Real Hardy Spaces Maximal Functions and Atoms — Examples

The support and size conditions hold. Since ${\mathbf{1}}_Q$ vanishes off $Q$, $\mathrm{supp}a\subseteq Q$, and $\mathrm{\mid }a\mathrm{\mid }=\mathrm{\mid }Q{\mathrm{\mid }}^{-1}{\mathbf{1}}_Q\le \mathrm{\mid }Q{\mathrm{\mid }}^{-1}$ pointwise.

The moment condition fails. By [F1], ${\int }_{{\mathbb{R}}^n}a=\mathrm{\mid }Q{\mathrm{\mid }}^{-1}\int {\mathbf{1}}_Q=\mathrm{\mid }Q{\mathrm{\mid }}^{-1}\mathrm{\mid }Q\mathrm{\mid }=1$, which is nonzero because $\mathrm{\mid }Q\mathrm{\mid }>0$. Hence the zeroth-moment requirement $\int a=0$ of [L1] fails, and $a$ is not a $(1,\mathrm{\infty },0)$-atom.

Conclusion. The function $a$ meets the support and size parts of the atom definition but not the cancellation part; therefore the support and size conditions alone do not suffice for the atom property, and the moment condition is a genuine part of $H^p$ atoms with a prescribed moment order.

$a$ is a $(1,\mathrm{\infty },0)$-atom. Clearly $\mathrm{supp}a\subseteq Q$ and $\mathrm{\mid }a\mathrm{\mid }\le \mathrm{\mid }Q{\mathrm{\mid }}^{-1}$ almost everywhere. For the zeroth moment, [F1] gives $\int a=\mathrm{\mid }Q{\mathrm{\mid }}^{-1}(\mathrm{\mid }Q\mathrm{\mid }\mathrm{/}2-\mathrm{\mid }Q\mathrm{\mid }\mathrm{/}2)=0$. Hence $a$ satisfies [L1] and is a $(1,\mathrm{\infty },0)$-atom.

$a$ is discontinuous across the cutting hyperplane. Fix $z\in H\cap \mathrm{int}Q$ and let $r>0$ be smaller than the distance from $z$ to $\mathrm{\partial }Q$; then $z+s{e}_0\in Q^{+}$ and $z-s{e}_0\in Q^{-}$ for every $0<s<r$, and $a(z+s{e}_0)=\mathrm{\mid }Q{\mathrm{\mid }}^{-1}$, $a(z-s{e}_0)=-\mathrm{\mid }Q{\mathrm{\mid }}^{-1}$. Both sequences tend to $z$, so continuity of $a$ at $z$ would force the two values to be equal: by [F2] applied to the neighbourhood $V=\{w:\mathrm{\mid }w-a(z)\mathrm{\mid }<\mathrm{\mid }Q{\mathrm{\mid }}^{-1}\mathrm{/}2\}$ of $a(z)$, every point of a sufficiently small neighbourhood $U$ of $z$ would satisfy $a(z+s{e}_0)\in V$ and $a(z-s{e}_0)\in V$, which is impossible because the two values differ by $2\mathrm{\mid }Q{\mathrm{\mid }}^{-1}>\mathrm{\mid }Q{\mathrm{\mid }}^{-1}$. Hence $a$ is discontinuous at every $z\in H\cap \mathrm{int}Q$, which is the relative interior of the cutting slice, and in particular is not continuous, hence not smooth.

Conclusion. Step 1.1 exhibits a $(1,\mathrm{\infty },0)$-atom and step 2.1 shows that it has a jump discontinuity on the cutting hyperplane; therefore the atomic size and cancellation conditions do not imply continuity or smoothness, and no regularity of atoms may be assumed in the atomic characterisation. In particular a proof producing $L^{\mathrm{\infty }}$ atoms with jumps is not deficient on that account.

The three atom properties hold. Since ${\mathbf{1}}_{Q^{\pm }}$ vanish off $Q$, $\mathrm{supp}a\subseteq Q$. Pointwise $\mathrm{\mid }a\mathrm{\mid }=\mathrm{\mid }Q{\mathrm{\mid }}^{-1}$ on $Q^{+}\cup Q^{-}$ and $a=0$ elsewhere, so $\mathrm{\mid }a\mathrm{\mid }\le \mathrm{\mid }Q{\mathrm{\mid }}^{-1}$ everywhere. Finally, by [F1], ${\int }_{{\mathbb{R}}^n}a=\mathrm{\mid }Q{\mathrm{\mid }}^{-1}(\int {\mathbf{1}}_{Q^{+}}-\int {\mathbf{1}}_{Q^{-}})=\mathrm{\mid }Q{\mathrm{\mid }}^{-1}(\mathrm{\mid }Q\mathrm{\mid }\mathrm{/}2-\mathrm{\mid }Q\mathrm{\mid }\mathrm{/}2)=0$. Hence $a$ is a $(1,\mathrm{\infty },0)$-atom.

