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Reverse Holder self-improvement for A_p weights

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $1 \leq p < \infty$ and $w \in A_{p}$ (Muckenhoupt A_p and A_1 weights). Fix $0 < α < 1$ and put $β : = 1 - \frac{(1 - α)^{p}}{K_{p}}, γ : = \frac{1}{2} \frac{- \log β}{\log (2^{n} α^{- 1})} > 0$ (the logarithm is the one of The logarithm to a positive base other than one, and the powers are those of Real powers for positive bases, with the zero-base positive-exponent convention). Then $(2^{n} α^{- 1})^{γ} β = β^{1 / 2} < 1,$ and for every axis-parallel cube $Q$ , $(⟨ w^{1 + γ} ⟩_{Q})^{1 / (1 + γ)} \leq C ⟨ w ⟩_{Q}, C : = (1 + \frac{(2^{n} α^{- 1})^{γ}}{1 - (2^{n} α^{- 1})^{γ} β})^{1 / (1 + γ)} < \infty .$ Thus $w$ satisfies a reverse Hölder inequality with exponent $1 + γ$ , and $γ$ and $C$ depend only on $n$ , $p$ , $K_{p}$ and the fixed $α$ .

Here $K_{p} = [w]_{A_{p}}$ for $p > 1$ , and $K_{1} = \sup_{Q} ⟨ w ⟩_{Q} / ({ess inf}_{Q} w)$ ; by The two defining forms of A_1 agree, $1 \leq K_{1} \leq c_{n} [w]_{A_{1}}$ . Thus the constants remain controlled by the stated $A_{p}$ data, including the ball-normalized endpoint characteristic.

Facts & Assumptions

Given: Countable Choice, $1 \leq p < \infty$ , $w \in A_{p}$ , $0 < α < 1$ , and the constants $β, γ$ of the Statement.

[F1]

$0 < w (Q) < \infty$ for every cube. For $p > 1$ , the defining product is at most $K_{p} = [w]_{A_{p}}$ and Hölder applies; for $p = 1$ , $K_{1}$ is the cube-average/essential-infimum characteristic above (Muckenhoupt A_p and A_1 weights, The two defining forms of A_1 agree, Holder's inequality for integrals, including the endpoint cases).

[F2]

For a cube $Q_{0}$ with $α_{0} = ⟨ w ⟩_{Q_{0}} > 0$ and $α_{k} = (2^{n} α^{- 1})^{k} α_{0}$ , the sets $U_{k}$ of the maximal dyadic subcubes with $⟨ w ⟩_{R} > α_{k}$ satisfy $U_{k + 1} \subseteq U_{k}$ , $w (U_{k}) \leq β^{k} w (Q_{0})$ and $w \leq α_{k}$ almost everywhere on $Q_{0} ∖ U_{k}$ (Distribution decay from maximal cubes for A_p weights).

[F3]

$a^{x} = \exp (x \log a)$ for $a > 0$ and real $x$ (Real powers for positive bases, with the zero-base positive-exponent convention), $\log_{b} x = \log x / \log b$ for $b > 0$ , $b \neq 1$ , $x > 0$ (The logarithm to a positive base other than one), and the elementary exponential identities $\exp (u + v) = \exp (u) \exp (v)$ , $\exp (- u) = 1 / \exp (u)$ follow from The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ and the inverse identities of The natural logarithm as the inverse of the exponential function.

[F4]

Integrals of nonnegative measurable functions over a measurable set are countably additive on disjoint measurable pieces, and monotone under inclusion (Countable additivity and continuity of finitely additive set functions).

Proof

technique · direct

1.1F1givenalgebra

For $p > 1$ , Hölder applied to $w^{1 / p} w^{- 1 / p}$ on a cube gives $1 \leq ⟨ w ⟩_{Q}^{1 / p} ⟨ w^{- 1 / (p - 1)} ⟩_{Q}^{(p - 1) / p} \leq K_{p}^{1 / p}$ . For $p = 1$ , $⟨ w ⟩_{Q} \geq {ess inf}_{Q} w$ gives $K_{1} \geq 1$ . Therefore $0 < (1 - α)^{p} < K_{p}$ and $β \in (0, 1)$ , so $γ > 0$ is well defined.

2.1F3step 1.1givenalgebra

By [F3], $\log β < 0$ and $(2^{n} α^{- 1})^{γ} = \exp (γ \log (2^{n} α^{- 1})) = \exp (\frac{1}{2} (- \log β)) = β^{- 1 / 2}$ ; therefore $(2^{n} α^{- 1})^{γ} β = β^{- 1 / 2} β = β^{1 / 2} < 1$ , and the geometric ratio $q : = (2^{n} α^{- 1})^{γ} β$ satisfies $0 < q < 1$ with $1 / (1 - q) < \infty$ .

3.1F2F4step 2.1givenalgebra

Fix a cube $Q_{0} = Q$ and apply [F2] with $α_{0} = ⟨ w ⟩_{Q_{0}}$ : the sets $Q_{0} ∖ U_{0}$ and $U_{k} ∖ U_{k + 1}$ , $k \geq 0$ , are disjoint and measurable and cover $Q_{0}$ up to a Lebesgue-null set: indeed $w (⋂_{k} U_{k}) \leq β^{k} w (Q_{0})$ for every $k$ , so the intersection is $w$ -null, hence Lebesgue-null, and $w \leq α_{0}$ a.e. on $Q_{0} ∖ U_{0}$ while $w \leq α_{k + 1}$ a.e. on $U_{k} ∖ U_{k + 1}$ because $U_{k} ∖ U_{k + 1} \subseteq Q_{0} ∖ U_{k + 1}$ . Hence, by countable additivity and monotonicity [F4], $\int_{Q_{0}} w^{1 + γ} d λ = \int_{Q_{0} ∖ U_{0}} w^{γ} w d λ + \sum_{k \geq 0} \int_{U_{k} ∖ U_{k + 1}} w^{γ} w d λ \leq α_{0}^{γ} w (Q_{0} ∖ U_{0}) + \sum_{k \geq 0} α_{k + 1}^{γ} w (U_{k}) \leq α_{0}^{γ} w (Q_{0}) [1 + (2^{n} α^{- 1})^{γ} \sum_{k \geq 0} q^{k}]$ , where we used $w (U_{k}) \leq β^{k} w (Q_{0})$ , $α_{k + 1}^{γ} = (2^{n} α^{- 1})^{(k + 1) γ} α_{0}^{γ}$ and $\sum_{k} q^{k} = 1 / (1 - q)$ from step 2.1.

4.1step 1.1step 2.1step 3.1givenalgebra∎

Dividing the display of step 3.1 by $∣ Q_{0} ∣$ and taking $(1 + γ)$ -th roots gives $(⟨ w^{1 + γ} ⟩_{Q_{0}})^{1 / (1 + γ)} \leq C ⟨ w ⟩_{Q_{0}}$ with $C = (1 + (2^{n} α^{- 1})^{γ} / (1 - q))^{1 / (1 + γ)} < \infty$ , which is the asserted reverse Hölder inequality; the constants $β, γ, C$ depend only on $n, p, K_{p}$ and $α$ .

Depends on

Used by

The Aₚ classes are open in the exponent Corollary
Aᵢnfinity weights satisfy power decay Lemma
The Aᵢnfinity power-decay characterisation Theorem

Dependency tree · two levels

57 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)