Weighted maximal operators of Fej\'er means of Walsh-Fourier series in the martingale Hardy space $H_{1/2}$

TL;DR

This paper establishes the boundedness of a weighted maximal operator of Fejér means of Walsh-Fourier series from the martingale Hardy space to Lebesgue space, including sharpness and new related results.

Contribution

It introduces a restricted weighted maximal operator for Walsh-Fourier Fejér means and proves its boundedness from $H_{1/2}$ to $L_{1/2}$, demonstrating sharpness and deriving new results.

Findings

Boundedness of the weighted maximal operator from $H_{1/2}$ to $L_{1/2}$.

Proof of the sharpness of the boundedness result.

Derivation of new and known results as corollaries.

Abstract

In this paper we derive the restricted weighted maximal operator, defined by ${\sup }_{k\in \mathbb{N}}\left(\left\vert \sigma _{k}F\right\vert/A^2_k\right)$ of Fej\'er means of Walsh-Fourier series and prove that the it is bounded from the martingale Hardy space $H_{1/2}(G)$ to the Lebesgue space $L_{1/2}(G).$ The sharpness of this result is also proved. As a consequence we obtain some new and and well-know results.

Equations

$${\sigma }^{\ast }f:=\underset{n\in \mathbb{N}}{\sup }\left|{\sigma }_{n}f\right|$$

$$\underset{n\in \mathbb{N}}{\sup }{\Vert {\sigma }_{n}F\Vert }_p=+\infty .$$

$$\underset{n\in \mathbb{N}}{\sup }\left|{\sigma }_{2^n}F\right|$$

$$n=\sum _{k=0}^{\infty }{n}_j{2}^j,n_j\in Z_2(j\in \mathbb{N}),$$

$$\left[n\right]:=\min \{j\in \mathbb{N},n_j\ne 0\},\left|n\right|:=\max \{j\in \mathbb{N},n_j\ne 0\},\rho \left(n\right)=\left|n\right|-\left[n\right]$$

$$V\left(n\right):=n_0+\stackrel{\infty }{\sum _{k=1}}\left|n_k-n_{k-1}\right|,\text{ for all }n\in \mathbb{N}.$$

$$\underset{k\in \mathbb{N}}{\sup }\rho \left(n_k\right)\le c<\infty ,$$

$${\tilde{\sigma }}^{\ast ,\nabla }F:=\underset{k\in \mathbb{N}}{\sup }\left|{\sigma }_{n_k}F\right|$$

$$\underset{k\in \mathbb{N}}{\sup }\rho \left(n_k\right)=\infty ,$$

$$\underset{k\in \mathbb{N}}{\sup }{\Vert {\sigma }_{n_k}F\Vert }_p=\infty .$$

$${\Vert {\sigma }_{n}F\Vert }_{H_{1/2}}\le c{V}^2\left(n\right){\Vert F\Vert }_{H_{1/2}}.$$

$$\underset{k\in \mathbb{N}}{\sup }V\left(n_k\right)=\infty$$

$$\overline{\underset{k\to \infty }{\lim }}\frac{V^2\left(n_k\right)}{{\Phi }_{n_k}}=\infty ,$$

$$\underset{k\in \mathbb{N}}{\sup }{\Vert \frac{{\sigma }_{n_k}F}{{\Phi }_{n_k}}\Vert }_{1/2}=\infty .$$

$$\underset{k\in \mathbb{N}}{\sup }V\left(n_k\right)<c<\infty .$$

$${\stackrel{\sim }{\sigma }}^{\ast }f:=\underset{n\in \mathbb{N}}{\sup }\frac{\left|{\sigma }_{n}f\right|}{{\log }^2\left(n+1\right)}$$

$$\underset{k\in \mathbb{N}}{\sup }{A}_{|n_k|}\le c<\infty ,$$

$$\underset{k\in \mathbb{N}}{\sup }{A}_{|n_k|}=\infty ,$$

$$\underset{k\in \mathbb{N}}{\sup }{\Vert {\sigma }_{n_k}F\Vert }_{1/2}=\infty .$$

