Timeline for symbol $m\in L^{\infty}$ implies any boundedness of a bilinear operator?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 23, 2016 at 3:14 | vote | accept | Tony B | ||
| Feb 18, 2016 at 14:39 | answer | added | ioannis.parissis | timeline score: 4 | |
| Dec 22, 2015 at 18:18 | comment | added | Matt Rosenzweig | 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. | |
| Jan 11, 2015 at 13:03 | comment | added | Joonas Ilmavirta | @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 $(x,x)\in\mathbb R^2$). | |
| Jan 11, 2015 at 10:52 | comment | added | Jochen Wengenroth | 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 $\hat f(\xi) \hat g(\eta)$. | |
| Jan 11, 2015 at 1:46 | history | asked | Tony B | CC BY-SA 3.0 |