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Real Hardy Spaces Maximal Functions and Atoms

1 · Prerequisites

2 · Summary

This page develops the real Hardy spaces $H^{p} (R^{n})$ for $0 < p < \infty$ . The space is defined by the radial maximal function of a fixed admissible Schwartz kernel $φ$ with $\int φ \neq 0$ , and the page compares that definition with the nontangential maximal functions of aperture $a \geq 1$ and with the grand maximal function of order $N$ built on the Schwartz test class $F_{N}$ . The atomic objects are the $(p, \infty, s)$ -atoms: functions supported in a nondegenerate axis-parallel cube, bounded by $∣ Q ∣^{- 1 / p}$ and with all moments of order at most $s = ⌊ n (1 / p - 1) ⌋$ vanishing.

The first part assembles the analytic tools. Dilations and their normalisations preserve Schwartz space with explicit seminorm identities, flat Schwartz functions exist with prescribed vanishing moments, Schwartz approximate identities converge in the tempered-distribution topology, and under Countable Choice, every tempered distribution has a unique local polynomial projection for a nonnegative smooth compactly supported weight of positive integral matching its moments through a prescribed order. Two covering lemmas are proved: a greedy countable cover of a proper open set by balls with controlled radii, sizes and overlap, and the all-generations dyadic Whitney decomposition with pairwise disjoint interiors, comparable touching cubes, a finite touching count and bounded overlap of the small dilates $R Q_{j}$ , $1 \leq R \leq 2$ . A smooth deconvolution along dyadic dilations then expresses an arbitrary Schwartz function as a rapidly convergent series in the negative dilates of the fixed kernel $φ$ , with Schwartz-norm coefficients decaying faster than any prescribed power.

The central theorem is the maximal-function characterisation: for each admissible kernel $φ$ there is a finite order threshold $N_{0} (n, p, φ)$ such that for every $N \geq N_{0} (n, p, φ)$ the conditions $M_{φ}^{0} f \in L^{p}$ , $M_{φ}^{*, a} f \in L^{p}$ and $M_{N} f \in L^{p}$ are equivalent, with equivalent extended quasi-norms. The threshold depends on the kernel through the deconvolution constants, while the thresholds recorded in the sources, $N \geq ⌊ n / p ⌋ + 1$ and $N > n / p + n + 1$ , refer to their own normalised grand maximal functions. The proof follows the truncation route: the truncated maximal functions $M^{ϵ, L}$ are finite and integrable for large $L$ , the grand truncated function is pointwise dominated by the truncated tangential one, the good-set bootstrap with the Hardy-Littlewood maximal theorem closes the a priori estimate, and monotone convergence as $ϵ ↓ 0$ removes the truncation. Borel measurability of all the maximal functions makes the $L^{p}$ statements meaningful. A flat compactly supported kernel and the approximate-identity limit yield the Calderon reproducing formula used later.

From the characterisation the page derives that $H^{p} = L^{p}$ with equivalent norms for $1 < p < \infty$ ; the inclusion $H^{p} \subseteq L^{p}$ uses the weak-star sequential compactness of the dual ball and is recorded as assuming the ultrafilter lemma, while the converse uses Countable Choice through the published maximal-function machinery. For $0 < p \leq 1$ the Calderon-Zygmund level decomposition of an $H^{p}$ distribution produces $(p, \infty, s)$ -atoms with summable $ℓ^{p}$ coefficients, $ℓ^{p}$ -sums of atoms converge in $S^{'}$ and in the $H^{p}$ quasi-norm, atoms have a uniform $H^{p}$ bound with a quantitative pairing estimate, and the resulting atomic characterisation identifies $H^{p}$ with the space of atomic sums and gives the two-sided quasi-norm equivalence. The same route gives the Fourier decay $∣ \hat{f} (ξ) ∣ \leq C ∥ f ∥_{H^{p}} ∣ ξ ∣^{n (1 / p - 1)}$ with a little- $o$ refinement, and hence the vanishing of all moments through order $⌊ n (1 / p - 1) ⌋$ for $H^{p}$ functions whose weighted moments through that order are absolutely integrable. Calderon-Zygmund operators with standard Holder kernels map $H^{1}$ boundedly into $L^{1}$ ; that theorem assumes Countable Choice, and the quasi-Banach remark records that $∥ \cdot ∥_{H^{p}}$ is only a quasi-norm for $p < 1$ .

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

$H^{p}$ atoms with a prescribed moment order

Definition

Let $n \geq 1$ , let $0 < p \leq 1$ and let $s \in N \cup {0}$ satisfy $s \geq ⌊ n (1 / p - 1) ⌋$ . A $(p, \infty, s)$ -atom is a measurable function $a ⁣ : R^{n} \to C$ for which there is a nondegenerate axis-parallel cube $Q$ (Axis-parallel rectangles in $R^{m}$ and their volume, so all $n$ side lengths are equal and positive) such that

$supp a \subseteq Q$ , where the support is the closure of ${a \neq 0}$ ;
$∣ a (x) ∣ \leq ∣ Q ∣^{- 1 / p}$ for almost every $x \in R^{n}$ ;
$\int_{R^{n}} a (x) x^{α} d x = 0$ for every multi-index $α$ with $∣ α ∣ \leq s$ ( $C^{k}$ maps and multi-index derivative notation in Euclidean space).

The exponent $1 / p \geq 1$ in the size bound and the order $s$ are part of the datum, not free parameters of the function: a $(p, \infty, s)$ -atom is also a $(p, \infty, s^{'})$ -atom for every $⌊ n (1 / p - 1) ⌋ \leq s^{'} \leq s$ , because the moment conditions for the smaller order are among those already imposed. The zero function satisfies all three conditions; zero terms may be omitted from atomic representations.

Under Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), every moment in condition 3 is an absolutely convergent Lebesgue integral; the supporting cube has its finite volume 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. Indeed $a$ vanishes a.e. off the bounded set $Q$ and $∣ a x^{α} ∣ \leq ∣ Q ∣^{- 1 / p} \sup_{x \in Q} ∣ x^{α} ∣$ a.e. on $Q$ , a bounded function on a set of finite measure; the monomial $x^{α}$ is continuous and hence Borel measurable by Continuous functions on Euclidean spaces are Borel measurable. Applying Arithmetic and lattice operations preserve measurability whenever they are defined to the real and imaginary parts of $a$ shows that $a x^{α}$ is measurable, so the integral is defined and finite. The a.e. bound in condition 2 is an essential supremum bound, $∥ a ∥_{\infty} \leq ∣ Q ∣^{- 1 / p}$ (The essential supremum of a measurable function with respect to a measure); replacing $a$ by another representative of its a.e. class preserves conditions 2 and 3 but can change the support in condition 1, so the support condition is imposed for the chosen representative.

The page fixes the order $s_{p} = ⌊ n (1 / p - 1) ⌋ .$ This is the least integer compatible with condition 3, and it is the order used by the sources: DKKP require moments through $n (p^{- 1} - 1)$ , Wang and Hiserote through $⌊ n (1 / p - 1) ⌋$ . The threshold moves exactly at the integers: $s_{p} = 0$ for $n / (n + 1) < p \leq 1$ , $s_{p} = 1$ for $n / (n + 2) < p \leq n / (n + 1)$ , and so on. For $p = 1$ the definition specialises to the classical $L^{\infty}$ atoms of $H^{1}$ : support in a cube, $∣ a ∣ \leq ∣ Q ∣^{- 1}$ a.e. and $\int a = 0$ (Conjugate exponents, including the endpoint conventions records the exponent convention used for the dual exponents invoked later on this page).

The ball-supported atoms of the sources differ from this convention only by fixed constants: a cube $Q$ containing a ball $B$ with $∣ B ∣ \leq ∣ Q ∣ \leq c_{n} ∣ B ∣$ carries the same conditions up to the dimensional factor $c_{n}^{1 / p}$ , and nondegeneracy rules out the degenerate cubes of zero volume that occur in moment conditions. The three conditions define the class without selecting representatives; the accompanying Lebesgue-integrability assertions use Countable Choice.

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

Schwartz functions with prescribed flatness of the Fourier transform at the origin

Statement

Assume Countable Choice. Let $n \geq 1$ . For every integer $m \geq 1$ there is a real, even function $φ \in C_{c}^{\infty} (R^{n})$ with $supp φ \subseteq B (0, 1), \int_{R^{n}} φ \neq 0, \int_{R^{n}} x^{α} φ (x) d x = 0 for 0 < ∣ α ∣ \leq m .$ Equivalently, under the Fourier convention of Fourier differentiation and multiplication identities on tempered distributions, $\hat{φ} (0) \neq 0$ and $\partial^{α} \hat{φ} (0) = 0$ for every multi-index with $0 < ∣ α ∣ \leq m$ . The construction is uniform in $m$ : a single one-dimensional finite-difference construction achieves every prescribed finite flatness order, and its tensor product is used.

Facts & Assumptions

Given: Countable Choice, an integer $n \geq 1$ and an integer $m \geq 1$ . The multi-index notation is that of $C^{k}$ maps and multi-index derivative notation in Euclidean space and the seminorms are those of Schwartz space and its seminorms.

[A1]

Countable Choice is assumed, in particular for the Fourier differentiation identity and the Lebesgue change-of-variables formula cited below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

There is a smooth bump: with $σ$ the standard smooth step function of The standard smooth step function, which is smooth, vanishes on $(- \infty, 0]$ and equals $1$ on $[1, \infty)$ , the function $θ (x) : = σ (\frac{1}{16} - x^{2})$ is smooth (composition of the smooth functions $σ$ and $x \mapsto \frac{1}{16} - x^{2}$ , The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ), even, positive on $(- 1 / 4, 1 / 4)$ (where $\frac{1}{16} - x^{2} \in (0, \frac{1}{16}]$ and $σ > 0$ on $(0, \infty)$ ) and supported in $[- 1 / 4, 1 / 4]$ (where $\frac{1}{16} - x^{2} \geq 0$ ).

[F2]

Newton-Leibniz with an interior derivative: if $G$ is continuous on $[a, b]$ , differentiable on $(a, b)$ and $G^{'} = g$ there with $g$ Riemann integrable, then $\int_{a}^{b} g = G (b) - G (a)$ (Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative). Applied inductively this gives, for $f \in C^{m} (R)$ and $h > 0$ , the iterated integral representation $Δ_{h}^{m} f (x) = \int_{[- h, h]^{m}} f^{(m)} (x + s_{1} + \dots + s_{m}) d s_{1} \dots d s_{m}, Δ_{h} f (x) = f (x + h) - f (x - h) .$ There is no factorial prefactor in this representation: each finite-difference factor introduces one integration over $[- h, h]$ . Any factorial below comes from evaluating the derivative $f^{(m)}$ , not from the integration formula.

[F3]

Under the Fourier convention of Fourier differentiation and multiplication identities on tempered distributions, for $φ \in S$ and every multi-index $α$ one has $\partial^{α} \hat{φ} (0) = (- 2 π i)^{∣ α ∣} \int_{R^{n}} x^{α} φ (x) d x$ ; equivalently, if all mixed moments $\int x^{α} φ$ , $0 < ∣ α ∣ \leq m$ , vanish then $\partial^{α} \hat{φ} (0) = 0$ for those $α$ , and conversely.

[F4]

Riemann Fubini on a product rectangle factors the integral of a continuous compactly supported tensor product; the coordinate dilation uses the change-of-variables formula (Riemann--Fubini on product rectangles, with lower and upper section integrals and content-zero exceptional sections, A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions).

Proof technique: finite differences of a bump in one dimension, then tensor product, dilation and normalisation.

Proof

technique · constructive

1.1givenconstruct

Reduction to even order. If $m$ is odd, replace it by $m + 1$ : a function whose moments vanish through order $m + 1$ also has all moments vanishing through order $m$ . We may therefore assume $m \geq 2$ is even, and we write $h = 1 / (8 m)$ . This uses no choice.

1.2F1givenalgebra

The one-dimensional construction. Let $θ$ be the even bump of [F1] and put $Θ (x) = θ (x + 1 / 2) - θ (x - 1 / 2)$ . Then $Θ \in C_{c}^{\infty} (R)$ is real, odd, and supported in $[- 3 / 4, - 1 / 4] \cup [1 / 4, 3 / 4]$ ; moreover $Θ (x) = - θ (x - 1 / 2) < 0$ on $(1 / 4, 3 / 4)$ and $Θ (x) = θ (x + 1 / 2) > 0$ on $(- 3 / 4, - 1 / 4)$ . Define $ϕ (x) = x^{- 1} Δ_{h}^{m} Θ (x)$ . Since $supp Δ_{h}^{m} Θ \subseteq [- 3 / 4 - m h, 3 / 4 + m h] = [- 7 / 8, 7 / 8]$ and is bounded away from $0$ , the factor $x^{- 1}$ is smooth on a neighbourhood of that support, so $ϕ \in C_{c}^{\infty} (R)$ with support in $[- 7 / 8, 7 / 8]$ . The reflection operator $(P f) (x) = f (- x)$ satisfies $P Δ_{h} = - Δ_{h} P$ ; since $Θ$ is odd and $m$ is even, $Δ_{h}^{m} Θ$ is odd, so $ϕ$ is even and real.

1.3A1algebra

The flatness of the one-dimensional Fourier transform. For $1 \leq ν \leq m$ , differentiating $\hat{ϕ} (ξ) = \int e^{- 2 π i ξ x} ϕ (x) d x$ under the integral sign and using the polynomial identity $Δ_{h}^{m} (x^{ν - 1}) = 0$ (the $m$ -th finite difference of a polynomial of degree $ν - 1 \leq m - 1$ vanishes) gives ${\hat{ϕ}}^{(ν)} (0) = (- 2 π i)^{ν} \int_{R} x^{ν - 1} Δ_{h}^{m} Θ (x) d x = (- 2 π i)^{ν} \int_{R} (Δ_{h}^{m} x^{ν - 1}) (x) Θ (x) d x = 0,$ where the middle equality is the self-adjointness $\int f (Δ_{h}^{m} g) = \int (Δ_{h}^{m} f) g$ of the even-order finite difference, which follows from the translation invariance of Lebesgue measure and $(Δ_{h})^{*} = - Δ_{h}$ .

1.4A1F2algebra

Nonvanishing of the mean. Using self-adjointness again, $\int_{R} ϕ = \int_{R} x^{- 1} Δ_{h}^{m} Θ (x) d x = \int_{R} (Δ_{h}^{m} x^{- 1}) (x) Θ (x) d x .$ On the support of $Θ$ one has $∣ x ∣ \geq 1 / 4$ , and all points $x + s_{1} + \dots + s_{m}$ in the iterated integral representation of [F2] stay on the same side of zero. Since $m$ is even, $(x^{- 1})^{(m)} = m! x^{- m - 1}$ has the sign of $x$ , while $Θ$ has the opposite sign on each of its two support components. Thus the integrand $(Δ_{h}^{m} x^{- 1}) Θ$ has one constant sign and there is no cancellation. By [F2], $Δ_{h}^{m} (x^{- 1}) (x) = m! \int_{[- h, h]^{m}} (x + s_{1} + \dots + s_{m})^{- m - 1} d s_{1} \dots d s_{m} .$ The integrand has a constant sign on this box, so its absolute value is the integral of the absolute value. The factor $m!$ comes from $(x^{- 1})^{(m)}$ ; the iterated integral contributes the box volume $(2 h)^{m}$ . Therefore $∣ Δ_{h}^{m} (x^{- 1}) (x) ∣ = m! \int_{[- h, h]^{m}} ∣ x + s_{1} + \dots + s_{m} ∣^{- m - 1} d s_{1} \dots d s_{m} \geq (2 h)^{m} m! (7 / 8)^{- m - 1}$ for every $x \in supp Θ$ , because $∣ x + s_{1} + \dots + s_{m} ∣ \leq 3 / 4 + m h = 7 / 8$ . Hence $∣ \int_{R} ϕ ∣ = \int_{R} ∣ Δ_{h}^{m} x^{- 1} ∣ ∣ Θ ∣ \geq (2 h)^{m} m! (7 / 8)^{- m - 1} \int_{R} ∣ Θ ∣ > 0,$ since $Θ$ is continuous and not identically zero.

2.1A1step 1.3step 1.4F3F4algebra

The tensor product and its moments. Put $ψ (x) = ϕ (x_{1}) ϕ (x_{2}) \dots ϕ (x_{n})$ and $Ψ (x) = ψ (λ x)$ with $λ = \sqrt{n} + 1 > 7 \sqrt{n} / 8$ ; then $Ψ \in C_{c}^{\infty} (R^{n})$ is real and even and $supp Ψ \subseteq [- 7 / (8 λ), 7 / (8 λ)]^{n} \subseteq B (0, 1)$ , since the euclidean circumradius of that cube is $(7 / 8) \sqrt{n} / λ < (7 / 8) \sqrt{n} / (7 \sqrt{n} / 8) = 1$ . For a multi-index $α$ with $0 < ∣ α ∣ \leq m$ the substitution $x = λ^{- 1} t$ gives the factorisation $\int_{R^{n}} x^{α} Ψ (x) d x = λ^{- n - ∣ α ∣} \prod_{j = 1}^{n} \int_{R} t^{α_{j}} ϕ (t) d t .$ If $α_{j} = 0$ the corresponding factor is $\int ϕ \neq 0$ by step 1.4; if $α_{j} \geq 1$ then $α_{j} \leq ∣ α ∣ \leq m$ , and $\int t^{α_{j}} ϕ (t) d t = 0$ by step 1.3 combined with [F3] applied in one dimension. Hence every factor with $α_{j} \geq 1$ vanishes and the product is zero.

3.1A1step 2.1F3discharge-construct∎

Normalisation and conclusion. Step 2.1 gives $\int Ψ = λ^{- n} (\int ϕ)^{n} \neq 0$ , so $φ = (\int_{R^{n}} Ψ)^{- 1} Ψ$ is real, even, smooth and compactly supported in $B (0, 1)$ , with $\int φ = 1$ and all moments $\int x^{α} φ$ , $0 < ∣ α ∣ \leq m$ , still vanishing. The Fourier form of the statement follows from the differentiation identity of [F3] and $\hat{Ψ} (0) = \int Ψ \neq 0$ , together with the linearity of the Fourier transform under the real scalar normalisation. This proves the lemma.

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Local polynomial projections matching moments through order $s$

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $Q \subseteq R^{n}$ be a nondegenerate axis-parallel cube with centre $c_{Q}$ , side length $ℓ (Q)$ and volume $∣ Q ∣$ in the sense of Axis-parallel rectangles in $R^{m}$ and their volume. Fix $λ > 1$ and write $Q^{*} : = {x \in R^{n} : ∥ x - c_{Q} ∥_{\infty} < λ ℓ (Q) / 2}$ for its open concentric dilation. Let $N \in N \cup {0}$ , and let $ω_{Q} \in C_{c}^{\infty} (R^{n})$ be real, nonnegative, with $\int_{R^{n}} ω_{Q} > 0$ and $supp ω_{Q} \subset Q^{*}$ . Then for every tempered distribution $f \in S^{'} (R^{n})$ there is a unique moment-matching polynomial $P_{Q}$ of total degree at most $N$ such that $⟨ f - P_{Q}, x^{α} ω_{Q} ⟩ = 0 for every multi-index ∣ α ∣ \leq N .$ The polynomial $P_{Q}$ depends only on the restriction of $f$ to an open neighbourhood of $supp ω_{Q}$ : for every open $U \supseteq supp ω_{Q}$ , if $g \in S^{'}$ agrees with $f$ as a distribution on $U$ , then $P_{Q}^{g} = P_{Q}^{f}$ . When $f$ is represented by a function in $L^{2} (ω_{Q} (x) d x)$ , this is the orthogonal projection of $f$ onto the polynomials of degree at most $N$ in that weighted inner product space, with inner product $(P, R)_{ω_{Q}} = \int_{R^{n}} P (x) \overline{R (x)} ω_{Q} (x) d x$ . This inner product is positive definite on the polynomial subspace because $ω_{Q} \geq 0$ and $\int ω_{Q} > 0$ .

Facts & Assumptions

Given: Countable Choice and a nondegenerate axis-parallel cube $Q$ , an integer $N \geq 0$ , a test function $ω_{Q}$ as in the statement, $f \in S^{'} (R^{n})$ , and multi-indices with the conventions of $C^{k}$ maps and multi-index derivative notation in Euclidean space.

[L1]

A nonnegative continuous function on an open set with positive integral is positive at some point, hence positive on a nonempty open subset of that set; a polynomial vanishing on a nonempty open set is zero: at an interior point all its partial derivatives vanish, and its finite expansion about that point, obtained by the binomial formula for each monomial, has precisely those derivatives as coefficients (The spaces $C_{c} (R^{n})$ and $C_{c}^{\infty} (R^{n})$ fixes the support convention).

[F1]

$x^{α} ω_{Q} \in C_{c}^{\infty} (R^{n}) \subseteq S (R^{n})$ , so the pairings $⟨ f, x^{α} ω_{Q} ⟩$ and, for polynomials $P$ , the regular-distribution pairings $⟨ P, x^{α} ω_{Q} ⟩ = \int P x^{α} ω_{Q}$ are defined; polynomials are locally integrable and $P x^{α} ω_{Q} \in S$ (Schwartz space and its seminorms, Tempered distribution, Regular distribution from a locally integrable function).

[F2]

The space $P_{N}$ of polynomials of total degree at most $N$ has finite dimension $d = (\binom{N + n}{n})$ , and the monomials $x^{α}$ , $∣ α ∣ \leq N$ , form a basis ( $C^{k}$ maps and multi-index derivative notation in Euclidean space). A finite-dimensional linear system with invertible matrix has a unique solution.

[F3]

If two distributions agree on an open set $U$ , then their pairings with every test function supported in $U$ agree: this is the definition of agreement of distributions on $U$ (Distribution). In particular, if $h \in C_{c}^{\infty} (R^{n})$ is supported in $U$ and $f = 0$ on $U$ , then $⟨ f, h ⟩ = 0$ .

[F4]

Since $ω_{Q}$ is measurable and nonnegative, $d μ_{Q} = ω_{Q} (x) d x$ is the measure with density $ω_{Q}$ relative to Lebesgue measure (The measure with density $f$ relative to $μ$ ).

[F5]

On the measure space $(R^{n}, μ_{Q})$ the pairing $([u], [v]) \mapsto \int u \overline{v} d μ_{Q}$ is the well-defined complex $L^{2}$ inner product (The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz).

Proof technique: positive-definite Gram matrix on the finite-dimensional polynomial space.

Proof

technique · direct

1.1L1F1F2algebra

The Gram matrix. Set $G_{α β} = \int_{R^{n}} x^{α + β} ω_{Q} (x) d x$ for $∣ α ∣, ∣ β ∣ \leq N$ . If $P = \sum_{∣ α ∣ \leq N} c_{α} x^{α}$ satisfies $\int ∣ P ∣^{2} ω_{Q} = 0$ , then $∣ P ∣^{2} ω_{Q} = 0$ Lebesgue-a.e.; since $∣ P ∣^{2}$ is continuous and $ω_{Q}$ is continuous and positive on a nonempty open set by [L1] (using $\int ω_{Q} > 0$ and $ω_{Q} \geq 0$ ), $P$ vanishes on that open set, hence $P = 0$ and all $c_{α} = 0$ by [L1] and [F2]. Writing $\int ∣ P ∣^{2} ω_{Q} = \sum_{α, β} \overline{c_{α}} c_{β} G_{α β}$ , positive definiteness follows, so $G$ is invertible.

2.1step 1.1F1F2F4F5algebra

Existence and uniqueness. The vector $b = (⟨ f, x^{α} ω_{Q} ⟩)_{∣ α ∣ \leq N} \in C^{d}$ is well defined by [F1], so [F2] gives a unique coefficient vector $c = G^{- 1} b$ and a polynomial $P_{Q} = \sum_{∣ α ∣ \leq N} c_{α} x^{α}$ with $⟨ P_{Q}, x^{β} ω_{Q} ⟩ = \sum_{α} c_{α} G_{α β} = b_{β} = ⟨ f, x^{β} ω_{Q} ⟩$ for every $∣ β ∣ \leq N$ ; that is, $⟨ f - P_{Q}, x^{β} ω_{Q} ⟩ = 0$ . If $f$ is represented by an element of $L^{2} (μ_{Q})$ , then for every polynomial $R$ of degree at most $N$ its conjugate is a linear combination of the real monomials $x^{α}$ , and the moment equations give $(f - P_{Q}, R)_{ω_{Q}} = \int (f - P_{Q}) \overline{R} d μ_{Q} = 0$ by [F4, F5]. Thus $P_{Q}$ is the orthogonal projection onto the polynomial subspace in the weighted $L^{2}$ inner product space. If $P^{'}, P^{''}$ both satisfy the moment equations, then $R = P^{'} - P^{''}$ satisfies $⟨ R, x^{α} ω_{Q} ⟩ = 0$ for $∣ α ∣ \leq N$ , so $\int ∣ R ∣^{2} ω_{Q} = \sum \overline{c_{α}} ⟨ R, x^{α} ω_{Q} ⟩ = 0$ with $c_{α}$ the coefficients of $R$ , because $\overline{R} = \sum \overline{c_{α}} x^{α}$ , and step 1.1 gives $R = 0$ . This proves existence and uniqueness.

3.1step 2.1F3given

Locality. Let $U$ be the given neighbourhood of $supp ω_{Q}$ on which $f$ and $g$ agree as distributions. For every $∣ α ∣ \leq N$ , the test function $x^{α} ω_{Q}$ is supported in $supp ω_{Q} \subset U$ , so [F3] gives $⟨ f - g, x^{α} ω_{Q} ⟩ = 0$ . Hence $f$ and $g$ produce the same vector $b$ in step 2.1, and therefore the same $P_{Q} = G^{- 1} b$ . This proves the locality statement.

4.1step 1.1step 2.1step 3.1∎

Conclusion. Step 1.1 shows that the Gram matrix is positive definite, step 2.1 constructs the unique moment-matching polynomial and identifies it as the weighted $L^{2}$ orthogonal projection when that interpretation applies, and step 3.1 records dependence only on the distribution near the support of the weight. This proves the lemma.

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Dilations and their normalisations preserve Schwartz space, with scaling identities

Statement

Let $n \geq 1$ , $φ \in S (R^{n})$ and $t > 0$ , and write $(D_{t} φ) (x) = φ (x / t), φ_{t} (x) = t^{- n} φ (x / t) .$ Then $D_{t} φ, φ_{t} \in S (R^{n})$ , with the seminorm identities $p_{α β} (D_{t} φ) = t^{∣ α ∣ - ∣ β ∣} p_{α β} (φ), p_{α β} (φ_{t}) = t^{∣ α ∣ - ∣ β ∣ - n} p_{α β} (φ)$ for all multi-indices $α, β$ (Schwartz space and its seminorms, $C^{k}$ maps and multi-index derivative notation in Euclidean space). Consequently each $D_{t}$ maps $S$ continuously into itself for the Schwartz topology (Schwartz topology and convergence).

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Then $\int_{R^{n}} φ_{t} (x) d x = \int_{R^{n}} φ (x) d x, \int_{R^{n}} ∣ x ∣^{m} ∣ φ_{t} (x) ∣ d x = t^{m} \int_{R^{n}} ∣ x ∣^{m} ∣ φ (x) ∣ d x$ for every integer $m \geq 0$ , both sides finite (Schwartz derivatives are integrable).

The identities are stated for $t > 0$ ; the normalisation is chosen so that the $L^{1}$ mass and the first moments scale by the powers $t^{m}$ , which is what the later approximate-identity argument consumes. The unnormalised dilation satisfies $D_{t} φ = t^{n} φ_{t}$ , and the factor $t^{- n}$ does not affect membership in $S$ , which is closed under nonzero scalar multiples.

Facts & Assumptions

Given: $n \geq 1$ , $φ \in S (R^{n})$ , $t > 0$ , and the seminorms, topology and partial derivatives of Schwartz space and its seminorms, Schwartz topology and convergence and $C^{k}$ maps and multi-index derivative notation in Euclidean space. Under countable choice, A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions gives the substitution formula for the $C^{1}$ diffeomorphism $T (y) = t y$ of $R^{n}$ with $∣ \det D T (y) ∣ = t^{n}$ , and Schwartz derivatives are integrable gives $x^{α} \partial^{β} φ \in L^{1}$ for all multi-indices.

[L1]

The $j$ -th partial derivative of a function $f$ at $x$ is $\partial_{j} f (x) = \lim_{h \to 0} (f (x + h e_{j}) - f (x)) / h$ , and partial derivatives of a Schwartz function exist and are continuous ( $C^{k}$ maps and multi-index derivative notation in Euclidean space, Schwartz space and its seminorms).

[F1]

$∣ x ∣^{m} \leq (1 + ∣ x ∣^{2})^{m / 2} \leq \sum_{∣ α ∣ \leq m} c_{α} ∣ x^{α} ∣$ with finite constants $c_{α}$ , and a nonnegative measurable function dominated by a finite sum of $L^{1}$ functions lies in $L^{1}$ (Schwartz derivatives are integrable).

Proof technique: direct computation with the chain rule along coordinate axes, then the change-of-variables formula.

Proof

technique · direct

1.1L1givenalgebra

Differentiation of a dilation. Let $f = φ \circ A$ with $A (x) = x / t$ , and fix $j \leq n$ and $x \in R^{n}$ . Writing $z = x / t$ and $s = h / t$ , the one-variable difference quotient of the map $h \mapsto f (x + h e_{j})$ equals $t^{- 1} (φ (z + s e_{j}) - φ (z)) / s$ , and $s \to 0$ exactly when $h \to 0$ , so the limit exists and equals $t^{- 1} \partial_{j} φ (z)$ by [L1]; there is no division by a vanishing quantity because $t > 0$ . Induction on $∣ β ∣$ , applying the same computation to the $C^{1}$ function $\partial^{γ} φ$ at the point $x$ with $γ$ the predecessor of $β$ , gives $\partial^{β} (D_{t} φ) (x) = t^{- ∣ β ∣} (\partial^{β} φ) (x / t) .$

1.2F1given

The scaling identities. By [F1] and [L1] the functions $φ$ , $∣ φ ∣$ , $x^{α} \partial^{β} φ$ and $∣ x ∣^{m} ∣ φ ∣$ are integrable, so the change-of-variables formula applies to them. Applying it to $φ$ with $T (y) = t y$ and $∣ \det D T ∣ = t^{n}$ gives $\int φ_{t} (x) d x = t^{- n} \int φ (x / t) d x = \int φ (y) d y,$ and applying it to the nonnegative integrable function $∣ x ∣^{m} ∣ φ (x) ∣$ gives $\int ∣ x ∣^{m} ∣ φ (x / t) ∣ d x = t^{n + m} \int ∣ y ∣^{m} ∣ φ (y) ∣ d y$ , whence the moment identity after multiplying by $t^{- n}$ . The factor $t^{- n} t^{n + m} = t^{m}$ is finite for every $m \geq 0$ and $t > 0$ .

2.1L1step 1.1givenalgebra

Membership and the seminorm identities. Substituting $y = x / t$ in $∣ x^{α} \partial^{β} (D_{t} φ) (x) ∣ = t^{- ∣ β ∣} ∣ x^{α} (\partial^{β} φ) (x / t) ∣$ gives $t^{∣ α ∣ - ∣ β ∣} ∣ y^{α} \partial^{β} φ (y) ∣$ , valid for every $x \in R^{n}$ ; taking suprema over $x$ is taking suprema over $y$ and proves $p_{α β} (D_{t} φ) = t^{∣ α ∣ - ∣ β ∣} p_{α β} (φ) < \infty$ . Multiplying by the scalar $t^{- n}$ proves the second identity and makes $D_{t} φ, φ_{t}$ elements of $S$ , since these are finite for all $α, β$ by [L1] and the given. For fixed $t$ the constants $t^{∣ α ∣ - ∣ β ∣ - n}$ are finite, so for every basic neighbourhood the finitely many relevant input seminorms of $φ$ control the output seminorms; this is continuity of $D_{t}$ at zero, hence everywhere by linearity.

3.1step 2.1step 1.2∎

Conclusion. Step 2.1 gives membership, the two seminorm identities, and continuity of $D_{t}$ on $S$ ; step 1.2 gives the integral and moment identities under countable choice, which is inherited from the substitution theorem. This proves the lemma.

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

Whitney decomposition of a proper open subset of Euclidean space

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ and let $Ω \subseteq R^{n}$ be nonempty, open and proper. Write $ℓ (Q) = 2^{- k} for the generation- k dyadic cube Q, d (Q) = \inf {∣ x - y ∣ : x \in Q, y \in R^{n} ∖ Ω}$ for the all-generations dyadic cubes of Dyadic cubes of all generations in R^n. Then there is a countable family $W = (Q_{j})_{j \in N}$ of dyadic cubes, with pairwise disjoint interiors, such that

$Ω = ⋃_{j} Q_{j}$ and $\sqrt{n} ℓ (Q_{j}) \leq d (Q_{j}) \leq 4 \sqrt{n} ℓ (Q_{j})$ for every $j$ ;
if $Q, Q^{'} \in W$ have intersecting closures, then $\frac{1}{5} ℓ (Q) \leq ℓ (Q^{'}) \leq 5 ℓ (Q)$ ;
every $Q \in W$ has intersecting closures with at most $K (n) : = 2 \cdot 2^{n} + 3^{n} + 4^{n} + 6^{n}$ cubes of $W$ (the source [K, Remark 1.11] records the sharper count $12^{n}$ for its construction; only finiteness of $K (n)$ is used below);
for every $1 \leq R \leq 2$ the dilated cubes $R Q_{j} : = {c_{j} + R (x - c_{j}) : x \in Q_{j}}$ , $c_{j}$ the centre of $Q_{j}$ , have overlap bounded by a constant $C_{n} < \infty$ depending only on $n$ (the bound is uniform in $R \in [1, 2]$ ).

In the diametral normalization the published form [W, Theorem 14.5] records $diam Q_{j} \leq dist (Q_{j}, Ω^{c}) \leq 4 diam Q_{j}$ for a family with the same covering and disjointness properties. The dilation restriction $R \leq 2$ is not a defect of the construction: for the Whitney family of $Ω = (0, \infty) \subseteq R$ the intervals $Q_{k} = (2^{k}, 2^{k + 1}]$ have $R$ -dilations containing the fixed point $x = 1$ for infinitely many $k$ as soon as $R \geq 3$ , so no bound uniform in $R$ can hold.

