Almost everywhere convergence and divergence of Cesàro means with varying parameters of Walsh–Fourier series

Abstract

In the present paper, we prove the almost everywhere convergence and divergence of subsequences of Cesàro means with zero tending parameters of Walsh–Fourier series.

1 Introduction

We denote the set of non-negative integers by $\mathbb {N}$. By a dyadic interval in $\mathbb {I}:=[0,1)$, we mean one of the form $I\left( l,k\right) :=\left[ \frac{l}{2^{k}},\frac{l+1}{2^{k}}\right)$ for some $k\in \mathbb {N}$, $0\le l<2^{k}$. Given $k\in \mathbb {N}$ and $x\in [0,1),$ let $I_{k}(x)$ denote the dyadic interval of length $2^{-k}$ which contains the point x. Also, use the notation $I_{n}:=I_{n}\left( 0\right) \left( n\in \mathbb {N}\right) ,\overline{I}_{k}\left( x\right) :=\mathbb {I}\backslash I_{k}\left( x\right)$. Let

$$\begin{aligned} x=\sum \limits _{n=0}^{\infty }x_{n}2^{-\left( n+1\right) } \end{aligned}$$

be the dyadic expansion of $x\in \mathbb {I}$, where $x_{n}=0$ or 1, and if x is a dyadic rational number, we choose the expansion which terminate in $0^{\prime }$s. We also use the following notation:

$$\begin{aligned} I_{k}(x)=I_{k}\left( x_{0},x_{1},...,x_{k-1}\right) . \end{aligned}$$

For any given $n\in \mathbb {N}$, it is possible to write n uniquely as

$$\begin{aligned} n=\sum \limits _{k=0}^{\infty }\varepsilon _{k}\left( n\right) 2^{k}, \end{aligned}$$

where $\varepsilon _{k}\left( n\right) =0$ or 1 for $k\in \mathbb {N}$. This expression will be called the binary expansion of n and the numbers $\varepsilon _{k}\left( n\right)$ will be called the binary coefficients of n. Denote for $1\le n\in \mathbb {N},\,\,$ $\left| n\right| :=\max \{j\in \mathbb {N}\mathbf {:}\varepsilon _{j}\left( n\right) \ne 0\}$, that is $2^{\left| n\right| }\le n<2^{\left| n\right| +1}.$

Set the definition of the nth $\left( n\in \mathbb {N}\right)$ Walsh–Paley function at point $x\in \mathbb {I}$ as

$$\begin{aligned} w_{n}\left( x\right) =\left( -1\right) ^{\sum \limits _{j=0}^{\infty }\varepsilon _{j}\left( n\right) x_{j}}. \end{aligned}$$

Denote by $\dotplus$ the logical addition on $\mathbb {I}$. That is, for any $x,y\in \mathbb {I}$ and $k,n\in \mathbb {N}$

$$\begin{aligned} x\dotplus y:=\sum \limits _{n=0}^{\infty }\left| x_{n}-y_{n}\right| 2^{-\left( n+1\right) }. \end{aligned}$$

Define the binary operator $\oplus :\mathbb {N}\mathbf {\times }\mathbb {N} \mathbf {\rightarrow }\mathbb {N}$ by

$$\begin{aligned} k\oplus n=\sum \limits _{i=0}^{\infty }\left| \varepsilon _{i}\left( k\right) -\varepsilon _{i}\left( n\right) \right| 2^{i}. \end{aligned}$$

(1)

It is well known (see, e.g., [13], p. 5) that

$$\begin{aligned} w_{m\oplus n}\left( x\right) =w_{m}\left( x\right) w_{n}\left( x\right) ,x\in [0,1),n,m\in \mathbb {N}. \end{aligned}$$

(2)

The Walsh–Dirichlet kernel is defined by

$$\begin{aligned} D_{n}\left( x\right) =\sum \limits _{k=0}^{n-1}w_{k}\left( x\right) . \end{aligned}$$

Set

$$\begin{aligned} D_{n}^{*}:=w_{n}D_{n}. \end{aligned}$$

Recall that [9, 13]

$$\begin{aligned} D_{2^{n}}\left( x\right) =2^{n}\chi _{I_{n}}\left( x\right) , \end{aligned}$$

(3)

where $\chi _{E}$ is the characteristic function of the set E.

Dyadic shift transformations of a function on the unit interval $\mathbb {I}$ will be denoted by $\tau _yf$ and it will be defined as

$$\begin{aligned} \left( \tau _{y}f\right) \left( x\right) :=f\left( x\dotplus y\right) { \ \ }\left( x\in \mathbb {I}\right) . \end{aligned}$$

The Fejér kernel of Walsh–Fourier series defined by

$$\begin{aligned} K_{n}\left( x\right) =\frac{1}{n}\sum \limits _{j=0}^{n-1}D_{j}\left( x\right) . \end{aligned}$$

The partial sums of the Walsh–Fourier series are defined as follows:

$$\begin{aligned} S_{m}(f,x)=\sum \limits _{j=0}^{m-1}\widehat{f}\left( j\right) w_{j}(x), \end{aligned}$$

where the number

$$\begin{aligned} \widehat{f}\left( j\right) =\int \limits _{\mathbb {I}}fw_{j} \end{aligned}$$

is said to be the jth Walsh–Fourier coefficient of the function f.

The space $L_{1}\left( \mathbb {I}\right)$ is defined by $\{f:\mathbb {I} \rightarrow \mathbb {R}:\Vert f\Vert _{1}<\infty \}$, where

$$\begin{aligned} \left\| f\right\| _{1}:=\int \limits _{\mathbb {I}}\left| f\left( x\right) \right| dx\,. \end{aligned}$$

The space weak-$L_{1}\left( \mathbb {I}\right)$ consists of all (Lebesgue) measurable functions f for which

$$\begin{aligned} \left\| f\right\| _{\text {weak}-L_{1}\left( \mathbb {I}\right) }:=\sup \limits _{\lambda>0}\lambda \text {mes}\left( \left| f\right| >\lambda \right) <+\infty . \end{aligned}$$

Let $f\in L_{1}\left( \mathbb {I}\right)$. Then, the maximal function given by

$$\begin{aligned} E^{*}\left( f, x\right) =\sup \limits _{n\in \mathbb {N}}\frac{1}{\left| I_{n}(x)\right| }\left| \int \limits _{I_{n}(x)}f\left( u\right) du\right| ,\,\,x\in \mathbb {I}. \end{aligned}$$

For each $n\in \mathbb {N}$, let $\mathcal {A}_{n}$ represent the $\sigma$ -algebra generated by the collection of dyadic intervals $\left\{ I\left( k,n\right) :k=0,1,...,2^{n}-1\right\}$. Thus, every element of $\mathcal {A}_{n}$ is a finite union of intervals of the form $\left[ k2^{-n},\left( k+1\right) 2^{-n}\right)$ or an empty set.

Let $L\left( \mathcal {A}_{n}\right)$ represent the collection of $\mathcal {A }_{n}$-measurable functions on $\mathbb {I}$. By the Paley Lemma ( see [13], Ch. 1, p. 12), $L\left( \mathcal {A}_{n}\right)$ coincides with the collection of Walsh polynomials of order less than $2^{n}$.

A sequence of functions $\left( f_{n}:n\in \mathbb {N}\right)$ is called a dyadic martingale if each $f_{n}$ belongs to $L\left( \mathcal {A}_{n}\right)$ and

$$\begin{aligned} \int \limits _{E}f_{n+1}=\int \limits _{E}f_{n}\left( E\in \mathcal {A}_{n},n\in \mathbb {N}\right) . \end{aligned}$$

It is clear that the $2^{n}$th partial sums of any Walsh series is a dyadic martingale. Conversely, it is easy to see that every dyadic martingale can be obtained in this way. Thus investigation of $2^{n}$th partial sums of Walsh series leads to the study of dyadic martingales. It is well known that $\left( f_{n}:n\in \mathbb {N}\right)$ is dyadic martingale if and only if $f_{n}\in L\left( \mathcal {A}_{n}\right)$ and

$$\begin{aligned} S_{2^{n}}\left( f_{n+1}\right) =f_{n}{ \ \ }\left( n\in \mathbb {N} \right) \text {.} \end{aligned}$$

A martingale $\left( f_{n}:n\in \mathbb {N}\right)$ will be called regular if there is an integrable function f, such that $f_{n}=S_{2^{n}}\left( f\right)$ for all $n\in \mathbb {N}$.

Let $\mathbf {A}$ denote the collection of sequences $\mathbf {\beta } :=\left\{ \beta _{n}:n\in \mathbb {N}\right\}$ which satisfy $\beta _{n}\in$ $L\left( \mathcal {A}_{n}\right)$ for $n\in \mathbb {N}$ and

$$\begin{aligned} \left\| \mathbf {\beta }\right\| :=\sup \limits _{n\in \mathbb {N} }\left\| \beta _{n}\right\| _{\infty }<\infty . \end{aligned}$$

For a given $\beta \in \mathbf {A}$ and $f\in L_{1}\left( \mathbb {I}\right)$, the martingale transform of f is defined by

$$\begin{aligned} \mathbf {T}\left( \mathbf {\beta }\right) f:=\sum \limits _{n=0}^{\infty }\beta _{n}\Delta _{n}f, \end{aligned}$$

where $\Delta _{n}f:=S_{2^{n+1}}\left( f\right) -S_{2^{n}}\left( f\right)$ for $n\in \mathbb {N}$. The maximal martingale transform is defined by

$$\begin{aligned} \mathbf {T}^{*}\left( \mathbf {\beta }\right) f:=\sup \limits _{N\in \mathbb { N}}\left| \sum \limits _{n=0}^{N}\beta _{n}\Delta _{n}f\right| . \end{aligned}$$

In fact, we will use the following theorem (see [13], Ch. 3, Theorem 4; see more details in [16]).

Theorem MT

There exists an absolute constant c, such that

$$\begin{aligned} \lambda \text {mes}\left( \left\{ \mathbf {T}^{*}\left( \mathbf {\beta } \right) f>\lambda \right\} \right) \le c\left\| \mathbf {\beta } \right\| \left\| f\right\| _{1} \end{aligned}$$

for all $f\in L_{1}\left( \mathbb {I}\right) ,\lambda >0,$ and $\mathbf {\beta }\in \mathbf {A}$.

