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?

$\begingroup$ Isn't $T$ the multiplier operator on $L^2(\mathbb R^2)= L^2(\mathbb R) \hat \otimes L^2(\mathbb R)$? Note that the Fourier tranform of $f\otimes g$ is $\hat f(\xi) \hat g(\eta)$. $\endgroup$– Jochen WengenrothJan 11, 2015 at 10:52

$\begingroup$ @JochenWengenroth, the operator $T$ is the restriction of the multiplier operator to the diagonal of $\mathbb R^2$ (if $x\in\mathbb R$ corresponds to $(x,x)\in\mathbb R^2$). $\endgroup$– Joonas IlmavirtaJan 11, 2015 at 13:03

$\begingroup$ If $m$ is a symbol of order $0$ in the sense that $$\nabla_{\xi}^{j}\nabla_{\eta}^{k}m(\xi,\eta)\lesssim_{j,k}(\xi+\eta)^{jk}$$ for all $j,k\geq 0$, then one has a special case of the CoifmanMeyer multiplier theorem, which says that $\T(f,g)\_{L^{r}}\lesssim_{p,q}\f\_{L^{p}}\g\_{L^{q}}$ for all $1<p,q<\infty$ satisfying $\frac{1}{p}+\frac{1}{q}=\frac{1}{r}$. Of course, the hypotheses here are much stronger than just $m$ is bounded. $\endgroup$– Matt RosenzweigDec 22, 2015 at 18:18
1 Answer
The answer is NO, in general. There is no such simple result in the bilinear case.
Look for example at:
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
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 $T$.