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John-Nirenberg stopping cubes have geometric decay

Statement

Assume Countable Choice. Let $b \in B M O (R^{n})$ , let $Q$ be an all-generations dyadic cube and let $s > ∥ b ∥_{B M O}$ . Then there are families of dyadic subcubes $Q_{j}^{(k)} \subseteq Q$ , $k \geq 1$ , pairwise disjoint for each fixed $k$ , such that, with $Q_{1}^{(0)} : = Q$ : (i) each $Q_{j}^{(k)}$ is contained in a unique level- $(k - 1)$ stopping cube. Its immediate dyadic parent is contained in that stopping cube but need not belong to the preceding stopping family; (ii) for every $k$ , $\sum_{j} ∣ Q_{j}^{(k)} ∣ \leq (∥ b ∥_{B M O} / s)^{k} ∣ Q ∣$ ; and (iii) for every $k \geq 1$ , $∣ b - b_{Q} ∣ \leq k 2^{n} s$ almost everywhere on $Q ∖ ⋃_{j} Q_{j}^{(k)}$ , and a fortiori $∣ b - b_{Q} ∣ \leq 2^{n k} s$ there, so ${x \in Q : ∣ b - b_{Q} ∣ > 2^{n k} s} \subseteq ⋃_{j} Q_{j}^{(k)}$ up to a Lebesgue-null set.

Facts & Assumptions

Given: Countable Choice, $b \in B M O (R^{n})$ , an all-generations dyadic cube $Q$ and a real number $s > ∥ b ∥_{B M O}$ .

[F1]

The all-generations dyadic cubes partition $R^{n}$ at each generation, have volumes $2^{- k n}$ , and for two dyadic cubes one contains the other or they are disjoint; the parent of a cube of generation $k$ is its unique ancestor of generation $k - 1$ , with volume $2^{n}$ times that of the cube (Dyadic cubes of all generations in R^n, All-generation dyadic cubes: partition, volume and nesting).

[F2]

For $f \in L^{1} (R^{n})$ and $λ > 0$ , the dyadic cubes $E$ with $∣ E ∣^{- 1} \int_{E} ∣ f ∣ > λ$ that are maximal under inclusion form a countable family of pairwise disjoint cubes; their union is exactly the dyadic maximal superlevel set ${M_{d} f > λ}$ ; each such $E$ satisfies $∣ E ∣^{- 1} \int_{E} ∣ f ∣ \leq 2^{n} λ$ ; and $\sum_{E} ∣ E ∣ \leq λ^{- 1} ∥ f ∥_{1}$ (Maximal dyadic cubes above a level).

[F3]

In the Calderon-Zygmund decomposition of $f \in L^{1}$ at height $λ > 0$ , the good part $g$ equals $f$ outside the union of those maximal cubes and satisfies $∣ g ∣ \leq 2^{n} λ$ almost everywhere (Calderón–Zygmund decomposition at height λ).

[F4]

Almost every point of $R^{n}$ is a Lebesgue point of a given $L_{l o c}^{1}$ function (Almost every point is a Lebesgue point of a locally integrable function).

[F5]

A countable union of at most countable sets is at most countable, and a countable union of Lebesgue-null subsets of $R^{n}$ is Lebesgue-null (Countable unions of at most countable sets, assuming ${A C}_{ω}$ , Subsets and countable unions of null subsets of $R^{m}$ are null).

[F6]

For every cube $E$ , $∣ E ∣^{- 1} \int_{E} ∣ b - b_{E} ∣ \leq ∥ b ∥_{B M O}$ (BMO seminorm and the quotient by constants).