The $H^1$ estimate. Apply [F2] with $p=1$ and $s=0$: since $a$ is a $(1,\mathrm{\infty },0)$-atom, $\mathrm{\parallel }a{\mathrm{\parallel }}_{H^1}\le C_1(n,1,0,N,\phi )$. This bound is uniform over the supporting cube and the position of the halving hyperplane, with the fixed kernel and order dependence shown.

Conclusion. The half-cube difference is a legitimate $(1,\mathrm{\infty },0)$-atom, and its fixed-kernel $H^1$ norm has the uniform bound stated above.

Kernel bounds. For the Hilbert kernel and $\mathrm{\mid }h\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }\mathrm{/}2$, $\mathrm{\mid }k(z-h)-k(z)\mathrm{\mid }=\mathrm{\mid }h\mathrm{\mid }\mathrm{/}(\pi \mathrm{\mid }z\mathrm{\mid }\mathrm{\mid }z-h\mathrm{\mid })\le (2\mathrm{/}\pi )\mathrm{\mid }h\mathrm{\mid }\mathrm{\mid }z{\mathrm{\mid }}^{-2}$. Its annular size constant is $A_1=2\log 2\mathrm{/}\pi$ and its cancellation constant is $A_3=0$ by oddness. For Riesz kernels [F1] gives $A_2^{\mathrm{\prime }}=c_n{2}^{n+1}(3n+4)$; polar coordinates give $A_1\le c_n\sigma (S^{n-1})\log 2$, and spherical cancellation gives $A_3=0$. Thus in either case $\mathrm{\mid }k(z-h)-k(z)\mathrm{\mid }\le A_2^{\mathrm{\prime }}\mathrm{\mid }h\mathrm{\mid }\mathrm{\mid }z{\mathrm{\mid }}^{-n-1}$ for $\mathrm{\mid }h\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }\mathrm{/}2$, with constants depending only on $n$.

Off-support representation. Each $T$ is skew-adjoint: this is [F1] for the Hilbert transform, and follows for Riesz transforms from Plancherel and $\overline{m_j}=-m_j$. Put $\mathrm{\Omega }={\mathbb{R}}^n\setminus Q$ and $g(x)={\int }_{Q}k(x-y)a(y)dy$ on $\mathrm{\Omega }$. For $\phi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ the supports of $a$ and $\phi$ have positive distance, so $T\phi (y)=\int k(y-x)\phi (x)dx$ on $Q$ by the Schwartz principal-value formulas [F1], and the double integral is absolutely integrable. Skew-adjointness, Fubini and the real odd kernel give $⟨Ta,\phi ⟩=-⟨a,T\phi ⟩={\int }_{\mathrm{\Omega }}g(x)\overline{\phi (x)}dx$. The function $g$ is locally bounded on $\mathrm{\Omega }$, since the kernel is bounded on each compact set separated from $Q$ and $a\in L^1$; also $Ta\in L^2\subset L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$. Therefore the injectivity of regular distributions gives $Ta(x)=g(x)$ almost everywhere on $\mathrm{\Omega }$.

Near estimate. Write $\mathrm{\ell }=\mathrm{\ell }(Q)$ and let $Q^{†}$ be the concentric cube of side length $2\sqrt{n}\mathrm{\ell }$. By [F3], $\mathrm{\mid }{Q}^{†}\mathrm{\mid }=(2\sqrt{n}{)}^n\mathrm{\mid }Q\mathrm{\mid }$. Since $a\in L^2$, Cauchy-Schwarz and the $L^2$ bound give ${\int }_{Q^{†}}\mathrm{\mid }Ta\mathrm{\mid }\le \mathrm{\mid }{Q}^{†}{\mathrm{\mid }}^{1\mathrm{/}2}\mathrm{\parallel }Ta{\mathrm{\parallel }}_{L^2}\le \mathrm{\mid }{Q}^{†}{\mathrm{\mid }}^{1\mathrm{/}2}B\mathrm{\parallel }a{\mathrm{\parallel }}_{L^2}\le (2\sqrt{n}{)}^{n\mathrm{/}2}B$, because $\mathrm{\parallel }a{\mathrm{\parallel }}_{L^2}\le \mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }Q{\mathrm{\mid }}^{1\mathrm{/}2}\le \mathrm{\mid }Q{\mathrm{\mid }}^{-1\mathrm{/}2}$.

