Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck pass
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Duality and nesting of the A_p classes

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (ACω)).

Let w be a weight on Rn (Weights, their associated measures, and the spaces L^p(w)). Then:

  1. For 1<p<∞, w∈Ap if and only if its reciprocal power w−1/(p−1) lies in Ap′, where 1/p+1/p′=1, and then [w−1/(p−1)]Ap′=[w]Ap1/(p−1) (Conjugate exponents, including the endpoint conventions).
  2. The classes are nested: Ap⊆Aq for 1<p<q<∞ with [w]Aq≤[w]Ap, and A1⊆Aq for every 1<q<∞ with [w]Aq≤Cn[w]A1 for a dimensional constant Cn.

Facts & Assumptions

Given: Countable Choice; A weight w and the characteristic constants [w]Ap=sup⁡Q⟨w⟩Q⟨w−1/(p−1)⟩Qp−1 of Muckenhoupt A_p and A_1 weights.

[F1]

The A1 condition is equivalent to the cube-average/essential-infimum form: there is a dimensional constant cn≥1 with ⟨w⟩Q≤cn[w]A1ess inf⁡Qw for every cube Q, and conversely the cube-average form with constant C′ gives [w]A1≤cnC′ (The two defining forms of A_1 agree).

[F2]

Hölder's inequality with conjugate exponents r,r′ and the generalized form for finitely many factors hold for nonnegative measurable functions; in particular ⟨gθ⟩Q≤⟨g⟩Qθ for 0<θ≤1 and nonnegative measurable g∈Lloc1 (Holder's inequality for integrals, including the endpoint cases, Generalized Holder inequality puts products into Lr).

Proof

technique · direct
1.1F2givenalgebra

Set v:=w−1/(p−1). Since p′−1=1/(p−1), one has v−1/(p′−1)=v−(p−1)=w, and therefore ⟨v⟩Q⟨v−1/(p′−1)⟩Qp′−1=⟨w−1/(p−1)⟩Q⟨w⟩Q1/(p−1)=(⟨w⟩Q⟨w−1/(p−1)⟩Qp−1)1/(p−1) for every cube Q. Taking suprema, the two suprema are finite simultaneously and [v]Ap′=[w]Ap1/(p−1); if w∈Ap, finiteness of its product and positivity of ⟨w⟩Q imply local integrability of v, so v is a weight. Conversely, if v∈Ap′, the displayed identity gives the Ap bound for the already given weight w.

1.2F2givenalgebra

For 1<p<q<∞ put θ:=(p−1)/(q−1)∈(0,1) and σ:=w−1/(p−1), so that w−1/(q−1)=σθ. By the power-mean inequality of [F2], ⟨σθ⟩Q≤⟨σ⟩Qθ for every cube Q; raising to the (q−1)-th power gives ⟨w−1/(q−1)⟩Qq−1≤⟨w−1/(p−1)⟩Qp−1.

1.3F1givenalgebra

For w∈A1 and every cube Q, put m=ess inf⁡Qw. By [F1], m≥⟨w⟩Q/(cn[w]A1)>0. For every q>1, w−1/(q−1)≤m−1/(q−1) a.e. on Q, so ⟨w⟩Q⟨w−1/(q−1)⟩Qq−1≤⟨w⟩Q/m≤cn[w]A1. Taking suprema proves A1⊆Aq with the stated bound for the entire range q>1.

2.1step 1.2givenalgebra

Combining step 1.2 with the definition, for every cube Q one has ⟨w⟩Q⟨w−1/(q−1)⟩Qq−1≤⟨w⟩Q⟨w−1/(p−1)⟩Qp−1≤[w]Ap; taking the supremum in Q gives [w]Aq≤[w]Ap and in particular Ap⊆Aq for 1<p<q<∞.

3.1step 1.1step 2.1step 1.3∎

Steps 2.1 and 1.3 are the two nesting assertions, and step 1.1 is the duality assertion together with the exact identity of characteristics; this proves the lemma with Cn=cn.

Depends on

Used by

Dependency tree · two levels

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Sources