Level decomposition of an $H^p$ distribution produces atoms

The level sets and scale shells. If $f=0$, the empty atomic family gives the conclusion, so assume $f\mathrm{\ne }0$. Put $M=M_{N}f$ and ${\mathrm{\Omega }}_r=\{M>2^r\}$. By [F1], [F2], and layer cake (summing nonnegative indicators using Monotone convergence for the integral), ${\sum }_{r\in \mathbb{Z}}{2}^{pr}\mathrm{\mid }{\mathrm{\Omega }}_r\mathrm{\mid }\le C_p\mathrm{\parallel }M{\mathrm{\parallel }}_p^p\le C\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}^p,$ so every ${\mathrm{\Omega }}_r$ has finite measure. For $k\in \mathbb{Z}$ set $a_k=2^{-k+1}$, $U_k^r=\{x\in {\mathrm{\Omega }}_r:d_r(x)>a_k\},V_k^r=\{x\in {\mathrm{\Omega }}_{r+1}:d_{r+1}(x)>a_k\},E_r^k=U_k^r\setminus V_k^r,$ where $d_r(x)=\mathrm{dist}(x,{\mathrm{\Omega }}_r^c)$. The sets $E_r^k$ are disjoint in $r$. At a fixed $k$, they cover every $y$ with $0<M(y)<\mathrm{\infty }$ for which ${\tilde{\psi }}_k\ast f(y)\mathrm{\ne }0$: indeed, with $\eta =4^n\tilde{\psi }(4\cdot )$ and $t_k=2^{-k+2}$ one has ${\eta }_{t_k}={\tilde{\psi }}_k$. After division by $P_N(\eta )$ this is a test in ${\mathcal{F}}_N$, so a nonzero value at $y$ gives a positive lower bound for $M(z)$ for all $\mathrm{\mid }z-y\mathrm{\mid }\le t_k$. Choosing $r$ below that bound puts $B(y,t_k)\subset {\mathrm{\Omega }}_r$, hence $y\in U_k^r$. Since $M\in L^p$ is finite a.e., for a.e. such $y$ it is outside ${\mathrm{\Omega }}_r$ for all sufficiently large $r$; the nested sets $U_k^r$ therefore give a unique last index $r$ with $y\in E_r^k$. Points where $M=\mathrm{\infty }$ form a null set, and where $M=0$ the displayed convolution is zero by the grand-maximal definition. Thus the $E_r^k$ partition the integrand ${\tilde{\psi }}_k\ast f$ up to a null set, which is all the integral and distributional identities below require. Finally, if ${\mathrm{\Omega }}_r\mathrm{\ne }\mathrm{\varnothing }$, choose $s_r$ so negative that ${\omega }_n{a}_k^n>\mathrm{\mid }{\mathrm{\Omega }}_r\mathrm{\mid }$ for $k<s_r$. Then $U_k^r=E_r^k=\mathrm{\varnothing }$ for $k<s_r$, since each point of $U_k^r$ would force a ball of radius $a_k$ inside ${\mathrm{\Omega }}_r$. If ${\mathrm{\Omega }}_r=\mathrm{\varnothing }$, set all its pieces to zero.

