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An $H^{1}$ atom need not be smooth or continuous

Statement refuted

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). The claim refuted is that every $(1, \infty, 0)$ -atom is continuous (or smooth). Let $Q \subseteq R^{n}$ be a nondegenerate closed axis-parallel cube and let $Q^{+}, Q^{-}$ be the two halves of $Q$ cut by a coordinate hyperplane through the centre $c_{Q}$ of $Q$ ; write $a = ∣ Q ∣^{- 1} (1_{Q^{+}} - 1_{Q^{-}}) .$ Then $a$ is a $(1, \infty, 0)$ -atom ( $H^{p}$ atoms with a prescribed moment order) but is discontinuous at every point of the relative interior of the cutting slice $Q \cap H$ , where $H$ is that hyperplane, so no continuity or smoothness may be assumed of a general atom.

Facts & Assumptions

Given: Countable Choice and $n \geq 1$ , a nondegenerate closed axis-parallel cube $Q$ with centre $c_{Q}$ and volume $∣ Q ∣$ , the halving hyperplane $H = {x : x_{0} = c_{Q, 0}}$ through the centre, and the sets $Q^{+} = Q \cap {x_{0} > c_{Q, 0}}$ , $Q^{-} = Q \cap {x_{0} < c_{Q, 0}}$ .

[L1]

A $(1, \infty, 0)$ -atom is a measurable $a$ with $supp a \subseteq Q$ , $∣ a ∣ \leq ∣ Q ∣^{- 1}$ a.e. and $\int a = 0$ ; no regularity is required ( $H^{p}$ atoms with a prescribed moment order).

[F1]

$Q^{\pm}$ are axis-parallel boxes each of volume $∣ Q ∣ / 2$ , and $Q$ is the disjoint union of $Q^{+}$ , $Q^{-}$ and $Q \cap H$ up to a Lebesgue-null set; consequently $\int_{Q^{\pm}} 1 = ∣ Q ∣ / 2$ (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).

[F2]

A function $g : R^{n} \to C$ is continuous at $z$ if and only if for every neighbourhood $V$ of $g (z)$ there is a neighbourhood $U$ of $z$ with $g [U] \subseteq V$ (Continuity of a map of topological spaces at a point and globally).

The witness is $a = ∣ Q ∣^{- 1} (1_{Q^{+}} - 1_{Q^{-}})$ .

Counterexample

technique · direct

1.1L1F1algebra

$a$ is a $(1, \infty, 0)$ -atom. Clearly $supp a \subseteq Q$ and $∣ a ∣ \leq ∣ Q ∣^{- 1}$ almost everywhere. For the zeroth moment, [F1] gives $\int a = ∣ Q ∣^{- 1} (∣ Q ∣ / 2 - ∣ Q ∣ / 2) = 0$ . Hence $a$ satisfies [L1] and is a $(1, \infty, 0)$ -atom.

2.1step 1.1F2givenalgebra

$a$ is discontinuous across the cutting hyperplane. Fix $z \in H \cap int Q$ and let $r > 0$ be smaller than the distance from $z$ to $\partial Q$ ; then $z + s e_{0} \in Q^{+}$ and $z - s e_{0} \in Q^{-}$ for every $0 < s < r$ , and $a (z + s e_{0}) = ∣ Q ∣^{- 1}$ , $a (z - s e_{0}) = - ∣ Q ∣^{- 1}$ . Both sequences tend to $z$ , so continuity of $a$ at $z$ would force the two values to be equal: by [F2] applied to the neighbourhood $V = {w : ∣ w - a (z) ∣ < ∣ Q ∣^{- 1} / 2}$ of $a (z)$ , every point of a sufficiently small neighbourhood $U$ of $z$ would satisfy $a (z + s e_{0}) \in V$ and $a (z - s e_{0}) \in V$ , which is impossible because the two values differ by $2 ∣ Q ∣^{- 1} > ∣ Q ∣^{- 1}$ . Hence $a$ is discontinuous at every $z \in H \cap int Q$ , which is the relative interior of the cutting slice, and in particular is not continuous, hence not smooth.

3.1step 1.1step 2.1∎

Conclusion. Step 1.1 exhibits a $(1, \infty, 0)$ -atom and step 2.1 shows that it has a jump discontinuity on the cutting hyperplane; therefore the atomic size and cancellation conditions do not imply continuity or smoothness, and no regularity of atoms may be assumed in the atomic characterisation. In particular a proof producing $L^{\infty}$ atoms with jumps is not deficient on that account.

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Sources

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

An H1H1 atom need not be smooth or continuous