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Source. Proposition 3.26, p. 17, of Cabrelli, Lacey, Molter and Pipher, Variations on the theme of Journé's lemma, in the edition named on the source card. The paper presents it as the lemma from the Appendix of S. H. Ferguson and M. T. Lacey, A characterization of product BMO by commutators, Acta Math. 189 (2002), 143--160.

Statement

Setting (p. 16). For a set $V$ containing the shadow and a rectangle $R\in\mathcal U$,

$$\operatorname{emb}(R,V)=\sup\{\mu\ge1:\mu R\subset V\},$$

with all sides of $R$ dilated by $\mu$ about its center.

Proposition 3.26 (p. 17). For each $0<\delta,\epsilon<1$ there is a constant $K_{\delta,\epsilon}$ such that for every collection $\mathcal U$ of rectangles whose shadow has finite measure in the plane there is a set $V\supset\operatorname{sh}(\mathcal U)$ with $\lvert V\rvert<(1+\delta)\lvert\operatorname{sh}(\mathcal U)\rvert$ such that for every collection $\mathcal U'\subset\mathcal U$,

$$\sum_{R\in\mathcal U'}\operatorname{emb}(R,V)^{-\epsilon}\lvert R\rvert \lesssim\lvert\operatorname{sh}(\mathcal U')\rvert\quad(3.27).$$

The implied constant depends only on $\epsilon$ and $\delta$. The statement introduces $K_{\delta,\epsilon}$ but writes the inequality with $\lesssim$.

Proof pointer

Pp. 17--18. The set $V$ is built from one-dimensional maximal functions on Christ's shifted dyadic grids (1.6), taken in one coordinate and then the other (3.28); the paper states $\lvert V\rvert<(1+K\delta\log\delta^{-1})\lvert\operatorname{sh}(\mathcal U)\rvert$ for this set, with its own $\delta=(1+2^{\mathsf d})^{-1}$, and takes it as $V$. The key property (3.29) is that a dyadic rectangle inside the first-stage set has its four $\delta$-shifted translates inside $V$. The estimate then follows the essentially-disjoint argument of Section 3.2, with scales separated by $10^6\mu\delta^{-1}$.

Dependencies

None in the corpus. Read depth: claims checked; the definition and statement were read clause by clause on pp. 16--17, the proof for structure only. Nothing here is independently reviewed.

Bears on

The paper names no Erdős problem.