For a linear multiplier operator $T(f)(x)=\int\_{\mathbb{R}} m(\xi)\hat{f}(\xi)e^{2\pi ix\xi}d\xi$, we know that $\|m\|\_{\infty}$ gives the operator norm of $T$ from $L^2$ to itself immediately. What about the bilinear case? Let $T(f,g)(x)=\int\_{\mathbb{R^2}} m(\xi,\eta)\hat{f}(\xi)\hat{g}(\eta)e^{2\pi ix(\xi+\eta)}d\xi d\eta$ be a bilinear multiplier operator. Can the knowledge of $\|m\|\_{\infty}$ provide any information about boundedness of $T$ in any sense?



symbol $m\in L^{\infty}$ implies any boundedness of a bilinear operator?
The answer is NO, in general. There is no such simple result in the bilinear case. 


Look for example at:


[L. Grafakos and N. Kalton, The Marcinkiewicz multiplier condition for bilinear operators. Studia Math. 146 (2001), 115–156.](http://arxiv.org/pdf/math/0010076v1.pdf)


There the authors show that there exist smooth multipliers whose derivatives of any order decay at infinity separately in $\xi$ and $\eta$, which do not map $L^{p\_1}\times L^{p\_2}$ into $L^{p\_3,\infty}$ where $p\_1,p\_2,p\_3$ are Hölder exponents.


Also take a loot at


[A. Benyi and R.H. Torres, Almost orthogonality and a class of bounded bilinear pseudodifferential operators. Math. Res. Lett. 11 (2004), 1-11.](http://faculty.wwu.edu/~benyia/papers/revised2.pdf)


In Proposition 1 the authors show (by contradiction) that the boundedness of derivatives of the symbols of all orders does not suffice for any boundedness result for $T$.


