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 $${T}^{\lambda}_{*}(f,g)(x):=\sup_{R>0}|{T}^{\lambda}_{R}(f,g)(x)|.$$