A Remark on Atomic Decompositions of Martingale Hardy’s Spaces

https://doi.org/10.1007/s12220-020-00449-4

A Remark on Atomic Decompositions of Martingale Hardy’s Spaces

Maciej Paluszynski¹ ·Jacek Zienkiewicz¹

Received: 12 March 2020 / Published online: 27 June 2020

© The Author(s) 2020

Abstract

We prove theorems and exhibit a counterexample concerning an atomic decomposition of martingaleH¹with atoms satisfying simultaneous cancellation condition (3).

Keywords Hardy space·Atomic decomposition·Dyadic martingale Mathematics Subject Classification 42B25·11P05

1 Introduction

Let f ∈ L¹(R). We say that f ∈H¹(dyadic martingale Hardy space) if

M f(x)= sup

I:x∈I

1

|I|

I

f∈L¹, f_H¹ = M fL¹, (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 aI is a (dyadic) atom, or anH¹- atom, associated with a dyadic intervalI if

• suppaI ⊂I,

• aIL^∞≤ _|¹_I|,

• aI =0.

Dedicated to Guido Weiss on the occasion of his 90th Birthday.

B

1 Instytut Matematyczny, Uniwersytet Wrocławski, pl. Grunwaldzki 2/4, 50-384 Wrocław, Poland

Any function f ∈H¹admits a decomposition

f =

I

λIaI, inH¹, (2)

whereaIare atoms and

I

|λI| ≤Cf_H¹.

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 intro- duction 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 weightsw1, w2onRand f ∈H¹satisfying 1≤wi ≤C, wi· f ∈H¹, i =1,2.

Can one obtain a decomposition (2) with atomsaI satisfying simultaneously

aI·w1=

aI ·w2=0? (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 ∈ H¹which admit decomposition (2) with atomsaI satisfying (3).

The Results

Given a weight functionw we call wH¹ the space {f : w· f ∈ H¹}with norm f_wH¹ = w· f_H¹. A functionais called awH¹atom ifw·ais an atom. Since the weightswwe consider are bounded and bounded away from 0, the only difference between an atom and awH¹atom is in the cancellation condition.

Theorem 1 There exist weights1≡w1≤w2≤c and a function f withw1f ∈H¹ andw2f ∈H¹, which does not have a joint atomic decomposition, that is it does not have a decomposition

f =

λQbQ,

|λQ| ≤Cmax

f_w1H¹,f_w2H¹

, (4)

with bQbeing bothw1H¹andw2H¹atoms.

Remark Theorem is stated forw1≡1. It is clear, that the same construction can be applied to the case of arbitraryw1∼1. That is for any weightw1∼1 there exists a w2satisfyingw1≤w2≤C, such that the theorem holds.

Proof Consider the interval[0,1]and arbitraryn ∈N. LetIkbe consecutive, adjacent intervals of length 2^−k

Ik =2^k−2

2^k ,2^k−1

2^k , k=1,2, . . . ,n.

Each of these intervals is an element of the standard dyadic grid. The left half ofIkis denoted byI_k⁺and the right half byI_k⁻:

I_k⁺=2^k−2

2^k ,2^k−3/2

2^k , I_k⁻=2^k−3/2

2^k ,2^k−1 2^k . We define:

aI_k(x)=

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩ 2^k

1+¹_nk−1

:x∈ I_k⁺,

−2^k

1+_n¹k−1

:x∈ I_k⁻,

0 :x∈/ Ik.

The functionsaIksatisfy:

• suppaIk ⊂Ik,

•

aIk =0,

• aIkL^∞ =2^k

1+¹_nk−1

=(1+¹_n)^k−1

|Ik| ≤ e

|Ik|, that is they aree- multiples ofH¹atoms. We let

f0= n k=1

aI_k,

and thus

f0_H¹ ≤ n k=1

aIk_H¹ ≤C n.