Facts & Assumptions

Given: $n \geq 1$ , a nonempty proper open set $Ω \subseteq R^{n}$ , Countable Choice, and the dyadic cubes of Dyadic cubes of all generations in R^n with the partition, volume and nesting properties of All-generation dyadic cubes: partition, volume and nesting.

[L1]

The dyadic cube $Q_{k, m}$ of generation $k$ is a half-open box of side $2^{- k}$ and volume $2^{- k n}$ ; cubes at one generation are pairwise disjoint and cover $R^{n}$ ; two dyadic cubes are disjoint or one contains the other (Dyadic cubes of all generations in R^n, All-generation dyadic cubes: partition, volume and nesting). The closed cube $\overline{Q}$ is contained in the closed ball of radius $\frac{\sqrt{n}}{2} ℓ (Q)$ about the centre $c_{Q}$ , and its diameter is $\sqrt{n} ℓ (Q)$ .

[L2]

For a nonempty set $A$ and $z \in R^{n}$ one has $∣ d (z) - d (z^{'}) ∣ \leq ∣ z - z^{'} ∣$ where $d (z) = dist (z, Ω^{c})$ ; in particular $d$ is continuous on $Ω$ and $d (Q) = \inf_{y \in Q} d (y)$ satisfies $d (Q) \leq d (y)$ for every $y \in Q$ . If $z$ lies in the closure of $Q$ , then $d (z) \geq d (Q)$ .

[L3]

The dilation volume identity $∣ r E ∣ = r^{n} ∣ E ∣$ holds for measurable $E$ and $r > 0$ (For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it).

Proof technique: the distance-compatible dyadic rule, then maximal elements and packing estimates.

Proof

technique · constructive

1.1L1givenalgebraconstruct

The distance rule. For $x \in Ω$ put $d (x) = dist (x, Ω^{c}) > 0$ ; the positivity uses that $Ω$ is open and $Ω^{c}$ closed. Since the dyadic numbers $2^{- k}$ , $k \in Z$ , partition $(0, \infty)$ into the intervals $(2^{- k - 1}, 2^{- k}]$ , there is exactly one $k (x) \in Z$ with $2^{- k (x) - 1} < d (x) / (4 \sqrt{n}) \leq 2^{- k (x)}$ , and then $d (x) / (4 \sqrt{n}) \leq ℓ (x) : = 2^{- k (x)} < d (x) / (2 \sqrt{n}) .$ Let $Q (x)$ be the unique dyadic cube of generation $k (x)$ containing $x$ , which exists because generation $k (x)$ partitions $R^{n}$ by [L1]. Put $G = {Q (x) : x \in Ω}$ .

2.1step 1.1L1L2algebra

Cubes of the rule are contained in $Ω$ with controlled distance. If $y \in Q (x)$ , then $∣ y - x ∣ \leq \sqrt{n} ℓ (x) < d (x) / 2$ , so $d (y) \geq d (x) - ∣ y - x ∣ > d (x) / 2 \geq \sqrt{n} ℓ (x)$ by [L2], and therefore $d (Q (x)) = \inf_{y \in Q (x)} d (y) \geq \sqrt{n} ℓ (x) > 0$ . In particular $Q (x) \cap Ω^{c} = \emptyset$ , that is, $Q (x) \subseteq Ω$ , and $Q (x) \in G$ satisfies the lower bound of claim 1. Thus every point of $Ω$ lies in an element of $G$ .

3.1step 1.1step 2.1L1given

Maximal elements. For each fixed $Q \in G$ , let $G_{Q} : = {Q^{'} \in G : Q \subseteq Q^{'}}$ . Choose the witness $x \in Ω$ with $Q = Q (x)$ supplied by the definition of $G$ . Every member $Q^{'}$ of this particular set $G_{Q}$ contains $x$ . By step 2.1, $\sqrt{n} ℓ (Q^{'}) \leq d (Q^{'}) \leq d (x)$ , while dyadic nesting gives $ℓ (Q^{'}) \geq ℓ (Q)$ . Thus its side length lies in the finite dyadic range $[ℓ (Q), d (x) / \sqrt{n}]$ ; at each generation there is only one dyadic cube containing this fixed $x$ . Therefore $G_{Q}$ is finite and nonempty. Its members are nested, so it has a unique largest member by side length, which is maximal in $G$ . Let $W$ be the set of all maximal elements of $G$ . Every $Q \in G$ lies in one of them by the preceding finite-superset argument; the set of all dyadic cubes is countable, hence so is $W$ . This construction uses only $G$ -supersets of each fixed cube, not a maximality principle over all cubes.

4.1step 2.1step 3.1L1

Covering, disjointness and the lower bound. Every $x \in Ω$ lies in some $Q (x) \in G$ , hence in a maximal element of $G$ containing it, so $Ω = ⋃_{Q \in W} Q$ . Dyadic cubes are nested or disjoint by [L1], and maximality of the elements of $W$ rules out proper inclusion, so distinct elements of $W$ are disjoint (their interiors are disjoint, indeed the cubes themselves are disjoint as half-open sets). Each $Q \in W$ belongs to $G$ , so $\sqrt{n} ℓ (Q) \leq d (Q)$ by step 2.1.

5.1step 1.1step 4.1L2algebra

The upper bound is built into the rule. Fix $Q \in W$ and $x \in Ω$ with $Q = Q (x)$ , which exists because $W \subseteq G$ . By definition of $ℓ (x)$ one has $d (x) \leq 4 \sqrt{n} ℓ (x)$ , and $d (Q) \leq d (x)$ because $x \in Q$ ; with the lower bound of step 4.1 this gives $\sqrt{n} ℓ (Q) \leq d (Q) \leq d (x) \leq 4 \sqrt{n} ℓ (Q)$ , which proves claim 1.

6.1step 4.1step 5.1L1L2algebra

Touching cubes have comparable sizes. Let $Q, Q^{'} \in W$ have intersecting closures and let $z$ be a common point. Then $d (z) \geq d (Q) \geq \sqrt{n} ℓ (Q)$ by [L2] and step 4.1. Choose $y \in Q^{'}$ with $d (y) < d (Q^{'}) + ε$ ; since $z \in \overline{Q^{'}}$ one has $∣ z - y ∣ \leq diam (Q^{'}) = \sqrt{n} ℓ (Q^{'})$ , so $d (z) \leq ∣ z - y ∣ + d (y) < \sqrt{n} ℓ (Q^{'}) + d (Q^{'}) + ε \leq \sqrt{n} ℓ (Q^{'}) + 4 \sqrt{n} ℓ (Q^{'}) + ε = 5 \sqrt{n} ℓ (Q^{'}) + ε,$ the last inequality by step 5.1 applied to $Q^{'}$ . Letting $ε ↓ 0$ gives $\sqrt{n} ℓ (Q) \leq 5 \sqrt{n} ℓ (Q^{'})$ , hence $ℓ (Q) \leq 5 ℓ (Q^{'})$ ; interchanging the roles of the two cubes gives $ℓ (Q^{'}) \leq 5 ℓ (Q)$ . This proves claim 2 with the stated factor $5$ .

6.2step 4.1step 5.1L2L3givenalgebra

Bounded overlap of small dilates. Fix $1 \leq R \leq 2$ and $x \in R^{n}$ , and let $J$ be the set of $j$ with $x \in R Q_{j}$ ; write $c_{j}$ for the centre, $ℓ_{j} = ℓ (Q_{j})$ and $B_{j} = B (c_{j}, ℓ_{j} / 2)$ . Distinct cubes of the family are disjoint and each contains $B_{j}$ , so the balls $B_{j}$ are pairwise disjoint. If $x \in R Q_{j}$ , then $x = c_{j} + R (y - c_{j})$ for some $y \in Q_{j}$ , so $dist (x, Q_{j}) \leq ∣ x - y ∣ = (R - 1) ∣ y - c_{j} ∣ \leq \frac{\sqrt{n}}{2} ℓ_{j}$ . Since $d$ is 1-Lipschitz and $d (z) \geq d (Q_{j})$ for $z \in Q_{j}$ , this gives $d (x) \geq d (Q_{j}) - dist (x, Q_{j}) \geq \frac{\sqrt{n}}{2} ℓ_{j} > 0$ . For the other direction, for each $δ > 0$ choose $u, v \in Q_{j}$ with $∣ x - u ∣ < dist (x, Q_{j}) + δ$ and $d (v) < d (Q_{j}) + δ$ . Lipschitz continuity and $∣ u - v ∣ \leq diam (Q_{j})$ give $d (x) \leq d (v) + ∣ x - v ∣ \leq d (Q_{j}) + dist (x, Q_{j}) + diam (Q_{j}) + 2 δ$ ; letting $δ ↓ 0$ yields $d (x) \leq d (Q_{j}) + dist (x, Q_{j}) + diam (Q_{j})$ . By step 5.1 and $diam (Q_{j}) = \sqrt{n} ℓ_{j}$ , $d (x) \leq 4 \sqrt{n} ℓ_{j} + \sqrt{n} ℓ_{j} + \frac{\sqrt{n}}{2} ℓ_{j} = \frac{11}{2} \sqrt{n} ℓ_{j},$ so $ℓ_{j} \geq 2 d (x) / (11 \sqrt{n})$ . The lower bound for $d (x)$ also gives $ℓ_{j} \leq 2 d (x) / \sqrt{n}$ , hence $∣ x - c_{j} ∣ \leq R \frac{\sqrt{n}}{2} ℓ_{j} \leq \sqrt{n} ℓ_{j} \leq 2 d (x)$ . The disjoint balls $B (c_{j}, d (x) / (11 \sqrt{n})) \subseteq B_{j}$ therefore have centres in $B (x, 3 d (x))$ and a common radius $d (x) / (11 \sqrt{n})$ . The packing estimate [L3] bounds their number by $(66 \sqrt{n} + 1)^{n}$ . Thus the dilated cubes have overlap at most $C_{n} = (66 \sqrt{n} + 1)^{n}$ , uniformly for $1 \leq R \leq 2$ .

7.1step 6.1L1algebra

Counting touching cubes. Fix $Q \in W$ with side $ℓ$ and generation $k$ . By step 6.1, a touching cube has side in $[ℓ / 5, 5 ℓ]$ , so its generation lies in ${k - 2, k - 1, k, k + 1, k + 2}$ . At generations $k, k + 1, k + 2$ , at most $3, 4, 6$ coordinate intervals, respectively, have closures meeting a given closed interval of length $ℓ$ ; hence the counts are at most $3^{n}, 4^{n}, 6^{n}$ . At each of the two coarser generations, the interval of $Q$ lies inside a single dyadic interval of that generation. Its closure can meet at most that interval and one adjacent interval, because $ℓ$ is strictly less than the coarse side length and all endpoints lie on the fine grid. Thus each coarser generation contributes at most $2^{n}$ cubes. Summing proves claim 3 with $K (n) = 2 \cdot 2^{n} + 3^{n} + 4^{n} + 6^{n}$ .

8.1step 4.1step 5.1step 6.1step 7.1step 6.2discharge-construct∎

Conclusion. Steps 4.1 and 5.1 provide a countable family of dyadic cubes with pairwise disjoint interiors, union $Ω$ and the two-sided distance estimate; step 6.1 gives the touching size comparison; step 7.1 gives the explicit touching count $K (n) = 2 \cdot 2^{n} + 3^{n} + 4^{n} + 6^{n}$ ; step 6.2 gives the bounded overlap of the dilations with $1 \leq R \leq 2$ . This proves the lemma.

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

Whitney-type ball cover with disjoint small balls and bounded overlap

Statement

Assume Countable Choice. Let $n \geq 1$ and let $Ω \subseteq R^{n}$ be nonempty, open and proper. Put $ρ (x) = dist (x, R^{n} ∖ Ω)$ for $x \in R^{n}$ . Then there is a countable family of points $ξ_{j} \in Ω$ , $j \in N$ , with $ρ_{j} : = ρ (ξ_{j})$ , such that

$Ω = ⋃_{j} B (ξ_{j}, ρ_{j} / 2)$ ;
the balls $B (ξ_{j}, ρ_{j} / 8)$ are pairwise disjoint (the source's maximal-selection construction records $ρ_{j} / 5$ , which the present choice-free greedy selection replaces by the fixed larger constant $8$ ; only the existence of a fixed constant matters below);
if $B (ξ_{j}, 3 ρ_{j} / 4) \cap B (ξ_{ν}, 3 ρ_{ν} / 4) \neq \emptyset$ , then $\frac{1}{7} ρ_{j} \leq ρ_{ν} \leq 7 ρ_{j}$ ;
for every $j$ at most $K (n) : = 785^{n}$ of the balls $B (ξ_{ν}, 3 ρ_{ν} / 4)$ meet $B (ξ_{j}, 3 ρ_{j} / 4)$ .

The family is the greedy subfamily of the countable rational grid ${B (q, ρ (q)) : q \in Ω \cap Q^{n}}$ : the grid is enumerated by restriction of a fixed enumeration of $Q^{n}$ , and the point $q^{(j)}$ is selected exactly when $B (q^{(j)}, ρ (q^{(j)}) / 8)$ meets none of the balls $B (q^{(i)}, ρ (q^{(i)}) / 8)$ with $i < j$ already selected. In particular no maximality principle and no choice beyond Countable Choice is used.

Facts & Assumptions

Given: $n \geq 1$ , $Ω$ nonempty, open and proper, and the distance function $ρ$ as in the statement.

[L1]

Since $Ω$ is proper, $A = R^{n} ∖ Ω$ is nonempty, so $ρ (x) = \inf_{a \in A} ∣ x - a ∣$ is finite and $∣ ρ (x) - ρ (y) ∣ \leq ∣ x - y ∣$ by $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz. If $x \in Ω$ , openness gives $r > 0$ with $B (x, r) \subseteq Ω$ , hence $ρ (x) \geq r > 0$ ; if $x \notin Ω$ , then $x \in A$ and $ρ (x) = 0$ . Here $B (x, r) = {y : ∣ y - x ∣ < r}$ as in Open ball, closed ball and sphere in a metric space.

[L2]

$Q^{n}$ is countable and dense in $R^{n}$ , so $Q^{n}$ admits a fixed enumeration $q^{(0)}, q^{(1)}, \dots$ and every nonempty open subset of $Ω$ contains a point of $Ω \cap Q^{n}$ ( $Q$ is countably infinite, Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable).

[L3]

For every $a \in R^{n}$ and $r > 0$ , $λ (B (a, r)) = c_{n} r^{n}$ with $c_{n} = ω_{n - 1} / n > 0$ ; this follows from the centred-ball formula and translation invariance. Lebesgue measure is finitely additive on disjoint measurable sets and monotone. Hence a finite family of pairwise disjoint open balls of common radius $r > 0$ with centres in a ball of radius $R$ has at most $(2 R / r + 1)^{n}$ members: they lie in the ball of radius $R + r$ , and comparing the volume of their union with that containing ball gives the bound (Sphere and ball measures scale in Rn, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume, Measures are monotone).

Proof technique: greedy selection on the countable rational grid, then the covering, comparison and packing estimates.

Proof

technique · constructive

1.1L1L2algebraconstruct

Grid covers $Ω$ . For $x \in Ω$ and $δ = ρ (x) / 64 > 0$ , density [L2] gives $q \in Ω \cap Q^{n}$ with $∣ x - q ∣ < δ$ . Then [L1] gives $ρ (q) \geq ρ (x) - ∣ x - q ∣ > (63 / 64) ρ (x)$ , so $ρ (q) > 0$ and $∣ x - q ∣ < ρ (x) / 64 < ρ (q) / 2$ ; hence $x \in B (q, ρ (q) / 2)$ . Thus ${B (q, ρ (q) / 2) : q \in Ω \cap Q^{n}}$ covers $Ω$ .

1.2L2given

Greedy selection. Enumerate $Ω \cap Q^{n}$ as $q^{(0)}, q^{(1)}, \dots$ by restriction of the fixed enumeration of $Q^{n}$ . Define $J \subseteq N$ recursively: $j \in J$ if and only if the ball $B (q^{(j)}, ρ (q^{(j)}) / 8)$ meets none of the balls $B (q^{(i)}, ρ (q^{(i)}) / 8)$ with $i < j$ , $i \in J$ ; the decision at step $j$ depends only on finitely many previous data, so this is a deterministic recursion requiring no choice. Writing $ξ_{j} = q^{(j)}$ and $ρ_{j} = ρ (ξ_{j})$ for $j \in J$ , the selected balls $B (ξ_{j}, ρ_{j} / 8)$ are pairwise disjoint by construction.

2.1step 1.1step 1.2L1L2algebra

Covering property. Let $x \in Ω$ and let $q = q^{(j)} \in Ω \cap Q^{n}$ satisfy $∣ x - q ∣ < ρ (x) / 64$ , which exists by density [L2]. If $j \in J$ , then $∣ x - ξ_{j} ∣ = ∣ x - q ∣ < ρ (x) / 64$ and $ρ_{j} = ρ (q) > 63 ρ (x) / 64$ , so $∣ x - ξ_{j} ∣ < ρ (x) / 64 < ρ_{j} / 63 < ρ_{j} / 2$ and $x \in B (ξ_{j}, ρ_{j} / 2)$ . If $j \notin J$ , then at step $j$ the ball $B (q, ρ (q) / 8)$ met some selected ball $B (ξ_{i}, ρ_{i} / 8)$ with $i < j$ , so $∣ q - ξ_{i} ∣ < \frac{ρ (q) + ρ_{i}}{8}, hence ρ (q) \leq ρ_{i} + ∣ q - ξ_{i} ∣ < ρ_{i} + \frac{ρ (q) + ρ_{i}}{8},$ which gives $\frac{7}{8} ρ (q) < \frac{9}{8} ρ_{i}$ , that is $ρ (q) < \frac{9}{7} ρ_{i}$ . Therefore $∣ q - ξ_{i} ∣ < \frac{1}{8} (\frac{9}{7} + 1) ρ_{i} = \frac{2}{7} ρ_{i} < \frac{1}{2} ρ_{i}$ , and, since $∣ x - q ∣ < ρ (x) / 64$ while $ρ (x) \leq ρ (q) + ∣ x - q ∣ < ρ (q) + ρ (x) / 64$ gives $ρ (x) < \frac{64}{63} ρ (q) < \frac{64}{63} \cdot \frac{9}{7} ρ_{i} = \frac{64}{49} ρ_{i}$ , we obtain $∣ x - ξ_{i} ∣ \leq ∣ x - q ∣ + ∣ q - ξ_{i} ∣ < \frac{ρ (x)}{64} + \frac{2}{7} ρ_{i} < \frac{1}{49} ρ_{i} + \frac{2}{7} ρ_{i} = \frac{15}{49} ρ_{i} < \frac{1}{2} ρ_{i} .$ Hence $x \in B (ξ_{i}, ρ_{i} / 2)$ in this case as well, which proves claim 1.

2.2step 1.2L1algebra

Comparison of meeting balls. Suppose $B (ξ_{j}, 3 ρ_{j} / 4) \cap B (ξ_{ν}, 3 ρ_{ν} / 4) \neq \emptyset$ . Then $∣ ξ_{j} - ξ_{ν} ∣ < \frac{3}{4} (ρ_{j} + ρ_{ν})$ and [L1] gives $ρ_{j} \leq ρ_{ν} + ∣ ξ_{j} - ξ_{ν} ∣ < ρ_{ν} + \frac{3}{4} ρ_{j} + \frac{3}{4} ρ_{ν}$ , hence $\frac{1}{4} ρ_{j} < \frac{7}{4} ρ_{ν}$ , that is $ρ_{j} < 7 ρ_{ν}$ ; interchanging $j, ν$ gives the reverse inequality. This proves claim 3.

3.1step 1.2step 2.2L3algebra

Bounded overlap. Fix $j$ and let $N$ be the set of $ν$ with $B (ξ_{ν}, 3 ρ_{ν} / 4) \cap B (ξ_{j}, 3 ρ_{j} / 4) \neq \emptyset$ . For $ν \in N$ step 2.2 gives $ρ_{ν} \leq 7 ρ_{j}$ , and $∣ ξ_{ν} - ξ_{j} ∣ < \frac{3}{4} (ρ_{j} + ρ_{ν}) \leq 6 ρ_{j}$ . The selected balls $B (ξ_{ν}, ρ_{ν} / 8)$ are pairwise disjoint, and their radii satisfy $ρ_{ν} / 8 \geq r : = ρ_{j} / 56$ . Thus the smaller balls $B (ξ_{ν}, r)$ , $ν \in N$ , remain pairwise disjoint. Their centres lie in $B (ξ_{j}, 6 ρ_{j})$ , so each smaller ball lies in $B (ξ_{j}, 6 ρ_{j} + r) \subset B (ξ_{j}, 7 ρ_{j})$ . For any finite subfamily, finite additivity and the ball-volume formula [L3] give $# F c_{n} r^{n} \leq c_{n} (7 ρ_{j})^{n},$ so $# F \leq (7 ρ_{j} / r)^{n} = 392^{n} \leq 785^{n}$ . Hence $N$ itself has at most $785^{n}$ members. This proves claim 4.

4.1step 1.1step 1.2step 2.1step 2.2step 3.1discharge-construct∎

Conclusion. Steps 1.1 and 1.2 provide a countable greedy family with covering property 1 and pairwise disjoint $B (ξ_{j}, ρ_{j} / 8)$ ; step 2.1 proves the covering property 1, step 2.2 gives the comparison property 3; step 3.1 gives the explicit finite overlap bound of property 4. This proves the lemma.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

Radial and nontangential maximal functions of a tempered distribution

Definition

Fix an integer $n \geq 1$ , work with $S (R^{n})$ and $S^{'} (R^{n})$ of Schwartz space and its seminorms and Tempered distribution, and let $φ \in S (R^{n})$ satisfy $\int_{R^{n}} φ \neq 0$ . For $t > 0$ write $φ_{t} (x) = t^{- n} φ (x / t), x \in R^{n} .$ For $f \in S^{'} (R^{n})$ define the radial maximal function $M_{φ}^{0} f (x) = \sup_{t > 0} ∣ (f * φ_{t}) (x) ∣, x \in R^{n},$ and, for an aperture $a \geq 1$ , the nontangential maximal function $M_{φ}^{*, a} f (x) = \sup_{t > 0} \sup_{∣ y - x ∣ \leq a t} ∣ (f * φ_{t}) (y) ∣, x \in R^{n} .$

Each convolution is the distributional convolution of Convolution of a tempered distribution with a schwartz function. It is well defined: $x \mapsto t^{- n} φ (x / t)$ again lies in $S (R^{n})$ by Dilations and their normalisations preserve Schwartz space, with scaling identities, so the pairing of $f$ with the reflected translate of $φ_{t}$ exists at every point. Its values are smooth and of polynomial growth by Tempered convolution is smooth with polynomial growth, so each function $x \mapsto (f * φ_{t}) (x)$ is finite at every point and the suprema displayed above are suprema of a nonempty family of real numbers; the radial case is the diagonal $y = x$ of the nontangential case, so $M_{φ}^{0} f \leq M_{φ}^{*, a} f$ for every $a \geq 1$ .

The aperture convention is $∣ y - x ∣ \leq a t$ with the closed cone, and the normalisation $t^{- n}$ is the one of the sources. Under Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), each $φ_{t}$ has the same integral as $φ$ by Dilations and their normalisations preserve Schwartz space, with scaling identities. This additional mass identity uses that supplier's stated choice premise. No measurability of $M_{φ}^{0} f$ or $M_{φ}^{*, a} f$ is asserted here, and the pointwise convolution and maximal-function definitions use no choice principle; the finiteness of the supremum at a point is not claimed, since the family ${(f * φ_{t}) (y)}$ need not be bounded a priori. Apertures $a > 1$ and the grand maximal function are treated in the following items.

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Schwartz approximate identities converge in the sense of tempered distributions

Statement

Assume Countable Choice. Let $n \geq 1$ , let $Φ \in S (R^{n})$ satisfy $\int_{R^{n}} Φ = 1$ , and for $t > 0$ write $Φ_{t} (x) = t^{- n} Φ (x / t)$ . Then for every $f \in S^{'} (R^{n})$ and every $ψ \in S (R^{n})$ , $⟨ Φ_{t} * f, ψ ⟩ ⟶ ⟨ f, ψ ⟩ (t ↓ 0),$ where the pairing is against the smooth convolution function of Convolution of a tempered distribution with a schwartz function; in other words $Φ_{t} * f \to f$ in $S^{'}$ as $t ↓ 0$ . Consequently for every sequence $t_{j} ↓ 0$ one has $Φ_{t_{j}} * f \to f$ in $S^{'}$ . If in addition $Φ * Φ$ denotes the everywhere-defined Schwartz convolution, then $Φ_{t} * Φ_{t} * f = (Φ * Φ)_{t} * f \to f$ in $S^{'}$ as $t ↓ 0$ , the dyadic instance being $Φ_{2^{- j}} * Φ_{2^{- j}} * f \to f$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $Φ \in S$ with $\int Φ = 1$ , $f \in S^{'}$ , $ψ \in S$ ; the seminorms and topology of Schwartz space and its seminorms and Schwartz topology and convergence; the convolution of Convolution of a tempered distribution with a schwartz function.

[F1]

For fixed $t > 0$ , $Φ_{t} \in S$ , $\int Φ_{t} = \int Φ = 1$ , and the translated and reflected family $x \mapsto Φ_{t} (x - \cdot)$ is smooth into $S$ : derivatives in $x$ correspond to derivatives of $Φ_{t}$ (Dilations and their normalisations preserve Schwartz space, with scaling identities, Schwartz parameter pairing and integral interchange).

[F2]

For $f \in S^{'}$ there are $C \geq 0$ and integers $N, M$ with $∣ ⟨ f, h ⟩ ∣ \leq C \max_{∣ α ∣ \leq N, ∣ β ∣ \leq M} p_{α β} (h)$ for all $h \in S$ (Finite seminorm bound characterizes tempered distributions).

[F3]

$C_{c}^{\infty} \subseteq S$ and Schwartz functions and all their polynomial multiples are integrable; in particular $\int ∣ Φ (w) ∣ (1 + ∣ w ∣)^{N + 1} d w < \infty$ for every $N$ (Schwartz derivatives are integrable).

[F4]

Substitution preserves Lebesgue integrals under Countable Choice, and $\int Φ = \int \overset{ˇ}{Φ}$ where $\overset{ˇ}{Φ} (u) = Φ (- u)$ (A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions).

Proof technique: reduce the distributional convergence to a Schwartz-norm estimate for the reflected approximate identity, then apply the finite-seminorm bound.

Proof

technique · direct

1.1F1F3F4algebra

The reflected approximate identity converges in Schwartz space. Put $\overset{ˇ}{Φ} (u) = Φ (- u)$ and $K_{t} = {\overset{ˇ}{Φ}}_{t} * ψ$ , so that $K_{t} (y) = \int_{R^{n}} Φ_{t} (x - y) ψ (x) d x = \int_{R^{n}} \overset{ˇ}{Φ} (w) ψ (y - t w) d w$ , the last form by the substitution $x = y - t w$ . Then $K_{t} - ψ = \int \overset{ˇ}{Φ} (w) (ψ (\cdot - t w) - ψ (\cdot)) d w$ because $\int \overset{ˇ}{Φ} = \int Φ = 1$ by [F1], [F4]. Fix multi-indices $α, β$ and $t \leq 1$ . Differentiating under the integral and applying the mean value theorem along the segment from $y$ to $y - t w$ gives $∣ \partial^{β} K_{t} (y) - \partial^{β} ψ (y) ∣ \leq \int_{R^{n}} ∣ \overset{ˇ}{Φ} (w) ∣ t ∣ w ∣ \int_{0}^{1} ∣ \nabla \partial^{β} ψ (y - s t w) ∣ d s d w .$ Since $∣ y^{α} ∣ \leq \sum_{γ \leq α} (\binom{α}{γ}) ∣ (y - s t w)^{γ} ∣ (t ∣ w ∣)^{∣ α ∣ - ∣ γ ∣} \leq (1 + t ∣ w ∣)^{∣ α ∣} \sum_{γ \leq α} (\binom{α}{γ}) ∣ (y - s t w)^{γ} ∣$ , taking the supremum in $y$ and using $t \leq 1$ yields $p_{α β} (K_{t} - ψ) \leq t C_{α} \int_{R^{n}} ∣ \overset{ˇ}{Φ} (w) ∣ (1 + ∣ w ∣)^{∣ α ∣ + 1} d w \cdot \max_{i \leq n} \max_{γ \leq α} p_{γ, β + e_{i}} (ψ),$ with $C_{α} = \sqrt{n} \sum_{γ \leq α} (\binom{α}{γ})$ , and the integral is finite by [F3]. Hence $p_{α β} (K_{t} - ψ) \to 0$ as $t ↓ 0$ : this is convergence in every Schwartz seminorm, i.e. $K_{t} \to ψ$ in $S$ .

2.1F1F2F3step 1.1given

The pairing identity. For every $t > 0$ , $⟨ Φ_{t} * f, ψ ⟩ = ⟨ f, K_{t} ⟩$ . Indeed, the parameter-pairing lemma applied to $H (x) = ψ (x) Φ_{t} (x - \cdot)$ and $u = f$ gives $⟨ f, \int H (x) d x ⟩ = \int ⟨ f, Φ_{t} (x - \cdot) ⟩ ψ (x) d x = \int (Φ_{t} * f) (x) ψ (x) d x$ , and $\int H (x) d x = K_{t}$ by the computation of step 1.1; the seminorm majorants required by that lemma are supplied by [F1] and [F3], since $∣ ψ (x) ∣ (1 + ∣ x ∣)^{N}$ is integrable for every $N$ .

3.1F1F2step 1.1step 2.1algebra∎

Conclusion. By step 1.1 $K_{t} \to ψ$ in $S$ , so the finite-seminorm bound of [F2] gives $⟨ f, K_{t} ⟩ \to ⟨ f, ψ ⟩$ ; step 2.1 identifies this with $⟨ Φ_{t} * f, ψ ⟩ \to ⟨ f, ψ ⟩$ as $t ↓ 0$ , which is convergence $Φ_{t} * f \to f$ in $S^{'}$ by the definition of that convergence. Sequences and the dyadic scale $t = 2^{- j}$ are instances. For the convolution form, $Φ * Φ \in S$ by Schwartz convolution and product laws, $\int (Φ * Φ) = (\int Φ)^{2} = 1$ , and the substitution $z = t w$ gives $(Φ * Φ)_{t} = Φ_{t} * Φ_{t}$ , so the statement applies to the Schwartz function $Φ * Φ$ . This proves the lemma.

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Deconvolution of a Schwartz function along the dyadic dilates of a fixed kernel

Statement

Assume Countable Choice. Let $n \geq 1$ and let $φ \in S (R^{n})$ satisfy $\int_{R^{n}} φ \neq 0$ . Then there is a constant $s_{0} > 0$ (depending only on $φ$ and $n$ ) such that for all integers $L, N > 0$ there exist $C > 0$ and $M > 0$ (depending on $φ, n, L, N$ but not on the input function) with the following property: for every $ψ \in S$ there are $η^{j} \in S$ , $j \geq 0$ , such that $ψ = \sum_{j = 0}^{\infty} η^{j} * φ_{s_{0} 2^{- j}} in S (R^{n}),$ and $∥ η^{j} ∥_{S_{N}} \leq C 2^{- j n L} ∥ ψ ∥_{S_{M}} (j \geq 0),$ where $φ_{t} (x) = t^{- n} φ (x / t)$ and $∥ h ∥_{S_{N}} : = \sup_{x \in R^{n}} \max (1, ∣ x ∣)^{N} \max_{∣ α ∣ \leq N} ∣ \partial^{α} h (x) ∣ .$ Any smaller positive value of $s_{0}$ also works, with the same conclusion and constants depending on the chosen value. The point of the estimate is that the coefficients $η^{j}$ become rapidly small in the strong Schwartz norm as $j \to \infty$ , uniformly in $ψ$ : this is what makes the deconvolution usable inside maximal-function estimates. Countable Choice is used through the Fourier automorphism, differentiation identities, and Schwartz convolution laws cited below.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $φ \in S$ with $\int φ \neq 0$ , and the seminorms and topology of Schwartz space and its seminorms, Schwartz topology and convergence, $C^{k}$ maps and multi-index derivative notation in Euclidean space.

[F1]

Fourier transformation is a topological automorphism of $S (R^{n})$ , with $\hat{φ_{t}} (ξ) = \hat{φ} (t ξ)$ for $t > 0$ (Fourier transform is a topological automorphism of Schwartz space, Fourier differentiation and multiplication identities on tempered distributions). In particular, the inverse transform of a compactly supported smooth function is Schwartz, and $F (f * g) = \hat{f} \hat{g}$ for Schwartz $f, g$ (Schwartz convolution and product laws).

[F2]

For every $m \in N$ there is $C_{m}$ with $∥ \hat{h} ∥_{S_{m}} \leq C_{m} ∥ h ∥_{S_{m + n + 1}}$ for all $h \in S$ : for multi-indices $∣ α ∣, ∣ β ∣ \leq m$ the identity $(2 π i)^{∣ β ∣} ξ^{β} \partial^{α} \hat{h} (ξ) = (- 2 π i)^{∣ α ∣} \int_{R^{n}} e^{- 2 π i x \cdot ξ} \partial^{β} (x^{α} h (x)) d x$ combined with the higher product rule and $\int (1 + ∣ x ∣)^{- n - 1} d x < \infty$ bounds $∣ ξ^{β} \partial^{α} \hat{h} (ξ) ∣$ by a finite sum of $S_{m + n + 1}$ seminorms of $h$ (Fourier differentiation and multiplication identities on tempered distributions, Basic operations are continuous on Schwartz space).

[F3]

Dilations act on $S$ with $p_{α β} (φ_{t}) = t^{∣ α ∣ - ∣ β ∣ - n} p_{α β} (φ)$ , so $φ_{t} \in S$ for every $t > 0$ (Dilations and their normalisations preserve Schwartz space, with scaling identities).

Proof technique: Fourier-side construction of a smooth dyadic partition and inversion of the symbol $\hat{φ}$ on the annuli where it does not vanish.

