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The A_p range of a power weight

Example

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

Fix 1<p<∞ and α∈R, and let w(x):=∣x∣α on Rn (the value at the origin being assigned arbitrarily, say w(0):=0). Then w is a weight exactly when α>−n, and for such α one has w∈Ap if and only if −n<α<n(p−1). In that open range the Ap characteristic is finite and bounded in terms of n,p,α; when α≥n(p−1) the reciprocal-power average diverges on cubes containing the origin; when α≤−n, the function is not a weight. Both thresholds are local integrability conditions at the origin. Moreover the associated measure is doubling for every α>−n.

Facts & Assumptions

Given: Countable Choice; n≥1, 1<p<∞, α∈R, and w=∣x∣α.

[F1]

w is a weight iff it is Lebesgue measurable, locally integrable and positive and finite a.e.; the Ap characteristic is sup⁡Q⟨w⟩Q⟨w−1/(p−1)⟩Qp−1 over cubes (Weights, their associated measures, and the spaces L^p(w), Muckenhoupt A_p and A_1 weights).

[F2]

Polar coordinates: ∫Rng(x) dx=∫0∞∫Sn−1g(rθ)rn−1dθ dr, so ∫B(0,R)∣x∣a dx=∣Sn−1∣Rn+a/(n+a) for a>−n and +∞ for a≤−n (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma, Continuity and derivatives of positive-base real powers, Comparison tests for improper integrals, Dominated convergence, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation).

[F3]

For a>−n, ∣x0∣≤3R implies B(x0,R)⊆B(0,4R), so [F2] bounds its ∣x∣a integral above by Cn,aRn+a. If a<0, then ∣x∣a≥(4R)a a.e. on this ball, giving a positive lower bound of that order. If a≥0, remove B(0,R/2); the remainder has measure at least (1−2−n)∣B(x0,R)∣ and ∣x∣a≥(R/2)a there, again giving a positive lower bound. For ∣x0∣≥3R, (2/3)∣x0∣≤∣x∣≤(4/3)∣x0∣ on the ball, so its integral is comparable to ∣x0∣aRn. Volume scaling is For a nonzero real c, dilation by c multiplies Lebesgue outer measure by ∣c∣n, and reflection in the origin preserves it, and ball/cube characteristic equivalence is Ball and cube maximal functions are pointwise comparable.

Verification

technique · direct
1.1F1F2F3givenalgebra

If α≤−n, then ∫B(0,1)∣x∣αdx=+∞ by [F2], so w is not locally integrable and hence not a weight. If α>−n, then w is locally integrable (apply [F2] on each ball, using [F3] to compare ∫B(x0,R)∣x∣α with the radial integral), it is positive and finite off the origin, and it is assigned the value 0 at the single point 0 of measure zero; hence w is a weight.

1.2F3givenalgebra

Let α>−n and let B=B(x0,R) be a ball with ∣x0∣≥3R. On B one has (2/3)∣x0∣≤∣x∣≤(4/3)∣x0∣, so both ∫B∣x∣αdx and ∫B∣x∣−α/(p−1)dx are comparable to ∣x0∣αRn and ∣x0∣−α/(p−1)Rn respectively, and the defining product is bounded by a constant depending only on n,α,p.

2.1F3step 1.2givenalgebra

Let α>−n and let B=B(x0,R) with ∣x0∣<3R. By [F3] the two integrals ∫B∣x∣αdx and ∫B∣x∣−α/(p−1)dx are comparable to Rn+α and Rn−α/(p−1) provided both exponents exceed −n; the normalized product ⟨w⟩B⟨w−1/(p−1)⟩Bp−1 is then comparable to RαR−α/(p−1)⋅(p−1)=1, uniformly in B. If −α/(p−1)≤−n, that is α≥n(p−1), the second exponent does not exceed −n and the corresponding integral over B(0,R) diverges, so the product is +∞ on the ball B(0,R).

3.1F2F3step 1.1step 1.2step 2.1givenalgebra∎

Steps 1.1–2.1 give the stated weight and Ap ranges for ball averages; the ball/cube comparison [F3] gives the same result for the defined cube characteristic. For doubling, if ∣x0∣<3R, the integral on B(x0,2R) is bounded above by the radial integral on B(0,5R), of order Rn+α, while [F3] bounds the integral on B(x0,R) below by a positive multiple of that order. If ∣x0∣≥3R, both balls have ∣x∣ comparable to ∣x0∣ (on the larger ball, ∣x0∣/3≤∣x∣≤5∣x0∣/3); their integrals are therefore comparable up to a fixed constant. Hence w(B(x0,2R))≤Cn,αw(B(x0,R)) for every α>−n.

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