Let us define the weightw2

w2(x)=

1+¹_n

:x∈ I_k⁻, k=1,2, . . . ,

1 :x∈/

I_k⁻. We obtain the decomposition

w2· f0=21_I⁺

1 −2ⁿ· 1+1

n n

1_In⁻+

n−1

k=1

bJ_k,

where

Jk =I_k⁻∪I_k⁺₊₁, and

bJ_k(x)=

⎧⎪

⎨

⎪⎩

−2^k 1+¹_nk

:x∈ I_k⁻, 2^k+1

1+¹_nk

:x∈ I_k⁺₊₁,

0 :x∈/ Jk.

Observe, thatbJ_kare 3e/2 multiples ofH¹atoms.

• suppbJ_k =Jk,

•

bJ_k = −¹₂(1+¹_n)^k+¹₂(1+¹_n)^k=0,

• bJ_kL∞=2^k+1(1+¹_n)^k = 3(1+_n¹)^k 2|Jk| ≤ 3e

2|Jk|. Note that

Jk =I_k⁻∪I_k⁺₊₁= 1− 3

2^k+1,1− 3 2^k+2 ⊂

1− 1

2^k−1,1 = ˜Jk,

where J˜kis an element of the standard dyadic grid, with length comparable to that of Jk

|Jk| = 3

2^k+2, | ˜Jk| = 1 2^k−1. Thus

n−1

k=1

bJk

H¹ ≤

n−1

k=1

bJk_H¹ ≤Cn.

To account for the remaining parts ofw2· f0we extend f0to[1,2]and modifyw2

there. Let

α=

f0w2=1 2 −1

2

1+1 n

n

, −1≤α≤ −1 2. Forx∈ [1,2]we let

w2(x)=

1 :x∈ [1,³₂], 1+_n¹

:x∈(³₂,2],

and

f(x)=

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

f0(x) :x∈ [0,1], 2αn :x∈ [1,³₂],

−2αn :x∈(³₂,2], 0 :x∈ [0,/ 2].

Then

f_H¹ ≤ f0_H¹ +2|α|n≤Cn. Moreoverw2· f decomposes as

w2· f =

n−1

k=1

bJ_k +A,

where

A=21_I⁺

1 −2ⁿ 1+¹_nn

1_In⁻+2αn1[1,3/2]−2αn 1+¹_n

1(3/2,2].

It is a straightforward argument to show thatA_H¹ ≤ Cn. Similar straightforward computation shows the same estimate holds for the classical, non-martingale Hardy’s space. We will comment on this later. Thus

w2· f_H¹ ≤

n−1

k=1

bJ_k

H¹+ A_H¹ ≤Cn. We have just shown

f_w1H¹ ≤Cn, f_w2H¹ ≤Cn. Consider the function

g(x)=n(w1(x)−w2(x))· f(x)

=

⎧⎪

⎨

⎪⎩

2^k(1+¹_n)^k−1 :x∈I_k⁻, k=1,2, . . . ,n 2αn :x∈(³₂,2],

0 : otherwise.

We will show, that

g_H¹ = MgL¹ ≥C n².

Let us considerx∈I_k⁻. Immediate dyadic parents ofI_k⁻are

Ik, [1−2⁻ⁱ,1], i =0,1, . . . ,k−1.

We compute the average ofgoverI = [1−2^−(k−1),1]. 1

|I|

I

g(y)d y=2^k−1 n i=k

I_i⁻

2ⁱ

1+1 n

i−1

d y

=2^k−1 n i=k

2⁻⁽ⁱ⁺¹⁾2ⁱ

1+1 n

i−1

=2^k−2

n−1

i=k−1

1+ 1

n i

=2^k−2n

1+1 n

n

− 1+1

n k−1

.

Thus

Mg(x)≥2^k−2n

1+1 n

n

− 1+1

n k−1

, x∈ I_k⁻.

IntegratingMgwe obtain

MgL¹ ≥ n k=1

|I_k⁻|2^k−2n

1+ 1 n

n

− 1+ 1

n k−1

=n² 8 .