The $\left( C,\alpha _{n}\right)$ means of the Walsh–Fourier series of the function f is given by

$$\begin{aligned} \sigma _{n}^{\alpha _{n}}(f,x)=\frac{1}{A_{n-1}^{\alpha _{n}}} \sum \limits _{j=1}^{n}A_{n-j}^{\alpha _{n}-1}S_{j}(f,x)=\frac{1}{ A_{n-1}^{\alpha _{n}}}\sum \limits _{j=0}^{n-1}A_{n-1-j}^{\alpha _{n}}\widehat{ f}\left( j\right) w_{j}\left( x\right) , \end{aligned}$$

where

$$\begin{aligned} A_{n}^{\alpha _{n}}:=\frac{(1+\alpha _{n})\dots (n+\alpha _{n})}{n!} \end{aligned}$$

for any $n\in \mathbb {N},\alpha _{n}\ne -1,-2,...$. It is known that [20]

$$\begin{aligned} A_{n}^{\alpha _{n}}=\sum \limits _{k=0}^{n}A_{k}^{\alpha _{n}-1},A_{n}^{\alpha _{n}-1}=\frac{\alpha _{n}}{\alpha _{n}+n}A_{n}^{\alpha _{n}}. \end{aligned}$$

(4)

The $\left( C,\alpha _{n}\right)$ kernel is defined by

$$\begin{aligned} K_{n}^{\alpha _{n}}=\frac{1}{A_{n-1}^{\alpha }}\sum \limits _{j=1}^{n}A_{n-j}^{\alpha _{n}-1}D_{j}=\frac{1}{A_{n-1}^{\alpha _{n}}} \sum \limits _{j=0}^{n-1}A_{n-j-1}^{\alpha _{n}}w_{j}. \end{aligned}$$

The idea of Cesàro means with variable parameters of numerical sequences is due to Kaplan [11] and the introduction of these $\left( C,\alpha _{n}\right)$ means of Fourier series is due to Akhobadze ( [1, 2]) who investigated the behavior of the $L_{1}$-norm convergence of $\sigma _{n}^{\alpha _{n}}\left( f\right) \rightarrow f$ for the trigonometric system.

The first result with respect to the a.e. convergence of the Walsh–Fejér means $\sigma _{n}^{\alpha _{n}}\left( f\right)$ for all integrable function f with constant sequence $\alpha _{n}=\alpha >0$ is due to Fine [4] (see also Weisz [17]). On the rate of convergence of Cesà ro means in this constant case, see the paper of Yano [19], Fridli [5]. Approximation properties of Cesàro means of negative order with constant sequence were investigated by the second author [8].

For $n:=\sum \nolimits _{i=0}^{\infty }\varepsilon _{i}\left( n\right) 2^{i}\left( \varepsilon _{i}\left( n\right) =0,1,i\in \mathbb {N}\right)$, set the two variable function

$$\begin{aligned} P\left( n,\alpha \right) :=\sum \nolimits _{i=0}^{\infty }\varepsilon _{i}\left( n\right) 2^{i\alpha _{n}}{ \ }\left( n\in \mathbb {N}\right) ,\alpha :=\left\{ \alpha _{n}:n\in \mathbb {N}\right\} . \end{aligned}$$

The function $P\left( n,\alpha \right)$ was introduced by Abu Joudeh and Gát in [10]. Also, set for sequence $\alpha :=\left\{ \alpha _{n}:n\in \mathbb {N}\right\}$ and positive reals K the subset of natural numbers

$$\begin{aligned} P_{K}\left( \alpha \right) :=\left\{ n\in \mathbb {N}:\frac{P\left( n,\alpha \right) }{n^{\alpha _{n}}}\le K\right\} . \end{aligned}$$

Under some conditions on $\left\{ \alpha _{n}:n\in \mathbb {N}\right\} ,$ Abu Joudeh and Gàt in [10] proved the almost everywhere convergence of the Cesàro $\left( C,\alpha _{n}\right)$ means of integrable functions. In particular, the following is proved.

Theorem JG

Suppose that $\alpha _{n}\in \left( 0,1\right)$. Let $f\in L_{1}\left( \mathbb {I}\right)$. Then, we have the almost everywhere convergence $\sigma _{n}^{\alpha _{n}}\left( f\right) \rightarrow f$ provided that $P_{K}\left( \alpha \right) \ni n\rightarrow \infty$.

The definition of the variation of an $n\in \mathbb {N}$ with binary coefficients

$$\begin{aligned} \left( \varepsilon _{k}\left( n\right) : k\in \mathbb {N}\right) \end{aligned}$$

was introduced in [13] by

$$\begin{aligned} V\left( n\right) :=\sum \limits _{i=0}^{\infty }\left| \varepsilon _{i}\left( n\right) -\varepsilon _{i+1}\left( n\right) \right| . \end{aligned}$$

In this paper, we define the weighted version of variation of an $n\in \mathbb {N}$ with binary coefficients $\left( \varepsilon _{k}\left( n\right) :k\in \mathbb {N}\right)$ by

$$\begin{aligned} V\left( n,\alpha \right) :=\sum \limits _{i=0}^{\infty }\left| \varepsilon _{i}\left( n\right) -\varepsilon _{i+1}\left( n\right) \right| 2^{i\alpha _{n}}{ \ }\left( n\in \mathbb {N}\right) . \end{aligned}$$

Set for sequence $\alpha :=\left\{ \alpha _{n}:n\in \mathbb {N}\right\}$ and positive reals K the subset of natural numbers

$$\begin{aligned} V_{K}\left( \alpha \right) :=\left\{ n\in \mathbb {N}:\frac{V\left( n,\alpha \right) }{n^{\alpha _{n}}}\le K<\infty \right\} . \end{aligned}$$

It is easy to see that $P_{K}\left( \alpha \right) \subsetneq V_{2K}\left( \alpha \right)$. On the other hand, if $\alpha _{n}\rightarrow 0$, then there exists K, such that $2^{n}-1\in V_{K}\left( \alpha \right)$ for all n, but there does not exist K, such that $2^{n}-1\in P_{K}\left( \alpha \right)$ for all n. In this paper, we are going to improve Theorem JG and to replace the condition $P_{K}\left( \alpha \right) \ni n\rightarrow \infty$ by the condition $V_{K}\left( \alpha \right) \ni n\rightarrow \infty$. In particular, the following will be proved.

Theorem 1.1

Suppose that $\alpha _{n}\in \left( 0,1\right)$. Let $f\in L_{1}\left( \mathbb {I}\right)$. Then, we have the almost everywhere convergence $\sigma _{n}^{\alpha _{n}}\left( f\right) \rightarrow f$ provided that $V_{K}\left( \alpha \right) \ni n\rightarrow \infty$.

From the proof of Theorem 1.1, we can obtain pointwise growth of Ces àro means with varying parameters of Walsh–Fourier series. The following is true.

Theorem 1.2

Let $f\in L_{1}\left( \mathbb {I}\right)$ and

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{V\left( n,\alpha \right) }{n^{\alpha _{n}}}=\infty . \end{aligned}$$

Then, we have the almost everywhere convergence

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{n^{\alpha _{n}}\sigma _{n}^{\alpha _{n}}\left( f,x\right) }{V\left( n,\alpha \right) }=0. \end{aligned}$$

Let $\lim \limits _{n\rightarrow \infty }\alpha _{n}=0$. We investigate two cases:

a) $\lim \limits _{n\rightarrow \infty }\left( \alpha _{n}\log n\right) >0$ and b) $\lim \limits _{n\rightarrow \infty }\left( \alpha _{n}\log n\right) =0$ . For case a), we have

$$\begin{aligned} \frac{V\left( n,\alpha \right) }{n^{\alpha _{n}}}\le \frac{c}{2^{|n|\alpha _{n}}}\sum \limits _{i=0}^{|n|}2^{i\alpha _{n}}\le c\alpha _{n}^{-1}, \end{aligned}$$

and for case b), we obtain

$$\begin{aligned} \frac{V\left( n,\alpha \right) }{n^{\alpha _{n}}}\le \frac{c}{2^{|n|\alpha _{n}}}\sum \limits _{i=0}^{|n|}2^{i\alpha _{n}}\le \frac{c|n|2^{|n|\alpha _{n}}}{2^{|n|\alpha _{n}}}\le c|n|. \end{aligned}$$

Hence, from Theorem 1.2, we get the following.

Corollary 1.3

Let $f\in L_{1}\left( \mathbb {I}\right)$ and

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\alpha _{n}=0. \end{aligned}$$

Then, we have the almost everywhere convergence:(a) If $\lim \limits _{n\rightarrow \infty }\left( \alpha _{n}\log n\right) >0$ , then $\lim \limits _{n\rightarrow \infty }\left( \alpha _{n}\sigma _{n}^{\alpha _{n}}\left( f,x\right) \right) =0;$ (b) If $\lim \limits _{n\rightarrow \infty }\left( \alpha _{n}\log n\right) =0$ , then $\lim \limits _{n\rightarrow \infty }\frac{\sigma _{n}^{\alpha _{n}}\left( f,x\right) }{\log n}=0.$

Theorem 1.4

Let $f\in L_{1}\left( \mathbb {I}\right)$ and $\alpha _{n}\in \left( 0,1\right)$. Then, the operator $\sigma _{n}^{\alpha _{n}}\left( f\right)$ is of weak type $\left( L_{1},L_{1}\right)$.

Theorem 1.4 imply

Corollary 1.5

Let $f\in L_{1}\left( \mathbb {I}\right)$ and $\alpha _{n}\in \left( 0,1\right)$. Then, $\sigma _{n}^{\alpha _{n}}\left( f\right) \rightarrow f$ in measure as $n\rightarrow \infty$.

Theorem 1.6

Let $f\in L_{1}\left( \mathbb {I}\right)$. Then, there exists a sequence $\mu _{j}\left( f\right)$, such that for each subsequence of natural numbers with $n_{j}\ge \mu _{j}\left( f\right)$, we have the a. e. relation

$$\begin{aligned} \sigma _{n_{j}}^{\alpha _{n_{j}}}\left( f\right) \rightarrow f\text {.} \end{aligned}$$

For the subsequence of the partial sums, we are going to prove the following.

