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Calderón–Zygmund Decomposition and Singular Integrals

1 · Prerequisites

2 · Summary

This page develops the Calderón–Zygmund decomposition and the mapping theory of singular integrals, and then applies it to the Hilbert and Riesz transforms and to Mihlin multipliers. Every argument that needs a choice principle assumes Countable Choice and the spending step is named.

A Calderón–Zygmund kernel is a locally integrable function off the origin with a finite annular $L^{1}$ mass and finite Hörmander integral; a standard $δ$ -Hölder kernel is the pointwise-smoothness special case, and the Hölder-to-Hörmander lemma converts the pointwise difference bound into the integral condition with the explicit constant $∣ S^{n - 1} ∣ 2^{- δ} δ^{- 1} A_{2}^{'}$ . The decomposition itself is driven by the half-open dyadic cubes of all integer generations: they partition $R^{n}$ at each generation, are nested or disjoint, and admit ancestors at every coarser generation. The maximal bad cubes at height $λ$ are pairwise disjoint, and their good and bad parts satisfy the $L^{1}$ , $L^{2}$ , $L^{\infty}$ and mean-zero bounds recorded in the decomposition lemma.

Combining the decomposition with Chebyshev's inequality and the off-support representation gives the weak $(1, 1)$ endpoint for a Calderón–Zygmund operator, and a two-level interpolation together with duality upgrades it to strong $L^{p}$ bounds for $1 < p < \infty$ with the constant $C_{n, p} (A_{2} + B) \max (p, (p - 1)^{- 1})$ . The same machinery applied to the maximal truncations, with Cotlar's inequality controlling the good part and a careful annulus analysis of the bad part, yields weak $(1, 1)$ and strong $L^{p}$ bounds for $T^{*}$ and $T^{* *}$ , and a density argument then gives almost-everywhere convergence of the principal-value truncations for the Hilbert and Riesz kernels. The kernels $1 / (π x)$ and $c_{n} x_{j} / ∣ x ∣^{n + 1}$ are verified to be standard $1$ -Hölder Calderón–Zygmund kernels, so the strict-range $L^{p}$ theory applies to them directly.

Finally, the dyadic pieces $m ζ (2^{- j} \cdot)$ of a Mihlin symbol are summed to produce an off-support kernel with annular and Hörmander bounds of size $C_{n} A$ , and the strict-range theorem then proves the Mihlin–Hörmander multiplier theorem on $L^{p}$ for $1 < p < \infty$ . The closing remark records the weak $(1, 1)$ endpoint for Mihlin multipliers themselves, using their Calderón–Zygmund operator representation. Weak bounds for maximal truncations require the stronger kernel hypotheses of the maximal-truncation theorem. The $L^{\infty} \to B M O$ endpoint belongs to the later BMO page, and no general strong $L^{1}$ or $L^{\infty}$ bound follows.

3 · Logical flowchart

4 · Definitions, theorems and proofs

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

Marcinkiewicz interpolation from weak (1,1) and strong (2,2)

Statement

Let $(X, A, μ)$ be a $σ$ -finite measure space, let $1 < p < 2$ , and let $T$ be a sublinear operator defined on $L^{1} (X) + L^{2} (X)$ and taking values in the measurable functions on $X$ , which is of weak type $(1, 1)$ with constant $A$ and of strong type $(2, 2)$ with constant $B$ . Then every $f \in L^{p} (X)$ satisfies $∥ T f ∥_{p} \leq [p (\frac{2 A}{p - 1} + \frac{4 B^{2}}{2 - p})]^{1 / p} ∥ f ∥_{p},$ so that $T$ is of strong type $(p, p)$ , with the stated constant.

Facts & Assumptions

Given: A $σ$ -finite measure space $(X, A, μ)$ ; an exponent $1 0$ .

[F1]

Sublinearity means $∣ T (a f + b g) ∣ \leq ∣ a ∣ ∣ T f ∣ + ∣ b ∣ ∣ T g ∣$ ; weak type $(1, 1)$ with constant $A$ means $μ ({∣ T g ∣ > s}) \leq A ∥ g ∥_{1} / s$ for all $g \in L^{1}$ and $s > 0$ ; strong type $(2, 2)$ with constant $B$ means $∥ T g ∥_{2} \leq B ∥ g ∥_{2}$ for all $g \in L^{2}$ (Sublinear operators and weak or strong type $(p, q)$ bounds).

[F2]

The distribution function of $∣ g ∣$ is $A_{g} (t) = μ ({∣ g ∣ > t})$ (The distribution function of absolute value), and for measurable $h \geq 0$ and $λ > 0$ one has $μ ({h \geq λ}) \leq λ^{- 1} \int h d μ$ (Chebyshev-Markov inequality for the integral).

[F3]

For measurable $g$ and $0 < q < \infty$ , $\int_{X} ∣ g ∣^{q} d μ = q \int_{0}^{\infty} t^{q - 1} A_{g} (t) d t$ , both sides possibly $+ \infty$ (For 0 < p < infinity, the layer-cake formula computes the integral of |f|^p from the distribution function).

[F4]

On a product of $σ$ -finite measure spaces, a nonnegative product-measurable function may be integrated in either order (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[F5]

Elements of $L^{p} (X)$ are a.e. equivalence classes, and $∥ \cdot ∥_{p}$ is the quotient norm (The space $L^{p} (μ)$ as the quotient by null functions).

Proof

technique · direct

1.1F5givenconstruct

Fix a representative of $f$ and, for each height $t > 0$ , split $f = f_{1}^{(t)} + f_{2}^{(t)}$ with $f_{1}^{(t)} : = f 1_{{∣ f ∣ > t}}$ and $f_{2}^{(t)} : = f 1_{{∣ f ∣ \leq t}}$ ; both are measurable, $∥ f_{1}^{(t)} ∥_{1} = \int_{{∣ f ∣ > t}} ∣ f ∣ d μ \leq t^{1 - p} ∥ f ∥_{p}^{p} < \infty$ and $∥ f_{2}^{(t)} ∥_{2}^{2} = \int_{{∣ f ∣ \leq t}} ∣ f ∣^{2} d μ \leq t^{2 - p} ∥ f ∥_{p}^{p} < \infty$ , so $f_{1}^{(t)} \in L^{1}$ , $f_{2}^{(t)} \in L^{2}$ , and both lie in $L^{1} + L^{2}$ ; moreover $∣ f_{2}^{(t)} ∣ \leq t$ and $f = f_{1}^{(t)} + f_{2}^{(t)}$ pointwise.

2.1F1F2step 1.1algebra

Sublinearity gives $∣ T f ∣ \leq ∣ T f_{1}^{(t)} ∣ + ∣ T f_{2}^{(t)} ∣$ pointwise, hence ${∣ T f ∣ > t} \subseteq {∣ T f_{1}^{(t)} ∣ > t / 2} \cup {∣ T f_{2}^{(t)} ∣ > t / 2}$ , and subadditivity of $μ$ together with [F1] applied to $f_{1}^{(t)}$ (weak $(1, 1)$ at level $t / 2$ ) and to $f_{2}^{(t)}$ (strong $(2, 2)$ , then [F2] applied to $∣ T f_{2}^{(t)} ∣^{2}$ at level $(t / 2)^{2}$ ) yields $A_{T f} (t) \leq \frac{2 A}{t} ∥ f_{1}^{(t)} ∥_{1} + \frac{4 B^{2}}{t^{2}} ∥ f_{2}^{(t)} ∥_{2}^{2} = \frac{2 A}{t} \int_{{∣ f ∣ > t}} ∣ f ∣ d μ + \frac{4 B^{2}}{t^{2}} \int_{{∣ f ∣ \leq t}} ∣ f ∣^{2} d μ .$

3.1F3F4step 2.1algebra

Tonelli's theorem applied to the nonnegative product-measurable function $(x, t) \mapsto t^{p - 2} ∣ f (x) ∣ 1_{{∣ f (x) ∣ > t}}$ on the $σ$ -finite product $X \times (0, \infty)$ converts the first term of the layer-cake integral into an $X$ -integral: $\int_{0}^{\infty} t^{p - 1} \frac{2 A}{t} (\int_{{∣ f ∣ > t}} ∣ f ∣ d μ) d t = 2 A \int_{X} ∣ f ∣ (\int_{0}^{∣ f ∣} t^{p - 2} d t) d μ = \frac{2 A}{p - 1} \int_{X} ∣ f ∣^{p} d μ,$ where the inner integral was evaluated as $∣ f ∣^{p - 1} / (p - 1)$ , legitimate because $p - 1 > 0$ , and the case $f (x) = 0$ contributes $0$ .

3.2F3F4step 2.1algebra

Likewise, Tonelli applied to $(x, t) \mapsto t^{p - 3} ∣ f (x) ∣^{2} 1_{{∣ f (x) ∣ \leq t}}$ gives, since $p - 2 < 0$ , $\int_{0}^{\infty} t^{p - 1} \frac{4 B^{2}}{t^{2}} (\int_{{∣ f ∣ \leq t}} ∣ f ∣^{2} d μ) d t = 4 B^{2} \int_{X} ∣ f ∣^{2} (\int_{∣ f ∣}^{\infty} t^{p - 3} d t) d μ = \frac{4 B^{2}}{2 - p} \int_{X} ∣ f ∣^{p} d μ,$ the inner integral being $∣ f ∣^{p - 2} / (2 - p)$ and the case $f (x) = 0$ contributing $0$ .

4.1F3step 2.1step 3.1step 3.2algebra∎

Combining the layer-cake identity [F3] for $g = T f$ with step 2.1 and steps 3.1 and 3.2 gives $∥ T f ∥_{p}^{p} = p \int_{0}^{\infty} t^{p - 1} A_{T f} (t) d t \leq p (\frac{2 A}{p - 1} + \frac{4 B^{2}}{2 - p}) ∥ f ∥_{p}^{p},$ because $t^{p - 1} A_{T f} (t) \leq t^{p - 1}$ times the two integrands integrated in steps 3.1 and 3.2; taking $p$ -th roots gives the asserted strong $(p, p)$ bound. Since $T f$ is determined a.e. by the class of $f$ , the bound descends to $L^{p} (X)$ by [F5].

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Calderón–Zygmund kernels and their associated operators

Definition

Fix an integer $n \geq 1$ ; Lebesgue measure, the Euclidean norm, and the complex test-function conventions are those of Complex Lp classes and Euclidean test-function conventions. A Calderón–Zygmund kernel with constants $A_{1}, A_{2}$ is a pair consisting of a measurable function $k : R^{n} ∖ {0} \to C$ that is integrable on compact subsets of $R^{n} ∖ {0}$ (A locally integrable function on $R^{n}$ ) and finite numbers $0 \leq A_{1}, A_{2} < \infty$ such that $\sup_{R > 0} \int_{R \leq ∣ x ∣ \leq 2 R} ∣ k (x) ∣ d x \leq A_{1} (1)$ and $\sup_{y \neq 0} \int_{∣ x ∣ \geq 2 ∣ y ∣} ∣ k (x - y) - k (x) ∣ d x \leq A_{2} . (2)$ Condition (1) is an annular size condition: it bounds the $L^{1}$ mass of every dyadic annulus $R \leq ∣ x ∣ \leq 2 R$ by $A_{1}$ , uniformly in the scale $R > 0$ . It is an integral, not a pointwise, bound: the pointwise estimate $∣ k (x) ∣ \leq A ∣ x ∣^{- n}$ implies (1) with $A_{1} = A ∣ S^{n - 1} ∣ \log 2$ , and not conversely. Since every compact subset of $R^{n} ∖ {0}$ lies in ${a \leq ∣ x ∣ \leq b}$ with $0 < a \leq b < \infty$ , and the latter is covered by the finitely many annuli $2^{j} a \leq ∣ x ∣ \leq 2^{j + 1} a$ for $0 \leq j \leq m$ with $2^{m} a \geq b$ , condition (1) also implies the local integrability listed above. Condition (2) is Hörmander's condition: an integral smoothness bound at scale $∣ y ∣$ . It is translation invariant, in that substituting $x - c$ for $x$ and leaving $y$ unchanged leaves the value of the integral unchanged, so it may be applied with the origin replaced by any centre $c$ .

A linear map $T : L^{2} (R^{n}) \to L^{2} (R^{n})$ with finite operator norm $B = ∥ T ∥_{L^{2} \to L^{2}}$ is a Calderón–Zygmund operator with kernel $k$ when for every compactly supported $f \in L^{2} (R^{n})$ the integral $\int_{R^{n}} k (x - y) f (y) d y$ converges absolutely for almost every $x \notin supp f$ and satisfies $T f (x) = \int_{R^{n}} k (x - y) f (y) d y for almost every x \notin supp f . (3)$ Here $supp f = ess supp f$ is the essential support defined in Complex Lp classes and Euclidean test-function conventions; compact support means that this closed set is compact. These conditions depend only on the almost-everywhere class of $f$ . Thus the only link between the operator and the kernel is the off-support representation (3): the action of $T$ on $L^{1}$ functions, on general bounded functions, or off the diagonal is not presupposed, and $T$ need not be convolution with any distribution. In the mean-zero applications below the absolute convergence in (3) is recovered from Hörmander's condition (2) by Tonelli's theorem; it is automatic whenever $k$ is locally square-integrable on $R^{n} ∖ {0}$ .

A principal-value distribution for $k$ is a tempered distribution $W$ on $R^{n}$ (Tempered distribution, Schwartz space and its seminorms) that agrees with $k$ on $R^{n} ∖ {0}$ , in the sense that $⟨ W, φ ⟩ = \int_{R^{n}} k (x) φ (x) d x$ for every $φ \in S (R^{n})$ supported in $R^{n} ∖ {0}$ , and for which some sequence $δ_{j} ↓ 0$ satisfies $⟨ W, φ ⟩ = \lim_{j \to \infty} \int_{∣ x ∣ \geq δ_{j}} k (x) φ (x) d x$ for every $φ \in S (R^{n})$ .

Neither the existence of a principal-value distribution nor the existence of any truncation limit is part of the definition of a Calderón–Zygmund kernel: a kernel may fail to have one, and the operator $T$ of (3) need not arise from one. Only the annular size condition (1), Hörmander's condition (2), the $L^{2}$ bound, and the off-support representation (3) are assumed. No choice principle is used in this definition.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Standard (Hölder) Calderón–Zygmund kernels

Definition

Let $k$ be a Calderón–Zygmund kernel with constants $A_{1}, A_{2}$ in the sense of Calderón–Zygmund kernels and their associated operators. Fix an exponent $0 < δ \leq 1$ . The kernel is standard $δ$ -Hölder with constant $A_{2}^{'} < \infty$ when $∣ k (x - y) - k (x) ∣ \leq A_{2}^{'} \frac{∣ y ∣^{δ}}{∣ x ∣^{n + δ}} whenever ∣ x ∣ \geq 2 ∣ y ∣ > 0. (1)$ Condition (1) is a pointwise estimate on the first difference of $k$ at the scale $∣ y ∣$ ; it is stated only on the regime $∣ x ∣ \geq 2 ∣ y ∣ > 0$ , where the two arguments $x$ and $x - y$ stay in the punctured space $R^{n} ∖ {0}$ and at comparable distance from the origin. The exponent is kept explicit and may be any number in $(0, 1]$ ; the constant $A_{2}^{'}$ may depend on $δ$ and on $k$ . A Calderón–Zygmund operator whose kernel is standard $δ$ -Hölder is called a standard-kernel Calderón–Zygmund operator.

The pointwise condition (1) is a sufficient hypothesis for Hörmander's integral condition (2) of the base definition: the next item proves that every standard $δ$ -Hölder kernel is a Calderón–Zygmund kernel in the sense of Calderón–Zygmund kernels and their associated operators, with the integral constant controlled by $A_{2}^{'}$ . No converse is claimed: a kernel satisfying Hörmander's integral condition need not satisfy the pointwise estimate (1), and the two hypotheses are recorded separately so that each theorem can invoke exactly the one it uses. Likewise no $L^{2}$ boundedness, no principal-value existence, and no cancellation of spherical means is asserted by this definition. No choice principle is used.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Standard Hölder kernels satisfy the Hörmander condition

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Every standard $δ$ -Hölder Calderón–Zygmund kernel with constant $A_{2}^{'}$ (Standard (Hölder) Calderón–Zygmund kernels) is a Calderón–Zygmund kernel in the sense of the base definition (Calderón–Zygmund kernels and their associated operators), and its Hörmander constant may be taken to be $A_{2} = ∣ S^{n - 1} ∣ 2^{- δ} δ^{- 1} A_{2}^{'} .$

Facts & Assumptions

Given: Countable Choice; a standard $δ$ -Hölder Calderón–Zygmund kernel $k$ with constant $A_{2}^{'}$ , $0 < δ \leq 1$ , and its a priori annular constant $A_{1}$ ; a vector $y \neq 0$ .

[F1]

$k$ is measurable on $R^{n} ∖ {0}$ , integrable on compact subsets of $R^{n} ∖ {0}$ , satisfies the annular bound with $A_{1}$ , and satisfies $∣ k (x - y) - k (x) ∣ \leq A_{2}^{'} ∣ y ∣^{δ} ∣ x ∣^{- n - δ}$ whenever $∣ x ∣ \geq 2 ∣ y ∣ > 0$ (Standard (Hölder) Calderón–Zygmund kernels, Calderón–Zygmund kernels and their associated operators).

[F2]

Polar coordinates: for a Borel function $g \geq 0$ on $R^{n}$ , $\int_{R^{n}} g d λ = \int_{0}^{\infty} \int_{S^{n - 1}} g (r ω) d σ (ω) r^{n - 1} d r$ , where $σ$ is the surface measure with $σ (S^{n - 1}) = ∣ S^{n - 1} ∣$ (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma); the integral over a smaller domain is at most the integral over a larger one (Measures are monotone).

Proof

technique · direct

1.1F1F2algebra

Fix $y \neq 0$ . By the pointwise Hölder bound of [F1] and monotonicity of the integral, $\int_{∣ x ∣ \geq 2 ∣ y ∣} ∣ k (x - y) - k (x) ∣ d x \leq A_{2}^{'} ∣ y ∣^{δ} \int_{∣ x ∣ \geq 2 ∣ y ∣} ∣ x ∣^{- n - δ} d x .$

1.2F2algebra

Polar coordinates evaluate the radial integral: substituting $x = r ω$ and using $∣ x ∣^{- n - δ} r^{n - 1} = r^{- 1 - δ}$ , $\int_{∣ x ∣ \geq 2 ∣ y ∣} ∣ x ∣^{- n - δ} d x = ∣ S^{n - 1} ∣ \int_{2 ∣ y ∣}^{\infty} r^{- 1 - δ} d r = ∣ S^{n - 1} ∣ \frac{(2 ∣ y ∣)^{- δ}}{δ},$ the last step being the elementary integral $\int_{a}^{\infty} r^{- 1 - δ} d r = a^{- δ} / δ$ for $a > 0$ and $δ > 0$ .

2.1F1step 1.1step 1.2algebra∎

Combining steps 1.1 and 1.2 gives $\int_{∣ x ∣ \geq 2 ∣ y ∣} ∣ k (x - y) - k (x) ∣ d x \leq A_{2}^{'} ∣ y ∣^{δ} ∣ S^{n - 1} ∣ (2 ∣ y ∣)^{- δ} δ^{- 1} = ∣ S^{n - 1} ∣ 2^{- δ} δ^{- 1} A_{2}^{'}$ ; taking the supremum over $y \neq 0$ shows that Hörmander's condition holds with $A_{2} = ∣ S^{n - 1} ∣ 2^{- δ} δ^{- 1} A_{2}^{'}$ , while the annular bound holds with $A_{1}$ by hypothesis. Hence $k$ is a Calderón–Zygmund kernel in the base sense with the stated Hörmander constant.

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

Dyadic cubes of all generations in R^n

Definition

Fix an integer $n \geq 1$ . Let $Z$ be the integers of The integers as equivalence classes of pairs of naturals with their order and ring operations, and read integer values inside $R$ along the canonical embedding. For $k \in Z$ and a function $m : n \to Z$ , the dyadic cube of generation $k$ and index $m$ is the half-open box (Half-open boxes in $R^{n}$ and their volume) $Q_{k, m} : = {x \in R^{n} : m_{i} 2^{- k} < x_{i} \leq (m_{i} + 1) 2^{- k} for every i < n},$ where $2^{- k}$ is the integer power of Integer powers $a^{m}$ and the products $m_{i} 2^{- k}$ are read in $R$ . The generation of the cube is $k$ and its side length is $2^{- k}$ . Since $2^{- k} > 0$ by Laws of integer exponents, the two endpoints of each coordinate interval satisfy $m_{i} 2^{- k} < (m_{i} + 1) 2^{- k}$ , so $Q_{k, m}$ is a nonempty half-open box with both parameters finite; assuming Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), its measure, computed from the half-open box measure theorem (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included), is the product of the $n$ equal side lengths $∣ Q_{k, m} ∣ = (2^{- k})^{n} = 2^{- k n} .$

Generations are indexed by all of $Z$ , not merely by the natural numbers: for $k > 0$ the side length $2^{- k}$ is smaller than $1$ , for $k = 0$ it is $1$ , and for $k < 0$ the side length is $2^{∣ k ∣} > 1$ , so cubes larger than the unit cube occur. The cubes with $k \geq 0$ are exactly the generation- $k$ dyadic cubes of the measure-theoretic convention Dyadic cubes of generation $k$ in $R^{n}$ , whose generation index is a natural number; that convention is bounded above in size by the unit cube $k = 0$ . Such a grid does not suffice for a stopping-time decomposition at a small height: a maximal bad cube can then be the unit cube while the height is far below its average. The all-generations family above is the grid used by the dyadic maximal function and by the Calderón–Zygmund decomposition on this page; the next item proves that it partitions $R^{n}$ at each generation, that each cube has exactly one ancestor of every coarser generation, and that two cubes are nested or disjoint. All parameters and the generation are determined by the cube as a set, by the parameter uniqueness recorded in Half-open boxes in $R^{n}$ and their volume. The set construction is choice-free; only the stated identification with Lebesgue measure uses Countable Choice.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

All-generation dyadic cubes: partition, volume and nesting

Statement

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

Let $n \geq 1$ and use the all-generations dyadic cubes of Dyadic cubes of all generations in R^n. Then:

For every $k \in Z$ the generation- $k$ dyadic cubes are pairwise disjoint and cover $R^{n}$ , and each has volume $∣ Q_{k, m} ∣ = 2^{- k n}$ .
Every dyadic cube $Q$ of generation $k$ has, for each $j < k$ , exactly one ancestor dyadic cube of generation $j$ containing $Q$ ; in particular the parent of $Q$ has generation $k - 1$ and volume $2^{n} ∣ Q ∣$ .
If dyadic cubes $Q, Q^{'}$ of generations $k \leq k^{'}$ intersect, then $Q^{'} \subseteq Q$ ; consequently two dyadic cubes are either disjoint or one contains the other, and cubes of one generation are equal or disjoint.

Facts & Assumptions

Given: An integer $n \geq 1$ ; dyadic cubes $Q = Q_{k, m}$ and $Q^{'} = Q_{k^{'}, m^{'}}$ of generations $k \leq k^{'}$ ; an ancestor generation $j < k$ ; Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) is assumed only in claim 1, for the identification of the box volume with Lebesgue measure.

[L1]

$Q_{k, m} = {x \in R^{n} : m_{i} 2^{- k} < x_{i} \leq (m_{i} + 1) 2^{- k} for every i < n}$ with $k \in Z$ , the side length is $2^{- k}$ , and every dyadic cube is nonempty (Dyadic cubes of all generations in R^n).

[L2]

$B (a, b) = {x \in R^{n} : a_{i} < x_{i} \leq b_{i} for every i < n}$ for real parameters; when $a_{i} < b_{i}$ for every $i < n$ , its box volume is $vol (B) = \prod_{i < n} (b_{i} - a_{i})$ . Empty boxes have volume zero (Half-open boxes in $R^{n}$ and their volume).

[F1]

For every real $t$ there is exactly one integer $m$ with $m \leq t < m + 1$ (Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ ).

[F2]

For $a \neq 0$ and integers $r, s$ one has $a^{r + s} = a^{r} a^{s}$ and $(a^{r})^{s} = a^{r s}$ ; in particular $2^{k} 2^{- k} = 1$ and $2^{k} > 0$ for every $k \in Z$ (Laws of integer exponents, Integer powers $a^{m}$ ).

[F3]

The order on $Z$ is total and compatible with addition, and $x \leq y$ implies $x + z \leq y + z$ (The integers form a totally ordered ring); the canonical embedding $N \to Z$ is injective, preserves the order, and has image exactly the nonnegative integers, so every positive integer is the image of a unique natural number $\geq 1$ (The naturals embed in the integers, The integers as equivalence classes of pairs of naturals).

