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ISSN 1402-1544 ISBN 978-91-7583-392-7 (print) ISBN 978-91-7583-393-4 (pdf) Luleå University of Technology 2015

George Tephnadze Martingale Hardy Spaces and Summability of the One Dimensional Vilenkin-Fourier Series

Department of Engineering Sciences and Mathematics Division of Mathematical Sciences

Martingale Hardy Spaces and Summability of the One Dimensional Vilenkin-Fourier Series

George Tephnadze

Martingale Hardy spaces and summability of the one dimensional Vilenkin-Fourier series

George Tephnadze

Department of Engineering Sciences and Mathematics Lule˚a University of Technology SE-971 87 Lule˚a, Sweden and Department of Mathematics Faculty of Exact and Natural Sciences Ivane Javakhishvili Tbilisi State University Chavchavadze str. 1, Tbilisi 0128 [email protected]

8 October, 2015

Printed by Luleå University of Technology, Graphic Production 2015 ISSN 1402-1544 ISBN 978-91-7583-392-7 (print) ISBN 978-91-7583-393-4 (pdf) Luleå 2015 www.ltu.se

August 30, 2015


3

C ONTENTS 1

2

Preliminaries 1.1

Basic notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11

1.2

Definition and examples of Nörlund means and its maximal operators . . .

14

1.3

Dirichlet kernels and Lebesgue constants with respect to Vilenkin systems .

18

1.4

Fejér Kernels with respect to Vilenkin systems . . . . . . . . . . . . . . .

21

1.5

Nörlund Kernels with respect to Vilenkin systems . . . . . . . . . . . . . .

27

1.6

Introduction to the theory of Martingale Hardy spaces . . . . . . . . . . .

39

1.7

Examples of p-atoms and Hp maringales . . . . . . . . . . . . . . . . . .

43

Fourier coefficients and partial sums of Vilenkin-Fourier series on Martingale Hardy spaces 55 2.1

Some classical results on the Vilenkin-Fourier coefficients and partial sums of Vilenkin-Fourier series . . . . . . . . . . . . . . . . . . . . . . . . . .

55

2.2

Estimations of the Vilenkin-Fourier coefficients on martingale Hardy spaces

57

2.3

Hardy and Paley type inequalities on martingale Hardy spaces . . . . . . .

59

2.4

Maximal operators of partial sums of Vilenkin-Fourier series on martingale Hardy spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

63

2.5

Norm convergence of partial sums of Vilenkin-Fourier series on martingale Hardy spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

73

2.6

Strong convergence of partial sums of Vilenkin-Fourier series on martingale Hardy spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

77

An application concerning estimations of Vilenkin-Fourier coefficients . .

80

2.7

3

11

Vilenkin-Fejér means on martingale Hardy spaces

83

3.1

Some classical results on Vilenkin-Fejér means . . . . . . . . . . . . . . .

83

3.2

Divergence of Vilenkin-Fejér means on martingale Hardy spaces . . . . . .

85

3.3

Maximal operators of Vilenkin-Fejér means on martingale Hardy spaces . .

88

3.4

Norm convergence of Vilenkin-Fejér means on martingale Hardy spaces . .

106

3.5

Strong convergence of Vilenkin-Fejér means on martingale Hardy spaces .

112

G.Tephnadze

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4 4

Contents

August 30, 2015

Vilenkin-Nörlund means on martingale Hardy spaces

121

4.1

Some classical results on Vilenkin-Nörlund means . . . . . . . . . . . . .

121

4.2

Maximal operators of Nörlund means on martingale Hardy spaces . . . . .

122

4.3

Strong convergence of Nörlund means on martingale Hardy spaces . . . . .

143

4.4

Maximal operators of Riesz and Nörlund logarithmic means on martingale Hardy spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

150

Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

155

4.5

,

Key words: Harmonic analysis, Vilenkin groups, Vilenkin systems, Lebesgue spaces, Weak-Lp spaces, modulus of continuity, Vilenkin-Fourier coefficients, partial sums of VilenkinFourier series, Lebesgue constants, Fejér means, Cesàro means, Nörlund means, Riesz and Nörlund logarithmic means, martingales, martingale Hardy spaces, maximal operators, strong convergence, inequalities, approximation.

6

Contents

August 30, 2015

Abstract The classical theory of Fourier series deals with decomposition of a function into sinusoidal waves. Unlike these continuous waves the Vilenkin (Walsh) functions are rectangular waves. Such waves have already been used frequently in the theory of signal transmission, multiplexing, filtering, image enhancement, codic theory, digital signal processing and pattern recognition. The development of the theory of Vilenkin-Fourier series has been strongly influenced by the classical theory of trigonometric series. Because of this it is inevitable to compare results of Vilenkin series to those on trigonometric series. There are many similarities between these theories, but there exist differences also. Much of these can be explained by modern abstract harmonic analysis, which studies orthonormal systems from the point of view of the structure of a topological group. In this PhD thesis we discuss, develop and apply this fascinating theory connected to modern harmonic analysis. In particular we make new estimations of Vilenkin-Fourier coefficients and prove some new results concerning boundedness of maximal operators of partial sums. Moreover, we derive necessary and sufficient conditions for the modulus of continuity so that norm convergence of the partial sums is valid and develop new methods to prove Hardy type inequalities for the partial sums with respect to the Vilenkin systems. We also do the similar investigation for the Fejér means. Furthermore, we investigate some Nörlund means but only in the case when their coefficients are monotone. Some well-know examples of Nörlund means are Fejér means, Cesàro means and Nörlund logarithmic means. In addition, we consider Riesz logarithmic means, which are not example of Nörlund means. It is also proved that these results are the best possible in a special sense. As applications both some well-known and new results are pointed out. This PhD is written as a monograph consisting of four Chapters: Preliminaries, Fourier coefficients and partial sums of Vilenkin-Fourier series on martingale Hardy spaces, VilenkinFejér means on martingale Hardy spaces, Vilenkin-Nörlund means on martingale Hardy spaces. It is based on 15 papers with the candidate as author or coauthor, but also some new results are presented for the first time. In Chapter 1 we first present some definitions and notations, which are crucial for our further investigations. After that we also define some summabilitity methods and remind about some classical facts and results. We investigate some well-known results and prove new estimates for the kernels of these summabilitity methods, which are very important to prove our main results. Moreover, we define martingale Hardy spaces and construct martingales, which help us to prove sharpness of our main results in the later chapters. Chapter 2 is devoted to present and prove some new and known results about VilenkinFourier coefficients and partial sums of martingales in Hardy spaces. First, we show that Fourier coefficients of martingales are not uniformly bounded when 0 < p < 1. By applying these results we prove some known Hardy and Paley type inequalities with a new method. After that we investigate partial sums with respect to Vilenkin systems and prove boundedness of maximal operators of partial sums. Moreover, we find necessary and sufficient ,

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7

conditions for the modulus of continuity for which norm convergence of partial sums hold and we present a new proof of a Hardy type inequality for it. In Chapter 3 we investigate some analogous problems concerning the partial sums of Fejér means. First we consider some weighted maximal operators of Fejér means and prove some boundedness results for them. After that we apply these results to find necessary and sufficient conditions for the modulus of continuity for which norm convergence of Fejér means hold. Finally, we prove some new Hardy type inequalities for Fejér means, which is a main part of this PhD thesis. We also prove sharpness of all our main results in this Chapter. In Chapter 4 we consider boundedness of maximal operators of Nörlund means. After that we prove some strong convergence theorems for these summablility methods. Since Fejér means, Cesàro means are well-know examples of Nörlund means some well-known and new results are pointed out. We also investigate Riesz and Nörlund logarithmic means simultaneously at the end of this chapter.

G.Tephnadze

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Contents

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Preface This PhD thesis is written as a monograph based on the following publications: [1] G. Tephnadze, The maximal operators of logarithmic means of one-dimensional Vilenkin-Fourier series, Acta Math. Acad. Paedagog. Nyházi. 27 (2011), no. 2, 245-256. [2] G. Tephnadze, Fejér means of Vilenkin-Fourier series. Studia Sci. Math. Hungar. 49 (2012), no. 1, 79-90. [3] G. Tephnadze, A note on the Fourier coefficients and partial sums of Vilenkin-Fourier series. Acta Math. Acad. Paedagog. Nyházi. 28 (2012), no. 2, 167-176. [4] G. Tephnadze, On the maximal operators of Vilenkin-Fejér means. Turkish J. Math. 37 (2013), no. 2, 308-318. [5] G. Tephnadze, On the maximal operators of Vilenkin-Fejér means on Hardy spaces. Math. Inequal. Appl. 16 (2013), no. 1, 301-312. [6] G. Tephnadze, On the Vilenkin-Fourier coefficients, Georgian Math. J. 20 (2013), no. 1, 169-177. [7] G. Tephnadze, On the partial sums of Vilenkin-Fourier series. J. Contemp. Math. Anal. 49 (2014), no. 1, 23-32. [8] G. Tephnadze, A note on the norm convergence by Vilenkin-Fejér means. Georgian Math. J. 21 (2014), no. 4, 511–517. [9] I. Blahota and G. Tephnadze, Strong convergence theorem for Vilenkin-Fejér means. Publ. Math. Debrecen 85 (2014), no. 1-2, 181-196. [10] L. E. Persson and G. Tephnadze, A note on Vilenkin-Fejér means, on the martingale Hardy space Hp , Bulletin of TICMI 18 (2014), no. 1, 55-64. [11] L. E. Persson, G. Tephnadze and P. Wall, Maximal operators of Vilenkin-Nörlund means, J. Fourier Anal. Appl. 21 (2015), no. 1, 76-94. [12] L. E. Persson, G. Tephnadze and P. Wall, Some new (Hp , Lp ) type inequalities of maximal operators of Vilenkin-Nörlund means with non-decreasing coefficients, J. Math. Inequal. (to appear). [13] L. E. Persson and G. Tephnadze, A sharp boundedness result concerning some maximal operators of Vilenkin-Fejér means, Mediterr. J. Math. (to appear). [14] I. Blahota, L. E. Persson and G. Tephnadze, On the Nörlund means of VilenkinFourier series, Czechoslovak Math. J. (to appear). [15] I. Blahota and G. Tephnadze, A note on maximal operators of Vilenkin-Nörlund means, Acta Math. Acad. Paedagog. Nyházi. (to appear). Remark: Also some new results which can not be found in these papers appear in this PhD thesis for the first time.

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August 30, 2015


9

Acknowledgement It is a pleasure to express my warmest thanks to my supervisors Professors Lars-Erik Persson, Ushangi Goginava and Peter Wall for the attention to my work, their valuable remarks and suggestions and for their constant support and help. I am also grateful to Doctors Yulia Koroleva, John Fabricius and Professors Niklas Grip, Maria Alessandra Ragusa and Natasha Samko for helping me with several practical things. Moreover, I appreciate very much the warm and friendly atmosphere at the Department of Mathematics at Lule˚a University of Technology, which helps me to do research more effectively. I thank my Georgian colleagues from Tbilisi State University for their interest in my research and for many fruitful discussions. In particular, I thank Professor Amiran Gogatishvili, for his support and his valuable remarks. Moreover, I am very grateful to Georgian Mathematical Union and its President Professor Roland Duduchava for all support they have given me. I also want to pronounce that the agreement about scientific collaboration and PhD education between Tbilisi State University and Lule˚a University of Technology has been very important. Especially, I express my deepest gratitude to Professor Ramaz Bochorishvili at Tbilisi State University and Professor Elisabet Kassfeldt at Lule˚a University of Technology for their creative and hard work to realize this important agreement. I thank Swedish Institute for economic support, which was very important to be able to travel and be accommodated in Sweden during this PhD program. Moreover, I thank Shota Rustaveli National Science Foundation for financial support. Finally, I thank my family for their love, understanding, patience and long lasting constant support.

G.Tephnadze

,

George Tephnadze

August 30, 2015

1

1.1

11


P RELIMINARIES

BASIC NOTATIONS

Denote by N+ the set of the positive integers, N := N+ ∪ {0}. Let m := (m0, m1 , . . .) be a sequence of positive integers not less than 2. Denote by Zmk := {0, 1, . . . , mk − 1} the additive group of integers modulo mk . Define the group Gm as the complete direct product of the groups Zmi with the product of the discrete topologies of Zmi . The direct product µ of the measures µk ({j}) := 1/mk

(j ∈ Zmk )

is the Haar measure on Gmk with µ (Gm ) = 1. In this paper we discuss bounded Vilenkin groups, i.e. the case when supn mn < ∞. The elements of Gm are represented by sequences x := (x0 , x1 , . . . , xj , . . .) xj ∈ Zmj . It is easy to give a base for the neighborhoods of Gm :

I0 (x) : = Gm , In (x) : = {y ∈ Gm | y0 = x0 , . . . , yn−1 = xn−1 } (x ∈ Gm , n ∈ N) . Let en := (0, . . . , 0, xn = 1, 0, . . .) ∈ Gm

(n ∈ N) .

If we define In := In (0) , for n ∈ N and In := Gm \ In , then ! ! N −1 N −2 N −1 −1 [ [ [ [ N[ k,l k,N IN = Is \Is+1 = IN IN , s=0

k=0 l=k+1

(1.1)

k=1

where

INk,l

 IN (0, . . . , 0, xk = 6 0, 0, ..., 0, xl 6= 0, xl+1 , . . . , xN −1 , . . .),    for k < l < N, := 6 0, xk+1 = 0, . . . , xN −1 = 0, xN , . . .),  IN (0, . . . , 0, xk =   for l = N. G.Tephnadze

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12

August 30, 2015

1. Preliminaries

The norm (or quasi-norm when 0 λ)1/p < +∞. λ>0

The norm of the space of continuous functions C(Gm ) is defined by kf kC := sup |f (x)| < c < ∞. x∈Gm

The best approximation of f ∈ Lp (Gm ) (1 ≤ p ∈ ∞) is defined as En (f, Lp ) := inf kf − ψkp , ψ∈Pn

where Pn is set of all Vilenkin polynomials of order less than n ∈ N. The modulus of continuity of f ∈ Lp (Gm ) and f ∈ C (Gm ) are defined by ωp and

ωC

1 ,f Mn

1 ,f Mn

:= sup kf (· − h) − f (·)kp h∈In

:= sup kf (· − h) − f (·)kC , h∈In

respectively. If we define the so-called generalized number system based on m in the following way : M0 := 1, Mk+1 := mk Mk

(k ∈ N),

then every n ∈ N can be uniquely expressed as n=

∞ X

nj Mj ,

j=0 0

where nj ∈ Zmj (j ∈ N+ ) and only a finite number of nj s differ from zero. Let hni := min{j ∈ N : nj 6= 0} and |n| := max{j ∈ N : nj 6= 0}, that is M|n| ≤ n ≤ M|n|+1 . Set d (n) := |n| − hni , for all n ∈ N. ,

August 30, 2015


For the natural number n = Lukomskii [38]) v (n) :=

P∞

∞ X

j=1

13

nj Mj , we define functions v and v ∗ by (for details see

|δj+1 − δj | + δ0 , v ∗ (n) :=

∞ X

j=1

δj∗ ,

j=1

where δj = sign (nj ) = sign ( nj ) and δj∗ = | nj − 1| δj and is the inverse operation for ak ⊕ bk := (ak + bk )modmk . Next, we introduce on Gm an orthonormal systems, which are called Vilenkin systems. At first, we define the complex-valued function rk (x) : Gm → C, the generalized Rademacher functions, by rk (x) := exp (2πixk /mk ) , i2 = −1, x ∈ Gm , k ∈ N . Now, define Vilenkin systems ψ := (ψn : n ∈ N) on Gm as: ψn (x) :=

∞ Y

rknk (x) ,

(n ∈ N) .

k=0

The Vilenkin systems are orthonormal and complete in L2 (Gm ) (see e.g. Vilenkin [77]). Specifically, we call this system the Walsh-Paley system when m ≡ 2. Next, we introduce some analogues of the usual definitions in Fourier-analysis. If f ∈ L1 (Gm ) we can define the Fourier coefficients, the partial sums of the Fourier series, the Dirichlet kernels with respect to Vilenkin systems in the usual manner: Z fb(n) :=

(n ∈ N) ,

f ψ n dµ, Gm

Sn f :=

n−1 X

fb(k) ψk ,

(n ∈ N+ ) ,

k=0

Dn :=

n−1 X

ψk ,

(n ∈ N+ ) ,

k=0

respectively. The n-th Lebesgue constant is defined in the following way: Ln := kDn k1 . G.Tephnadze

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14 1.2

August 30, 2015

1. Preliminaries

D EFINITION

AND EXAMPLES OF

¨ N ORLUND

MEANS AND ITS MAXIMAL OPERA -

TORS

Let {qk : k ∈ N} be a sequence of nonnegative numbers. The n-th Nörlund means for the Fourier series of f is defined by tn f :=

n 1 X qn−k Sk f, Qn k=1

(1.2)

where Qn :=

n−1 X

qk .

k=0

We always assume that q0 > 0 and lim Qn = ∞.

n→∞

In this case it is well-known that the summability method generated by {qk : k ≥ 0} is regular if and only if qn−1 lim = 0. (1.3) n→∞ Qn Concerning this fact and related basic results, we refer to [40]. The next remark is due to Persson, Tephnadze and Wall [51]: Remark 1.1 a) Let the sequence {qk : k ∈ N} be non-increasing. Then the summability method generated by {qk : k ∈ N} is regular. b) Let the sequence {qk : k ∈ N} be non-decreasing. Then the summability method generated by {qk : k ∈ N} is not always regular. Proof: Let the sequence {qk : k ∈ N} be non-increasing. Then qn−1 qn−1 1 ≤ = → 0, when n → ∞. Qn nqn−1 n According to (1.3) we conclude that in this case the summability method is regular. Now, we prove part b) of theorem and construct Nörlund mean, with non-decreasing coefficients {qk : k ∈ N}, which is not regular. Let {qk = 2k : k ∈ N}. Then Qn =

n−1 X 2k = 2n − 1 ≤ 2n k=0 ,

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and


15

qn−1 2n−1 2n−1 1 = n ≥ n = 9 0, when n → ∞. Qn 2 −1 2 2

By using again (1.3) we obtain that when the sequence {qk : k ∈ N} is non-decreasing, then the summability method is not always regular. The proof is complete. Let tn be Nörlund means with monotone and bounded sequence {qk : k ∈ N}, such that q := lim qn > c > 0. n→∞

Then, if the sequence {qk : k ∈ N} is non-decreasing, we get that nq0 ≤ Qn ≤ nq. In the case when the sequence {qk : k ∈ N} is non-increasing, then nq ≤ Qn ≤ nq0 . In both cases we can conclude that 1 qn−1 =O , when n → ∞. Qn n

(1.4)

(1.5)

In the special case when {qk = 1 : k ∈ N}, we get Fejér means n

σn f :=

1X Sk f . n k=1

The (C, α)-means (Cesàro means) of the Vilenkin-Fourier series are defined by σnα f := where Aα0 := 0, Aαn :=

n 1 X α−1 A Sk f, Aαn k=1 n−k

(α + 1) ... (α + n) , α 6= −1, −2, ... n!

It is well-known that (see e.g. Zygmund [88]) Aαn =

n X Aα−1 n−k ,

(1.6)

k=0

Aαn − Aαn−1 = Aα−1 n , G.Tephnadze

Aαn v nα .

(1.7) ,

16

August 30, 2015

1. Preliminaries

In the literature, there is the notion of Riesz means ((R, α)-means) of the Fourier series. Let βnα denote the Nörlund mean, where q0 = 1, qk = k α−1 : k ∈ N+ , that is βnα f :=

n 1 X (n − k)α−1 Sk f, 0 < α < 1. Qn k=1

It is obvious that |qn − qn+1 | = O (1) , when n → ∞. nα−2 and

q0 =O Qn

1 nα

(1.8)

, when n → ∞.

(1.9)

The n-th Nörlund logarithmic mean Ln and the Riesz logarithmic mean Rn are defined by Ln f :=

n−1 n−1 1 X Sk f 1 X Sk f , Rn f := , ln k=1 n − k ln k=1 k

respectively, where ln :=

n−1 X 1 k=1

k

.

Up to now we have considered Nörlund mean in the case when the sequence {qk : k ∈ N} is bounded but now we consider Nörlund summabilities with unbounded sequence {qk : k ∈ N}. Let α ∈ R+ , β ∈ N+ and β times

log

(β)

z }| { x := log ... logx.

If we define the sequence {qk : k ∈ N} by n o q0 = 0 and qk = log(β) k α : k ∈ N+ , then we get the class of Nörlund means with non-decreasing coefficients: κα,β n f :=

n 1 X (β) log (n − k)α Sk f. Qn k=1

First we note that κα,β n are well-defined for every n ∈ N+ , if we rewrite them as: κα,β n f =

n X log(β) (n − k)α k=1

Qn

Sk f.

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August 30, 2015


17

It is obvious that nα n log(β) α ≤ Qn ≤ n log(β) nα . 2 2 It follows that qn−1 c log(β) (n − 1)α =O ≤ Qn n log(β) nα

1 → 0, when n → ∞. n

(1.10)

For the function f we consider the following maximal operators: S ∗ f := sup |Sn f | ,

t∗ f := sup |tn f | ,

σ ∗ f := sup |σn f | ,

n∈N

n∈N

n∈N

σ α,∗ f := sup |σnα f | , R∗ f := sup |Rn f | , L∗ f := sup |Ln f | , n∈N

n∈N

n∈N

α,∗ κα,β,∗ f := sup κα,β f := sup |βnα f | . n f , β n∈N

n∈N

We also consider the following weighted maximal operators: |tn f |

∼∗

t p,α f := sup n∈N

(n + 1)1/p−1−α e t∗α f := sup n∈N

∼ α,∗

σ p f := sup n∈N

∼ α,∗

β p f := sup n∈N

(0 < p < 1/ (1 + α) , 0 < α ≤ 1),

,

|tn f | , log1+α (n + 1)

|σnα f | (n + 1)1/p−1−α |βnα f | (n + 1)1/p−1−α

(0 < α ≤ 1) ,

,

(0 < p < 1/ (1 + α) , 0 < α ≤ 1),

,

(0 < p < 1/ (1 + α) , 0 < α ≤ 1),

|σnα f | |βnα f | , βeα,∗ f := sup , (0 < α < 1) , 1+α (n + 1) (n + 1) n∈N log n∈N log α,β κn f |σn | ∗ α,β,∗ σ ep := sup , κ ep f := sup , (0 < p < 1/2) , 1/p−2 1/p−2 n∈N (n + 1) n∈N (n + 1)

σ eα,∗ f := sup

1+α

and σ e∗ f := sup n∈N

|σn f | , log2 (n + 1)

κ eα,β,∗ f := sup

G.Tephnadze

n∈N

α,β κn f log2 (n + 1)

.

