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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.

Depends on

Used by

Dependency tree · two levels

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

Sources

Mark Williams, Notes on Harmonic Analysis (January 11, 2022) (standard reference, not scraped)