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BMO, John-Nirenberg, and H1 Duality

1 · Prerequisites

2 · Summary

This page develops the space $B M O (R^{n})$ of bounded mean oscillation modulo constants, the John-Nirenberg exponential inequality, and the duality $(H^{1} (R^{n}))^{*} ≅ B M O (R^{n}) / C$ of the real Hardy space $H^{1}$ of the companion FR-9 page. The definition fixes the cube-based seminorm and the quotient by the constants, with the mean recorded as the optimal constant up to the factor $2$ ; the zero seminorm is exactly the class of the constants. Two elementary tools are developed first: nested-cube averages grow at most logarithmically, and range truncations preserve the seminorm up to the constants $3 / 2$ , $9 / 4$ and $9 / 2$ that the Banach-Alaoglu step needs.

The John-Nirenberg theory is proved by recursive stopping on the all-generations dyadic grid at a fixed height $s > ∥ b ∥_{B M O}$ : the generation- $k$ cubes decay geometrically in measure, and off them the oscillation is at most $k 2^{n} s$ , which integrates to the exponential tail bound and, through layer cake, to the equivalence of the $L^{q}$ oscillation norms for every $1 \leq q < \infty$ . A separate local/far kernel decomposition, together with the $L^{2}$ bound, proves the endpoint result at the other end of the scale: every Calderon-Zygmund operator with a standard Holder kernel maps $L^{\infty}$ into $B M O$ modulo constants, and the Hilbert and Riesz transforms are verified to be such operators, closing the $L^{\infty}$ endpoint left open by the preceding pages.

The second half of the page proves the duality. Atoms pair boundedly with BMO functions, $L^{2}$ -normalised atoms and mean-zero $L^{2}$ functions on a cube embed continuously into $H^{1}$ , and finite atomic sums are dense. A bounded functional on $H^{1}$ is then represented on each local mean-zero $L^{2}$ space by Riesz representation, the local representatives glue to a BMO function, and their uniform oscillation bound gives surjectivity; injectivity of the atom pairing gives uniqueness of the class. Hence $(H^{1})^{*}$ is isomorphic to $B M O / C$ with equivalent norms. The proof assumes the Axiom of Choice, used for the weak-star cluster point of the truncated functionals in the dual of $H^{1}$ , and implying the Countable Choice assumptions of its suppliers; the earlier items assume only Countable Choice or none. The companion page carries the logarithm example, the strictness of the $L^{\infty}$ inclusion, the failure of global integrability, the worked tail integration, and a recorded two-grid dyadic result.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

BMO seminorm and the quotient by constants

Definition

Complex scalars, Lebesgue measure on $R^{n}$ , and the complex $L^{p}$ and test-function conventions of Complex Lp classes and Euclidean test-function conventions. A cube is a nondegenerate axis-parallel cube (Axis-parallel rectangles in $R^{m}$ and their volume). For $b \in L_{l o c}^{1} (R^{n})$ and a cube $Q$ put $b_{Q} : = ∣ Q ∣^{- 1} \int_{Q} b$ , so that $b_{Q}$ is the mean of $b$ over $Q$ , and define the $B M O$ seminorm $∥ b ∥_{B M O} : = \sup_{Q} ∣ Q ∣^{- 1} \int_{Q} ∣ b - b_{Q} ∣ \in [0, \infty]$ , the supremum over all cubes, and $B M O (R^{n}) : = {b \in L_{l o c}^{1} : ∥ b ∥_{B M O} < \infty}$ . The mean is optimal up to a factor of $2$ in the average: $∣ Q ∣^{- 1} \int_{Q} ∣ b - b_{Q} ∣ \leq 2 \inf_{c} ∣ Q ∣^{- 1} \int_{Q} ∣ b - c ∣ \leq 2 ∣ Q ∣^{- 1} \int_{Q} ∣ b - b_{Q} ∣$ where the infimum is over $c \in C$ . Moreover $∥ b ∥_{B M O} = 0$ exactly when $b$ is constant almost everywhere, so the page works with the quotient $B M O (R^{n}) / C$ of equivalence classes modulo constants; a class is written $b \in B M O / C$ and the zero class is the class of the constants. Replacing cubes by balls in the supremum defines an equivalent seminorm with comparison constants depending only on $n$ .

These assertions follow directly from the averages. For every $c \in C$ , $∣ b_{Q} - c ∣ \leq ∣ Q ∣^{- 1} \int_{Q} ∣ b - c ∣$ , so the triangle inequality gives the factor- $2$ upper bound; taking $c = b_{Q}$ gives the other inequality. For locally integrable functions $b, d$ , homogeneity and $(b + d)_{Q} = b_{Q} + d_{Q}$ give the seminorm laws. If the seminorm is zero, $b = b_{Q_{m}}$ almost everywhere on each $Q_{m} = [- m, m]^{n}$ , $m \geq 1$ . The constants agree on their positive-measure overlaps, and the exceptional set is the union of the explicit measurable null sets $Q_{m} \cap {b \neq b_{Q_{1}}}$ , hence is null by countable additivity of Lebesgue measure. Conversely an almost-everywhere constant has zero oscillation. Thus the quotient also identifies almost-everywhere equal representatives. Finally, if $E \subseteq F$ are measurable sets of finite positive measure and $b \in L^{1} (F)$ , then $∣ E ∣^{- 1} \int_{E} ∣ b - b_{E} ∣ \leq 2 (∣ F ∣ / ∣ E ∣) ∣ F ∣^{- 1} \int_{F} ∣ b - b_{F} ∣$ . Every ball is contained in a concentric cube of comparable volume, and every cube is contained in a concentric ball of comparable volume. Applying this inequality in both directions proves the cube/ball comparison.

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

The Hilbert and Riesz transforms are Calderon-Zygmund operators

Statement

Assume Countable Choice. Let $H$ be the Hilbert transform of the line and, for $1 \leq j \leq n$ , let $R_{j}$ be the Riesz transform of Riesz transforms on Euclidean space. Then $H$ and each $R_{j}$ are Calderon-Zygmund operators with the kernels $k (x) = 1 / (π x)$ and $K_{j} (x) = c_{n} x_{j} / ∣ x ∣^{n + 1}$ : they are $L^{2}$ -bounded with norm at most $1$ , and for every compactly supported $f \in L^{2}$ and almost every $x \notin supp f$ one has $T f (x) = \int k (x - y) f (y) d y$ with the kernel of $T$ . Moreover both kernels are standard $1$ -Holder Calderon-Zygmund kernels with the published constants ( $A_{2}^{'} = 2 / π$ and $C_{n} = c_{n} 2^{n + 1} (3 n + 4)$ , respectively).

Facts & Assumptions

Given: Countable Choice, the Hilbert transform $H$ on $R$ with kernel $k (x) = 1 / (π x)$ , and the Riesz transforms $R_{1}, \dots, R_{n}$ on $R^{n}$ with kernels $K_{j} (x) = c_{n} x_{j} / ∣ x ∣^{n + 1}$ .

[F1]

The Hilbert transform is a well-defined $L^{2}$ operator with $∥ H f ∥_{2} = ∥ f ∥_{2}$ and $H^{2} f = - f$ ; it is skew-adjoint, $⟨ H f, g ⟩ = - ⟨ f, H g ⟩$ ; on Schwartz functions $H f = W * f$ , the principal value $\lim_{ε ↓ 0} H_{ε} f (x)$ exists at every $x$ and equals $(W * f) (x)$ , with $F (H f) = - i sgn (ξ) \hat{f} (ξ)$ (The Hilbert transform is an L2 isometry and squares to minus the identity, The Hilbert transform is skew-adjoint on L2, The Hilbert transform is the tempered convolution with pv(1/(pi x)) and has signum Fourier multiplier, Truncated Hilbert transform and principal value).

[F2]

The Riesz transform $R_{j}$ has multiplier $m_{j} (ξ) = - i ξ_{j} / ∣ ξ ∣$ for $ξ \neq 0$ (and $m_{j} (0) = 0$ ), acts on Schwartz functions, and satisfies $∥ R_{j} f ∥_{2} \leq ∥ f ∥_{2}$ for all $f \in L^{2}$ ; its kernel $K_{j} (x) = c_{n} x_{j} / ∣ x ∣^{n + 1}$ obeys $∣ K_{j} (x) ∣ \leq c_{n} ∣ x ∣^{- n}$ and $∣ K_{j} (x - h) - K_{j} (x) ∣ \leq C_{n} ∣ h ∣ ∣ x ∣^{- (n + 1)}$ whenever $x \neq 0$ and $∣ h ∣ \leq ∣ x ∣ / 2$ , with $C_{n} = c_{n} 2^{n + 1} (3 n + 4)$ ; and for every Schwartz function $f$ the truncated integrals $\int_{∣ y ∣ > ε} K_{j} (y) f (x - y) d y$ converge as $ε ↓ 0$ , for every $x$ , to a continuous representative of the $L^{2}$ class $R_{j} f$ (Riesz transforms on Euclidean space, Riesz transforms are L2 contractions and square to minus the identity in sum, Riesz kernel size, difference and spherical-cancellation bounds, The Riesz transform is the principal value of its kernel, with the matching constant).

[F3]

Polar coordinates under Countable Choice: for every Borel measurable nonnegative $g$ on $R^{n}$ , $\int_{R^{n}} g (x) d x = \int_{0}^{\infty} \int_{S^{n - 1}} g (r ω) r^{n - 1} d σ (ω) d r,$ where $σ (S^{n - 1}) = ∣ S^{n - 1} ∣$ (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma).

[F4]

The Fourier transform extends to a unitary $F_{2} : L^{2} (R^{n}) \to L^{2} (R^{n})$ preserving the first-variable-linear inner product (Plancherel theorem).

[F5]

The Fubini theorem applies to $L^{1}$ functions on products of $σ$ -finite measure spaces (Fubini's theorem for L^1 functions on a sigma-finite product).

[F6]

The map sending a locally integrable function to its regular distribution is injective on $L_{l o c}^{1} (Ω)$ modulo almost-everywhere equality, for every open $Ω$ (Locally integrable functions embed in distributions).

[F7]

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

[F9]

The pointwise first-difference estimate defines the standard $1$ -Holder kernel condition once the base Calderon-Zygmund conditions have been verified (Standard (Hölder) Calderón–Zygmund kernels).

[F8]

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

Proof

technique · direct

1.1F1F7F8algebra

The Hilbert kernel $k (x) = 1 / (π x)$ is odd, continuous on $R ∖ {0}$ and hence locally integrable there, and for every $R > 0$ the annular integral is $\int_{R \leq ∣ x ∣ \leq 2 R} ∣ k (x) ∣ d x = \frac{1}{π} \int_{R \leq ∣ x ∣ \leq 2 R} \frac{d x}{∣ x ∣} = \frac{2 \ln 2}{π}$ , so condition (1) of [F7] holds with $A_{1} = 2 \ln 2 / π$ . If $∣ x ∣ \geq 2 ∣ h ∣ > 0$ then $∣ x - h ∣ \geq ∣ x ∣ / 2$ , so $∣ k (x - h) - k (x) ∣ = \frac{∣ h ∣}{π ∣ x ∣ ∣ x - h ∣} \leq \frac{2}{π} \frac{∣ h ∣}{∣ x ∣^{2}}$ ; this gives the required pointwise first-difference bound with constant $A_{2}^{'} = 2 / π$ . The standing Countable Choice hypothesis [F8] is inherited with the $L^{2}$ Hilbert theory of [F1], which is stated under it, and is used nowhere else in this step.

1.2F2F7F8algebra

The Riesz kernel $K_{j} (x) = c_{n} x_{j} / ∣ x ∣^{n + 1}$ is smooth and odd on $R^{n} ∖ {0}$ , hence locally integrable there, and $∣ K_{j} (x) ∣ \leq c_{n} ∣ x ∣^{- n}$ gives $\int_{R \leq ∣ x ∣ \leq 2 R} ∣ K_{j} (x) ∣ d x \leq c_{n} ∣ S^{n - 1} ∣ \ln 2$ for every $R > 0$ , so condition (1) of [F7] holds with $A_{1} = c_{n} ∣ S^{n - 1} ∣ \ln 2$ ; [F2] gives the pointwise first-difference bound with $A_{2}^{'} = C_{n} = c_{n} 2^{n + 1} (3 n + 4)$ . The standing Countable Choice hypothesis [F8] is inherited here with the Riesz theory of [F2], which is stated under it.

1.3F1F2F4algebra

The operators are $L^{2}$ -bounded with norm at most $1$ : $∥ H f ∥_{2} = ∥ f ∥_{2}$ and $∥ R_{j} f ∥_{2} \leq ∥ f ∥_{2}$ by [F1] and [F2]. The Hilbert transform is skew-adjoint by [F1]. For the Riesz transforms, Plancherel [F4] and the multiplier representation give, for $f, g \in L^{2} (R^{n})$ , $⟨ R_{j} f, g ⟩ = \int_{R^{n}} m_{j} \hat{f} \overline{\hat{g}} d ξ = \int_{R^{n}} \hat{f} \overline{(- m_{j}) \hat{g}} d ξ = ⟨ f, - R_{j} g ⟩$ , because $\overline{m_{j}} = - m_{j}$ on the purely imaginary symbol; hence $R_{j}^{*} = - R_{j}$ .

2.1F3step 1.1step 1.2F7F9algebra

Both kernels satisfy the annular size condition (1) of [F7] by steps 1.1 and 1.2. For either kernel $q$ and every $h \neq 0$ , the pointwise difference bound in step 1.1 or 1.2 gives $∣ q (x - h) - q (x) ∣ \leq A_{2}^{'} ∣ h ∣ ∣ x ∣^{- n - 1}$ for $∣ x ∣ \geq 2 ∣ h ∣$ . By polar coordinates [F3], $\int_{∣ x ∣ \geq 2 ∣ h ∣} ∣ q (x - h) - q (x) ∣ d x \leq A_{2}^{'} ∣ h ∣ \int_{∣ x ∣ \geq 2 ∣ h ∣} ∣ x ∣^{- n - 1} d x = A_{2}^{'} ∣ h ∣ ∣ S^{n - 1} ∣ \int_{2 ∣ h ∣}^{\infty} r^{- 2} d r = \frac{∣ S^{n - 1} ∣}{2} A_{2}^{'} .$ Thus condition (2) of [F7] holds directly with $A_{2} = ∣ S^{n - 1} ∣ A_{2}^{'} / 2$ , so both kernels are Calderon-Zygmund kernels in the base sense. Since condition (2) is now established, their pointwise first-difference bounds make them standard $1$ -Holder kernels by Standard (Hölder) Calderón–Zygmund kernels, with constants $2 / π$ and $C_{n}$ .

2.2F1F2F5step 1.3algebra

Fix a compactly supported $f \in L^{2} (R^{n})$ and put $Ω = R^{n} ∖ supp f$ ; fix also $φ \in C_{c}^{\infty} (Ω)$ . Then $supp f$ and $supp φ$ are disjoint compact sets, so their distance is positive and the kernel is bounded on $supp f \times supp φ$ ; since $f$ is integrable on its compact support by Cauchy-Schwarz, the double integral below is absolutely convergent. For $x \in supp f$ one has $x \notin supp φ$ , and the truncated integrals in the principal-value formulas [F1] and [F2] converge to the full absolutely convergent integral $\int k (x - y) φ (y) d y$ , so $T φ (x) = \int k (x - y) φ (y) d y$ there. Skew-adjointness from step 1.3 therefore gives $⟨ T f, φ ⟩ = - ⟨ f, T φ ⟩ = - \int_{supp f} f (x) \overline{\int k (x - y) φ (y) d y} d x$ , and since the kernels are real-valued this equals $- \iint f (x) k (x - y) \overline{φ (y)} d y d x$ . Substituting $k (x - y) = - k (y - x)$ by oddness and applying Fubini [F5] yields $⟨ T f, φ ⟩ = \int \overline{φ (y)} g (y) d y$ with $g (y) : = \int k (y - x) f (x) d x$ .

3.1step 2.2F6

For every $y \in Ω$ the integral defining $g (y)$ is absolutely convergent: $x \mapsto k (y - x)$ is bounded on the compact set $supp f$ , and $\int_{supp f} ∣ f (x) ∣ d x \leq ∣ supp f ∣^{1 / 2} ∥ f ∥_{2} < \infty$ . Hence $g$ is locally integrable on the open set $Ω$ , and $T f ∣_{Ω}$ is locally integrable as well because $T f \in L^{2} (R^{n})$ ; step 2.2 says that the regular distributions of $T f ∣_{Ω}$ and $g ∣_{Ω}$ agree on every test function supported in $Ω$ , so the injectivity of [F6], applied on $Ω$ , gives $T f = g$ almost everywhere on $Ω$ .

