A Remark on Atomic Decompositions of Martingale Hardy’s Spaces

Abstract

We prove theorems and exhibit a counterexample concerning an atomic decomposition of martingale ${{\mathbb {H}}^1}$ with atoms satisfying simultaneous cancellation condition (3).

1 Introduction

Let $f\in L^1({\mathbb {R}})$. We say that $f\in {{\mathbb {H}}^1}$ (dyadic martingale Hardy space) if

$$\begin{aligned} Mf(x)=\sup _{I\,:x\in I}\,\frac{1}{|I|}\Big |\int _I f\Big |\in L^1,\qquad \Vert f\Vert _{{\mathbb {H}}^1}=\Vert Mf\Vert _{L^1}, \end{aligned}$$

(1)

where the supremum is taken over all dyadic intervals I containing x. The basic property of Hardy spaces is the so called atomic decomposition. We say that a function $a_I$ is a (dyadic) atom, or an ${{\mathbb {H}}^1}$ - atom, associated with a dyadic interval I if

• $\mathrm{supp}\, a_I\subset I$,

• $\Vert a_I\Vert _{L^\infty }\le \frac{1}{|I|}$,

• $\int a_I=0$.

Any function $f\in {{\mathbb {H}}^1}$ admits a decomposition

$$\begin{aligned} f=\sum _I\lambda _I\,a_I,\qquad \text {in }{{\mathbb {H}}^1}, \end{aligned}$$

(2)

where $a_I$ are atoms and

$$\begin{aligned} \sum _I|\lambda _I|\le C\Vert f\Vert _{{\mathbb {H}}^1}. \end{aligned}$$

The martingale Hardy spaces were first introduced in [3], with atomic decomposition implicitly appearing in [4], and the explicit proof appearing in [1]. More on the subject of atomic decompositions in the martingale setting can be found in [8]. Classical introduction to Hardy spaces can be found in [7]. Atomic decompositions in the classical case were developed in [2] and [5]. In this note we address the following question: Suppose we are given two weights $w_1,w_2$ on ${\mathbb {R}}$ and $f\in {{\mathbb {H}}^1}$ satisfying

$$\begin{aligned} 1\le w_i\le C,\qquad w_i\cdot f\in {{\mathbb {H}}^1},\ i=1,2. \end{aligned}$$

Can one obtain a decomposition (2) with atoms $a_I$ satisfying simultaneously

$$\begin{aligned} \int a_I\cdot w_1=\int a_I\cdot w_2=0\mathrm {?} \end{aligned}$$

(3)

One of the main results of this note is a negative answer to this natural question.

We also give a maximal function characterization of those $f\in {{\mathbb {H}}^1}$ which admit decomposition (2) with atoms $a_I$ satisfying (3).

2 The Results

Given a weight function w we call $w{{\mathbb {H}}^1}$ the space $\{f: w\cdot f\in {{\mathbb {H}}^1}\}$ with norm $\Vert f\Vert _{w{{\mathbb {H}}^1}}=\Vert w\cdot f\Vert _{{\mathbb {H}}^1}$. A function a is called a $w{{\mathbb {H}}^1}$ atom if $w\cdot a$ is an atom. Since the weights w we consider are bounded and bounded away from 0, the only difference between an atom and a $w{{\mathbb {H}}^1}$ atom is in the cancellation condition.

Theorem 1

There exist weights $1\equiv w_1\le w_2\le c$ and a function f with $w_1f\in {{\mathbb {H}}^1}$ and $w_2f\in {{\mathbb {H}}^1}$, which does not have a joint atomic decomposition, that is it does not have a decomposition

$$\begin{aligned} f=\sum \lambda _Qb_Q,\qquad \sum |\lambda _Q|\le C\max \big \{\Vert f\Vert _{w_1{{\mathbb {H}}^1}},\Vert f\Vert _{w_2{{\mathbb {H}}^1}}\big \}, \end{aligned}$$

(4)

with $b_Q$ being both $w_1{{\mathbb {H}}^1}$ and $w_2{{\mathbb {H}}^1}$ atoms.

Remark

Theorem is stated for $w_1\equiv 1$. It is clear, that the same construction can be applied to the case of arbitrary $w_1\sim 1$. That is for any weight $w_1\sim 1$ there exists a $w_2$ satisfying $w_1\le w_2\le C$, such that the theorem holds.

