# On weighted integrability of functions defined by trigonometric series with p-bounded variation coefficients

Open Access
Research

## Abstract

In this paper we introduce new classes of p-bounded variation sequences and give a sufficient and necessary condition for weighted integrability of trigonometric series with coefficients belonging to these classes. This is a generalization of the results obtained by the first author [J. Inequal. Appl. 2010:1–19, 2010] and Dyachenko and Tikhonov [Stud. Math. 193(3):285–306, 2009].

## Keywords

Trigonometric series Sequences of p-bounded variation Weighted $$L^{p}$$ integrability

## MSC

42A16 42A20 42A32 41A50

## 1 Introduction

Let $$L^{s}$$, $$1 \leqslant s< \infty$$, be the space of all s-power integrable functions f of period 2π with the norm
$$\Vert f \Vert _{L^{s}}= \biggl( \int _{-\pi }^{\pi } \bigl\vert f(x) \bigr\vert ^{s}\,dx \biggr) ^{\frac{1}{s}}.$$
Write
$$f(x)=\sum_{k=1}^{\infty } a_{k}\cos kx, \qquad g(x)=\sum_{k=1}^{\infty } a_{k}\sin kx$$
for those x, where the above series converge.

Denote by ϕ and $$\lambda _{n}$$ either f or g and either $$a_{n}$$ and $$b_{n}$$, respectively.

Let $$\triangle _{r}a_{n}=a_{n}-a_{n+r}$$ for a sequence of complex numbers $$(a_{n})$$ and $$r\in \mathbb{N}$$.

### Theorem 1

Let a nonnegative sequence$$( \lambda _{n} ) \in \Re$$, $$1< s<\infty$$and$$1-s<\alpha <1$$. Then
$$x^{-\alpha } \vert \phi \vert ^{s}\in L^{1} \quad \Longleftrightarrow\quad \sum_{n=1}^{\infty }n^{\alpha +s-2} \lambda _{n}^{s}< \infty .$$
This theorem was proved for $$\Re =DS$$, where DS denotes all decreasing sequences, in [1, 5, 14], and . Later, Theorem 1 was showed in  for
$$\Re =\overline{GM} ( {}_{1}\beta ) := \Biggl\{ ( a _{n} ) \subset \mathbb{C} :\sum_{k=n}^{\infty } \vert \Delta _{1}a_{k} \vert \leqslant C\cdot {}_{1}\beta _{n} \Biggr\} ,$$
and in  for
$$\Re =GM ( {}_{1}\beta ) := \Biggl\{ ( a_{n} ) \subset \mathbb{C} :\sum_{k=n}^{2n-1} \vert \Delta _{1}a _{k} \vert \leqslant C\cdot {}_{1}\beta _{n} \Biggr\} ,$$
where $${}_{1}\beta _{n}= \vert a_{n} \vert$$; C here and throughout the paper denotes a positive constant.
The proof in the case of class
$$\Re =GM ( {}_{2}\beta ) := \Biggl\{ ( a_{n} ) \subset \mathbb{C} :\sum_{k=n}^{2n-1} \vert \Delta _{1}a _{k} \vert \leqslant C\cdot {}_{2}\beta _{n} \Biggr\} ,$$
where $${}_{2}\beta _{n}=\sum_{k= [ n/c ] }^{ [ cn ] }\frac{ \vert a_{k} \vert }{k}$$ for some $$c>1$$, is included in .
In  Dyachenko and Tikhonov extended this theorem to the class
$$\Re =\overline{GM} \bigl( {}_{3}\beta (\theta ) \bigr) := \Biggl\{ ( a_{n} ) \subset \mathbb{C} :\sum_{k=n}^{\infty } \vert \Delta _{1}a_{k} \vert \leqslant C\cdot {}_{3}\beta _{n}( \theta ) \Biggr\} ,$$
where $${}_{3}\beta _{n}(\theta )=n^{\theta -1}\sum_{k= [ n/c ] }^{\infty }\frac{ \vert a_{k} \vert }{k ^{\theta }}<\infty$$ for some $$c>1$$ and $$\theta \in (0,1]$$.
From the articles of Dyachenko and Tikhonov  and Leindler , it is well known that
\begin{aligned} DS \varsubsetneq &\overline{GM}({}_{1}\beta )\varsubsetneq GM({}_{1} \beta )\varsubsetneq GM({}_{2}\beta ) \\ \varsubsetneq &\overline{GM} \bigl( {}_{3}\beta (1) \bigr) \subseteq \overline{GM} \bigl( {}_{3}\beta (\theta _{2}) \bigr) \subseteq \overline{GM} \bigl( {}_{3}\beta (\theta _{1}) \bigr) , \end{aligned}
(1)
for $$0<\theta _{1}\leqslant \theta _{2}\leqslant 1$$.

Further, Szal defined a new class of sequences in the following way (see ):

### Definition 1

Let $$\beta := ( \beta _{n} )$$ be a nonnegative sequence and r a natural number. The sequence of complex numbers $$a:= ( a _{n} ) \in \overline{GM} ( \beta ,r )$$ if the relation
$$\sum_{k=n}^{\infty } \vert \triangle _{r}a_{n} \vert \leqslant C\beta _{n}$$
holds for all $$n\in \mathbb{N}$$.
Moreover, from  we know that
$$\overline{GM}\bigl({}_{3}\beta (\theta ),r_{1}\bigr) \subsetneq \overline{GM}\bigl({}_{3} \beta (\theta ),r_{2} \bigr),$$
(2)
where $$r_{1}< r_{2}$$, $$\theta \in (0,1]$$ and $$r_{1}\mid r_{2}$$.
Let $$r\in \mathbb{N}$$ and $$\alpha \in \mathbb{R}$$. We define on the interval $$[ -\pi ,\pi ]$$ an even function $$\omega _{\alpha ,r}$$, which is given on the interval $$[ 0,\pi ]$$ by the formula
$$\omega _{\alpha ,r} ( x ) :=\textstyle\begin{cases} ( x-\frac{2l\pi }{r} ) ^{-\alpha }\quad \text{for }x \in (\frac{ 2l\pi }{r},\frac{ ( 2l+1 ) \pi }{r}] \text{ and }l\in U_{1,} \\ ( \frac{2 ( l+1 ) \pi }{r}-x ) ^{-\alpha }\quad \text{for }x\in ( \frac{ ( 2l+1 ) \pi }{r},\frac{2 ( l+1 ) \pi }{r} ) \text{ and }l\in U_{2}, \\ 0\quad \text{for }x=\frac{2l\pi }{r}\text{ and }l\in U_{3}, \end{cases}$$
where $$U_{1}= \{ 0,1,\dots ,[r/2] \}$$ if r is an odd number and $$U_{1}= \{ 0,1,\dots , [ r/2 ] -1 \}$$ if r is an even number, $$U_{2}= \{ 0,1,\dots , [ r/2 ] -1 \}$$ for $$r\geqslant 2$$, and $$U_{3}= \{ 0,1, \dots , [ r/2 ] \}$$ for $$r\geqslant 1$$.

Theorem 1 was generalized for the class $$\overline{GM}({}_{3}\beta ( \theta ),r)$$, where $$r\in \mathbb{N}$$ and $$\theta \in (0,1]$$, in . We can formulate this result in the following way.

