A PROOF OF THE DYADIC JOURNÉ LEMMA

TERENCE TAO

Abstract. We give a proof of Journé’s lemma in the dyadic setting.

1. Introduction

Let {R} be a collection of dyadic rectangles in R^{2} which are non-nested in the sense that R ⊄ R′ for any distinct R, R′ ∈ {R}. Let Ω denote the set

Ω = \{M_{S}(χ_{⋃_{R} R}) ≥ \frac{1}{4}\}

where M_{S} is the strong maximal function. Since M_{S} is bounded on L^{2}, we see from Chebyshev’s inequality that

|Ω| ∼ | ⋃_{R} R|.

If the R were all disjoint, we would of course have

∑_{R} |R| = | ⋃_{R} R|.

Unfortunately, this estimate is not correct, however we have the substitute

∑_{R} |R|(\frac{|R|}{|R^{∗}|})^{ε} ≲ | ⋃_{R} R|

for any ε > 0 (with the implicit constant depending on ε of course), where R^{∗} is the largest dyadic rectangle in Ω which contains R (of course, R^{∗} will almost certainly not be in {R}). The ratio |R^{∗}|/|R| measures how deeply R is embedded inside Ω, and the above estimate (which is known as Journé’s lemma) states that the shallowly embedded rectangles R are essentially disjoint.

We now prove this lemma, by first introducing some notation. Let π_{1}(R) and π_{2}(R) denote the horizontal and vertical projections of R, hence R = π_{1}(R)×π_{2}(R). We introduce the Fefferman ordering on dyadic rectangles, defining R < R′ if π_{1}(R) ⊂ π_{1}(R′) and π_{2}(R) ⊃ π_{2}(R′). (Note that this breaks the symmetry between the horizontal and vertical co-ordinates). For each x ∈ ⋃_{R} R, we define R_{max}(x) to be the maximal rectangle in {R} (with respect to the Fefferman ordering) which contains x.

We now make the crucial observation that

|\{x ∈ R : R_{max}(x) > 2R^{∗}\}| ≤ \frac{1}{2}|R|

for all R ∈ R, where 2R^{∗} is the parent of R^{∗}. For if this were not the case, and over half of the points in R satisfied R_{max}(x) > 2R^{∗}, then it is clear that 2R^{∗} is at least one-quarter filled with elements from ⋃_{x∈R} R_{max}(x) ⊂ ⋃_{R} R, contradicting the maximality of R^{∗}.

1991 Mathematics Subject Classification. 42B.

From the above observation we thus see that

∑_{R} |R|(\frac{|R|}{|R^{∗}|})^{ε} ≲ ∑_{R} ∫_{x∈R:R_{max}(x)>2R^{∗}}(\frac{|R|}{|R^{∗}|})^{ε} dx.

Since R_{max}(x) > 2R^{∗}, we have

\frac{|R|}{|R^{∗}|} ≲ \frac{|π_{1}(R)|}{|π_{1}(R_{max}(x))|}.

Thus we have

∑_{R} |R|(\frac{|R|}{|R^{∗}|})^{ε} ≲ ∑_{R} ∫_{R}(\frac{|π_{1}(R)|}{|π(R_{max}(x))})^{ε} dx.

By Fubini we thus have

∑_{R} |R|(\frac{|R|}{|R^{∗}|})^{ε} ≲ ∫_{⋃_{R} R} ∑_{Rnix}(\frac{|π_{1}(R)|}{|π_{1}(R_{max}(x))})^{ε} dx.

However, from the non-nestedness of the R we thus see that the dyadic numbers |π_{1}(R)| are never repeated as R ranges over all rectangles containing x. Thus the inner sum is convergent, and the claim follows.

References

[1] J.L. Journé, A covering lemma for product spaces, Proc. Amer. Math. Soc. 96 (1986), 593– 598.

Department of Mathematics, UCLA, Los Angeles, CA 90024

   E-mail address: tao@math.ucla.edu