Theorem 1.7

For each sequence of natural numbers $\nu _{j}\uparrow \infty$, there exists a function $f\in L_{1}\left( \mathbb {I}\right)$ and an another sequence of natural numbers with $N_{j}\ge \nu _{j}$ for which we have the everywhere divergence of $S_{N_{j}}\left( f\right)$.

The a. e. divergence of Cesàro means with varying parameters of Walsh–Fourier series was investigated by Tetunashvili [14]. In particular, the following is proved: Assume that $\left\{ \alpha _{n}\right\}$ is such that for a positive number $n_{0}$, we have

$$\begin{aligned} \alpha _{n}\le \frac{c}{\log _{2}n},0\le c<1,n>n_{0}\text {.} \end{aligned}$$

(5)

Then, there exists such a function f that the sequence $\sigma _{n}^{\alpha _{n}}\left( f\right)$ diverges everywhere unboundedly.

In this paper, we improve this theorem of Tetunashvili (5) in a way that we enlarge the set of sequences $(\alpha _n)$ for which we have divergence results of the Cesàro means with variable parameters. In particular, the following is true.

Theorem 1.8

Assume that $\left\{ \alpha _{n}\right\}$ is such that for some positive integer $n_{0}$, we have

$$\begin{aligned} \frac{c_{1}}{\log _{2}n}\le \alpha _{n}\le \frac{c_{0}\log _{2}\log _{2}n}{ \log _{2}n},0\le c_{0}<\frac{1}{2},n>n_{0}\text {.} \end{aligned}$$

Then, there exists a integrable function f that the sequence $\sigma _{n}^{\alpha _{n}}\left( f\right)$ diverges almost everywhere unboundedly.

The boundedness of maximal operators of subsequences of $\left( C,\alpha _{n}\right) -$ means of partial sums of Walsh–Fourier series from the Hardy space $H_{p}$ into the space $L_{p}$ is studied in [7]. In particular, the following is proved.

Theorem GG

Let $p>0$. Then, there exists a positive constant $c_{p}$, such that

$$\begin{aligned} \left\| \sup \limits _{N\in \mathbb {N}}\left| f*\left| K_{2^{N}}^{\alpha _{N}}\right| \right| \right\| _{p}\le c_{p}\left\| f\right\| _{H_{p}}{ \ \ }\left( f\in H_{p}\right) . \end{aligned}$$

Weisz [18] generalized Theorem GG for both the Cesàro and Riesz means by taking the supremum over all indices $n\in \mathbb {N}_{v}$. Here, $\mathbb {N}_{v}$ denotes the set of all $n=2^{n_{1}}+\cdots +2^{n_{v}}$ with a fixed parameter v. In particular, the following is proved.

Theorem W

(Weisz [18]) Let $p>0$. Then, there exists a positive constant $c_{p}$, such that

$$\begin{aligned} \left\| \sup \limits _{n\in P_{K}\left( \alpha \right) }\left| f*K_{n}^{\alpha _{n}}\right| \right\| _{p}\le c_{p}\left\| \left| f\right| \right\| _{H_{p}}{ \ \ }\left( \left| f\right| \in H_{p}\right) . \end{aligned}$$

2 Auxiliary results

We shall need the following.

Lemma 2.1

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

$$\begin{aligned}&c_{1}\left( 1+\alpha _{n}\right) \left( 2+\alpha _{n}\right) k^{\alpha _{n}}<A_{k}^{\alpha _{n}}<c_{2}\left( 1+\alpha _{n}\right) \left( 2+\alpha _{n}\right) k^{\alpha _{n}}, -2<\alpha _{n}<-1;\\&c_{1}\left( 1+\alpha _{n}\right) k^{\alpha _{n}}<A_{k}^{\alpha _{n}}<c_{2}\left( 1+\alpha _{n}\right) k^{\alpha _{n}}, -1<\alpha _{n}<0;\\&c_{1}\left( d\right) k^{\alpha _{n}}<A_{k}^{\alpha _{n}}<c_{2}\left( d\right) k^{\alpha _{n}}, 0<\alpha _{n}\le d\text {.} \end{aligned}$$

The proof can be found in the paper of Akhobadze [1].

Set

$$\begin{aligned} n^{\left( s\right) }:=\sum \limits _{j=s}^{\infty }\varepsilon _{j}\left( n\right) 2^{j}, n_{\left( s\right) }=n-n^{\left( s+1\right) }=\sum \limits _{j=0}^{s}\varepsilon _{j}\left( n\right) 2^{j}. \end{aligned}$$

Lemma 2.2

Let $\alpha _{n}\in \left( 0,1\right) , 1\le n\in \mathbb {N}$. Then, we have

$$\begin{aligned} K_{n}^{\alpha _{n}}&=\frac{1}{A_{n-1}^{\alpha _{n}}}\sum \limits _{s=0}^{|n|}\varepsilon _{s}\left( n\right) w_{n^{\left( s\right) }-1}\sum \limits _{j=1}^{2^{s}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}-2}jK_{j} \nonumber \\&\quad -\frac{1}{A_{n-1}^{\alpha _{n}}}\sum \limits _{s=0}^{|n|}\varepsilon _{s}\left( n\right) w_{n^{\left( s\right) }-1}A_{n_{\left( s\right) }-1}^{\alpha _{n}-1}2^{s}K_{2^{s}} \nonumber \\&\quad +\frac{1}{A_{n-1}^{\alpha _{n}}}\sum \limits _{s=0}^{|n|}\varepsilon _{s}\left( n\right) w_{n^{\left( s\right) }-1}A_{n_{\left( s\right) }-1}^{\alpha _{n}}D_{2^{s}} \nonumber \\&=:T_{n}^{\left( 1\right) }+T_{n}^{\left( 2\right) }+T_{n}^{\left( 3\right) }. \end{aligned}$$

(6)

Proof of Lemma 2.2

We can write

$$\begin{aligned} A_{n-1}^{\alpha _{n}}K_{n}^{\alpha _{n}}= & {} \sum \limits _{j=0}^{n-1}A_{n-j-1}^{\alpha _{n}}w_{j}=\sum \limits _{s=0}^{|n|}\varepsilon _{s}\left( n\right) \sum \limits _{j=n^{\left( s+1\right) }}^{n^{\left( s\right) }-1}A_{n-j-1}^{\alpha _{n}}w_{j} \\= & {} \sum \limits _{s=0}^{|n|}\varepsilon _{s}\left( n\right) \sum \limits _{j=0}^{2^{s}-1}A_{n_{\left( s\right) }-j-1}^{\alpha _{n}}w_{j+n^{\left( s+1\right) }} \\= & {} \sum \limits _{s=0}^{|n|}\varepsilon _{s}\left( n\right) w_{n^{\left( s+1\right) }}\sum \limits _{j=0}^{2^{s}-1}A_{n_{\left( s\right) }-j-1}^{\alpha _{n}}w_{j}. \end{aligned}$$

Since

$$\begin{aligned} n_{\left( s\right) }-j-1=n_{\left( s-1\right) }+2^{s}-1-j, \varepsilon _{s}\left( n\right) =1, \end{aligned}$$

(otherwise nothing to be investigated here) and

$$\begin{aligned} 2^{s}-1-j=\left( 2^{s}-1\right) \oplus j \end{aligned}$$

from (2), we obtain

$$\begin{aligned} A_{n-1}^{\alpha _{n}}K_{n}^{\alpha _{n}}= & {} \sum \limits _{s=0}^{|n|}\varepsilon _{s}\left( n\right) w_{n^{\left( s+1\right) }}\sum \limits _{j=0}^{2^{s}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}}w_{\left( 2^{s}-1\right) \oplus j} \nonumber \\= & {} \sum \limits _{s=0}^{|n|}\varepsilon _{s}\left( n\right) w_{n^{\left( s\right) }-1}\sum \limits _{j=0}^{2^{s}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}}w_{j}. \end{aligned}$$

(7)

Applying Abel’s transformation (twice), we get

$$\begin{aligned}&\sum \limits _{j=0}^{2^{s}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}}w_{j} =\sum \limits _{j=1}^{2^{s}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}-2}jK_{j}-A_{n_{\left( s\right) }-1}^{\alpha _{n}-1}2^{s}K_{2^{s}}+A_{n_{\left( s\right) }-1}^{\alpha _{n}}D_{2^{s}}. \end{aligned}$$

Hence, from (7), we conclude (6). $\square$

From (4), we can write

$$\begin{aligned} \left| T_{n}^{\left( 1\right) }\right| \le \frac{2}{A_{n-1}^{\alpha _{n}}}\sum \limits _{s=0}^{|n|}\varepsilon _{s}\left( n\right) \sum \limits _{j=1}^{2^{s}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}-1}\left| K_{j}\right| :=\widetilde{T}_{n}^{\left( 1\right) }. \end{aligned}$$

Lemma 2.3

Let $\alpha _{n}\in \left( 0,1\right) ,n\in \mathbb {N}$ and $f\in L_{1}\left( \mathbb {I}\right)$, such that $\mathop {supp}\left( f\right) \subset I_{N}\left( u^{\prime }\right) ,\int \limits _{I_{N}\left( u^{\prime }\right) }f=0$ for some dyadic interval $I_{N}\left( u^{\prime }\right)$. Then, we have

$$\begin{aligned} \int \limits _{\overline{I}_{N}\left( u^{\prime }\right) }\sup \limits _{n\in \mathbb {N}}\left| f*\widetilde{T}_{n}^{\left( 1\right) }\right| \le c\left\| f\right\| _{1}. \end{aligned}$$