[F4]

Finite products are defined by the recursion $Π_{0} = 1$ , $Π_{σ (n)} = Π_{n} \cdot a_{n}$ , and $\prod_{i < n} (a_{i} b_{i}) = (\prod_{i < n} a_{i}) (\prod_{i < n} b_{i})$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[F5]

Every half-open box $B (a, b)$ with real parameters satisfying $a_{i} \leq b_{i}$ for every $i < n$ is Lebesgue measurable with $λ_{n} (B (a, b)) = \prod_{i < n} (b_{i} - a_{i})$ (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).

Proof

technique · direct

1.1F1F3algebra

For a real $t$ there is exactly one integer $m$ with $m < t \leq m + 1$ : applying [F1] to $- t$ gives a unique $p$ with $p \leq - t < p + 1$ , and $m : = - p - 1$ satisfies $m < t \leq m + 1$ ; uniqueness follows because any integer $m^{'}$ with $m^{'} < t \leq m^{'} + 1$ gives $- m^{'} - 1 \leq - t < - m^{'}$ , so $- m^{'} - 1 = p$ and $m^{'} = m$ .

1.2F2givenalgebra

For $k \in Z$ and real $x_{i}$ , the condition $m_{i} 2^{- k} < x_{i} \leq (m_{i} + 1) 2^{- k}$ is equivalent to $m_{i} < 2^{k} x_{i} \leq m_{i} + 1$ , because $2^{k} > 0$ and $2^{k} \cdot 2^{- k} = 1$ by [F2]; multiplying the chain by $2^{k}$ preserves the two inequalities.

1.3F3algebra

For integers $u < v$ one has $u + 1 \leq v$ : by [F3] the positive integer $v - u$ is the image of a natural number $d \neq 0$ , and every nonzero natural number satisfies $1 \leq d$ (its predecessor is a natural number), so $v - u \geq 1$ . Consequently, if integers $A < B$ and $C$ , and a real $t$ , satisfy $A < t \leq B$ and $C < t \leq C + 1$ , then $A \leq C$ and $C + 1 \leq B$ : if $C < A$ then $C + 1 \leq A$ and $t \leq C + 1 \leq A < t$ , a contradiction, and if $B < C + 1$ then $B \leq C < t$ , contradicting $t \leq B$ .

1.4L2F2F4F5algebra

The box $Q_{k, m}$ has Lebesgue measure $∣ Q_{k, m} ∣ = \prod_{i < n} ((m_{i} + 1) 2^{- k} - m_{i} 2^{- k}) = \prod_{i < n} 2^{- k} = (2^{- k})^{n} = 2^{- k n}$ , the last two equalities by the finite-product recursion and the power laws; here $∣ Q ∣$ denotes Lebesgue measure, identified with the box volume by [F5].

2.1L1step 1.1step 1.2step 1.4

Given $x \in R^{n}$ , step 1.1 applied in each coordinate to the real $2^{k} x_{i}$ produces exactly one integer $m_{i}$ with $m_{i} < 2^{k} x_{i} \leq m_{i} + 1$ ; by step 1.2 the function $m$ is the unique index of a generation- $k$ dyadic cube containing $x$ . Hence the generation- $k$ cubes cover $R^{n}$ and no two distinct ones share a point, and by step 1.4 each has volume $2^{- k n}$ .

2.2L1L2F2step 1.3algebra

Put $d : = k^{'} - k \geq 0$ and $M_{i} : = m_{i} 2^{d} \in Z$ ; by [F2], $m_{i} 2^{- k} = M_{i} 2^{- k^{'}}$ and $(m_{i} + 1) 2^{- k} = (M_{i} + 2^{d}) 2^{- k^{'}}$ , so in coordinate $i$ the cube $Q$ is cut out by $M_{i} 2^{- k^{'}} < x_{i} \leq (M_{i} + 2^{d}) 2^{- k^{'}}$ while $Q^{'}$ is cut out by $m_{i}^{'} 2^{- k^{'}} < x_{i} \leq (m_{i}^{'} + 1) 2^{- k^{'}}$ . If $x \in Q \cap Q^{'}$ , step 1.3 with $A : = M_{i}$ , $B : = M_{i} + 2^{d}$ , $C : = m_{i}^{'}$ and $t : = 2^{k^{'}} x_{i}$ gives $M_{i} \leq m_{i}^{'}$ and $m_{i}^{'} + 1 \leq M_{i} + 2^{d}$ in every coordinate, so $Q^{'} \subseteq Q$ .

3.1L1F2step 1.4step 2.1step 2.2algebra

Fix $j < k$ and take the upper corner $x_{i} = (m_{i} + 1) 2^{- k}$ of $Q$ ; the half-open convention places $x$ in $Q$ . By step 2.1 there is exactly one generation- $j$ cube $P$ containing $x$ . Since $P$ and $Q$ intersect and $j < k$ , step 2.2 gives $Q \subseteq P$ . If $P^{'}$ is another generation- $j$ cube containing $Q$ , it contains $x$ , hence $P^{'} = P$ by step 2.1. This proves unique ancestry without any erroneous scaling of the integer index. For the parent $j = k - 1$ , step 1.4 gives $∣ P ∣ = 2^{- (k - 1) n} = 2^{n} 2^{- k n} = 2^{n} ∣ Q ∣$ .

4.1step 2.1step 2.2step 3.1∎

Claim 1 is steps 2.1 and 1.4, claim 2 is step 3.1, and claim 3 is step 2.2 together with its same-generation special case; this proves the lemma.

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

Maximal dyadic cubes above a level

Statement

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

Let $f \in L^{1} (R^{n})$ and $λ > 0$ , and let the dyadic cubes be the all-generations cubes of Dyadic cubes of all generations in R^n. The dyadic cubes $Q$ with average $∣ Q ∣^{- 1} \int_{Q} ∣ f ∣ > λ$ that are maximal under inclusion form a countable family of pairwise disjoint cubes; their union is exactly the dyadic maximal superlevel set ${M_{d} f > λ}$ , where $M_{d} f (x) = \sup {∣ Q ∣^{- 1} \int_{Q} ∣ f ∣ : Q dyadic, x \in Q}$ over all generations; each such $Q$ satisfies $∣ Q ∣^{- 1} \int_{Q} ∣ f ∣ \leq 2^{n} λ$ ; and $\sum_{Q} ∣ Q ∣ \leq λ^{- 1} ∥ f ∥_{1}$ .

Facts & Assumptions

Given: $f \in L^{1} (R^{n})$ and $λ > 0$ ; a dyadic cube $Q$ of generation $k$ with centre-related index $m$ ; the all-generations dyadic grid of Dyadic cubes of all generations in R^n; two dyadic cubes $Q, Q^{'}$ of generations $k^{'} \geq k$ .

[F1]

For every generation $k \in Z$ the generation- $k$ cubes are pairwise disjoint with union $R^{n}$ and volume $2^{- k n}$ ; every dyadic cube $Q$ of generation $k$ has for each $j < k$ exactly one ancestor $A_{j} (Q)$ of generation $j$ containing it, and the parent $A_{k - 1} (Q)$ has volume $2^{n} ∣ Q ∣$ ; and if two dyadic cubes intersect then one contains the other (All-generation dyadic cubes: partition, volume and nesting).

[F2]

$\int_{Q} ∣ f ∣ d λ \leq ∥ f ∥_{1} < \infty$ for every measurable $Q$ , and every dyadic cube has finite volume (The class $L^{1} (μ)$ of integrable functions, Dyadic cubes of all generations in R^n).

[F3]

The set $Z \times Z^{n}$ is countable, and every subset of a countable set is countable (Finite, countably infinite, countable, uncountable, Every subset of an at most countable set is at most countable); finite and countable sums of nonnegative extended reals are defined by the usual supremum over finite partial sums (Series in the nonnegative extended real line).

Proof

technique · direct

1.1F2givenalgebra

Call a dyadic cube bad when $∣ Q ∣^{- 1} \int_{Q} ∣ f ∣ > λ$ . Every bad cube satisfies $0 \leq λ ∣ Q ∣ < \int_{Q} ∣ f ∣ \leq ∥ f ∥_{1}$ , so $∣ Q ∣ < λ^{- 1} ∥ f ∥_{1}$ ; in particular there is a scale above which no bad cube lives.

1.2F1F3givenalgebra

A bad cube is maximal exactly when none of its strictly larger ancestors is bad: by [F1] any intersecting cube is nested, and any containing cube of coarser generation is the unique ancestor of that generation. Every maximal bad cube therefore has a good parent. A good parent alone need not imply maximality; coarser ancestors must also be excluded. Distinct maximal bad cubes are disjoint, since nesting would otherwise make one a strictly larger bad cube containing the other. The family is countable because it is a subset of the dyadic grid parameterized by $Z \times Z^{n}$ ; no selection is required.

2.1F1step 1.1step 1.2algebra

Every bad cube is contained in a maximal bad cube. Let $Q$ be bad of generation $k$ , and let $J : = {j \leq k : A_{j} (Q) is bad} \subseteq Z$ , where $A_{j} (Q)$ is the unique generation- $j$ ancestor of $Q$ from [F1]. The set $J$ is nonempty because $k \in J$ , and it is bounded below: if $j \in J$ then $∣ A_{j} (Q) ∣ = 2^{- j n} < λ^{- 1} ∥ f ∥_{1}$ by step 1.1, so $2^{- j} < (λ^{- 1} ∥ f ∥_{1})^{1 / n}$ and $j$ exceeds a fixed bound. A nonempty subset of $Z$ that is bounded below has a least element $j_{0}$ ; the ancestor $A_{j_{0}} (Q)$ is bad by definition, and every strictly larger ancestor has generation $j < j_{0}$ and is not bad by minimality. Thus $A_{j_{0}} (Q)$ is maximal by step 1.2, and contains $Q$ .

2.2F1F2F3step 1.2algebra

Let $Q$ be a maximal bad cube and $P$ its parent; by step 1.2 the cube $P$ is good, that is, $∣ P ∣^{- 1} \int_{P} ∣ f ∣ \leq λ$ . Since $Q \subseteq P$ and $∣ P ∣ = 2^{n} ∣ Q ∣$ by [F1], $\int_{Q} ∣ f ∣ \leq \int_{P} ∣ f ∣ \leq λ ∣ P ∣ = 2^{n} λ ∣ Q ∣$ , so the average of every maximal bad cube is at most $2^{n} λ$ . For the sum, the maximal bad cubes are pairwise disjoint by step 1.2, so with disjoint additivity and monotonicity of the integral, $\sum_{Q} λ ∣ Q ∣ \leq \sum_{Q} \int_{Q} ∣ f ∣ = \int_{⋃ Q} ∣ f ∣ \leq ∥ f ∥_{1},$ because each maximal bad cube has average $> λ$ ; hence $\sum_{Q} ∣ Q ∣ \leq λ^{- 1} ∥ f ∥_{1}$ , the sums being understood as suprema of finite partial sums over the countable family.

3.1F1step 2.1algebra

The union of the maximal bad cubes is ${M_{d} f > λ}$ . If $x$ lies in a maximal bad cube $Q$ , then [F1] gives $M_{d} f (x) \geq ∣ Q ∣^{- 1} \int_{Q} ∣ f ∣ > λ$ . Conversely, if $M_{d} f (x) > λ$ , then by definition of the supremum over a nonempty set of real numbers there is a dyadic cube $Q ∋ x$ with $∣ Q ∣^{- 1} \int_{Q} ∣ f ∣ > λ$ , i.e. $Q$ is bad; step 2.1 provides a maximal bad cube containing $Q$ , hence containing $x$ .

4.1step 1.2step 2.1step 2.2step 3.1∎

Steps 1.2 and 2.1 give the countable pairwise disjoint maximal family with the containment property, step 3.1 identifies its union with ${M_{d} f > λ}$ , and step 2.2 gives both the average bound and the sum bound. This proves the lemma.

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

Radially decreasing kernels are dominated by the maximal function

Statement

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

Let $ω \geq 0$ be a measurable, radially nonincreasing, integrable function on $R^{n}$ : that is, $ω (x) = ω (y)$ whenever $∣ x ∣ = ∣ y ∣$ and $ω (x) \geq ω (y)$ whenever $∣ x ∣ \leq ∣ y ∣$ . Let $f \in L_{l o c}^{1} (R^{n})$ . Then for every $x \in R^{n}$ , $\int_{R^{n}} ∣ f (x - y) ∣ ω (y) d y \leq ∥ ω ∥_{1} M f (x),$ where $M$ is the centered Hardy–Littlewood maximal operator and both sides may be $+ \infty$ .

Facts & Assumptions

Given: Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), which is assumed both by the definition of the maximal function [F1] and by the scaling identity [F5]; a radially nonincreasing integrable $ω \geq 0$ ; a function $f \in L_{l o c}^{1} (R^{n})$ ; a point $x \in R^{n}$ ; a height $t > 0$ .

[F1]

$M f (x) = \sup_{r > 0} λ (B (x, r))^{- 1} \int_{B (x, r)} ∣ f ∣ d λ$ , with values in $[0, \infty]$ (The centered and uncentered Hardy-Littlewood maximal functions); consequently $\int_{B (x, r)} ∣ f ∣ d λ \leq λ (B (x, r)) M f (x)$ for every $r > 0$ whenever $M f (x) < \infty$ .

[F2]

For measurable $A \subseteq B$ one has $μ (A) \leq μ (B)$ (Measures are monotone), and every Euclidean ball is Lebesgue measurable with $0 < λ (B (x, r)) < \infty$ (Euclidean balls have positive finite Lebesgue measure).

[F3]

For measurable $g$ and $0 < q < \infty$ , $\int ∣ g ∣^{q} d λ = q \int_{0}^{\infty} t^{q - 1} λ ({∣ g ∣ > t}) d t$ , both sides possibly $+ \infty$ ; in particular the case $q = 1$ computes $∥ ω ∥_{1}$ (For 0 < p < infinity, the layer-cake formula computes the integral of |f|^p from the distribution function).

[F4]

[F5]

For nonzero real $c$ and Lebesgue measurable $E$ , $λ (c E) = ∣ c ∣^{n} λ (E)$ (For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it); in particular $λ (B (0, r)) = r^{n} λ (B (0, 1))$ for $r > 0$ , and $t \mapsto t^{n}$ is continuous.

Proof

technique · direct

1.1F1givenconstruct

Fix $x$ and put $g (y) : = ∣ f (x - y) ∣$ for $y \in R^{n}$ ; then $g \geq 0$ is measurable and locally integrable, $\int ∣ f (x - y) ∣ ω (y) d y = \int g ω d λ$ , and substitution $z = x - y$ (with $B (x, r) = x - B (0, r)$ and translation invariance of $λ$ ) gives $\sup_{r > 0} λ (B (0, r))^{- 1} \int_{B (0, r)} g d λ = \sup_{r > 0} λ (B (x, r))^{- 1} \int_{B (x, r)} ∣ f ∣ d λ = M f (x)$ , that is, $M g (0) = M f (x)$ ; if $∥ ω ∥_{1} = 0$ , the nonnegative integrand vanishes almost everywhere and both sides are zero (with the usual zero-times-infinity convention). Otherwise the case $M f (x) = + \infty$ makes the desired inequality trivial, so assume $M f (x) < \infty$ and fix a height $t > 0$ .

1.2F1F2F5givenalgebra

For $t > 0$ put $S_{t} : = {ω > t}$ and $r_{t} : = \sup {∣ y ∣ : y \in S_{t}} \in [0, \infty]$ , using $r_{t} = 0$ when $S_{t} = \emptyset$ . Then $r_{t} < \infty$ : if $r_{t} = \infty$ , then for every $ρ > 0$ radial monotonicity and the definition of the supremum give $B (0, ρ) \subseteq S_{t}$ , so $∥ ω ∥_{1} \geq t λ (B (0, ρ))$ for all $ρ$ , which is impossible because $λ (B (0, ρ)) \to \infty$ as $ρ \to \infty$ . Moreover $B (0, r_{t}) \subseteq S_{t} \subseteq B (0, r_{t} + ε)$ for every $ε > 0$ : the first inclusion uses that $∣ z ∣ < r_{t}$ provides $y \in S_{t}$ with $∣ y ∣ > ∣ z ∣$ and then $ω (z) \geq ω (y) > t$ , and the second uses $∣ y ∣ \leq r_{t}$ for $y \in S_{t}$ . Consequently, by monotonicity [F2] and the scaling identity [F5], $λ (S_{t}) \leq λ (B (0, r_{t} + ε)) = (r_{t} + ε)^{n} λ (B (0, 1)) (ε > 0), λ (B (0, r_{t})) = r_{t}^{n} λ (B (0, 1)),$ so letting $ε ↓ 0$ along $ε = 1 / k$ and using continuity of $t \mapsto t^{n}$ yields $λ (S_{t}) = λ (B (0, r_{t}))$ .

2.1F1F2F5step 1.1step 1.2algebra

For every $ρ > 0$ one has $\int_{B (0, ρ)} g d λ \leq λ (B (0, ρ)) M g (0) = λ (B (0, ρ)) M f (x)$ by [F1] and step 1.1, hence for every $ε > 0$ the inclusions of step 1.2 give $\int_{S_{t}} g d λ \leq \int_{B (0, r_{t} + ε)} g d λ \leq (r_{t} + ε)^{n} λ (B (0, 1)) M f (x);$ letting $ε ↓ 0$ as in step 1.2 gives $\int_{S_{t}} g d λ \leq λ (B (0, r_{t})) M f (x)$ .

2.2F3step 1.2algebra

The layer-cake identity [F3] applied to $ω$ with $q = 1$ , together with $λ (S_{t}) = λ (B (0, r_{t}))$ from step 1.2, gives $∥ ω ∥_{1} = \int_{0}^{\infty} λ (S_{t}) d t = \int_{0}^{\infty} λ (B (0, r_{t})) d t$ .

3.1F4step 2.1step 2.2algebra∎

The pointwise identity $ω (y) = \int_{0}^{\infty} 1_{S_{t}} (y) d t$ for $y \in R^{n}$ , Tonelli's theorem [F4] applied to the nonnegative product-measurable integrand $(y, t) \mapsto g (y) 1_{S_{t}} (y)$ , and steps 2.1 and 2.2 give $\int g ω d λ = \int_{0}^{\infty} ⁣ ⁣ \int_{S_{t}} g d λ d t \leq \int_{0}^{\infty} λ (B (0, r_{t})) M f (x) d t = M f (x) \int_{0}^{\infty} λ (B (0, r_{t})) d t = M f (x) ∥ ω ∥_{1},$ which is the asserted inequality; the case $M f (x) = + \infty$ was already trivial in step 1.1.

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

Calderón–Zygmund decomposition at height λ

Statement

Assume Countable Choice. Let $f \in L^{1} (R^{n})$ and $λ > 0$ . Then $f = g + \sum_{j} b_{j}$ almost everywhere, where the cubes $Q_{j}$ are the maximal all-generation dyadic cubes of Maximal dyadic cubes above a level, $b_{j} = (f - ∣ Q_{j} ∣^{- 1} \int_{Q_{j}} f) 1_{Q_{j}}$ satisfies $\int b_{j} = 0$ and $∥ b_{j} ∥_{1} \leq 2^{n + 1} λ ∣ Q_{j} ∣$ , and $g = f outside ⋃_{j} Q_{j}, g = ∣ Q_{j} ∣^{- 1} \int_{Q_{j}} f on Q_{j}$ satisfies $∥ g ∥_{1} \leq ∥ f ∥_{1}$ , $∣ g ∣ \leq 2^{n} λ$ almost everywhere, and $∥ g ∥_{2}^{2} \leq 2^{n} λ ∥ f ∥_{1}$ ; moreover $\sum_{j} ∣ Q_{j} ∣ \leq λ^{- 1} ∥ f ∥_{1}$ .

Facts & Assumptions

Given: $f \in L^{1} (R^{n})$ and $λ > 0$ ; the maximal bad dyadic cubes $Q_{j}$ of the previous lemma, pairwise disjoint with $\sum_{j} ∣ Q_{j} ∣ \leq λ^{- 1} ∥ f ∥_{1}$ and $∣ Q_{j} ∣^{- 1} \int_{Q_{j}} ∣ f ∣ \leq 2^{n} λ$ ; the functions $g$ and $b_{j}$ defined above.

[F1]

The maximal bad cubes $Q_{j}$ (cubes with $∣ Q ∣^{- 1} \int_{Q} ∣ f ∣ > λ$ , maximal under inclusion) are countable, pairwise disjoint, have union ${M_{d} f > λ}$ , and satisfy $∣ Q_{j} ∣^{- 1} \int_{Q_{j}} ∣ f ∣ \leq 2^{n} λ$ and $\sum_{j} ∣ Q_{j} ∣ \leq λ^{- 1} ∥ f ∥_{1}$ (Maximal dyadic cubes above a level).

[F2]

A family $(E_{r})_{r > 0}$ shrinks nicely to $x$ with constant $α$ when $E_{r} \subseteq B (x, r)$ and $λ (E_{r}) \geq α λ (B (x, r))$ ; if for each $x$ in a set $A$ such a family is given, then for almost every $x \in A$ the averages of an $L_{l o c}^{1}$ function over $E_{r}$ converge to the function value as $r \to 0^{+}$ (Differentiation holds along families shrinking nicely, Almost every point is a Lebesgue point of a locally integrable function).

Proof

technique · direct

1.1F1givenalgebra

The functions $g$ and $b_{j}$ are measurable, $b_{j}$ is supported in $Q_{j}$ , and $f = g + \sum_{j} b_{j}$ everywhere: on $Q_{j}$ the sum $g + \sum_{k} b_{k}$ has the single nonzero term $g + b_{j} = ∣ Q_{j} ∣^{- 1} \int_{Q_{j}} f + f - ∣ Q_{j} ∣^{- 1} \int_{Q_{j}} f = f$ by disjointness of the $Q_{k}$ , while off $⋃_{k} Q_{k}$ one has $g = f$ and every $b_{k} = 0$ . Moreover $\int b_{j} = \int_{Q_{j}} f - ∣ Q_{j} ∣^{- 1} \int_{Q_{j}} f ∣ Q_{j} ∣ = 0$ and $∥ b_{j} ∥_{1} \leq \int_{Q_{j}} ∣ f ∣ + ∣ Q_{j} ∣^{- 1} ∣ \int_{Q_{j}} f ∣ ∣ Q_{j} ∣ \leq 2 \int_{Q_{j}} ∣ f ∣ \leq 2 \cdot 2^{n} λ ∣ Q_{j} ∣ = 2^{n + 1} λ ∣ Q_{j} ∣$ .

2.1F1F2step 1.1algebra

On each bad cube, $∣ g ∣ = ∣ Q_{j} ∣^{- 1} ∣ \int_{Q_{j}} f ∣ \leq ∣ Q_{j} ∣^{- 1} \int_{Q_{j}} ∣ f ∣ \leq 2^{n} λ$ , so $∣ g ∣ \leq 2^{n} λ$ on $⋃_{j} Q_{j}$ ; off $⋃_{j} Q_{j}$ one has $g = f$ and all dyadic cubes through $x$ are good, so the averages of $∣ f ∣$ over those cubes are at most $λ$ . These cubes, indexed by generation and assigned to the parameter $r = 2 \sqrt{n} 2^{- k}$ for the generation- $k$ cube through $x$ and extended constantly on $[2 \sqrt{n} 2^{- k}, 2 \sqrt{n} 2^{- k + 1})$ , shrink nicely to $x$ with a dimensional constant: each lies in $B (x, r)$ and has measure $2^{- k n} = c_{n} λ (B (x, 2 \sqrt{n} 2^{- k}))$ . Hence [F2] gives $∣ f (x) ∣ \leq λ$ for almost every $x \notin ⋃_{j} Q_{j}$ , and since $g = f$ there, $∣ g ∣ \leq \max (2^{n} λ, λ) = 2^{n} λ$ almost everywhere.