Proof

technique · constructive

1.1givenF1construct

Normalisation and scaling of the kernel. The construction below is uniform in the scale: for a fixed parameter $s_{0} > 0$ the annuli ${∣ ξ ∣ ≍ s_{0}^{- 1} 2^{j}}$ play the role of the annuli ${∣ ξ ∣ ≍ 2^{j}}$ at $s_{0} = 1$ , and every estimate keeps the same form with constants depending on $s_{0}$ ; in particular the same argument run at a smaller parameter gives the statement for every smaller scale, the constants changing by a fixed factor. Since $\hat{φ} (0) = \int φ \neq 0$ and $\hat{φ}$ is continuous, after multiplying $φ$ by the nonzero complex multiple $(1 / \int φ)$ and then replacing it by a suitable positive dilation $δ^{n} φ (δ \cdot) = φ_{1 / δ}$ , we may assume $\int_{R^{n}} φ = 1, ∣ \hat{φ} (ξ) ∣ \geq \frac{1}{2} for ∣ ξ ∣ \leq 2.$ We prove the lemma in this normalisation with $s_{0} = 1$ ; undoing the dilation replaces the scale $1$ by the fixed positive number $1 / δ$ and does not change the form of the estimates.

2.1step 1.1givenalgebra

A smooth dyadic partition of unity. With the smooth step $σ$ of The standard smooth step function, take $ζ (ξ) = σ ((9 / 4 - ∣ ξ ∣^{2}) / (5 / 4))$ ; it equals one on $B (0, 1)$ and its support is contained in the closed ball of radius $3 / 2$ , hence in $B (0, 2)$ , and put $ζ_{0} = ζ$ and $ζ_{j} (ξ) = ζ (2^{- j} ξ) - ζ (2^{- j + 1} ξ)$ for $j \geq 1$ . Then for every $J \geq 0$ , $\sum_{j = 0}^{J} ζ_{j} (ξ) = ζ (2^{- J} ξ),$ so $\sum_{j \geq 0} ζ_{j} (ξ) = 1$ for every $ξ$ : at $ξ = 0$ both sides equal one because $ζ (0) = 1$ , and for $ξ \neq 0$ the limit $ζ (2^{- J} ξ) \to ζ (0) = 1$ as $J \to \infty$ gives the identity. If $ζ_{j} (ξ) \neq 0$ , then $2^{- j} ξ \in supp ζ$ or $2^{- j + 1} ξ \in supp ζ$ , so $∣ 2^{- j} ξ ∣ \leq 2$ in either case; by step 1.1, $∣ \hat{φ} (2^{- j} ξ) ∣ \geq \frac{1}{2}$ on $supp ζ_{j}$ .

3.1F1F3step 2.1givenalgebra

The deconvolution coefficients and convergence. For $ψ \in S$ and $j \geq 0$ define the compactly supported smooth function $\hat{η^{j}} (ξ) = \frac{ζ_{j} (ξ)}{\hat{φ} (2^{- j} ξ)} \hat{ψ} (ξ) .$ The quotient is well defined and smooth on a neighbourhood of $supp ζ_{j}$ by step 2.1, and $\hat{η^{j}} \in C_{c}^{\infty} (R^{n}) \subseteq S$ ; hence $η^{j} : = F^{- 1} \hat{η^{j}} \in S$ by [F1]. On the Fourier side, for every $J \geq 0$ , $F (\sum_{j = 0}^{J} η^{j} * φ_{2^{- j}}) (ξ) = \hat{ψ} (ξ) \sum_{j = 0}^{J} ζ_{j} (ξ) = \hat{ψ} (ξ) ζ (2^{- J} ξ),$ where we used $\hat{φ_{2^{- j}}} (ξ) = \hat{φ} (2^{- j} ξ)$ from [F1]. To prove convergence in $S$ , put $χ_{J} (ξ) = ζ (2^{- J} ξ) - 1$ . The multiplier $χ_{J}$ itself is not Schwartz and does not converge to zero in $S$ ; instead we prove $\hat{ψ} χ_{J} \to 0$ in every Schwartz seminorm. Fix multi-indices $α, β$ . Leibniz's rule writes $\partial^{β} (\hat{ψ} χ_{J})$ as a finite sum of terms $C_{γ, β} (\partial^{γ} \hat{ψ}) (\partial^{β - γ} χ_{J})$ , $γ \leq β$ . For the term with $γ = β$ , the cutoff is undifferentiated: $χ_{J} = 0$ on $∣ ξ ∣ \leq 2^{J}$ because $ζ = 1$ on $B (0, 1)$ , and $∣ χ_{J} ∣ \leq 1 + ∥ ζ ∥_{\infty}$ everywhere. Thus its $p_{α β}$ contribution is bounded by $(1 + ∥ ζ ∥_{\infty}) \sup_{∣ ξ ∣ \geq 2^{J}} ∣ ξ^{α} \partial^{β} \hat{ψ} (ξ) ∣$ , which tends to zero by Schwartz decay. For every term with $γ < β$ , let $δ = β - γ \neq 0$ . The chain rule gives $\partial^{δ} χ_{J} (ξ) = 2^{- J ∣ δ ∣} (\partial^{δ} ζ) (2^{- J} ξ)$ , supported in the annulus $2^{J} \leq ∣ ξ ∣ \leq 2^{J + 1}$ since $ζ$ is constant on $B (0, 1)$ and vanishes outside $B (0, 2)$ . Its contribution is therefore bounded by $C_{δ} 2^{- J ∣ δ ∣} \sup_{∣ ξ ∣ \geq 2^{J}} ∣ ξ^{α} \partial^{γ} \hat{ψ} (ξ) ∣$ , which also tends to zero. There are only finitely many terms for each $α, β$ , so $p_{α β} (\hat{ψ} χ_{J}) \to 0$ . This proves $\hat{ψ} ζ (2^{- J} \cdot) \to \hat{ψ}$ in $S$ . Since Fourier transformation is a homeomorphism of $S$ [F1], the partial sums converge to $ψ$ in $S$ , and (with $\sum_{j \geq 0}$ denoting that limit) $ψ = \sum_{j \geq 0} η^{j} * φ_{2^{- j}}$ .

4.1step 2.1step 3.1F1F2algebra

The rapid norm decay. Fix $L, N > 0$ ; all constants below depend on $φ, n, L, N$ only. For $j \geq 1$ , on $supp ζ_{j}$ one has $(1 + ∣ ξ ∣) ≍ 2^{j}$ : if $ζ_{j} (ξ) \neq 0$ then $2^{- j} ∣ ξ ∣ \leq 2$ , while $2^{- j + 1} ∣ ξ ∣ \geq 1$ because otherwise both $ζ (2^{- j} ξ)$ and $ζ (2^{- j + 1} ξ)$ would equal one, so $\frac{1}{2} \cdot 2^{j} \leq 1 + ∣ ξ ∣ \leq 3 \cdot 2^{j}$ . For $j = 0$ , the support lies in a fixed ball and all the following estimates hold by enlarging the constant, since the target factor is $2^{- 0 n L} = 1$ . The quotient $ξ \mapsto ζ_{j} (ξ) / \hat{φ} (2^{- j} ξ)$ is smooth on the neighbourhood ${∣ ξ ∣ < 2^{j + 1}}$ of $supp ζ_{j}$ , and its derivatives of order at most $N + n + 1$ are bounded by a constant $C_{0} = C_{0} (n, N, φ)$ independent of $j$ : the chain rule contributes the factors $2^{- j ∣ β ∣} \leq 1$ to the derivatives both of $ζ_{j}$ and of the composition of $1 / \hat{φ}$ with $ξ \mapsto 2^{- j} ξ$ , and $1 / \hat{φ}$ is smooth with bounded derivatives on the fixed ball $∣ z ∣ \leq 2$ , where $∣ \hat{φ} ∣ \geq \frac{1}{2}$ by step 1.1. Multiplying by $\hat{ψ}$ with the higher product rule, $∣ \partial^{α} \hat{η^{j}} (ξ) ∣ \leq C_{1} \max_{∣ β ∣ \leq N + n + 1} ∣ \partial^{β} \hat{ψ} (ξ) ∣, ∣ α ∣ \leq N + n + 1, ξ \in supp ζ_{j} .$ Multiplying by $(1 + ∣ ξ ∣)^{N + n + 1}$ and using the definition of the $S_{M}$ norm with $M \geq N + n + 1$ together with the lower bound just proved, $∥ \hat{η^{j}} ∥_{S_{N + n + 1}} \leq C_{1} \sup_{ξ \in supp ζ_{j}} (1 + ∣ ξ ∣)^{N + n + 1 - M} ∥ \hat{ψ} ∥_{S_{M}} \leq C_{2} 2^{- j (M - N - n - 1)} ∥ \hat{ψ} ∥_{S_{M}},$ so with $M \geq N + n + 1 + n L$ the right-hand side is at most $C_{2} 2^{- j n L} ∥ \hat{ψ} ∥_{S_{M}}$ . Finally [F2] applied with the roles of a function and its transform interchanged gives $∥ η^{j} ∥_{S_{N}} \leq C_{3} ∥ \hat{η^{j}} ∥_{S_{N + n + 1}}$ (the Fourier transform of $\hat{η^{j}}$ is $η^{j} (- \cdot)$ , which has the same seminorms), and [F2] gives $∥ \hat{ψ} ∥_{S_{M}} \leq C_{M} ∥ ψ ∥_{S_{M + n + 1}}$ . Hence $∥ η^{j} ∥_{S_{N}} \leq C 2^{- j n L} ∥ ψ ∥_{S_{M + n + 1}}$ with $C = C_{2} C_{3} C_{M}$ independent of $ψ$ and $j$ , so the statement holds with $M$ replaced by $M + n + 1$ .

5.1step 1.1step 2.1step 3.1step 4.1discharge-construct∎

Conclusion. Steps 2.1 and 3.1 construct $η^{j} \in S$ with $ψ = \sum_{j \geq 0} η^{j} * φ_{2^{- j}}$ in $S$ , and step 4.1 gives the estimate $∥ η^{j} ∥_{S_{N}} \leq C 2^{- j n L} ∥ ψ ∥_{S_{M}}$ with constants independent of $ψ$ . Undoing the normalisation of step 1.1 replaces the scale family $2^{- j}$ by $s_{0} 2^{- j}$ with the fixed $s_{0} = 1 / δ$ and does not affect the convergence or the estimates, and the same construction run at a smaller parameter gives the statement there. This proves the lemma.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Grand maximal test class of order N and the grand maximal function

Definition

Fix an integer $n \geq 1$ and let $S (R^{n})$ carry the seminorms and topology of Schwartz space and its seminorms and Schwartz topology and convergence, with multi-indices as in $C^{k}$ maps and multi-index derivative notation in Euclidean space. For each integer $N \geq 1$ and $φ \in S (R^{n})$ define the Schwartz test seminorm of order $N$ $P_{N} (φ) = \sup_{x \in R^{n}} (1 + ∣ x ∣)^{N} \max_{∣ α ∣ \leq N + 1} ∣ \partial^{α} φ (x) ∣$ and the grand maximal test class of order $N$ $F_{N} = {φ \in S (R^{n}) : P_{N} (φ) \leq 1} .$

For $f \in S^{'} (R^{n})$ the grand maximal function of order $N$ is $M_{N} f (x) = \sup_{φ \in F_{N}} \sup_{t > 0} \sup_{∣ y - x ∣ \leq t} ∣ (f * φ_{t}) (y) ∣, x \in R^{n},$ where $φ_{t} (u) = t^{- n} φ (u / t)$ and convolution is the distributional convolution of Convolution of a tempered distribution with a schwartz function. The dilated test $φ_{t}$ is Schwartz for every $t > 0$ (Dilations and their normalisations preserve Schwartz space, with scaling identities), so each displayed convolution is defined even when $\int φ = 0$ . Thus the full class $F_{N}$ , including its zero-integral tests, is used. Each convolution has a finite scalar value and the supremum is a well-defined function with values in $[0, \infty]$ ; the value $+ \infty$ is allowed and no measurability is asserted here. The class $F_{N}$ contains the zero function, is symmetric under $φ \mapsto - φ$ and under complex conjugation, and is nonempty for every $N$ . No choice principle is used in this definition.

The order $N$ is a parameter. The characterisation theorem on this page fixes a finite admissible order $N_{0} (n, p, φ) < \infty$ depending only on $n$ , $p$ and the fixed kernel $φ$ and works for every $N \geq N_{0} (n, p, φ)$ ; the value of $N_{0}$ is whatever the accumulated comparison estimates of that proof require, and its existence, not an explicit formula, is what the page uses. The sources record the explicit sufficient choices $N \geq ⌊ n / p ⌋ + 1$ for the nontangential class $F_{N}$ in [DKKP, Proposition 1, p. 60], $N > 1 + n / p$ for the radial class $B_{N}$ (with derivatives through order $N$ ) in [MSV, section 1, p. 16], and $N > n / p + n + 1$ in [CUW, Theorem 3.1, p. 8]; these recorded choices are not used as the definition of $N_{0}$ below. If $N^{'} \geq N$ then $(1 + ∣ x ∣)^{N^{'}} \max_{∣ α ∣ \leq N^{'} + 1} ∣ \partial^{α} φ (x) ∣ \geq (1 + ∣ x ∣)^{N} \max_{∣ α ∣ \leq N + 1} ∣ \partial^{α} φ (x) ∣$ pointwise, so $F_{N^{'}} \subseteq F_{N}$ and hence $M_{N^{'}} f \leq M_{N} f pointwise on R^{n}$ for every $f \in S^{'}$ : the grand maximal functions are monotone in the order. The aperture is fixed to one; the comparison with larger apertures is the subject of the domination lemma on this page.

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The tangential maximal function is controlled by the aperture-one nontangential maximal function in $L^{p}$

Statement

Assume Countable Choice. Let $n \geq 1$ , $0 < q 0} \sup_{y \in R^{n}} \frac{∣ (f * φ_{t}) (x - y) ∣}{(1 + ∣ y ∣ / t)^{T}}, x \in R^{n} .$ For a nonnegative Borel function $g$ define the extended centered average $\tilde{M} g (x) : = \sup_{r > 0} \frac{1}{λ (B (x, r))} \int_{B (x, r)} g,$ where the nonnegative Lebesgue integral may be $+ \infty$ . If $g \in L_{l o c}^{1}$ , then $\tilde{M} g$ agrees with the centered Hardy-Littlewood maximal function $M g$ of The centered and uncentered Hardy-Littlewood maximal functions. The aperture-one nontangential maximal function $M_{φ}^{*, 1} f$ is the one of Radial and nontangential maximal functions of a tempered distribution. Then, pointwise, $M_{φ, T} f (x)^{q} \leq \tilde{M} ((M_{φ}^{*, 1} f)^{q}) (x) .$ If $∥ M_{φ}^{*, 1} f ∥_{L^{p}} < \infty$ , then $∥ M_{φ, T} f ∥_{L^{p}} \leq C_{n, p, q} ∥ M_{φ}^{*, 1} f ∥_{L^{p}},$ with $C_{n, p, q}$ depending only on $n, p, q$ . The norm inequality also holds in the extended sense when the right-hand side is infinite, in which case it is trivial.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $0 < q < p < \infty$ , $T = n / q$ , $φ \in S$ with $\int φ \neq 0$ , $f \in S^{'}$ and $x \in R^{n}$ .

[L1]

For every $a \in R^{n}$ and $r > 0$ , $λ (B (a, r)) = c_{n} r^{n}$ with $c_{n} = ω_{n - 1} / n > 0$ : the centred-ball formula is Sphere and ball measures scale in Rn and translation invariance is Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation. Hence $B (x - y, t) \subseteq B (x, ∣ y ∣ + t)$ and the volume ratio is $(1 + ∣ y ∣ / t)^{n}$ .

[F1]

For every $z \in B (x - y, t)$ one has $∣ (f * φ_{t}) (x - y) ∣ \leq M_{φ}^{*, 1} f (z)$ , because $∣ z - (x - y) ∣ < t$ and the supremum defining $M_{φ}^{*, 1} f (z)$ runs over $t > 0$ and all points within distance $t$ of $z$ (Radial and nontangential maximal functions of a tempered distribution).

[F2]

The centered Hardy-Littlewood maximal operator satisfies the strong $L^{r}$ bound $∥ M g ∥_{r} \leq C_{n, r} ∥ g ∥_{r}$ for $1 < r < \infty$ (The centered maximal operator is bounded on $L^{p} (R^{n})$ for $1 < p < \infty$ , The centered and uncentered Hardy-Littlewood maximal functions).

[F3]

The tangential maximal function is Borel because it is a supremum, over fixed $t, y$ , of continuous functions of $x$ . The aperture-one nontangential maximal function is Borel for $f \in S^{'}$ (Measurability and lower semicontinuity of the smooth maximal functions). Also, for every $r > 1$ , $L^{r} (R^{n}) \subset L_{l o c}^{1} (R^{n})$ : for compact $K$ , Holder gives $\int_{K} ∣ g ∣ \leq λ (K)^{1 - 1 / r} ∥ g ∥_{r}$ , and bounded sets have finite measure (Holder's inequality for integrals, including the endpoint cases, Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure).

Proof technique: local averaging over the ball of radius $t$ and the Hardy-Littlewood maximal bound.

Proof

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1.1L1F1F3algebra

Pointwise bound. Fix $x, y, t$ and put $g = (M_{φ}^{*, 1} f)^{q}$ , which is nonnegative Borel by [F3]. By [F1], $∣ (f * φ_{t}) (x - y) ∣^{q} \leq g (z)$ for every $z \in B (x - y, t)$ . Averaging (allowing an infinite integral) and enlarging to $B (x, ∣ y ∣ + t)$ gives $∣ (f * φ_{t}) (x - y) ∣^{q} \leq \frac{1}{λ (B (x - y, t))} \int_{B (x - y, t)} g \leq \frac{λ (B (x, ∣ y ∣ + t))}{λ (B (x - y, t))} \cdot \frac{1}{λ (B (x, ∣ y ∣ + t))} \int_{B (x, ∣ y ∣ + t)} g \leq (1 + ∣ y ∣ / t)^{n} \tilde{M} g (x),$ by [L1]. Since $T q = n$ , division by $(1 + ∣ y ∣ / t)^{T q}$ and taking the supremum over $t, y$ proves the pointwise inequality.

2.1step 1.1F2F3algebra

$L^{p}$ bound. If $∥ M_{φ}^{*, 1} f ∥_{L^{p}} = \infty$ , the asserted norm inequality is trivial. Otherwise [F3] gives that $g = (M_{φ}^{*, 1} f)^{q}$ is Borel and belongs to $L^{p / q}$ . Since $p / q > 1$ , [F3] also gives $g \in L_{l o c}^{1}$ , so $\tilde{M} g = M g$ . Apply [F2] with $r = p / q$ and use step 1.1: $∥ M_{φ, T} f ∥_{L^{p}}^{q} = ∥ (M_{φ, T} f)^{q} ∥_{L^{p / q}} \leq ∥ M g ∥_{L^{p / q}} \leq C_{n, p, q} ∥ g ∥_{L^{p / q}} = C_{n, p, q} ∥ M_{φ}^{*, 1} f ∥_{L^{p}}^{q} .$ Taking $q$ -th roots proves the estimate (with the constant renamed).

3.1step 1.1step 2.1∎

Conclusion. Step 1.1 gives pointwise domination by the extended centered average, which agrees with the ordinary maximal operator on the locally integrable input in step 2.1; the strong $L^{p / q}$ bound then proves the norm estimate.

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The grand maximal function is pointwise dominated by a tangential maximal function

Statement

Assume Countable Choice. Let $n \geq 1$ , $T > 0$ and let $φ \in S (R^{n})$ with $\int φ \neq 0$ . Then there are $N = N (n, φ, T)$ and $C = C (n, φ, T) < \infty$ such that for every $f \in S^{'} (R^{n})$ and every $x \in R^{n}$ , $M_{N} f (x) \leq C M_{φ, T} f (x),$ where $M_{N}$ is the grand maximal function of Grand maximal test class of order N and the grand maximal function and $M_{φ, T}$ is the tangential maximal function $M_{φ, T} f (x) = \sup_{t > 0} \sup_{y} ∣ (f * φ_{t}) (x - y) ∣ (1 + ∣ y ∣ / t)^{- T}$ . Consequently $∥ M_{N} f ∥_{L^{p}} \leq C ∥ M_{φ, T} f ∥_{L^{p}}$ for every $0 < p \leq \infty$ , both sides extended values.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $T > 0$ , $φ \in S$ with $\int φ \neq 0$ , $f \in S^{'}$ .

[F1]

Deconvolution: for every $ψ \in S$ and every choice of positive integer parameters $L^{'}, N^{'}$ there are $C_{d e c}, M_{d e c}, s_{0}$ and $η^{j} \in S$ with $ψ = \sum_{j \geq 0} η^{j} * φ_{s_{0} 2^{- j}}$ in $S$ and $∥ η^{j} ∥_{S_{N^{'}}} \leq C_{d e c} 2^{- j n L^{'}} ∥ ψ ∥_{S_{M_{d e c}}}$ (Deconvolution of a Schwartz function along the dyadic dilates of a fixed kernel); here $S_{M}$ is the norm $h \mapsto \sup_{w} (1 + ∣ w ∣)^{M} \max_{∣ α ∣ \leq M} ∣ \partial^{α} h (w) ∣$ . The supplier's weight $\max (1, ∣ w ∣)^{M}$ and this weight satisfy $\max (1, ∣ w ∣)^{M} \leq (1 + ∣ w ∣)^{M} \leq 2^{M} \max (1, ∣ w ∣)^{M}$ ; absorb $2^{N^{'}}$ in $C_{d e c}$ .

[F2]

If $Ψ \in S$ and $t > 0$ , then $(f * Ψ_{t}) (x) = \int_{R^{n}} (f * φ_{s t}) (x - w) η_{t} (w) d w$ whenever $Ψ = η * φ_{s}$ in the sense of the decomposition of [F1]: apply Schwartz parameter pairing and integral interchange to $H (w) (z) = η_{t} (w) φ_{s t} (x - w - z)$ . Each Schwartz seminorm of this family is bounded by $C (1 + ∣ w ∣)^{m} ∣ η_{t} (w) ∣$ , an integrable function because $η_{t}$ is Schwartz. The interchange proves the identity. Applying continuity of $f$ to the reflected translate of each partial sum in [F1] also justifies passage to the series; the subsequent nonnegative estimates apply to finite sums first, then to their limit.

[F3]

The translated kernel $ψ^{v} (w) = ψ (w + v)$ satisfies $S_{M} (ψ^{v}) \leq 2^{M} S_{M} (ψ)$ for $∣ v ∣ \leq 1$ , and if $P_{N} (ψ) \leq 1$ then $S_{M} (ψ) \leq 1$ for $M \leq N$ (Grand maximal test class of order N and the grand maximal function, Schwartz space and its seminorms).

Proof technique: insert the dyadic deconvolution identity and sum the rapidly decaying coefficients.

Proof

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Single-kernel estimate. Choose integers $L^{'} > T$ and $N^{'} > T + n$ (for example, $L^{'} = ⌊ T ⌋ + 1$ and $N^{'} = ⌊ T + n ⌋ + 1$ ), and apply [F1] with these parameters, obtaining $C_{d e c}, M_{d e c}, s_{0}$ . By the smaller-scale clause of [F1], decrease $s_{0}$ if necessary so that $s_{0} \leq 1$ . Take an integer $N \geq \max (M_{d e c}, N^{'})$ . Fix $ψ \in F_{N}$ , $t > 0$ and $x$ ; write the deconvolution of [F1] and set $s = s_{0} 2^{- j}$ . If $M_{φ, T} f (x) = + \infty$ , the desired estimate is immediate. Otherwise, by [F2], $∣ (f * ψ_{t}) (x) ∣ \leq \sum_{j \geq 0} \int_{R^{n}} ∣ (f * φ_{s t}) (x - w) ∣ ∣ η_{t}^{j} (w) ∣ d w \leq M_{φ, T} f (x) \sum_{j \geq 0} \int_{R^{n}} (1 + \frac{∣ w ∣}{s t})^{T} ∣ η_{t}^{j} (w) ∣ d w,$ where we used the definition of $M_{φ, T} f (x)$ with the displacement $w$ at scale $s t$ . Substituting $w = t u$ and using $η_{t}^{j} (t u) = t^{- n} η^{j} (u)$ gives $\int_{R^{n}} (1 + \frac{∣ w ∣}{s t})^{T} ∣ η_{t}^{j} (w) ∣ d w = \int_{R^{n}} (1 + \frac{∣ u ∣}{s})^{T} ∣ η^{j} (u) ∣ d u \leq s^{- T} \int_{R^{n}} (1 + ∣ u ∣)^{T} ∣ η^{j} (u) ∣ d u \leq C_{T, N^{'}} s^{- T} ∥ η^{j} ∥_{S_{N^{'}}} \leq C^{'} 2^{j T} 2^{- j n L^{'}} ∥ ψ ∥_{S_{M_{d e c}}},$ since $s \leq 1$ , $N^{'} > T + n$ makes $(1 + ∣ u ∣)^{T - N^{'}}$ integrable, and [F1] gives $∥ η^{j} ∥_{S_{N^{'}}} \leq C 2^{- j n L^{'}} ∥ ψ ∥_{S_{M_{d e c}}}$ . Also $∥ ψ ∥_{S_{M_{d e c}}} \leq 1$ by the choice of $N$ . The series converges because $n L^{'} > T$ , so $∣ (f * ψ_{t}) (x) ∣ \leq C^{''} M_{φ, T} f (x)$ with $C^{''}$ independent of $ψ, t, x$ .

2.1step 1.1F3givenalgebra

Aperture. For $t > 0$ and $y$ with $∣ x - y ∣ \leq t$ write $v = (y - x) / t$ , so $∣ v ∣ \leq 1$ , and let $ψ^{v} (w) = ψ (w + v)$ ; then $(f * ψ_{t}) (y) = (f * ψ_{t}^{v}) (x)$ : both equal $t^{- n} ⟨ f_{w}, ψ ((y - w) / t) ⟩$ and $t^{- n} ⟨ f_{w}, ψ ((x - w) / t + v) ⟩ = t^{- n} ⟨ f_{w}, ψ ((y - w) / t) ⟩$ . By [F3], $S_{M_{d e c}} (ψ^{v}) \leq 2^{M_{d e c}} S_{M_{d e c}} (ψ) \leq 2^{M_{d e c}}$ , so the kernel $Ψ = ψ^{v} / 2^{M_{d e c}}$ satisfies $∥ Ψ ∥_{S_{M_{d e c}}} \leq 1$ ; the argument of step 1.1 uses only this bound on the kernel and the deconvolution of [F1] with the kernel $φ$ , so it applies verbatim to $Ψ$ and gives $∣ (f * Ψ_{t}) (x) ∣ \leq C^{''} M_{φ, T} f (x)$ . Since $(ψ^{v} / 2^{M_{d e c}})_{t} = ψ_{t}^{v} / 2^{M_{d e c}}$ and $(f * ψ_{t}) (y) = (f * ψ_{t}^{v}) (x)$ , multiplying by $2^{M_{d e c}}$ gives $∣ (f * ψ_{t}) (y) ∣ \leq C^{''} 2^{M_{d e c}} M_{φ, T} f (x)$ . Taking the supremum over $ψ \in F_{N}$ , $t > 0$ and $∣ x - y ∣ \leq t$ gives $M_{N} f (x) \leq C M_{φ, T} f (x)$ . The $L^{p}$ statement follows by monotonicity of the integral; for $p = \infty$ it is the pointwise bound.

3.1step 1.1step 2.1∎

Conclusion. Step 1.1 controls a single test kernel by the tangential maximal function with a rapidly convergent deconvolution expansion, and step 2.1 removes the aperture restriction by translating the kernel; the class $F_{N}$ is then dominated pointwise. This proves the lemma.

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Measurability and lower semicontinuity of the smooth maximal functions

Statement

Let $n \geq 1$ , $f \in S^{'} (R^{n})$ and $φ \in S (R^{n})$ . Then the function $(y, t) \mapsto (f * φ_{t}) (y)$ is continuous on $R^{n} \times (0, \infty)$ , where $φ_{t} (x) = t^{- n} φ (x / t)$ and the convolution is the distributional convolution of Convolution of a tempered distribution with a schwartz function. If $\int_{R^{n}} φ \neq 0$ , the radial maximal function $M_{φ}^{0} f$ and every nontangential maximal function $M_{φ}^{*, a} f$ with $a \geq 1$ (Radial and nontangential maximal functions of a tempered distribution) are Borel measurable as $[0, \infty]$ -valued functions and may be identically $+ \infty$ . For every integer $N \geq 1$ , the grand maximal function $M_{N} f$ of Grand maximal test class of order N and the grand maximal function is also Borel measurable, may be identically $+ \infty$ , and its definition imposes no integral condition on the tests in $F_{N}$ . Moreover, when $\int φ \neq 0$ , $M_{φ}^{0} f$ and every $M_{φ}^{*, a} f$ are lower semicontinuous, and every $M_{N} f$ is lower semicontinuous for all $N \geq 1$ . Thus their strict superlevel sets are open; in particular $Ω_{r} = {M_{N} f > 2^{r}}$ , $r \in Z$ , used in the level decomposition are open.

Facts & Assumptions

Given: $n \geq 1$ , $f \in S^{'}$ , $φ \in S$ and an integer $N \geq 1$ . For the radial and nontangential conclusions, also assume $\int φ \neq 0$ and an aperture $a \geq 1$ .

[F1]

For every fixed $t > 0$ the function $x \mapsto (f * φ_{t}) (x)$ is smooth (Tempered convolution is smooth with polynomial growth); the convolution is $(f * φ_{t}) (x) = ⟨ f_{y}, φ_{t} (x - y) ⟩$ (Convolution of a tempered distribution with a schwartz function).

[F2]

Translations and dilations preserve $S$ continuously: $h \mapsto h (\cdot - c)$ is continuous in every seminorm for fixed $c$ , and the seminorms of $φ_{t}$ are $p_{α β} (φ_{t}) = t^{∣ α ∣ - ∣ β ∣ - n} p_{α β} (φ)$ (Basic operations are continuous on Schwartz space, Dilations and their normalisations preserve Schwartz space, with scaling identities, Schwartz space and its seminorms).

[F3]

A tempered distribution is continuous on $S$ , so convergence in every seminorm implies convergence of the pairings; this is the definition of tempered distribution and of the Schwartz topology (Tempered distribution, Schwartz topology and convergence).

Proof technique: direct seminorm estimates for the parameter family, then lower semicontinuity of suprema.

Proof

technique · direct

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Continuity of the parameter family in Schwartz space. Fix $(y_{0}, t_{0}) \in R^{n} \times (0, \infty)$ and a compact interval $[a, b] \subseteq (0, \infty)$ containing $t_{0}$ in its interior. For $h_{y, t} (z) = φ_{t} (y - z)$ and multi-indices $α, β$ , a first-order Taylor expansion of $\partial^{β} φ$ along the segment from $(y_{0} - z) / t$ to $(y - z) / t$ gives, for $t \in [a, b]$ and $ε : = ∣ y - y_{0} ∣ / a \leq 1$ , $p_{α β} (h_{y, t} - h_{y_{0}, t}) \leq C_{a, b} (1 + ∣ y_{0} ∣)^{∣ α ∣} \frac{∣ y - y_{0} ∣}{a} \sum_{∣ γ ∣ \leq ∣ α ∣} \sum_{∣ e ∣ = 1} p_{γ, β + e} (φ) .$ For the scale variation put $u = (y_{0} - z) / t$ . Differentiating $\partial_{z}^{β} h_{y_{0}, t} (z) = (- 1)^{∣ β ∣} t^{- n - ∣ β ∣} \partial^{β} φ (u)$ gives $\partial_{t} \partial_{z}^{β} h_{y_{0}, t} (z) = (- 1)^{∣ β ∣ + 1} t^{- n - ∣ β ∣ - 1} ((n + ∣ β ∣) \partial^{β} φ (u) + u \cdot \nabla \partial^{β} φ (u)) .$ Since $z = y_{0} - t u$ and $t \in [a, b]$ , $∣ z ∣^{∣ α ∣} \leq C_{a, b, α} (1 + ∣ y_{0} ∣)^{∣ α ∣} (1 + ∣ u ∣)^{∣ α ∣}$ . The factor $u$ in the scale derivative therefore requires one additional polynomial weight, and the mean-value estimate gives $p_{α β} (h_{y_{0}, t} - h_{y_{0}, t_{0}}) \leq C_{a, b, n, α, β} (1 + ∣ y_{0} ∣)^{∣ α ∣} ∣ t - t_{0} ∣ (\sum_{∣ γ ∣ \leq ∣ α ∣} p_{γ, β} (φ) + \sum_{∣ γ ∣ \leq ∣ α ∣ + 1} \sum_{∣ e ∣ = 1} p_{γ, β + e} (φ)) .$ The two estimates tend to $0$ as $(y, t) \to (y_{0}, t_{0})$ ; hence $(y, t) \mapsto h_{y, t}$ is continuous from $R^{n} \times (0, \infty)$ into $S$ .

2.1step 1.1F1F3

Joint continuity of the convolution. Since $(f * φ_{t}) (y) = ⟨ f, h_{y, t} ⟩$ by [F1] and $h_{y, t} \to h_{y_{0}, t_{0}}$ in $S$ by step 1.1, the continuity of $f$ on $S$ [F3] gives $(f * φ_{t}) (y) \to (f * φ_{t_{0}}) (y_{0})$ as $(y, t) \to (y_{0}, t_{0})$ . This proves the first clause, and in particular each function $y \mapsto (f * φ_{t}) (y)$ is continuous on $R^{n}$ for every fixed $t$ .