Local bounds and the two-boundary scale split. For every measurable $A\subseteq E_r^k$, $$\begin{array}{cccc}& {\parallel {\int }_A{\psi }_k(\cdot -y)({\tilde{\psi }}_k\ast f)(y)dy\parallel }_{\mathrm{\infty }}\le C{2}^r,{\parallel {\int }_A{\mathrm{\Phi }}_m(\cdot -y)({\mathrm{\Phi }}_m\ast f)(y)dy\parallel }_{\mathrm{\infty }}\le C{2}^r(m=k,k+1)\mathrm{.}& & \text{(1)}\end{array}$$ To see this, if $y\in E_r^k$ is outside ${\mathrm{\Omega }}_{r+1}$, use $z=y$. Otherwise $d_{r+1}(y)\le a_k<d_r(y)$; choose $z\in {\mathrm{\Omega }}_{r+1}^c$ with $\mathrm{\mid }z-y\mathrm{\mid }<d_{r+1}(y)+ϵ<d_r(y)$ and $\mathrm{\mid }z-y\mathrm{\mid }<2a_k$. Then $z\in {\mathrm{\Omega }}_r\setminus {\mathrm{\Omega }}_{r+1}$ and $M(z)\le 2^{r+1}$. Each inner kernel here is exactly a grand-maximal test dilation at cone scale $2a_k$: use $4^n\tilde{\psi }(4\cdot )$ for ${\tilde{\psi }}_k$, $4^n\mathrm{\Phi }(4\cdot )$ for ${\mathrm{\Phi }}_k$, and $8^n\mathrm{\Phi }(8\cdot )$ for ${\mathrm{\Phi }}_{k+1}$. The cone definition of $M_N$ therefore gives $\mathrm{\mid }{\tilde{\psi }}_k\ast f(y)\mathrm{\mid }+\mathrm{\mid }{\mathrm{\Phi }}_m\ast f(y)\mathrm{\mid }\le C{2}^r$. Multiplication by the fixed $L^1$ norms of the outer kernels ${\psi }_k$ and ${\mathrm{\Phi }}_m$ proves (1). For $s_r\le q\le m$ write $F_{r,q,m}(x)={\sum }_{k=q}^m{\int }_{E_r^k}{\psi }_k(x-y)({\tilde{\psi }}_k\ast f)(y)dy\mathrm{.}$ Its $L^{\mathrm{\infty }}$ norm is at most $C{2}^r$, uniformly in $r,q,m$. Here are the scale details. If a summand can be nonzero at $x$, then $x\in {\mathrm{\Omega }}_r$. Choose $\mathrm{\ell }$ with $x\in U_{\mathrm{\ell }+1}^r\setminus U_{\mathrm{\ell }}^r$, so $2^{-\mathrm{\ell }}<d_r(x)\le 2^{-\mathrm{\ell }+1}$. If also $x\in {\mathrm{\Omega }}_{r+1}$, choose $\nu$ with $x\in V_{\nu +1}^r\setminus V_{\nu }^r$; since $d_r(x)\ge d_{r+1}(x)$, $\nu \ge \mathrm{\ell }$. For $\nu \ge \mathrm{\ell }+3$, the Lipschitz property of distance gives $B(x,2^{-k})\cap E_r^k=\mathrm{\varnothing }(k\le \mathrm{\ell }-1\text{ or }k\ge \nu +2),B(x,2^{-k})\subset E_r^k(\mathrm{\ell }+2\le k\le \nu -1)\mathrm{.}$ Thus, in any finite interval $[q,m]$, at most four non-full boundary terms remain; the consecutive full terms $k=u,\dots ,v$ telescope to ${\sum }_{k=u}^v{\psi }_k\ast {\tilde{\psi }}_k\ast f(x)={\mathrm{\Phi }}_{v+1}\ast {\mathrm{\Phi }}_{v+1}\ast f(x)-{\mathrm{\Phi }}_u\ast {\mathrm{\Phi }}_u\ast f(x)\mathrm{.}$ The support balls of the two endpoint convolutions lie in $E_r^v$ and $E_r^u$, respectively, so (1) bounds both endpoints by $C{2}^r$. If $\mathrm{\ell }\le \nu \le \mathrm{\ell }+2$, only the at most four indices $\mathrm{\ell },\dots ,\nu +1$ can contribute, and (1) applies directly. If $x\in {\mathrm{\Omega }}_r\setminus {\mathrm{\Omega }}_{r+1}$, choose $\mathrm{\ell }$ as above. There is no interaction for $k\le \mathrm{\ell }-1$, while $B(x,2^{-k})\subset E_r^k$ for every $k\ge \mathrm{\ell }+2$. The latter tail in any finite interval telescopes, with both endpoint convolutions localized in the corresponding $E_r^k$ and bounded by (1); only the two scales $\mathrm{\ell },\mathrm{\ell }+1$ are left. These cases prove the uniform bound for every finite partial sum. For $\chi \in \mathcal{S}$, the moment estimate [F7] and [F4] give, for $k\ge 0$, ${\sum }_r\mathrm{\mid }⟨g_{rk},\chi ⟩\mathrm{\mid }\le C_{\chi }{2}^{-k(K+1-n\mathrm{/}p)}\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p},g_{rk}(x)={\int }_{E_r^k}{\psi }_k(x-y)({\tilde{\psi }}_k\ast f)(y)dy\mathrm{.}$ For $k<0$, the same sum is at most $C_{\chi }{2}^{kn\mathrm{/}p}\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}$, using $\mathrm{\parallel }{\psi }_k{\mathrm{\parallel }}_1=\mathrm{\parallel }\psi {\mathrm{\parallel }}_1$ and $\mathrm{\parallel }{\tilde{\psi }}_k\ast f{\mathrm{\parallel }}_{\mathrm{\infty }}\le C{2}^{kn\mathrm{/}p}\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}$. The first bound is summable because $K\ge n\mathrm{/}p$, and for each fixed $r$ only finitely many negative $k$ occur because $k\ge s_r$. Thus $F_r={\sum }_{k\ge s_r}{g}_{rk}$ converges in ${\mathcal{S}}^{\mathrm{\prime }}$. The uniform bounds for finite partial sums imply $\mathrm{\mid }⟨F_r,\chi ⟩\mathrm{\mid }\le C{2}^r\mathrm{\parallel }\chi {\mathrm{\parallel }}_1$; by the bounded-distribution representation [F9], $F_r$ is represented by an $L^{\mathrm{\infty }}$ function with $\mathrm{\parallel }{F}_r{\mathrm{\parallel }}_{\mathrm{\infty }}\le C{2}^r$. No pointwise convergence of the infinite scale series is needed.