### Theorem 2

([9, Theorem 5])

Let a nonnegative sequence$$( \lambda _{n} ) \in \overline{GM} ( {}_{3}\beta (\theta ),r )$$, where$$r\in \mathbb{N}$$, $$\theta \in (0,1]$$and$$1\leqslant s<\infty$$. If
$$1-\theta s< \alpha < 1,$$
then$$\omega _{\alpha ,r} \vert \phi \vert ^{s}\in L^{1}$$if and only if
$$\sum_{n=1}^{\infty }n^{\alpha +s-2} \vert \lambda _{n} \vert ^{s}< \infty .$$

Now, we define new classes of sequences.

### Definition 2

Let $$\beta := ( \beta _{n} )$$ be a nonnegative sequence, p a positive real number, $$r\in \mathbb{N}$$. One says that a sequence $$a=(a_{n})$$ of complex numbers belongs to $$GM(p,\beta ,r)$$ if the relation
$$\Biggl( \sum_{k=n}^{2n-1} \vert \triangle _{r}a_{k} \vert ^{p} \Biggr) ^{ \frac{1}{p}}\leqslant C\beta _{n}$$
holds for all $$n\in \mathbb{N}$$.
Moreover, we say that a sequence $$(a_{n})\in \overline{GM}(p,\beta ,r)$$ if the relation
$$\Biggl( \sum_{k=n}^{\infty } \vert \triangle _{r}a_{k} \vert ^{p} \Biggr) ^{\frac{1}{p}}\leqslant C\beta _{n}$$
holds for all $$n\in \mathbb{N}$$.

The class $$GM(p,\beta ,1)$$ was defined by Tikhonov and Liflyand in .

In this paper we present some properties of the classes $$\overline{GM} ( p,{}_{3}\beta (\theta ),r )$$ and $$GM ( p, {}_{3}\beta (\theta ),r )$$. Moreover, we will generalize Theorem 2 for the class $$GM ( p,{}_{3}\beta (\theta ),r )$$ with $$0<\theta < \frac{1}{s}$$ and $$r\in \mathbb{N}$$.

We will write $$I_{1}\ll I_{2}$$ if there exists a positive constant C such that $$I_{1}\leqslant CI_{2}$$.

## 2 Main results

We formulate our results as follows:

### Theorem 3

Let$$r\in \mathbb{N}$$, $$\theta \in (0,1)$$, andpbe a positive real number. Then
\begin{aligned}& \overline{GM} \bigl( p,{}_{3}\beta (\theta ),r \bigr) =GM \bigl( p,{}_{3} \beta (\theta ),r \bigr) \quad \textit{and} \\& \overline{GM} \bigl( p,{}_{3}\beta (1),r \bigr) \subseteq GM \bigl( p,{}_{3} \beta (1),r \bigr) . \end{aligned}

### Theorem 4

Let$$r\in \mathbb{N}$$, $$\theta \in (0,1)$$, and$$p_{1}$$, $$p_{2}$$be two positive real numbers such that$$0< p_{1}<p_{2}$$. Then
$$GM \bigl( p_{1},{}_{3}\beta (\theta ),r \bigr) \varsubsetneq GM \bigl( p_{2},{}_{3}\beta (\theta ),r \bigr) .$$

### Theorem 5

Let$$r_{1},r_{2}\in \mathbb{N}$$, $$r_{1}< r_{2}$$, $$\theta \in (0,1]$$and$$p\geqslant 1$$. If$$r_{1}|r_{2}$$, then
$$GM \bigl( p,{}_{3}\beta (\theta ),r_{1} \bigr) \varsubsetneq GM \bigl( p,{}_{3}\beta (\theta ),r_{2} \bigr) .$$

### Theorem 6

Let$$(b_{n})\in GM ( p,{}_{3}\beta (\theta ),r )$$, where$$r\in \mathbb{N}$$, $$p\geqslant 1$$, $$0<\theta <\frac{1}{p}$$and$$1\leqslant s<\infty$$. If
$$1-\theta s-s+\frac{s}{p}< \alpha < 1$$
and
$$\sum_{n=1}^{\infty }n^{\alpha -2-\frac{s}{p}+2s} \vert b _{n} \vert ^{s}< \infty$$
then$$\omega _{\alpha .r}|\phi |^{s}\in L^{1}$$.

### Theorem 7

Let a nonnegative sequence$$(b_{n})$$belong to$$GM ( p,{}_{3} \beta (\theta ),r )$$, where$$r\in \mathbb{N}$$, $$p\geqslant 1$$, $$0<\theta <\frac{1}{p}$$and$$1\leqslant s<\infty$$. If
$$1-\theta s< \alpha < 1+s$$
and$$\omega _{\alpha .r}|\phi |^{s}\in L^{1}$$then
$$\sum_{n=1}^{\infty }n^{\alpha -2+\frac{s}{p}}b_{n}^{s}< \infty .$$

### Remark 1

If we take $$p=1$$, then the result of Szal  (Theorem 2) follows from our Theorem 6 and 7. Moreover, by the embedding relations (1) and (2), we can also derive from Theorem 6 and 7 the result of Dyachenko and Tikhonov  and all the results mentioned before.

## 3 Auxiliary results

For $$n\in \mathbb{N}$$ and $$k=0,1,2,\ldots$$ , denote by
\begin{aligned}& D_{k,r}(x)=\frac{\sin (k+r/2)x}{2\sin (rx/2)}, \\& \overset{\sim }{D}_{k,r}(x)=\frac{\cos (k+r/2)x}{2\sin (rx/2)} \end{aligned}
the Dirichlet-type kernels.

### Lemma 1

([10, Lemma 3.1] and [11, Lemma 17])

Let$$r\in \mathbb{N}$$, $$l\in \mathbb{Z}$$, and$$(a_{n})\subset \mathbb{C}$$. If$$x\neq \frac{2l\pi }{r}$$, then for all$$m\geqslant n$$
\begin{aligned}& \sum_{k=n}^{m}a_{k}\cos kx=\sum_{k=n}^{m}\triangle _{r}a _{k}D_{k,r}(x)-\sum _{k=m+1}^{m+r}a_{k}D_{k,-r}(x)+ \sum_{k=n}^{n+r-1}a_{k}D_{k,-r}(x), \\& \sum_{k=n}^{m}a_{k}\sin kx=\sum_{k=m+1}^{m+r}a_{k} \overset{\sim }{D}_{k,-r}(x)-\sum_{k=n}^{n+r-1}a_{k} \overset{\sim }{D}_{k,-r}(x)-\sum_{k=n}^{m} \triangle _{r}a_{k} \overset{\sim }{D}_{k,r}(x). \end{aligned}

### Lemma 2

([6, Corollary 1])

Let$$p\geqslant 1,\gamma _{n}>0$$, and$$a_{n}\geqslant 0$$for$$n\in \mathbb{N}$$. Then
\begin{aligned}& \sum_{n=1}^{\infty }\gamma _{n} \Biggl( \sum_{k=1}^{n}a _{k} \Biggr) ^{p}\leqslant p^{p}\sum _{n=1}^{\infty }\gamma _{n} ^{1-p}a_{n}^{p} \Biggl( \sum _{k=n}^{\infty }\gamma _{k} \Biggr) ^{p}, \\& \sum_{n=1}^{\infty }\gamma _{n} \Biggl( \sum_{k=n}^{ \infty }a_{k} \Biggr) ^{p}\leqslant p^{p}\sum _{n=1}^{\infty } \gamma _{n}^{1-p}a_{n}^{p} \Biggl( \sum_{k=1}^{n}\gamma _{k} \Biggr) ^{p}. \end{aligned}