4.1step 2.1step 1.3step 3.1F7∎

By step 2.1 the kernels $k$ and $K_{j}$ are Calderon-Zygmund kernels, by step 1.3 the operators $H$ and $R_{j}$ are $L^{2}$ -bounded with norm at most $1$ and skew-adjoint, and by step 3.1 the off-support representation (3) of [F7] holds: for almost every $x \notin supp f$ , $T f (x) = \int k (x - y) f (y) d y$ . Therefore $H$ and $R_{j}$ are Calderon-Zygmund operators with the kernels $k$ and $K_{j}$ , and the standard $1$ -Holder constants are $A_{2}^{'} = 2 / π$ and $C_{n} = c_{n} 2^{n + 1} (3 n + 4)$ .

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

L-infinity embeds continuously into BMO modulo constants

Statement

Every $b \in L^{\infty} (R^{n})$ lies in $B M O (R^{n})$ with $∥ b ∥_{B M O} \leq 2 ∥ b ∥_{L^{\infty}}$ , so the class map $L^{\infty} (R^{n}) / C \to B M O (R^{n}) / C$ is a continuous injection. Whether the injection is surjective is not claimed here; the companion examples page records an unbounded BMO function, which exhibits strictness independently.

Facts & Assumptions

Given: A bounded function $b \in L^{\infty} (R^{n})$ and a cube $Q$ , with the cube convention, the mean $b_{Q}$ and the seminorm and quotient of BMO seminorm and the quotient by constants.

[F1]

The mean is defined by $b_{Q} = ∣ Q ∣^{- 1} \int_{Q} b$ , and the seminorm is $∥ b ∥_{B M O} = \sup_{Q} ∣ Q ∣^{- 1} \int_{Q} ∣ b - b_{Q} ∣$ ; the zero-seminorm class is exactly the class of functions constant almost everywhere (BMO seminorm and the quotient by constants).

[F2]

On a cube of finite volume, a bounded function is integrable and $∣ b_{Q} ∣ \leq ∥ b ∥_{L^{\infty}}$ (BMO seminorm and the quotient by constants).

Proof

technique · direct

1.1F1F2algebra

For every cube $Q$ the triangle inequality and [F2] give $∣ Q ∣^{- 1} \int_{Q} ∣ b - b_{Q} ∣ \leq ∣ Q ∣^{- 1} \int_{Q} ∣ b ∣ + ∣ b_{Q} ∣ \leq 2 ∥ b ∥_{L^{\infty}}$ , so $b$ is locally integrable and $∥ b ∥_{B M O} \leq 2 ∥ b ∥_{L^{\infty}}$ ; in particular $L^{\infty} (R^{n}) \subseteq B M O (R^{n})$ .

1.2F1

Adding a constant $c$ to a representative gives $(b + c)_{Q} = b_{Q} + c$ by linearity of the integral, so the class map is well defined on $L^{\infty} (R^{n}) / C$ : both quotients identify functions whose difference is almost everywhere constant. If two classes $b_{1}, b_{2} \in L^{\infty} (R^{n}) / C$ have the same image in $B M O (R^{n}) / C$ , then $b_{1} - b_{2}$ is almost everywhere constant by the definition of that quotient as the quotient of $B M O$ by the constants, so $b_{1}$ and $b_{2}$ already represent the same class in $L^{\infty} (R^{n}) / C$ ; that is injectivity.

2.1step 1.1step 1.2∎

By invariance of the BMO seminorm under constants and step 1.1, for every $c \in C$ one has $∥ b ∥_{B M O} = ∥ b - c ∥_{B M O} \leq 2 ∥ b - c ∥_{L^{\infty}}$ . Taking the infimum over $c$ gives $∥ [b] ∥_{B M O / C} \leq 2 ∥ [b] ∥_{L^{\infty} / C}$ for the quotient norms, so the induced class map is continuous; step 1.2 gives injectivity. Both statements are choice-free: only linearity of the integral and the definition of the seminorm are used.

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

BMO averages on nested cubes grow at most logarithmically

Statement

There is $C_{n} < \infty$ such that for every $b \in B M O (R^{n})$ and all cubes $Q \subseteq R$ , writing $ℓ (Q)$ for the side length, $∣ b_{Q} - b_{R} ∣ \leq C_{n} (1 + \log_{2} (ℓ (R) / ℓ (Q))) ∥ b ∥_{B M O}$ .

Facts & Assumptions

Given: $b \in B M O (R^{n})$ and cubes $Q \subseteq R$ with side lengths $ℓ (Q) \leq ℓ (R)$ , together with the mean $b_{E} = ∣ E ∣^{- 1} \int_{E} b$ , the seminorm and the cube and volume conventions of BMO seminorm and the quotient by constants and Axis-parallel rectangles in $R^{m}$ and their volume.

[F1]

For every cube $E$ , $∣ E ∣^{- 1} \int_{E} ∣ b - b_{E} ∣ \leq ∥ b ∥_{B M O}$ , and for cubes $E \subseteq F$ one has $∣ b_{E} - b_{F} ∣ \leq ∣ E ∣^{- 1} \int_{E} ∣ b - b_{F} ∣ \leq (∣ F ∣ / ∣ E ∣) ∥ b ∥_{B M O}$ (BMO seminorm and the quotient by constants).

[F2]

A cube is a nondegenerate axis-parallel cube; the concentric cube with side length $2^{k} ℓ (Q)$ has volume $2^{k n} ∣ Q ∣$ , and a cube whose side length is at most that of $R$ and which is contained in $R$ has volume at most $∣ R ∣$ (Axis-parallel rectangles in $R^{m}$ and their volume).

Proof

technique · direct

1.1F2algebra

If $Q = R$ the claim is trivial, so assume $Q ⊊ R$ . Let $m \geq 0$ be the least integer such that $2^{m} ℓ (Q) \geq 2 ℓ (R)$ , and let $R^{'}$ be the cube concentric with $Q$ of side length $2^{m} ℓ (Q)$ . The centres of $Q$ and $R$ both lie in $R$ , so their coordinatewise distance is at most $ℓ (R) / 2$ ; every point of $R$ is therefore within coordinate distance $ℓ (R)$ of the centre of $Q$ , and $R \subseteq R^{'}$ . Since $Q ⊊ R$ , $ℓ (Q) < ℓ (R)$ , so $m \geq 1$ . Minimality gives $2^{m - 1} ℓ (Q) < 2 ℓ (R)$ and hence $2^{m} ℓ (Q) < 4 ℓ (R)$ . Therefore $∣ R^{'} ∣ = 2^{m n} ∣ Q ∣ < 4^{n} ∣ R ∣$ .

2.1step 1.1F1

Let $Q = Q_{0} \subseteq Q_{1} \subseteq \dots \subseteq Q_{m} = R^{'}$ be the concentric cubes of side lengths $2^{k} ℓ (Q)$ . Since $∣ Q_{k + 1} ∣ = 2^{n} ∣ Q_{k} ∣$ , [F1] gives $∣ b_{Q_{k}} - b_{Q_{k + 1}} ∣ \leq ∣ Q_{k} ∣^{- 1} \int_{Q_{k}} ∣ b - b_{Q_{k + 1}} ∣ \leq 2^{n} ∥ b ∥_{B M O}$ for each $k < m$ , and the triangle inequality over the $m$ steps gives $∣ b_{Q} - b_{R^{'}} ∣ \leq m 2^{n} ∥ b ∥_{B M O}$ .

2.2step 1.1F1

Since $R \subseteq R^{'}$ , [F1] and step 1.1 give $∣ b_{R^{'}} - b_{R} ∣ \leq ∣ R ∣^{- 1} \int_{R} ∣ b - b_{R^{'}} ∣ \leq (∣ R^{'} ∣ / ∣ R ∣) ∥ b ∥_{B M O} \leq 4^{n} ∥ b ∥_{B M O}$ .

3.1step 2.1step 2.2algebra∎

Combining steps 2.1 and 2.2, $∣ b_{Q} - b_{R} ∣ \leq (2^{n} m + 4^{n}) ∥ b ∥_{B M O}$ . From $2^{m} ℓ (Q) < 4 ℓ (R)$ we get $m < 2 + \log_{2} (ℓ (R) / ℓ (Q))$ , hence $m \leq 2 + \log_{2} (ℓ (R) / ℓ (Q)) \leq 2 (1 + \log_{2} (ℓ (R) / ℓ (Q)))$ . Therefore $∣ b_{Q} - b_{R} ∣ \leq (2^{n + 1} + 4^{n}) (1 + \log_{2} (ℓ (R) / ℓ (Q))) ∥ b ∥_{B M O}$ , which is the claim with $C_{n} = 2^{n + 1} + 4^{n}$ .

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

BMO functions pair uniformly with H1 atoms

Statement

Let $b \in B M O (R^{n})$ and let $a$ be an $H^{1}$ atom supported in a cube $Q$ , i.e. a $(1, \infty, 0)$ -atom in the sense of $H^{p}$ atoms with a prescribed moment order: $a$ vanishes off $Q$ , $∣ a ∣ \leq ∣ Q ∣^{- 1}$ almost everywhere and $\int a = 0$ . Then the integral $\int a b$ converges absolutely and $∣ \int a b ∣ \leq ∥ b ∥_{B M O}$ , with the bound independent of the atom and of $Q$ .

Facts & Assumptions

Given: $b \in B M O (R^{n})$ and a $(1, \infty, 0)$ -atom $a$ supported in a cube $Q$ with $∣ a ∣ \leq ∣ Q ∣^{- 1}$ almost everywhere and $\int a = 0$ .

[F1]

The atom hypotheses are $supp a \subseteq Q$ , $∣ a ∣ \leq ∣ Q ∣^{- 1}$ almost everywhere and $\int_{R^{n}} a = 0$ ; an atom is bounded and compactly supported, hence locally integrable ( $H^{p}$ atoms with a prescribed moment order).

[F2]

$b \in L_{l o c}^{1} (R^{n})$ , $b_{Q} = ∣ Q ∣^{- 1} \int_{Q} b$ and $∣ Q ∣^{- 1} \int_{Q} ∣ b - b_{Q} ∣ \leq ∥ b ∥_{B M O}$ (BMO seminorm and the quotient by constants, A locally integrable function on $R^{n}$ ).

Proof

technique · direct

1.1F1F2

The product $a b$ is integrable: $a$ vanishes off the cube $Q$ and $∣ a ∣ \leq ∣ Q ∣^{- 1}$ almost everywhere by [F1], while $\int_{Q} ∣ b ∣ < \infty$ by [F2], so $\int_{R^{n}} ∣ a ∣ ∣ b ∣ \leq ∣ Q ∣^{- 1} \int_{Q} ∣ b ∣ < \infty$ .

2.1F1F2step 1.1algebra

Because $\int a = 0$ by [F1], subtracting the constant $b_{Q}$ changes nothing: $\int a b = \int a (b - b_{Q})$ , and step 1.1 makes this integral absolutely convergent. Hence $∣ \int a b ∣ \leq ∥ a ∥_{L^{\infty}} \int_{Q} ∣ b - b_{Q} ∣ \leq ∣ Q ∣^{- 1} \cdot ∣ Q ∣ \cdot ∥ b ∥_{B M O} = ∥ b ∥_{B M O}$ , where the middle inequality uses the almost-everywhere bound on $a$ and the last one the mean-oscillation bound of [F2].

3.1step 2.1∎

The estimate of step 2.1 depends only on $∥ b ∥_{B M O}$ , with no reference to the particular atom or cube, so the pairing is bounded uniformly over all $H^{1}$ atoms. No choice principle is used.

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

John-Nirenberg stopping cubes have geometric decay

Statement

Assume Countable Choice. Let $b \in B M O (R^{n})$ , let $Q$ be an all-generations dyadic cube and let $s > ∥ b ∥_{B M O}$ . Then there are families of dyadic subcubes $Q_{j}^{(k)} \subseteq Q$ , $k \geq 1$ , pairwise disjoint for each fixed $k$ , such that, with $Q_{1}^{(0)} : = Q$ : (i) each $Q_{j}^{(k)}$ is contained in a unique level- $(k - 1)$ stopping cube. Its immediate dyadic parent is contained in that stopping cube but need not belong to the preceding stopping family; (ii) for every $k$ , $\sum_{j} ∣ Q_{j}^{(k)} ∣ \leq (∥ b ∥_{B M O} / s)^{k} ∣ Q ∣$ ; and (iii) for every $k \geq 1$ , $∣ b - b_{Q} ∣ \leq k 2^{n} s$ almost everywhere on $Q ∖ ⋃_{j} Q_{j}^{(k)}$ , and a fortiori $∣ b - b_{Q} ∣ \leq 2^{n k} s$ there, so ${x \in Q : ∣ b - b_{Q} ∣ > 2^{n k} s} \subseteq ⋃_{j} Q_{j}^{(k)}$ up to a Lebesgue-null set.

Facts & Assumptions

Given: Countable Choice, $b \in B M O (R^{n})$ , an all-generations dyadic cube $Q$ and a real number $s > ∥ b ∥_{B M O}$ .

[F1]

The all-generations dyadic cubes partition $R^{n}$ at each generation, have volumes $2^{- k n}$ , and for two dyadic cubes one contains the other or they are disjoint; the parent of a cube of generation $k$ is its unique ancestor of generation $k - 1$ , with volume $2^{n}$ times that of the cube (Dyadic cubes of all generations in R^n, All-generation dyadic cubes: partition, volume and nesting).

[F2]

For $f \in L^{1} (R^{n})$ and $λ > 0$ , the dyadic cubes $E$ with $∣ E ∣^{- 1} \int_{E} ∣ f ∣ > λ$ that are maximal under inclusion form a countable family of pairwise disjoint cubes; their union is exactly the dyadic maximal superlevel set ${M_{d} f > λ}$ ; each such $E$ satisfies $∣ E ∣^{- 1} \int_{E} ∣ f ∣ \leq 2^{n} λ$ ; and $\sum_{E} ∣ E ∣ \leq λ^{- 1} ∥ f ∥_{1}$ (Maximal dyadic cubes above a level).

[F3]

In the Calderon-Zygmund decomposition of $f \in L^{1}$ at height $λ > 0$ , the good part $g$ equals $f$ outside the union of those maximal cubes and satisfies $∣ g ∣ \leq 2^{n} λ$ almost everywhere (Calderón–Zygmund decomposition at height λ).

[F4]

Almost every point of $R^{n}$ is a Lebesgue point of a given $L_{l o c}^{1}$ function (Almost every point is a Lebesgue point of a locally integrable function).

[F5]

A countable union of at most countable sets is at most countable, and a countable union of Lebesgue-null subsets of $R^{n}$ is Lebesgue-null (Countable unions of at most countable sets, assuming ${A C}_{ω}$ , Subsets and countable unions of null subsets of $R^{m}$ are null).

[F6]

For every cube $E$ , $∣ E ∣^{- 1} \int_{E} ∣ b - b_{E} ∣ \leq ∥ b ∥_{B M O}$ (BMO seminorm and the quotient by constants).

Proof

technique · direct

1.1F2F3F6algebra

Put $F = (b - b_{Q}) 1_{Q}$ . Then $F \in L^{1} (R^{n})$ with $\int_{R^{n}} ∣ F ∣ = \int_{Q} ∣ b - b_{Q} ∣ \leq ∣ Q ∣ ∥ b ∥_{B M O} < s ∣ Q ∣$ by [F6] and the hypothesis on $s$ . Let $Q^{(1)}$ be the family of dyadic cubes $E$ with $∣ E ∣^{- 1} \int_{E} ∣ F ∣ > s$ maximal under inclusion. By [F2] applied to $F$ and $s$ , the family $Q^{(1)}$ is countable and pairwise disjoint, each $E \in Q^{(1)}$ satisfies $s < ∣ E ∣^{- 1} \int_{E} ∣ F ∣ \leq 2^{n} s$ , and $\sum_{E \in Q^{(1)}} ∣ E ∣ \leq s^{- 1} \int_{R^{n}} ∣ F ∣ \leq (∥ b ∥_{B M O} / s) ∣ Q ∣$ . Every $E \in Q^{(1)}$ is a proper dyadic subcube of $Q$ : it meets $Q$ because its average of $∣ F ∣$ is positive, and if $Q ⊊ E$ then $∣ E ∣^{- 1} \int_{E} ∣ F ∣ \leq (∣ Q ∣ / ∣ E ∣) ∥ b ∥_{B M O} < s$ , a contradiction, while $E = Q$ is excluded by the same average bound. By [F3] the good part of the decomposition of $F$ equals $F$ off $⋃_{E \in Q^{(1)}} E$ , so $∣ F ∣ \leq 2^{n} s$ almost everywhere there.

