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Power decay implies doubling

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Let $w$ be a weight on $R^{n}$ (Weights, their associated measures, and the spaces L^p(w)) and suppose there are constants $C, δ > 0$ such that $\frac{w (E)}{w (Q)} \leq C (\frac{∣ E ∣}{∣ Q ∣})^{δ}$ for every axis-parallel cube $Q$ and every measurable $E \subseteq Q$ . Then the measure $w d λ$ is doubling, with a constant depending only on $n$ , $C$ and $δ$ : there is $C^{'} = C^{'} (n, C, δ) < \infty$ with $w (B (x, 2 r)) \leq C^{'} w (B (x, r))$ for all $x \in R^{n}$ and $r > 0$ .

Facts & Assumptions

Given: Countable Choice; A weight $w$ and constants $C, δ > 0$ with the displayed power decay property.

[F1]

$w > 0$ and $w < \infty$ Lebesgue-a.e., $w \in L_{l o c}^{1}$ , and $E \mapsto w (E) = \int_{E} w d λ$ is a locally finite measure with the $w$ -null sets equal to the Lebesgue-null sets (Weights, their associated measures, and the spaces L^p(w)).

[F2]

$Q (x, ρ) = \prod_{i} (x_{i} - ρ, x_{i} + ρ)$ has side $2 ρ$ and Lebesgue measure $(2 ρ)^{n}$ ; if $E \subseteq F$ are measurable then $w (E) \leq w (F)$ , and for nested cubes $∣ Q (x, ρ^{'}) ∣ = ρ^{' n} ρ^{- n} ∣ Q (x, ρ) ∣$ (Axis-parallel cubes, their averages, and cube maximal functions, For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it).

[F3]

A ball $B (x, 2 r)$ is contained in the cube $Q (x, 4 r)$ , and a cube $Q (x, ρ)$ is contained in $B (x, \sqrt{n} ρ)$ (Ball and cube maximal functions are pointwise comparable).

Proof

technique · direct

1.1F1F2givenalgebra

Choose $β \in (0, 1)$ with $C β^{δ} \leq \frac{1}{2}$ , for instance $β : = \min {\frac{1}{2}, (2 C)^{- 1 / δ}}$ , and put $α : = 1 - C β^{δ} \geq \frac{1}{2}$ . If $E \subseteq Q$ is measurable with $∣ E ∣ \geq (1 - β) ∣ Q ∣$ , then $∣ Q ∖ E ∣ \leq β ∣ Q ∣$ , so the power decay applied to $Q ∖ E$ gives $w (Q ∖ E) \leq C β^{δ} w (Q)$ and hence $w (E) = w (Q) - w (Q ∖ E) \geq α w (Q) > 0$ . Both $β$ and $α$ depend only on $C$ and $δ$ .

2.1F2step 1.1givenalgebra

Fix $x \in R^{n}$ and $r > 0$ , let $l_{1} : = 4 r$ and $l_{j + 1} : = (1 - β)^{1 / n} l_{j}$ , and let $k \geq 1$ be the least integer with $l_{k} \leq r / \sqrt{n}$ ; then $k$ depends only on $n$ and $β$ , hence only on $n, C, δ$ . By [F2] one has $∣ Q (x, l_{j + 1}) ∣ = (1 - β) ∣ Q (x, l_{j}) ∣$ , so the shell $Q (x, l_{j}) ∖ Q (x, l_{j + 1})$ has measure $β ∣ Q (x, l_{j}) ∣$ and step 1.1 applied to $E = Q (x, l_{j + 1})$ (whose complement in $Q (x, l_{j})$ has measure $β ∣ Q (x, l_{j}) ∣$ ) yields $w (Q (x, l_{j + 1})) \geq α w (Q (x, l_{j}))$ for every $j < k$ . Iterating, $w (Q (x, 4 r)) \leq α^{- (k - 1)} w (Q (x, l_{k}))$ .

3.1F1F3step 2.1algebra∎

Since $l_{k} \leq r / \sqrt{n}$ , the cube $Q (x, l_{k})$ is contained in $B (x, r)$ by [F3], so [F1] gives $w (Q (x, l_{k})) \leq w (B (x, r))$ ; and $B (x, 2 r) \subseteq Q (x, 4 r)$ by [F3], so $w (B (x, 2 r)) \leq w (Q (x, 4 r)) \leq α^{- (k - 1)} w (B (x, r))$ . Thus $w d λ$ is doubling with $C^{'} = α^{- (k - 1)}$ , a constant depending only on $n$ , $C$ and $δ$ ; no property of $x$ or $r$ entered beyond the display, and the degenerate case $r > 0$ is the only case needed since doubling is asserted for positive radii.

Depends on

Used by

Power decay implies membership in some Aₚ Lemma

Dependency tree · two levels

46 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

Loukas Grafakos, Classical Fourier Analysis, 3rd ed. (Springer GTM 249, 2014) (standard reference, not scraped)
Juha Kinnunen, Harmonic Analysis (Aalto University lecture notes) (standard reference, not scraped)