Weighted and Ergodic Theorems for Series of Differences of Convolutions

J Fourier Anal Appl (2008) 14: 39–59DOI 10.1007/s00041-007-9005-x

Weighted and Ergodic Theorems for Seriesof Differences of Convolutions

María Lorente · Alberto de la Torre

Received: 13 September 2006 / Revised: 04 July 2007 / Published online: 23 January 2008© Birkhäuser Boston 2008

Abstract We extend some results of C. Demeter about the boundedness of series ofdifference of convolutions to the setting of one-sided Ap weights. We need to usea different approach. To be precise, we prove that the kernel of the associated op-erator satisfies a generalized Hörmander condition. The weighted inequalities allowus to transfer the result to the ergodic case, when the operator is induced by a meanbounded, invertible, positive groups.

Keywords Singular integral · Weights · Ergodic averages · Maximal operator

Mathematics Subject Classification (2000) 42B20 · 42B25

1 Introduction

Let f be a locally integrable function defined on R. For each integer k let us denote

by Dk the average Dkf (x) = 12k

∫ x+2k

xf (y) dy. In [7] Jones and Rosenblatt have

proved that if we identify Lp of the torus T with the periodic extensions of functionsf ∈ Lp(0,1), and {vk} is a bounded sequence, then the operator

Tf (x) =∑

k≤0

vk(Dk − Dk−1)f (x)

The authors were supported by MEC Grant MTM2005-08350-C03-02 and Junta de Andalucía GrantFQM354.

M. Lorente (�) · A. de la TorreDepartamento de Análisis Matemático, Facultad de Ciencias, Universidad de Málaga, 29071 Málaga,Spaine-mail: [email protected]

A. de la Torree-mail: [email protected]

40 J Fourier Anal Appl (2008) 14: 39–59

is bounded in Lp(T), p > 1 and of weak type (1,1). In [2], using a different ap-proach, the result is extended to the context of Lp(w dx), where w is a one-sided Ap

weight, and the index k takes values in Z. Recently C. Demeter [4] has studied a sim-

ilar operator but in his case f is not periodic and Dkf (x) = ∫ 2k+1

2k f (x − y)ϕ(y) dy,where ϕ is a function defined on (0,∞) such that:

(a) ϕ is nonincreasing.(b) xϕ(x) is bounded.

(c)∑

k≤0 |sk − sk+1| < C, where sk = ∫ 2k+1

2k ϕ.

Observe that the D′ks are convolutions with truncations of a fix function but they are

not averages. Condition (c) is a substitute for that. It tells us that, although the sk arenot all identical, the differences are small. It can be seen that the theorem in [4] givesthe one in [7] as a particular case. Demeter’s proof is based on Fourier Transformestimates and a theorem of Duoandikoetxea and Rubio de Francia [5], that has noanalogue with weights.

In this article we use the methods of [2] and [8], to extend the results of [4] to thecontext of one-sided Ap weights. In order to obtain a weighted theory we need toimpose an additional condition on ϕ. The reason is that, although, as we will showbelow, conditions (a), (b) imply that the kernel of the corresponding singular integralsatisfies the H1 Hörmander condition, it is known that, in general, H1 is not enough toobtain a weighted theory [9]. The condition we impose is so mild that it is satisfied byany function ϕ, differentiable on every interval of the form [2k,2k+1) and such that|ϕ′(y)| ≤ C

|y|1+α for some α > 0. This includes all the examples in [4]. In this waywe give a result that extends those in the above mentioned articles. As an applicationof our theorem we prove that if (X,μ) is a sigma-finite measure space and St is astrongly measurable, positive, mean bounded group acting on Lp(X,μ), then we cantransfer the result to the ergodic version of T . We point out that the boundednessin Lp(R), without weights, only allows to transfer results for operators induced bygroups of measure preserving transformations in (X,μ).

The article is organized as follows: In Sect. 2 we introduce notation, recall somefacts about one-sided weights and state the main results. In Sect. 3 we give the proofsin the real variable case and point out how the result can be extended to the Orliczspace setting and finally in Sect. 4 we give the proof of the ergodic case.

2 Notation and Main Results

The one-sided Hardy-Littlewood maximal operators are defined as

M+f (x) = suph>0

1

h

∫ x+h

x

|f | and M−f (x) = suph>0

1

h

∫ x

x−h

|f |.

The one-sided Ap weights, 1 ≤ p < ∞, were introduced by E. Swayer [15]. They aredefined as follows:

We say that w ∈ A+1 if there exists C > 0 such that

M−w(x) ≤ Cw(x), a.e. (A+1 )

J Fourier Anal Appl (2008) 14: 39–59 41

We say that w ∈ A+p , p > 1, if there exists C > 0 such that for any a < b < c,

∫ b

a

w

(∫ c

b

w−p′p

) p

p′≤ C(c − a)p, (A+

p )

where p + p′ = pp′.The following results about these weights can be found in [15] and [10].

(1) The operator M+ is of weak type (1,1) with respect to the measure w(x)dx ifand only if w ∈ A+

1 .(2) The operator M+ is a bounded operator on Lp(w), p > 1, if and only if, w ∈ A+

p .(3) If w ∈ A+

1 , then there exists r > 1 so that wr ∈ A+1 .

(4) If w ∈ A+p , p > 1, then there exists 1 < t < p so that w ∈ A+

t .(5) One-sided doubling condition. If w ∈ A+

p for some p, 1 ≤ p, then there exists C

such that for any a ∈ R and any h > 0,∫(a,a+2h)

w ≤ C∫(a+h,a+2h)

w.

There is also a notion of A+∞ : A+∞ = ∪p≥1A+p .

The properties of this class can be seen in [11] and [12]. Of course there are sim-ilar definitions and theorems for A−

p , 1 ≤ p ≤ ∞, that are obtained reversing theorientation of R.