3.1F1step 1.1step 2.1algebra∎

From step 2.1, $∥ g ∥_{1} \leq \int_{⋃ Q_{j}} ∣ g ∣ + \int_{R^{n} ∖ ⋃ Q_{j}} ∣ f ∣ \leq \sum_{j} \int_{Q_{j}} ∣ f ∣ + \int_{R^{n} ∖ ⋃ Q_{j}} ∣ f ∣ = ∥ f ∥_{1}$ , and $∥ g ∥_{2}^{2} = \int ∣ g ∣^{2} \leq ∥ g ∥_{\infty} \int ∣ g ∣ \leq 2^{n} λ ∥ f ∥_{1}$ . Together with step 1.1 and the bounds $\sum_{j} ∣ Q_{j} ∣ \leq λ^{- 1} ∥ f ∥_{1}$ and $∣ Q_{j} ∣^{- 1} \int_{Q_{j}} ∣ f ∣ \leq 2^{n} λ$ from [F1], this is the asserted decomposition.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The good part has controlled L2 image

Statement

Assume Countable Choice. Let $T$ be a Calderón–Zygmund operator with kernel constants $A_{1}, A_{2}$ and $L^{2}$ norm $B$ , and let $f = g + \sum_{j} b_{j}$ be the Calderón–Zygmund decomposition of $f \in L^{1} (R^{n})$ at height $λ > 0$ (Calderón–Zygmund decomposition at height λ). Then $∣ {∣ T g ∣ > λ / 2} ∣ \leq 4 B^{2} 2^{n} λ^{- 1} ∥ f ∥_{1} .$

Facts & Assumptions

Given: $f \in L^{1}$ , $λ > 0$ , its Calderón–Zygmund decomposition with good part $g$ at height $λ$ ; a Calderón–Zygmund operator $T$ with $L^{2}$ norm bound $B$ .

[F1]

The good part satisfies $g \in L^{2} (R^{n})$ with $∥ g ∥_{2}^{2} \leq 2^{n} λ ∥ f ∥_{1}$ (Calderón–Zygmund decomposition at height λ).

[F2]

$T : L^{2} \to L^{2}$ is linear with $∥ T h ∥_{2} \leq B ∥ h ∥_{2}$ for every $h \in L^{2}$ (Calderón–Zygmund kernels and their associated operators).

[F3]

For a measurable $u$ and $μ > 0$ , $μ ({∣ u ∣ \geq μ}) \leq μ^{- 2} \int ∣ u ∣^{2} d μ$ ; more precisely $λ ({∣ u ∣ > μ}) \leq μ^{- 2} \int ∣ u ∣^{2} d λ$ by Chebyshev's inequality applied to $∣ u ∣^{2}$ at level $μ^{2}$ (Chebyshev-Markov inequality for the integral).

Proof

technique · direct

1.1F1F2given

Since $g \in L^{2}$ by [F1], linearity of $T$ gives $T g \in L^{2}$ with $∥ T g ∥_{2} \leq B ∥ g ∥_{2}$ .

2.1F1F3step 1.1algebra∎

Chebyshev's inequality at level $(λ / 2)^{2}$ applied to $∣ T g ∣^{2}$ , followed by step 1.1 and the bound of [F1], gives $∣ {∣ T g ∣ > λ / 2} ∣ \leq \frac{4}{λ^{2}} ∥ T g ∥_{2}^{2} \leq \frac{4 B^{2}}{λ^{2}} ∥ g ∥_{2}^{2} \leq \frac{4 B^{2}}{λ^{2}} 2^{n} λ ∥ f ∥_{1} = 4 B^{2} 2^{n} λ^{- 1} ∥ f ∥_{1},$ which is the asserted estimate.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The bad part is integrable away from expanded cubes

Statement

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

Let $T$ be a Calderón–Zygmund operator with kernel constants $A_{1}, A_{2}$ (Calderón–Zygmund kernels and their associated operators), let $Q$ be a dyadic cube (Dyadic cubes of all generations in R^n) with centre $c_{Q}$ , and let $b_{Q} \in L^{2} (R^{n})$ be supported in $Q$ with $\int b_{Q} = 0$ . If $Q^{*}$ is the cube concentric with $Q$ whose side length is $2 \sqrt{n}$ times the side length of $Q$ , then $\int_{R^{n} ∖ Q^{*}} ∣ T b_{Q} (x) ∣ d x \leq A_{2} ∥ b_{Q} ∥_{1} .$

Facts & Assumptions

Given: A Calderón–Zygmund operator $T$ with kernel $k$ and constants $A_{1}, A_{2}$ ; a dyadic cube $Q$ of side length $ℓ = 2^{- k}$ with centre $c_{Q}$ ; the concentric cube $Q^{*}$ of side length $2 \sqrt{n} ℓ$ ; a function $b_{Q} \in L^{2}$ supported in $Q$ with $\int b_{Q} = 0$ .

[F1]

$T$ is linear and $L^{2}$ -bounded, and for every compactly supported $f \in L^{2}$ one has $T f (x) = \int k (x - y) f (y) d y$ for almost every $x \notin supp f$ , the integral converging absolutely there; the Hörmander condition reads $\sup_{y \neq 0} \int_{∣ x ∣ \geq 2 ∣ y ∣} ∣ k (x - y) - k (x) ∣ d x \leq A_{2}$ and is invariant under replacing the origin by any centre (Calderón–Zygmund kernels and their associated operators).

[F2]

$Q \subseteq {x : ∣ x_{i} - c_{Q, i} ∣ \leq ℓ / 2 for every i}$ and $Q^{*} = {x : ∣ x_{i} - c_{Q, i} ∣ \leq \sqrt{n} ℓ for every i}$ in the notation of Dyadic cubes of all generations in R^n; the Euclidean and supremum norms on $R^{n}$ satisfy $∥ v ∥_{\infty} \leq ∥ v ∥_{2} \leq \sqrt{n} ∥ v ∥_{\infty}$ .

[F3]

On a product of $σ$ -finite measure spaces a nonnegative product-measurable function may be integrated in either order, both integrals possibly $+ \infty$ (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

Proof

technique · direct

1.1F2givenalgebra

If $y \in Q$ and $x \notin Q^{*}$ , then $∥ x - c_{Q} ∥_{\infty} > \sqrt{n} ℓ$ and $∥ y - c_{Q} ∥_{\infty} \leq ℓ / 2$ , so by [F2] $∥ x - c_{Q} ∥_{2} \geq ∥ x - c_{Q} ∥_{\infty} > \sqrt{n} ℓ \geq 2 ∥ y - c_{Q} ∥_{2}$ ; hence $∣ x - c_{Q} ∣ \geq 2 ∣ y - c_{Q} ∣$ , and in particular $x \neq c_{Q}$ and $x \neq y$ .

2.1F1givenalgebra

For almost every $x \notin Q^{*}$ one has, using [F1] and the mean-zero condition, $T b_{Q} (x) = \int k (x - y) b_{Q} (y) d y = \int_{R^{n}} [k (x - y) - k (x - c_{Q})] b_{Q} (y) d y,$ because $b_{Q}$ vanishes off $Q$ , the subtracted term is the constant $k (x - c_{Q})$ times $\int b_{Q} = 0$ , and $x - c_{Q} \neq 0$ by step 1.1.

3.1F3step 2.1algebra

By step 2.1 and nonnegativity, for almost every $x \notin Q^{*}$ , $∣ T b_{Q} (x) ∣ \leq \int_{Q} ∣ k (x - y) - k (x - c_{Q}) ∣ ∣ b_{Q} (y) ∣ d y .$ Integrating this inequality over $R^{n} ∖ Q^{*}$ , whose complement has finite measure at every scale and which is $σ$ -finite, and applying Tonelli's theorem [F3] to the nonnegative product-measurable integrand gives $\int_{R^{n} ∖ Q^{*}} ∣ T b_{Q} ∣ \leq \int_{Q} ∣ b_{Q} (y) ∣ (\int_{R^{n} ∖ Q^{*}} ∣ k (x - y) - k (x - c_{Q}) ∣ d x) d y .$

4.1F1step 1.1step 3.1algebra∎

For $y = c_{Q}$ the difference integrand is zero. For every other $y \in Q$ the inner integral is at most $A_{2}$ : by step 1.1 the domain $R^{n} ∖ Q^{*}$ is contained in ${x : ∣ x - c_{Q} ∣ \geq 2 ∣ y - c_{Q} ∣}$ , and the change of variables $u = x - c_{Q}$ , $v = y - c_{Q}$ turns the integral over that larger set into $\int_{∣ u ∣ \geq 2 ∣ v ∣} ∣ k (u - v) - k (u) ∣ d u \leq A_{2}$ by the translation-invariant Hörmander condition of [F1]. Substituting into step 3.1 yields the asserted bound $\int_{R^{n} ∖ Q^{*}} ∣ T b_{Q} ∣ \leq A_{2} \int_{Q} ∣ b_{Q} ∣ = A_{2} ∥ b_{Q} ∥_{1}$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Calderón–Zygmund operators are of weak type (1,1)

Statement

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

Let $T$ be a Calderón–Zygmund operator with kernel constants $A_{1}, A_{2}$ and $L^{2}$ norm $B$ . Then for every $f \in L^{1} (R^{n})$ and every $λ > 0$ , $∣ {∣ T f ∣ > λ} ∣ \leq C_{n} (A_{2} + B) λ^{- 1} ∥ f ∥_{1}$ with a dimensional constant $C_{n}$ independent of $T$ , $f$ and $λ$ ; equivalently, $T$ extends to a bounded operator $L^{1} (R^{n}) \to L^{1, \infty} (R^{n})$ .

Facts & Assumptions

Given: A Calderón–Zygmund operator $T$ with kernel constants $A_{1}, A_{2}$ and $L^{2}$ norm bound $B$ ; $f \in L^{1} (R^{n})$ and $λ > 0$ ; a constant $γ > 0$ to be fixed; Countable Choice.

[F1]

$T$ is linear, $L^{2}$ -bounded with $∥ T h ∥_{2} \leq B ∥ h ∥_{2}$ , and has the off-support kernel representation with constants $A_{1}, A_{2}$ (Calderón–Zygmund kernels and their associated operators); weak type $(1, 1)$ with constant $A$ means exactly the inequality $μ ({∣ T h ∣ > λ}) \leq A λ^{- 1} ∥ h ∥_{1}$ for all $h \in L^{1}$ and $λ > 0$ (Sublinear operators and weak or strong type $(p, q)$ bounds).

[F2]

The Calderón–Zygmund decomposition of $h \in L^{1}$ at height $μ > 0$ writes $h = g + \sum_{j} b_{j}$ a.e. with $∥ g ∥_{1} \leq ∥ h ∥_{1}$ , $∥ g ∥_{2}^{2} \leq 2^{n} μ ∥ h ∥_{1}$ , $∣ g ∣ \leq 2^{n} μ$ a.e., $\int b_{j} = 0$ , $∥ b_{j} ∥_{1} \leq 2^{n + 1} μ ∣ Q_{j} ∣$ and $\sum_{j} ∣ Q_{j} ∣ \leq μ^{- 1} ∥ h ∥_{1}$ (Calderón–Zygmund decomposition at height λ); the good part satisfies $∣ {∣ T g ∣ > μ / 2} ∣ \leq 4 B^{2} 2^{n} μ^{- 1} ∥ h ∥_{1}$ (The good part has controlled L2 image). If additionally $b_{j} \in L^{2}$ , then for the dilated cube $Q_{j}^{*}$ of side $2 \sqrt{n}$ times that of $Q_{j}$ one has $\int_{R^{n} ∖ Q_{j}^{*}} ∣ T b_{j} ∣ \leq A_{2} ∥ b_{j} ∥_{1}$ (The bad part is integrable away from expanded cubes); step 1.1 verifies this additional hypothesis before the estimate is used.

[F3]

Chebyshev: $μ ({∣ u ∣ \geq t}) \leq t^{- 1} \int ∣ u ∣$ for nonnegative measurable $∣ u ∣$ (Chebyshev-Markov inequality for the integral); dilation: $λ (r E) = r^{n} λ (E)$ for measurable $E$ (For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it); Tonelli applies to nonnegative product-measurable integrands over $σ$ -finite products (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product); the $L^{1}$ convention is The class $L^{1} (μ)$ of integrable functions and Countable Choice is The Axiom of Countable Choice ( ${A C}_{ω}$ ).

[F4]

For nonnegative measurable $v_{r}$ , $\int {lim inf}_{r} v_{r} \leq {lim inf}_{r} \int v_{r}$ (Fatou's lemma).

Proof

technique · direct

1.1F1F2F3algebra

Assume first $f \in L^{1} \cap L^{2}$ and $B > 0$ , and apply the decomposition [F2] at height $μ = γ λ$ with $γ : = 2^{- (n + 1)} B^{- 1}$ ; write $f = g + b$ , $b = \sum_{j} b_{j}$ . The series $\sum_{j} b_{j}$ converges in $L^{2}$ : the $b_{j}$ are supported on the pairwise disjoint cubes $Q_{j}$ , and $∥ b_{j} ∥_{2}^{2} \leq 2 \int_{Q_{j}} ∣ f ∣^{2} + 2 (2^{n} γ λ)^{2} ∣ Q_{j} ∣$ by $b_{j} = (f - ∣ Q_{j} ∣^{- 1} \int_{Q_{j}} f) 1_{Q_{j}}$ and $∣ ∣ Q_{j} ∣^{- 1} \int_{Q_{j}} f ∣ \leq 2^{n} γ λ$ (the value of $∣ g ∣$ on $Q_{j}$ ), so $\sum_{j} ∥ b_{j} ∥_{2}^{2} \leq 2 ∥ f ∥_{2}^{2} + 2 \cdot 4^{n} γ λ ∥ f ∥_{1} < \infty$ because $f \in L^{1} \cap L^{2}$ and $\sum_{j} ∣ Q_{j} ∣ \leq (γ λ)^{- 1} ∥ f ∥_{1}$ ; hence $b \in L^{2}$ and, by linearity and $L^{2}$ -continuity of $T$ in [F1], $T f = T g + \sum_{j} T b_{j}$ as $L^{2}$ classes, so $∣ T f ∣ \leq ∣ T g ∣ + \sum_{j} ∣ T b_{j} ∣$ almost everywhere: choose image partial sums whose squared $L^{2}$ errors relative to $T b$ are at most $2^{- 3 r}$ . By Chebyshev the sets where the errors exceed $2^{- r}$ have measures at most $2^{- r}$ ; their tail unions have measures tending to zero, so this subsequence converges almost everywhere, and the finite triangle inequalities pass to the limit. The good part is controlled at the target level $λ / 2$ directly: Chebyshev's inequality for $∣ T g ∣^{2}$ at level $(λ / 2)^{2}$ , the $L^{2}$ bound $∥ T g ∥_{2} \leq B ∥ g ∥_{2}$ of [F1] and the decomposition bound $∥ g ∥_{2}^{2} \leq 2^{n} γ λ ∥ f ∥_{1}$ at height $μ = γ λ$ from [F2] give $∣ {∣ T g ∣ > λ / 2} ∣ \leq \frac{4}{λ^{2}} ∥ T g ∥_{2}^{2} \leq \frac{4 B^{2}}{λ^{2}} 2^{n} γ λ ∥ f ∥_{1} = 4 B^{2} 2^{n} γ λ^{- 1} ∥ f ∥_{1} = 2 B λ^{- 1} ∥ f ∥_{1},$ the last equality by the choice $γ = 2^{- (n + 1)} B^{- 1}$ .

1.2F2F3algebra

The dilated cubes satisfy $∣ Q_{j}^{*} ∣ = (2 \sqrt{n})^{n} ∣ Q_{j} ∣$ by the dilation identity [F3] applied to the concentric dilation of $Q_{j}$ , so the union bound and the decomposition's summability give $∣ ⋃_{j} Q_{j}^{*} ∣ \leq \sum_{j} ∣ Q_{j}^{*} ∣ \leq (2 \sqrt{n})^{n} (γ λ)^{- 1} ∥ f ∥_{1} = 2^{n + 1} n^{n / 2} 2^{n} B λ^{- 1} ∥ f ∥_{1} .$

1.3F2F3algebra

On the complement, Tonelli's theorem for the nonnegative series and the bad-part bound of [F2] give $\int_{R^{n} ∖ ⋃_{j} Q_{j}^{*}} \sum_{j} ∣ T b_{j} ∣ d x = \sum_{j} \int_{R^{n} ∖ ⋃_{k} Q_{k}^{*}} ∣ T b_{j} ∣ d x \leq \sum_{j} \int_{R^{n} ∖ Q_{j}^{*}} ∣ T b_{j} ∣ d x \leq A_{2} \sum_{j} ∥ b_{j} ∥_{1},$ and the decomposition's bounds $∥ b_{j} ∥_{1} \leq 2^{n + 1} γ λ ∣ Q_{j} ∣$ and $\sum_{j} ∣ Q_{j} ∣ \leq (γ λ)^{- 1} ∥ f ∥_{1}$ show this is at most $2^{n + 1} γ λ (γ λ)^{- 1} A_{2} ∥ f ∥_{1} = 2^{n + 1} A_{2} ∥ f ∥_{1}$ ; hence Chebyshev [F3] at level $λ / 2$ yields $∣ {x \notin ⋃_{j} Q_{j}^{*} : \sum_{j} ∣ T b_{j} ∣ > λ / 2} ∣ \leq 2^{n + 2} A_{2} λ^{- 1} ∥ f ∥_{1}$ .

2.1step 1.1step 1.2step 1.3algebra

Combining step 1.1 (which supplies the almost-everywhere inequality and the good-part estimate), step 1.2 and step 1.3, $∣ {∣ T f ∣ > λ} ∣ \leq ∣ {∣ T g ∣ > λ / 2} ∣ + ∣ ⋃_{j} Q_{j}^{*} ∣ + ∣ {x outside the cubes : \sum_{j} ∣ T b_{j} ∣ > λ / 2} ∣ \leq C_{n} (A_{2} + B) λ^{- 1} ∥ f ∥_{1},$ where $C_{n} : = \max {2 + 2^{2 n + 1} n^{n / 2}, 2^{n + 2}}$ is the maximum of the three dimensional constants collected from steps 1.1–1.3; this is the assertion for $f \in L^{1} \cap L^{2}$ .

3.1F1F4step 2.1algebra∎

Extension to all $f \in L^{1}$ and $B \geq 0$ . If $B = 0$ , extend the zero operator on $L^{2}$ by zero on $L^{1}$ . Otherwise put $D = C_{n} (A_{2} + B)$ and $f_{m} = f 1_{B (0, m)} 1_{{∣ f ∣ \leq m}} \in L^{1} \cap L^{2}$ . The integrable tails show $∥ f_{m} - f ∥_{1} \to 0$ , so step 2.1 applied to differences makes $u_{m} = T f_{m}$ Cauchy in measure. Select increasing $m_{r}$ such that $∣ {∣ u_{m_{r + 1}} - u_{m_{r}} ∣ > 2^{- r}} ∣ \leq 2^{- r}$ . The measure of the union of these exceptional sets for $r \geq R$ is at most $\sum_{r \geq R} 2^{- r} \to 0$ ; outside their null limsup the successive differences are eventually bounded by $2^{- r}$ , so $u_{m_{r}}$ converges to a finite measurable limit, denoted $T f$ . For each $λ > 0$ , $1_{{∣ T f ∣ > λ}} \leq {lim inf}_{r} 1_{{∣ T f_{m_{r}} ∣ > λ}}$ almost everywhere. Fatou's lemma [F4] and step 2.1 yield $∣ {∣ T f ∣ > λ} ∣ \leq {lim inf}_{r} ∣ {∣ T f_{m_{r}} ∣ > λ} ∣ \leq D λ^{- 1} ∥ f ∥_{1}$ . The same difference estimate implies uniqueness of limits in measure and independence of the chosen $L^{1} \cap L^{2}$ approximants; it also proves linearity by approximating two inputs and their linear combination. For $f \in L^{1} \cap L^{2}$ these truncations converge in $L^{2}$ , so $T f$ agrees with the original operator. Thus the compatible linear extension satisfies the required weak $(1, 1)$ bound on all of $L^{1}$ .

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The Lp range: interpolation below two and adjoint duality above two

Statement

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

Let $T$ be a linear operator defined on $L^{1} (R^{n}) + L^{2} (R^{n})$ that is bounded on $L^{2} (R^{n})$ with norm $B$ , let $T^{*}$ denote its adjoint with respect to the $L^{2}$ pairing, and suppose that $T$ and $T^{*}$ both satisfy the weak $(1, 1)$ bound $∣ {∣ T f ∣ > λ} ∣ \leq A λ^{- 1} ∥ f ∥_{1}$ for every $f \in L^{1} (R^{n})$ and $λ > 0$ (and likewise for a compatible linear extension of $T^{*}$ to $L^{1} + L^{2}$ ). Then for every $1 < p < \infty$ , $p \neq 2$ , $T$ is bounded on $L^{p} (R^{n})$ : for $1 < p < 2$ the operator norm is at most $C_{p} : = [p (\frac{2 A}{p - 1} + \frac{4 B^{2}}{2 - p})]^{1 / p},$ and for $2 < p < \infty$ it is at most $C_{p^{'}}$ , the same expression evaluated at the conjugate exponent $p^{'} = p / (p - 1) \in (1, 2)$ .

Facts & Assumptions

Given: A linear operator $T$ on $L^{1} + L^{2}$ , bounded on $L^{2}$ with norm $B$ ; its $L^{2}$ adjoint $T^{*}$ ; weak $(1, 1)$ bounds with constant $A$ for both $T$ and $T^{*}$ ; an exponent $1 < p < \infty$ , $p \neq 2$ , with conjugate $p^{'}$ ; Countable Choice.

[F1]

$T^{*}$ is the bounded $L^{2}$ adjoint: $⟨ T f, g ⟩ = ⟨ f, T^{*} g ⟩$ for all $f, g \in L^{2} (R^{n}; C)$ with the first-variable-linear pairing $⟨ u, v ⟩ = \int u \overline{v}$ , and $T^{*}$ is likewise bounded on $L^{2}$ with norm $B$ (The Hilbert-space adjoint of a bounded operator, Complex Lp classes and Euclidean test-function conventions).

[F2]

Let $(X, μ)$ be $σ$ -finite, $1 < q < 2$ , and let $S$ be a sublinear operator on $L^{1} (X) + L^{2} (X)$ , weak $(1, 1)$ with constant $A$ and strong $(2, 2)$ with constant $B$ . Then $∥ S f ∥_{q} \leq [q (2 A / (q - 1) + 4 B^{2} / (2 - q))]^{1 / q} ∥ f ∥_{q}$ for every $f \in L^{q} (X)$ (Marcinkiewicz interpolation from weak (1,1) and strong (2,2)).

[F3]

Complex finite simple functions, and under Countable Choice also $C_{c}^{\infty} (R^{n}; C)$ , are dense in $L^{p} (R^{n}; C)$ for every $1 \leq p < \infty$ (Complex finite-simple and smooth compact-support density for finite p, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F4]

For $1 \leq p < \infty$ and conjugate $q$ , $∥ f ∥_{p} = \sup {∣ \int f g d μ ∣ : g \in L^{q}, ∥ g ∥_{q} \leq 1}$ (The $L^{p}$ norm is the supremum of pairings against unit $L^{q}$ functions); the Hölder and Minkowski inequalities for the complex $L^{p}$ spaces are recorded in Complex Holder, Minkowski, and the quotient norm, and the integral conventions in The class $L^{1} (μ)$ of integrable functions. Monotone convergence is Monotone convergence for the integral.

Proof

technique · direct

1.1F2given

Let $1 < q < 2$ and $f \in L^{q} \cap L^{2}$ . Then $f \in L^{1} + L^{2}$ and $T$ is defined at $f$ ; sublinearity is automatic for the linear $T$ , the weak $(1, 1)$ and strong $(2, 2)$ hypotheses are those assumed, so [F2] applies with $(X, μ) = (R^{n}, λ)$ and gives $∥ T f ∥_{q} \leq C_{q} ∥ f ∥_{q}$ with $C_{q} = [q (2 A / (q - 1) + 4 B^{2} / (2 - q))]^{1 / q}$ .

2.1F3step 1.1algebra

Consequently, for every $1 < q < 2$ the operator $T$ has a unique bounded extension to all of $L^{q} (R^{n}; C)$ with norm at most $C_{q}$ : since $C_{c}^{\infty} (R^{n}; C) \subseteq L^{q} \cap L^{2}$ is dense in $L^{q}$ by [F3], step 1.1 applied to differences of test functions shows that $g \mapsto T g$ is uniformly continuous on this dense subspace, so it extends uniquely to the closure with the same bound.

3.1F1step 2.1given

The adjoint $T^{*}$ is a bounded linear operator on $L^{2}$ with norm $B$ by [F1]; it satisfies the weak $(1, 1)$ bound with constant $A$ by hypothesis, and it is linear, hence sublinear. Therefore step 2.1 applies to $T^{*}$ at the exponent $p^{'} \in (1, 2)$ whenever $2 < p < \infty$ : for every $h \in L^{p^{'}} (R^{n}; C)$ , $∥ T^{*} h ∥_{p^{'}} \leq C_{p^{'}} ∥ h ∥_{p^{'}}$ .