,

18 1.3

August 30, 2015

1. Preliminaries

D IRICHLET

KERNELS AND

L EBESGUE

CONSTANTS WITH RESPECT TO

V ILENKIN

SYSTEMS

It is easy to see that Z Sn f (x) =

f (t)

n−1 X

Gm

ψk (x − t) dµ (t)

k=0

Z f (t) Dn (x − t) dµ (t) = (f ∗ Dn ) (x) .

= Gm

The next well-known identities with respect to Dirichlet kernels (see Lemmas 1.2 and 1.3, Corollaries 1.4 and 1.5) will be used many time in the proofs of our main results. The first equality can be found in Vilenkin [77] and the second identity in Gàt and Goginava [19]: Lemma 1.2 Let n ∈ N, 1 ≤ sn ≤ mn − 1. Then j ≤ (mn − 1) Mn

Dj+Mn = DMn + ψMn Dj = DMn + rn Dj ,

(1.11)

and Dsn Mn −j = Dsn Mn − ψsn Mn −1 Dj , j < Mn .

(1.12)

Lemma 1.3 Let n ∈ N and 1 ≤ sn ≤ mn − 1. Then Dsn Mn = DMn

sX n −1

ψkMn = DMn

sX n −1

and

 Dn = ψn 

∞ X

for n =

P∞

i=0

(1.13)



mj −1

DMj

j=0

rnk

k=0

k=0

X

rjk  ,

(1.14)

k=mj −nj

ni Mi .

Corollary 1.4 Let n ∈ N. Then DMn+1 =

n Y

m k −1 X

k=0

s=0

! rks

.

Corollary 1.5 Let n ∈ N. Then DMn (x) =

Mn , x ∈ In , 0, x∈ / In . ,

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19

We also need the following estimate (see Tephnadze [71]): Lemma 1.6 Let x ∈ Is \Is+1 , s = 0, ..., N − 1. Then Z cMs , |Dn (x − t)| dµ (t) ≤ MN IN where c is an absolute constant. Proof: By combining (1.14) in Lemma 1.3 and Corollary 1.5 we have that |Dn (x)| ≤

l X

nj DMj (x) =

j=0

l X

nj Mj ≤ cMl .

j=0

Since t ∈ IN and x ∈ Is \Is+1 , s = 0, ..., N − 1, we obtain that x − t ∈ Is \Is+1 . By using the estimate above we get that |Dn (x − t)| ≤ cMs and

Z |Dn (x − t)| dµ (t) ≤ IN

cMs . MN

The proof is complete. To study the Dirichlet kernels we need an estimate of some sums of Rademacher functions. This Lemma can be found in Persson and Tephnadze [53]. Lemma 1.7 Let n ∈ N, and xn = 1. Then n −1 sX u rn (x) ≥ 1, for some 1 ≤ sn ≤ mn − 1 u=0 and

s −1 n X u rn (x) ≥ 1, for some 2 ≤ sn ≤ mn − 1. u=1

Proof: Let xn = 1. Then we readily get that n −1 sX rsn (x) − 1 u rn (x) = n u=0 rn (x) − 1 =

sin (πsn xn /mn ) sin (πsn /mn ) = ≥ 1. sin (πxn /mn ) sin (π/mn ) G.Tephnadze

,

20

August 30, 2015

1. Preliminaries

Analogously, we can prove that n −1 sX rsn −1 (x) − 1 sin (π (sn − 1) /mn ) u = ≥ 1. rn (x) = rn (x) n u=1 rn (x) − 1 sin (π/mn ) The proof is complete. The next Lemma can be found in Tephnadze [71]: Lemma 1.8 Let n ∈ N, |n| = 6 hni and x ∈ Ihni \Ihni+1 . Then |Dn | = Dn−M|n| ≥ Mhni . Proof: Let x ∈ Ihni \Ihni+1 . Since |n|−1

n = nhni Mhni +

X

nj Mj + n|n| M|n|

j=hni

and

|n|−1

n − M|n| = nhni Mhni +

X

nj Mj + n|n| − 1 M|n| ,

j=hni

if we apply Lemma 1.7, Corollary 1.5 and (1.14) in Lemma 1.3 we can conclude that Dn−M|n| mhni −1 mj −1 |n| X X X s s ψhni r rhni − D ≥ ψhni DMhni M j j s=mj −nj s=mhni −nhni j=hni+1 mhni −1 nhni −1 X X mhni −αhni s s = DMhni rhni = DMhni rhni rhni s=mhni −nhni s=0 nX −1 hni s = DMhni rhni ≥ DMhni ≥ Mhni . s=0 Analogously we can show that nhni X s |Dn | = DMhni rhni ≥ Mhni s=0

and the proof is complete. The next Lemma can be found in Lukomskii [38]: ,

August 30, 2015


Lemma 1.9 Let n =

P∞

i=1

21

ni Mi . Then

1 1 1 3 v (n) + v ∗ (n) + ≤ Ln ≤ v (n) + 4v ∗ (n) − 1, 4λ λ 2λ 2 where λ := supn∈N mn . The next result for the Walsh system can be found in the book [55] and for bounded Vilenkin systems in the book [2].

Corollary 1.10 Let qnk = M2nk + M2nk −2 + M2 + M0 . Then

nk ≤ Dqnk 1 ≤ λnk , 2λ where λ := supn∈N mn . Proof: The proof readily follows by just using Theorem 1.9 and the following identity v (qnk ) = 2nk . Thus, we leave out the details. 1.4

F EJ E´ R K ERNELS WITH RESPECT TO V ILENKIN SYSTEMS

It is obvious that n−1

σn f (x) =

1X (Dk ∗ f ) (x) n k=0

Z = (f ∗ Kn ) (x) =

f (t) Kn (x − t) dµ (t) . Gm

We frequently use the following well-known result, which was proved in Gát [17]: Lemma 1.11 Let n > t, t, n ∈ N. Then

KMn (x) =

 

Mt , 1−rt (x) Mn +1 , 2



0,

x ∈ It \It+1 , x − xt et ∈ In , x ∈ In , otherwise.

The proof of the next lemma can easily be done by using last lemma (c.f. also the book [2] and Tephnadze [67, 68]:

G.Tephnadze

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22

August 30, 2015

1. Preliminaries

Lemma 1.12 Let n ∈ N and x ∈ INk,l , where k < l. Then KMn (x) = 0, if n > l.

(1.15)

|KMn (x)| ≤ cMk .

(1.16)

and Moreover, Z |KMn | dµ ≤ c < ∞,

(1.17)

Gm

where c is an absolute constant. We also need the following useful result: Lemma 1.13 Let t, sn , n ∈ N, and 1 ≤ sn ≤ mn − 1. Then

sn Mn Ksn Mn =

sX n −1

l−1 X

l=0

i=0

and |Ksn Mn (x)| ≥

sX n −1

! rni

Mn DMn +

! rnl

Mn KMn

(1.18)

l=0

Mn , for x ∈ In+1 (en−1 + en ) . 2πsn

(1.19)

Moreover, if x ∈ It \It+1 , x − xt et ∈ / In and n > t, then Ksn Mn (x) = 0. Remark 1.14 This result was proved by Blahota and Tephnadze [5], but here we will give a completely different and simpler proof. Proof: We can write that sn Mn Ksn Mn =

sX n −1 (l+1)M Xn −1 l=0

=

sX n −1 (l+1)M Xn −1 l=0

Dk =

k=lMn

Dk

(1.20)

k=lMn

sX n −1 M n −1 X l=0

Dk+lMn .

k=0

Let 0 ≤ k < Mn . Then, by using (1.13) in Lemma 1.2 we have that Dk+lMn =

lM n −1 X m=0

ψm +

lMX n +k−1

ψm

m=lMn ,

August 30, 2015


= DlMn +

k−1 X

ψm+lMn = DlMn + rnl

m=0

ψm

m=0

l−1 X

=

k−1 X

23

! rni

DMn + rnl Dk .

i=0

According to (1.20) we readily get that sn Mn Ksn Mn =

sX n −1 M n −1 X l=0

=

sX n −1 M n −1 X l=0

=

=

l−1 X

l−1 X

l=0

i=0

sX n −1

l−1 X

l=0

i=0

!

! rni

DMn +

rnl Dk

i=0

k=0

sX n −1

Dk+lMn

k=0

! rni

Mn DMn +

sX n −1

rnl

! M −1 n X

l=0

! rni

Mn DMn +

sX n −1

Dk

k=0

! rnl

Mn KMn .

l=0

The first part of the Lemma is proved. Now, let t, sn , n ∈ N, n > t, x ∈ It \It+1 . If x−xt et ∈ / In , then by combining Corollary 1.5 and Lemma 1.11 we obtain that DMn (x) = KMn (x) = 0.

(1.21)

Let x ∈ In+1 (en−1 + en ) . By Lemma 1.11 we have that |KMn (x)| =

Mn−1 Mn−1 = . |1 − rn−1 (x)| 2 sin π/mn−1

By combining Lemmas 1.7 and 1.11 and the first part of Lemma 1.13 we immediately get that ! n −1 sX l |sn Mn Ksn Mn (x)| = rn (x) Mn KMn (x) l=0

Mn Mn−1 Mn Mn−1 mn−1 M2 Mn Mn−1 = ≥ ≥ n. |1 − rn (x)| 2 sin π/mn−1 2π 2π Now, let t, sn , n ∈ N, n > t, x ∈ It \It+1 . If x − xt et ∈ / In , then, by using the first part of Lemma 1.13, with identities (1.21) we immediately get that Ksn Mn (x) = 0. The proof is complete. In the same paper Blahota and Tephnadze [5] also proved the following result: G.Tephnadze

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24

August 30, 2015

1. Preliminaries

P Lemma 1.15 Let n = ri=1 sni Mni , where n1 > n2 > · · · > nr ≥ 0 and 1 ≤ sni < mni P for all 1 ≤ i ≤ r as well as n(k) = n − ki=1 sni Mni , where 0 < k ≤ r. Then ! ! r−1 k−1 r k−1 X Y sn X Y sn j j rnj n(k) Dsnk Mnk . nKn = rnj snk Mnk Ksnk Mnk + j=1

k=1

j=1

k=1

Proof: Let k, n ∈ N, 0 ≤ k < Mn . If we use the identity (1.11) in Lemma 1.2 we readily get that s n1 M n1 n n X X X nKn = Dk = Dk + Dk k=1

k=1

= sn1 Mn1 Ksn1 Mn1 +

k=sn1 Mn1 +1 n(1) X

Dk+sn1 Mn1

k=1 (1)

= sn1 Mn1 Ksn1 Mn1 +

n X

s

Dsn1 Mn1 + rnn11 Dk

k=1 s

(1)

= sn1 Mn1 Ksn1 Mn1 + n Dsn1 Mn1 + rnn11 n(1) Kn(1) . If we calculate n(1) Kn(1) in similar way, we get that s

n(1) Kn(1) = sn2 Mn2 Ksn2 Mn2 + n(2) Dsn2 Mn2 + rnn22 n(2) Kn(2) , so

s

nKn = sn1 Mn1 Ksn1 Mn1 + rnn11 sn2 Mn2 Ksn2 Mn2 s

s

s

+rnn11 rnn22 n(2) Kn(2) + n(1) Dsn1 Mn1 + rnn11 n(2) Dsn2 Mn2 . By using this method successively with n(2) Kn(2) , . . . , n(r−1) Kn(r−1) , we obtain that ! r k−1 X Y sn j nKn = rnj snk Mnk Ksnk Mnk k=1

+

r Y

sn rnj j

j=1

! (r)

n Kn(r) +

j=1

r−1 k−1 X Y k=1

sn rnj j

! n(k) Dsnk Mnk .

j=1

Since n(r) = 0 we conclude that the proof is complete. Corollary 1.16 Let n ∈ N. Then n |Kn | ≤ c

|n| X l=hni

Ml |KMl | ≤ c

|n| X

Ml |KMl |

(1.22)

l=0

,

August 30, 2015


and

25

Z |Kn | dµ ≤ c < ∞,

sup n

(1.23)

Gm

where c is an absolute constant. Remark 1.17 Corollary 1.16 is known (see the book [2]), but it is also a simple consequence of Lemmas 1.12 and 1.15. The next lemma can be found as a part of a more general result by Persson and Tephnadze [53]: Lemma 1.18 Let n ∈ N, hni = 6 |n| and x ∈ Ihni+1 ehni−1 + ehni . Then M2 hni , |nKn | = n − M|n| K−M|n| ≥ 2πλ where λ := sup mn . hni−1,hni Proof: Let x ∈ Ihni+1 . Since |n|−1

n = nhni Mhni +

X

nj Mj + n|n| M|n|

j=hni

and

|n|−1

n − M|n| = nhni Mhni +

X

nj Mj + n|n| − 1 M|n| ,

j=hni

if we combine (1.13), (1.19) and invoke Corollary 1.5, Lemmas 1.11 and 1.15 we obtain that n |Kn | = n − M|n| Kn−M|n|   hni−1 Y j   = ψMn shni Mhni Kshni Mhni = shni Mhni Kshni Mhni j j=1   ! shni −1 l−1 shni −1 X X X i l = rhni Mhni DMhni +  rhni  Mhni KMhni l=0 i=0 l=0   shni −1 M2 X l hni . =  rhni  Mhni KMhni ≥ Mhni KMhni ≥ 2πλ l=0 The proof is complete. The next result is proved in Blahota, Gát and Goginava [9] and [10], (see also Tephnadze [70]): G.Tephnadze

,

26

August 30, 2015

1. Preliminaries

Lemma 1.19 Let 2 < n ∈ N+ , k ≤ s < n and qn = M2n + M2n−2 + ... + M2 + M0 . Then M2k M2s , qn−1 Kqn−1 (x) ≥ 8 for x ∈ I2n (0, ..., x2k 6= 0, 0, ..., 0, x2s 6= 0, x2s+1 , ..., x2n−1 ) , k = 0, 1, ..., n − 3, s = k + 2, k + 3, ..., n − 1. The next lemma can be found in Blahota and Tephnadze [5] (see also Tephnadze [67] and [68]): Lemma 1.20 Let x ∈ INk,l , k = 0, . . . , N − 2, l = k + 1, . . . , N − 1. Then Z cMl Mk . |Kn (x − t)| dµ (t) ≤ nMN IN Let x ∈ INk,N , k = 0, . . . , N − 1. Then Z cMk |Kn (x − t)| dµ (t) ≤ , MN IN where c is an absolute constant. Proof: Let x ∈ INk,l , for 0 ≤ k < l ≤ N − 1 and t ∈ IN . Since x − t ∈ INk,l and n ≥ MN , by combining Lemma 1.11 and (1.22) in Corollary 1.16 we obtain that n |Kn (x)| ≤c

Z l X Mi i=0

and

|KMi (x − t)| dµ (t) ≤ c

IN

l X

Mi Mk ≤ cMk Ml

i=0

Z |Kn (x − t)| dµ (t) ≤ IN

cMk Ml . nMN

(1.24)

Let x ∈ INk,N . Then by applying Lemma 1.11 and (1.22) in Corollary 1.16, we have that Z n |Kn (x − t)| dµ (t) ≤ IN

|n| X i=0

Z |KMi (x − t)| dµ (t) .

Mi

(1.25)

IN

Let x = 0, . . . , 0, xk 6= 0, . . . , xN −1 = 0, xN , xN +1 , xq , ..., x|n|−1 , . . . , t = 0, . . . , 0, xN , . . . , xq−1 , tq 6= xq , tq+1 , . . . , t|n|−1 , . . . , q = N, . . . , |n| − 1. ,

August 30, 2015


By using Lemmas 1.11 and 1.12 in (1.25) it is easy to see that Z |Kn (x − t)| dµ (t)

27

(1.26)

IN q−1

≤

cX Mi n i=0

Z Mk dµ (t) ≤ IN

cMk cMk Mq ≤ . nMN MN

Let (

x = 0, . . . , 0, xm 6= 0, 0, . . . , 0, xN, xN +1 , xq , . . . , x|n|−1 , . . . ,

t = 0, , . . . , xN = 0, . . . , x|n|−1 , . . . .

If we apply again Lemmas 1.11 and 1.12 in (1.25) we obtain that Z Z |n|−1 cX cMk |Kn (x − t)| dµ (t) ≤ . Mi Mk dµ (t) ≤ n i=0 MN IN IN

(1.27)

By combining (1.24), (1.26) and (1.27) we can complete the proof. Also the next lemma is due to Tephnadze [67, 68], but it is also a simple consequence of Lemma 1.20. Lemma 1.21 Let x ∈ INk,l , k = 0, . . . , N − 1, l = k + 1, . . . , N. Then Z cMl Mk |Kn (x − t)| dµ (t) ≤ , for n ≥ MN , MN2 IN where c is an absolute constant. Proof: Since n ≥ MN if we apply Lemma 1.20 we immediately get a proof of this estimate. 1.5

¨ N ORLUND K ERNELS WITH RESPECT TO V ILENKIN SYSTEMS

A representation Z f (t) Fn (x − t) dµ (t)

tn f (x) = G

plays a central role in the sequel, where Fn :=

n 1 X qn−k Dk Qn k=1

is the so-called Nörlund kernel. In this section we study Nörlund kernels with respect to Vilenkin systems. The next results (Lemmas 1.22-1.27) are due to Persson, Tephnadze and Wall [51]: G.Tephnadze

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28

August 30, 2015

1. Preliminaries

Lemma 1.22 Let {qk : k ∈ N} be a sequence of non-decreasing numbers, satisfying the condition 1 qn−1 =O , when n → ∞. (1.28) Qn n Then

  |n|  c X |Fn | ≤ Mj KMj ,  n  j=0

where c is an absolute constant. Proof: First, we invoke Abel transformation to obtain the following identities Qn :=

n−1 n n−1 X X X qj = qn−j · 1 = (qn−j − qn−j−1 ) j + q0 n j=0

j=1

and

n−1 X

1 Fn = Qn

(1.29)

j=1

! (qn−j − qn−j−1 ) jKj + q0 nKn

.

(1.30)

j=1

Let the sequence {qk : k ∈ N} be non-decreasing. Then, by using (1.28), we get that ! n−1 1 X |qn−j − qn−j−1 | + q0 Qn j=1 1 ≤ Qn

n−1 X

! (qn−j − qn−j−1 ) + q0

≤

j=1

qn−1 c ≤ . Qn n

Under condition (1.28) if we apply (1.22) in Corollary 1.16 and use the equalities (1.29) and (1.30) we immediately get that |Fn | ≤

=

1 Qn 1 Qn ≤

n−1 X

!! |qn−j − qn−j−1 | + q0

j=1 n−1 X j=1

Mi |KMi |

i=0

!! (qn−j − qn−j−1 ) + q0

|n| X

|n| X

Mi |KMi |

i=0

|n| |n| qn−1 X cX Mi |KMi | ≤ Mi |KMi | . Qn i=0 n i=0

The proof is complete by just combining the estimates above. ,

August 30, 2015


29

Corollary 1.23 Let {qk : k ∈ N} be a sequence of non-decreasing numbers. Then Z sup |Fn | dµ ≤ c < ∞, n

Gm

where c is an absolute constant. Proof: If we apply (1.23) in Corollary 1.16 and invoke the identities (1.29) and (1.30) we readily get the proof. So, we leave out the details. Lemma 1.24 Let n ≥ MN and {qk : k ∈ N} be a sequence of non-decreasing numbers. Then   |n| n   1 X X c qn−j Dj ≤ Mj KMj , MN   Qn j=0 j=M N

where c is an absolute constant. Proof: Let MN ≤ j ≤ n. By using (1.22) in Corollary 1.16 we get that |Kj | ≤

|j| |n| 1X 1 X Ml |KMl | ≤ Ml |KMl | . j l=0 MN l=0

Let the sequence {qk : k ∈ N} be non-decreasing. Then n−1 X

|qn−j − qn−j−1 | j + q0 n ≤

n−1 X

|qn−j − qn−j−1 | j + q0 n

j=0

j=MN

=

n−1 X

(qn−j − qn−j−1 ) j + q0 n = Qn .

j=1

By using Abel transformation we can write that n 1 X qn−j Dj Qn j=MN ! n−1 1 X = (qn−j − qn−j−1 ) jKj + q0 nKn Qn j=MN !! |n| n−1 X 1 X 1 |qn−j − qn−j−1 | j + q0 n Mi |KMi | Qn j=M MN i=0 N

≤

|n| 1 X Mi |KMi | . MN i=0

The proof is complete. G.Tephnadze

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30

August 30, 2015

1. Preliminaries

Lemma 1.25 Let x ∈ INk,l , k = 0, . . . , N − 2, l = k + 1, . . . , N − 1 and {qk : k ∈ N} be a sequence of non-decreasing numbers, satisfying condition (1.28). Then Z cMl Mk . |Fn (x − t)| dµ (t) ≤ nMN IN Let x ∈ INk,N , k = 0, . . . , N − 1. Then Z cMk |Fn (x − t)| dµ (t) ≤ . MN IN Here c is an absolute constant.

INk,l .

Proof: Let x ∈ INk,l , for 0 ≤ k < l ≤ N − 1 and t ∈ IN . First, we observe that x − t ∈ Next, we apply Lemma 1.22 and invoke (1.15) and (1.16) in Lemma 1.12 to obtain that Z |Fn (x − t)| dµ (t) (1.31) IN |n|

≤

≤

cX Mi n i=0 c n

Z IN

Z |KMi (x − t)| dµ (t) IN

l X cMk Ml Mi Mk dµ (t) ≤ nMN i=0

and the first estimate is proved. Now, let x ∈ INk,N . Since x − t ∈ INk,N for t ∈ IN , by combining Lemmas 1.11 and 1.22 with (1.15) and (1.16) in 1.12, we have that Z |Fn (x − t)| dµ (t) (1.32) IN |n|

≤

cX Mi n i=0 |n|−1

Z

cX ≤ Mi n i=0

|KMi (x − t)| dµ (t) IN

Z Mk dµ (t) ≤ IN

cMk . MN

By combining (1.31) and (1.32) we complete the proof. Lemma 1.26 Let n ≥ MN , x ∈ INk,l , k = 0, . . . , N − 1, l = k + 1, . . . , N and {qk : k ∈ N} be a sequence of non-decreasing sequence, satisfying condition (1.28). Then Z cMl Mk |Fn (x − t)| dµ (t) ≤ , MN2 IN where c is an absolute constant. ,

August 30, 2015


31

Proof: Since n ≥ MN if we apply Lemma 1.25 we immediately get the proof. Next, we state analogical estimate, but now without any restriction like (1.28): Lemma 1.27 Let x ∈ INk,l , k = 0, . . . , N − 1, l = k + 1, . . . , N and {qk : k ∈ N} be a sequence of non-decreasing sequence. Then Z n cMl Mk 1 X qn−j Dj (x − t) dµ (t) ≤ , Q MN2 n IN j=M N

where c is an absolute constant. Proof: Let x ∈ INk,l , for 0 ≤ k < l ≤ N − 1 and t ∈ IN . Since x − t ∈ INk,l and n ≥ MN , if we combine Lemmas 1.11 and 1.24 and invoke (1.15) and (1.16) in Lemma 1.12 we readily obtain that Z n 1 X qn−j Dj (x − t) dµ (t) (1.33) IN Q n j=MN

|n|

≤

≤

c X Mi MN i=0

Z

l

Z

c X Mi MN i=0

|KMi (x − t)| dµ (t) IN

Mk dµ (t) ≤ IN

cMk Ml MN2

and the first estimate is proved. Now, let x ∈ INk,N . Since x − t ∈ INk,N for t ∈ IN , if we apply again Lemmas 1.11, 1.22 and (1.15) and (1.16) in Lemma 1.12 we can conclude that Z n 1 X qn−j Dj (x − t) dµ (t) (1.34) Q n IN j=MN

|n|

≤

c X Mi MN i=0

Z |KMi (x − t)| dµ (t) IN

Z |n|−1 cMk c X ≤ Mi Mk dµ (t) ≤ . MN i=0 MN IN By combining (1.33) and (1.34) we complete the proof. The next results (Lemmas 1.28-1.34) are due to Blahota, Persson and Tephnadze [7]: G.Tephnadze

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32

August 30, 2015

1. Preliminaries

Lemma 1.28 Let sn Mn < r ≤ (sn + 1) Mn , where 1 ≤ sn ≤ mn − 1. Then Qr Fr = Qr Dsn Mn − ψsn Mn −1

snX Mn −2

(qr−sn Mn +l − qr−sn Mn +l+1 ) lKl

l=1

−ψsn Mn −1 (sn Mn − 1) qr−1 Ksn Mn −1 + ψsn Mn Qr−sn Mn Fr−sn Mn . Remark 1.29 We note that Lemma 1.28 is true for every Nörlund mean, without any restriction on the generative sequence {qk : k ∈ N}. Proof: Let sn Mn < r ≤ (sn + 1) Mn , where 1 ≤ sn ≤ mn − 1. It is easy to see that Qr Fr =

sX r n Mn X qr−l Dl + qr−k Dk =

qr−l Dl := I + II.

