Pipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

✓ 25 results · all verified · 5 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 20 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Muckenhoupt Weights and Weighted Estimates

1 · Prerequisites

2 · Summary

This page develops the Muckenhoupt weighted theory that upgrades the unweighted maximal and singular-integral estimates of the Fourier analysis track to measures $w d λ$ . It begins with the conventions: a weight is a locally integrable almost-everywhere positive function, its associated measure is a locally finite regular Borel measure, $L^{p} (w)$ is the corresponding weighted space, and cube averages and the centred and uncentred maximal functions are compared across cubes and balls up to dimensional constants.

The core of the page is the $A_{p}$ hierarchy. The characteristic $[w]_{A_{p}} = \sup_{Q} ⟨ w ⟩_{Q} ⟨ w^{- 1 / (p - 1)} ⟩_{Q}^{p - 1}$ is shown to be equivalent in its cube and ball forms; the endpoint class $A_{1}$ is controlled by $M^{*} w \leq C w$ and by the cube-average/essential-infimum form. The classes are nested and closed under the dual weight $w^{- 1 / (p - 1)} \in A_{p^{'}}$ with the exact characteristic identity; they are doubling; and their weighted averages dominate unweighted averages in the density-to-mass form. Maximal dyadic subcubes at a height give the distribution decay of the weight averages that drives the reverse Hölder self-improvement and the openness of the $A_{p}$ range in the exponent.

The second half defines $A_{\infty} = ⋃_{1 \leq p < \infty} A_{p}$ and proves the equivalence of $A_{\infty}$ membership, power decay $w (E) / w (Q) \leq C (∣ E ∣ / ∣ Q ∣)^{δ}$ and a reverse Hölder inequality. It also proves the weighted maximal theory for doubling weights: the weak $(1, 1)$ bound under $A_{1}$ , Marcinkiewicz interpolation, and the characterisation of $A_{p}$ by the boundedness of the Hardy–Littlewood maximal operator on $L^{p} (w)$ .

The final block assembles the good- $λ$ machinery — a Whitney-type dyadic covering of proper open sets, annulus far-field estimates, finiteness of the kernel tails of weighted $L^{p}$ functions, and the unweighted and weighted local good- $λ$ inequalities — and proves the weighted Calderón–Zygmund theorem for maximal truncations: strong $L^{p} (w)$ bounds for $w \in A_{p}$ , the weak $(1, 1)$ endpoint for $w \in A_{1}$ , and the conditional clause upgrading almost-everywhere convergence on a dense subspace to all of $L^{p} (w)$ . The Hilbert and Riesz transforms are the concrete corollary, and a closing remark explains why the endpoints are not obtained by setting $p = 1$ .

Every statement that needs a choice principle names it: the differentiation lemma for doubling weights, the reverse Hölder lemma, the power-decay converse chain, the $A_{\infty}$ characterisation and the weighted Calderón–Zygmund/Hilbert–Riesz results assume Dependent Choice, which supplies the density of $C_{c}^{\infty}$ in $L^{p} (w)$ and the almost-everywhere differentiation inputs; the remaining items use at most Countable Choice.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Weights, their associated measures, and the spaces L^p(w)

Definition

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), the principle used by the completeness statement below.

Weights. A weight on $R^{n}$ is a Lebesgue measurable function $w : R^{n} \to [0, \infty]$ such that $\int_{B (x, r)} w d λ < \infty$ for every Euclidean ball with $r > 0$ , and $0 < w (x) < \infty$ for Lebesgue-almost every $x$ (Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ , Measure-null sets and almost-everywhere statements relative to a measure). Thus a weight may vanish or be infinite only on a Lebesgue null set.

Set $\tilde{w} (x) = w (x)$ where $0 < w (x) < \infty$ and $\tilde{w} (x) = 1$ otherwise. This positive finite-valued representative belongs to $L_{l o c}^{1} (R^{n})$ in the precise sense of A locally integrable function on $R^{n}$ . In all finite-valued function interfaces and reciprocal powers below, use this representative and denote it again by $w$ . Its associated measure, cube integrals and almost-everywhere assertions agree with those of the original extended-valued weight.

The $w$ -measure. For a Lebesgue measurable set $E$ the $w$ -measure of $E$ is $w (E) : = \int_{E} w d λ .$ The map $E \mapsto w (E)$ is countably additive by the indefinite-integral theorem (The indefinite integral of a nonnegative measurable function is a measure), so it is a measure on the Lebesgue $σ$ -algebra; its restriction to the Borel sets is a Borel measure. It is finite on bounded sets: a bounded $E$ lies in some ball $B$ , and monotonicity of the integral together with $w \in L_{l o c}^{1}$ gives $w (E) \leq \int_{B} w d λ < \infty$ . It is therefore a locally finite (equivalently, Radon) Borel measure: $R^{n}$ is locally compact and $σ$ -compact ( $R^{n}$ is locally compact and $σ$ -compact), so every open subset is $σ$ -compact (for a proper open $U$ , use, for integers $m \geq 1$ , $K_{m} = {x : ∣ x ∣ \leq m, dist (x, U^{c}) \geq 1 / m}$ : distance to the closed complement is continuous by the triangle inequality, so $K_{m}$ is closed and bounded, hence compact by Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, and $U = ⋃_{m} K_{m}$ ; for $U = R^{n}$ use closed balls), and Sigma-compact open sets make locally finite Borel measures regular makes the measure regular. Since $R^{n}$ is $σ$ -compact, $w$ is $σ$ -finite as well.

Because $0 < w < \infty$ almost everywhere, a Lebesgue measurable set is $w$ -null exactly when it is Lebesgue null: the integral of $w$ over a Lebesgue null set vanishes, and conversely $\int_{N} w d λ = 0$ forces $w = 0$ a.e. on $N$ , hence $λ (N) = 0$ since $w > 0$ a.e.

The spaces $L^{p} (w)$ . Fix $1 \leq p < \infty$ . For a measurable $f$ the weighted $L^{p}$ functional is $∥ f ∥_{L^{p} (w)} : = (\int_{R^{n}} ∣ f ∣^{p} w d λ)^{1 / p} \in [0, \infty],$ the value $+ \infty$ being assigned when the integral diverges. Since $\int ∣ f ∣^{p} w d λ = \int ∣ f ∣^{p} d (w λ)$ and $w λ$ is a measure, this is the $L^{p}$ functional of the measure space $(R^{n}, w λ)$ in the sense of The function space $L^{p} (μ)$ for $0 < p < \infty$ , and $L^{p} (w)$ is the corresponding quotient of the class of measurable $f$ with $∥ f ∥_{L^{p} (w)} < \infty$ by the functions that vanish $w$ -almost everywhere (The space $L^{p} (μ)$ as the quotient by null functions); complex-valued $f$ are admitted under the componentwise conventions of Complex Lp classes and Euclidean test-function conventions. Equivalently, and this is how the norm is used below, $∥ f ∥_{L^{p} (w)} = ∥ ∣ f ∣ ∥_{L^{p} (w)}$ and the quotient norm is well defined (The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$ ); the real space is complete by Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ applied to $w λ$ . For complex functions, real and imaginary component projections contract the norm and recombination has norm at most the sum of the component norms (Complex Holder, Minkowski, and the quotient norm). A complex Cauchy sequence therefore has two real Cauchy components with limits, whose recombination is its complex norm limit; hence the complex space is Banach as well. Because the $w$ -null sets are exactly the Lebesgue null sets, membership of $L^{p} (w)$ and equality in $L^{p} (w)$ are determined by the same negligible sets as in unweighted measure theory.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Axis-parallel cubes, their averages, and cube maximal functions

Definition

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), the principle already assumed by the published ball-based maximal functions.

For $x \in R^{n}$ and $r > 0$ let $Q (x, r) : = \prod_{i < n} (x_{i} - r, x_{i} + r)$ be the open axis-parallel cube with centre $x$ and side length $2 r$ , so that $Q (x, r) = {y \in R^{n} : ∣ y_{i} - x_{i} ∣ < r for every i < n}$ . Its half-open counterpart is a box in the sense of Half-open boxes in $R^{n}$ and their volume with the same real endpoints $a_{i} = x_{i} - r$ and $b_{i} = x_{i} + r$ , and the box-measure theorem (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included) gives $∣ Q (x, r) ∣ = (2 r)^{n}$ ; the face convention is immaterial because all boxes with the same endpoints have the same Lebesgue measure. For $λ > 0$ , $λ Q$ denotes the cube concentric with $Q$ whose side length is $λ$ times that of $Q$ ; thus $λ Q (x, r) = Q (x, λ r)$ and $∣ λ Q (x, r) ∣ = (2 λ r)^{n} = λ^{n} ∣ Q (x, r) ∣$ by the same box-measure computation.

For $f \in L_{l o c}^{1} (R^{n})$ (A locally integrable function on $R^{n}$ ) and an axis-parallel cube $Q$ the average of $f$ over $Q$ is the finite number $⟨ f ⟩_{Q} : = ∣ Q ∣^{- 1} \int_{Q} f d λ$ , and $⟨ ∣ f ∣ ⟩_{Q}$ is the average of the nonnegative function $∣ f ∣$ . For any nonnegative measurable $g$ , the same notation $⟨ g ⟩_{Q} : = ∣ Q ∣^{- 1} \int_{Q} g d λ$ denotes an extended average in $[0, \infty]$ , with value $+ \infty$ when the integral diverges. The centred and uncentred cube maximal functions of $f$ are $M_{c} f (x) : = \sup_{r > 0} ⟨ ∣ f ∣ ⟩_{Q (x, r)}, M_{c}^{*} f (x) : = \sup_{Q ∋ x} ⟨ ∣ f ∣ ⟩_{Q},$ the second supremum taken over all axis-parallel cubes $Q = Q (y, r)$ with $y \in R^{n}$ , $r > 0$ that contain $x$ . Both functions take values in $[0, \infty]$ .

These are the cube analogues of the published centred and uncentred ball-based Hardy-Littlewood maximal functions $M f$ and $M^{*} f$ (The centered and uncentered Hardy-Littlewood maximal functions), whose averages are formed with the ball average operator (The average of a locally integrable function over a Euclidean ball). Every Euclidean ball between the inscribed and circumscribed cube of a fixed cube has comparable volume, so the ball and cube maximal functions are pointwise comparable by a constant depending only on $n$ ; that comparison is proved on this page. The two functions $M_{c} f$ and $M_{c}^{*} f$ are themselves pointwise comparable by $2^{n}$ , since the centred cube $Q (x, r)$ is among the cubes containing $x$ , and every cube $Q (y, r) ∋ x$ lies in $Q (x, 2 r)$ , whose volume is $2^{n} ∣ Q (y, r) ∣$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Ball and cube maximal functions are pointwise comparable

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). There is a constant $C_{n} < \infty$ , depending only on the dimension, such that every Euclidean ball $B$ contains an axis-parallel cube $Q_{1}$ and is contained in an axis-parallel cube $Q_{2}$ with $∣ Q_{2} ∣ \leq C_{n} ∣ Q_{1} ∣$ ; explicitly, for $B = B (x, r)$ one may take $Q_{1} = Q (x, r / \sqrt{n})$ and $Q_{2} = Q (x, r)$ and $C_{n} = n^{n / 2}$ . Consequently, for every $f \in L_{l o c}^{1} (R^{n})$ and every $x \in R^{n}$ , $M^{*} f (x) \leq C_{n} M_{c}^{*} f (x), M_{c}^{*} f (x) \leq C_{n} M^{*} f (x),$ and the centred versions satisfy the same two-sided comparison with a constant that is a dimensional power of $C_{n}$ (with $C_{n} = \max {2^{n} / v_{n}, v_{n} n^{n / 2} / 2^{n}}$ and $v_{n} = λ (B (0, 1))$ for a comparison constant; the displayed $C_{n} = n^{n / 2}$ is the one for the pure cube sandwich, up to dimensional factors). Hence the cube-based and ball-based Muckenhoupt characteristics $\sup_{Q} ⟨ w ⟩_{Q} ⟨ w^{- 1 / (p - 1)} ⟩_{Q}^{p - 1}$ and $\sup_{B} ⟨ w ⟩_{B} ⟨ w^{- 1 / (p - 1)} ⟩_{B}^{p - 1}$ differ by at most a dimensional power of such a $C_{n}$ , and boundedness of the ball maximal operator on $L^{p} (w)$ is equivalent to boundedness of any cube maximal function.

Facts & Assumptions

Given: Countable Choice, a locally integrable $f$ , and points and radii as below.

[F1]

$Q (x, r) = \prod_{i} (x_{i} - r, x_{i} + r)$ has side $2 r$ , Lebesgue measure $(2 r)^{n}$ , its dilates satisfy $λ Q (x, r) = Q (x, λ r)$ , and $⟨ f ⟩_{Q} = ∣ Q ∣^{- 1} \int_{Q} f d λ$ (Axis-parallel cubes, their averages, and cube maximal functions).

[F2]

$M f (x) = \sup_{r > 0} λ (B (x, r))^{- 1} \int_{B (x, r)} ∣ f ∣$ and $M^{*} f (x) = \sup_{B ∋ x} λ (B)^{- 1} \int_{B} ∣ f ∣$ , the second supremum over all Euclidean balls containing $x$ (The centered and uncentered Hardy-Littlewood maximal functions).

[F3]

Every ball $B (x, r)$ is Lebesgue measurable with $0 < λ (B (x, r)) < \infty$ , and $λ (B (x, r)) = v_{n} r^{n}$ with $v_{n} = λ (B (0, 1)) \in (0, \infty)$ (Euclidean balls have positive finite Lebesgue measure, For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation), since $B (x, r) = x + r B (0, 1)$ .

[F4]

For a nonnegative measurable $g$ the set function $A \mapsto \int_{A} g d λ$ is a measure (The indefinite integral of a nonnegative measurable function is a measure), so $A \subseteq B$ measurable implies $\int_{A} g d λ \leq \int_{B} g d λ$ (Measures are monotone).

Proof

technique · direct

1.1F1F3algebra

Sandwiches: for $B = B (x, r)$ one has $Q (x, r / \sqrt{n}) \subseteq B (x, r)$ and $B (x, r) \subseteq Q (x, r)$ , because $∣ y_{i} - x_{i} ∣ < r / \sqrt{n}$ for all $i$ gives $∣ y - x ∣ < \sqrt{n} (r / \sqrt{n}) = r$ , and $∣ y - x ∣ < r$ gives $∣ y_{i} - x_{i} ∣ < r$ for every $i$ . By [F1] and [F3], $∣ Q (x, r) ∣ = (2 r)^{n}$ and $∣ Q (x, r / \sqrt{n}) ∣ = (2 r / \sqrt{n})^{n}$ , so the ratio is $n^{n / 2}$ ; also $∣ Q (x, r) ∣ / ∣ B (x, r) ∣ = 2^{n} / v_{n}$ and $∣ B (x, r) ∣ / ∣ Q (x, r) ∣ = v_{n} / 2^{n}$ .

2.1F1F2F4step 1.1algebra

First comparison: fix $f \in L_{l o c}^{1}$ and $x$ . For every $r > 0$ , step 1.1 and [F4] give $\int_{B (x, r)} ∣ f ∣ \leq \int_{Q (x, r)} ∣ f ∣$ , hence $λ (B (x, r))^{- 1} \int_{B (x, r)} ∣ f ∣ \leq (∣ Q (x, r) ∣ / λ (B (x, r))) ⟨ ∣ f ∣ ⟩_{Q (x, r)} = (2^{n} / v_{n}) ⟨ ∣ f ∣ ⟩_{Q (x, r)} \leq (2^{n} / v_{n}) M_{c} f (x)$ ; taking the supremum over $r$ gives $M f (x) \leq (2^{n} / v_{n}) M_{c} f (x)$ , and the same computation with a ball containing $x$ in place of the centred ball gives $M^{*} f (x) \leq (2^{n} / v_{n}) M_{c}^{*} f (x)$ .

2.2F1F2F3F4step 1.1algebra

Second comparison: let $Q = Q (y, r)$ contain $x$ . Then $Q \subseteq B (y, \sqrt{n} r)$ , and $x \in B (y, \sqrt{n} r)$ , so by [F2] $⟨ ∣ f ∣ ⟩_{B (y, \sqrt{n} r)} \leq M^{*} f (x)$ ; moreover $Q \subseteq B (y, \sqrt{n} r)$ and [F4] give $\int_{Q} ∣ f ∣ \leq \int_{B (y, \sqrt{n} r)} ∣ f ∣$ , hence $⟨ ∣ f ∣ ⟩_{Q} \leq (λ (B (y, \sqrt{n} r)) / ∣ Q ∣) ⟨ ∣ f ∣ ⟩_{B (y, \sqrt{n} r)} \leq (v_{n} n^{n / 2} / 2^{n}) M^{*} f (x)$ by [F3]. Taking the supremum over all $Q ∋ x$ gives $M_{c}^{*} f (x) \leq (v_{n} n^{n / 2} / 2^{n}) M^{*} f (x)$ ; for the centred version the same computation with $Q = Q (x, r)$ and $B (x, \sqrt{n} r)$ gives $M_{c} f (x) \leq (v_{n} n^{n / 2} / 2^{n}) M f (x)$ .

3.1F1F2F4step 1.1step 2.1step 2.2algebra∎

Both assertions of the Statement now follow with $C_{n} = \max {2^{n} / v_{n}, v_{n} n^{n / 2} / 2^{n}}$ , which is finite because $v_{n} \in (0, \infty)$ : the two-sided pointwise bounds are steps 2.1 and 2.2 (the displayed $C_{n} = n^{n / 2}$ sandwich constant appears here only through dimensional factors). For the characteristic comparison, let $w$ be a weight, $1 < p < \infty$ and $σ = w^{- 1 / (p - 1)}$ ; for a cube $Q$ let $B_{Q}$ be a ball with $Q \subseteq B_{Q}$ and $∣ B_{Q} ∣ \leq C_{n} ∣ Q ∣$ , and note that [F4] applied to $g = w$ and $g = σ$ gives $⟨ w ⟩_{Q} ⟨ σ ⟩_{Q}^{p - 1} \leq (∣ B_{Q} ∣ / ∣ Q ∣)^{p} ⟨ w ⟩_{B_{Q}} ⟨ σ ⟩_{B_{Q}}^{p - 1} \leq C_{n}^{p} \sup_{B} ⟨ w ⟩_{B} ⟨ σ ⟩_{B}^{p - 1}$ . Conversely every ball $B$ is contained in a cube $Q_{B}$ with $∣ Q_{B} ∣ \leq C_{n} ∣ B ∣$ , so monotonicity of both integrals bounds the ball product by $C_{n}^{p}$ times the cube product; the two suprema therefore differ by at most the dimensional factor $C_{n}^{p}$ . Finally, because all four maximal functions are pointwise comparable in pairs by constants independent of $f$ , if one of them has finite $L^{p} (w)$ norm for every $f$ then so do the others, with norms bounded by the corresponding dimensional multiples; this is the asserted equivalence of boundedness.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Muckenhoupt A_p and A_1 weights

Definition

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

Fix $n \geq 1$ . All averages and maximal functions below are the cube-based ones of Axis-parallel cubes, their averages, and cube maximal functions. Nonnegative measurable averages are extended integrals in $[0, \infty]$ ; in particular the reciprocal-weight average may be infinite before $A_{p}$ membership is established. Use the positive finite representative of Weights, their associated measures, and the spaces L^p(w) for reciprocal powers.

The class $A_{p}$ , $1 < p < \infty$ . A weight $w$ (Weights, their associated measures, and the spaces L^p(w)) belongs to $A_{p}$ when $[w]_{A_{p}} : = \sup_{Q} ⟨ w ⟩_{Q} ⟨ w^{- 1 / (p - 1)} ⟩_{Q}^{p - 1} < \infty,$ the supremum over all axis-parallel cubes $Q$ , and $[w]_{A_{p}}$ is the $A_{p}$ characteristic of $w$ . The supremum over all Euclidean balls instead differs from the cube supremum by at most a dimensional factor: the two sandwiches of Ball and cube maximal functions are pointwise comparable compare $w$ - and $w^{- 1 / (p - 1)}$ -averages over nested balls and cubes, so $\sup_{B} ⟨ w ⟩_{B} ⟨ w^{- 1 / (p - 1)} ⟩_{B}^{p - 1}$ is finite exactly when $[w]_{A_{p}}$ is, and the two numbers are bounded by dimensional powers of one another. If $[w]_{A_{p}} < \infty$ , then $⟨ w^{- 1 / (p - 1)} ⟩_{Q} < \infty$ and $⟨ w ⟩_{Q} > 0$ for every cube $Q$ (both are finite or nonzero because $w > 0$ a.e. and $[w]_{A_{p}}$ is finite), so $w^{- 1 / (p - 1)} \in L_{l o c}^{1}$ and every cube satisfies $0 < w (Q) < \infty$ and $0 < \int_{Q} w^{- 1 / (p - 1)} d λ < \infty$ .

The class $A_{1}$ . A weight $w$ belongs to $A_{1}$ when $M^{*} w \leq C w$ almost everywhere for some constant $C < \infty$ , where $M^{*}$ is the uncentred ball maximal function of The centered and uncentered Hardy-Littlewood maximal functions; by the ball-cube comparison this is equivalent to the same pointwise bound for the uncentred cube maximal function $M_{c}^{*}$ , and $M_{c} w \leq M_{c}^{*} w \leq 2^{n} M_{c} w$ pointwise (an uncentred cube $Q ∋ x$ of side $ℓ$ is contained in the centred cube $Q (x, ℓ)$ of $2^{n}$ -fold volume), so the centred form is equivalent as well. For an $A_{1}$ weight the infimum $a$ of admissible constants is attained: choose admissible $C_{j} < a + 1 / (j + 1)$ , discard their countably many exceptional null sets, and pass to the limit in $M^{*} w \leq C_{j} w$ . Thus the $A_{1}$ characteristic $[w]_{A_{1}}$ is defined as the least such $C$ , and $M^{*} w \leq [w]_{A_{1}} w$ holds almost everywhere. The class $A_{1}$ is not obtained by substituting $p = 1$ into the $A_{p}$ formula: the reciprocal power $w^{- 1 / (p - 1)}$ has no finite-exponent analogue, and the equivalent cube-average/essential-infimum condition $⟨ w ⟩_{Q} \leq c_{n} [w]_{A_{1}} {ess inf}_{Q} w$ , with a dimensional factor $c_{n}$ is a separate lemma on this page. Here ${ess inf}_{Q} w : = \sup {a \geq 0 : w \geq a a.e. on Q}$ , the lower-bound analogue of The essential supremum of a measurable function with respect to a measure. It is finite because $w$ is integrable on $Q$ , and $w$ is at least this supremum a.e.: take a sequence of admissible bounds tending to it and discard their countable union of null exceptional sets.