### Lemma 3

([4, Theorem 19])

If$$a_{n}\geqslant 0$$for$$n\in \mathbb{N}$$and$$0< p_{1}\leqslant p_{2}<\infty$$, then
$$\Biggl( \sum_{n=1}^{\infty }a_{n}^{p_{2}} \Biggr) ^{\frac{1}{p _{2}}}\leqslant \Biggl( \sum_{n=1}^{\infty }a_{n}^{p_{1}} \Biggr) ^{\frac{1}{p_{1}}}.$$

### Lemma 4

()

Let$$a_{k}\geqslant 0$$for$$k\in \mathbb{N}$$and$$p\geqslant 1$$. Then
$$\Biggl( \frac{1}{n}\sum_{k=n}^{2n-1}a_{k}^{p} \Biggr) ^{ \frac{1}{p}}\geqslant \frac{1}{n}\sum _{k=n}^{2n-1}a_{k}.$$

### Lemma 5

Let$$(a_{k})\subset \mathbb{C}$$, $$p\geqslant 1$$, $$r,n\in \mathbb{N}$$and$$d\in \mathbb{N} _{0}=\mathbb{N}$$$$\cup \{ 0 \}$$. Then
$$\sum_{k=2^{d+1}(n+1)}^{2^{d+1}(n+1)+r-1} \vert a_{k} \vert \leq \sum_{k=2^{d}(n+1)}^{2^{d}(n+1)+r-1} \vert a_{k} \vert + \bigl[ 2^{d}(n+1) \bigr] ^{1-\frac{1}{p}} \Biggl( \sum_{k=2^{d}(n+1)} ^{2^{d+1}(n+1)-1} \vert \triangle _{r}a_{k} \vert ^{p} \Biggr) ^{\frac{1}{p}}.$$

### Proof

From Lemma 4 we have
\begin{aligned} \Biggl(\sum_{k=2^{d}(n+1)}^{2^{d+1}(n+1)-1} \vert \triangle _{r}a_{k} \vert ^{p} \Biggr) ^{\frac{1}{p}} =& \bigl[2^{d}(n+1) \bigr]^{\frac{1}{p}} \Biggl(\frac{1}{2^{d}(n+1)} \sum_{k=2^{d}(n+1)}^{2^{d+1}(n+1)-1} \vert \triangle _{r}a_{k} \vert ^{p} \Biggr) ^{\frac{1}{p}} \\ \geqslant& \bigl[2^{d}(n+1) \bigr]^{\frac{1}{p}} \frac{1}{2^{d}(n+1)} \sum_{k=2^{d}(n+1)}^{2^{d+1}(n+1)-1} \vert \triangle _{r}a_{k} \vert \\ \geqslant& \bigl[2^{d}(n+1) \bigr]^{\frac{1}{p}-1} \Biggl(\sum _{k=2^{d+1}(n+1)}^{2^{d+1}(n+1)+r-1} \vert a_{k} \vert -\sum_{k=2^{d}(n+1)}^{2^{d}(n+1)+r-1} \vert a_{k} \vert \Biggr). \end{aligned}
Hence
$$\sum_{k=2^{d+1}(n+1)}^{2^{d+1}(n+1)+r-1} \vert a_{k} \vert \leqslant \sum_{k=2^{d}(n+1)}^{2^{d}(n+1)+r-1} \vert a_{k} \vert + \bigl[2^{d}(n+1) \bigr] ^{1-\frac{1}{p}} \Biggl(\sum_{k=2^{d}(n+1)}^{2^{d+1}(n+1)-1} \vert \triangle _{r}a_{k} \vert ^{p} \Biggr)^{\frac{1}{p}}$$
and this ends our proof. □

### Lemma 6

Let$$(a_{k})\in GM ( p,{}_{3}\beta (\theta ),r )$$, $$p\geqslant 1$$, $$r\in \mathbb{N}$$, $$d\in \mathbb{N} _{0}$$, and$$0<\theta <\frac{1}{p}$$. Then
$$\sum_{k=2^{d}(n+1)}^{2^{d}(n+1)+r-1} \vert a_{k} \vert \leqslant C\frac{1}{1-2^{ \theta -\frac{1}{p}}} \bigl( 2^{d}(n+1) \bigr) ^{\theta -\frac{1}{p}} \sum _{k= [ \frac{2^{d}(n+1)}{c} ] }^{\infty }\frac{ \vert a _{k} \vert }{k^{\theta }}.$$

### Proof

We have
$$\sum_{k=2^{d}(n+1)}^{2^{d}(n+1)+r-1} \vert a_{k} \vert \leqslant \sum_{j=0}^{\infty } \sum_{k=2^{j}2^{d}(n+1)}^{2^{j+1}2^{d}(n+1)-1} \vert \triangle _{r}a_{k} \vert .$$
Using Hölder inequality with $$p>1$$, we get
\begin{aligned}& \sum_{j=0}^{\infty }\sum _{k=2^{j}2^{d}(n+1)}^{2^{j+1}2^{d}(n+1)-1} \vert \triangle _{r}a_{k} \vert \\& \quad \leqslant \sum_{j=0} ^{\infty } \Biggl[ \Biggl( \sum_{k=2^{j}2^{d}(n+1)}^{2^{j+1}2^{d}(n+1)-1} \vert \triangle _{r}a_{k} \vert ^{p} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{k=2^{j}2^{d}(n+1)}^{2^{j+1}2^{d}(n+1)-1}1^{ \frac{p}{p-1}} \Biggr) ^{1-\frac{1}{p}} \Biggr] \\& \quad \leqslant C\sum_{j=0}^{\infty } \bigl( 2^{j}2^{d}(n+1) \bigr) ^{1-\frac{1}{p}} \bigl( 2^{j}2^{d}(n+1) \bigr) ^{\theta -1}\sum _{k= [ \frac{2^{j}2^{d}(n+1)}{c} ] }^{\infty }\frac{ \vert a _{k} \vert }{k^{\theta }} \\& \quad \leqslant C \bigl(2^{d}(n+1) \bigr)^{\theta -\frac{1}{p}}\sum _{k= [\frac{2^{d}(n+1)}{c} ]}^{\infty }\frac{ \vert a_{k} \vert }{k ^{\theta }} \sum _{j=0}^{\infty } \bigl(2^{\theta -\frac{1}{p}} \bigr) ^{j}. \end{aligned}
When $$p=1$$, we have
\begin{aligned} \sum_{j=0}^{\infty }\sum _{k=2^{j}2^{d}(n+1)}^{2^{j+1}2^{d}(n+1)-1} \vert \triangle _{r}a_{k} \vert \leqslant& C\sum _{j=0}^{ \infty } \bigl(2^{j}2^{d}(n+1) \bigr)^{\theta -1}\sum_{k= [\frac{2^{j}2^{d}(n+1)}{c} ]}^{\infty } \frac{ \vert a _{k} \vert }{k^{\theta }} \\ \leqslant& C \bigl(2^{d}(n+1) \bigr)^{\theta -1}\sum _{k= [\frac{2^{d}(n+1)}{c} ]}^{\infty }\frac{ \vert a_{k} \vert }{k ^{\theta }}\sum _{j=0}^{\infty } \bigl(2^{\theta -1} \bigr) ^{j}. \end{aligned}
If $$\theta -\frac{1}{p} <0$$, then
$$\sum_{k=2^{d}(n+1)}^{2^{d}(n+1)+r-1} \vert a_{k} \vert \leqslant C\frac{1}{1-2^{ \theta -\frac{1}{p}}} \bigl(2^{d}(n+1) \bigr)^{\theta -\frac{1}{p}} \sum _{k= [\frac{2^{d}(n+1)}{c} ]}^{\infty }\frac{ \vert a _{k} \vert }{k^{\theta }}$$
and our proof is complete. □

