Atoms have uniformly bounded $H^p$ quasi-norm and uniformly bounded test pairings

Near estimate. For $x\in {\mathbb{R}}^n$, $\mathrm{\Psi }\in {\mathcal{F}}_N$, $t>0$, and every convolution centre $y$ with $\mathrm{\mid }y-x\mathrm{\mid }\le t$, the atom bound gives $\mathrm{\mid }(a\ast {\mathrm{\Psi }}_t)(y)\mathrm{\mid }\le \mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_Q\mathrm{\mid }{\mathrm{\Psi }}_t(y-z)\mathrm{\mid }dz\le \mathrm{\mid }Q{\mathrm{\mid }}^{-1\mathrm{/}p}\mathrm{\parallel }{\mathrm{\Psi }}_t{\mathrm{\parallel }}_1=\mathrm{\mid }Q{\mathrm{\mid }}^{-1\mathrm{/}p}\mathrm{\parallel }\mathrm{\Psi }{\mathrm{\parallel }}_1\le C\mathrm{\mid }Q{\mathrm{\mid }}^{-1\mathrm{/}p}$. The last constant is uniform over $\mathrm{\Psi }\in {\mathcal{F}}_N$ by [F2] and $N\ge n+1$. By [F8], $\mathrm{\parallel }{\mathrm{\Psi }}_t{\mathrm{\parallel }}_1=\mathrm{\parallel }\mathrm{\Psi }{\mathrm{\parallel }}_1$. Taking the suprema over $\mathrm{\Psi }$, $t$, and all $y$ with $\mathrm{\mid }y-x\mathrm{\mid }\le t$ proves this bound for every $x$. Let $Q^{\ast }$ be the concentric closed cube of side $4\sqrt{n}\mathrm{\ell }$. By [F5], $\mathrm{\mid }{Q}^{\ast }\mathrm{\mid }=(4\sqrt{n}{)}^n\mathrm{\mid }Q\mathrm{\mid }$, and hence ${\int }_{Q^{\ast }}(M_{N}a{)}^p\le C^p\mathrm{\mid }Q{\mathrm{\mid }}^{-1}\mathrm{\mid }{Q}^{\ast }\mathrm{\mid }\le C^p(4\sqrt{n}{)}^n\mathrm{.}$

Far estimate. Set $m:=n+s+1$. If $x\mathrm{\notin }{Q}^{\ast }$, put $r:=\mathrm{\mid }x-c_Q\mathrm{\mid }>2\sqrt{n}\mathrm{\ell }$. Fix $\mathrm{\Psi }\in {\mathcal{F}}_N$, $t>0$, and any $y$ with $\mathrm{\mid }y-x\mathrm{\mid }\le t$; all estimates below are uniform in this $y$, so taking the suprema over $y,t,\mathrm{\Psi }$ at the end gives the estimate for $M_{N}a(x)$. For $z\in Q$, $\mathrm{\mid }z-c_Q\mathrm{\mid }\le \sqrt{n}\mathrm{\ell }\mathrm{/}2$. When $0<t\le \mathrm{\ell }$, the triangle inequality gives $\mathrm{\mid }y-z\mathrm{\mid }\ge r-t-\sqrt{n}\mathrm{\ell }\mathrm{/}2\ge r\mathrm{/}4$. Thus [F2] and $N\ge m$ imply $\mathrm{\mid }(a\ast {\mathrm{\Psi }}_t)(y)\mathrm{\mid }\le \mathrm{\mid }Q{\mathrm{\mid }}^{-1\mathrm{/}p}\mathrm{\mid }Q\mathrm{\mid }{t}^{-n}(1+r\mathrm{/}(4t){)}^{-m}\le C\mathrm{\mid }Q{\mathrm{\mid }}^{-1\mathrm{/}p}(\mathrm{\ell }\mathrm{/}r{)}^m(t\mathrm{/}\mathrm{\ell }{)}^{s+1}\le C\mathrm{\mid }Q{\mathrm{\mid }}^{-1\mathrm{/}p}(\mathrm{\ell }\mathrm{/}r{)}^m$, using $\mathrm{\mid }Q\mathrm{\mid }={\mathrm{\ell }}^n$ and $m-n=s+1$. When $t\ge \mathrm{\ell }$, expand the complex function $z\mapsto {\mathrm{\Psi }}_t(y-z)$ about $c_Q$ through degree $s$, applying [F4] to its real and imaginary parts. The Taylor polynomial integrates to zero against $a$ because each $(z-c_Q{)}^{\alpha }$, $\mathrm{\mid }\alpha \mathrm{\mid }\le s$, is a linear combination of monomials $z^{\beta }$ of degree at most $s$, whose moments vanish by [L1]. For every remainder point $\zeta$ on the segment from $c_Q$ to $z\in Q$, the cone condition and $t\ge \mathrm{\ell }$ give $r\le \mathrm{\mid }x-y\mathrm{\mid }+\mathrm{\mid }y-\zeta \mathrm{\mid }+\mathrm{\mid }\zeta -c_Q\mathrm{\mid }\le t+\mathrm{\mid }y-\zeta \mathrm{\mid }+\frac{\sqrt{n}}{2}\mathrm{\ell }\le (1+\sqrt{n}\mathrm{/}2)(t+\mathrm{\mid }y-\zeta \mathrm{\mid })$. If $\mathrm{\mid }\beta \mathrm{\mid }=s+1$, [F2] yields $\mathrm{\mid }{\mathrm{\partial }}^{\beta }{\mathrm{\Psi }}_t(y-\zeta )\mathrm{\mid }\le t^{-n-s-1}(1+\mathrm{\mid }y-\zeta \mathrm{\mid }\mathrm{/}t{)}^{-m}=(t+\mathrm{\mid }y-\zeta \mathrm{\mid }{)}^{-m}\le (1+\sqrt{n}\mathrm{/}2{)}^m{r}^{-m}$. The Taylor remainder and $\mathrm{\mid }z-c_Q\mathrm{\mid }\le \sqrt{n}\mathrm{\ell }\mathrm{/}2$ now give $\mathrm{\mid }(a\ast {\mathrm{\Psi }}_t)(y)\mathrm{\mid }\le C\mathrm{\mid }Q{\mathrm{\mid }}^{1-1\mathrm{/}p}{\mathrm{\ell }}^{s+1}{r}^{-m}=C\mathrm{\mid }Q{\mathrm{\mid }}^{-1\mathrm{/}p}(\mathrm{\ell }\mathrm{/}r{)}^m$. These bounds hold for every $\mathrm{\Psi },t,y$ in the grand-maximal supremum. Therefore $M_{N}a(x)\le C\mathrm{\mid }Q{\mathrm{\mid }}^{-1\mathrm{/}p}(1+r\mathrm{/}\mathrm{\ell }{)}^{-m}$. Since $s\ge \lfloor n(1\mathrm{/}p-1)\rfloor$, $pm>n$. Cover the far region by shells $E_j=\{x:2^j{R}_0\le \mathrm{\mid }x-c_Q\mathrm{\mid }<2^{j+1}{R}_0\}$, $j\ge 0$, with $R_0=2\sqrt{n}\mathrm{\ell }$. Each is contained in a concentric closed cube of side $2^{j+2}{R}_0$, so [F5] gives $\mathrm{\mid }{E}_j\mathrm{\mid }\le C_n{2}^{jn}\mathrm{\mid }Q\mathrm{\mid }$. By [F7] and the pointwise bound, ${\int }_{{\mathbb{R}}^n\setminus Q^{\ast }}(M_{N}a{)}^p\le {\sum }_{j\ge 0}C\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{2}^{-jmp}\mathrm{\mid }{E}_j\mathrm{\mid }\le C{\sum }_{j\ge 0}{2}^{-j(pm-n)}<\mathrm{\infty }\mathrm{.}$