Invariance. Translations, positive isotropic dilations and positive scalar multiples preserve both classes with the same characteristic: for $τ_{z} w (x) : = w (x - z)$ one has $⟨ τ_{z} w ⟩_{Q} = ⟨ w ⟩_{Q - z}$ and, with $σ = w^{- 1 / (p - 1)}$ , $⟨ τ_{z} σ ⟩_{Q}^{p - 1} = ⟨ σ ⟩_{Q - z}^{p - 1}$ by the $C^{1}$ change-of-variables theorem applied to the translation $x \mapsto x - z$ (A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions), which carries cubes to cubes of the same side length; for $δ_{λ} w (x) : = w (λ x)$ with $λ > 0$ the determinant $λ^{n}$ introduced by the substitution $x \mapsto λ x$ cancels between the two averages, since $δ_{λ} σ = (δ_{λ} w)^{- 1 / (p - 1)}$ ; and $⟨ c w ⟩_{Q} ⟨ (c w)^{- 1 / (p - 1)} ⟩_{Q}^{p - 1} = ⟨ w ⟩_{Q} ⟨ w^{- 1 / (p - 1)} ⟩_{Q}^{p - 1}$ for $c > 0$ . The same computations apply verbatim to the condition $M^{*} w \leq C w$ a.e.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

The two defining forms of A_1 agree

Statement

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

Let $w$ be a weight on $R^{n}$ (Weights, their associated measures, and the spaces L^p(w)). Then the following are equivalent:

$M w \leq C w$ almost everywhere for some constant $C < \infty$ , where $M$ is the centred ball maximal function (The centered and uncentered Hardy-Littlewood maximal functions);
$\sup_{Q} ⟨ w ⟩_{Q} ({ess inf}_{Q} w)^{- 1} = : C^{'} < \infty$ , the supremum over axis-parallel cubes, where the essential infimum is defined in Muckenhoupt A_p and A_1 weights.

Moreover the least constants satisfy $C^{'} \leq C_{n} C$ and $C \leq C_{n} C^{'}$ for a dimensional constant $C_{n}$ , and the same equivalence holds with the uncentred maximal function $M^{*}$ in place of $M$ . In particular the cube-average/ essential-infimum condition may be used as an equivalent definition of $A_{1}$ with a characteristic changed only by a dimensional factor; a positive average divided by a zero essential infimum is interpreted as $+ \infty$ ; the class $A_{1}$ itself is the one of Muckenhoupt A_p and A_1 weights.

Facts & Assumptions

Given: Countable Choice; A weight $w$ , the centred and uncentred ball maximal functions $M$ and $M^{*}$ , and the cube averages $⟨ w ⟩_{Q}$ .

[F1]

$w > 0$ and $w < \infty$ Lebesgue-a.e., and for every cube $Q$ one has $0 < w (Q) < \infty$ (Weights, their associated measures, and the spaces L^p(w)).

[F2]

For every ball $B = B (x, r)$ there is an axis-parallel cube $Q \supseteq B$ with $∣ Q ∣ \leq C_{n} ∣ B ∣$ , and for every cube $Q$ there is a ball $B \supseteq Q$ with $∣ B ∣ \leq C_{n} ∣ Q ∣$ ; consequently, if $E \subseteq F$ and $∣ F ∣ \leq C_{n} ∣ E ∣$ , then $⟨ w ⟩_{E} \leq C_{n} ⟨ w ⟩_{F}$ by nonnegativity (Ball and cube maximal functions are pointwise comparable).

[F3]

For a nonnegative function $g$ the set where $g$ does not satisfy a pointwise inequality of the form $g \leq c$ a.e. is contained in a null set, and countable unions of null sets are null (Measure-null sets and almost-everywhere statements relative to a measure).

[F4]

$Q^{n}$ is countable and dense in $R^{n}$ , so cubes with rational centre and rational side length approximate any given cube from outside with volume comparable by a fixed factor ( $Q^{n}$ is a countable dense subset of $R^{n}$ , and rational open boxes form a countable basis).

Proof

technique · direct

1.1F1F2givenalgebra

Assume (1), with constant $C$ . For each cube $Q$ of side $ℓ$ and each $x \in Q$ , the centred ball $B (x, \sqrt{n} ℓ)$ contains $Q$ and has volume at most a dimensional multiple of $∣ Q ∣$ . Thus $⟨ w ⟩_{Q} \leq C_{n} M w (x) \leq C_{n} C w (x)$ for almost every $x \in Q$ . Taking the essential infimum and then the supremum in $Q$ gives $C^{'} \leq C_{n} C$ . If the hypothesis instead uses $M^{*}$ , the same estimate holds since $M \leq M^{*}$ .

1.2F1F2F3F4givenchoose

Assume (2). For each rational-centred, rational-sided cube $Q$ , the set $N (Q) = {x \in Q : ⟨ w ⟩_{Q} > C^{'} w (x)}$ is null. Their union $N$ is null by [F3, F4]. For $x \notin N$ and any ball $B ∋ x$ , choose a rational cube $Q \supseteq B$ with $∣ Q ∣ \leq C_{n} ∣ B ∣$ . Then $⟨ w ⟩_{B} \leq C_{n} ⟨ w ⟩_{Q} \leq C_{n} C^{'} w (x)$ . Taking the supremum over these balls gives $M^{*} w (x) \leq C_{n} C^{'} w (x)$ , hence also $M w (x) \leq C_{n} C^{'} w (x)$ .

2.1step 1.1step 1.2algebra∎

Steps 1.1 and 1.2 prove the equivalence together with the comparable bounds $C^{'} \leq C_{n} C$ and $C \leq C_{n} C^{'}$ for one and the same dimensional constant $C_{n}$ (renaming constants if necessary), and each direction was proved both for $M$ and for $M^{*}$ , so the centred and uncentred forms of condition (1) are equivalent to (2). Therefore the cube-average/essential-infimum condition defines the same class as $M w \leq C w$ a.e., with characteristic changed only by dimensional factors.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Duality and nesting of the A_p classes

Statement

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

Let $w$ be a weight on $R^{n}$ (Weights, their associated measures, and the spaces L^p(w)). Then:

For $1 < p < \infty$ , $w \in A_{p}$ if and only if its reciprocal power $w^{- 1 / (p - 1)}$ lies in $A_{p^{'}}$ , where $1 / p + 1 / p^{'} = 1$ , and then $[w^{- 1 / (p - 1)}]_{A_{p^{'}}} = [w]_{A_{p}}^{1 / (p - 1)}$ (Conjugate exponents, including the endpoint conventions).
The classes are nested: $A_{p} \subseteq A_{q}$ for $1 < p < q < \infty$ with $[w]_{A_{q}} \leq [w]_{A_{p}}$ , and $A_{1} \subseteq A_{q}$ for every $1 < q < \infty$ with $[w]_{A_{q}} \leq C_{n} [w]_{A_{1}}$ for a dimensional constant $C_{n}$ .

Facts & Assumptions

Given: Countable Choice; A weight $w$ and the characteristic constants $[w]_{A_{p}} = \sup_{Q} ⟨ w ⟩_{Q} ⟨ w^{- 1 / (p - 1)} ⟩_{Q}^{p - 1}$ of Muckenhoupt A_p and A_1 weights.

[F1]

The $A_{1}$ condition is equivalent to the cube-average/essential-infimum form: there is a dimensional constant $c_{n} \geq 1$ with $⟨ w ⟩_{Q} \leq c_{n} [w]_{A_{1}} {ess inf}_{Q} w$ for every cube $Q$ , and conversely the cube-average form with constant $C^{'}$ gives $[w]_{A_{1}} \leq c_{n} C^{'}$ (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 $⟨ g^{θ} ⟩_{Q} \leq ⟨ g ⟩_{Q}^{θ}$ for $0 < θ \leq 1$ and nonnegative measurable $g \in L_{l o c}^{1}$ (Holder's inequality for integrals, including the endpoint cases, Generalized Holder inequality puts products into $L^{r}$ ).

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 $⟨ v ⟩_{Q} ⟨ v^{- 1 / (p^{'} - 1)} ⟩_{Q}^{p^{'} - 1} = ⟨ w^{- 1 / (p - 1)} ⟩_{Q} ⟨ w ⟩_{Q}^{1 / (p - 1)} = (⟨ w ⟩_{Q} ⟨ w^{- 1 / (p - 1)} ⟩_{Q}^{p - 1})^{1 / (p - 1)}$ for every cube $Q$ . Taking suprema, the two suprema are finite simultaneously and $[v]_{A_{p^{'}}} = [w]_{A_{p}}^{1 / (p - 1)}$ ; if $w \in A_{p}$ , finiteness of its product and positivity of $⟨ w ⟩_{Q}$ imply local integrability of $v$ , so $v$ is a weight. Conversely, if $v \in A_{p^{'}}$ , the displayed identity gives the $A_{p}$ bound for the already given weight $w$ .

1.2F2givenalgebra

For $1 < p < q < \infty$ put $θ : = (p - 1) / (q - 1) \in (0, 1)$ and $σ : = w^{- 1 / (p - 1)}$ , so that $w^{- 1 / (q - 1)} = σ^{θ}$ . By the power-mean inequality of [F2], $⟨ σ^{θ} ⟩_{Q} \leq ⟨ σ ⟩_{Q}^{θ}$ for every cube $Q$ ; raising to the $(q - 1)$ -th power gives $⟨ w^{- 1 / (q - 1)} ⟩_{Q}^{q - 1} \leq ⟨ w^{- 1 / (p - 1)} ⟩_{Q}^{p - 1}$ .

1.3F1givenalgebra

For $w \in A_{1}$ and every cube $Q$ , put $m = {ess inf}_{Q} w$ . By [F1], $m \geq ⟨ w ⟩_{Q} / (c_{n} [w]_{A_{1}}) > 0$ . For every $q > 1$ , $w^{- 1 / (q - 1)} \leq m^{- 1 / (q - 1)}$ a.e. on $Q$ , so $⟨ w ⟩_{Q} ⟨ w^{- 1 / (q - 1)} ⟩_{Q}^{q - 1} \leq ⟨ w ⟩_{Q} / m \leq c_{n} [w]_{A_{1}}$ . Taking suprema proves $A_{1} \subseteq A_{q}$ 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 $⟨ w ⟩_{Q} ⟨ w^{- 1 / (q - 1)} ⟩_{Q}^{q - 1} \leq ⟨ w ⟩_{Q} ⟨ w^{- 1 / (p - 1)} ⟩_{Q}^{p - 1} \leq [w]_{A_{p}}$ ; taking the supremum in $Q$ gives $[w]_{A_{q}} \leq [w]_{A_{p}}$ and in particular $A_{p} \subseteq A_{q}$ for $1 < p < q < \infty$ .

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 $C_{n} = c_{n}$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

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].

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

A_p weights are doubling

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). Then the measure $w d λ$ is doubling: for every axis-parallel cube $Q$ and every $λ > 1$ , $w (λ Q) \leq λ^{n p} K_{p} w (Q),$ and for every ball $B (x, r)$ , $w (B (x, 2 r)) \leq (4 \sqrt{n})^{n p} K_{p} w (B (x, r)) .$ All constants depend only on $n$ , $p$ and $K_{p}$ , never on the particular cube, ball, or point.

Here $K_{p} = [w]_{A_{p}}$ for $p > 1$ , and $K_{1} = \sup_{Q} ⟨ w ⟩_{Q} / ({ess inf}_{Q} w)$ ; by The two defining forms of A_1 agree, $1 \leq K_{1} \leq c_{n} [w]_{A_{1}}$ . Thus the constants remain controlled by the stated $A_{p}$ data, including the ball-normalized endpoint characteristic.

Facts & Assumptions

Given: Countable Choice; $1 \leq p < \infty$ , $w \in A_{p}$ , a cube $Q$ , $λ > 1$ , and a ball $B (x, r)$ .

[F1]

For $w \in A_{p}$ , every cube satisfies $0 < w (Q) < \infty$ and $λ Q$ is an axis-parallel cube with $∣ λ Q ∣ = λ^{n} ∣ Q ∣$ (Axis-parallel cubes, their averages, and cube maximal functions, Muckenhoupt A_p and A_1 weights).

[F2]

For $w \in A_{p}$ , every cube $R$ and every nonnegative measurable $f$ obey $⟨ f ⟩_{R} \leq K_{p}^{1 / p} (w (R)^{- 1} \int_{R} f^{p} w)^{1 / p}$ for $p > 1$ ; for $p = 1$ , $w \geq ⟨ w ⟩_{R} / K_{1}$ a.e. gives the same bound with $K_{1}$ . (Weighted average comparison and the density-to-mass estimate for A_p weights, The two defining forms of A_1 agree).

[F3]

For every ball $B (x, r)$ one has $Q (x, r / \sqrt{n}) \subseteq B (x, r) \subseteq B (x, 2 r) \subseteq Q (x, 4 r)$ , and $Q (x, 4 r) = (4 \sqrt{n}) Q (x, r / \sqrt{n})$ (Ball and cube maximal functions are pointwise comparable, Axis-parallel cubes, their averages, and cube maximal functions), while $w d λ$ is a measure, so $A \subseteq B$ implies $w (A) \leq w (B)$ (Weights, their associated measures, and the spaces L^p(w)).

Proof

technique · direct

1.1F1F2givenalgebra

Apply [F2] on the cube $λ Q$ to $f : = 1_{Q}$ : then $⟨ f ⟩_{λ Q} = ∣ Q ∣ / ∣ λ Q ∣ = λ^{- n}$ by [F1] and $\int_{λ Q} f^{p} w d λ = w (Q)$ , so $λ^{- n} \leq K_{p}^{1 / p} (w (Q) / w (λ Q))^{1 / p}$ . Raising to the $p$ -th power and rearranging using $0 < w (Q), w (λ Q) < \infty$ gives $w (λ Q) \leq λ^{n p} K_{p} w (Q)$ , which is the cube assertion.

2.1F1F3step 1.1givenalgebra

For the ball assertion apply step 1.1 to the cube $Q_{0} : = Q (x, r / \sqrt{n})$ and the dilation factor $λ_{0} : = 4 \sqrt{n} > 1$ : by [F3], $B (x, 2 r) \subseteq Q (x, 4 r) = λ_{0} Q_{0}$ , and $Q_{0} \subseteq B (x, r)$ , so $w (B (x, 2 r)) \leq w (λ_{0} Q_{0}) \leq λ_{0}^{n p} K_{p} w (Q_{0}) \leq (4 \sqrt{n})^{n p} K_{p} w (B (x, r))$ .

3.1step 1.1step 2.1∎

Step 1.1 is the cube form with constant $λ^{n p} K_{p}$ and step 2.1 the ball form with constant $(4 \sqrt{n})^{n p} K_{p}$ ; both depend only on $n, p, K_{p}$ and the stated dilation, and no property of $Q$ , $x$ or $r$ entered otherwise. This is the asserted doubling property.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Maximal dyadic subcubes of a cube at a height

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ , let $Q_{0}$ be an axis-parallel cube with side length $ℓ > 0$ and centre $x_{0}$ , and let $Φ (y) : = x_{0} + ℓ (y - \frac{1}{2} (1, \dots, 1))$ be the unique translation-dilation carrying $(0, 1]^{n}$ onto the half-open box $R_{0}$ with the same centre and side length as $Q_{0}$ ; $R_{0}$ and $Q_{0}$ differ by a Lebesgue-null set. The dyadic subcubes of $Q_{0}$ are the images $Φ (D)$ of the dyadic cubes $D \subseteq (0, 1]^{n}$ of the all-generations grid (Dyadic cubes of all generations in R^n).

When forming averages over half-open descendants, extend functions on $Q_{0}$ by zero on $R_{0} ∖ Q_{0}$ ; all such boundary changes are null. Here a dyadic subcube of $Q_{0}$ means a descendant of $R_{0}$ , rather than literal inclusion in the open cube.

Let $f \geq 0$ satisfy $f \in L^{1} (Q_{0})$ , and let $α \geq ⟨ f ⟩_{Q_{0}}$ with $α > 0$ . Then the dyadic subcubes $R \subseteq Q_{0}$ with $⟨ f ⟩_{R} > α$ that are maximal under inclusion are pairwise disjoint and at most countable, their union equals ${M_{d, Q_{0}} f > α} = {x \in R_{0} : \sup_{R ∋ x} ⟨ f ⟩_{R} > α}$ up to a Lebesgue-null set, where the supremum is over the dyadic subcubes of $Q_{0}$ containing $x$ and $R_{0}$ is the half-open box above; since $Q_{0} ∖ R_{0}$ is Lebesgue null, this is the same as the corresponding set with $Q_{0}$ in place of $R_{0}$ , up to a null set. Each such maximal $R$ satisfies $⟨ f ⟩_{R} \leq 2^{n} α$ ; and $\sum_{R} ∣ R ∣ \leq α^{- 1} \int_{Q_{0}} f d λ$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , the cube $Q_{0}$ and its dyadic subcubes via $Φ$ , a nonnegative $f \in L^{1} (Q_{0})$ , and $α \geq ⟨ f ⟩_{Q_{0}}$ and $α > 0$ .

[F1]

For dyadic cubes $D, D^{'}$ of generations $k \leq k^{'}$ with $D \cap D^{'} \neq \emptyset$ one has $D^{'} \subseteq D$ ; every dyadic cube of generation $k$ has a unique parent of generation $k - 1$ containing it, of volume $2^{n}$ times its own; and two dyadic cubes are disjoint or one contains the other (All-generation dyadic cubes: partition, volume and nesting, Dyadic cubes of all generations in R^n).

[F2]

A generation- $k$ dyadic cube $D \subseteq (0, 1]^{n}$ has centre $c_{D}$ and side $2^{- k}$ . Its image $Φ (D)$ is the half-open box with centre $Φ (c_{D})$ and side $2^{- k} ℓ$ , hence $∣ Φ (D) ∣ = ℓ^{n} ∣ D ∣$ by A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included. The map $Φ$ is bijective, so it preserves inclusion and disjointness; the images of the generation- $k$ descendants partition $R_{0}$ for each $k \geq 0$ .

[F3]

The set of all dyadic cubes is at most countable: the parameters $(k, m)$ inject into $Q^{n + 1}$ , which is at most countable ( $Q^{n}$ is a countable dense subset of $R^{n}$ , and rational open boxes form a countable basis, Every subset of an at most countable set is at most countable).

[F4]

For nonnegative measurable functions and measurable sets, integrals are monotone in the set (The indefinite integral of a nonnegative measurable function is a measure, Measures are monotone) and finite on $Q_{0}$ by the hypothesis $f \in L^{1} (Q_{0})$ .

Proof

technique · direct

1.1F1F2F4givenalgebra

Call a dyadic subcube $R$ of $Q_{0}$ bad when $⟨ f ⟩_{R} > α$ . Every bad $R$ satisfies $α ∣ R ∣ < \int_{R} f d λ \leq \int_{Q_{0}} f d λ < \infty$ by [F4], so $∣ R ∣ < α^{- 1} \int_{Q_{0}} f$ ; since the ancestors of a subcube have volumes $ℓ^{n} 2^{- k n}$ growing by the factor $2^{n}$ from generation to generation, only finitely many ancestors of a given bad cube can be bad. The top cube $Q_{0}$ is not bad because $α \geq ⟨ f ⟩_{Q_{0}}$ , and its subcube family is identified with the all-generations dyadic cubes inside $(0, 1]^{n}$ , so the ancestors of any bad subcube that lie inside $(0, 1]^{n}$ form a finite chain starting at the bad cube; a maximal bad subcube containing it is therefore obtained by taking the last bad member of that chain.

2.1F1F2F3step 1.1

The maximal bad subcubes are pairwise disjoint: if two of them meet, [F1] and injectivity of $Φ$ make one contain the other, and maximality forces equality. They are at most countable because they are images under the fixed map $Φ$ of a subfamily of the at most countable dyadic grid [F3].

3.1F1step 1.1step 2.1

The union of the maximal bad subcubes is exactly ${M_{d, Q_{0}} f > α}$ : if $x$ lies in a maximal bad $R$ , then $M_{d, Q_{0}} f (x) \geq ⟨ f ⟩_{R} > α$ ; conversely, if $M_{d, Q_{0}} f (x) > α$ then some dyadic subcube $R ∋ x$ is bad, and step 1.1 contains it in a maximal bad subcube $R^{'}$ , which also contains $x$ since $R \cap R^{'} \neq \emptyset$ and dyadic subcubes are nested [F1]. This is an equality of sets, hence a fortiori equality up to a null set.

