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A_infinity weights satisfy power decay

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $w \in A_{\infty}$ (The Muckenhoupt A_infinity class). Then there are $C, δ > 0$ , depending only on $n$ , on a witnessing exponent $p$ and on $[w]_{A_{p}}$ , such that $\frac{w (E)}{w (Q)} \leq C (\frac{∣ E ∣}{∣ Q ∣})^{δ}$ for every axis-parallel cube $Q$ and every measurable $E \subseteq Q$ . Explicitly one may take $δ = γ / (1 + γ)$ and $C = C_{1}$ , where $1 + γ$ and $C_{1}$ are the reverse Hölder exponent and constant supplied for a witnessing $A_{p}$ weight.

Facts & Assumptions

Given: Countable Choice, $w \in A_{\infty}$ , a witnessing exponent $p$ with $w \in A_{p}$ , a reverse Hölder pair $(1 + γ, C_{1})$ , a cube $Q$ and a measurable $E \subseteq Q$ .

[F1]

$w \in A_{\infty}$ means $w \in A_{p}$ for some $1 \leq p < \infty$ , and then $0 < w (Q) < \infty$ for every cube $Q$ (The Muckenhoupt A_infinity class).

[F2]

The reverse Hölder theorem applied to a fixed $0 < α < 1$ gives $γ > 0$ and $C_{1} < \infty$ , depending only on $n, p, [w]_{A_{p}}$ and $α$ , with $(⟨ w^{1 + γ} ⟩_{Q})^{1 / (1 + γ)} \leq C_{1} ⟨ w ⟩_{Q}$ for every cube (Reverse Holder self-improvement for A_p weights).

[F3]

Hölder's inequality: for a measurable set $E$ , $\int_{E} w d λ \leq (\int_{Q} w^{1 + γ} d λ)^{1 / (1 + γ)} ∣ E ∣^{γ / (1 + γ)}$ (the exponents $1 + γ$ and $(1 + γ) / γ$ are conjugate) (Holder's inequality for integrals, including the endpoint cases, Real powers for positive bases, with the zero-base positive-exponent convention).

Proof

technique · direct

1.1F1F2given

Since $w \in A_{\infty}$ , there is a witnessing exponent $p$ with $w \in A_{p}$ by [F1]; applying [F2] produces $γ > 0$ and $C_{1}$ with the normalized reverse Hölder inequality.

2.1F2F3step 1.1givenalgebra

For the cube $Q$ and measurable $E \subseteq Q$ , Hölder's inequality [F3] gives $w (E) \leq (\int_{Q} w^{1 + γ})^{1 / (1 + γ)} ∣ E ∣^{γ / (1 + γ)}$ ; the reverse Hölder inequality bounds the first factor by $C_{1} ⟨ w ⟩_{Q} ∣ Q ∣^{1 / (1 + γ)} = C_{1} w (Q) ∣ Q ∣^{- γ / (1 + γ)}$ , so $w (E) \leq C_{1} w (Q) (∣ E ∣ / ∣ Q ∣)^{γ / (1 + γ)}$ .

3.1F1step 2.1givenalgebra∎

Dividing by $w (Q) \in (0, \infty)$ gives $w (E) / w (Q) \leq C (∣ E ∣ / ∣ Q ∣)^{δ}$ with $C = C_{1}$ and $δ = γ / (1 + γ) > 0$ , and both constants depend only on $n$ , the witnessing exponent $p$ and $[w]_{A_{p}}$ (and on the auxiliary $α$ , which is fixed in the construction).

Depends on

Used by

Dependency tree · two levels

32 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)