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Weighted average comparison and the density-to-mass estimate 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). For every axis-parallel cube $Q$ and every nonnegative measurable $f$ on $Q$ , $⟨ f ⟩_{Q} \leq [w]_{A_{p}}^{1 / p} (\frac{1}{w (Q)} \int_{Q} f^{p} w d λ)^{1 / p} (1 < p < \infty),$ and for $p = 1$ the same inequality holds with $[w]_{A_{p}}^{1 / p}$ replaced by $c_{n} [w]_{A_{1}}$ , where $c_{n}$ is the dimensional constant of The two defining forms of A_1 agree (with the cube-average normalization of the $A_{1}$ characteristic the factor is exactly $[w]_{A_{1}}$ ). In particular every $f \in L^{p} (w)$ is locally integrable. Consequently, for every measurable $E \subseteq Q$ with $1 < p < \infty$ , $(\frac{∣ E ∣}{∣ Q ∣})^{p} \leq [w]_{A_{p}} \frac{w (E)}{w (Q)};$ equivalently, if $S \subseteq Q$ is measurable and $∣ S ∣ \leq α ∣ Q ∣$ for some $0 < α < 1$ , then $w (S) \leq (1 - \frac{(1 - α)^{p}}{[w]_{A_{p}}}) w (Q) .$

Facts & Assumptions

Given: Countable Choice; $1 \leq p < \infty$ , $w \in A_{p}$ , a cube $Q$ , a nonnegative measurable $f$ on $Q$ , and a measurable $E \subseteq Q$ .

[F1]

For $1 < p < \infty$ the $A_{p}$ characteristic is $[w]_{A_{p}} = \sup_{Q} ⟨ w ⟩_{Q} ⟨ w^{- 1 / (p - 1)} ⟩_{Q}^{p - 1} < \infty$ , and $w^{- 1 / (p - 1)} \in L_{l o c}^{1}$ with $0 < w (Q) < \infty$ for every cube; for $p = 1$ the equivalent cube-average/essential-infimum form gives $⟨ w ⟩_{Q} \leq c_{n} [w]_{A_{1}} {ess inf}_{Q} w$ (Muckenhoupt A_p and A_1 weights, The two defining forms of A_1 agree).

[F2]

Hölder's inequality: for finite-valued nonnegative measurable $g, h$ on $Q$ with $\int_{Q} g^{r} < \infty$ and $\int_{Q} h^{r^{'}} < \infty$ , and conjugate finite exponents $r, r^{'} > 1$ , $⟨ g h ⟩_{Q} \leq ⟨ g^{r} ⟩_{Q}^{1 / r} ⟨ h^{r^{'}} ⟩_{Q}^{1 / r^{'}}$ (Holder's inequality for integrals, including the endpoint cases).

[F3]

$f \in L^{p} (w)$ means $\int ∣ f ∣^{p} w d λ < \infty$ , with $∣ f ∣$ measurable when $f$ is; the integral over a set is additive and monotone (Weights, their associated measures, and the spaces L^p(w), The function space $L^{p} (μ)$ for $0 < p < \infty$ ).

Proof

technique · direct

1.1F1F2givenalgebra

Let $1 < p < \infty$ and use the positive finite representative of $w$ . If $\int_{Q} f^{p} w = + \infty$ , the claimed inequality is immediate in the extended order because its right-hand side is $+ \infty$ . Otherwise $f$ is finite a.e.; replace its infinite values on a null set by zero if needed. The functions $f w^{1 / p}$ and $w^{- 1 / p}$ have finite $p$ - and $p^{'}$ -integrals respectively, the latter by [F1]. Hölder's inequality with exponents $p$ and $p^{'}$ applied to $∣ f ∣ w^{1 / p}$ and $w^{- 1 / p} = w^{- 1 / (p - 1) \cdot (p - 1) / p}$ gives $⟨ f ⟩_{Q} \leq ⟨ f^{p} w ⟩_{Q}^{1 / p} ⟨ w^{- 1 / (p - 1)} ⟩_{Q}^{(p - 1) / p}$ . Since $⟨ w ⟩_{Q} ⟨ w^{- 1 / (p - 1)} ⟩_{Q}^{p - 1} \leq [w]_{A_{p}}$ , one has $⟨ w^{- 1 / (p - 1)} ⟩_{Q}^{(p - 1) / p} \leq [w]_{A_{p}}^{1 / p} ⟨ w ⟩_{Q}^{- 1 / p} = [w]_{A_{p}}^{1 / p} (∣ Q ∣ / w (Q))^{1 / p}$ . Substituting and using $⟨ f^{p} w ⟩_{Q} = ∣ Q ∣^{- 1} \int_{Q} f^{p} w$ yields $⟨ f ⟩_{Q} \leq [w]_{A_{p}}^{1 / p} (w (Q)^{- 1} \int_{Q} f^{p} w)^{1 / p}$ .

1.2F1givenalgebra

Let $p = 1$ . Since $0 < w (Q) < \infty$ and [F1] imply ${ess inf}_{Q} w > 0$ , and $w \geq {ess inf}_{Q} w$ a.e. on $Q$ , one has $\int_{Q} f = \int_{Q} (f w) / w \leq ({ess inf}_{Q} w)^{- 1} \int_{Q} f w$ , and [F1] gives $({ess inf}_{Q} w)^{- 1} \leq c_{n} [w]_{A_{1}} ⟨ w ⟩_{Q}^{- 1} = c_{n} [w]_{A_{1}} ∣ Q ∣ / w (Q)$ ; dividing by $∣ Q ∣$ gives $⟨ f ⟩_{Q} \leq c_{n} [w]_{A_{1}} w (Q)^{- 1} \int_{Q} f w$ .

2.1F1step 1.1givenalgebra

Density-to-mass. Apply step 1.1 with $f = 1_{E}$ (for $1 < p < \infty$ ): $⟨ 1_{E} ⟩_{Q} = ∣ E ∣ / ∣ Q ∣$ and $\int_{Q} 1_{E} w = w (E)$ , so $(∣ E ∣ / ∣ Q ∣)^{p} \leq [w]_{A_{p}} w (E) / w (Q)$ , which is the first display. Applying it to $E = Q ∖ S$ when $∣ S ∣ \leq α ∣ Q ∣$ gives $∣ Q ∖ S ∣ \geq (1 - α) ∣ Q ∣$ , hence $(1 - α)^{p} \leq [w]_{A_{p}} w (Q ∖ S) / w (Q)$ and therefore $w (S) = w (Q) - w (Q ∖ S) \leq (1 - (1 - α)^{p} / [w]_{A_{p}}) w (Q)$ , the equivalent form.

3.1F1F3step 1.1step 1.2givenalgebra∎

Local integrability. If $f \in L^{p} (w)$ and $Q$ is a cube, then steps 1.1 and 1.2 applied to $∣ f ∣$ give $⟨ ∣ f ∣ ⟩_{Q} \leq [w]_{A_{p}}^{1 / p} (w (Q)^{- 1} \int_{Q} ∣ f ∣^{p} w)^{1 / p} \leq [w]_{A_{p}}^{1 / p} w (Q)^{- 1 / p} ∥ f ∥_{L^{p} (w)} < \infty$ for $1 < p < \infty$ , and the $p = 1$ analogue holds with $c_{n} [w]_{A_{1}}$ ; hence $f$ is integrable over every cube, i.e. locally integrable. This uses that $0 < w (Q) < \infty$ for every cube from [F1].

Depends on

Used by

Dependency tree · two levels

41 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)