Abstract
AbstractLet $\mathscr{R}$ be a collection of disjoint dyadic rectangles $R$, let $\pi _R$ denote the non-smooth bilinear projection onto $R$and let $r>2$. We show that the bilinear Rubio de Francia operator associated with $\mathscr{R}$ given by $$\begin{equation*} f,g \mapsto \left(\sum_{R\in\mathscr{R}} |\pi_{R} (f,g)|^r \right)^{1/r} \end{equation*}$$ is $L^p \times L^q \rightarrow L^s$ bounded whenever $1/p + 1/q = 1/s$, $r^{\prime}<p,q<r$. This extends from squares to rectangles a previous result by the same authors in [7], and as a corollary extends in the same way a previous result from [2] for smooth projections, albeit in a reduced range.
Highlights
In this paper we present an improvement over the results of Benea and the first author in [2] and the results of [7] by the same authors
Linear variants of the operator TRr are known in the literature as Rubio de Francia operators
Let us briefly explain the structure of the proof by analogy with the linear case of Rubio de Francia operators RdFrI, where I “ tIn, n P Zu is a given collection of disjoint intervals
Summary
In this paper we present an improvement over the results of Benea and the first author in [2] and the results of [7] by the same authors. In this paper we will extend the result above to the case where the collection consists of arbitrary dyadic rectangles instead, again taking the frequency projections to be non-smooth. A standard argument shows that we have as a corollary that the same result holds for smooth frequency projections instead, but this time with the added benefit of allowing arbitrary disjoint rectangles, that is with sides not necessarily dyadic in any way (but still parallel to the axes, ). Let us briefly explain the structure of the proof by analogy with the linear case of Rubio de Francia operators RdFrI , where I “ tIn, n P Zu is a given collection of disjoint intervals.
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