4.1F1F2F4step 3.1algebra∎

For a maximal bad subcube $R$ : if $R \neq Q_{0}$ its parent $P = Φ (D^{'})$ exists with $Φ^{- 1} (R) ⊊ D^{'} \subseteq (0, 1]^{n}$ and $∣ P ∣ = 2^{n} ∣ R ∣$ by [F1] and [F2]; maximality makes $P$ good, so $\int_{R} f \leq \int_{P} f \leq α ∣ P ∣ = 2^{n} α ∣ R ∣$ by [F4], and hence $⟨ f ⟩_{R} \leq 2^{n} α$ . If $R = Q_{0}$ then $⟨ f ⟩_{Q_{0}} \leq α \leq 2^{n} α$ as well. Finally, pairwise disjointness gives $\sum_{R} ∣ R ∣ = ∣ ⋃_{R} R ∣ \leq ∣ {M_{d, Q_{0}} f > α} ∣$ , and on each bad $R$ one has $α ∣ R ∣ < \int_{R} f$ , so summing over the at most countable disjoint family and using $f \geq 0$ yields $\sum_{R} ∣ R ∣ \leq α^{- 1} \sum_{R} \int_{R} f \leq α^{- 1} \int_{Q_{0}} f$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Distribution decay from maximal cubes for A_p weights

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $1 \leq p < \infty$ , let $w \in A_{p}$ (Muckenhoupt A_p and A_1 weights), let $Q_{0}$ be an axis-parallel cube with $α_{0} : = ⟨ w ⟩_{Q_{0}} > 0$ , and fix $0 < α < 1$ . Put $α_{k} : = (2^{n} α^{- 1})^{k} α_{0}$ and let $U_{k}$ be the union of the maximal dyadic subcubes $R \subseteq Q_{0}$ (in the sense of Maximal dyadic subcubes of a cube at a height) with $⟨ w ⟩_{R} > α_{k}$ , with $U_{k} = \emptyset$ when there is none. Then $U_{k + 1} \subseteq U_{k}$ , $∣ U_{k + 1} ∣ \leq α ∣ U_{k} ∣$ and $∣ U_{k} ∣ \leq α^{k} ∣ Q_{0} ∣$ , and with $β : = 1 - \frac{(1 - α)^{p}}{K_{p}}$ one has $w (U_{k + 1}) \leq β w (U_{k})$ and $w (U_{k}) \leq β^{k} w (Q_{0})$ ; moreover $w \leq α_{k}$ almost everywhere on $Q_{0} ∖ U_{k}$ .

Facts & Assumptions

Given: Countable Choice, $1 \leq p < \infty$ , $w \in A_{p}$ , the cube $Q_{0}$ with $α_{0} = ⟨ w ⟩_{Q_{0}} > 0$ , $0 < α < 1$ , the levels $α_{k}$ and the sets $U_{k}$ .

[F1]

For every $k \geq 0$ one has $α_{k} \geq α_{0}$ because $2^{n} α^{- 1} > 1$ , so the subcube lemma applies at the height $α_{k}$ : the maximal dyadic subcubes $R \subseteq Q_{0}$ with $⟨ w ⟩_{R} > α_{k}$ are pairwise disjoint, at most countable, their union equals ${M_{d, Q_{0}} w > α_{k}}$ up to a null set, and each of them satisfies $⟨ w ⟩_{R} \leq 2^{n} α_{k}$ (Maximal dyadic subcubes of a cube at a height).

[F2]

Since $w > 0$ a.e. and $w (Q_{0}) > 0$ , all the sets $U_{k}$ and their intersections with the maximal cubes are measurable, and $w$ is finite on $Q_{0}$ (Muckenhoupt A_p and A_1 weights).

[F3]

Density-to-mass: for $1 \leq p < \infty$ , $w \in A_{p}$ , a cube $Q$ and a measurable $S \subseteq Q$ with $∣ S ∣ \leq α ∣ Q ∣$ one has $w (S) \leq β w (Q)$ with $β = 1 - (1 - α)^{p} / K_{p}$ for $p > 1$ ; for $p = 1$ use $w \geq ⟨ w ⟩_{Q} / K_{1}$ a.e. to get $w (Q ∖ S) \geq (1 - α) w (Q) / K_{1}$ . (Weighted average comparison and the density-to-mass estimate for A_p weights, The two defining forms of A_1 agree).

[F4]

For a locally integrable function and almost every point, the averages over a family of sets shrinking nicely to the point converge to the value of the function (Lebesgue differentiation theorem on $R^{n}$ , Differentiation holds along families shrinking nicely); the dyadic subcubes of $Q_{0}$ containing a point of $Q_{0}$ contain cubes of arbitrarily small side length, and such a cube $R$ satisfies $R \subseteq B (x, \sqrt{n} ℓ (R))$ , so the family shrinks nicely.

Proof

technique · direct

1.1F1F2givenalgebra

Since $α_{k + 1} > α_{k}$ , every dyadic subcube counted in step [F1] at level $α_{k + 1}$ has average exceeding $α_{k}$ as well, so it is contained in a maximal subcube at level $α_{k}$ ; hence $U_{k + 1} \subseteq U_{k}$ . For a maximal level- $α_{k}$ cube $R$ , the set $S : = R \cap U_{k + 1}$ is contained in $R$ and measurable, and $α_{k + 1} ∣ S ∣ \leq \int_{S} w d λ \leq \int_{R} w d λ \leq 2^{n} α_{k} ∣ R ∣$ by [F1]; since $α_{k + 1} = 2^{n} α^{- 1} α_{k}$ , this gives $∣ S ∣ \leq α ∣ R ∣$ . Summing over the pairwise disjoint maximal level- $α_{k}$ cubes gives $∣ U_{k + 1} ∣ \leq α ∣ U_{k} ∣$ , and iterating with $U_{0} \subseteq Q_{0}$ gives $∣ U_{k} ∣ \leq α^{k} ∣ Q_{0} ∣$ .

2.1F2F3step 1.1givenalgebra

With the same set $S = R \cap U_{k + 1}$ of step 1.1 we have $∣ S ∣ \leq α ∣ R ∣$ , so the density-to-mass estimate [F3] applies: $w (S) \leq β w (R)$ with $β = 1 - (1 - α)^{p} / K_{p}$ . Summing over the pairwise disjoint maximal level- $α_{k}$ cubes, whose union is $U_{k}$ , gives $w (U_{k + 1}) \leq β w (U_{k})$ ; iterating with $U_{0} \subseteq Q_{0}$ gives $w (U_{k}) \leq β^{k} w (Q_{0})$ .

2.2F1F4step 1.1givenalgebra

Almost everywhere bound. Fix $x \in Q_{0} ∖ U_{k}$ outside the null sets of [F4] and outside the null set on which the union of the maximal subcubes differs from ${M_{d, Q_{0}} w > α_{k}}$ . Then no dyadic subcube $R \subseteq Q_{0}$ containing $x$ has $⟨ w ⟩_{R} > α_{k}$ , for such an $R$ would lie in a maximal subcube counted at level $α_{k}$ and hence in $U_{k}$ . The dyadic subcubes of $Q_{0}$ containing $x$ shrink nicely to $x$ by [F4], so their averages of $w$ converge to $w (x)$ ; since every such average is at most $α_{k}$ , the limit satisfies $w (x) \leq α_{k}$ . Thus $w \leq α_{k}$ almost everywhere on $Q_{0} ∖ U_{k}$ .

3.1step 1.1step 2.1step 2.2∎

Steps 1.1, 2.1 and 2.2 are exactly the assertions of the Statement, namely nesting and the two chains of measure bounds together with the almost everywhere bound.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Reverse Holder self-improvement 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). Fix $0 < α < 1$ and put $β : = 1 - \frac{(1 - α)^{p}}{K_{p}}, γ : = \frac{1}{2} \frac{- \log β}{\log (2^{n} α^{- 1})} > 0$ (the logarithm is the one of The logarithm to a positive base other than one, and the powers are those of Real powers for positive bases, with the zero-base positive-exponent convention). Then $(2^{n} α^{- 1})^{γ} β = β^{1 / 2} < 1,$ and for every axis-parallel cube $Q$ , $(⟨ w^{1 + γ} ⟩_{Q})^{1 / (1 + γ)} \leq C ⟨ w ⟩_{Q}, C : = (1 + \frac{(2^{n} α^{- 1})^{γ}}{1 - (2^{n} α^{- 1})^{γ} β})^{1 / (1 + γ)} < \infty .$ Thus $w$ satisfies a reverse Hölder inequality with exponent $1 + γ$ , and $γ$ and $C$ depend only on $n$ , $p$ , $K_{p}$ and the fixed $α$ .

Facts & Assumptions

Given: Countable Choice, $1 \leq p < \infty$ , $w \in A_{p}$ , $0 < α < 1$ , and the constants $β, γ$ of the Statement.

[F1]

$0 < w (Q) < \infty$ for every cube. For $p > 1$ , the defining product is at most $K_{p} = [w]_{A_{p}}$ and Hölder applies; for $p = 1$ , $K_{1}$ is the cube-average/essential-infimum characteristic above (Muckenhoupt A_p and A_1 weights, The two defining forms of A_1 agree, Holder's inequality for integrals, including the endpoint cases).

[F2]

For a cube $Q_{0}$ with $α_{0} = ⟨ w ⟩_{Q_{0}} > 0$ and $α_{k} = (2^{n} α^{- 1})^{k} α_{0}$ , the sets $U_{k}$ of the maximal dyadic subcubes with $⟨ w ⟩_{R} > α_{k}$ satisfy $U_{k + 1} \subseteq U_{k}$ , $w (U_{k}) \leq β^{k} w (Q_{0})$ and $w \leq α_{k}$ almost everywhere on $Q_{0} ∖ U_{k}$ (Distribution decay from maximal cubes for A_p weights).

[F3]

$a^{x} = \exp (x \log a)$ for $a > 0$ and real $x$ (Real powers for positive bases, with the zero-base positive-exponent convention), $\log_{b} x = \log x / \log b$ for $b > 0$ , $b \neq 1$ , $x > 0$ (The logarithm to a positive base other than one), and the elementary exponential identities $\exp (u + v) = \exp (u) \exp (v)$ , $\exp (- u) = 1 / \exp (u)$ follow from The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ and the inverse identities of The natural logarithm as the inverse of the exponential function.

[F4]

Integrals of nonnegative measurable functions over a measurable set are countably additive on disjoint measurable pieces, and monotone under inclusion (Countable additivity and continuity of finitely additive set functions).

Proof

technique · direct

1.1F1givenalgebra

For $p > 1$ , Hölder applied to $w^{1 / p} w^{- 1 / p}$ on a cube gives $1 \leq ⟨ w ⟩_{Q}^{1 / p} ⟨ w^{- 1 / (p - 1)} ⟩_{Q}^{(p - 1) / p} \leq K_{p}^{1 / p}$ . For $p = 1$ , $⟨ w ⟩_{Q} \geq {ess inf}_{Q} w$ gives $K_{1} \geq 1$ . Therefore $0 < (1 - α)^{p} < K_{p}$ and $β \in (0, 1)$ , so $γ > 0$ is well defined.

2.1F3step 1.1givenalgebra

By [F3], $\log β < 0$ and $(2^{n} α^{- 1})^{γ} = \exp (γ \log (2^{n} α^{- 1})) = \exp (\frac{1}{2} (- \log β)) = β^{- 1 / 2}$ ; therefore $(2^{n} α^{- 1})^{γ} β = β^{- 1 / 2} β = β^{1 / 2} < 1$ , and the geometric ratio $q : = (2^{n} α^{- 1})^{γ} β$ satisfies $0 < q < 1$ with $1 / (1 - q) < \infty$ .

3.1F2F4step 2.1givenalgebra

Fix a cube $Q_{0} = Q$ and apply [F2] with $α_{0} = ⟨ w ⟩_{Q_{0}}$ : the sets $Q_{0} ∖ U_{0}$ and $U_{k} ∖ U_{k + 1}$ , $k \geq 0$ , are disjoint and measurable and cover $Q_{0}$ up to a Lebesgue-null set: indeed $w (⋂_{k} U_{k}) \leq β^{k} w (Q_{0})$ for every $k$ , so the intersection is $w$ -null, hence Lebesgue-null, and $w \leq α_{0}$ a.e. on $Q_{0} ∖ U_{0}$ while $w \leq α_{k + 1}$ a.e. on $U_{k} ∖ U_{k + 1}$ because $U_{k} ∖ U_{k + 1} \subseteq Q_{0} ∖ U_{k + 1}$ . Hence, by countable additivity and monotonicity [F4], $\int_{Q_{0}} w^{1 + γ} d λ = \int_{Q_{0} ∖ U_{0}} w^{γ} w d λ + \sum_{k \geq 0} \int_{U_{k} ∖ U_{k + 1}} w^{γ} w d λ \leq α_{0}^{γ} w (Q_{0} ∖ U_{0}) + \sum_{k \geq 0} α_{k + 1}^{γ} w (U_{k}) \leq α_{0}^{γ} w (Q_{0}) [1 + (2^{n} α^{- 1})^{γ} \sum_{k \geq 0} q^{k}]$ , where we used $w (U_{k}) \leq β^{k} w (Q_{0})$ , $α_{k + 1}^{γ} = (2^{n} α^{- 1})^{(k + 1) γ} α_{0}^{γ}$ and $\sum_{k} q^{k} = 1 / (1 - q)$ from step 2.1.

4.1step 1.1step 2.1step 3.1givenalgebra∎

Dividing the display of step 3.1 by $∣ Q_{0} ∣$ and taking $(1 + γ)$ -th roots gives $(⟨ w^{1 + γ} ⟩_{Q_{0}})^{1 / (1 + γ)} \leq C ⟨ w ⟩_{Q_{0}}$ with $C = (1 + (2^{n} α^{- 1})^{γ} / (1 - q))^{1 / (1 + γ)} < \infty$ , which is the asserted reverse Hölder inequality; the constants $β, γ, C$ depend only on $n, p, K_{p}$ and $α$ .

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

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 $α$ ).

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

The Muckenhoupt A_infinity class

Definition

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

The Muckenhoupt $A_{\infty}$ class is the union of the finite-exponent classes: $A_{\infty} : = ⋃_{1 \leq p < \infty} A_{p},$ where $A_{p}$ and $A_{1}$ are the classes of Muckenhoupt A_p and A_1 weights. Thus a weight $w$ belongs to $A_{\infty}$ exactly when $w \in A_{p}$ for some finite $p$ , and by the nesting property of Duality and nesting of the A_p classes the witnessing exponent may be replaced by any larger one: if $w \in A_{p}$ then $w \in A_{q}$ for every $q > p$ . The class is not defined by a single limit formula in $p$ ; its equivalence with the power-decay condition $w (E) / w (Q) \leq C (∣ E ∣ / ∣ Q ∣)^{δ}$ and with the reverse Hölder property is a theorem proved separately on this page.

Every $A_{\infty}$ weight is doubling: choose a witnessing exponent $p > 1$ using the nesting lemma. Then then A_p weights are doubling gives $w (λ Q) \leq λ^{n p} [w]_{A_{p}} w (Q)$ for every cube and $λ > 1$ , and the corresponding ball bound with a constant depending only on the indicated data. No new choice principle is used in the definition itself.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Weighted weak (1,1) bound for the maximal function under A_1

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $w \in A_{1}$ (Muckenhoupt A_p and A_1 weights) and $f \in L^{1} (w)$ (so $f \in L_{l o c}^{1} (λ)$ and the maximal functions of The centered and uncentered Hardy-Littlewood maximal functions are defined). Then for every $λ > 0$ , $w ({M f > λ}) \leq 5^{n} [w]_{A_{1}} λ^{- 1} \int_{R^{n}} ∣ f ∣ w d λ,$ and the uncentred maximal function satisfies the same estimate with constant $2^{n} 5^{n} [w]_{A_{1}}$ .

Facts & Assumptions

Given: Countable Choice, $w \in A_{1}$ , $f \in L^{1} (w)$ and $λ > 0$ .

[F1]

$M^{*} w \leq [w]_{A_{1}} w$ almost everywhere, and $w d λ$ is a locally finite regular Borel (Radon) measure; the cube-average/essential-infimum form of the $A_{1}$ condition is equivalent to this pointwise form (Muckenhoupt A_p and A_1 weights, The two defining forms of A_1 agree, Sigma-compact open sets make locally finite Borel measures regular, Radon measure on an LCH space).

[F2]

$f \in L^{1} (w)$ implies $f \in L_{l o c}^{1} (λ)$ by the weighted average comparison at $p = 1$ , so every ball average of $∣ f ∣$ is finite (Weighted average comparison and the density-to-mass estimate for A_p weights), and for every $r > 0$ the function $x \mapsto A_{r} ∣ f ∣ (x)$ is continuous in the centre and radius (Ball averages vary continuously with the centre and radius).

[F3]

Fivefold Vitali covering: for a finite family of balls $B_{1}, \dots, B_{m}$ there is a pairwise disjoint subfamily $B_{i_{1}}, \dots, B_{i_{ℓ}}$ with $⋃_{j} B_{j} \subseteq ⋃_{k} 5 B_{i_{k}}$ (Vitali covering lemma for Euclidean balls with fivefold dilates).

[F4]

Inner regularity: for the Radon measure $w d λ$ and a Borel set $E$ , $w (E) = \sup {w (K) : K \subseteq E compact}$ (Radon measure on an LCH space, Sigma-compact open sets make locally finite Borel measures regular).

Proof

technique · direct

1.1F2F4given

The level set $E_{λ} : = {M f > λ}$ is open: $M f$ is the supremum of the functions $x \mapsto A_{r} ∣ f ∣ (x)$ , $r > 0$ , each continuous by [F2], so $M f$ is lower semicontinuous. By [F4] its $w$ -measure is the supremum of $w (K)$ over compact $K \subseteq E_{λ}$ .

1.2F2F3givenchoose

Let $K \subseteq E_{λ}$ be compact. Each $x \in K$ has a ball $B_{x} ∋ x$ with $A_{r_{x}} ∣ f ∣ (x) > λ$ , i.e. $\int_{B_{x}} ∣ f ∣ d λ > λ ∣ B_{x} ∣$ ; finitely many of the open balls $B_{x}$ cover $K$ , and [F3] supplies pairwise disjoint balls $B_{x_{1}}, \dots, B_{x_{ℓ}}$ from that finite cover with $K \subseteq ⋃_{j} 5 B_{x_{j}}$ and $\int_{B_{x_{j}}} ∣ f ∣ > λ ∣ B_{x_{j}} ∣$ for every $j$ .

2.1F1step 1.2givenalgebra

For each ball $B_{j}$ of step 1.2 and each $y \in B_{j}$ one has $M^{*} w (y) \geq ∣ 5 B_{j} ∣^{- 1} \int_{5 B_{j}} w = w (5 B_{j}) / (5^{n} ∣ B_{j} ∣)$ because $B_{j} \subseteq 5 B_{j}$ ; integrating over $y \in B_{j}$ against $∣ f ∣$ gives $\int_{B_{j}} ∣ f (y) ∣ M^{*} w (y) d y \geq (w (5 B_{j}) / (5^{n} ∣ B_{j} ∣)) \int_{B_{j}} ∣ f ∣$ , and since $\int_{B_{j}} ∣ f ∣ > λ ∣ B_{j} ∣$ we get $w (5 B_{j}) \leq 5^{n} λ^{- 1} \int_{B_{j}} ∣ f (y) ∣ M^{*} w (y) d y$ .

3.1F1step 2.1givenalgebra

Summing over the pairwise disjoint $B_{j}$ and using $M^{*} w \leq [w]_{A_{1}} w$ almost everywhere from [F1], $w (K) \leq \sum_{j} w (5 B_{j}) \leq 5^{n} λ^{- 1} \sum_{j} \int_{B_{j}} ∣ f ∣ M^{*} w d λ \leq 5^{n} λ^{- 1} \int_{R^{n}} ∣ f ∣ M^{*} w d λ \leq 5^{n} [w]_{A_{1}} λ^{- 1} \int_{R^{n}} ∣ f ∣ w d λ$ .

4.1step 1.1step 3.1givenalgebra∎

Taking the supremum over compact $K \subseteq E_{λ}$ in step 3.1 and using the inner regularity of step 1.1 gives $w ({M f > λ}) \leq 5^{n} [w]_{A_{1}} λ^{- 1} \int ∣ f ∣ w d λ$ . For the uncentred maximal function, every ball $B = B (y, r) ∋ x$ satisfies $B \subseteq B (x, 2 r)$ and $∣ B (x, 2 r) ∣ = 2^{n} ∣ B ∣$ , so $⟨ ∣ f ∣ ⟩_{B} \leq 2^{n} ⟨ ∣ f ∣ ⟩_{B (x, 2 r)} \leq 2^{n} M f (x)$ and hence $M^{*} f \leq 2^{n} M f$ pointwise; consequently ${M^{*} f > λ} \subseteq {M f > λ / 2^{n}}$ and the centred estimate gives $w ({M^{*} f > λ}) \leq 2^{n} 5^{n} [w]_{A_{1}} λ^{- 1} \int ∣ f ∣ w d λ$ .

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

The weighted maximal function of a doubling weight

Definition

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

Let $v$ be a weight (Weights, their associated measures, and the spaces L^p(w)) whose measure $v d λ$ is doubling with constant $c_{v}$ : $v (B (x, 2 r)) \leq c_{v} v (B (x, r)) (x \in R^{n}, r > 0) .$

For $f \in L_{l o c}^{1} (v)$ the weighted maximal function is $M^{v} f (x) : = \sup_{r > 0} \frac{1}{v (B (x, r))} \int_{B (x, r)} ∣ f ∣ v d λ,$ and its cube analogue is $M_{c}^{v} f (x) : = \sup_{Q ∋ x} \frac{1}{v (Q)} \int_{Q} ∣ f ∣ v d λ,$ the supremum over the axis-parallel cubes of Axis-parallel cubes, their averages, and cube maximal functions containing $x$ . The weighted averages are finite because $f \in L_{l o c}^{1} (v)$ and $0 < v (Q) < \infty$ for bounded $Q$ .

Ball-cube comparability. Balls and cubes of comparable size have $v$ -measure comparable by a constant depending only on $n$ and $c_{v}$ : for a cube $Q$ with centre $y$ and side length $2 r$ one has $Q \subseteq B (y, \sqrt{n} r) \subseteq Q (y, \sqrt{n} r)$ , and iterating the doubling inequality a number of times depending only on $n$ gives $v (Q) \leq v (B (y, \sqrt{n} r)) \leq C (n, c_{v}) v (B (y, r)) \leq C (n, c_{v}) v (Q)$ , while $B (x, r) \subseteq Q (x, r) \subseteq B (x, \sqrt{n} r)$ gives the same comparison for balls. Consequently $M^{v} f \leq C (n, c_{v}) M_{c}^{v} f$ and $M_{c}^{v} f \leq C (n, c_{v}) M^{v} f$ pointwise: a cube $Q = Q (y, r) ∋ x$ is contained in $B (x, 2 \sqrt{n} r) \subseteq B (y, 3 \sqrt{n} r)$ , whose $v$ -measure is at most a dimensional number of doublings times $v (B (y, r)) \leq v (Q)$ ; conversely each centred ball $B (x, r)$ lies in $Q (x, r)$ of comparable $v$ -measure. These containments compare each average to one in the appropriate supremum (Ball and cube maximal functions are pointwise comparable provides the unweighted geometric sandwich, and the doubling of $v d λ$ converts it to a $v$ -measure comparison).