(1.35)

l=sn Mn +1

l=1

k=1

r X

We apply (1.12) in Lemma 1.2 and invoke Abel transformation to obtain that I=

snX Mn −1

qr−sn Mn +l Dsn Mn −l

(1.36)

l=0

=

snX Mn −1

qr−sn Mn +l Dsn Mn −l + qr−sn Mn Dsn Mn

l=1

= Dsn Mn

snX Mn −1

snX Mn −1

l=0

l=1

qr−sn Mn +l − ψsn Mn −1

qr−sn Mn +l Dl

= (Qr − Qr−sn Mn ) Dsn Mn −ψsn Mn −1

snX Mn −2

(qr−sn Mn +l − qr−sn Mn +l+1 ) lKl

l=1

−ψsn Mn −1 qr−1 (sn Mn − 1) Ksn Mn −1 . By using (1.11) in Lemma 1.2 we can rewrite II as II =

r−s n Mn X

qr−sn Mn −l Dl+sn Mn

(1.37)

l=1

= Qr−sn Mn Dsn Mn + ψsn Mn Qr−sn Mn Fr−sn Mn . The proof is complete by just combining (1.35-1.37). ,

August 30, 2015


33

Lemma 1.30 Let {qk : k ∈ N} be a sequence of non-increasing numbers satisfying the condition 1 1 =O , when n → ∞. (1.38) Qn n Then   |n|   X cα |Fn | ≤ Mj KMj ,  n  j=0

where c is an absolute constant. Proof: Let the sequence {qk : k ∈ N} be non-increasing satisfying condition (1.38). Then ! n−1 1 X |qn−j − qn−j−1 | + q0 Qn j=1 ! n−1 1 X ≤ − (qn−j − qn−j−1 ) + q0 Qn j=1 ≤

2q0 − qn−1 2q0 c ≤ ≤ . Qn Qn n

If we apply (1.22) in Corollary 1.16 and invoke equalities (1.29) and (1.30) we immediately get that !! |n| n−1 X 1 X |Fn | ≤ |qn−j − qn−j−1 | + q0 Mi |KMi | Qn j=1 i=0 =

1 Qn ≤

n−1 X

!! − (qn−j − qn−j−1 ) + q0

j=1

|n| X

Mi |KMi |

i=0

|n| |n| 2q0 − qn−1 X 2q0 X Mi |KMi | ≤ Mi |KMi | Qn Qn i=0 i=0 |n|

≤

cX Mi |KMi | . n i=0

The proof is complete by combining the estimates above. Corollary 1.31 Let {qk : k ∈ N} be a sequence of non-increasing numbers satisfying condition (1.38). Then Z sup |Fn | dµ ≤ cα < ∞, n

Gm

where cα is an absolute constant depending only on α. G.Tephnadze

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34

August 30, 2015

1. Preliminaries

Proof: By applying Lemma 1.30 we readily get the proof. So, we leave out the details.

Lemma 1.32 Let {qk : k ∈ N} be a sequence of non-increasing numbers, 0 < α < 1, and 1 1 =O , when n → ∞ (1.39) Qn nα and

qn − qn+1 = O

Then

1

n2−α

, when n → ∞.

(1.40)

  |n|  cα X α |Fn | ≤ α M KMj ,  n  j=0 j

where cα is an absolute constant depending only on α. Proof: According to the fact that {qk : k ∈ N} is a sequence of non-negative and non-increasing numbers we have two cases: 1. limk→∞ qk ≥ c > 0, 2. limk→∞ qk = 0, In the first case we obtain that (1.4) and (1.5) are satisfied. Since the case n = O (1) , when n → ∞, Qn have already been considered in Lemma 1.22, we can exclude it. Hence, we may assume that qn = o(1),

when n → ∞.

(1.41)

By combining (1.40) and (1.41) we immediately get that qn =

∞ X

(ql − ql+1 ) ≤

l=n

and Qn =

∞ X c c ≤ 1−α 2−α l n l=n

n−1 n X X c ql ≤ ≤ cnα . 1−α l l=0 l=1

Let Mn < k ≤ Mn+1 . It is easy to see that Qk |DsMn | ≤ cMnα |DsMn |

(1.42) ,

August 30, 2015

35


and (sMn − 1) qk−1 |KsMn −1 | ≤ ck

α−1

Mn |KsMn −1 | ≤

cMnα

(1.43)

|KsMn −1 | .

Let n = sn1 Mn1 + sn2 Mn2 + · · · + snr Mnr , n1 > n2 > · · · > nr , and n(k) = snk+1 Mnk+1 + · · · + snr Mnr , 1 ≤ snl ≤ ml − 1, l = 1, . . . , r. By combining (1.42), (1.43) and Lemma 1.28 we have that |Qn Fn |  ≤ cα Mnα1 Dsn1 Mn1 +

 α−2 (1) n +l |lKl | + Mnα1 Ksn1 Mn1 −1  .

sn1 Mn1 −1

X l=1

+cα |Qn(1) Fn(1) | By repeating this process r times we get that |Qn Fn | ≤ cα

r X



snk Mnk −1

Mnα

k

Dsn

k

Mnk

+

X

n(k) + l

α−2

|lKl | + Mnαk

Ksn

k

M nk

  −1

l=1

k=1

:= I + II + III. We combine Corollary 1.5 and Lemma 1.11 and invoke (1.13) in Lemma 1.3 to obtain that I ≤ cα

|n| X

Mkα |Dsk Mk |

k=1 |n|

≤ cα

X

Mkα |DMk | ≤ cα

k=1

and III ≤ cα

r X

|n| X

Mkα |KMk |

k=1

Mn Ksn Mn − Mn Dsn Mn Mnα−1 k k k k k k k

k=1

≤ cα

r X

r X Mnαk Ksnk Mnk + cα Mnαk Dsnk Mnk

k=1

k=1

≤ cα

r X

Mkα |KMk | .

k=1 G.Tephnadze

,

36

August 30, 2015

1. Preliminaries

Next, we can rewrite II as snk+1 Mnk+1 −1

II = cα

+cα

r X

X

k=1

l=1

r X

n(k) + l

snk Mnk −1

X

n(k) + l

α−2

α−2

|lKl |

|lKl |

k=1 l=snk+1 Mnk+1

:= II1 + II2 . For II1 we find that II1 ≤ cα

r X

snk+1 Mnk+1 −1

X

α−2 sα−2 nk+1 Mnk+1

≤ cα

n1 X

Mkα−2

k=1

= cα

n1 X

M k −1 X

|lKl |

l=1

Mkα−2

k M i −1 X X

|lKl |

i=1 l=Mi−1

k=1

≤ cα

|lKl |

l=1

k=1

n1 k i X X X Mkα−2 Mi Mj KMj i=1

k=1

≤ cα

n1 X

Mkα−1

n1 X

k X

n1 X Mj |KMj | Mkα−1

j=0

≤ cα

Mj |KMj |

j=0

k=0

= cα

j=0

k=j n1 X

Mjα KMj .

j=0

Moreover, II2 ≤ cα

r X

snk Mnk −1

X

lα−2 |lKl |

k=1 l=snk+1 Mnk+1

≤ cα

nk r X X

Mi+1 −1

Miα−2

k=1 i=nk+1 +1

X

|lKl |

l=Mi ,

August 30, 2015


≤ cα

nk r X X

Miα−2 Mi

k=1 i=nk+1 +1

= cα

r X

nk X

≤ cα

Miα−1

Miα−1

i X

Mj KMj

j=0 i X

Mj KMj

j=0

i=1

≤ cα

Mj KMj

j=0

k=1 i=nk+1 +1 n1 X

i X

37

n1 X

Mjα KMj .

j=0

The proof is complete by just combining the estimates above. Corollary 1.33 Let 0 < α ≤ 1 and {qk : k ∈ N} be a sequence of non-increasing numbers satisfying the conditions (1.39) and (1.40). Then Z sup |Fn | dµ ≤ cα < ∞, n

Gm

where cα is an absolute constant depending only on α. Proof: If we use Lemma 1.32 we readily get the proof. Thus, we leave out the details. Lemma 1.34 Let 0 < α ≤ 1 and {qk : k ∈ N} be a sequence of non-increasing numbers satisfying the conditions (1.39) and (1.40). Then Z cα Mlα Mk |Fm (x − t)| dµ (t) ≤ , for x ∈ INk,l , mα MN IN where k = 0, . . . , N − 2, l = k + 2, . . . , N − 1. Moreover, Z cα Mk |Fm (x − t)| dµ (t) ≤ , for x ∈ INk,N , k = 0, . . . , N − 1. MN IN Here cα is an absolute constant depending only on α. Proof: Let x ∈ INk,l , where k = 0, . . . , N − 2, l = k + 2, . . . , N − 1. Since x − t ∈ INk,l , for t ∈ IN if we apply Lemma 1.32 with (1.15) and (1.16) in Lemma 1.12 we can conclude that l cα X α |Fm (x − t)| ≤ α M |KMi (x − t)| (1.44) m i=0 i ≤

l cα X α cα Mlα Mk Mi Mk ≤ . α m i=0 mα G.Tephnadze

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38

August 30, 2015

1. Preliminaries

Let x ∈ INk,l , for some 0 ≤ k < l ≤ N − 1. Since x − t ∈ INk,l , for t ∈ IN and m ≥ MN from (1.44) we readily obtain that Z cα Mlα Mk |Fm (x − t)| dµ (t) ≤ . (1.45) mα MN IN Let x ∈ INk,N , k = 0, . . . , N − 1. Then, by applying Lemma 1.32, we have that Z |Fm (x − t)| dµ (t)

(1.46)

IN

≤

Z |m| cα X α M |KMi (x − t)| dµ (t) . mα i=0 i IN

Let x ∈ INk,N , k = 0, . . . , N − 1, t ∈ IN and xq 6= tq , where N ≤ q ≤ |m| − 1. By using Lemma 1.11 and estimate (1.46) we get that Z |Fm (x − t)| dµ (t) (1.47) IN q−1

≤

cα X α M mα i=0 i

Z Mk dµ (t) ≤ IN

cα Mk Mqα cα Mk ≤ . mα MN MN

Let x ∈ INk,N , k = 0, . . . , N − 1, t ∈ IN and xN = tN , . . . , x|m|−1 = t|m|−1 . By again applying Lemma 1.11 and estimate (1.46) we have that Z |Fm (x − t)| dµ (t) ≤ IN

Z |m|−1 cα X α cα Mk M Mk dµ (t) ≤ . i α m i=0 MN IN

(1.48)

The proof follows by combining (1.45), (1.47) and (1.48). Corollary 1.35 Let 0 < α < 1, m ≥ MN and {qk : k ∈ N} be a sequence of non-increasing numbers satisfying the conditions (1.39) and (1.40). Then there exists an absolute constant cα , depending only on α, such that Z cα Mlα Mk |Fm (x − t)| dµ (t) ≤ , for x ∈ INk,l MN1+α IN where k = 0, . . . , N − 1, l = k + 2, . . . , N . Proof: The proof readily follows Lemma 1.34 if we use additional condition n ≥ MN . Remark 1.36 For some sequences {qk : k ∈ N} of non-increasing numbers conditions (1.39) and (1.40) can be true or false independently.

,

August 30, 2015

1.6


39

I NTRODUCTION TO THE THEORY OF M ARTINGALE H ARDY SPACES

The σ-algebra generated by the intervals {In (x) : x ∈ Gm } will be denoted by zn (n ∈ N) . A sequence f = f (n) : n ∈ N of integrable functions f (n) is said to be a martingale with respect to the σ-algebras zn (n ∈ N) if (for details see e.g. Weisz [83]) 1) fn is zn measurable for all n ∈ N, 2) SMn fm = fn for all n ≤ m. The martingale f = f (n) , n ∈ N is said to be Lp -bounded (0 < p ≤ ∞) if f (n) ∈ Lp and kf kp := sup kfn kp < ∞. n∈N

If f ∈ L1 (Gm ) , then it is easy to show that the sequence F = (En f : n ∈ N) is a martingale. This type of martingales is called regular. If 1 ≤ p ≤ ∞ and f ∈ Lp (Gm ) then f = f (n) , n ∈ N is Lp -bounded and lim kEn f − f kp = 0

n→∞

and consequently kF kp = kf kp (see [46]). The converse of the latest statement holds also if 1 < p ≤ ∞ (see [46]): for an arbitrary Lp -bounded martingale f = f (n) , n ∈ N there exists a function f ∈ Lp (Gm ) for which f (n) = En f. If p = 1, then there exists a function f ∈ L1 (Gm ) of the preceding type if and only if f is uniformly integrable (see [46]), namely, if Z lim sup |fn (x)| dµ (x) = 0. y→∞ n∈N

{|fn |>y}

Thus the map f → f := (En f : n ∈ N) is isometric from Lp onto the space of Lp bounded martingales when 1 < p ≤ ∞. Consequently, these two spaces can be identified with each other. Similarly, the space L1 (Gm ) can be identified with the space of uniformly integrable martingales. Analogously, the martingale f = f (n) , n ∈ N is said to be weak − Lp -bounded (0 < p ≤ ∞) if f (n) ∈ Lp and kf kweak−Lp := sup kfn kweak−Lp < ∞. n∈N

The maximal function of a martingale f is defined by f ∗ := sup f (n) . n∈N G.Tephnadze

,

40

August 30, 2015

1. Preliminaries

In the case f ∈ L1 (Gm ), the maximal functions are also given by Z 1 ∗ f (x) := sup f (u) dµ (u) . n∈N |In (x)| In (x) For 0 < p < ∞ the Hardy martingale spaces Hp consist of all martingales for which kf kHp := kf ∗ kp < ∞. If f = f (n) : n ∈ N is a martingale, then the Vilenkin-Fourier coefficients must be defined in a slightly different manner: Z fb(i) := lim f (k) ψ i dµ. k→∞

Gm

The next Lemma can be found in [84] (see also book [55]): Lemma 1.37 If f ∈ L1 , then the sequence F := (SMn f : n ∈ N) is a martingale and

kF kHp ∼ sup |SMn f |

. n∈N

p

Moreover, if F := (SMn f : n ∈ N) is a regular martingale generated by f ∈ L1 , then

Z Fb (k) =

f (x) ψk (x) dµ (x) = fb(k) ,

k ∈ N.

G

A bounded measurable function a is a p-atom if there exist an interval I such that Z adµ = 0, kak∞ ≤ µ (I)−1/p , supp (a) ⊂ I. I

Next, we note that the Hardy martingale spaces Hp (Gm ) for 0 < p ≤ 1 have atomic characterizations (see e.g. Weisz [83, 84]): Lemma 1.38 A martingale f = f (n) : n ∈ N is in Hp (0 < p ≤ 1) if and only if there exist a sequence (ak , k ∈ N) of p-atoms and a sequence (µk : k ∈ N) of real numbers such that, for every n ∈ N, ∞ X µk SMn ak = f (n) , a.e., (1.49) k=0

where

∞ X

|µk |p < ∞.

k=0 ,

August 30, 2015


41

Moreover, kf kHp v inf

∞ X

!1/p p

|µk |

,

k=0

where the infimum is taken over all decomposition of f = f (n) : n ∈ N of the form (1.49). By using atomic characterization it can be easily proved that the following Lemmas hold (see e.g. Weisz [84]): Lemma 1.39 Suppose that an operator T is sub-linear and for some 0 < p ≤ 1 Z |T a|p dµ ≤ cp < ∞ −

I

for every p-atom a, where I denotes the support of the atom. If T is bounded from L∞ to L∞ , then kT f kp ≤ cp kf kHp . Moreover, if p < 1, then we have the weak (1,1) type estimate λµ {x ∈ Gm : |T f (x)| > λ} ≤ kf k1 for all f ∈ L1 . Lemma 1.40 Suppose that an operator T is sub-linear and for some 0 λ ≤ cp < +∞ λ>0

for every p-atom a, where I denote the support of the atom. If T is bounded from L∞ to L∞ , then kT f kweak−Lp ≤ cp kf kHp . Moreover, if p < 1, then λµ {x ∈ Gm : |T f (x)| > λ} ≤ kf k1 , for all f ∈ L1 . The concept of modulus of continuity in Hp (p > 0) is defined by 1 , f := kf − SMn f kHp . ωHp Mn We need to understand the meaning of the expression f − SMn f where f is a martingale and SMn f is function. So, we give an explanation in the following remark: G.Tephnadze

,

42

August 30, 2015

1. Preliminaries

Remark 1.41 Let 0 < p ≤ 1. Since SMn f = f (n) , for f = f (n) : n ∈ N ∈ Hp and SMk f (n) : k ∈ N = (SMk SMn f, k ∈ N) = SM0 f, . . . , SMn−1 f, SMn f, SMn f, . . . = f (0) , . . . , f (n−1) , f (n) , f (n) , . . .

we obtain that f − SMn f = f (k) − SMk f : k ∈ N is a martingale, for which (k)

(f − SMn f )

=

0, k = 0, . . . . , n, f (k) − f (n) , k ≥ n + 1,

(1.50)

Watari [79] showed that there are strong connections between ωp

1 ,f Mn

, EMn (Lp , f ) and kf − SMn f kp , p ≥ 1, n ∈ N.

In particular, 1 ωp 2

1 ,f Mn

≤ kf − SMn f kp ≤ ωp

1 ,f Mn

(1.51)

and 1 kf − SMn f kp ≤ EMn (Lp , f ) ≤ kf − SMn f kp . 2 Remark 1.42 Since kf kHp ∼ kf kp , when p > 1, by applying (1.51), we obtain that ωHp

1 ,f Mn

∼ ωp

1 ,f Mn

.

,

August 30, 2015

1.7


E XAMPLES OF p- ATOMS AND Hp

43

MARINGALES

The next two Examples can be found in the papers [9] and [10] by Blahota, Gát and Goginava (see also : Example 1.43 Let 0 < p ≤ 1 and λ = supn mn . Then the function 1/p−1

ak :=

Mαk λ

DMαk +1 − DMαk

is a p-atom. Moreover, kak kHp ≤ 1. Proof: Here we present the proof from Tephnadze [67] and [68]. Since Z ak dµ = 0 supp(ak ) = Iαk , Iαk

and

1/p−1

Mαk λ we conclude that ak is a p-atom.

Mαk +1 ≤ Mα1/p = (supp ak )−1/p , k

kak k∞ ≤

Moreover, by using the orthonormality of Vilenkin functions, we find that 0, n = 0, ..., αk , SMn DMαk +1 − DMαk = DMαk +1 − DMαk , n ≥ αk + 1, and

sup SMn DMαk +1 (x) − DMαk (x) n∈N

= DMαk +1 (x) − DMαk (x) , for all x ∈ Gm . If we invoke Lemma 1.37 we obtain that 1/p−1

Mαk λ

kak kHp =

1/p−1

=

=

1/p−1 Mαk

λ

Mαk λ

Z Iα \ Iα k

Hp

DMαk +1 − DMαk

p

Mαpk dµ k +1

DMαk +1 − DMαk

!1/p

Z

p

(Mαk +1 − Mαk ) dµ

+ Iα

G.Tephnadze

k +1

,

44

August 30, 2015

1. Preliminaries 1/p−1

=

Mαk λ

m αk − 1 p (mαk − 1)p p Mαk + Mαk Mαk +1 Mαk +1

1/p

1/p−1

≤

Mαk λ

≤ 1. · Mα1−1/p k

The proof is complete. Example 1.44 Let 0 < p ≤ 1 and fk = DM2nk +1 − DM2n . k

Then

fbk (i) =

and

1, i = M2nk , ..., M2nk +1 − 1, 0, otherwise,

  Di − DM2nk , i = M2nk + 1, ..., M2nk +1 − 1, Si fk = f , i ≥ M2nk +1 ,  k 0, otherwise.