$${\sup }_{k\in \mathbb{N}}\left(\left|{\sigma }_{k}F\right|/A_k^2\right)$$

$$I_0\left(x\right):=G,I_n(x):=\{y\in G:y_0=x_0,...,y_{n-1}=x_{n-1}\}(n\in \mathbb{N}).$$

$$\overline{I_M}=\underset{i=0}{\bigcup ^{M-1}}{I}_i\backslash I_{i+1}=\left(\stackrel{M-2}{\bigcup _{k=0}}\stackrel{M-1}{\bigcup _{l=k+1}}{I}_{l+1}\left(e_k+e_l\right)\right)\bigcup \left(\underset{k=0}{\bigcup ^{M-1}}{I}_M\left(e_k\right)\right).$$

$$n^{\left(i\right)}=2^{n_{i+1}}+...+2^{n_r},i=1,...,r.$$

$$A_s$$

$$card(A_s):=|A_s|.$$

$$V(n_{s_j})<c<\infty ,\text{for each}j=1,2,\dots ,r.$$

$$[u_1^s,u_1^s],[u_1^s,u_2^s],...,[u_1^s,u_{r_s^3}^s],[u_2^s,u_2^s],[u_2^s,u_3^s],...,[u_2^s,u_{r_s^3}^s]\dots [u_{r_s^3}^s,u_{r_s^3}^s],$$

$$s_r<\infty ,\text{for all}s\in \mathbb{N}.$$

$${\Vert f\Vert }_p^p:={\int }_G{\left|f\right|}^{p}d\mu \text{and}{\Vert f\Vert }_{\text{weak}-L_p}^p:=\underset{\lambda >0}{\sup }{\lambda }^p\mu \left(f>\lambda \right).$$

$$r_k\left(x\right):={\left(-1\right)}^{x_k}\left(x\in G,k\in \mathbb{N}\right).$$

Full text

The research was supported by Shota Rustaveli National Science Foundation grant no. PHDF-21-1702.

2020 Mathematics Subject Classification. 42C10.

Key words and phrases: Walsh system, Fejér means, martingale Hardy space, maximal operator.

1. INTRODUCTION

In the one-dimensional case the weak (1,1)-type inequality for the maximal operator of Fejér means with respect to Walsh system, defined by

[TABLE]

was investigated in Schipp [15] and Pál, Simon [10] (see also [1], [9] and [12]). Fujii [3] and Simon [17] verified that $\sigma^{\ast}$ is bounded from $H_{1}(G)$ to $L_{1}(G)$. Weisz [23] generalized this result and proved boundedness of $\sigma^{\ast}$ from the martingale space $H_{p}$ to the Lebesgue space $L_{p}$ for $p>1/2$. Simon [16] gave a counterexample, which shows that boundedness does not hold for $0<p<1/2.$ A counterexample for $p=1/2$ was given by Goginava [5]. Moreover, [6] (see also [19]) he proved that there exists a martingale $F\in H_{p}(G)$ $\left(0<p\leq 1/2\right),$ such that

[TABLE]

Weisz [26] proved that the maximal operator $\sigma^{\ast}$ of the Fejér means is bounded from the Hardy space $H_{1/2}$ to the space $weak-L_{1/2}$. Weisz [25] (see also [24]) also proved that for any $F\in H_{p}(G)$ $(p>0),$ the maximal operator

[TABLE]

is bounded from the Hardy space $H_{p}(G)$ to the space $L_{p}(G)$, but (for details see [13]) it is not bounded from the hardy space $H_{p}(G)$ to the space $H_{p}(G).$

To study convergence of subsequences of Fejér means and their restricted maximal operators on the martingale Hardy spaces $H_{p}(G)$ for $0<p\leq 1/2,$ the central role is played by the fact that any natural number $n\in\mathbb{N}$ can be uniquely expression as

[TABLE]

where only a finite numbers of $n_{j}$ differ from zero and their important characters $\left[n\right],$ $\left|n\right|,$ $\rho\left(n\right)$ and $V(n)$ are defined by