Measurability. For each fixed $r > 0$ the function $x \mapsto v (B (x, r))^{- 1} \int_{B (x, r)} ∣ f ∣ v d λ$ is continuous: the numerator is continuous in the centre by dominated convergence with dominating function $∣ f ∣ v$ over a fixed bounded ball containing all the translates, and the denominator $x \mapsto v (B (x, r))$ is continuous by dominated convergence with dominating function $v$ over such a ball; the denominator is positive (Dominated convergence, and Ball averages vary continuously with the centre and radius for the unweighted averages that underlie the same argument). The same dominated-convergence argument gives continuity in $r$ , so the supremum over positive radii equals that over positive rational radii. Hence $M^{v} f$ is Borel measurable (Sequential suprema, infima, limsup, liminf, and pointwise limits of measurable functions are measurable); the same holds for $M_{c}^{v} f$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

The weighted maximal function of a doubling weight is weak (1,1)

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $v$ be a weight on $R^{n}$ whose measure $v d λ$ is doubling with constant $c_{v}$ , and let $M^{v}$ be the weighted maximal function of The weighted maximal function of a doubling weight. Then there is $C (n, c_{v}) < \infty$ such that for every $f \in L^{1} (v)$ and every $λ > 0$ , $v ({M^{v} f > λ}) \leq C (n, c_{v}) λ^{- 1} \int_{R^{n}} ∣ f ∣ v d λ .$ Consequently $M^{v}$ is of strong type $(q, q)$ with respect to $v d λ$ for every $1 < q < \infty$ (Sublinear operators and weak or strong type $(p, q)$ bounds), with norm depending only on $n$ , $q$ and the doubling constant $c_{v}$ .

Facts & Assumptions

Given: Countable Choice, a weight $v$ with $v d λ$ doubling of constant $c_{v}$ , the weighted maximal function $M^{v}$ , $f \in L^{1} (v)$ and $λ > 0$ .

[F1]

$M^{v} f (x) = \sup_{r > 0} v (B (x, r))^{- 1} \int_{B (x, r)} ∣ f ∣ v d λ$ ; for each fixed $r$ the function $x \mapsto v (B (x, r))^{- 1} \int_{B (x, r)} ∣ f ∣ v d λ$ is continuous, and $v d λ$ is a locally finite regular Borel measure whose level sets are Borel (The weighted maximal function of a doubling weight, Weights, their associated measures, and the spaces L^p(w), Sigma-compact open sets make locally finite Borel measures regular).

[F2]

Doubling: $v (B (x, 2 r)) \leq c_{v} v (B (x, r))$ , so iterating gives $v (5 B) \leq c_{v}^{3} v (B)$ for every ball $B$ , since $5 r \leq 8 r$ .

[F3]

Fivefold Vitali covering: a finite family of balls $B_{1}, \dots, B_{m}$ has a pairwise disjoint subfamily $B_{i_{1}}, \dots, B_{i_{ℓ}}$ with $⋃_{j} B_{j} \subseteq ⋃_{k} 5 B_{i_{k}}$ (Vitali covering lemma for Euclidean balls with fivefold dilates).

[F4]

Inner regularity: for the regular Borel measure $v d λ$ and a Borel set $E$ , $v (E) = \sup {v (K) : K \subseteq E compact}$ (Sigma-compact open sets make locally finite Borel measures regular).

[F5]

Marcinkiewicz interpolation: a sublinear operator that is weak $(1, 1)$ with constant $A$ and strong $(\infty, \infty)$ with constant $B$ is strong $(p, p)$ for every $1 < p < \infty$ with norm at most $2 (A p / (p - 1))^{1 / p} B^{1 - 1 / p}$ (Marcinkiewicz interpolation from weak $(1, 1)$ and strong $(\infty, \infty)$ , Sublinear operators and weak or strong type $(p, q)$ bounds).

Proof

technique · direct

1.1F1F4given

The level set $E_{λ} : = {M^{v} f > λ}$ is open, being the union over $r > 0$ of the open sets ${x : v (B (x, r))^{- 1} \int_{B (x, r)} ∣ f ∣ v d λ > λ}$ , which are open because the displayed functions are continuous by [F1]. By [F4] its $v$ -measure is the supremum of $v (K)$ over compact $K \subseteq E_{λ}$ .

1.2F1F3givenchoose

Let $K \subseteq E_{λ}$ be compact. Each $x \in K$ admits $r_{x} > 0$ with $\int_{B (x, r_{x})} ∣ f ∣ v d λ > λ v (B (x, r_{x}))$ ; finitely many of these open balls cover $K$ , and [F3] selects pairwise disjoint balls $B_{1}, \dots, B_{ℓ}$ among them with $K \subseteq ⋃_{j} 5 B_{j}$ and $\int_{B_{j}} ∣ f ∣ v > λ v (B_{j})$ for every $j$ .

2.1F2step 1.2givenalgebra

By [F2] and the selection of step 1.2, $v (K) \leq \sum_{j} v (5 B_{j}) \leq c_{v}^{3} \sum_{j} v (B_{j}) \leq c_{v}^{3} λ^{- 1} \sum_{j} \int_{B_{j}} ∣ f ∣ v d λ \leq c_{v}^{3} λ^{- 1} \int_{R^{n}} ∣ f ∣ v d λ$ , where the last inequality uses the pairwise disjointness of the $B_{j}$ .

3.1step 1.1step 2.1givenalgebra

Taking the supremum over compact $K \subseteq E_{λ}$ in step 2.1 and using the inner regularity of step 1.1 gives $v ({M^{v} f > λ}) \leq c_{v}^{3} λ^{- 1} \int ∣ f ∣ v d λ$ , which is the asserted weak $(1, 1)$ bound with $C (n, c_{v}) = c_{v}^{3}$ .

4.1F5step 3.1givenalgebra∎

$M^{v}$ is sublinear and homogeneous; besides the weak $(1, 1)$ bound of step 3.1 with constant $A = c_{v}^{3}$ , it satisfies the trivial strong $(\infty, \infty)$ bound with constant $B = 1$ with respect to the measure $v d λ$ . Hence [F5] applies and gives, for every $1 < q < \infty$ , $∥ M^{v} f ∥_{L^{q} (v)} \leq 2 (A q / (q - 1))^{1 / q} ∥ f ∥_{L^{q} (v)}$ for all $f \in L^{q} (v)$ , a constant depending only on $n$ , $q$ and $c_{v}$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

The Hardy-Littlewood maximal operator characterises A_p

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $1 < p < \infty$ and let $w$ be a weight (Weights, their associated measures, and the spaces L^p(w)). Then the centred maximal operator $M$ is bounded on $L^{p} (w)$ if and only if $w \in A_{p}$ (Muckenhoupt A_p and A_1 weights); equivalently the uncentred and the cube maximal operators are bounded on $L^{p} (w)$ exactly for $w \in A_{p}$ . If $w \in A_{p}$ then $∥ M f ∥_{L^{p} (w)} \leq C_{n, p} [w]_{A_{p}}^{1 / (p - 1)} ∥ f ∥_{L^{p} (w)} (f \in L^{p} (w)),$ and conversely if $M$ is of strong type $(p, p)$ with respect to $w d λ$ then $w \in A_{p}$ .

Facts & Assumptions

Given: Countable Choice, $1 < p < \infty$ , a weight $w$ , and the exponent $p^{'} = p / (p - 1)$ ; for the sufficiency direction also $w \in A_{p}$ and the dual weight $σ : = w^{- 1 / (p - 1)}$ .

[F1]

$A_{p}$ weights are doubling, $σ \in A_{p^{'}}$ with $[σ]_{A_{p^{'}}} = [w]_{A_{p}}^{1 / (p - 1)}$ , and for every cube $Q$ $⟨ w ⟩_{Q} ⟨ σ ⟩_{Q}^{p - 1} \leq [w]_{A_{p}}$ (Duality and nesting of the A_p classes, A_p weights are doubling, Muckenhoupt A_p and A_1 weights).

[F2]

Marcinkiewicz interpolation gives $∥ S u ∥_{L^{q} (μ)} \leq a_{q} ∥ u ∥_{L^{q} (μ)}$ for $q > 1$ whenever $S$ is sublinear, weak $(1, 1)$ with constant one and bounded on $L^{\infty}$ with constant one; here $a_{q} = 2 (q / (q - 1))^{1 / q}$ (Marcinkiewicz interpolation from weak $(1, 1)$ and strong $(\infty, \infty)$ ).

[F3]

Dyadic cubes partition each generation and are nested or disjoint; all face conventions have the same volume and null boundaries (All-generation dyadic cubes: partition, volume and nesting, A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included). The shifted grids used below have the same properties by the explicit boundary calculation in Proof 1.1.

[F4]

Chebyshev's inequality and monotone convergence (Chebyshev-Markov inequality for the integral, Monotone convergence for the integral); the ball and cube maximal operators are pointwise comparable by a dimensional constant, as are their centred and uncentred versions (Ball and cube maximal functions are pointwise comparable).

Proof

technique · direct

1.1F3givenconstructalgebra

Finite dyadic covering. For $a \in {0, 1 / 3, 2 / 3}^{n}$ use the grid $D^{a}$ of half-open cubes $2^{- k} (m + (- 1)^{k} a + (0, 1]^{n})$ , $k \in Z$ , $m \in Z^{n}$ . The boundary offset of a parent, in child units, differs from the child offset by $- 3 (- 1)^{k} a \in Z^{n}$ , so the grids are nested and partition each generation. Given an open cube $Q$ of side $ℓ$ , choose a scale $L = 2^{- k}$ with $3 ℓ < L \leq 6 ℓ$ . In each coordinate the boundary sets of the three offsets are separated by $L / 3$ , so an interval of length $ℓ$ meets boundaries from at most one offset. Choose an offset avoiding its closure, coordinate by coordinate; then $Q$ is contained in one $R \in D^{a}$ of side $L$ , with $∣ R ∣ \leq 6^{n} ∣ Q ∣$ . Consequently $M_{c}^{*} f \leq 6^{n} \max_{a} M_{D^{a}} f$ , where $M_{D} f = \sup_{R \in D, x \in R} ∣ R ∣^{- 1} \int_{R} ∣ f ∣$ .

1.2F2F3givenalgebra

Uniform weighted dyadic bounds. For any locally finite measure $μ = v d λ$ with $v$ a weight, set $M_{D}^{μ} u (x) = \sup_{R \in D, x \in R} μ (R)^{- 1} \int_{R} ∣ u ∣ d μ$ . Restrict first to generations $- N \leq k \leq N$ . The maximal bad cubes at height $t > 0$ exist because ancestor chains are finite; they are countable and disjoint, and cover the restricted superlevel set, so its $μ$ -measure is at most $t^{- 1} \int ∣ u ∣ d μ$ . Letting $N \to \infty$ gives weak $(1, 1)$ with constant one. The operator is sublinear and bounded on $L^{\infty} (μ)$ with constant one, so [F2] gives $∥ M_{D}^{μ} u ∥_{L^{q} (μ)} \leq a_{q} ∥ u ∥_{L^{q} (μ)}$ , independently of $v$ and its doubling constant. Measurability follows from the countable grid.

1.3F1F3givenalgebra

Sufficiency, pointwise estimate. Suppose $w \in A_{p}$ , let $σ = w^{- 1 / (p - 1)}$ , and fix one grid. Set $u = ∣ f ∣ σ^{- 1}$ , $F = M_{D}^{σ} u$ and $h = F^{p - 1} w^{- 1}$ , with powers assigned arbitrarily on the common null set where $w$ or $σ$ is not positive finite. For every grid cube $R$ and almost every $y \in R$ , $F (y) \geq σ (R)^{- 1} \int_{R} ∣ f ∣$ . Thus $⟨ ∣ f ∣ ⟩_{R} \leq (σ (R) / ∣ R ∣) F (y)$ . Raise to $p - 1$ and integrate with respect to Lebesgue measure over $R$ to get $(⟨ ∣ f ∣ ⟩_{R})^{p - 1} \leq (σ (R) / ∣ R ∣)^{p - 1} ∣ R ∣^{- 1} \int_{R} F^{p - 1} \leq [w]_{A_{p}} w (R)^{- 1} \int_{R} h w$ . The $A_{p}$ bound holds also for half-open cubes because their boundaries are null. Taking suprema at $x$ yields $M_{D} f (x) \leq [w]_{A_{p}}^{1 / (p - 1)} (M_{D}^{w} h (x))^{1 / (p - 1)}$ .

2.1F4step 1.1step 1.2step 1.3givenalgebra

Sufficiency, norm estimate. We have $∥ u ∥_{L^{p} (σ)} = ∥ f ∥_{L^{p} (w)}$ and $h^{p^{'}} w = F^{p} σ$ . Step 1.2 therefore ensures $F \in L^{p} (σ)$ and $h \in L^{p^{'}} (w)$ , and step 1.3 gives $∥ M_{D} f ∥_{L^{p} (w)} \leq [w]_{A_{p}}^{1 / (p - 1)} a_{p^{'}}^{1 / (p - 1)} ∥ h ∥_{L^{p^{'}} (w)}^{1 / (p - 1)} \leq [w]_{A_{p}}^{1 / (p - 1)} a_{p^{'}}^{1 / (p - 1)} a_{p} ∥ f ∥_{L^{p} (w)}$ . By step 1.1, the maximum over the $3^{n}$ grids has norm at most $3^{n / p}$ times this bound. The ball/cube comparisons [F4] give the displayed estimate for all the stated maximal functions, with a constant depending only on $n, p$ .

3.1F4step 2.1givenalgebra∎

Necessity. Suppose $∥ M f ∥_{L^{p} (w)} \leq C ∥ f ∥_{L^{p} (w)}$ for locally integrable inputs in $L^{p} (w)$ . By [F4], $M_{c}^{*}$ has bound $C_{0} = C_{n} C$ . For a cube $Q$ put $f_{ε} = 1_{Q} (w + ε)^{- 1 / (p - 1)}$ , setting it to zero on the null set of exceptional weight values. It is bounded, compactly supported, and belongs to $L^{p} (w)$ . On $Q$ , $M_{c}^{*} f_{ε} \geq ⟨ f_{ε} ⟩_{Q}$ . Since $f_{ε}^{p} w \leq f_{ε}$ , the norm bound implies $w (Q) ⟨ f_{ε} ⟩_{Q}^{p} \leq C_{0}^{p} \int_{Q} f_{ε}$ , hence $⟨ w ⟩_{Q} ⟨ f_{ε} ⟩_{Q}^{p - 1} \leq C_{0}^{p}$ . Let $ε = 1 / j ↓ 0$ and use monotone convergence [F4] to obtain $⟨ w ⟩_{Q} ⟨ w^{- 1 / (p - 1)} ⟩_{Q}^{p - 1} \leq C_{0}^{p}$ for every cube. Thus $w \in A_{p}$ , completing both directions.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

A_infinity weights satisfy power decay

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $w \in A_{\infty}$ (The Muckenhoupt A_infinity class). Then there are $C, δ > 0$ , depending only on $n$ , on a witnessing exponent $p$ and on $[w]_{A_{p}}$ , such that $\frac{w (E)}{w (Q)} \leq C (\frac{∣ E ∣}{∣ Q ∣})^{δ}$ for every axis-parallel cube $Q$ and every measurable $E \subseteq Q$ . Explicitly one may take $δ = γ / (1 + γ)$ and $C = C_{1}$ , where $1 + γ$ and $C_{1}$ are the reverse Hölder exponent and constant supplied for a witnessing $A_{p}$ weight.

Facts & Assumptions

Given: Countable Choice, $w \in A_{\infty}$ , a witnessing exponent $p$ with $w \in A_{p}$ , a reverse Hölder pair $(1 + γ, C_{1})$ , a cube $Q$ and a measurable $E \subseteq Q$ .

[F1]

$w \in A_{\infty}$ means $w \in A_{p}$ for some $1 \leq p < \infty$ , and then $0 < w (Q) < \infty$ for every cube $Q$ (The Muckenhoupt A_infinity class).

[F2]

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

[F3]

Hölder's inequality: for a measurable set $E$ , $\int_{E} w d λ \leq (\int_{Q} w^{1 + γ} d λ)^{1 / (1 + γ)} ∣ E ∣^{γ / (1 + γ)}$ (the exponents $1 + γ$ and $(1 + γ) / γ$ are conjugate) (Holder's inequality for integrals, including the endpoint cases, Real powers for positive bases, with the zero-base positive-exponent convention).

Proof

technique · direct

1.1F1F2given

Since $w \in A_{\infty}$ , there is a witnessing exponent $p$ with $w \in A_{p}$ by [F1]; applying [F2] produces $γ > 0$ and $C_{1}$ with the normalized reverse Hölder inequality.

2.1F2F3step 1.1givenalgebra

For the cube $Q$ and measurable $E \subseteq Q$ , Hölder's inequality [F3] gives $w (E) \leq (\int_{Q} w^{1 + γ})^{1 / (1 + γ)} ∣ E ∣^{γ / (1 + γ)}$ ; the reverse Hölder inequality bounds the first factor by $C_{1} ⟨ w ⟩_{Q} ∣ Q ∣^{1 / (1 + γ)} = C_{1} w (Q) ∣ Q ∣^{- γ / (1 + γ)}$ , so $w (E) \leq C_{1} w (Q) (∣ E ∣ / ∣ Q ∣)^{γ / (1 + γ)}$ .

3.1F1step 2.1givenalgebra∎

Dividing by $w (Q) \in (0, \infty)$ gives $w (E) / w (Q) \leq C (∣ E ∣ / ∣ Q ∣)^{δ}$ with $C = C_{1}$ and $δ = γ / (1 + γ) > 0$ , and both constants depend only on $n$ , the witnessing exponent $p$ and $[w]_{A_{p}}$ (and on the auxiliary $α$ , which is fixed in the construction).

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Power decay implies doubling

Statement

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

Let $w$ be a weight on $R^{n}$ (Weights, their associated measures, and the spaces L^p(w)) and suppose there are constants $C, δ > 0$ such that $\frac{w (E)}{w (Q)} \leq C (\frac{∣ E ∣}{∣ Q ∣})^{δ}$ for every axis-parallel cube $Q$ and every measurable $E \subseteq Q$ . Then the measure $w d λ$ is doubling, with a constant depending only on $n$ , $C$ and $δ$ : there is $C^{'} = C^{'} (n, C, δ) < \infty$ with $w (B (x, 2 r)) \leq C^{'} w (B (x, r))$ for all $x \in R^{n}$ and $r > 0$ .

Facts & Assumptions

Given: Countable Choice; A weight $w$ and constants $C, δ > 0$ with the displayed power decay property.

[F1]

$w > 0$ and $w < \infty$ Lebesgue-a.e., $w \in L_{l o c}^{1}$ , and $E \mapsto w (E) = \int_{E} w d λ$ is a locally finite measure with the $w$ -null sets equal to the Lebesgue-null sets (Weights, their associated measures, and the spaces L^p(w)).

[F2]

$Q (x, ρ) = \prod_{i} (x_{i} - ρ, x_{i} + ρ)$ has side $2 ρ$ and Lebesgue measure $(2 ρ)^{n}$ ; if $E \subseteq F$ are measurable then $w (E) \leq w (F)$ , and for nested cubes $∣ Q (x, ρ^{'}) ∣ = ρ^{' n} ρ^{- n} ∣ Q (x, ρ) ∣$ (Axis-parallel cubes, their averages, and cube maximal functions, For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it).

[F3]

A ball $B (x, 2 r)$ is contained in the cube $Q (x, 4 r)$ , and a cube $Q (x, ρ)$ is contained in $B (x, \sqrt{n} ρ)$ (Ball and cube maximal functions are pointwise comparable).

Proof

technique · direct

1.1F1F2givenalgebra

Choose $β \in (0, 1)$ with $C β^{δ} \leq \frac{1}{2}$ , for instance $β : = \min {\frac{1}{2}, (2 C)^{- 1 / δ}}$ , and put $α : = 1 - C β^{δ} \geq \frac{1}{2}$ . If $E \subseteq Q$ is measurable with $∣ E ∣ \geq (1 - β) ∣ Q ∣$ , then $∣ Q ∖ E ∣ \leq β ∣ Q ∣$ , so the power decay applied to $Q ∖ E$ gives $w (Q ∖ E) \leq C β^{δ} w (Q)$ and hence $w (E) = w (Q) - w (Q ∖ E) \geq α w (Q) > 0$ . Both $β$ and $α$ depend only on $C$ and $δ$ .

2.1F2step 1.1givenalgebra

Fix $x \in R^{n}$ and $r > 0$ , let $l_{1} : = 4 r$ and $l_{j + 1} : = (1 - β)^{1 / n} l_{j}$ , and let $k \geq 1$ be the least integer with $l_{k} \leq r / \sqrt{n}$ ; then $k$ depends only on $n$ and $β$ , hence only on $n, C, δ$ . By [F2] one has $∣ Q (x, l_{j + 1}) ∣ = (1 - β) ∣ Q (x, l_{j}) ∣$ , so the shell $Q (x, l_{j}) ∖ Q (x, l_{j + 1})$ has measure $β ∣ Q (x, l_{j}) ∣$ and step 1.1 applied to $E = Q (x, l_{j + 1})$ (whose complement in $Q (x, l_{j})$ has measure $β ∣ Q (x, l_{j}) ∣$ ) yields $w (Q (x, l_{j + 1})) \geq α w (Q (x, l_{j}))$ for every $j < k$ . Iterating, $w (Q (x, 4 r)) \leq α^{- (k - 1)} w (Q (x, l_{k}))$ .