(1.52)

(1.53)

Moreover, 1−1/p kfk kHp ≤ λM2n , k

(1.54)

where λ = supn mn . Proof: The proof follows by using Example 1.43 in the case when αk = 2nk . We leave out the details. The next three examples of regular martingales will be used frequently and it can be found in Persson and Tephnadze [52]: Example 1.45 Let Mk ≤ n < Mk+1 and Sn f be the n-th partial sum with respect to Vilenkin systems, where f ∈ Hp for some 0 < p ≤ 1. Then Sn f ∈ L1 for every fixed n ∈ N and

kSn f kHp ≤ sup |S f |

0≤l≤k Ml + kSn f kp p

e∗ ≤ S# f + kSn f kp . p

Proof: We consider the following martingale f# = (SMl Sn , l ∈ N) f# := (SMk Sn f, k ≥ 1) = (SM0 , ..., SMk f, ..., Sn f, ..., Sn f, ..) . It immediately follows that

∗ e#

+ kSn f k ≤ kSn f kHp ≤ sup |S f | S f

+ kSn f kp . M l p

0≤l≤k

p p

The proof is complete. ,

August 30, 2015


45

Example 1.46 Let Mk ≤ n < Mk+1 and σn f be n-th Fejér means with respect to Vilenkin systems, where f ∈ Hp for some 0 < p ≤ 1. Then σn f ∈ L1 for every fixed n ∈ N and

∗

e# f + kσn f k . kσn f kHp ≤ sup |σMl f | p

+ kσn f kp ≤ σ p 0≤l≤k

p

Proof: We consider the following martingale f# = (SMk σn f, k ≥ 1) =

Mk σMk f M0 σM0 , ..., , ..., σn f, ..., σn f, ... . n n

and if we follow analogous steps as in Example 1.45 we readily can complete the proof. Thus, we leave out the details. The next five Examples of martingales will be used many times to prove sharpness of our main results (c.f. the papers [66], [69], [71], [72], [73] by Tephnadze). Example 1.47 Let 0 < p ≤ 1, λ = supn mn , {λk : k ∈ N} be a sequence of real numbers R, such that ∞ X |λk |p ≤ cp < ∞ (1.55) k=0

and {ak : k ∈ N} be a sequence of p-atoms, defined by 1/p−1

ak :=

M2αk λ

DM2αk +1 − DM2α

.

k

(1.56)

Then f = f (n) : n ∈ N , where X

f (n) :=

λk ak

{k: 2αk 0 and f ∈ Hp . Then the maximal operator sup SMn +Mn−1 f n∈N+

is bounded from the Hardy space Hp to the space Lp . Corollary 2.15 Let p > 0 and f ∈ Hp . Then the maximal operator sup |SMn +1 f | n∈N+

is not bounded from the Hardy space Hp to the space Lp .

,

August 30, 2015

2.5


73

N ORM CONVERGENCE OF PARTIAL SUMS OF V ILENKIN -F OURIER SERIES ON MAR H ARDY SPACES

TINGALE

By applying Corollaries 2.5 and 2.9 we find necessary and sufficient conditions for the modulus of continuity of martingale Hardy spaces Hp , for which the partial sums of VilenkinFourier series convergence in Lp -norm. We also study sharpness of these results. All results in this section can be found in Tephnadze [72]. Theorem 2.16 Let 0 < p < 1 and f ∈ Hp . Then there exists an absolute constant cp depending only on p such that kSn f kHp ≤ cp n1/p−1 kf kHp . Remark 2.17 We note that the asymptotic behaviour of the sequence n1/p−1 : n ∈ N in Theorem 2.16 can not be improved (c.f. part b) of Theorem 2.4). Proof: According to Corollaries 2.5 and 2.13 if we invoke Example 1.45 we can conclude that

kSn f kHp ≤ S # f p + kSn f kp (2.28) ≤ kf kHp + cp n1/p−1 kf kHp ≤ cp n1/p−1 kf kHp . The proof is complete. Theorem 2.18 Let f ∈ H1 . Then there exists an absolute constant c such that kSn f kH1 ≤ c log n kf kH1 . Remark 2.19 We note that the asymptotic behaviour of the sequence {log n : n ∈ N} in Theorem 2.18 can not be improved (c.f. part b) of Theorem 2.8). Proof: According to Corollaries 2.9 and 2.13 if we invoke Example 1.45 we can write that

kSn f kH1 ≤ S # f 1 + kSn f k1 ≤ kf kH1 + c log n kf kH1 ≤ c log n kf kH1 . The proof is complete. G.Tephnadze

,

74

2. Fourier coefficients and partial sums of Vilenkin-Fourier series on Martingale Hardy spaces

August 30, 2015

Theorem 2.20 Let p > 0 and f ∈ Hp . Then there exists an absolute constant cp depending only on p such that kSMn f kHp ≤ cp kf kHp . Proof: In the view of Corollary 2.13 and Example 1.45 we can conclude that

kSMn f kHp ≤ sup |SMl f |

≤ kf kHp . 0≤l≤n

p

The proof is complete. Theorem 2.21 Let 0 < p < 1, f ∈ Hp and Mk < n ≤ Mk+1 . Then there is an absolute constant cp depending only on p such that 1 kSn f − f kHp ≤ cp n1/p−1 ωHp ,f . Mk Proof: Let 0 < p < 1 and Mk < n ≤ Mk+1 . By using Corollary 2.5 we immediately get that kSn f − f kpHp ≤ kSn f − SMk f kpHp + kSMk f − f kpHp = kSn (SMk f − f )kpHp + kSMk f − f kpHp ≤ cp n1−p + 1 kSMk f − f kpHp cp n1−p + 1 kSMk f − f kpHp 1 p , f . ≤ cp n1−p ωH p Mk The proof is complete. Theorem 2.22 Let f ∈ H1 and Mk < n ≤ Mk+1 . Then there is an absolute constant c such that 1 ,f . kSn f − f kH1 ≤ c lg nωH1 Mk Proof: Let Mk < n ≤ Mk+1 . Then, by using Corollary 2.9, we immediately get that kSn f − f kH1 ≤ kSn (SMk f − f )kH1 + kSMk f − f kH1 ≤ c (log n + 1) kSMk f − f kH1 1 ≤ c log nωH1 ,f . Mk The proof is complete. ,

August 30, 2015


75

Theorem 2.23 a) Let 0 < p < 1, f ∈ Hp and 1 1 ,f = o ωHp , when n → ∞. 1/p−1 Mn Mn Then kSk f − f kHp → 0, when

k → ∞.

b) For every 0 < p < 1 there exists a martingale f ∈ Hp , for which 1 1 ωHp ,f = O , when n → ∞ 1/p−1 Mn Mn and kSk f − f kweak−Lp 9 0, when k → ∞. Proof: Let 0 < p < 1, f ∈ Hp and 1 1 ωHp , when n → ∞. ,f = o 1/p−1 Mn Mn By using Theorem 2.21 we immediately get that kSn f − f kHp → 0, when n → ∞. The proof of part a) is complete. Let f = f (n) : n ∈ N be the martingale defined in Example 1.47, where λk =

λ 1/p−1 M2αk

Since

, where λ = sup mn .

(2.29)

n∈N

∞ X λp 1−p < c < ∞ M2α k k=0

we conclude that f ∈ Hp . Moreover, according to (1.58) we find that ! ∞ X 1 1 ωHp ( , f) = O 1/p−1 Mn k=n Mk =O

1 1/p−1

, when n → ∞.

Mn

By now using (1.57) with λk defined by (2.29) we readily see that G.Tephnadze

,

76


  1, j ∈ {M2αk , ..., M2αk +1 − 1} , k ∈ N+ , ∞ S fb(j) = / {M2αk , ..., M2αk +1 − 1} .  0, j ∈

August 30, 2015

(2.30)

k=1

According to (2.30) and Corollary 1.5 we can establish that lim sup kSM2αk +1 f − f kweak−Lp k→∞

≥ lim sup kSM2αk f + fb(M2αk )ψM2αk − f kweak−Lp k→∞

≥ lim sup kψM2αk kweak−Lp − kf − SM2αk f kweak−Lp k→∞

≥ lim sup (1 − o (1)) = 1. k→∞

This completes the proof of the part b) and the proof is complete. Theorem 2.24 a) Let f ∈ H1 and 1 1 ωH1 ,f = o , when n → ∞. Mn n Then kSk f − f kH1 → 0, when k → ∞. b) There exists a martingale f ∈ H1 , for which 1 1 ,f = O ωH1 , when n → ∞ M2Mn Mn and kSk f − f k1 9 0, when k → ∞. Proof: Let f ∈ H1 and ωH1

1 ,f Mn

=o

1 , when n → ∞. n

By using Corollary 2.9 we immediately get that kSn f − f kH1 → ∞, when n → ∞. For the proof of part b) we use the martingale defined in Example 1.52, for p = 1. Then, f ∈ H1 and by applying (1.86) we can conclude that 1 1 ωH1 ( , f) = O , when n → ∞. Mn n ,

August 30, 2015

77


By combining (1.85) and (1.87) with Corollaries 1.10 and 1.5 we find that lim sup kSqMk f − f k1 k→∞

= lim sup k k→∞

ψMMk DqMk M2k

+ SMMk f − f k1

1 kDqMk k1 + kSMMk f − f k1 M2k k→∞ ! ∞ X 1 1 ≥ c − lim sup − = c > 0. M2i M2k k→∞ i=k+1

− lim sup

The proof is complete. 2.6

S TRONG

CONVERGENCE OF PARTIAL SUMS OF

MARTINGALE

V ILENKIN -F OURIER

SERIES ON

H ARDY SPACES

In this section we prove a Hardy type inequality for partial sums with respect to Vilenkin systems. This result was first proved by Simon [60]. Here we use some new estimations and present a simpler proof, which is due to in Tephnadze [72]. We also show sharpness of this result in the special sense (see Tephnadze [69]). Theorem 2.25 (Simon [60]) a) Let 0 < p < 1 and f ∈ Hp . Then there is an absolute constant cp , depending only on p, such that ∞ X kSk f kpp k=1

k 2−p

≤ cp kf kpHp .

b) Let 0 < p < 1 and {Φn : n ∈ N} be any non-decreasing sequence satisfying the condition lim Φn = +∞.

(2.31)

n→∞

Then there exists a martingale f ∈ Hp such that ∞ X kSk f kpweak−Lp Φk k=1

k 2−p

= ∞.

Proof: Let 0 < p < 1. By applying (1.1) with (2.11) in Theorem 2.4 we have that ∞ X kSk akpp k 2−p k=M N

G.Tephnadze

,

78


≤

Z ∞ X 1 k IN k=M

Sk a (x) p k 1/p−1 dµ (x)

N

∞ X

August 30, 2015

N −1 Z X

Sk a (x) p 1/p−1 dµ (x) Is \Is+1 k k=MN ∞ N −1 Z M 1/p−1 M p X 1X N s ≤c dµ k k 1/p−1 Is \Is+1 =

1 k s=0

k=MN

s=0

≤ cp MN1−p

N −1 Z 1 X

∞ X k=MN

≤ cp MN1−p

k 2−p s=0 N −1 1 X

∞ X k=MN

+cp MN1−p

∞ X k=MN

k 2−p s=0

Msp

Is \Is+1

Msp−1 dµ

1 ≤ cp < ∞. k 2−p

This completes the proof of the part a). Next, we note that under condition (2.31) there exists an increasing sequence {αk ≥ 2 : k ∈ N+ } of positive integers such that ∞ X 1 < ∞. (2.32) p/4 Φ M k=1 2αk

Let f = f (k) : k ∈ N be the martingale defined in Example 1.51. By using (2.32) we conclude that the martingale f ∈ Hp . Let Mαk ≤ j < Mαk +1 . By using (1.84) if we repeat the steps which leads to (1.70) 1/4 obtained by l = k in the case when λk = 1/ΦM2αη we obtain that Sj f =

1/p−1 k−1 X M2αη 1/4 η=0 ΦM2αη

DM2αη +1 − DM2αη

1/p−1

+

M2αk ψM2αk Dj−M2α

k

1/4

ΦM2α

k

:= I + II. We calculate each term separately. By using Corollary 1.5 with condition αn ≥ 2 (n ∈ N) for I we find that DMαn = 0, for x ∈ Gm \I1 ,

August 30, 2015


79

and for x ∈ Gm \I1 .

I = 0,

Denote by N0 the subset of positive integers N+ , for which hni = 0. Then every n ∈ N0 , Mk < n < Mk+1 (k > 1) can be written as n = n0 M0 +

k X

nj Mj ,

j=1

where n0 ∈ {1, ..., m0 − 1} and nj ∈ {0, ..., mj − 1} , (j ∈ N+ ). Let j ∈ N0 and x ∈ Gm \I1 = I0 \I1 . According to Lemma 1.8 we find that Dα −Mα ≥ cM0 ≥ c > 0 k k and

1/p−1

|II| =

M2αk 1/4

ΦM2α

k

M 1/p−1 2α (x) Dj−M2α = 1/4k . k ΦM2α k

It follows that 1/p−1

|Sj f (x)| = |II| =

M2αk

for x ∈ Gm \I1 ,

,

1/4

ΦM2α

k

and kSj f kweak−Lp 

1/p−1

≥

M2αk

µ x ∈ Gm \I1 : |Sj f (x)| >

1/4

2ΦM2α

k

=

1/p−1 M2αk 1/4 2ΦM2α k

|Gm \I1 | ≥

(2.33) 1/p−1

1/p

1/4



M2αk

2ΦM2α

k

1/p−1 cM2αk . 1/4 ΦM2α k

Since X

1 ≥ cMk ,

{n∈N0 :Mk ≤n≤Mk+1 ,}

where c is an absolute constant, by applying (2.33) we obtain that M2αk +1 −1

X

kSj f kpweak−Lp Φj j 2−p

j=1 M2αk +1 −1

≥

X j=M2αk

kSj f kpweak−Lp Φj j 2−p

G.Tephnadze

,

80


kSj f kpweak−Lp

X

≥ ΦM2αk

j 2−p

{j∈N0 :M2αk ≤j≤M2αk +1 } ≥ cΦM2αk

1−p M2α k

X

p/4 ΦM2α k

{j∈N0 :M2αk ≤j≤M2αk +1 }

1 j 2−p 1

X

3/4

≥ cΦMα Mα1−p k

August 30, 2015

k

{j∈N0 :M2αk ≤j≤M2αk +1 }

2−p M2α k +1

3/4

≥c

ΦM2α

X

k

M2αk +1

1

{j∈N0 :M2αk ≤j≤M2αk +1 }

3/4

≥ cΦM2α → ∞, when k

k → ∞.

Hence, also the part b) is proved so the proof is complete. 2.7

A N APPLICATION CONCERNING ESTIMATIONS OF V ILENKIN -F OURIER COEFFICIENTS

The following inequalities follow from our results: Corollary 2.26 Let 0 < p < 1 and f ∈ Hp . Then there exists an absolute constant cp , depending only on p, such that b f (n) ≤ cp n1/p−1 kf kHp , p ∞ fb (k) X k=1

k 2−p

≤ cp kf kpHp , (f ∈ Hp , 0 < p < 1)

and ∞ X k=1

2−2/p Mk

m k −1 X

2 b f (jMk )

!1/2 ≤ cp kf kHp .

j=1

Remark 2.27 The first inequality was first proved by Tephnadze [73], the second by Weisz [81, 83] and the third also by Weisz [62]. The proofs are different and simpler then the original ones. ,

August 30, 2015


Proof: Let 0 < p < 1. Since b f (n) = |Sn+1 f − Sn f | ≤ |Sn+1 f | + |Sn f | and

81

(2.34)

b f (n) (n + 1)1/p−1 ≤

|Sn+1 f | (n + 1)1/p−1

+

|Sn f | . n1/p−1

By using part a) of Theorem 2.4 we can write that b f (n) (n + 1)1/p−1

S f

Sn f

n+1

≤

+ 1−p

(n + 1)1/p−1 n p p

≤ cp kf kHp . It follows that

b f (n) ≤ cp n1/p−1 kf kHp

and the first inequality is proved. To prove the second inequality we use (2.34) again. We find that b f (n) (n + 1)2/p−1 ≤

|Sn+1 f | (n + 1)

2/p−1

+

(2.35)

|Sn f | . n2/p−1

By combining (2.35) and Theorem 2.4 we get that p ∞ fb (n) X n=1

n2−p

 

p

S f

S f

n+1 n

  ≤

+

n2/p−1 2/p−1

(n + 1) p n=1 p ! ∞ X kSn+1 f kpp kSn f kpp ≤ + n2−p (n + 1)2−p n=1 ∞ X

G.Tephnadze

,

82


August 30, 2015

∞ X kSn f kpp ≤2 ≤ cp kf kpHp . 2−p n n=1

and also the second inequality is proved. The third inequality can be proved analogously so we leave the details. The proof is complete.

,

August 30, 2015

3


83

´ MEANS ON MARTINGALE H ARDY SPACES V ILENKIN -F EJ ER

S OME CLASSICAL RESULTS ON V ILENKIN -F EJ E´ R MEANS

3.1

In the one-dimensional case Yano [78] proved that kKn k ≤ 2 (n ∈ N). Consequently, kσn f − f kp → 0,

when

n → ∞, (f ∈ Lp , 1 ≤ p ≤ ∞).

However (see [39, 55]) the rate of convergence can not be better then O (n−1 ) (n → ∞) for non-constant functions. a.e, if f ∈ Lp , 1 ≤ p ≤ ∞ and 1 kσMn f − f kp = o , when n → ∞, Mn then f is a constant function. Fridli [14] used dyadic modulus of continuity to characterize the set of functions in the space Lp , whose Vilenkin-Fejér means converge at a given rate. It is also known that (see e.g books [2] and [55]) kσn f − f kp N −1 X Ms 1 1 , f + cp ωp , f , (1 ≤ p ≤ ∞, n ∈ N) . ≤ cp ω p MN MN Ms s=0 By applying this estimate, we immediately obtain that if f ∈ lip (α, p) , i.e., 1 1 ωp ,f = O , n → ∞, Mn Mnα then

  O 1 , if α > 1,    MN O MNN , if α = 1, kσn f − f kp =     O 1α , if α < 1. M N

Weisz [85] considered the norm convergence of Fejér means of Vilenkin-Fourier series and proved that kσk f kp ≤ cp kf kHp , p > 1/2 and f ∈ Hp . (3.1) This result implies that p n 1 X kσk f kp ≤ cp kf kpHp , n2p−1 k=1 k 2−2p G.Tephnadze

(1/2 < p < ∞) .

,

84

August 30, 2015

3. Vilenkin-Fejér means on martingale Hardy spaces

If (3.1) hold for 0 1/2 in (3.1) is essential. In particular, is was proved that there exists a martingale f ∈ H1/2 such that sup kσn f k1/2 = +∞. n∈N

For Vilenkin systems in [70] it was proved that (3.2) holds, though inequality (3.1) is not true for 0 λ) ≤

c kf k1 , λ

(f ∈ L1 ,

λ > 0)

can be found in Zygmund [88] for the trigonometric series, in Schipp [54] for Walsh series and in Pál, Simon [49] for bounded Vilenkin series. Fujji [16] and Simon [57] verified that σ ∗ is bounded from H1 to L1 . Weisz [85] generalized this result and proved the boundedness of σ ∗ from the martingale space Hp to the Lebesgue space Lp for p > 1/2. Simon [56] gave a counterexample, which shows that boundedness does not hold for 0 < p < 1/2. The counterexample for p = 1/2 due to Goginava [27], (see also [9] and [10]). Weisz [86] proved that σ ∗ is bounded from the Hardy space H1/2 to the space weak − L1/2 . In [67] and [68] (for Walsh system see [28]) it was proved that the maximal operator σ ep∗ with respect to Vilenkin systems defined by |σn | , σ ep∗ := sup 1/p−2 n∈N (n + 1) log2[1/2+p] (n + 1) where 0 < p ≤ 1/2 and [1/2 + p] denotes integer part of 1/2 + p, is bounded from the Hardy space Hp to the Lebesgue space Lp . Moreover, the order of deviant behavior of the n-th Fejér mean was given exactly. As a corollary we get that kσn f kp ≤ cp (n + 1)1/p−2 log2[1/2+p] (n + 1) kf kHp . For Walsh-Kaczmarz system analogical theorems were proved in [33] and [69]. For the one-dimensional Vilenkin-Fourier series Weisz [85] proved that the maximal operator σ # f = sup |σMn f | n∈N

is bounded from the martingale Hardy space Hp to the Lebesgue space Lp for p > 0. For the Walsh-Fourier series Goginava [30] proved that the operator |σ2n f | is not bounded from the space Hp to the space Hp , for 0 < p ≤ 1. ,

August 30, 2015

3.2


85

D IVERGENCE OF V ILENKIN -F EJ E´ R MEANS ON MARTINGALE H ARDY SPACES

Theorem 3.1 a)There exist a martingale f ∈ H1/2 such that sup kσn f k1/2 = +∞. n∈N

b) Let 0 < p < 1/2. There exist a martingale f ∈ Hp , such that sup kσn f kweak−Lp = +∞. n∈N

Remark 3.2 This result for p = 1/2 can be found in Tephnadze [66]. By interpolation automatically follows the second part of this Theorem. Here, we make a more concrete proof of this fact by even constructing a martingale f ∈ Hp , for which the Vilenkin-Fejér means are not bounded in the space weak − Lp . Proof: Let f = f (n) : n ∈ N be the martingale defined in Example 1.49 in the case when p = q = 1/2. We can write that M2α 1 1 Xk Sj f + σqαk f = qαk j=1 qαk

qαk X

Sj f

j=M2αk +1

:= I + II. According to (1.76) in Example 1.49 we can conclude that |I| ≤

≤

M2α 1 Xk |Si f | qαk j=1

2 2λM2α k−1 M2αk 1/2

qαk

αk−1

≤

(3.3)

2 2λM2α k−1 1/2

.

αk−1

By applying (1.75) obtained by letting l = k we can rewrite II as II =

+

(qαk − M2αk ) SM2αk qαk

M2αk ψM2αk

qαk X

1/2

α k qαk

Dj−M2α

k

j=M2αk +1

:= II1 + II2 . G.Tephnadze

,

86

August 30, 2015


According to (1.76) for j = M2αk and p = 1/2 we find that 2λM 2 2αk−1 |II1 | ≤ SM2αk ≤ . 1/2 αk−1

(3.4)

In view of (3.3) and (3.4) we invoke estimate (1.73) for p = q = 1/2 to conclude that 2 2λM2α k−1

|II1 | ≤ and

1/2

αk−1

≤

1/2

αk−1 Let 2η,2(η+1)

x ∈ I2αk

, η=

Mα2k 3/2

16αk

2 2λM2α k−1

|I| ≤

≤

Mα2k 3/2

.

16αk hα i k

2

+ 1, ..., αk − 3.

By applying Lemma 1.19 we have that 2 4qαk −1 Kqαk −1 (x) ≥ M2η M2(η+1) ≥ M2η . Hence, for II2 we readily get that qαk −1 X M2αk |II2 | = 1/2 Dj ψM2αk α qα k

k

j=1

M2αk qαk −1 Kqαk −1 1/2 qαk α k 2 qαk −1 Kqαk −1 M2η ≥ ≥ . 1/2 1/2 2αk 8αk =

Since M2η ≥ Mαk we obtain that σqα f (x) ≥ |II2 | − (|I| + |II1 |) k ≥

1 1/2

8αk

(3.5)

Mα2k 2 M2η − . 2αk

From (3.5) it follows that 2 cM2η 2η,2(η+1) σqα f (x) ≥ , x ∈ I2αk , 1/2 k αk ,

August 30, 2015


where η=

hα i k

2

87

+ 1, ..., αk − 3,

Thus, Z Gm

≥

αX k −3

m2αk −1 −1 Z

m2η+3 −1

X

x2α

k −1

=0

X

1/4 αk η=[αk /2]+1 x2η+3 =0 αX k −3

c

≥

η=[αk /2]+1

≥

...

m2αk −1 −1 x2α

k −1

1/4 αk η=[αk /2]+1

3/4

≥ αk

1

1/4

αk

=0

M2η M2η+2

αX k −3

c

≥

2η,2(η+1) I2αk M2η

X

αX k −3

c

σqα f (x) 1/2 dµ (x) k

m2η+3... m2αk −1 M2η M2αk

1/4

αk

2η,2(η+1) I2α k

m2η+3 −1

αX k −3

c

X

...

