---
tag: AQ15060AAA
kind: statement
role: theorem
name: Atomic decomposition of Hardy spaces
msc: [42B30]
uses: [75WGJ014GR, SASP14MA1Z, 2Z2QVWZ262, FBERT0KQXN]
about: [75WGJ014GR]
sources:
  - url: https://en.wikipedia.org/wiki/Hardy_space
---

## Statement

Let $n \ge 1$.
For a ball $B$, let $\abs{B}$ be its [[Lebesgue measure|2Z2QVWZ262]].
An *atom* is a measurable function $a$ on $\R^n$ with a ball $B$ such that $a = 0$ outside $B$, $\abs{a} \le 1/\abs{B}$, and $\int_B a \, dx = 0$.
Let $H^1(\R^n)$ be the [[real Hardy space|75WGJ014GR]].
Then a measurable $f$ lies in $H^1(\R^n)$ exactly when there are atoms $a_k$ and complex numbers $\lambda_k$ with
$$\sum_{k} \abs{\lambda_k} < \infty \quad \text{and} \quad f = \sum_k \lambda_k a_k,$$
the series converging in $L^1(\R^n)$.
Moreover there are constants $0 < c \le C$, depending only on $n$, such that
$$c \norm{f}_{H^1} \le \inf \sum_k \abs{\lambda_k} \le C \norm{f}_{H^1},$$
where the infimum is over all such representations of $f$.

## 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|FBERT0KQXN]].
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.

## Explanations

### L1 human

Every function in the Hardy space $H^1$ can be built from simple pieces, called atoms.
An atom lives on a ball, is not too large there, and has integral zero.
The theorem says that $H^1$ consists exactly of absolutely summable combinations of atoms.
So, to show that an operator behaves well on $H^1$, it is often enough to check it on a single atom.

### L2 human

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.