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A normalised cube indicator is not an $H^{1}$ atom

Statement refuted

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). The claim refuted is that the support and $L^{\infty}$ size conditions alone characterise $(1, \infty, 0)$ -atoms, in particular that the normalised cube indicator is an atom. Let $Q \subseteq R^{n}$ be a nondegenerate axis-parallel cube and put $a = ∣ Q ∣^{- 1} 1_{Q}$ . Then $supp a \subseteq Q$ and $∣ a ∣ = ∣ Q ∣^{- 1} 1_{Q} \leq ∣ Q ∣^{- 1}$ pointwise, so $a$ satisfies the support and size conditions of a $(1, \infty, 0)$ -atom ( $H^{p}$ atoms with a prescribed moment order); but $\int_{R^{n}} a = ∣ Q ∣^{- 1} ∣ Q ∣ = 1 \neq 0$ , so the zeroth-moment condition fails and $a$ is not an atom. This refutes only the atom property; it does not by itself prove $∣ Q ∣^{- 1} 1_{Q} \notin H^{1}$ , which is the separate and stronger counterexample on this page and requires the vanishing-moment corollary.

Facts & Assumptions

Given: Countable Choice and $n \geq 1$ , a nondegenerate closed axis-parallel cube $Q$ with volume $∣ Q ∣$ in the sense of Axis-parallel rectangles in $R^{m}$ and their volume, and the function $a = ∣ Q ∣^{- 1} 1_{Q}$ .

[L1]

A $(1, \infty, 0)$ -atom is a measurable $a$ with $supp a \subseteq Q$ , $∣ a ∣ \leq ∣ Q ∣^{- 1}$ almost everywhere, and $\int_{R^{n}} a = 0$ ( $H^{p}$ atoms with a prescribed moment order).

[F1]

$∣ Q ∣ > 0$ and $\int_{R^{n}} 1_{Q} = ∣ Q ∣$ (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included, Axis-parallel rectangles in $R^{m}$ and their volume); $L^{1}$ is the quotient by almost-everywhere null functions (The space $L^{p} (μ)$ as the quotient by null functions).

Counterexample

The witness is the pair $(Q, a)$ with $a = ∣ Q ∣^{- 1} 1_{Q}$ .

1.1L1given

The support and size conditions hold. Since $1_{Q}$ vanishes off $Q$ , $supp a \subseteq Q$ , and $∣ a ∣ = ∣ Q ∣^{- 1} 1_{Q} \leq ∣ Q ∣^{- 1}$ pointwise.

1.2L1F1algebra

The moment condition fails. By [F1], $\int_{R^{n}} a = ∣ Q ∣^{- 1} \int 1_{Q} = ∣ Q ∣^{- 1} ∣ Q ∣ = 1$ , which is nonzero because $∣ Q ∣ > 0$ . Hence the zeroth-moment requirement $\int a = 0$ of [L1] fails, and $a$ is not a $(1, \infty, 0)$ -atom.

2.1step 1.1step 1.2∎

Conclusion. The function $a$ meets the support and size parts of the atom definition but not the cancellation part; therefore the support and size conditions alone do not suffice for the atom property, and the moment condition is a genuine part of $H^{p}$ atoms with a prescribed moment order.

Depends on

Used by

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Dependency tree · two levels

36 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

Stefano Meda, Peter Sjogren, Maria Vallarino, Atomic decompositions and operators on Hardy spaces, Revista de la Union Matematica Argentina 50 (2009), no. 2, 15-22 (standard reference, not scraped)
Li-An Daniel Wang, Multiplier Theorems on Anisotropic Hardy Spaces (PhD dissertation, University of Oregon, 2012) (standard reference, not scraped)

A normalised cube indicator is not an H1H1 atom