Whitney localization, overlap, and atoms. Fix a nonempty ${\mathrm{\Omega }}_r$ and take the cover [F5], writing $B_j=B({\xi }_j,{\rho }_j\mathrm{/}2)$ and $W_j=B({\xi }_j,3{\rho }_j\mathrm{/}4)$. Then $B_j\subset W_j$, so the $B_j$ cover ${\mathrm{\Omega }}_r$ with multiplicity at most the exact constant $K(n)={785}^n$ from [F5]. For each $j$, let $k_0(j)$ be the unique integer with $2^{-k_0(j)}<{\rho }_j\le 2^{-k_0(j)+1}$. For $k\ge s_r$, put $N_{j,k}=\{y:\mathrm{dist}(y,B_j)<a_k\}$ and $A_{rj}^k=\{\begin{array}{ll}{E}_r^k\cap N_{j,k},& k\ge k_0(j),\\ \mathrm{\varnothing },& k<k_0(j)\mathrm{.}\end{array}$ Put $R_{rj}^k=A_{rj}^k\setminus {\bigcup }_{m>j}{A}_{rm}^k$ and $g_{rjk}(x)={\int }_{R_{rj}^k}{\psi }_k(x-y)({\tilde{\psi }}_k\ast f)(y)dy$. These sets are disjoint and cover $E_r^k$: if $y\in E_r^k$, choose a covering ball $B_j$ containing $y$. Then $d_r(y)>a_k=2\cdot 2^{-k}$ and $d_r(y)\le d_r({\xi }_j)+\mathrm{\mid }y-{\xi }_j\mathrm{\mid }<3{\rho }_j\mathrm{/}2$, so $2^{-k}<3{\rho }_j\mathrm{/}4<{\rho }_j$ and $k\ge k_0(j)$. Hence $y\in A_{rj}^k$, and the greatest eligible index assigns it to exactly one $R_{rj}^k$, once the finite overlap below is established. The actual localization neighborhoods $N_{j,k}$ with $k\ge k_0(j)$ have uniformly finite overlap. For such an index $2^{-k}<{\rho }_j$, so $a_k<2{\rho }_j$ and $\mathrm{\mid }y-{\xi }_j\mathrm{\mid }<{\rho }_j\mathrm{/}2+a_k<5{\rho }_j\mathrm{/}2$ whenever $y\in N_{j,k}$. If two eligible neighborhoods meet, the 1-Lipschitz property of $d_r$ gives, for $R=\max ({\rho }_j,{\rho }_m)$ and $r_0=\min ({\rho }_j,{\rho }_m)$, $R-r_0\le \mathrm{\mid }{\xi }_j-{\xi }_m\mathrm{\mid }<\frac{1}{2}(R+r_0)+4r_0,$ so $R<11r_0$. For any finite collection of eligible neighborhoods containing one point $y$, the disjoint balls $B({\xi }_j,{\rho }_j\mathrm{/}8)$ therefore have radii at least $R\mathrm{/}88$ and lie in $B(y,21R\mathrm{/}8)$. Comparing volumes gives multiplicity at most ${231}^n$ at every fixed scale. If $A_{rj}^k\mathrm{\ne }\mathrm{\varnothing }$, then $k\ge k_0(j)$ and the support of ${\psi }_k\ast [({\tilde{\psi }}_k\ast f)1_{R_{rj}^k}]$ lies in the ball of radius ${\rho }_j\mathrm{/}2+3\cdot 2^{-k}<7{\rho }_j\mathrm{/}2$ about ${\xi }_j$, hence in $7B_j$. It remains to prove a uniform $L^{\mathrm{\infty }}$ estimate for $F_{B_j}:={\sum }_{k\ge s_r}{\int }_{R_{rj}^k}{\psi }_k(\cdot -y)({\tilde{\psi }}_k\ast f)(y)dy$. We use the following explicit localization estimate. For any set $S\subseteq {\mathbb{R}}^n$ and integers $s_r\le q\le m$, replacing $E_r^k$ by $E_r^k\cap \{y:\mathrm{dist}(y,S)<a_k\}$ in $F_{r,q,m}$ still gives an $L^{\mathrm{\infty }}$ norm at most $C{2}^r$. If $S=\mathrm{\varnothing }$, the sum is zero. If $S\mathrm{\ne }\mathrm{\varnothing }$ and $\mathrm{dist}(x,S)=0$, all kernel-support balls $B(x,2^{-k})$ lie in these neighborhoods and the sum is the full bounded partial sum. If $\delta =\mathrm{dist}(x,S)>0$, choose $\mathrm{\ell }$ with $2^{-\mathrm{\ell }}<\delta \le 2^{-\mathrm{\ell }+1}$. Then $B(x,2^{-k})$ lies in the neighborhood for $k\le \mathrm{\ell }-1$ and is disjoint from it for $k\ge \mathrm{\ell }+2$, leaving only two boundary scales, each bounded by (1). The same estimate holds for the infinite tail: the preceding absolute pairing bounds give distributional convergence, and the uniform finite-sum bounds pass