3.1step 2.1givenalgebra

Lower semicontinuity. Let $λ > 0$ . For the radial and nontangential functions assume $\int φ \neq 0$ , and suppose $M_{φ}^{*, a} f (x_{0}) > λ$ . Since by step 2.1 the function $(y, t) \mapsto ∣ (f * φ_{t}) (y) ∣$ is continuous and the closed cone ${∣ y - x_{0} ∣ \leq a t}$ is the closure of the open cone ${∣ y - x_{0} ∣ < a t}$ , the supremum over the open cone equals the supremum over the closed one: a witness in the closed cone with value $> λ$ can be moved slightly along the segment towards $x_{0}$ to a witness with $∣ y - x_{0} ∣ < a t$ and value still $> λ$ . Fix such $t > 0$ , $y$ with $∣ y - x_{0} ∣ < a t$ and $∣ (f * φ_{t}) (y) ∣ > λ$ ; by step 2.1 there is a neighbourhood $U$ of $y$ on which $∣ (f * φ_{t}) ∣ > λ$ . The set of $x$ with $∣ y - x ∣ < a t$ is open and contains $x_{0}$ , so $U^{'} = {x : ∣ y - x ∣ < a t}$ is a neighbourhood of $x_{0}$ on which $M_{φ}^{*, a} f (x) \geq ∣ (f * φ_{t}) (y) ∣ > λ$ for every $y \in U \cap$ the ball of radius $a t$ centred at $x$ : more precisely, for $x \in U^{'}$ choose $y^{'} \in U \subset B (x, a t)$ (possible because $U$ is a neighbourhood of $y$ and $∣ y - x ∣ < a t$ , so $U \cap B (x, a t) \neq \emptyset$ ), and then $M_{φ}^{*, a} f (x) > λ$ . Hence ${M_{φ}^{*, a} f > λ}$ is open and $M_{φ}^{*, a} f$ is lower semicontinuous. The radial case is identical with $y = x$ and the value $∣ (f * φ_{t}) (x_{0}) ∣ > λ$ : step 2.1 gives a neighbourhood of $x_{0}$ on which $∣ (f * φ_{t}) ∣ > λ$ , so $M_{φ}^{0} f > λ$ there. For the grand maximal function, with no integral restriction on $ψ \in F_{N}$ , if $M_{N} f (x_{0}) > λ$ , the direct definition gives $ψ \in F_{N}$ , $t > 0$ and $y$ with $∣ y - x_{0} ∣ \leq t$ and $∣ (f * ψ_{t}) (y) ∣ > λ$ . If $∣ y - x_{0} ∣ = t$ , continuity from step 2.1 lets us move $y$ slightly toward $x_{0}$ while keeping the value above $λ$ , so we may assume $∣ y - x_{0} ∣ < t$ . Then $U = {x : ∣ y - x ∣ < t}$ is an open neighbourhood of $x_{0}$ , and for every $x \in U$ the same $ψ, t, y$ is admissible in the defining supremum, giving $M_{N} f (x) > λ$ . Thus ${M_{N} f > λ}$ is open and $M_{N} f$ is lower semicontinuous.

4.1step 3.1∎

Measurability. An extended-real lower semicontinuous function is Borel: for each real $λ$ the set ${g > λ}$ is open, hence Borel, and the Borel structure of $[0, \infty]$ is generated by the open (or by the intervals $(λ, \infty]$ and $[0, λ)$ ) sets. Applying this to $M_{φ}^{0} f$ , $M_{φ}^{*, a} f$ and $M_{N} f$ by step 3.1 gives the stated Borel measurability, with values in $[0, \infty]$ ; the value $+ \infty$ is not excluded, and if it occurs it occurs on a measurable set. This proves the lemma.

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Truncated maximal functions: finiteness, comparison estimates and the good-set bound

Statement

Assume Countable Choice. Let $n \geq 1$ , $0 0$ , $T > 0$ and $f \in S^{'} (R^{n})$ define the truncated maximal functions $M_{φ, 0}^{ϵ, L} f (x) = \sup_{0 < t < 1 / ϵ} \frac{∣ (f * φ_{t}) (x) ∣ t^{L}}{(t + ϵ + ϵ ∣ x ∣)^{L}},$ $M_{φ, 1}^{ϵ, L} f (x) = \sup_{0 < t < 1 / ϵ} \sup_{∣ x - y ∣ < t} \frac{∣ (f * φ_{t}) (y) ∣ t^{L}}{(t + ϵ + ϵ ∣ y ∣)^{L}},$ $M_{φ, T}^{ϵ, L} f (x) = \sup_{0 < t < 1 / ϵ} \sup_{y \in R^{n}} \frac{∣ (f * φ_{t}) (x - y) ∣}{(1 + ∣ y ∣ / t)^{T}} \frac{t^{L}}{(t + ϵ + ϵ ∣ x - y ∣)^{L}},$ $M_{N}^{ϵ, L} f (x) = \sup_{ψ \in F_{N}} M_{ψ, 1}^{ϵ, L} f (x),$ where $M_{φ}^{0} f$ , $M_{φ}^{*, 1} f$ and $F_{N}$ are those of Radial and nontangential maximal functions of a tempered distribution and Grand maximal test class of order N and the grand maximal function. The displayed radial and aperture-one truncation formulas also apply to every Schwartz test $ψ$ , including tests of zero integral; the nonzero-integral hypothesis is needed only for estimates involving the fixed comparison kernel $φ$ . Then:

For a nonnegative Borel function $g$ , write $\tilde{M} g (x)$ for the supremum of its centered ball averages, with the nonnegative Lebesgue integral allowed to equal $+ \infty$ . If $g \in L_{l o c}^{1}$ , then $\tilde{M} g = M g$ for the centered maximal operator of The centered and uncentered Hardy-Littlewood maximal functions.

For every $f \in S^{'}$ and every $0 < p < \infty$ there is $L_{0} = L_{0} (f, n, φ, p) < \infty$ such that for every $L \geq L_{0}$ and every $0 < ϵ \leq 1 / 2$ the function $M_{φ, 1}^{ϵ, L} f$ belongs to $L^{p} (R^{n})$ and satisfies $M_{φ, 1}^{ϵ, L} f (x) \leq C ϵ^{- K} (1 + ∣ x ∣)^{- M}$ with some finite $C, K, M > 0$ depending on $f, L, p$ (so the quantity $∥ M_{φ, 1}^{ϵ, L} f ∥_{p}$ is finite).
For every $T > 0$ and $L > 0$ there is $N_{1}$ such that for every $N \geq N_{1}$ , every $0 < ϵ \leq 1 / 2$ , every $f \in S^{'}$ and every $x$ , $M_{N}^{ϵ, L} f (x) \leq C_{1} M_{φ, T}^{ϵ, L} f (x)$ with $C_{1} = C_{1} (n, φ, T, L)$ independent of $ϵ$ and $f$ .
For every $T > 0$ , setting $q = n / T$ and assuming $0 < q < p$ , one has for every $0 < ϵ \leq 1 / 2$ , $L > 0$ and $f \in S^{'}$ $M_{φ, T}^{ϵ, L} f (x)^{q} \leq \tilde{M} ((M_{φ, 1}^{ϵ, L} f)^{q}) (x) (x \in R^{n}), ∥ M_{φ, T}^{ϵ, L} f ∥_{L^{p}} \leq C_{2} ∥ M_{φ, 1}^{ϵ, L} f ∥_{L^{p}},$ where $\tilde{M}$ is the extended centered average defined above and $C_{2} = C_{2} (n, p, q)$ . The norm inequality is interpreted in the extended sense if its right-hand side is infinite.
For every $p_{0} > 0$ , $T > 0$ , $L > 0$ and $λ > 0$ there is $N_{2}$ such that for every $N \geq N_{2}$ there is $C_{3} = C_{3} (n, φ, p_{0}, T, L, λ, N) < \infty$ with the property that for every $0 < ϵ \leq 1 / 2$ , every $f \in S^{'}$ and every $x$ satisfying $M_{N}^{ϵ, L} f (x) < λ M_{φ, 1}^{ϵ, L} f (x)$ , $M_{φ, 1}^{ϵ, L} f (x) \leq C_{3} \tilde{M} ((M_{φ}^{0} f)^{p_{0}}) (x)^{1 / p_{0}} .$
(Untruncated good-set estimate.) For every $p_{0} > 0$ and $λ > 0$ there is $N_{3}$ such that for every $N \geq N_{3}$ there is $C_{4} = C_{4} (n, φ, p_{0}, λ, N) < \infty$ with the property that for every $f \in S^{'}$ and every $x$ with $M_{N} f (x) \leq λ M_{φ}^{*, 1} f (x) < \infty$ , $M_{φ}^{*, 1} f (x) \leq C_{4} \tilde{M} ((M_{φ}^{0} f)^{p_{0}}) (x)^{1 / p_{0}} .$ The finiteness $M_{φ}^{*, 1} f (x) < \infty$ is part of the hypothesis: no claim is made at points where $M_{φ}^{*, 1} f (x) = + \infty$ . For every $f$ with $M_{φ}^{*, 1} f$ finite a.e. the estimate therefore holds a.e. on the set $F = {M_{N} f \leq λ M_{φ}^{*, 1} f}$ .

The point of the truncation is that $M_{φ, 1}^{ϵ, L} f$ is finite and integrable, so the good-set argument of the last item can be run without an a priori finiteness assumption on $M_{φ}^{*, 1} f$ ; as $ϵ ↓ 0$ the truncated functions increase pointwise to the untruncated ones. Countable Choice is assumed through dyadic deconvolution and the measure-theoretic estimates used below.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $0 0$ , $f \in S^{'}$ ; the maximal functions of Radial and nontangential maximal functions of a tempered distribution and Grand maximal test class of order N and the grand maximal function.

[F1]

Finite-seminorm bound: there are integers $N_{0}, M \geq 0$ and $C$ with $∣ ⟨ f, h ⟩ ∣ \leq C \max_{∣ α ∣ \leq N_{0}, ∣ β ∣ \leq M} \sup_{z} ∣ z^{α} \partial^{β} h (z) ∣$ for all $h \in S$ . Applying this to $h (z) = φ_{t} (y - z)$ gives $∣ (f * φ_{t}) (y) ∣ \leq C_{φ} t^{- (n + M)} (1 + ∣ y ∣)^{N_{0}}$ when $0 < t \leq 1$ and $∣ (f * φ_{t}) (y) ∣ \leq C_{φ} t^{N_{0} - n} (1 + ∣ y ∣)^{N_{0}}$ when $t \geq 1$ : for small scales the largest derivative seminorm is bounded by $t^{- n - M}$ , while for large scales it is bounded by $t^{- n}$ and the polynomial weight contributes at most $t^{N_{0}}$ (Finite seminorm bound characterizes tempered distributions, Schwartz space and its seminorms).

[F2]

The centered Hardy-Littlewood maximal operator is defined for $L_{l o c}^{1}$ inputs and satisfies $∥ M g ∥_{L^{r}} \leq C_{n, r} ∥ g ∥_{L^{r}}$ for $1 < r < \infty$ ; also $M (h_{1} + h_{2}) \leq M h_{1} + M h_{2}$ (The centered and uncentered Hardy-Littlewood maximal functions, The centered maximal operator is bounded on $L^{p} (R^{n})$ for $1 < p < \infty$ ).

[F3]

The deconvolution lemma: for $φ$ with $\int φ \neq 0$ and every pair of positive integers $L^{'}, N^{'}$ , there are $C_{d e c}, M_{d e c}, s_{0} > 0$ such that for every $ψ \in S$ there are $η^{j} \in S$ with $ψ = \sum_{j \geq 0} η^{j} * φ_{s_{0} 2^{- j}}$ in $S$ and $∥ η^{j} ∥_{S_{N^{'}}} \leq C_{d e c} 2^{- j n L^{'}} ∥ ψ ∥_{S_{M_{d e c}}}$ Any smaller positive value of $s_{0}$ also works, with constants depending on that value (Deconvolution of a Schwartz function along the dyadic dilates of a fixed kernel).

[F4]

For every $a \in R^{n}$ and $r > 0$ , $λ (B (a, r)) = c_{n} r^{n}$ with $c_{n} = ω_{n - 1} / n > 0$ ; this follows from the centred-ball formula and translation invariance. Consequently, $B (x - y, t) \subseteq B (x, ∣ y ∣ + t)$ and $λ (B (x, ∣ y ∣ + t)) / λ (B (x - y, t)) = (1 + ∣ y ∣ / t)^{n}$ (Sphere and ball measures scale in Rn, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, Open ball, closed ball and sphere in a metric space).

[F5]

Translate bound for the test seminorm: for $ψ \in S$ and $h \in R^{n}$ one has $P_{N} (ψ (\cdot + h)) \leq (1 + ∣ h ∣)^{N} P_{N} (ψ)$ , because $(1 + ∣ w ∣) \leq (1 + ∣ h ∣) (1 + ∣ w + h ∣)$ pointwise; hence for the translated derivative kernel $Ψ^{z} (w) = (\partial_{j} φ) (w + (z - x) / t)$ and $∣ (z - x) / t ∣ \leq r + 1$ with $r \leq 1$ one has $P_{N} (Ψ^{z}) \leq 3^{N} P_{N} (\partial_{j} φ) \leq 3^{N} P_{N + 1} (φ) < \infty$ . Combining the componentwise bounds for $0 \leq j < n$ gives $∣ \nabla (f * φ_{t}) (z) ∣ \leq \sqrt{n}$ times this bound after applying the grand maximal estimate; write $c_{N, φ} : = \sqrt{n} 3^{N} P_{N + 1} (φ)$ for the resulting gradient constant (Grand maximal test class of order N and the grand maximal function, Schwartz space and its seminorms).

[F6]

For every $r > 1$ , $L^{r} (R^{n}) \subset L_{l o c}^{1} (R^{n})$ : if $K$ is compact, it is bounded and has finite measure, and Holder gives $\int_{K} ∣ g ∣ \leq λ (K)^{1 - 1 / r} ∥ g ∥_{L^{r}}$ (Holder's inequality for integrals, including the endpoint cases, Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure).

[F7]

The radial and aperture-one nontangential maximal functions are Borel measurable for every $f \in S^{'}$ (Measurability and lower semicontinuity of the smooth maximal functions). For fixed $ϵ, L$ , the truncated aperture-one function is Borel: each strict superlevel set is the union of the open balls $B (y, t)$ indexed by the admissible witnesses whose weighted convolution value exceeds that level. The truncated grand maximal function is Borel by the same open-ball superlevel argument, with tests also indexed over $F_{N}$ . The tangential function is Borel because it is a supremum, over fixed witnesses, of continuous functions of $x$ ; smoothness of each convolution is Tempered convolution is smooth with polynomial growth.

Proof technique: weighted distance estimates, the dyadic deconvolution comparison and a mean-value argument on the good set.

Proof

technique · direct

1.1F1F7givenalgebra

Finiteness bound. Let $N_{0}, M, C_{φ}$ be as in [F1], and choose $L_{0} = L_{0} (f, n, φ, p)$ so large that $L_{0} > n + M$ and $L_{0} > N_{0} + n / p$ . Fix $L \geq L_{0}$ . For a witness $(y, t)$ with $0 < t < 1 / ϵ$ and $∣ x - y ∣ < t$ , write $W = \frac{∣ (f * φ_{t}) (y) ∣ t^{L}}{(t + ϵ + ϵ ∣ y ∣)^{L}}$ . If $t \leq 1$ , [F1] and $t + ϵ + ϵ ∣ y ∣ \geq ϵ (1 + ∣ y ∣)$ give $W \leq C_{φ} t^{L - (n + M)} ϵ^{- L} (1 + ∣ y ∣)^{N_{0} - L} \leq C_{φ} ϵ^{- L} (1 + ∣ y ∣)^{N_{0} - L},$ because $L > n + M$ . If $t \geq 1$ , the large-scale bound in [F1] gives $W \leq C_{φ} t^{L + N_{0} - n} ϵ^{- L} (1 + ∣ y ∣)^{N_{0} - L} \leq C_{φ} ϵ^{- (L + \max {0, L + N_{0} - n})} (1 + ∣ y ∣)^{N_{0} - L},$ because when $L + N_{0} - n \geq 0$ one has $t^{L + N_{0} - n} \leq ϵ^{- (L + N_{0} - n)}$ , while when $L + N_{0} - n < 0$ the factor is at most $1$ . Thus in both cases $W \leq C^{'} ϵ^{- K} (1 + ∣ y ∣)^{- (L - N_{0})}$ for some finite $K$ . If $∣ x ∣ \geq 2 / ϵ$ , then $∣ y ∣ \geq ∣ x ∣ - t \geq ∣ x ∣ / 2$ , so this is at most $C^{''} ϵ^{- K} (1 + ∣ x ∣)^{- (L - N_{0})}$ . If $∣ x ∣ \leq 2 / ϵ$ , then $1 + ∣ x ∣ \leq 1 + 2 / ϵ \leq 3 / ϵ$ ; absorbing the resulting factor $ϵ^{- (L - N_{0})}$ gives the same spatial-decay form with a possibly larger finite power of $ϵ^{- 1}$ . Taking the supremum over witnesses yields $M_{φ, 1}^{ϵ, L} f (x) \leq C ϵ^{- K^{'}} (1 + ∣ x ∣)^{- (L - N_{0})} .$ Because $L > N_{0} + n / p$ , this bound belongs to $L^{p} (R^{n})$ , uniformly for each fixed $0 < ϵ \leq 1 / 2$ . By [F7] the maximal function is Borel, so its $L^{p}$ norm is defined.

1.2F2F4F6F7algebra

Tangential dominated by aperture one. Fix $x, y$ and $0 < t < 1 / ϵ$ , and put $q = n / T > 0$ and $g = (M_{φ, 1}^{ϵ, L} f)^{q}$ , which is nonnegative Borel by [F7]. For every $z \in B (x - y, t)$ the definition gives $\frac{∣ (f * φ_{t}) (x - y) ∣ t^{L}}{(t + ϵ + ϵ ∣ x - y ∣)^{L}} \leq M_{φ, 1}^{ϵ, L} f (z)$ . Raising to the $q$ -th power, averaging with the integral allowed to be infinite, and enlarging the ball using [F4] gives $∣ (f * φ_{t}) (x - y) ∣^{q} \frac{t^{L q}}{(t + ϵ + ϵ ∣ x - y ∣)^{L q}} \leq \frac{1}{λ (B (x - y, t))} \int_{B (x - y, t)} g \leq (1 + \frac{∣ y ∣}{t})^{n} \tilde{M} g (x) .$ Since $n = T q$ , division by $(1 + ∣ y ∣ / t)^{T q}$ and taking the supremum over $y, t$ proves the pointwise estimate for every $L > 0$ . For the norm estimate assume $q < p$ . If $∥ M_{φ, 1}^{ϵ, L} f ∥_{p} = \infty$ , the extended norm inequality is trivial. Otherwise $g \in L^{p / q}$ , and [F6] gives $g \in L_{l o c}^{1}$ , so $\tilde{M} g = M g$ . Applying [F2] with $r = p / q > 1$ yields $∥ M_{φ, T}^{ϵ, L} f ∥_{p}^{q} \leq ∥ M g ∥_{p / q} \leq C_{n, p, q} ∥ g ∥_{p / q} = C_{n, p, q} ∥ M_{φ, 1}^{ϵ, L} f ∥_{p}^{q} .$ Taking $q$ -th roots and renaming the constant proves assertion 3.

1.3F3F5algebra

Truncated grand maximal dominated by the truncated tangential maximal function. Fix $T > 0$ , $L > 0$ and choose integers $L^{'} > L + T$ and $N^{'} > L + T + n$ , for example $L^{'} = ⌊ L + T ⌋ + 1$ and $N^{'} = ⌊ L + T + n ⌋ + 1$ . Apply [F3] with these parameters, obtaining $C_{d e c}, M_{d e c}, s_{0}$ ; by the scaling clause of [F3] we may assume $s_{0} \leq 1$ , and we take an integer $N \geq \max (M_{d e c}, N^{'})$ so that $s t \leq t < 1 / ϵ$ for every $j$ and $∥ ψ ∥_{S_{M_{d e c}}} \leq 1$ for $ψ \in F_{N}$ . Write $M_{*} = M_{d e c}$ . For $ψ \in F_{N}$ , $t \in (0, 1 / ϵ)$ and $x$ write the deconvolution $ψ = \sum_{j} η^{j} * φ_{s_{0} 2^{- j}}$ and set $s = s_{0} 2^{- j}$ . If $M_{φ, T}^{ϵ, L} f (x) = + \infty$ , assertion 2 is immediate. Otherwise associativity follows from Schwartz parameter pairing and integral interchange applied to $H (w) (z) = η_{t}^{j} (w) φ_{s t} (x - w - z)$ : each seminorm is bounded by an integrable polynomial weight times $∣ η_{t}^{j} (w) ∣$ . Continuity of $f$ passes the Schwartz deconvolution partial sums to the scalar limit. Thus, with $w$ the integration variable, $∣ (f * ψ_{t}) (x) ∣ \leq \sum_{j} \int_{R^{n}} ∣ (f * φ_{s t}) (x - w) ∣ ∣ η_{t}^{j} (w) ∣ d w .$ By definition of $M_{φ, T}^{ϵ, L} f (x)$ , for every $w$ and $s, t > 0$ with $s t < 1 / ϵ$ , $∣ (f * φ_{s t}) (x - w) ∣ \leq M_{φ, T}^{ϵ, L} f (x) (1 + \frac{∣ w ∣}{s t})^{T} \frac{(s t + ϵ + ϵ ∣ x - w ∣)^{L}}{(s t)^{L}} .$ Multiplying by $t^{L} / (t + ϵ + ϵ ∣ x ∣)^{L}$ and using the triangle inequality $∣ x - w ∣ \leq ∣ x ∣ + ∣ w ∣$ , together with $ϵ / (t + ϵ) \leq 1 / t$ (valid since $ϵ \leq 1$ and $t \leq 1 / ϵ$ ), gives $(\frac{s t + ϵ + ϵ ∣ x - w ∣}{s t})^{L} (\frac{t}{t + ϵ + ϵ ∣ x ∣})^{L} \leq (\frac{1}{s} + \frac{∣ w ∣}{s t})^{L} .$ Therefore, substituting $w = t u$ and using $N^{'} > L + T + n$ so $(1 + ∣ u ∣)^{L + T - N^{'}}$ is integrable, $∣ (f * ψ_{t}) (x) ∣ \frac{t^{L}}{(t + ϵ + ϵ ∣ x ∣)^{L}} \leq M_{φ, T}^{ϵ, L} f (x) \sum_{j} \int_{R^{n}} (1 + \frac{∣ u ∣}{s})^{T} (\frac{1}{s} + \frac{∣ u ∣}{s})^{L} ∣ η^{j} (u) ∣ d u \leq M_{φ, T}^{ϵ, L} f (x) \sum_{j} C_{T, L, N^{'}} s^{- (T + L)} ∥ η^{j} ∥_{S_{N^{'}}} \leq C^{'} M_{φ, T}^{ϵ, L} f (x) \sum_{j} 2^{j (T + L)} 2^{- j n L^{'}} = C_{r a d} M_{φ, T}^{ϵ, L} f (x),$ since $s \leq 1$ , [F3] gives $∥ η^{j} ∥_{S_{N^{'}}} \leq C 2^{- j n L^{'}} ∥ ψ ∥_{S_{M_{d e c}}}$ , and $n L^{'} > T + L$ makes the geometric series converge. The constants are independent of $N$ , $ϵ$ , $ψ$ , $t$ and $x$ ; the constant is independent of $ϵ$ because no weight with $ϵ$ remains. The same calculation for an arbitrary Schwartz test $θ$ retains the factor $∥ θ ∥_{S_{M_{*}}}$ on the right. For a cone witness $∣ z - x ∣ < t$ and $ψ \in F_{N}$ , put $h = (z - x) / t$ and $θ (w) = ψ (w + h)$ , so $(f * θ_{t}) (x) = (f * ψ_{t}) (z)$ . Since $∣ h ∣ < 1$ and $N \geq M_{*}$ , the weighted derivative inequality gives $∥ θ ∥_{S_{M_{*}}} \leq 2^{M_{*}}$ , independently of $N$ . Also $t + ϵ + ϵ ∣ x ∣ \leq 2 (t + ϵ + ϵ ∣ z ∣)$ since $∣ x - z ∣ < t$ and $ϵ \leq 1$ . Thus $∣ (f * ψ_{t}) (z) ∣ \frac{t^{L}}{(t + ϵ + ϵ ∣ z ∣)^{L}} \leq 2^{L + M_{*}} C M_{φ, T}^{ϵ, L} f (x) .$ Taking the supremum over these cone witnesses and tests proves assertion 2, with $C_{1} = C_{1} (n, φ, T, L)$ independent of $N$ and $ϵ$ .

1.4F5F7F4algebra

Good-set bound. Fix $p_{0} > 0$ , $λ > 0$ , $N$ large enough for the estimate below, $0 < ϵ \leq 1 / 2$ , and $x$ with $M_{N}^{ϵ, L} f (x) < λ M_{φ, 1}^{ϵ, L} f (x)$ . This condition forces $M_{φ, 1}^{ϵ, L} f (x) < \infty$ : put $A = P_{N} (φ) > 0$ and $φ_{0} = φ / A \in F_{N}$ . For every aperture-one witness $(t, y)$ at $x$ , $v = (x - y) / t$ has $∣ v ∣ < 1$ , and the translated kernel $Ψ (w) = φ_{0} (w - v)$ satisfies $P_{N} (Ψ) \leq 2^{N}$ by [F5]. Thus $Ψ / 2^{N} \in F_{N}$ , $(f * Ψ_{t}) (x) = (f * (φ_{0})_{t}) (y)$ , and the denominator comparison $t + ϵ + ϵ ∣ x ∣ \leq 2 (t + ϵ + ϵ ∣ y ∣)$ gives $M_{N}^{ϵ, L} f (x) \geq 2^{- N - L} A^{- 1} M_{φ, 1}^{ϵ, L} f (x) .$ In particular an infinite right-hand side would force an infinite left-hand side, contrary to the strict good-set inequality. If the aperture-one quantity is positive, its finiteness lets us choose $t \in (0, 1 / ϵ)$ and $y$ with $∣ x - y ∣ < t$ such that $M_{φ, 1}^{ϵ, L} f (x) \leq 2 ∣ h (y) ∣ \frac{t^{L}}{(t + ϵ + ϵ ∣ y ∣)^{L}}, h = f * φ_{t} .$ If the aperture-one quantity is zero, the strict good-set inequality is impossible because $M_{N}^{ϵ, L} f (x) \geq 0$ . With $c = 2^{L} c_{N, φ}$ , [F5] gives $t \sup_{∣ z - y ∣ < r t} ∣ \nabla h (z) ∣ \leq c M_{N}^{ϵ, L} f (x) \frac{(t + ϵ + ϵ ∣ y ∣)^{L}}{t^{L}} (0 < r \leq 1) .$ Indeed, $t \partial_{j} h (z) = (f * Ψ_{t}^{z}) (x)$ for the translated derivative test function $Ψ^{z}$ , whose translation length is at most $2$ ; its $P_{N}$ seminorm is at most $c_{N, φ}$ . Also $t + ϵ + ϵ ∣ x ∣ \leq 2 (t + ϵ + ϵ ∣ y ∣)$ because $∣ x - y ∣ < t$ and $ϵ \leq 1$ . The mean value theorem and the good-set hypothesis now give $∣ h (x^{'}) - h (y) ∣ \leq c r λ M_{φ, 1}^{ϵ, L} f (x) \frac{(t + ϵ + ϵ ∣ y ∣)^{L}}{t^{L}} (x^{'} \in B (y, r t)) .$ Choose $0 < r \leq 1$ so $c r λ \leq 1 / 4$ . The saturation estimate yields $∣ h (x^{'}) ∣ \geq \frac{1}{4} M_{φ, 1}^{ϵ, L} f (x) (t + ϵ + ϵ ∣ y ∣)^{L} / t^{L} \geq \frac{1}{4} M_{φ, 1}^{ϵ, L} f (x)$ on $B (y, r t)$ , and $∣ h (x^{'}) ∣ \leq M_{φ}^{0} f (x^{'})$ . Thus, using the ball inclusion and [F4], $M_{φ, 1}^{ϵ, L} f (x)^{p_{0}} \leq 4^{p_{0}} (\frac{1 + r}{r})^{n} \frac{1}{λ (B (x, (1 + r) t))} \int_{B (x, (1 + r) t)} (M_{φ}^{0} f)^{p_{0}} .$ By [F7], $(M_{φ}^{0} f)^{p_{0}}$ is Borel, so the extended average defined before assertion 1 applies even if it is not locally integrable. The displayed average is at most $\tilde{M} ((M_{φ}^{0} f)^{p_{0}}) (x)$ , so assertion 4 follows with $C_{3} = 4 ((1 + r) / r)^{n / p_{0}}$ .

1.5givenF7algebra

Removing truncation. For fixed $L, T, N$ and every fixed admissible witness, the weight $t^{L} / (t + ϵ + ϵ ∣ z ∣)^{L}$ increases to $1$ as $ϵ ↓ 0$ , and the permitted scale range increases to all $t > 0$ . Hence the radial, aperture-one and tangential truncated functions increase to their corresponding untruncated suprema. The same argument, also taking the supremum over $ψ \in F_{N}$ , gives $M_{N}^{ϵ, L} f (x) ↑ M_{N} f (x)$ . The strict cone $∣ z - x ∣ < t$ has the same supremum as the closed cone $∣ z - x ∣ \leq t$ , because each convolution is continuous and every boundary point is a limit of interior points at fixed $t$ ; therefore this limit is precisely the supplied nontangential $M_{N}$ .

2.1step 1.4F5F4algebra

Untruncated good-set estimate. Let $x$ satisfy $M_{N} f (x) \leq λ M_{φ}^{*, 1} f (x) < \infty$ , with $N \geq 1$ . If $M_{φ}^{*, 1} f (x) = 0$ , the asserted bound is immediate; otherwise, by definition of the supremum choose $t > 0, y$ with $∣ x - y ∣ < t$ and $∣ h (y) ∣ \geq \frac{1}{2} M_{φ}^{*, 1} f (x)$ , where $h = f * φ_{t}$ . As in step 1.4 but without truncation weights, $t \sup_{∣ z - y ∣ < r t} ∣ \nabla h (z) ∣ \leq c_{N, φ} M_{N} f (x)$ by [F5]. Choose $r \leq 1$ so $c_{N, φ} r λ \leq 1 / 4$ . The mean value theorem gives $∣ h (z) ∣ \geq \frac{1}{4} M_{φ}^{*, 1} f (x)$ on $B (y, r t)$ , and $∣ h (z) ∣ \leq M_{φ}^{0} f (z)$ . Using [F4], the ball inclusion and the extended average $\tilde{M}$ defined in step 1.4 yields $M_{φ}^{*, 1} f (x)^{p_{0}} \leq 4^{p_{0}} (\frac{1 + r}{r})^{n} \tilde{M} ((M_{φ}^{0} f)^{p_{0}}) (x) .$ This is assertion 5 with $N_{3} = 1$ and $C_{4} = 4 ((1 + r) / r)^{n / p_{0}}$ .

3.1step 1.1step 1.3step 1.2step 1.4step 2.1step 1.5∎

Conclusion. Step 1.1 gives the finiteness and pointwise decay of the aperture-one truncated function; step 1.3 gives the pointwise comparison of the truncated grand maximal function with the truncated tangential one, with constants independent of $ϵ$ ; step 1.2 gives the tangential-to-aperture-one comparison via the Hardy-Littlewood maximal operator; step 1.4 gives the truncated good-set bound and step 2.1 the untruncated one. Step 1.5 proves the stated monotone limits.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

The real Hardy space $H^{p}$ defined by a radial maximal function

Definition

Fix $n \geq 1$ , $0 < p < \infty$ and an admissible kernel $φ \in S (R^{n})$ with $\int_{R^{n}} φ \neq 0$ , and let $M_{φ}^{0} f$ be the radial maximal function of Radial and nontangential maximal functions of a tempered distribution. The real Hardy space is $H^{p} (R^{n}) = {f \in S^{'} (R^{n}) : M_{φ}^{0} f \in L^{p} (R^{n})},$ with the functional $∥ f ∥_{H^{p}} : = ∥ M_{φ}^{0} f ∥_{L^{p} (R^{n})} .$ Here $L^{p}$ is the quotient by almost-everywhere null functions of The space $L^{p} (μ)$ as the quotient by null functions with the complex scalar conventions of Complex Lp classes and Euclidean test-function conventions, and $M_{φ}^{0} f$ is the Borel measurable extended-real function supplied by Measurability and lower semicontinuity of the smooth maximal functions; the membership condition includes that $M_{φ}^{0} f$ is finite almost everywhere and that its class lies in $L^{p}$ . Since $M_{φ}^{0} f \geq 0$ , the functional takes values in $[0, \infty]$ and is finite on $H^{p}$ ; the zero distribution lies in $H^{p}$ with $∥ 0 ∥_{H^{p}} = 0$ . Under Countable Choice, for $p \geq 1$ the functional $∥ \cdot ∥_{H^{p}}$ is a norm. For $0 < p < 1$ it is $p$ -subadditive. No Banach-space duality of $H^{p}$ with a normed dual is asserted here below $p = 1$ . The kernel $φ$ is held fixed in the definition; the maximal-characterisation theorem on this page shows that different admissible kernels give the same space with equivalent quasi-norms, so that the notation $H^{p} (R^{n})$ does not depend on the choice up to equivalence. No choice principle is used in the definition itself.

Norm properties

The set defining $H^{p}$ is specified without a choice principle. Under Countable Choice, positive definiteness holds for every $p > 0$ . If $∥ f ∥_{H^{p}} = 0$ , then $M_{φ}^{0} f = 0$ almost everywhere. Lower semicontinuity of the radial maximal function makes it identically zero: if it has a positive value, one of its open strict superlevel sets contains a Euclidean ball, which has positive measure under Countable Choice by Euclidean balls have positive finite Lebesgue measure. Thus $f * φ_{t} = 0$ for every $t > 0$ at every point. Apply Schwartz approximate identities converge in the sense of tempered distributions to $Φ = φ / (\int φ)$ to conclude $f = 0$ in $S^{'}$ . The lower-semicontinuity input is Measurability and lower semicontinuity of the smooth maximal functions, and the approximate-identity limit is as stated. For $p \geq 1$ , homogeneity and the triangle inequality follow from $M_{φ}^{0} (f + g) \leq M_{φ}^{0} f + M_{φ}^{0} g$ and the complex $L^{p}$ norm properties Complex Holder, Minkowski, and the quotient norm, so the functional is a norm.

For $0 < p < 1$ , pointwise sublinearity and $(u + v)^{p} \leq u^{p} + v^{p}$ for $u, v \geq 0$ give $∥ f + g ∥_{H^{p}}^{p} \leq ∥ f ∥_{H^{p}}^{p} + ∥ g ∥_{H^{p}}^{p}$ . The $H^{p}$ membership definition itself uses no choice principle.