3.1F1F3step 2.1algebra∎

Since $l_{k} \leq r / \sqrt{n}$ , the cube $Q (x, l_{k})$ is contained in $B (x, r)$ by [F3], so [F1] gives $w (Q (x, l_{k})) \leq w (B (x, r))$ ; and $B (x, 2 r) \subseteq Q (x, 4 r)$ by [F3], so $w (B (x, 2 r)) \leq w (Q (x, 4 r)) \leq α^{- (k - 1)} w (B (x, r))$ . Thus $w d λ$ is doubling with $C^{'} = α^{- (k - 1)}$ , a constant depending only on $n$ , $C$ and $δ$ ; no property of $x$ or $r$ entered beyond the display, and the degenerate case $r > 0$ is the only case needed since doubling is asserted for positive radii.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Differentiation of L-one functions for a doubling weight

Statement

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain), hence Countable Choice (Dependent choice implies countable choice). Let $v$ be a weight on $R^{n}$ whose measure $v d λ$ is doubling, and let $h \in L_{l o c}^{1} (v)$ (Weights, their associated measures, and the spaces L^p(w)). Then for $v$ -almost every $x \in R^{n}$ , $\lim_{r \to 0^{+}} \frac{1}{v (B (x, r))} \int_{B (x, r)} h v d λ = h (x);$ more generally the same limit holds along any family of balls or cubes shrinking nicely to $x$ (A family shrinking nicely to a point) with $v$ -measure comparable to the corresponding ball.

Facts & Assumptions

Given: Dependent Choice, a weight $v$ with $v d λ$ doubling, and $h \in L_{l o c}^{1} (v)$ .

[F1]

$M^{v} h (x) = \sup_{r > 0} v (B (x, r))^{- 1} \int_{B (x, r)} ∣ h ∣ v d λ$ is the weighted maximal function of The weighted maximal function of a doubling weight, and it obeys the weak $(1, 1)$ bound $v ({M^{v} g > η}) \leq C (n, c_{v}) η^{- 1} \int ∣ g ∣ v d λ$ for every $g \in L^{1} (v)$ (The weighted maximal function of a doubling weight is weak (1,1)).

[F2]

$C_{c} (R^{n})$ is dense in $L^{1} (v d λ)$ because $v d λ$ is a Radon measure (C_c(X) is dense in L^p(mu) for a Radon measure), with complex density obtained by approximating the two real components separately; a function $g \in C_{c}$ is continuous, so for every $x$ and every family shrinking nicely to $x$ the weighted averages of $g$ converge to $g (x)$ (the averages of $∣ g (\cdot) - g (x) ∣$ are bounded by the maximum of $∣ g - g (x) ∣$ over the shrinking sets, which tends to $0$ ).

[F3]

Chebyshev's inequality: for a nonnegative measurable $G$ and $η > 0$ , $v ({G > η}) \leq η^{- 1} \int G v d λ$ (Chebyshev-Markov inequality for the integral).

[F4]

Dominated convergence gives continuity in radius for local weighted integrals, and Lebesgue-measurable functions have Borel representatives: apply A function measurable for a completion is almost everywhere equal to one measurable for the original sigma-algebra componentwise using $L (R^{n})$ is exactly the completion of the restriction of $λ_{n}$ to the Borel sets, setting infinite values on null sets to zero. These representatives give jointly Borel integrands by composition with continuous maps; Tonelli gives measurable section integrals (Dominated convergence, Borel representatives make the convolution integrand Borel measurable, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

Proof

technique · direct

1.1F1F2F4givenalgebra

First suppose $h \in L^{1} (v)$ globally and put $μ = v d λ$ . For $g \in C_{c}$ and $G = ∣ h - g ∣$ , the centred weighted oscillation $D h (x) : = {lim sup}_{r ↓ 0} μ (B (x, r))^{- 1} \int_{B (x, r)} ∣ h (y) - h (x) ∣ d μ (y)$ is at most $M^{v} G (x) + G (x)$ , since the corresponding oscillation of continuous $g$ tends to zero. These limsups are measurable: for each fixed $x$ the integrals are continuous in positive radius by dominated convergence, hence rational radii suffice; joint measurability follows after choosing Borel representatives, and changing them on a null set does not change the a.e. assertion.

2.1F1F2F3step 1.1givenalgebra

For $η > 0$ , ${D h > η} \subseteq {M^{v} G > η / 2} \cup {G > η / 2}$ . The weak bound and Chebyshev give $μ ({D h > η}) \leq 2 (C (n, c_{v}) + 1) η^{- 1} ∥ h - g ∥_{L^{1} (v)}$ . Taking the infimum over $g \in C_{c}$ using [F2] makes this measure zero. The countable union over $η = 1 / k$ is null, so the centred absolute oscillation tends to zero a.e. For local $h$ , apply this result to $h_{m} = h 1_{B (0, m + 1)} \in L^{1} (v)$ ; on $B (0, m)$ , sufficiently small centred balls see $h = h_{m}$ . The countable union of these exceptional sets is null and $⋃_{m} B (0, m) = R^{n}$ , proving the same conclusion for every local input.

3.1step 2.1givenalgebra∎

Outside the null set of step 2.1, the absolute oscillation over any shrinking set $E_{r} \subseteq B (x, ρ_{r})$ with $ρ_{r} \to 0$ and $μ (B (x, ρ_{r})) \leq C_{x} μ (E_{r})$ is bounded by $C_{x}$ times the centred oscillation. It therefore tends to zero, and the modulus of the difference between the average of $h$ and $h (x)$ is at most this oscillation. Balls and cubes shrinking nicely with the stipulated weighted comparability meet these hypotheses. This proves all the stated limits, on a common full-measure set.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Reverse Holder from a distribution estimate for a doubling weight

Statement

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain). Let $v$ be a weight on $R^{n}$ whose measure $v d λ$ is doubling (The weighted maximal function of a doubling weight), and let $h \geq 0$ be measurable with $h v \in L_{l o c}^{1} (λ)$ . Suppose there are $0 < α, β < 1$ such that for every cube $Q$ and every measurable $S \subseteq Q$ , $v (S) \leq α v (Q) ⟹ \int_{S} h v d λ \leq β \int_{Q} h v d λ .$ Then there are $q > 1$ and $c < \infty$ , depending only on $n$ , the doubling constant of $v$ , $α$ and $β$ , such that $(\frac{1}{v (Q)} \int_{Q} h^{q} v d λ)^{1 / q} \leq \frac{c}{v (Q)} \int_{Q} h v d λ$ for every cube $Q$ .

Facts & Assumptions

Given: Dependent Choice, a weight $v$ with $μ : = v d λ$ doubling, a nonnegative measurable $h$ with $h \in L_{l o c}^{1} (μ)$ , constants $0 < α, β < 1$ , a cube $Q_{0}$ , and the levels $α_{k} : = (C_{n} α^{- 1})^{k} α_{0}$ with $α_{0} : = μ (Q_{0})^{- 1} \int_{Q_{0}} h d μ > 0$ .

[F1]

$μ$ is a locally finite measure with $0 < μ (Q) < \infty$ for every cube; cubes and balls of comparable size have comparable $μ$ -measure, with a constant depending only on $n$ and the doubling constant of $v$ (The weighted maximal function of a doubling weight, Weights, their associated measures, and the spaces L^p(w)).

[F2]

Inside $Q_{0}$ the dyadic subcubes form a family with a top element $Q_{0}$ in which every proper descendant has a parent inside $Q_{0}$ and two cubes are nested or disjoint (Maximal dyadic subcubes of a cube at a height).

[F3]

Differentiation for the doubling weight $v$ : for $μ$ -almost every point, the $μ$ -averages of an $L_{l o c}^{1} (μ)$ function over the dyadic subcubes shrinking nicely to the point converge to the value of the function (Differentiation of L-one functions for a doubling weight).

[F4]

$μ$ is countably additive on disjoint measurable pieces, and the layered integral identity and Fubini/Tonelli are available (Countable additivity and continuity of finitely additive set functions, Fubini's theorem for L^1 functions on a sigma-finite product, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, For 0 < p < infinity, the layer-cake formula computes the integral of |f|^p from the distribution function).

Proof

technique · direct

1.1F1F2givenalgebra

Since $α_{k} \geq α_{0}$ for every $k \geq 0$ and the top cube $Q_{0}$ has $μ$ -average $α_{0}$ , the cube $Q_{0}$ is not bad at any level $k$ . For every bad subcube $R$ the ancestors of $R$ inside $Q_{0}$ form a finite chain ending at $Q_{0}$ ; hence there is a topmost bad ancestor, and the family of maximal bad subcubes (those with no bad proper ancestor inside $Q_{0}$ ) is well defined, pairwise disjoint and at most countable. Let $U_{k}$ be their union.

2.1F1F2F3step 1.1givenalgebra

Let $R$ be a maximal bad subcube at level $k$ with parent $P$ : then $⟨ h ⟩_{P}^{μ} \leq α_{k}$ by maximality, and $μ (P) \leq C_{n} μ (R)$ by [F1] since $P$ is a cube of twice the side length containing $R$ ; hence $\int_{R} h d μ \leq \int_{P} h d μ \leq C_{n} α_{k} μ (R)$ , that is, $⟨ h ⟩_{R}^{μ} \leq C_{n} α_{k}$ . Moreover $U_{k + 1} \subseteq U_{k}$ by the same parent argument, and $h \leq α_{k}$ $μ$ -almost everywhere on $Q_{0} ∖ U_{k}$ : for a point $x$ outside $U_{k}$ and outside the $μ$ -null exceptional set of the differentiation lemma, no dyadic subcube $R ∋ x$ has $⟨ h ⟩_{R}^{μ} > α_{k}$ , since such an $R$ would lie in a maximal bad cube containing $x$ ; the subcubes containing $x$ shrink nicely to $x$ , so their $μ$ -averages converge to $h (x)$ by that lemma, and the limit satisfies $h (x) \leq α_{k}$ .

3.1F1F4step 1.1step 2.1givenalgebra

Decay of the integrals. For a maximal bad subcube $R$ at level $k$ , the set $S : = R \cap U_{k + 1}$ is measurable and contained in $R$ ; since $U_{k + 1}$ is the disjoint union of its maximal bad subcubes, on each of which the $μ$ -average of $h$ exceeds $α_{k + 1}$ , $α_{k + 1} μ (S) \leq \int_{S} h d μ \leq \int_{R} h d μ \leq C_{n} α_{k} μ (R)$ , so $μ (S) \leq α μ (R)$ because $C_{n} α_{k} = α α_{k + 1}$ . The hypothesis applied to $S \subseteq R$ therefore gives $\int_{S} h d μ \leq β \int_{R} h d μ$ ; summing over the pairwise disjoint maximal cubes at level $k$ yields $\int_{U_{k + 1}} h d μ \leq β \int_{U_{k}} h d μ$ , hence $\int_{U_{k}} h d μ \leq β^{k} \int_{Q_{0}} h d μ$ by iteration.

4.1F4step 2.1step 3.1givenalgebra

Integral bound. If $\int_{Q_{0}} h d μ = 0$ , then $h = 0$ $μ$ -a.e. on $Q_{0}$ and the conclusion is immediate. Otherwise $α_{0} > 0$ as above. Summing the bounds $μ (R \cap U_{k + 1}) \leq α μ (R)$ from step 3.1 shows $μ (U_{k + 1}) \leq α μ (U_{k})$ ; hence $μ (⋂_{k} U_{k}) = 0$ . The sets $Q_{0} ∖ U_{0}$ and $U_{k} ∖ U_{k + 1}$ are disjoint measurable pieces covering $Q_{0}$ up to a $μ$ -null set; by step 2.1, $h \leq α_{0}$ on $Q_{0} ∖ U_{0}$ and $h \leq α_{k + 1}$ on $U_{k} ∖ U_{k + 1}$ , all $μ$ -a.e. Hence, for every $γ > 0$ , $\int_{Q_{0}} h^{1 + γ} d μ \leq α_{0}^{γ} \int_{Q_{0} ∖ U_{0}} h d μ + \sum_{k \geq 0} α_{k + 1}^{γ} \int_{U_{k}} h d μ \leq α_{0}^{γ} (\int_{Q_{0}} h d μ) [1 + (C_{n} α^{- 1})^{γ} \sum_{k \geq 0} ((C_{n} α^{- 1})^{γ} β)^{k}]$ . Choose $γ > 0$ so small that $r : = (C_{n} α^{- 1})^{γ} β < 1$ ; then the geometric series converges.

5.1step 3.1step 4.1givenalgebra∎

Dividing the display of step 4.1 by $μ (Q_{0})$ and using $α_{0} = μ (Q_{0})^{- 1} \int_{Q_{0}} h d μ$ gives $(μ (Q_{0})^{- 1} \int_{Q_{0}} h^{1 + γ} d μ)^{1 / (1 + γ)} \leq c μ (Q_{0})^{- 1} \int_{Q_{0}} h d μ$ with $c = (1 + (C_{n} α^{- 1})^{γ} / (1 - r))^{1 / (1 + γ)} < \infty$ ; since $Q_{0}$ was arbitrary, the reverse Hölder inequality holds with $q = 1 + γ > 1$ and this $c$ , both depending only on $n$ , the doubling constant of $v$ , $α$ and $β$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Power decay implies membership in some A_p

Statement

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain). Let $w$ be a weight on $R^{n}$ (Weights, their associated measures, and the spaces L^p(w)) and suppose there are constants $C, δ > 0$ with $\frac{w (E)}{w (Q)} \leq C (\frac{∣ E ∣}{∣ Q ∣})^{δ}$ for every axis-parallel cube $Q$ and every measurable $E \subseteq Q$ . Then $w \in A_{p}$ (Muckenhoupt A_p and A_1 weights) for $p = 1 + 1 / γ < \infty$ , where $γ > 0$ and the resulting bound on $[w]_{A_{p}}$ depend only on $n, C, δ$ ; consequently $w \in A_{\infty}$ (The Muckenhoupt A_infinity class).

Facts & Assumptions

Given: Dependent Choice, a weight $w$ , constants $C, δ > 0$ with the displayed power decay, and a cube $Q$ .

[F1]

$w d λ$ is a locally finite measure with $0 < w (Q) < \infty$ for every cube $Q$ ; subsets have smaller measure, and $w (\emptyset) = 0$ (Weights, their associated measures, and the spaces L^p(w)).

[F2]

Power decay implies that $w d λ$ is doubling, with a doubling constant depending only on $n, C, δ$ (Power decay implies doubling).

[F3]

Reverse Hölder from a distribution estimate: if $μ$ is a doubling measure of the form $v d λ$ and $h \geq 0$ is measurable with $h v \in L_{l o c}^{1} (λ)$ and with $μ (S) \leq α μ (Q) \Rightarrow \int_{S} h d μ \leq β \int_{Q} h d μ$ for some $0 < α, β < 1$ and every cube $Q$ and measurable $S \subseteq Q$ , then there are $q > 1$ and $c < \infty$ , depending only on $n$ , the doubling constant of $μ$ , $α$ and $β$ , with $(μ (Q)^{- 1} \int_{Q} h^{q} d μ)^{1 / q} \leq c μ (Q)^{- 1} \int_{Q} h d μ$ for every cube $Q$ (Reverse Holder from a distribution estimate for a doubling weight, The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

[F4]

For real exponents, $1 - q = - (q - 1)$ and $p = q / (q - 1) = 1 + 1 / (q - 1)$ satisfies $p - 1 = 1 / (q - 1)$ and $p^{'} = q$ (Real powers for positive bases, with the zero-base positive-exponent convention).

Proof

technique · direct

1.1F1givenalgebra

The density implication. Put $β : = 1 - \min {\frac{1}{2}, (2 C)^{- 1 / δ}} \in (0, 1)$ and let $S \subseteq Q$ be measurable with $w (S) \leq \frac{1}{2} w (Q)$ . Then $∣ S ∣ \leq β ∣ Q ∣$ : otherwise $∣ Q ∖ S ∣ < (1 - β) ∣ Q ∣ \leq (2 C)^{- 1 / δ} ∣ Q ∣$ , so power decay applied to the complement gives $w (Q ∖ S) < \frac{1}{2} w (Q)$ , whence $w (S) > \frac{1}{2} w (Q)$ , a contradiction. Equivalently, writing $μ : = w d λ$ and $h : = w^{- 1}$ , the hypothesis of [F3] holds with $α = \frac{1}{2}$ and this $β$ : $μ (S) \leq α μ (Q)$ implies $\int_{S} h d μ = ∣ S ∣ \leq β ∣ Q ∣ = β \int_{Q} h d μ$ .

2.1F2F3step 1.1givenalgebra

Applying the reverse Hölder lemma. The measure $μ = w d λ$ is doubling by [F2], and $h = w^{- 1} \geq 0$ has $\int_{Q} h d μ = ∣ Q ∣ < \infty$ for every cube, so $h w \in L_{l o c}^{1} (λ)$ ; step 1.1 supplies the density implication with $α = \frac{1}{2}$ and $β < 1$ . By [F3] there are $q > 1$ and $c < \infty$ , depending only on $n$ , the doubling constant of $w$ , and hence only on $n, C, δ$ , such that, for every cube $Q$ , $(w (Q)^{- 1} \int_{Q} w^{1 - q} d λ)^{1 / q} \leq c ∣ Q ∣ / w (Q)$ , since $h^{q} d μ = w^{- q} \cdot w d λ = w^{1 - q} d λ$ and $\int_{Q} h d μ = ∣ Q ∣$ .

3.1F4step 2.1givenalgebra∎

From the reverse Hölder estimate to $A_{p}$ . Put $γ : = q - 1 > 0$ and $p : = q / (q - 1) = 1 + 1 / γ$ , so that $p - 1 = 1 / γ$ and $w^{- 1 / (p - 1)} = w^{- γ}$ ; write $I : = w (Q)^{- 1} \int_{Q} w^{- γ} d λ = ⟨ w^{- γ} ⟩_{Q} / ⟨ w ⟩_{Q}$ . Step 2.1 reads $I^{1 / q} \leq c ⟨ w ⟩_{Q}^{- 1}$ , hence $I^{1 / γ} \leq (c ⟨ w ⟩_{Q}^{- 1})^{q / γ} = c^{p} ⟨ w ⟩_{Q}^{- p}$ by [F4], since $q / γ = q / (q - 1) = p$ . Multiplying by $⟨ w ⟩_{Q}^{p}$ and using $⟨ w ⟩_{Q}^{p} I^{1 / γ} = ⟨ w ⟩_{Q}^{1 + 1 / γ} (⟨ w^{- γ} ⟩_{Q} / ⟨ w ⟩_{Q})^{1 / γ} = ⟨ w ⟩_{Q} ⟨ w^{- γ} ⟩_{Q}^{1 / γ}$ gives $⟨ w ⟩_{Q} ⟨ w^{- 1 / (p - 1)} ⟩_{Q}^{p - 1} = ⟨ w ⟩_{Q} ⟨ w^{- γ} ⟩_{Q}^{1 / γ} \leq c^{p}$ for every cube $Q$ . Therefore $[w]_{A_{p}} \leq c^{p} < \infty$ , that is, $w \in A_{p}$ , and $w \in A_{\infty}$ by the definition of the latter as the union of the finite-exponent classes.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

The A_infinity power-decay characterisation

Statement

$w \in A_{\infty}$ (The Muckenhoupt A_infinity class), that is, $w \in A_{p}$ (Muckenhoupt A_p and A_1 weights) for some $1 \leq p < \infty$ ;
there are $C, δ > 0$ with $w (E) / w (Q) \leq C (∣ E ∣ / ∣ Q ∣)^{δ}$ for every axis-parallel cube $Q$ and every measurable $E \subseteq Q$ ;
$w$ satisfies a reverse Hölder inequality: there are $γ > 0$ and $C < \infty$ with $(⟨ w^{1 + γ} ⟩_{Q})^{1 / (1 + γ)} \leq C ⟨ w ⟩_{Q}$ for every axis-parallel cube $Q$ .

All constants in each condition depend only on $n$ and on the constants appearing in the assumed condition.

Facts & Assumptions

Given: Dependent Choice; a weight $w$ ; the three conditions (i), (ii), (iii) of the statement.

[F1]

(i) implies (ii): for $w \in A_{\infty}$ there are $C, δ > 0$ , depending only on $n$ , a witnessing exponent $p$ and $[w]_{A_{p}}$ , with $w (E) / w (Q) \leq C (∣ E ∣ / ∣ Q ∣)^{δ}$ for every cube $Q$ and measurable $E \subseteq Q$ (A_infinity weights satisfy power decay).

[F2]

(ii) implies (i): power decay with constants $C, δ$ forces $w \in A_{p}$ for $p = 1 + 1 / γ < \infty$ with $γ > 0$ and the bound on $[w]_{A_{p}}$ depending only on $n, C, δ$ , hence $w \in A_{\infty}$ (Power decay implies membership in some A_p).

[F3]

(i) implies (iii): for $1 \leq p < \infty$ and $w \in A_{p}$ there are $γ > 0$ and $C < \infty$ , depending only on $n$ , $p$ and $[w]_{A_{p}}$ , with $(⟨ w^{1 + γ} ⟩_{Q})^{1 / (1 + γ)} \leq C ⟨ w ⟩_{Q}$ for every cube $Q$ (Reverse Holder self-improvement for A_p weights); this is applied to a witnessing exponent of $A_{\infty}$ .

[F4]

Hölder's inequality for the conjugate exponents $1 + γ$ and $(1 + γ) / γ$ : $\int_{E} w d λ \leq (\int_{Q} w^{1 + γ} d λ)^{1 / (1 + γ)} ∣ E ∣^{γ / (1 + γ)}$ for measurable $E \subseteq Q$ ; all cube averages are those of Muckenhoupt A_p and A_1 weights and $(1 + γ) / γ > 1$ is a real exponent (Holder's inequality for integrals, including the endpoint cases, Real powers for positive bases, with the zero-base positive-exponent convention).

Proof

technique · direct

1.1F1given

(i) implies (ii). If $w \in A_{\infty}$ , then $w \in A_{p}$ for a witnessing $1 \leq p < \infty$ , and [F1] supplies $C, δ > 0$ , depending only on $n$ , $p$ and $[w]_{A_{p}}$ , with $w (E) / w (Q) \leq C (∣ E ∣ / ∣ Q ∣)^{δ}$ for every cube $Q$ and measurable $E \subseteq Q$ .

1.2F2given

(ii) implies (i). If power decay holds with constants $C, δ$ , then [F2] gives $γ > 0$ and $p = 1 + 1 / γ < \infty$ with $w \in A_{p}$ and $[w]_{A_{p}}$ bounded in terms of $n, C, δ$ ; by the definition of $A_{\infty}$ as the union of the classes $A_{p}$ , $1 \leq p < \infty$ , this is (i).