2.1step 1.1F1F2F3F5algebra

Recursion. Let $Q^{(k)}$ be a countable pairwise disjoint family of proper dyadic subcubes of $Q$ , and for each $P \in Q^{(k)}$ put $F_{P} = (b - b_{P}) 1_{P}$ and let $Q_{P}$ be the family of maximal dyadic cubes $E$ with $∣ E ∣^{- 1} \int_{E} ∣ F_{P} ∣ > s$ . The argument of step 1.1, with $Q$ replaced by $P$ and $F$ by $F_{P}$ , shows that every $E \in Q_{P}$ is a proper dyadic subcube of $P$ , that $Q_{P}$ is countable and pairwise disjoint, that $s < ∣ E ∣^{- 1} \int_{E} ∣ F_{P} ∣ \leq 2^{n} s$ and $\sum_{E \in Q_{P}} ∣ E ∣ \leq (∥ b ∥_{B M O} / s) ∣ P ∣$ , and that the good part of the decomposition of $F_{P}$ satisfies $∣ F_{P} ∣ \leq 2^{n} s$ almost everywhere off $⋃_{E \in Q_{P}} E$ . Each selected $E$ has a unique stopping parent $P \in Q^{(k)}$ . Its immediate dyadic parent $R$ is contained in $P$ : since $R$ and $P$ both contain $E$ , [F1] makes them nested; if $P ⊊ R$ , the generations of $R$ and $E$ differ by one and $E ⊊ P ⊊ R$ is impossible, so $R \subseteq P$ . The dyadic parent has the exact volume ratio $∣ R ∣ = 2^{n} ∣ E ∣$ by [F1]. Maximality gives $∣ R ∣^{- 1} \int_{R} ∣ F_{P} ∣ \leq s$ , and hence $∣ E ∣^{- 1} \int_{E} ∣ F_{P} ∣ \leq 2^{n} s$ . Thus the stopping parent is $P$ , while the immediate dyadic parent $R$ need only be contained in $P$ and need not itself belong to the preceding stopping family. Define $Q^{(k + 1)} = ⋃_{P \in Q^{(k)}} Q_{P}$ ; indexed by pairs it is countable by [F5], its members are pairwise disjoint because distinct cubes $P$ are disjoint and each $Q_{P}$ is a pairwise disjoint family of subcubes of its $P$ , and each member is a proper dyadic subcube of a unique $P \in Q^{(k)}$ , so its interior is contained in that cube.

3.1step 1.1step 2.1

By induction on $k \geq 1$ , $\sum_{E \in Q^{(k)}} ∣ E ∣ \leq (∥ b ∥_{B M O} / s)^{k} ∣ Q ∣$ : step 1.1 is the case $k = 1$ , and the ratio estimate of step 2.1 gives $\sum_{E \in Q^{(k + 1)}} ∣ E ∣ = \sum_{P \in Q^{(k)}} \sum_{E \in Q_{P}} ∣ E ∣ \leq (∥ b ∥_{B M O} / s) \sum_{P \in Q^{(k)}} ∣ P ∣$ , which is the induction step.

3.2step 1.1step 2.1F2F3F4F5algebra

Fix $k \geq 1$ and discard the exceptional Lebesgue-null sets of steps 1.1 and 2.1 together with the null sets of non-Lebesgue points of the countably many functions $F_{P}$ for $P$ of level at most $k - 1$ ; by [F4] and [F5] the discarded set is Lebesgue-null. Let $x \in Q ∖ ⋃_{E \in Q^{(k)}} E$ lie outside it, and let $0 \leq l \leq k - 1$ be the largest integer such that $x$ lies in some cube $P_{l}$ of level $l$ , where $P_{0} : = Q$ ; such an $l$ exists because $x \in Q$ , and $P_{l}$ is unique because the families at each level are pairwise disjoint. Since $x$ lies in no cube of level $l + 1$ , the good-part bound of [F3] for the ambient cube $P_{l}$ gives $∣ b (x) - b_{P_{l}} ∣ = ∣ F_{P_{l}} (x) ∣ \leq 2^{n} s$ ; moreover, because $x$ is a Lebesgue point of $F_{P_{l}}$ , the dyadic cubes of generation $m$ containing $x$ lie in $B (x, \sqrt{n} 2^{- m})$ with volume comparable to $2^{- m n}$ , so their averages of $F_{P_{l}}$ converge to $F_{P_{l}} (x)$ , giving $∣ F_{P_{l}} (x) ∣ \leq M_{d} F_{P_{l}} (x) \leq s$ by [F2] and [F4]. For $l \geq 1$ , $∣ b_{P_{l}} - b_{Q} ∣ \leq \sum_{i = 1}^{l} ∣ b_{P_{i}} - b_{P_{i - 1}} ∣$ with $∣ b_{P_{i}} - b_{P_{i - 1}} ∣ \leq ∣ P_{i} ∣^{- 1} \int_{P_{i}} ∣ b - b_{P_{i - 1}} ∣ = ∣ P_{i} ∣^{- 1} \int_{P_{i}} ∣ F_{P_{i - 1}} ∣ \leq 2^{n} s$ , because $P_{i}$ was selected inside the ambient cube $P_{i - 1}$ . Hence $∣ b (x) - b_{Q} ∣ \leq s + l 2^{n} s \leq (l + 1) 2^{n} s \leq k 2^{n} s \leq 2^{n k} s$ , using $l + 1 \leq k$ and $k \leq 2^{n (k - 1)}$ for $k \geq 1$ , $n \geq 1$ . The complement of $⋃_{E \in Q^{(k)}} E$ inside $Q$ therefore satisfies the asserted almost-everywhere bound, its exceptional set being a countable union of Lebesgue-null sets by [F5].

4.1step 1.1step 2.1step 3.1step 3.2∎

Steps 1.1 and 2.1 construct, for every $k \geq 1$ , a countable pairwise disjoint family $Q^{(k)}$ of proper dyadic subcubes of $Q$ , each with a unique stopping parent in the preceding family and with its immediate dyadic parent contained in that stopping parent, which is (i); step 3.1 is the geometric-decay estimate (ii); and step 3.2 is the level bound (iii). The construction applies the published maximal-cube and decomposition lemmas countably many times, which is exactly where Countable Choice is used, together with the countable-union facts in [F5].

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

Range truncations preserve the BMO seminorm up to a constant

Statement

For every real-valued $b \in B M O (R^{n})$ and every $c \in R$ the one-sided truncations satisfy $∥ \max (b, c) ∥_{B M O} \leq \frac{3}{2} ∥ b ∥_{B M O}$ and $∥ \min (b, c) ∥_{B M O} \leq \frac{3}{2} ∥ b ∥_{B M O}$ ; consequently for $- \infty < L \leq K < \infty$ the two-sided truncation $b_{[L, K]} : = \min (K, \max (L, b))$ satisfies $∥ b_{[L, K]} ∥_{B M O} \leq \frac{9}{4} ∥ b ∥_{B M O}$ . For a complex-valued $b$ the componentwise truncation $b_{M} : = \min (M, \max (- M, Re b)) + i \min (M, \max (- M, Im b))$ satisfies $∣ b_{M} ∣ \leq ∣ b ∣$ , $b_{M} \to b$ pointwise as $M \to \infty$ , and $∥ b_{M} ∥_{B M O} \leq \frac{9}{2} ∥ b ∥_{B M O}$ .

Facts & Assumptions

Given: A function $b \in B M O (R^{n})$ , a cube $Q$ and constants $c \in R$ , $- \infty < L \leq K < \infty$ and $M > 0$ , with the mean and seminorm of BMO seminorm and the quotient by constants.

[F1]

The mean is $b_{Q} = ∣ Q ∣^{- 1} \int_{Q} b$ and $∥ b ∥_{B M O} = \sup_{Q} ∣ Q ∣^{- 1} \int_{Q} ∣ b - b_{Q} ∣$ ; for every constant $d$ one has $(b + d)_{Q} = b_{Q} + d$ and hence $∥ b + d ∥_{B M O} = ∥ b ∥_{B M O}$ (BMO seminorm and the quotient by constants).

[F2]

The reverse triangle inequality gives $∣ ∣ u ∣ - ∣ v ∣ ∣ \leq ∣ u - v ∣$ for real or complex numbers $u, v$ , and for real $b, c$ the maximum and minimum decompose as $\max (b, c) = c + \frac{1}{2} ((b - c) + ∣ b - c ∣)$ and $\min (b, c) = c + \frac{1}{2} ((b - c) - ∣ b - c ∣)$ .

Proof

technique · direct

1.1F1

For a locally integrable $g$ , a cube $Q$ and a constant $d$ , one has $∣ g_{Q} - d ∣ \leq ∣ Q ∣^{- 1} \int_{Q} ∣ g - d ∣$ , so $∣ Q ∣^{- 1} \int_{Q} ∣ g - g_{Q} ∣ \leq ∣ Q ∣^{- 1} \int_{Q} ∣ g - d ∣ + ∣ g_{Q} - d ∣ \leq 2 ∣ Q ∣^{- 1} \int_{Q} ∣ g - d ∣$ ; taking the supremum over $Q$ gives $∥ g ∥_{B M O} \leq 2 \sup_{Q} ∣ Q ∣^{- 1} \int_{Q} ∣ g - d_{Q} ∣$ for any choice of constants $d_{Q}$ .

2.1F2step 1.1

Let $b$ be real-valued, fix $c \in R$ and put $g = ∣ b - c ∣$ . For every cube $Q$ , [F2] with $d = ∣ b_{Q} - c ∣$ gives $∣ ∣ b - c ∣ - ∣ b_{Q} - c ∣ ∣ \leq ∣ b - b_{Q} ∣$ pointwise on $Q$ , so choosing $d_{Q} = ∣ b_{Q} - c ∣$ in step 1.1 yields $∥ ∣ b - c ∣ ∥_{B M O} \leq 2 ∥ b ∥_{B M O}$ .

3.1F1step 2.1algebra

By [F2], $\max (b, c) = c + \frac{1}{2} (b - c) + \frac{1}{2} ∣ b - c ∣$ and $\min (b, c) = c + \frac{1}{2} (b - c) - \frac{1}{2} ∣ b - c ∣$ ; translation by constants leaves the seminorm unchanged by [F1] and the triangle inequality for the supremum gives $∥ \max (b, c) ∥_{B M O} \leq \frac{1}{2} ∥ b - c ∥_{B M O} + \frac{1}{2} ∥ ∣ b - c ∣ ∥_{B M O} \leq \frac{1}{2} ∥ b ∥_{B M O} + ∥ b ∥_{B M O} = \frac{3}{2} ∥ b ∥_{B M O}$ , and the same computation applies to $\min (b, c)$ .

4.1step 3.1algebra

If $- \infty < L \leq K < \infty$ , then $b_{[L, K]} = \min (K, \max (L, b))$ and applying step 3.1 twice gives $∥ b_{[L, K]} ∥_{B M O} \leq \frac{3}{2} ∥ \max (L, b) ∥_{B M O} \leq \frac{9}{4} ∥ b ∥_{B M O}$ .

5.1step 4.1algebra

Let $b = u + i v$ be complex-valued and put $u_{M} = \min (M, \max (- M, u))$ and $v_{M} = \min (M, \max (- M, v))$ , so $b_{M} = u_{M} + i v_{M}$ . Then $∣ u_{M} ∣ \leq ∣ u ∣$ and $∣ v_{M} ∣ \leq ∣ v ∣$ , hence $∣ b_{M} ∣ = (u_{M}^{2} + v_{M}^{2})^{1 / 2} \leq (u^{2} + v^{2})^{1 / 2} = ∣ b ∣$ , and $u_{M} \to u$ , $v_{M} \to v$ pointwise as $M \to \infty$ . Moreover $∥ Re b ∥_{B M O} \leq ∥ b ∥_{B M O}$ and $∥ Im b ∥_{B M O} \leq ∥ b ∥_{B M O}$ , because $∣ Re (b - b_{Q}) ∣ \leq ∣ b - b_{Q} ∣$ and similarly for the imaginary part; step 4.1 applied to $u$ and $v$ gives $∥ b_{M} ∥_{B M O} \leq ∥ u_{M} ∥_{B M O} + ∥ v_{M} ∥_{B M O} \leq \frac{9}{4} (∥ u ∥_{B M O} + ∥ v ∥_{B M O}) \leq \frac{9}{2} ∥ b ∥_{B M O}$ .

6.1step 3.1step 4.1step 5.1∎

Steps 3.1, 4.1 and 5.1 are the three assertions of the statement. No choice principle is used: only linearity of the integral, the definition of the seminorm and elementary real and complex inequalities.

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

Calderon-Zygmund operators map L-infinity to BMO

Statement

Assume Countable Choice. Let $T$ be a Calderon-Zygmund operator with kernel $k$ in the sense of Calderón–Zygmund kernels and their associated operators: $k$ satisfies the annular size bound with constant $A_{1}$ and Hormander's condition with constant $A_{2}$ , $T$ is $L^{2}$ -bounded with norm $B$ and satisfies the off-support representation (3) of that definition. Assume moreover that $k$ is standard $δ$ -Holder with constant $A_{2}^{'}$ for some $0 < δ \leq 1$ (Standard (Hölder) Calderón–Zygmund kernels). Fix $λ_{n} : = 1 + 4 \sqrt{n}$ , chosen so that $∣ a - y ∣ \geq 2 \sqrt{n} ℓ (Q) \geq 2 ∣ x - a ∣$ whenever $x, a \in Q$ and $y \notin λ_{n} Q$ . For a cube $Q$ and a point $a \in Q$ put, for $x \in Q$ , $(T b)_{Q} (x) : = T (b 1_{λ_{n} Q}) (x) + \int_{R^{n} ∖ λ_{n} Q} [k (x - y) - k (a - y)] b (y) d y$ , where $λ_{n} Q$ is the concentric cube with side length $λ_{n} ℓ (Q)$ and $b \in L^{\infty} (R^{n})$ . Then: (i) the tail integral converges absolutely for every $x \in Q$ ; (ii) for nested cubes $Q \subseteq R$ the difference $(T b)_{Q} - (T b)_{R}$ is almost everywhere constant on $Q$ , so the localisations define a class $T b \in B M O (R^{n}) / C$ ; (iii) $∥ T b ∥_{B M O} \leq C_{n, δ} (A_{2}^{'} + B) ∥ b ∥_{L^{\infty}}$ ; and (iv) if in addition $b \in L^{2} (R^{n})$ , then the class $T b$ is the class of the $L^{2}$ function $T b$ defined by the operator, modulo constants.

Facts & Assumptions

Given: Countable Choice, a Calderon-Zygmund operator $T$ with kernel $k$ and constants $A_{1}, A_{2}, B$ , standard $δ$ -Holder with constant $A_{2}^{'}$ , a bounded function $b \in L^{\infty} (R^{n})$ , cubes $Q \subseteq R$ and points $x, a \in Q$ .

[F1]

The kernel satisfies $\sup_{R > 0} \int_{R \leq ∣ x ∣ \leq 2 R} ∣ k (x) ∣ d x \leq A_{1}$ and $\sup_{v \neq 0} \int_{∣ z ∣ \geq 2 ∣ v ∣} ∣ k (z - v) - k (z) ∣ d z \leq A_{2}$ ; $T$ is $L^{2}$ -bounded with norm $B$ and, for every compactly supported $f \in L^{2}$ , $T f (x) = \int_{R^{n}} k (x - y) f (y) d y$ for almost every $x \notin supp f$ , the integral converging absolutely (Calderón–Zygmund kernels and their associated operators).

[F2]

The kernel is standard $δ$ -Holder with constant $A_{2}^{'}$ : $∣ k (z - v) - k (z) ∣ \leq A_{2}^{'} ∣ v ∣^{δ} ∣ z ∣^{- n - δ}$ whenever $∣ z ∣ \geq 2 ∣ v ∣ > 0$ (Standard (Hölder) Calderón–Zygmund kernels).

