A normalised mean-zero $H^1$ atom

Given: Countable Choice, $n\ge 1$, the fixed admissible kernel $\phi$ and order $N$, a nondegenerate closed axis-parallel cube $Q$ with volume $\mathrm{\mid }Q\mathrm{\mid }$ and centre $c_Q$, the halving hyperplane $\{x_0=c_{Q,0}\}$, and the sets $Q^{+}=Q\cap \{x_0>c_{Q,0}\}$, $Q^{-}=Q\cap \{x_0<c_{Q,0}\}$.

A $(1,\mathrm{\infty },0)$-atom is a measurable $a$ with $\mathrm{supp}a\subseteq Q$, $\mathrm{\mid }a\mathrm{\mid }\le \mathrm{\mid }Q{\mathrm{\mid }}^{-1}$ a.e. and $\int a=0$ ($H^p$ atoms with a prescribed moment order).

Under Countable Choice, $Q^{\pm }$ are measurable axis-parallel boxes of measure $\mathrm{\mid }Q\mathrm{\mid }\mathrm{/}2$, so ${\int }_{Q^{\pm }}\mathbf{1}=\mathrm{\mid }Q\mathrm{\mid }\mathrm{/}2$; $\mathrm{\mid }Q\mathrm{\mid }>0$ (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$), Axis-parallel rectangles in ${\mathbb{R}}^m$ and their volume, 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).

For the fixed kernel $\phi$, $\mathrm{\parallel }a{\mathrm{\parallel }}_{H^p}:=\mathrm{\parallel }{M}_{\phi }^{0}a{\mathrm{\parallel }}_{L^p}$ (The real Hardy space $H^p$ defined by a radial maximal function); under Countable Choice and for an admissible order $N$, every $(p,\mathrm{\infty },s)$-atom satisfies $\mathrm{\parallel }{M}_{N}a{\mathrm{\parallel }}_{L^p}\le C_0(n,p,s)$ and $\mathrm{\parallel }a{\mathrm{\parallel }}_{H^p}\le C_1(n,p,s,N,\phi )$, uniformly in its supporting cube (Atoms have uniformly bounded $H^p$ quasi-norm and uniformly bounded test pairings).

Proof technique: direct verification of the three defining properties, then the uniform atom bound.