η=[αk /2]+1 x2η+3 =0

≥

σqα f (x) 1/2 dµ (x) k

η=[αk /2]+1

→ ∞, when k −→ ∞.

The proof of part a) is complete so we turn to the proof of part b). Let 0 < p < 1/2. Let f = f (n) : n ∈ N be the martingale defined in Example 1.49 in the case when 0 < p < q = 1/2. We can write that σM2αk +1 f =

M2αk

1

X

M2αk + 1

Sj f +

j=0

SM2αk +1 f M2αk + 1

:= III + IV. We combine (1.76) and (1.77) and invoke (1.73) in the case when p < q = 1/2 to obtain the following estimates: 1/p M2αk 2λM2αk−1 |III| ≤ 1/2 M2αk + 1 αk−1 1/p

≤

2λM2αk−1 1/2

1/p−2

≤

αk−1 G.Tephnadze

Mαk

3/2

16αk

,

88

August 30, 2015


and |IV | ≥

SM2αk +1 f M2αk + 1

1/p−2

≥

M2αk 2αk

Let x ∈ Gm . We conclude that σM2αk +1 f (x) ≥ |IV | − |III| 1/p−2

≥

M2αk

1/2

2αk

1/p−2

1/p−2

−

Mαk

3/2

≥

M2αk

16αk

1/2

.

4αk

It follows that 1/p−2

M2αk

1/2

4αk

(

µ x ∈ Gm : σM2α

M 1/p−2 2αk +1 f (x) ≥ 1/2 k 4αk

)!1/p

1/p−2

≥

M2αk

1/2

4αk

→ ∞, when k → ∞.

Hence, also part b) is proved so the proof is complete. 3.3

M AXIMAL OPERATORS OF V ILENKIN -F EJ E´ R MEANS ON MARTINGALE H ARDY SPACES

In this subsection we consider weighted Maximal operators of Fejér means of VilenkinFourier series and prove (Hp , Lp ) and (Hp , weak − Lp ) type inequalities. In all cases we also show sharpness in a special sense. First we note the following consequence of Theorem 3.1: Corollary 3.3 a) There exist a martingale f ∈ H1/2 such that kσ ∗ f k1/2 = +∞. b) Let 0 < p < 1/2. There exist a martingale f ∈ Hp such that kσ ∗ f kweak−Lp = +∞. The next theorem can be found in Tephnadze [68], but here we will give a simpler proof of part b). Theorem 3.4 a) Let 0 < p < 1/2. Then the maximal operator σ ep∗ f := sup n∈N

|σn f | (n + 1)1/p−2 ,

August 30, 2015


89

is bounded from the Hardy martingale space Hp to the Lebesgue space Lp . b) Let {Φn : n ∈ N} be any non-decreasing sequence satisfying the condition (n + 1)1/p−2 = +∞. n→∞ Φn lim

Then sup

σM2n +1 fk k

ΦM2n +1 k

weal−Lp

kfk kHp

k∈N

(3.6)

= ∞.

Proof: First place we note that σn is bounded from L∞ to L∞ (see (1.23) in Corollary 1.16). Hence, by Lemma 1.39 the proof of Theorem 3.4 will be complete if we show that !p Z |σn a| dµ ≤ c < ∞ sup 1/p−2 n∈N (n + 1) IN

for every p-atom a, where I denotes the support of the atom. Let a be an arbitrary p-atom with support I and µ (I) = MN−1 . We may assume that I = IN . It is easy to see that σn (a) = 0 when n ≤ MN . Therefore we can suppose that n > MN . 1/p

Since kak∞ ≤ MN it follows that |σn (a)|

≤

1 (n + 1)1/p−2 ≤

(n + 1)1/p−2 Z |a (t)| |Kn (x − t)| dµ (t) IN

kak∞

Z

1/p

≤

|Kn (x − t)| dµ (t)

(n + 1)1/p−2

IN

Z

cMN

(n + 1)1/p−2

|Kn (x − t)| dµ (t) . IN

Let x ∈ INk,l , 0 ≤ k < l ≤ N. From Corollary 1.21 we can deduce that |σn (a)| (n + 1)1/p−2

(3.7)

1/p

≤

cp MN Ml Mk = cp Ml Mk . 1/p−2 M 2 MN N G.Tephnadze

,

90


August 30, 2015

The expression on the right-hand side of (3.7) does not depend on n. Thus, ∗ σ ep a (x) ≤ cp Ml Mk , for x ∈ INk,l , 0 ≤ k < l ≤ N.

(3.8)

By using (3.8) with identity (1.1) we obtain that Z ∗ p σ ep a (x) dµ (x) IN

=

mj−1

−1 N −2 N X X

Z

X

k=0 l=k+1 xj =0,j∈{l+1,...,N −1} N −1 Z X

+

k,N IN

k=0

≤ cp

∗ p σ ep a (x) dµ (x)

∗ p σ ep a (x) dµ (x)

−1 N −2 N X X

ml+1 ...mN −1 p p Ml Mk MN k=0 l=k+1 +cp

≤ cp

k,l IN

N −1 X

1 MNp Mkp M N k=0

N −1 X Mkp Mlp Mkp + cp Ml MN1−p k=0 k=0 l=k+1

N −2 N −1 X X

:= I + II. We estimate each term separately. I = cp

−1 N −2 N X X

≤ cp

≤ cp

Mlp Mkp Ml2p

(3.9)

N −2 N −1 X X

1 1−2p M l k=0 l=k+1

N −2 N −1 X X k=0 l=k+1

≤ cp

1

Ml1−2p k=0 l=k+1

1 2l(1−2p)

N −2 X

1 < cp < ∞. k(1−2p) 2 k=0

It is obvious that II ≤

N −1 cp X Mkp p MN1−2p k=0 MN

(3.10)

,

August 30, 2015

91


≤

cp < c < ∞. MN1−2p

The proof of the part a) is complete by combining the estimates above. Let 0 < p < 1/2. Under condition (3.6) there exists an increasing sequence of positive integers {λk : k ∈ N} such that 1/p−2

λk = ∞. k→∞ Φλk lim

It is evident that for every λk there exists a positive integers m,k such that q 2qm, . Since {Φn : n ∈ N} is a non-decreasing function we have that

m

0 k

< λk
MN . G.Tephnadze

,

94

August 30, 2015


Since kak∞ ≤ MN2 we obtain that

≤

1 log2 (n + 1) ≤ ≤

|σn (a)| log2 (n + 1) Z |a (t)| |Kn (x − t)| dµ (t) IN

kak∞ log2 (n + 1)

Z

MN2 2 log (n + 1)

Z

|Kn (x − t)| dµ (t) IN

|Kn (x − t)| dµ (t) . IN

Let x ∈ INk,l , 0 ≤ k < l ≤ N. Then, from Corollary 1.21 it follows that |σn (a)| log2 (n + 1)

(3.12)

cMl Mk cMN2 Ml Mk = . 2 2 N MN N2 The expression on the right-hand side of (3.12) does not depend on n. Therefore, ≤

|e σ ∗ a (x)| ≤

cMl Mk , for x ∈ INk,l , 0 ≤ k < l ≤ N. N2

(3.13)

By applying (3.13) with identity (1.1) we obtain that Z |e σ ∗ a (x)|1/2 dµ (x) IN

=

mj −1

N −2 N −1 X X

Z

X

k=0 l=k+1 xj =0,j∈{l+1,...,N −1} N −1 Z X

+

k=0

≤c

k,l IN

∗ 1/2 ∼ σ a (x) dµ (x)

|e σ ∗ a (x)|1/2 dµ (x)

k,N IN

N −2 N −1 X X

1/2

1/2

ml+1 . . . mN −1 Ml Mk MN N k=0 l=k+1

+c

N −1 X

1/2

1/2

1 MN Mk M N N k=0

:= I + II.

We estimate each term separately. I≤

N −2 N −1 N −2 1/2 c X X Mk cX ≤ 1 ≤ c < ∞. N k=0 l=k+1 Ml1/2 N k=0 ,

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95

Moreover, N −1 X

1

II ≤

1/2

Mk

1/2

MN N

≤

k=0

c < c < ∞. N

The proof of part a) is complete. Let {λk : k ∈ N} be an increasing sequence of positive integers such that log2 (λk + 1) = ∞. k→∞ Φ λk lim

It is evident that for every λk there exists a positive integers m,k such that qm0 ≤ λk < k qm0k +1 < c qm0 . Since Φn is a non-decreasing function we have that k

0 2 mk log2 (λk + 1) lim ≥ c lim = ∞. k→∞ Φqm, k→∞ Φλk k

Let {nk : k ∈ N} ⊂ {m0k : k ∈ N} be a subsequence of positive numbers N+ such that n2k =∞ k→∞ Φqn k lim

and let fk be the atom defined in Example 1.44. We combine (1.52) and (1.53) in Example 1.44 and invoke (1.11) in Lemma 1.2 to obtain that σqn fk k

Φqnk =

1

qn k X Sj fk

Φqnk qnk j=1 qnk X 1 = S f j k Φqnk qnk j=M +1 2nk qnk X 1 = D − D j M2n k Φqnk qnk j=M +1 2nk qn −1 k X 1 Dj+M2n − DM2n = k k Φqnk qnk j=1 qn −1 k X 1 = Dj Φqnk qnk j=1

G.Tephnadze

,

96

August 30, 2015


1 qnk −1 Kqnk −1 . Φqnk qnk

=

2s,2l Let x ∈ I2n . Then, in view of Lemma 1.19, we can conclude that k

σqn fk (x) cM2s M2l k ≥ . Φqnk M2nk Φqnk Hence, σ f 1/2 qnk k dµ Gm Φqnk

Z

≥

m2nk −1

nX 2l+1 k −3 n k −1 m X X

...

s=0 l=s+1x2l+1=0

≥c

σ f 1/2 qnk k dµ 2s,2l Φ qnk I2n

Z

X

x2nk −1 =0

k

nX k −3 n k −1 X

1/2 1/2 m2l+1 ...m2nk −1 M2s M2l 1/2 1/2 M2nk Φqn M2n s=0 l=s+1 k

≥c

k

nX k −3 n k −1 X

1/2 M2s 1/2 1/2 1/2 s=0 l=s+1 M2l M2nk Φqnk

cnk

≥

1/2

1/2

.

M2nk Φqnk From (1.54) in Example 1.44 we have that

σ f 1/2 2 q nk k dµ Gm Φqn

R

k

kfk kH1/ 2 ≥ ≥

cn2k M2nk M2nk Φqnk

cn2k → ∞, Φqnk

when

k → ∞.

Thus, also part b) is proved so the proof is complete. We also point out the following consequence of Theorem 3.8, which we need later on. Corollary 3.9 Let f ∈ H1/2 . Then there exists an absolute constant c such that kσn f k1/2 ≤ c log2 (n + 1) kf kH1/2 , n ∈ N+ . ,

August 30, 2015


97

Proof: According to part a) of Theorem 3.8 we readily conclude that

σn f

log2 (n + 1) 1/2

≤

sup n∈N

|σn f |

≤ c kf k H1/2 , n ∈ N+ . log2 (n + 1) 1/2

The proof is complete. We also mention the following corollaries: Corollary 3.10 Let {Φn : n ∈ N} be any non-decreasing sequence, satisfying the condition (3.11). Then there exists a martingale f ∈ H1/2 , such that

σn f

= ∞. sup Φn 1/2 n∈N Corollary 3.11 Let {Φn : n ∈ N} be any non-decreasing sequence, satisfying the condition (3.11). Then the following maximal operator sup n∈N

|σn f | Φn

is not bounded from the Hardy space H1/2 to the Lebesgue space L1/2 . The next results was proved by Tephnadze [53]. Theorem 3.12 a) Let 0 < p ≤ 1/2 and {nk : k ∈ N} be a subsequence of positive numbers such that sup ρ (nk ) = κ < c < ∞. (3.14) k∈N

Then the maximal operator σ e∗,M f := sup |σnk f | k∈N

is bounded from the Hardy space Hp to the Lebesgue space Lp . b) Let 0 < p < 1/2 and {nk : k ∈ N} be a subsequence of positive numbers satisfying the condition sup ρ (nk ) = ∞. (3.15) k∈N

Then there exists an martingale f ∈ Hp such that sup kσnk f kweak−Lp = ∞,

(0 < p < 1/2) .

k∈N

G.Tephnadze

,

98


August 30, 2015

Proof: Since σ e∗,M is bounded from L∞ to L∞ , by using Lemma 1.39, we obtain that the proof of part a) is complete if we show that Z |e σ ∗,M a (x)| < c < ∞, IN

for every p-atom a with support IN and µ (IN ) = MN−1 . Analogously to Theorem 3.8 we may assume that nk > MN . 1/p

Since kak∞ ≤ MN we find that Z |σnk (a)| ≤

|a (t)| |Knk (x − t)| dµ (t)

(3.16)

IN

Z |Knk (x − t)| dµ (t)

≤ kak∞ IN 1/p

Z |Knk (x − t)| dµ (t) .

≤ MN

IN

Let x ∈ INi,j and i < j < hnk i . Then x − t ∈ INi,j for t ∈ IN and, according to Lemma 1.11 we obtain that |KMl (x − t)| = 0, for all hnk i ≤ l ≤ |nk | . By applying (3.16) and (1.22) in Lemma 1.16 we get that |σnk a (x)| ≤

1/p MN

|nk | Z X l=hnk i

|KMl (x − t)| dµ (t) = 0,

IN

for x ∈ INi,j , 0 ≤ i < j < hnk i ≤ l ≤ |nk | . Let x ∈ INi,j , where hnk i ≤ j ≤ N. Then, in view of Corollary 1.21, we have that Z cMi Mj |Knk (x − t)| dµ (t) ≤ . MN2 IN By using again (3.16) we obtain that 1/p−2

|σnk a (x)| ≤ cp MN

Mi Mj .

Set % := min hαk i . k∈N

Then |e σ ∗,M a (x)| = 0, for x ∈ INi,j , 0 ≤ i < j ≤ %

(3.17) ,

August 30, 2015


99

and 1/p−2

|e σ ∗,M a (x)| ≤ cp MN

Mi Mj , for x ∈ INi,j , i < % ≤ j ≤ N − 1.

(3.18)

Analogously to (2.24) we can conclude that

and

N −%≤κ

(3.19)

MN1−2p ≤ λ(N −%)(1−2p) ≤ λκ(1−2p) < c < ∞, M%1−2p

(3.20)

where λ = supk mk . Let p = 1/2. By combining (3.17)-(3.19) with (1.1) we get that Z 1/2 |e σ ∗,M a| dµ IN

=

N −2 N −1 X X

Z

N −1 Z X

1/2

|e σ ∗,M a|

+

%−1 N −1 Z X X

≤ Z

i=% j=i+1

1/2

|e σ ∗,M a|

1/2

|e σ ∗,M a|

dµ +

i,j IN

% X

1/2

Mi

i=%

≤ cM%1/2

N −1 X

1/2

Mi

i=0 N −2 X

N −1 X

1/2 M%

+c

1/2

|e σ ∗,M a|

dµ

i,N IN

1

1/2 j=%+1 Mj

1

+ cp

1/2 j=i+1 Mj

1

dµ

N −1 Z X i=0

≤c

+

dµ

i,j IN

i=0 j=%

+

dµ

i,N IN

i=0

N −2 N −1 X X

1/2

|e σ ∗,M a|

i,j IN

i=0 j=i+1

N −2 X

Mi

i=0

MN

1/2

Mi

i=%

N −1 X

1/2 1/2

1 1/2

+c

Mi

≤ N − % + c ≤ c < ∞. Let 0 < p < 1/2. By combining (3.17), (3.18) and (3.20) with (1.1) we have that Z p |e σ ∗,M a| dµ IN G.Tephnadze

,

100

August 30, 2015


=

N −2 N −1 X X

Z

p

|e σ ∗,M a| dµ

i,j IN

i=0 j=i+1 N −1 Z X

p

|e σ ∗,M a| dµ

+

k,N IN

i=0

%−1 N −1 Z

≤

XX i=0 j=%

N −2 N −1 X X

Z

p

|e σ ∗,M a| dµ +

+

i=% j=i+1

p

|e σ ∗,M a| dµ

i,j IN

i,j IN

i=0 %

≤ cp MN1−2p

X

N −2 X

N −1 X

Mip

i=0

+MN1−2p

Mip

i=%

≤

cp MN1−2p M%1−2p

N −1 Z X

p

|e σ ∗,M a| dµ

i,N IN

N −1 X

1 Mj1−p j=%+1

N −1 X 1 Mip + c p MNp Mj1−p j=i+1 i=0

+ cp ≤ cp λ(|nk |−hnk i)(1−2p) + cp < ∞.

The proof of part a) is complete. Let {nk : k ∈ N} be a sequence of positive numbers satisfying condition (3.15). Then sup k∈N

M|nk | = ∞. Mhnk i

(3.21)

Under condition (3.21) there exists a sequence {αk : k ∈ N} ⊂ {nk : k ∈ N} such that (1−2p)/2 ∞ X Mhα i k

(1−2p)/2 k=0 M|αk |

< c < ∞.

Let f = f (n) : n ∈ N be the martingae defined in Example 1.48, where (1/p−2)/2

λk =

λMhαk i

(1/p−2)/2

M|αk |

,

λ = sup mn .

(3.22)

n∈N

By applying (1.38) we can conclude that f ∈ Hp . By now using (1.68) with λk defined by (3.22) we obtain that fb(j)

(3.23) ,

August 30, 2015


101

 1/2p (1/p−2)/2  M|αk | Mhαk i , j ∈ M|αk | , ..., M|αk |+1 − 1 , k ∈ N+ , ∞ S =  0, j∈ / M|αk | , ..., M|αk |+1 − 1 . k=0

Therefore, M

σαk f =

|αk | 1 X 1 Sj f + αk j=1 αk

αk X

Sj f

j=M|α | +1 k

:= I + II. Let M|αk | < j ≤ αk . Then, according to (3.23) we have that 1/2p

(1/p−2)/2

Sj f = SM|α | f + M|αk | Mhαk i

Dj−M|α | . k

k

(3.24)

By applying (3.24) we can rewrite II as II = 1/2p

+

αk − M|αk | SM|α | f k αk

(1/p−2)/2

M|αk | Mhαk i

ψM|α

αk X

k|

αk

Dj−M|α

k|

j=M|α | +1 k

:= II1 + II2 . In view of Corollary 2.13 for II1 we find that kII1 kpweak−Lp ≤

αk − M|αk | αk ≤

p

p

SM|α | f

cp kf kpHp

k

weak−Lp

< ∞.

By using part a) of Theorem 3.12 for I we obtain that

p M|αk | p

p kIkweak−Lp =

σM|α | f k αk weak−Lp ≤ c kf kpHp ≤ cp < ∞. Under condition (3.15) we can conclude that

hαk i = 6 |αk | and αk − M|αk | = hαk i . G.Tephnadze

,

102

August 30, 2015


k i−1,hαk i . According to Lemma 1.18 we get that Let x ∈ I hα α +1

h

ki

(1/p−2)/2 αk −M|αk |

1/2p

|II2 | = 1/2p

M|αk | Mhαk i αk

X j=1

Dj

(1/p−2)/2

M|αk | Mhαk i

αk − M|αk | Kαk −M|α | k αk (1/p−2)/2 1/2p−1 αk − M|αk | Kαk −M|α | ≥ cM|αk | Mhαk i =

k

≥

(1/p+2)/2 1/2p−1 . cM|αk | Mhαk i

It follows that kII2 kpweak−Lp o p n (1/p+2)/2 (1/p−2)/2 (1/p+2)/2 (1/p−2)/2 − cp Mhαk i Mhαk i µ x ∈ Gm : |II2 | ≥ cp M|αk | ≥ cp M|αk | n o 1/2+p 1/2−p k i−1,hαk i ≥ cp M|αk | Mhαk i µ I hα α +1 h

≥

1/2−p cp M|αk | 1/2−p Mhαk i

ki

→ ∞, when k → ∞.

By now combining the estimates above we can conclude that kσαk f kpweak−Lp ≥ kII2 kpweak−Lp − kII1 kpweak−Lp − kIkpweak−Lp ≥

1 kII2 kpweak−Lp 2

1/2−p

≥

cp M|αk |

1/2−p

Mhαk i

→ ∞, when k → ∞.

The proof is complete. From the part a) of Theorem 3.12 follows immediately the following well-known results of Weisz [85]: Corollary 3.13 Let p > 0. Then the maximal operator σ # f := sup |σMn f | n∈N

is bounded from the Hardy space Hp to the Lebesgue space Lp . ,

August 30, 2015


103

Proof: It is obvious that |Mn | = [Mn ] = n and sup ρ (Mn ) = 0 < c < ∞. n∈N

It follows that condition (3.14) is satisfied and the proof is complete by just applying part a) of Theorem 3.12. Our next results reads: Theorem 3.14 Let 0 < p ≤ 1. Then the operator |σMn f | is not bounded from the martingale Hardy space Hp to the martingale Hardy space Hp . Remark 3.15 This result for the Walsh system can be found in Goginava [30] and for bounded Vilenkin systems in the paper of Persson and Tephnadze [52]. Proof: Let fk be the martingale from Example 1.44. By combining (1.11) in Lemma 1.2 and (1.52), (1.53) in Example 1.44 we find that σM2nk +1 fk =

=

M2nk +1

1

X

M2nk +1

Sj fk

j=1 M2nk +1

1

X

M2nk +1 j=M

Sj fk

2nk +1

=

1

M2nk +1

=

X

M2nk +1 j=M 1

=

M2nk +1

Dj − DM2nk

2nk +1

M2nk

X

Dj+M2nk − DM2nk

j=1

M2n ψM2nk Xk ψM2nk Dj = KM2nk . M2nk +1 j=1 m2nk

It is evident that Z = Gm

SMN σM2nk +1 fk (x) σM2nk fk (t) DMN (x − t) dµ (t) G.Tephnadze

,

104

August 30, 2015


Z

σM2nk fk (t) DMN (x − t) dµ (t)

≥ I2nk

Z ≥c I2nk

KM2nk +1 (t) DMN (x − t) dµ (t) Z

≥ cM2nk

DMN (x − t) dµ (t) . I2nk

Since ψj (t) = 1, for t ∈ I2nk , j = 0, 1, ..., M2nk − 1 we obtain that

SMN σM2nk +1 fk (x) ≥ cM2nk

1 DMN (x) , N = 0, 1, ..., MN − 1 M2nk

and sup 1≤N 0. M2αk + 1

Hence also part b) is proved so the proof is complete. Theorem 3.24 a) Let f ∈ H1/2 and 1 1 ωH1/2 ,f = o , when n → ∞. Mn n2

(3.34)

Then kσn f − f kH1/2 → 0, when n → ∞. b) There exists a martingale f ∈ H1/2 for which 1 1 ωH1/2 ,f = O , when n → ∞ Mn n2 and kσn f − f k1/2 9 0, when n → ∞. G.Tephnadze

,

110

August 30, 2015


Proof: Let f ∈ H1/2 and MN < n ≤ MN +1 . By using identity (3.30) we find that 1/2

kσn f − f kH1/2 ≤ kσn f − σn SMN f k1/2 H1/2 1/2 + kσn SMN f − SMN f k1/2 H1/2 + kSMN f − f kH1/2 1/2 1/2 = kσn (SMN f − f )k1/2 H1/2 + kSMN f − f kH1/2 + kσn SMN f − SMN f kH1/2 1/2 1 2 , f + kσn SMN f − SMN f k1/2 ≤ c log n + 1 ωH1/2 H1/2 . MN

Let p > 0 and f ∈ Hp . By combining (3.31), (3.32) and equality (3.30) we have that kσn SMN f − SMN f k1/2 H1/2 =

(3.35)

MN kSMN (σMN f − f )k1/2 H1/2 → 0 when k → ∞. n

It follows that under condition (3.34) we obtain that kσn f − f kH1/2 → 0, when n → ∞. and the proof of part a) is complete. Let f = f (n) : n ∈ N be a martingale from the Lemma 1.52 where p = 1/2. Then f ∈ H1/2 and 1 1 ωH1/2 ( , f) = O , when n → ∞. Mn n2 Hence, σqMk f − f =

M2Mk σM2Mk f qMk −

+

1 qMk

(3.36) qMk X

Sj f

j=M2Mk +1

M2Mk f qM −1 f − k . qMk qMk

Let M2Mk < j ≤ qMk . By combining (1.85) and (1.87) we have that qMk X

1 qMk

Sj f

j=M2M +1 k

=

qMk −1 SM2Mk f qMk

+

M2Mk ψM2M

k

qMk Mk2

qMk −1

X

Dj

j=1 ,

August 30, 2015


=

qMk −1 SM2Mk f

+

qMk

M2Mk ψM2M qMk −1 KqMk −1 k

qMk Mk2

111

.