In fact, we can show a stronger estimate, namelyg_H¹ ≥Cn², 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

1[−1,1]≤≤1[−3/2,3/2]. Then

t ∗g(x)= t∗

n k=1

2^k

1+ 1 n

k−1

1I_k⁻

(x)+

t∗(2αn1(3/2,2]) (x)

= t∗F

(x)+ t ∗G

(x),

=1 t

x+³2t x−³₂t

x−y t

F(y)d y+1 t

x+³2t x−³₂t

x−y t

G(y)d y

where

suppF= n k=n/2

I_k⁻⊂ [1−3·2^−n/2−1,1], suppG=(3/2,2].

Now, takex ∈ [0,1−3·2^−n/2](n ≥ 4), andt = 1−x. Then the second integral vanishes due to disjoint supports. Thus

t∗g(x)= 1 t

_x+³

2t

x−³2t x−y t

F(y)d y≥ 1 t

_x+t

x−t

F(y)d y

Sincet =1−x, we havex+t =1 andx−t =2x−1≤ 1−3·2^−n/2−1, so the integration interval covers the entire support ofF. Thus, in the chosen range ofx

Mg(x)≥1−x∗g(x)

≥ 1 1−x

₁

2x−1

F(y)d y

= 1 1−x

n k=n/2

1+1

n k−1

·1 2

≥ 1 1−x ·n

4. Consequently,

MgL¹ ≥

_1−3·2−n/2

0

|Mg(x)|d x

≥ n 4

_1−3·2−n/2

0

d x 1−x

=n 4 ·n

2 log 2−log 3

≥Cn².

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

f =

Q

λQbQ,

wherebQare bothw1H¹andw2H¹atoms, as in the statement of the theorem. Then, by estimates above

Q

|λQ| ≤Cn.

This would imply

g_H¹ =n(w1−w2)

Q

λQ·bQ

H¹

≤

Q

|λQ| · n(w1−w2)bQ_H¹

≤Cn, since eachn(w1−w2)bQ is anH¹atom:

• suppn(w1−w2)bQ ⊂Q,

•

n(w1−w2)bQ =0,

• n(w1−w2)bQL^∞≤ bQL^∞ ≤ _|Q¹_|.

Thus, since the constants are independent ofn, we have obtained a contradiction n²g_H¹ n.

We call just constructed function fn, and the weightwⁿ. Both are localized on[0,2].

It is now routine to appropriatelyH¹-scale and shift thus constructed fn’s, together withwⁿ’s (both operations necessarily dyadic), so they are all localized within[0,1], with disjoint supports. The sum ofn⁻³² fn’s over a dyadicn’s, together with weight w, being the sum ofwⁿ’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 weightwfrom Theorem1, very much in the spirit of tweaks known from the theory of Cauchy Integral. LetI ⊂ [0,1]

be dyadic. We letwⁿbe given by (5) below (modified weight from the above theorem), and denote bywⁿ_I this weight re-scaled and translated toI. Suppose{nk}is a sequence of naturals increasing to infinity sufficiently fast. We construct a sequence of weights ω^k.

(i) We putω¹=wⁿ_[0¹_,1].

(ii) Assumeω¹, . . . , ω^khave already been constructed. Let Ik,j, j =1, . . . ,lk be the maximal dyadic intervals on whichω^k =ck,j is constant. Then, forx∈ Ik,j

we putω^k+1(x)=ck,jwⁿ_Ik^k_,⁺_j¹(x). Observe that by construction

jIk,j = [0,1]

(iii) We putw(x)=limk→∞ω^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 condi- tion. From now on we fixnand the weightsw1≡1 andw2=won[0,1]constructed in the proof of Theorem1. Let us recall

w(x)=

1+¹_n

:x∈ I_k⁻, k=1,2, . . . ,

1 :x∈/

I_k⁻, (5)

(we do no restrictkto be≤ n, thus suppw = [0,1]). We will state a quantitative version of our result for these weights. The argument clearly extends to any pair w1, w2satisfying condition (6) below, withw =w2w⁻₁¹. Typical examples of such weights are those defined by lacunary Fourier series or positive polynomials. See Corollary following Theorem2.