Proof

Consider the interval [0, 1] and arbitrary $n\in {\mathbb {N}}$. Let $I_k$ be consecutive, adjacent intervals of length $2^{-k}$

$$\begin{aligned} I_k=\Big [\frac{2^k-2}{2^k},\frac{2^k-1}{2^k}\Big ],\quad k=1,2,\dots ,n. \end{aligned}$$

Each of these intervals is an element of the standard dyadic grid. The left half of $I_k$ is denoted by $I_k^+$ and the right half by $I_k^-$:

$$\begin{aligned} I_k^+=\Big [\frac{2^k-2}{2^k},\frac{2^k-3/2}{2^k}\Big ],\qquad I_k^-=\Big [\frac{2^k-3/2}{2^k},\frac{2^k-1}{2^k}\Big ]. \end{aligned}$$

We define:

$$\begin{aligned} a_{I_k}(x)={\left\{ \begin{array}{ll}2^k\Big (1+\frac{1}{n}\Big )^{k-1}&{}:x\in I_k^+,\\ -2^k\Big (1+\frac{1}{n}\Big )^{k-1}&{}:x\in I_k^-,\\ 0&{}:x\notin I_k. \end{array}\right. } \end{aligned}$$

The functions $a_{I_k}$ satisfy:

• $\mathrm{supp}\ a_{I_k}\subset I_k$,

• $\int a_{I_k}=0$,

• $\Vert a_{I_k}\Vert _{L^\infty }=2^k\big (1+\frac{1}{n}\big )^{k-1}=\displaystyle \frac{(1+\frac{1}{n})^{k-1}}{|I_k|}\le \frac{e}{|I_k|}$,

that is they are e - multiples of ${{\mathbb {H}}^1}$ atoms. We let

$$\begin{aligned} f_0=\sum _{k=1}^n a_{I_k}, \end{aligned}$$

and thus

$$\begin{aligned} \Vert f_0\Vert _{{{\mathbb {H}}^1}}\le \sum _{k=1}^n\Vert a_{I_k}\Vert _{{{\mathbb {H}}^1}}\le C\,n. \end{aligned}$$

Let us define the weight $w_2$

$$\begin{aligned} w_2(x)={\left\{ \begin{array}{ll} \big (1+\frac{1}{n}\big )&{}: x\in I_k^-,\quad k=1,2,\dots ,\\ 1&{}: x\notin \bigcup I_k^-. \end{array}\right. } \end{aligned}$$

We obtain the decomposition

$$\begin{aligned} w_2\cdot f_0=2\mathbb {1}_{I_1^+}-2^n\cdot \Big (1+\frac{1}{n}\Big )^n\mathbb {1}_{I_n^-}+\sum _{k=1}^{n-1}b_{J_k}, \end{aligned}$$

where

$$\begin{aligned} J_k=I_k^-\cup I_{k+1}^+, \end{aligned}$$

and

$$\begin{aligned} b_{J_k}(x)={\left\{ \begin{array}{ll}-2^k\,\big (1+\frac{1}{n}\big )^k&{}: x\in I_k^-,\\ 2^{k+1}\,\big (1+\frac{1}{n}\big )^k&{}: x\in I_{k+1}^+,\\ 0&{}: x\notin J_k. \end{array}\right. } \end{aligned}$$

Observe, that $b_{J_k}$ are 3e/2 multiples of ${{\mathbb {H}}^1}$ atoms.

• $\mathrm{supp}\ b_{J_k}=J_k$,

• $\int b_{J_k}=-\frac{1}{2}(1+\frac{1}{n})^k+\frac{1}{2}(1+\frac{1}{n})^k=0$,

• $\Vert b_{J_k}\Vert _{L^\infty }=2^{k+1}(1+\frac{1}{n})^k=\displaystyle \frac{3(1+\frac{1}{n})^k}{2|J_k|}\le \frac{3e}{2|J_k|}$.

Note that

$$\begin{aligned} J_k=I_k^-\cup I_{k+1}^+=\Big [1-\frac{3}{2^{k+1}},1-\frac{3}{2^{k+2}}\Big ]\subset \Big [1-\frac{1}{2^{k-1}},1\Big ]={\tilde{J}}_k, \end{aligned}$$

where ${\tilde{J}}_k$ is an element of the standard dyadic grid, with length comparable to that of $J_k$

$$\begin{aligned} |J_k|=\frac{3}{2^{k+2}},\qquad |{\tilde{J}}_k|=\frac{1}{2^{k-1}}. \end{aligned}$$

Thus

$$\begin{aligned} \Big \Vert \sum _{k=1}^{n-1}b_{J_k}\Big \Vert _{{{\mathbb {H}}^1}}\le \sum _{k=1}^{n-1}\big \Vert b_{J_k}\Vert _{{{\mathbb {H}}^1}}\le Cn. \end{aligned}$$