[TABLE]

and

[TABLE]

Persson and Tephnadze [11] generalized this result and proved that if $0<p\leq 1/2$ and $\left\{n_{k}:k\geq 0\right\}$ is a sequence of positive numbers, such that

[TABLE]

then the maximal operator $\widetilde{\sigma}^{\ast,\nabla},$ defined by

[TABLE]

is bounded from the Hardy space $H_{p}(G)$ to the Lebesgue space $L_{p}(G).$ Moreover, if $0<p<1/2$ and $\left\{n_{k}:k\geq 0\right\}$ is a sequence of positive numbers, such that

[TABLE]

then there exists a martingale $F\in H_{p}(G)$ such that

[TABLE]

In [22] it was proved that if $F\in H_{1/2}(G),$ then there exists an absolute constant $c,$ such that

[TABLE]

Moreover, if $\left\{n_{k}:k\geq 0\right\}$ is subsequence of positive integers $\mathbb{N}_{+},$ such that

[TABLE]

and $\{\varphi_{n}\}$ is any nondecreasing, nonnegative function, satisfying conditions $\Phi_{n}\uparrow\infty$ and

[TABLE]

then there exists a martingale $F\in H_{1/2}(G),$ such that

[TABLE]

It follows that if $f\in H_{1/2}(G)$ and $\left\{n_{k}:k\geq 0\right\}$ is any sequence of positive numbers, then $\sigma_{n_{k}}f$ are bounded from the Hardy space $H_{1/2}(G)$ to the space $H_{1/2}(G)$ if and only if, for some $c,$

[TABLE]

In [20] it was proved that the weighted maximal operator

[TABLE]

is bounded from the Hardy space $H_{1/2}\left(G\right)$ to the space $L_{1/2}\left(G\right).$ Moreover, it was also proved that the rate of denominator $\{\log^{2}(n+1)\}$ can not be improved.

Baramidze and Tephnadze [2] proved that if $\left\{n_{k}:k\geq 0\right\}$ be a sequence of positive numbers, such that

[TABLE]

then the maximal operator (1) is bounded from the Hardy space $H_{1/2}(G)$ to the space $L_{1/2}(G).$ Moreover, if $\left\{n_{k}:k\geq 0\right\}$ is a sequence of positive numbers, such that

[TABLE]

then there exists a martingale $F\in H_{p}(G)$ such that

[TABLE]

In this paper we investigate the restricted weighted maximal operator, defined by

[TABLE]

of Fejér means of Walsh-Fourier series and prove that the it is bounded from the martingale Hardy space $H_{1/2}(G)$ to the Lebesgue space $L_{1/2}(G).$ As a consequence we obtain some new and and well-know results.

This paper is organized as follows: In order not to disturb our discussions later on some definitions and notations are presented in Section 2. The main result and some of its consequences can be found in Section 3. For the proof of the main result we need some auxiliary statements. These results are presented in Section 4. The detailed proofs are given in Section 5.

2. Definitions and Notations

Let $\mathbb{N}_{+}$ denote the set of the positive integers, $\mathbb{N}:=\mathbb{N}_{+}\cup\{0\}.$ Denote by $Z_{2}$ the discrete cyclic group of order 2, that is $Z_{2}:=\{0,1\},$ where the group operation is the modulo 2 addition and every subset is open. The Haar measure on $Z_{2}$ is given so that the measure of a singleton is 1/2.