Let us recall that we are working in the setting of the standard dyadic martingale onR. Our aim is to define a maximal function which would characterize an atomic space with atoms simultaneously orthogonal to both 1 andw. We put

I(x)=1I(x)(βI −w(x)) with the constantβI chosen so that

I =0. Obviously, 1I(x)

|I|^1/2, I(x) IL²(I)

are orthonormal functions in L²(I), obtained by Gramm-Schmidt orthogonalization of weights1I and1I ·wonI. We define the following maximal operator

Mf(x)= sup

I⊂[0,1]−dyad.

α∈R

1I(x)

|I| · α−wL^∞(I)

I

f ·(α−w).

It is immediate thatMf ≤C MH L f, whereMH L f is the standard Hardy-Littlewood maximal function on[0,1]. We will use the following

Lemma Let I ⊂ [0,1]. Then, for some constant C independent on n, we have I(x)

I²_L2

I

f ·I≤CMf(x)

Proof The lemma follows immediately from the following condition satisfied by the weightα−w: there is the constantCindependent ofαandnsuch that for any dyadic intervalI we have

|I| · α−w²L^∞(I)≤Cα−w²_L2 (6) To see this, supposeI ⊂ [0,1]is a dyadic interval. Since

k≥1Ik = [0,1)one of the following cases has to hold.

(i) There exists aksuch thatI ⊂Ik properly. ThenI ⊂I_k⁻or I ⊂I_k⁺andα−w is constant onI.

(ii) There exists aksuch thatI =Ik. Thenα−wis constant on both I_k⁻,I_k⁺. (iii) There exists aksuch thatIk ⊂ I properly. We denote byk0minimal suchk.

LetJ^#denote the immediate dyadic parent of J. ThenI_k^#

0 = [^2k₂⁰k⁻0²,1] ⊂ I. If I_k^#

0 ⊂ I properly, thanIk0−1 ⊂(I_k^#₀)^# ⊂ I, contradicting the definition ofk0. HenceI =I_k^#

0,Ik0 ⊂I, 2|Ik0| = |I|andα−wtakes exactly two values onI. We note that both values are taken onI_k⁻

0,I_k⁺

0.

To summarize,I either is contained within someIk, or contains a number of Ik’s in their entirety. In either case the functionα−wonIis 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 Theorem1. We recall that we work with the weightwconstructed for a fixedn. Theorem 2 IfMf ∈ L¹([0,1]),supp f ⊂ [0,1],

f =

wf =0, then f admits decomposition

f =

I−dyad.

λIaI,

where, for some constant C independent on n

I

|λI| ≤CMf

L¹, and aIare atoms, satisfying double cancellation condition

aI =

aIw=0.

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 operatorM^∗, playing the role of the classical grand maximal operator.

M^∗f(x)= sup

I⊂[0,1],I-dyad.

α∈R

110I(x)

|I| · α−wL^∞(I)

I

f ·(α−w). It is an easy consequence of the definition ofMandM^∗that

{x:M^∗f(x) > λ}≤10{x:Mf(x) > λ}.

If x ∈ {x : M^∗f(x) > λ} than x ∈ 10I for some dyadic interval I such that I ⊂ {x:Mf(x) > λ}. Now, if we write

{x:Mf(x) > λ} = I˜

whereI˜are maximal dyadic. Thus eachI ⊂ ˜I for someI˜and consequently {x:M^∗f(x) > λ}≤10{x:Mf(x) > λ}.

The immediate corollary is

M^∗fL¹ ≤CMfL¹.