4.1F1F4step 3.1algebra

Let $2 < p < \infty$ , $f \in L^{p} \cap L^{2}$ and $g \in L^{p^{'}} \cap L^{2}$ with $∥ g ∥_{p^{'}} \leq 1$ . Using $\overline{g} \in L^{p^{'}} \cap L^{2}$ and the adjoint identity of [F1] applied to the pair $(T f, \overline{g})$ , $∣ \int_{R^{n}} (T f) g d λ ∣ = ∣ ⟨ T f, \overline{g} ⟩ ∣ = ∣ ⟨ f, T^{*} \overline{g} ⟩ ∣ \leq ∥ f ∥_{p} ∥ T^{*} \overline{g} ∥_{p^{'}} \leq C_{p^{'}} ∥ f ∥_{p},$ where the first inequality is Hölder's inequality [F4] and the second is step 3.1 applied to $h = \overline{g}$ , which has $∥ \overline{g} ∥_{p^{'}} = ∥ g ∥_{p^{'}} \leq 1$ . To establish $T f \in L^{p}$ before using norm recovery, put $u = T f \in L^{2}$ , $E_{N} = B (0, N) \cap {∣ u ∣ \leq N}$ and $a_{N} = (\int_{E_{N}} ∣ u ∣^{p})^{1 / p}$ . If $a_{N} > 0$ , take $g_{N} = 1_{E_{N}} ∣ u ∣^{p - 1} θ_{u} / a_{N}^{p - 1}$ , with $θ_{u} = \overline{u} / ∣ u ∣$ on $u \neq 0$ and zero otherwise. This test is bounded on a finite-measure set, hence lies in $L^{p^{'}} \cap L^{2}$ , and satisfies $∥ g_{N} ∥_{p^{'}} = 1$ , $\int u g_{N} = a_{N}$ . The preceding pairing bound gives $a_{N} \leq C_{p^{'}} ∥ f ∥_{p}$ ; if $a_{N} = 0$ the same inequality is immediate. Since $E_{N}$ increases to a full-measure set, monotone convergence gives $∥ T f ∥_{p} \leq C_{p^{'}} ∥ f ∥_{p}$ .

5.1F3step 2.1step 4.1∎

If $2 < p < \infty$ , step 4.1 and the density [F3] of $C_{c}^{\infty} \subseteq L^{p} \cap L^{2}$ in $L^{p}$ extend the bound to all $f \in L^{p}$ with the same constant $C_{p^{'}}$ , exactly as in step 2.1. Together with step 2.1 for $1 < p < 2$ , this proves the asserted bounds for every $1 < p < \infty$ , $p \neq 2$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Calderón–Zygmund operators are bounded on Lp

Statement

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

Let $T$ be a Calderón–Zygmund operator with kernel constants $A_{1}, A_{2}$ and $L^{2}$ norm $B$ . Then for every $1 < p < \infty$ , $T$ extends uniquely to a bounded operator on $L^{p} (R^{n})$ with $∥ T f ∥_{p} \leq C_{n, p} (A_{2} + B) \max (p, (p - 1)^{- 1}) ∥ f ∥_{p},$ where $C_{n, p}$ depends only on $n$ and $p$ . In fact $C_{n, p}$ may be chosen to be a dimensional constant $C_{n}$ independent of $p$ .

Facts & Assumptions

Given: A Calderón–Zygmund operator $T$ with kernel $k$ , constants $A_{1}, A_{2}$ and $L^{2}$ norm bound $B$ ; an exponent $1 0$ ; Countable Choice.

[F1]

$T$ is linear, $L^{2}$ -bounded with $∥ T h ∥_{2} \leq B ∥ h ∥_{2}$ , and satisfies the off-support representation with kernel $k$ off the support of compactly supported $L^{2}$ inputs; the kernel obeys the annular bound $A_{1}$ and Hörmander's condition $A_{2}$ , the latter invariant under reflection: $k^{*} (x) : = \overline{k (- x)}$ also satisfies both bounds with the same constants (Calderón–Zygmund kernels and their associated operators).

[F2]

$T$ is of weak type $(1, 1)$ with constant $C_{n} (A_{2} + B)$ : $∣ {∣ T h ∣ > λ} ∣ \leq C_{n} (A_{2} + B) λ^{- 1} ∥ h ∥_{1}$ for all $h \in L^{1}$ , and $∥ T h ∥_{2} \leq B ∥ h ∥_{2}$ for $h \in L^{2}$ (Calderón–Zygmund operators are of weak type (1,1)); weak and strong type are as in Sublinear operators and weak or strong type $(p, q)$ bounds.

[F3]

Chebyshev: $∣ {∣ u ∣ > t} ∣ \leq t^{- 2} \int ∣ u ∣^{2}$ for measurable $u$ ; layer cake: $∥ g ∥_{q}^{q} = q \int_{0}^{\infty} t^{q - 1} ∣ {∣ g ∣ > t} ∣ d t$ ; Fubini applies to absolutely integrable complex kernels (Fubini's theorem for L^1 functions on a sigma-finite product); Tonelli applies to nonnegative product-measurable integrands on $σ$ -finite products (Chebyshev-Markov inequality for the integral, For 0 < p < infinity, the layer-cake formula computes the integral of |f|^p from the distribution function, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product); the $L^{p}$ and $L^{p^{'}}$ norms are the quotient norms of The space $L^{p} (μ)$ as the quotient by null functions, Hölder's inequality is Complex Holder, Minkowski, and the quotient norm, and $C_{c}^{\infty}$ is dense in $L^{p}$ for finite $p$ (Complex finite-simple and smooth compact-support density for finite p).

[F4]

For $1 \leq p < \infty$ with conjugate $q$ , $∥ f ∥_{p} = \sup {∣ \int f g ∣ : g \in L^{q}, ∥ g ∥_{q} \leq 1}$ (The $L^{p}$ norm is the supremum of pairings against unit $L^{q}$ functions); the $L^{2}$ adjoint satisfies $⟨ T h, g ⟩ = ⟨ h, T^{*} g ⟩$ for the first-variable-linear pairing (The Hilbert-space adjoint of a bounded operator); Monotone convergence is Monotone convergence for the integral; the interpolation-and-duality lemma with explicit constants is The Lp range: interpolation below two and adjoint duality above two.

Proof

technique · direct

1.1F1F2F4givenalgebra

The adjoint kernel $k^{*} (x) = \overline{k (- x)}$ satisfies the annular bound and Hörmander's condition with the constants $A_{1}, A_{2}$ : the annular integral of $∣ k^{*} ∣$ is that of $∣ k ∣$ under the reflection $x \mapsto - x$ , and $∣ k^{*} (x - y) - k^{*} (x) ∣ = ∣ k (y - x) - k (- x) ∣ = ∣ \overline{k (y - x)} - \overline{k (- x)} ∣$ , whose integral over $∣ x ∣ \geq 2 ∣ y ∣$ equals $\int_{∣ u ∣ \geq 2 ∣ y ∣} ∣ k (u + y) - k (u) ∣ d u \leq A_{2}$ by the substitution $u = - x$ and the Hörmander condition applied to $- y$ . The off-support representation for $T^{*}$ also follows from that of $T$ . For compactly supported $h \in L^{2}$ and a compact set $E$ disjoint from its support, $u (y) = \int \overline{k (x - y)} h (x) d x$ is absolutely convergent for almost every $y \in E$ : integrating its absolute majorant over $E$ is bounded by $∥ h ∥_{1} \int_{E - supp h} ∣ k (- z) ∣ d z < \infty$ , since the difference set is compact and avoids zero. For a bounded test $g$ supported in $E$ , Fubini in the kernel formula for $T g$ gives $⟨ T g, h ⟩ = ⟨ g, u ⟩$ ; absolute integrability follows from the same majorant times $∥ g ∥_{\infty}$ . The adjoint identity then says $T^{*} h = u$ almost everywhere on $E$ , since both are locally integrable and agree against all such tests. Exhausting the complement of the support by countably many compact sets proves exactly the required representation with $k^{*}$ . Hence $T^{*}$ , which is $L^{2}$ -bounded with norm $B$ by [F4], is again a Calderón–Zygmund operator with constants $A_{1}, A_{2}, B$ , and by [F2] both $T$ and $T^{*}$ are weak $(1, 1)$ with constant $A : = C_{n} (A_{2} + B)$ .

1.2F1F2F3algebra

Sharp two-level interpolation. Let $S$ be any linear operator defined on $L^{1} + L^{2}$ , weak $(1, 1)$ with constant $A$ and $L^{2}$ -bounded with constant $B$ , and let $1 < q < 2$ . For $f \in L^{q}$ and any $δ > 0$ put $f_{0} : = f 1_{{∣ f ∣ > δ t}}$ and $f_{1} : = f 1_{{∣ f ∣ \leq δ t}}$ at height $t > 0$ ; then $f_{0} \in L^{1}$ and $f_{1} \in L^{2}$ , since $∥ f_{0} ∥_{1} \leq (δ t)^{1 - q} ∥ f ∥_{q}^{q}$ and $∥ f_{1} ∥_{2}^{2} \leq (δ t)^{2 - q} ∥ f ∥_{q}^{q}$ . Hence ${∣ S f ∣ > t} \subseteq {∣ S f_{0} ∣ > t / 2} \cup {∣ S f_{1} ∣ > t / 2}$ and the weak $(1, 1)$ bound for $S f_{0}$ together with Chebyshev and the $L^{2}$ bound for $S f_{1}$ give $∣ {∣ S f ∣ > t} ∣ \leq \frac{2 A}{t} \int_{{∣ f ∣ > δ t}} ∣ f ∣ + \frac{4 B^{2}}{t^{2}} \int_{{∣ f ∣ \leq δ t}} ∣ f ∣^{2}$ . Integrating $t^{q - 1} ∣ {∣ S f ∣ > t} ∣$ over $(0, \infty)$ and exchanging the integrals by Tonelli gives $∥ S f ∥_{q}^{q} \leq q [\frac{2 A δ^{1 - q}}{q - 1} + \frac{4 B^{2} δ^{2 - q}}{2 - q}] ∥ f ∥_{q}^{q}$ , because $\int_{0}^{∣ f ∣ / δ} t^{q - 2} d t = \frac{(∣ f ∣ / δ)^{q - 1}}{q - 1}$ and $\int_{∣ f ∣ / δ}^{\infty} t^{q - 3} d t = \frac{(∣ f ∣ / δ)^{q - 2}}{2 - q}$ ; choosing $δ : = A / (2 B^{2})$ when $A, B > 0$ balances the two terms at a constant multiple of $A^{2 - q} B^{2 q - 2}$ , so that $∥ S f ∥_{q} \leq C_{q} A^{\frac{2}{q} - 1} B^{2 - \frac{2}{q}} ∥ f ∥_{q} \leq C_{q} (A + B) ∥ f ∥_{q},$ the last inequality because the exponents $\frac{2}{q} - 1$ and $2 - \frac{2}{q}$ are nonnegative and sum to one, so the weighted geometric mean is at most the sum; if $A = 0$ , let $δ ↓ 0$ , and if $B = 0$ , let $δ \to \infty$ in the preceding inequality, obtaining $S f = 0$ in either case.

2.1F2step 1.1step 1.2algebra

The case $1 < p < 2$ : by step 1.1 the operator $T$ has weak $(1, 1)$ constant $A = C_{n} (A_{2} + B)$ and $L^{2}$ norm $B$ , so step 1.2 with $q = p$ gives $∥ T f ∥_{p} \leq C_{p} A^{2 / p - 1} B^{2 - 2 / p} ∥ f ∥_{p} \leq C_{n, p} (A_{2} + B) ∥ f ∥_{p} \leq C_{n, p} (A_{2} + B) \max (p, (p - 1)^{- 1}) ∥ f ∥_{p}$ for every $f \in L^{p}$ (which lies in $L^{1} + L^{2}$ by the canonical split of step 1.2, so $T f$ is defined).

2.2F3F4step 1.1step 1.2algebra

The case $2 < p < \infty$ : apply step 1.2 with $q = p^{'}$ to the adjoint $T^{*}$ , which by step 1.1 has weak $(1, 1)$ constant $A$ and $L^{2}$ norm $B$ : for every $g \in L^{p^{'}}$ , $∥ T^{*} g ∥_{p^{'}} \leq C_{p^{'}} (A_{2} + B) ∥ g ∥_{p^{'}}$ . Then for $f \in L^{p} \cap L^{2}$ and $g \in L^{p^{'}} \cap L^{2}$ with $∥ g ∥_{p^{'}} \leq 1$ , the adjoint identity and Hölder's inequality give $∣ \int (T f) g ∣ = ∣ ⟨ f, T^{*} \overline{g} ⟩ ∣ \leq ∥ f ∥_{p} ∥ T^{*} \overline{g} ∥_{p^{'}} \leq C_{p^{'}} (A_{2} + B) ∥ f ∥_{p}$ , To establish $T f \in L^{p}$ before using norm recovery, put $u = T f \in L^{2}$ , $E_{N} = B (0, N) \cap {∣ u ∣ \leq N}$ and $a_{N} = (\int_{E_{N}} ∣ u ∣^{p})^{1 / p}$ . If $a_{N} > 0$ , take $g_{N} = 1_{E_{N}} ∣ u ∣^{p - 1} θ_{u} / a_{N}^{p - 1}$ , with $θ_{u} = \overline{u} / ∣ u ∣$ on $u \neq 0$ and zero otherwise. This test is bounded on a finite-measure set, hence lies in $L^{p^{'}} \cap L^{2}$ , and satisfies $∥ g_{N} ∥_{p^{'}} = 1$ , $\int u g_{N} = a_{N}$ . The preceding pairing bound gives $a_{N} \leq C_{p^{'}} (A_{2} + B) ∥ f ∥_{p}$ ; if $a_{N} = 0$ the same inequality is immediate. Since $E_{N}$ increases to a full-measure set, monotone convergence gives $∥ T f ∥_{p} \leq C_{p^{'}} (A_{2} + B) ∥ f ∥_{p}$ ; since $C_{c}^{\infty} \subseteq L^{p} \cap L^{2}$ is dense in $L^{p}$ , this bound extends uniquely to all of $L^{p}$ , and $C_{p^{'}} (A_{2} + B) \leq C_{n, p} (A_{2} + B) \max (p, (p - 1)^{- 1})$ .

3.1F2F3F4step 1.1step 1.2step 2.2algebra

The exponent dependence can be made dimension-only. Put $D = A_{2} + B$ . If $D = 0$ , then $B = 0$ and $T = 0$ . Otherwise normalize $S = T / D$ . Step 1.1 and the weak endpoint give weak $(1, 1)$ constants at most $a_{n} = C_{n}$ for both $S$ and $S^{*}$ , and their $L^{2}$ norms are at most one. The interpolation-and-duality lemma [F4] at $q = 3 / 2$ and its conjugate $3$ gives $∥ S ∥_{3 \to 3}, ∥ S^{*} ∥_{3 \to 3} \leq d_{n}$ with $d_{n}$ depending only on $n$ . For $1 t}} + f 1_{{∣ f ∣ \leq t}}$ . Weak $(1, 1)$ on the first part and Chebyshev with the strong $(3, 3)$ bound on the second yield $∣ {∣ S f ∣ > t} ∣ \leq 2 a_{n} t^{- 1} \int_{∣ f ∣ > t} ∣ f ∣ + 8 d_{n}^{3} t^{- 3} \int_{∣ f ∣ \leq t} ∣ f ∣^{3}$ . Layer cake and Tonelli, as in step 1.2, give $∥ S f ∥_{p}^{p} \leq p [2 a_{n} / (p - 1) + 8 d_{n}^{3} / (3 - p)] ∥ f ∥_{p}^{p} \leq K_{n} (p - 1)^{- 1} ∥ f ∥_{p}^{p}$ , with $K_{n} \geq 1$ independent of $p$ . The compatible $L^{1}$ and $L^{3}$ actions agree on their intersection by approximation with bounded compact-support functions in both norms. Thus $∥ S ∥_{p \to p} \leq K_{n} / (p - 1)$ for $1 2$ . Consequently $∥ T ∥_{p \to p} \leq K_{n} D \max (p, (p - 1)^{- 1})$ throughout the strict range.

4.1F3step 2.1step 2.2step 3.1∎

Steps 2.1 and 2.2 prove boundedness and uniqueness in the two open ranges; the given $L^{2}$ bound handles $p = 2$ . Step 3.1 also proves the stronger bound with a dimensional constant independent of $p$ , and hence the stated estimate.

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Maximal truncated singular integrals

Definition

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

Fix an integer $n \geq 1$ and let $k : R^{n} ∖ {0} \to C$ be a measurable function that is integrable on compact subsets of $R^{n} ∖ {0}$ and satisfies the pointwise size bound $∣ k (x) ∣ \leq A_{1} ∣ x ∣^{- n} (x \neq 0) (1)$ for some finite constant $0 \leq A_{1} < \infty$ . With the complex $L^{p}$ conventions of Complex Lp classes and Euclidean test-function conventions, fix $1 \leq p < \infty$ and $f \in L^{p} (R^{n})$ .

For $0 < ε < \infty$ define the truncated singular integral $T_{ε} f (x) : = \int_{∣ y ∣ > ε} k (y) f (x - y) d y, x \in R^{n},$ and for $0 < ε < N < \infty$ define the doubly truncated singular integral $T^{(ε, N)} f (x) : = \int_{ε < ∣ y ∣ < N} k (y) f (x - y) d y, x \in R^{n} .$ Both integrals converge absolutely for every $x$ . Indeed, the truncated kernel $k_{ε} : = k 1_{{∣ \cdot ∣ > ε}}$ satisfies $∥ k_{ε} ∥_{p^{'}}^{p^{'}} = \int_{∣ y ∣ > ε} ∣ k (y) ∣^{p^{'}} d y \leq A_{1}^{p^{'}} \int_{∣ y ∣ > ε} ∣ y ∣^{- n p^{'}} d y = A_{1}^{p^{'}} ∣ S^{n - 1} ∣ \frac{ε^{n - n p^{'}}}{n p^{'} - n} < \infty$ by Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma for $1 < p < \infty$ (the case $p = 1$ , $p^{'} = \infty$ , uses $∥ k_{ε} ∥_{\infty} \leq A_{1} ε^{- n}$ ), and likewise $∥ k 1_{{ε < ∣ \cdot ∣ < N}} ∥_{p^{'}} < \infty$ ; Hölder's inequality (Complex Holder, Minkowski, and the quotient norm) therefore gives $∣ T_{ε} f (x) ∣ \leq ∥ f ∥_{p} ∥ k_{ε} ∥_{p^{'}} < \infty$ and $∣ T^{(ε, N)} f (x) ∣ \leq ∥ f ∥_{p} ∥ k 1_{{ε < ∣ \cdot ∣ < N}} ∥_{p^{'}} < \infty$ at every $x$ . The maximal truncated singular integral and the doubly truncated maximal singular integral are $T^{*} f (x) : = \sup_{ε > 0} ∣ T_{ε} f (x) ∣, T^{* *} f (x) : = \sup_{0 < ε < N < \infty} ∣ T^{(ε, N)} f (x) ∣,$ with values in $[0, \infty]$ .

Under the size bound (1) the two maximal operators are pointwise comparable: $T^{*} f \leq T^{* *} f \leq 2 T^{*} f pointwise on R^{n} . (2)$ For the left inequality, fix $ε > 0$ and $x$ : dominated convergence (Dominated convergence) applied to the absolutely convergent integral defining $T_{ε} f (x)$ gives $T_{ε} f (x) = \lim_{N \to \infty} T^{(ε, N)} f (x)$ , so $∣ T_{ε} f (x) ∣ \leq T^{* *} f (x)$ ; take the supremum in $ε$ . For the right inequality, split the defining integral of $T^{(ε, N)}$ at $N$ to get $T^{(ε, N)} f (x) = T_{ε} f (x) - T_{N} f (x)$ and hence $∣ T^{(ε, N)} f (x) ∣ \leq ∣ T_{ε} f (x) ∣ + ∣ T_{N} f (x) ∣ \leq 2 T^{*} f (x)$ ; take the supremum in $ε < N$ . Thus $T^{*}$ and $T^{* *}$ have the same finiteness set and the same boundedness properties. The definition itself asserts neither an upper truncation for $T^{*}$ nor the existence of a principal-value limit $\lim_{ε ↓ 0} T_{ε} f (x)$ ; the doubly truncated form is the primitive object, because it is defined from the kernel alone. Here the pointwise size bound (1) is a stronger hypothesis than the annular size condition of the Calderón–Zygmund kernel definition: (1) implies $\int_{R \leq ∣ x ∣ \leq 2 R} ∣ k ∣ \leq A_{1} ∣ S^{n - 1} ∣ \log 2$ , while the annular condition does not by itself prevent pointwise spikes. Countable Choice is inherited from the polar-coordinate evaluation; the truncations themselves require no selection.

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Cotlar's inequality for maximal truncations

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $k : R^{n} ∖ {0} \to C$ satisfy the pointwise size bound $∣ k (x) ∣ \leq A_{1} ∣ x ∣^{- n}$ , the $δ$ -Hölder smoothness bound $∣ k (x - y) - k (x) ∣ \leq A_{2}^{'} ∣ y ∣^{δ} ∣ x ∣^{- n - δ}$ for $∣ x ∣ \geq 2 ∣ y ∣ > 0$ with $0 < δ \leq 1$ , 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 extending $k$ (Calderón–Zygmund kernels and their associated operators) and let $T$ be the convolution operator with $W$ , bounded on $L^{2} (R^{n})$ . Then for every $f \in S (R^{n})$ and almost every $x$ , $T^{*} f (x) \leq M (T f) (x) + C_{n, δ} (A_{1} + A_{2}^{'} + A_{3}) M f (x),$ where $M$ is the centered Hardy–Littlewood maximal operator of The centered and uncentered Hardy-Littlewood maximal functions and $T^{*}$ is the maximal truncated operator of Maximal truncated singular integrals.

Facts & Assumptions

Given: Countable Choice; $n \geq 1$ ; $0 < δ \leq 1$ ; a kernel $k$ with the size, Hölder and cancellation bounds; a principal-value distribution $W$ extending $k$ ; the convolution operator $T$ with $W$ , bounded on $L^{2}$ ; a Schwartz function $f$ ; a point $x \in R^{n}$ ; a scale $ε > 0$ ; the canonical dimension-dependent nonnegative radially nonincreasing $φ \in C_{c}^{\infty} (R^{n})$ with $\int φ = 1$ and $supp φ \subseteq B (0, 1 / 2)$ , and its mollifiers $φ_{ε} (y) = ε^{- n} φ (y / ε)$ (The mollifier family generated by a unit-mass smooth bump; the approximate-identity properties are recorded in A unit-mass smooth bump generates an $L^{1}$ approximate identity).

[F1]

$T_{ε} f (x) = \int_{∣ y ∣ \geq ε} k (y) f (x - y) d y$ is absolutely convergent and $T^{*} f (x) = \sup_{ε > 0} ∣ T_{ε} f (x) ∣$ (Maximal truncated singular integrals).

[F2]

For $u \in S^{'}$ and $ψ \in S$ , $(u * ψ) (x) = ⟨ u_{y}, ψ (x - y) ⟩$ defines a smooth function of polynomial growth, and $T f = W * f$ is the convolution of the tempered distribution $W$ with the Schwartz function $f$ (Convolution of a tempered distribution with a schwartz function, Tempered convolution is smooth with polynomial growth).

[F3]

A principal-value distribution $W$ for $k$ has a sequence $δ_{j} ↓ 0$ such that it satisfies $⟨ W, ψ ⟩ = \lim_{j \to \infty} \int_{∣ z ∣ \geq δ_{j}} k (z) ψ (z) d z$ for every $ψ \in S (R^{n})$ (Calderón–Zygmund kernels and their associated operators).

[F4]

If $ω \geq 0$ is measurable, radially nonincreasing and integrable and $g \in L_{l o c}^{1}$ , then $\int ∣ g (x - y) ∣ ω (y) d y \leq ∥ ω ∥_{1} M g (x)$ for every $x$ (Radially decreasing kernels are dominated by the maximal function).

[F5]

Convolution of functions $g, h$ on $R^{n}$ is $g * h (x) = \int g (x - y) h (y) d y$ , whenever the integral converges absolutely (Convolution of two functions on $R^{n}$ ); on $σ$ -finite products a nonnegative product-measurable integrand may be integrated in either order (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product); the Schwartz conventions are those of Schwartz space and its seminorms.

