# symbol $m\in L^{\infty}$ implies any boundedness of a bilinear operator?

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?

• 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 $\hat f(\xi) \hat g(\eta)$. – Jochen Wengenroth Jan 11 '15 at 10:52
• @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 $(x,x)\in\mathbb R^2$). – Joonas Ilmavirta Jan 11 '15 at 13:03
• If $m$ is a symbol of order $0$ in the sense that $$|\nabla_{\xi}^{j}\nabla_{\eta}^{k}m(\xi,\eta)\lesssim_{j,k}(|\xi|+|\eta|)^{-j-k}$$ for all $j,k\geq 0$, then one has a special case of the Coifman-Meyer 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. – Matt Rosenzweig Dec 22 '15 at 18:18

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.

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.

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 $T$.