Define the group $G$ as the complete direct product of the group $Z_{2},$ with the product of the discrete topologies of $Z_{2}$. The elements of $G$ are represented by sequences $x:=(x_{0},x_{1},...,x_{j},...),$ where $x_{k}=0\vee 1.$

It is easy to give a base for the neighborhood of $x\in G$

[TABLE]

Denote $I_{n}:=I_{n}\left(0\right),$ $\overline{I_{n}}:=G$ $\backslash$ $I_{n}$ and $e_{n}:=\left(0,...,0,x_{n}=1,0,...\right)\in G,$ for $n\in\mathbb{N}$. Then it is easy to prove that

[TABLE]

If $n\in\mathbb{N},$ then every $n$ can be uniquely expressed as $n=\sum_{j=0}^{\infty}n_{j}2^{j},$ where $n_{j}\in Z_{2}$ $~{}(j\in\mathbb{N})$ and only a finite numbers of $n_{j}$ differ from zero. Every $n\in\mathbb{N}$ can be also represented as $n=\sum_{i=1}^{r}2^{n_{i}},n_{1}>n_{2}>...n_{r}\geq 0.$ For such a representation of $n\in\mathbb{N},$ we denote numbers

[TABLE]

Let $\begin{matrix}2^{s}\leq{{n}_{{{s}_{1}}}}\leq{{n}_{{{s}_{2}}}}\leq...\leq{{n}_{{{s}_{r}}}}\leq{2^{s+1}},\ s\in\mathbb{N}.\\ \end{matrix}$ For such ${{n}_{{{s}_{j}}}},$ which can be written as ${{n}_{{{s}_{j}}}}=\sum\limits_{i=1}^{{{r}_{{{s}_{j}}}}}{\sum\limits_{k=l_{i}^{{{s}_{j}}}}^{t_{i}^{{{s}_{j}}}}{2^{k}}},$ where $0\leq l_{1}^{{{s}_{j}}}\leq t_{1}^{{{s}_{j}}}\leq l_{2}^{{{s}_{j}}}-2<l_{2}^{{{s}_{j}}}\leq t_{2}^{{{s}_{j}}}\leq...\leq l_{{{r}_{j}}}^{{{s}_{j}}}-2<l_{{{r}_{s_{j}}}}^{{{s}_{j}}}\leq t_{{{r}_{s_{j}}}}^{{{s}_{j}}},$ we denote

[TABLE]

where $u_{1}^{s}<u_{2}^{s}<...<u_{r^{3}_{s}}^{s}.$ We note that $t_{{{r}_{s_{j}}}}^{{{s}_{j}}}=s\in{{A}_{s}},\ \ \text{ for }\ \ j=1,2,...,r.$

Let us denote the cardinality of the set $A_{s}$ by $|A_{s}|$, that is

[TABLE]

It is evident that $|A_{s}|=r^{3}_{s}\leq r_{s}^{1}+r_{s}^{2}.$ Moreover, $r^{s}_{3}=card(A_{s})<\infty$ if and only if $r_{s}^{1}<\infty\ \ \text{and}\ \ r_{s}^{2}<\infty.$ We note that if $|A_{s}|<\infty,$ then each ${{n}_{{{s}_{j}}}}$ has bounded variation

[TABLE]

and   $r_{s}^{1}<\infty,\ r_{s}^{2}<\infty\ \text{and}\ r_{s}^{3}<\infty.$ Therefore, if we consider blocks (intervals)

[TABLE]

then it is easy to see that it contains $\left({{r_{s}^{3}}{(r_{s}^{3}+1)}}\right)/{2}$ different blocks. Therefore, the dyadic representation of different natural numbers, which contains blocks from (4), can be at most $2^{\frac{{r_{s}^{3}}{(r_{s}^{3}+1)}}{2}}-1,$ which is finite number and the set $\{{{n}_{{{s}_{1}}}},{{n}_{{{s}_{2}}}},...{{n}_{{{s}_{r}}}}\}$ is finite for all $s\in\mathbb{N}_{+},$ from which it follows that

[TABLE]

Summing up, we can conclude that $\sup_{s\in\mathbb{N}}|A_{s}|<\infty$ if and only if the set $\{{{n}_{{{s}_{1}}}},{{n}_{{{s}_{2}}}},...,{{n}_{{{s}_{r}}}}\}$ is finite for all $s\in\mathbb{N}_{+}$ and each ${{n}_{{{s}_{j}}}}$ has bounded variation, that is, conditions (3) and (5) are fulfilled.