[F6]

Exponentials dominate every fixed polynomial at positive infinity, and Euclidean balls of positive radius have positive finite Lebesgue measure. (The exponential dominates every fixed nonnegative integer power at $+ \infty$ , Euclidean balls have positive finite Lebesgue measure)

Proof

1.1F2F3F6givenconstruct

Fix the auxiliary bump once as a function of dimension only: let $a (t) = e^{- 1 / t}$ for $t > 0$ and $a (t) = 0$ otherwise, put $ρ (y) = a (1 - 16 ∣ y ∣^{2})$ and $φ (y) = c_{n} ρ (y)$ with $c_{n} = (\int ρ)^{- 1}$ . The derivatives of $a$ on $t > 0$ have the form $P_{k} (1 / t) e^{- 1 / t}$ , with $P_{k + 1} (s) = s^{2} (P_{k} (s) - P_{k}^{'} (s))$ ; [F6] makes each derivative and its difference quotient tend to zero at $t = 0$ , proving smoothness across that point. Thus $ρ$ is smooth, supported in the closed radius- $1 / 4$ ball, nonnegative and radially nonincreasing since $a^{'} (t) \geq 0$ . Its mass is finite by boundedness and compact support, and positive since it is bounded below by $a (3 / 4) > 0$ on the radius- $1 / 8$ ball, which has positive measure by [F6]. This gives the required unit-mass bump with support inside $B (0, 1 / 2)$ . All its derivative bounds and $c_{n}$ depend only on $n$ . For fixed $ε > 0$ put $k^{(ε)} = k 1_{{∣ \cdot ∣ \geq ε}}$ and $R_{ε} = k^{(ε)} - W * φ_{ε}$ . By [F2,F3], $W * φ_{ε}$ is smooth and equals the principal-value limit $\lim_{j \to \infty} \int_{∣ z ∣ \geq δ_{j}} k (z) φ_{ε} (x - z) d z$ . This need not be an absolutely convergent integral near $z = 0$ . If $∣ x ∣ \geq 2 ε$ , the support condition $∣ x - z ∣ \leq ε / 2$ implies $∣ z ∣ \geq 3 ε / 2$ , so in that region the same formula is an ordinary absolutely convergent integral. Near the origin retain the principal-value limit and use the cancellation estimate in the next step.

1.2F3givenalgebra

Case $∣ x ∣ < 2 ε$ . Write $(W * φ_{ε}) (x) = \lim_{j \to \infty} (I_{1}^{(δ)} + I_{2}^{(δ)} + I_{3}^{(δ)})$ along $δ = δ_{j}$ for all sufficiently large $j$ such that $0 < δ_{j} < ε / 4$ , with the three pieces obtained by inserting $φ_{ε} (y) = φ_{ε} (x) + (φ_{ε} (y) - φ_{ε} (x))$ and splitting at $∣ x - y ∣ = ε / 4$ : $I_{1}^{(δ)} = \int_{∣ x - y ∣ > ε / 4} k (x - y) φ_{ε} (y) d y$ , $I_{2}^{(δ)} = \int_{δ \leq ∣ x - y ∣ \leq ε / 4} k (x - y) (φ_{ε} (y) - φ_{ε} (x)) d y$ , and $I_{3}^{(δ)} = φ_{ε} (x) \int_{δ \leq ∣ x - y ∣ \leq ε / 4} k (x - y) d y$ ; the three pieces are absolutely convergent and their sum is the truncation $\int_{∣ x - y ∣ \geq δ} k (x - y) φ_{ε} (y) d y$ . Here $∣ I_{1}^{(δ)} ∣ \leq A_{1} (4 / ε)^{n} \int φ_{ε} = A_{1} 4^{n} ε^{- n}$ because $∣ k (x - y) ∣ \leq A_{1} ∣ x - y ∣^{- n}$ on the domain; $∣ I_{2}^{(δ)} ∣ \leq ∥ \nabla φ_{ε} ∥_{\infty} A_{1} \int_{∣ x - y ∣ \leq ε / 4} ∣ x - y ∣^{- n} ∣ x - y ∣ d y \leq C_{φ} A_{1} ε^{- n}$ by the mean value theorem, the bound $∣ k ∣ \leq A_{1} ∣ \cdot ∣^{- n}$ and polar coordinates of the punctured ball; and $∣ I_{3}^{(δ)} ∣ \leq ∥ φ_{ε} ∥_{\infty} A_{3} \leq C_{φ} A_{3} ε^{- n}$ by the cancellation bound after the substitution $z = x - y$ . Finally $∣ k^{(ε)} (x) ∣ \leq A_{1} ∣ x ∣^{- n} 1_{{∣ x ∣ \geq ε}} \leq A_{1} ε^{- n}$ . Hence $∣ R_{ε} (x) ∣ \leq C_{n, φ} (A_{1} + A_{3}) ε^{- n} \leq C_{n, δ, φ} (A_{1} + A_{2}^{'} + A_{3}) ε^{δ} (ε + ∣ x ∣)^{- n - δ}$ , the last inequality because $∣ x ∣ < 2 ε$ gives $(ε + ∣ x ∣)^{- n - δ} \geq (3 ε)^{- n - δ}$ .

2.1F3givenalgebra

Case $∣ x ∣ \geq 2 ε$ of the error bound. Since $φ_{ε}$ is supported in $∣ y ∣ \leq ε / 2$ , for $∣ x ∣ \geq 2 ε$ one has $∣ x ∣ \geq 2 ∣ y ∣$ on the support, so $k^{(ε)} (x) = k (x)$ and, substituting $z = x - y$ in the formula of step 1.1, $R_{ε} (x) = k (x) - \int k (x - y) φ_{ε} (y) d y = \int [k (x) - k (x - y)] φ_{ε} (y) d y,$ whence the Hölder bound gives $∣ R_{ε} (x) ∣ \leq A_{2}^{'} \int ∣ y ∣^{δ} φ_{ε} (y) d y ∣ x ∣^{- n - δ} = A_{2}^{'} C_{φ} ε^{δ} ∣ x ∣^{- n - δ} \leq A_{2}^{'} C_{φ} 2^{n + δ} ε^{δ} (ε + ∣ x ∣)^{- n - δ}$ .

3.1F1F2F5step 1.2step 2.1algebra

The convolution identity is $T_{ε} f = ((T f) * φ_{ε}) + f * R_{ε}$ . Indeed, by the bounds in steps 1.2 and 2.1, $R_{ε}$ is integrable and its convolution with $f$ converges absolutely; $k^{(ε)} * f = T_{ε} f$ also converges absolutely by the size bound and Schwartz decay. For the remaining term use the distribution pairing rather than interchange nonabsolute kernel integrals. The compactly supported integral $\int φ_{ε} (y) f (x - y - \cdot) d y$ converges in every Schwartz seminorm: all derivatives of $f$ decay rapidly, uniformly over $y$ in the fixed compact support. Continuity of the tempered distribution $W$ therefore permits its pairing to pass through that integral. This gives $(W * f) * φ_{ε} = W * (f * φ_{ε})$ . Applying the same argument to $\int f (x - z) φ_{ε} (z - \cdot) d z$ gives $f * (W * φ_{ε}) = W * (f * φ_{ε})$ : its Schwartz seminorms are bounded by integrals of $∣ f (x - z) ∣ (1 + ∣ z ∣)^{N}$ , finite for every $N$ . Hence $f * (k^{(ε)} - R_{ε}) = (W * f) * φ_{ε}$ , proving the identity.

3.2step 2.1step 1.2algebra

Steps 2.1 and 1.2 together show that for every $ε > 0$ and every $x \in R^{n}$ , $∣ R_{ε} (x) ∣ \leq C_{n, δ} (A_{1} + A_{2}^{'} + A_{3}) ε^{δ} (ε + ∣ x ∣)^{- n - δ} = C_{n, δ} (A_{1} + A_{2}^{'} + A_{3}) ω_{ε} (x),$ where $ω (x) : = (1 + ∣ x ∣)^{- n - δ}$ , $ω_{ε} (y) = ε^{- n} ω (y / ε)$ , and $ω$ is nonnegative, radially nonincreasing and integrable; note $ε^{δ} (ε + ∣ x ∣)^{- n - δ} = ε^{- n} (1 + ∣ x ∣ / ε)^{- n - δ}$ .

4.1F4step 3.1givenalgebra

First term bound: $∣ ((T f) * φ_{ε}) (x) ∣ \leq \int ∣ T f (x - y) ∣ φ_{ε} (y) d y \leq ∥ φ_{ε} ∥_{1} M (T f) (x) = M (T f) (x)$ , using that $φ_{ε}$ is radially nonincreasing with $\int φ_{ε} = 1$ and applying the domination lemma [F4] with $g = T f$ (a smooth function of polynomial growth, hence locally integrable).

4.2F4step 3.2givenalgebra

Second term bound: $∣ (f * R_{ε}) (x) ∣ \leq \int ∣ f (x - y) ∣ ∣ R_{ε} (y) ∣ d y \leq C_{n, δ} (A_{1} + A_{2}^{'} + A_{3}) \int ∣ f (x - y) ∣ ω_{ε} (y) d y \leq C_{n, δ} (A_{1} + A_{2}^{'} + A_{3}) ∥ ω ∥_{1} M f (x)$ by step 3.2 and the domination lemma [F4] applied to $ω_{ε}$ , which is radially nonincreasing, integrable with $∥ ω_{ε} ∥_{1} = ∥ ω ∥_{1}$ .

5.1F1step 4.1step 4.2∎

For every $ε > 0$ and every $x$ , steps 3.1, 4.1 and 4.2 give $∣ T_{ε} f (x) ∣ \leq M (T f) (x) + C_{n, δ} (A_{1} + A_{2}^{'} + A_{3}) ∥ ω ∥_{1} M f (x)$ ; taking the supremum over $ε > 0$ and using [F1] yields the asserted inequality with $C_{n, δ}$ redefined to absorb $∥ ω ∥_{1} = \int (1 + ∣ y ∣)^{- n - δ} d y < \infty$ , in particular for almost every $x$ .

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Maximal truncations: weak (1,1) and strong Lp bounds

Statement

Assume Countable Choice. Let $n \geq 1$ , $0 < δ \leq 1$ , and let $k$ , $W$ , $T$ satisfy the hypotheses of Cotlar's inequality for maximal truncations: $k$ is measurable and locally integrable on $R^{n} ∖ {0}$ with $∣ k (x) ∣ \leq A_{1} ∣ x ∣^{- n}, ∣ k (x - y) - k (x) ∣ \leq A_{2}^{'} ∣ y ∣^{δ} ∣ x ∣^{- n - δ} (∣ x ∣ \geq 2 ∣ y ∣ > 0),$ and cancellation $\sup_{0 < r < R} ∣ \int_{r < ∣ x ∣ < R} k (x) d x ∣ \leq A_{3}$ ; $W$ is a principal-value distribution for $k$ ; and $T$ , the convolution operator with $W$ , is $L^{2}$ -bounded with norm $B$ and satisfies the off-support representation (3) of Calderón–Zygmund kernels and their associated operators with kernel $k$ (so $T$ is also a Calderón–Zygmund operator with kernel $k$ ). Then $T^{* *}$ and $T^{*}$ are of weak type $(1, 1)$ : there is a constant $C_{n, δ}$ , depending only on $n$ and on the fixed exponent $δ$ , with $∣ {∣ T^{* *} f ∣ > λ} ∣ \leq C_{n, δ} (A_{1} + A_{2}^{'} + A_{3} + B) λ^{- 1} ∥ f ∥_{1} (f \in L^{1} (R^{n}), λ > 0),$ and for every $1 < p < \infty$ there is a constant $C_{n, p, δ}$ with $∥ T^{* *} f ∥_{p} \leq C_{n, p, δ} (A_{1} + A_{2}^{'} + A_{3} + B) \max (p, (p - 1)^{- 1}) ∥ f ∥_{p} (f \in L^{p} (R^{n})) .$ The same bounds hold for $T^{*}$ , which satisfies $T^{*} f \leq T^{* *} f$ pointwise. The proof consumes the Hörmander constant $A_{2} = ∣ S^{n - 1} ∣ 2^{- δ} δ^{- 1} A_{2}^{'}$ supplied by the standard $δ$ -Hölder bound; since $A_{2} \leq C_{n, δ} A_{2}^{'}$ , this is why the constants depend on the fixed exponent $δ$ .

Facts & Assumptions

Given: Countable Choice; $n \geq 1$ , $0 < δ \leq 1$ , finite constants $A_{1}, A_{2}^{'}, A_{3}, B \geq 0$ ; the kernel $k$ , principal-value distribution $W$ and $L^{2}$ -bounded convolution operator $T$ with off-support representation as in the statement; $f \in L^{1} (R^{n})$ and $λ > 0$ ; a height $μ = γ λ$ with $γ > 0$ to be fixed; the centered Hardy–Littlewood maximal operator $M$ .

[F1]

For $1 \leq p < \infty$ and $g \in L^{p}$ the integrals defining $T_{ε} g$ and $T^{(ε, N)} g$ converge absolutely at every point, $T^{*} g = \sup_{ε > 0} ∣ T_{ε} g ∣$ , $T^{* *} g = \sup_{0 < ε < N < \infty} ∣ T^{(ε, N)} g ∣$ , and $T^{*} g \leq T^{* *} g \leq 2 T^{*} g$ pointwise (Maximal truncated singular integrals).

[F2]

Cotlar's inequality: for every $u \in S (R^{n})$ and almost every $x$ , $T^{*} u (x) \leq M (T u) (x) + C_{0} (A_{1} + A_{2}^{'} + A_{3}) M u (x)$ with a constant $C_{0} = C_{n, δ}$ (Cotlar's inequality for maximal truncations).

[F3]

Calderón–Zygmund decomposition at height $μ$ : $f = g + \sum_{j} b_{j}$ almost everywhere, with the maximal dyadic cubes $Q_{j}$ pairwise disjoint, $\sum_{j} ∣ Q_{j} ∣ \leq μ^{- 1} ∥ f ∥_{1}$ , $b_{j}$ supported in $Q_{j}$ with $\int b_{j} = 0$ and $∥ b_{j} ∥_{1} \leq 2^{n + 1} μ ∣ Q_{j} ∣$ , and the good part satisfying $∥ g ∥_{1} \leq ∥ f ∥_{1}$ , $∣ g ∣ \leq 2^{n} μ$ and $∥ g ∥_{2}^{2} \leq 2^{n} μ ∥ f ∥_{1}$ (Calderón–Zygmund decomposition at height λ).

[F4]

The pointwise size bound gives the annular condition with $A_{1} ∣ S^{n - 1} ∣ \log 2$ , and the standard $δ$ -Hölder bound gives Hörmander's condition with $A_{2} = ∣ S^{n - 1} ∣ 2^{- δ} δ^{- 1} A_{2}^{'}$ ; hence $k$ is a Calderón–Zygmund kernel in the base sense and, since $T$ is a Calderón–Zygmund operator with kernel $k$ , $T$ extends uniquely to a bounded operator on $L^{p}$ for $1 < p < \infty$ with $∥ T u ∥_{p} \leq C_{n, p} (A_{2} + B) \max (p, (p - 1)^{- 1}) ∥ u ∥_{p}$ (Standard Hölder kernels satisfy the Hörmander condition, Standard (Hölder) Calderón–Zygmund kernels, Calderón–Zygmund operators are bounded on Lp).

[F5]

$M$ is the centered Hardy–Littlewood maximal operator: $∣ {M u > t} ∣ \leq C_{n} t^{- 1} ∥ u ∥_{1}$ for $u \in L^{1}$ , and $∥ M u ∥_{2} \leq C_{M} ∥ u ∥_{2}$ , $∥ M u ∥_{p} \leq C_{n, p} \max (p, (p - 1)^{- 1}) ∥ u ∥_{p}$ for $1 < p < \infty$ (The centered Hardy-Littlewood maximal operator is weak type $(1, 1)$ , The centered maximal operator is bounded on $L^{p} (R^{n})$ for $1 < p < \infty$ , The centered and uncentered Hardy-Littlewood maximal functions).

[F6]

Chebyshev's inequality; $∣ r E ∣ = r^{n} ∣ E ∣$ for measurable $E$ ; an $L^{q}$ -convergent sequence has an almost-everywhere convergent subsequence; Tonelli's theorem applies to nonnegative product-measurable integrands; the $L^{1}$ convention is The class $L^{1} (μ)$ of integrable functions and Countable Choice is The Axiom of Countable Choice ( ${A C}_{ω}$ ) (Chebyshev-Markov inequality for the integral, For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it, Assuming Countable Choice, $L^{p}$ -convergent sequences have almost-everywhere convergent subsequences, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[F7]

Hölder holds for complex $L^{p}$ functions; $C_{c}^{\infty}$ is dense in finite-exponent Euclidean $L^{p}$ under Countable Choice; dominated convergence applies under an integrable majorant, and Fatou applies to nonnegative measurable functions. (Complex Holder, Minkowski, and the quotient norm, Complex finite-simple and smooth compact-support density for finite p, Dominated convergence, Fatou's lemma)

Proof

technique · direct

1.1F1F2F5F6F7given

Cotlar's inequality extends to $L^{2}$ inputs. Choose $h_{m} \in C_{c}^{\infty}$ with $h_{m} \to h$ in $L^{2}$ by [F7]. For each $t > 0$ , the size bound makes $k_{t} = k 1_{{∣ \cdot ∣ > t}} \in L^{2}$ , so Hölder gives $∣ T_{t} (h_{m} - h) (x) ∣ \leq ∥ k_{t} ∥_{2} ∥ h_{m} - h ∥_{2} \to 0$ for every $x$ . Sublinearity and the $L^{2}$ bound of $M$ imply $M (T h_{m}) \to M (T h)$ and $M h_{m} \to M h$ in $L^{2}$ ; [F6] gives a common almost-everywhere convergent subsequence for these two families. Outside the countable union of the exceptional sets for [F2], pass its bound for each $h_{m}$ to the limit at every $t > 0$ , then take the supremum to get $T^{*} h \leq M (T h) + C_{0} (A_{1} + A_{2}^{'} + A_{3}) M h$ almost everywhere. For every finite-exponent input, dominated convergence shows that $t \mapsto T_{t} h (x)$ is continuous on $(0, \infty)$ for every $x$ : nearby truncations are dominated by $∣ k (z) h (x - z) ∣ 1_{{∣ z ∣ > t / 2}}$ , integrable by [F1]. Thus both maximal suprema can be taken over rational parameters and are measurable.

1.2F3F6algebra

Geometry of the dilated cubes. For each maximal cube $Q_{j}$ with centre $c_{j}$ and side length $ℓ_{j}$ , let $Q_{j}^{*}$ be the cube concentric with $Q_{j}$ and with side length $5 \sqrt{n} ℓ_{j}$ ; then $∣ Q_{j}^{*} ∣ = (5 \sqrt{n})^{n} ∣ Q_{j} ∣$ by the dilation identity [F6]. If $x \notin ⋃_{k} Q_{k}^{*}$ and $y \in Q_{j}$ , then $∣ x - c_{j} ∣_{\infty} > \frac{5 \sqrt{n}}{2} ℓ_{j}$ , so $∣ x - c_{j} ∣ \geq \frac{5 \sqrt{n}}{2} ℓ_{j} > 2 \cdot \frac{\sqrt{n}}{2} ℓ_{j} \geq 2 ∣ y - c_{j} ∣$ ; in particular $∣ x - y ∣ \geq ∣ x - c_{j} ∣ - ∣ y - c_{j} ∣ \geq \frac{\sqrt{n}}{2} ℓ_{j} > 0$ . Moreover, if $t > 0$ and $j$ belongs to $J_{3} (x, t) : = {j : \exists y_{0} \in Q_{j}, ∣ x - y_{0} ∣ = t}$ , then $t \geq ∣ x - c_{j} ∣ - ∣ y_{0} - c_{j} ∣ \geq (\frac{5 \sqrt{n}}{2} - \frac{\sqrt{n}}{2}) ℓ_{j} = 2 \sqrt{n} ℓ_{j}$ , and consequently every $y \in Q_{j}$ satisfies $\frac{t}{2} \leq t - \sqrt{n} ℓ_{j} \leq ∣ x - y ∣ \leq t + \sqrt{n} ℓ_{j} \leq \frac{3 t}{2}$ ; hence $⋃_{j \in J_{3} (x, t)} Q_{j} \subseteq B (x, \frac{3 t}{2}) ∖ B (x, \frac{t}{2})$ .

2.1F1F3step 1.2algebra

Splitting the truncated bad part. Fix $x \notin ⋃_{k} Q_{k}^{*}$ and $t > 0$ , and split the indices into $J_{1}, J_{2}, J_{3}$ according to whether $∣ x - y ∣ < t$ for all $y \in Q_{j}$ , $∣ x - y ∣ > t$ for all $y \in Q_{j}$ , or $∣ x - y ∣ = t$ for some $y \in Q_{j}$ . Each index lies in exactly one of the three classes because $y \mapsto ∣ x - y ∣$ is continuous on the connected cube. For $j \in J_{1}$ the integrand of $T_{t} b (x)$ vanishes on $Q_{j}$ ; for $j \in J_{2}$ one has $∣ k (x - y) ∣ \leq A_{1} t^{- n}$ on $Q_{j}$ ; for $j \in J_{3}$ step 1.2 gives $∣ k (x - y) ∣ \leq A_{1} (2 / t)^{n}$ on $Q_{j}$ . Hence $\sum_{j} \int_{Q_{j} \cap {∣ x - y ∣ > t}} ∣ k (x - y) ∣ ∣ b_{j} (y) ∣ d y \leq \sum_{j \in J_{2} \cup J_{3}} \int_{Q_{j}} ∣ k (x - y) ∣ ∣ b_{j} (y) ∣ d y \leq \max {A_{1} t^{- n}, A_{1} (2 / t)^{n}} \sum_{j} ∥ b_{j} ∥_{1} < \infty$ by [F3], so the series converges absolutely and $T_{t} b (x) = \sum_{j} \int_{Q_{j} \cap {∣ x - y ∣ > t}} k (x - y) b_{j} (y) d y$ . For $j \in J_{2}$ the truncation is inactive on $Q_{j}$ and the mean-zero property of $b_{j}$ gives $∣ \int_{Q_{j}} k (x - y) b_{j} (y) d y ∣ = ∣ \int_{Q_{j}} [k (x - y) - k (x - c_{j})] b_{j} (y) d y ∣$ . For $j \in J_{3}$ , put $c_{j} (t) : = ∣ Q_{j} ∣^{- 1} \int_{Q_{j}} b_{j} (y) 1_{{∣ x - y ∣ > t}} (y) d y$ ; then $∣ c_{j} (t) ∣ \leq ∣ Q_{j} ∣^{- 1} ∥ b_{j} ∥_{1} \leq 2^{n + 1} μ$ and, since $\int_{Q_{j}} (b_{j} 1_{{∣ x - y ∣ > t}} - c_{j} (t)) = 0$ , $\int_{Q_{j}} k (x - y) b_{j} (y) 1_{{∣ x - y ∣ > t}} (y) d y = \int_{Q_{j}} [k (x - y) - k (x - c_{j})] (b_{j} 1_{{∣ x - y ∣ > t}} - c_{j} (t)) d y + c_{j} (t) \int_{Q_{j}} k (x - y) d y .$ Summing, using $∣ b_{j} 1 - c_{j} (t) ∣ \leq ∣ b_{j} ∣ + 2^{n + 1} μ$ on $Q_{j}$ , and using step 1.2 with $\int_{{∣ z ∣ \in [t / 2, 3 t / 2]}} ∣ k (z) ∣ d z \leq A_{1} ∣ S^{n - 1} ∣ \log 3$ , yields $\sup_{t > 0} ∣ T_{t} b (x) ∣ \leq 2 E_{1} (x) + 2^{n + 1} μ E_{2} (x) + C_{n} μ A_{1},$ where $E_{1} (x) : = \sum_{j} \int_{Q_{j}} ∣ k (x - y) - k (x - c_{j}) ∣ ∣ b_{j} (y) ∣ d y$ and $E_{2} (x) : = \sum_{j} \int_{Q_{j}} ∣ k (x - y) - k (x - c_{j}) ∣ d y$ : the $J_{2}$ sum and the first $J_{3}$ sum together contribute at most $E_{1} + 2^{n + 1} μ E_{2}$ (both $E_{1}$ and $E_{2}$ majorize their sub-sums over $J_{2}$ and $J_{3}$ ; if one of them is infinite the displayed inequality is trivial), while $\sum_{j \in J_{3}} ∣ c_{j} (t) ∣ \int_{Q_{j}} ∣ k (x - y) ∣ d y \leq 2^{n + 1} μ \int_{{∣ z ∣ \in [t / 2, 3 t / 2]}} ∣ k (z) ∣ d z$ by the containment of step 1.2 and the disjointness of the cubes. Since $T^{(ε, N)} b = T_{ε} b - T_{N} b$ , we obtain $T^{* *} b (x) \leq 2 \sup_{t > 0} ∣ T_{t} b (x) ∣ \leq 4 E_{1} (x) + 2^{n + 2} μ E_{2} (x) + C_{n} μ A_{1}$ at every such $x$ (with $C_{n}$ denoting a dimensional constant, as everywhere).