To account for the remaining parts of $w_2\cdot f_0$ we extend $f_0$ to [1, 2] and modify $w_2$ there. Let

$$\begin{aligned} \alpha = \int f_0w_2=\frac{1}{2} - \frac{1}{2}\Big (1+\frac{1}{n}\Big )^n,\qquad -1\le \alpha \le -\frac{1}{2}. \end{aligned}$$

For $x\in [1,2]$ we let

$$\begin{aligned} w_2(x)={\left\{ \begin{array}{ll}1&{}:x\in [1,\frac{3}{2}],\\ \big (1+\frac{1}{n}\big ) &{}: x\in (\frac{3}{2},2], \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} f(x)={\left\{ \begin{array}{ll}f_0(x)&{}:x\in [0,1],\\ 2\alpha n&{}: x\in [1,\frac{3}{2}],\\ -2\alpha n&{}: x\in (\frac{3}{2},2],\\ 0&{}: x\notin [0,2]. \end{array}\right. } \end{aligned}$$

Then

$$\begin{aligned} \Vert f\Vert _{{{\mathbb {H}}^1}}\le \Vert f_0\Vert _{{{\mathbb {H}}^1}}+2|\alpha |n\le C n. \end{aligned}$$

Moreover $w_2\cdot f$ decomposes as

$$\begin{aligned} w_2\cdot f=\sum _{k=1}^{n-1} b_{J_k} + A, \end{aligned}$$

where

$$\begin{aligned} A=2\mathbb {1}_{I_1^+}-2^n\big (1+\textstyle \frac{1}{n}\big )^n\mathbb {1}_{I_n^-}+2\alpha n\mathbb {1}_{[1,3/2]}-2\alpha n\big (1+\frac{1}{n}\big )\mathbb {1}_{(3/2,2]}. \end{aligned}$$

It is a straightforward argument to show that $\Vert A\Vert _{{{\mathbb {H}}^1}}\le Cn$. Similar straightforward computation shows the same estimate holds for the classical, non-martingale Hardy’s space. We will comment on this later. Thus

$$\begin{aligned} \Vert w_2\cdot f\Vert _{{{\mathbb {H}}^1}}\le \Big \Vert \sum _{k=1}^{n-1}b_{J_k}\Big \Vert _{{{\mathbb {H}}^1}}+\Vert A\Vert _{{{\mathbb {H}}^1}}\le C n. \end{aligned}$$

We have just shown

$$\begin{aligned} \Vert f\Vert _{w_1{{\mathbb {H}}^1}}\le C n, \qquad \Vert f\Vert _{w_2{{\mathbb {H}}^1}}\le C n. \end{aligned}$$

Consider the function

$$\begin{aligned} g(x)&=n(w_1(x)-w_2(x))\cdot f(x)\\&={\left\{ \begin{array}{ll} 2^k(1+\frac{1}{n})^{k-1}&{}: x\in I_k^-,\quad k=1,2,\dots ,n\\ 2\alpha n &{}:x\in (\frac{3}{2},2],\\ 0&{}: \text { otherwise}. \end{array}\right. } \end{aligned}$$

We will show, that

$$\begin{aligned} \Vert g\Vert _{{{\mathbb {H}}^1}}=\Vert Mg\Vert _{L^1}\ge C\,n^2. \end{aligned}$$

Let us consider $x\in I_k^-$. Immediate dyadic parents of $I_k^-$ are

$$\begin{aligned} I_k,\quad [1-2^{-i},1],\ i=0,1,\dots ,k-1. \end{aligned}$$

We compute the average of g over $I=[1-2^{-(k-1)},1]$.

$$\begin{aligned} \frac{1}{|I|}\int _Ig(y)\,dy&=2^{k-1}\sum _{i=k}^n\int _{I_i^-}2^i\Big (1+\frac{1}{n}\Big )^{i-1}\,dy\\&=2^{k-1}\sum _{i=k}^n2^{-(i+1)}2^i\Big (1+\frac{1}{n}\Big )^{i-1}\\&=2^{k-2}\sum _{i=k-1}^{n-1}\Big (1+\frac{1}{n}\Big )^i\\&=2^{k-2}n\,\Big (\Big (1+\frac{1}{n}\Big )^{n}-\Big (1+\frac{1}{n}\Big )^{k-1}\Big ). \end{aligned}$$

Thus

$$\begin{aligned} Mg(x)\ge 2^{k-2}n\,\Big (\Big (1+\frac{1}{n}\Big )^{n}-\Big (1+\frac{1}{n}\Big )^{k-1}\Big ),\qquad x\in I_k^-. \end{aligned}$$