The norms (or quasi-norm) of the spaces $L_{p}$ and $\text{weak}-L_{p},\left(0<p<\infty\right)$ are respectively defined by

[TABLE]

The $k$-th Rademacher function is defined by

[TABLE]

Now, define the Walsh system $w:=(w_{n}:n\in\mathbb{N})$ on $G$ as:

[TABLE]

The Walsh system is orthonormal and complete in $L_{2}\left(G\right)$ (see [14]).

If $f\in L_{1}\left(G\right),$ then we can define the Fourier coefficients, partial sums of Fourier series, Fejér means, Dirichlet and Fejér kernels in the usual manner:

[TABLE]

Recall that (see [8] and [14])

[TABLE]

Let $n=\sum_{i=1}^{r}2^{n_{i}},\ \ \ n_{1}>n_{2}>...>n_{r}\geq 0.$ Then (see [4] and [14])

[TABLE]

The $\sigma$-algebra, generated by the intervals $\left\{I_{n}\left(x\right):x\in G\right\}$ will be denoted by $\zeta_{n}$ $\left(n\in\mathbb{N}\right).$ Denote by $F=\left(F_{n},n\in\mathbb{N}\right)$ a martingale with respect to $\zeta_{n}$ $\left(n\in\mathbb{N}\right)$ (for details see e.g. [24]). The maximal function $F^{\ast}$ of a martingale $F$ is defined by $F^{\ast}:=\sup_{n\in\mathbb{N}}\left|F_{n}\right|.$

In the case $f\in L_{1}\left(G\right),$ the maximal functions $f^{\ast}$ are also given by

[TABLE]

For $0<p<\infty,$ the Hardy martingale spaces $H_{p}\left(G\right)$ consist of all martingales, for which

[TABLE]

A bounded measurable function $a$ is a $p$-atom, if there exists an interval $I,$ such that

[TABLE]

It is easy to check that for every martingale $F=\left(F_{n},n\in\mathbb{N}\right)$ and every $k\in\mathbb{N}$ the limit

[TABLE]

exists and it is called the $k$-th Walsh-Fourier coefficients of $F.$

The Walsh-Fourier coefficients of $f\in L_{1}\left(G\right)$ are the same as those of the martingale $\left(S_{2^{n}}f,n\in\mathbb{N}\right)$ obtained from $f$.

3. The Main Result and its Consequences

Our main result reads:

Theorem 1**.**

a) Let $f\in{{H}_{1/2}}\left(G\right)$ and $\left\{n_{k}:k\geq 0\right\}$ is a sequence of positive numbers. Then the weighted maximal operator $\widetilde{\sigma}^{\ast,\nabla},$ defined by

[TABLE]

is bounded from the Hardy space ${{H}_{1/2}}$ to the Lebesgue space ${{L}_{1/2}}$.

b) (Sharpness) Let

[TABLE]

and $\{\varphi_{n}\}$ is a nondecreasing sequence satisfying the condition

[TABLE]

Then there exists a martingale $f\in{{H}_{1/2}}\left(G\right),$ such that the maximal operator

[TABLE]

is not bounded from the Hardy space ${{H}_{1/2}}(G)$ to the Lebesgue space ${{L}_{1/2}}(G).$

Corollary 1**.**

The maximal operator $\overset{\sim}{\sigma}^{*},$ defined by

[TABLE]

is bounded from the Hardy space $H_{1/2}\left(G\right)$ to the space $L_{1/2}\left(G\right).$

b) Let $\{\varphi_{n}\}$ is a nondecreasing sequence satisfying the condition

[TABLE]

Then there exists a martingale $f\in H_{1/2},$ such that the maximal operator

[TABLE]

is not bounded from the Hardy space ${{H}_{1/2}}(G)$ to the Lebesgue space ${{L}_{1/2}}(G).$

In order to be able to compare with some other results in the literature (see Remark 1) we also state the following:

Corollary 2**.**

Let $f\in{{H}_{1/2}}\left(G\right)$. Then the restricted maximal operators $\widetilde{\sigma}^{\ast,\nabla}_{i},\ i=1,2,3,$ defined by

[TABLE]

[TABLE]

[TABLE]

where $[n]$ denotes integer part of $n$, all are bounded from the Hardy space ${{H}_{1/2}}(G)$ to the Lebesgue space ${{L}_{1/2}}(G)$.