For the dyadic interval I we denote by PI(f)the orthonormal projection of f onto the space spanned by the weights1I and1I ·w

PI(f)(x)= 1I(x)

|I|

I

f(y)d y+ I(x) I²_L2

I

f(y)I(y)d y. (7)

We observe that

PI(f),1 = f ·1I,1, and PI(f), w = f1I, w,

directly by the definition (7). Denote by As the set {M^∗f(x) > 2^s} and let A1=

r1Ir1 be the Whitney decomposition. Since by the construction As+1 ⊂ As

so we can choose the Whitney decomposition A2 =

r1,r2 Ir1,r2 in such a way that Ir1,r2 ⊂ Ir1. We continue this process, obtaining of a tree of dyadic intervals {Ir1,· · ·,Ir1,r2,···,rs,· · · }. We write

f(x)=

s

r₁,r₂,···,r_s

1I_{r1,r2,···,rs}(x)−

r_s+1

1I_{r1,r2,···,rs,rs+1}(x)

· f(x)−

−PI_{r1,r2,···,rs}(f)(x)+

r_s+1

PI_{r1,r2,···,rs,rs+1}(f)(x)

Each component

bIr1,r2,···,rs(x)=

1Ir1,r2,···,rs(x)−

rs+1

1Ir1,r2,···,rs,rs+1(x)

· f(x)−

−PIr1,r2,···,rs(f)(x)+

rs+1

PIr1,r2,···,rs,rs+1(f)(x)

satisfies

|bI_{r1,r2,···,rs}(x)| ≤C2^s

s

r₁,r₂,···,r_s

2^s|Ir₁,r₂,···,r_s| ≤CM^∗fL¹

bIr1,r2,···,rs,1I =0, bIr1,r2,···,rs, I =0. (8)

This means that

aIr1,r2,···,rs(x)= bIr1,r2,···,rs(x) 2^s|Ir₁,r₂,···,r_s| are double cancellation atoms and

f(x)=

λIr1,r2,···,rsaIr1,r2,···,rs(x), where

|λIr1,r2,···,rs| ≤CM^∗fL¹.

The theorem follows.

We leave it to the reader to extend Theorem2to any pairw1, w2having the property thatw=w2w₁⁻¹satisfies (6).

Corollary Assume thatw1, w2are polynomials satisfying1≤w1, w2≤C on[0,1]

and that f ∈H¹,supp f ⊂ [0,1]is simultaneously orthogonal tow1andw2. Then f admits an atomic decomposition with atoms simultaneously orthogonal tow1, w2. Proof The proof is based on standard norm-comparison properties for polynomials . We need to check the assumption of the Theorem2for polynomialsw1andw2.

If J ⊂ [0,1]is an interval,|J| = R andw is a polynomial of degree d (here w=αw1−w2), then

sup

x∈J

|w(x)| ≈ d k=0

R^k|w^(k)(a)| (9)

The implied constants in (9) do not depend onRanda∈ J. Hence, for anyI ⊂J, we have sup_x∈J|w(x)| ≥C Rsup_x∈I|w⁽¹⁾(x)|again with the constantCindependent of I andJ. With this observation the estimate forMa,abeing a classicalH¹atom, is an application of the standard cancellation argument. We note, that the proof of the corollary can be made independent of Theorem2. We leave details to the reader.

We present one more result.

Theorem 3 Suppose f ∈w1H¹∩w2H¹. Then there exist coefficients{λr1,r2,...,rs :ri ∈ Ji−finite, s∈N}and a tree of dyadic intervals{Ir1,r2,...,rs :ri ∈ Ji −finite, s∈N}, with inclusions

Ir1 ⊃Ir1,r2 ⊃ · · · ⊃Ir1,r2,...,rs ⊃. . . such that f admits atomic decompositions

f =

ri∈Ii,i=1,...s s∈N

λr₁,r₂,...,r_sbr^j₁,r₂,...,r_s, j =1,2,

where the atoms are given by br^j1,r2,...,rs(x)= 1

λr₁,r₂,...,r_s

f ·1I_r1,...,rs\

rs+1I_r1,...,rs+1(x)

− fI_r1,...,rs,wj +

r_s+1

1I_r1,...,rs+1(x)fI_r1,...,rs+1,wj

,

and

fI,w = 1

|I|

I

f ·w.

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 ofM^∗₁(w1· f)+M^∗₁(w2· f), forM^∗₁being the dyadic grand maximal operator

M^∗f(x)= sup

I⊂[0,1],I-dyad.

α∈R

110I(x)

|I|

I

f. We leave details to the reader.

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