## 4 Proofs

### 4.1 Proof of Theorem 3

Let $$(a_{n})\in GM ( p,{}_{3}\beta (\theta ),r )$$, where $$p>0$$, $$r\in \mathbb{N,}$$ and $$\theta \in (0,1)$$. Then
\begin{aligned} \Biggl( \sum_{k=n}^{\infty } \vert \triangle _{r}a_{k} \vert ^{p} \Biggr) ^{\frac{1}{p}} =& \Biggl( \sum_{d=0}^{\infty } \sum_{k=2^{d}n} ^{2^{d+1}n-1} \vert \triangle _{r}a_{k} \vert ^{p} \Biggr) ^{\frac{1}{p}} \\ \leqslant& \Biggl( \sum_{d=0}^{\infty } \Biggl( C\bigl(2^{d}n\bigr)^{\theta -1} \sum _{k= [ \frac{2^{d}n}{c} ] }^{\infty }\frac{ \vert a _{k} \vert }{k^{\theta }} \Biggr) ^{p} \Biggr) ^{\frac{1}{p}} \\ \leqslant& Cn^{\theta -1}\sum_{k= [ \frac{n}{c} ] } ^{\infty }\frac{ \vert a_{k} \vert }{k^{\theta }} \Biggl( \sum_{d=0}^{ \infty } \bigl( 2^{(\theta -1)p} \bigr) ^{d} \Biggr) ^{\frac{1}{p}}. \end{aligned}
If $$0<\theta <1$$ then $$(\theta -1)p<0$$, and we have
$$\Biggl( \sum_{k=n}^{\infty } \vert \triangle _{r}a_{k} \vert ^{p} \Biggr) ^{\frac{1}{p}}\leqslant C \Biggl( \frac{1}{1-2^{ (\theta -1)p}} \Biggr) ^{\frac{1}{p}} n^{\theta -1}\sum_{k= [ \frac{n}{c} ] }^{\infty } \frac{ \vert a_{k} \vert }{k ^{\theta }}.$$
So $$(a_{n})\in \overline{GM} ( p,{}_{3}\beta (\theta ),r )$$.
Now we assume $$(a_{n})\in \overline{GM} ( p,{}_{3}\beta (1),r )$$, $$p>0$$, $$r\in \mathbb{N}$$. We have
$$\Biggl( \sum_{k=n}^{2n-1} \vert \triangle _{r}a_{k} \vert ^{p} \Biggr) ^{ \frac{1}{p}}\leqslant \Biggl( \sum_{k=n}^{\infty } \vert \triangle _{r}a _{k} \vert ^{p} \Biggr) ^{\frac{1}{p}}\leqslant Cn^{\theta -1}\sum _{k= [ \frac{n}{c} ] }^{\infty }\frac{ \vert a_{k} \vert }{k ^{\theta }}.$$
This means $$(a_{n})\in GM ( p,{}_{3}\beta (1),r )$$. □

### 4.2 Proof of Theorem 4

Let $$r\in \mathbb{N}$$, $$\theta \in (0,1]$$, $$0< p_{1}\leqslant p_{2}$$, and $$(a_{n})\in GM ( p_{1},{}_{3}\beta (\theta ),r )$$. We will show that $$GM ( p_{1},{}_{3}\beta (\theta ),r ) \subseteq GM ( p_{2},{}_{3}\beta (\theta ),r )$$. Using Lemma 3, we have
$$\Biggl( \sum_{k=n}^{2n-1} \vert \triangle _{r}a_{k} \vert ^{p_{2}} \Biggr) ^{\frac{1}{p_{2}}}\leqslant \Biggl( \sum_{k=n}^{2n-1} \vert \triangle _{r}a _{k} \vert ^{p_{1}} \Biggr) ^{\frac{1}{p_{1}}}\leqslant cn^{\theta -1} \sum _{k=n}^{\infty }\frac{ \vert a_{k} \vert }{k^{\theta }}.$$
This means that $$(a_{n})\in GM ( p_{2},{}_{3}\beta (\theta ),r )$$.
Now we will show that $$GM ( p_{1},{}_{3}\beta (\theta ),r ) \neq GM ( p_{2},{}_{3}\beta (\theta ),r )$$ for $$0< p_{1}< p _{2}$$. Let
\begin{aligned}& a_{n} =\textstyle\begin{cases} \frac{1}{n^{2}},\quad \text{when } 2r\nmid n, \\ \frac{1}{(n-r)^{2}}+\frac{1}{n^{2}n^{\frac{1}{p_{2}}}},\quad \text{when } 2r | n. \end{cases}\displaystyle \end{aligned}
We prove that $$(a_{n}) \in GM (p_{2},{}_{3}\beta (\theta ),r )$$. Suppose
\begin{aligned}& A_{n} = \{ k \in \mathbb{N} : n\leqslant k \leqslant 2n-1\text{ and }2r | k \}, \\& B_{n} = \{ k \in \mathbb{N} : n\leqslant k\leqslant 2n-1, 2r \nmid k \text{ and }2r \nmid k+r \}, \\& C_{n} = \{ k \in \mathbb{N} : n\leqslant k\leqslant 2n-1, 2r \nmid k\text{ and }2r | k+r \}. \end{aligned}
Then
\begin{aligned}& \Biggl( \sum_{k=n}^{2n-1} \vert a_{k}-a_{k+r} \vert ^{p_{2}} \Biggr)^{\frac{1}{p _{2}}} \\& \quad = \biggl( \sum_{k \in A_{n}} \biggl\vert \frac{1}{(k-r)^{2}} + \frac{1}{k^{2}k^{\frac{1}{p_{2}}}} - \frac{1}{(k+r)^{2}} \biggr\vert ^{p_{2}} \\& \qquad {} +\sum_{k \in B_{n}} \biggl\vert \frac{1}{k^{2}} - \frac{1}{(k+r)^{2}} \biggr\vert ^{p_{2}} + \sum_{k \in C_{n}} \biggl\vert \frac{1}{k^{2}} - \frac{1}{k^{2}} - \frac{1}{(k+r)^{2}(k+r)^{\frac{1}{p _{2}}}} \biggr\vert ^{p_{2}} \biggr)^{\frac{1}{p_{2}}} \\& \quad \leqslant \biggl( \sum_{k \in A_{n}} \biggl( \frac{4kr}{ \frac{1}{4} k^{2}k^{2}} + \frac{1}{k^{2+\frac{1}{p_{2}}}} \biggr) ^{p_{2}} + \sum _{k \in B_{n}} \biggl( \frac{2kr + r^{2}}{k^{2}(k+r)^{2}} \biggr)^{p_{2}} + \sum_{k \in C_{n}} \biggl( \frac{1}{(k+r)^{2+{\frac{1}{p_{2}}}}} \biggr) ^{p_{2}} \biggr)^{\frac{1}{p_{2}}} \\& \quad \leqslant (16r+1) \Biggl( \sum_{k=n}^{2n-1} \biggl( \frac{1}{ k ^{2+\frac{1}{p_{2}}}} \biggr)^{p_{2}} \Biggr)^{\frac{1}{p_{2}}} \leqslant \frac{17r}{n^{2}}. \end{aligned}
Moreover,
$$\frac{17r}{n^{2}} \leqslant 2^{2+\theta }17r \Biggl(n^{\theta -1} \sum_{k=n}^{2n-1}\frac{1}{k^{2}} \frac{1}{k^{\theta }} \Biggr) \leqslant 2^{2+\theta }17r n^{\theta -1} \sum_{k= [ \frac{n}{c} ]}^{\infty }\frac{ \vert a _{k} \vert }{k^{\theta }}.$$
This means $$(a_{n})\in GM ( p_{2},{}_{3}\beta (\theta ),r )$$. We will show that $$(a_{n})\notin GM ( p_{1},{}_{3}\beta (\theta ),r )$$. We have
$$\Biggl( \sum_{k=n}^{2n-1} \vert a_{k}-a_{k+r} \vert ^{p_{1}} \Biggr) ^{\frac{1}{p _{1}}}\geqslant \biggl( \sum_{k\in C_{n}} \frac{1}{(k+r)^{2p _{1}+\frac{p_{1}}{p_{2}}}} \biggr) ^{\frac{1}{p_{1}}} \geqslant \frac{1}{(4r)^{2+\frac{1}{p _{2}}+\frac{2}{p_{1}}}} \frac{n^{\frac{1}{p_{1}}}}{n^{2+ \frac{1}{p_{2}}}}.$$
Let
\begin{aligned}& D_{n}=\biggl\{ k\in \mathbb{N} : \biggl[ \frac{n}{c} \biggr] \leqslant k\text{ and }2r|k\biggr\} , \\& E_{n}=\biggl\{ k\in \mathbb{N} : \biggl[ \frac{n}{c} \biggr] \leqslant k\text{ and }2r\nmid k\biggr\} . \end{aligned}
On the other hand, we get
\begin{aligned} n^{\theta -1}\sum_{k= [ \frac{n}{c} ] }^{\infty } \frac{a _{k}}{k^{\theta }} =&n^{\theta -1} \biggl( \sum _{k\in D_{n}}\frac{1}{k ^{2}k^{\theta }}+\sum _{k\in E_{n}} \biggl( \frac{1}{(k-r)^{2}}+\frac{1}{k ^{2+\frac{1}{p_{2}}}} \biggr) \frac{1}{k^{\theta }} \biggr) \\ \leqslant& 5n^{\theta -1}\sum_{k= [ \frac{n}{c} ] } ^{\infty }\frac{1}{k^{2+\theta }}\ll n^{-2}. \end{aligned}
Therefore the inequality
$$\Biggl( \sum_{k=n}^{2n-1} \vert \triangle _{r}a_{k} \vert ^{p_{1}} \Biggr) ^{\frac{1}{p_{1}}}\leqslant Cn^{\theta -1}\sum_{k= [ \frac{n}{c} ] }^{\infty } \frac{a_{k}}{k^{ \theta }}$$
cannot be satisfied because $$n^{\frac{1}{p_{1}}-\frac{1}{p_{2}}} \rightarrow \infty$$ as $$n\rightarrow \infty$$. □