Integrating Mg we obtain

$$\begin{aligned} \Vert Mg\Vert _{L^1}\ge \sum _{k=1}^n|I_k^-|2^{k-2}n\,\Big (\Big (1+\frac{1}{n}\Big )^{n}-\Big (1+\frac{1}{n}\Big )^{k-1}\Big )=\frac{n^2}{8}. \end{aligned}$$

In fact, we can show a stronger estimate, namely $\Vert g\Vert _{{{\mathbb {H}}^1}}\ge Cn^2$, where the norm is in the classical Hardy’s space. We will comment on that in a remark below. To see this stronger estimate, let us fix a test function

$$\begin{aligned} \mathbb {1}_{[-1,1]}\le \Phi \le \mathbb {1}_{[-3/2,3/2]}. \end{aligned}$$

Then

$$\begin{aligned} \Phi _t*g(x)&=\Big (\Phi _t*\sum _{k=1}^n 2^k\Big (1+\frac{1}{n}\Big )^{k-1}\mathbb {1}_{I_k^-}\Big )(x)+\Big (\Phi _t*(2\alpha n\mathbb {1}_{(3/2,2]})\Big )(x)\\&=\Big (\Phi _t*F\Big )(x)+\Big (\Phi _t*G\Big )(x),\\&=\frac{1}{t}\int _{x-\frac{3}{2}t}^{x+\frac{3}{2}t}\Phi \Big (\frac{x-y}{t}\Big )F(y)\,dy+ \frac{1}{t}\int _{x-\frac{3}{2}t}^{x+\frac{3}{2}t}\Phi \Big (\frac{x-y}{t}\Big )G(y)\,dy \end{aligned}$$

where

$$\begin{aligned} \mathrm{supp}\ F=\bigcup _{k=n/2}^nI_k^-\subset [1-3\cdot 2^{-n/2-1},1],\quad \mathrm{supp}\ G=(3/2,2]. \end{aligned}$$

Now, take $x\in [0,1-3\cdot 2^{-n/2}]$ ($n\ge 4$), and $t=1-x$. Then the second integral vanishes due to disjoint supports. Thus

$$\begin{aligned} \Phi _t*g(x)=\frac{1}{t}\int _{x-\frac{3}{2}t}^{x+\frac{3}{2}t}\Phi \Big (\frac{x-y}{t}\Big )F(y)\,dy\ge \frac{1}{t}\int _{x-t}^{x+t}F(y)\,dy \end{aligned}$$

Since $t=1-x$, we have $x+t=1$ and $x-t=2x-1\le 1-3\cdot 2^{-n/2-1}$, so the integration interval covers the entire support of F. Thus, in the chosen range of x

$$\begin{aligned} \big |Mg(x)\big |&\ge \Big |\Phi _{1-x}*g(x)\Big |\\&\ge \frac{1}{1-x}\int _{2x-1}^1F(y)\,dy\\&=\frac{1}{1-x}\sum _{k=n/2}^n\Big (1+\frac{1}{n}\Big )^{k-1}\cdot \frac{1}{2}\\&\ge \frac{1}{1-x}\cdot \frac{n}{4}. \end{aligned}$$

Consequently,

$$\begin{aligned} \Vert Mg\Vert _{L^1}&\ge \int _0^{1-3\cdot 2^{-n/2}}|Mg(x)|\,dx\\&\ge \frac{n}{4}\int _0^{1-3\cdot 2^{-n/2}}\frac{dx}{1-x}\\&=\frac{n}{4}\cdot \Big (\frac{n}{2}\log 2-\log 3\Big )\\&\ge C n^2. \end{aligned}$$

We continue with the proof of the theorem. Suppose f does have a decomposition

$$\begin{aligned} f=\sum _Q\lambda _Qb_Q, \end{aligned}$$

where $b_Q$ are both $w_1{{\mathbb {H}}^1}$ and $w_2{{\mathbb {H}}^1}$ atoms, as in the statement of the theorem. Then, by estimates above

$$\begin{aligned} \sum _Q|\lambda _Q|\le C n. \end{aligned}$$

This would imply

$$\begin{aligned} \Vert g\Vert _{{{\mathbb {H}}^1}}&=\Big \Vert n(w_1-w_2)\sum _Q\lambda _Q\cdot b_Q\Big \Vert _{{{\mathbb {H}}^1}}\\&\le \sum _Q|\lambda _Q|\cdot \Vert n(w_1-w_2)b_Q\Vert _{{{\mathbb {H}}^1}}\\&\le C n, \end{aligned}$$

since each $n(w_1-w_2)b_Q$ is an ${{\mathbb {H}}^1}$ atom:

• $\mathrm{supp}\ n(w_1-w_2)b_Q\subset Q$,

• $\int n(w_1-w_2)b_Q=0$,

• $\Vert n(w_1-w_2)b_Q\Vert _{L^\infty }\le \Vert b_Q\Vert _{L^\infty }\le \frac{1}{|Q|}$.

