任意正方形上的双线性 Rubio de Francia 不等式
经典分析与常微分方程
2016-02-08 v1
摘要
我们证明了与频率平面上任意正方形集合(边平行于坐标轴)相关的光滑双线性 Rubio de Francia 算子的有界性( f , g ) ↦ ( ∑ ω ∈ Ω ∣ ∫ R 2 f ^ ( ξ ) g ^ ( η ) Φ ω ( ξ , η ) e 2 π i x ( ξ + η ) d ξ d η ∣ r ) 1 / r , \left(f, g \right)\mapsto \left( \sum_{\omega \in \Omega}\left| \int_{\mathbb{R}^2} \hat{f}(\xi) \hat{g}(\eta) \Phi_{\omega}(\xi, \eta) e^{2 \pi i x\left(\xi+\eta \right)} d \xi d \eta\right|^r \right)^{1/r}, ( f , g ) ↦ ( ω ∈ Ω ∑ ∫ R 2 f ^ ( ξ ) g ^ ( η ) Φ ω ( ξ , η ) e 2 π i x ( ξ + η ) d ξ d η r ) 1/ r , 条件是 r \textgreater 2 r\textgreater{}2 r \textgreater 2 。更确切地说,我们表明上述算子映射 L p × L q → L s L^p \times L^q \to L^s L p × L q → L s ,只要 p , q , s ′ p, q, s' p , q , s ′ 处于“局部 L r ′ L^{r'} L r ′ ”范围内,即 1 p + 1 q + 1 s ′ = 1 \displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{s'}=1 p 1 + q 1 + s ′ 1 = 1 ,0 ≤ 1 p , 1 q \textless 1 r ′ \displaystyle0 \leq \frac{1}{p}, \frac{1}{q} \textless{\frac{1}{r'}} 0 ≤ p 1 , q 1 \textless r ′ 1 ,且 1 s ′ \textless 1 r ′ \displaystyle\frac{1}{s'}\textless{\frac{1}{r'}} s ′ 1 \textless r ′ 1 。注意我们允许 s ′ s' s ′ 取负值,其对应于拟 Banach 空间 L s L^s L s 。
引用
@article{arxiv.1602.01948,
title = {A bilinear Rubio de Francia inequality for arbitrary squares},
author = {Cristina Benea and Frederic Bernicot},
journal= {arXiv preprint arXiv:1602.01948},
year = {2016}
}