On the bilinear cone multiplier
On the bilinear cone multiplier
For $f,g \in \mathscr{S}(\R^n), n\geq 3$, consider the bilinear cone multiplier operator defined by \[{T}^{\lambda}_{R}(f,g)(x):=\int_{\mathbb{R}^{2n}}m^{\lambda}\left(\frac{\xi'}{R\xi_n},\frac{\eta'}{R\eta_n}\right)\hat{f}(\xi)\hat{g}(\eta)e^{2\pi\iota x\cdot(\xi+\eta)}~d\xi d\eta,\] where $\lambda>0, R>0$ and \[m^{\lambda}\left(\frac{\xi'}{R\xi_n},\frac{\eta'}{R\eta_n}\right)=\Big(1-\frac{|\xi'|^2}{R^2\xi^2_n}-\frac{|\eta'|^2}{R^2\eta^2_n}\Big)^{\lambda}_{+}\varphi(\xi_n)\varphi(\eta_n),\] $(\xi',\xi_n), (\eta',\eta_n)\in\mathbb{R}^{n-1}\times \mathbb{R}$ and $\varphi\in C_{c}^{\infty}([\frac{1}{2},2])$. We investigate the problem of pointwise almost everywhere convergence of ${T}^{\lambda}_{R}(f,g)(x)$ as $R\rightarrow \infty$ for $(f,g)\in L^{p_1}\times L^{p_2}$ for a wide range of exponents $p_1, p_2$ satisfying the H\"{o}lder relation $\frac{1}{p_1}+\frac{1}{p_2}=\frac{1}{p}$. This assertion is proved by establishing suitable weighted $L^{2}\times L^{2}\rightarrow L^{1}$--estimates of the maximal bilinear cone multiplier operator \[{T}^{\lambda}_{*}(f,g)(x):=\sup_{R>0}|{T}^{\lambda}_{R}(f,g)(x)|.\]
Saurabh Shrivastava、Kalachand Shuin
数学
Saurabh Shrivastava,Kalachand Shuin.On the bilinear cone multiplier[EB/OL].(2025-05-19)[2025-07-01].https://arxiv.org/abs/2505.13108.点此复制
评论