Atomic decomposition of Hardy spaces

Remarks

Each atom has $\norm{a}_{L^1} \le 1$, and a computation with the Riesz transforms shows $\norm{a}_{H^1} \le C$; this gives the easy direction. The hard direction starts from the maximal function characterization of $H^1$. For each $k \in \Z$, the level set $\set{f^* > 2^k}$ of a suitable maximal function $f^*$ is decomposed into Whitney cubes. A Calderón–Zygmund decomposition at height $2^k$ produces functions supported on these cubes. The differences between consecutive levels split into pieces that live on slightly enlarged Whitney cubes $Q$, have integral zero and are bounded by $C 2^k$. Each such piece is a multiple $\lambda$ of an atom with $\abs{\lambda} \le C 2^k \abs{Q}$, and summing over $k$ gives $\sum \abs{\lambda} \le C \sum_k 2^k \abs{\set{f^* > 2^k}} \le C \int f^* \le C \norm{f}_{H^1}$.

The theorem is due to Coifman for $n = 1$ and to Latter in all dimensions. It reduces many questions about $H^1$ to atoms. For example, let $T$ be a Calderón–Zygmund operator and $a$ an atom on a ball $B$. Then $Ta$ is controlled on the double ball $2B$ by the $L^2$ bound of $T$. Off $2B$ it is controlled by the cancellation of $a$ and the smoothness of the kernel. This gives $\norm{Ta}_{L^1} \le C$ for all atoms, and hence $T$ is bounded from $H^1$ to $L^1$. Care is needed in such arguments, since for a general operator a uniform bound on atoms does not by itself imply boundedness on all of $H^1$. For Calderón–Zygmund operators the passage from atoms to $H^1$ can be justified.

Going deeper

For a student

The atomic decomposition is the most used description of $H^1(\R^n)$. It gives a short proof of the duality between $H^1$ and BMO: pairing an atom $a$ on a ball $B$ with a BMO function $b$ gives $\abs{\int a b} = \abs{\int_B a (b - b_B)} \le \frac{1}{\abs{B}}\int_B \abs{b - b_B}$. It extends to the spaces $H^p(\R^n)$ for $0 < p < 1$, where atoms must have vanishing moments up to an order depending on $p$. Similar decompositions exist on spaces of homogeneous type, where the Fourier transform is not available.