If 1 < s < ∞ the operators (M−|f |s) 1s , (M+|f |s) 1

s will be denoted byM−

s f, M+s f .

Our last definition is the Hr -Hörmander condition. See [14].

Definition 1 Let 1 ≤ r < ∞, we say that the function K satisfies the Hr -Hörmandercondition, or that K ∈ Hr , if there are numbers cr > 1 and Cr > 0, such that for anyx ∈ R and R > cr |x|,

∞∑

m=1

2mR

(1

2mR

∫

2mR<|y|≤2m+1R

|K(y − x) − K(y)|r dy

) 1r ≤ Cr. (2.1)

There is a similar definition for vector valued kernels K .

Definition 2 Let 1 ≤ r < ∞, we say that the kernel K , taking values in the Ba-nach space X, satisfies the Hr -Hörmander condition, if there are numbers cr > 1 andCr > 0, such that for any x ∈ R and R > cr |x|,

∞∑

m=1

2mR

(1

2mR

∫

2mR<|y|≤2m+1R

‖K(y − x) − K(y)‖rX dy

) 1r ≤ Cr. (2.2)

The following theorem is proved in [8].

Theorem 1 Let K be a kernel with support on (−∞,0), taking values in the Banachspace X, such that the operator Tf (x) = (K ∗ f )(x) is bounded from Lq(R) to


LqX(R) for some 1 < q < ∞ i.e.,

∫

R

‖Tf (x)‖qX dx ≤ C

∫

R

|f (x)|q dx.

If K satisfies Hr for some 1 < r < ∞ and r ′ is the conjugate exponent of r , then forany w ∈ A+∞ and any 0 < p < ∞, there exists C > 0 so that

∫

R

‖Tf (x)‖pXw(x)dx ≤ C

∫

R

∣∣M+

r ′ f (x)∣∣pw(x)dx,

whenever the left-hand side is finite.

Since our kernels will be supported in (0,∞) we will use this theorem changingthe orientation of R i.e., with w ∈ A−∞.

We now state our main results. Let ϕ be a locally integrable function supportedon (0,∞). Denote by Ik the interval [2k,2k+1), define sk = ∫

Ikϕ, and assume ϕ

satisfies (a), (b), and (c’) where

(c’)∑∞

−∞ |sk − sk+1| < ∞.

For each k ∈ Z define Dkf (x) = ∫Ik

f (x − y)ϕ(y) dy. Let {vk}k∈Z be a boundedsequence and (n,m) ∈ N × N. We will study the linear operators

Tn,mf (x) =m∑

k=−n

vk(Dk − Dk+1)f (x)

and the corresponding maximal operator T ∗f (x)=supN≥0 T ∗Nf (x), where T ∗

Nf (x) =supn,m≤N |Tn,mf (x)|. Our operators include the case where k moves only in the neg-ative integers or only in N, choosing vk = 0 for k ∈ N or −k ∈ N. It is clear that theoperators Tn,m are bounded in Lp(R) because they are convolution operators withnice kernels. Our main result is that they are uniformly bounded. Furthermore themaximal operator T ∗ is bounded in Lp(w), p > 1, for any w ∈ A−

p and of weak type

(1,1) if w ∈ A−1 .

Theorem 2 If ϕ satisfies Hr for some r > 1 then for any 0 < p < ∞ and any w ∈A−∞, there exists C such that

∫|Tn,mf (x)|pw(x)dx ≤ C

∫M−

r ′ f (x)pw(x)dx,

for any f for which the left-hand side is finite.

Theorem 3 If ϕ satisfies Hr for some r > 1 then for any 0 < p < ∞ and any w ∈A−∞, there exists C such that

∫T ∗f (x)pw(x)dx ≤ C

∫M−

r ′ f (x)pw(x)dx,



Theorem 4 If ϕ satisfies Hr for r ≥ 1, p > r ′ and w ∈ A−p

r′, then there exists C such

that for any f ∈ Lp(w),∫

T ∗f (x)pw(x)dx ≤ C

∫|f (x)|pw(x)dx.

Theorem 5 Let 1 < p < ∞ and w ∈ A−p . If ϕ satisfies Hr for all r ≥ 1, then there

exists C such that for any f ∈ Lp(w),∫

T ∗f (x)pw(x)dx ≤ C

∫|f (x)|pw(x)dx.

Theorem 6 If w ∈ A−1 and ϕ satisfies Hr for all r ≥ 1, then there exists C such that

for any d > 0 and any f ∈ L1(w),∫

{x:|T ∗f (x)|>d}w(x)dx ≤ C

d

∫|f (x)|w(x)dx.

In order to state our ergodic theorem we need some notation. Let (X,σ,μ) bea sigma finite measure space. Let us fix 1 < p < ∞ and let St a positive, stronglycontinuous, group of operators acting on Lp(X) that is mean bounded, i.e., ifAεf (x) = 1

ε

∫ ε

0 Stf (x) dt , then there exists C so that

‖Aεf ‖Lp(X) ≤ C‖f ‖Lp(X),

for any ε ∈ R+ and any f ∈ Lp(X). Let us fix a bounded sequence {vk}, a

function ϕ that satisfies (a), (b), (c’), and Hr for some r > p′ and let us de-

fine Dkf (x) = ∫ 2k+1

2k Stf (x)ϕ(t) dt . The ergodic version of our operators areτn,mf (x) = ∑m

k=−n vk(Dk − Dk+1)f (x), τ ∗Nf (x) = supn,m≤N |τn,mf (x)| and

τ ∗f (x) = supN∈N |τ ∗Nf (x)|.

Theorem 7 There exists C so that

∥∥τ ∗f

∥∥

Lp(X)≤ C‖f ‖Lp(X).