[F3]

A cube of side length $ℓ$ has diameter at most $\sqrt{n} ℓ$ : its points lie in an axis-parallel box with side lengths $ℓ$ , so coordinatewise $∣ x_{i} - a_{i} ∣ \leq ℓ$ and $∣ x - a ∣ \leq \sqrt{n} ℓ$ ; the concentric cube of side $λ ℓ$ has volume $λ^{n}$ times the volume, and containment of cubes is preserved under concentric dilation (Axis-parallel rectangles in $R^{m}$ and their volume).

[F4]

The Cauchy-Schwarz inequality gives $\int_{E} ∣ f g ∣ \leq ∥ f ∥_{L^{2} (E)} ∥ g ∥_{L^{2} (E)}$ on a finite-measure set $E$ , and $L^{2}$ is the quotient of measurable functions modulo almost-everywhere equality, with $∥ f ∥_{2} = (\int ∣ f ∣^{2})^{1 / 2}$ (Cauchy-Schwarz inequality for $L^{2}$ , The space $L^{p} (μ)$ as the quotient by null functions).

[F5]

Polar coordinates: for Borel measurable $F \geq 0$ , $\int_{R^{n}} F d x = \int_{0}^{\infty} \int_{S^{n - 1}} F (r ω) r^{n - 1} d σ (ω) d r$ , with $σ$ the finite Borel surface measure on $S^{n - 1}$ (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma, The polar surface set function on the unit sphere).

[F6]

Fubini applies to $L^{1}$ functions on products of $σ$ -finite measure spaces, and dominated convergence applies to pointwise convergent measurable functions dominated by one integrable function (Fubini's theorem for L^1 functions on a sigma-finite product, Dominated convergence).

[F7]

Countable unions of Lebesgue-null sets are Lebesgue-null, and Countable Choice permits the countably many selections made below (Subsets and countable unions of null subsets of $R^{m}$ are null, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F8]

The seminorm is $∥ u ∥_{B M O} = \sup_{Q} ∣ Q ∣^{- 1} \int_{Q} ∣ u - u_{Q} ∣$ , and the quotient $B M O (R^{n}) / C$ identifies functions differing by an almost-everywhere constant (BMO seminorm and the quotient by constants).

[F9]

If a sequence converges in $L^{2}$ , it has a subsequence of measurable representatives converging almost everywhere to a representative of the limit under Countable Choice (Assuming Countable Choice, $L^{p}$ -convergent sequences have almost-everywhere convergent subsequences).

Proof

technique · direct

1.1F1F3F4

The local term obeys $∣ Q ∣^{- 1} \int_{Q} ∣ T (b 1_{λ_{n} Q}) ∣ \leq ∣ Q ∣^{- 1 / 2} ∥ T (b 1_{λ_{n} Q}) ∥_{2} \leq ∣ Q ∣^{- 1 / 2} B ∥ b 1_{λ_{n} Q} ∥_{2} \leq ∣ Q ∣^{- 1 / 2} B ∥ b ∥_{L^{\infty}} ∣ λ_{n} Q ∣^{1 / 2} = λ_{n}^{n / 2} B ∥ b ∥_{L^{\infty}}$ , by Cauchy-Schwarz [F4], the $L^{2}$ bound [F1] and $∣ λ_{n} Q ∣ = λ_{n}^{n} ∣ Q ∣$ [F3].

1.2F3

For $x, a \in Q$ one has $∣ x - a ∣ \leq \sqrt{n} ℓ (Q)$ ; for $y \notin λ_{n} Q$ the coordinatewise distance from $a$ to the complement of the concentric cube $λ_{n} Q$ is at least $\frac{λ_{n} - 1}{2} ℓ (Q) = 2 \sqrt{n} ℓ (Q)$ , so $∣ a - y ∣ \geq 2 \sqrt{n} ℓ (Q) \geq 2 ∣ x - a ∣$ ; and if $Q \subseteq R$ , $a \in Q \subseteq R$ and $y \notin λ_{n} R$ , the same computation with $ℓ (R)$ gives $∣ a - y ∣ \geq 2 \sqrt{n} ℓ (R) \geq 2 ∣ x - a ∣$ .

2.1F2F5step 1.2algebra

The tail integral converges absolutely for every $x \in Q$ . If $x = a$ , its integrand is zero. Otherwise, by [F2] and step 1.2, $∣ k (x - y) - k (a - y) ∣ \leq A_{2}^{'} ∣ x - a ∣^{δ} ∣ a - y ∣^{- n - δ} \leq A_{2}^{'} (\sqrt{n} ℓ (Q))^{δ} ∣ a - y ∣^{- n - δ}$ for $y \notin λ_{n} Q$ , and by [F5] the tail power integral is $\int_{R^{n} ∖ λ_{n} Q} ∣ a - y ∣^{- n - δ} d y \leq \int_{∣ z ∣ \geq 2 \sqrt{n} ℓ (Q)} ∣ z ∣^{- n - δ} d z = σ (S^{n - 1}) δ^{- 1} (2 \sqrt{n} ℓ (Q))^{- δ}$ . Hence $∣ \int_{R^{n} ∖ λ_{n} Q} [k (x - y) - k (a - y)] b (y) d y ∣ \leq A_{2}^{'} σ (S^{n - 1}) δ^{- 1} 2^{- δ} ∥ b ∥_{L^{\infty}}$ , a bound independent of $x$ and $Q$ ; this is (i), and the tail is a bounded measurable function of $x$ : [F2] implies continuity of $k$ away from zero, while its displayed bound gives an integrable majorant uniformly on $Q$ , so dominated convergence gives continuity of the tail there.

2.2F1step 1.2algebra

Nested consistency. Let $Q \subseteq R$ be cubes with points $a_{Q} \in Q$ and $a_{R} \in R$ , and let $x \in Q$ . Since $λ_{n} Q \subseteq λ_{n} R$ , splitting the complement of $λ_{n} Q$ into $λ_{n} R ∖ λ_{n} Q$ and $R^{n} ∖ λ_{n} R$ and using linearity of $T$ gives $(T b)_{Q} (x) - (T b)_{R} (x) = - T (b 1_{λ_{n} R ∖ λ_{n} Q}) (x) + \int_{λ_{n} R ∖ λ_{n} Q} [k (x - y) - k (a_{Q} - y)] b (y) d y + \int_{R^{n} ∖ λ_{n} R} [k (a_{R} - y) - k (a_{Q} - y)] b (y) d y$ . The function $b 1_{λ_{n} R ∖ λ_{n} Q}$ is compactly supported $L^{2}$ with support disjoint from $Q$ , so the off-support representation [F1] gives $T (b 1_{λ_{n} R ∖ λ_{n} Q}) (x) = \int_{λ_{n} R ∖ λ_{n} Q} k (x - y) b (y) d y$ for almost every $x \in Q$ ; the first integral over $λ_{n} R ∖ λ_{n} Q$ converges absolutely because that region lies in a bounded annulus about $a_{Q}$ on which the annular bound of [F1] controls $k$ . If $a_{R} = a_{Q}$ , the last integral is zero; otherwise it converges absolutely by Hormander's condition [F1] applied at centre $a_{R}$ with nonzero translation $v = a_{R} - a_{Q}$ , since $∣ a_{R} - y ∣ \geq 2 \sqrt{n} ℓ (R) \geq 2 ∣ a_{R} - a_{Q} ∣$ for $y \notin λ_{n} R$ by step 1.2. The resulting expression is independent of $x$ , so $(T b)_{Q} - (T b)_{R}$ is almost everywhere constant on $Q$ , which is (ii).

3.1step 1.1step 2.1F8algebra

Combining steps 1.1 and 2.1, $∣ Q ∣^{- 1} \int_{Q} ∣ (T b)_{Q} ∣ \leq (λ_{n}^{n / 2} B + A_{2}^{'} σ (S^{n - 1}) δ^{- 1} 2^{- δ}) ∥ b ∥_{L^{\infty}}$ for every cube $Q$ , hence by the optimal-constant bound of [F8] with $c = 0$ the mean oscillation of $(T b)_{Q}$ over $Q$ is at most $2 (λ_{n}^{n / 2} B + A_{2}^{'} σ (S^{n - 1}) δ^{- 1} 2^{- δ}) ∥ b ∥_{L^{\infty}}$ .

3.2step 2.2F7

Coherence and gluing. If $Q_{1}, Q_{2}$ are cubes, choose a cube $R$ containing both; step 2.2 shows that each $(T b)_{Q_{i}} - (T b)_{R}$ is constant almost everywhere on $Q_{i}$ , so $(T b)_{Q_{1}} - (T b)_{Q_{2}}$ is constant almost everywhere on $Q_{1} \cap Q_{2}$ . Let $Q_{k} = [- k, k]^{n}$ for $k \geq 1$ and choose representatives of the countably many localisations, which Countable Choice permits [F7]. Define constants $c_{k}$ inductively by $c_{1} = 0$ and $c_{k + 1} = c_{k} - [(T b)_{Q_{k + 1}} - (T b)_{Q_{k}}]$ , the bracket being the constant of step 2.2 on $Q_{k}$ ; then $u_{k} : = (T b)_{Q_{k}} + c_{k}$ satisfies $u_{k + 1} = u_{k}$ almost everywhere on $Q_{k}$ . Removing the countable union of the exceptional null sets, which is null by [F7], define $u (x) : = u_{k} (x)$ for $x \in Q_{k}$ outside that null set, and set $u = 0$ on the null set; this is well defined and locally integrable, and for every cube $Q$ , choosing $k$ with $Q \subseteq Q_{k}$ , the function $u - (T b)_{Q} = [u - (T b)_{Q_{k}}] + [(T b)_{Q_{k}} - (T b)_{Q}]$ is almost everywhere constant on $Q$ by the construction and step 2.2. Two such global representatives differ by constants on the nested $Q_{k}$ ; these constants agree on their positive-measure overlaps, so the global class is unique. Linearity follows from linearity of each localisation and this uniqueness.

4.1step 3.1step 3.2F8algebra

By step 3.2, $u - (T b)_{Q}$ is constant almost everywhere on every cube $Q$ , so the mean oscillation of $u$ over $Q$ equals that of $(T b)_{Q}$ ; by step 3.1, $∣ Q ∣^{- 1} \int_{Q} ∣ u - u_{Q} ∣ \leq 2 (λ_{n}^{n / 2} B + A_{2}^{'} σ (S^{n - 1}) δ^{- 1} 2^{- δ}) ∥ b ∥_{L^{\infty}}$ for every cube. Hence $u \in B M O (R^{n})$ with $∥ u ∥_{B M O} \leq C_{n, δ} (A_{2}^{'} + B) ∥ b ∥_{L^{\infty}}$ for $C_{n, δ} : = 2 \max (λ_{n}^{n / 2}, σ (S^{n - 1}) δ^{- 1} 2^{- δ})$ , and its class modulo constants is the class $T b$ of the statement, which is (iii).

4.2F1F2F4F5F6F7F9step 1.2step 3.2

Suppose $b \in L^{2} (R^{n})$ ; fix a cube $Q$ and $a \in Q$ and write $c : = R^{n} ∖ λ_{n} Q$ and $Ψ (x) : = \int_{c} [k (x - y) - k (a - y)] b (y) d y$ for the tail of the localisation. Let $b_{N} : = b 1_{c \cap B (0, N)}$ for $N \geq 1$ . Each $b_{N}$ is compactly supported $L^{2}$ with support disjoint from $Q$ , so [F1] gives $T b_{N} (x) = \int_{c} k (x - y) b_{N} (y) d y$ for almost every $x \in Q$ ; since $b_{N} \to b 1_{c}$ in $L^{2}$ , boundedness of $T$ gives $T b_{N} \to T (b 1_{c})$ in $L^{2}$ . By [F9] choose a subsequence whose representatives converge almost everywhere to a representative of $T (b 1_{c})$ , and intersect this full-measure set with the full-measure set where the countably many off-support identities hold. For $x = x^{'}$ the kernel difference below is zero. For distinct $x, x^{'} \in Q$ outside the exceptional null set and $y \in c$ , one has $∣ x^{'} - y ∣ \geq 2 \sqrt{n} ℓ (Q) \geq 2 ∣ x - x^{'} ∣$ by step 1.2, so $∣ k (x - y) - k (x^{'} - y) ∣ \leq A_{2}^{'} ∣ x - x^{'} ∣^{δ} ∣ x^{'} - y ∣^{- n - δ}$ by [F2]; the function $y \mapsto A_{2}^{'} ∣ x - x^{'} ∣^{δ} ∣ x^{'} - y ∣^{- n - δ} ∣ b (y) ∣$ is integrable on $c$ by Cauchy-Schwarz [F4] and [F5], so dominated convergence [F6] gives along that subsequence $\lim_{j} [T b_{N_{j}} (x) - T b_{N_{j}} (x^{'})] = \int_{c} [k (x - y) - k (x^{'} - y)] b (y) d y = Ψ (x) - Ψ (x^{'})$ . Hence $T (b 1_{c}) - Ψ$ has equal values at almost every pair of points of $Q$ , so by Fubini [F6] it is almost everywhere constant on $Q$ . Since $T b = T (b 1_{λ_{n} Q}) + T (b 1_{c})$ in $L^{2}$ and $(T b)_{Q} = T (b 1_{λ_{n} Q}) + Ψ$ , the difference $T b - (T b)_{Q}$ is almost everywhere constant on $Q$ ; that is (iv), and it identifies the $L^{2}$ class of $T b$ with the localisation class of step 3.2 modulo constants.

5.1step 2.1step 2.2step 3.1step 3.2step 4.1step 4.2F9∎

Steps 2.1, 2.2, 4.1 and 4.2 prove (i), (ii), (iii) and (iv) respectively, and steps 3.1 and 3.2 supply the global representative $u$ used in (iii). The argument uses Countable Choice exactly in the countably many applications of the off-support representation and in the selection of the representatives and subsequence in steps 3.2 and 4.2, and it uses no other choice principle.

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The Hilbert and Riesz transforms map L-infinity to BMO

Statement

Assume Countable Choice. The Hilbert transform $H$ on $R$ and the Riesz transforms $R_{1}, \dots, R_{n}$ on $R^{n}$ extend to bounded maps $L^{\infty} (R^{d}) \to B M O (R^{d}) / C$ , where $d = 1$ for $T = H$ and $d = n$ for $T = R_{j}$ : for every $b \in L^{\infty} (R^{d})$ the class of $T b$ is well defined modulo constants, agrees with the $L^{2}$ action of $T$ when $b \in L^{2}$ , and satisfies $∥ T b ∥_{B M O} \leq C_{d} ∥ b ∥_{L^{\infty}}$ with a dimensional constant.

Facts & Assumptions

Given: Countable Choice, the Hilbert transform $H$ and the Riesz transforms $R_{1}, \dots, R_{n}$ , and a bounded function $b \in L^{\infty} (R^{d})$ , with $d = 1$ for $H$ and $d = n$ for $R_{j}$ .

[F1]

The Hilbert transform and each Riesz transform are Calderon-Zygmund operators whose kernels are standard $1$ -Holder, with $L^{2}$ operator norm at most $1$ ; the Hilbert transform is the case $n = 1$ with kernel $1 / (π x)$ and the Riesz transforms have kernels $c_{n} x_{j} / ∣ x ∣^{n + 1}$ (The Hilbert and Riesz transforms are Calderon-Zygmund operators).

[F2]

Every Calderon-Zygmund operator with a standard $δ$ -Holder kernel and $L^{2}$ norm at most $1$ maps $L^{\infty}$ into $B M O (R^{n}) / C$ : the class of $T b$ is well defined modulo constants, agrees with the $L^{2}$ action when $b \in L^{2}$ , and has $∥ T b ∥_{B M O} \leq C_{n, δ} (A_{2}^{'} + 1) ∥ b ∥_{L^{\infty}}$ (Calderon-Zygmund operators map L-infinity to BMO).

Proof

technique · direct

1.1F1

The hypotheses of [F2] are satisfied with $δ = 1$ : [F1] supplies the Calderon-Zygmund operator, the standard $1$ -Holder kernel with its constant, and the $L^{2}$ bound by $1$ ; for the Hilbert transform this is the case $n = 1$ and for each Riesz transform the case of the corresponding kernel $c_{n} x_{j} / ∣ x ∣^{n + 1}$ .