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

The grand maximal function dominates every admissible radial and nontangential maximal function

Statement

Let $n \geq 1$ , $f \in S^{'} (R^{n})$ , $φ \in S (R^{n})$ with $\int_{R^{n}} φ \neq 0$ , $a \geq 1$ , and an integer $N \geq 1$ . Then, with $P_{N}$ and $F_{N}$ as in Grand maximal test class of order N and the grand maximal function and the maximal functions of Radial and nontangential maximal functions of a tempered distribution, $M_{φ}^{*, a} f (x) \leq (1 + a)^{N} P_{N} (φ) M_{N} f (x) for every x \in R^{n},$ and in particular $M_{φ}^{0} f (x) \leq 2^{N} P_{N} (φ) M_{N} f (x), ∥ M_{φ}^{0} f ∥_{L^{p}} \leq 2^{N} P_{N} (φ) ∥ M_{N} f ∥_{L^{p}}$ whenever $0 < p \leq \infty$ and the right-hand side is finite. The constants $(1 + a)^{N}$ differ from the source's sharper $a^{N}$ but are equivalent for fixed $a, N$ and are the ones produced by the elementary translate estimate below. Consequently every admissible radial maximal function is pointwise dominated by the grand maximal function of every sufficiently large order, and the space defined by the radial maximal function of one kernel contains the space defined by $M_{N}$ .

Facts & Assumptions

Given: $n \geq 1$ , $φ \in S$ with $\int φ \neq 0$ , $a \geq 1$ , an integer $N \geq 1$ , $f \in S^{'}$ , and a point $x \in R^{n}$ .

[F1]

$(f * Ψ_{t}) (y) = ⟨ f_{z}, Ψ_{t} (y - z) ⟩$ for $Ψ \in S$ and $t > 0$ , and $M_{N} f (x) = \sup_{Ψ \in F_{N}} \sup_{t > 0} \sup_{∣ y - x ∣ \leq t} ∣ (f * Ψ_{t}) (y) ∣$ (Convolution of a tempered distribution with a schwartz function, Grand maximal test class of order N and the grand maximal function).

[F2]

The test seminorm satisfies $P_{N} (Ψ) \leq 1$ exactly for $Ψ \in F_{N}$ , and for every nonzero $Ψ$ , $P_{N} (Ψ / P_{N} (Ψ)) = 1$ ; $P_{N} (Ψ) = \sup_{w} (1 + ∣ w ∣)^{N} \max_{∣ α ∣ \leq N + 1} ∣ \partial^{α} Ψ (w) ∣$ (Grand maximal test class of order N and the grand maximal function, Schwartz space and its seminorms).

[F3]

If $∣ y - x ∣ \leq a t$ then $y = x + t γ$ with $∣ γ ∣ \leq a$ ; the translation identity $φ_{t} (y - z) = G_{t} (x - z)$ for $G (w) = φ (w + γ)$ holds for every $z$ , as both sides equal $t^{- n} φ ((x - z) / t + γ)$ (Dilations and their normalisations preserve Schwartz space, with scaling identities).

[F4]

Maximal functions are Borel measurable, so the $L^{p}$ statement is meaningful (Measurability and lower semicontinuity of the smooth maximal functions).

Proof technique: translate the kernel into the aperture-one cone at the base point, using the translate bound for $P_{N}$ .

Proof

technique · direct

1.1F2F3givenalgebra

The translate bound. Fix $γ \in R^{n}$ and put $G (w) = φ (w + γ)$ . Then $G \in S (R^{n})$ and $P_{N} (G) \leq (1 + ∣ γ ∣)^{N} P_{N} (φ)$ : indeed $\partial^{α} G (w) = (\partial^{α} φ) (w + γ)$ and $(1 + ∣ w ∣) \leq (1 + ∣ γ ∣) (1 + ∣ w + γ ∣)$ , so $(1 + ∣ w ∣)^{N} \max_{∣ α ∣ \leq N + 1} ∣ \partial^{α} φ (w + γ) ∣ \leq (1 + ∣ γ ∣)^{N} P_{N} (φ)$ pointwise in $w$ , and taking the supremum proves the claim. If $P_{N} (G) = 0$ then $G = 0$ and $f * G_{t} = 0$ ; otherwise $G / P_{N} (G) \in F_{N}$ by [F2], and $t^{- n}$ normalisation is the same for $G$ .

2.1step 1.1F1F3algebra

Pointwise domination. Fix $t > 0$ and $y$ with $∣ y - x ∣ \leq a t$ , and write $y = x + t γ$ with $∣ γ ∣ \leq a$ . By [F3], $(f * φ_{t}) (y) = (f * G_{t}) (x)$ for $G = φ (\cdot + γ)$ . Taking absolute values and applying the definition of $M_{N}$ through [F1] and step 1.1, $∣ (f * φ_{t}) (y) ∣ = ∣ (f * G_{t}) (x) ∣ \leq P_{N} (G) M_{N} f (x) \leq (1 + a)^{N} P_{N} (φ) M_{N} f (x) .$ Taking the supremum over all such $t, y$ gives $M_{φ}^{*, a} f (x) \leq (1 + a)^{N} P_{N} (φ) M_{N} f (x)$ .

3.1step 2.1F1F4∎

Radial case and $L^{p}$ consequence. Since $M_{φ}^{0} f (x) = \sup_{t > 0} ∣ (f * φ_{t}) (x) ∣$ is the diagonal $y = x$ instance of the aperture-one supremum, $M_{φ}^{0} f (x) \leq M_{φ}^{*, 1} f (x) \leq 2^{N} P_{N} (φ) M_{N} f (x)$ by step 2.1 with $a = 1$ . If $∥ M_{N} f ∥_{p} < \infty$ , the pointwise inequality and the Borel measurability of [F4] give $∥ M_{φ}^{0} f ∥_{p} \leq 2^{N} P_{N} (φ) ∥ M_{N} f ∥_{p}$ for every $0 < p \leq \infty$ by monotonicity of the integral. This proves the lemma.

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

Atoms have uniformly bounded $H^{p}$ quasi-norm and uniformly bounded test pairings

Statement

Assume Countable Choice. Let $n \geq 1$ , $0 < p \leq 1$ , $s \geq ⌊ n (1 / p - 1) ⌋$ , fix the admissible kernel $φ$ defining $H^{p}$ , and let $N \geq \max (N_{0} (n, p, φ), n + s + 1)$ be an admissible grand-maximal order. There are constants $C_{0} = C_{0} (n, p, s) < \infty$ and $C_{1} = C_{1} (n, p, s, N, φ) < \infty$ such that every $(p, \infty, s)$ -atom $a$ supported in an axis-parallel cube $Q$ ( $H^{p}$ atoms with a prescribed moment order) satisfies

$∥ M_{N} a ∥_{L^{p}} \leq C_{0}$ and hence $∥ a ∥_{H^{p}} \leq C_{1}$ ;
for every $ψ \in S (R^{n})$ , $∣ ⟨ a, ψ ⟩ ∣ \leq C_{0} \min (∣ Q ∣^{1 - 1 / p + (s + 1) / n}, ∣ Q ∣^{1 - 1 / p}) \max (∥ ψ ∥_{L^{\infty} (Q)}, ∥ ψ ∥_{C^{s + 1} (Q)}),$ where $∥ ψ ∥_{C^{s + 1} (Q)} = \max_{∣ β ∣ \leq s + 1} \sup_{Q} ∣ \partial^{β} ψ ∣$ ;
in particular $\sup_{j} ∣ ⟨ a_{j}, ψ ⟩ ∣ < \infty$ for every fixed $ψ \in S$ and every family $(a_{j})$ of such atoms.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $0 < p \leq 1$ , $s \geq ⌊ n (1 / p - 1) ⌋$ , the fixed admissible kernel $φ$ , an admissible order $N \geq \max (N_{0} (n, p, φ), n + s + 1)$ , and a $(p, \infty, s)$ -atom $a$ supported in a cube $Q$ with centre $c_{Q}$ and side length $ℓ = ℓ (Q)$ .

[L1]

$a$ is measurable, $supp a \subseteq Q$ , $∣ a ∣ \leq ∣ Q ∣^{- 1 / p}$ a.e., and $\int a (y) y^{α} d y = 0$ for every multi-index $∣ α ∣ \leq s$ ( $H^{p}$ atoms with a prescribed moment order).

[F1]

The cube $Q$ has side length $ℓ$ , is contained in the closed ball $\overline{B (c_{Q}, \sqrt{n} ℓ / 2)}$ , and $∣ y - c_{Q} ∣ \leq \sqrt{n} ℓ / 2$ for $y \in Q$ (Axis-parallel rectangles in $R^{m}$ and their volume).

[F2]

For $Ψ \in F_{N}$ , $Ψ_{t} (w) = t^{- n} Ψ (w / t)$ and each derivative through order $N + 1$ satisfies $∣ \partial^{β} Ψ (u) ∣ \leq (1 + ∣ u ∣)^{- N}$ ; in particular $∣ Ψ (u) ∣ \leq (1 + ∣ u ∣)^{- N}$ and $∥ Ψ ∥_{1} \leq C_{n}$ because $N \geq n + 1$ (Grand maximal test class of order N and the grand maximal function, Schwartz space and its seminorms).

[F3]

Domination: $M_{φ}^{0} a \leq 2^{N} P_{N} (φ) M_{N} a$ for the fixed admissible kernel $φ$ , so $∥ a ∥_{H^{p}} \leq 2^{N} P_{N} (φ) ∥ M_{N} a ∥_{L^{p}}$ (The grand maximal function dominates every admissible radial and nontangential maximal function).

[F4]

Taylor remainder: for real $G \in C^{s + 1} (R^{n})$ , the multivariable Lagrange formula gives $G (y) = T_{s} G (c; y - c) + R_{G} (y)$ and $∣ R_{G} (y) ∣ \leq C_{n, s} ∣ y - c ∣^{s + 1} \max_{∣ β ∣ = s + 1} \sup_{z \in [c, y]} ∣ \partial^{β} G (z) ∣$ (Multivariable Taylor formula with a Lagrange remainder along a line segment, $C^{k}$ maps and multi-index derivative notation in Euclidean space). For complex $G$ , apply the real formula to $Re G$ and $Im G$ and add the two remainder bounds; each component derivative is bounded by $∣ \partial^{β} G ∣$ (Complex Lp classes and Euclidean test-function conventions).

[F5]

Under Countable Choice, a closed axis-parallel box is Lebesgue measurable with measure equal to the product of its side lengths, and Lebesgue measure is monotone (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, Measures are monotone, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F6]

The grand maximal function $M_{N} a$ is Borel measurable (Measurability and lower semicontinuity of the smooth maximal functions).

[F7]

For nonnegative measurable functions, integration over an increasing union of measurable sets is the limit of the integrals over the finite unions (Monotone convergence for the integral).

[F8]

Normalized dilation preserves the $L^{1}$ norm: $∥ Ψ_{t} ∥_{1} = ∥ Ψ ∥_{1}$ by A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions applied to $T (w) = t w$ and the integrand $∣ Ψ ∣$ .

[F9]

The tempered-distribution test pairing is bilinear: $⟨ u, ψ ⟩ = u (ψ)$ with no conjugation of $ψ$ (Tempered distribution).

Proof technique: near/far splitting with the Taylor remainder and the moment conditions.

Proof

technique · direct

1.1L1F1F2F5F6F8algebra

Near estimate. For $x \in R^{n}$ , $Ψ \in F_{N}$ , $t > 0$ , and every convolution centre $y$ with $∣ y - x ∣ \leq t$ , the atom bound gives $∣ (a * Ψ_{t}) (y) ∣ \leq ∥ a ∥_{\infty} \int_{Q} ∣ Ψ_{t} (y - z) ∣ d z \leq ∣ Q ∣^{- 1 / p} ∥ Ψ_{t} ∥_{1} = ∣ Q ∣^{- 1 / p} ∥ Ψ ∥_{1} \leq C ∣ Q ∣^{- 1 / p}$ . The last constant is uniform over $Ψ \in F_{N}$ by [F2] and $N \geq n + 1$ . By [F8], $∥ Ψ_{t} ∥_{1} = ∥ Ψ ∥_{1}$ . Taking the suprema over $Ψ$ , $t$ , and all $y$ with $∣ y - x ∣ \leq t$ proves this bound for every $x$ . Let $Q^{*}$ be the concentric closed cube of side $4 \sqrt{n} ℓ$ . By [F5], $∣ Q^{*} ∣ = (4 \sqrt{n})^{n} ∣ Q ∣$ , and hence $\int_{Q^{*}} (M_{N} a)^{p} \leq C^{p} ∣ Q ∣^{- 1} ∣ Q^{*} ∣ \leq C^{p} (4 \sqrt{n})^{n} .$

2.1step 1.1L1F1F2F4F5F7algebra

Far estimate. Set $m : = n + s + 1$ . If $x \notin Q^{*}$ , put $r : = ∣ x - c_{Q} ∣ > 2 \sqrt{n} ℓ$ . Fix $Ψ \in F_{N}$ , $t > 0$ , and any $y$ with $∣ y - x ∣ \leq t$ ; all estimates below are uniform in this $y$ , so taking the suprema over $y, t, Ψ$ at the end gives the estimate for $M_{N} a (x)$ . For $z \in Q$ , $∣ z - c_{Q} ∣ \leq \sqrt{n} ℓ / 2$ . When $0 < t \leq ℓ$ , the triangle inequality gives $∣ y - z ∣ \geq r - t - \sqrt{n} ℓ / 2 \geq r / 4$ . Thus [F2] and $N \geq m$ imply $∣ (a * Ψ_{t}) (y) ∣ \leq ∣ Q ∣^{- 1 / p} ∣ Q ∣ t^{- n} (1 + r / (4 t))^{- m} \leq C ∣ Q ∣^{- 1 / p} (ℓ / r)^{m} (t / ℓ)^{s + 1} \leq C ∣ Q ∣^{- 1 / p} (ℓ / r)^{m}$ , using $∣ Q ∣ = ℓ^{n}$ and $m - n = s + 1$ . When $t \geq ℓ$ , expand the complex function $z \mapsto Ψ_{t} (y - z)$ about $c_{Q}$ through degree $s$ , applying [F4] to its real and imaginary parts. The Taylor polynomial integrates to zero against $a$ because each $(z - c_{Q})^{α}$ , $∣ α ∣ \leq s$ , is a linear combination of monomials $z^{β}$ of degree at most $s$ , whose moments vanish by [L1]. For every remainder point $ζ$ on the segment from $c_{Q}$ to $z \in Q$ , the cone condition and $t \geq ℓ$ give $r \leq ∣ x - y ∣ + ∣ y - ζ ∣ + ∣ ζ - c_{Q} ∣ \leq t + ∣ y - ζ ∣ + \frac{\sqrt{n}}{2} ℓ \leq (1 + \sqrt{n} / 2) (t + ∣ y - ζ ∣)$ . If $∣ β ∣ = s + 1$ , [F2] yields $∣ \partial^{β} Ψ_{t} (y - ζ) ∣ \leq t^{- n - s - 1} (1 + ∣ y - ζ ∣ / t)^{- m} = (t + ∣ y - ζ ∣)^{- m} \leq (1 + \sqrt{n} / 2)^{m} r^{- m}$ . The Taylor remainder and $∣ z - c_{Q} ∣ \leq \sqrt{n} ℓ / 2$ now give $∣ (a * Ψ_{t}) (y) ∣ \leq C ∣ Q ∣^{1 - 1 / p} ℓ^{s + 1} r^{- m} = C ∣ Q ∣^{- 1 / p} (ℓ / r)^{m}$ . These bounds hold for every $Ψ, t, y$ in the grand-maximal supremum. Therefore $M_{N} a (x) \leq C ∣ Q ∣^{- 1 / p} (1 + r / ℓ)^{- m}$ . Since $s \geq ⌊ n (1 / p - 1) ⌋$ , $p m > n$ . Cover the far region by shells $E_{j} = {x : 2^{j} R_{0} \leq ∣ x - c_{Q} ∣ < 2^{j + 1} R_{0}}$ , $j \geq 0$ , with $R_{0} = 2 \sqrt{n} ℓ$ . Each is contained in a concentric closed cube of side $2^{j + 2} R_{0}$ , so [F5] gives $∣ E_{j} ∣ \leq C_{n} 2^{j n} ∣ Q ∣$ . By [F7] and the pointwise bound, $\int_{R^{n} ∖ Q^{*}} (M_{N} a)^{p} \leq \sum_{j \geq 0} C ∣ Q ∣^{- 1} 2^{- j m p} ∣ E_{j} ∣ \leq C \sum_{j \geq 0} 2^{- j (p m - n)} < \infty .$

3.1step 1.1step 2.1F3F6algebra

Conclusion of (a). Steps 1.1 and 2.1 give $\int_{R^{n}} (M_{N} a)^{p} \leq C_{0}^{p}$ after enlarging $C_{0} = C_{0} (n, p, s)$ , independently of $Q$ , $a$ and admissible $N$ ; measurability is [F6]. Hence $∥ M_{N} a ∥_{L^{p}} \leq C_{0}$ , and [F3] gives $∥ a ∥_{H^{p}} \leq 2^{N} P_{N} (φ) C_{0} = : C_{1} (n, p, s, N, φ)$ . This proves assertion 1.

4.1step 3.1L1F4F5F9algebra

Pairing bounds. The atom function induces a tempered distribution by $∣ \int_{Q} a (y) ψ (y) d y ∣ \leq ∣ Q ∣^{1 - 1 / p} p_{00} (ψ)$ , and [F9] fixes the bilinear convention. Thus for a complex test $ψ$ , $⟨ a, ψ ⟩ = \int_{Q} a (y) ψ (y) d y$ ; this integral is absolutely convergent by [L1] and boundedness of $ψ$ on $Q$ . The plain estimate is $∣ ⟨ a, ψ ⟩ ∣ \leq ∣ Q ∣^{1 - 1 / p} ∥ ψ ∥_{L^{\infty} (Q)}$ . For the Taylor estimate, apply [F4] separately to $Re ψ$ and $Im ψ$ and use the same moment cancellation as in step 2.1. The combined remainder obeys $\sup_{Q} ∣ R_{ψ} ∣ \leq C_{n, s} ℓ^{s + 1} ∥ ψ ∥_{C^{s + 1} (Q)}$ , hence $∣ ⟨ a, ψ ⟩ ∣ = ∣ \int_{Q} a (y) R_{ψ} (y) d y ∣ \leq C_{n, s} ∣ Q ∣^{1 - 1 / p + (s + 1) / n} ∥ ψ ∥_{C^{s + 1} (Q)} .$ Taking the smaller of the plain and Taylor bounds proves assertion 2 after enlarging $C_{0}$ . Put $X = n (1 / p - 1) \geq 0$ : then $1 - 1 / p + (s + 1) / n = (s + 1 - X) / n > 0$ , whereas $1 - 1 / p \leq 0$ (including equality when $p = 1$ ). Thus the two powers have one positive and one nonpositive exponent, and $\min (∣ Q ∣^{1 - 1 / p + (s + 1) / n}, ∣ Q ∣^{1 - 1 / p}) \leq 1$ for all $∣ Q ∣ > 0$ . Since a Schwartz test and its derivatives through order $s + 1$ are bounded globally, assertion 3 follows uniformly over every family of atoms.

5.1step 1.1step 2.1step 3.1step 4.1F5F6∎

Conclusion. Steps 1.1 and 2.1 give the uniform grand-maximal estimate, [F3] gives the kernel/order-dependent $H^{p}$ bound, and step 4.1 proves the uniform pairing estimates. Countable Choice is used for the explicit box measures and maximal-function measurability in [F5]--[F6]. This proves the lemma.

LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passOpen item page →

Calderon reproducing pair and the telescoping identity in $S^{'}$

Statement

Assume Countable Choice. Let $K \geq 0$ and let $φ \in C_{c}^{\infty} (R^{n})$ satisfy $supp φ \subseteq B (0, 1)$ , $\hat{φ} (0) = 1$ and $\partial^{α} \hat{φ} (0) = 0$ for $0 < ∣ α ∣ \leq K$ (such a $φ$ exists by Schwartz functions with prescribed flatness of the Fourier transform at the origin). Put $ψ = 2^{n} φ (2 \cdot) - φ$ and $\tilde{ψ} = 2^{n} φ (2 \cdot) + φ$ , and for $k \in Z$ write $h_{k} (x) = 2^{k n} h (2^{k} x)$ . Then $supp ψ_{k} \subseteq B (0, 2^{- k}), supp {\tilde{ψ}}_{k} \subseteq B (0, 2^{- k}), \int_{R^{n}} ψ (x) x^{α} d x = 0 (∣ α ∣ \leq K),$ and for every $f \in S^{'} (R^{n})$ and every $j \in Z$ the identity $f = φ_{j} * φ_{j} * f + \sum_{k \geq j} ψ_{k} * {\tilde{ψ}}_{k} * f in S^{'} (R^{n})$ holds, the series being the limit of its partial sums in $S^{'}$ . If $f \in H^{p} (R^{n})$ for some $0 < p < \infty$ (with the fixed admissible kernel and the space of The real Hardy space $H^{p}$ defined by a radial maximal function), then also $f = \sum_{k \in Z} ψ_{k} * {\tilde{ψ}}_{k} * f in S^{'} .$ The absolute convergence of the scalar series $\sum_{k \geq j} ∣ ⟨ ψ_{k} * {\tilde{ψ}}_{k} * f, χ ⟩ ∣$ for every test function $χ$ is proved where it is consumed, in the level-decomposition item, whose quantitative hypotheses are available there. The two-sided identity can fail for general $f \in S^{'}$ : for the constant function $f = 1$ one has $φ_{j} * φ_{j} * f = f \neq 0$ for every $j$ , while $ψ_{k} * {\tilde{ψ}}_{k} * f = 0$ for every $k$ because $\int ψ = 0$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $K \geq 0$ , $φ, ψ, \tilde{ψ}$ as in the statement; the convolutions and dilations of Convolution of a tempered distribution with a schwartz function, Dilations and their normalisations preserve Schwartz space, with scaling identities, Schwartz space and its seminorms and Schwartz topology and convergence.

[F1]

The Fourier identity $\partial^{α} \hat{φ} (0) = (- 2 π i)^{∣ α ∣} \int_{R^{n}} x^{α} φ (x) d x$ holds, so the moment conditions on $φ$ at the origin are equivalent to the vanishing of the positive-order moments of $φ$ ; the mean of $φ$ is one, and the mean of $ψ$ is zero (Fourier differentiation and multiplication identities on tempered distributions).

[F2]

For $Φ = φ * φ \in S$ one has $\int Φ = 1$ and $Φ_{t} = φ_{t} * φ_{t}$ , so $Φ_{t} * f \to f$ in $S^{'}$ as $t ↓ 0$ (Schwartz approximate identities converge in the sense of tempered distributions).

[F3]

By kernel independence in Maximal-function characterisations of real Hardy spaces, membership in $H^{p}$ gives integrability for the reproducing kernel $φ$ , even when a different admissible kernel defines the given quasi-norm. Thus for $f \in H^{p}$ the radial maximal function $g = M_{φ}^{0} f$ belongs to $L^{p}$ . When $p \geq 1$ , $h = φ_{j} * f$ satisfies $∣ h ∣ \leq g$ and hence $∥ h ∥_{p} \leq ∥ g ∥_{p}$ ; the regular-distribution convolution formula and Hölder give $∥ φ_{j} * h ∥_{\infty} \leq ∥ φ_{j} ∥_{p^{'}} ∥ h ∥_{p}$ (The real Hardy space $H^{p}$ defined by a radial maximal function, Tempered distribution, Convolution of a tempered distribution with a schwartz function, Schwartz space and its seminorms, Holder's inequality for integrals, including the endpoint cases).

[F4]

Schwartz functions and their polynomial multiples are integrable, so $x^{α} ψ \in L^{1}$ and $\int_{R^{n}} (1 + ∣ x ∣)^{N} ∣ ψ ∣ < \infty$ for every $N$ (Schwartz derivatives are integrable).

[F5]

For $f \in H^{p}$ , the maximal-characterisation theorem supplies an admissible integer order $N$ with $G = M_{N} f \in L^{p}$ . Grand-maximal domination gives $∣ (f * φ_{t}) (y) ∣ \leq 3^{N} P_{N} (φ) G (x)$ whenever $∣ y - x ∣ \leq 2 t$ (Maximal-function characterisations of real Hardy spaces, The grand maximal function dominates every admissible radial and nontangential maximal function).

[F6]

Under Countable Choice, translation invariance and the ball-volume formula give $λ (B (x, R)) = c_{n} R^{n}$ for every $x$ and $R > 0$ , while dilation gives $∥ φ_{t} ∥_{L^{1}} = ∥ φ ∥_{L^{1}}$ and, for $1 \leq q < \infty$ , $∥ φ_{j} ∥_{q} = 2^{j n (1 - 1 / q)} ∥ φ ∥_{q}$ ; for $q = \infty$ , $∥ φ_{j} ∥_{\infty} = 2^{j n} ∥ φ ∥_{\infty}$ . Indeed, for finite $q$ , $\int ∣ φ_{j} (x) ∣^{q} d x = 2^{j n q} 2^{- j n} \int ∣ φ (u) ∣^{q} d u$ under $u = 2^{j} x$ (Sphere and ball measures scale in Rn, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, Dilations and their normalisations preserve Schwartz space, with scaling identities, For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it).

[F7]

The grand maximal function $M_{N} f$ is Borel measurable, so its strict superlevel sets are measurable (Measurability and lower semicontinuity of the smooth maximal functions).

[F8]

If $G \geq 0$ is measurable and $t > 0$ , then $λ ({G \geq t}) \leq t^{- 1} \int G$ by the Chebyshev-Markov inequality (Chebyshev-Markov inequality for the integral).

Proof technique: telescoping of the two-scale identity, an elementary limit at $- \infty$ , and a moment-Taylor estimate for the absolute convergence.

Proof

technique · constructive

1.1F1F4algebraconstruct

Support and moments. Since $supp φ \subseteq B (0, 1)$ , both $supp (2^{n} φ (2 \cdot))$ and $supp φ$ lie in $B (0, 1)$ ; hence $supp ψ, supp \tilde{ψ} \subseteq B (0, 1)$ and, after dilation, $supp ψ_{k}, supp {\tilde{ψ}}_{k} \subseteq B (0, 2^{- k})$ for every $k$ . For $0 < ∣ α ∣ \leq K$ one has $\int x^{α} ψ (x) d x = 2^{- ∣ α ∣} \int x^{α} φ (x) d x - \int x^{α} φ (x) d x = (2^{- ∣ α ∣} - 1) \int x^{α} φ = 0$ by [F1] and the flatness of $\hat{φ}$ ; the case $α = 0$ gives $\int ψ = 0$ as well.

2.1F2step 1.1algebra

Telescoping. For every $k \in Z$ the definitions give $ψ_{k} = φ_{k + 1} - φ_{k}$ and ${\tilde{ψ}}_{k} = φ_{k + 1} + φ_{k}$ : indeed $h = 2^{n} φ (2 \cdot)$ has $h_{k} = φ_{k + 1}$ and $φ_{k} = φ_{k}$ . Therefore $ψ_{k} * {\tilde{ψ}}_{k} * f = (φ_{k + 1} * φ_{k + 1} - φ_{k} * φ_{k}) * f$ , and the finite sums telescope: $φ_{j} * φ_{j} * f + \sum_{k = j}^{N} ψ_{k} * {\tilde{ψ}}_{k} * f = φ_{N + 1} * φ_{N + 1} * f = (φ * φ)_{N + 1} * f$ for every $N \geq j$ . By [F2], $(φ * φ)_{N + 1} * f \to f$ in $S^{'}$ as $N \to \infty$ , so the partial sums converge to $f - φ_{j} * φ_{j} * f$ and the displayed one-sided identity holds for every $f \in S^{'}$ and $j \in Z$ .

3.1F3F4F5F6F7F8step 2.1givenalgebra

The two-sided identity for $H^{p}$ elements. Let $f \in H^{p}$ and set $g = M_{φ}^{0} f \in L^{p}$ , so $∣ φ_{j} * f ∣ \leq g$ pointwise for every $j$ . If $0 2^{r}}$ ; by [F7] it is measurable, and $Ω_{r} \subseteq {G^{p} \geq 2^{r p}}$ , so [F8] applied to $G^{p}$ at threshold $2^{r p}$ gives $λ (Ω_{r}) \leq 2^{- r p} ∥ G ∥_{p}^{p} < \infty$ . Fix $r$ . By [F6], $λ (B (y, 2^{- j + 1})) = c_{n} 2^{(- j + 1) n} \to \infty$ as $j \to - \infty$ , uniformly in $y$ . Thus for all sufficiently negative $j$ and every $y \in R^{n}$ there is $x \in B (y, 2^{- j + 1}) ∖ Ω_{r}$ ; otherwise this ball would be contained in $Ω_{r}$ and have measure at most $λ (Ω_{r})$ . Then $∣ y - x ∣ < 2 \cdot 2^{- j}$ , so [F5] gives $∣ (φ_{j} * f) (y) ∣ \leq C_{N} G (x) \leq C_{N} 2^{r}$ . Hence $∥ φ_{j} * f ∥_{\infty} \leq C_{N} 2^{r}$ , and [F6] gives $∥ φ_{j} * φ_{j} * f ∥_{\infty} \leq ∥ φ_{j} ∥_{1} ∥ φ_{j} * f ∥_{\infty} \leq ∥ φ ∥_{1} C_{N} 2^{r} .$ Given $ε > 0$ , choose $r$ so negative that the right-hand side is less than $ε$ , and then choose $j$ sufficiently negative. Thus $φ_{j} * φ_{j} * f \to 0$ uniformly, hence in $S^{'}$ because every Schwartz test function is integrable by [F4]. If $p \geq 1$ , Hölder instead gives $∣ φ_{j} * φ_{j} * f (x) ∣ \leq ∥ φ_{j} ∥_{p^{'}} ∥ φ_{j} * f ∥_{p} \leq 2^{j n / p} ∥ φ ∥_{p^{'}} ∥ g ∥_{p} ⟶ 0$ as $j \to - \infty$ . For $p > 1$ , [F6] gives this norm scaling since $\int ∣ φ_{j} ∣^{p^{'}} = 2^{j n p^{'}} 2^{- j n} \int ∣ φ (u) ∣^{p^{'}} d u$ , so $n (1 - 1 / p^{'}) = n / p$ ; for $p = 1$ it is the supremum scaling $∥ φ_{j} ∥_{\infty} = 2^{j n} ∥ φ ∥_{\infty}$ . The other bound uses $∣ φ_{j} * f ∣ \leq g$ and [F3]. Hence in every case $φ_{j} * φ_{j} * f \to 0$ in $S^{'}$ . Passing to the limit $j \to - \infty$ in the one-sided identity of step 2.1 (with the series understood as $\lim_{j \to - \infty} \sum_{k \geq j}$ , whose partial sums are $φ_{N + 1} * φ_{N + 1} * f - φ_{j} * φ_{j} * f$ ) gives $f = \sum_{k \in Z} ψ_{k} * {\tilde{ψ}}_{k} * f$ in $S^{'}$ .

4.1step 1.1step 2.1step 3.1discharge-construct∎

Conclusion. Step 1.1 gives the support and moment properties of $ψ$ ; step 2.1 gives the one-sided telescoping identity for every tempered distribution; step 3.1 gives the two-sided identity for $H^{p}$ elements. This proves the lemma.

TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passOpen item page →

Maximal-function characterisations of real Hardy spaces

Statement

Assume Countable Choice. Let $n \geq 1$ , $0 < p < \infty$ and let $φ \in S (R^{n})$ with $\int φ \neq 0$ . Then there is $N_{0} (n, p, φ) < \infty$ , depending only on $n$ , $p$ and the fixed kernel $φ$ , such that for every $N \geq N_{0} (n, p, φ)$ and every $f \in S^{'} (R^{n})$ the following assertions are equivalent:

$M_{φ}^{0} f \in L^{p} (R^{n})$ ;
$M_{φ}^{*, a} f \in L^{p} (R^{n})$ for some $a \geq 1$ (equivalently, for every $a \geq 1$ );
$M_{N} f \in L^{p} (R^{n})$ .

Here $M_{φ}^{0} f$ , $M_{φ}^{*, a} f$ are the maximal functions of Radial and nontangential maximal functions of a tempered distribution and $M_{N}$ is the grand maximal function of Grand maximal test class of order N and the grand maximal function. Moreover the extended quantities $∥ M_{φ}^{0} f ∥_{L^{p}}$ , $∥ M_{φ}^{*, a} f ∥_{L^{p}}$ and $∥ M_{N} f ∥_{L^{p}}$ are finite exactly on the common set of $f$ satisfying 1-3, and on that set they are equivalent: $∥ M_{φ}^{0} f ∥_{L^{p}} \leq ∥ M_{φ}^{*, a} f ∥_{L^{p}} \leq (1 + a)^{N} P_{N} (φ) ∥ M_{N} f ∥_{L^{p}}, ∥ M_{N} f ∥_{L^{p}} \leq C ∥ M_{φ}^{0} f ∥_{L^{p}},$ with $C = C (n, p, N, φ) < \infty$ depending only on $n, p, N$ and finitely many Schwartz seminorms of $φ$ together with quantitative nonvanishing data for $\hat{φ}$ near zero (as used in the deconvolution lemma). Consequently the space $H^{p} (R^{n})$ of The real Hardy space $H^{p}$ defined by a radial maximal function does not depend on the choice of admissible $φ$ , and for each admissible $φ$ the grand maximal function may be used to define the same space with an equivalent quasi-norm for every order $N \geq N_{0} (n, p, φ)$ ; for two admissible kernels $φ, ψ$ the two radial definitions agree because both are equivalent to $M_{N}$ for every $N \geq \max (N_{0} (n, p, φ), N_{0} (n, p, ψ))$ . The recorded admissible thresholds of the sources are $N \geq ⌊ n / p ⌋ + 1$ for the nontangential class $F_{N}$ with derivatives through $N + 1$ [DKKP], $N > 1 + n / p$ for the radial class $B_{N}$ with derivatives through $N$ [MSV, section 1, p. 16, for $0 n / p + n + 1$ [CUW], stated there for the grand maximal functions normalised by the test classes of those papers; the proof below uses an unspecified finite $N_{0} (n, p, φ)$ that is at least as large as the order thresholds consumed by the finitely many comparison estimates for the fixed kernel $φ$ (the deconvolution constants of the comparison lemmas depend on the kernel, so the order threshold asserted here depends on $φ$ as well as on $n$ and $p$ ), and the existence of such a finite threshold is what is asserted.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $0 < p < \infty$ , $φ \in S$ with $\int φ \neq 0$ , $f \in S^{'}$ , and a fixed order $N$ .