Proof of Lemma 2.3

Let $n\le 2^{N}$. From the condition of the lemma, it is easy to see that $f*\widetilde{T}_{n}^{\left( 1\right) }=0$. Hence, we can suppose that $n>2^{N}$. Without lost of generality, we may assume that $u^{\prime }=0$. It is easy to see that

$$\begin{aligned} f*\left( \frac{\widetilde{T}_{n}^{\left( 1\right) }}{2}\right)= & {} f*\left( \frac{1}{A_{n-1}^{\alpha _{n}}}\sum \limits _{s=0}^{|n|}\varepsilon _{s}\left( n\right) \sum \limits _{j=1}^{2^{s}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}-1}\left| K_{j}\right| \right) \\= & {} \int \limits _{I_{N}}f\left( u\right) \frac{1}{A_{n-1}^{\alpha _{n}}} \sum \limits _{s=0}^{|n|}\varepsilon _{s}\left( n\right) \sum \limits _{j=1}^{2^{s}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}-1}\left| K_{j}\left( x\dotplus u\right) \right| du \\= & {} \int \limits _{I_{N}}f\left( u\right) \frac{1}{A_{n-1}^{\alpha _{n}}} \sum \limits _{s=0}^{N}\varepsilon _{s}\left( n\right) \sum \limits _{j=1}^{2^{s}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}-1}\left| K_{j}\left( x\dotplus u\right) \right| du \\&+\int \limits _{I_{N}}f\left( u\right) \frac{1}{A_{n-1}^{\alpha _{n}}} \sum \limits _{s=N+1}^{|n|}\varepsilon _{s}\left( n\right) \sum \limits _{j=1}^{2^{N}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}-1}\left| K_{j}\left( x\dotplus u\right) \right| du \\&+\int \limits _{I_{N}}f\left( u\right) \frac{1}{A_{n-1}^{\alpha _{n}}} \sum \limits _{s=N+1}^{|n|}\varepsilon _{s}\left( n\right) \sum \limits _{j=2^{N}}^{2^{s}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}-1}\left| K_{j}\left( x\dotplus u\right) \right| du \\= & {} \frac{1}{A_{n-1}^{\alpha _{n}}}\sum \limits _{s=0}^{N}\varepsilon _{s}\left( n\right) \sum \limits _{j=1}^{2^{s}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}-1}\left| K_{j}\left( x\right) \right| \int \limits _{I_{N}}f\left( u\right) du\\&+\frac{1}{A_{n-1}^{\alpha _{n}}}\sum \limits _{s=N+1}^{|n|}\varepsilon _{s}\left( n\right) \sum \limits _{j=1}^{2^{N}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}-1}\left| K_{j}\left( x\right) \right| \int \limits _{I_{N}}f\left( u\right) du \\&+\int \limits _{I_{N}}f\left( u\right) \frac{1}{A_{n-1}^{\alpha _{n}}} \sum \limits _{s=N+1}^{|n|}\varepsilon _{s}\left( n\right) \sum \limits _{j=2^{N}}^{2^{s}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}-1}\left| K_{j}\left( x\dotplus u\right) \right| du \\= & {} \int \limits _{I_{N}}f\left( u\right) \frac{1}{A_{n-1}^{\alpha _{n}}} \sum \limits _{s=N+1}^{|n|}\varepsilon _{s}\left( n\right) \sum \limits _{j=2^{N}}^{2^{s}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}-1}\left| K_{j}\left( x\dotplus u\right) \right| du. \end{aligned}$$

It is easy to see from (4) and Lemma 2.1 that

$$\begin{aligned}&\frac{1}{A_{n-1}^{\alpha _{n}}}\sum \limits _{s=1}^{\left| n\right| }\varepsilon _{s}\left( n\right) \sum \limits _{j=1}^{2^{s}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}-1} \nonumber \\= & {} \frac{1}{A_{n-1}^{\alpha _{n}}}\sum \limits _{s=1}^{\left| n\right| }\varepsilon _{s}\left( n\right) \sum \limits _{j=n_{\left( s-1\right) }+1}^{n_{\left( s\right) }-1}A_{j}^{\alpha _{n}-1} \nonumber \\= & {} \frac{1}{A_{n-1}^{\alpha _{n}}}\sum \limits _{s=1}^{\left| n\right| }\varepsilon _{s}\left( n\right) \left( A_{n_{\left( s\right) }-1}^{\alpha _{n}}-A_{n_{\left( s-1\right) }}^{\alpha _{n}}\right) \nonumber \\\le & {} \frac{1}{A_{n-1}^{\alpha _{n}}}\sum \limits _{s=1}^{\left| n\right| }\left( A_{n_{\left( s\right) }}^{\alpha _{n}}-A_{n_{\left( s-1\right) }}^{\alpha _{n}}\right) \nonumber \\< & {} \frac{A_{n\left( \left| n\right| \right) }}{A_{n-1}^{\alpha _{n}}} \le c. \end{aligned}$$

(8)

Set

$$\begin{aligned} K_{2^{N}}^{*}:=\sup \limits _{n\ge 2^{N}}\left| K_{n}\right| . \end{aligned}$$

It is proved in [6] that

$$\begin{aligned} \int \limits _{\overline{I}_{N}}K_{2^{N}}^{*}\le c<\infty ,N\in \mathbb {N} . \end{aligned}$$

Then, from (8), we have

$$\begin{aligned}&\int \limits _{\overline{I}_{N}}\sup \limits _{n\ge 2^{N}}\frac{1}{ A_{n-1}^{\alpha _{n}}}\sum \limits _{s=N+1}^{|n|}\sum \limits _{j=2^{N}}^{2^{s}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}-1}\left| K_{j}\left( u\right) \right| du \\&\quad \le \sup \limits _{n\ge 2^{N}}\frac{1}{A_{n-1}^{\alpha _{n}}} \sum \limits _{s=N+1}^{|n|}\sum \limits _{j=2^{N}}^{2^{s}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}-1}\int \limits _{\overline{I}_{N}}K_{2^{N}}^{*} \\&\quad \le c\sup \limits _{n\ge 2^{N}}\frac{1}{A_{n-1}^{\alpha _{n}}} \sum \limits _{s=N+1}^{|n|}\sum \limits _{j=2^{N}}^{2^{s}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}-1}\le c<\infty . \end{aligned}$$

Consequently

$$\begin{aligned}&\int \limits _{\overline{I}_{N}}\sup \limits _{n\in \mathbb {N}}\left| f*\widetilde{T}_{n}^{\left( 1\right) }\right| \\&\quad = \int \limits _{\overline{I}_{N}}\sup \limits _{n\ge 2^{N}}\left| \int \limits _{I_{N}}f\left( u\right) \frac{1}{A_{n-1}^{\alpha _{n}}} \sum \limits _{s=N+1}^{|n|}\varepsilon _{s}\left( n\right) \sum \limits _{j=2^{N}}^{2^{s}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}-1}\left| K_{j}\left( x\dotplus u\right) \right| du\right| \\&\quad \le \int \limits _{\overline{I}_{N}}\left( \int \limits _{I_{N}}\left| f\left( u\right) \right| \sup \limits _{n\ge 2^{N}}\frac{1}{ A_{n-1}^{\alpha _{n}}}\sum \limits _{s=N+1}^{|n|}\sum \limits _{j=2^{N}}^{2^{s}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}-1}\left| K_{j}\left( x\dotplus u\right) \right| du\right) dx \\&\quad =\int \limits _{I_{N}}\left| f\left( u\right) \right| \left( \int \limits _{\overline{I}_{N}}\sup \limits _{n\ge 2^{N}}\frac{1}{ A_{n-1}^{\alpha _{n}}}\sum \limits _{s=N+1}^{|n|}\sum \limits _{j=2^{N}}^{2^{s}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}-1}\left| K_{j}\left( x\dotplus u\right) \right| dx\right) du \\&\quad \le c\left\| f\right\| _{1}\text {.} \end{aligned}$$

This completes the proof of Lemma 2.3. $\square$

Lemma 2.4

The operator $\sup \limits _{n\in \mathbb {N}}\left| f*\widetilde{T}_{n}^{\left( 1\right) }\right|$ is of type $\left( L_{\infty },L_{\infty }\right)$.

Proof of Lemma 2.4

Since (see [13]) $\sup \limits _{n}\left\| K_{n}\right\| _1 <2$ from (8) (or even see [15] $\sup \limits _{n}\left\| K_{n}\right\| _1 \le 17/15$), we have

$$\begin{aligned} \sup \limits _{n\in \mathbb {N}}\left\| \widetilde{T}_{n}^{\left( 1\right) }\right\| _{1}\le \sup \limits _{n\in \mathbb {N}}\frac{1}{A_{n-1}^{\alpha _{n}}}\sum \limits _{s=0}^{|n|}\varepsilon _{s}\left( n\right) \sum \limits _{j=1}^{2^{s}-1}A_{n_{\left( s-1\right) }+j}^{\alpha _{n}-1}\le c<\infty , \end{aligned}$$

which implies the boundedness of operator $\sup \limits _{n\in \mathbb {N} }\left| f*\widetilde{T}_{n}^{\left( 1\right) }\right|$ from the space $L_{\infty }$ to the space $L_{\infty }$. $\square$

Combine Lemmas 2.3 and 2.4 to have the following.

Lemma 2.5

The operator $\sup \limits _{n\in \mathbb {N}}\left| f*\widetilde{T}_{n}^{\left( 1\right) }\right|$ is of weak type $\left( L_{1},L_{1}\right)$.

Since

$$\begin{aligned} \left| f*T_{n}^{\left( 1\right) }\right| \le \left| f\right| *\widetilde{T}_{n}^{\left( 1\right) }, \end{aligned}$$

from Lemma 2.5, we obtain

Lemma 2.6

The operator $\sup \limits _{n\in \mathbb {N}}\left| f*T_{n}^{\left( 1\right) }\right|$ is of weak type $\left( L_{1},L_{1}\right)$.

Analogously, we can prove

Lemma 2.7

The operator $\sup \limits _{n\in \mathbb {N}}\left| f*T_{n}^{\left( 2\right) }\right|$ is of weak type $\left( L_{1},L_{1}\right)$.