2.1step 1.1F2

Applying [F2] to the Hilbert transform and to each Riesz transform gives, for every $b \in L^{\infty} (R^{d})$ in the corresponding dimension, a well-defined class $T b$ modulo constants with $∥ T b ∥_{B M O} \leq C_{d, 1} (A_{2}^{'} + 1) ∥ b ∥_{L^{\infty}}$ , and when $b \in L^{2}$ that class is the class of the $L^{2}$ function $T b$ ; the constants $A_{2}^{'}$ and hence $C_{d, 1} (A_{2}^{'} + 1)$ depend only on $d$ .

3.1step 2.1∎

The assertions of the statement are exactly those of step 2.1 for $T = H$ and $T = R_{j}$ , with $C_{d} : = C_{d, 1} (A_{2}^{'} + 1)$ . Countable Choice is inherited from both [F1] and [F2], including the endpoint gluing and $L^{2}$ consistency argument.

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John-Nirenberg exponential inequality

Statement

Assume Countable Choice. There are constants $c_{n}, C_{n} \in (0, \infty)$ such that for every $b \in B M O (R^{n})$ , every cube $Q$ and every $λ > 0$ , $∣ {x \in Q : ∣ b (x) - b_{Q} ∣ > λ} ∣ \leq C_{n} ∣ Q ∣ \exp (- c_{n} λ / ∥ b ∥_{B M O})$ . If $∥ b ∥_{B M O} = 0$ then $b$ is constant almost everywhere and the left-hand side is $0$ for every $λ > 0$ , which is the interpretation used throughout.

Facts & Assumptions

Given: Countable Choice, $b \in B M O (R^{n})$ , a cube $Q$ and a level $λ > 0$ .

[F1]

The seminorm is $∥ b ∥_{B M O} = \sup_{E} ∣ E ∣^{- 1} \int_{E} ∣ b - b_{E} ∣$ and it vanishes exactly on the almost-everywhere constants (BMO seminorm and the quotient by constants).

[F2]

For every $s > ∥ b ∥_{B M O}$ and every all-generations dyadic cube $Q^{'}$ there are pairwise disjoint dyadic subcubes $Q_{j}^{' (k)}$ such that $\sum_{j} ∣ Q_{j}^{' (k)} ∣ \leq (∥ b ∥_{B M O} / s)^{k} ∣ Q^{'} ∣$ and $∣ b - b_{Q^{'}} ∣ \leq k 2^{n} s$ almost everywhere on $Q^{'} ∖ ⋃_{j} Q_{j}^{' (k)}$ for every $k \geq 1$ (John-Nirenberg stopping cubes have geometric decay).

[F3]

The all-generations dyadic cubes of Dyadic cubes of all generations in R^n partition $R^{n}$ at each generation, with parent and nesting as in All-generation dyadic cubes: partition, volume and nesting: every point lies in exactly one cube of each generation, and a dyadic cube of a coarser generation containing a point contains every finer dyadic cube through that point.

Proof

technique · direct

1.1F1

If $∥ b ∥_{B M O} = 0$ , then $b$ is constant almost everywhere by [F1], so $∣ b - b_{Q} ∣ = 0$ almost everywhere for every cube $Q$ and the asserted left-hand side vanishes for every $λ > 0$ ; this is the interpretation required by the statement.

1.2F3algebra

Covering by dyadic cubes. Let $m$ be the greatest integer with $2^{- m} \geq ℓ (Q)$ and put $S : = 2^{- m}$ , so that $ℓ (Q) \leq S < 2 ℓ (Q)$ . Every coordinate interval of $Q$ has length at most $S$ and therefore meets at most two of the generation- $m$ dyadic intervals, so $Q$ is contained in the union of the $N \leq 2^{n}$ generation- $m$ dyadic cubes $Q_{1}, \dots, Q_{N}$ that meet it. Each $Q_{j}$ has side $S$ , so $∣ Q ∣ = ℓ (Q)^{n} \leq S^{n} = ∣ Q_{j} ∣ < 2^{n} ∣ Q ∣$ .

1.3algebra

Small levels. If $0 < λ \leq Λ_{n} ∥ b ∥_{B M O}$ , then for every $γ > 0$ one has $∣ {x \in Q : ∣ b - b_{Q} ∣ > λ} ∣ \leq ∣ Q ∣ = e^{γ Λ_{n}} ∣ Q ∣ e^{- γ Λ_{n}} \leq e^{γ Λ_{n}} ∣ Q ∣ e^{- γ λ / ∥ b ∥_{B M O}}$ , because $λ / ∥ b ∥_{B M O} \leq Λ_{n}$ .

2.1step 1.2F1algebra

Comparison of the means. Fix $j$ and a point $p \in Q \cap Q_{j}$ ; both cubes lie in the cube $R_{j}$ centred at $p$ of side $4 \sqrt{n} S$ , because their diameters are $\sqrt{n} ℓ (Q) \leq \sqrt{n} S$ and $\sqrt{n} S$ respectively. By [F1], for cubes $E \subseteq R$ one has $∣ b_{E} - b_{R} ∣ \leq ∣ E ∣^{- 1} \int_{E} ∣ b - b_{R} ∣ \leq (∣ R ∣ / ∣ E ∣) ∣ R ∣^{- 1} \int_{R} ∣ b - b_{R} ∣ \leq (∣ R ∣ / ∣ E ∣) ∥ b ∥_{B M O}$ . Since $∣ R_{j} ∣ = (4 \sqrt{n})^{n} ∣ Q_{j} ∣$ and $∣ R_{j} ∣ = (4 \sqrt{n} S / ℓ (Q))^{n} ∣ Q ∣ < (8 \sqrt{n})^{n} ∣ Q ∣$ , applying this with $E = Q_{j}$ and with $E = Q$ gives $∣ b_{Q_{j}} - b_{Q} ∣ \leq 2 (8 \sqrt{n})^{n} ∥ b ∥_{B M O} = : B_{n} ∥ b ∥_{B M O}$ .

3.1step 1.2step 2.1F2algebra

Large levels. Suppose $∥ b ∥_{B M O} > 0$ and $λ > Λ_{n} ∥ b ∥_{B M O}$ with $Λ_{n} : = \max {2 B_{n}, 2^{2 n + 3}}$ , and put $s : = 2^{n + 1} ∥ b ∥_{B M O} > ∥ b ∥_{B M O}$ , $λ^{'} : = λ - B_{n} ∥ b ∥_{B M O} > λ / 2$ , and $k : = ⌊ λ^{'} / (2^{n} s) ⌋$ , where $2^{n} s = 2^{2 n + 1} ∥ b ∥_{B M O}$ . Then $k \geq 1$ , $k 2^{n} s \leq λ^{'}$ , and $k > λ / (2^{2 n + 3} ∥ b ∥_{B M O})$ because $k > λ^{'} / (2^{2 n + 1} ∥ b ∥_{B M O}) - 1 > λ / (2^{2 n + 2} ∥ b ∥_{B M O}) - 1 \geq λ / (2^{2 n + 3} ∥ b ∥_{B M O})$ as $λ > 2^{2 n + 3} ∥ b ∥_{B M O}$ . Apply [F2] on each dyadic cube $Q_{j}$ at height $s$ : the level- $k$ stopping families are pairwise disjoint with $\sum_{i} ∣ Q_{j, i}^{(k)} ∣ \leq (∥ b ∥_{B M O} / s)^{k} ∣ Q_{j} ∣ = 2^{- (n + 1) k} ∣ Q_{j} ∣$ , and $∣ b - b_{Q_{j}} ∣ \leq k 2^{n} s \leq λ^{'}$ almost everywhere off their union, so step 2.1 gives $∣ b - b_{Q} ∣ \leq λ$ there. Hence ${x \in Q_{j} : ∣ b - b_{Q} ∣ > λ}$ is contained in $⋃_{i} Q_{j, i}^{(k)}$ up to a Lebesgue-null set and has measure at most $2^{- (n + 1) k} ∣ Q_{j} ∣ \leq 2^{- (n + 1) k} 2^{n} ∣ Q ∣$ . Summing over the $N \leq 2^{n}$ cubes of step 1.2 gives $∣ {x \in Q : ∣ b - b_{Q} ∣ > λ} ∣ \leq 2^{2 n} 2^{- (n + 1) k} ∣ Q ∣ \leq 2^{2 n} ∣ Q ∣ e^{- k \ln 2} \leq 2^{2 n} ∣ Q ∣ e^{- γ λ / ∥ b ∥_{B M O}}$ with $γ : = \ln 2 / 2^{2 n + 3}$ .

4.1step 1.1step 3.1step 1.3F2∎

Assembly. Take $γ : = \ln 2 / 2^{2 n + 3}$ , $B_{n} : = 2 (8 \sqrt{n})^{n}$ , $Λ_{n} : = \max {2 B_{n}, 2^{2 n + 3}}$ and $C_{n} : = \max {2^{2 n}, e^{γ Λ_{n}}}$ . Steps 3.1 and 1.3 give $∣ {x \in Q : ∣ b (x) - b_{Q} ∣ > λ} ∣ \leq C_{n} ∣ Q ∣ e^{- γ λ / ∥ b ∥_{B M O}}$ for every $λ > 0$ when $∥ b ∥_{B M O} > 0$ , covering the large levels and the small levels respectively, and step 1.1 gives the bound when $∥ b ∥_{B M O} = 0$ . Both constants depend only on $n$ , and the stopping construction inside [F2] is the only place Countable Choice is used.

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BMO oscillation norms in Lq are equivalent

Statement

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

Let $1 \leq q < \infty$ and for $b \in L_{l o c}^{1} (R^{n})$ put $∥ b ∥_{B M O, q} : = \sup_{Q} (∣ Q ∣^{- 1} \int_{Q} ∣ b - b_{Q} ∣^{q})^{1 / q} \in [0, \infty]$ , the supremum over all cubes. Then there are constants $0 < c_{n, q} \leq C_{n, q} < \infty$ such that $c_{n, q} ∥ b ∥_{B M O} \leq ∥ b ∥_{B M O, q} \leq C_{n, q} ∥ b ∥_{B M O}$ for every $b \in B M O (R^{n})$ ; indeed $(∣ Q ∣^{- 1} \int_{Q} ∣ b - b_{Q} ∣^{q})^{1 / q} \leq C_{n, q} ∥ b ∥_{B M O}$ for every cube $Q$ . In particular every $B M O$ function lies in $L_{l o c}^{q} (R^{n})$ .

Facts & Assumptions

Given: Countable Choice, $1 \leq q < \infty$ , a function $b \in L_{l o c}^{1} (R^{n})$ , a cube $Q$ , and the mean and seminorm of BMO seminorm and the quotient by constants.

[F1]

The mean is $b_{Q} = ∣ Q ∣^{- 1} \int_{Q} b$ and $∥ b ∥_{B M O} = \sup_{Q} ∣ Q ∣^{- 1} \int_{Q} ∣ b - b_{Q} ∣$ ; the seminorm vanishes exactly on the almost-everywhere constants (BMO seminorm and the quotient by constants).

[F2]

Holder's inequality on the finite-measure cube $Q$ , applied to the nonnegative functions $∣ b - b_{Q} ∣$ and $1$ , gives $∣ Q ∣^{- 1} \int_{Q} ∣ b - b_{Q} ∣ \leq (∣ Q ∣^{- 1} \int_{Q} ∣ b - b_{Q} ∣^{q})^{1 / q}$ for $1 \leq q < \infty$ (Holder's inequality for integrals, including the endpoint cases); if the right-hand side is infinite the inequality is immediate, and $q = 1$ is equality.

[F3]

The layer-cake formula gives $∣ Q ∣^{- 1} \int_{Q} ∣ b - b_{Q} ∣^{q} = q \int_{0}^{\infty} λ^{q - 1} ∣ Q ∣^{- 1} ∣ {x \in Q : ∣ b - b_{Q} ∣ > λ} ∣ d λ$ (For 0 < p < infinity, the layer-cake formula computes the integral of |f|^p from the distribution function).

[F4]

There are constants $c_{n}, C_{n} \in (0, \infty)$ with $∣ {x \in Q : ∣ b - b_{Q} ∣ > λ} ∣ \leq C_{n} ∣ Q ∣ e^{- c_{n} λ / ∥ b ∥_{B M O}}$ for every $λ > 0$ , the zero-seminorm case giving the value $0$ (John-Nirenberg exponential inequality).

Proof

technique · direct

1.1F1F2

Lower bound. By [F2], $∣ Q ∣^{- 1} \int_{Q} ∣ b - b_{Q} ∣ \leq (∣ Q ∣^{- 1} \int_{Q} ∣ b - b_{Q} ∣^{q})^{1 / q}$ for every cube $Q$ ; taking the supremum over all cubes gives $∥ b ∥_{B M O} \leq ∥ b ∥_{B M O, q}$ , so $c_{n, q} = 1$ is admissible.

1.2F3F4F1algebra

Upper bound for a fixed cube. If $∥ b ∥_{B M O} > 0$ , [F4] inserted into the layer-cake formula [F3] gives $∣ Q ∣^{- 1} \int_{Q} ∣ b - b_{Q} ∣^{q} \leq q C_{n} \int_{0}^{\infty} λ^{q - 1} e^{- c_{n} λ / ∥ b ∥_{B M O}} d λ = q C_{n} ∥ b ∥_{B M O}^{q} \int_{0}^{\infty} μ^{q - 1} e^{- c_{n} μ} d μ$ after the substitution $λ = μ ∥ b ∥_{B M O}$ ; the last integral is finite because $μ^{q - 1} e^{- c_{n} μ}$ is integrable on $(0, \infty)$ . If $∥ b ∥_{B M O} = 0$ the left-hand side is $0$ by [F1], so the same estimate holds with either side zero.

2.1step 1.2step 1.1algebra

With $K_{n, q} : = (q C_{n} \int_{0}^{\infty} μ^{q - 1} e^{- c_{n} μ} d μ)^{1 / q} < \infty$ , step 1.2 gives $(∣ Q ∣^{- 1} \int_{Q} ∣ b - b_{Q} ∣^{q})^{1 / q} \leq K_{n, q} ∥ b ∥_{B M O}$ for every cube $Q$ , and taking the supremum over $Q$ gives $∥ b ∥_{B M O, q} \leq K_{n, q} ∥ b ∥_{B M O}$ ; so $C_{n, q} = K_{n, q}$ is admissible. Both constants depend only on $n$ and $q$ .

3.1step 2.1F1∎

Finally, for $b \in B M O (R^{n})$ and a cube $Q$ one has $∣ b ∣ \leq ∣ b - b_{Q} ∣ + ∣ b_{Q} ∣$ , hence $\int_{Q} ∣ b ∣^{q} \leq 2^{q - 1} (\int_{Q} ∣ b - b_{Q} ∣^{q} + ∣ Q ∣ ∣ b_{Q} ∣^{q}) < \infty$ by step 2.1 and the finiteness of the local mean $b_{Q}$ ; every compact set is covered by finitely many cubes, so $b \in L_{l o c}^{q} (R^{n})$ .

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L2-normalised H1 atoms have uniformly bounded H1 norm

Statement

Assume Countable Choice. Fix an admissible kernel $φ$ with $\int φ \neq 0$ and an auxiliary integer order $\tilde{N} \geq \max {N_{0} (n, 1, φ), n + 1}$ for the grand-maximal characterisation. Let $a$ be a $(1, 2)$ -atom supported in a cube $Q$ : $a$ vanishes off $Q$ , $\int a = 0$ and $(∣ Q ∣^{- 1} \int_{Q} ∣ a ∣^{2})^{1 / 2} \leq ∣ Q ∣^{- 1}$ . Then $∥ a ∥_{H^{1}} = ∥ M_{φ}^{0} a ∥_{L^{1}} \leq C_{n, \tilde{N}, φ}$ , with the constant independent of $Q$ and the atom.

Facts & Assumptions

Given: Countable Choice, the fixed admissible kernel $φ$ and auxiliary integer order $\tilde{N}$ from the statement; also let $a$ be a $(1, 2)$ -atom supported in a cube $Q$ with centre $x_{Q}$ and side $ℓ (Q)$ , and the functionals and spaces of The real Hardy space $H^{p}$ defined by a radial maximal function and Grand maximal test class of order N and the grand maximal function.