1.3F3given

(i) implies (iii). Let $w \in A_{p}$ with witnessing exponent $p$ , which may be taken finite by (i). By [F3] there are $γ > 0$ , $C < \infty$ , depending only on $n$ , $p$ and $[w]_{A_{p}}$ , with $(⟨ w^{1 + γ} ⟩_{Q})^{1 / (1 + γ)} \leq C ⟨ w ⟩_{Q}$ for every cube $Q$ , which is condition (iii).

1.4F4givenalgebra

(iii) implies (ii). Suppose (iii) holds with $γ > 0$ and $C$ . For a cube $Q$ and measurable $E \subseteq Q$ , [F4] gives $\int_{E} w d λ \leq (\int_{Q} w^{1 + γ} d λ)^{1 / (1 + γ)} ∣ E ∣^{γ / (1 + γ)}$ ; substituting $\int_{Q} w^{1 + γ} d λ = ∣ Q ∣ ⟨ w^{1 + γ} ⟩_{Q} \leq C^{1 + γ} w (Q)^{1 + γ} ∣ Q ∣^{- γ}$ (the $(1 + γ)$ -th power of the reverse Hölder inequality, since $⟨ w ⟩_{Q} = w (Q) / ∣ Q ∣$ ) yields $w (E) \leq C w (Q) (∣ E ∣ / ∣ Q ∣)^{γ / (1 + γ)}$ , that is, (ii) with this same $C$ and $δ = γ / (1 + γ)$ .

2.1step 1.1step 1.2step 1.3step 1.4∎

The cycles (i) $\Rightarrow$ (ii) $\Rightarrow$ (i) of steps 1.1 and 1.2 and (i) $\Rightarrow$ (iii) $\Rightarrow$ (ii) of steps 1.3 and 1.4 exhibit each of the three conditions as equivalent to the others; the constants recorded in steps 1.1, 1.2, 1.3 and 1.4 depend only on $n$ and on the constants appearing in the assumed condition, as claimed.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Maximal dyadic cubes covering a proper open set

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ and let $Ω \subseteq R^{n}$ be a nonempty, open and proper subset of $R^{n}$ . Use the all-generations dyadic cubes of Dyadic cubes of all generations in R^n; for a dyadic cube $Q$ and $λ > 0$ let $λ Q$ denote the concentric cube with side length $λ$ times that of $Q$ , and write $ℓ (Q)$ for the side length. Put $F : = {Q dyadic : 5 \sqrt{n} Q \subseteq Ω} .$ Then:

$F$ possesses maximal elements, i.e. cubes of $F$ that are contained in no strictly larger cube of $F$ .
The maximal elements of $F$ are pairwise disjoint, they are at most countable, and their union is exactly $Ω$ .
If $Q$ is a maximal element of $F$ and $P$ is its dyadic parent, then $P \subseteq 3 Q$ and $P \notin F$ , so some $y \in 5 \sqrt{n} P$ satisfies $y \notin Ω$ ; every such $y$ obeys $∣ x - y ∣ \leq 6 n \sqrt{n} ℓ (Q)$ for all $x \in Q$ . In particular $dist (Q, Ω^{c}) \leq 6 n \sqrt{n} ℓ (Q)$ while $5 \sqrt{n} Q \subseteq Ω$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , a nonempty open proper $Ω$ , and the family $F$ of the Statement.

[F1]

A dyadic cube $Q_{k, m}$ with $k \in Z$ , $m \in Z^{n}$ is the half-open box ${x : m_{i} 2^{- k} < x_{i} \leq (m_{i} + 1) 2^{- k} for all i}$ of side $ℓ (Q_{k, m}) = 2^{- k}$ , it contains its centre $c (Q_{k, m}) = ((m_{i} + \frac{1}{2}) 2^{- k})_{i}$ , and two dyadic cubes of generations $k \leq k^{'}$ that meet satisfy $Q_{k^{'}, m^{'}} \subseteq Q_{k, m}$ (Dyadic cubes of all generations in R^n, All-generation dyadic cubes: partition, volume and nesting).

[F2]

A subset of $R^{n}$ is open in the metric topology when every point of it has a Euclidean ball around it contained in it (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space), and the Euclidean, $ℓ^{1}$ and $ℓ^{\infty}$ data satisfy $d_{2} (x, y) = ∥ x - y ∥_{2} \leq ∥ x - y ∥_{1} \leq n d_{\infty} (x, y)$ (The finite and reverse triangle inequalities for a norm; and for $n \geq 1$ every norm $N$ on $R^{n}$ satisfies $N (x) \leq C ∥ x ∥_{1}$ and is Lipschitz, hence continuous, for $d_{2}$ ).

[F3]

The set $Q^{n}$ is at most countable ( $Q^{n}$ is a countable dense subset of $R^{n}$ , and rational open boxes form a countable basis) and a subset of an at most countable set is at most countable (Every subset of an at most countable set is at most countable).

[F4]

For every real $M$ there is a natural number $k$ with $M < k$ (Every complete ordered field is Archimedean).

Proof

technique · direct

1.1F1F2givenchoose

$F$ is nonempty and covers $Ω$ locally: fix $x \in Ω$ ; by [F2] there is $r > 0$ with $B_{2} (x, r) \subseteq Ω$ . Choose $k$ with $3 n \sqrt{n} 2^{- k} < r$ , let $Q$ be the generation- $k$ dyadic cube containing $x$ , and let $y \in 5 \sqrt{n} Q$ . Since $x, y \in 5 \sqrt{n} Q$ and $5 \sqrt{n} Q$ has side $5 \sqrt{n} 2^{- k}$ , [F1] gives $d_{\infty} (x, c (Q)) \leq \frac{1}{2} 2^{- k}$ and $d_{\infty} (c (Q), y) \leq \frac{5}{2} \sqrt{n} 2^{- k}$ , so by [F2] $d_{2} (x, y) \leq d_{2} (x, c (Q)) + d_{2} (c (Q), y) \leq n (\frac{1}{2} + \frac{5}{2} \sqrt{n}) 2^{- k} \leq 3 n \sqrt{n} 2^{- k} < r$ ; hence $y \in B_{2} (x, r) \subseteq Ω$ . Thus $5 \sqrt{n} Q \subseteq Ω$ and $Q \in F$ contains $x$ .

2.1F1F2F4step 1.1algebra

Maximal elements exist: let $Q \in F$ and $w \notin Ω$ , which exists because $Ω$ is proper, and fix $x \in Q$ . If an ancestor $A$ of $Q$ of side $s$ lies in $F$ , then $5 \sqrt{n} A \subseteq Ω$ , so $w \notin 5 \sqrt{n} A$ ; now $x \in Q \subseteq A$ gives $d_{\infty} (x, c (A)) \leq s / 2$ and every point of $5 \sqrt{n} A$ is within $ℓ^{\infty}$ -distance $(5 \sqrt{n} / 2) s$ of $c (A)$ , so [F2] gives $d_{2} (x, c (A)) \leq n s / 2$ and, for $z = w$ first, $d_{2} (w, x) \geq d_{\infty} (w, x) \geq \frac{1}{2} (5 \sqrt{n} - 1) s$ , by the reverse triangle inequality in $d_{\infty}$ ; that is, $s \leq 2 d_{2} (w, x) / (5 \sqrt{n} - 1) = : M$ . The ancestors of $Q$ have side lengths $2^{- j}$ with $j \leq k$ increasing as $j$ decreases, so by [F4] only finitely many of them have side $s \leq M$ ; hence only finitely many ancestors of $Q$ lie in $F$ , and among those finitely many there is one of least generation, which is a maximal element of $F$ containing $Q$ . Taking $Q$ arbitrary shows that every cube of $F$ lies below a maximal element, and in particular maximal elements exist.

3.1F1F3step 2.1algebra

The maximal elements are pairwise disjoint: if $Q, Q^{'}$ are maximal and meet, then by [F1] one contains the other, and maximality forces $Q = Q^{'}$ . They are at most countable: the map sending a dyadic cube to its centre is injective on any family of pairwise disjoint cubes (a cube contains its own centre), its values are points of $Q^{n}$ because $c (Q_{k, m})_{i} = (m_{i} + \frac{1}{2}) 2^{- k} \in Q$ , and $Q^{n}$ is at most countable, so [F3] makes the family at most countable.

3.2step 1.1step 2.1given

Their union is exactly $Ω$ : each maximal element lies in $F$ , hence is contained in $Ω$ , so the union is a subset of $Ω$ ; conversely, for $x \in Ω$ step 1.1 supplies $Q \in F$ with $x \in Q$ , and step 2.1 supplies a maximal element containing $Q$ , hence containing $x$ . Thus $Ω = ⋃ {Q : Q maximal in F}$ .

4.1F1F2step 2.1algebra∎

Let $Q$ be maximal in $F$ and let $P$ be its dyadic parent: $P$ has side $2 ℓ (Q)$ , its centre differs from $c (Q)$ by at most $\frac{1}{2} ℓ (Q)$ in each coordinate, so $P \subseteq 3 Q$ . Maximality gives $P \notin F$ , that is, $5 \sqrt{n} P \subseteq̸ Ω$ , so there is $y \in 5 \sqrt{n} P$ with $y \notin Ω$ ; every point of $P$ is within $ℓ^{\infty}$ -distance $ℓ (Q)$ and every point of $5 \sqrt{n} P$ within $ℓ^{\infty}$ -distance $5 \sqrt{n} ℓ (Q)$ of the centre of $P$ , so [F2] bounds the $d_{2}$ -distance between any $x \in Q \subseteq P$ and $y$ by $n ℓ (Q) + 5 n \sqrt{n} ℓ (Q) \leq 6 n \sqrt{n} ℓ (Q)$ ; in particular $dist (Q, Ω^{c}) \leq 6 n \sqrt{n} ℓ (Q)$ while $5 \sqrt{n} Q \subseteq Ω$ .

Scaffold repair recorded. The scaffolded form of this lemma asked for the dyadic cubes contained in $Ω$ that are maximal under inclusion, with the parent of a maximal cube not contained in $Ω$ . That form is false: for the nonempty open proper set $Ω = (0, \infty)^{n}$ and the all-generations grid, every dyadic cube contained in $Ω$ is contained in a strictly larger ancestor also contained in $Ω$ (the ancestors of the cube $(0, 2^{- k}]^{n}$ are $(0, 2^{- k + 1}]^{n}, \dots$ , all inside $Ω$ ), so maximal elements do not exist at all; with the bounded grid of generations $k \geq 0$ taken instead, a generation- $0$ maximal cube can be at distance far exceeding a multiple of its side length from $Ω^{c}$ , so no point $y \in Ω^{c}$ can be found near it. The version proved above is the Whitney-type statement actually needed by the good- $λ$ estimate: the cubes are maximal in the family adapted to $Ω$ ( $5 \sqrt{n} Q \subseteq Ω$ ), and they retain the near-boundary point $y$ with the uniform bound $∣ x - y ∣ \leq 6 n \sqrt{n} ℓ (Q)$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Dyadic annulus far-field estimates for the maximal function

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $f \in L_{l o c}^{1} (R^{n})$ , $z \in R^{n}$ , $r > 0$ and $δ > 0$ . Then $\int_{∣ t - z ∣ \geq r} ∣ f (t) ∣ ∣ t - z ∣^{- n - δ} d t \leq C_{n, δ} r^{- δ} M f (z),$ where $M$ is the centred Hardy-Littlewood maximal function (The centered and uncentered Hardy-Littlewood maximal functions) and $C_{n, δ} = 2^{2 n} (1 - 2^{- δ})^{- 1}$ depends only on $n$ and $δ$ .

Facts & Assumptions

Given: Countable Choice, $f \in L_{l o c}^{1} (R^{n})$ , $z \in R^{n}$ , $r > 0$ and $δ > 0$ .

[F1]

For every locally integrable $f$ one has $M f (z) = \sup_{ρ > 0} λ (B (z, ρ))^{- 1} \int_{B (z, ρ)} ∣ f ∣ d λ$ , and each average is finite because $f$ is integrable over balls (The centered and uncentered Hardy-Littlewood maximal functions, A locally integrable function on $R^{n}$ ).

[F2]

Every ball $B (x, ρ)$ is Lebesgue measurable with $0 < λ (B (x, ρ)) < \infty$ (Euclidean balls have positive finite Lebesgue measure), the dilates of the unit ball satisfy $λ (B (0, ρ)) = v_{n} ρ^{n}$ with $v_{n} = λ (B (0, 1)) \in (0, \infty)$ (For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it), and a ball is contained in the axis-parallel cube $Q (z, ρ) = \prod_{i} (z_{i} - ρ, z_{i} + ρ)$ of measure $(2 ρ)^{n}$ (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

[F3]

For a sequence of nonnegative measurable functions increasing to $g$ , the integrals increase to $\int g$ (Monotone convergence for the integral).

[F4]

If $A \subseteq B$ are measurable then $\int_{A} ∣ f ∣ \leq \int_{B} ∣ f ∣$ (Measures are monotone).

Proof

technique · Decompose the far field into dyadic annuli, bound each annulus by the maximal function through its measure, and sum the resulting geometric series

1.1F1F2F3given

The sets $A_{k} = {t : 2^{k} r \leq ∣ t - z ∣ < 2^{k + 1} r}$ , $k \geq 0$ , are pairwise disjoint measurable sets whose union is ${∣ t - z ∣ \geq r}$ . Each partial sum $\sum_{k < K} ∣ f ∣ ∣ t - z ∣^{- n - δ} 1_{A_{k}}$ increases with $K$ to $∣ f (t) ∣ ∣ t - z ∣^{- n - δ} 1_{{∣ t - z ∣ \geq r}}$ , so monotone convergence [F3] gives $\int_{∣ t - z ∣ \geq r} ∣ f (t) ∣ ∣ t - z ∣^{- n - δ} d t = \sum_{k \geq 0} \int_{A_{k}} ∣ f (t) ∣ ∣ t - z ∣^{- n - δ} d t$ , and every term is finite because $(2^{k} r)^{- n - δ} \int_{B (z, 2^{k + 1} r)} ∣ f ∣ < \infty$ by [F1] and [F2].

2.1F1F2F4step 1.1algebra

For $t \in A_{k}$ one has $∣ t - z ∣^{- n - δ} \leq (2^{k} r)^{- n - δ}$ , and $A_{k} \subseteq B (z, 2^{k + 1} r) \subseteq Q (z, 2^{k + 1} r)$ , so [F4] and [F2] give $\int_{A_{k}} ∣ f (t) ∣ ∣ t - z ∣^{- n - δ} d t \leq (2^{k} r)^{- n - δ} \int_{B (z, 2^{k + 1} r)} ∣ f ∣ \leq (2^{k} r)^{- n - δ} λ (B (z, 2^{k + 1} r)) M f (z) \leq (2^{k} r)^{- n - δ} (2^{k + 2} r)^{n} M f (z) = 2^{2 n} 2^{- k δ} r^{- δ} M f (z)$ .

3.1step 1.1step 2.1algebra∎

Summing the geometric series in step 2.1 with ratio $2^{- δ} < 1$ gives $\int_{∣ t - z ∣ \geq r} ∣ f (t) ∣ ∣ t - z ∣^{- n - δ} d t \leq 2^{2 n} (1 - 2^{- δ})^{- 1} r^{- δ} M f (z)$ , which is the asserted inequality with $C_{n, δ} = 2^{2 n} (1 - 2^{- δ})^{- 1}$ .

Why the exponent range is $δ > 0$ . The endpoint $δ = 0$ is not available: for the locally integrable function $f = 1_{B (0, R)}$ , the point $z = 0$ and $r = 1$ one has $M f (0) = 1$ , while $\int_{∣ t ∣ \geq 1} ∣ f (t) ∣ ∣ t ∣^{- n} d t = \int_{1 \leq ∣ t ∣ \leq R} ∣ t ∣^{- n} d t = ∣ S^{n - 1} ∣ \log R$ grows without bound as $R \to \infty$ , by Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma and The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t. Hence no constant independent of $R$ can bound that integral by $M f (0)$ ; the divergence of $\sum_{k \geq 0} 2^{- k δ}$ at $δ = 0$ is not removable, and only exponents $δ > 0$ occur in the Hölder estimates for standard kernels used on this page.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Kernel tail integrals of weighted L-p functions are finite

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $1 \leq p < \infty$ , $w \in A_{p}$ (Muckenhoupt A_p and A_1 weights), and let $k : R^{n} ∖ {0} \to C$ be measurable and satisfy the pointwise size bound $∣ k (y) ∣ \leq A_{1} ∣ y ∣^{- n}$ for $y \neq 0$ . Then for every $f \in L^{p} (w)$ , every $x \in R^{n}$ and every $ε > 0$ , $\int_{∣ y ∣ \geq ε} ∣ k (y) ∣ ∣ f (x - y) ∣ d y \leq C (n, p, [w]_{A_{p}}) A_{1} w (Q (x, ε))^{- 1 / p} ∥ f ∥_{L^{p} (w)},$ a finite bound depending only on the stated data and on the cube $Q (x, ε)$ . Consequently the truncated singular integrals $T_{ε} f$ and $T^{(ε, N)} f$ (Maximal truncated singular integrals) are defined at every point by absolutely convergent integrals, and they are Borel measurable functions of the centre $x$ .

Facts & Assumptions

Given: Countable Choice, $1 \leq p < \infty$ , $w \in A_{p}$ , the size bound $∣ k (y) ∣ \leq A_{1} ∣ y ∣^{- n}$ , $f \in L^{p} (w)$ , $x \in R^{n}$ and $ε > 0$ .

[F1]

Write $D_{p} = [w]_{A_{p}}^{1 / p}$ for $p > 1$ and $D_{1} = c_{n} [w]_{A_{1}}$ . Weighted average comparison: for every cube $Q$ and nonnegative measurable $g$ , $⟨ g ⟩_{Q} \leq D_{p} (w (Q)^{- 1} \int_{Q} g^{p} w)^{1 / p}$ , so $\int_{Q} ∣ f ∣ \leq D_{p} ∣ Q ∣ w (Q)^{- 1 / p} ∥ f ∥_{L^{p} (w)}$ (Weighted average comparison and the density-to-mass estimate for A_p weights).

[F2]

Power decay: since $w \in A_{\infty}$ (The Muckenhoupt A_infinity class), there are $C, δ > 0$ , depending only on $n$ and the data of a witnessing exponent, with $w (E) / w (Q) \leq C (∣ E ∣ / ∣ Q ∣)^{δ}$ for measurable $E \subseteq Q$ (A_infinity weights satisfy power decay).

[F3]

$w d λ$ is a locally finite measure and $0 < w (Q) < \infty$ for every cube $Q$ (Weights, their associated measures, and the spaces L^p(w), Muckenhoupt A_p and A_1 weights).

[F4]

Monotone convergence passes through increasing nonnegative sums (Monotone convergence for the integral). For a jointly measurable nonnegative integrand, the integral over a product space may be computed by iterated integrals (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product), and $∥ f ∥_{L^{p} (w)}^{p} = \int ∣ f ∣^{p} w d λ$ (Weights, their associated measures, and the spaces L^p(w)).

[F5]

Translations are norm-continuous in complex $L^{1}$ (Complex translation, convolution, approximate identities, and mollification); dominated convergence applies under an integrable majorant (Dominated convergence).

Proof

technique · direct

1.1F4givenalgebra

Cover ${∣ y ∣ \geq ε}$ by the annuli $A_{j} = {2^{j} ε \leq ∣ y ∣ < 2^{j + 1} ε}$ , $j \geq 0$ , which are pairwise disjoint with union ${∣ y ∣ \geq ε}$ and each contained in the ball $B (x, 2^{j + 2} ε)$ after the substitution $y \mapsto x - y$ . Hence $\int_{∣ y ∣ \geq ε} ∣ k (y) ∣ ∣ f (x - y) ∣ d y \leq \sum_{j \geq 0} (2^{j} ε)^{- n} A_{1} \int_{A_{j}} ∣ f (x - y) ∣ d y \leq \sum_{j \geq 0} (2^{j} ε)^{- n} A_{1} \int_{Q (x, 2^{j + 2} ε)} ∣ f ∣$ by monotone convergence and monotonicity of the integral.

1.2F1F2F3givenalgebra

For each $j \geq 0$ put $Q_{j} : = Q (x, 2^{j + 2} ε)$ and $R : = Q (x, ε)$ . By [F1], $\int_{Q_{j}} ∣ f ∣ \leq D_{p} ∣ Q_{j} ∣ w (Q_{j})^{- 1 / p} ∥ f ∥_{L^{p} (w)}$ , while $(2^{j} ε)^{- n} ∣ Q_{j} ∣ = (2^{j} ε)^{- n} (2^{j + 3} ε)^{n} = 2^{3 n}$ is a dimensional constant. By [F2] applied to $R \subseteq Q_{j}$ , $w (R) / w (Q_{j}) \leq C (∣ R ∣ / ∣ Q_{j} ∣)^{δ} = C 2^{- (j + 2) n δ}$ , so $w (Q_{j})^{- 1 / p} \leq C^{1 / p} 2^{- (j + 2) n δ / p} w (R)^{- 1 / p}$ .

2.1F2step 1.1step 1.2givenalgebra

Substituting the bounds of step 1.2 into step 1.1 gives $\int_{∣ y ∣ \geq ε} ∣ k (y) ∣ ∣ f (x - y) ∣ d y \leq 2^{3 n} D_{p} C^{1 / p} A_{1} ∥ f ∥_{L^{p} (w)} w (R)^{- 1 / p} \sum_{j \geq 0} 2^{- (j + 2) n δ / p}$ , and the geometric series converges because $δ > 0$ ; this is the asserted finite bound with $C (n, p, [w]_{A_{p}}) = 2^{3 n} D_{p} C^{1 / p} (1 - 2^{- n δ / p})^{- 1} 2^{- 2 n δ / p}$ .

