Consider a general bilinear multiplier operator:
$$
T(f,g)(n)=\int\_{\Pi}\int\_{\Pi}\hat{f}(\xi)\hat{g}(\eta)e^{2\pi i(\xi+\eta)n}m(\xi,\eta)d\xi d\eta,
$$
where $\Pi$ is the torus, $n\in\mathbb{Z}$, $m$ is periodic in both variables, $f$ and $g$ are compactly supported functions defined on $\mathbb{Z}$ and $\hat{f}(\xi):=\sum\_nf(n)e^{-2\pi in\xi}$.


**Prove or disprove:** $\|T(f,g)\|\_{L^2(\mathbb{Z})}\le C\|m\|\_{L^2(\Pi^2)}\|f\|\_{L^2(\mathbb{Z})}\|g\|\_{L^2(\mathbb{Z})}$.


The motivation of my question is the following: It is well-known that in the linear case the $L^2$-norm of a multiplier operator is just the $L^{\infty}$-norm of the multiplier. A bilinear analogue of this fact is that the $L^2\times L^2\to L^2$ norm of a bilinear multiplier operator is at most the $L^{\infty}$-norm of the multiplier (which is sometimes called "symbol"). It is natural to ask if we can get a better estimate, say $L^2$-norm of the multipler (instead of $L^{\infty}$-norm) in the bilinear setting.



The $L^2\times L^2\to L^2$ norm of the bilinear multiplier operator
Here is an example where $T(f,g)\notin l^2$ while $m\in L^2$ and $f=g\in l^2$:


Let $a(n)=|n|^{-1+s}$ with $\frac38<s<\frac12$ (so that $a\in l^2$), $m(\xi,\eta)=\hat{a}(\xi)\hat{a}(\eta)$. Take also $f=g=a$.


One has $\hat{a}(\xi)\propto |\xi|^{-s}$ near $0$. Then $\hat{a}(\xi)\propto |\xi|^{-2s}$, so that $(a \* a)(n)\propto n^{-1+2s}$ for large $n$. So $T(f,g)^2=(a\* a)^2\notin l^2$ because $-2+4s>-\frac12$.


