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A compactly supported $L^{1}$ function of nonzero integral is not in $H^{1}$

Statement refuted

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). The claim refuted is that the size and compact support of an $L^{1}$ function suffice for membership in $H^{1}$ . Let $f \in L^{1} (R^{n})$ be compactly supported with $\int_{R^{n}} f \neq 0$ ; for instance $f = 1_{Q}$ for a nondegenerate cube $Q$ . Then $f \notin H^{1} (R^{n})$ . Consequently $H^{1} (R^{n}) ⊊ L^{1} (R^{n})$ : the atomic characterisation gives the inclusion, and $1_{Q}$ is in $L^{1}$ but outside $H^{1}$ . In particular no compactly supported integrable function of nonzero integral is an $H^{1}$ atom or a finite sum of atoms.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , a compactly supported $f \in L^{1} (R^{n})$ with $\int_{R^{n}} f \neq 0$ , and the space $H^{1}$ of The real Hardy space $H^{p}$ defined by a radial maximal function.

[A1]

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

[L1]

Vanishing moments: if $g \in H^{1}$ is represented by a locally integrable function with $g \in L^{1}$ , then $\int_{R^{n}} g = 0$ (Weighted-integrable $H^{p}$ functions have vanishing moments in the atomic range with $p = 1$ , $s = 0$ ).

[F1]

For a nondegenerate cube $Q$ , $1_{Q} \in L_{c}^{1}$ with $\int 1_{Q} = ∣ Q ∣ > 0$ (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]

Every $g \in H^{1}$ has an atomic representation $g = \sum_{j} λ_{j} a_{j}$ in $S^{'}$ with $\sum_{j} ∣ λ_{j} ∣ < \infty$ ; the $(1, \infty, 0)$ -atoms obey $∥ a_{j} ∥_{1} \leq 1$ (Atomic characterisation of real $H^{p}$ for $0 < p \leq 1$ , $H^{p}$ atoms with a prescribed moment order). Complex $L^{1}$ is complete under Countable Choice (Complex Lp completeness and almost-everywhere subsequences).

[F3]

The integral pairing satisfies $∣ \int u v ∣ \leq ∥ u ∥_{1} ∥ v ∥_{\infty}$ for $u \in L^{1}$ and $v \in L^{\infty}$ (Holder's inequality for integrals, including the endpoint cases).

The witness is a compactly supported $f \in L^{1}$ with $\int f \neq 0$ , for instance $f = 1_{Q}$ .

Counterexample

technique · direct

1.1F1given

The witness is admissible. For $f = 1_{Q}$ with $Q$ nondegenerate, $f$ is compactly supported and integrable, and $\int f = ∣ Q ∣ > 0$ by [F1]; more generally the assumed $f$ is itself compactly supported in $L^{1}$ with nonzero integral.

1.2A1F2F3algebra

Every $H^{1}$ element has an $L^{1}$ representative. For $g \in H^{1}$ , take the representation of [F2]. Its partial sums $S_{N} = \sum_{j \leq N} λ_{j} a_{j}$ are Cauchy in $L^{1}$ , since $∥ S_{M} - S_{N} ∥_{1} \leq \sum_{N < j \leq M} ∣ λ_{j} ∣$ . By completeness they converge to $h \in L^{1}$ . For every Schwartz test $ψ$ , [F3] gives $∣ \int (S_{N} - h) ψ ∣ \leq ∥ S_{N} - h ∥_{1} ∥ ψ ∥_{\infty} \to 0$ , whereas the atomic series converges to $g$ in $S^{'}$ . Thus $g$ is the regular distribution of $h$ , proving $H^{1} \subseteq L^{1}$ .

2.1L1step 1.1step 1.2given

Nonzero integral excludes $H^{1}$ . Suppose $f \in H^{1}$ . Since $f$ is (represented by) an $L^{1}$ function, [L1] forces $\int f = 0$ , contradicting the hypothesis $\int f \neq 0$ . Hence $f \notin H^{1}$ ; specializing to $1_{Q}$ gives $1_{Q} \notin H^{1}$ with $1_{Q} \in L^{1}$ , so $H^{1} ⊊ L^{1}$ .

3.1L1step 2.1

Consequences for atoms. Every $(1, \infty, 0)$ -atom has integral zero by definition, so a compactly supported integrable function with nonzero integral is not an atom; and since finite sums of atoms have zero integral as well, such a function is not a finite sum of atoms either, in accordance with its exclusion from $H^{1}$ .

4.1step 1.1step 1.2step 2.1step 3.1∎

Conclusion. The compactly supported $L^{1}$ function of nonzero integral is a witness that $L^{1} ⊈ H^{1}$ ; together with step 1.2 this proves $H^{1} ⊊ L^{1}$ , and refutes the claimed sufficiency of compact support and integrability.

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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)

A compactly supported L1L1 function of nonzero integral is not in H1H1