Thus, since the constants are independent of n, we have obtained a contradiction

$$\begin{aligned} n^2\lesssim \Vert g\Vert _{{{\mathbb {H}}^1}}\lesssim n. \end{aligned}$$

We call just constructed function $f_n$, and the weight $w^n$. Both are localized on [0, 2]. It is now routine to appropriately ${{\mathbb {H}}^1}$-scale and shift thus constructed $f_n$’s, together with $w^n$’s (both operations necessarily dyadic), so they are all localized within [0, 1], with disjoint supports. The sum of ${n^{-\frac{3}{2}}}f_n$’s over a dyadic n’s, together with weight w, being the sum of $w^n$’s is the required example for which the condition (4) cannot hold. This completes the proof. $\square$

Remark

Observe, that the above theorem is also valid in the case of classical Hardy’s space.

We point to another possible construction of the weight w from Theorem 1, very much in the spirit of tweaks known from the theory of Cauchy Integral. Let $I\subset [0,1]$ be dyadic. We let $w^n$ be given by (5) below (modified weight from the above theorem), and denote by $w^n_I$ this weight re-scaled and translated to I. Suppose $\{n_k\}$ is a sequence of naturals increasing to infinity sufficiently fast. We construct a sequence of weights $\omega ^k$.

• (i) We put $\omega ^1=w^{n_1}_{[0,1]}$.

• (ii) Assume $\omega ^1,\dots ,\omega ^k$ have already been constructed. Let $I_{k,j},\ j=1,\dots ,l_k$ be the maximal dyadic intervals on which $\omega ^k=c_{k,j}$ is constant. Then, for $x\in I_{k,j}$ we put $\omega ^{k+1}(x)=c_{k,j}w^{n_{k+1}}_{I_{k,j}}(x)$. Observe that by construction $\bigcup _j I_{k,j}=[0,1]$

• (iii) We put $w(x)=\lim _{k\rightarrow \infty }\omega ^k(x)$.

The weight obviously satisfies requirements of the Theorem 1 (together with the function f, which consists of parts constructed in the proof, but summed differently). We also point out (we leave the proof to the reader), that it satisfies the condition (6) below.

We will now prove a maximal function characterization of those functions on [0, 1] that do admit atomic decomposition with atoms satisfying double cancellation condition. From now on we fix n and the weights $w_1\equiv 1$ and $w_2=w$ on [0, 1] constructed in the proof of Theorem 1. Let us recall

$$\begin{aligned} w(x)={\left\{ \begin{array}{ll}\big (1+\frac{1}{n}\big )&{}: x\in I_k^-,\quad k=1,2,\dots ,\\ 1&{}: x\notin \bigcup I_k^-, \end{array}\right. } \end{aligned}$$

(5)

(we do no restrict k to be $\le n$, thus $\mathrm{supp }\,w=[0,1]$). We will state a quantitative version of our result for these weights. The argument clearly extends to any pair $w_1,w_2$ satisfying condition (6) below, with $w={w_2}{w_1^{-1}}$. Typical examples of such weights are those defined by lacunary Fourier series or positive polynomials. See Corollary following Theorem 2.

Let us recall that we are working in the setting of the standard dyadic martingale on ${\mathbb {R}}$. Our aim is to define a maximal function which would characterize an atomic space with atoms simultaneously orthogonal to both 1 and w. We put

$$\begin{aligned} \Delta _I(x)=\mathbb {1}_I(x)(\beta _I- w(x)) \end{aligned}$$

with the constant $\beta _I$ chosen so that $\int \Delta _I=0$. Obviously,

$$\begin{aligned} \frac{\mathbb {1}_I(x)}{|I|^{1/2}},\qquad \frac{\Delta _I(x)}{\Vert \Delta _I\Vert _{L^2(I)}} \end{aligned}$$

are orthonormal functions in $L^2(I)$, obtained by Gramm-Schmidt orthogonalization of weights $\mathbb {1}_I$ and $\mathbb {1}_I\cdot w$ on I. We define the following maximal operator

$$\begin{aligned} {\mathcal {M}} f(x)=\sup _{\genfrac{}{}{0.0pt}{}{I\subset [0,1]-\text {dyad.}}{\alpha \in {\mathbb {R}}}}\frac{\mathbb {1}_{I}(x)}{|I|\cdot \Vert \alpha - w\Vert _{L^\infty (I)}} \Big |\int _{I}f\cdot (\alpha -w)\Big |. \end{aligned}$$