Remark 1**.**

In [11] it was proved that if $0<p<1/2,$ then the restricted maximal operators $\widetilde{\sigma}^{\ast,\nabla}_{2}$ and $\widetilde{\sigma}^{\ast,\nabla}_{3}$ defined by (9) and (10), are not bounded from the Hardy space ${{H}_{p}}(G)$ to the Lebesgue space $weak-{{L}_{p}}(G)$.

On the other hand, in [11] it was proved that if $0<p\leq 1/2$, then the restricted maximal operator $\widetilde{\sigma}^{\ast,\nabla}_{1}$ defined by (8) is bounded from the Hardy space ${{H}_{p}}(G)$ to the Lebesgue space $L_{p}(G).$

Corollary 3**.**

Let $f\in{{H}_{1/2}}\left(G\right)$ and $\left\{A^{s}_{k}:0\leq k\leq s-1\right\}$ be a sequence of positive numbers, defined by

[TABLE]

Then the restricted maximal operator $\widetilde{\sigma}^{\ast,\nabla}_{4},$ defined by

[TABLE]

is bounded from the Hardy space ${{H}_{1/2}}$ to the Lebesgue space ${{L}_{1/2}}.$

Corollary 4**.**

Let $f\in{{H}_{1/2}}\left(G\right)$ and $\left\{A^{s}_{k}:0\leq k\leq s-1\right\}$ be a sequence of positive numbers, defined by

[TABLE]

Then the restricted maximal operator $\widetilde{\sigma}^{\ast,\nabla}_{5},$ defined by

[TABLE]

is bounded from the Hardy space ${{H}_{1/2}}(G)$ to the Lebesgue space ${{L}_{1/2}}(G).$

Remark 2**.**

Let $f\in{{H}_{1/2}}\left(G\right)$ and $\left\{\alpha^{s}_{k}:0\leq k\leq s-1\right\}$ and $\left\{\beta^{s}_{k}:0\leq k\leq s-1\right\}$ are sequences of positive numbers, defined in Corollaries 3 and 4. Then there exist absolute constants $c_{1}$ and $c_{2}$ such that

[TABLE]

for any $0\leq k\leq s-1\ \text{and}\ s\in\mathbb{N}.$

We note that these results was proved in [22] and follow the facts that

[TABLE]

Corollary 5**.**

Let $f\in{{H}_{1/2}}\left(G\right)$ and $\left\{A^{s}_{k}:0\leq k\leq s-1\right\}$ be a sequence of positive numbers, defined by

[TABLE]

where $[n]$ denotes the integer part of $n.$ Then the restricted maximal operator $\widetilde{\sigma}^{\ast,\nabla}_{6},$ defined by

[TABLE]

is bounded from the Hardy space ${{H}_{1/2}}(G)$ to the Lebesgue space ${{L}_{1/2}}(G).$

Corollary 6**.**

Let $f\in{{H}_{1/2}}\left(G\right)$ and $\left\{A^{s}_{k}:0\leq k\leq s-1\right\}$ be a sequence of positive numbers, defined by

[TABLE]

Then the restricted maximal operator $\widetilde{\sigma}^{\ast,\nabla}_{7},$ defined by

[TABLE]

is bounded from the Hardy space ${{H}_{1/2}}(G)$ to the Lebesgue space ${{L}_{1/2}}(G).$

4. Auxiliary Lemmas

Lemma 1** (Weisz [25] and Simon [18]).**

A martingale $F=\left(F_{n},\text{ }n\in\mathbb{N}\right)$ is in $H_{p}\left(0<p\leq 1\right)$ if and only if there exists a sequence $\left(a_{k},k\in\mathbb{N}\right)$ of p-atoms and a sequence $\left(\mu_{k},k\in\mathbb{N}\right)$ of a real numbers, such that for every $n\in\mathbb{N}$

[TABLE]

Moreover, $\left\|F\right\|_{H_{p}}\backsim\inf\left(\sum_{k=0}^{\infty}\left|\mu_{k}\right|^{p}\right)^{1/p},$ where the infimum is taken over all decomposition of $F$ of the form (11).