3 Proofs of main results

Proof of Theorem 1.1

We have

$$\begin{aligned} w_{n}T_{n}^{\left( 3\right) }=\frac{1}{A_{n-1}^{\alpha _{n}}} \sum \limits _{s=0}^{|n|}\varepsilon _{s}\left( n\right) w_{n}w_{n^{\left( s\right) }-1}A_{n_{\left( s\right) }-1}^{\alpha _{n}}D_{2^{s}}. \end{aligned}$$

From (2), we get ($\varepsilon _{s}\left( n\right) =1$; otherwise, there is nothing to be discussed here)

$$\begin{aligned} w_{n}w_{n^{\left( s\right) }-1}= & {} w_{n}w_{n^{\left( s+1\right) }+2^{s}-1}=w_{n}w_{n^{\left( s+1\right) }}w_{2^{s}-1} \\= & {} w_{n\oplus n^{\left( s+1\right) }}w_{2^{s}-1}=w_{n_{\left( s\right) }}w_{2^{s}-1} \\= & {} w_{2^{s}}w_{n_{\left( s-1\right) }}w_{2^{s}-1}=w_{2^{s}}w_{n_{\left( s-1\right) \oplus \left( 2^{s}-1\right) }}. \end{aligned}$$

Since $n_{\left( s-1\right) \oplus \left( 2^{s}-1\right) }<2^{s}$ from (3), we have

$$\begin{aligned} D_{2^{s}}w_{n_{\left( s-1\right) \oplus \left( 2^{s}-1\right) }}=D_{2^{s}}. \end{aligned}$$

Consequently

$$\begin{aligned} w_{n}T_{n}^{\left( 3\right) }= & {} \frac{1}{A_{n-1}^{\alpha _{n}}} \sum \limits _{s=0}^{|n|}\varepsilon _{s}\left( n\right) A_{n_{\left( s\right) }-1}^{\alpha _{n}}w_{2^{s}}D_{2^{s}} \nonumber \\= & {} \frac{1}{A_{n-1}^{\alpha _{n}}}\sum \limits _{s=0}^{|n|}\varepsilon _{s}\left( n\right) A_{n_{\left( s\right) }-1}^{\alpha _{n}}\left( D_{2^{s+1}}-D_{2^{s}}\right) \nonumber \\= & {} \frac{1}{A_{n-1}^{\alpha _{n}}}\sum \limits _{s=1}^{|n|} \left( \varepsilon _{s-1}\left( n\right) -\varepsilon _{s}\left( n\right) \right) A_{n_{\left( s-1\right) }-1}^{\alpha _{n}}D_{2^{s}} \nonumber \\&+\frac{1}{A_{n-1}^{\alpha _{n}}}\sum \limits _{s=1}^{|n|}\varepsilon _{s}\left( n\right) \left( A_{n_{\left( s-1\right) }-1}^{\alpha _{n}}-A_{n_{\left( s\right) }-1}^{\alpha _{n}}\right) D_{2^{s}} \nonumber \\&+\frac{1}{A_{n-1}^{\alpha _{n}}}\varepsilon _{|n|}\left( n\right) A_{n_{\left( |n|\right) }-1}^{\alpha _{n}}D_{2^{|n|+1}} \nonumber \\&-\frac{1}{A_{n-1}^{\alpha _{n}}}\varepsilon _{0}\left( n\right) A_{n_{\left( 0\right) }-1}^{\alpha _{n}}D_{1} \nonumber \\&=: T_{n}^{\left( 31\right) }+T_{n}^{\left( 32\right) }+T_{n}^{\left( 33\right) }+T_{n}^{\left( 34\right) }. \end{aligned}$$

(9)

From the condition of Theorem 1.1, we can write

$$\begin{aligned}&\sup \limits _{n\in \mathbb {N}}\left( \left| f\right| *\left| T_{n}^{\left( 3\right) }\right| \right) \nonumber \\&\quad =\sup \limits _{n\in \mathbb {N}}\left( \left| f\right| *\left| w_{n}T_{n}^{\left( 3\right) }\right| \right) \nonumber \\&\quad \le \sup \limits _{n\in \mathbb {N}}\left( \left| f\right| *\left| w_{n}T_{n}^{\left( 31\right) }\right| \right) +\sup \limits _{n\in \mathbb {N}}\left( \left| f\right| *\left| w_{n}T_{n}^{\left( 32\right) }\right| \right) \nonumber \\&\qquad +\sup \limits _{n\in \mathbb {N}}\left( \left| f\right| *\left| w_{n}T_{n}^{\left( 33\right) }\right| \right) +\sup \limits _{n\in \mathbb {N}}\left( \left| f\right| *\left| w_{n}T_{n}^{\left( 34\right) }\right| \right) \nonumber \\&\quad \le cE^{*}\left( x,\left| f\right| \right) \frac{1}{n^{\alpha _{n}}}\sum \limits _{s=1}^{|n|}\left| \varepsilon _{s-1}\left( n\right) -\varepsilon _{s}\left( n\right) \right| 2^{s\alpha _{n}} \nonumber \\&\qquad +cE^{*}\left( x,\left| f\right| \right) \frac{1}{ A_{n-1}^{\alpha _{n}}}\sum \limits _{s=1}^{|n|}\left( A_{n_{\left( s-1\right) }-1}^{\alpha _{n}}-A_{n_{\left( s\right) }-1}^{\alpha _{n}}\right) \nonumber \\&\qquad +cE^{*}\left( x,\left| f\right| \right) \nonumber \\&\quad \le c_{K}E^{*}\left( x,\left| f\right| \right) . \end{aligned}$$

(10)

Since the operator $E^{*}\left( x,\left| f\right| \right)$ is of weak type $\left( L_{1},L_{1}\right)$, we obtain that

$$\begin{aligned} \left\| \sup \limits _{n\in V_{K}\left( \alpha \right) }\left( \left| f\right| *\left| T_{n}^{\left( 3\right) }\right| \right) \right\| _{weak-L_{1}}\le c_{K}\left\| f\right\| _{1}\text {.} \end{aligned}$$

(11)

Combining Lemmas 2.6, 2.7, estimation (11) from (6) we conclude that

$$\begin{aligned} \left\| \sup \limits _{n\in V_{K}\left( \alpha \right) }\left| \sigma _{n}^{\alpha _{n}}f\right| \right\| _{weak-L_{1}}\le c_K\left\| f\right\| _{1}\text {.} \end{aligned}$$

(12)

Using the standard argument of Marcinkiewicz and Zygmund [12] from the estimation (12), we obtain the validity of Theorem  1.1. $\square$

Proof of Theorem 1.2

From (6), we have

$$\begin{aligned} \frac{n^{\alpha _{n}}\left( f*K_{n}^{\alpha _{n}}\right) }{V\left( n,\alpha \right) }=\frac{n^{\alpha _{n}}\left( f*T_{n}^{\left( 1\right) }\right) }{V\left( n,\alpha \right) }+\frac{n^{\alpha _{n}}\left( f*T_{n}^{\left( 2\right) }\right) }{V\left( n,\alpha \right) }+\frac{n^{\alpha _{n}}\left( f*T_{n}^{\left( 3\right) }\right) }{V\left( n,\alpha \right) }. \end{aligned}$$

(13)

Lemmas 2.6 and 2.7 imply that

$$\begin{aligned} \sup \limits _{n}\left| f*T_{n}^{\left( l\right) }\right| <\infty \ \ \text {a. e. for }f\in L_{1}\left( \mathbb {I}\right) ,l=1,2. \end{aligned}$$

Hence

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{n^{\alpha _{n}}\left( f*T_{n}^{\left( l\right) }\right) }{V\left( n,\alpha \right) }=0\text { a. e. }l=1,2. \end{aligned}$$

(14)

Using estimation (10), we have

$$\begin{aligned} \sup \limits _{n}\frac{n^{\alpha _{n}}\left( f*T_{n}^{\left( 3\right) }\right) }{V\left( n,\alpha \right) }\le cE^{*}\left( x,\left| f\right| \right) . \end{aligned}$$

Since the operator $E^{*}\left( x,\left| f\right| \right)$ is of weak type $\left( L_{1},L_{1}\right)$, we obtain that the maximal operator

$$\begin{aligned} \sup \limits _{n}\frac{n^{\alpha _{n}}\left( f*T_{n}^{\left( 3\right) }\right) }{V\left( n,\alpha \right) } \end{aligned}$$

is of weak type $\left( L_{1},L_{1}\right)$. It is clear that

$$\begin{aligned} \frac{n^{\alpha _{n}}\left( W*T_{n}^{\left( 3\right) }\right) }{V\left( n,\alpha \right) }\rightarrow 0\quad \text { as }n\rightarrow \infty \end{aligned}$$

for every Walsh polynomial W. By the well-known density argument, we conclude that

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{n^{\alpha _{n}}\left( f*T_{n}^{\left( 3\right) }\right) }{V\left( n,\alpha \right) }=0\quad \text { a. e.} \end{aligned}$$

(15)

Combining (13)–(15), we conclude the proof of Theorem 1.2. $\square$

Proof of Theorem 1.4

From (6), we have

$$\begin{aligned} \sigma _{n}^{\alpha _{n}}\left( f\right) =f*K_{n}^{\alpha _{n}}=f*T_{n}^{\left( 1\right) }+f*T_{n}^{\left( 2\right) }+f*T_{n}^{\left( 3\right) }. \end{aligned}$$

(16)

Applying Lemmas 2.6 and 2.7, we conclude that the operators $f*T_{n}^{\left( l\right) },l=1,2$ are of weak type $\left( L_{1},L_{1}\right)$. Now, we consider the operator $f*T_{n}^{\left( 3\right) }$. From (9), we have

$$\begin{aligned}&f*T_{n}^{\left( 3\right) } \nonumber \\&\quad =w_{n}\left( \left( fw_{n}\right) *w_{n}T_{n}^{\left( 3\right) }\right) \nonumber \\&\quad =\frac{w_{n}}{A_{n-1}^{\alpha _{n}}}\sum \limits _{s=0}^{\infty }\varepsilon _{s}\left( n\right) A_{n_{\left( s\right) -1}}^{\alpha _{n}}\left( \left( fw_{n}\right) *\left( D_{2^{s+1}}-D_{2^{s}}\right) \right) \nonumber \\&\quad = w_{n}\sum \limits _{s=0}^{\infty }\frac{\varepsilon _{s}\left( n\right) A_{n_{\left( s\right) -1}}^{\alpha _{n}}}{A_{n-1}^{\alpha _{n}}}\Delta _{s}\left( fw_{n}\right) \nonumber \\&\quad = \mathbf {T}\left( \mathbf {\beta }\right) \left( fw_{n}\right) , \end{aligned}$$

