$H^p$ atoms with a prescribed moment order

Let $n\ge 1$, let $0<p\le 1$ and let $s\in \mathbb{N}\cup \{0\}$ satisfy $s\ge \lfloor n(1\mathrm{/}p-1)\rfloor$. A $(p,\mathrm{\infty },s)$-atom is a measurable function $a\text{\hspace{0.17em}⁣}:{\mathbb{R}}^n\to \mathbb{C}$ for which there is a nondegenerate axis-parallel cube $Q$ (Axis-parallel rectangles in ${\mathbb{R}}^m$ and their volume, so all $n$ side lengths are equal and positive) such that

1. $\mathrm{supp}a\subseteq Q$, where the support is the closure of $\{a\mathrm{\ne }0\}$;
2. $\mathrm{\mid }a(x)\mathrm{\mid }\le \mathrm{\mid }Q{\mathrm{\mid }}^{-1\mathrm{/}p}$ for almost every $x\in {\mathbb{R}}^n$;
3. ${\displaystyle {\int }_{{\mathbb{R}}^n}a(x)x^{\alpha }dx=0}$ for every multi-index $\alpha$ with $\mathrm{\mid }\alpha \mathrm{\mid }\le s$ ($C^k$ maps and multi-index derivative notation in Euclidean space).

The exponent $1\mathrm{/}p\ge 1$ in the size bound and the order $s$ are part of the datum, not free parameters of the function: a $(p,\mathrm{\infty },s)$-atom is also a $(p,\mathrm{\infty },s^{\mathrm{\prime }})$-atom for every $\lfloor n(1\mathrm{/}p-1)\rfloor \le s^{\mathrm{\prime }}\le s$, because the moment conditions for the smaller order are among those already imposed. The zero function satisfies all three conditions; zero terms may be omitted from atomic representations.

Under Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)), every moment in condition 3 is an absolutely convergent Lebesgue integral; the supporting cube has its finite volume by A box in ${\mathbb{R}}^n$ with parameters $a_i\le b_i$ is Lebesgue measurable of measure ${\prod }_{i<n}(b_i-a_i)$, whichever of its faces are included. Indeed $a$ vanishes a.e. off the bounded set $Q$ and $\mathrm{\mid }a{x}^{\alpha }\mathrm{\mid }\le \mathrm{\mid }Q{\mathrm{\mid }}^{-1\mathrm{/}p}{\sup }_{x\in Q}\mathrm{\mid }{x}^{\alpha }\mathrm{\mid }$ a.e. on $Q$, a bounded function on a set of finite measure; the monomial $x^{\alpha }$ is continuous and hence Borel measurable by Continuous functions on Euclidean spaces are Borel measurable. Applying Arithmetic and lattice operations preserve measurability whenever they are defined to the real and imaginary parts of $a$ shows that $a{x}^{\alpha }$ is measurable, so the integral is defined and finite. The a.e. bound in condition 2 is an essential supremum bound, $\mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\mid }Q{\mathrm{\mid }}^{-1\mathrm{/}p}$ (The essential supremum of a measurable function with respect to a measure); replacing $a$ by another representative of its a.e. class preserves conditions 2 and 3 but can change the support in condition 1, so the support condition is imposed for the chosen representative.

The page fixes the order $s_p=\lfloor n(1\mathrm{/}p-1)\rfloor \mathrm{.}$ This is the least integer compatible with condition 3, and it is the order used by the sources: DKKP require moments through $n(p^{-1}-1)$, Wang and Hiserote through $\lfloor n(1\mathrm{/}p-1)\rfloor$. The threshold moves exactly at the integers: $s_p=0$ for $n\mathrm{/}(n+1)<p\le 1$, $s_p=1$ for $n\mathrm{/}(n+2)<p\le n\mathrm{/}(n+1)$, and so on. For $p=1$ the definition specialises to the classical $L^{\mathrm{\infty }}$ atoms of $H^1$: support in a cube, $\mathrm{\mid }a\mathrm{\mid }\le \mathrm{\mid }Q{\mathrm{\mid }}^{-1}$ a.e. and $\int a=0$ (Conjugate exponents, including the endpoint conventions records the exponent convention used for the dual exponents invoked later on this page).

The ball-supported atoms of the sources differ from this convention only by fixed constants: a cube $Q$ containing a ball $B$ with $\mathrm{\mid }B\mathrm{\mid }\le \mathrm{\mid }Q\mathrm{\mid }\le c_n\mathrm{\mid }B\mathrm{\mid }$ carries the same conditions up to the dimensional factor $c_n^{1\mathrm{/}p}$, and nondegeneracy rules out the degenerate cubes of zero volume that occur in moment conditions. The three conditions define the class without selecting representatives; the accompanying Lebesgue-integrability assertions use Countable Choice.