e^{-2\pi i m_2 \xi} + \sum_{n\in F}c_{mn}e^{2\pi i n_1 x}e^{-2\pi i n_2 \xi}, $$ where $m = (m_1,m_2)$ and $n = (n_1,n_2)$. Note that the coefficients $c_{mn}$ depend on both $m$ and $n$.
Conjecture. Let $g\in L^2(Q)$ such that $\frac{1}{g} \not\in L^2(Q)$. If $\frac{p_m}{g}\in L^2(Q)$ for all $m\not\in F$ then the following holds:
- $g$ has finitely many zeros in $Q$.
- Each zero of $g$, namely, $(x_i,\xi_i)$ is a weak $L^2$-zero of $g$.
Ideas. So I previously posted a question on Stack Exchange regarding the idea that $g$ has finitely many zeros. Specifically, the post is Zero set of double trigonometric polynomials.. I came up with a proof, but have yet to post it. I'm now stuck trying to prove the weak $L^2$-zero conclusion. The main issue is that the zero set of each $p_m$ is infinite.
However, one can show that the set of functions $\{p_m\}_{m\not\in F}$ has only finitely many zeros. Therefore, my current idea is to focus my attention on the following function: $$ f(x,\xi) = \sum_{m\not\in F}d_{m}|p_m|^2. $$ Now, let $d_m > 0$ and the coefficients $d_m$ decay faster than any polynomial. In other words, let $I$ be some enumeration of $\mathbf{Z}^2\setminus F$, then $d_i = \mathcal{O}(i^{-k})$ for all $k \in \mathbf{N}$. Then one can show that $f$ is infinitely differentiable, is non-negative, and possesses finitely many zeros in $Q$. Furthermore, with adequate coefficients $d_m$ one can also show that $\frac{f}{g}\in L^2(Q)$.
But, that is as far as I've been able to go. I would like to be able to show that if $f(x_0,\xi_0) = 0$, then $$ \frac{\rho_{x_0,\xi_0}^k}{f}\chi_{B_{\delta}} $$ is bounded for some $k\in \mathbf{N}$. However, I've not been able to deduce nor find any sort of lower bound inequality regarding a sum of double trigonometric polynomials. Any advice or counterexamples or anything would be appreciated.
As a side note, I can prove the following lemma directly. However, it does not directly utilize the sum of trigonometric polynomials idea.
Lemma. Let $F \subseteq \mathbf{Z}^2$ such that $|F|=1$. Let $g\in L^2(Q)$ such that $\frac{1}{g} \not\in L^2(Q)$. If $\frac{p_m}{g}\in L^2(Q)$ for all $m\not\in F$ then the following holds:
- $g$ has one zero in $Q$, specifically, $(x_0,\xi_0)$.
- $(x_0,\xi_0)$ is a weak $L^2$-zero of $g$.
Proof Sketch. $g$ must contain a zero that coincides with $p_m$ for all $m\not\in F$, otherwise $\frac{1}{g} \in L^2(Q)$ or $\frac{p_m}{g}\not\in L^2(Q)$. Let $F = {(a,b)}$. Additionally, note that the functions $p_m$ must have a shared zero. Let's call it $(x_0, \xi_0)$. Therefore, we have that $$ e^{2\pi i m_1 x_0}e^{-2\pi i m_2 \xi_0} + c_{m(a,b)}e^{2 \pi i a x_0}e^{-2\pi i b \xi_0} =0. $$ Hence, $c_m = -e^{2\pi i (m_1-a)x_0}e^{-2\pi i (m_2-b)\xi_0)}$. Therefore, we have the following simplification, $$ \begin{align*} |p_m(x,\xi)| &= |e^{2\pi i m_1 x}e^{-2\pi i m_2 \xi} -e^{2\pi i (m_1-a)x_0}e^{-2\pi i (m_2-b)\xi_0)}e^{2 \pi i a x}e^{-2\pi i b \xi}| \\ &= |e^{2\pi i (m_1-a)x_0}e^{-2\pi i (m_2-b)\xi_0)}e^{2 \pi i a x}e^{-2\pi i b \xi}|\cdot |e^{2\pi i (m_1-a)(x-x_0)}e^{-2\pi i (m_2-b)(\xi-\xi_0)}-1| \\ &= |e^{2\pi i (m_1-a)(x-x_0)}e^{-2\pi i (m_2-b)(\xi-\xi_0)}-1|. \end{align*} $$ Now, choose $m = (m_1,m_2) = (a+1,b)$. Therefore, for this specific $m$ we have that $$ |p_m| = |e^{2\pi i (x-x_0) - 1}| \geq C |x-x_0| $$ for $x$ sufficiently close to $x_0$. Additionally, one could also choose $m' = (m_1,m_2) = (a,b-1)$. And for this $m'$ we have that
$$ |p_{m'}| = |e^{2\pi i (\xi-\xi_0) - 1}| \geq C' |\xi-\xi_0| $$ for $\xi$ sufficiently close to $x_0$. Also observe that $|p_m|$ and $|p_{m'}|$ have perpendicular zero sets and also frequency one. Hence, intersect once. Therefore, we can conclude that $$ \frac{\rho_{x_0,\xi_0}}{g}\chi_{B} \leq \frac{|p_m|}{g} + \frac{|p_{m'}|}{g} \in L^2(Q); $$ for some open ball $B$ and positive constant $c$.
However, I have yet to be able to generalize this result to arbitrary $F$. Again, any advice/direction would be appreciated.