3 Proof of the Theorems in R

In order to prove the boundedness of the operators Tn,m we will consider them assingular integrals given by convolution with the kernels

Kn,m(y) =m∑

k=−n

vk(χIk− χIk+1)(y)ϕ(y)

= v−nχI−n(y)ϕ(y) +m∑

k=−n+1

(vk − vk−1)χIk(y)ϕ(y) − vmχIm+1(y)ϕ(y).


The classical assumption on the kernel of a singular integral, if one wants to have aweighted theory is a Lipschitz condition, i.e., there exists α > 0, c > 1 and C > 0 sothat

|K(y − x) − K(y)| ≤ C|x|α

|y|1+α, |y| > c|x|.

The problem is that our kernel is not even continuous. But Theorem 1 tells us thatthe natural substitutes for Lipschitz are the Hr conditions. Our first result is that theKn,m satisfy Hr for finite r if the function ϕ does.

Theorem 8 Let 1 ≤ r < ∞. If ϕ satisfies Hr so does Kn,m with a constant indepen-dent of n and m.

Proof Using that our kernel has support in (0,∞), it is not difficult to check (see [2])that it is enough to prove the following:

For 0 < x < 2i

∑

j>i

2j

(1

2j

∫

2j <y≤2j+1|Kn,m(y − x) − Kn,m(y)|r dy

) 1r ≤ Cr. (3.1)

Therefore, we will fix 0 < x < 2i and estimate Kn,m(y − x) − Kn,m(y) for y ∈ Ij ,where i < j . Keeping in mind the expression for Kn,m evaluated in y − x and ob-serving that y − x ∈ Ik if and only if y ∈ [2k + x,2k+1 + x) = Jk,x , we get

Kn,m(y −x) =[

v−nχJ−n,x (y)+m∑

k=−n+1

(vk −vk−1)χJk,x(y)−vmχJm+1,x

(y)

]

ϕ(y−x).

Since i < j and 0 < x < 2i we have that, for any k,

χJk,x(y) = χIk

(y) − χ[2k,2k+x)(y) + χ[2k+1,2k+1+x)(y)

and therefore,

χJk,x(y)ϕ(y − x) − χIk

(y)ϕ(y) =χIk(y)(ϕ(y − x) − ϕ(y))

− χ[2k,2k+x)(y)ϕ(y − x)

+ χ[2k+1,2k+1+x)(y)ϕ(y−x).


As a consequence,

Kn,m(y − x) − Kn,m(y) = v−nχI−n(y)(ϕ(y − x) − ϕ(y))

+ v−nχ[2−n,2−n+x)(y)ϕ(y − x) + v−nχ[2−n+1,2−n+1+x)(y)ϕ(y − x)

+m∑

k=−n+1

(vk − vk−1)[χIk

(y)(ϕ(y − x) − ϕ(y)) − χ[2k,2k+x)(y)ϕ(y − x)

+ χ[2k+1,2k+1+x)(y)ϕ(y − x)] − vmχIm+1(y)(ϕ(y − x) − ϕ(y))

+ vmχ[2m+1,2m+1+x)(y)ϕ(y − x) + vmχ[2m+2,2m+2+x)(y)ϕ(y − x).

Then, for each y ∈ Ij , there are at most three non zero summands in the above sum.This and the fact that the sequence {vj } is bounded give

∑

j>i

2j

(1

2j

∫

2j <y≤2j+1|Kn,m(y − x) − Kn,m(y)|r dy

) 1r

≤ C∑

j>i

2j

(1

2j

∫

2j <y≤2j+1|ϕ(y − x) − ϕ(y)|r dy

) 1r

+ C∑

j>i

2j

(1

2j

∫

2j <y≤2j +x

|ϕ(y − x)|r dy

) 1r ≡ A + B. (3.2)

The term B is easy to estimate. We use that if y ∈ Ij , then y − x � 2j and thatϕ(z)z ≤ C, to obtain

∫ 2j +x

2j

ϕ(y − x)r dy ≤ C2i2−jr ,

and then

B ≤ C∑

j>i

2i−jr ≤ C, (3.3)

where C does not depend on i. Since the statement: There exists C so that A < C, isnothing but condition Hr for the function ϕ the theorem follows. �

Remark 1 It is clear that (3.2) implies that the theorem can be improved to: If ϕ sat-isfies Hr then the vector valued kernel K∗(y) ≡ {Kn,m(y)} satisfies that there existsCr such that for any i ∈ Z and any x ∈ (0,2i )

∑

j>i

2j

(1

2j

∫

2j <y≤2j+1

∥∥K∗(y − x) − K∗(y)

∥∥r

l∞ dy

) 1r ≤ Cr, (3.4)

and therefore belongs to Hr .


Our next result is that to obtain Kn,m ∈ H1 we need only (a) and (b).

Theorem 9 If ϕ satisfies (a) and (b) then ϕ, and therefore Kn,m, satisfy H1.

Proof We must prove that with x, i and j as above,∑

j>i

∫Ij

|ϕ(y − x)−ϕ(y)|dy ≤ C. If we observe that because ϕ is nonincreasing the term ϕ(y −x)−ϕ(y)

is always positive and that∫Ij

ϕ(y − x)dy = ∫ 2j+1−x

2j −xϕ(y) dy we have:

∑

j>i

∫

Ij

|ϕ(y − x) − ϕ(y)|dy =∫ 2i+1

2i+1−x

ϕ ≤ Cx

2i+1≤ C.

The standard Calderón-Zygmund theory of singular integrals [6], says that if a kernelsatisfies H1 and has bounded Fourier Transform then the corresponding operator isbounded in Lp(R), 1 < p, and of weak type (1,1) with respect to Lebesgue measurewith constants that depend only on p, the H1 constant and the L∞ norm of the FourierTransform of the kernel. Our next step is therefore to study the Fourier Transform ofKn,m. �

Theorem 10 There exists C such that for any (n,m) ∈ N × N and ξ ∈ R,

|Kn,m (ξ)| ≤ C.