2.2F3F4F6step 1.2

Integrating $E_{1}$ and $E_{2}$ off the dilated cubes. By step 1.2, for $x \notin Q_{j}^{*}$ and $y \in Q_{j}$ one has $∣ x - c_{j} ∣ \geq 2 ∣ y - c_{j} ∣$ , so in $\int_{(⋃_{k} Q_{k}^{*})^{c}} E_{1} \leq \sum_{j} \int_{Q_{j}} ∣ b_{j} (y) ∣ (\int_{∣ x - c_{j} ∣ \geq 2 ∣ y - c_{j} ∣} ∣ k (x - y) - k (x - c_{j}) ∣ d x) d y$ the inner integral is at most $A_{2}$ by Hörmander's condition, giving $\int_{(⋃_{k} Q_{k}^{*})^{c}} E_{1} \leq A_{2} \sum_{j} ∥ b_{j} ∥_{1} \leq A_{2} 2^{n + 1} μ \sum_{j} ∣ Q_{j} ∣ \leq 2^{n + 1} A_{2} ∥ f ∥_{1}$ ; similarly $\int_{(⋃_{k} Q_{k}^{*})^{c}} E_{2} \leq A_{2} \sum_{j} ∣ Q_{j} ∣ \leq A_{2} μ^{- 1} ∥ f ∥_{1}$ . Both interchanges are Tonelli's theorem applied to nonnegative product-measurable integrands, and $A_{2} = ∣ S^{n - 1} ∣ 2^{- δ} δ^{- 1} A_{2}^{'}$ by [F4].

3.1F6step 2.1step 2.2algebra

The bad part is controlled off the cubes. Choose $γ : = (K_{n} (A_{1} + A_{2}^{'} + A_{3} + B))^{- 1}$ with a dimensional constant $K_{n}$ large enough that the last term of step 2.1 satisfies $C_{n} μ A_{1} = C_{n} γ λ A_{1} \leq λ / 3$ ; if $A_{1} + A_{2}^{'} + A_{3} + B = 0$ then $k = 0$ and $T^{* *} = 0$ , so the theorem is trivial, and otherwise $γ > 0$ is well defined. Then $λ / 2 = λ / 3 + λ / 12 + λ / 12$ and step 2.1 give ${x \notin ⋃_{k} Q_{k}^{*} : T^{* *} b (x) > λ / 2} \subseteq {4 E_{1} > λ / 12} \cup {2^{n + 2} μ E_{2} > λ / 12}$ , so by Chebyshev's inequality and step 2.2, $∣ {x \notin ⋃_{k} Q_{k}^{*} : T^{* *} b > λ / 2} ∣ \leq \frac{48}{λ} \int E_{1} + \frac{12 \cdot 2^{n + 2} μ}{λ} \int E_{2} \leq C_{n} A_{2} λ^{- 1} ∥ f ∥_{1} \leq C_{n, δ} A_{2}^{'} λ^{- 1} ∥ f ∥_{1} .$

4.1F3F5F6step 1.1

The good part. By step 1.1 and [F1], $T^{* *} g \leq 2 T^{*} g \leq 2 M (T g) + 2 C_{0} (A_{1} + A_{2}^{'} + A_{3}) M g$ almost everywhere, and $∥ T g ∥_{2} \leq B ∥ g ∥_{2}$ by the $L^{2}$ bound of $T$ . Chebyshev's inequality, the $L^{2}$ bound of $M$ and $∥ g ∥_{1} \leq ∥ f ∥_{1}$ give $∣ {T^{* *} g > λ / 2} ∣ \leq ∣ {M (T g) > λ / 8} ∣ + ∣ {M g > λ / (8 C_{0} (A_{1} + A_{2}^{'} + A_{3}))} ∣ \leq \frac{64}{λ^{2}} ∥ M (T g) ∥_{2}^{2} + \frac{8 C_{n} C_{0} (A_{1} + A_{2}^{'} + A_{3})}{λ} ∥ f ∥_{1} \leq \frac{64 C_{M}^{2} B^{2}}{λ^{2}} ∥ g ∥_{2}^{2} + \frac{8 C_{n} C_{0} (A_{1} + A_{2}^{'} + A_{3})}{λ} ∥ f ∥_{1} \leq C_{n, δ} (A_{1} + A_{2}^{'} + A_{3} + B) λ^{- 1} ∥ f ∥_{1},$ where the last step uses $∥ g ∥_{2}^{2} \leq 2^{n} μ ∥ f ∥_{1} = 2^{n} γ λ ∥ f ∥_{1}$ and $γ B^{2} \leq B / K_{n} \leq (A_{1} + A_{2}^{'} + A_{3} + B) / K_{n}$ (in the degenerate case of step 3.1 the bound is trivial).

5.1F3step 1.2step 3.1step 4.1algebra

Weak $(1, 1)$ for $f \in L^{1} \cap L^{2}$ . Subadditivity of the supremum gives $T^{* *} f \leq T^{* *} g + T^{* *} b$ pointwise, so $∣ {∣ T^{* *} f ∣ > λ} ∣ \leq ∣ {T^{* *} g > λ / 2} ∣ + ∣ ⋃_{k} Q_{k}^{*} ∣ + ∣ {x \notin ⋃_{k} Q_{k}^{*} : T^{* *} b > λ / 2} ∣ \leq C_{n, δ} (A_{1} + A_{2}^{'} + A_{3} + B) λ^{- 1} ∥ f ∥_{1},$ because the union of cubes has measure at most $(5 \sqrt{n})^{n} \sum_{j} ∣ Q_{j} ∣ \leq (5 \sqrt{n})^{n} μ^{- 1} ∥ f ∥_{1} \leq (5 \sqrt{n})^{n} K_{n} (A_{1} + A_{2}^{'} + A_{3} + B) λ^{- 1} ∥ f ∥_{1}$ by steps 1.2 and 3.1 and [F3], while steps 3.1 and 4.1 bound the other two terms by dimensional multiples of $(A_{1} + A_{2}^{'} + A_{3} + B) λ^{- 1} ∥ f ∥_{1}$ .

6.1F1F7step 5.1algebra

For general $f \in L^{1}$ , put $f_{m} = f 1_{B (0, m)} 1_{{∣ f ∣ \leq m}} \in L^{1} \cap L^{2}$ . Then $f_{m} \to f$ in $L^{1}$ and $∥ f_{m} ∥_{1} \leq ∥ f ∥_{1}$ . For every $0 < ε < N$ , the size bound gives $∣ T^{(ε, N)} (f_{m} - f) (x) ∣ \leq A_{1} ε^{- n} ∥ f_{m} - f ∥_{1} \to 0$ at every $x$ . Therefore $T^{* *} f (x) \leq {lim inf}_{m} T^{* *} f_{m} (x)$ : each fixed truncation is bounded by this liminf, and then one takes its supremum. Fatou [F7] applied to superlevel indicators and step 5.1 give the weak $(1, 1)$ bound; $T^{*} \leq T^{* *}$ transfers it to $T^{*}$ . No subsequence selection is needed here.

7.1F1F4F5F7step 1.1step 6.1algebra

For $1 < p < \infty$ and $f \in L^{p} \cap L^{2}$ , step 1.1 gives $T^{* *} f \leq 2 M (T f) + 2 C_{0} (A_{1} + A_{2}^{'} + A_{3}) M f$ almost everywhere. The strong $L^{p}$ bounds [F4,F5] and $A_{2} \leq C_{n, δ} A_{2}^{'}$ yield $∥ T^{* *} f ∥_{p} \leq C_{n, p, δ} (A_{1} + A_{2}^{'} + A_{3} + B) \max (p, (p - 1)^{- 1}) ∥ f ∥_{p}$ ; additional factors depending on $p$ are included in $C_{n, p, δ}$ , as allowed by the statement. For general $f \in L^{p}$ , the same bounded compact-support approximants $f_{m}$ converge in $L^{p}$ and satisfy $∥ f_{m} ∥_{p} \leq ∥ f ∥_{p}$ . Every doubly truncated kernel lies in $L^{p^{'}}$ , so Hölder gives $T^{(ε, N)} f_{m} (x) \to T^{(ε, N)} f (x)$ at every $x$ for every parameter pair. Hence $T^{* *} f \leq {lim inf}_{m} T^{* *} f_{m}$ pointwise, and Fatou [F7] applied to the $p$ th powers extends the bound to all $L^{p}$ . The comparison $T^{*} \leq T^{* *}$ gives its bound too.

8.1F1step 5.1step 6.1step 7.1∎

Steps 5.1 and 6.1 give the weak $(1, 1)$ bound for $T^{* *}$ , and step 7.1 gives the strong $L^{p}$ bounds for $T^{* *}$ ; the pointwise comparison $T^{*} \leq T^{* *}$ of [F1] transfers both to $T^{*}$ . This proves the theorem.

CorollaryStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Almost-everywhere convergence of principal-value truncations

Statement

Assume Countable Choice. Fix $1 \leq p < \infty$ and let $k$ and $T$ be as in Maximal truncations: weak (1,1) and strong Lp bounds: $k$ satisfies the pointwise size bound with constant $A_{1}$ , the standard $δ$ -Hölder bound with constant $A_{2}^{'}$ and the cancellation bound $A_{3}$ , and $T$ is the associated $L^{2}$ -bounded operator with off-support representation and $L^{2}$ norm $B$ . Let $D \subseteq L^{p} (R^{n}; C)$ be dense in $L^{p} (R^{n}; C)$ and suppose that for every $g \in D$ the limit $\lim_{ε ↓ 0} T_{ε} g (x)$ exists for almost every $x \in R^{n}$ . Then for every $f \in L^{p} (R^{n}; C)$ the limit $\lim_{ε ↓ 0} T_{ε} f (x)$ exists for almost every $x \in R^{n}$ .

In particular, for the Hilbert kernel $1 / (π x)$ on $R$ and the Riesz kernels $c_{n} x_{j} / ∣ x ∣^{n + 1}$ on $R^{n}$ the dense class $D = S (R^{n})$ satisfies the hypothesis, by the published principal-value formulas for Schwartz functions.

Facts & Assumptions

Given: Countable Choice; $1 \leq p < \infty$ ; $k$ , $T$ as in the statement; a dense subspace $D \subseteq L^{p}$ such that $\lim_{ε ↓ 0} T_{ε} g (x)$ exists a.e. for every $g \in D$ ; a function $f \in L^{p}$ and $λ > 0$ .

[F1]

For $f \in L^{p}$ the truncations $T_{ε} f$ , $ε > 0$ , are defined pointwise by absolutely convergent integrals and $T^{*} f = \sup_{ε > 0} ∣ T_{ε} f ∣$ (Maximal truncated singular integrals); $T^{*} f \leq T^{* *} f$ , and the maximal truncation $T^{* *}$ satisfies $∣ {∣ T^{* *} h ∣ > λ} ∣ \leq C_{n, δ} (A_{1} + A_{2}^{'} + A_{3} + B) λ^{- 1} ∥ h ∥_{1}$ for $h \in L^{1}$ and $∥ T^{* *} h ∥_{p} \leq C_{n, p, δ} (A_{1} + A_{2}^{'} + A_{3} + B) \max (p, (p - 1)^{- 1}) ∥ h ∥_{p}$ for $1 < p < \infty$ , hence the same bounds hold for $T^{*}$ (Maximal truncations: weak (1,1) and strong Lp bounds).

[F2]

Chebyshev's inequality: for measurable $u$ and $t > 0$ , $∣ {∣ u ∣ > t} ∣ \leq t^{- p} \int ∣ u ∣^{p}$ when $u \in L^{p}$ (Chebyshev-Markov inequality for the integral); $C_{c}^{\infty} (R^{n})$ — and hence its superset $S (R^{n})$ — is dense in $L^{p} (R^{n}; C)$ for $1 \leq p < \infty$ (Complex finite-simple and smooth compact-support density for finite p); the $L^{p}$ conventions are those of the maximal-truncation theorem and Countable Choice is The Axiom of Countable Choice ( ${A C}_{ω}$ ).

[F3]

For the Hilbert and Riesz kernels the principal-value truncations converge on every Schwartz input and identify the $L^{2}$ operators. Their kernel size, first-difference, cancellation, $L^{2}$ and off-support operator conditions are proved in the corresponding items. (The Hilbert transform is the tempered convolution with pv(1/(pi x)) and has signum Fourier multiplier, The Riesz transform is the principal value of its kernel, with the matching constant, Truncated Hilbert transform and principal value, Riesz transforms on Euclidean space, The Hilbert transform is bounded on Lp, The Riesz transforms are bounded on Lp, Riesz kernel size, difference and spherical-cancellation bounds)

Proof

technique · direct

1.1F1givenalgebra

Oscillation bound. Put $H_{f} (x) : = {lim sup}_{ε, θ ↓ 0} ∣ T_{ε} f (x) - T_{θ} f (x) ∣$ . For every $g \in D$ , on the full-measure set where $(T_{ε} g (x))_{ε}$ converges, the triangle inequality gives $∣ T_{ε} f - T_{θ} f ∣ \leq ∣ T_{ε} (f - g) ∣ + ∣ T_{θ} (f - g) ∣ + ∣ T_{ε} g - T_{θ} g ∣ \leq 2 T^{*} (f - g) (x) + ∣ T_{ε} g - T_{θ} g ∣$ , and the last term tends to $0$ as $ε, θ ↓ 0$ ; hence $H_{f} \leq 2 T^{*} (f - g)$ almost everywhere.

2.1F1step 1.1algebra

The case $p = 1$ . Taking $g \in D$ and using step 1.1 and the weak $(1, 1)$ bound of [F1] for $f - g \in L^{1}$ , $∣ {H_{f} > λ} ∣ \leq ∣ {T^{*} (f - g) > λ / 2} ∣ \leq 2 C_{n, δ} (A_{1} + A_{2}^{'} + A_{3} + B) λ^{- 1} ∥ f - g ∥_{1}$ for every $λ > 0$ ; since $D$ is dense in $L^{1}$ , the infimum over $g \in D$ gives $∣ {H_{f} > λ} ∣ = 0$ for every $λ > 0$ , hence $H_{f} = 0$ almost everywhere. Thus $(T_{ε} f (x))_{ε > 0}$ is a Cauchy family as $ε ↓ 0$ for almost every $x$ , so its limit exists almost everywhere.

2.2F1F2step 1.1algebra

The case $1 λ} ∣ \leq ∣ {T^{*} (f - g) > λ / 2} ∣ \leq (2 / λ)^{p} ∥ T^{*} (f - g) ∥_{p}^{p} \leq (2 C_{n, p, δ} (A_{1} + A_{2}^{'} + A_{3} + B) \max (p, (p - 1)^{- 1}) / λ)^{p} ∥ f - g ∥_{p}^{p},$ and letting $g \to f$ in $L^{p}$ through $D$ gives $∣ {H_{f} > λ} ∣ = 0$ for every $λ > 0$ , hence $H_{f} = 0$ almost everywhere and the limit exists almost everywhere.

3.1F2F3step 2.1step 2.2∎

The Hilbert and Riesz kernels have the size, Hölder, spherical cancellation, $L^{2}$ bound and off-support representation in [F3]. Zero spherical means give the annular cancellation bound $A_{3} = 0$ . Their principal-value distributions are defined by subtracting a test's value at zero on $∣ y ∣ < 1$ ; the size estimate makes the resulting integrand integrable, bounded by $C ∣ y ∣^{1 - n}$ there, and Schwartz decay controls infinity. Thus they satisfy the maximal theorem's hypotheses. The dense class $S$ has convergence at every point by [F3], and is dense in each finite-exponent $L^{p}$ by [F2]. Steps 2.1 and 2.2 therefore give the asserted almost-everywhere convergence for every $f \in L^{p}$ .

CorollaryStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The Hilbert transform is bounded on Lp

Statement

Assume Countable Choice. The Hilbert transform $H$ of the $L^{2}$ multiplier definition extends uniquely to a bounded operator on $L^{p} (R; C)$ for every $1 < p < \infty$ , with norm at most $C_{p}$ , a constant depending only on $p$ : more explicitly the kernel $k (x) = 1 / (π x)$ is a standard $1$ -Hölder Calderón–Zygmund kernel with $A_{2}^{'} = 2 / π$ and annular constant $A_{1} = 2 \log 2 / π$ , so that $∥ H f ∥_{p} \leq C_{p} (1 + \frac{2}{π}) \max (p, (p - 1)^{- 1}) ∥ f ∥_{p} (f \in L^{p} (R; C))$ for a numerical constant $C_{p}$ .

Facts & Assumptions

Given: Countable Choice; the kernel $k (x) = 1 / (π x)$ on $R ∖ {0}$ ; the operator $H$ of The Hilbert transform is the tempered convolution with pv(1/(pi x)) and has signum Fourier multiplier and The Hilbert transform is an L2 isometry and squares to minus the identity; a compactly supported $f \in L^{2} (R; C)$ ; a test function $φ \in C_{c}^{\infty} (R)$ supported off $supp f$ .

[F1]

On Schwartz functions $H$ is the principal-value operator with kernel $k$ : for every Schwartz $g$ the limits $\lim_{ε ↓ 0} H_{ε} g (x)$ exist at every $x$ and equal $(W * g) (x)$ for the tempered distribution $W = p v 1 / (π x)$ , and $F (H g) = - i sgn (ξ) \hat{g} (ξ)$ (The Hilbert transform is the tempered convolution with pv(1/(pi x)) and has signum Fourier multiplier, Truncated Hilbert transform and principal value).

[F2]

$H$ has a unique extension to an $L^{2}$ -bounded operator with $∥ H g ∥_{2} = ∥ g ∥_{2}$ for all $g \in L^{2}$ and $H^{2} = - I$ ; it is skew-adjoint, $H^{*} = - H$ for the first-variable-linear pairing $⟨ u, v ⟩ = \int u \overline{v}$ , so $⟨ H f, φ ⟩ = ⟨ f, H^{*} φ ⟩ = - ⟨ f, H φ ⟩$ (The Hilbert transform is an L2 isometry and squares to minus the identity, The Hilbert transform is skew-adjoint on L2, The Hilbert-space adjoint of a bounded operator).

[F3]

A measurable kernel with pointwise bound $∣ k ∣ \leq c ∣ \cdot ∣^{- n}$ satisfies the annular condition of the base definition with $A_{1} = c ∣ S^{n - 1} ∣ \log 2$ ; a standard $δ$ -Hölder kernel is a Calderón–Zygmund kernel with Hörmander constant $A_{2} = ∣ S^{n - 1} ∣ 2^{- δ} δ^{- 1} A_{2}^{'}$ ; and a Calderón–Zygmund operator with constants $A_{1}, A_{2}$ and $L^{2}$ norm $B$ extends uniquely to a bounded operator on $L^{p}$ for $1 < p < \infty$ with $∥ T g ∥_{p} \leq C_{n, p} (A_{2} + B) \max (p, (p - 1)^{- 1}) ∥ g ∥_{p}$ (Calderón–Zygmund kernels and their associated operators, Standard (Hölder) Calderón–Zygmund kernels, Standard Hölder kernels satisfy the Hörmander condition, Calderón–Zygmund operators are bounded on Lp).

[F4]

Fubini interchanges absolutely integrable complex double integrals, and locally integrable functions with equal distribution pairings agree almost everywhere. (Fubini's theorem for L^1 functions on a sigma-finite product, Locally integrable functions embed in distributions)

[F5]

Under Countable Choice, polar coordinates integrate every nonnegative Borel function against $r^{n - 1} d r d σ$ ; in dimension one $S^{0} = {- 1, 1}$ has counting measure. (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma)

Proof

technique · direct

1.1givenalgebra

Size and smoothness of the kernel. $∣ k (x) ∣ = 1 / (π ∣ x ∣)$ for $x \neq 0$ , which is the pointwise size bound with constant $1 / π$ ; and for $∣ x ∣ \geq 2 ∣ y ∣ > 0$ the difference is $∣ k (x - y) - k (x) ∣ = \frac{1}{π} ∣ \frac{1}{x - y} - \frac{1}{x} ∣ = \frac{∣ y ∣}{π ∣ x ∣ ∣ x - y ∣} \leq \frac{2}{π} \frac{∣ y ∣}{∣ x ∣^{2}}$ , because $∣ x - y ∣ \geq ∣ x ∣ - ∣ y ∣ \geq ∣ x ∣ / 2$ . Hence $k$ satisfies the standard $1$ -Hölder condition with constant $A_{2}^{'} = 2 / π$ .

2.1F3F5step 1.1algebra

The smooth kernel is measurable and locally integrable away from zero. Its oddness gives $k (r) + k (- r) = 0$ , and [F5] gives $\int_{R \leq ∣ x ∣ \leq 2 R} ∣ k (x) ∣ d x = (2 / π) \int_{R}^{2 R} d r / r = 2 \log 2 / π$ . For each $y \neq 0$ , directly integrating the estimate of step 1.1 gives $\int_{∣ x ∣ \geq 2 ∣ y ∣} ∣ k (x - y) - k (x) ∣ d x \leq (2 / π) ∣ y ∣ 2 \int_{2 ∣ y ∣}^{\infty} r^{- 2} d r = 2 / π$ . Thus the base kernel conditions hold with $A_{1} = 2 \log 2 / π$ and $A_{2} = 2 / π$ ; together with step 1.1 they establish standard $1$ -Hölder status with $A_{2}^{'} = 2 / π$ .

3.1F1F2step 2.1algebraF4

$H$ is a Calderón–Zygmund operator with kernel $k$ and $L^{2}$ norm $B = 1$ . It is $L^{2}$ -bounded with norm one by [F2], so it remains to prove the off-support representation. Let $f \in L^{2}$ be compactly supported, let $φ \in C_{c}^{\infty}$ be supported off $supp f$ , and put $d : = dist (supp f, supp φ) > 0$ . By [F2] and the reality of $k$ , $⟨ H f, φ ⟩ = - \int f (y) \overline{H φ (y)} d y = - \frac{1}{π} \iint \frac{f (y) \overline{φ (x)}}{y - x} d x d y,$ where the inner limit defining $\overline{H φ (y)} = \frac{1}{π} p . v . \int \frac{\overline{φ (x)}}{y - x} d x$ is an absolutely convergent integral because $∣ x - y ∣ \geq d > 0$ on $supp f \times supp φ$ ; the double integral is absolutely convergent over the bounded supports, so Fubini's theorem may be applied and the sign of the denominator changed: $⟨ H f, φ ⟩ = \frac{1}{π} \iint \frac{f (y) \overline{φ (x)}}{x - y} d x d y = \int (\int k (x - y) f (y) d y) \overline{φ (x)} d x,$ the last equality by the definition $k (x - y) = 1 / (π (x - y))$ and Fubini. Since the pairing against every test function supported off $supp f$ determines the $L^{2}$ class off that support, the $L^{2}$ function $H f$ agrees almost everywhere off $supp f$ with the locally integrable function $x \mapsto \int k (x - y) f (y) d y$ , which is the representation (3) required of a Calderón–Zygmund operator.

4.1F3step 2.1step 3.1∎

Applying the strict-range theorem [F3] to the Calderón–Zygmund operator $H$ with constants $A_{1} = 2 \log 2 / π$ , $A_{2} = 2 / π$ and $B = 1$ yields a unique bounded extension of $H$ to $L^{p} (R; C)$ for every $1 < p < \infty$ with $∥ H g ∥_{p} \leq C_{1, p} (1 + 2 / π) \max (p, (p - 1)^{- 1}) ∥ g ∥_{p}$ ; since the dimension is one, $C_{1, p}$ depends only on $p$ . This is the assertion.

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The Riesz transforms are bounded on Lp

Statement

Assume Countable Choice. Let $n \geq 1$ and $1 \leq j \leq n$ . The $j$ -th Riesz transform $R_{j}$ of Riesz transforms on Euclidean space extends uniquely to a bounded operator on $L^{p} (R^{n}; C)$ for every $1 < p < \infty$ , with norm at most $C_{n, p}$ , a constant depending only on $n$ and $p$ : the Riesz kernel $K_{j} (x) = c_{n} x_{j} / ∣ x ∣^{n + 1}$ is a standard $1$ -Hölder Calderón–Zygmund kernel with constant $C_{n} = c_{n} 2^{n + 1} (3 n + 4)$ , so that $∥ R_{j} f ∥_{p} \leq C_{n, p} (1 + ∣ S^{n - 1} ∣ 2^{- 1} C_{n}) \max (p, (p - 1)^{- 1}) ∥ f ∥_{p} (f \in L^{p} (R^{n}; C)) .$

Facts & Assumptions

Given: Countable Choice; the dimension $n \geq 1$ and index $1 \leq j \leq n$ ; the Riesz kernel $K_{j} (x) = c_{n} x_{j} / ∣ x ∣^{n + 1}$ and operator $R_{j}$ of Riesz transforms on Euclidean space; a compactly supported $f \in L^{2} (R^{n}; C)$ ; a test function $φ \in C_{c}^{\infty} (R^{n})$ supported off $supp f$ .