By using (3.36) we get that 1/2

kσqMk f − f k1/2 ≥

−

c 1/2 kqMk −1 KqMk −1 k1/2 Mk

M2Mk qMk

1/2

qMk −1 qMk

1/2

−

(3.37)

1/2

kσM2M f − f k1/2 k

1/2

kSM2Mk f − f k1/2 .

Let 2s,2η x ∈ I2M , k

s = η + 2, η + 3, Mk − 2.

By applying Lemma 1.19 we find that M M 2η 2s qMk −1 KqMk −1 (x) ≥ . 4 Hence, Z Gm

≥c

M k −4 M k −2 m2s+1 X X X−1

1/2 qMk −1 KqMk −1 dµ m2Mk −1 −1 Z

...

η=1 s=η+2 x2s+1 =0

≥c

X

x2α

k −1

=0

2s,2η I2M

(3.38)

1/2 qMk −1 KqMk −1 dµ

k

M k −4 M k −2 X X η=1

1 1/2 1/2 M2s M2η ≥ cMk . 2M 2s s=η+2

By combining (3.37) and (3.38) we obtain that lim supkσqMk f − f k1/2 ≥ c > 0. k→∞

Thus also the part b) is proved so the proof is complete. G.Tephnadze

,

112 3.5


S TRONG

CONVERGENCE OF

V ILENKIN -F EJ E´ R

August 30, 2015

MEANS ON MARTINGALE

H ARDY

SPACES

The first result in this section is due to Blahota and Tephnadze [5]: Theorem 3.25 a) Let 0 < p < 1/2 and f ∈ Hp . Then there exists an absolute constant cp , depending only on p, such that ∞ X kσk f kpp k=1

k 2−2p

≤ cp kf kpHp .

b) Let 0 < p < 1/2 and {Φk : k ∈ N} be any non-decreasing sequence satisfying the conditions Φn ↑ ∞ and k 2−2p lim = ∞. (3.39) k→∞ Φk Then there exists a martingale f ∈ Hp such that ∞ X kσk f kpweak−Lp

Φk

k=1

= ∞.

Proof: According to Lemma 1.38 it suffices to show that ∞ X kσm akpp m=1

m2−2p

≤c MN . Let x ∈ IN . Since σn is bounded from L∞ to L∞ (see (1.23) in Corollary 1.16) and 1/p kak∞ ≤ MN we obtain that Z kakp∞ |σm a|p dµ ≤ c ≤ c < ∞, 0 < p < 1/2. MN IN Hence,

∞ X

R IN

|σm a|p dµ m2−2p

m=1

≤

∞ X m=1

1 m2−2p

≤ c < ∞.

It is easy to see that |σm a (x)| Z ≤

|a (t)| |Km (x − t)| dµ (t) IN ,

August 30, 2015

113


Z |Km (x − t)| dµ (t)

≤ ka (x)k∞ IN 1/p

Z

≤ MN

|Km (x − t)| dµ (t) . IN

Let x ∈ INk,l , 0 ≤ k < l ≤ N. Then, from Corollary 1.21 it follows that 1/p−2

|σm a (x)| ≤ cp Ml Mk MN

that

.

(3.40)

By combining (1.1) and (3.40) if we invoke also estimates (3.9) and (3.10) we obtain Z |σm a (x)|p dµ (x) (3.41) IN

=

mj−1

N −2 N −1 X X

Z

X

k=0 l=k+1 xj =0, j∈{l+1,...,N −1}

+cp

N −1 Z X k,N IN

k=0

≤ cp

k,l IN

|σm a (x)|p dµ (x)

|σm a (x)|p dµ (x)

−1 N −2 N X X

ml+1 · · · mN −1 p p 1−2p Ml Mk MN MN k=0 l=k+1 +cp

≤ cp MN1−2p

N −1 X

1 Mkp MN1−p M N k=0

N −1 X Mkp Mlp Mkp + cp Ml MNp k=0 k=0 l=k+1

N −2 N −1 X X

≤ cp MN1−2p . Let 0 < p < 1/2. By using (3.41) we get that R ∞ X |σm a (x)|p dµ (x) IN m2−2p

m=MN +1 ∞ X

≤

m=MN

cMN1−2p < c < ∞. m2−p +1

and the proof of part a) is complete. Under condition (3.39) there exists an increasing numbers {αk : k ∈ N}, such that lim

2−2p cM|α k |+1

k→∞ ΦM |αk |+1 G.Tephnadze

= ∞.

,

114

August 30, 2015


There exists a sequence {αk : k ∈ N} ⊂ {nk : k ∈ N} such that |αk | > 2, for k ∈ N and

1/2

∞ Φ X M|αη |+1

(3.42)

1−p M|α η|

η=0

1/2

=

m1−p |αη |

∞ Φ X M|αη |+1 η=0

1−p M|α η |+1

< c < ∞.

Let f be the martingale defined in Example 1.48 in the case when 1/2p

λk =

λΦM α

| k |+1 . 1/p−1

(3.43)

M|αk |

By applying (3.42) we can conclude that f ∈ Hp . By now using (3.43) with λk defined by (1.68) we readily get that fb(j)

(3.44)

 1/2p   ΦM|α |+1 , j ∈ M|αk | , . . . , M|αk |+1 − 1 , k ∈ N, k ∞ = S  j∈ / M|αk | , . . . , M|αk |+1 − 1 .  0, k=0

Hence, M

|αk | 1 X 1 σαk f = Sj f + αk j=1 αk

αk X

Sj f

(3.45)

j=M|α | +1 k

:= III + IV. Let M|nk | < j < αk . According to (3.44) and (1.69), we can write that Sj f =

k−1 X

1/2p

ΦM α |

η |+1

(3.46)

DM|αη |+1 − DM|αη |

η=0 1/2p

+ΦM α |

k |+1

ψM|α | Dj−M|α | . k

k

By using (3.46) in IV we obtain that IV =

k−1 αk − M|nk | X 1/2p ΦM α +1 DM|αη |+1 − DM|αη | | η| αk η=0

(3.47)

,

August 30, 2015

Martingale Hardy spaces and summability of the one dimensional Vilenkin-Fourier series 1/2p

+

ΦM α |

k |+1

ψM|α

αk

αk X

k|

Dj−M|α

115

k|

j=M|α | +1 k

:= IV1 + IV2 . We calculate each term separately. First we define a set of positive numbers n ∈ N1, , for which hni = 1, that is,   |n|   X N1 := n ∈ N : n = n1 M1 + ni Mi ,   i=2

where n1 ∈ {1, . . . , m1 − 1} and ni ∈ {0, . . . , mi − 1} , for i ≥ 2. Let αk ∈ N1 and x ∈ I20,1 . Since αk − M|αk | ∈ N2 and |αk | = 6 hαk i for every |αk | > 2, by applying Lemma 1.18, we find that 1/2p |αk | cp ΦM α +1 αk −M X | k| |IV2 | = D (x) j αk j=1 1/2p

=

cp ΦM α |

k |+1

αk

αk − M|αk | Kαk −M|α | (x) k

≥

1/2p cp ΦM α +1 | k|

αk

.

Let x ∈ I20,1 , n ≥ 2 and 1 ≤ sn ≤ mn − 1. By combining Corollary 1.5, Lemma 1.11 and (1.18) in Lemma 1.13 we have that Ksn Mn (x) = DMn (x) = 0,

for n ≥ 2.

Since |αk | > 2, k ∈ N, we obtain that IV1 = 0, for x ∈ I20,1 .

(3.48)

Moreover, if we invoke (1.69) and (1.70) with (3.43) we get that  1/2p Φ M|αs |+1 ψM|αs | Dj−M|α | ,   s  M s ∈ N + |αs | < j ≤ M|αs |+1 , Sj f =  0,   M|αs |+1 < j ≤ M|αs+1 | , s ∈ N+ and

M

|αη |+1 k−1 X 1 X 1/2p III = Φ M|αη |+1 ψM|αη | Dv−M|αη | n η=0 v=M +1

(3.49)

|αη |

G.Tephnadze

,

116

August 30, 2015


k−1

1 X 1/2p M|αη |+1 ψM|αη | Φ = n η=0 =

m|αη | −1 M|αη |

X

Dv−M|αη |

v=1

k−1 1 X 1/2p M|αη |+1 m|αη | − 1 M|αη | Km −1M (x) = 0. Φ n η=0 |αη | |αη |

Let 0 < p < 1/2, n ∈ N2 and M|αk | < n < M|αk |+1 . By combining (3.45)-(3.49) we find that kσn f kpweak−Lp   1/2 1/2p cp ΦM α +1  cp ΦM α +1  | k| | k| ≥ µ x ∈ I20,1 : |IV2 | ≥   αkp αk 1/2

1/2

≥

cp ΦM α

| k |+1 0,1 µ I2 αkp

≥

cp ΦM α |

k |+1

p M|α k |+1

.

Since X

1 ≥ cMk ,

{n∈N1 :Mk ≤n≤Mk+1 }

we obtain that

∞ X kσn f kpweak−Lp

Φn

n=1

kσn f kpweak−Lp

X

≥

n o n∈N1 : M|α | MN . Hence, tn (a) =

n 1 X qn−k Sk (a) Qn k=1

(4.7)

n 1 X qn−k Sk (a) Qn k=M N Z n 1 X qn−k a (x) Dk (x − t) dµ (t) . Qn k=M IN

=

N

Since kak∞ ≤

1/p MN

it follows that

|tn (a)| (n + 1)1/p−2

≤

kak∞

Z

n 1 X qn−k Dk (x − t) dµ (t) Qn

(n + 1)1/p−2 IN k=MN Z n 1/p 1 X MN ≤ qn−k Dk (x − t) dµ (t) . 1/p−2 Q n (n + 1) IN k=M N

Let x ∈

INk,l ,

0 ≤ k < l ≤ N. From Lemma 1.27 we can deduce that |tn (a)| (n + 1)

1/p−2

1/p

≤

cp MN Ml Mk = cp Ml Mk . 1/p−2 M 2 MN N

(4.8)

,

August 30, 2015


The expression on the right-hand side of (4.8) does not depend on n. Therefore, ∗ e tp,1 a (x) ≤ cp Ml Mk , for x ∈ INk,l , 0 ≤ k < l ≤ N.

125

(4.9)

By combining (1.1) with (4.9) we obtain that Z ∗ p e tp,1 a (x) dµ (x) IN

=

mj−1

N −2 N −1 X X

Z

X

k=0 l=k+1 xj =0,j∈{l+1,...,N −1} N −1 Z X

+

k=0

≤ cp

k,N IN

k,l IN

∗ p e tp,1 a (x) dµ (x)

∗ p e tp,1 a (x) dµ (x)

N −2 N −1 X X

N −1 X ml+1 ...mN −1 p p 1 Ml Mk + cp MNp Mkp M M N N k=0 l=k+1 k=0

≤ cp

−1 N −2 N X X

N −1 X Mlp Mkp Mkp < c < ∞. + cp Ml MN1−p k=0 l=k+1 k=0

The proof is complete by using this estimate and Lemma 1.39. Theorem 4.4 Let the sequence {qk : k ∈ N} be non-decreasing. Then the maximal operator ∼∗

t 1 f := sup n∈N

|tn f | log2 (n + 1)

is bounded from the Hardy space H1/2 to the Lebesgue space L1/2 . Remark 4.5 Since the Fejér means are examples of Nörlund means with non-decreasing sequence {qk : k ∈ N} we immediately obtain from part b) of Theorem 3.8 that the asymptotic behaviour of the sequence of weights 1/ log2 (n + 1) : n ∈ N in Nörlund means in Theorem 4.4 can not be improved. Proof: The idea of proof is similar as that of part a) of Theorem 3.8, but since this case is more general we give the details. Analogously to Theorem 4.2 we may assume that n > MN and a be a p-atom with support I = IN . Since kak∞ ≤ MN2 if we apply (4.7) we obtain that |tn (a)| log2 (n + 1) G.Tephnadze

,

126

August 30, 2015

4. Vilenkin-Nörlund means on martingale Hardy spaces

n 1 X qn−k Dk (x − t) dµ (t) IN Qn k=M N Z n MN2 1 X ≤ qn−k Dk (x − t) dµ (t) . 2 log (n + 1) IN Qn k=M kak∞ ≤ log2 (n + 1)

Z

N

Let x ∈ INk,l , 0 ≤ k < l ≤ N. Then, from Lemma 1.27 it follows that Ml Mk cMl Mk |tn (a)| cMN2 = ≤ 2 2 2 M log (n + 1) log (n + 1) log2 (n + 1) N and

cM M e∗ l k . t1 a (x) ≤ N2

(4.10)

By combining (1.1) with (4.10) we find that Z 1/2 e∗ t1 a (x) dµ (x) IN

=

mj −1

−1 N −2 N X X

Z

X

k,l IN

k=0 l=k+1 xj =0,j∈{l+1,...,N −1} N −1 Z X

+

k,N IN

k=0

≤c

∼∗ 1/2 t1 a (x) dµ (x)

1/2 e∗ t1 a (x) dµ (x)

N −2 N −1 X X

1/2

1/2

ml+1 ...mN −1 Ml Mk MN N k=0 l=k+1 +c

N −1 X

≤c

+c

1/2

1/2

1 MN Mk MN N k=0

N −1 X

N −2 N −1 X X

1/2

Mk

1/2 k=0 l=k+1 N Ml

1

1/2

Mk 1/2 N k=0 MN

≤ c < ∞.

The proof is complete. The next results deal with a Nörlund means with non-increasing sequence {qk : k ∈ N} . We begin by stating a divergence result for all such summability methods when 0 < p < 1/2, which can be found in Persson, Tephnade and Wall [50]. ,

August 30, 2015


127

Theorem 4.6 Let 0 < p < 1/2. Then, for all Nörlund means with non-increasing sequence {qk : k ∈ N} there exists a martingale f ∈ Hp such that sup ktn f kweak−Lp = ∞. n∈N

Proof: For the proof we use the martingale defined in Example 1.49 (see also the part b) of Theorem 4.1). It is obvious that for every non-increasing sequence {qk : k ∈ N} it automatically holds that

1 q0 ≥ . QM2αk +1 M2αk + 1 Since tM2αk +1 f = +

M2αk

1

X

QM2αk +1

qM2αk +1−j Sj f

j=0

q0 SM +1 f := I + II. QM2αk +1 2αk

by combining (4.3) and (4.4) we see that 1/p−2 M2αk . tM2αk +1 f ≥ |II| − |I| ≥ 1/2 8αk

Analogously to (4.5) and (4.6) we get that

sup tM2αk +1 f k∈N

= ∞.

weak−Lp

The proof is complete. Corollary 4.7 Let 0 < p < 1/2 and tn be Nörlund means with non-increasing sequence {qk : k ∈ N}. Then the maximal operator t∗ is not bounded from the martingale Hardy space Hp to the space weak − Lp , that is there exists a martingale f ∈ Hp , such that sup kt∗ f kweak−Lp = ∞. n∈N

Next, we present necessary condition for the Nörlund means with non-increasing sequence {qk : k ∈ N}, when 1/2 ≤ p < 1, which can be found in Persson, Tephnade and Wall [50]. G.Tephnadze

,

128

August 30, 2015


Theorem 4.8 a) Let 0 0, 0 < α ≤ 1. n→∞ Qn lim

(4.11)

Then there exists a martingale f ∈ Hp such that sup ktn f kweak−Lp = ∞.

(4.12)

n∈N

b) Let {qk : k ∈ N} be a non-increasing sequence satisfying the condition nα = ∞, n→∞ Qn lim

(0 < α ≤ 1) .

(4.13)

Then there exists an martingale f ∈ H1/(1+α) , such that sup ktn f kweak−L1/(1+α) = ∞.

(4.14)

n∈N

Proof: Under condition (4.11) there exists an increasing sequence {αk : k ∈ N} of positive integers such that α M2α k ≥ c, k ∈ N QM2αk +1 and the estimates (1.71)-(1.73) are satisfied. To prove part a) we use the martingale defined in Example 1.49 in the case when 0 < p < q = 1/ (1 + α) . Since tM2αk +1 f =

M2αk

1

X

QM2αk +1

qM2αk +1−j Sj f +

j=0

1 SM +1 f QM2αk +1 2αk

:= I + II, by combining (4.3) and (4.4), we get that tM2αk +1 f ≥ |II| − |I| 1/p−1

=

M2αk

1/2

4αk that

1 QM2αk +1

1/p

−

2λMαk−1 1/2

.

αk−1

Without lost the generality we may assume that c = 1 in (4.11). By using (1.73) we find tM2αk +1 f ,

August 30, 2015

Martingale Hardy spaces and summability of the one dimensional Vilenkin-Fourier series 1/p−1−α

≥

M2αk

1/2

2λMαk−1

4αk M2αk

1/2

M2αk

1/2

8αk

1/p−1−α

1/p−1−α

−

λMαk

≥

3/2

4αk and

1/2

≥

αk−1

1/p−1−α

≥

1/p−1−α

1/p

−

M2αk

1/2

16αk

8αk

(

M 1/p−1−α 2αk · µ x ∈ Gm : tM2αk +1 f ≥ 8αk

1/p−1−α

M2αk 8αk

129

)

1/p−1−α

=

M2αk 8αk

→ ∞, when k → ∞.

Which means that 4.12 holds and part (a) is proved. Under condition (4.13) there exists an increasing sequence {αk : k ∈ N} which satisfies the conditions ∞ Q1/2(1+α) X M2αk +1 ≤ c < ∞, (4.15) α/2(1+α) M 2αk k=0 k−1 X

α/2+1

QM2αη +1 M2αη

α/2+1

≤ QM2αk +1 M2αk

(4.16)

η=0

and

α/2

α/2+1

32λQM2αk−1 +1 M2αk−1
0 (4.22) n→∞ Qn and |qn − qn+1 | ≥ cα nα−2 , n ∈ N. (4.23) Then there exists a martingale f ∈ H1/(1+α) such that sup ktn f k1/(1+α) = ∞. n∈N

Remark 4.12 Part a) of this result can be found in [50], but below we give a different proof. Part b) of the theorem is new. G.Tephnadze

,

132


August 30, 2015

Proof: By Lemma 1.40 the proof of part a) is complete if we show that tµ x ∈ IN : t∗ f ≥ t1+α ≤ c < ∞, t ≥ 0, for every 1/ (1 + α)-atom a. We may assume that a is an arbitrary 1/ (1 + α)-atom with support I, µ (I) = MN−1 and I = IN . It is easy to see that tm (a) = 0, when m ≤ MN . Therefore we can suppose that m > MN . Let x ∈ IN . Since tm is bounded from L∞ to L∞ (the boundedness follows from Corol1/(1+α) lary 1.33) and kak∞ ≤ cMN we obtain that Z |tm a (x)| ≤ |a (t)| |Fm (x − t)| dµ (t) IN

Z ≤ kak∞

|Fm (x − t)| dµ (t) IN

≤ MN1+α

Z |Fm (x − t)| dµ (t) . IN

Let x ∈ INk,l , 0 ≤ k < l ≤ N and m > MN . From Corollary 1.35 we get that |tm a (x)| ≤ cα Mk Mlα .

(4.24)

The expression on the right-hand side of (4.24) does not depend on m. Hence, |t∗ a (x)| ≤ cα Mk Mlα

(4.25)

Let n ≥ N. According to (4.25) we conclude that µ x ∈ IN : t∗ f ≥ cα Mn1+α = 0. Thus, we can suppose that 0 < n < N. Let λ = supn mn and [x] denotes the integer part of x. It is obvious that for fixed λ there exists a positive number θ so that λ1/θ ≤ 2. Then, for every k < n, it yields that α Mk Mn+[(n−k)/θ] ≤ Mk Mnα λ[(n−k−1)/θ]

≤ Mk Mnα λ1/θ

n−k−1

≤ Mk Mnα 2n−k−1

≤ Mk Mnα mk mk+1 ...mn−1 ≤ Mn1+α . It is obvious that if n + [(n − k − 1) /λ] > N , for some k < l < N we readily get that α cα Mk Mlα ≤ cα Mk MNα ≤ cα Mk Mn+[(n−k−1)/λ] ≤ cα Mn1+α .