Proof We rewrite Kn,m as Ln,m + Hn,m where

Ln,m =m∑

−n

vk

(sk+1

skχIk

− χIk+1

)

(y)ϕ(y).

It is easy to see that Hn,m ∈ L1 with norm bounded by∑ |sk − sk+1|. This implies

that ‖Hn,m(ξ)‖∞ ≤ ∑ |sk − sk+1| ≤ C. It is enough then, to consider Ln,m. Observethat Ln,m is sum of terms of the form Ak(y) = vk(

sk+1sk

χIk− χIk+1)(y)ϕ(y) and for

each k, Ak has average 0 and it is supported on Ik ∪ Ik+1. We will prove that thereexists C independent of k such that

(1) |Ak(ξ)| ≤ C|ξ |2k, ξ ∈ R.(2) |Ak(ξ)| ≤ C 1

|ξ |2k , ξ �= 0.

Once we have proved this we have

∣∣Ln,m (ξ)

∣∣ =

∑

{k:2k≥|ξ |−1}

∣∣Ak(ξ)

∣∣ +

∑

{k:2k<|ξ |−1}

∣∣Ak(ξ)

∣∣.

Using the first estimate on the second term of the sum and the second on the other weare finished.

To obtain our first estimate we use that Ak has average 0. From properties (a)and (b) it follows that sk+1 ≤ 2sk and then |Ak(x)x| ≤ C, for some C that does not


depend on k. Now

∣∣Ak(ξ)

∣∣ ≤ C

∫

Ik∪Ik+1

∣∣e−ixξ − 1

∣∣Ak(x)dx ≤ C

∫

Ik∪Ik+1

|ξ |dx ≤ C|ξ |2k.

For the second we have,

∣∣Ak(ξ)

∣∣ ≤ 2

∣∣∣∣

∫

Ik

e−ixξ ϕ(x) dx

∣∣∣∣ +

∣∣∣∣

∫

Ik+1

e−ixξ ϕ(x) dx

∣∣∣∣ ,

and it is enough to control one of them. We will use integration by parts. Let usassume ϕ is absolutely continuous. We have

∣∣∣∣

∫

Ik

e−ixξ ϕ(x) dx

∣∣∣∣ =

∣∣∣∣∣

[ϕ(x)e−ixξ

−iξ

]2k+1

2k

−∫

Ik

ϕ′(x)e−ixξ

−iξdx

∣∣∣∣∣

≤ Cϕ(2k)

|ξ | + 1

|ξ |∫ 2k+1

2k

∣∣ϕ′(x)

∣∣dx.

But ϕ is nonincreasing which means ϕ′(x) ≤ 0 and then we can dominate our lastterm by

C

(ϕ(2k)

|ξ | + ϕ(2k+1

)

|ξ |

)

≤ C

|ξ |2k.

If the function ϕ is not absolutely continuous we use an approximation argument. Letus fix a C1 function τ , with support on (−1,1), which is even, decreasing in (0,1)

and has integral 1.We need to prove that (χIk

ϕ)(ξ) ≤ C|ξ |2k . For each ε > 0 define τε(x) = 1

ετ ( x

ε).

Let us write

(χIkϕ)(ξ) = (χIk

ϕ − χIk(ϕ ∗ τε)) (ξ) + (χIk

(ϕ ∗ τε)) (ξ).

If ε < 2k−2, the first term is easily controlled since

∥∥(χIk

ϕ − χIk(ϕ ∗ τε))

∥∥∞ ≤ ‖χIk

ϕ − χIk(ϕ ∗ τε)‖1

≤ ‖χ[2k−1,2k+2)ϕ − (χ[2k−1,2k+2)ϕ) ∗ τε‖1,

which tends to 0 because χ[2k−1,2k+2)ϕ ∈ L1(R). It will be enough to prove that foreach k there exist an εk and a constant C, that does not depend on k, such that forevery ε < εk and any ξ ∈ R,

∣∣(χIk

(ϕ ∗ τε)) (ξ)∣∣ ≤ C

|ξ |2k.


Let us take any ε < 2k−2. As before, using integration by parts we get

∣∣(χIk

(ϕ ∗ τε))) (ξ)∣∣ ≤ |ϕ ∗ τε(2k)| + |ϕ ∗ τε(2k+1)|

|ξ |

+ 1

|ξ |∫ 2k+1

2k

∣∣(ϕ ∗ τε)

′(x)∣∣dx.

We now observe that (ϕ ∗ τε)′(x) ≤ 0 for any x. Indeed, (ϕ ∗ τε)

′(x) = ∫ ε

−εϕ(x − y)

τ ′ε(y) dy. Writing the integral as the sum of the integrals in (−ε,0) and (0, ε), chang-

ing in the first integral y into −y and using that τ ′ε is an odd function we have

(ϕ ∗ τε)′(x) =

∫ ε

0(ϕ(x − y) − ϕ(x + y))τ ′

ε(y) dy.

This is negative because ϕ is nonincreasing and τ ′ε(y) ≤ 0 for y > 0. This means that

we can write

∫ 2k+1

2k

∣∣(ϕ ∗ τε)

′(x)∣∣dx = (ϕ ∗ τε)

(2k

) − (ϕ ∗ τε)(2k+1).

But

(ϕ ∗ τε)(2k

) =∫ ε

−ε

ϕ(2k − y

)τε(y) dy ≤ ϕ

(2k−1)

∫ ε

−ε

τε(y) dy

≤ C

2k

∫ ε

−ε

τε(y) dy = C

2k.

Obviously the same argument works for the point 2k+1.We have then proved that our operators Tn,m are uniformly bounded in L2. Since

the kernels Kn,m satisfy H1 uniformly, the standard theory of singular integrals givesus that, for any 1 < p, Tn,m are bounded operators in Lp with a constant independentof n and m. Now Theorem 2 follows using Theorem 1.