### 4.3 Proof of Theorem 5

Let $$r_{1},r_{2}\in \mathbb{N}$$, $$r_{1}\leqslant r_{2}$$, $$r_{1}|r_{2}$$, $$p\geqslant 1$$ and $$(a_{n})\in GM ( p,{}_{3}\beta (\theta ),r_{1} )$$.

If $$r_{1}|r_{2}$$, then $$r_{2}=\alpha r_{1}$$, where $$\alpha \in \mathbb{N}$$. Using Hölder inequality with $$p>1$$, we have
\begin{aligned}& \Biggl( \sum_{k=n}^{2n-1} \vert a_{k}-a_{k+r_{2}} \vert ^{p} \Biggr) ^{ \frac{1}{p}} \\& \quad = \Biggl( \sum_{k=n}^{2n-1} \Biggl\vert \sum_{l=0}^{\alpha -1} ( a_{k+lr_{1}}-a_{k+(l+1)r_{1}} ) \Biggr\vert ^{p} \Biggr) ^{\frac{1}{p}} \\& \quad \leqslant \Biggl(\sum_{k=n}^{2n-1} \Biggl(\sum_{l=0}^{ \alpha -1} \vert a_{k+lr_{1}}-a_{k+(l+1)r_{1}} \vert \Biggr)^{p} \Biggr)^{\frac{1}{p}} \\& \quad \leqslant \Biggl( \sum_{k=n}^{2n-1} \Biggl( \Biggl( \sum_{l=0}^{\alpha -1} \vert a_{k+lr_{1}}-a_{k+(l+1)r_{1}} \vert ^{p} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{l=0}^{\alpha -1}1^{ \frac{p}{p-1}} \Biggr) ^{1-\frac{1}{p}} \Biggr) ^{p} \Biggr) ^{ \frac{1}{p}} \\& \quad \leqslant \alpha ^{1-\frac{1}{p}} \Biggl( \sum _{k=n}^{2n-1} \Biggl( \sum _{l=0}^{\alpha -1} \vert a_{k+lr_{1}}-a_{k+(l+1)r_{1}} \vert ^{p} \Biggr) \Biggr) ^{\frac{1}{p}} \\& \quad \leqslant \alpha ^{1-\frac{1}{p}} \Biggl( \sum _{l=0}^{\alpha -1} \Biggl( C(n+lr_{1})^{\theta -1} \sum_{k= [ \frac{n+lr_{1}}{c} ] }^{\infty }\frac{ \vert a_{k} \vert }{k ^{\theta }} \Biggr) ^{p} \Biggr) ^{\frac{1}{p}} \\& \quad \leqslant \alpha Cn ^{\theta -1}\sum_{k= [ \frac{n}{c} ] }^{\infty } \frac{ \vert a _{k} \vert }{k^{\theta }}. \end{aligned}
If $$p=1$$ then
\begin{aligned} \sum_{k=n}^{2n} \vert a_{k}-a_{k+r_{2}} \vert \leqslant& \sum _{k=n}^{2n-1}\sum_{l=0}^{\alpha -1} \vert a _{k+lr_{1}}-a_{k+(l-1)r_{1}} \vert \\ \leqslant& C\sum_{l=0}^{\alpha -1} ( n+lr_{1} ) ^{ \theta -1}\sum_{k= [ \frac{n+lr_{1}}{c} ] }^{\infty } \frac{ \vert a_{k} \vert }{k^{\theta }}\leqslant \alpha Cn^{\theta -1}\sum _{k= [ \frac{n}{c} ] }^{\infty }\frac{ \vert a_{k} \vert }{k ^{\theta }}. \end{aligned}
Hence $$(a_{n})\in GM ( p,{}_{3}\beta (\theta ),r_{2} )$$.