It follows that for such k < l < N we have the following estimate |tm a (x)| ≤ cα Mn1+α , for x ∈ INk,l ,

August 30, 2015


and

133

o n µ x ∈ INk,l : t∗ f ≥ cα Mn1+α = 0. Therefore, we may assume that n + [(n − k − 1) /λ] ≤ N. By combining (1.1) and (4.25) we obtain that µ x ∈ IN : t∗ f ≥ cα Mn1+α ≤

N −1 X

mj−1

N X

k,l IN

X

k=n l=k+1 xj =0, j∈{l+1,...,N −1} n X +

mj−1

N X

k,l IN

X

k=0 l=n+[(n−k−1)/λ] xj =0, j∈{l+1,...,N −1}

≤

N −1 X

n N X X 1 + Ml k=0 k=n l=k+1

N X l=n+[(n−k)/θ]

1 c ≤ . Ml Mn

Hence, sup Mn µ x ∈ IN : t∗ f ≥ Mn1+α ≤ c < ∞. n∈N

Part a) is proved. Under condition (4.22) there exists an increasing sequence {αk : k ∈ N} of positive integers such that α M2α k +1 > cα > 0, k ∈ N, (4.26) QM2αk +1 and estimates (1.71)-(1.73) are satisfied. To prove part b) of Theorem 4.8 we use the martingale defined in Example 1.49 in the case when p = q = 1/ (1 + α) . In particular, as we proved there, the martingale belongs to the space H1/(1+α) . Moreover,

fb(j) =

α M2α

  

k 1/2

αk

  0,

, j ∈ {M2αk , ..., M2αk +1 − 1} , k ∈ N, ∞ S j∈ / {M2αk , ..., M2αk +1 − 1} . k=1

We can write that tM2αk +M2s f =

+

1 QM2αk +M2s

M2αk

1

X

QM2αk +M2s

qM2αk +M2s −j Sj f

j=0

M2αk +M2s

X

qM2αk +M2s −j Sj f := I + II.

j=M2αk +1 G.Tephnadze

,

134

August 30, 2015


According to (1.76) we can conclude that |I| ≤

≤

M2αk

1

X

QM2αk +M2s

qM2αk +M2s −j |Sj f | M2αk

α+1 2λM2α k−1

1

1/2 αk−1

QM2αk +M2s

≤

(4.27)

j=0

X

α+1 2λM2α k−1 1/2 αk−1

≤

qM2αk +M2s −j

j=0

Mααk 3/2

.

16αk

Let x ∈ Is /Is+1 and M2αk + 1 ≤ j ≤ M2αk + M2s . In view of the second inequality of (1.75) in the case when l = k and p = 1/ (1 + α) we can write that Sj f = SM2α f +

α M2α ψ Dj−M2α k M2α k

1/2

k

.

αk

k

Hence, it yields that II =

M2αk +M2s

1

X

QM2αk +M2s

+

qM2αk +M2s −j

α Dj−M2α M2α ψ k M2α k

k

1/2

αk

j=M2αk +1 M2αk +M2s

1

X

QM2αk +M2s

qM2αk +M2s −j SM2α f k

j=M2αk +1

:= II1 + II2 . By using again (1.76) we get that |II2 | ≤

M2αk +M2s

α+1 2λM2α k−1

1

1/2 αk−1

QM2αk +M2s

≤

α+1 2λM2α k−1 1/2 αk−1

X

≤

qM2αk +M2s −j

j=M2αk +1

Mαα+1 k 1/2

.

16αk

Let x ∈ Is /Is+1 , for [αk /2] < s ≤ αk . Then, according to (4.26) we find that M2s ψM M α X 1 2αk 2αk |II1 | = qM2s −j Dj 1/2 QM2αk +M2s α j=1

k

=

α M2α k 1/2

αk QM2αk +M2s

M 2s X qM2s −j j j=1

,

August 30, 2015


135

M 2s X ≥ 1/2 qM2s −j j αk QM2αk +1 j=1 M 2s c X qM2s −j j . ≥ 1/2 α λ α α M2α k

j=1

k

We invoke Abel transformation and apply (4.23) to get that M 2s X qM2s −j j j=1

M 2s X j (j + 1) (qM2s −j − qM2s −j−1 ) = 2 j=1

≥

2 cα M2s

M2s X

1/2 αk

j=[M2s /2]

|qM2s −j − qM2s −j−1 | j 2

M2s X

2 ≥ cα M2s

|qM2s −j − qM2s −j−1 |

j=[M2s /2] [M2s /2] 2 ≥ cα M2s

X

|qj − qj+1 |

j=1 [M2s /2] 2 ≥ cα M2s

X j=1

1 α−1 2 α+1 ≥ cα M2s M2s ≥ cα M2s . j α−2

Hence, M 2s c M α+1 cα X α 2s |II1 | ≥ 1/2 qM2s −j j ≥ . 1/2 αk j=1 αk By now using the estimates above we obtain that Z 1/(1+α) dµ ≥ |II1 | − |II2 | − |I| tM2αk +M2s f

(4.28)

Gm

≥

1+α cα M2s 1/2

−

αk

4λMαα+1 k 3/2

≥

1+α cα M2s

αk

1/2

.

αk

By combining (4.27) and (4.28) we find that Z |t∗ f |1/(1+α) dµ Gm G.Tephnadze

,

136

August 30, 2015


≥

αX k −1

Z Is /Is+1

s=[αk /2]

≥ cα

αX k −1

1/(1+α) dµ tM2αk +M2s f

M2s

1/2(1+α) s=[αk /2] M2s αk

cα αk 1/2(1+α) αk

1

1/2(1+α) s=[αk /2] αk

1/2(1+α)

αk ≥

αX k −1

αX k −1

cα

≥

≥ cα

1

s=[αk /2] 1/2

≥ cα α k

→ ∞, when k → ∞.

The proof is complete. Our next result reads: Theorem 4.13 Let f ∈ Hp , where 0 < p < 1/ ()1 + α) for some 0 < α ≤ 1, and {qk : k ∈ N} be a sequence of non-increasing numbers satisfying conditions (1.39) and (1.40). Then the maximal operator ∼∗ |tn f | t p,α := (n + 1)1/p−1−α is bounded from the martingale Hardy space Hp to the Lebesgue space Lp . b) Let {Φn : n ∈ N+ } be any non-decreasing sequence, satisfying the condition (n + 1)1/p−1−α = +∞. n→∞ Φn lim

(4.29)

Then there exists Nörlund means with non-increasing sequence {qk : k ∈ N} satisfying the conditions (4.22) and (4.23) such that

tM2n +1 fk k

ΦM2n +1 k weal−Lp = ∞. sup kfk kHp k∈N Remark 4.14 Part a) can be found in the paper Blahota and Tephnadze [6], while part b) has not been stated before for such a general case. Proof: Since the Nörlund means tn are bounded from L∞ to L∞ (the boundedness follows from Corollary 1.33), according to Lemma 1.39 it actually suffices to show that Z ∗ p ∼ t p,α a dµ < c IN

,

August 30, 2015


137

for some constant c and every p-atom a. We may assume that a is an arbitrary p-atom with support I, µ (I) = MN−1 and I = IN . It is easy to see that Sn (a) = tn (a) = 0 when n ≤ MN . Therefore we can suppose that n > MN . 1/p

Let x ∈ IN . Since kak∞ ≤ MN we obtain that Z |tn a (x)| ≤ |a (t)| |Fn (x − t)| dµ (t) IN

Z |Fn (x − t)| dµ (t) ≤

≤ kak∞

1/p MN

Z |Fn (x − t)| dµ (t) .

IN

IN

Let x ∈ INk,l , 0 ≤ k < l < N. Then, from Lemma 1.34 we get that 1/p−1

cα,p MN

|tn a (x)| ≤

Mlα Mk

nα

.

(4.30)

Let x ∈ INk,N , 0 ≤ k < N. Then, according to Lemma 1.34, we have that 1/p−1

|tn a (x)| ≤ cα,p MN

Mk .

(4.31)

Let x ∈ INk,l , 0 ≤ k < l ≤ N. Since n > MN we can conclude that |tn a (x)| ≤ cα,p Mlα Mk . n1/p−1−α

(4.32)

The expression on the right-hand side of (4.32) does not depend on n. Hence, we can conclude that ∼ ∗ (4.33) t p,α a (x) ≤ cα,p Mlα Mk for x ∈ INk,l , 0 ≤ k < l ≤ N. By combining (1.1) and (4.33) we obtain that Z ∗ p ∼ t p,α a dµ IN

=

mj −1

N −2 N −1 X X

Z

X

k=0 l=k+1 xj =0,j∈{l+1,...,N −1} N −1 Z X

+

k=0

≤ cα,p

k,N IN

k,l IN

∼∗ p t p,α a dµ

∼ ∗ p t p,α a dµ

N −2 N −1 X X

N −1 X 1 1 Mlαp Mkp + cα,p MN1−p Mkp M M l N k=0 l=k+1 k=0 G.Tephnadze

,

138

August 30, 2015


≤ cα,p

N −2 X

Mkp

k=0

N −1 X

1

Ml1−αp l=k+1

+ cα,p

N −1 X

Mkp ≤ cα,p < ∞. MNp k=0

The proof of part a) is complete. Let 0 < p < 1/ (1 + α) . Under condition (4.29) there exists positive integers nk such that

(M2nk + 1)1/p−1−α = ∞, 0 < p < 1/ (1 + α) . k→∞ ΦM2nk +1 lim

To prove part b) we apply the p-atoms defined in Example 1.44. Under conditions (1.39) and (1.40), if we invoke (1.52) and (1.53) we find that tM2nk +1 fk SM2nk +1 = ΦM2nk +1 QM2nk +1 ΦM2nk +1 q0 DM2nk +1 − DM2nk q0 ψM2nk cα = = = α . QM2nk +1 ΦM2nk +1 QM2nk +1 ΦM2nk +1 M2nk ΦM2nk +1 Hence, µ

 

x ∈ Gm :



tM2nk +1 fk (x) ΦM2nk +1

  cα ≥ α = 1. M2nk ΦM2nk +1 

By combining (1.54) and (4.34) we have that SM2n +1 fk (x) cα k µ x ∈ G : ≥ α m ΦM M ΦM +1 +1 2nk

2nk

2nk

q0 α Φ M2n M2n +1 k k

(4.34)

1/p

kfk kHp 1/p−1−α

cα M2nk ≥ ΦM2nk +1

≥

1/p−1−α cα M2nk + 1 ΦM2nk +1

→ ∞, when k → ∞.

The proof is complete. Corollary 4.15 Let 0 < p < 1/ (1 + α) and f ∈ Hp . Then there exists an absolute constant cp,α , depending only on p and α, such that ktn f kp ≤ cp,α (n + 1)1/p−1−α kf kHp , n ∈ N+ . Proof: According to part a) of Theorem 4.13 we conclude that

tn f |tn f |

≤ sup

(n + 1)1/p−1−α

n∈N (n + 1)1/p−1−α p p ,

August 30, 2015

139


≤ cp,α kf kHp , n ∈ N+ . The proof is complete. Corollary 4.16 Let {Φn : n ∈ N} be any non-decreasing sequence satisfying the condition (4.29). Then there exists a martingale f ∈ Hp such that

tn f

sup = ∞.

n∈N Φn weak−Lp

Corollary 4.17 Let {Φn : n ∈ N} be any non-decreasing sequence satisfying the condition (4.29). Then the maximal operator |tn f | sup n∈N Φn is not bounded from the Hardy space Hp to the space weak − Lp . We now formulate our final result in this section. Theorem 4.18 Let f ∈ H1/(1+α) , where 0 < α ≤ 1 and {qk : k ∈ N} be a sequence of nonincreasing numbers satisfying the conditions (1.39) and (1.40). Then there exists an absolute constant cα depending only on α such that the maximal operator ∼∗

t α :=

log

|tn f | 1+α

(n + 1)

is bounded from the martingale Hardy space H1/(1+α) to the Lebesgue space L1/(1+α) . b) Let {Φn : n ∈ N+ } be any non-decreasing sequence satisfying the condition log1+α (n + 1) = +∞. n→∞ Φn lim

(4.35)

Then there exists Nörlund means with non-increasing sequence {qk : k ∈ N} satisfying the conditions (4.22) and (4.23) such that

supn tΦn fnk 1/(1+α) sup = ∞. kf kH1/(1+α) k∈N

Remark 4.19 Part a) of this result can be found in the paper Bhahota and Tephnadze [7]. Part b) has not been stated before for such a general case. G.Tephnadze

,

140


August 30, 2015

Proof: According to Lemma 1.39 the proof of part a) will be complete if we show that Z ∗ 1/(1+α) ∼ dµ < ∞ t α a IN

for every 1/ (1 + α)-atom a. We may assume that a is an arbitrary 1/ (1 + α)-atom with support I, µ (I) = MN−1 and I = IN . It is easy to see that tm (a) = 0, when m ≤ MN . Therefore we can suppose that m > MN . Let x ∈ IN . Since tm is bounded from L∞ to L∞ (the boundedness follows from Corol1/(1+α) lary 1.33) and kak∞ ≤ MN we obtain that Z |a (t)| |Fm (x − t)| dµ (t) |tm a (x)| ≤ IN

Z ≤ ka (x)k∞

|Fm (x − t)| dµ (t) IN

≤ MN1+α

Z |Fm (x − t)| dµ (t) . IN

Let x ∈ INk,l , 0 ≤ k < l < N. From Lemma 1.34 we get that |tm a (x)| ≤

cα Mk Mlα MNα . mα

(4.36)

Let x ∈ INk,N , 0 ≤ k < N. In the view of Lemma 1.34 we have that |tm a (x)| ≤ cα Mk MNα .

(4.37)

Let x ∈ INk,l , 0 ≤ k < l ≤ N. Since n > MN we can conclude that |tn a (x)| cα Mk Mlα ≤ . 1+α N 1+α log n The expression on the right-hand side of (4.38) does not depend on n. Hence, ∼∗ c M Mα α k N , tα a (x) ≤ N 1+α

(4.38)

(4.39)

for x ∈ INk,l , 0 ≤ k < l ≤ N. According to (1.1) and (4.39) we obtain that Z ∗ 1/(1+α) ∼ dµ tα a IN

=

N −2 N −1 X X

mj −1

X

k=0 l=k+1 xj =0,j∈{l+1,...,N −1}

Z k,l IN

∼ ∗ 1/(1+α) dµ tα a ,

August 30, 2015

Martingale Hardy spaces and summability of the one dimensional Vilenkin-Fourier series N −1 Z X

+

k=0

≤

k,N IN

141

∼ ∗ 1/(1+α) dµ t α a

N −2 N −1 cα X X 1 α/(1+α) 1/(1+α) M Mk N k=0 l=k+1 Ml l N −1

+

≤

cα X 1 α/(1+α) 1/(1+α) M Mk N k=0 MN N N −2 N −1 α/(1+α) 1/(1+α) cα X X Ml Mk N k=0 l=k+1 Ml N −1

+

1/(1+α)

cα X Mk ≤ cα < ∞. N k=0 MN1/(1+α)

The proof of part a) is complete. Under condition (4.35) there exists a positive integers m,k such that M2m,k +1 ≤ λk < 2M2m,k +1 . Since Φn is a non-decreasing function we have that 0 1+α mk log1+α (λk ) lim = ∞. ≥ c lim k→∞ ΦM2m, +1 k→∞ ΦM2λ +1 k k

Let {nk : k ∈ N+ } ⊂ {m0k : k ∈ N+ } be a sequence of positive numbers such that n1+α k = ∞. k→∞ ΦM2n +1 k lim

To prove part b) we use the 1/ (1 + α)-atoms defined in Example 1.44. If we apply (1.11) in Lemma 1.2 with (1.52) and (1.53) and invoke Abel transformation we get that M2nX +M tM2nk +M2s fk k 2s 1 = qM +M −j Dj − DM2nk ΦM2nk +M2s ΦM2nk +M2s QM2nk +M2s j=M +1 2nk 2s 2nk M 2s X qM2s −j Dj+M2nk − DM2nk j=1 M2s X 1 = qM2s −j Dj ψM2nk ΦM2nk +M2s QM2nk +M2s j=1 M 2s X 1 = qM2s −j j . ΦM2nk +M2s QM2nk +M2s

1 = ΦM2nk +M2s QM2nk +M2s

j=1

G.Tephnadze

,

142

August 30, 2015


Let x ∈ I2s /I2s+1 , s = [nk /2] , ..., nk . If we again use abel transformation, then under the conditions (4.22) and (4.23) we find that M 2s X tM2nk +M2s fk c 2 (q − q ) j ≥ M2s −j M2s −j−1 α ΦM2nk +M2s ΦM2nk +M2s M2n k j=1 ≥

c

M2s X

α ΦM2nk +M2s M2n k

j=[M2s ]/2

≥

2 cM2s

M2s X


j=[M2s ]/2

X


|qj − qj+1 |

j=1

[M2s ]/2

2 cM2s

X


≥

|qM2s −j − qM2s −j−1 |

[M2s ]/2

2 cM2s

≥

≥

|qM2s −j − qM2s −j−1 | j 2

j α−2 ≥

j=1

α+1 cM2s α ΦM2nk +M2s M2n k

≥

α−1 2 cM2s M2s α ΦM2nk +M2s M2n k

α+1 cM2s . α ΦM2nk +1 M2n k

Hence, tn fk 1/(1+α) dµ sup Φn n Gm nX k −1 Z tM +M fk 1/(1+α) 2nk 2s dµ I2s /I2s+1 ΦM2nk +M2s Z

≥

s=[nk /2]

≥

c

Z

1/(1+α) ΦM2n +1 s=[nk /2] k

≥

≥

nX k −1

c

c

I2s /I2s+1

nX k −1

M2s dµ α/(1+α) M2nk

α+1 M2s

1/(1+α) α/(1+α) ΦM2n +1 s=[nk /2] M2nk k nX k −1

1

1/(1+α) α/(1+α) ΦM2n +1 s=[nk /2] M2nk k

≥

1 α+1 M2s

cnk . α/(1+α) 1/(1+α) M2nk ΦM2n +1 k

Therefore, by also using (1.54) for p = 1/ (1 + α) we have that 1+α R tn fk 1/(1+α) dµ sup n Φn Gm kfk kH1/(1+α) ,

August 30, 2015


≥

cn1+α cn1+α α k k M ≥ → ∞, when 2nk α Φ M2n ΦM2nk +1 M2nk +1

143

k → ∞.

k

The proof is complete. Corollary 4.20 Let {Φn : n ∈ N+ } be any non-decreasing sequence satisfying the condition (4.35). Then the following maximal operator tn f sup Φn n is not bounded from the Hardy space H1/(1+α) to the space L1/(1+α) .

4.3

¨ S TRONG CONVERGENCE OF N ORLUND MEANS ON MARTINGALE H ARDY SPACES

The first result in this section is due to Persson, Tephnadze and Wall [51]. Theorem 4.21 Let 0 < p < 1/2, f ∈ Hp and the sequence {qk : k ∈ N} be non-decreasing. Then there exists an absolute constant cp depending only on p such that

∞ X ktk f kpp

k 2−2p

k=1

≤ cp kf kpHp .

Remark 4.22 Since the Fejér means are examples of Nörlund means with non-decreasing sequence {qk : k ∈ N} we immediately obtain from part b) of Theorem 3.25 that the asymptotic behaviour of the sequence of weights

1/k 2−2p : k ∈ N

in Nörlund means in Theorem 4.21 can not be improved. Proof: The proof is similar to that of part a) of Theorem 3.25, but since this case is more general we give the details. By Lemma 1.38 the proof is complete if we show that ∞ X ktm akpp m=1

m2−2p

≤ cp

(4.40)

for every p-atom a with support I, µ (I) = MN−1 . We may assume that I = IN . It is easy to see that Sn (a) = tn (a) = 0, when n ≤ MN . Therefore, we can suppose that n > MN . G.Tephnadze

,

144

August 30, 2015


Let x ∈ IN . Since Nörlund means tn with non-decreasing sequence {qk : k ∈ N} are 1/p bounded from L∞ to L∞ (the boundedness follows from Corollary 1.23) and kak∞ ≤ MN we obtain that Z |tm a|p dµ IN

≤

kakp∞

≤ 1, 0 < p ≤ 1/2.

MN

Hence,

∞ X

R IN

(4.41)

m2−2p

m=1

≤

|tm a|p dµ

∞ X

1 ≤ c < ∞. 2−2p m k=1

It is easy to see that |tn a (x)| Z n 1 X = a (t) qn−k Dk (x − t) dµ (t) IN Qn k=M N Z n 1 X |a (t)| ≤ qn−k Dk (x − t) dµ (t) Qn IN k=MN Z n 1 X qn−k Dk (x − t) dµ (t) ≤ kak∞ IN Qn k=M N Z n 1 X 1/p ≤ MN qn−k Dk (x − t) dµ (t) . IN Qn k=MN

Let x ∈ INk,l , 0 ≤ k < l ≤ N. Then, in the view of Lemma 1.27, we get that 1/p−2

|tm a (x)| ≤ cp Ml Mk MN

, for 0 < p < 1/2.

According to (1.1) with (4.42) we find that Z |tm a|p dµ

(4.42)

(4.43)

IN

=

N −2 N −1 X X

mj−1

X

k=0 l=k+1 xj =0, j∈{l+1,...,N −1}

Z k,l IN

|tm a|p dµ

,

August 30, 2015

Martingale Hardy spaces and summability of the one dimensional Vilenkin-Fourier series N −1 Z X

+

k,N IN

k=0

≤ cp

145

|tm a|p dµ

N −2 N −1 X X

ml+1 · · · mN −1 p p 1−2p Ml Mk MN MN k=0 l=k+1 +cp

N −1 X

1 Mkp MN1−p M N k=0

≤ cp MN1−2p

+cp

N −2 N −1 X X

Mlp Mkp Ml k=0 l=k+1

N −1 X

Mkp 1−2p . p ≤ cMN M N k=0

Moreover, according to (4.43), we get that R ∞ X |t a|p dµ IN m m=MN +1

≤ cp

∞ X m=MN

m2−2p

MN1−2p < c < ∞, (0 < p < 1/2) . m2−2p +1

Now, by combining this estimate with (4.41) we obtain (4.40) so the proof is complete. Also the next theorem is proved in Persson, Tephnadze and Wall [51]. Theorem 4.23 Let f ∈ H1/2 and the sequence {qk : k ∈ N} be non-decreasing satisfying condition (1.28). Then there exists an absolute constant c, such that 1/2 n 1 X ktk f k1/2 1/2 ≤ c kf kH1/2 . log n k=1 k

Proof: The proof is similar to that of part a) of Theorem 3.28, but since this case is more general we present the details. According to Lemma 1.38 it suffices to show that 1/2 n 1 X ktm ak1/2 ≤c log n m=1 m

for every p-atom a with support I, µ (I) = MN−1 . We may assume that I = IN . It is easy to see that Sn (a) = tn (a) = 0 when n ≤ MN . Therefore we can suppose that n > MN . G.Tephnadze

,

146

August 30, 2015


Let x ∈ IN . Since tn is bounded from L∞ to L∞ (the boundedness follows from Corollary 1.23) and kak∞ ≤ MN2 we obtain that Z

kak1/2 ∞ ≤ c < ∞. MN

|tm a|1/2 dµ ≤

IN

Hence, n 1 X log n m=1

≤

R IN

|tm a|1/2 dµ

(4.44)

m

n 1 X1 ≤ c < ∞, n ∈ N. log n m=1 m

It is easy to see that Z |tm a (x)| ≤

|a (t)| |Fm (x − t)| dµ (t) IN

Z |Fm (x − t)| dµ (t)

≤ kak∞ IN

≤ MN2

Z |Fm (x − t)| dµ (t) . IN

Let x ∈ INk,l , 0 ≤ k < l < N. Then, in the view of Lemma 1.25, we get that |tm a (x)| ≤

cMl Mk MN . m

(4.45)

Let x ∈ INk,N . Then, according to Lemma 1.25, we find that |tm a (x)| ≤ cMk MN .