[F1]

Pointwise domination by the grand maximal function: $M_{φ}^{0} f \leq M_{φ}^{*, 1} f \leq 2^{N} P_{N} (φ) M_{N} f$ and $M_{φ}^{*, a} f \leq (1 + a)^{N} P_{N} (φ) M_{N} f$ for every $a \geq 1$ , provided $N$ is the order of the grand maximal function (The grand maximal function dominates every admissible radial and nontangential maximal function).

[F2]

Tangential comparison: for $0 < q < p$ and $T = n / q$ , $∥ M_{φ, T} f ∥_{L^{p}} \leq C_{n, p, q} ∥ M_{φ}^{*, 1} f ∥_{L^{p}}$ (The tangential maximal function is controlled by the aperture-one nontangential maximal function in $L^{p}$ ).

[F3]

Grand dominated by tangential: for every $T > 0$ there are $N_{2} (n, φ, T)$ and $C_{0}$ with $M_{N} f \leq C_{0} M_{φ, T} f$ whenever $N \geq N_{2}$ (The grand maximal function is pointwise dominated by a tangential maximal function).

[F4]

Good-set estimates: assertion 5 of Truncated maximal functions: finiteness, comparison estimates and the good-set bound uses the extended centered average $\tilde{M}$ of that item for general nonnegative Borel inputs. When $M_{φ}^{0} f \in L^{p}$ and $q_{0} = p / 2$ , the input $(M_{φ}^{0} f)^{q_{0}}$ belongs to $L^{2} \subset L_{l o c}^{1}$ , so $\tilde{M} = M$ and the standard maximal operator can be used. Assertions 1-4 of the same lemma provide the truncated functions $M_{φ, j}^{ϵ, L}$ and their finiteness, grand/tangential comparison, tangential/aperture-one comparison and good-set bound.

[F5]

Hardy-Littlewood boundedness: $∥ M g ∥_{L^{r}} \leq C_{n, r} ∥ g ∥_{L^{r}}$ for $1 < r < \infty$ (The centered maximal operator is bounded on $L^{p} (R^{n})$ for $1 < p < \infty$ ).

[F6]

The maximal functions are Borel measurable, so all $L^{p}$ expressions are meaningful with values in $[0, \infty]$ (Measurability and lower semicontinuity of the smooth maximal functions).

[F7]

$L^{2} (R^{n}) \subset L_{l o c}^{1} (R^{n})$ : on each compact $K$ , Holder gives $\int_{K} ∣ g ∣ \leq λ (K)^{1 / 2} ∥ g ∥_{2}$ , and compact sets have finite measure because they are bounded (Holder's inequality for integrals, including the endpoint cases, Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure).

[F8]

If nonnegative measurable functions $h_{m}$ increase pointwise to $h$ , then their integrals increase to $\int h$ by the monotone convergence theorem (Monotone convergence for the integral).

Proof technique: a priori good-set argument, then removal of the a priori finiteness by truncation, then the pointwise domination for the converse.

Proof

technique · direct

1.1F1F2F3F4F5F6F7algebra

A priori estimate. Assume $M_{φ}^{*, 1} f \in L^{p}$ and let $q_{0} = p / 2$ , $T = 2 n / p$ , so that $T = n / q_{0}$ and $q_{0} < p$ . Let $N_{2} = N_{2} (n, φ, T)$ and $C_{0} = C_{0} (n, φ, T)$ be the constants of [F3], let $C_{T} = C_{n, p, q_{0}}$ be the norm constant from [F2], and let $C_{H} = C_{n, 2}^{1 / q_{0}}$ be the powered Hardy-Littlewood constant from [F5] at $r = 2$ . Put $C_{1} = C_{0} C_{T}$ and $λ = 2^{1 / p} C_{1}$ , so that $∥ M_{N} f ∥_{L^{p}} \leq C_{1} ∥ M_{φ}^{*, 1} f ∥_{L^{p}}$ by [F2], [F3] for every $N \geq N_{2}$ . Let $N_{3} = N_{3} (n, φ, q_{0}, λ)$ be the threshold of [F4, assertion 5]; fix $N \geq \max (N_{2}, N_{3})$ and let $C_{4} = C_{4} (n, φ, q_{0}, λ, N)$ be the constant of [F4, assertion 5] at this order. Set $F = {x : M_{N} f (x) \leq λ M_{φ}^{*, 1} f (x)}$ . On $F^{c}$ one has $M_{φ}^{*, 1} f \leq λ^{- 1} M_{N} f$ pointwise, hence $∥ M_{φ}^{*, 1} f χ_{F^{c}} ∥_{L^{p}}^{p} \leq λ^{- p} ∥ M_{N} f ∥_{L^{p}}^{p} \leq (C_{1} / λ)^{p} ∥ M_{φ}^{*, 1} f ∥_{L^{p}}^{p} = \frac{1}{2} ∥ M_{φ}^{*, 1} f ∥_{L^{p}}^{p}$ . On $F$ , [F4, assertion 5] gives $M_{φ}^{*, 1} f \leq C_{4} \tilde{M} ((M_{φ}^{0} f)^{q_{0}})^{1 / q_{0}}$ at every point at which $M_{φ}^{*, 1} f$ is finite, hence almost everywhere. Since [F1] gives $M_{φ}^{0} f \leq M_{φ}^{*, 1} f$ , we have $(M_{φ}^{0} f)^{q_{0}} \in L^{2}$ ; [F7] gives its local integrability and hence $\tilde{M} = M$ on this input. Integrating, $∥ M_{φ}^{*, 1} f χ_{F} ∥_{L^{p}} \leq C_{4} ∥ M ((M_{φ}^{0} f)^{q_{0}})^{1 / q_{0}} ∥_{L^{p}} = C_{4} ∥ M ((M_{φ}^{0} f)^{q_{0}}) ∥_{L^{p / q_{0}}}^{1 / q_{0}} \leq C_{4} C_{H} ∥ (M_{φ}^{0} f)^{q_{0}} ∥_{L^{p / q_{0}}}^{1 / q_{0}} = C_{4} C_{H} ∥ M_{φ}^{0} f ∥_{L^{p}}$ by [F5] with $r = p / q_{0} = 2 > 1$ . Combining the two pieces, $∥ M_{φ}^{*, 1} f ∥_{L^{p}}^{p} \leq (C_{4} C_{H})^{p} ∥ M_{φ}^{0} f ∥_{L^{p}}^{p} + \frac{1}{2} ∥ M_{φ}^{*, 1} f ∥_{L^{p}}^{p}$ , so $∥ M_{φ}^{*, 1} f ∥_{L^{p}} \leq 2^{1 / p} C_{4} C_{H} ∥ M_{φ}^{0} f ∥_{L^{p}}$ ; this is the a priori estimate, with a constant independent of $f$ at each fixed order $N \geq \max (N_{2}, N_{3})$ .

2.1F4F5F6F7F8step 1.1algebra

Finiteness of $M_{φ}^{*, 1} f$ when $M_{φ}^{0} f \in L^{p}$ . Let $f$ be arbitrary with $M_{φ}^{0} f \in L^{p}$ . Choose $L \geq L_{0} (f, n, φ, p)$ as in assertion 1 of [F4] for the fixed exponent $p$ of the theorem and then $N^{'} = N^{'} (n, φ, T, L)$ as in assertions 2-4 of [F4], where $T = 2 n / p$ and $q_{0} = p / 2$ are as above (note that $N^{'}$ may be much larger than the theorem's fixed $N$ ; it is used only to prove finiteness). With $0 < ϵ \leq 1 / 2$ set $M_{1} = M_{φ, 1}^{ϵ, L} f$ , $M_{0} = M_{φ, 0}^{ϵ, L} f$ , $M_{T} = M_{φ, T}^{ϵ, L} f$ , $M_{G} = M_{N^{'}}^{ϵ, L} f$ and $F_{ϵ} = {x : M_{G} (x) < λ M_{1} (x)}$ with $λ = 2^{1 / p} C_{1}^{'}$ , $C_{1}^{'} = C_{1}^{'} (n, φ, T, L)$ the product of the constants in assertions 2 and 3 of [F4]. Assertion 1 of [F4] gives $∥ M_{1} ∥_{L^{p}} < \infty$ . On $F_{ϵ}^{c}$ one has $M_{1} \leq λ^{- 1} M_{G}$ , so $∥ M_{1} χ_{F_{ϵ}^{c}} ∥_{L^{p}}^{p} \leq λ^{- p} ∥ M_{G} ∥_{L^{p}}^{p} \leq (C_{1}^{'} / λ)^{p} ∥ M_{1} ∥_{L^{p}}^{p} = \frac{1}{2} ∥ M_{1} ∥_{L^{p}}^{p}$ . On $F_{ϵ}$ , assertion 4 of [F4] gives $M_{1} \leq C_{3} \tilde{M} ((M_{φ}^{0} f)^{q_{0}})^{1 / q_{0}}$ with $C_{3} = C_{3} (n, φ, q_{0}, T, L, λ, N^{'})$ . Since $(M_{φ}^{0} f)^{q_{0}} \in L^{2}$ , [F7] gives $\tilde{M} = M$ on this input; integrating as in step 1.1 and using [F5] gives $∥ M_{1} χ_{F_{ϵ}} ∥_{L^{p}} \leq C^{''} ∥ M_{φ}^{0} f ∥_{L^{p}}$ with $C^{''}$ independent of $ϵ$ (but depending on the fixed $L$ and hence on $f$ ). Hence $∥ M_{φ, 1}^{ϵ, L} f ∥_{L^{p}} \leq 2^{1 / p} C^{''} ∥ M_{φ}^{0} f ∥_{L^{p}}$ for every $0 < ϵ \leq 1 / 2$ . Take $ϵ_{m} = 1 / (m + 2)$ for $m \geq 0$ . Then the weights $t^{L} / (t + ϵ_{m} + ϵ_{m} ∣ y ∣)^{L}$ increase pointwise to $1$ and the ranges $0 < t < 1 / ϵ_{m}$ increase to $(0, \infty)$ . For each fixed witness $(t, y)$ , eventually $t < 1 / ϵ_{m}$ and its weight tends to $1$ , so $M_{φ, 1}^{ϵ_{m}, L} f ↗ M_{φ}^{*, 1} f$ pointwise. By [F8], monotone convergence gives $∥ M_{φ}^{*, 1} f ∥_{L^{p}} \leq 2^{1 / p} C^{''} ∥ M_{φ}^{0} f ∥_{L^{p}} < \infty$ .

3.1step 1.1step 2.1F2F3F5algebra

The implication 1 $\Rightarrow$ 3 and the norm bound. Let $f$ satisfy 1. By step 2.1, $M_{φ}^{*, 1} f \in L^{p}$ , so the a priori estimate of step 1.1 applies at every order $N \geq N_{0} (n, p, φ) : = \max (N_{2} (n, φ, 2 n / p), N_{3} (n, φ, p / 2, 2^{1 / p} C_{1}))$ and gives $∥ M_{φ}^{*, 1} f ∥_{L^{p}} \leq C ∥ M_{φ}^{0} f ∥_{L^{p}}$ with $C = 2^{1 / p} C_{4} (n, φ, p / 2, 2^{1 / p} C_{1}, N) C_{H}$ independent of $f$ , where $C_{H} = C_{n, 2}^{1 / q_{0}}$ is the constant from [F5] at $r = 2$ ; then [F2], [F3] give $∥ M_{N} f ∥_{L^{p}} \leq C_{1} ∥ M_{φ}^{*, 1} f ∥_{L^{p}} \leq C_{1} C ∥ M_{φ}^{0} f ∥_{L^{p}}$ for the same orders $N \geq N_{0} (n, p, φ)$ , since the estimates of steps 1.1 and 2.1 hold with the stated constants for every such $N$ . Thus 1 implies 3 for every $N \geq N_{0} (n, p, φ)$ , with the stated norm bound.

4.1step 3.1F1F6∎

The remaining implications. If 3 holds, then [F1] gives $M_{φ}^{0} f \leq 2^{N} P_{N} (φ) M_{N} f \in L^{p}$ and $M_{φ}^{*, a} f \leq (1 + a)^{N} P_{N} (φ) M_{N} f \in L^{p}$ for every $a \geq 1$ , so 3 implies 1 and 2 for every aperture, with the displayed bounds (the first inequality $∥ M^{0} ∥ \leq ∥ M^{*, a} ∥$ is pointwise since $M_{φ}^{0} f \leq M_{φ}^{*, a} f$ ). If 2 holds for some $a$ , then $M_{φ}^{0} f \leq M_{φ}^{*, a} f \in L^{p}$ pointwise, so 2 implies 1. Hence all three assertions are equivalent, the quantities are finite exactly on the common set, and the displayed equivalence of extended norms holds with constants depending only on $n, p, N$ and the kernel data used in the deconvolution comparison and in $P_{N} (φ)$ . The last sentence about the kernel-independence of $H^{p}$ follows by applying the equivalence to two admissible kernels $φ, ψ$ and a common order $N \geq \max (N_{0} (n, p, φ), N_{0} (n, p, ψ))$ . This proves the theorem.

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

$H^{p}$ equals $L^{p}$ with equivalent norms for $1 < p < \infty$

Statement

Assume Countable Choice and the ultrafilter lemma used in the $H^{p} \subseteq L^{p}$ part of the proof. Let $n \geq 1$ and $1 < p < \infty$ , and fix an admissible kernel $φ \in S (R^{n})$ with $\int φ \neq 0$ as in The real Hardy space $H^{p}$ defined by a radial maximal function. Then $f \in H^{p} (R^{n})$ if and only if $f$ is (represented by) a function of $L^{p} (R^{n})$ , the two classes coincide, and $∥ f ∥_{H^{p}} \leq C_{1} ∥ f ∥_{L^{p}}, ∥ f ∥_{L^{p}} \leq C_{2} ∥ f ∥_{H^{p}}, f \in H^{p},$ with constants depending on $n, p$ , finitely many Schwartz seminorms of $φ$ , and $∣ \int φ ∣^{- 1}$ . One may take $C_{2} = ∣ \int φ ∣^{- 1}$ . The proof of the inclusion $H^{p} \subseteq L^{p}$ assumes the ultrafilter lemma (a consequence of the Axiom of Choice, The Axiom of Choice) through the weak-star sequential compactness of the dual ball; the inclusion $L^{p} \subseteq H^{p}$ is choice-free beyond the published maximal-function machinery. In particular the scale $H^{p}$ is new only for $0 < p \leq 1$ .

Facts & Assumptions

Given: Countable Choice and the ultrafilter lemma, $n \geq 1$ , $1 < p < \infty$ , an admissible kernel $φ$ , and $f \in S^{'}$ .

[F2]

If $0 \leq Φ \in L^{1} (R^{n})$ is radially nonincreasing, then the associated maximal operator is dominated by the centered Hardy-Littlewood maximal operator: $(Φ * ∣ u ∣) (x) \leq ∥ Φ ∥_{1} M u (x)$ (Radially decreasing kernels are dominated by the maximal function).

[F3]

The centered Hardy-Littlewood maximal operator satisfies the strong $L^{p}$ bound $∥ M u ∥_{L^{p}} \leq C_{n, p} ∥ u ∥_{L^{p}}$ , $1 < p < \infty$ (The centered maximal operator is bounded on $L^{p} (R^{n})$ for $1 < p < \infty$ , The centered and uncentered Hardy-Littlewood maximal functions).

[F4]

Let $p_{α, 0} (φ) = \sup_{x} ∣ x^{α} φ (x) ∣$ be the Schwartz seminorms of Schwartz space and its seminorms, and set $C_{φ} : = 2^{n} (1 + \sqrt{n})^{n + 1} (p_{0, 0} (φ) + \sum_{i = 1}^{n} p_{(n + 1) e_{i}, 0} (φ)) < \infty$ . Since $∣ x ∣ \leq \sqrt{n} \max_{i} ∣ x_{i} ∣$ , the elementary inequality $(1 + r)^{n + 1} \leq 2^{n} (1 + r^{n + 1})$ shows $(1 + ∣ x ∣)^{n + 1} ∣ φ (x) ∣ \leq C_{φ}$ . Thus $G (x) : = C_{φ} (1 + ∣ x ∣)^{- n - 1}$ is a radially nonincreasing integrable pointwise majorant of $∣ φ ∣$ . If $u \in L^{p}$ , then $∣ (u * φ_{t}) (x) ∣ \leq (∣ u ∣ * ∣ φ ∣_{t}) (x)$ ; normalised dilations $G_{t}$ are radially nonincreasing with $∥ G_{t} ∥_{1} = ∥ G ∥_{1}$ , so $∣ u ∣ * ∣ φ ∣_{t} \leq ∣ u ∣ * G_{t} \leq ∥ G ∥_{1} M u$ (Radially decreasing kernels are dominated by the maximal function).

[F5]

For $1 < p < \infty$ complex $L^{p}$ is the dual of complex $L^{p^{'}}$ by Complex Lp duality from real Lp duality. Separability first applies to the Borel restriction: rational boxes countably generate it (For n at least one, open sets, closed sets, compact sets, open balls, boxes, rational open boxes, and rational half-open boxes generate the Borel sigma-algebra on R^n), and bounded cubes give sigma-finiteness. Completion does not change $L^{p^{'}}$ : $L (R^{n})$ is exactly the completion of the restriction of $λ_{n}$ to the Borel sets replaces each measurable set in a sequence of simple approximants by a Borel set modulo a null set; the countable union of these exceptions is null, yielding a Borel representative. Separability on the Borel restriction therefore gives separability of Lebesgue $L^{p^{'}}$ ; hence the unit ball of $L^{p}$ is weak-star sequentially compact, the weak-star topology on norm-bounded sets is metrised by a countable dense set, and the dual norm is weak-star lower semicontinuous (For $1 < p < \infty$ , the same representation theorem holds on arbitrary measure spaces, If $μ$ is sigma-finite and $A$ is countably generated, then $L^{p} (μ)$ is separable for $1 \leq p < \infty$ , A separable predual has weak-star sequentially compact dual ball, Conjugate exponents, including the endpoint conventions).

Proof technique: direct domination by the Hardy-Littlewood maximal function, then weak-star sequential compactness for the reverse inclusion.

Proof

technique · direct

1.1F2F3F4algebra

$L^{p} \subset H^{p}$ . Let $f \in L^{p}$ and use the radially nonincreasing integrable majorant $G$ from [F4]. For every $t > 0$ the normalised kernel $G_{t}$ is again radially nonincreasing with $∥ G_{t} ∥_{1} = ∥ G ∥_{1}$ , and $∣ f * φ_{t} ∣ \leq ∣ f ∣ * ∣ φ ∣_{t} \leq ∣ f ∣ * G_{t} \leq ∥ G ∥_{1} M f$ pointwise by [F2]. Hence $M_{φ}^{0} f \leq ∥ G ∥_{1} M f$ , and [F3] gives $∥ f ∥_{H^{p}} = ∥ M_{φ}^{0} f ∥_{L^{p}} \leq ∥ G ∥_{1} C_{n, p} ∥ f ∥_{L^{p}}$ , so $f \in H^{p}$ with the stated bound.

1.2F5givenalgebra

$H^{p} \subset L^{p}$ . Let $f \in H^{p}$ and set $Φ = φ / \int φ$ . Then $\int Φ = 1$ and $M_{Φ}^{0} f = ∣ \int φ ∣^{- 1} M_{φ}^{0} f \in L^{p}$ exactly by linearity. The functions $u_{t} = f * Φ_{t}$ , $0 < t < 1$ , satisfy $∣ u_{t} ∣ \leq M_{Φ}^{0} f$ pointwise, hence form a bounded family in $L^{p}$ . By Schwartz approximate identities converge in the sense of tempered distributions, $u_{t} \to f$ in $S^{'}$ as $t ↓ 0$ : for $ψ \in S$ , $⟨ u_{t}, ψ ⟩ = ⟨ f, {\overset{ˇ}{Φ}}_{t} * ψ ⟩ \to ⟨ f, ψ ⟩$ . The space $L^{p^{'}}$ is separable for $1 < p < \infty$ , so the unit ball of its dual $L^{p}$ is weak-star sequentially compact, and the bounded sequence $u_{t_{k}}$ over a fixed sequence $t_{k} ↓ 0$ has a subsequence $(u_{t_{k_{ℓ}}})$ converging weak-star to some $v \in L^{p}$ . By weak-star lower semicontinuity of the norm, $∥ v ∥_{L^{p}} \leq {lim inf}_{ℓ} ∥ u_{t_{k_{ℓ}}} ∥_{L^{p}} \leq ∥ M_{Φ}^{0} f ∥_{L^{p}} \leq ∣ \int φ ∣^{- 1} ∥ f ∥_{H^{p}}$ . For every $ψ \in S \subset L^{p^{'}}$ one has $⟨ v, ψ ⟩ = \lim_{ℓ} ⟨ u_{t_{k_{ℓ}}}, ψ ⟩ = ⟨ f, ψ ⟩$ , since $u_{t} \to f$ in $S^{'}$ ; since equality of tempered distributions is tested against $S$ , the distribution $f$ is represented by the $L^{p}$ function $v$ . Hence $f \in L^{p}$ with $∥ f ∥_{L^{p}} \leq ∣ \int φ ∣^{- 1} ∥ f ∥_{H^{p}}$ .

2.1step 1.1step 1.2∎

Conclusion. Steps 1.1 and 1.2 show that $H^{p}$ and $L^{p}$ have the same elements and equivalent (quasi-)norms for $1 < p < \infty$ .

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

$ℓ^{p}$ sums of atoms converge in $S^{'}$ and in $H^{p}$

Statement

Assume Countable Choice. Let $n \geq 1$ , $0 < p \leq 1$ , $s \geq ⌊ n (1 / p - 1) ⌋$ , fix the admissible kernel $φ$ defining $H^{p}$ , and let $N$ be an admissible order for the grand maximal function with $N \geq \max (N_{0} (n, p, φ), n + s + 1)$ . Let $(a_{j})$ be a sequence of $(p, \infty, s)$ -atoms and $(λ_{j}) \in ℓ^{p}$ . Then the series $\sum_{j} λ_{j} a_{j}$ converges absolutely in $S^{'} (R^{n})$ to an element $g \in S^{'} (R^{n})$ ; the partial sums converge to $g$ in the $H^{p}$ quasi-norm of The real Hardy space $H^{p}$ defined by a radial maximal function; $g \in H^{p}$ ; and with $C_{p} = C (n, p, s, N, φ) < \infty$ independent of the atoms and coefficients, $∥ g ∥_{H^{p}} \leq C_{p} (\sum_{j} ∣ λ_{j} ∣^{p})^{1 / p}, ∥ g - \sum_{j \leq J} λ_{j} a_{j} ∥_{H^{p}} \leq C_{p} (\sum_{j > J} ∣ λ_{j} ∣^{p})^{1 / p} .$

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $0 < p \leq 1$ , $s \geq ⌊ n (1 / p - 1) ⌋$ , the fixed admissible kernel $φ$ , an admissible order $N \geq \max (N_{0} (n, p, φ), n + s + 1)$ , atoms $a_{j}$ , coefficients $(λ_{j}) \in ℓ^{p}$ .

[F1]

Uniform atom bound: there is $C = C (n, p, s)$ with $∥ M_{N} a_{j} ∥_{L^{p}} \leq C$ and $∣ ⟨ a_{j}, ψ ⟩ ∣ \leq C ∥ ψ ∥_{C^{s + 1} (R^{n})}, ∥ ψ ∥_{C^{s + 1} (R^{n})} : = \max_{∣ β ∣ \leq s + 1} \sup_{x \in R^{n}} ∣ \partial^{β} ψ (x) ∣,$ for every $ψ \in S$ . The pairing estimate follows from assertion 2 of Atoms have uniformly bounded $H^{p}$ quasi-norm and uniformly bounded test pairings because its minimum cube-volume factor is at most one; in particular this is a uniform bound by a continuous Schwartz seminorm.

[F2]

Domination: $M_{φ}^{0} g \leq 2^{N} P_{N} (φ) M_{N} g$ for every $g \in S^{'}$ (The grand maximal function dominates every admissible radial and nontangential maximal function).

[F3]

Convergence in $S^{'}$ of the partial sums implies pointwise convergence of the convolutions: if $g_{J} \to g$ in $S^{'}$ , then $(g_{J} * φ_{t}) (y) \to (g * φ_{t}) (y)$ for every $t > 0$ and $y$ , since $(g_{J} * φ_{t}) (y) = ⟨ g_{J}, φ_{t} (y - \cdot) ⟩$ and $φ_{t} (y - \cdot) \in S$ (Convolution of a tempered distribution with a schwartz function, Tempered distribution).

[F4]

$M_{N}$ is Borel measurable and the $p$ -th power inequality $∣ \sum_{j} z_{j} ∣^{p} \leq \sum_{j} ∣ z_{j} ∣^{p}$ holds for $0 < p \leq 1$ ; monotone convergence applies to the nonnegative measurable partial sums (Measurability and lower semicontinuity of the smooth maximal functions, Monotone convergence for the integral).

Proof technique: absolute convergence of pairings, monotone maximal control and monotone convergence.

Proof

technique · direct

1.1F1givenalgebra

Absolute convergence in $S^{'}$ . Fix $ψ \in S$ . By [F1], $∣ λ_{j} ⟨ a_{j}, ψ ⟩ ∣ \leq C ∥ ψ ∥_{C^{s + 1} (R^{n})} ∣ λ_{j} ∣$ , and $\sum_{j} ∣ λ_{j} ∣ < \infty$ because $(λ_{j}) \in ℓ^{p}$ and $p \leq 1$ . Thus the scalar series $\sum_{j} λ_{j} ⟨ a_{j}, ψ ⟩$ converges absolutely for every $ψ$ . Its limit defines a linear functional $g$ satisfying $∣ ⟨ g, ψ ⟩ ∣ \leq C ∥ ψ ∥_{C^{s + 1} (R^{n})} \sum_{j} ∣ λ_{j} ∣$ ; this continuous-seminorm bound proves $g \in S^{'}$ , and the partial sums converge to $g$ on every Schwartz test.

1.2F3givenalgebra

Maximal control of the sum. For every $J$ put $g_{J} = \sum_{j \leq J} λ_{j} a_{j}$ . For fixed $Ψ \in F_{N}$ , $t > 0$ and $y$ with $∣ y - x ∣ \leq t$ , [F3] gives $(g * Ψ_{t}) (y) = \lim_{J} (g_{J} * Ψ_{t}) (y)$ , so $∣ (g * Ψ_{t}) (y) ∣ \leq {lim sup}_{J} \sum_{j \leq J} ∣ λ_{j} ∣ ∣ (a_{j} * Ψ_{t}) (y) ∣ \leq \sum_{j} ∣ λ_{j} ∣ ∣ (a_{j} * Ψ_{t}) (y) ∣$ . Taking the defining suprema and using $∣ (a_{j} * Ψ_{t}) (y) ∣ \leq M_{N} a_{j} (x)$ for each such $Ψ, t, y$ gives $M_{N} g (x) = \sup_{Ψ \in F_{N}} \sup_{t > 0} \sup_{∣ y - x ∣ \leq t} ∣ (g * Ψ_{t}) (y) ∣ \leq \sum_{j} ∣ λ_{j} ∣ M_{N} a_{j} (x)$ pointwise.

2.1step 1.1step 1.2F1F2F4algebra

$H^{p}$ bound and convergence. By [F4] and the $p$ -power inequality for finite sums, letting the number of terms increase in the display of step 1.2 gives $(M_{N} g)^{p} \leq \sum_{j} ∣ λ_{j} ∣^{p} (M_{N} a_{j})^{p} .$ The nonnegative partial sums on the right are measurable; monotone convergence and [F1] therefore give $∥ M_{N} g ∥_{L^{p}}^{p} \leq C^{p} \sum_{j} ∣ λ_{j} ∣^{p}$ , so $∥ g ∥_{H^{p}} \leq 2^{N} P_{N} (φ) C (\sum_{j} ∣ λ_{j} ∣^{p})^{1 / p}$ by [F2]. Applying the same argument to the tail $g - \sum_{j \leq J} λ_{j} a_{j} = \sum_{j > J} λ_{j} a_{j}$ gives the stated tail bound; in particular the partial sums converge to $g$ in the $H^{p}$ quasi-norm.

3.1step 1.1step 1.2step 2.1∎

Conclusion. Steps 1.1 and 1.2 establish the absolute convergence in $S^{'}$ , and step 2.1 establishes the membership $g \in H^{p}$ , the quasi-norm bound and the tail bound. This proves the lemma.

LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passOpen item page →

Level decomposition of an $H^{p}$ distribution produces atoms

Statement

Assume Countable Choice. Fix the admissible kernel $φ$ defining the $H^{p}$ quasi-norm. Let $n \geq 1$ , $0 < p \leq 1$ , and let $f \in H^{p} (R^{n})$ with the space and quasi-norm of The real Hardy space $H^{p}$ defined by a radial maximal function. Put $s = ⌊ n (1 / p - 1) ⌋$ , let $K \geq n / p$ be an integer and let $Φ, ψ, \tilde{ψ}$ be the Calderon reproducing pair of flatness $K$ from Calderon reproducing pair and the telescoping identity in $S^{'}$ . Fix an admissible grand-maximal order $N \geq \max {N_{0} (n, p, φ), n, ⌊ n / p ⌋ + 1}$ . For $r \in Z$ put $Ω_{r} = {x : M_{N} f (x) > 2^{r}}$ . Then there are a countable family $(a_{B})$ of $(p, \infty, s)$ -atoms and positive coefficients $λ_{B}$ such that

$\sum_{B} λ_{B}^{p} \leq C ∥ f ∥_{H^{p}}^{p}$ with $C = C (n, p, N, K, φ, Φ) < \infty$ ;
$f = \sum_{B} λ_{B} a_{B}$ with convergence in $S^{'} (R^{n})$ ;
each atom $a_{B}$ is supported in a fixed dilation $B^{⋆} = 7 B$ of a ball $B = B (ξ, ρ (ξ) / 2)$ of the Whitney-type ball cover of $Ω_{r}$ for the corresponding level $r$ , with $∥ a_{B} ∥_{L^{\infty}} \leq ∣ B^{⋆} ∣^{- 1 / p}$ and vanishing moments through order $s$ , and the balls $B$ cover $Ω_{r}$ with multiplicity at most $K (n) = 785^{n}$ ;
the centres and radii are those of Whitney-type ball cover with disjoint small balls and bounded overlap applied to each nonempty $Ω_{r}$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $0 < p \leq 1$ , admissible $N \geq \max (N_{0} (n, p, φ), \max (n, ⌊ n / p ⌋ + 1))$ , $f \in H^{p}$ , $s = ⌊ n (1 / p - 1) ⌋$ , an integer $K \geq n / p$ , and the fixed reproducing pair $Φ, ψ, \tilde{ψ}$ with $ψ_{k} (x) = 2^{k n} ψ (2^{k} x)$ , ${\tilde{ψ}}_{k} (x) = 2^{k n} \tilde{ψ} (2^{k} x)$ .

[F1]

Maximal characterisation: $∥ M_{N} f ∥_{L^{p}} \leq C_{0} (n, p, N, φ) ∥ f ∥_{H^{p}}$ and $M_{N} f$ is Borel and lower semicontinuous, so each $Ω_{r}$ is open (Maximal-function characterisations of real Hardy spaces, Measurability and lower semicontinuity of the smooth maximal functions).

[F2]

Layer-cake bound: $\sum_{r \in Z} 2^{p r} ∣ Ω_{r} ∣ \leq C_{p} ∥ M_{N} f ∥_{L^{p}}^{p}$ , since $\sum_{r} 2^{p r} ∣ Ω_{r} ∣ = \sum_{ν} ∣ Ω_{ν} ∖ Ω_{ν + 1} ∣ \sum_{r \leq ν} 2^{p r} \leq C_{p} \sum_{ν} 2^{p ν} ∣ Ω_{ν} ∖ Ω_{ν + 1} ∣ \leq C_{p} \int (M_{N} f)^{p}$ (Measurability and lower semicontinuity of the smooth maximal functions, The real Hardy space $H^{p}$ defined by a radial maximal function).

[F3]

Reproducing identity: $f = \sum_{k \in Z} ψ_{k} * {\tilde{ψ}}_{k} * f$ in $S^{'}$ , $supp ψ_{k} \subseteq B (0, 2^{- k})$ , and $\int ψ (x) x^{α} d x = 0$ for $∣ α ∣ \leq K$ ; the identity and its justification are in Calderon reproducing pair and the telescoping identity in $S^{'}$ .

[F4]

For every $k \in Z$ one has $∥ {\tilde{ψ}}_{k} * f ∥_{L^{\infty}} \leq c 2^{k n / p} ∥ f ∥_{H^{p}}$ : indeed $∣ {\tilde{ψ}}_{k} * f (x) ∣^{p} \leq \inf_{∣ x - y ∣ \leq 2^{- k}} \sup_{∣ y - z ∣ \leq 2^{- k}} ∣ {\tilde{ψ}}_{k} * f (z) ∣^{p} \leq c \inf_{∣ x - y ∣ \leq 2^{- k}} M_{N} f (y)^{p} \leq c ∣ B (x, 2^{- k}) ∣^{- 1} \int_{B (x, 2^{- k})} (M_{N} f)^{p} \leq c 2^{k n} ∥ M_{N} f ∥_{p}^{p}$ (Grand maximal test class of order N and the grand maximal function, [F1]).

[F5]

Whitney-type ball cover of each nonempty $Ω_{r}$ : points $ξ_{j}$ , radii $ρ_{j} = ρ (ξ_{j})$ , pairwise disjoint balls $B (ξ_{j}, ρ_{j} / 8)$ with $⋃_{j} B (ξ_{j}, ρ_{j} / 2) = Ω_{r}$ , comparison $ρ_{j} / 7 \leq ρ_{ν} \leq 7 ρ_{j}$ for meeting $3 ρ / 4$ -balls and bounded overlap $K (n)$ of the dilated balls $B (ξ_{j}, 3 ρ_{j} / 4)$ (Whitney-type ball cover with disjoint small balls and bounded overlap).

[F6]

Every $x$ with $M_{N} f (x) > 2^{r}$ lies in $Ω_{r}$ ; the sets $Ω_{r}$ decrease in $r$ and are open, so $dist (\cdot, Ω_{r}^{c})$ is continuous and positive on $Ω_{r}$ (Measurability and lower semicontinuity of the smooth maximal functions).