[F1]

The test class is $F_{\tilde{N}} = {ψ \in S : P_{\tilde{N}} (ψ) \leq 1}$ with $P_{\tilde{N}} (ψ) = \sup_{x} (1 + ∣ x ∣)^{\tilde{N}} \max_{∣ α ∣ \leq \tilde{N} + 1} ∣ \partial^{α} ψ (x) ∣$ , $M_{\tilde{N}} f (x) = \sup_{ψ \in F_{\tilde{N}}} \sup_{t > 0} \sup_{∣ y - x ∣ \leq t} ∣ (f * ψ_{t}) (y) ∣$ with $ψ_{t} (u) = t^{- n} ψ (u / t)$ ; in particular $∣ ψ ∣ \leq 1$ and, from the componentwise derivative bounds in the seminorm, $∣ \nabla ψ (u) ∣ \leq \sqrt{n} (1 + ∣ u ∣)^{- \tilde{N}}$ for every $ψ \in F_{\tilde{N}}$ (Grand maximal test class of order N and the grand maximal function).

[F2]

Every $L^{2}$ class has a representative defining a tempered distribution, and for the regular distribution of a locally integrable compactly supported $a$ the convolution is $(a * ψ_{t}) (y) = \int a (z) ψ_{t} (y - z) d z$ (Polynomial growth functions define tempered distributions, Convolution of a tempered distribution with a schwartz function).

[F3]

The stated atom hypotheses give $supp a \subseteq Q$ , $\int a = 0$ and $∥ a ∥_{L^{2}} \leq ∣ Q ∣^{- 1 / 2}$ ; hence $∥ a ∥_{L^{1}} \leq ∣ Q ∣^{1 / 2} ∥ a ∥_{L^{2}} \leq 1$ by Cauchy-Schwarz on the finite-measure cube (Cauchy-Schwarz inequality for $L^{2}$ ).

[F4]

Once $M_{\tilde{N}} a \in L^{1}$ has been established, the maximal-function characterisation gives $a \in H^{1}$ and $∥ M_{φ}^{0} a ∥_{L^{1}} \leq C_{n, \tilde{N}, φ} ∥ M_{\tilde{N}} a ∥_{L^{1}}$ and $M_{\tilde{N}} a$ is Borel measurable (Maximal-function characterisations of real Hardy spaces, Measurability and lower semicontinuity of the smooth maximal functions).

[F5]

The centered Hardy-Littlewood maximal operator satisfies $∥ M f ∥_{L^{2}} \leq C_{n, 2} ∥ f ∥_{L^{2}}$ and $M f$ is Borel measurable for $f \in L_{l o c}^{1}$ (The centered maximal operator is bounded on $L^{p} (R^{n})$ for $1 < p < \infty$ , The centered Hardy-Littlewood maximal function is Borel measurable, The centered and uncentered Hardy-Littlewood maximal functions).

[F6]

A Euclidean ball of radius $R$ is contained in the axis-parallel box of side $2 R$ with the same centre; under Countable Choice that box has Lebesgue measure $(2 R)^{n}$ , and Lebesgue measure is monotone (Axis-parallel rectangles in $R^{m}$ and their volume, A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included, Measures are monotone).

[F7]

Under Countable Choice, $λ (B (x, r)) = c_{n} r^{n}$ for a finite positive constant $c_{n}$ depending only on dimension (Sphere and ball measures scale in Rn).

Proof

technique · direct

1.1F1F2F3F5F7algebra

Since $a \in L^{2}$ is supported in the finite-measure cube $Q$ , [F2] identifies its regular distribution with a tempered distribution; [F3] gives $∥ a ∥_{L^{1}} \leq 1$ . For every $ψ \in F_{\tilde{N}}$ , $∣ ψ (u) ∣ \leq (1 + ∣ u ∣)^{- \tilde{N}}$ . Fix $t > 0$ and $∣ y - x ∣ \leq t$ . Split the convolution integral into $∣ y - z ∣ < t$ and the annuli $2^{k - 1} t \leq ∣ y - z ∣ < 2^{k} t$ for $k \geq 1$ . The ball inclusions $B (y, t) \subseteq B (x, 2 t)$ and $B (y, 2^{k} t) \subseteq B (x, 2^{k + 1} t)$ , together with [F5] and [F7], give $∣ (a * ψ_{t}) (y) ∣ \leq t^{- n} \int_{B (y, t)} ∣ a (z) ∣ d z + \sum_{k \geq 1} 2^{- (k - 1) \tilde{N}} t^{- n} \int_{B (y, 2^{k} t)} ∣ a (z) ∣ d z$ $\leq c_{n} 2^{n} M (∣ a ∣) (x) + c_{n} \sum_{k \geq 1} 2^{- (k - 1) \tilde{N}} 2^{(k + 1) n} M (∣ a ∣) (x) \leq C_{n, \tilde{N}} M (∣ a ∣) (x),$ since $\tilde{N} \geq n + 1$ makes the geometric series converge. Taking suprema gives $M_{\tilde{N}} a \leq C_{n, \tilde{N}} M (∣ a ∣)$ pointwise.

2.1step 1.1F3F5F6

Near region. Put $ρ_{n} : = 4 \sqrt{n}$ and $Q^{*} : = {x : ∣ x - x_{Q} ∣ \leq ρ_{n} ℓ (Q)}$ . By [F6], $∣ Q^{*} ∣ \leq (2 ρ_{n} ℓ (Q))^{n} = C_{n}^{'} ∣ Q ∣$ , since the ball is contained in the corresponding axis-parallel cube. Cauchy-Schwarz together with the $L^{2}$ bound of [F5] and step 1.1 gives $\int_{Q^{*}} ∣ M_{\tilde{N}} a ∣ \leq ∣ Q^{*} ∣^{1 / 2} ∥ M_{\tilde{N}} a ∥_{2} \leq ∣ Q^{*} ∣^{1 / 2} C_{n, \tilde{N}} ∥ M (∣ a ∣) ∥_{2} \leq C_{n, 2}^{'} ∣ Q^{*} ∣^{1 / 2} ∥ a ∥_{2} \leq C_{n}$ .

2.2step 1.1F1F3F2

Far region, pointwise. Assume $r : = ∣ x - x_{Q} ∣ \geq ρ_{n} ℓ (Q)$ ; then every $z \in Q$ satisfies $∣ z - x_{Q} ∣ \leq \sqrt{n} ℓ (Q) \leq r / 4$ , so for $∣ y - x ∣ \leq t$ and $w$ on the segment between $y - z$ and $y - x_{Q}$ one has $∣ w ∣ \geq r - t - r / 4$ . If $t \leq r / 2$ , then $∣ w ∣ \geq r / 4$ and $∣ ψ_{t} (y - z) - ψ_{t} (y - x_{Q}) ∣ \leq \sqrt{n} ℓ (Q) \sup ∣ \nabla ψ_{t} ∣$ with $∣ \nabla ψ_{t} (w) ∣ = t^{- n - 1} ∣ \nabla ψ (w / t) ∣ \leq \sqrt{n} t^{- n - 1} (t / ∣ w ∣)^{\tilde{N}} \leq C_{\tilde{N}, n} r^{- n - 1}$ because $\tilde{N} \geq n + 1$ and $t \leq r / 2$ ; if $t > r / 2$ , the same difference is at most $\sqrt{n} ℓ (Q) \cdot \sqrt{n} (r / 2)^{- n - 1} \leq C_{n} ℓ (Q) r^{- n - 1}$ by [F1]. Using $\int a = 0$ to write $(a * ψ_{t}) (y) = \int a (z) [ψ_{t} (y - z) - ψ_{t} (y - x_{Q})] d z$ and $∥ a ∥_{L^{1}} \leq 1$ from [F3], both cases give $∣ (a * ψ_{t}) (y) ∣ \leq C_{\tilde{N}, n} ℓ (Q) r^{- n - 1}$ , and taking suprema over $ψ \in F_{\tilde{N}}$ , $t > 0$ and $∣ y - x ∣ \leq t$ yields $M_{\tilde{N}} a (x) \leq C_{\tilde{N}, n} ℓ (Q) r^{- n - 1}$ .

3.1step 2.2algebra

Far region, integration. Covering ${r \geq ρ_{n} ℓ (Q)}$ by the shells ${2^{j} ρ_{n} ℓ (Q) \leq r < 2^{j + 1} ρ_{n} ℓ (Q)}$ , each contained in a cube of side $4 \cdot 2^{j} ρ_{n} ℓ (Q)$ , and using step 2.2 gives $\int_{r \geq ρ_{n} ℓ (Q)} M_{\tilde{N}} a \leq C_{\tilde{N}, n} ℓ (Q) \sum_{j \geq 0} (2^{j} ρ_{n} ℓ (Q))^{- n - 1} (4 \cdot 2^{j} ρ_{n} ℓ (Q))^{n} = C_{\tilde{N}, n}^{'}$ .

4.1step 2.1step 3.1F4∎

Combining steps 2.1 and 3.1, $∥ M_{\tilde{N}} a ∥_{L^{1}} = \int_{Q^{*}} ∣ M_{\tilde{N}} a ∣ + \int_{R^{n} ∖ Q^{*}} ∣ M_{\tilde{N}} a ∣ \leq C_{n, \tilde{N}} < \infty$ , so [F4] gives $∥ a ∥_{H^{1}} = ∥ M_{φ}^{0} a ∥_{L^{1}} \leq C_{n, \tilde{N}, φ} ∥ M_{\tilde{N}} a ∥_{L^{1}} \leq C_{n, \tilde{N}, φ}$ , independent of $Q$ and of the atom. Countable Choice is inherited from the maximal-function and measure suppliers [F4]-[F7].

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

Mean-zero L2 functions on a cube embed continuously into H1

Statement

Assume Countable Choice, and fix the admissible kernel $φ$ and auxiliary order $\tilde{N}$ of L2-normalised H1 atoms have uniformly bounded H1 norm. There is $C_{n, \tilde{N}, φ} < \infty$ such that for every cube $Q$ and every $f \in L^{2} (R^{n})$ with $supp f \subseteq Q$ and $\int_{Q} f = 0$ , the class of $f$ lies in $H^{1} (R^{n})$ and $∥ f ∥_{H^{1}} \leq C_{n, \tilde{N}, φ} ∣ Q ∣^{1 / 2} ∥ f ∥_{L^{2}}$ .

Facts & Assumptions

Given: Countable Choice, the fixed $φ$ and $\tilde{N}$ , a cube $Q$ of finite positive volume $∣ Q ∣$ (Axis-parallel rectangles in $R^{m}$ and their volume) and a function $f \in L^{2} (R^{n})$ with $supp f \subseteq Q$ and $\int_{Q} f = 0$ .

[F1]

The $H^{1}$ functional is $∥ g ∥_{H^{1}} = ∥ M_{φ}^{0} g ∥_{L^{1}}$ for an admissible Schwartz function $φ$ with $\int φ \neq 0$ , defined for tempered distributions (The real Hardy space $H^{p}$ defined by a radial maximal function), and every $L^{2}$ class has a representative defining a tempered distribution (Polynomial growth functions define tempered distributions).

[F2]

If $a$ vanishes off a cube $Q$ , has $\int a = 0$ and $(∣ Q ∣^{- 1} \int_{Q} ∣ a ∣^{2})^{1 / 2} \leq ∣ Q ∣^{- 1}$ , then $∥ a ∥_{H^{1}} \leq C_{n, \tilde{N}, φ}$ with this constant independent of $Q$ and $a$ (L2-normalised H1 atoms have uniformly bounded H1 norm).

[F3]

For $λ \in C$ and a tempered distribution $g$ , $M_{φ}^{0} (λ g) = ∣ λ ∣ M_{φ}^{0} g$ pointwise, because $(g * φ_{t}) (y) = ⟨ g, φ_{t} (y - \cdot) ⟩$ is complex-linear in $g$ ; hence $∥ λ g ∥_{H^{1}} = ∣ λ ∣ ∥ g ∥_{H^{1}}$ (The real Hardy space $H^{p}$ defined by a radial maximal function).

[F4]

The $L^{2}$ classes of The space $L^{p} (μ)$ as the quotient by null functions are used, so all statements below are insensitive to changes on null sets.

Proof

technique · direct

1.1F1

If $f = 0$ almost everywhere, then $f$ is the zero distribution and $∥ f ∥_{H^{1}} = 0 \leq C_{n, \tilde{N}, φ} ∣ Q ∣^{1 / 2} ∥ f ∥_{L^{2}} = 0$ .

1.2F1F2F4

Assume $f \neq 0$ in $L^{2}$ , so $∥ f ∥_{L^{2}} > 0$ because $Q$ has finite measure, and put $a : = f / (∣ Q ∣^{1 / 2} ∥ f ∥_{L^{2}})$ . Then $a$ vanishes off $Q$ and $\int a = 0$ ; moreover $∣ a ∣^{2} = ∣ f ∣^{2} / (∣ Q ∣ ∥ f ∥_{L^{2}}^{2})$ , so $\int_{Q} ∣ a ∣^{2} = ∣ Q ∣^{- 1}$ and $(∣ Q ∣^{- 1} \int_{Q} ∣ a ∣^{2})^{1 / 2} = ∣ Q ∣^{- 1}$ : the function $a$ is a $(1, 2)$ -atom. By [F2], $∥ a ∥_{H^{1}} \leq C_{n, \tilde{N}, φ}$ .

2.1step 1.1step 1.2F3∎

Since $f = ∣ Q ∣^{1 / 2} ∥ f ∥_{L^{2}} a$ , the homogeneity [F3] gives $∥ f ∥_{H^{1}} = ∣ Q ∣^{1 / 2} ∥ f ∥_{L^{2}} ∥ a ∥_{H^{1}} \leq C_{n, \tilde{N}, φ} ∣ Q ∣^{1 / 2} ∥ f ∥_{L^{2}}$ ; together with the trivial case of step 1.1 this is the asserted estimate, and the class of $f$ lies in $H^{1}$ because $a$ does and multiplication by a scalar preserves the space.

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

Finite atomic sums are dense in H1

Statement

Assume Countable Choice. The finite linear combinations of $H^{1}$ atoms are dense in $H^{1} (R^{n})$ : for every $f \in H^{1} (R^{n})$ and every $ε > 0$ there are finitely many atoms $a_{1}, \dots, a_{N}$ and coefficients $λ_{1}, \dots, λ_{N}$ with $∥ f - \sum_{1 \leq j \leq N} λ_{j} a_{j} ∥_{H^{1}} < ε$ .

Facts & Assumptions

Given: Countable Choice, $f \in H^{1} (R^{n})$ and $ε > 0$ , with the $(1, \infty, 0)$ -atoms of $H^{p}$ atoms with a prescribed moment order and the $H^{1}$ functional of The real Hardy space $H^{p}$ defined by a radial maximal function.

[F1]

The atomic characterisation of $H^{1}$ gives a sequence $(λ_{j}) \in ℓ^{1}$ and $(1, \infty, 0)$ -atoms $a_{j}$ , reindexed by $j \geq 1$ , with $f = \sum_{j} λ_{j} a_{j}$ converging in $S^{'}$ , with $\sum_{j} ∣ λ_{j} ∣ \leq C ∥ f ∥_{H^{1}}$ for a suitable constant; moreover every such series converges also in the $H^{1}$ quasi-norm, which for $p = 1$ is the norm $∥ \cdot ∥_{H^{1}}$ (Atomic characterisation of real $H^{p}$ for $0 < p \leq 1$ , The real Hardy space $H^{p}$ defined by a radial maximal function).

[F2]

For an $ℓ^{1}$ sum of atoms indexed by $j \geq 1$ , the partial sums converge to the sum in the $H^{1}$ quasi-norm and the tail bound $∥ g - \sum_{1 \leq j \leq N} λ_{j} a_{j} ∥_{H^{1}} \leq C (\sum_{j > N} ∣ λ_{j} ∣)$ holds ( $ℓ^{p}$ sums of atoms converge in $S^{'}$ and in $H^{p}$ with $p = 1$ ).

Proof

technique · direct

1.1F1

By [F1] fix $ℓ^{1}$ -coefficients $(λ_{j})$ and atoms $(a_{j})$ , indexing both sequences by $j \geq 1$ , with $f = \sum_{j} λ_{j} a_{j}$ converging in $S^{'}$ and with $\sum_{j} ∣ λ_{j} ∣ \leq C ∥ f ∥_{H^{1}}$ ; by the same item the partial sums $S_{N} : = \sum_{1 \leq j \leq N} λ_{j} a_{j}$ converge to $f$ in the $H^{1}$ norm.