For the maximal operator T ∗ we can not use the Fourier Transform argument. Thealternative is to control it by operators that we know are bounded. Observe that ouroperators Tn,m are one-sided in the sense that Tn,mf (x) = Tn,m(f χ(−∞,x)), for any x

and f . It is then natural to try and control them with operators that are also one-sided,namely the one-sided Hardy-Littlewood maximal operators M−

s . �

Theorem 11 Assume that the kernels Kn,m satisfy Hr for some r > 1 with constantindependent of (n,m). Then there exists a constant C such that for any N ≥ 0,

T ∗Nf (x) ≤ C

(M−(TN,Nf )(x) + M−

r ′ f (x)).

Proof Since we are dealing with translation invariant operators, it is enough to proveour inequality at x = 0. We will prove that if n ≤ N and m ≤ N ,

|Tn,mf (0)| ≤ C(M−(TN,Nf )(0) + M−

r ′ f (0)),


with a constant independent of n,m, and N .

Tn,mf (0) =m∑

−n

vk

(∫ 2k+1

2k

f (−y)ϕ(y) dy −∫ 2k+2

2k+1f (−y)ϕ(y) dy

)

=m∑

−n

vk

(∫ −2k

−2k+1f (y)ϕ(−y)dy −

∫ −2k+1

−2k+2f (y)ϕ(−y)dy

)

.

This means that

Tn,mf (0) = Tn,m(f χ(−2m+2,−2−n))(0)

and we can write

Tn,mf (0) = Tn,m(f χ(−2−n+1,−2−n))(0) + Tn,m(f χ(−2m+2,−2−n+1))(0)

≡ Tn,mf1(0) + Tn,mf2(0).

Now

|Tn,mf1(0)| =∣∣∣∣v−n

∫ −2−n

−2−n+1f (y)ϕ(−y)dy

∣∣∣∣

≤ C2n

∫ −2−n

−2−n+1|f (y)|dy ≤ CM−(f )(0).

For f2 we proceed as follows:

|Tn,mf2(0)| ≤ C2n

∫ −2−n−2

−2−n−1|TN,Nf (x) − Tn,m(f2)(0)|dx

+ C2n

∫ −2−n−2

−2−n−1|TN,Nf (x)|dx

≤ C2n

∫ −2−n−2

−2−n−1|TN,Nf (x) − Tn,m(f2)(0)|dx

+ CM−(TN,Nf )(0).

We need to control

2n

∫ −2−n−2

−2−n−1|TN,Nf (x) − Tn,m(f2)(0)|dx.


Let us write TN,Nf (x) − Tn,mf2(0) = TN,Nf1(x) + TN,Nf2(x) − Tn,mf2(x) +Tn,mf2(x) − Tn,mf2(0). Then

2n

∫ −2−n−2

−2−n−1|TN,Nf1| ≤ C

(

2n

∫ −2−n−2

−2−n−1|TN,Nf1|r ′

) 1r′

≤ C

(

2n

∫

R

|f1|r ′) 1

r′

= C

(

2n

∫ −2−n

−2−n+1|f |r ′

) 1r′

≤ CM−r ′ f (0).

Next

TN,Nf2(x) − Tn,mf2(x) =−n−1∑

−N

vk(Dk − Dk+1)f2(x)

+N∑

m+1

vk(Dk − Dk+1)f2(x).

But if x ∈ (−2−n−1,−2−n−2) and y ≤ 2−n−1, then x − y > −2−n and is not in the

support of f2. This means that∫ 2k+1

2k f2(x−y)ϕ(y) dy = 0 for k ≤ −n−2. Therefore,in the first sum above there are at most 2 terms and each of them, in absolute value,is dominated again by CM−f (0). Now if y > 2m+2, x ∈ (−2−n−1,−2−n−2), thenx − y < −2m+2 which is not in the support of f2. Therefore,

∣∣∣∣

N∑

m+1

vk(Dk − Dk+1)f2(x)

∣∣∣∣ = |vm+1(Dm+1 − Dm+2)f (x)| ≤ CM−f (0).

We have thus proved that

|Tn,mf2(0)| ≤ C

(

M−(TN,Nf )(0) + M−r ′ f (0)

+ 2n

∫ −2−n−2

−2−n−1|Tn,mf2(x) − Tn,mf2(0)|dx

)

.

So far we have not made use of the assumption that Kn,m ∈ Hr . We will use it toprove that

|Tn,mf2(x) − Tn,mf2(0)| ≤ CM−r ′ f (0).

Indeed,

|Tn,mf2(x) − Tn,mf2(0)|

=∣∣∣∣

∫ −2−n+1

−2m+2(Kn,m(x − y) − Kn,m(−y))f2(y) dy

∣∣∣∣


=∣∣∣∣

m+1∑

k=−n+1

∫ −2k

−2k+1(Kn,m(x − y) − Kn,m(−y))f (y) dy

∣∣∣∣

≤m+1∑

k=−n+1

2k

(1

2k

∫ −2k

−2k+1|Kn,m(x − y) − Kn,m(−y)|r dy

) 1r

×(

1

2k

∫ −2k

−2k+1|f (y)|r ′

dy

) 1r′

≤ CM−r ′ f (0)

m+1∑

k=−n+1

2k

(1

2k

∫ −2k

−2k+1|Kn,m(x − y) − Kn,m(−y)|r dy

) 1r

≤ CM−r ′ f (0)

m+1∑

k=−n+1

2k

(1

2k

∫ 2k+1

2k

|Kn,m(y − (−x)) − Kn,m(y) dy|r) 1

r

≤ CM−r ′ f (0).

�

Proof of Theorem 3 Let us fix some q > r ′. The preceding theorem gives that foreach N , T ∗

N is bounded in Lq(R) with a constant independent of N . By monotoneconvergence T ∗ is bounded in Lq(R). Now the operator T ∗ can be written as

T ∗f (x) = ∥∥K∗f (x)

∥∥

l∞ .

