Some new restricted maximal operators of Fej\'er means of Walsh-Fourier series in the space $H_{1/2}$

TL;DR

This paper identifies the largest subset of natural numbers for which a restricted maximal operator of Fejér means of Walsh-Fourier series remains bounded from the Hardy space $H_{1/2}$ to $L_{1/2}$, and proves the result's optimality.

Contribution

It determines the maximal subspace of natural numbers ensuring boundedness of the restricted maximal operator in Walsh-Fourier series, establishing the sharpness of this characterization.

Findings

Identified the maximal subset of natural numbers for boundedness.

Proved the boundedness of the restricted maximal operator.

Established the sharpness of the main result.

Abstract

In this paper we derive the maximal subspace of natural numbers $\left\{n_{k}:k\geq 0\right\}$, such that the restricted maximal operator, defined by $\sup_{k\in \mathbb{N}}\left\vert \sigma_{n_{k}}F \right\vert$ on this subspace of Fej\'er means of Walsh-Fourier series is bounded from the martingale Hardy space $H_{1/2}$ to the Lebesgue space $L_{1/2}$. The sharpness of this result is also proved.

Equations

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

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

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

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

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

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

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

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

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

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

$$\rho \left(2^k+1\right)=k\to \infty ,\text{as}k\to \infty ,$$

$$\underset{k\in \mathbb{N}}{\sup }\left|{\sigma }_{2^k+1}F\right|$$

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

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

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

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

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

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

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

$$\left[n\right]:=\min \{j\in \mathbb{N},n_j\ne 0\}\text{ and }\left|n\right|:=\max \{j\in \mathbb{N},n_j\ne 0\},$$

$$d\left(n\right)=\left|n\right|-\left[n\right],\text{ for all }n\in \mathbb{N}.$$

$$V\left(n\right):=n_0+\sum _{k=1}^{\infty }\left|n_k-n_{k-1}\right|.$$

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

$$n_{s_j}=\sum _{i=1}^{r_{s_j}}\sum _{k=l_i^{s_j}}^{t_i^{s_j}}{2}^k,$$

$$A_s$$

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

$$|A_s|=r_s^3\le r_s^1+r_s^2.$$

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

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

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

$$\dots$$

Full text

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

2020 Mathematics Subject Classification. 26015, 42C10, 42B30.

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

1. INTRODUCTION

All symbols used in this introduction can be found in Section 2.

In the one-dimensional case, the weak (1,1)-type inequality for the maximal operator $\sigma^{\ast}$ of Fejér means $\sigma_{n}$ with respect to the Walsh system, defined by

[TABLE]

was investigated in Schipp [19] and Pál, Simon [12] (see also [1], [10] and [16]). Fujii [3] and Simon [21] proved that $\sigma^{\ast}$ is bounded from $H_{1}$ to $L_{1}$. Weisz [32] 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 [20] 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, in [6] (see also [23]) he proved that there exists a martingale $F\in H_{p}$ $\left(0<p\leq 1/2\right),$ such that

[TABLE]

Weisz [35] proved that the maximal operator $\sigma^{\ast}$ of the Fejér means is bounded from the Hardy space $H_{1/2}$ to the space $weak-L_{1/2}$.

The boundedness of weighted maximal operators are considered in [24], [25]. The results for summability of Fejér means of Walsh-Fourier series can be found in [2], [9], [11], [14], [28], [32].

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]

Weisz [34] (see also [33]) also proved that for any $F\in H_{p}(G)$ $(p>0),$ the maximal operator

[TABLE]

is bounded from the Hardy space $H_{p}$ to the Lebesgue space $L_{p}$, but (for details see [17]) it is not bounded from the Hardy space $H_{p}$ to the space $H_{p}.$ Persson and Tephnadze [15] 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 $H_{p}$ to the Lebesgue space $L_{p}.$ 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 $F\in H_{p}$ such that

[TABLE]