2.1step 1.1F2

Since $∥ f - S_{N} ∥_{H^{1}} \to 0$ as $N \to \infty$ by step 1.1 and $ε > 0$ , there is $N \geq 1$ with $∥ f - S_{N} ∥_{H^{1}} < ε$ ; the sum $S_{N}$ is a finite linear combination of the atoms $a_{1}, \dots, a_{N}$ with coefficients $λ_{1}, \dots, λ_{N}$ , and [F2] gives the same conclusion with the explicit tail bound.

3.1step 2.1∎

Thus for every $f \in H^{1} (R^{n})$ and every $ε > 0$ there is a finite atomic sum within $ε$ in the $H^{1}$ norm, which is density. Countable Choice is inherited from the atomic characterisation.

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

Bounded H1 functionals have compatible local L2 representatives

Statement

Assume Countable Choice and fix the kernel $φ$ and auxiliary order $\tilde{N}$ of Mean-zero L2 functions on a cube embed continuously into H1. Let $Λ \in (H^{1} (R^{n}))^{*}$ and let $L_{0}^{2} (Q)$ be the closed subspace of $L^{2} (R^{n})$ of functions supported in the cube $Q$ with $\int_{Q} f = 0$ (equivalently, mean $0$ when $∣ Q ∣ > 0$ ). For every cube $Q$ there is a unique $u_{Q} \in L_{0}^{2} (Q)$ with $Λ (f) = \int_{Q} u_{Q} f$ for every $f \in L_{0}^{2} (Q)$ ; moreover $Q \subseteq R$ implies that $u_{Q} - u_{R}$ is almost everywhere constant on $Q$ . Consequently there are a locally integrable function $u$ on $R^{n}$ , unique up to additive constants, and for every cube $Q$ a constant $c_{Q}$ with $u - u_{Q} = c_{Q}$ almost everywhere on $Q$ .

Facts & Assumptions

Given: Countable Choice, a bounded linear functional $Λ \in (H^{1} (R^{n}))^{*}$ and cubes $Q \subseteq R$ , with the complex $L^{2}$ space and its integral pairing of $L^{2}$ with the integral pairing is a Hilbert space and The space $L^{p} (μ)$ as the quotient by null functions.

[F1]

For every cube $Q$ and every $f \in L^{2} (R^{n})$ with $supp f \subseteq Q$ and $\int_{Q} f = 0$ one has $∥ f ∥_{H^{1}} \leq C_{n, \tilde{N}, φ} ∣ Q ∣^{1 / 2} ∥ f ∥_{L^{2}}$ for the fixed kernel and order of (Mean-zero L2 functions on a cube embed continuously into H1).

[F2]

On complex $L^{2} (R^{n})$ the form $⟨ f, g ⟩ = \int f \overline{g}$ is a Hilbert-space inner product, and Riesz representation holds: for every bounded linear functional $λ$ on a closed subspace $H_{0}$ there is a unique $y \in H_{0}$ with $λ (f) = ⟨ f, y ⟩$ for all $f \in H_{0}$ and $∥ λ ∥ = ∥ y ∥$ ( $L^{2}$ with the integral pairing is a Hilbert space, Riesz representation for Hilbert spaces).

[F4]

Cauchy-Schwarz gives $∣ \int_{Q} f ∣ \leq ∣ Q ∣^{1 / 2} ∥ f ∥_{2}$ (Cauchy-Schwarz inequality for $L^{2}$ ).

[F3]

A countable union of Lebesgue-null sets is Lebesgue-null (Subsets and countable unions of null subsets of $R^{m}$ are null), and Axis-parallel rectangles in $R^{m}$ and their volume supplies the cube conventions of the chain $Q_{k} = [- k, k]^{n}$ below.

Proof

technique · direct

1.1F1F2F4

For a cube $Q$ , the set $L_{0}^{2} (Q)$ is the kernel of the continuous linear functional $f \mapsto \int_{Q} f$ on the closed subspace ${f \in L^{2} : f = 0 a.e. off Q}$ (closedness follows from $∥ f 1_{Q^{c}} ∥_{2} \leq ∥ f - f_{m} ∥_{2}$ for a supported approximating sequence $f_{m}$ , and continuity of the integral from [F4]), hence a closed subspace of the Hilbert space $L^{2} (R^{n})$ ; and for $f \in L_{0}^{2} (Q)$ the boundedness of $Λ$ and [F1] give $∣ Λ (f) ∣ \leq ∥ Λ ∥ ∥ f ∥_{H^{1}} \leq C_{n, \tilde{N}, φ} ∣ Q ∣^{1 / 2} ∥ Λ ∥ ∥ f ∥_{L^{2}}$ , so $Λ ∣_{L_{0}^{2} (Q)}$ is bounded for the $L^{2}$ norm.

2.1step 1.1F2

By [F2] applied to the closed subspace $L_{0}^{2} (Q)$ and the bounded functional $Λ ∣_{L_{0}^{2} (Q)}$ , there is a unique $y_{Q} \in L_{0}^{2} (Q)$ with $Λ (f) = ⟨ f, y_{Q} ⟩$ for $f \in L_{0}^{2} (Q)$ ; setting $u_{Q} : = \overline{y_{Q}}$ , which still lies in $L_{0}^{2} (Q)$ because conjugation preserves supports and means, gives $Λ (f) = \int_{Q} u_{Q} f$ for every $f \in L_{0}^{2} (Q)$ , and $u_{Q}$ is unique with this property.

3.1step 2.1F2

Nested compatibility. If $∣ Q ∣ = 0$ , then $L_{0}^{2} (Q) = {0}$ as an almost-everywhere quotient, and constancy almost everywhere on $Q$ is vacuous. Assume now $∣ Q ∣ > 0$ . If $Q \subseteq R$ then $L_{0}^{2} (Q) \subseteq L_{0}^{2} (R)$ , and for $f \in L_{0}^{2} (Q)$ step 2.1 gives $\int_{Q} u_{Q} f = Λ (f) = \int_{R} u_{R} f = \int_{Q} u_{R} f$ . Put $h : = u_{Q} - u_{R}$ and apply this identity with $f : = \overline{h - {mean}_{Q} (h)} 1_{Q}$ , which belongs to $L_{0}^{2} (Q)$ . Then $\int_{Q} ∣ h - {mean}_{Q} (h) ∣^{2} = 0$ , so $u_{Q} - u_{R}$ equals the constant ${mean}_{Q} (u_{Q} - u_{R})$ almost everywhere on $Q$ .

4.1step 3.1F3

Gluing. Let $Q_{k} = [- k, k]^{n}$ for $k \geq 1$ and use Countable Choice to select measurable representatives of their $u_{Q_{k}}$ . By step 3.1 the difference $(u_{Q_{k + 1}} - u_{Q_{k}})$ is almost everywhere constant $a_{k}$ on $Q_{k}$ ; define $c_{1} : = 0$ and $c_{k + 1} : = c_{k} - a_{k}$ , so that $u_{Q_{k + 1}} + c_{k + 1} = u_{Q_{k}} + c_{k}$ almost everywhere on $Q_{k}$ . Removing the countable union of the exceptional null sets, which is null by [F3], define $u (x) : = u_{Q_{k}} (x) + c_{k}$ for $x \in Q_{k}$ outside that null set, and set $u = 0$ on the null set; this is well defined, locally integrable, and for every cube $Q$ , choosing $k$ with $Q \subseteq Q_{k}$ , the function $u - u_{Q} = [u - (u_{Q_{k}} + c_{k})] + [(u_{Q_{k}} + c_{k}) - u_{Q}]$ is almost everywhere constant on $Q$ by the construction and step 3.1. If $u^{'}$ is another such function, then on each $Q_{k}$ the difference $u - u^{'}$ is constant almost everywhere, and the constants agree on the positive-measure overlap $Q_{k} \cap Q_{k + 1} = Q_{k}$ , so $u - u^{'}$ is almost everywhere equal to a single constant on $⋃_{k} Q_{k} = R^{n}$ : uniqueness up to additive constants.

5.1step 2.1step 3.1step 4.1∎

Steps 2.1, 3.1 and 4.1 prove the existence and uniqueness of each $u_{Q}$ , the nested constancy, and the existence of the global representative $u$ with constants $c_{Q}$ , which is the statement. Countable Choice is used for the countably many representations and the countable union in step 4.1.

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

Bounded BMO functions dualise H1 boundedly

Statement

Assume Countable Choice, fix the $H^{1}$ kernel $φ$ and admissible atomic order $\tilde{N}$ used in Atomic characterisation of real $H^{p}$ for $0 < p \leq 1$ . There is $C_{n, \tilde{N}, φ} < \infty$ such that for every $b \in L^{\infty} (R^{n}) \cap B M O (R^{n})$ and every $f \in H^{1} (R^{n})$ the integral $\int f b$ converges absolutely and $∣ \int f b ∣ \leq C_{n, \tilde{N}, φ} ∥ b ∥_{B M O} ∥ f ∥_{H^{1}}$ .

Facts & Assumptions

Given: Countable Choice, the fixed $φ, \tilde{N}$ , $b \in L^{\infty} (R^{n}) \cap B M O (R^{n})$ and $f \in H^{1} (R^{n})$ .

[F1]

The atomic characterisation gives $(λ_{j}) \in ℓ^{1}$ and $(1, \infty, 0)$ -atoms $a_{j}$ with $f = \sum_{j} λ_{j} a_{j}$ in $S^{'}$ and with the partial sums $S_{N} = \sum_{j \leq N} λ_{j} a_{j}$ converging to $f$ in the $H^{1}$ norm; the coefficients satisfy $\sum_{j} ∣ λ_{j} ∣ \leq C_{n, \tilde{N}, φ} ∥ f ∥_{H^{1}}$ (Atomic characterisation of real $H^{p}$ for $0 < p \leq 1$ , $ℓ^{p}$ sums of atoms converge in $S^{'}$ and in $H^{p}$ ).

[F2]

Each atom is bounded, compactly supported and has $∥ a_{j} ∥_{L^{1}} \leq 1$ ; the pairing with $b$ satisfies $∣ \int a_{j} b ∣ \leq ∥ b ∥_{B M O}$ ( $H^{p}$ atoms with a prescribed moment order, BMO functions pair uniformly with H1 atoms).

[F3]

Complex $L^{1}$ is complete, and on any measure space the pairing of an $L^{1}$ function with an $L^{\infty}$ function obeys $\int ∣ g h ∣ \leq ∥ g ∥_{L^{1}} ∥ h ∥_{L^{\infty}}$ (Complex Lp completeness and almost-everywhere subsequences, Holder's inequality for integrals, including the endpoint cases, The space $L^{p} (μ)$ as the quotient by null functions).

Proof

technique · direct

1.1F1F2F3

By [F1] fix a representation $f = \sum_{j} λ_{j} a_{j}$ with $\sum_{j} ∣ λ_{j} ∣ \leq C_{n, \tilde{N}, φ} ∥ f ∥_{H^{1}}$ . The partial sums are $L^{1}$ functions with $∥ S_{N} ∥_{L^{1}} \leq \sum_{j \leq N} ∣ λ_{j} ∣ ∥ a_{j} ∥_{L^{1}} \leq \sum_{j \leq N} ∣ λ_{j} ∣$ by [F2]; they therefore form a Cauchy sequence in $L^{1}$ , and by completeness [F3] converge in $L^{1}$ to some $g \in L^{1}$ with $∥ g ∥_{L^{1}} \leq \sum_{j} ∣ λ_{j} ∣$ . For every test function $ψ$ one has $⟨ f, ψ ⟩ = \lim_{N} ⟨ S_{N}, ψ ⟩ = \lim_{N} \int S_{N} ψ = \int g ψ$ by [F3] and the $S^{'}$ -convergence of the partial sums; hence $f$ is represented by the $L^{1}$ function $g$ , and $\int f b = \int g b$ converges absolutely with $\int ∣ g b ∣ \leq ∥ g ∥_{L^{1}} ∥ b ∥_{L^{\infty}} < \infty$ .

2.1step 1.1F2F3

Since $S_{N} \to g$ in $L^{1}$ and $b \in L^{\infty}$ , [F3] gives $\int S_{N} b \to \int g b$ ; and $∣ \int S_{N} b ∣ = ∣ \sum_{j \leq N} λ_{j} \int a_{j} b ∣ \leq \sum_{j \leq N} ∣ λ_{j} ∣ ∥ b ∥_{B M O} \leq C_{n, \tilde{N}, φ} ∥ b ∥_{B M O} ∥ f ∥_{H^{1}}$ by [F2]. Passing to the limit gives $∣ \int f b ∣ = ∣ \int g b ∣ \leq C_{n, \tilde{N}, φ} ∥ b ∥_{B M O} ∥ f ∥_{H^{1}}$ .

3.1step 1.1step 2.1∎

Step 2.1 is the asserted bound with constant $C_{n, \tilde{N}, φ}$ , and step 1.1 is the asserted absolute convergence. Countable Choice is inherited from the suppliers.

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

The dual representative has uniformly bounded BMO oscillation

Statement

Assume Countable Choice, fix the $H^{1}$ kernel $φ$ and auxiliary order $\tilde{N}$ of Mean-zero L2 functions on a cube embed continuously into H1, let $Λ \in (H^{1} (R^{n}))^{*}$ and let $u$ be the representative of the preceding lemma. Then $u \in B M O (R^{n})$ and $∥ u ∥_{B M O} \leq C_{n, \tilde{N}, φ} ∥ Λ ∥$ , with the constant independent of $Λ$ .

Facts & Assumptions

Given: Countable Choice, the fixed $φ, \tilde{N}$ , $Λ \in (H^{1} (R^{n}))^{*}$ , the locally integrable representative $u$ and the local representatives $u_{Q}$ of Bounded H1 functionals have compatible local L2 representatives, and a nondegenerate cube $Q$ .

[F1]

For every nondegenerate cube $Q$ the representative $u_{Q} \in L_{0}^{2} (Q)$ has mean $0$ on $Q$ , satisfies $Λ (f) = \int_{Q} u_{Q} f$ for $f \in L_{0}^{2} (Q)$ , and $u - u_{Q}$ is almost everywhere constant on $Q$ (Bounded H1 functionals have compatible local L2 representatives).

[F2]

The restriction of $Λ$ to $L_{0}^{2} (Q)$ obeys $∣ Λ (f) ∣ \leq C_{n, \tilde{N}, φ} ∣ Q ∣^{1 / 2} ∥ Λ ∥ ∥ f ∥_{L^{2}}$ for $f \in L_{0}^{2} (Q)$ : this is the mean-zero embedding $∥ f ∥_{H^{1}} \leq C_{n, \tilde{N}, φ} ∣ Q ∣^{1 / 2} ∥ f ∥_{L^{2}}$ composed with $∣ Λ (f) ∣ \leq ∥ Λ ∥ ∥ f ∥_{H^{1}}$ (Mean-zero L2 functions on a cube embed continuously into H1).

[F3]

Riesz representation is an isometry: the representing vector $u_{Q}$ of $Λ ∣_{L_{0}^{2} (Q)}$ has $∥ u_{Q} ∥_{L^{2}} = ∥ Λ ∣_{L_{0}^{2} (Q)} ∥$ , the operator norm on the subspace (Riesz representation for Hilbert spaces).

[F4]

On the positive finite-measure cube $Q$ , $∣ Q ∣^{- 1} \int_{Q} ∣ u_{Q} ∣ \leq ∣ Q ∣^{- 1 / 2} ∥ u_{Q} ∥_{L^{2}}$ (Cauchy-Schwarz inequality for $L^{2}$ ).

Proof

technique · direct

1.1F1

By [F1], $u - u_{Q}$ equals a constant $c$ almost everywhere on $Q$ , and $u_{Q}$ has mean $0$ on $Q$ ; hence ${mean}_{Q} (u) = c$ and $u - {mean}_{Q} (u) = u_{Q}$ almost everywhere on $Q$ . Therefore the mean oscillation of $u$ over $Q$ is $∣ Q ∣^{- 1} \int_{Q} ∣ u_{Q} ∣$ .

1.2F2F3

By [F2] and [F3], $∥ u_{Q} ∥_{L^{2}} \leq C_{n, \tilde{N}, φ} ∣ Q ∣^{1 / 2} ∥ Λ ∥$ .