[F1]

$∣ K_{j} (x) ∣ \leq c_{n} ∣ x ∣^{- n}$ for $x \neq 0$ ; $∣ K_{j} (x - h) - K_{j} (x) ∣ \leq C_{n} ∣ h ∣ ∣ x ∣^{- (n + 1)}$ with $C_{n} = c_{n} 2^{n + 1} (3 n + 4)$ whenever $∣ h ∣ \leq ∣ x ∣ / 2$ ; and $\int_{S^{n - 1}} K_{j} (r ω) d σ (ω) = 0$ for every $r > 0$ (Riesz kernel size, difference and spherical-cancellation bounds).

[F2]

For every Schwartz function $g$ the truncated integrals $\int_{∣ y ∣ > ε} K_{j} (y) g (x - y) d y$ converge as $ε ↓ 0$ for every $x$ , and the limit is a continuous representative of the $L^{2}$ class $R_{j} g$ (The Riesz transform is the principal value of its kernel, with the matching constant).

[F3]

$R_{j}$ is the $L^{2}$ Fourier multiplier with symbol $m_{j} (ξ) = - i ξ_{j} / ∣ ξ ∣$ for $ξ \neq 0$ and $m_{j} (0) = 0$ , is bounded with $∥ R_{j} g ∥_{2} \leq ∥ g ∥_{2}$ for all $g \in L^{2}$ , and satisfies $⟨ R_{j} f, g ⟩ = \int m_{j} \hat{f} \overline{\hat{g}}$ for the first-variable-linear pairing $⟨ u, v ⟩ = \int u \overline{v}$ (Riesz transforms on Euclidean space, Riesz transforms are L2 contractions and square to minus the identity in sum, Plancherel theorem, The Hilbert-space adjoint of a bounded operator).

[F4]

A pointwise bound $∣ k ∣ \leq c ∣ \cdot ∣^{- n}$ implies the annular condition with $A_{1} = c ∣ S^{n - 1} ∣ \log 2$ ; a standard $δ$ -Hölder kernel with constant $A_{2}^{'}$ is a Calderón–Zygmund kernel with Hörmander constant $A_{2} = ∣ S^{n - 1} ∣ 2^{- δ} δ^{- 1} A_{2}^{'}$ ; and a Calderón–Zygmund operator with constants $A_{1}, A_{2}$ and $L^{2}$ norm $B$ extends uniquely to a bounded operator on $L^{p}$ for $1 < p < \infty$ with $∥ T g ∥_{p} \leq C_{n, p} (A_{2} + B) \max (p, (p - 1)^{- 1}) ∥ g ∥_{p}$ (Calderón–Zygmund kernels and their associated operators, Standard (Hölder) Calderón–Zygmund kernels, Standard Hölder kernels satisfy the Hörmander condition, Calderón–Zygmund operators are bounded on Lp).

[F5]

[F6]

Under Countable Choice, for every nonnegative Borel function $g$ , $\int_{R^{n}} g (x) d x = \int_{0}^{\infty} \int_{S^{n - 1}} g (r ω) r^{n - 1} d σ (ω) d r$ , with $σ$ a finite Borel measure. (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma)

Proof

technique · direct

1.1F1given

The published estimates give the size bound $∣ K_{j} (x) ∣ \leq c_{n} ∣ x ∣^{- n}$ , the first-difference bound $∣ K_{j} (x - h) - K_{j} (x) ∣ \leq C_{n} ∣ h ∣ ∣ x ∣^{- (n + 1)}$ for $∣ h ∣ \leq ∣ x ∣ / 2$ , and the vanishing of every spherical mean. The explicit kernel is smooth on the punctured space, hence Borel measurable and locally integrable there.

1.2F3givenalgebra

$R_{j}$ is $L^{2}$ -bounded with $∥ R_{j} g ∥_{2} \leq ∥ g ∥_{2}$ , since $∣ m_{j} ∣ \leq 1$ and the $L^{2}$ multiplier bound gives $∥ R_{j} ∥ \leq 1$ ; hence $B = 1$ is an admissible $L^{2}$ norm bound. Moreover $R_{j}$ is skew-adjoint: for $f, g \in L^{2}$ , $⟨ R_{j} f, g ⟩ = \int m_{j} \hat{f} \overline{\hat{g}} = \int \hat{f} \overline{(- m_{j}) \hat{g}} = ⟨ f, - R_{j} g ⟩$ , because $\overline{m_{j}} = - m_{j}$ for the purely imaginary symbol; that is, $R_{j}^{*} = - R_{j}$ .

2.1F1F4F6step 1.1algebra

By [F6], the size bound gives $\int_{R \leq ∣ x ∣ \leq 2 R} ∣ K_{j} (x) ∣ d x \leq c_{n} ∣ S^{n - 1} ∣ \int_{R}^{2 R} d r / r = c_{n} ∣ S^{n - 1} ∣ \log 2$ . For every $h \neq 0$ , directly integrating the difference bound gives $\int_{∣ x ∣ \geq 2 ∣ h ∣} ∣ K_{j} (x - h) - K_{j} (x) ∣ d x \leq C_{n} ∣ h ∣ ∣ S^{n - 1} ∣ \int_{2 ∣ h ∣}^{\infty} r^{- 2} d r = ∣ S^{n - 1} ∣ C_{n} / 2$ . Thus $K_{j}$ is a base Calderón–Zygmund kernel with $A_{1} = c_{n} ∣ S^{n - 1} ∣ \log 2$ and $A_{2} = ∣ S^{n - 1} ∣ C_{n} / 2$ ; its first-difference estimate now establishes standard $1$ -Hölder status with $A_{2}^{'} = C_{n}$ .

2.2F2F3step 1.2algebraF5

Off-support representation: let $f \in L^{2}$ be compactly supported, let $φ \in C_{c}^{\infty}$ be supported off $supp f$ , and put $d : = dist (supp f, supp φ) > 0$ . Using the adjoint identity of [F3], skew-adjointness from step 1.2 and the Schwartz principal-value formula [F2], and writing $K_{j}$ for the real-valued kernel, $⟨ R_{j} f, φ ⟩ = ⟨ f, R_{j}^{*} φ ⟩ = - \int f (y) \overline{R_{j} φ (y)} d y = - \iint K_{j} (y - x) f (y) \overline{φ (x)} d x d y = \iint K_{j} (x - y) f (y) \overline{φ (x)} d x d y,$ where $\overline{R_{j} φ (y)} = \lim_{ε ↓ 0} \int_{∣ y - x ∣ > ε} K_{j} (y - x) \overline{φ (x)} d x$ is an absolutely convergent integral on the two supports (there $∣ x - y ∣ \geq d > 0$ , so the limit may be taken inside the $y$ -integration), Fubini applies over the bounded supports, and the last step uses the oddness $K_{j} (- z) = - K_{j} (z)$ . Since this holds for every test function supported off $supp f$ , the $L^{2}$ class $R_{j} f$ agrees almost everywhere off $supp f$ with the locally integrable function $x \mapsto \int K_{j} (x - y) f (y) d y$ ; this is the off-support representation (3) required of a Calderón–Zygmund operator.

3.1F4step 2.1step 1.2step 2.2∎

By steps 2.1, 1.2 and 2.2 the operator $R_{j}$ is a standard-kernel Calderón–Zygmund operator with annular constant $c_{n} ∣ S^{n - 1} ∣ \log 2$ , Hörmander constant $∣ S^{n - 1} ∣ 2^{- 1} C_{n}$ and $L^{2}$ norm $B = 1$ ; the strict-range theorem [F4] therefore gives its unique extension to a bounded operator on $L^{p} (R^{n}; C)$ , $1 < p < \infty$ , with $∥ R_{j} f ∥_{p} \leq C_{n, p} (1 + ∣ S^{n - 1} ∣ 2^{- 1} C_{n}) \max (p, (p - 1)^{- 1}) ∥ f ∥_{p}$ . This is the assertion.

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Dyadic Mihlin pieces: uniform L1 and first-difference bounds

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ , put $n_{0} : = ⌊ n / 2 ⌋ + 1$ , and let $χ \in C_{c}^{\infty} (R^{n})$ be the specific radially nonincreasing smooth cutoff constructed in Explicit compactly supported smooth cutoffs, with $0 \leq χ \leq 1$ , $χ = 1$ on $∣ ξ ∣ \leq 1$ and $χ = 0$ on $∣ ξ ∣ \geq 2$ . Put $ζ (ξ) : = χ (ξ) - χ (2 ξ) .$ Then $ζ$ is supported in the annulus $1 / 2 \leq ∣ ξ ∣ \leq 2$ , satisfies $0 \leq ζ \leq 1$ , and $\sum_{j \in Z} ζ (2^{- j} ξ) = 1$ for every $ξ \neq 0$ . Let $m$ be a Mihlin symbol with constants $C_{α}$ as in Mihlin smoothness convention above half the dimension, put $m_{j} (ξ) : = m (ξ) ζ (2^{- j} ξ), K_{j} : = F^{- 1} (u_{m_{j}}),$ where $u_{m_{j}}$ is the regular tempered distribution of $m_{j}$ and $F^{- 1}$ is the inverse Fourier transform of Fourier transform of a tempered distribution, and let $A$ be any finite quantity with $A \geq \max {∥ m ∥_{\infty}, \max_{∣ α ∣ \leq n_{0}} C_{α}}$ . Then $K_{j}$ is (the regular distribution of) an $L^{2}$ function, and there is a constant $C_{n}$ , depending only on $n$ and on the fixed cutoff $χ$ , such that $\sup_{j \in Z} \int_{R^{n}} ∣ K_{j} (x) ∣ (1 + 2^{j} ∣ x ∣)^{1 / 4} d x \leq C_{n} A (1)$ and $\sup_{j \in Z} 2^{- j} \int_{R^{n}} ∣ \nabla K_{j} (x) ∣ (1 + 2^{j} ∣ x ∣)^{1 / 4} d x \leq C_{n} A . (2)$

Facts & Assumptions

Given: Countable Choice; an integer $n \geq 1$ ; the smooth step $χ$ and the Mihlin symbol $m$ with its constants $C_{α}$ ; the derived objects $ζ$ , $m_{j}$ , $K_{j}$ ; a finite quantity $A \geq \max {∥ m ∥_{\infty}, \max_{∣ α ∣ \leq n_{0}} C_{α}}$ , where $n_{0} = ⌊ n / 2 ⌋ + 1$ .

[F1]

$m$ agrees almost everywhere with a function $m_{0} \in C^{n_{0}} (R^{n} ∖ {0})$ satisfying $∣ \partial^{α} m_{0} (ξ) ∣ \leq C_{α} ∣ ξ ∣^{- ∣ α ∣}$ for $∣ α ∣ \leq n_{0}$ and $ξ \neq 0$ , and $∥ m ∥_{\infty} \leq C_{0}$ (Mihlin smoothness convention above half the dimension).

[F2]

$χ \in C_{c}^{\infty} (R^{n})$ obeys $0 \leq χ \leq 1$ , $χ = 1$ on $∣ ξ ∣ \leq 1$ , $χ = 0$ on $∣ ξ ∣ \geq 2$ (Explicit compactly supported smooth cutoffs).

[F3]

For an $L^{2}$ class $h$ with corresponding regular distribution $u_{h}$ , the transform $F^{- 1} (u_{h})$ is the regular distribution of the inverse Plancherel transform $F_{2}^{- 1} h$ , so $F^{- 1} (u_{h}) = u_{F_{2}^{- 1} h}$ (Fourier transform agrees with l one and plancherel transforms), and Plancherel's isometry gives $∥ u_{F_{2}^{- 1} h} ∥_{2} = ∥ h ∥_{2}$ (Plancherel theorem).

[F4]

For every tempered distribution $u$ and multi-index $γ$ , $F (x^{γ} u) = (- 1 / (2 π i))^{∣ γ ∣} \partial^{γ} F u$ and $F (\partial^{γ} u) = (2 π i ξ)^{γ} F u$ in $S^{'} (R^{n})$ (Fourier differentiation and multiplication identities on tempered distributions). The transform conventions are those of Fourier transform of a tempered distribution and Schwartz space and its seminorms.

[F5]

Integral Cauchy–Schwarz is the $p = q = 2$ case of Hölder: $\int ∣ u v ∣ \leq ∥ u ∥_{2} ∥ v ∥_{2}$ . (Holder's inequality for integrals, including the endpoint cases)

[F6]

Fubini interchanges absolutely integrable complex double integrals; under the assumed Countable Choice, locally integrable functions have equal regular distributions exactly when they agree almost everywhere. Distributional derivatives on Schwartz tests satisfy $⟨ \partial^{β} u, φ ⟩ = (- 1)^{∣ β ∣} ⟨ u, \partial^{β} φ ⟩$ . (Fubini's theorem for L^1 functions on a sigma-finite product, Locally integrable functions embed in distributions, Differentiation and polynomial multiplication preserve tempered distributions)

Proof

technique · direct

1.1F2givenalgebra

The difference $ζ = χ (ξ) - χ (2 ξ)$ vanishes for $∣ ξ ∣ \leq 1 / 2$ , since there $χ (ξ) = χ (2 ξ) = 1$ , and vanishes for $∣ ξ ∣ \geq 2$ , since there $χ (ξ) = χ (2 ξ) = 0$ ; hence $ζ$ is supported in the annulus $1 / 2 \leq ∣ ξ ∣ \leq 2$ , and $0 \leq ζ \leq 1$ for the fixed smooth-step cutoff: its construction is $χ (ξ) = σ ((4 - ∣ ξ ∣^{2}) / 3)$ , $σ (t) = a (t) / (a (t) + a (1 - t))$ , with $a (t) = e^{- 1 / t}$ for $t > 0$ and $a (t) = 0$ otherwise. On $0 < t < 1$ one has $σ^{'} (t) = (a^{'} (t) a (1 - t) + a (t) a^{'} (1 - t)) / (a (t) + a (1 - t))^{2} \geq 0$ ; on the constant regions its derivative is zero. Thus $χ$ decreases with radius and $χ (ξ) \geq χ (2 ξ)$ , proving the asserted nonnegativity. For every $N \geq 1$ the sum telescopes: $\sum_{j = - N}^{N} ζ (2^{- j} ξ) = \sum_{j = - N}^{N} [χ (2^{- j} ξ) - χ (2^{1 - j} ξ)] = χ (2^{- N} ξ) - χ (2^{N + 1} ξ)$ . For $ξ \neq 0$ one has $2^{- N} ∣ ξ ∣ \leq 1$ and $2^{N + 1} ∣ ξ ∣ \geq 2$ for all large $N$ , so the last expression equals $1 - 0 = 1$ there; this gives the asserted partition of unity.

1.2F1F3F6givenconstruct

For every $j$ the function $m_{j} = m ζ (2^{- j} \cdot)$ is supported in the annulus $2^{j - 1} \leq ∣ ξ ∣ \leq 2^{j + 1}$ , where it agrees almost everywhere with $m_{0} (ξ) ζ (2^{- j} ξ)$ ; we use this representative in the derivative estimates. Since $m_{0}$ is $C^{n_{0}}$ on that annulus and $ζ$ is compactly supported and smooth, $m_{j}$ is represented by a compactly supported $C^{n_{0}}$ function, so $m_{j} \in L^{1} \cap L^{2}$ and $u_{m_{j}}$ is a well-defined regular tempered distribution. By [F3] the object $K_{j} = F^{- 1} (u_{m_{j}})$ is the regular distribution of the $L^{2}$ function $F_{2}^{- 1} m_{j}$ ; we use $K_{j}$ to denote that $L^{2}$ class, so that $F K_{j} = u_{m_{j}}$ and $∥ K_{j} ∥_{2} \leq ∥ m_{j} ∥_{2}$ . It has the smooth integral representative $G_{j} (x) = \int m_{j} (ξ) e^{2 π i x \cdot ξ} d ξ$ : for every Schwartz test $φ$ , Fubini applies with absolute bound $∥ m_{j} ∥_{1} ∥ φ ∥_{1}$ , giving $\int G_{j} φ = \int m_{j} F^{- 1} φ = ⟨ F^{- 1} u_{m_{j}}, φ ⟩$ . Thus [F6] identifies $G_{j}$ with the $L^{2}$ class. Every $ξ^{β} m_{j}$ is integrable on the fixed compact frequency support. Put $G_{j, β} (x) = \int (2 π i ξ)^{β} m_{j} (ξ) e^{2 π i x \cdot ξ} d ξ$ . The bounds $∣ e^{i t} - 1 ∣ \leq ∣ t ∣$ and $∣ e^{i t} - 1 - i t ∣ \leq t^{2} / 2$ give continuity of $G_{j, β}$ and a coordinate difference-quotient remainder bounded uniformly in $x$ by $C_{j, β} ∣ h ∣ ∥ m_{j} ∥_{1}$ as $h \to 0$ . Hence $\partial_{r} G_{j, β} = G_{j, β + e_{r}}$ , proving $G_{j} \in C^{\infty}$ . These derivatives are bounded; repeated integration by parts against rapidly decaying Schwartz tests therefore has no boundary term and identifies each classical derivative with its regular distributional derivative as defined in [F6]. We henceforth use this smooth representative for $K_{j}$ and its gradients.

2.1F3F4F6step 1.2algebra

Claim: for every multi-index $γ$ with $∣ γ ∣ \leq n_{0}$ , the product $x^{γ} K_{j}$ is (the regular distribution of) an $L^{2}$ function and $∥ x^{γ} K_{j} ∥_{2} = (2 π)^{- ∣ γ ∣} ∥ \partial^{γ} m_{j} ∥_{2} .$ Indeed, applying the first identity of [F4] to $u = K_{j}$ and using $F K_{j} = u_{m_{j}}$ gives $F (x^{γ} K_{j}) = (- 1 / (2 π i))^{∣ γ ∣} \partial^{γ} u_{m_{j}} = c_{γ} u_{\partial^{γ} m_{j}}$ for the scalar $c_{γ} = (- 1 / (2 π i))^{∣ γ ∣}$ , the last equality because $\partial^{γ} m_{j}$ is continuous and compactly supported, hence a regular distribution, and differentiation of a regular distribution of a $C^{∣ γ ∣}$ function is the regular distribution of its classical derivative. Since $\partial^{γ} m_{j} \in C_{c} \subset L^{2}$ , [F3] applied to $h = c_{γ} \partial^{γ} m_{j}$ identifies $x^{γ} K_{j}$ with the regular distribution of $F_{2}^{- 1} (c_{γ} \partial^{γ} m_{j})$ , and Plancherel gives $∥ x^{γ} K_{j} ∥_{2} = ∥ c_{γ} \partial^{γ} m_{j} ∥_{2} = (2 π)^{- ∣ γ ∣} ∥ \partial^{γ} m_{j} ∥_{2}$ .

2.2F1step 1.2algebra

Claim: there is $C_{n, χ}$ with $∥ \partial^{γ} m_{j} ∥_{2} \leq C_{n, χ} A 2^{j (n / 2 - ∣ γ ∣)}$ for all $j \in Z$ and all $∣ γ ∣ \leq n_{0}$ . Leibniz's rule on $m_{j} = m_{0} ζ (2^{- j} \cdot)$ gives $\partial^{γ} m_{j} = \sum_{δ \leq γ} C_{δ, γ} \partial^{γ - δ} (ζ (2^{- j} \cdot)) \partial^{δ} m_{0}$ ; the chain rule bounds the factor by $2^{- j ∣ γ - δ ∣} ∥ \partial^{γ - δ} ζ ∥_{\infty}$ , and on the support of $m_{j}$ one has $∣ \partial^{δ} m_{0} (ξ) ∣ \leq C_{δ} ∣ ξ ∣^{- ∣ δ ∣} \leq C_{δ} 2^{- j ∣ δ ∣} \cdot 2^{∣ δ ∣}$ since $∣ ξ ∣ \geq 2^{j - 1}$ . Taking $L^{2}$ norms and bounding the support measure by $∣ S^{n - 1} ∣ (2^{n} - 2^{- n}) 2^{j n}$ yields $∥ \partial^{γ} m_{j} ∥_{2} \leq \sum_{δ \leq γ} C_{δ, γ} 2^{- j ∣ γ - δ ∣} ∥ \partial^{γ - δ} ζ ∥_{\infty} C_{δ} 2^{- j ∣ δ ∣} 2^{j n / 2} \cdot c_{n}$ , that is, $C_{n, χ} (\max_{∣ δ ∣ \leq n_{0}} C_{δ}) 2^{j (n / 2 - ∣ γ ∣)} \leq C_{n, χ} A 2^{j (n / 2 - ∣ γ ∣)}$ , because $∣ γ - δ ∣ + ∣ δ ∣ = ∣ γ ∣$ for $δ \leq γ$ .

3.1F1F5givenstep 2.1step 2.2algebra

Proof of (1). Fix $j$ and write $w (x) : = (1 + 2^{j} ∣ x ∣)^{1 / 4}$ , $W (x) : = (1 + 2^{j} ∣ x ∣)^{n_{0}}$ . Since $- 2 n_{0} + 1 / 2 < - n$ , the substitution $u = 2^{j} x$ gives $\int_{R^{n}} W (x)^{- 2} w (x)^{2} d x = \int_{R^{n}} (1 + 2^{j} ∣ x ∣)^{- 2 n_{0} + 1 / 2} d x = 2^{- j n} c_{n}$ for a constant $c_{n} = \int (1 + ∣ u ∣)^{- 2 n_{0} + 1 / 2} d u$ . Cauchy–Schwarz and the elementary bound $W (x) \leq C (n) \sum_{∣ γ ∣ \leq n_{0}} 2^{j ∣ γ ∣} ∣ x^{γ} ∣$ give $\int ∣ K_{j} ∣ w \leq (\int W^{2} ∣ K_{j} ∣^{2})^{1 / 2} (\int W^{- 2} w^{2})^{1 / 2} \leq C (n) \sum_{∣ γ ∣ \leq n_{0}} 2^{j ∣ γ ∣} ∥ x^{γ} K_{j} ∥_{2} \cdot 2^{- j n / 2} c_{n}^{1 / 2}$ . By steps 2.1 and 2.2 this is at most $C_{n, χ} A \sum_{∣ γ ∣ \leq n_{0}} 2^{j ∣ γ ∣} 2^{j (n / 2 - ∣ γ ∣)} 2^{- j n / 2} = C_{n, χ}^{'} A$ , uniformly in $j$ .

4.1F4step 2.1step 2.2step 3.1algebra

Proof of (2), one coordinate at a time. Fix $r \leq n$ and put $ζ_{r} (ξ) : = ξ_{r} ζ (ξ)$ and ${\tilde{m}}_{j} (ξ) : = m (ξ) ζ_{r} (2^{- j} ξ) = 2^{- j} ξ_{r} m_{j} (ξ)$ , so that ${\tilde{m}}_{j}$ is again compactly supported and $C^{n_{0}}$ . The second identity of [F4] gives $F (\partial_{r} K_{j}) = (2 π i ξ_{r}) F K_{j} = 2 π i u_{ξ_{r} m_{j}}$ , hence $F (2^{- j} \partial_{r} K_{j}) = 2 π i u_{{\tilde{m}}_{j}}$ . Identifying $2^{- j} \partial_{r} K_{j}$ with the regular distribution of $F_{2}^{- 1} (2 π i {\tilde{m}}_{j})$ as in step 1.2 and repeating steps 2.1, 2.2 and 3.1 with the fixed cutoff $ζ_{r}$ in place of $ζ$ (whose support and derivatives are again bounded by constants $C_{n, χ}$ ) yields $\sup_{j} \int ∣ 2^{- j} \partial_{r} K_{j} ∣ w \leq C_{n, χ} A$ . Summing these $n$ estimates over $r \leq n$ and using $∣ \nabla K_{j} ∣ \leq \sum_{r} ∣ \partial_{r} K_{j} ∣$ gives (2).

5.1step 3.1step 4.1∎

Steps 3.1 and 4.1 are exactly the two asserted estimates, with constants depending only on $n$ and the fixed cutoff $χ$ ; the auxiliary claim of step 1.2 supplies the $L^{2}$ reading of $K_{j}$ used throughout. This proves the lemma.