Lemma 2** (Weisz [24]).**

Suppose that an operator $T$ is $\sigma$-linear and

[TABLE]

for every $p$-atom $a$, where $I$ denote the support of the atom. If $T$ is bounded from $L_{\infty\text{ }}$ to $L_{\infty},$ then

[TABLE]

Lemma 3** (see e.g. [8], [14]).**

Let $t,n\in\mathbb{N}.$ Then

[TABLE]

Lemma 4** (see e.g. [13], [22]).**

Let

[TABLE]

Then

[TABLE]

Lemma 5** (see e.g. [13], [22]).**

Let

[TABLE]

Then, for any $i=1,2,...,s,$

[TABLE]

and

[TABLE]

where $I_{1}\left(e_{-1}+e_{0}\right)=I_{2}\left(e_{0}+e_{1}\right).$

5. Proof of the Theorem 1

Proof.

Since $\sigma_{n}$ is bounded from $L_{\infty}$ to $L_{\infty},$ by Lemma 2, the proof of theorem 1 will be complete, if we prove that

[TABLE]

for every 1/2-atom $a.$ We may assume that $a$ is an arbitrary $1/2$-atom, with support$\ I,$ $\mu\left(I\right)=2^{-M}$ and $I=I_{M}.$ It is easy to see that $\sigma_{n}\left(a\right)=0,\ \text{ when }\ n<2^{M}.$ Therefore, we can suppose that $n_{s_{k}}\geq 2^{M}.$ Let $x\in I_{M}$ and $2^{s}\leq n_{s_{k}}<2^{s+1}$ for some $n_{s_{k}}\geq 2^{M}.$ Since $\left\|a\right\|_{\infty}\leq 2^{2M}$ and $\left|s_{k}\right|=s,$ by using Lemma 4 we obtain that

[TABLE]

If we denote by

[TABLE]

from (5) we can conclude that

[TABLE]

Hence,

[TABLE]

Since

[TABLE]

we obtain that Theorem 1 is proved if we can prove that

[TABLE]

and

[TABLE]

for all

[TABLE]

Indeed, if (14) and (15) hold, from (5) we get that

[TABLE]

It remains to prove (14) and (15). Let $t\in I_{M}\ \text{ and }\ x\in I_{l+1}\left(e_{k}+e_{l}\right).$ If $0\leq k<l<\alpha^{s}_{A}\leq M\ \ \ \text{or }\ \ \ 0\leq k<l<M<\alpha^{s}_{A},$ then $x+t\in I_{l+1}\left(e_{k}+e_{l}\right)$ and if we apply Lemma 3 we obtain that

[TABLE]

Let $0\leq k<\alpha^{s}_{A}\leq l<M.$ Then $x+t\in I_{l+1}\left(e_{k}+e_{l}\right)$ and if we use Lemma 3 we get that

[TABLE]

so that

[TABLE]

Analogously to (17) we can prove that if $0\leq\alpha^{s}_{A}\leq k<l<M$, then

[TABLE]

so that

[TABLE]

Let $t\in I_{M}\ \text{ and }\ x\in I_{M}(e_{k}).$ If $0\leq k<\alpha^{s}_{A}\leq M$ or $0\leq k<M\leq\alpha^{s}_{A}.$ Since $x+t\in x\in I_{M}(e_{k})$ and if we apply Lemma 3, then we find that

[TABLE]

so that

[TABLE]

Let $0\leq\alpha^{s}_{A}\leq k<M.$ Since $x+t\in x\in I_{M}(e_{k})$ and if we apply Lemma 3 we obtain that

[TABLE]

and

[TABLE]

Let $0\leq\alpha^{s}_{A}<M,A=1,...,s.$ By combining (2) with (16)-(20) we get that

[TABLE]

Analogously we can prove that (15) holds also for the case $\alpha^{s}_{A}\geq M.$ Hence, (15) holds and it remains to prove (14).