From this fact it follows that if $0<p<1/2,$ $f\in H_{p}$ and $\left\{n_{k}:k\geq 0\right\}$ is any sequence of positive numbers, then the maximal operator, defined by (2) is bounded from the Hardy space $H_{p}$ to the space $L_{p}$ if and only if the condition (1) is fulfilled. Moreover, if $0<p<1/2,$ $f\in H_{p}$ and $\left\{n_{k}:k\geq 0\right\}$ is any sequence of positive numbers, then $\sigma_{n_{k}}f$ are uniformly bounded from the Hardy space $H_{p}$ to the Lebesgue space $L_{p}$ if and only if the condition (1) is fulfilled. Moreover, condition (1) is necessary and sufficient condition for the boundedness of subsequence $\sigma_{n_{k}}f$ from the Hardy space $H_{p}$ to the Hardy space $H_{p}.$

It is easy to prove that condition (1) also implies that the maximal operator (2) is bounded from the Hardy space $H_{1/2}$ to the Lebesgue space $L_{1/2}.$ Moreover, for the subsequence $\{n_{k}=2^{k}+1\},$ we have that

[TABLE]

but the maximal operator

[TABLE]

is bounded from the Hardy space $H_{1/2}$ to the Lebesgue space $L_{1/2}.$

In [30] it was proved that if $F\in H_{1/2},$ then there exists an absolute constant $c,$ such that

[TABLE]

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

[TABLE]

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

[TABLE]

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

[TABLE]

It follows that if $f\in H_{1/2}$ 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 $H_{1/2}$ to the space $H_{1/2}$ if and only if, for some $c,$

[TABLE]

Convergence of subsequences of partial sums and Fejér means of Walsh-Fourier series can be found in [13], [26], [27] and [31].

In this paper we complement the reported research above by investigating the limit case $p=1/2.$ In particular, we derive the maximal subspace of natural numbers $\left\{n_{k}:k\geq 0\right\}$, such that restricted maximal operator, defined by ${\sup}_{k\in\mathbb{N}}\left|\sigma_{n_{k}}F\right|$ on this subspace of Fejér means of Walsh-Fourier series is bounded from the martingale Hardy space $H_{1/2}$ to the Lebesgue space $L_{1/2}.$

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, some of them are new and of independent interest (see Propositions 1 and 2). 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.

Let

[TABLE]

that is $2^{\left|n\right|}\leq n<2^{\left|n\right|+1}.$ Set

[TABLE]

Define the variation $V(n)$ of $n\in\mathbb{N},$ with binary coefficients $\left(n_{k},\ k\in\mathbb{N}\right),$ by

[TABLE]

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

[TABLE]

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 define

[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

[TABLE]

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 $\frac{{r_{s}^{3}}{(r_{s}^{3}+1)}}{2}$ different blocks. Therefore, the dyadic representation of different natural numbers, which contains blocks from (6), can be at most $2^{\frac{{r_{s}^{3}}{(r_{s}^{3}+1)}}{2}}-1,$ which is a 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 (5) and (7) are fullfiled.

The norms (or quasi-norm) of the spaces $L_{p}(G)$ and $L_{p,\infty}\left(G\right),\ \left(0<p<\infty\right)$ 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 [18]).

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 [18])

[TABLE]

Let $n=\sum_{i=1}^{r}2^{n_{i}},\ \ \ n_{1}>n_{2}>...>n_{r}\geq 0.$ Then (see [8] and [18])

[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. [33]). The maximal function $F^{\ast}$ of a martingale $F$ is defined by

[TABLE]

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 $\left\{n_{k}:k\geq 0\right\}$ be a sequence of positive numbers and let $\left\{n_{s_{i}}:\ 1\leq i\leq r\right\}\subset\left\{n_{k}:\ k\geq 0\right\}$ be numbers such that $2^{s}\leq{{n}_{{{s}_{1}}}}\leq{{n}_{{{s}_{2}}}}\leq...\leq{{n}_{{{s}_{r}}}}\leq{2^{s+1}},\ s\in\mathbb{N}.$ If the sets ${{A}_{s}},$ defined by (2), are finite for all $s\in\mathbb{N},$ that is the cardinality of the sets ${{A}_{s}}$ are finite:

[TABLE]

then the restricted 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]

Then there exists a martingale $f\in{{H}_{1/2}}\left(G\right),$ such that the maximal operator, defined by (10), is not bounded from the Hardy space ${{H}_{1/2}}$ to the Lebesgue space ${{L}_{1/2}}.$

In particular, Theorem 1 implies the following optimal characterization:

Corollary 1**.**

Let $F\in{{H}_{1/2}}\left(G\right)$ and $\left\{n_{k}:k\geq 0\right\}$ be a sequence of positive numbers. Then the restricted maximal operator $\widetilde{\sigma}^{\ast,\nabla},$ defined by (10), is bounded from the Hardy space ${{H}_{1/2}}$ to the Lebesgue space ${{L}_{1/2}}$ if and only if any sequence of positive numbers $\left\{n_{k}:k\geq 0\right\}$ which satisfies $n_{k}\in[2^{s},2^{s+1}),$ is finite for each $s\in\mathbb{N}_{+}$ and each $\left\{n_{k}:k\geq 0\right\}$ has bounded variation, i.e.

[TABLE]

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 $\widetilde{\sigma}_{i}^{\ast,\nabla},\ i=1,2,3,$ defined by

[TABLE]

[TABLE]

[TABLE]

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

Remark 1**.**

In [15] 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 (12) and (13), are not bounded from the Hardy space ${{H}_{p}}$ to the Lebesgue space $weak-{{L}_{p}}$.

On the other hand, in [15] it was proved that if $0<p\leq 1/2$, then the restricted maximal operator defined by $\widetilde{\sigma}^{\ast,\nabla}_{1}$ is bounded from the Hardy space ${{H}_{p}}$ to the Lebesgue space $L_{p}.$

We also have the following related consequence of Theorem 1:

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

Remark 2**.**

We note that in this case $V(\alpha_{k}^{s})\leq 6$ for any $0\leq k\leq s-1$ and $s\in\mathbb{N},$

[TABLE]

i.e. it contains only two elements, the set

[TABLE]

and

[TABLE]

Finally, we state the following related result, which is connected to the research in [30] (see Remark 4):

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

Remark 3**.**

We note that now $V(\beta_{k}^{s})\leq 4,$ for any $0\leq k\leq s-1$ and $s\in\mathbb{N},$

[TABLE]

and

[TABLE]

Remark 4**.**

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

[TABLE]

and

[TABLE]

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

We note that these results were proved in [30] and they follow, respectively, from the facts that

[TABLE]

4. Auxiliary Lemmas and Propositions

Lemma 1** (Weisz [34] (see also Simon [22])).**

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,

[TABLE]

where the infimum is taken over all decomposition of $F$ of the form (14).

Lemma 2** (Weisz [33]).**

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. [4]).**

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

[TABLE]

Lemma 4** (see e.g. [7], [25]).**

Let $n\geq 2^{M}$ and

[TABLE]

Then

[TABLE]

Lemma 5** (see [29]).**

Let

[TABLE]

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

[TABLE]

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

We also need the following new statement of independent interest:

Proposition 1**.**

Let

[TABLE]

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

[TABLE]

Proof.

It is evident that we always have that $t_{i}+2\leq l_{i+1}.$ If $t_{i}+2=l_{i+1},$ then $E_{t_{i}}=I_{t_{i}+3}\left(e_{t_{i}+1}+e_{t_{i}+2}\right)=I_{l_{i+1}+1}\left(e_{l_{i+1}-1}+e_{l_{i+1}}\right)=E_{l_{i+1}}$ and if we apply Lemma 5 we find that

[TABLE]

Let $t_{i}+2<l_{i+1}.$ By combining (8) and Lemma 3, for any $n\geq t_{i}+3$ we get that

[TABLE]

From (9), for $x\in E_{t_{i}}$ we can conclude that

[TABLE]

[TABLE]

Suppose that $l_{i}<t_{i}.$ Since

[TABLE]

for $x\in E_{t_{i}}$ we find that

[TABLE]

Moreover, by combining (8) and Lemma 3 we get that

[TABLE]

For $II$ we have that

[TABLE]

Moreover, $III$ can be estimated as follows

[TABLE]

By combining (17)-(19) and putting them into (4) we obtain that

[TABLE]

If ${t_{i}}=l_{i}$ we get that $t_{i-1}\leq l_{i}-2={t_{i}}-2$. Hence, by using (15) we find that

[TABLE]

By simple calculations we get that

[TABLE]

[TABLE]

and

[TABLE]

We insert (22)-(24) into (21) and find that

[TABLE]

The proof is complete by just combining the estimates (20) and (25). ∎

Corollary 5**.**

Let

[TABLE]

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

[TABLE]

and

[TABLE]

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

Our second auxiliary result of independent interest is the following:

Proposition 2**.**

Let

[TABLE]

Then

[TABLE]

Proof.