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Dyadic Mihlin pieces sum to an off-support kernel representation

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). In the setting of Dyadic Mihlin pieces: uniform L1 and first-difference bounds — $m$ a Mihlin symbol with constants $C_{α}$ (Mihlin smoothness convention above half the dimension), $ζ$ the annulus cutoff with $\sum_{j \in Z} ζ (2^{- j} ξ) = 1$ for $ξ \neq 0$ , $m_{j} = m ζ (2^{- j} \cdot)$ , $K_{j} = F^{- 1} (u_{m_{j}})$ — the following hold, with $W : = m^{\lor}$ and $A \geq \max {∥ m ∥_{\infty}, \max_{∣ α ∣ \leq n_{0}} C_{α}}$ :

$\sum_{∣ j ∣ \leq N} K_{j} \to W$ in $S^{'} (R^{n})$ as $N \to \infty$ ;
the series $\sum_{j \in Z} K_{j} (x)$ converges for almost every $x \in R^{n} ∖ {0}$ to a function $k$ that coincides with $W$ on $R^{n} ∖ {0}$ ; and
that function $k$ satisfies the annular size bound $\sup_{δ > 0} \int_{δ \leq ∣ x ∣ \leq 2 δ} ∣ k (x) ∣ d x \leq C_{n} A$ and Hörmander's condition $\sup_{y \neq 0} \int_{∣ x ∣ \geq 2 ∣ y ∣} ∣ k (x - y) - k (x) ∣ d x \leq C_{n} A$ .

Facts & Assumptions

Given: Countable Choice; the Mihlin symbol $m$ with constants $C_{α}$ and $q = n_{0} = ⌊ n / 2 ⌋ + 1$ ; the cutoff $ζ$ and the pieces $m_{j}, K_{j}$ ; a finite $A \geq \max {∥ m ∥_{\infty}, \max_{∣ α ∣ \leq n_{0}} C_{α}}$ ; a scale $δ > 0$ ; a dyadic integer $k$ and a vector $y \neq 0$ .

[F1]

$S^{'}$ is endowed with the pairing $⟨ u, φ ⟩$ ; the Fourier transform $F$ is a linear automorphism of $S^{'}$ with inverse $F^{- 1}$ , and $F$ maps $S$ into $S$ ; and for every $ψ \in S$ the regular distribution $u_{m_{j}}$ of the $L^{1}$ function $m_{j}$ satisfies $⟨ u_{m_{j}}, ψ ⟩ = \int m_{j} ψ$ . Hence $⟨ K_{j}, φ ⟩ = ⟨ F^{- 1} u_{m_{j}}, φ ⟩ = ⟨ u_{m_{j}}, F^{- 1} φ ⟩ = \int m_{j} (ξ) (F^{- 1} φ) (ξ) d ξ$ for $φ \in S$ (Fourier transform of a tempered distribution).

[F2]

$m$ agrees with a $C^{n_{0}}$ function off the origin, $∣ m ∣ \leq ∥ m ∥_{\infty} \leq A$ almost everywhere, and the cutoff $ζ$ is nonnegative, supported in $1 / 2 \leq ∣ ξ ∣ \leq 2$ , bounded by $1$ , with $\sum_{j \in Z} ζ (2^{- j} ξ) = 1$ for $ξ \neq 0$ ; consequently $\sum_{∣ j ∣ \leq N} m_{j} (ξ) = m (ξ) \sum_{∣ j ∣ \leq N} ζ (2^{- j} ξ) \to m (ξ)$ for every $ξ \neq 0$ , with $∣ \sum_{∣ j ∣ \leq N} m_{j} ∣ \leq ∥ m ∥_{\infty}$ (Mihlin smoothness convention above half the dimension, Dyadic Mihlin pieces: uniform L1 and first-difference bounds).

[F3]

$\sup_{j} \int ∣ K_{j} (x) ∣ (1 + 2^{j} ∣ x ∣)^{1 / 4} d x \leq C_{n} A$ and $\sup_{j} 2^{- j} \int ∣ \nabla K_{j} (x) ∣ (1 + 2^{j} ∣ x ∣)^{1 / 4} d x \leq C_{n} A$ (Dyadic Mihlin pieces: uniform L1 and first-difference bounds).

[F4]

Dominated convergence permits passage to an almost-everywhere limit under an integrable majorant, and Tonelli permits interchange of nonnegative sums and integrals on the sigma-finite Euclidean product. (Dominated convergence, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product)

Proof

technique · direct

1.1F1F2F4algebra

Transposition of the inverse Schwartz transform is the inverse distribution transform: composing either way tests against $F F^{- 1} φ = φ$ . Thus the inverse in [F1] uses $F^{- 1}$ on tests. The partial sums converge in $S^{'}$ : for $φ \in S$ , using [F1] and [F2], $⟨ \sum_{∣ j ∣ \leq N} K_{j}, φ ⟩ = \int_{R^{n}} (\sum_{∣ j ∣ \leq N} m_{j} (ξ)) (F^{- 1} φ) (ξ) d ξ ⟶ \int_{R^{n}} m (ξ) (F^{- 1} φ) (ξ) d ξ = ⟨ W, φ ⟩,$ by dominated convergence with majorant $∥ m ∥_{\infty} ∣ F^{- 1} φ ∣ \in L^{1}$ and pointwise convergence $\sum_{∣ j ∣ \leq N} m_{j} (ξ) = m (ξ) \sum_{∣ j ∣ \leq N} ζ (2^{- j} ξ) \to m (ξ)$ for $ξ \neq 0$ ; the limit pairing is $⟨ m^{\lor}, φ ⟩ = ⟨ W, φ ⟩$ .

1.2F2F3givenalgebra

Two elementary estimates. First, from $K_{j} (x) = \int m_{j} (ξ) e^{2 π i x \cdot ξ} d ξ$ , which is absolutely convergent because $m_{j}$ is supported in the annulus $2^{j - 1} \leq ∣ ξ ∣ \leq 2^{j + 1}$ and bounded by $∥ m ∥_{\infty}$ , one has the pointwise bound $∣ K_{j} (x) ∣ \leq 2^{j n} C_{ζ} A$ for every $j$ and every $x$ . Second, for every $δ > 0$ and $j > 0$ , [F3] gives $\int_{∣ x ∣ \geq δ} ∣ K_{j} (x) ∣ d x \leq C_{n} A (1 + 2^{j} δ)^{- 1 / 4}$ , so $\sum_{j > 0} \int_{∣ x ∣ \geq δ} ∣ K_{j} ∣$ is finite for every fixed $δ > 0$ , and for $j \leq 0$ the pointwise bound gives $\int_{δ \leq ∣ x ∣ \leq 2 δ} ∣ K_{j} ∣ \leq 2^{j n} C_{ζ} A ∣ B (0, 2 δ) ∣ \leq C_{n} A δ^{n} 2^{j n}$ , whose sum over $j \leq 0$ is at most $C_{n} A δ^{n}$ , finite for every fixed $δ$ .

2.1F4step 1.2algebra

Almost everywhere convergence. By step 1.2, $\sum_{j \leq 0} ∣ K_{j} (x) ∣ < \infty$ for every $x$ ; for each fixed $δ > 0$ , $\sum_{j > 0} \int_{∣ x ∣ \geq δ} ∣ K_{j} ∣ < \infty$ implies $\sum_{j > 0} ∣ K_{j} (x) ∣ < \infty$ for almost every $x$ with $∣ x ∣ \geq δ$ . Taking the union of the exceptional sets over $δ = 1 / m$ , $m \geq 1$ , the series $\sum_{j \in Z} K_{j} (x)$ converges absolutely for almost every $x \in R^{n} ∖ {0}$ ; denote its sum by $k (x)$ , a measurable function on $R^{n} ∖ {0}$ .

3.1F4step 1.1step 1.2step 2.1algebra

On every compact $K \subset R^{n} ∖ {0}$ the dominating function $\sum_{j} ∣ K_{j} ∣$ is integrable: $K$ lies in some annulus $δ \leq ∣ x ∣ \leq 2^{m} δ$ , and the two estimates of step 1.2 (the second applied after covering the outer annulus by finitely many dyadic annuli of the same type) give $\int_{K} \sum_{j} ∣ K_{j} ∣ < \infty$ . Hence for $φ \in C_{c}^{\infty} (R^{n} ∖ {0})$ , dominated convergence with the partial sums bounded by $\sum_{j} ∣ K_{j} ∣$ gives $⟨ W, φ ⟩ = \lim_{N} ⟨ \sum_{∣ j ∣ \leq N} K_{j}, φ ⟩ = \int k φ$ by step 1.1, so $k$ coincides with $W$ on $R^{n} ∖ {0}$ .

3.2F3step 1.2step 2.1algebra

Annular size bound. Fix $δ > 0$ . Split $\sum_{j} \int_{δ \leq ∣ x ∣ \leq 2 δ} ∣ K_{j} ∣$ according to whether $2^{j} δ > 1$ . For $2^{j} δ \leq 1$ the pointwise bound of step 1.2 gives $\int_{δ \leq ∣ x ∣ \leq 2 δ} ∣ K_{j} ∣ \leq 2^{j n} C_{ζ} A ∣ B (0, 2 δ) ∣$ , so the sum over these $j$ is at most $C_{n} A δ^{n} \sum_{2^{j} \leq 1 / δ} 2^{j n} \leq C_{n} A$ . For $2^{j} δ > 1$ , $\int_{δ \leq ∣ x ∣ \leq 2 δ} ∣ K_{j} ∣ \leq (2^{j} δ)^{- 1 / 4} \int ∣ K_{j} ∣ (1 + 2^{j} ∣ x ∣)^{1 / 4} \leq C_{n} A (2^{j} δ)^{- 1 / 4}$ by [F3], Put $j_{0} = \min {j \in Z : 2^{j} δ > 1}$ , so $1 < 2^{j_{0}} δ \leq 2$ by minimality. The high-frequency sum is therefore bounded by $C_{n} A \sum_{j \geq j_{0}} (2^{j} δ)^{- 1 / 4} = C_{n} A (2^{j_{0}} δ)^{- 1 / 4} / (1 - 2^{- 1 / 4}) \leq C_{n} A / (1 - 2^{- 1 / 4})$ , independent of $δ$ . Hence $\int_{δ \leq ∣ x ∣ \leq 2 δ} ∣ k ∣ \leq \sum_{j} \int_{δ \leq ∣ x ∣ \leq 2 δ} ∣ K_{j} ∣ \leq C_{n} A$ , uniformly in $δ$ .

3.3F3step 1.2step 2.1algebra

Hörmander's condition. Fix $y \neq 0$ and choose $k \in Z$ with $2^{- k} \leq ∣ y ∣ \leq 2^{1 - k}$ . For $j > k$ the triangle inequality and [F3] give $\int_{∣ x ∣ \geq 2 ∣ y ∣} ∣ K_{j} (x - y) - K_{j} (x) ∣ d x \leq 2 \int_{∣ x ∣ \geq ∣ y ∣} ∣ K_{j} (x) ∣ d x \leq 2 C_{n} A (1 + 2^{j} ∣ y ∣)^{- 1 / 4}$ , and summing over $j > k$ yields at most $C_{n} A$ , since $2^{j} ∣ y ∣ \geq 2^{j - k}$ . For $j \leq k$ , the mean value theorem and translation of the integral give $\int_{∣ x ∣ \geq 2 ∣ y ∣} ∣ K_{j} (x - y) - K_{j} (x) ∣ d x \leq ∣ y ∣ \int_{R^{n}} ∣ \nabla K_{j} (u) ∣ d u \leq C_{n} A ∣ y ∣ 2^{j}$ by [F3]. Hence the sum over $j \leq k$ is bounded by $C_{n} A ∣ y ∣ \sum_{j \leq k} 2^{j} = 2 C_{n} A ∣ y ∣ 2^{k} \leq 4 C_{n} A$ , using the upper dyadic inequality $∣ y ∣ \leq 2^{1 - k}$ . Summing in $j$ yields the asserted Hörmander bound, uniformly in $y \neq 0$ .

4.1step 1.1step 3.1step 3.2step 3.3∎

Steps 1.1, 3.1, 3.2 and 3.3 are the four assertions: $S^{'}$ convergence, the a.e. convergent series defining $k$ , its coincidence with $W$ off the origin, and the two kernel bounds with constant $C_{n} A$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The Mihlin–Hörmander Fourier multiplier theorem

Statement

Assume Countable Choice. Let $n \geq 1$ , put $q : = ⌊ n / 2 ⌋ + 1$ , and let $m$ be a Mihlin symbol with constants $C_{α}$ , $∣ α ∣ \leq q$ (Mihlin smoothness convention above half the dimension); put $A : = \max_{∣ α ∣ \leq q} C_{α}$ , so that $∥ m ∥_{\infty} \leq C_{0} \leq A$ . Then $m$ is an $L^{p}$ Fourier multiplier for every $1 < p < \infty$ (Lp Fourier multiplier and its norm), and there is a constant $C_{n}$ , depending only on $n$ , with $∥ m ∥_{M_{p}} \leq C_{n} \max (p, (p - 1)^{- 1}) (A + ∥ m ∥_{\infty}) (1 < p < \infty) .$ Equivalently, the operator $T_{m}$ of Translation-invariant Fourier multiplier on the Schwartz core satisfies $∥ T_{m} f ∥_{p} \leq C_{n} \max (p, (p - 1)^{- 1}) (A + ∥ m ∥_{\infty}) ∥ f ∥_{p}$ for every $f \in L^{p} (R^{n}; C)$ . Only strict-range bounds are asserted: no endpoint $p = 1$ or $p = \infty$ claim is made.

Facts & Assumptions

Given: Countable Choice; $n \geq 1$ , $q = n_{0} = ⌊ n / 2 ⌋ + 1$ , the Mihlin symbol $m$ with constants $C_{α}$ and the constant $A = \max_{∣ α ∣ \leq q} C_{α}$ ; a compactly supported $f \in L^{2} (R^{n}; C)$ ; the mollifier family $φ_{ε}$ generated by a unit-mass $φ \in C_{c}^{\infty}$ (The mollifier family generated by a unit-mass smooth bump, A unit-mass smooth bump generates an $L^{1}$ approximate identity).

[F1]

In the dyadic-piece setting attached to $m$ — the annulus cutoff $ζ$ , the symbols $m_{j} = m ζ (2^{- j} \cdot)$ and the functions $K_{j} = F^{- 1} (u_{m_{j}})$ — one has $\sup_{j} \int ∣ K_{j} (x) ∣ (1 + 2^{j} ∣ x ∣)^{1 / 4} d x \leq C_{n} A$ and $\sup_{j} 2^{- j} \int ∣ \nabla K_{j} (x) ∣ (1 + 2^{j} ∣ x ∣)^{1 / 4} d x \leq C_{n} A$ whenever $A \geq \max {∥ m ∥_{\infty}, \max_{∣ α ∣ \leq n_{0}} C_{α}}$ (Mihlin smoothness convention above half the dimension, Dyadic Mihlin pieces: uniform L1 and first-difference bounds).

[F2]

With $W : = F^{- 1} (u_{m})$ , a tempered distribution extending the off-origin kernel (no principal-value representation is asserted): the partial sums $\sum_{∣ j ∣ \leq N} K_{j}$ converge to $W$ in $S^{'} (R^{n})$ ; the series $\sum_{j \in Z} K_{j} (x)$ converges for almost every $x \in R^{n} ∖ {0}$ to a function $k$ with $k = W$ on $R^{n} ∖ {0}$ (that is, $⟨ W, ψ ⟩ = \int k ψ$ for every $ψ \in C_{c}^{\infty} (R^{n} ∖ {0})$ ); and $k$ satisfies the annular bound $\sup_{δ > 0} \int_{δ \leq ∣ x ∣ \leq 2 δ} ∣ k ∣ \leq C_{n} A$ and Hörmander's condition $\sup_{y \neq 0} \int_{∣ x ∣ \geq 2 ∣ y ∣} ∣ k (x - y) - k (x) ∣ d x \leq C_{n} A$ (Dyadic Mihlin pieces sum to an off-support kernel representation).

[F3]

$S (R^{n}) \subseteq D_{m}$ and, as $L^{2}$ classes, $T_{m} f = F_{2}^{- 1} (m F_{2} f)$ for $f \in S$ , with $∥ T_{m} f ∥_{2} \leq ∥ m ∥_{\infty} ∥ f ∥_{2}$ ; the Schwartz-core action extends uniquely to the bounded $L^{2}$ operator $T_{m} = F_{2}^{- 1} M_{m} F_{2}$ of norm $∥ m ∥_{\infty}$ (Exact L2 Fourier multiplier norm, Translation-invariant Fourier multiplier on the Schwartz core).

[F4]

For $u \in S^{'}$ and Schwartz $f$ , $F (u * f) = (F u) (F f)$ ; the convolution $u * f$ is the smooth function of polynomial growth $x \mapsto ⟨ u_{y}, f (x - y) ⟩$ (Fourier transform converts allowed tempered convolutions to products, Tempered convolution is smooth with polynomial growth, Convolution of a tempered distribution with a schwartz function).

[F5]

A Calderón–Zygmund kernel in the base sense with constants $A_{1}, A_{2}$ and its $L^{2}$ -bounded operator with norm $B$ satisfy $∥ T g ∥_{p} \leq C_{n} (A_{2} + B) \max (p, (p - 1)^{- 1}) ∥ g ∥_{p}$ for $1 < p < \infty$ (Calderón–Zygmund kernels and their associated operators, Calderón–Zygmund operators are bounded on Lp); an $L^{1}$ approximate identity converges in $L^{p}$ : $∥ g * φ_{ε} - g ∥_{p} \to 0$ for $g \in L^{p}$ , $1 \leq p < \infty$ (Every $L^{1}$ approximate identity converges to the identity in $L^{p}$ for $1 \leq p < \infty$ ); and $g * φ_{ε}$ is smooth for locally integrable $g$ (Convolution with a mollifier is smooth, and derivatives pass under the integral sign).

[F6]

Tonelli interchanges nonnegative product integrals; dominated convergence applies under an integrable majorant; locally integrable functions are determined almost everywhere by their distribution pairings. (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, Dominated convergence, Locally integrable functions embed in distributions)

Proof

technique · direct

1.1F1F2given

The dyadic pieces and the kernel. The hypotheses of [F1] are satisfied, so with $ζ$ , $m_{j}$ , $K_{j}$ as there the weighted bounds hold; [F2] then supplies the tempered distribution $W = F^{- 1} (u_{m})$ , the almost-everywhere convergent series $k = \sum_{j} K_{j}$ on $R^{n} ∖ {0}$ coinciding with $W$ off the origin, and the annular and Hörmander bounds $\leq C_{n} A$ for $k$ .

1.2F1F2F4F6givenalgebra

On Schwartz inputs, [F4] gives $F (W * f) = u_{m \hat{f}}$ , so $W * f = T_{m} f$ . Suppose $x \notin supp f$ and put $ψ (y) = f (x - y)$ , which vanishes near zero. Cut $ψ$ off at infinity using the fixed cutoff $χ_{R}$ of [F1]. Then $χ_{R} ψ \to ψ$ in Schwartz space, and [F2] gives $⟨ W, χ_{R} ψ ⟩ = \int k χ_{R} ψ$ . The annular bound makes $\int ∣ k ψ ∣$ finite: on $2^{j} d \leq ∣ y ∣ \leq 2^{j + 1} d$ , its contribution is at most $C_{n} A \sup_{∣ y ∣ \geq 2^{j} d} ∣ ψ (y) ∣$ , summable for $j \geq 0$ by Schwartz decay, where $d > 0$ is smaller than the distance from zero to the support of $ψ$ . Dominated convergence [F6] gives $T_{m} f (x) = \int k (y) f (x - y) d y$ . Only annular size and local integrability are used.

2.1F2F3F5F6step 1.2algebra

For compactly supported $f \in L^{2}$ , its mollifications $f_{ε} = f * φ_{ε}$ are smooth with uniformly bounded compact support and converge to $f$ in $L^{1}$ and $L^{2}$ by [F5]. Fix a compact $E$ disjoint from $supp f$ ; for sufficiently small $ε$ all these supports lie in a fixed compact set $S$ disjoint from $E$ . The compact difference set $E - S$ avoids zero. Tonelli and [F2] give $\int_{E} \int_{S} ∣ k (x - y) ∣ ∣ f_{ε} (y) - f (y) ∣ d y d x \leq ∥ f_{ε} - f ∥_{1} \int_{E - S} ∣ k (z) ∣ d z \to 0$ . The same estimate with $f$ proves absolute convergence of its kernel integral almost everywhere on $E$ . Step 1.2 identifies $T_{m} f_{ε}$ there with the kernel integral of $f_{ε}$ . By [F3], $T_{m} f_{ε} \to T_{m} f$ in $L^{2}$ , hence in $L^{1} (E)$ by Cauchy–Schwarz. Uniqueness of the $L^{1} (E)$ limit identifies $T_{m} f$ with the kernel integral of $f$ . A countable compact exhaustion proves the off-support representation on $(supp f)^{c}$ .

3.1F3F5step 1.1step 2.1algebra

The kernel $k$ has annular and Hörmander constants at most $C_{n} A$ , and $T_{m}$ has $L^{2}$ norm $B = ∥ m ∥_{\infty}$ . Step 2.1 proves its off-support representation, so $T_{m}$ is a Calderón–Zygmund operator. Apply the dimension-only estimate proved in the quantitative argument of Calderón–Zygmund operators are bounded on Lp to get $∥ T_{m} g ∥_{p} \leq C_{n} \max (p, (p - 1)^{- 1}) (A + ∥ m ∥_{\infty}) ∥ g ∥_{p}$ . The constant has no unrecorded dependence on $p$ .

4.1F3F5step 3.1∎

Consequently $m$ is an $L^{p}$ Fourier multiplier in the sense of the multiplier definition: [F3] gives $S \subseteq D_{m}$ , for $f \in S$ the distribution $T_{m} f$ is the regular distribution of the $L^{2}$ class $F_{2}^{- 1} (m F_{2} f)$ , which is the class assigned to $f$ by the $L^{p}$ extension of step 3.1 (the two extensions of the Schwartz-core action agree on the dense subspace $S$ ), and step 3.1 is exactly the required norm bound. Hence $∥ m ∥_{M_{p}} \leq C_{n} \max (p, (p - 1)^{- 1}) (A + ∥ m ∥_{\infty})$ for every $1 < p < \infty$ , which is the assertion.

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Endpoint targets: weak (1,1) here, L∞ to BMO later; strong endpoints fail in general

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). For an $L^{2}$ -bounded Calderón–Zygmund operator the two endpoint-adjacent facts established on this page are the weak $(1, 1)$ estimate of Calderón–Zygmund operators are of weak type (1,1) and the strong $L^{p}$ bounds for $1 < p < \infty$ of Calderón–Zygmund operators are bounded on Lp. The following claims are deliberately not made here, and the reader should not read the page as asserting or refuting them.

First, no strong type $(1, 1)$ bound and no bounded action on $L^{\infty} (R^{n})$ is claimed; the weak $(1, 1)$ estimate is the endpoint substitute for the former, and the companion examples page exhibits the Hilbert transform of an interval indicator as a counterexample to compatible strong $L^{1}$ and $L^{\infty}$ conclusions, for the Hilbert transform and hence for the class of Calderón–Zygmund operators.

Second, the $L^{\infty} \to B M O$ endpoint is the subject of the later BMO page of this track, where bounded mean oscillation is defined and the endpoint estimate is proved; neither the statement nor the proof of that estimate is used here, and no $L^{\infty}$ conclusion is available from it.

The strict-range bounds are stated with the exponent range $1 < p < \infty$ only; the constants blow up as $p ↓ 1$ and as $p \to \infty$ in the estimates recorded above, consistently with the two refuted endpoints.

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Mihlin endpoints: weak (1,1), but no general strong endpoint bounds

Statement

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). The displayed estimate in The Mihlin–Hörmander Fourier multiplier theorem gives strong $L^{p}$ bounds for $1 < p < \infty$ . Its proof also identifies $T_{m}$ as a Calderón–Zygmund operator with Hörmander constant at most $C_{n} A$ and $L^{2}$ norm $∥ m ∥_{\infty}$ . Consequently Calderón–Zygmund operators are of weak type (1,1) gives the weak endpoint $∣ {∣ T_{m} f ∣ > λ} ∣ \leq C_{n} (A + ∥ m ∥_{\infty}) λ^{- 1} ∥ f ∥_{1} (f \in L^{1}, λ > 0),$ where $A$ is the derivative-bound constant in the Mihlin theorem. This is a bound for the multiplier operator; it does not require the stronger principal-value and pointwise kernel hypotheses used for maximal truncations.

Remarks

No general strong $L^{1}$ or $L^{\infty}$ bound follows. In dimension one, $m (ξ) = - i sgn ξ$ satisfies the Mihlin hypotheses despite its jump at the origin, and its Hilbert transform has the weak endpoint above while the interval-indicator counterexamples refute compatible strong $L^{1}$ and $L^{\infty}$ bounds (Strong type (1,1) fails for the Hilbert transform ↗, Calderón–Zygmund operators need not map L∞ to L∞ ↗). A jump at a nonzero frequency violates the Mihlin smoothness hypothesis; boundedness of a symbol alone does not imply a strong $L^{1}$ bound.

5 · Examples, counterexamples and false statements

None yet.