A Remark on Atomic Decompositions of Martingale Hardy’s Spaces

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


Introduction
Let f ∈ L 1 (R). We say that f ∈ H 1 (dyadic martingale Hardy space) if 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 H 1 -atom, associated with a dyadic interval I if where a I are atoms and 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 R and f ∈ H 1 satisfying Can one obtain a decomposition (2) with atoms a I satisfying simultaneously 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 ∈ H 1 which admit decomposition (2) with atoms a I satisfying (3).

The Results
Given a weight function w we call wH 1 A function a is called a wH 1 atom if w · 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 wH 1 atom is in the cancellation condition.
Theorem 1 There exist weights 1 ≡ w 1 ≤ w 2 ≤ c and a function f with w 1 f ∈ H 1 and w 2 f ∈ H 1 , which does not have a joint atomic decomposition, that is it does not have a decomposition with b Q being both w 1 H 1 and w 2 H 1 atoms.
Remark Theorem is stated for w 1 ≡ 1. It is clear, that the same construction can be applied to the case of arbitrary w 1 ∼ 1. That is for any weight w 1 ∼ 1 there exists a w 2 satisfying w 1 ≤ w 2 ≤ C, such that the theorem holds.
Proof Consider the interval [0, 1] and arbitrary n ∈ N. Let I k be consecutive, adjacent intervals of length 2 −k 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 : We define: The functions a I k satisfy: • supp a I k ⊂ I k , • a I k = 0, that is they are e -multiples of H 1 atoms. We let and thus Let us define the weight w 2 We obtain the decomposition Observe, that b J k are 3e/2 multiples of H 1 atoms.
Note that whereJ k is an element of the standard dyadic grid, with length comparable to that of J k To account for the remaining parts of w 2 · f 0 we extend f 0 to [1,2] and modify w 2 there. Let Then Moreover w 2 · f decomposes as It is a straightforward argument to show that A H 1 ≤ Cn. Similar straightforward computation shows the same estimate holds for the classical, non-martingale Hardy's space. We will comment on this later. Thus We have just shown Consider the function : otherwise.
We will show, that Let us consider x ∈ I − k . Immediate dyadic parents of I − k are We compute the average of g over Integrating Mg we obtain In fact, we can show a stronger estimate, namely g H 1 ≥ 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 Then Then the second integral vanishes due to disjoint supports. Thus Consequently, We continue with the proof of the theorem. Suppose f does have a decomposition where b Q are both w 1 H 1 and w 2 H 1 atoms, as in the statement of the theorem. Then, by estimates above This would imply Thus, since the constants are independent of n, we have obtained a contradiction We call just constructed function f n , and the weight w n . Both are localized on [0, 2]. It is now routine to appropriately 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 − 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.
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 ⊂ [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 ω k .
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 ≡ 1 and w 2 = w on [0, 1] constructed in the proof of Theorem 1. Let us recall (we do no restrict k to be ≤ n, thus 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 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 with the constant β I chosen so that I = 0. Obviously, are orthonormal functions in L 2 (I ), obtained by Gramm-Schmidt orthogonalization of weights 1 I and 1 I · w on I . We define the following maximal operator Proof The lemma follows immediately from the following condition satisfied by the weight α − w: there is the constant C independent of α and n such that for any dyadic interval I we have To see this, suppose I ⊂ [0, 1] is a dyadic interval. Since k≥1 I k = [0, 1) one of the following cases has to hold.
(i) There exists a k such that I ⊂ I k properly. Then I ⊂ I − k or I ⊂ I + k and α − w is constant on I . (ii) There exists a k such that I = I k . Then α − w is constant on both I − k , I + k . (iii) There exists a k such that I k ⊂ I properly. We denote by k 0 minimal such k.
Let J # denote the immediate dyadic parent of J . Then contradicting the definition of k 0 . Hence I = I # k 0 , I k 0 ⊂ I , 2|I k 0 | = |I | and α − 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 α − 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.
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. 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 M * , playing the role of the classical grand maximal operator. The immediate corollary is For the dyadic interval I we denote by P I ( f ) the orthonormal projection of f onto the space spanned by the weights 1 I and 1 I · w We observe that directly by the definition (7). Denote by A s the set {M * f (x) > 2 s } and let A 1 = r 1 I r 1 be the Whitney decomposition. Since by the construction A s+1 ⊂ A s so we can choose the Whitney decomposition A 2 = r 1 ,r 2 I r 1 ,r 2 in such a way that I r 1 ,r 2 ⊂ I r 1 . We continue this process, obtaining of a tree of dyadic intervals {I r 1 , · · · , I r 1 ,r 2 ,··· ,r s , · · · }. We write The theorem follows.
We leave it to the reader to extend Theorem 2 to any pair w 1 , w 2 having the property that w = w 2 w −1 1 satisfies (6). Corollary Assume that w 1 , w 2 are polynomials satisfying 1 ≤ w 1 , w 2 ≤ C on [0, 1] and that f ∈ H 1 , supp f ⊂ [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 ⊂ [0, 1] is an interval, |J | = R and w is a polynomial of degree d (here w = αw 1 − w 2 ), then sup x∈J |w(x)| ≈ d k=0 R k |w (k) (a)| The implied constants in (9) do not depend on R and a ∈ J . Hence, for any I ⊂ J , we have sup x∈J |w(x)| ≥ C R sup x∈I |w (1) (x)| again with the constant C independent of I and J . With this observation the estimate for Ma, a being a classical 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.
We present one more result.