Proof

technique · direct

1.1F2F3F6algebra

Put $F = (b - b_{Q}) 1_{Q}$ . Then $F \in L^{1} (R^{n})$ with $\int_{R^{n}} ∣ F ∣ = \int_{Q} ∣ b - b_{Q} ∣ \leq ∣ Q ∣ ∥ b ∥_{B M O} < s ∣ Q ∣$ by [F6] and the hypothesis on $s$ . Let $Q^{(1)}$ be the family of dyadic cubes $E$ with $∣ E ∣^{- 1} \int_{E} ∣ F ∣ > s$ maximal under inclusion. By [F2] applied to $F$ and $s$ , the family $Q^{(1)}$ is countable and pairwise disjoint, each $E \in Q^{(1)}$ satisfies $s < ∣ E ∣^{- 1} \int_{E} ∣ F ∣ \leq 2^{n} s$ , and $\sum_{E \in Q^{(1)}} ∣ E ∣ \leq s^{- 1} \int_{R^{n}} ∣ F ∣ \leq (∥ b ∥_{B M O} / s) ∣ Q ∣$ . Every $E \in Q^{(1)}$ is a proper dyadic subcube of $Q$ : it meets $Q$ because its average of $∣ F ∣$ is positive, and if $Q ⊊ E$ then $∣ E ∣^{- 1} \int_{E} ∣ F ∣ \leq (∣ Q ∣ / ∣ E ∣) ∥ b ∥_{B M O} < s$ , a contradiction, while $E = Q$ is excluded by the same average bound. By [F3] the good part of the decomposition of $F$ equals $F$ off $⋃_{E \in Q^{(1)}} E$ , so $∣ F ∣ \leq 2^{n} s$ almost everywhere there.

2.1step 1.1F1F2F3F5algebra

Recursion. Let $Q^{(k)}$ be a countable pairwise disjoint family of proper dyadic subcubes of $Q$ , and for each $P \in Q^{(k)}$ put $F_{P} = (b - b_{P}) 1_{P}$ and let $Q_{P}$ be the family of maximal dyadic cubes $E$ with $∣ E ∣^{- 1} \int_{E} ∣ F_{P} ∣ > s$ . The argument of step 1.1, with $Q$ replaced by $P$ and $F$ by $F_{P}$ , shows that every $E \in Q_{P}$ is a proper dyadic subcube of $P$ , that $Q_{P}$ is countable and pairwise disjoint, that $s < ∣ E ∣^{- 1} \int_{E} ∣ F_{P} ∣ \leq 2^{n} s$ and $\sum_{E \in Q_{P}} ∣ E ∣ \leq (∥ b ∥_{B M O} / s) ∣ P ∣$ , and that the good part of the decomposition of $F_{P}$ satisfies $∣ F_{P} ∣ \leq 2^{n} s$ almost everywhere off $⋃_{E \in Q_{P}} E$ . Each selected $E$ has a unique stopping parent $P \in Q^{(k)}$ . Its immediate dyadic parent $R$ is contained in $P$ : since $R$ and $P$ both contain $E$ , [F1] makes them nested; if $P ⊊ R$ , the generations of $R$ and $E$ differ by one and $E ⊊ P ⊊ R$ is impossible, so $R \subseteq P$ . The dyadic parent has the exact volume ratio $∣ R ∣ = 2^{n} ∣ E ∣$ by [F1]. Maximality gives $∣ R ∣^{- 1} \int_{R} ∣ F_{P} ∣ \leq s$ , and hence $∣ E ∣^{- 1} \int_{E} ∣ F_{P} ∣ \leq 2^{n} s$ . Thus the stopping parent is $P$ , while the immediate dyadic parent $R$ need only be contained in $P$ and need not itself belong to the preceding stopping family. Define $Q^{(k + 1)} = ⋃_{P \in Q^{(k)}} Q_{P}$ ; indexed by pairs it is countable by [F5], its members are pairwise disjoint because distinct cubes $P$ are disjoint and each $Q_{P}$ is a pairwise disjoint family of subcubes of its $P$ , and each member is a proper dyadic subcube of a unique $P \in Q^{(k)}$ , so its interior is contained in that cube.

3.1step 1.1step 2.1

By induction on $k \geq 1$ , $\sum_{E \in Q^{(k)}} ∣ E ∣ \leq (∥ b ∥_{B M O} / s)^{k} ∣ Q ∣$ : step 1.1 is the case $k = 1$ , and the ratio estimate of step 2.1 gives $\sum_{E \in Q^{(k + 1)}} ∣ E ∣ = \sum_{P \in Q^{(k)}} \sum_{E \in Q_{P}} ∣ E ∣ \leq (∥ b ∥_{B M O} / s) \sum_{P \in Q^{(k)}} ∣ P ∣$ , which is the induction step.