(17)

where

$$\begin{aligned} \mathbf {\beta :=}\left( \frac{\varepsilon _{0}\left( n\right) A_{n_{\left( 0\right) -1}}^{\alpha _{n}}}{A_{n-1}^{\alpha _{n}}},...,\frac{\varepsilon _{|n|}\left( n\right) A_{n_{\left( |n|\right) -1}}^{\alpha _{n}}}{ A_{n-1}^{\alpha _{n}}},0,...\right) . \end{aligned}$$

Since $\left\| \mathbf {\beta }\right\| \le 1$ from Theorem MT, we get that the operator $\left| \left( fw_{n}\right) *w_{n}T_{n}^{\left( 3\right) }\right|$ is of weak type $\left( L_{1},L_{1}\right)$. Consequently

$$\begin{aligned} \left\| f*T_{n}^{\left( 3\right) }\right\| _{weak-L_{1}\left( \mathbb {I}\right) }\le c\left\| f\right\| _{1}. \end{aligned}$$

(18)

From (16), we complete the proof of Theorem 1.4. $\square$

Proof of Theorem 1.7

Basically, we use the method of Schipp (see [13], Ch. 4, Theorem 12) with some necessary modifications. For natural numbers n, k, set

$$\begin{aligned}&i_{n}:=\sum _{k=1}^{\lfloor n/2\rfloor }2^{2k-1}<2^{n},\quad g_{n}:= \mathop {sgn}D_{i_{n}}, \\&R_{k}^{(n)}:=r_{2^{n}+k}\tau _{k/2^{n}}g_{n},\quad Q_{n}:=\prod _{k=0}^{2^{n}-1}\left( 1+R_{k}^{(n)}\right) . \end{aligned}$$

Then, in the sequel, we prove

$$\begin{aligned} S_{2^{2^{n}+k}+i_{n}}\left( Q_{n},x\right) -S_{2^{2^{n}+k}}\left( Q_{n},x\right) =r_{2^{n}+k}(x)g_{n}(x\dotplus k/2^{n}). \end{aligned}$$

(19)

Since $Q_{n}$ is the sum of the product of terms $R_{k}^{(n)}$, then we have to check $R:=R_{l_{1}}^{(n)}\dots R_{l_{s}}^{(n)}$ for $l_{1}<\dots <l_{s}$ and let the empty product be 1. If the case is the latter, i.e., $R=1$, then the left-hand side of (19) is zero. Therefore, suppose that we are checking not the empty product. Then

$$\begin{aligned} R=R_{l_{1}}^{(n)}\dots R_{l_{s}}^{(n)}=r_{2^{n}+l_{1}}\dots r_{2^{n}+l_{s}}\left( \tau _{l_{1}2^{-n}}g_{n}\dots \tau _{l_{s}2^{-n}}g_{n}\right) =:r_{2^{n}+l_{1}}\dots r_{2^{n}+l_{s}}h, \end{aligned}$$

where function h is $\mathcal {A}_{n}$ measurable. Therefore, in the case of $k<l_{s}$, we have

$$\begin{aligned} S_{2^{2^{n}+k}}\left( R\right) =0. \end{aligned}$$

Besides, in the case of $k>l_{s}$, we have

$$\begin{aligned} S_{2^{2^{n}+k}}\left( R\right) =R. \end{aligned}$$

That is, in both cases, the left-hand side of (19) is

$$\begin{aligned} S_{2^{2^{n}+k}+i_{n}}\left( R,x\right) -S_{2^{2^{n}+k}}\left( R,x\right) =r_{2^{n}+k}(x)S_{i_{n}}(Rr_{2^{n}+k},x), \end{aligned}$$

(20)

which can be different from zero only in the case when $s=1$ and $l_{s}=k$. In this situation, it is exactly

$$\begin{aligned} r_{2^{n}+k}S_{i_{n}}(R_{k}^{(n)}r_{2^{n}+k})=r_{2^{n}+k}S_{i_{n}}(\tau _{k2^{-n}}g_{n})=r_{2^{n}+k}\tau _{k2^{-n}}g_{n}=R_{k}^{(n)}. \end{aligned}$$

(21)

Just add a few details to equality (21): Let $a=2\lfloor n/2\rfloor -1$ . Then, $i_{n}=2^{1}+2^{3}+\dots 2^{a}$. It is easy to have that

$$\begin{aligned} g_{n}(x)= {\left\{ \begin{array}{ll} 1, &{} \text{ if } x\in I_{a}, \\ r_{a}(x), &{} \text{ if } x\in I_{a-2}{\setminus } I_{a}, \\ r_{a}(x)r_{a-2}(x), &{} \text{ if } x\in I_{a-4}{\setminus } I_{a-2}, \\ \dots , &{} \\ r_{a}(x)r_{a-2}(x)\cdots r_{3}(x), &{} \text{ if } x\in I_{1}{\setminus } I_{3}, \\ 0, &{} \text{ if } x\in \mathbb {I}{\setminus } I_{1}. \end{array}\right. } \end{aligned}$$

Let $e_{i}=1/2^{i+1}$. It gives that $g_{n}$ is the sum of functions $g_{n,\epsilon }$

$$\begin{aligned} \begin{aligned} g_{n,\epsilon }(x)&:=\frac{1}{2^{a}}D_{2^{a}}(x)+\frac{1}{2^{a}} D_{2^{a}}(x+\epsilon _{a-2}e_{a-2}+\epsilon _{a-1}e_{a-1})r_{a}(x) \\&\quad +\frac{1}{2^{a-2}}D_{2^{a-2}}(x+\epsilon _{a-4}e_{a-4}+\epsilon _{a-3}e_{a-3})r_{a}(x)r_{a-2}(x) \\&\quad +\dots +\frac{1}{2^{3}}D_{2^{3}}(x+\epsilon _{1}e_{1}+\epsilon _{2}e_{2})r_{a}(x)r_{a-2}(x)\cdots r_{3}(x), \end{aligned} \end{aligned}$$

where each $\epsilon _{i}$ is either 0 or 1, but $\epsilon _{a-2}+\epsilon _{a-1},\epsilon _{a-4}+\epsilon _{a-3},\dots ,\epsilon _{1}+\epsilon _{2}\not =0$ and we do the summing with respect to $\epsilon$. That is, $g_{n}=\sum _{\epsilon }g_{n,\epsilon }$. Then, for any of the addends of type $g_{n,\epsilon }$, we have

$$\begin{aligned} \begin{aligned}&S_{i_{n}}\left( r_{a}\cdots r_{a-2i}D_{2^{a-2i}}(\cdot +\epsilon _{a-2i-2}e_{a-2i-2}+\epsilon _{a-2i-1}e_{a-2i-1})\right) \\&\quad =r_{a}\cdots r_{a-2i}D_{2^{a-2i}}(\cdot +\epsilon _{a-2i-2}e_{a-2i-2}+\epsilon _{a-2i-1}e_{a-2i-1}), \end{aligned} \end{aligned}$$

and consequently, $S_{i_{n}}g_{n}=g_{n}$. In other words, (19) is proved. Let $n_{m}\in \mathbb {N},x\in \mathbb {I}$ be arbitrary and suppose that $n_{m}$ is a cube and $n_{m}\ge \nu _{2m+1}$. Then, there exists one $k\in \left\{ 0,1,...,2^{n_{m}}-1\right\}$, such that

$$\begin{aligned} x\dotplus k2^{-n_{m}}\in I_{n_{m}}. \end{aligned}$$

(22)

Set

$$\begin{aligned} N_{2m}:=2^{2^{n_{m}}+k},N_{2m+1}:=2^{2^{n_{m}}+k}+i_{n_{m}},m=1,2,.... \end{aligned}$$

It is easy to see that

$$\begin{aligned} N_{2m}\ge 2^{2^{n_{m}}}>n_{m}\ge v_{2m+1}>v_{2m} \end{aligned}$$

and

$$\begin{aligned} N_{2m+1}\ge 2^{2^{n_{m}}}>n_{m}\ge v_{2m+1}. \end{aligned}$$

Hence

$$\begin{aligned} N_{j}\ge v_{j},j=1,2,.... \end{aligned}$$

Let

$$\begin{aligned} f\left( x\right) :=\sum \limits _{m=1}^{\infty }\frac{Q_{n_{m}}\left( x\right) }{\root 3 \of {n_{m}^{2}}}. \end{aligned}$$

Since $\Vert Q_{n}\Vert _{1}=1$ (see [13, ch. 4, Theorem 12]), then $f\in L_{1}\left( \mathbb {I}\right)$. From the definition of function $Q_{n}$, it follows for its spectrum:

$$\begin{aligned} \text {sp}(Q_{n_{j}})\subset \left[ 0,2^{2^{n_{j}+1}}\right) , \end{aligned}$$

and since

$$\begin{aligned} N_{2m}\ge 2^{2^{n_{m}}}\ge 2^{2^{n_{j}+1}}\left( j<m\right) , \end{aligned}$$

we obtain

$$\begin{aligned} S_{N_{2m+1}}\left( Q_{n_{j}},x\right) -S_{N_{2m}}\left( Q_{n_{j}},x\right) =0,j<m. \end{aligned}$$

(23)

On the other hand, check the same difference of partial sums for $Q_{n_{j}}$ ($j>m$). Let again $R:=R_{l_{1}}^{(n_{j})}\dots R_{l_{s}}^{(n_{j})}$ be different from the empty product. Then

$$\begin{aligned}&\left| S_{N_{2m+1}}\left( Q_{n_{j}},x\right) -S_{N_{2m}}\left( Q_{n_{j}},x\right) \right| \nonumber \\&\quad =\left| S_{i_{n_{m}}}\left( R\cdot r_{2^{n_{m}}+k}\right) \right| \nonumber \\&\quad = \left| S_{i_{n_{m}}}\left( r_{2^{n_{j}}+l_{1}}\cdots r_{2^{n_{j}}+l_{s}}\cdot r_{2^{n_{m}}+k}\cdot h\right) \right| =0, \end{aligned}$$

(24)

because the function h is $\mathcal {A}_{n_{j}}$ measurable.