to the limit by the same bounded-distribution representation [F9] used in step 2.1. Let $J_j$ be all later indices $m>j$ for which $W_m\cap W_j\mathrm{\ne }\mathrm{\varnothing }$. By [F5], $\mathrm{\#}{J}_j\le K(n)$ and ${\rho }_m\ge {\rho }_j\mathrm{/}7$. Put ${\rho }_{\ast }=\min (\{{\rho }_j\mathrm{/}7\}\cup \{{\rho }_m:m\in J_j\})={\rho }_j\mathrm{/}7$, let $k_1$ be the least integer with $a_{k_1}<{\rho }_{\ast }\mathrm{/}4$, and set $S={\bigcup }_{m\in J_j}{B}_m$. Then ${\rho }_j\mathrm{/}7\le {\rho }_{\ast }\le {\rho }_j$, while ${\rho }_j\mathrm{/}2\le 2^{-k_0}<{\rho }_j$ and ${\rho }_{\ast }\mathrm{/}16\le 2^{-k_1}<{\rho }_{\ast }\mathrm{/}8$. Hence $k_0<k_1$ and $2^{k_1-k_0}=2^{-k_0}\mathrm{/}{2}^{-k_1}<112<2^7$, so $k_1-k_0\le 7$. For every $k\ge k_1$ each neighborhood $N_{m,k}$ with $m\in J_j\cup \{j\}$ lies in $W_m$. To exclude other later indices without presupposing their radii, a meeting point gives $\mathrm{\mid }{\xi }_m-{\xi }_j\mathrm{\mid }<({\rho }_m+{\rho }_j)\mathrm{/}2+2a_k$. Here $a_k<{\rho }_{\ast }\mathrm{/}4\le {\rho }_j\mathrm{/}28$, so this distance is less than $3({\rho }_m+{\rho }_j)\mathrm{/}4$. Thus $W_m\cap W_j\mathrm{\ne }\mathrm{\varnothing }$ and $m\in J_j$. All indices in $J_j$ are eligible at these scales because $2^{-k}<{\rho }_{\ast }\mathrm{/}8\le {\rho }_m\mathrm{/}8$. Consequently the high-scale part is exactly the difference of the two localized sums associated with $S\cup B_j$ and $S$, each bounded by the localization estimate. There are at most seven lower scales $k_0\le k<k_1$, each bounded by (1). Thus $\mathrm{\parallel }{F}_{B_j}{\mathrm{\parallel }}_{\mathrm{\infty }}\le C_{\mathrm{♯}}{2}^r$ uniformly in $r,j$. Absolute convergence against Schwartz tests follows by summing [F7] over the disjoint sets $R_{rj}^k$; for $k<0$ the bound is $C_{\chi }{2}^{kn\mathrm{/}p}\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}$ as above. Each summand is compactly supported in $7B_j$ and has zero moments through order $K$ by the moments of $\psi$. Choose the representative of $F_{B_j}$ to vanish outside its compact support. That support lies in the closed ball of radius ${\rho }_j\mathrm{/}2+3\cdot 2^{-k_0(j)}<7{\rho }_j\mathrm{/}2$, so it is contained in the open ball $7B_j$. For fixed $j,k$, $R_{rj}^k\subset N_{j,k}$ is bounded, has finite measure, and ${\tilde{\psi }}_k\ast f$ is bounded by [F4], so absolute integrability and Tonelli's theorem for nonnegative measurable functions on a sigma-finite product applied to the absolute values justify each moment integral. The distributional limit is supported in $7B_j$ and has the same moments, by testing against a smooth compactly supported test equal to each monomial on a neighborhood of the closed ball $\overline{7B_j}$ (construct the cutoff from The standard smooth step function). Finally, the sets $R_{rj}^k$ partition the $E_r^k$, and the absolute pairing bounds summed over all $r,j,k$ show $f={\sum }_{r,j}{F}_{B_j}\text{in }{\mathcal{S}}^{\mathrm{\prime }}\mathrm{.}$ Indeed, for $k\ge 0$ the full sum of absolute pairings is bounded by $C_{\chi }{2}^{-k(K+1-n\mathrm{/}p)}\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}$ using [F7]; for $k<0$ it is bounded by $C_{\chi }{2}^{kn\mathrm{/}p}\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}$. Sum in $k$ and apply the Calderon reproducing identity [F3].