3.2step 1.1step 2.1F2F3F4F5algebra

Fix $k \geq 1$ and discard the exceptional Lebesgue-null sets of steps 1.1 and 2.1 together with the null sets of non-Lebesgue points of the countably many functions $F_{P}$ for $P$ of level at most $k - 1$ ; by [F4] and [F5] the discarded set is Lebesgue-null. Let $x \in Q ∖ ⋃_{E \in Q^{(k)}} E$ lie outside it, and let $0 \leq l \leq k - 1$ be the largest integer such that $x$ lies in some cube $P_{l}$ of level $l$ , where $P_{0} : = Q$ ; such an $l$ exists because $x \in Q$ , and $P_{l}$ is unique because the families at each level are pairwise disjoint. Since $x$ lies in no cube of level $l + 1$ , the good-part bound of [F3] for the ambient cube $P_{l}$ gives $∣ b (x) - b_{P_{l}} ∣ = ∣ F_{P_{l}} (x) ∣ \leq 2^{n} s$ ; moreover, because $x$ is a Lebesgue point of $F_{P_{l}}$ , the dyadic cubes of generation $m$ containing $x$ lie in $B (x, \sqrt{n} 2^{- m})$ with volume comparable to $2^{- m n}$ , so their averages of $F_{P_{l}}$ converge to $F_{P_{l}} (x)$ , giving $∣ F_{P_{l}} (x) ∣ \leq M_{d} F_{P_{l}} (x) \leq s$ by [F2] and [F4]. For $l \geq 1$ , $∣ b_{P_{l}} - b_{Q} ∣ \leq \sum_{i = 1}^{l} ∣ b_{P_{i}} - b_{P_{i - 1}} ∣$ with $∣ b_{P_{i}} - b_{P_{i - 1}} ∣ \leq ∣ P_{i} ∣^{- 1} \int_{P_{i}} ∣ b - b_{P_{i - 1}} ∣ = ∣ P_{i} ∣^{- 1} \int_{P_{i}} ∣ F_{P_{i - 1}} ∣ \leq 2^{n} s$ , because $P_{i}$ was selected inside the ambient cube $P_{i - 1}$ . Hence $∣ b (x) - b_{Q} ∣ \leq s + l 2^{n} s \leq (l + 1) 2^{n} s \leq k 2^{n} s \leq 2^{n k} s$ , using $l + 1 \leq k$ and $k \leq 2^{n (k - 1)}$ for $k \geq 1$ , $n \geq 1$ . The complement of $⋃_{E \in Q^{(k)}} E$ inside $Q$ therefore satisfies the asserted almost-everywhere bound, its exceptional set being a countable union of Lebesgue-null sets by [F5].

4.1step 1.1step 2.1step 3.1step 3.2∎

Steps 1.1 and 2.1 construct, for every $k \geq 1$ , a countable pairwise disjoint family $Q^{(k)}$ of proper dyadic subcubes of $Q$ , each with a unique stopping parent in the preceding family and with its immediate dyadic parent contained in that stopping parent, which is (i); step 3.1 is the geometric-decay estimate (ii); and step 3.2 is the level bound (iii). The construction applies the published maximal-cube and decomposition lemmas countably many times, which is exactly where Countable Choice is used, together with the countable-union facts in [F5].

Depends on

Used by

John-Nirenberg exponential inequality Theorem

Dependency tree · two levels

47 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Juha Kinnunen, Harmonic Analysis (Aalto University lecture notes) (standard reference, not scraped)
Mark Williams, Notes on Harmonic Analysis (January 11, 2022) (standard reference, not scraped)
Brooke Wilson, Math 581A Classical and Multilinear Harmonic Analysis (University of Washington, Fall 2024), lecture 18 (standard reference, not scraped)