Now, we will show that $$GM ( p,{}_{3}\beta (\theta ),r_{1} ) \varsubsetneq GM ( p,{}_{3}\beta (\theta ),r_{2} )$$, when $$r_{1}< r_{2}$$. Let $$a_{n}=\frac{2+\alpha _{n}}{n^{2}}$$, where $$\alpha _{n}= \bigl\{ \scriptsize{ \begin{array}{l@{\quad}l} -1,&\text{when }r_{1}|n, \\ 1,&\text{when }r_{1}\nmid n. \end{array} }$$

We will prove that $$(a_{n})\in GM ( p,{}_{3}\beta (\theta ),r_{2} )$$ and $$(a_{n})\notin GM ( p,{}_{3} \beta (\theta ),r_{1} )$$. Let
\begin{aligned}& A_{n}:=\{k\in \mathbb{N} :n\leqslant k\leqslant 2n-1\text{ and }r_{2}|k \}, \\& B_{n}:=\{k\in \mathbb{N} :n\leqslant k\leqslant 2n-1\text{ and }r_{2} \nmid k\}. \end{aligned}
Then using Lemma 3 for $$p\geqslant 1$$, we have
\begin{aligned} \Biggl( \sum_{k=n}^{2n-1} \vert a_{k}-a_{k+r_{2}} \vert ^{p} \Biggr) ^{ \frac{1}{p}} =& \biggl( \biggl( \sum_{k\in A_{n}}+ \sum_{k\in B_{n}} \biggr) \vert a_{k}-a_{k+r_{2}} \vert ^{p} \biggr) ^{ \frac{1}{p}} \\ =& \biggl( \sum_{k\in A_{n}} \biggl\vert \frac{1}{k^{2}}-\frac{1}{(k+r _{2})^{2}} \biggr\vert ^{p}+\sum _{k\in B_{n}} \biggl\vert \frac{3}{k ^{2}}- \frac{3}{(k+r_{2})^{2}} \biggr\vert ^{p} \biggr) ^{\frac{1}{p}} \\ \leqslant& \Biggl( 3^{p}\sum_{k=n}^{2n-1} \biggl\vert \frac{(k+r _{2})^{2}-k^{2}}{(k+r_{2})^{2}k^{2}} \biggr\vert ^{p} \Biggr) ^{ \frac{1}{p}} \\ =&3 \Biggl( \sum_{k=n}^{2n-1} \biggl\vert \frac{2r_{2}k+r_{2} ^{2}}{(k+r_{2})^{2}k^{2}} \biggr\vert ^{p} \Biggr) ^{\frac{1}{p}} \\ \leqslant& 6r_{2} \Biggl( \sum_{k=n}^{2n-1} \biggl( \frac{1}{k ^{3}} \biggr) ^{p} \Biggr) ^{\frac{1}{p}} \leqslant 6r_{2}\sum_{k=n}^{2n-1} \frac{1}{k^{3}}\leqslant \frac{6r_{2}}{n}\sum _{k=n} ^{2n-1}\frac{1}{k^{2}}. \end{aligned}
Moreover,
\begin{aligned} \frac{6r_{2}}{n}\sum_{k=n}^{2n-1} \frac{1}{k^{2}} =&6r_{2}n^{ \theta -1}\frac{1}{(2n)^{\theta }}2^{\theta } \sum_{k=n}^{2n-1}\frac{1}{k ^{2}} \\ \leqslant& 6r_{2}2^{\theta }n^{\theta -1}\sum _{k=n}^{2n-1}\frac{1}{k ^{2}} \frac{1}{k^{\theta }} \\ \leqslant& 6r_{2}2^{\theta }n^{\theta -1} \sum _{k=n}^{2n-1}\frac{a_{k}}{k^{\theta }} \leqslant 6r_{2}2^{ \theta }n^{\theta -1}\sum _{k= [ \frac{n}{c} ] }^{ \infty }\frac{a_{k}}{k^{\theta }}. \end{aligned}
It means that $$(a_{n})\in GM ( p,{}_{3}\beta (\theta ),r_{2} )$$. Furthermore,
\begin{aligned} \Biggl( \sum_{k=n}^{2n-1} \vert a_{k}-a_{k+r_{1}} \vert ^{p} \Biggr) ^{ \frac{1}{p}} \geqslant& \biggl( \sum_{k\in A_{n}} \vert a_{k}-a_{k+r _{1}} \vert ^{p} \biggr) ^{\frac{1}{p}}\geqslant \biggl( \sum_{k\in A_{n}} \biggl\vert \frac{1}{k^{3}}-\frac{3}{(k+r_{1})^{2}} \biggr\vert ^{p} \biggr) ^{\frac{1}{p}} \\ =& \biggl( \sum_{k\in A_{n}} \biggl\vert \frac{(k+r_{1})^{2}-3k^{2}}{(k+r _{1})^{2}k^{2}} \biggr\vert ^{p} \biggr) ^{\frac{1}{p}}= \biggl( \sum_{k\in A_{n}} \biggl\vert \frac{-2k^{2}+2kr_{1}+r_{1}^{2}}{(k+r _{1})^{2}k^{2}} \biggr\vert ^{p} \biggr) ^{\frac{1}{p}}. \end{aligned}
If $$n\geqslant 5r_{1}$$, then $$2n^{2}-2nr_{1}-r_{1}^{2}\geqslant (n+r _{1})^{2}$$. Whence for $$n\geqslant 5r_{1}$$,
\begin{aligned} \Biggl( \sum_{k=n}^{2n-1} \vert a_{k}-a_{k+r_{1}} \vert ^{p} \Biggr) ^{ \frac{1}{p}} \geqslant& \biggl( \sum_{k\in A_{n}} \biggl( \frac{2k ^{2}-2kr_{1}-r_{1}^{2}}{(k+r_{1})^{2}k^{2}} \biggr) ^{p} \biggr) ^{ \frac{1}{p}} \\ \geqslant& \frac{1}{(2n)^{2}} \biggl( \frac{n}{2r_{1}} \biggr) ^{ \frac{1}{p}}=\frac{1}{2^{2+\frac{1}{p}}r_{1}}n^{-2+\frac{1}{p}}. \end{aligned}
On the other hand,
$$n^{\theta -1}\sum_{k= [ \frac{n}{c} ] }^{\infty } \frac{a _{k}}{k^{\theta }}\leqslant n^{\theta -1}\sum _{k= [ \frac{n}{c} ] }^{\infty }\frac{3}{k^{2}} \frac{1}{k ^{\theta }}\leqslant 3n^{\theta -1}\sum _{k= [ \frac{n}{c} ] }^{\infty }\frac{1}{k^{2+ \theta }}\ll n^{-2}.$$
Therefore, the inequality
$$\Biggl( \sum_{k=n}^{2n-1} \vert \triangle _{r_{1}}a_{k} \vert ^{p} \Biggr) ^{\frac{1}{p}}\leqslant Cn^{\theta -1}\sum_{k= [ \frac{n}{c} ] }^{\infty } \frac{a_{k}}{k^{ \theta }}$$
cannot be satisfied because $$n^{\frac{1}{p}}\rightarrow \infty$$ as $$n\rightarrow \infty$$. □