We have already shown that if ϕ ∈ Hr then K∗ ∈ Hr therefore we can applyTheorem 1 with X = l∞. Using an approximation argument we may assume∫

T ∗f (x)pw(x)dx is finite, and obtain that if 0 < p < ∞ and w ∈ A−∞,

∫T ∗f (x)pw(x)dx ≤ C

∫M−

r ′ f (x)pw(x)dx.

�

Proof of Theorem 4 Just use Theorem 3 and w ∈ A−p

r′. �

Proof of Theorem 5 Let p ∈ (1,∞) and w ∈ A−p , we know that there exists t ∈ (1,p)

such that w ∈ A−t . Choosing r so that r ′ = p

twe get,

∫(T ∗f )pw ≤ C

∫(M−

r ′ f)p

w = C

∫(M−|f |r ′)t

w ≤ C

∫|f |pw.

�


Theorem 12 If the vector valued kernel K∗ satisfies (3.1) for every 1 ≤ r < ∞, thenfor any w ∈ A−

1 there exists a constant C such that

w{x : ∣∣T ∗f (x)

∣∣ > d

} ≤ C

d

∫|f (x)|w(x)dx.

The proof uses the following lemma which is of independent interest.

Lemma 1 Let B be a function supported on an interval I = (a, b) and such that∫IB = 0 and let w ∈ A−

1 . If the vector valued kernel K∗ satisfies (3.1) for every1 ≤ r < ∞, then there exists C such that

∫

x>b+(b−a)

T ∗(B)(x)w(x)dx ≤ C

∫

I

|B(x)|w(x)dx.

Proof We may and do assume that I = (0,2i ). Let us fix j > i. We know that thereexists r > 1 so that wr ∈ A−

1 . For this r we have:

∫

y>2i+1

∣∣T ∗B(y)

∣∣w(y)dy

=∑

j>i

∫ 2j+1

2j

∥∥∥∥

{∫

I

(Kn,m(y − x) − Kn,m(y))B(x)dx

}

n,m

∥∥∥∥

l∞w(y)dy

≤∑

j>i

2j

∫

I

|B(x)|(

1

2j

∫ 2j+1

2j

∥∥{Kn,m(y − x) − Kn,m(y)}n,m

∥∥r ′

l∞ dy

) 1r′

×(

1

2j

∫ 2j+1

2j

w(y)r dy

) 1r

dx.

Now wr ∈ A−1 tells us that

( 12j

∫ 2j+1

2j wr(y) dy) 1

r ≤ Cw(x) for a.e. x ∈ I .Therefore,

∫

y>2i+1

∣∣T ∗B(y)

∣∣w(y)dy

≤∫

I

∑

j>i

2j

(1

2j

∫

2j <y≤2j+1‖{Kn,m(y − x) − Kn,m(y)}n,m‖r ′

l∞ dy

) 1r′

× |B(x)|w(x)dx

and the lemma follows. �

Proof of Theorem 12 The argument is a one-sided version of standard Calderón-Zygmund theory. If we look at a similar result in [2] (Theorem 7.4), we see that oneonly needs boundedness of the operator in Lp(w) for some p > 1 and the lemma


above. We give the proof just to make the article self contained. Assume f ≥ 0. LetOd = {M−f (x) > d}. It is well known that then d = 1

|Ii |∫Ii

f ≡ fIi, where Ii are

the connected components of Od . We decompose f in the usual fashion f = g + b

where b = ∑(f − fIi

)χIi≡ ∑

bi . Observe that each bi is supported on an intervaland has average 0 and that g(x) ≤ d for a.e. x. Then

∫g(y)w(y)dy ≤

∫

{y /∈Od }f (y)w(y)dy +

∑∫

Ii

w(y) dyfIi

=∫

{y /∈Od }f (y)w(y)dy + d

∫

Od

w(y)dy ≤ C

∫f (y)w(y)dy

because the operator M− is of weak type (1,1) with respect to the measure w(y)dy.Let Pd be the union of 2Ii where if Ii = (a, b) then 2Ii = (a, b + (b − a)). Observethat we have duplicated the interval in one direction only. The weights w ∈ A−

1 do nothave, in general, the doubling property but they have a one-sided doubling propertyi.e., w(2I ) ≤ Cw(I) and therefore w(Pd) ≤ C

d

∫f w. Now we write

w{x : T ∗f (x) > d

} ≤ w

{

x : T ∗g(x) >d

2

}

+ w(Pd)

+ w

{

x /∈ Pd : T ∗b(x) >d

2

}

≡ I + II + III.

We have already controlled II . For I we use that T ∗ is bounded in Lp(w), p > 1,because A−

1 ⊂ A−p for any p > 1 and we obtain,

I ≤ C

dp

∫T ∗g(y)pw(y)dy

≤ C

dp

∫|g(y)|pw(y)

≤ C

d

∫|g(y)|w(y)dy

≤ C

d

∫|f (y)|w(y)dy,

because |g| ≤ d and we have already seen that

∫|g(y)|w(y)dy ≤ C

∫|f (y)|w(y)dy.

For III we use Remark 1 and the Lemma 1 to obtain,

III ≤ C

d

∑∫

R−2Ii

T ∗bi(y)w(y)dy ≤ C

d

∑∫

Ii

|bi(y)|w(y)dy.


Since the intervals are pairwise disjoint and on each of them b = bi , we can dominatethe last term by

C

d

∫|b(y)|w(y)dy = C

d

∫|f (y) − g(y)|w(y)dy ≤ C

d

∫|f (y)|w(y)dy.

�

3.1 Results for Orlicz Spaces

One of the main ingredients in the proofs above is Theorem 1 from which we getTheorems 2 and 3. Since Theorem 1 has an Orlicz space version, it is natural to useit to prove the corresponding version of Theorem 3.

