$\textbf{Set up.}$ Let $Q = [0,1]\times [0,1]$. Let $m = (m_1,m_2)\in \mathbf{Z}^2$. Let $F\subseteq \mathbf{Z}^2$ be a finite set. Define the double trigonometric polynomial $$ p_m(x,\xi) = e^{2\pi i m_1 x}e^{-2\pi i m_2 \xi}+\sum_{n\in F}c_n e^{2\pi i n_1 x}e^{-2\pi i n_2 \xi}. $$ Let $g\in L^2(Q)$ such that $\frac{1}{g}\not\in L^2(Q)$. Moreover, define $$ d(x,\xi) = \sqrt{x^2+\xi^2}. $$
$\textbf{Question/Conjecture.}$ Suppose that $\frac{p_m}{g}\in L^{\infty}(Q)$ for all $m\not\in F$. Then, $g$ has finitely many zeros $\{(x_i,\xi_i)\}_{i=1}^k$ (proven elsewhere). Additionally, there exists constants $\delta_k > 0$, $c_k$, and $r_k > 0$ such that for almost all $(x,\xi)\in B_{\delta_k}$, where $B_{\delta_k}$ is the ball of radius $\delta_k$ centered at $(x_k,\xi_k)$ $$ |g(x,\xi)| \leq c_k\, d(x-x_k,\xi-\xi_k)^{r_k}. $$
$\textbf{Ideas.}$ This is almost like a reverse Lojasiewicz's Inequality, however, $g$ may not be real analytic. The boundedness assumption on $\frac{p_m}{g}$ helps with a lower bound, but I'm not sure if an upper bound could be derived. Any help would be appreciated, or a counter example.