It is immediate that $\mathcal Mf\le C M_{HL}f$, where $M_{HL}f$ is the standard Hardy-Littlewood maximal function on [0, 1]. We will use the following

Lemma

Let $I\subset [0,1]$. Then, for some constant C independent on n, we have

$$\begin{aligned} \Big |\frac{\Delta _I(x)}{\Vert \Delta _I\Vert _{L^2}^2}\int _{I}f\cdot \Delta _I\Big |\le C\mathcal Mf(x) \end{aligned}$$

Proof

The lemma follows immediately from the following condition satisfied by the weight $\alpha -w$: there is the constant C independent of $\alpha$ and n such that for any dyadic interval I we have

$$\begin{aligned} |I|\cdot \Vert \alpha -w\Vert _{L^\infty (I)}^2\le C\Vert \alpha - w\Vert _{L^2}^2 \end{aligned}$$

(6)

To see this, suppose $I\subset [0,1]$ is a dyadic interval. Since $\bigcup _{k\ge 1} I_k=[0,1)$ one of the following cases has to hold.

• (i) There exists a k such that $I\subset I_k$ properly. Then $I\subset I^-_k$ or $I\subset I^+_k$ and $\alpha -w$ is constant on I.

• (ii) There exists a k such that $I= I_k$. Then $\alpha -w$ is constant on both $I^-_k, I^+_k$.

• (iii) There exists a k such that $I_k\subset I$ properly. We denote by $k_0$ minimal such k. Let $J^\#$ denote the immediate dyadic parent of J. Then $I^\#_{k_0}=[\frac{2^{k_0}-2}{2^{k_0}},1]\subset I$. If $I^\#_{k_0}\subset I$ properly, than $I_{k_0-1}\subset (I^\#_{k_0})^\#\subset I$, contradicting the definition of $k_0$. Hence $I=I^\#_{k_0}$, $I_{k_0}\subset I$, $2|I_{k_0}|=|I|$ and $\alpha -w$ takes exactly two values on I. We note that both values are taken on $I^-_{k_0}, I^+_{k_0}$.

To summarize, I either is contained within some $I_k$, or contains a number of $I_k$’s in their entirety. In either case the function $\alpha - w$ on I is constant, or assumes exactly 2 values, spread over sets of equal measure. In any case, the norm equivalence condition (6) is immediate. $\square$

The following two theorems have motivated the construction of the counterexample in Theorem 1. We recall that we work with the weight w constructed for a fixed n.

Theorem 2

If $\mathcal Mf\in L^1([0,1])$, $\mathrm{supp\,}f\subset [0,1]$, $\int f=\int wf =0$, then f admits decomposition

$$\begin{aligned} f=\sum _{I-\text {dyad.}}\lambda _I\,a_I, \end{aligned}$$

where, for some constant C independent on n

$$\begin{aligned} \sum _I |\lambda _I|\le C\big \Vert \mathcal Mf\big \Vert _{L^1}, \end{aligned}$$

and $a_I$ are atoms, satisfying double cancellation condition

$$\begin{aligned} \int a_I=\int a_I\,w=0. \end{aligned}$$

Proof

We note that the argument we use in the proof is standard, the only difference is in the cancellation statements involved. We begin with a definition of an auxiliary maximal operator ${\mathcal {M}}^*$, playing the role of the classical grand maximal operator.

$$\begin{aligned} {\mathcal {M}}^*f(x)=\sup _{\genfrac{}{}{0.0pt}{}{I\subset [0,1],I\text {-dyad.}}{\alpha \in {\mathbb {R}}}}\frac{\mathbb {1}_{10I}(x)}{|I|\cdot \Vert \alpha - w\Vert _{L^\infty (I)}}\Big |\int _If\cdot (\alpha - w)\Big |. \end{aligned}$$

It is an easy consequence of the definition of ${\mathcal {M}}$ and ${\mathcal {M}}^*$ that

$$\begin{aligned} \big |\{x:{\mathcal {M}}^*f(x)>\lambda \}\big |\le 10\big |\{x:\mathcal Mf(x)>\lambda \}\big |. \end{aligned}$$