Conclusion of (a). Steps 1.1 and 2.1 give ${\int }_{{\mathbb{R}}^n}(M_{N}a{)}^p\le C_0^p$ after enlarging $C_0=C_0(n,p,s)$, independently of $Q$, $a$ and admissible $N$; measurability is [F6]. Hence $\mathrm{\parallel }{M}_{N}a{\mathrm{\parallel }}_{L^p}\le C_0$, and [F3] gives $\mathrm{\parallel }a{\mathrm{\parallel }}_{H^p}\le 2^N{P}_N(\phi )C_0=:C_1(n,p,s,N,\phi )$. This proves assertion 1.

Pairing bounds. The atom function induces a tempered distribution by $\mathrm{\mid }{\int }_{Q}a(y)\psi (y)dy\mathrm{\mid }\le \mathrm{\mid }Q{\mathrm{\mid }}^{1-1\mathrm{/}p}{p}_{00}(\psi )$, and [F9] fixes the bilinear convention. Thus for a complex test $\psi$, $⟨a,\psi ⟩={\int }_{Q}a(y)\psi (y)dy$; this integral is absolutely convergent by [L1] and boundedness of $\psi$ on $Q$. The plain estimate is $\mathrm{\mid }⟨a,\psi ⟩\mathrm{\mid }\le \mathrm{\mid }Q{\mathrm{\mid }}^{1-1\mathrm{/}p}\mathrm{\parallel }\psi {\mathrm{\parallel }}_{L^{\mathrm{\infty }}(Q)}$. For the Taylor estimate, apply [F4] separately to $\mathrm{Re}\psi$ and $\mathrm{Im}\psi$ and use the same moment cancellation as in step 2.1. The combined remainder obeys ${\sup }_Q\mathrm{\mid }{R}_{\psi }\mathrm{\mid }\le C_{n,s}{\mathrm{\ell }}^{s+1}\mathrm{\parallel }\psi {\mathrm{\parallel }}_{C^{s+1}(Q)}$, hence $\mathrm{\mid }⟨a,\psi ⟩\mathrm{\mid }=\mathrm{\mid }{\int }_{Q}a(y)R_{\psi }(y)dy\mathrm{\mid }\le C_{n,s}\mathrm{\mid }Q{\mathrm{\mid }}^{1-1\mathrm{/}p+(s+1)\mathrm{/}n}\mathrm{\parallel }\psi {\mathrm{\parallel }}_{C^{s+1}(Q)}\mathrm{.}$ Taking the smaller of the plain and Taylor bounds proves assertion 2 after enlarging $C_0$. Put $X=n(1\mathrm{/}p-1)\ge 0$: then $1-1\mathrm{/}p+(s+1)\mathrm{/}n=(s+1-X)\mathrm{/}n>0$, whereas $1-1\mathrm{/}p\le 0$ (including equality when $p=1$). Thus the two powers have one positive and one nonpositive exponent, and $\min (\mathrm{\mid }Q{\mathrm{\mid }}^{1-1\mathrm{/}p+(s+1)\mathrm{/}n},\mathrm{\mid }Q{\mathrm{\mid }}^{1-1\mathrm{/}p})\le 1$ for all $\mathrm{\mid }Q\mathrm{\mid }>0$. Since a Schwartz test and its derivatives through order $s+1$ are bounded globally, assertion 3 follows uniformly over every family of atoms.

Conclusion. Steps 1.1 and 2.1 give the uniform grand-maximal estimate, [F3] gives the kernel/order-dependent $H^p$ bound, and step 4.1 proves the uniform pairing estimates. Countable Choice is used for the explicit box measures and maximal-function measurability in [F5]--[F6]. This proves the lemma.