Atom normalization. Choose an axis-parallel cube $Q_j$ centered at ${\xi }_j$ with side length $7{\rho }_j$; it contains $7B_j$, and $\mathrm{\mid }{Q}_j\mathrm{\mid }\mathrm{/}\mathrm{\mid }7B_j\mathrm{\mid }=2^n\mathrm{/}{\omega }_n$. Enlarge $C_{\mathrm{♯}}$ if needed so that $C_{\mathrm{♯}}\ge C_0(2^n\mathrm{/}{\omega }_n{)}^{1\mathrm{/}p}$, where $C_0$ is the constant in the preceding $L^{\mathrm{\infty }}$ estimate. Put $a_{rj}=C_{\mathrm{♯}}^{-1}\mathrm{\mid }7B_j{\mathrm{\mid }}^{-1\mathrm{/}p}{2}^{-r}{F}_{B_j},{\lambda }_{rj}=C_{\mathrm{♯}}\mathrm{\mid }7B_j{\mathrm{\mid }}^{1\mathrm{/}p}{2}^r\mathrm{.}$ Then $\mathrm{supp}{a}_{rj}\subseteq 7B_j$, $\mathrm{\parallel }{a}_{rj}{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\mid }7B_j{\mathrm{\mid }}^{-1\mathrm{/}p}$, and the choice of $C_{\mathrm{♯}}$ gives $\mathrm{\parallel }{a}_{rj}{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\mid }{Q}_j{\mathrm{\mid }}^{-1\mathrm{/}p}$. Its moments vanish through order $K$, hence through $s$, because $s=\lfloor n(1\mathrm{/}p-1)\rfloor <n\mathrm{/}p\le K$. Thus $a_{rj}$ is a $(p,\mathrm{\infty },s)$-atom in the cube-supported convention of $H^p$ atoms with a prescribed moment order, and $f={\sum }_{r,j}{\lambda }_{rj}{a}_{rj}$ in ${\mathcal{S}}^{\mathrm{\prime }}$.

Coefficient bound. Since $\mathrm{\mid }7B_j\mathrm{\mid }=7^n\mathrm{\mid }{B}_j\mathrm{\mid }$ and $B_j\subset W_j$ with multiplicity at most $K(n)$, ${\sum }_{r,j}{\lambda }_{rj}^p=C_{\mathrm{♯}}^p{7}^n{\sum }_r{2}^{pr}{\sum }_j\mathrm{\mid }{B}_j\mathrm{\mid }\le C_{\mathrm{♯}}^p{7}^{n}K(n){\sum }_r{2}^{pr}\mathrm{\mid }{\mathrm{\Omega }}_r\mathrm{\mid }\le C\mathrm{\parallel }f{\mathrm{\parallel }}_{H^p}^p$ by the layer-cake estimate and [F1]. This proves the coefficient bound.

Conclusion. The zero case was handled at the start. For $f\mathrm{\ne }0$, steps 3.1 and 4.1 produce the atoms and coefficients, step 5.1 gives the ${\mathrm{\ell }}^p$ estimate, and the absolutely convergent distributional sum in step 3.1 equals $f$. Hence all four claims hold.