A bilinear Rubio de Francia inequality for arbitrary rectangles

TL;DR

This paper proves a new boundedness result for a bilinear Rubio de Francia operator associated with arbitrary rectangles, extending previous results from squares and smooth projections to a broader class of rectangles.

Contribution

It establishes $L^p imes L^q ightarrow L^s$ bounds for a bilinear operator over arbitrary rectangles, extending prior work from squares and smooth projections.

Findings

Extends boundedness results from squares to arbitrary rectangles.

Generalizes previous results to non-smooth bilinear projections.

Provides a broader range of $p,q$ for which the operator is bounded.

Abstract

Let $\mathscr{R}$ be a collection of disjoint dyadic rectangles $R$ with sides parallel to the axes, let $\pi_R$ denote the non-smooth bilinear projection onto $R$ \[ \pi_R (f,g)(x):=\iint \mathbf{1}_{R}(\xi,\eta) \widehat{f}(\xi) \widehat{g}(\eta) e^{2\pi i (\xi + \eta) x} d\xi d\eta \] and let $r>2$. We show that the bilinear Rubio de Francia operator associated to $\mathscr{R}$ given by \[ f,g \mapsto \Big(\sum_{R\in\mathscr{R}} |\pi_{R} (f,g)|^r \Big)^{1/r} \] is $L^p \times L^q \rightarrow L^s$ bounded whenever $1/p + 1/q = 1/s$, $r'<p,q<r$. This extends from squares to rectangles a previous result by the same authors, and as a corollary extends in the same way a previous result from Benea and the first author for smooth projections, albeit in a reduced range.