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A power weight fails at both A_p endpoints

Statement refuted

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Fix $1 < p < \infty$ . The claims that the power weight $∣ x ∣^{α}$ is in $A_{p}$ also at the endpoints of its admissible interval are false:

at the upper endpoint $α = n (p - 1)$ the function $∣ x ∣^{α}$ is a weight but not an $A_{p}$ weight;
at the lower endpoint $α = - n$ the function $∣ x ∣^{α}$ is not even locally integrable, so it is not a weight.

Hence the admissible interval $- n < α < n (p - 1)$ is open at both ends and cannot be enlarged.

Facts & Assumptions

Given: Countable Choice; $n \geq 1$ , $1 < p < \infty$ , the power function $∣ x ∣^{α}$ and a radius $R > 0$ .

[F1]

For $α > - n$ the function $∣ x ∣^{α}$ is a weight, and the $A_{p}$ characteristic is the supremum of $⟨ w ⟩_{Q} ⟨ w^{- 1 / (p - 1)} ⟩_{Q}^{p - 1}$ over cubes (The A_p range of a power weight, Muckenhoupt A_p and A_1 weights, Weights, their associated measures, and the spaces L^p(w)).

[F2]

Polar coordinates give $\int_{B (0, R)} ∣ x ∣^{a} d x = ∣ S^{n - 1} ∣ R^{n + a} / (n + a)$ for $a > - n$ and $+ \infty$ for $a \leq - n$ , and $\int_{0}^{R} r^{- 1} d r = + \infty$ for every $R > 0$ (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma, Continuity and derivatives of positive-base real powers, The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t, Comparison tests for improper integrals).

Counterexample

technique · direct

1.1F1F2givenalgebra

At the upper endpoint: for $α = n (p - 1)$ one has $- α / (p - 1) = - n$ , so polar coordinates give $\int_{B (0, R)} ∣ x ∣^{- α / (p - 1)} d x = \int_{B (0, R)} ∣ x ∣^{- n} d x = ∣ S^{n - 1} ∣ \int_{0}^{R} r^{- 1} d r = + \infty$ for every $R > 0$ by the logarithmic divergence of $\int_{0}^{1} r^{- 1} d r$ . The second factor of the defining product $⟨ ∣ x ∣^{α} ⟩_{Q} ⟨ ∣ x ∣^{- α / (p - 1)} ⟩_{Q}^{p - 1}$ is therefore $+ \infty$ on the cube $Q = Q (0, R)$ containing $B (0, R)$ , so the defining supremum is $+ \infty$ and $∣ x ∣^{n (p - 1)}$ is not in $A_{p}$ , while it is locally integrable and hence a weight.

1.2F2givenalgebra

At the lower endpoint: for $α = - n$ polar coordinates give $\int_{B (0, R)} ∣ x ∣^{- n} d x = ∣ S^{n - 1} ∣ \int_{0}^{R} r^{- 1} d r = + \infty$ , so $∣ x ∣^{- n} \notin L_{l o c}^{1} (R^{n})$ and no $A_{p}$ membership is defined; this is the same logarithmic divergence of the radial integral $\int_{0}^{R} r^{- 1} d r$ .

2.1step 1.1step 1.2given∎

Steps 1.1 and 1.2 show the failure at both endpoints, so the range $- n < α < n (p - 1)$ determined in The A_p range of a power weight is exactly the open admissible interval and cannot be enlarged.

Depends on

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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)