If $x\in \{x:{\mathcal {M}}^*f(x)>\lambda \}$ than $x\in 10I$ for some dyadic interval I such that $I\subset \{x:\mathcal Mf(x)>\lambda \}$. Now, if we write

$$\begin{aligned} \{x:\mathcal Mf(x)>\lambda \}=\bigcup {{\tilde{I}}} \end{aligned}$$

where ${\tilde{I}}$ are maximal dyadic. Thus each $I\subset {\tilde{I}}$ for some ${\tilde{I}}$ and consequently

$$\begin{aligned} \big |\{x:{\mathcal {M}}^*f(x)>\lambda \}\big |\le 10\big |\{x:\mathcal Mf(x)>\lambda \}\big |. \end{aligned}$$

The immediate corollary is

$$\begin{aligned} \Vert {\mathcal {M}}^*f\Vert _{L^1}\le C\Vert \mathcal Mf\Vert _{L^1}. \end{aligned}$$

For the dyadic interval I we denote by $P_I(f)$ the orthonormal projection of f onto the space spanned by the weights $\mathbb {1}_I$ and $\mathbb {1}_I\cdot w$

$$\begin{aligned} P_I(f)(x)=\frac{\mathbb {1}_I(x)}{|I|}\int _If(y)\,dy+ \frac{\Delta _I(x)}{\Vert \Delta _I\Vert _{L^2}^2}\int _If(y)\Delta _I(y)\,dy. \end{aligned}$$

(7)

We observe that

$$\begin{aligned} \langle P_I(f),1\rangle =\langle f\cdot \mathbb {1}_I,1\rangle ,\quad \text {and}\quad \langle P_I(f),w\rangle =\langle f\mathbb {1}_I,w\rangle , \end{aligned}$$

directly by the definition (7). Denote by $A_s$ the set $\{{\mathcal {M}}^*f(x)>2^s\}$ and let $A_1=\bigcup _{r_1}I_{r_1}$ be the Whitney decomposition. Since by the construction $A_{s+1}\subset A_s$ so we can choose the Whitney decomposition $A_2=\bigcup _{r_1,r_2}I_{r_1,r_2}$ in such a way that $I_{r_1,r_2}\subset I_{r_1}$. We continue this process, obtaining of a tree of dyadic intervals $\{I_{r_1}, \cdots , I_{r_1,r_2,\cdots ,r_s}, \cdots \}$. We write

$$\begin{aligned} f(x)&=\sum _s \sum _{{r_1,r_2,\cdots ,r_s}}\bigg (\Big ( \mathbb {1}_{I_{r_1,r_2,\cdots ,r_s}}(x)- \sum _{r_{s+1}}\mathbb {1}_{I_{r_1,r_2,\cdots ,r_s,r_{s+1}}}(x)\Big )\cdot f(x)-\\&\qquad - P_{ I_{r_1,r_2,\cdots ,r_s}}(f)(x)+\sum _{r_{s+1}} P_{ I_{r_1,r_2,\cdots ,r_s,r_{s+1}}}(f)(x)\bigg ) \end{aligned}$$

Each component

$$\begin{aligned} b_{ I_{r_1,r_2,\cdots ,r_s}}(x)&=\Big ( \mathbb {1}_{I_{r_1,r_2,\cdots ,r_s}}(x)- \sum _{r_{s+1}}\mathbb {1}_{I_{r_1,r_2,\cdots ,r_s,r_{s+1}}}(x)\Big )\cdot f(x)-\\&\qquad - P_{ I_{r_1,r_2,\cdots ,r_s}}(f)(x)+\sum _{r_{s+1}} P_{ I_{r_1,r_2,\cdots ,r_s,r_{s+1}}}(f)(x) \end{aligned}$$

satisfies

$$\begin{aligned}&|b_{ I_{r_1,r_2,\cdots ,r_s}}(x)|\le C2^s\nonumber \\&\quad \sum _s \sum _{r_1,r_2,\cdots ,r_s} 2^s|I_{r_1,r_2,\cdots ,r_s}|\le C\Vert {\mathcal {M}}^*f\Vert _{L^1}\nonumber \\&\quad \langle b_{ I_{r_1,r_2,\cdots ,r_s}},\mathbb {1}_I\rangle =0,\qquad \langle b_{I_{r_1,r_2,\cdots ,r_s}},\Delta _I\rangle =0. \end{aligned}$$

(8)

This means that

$$\begin{aligned} a_{I_{r_1,r_2,\cdots ,r_s}}(x)=\frac{b_{I_{r_1,r_2,\cdots ,r_s}}(x)}{2^s|I_{r_1,r_2,\cdots ,r_s}|} \end{aligned}$$

are double cancellation atoms and

$$\begin{aligned} f(x)=\sum \lambda _{I_{r_1,r_2,\cdots ,r_s}}a_{I_{r_1,r_2,\cdots ,r_s}}(x), \end{aligned}$$

where

$$\begin{aligned} \sum |\lambda _{I_{r_1,r_2,\cdots ,r_s}}|\le C\Vert {\mathcal {M}}^*f\Vert _{L^1}. \end{aligned}$$

The theorem follows. $\square$

We leave it to the reader to extend Theorem 2 to any pair $w_1,w_2$ having the property that $w=w_2w_1^{-1}$ satisfies (6).

