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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.

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  • $\begingroup$ Consider what happens when you set $m=\hat f\hat g$. $\endgroup$ – Michael Renardy Jul 10 '15 at 8:09
  • $\begingroup$ Compactly supported functions are dense in $\ell^2(\mathbb Z)$, so your estimate must hold for all $f,g\in\ell^2$ if it holds for compactly supported ones. $\endgroup$ – Joonas Ilmavirta Jul 10 '15 at 11:05
  • $\begingroup$ @MichaelRenardy Then the operator becomes $T(f,g)(n)=f*f(n)g*g(n)$ $\endgroup$ – Tony B Jul 10 '15 at 12:45
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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 $-2+4s>-\frac12$.

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