From (20), (21), (22), (23), and (24), we obtain

$$\begin{aligned}&\left| S_{N_{2m+1}}\left( f,x\right) -S_{N_{2m}}\left( f,x\right) \right| \\&\quad =\frac{1}{\root 3 \of {n_{m}^{2}}}\left| S_{N_{2m+1}}\left( Q_{n_{m}},x\right) -S_{N_{2m}}\left( Q_{n_{m}},x\right) \right| \\&\quad =\frac{1}{\root 3 \of {n_{m}^{2}}}\left| S_{i_{n_{m}}}(\tau _{k2^{-n_{m}}}g_{n_{m}},x)\right| \\&\quad = \frac{1}{\root 3 \of {n_{m}^{2}}}\left| \left( \tau _{k2^{-n_{m}}}g_{n_{m}}\right) *D_{i_{n_{m}}}\left( x\right) \right| \\&\quad = \frac{1}{\root 3 \of {n_{m}^{2}}}\left| g_{n_{m}}*\left( \tau _{k2^{-n_{m}}}D_{i_{n_{m}}}\left( x\right) \right) \right| \\&\quad = \frac{1}{\root 3 \of {n_{m}^{2}}}\left| g_{n_{m}}*D_{i_{n_{m}}}\left( 0\right) \right| \\&\quad = \frac{1}{\root 3 \of {n_{m}^{2}}}\left\| D_{i_{n_{m}}}\right\| _{1}\ge cn_{m}^{1/3}. \end{aligned}$$

It means that for every $x\in \mathbb {I}$, we have

$$\begin{aligned} \sup \limits _{m}\left| S_{N_{2m+1}}\left( f,x\right) -S_{N_{2m}}\left( f,x\right) \right| =\infty , \end{aligned}$$

provided that $N_{m}\ge \nu _{m}$. This completes the proof of Theorem  1.7. $\square$

Proof of Theorem 1.8

During the proof, we apply some idea of Bochkarev [3]. Consider the function $W_{N}\left( t\right)$ defined by

$$\begin{aligned}&W_{N}\left( t\right) \\&\quad : =\left\{ \begin{array}{l} \frac{2^{N}}{\sqrt{N}}\sum \limits _{j=2N}^{3N-1}w_{2^{j}}\left( t\right) ,t\in \bigcup \limits _{y_{0}=0}^{1}\cdots \bigcup \limits _{y_{3N-1}=0}^{1}I_{4N}\left( y_{0},...,y_{3N-1},y_{2N},...,y_{3N-1}\right) \\ 0,\text {otherwise} \end{array} .\right. \end{aligned}$$

Set

$$\begin{aligned} n\left( N,x\right) =\sum \limits _{j=2N}^{3N-1}\varepsilon _{j}\left( x\right) 2^{j}+\sum \limits _{j=3N}^{4N-1}\varepsilon _{j-N}\left( x\right) 2^{j}, \end{aligned}$$

(25)

where $\varepsilon _{j}\left( x\right) =0,1$ which will be defined below. We suppose that

$$\begin{aligned} x\in I_{3N+1}\left( x_{0},...,x_{3N-1},1-x_{2N}\right) . \end{aligned}$$

Denote

$$\begin{aligned}&E_{N}^{\prime }:=\bigcup \limits _{x_{0}=0}^{1}\cdots \bigcup \limits _{x_{3N-1}=0}^{1}I_{3N+1}\left( x_{0},...,x_{3N-1},1-x_{2N}\right) , \\&E^{\prime }:=\bigcap \limits _{k=1}^{\infty }\bigcup \limits _{N=k}^{\infty }E_{N}^{\prime }. \end{aligned}$$

It is easy to see that

$$\begin{aligned} \text {mes}\left( E^{\prime }\right) =1. \end{aligned}$$

and

$$\begin{aligned} I_{3N+1}\left( x_{0},...,x_{3N-1},1-x_{2N}\right) \cap I_{4N}\left( y_{0},...,y_{3N-1},y_{2N},...,y_{3N-1}\right) =\emptyset . \end{aligned}$$

Let $\left\{ N_{v}\right\}$ be a subsequence for which $x\in E_{N_{v}}^{\prime },v=1,2,....$. Without lost of generality, we can suppose that $N_{v}^{\prime }=N$. Since

$$\begin{aligned} \left( x\dotplus t\right) _{2N}\dotplus \left( x\dotplus t\right) _{3N}=1,t\in \mathop {supp}\left( W_{N}\right) ,x\in E^{\prime }, \end{aligned}$$

then from (3) and (9), we have (for the sake of brevity $A_{n(N,x)-1}^{\alpha _{n(N,x)}}$ will be denoted as $A_{n(N,x)-1}^{\alpha _{n}}$ which will not cause misunderstand)

$$\begin{aligned}&W_{N}*T_{n\left( N,x\right) }^{\left( 3\right) } \\&\quad =\frac{1}{A_{n\left( N,x\right) -1}^{\alpha _{n}}}\sum \limits _{j=2N}^{3N} \varepsilon _{j}\left( x\right) A_{n_{\left( j\right) }\left( N,x\right) -1}^{\alpha _{n}} \\&\qquad \times \int \limits _{\mathbb {I}}W_{N}\left( t\right) w_{n\left( N,x\right) }\left( x\dotplus t\right) D_{2^{j}}^{*}\left( x\dotplus t\right) dt. \end{aligned}$$

Set

$$\begin{aligned} q\left( N,x\right) :=\sum \limits _{j=3N}^{4N-1}\varepsilon _{j-N}\left( x\right) 2^{j}. \end{aligned}$$

Then, we can write

$$\begin{aligned} w_{n\left( N,x\right) }\left( t\right) =w_{n\left( N,x\right) -q\left( Nx\right) }\left( t\right) w_{q\left( N,x\right) }\left( t\right) =1,t\in \mathop {supp}\left( W_{N}\right) . \end{aligned}$$

Consequently

$$\begin{aligned}&W_{N}*T_{n\left( N,x\right) }^{\left( 3\right) } \\&\quad =\frac{w_{n\left( N,x\right) }\left( x\right) }{A_{n\left( N,x\right) -1}^{\alpha _{n}}}\sum \limits _{j=2N}^{3N}\varepsilon _{j}\left( x\right) A_{n_{\left( j\right) }\left( N,x\right) -1}^{\alpha _{n}}\frac{2^{N}}{\sqrt{ N}} \\&\qquad \times \bigcup \limits _{y_{0}=0}^{1}\cdots \bigcup \limits _{y_{3N-1}=0}^{1}\int \limits _{I_{4N}\left( y_{0},...,y_{3N-1},y_{2N},...,y_{3N-1}\right) }\left( \sum \limits _{i=2N}^{3N-1}w_{2^{i}}\left( t\right) \right) D_{2^{j}}^{*}\left( x\dotplus t\right) dt \\&\quad = \frac{w_{n\left( N,x\right) }\left( x\right) }{\sqrt{N}A_{n\left( N,x\right) -1}^{\alpha _{n}}}\sum \limits _{j=2N}^{3N}\varepsilon _{j}\left( x\right) A_{n_{\left( j\right) }\left( N,x\right) -1}^{\alpha _{n}}\\&\qquad \times \bigcup \limits _{y_{0}=0}^{1}\cdots \bigcup \limits _{y_{3N-1}=0}^{1}\int \limits _{I_{3N}\left( y_{0},...,y_{3N-1}\right) }\left( \sum \limits _{i=2N}^{3N-1}w_{2^{i}}\left( t\right) \right) D_{2^{j}}^{*}\left( x\dotplus t\right) dt \\&\quad =\frac{w_{n\left( N,x\right) }\left( x\right) }{\sqrt{N}A_{n\left( N,x\right) -1}^{\alpha _{n}}}\sum \limits _{j=2N}^{3N-1}\varepsilon _{j}\left( x\right) A_{n_{\left( j\right) }\left( N,x\right) -1 }^{\alpha _{n}}\int \limits _{\mathbb {I}}\left( \sum \limits _{i=2N}^{3N-1}w_{2^{i}}\left( t\right) \right) D_{2^{j}}^{*}\left( x\dotplus t\right) dt \\&\quad =\frac{w_{n\left( N,x\right) }\left( x\right) }{\sqrt{N}A_{n\left( N,x\right) -1}^{\alpha _{n}}}\sum \limits _{j=2N}^{3N-1}\varepsilon _{j}\left( x\right) A_{n_{\left( j\right) }\left( N,x\right) -1 }^{\alpha _{n}}w_{2^{j}}\left( x\right) . \end{aligned}$$

Two cases are possible:

• (a)
  $$\begin{aligned} \sum \limits _{k=2N}^{3N-1}x_{k}<\frac{N}{3}; \end{aligned}$$
• (b)
  $$\begin{aligned} \sum \limits _{k=2N}^{3N-1}x_{k}\ge \frac{N}{3}. \end{aligned}$$

  First, we consider the case a) and let us define digits $\varepsilon _{k}\left( x\right)$ by $\varepsilon _{k}\left( x\right) =1-x_{k}$. Then, we can write

  $$\begin{aligned} \left| W_{N}*T_{n\left( N,x\right) }^{\left( 3\right) }\right| \ge \frac{c}{\sqrt{N}2^{4N\alpha _{n}}}\sum \limits _{2N\le j\le 2N+\left( 2N\right) /3}2^{j\alpha _{n}}\ge \frac{c}{\sqrt{N}2^{2N\alpha _{n}}\alpha _{n}}. \end{aligned}$$

  Since

  $$\begin{aligned} \alpha _{n}\le \frac{c_{0}\log \log n\left( N,x\right) }{\log \left( N,x\right) }\le \frac{c_{0}\log \left( 4N\right) }{2N}\text { }\left( n>n_{0}\right) \text {,} \end{aligned}$$

  we obtain

  $$\begin{aligned} \left| W_{N}*T_{n\left( N,x\right) }^{\left( 3\right) }\right| \ge \frac{cN^{1/2-c_{0}}}{\log \left( 4N\right) }. \end{aligned}$$

  (26)

  Now, we consider the case b). The digits $\varepsilon _{k}\left( x\right)$ define by $\varepsilon _{k}\left( x\right) =x_{k}.$ Analogously, we can prove the validity of estimation (26).