Corollary

Assume that $w_1, w_2$ are polynomials satisfying $1\le w_1, w_2 \le C$ on [0, 1] and that $f\in {{\mathbb {H}}^1}, \mathrm{supp}\, f\subset [0,1]$ is simultaneously orthogonal to $w_1$ and $w_2$. Then f admits an atomic decomposition with atoms simultaneously orthogonal to $w_1, w_2$.

Proof

The proof is based on standard norm-comparison properties for polynomials . We need to check the assumption of the Theorem 2 for polynomials $w_1$ and $w_2$.

If $J\subset [0,1]$ is an interval, $|J|=R$ and w is a polynomial of degree d (here $w=\alpha w_1-w_2$), then

$$\begin{aligned} \sup _{x\in J} |w(x)| \approx \sum _{k=0}^{d} R^k|w^{(k)}(a)| \end{aligned}$$

(9)

The implied constants in (9) do not depend on R and $a\in J$. Hence, for any $I\subset J$, we have $\sup _{x\in J} |w(x)|\ge CR\sup _{x\in I} |w^{(1)}(x)|$ again with the constant C independent of I and J. With this observation the estimate for ${\mathcal {M}} a$, a being a classical ${{\mathbb {H}}^1}$ atom, is an application of the standard cancellation argument. We note, that the proof of the corollary can be made independent of Theorem 2. We leave details to the reader. $\square$

We present one more result.

Theorem 3

Suppose $f\in w_1{{\mathbb {H}}^1}\cap w_2{{\mathbb {H}}^1}$. Then there exist coefficients $\{\lambda _{r_1,r_2,\dots ,r_s}:r_i\in J_i-\text {finite, }s\in {\mathbb {N}}\}$ and a tree of dyadic intervals $\{I_{r_1,r_2,\dots ,r_s}:r_i\in J_i-\text {finite, }s\in {\mathbb {N}}\}$, with inclusions

$$\begin{aligned} I_{r_1}\supset I_{r_1,r_2}\supset \dots \supset I_{r_1,r_2,\dots , r_s}\supset \dots \end{aligned}$$

such that f admits atomic decompositions

$$\begin{aligned} f=\sum _{\genfrac{}{}{0.0pt}{}{r_i\in I_i,i=1,\dots s}{s\in {\mathbb {N}}}} \lambda _{r_1,r_2,\dots ,r_s}b^j_{r_1,r_2,\dots ,r_s},\qquad j=1,2, \end{aligned}$$

where the atoms are given by

$$\begin{aligned} b^j_{r_1,r_2,\dots ,r_s}(x)&=\frac{1}{\lambda _{r_1,r_2,\dots ,r_s}}\Big (f\cdot \mathbb {1}_{I_{r_1,\dots ,r_s}\setminus \bigcup _{r_{s+1}} I_{r_1,\dots ,r_{s+1}}}(x) \\&\quad -\langle f\rangle _{I_{r_1,\dots ,r_s},w_j}+\sum _{r_{s+1}}\mathbb {1}_{I_{r_1,\dots ,r_{s+1}}}(x)\langle f\rangle _{I_{r_1,\dots ,r_{s+1}},w_j}\Big ), \end{aligned}$$

and

$$\begin{aligned} \langle f\rangle _{I,w}=\frac{1}{|I|}\int _If\cdot w. \end{aligned}$$

We omit the proof which is similar to the classical case of dyadic atomic decomposition. The argument yielding Theorem 2 can be adapted here as well. The only change required is to consider level sets of ${\mathcal {M}}_1^*(w_1\cdot f)+{\mathcal {M}}_1^*(w_2\cdot f)$, for ${\mathcal {M}}_1^*$ being the dyadic grand maximal operator

$$\begin{aligned} {\mathcal {M}}^*f(x)=\sup _{\genfrac{}{}{0.0pt}{}{I\subset [0,1],I\text {-dyad.}}{\alpha \in {\mathbb {R}}}}\frac{\mathbb {1}_{10I}(x)}{|I|} \Big |\int _If\Big |. \end{aligned}$$

We leave details to the reader.