2.1step 1.1step 1.2F4∎

Combining steps 1.1 and 1.2 with the Cauchy-Schwarz bound [F4] gives $∣ Q ∣^{- 1} \int_{Q} ∣ u - {mean}_{Q} (u) ∣ = ∣ Q ∣^{- 1} \int_{Q} ∣ u_{Q} ∣ \leq ∣ Q ∣^{- 1 / 2} ∥ u_{Q} ∥_{L^{2}} \leq C_{n, \tilde{N}, φ} ∥ Λ ∥$ for every nondegenerate cube $Q$ . Taking the supremum over $Q$ shows $u \in B M O (R^{n})$ with $∥ u ∥_{B M O} \leq C_{n, \tilde{N}, φ} ∥ Λ ∥$ .

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

BMO classes define bounded functionals on H1

Statement

Assume the Axiom of Choice, with the fixed $H^{1}$ kernel $φ$ and admissible atomic order $\tilde{N}$ of Atomic characterisation of real $H^{p}$ for $0 < p \leq 1$ . For every $b \in B M O (R^{n})$ there is a unique bounded linear functional $Λ_{b} \in (H^{1} (R^{n}))^{*}$ with $Λ_{b} (a) = \int a b$ for every $H^{1}$ atom $a$ , and $∥ Λ_{b} ∥ \leq C_{n, \tilde{N}, φ} ∥ b ∥_{B M O}$ with the constant independent of $b$ . The map $b \mapsto Λ_{b}$ is linear, annihilates constants, and therefore factors through $B M O (R^{n}) / C$ ; on every finite sum of atoms $g$ one has $Λ_{b} (g) = \int g b$ .

Facts & Assumptions

Given: The Axiom of Choice, the fixed $φ, \tilde{N}$ , a function $b \in B M O (R^{n})$ , the $(1, \infty, 0)$ -atoms of $H^{p}$ atoms with a prescribed moment order, and the space $H^{1} (R^{n})$ with its atoms.

[F1]

The atom pairing is bounded: for every atom $a$ the integral $\int a b$ converges absolutely and $∣ \int a b ∣ \leq ∥ b ∥_{B M O}$ (BMO functions pair uniformly with H1 atoms, $H^{p}$ atoms with a prescribed moment order); in particular $\int a = 0$ and atoms are bounded with compact support.

[F2]

If $b \in L^{\infty} (R^{n}) \cap B M O (R^{n})$ then for every $f \in H^{1} (R^{n})$ the integral $\int f b$ converges absolutely and $∣ \int f b ∣ \leq C_{n, \tilde{N}, φ} ∥ b ∥_{B M O} ∥ f ∥_{H^{1}}$ (Bounded BMO functions dualise H1 boundedly).

[F3]

Every $ℓ^{1}$ sum of atoms lies in $H^{1}$ : if $g = \sum_{j \leq N} λ_{j} a_{j}$ is a finite atomic sum then $∥ g ∥_{H^{1}} \leq C \sum_{j \leq N} ∣ λ_{j} ∣$ , and the finite atomic sums are dense in $H^{1}$ (Atomic characterisation of real $H^{p}$ for $0 < p \leq 1$ , Finite atomic sums are dense in H1).

[F4]

The componentwise truncations $b_{M}$ of $b$ satisfy $b_{M} \in L^{\infty}$ , $∣ b_{M} ∣ \leq ∣ b ∣$ , $b_{M} \to b$ pointwise and $∥ b_{M} ∥_{B M O} \leq \frac{9}{2} ∥ b ∥_{B M O}$ (Range truncations preserve the BMO seminorm up to a constant).

[F5]

The Axiom of Choice implies the ultrafilter lemma (The Axiom of Choice, The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter); under the ultrafilter lemma the closed dual ball of a normed space is weak-star compact (Banach–Alaoglu) and every net in a compact space has a cluster point (Assuming the ultrafilter lemma, compactness is equivalent to every net having a cluster point, every net having a convergent subnet, every filter having a cluster point, and every ultrafilter converging, Convergence and cluster points of a net in a topological space); the evaluations $Λ \mapsto Λ (a)$ are continuous for the weak-star topology (The weak-star topology from finite evaluations).

[F6]

If $b_{M} \to b$ pointwise with $∣ b_{M} ∣ \leq ∣ b ∣$ and $a$ is an atom, then $\int a b_{M} \to \int a b$ by dominated convergence, the dominating function $∣ a ∣ ∣ b ∣$ being integrable because $a$ is bounded with compact support and $b \in L_{l o c}^{1}$ (Dominated convergence, BMO seminorm and the quotient by constants).

[F7]

A locally integrable function whose regular distribution vanishes is zero almost everywhere (Locally integrable functions embed in distributions).

Proof

technique · direct

1.1F1F2F3F7

The bounded case. Let $b \in L^{\infty} \cap B M O$ and let $F$ be the linear span of the atoms, viewed as a subspace of $H^{1}$ by [F3]. Every $g \in F$ is a finite sum of bounded compactly supported atoms, hence an $L^{1}$ function with $g \in H^{1}$ , so $\int g b$ converges absolutely and $∣ \int g b ∣ \leq C_{n, \tilde{N}, φ} ∥ b ∥_{B M O} ∥ g ∥_{H^{1}}$ by [F2]. The value $\int g b$ depends only on the element $g \in H^{1}$ : if two finite sums $g, g^{'}$ represent the same element, then the locally integrable function $g - g^{'}$ has zero regular distribution, so $g = g^{'}$ almost everywhere by [F7] and the two integrals agree. Thus $g \mapsto \int g b$ is a well-defined linear functional on $F$ , bounded by $C_{n, \tilde{N}, φ} ∥ b ∥_{B M O}$ , and it extends uniquely to a bounded $Λ_{b} \in (H^{1})^{*}$ by density [F3]; the extension is the unique bounded functional whose value at every atom $a$ is $\int a b$ , since two such functionals agree on $F$ and $F$ is dense.

2.1step 1.1F4F5F6

The general case, existence of a cluster point. For general $b \in B M O$ let $b_{M}$ be the componentwise truncation of [F4]; step 1.1 gives bounded functionals $Λ_{b_{M}}$ with $∥ Λ_{b_{M}} ∥ \leq \frac{9}{2} C_{n, \tilde{N}, φ} ∥ b ∥_{B M O}$ . The Axiom of Choice yields the ultrafilter lemma [F5], so the closed ball of radius $\frac{9}{2} C_{n, \tilde{N}, φ} ∥ b ∥_{B M O}$ in $(H^{1})^{*}$ is weak-star compact [F5]; by the compactness characterization [F5] the sequence, viewed as a net, $(Λ_{b_{M}})_{M \in N, M \geq 1}$ has a weak-star cluster point $Λ$ . For every atom $a$ the evaluations converge: $Λ_{b_{M}} (a) = \int a b_{M} \to \int a b$ by [F6]. Evaluation at $a$ is weak-star continuous [F5], so $Λ (a)$ is a cluster point of the convergent net $(Λ_{b_{M}} (a))_{M}$ in $C$ and therefore equals its limit, $Λ (a) = \int a b$ .

3.1step 2.1F3

Uniqueness and the norm bound. If $Λ, Λ^{'}$ are bounded functionals with $Λ (a) = Λ^{'} (a) = \int a b$ for every atom $a$ , then by linearity they agree on the span $F$ of the atoms and hence, by density [F3] and continuity, on all of $H^{1}$ ; so the functional of step 2.1 is the unique bounded functional with the required atom values, and $∥ Λ ∥ \leq \frac{9}{2} C_{n, \tilde{N}, φ} ∥ b ∥_{B M O}$ .

4.1step 1.1step 3.1F1algebra

Linearity, constants and finite sums. For $b, b^{'} \in B M O$ and $λ \in C$ , the functionals $Λ_{b + b^{'}}$ and $Λ_{b} + Λ_{b^{'}}$ both assign to every atom $a$ the value $\int a (b + b^{'}) = \int a b + \int a b^{'}$ , so they are equal by the uniqueness of step 3.1; the same argument gives $Λ_{λ b} = λ Λ_{b}$ . If $b$ is constant almost everywhere, then $\int a b = 0$ for every atom because $\int a = 0$ by [F1], so $Λ_{b} = 0$ by uniqueness. Hence $b \mapsto Λ_{b}$ is linear with image of the constants in the zero functional, so it factors through $B M O (R^{n}) / C$ . Finally, for a finite atomic sum $g = \sum_{j \leq N} λ_{j} a_{j}$ , linearity and step 1.1 give $Λ_{b} (g) = \sum_{j \leq N} λ_{j} \int a_{j} b = \int g b$ .

5.1step 1.1step 2.1step 3.1step 4.1∎

Steps 1.1 and 2.1 construct, for every $b \in B M O (R^{n})$ , a bounded functional with the required atom values, step 3.1 gives uniqueness and the bound $∥ Λ_{b} ∥ \leq \frac{9}{2} C_{n, \tilde{N}, φ} ∥ b ∥_{B M O}$ , and step 4.1 gives linearity, the annihilation of constants, the factorisation through the quotient and the finite-sum identity. The Axiom of Choice is spent exactly at the ultrafilter lemma and the Banach-Alaoglu cluster point in step 2.1.

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

BMO classes are determined by their pairings with H1 atoms

Statement

Let $b \in B M O (R^{n})$ and suppose $\int a b = 0$ for every $H^{1}$ atom $a$ . Then $b$ is constant almost everywhere. Equivalently, the evaluation map on atoms is injective on $B M O / C$ .

Facts & Assumptions

Given: $b \in B M O (R^{n})$ with $\int a b = 0$ for every $(1, \infty, 0)$ -atom $a$ , and a cube $Q$ .

[F1]

A bounded mean-zero function $g$ vanishing off $Q$ is a scalar multiple of an atom: for $∥ g ∥_{\infty} > 0$ , divide by $∣ Q ∣ ∥ g ∥_{\infty}$ and choose the representative vanishing off the closed cube $Q$ ; a zero $L^{\infty}$ class has zero pairing ( $H^{p}$ atoms with a prescribed moment order).

[F2]

$b$ is locally integrable, $b_{Q} = ∣ Q ∣^{- 1} \int_{Q} b$ , and the BMO seminorm vanishes exactly on the almost-everywhere constants (BMO seminorm and the quotient by constants).

Proof

technique · direct

1.1F1F2givenconstruct

Define $h (x) = \overline{b (x) - b_{Q}} / ∣ b (x) - b_{Q} ∣$ on $Q$ where $b (x) \neq b_{Q}$ , and $h (x) = 0$ elsewhere. Then $∣ h ∣ \leq 1$ , and $g : = (h - h_{Q}) 1_{Q}$ is bounded, supported in $Q$ and mean zero. By [F1] and the hypothesis, $\int g b = 0$ , including the zero-class case. All products are integrable because $b \in L^{1} (Q)$ and $g$ is bounded.

2.1step 1.1F1F2algebra∎

Using $\int_{Q} (b - b_{Q}) = 0$ and the definition of $h$ , we obtain $0 = \int_{Q} g b = \int_{Q} g (b - b_{Q}) = \int_{Q} h (b - b_{Q}) = \int_{Q} ∣ b - b_{Q} ∣$ . Since $Q$ was arbitrary, the BMO seminorm is zero, so $b$ is constant almost everywhere by [F2]. Applying this to $b - b^{'}$ proves injectivity on classes with equal atom pairings; conversely constants pair to zero by atom cancellation. No choice principle or local $L^{2}$ estimate is used.

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

Real H1-BMO duality

Statement

Assume the Axiom of Choice, with the fixed $H^{1}$ kernel $φ$ and auxiliary order $\tilde{N}$ used in Mean-zero L2 functions on a cube embed continuously into H1. The map $Φ : B M O (R^{n}) / C \to (H^{1} (R^{n}))^{*}$ , $b \mapsto Λ_{b}$ of the preceding theorem is a linear bijection, and there are constants $0 < c_{n, \tilde{N}, φ} \leq C_{n, \tilde{N}, φ} < \infty$ with $c_{n, \tilde{N}, φ} ∥ b ∥_{B M O} \leq ∥ Λ_{b} ∥ \leq C_{n, \tilde{N}, φ} ∥ b ∥_{B M O}$ for every class $b$ . Thus $(H^{1} (R^{n}))^{*}$ is isomorphic to $B M O (R^{n}) / C$ with equivalent norms.

Facts & Assumptions

Given: The Axiom of Choice and a bounded functional $Λ \in (H^{1} (R^{n}))^{*}$ , with the map $Φ$ of BMO classes define bounded functionals on H1.

[F1]

The map $Φ$ is linear and bounded: for every $b \in B M O (R^{n})$ the functional $Λ_{b}$ satisfies $Λ_{b} (a) = \int a b$ on atoms and $∥ Λ_{b} ∥ \leq C_{n, \tilde{N}, φ}^{+} ∥ b ∥_{B M O}$ , and $Λ_{b} = 0$ for constant $b$ (BMO classes define bounded functionals on H1).

[F2]

The atom pairings determine the class: if $\int a b = 0$ for every atom $a$ , then $b$ is constant almost everywhere; equivalently $Φ$ is injective (BMO classes are determined by their pairings with H1 atoms).

[F3]

For every $Λ \in (H^{1})^{*}$ the preceding local representatives produce a locally integrable $u$ with $u - u_{Q}$ constant almost everywhere on every cube $Q$ , where $u_{Q} \in L_{0}^{2} (Q)$ represents $Λ ∣_{L_{0}^{2} (Q)}$ , and $∥ u ∥_{B M O} \leq C_{n, \tilde{N}, φ}^{-} ∥ Λ ∥$ (Bounded H1 functionals have compatible local L2 representatives, The dual representative has uniformly bounded BMO oscillation).

[F4]

The finite atomic sums are dense in $H^{1}$ (Finite atomic sums are dense in H1), and two bounded functionals agreeing on a dense subspace agree everywhere.

Proof

technique · direct

1.1F1F2

The map $Φ$ is linear by [F1]; it is bounded with $∥ Λ_{b} ∥ \leq C_{n, \tilde{N}, φ}^{+} ∥ b ∥_{B M O}$ by [F1]; and it is injective because a class in its kernel has vanishing pairings with all atoms and is therefore the class of the constants by [F2].

2.1step 1.1F1F3F4

Surjectivity. Let $Λ \in (H^{1})^{*}$ and let $u$ be the representative of [F3], so that $∥ u ∥_{B M O} \leq C_{n, \tilde{N}, φ}^{-} ∥ Λ ∥$ . For an atom $a$ supported in a cube $Q$ one has $a \in L_{0}^{2} (Q)$ , hence $Λ (a) = \int_{Q} u_{Q} a = \int_{Q} (u - c_{Q}) a = \int u a$ because $\int a = 0$ and $u - u_{Q} = c_{Q}$ almost everywhere on $Q$ ; meanwhile $Λ_{u} (a) = \int a u$ by the definition of $Φ$ [F1]. Thus $Λ$ and $Λ_{u}$ agree on every atom, hence on every finite atomic sum by linearity, and therefore on all of $H^{1}$ by density and continuity [F4]; that is, $Λ = Φ (u)$ and $Φ$ is surjective.

3.1step 1.1step 2.1F2

The reverse norm bound. Given a class $b$ , apply step 2.1 to $Λ_{b}$ : there is $u$ with $Λ_{b} = Φ (u)$ and $∥ u ∥_{B M O} \leq C_{n, \tilde{N}, φ}^{-} ∥ Λ_{b} ∥$ . By injectivity of $Φ$ from step 1.1 and [F2], $b - u$ is constant almost everywhere, so $∥ b ∥_{B M O} = ∥ u ∥_{B M O} \leq C_{n, \tilde{N}, φ}^{-} ∥ Λ_{b} ∥$ ; combined with step 1.1 this gives $c_{n, \tilde{N}, φ} ∥ b ∥_{B M O} \leq ∥ Λ_{b} ∥ \leq C_{n, \tilde{N}, φ} ∥ b ∥_{B M O}$ with $c_{n, \tilde{N}, φ} : = 1 / C_{n, \tilde{N}, φ}^{-}$ and $C_{n, \tilde{N}, φ} : = C_{n, \tilde{N}, φ}^{+}$ .

4.1step 1.1step 2.1step 3.1∎

Steps 1.1, 2.1 and 3.1 show that $Φ$ is a linear bijection with the two-sided norm bound, so $(H^{1} (R^{n}))^{*}$ is isomorphic to $B M O (R^{n}) / C$ with equivalent norms. The Axiom of Choice is inherited from the construction of $Λ_{b}$ and from the local representatives.

5 · Examples, counterexamples and false statements

None yet.