# the (2,2,1) boundedness of a “product” operator

Let $\{E_j\}_{j\in\mathbb{Z}}$ and $\{F_k\}_{k\in\mathbb{Z}}$ be two collections of pairwise disjoint sets in $\mathbb{R}$. Let $C(j,k)$ be a bounded function (e.g. $|C(j,k)|<1$) defined on $\mathbb{Z}^2$. For Schwartz functions $f$ and $g$, consider the operator $$T(f,g)=\sum_{j\in \mathbb{Z}}\sum_{k\in \mathbb{Z}}C(j,k)f_j g_k$$ where $f_j$ and $g_k$ are defined by $\hat{f_j}=\hat{f}1_{E_j}$ and $\hat{g_k}=\hat{g}1_{F_k}$.

It seems that $T$ is similar to the product operator $fg$ or a paraproduct. So we may have $\|T(f,g)\|_1\le C\|f\|_2\|g\|_2$ for some absolute constant $C$ and I'm trying to prove it. If the function $C(j,k)$ is independent of one of $j$ and $k$, then I can write the double sum as a product of two single sums and proceed with Cauchy-Schwartz and Plancherel. The real difficulty is that $C(j,k)$ depends on both $j$ and $k$. Any ideas?