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The A_p classes are open in the exponent

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $1 < p < \infty$ and $w \in A_{p}$ (Muckenhoupt A_p and A_1 weights). Then there is $ε = ε (n, p, [w]_{A_{p}}) > 0$ with $w \in A_{p - ε}$ and $[w]_{A_{p - ε}}$ bounded in terms of $n, p, [w]_{A_{p}}$ . More precisely, if the dual weight $v : = w^{1 / (1 - p)} = w^{- 1 / (p - 1)} \in A_{p^{'}}$ satisfies a reverse Hölder inequality with exponent $q > 1$ and constant $c$ (as supplied by the previous theorem), then one may take $p - ε = 1 + \frac{p - 1}{q} < p, [w]_{A_{p - ε}} \leq c^{p - 1} [w]_{A_{p}} .$ Hence $A_{p} = ⋃_{1 < r < p} A_{r}$ .

Facts & Assumptions

Given: Countable Choice, $1 < p < \infty$ , $w \in A_{p}$ , the dual weight $v = w^{- 1 / (p - 1)}$ , and a reverse Hölder pair $(q, c)$ for $v$ .

[F1]

Duality: $v \in A_{p^{'}}$ with $[v]_{A_{p^{'}}} = [w]_{A_{p}}^{1 / (p - 1)}$ , where $p^{'}$ is the conjugate exponent of $p$ (Duality and nesting of the A_p classes, Conjugate exponents, including the endpoint conventions).

[F2]

The reverse Hölder theorem applied to $v$ with a fixed $0 < α < 1$ supplies $q = 1 + γ > 1$ and $c = C < \infty$ , depending only on $n, p^{'}, [v]_{A_{p^{'}}}$ and $α$ , with $(⟨ v^{q} ⟩_{Q})^{1 / q} \leq c ⟨ v ⟩_{Q}$ for every cube $Q$ (Reverse Holder self-improvement for A_p weights).

[F3]

The $A_{p}$ condition is the finiteness of $⟨ w ⟩_{Q} ⟨ w^{- 1 / (p - 1)} ⟩_{Q}^{p - 1}$ uniformly in $Q$ ; all averages are nonnegative and the exponents combine by the usual power laws (Muckenhoupt A_p and A_1 weights, Holder's inequality for integrals, including the endpoint cases).

Proof

technique · direct

1.1F1F2givenalgebra

By step [F1], $v \in A_{p^{'}}$ with $[v]_{A_{p^{'}}} = [w]_{A_{p}}^{1 / (p - 1)}$ , so [F2] applies to $v$ and yields $q > 1$ , $c < \infty$ with $(⟨ v^{q} ⟩_{Q})^{1 / q} \leq c ⟨ v ⟩_{Q}$ for every cube $Q$ . Define $ε$ by $1 / (p - ε - 1) = q / (p - 1)$ , i.e. $p - ε = 1 + (p - 1) / q$ ; since $q > 1$ one has $p - ε < p$ and $p - ε > 1$ .

2.1F2step 1.1givenalgebra

With this choice, $1 / (1 - (p - ε)) = - 1 / (p - ε - 1) = - q / (p - 1)$ , so $w^{1 / (1 - (p - ε))} = w^{- q / (p - 1)} = (w^{- 1 / (p - 1)})^{q} = v^{q}$ . Therefore, for every cube $Q$ , $(⟨ w^{1 / (1 - (p - ε))} ⟩_{Q})^{p - ε - 1} = (⟨ v^{q} ⟩_{Q})^{(p - 1) / q} \leq (c ⟨ v ⟩_{Q})^{p - 1} = c^{p - 1} ⟨ v ⟩_{Q}^{p - 1}$ by the reverse Hölder inequality and the exponent identity.

3.1F3step 2.1givenalgebra

Multiplying the estimate of step 2.1 by $⟨ w ⟩_{Q}$ and inserting the $A_{p}$ bound $⟨ w ⟩_{Q} ⟨ v ⟩_{Q}^{p - 1} \leq [w]_{A_{p}}$ valid for every cube $Q$ gives $⟨ w ⟩_{Q} (⟨ w^{1 / (1 - (p - ε))} ⟩_{Q})^{p - ε - 1} \leq c^{p - 1} [w]_{A_{p}}$ for every cube; this is exactly the $A_{p - ε}$ condition with characteristic at most $c^{p - 1} [w]_{A_{p}}$ .

4.1F1step 1.1step 3.1givenalgebra∎

The monotonicity part of [F1] gives $A_{r} \subseteq A_{p}$ for every $1 < r < p$ ; step 1.1 and step 3.1 exhibit, for each $w \in A_{p}$ , an exponent $p - ε < p$ with $w \in A_{p - ε}$ , so every element of $A_{p}$ lies in some $A_{r}$ with $r < p$ . Hence $A_{p} = ⋃_{1 < r < p} A_{r}$ , and $ε$ and the bound on $[w]_{A_{p - ε}}$ depend only on $n, p, [w]_{A_{p}}$ (through $[v]_{A_{p^{'}}}$ and the fixed $α$ ).

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)