Far estimate. For $y\in Q$ one has $\mathrm{\mid }y-c_Q\mathrm{\mid }\le \sqrt{n}\mathrm{\ell }\mathrm{/}2$, while $x\mathrm{\notin }{Q}^{†}$ gives $\mathrm{\mid }x-c_Q\mathrm{\mid }\ge \sqrt{n}\mathrm{\ell }$. By 1.2 and $\int a=0$, $Ta(x)={\int }_Q[k(x-y)-k(x-c_Q)]a(y)dy$ almost everywhere there. For $h=y-c_Q\mathrm{\ne }0$, step 1.1 and polar coordinates give ${\int }_{\mathrm{\mid }z\mathrm{\mid }\ge 2\mathrm{\mid }h\mathrm{\mid }}\mathrm{\mid }k(z-h)-k(z)\mathrm{\mid }dz\le A_2^{\mathrm{\prime }}\mathrm{\mid }h\mathrm{\mid }\sigma (S^{n-1})\mathrm{/}(2\mathrm{\mid }h\mathrm{\mid })=\sigma (S^{n-1})A_2^{\mathrm{\prime }}\mathrm{/}2$; for $h=0$ the difference is identically zero. Tonelli consequently gives ${\int }_{(Q^{†}{)}^c}\mathrm{\mid }Ta\mathrm{\mid }\le {\int }_Q\mathrm{\mid }a(y)\mathrm{\mid }{\int }_{\mathrm{\mid }z\mathrm{\mid }\ge 2\mathrm{\mid }y-c_Q\mathrm{\mid }}\mathrm{\mid }k(z-(y-c_Q))-k(z)\mathrm{\mid }dzdy\le \sigma (S^{n-1})A_2^{\mathrm{\prime }}\mathrm{\parallel }a{\mathrm{\parallel }}_1\mathrm{/}2\le \sigma (S^{n-1})A_2^{\mathrm{\prime }}\mathrm{/}2$.

Conclusion. Steps 1.3 and 2.1 give $\mathrm{\parallel }Ta{\mathrm{\parallel }}_1\le (2\sqrt{n}{)}^{n\mathrm{/}2}B+\sigma (S^{n-1})A_2^{\mathrm{\prime }}\mathrm{/}2\le C_n(A_1+A_2^{\mathrm{\prime }}+A_3+B)$ for every component. The atom is in $H^1$ by [F2], and these estimates prove directly that its $L^2$ transform is integrable, without requiring smoothness of the atom.

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

Every $H^1$ element has an $L^1$ representative. For $g\in H^1$, take the representation of [F2]. Its partial sums $S_N={\sum }_{j\le N}{\lambda }_j{a}_j$ are Cauchy in $L^1$, since $\mathrm{\parallel }{S}_M-S_N{\mathrm{\parallel }}_1\le {\sum }_{N<j\le M}\mathrm{\mid }{\lambda }_j\mathrm{\mid }$. By completeness they converge to $h\in L^1$. For every Schwartz test $\psi$, [F3] gives $\mathrm{\mid }\int (S_N-h)\psi \mathrm{\mid }\le \mathrm{\parallel }{S}_N-h{\mathrm{\parallel }}_1\mathrm{\parallel }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$, whereas the atomic series converges to $g$ in ${\mathcal{S}}^{\mathrm{\prime }}$. Thus $g$ is the regular distribution of $h$, proving $H^1\subseteq L^1$.

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 $\int f\mathrm{\ne }0$. Hence $f\mathrm{\notin }{H}^1$; specializing to ${\mathbf{1}}_Q$ gives ${\mathbf{1}}_Q\mathrm{\notin }{H}^1$ with ${\mathbf{1}}_Q\in L^1$, so $H^1⊊L^1$.

Consequences for atoms. Every $(1,\mathrm{\infty },0)$-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 $H^1$.

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.