Set

$$\begin{aligned} \gamma _{N}:=\frac{N^{1/2-c_{0}}}{\log \left( 4N\right) }. \end{aligned}$$

Let $\left\{ N_{v}:v\ge 1\right\}$ be a subsequence for which

$$\begin{aligned}&x\in E_{N_{v}},v=1,2,..., \nonumber \\&\quad N_{v+1}\ge 2N_{v}, \end{aligned}$$

(27)

$$\begin{aligned}&\quad \gamma _{N_{v}}\ge v^{4}, \end{aligned}$$

(28)

$$\begin{aligned}&\quad \sum \limits _{j=1}^{v-1}\frac{2^{N_{j}}\sqrt{N_{j}}}{\sqrt{\gamma _{N_{j}}}}< \frac{\sqrt{\gamma _{N_{v}}}}{v}. \end{aligned}$$

(29)

Let

$$\begin{aligned} f\left( t\right) :=\sum \limits _{j=1}^{\infty }\frac{W_{N_{j}}\left( t\right) }{\sqrt{\gamma _{N_{j}}}}. \end{aligned}$$

It is easy to show that

$$\begin{aligned}&\left\| W_{N}\right\| _{1}=\frac{2^{N}}{\sqrt{N}}\bigcup \limits _{x_{0}=0}^{1}\cdots \bigcup \limits _{x_{3N-1}=0}^{1}\int \limits _{I_{4N}\left( x_{0},...,x_{3N-1},x_{2N},...,x_{3N-1}\right) }\left| \sum \limits _{j=2N}^{3N-1}w_{2^{j}}\left( t\right) \right| dt \\&\quad =\frac{1}{\sqrt{N}}\bigcup \limits _{x_{0}=0}^{1}\cdots \bigcup \limits _{x_{3N-1}=0}^{1}\int \limits _{I_{3N}\left( x_{0},...,x_{3N-1}\right) }\left| \sum \limits _{j=2N}^{3N-1}w_{2^{j}}\left( t\right) \right| dt \\&\quad =\frac{1}{\sqrt{N}}\int \limits _{\mathbb {I}}\left| \sum \limits _{j=2N}^{3N-1}w_{2^{j}}\left( t\right) \right| dt\le \frac{1 }{\sqrt{N}}\left( \int \limits _{I}\left| \sum \limits _{j=2N}^{3N-1}w_{2^{j}}\left( t\right) \right| ^{2}dt\right) ^{1/2}=1. \end{aligned}$$

Then, from (28), we conclude that $f\in L_{1}\left( \mathbb {I}\right)$.

It is easy to see that

$$\begin{aligned} f*T_{n\left( N_{v},x\right) }^{\left( 3\right) }=\sum \limits _{j=1}^{v-1} \frac{1}{\sqrt{\gamma _{N_{j}}}}\left( W_{N_{j}}*T_{n\left( N_{v},x\right) }^{\left( 3\right) }\right) +\frac{1}{\sqrt{\gamma _{N_{v}}}} \left( W_{N_{v}}*T_{n\left( N_{v},x\right) }^{\left( 3\right) }\right) . \end{aligned}$$

(30)

We can write (see (6) and (25))

$$\begin{aligned}&W_{N_{j}}*T_{n\left( N_{v},x\right) }^{\left( 3\right) } \nonumber \\&\quad = \frac{1}{A_{n\left( N_{v},x\right) -1}^{\alpha _{n}}}\sum \limits _{k=2N_{v}}^{3N_{v}-1}\varepsilon _{k}\left( x\right) A_{n_{\left( k\right) }\left( N_{v},x\right) -1}^{\alpha _{n}}\left( W_{N_{j}}*\left( w_{n^{\left( k\right) }\left( N_{v},x\right) -1}D_{2^{k}}\right) \right) \nonumber \\&\qquad + \frac{1}{A_{n\left( N_{v},x\right) -1}^{\alpha _{n}}}\sum \limits _{k=3N_{v}}^{4N_{v}-1}\varepsilon _{k-N_{v}}\left( x\right) A_{n_{\left( k\right) }\left( N_{v},x\right) -1}^{\alpha _{n}}\left( W_{N_{j}}*\left( w_{n^{\left( k\right) }\left( N_{v},x\right) -1}D_{2^{k}}\right) \right) . \end{aligned}$$

(31)

Let

$$\begin{aligned} n^{\left( k\right) }\left( N_{v},x\right) -1=2^{k}-1+n^{\left( k+1\right) }\left( N_{v},x\right) . \end{aligned}$$

Suppose that $n^{\left( k+1\right) }\left( N_{v},x\right) \ne 0$. Then, it is easy to see that

$$\begin{aligned} W_{N_{j}}*\left( w_{n^{\left( k\right) }\left( N_{v},x\right) -1}D_{2^{k}}\right) =0,j<v,2N_{v}\le k<3N_{v}. \end{aligned}$$

Hence, we can suppose that there exists $k_{0}\in \left\{ 2N_{v},...,3N_{v}-1\right\}$, such that $n^{\left( k_{0}+1\right) }\left( N_{v},x\right) =0$ and $\varepsilon _{k_{0}}\left( x\right) =1.$ Since $n^{\left( k_{0}\right) }\left( N_{v},x\right) \ne 0$, we conclude that

$$\begin{aligned} W_{N_{j}}*\left( w_{n^{\left( k\right) }\left( N_{v},x\right) -1}D_{2^{k}}\right) =0 \end{aligned}$$

when $k<k_{0}$. Consequently, we have $\left( w_{-1}=0\right)$

$$\begin{aligned}&\frac{1}{A_{n\left( N_{v},x\right) -1}^{\alpha _{n}}}\sum \limits _{k=2N_{v}}^{3N_{v}-1}\varepsilon _{k}\left( x\right) A_{n_{\left( k\right) }\left( N_{v},x\right) -1}^{\alpha _{n}}\left( W_{N_{j}}*\left( w_{n^{\left( k\right) }\left( N_{v},x\right) -1}D_{2^{k}}\right) \right) \nonumber \\&\quad =\frac{A_{n_{\left( k_{0}\right) }\left( N_{v},x\right) -1}^{\alpha _{n}}}{ A_{n\left( N_{v},x\right) -1}^{\alpha _{n}}}\left( W_{N_{j}}*\left( w_{2^{k_{0}}-1}D_{2^{k_{0}}}\right) \right) \nonumber \\&\quad =\frac{A_{n_{\left( k_{0}\right) }\left( N_{v},x\right) -1}^{\alpha _{n}}}{ A_{n\left( N_{v},x\right) -1}^{\alpha _{n}}}\left( W_{N_{j}}*D_{2^{k_{0}}}\right) \nonumber \\&\quad =\frac{A_{n_{\left( k_{0}\right) }\left( N_{v},x\right) -1}^{\alpha _{n}}}{ A_{n\left( N_{v},x\right) -1}^{\alpha _{n}}}S_{2^{k_{0}}}\left( W_{N_{j}}\right) \nonumber \\&\quad =\frac{A_{n_{\left( k_{0}\right) }\left( N_{v},x\right) -1}^{\alpha _{n}}}{ A_{n\left( N_{v},x\right) -1}^{\alpha _{n}}}W_{N_{j}}. \end{aligned}$$

(32)

Analogously, we can prove that

$$\begin{aligned}&\frac{1}{A_{n\left( N_{v},x\right) }^{\alpha _{n}}}\sum \limits _{k=3N_{v}}^{4N_{v}-1}\varepsilon _{k-N_{v}}\left( x\right) A_{n_{\left( k\right) }\left( N_{v},x\right) -1}^{\alpha _{n}}\left( W_{N_{j}}*\left( w_{n^{\left( k\right) }\left( N_{v},x\right) -1}D_{2^{k}}\right) \right) \nonumber \\&\quad =\frac{A_{n_{\left( k_{0}\right) }\left( N_{v},x\right) -1}^{\alpha _{n}}}{ A_{n\left( N_{v},x\right) -1}^{\alpha _{n}}}W_{N_{j}}. \end{aligned}$$

(33)

Combining (31)–(33) from (29), we get

$$\begin{aligned}&\left| \sum \limits _{j=1}^{v-1}\frac{1}{\sqrt{\gamma _{N_{j}}}}\left( W_{N_{j}}*T_{n\left( N_{v},x\right) }^{\left( 3\right) }\right) \right| \nonumber \\&\quad \le \sum \limits _{j=1}^{v-1}\frac{\left| W_{N_{j}}\right| }{\sqrt{ \gamma _{N_{j}}}}\le \sum \limits _{j=1}^{v-1}\frac{2^{N_{j}}\sqrt{N_{j}}}{ \sqrt{\gamma _{N_{j}}}}<\frac{\sqrt{\gamma _{N_{v}}}}{v}. \end{aligned}$$

(34)

From (26), (30), and (34), we conclude that $\left( x\in E^{\prime }\right)$

$$\begin{aligned} \left| f*T_{n\left( N_{v},x\right) }^{\left( 3\right) }\right| \ge c\sqrt{\gamma _{N_{v}}}\rightarrow \infty \text { as }v\rightarrow \infty . \end{aligned}$$

(35)

From (6), we can write

$$\begin{aligned} f*K_{n\left( N_{v},x\right) }^{\alpha _{n}}=f*T_{n\left( N_{v},x\right) }^{\left( 1\right) }+f*T_{n\left( N_{v},x\right) }^{\left( 2\right) }+f*T_{n\left( N_{v},x\right) }^{\left( 3\right) }. \end{aligned}$$

(36)

Lemmas 2.6 and 2.7 imply that

$$\begin{aligned} \sup \limits _{n}\left| f*T_{n}^{\left( l\right) }\right| <\infty \text { a. e. for }f\in L_{1}\left( \mathbb {I}\right) ,l=1,2. \end{aligned}$$

(37)

Let $E_{0}$ be the set for which (37) does not hold. Denote $E:=E^{\prime }\backslash E_{0}$. Then, it is evident that mes$\left( E\right) = 1$. Let $x\in E$. Then, (35)–(37) imply that

$$\begin{aligned} \sup \limits _{n}\left| \sigma _{n}^{\alpha _{n}}\left( f,x\right) \right| =\infty { \ \ \ }\left( x\in E\right) . \end{aligned}$$

Theorem 1.8 is proved. $\square$