[F7]

Moment-tail estimate: if $Ψ \in C_{c}^{\infty}$ with $supp Ψ \subseteq B (0, 1)$ and $\int Ψ (z) z^{α} d z = 0$ for $∣ α ∣ \leq K$ , then for every $χ \in S$ and $σ > n$ there is $c_{σ}$ with $∣ \int_{R^{n}} Ψ_{k} (x - y) χ (x) d x ∣ \leq c_{σ} 2^{- k (K + 1)} (1 + ∣ y ∣)^{- σ}$ for $k \geq 0$ . This is the Taylor estimate (using Multivariable Taylor formula with a Lagrange remainder along a line segment separately on real and imaginary parts): expand $χ$ about $y$ , use the vanishing moments, bound the remainder by $C 2^{- k (K + 1)} (1 + ∣ y ∣)^{- σ}$ with the Schwartz decay of $χ$ and of its derivatives (Schwartz space and its seminorms, Dilations and their normalisations preserve Schwartz space, with scaling identities).

[F8]

Ball volumes: with $ω_{n} = ∣ B (0, 1) ∣$ , one has $0 < ω_{n} < \infty$ and $∣ B (x, R) ∣ = ω_{n} R^{n}$ for $R > 0$ , by Euclidean balls have positive finite Lebesgue measure and For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it.

[F9]

A tempered distribution $u$ satisfying $∣ ⟨ u, χ ⟩ ∣ \leq D ∥ χ ∥_{1}$ for all Schwartz tests is represented by an $L^{\infty}$ function bounded by $D$ . Indeed, density of complex $C_{c}^{\infty}$ in $L^{1}$ (Complex finite-simple and smooth compact-support density for finite p) uniquely extends $u$ to a bounded complex-linear functional on $L^{1}$ . Restrict it to $L^{2} ([- m, m]^{n})$ , extending inputs by zero; $∥ h ∥_{1} \leq (2 m)^{n / 2} ∥ h ∥_{2}$ . Apply For $1 < p < \infty$ , the same representation theorem holds on arbitrary measure spaces at exponent two separately to the real and imaginary parts on real inputs, then combine by complex linearity, to obtain a complex density $g_{m}$ . Testing with $\overline{g_{m}} / ∣ g_{m} ∣$ times indicators of measurable subsets (zero where $g_{m} = 0$ ) gives $\int_{E} ∣ g_{m} ∣ \leq D ∣ E ∣$ , hence $∣ g_{m} ∣ \leq D$ a.e. Uniqueness of the $L^{2}$ densities makes them agree on nested cubes; choosing representatives under Countable Choice and discarding their countable union of disagreement null sets glues a globally bounded density. Density and truncation recover the functional on all of $L^{1}$ . All hypotheses of these suppliers hold under the assumed Countable Choice.

Proof technique: level-set decomposition, telescoping cancellation, Whitney covering and atom normalisation.

Proof

technique · constructive

1.1F1F2F4F6F8algebraconstruct

The level sets and scale shells. If $f = 0$ , the empty atomic family gives the conclusion, so assume $f \neq 0$ . Put $M = M_{N} f$ and $Ω_{r} = {M > 2^{r}}$ . By [F1], [F2], and layer cake (summing nonnegative indicators using Monotone convergence for the integral), $\sum_{r \in Z} 2^{p r} ∣ Ω_{r} ∣ \leq C_{p} ∥ M ∥_{p}^{p} \leq C ∥ f ∥_{H^{p}}^{p},$ so every $Ω_{r}$ has finite measure. For $k \in Z$ set $a_{k} = 2^{- k + 1}$ , $U_{k}^{r} = {x \in Ω_{r} : d_{r} (x) > a_{k}}, V_{k}^{r} = {x \in Ω_{r + 1} : d_{r + 1} (x) > a_{k}}, E_{r}^{k} = U_{k}^{r} ∖ V_{k}^{r},$ where $d_{r} (x) = dist (x, Ω_{r}^{c})$ . The sets $E_{r}^{k}$ are disjoint in $r$ . At a fixed $k$ , they cover every $y$ with $0 < M (y) < \infty$ for which ${\tilde{ψ}}_{k} * f (y) \neq 0$ : indeed, with $η = 4^{n} \tilde{ψ} (4 \cdot)$ and $t_{k} = 2^{- k + 2}$ one has $η_{t_{k}} = {\tilde{ψ}}_{k}$ . After division by $P_{N} (η)$ this is a test in $F_{N}$ , so a nonzero value at $y$ gives a positive lower bound for $M (z)$ for all $∣ z - y ∣ \leq t_{k}$ . Choosing $r$ below that bound puts $B (y, t_{k}) \subset Ω_{r}$ , hence $y \in U_{k}^{r}$ . Since $M \in L^{p}$ is finite a.e., for a.e. such $y$ it is outside $Ω_{r}$ for all sufficiently large $r$ ; the nested sets $U_{k}^{r}$ therefore give a unique last index $r$ with $y \in E_{r}^{k}$ . Points where $M = \infty$ form a null set, and where $M = 0$ the displayed convolution is zero by the grand-maximal definition. Thus the $E_{r}^{k}$ partition the integrand ${\tilde{ψ}}_{k} * f$ up to a null set, which is all the integral and distributional identities below require. Finally, if $Ω_{r} \neq \emptyset$ , choose $s_{r}$ so negative that $ω_{n} a_{k}^{n} > ∣ Ω_{r} ∣$ for $k < s_{r}$ . Then $U_{k}^{r} = E_{r}^{k} = \emptyset$ for $k < s_{r}$ , since each point of $U_{k}^{r}$ would force a ball of radius $a_{k}$ inside $Ω_{r}$ . If $Ω_{r} = \emptyset$ , set all its pieces to zero.

2.1step 1.1F3F4F6F7F9algebra

Local bounds and the two-boundary scale split. For every measurable $A \subseteq E_{r}^{k}$ , $\begin{matrix} {∥ \int_{A} ψ_{k} (\cdot - y) ({\tilde{ψ}}_{k} * f) (y) d y ∥}_{\infty} \leq C 2^{r}, {∥ \int_{A} Φ_{m} (\cdot - y) (Φ_{m} * f) (y) d y ∥}_{\infty} \leq C 2^{r} (m = k, k + 1) . & (1) \end{matrix}$ To see this, if $y \in E_{r}^{k}$ is outside $Ω_{r + 1}$ , use $z = y$ . Otherwise $d_{r + 1} (y) \leq a_{k} < d_{r} (y)$ ; choose $z \in Ω_{r + 1}^{c}$ with $∣ z - y ∣ < d_{r + 1} (y) + ϵ < d_{r} (y)$ and $∣ z - y ∣ < 2 a_{k}$ . Then $z \in Ω_{r} ∖ Ω_{r + 1}$ and $M (z) \leq 2^{r + 1}$ . Each inner kernel here is exactly a grand-maximal test dilation at cone scale $2 a_{k}$ : use $4^{n} \tilde{ψ} (4 \cdot)$ for ${\tilde{ψ}}_{k}$ , $4^{n} Φ (4 \cdot)$ for $Φ_{k}$ , and $8^{n} Φ (8 \cdot)$ for $Φ_{k + 1}$ . The cone definition of $M_{N}$ therefore gives $∣ {\tilde{ψ}}_{k} * f (y) ∣ + ∣ Φ_{m} * f (y) ∣ \leq C 2^{r}$ . Multiplication by the fixed $L^{1}$ norms of the outer kernels $ψ_{k}$ and $Φ_{m}$ proves (1). For $s_{r} \leq q \leq m$ write $F_{r, q, m} (x) = \sum_{k = q}^{m} \int_{E_{r}^{k}} ψ_{k} (x - y) ({\tilde{ψ}}_{k} * f) (y) d y .$ Its $L^{\infty}$ norm is at most $C 2^{r}$ , uniformly in $r, q, m$ . Here are the scale details. If a summand can be nonzero at $x$ , then $x \in Ω_{r}$ . Choose $ℓ$ with $x \in U_{ℓ + 1}^{r} ∖ U_{ℓ}^{r}$ , so $2^{- ℓ} < d_{r} (x) \leq 2^{- ℓ + 1}$ . If also $x \in Ω_{r + 1}$ , choose $ν$ with $x \in V_{ν + 1}^{r} ∖ V_{ν}^{r}$ ; since $d_{r} (x) \geq d_{r + 1} (x)$ , $ν \geq ℓ$ . For $ν \geq ℓ + 3$ , the Lipschitz property of distance gives $B (x, 2^{- k}) \cap E_{r}^{k} = \emptyset (k \leq ℓ - 1 or k \geq ν + 2), B (x, 2^{- k}) \subset E_{r}^{k} (ℓ + 2 \leq k \leq ν - 1) .$ Thus, in any finite interval $[q, m]$ , at most four non-full boundary terms remain; the consecutive full terms $k = u, \dots, v$ telescope to $\sum_{k = u}^{v} ψ_{k} * {\tilde{ψ}}_{k} * f (x) = Φ_{v + 1} * Φ_{v + 1} * f (x) - Φ_{u} * Φ_{u} * f (x) .$ The support balls of the two endpoint convolutions lie in $E_{r}^{v}$ and $E_{r}^{u}$ , respectively, so (1) bounds both endpoints by $C 2^{r}$ . If $ℓ \leq ν \leq ℓ + 2$ , only the at most four indices $ℓ, \dots, ν + 1$ can contribute, and (1) applies directly. If $x \in Ω_{r} ∖ Ω_{r + 1}$ , choose $ℓ$ as above. There is no interaction for $k \leq ℓ - 1$ , while $B (x, 2^{- k}) \subset E_{r}^{k}$ for every $k \geq ℓ + 2$ . The latter tail in any finite interval telescopes, with both endpoint convolutions localized in the corresponding $E_{r}^{k}$ and bounded by (1); only the two scales $ℓ, ℓ + 1$ are left. These cases prove the uniform bound for every finite partial sum. For $χ \in S$ , the moment estimate [F7] and [F4] give, for $k \geq 0$ , $\sum_{r} ∣ ⟨ g_{r k}, χ ⟩ ∣ \leq C_{χ} 2^{- k (K + 1 - n / p)} ∥ f ∥_{H^{p}}, g_{r k} (x) = \int_{E_{r}^{k}} ψ_{k} (x - y) ({\tilde{ψ}}_{k} * f) (y) d y .$ For $k < 0$ , the same sum is at most $C_{χ} 2^{k n / p} ∥ f ∥_{H^{p}}$ , using $∥ ψ_{k} ∥_{1} = ∥ ψ ∥_{1}$ and $∥ {\tilde{ψ}}_{k} * f ∥_{\infty} \leq C 2^{k n / p} ∥ f ∥_{H^{p}}$ . The first bound is summable because $K \geq n / p$ , and for each fixed $r$ only finitely many negative $k$ occur because $k \geq s_{r}$ . Thus $F_{r} = \sum_{k \geq s_{r}} g_{r k}$ converges in $S^{'}$ . The uniform bounds for finite partial sums imply $∣ ⟨ F_{r}, χ ⟩ ∣ \leq C 2^{r} ∥ χ ∥_{1}$ ; by the bounded-distribution representation [F9], $F_{r}$ is represented by an $L^{\infty}$ function with $∥ F_{r} ∥_{\infty} \leq C 2^{r}$ . No pointwise convergence of the infinite scale series is needed.

3.1step 2.1F3F4F5F7F8F9algebra

Whitney localization, overlap, and atoms. Fix a nonempty $Ω_{r}$ and take the cover [F5], writing $B_{j} = B (ξ_{j}, ρ_{j} / 2)$ and $W_{j} = B (ξ_{j}, 3 ρ_{j} / 4)$ . Then $B_{j} \subset W_{j}$ , so the $B_{j}$ cover $Ω_{r}$ with multiplicity at most the exact constant $K (n) = 785^{n}$ from [F5]. For each $j$ , let $k_{0} (j)$ be the unique integer with $2^{- k_{0} (j)} < ρ_{j} \leq 2^{- k_{0} (j) + 1}$ . For $k \geq s_{r}$ , put $N_{j, k} = {y : dist (y, B_{j}) < a_{k}}$ and $A_{r j}^{k} = {\begin{cases} E_{r}^{k} \cap N_{j, k}, & k \geq k_{0} (j), \\ \emptyset, & k < k_{0} (j) . \end{cases}$ Put $R_{r j}^{k} = A_{r j}^{k} ∖ ⋃_{m > j} A_{r m}^{k}$ and $g_{r j k} (x) = \int_{R_{r j}^{k}} ψ_{k} (x - y) ({\tilde{ψ}}_{k} * f) (y) d y$ . These sets are disjoint and cover $E_{r}^{k}$ : if $y \in E_{r}^{k}$ , choose a covering ball $B_{j}$ containing $y$ . Then $d_{r} (y) > a_{k} = 2 \cdot 2^{- k}$ and $d_{r} (y) \leq d_{r} (ξ_{j}) + ∣ y - ξ_{j} ∣ < 3 ρ_{j} / 2$ , so $2^{- k} < 3 ρ_{j} / 4 < ρ_{j}$ and $k \geq k_{0} (j)$ . Hence $y \in A_{r j}^{k}$ , and the greatest eligible index assigns it to exactly one $R_{r j}^{k}$ , once the finite overlap below is established. The actual localization neighborhoods $N_{j, k}$ with $k \geq k_{0} (j)$ have uniformly finite overlap. For such an index $2^{- k} < ρ_{j}$ , so $a_{k} < 2 ρ_{j}$ and $∣ y - ξ_{j} ∣ < ρ_{j} / 2 + a_{k} < 5 ρ_{j} / 2$ whenever $y \in N_{j, k}$ . If two eligible neighborhoods meet, the 1-Lipschitz property of $d_{r}$ gives, for $R = \max (ρ_{j}, ρ_{m})$ and $r_{0} = \min (ρ_{j}, ρ_{m})$ , $R - r_{0} \leq ∣ ξ_{j} - ξ_{m} ∣ < \frac{1}{2} (R + r_{0}) + 4 r_{0},$ so $R < 11 r_{0}$ . For any finite collection of eligible neighborhoods containing one point $y$ , the disjoint balls $B (ξ_{j}, ρ_{j} / 8)$ therefore have radii at least $R / 88$ and lie in $B (y, 21 R / 8)$ . Comparing volumes gives multiplicity at most $231^{n}$ at every fixed scale. If $A_{r j}^{k} \neq \emptyset$ , then $k \geq k_{0} (j)$ and the support of $ψ_{k} * [({\tilde{ψ}}_{k} * f) 1_{R_{r j}^{k}}]$ lies in the ball of radius $ρ_{j} / 2 + 3 \cdot 2^{- k} < 7 ρ_{j} / 2$ about $ξ_{j}$ , hence in $7 B_{j}$ . It remains to prove a uniform $L^{\infty}$ estimate for $F_{B_{j}} : = \sum_{k \geq s_{r}} \int_{R_{r j}^{k}} ψ_{k} (\cdot - y) ({\tilde{ψ}}_{k} * f) (y) d y$ . We use the following explicit localization estimate. For any set $S \subseteq R^{n}$ and integers $s_{r} \leq q \leq m$ , replacing $E_{r}^{k}$ by $E_{r}^{k} \cap {y : dist (y, S) < a_{k}}$ in $F_{r, q, m}$ still gives an $L^{\infty}$ norm at most $C 2^{r}$ . If $S = \emptyset$ , the sum is zero. If $S \neq \emptyset$ and $dist (x, S) = 0$ , all kernel-support balls $B (x, 2^{- k})$ lie in these neighborhoods and the sum is the full bounded partial sum. If $δ = dist (x, S) > 0$ , choose $ℓ$ with $2^{- ℓ} < δ \leq 2^{- ℓ + 1}$ . Then $B (x, 2^{- k})$ lies in the neighborhood for $k \leq ℓ - 1$ and is disjoint from it for $k \geq ℓ + 2$ , leaving only two boundary scales, each bounded by (1). The same estimate holds for the infinite tail: the preceding absolute pairing bounds give distributional convergence, and the uniform finite-sum bounds pass to the limit by the same bounded-distribution representation [F9] used in step 2.1. Let $J_{j}$ be all later indices $m > j$ for which $W_{m} \cap W_{j} \neq \emptyset$ . By [F5], $# J_{j} \leq K (n)$ and $ρ_{m} \geq ρ_{j} / 7$ . Put $ρ_{*} = \min ({ρ_{j} / 7} \cup {ρ_{m} : m \in J_{j}}) = ρ_{j} / 7$ , let $k_{1}$ be the least integer with $a_{k_{1}} < ρ_{*} / 4$ , and set $S = ⋃_{m \in J_{j}} B_{m}$ . Then $ρ_{j} / 7 \leq ρ_{*} \leq ρ_{j}$ , while $ρ_{j} / 2 \leq 2^{- k_{0}} < ρ_{j}$ and $ρ_{*} / 16 \leq 2^{- k_{1}} < ρ_{*} / 8$ . Hence $k_{0} < k_{1}$ and $2^{k_{1} - k_{0}} = 2^{- k_{0}} / 2^{- k_{1}} < 112 < 2^{7}$ , so $k_{1} - k_{0} \leq 7$ . For every $k \geq k_{1}$ each neighborhood $N_{m, k}$ with $m \in J_{j} \cup {j}$ lies in $W_{m}$ . To exclude other later indices without presupposing their radii, a meeting point gives $∣ ξ_{m} - ξ_{j} ∣ < (ρ_{m} + ρ_{j}) / 2 + 2 a_{k}$ . Here $a_{k} < ρ_{*} / 4 \leq ρ_{j} / 28$ , so this distance is less than $3 (ρ_{m} + ρ_{j}) / 4$ . Thus $W_{m} \cap W_{j} \neq \emptyset$ and $m \in J_{j}$ . All indices in $J_{j}$ are eligible at these scales because $2^{- k} < ρ_{*} / 8 \leq ρ_{m} / 8$ . Consequently the high-scale part is exactly the difference of the two localized sums associated with $S \cup B_{j}$ and $S$ , each bounded by the localization estimate. There are at most seven lower scales $k_{0} \leq k < k_{1}$ , each bounded by (1). Thus $∥ F_{B_{j}} ∥_{\infty} \leq C_{♯} 2^{r}$ uniformly in $r, j$ . Absolute convergence against Schwartz tests follows by summing [F7] over the disjoint sets $R_{r j}^{k}$ ; for $k < 0$ the bound is $C_{χ} 2^{k n / p} ∥ f ∥_{H^{p}}$ as above. Each summand is compactly supported in $7 B_{j}$ and has zero moments through order $K$ by the moments of $ψ$ . Choose the representative of $F_{B_{j}}$ to vanish outside its compact support. That support lies in the closed ball of radius $ρ_{j} / 2 + 3 \cdot 2^{- k_{0} (j)} < 7 ρ_{j} / 2$ , so it is contained in the open ball $7 B_{j}$ . For fixed $j, k$ , $R_{r j}^{k} \subset N_{j, k}$ is bounded, has finite measure, and ${\tilde{ψ}}_{k} * f$ is bounded by [F4], so absolute integrability and Tonelli's theorem for nonnegative measurable functions on a sigma-finite product applied to the absolute values justify each moment integral. The distributional limit is supported in $7 B_{j}$ and has the same moments, by testing against a smooth compactly supported test equal to each monomial on a neighborhood of the closed ball $\overline{7 B_{j}}$ (construct the cutoff from The standard smooth step function). Finally, the sets $R_{r j}^{k}$ partition the $E_{r}^{k}$ , and the absolute pairing bounds summed over all $r, j, k$ show $f = \sum_{r, j} F_{B_{j}} in S^{'} .$ Indeed, for $k \geq 0$ the full sum of absolute pairings is bounded by $C_{χ} 2^{- k (K + 1 - n / p)} ∥ f ∥_{H^{p}}$ using [F7]; for $k < 0$ it is bounded by $C_{χ} 2^{k n / p} ∥ f ∥_{H^{p}}$ . Sum in $k$ and apply the Calderon reproducing identity [F3].

4.1step 3.1F8algebra

Atom normalization. Choose an axis-parallel cube $Q_{j}$ centered at $ξ_{j}$ with side length $7 ρ_{j}$ ; it contains $7 B_{j}$ , and $∣ Q_{j} ∣ / ∣ 7 B_{j} ∣ = 2^{n} / ω_{n}$ . Enlarge $C_{♯}$ if needed so that $C_{♯} \geq C_{0} (2^{n} / ω_{n})^{1 / p}$ , where $C_{0}$ is the constant in the preceding $L^{\infty}$ estimate. Put $a_{r j} = C_{♯}^{- 1} ∣ 7 B_{j} ∣^{- 1 / p} 2^{- r} F_{B_{j}}, λ_{r j} = C_{♯} ∣ 7 B_{j} ∣^{1 / p} 2^{r} .$ Then $supp a_{r j} \subseteq 7 B_{j}$ , $∥ a_{r j} ∥_{\infty} \leq ∣ 7 B_{j} ∣^{- 1 / p}$ , and the choice of $C_{♯}$ gives $∥ a_{r j} ∥_{\infty} \leq ∣ Q_{j} ∣^{- 1 / p}$ . Its moments vanish through order $K$ , hence through $s$ , because $s = ⌊ n (1 / p - 1) ⌋ < n / p \leq K$ . Thus $a_{r j}$ is a $(p, \infty, s)$ -atom in the cube-supported convention of $H^{p}$ atoms with a prescribed moment order, and $f = \sum_{r, j} λ_{r j} a_{r j}$ in $S^{'}$ .

5.1step 4.1F1F2F5F8algebra

Coefficient bound. Since $∣ 7 B_{j} ∣ = 7^{n} ∣ B_{j} ∣$ and $B_{j} \subset W_{j}$ with multiplicity at most $K (n)$ , $\sum_{r, j} λ_{r j}^{p} = C_{♯}^{p} 7^{n} \sum_{r} 2^{p r} \sum_{j} ∣ B_{j} ∣ \leq C_{♯}^{p} 7^{n} K (n) \sum_{r} 2^{p r} ∣ Ω_{r} ∣ \leq C ∥ f ∥_{H^{p}}^{p}$ by the layer-cake estimate and [F1]. This proves the coefficient bound.

6.1step 1.1step 3.1step 4.1step 5.1discharge-construct∎

Conclusion. The zero case was handled at the start. For $f \neq 0$ , steps 3.1 and 4.1 produce the atoms and coefficients, step 5.1 gives the $ℓ^{p}$ estimate, and the absolutely convergent distributional sum in step 3.1 equals $f$ . Hence all four claims hold.

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Atomic characterisation of real $H^{p}$ for $0 < p \leq 1$

Statement

Assume Countable Choice. Let $n \geq 1$ , $0 < p \leq 1$ , fix the admissible kernel $φ$ defining $H^{p}$ , and set $s = ⌊ n (1 / p - 1) ⌋$ . Fix an integer $K \geq n / p$ and the associated reproducing pair from Calderon reproducing pair and the telescoping identity in $S^{'}$ . Fix an admissible grand-maximal order $N \geq \max {N_{0} (n, p, φ), n + s + 1}$ as in the two cited lemmas. For $f \in S^{'} (R^{n})$ the following are equivalent:

$f \in H^{p} (R^{n})$ in the sense of The real Hardy space $H^{p}$ defined by a radial maximal function;
there exist a sequence $(λ_{j}) \in ℓ^{p}$ and a sequence $(a_{j})$ of $(p, \infty, s)$ -atoms ( $H^{p}$ atoms with a prescribed moment order) with $f = \sum_{j} λ_{j} a_{j}$ converging in $S^{'} (R^{n})$ .

In that case $∥ f ∥_{H^{p}} ≍_{n, p, N, K, φ} \inf (\sum_{j} ∣ λ_{j} ∣^{p})^{1 / p},$ the infimum being taken over all atomic representations of $f$ , and every such series converges also in the $H^{p}$ quasi-norm.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $0 < p \leq 1$ , the fixed kernel $φ$ and reproducing order $K \geq n / p$ , $s = ⌊ n (1 / p - 1) ⌋$ , an admissible order $N \geq \max {N_{0} (n, p, φ), n + s + 1}$ as in the two cited lemmas, and $f \in S^{'}$ .

[F1]

Level decomposition: if $f \in H^{p}$ then there are $(p, \infty, s)$ -atoms $a_{B}$ and coefficients $λ_{B} > 0$ with $f = \sum_{B} λ_{B} a_{B}$ in $S^{'}$ and $\sum_{B} λ_{B}^{p} \leq C_{1} ∥ f ∥_{H^{p}}^{p}$ , where $C_{1} = C (n, p, N, K, φ, Φ)$ includes the auxiliary flat reproducing kernel $Φ$ (Level decomposition of an $H^{p}$ distribution produces atoms). For the norm bound, use Countable Choice over the integer pairs $(n, K)$ to fix once one admissible kernel $Φ_{n, K}$ from Calderon reproducing pair and the telescoping identity in $S^{'}$ . With this fixed family, $C (n, p, N, K, φ, Φ_{n, K})$ is a function of $n, p, N, K, φ$ . No uniformity over all admissible reproducing kernels is asserted or needed: the atomic class and the infimum over representations do not depend on the auxiliary kernel.

[F2]

$ℓ^{p}$ sums: if $(a_{j})$ are $(p, \infty, s)$ -atoms and $(λ_{j}) \in ℓ^{p}$ , then $g = \sum_{j} λ_{j} a_{j}$ converges absolutely in $S^{'}$ , lies in $H^{p}$ and satisfies $∥ g ∥_{H^{p}} \leq C_{2} (\sum_{j} ∣ λ_{j} ∣^{p})^{1 / p}$ with $C_{2}$ depending on $n, p, s, N, φ$ ; the tail bound of that lemma gives convergence in the $H^{p}$ quasi-norm ( $ℓ^{p}$ sums of atoms converge in $S^{'}$ and in $H^{p}$ ).

Proof technique: the two implications supplied by the level decomposition and the $ℓ^{p}$ -summation lemma, then the infimum.

Proof

technique · direct

1.1F2algebra

2 $\Rightarrow$ 1 and the upper norm bound. Let $f = \sum_{j} λ_{j} a_{j}$ with $(λ_{j}) \in ℓ^{p}$ and atoms $a_{j}$ . By [F2], $f \in H^{p}$ and $∥ f ∥_{H^{p}} \leq C_{2} (\sum_{j} ∣ λ_{j} ∣^{p})^{1 / p}$ ; taking the infimum over all representations gives the inequality $∥ f ∥_{H^{p}} \leq C_{2} \inf (\sum_{j} ∣ λ_{j} ∣^{p})^{1 / p}$ .

1.2F1algebra

1 $\Rightarrow$ 2 and the lower norm bound. Let $f \in H^{p}$ and apply [F1] using the auxiliary kernel $Φ_{n, K}$ fixed there, obtaining $f = \sum_{B} λ_{B} a_{B}$ . Then $(λ_{B}) \in ℓ^{p}$ with $\sum_{B} λ_{B}^{p} \leq C_{1} ∥ f ∥_{H^{p}}^{p}$ , so $f$ has an atomic representation and $\inf (\sum_{j} ∣ λ_{j} ∣^{p})^{1 / p} \leq C_{1}^{1 / p} ∥ f ∥_{H^{p}}$ .

2.1F2step 1.2

$H^{p}$ convergence. If $f = \sum_{j} λ_{j} a_{j}$ with $(λ_{j}) \in ℓ^{p}$ , the tail estimate of [F2] applied to the partial sums gives $∥ f - \sum_{j \leq J} λ_{j} a_{j} ∥_{H^{p}} \leq C_{2} (\sum_{j > J} ∣ λ_{j} ∣^{p})^{1 / p} \to 0$ ; hence the series converges in the $H^{p}$ quasi-norm. This applies in particular to the level-decomposition representation of [F1] and to any atomic representation of $f \in H^{p}$ .

3.1step 1.1step 1.2step 2.1∎

Conclusion. Steps 1.1-1.2 prove the equivalence and the two-sided norm bound, and step 2.1 gives the quasi-norm convergence. This proves the theorem.

RemarkRemark: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

For $0 < p < 1$ the $H^{p}$ functional is a quasi-norm, and $H^{p}$ is a quasi-Banach space

Statement

Assume Countable Choice. Fix $n \geq 1$ , $0 < p < 1$ and an admissible kernel $φ \in S (R^{n})$ with $\int φ \neq 0$ as in The real Hardy space $H^{p}$ defined by a radial maximal function. Its functional is $p$ -subadditive, $∥ f + g ∥_{H^{p}}^{p} \leq ∥ f ∥_{H^{p}}^{p} + ∥ g ∥_{H^{p}}^{p},$ and fails the ordinary triangle inequality for a pair of elements of this same $H^{p}$ . It also satisfies $∥ f + g ∥_{H^{p}} \leq 2^{1 / p - 1} (∥ f ∥_{H^{p}} + ∥ g ∥_{H^{p}}) .$ Consequently $d (f, g) = ∥ f - g ∥_{H^{p}}^{p}$ is a translation-invariant metric on $H^{p}$ under which $H^{p}$ is complete. Thus $H^{p}$ is a quasi-Banach space. No statement is made identifying $H^{p}$ with the dual of, or a dual of, a Banach space when $p < 1$ , and no Banach-space duality theorem is applied to $H^{p}$ below $p = 1$ on this page; the only duality statement here is $H^{p} = L^{p}$ for $p > 1$ .

Remarks

The maximal operator is pointwise sublinear: $M_{φ}^{0} (f + g) \leq M_{φ}^{0} f + M_{φ}^{0} g$ . Since $(u + v)^{p} \leq u^{p} + v^{p}$ for $u, v \geq 0$ and $0 < p < 1$ , integration gives the stated $p$ -subadditivity. The displayed quasi-triangle inequality follows as well because concavity of $r \mapsto r^{p}$ gives $a^{p} + b^{p} \leq 2^{1 - p} (a + b)^{p}$ for $a, b \geq 0$ ; take $p$ -th roots after $∥ f + g ∥_{H^{p}}^{p} \leq ∥ f ∥_{H^{p}}^{p} + ∥ g ∥_{H^{p}}^{p}$ .

Here is an $H^{p}$ -specific witness that the ordinary triangle inequality fails; the proof does not use the later uniform atom estimate. Put $s = ⌊ n (1 / p - 1) ⌋$ and $θ (u) = {\begin{cases} e^{- 1 / (1 - u^{2})}, & ∣ u ∣ < 1, \\ 0, & ∣ u ∣ \geq 1, \end{cases} ρ (x) = \prod_{j = 1}^{n} θ (x_{j}), a = \partial_{1}^{s + 1} ρ .$ Here $θ (u) = β (1 - u^{2})$ for the standard flat function of The standard flat function; The standard flat function is smooth and flat at zero establishes smoothness through the endpoints. Thus $a$ is a smooth function on $R^{n}$ supported in $[- 1, 1]^{n}$ . It is nonzero: otherwise each one-variable section of $ρ$ would have $(s + 1)$ st derivative zero, hence would be a polynomial of degree at most $s$ by repeated Newton-Leibniz, impossible for its nonzero compact support. Repeated one-variable integration by parts (obtained from the product rule and Newton–Leibniz needs only continuity on $[a, b]$ , differentiability on $(a, b)$ , and a Riemann-integrable extension of the interior derivative) has no boundary terms and gives $\int x^{α} a (x) d x = 0$ for every multi-index $∣ α ∣ \leq s$ : indeed $α_{1} \leq s$ , so $\partial_{1}^{s + 1} x^{α} = 0$ . Let $m = M_{φ}^{0} a$ . For every $x$ , the convolution estimate gives $m (x) \leq ∥ a ∥_{\infty} ∥ φ ∥_{1}$ . For $∣ x ∣ > 2 \sqrt{n}$ , Taylor's formula in the variable $y$ , with the cancellation moments through order $s$ removed, gives for every $t > 0$ $∣ (a * φ_{t}) (x) ∣ \leq C_{n, s} \int ∣ a (y) ∣ ∣ y ∣^{s + 1} \max_{∣ α ∣ = s + 1} \sup_{0 \leq τ \leq 1} ∣ \partial^{α} φ_{t} (x - τ y) ∣ d y .$ Writing $q = n + s + 1$ , Schwartz decay says for any $r_{0} > q$ and $∣ α ∣ = s + 1$ , $∣ \partial^{α} φ_{t} (w) ∣ \leq C_{r_{0}, α} t^{- q} (1 + ∣ w ∣ / t)^{- r_{0}} \leq C_{r_{0}, α}^{'} ∣ w ∣^{- q} (w \neq 0, t > 0) .$ Since $∣ y ∣ \leq \sqrt{n}$ on the support of $a$ , this proves $m (x) \leq C ∣ x ∣^{- q}$ for $∣ x ∣ > 2 \sqrt{n}$ . The choice of $s$ gives $p q = p (n + s + 1) > n$ , so $m \in L^{p}$ and $a \in H^{p}$ .

Also $m$ is positive on a nonempty open set. To see this, normalize $Φ = φ / (\int φ)$ . By Schwartz approximate identities converge in the sense of tempered distributions, $Φ_{t} * a \to a$ in $S^{'}$ , so some $t_{0} > 0$ has $a * φ_{t_{0}} \equiv̸ 0$ ; otherwise the distributional limit would be zero. This convolution is continuous, so its absolute value, and hence $m$ , is positive on a nonempty open set. Thus $A : = \int_{R^{n}} m (x)^{p} d x = ∥ a ∥_{H^{p}}^{p}$ is finite and positive.

For $y \in R^{n}$ put $a_{y} (x) = a (x - y)$ and $m_{y} (x) = m (x - y)$ . Translation invariance of the convolution and Lebesgue measure gives $M_{φ}^{0} a_{y} = m_{y}$ and $∥ a_{y} ∥_{H^{p}} = ∥ a ∥_{H^{p}}$ . Pointwise sublinearity gives $M_{φ}^{0} (a + a_{y}) \leq m + m_{y}$ , so $a + a_{y} \in H^{p}$ ; the reverse triangle inequality for this sublinear maximal operator gives $M_{φ}^{0} (a + a_{y}) (x) \geq ∣ m (x) - m_{y} (x) ∣ .$ For $Q_{R} = [- R, R]^{n}$ , choose $R$ with $\int_{Q_{R}} m^{p} > 2^{p - 1} A$ , possible since $2^{p - 1} < 1$ and $m^{p} \in L^{1}$ . Take $y = T e_{1}$ with $T > 2 R$ ; then $Q_{R}$ and $Q_{R} + y$ are disjoint. The scalar inequality $∣ u - v ∣^{p} \geq u^{p} - v^{p}$ for $u, v \geq 0$ , applied with the local copy as $u$ on each cube, gives $\int_{R^{n}} ∣ m - m_{y} ∣^{p} d x \geq 2 \int_{Q_{R}} m^{p} d x - \int_{Q_{R}} m_{y}^{p} d x - \int_{Q_{R} + y} m^{p} d x .$ As $y \to \infty$ along a coordinate ray, the last two integrals tend to zero because $m^{p} \in L^{1}$ . The first term is strictly larger than $2^{p} A$ . Therefore for all sufficiently large such $y$ , $∥ a + a_{y} ∥_{H^{p}}^{p} \geq \int ∣ m - m_{y} ∣^{p} > 2^{p} A = (∥ a ∥_{H^{p}} + ∥ a_{y} ∥_{H^{p}})^{p},$ which contradicts the ordinary triangle inequality. This proves the claimed failure within the radial-maximal definition of $H^{p}$ .