Now, prove boundedness of $II_{l^{s}_{A}}^{2}$. Let $t\in I_{M}\ \text{ and }\ x\in I_{i}\backslash I_{i+1}.$  If $i\leq l^{s}_{A}-1,$ since $x+t\in I_{i}\backslash I_{i+1},$ by using (6) we have that

[TABLE]

If $l^{s}_{A}\leq i<M,$ by using (6) we obtain that

[TABLE]

Let $0\leq l^{s}_{A}<M.$ By combining (2), (21) and (22) we get that

[TABLE]

If $M\leq l^{s}_{A},$ then $i<M\leq l^{s}_{A}$ and apply (21) we get that

[TABLE]

and also (14) is proved by just combining (5) and (24) so part a) is complete and we turn to the proof of b).

Under condition (7), there exists an increasing sequence $\left\{\alpha_{k}:\text{ }k\geq 0\right\}\subset\left\{n_{k}\text{ }:k\geq 0\right\}$ of positive integers, such that

[TABLE]

and

[TABLE]

Let

[TABLE]

where

[TABLE]

Since

[TABLE]

and

[TABLE]

if we apply Lemma 1 and (25) we can conclude that $F=\left(F_{1},F_{2},...\right)\in H_{1/2}.$

It is easy to prove that

[TABLE]

Let $2^{\left|\alpha_{k}\right|}<j<\alpha_{k}.$ By using (26) we get that

[TABLE]

Let  $\begin{matrix}2^{\left|\alpha_{k}\right|}\leq{{\alpha}_{{{s}_{n}}}}\leq{2^{{\left|\alpha_{k}\right|}+1}}.\end{matrix}$  Then, by using (27) we find that

[TABLE]

Since

[TABLE]

we obtain that

[TABLE]

By combining well-known estimates (see [13])

[TABLE]

we obtain that

[TABLE]

Let  $\begin{matrix}2^{\left|\alpha_{k}\right|}\leq{{\alpha}_{{{s}_{1}}}}\leq{{\alpha}_{{{s}_{2}}}}\leq...\leq{{\alpha}_{{{s}_{r}}}}\leq{2^{{\left|\alpha_{k}\right|}+1}}\end{matrix}$  be natural numbers which generates the set

[TABLE]

and choose number $\alpha_{s_{n}}=\sum_{i=1}^{r_{n}}\sum_{k=l_{i}^{n}}^{t_{i}^{n}}2^{k},$ where

[TABLE]

for some $1\leq n\leq r,$ such that $l^{\left|\alpha_{k}\right|}_{u}=l_{i},\ \text{for some}\ 1\leq u\leq r^{1}_{\left|\alpha_{k}\right|},1\leq i\leq r_{\left|\alpha_{k}\right|}^{1}.$

Since $\mu\left\{E_{l_{i}}\right\}\geq 1/2^{l_{i}+1},$ by using Lemma 5 we get that

[TABLE]

On the other hand, we can also choose number $\alpha_{s_{n}},$ for some $1\leq n\leq r,$ such that $t^{\left|\alpha_{k}\right|}_{u}=t_{i},\ \text{for some}\ 1\leq u\leq r^{2}_{\left|\alpha_{k}\right|},\ \ 1\leq i\leq r_{\left|\alpha_{k}\right|}^{2}.$ According the fact that $\mu\left\{E_{t_{i}}\right\}\geq 1/2^{t_{i}+3},$ by using again Lemma 5 for some $\alpha_{k}$ and $1\leq i\leq r_{s}^{2}$ we also get that

[TABLE]

By combining (5)-(5) and Lemma 5 for sufficiently big $\alpha_{k}$ we obtain that

[TABLE]

so also Part b) id proved and the proof is complete. ∎