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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.

Depends on

Used by

Muckenhoupt Aₚ and A₁ weights Definition
The weighted maximal function of a doubling weight Definition
The Aₚ range of a power weight Example
Aₚ weights are doubling Lemma
Power decay implies doubling Lemma
The two defining forms of A₁ agree Lemma
Unweighted local good-lambda estimate for maximal truncations Lemma
The Hardy-Littlewood maximal operator characterises Aₚ Theorem

Dependency tree · two levels

47 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Loukas Grafakos, Classical Fourier Analysis, 3rd ed. (Springer GTM 249, 2014) (standard reference, not scraped)
Juha Kinnunen, Harmonic Analysis (Aalto University lecture notes) (standard reference, not scraped)