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On the bilinear cone multiplier (2505.13108v1)

Published 19 May 2025 in math.CA

Abstract: For $f,g \in \mathscr{S}(\Rn), n\geq 3$, 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}{R2\xi2_n}-\frac{|\eta'|2}{R2\eta2_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)|.]

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