3.1F5step 2.1givenalgebra∎

The estimate proves absolute convergence of every truncation. For fixed $0 < ε < N$ , the annular kernel is bounded and compactly supported. Near a fixed $x_{0}$ , truncate $f$ to a bounded ball containing all arguments $x - y$ under consideration, obtaining an $L^{1}$ function $f_{0}$ . Then $∣ T^{(ε, N)} f (x + h) - T^{(ε, N)} f (x) ∣ \leq A_{1} ε^{- n} ∥ f_{0} (\cdot + h) - f_{0} ∥_{1} \to 0$ by [F5]. Thus each finite truncation is continuous and Borel. Dominated convergence gives $T_{ε} f = \lim_{N \to \infty} T^{(ε, N)} f$ , which is Borel. Continuity of the defining integrals in the cutoff radii follows from absolute integrability and null spherical boundaries, so rational cutoffs suffice in both maximal suprema; these maximal functions are Borel as well.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Unweighted local good-lambda estimate for maximal truncations

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ , $0 < δ \leq 1$ , and let $k$ (pointwise size $A_{1}$ , standard $δ$ -Hölder $A_{2}^{'}$ , cancellation $A_{3}$ ), a principal-value distribution $W$ for $k$ , and the associated $L^{2}$ -bounded convolution operator $T$ with norm $B$ be as in the published maximal-truncation theorem, with truncations $T_{ε}$ , $T^{(ε, N)}$ and maximal truncations $T^{*}, T^{* *}$ (Maximal truncated singular integrals, Standard (Hölder) Calderón–Zygmund kernels, Calderón–Zygmund kernels and their associated operators). Let $f \in L_{l o c}^{1} (R^{n})$ be such that $\int_{∣ x - y ∣ \geq ε} ∣ f (y) ∣ ∣ x - y ∣^{- n} d y < \infty$ for every $x$ and $ε > 0$ , so that $T_{ε} f$ , $T^{(ε, N)} f$ , $T^{*} f$ and $T^{* *} f$ are defined at every point, and let $λ > 0$ be such that $Ω_{λ} = {T^{* *} f > λ}$ is a proper open set. Then there are constants $γ_{0} = c_{0} (n, δ) (A_{1} + A_{2}^{'} + A_{3})^{- 1}$ and $C_{n}$ , depending only on $n$ and $δ$ , such that for every $0 < γ < γ_{0}$ , $∣ {T^{* *} f > 2 λ} \cap {M f \leq γ λ} ∣ \leq C_{n} γ (A_{1} + A_{2}^{'} + A_{3} + B) ∣ {T^{* *} f > λ} ∣,$ The same inequality holds with $T^{*}$ throughout whenever ${T^{*} f > λ}$ is a proper open set. More precisely, for either $U = T^{* *}$ or $U = T^{*}$ under its stipulated level-set hypothesis, each Whitney cube $Q_{j}$ used in the proof satisfies $∣ Q_{j} \cap {U f > 2 λ} \cap {M f \leq γ λ} ∣ \leq C_{n} γ (A_{1} + A_{2}^{'} + A_{3} + B) ∣ Q_{j} ∣ .$ If $A_{1} + A_{2}^{'} + A_{3} = 0$ , interpret $γ_{0} = + \infty$ ; then the kernel and both maximal truncations vanish.

Facts & Assumptions

Given: Countable Choice; $n \geq 1$ , $0 < δ \leq 1$ , constants $A_{1}, A_{2}^{'}, A_{3}, B$ ; the kernel $k$ , distribution $W$ , operator $T$ and function $f$ of the Statement; $λ > 0$ with $Ω_{λ}$ a proper open set; $0 < γ < γ_{0}$ with $γ_{0}$ fixed in step 3.1; the centred maximal function $M$ (The centered and uncentered Hardy-Littlewood maximal functions).

[F1]

$∣ k (y) ∣ \leq A_{1} ∣ y ∣^{- n}$ for $y \neq 0$ and $∣ k (x - y) - k (x) ∣ \leq A_{2}^{'} ∣ y ∣^{δ} ∣ x ∣^{- n - δ}$ whenever $∣ x ∣ \geq 2 ∣ y ∣ > 0$ (Maximal truncated singular integrals, Standard (Hölder) Calderón–Zygmund kernels), and for $h \in L^{1} (R^{n})$ and $μ > 0$ one has $∣ {T^{* *} h > μ} ∣ \leq C_{n, δ} (A_{1} + A_{2}^{'} + A_{3} + B) μ^{- 1} ∥ h ∥_{1}$ , with $T^{*} h \leq T^{* *} h \leq 2 T^{*} h$ pointwise (Maximal truncations: weak (1,1) and strong Lp bounds).

[F2]

For a nonempty open proper $Ω \subseteq R^{n}$ the Whitney family of Maximal dyadic cubes covering a proper open set consists of pairwise disjoint dyadic cubes $Q_{j}$ , at most countable with $⋃_{j} Q_{j} = Ω$ , and each $Q_{j}$ comes with $y_{j} \notin Ω$ satisfying $∣ x - y_{j} ∣ \leq 6 n \sqrt{n} ℓ (Q_{j})$ for all $x \in Q_{j}$ .

[F3]

For $g \in L_{l o c}^{1} (R^{n})$ , $z \in R^{n}$ , $r > 0$ and $δ > 0$ , $\int_{∣ t - z ∣ \geq r} ∣ g (t) ∣ ∣ t - z ∣^{- n - δ} d t \leq C_{n, δ} r^{- δ} M g (z)$ (Dyadic annulus far-field estimates for the maximal function).

[F4]

For every cube $Q$ and $z \in Q$ one has $∣ Q ∣^{- 1} \int_{Q} ∣ g ∣ \leq C_{n} M g (z)$ ; in particular if $M g (z) \leq γ λ$ then $\int_{Q} ∣ g ∣ \leq C_{n} ∣ Q ∣ γ λ$ (Ball and cube maximal functions are pointwise comparable).

[F5]

Lebesgue measure (and any measure) is countably additive on pairwise disjoint measurable sets, so for pairwise disjoint measurable sets $A_{j}$ one has $∣ ⋃_{j} A_{j} ∣ = \sum_{j} ∣ A_{j} ∣$ (Countable additivity and continuity of finitely additive set functions).

Proof

technique · direct

1.1F2givenchoosealgebra

Whitney geometry. Treat either $U = T^{* *}$ or $U = T^{*}$ , with its own proper open set $Ω_{λ} = {U f > λ}$ ; if this set is empty the assertion is immediate. Apply [F2] to obtain disjoint cubes $Q_{j}$ covering it and points $y_{j} \notin Ω_{λ}$ with $∣ x - y_{j} ∣ \leq d_{n} ℓ (Q_{j})$ for $x \in Q_{j}$ , where $d_{n} = 6 n \sqrt{n}$ . Put $L_{j} = ℓ (Q_{j})$ , $Λ = 24 n \sqrt{n}$ and $V_{j} = Λ^{2} Q_{j}$ . Choose $z_{j} \in Q_{j}$ with $M f (z_{j}) \leq γ λ$ whenever such a point exists. For $t \notin V_{j}$ and $x, z_{j} \in Q_{j}$ , $∣ t - y_{j} ∣ \geq (Λ^{2} / 2 - d_{n} - 1 / 2) L_{j} = : b_{n} L_{j}$ , with $b_{n} > 2 d_{n}$ ; the distances $∣ t - x ∣, ∣ t - y_{j} ∣, ∣ t - z_{j} ∣$ are mutually comparable. Also $V_{j} \subseteq B (y_{j}, R_{j})$ with $R_{j} \leq C_{n} L_{j}$ . These follow from coordinate bounds and the triangle inequality.

2.1F1F4step 1.1givenalgebra

Local part. Put $u_{j} = f 1_{V_{j}}$ and $g_{j} = f - u_{j}$ . By [F4], $∥ u_{j} ∥_{1} \leq C_{n} ∣ V_{j} ∣ M f (z_{j}) \leq C_{n} γ λ ∣ Q_{j} ∣$ . The weak $(1, 1)$ estimate [F1] for either maximal truncation therefore gives $∣ {U u_{j} > λ / 2} ∣ \leq C_{n, δ} (A_{1} + A_{2}^{'} + A_{3} + B) γ ∣ Q_{j} ∣$ . Cubes with no $z_{j}$ contribute nothing to the target set.

2.2F1F3step 1.1givenalgebra

Uniform difference of far truncations. Fix $x \in Q_{j}$ and $y = y_{j}$ . On the common part of the cutoff domains, Hölder smoothness bounds the integral of the kernel difference by $A_{2}^{'} ∣ x - y ∣^{δ} \int_{V_{j}^{c}} ∣ f (t) ∣ ∣ t - y ∣^{- n - δ} d t \leq C_{n, δ} A_{2}^{'} M f (z_{j})$ : compare $∣ t - y ∣$ with $∣ t - z_{j} ∣$ , use $∣ t - z_{j} ∣ \geq c_{n} L_{j}$ , and apply [F3] at $z_{j}$ . For each cutoff radius $a$ (the lower radius $ε$ and, for double truncations, also the upper radius $N$ ), a mismatch lies where one of $∣ t - x ∣, ∣ t - y ∣$ is at most $a$ and the other is greater than $a$ . Their difference is at most $∣ x - y ∣$ . If the mismatch meets $V_{j}^{c}$ , the geometry implies $a \geq c_{n} L_{j}$ , both distances are comparable to $a$ , and $∣ t - z_{j} ∣ \leq C_{n} a$ . Thus the kernel contributing on either mismatch is at most $C_{n} A_{1} a^{- n}$ and its integral is bounded by $C_{n} A_{1} a^{- n} \int_{B (z_{j}, C_{n} a)} ∣ f ∣ \leq C_{n} A_{1} M f (z_{j})$ . There are at most four mismatch pieces. It follows uniformly in all admissible cutoffs that the two far truncations at $x$ and $y_{j}$ differ by at most $C_{n, δ} (A_{1} + A_{2}^{'}) M f (z_{j})$ . For $T^{*}$ use only the lower cutoff.

2.3F1F4step 1.1givenalgebra

Far truncations at the boundary point. Choose $R_{j}$ with $V_{j} \subseteq B (y_{j}, R_{j})$ and $R_{j} \leq C_{n} L_{j}$ . If $ε \geq R_{j}$ , the far truncation equals the corresponding truncation of $f$ , bounded by $U f (y_{j}) \leq λ$ . If $ε < R_{j} < N$ , split the far integral at $R_{j}$ ; the part beyond $R_{j}$ is the corresponding truncation of $f$ and is at most $λ$ , while the part below $R_{j}$ is bounded by $A_{1} (b_{n} L_{j})^{- n} \int_{B (z_{j}, C_{n} L_{j})} ∣ f ∣ \leq C_{n} A_{1} M f (z_{j})$ because $g_{j}$ vanishes within distance $b_{n} L_{j}$ of $y_{j}$ . If $N \leq R_{j}$ , only this latter bound is needed. For single truncations the same proof uses $N = \infty$ . Consequently $U g_{j} (y_{j}) \leq λ + C_{n} A_{1} M f (z_{j})$ .

3.1F5step 2.1step 2.2step 2.3givenalgebra∎

Combining steps 2.2 and 2.3 gives $U g_{j} (x) \leq λ + C_{n, δ} (A_{1} + A_{2}^{'}) γ λ$ on $Q_{j}$ . Choose the dimensional $c_{0}$ in $γ_{0} = c_{0} / (A_{1} + A_{2}^{'} + A_{3})$ so that this is at most $3 λ / 2$ for $γ < γ_{0}$ . Subadditivity then implies $Q_{j} \cap {U f > 2 λ} \cap {M f \leq γ λ} \subseteq {U u_{j} > λ / 2}$ . Apply step 2.1 and sum over the disjoint Whitney cubes using [F5] to obtain $∣ {U f > 2 λ} \cap {M f \leq γ λ} ∣ \leq C_{n, δ} γ (A_{1} + A_{2}^{'} + A_{3} + B) ∣ {U f > λ} ∣$ . The proof applies separately to both $U$ , establishing the two asserted estimates.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Weighted good-lambda inequality for maximal truncations

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $w \in A_{\infty}$ (The Muckenhoupt A_infinity class), let $T$ be a standard-kernel Calderón–Zygmund operator as in the unweighted local estimate, let $λ > 0$ , and let $f \in L_{l o c}^{1} (R^{n})$ with the defining finiteness property and with ${T^{* *} f > λ}$ a proper open set. Then there are $γ_{0} > 0$ , $C < \infty$ and $δ^{'} > 0$ , depending only on $n$ , the Hölder exponent $δ$ , a witnessing finite exponent $p$ and $[w]_{A_{p}}$ and the constants $A_{1}, A_{2}^{'}, A_{3}, B$ , such that for every $0 < γ < γ_{0}$ , $w ({T^{* *} f > 2 λ} \cap {M f \leq γ λ}) \leq C γ^{δ^{'}} w ({T^{* *} f > λ}) .$ The same holds with $T^{*}$ throughout when its own level set ${T^{*} f > λ}$ is a proper open set.

Facts & Assumptions

Given: Countable Choice, $w \in A_{\infty}$ , the operator data $A_{1}, A_{2}^{'}, A_{3}, B, δ$ , the function $f$ and the height $λ$ .

[F1]

Unweighted local estimate: with $Ω_{λ} = {T^{* *} f > λ}$ decomposed into the Whitney cubes $Q_{j}$ , for $γ < γ_{0}^{u n w}$ the set $E_{j} : = Q_{j} \cap {T^{* *} f > 2 λ} \cap {M f \leq γ λ}$ satisfies $∣ E_{j} ∣ \leq C_{n} γ (A_{1} + A_{2}^{'} + A_{3} + B) ∣ Q_{j} ∣$ ; the cubes $Q_{j}$ are pairwise disjoint with union $Ω_{λ}$ (Unweighted local good-lambda estimate for maximal truncations).

[F2]

Power decay: since $w \in A_{\infty}$ , there are $C_{w}, η > 0$ , depending only on $n$ and the $A_{\infty}$ data, such that $w (E) / w (Q) \leq C_{w} (∣ E ∣ / ∣ Q ∣)^{η}$ for every cube $Q$ and measurable $E \subseteq Q$ (A_infinity weights satisfy power decay); in particular $0 < w (Q) < \infty$ for every cube.

[F3]

Lebesgue measure and $w d λ$ are countably additive on pairwise disjoint measurable sets (Countable additivity and continuity of finitely additive set functions).

Proof

technique · direct

1.1F2givenalgebra

Keep the Whitney decomposition of $Ω_{λ}$ supplied by [F1] and let $E_{j}$ be as there. Each $E_{j}$ is a measurable subset of $Q_{j}$ with $∣ E_{j} ∣ \leq (C_{n} γ (A_{1} + A_{2}^{'} + A_{3} + B)) ∣ Q_{j} ∣$ , so the power decay [F2] gives $w (E_{j}) \leq C_{w} (C_{n} γ (A_{1} + A_{2}^{'} + A_{3} + B))^{η} w (Q_{j})$ .

2.1F1F3step 1.1givenalgebra

Summing step 1.1 over the pairwise disjoint $Q_{j}$ , whose union is $Ω_{λ}$ by [F1], and using countable additivity [F3]: $w ({T^{* *} f > 2 λ} \cap {M f \leq γ λ}) = \sum_{j} w (E_{j}) \leq C_{w} C_{n}^{η} (A_{1} + A_{2}^{'} + A_{3} + B)^{η} γ^{η} \sum_{j} w (Q_{j}) = C_{w} C_{n}^{η} (A_{1} + A_{2}^{'} + A_{3} + B)^{η} γ^{η} w (Ω_{λ})$ .

3.1F1step 2.1givenalgebra∎

The display of step 2.1 is the claimed inequality with $δ^{'} = η$ and $C = C_{w} C_{n}^{η} (A_{1} + A_{2}^{'} + A_{3} + B)^{η}$ , valid for every $0 < γ < γ_{0}$ with $γ_{0} = γ_{0}^{u n w}$ from [F1]; the same argument applies with $T^{*}$ in place of $T^{* *}$ by applying the unweighted lemma to its own level set and Whitney decomposition.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Weighted L-p bounds for standard Calderon-Zygmund maximal truncations

Statement

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain); this supplies Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Dependent choice implies countable choice), which the published truncation definitions use. Let $1 < p < \infty$ , let $w \in A_{p}$ (Muckenhoupt A_p and A_1 weights), and let $k$ (pointwise size $A_{1}$ , standard $δ$ -Hölder $A_{2}^{'}$ , cancellation $A_{3}$ ), a principal-value distribution $W$ for $k$ , and the associated $L^{2}$ -bounded convolution operator $T$ with norm $B$ be as in the published maximal-truncation theorem (Maximal truncations: weak (1,1) and strong Lp bounds, Maximal truncated singular integrals, Standard (Hölder) Calderón–Zygmund kernels, Calderón–Zygmund kernels and their associated operators). If $A_{1} + A_{2}^{'} + A_{3} + B = 0$ , the kernel and both maximal truncations vanish; otherwise the constants below are obtained by the indicated homogeneous bounds. Then for every $f \in L^{p} (w)$ the maximal truncations are finite almost everywhere and $∥ T^{* *} f ∥_{L^{p} (w)} \leq C (A_{1} + A_{2}^{'} + A_{3} + B) ∥ f ∥_{L^{p} (w)}, C = C (n, p, δ, [w]_{A_{p}}),$ and the same bound holds for $T^{*}$ . Moreover, for $w \in A_{1}$ the weak type $(1, 1)$ bound (Sublinear operators and weak or strong type $(p, q)$ bounds) $w ({T^{* *} f > λ}) \leq C (A_{1} + A_{2}^{'} + A_{3} + B) λ^{- 1} ∥ f ∥_{L^{1} (w)}$ holds with $C = C (n, δ, [w]_{A_{1}})$ . Finally, if the principal-value truncations $T_{ε} f$ converge almost everywhere to a measurable $T f$ for every $f$ in a dense subspace of $L^{p} (w)$ , then the limit exists almost everywhere for every $f \in L^{p} (w)$ and satisfies the same $L^{p} (w)$ bound.

Facts & Assumptions

Given: Dependent Choice; the kernel data $A_{1}, A_{2}^{'}, A_{3}, B, δ$ ; a weight $w$ ; $1 < p < \infty$ ; and, when the weak endpoint is treated, $w \in A_{1}$ .

[F1]

For every $f \in L^{p} (w)$ , every $x$ and every $ε > 0$ the truncated integrals are defined by absolutely convergent integrals, $\int_{∣ y ∣ \geq ε} ∣ k (y) ∣ ∣ f (x - y) ∣ d y \leq C (n, p, [w]_{A_{p}}) A_{1} w (Q (x, ε))^{- 1 / p} ∥ f ∥_{L^{p} (w)}$ , and the maximal truncations are Borel measurable functions of the centre. Applying the same lemma to the auxiliary kernel $∣ y ∣^{- n}$ also establishes $\int_{∣ y ∣ \geq ε} ∣ f (x - y) ∣ ∣ y ∣^{- n} d y < \infty$ , the finiteness hypothesis of the good- $λ$ lemma (Kernel tail integrals of weighted L-p functions are finite).

[F2]

Quantitative weighted good- $λ$ . Put $S = A_{1} + A_{2}^{'} + A_{3} + B$ . When $A_{1} + A_{2}^{'} + A_{3} > 0$ , the local estimate of Unweighted local good-lambda estimate for maximal truncations gives, in each Whitney cube $Q_{j}$ , $∣ E_{j} ∣ \leq C_{n, δ} γ S ∣ Q_{j} ∣$ for $0 < γ < γ_{0} = c_{0} (n, δ) / (A_{1} + A_{2}^{'} + A_{3})$ . For $w \in A_{\infty}$ (The Muckenhoupt A_infinity class), A_infinity weights satisfy power decay gives $w (E_{j}) \leq C_{w} (∣ E_{j} ∣ / ∣ Q_{j} ∣)^{η} w (Q_{j})$ , with $C_{w}, η > 0$ depending only on $n$ and a witnessing exponent and characteristic. Summing over the disjoint Whitney cubes, as in Weighted good-lambda inequality for maximal truncations, yields the good- $λ$ bound with $δ^{'} = η$ and $C_{1} = C_{w} C_{n, δ}^{η} S^{η}$ . This applies to $T^{* *}$ when its level set is proper and open, and separately to $T^{*}$ under the corresponding hypothesis. If $A_{1} + A_{2}^{'} + A_{3} = 0$ , the kernel and its truncations vanish and no absorption is needed.

[F3]

Maximal-function bounds: for $w \in A_{p}$ and $f \in L^{p} (w)$ , $∥ M f ∥_{L^{p} (w)} \leq C_{n, p} [w]_{A_{p}}^{1 / (p - 1)} ∥ f ∥_{L^{p} (w)}$ (The Hardy-Littlewood maximal operator characterises A_p); for $w \in A_{1}$ and $f \in L^{1} (w)$ , $w ({M f > λ}) \leq 5^{n} [w]_{A_{1}} λ^{- 1} ∥ f ∥_{L^{1} (w)}$ (Weighted weak (1,1) bound for the maximal function under A_1); and $M (1_{B (0, 1)}) (x) \geq 2^{- n} (1 + ∣ x ∣)^{- n}$ for every $x$ , because $B (0, 1) \subseteq B (x, 2 (1 + ∣ x ∣))$ (The centered and uncentered Hardy-Littlewood maximal functions).

[F4]

Layer cake: for measurable $g \geq 0$ and $0 < p < \infty$ , $\int g^{p} w d λ = \int_{0}^{\infty} p λ^{p - 1} w ({g > λ}) d λ$ , both sides allowed to be $+ \infty$ ; Fatou's lemma holds for nonnegative measurable functions (For 0 < p < infinity, the layer-cake formula computes the integral of |f|^p from the distribution function, Fatou's lemma).

[F5]

$w d λ$ is Radon. Under DC, $C_{c}$ is dense in real $L^{p} (w)$ by C_c(X) is dense in L^p(mu) for a Radon measure; apply it to each component for complex inputs. To obtain smooth density, choose a nonnegative smooth bump equal to one on a small ball and supported in a larger ball by A smooth bump between concentric Euclidean balls, and divide by its positive finite Lebesgue integral to get a unit-mass $ρ$ . For each $h \in C_{c}$ , the functions $h * ρ_{ε}$ are smooth, compactly supported in one fixed bounded ball for $ε \leq 1$ , and converge uniformly to $h$ (Complex translation, convolution, approximate identities, and mollification, Statement and Proof 1.3–1.4, 2.2, 5.1). Their $L^{p} (w)$ error is at most the uniform error times the finite $w$ -measure of that ball to the power $1 / p$ . Thus $C_{c}^{\infty}$ is dense for every finite $p$ , including $p = 1$ .

