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Calderón–Zygmund decomposition of an interval indicator

Example

Assume Countable Choice. Take $f = 1_{(0, 1]}$ on $R$ with the half-open all-generation dyadic grid $(a, b]$ of Dyadic cubes of all generations in R^n and $λ = 1 / 4$ . The unique maximal dyadic interval with average of $∣ f ∣$ above $λ$ is $(0, 2]$ , whose average is $1 / 2 = 2 λ$ ; the bad part is $b = \frac{1}{2} 1_{(0, 1]} - \frac{1}{2} 1_{(1, 2]}$ and the good part is $g = \frac{1}{2} 1_{(0, 2]}$ , so $∣ g ∣ = 2 λ$ on $(0, 2]$ , $\int b = 0$ and $\sum ∣ Q ∣ = 2 \leq λ^{- 1} ∥ f ∥_{1} = 4$ . The parent $(0, 4]$ of the maximal bad interval has average exactly $λ$ and is therefore good, which is how the $2 λ$ average bound is attained.

Facts & Assumptions

Given: Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )); the function $f = 1_{(0, 1]}$ on $R$ ; the all-generation half-open dyadic intervals $(m 2^{- k}, (m + 1) 2^{- k}]$ of Dyadic cubes of all generations in R^n; the height $λ = 1 / 4$ ; the decomposition of Calderón–Zygmund decomposition at height λ.

[F1]

The dyadic intervals of all generations are nested or disjoint, at each generation they partition $R$ , and the interval $(m 2^{- k}, (m + 1) 2^{- k}]$ has length $2^{- k}$ (All-generation dyadic cubes: partition, volume and nesting).

[F2]

Decomposition at height $λ$ : for $f \in L^{1} (R)$ the maximal bad dyadic intervals $Q$ , those with $∣ Q ∣^{- 1} \int_{Q} ∣ f ∣ > λ$ that are maximal under inclusion, are pairwise disjoint, and with $b = \sum_{j} (f - ∣ Q_{j} ∣^{- 1} \int_{Q_{j}} f) 1_{Q_{j}}$ and $g = f - b$ one has $\int b = 0$ , $∣ g ∣ \leq 2^{n} λ = 2 λ$ on the bad intervals, and $\sum_{j} ∣ Q_{j} ∣ \leq λ^{- 1} ∥ f ∥_{1}$ (Calderón–Zygmund decomposition at height λ).

Verification

technique · direct

1.1F1F2givenalgebra

By [F1], a dyadic interval meeting $(0, 1]$ is either contained in it or contains it. The former have average $1$ . A containing interval of length $2^{a}$ , $a \geq 0$ , has average $2^{- a}$ , exceeding $λ = 1 / 4$ exactly when $a = 0$ or $a = 1$ . The unique ancestors of these lengths are $(0, 1]$ and $(0, 2]$ , while $(0, 4]$ has average exactly $λ$ and every coarser ancestor has smaller average. Intervals disjoint from $(0, 1]$ have average zero. Thus all bad intervals are contained in $(0, 2]$ , which is itself bad and is the unique maximal bad interval, with average $1 / 2 = 2 λ$ .

2.1F2step 1.1givenalgebra

Reading off the formulas of the decomposition [F2] for the single maximal bad interval $Q = (0, 2]$ : $∣ Q ∣ = 2$ , $∣ Q ∣^{- 1} \int_{Q} f = 1 / 2$ , so $b = f - \frac{1}{2} 1_{(0, 2]} = \frac{1}{2} 1_{(0, 1]} - \frac{1}{2} 1_{(1, 2]}$ and $g = \frac{1}{2} 1_{(0, 2]}$ .

3.1step 1.1step 2.1algebra∎

The checks: $f = 1_{(0, 1]}$ has $∥ f ∥_{1} = 1$ and $λ^{- 1} ∥ f ∥_{1} = 4$ ; $\sum ∣ Q ∣ = ∣ (0, 2] ∣ = 2 \leq 4$ ; $∣ g ∣ = 1 / 2 = 2 λ$ on its support; and $\int b = \frac{1}{2} λ ((0, 1]) - \frac{1}{2} λ ((1, 2]) = \frac{1}{2} - \frac{1}{2} = 0$ . This is the asserted finite verification of maximality, of the $2 λ$ average bound, and of the parent-good property.

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Sources

Loukas Grafakos, Classical Fourier Analysis, third edition (standard reference, not scraped)