### 4.4 Proof of Theorem 6

We prove the theorem for the case when $$\phi (x)=g(x)$$. We have
$$\bigl\Vert \omega _{\alpha ,r} \vert g \vert ^{s} \bigr\Vert _{L^{1}}=2 \int _{0}^{\pi }\omega _{\alpha ,r} ( x ) \bigl\vert g ( x ) \bigr\vert ^{s}\,dx.$$
For an odd r,
\begin{aligned} \int _{0}^{\pi }\omega _{\alpha ,r} ( x ) \bigl\vert g ( x ) \bigr\vert ^{s}\,dx =&\sum _{l=0}^{ [ r/2 ] } \int _{\frac{2l\pi }{r}}^{\frac{2l\pi }{r}+\frac{ \pi }{r}}\omega _{\alpha ,r} ( x ) \Biggl\vert \sum_{k=1}^{\infty }b_{k} \sin kx \Biggr\vert ^{s}\,dx \\ &{}+\sum_{l=0}^{ [ r/2 ] -1} \int _{\frac{2l\pi }{r}+\frac{\pi }{r}}^{\frac{2 ( l+1 ) \pi }{r}}\omega _{\alpha ,r} ( x ) \Biggl\vert \sum_{k=1}^{\infty }b_{k} \sin kx \Biggr\vert ^{s}\,dx \end{aligned}
(for $$r=1$$ the last sum should be omitted), and for an even r,
$$\int _{0}^{\pi }\omega _{\alpha ,r} ( x ) \bigl\vert g ( x ) \bigr\vert ^{s}\,dx=\sum _{l=0}^{ [ r/2 ] } \biggl( \int _{\frac{2l\pi }{r}}^{\frac{2l\pi }{r}+\frac{ \pi }{r}}+ \int _{\frac{2l\pi }{r}+\frac{\pi }{r}}^{\frac{2 ( l+1 ) \pi }{r}} \biggr) \omega _{\alpha ,r} ( x ) \Biggl\vert \sum_{k=1}^{\infty }b_{k} \sin kx \Biggr\vert ^{s}\,dx.$$
Now, we estimate the following integral:
\begin{aligned} \int _{\frac{2l\pi }{r}}^{\frac{2l\pi }{r}+\frac{\pi }{r}} \omega _{\alpha ,r} ( x ) \Biggl\vert \sum_{k=1}^{ \infty }b_{k} \sin kx \Biggr\vert ^{s}\,dx \ll& \Biggl( \int _{\frac{2l\pi }{r}}^{\frac{2l\pi }{r}+\frac{\pi }{r}} \omega _{\alpha ,r} ( x ) \Biggl\vert \sum_{k=1}^{n}b _{k}\sin kx \Biggr\vert ^{s}\,dx \\ &{}+ \int _{\frac{2l\pi }{r}}^{\frac{2l\pi }{r}+\frac{ \pi }{r}}\omega _{\alpha ,r} ( x ) \Biggl\vert \sum_{k=n+1}^{\infty }b_{k} \sin kx \Biggr\vert ^{s}\,dx \Biggr) \\ :=&I _{1}+I_{2}. \end{aligned}
By Lemma 2, for $$\alpha <1$$, we have
\begin{aligned} I_{1} =&\sum_{n=r}^{\infty } \int _{\frac{2l\pi }{r}+\frac{\pi }{n+1}}^{\frac{2l\pi }{r}+\frac{ \pi }{n}} \biggl( x-\frac{2l\pi }{r} \biggr) ^{-\alpha } \Biggl\vert \sum_{k=1}^{n}b_{k} \sin kx \Biggr\vert ^{s}\,dx \\ \ll& \sum_{n=r}^{\infty }n^{\alpha -2} \Biggl( \sum_{k=1} ^{n} \vert b_{k} \vert \Biggr) ^{s} \\ \leqslant& \sum _{n=1}^{\infty }n^{\alpha -2-\frac{s}{p}+2s} \vert b_{n} \vert ^{s}. \end{aligned}
(3)
Using Lemma 1 when $$m\rightarrow \infty$$ and the inequality
$$\frac{r}{\pi }x-2l\leqslant \biggl\vert \sin \frac{rx}{2} \biggr\vert \quad \text{for }x\in \biggl( \frac{2l\pi }{r}, \frac{2l\pi }{r}+\frac{ \pi }{r} \biggr) ,$$
we get
\begin{aligned} I_{2} =&\sum_{n=r}^{\infty } \int _{2l\pi /r+\pi /(n+1)} ^{2l\pi /r+\pi /n} \biggl( x-\frac{2l\pi }{r} \biggr) ^{-\alpha } \Biggl\vert \sum_{k=n+1}^{\infty }b_{k} \sin kx \Biggr\vert ^{s}\,dx \\ \ll& \sum_{n=r}^{\infty }n^{\alpha } \int _{2l\pi /r+\pi /(n+1)}^{2l\pi /r+\pi /n}\Biggl\vert \sum_{d=0}^{\infty } \Biggl(\sum_{k=2^{d+1}(n+1)} ^{2^{d+1}(n+1)-1+r}b_{k} \overset{\sim }{D}_{k,-r}(x) \\ &{}-\sum_{k=2^{d}(n+1)}^{2^{d}(n+1)+r-1}b_{k} \overset{\sim }{D}_{k,-r}(x)- \sum_{k=2^{d}(n+1)}^{2^{d+1}(n+1)-1} \triangle _{r} b_{k} \overset{\sim }{D}_{k,r}(x) \Biggr) \Biggr\vert ^{s}\,dx \\ \leqslant& \sum_{n=r}^{\infty }n^{\alpha } \int _{2l\pi /r+\pi /(n+1)}^{2l\pi /r+\pi /n} \frac{1}{(rx/\pi -2l)^{s}} \\ &{} \times \Biggl( \sum_{d=0} ^{\infty } \Biggl( \sum _{k=2^{d+1}(n+1)}^{2^{d+1}(n+1)-1+r} \vert b _{k} \vert +\sum_{k=2^{d}(n+1)}^{2^{d}(n+1)+r-1} \vert b_{k} \vert +\sum_{k=2^{d}(n+1)}^{2^{d+1}(n+1)-1} \vert \triangle _{r}b_{k} \vert \Biggr) \Biggr) ^{s}\,dx \\ \ll& \sum_{n=r}^{\infty }n^{\alpha +s-2} \Biggl( \sum_{d=0} ^{\infty } \Biggl( \sum _{k=2^{d+1}(n+1)}^{2^{d+1}(n+1)-1+r} \vert b _{k} \vert +\sum_{k=2^{d}(n+1)}^{2^{d}(n+1)+r-1} \vert b_{k} \vert +\sum_{k=2^{d}(n+1)}^{2^{d+1}(n+1)-1} \vert \triangle _{r}b_{k} \vert \Biggr) \Biggr) ^{s}. \end{aligned}
Further by Hölder inequality with $$p>1$$, we get