Completeness: a complete argument is sketched here for the record. Let $(f_{k})$ be Cauchy for $d$ . Passing to a subsequence, assume $∥ f_{k + 1} - f_{k} ∥_{H^{p}}^{p} \leq 2^{- k}$ , and set $g_{k} = f_{k + 1} - f_{k}$ . By Atomic characterisation of real $H^{p}$ for $0 < p \leq 1$ each $g_{k}$ has an atomic representation $g_{k} = \sum_{j} λ_{k, j} a_{k, j}$ with $\sum_{j} ∣ λ_{k, j} ∣ \leq C ∥ g_{k} ∥_{H^{p}} \leq C 2^{- k / p}$ ; the pairing bound for atoms Atoms have uniformly bounded $H^{p}$ quasi-norm and uniformly bounded test pairings then gives, for every $ψ \in S$ , $∣ ⟨ g_{k}, ψ ⟩ ∣ \leq C (ψ) \sum_{j} ∣ λ_{k, j} ∣ \leq C^{'} (ψ) 2^{- k / p}$ , so $\sum_{k} g_{k}$ defines a continuous linear functional $h$ bounded by a fixed Schwartz seminorm times $\sum_{k} 2^{- k / p}$ , and converges to $h$ in $S^{'}$ ; put $f = f_{0} + h$ . The same bound applied to the tails shows that $f_{k} \to f$ in $S^{'}$ . For fixed $φ$ and $(t, x)$ , the convolutions $(f_{k} * φ_{t}) (x)$ converge to $(f * φ_{t}) (x)$ ; taking the supremum after pointwise convergence of each convolution gives $M_{φ}^{0} (f - f_{l}) \leq {lim inf}_{k} M_{φ}^{0} (f_{k} - f_{l})$ . Hence Fatou's lemma applied to the measurable functions $∣ M_{φ}^{0} (f_{k} - f_{l}) ∣^{p}$ gives $∥ f - f_{l} ∥_{H^{p}}^{p} \leq {lim inf}_{k} ∥ f_{k} - f_{l} ∥_{H^{p}}^{p}$ , which tends to $0$ as $l \to \infty$ ; hence $f_{l} \to f$ in $H^{p}$ and $f \in H^{p}$ . The metric $d$ is translation invariant because $M_{φ}^{0} ((f + h) - (g + h)) = M_{φ}^{0} (f - g)$ . No Banach duality is used in this argument, and the completion obtained is the space $H^{p}$ itself.

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Calderon-Zygmund operators map $H^{1}$ boundedly into $L^{1}$

Statement

Assume Countable Choice. Let $n \geq 1$ , $0 < δ \leq 1$ , fix the kernel $φ$ defining $H^{1}$ and an admissible order $N$ for its atomic characterisation, and let $k : R^{n} ∖ {0} \to C$ satisfy the pointwise size bound $∣ k (x) ∣ \leq A_{1} ∣ x ∣^{- n}$ , the standard $δ$ -Holder bound $∣ k (x - y) - k (x) ∣ \leq A_{2}^{'} ∣ y ∣^{δ} ∣ x ∣^{- n - δ}$ for $∣ x ∣ \geq 2 ∣ y ∣ > 0$ , and the cancellation bound $\sup_{0 < r < R} ∣ \int_{r < ∣ x ∣ < R} k (x) d x ∣ \leq A_{3}$ . Let $W$ be a principal-value distribution for $k$ and let $T$ be the convolution operator with $W$ , assumed $L^{2}$ -bounded with norm $B$ and satisfying the off-support representation of Calderón–Zygmund kernels and their associated operators with kernel $k$ . Then $T$ has a unique extension to a bounded linear operator $H^{1} (R^{n}) \to L^{1} (R^{n})$ , and there is $C = C_{n, δ, N, φ}$ with $∥ T f ∥_{L^{1}} \leq C (A_{1} + A_{2}^{'} + A_{3} + B) ∥ f ∥_{H^{1}} (f \in H^{1}) .$ The extension agrees with the given $L^{2}$ operator on $L^{2} \cap H^{1}$ , and for any fixed sequence $δ_{j} ↓ 0$ realizing $W$ in Calderón–Zygmund kernels and their associated operators, its values are $\lim_{j} T_{δ_{j}} f$ almost everywhere. A full limit as $ε ↓ 0$ requires the additional hypothesis that the defining principal-value integrals converge along all radii; sequence-based principal-value existence alone does not imply this.

Facts & Assumptions

Given: Countable Choice, a fixed sequence $δ_{j}$ realizing $W$ , $n \geq 1$ , $0 < δ \leq 1$ , the fixed $H^{1}$ kernel $φ$ and atomic order $N$ , the kernel $k$ , the principal-value distribution $W$ , the operator $T$ and the constants as in the statement.

[F1]

The Holder bound makes $k$ continuous at each nonzero point: take $y \to 0$ with $2 ∣ y ∣ \leq ∣ x ∣$ in the stated difference bound. Thus $k$ is Borel, and its size bound gives integrability on compact sets away from zero. Truncations: for $f \in L^{p}$ , $1 \leq p < \infty$ , and $0 < ε < \infty$ , $T_{ε} f (x) = \int_{∣ y ∣ > ε} k (y) f (x - y) d y$ converges absolutely at every $x$ ; the maximal truncations obey the weak $(1, 1)$ bound $∣ {T^{*} f > λ} ∣ \leq C_{n, δ} (A_{1} + A_{2}^{'} + A_{3} + B) λ^{- 1} ∥ f ∥_{1}$ for $f \in L^{1}$ and the strong $L^{p}$ bounds for $1 < p < \infty$ (Maximal truncated singular integrals, Maximal truncations: weak (1,1) and strong Lp bounds, Standard Hölder kernels satisfy the Hörmander condition, Standard (Hölder) Calderón–Zygmund kernels).

[F2]

Fix a sequence $δ_{j} ↓ 0$ realizing $W$ . For $g \in C_{c}^{\infty}$ , the definition of $W$ applied to the Schwartz test $g (x - \cdot)$ gives $T_{δ_{j}} g (x) \to (W * g) (x)$ at every $x$ . This extends to a.e. sequential convergence for every $f \in L^{1}$ : for $g$ approximating $f$ in $L^{1}$ , the tail oscillation of $(T_{δ_{j}} f)$ is at most $2 T^{*} (f - g)$ . For every $η > 0$ , [F1] therefore bounds the measure of the set where that oscillation exceeds $η$ by $2 C η^{- 1} ∥ f - g ∥_{1}$ . Density (Complex finite-simple and smooth compact-support density for finite p) makes this zero. Taking a countable sequence of $η$ shows that the scalar sequence is Cauchy, hence convergent, a.e. The same argument uses the strong $L^{2}$ bound for $f \in L^{2}$ . On $C_{c}^{\infty}$ , dominated convergence with majorant $T^{*} g \in L^{2}$ gives $T_{δ_{j}} g \to T g$ in $L^{2}$ ; density and the uniform $L^{2}$ bound of $T^{*}$ extend this to all $L^{2}$ . If the principal-value integrals converge along all radii on Schwartz tests, the identical oscillation argument over $0 < ε < η$ gives the full a.e. limit. The measurable suprema can be reduced to rational radii by absolute convergence away from zero.

[F3]

Atomic characterisation: every $f \in H^{1}$ has a representation $f = \sum_{j} λ_{j} a_{j}$ in $S^{'}$ with $(1, \infty, 0)$ -atoms $a_{j}$ and $(λ_{j}) \in ℓ^{1}$ ; the series also converges in the $H^{1}$ norm and one may choose $\sum_{j} ∣ λ_{j} ∣ \leq C_{n, N, φ} ∥ f ∥_{H^{1}}$ (Atomic characterisation of real $H^{p}$ for $0 < p \leq 1$ ). A $(1, \infty, 0)$ -atom is supported in a cube $Q$ , satisfies $∣ a ∣ \leq ∣ Q ∣^{- 1}$ and $\int a = 0$ ( $H^{p}$ atoms with a prescribed moment order).

[F4]

Complex $L^{1}$ is complete under Countable Choice (Complex Lp completeness and almost-everywhere subsequences). Norm convergence implies convergence in measure (Convergence in $L^{p}$ implies convergence in measure). A weak $(1, 1)$ difference estimate also gives convergence in measure directly, since $∣ {∣ u_{j} - u ∣ > η} ∣ \leq C η^{- 1} ∥ f_{j} - f ∥_{1} \to 0$ . Limits in measure are unique: ${∣ u - v ∣ > η}$ lies in the union of the two error sets at threshold $η / 2$ , whose measures tend to zero.

[F5]

Under Countable Choice, an $L^{2}$ -norm convergent sequence has a subsequence of representatives converging almost everywhere to a representative of its limit (Assuming Countable Choice, $L^{p}$ -convergent sequences have almost-everywhere convergent subsequences).

[F6]

Tonelli's theorem permits interchanging the integrals of nonnegative measurable functions on sigma-finite product measure spaces (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[F7]

The Calderon-Zygmund kernel and operator conventions are those of Calderón–Zygmund kernels and their associated operators: conditions (1) and (2) are the annular size and Hormander conditions, and condition (3) is the off-support representation by the kernel.

[F8]

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

[F9]

Under Countable Choice, a closed cube of side length $L$ in $R^{n}$ is Lebesgue measurable and has measure $L^{n}$ : its volume is the product of its side lengths (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, Axis-parallel rectangles in $R^{m}$ and their volume).

Proof technique: the near/far atom estimate, the a.e. sequential limit of the truncations, and summation over an atomic representation.

Proof

technique · direct

1.1F1F3F6F7F9givenalgebra

Atom estimate. Let $a$ be a $(1, \infty, 0)$ -atom supported in a cube $Q$ of side length $ℓ (Q)$ and centre $c_{Q}$ , and let $Q^{†}$ be the concentric cube of side length $2 \sqrt{n} ℓ (Q)$ . Since $a \in L^{2}$ and $T$ is $L^{2}$ -bounded, Cauchy-Schwarz and [F9] give $\int_{Q^{†}} ∣ T a ∣ \leq ∣ Q^{†} ∣^{1 / 2} ∥ T a ∥_{L^{2}} \leq ∣ Q^{†} ∣^{1 / 2} B ∥ a ∥_{L^{2}} \leq (2 \sqrt{n})^{n / 2} B,$ using $∥ a ∥_{2} \leq ∣ Q ∣^{- 1 / 2}$ . Outside $Q^{†}$ , for almost every $x$ the off-support representation gives $T a (x) = \int_{Q} k (x - y) a (y) d y$ , and the mean-zero property rewrites this as $\int_{Q} [k (x - y) - k (x - c_{Q})] a (y) d y$ . For every $y \in Q$ one has $∣ y - c_{Q} ∣ \leq \sqrt{n} ℓ (Q) / 2$ , while $x \notin Q^{†}$ implies $∣ x - c_{Q} ∣ \geq \sqrt{n} ℓ (Q)$ ; hence $∣ x - c_{Q} ∣ \geq 2 ∣ y - c_{Q} ∣$ . The standard H"older bound and the H"ormander condition therefore give $\int_{(Q^{†})^{c}} ∣ T a ∣ \leq \int_{Q} ∣ a (y) ∣ \int_{∣ x - c_{Q} ∣ \geq 2 ∣ y - c_{Q} ∣} ∣ k (x - y) - k (x - c_{Q}) ∣ d x d y \leq A_{2} ∥ a ∥_{1} \leq C_{n, δ} A_{2}^{'},$ where $A_{2} \leq C_{n, δ} A_{2}^{'}$ by [F1] and $∥ a ∥_{1} \leq 1$ . Thus $∥ T a ∥_{L^{1}} \leq (2 \sqrt{n})^{n / 2} B + C_{n, δ} A_{2}^{'}$ .

1.2F1F2algebra

The almost-everywhere limit extension. Define $\tilde{T} f (x) = \lim_{j} T_{δ_{j}} f (x)$ for $f \in L^{1}$ , which exists almost everywhere by [F2]. Then $∣ \tilde{T} f ∣ \leq T^{*} f$ pointwise, so $\tilde{T}$ is linear on $L^{1}$ (limits of linear expressions) and $∥ \tilde{T} f ∥_{L^{1, \infty}} \leq C_{n, δ} (A_{1} + A_{2}^{'} + A_{3} + B) ∥ f ∥_{1}$ by the weak $(1, 1)$ bound of [F1].

2.1step 1.1F2F3F5

Identification on atoms. Let $a$ be a $(1, \infty, 0)$ -atom, so $a \in L^{1} \cap L^{2}$ . By [F2], the truncations $T_{δ_{j}} a$ converge almost everywhere to $\tilde{T} a$ as $j \to \infty$ . The $L^{2}$ convergence in [F2] and the subsequence principle [F5] give a subsequence $T_{δ_{j_{ℓ}}} a \to T a$ almost everywhere. On the intersection of these two full-measure sets, this subsequence converges to both limits, so $\tilde{T} a = T a$ almost everywhere. Step 1.1 therefore gives $∥ \tilde{T} a ∥_{L^{1}} \leq C_{n, δ} (A_{2}^{'} + B)$ .

3.1step 1.2step 2.1F3F4algebra

Summation. Let $f \in H^{1}$ and let $f = \sum_{j} λ_{j} a_{j}$ be the representation of [F3] with $\sum_{j} ∣ λ_{j} ∣ \leq C_{n, N, φ} ∥ f ∥_{H^{1}}$ . Since $∥ a_{j} ∥_{L^{1}} \leq 1$ , the series converges absolutely in $L^{1}$ to $f$ , so $f \in L^{1}$ and the partial sums $g_{J} = \sum_{j \leq J} λ_{j} a_{j}$ satisfy $∥ f - g_{J} ∥_{L^{1}} \leq \sum_{j > J} ∣ λ_{j} ∣ \to 0$ . By step 1.2 and linearity, $\tilde{T} g_{J} = \sum_{j \leq J} λ_{j} \tilde{T} a_{j}$ , and by the $L^{1, \infty}$ bound $\tilde{T} g_{J} \to \tilde{T} f$ in measure; on the other hand step 2.1 gives $\sum_{j} ∣ λ_{j} ∣ ∥ \tilde{T} a_{j} ∥_{L^{1}} \leq C_{n, δ} (A_{2}^{'} + B) \sum_{j} ∣ λ_{j} ∣ < \infty$ , so $\tilde{T} g_{J}$ converges absolutely in $L^{1}$ . The $L^{1}$ limit is also a limit in measure, so it equals $\tilde{T} f$ a.e., and $∥ \tilde{T} f ∥_{L^{1}} \leq \sum_{j} ∣ λ_{j} ∣ ∥ \tilde{T} a_{j} ∥_{L^{1}} \leq C_{n, δ, N, φ} (A_{2}^{'} + B) ∥ f ∥_{H^{1}} \leq C (A_{1} + A_{2}^{'} + A_{3} + B) ∥ f ∥_{H^{1}}$ after enlarging the constant.

4.1step 1.2step 3.1F2F3F4F5algebra

Agreement and uniqueness. Let $f \in L^{2} \cap H^{1}$ . By [F2], $T_{δ_{j}} f \to T f$ in $L^{2}$ , so a subsequence converges almost everywhere to $T f$ ; by [F2] the sequential limit $\lim_{j} T_{δ_{j}} f = \tilde{T} f$ exists almost everywhere, hence $\tilde{T} f = T f$ a.e. Thus the bounded operator $\tilde{T} : H^{1} \to L^{1}$ extends the given $L^{2}$ operator on the dense subspace $L^{2} \cap H^{1}$ of $H^{1}$ (dense because finite atomic sums lie there and approximate every $H^{1}$ element in the $H^{1}$ quasi-norm by [F3]). Any two bounded extensions with the same bound agree on the dense subspace and hence everywhere, so the extension is unique.

5.1step 1.1step 1.2step 2.1step 3.1step 4.1F8∎

Conclusion. Steps 1.1 and 1.2 give the atom estimate and construct the extension as the almost-everywhere sequential limit of the truncations, step 2.1 identifies it with $T$ on atoms, step 3.1 bounds it on $H^{1}$ by summation over the atomic representation, and step 4.1 proves agreement with the $L^{2}$ operator and uniqueness. Countable Choice is used through the cited subsequence and atomic-representation results. This proves the theorem.

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

Fourier transform decay of real $H^{p}$ elements

Statement

Assume Countable Choice. Let $n \geq 1$ , $0 < p \leq 1$ , fix the admissible kernel $φ$ defining $H^{p}$ , and set $s = ⌊ n (1 / p - 1) ⌋$ . Fix an integer $K \geq n / p$ and the associated reproducing pair of the atomic decomposition, and an admissible grand-maximal order $N \geq \max {N_{0} (n, p, φ), n + s + 1}$ . There is $C = C (n, p, N, K, φ) < \infty$ such that every $f \in H^{p} (R^{n})$ has a Fourier transform that is a continuous function on $R^{n} ∖ {0}$ and satisfies $∣ \hat{f} (ξ) ∣ \leq C ∥ f ∥_{H^{p}} ∣ ξ ∣^{n (1 / p - 1)}, ξ \neq 0,$ and moreover $\lim_{ξ \to 0} \frac{∣ \hat{f} (ξ) ∣}{∣ ξ ∣^{n (1 / p - 1)}} = 0.$ Here $\hat{f}$ is the tempered-distribution Fourier transform (Fourier transform of a tempered distribution), identified with a continuous function off the origin by the estimate.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $0 < p \leq 1$ , the fixed kernel $φ$ , reproducing order $K$ and order $N$ , $s = ⌊ n (1 / p - 1) ⌋$ , $f \in H^{p}$ , and multi-indices as in $C^{k}$ maps and multi-index derivative notation in Euclidean space.

[F1]

Atomic characterisation: $f = \sum_{j} λ_{j} a_{j}$ in $S^{'}$ with $(p, \infty, s)$ -atoms $a_{j}$ and $(λ_{j}) \in ℓ^{p}$ ; the representation may be chosen with $\sum_{j} ∣ λ_{j} ∣ \leq C_{n, p, N, K, φ} ∥ f ∥_{H^{p}}$ (Atomic characterisation of real $H^{p}$ for $0 < p \leq 1$ ).

[F2]

For an $L^{1}$ atom the distributional transform agrees with the integral transform: the absolute double integral against a Schwartz test $χ$ is bounded by $∥ a ∥_{1} ∥ χ ∥_{1}$ , so Fubini identifies $⟨ a, \hat{χ} ⟩$ with $\int \hat{a} χ$ (Fubini's theorem for L^1 functions on a sigma-finite product). Fourier transform is continuous on $S^{'}$ : if $g_{J} \to g$ in $S^{'}$ then $\hat{g_{J}} \to \hat{g}$ in $S^{'}$ (Fourier transform of a tempered distribution).

[F3]

For an atom $a$ supported in a cube $Q$ with centre $c_{Q}$ , $∥ a ∥_{L^{\infty}} \leq ∣ Q ∣^{- 1 / p}$ and moments vanishing through order $s$ ( $H^{p}$ atoms with a prescribed moment order), the Taylor expansion of $x \mapsto e^{- 2 π i x \cdot ξ}$ about $c_{Q}$ through order $s$ gives $∣ \hat{a} (ξ) ∣ \leq C \min (∣ Q ∣^{1 - 1 / p}, ∣ ξ ∣^{s + 1} ∣ Q ∣^{1 - 1 / p + (s + 1) / n}) (ξ \neq 0),$ with $C = C (n, s)$ : the first bound is $∥ a ∥_{1} \leq ∥ a ∥_{\infty} ∣ Q ∣ \leq ∣ Q ∣^{1 - 1 / p}$ , and the second uses the vanishing moments, the Taylor remainder bound $∣ \partial^{β} e^{- 2 π i x \cdot ξ} ∣ \leq (2 π ∣ ξ ∣)^{∣ β ∣}$ and $\int_{Q} ∣ x - c_{Q} ∣^{s + 1} d x \leq C ℓ (Q)^{s + 1 + n}$ . Consequently $∣ \hat{a} (ξ) ∣ \leq C^{'} ∣ ξ ∣^{n (1 / p - 1)}$ for $ξ \neq 0$ (split at $∣ ξ ∣ ℓ (Q) ≍ 1$ and use $s + 1 > n (1 / p - 1)$ ) and $∣ \hat{a} (ξ) ∣ / ∣ ξ ∣^{n (1 / p - 1)} \to 0$ as $ξ \to 0$ for each fixed atom. [def-multidimensional-rectangle-and-volume, def-ck-and-multi-index-notation-in-several-variables, algebra]

Proof technique: the atomic representation, termwise Fourier transformation and dominated summation.

Proof

technique · direct

1.1F1F2F3algebra

Continuity and decay off the origin. Let $f = \sum_{j} λ_{j} a_{j}$ be the representation of [F1]. By [F2], $\hat{f} = \sum_{j} λ_{j} \hat{a_{j}}$ in $S^{'}$ ; since each $\hat{a_{j}}$ is a continuous function (the atoms are integrable) and, by [F3], $∣ λ_{j} \hat{a_{j}} (ξ) ∣ \leq C ∣ λ_{j} ∣ ∣ ξ ∣^{n (1 / p - 1)}$ for $ξ \neq 0$ , the numerical series $\sum_{j} λ_{j} \hat{a_{j}}$ converges absolutely and locally uniformly on $R^{n} ∖ {0}$ . Its sum is therefore a continuous function off the origin and agrees with $\hat{f}$ there as a distribution. There is no additional distribution supported at the origin: define the sum to be zero there. The uniform atom bound holds globally after this assignment, and $∣ ξ ∣^{n (1 / p - 1)} ∣ χ (ξ) ∣$ is integrable for every Schwartz test $χ$ . Dominated convergence therefore identifies the regular distribution of this sum with the distributional limit of the transformed partial sums on all of $R^{n}$ ; this identifies $\hat{f}$ with that continuous function on $R^{n} ∖ {0}$ and gives $∣ \hat{f} (ξ) ∣ \leq C (\sum_{j} ∣ λ_{j} ∣) ∣ ξ ∣^{n (1 / p - 1)} \leq C^{'} ∥ f ∥_{H^{p}} ∣ ξ ∣^{n (1 / p - 1)}$ .

2.1step 1.1F1F3algebra

The little- $o$ statement. Fix $η > 0$ . Choose $J$ so large that $C \sum_{j > J} ∣ λ_{j} ∣ < η / 2$ , possible because $(λ_{j}) \in ℓ^{p} \subseteq ℓ^{1}$ ; then by [F3] the tail satisfies $\sum_{j > J} ∣ λ_{j} ∣ ∣ \hat{a_{j}} (ξ) ∣ \leq (η / 2) ∣ ξ ∣^{n (1 / p - 1)}$ for every $ξ \neq 0$ . The finite sum $\sum_{j \leq J} λ_{j} \hat{a_{j}}$ is a finite combination of continuous functions each vanishing faster than $∣ ξ ∣^{n (1 / p - 1)}$ at the origin, so there is $δ > 0$ with $∣ \sum_{j \leq J} λ_{j} \hat{a_{j}} (ξ) ∣ < (η / 2) ∣ ξ ∣^{n (1 / p - 1)}$ for $0 < ∣ ξ ∣ < δ$ . Hence $∣ \hat{f} (ξ) ∣ \leq η ∣ ξ ∣^{n (1 / p - 1)}$ for $0 < ∣ ξ ∣ < δ$ , which is the stated little- $o$ relation since $η > 0$ was arbitrary.

3.1step 1.1step 2.1∎

Conclusion. Steps 1.1 and 2.1 give the identification of $\hat{f}$ with a continuous function off the origin, the decay estimate and the little- $o$ refinement. This proves the theorem.

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

Weighted-integrable $H^{p}$ functions have vanishing moments in the atomic range

Statement

Assume Countable Choice. Let $n \geq 1$ , $0 < p \leq 1$ and $s = ⌊ n (1 / p - 1) ⌋$ . Suppose $f \in H^{p} (R^{n})$ is represented by a locally integrable function and assume additionally that $x^{α} f \in L^{1} (R^{n})$ for every multi-index $∣ α ∣ \leq s$ . Then $\int_{R^{n}} f (x) x^{α} d x = 0 (∣ α ∣ \leq s) .$ In particular every compactly supported $L^{1}$ function $f \in H^{1}$ satisfies $\int_{R^{n}} f = 0$ , and no compactly supported integrable function of nonzero integral lies in $H^{1}$ .

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $0 < p \leq 1$ , $s = ⌊ n (1 / p - 1) ⌋$ , $f \in H^{p} \cap L_{l o c}^{1}$ with $x^{α} f \in L^{1}$ for $∣ α ∣ \leq s$ .

[F1]

Fourier decay: for the fixed kernel, reproducing order and grand-maximal order of the Fourier-decay theorem, every $f \in H^{p}$ has $\hat{f}$ continuous on $R^{n} ∖ {0}$ with $∣ \hat{f} (ξ) ∣ \leq C_{n, p, N, K, φ} ∥ f ∥_{H^{p}} ∣ ξ ∣^{n (1 / p - 1)}$ and $\hat{f} (ξ) = o (∣ ξ ∣^{n (1 / p - 1)})$ as $ξ \to 0$ (Fourier transform decay of real $H^{p}$ elements).

[F2]

If $f \in L^{1}$ then $\hat{f}$ is bounded and uniformly continuous on $R^{n}$ , and $\hat{f} (ξ) = \int f (x) e^{- 2 π i x \cdot ξ} d x$ ; if moreover $x^{α} f \in L^{1}$ , then $\partial^{α} \hat{f} (ξ) = \int (- 2 π i x)^{α} f (x) e^{- 2 π i x \cdot ξ} d x$ , so $\partial^{α} \hat{f} (0) = (- 2 π i)^{∣ α ∣} \int x^{α} f$ (The L1 transform is bounded and uniformly continuous, Fourier differentiation and multiplication identities on tempered distributions).

[F3]

If $m \geq 1$ , the Peano Taylor formula applies to every real $C^{m}$ function near $0$ : $g (h) = T_{m} g (0; h) + o (∣ h ∣^{m})$ (Multivariable Taylor formula with $o (∥ h ∥^{k})$ remainder). For a complex-valued function, apply this to its real and imaginary parts and combine the two expansions.

[F4]

At $p = 1$ , Atomic characterisation of real $H^{p}$ for $0 < p \leq 1$ gives $f = \sum_{j} λ_{j} a_{j}$ in $S^{'}$ with $\sum_{j} ∣ λ_{j} ∣ < \infty$ . The size/support conditions of $H^{p}$ atoms with a prescribed moment order give $∥ a_{j} ∥_{1} \leq 1$ . Thus the partial sums converge in complex $L^{1}$ by Complex Lp completeness and almost-everywhere subsequences, and their $L^{1}$ limit has the same distributional limit since $∣ ⟨ h, χ ⟩ ∣ \leq ∥ h ∥_{1} ∥ χ ∥_{\infty}$ . Injectivity of Locally integrable functions embed in distributions identifies it a.e. with the given locally integrable representative of $f$ . Hence that representative belongs to $L^{1}$ .

Proof technique: the little- $o$ Fourier decay against the Taylor expansion of $\hat{f}$ at the origin.

Proof

technique · direct

1.1F2given

Smoothness of $\hat{f}$ at the origin. For each coordinate, the exponential difference quotient is bounded by $2 π ∣ x_{j} ∣$ , since $∣ e^{i u} - 1 ∣ \leq ∣ u ∣$ . Iterating dominated convergence with the assumed integrable functions $∣ x^{α} f ∣$ proves the derivative formula in [F2]; dominated convergence applied to each derivative integrand proves its continuity. Since $x^{α} f \in L^{1}$ for $∣ α ∣ \leq s$ , [F2] gives that $\hat{f}$ is $s$ times continuously differentiable near the origin and that $\partial^{α} \hat{f} (0)$ is the Fourier transform of $(- 2 π i x)^{α} f$ at the origin.

2.1F1F2F3step 1.1algebra

A nonvanishing lowest derivative contradicts the little- $o$ decay. Suppose some $\partial^{α} \hat{f} (0) \neq 0$ with $∣ α ∣ \leq s$ , and choose such an $α$ of minimal total degree $m$ . Put $γ = n (1 / p - 1) \geq 0$ . If $m = 0$ , then $\hat{f} (0) \neq 0$ ; choose any unit vector $η$ . Continuity from [F2] gives $∣ \hat{f} (t η) ∣ \geq ∣ \hat{f} (0) ∣ / 2$ for all sufficiently small $t > 0$ , contradicting [F1], which says $\hat{f} (t η) = o (t^{γ})$ and hence tends to zero. If $m \geq 1$ , every derivative of order below $m$ vanishes. By [F2], $\hat{f}$ is $C^{m}$ near $0$ , so [F3] applied to its real and imaginary parts gives, for fixed $η \in R^{n}$ , $\hat{f} (t η) = t^{m} Q (η) + o (t^{m}) (t ↓ 0), Q (η) = \sum_{∣ β ∣ = m} \frac{\partial^{β} \hat{f} (0)}{β!} η^{β} .$ This complex homogeneous polynomial is not identically zero, so choose a unit vector $η$ with $Q (η) \neq 0$ . Then $∣ \hat{f} (t η) ∣ \geq c t^{m}$ for all sufficiently small $t > 0$ . Since $m \leq s = ⌊ γ ⌋ \leq γ$ , one has $t^{m} \geq t^{γ}$ for $0 < t \leq 1$ , contradicting [F1]. Thus every $\partial^{α} \hat{f} (0)$ with $∣ α ∣ \leq s$ vanishes.

3.1step 2.1F2algebra

Conclusion. By [F2], $\partial^{α} \hat{f} (0) = (- 2 π i)^{∣ α ∣} \int x^{α} f$ for every $∣ α ∣ \leq s$ ; step 2.1 shows these derivatives all vanish, so $\int x^{α} f = 0$ for $∣ α ∣ \leq s$ . For $p = 1$ one has $s = 0$ , so the integral of $f$ vanishes; applying this to a compactly supported $L^{1}$ function $f \in H^{1}$ gives $\int f = 0$ , and a compactly supported $L^{1}$ function with $\int f \neq 0$ cannot be in $H^{1}$ .

4.1step 3.1F4given∎

Remark on the hypothesis. For $p = 1$ every $H^{1}$ function that is a locally integrable function automatically has $f \in L^{1}$ by [F4], so the "compactly supported $L^{1}$ " formulation is a special case; for $p < 1$ the hypothesis $x^{α} f \in L^{1}$ is a genuine additional assumption. This corollary proves the stated vanishing moments and no more.

5 · Examples, counterexamples and false statements

None yet.

Real Hardy Spaces Maximal Functions and Atoms

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

HpHp atoms with a prescribed moment order

Definition

Schwartz functions with prescribed flatness of the Fourier transform at the origin

Statement

Facts & Assumptions

Proof

Local polynomial projections matching moments through order ss

Statement

Facts & Assumptions

Proof

Dilations and their normalisations preserve Schwartz space, with scaling identities

Statement

Facts & Assumptions

Proof

Whitney decomposition of a proper open subset of Euclidean space

Statement

Facts & Assumptions

Proof

Whitney-type ball cover with disjoint small balls and bounded overlap

Statement

Facts & Assumptions

Proof

Radial and nontangential maximal functions of a tempered distribution

Definition

Schwartz approximate identities converge in the sense of tempered distributions

Statement

Facts & Assumptions

Proof

Deconvolution of a Schwartz function along the dyadic dilates of a fixed kernel

Statement

Facts & Assumptions

Proof

Grand maximal test class of order N and the grand maximal function

Definition

The tangential maximal function is controlled by the aperture-one nontangential maximal function in LpLp

Statement

Facts & Assumptions

Proof

The grand maximal function is pointwise dominated by a tangential maximal function

Statement

Facts & Assumptions

Proof

Measurability and lower semicontinuity of the smooth maximal functions

Statement

Facts & Assumptions

Proof

Truncated maximal functions: finiteness, comparison estimates and the good-set bound

Statement

Facts & Assumptions

Proof

The real Hardy space HpHp defined by a radial maximal function

Definition

Norm properties

The grand maximal function dominates every admissible radial and nontangential maximal function

Statement

Facts & Assumptions

Proof

Atoms have uniformly bounded HpHp quasi-norm and uniformly bounded test pairings

Statement

Facts & Assumptions

Proof

Calderon reproducing pair and the telescoping identity in S′S′

Statement

Facts & Assumptions

Proof

Maximal-function characterisations of real Hardy spaces

Statement

Facts & Assumptions

Proof

HpHp equals LpLp with equivalent norms for 1<p<∞1<p<∞

Statement

Facts & Assumptions

Proof

ℓpℓp sums of atoms converge in S′S′ and in HpHp

Statement

$H^{p}$ atoms with a prescribed moment order

Local polynomial projections matching moments through order $s$

The tangential maximal function is controlled by the aperture-one nontangential maximal function in $L^{p}$

The real Hardy space $H^{p}$ defined by a radial maximal function

Atoms have uniformly bounded $H^{p}$ quasi-norm and uniformly bounded test pairings

Calderon reproducing pair and the telescoping identity in $S^{'}$

$H^{p}$ equals $L^{p}$ with equivalent norms for $1 < p < \infty$

$ℓ^{p}$ sums of atoms converge in $S^{'}$ and in $H^{p}$

Level decomposition of an $H^{p}$ distribution produces atoms

Atomic characterisation of real $H^{p}$ for $0 < p \leq 1$

For $0 < p < 1$ the $H^{p}$ functional is a quasi-norm, and $H^{p}$ is a quasi-Banach space

Calderon-Zygmund operators map $H^{1}$ boundedly into $L^{1}$

Fourier transform decay of real $H^{p}$ elements

Weighted-integrable $H^{p}$ functions have vanishing moments in the atomic range