(4.46)

By combining (4.45) and (4.46) with (1.1) we can conclude that Z |tm a (x)|1/2 dµ (x) IN

≤c

N −2 N −1 X X

1/2

1/2

1/2

ml+1 · · · mN −1 Ml Mk MN MN m1/2 k=0 l=k+1 +c 1/2

≤ cMN

N −1 X

1 1/2 1/2 Mk MN M N k=0

N −2 N −1 X X

N −1 1/2 1/2 1/2 X Ml Mk Mk + c 1/2 m1/2 Ml k=0 l=k+1 k=0 MN ,

August 30, 2015


147

1/2

≤

cMN N + c. m1/2

It follows that n X 1 log n m=M +1

R IN

|tm a (x)|1/2 dµ (x) m

(4.47)

N

n X 1 ≤ log n m=M +1

1/2

cMN N c + m3/2 m

! < c < ∞.

N

The proof is complete by just combining (4.44) and (4.47). Next, we investigate Nörlund means with non-increasing sequence {qk : k ∈ N}. At first we consider the case 0 < p < 1/ (1 + α) where 0 < α < 1. For details see the paper of Blahota and Tephnadze [6]. Theorem 4.24 Let f ∈ Hp , where 0 < p < 1/ (1 + α) , 0 < α ≤ 1 and {qk : k ∈ N}, be a sequence of non-increasing numbers satisfying the conditions (1.39) and (1.40). Then there exists an absolute constant cα,p , depending only on α and p such that ∞ X ktk f kpHp

k 2−(1+α)p k=1

≤ cα,p kf kpHp .

Proof: By Lemma 1.38 the it suffices to show that ∞ X ktm akpp ≤ cα,p < ∞ m2−(1+α)p m=1

for every p-atom a. Analogously to the proofs of the previous theorems we may assume that a be an arbitrary p-atom with support I, µ (I) = MN−1 and I = IN and m > MN . Let x ∈ IN . Since tm is bounded from L∞ to L∞ (the boundedness follows from Corol1/p lary 1.33) and kak∞ ≤ MN we obtain that Z |tm a|p dµ ≤ kakp∞ MN−1 ≤ 1. IN

Hence ∞ X m=MN

≤

R IN

|tm a|1/(1+α) dµ m2−(1+α)p

∞ X

1 ≤ cα,p < ∞. 2−(1+α)p m m=1 G.Tephnadze

,

148

August 30, 2015


According to (1.1) and (4.30)-(4.31) we can conclude that R ∞ X |t a|p dµ IN m m2−(1+α)p

m=MN +1

=

mj −1

N −2 N −1 X X

R

X

k=0 l=k+1 xj =0,j∈{l+1,...,N −1}

+

n N −1 X X

R

k,N IN

m2−(1+α)p

|tm a|p dµ

N −2 N −1 N −1 MN1−p X X Mlpα Mkp MN1−p X Mkp + m2−p k=0 l=k+1 Ml m2−(1+α)p k=0 MN

∞ X m=MN +1

Since

|tm a|p dµ

m2−(1+α)p

m=MN +1 k=0

≤ cα,p

k,l IN

! .

−1 N −2 N X X

N −2 N −1 X X Mlpα Mkp 1 Mkp ≤ 1−pα Ml Ml k=0 l=k+1 k=0 l=k+1

N −2 X

≤

1

p 1−pα Mk ≤

Mk k=0

and

N −2 X

1 1−p(α+1)

s. Then we can write that qs

αk Sj f 1 X Rqαs k f = s lqαk j=1 j

qs

M2α αk X 1 Xk Sj f 1 Sj f = + := I + II. lqαs k j=1 j lqαs k j=M +1 j 2αk

According to (1.76) we have that |I| ≤

1

M2αk −1

X |Sj f (x)| j j=1

l

s qα k

M2αk −1 1/p 1 1 2λM2αk−1 X ≤ 1/2 αk αk−1 j j=1 1/p

≤

2λM2αk−1 1/2 αk−1

1/p

≤

2λMαk 3/2 αk

.

Let M2αk ≤ j ≤ qαs k . According to the second inequality of (1.75) in the case when l = k, we deduce that 1/p−1

Sj f = SM2α f +

M2αk ψM2α Dj−M2α

k

G.Tephnadze

k

1/2

αk

k

.

(4.49)

,

152

August 30, 2015


Hence, we can rewrite II as qs

II =

αk X SM2α f

1

k

lqαs k j=M

2αk

qs

1/p−1

+

αk X Dj−M2α

1 M2αk ψM2αk 1/2

lqαs k

αk

(4.50)

j

k

j

j=M2αk

:= II1 + II2 . In view of (1.76) we find that s

qα k X 1 |II1 | ≤ SM2α f k lqαs k j=M j

1

(4.51)

2αk

2λM 1/p 2αk−1 ≤ SM2α f ≤ . 1/2 k αk−1 Let 0 < p ≤ 1/2, x ∈ I2s \I2s+1 for s = [2αk /3] , . . . , αk . Since MX 2s −1

Dj (x)

j=0

j+M2αk

≥ we obtain that

j

j=0

j+M2αk

j

j=0

2M2αk

1 M2αk lqαs k

MX 2s −1

MX 2s −1

1/p−1

|II2 | =

≥

1/2

αk

≥

MX 2s −1 Dj ψM2α k j+M2αk

(4.52)

j=0

1/p−1

≥

2 cM2s M2αk

1/p−2

2 2 cM2αk M2s c M2αk M2s ≥ . 1/2 3/2 αk αk M2αk αk

By combining (1.71)-(1.73) with (4.50)-(4.52) we get that Rqαs k f = |II2 − I − II1 | 1/p

≥ |II2 | − |I| − |II1 | ≥ |II2 | − 1/p−2

≥

2 cM2αk M2s 3/2 αk

2λMαk 3/2

αk

1/p

−

cMαk 3/2 αk

,

August 30, 2015


153

1/p−2

≥

2 cM2αk M2s 3/2

.

αk Hence,

Z

|R∗ f (x)|p dµ (x)

Gm αk X

≥ cp

Z

p Rqαs k f (x) dµ (x)

s=[2αk /3] I \I 2s 2s+1 αk X

≥ cp

1−2p 2p cM2α M2s k

Z

3p/2

αk X

≥ cp

1−2p 2p−1 M2α M2s k 3p/2

αk

s=[2αk /3]

(

cp 2αk (1−2p)/3 3p/2

≥

αk 1/4 cαk ,

dµ (x)

αk

s=[2αk /3]I \I 2s 2s+1

, when 0 < p < 1/2,

→ ∞, when k → ∞.

when p = 1/2,

The proof is complete. Theorem 4.27 Let 0 < p ≤ 1. Then there exists a martingale f ∈ Hp such that kL∗ f kp = +∞. Proof: We write that

qs

αk 1 X Sj f s Lqαk f = lqαk ,s j=1 qαs k − j

=

1 lqαs k

M2αk −1

X j=1

(4.53)

qs

αk Sj f 1 X Sj f + := III + IV. s s s qαk − j qαk j=M qαk − j 2αk

In the view of (1.76) for III we get the following estimate: M

2αk−1 1/p 1/p 1/p M2αk−1 1 X 1 M2αk−1 Mαk |III| ≤ ≤ ≤ . 1/2 1/2 3/2 αk j=0 qαs k − j αk−1 αk−1 αk

(4.54)

Moreover, according to the second inequality of (1.75) in the case when l = k, (see also (4.49)) we can rewrite IV as qs

IV =

1 lqαs k

αk X

j=M2αk

G.Tephnadze

SM2α f k

qαk ,s − j

(4.55)

,

154

August 30, 2015

4. Vilenkin-Nörlund means on martingale Hardy spaces qs

1/p−1

+

αk X Dj−M2α

1 M2αk ψM2αk 1/2

lqαs k

αk

j=M2αk

k

:= IV1 + IV2 .

qαs k − j

By now applying (1.76) again we have that 1/p

1/p

2λM2αk−1

|IV1 | ≤

≤

1/2 αk−1

2λMαk

.

3/2 αk

(4.56)

Let x ∈ I2s \I2s+1 , s = [2αk /3] , ..., αk . Since MX 2s −1 j=0

MX MX MX 2s −1 2s −1 2s −1 j M2s M2s − j Dj = = − −j −j −j M2s − j j=0 M2s j=0 M2s j=0 M2s

= cp sM2s − M2s ≥ cp sM2s , we obtain that |IV2 | =

1/p−1 MX 2s −1

1 M2αk

lqαk ,s αk1/2

j=0

Dj M2s − j

(4.57)

1/p−1

≥

cp M2αk

x ∈ I2s /I2s+1 .

sM2s ,

3/2

αk

By combining (1.73) with (4.53)-(4.57) for x ∈ I2s \I2s+1 , s = [2αk /3] , ..., αk and 0 < p ≤ 1 we get that Lqαs k f (x) = |III + IV1 + IV2 | ≥ |IV2 | − |III| − |IV1 | 1/p−1

≥ Consequently, Z

cp M2αk

sM2s −

3/2 αk

p

∗

|L f (x)| dµ (x) ≥ Gm

≥

≥ cp

Z I2s \I2s+1

s=[2αk /3]

( ≥

αk X

Z

s=[2αk /3] αk X

cp 2αk (1−p)/3 p/2

αk 1/2 cαk ,

≥

3/2 αk

s=[2αk /3] αk X

1/p−1

2λMαk

I2s \I2s+1

cp M2αk 3/2 αk

Z

sM2s .

|L∗ f (x)|p dµ (x)

I2s \I2s+1

p Lqαs k f (x) dµ (x)

1−p M2α p k−1 p s M2s dµ 3p/2 αk

, when 0 < p < 1,

≥ cp

1−p αk X M2α k−1 s=[2αk /3]

p/2 αk

p−1 M2s

→ ∞, when k → ∞.

when p = 1,

The proof is complete. ,

August 30, 2015

4.5


155

A PPLICATIONS

First we consider Nörlund means tn with monotone and bounded sequence {qk : k ∈ N} . The results in our previous sections in particular imply the following results: Theorem 4.28 a) Let f ∈ H1/2 and tn be Nörlund means with monotone and bounded sequence {qk : k ∈ N} . Then there exists an absolute constant c such that kt∗ f kweak−L1/2 ≤ c kf kH1/2 . b) There exists a martingale f ∈ H1/2 , such that kt∗ f k1/2 = ∞. Theorem 4.29 a) Let p > 1/2, f ∈ Hp and tn be Nörlund means with monotone and bounded sequence {qk : k ∈ N} . Then there exists an absolute constant cp depending only on p such that

∼∗

tp,1 f ≤ cp kf kHp . p

b) Let ϕ : N+ → [1, ∞) be a non-decreasing function satisfying the condition (3.6). Then the following maximal operator |tn f | sup n∈N Φn is not bounded from the Hardy space Hp to the space weak − Lp . Theorem 4.30 a) Let f ∈ H1/2 and tn be Nörlund means with monotone and bounded sequence {qk : k ∈ N} . Then there exists an absolute constant c such that

∼∗

t1 f ≤ c kf kH1/2 . 1/2

b) Let ϕ : N+ → [1, ∞) be a non-decreasing function satisfying the condition (3.11). Then the following maximal operator |Bn f | sup n∈N Φn is not bounded from the Hardy space H1/2 to the Lebesgue space L1/2 . Theorem 4.31 Let 0 < p < 1/2, f ∈ Hp and tn be Nörlund means with monotone and bounded sequence {qk : k ∈ N} . Then there exists an absolute constant cp depending only on p such that ∞ X ktm f kpp ≤ cp kf kpHp . 2−2p m m=1 G.Tephnadze

,

156


August 30, 2015

Theorem 4.32 Let f ∈ H1/2 and tn be Nörlund means with monotone and bounded sequence {qk : k ∈ N} . Then there exists an absolute constant c such that 1/2 n 1 X ktm f k1/2 1/2 ≤ c kf kH1/2 , n = 2, 3, . . . . log n m=1 m

Next we remark that σn are Nörlund means with monotone and bounded sequence {qk : k ∈ N} and κα,β are concrete examples of Nörlund means with non-decreasing and n unbounded sequence {qk : k ∈ N} but they have similar boundedness properties. The results in our previous sections in particular imply the following results: Theorem 4.33 a) Let f ∈ H1/2 . Then there exists absolute constant c such that kσ ∗ f kweak−L1/2 ≤ c kf kH1/2 and

α,β,∗

κ f weak−L

1/2

≤ c kf kH1/2 .

b) There exists a martingale f ∈ H1/2 such that kσ ∗ f k1/2 = ∞ and

α,β,∗

κ f 1/2 = ∞. Theorem 4.34 a) Let 0 < p < 1/2 and f ∈ Hp . Then there exists an absolute constant cp depending only on p such that

∗

∼

σp f ≤ cp kf kHp p

and

α,β,∗

κ ep f p ≤ cp kf kHp . b) Let ϕ : N+ → [1, ∞) be a non-decreasing function satisfying the condition (3.6). Then the following maximal operators α,β κ |σn f | and sup n sup n∈N Φn n∈N Φn are not bounded from the Hardy space Hp to the space weak − Lp . Theorem 4.35 a) Let f ∈ H1/2 . Then there exists an absolute constant c such that

∗

∼

σ f ≤ c kf kH1/2 1/2

,

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and

α,β,∗

κ e f 1/2 ≤ c kf kH1/2 . b) Let ϕ : N+ → [1, ∞) be a nondecreasing function satisfying the condition (3.11). Then the following maximal operators α,β κ |σn f | and sup n sup n∈N Φn n∈N Φn are not bounded from the Hardy space H1/2 to the Lebesgue space L1/2 . Theorem 4.36 Let 0 < p < 1/2 and f ∈ Hp . Then there exists an absolute constant cp depending only on p such that ∞ X kσm f kpp m=1

and

m2−2p

∞ α,β p X κm f p m=1

m2−2p

≤ cp kf kpHp

≤ cp kf kpHp .

Theorem 4.37 Let f ∈ H1/2 . Then there exists an absolute constant c such that 1/2 n 1 X kσm f k1/2 1/2 ≤ c kf kH1/2 , n = 2, 3, . . . , log n m=1 m

and

n α,β 1/2 1 X κm f 1/2 1/2 ≤ c kf kH1/2 , n = 2, 3, . . . . log n m=1 m

Furthemore, we note that it is obvious that Cesàro (C, α) and Riesz (R, α) means are examples of Nörlund means with non-increasing sequence satisfying the conditions (1.39) and (1.40). It follows that all results concerning such summability methods are true also for the Cesàro (C, α) and Riesz (R, α) means. Hence, the results in our previous sections in particular imply the following results: Theorem 4.38 a) Let 0 < p < 1/2. Then there exists a martingale f ∈ Hp such that sup kσnα f kweak−Lp = ∞ n∈N

and sup kβnα f kweak−Lp = ∞. n∈N

G.Tephnadze

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Theorem 4.39 a) Let 0 < α ≤ 1 and f ∈ H1/(1+α) . Then there exists absolute constant cα depending only on α such that kσ α,∗ f kweak−L1/(1+α) ≤ cα kf kH1/(1+α) and kβ α,∗ f kweak−L1/(1+α) ≤ cα kf kH1/(1+α) . b) Let 0 < α ≤ 1. Then there exists a martingale f ∈ H1/(1+α) such that kσ α,∗ f k1/(1+α) = ∞ and kβ α,∗ f k1/(1+α) = ∞. Theorem 4.40 a) Let 0 < α ≤ 1, 0 < p < 1/ (1 + α) and f ∈ Hp . Then there exists an absolute constant cα,p depending only on α and p such that

α,∗

∼

σp f ≤ cα,p kf kHp p

and

α,∗

∼

βp f ≤ cα,p kf k . Hp

p

b) Let 0 < α ≤ 1 and ϕ : N+ → [1, ∞) be a non-decreasing function satisfying the condition (4.29). Then the following maximal operators sup n∈N

|σnα f | Φn

and

sup n∈N

|βnα f | Φn

are not bounded from the Hardy space Hp to the space weak − Lp . Theorem 4.41 a) Let 0 < α ≤ 1 and f ∈ H1/(1+α) . Then there exists an absolute constant cα depending only on α such that

α,∗

∼

≤ cα kf kH1/(1+α)

σ f 1/(1+α)

and

α,∗

∼

β f

1/(1+α)

≤ cα kf kH1/(1+α) .

b) Let 0 < α ≤ 1 and ϕ : N+ → [1, ∞) be a nondecreasing function satisfying the condition (4.35). Then the following maximal operators sup n∈N

|σnα f | and Φn

sup n∈N

|βnα f | Φn

are not bounded from the Hardy space H1/(1+α) to the Lebesgue space L1/(1+α) . ,

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Theorem 4.42 Let 0 < α ≤ 1, 0 < p < 1/ (1 + α) and f ∈ Hp . Then there exists an absolute constant cα,p depending only on α and p such that ∞ α X akpp kσm ≤ cα,p kf kpHp (1+α)(1−p) m m=1

and

∞ α X kβm akpp ≤ cα,p kf kpHp . (1+α)(1−p) m m=1

Theorem 4.43 Let 0 < α ≤ 1 and f ∈ H1/(1+α) . Then there exists an absolute constant cα depending only on α such that 1/(1+α) n α 1 X kσm f k1/(1+α) 1/(1+α) ≤ cα kf kH1/(1+α) log n m=1 m

and

1/(1+α) n α 1 X kβm f k1/(1+α) 1/(1+α) ≤ cα kf kH1/(1+α) . log n m=1 m

Finally, we present some applications concerning almost everywhere convergence of some summability methods. To study almost everywhere convergence of some summability methods is one of the most difficult topics in Fourier analysis. They involve techniques from function theory and Hardy spaces. In most applications the a.e. convergence of {Tn : n ∈ N} can be established for f in some dense class of L1 (Gm ) . In particular, the following result play an important role for studying this type of questions (see e.g. the books [35], [55] and [88]). Lemma 4.44 Let f ∈ L1 and Tn : L1 → L1 be some sub-linear operators and T ∗ := sup |Tn | . n∈N

If Tn f → f a.e. for every f ∈ S where the set S is dense in the space L1 and the maximal operator T ∗ is bounded from the space L1 to the space weak − L1 , that is sup λµ {x ∈ Gm : |T ∗ f (x)| > λ} ≤ kf k1 , λ>0

then Tn f → f, a.e. for every f ∈ L1 (Gm ) .

G.Tephnadze

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August 30, 2015

Remark 4.45 Since the Vilenkin function ψm is constant on In (x) for every x ∈ Gm and 0 ≤ m < Mn , it is clear that each Vilenkin function is a complex-valued step function, that is, it is a finite linear combination of the characteristic functions 1, x ∈ E, χ (E) = 0, x ∈ / E. On the other hand, notice that, by Corollary 1.5 (Paley lemma), it yields that χ (In (t)) (x) =

Mn −1 1 X ψj (x − t) , x ∈ In (t), Mn j=0

for each x, t ∈ Gm and n ∈ N. Thus each step function is a Vilenkin polynomial. Consequently, we obtain that the collection of step functions coincides with a collection of Vilenkin polynomials P. Since the Lebesgue measure is regular it follows from Lusin theorem that given f ∈ L1 there exist Vilenkin polynomials P1 , P2 ..., such that Pn → f a.e. when n → ∞. This means that the Vilenkin polynomials are dense in the space L1 . By using Lemma 4.44, remarks 4.45 and the results in our previous sections we in particular obtain the following a.e. convergence results. Theorem 4.46 Let f ∈ L1 and tn be the regular Nörlund means with non-decreasing sequence {qk : k ∈ N}. Then tn f → f a.e., when n → ∞. Proof: Since Sn P = P, for every P ∈ P we obtain that tn P → P a.e., when

n → ∞,

for every regular Nörlund mean with non-decreasing sequence {qk : k ∈ N}, where P ∈ P is dense in the space L1 (see Remark 4.45). On the other hand, by combining Lemma 1.40 and Theorem 4.1 we obtain that the maximal operator of Nörlund means, with non-decreasing sequence {qk : k ∈ N} satisfying conditions (1.38) is bounded from the space L1 to the space weak − L1 , that is, sup λµ {x ∈ Gm : |t∗ f (x)| > λ} ≤ kf k1 . λ>0

According to Lemma 4.44 we obtain that under condition (1.38) we have a.e. convergence tn f → f a.e., when n → ∞. The proof is complete. ,

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Corollary 4.47 Let f ∈ L1 . Then σn f → f

a.e.,

when n → ∞,

and κα,β n f → f

a.e., when n → ∞.

Proof: Since σn and κn are regular Nörlund means with non-decreasing sequence {qk : k ∈ N} (see (1.5) and (1.10)) the proof is complete by just using Theorem 4.46. Theorem 4.48 Let f ∈ L1 and tn be Nörlund means with non-increasing sequence {qk : k ∈ N} satisfying conditions (1.39) and (1.40). Then tn f → f a.e., when n → ∞. Proof: By using part b) of Theorem 1.1 we get that every Nörlund mean, with nonincreasing sequence {qk : k ∈ N} is regular. Since Sn P = P, for every P ∈ P we obtain that tn P → P, when n → ∞, where P ∈ P is dense in the space L1 (see Remark 4.45). On the other hand, by combining Lemma 1.40 and part a) of Theorem 4.11 we obtain that the maximal operator of Nörlund means with non-increasing sequence {qk : k ∈ N} satisfying conditions (1.39) and (1.40) is bounded from the space L1 to the space weak − L1 , that is sup λµ {x ∈ Gm : |t∗ f (x)| > λ} ≤ kf k1 . λ>0

According to Lemma 4.44 we obtain that under conditions (1.39) and (1.40) we have a.e. convergence tn f → f a.e., when n → ∞. The proof is complete. Corollary 4.49 Let f ∈ L1 . Then σnα f → f

a.e., when n → ∞, when 0 < α < 1.

and βnα f → f

a.e., when n → ∞,

when 0 < α < 1.

Proof: By combining (1.6)-(1.7) with (1.8)-(1.9) we obtain that the Nörlund means σnα and βnα , with non-increasing sequence {qk : k ∈ N} satisfy conditions (1.39) and (1.40). Hence, the proof is complete by just using Theorem 4.48. G.Tephnadze

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Corollary 4.50 Let f ∈ L1 and tn be Nörlund means with monotone and bounded sequence {qk : k ∈ N}. Then tn f → f a.e., when n → ∞. Proof: The proof is analogously to the proofs of Theorems 4.46 and 4.48 so we leave out the details. Theorem 4.51 Let f ∈ L1 . Then Rn f → f,

a.e.

when n → ∞.

Proof: The proof is analogously to the proofs of Theorems 4.46 and 4.48 so we leave out the details.

,

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,