Let $n=\sum_{i=1}^{s}2^{n_{i}},\ \ \ n_{1}>n_{2}>...>n_{r}\geq 0.$ By using (9) we get that

[TABLE]

For $I_{1}$ we have that

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Since $2^{n}-1=\sum_{k=0}^{n-1}2^{k}$ and

[TABLE]

we obtain that

[TABLE]

[TABLE]

Thus, by using the estimates

[TABLE]

we can conclude that

[TABLE]

Let $l_{j}<n_{A}\leq t_{j},$ for some $j=1,...,s.$ Then

[TABLE]

If $l_{j}=n_{A}$ for some $j=1,...,s,$ then

[TABLE]

Hence,

[TABLE]

Finally, we use the estimates (27)-(28) in (4) and the proof is complete. ∎

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

[TABLE]

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

[TABLE]

by using Proposition 2 we obtain that

[TABLE]

If we denote by

[TABLE]

from (5) we can conclude that

[TABLE]

and

[TABLE]

Hence,

[TABLE]

Since $\sup_{s\in\mathbb{N}}r_{s}^{1}<r<\infty,\ \ \ \sup_{s\in\mathbb{N}}r_{s}^{2}<r<\infty,$ we obtain that (29) holds so that Theorem 1 a) is proved if we can prove that

[TABLE]

and

[TABLE]

for all $\alpha^{s}_{A}=l^{s}_{A},\ \ \ A=1,...,r_{s}^{1}$ and $\alpha^{s}_{A}=t^{s}_{A},\ \ \ A=1,...,r_{s}^{2}$. Indeed, if (32) and (33) hold, from (5) we get that

[TABLE]

It remains to prove (32) and (33). Let

[TABLE]

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

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so that

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Analogously to (35) we can prove that if $0\leq\alpha^{s}_{A}\leq k<l<M$, then

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so that

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Let

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Let $0\leq k<\alpha^{s}_{A}\leq M$ 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 we obtain that

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and

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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, then we find that

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and

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Let $0\leq\alpha^{s}_{A}<M.$ By combining (3) with (34)-(38) for any $A=1,...,s$ we have that

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Analogously we can prove that (33) holds also for the case $\alpha^{s}_{A}\geq M.$ Hence, (33) holds and it remains to prove (32).

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 (8) we have that

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If $l^{s}_{A}\leq i<M,$ then by using (8) we obtain that

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Let $0\leq l^{s}_{A}<M.$ By combining (3), (39) and (40) we get that

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If $M\leq l^{s}_{A},$ then $i<M\leq l^{s}_{A}$ and apply (39) we get that

[TABLE]

and also (32) is proved by just combining (5) and (42) so part a) is complete and we turn to the proof of b).

Under condition (11), 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]

Let

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where

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Since   \text{supp}(a_{k})=I_{\left|\alpha_{k}\right|},\ \ \ \left\|a_{k}\right\|_{\infty}\leq 2^{2\left|\alpha_{k}\right|}=\mu(\text{supp }a_{k})^{-2}\ \ and

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if we apply Lemma 1 and (43) we can conclude that $F=\left(F_{1},F_{2},...\right)\in H_{1/2}.$

It is easy to prove that

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Let $2^{\left|\alpha_{k}\right|}<j<\alpha_{k}.$ By using (44) we get that

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Let  $\begin{matrix}2^{\left|\alpha_{k}\right|}\leq{{\alpha}_{{{s}_{n}}}}\leq{2^{{\left|\alpha_{k}\right|}+1}}.\end{matrix}$  Then, by using (45) we find that

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Since

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we obtain that

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By combining the well-known estimates (see [17])

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we obtain that

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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

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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 Corollary 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 to the fact that $\mu\left\{E_{t_{i}}\right\}\geq 1/2^{t_{i}+3},$ by using again Corollary 5 for some $\alpha_{k}$ and $1\leq i\leq r_{s}^{2}$ we also get that

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By combining (5)-(5) with Proposition 1 for sufficiently big $\alpha_{k}$ we get that

[TABLE]

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