We recall some basic definitions and refer the reader to [1] for details. A functionB : [0,∞) → [0,∞) is a Young function if it is continuous, convex, increasing andsatisfies B(0) = 0 and B(t) → ∞ as t → ∞. The Luxemburg norm of a function f ,given by B is

||f ||B = inf

{

λ > 0 :∫

B

( |f |λ

)

≤ 1

}

,

and so the B-average of f over I is

||f ||B,I = inf

{

λ > 0 : 1

|I |∫

I

B

( |f |λ

)

≤ 1

}

.

We will denote by B the complementary function associated to B (see, [1]). Thegeneralization of Hölder’s inequality is the following:

1

|I |∫

I

|f g| ≤ 2||f ||B,I ||g||B,I .

Definition 3 For each locally integrable function f , the one-sided maximal operatorsassociated to the Young function B are defined by

M+B f (x) = sup

x<b

‖f ‖B,(x,b) and M−B f (x) = sup

a<x‖f ‖B,(a,x).

The Orlicz version of Hr is the following.

Definition 4 Let A be a Young function, we say that the kernel K , taking valuesin the Banach space X, satisfies the HA-Hörmander condition, if there are numberscA > 1 and CA > 0, such that for any x ∈ R and R > cA|x|,

∞∑

m=1

2mR∥∥‖K(x − ·) − K(·)‖Xχ(2mR<|y|≤2m+1R)

∥∥

A,B(0,2m+1R)≤ CA.

The following is proved in [8].

Theorem 13 Let K be a kernel with support on (−∞,0), taking values in the Banachspace X, such that the operator Tf (x) = (K ∗ f )(x) is bounded from Lq(R) to


LqX(R) for some 1 < q < ∞, i.e.,

∫

R

‖Tf (x)‖qX dx ≤ C

∫

R

|f (x)|q dx.

If A is a Young function and K ∈ HA then for any 0 < p < ∞ and any w ∈ A+∞, thereexists C so that

∫

R

‖Tf ‖pXw ≤ C

∫

R

(M+

Af

)pw,


As in the Hr case, the kernel K is as good as ϕ.

Theorem 14 Let A be a Young function such that∑ 1

A−1(2i )< ∞. If ϕ satisfies HA

so does Kn,m with a constant independent of n and m.

Proof The proof is like the one in the Hr case but now the term in (3.2) is

∑

j>i

2j‖ϕ(y − x)χ(2j ,2j +x)‖A,(2j ,2j+1),

where 0 < x < 2i . We dominate |ϕ(y − x)| by C|y − x|−1 that cancels the2j and it is enough control ‖χ(2j ,2j +x)‖A,(2j ,2j+1) but it is easy to see that

‖χ(2j ,2j +x)‖A,(2j ,2j+1) ≤ C 1A−1(2j−i )

. �

Theorem 15 Let us assume that ϕ satisfies HA for some Young function A such that∑ 1A−1(2i )

< ∞. Then for any 0 < p < ∞ and any w ∈ A−∞, there exists C such that

∫|Tn,mf (x)|pw(x)dx ≤ C

∫M−

Af (x)pw(x)dx,


Proof We know from the preceding theorem that the Kn,m satisfy HA with constantindependent of n,m. The result follows from Theorem 13. �

Theorem 16 If A is a Young function such that∑ 1

A−1(2i )< ∞, ϕ satisfies HA and

M−A

is a bounded operator in Ls for some s, 1 < s < ∞, then for any w ∈ A−∞, thereexists C such that

∫ ∣∣T ∗f (x)

∣∣pw(x)dx ≤ C

∫M−

Af (x)pw(x)dx,



Proof It is enough to prove that for any 1 < t < s there exists C such that

T ∗Nf (x) ≤ C

(M−(TN,Nf )(x) + M−

t f (x) + M−A

f (x)).

Looking back at the proof in the Hr setting, we are led to estimate

Tn,mf (0) = Tn,m(f χ(−2−n+1,−2−n))(0) + Tn,m(f χ(−2m+2,−2−n+1))(0)

≡ Tn,mf1(0) + Tn,mf2(0).

As before

|Tn,mf1(0)| ≤ CM−(f )(0)

and

|Tn,mf2(0)| ≤ C2n

∫ −2−n−2

−2−n−1|TN,Nf (x) − Tn,m(f2)(0)|dx

+ CM−(TN,Nf )(0).

For the first term we write, as in the Hr case,

TN,Nf (x) − Tn,mf2(0) = TN,Nf1(x) + TN,Nf2(x) − Tn,mf2(x)

+ Tn,mf2(x) − Tn,mf2(0).

TN,Nf1(x) and TN,Nf2(x) − Tn,mf2(x) are dominated by CM−t f (0), using that

the TN,N are uniformly bounded in Lt . The Hr condition is needed in the term|Tn,mf2(x) − Tn,mf2(0)|. If we repeat the argument of Theorem 11 using the Or-licz version of Hölder’s inequality it is easy to see that this is less than a constanttimes M−

A. Now we know that the kernel K∗ satisfies HA and the preceding inequal-

ity tells us that the T ∗N are uniformly bounded in Ls so we can apply Theorem 13 and

finish the proof. �

4 The Ergodic Case

Once we have a weighted theory there is a way to extend the result to the meanbounded ergodic setting. We include it because it is not widely known and also toillustrate the interest of having a weighted theory. The argument was used for thefirst time in [13] to obtain a dominated ergodic estimate for mean bounded positiveoperators. The idea is that, once we have a weighted theory, we can use the CalderónTransference Method in the setting of mean bounded groups of operators. In orderto extend Calderón method from the measure preserving transformation case to thesetting of mean bounded group of operators, we use the weighted result, the fact thatfor invertible groups of positive operators there is a “change of variables” formula inLp and an idea of José Luis Rubio de Francia on his famous proof of the factorizationof Ap weights. We summarize the steps in the following.