[F6]

Chebyshev's inequality and the dominated convergence theorem (Chebyshev-Markov inequality for the integral, Dominated convergence).

[F7]

$T_{ε}$ and $T^{(ε, N)}$ are the truncations, $T^{*} = \sup_{ε > 0} ∣ T_{ε} ∣$ , $T^{* *} = \sup_{0 < ε < N} ∣ T^{(ε, N)} ∣$ , and $T^{*} \leq T^{* *} \leq 2 T^{*}$ pointwise; the kernel obeys $∣ k (y) ∣ \leq A_{1} ∣ y ∣^{- n}$ and the cancellation bound $\sup_{0 < r < R} ∣ \int_{r < ∣ y ∣ < R} k (y) d y ∣ \leq A_{3}$ ; and $T$ is the convolution operator with $W$ , $L^{2}$ -bounded with norm $B$ , satisfying the off-support representation with kernel $k$ (Maximal truncated singular integrals, Standard (Hölder) Calderón–Zygmund kernels, Calderón–Zygmund kernels and their associated operators).

[F8]

Polar coordinates integrate radial nonnegative kernels (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma). Applying the one-variable mean value theorem along coordinate line segments, componentwise, gives $∣ g (x - z) - g (x) ∣ \leq L_{g} ∣ z ∣$ for a smooth compactly supported $g$ and a finite $L_{g}$ determined by its bounded first derivatives (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

Proof

technique · direct

1.1F1F3F6F7F8givenalgebra

A priori finiteness on the dense class. Fix $g \in C_{c}^{\infty}$ supported in $B (0, R_{g})$ . In a double truncation, split at radius one. On the part below one, subtract $g (x)$ : [F8] bounds the resulting integral by $A_{1} L_{g} \int_{∣ z ∣ < 1} ∣ z ∣^{1 - n} d z = A_{1} L_{g} ∣ S^{n - 1} ∣$ , while cancellation bounds the constant term by $A_{3} ∥ g ∥_{\infty}$ . On the part above one, use the original integrand and $∣ k (z) ∣ \leq A_{1}$ to obtain the bound $A_{1} ∥ g ∥_{1}$ . Hence $T^{* *} g$ is uniformly bounded. If $∣ x ∣ > 2 R_{g} + 1$ , the original size estimate gives $T^{* *} g (x) \leq 2^{n} A_{1} ∥ g ∥_{1} ∣ x ∣^{- n}$ , so $T^{* *} g \leq C_{g} (1 + ∣ x ∣)^{- n}$ . Each finite truncation is continuous by dominated convergence, so its supremum has open level sets; the decay makes these bounded and proper. By [F3], $T^{* *} g \leq 2^{n} C_{g} M (1_{B (0, 1)})$ . The maximal bounds in [F3] therefore give finite $L^{p} (w)$ norm for $w \in A_{p}$ , $p > 1$ , and finite weak $L^{1} (w)$ norm for $w \in A_{1}$ .

2.1F2F3F4step 1.1givenalgebra

Fix $p > 1$ , $w \in A_{p}$ and $g \in C_{c}^{\infty}$ , with $A_{1} + A_{2}^{'} + A_{3} > 0$ ; if this sum vanishes the truncations are zero and the strong bound is immediate. Write $D (μ) = w ({T^{* *} g > μ})$ . For every $0 < γ < γ_{0}$ , splitting the level set and applying [F2] gives $D (2 μ) \leq C_{1} γ^{δ^{'}} D (μ) + w ({M g > γ μ})$ . By layer cake [F4], $∥ T^{* *} g ∥_{L^{p} (w)}^{p} = 2^{p} \int_{0}^{\infty} p μ^{p - 1} D (2 μ) d μ$ . The second distribution term integrates exactly to $γ^{- p} ∥ M g ∥_{L^{p} (w)}^{p}$ by the substitution $t = γ μ$ ; it is finite by [F3].

2.2F2F3step 1.1givenalgebra

Weak endpoint on the dense class. Let $w \in A_{1}$ and $g \in C_{c}^{\infty}$ . If $A_{1} + A_{2}^{'} + A_{3} = 0$ the kernel vanishes and the bound is immediate. Otherwise, since $A_{1} \subseteq A_{\infty}$ , [F2] applies; for every $λ > 0$ the same splitting with the weak bound of [F3] in place of the strong one gives $w ({T^{* *} g > 2 λ}) \leq C_{1} γ^{δ^{'}} w ({T^{* *} g > λ}) + 5^{n} [w]_{A_{1}} (γ λ)^{- 1} ∥ g ∥_{L^{1} (w)}$ . Multiplying by $2 λ$ , taking the supremum over $λ > 0$ (finite by step 1.1) and choosing $γ = \frac{1}{2} \min {γ_{0}, (4 C_{1})^{- 1 / δ^{'}}}$ gives $2 C_{1} γ^{δ^{'}} \leq \frac{1}{2}$ and $γ^{- 1} \leq C (n, δ, [w]_{A_{1}}) S$ , and yields $∥ T^{* *} g ∥_{L^{1, \infty} (w)} \leq C (n, δ, [w]_{A_{1}}) (A_{1} + A_{2}^{'} + A_{3} + B) ∥ g ∥_{L^{1} (w)}$ ; the weak $(1, 1)$ bound for $T^{*}$ follows from $T^{*} \leq T^{* *}$ .

3.1F2F3F7step 1.1step 2.1givenalgebra

If $A_{1} + A_{2}^{'} + A_{3} = 0$ , both maximal truncations vanish and the bounds are immediate. Otherwise, substituting step 2.1 into the layer-cake identity and using the finiteness of $∥ T^{* *} g ∥_{L^{p} (w)}$ from step 1.1 gives $∥ T^{* *} g ∥_{L^{p} (w)}^{p} \leq 2^{p} C_{1} γ^{δ^{'}} ∥ T^{* *} g ∥_{L^{p} (w)}^{p} + 2^{p} (C_{n, p} [w]_{A_{p}}^{1 / (p - 1)})^{p} γ^{- p} ∥ g ∥_{L^{p} (w)}^{p}$ ; choosing $γ = \frac{1}{2} \min {γ_{0}, (2^{p + 1} C_{1})^{- 1 / δ^{'}}}$ gives $2^{p} C_{1} γ^{δ^{'}} \leq \frac{1}{2}$ , and since $γ_{0} = c_{0} (A_{1} + A_{2}^{'} + A_{3})^{- 1}$ and $C_{1} \leq C_{w} C_{n}^{δ^{'}} (A_{1} + A_{2}^{'} + A_{3} + B)^{δ^{'}}$ one has $γ^{- 1} \leq C^{'} (n, p, δ, [w]_{A_{p}}) (A_{1} + A_{2}^{'} + A_{3} + B)$ ; hence $∥ T^{* *} g ∥_{L^{p} (w)} \leq C (n, p, δ, [w]_{A_{p}}) (A_{1} + A_{2}^{'} + A_{3} + B) ∥ g ∥_{L^{p} (w)}$ , and $T^{*}$ inherits the bound because $T^{*} \leq T^{* *} \leq 2 T^{*}$ by [F7].

4.1F1F4F5step 3.1step 2.2givenalgebra

Extension to $f \in L^{p} (w)$ . Let $w \in A_{p}$ , $1 < p < \infty$ , $f \in L^{p} (w)$ , and choose $g_{m} \in C_{c}^{\infty}$ with $∥ g_{m} - f ∥_{L^{p} (w)} \to 0$ by [F5]. For every $x$ and every fixed pair $0 < ε < N$ , [F1] gives $∣ T^{(ε, N)} f (x) - T^{(ε, N)} g_{m} (x) ∣ \leq C (n, p, [w]_{A_{p}}) A_{1} w (Q (x, ε))^{- 1 / p} ∥ f - g_{m} ∥_{L^{p} (w)} \to 0$ , so $∣ T^{(ε, N)} f (x) ∣ = \lim_{m} ∣ T^{(ε, N)} g_{m} (x) ∣ \leq {lim inf}_{m} T^{* *} g_{m} (x)$ for each fixed pair and, taking the supremum over all pairs, $T^{* *} f \leq {lim inf}_{m} T^{* *} g_{m}$ pointwise. Fatou's lemma [F4] and step 3.1 then give $∥ T^{* *} f ∥_{L^{p} (w)}^{p} \leq {lim inf}_{m} ∥ T^{* *} g_{m} ∥_{L^{p} (w)}^{p} \leq C (n, p, δ, [w]_{A_{p}})^{p} (A_{1} + A_{2}^{'} + A_{3} + B)^{p} ∥ f ∥_{L^{p} (w)}^{p}$ , in particular $T^{* *} f < \infty$ $w$ -almost everywhere, and $T^{*} f \leq T^{* *} f \leq 2 T^{*} f$ carries the same conclusions. If instead $w \in A_{1}$ and $f \in L^{1} (w)$ , the same approximation gives $T^{* *} f \leq {lim inf}_{m} T^{* *} g_{m}$ pointwise, hence ${T^{* *} f > λ} \subseteq {lim inf}_{m} {T^{* *} g_{m} > λ}$ for every $λ > 0$ and $w ({T^{* *} f > λ}) \leq {lim inf}_{m} w ({T^{* *} g_{m} > λ}) \leq C (n, δ, [w]_{A_{1}}) (A_{1} + A_{2}^{'} + A_{3} + B) λ^{- 1} ∥ f ∥_{L^{1} (w)}$ by step 2.2, so again $T^{* *} f < \infty$ $w$ -a.e.; $T^{*}$ inherits both bounds by the pointwise comparison.

5.1F1F6step 4.1givenalgebra∎

Final clause. Let $D \subseteq L^{p} (w)$ be a dense subspace on which $T_{ε}$ converges almost everywhere as $ε ↓ 0$ , and let $f \in L^{p} (w)$ . For $g \in D$ and $ε, ε^{'} > 0$ one has $∣ T_{ε} f - T_{ε^{'}} f ∣ \leq 2 T^{*} (f - g) + ∣ T_{ε} g - T_{ε^{'}} g ∣$ pointwise, so for every $η > 0$ the set where ${lim sup}_{ε, ε^{'} ↓ 0} ∣ T_{ε} f - T_{ε^{'}} f ∣ > 4 η$ has $w$ -measure at most $w ({T^{*} (f - g) > η}) \leq η^{- p} ∥ T^{*} (f - g) ∥_{L^{p} (w)}^{p} \leq C (n, p, δ, [w]_{A_{p}})^{p} (A_{1} + A_{2}^{'} + A_{3} + B)^{p} η^{- p} ∥ f - g ∥_{L^{p} (w)}^{p}$ by Chebyshev [F6] and step 4.1. For every $η > 0$ , take the infimum of this bound over $g \in D$ ; density makes the infimum zero, so the exceptional set is null; intersecting the resulting full-measure sets over $η = 1 / m$ , $m \geq 1$ , shows that $(T_{ε} f)$ is $w$ -almost everywhere Cauchy as $ε ↓ 0$ , so the limit $T f$ exists $w$ -a.e. and is measurable as an a.e. limit of measurable functions [F1]. It obeys $∣ T f ∣ \leq T^{*} f$ pointwise, hence $∥ T f ∥_{L^{p} (w)} \leq ∥ T^{*} f ∥_{L^{p} (w)} \leq C (n, p, δ, [w]_{A_{p}}) (A_{1} + A_{2}^{'} + A_{3} + B) ∥ f ∥_{L^{p} (w)}$ by step 4.1.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Hilbert and Riesz transforms are bounded on weighted L-p

Statement

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain); this supplies Countable Choice (Dependent choice implies countable choice) for the truncation definitions. Let $1 < p < \infty$ and $w \in A_{p}$ (Muckenhoupt A_p and A_1 weights). Then the Hilbert transform $H$ on $R$ (Truncated Hilbert transform and principal value) and each Riesz transform $R_{j}$ on $R^{n}$ (Riesz transforms on Euclidean space) extend boundedly to $L^{p} (w)$ (Weights, their associated measures, and the spaces L^p(w)), with norms bounded by $C (n, p, [w]_{A_{p}}) (A_{1} + A_{2}^{'} + A_{3} + B)$ , where the kernel constants are those recorded for the kernels $1 / (π x)$ and $c_{n} x_{j} / ∣ x ∣^{n + 1}$ : for the Hilbert kernel $A_{1} = 1 / π$ , $A_{2}^{'} = 2 / π$ , $A_{3} = 0$ and $B = 1$ , and for each Riesz kernel $A_{1} = c_{n}$ , $A_{2}^{'} = c_{n} 2^{n + 1} (3 n + 4)$ , $A_{3} = 0$ and $B = 1$ . The maximal truncations of these transforms obey the same bound, and for $w \in A_{1}$ they satisfy the weighted weak $(1, 1)$ estimate with the same structure of constants.

Facts & Assumptions

Given: Dependent Choice; $1 < p < \infty$ ; $w \in A_{p}$ , and in the endpoint discussion $w \in A_{1}$ ; the Hilbert transform $H$ on $R$ and the Riesz transforms $R_{j}$ , $1 \leq j \leq n$ , on $R^{n}$ .

[F1]

The Hilbert kernel $k (x) = 1 / (π x)$ satisfies $∣ k (x) ∣ = 1 / (π ∣ x ∣)$ for $x \neq 0$ and $∣ k (x - y) - k (x) ∣ \leq (2 / π) ∣ y ∣ ∣ x ∣^{- 2}$ whenever $∣ x ∣ \geq 2 ∣ y ∣ > 0$ , is odd, and is recorded as a standard $1$ -Hölder Calderón–Zygmund kernel with these constants; $H$ is the $L^{2}$ -bounded convolution operator with norm one for the principal-value distribution $W = p v 1 / (π x)$ , and the truncations $H_{ε} g$ converge at every point of every Schwartz input to the corresponding value of the $L^{2}$ class (The Hilbert transform is bounded on Lp, The Hilbert transform is the tempered convolution with pv(1/(pi x)) and has signum Fourier multiplier, Truncated Hilbert transform and principal value).

[F2]

Each Riesz kernel $K_{j} (x) = c_{n} x_{j} / ∣ x ∣^{n + 1}$ satisfies $∣ K_{j} (x) ∣ \leq c_{n} ∣ x ∣^{- n}$ , $∣ K_{j} (x - h) - K_{j} (x) ∣ \leq C_{n} ∣ h ∣ ∣ x ∣^{- (n + 1)}$ with $C_{n} = c_{n} 2^{n + 1} (3 n + 4)$ whenever $∣ h ∣ \leq ∣ x ∣ / 2$ , and $\int_{S^{n - 1}} K_{j} (r ω) d σ (ω) = 0$ for every $r > 0$ ; $R_{j}$ is the $L^{2}$ -bounded Fourier multiplier with symbol $- i ξ_{j} / ∣ ξ ∣$ and operator norm at most one, and its truncations converge at every point of every Schwartz input to the corresponding value of the $L^{2}$ class (Riesz kernel size, difference and spherical-cancellation bounds, The Riesz transforms are bounded on Lp, The Riesz transform is the principal value of its kernel, with the matching constant, Riesz transforms on Euclidean space).

[F3]

Weighted Calderón–Zygmund theorem: under the standing Dependent Choice hypothesis, for $1 < p < \infty$ , $w \in A_{p}$ , and any kernel data $A_{1}, A_{2}^{'}, A_{3}, B$ as in that theorem, every $f \in L^{p} (w)$ has $∥ T^{* *} f ∥_{L^{p} (w)} \leq C (n, p, δ, [w]_{A_{p}}) (A_{1} + A_{2}^{'} + A_{3} + B) ∥ f ∥_{L^{p} (w)}$ with the same bound for $T^{*}$ ; if $w \in A_{1}$ the weak $(1, 1)$ bound $w ({T^{* *} f > λ}) \leq C (n, δ, [w]_{A_{1}}) (A_{1} + A_{2}^{'} + A_{3} + B) λ^{- 1} ∥ f ∥_{L^{1} (w)}$ holds; and if the truncations converge almost everywhere to a measurable limit on a dense subspace of $L^{p} (w)$ , then the limit exists almost everywhere for every $f \in L^{p} (w)$ and satisfies the same $L^{p} (w)$ bound (Weighted L-p bounds for standard Calderon-Zygmund maximal truncations).

[F4]

$w d λ$ is a Radon measure and $C_{c}^{\infty} (R^{n})$ is dense in $L^{p} (w d λ)$ for $1 \leq p < \infty$ under Dependent Choice (Weights, their associated measures, and the spaces L^p(w), Weighted L-p bounds for standard Calderon-Zygmund maximal truncations, Facts [F5], which combines Radon $C_{c}$ density with uniform compact-support smoothing).

[F5]

The principal-value truncations of the Hilbert and Riesz transforms converge almost everywhere on the dense class $S (R^{n})$ of Schwartz functions; indeed the published convergence theorem applies to these kernels with that dense class (Almost-everywhere convergence of principal-value truncations).

Proof

technique · direct

1.1F1F2givenalgebra

Kernel data in normalized form. For the Hilbert kernel, $∣ k (x) ∣ = π^{- 1} ∣ x ∣^{- 1}$ gives the pointwise size constant $A_{1} = 1 / π$ , the difference estimate of [F1] is the standard $1$ -Hölder condition with $A_{2}^{'} = 2 / π$ , and oddness gives $\int_{r < ∣ x ∣ < R} k (x) d x = 0$ for all $0 < r < R$ , so the cancellation constant is $A_{3} = 0$ ; the $L^{2}$ norm bound is $B = 1$ , and $H$ is the convolution operator with the principal-value distribution of [F1] satisfying the off-support representation with kernel $k$ . For each Riesz kernel, [F2] gives the size constant $A_{1} = c_{n}$ , the standard $1$ -Hölder constant $A_{2}^{'} = C_{n} = c_{n} 2^{n + 1} (3 n + 4)$ , vanishing annulus integrals and hence $A_{3} = 0$ , and $B = 1$ , together with the principal value on Schwartz functions. Hence both families meet the hypotheses of the weighted theorem [F3] with the constants displayed in the statement.

1.2F1F2F3givenalgebra

Weighted bounds for the maximal truncations. By [F3] applied to the Hilbert kernel and to each Riesz kernel: for $f \in L^{p} (w)$ one has $∥ H^{* *} f ∥_{L^{p} (w)} \leq C (n, p, [w]_{A_{p}}) (1 / π + 2 / π + 0 + 1) ∥ f ∥_{L^{p} (w)}$ and $∥ R_{j}^{* *} f ∥_{L^{p} (w)} \leq C (n, p, [w]_{A_{p}}) (c_{n} + C_{n} + 0 + 1) ∥ f ∥_{L^{p} (w)}$ , with the same bounds for $H^{*}$ and $R_{j}^{*}$ since $T^{*} \leq T^{* *} \leq 2 T^{*}$ ; for $w \in A_{1}$ the theorem gives the weighted weak $(1, 1)$ bounds for the maximal truncations with the corresponding constants.

2.1F3F4F5step 1.2givenalgebra∎

Almost-everywhere convergence and the bounded extension. Let $1 < p < \infty$ and $w \in A_{p}$ . The subspace $C_{c}^{\infty} (R^{n})$ is dense in $L^{p} (w)$ by [F4], and for every $g \in C_{c}^{\infty} (R^{n})$ — a Schwartz function — the truncations converge almost everywhere by [F5]. The final clause of [F3] applied to the Hilbert kernel and to each Riesz kernel therefore gives, for every $f \in L^{p} (w)$ , an almost-everywhere limit $H f$ or $R_{j} f$ satisfying the displayed $L^{p} (w)$ bounds; these limits define the stated bounded extensions. For $w \in A_{1}$ and $f \in L^{1} (w)$ the same closure argument applies with the weak $(1, 1)$ bound in place of the strong bound: for $g \in C_{c}^{\infty} (R^{n})$ , ${lim sup}_{ε, ε^{'} ↓ 0} ∣ T_{ε} f - T_{ε^{'}} f ∣ \leq 2 T^{*} (f - g)$ off the null set where the truncations of $g$ converge, and $w ({T^{*} (f - g) > η}) \leq C (n, δ, [w]_{A_{1}}) (A_{1} + A_{2}^{'} + A_{3} + B) η^{- 1} ∥ f - g ∥_{L^{1} (w)}$ tends to zero by density [F4], so the truncations converge $w$ -almost everywhere and the limit obeys $∣ T f ∣ \leq T^{*} f$ .

RemarkRemark: Literature-sourcedProof: Not applicableOpen item page →

Weighted endpoints are not obtained by setting p equal to one

Remark

The strong weighted estimates of this page, which are stated for $1 < p < \infty$ , are not the instance $p = 1$ of an $A_{p}$ theory, in three distinct senses.

First, the $A_{p}$ condition itself degenerates at $p = 1$ : the factor $w^{- 1 / (p - 1)}$ in $[w]_{A_{p}}$ is undefined, and the class $A_{1}$ is defined instead by the pointwise bound $M^{*} w \leq C w$ almost everywhere (Muckenhoupt A_p and A_1 weights). The correct replacement for the maximal function is a weak-type estimate, $w ({M f > λ}) \leq 5^{n} [w]_{A_{1}} λ^{- 1} \int ∣ f ∣ w d λ$ (Weighted weak (1,1) bound for the maximal function under A_1), and the maximal characterisation of $A_{p}$ is likewise a statement about the strict range $1 < p < \infty$ (The Hardy-Littlewood maximal operator characterises A_p).

Second, the weighted Calderón–Zygmund theorem attaches the weak $(1, 1)$ bound to the endpoint $w \in A_{1}$ , not a strong $L^{1} (w)$ bound (Weighted L-p bounds for standard Calderon-Zygmund maximal truncations); its constant depends on the $A_{1}$ characteristic and is not universal, exactly as in the maximal case.

Third, the obstruction is not an artefact of the weights: already for the Lebesgue weight $w = 1$ , which lies in $A_{1}$ , the Hardy–Littlewood maximal operator is not of strong type $(1, 1)$ (The Hardy-Littlewood maximal operator is not strong type $(1, 1)$ ). Hence no strong $L^{1} (w)$ endpoint can be expected for all $w \in A_{1}$ , and the strict range $1 < p < \infty$ in the strong weighted theorems is essential.

5 · Examples, counterexamples and false statements

None yet.