Remark 2 [Change of variables formula (see, [3])]


(1) For each t ∈ R there exists a function Ht such that for any f ∈ Lp(X),

∫

X

|Stf (x)|pHt(x) dμ(x) =∫

X

|f (x)|p dμ(x).

(2) Furthermore, if (St )∗ denotes the adjoint of St , then

Ht = (S∗−t g

p)(

Stgp′)1−p

for any function g ≥ 0, g ∈ Lpp′(dμ).

Remark 3 For any t ∈ R, x ∈ X, N ∈ N and f ∈ Lp(X) we have

St

(τ ∗Nf

)(x) = τ ∗

N(Stf )(x).

Proof Since the St are linear and positive, we have that for any n,m < N

|τn,m(Stf )(x)| ≤ St (|τn,mf |)(x) ≤ St

(τ ∗Nf

)(x).

Therefore, τ ∗N(Stf )(x) ≤ St (τ

∗Nf )(x). Since a similar inequality holds for S−t , we

may write

St

(τ ∗Nf

)(x) = St

(τ ∗N

(S−t Stf

)(x)

) ≤ St

(S−t

(τ ∗N(Stf )(x)

) = τ ∗N(Stf )(x).

�

Remark 4 The functions t → Ht(x) satisfy condition A−p with a constant indepen-

dent of x ∈ X.

Proof The proof is a continuous version of the original argument in [13]. The detailscan be found in [3]. It is here that the Rubio de Francia method of factorizing Ap

weights play an essential role. For those familiar with the theory of Ap weights it isclear that (2) would be the proof that Ht is in A+

p if we knew that (S−t )∗gp and Stg

p′

are A+1 and A−

1 weights, respectively. This does not need to be the case but, playingwith the fact that g is arbitrary, the difficulty can be overcame using the Rubio deFrancia algorithm. The following identity is the final ingredient in the proof. �

Remark 5 τ ∗N(Stf )(x) = T ∗

Nfx(t), where fx(t) is the real variable function t →Stf (x).


Proof of Theorem 7 For any R > 0 we have

∫

X

(τ ∗Nf (x)

)pdμ(x) = 1

R

∫ R

0

∫

X

∣∣Stτ

∗Nf (x)

∣∣pHt(x) dμ(x)dt

≤∫

X

1

R

∫ R

0

∣∣T ∗

N(fxχ[0,R+N ])(t)∣∣pHt(x) dt dμ(x)

≤ C

∫

X

1

R

∫ R+N

0|fx(t)|pHt(x) dt dμ(x)

= C1

R

∫ R+N

0

∫

X

|Stf (x)|pHt(x) dμ(x)dt

= C1

R

∫ R+N

0

∫

X

|f (x)|p dμ(x)dt

= CR + N

R

∫

X

|f (x)|p dμ(x).

Letting, first R, and then N go to infinity we obtain∫

X

(τ ∗f (x)

)pdμ(x) ≤ C

∫

X

|f (x)|p dμ(x).

�

Acknowledgements We want to thank the referee for his helpful comments and suggestions.

References

1. Bennett, C., Sharpley, R.: Interpolation of Operators. Academic Press, New York (1998)2. Bernardis, A.L., Lorente, M., Martín-Reyes, F.J., Martínez, M.T., de la Torre, A., Torrea, J.L.: Differ-

ential transforms in weighted spaces. J. Fourier Anal. Appl. 12(1), 83–103 (2006)3. Bernardis, A.L., Lorente, M., Martín-Reyes, F.J., Martínez, M.T., de la Torre, A.: Differences of er-

godic averages for Cesàro bounded operators. Q. J. Math. 58, 137–150 (2007)4. Demeter, C.: Almost everywhere convergence of series in L1. Proc. Am. Math. Soc. 133(8), 2319–

2326 (2005)5. Duoandikoetxea, J., Rubio de Francia, J.L.: Maximal and singular integral operators via Fourier trans-

form estimates. Invent. Math. 84, 541–566 (1986)6. García-Cuerva, J., Rubio de Francia, J.L.: Weighted Norm Inequalities and Related Topics. North

Holland Mathematics Studies, vol. 116 (1985)7. Jones, R.L., Rosenblatt, J.: Differential and ergodic transforms. Math. Ann. 323, 525–546 (2002)8. Lorente, M., Riveros, M.S., de la Torre, A.: Weighted estimates for singular integral Operators satis-

fying Hörmander’s conditions of Young type. J. Fourier Anal. Appl. 11(5), 497–509 (2005)9. Martell, J.M., Pérez, C., Trujillo-González, R.: Lack of natural weighted estimates for some singular

integral operators. Trans. Am. Math. Soc. 357(1), 385–396 (2005)10. Martín-Reyes, F.J., Ortega, P., de la Torre, A.: Weighted inequalities for one-sided maximal functions.

Trans. Am. Math. Soc. 319(2), 517–534 (1990)11. Martín-Reyes, F.J., Pick, L., de la Torre, A.: A+∞ condition. Can. J. Math. 45, 1231–1244 (1993)12. Martín-Reyes, F.J., de la Torre, A.: One sided BMO spaces. J. Lond. Math. Soc. 2(49), 529–542

(1994)


13. Martín-Reyes, F.J., de la Torre, A.: The dominated ergodic theorem for mean bounded invertible,positive operators. Proc. Am. Math. Soc. 104, 69–75 (1988)

14. Rubio de Francia, J.L., Ruiz, F.J., Torrea, J.L.: Calderón-Zygmund theory for vector-valued functions.Adv. Math. 62, 7–48 (1986)

15. Sawyer, E.: Weighted inequalities for the one-sided Hardy-Littlewood maximal functions. Trans. Am.Math. Soc. 297, 53–61 (1986)

Weighted and Ergodic Theorems for Series of Differences of Convolutions

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