\begin{aligned} I_{2} \ll& \sum_{n=r}^{\infty }n^{\alpha +s-2} \Biggl(\sum_{d=0}^{\infty } \Biggl[ \Biggl( \sum_{k=2^{d}(n+1)} ^{2^{d+1}(n+1)-1} \vert \triangle _{r}b_{k} \vert ^{p} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{k=2^{d}(n+1)}^{2^{d+1}(n+1)-1}1 \Biggr) ^{1-\frac{1}{p}} \\ &{}+ \sum_{k=2^{d+1}(n+1)}^{2^{d+1}(n+1)-1+r} \vert b_{k} \vert +\sum_{k=2^{d}(n+1)}^{2^{d}(n+1)+r-1} \vert b_{k} \vert \Biggr] \Biggr)^{s} \\ \leqslant& \sum_{n=r}^{\infty }n^{\alpha +s-2} \Biggl( \sum_{d=0}^{\infty } \Biggl[ \Biggl( \sum_{k=2^{d}(n+1)}^{2^{d+1}(n+1)-1} \vert \triangle _{r} b_{k} \vert ^{p} \Biggr)^{\frac{1}{p}} \bigl(2^{d}(n+1) \bigr)^{1-\frac{1}{p}} \\ &{}+ \sum_{k=2^{d+1}(n+1)}^{2^{d+1}(n+1)-1+r} \vert b_{k} \vert +\sum_{k=2^{d}(n+1)}^{2^{d}(n+1)+r-1} \vert b_{k} \vert \Biggr] \Biggr)^{s}. \end{aligned}
Applying Lemma 5, we have
$$I_{2} \ll \sum_{n=r}^{\infty }n^{\alpha +s-2} \Biggl( \sum_{d=0}^{\infty } \Biggl[ \bigl( 2^{d}(n+1) \bigr) ^{1- \frac{1}{p}} \bigl( 2^{d}(n+1) \bigr) ^{\theta -1}\sum_{k=[\frac{2^{d}(n+1)}{c}]}^{\infty } \frac{ \vert b_{k} \vert }{k^{\theta }}+\sum_{k=2^{d}(n+1)}^{2^{d}(n+1)+r-1} \vert b_{k} \vert \Biggr] \Biggr) ^{s}.$$
From Lemma 6, we get
\begin{aligned} I_{2} \ll& \sum_{n=r}^{\infty }n^{\alpha +s-2}\Biggl( \sum_{d=0}^{\infty } \Biggl[ \bigl( 2^{d}(n+1) \bigr) ^{\theta -\frac{1}{p}}\sum _{k=[\frac{2^{d}(n+1)}{c}]}^{\infty }\frac{ \vert b_{k} \vert }{k^{\theta }} \\ &{}+ \frac{1}{1-2^{\theta -\frac{1}{p}}} \bigl( 2^{d}(n+1) \bigr) ^{\theta -\frac{1}{p}}\sum _{k= [ \frac{2^{d}(n+1)}{c} ] }^{\infty }\frac{ \vert b _{k} \vert }{k^{\theta }} \Biggr] \Biggr) ^{s} \\ \ll& \sum_{n=r}^{\infty }n^{\alpha +s-2+\theta s-\frac{s}{p}} \Biggl( \sum_{d=0}^{\infty } \bigl( 2^{d} \bigr) ^{\theta - \frac{1}{p}}\sum_{k=[\frac{2^{d}(n+1)}{c}]}^{\infty } \frac{ \vert b _{k} \vert }{k^{\theta }} \Biggr) ^{s}. \end{aligned}
If $$\theta -\frac{1}{p} <0$$, then
\begin{aligned} I_{2} \ll& \sum_{n=r}^{\infty }n^{\alpha +s-2+\theta s- \frac{s}{p}} \Biggl( \sum_{k=[\frac{n+1}{c}]}^{\infty } \frac{ \vert b _{k} \vert }{k^{\theta }} \Biggr) ^{s} \\ \ll& \sum_{n=r}^{\infty }n^{\alpha +s-2-\frac{s}{p}+\theta s} \Biggl( \sum_{k=[\frac{n}{c}]}^{n} \frac{ \vert b_{k} \vert }{k^{\theta }} \Biggr) ^{s}+\sum_{n=r}^{\infty }n^{\alpha +s-2-\frac{s}{p}+\theta s} \Biggl( \sum_{k=n}^{\infty } \frac{ \vert b_{k} \vert }{k^{\theta }} \Biggr) ^{s} \\ \leqslant& \sum_{n=1}^{\infty }n^{\alpha -2-\frac{s}{p}} \Biggl( \sum_{k=1}^{n}k \vert b_{k} \vert \Biggr) ^{s}+\sum _{n=1}^{\infty }n^{\alpha +s-2-\frac{s}{p}+\theta s} \Biggl( \sum _{k=n}^{ \infty }\frac{ \vert b_{k} \vert }{k^{\theta }} \Biggr) ^{s}. \end{aligned}
Now, we use Lemma 2 and get
\begin{aligned} I_{2} \ll& \sum_{n=1}^{\infty } \bigl( n^{\alpha -2-\frac{s}{p}} \bigr) ^{1-s}\bigl(n \vert b_{n} \vert \bigr)^{s} \Biggl( \sum _{k=n}^{\infty }k^{\alpha -2-\frac{s}{p}} \Biggr) ^{s} \\ &{}+\sum_{n=1}^{\infty } \bigl( n^{\alpha +s-2-\frac{s}{p}+\theta s} \bigr) ^{1-s} \biggl( \frac{ \vert b_{n} \vert }{n^{ \theta }} \biggr) ^{s} \Biggl( \sum_{k=1}^{n}k^{\alpha +s-2- \frac{s}{p}+\theta s} \Biggr) ^{s}. \end{aligned}
For $$1+\frac{s}{p}-\theta s-s<\alpha <1+\frac{s}{p}$$, we have
$$I_{2}\ll \sum_{n=1}^{\infty }n^{\alpha -2-\frac{s}{p}+2s} \vert b _{n} \vert ^{s}.$$
(4)
Now, we estimate the following integral:
\begin{aligned} \int _{\frac{2l\pi }{r}+\frac{\pi }{r}}^{\frac{2 ( l+1 ) \pi }{r}}\omega _{\alpha ,r} ( x ) \Biggl\vert \sum_{k=1}^{\infty }b_{k} \sin kx \Biggr\vert ^{s}\,dx \ll& \int _{\frac{2 ( l+1 ) \pi }{r}-\frac{\pi }{r}}^{\frac{2 ( l+1 ) \pi }{r}}\omega _{\alpha ,r} ( x ) \Biggl\vert \sum_{k=1}^{n}b_{k} \sin kx \Biggr\vert ^{s}\,dx \\ &{}+ \int _{\frac{2 ( l+1 ) \pi }{r}-\frac{\pi }{r}}^{\frac{2 ( l+1 ) \pi }{r}}\omega _{\alpha ,r} ( x ) \Biggl\vert \sum_{k=n+1}^{\infty }b_{k} \sin kx \Biggr\vert ^{s}\,dx \\ :=&I_{3}+I_{4}. \end{aligned}
By Lemma 2, for $$\alpha <1$$, we have
\begin{aligned} I_{3} =&\sum_{n=r}^{\infty } \int _{\frac{2 ( l+1 ) \pi }{r}-\frac{\pi }{n}}^{\frac{2 ( l+1 ) \pi }{r}-\frac{\pi }{n+1}} \biggl( \frac{2 ( l+1 ) \pi }{r}-x \biggr) ^{-\alpha } \Biggl\vert \sum_{k=1} ^{n}b_{k}\sin kx \Biggr\vert ^{s}\,dx \\ \ll& \sum_{n=1}^{\infty }n^{\alpha -2} \Biggl( \sum_{k=1} ^{n} \vert b_{k} \vert \Biggr) ^{s} \\ \ll& \sum _{n=1} ^{\infty }n^{\alpha +s-2} \vert b_{n} \vert ^{s} \leqslant \sum _{n=1}^{\infty }n^{\alpha -2-\frac{s}{p}+2s} \vert b _{n} \vert ^{s}. \end{aligned}
(5)
Using Lemma 1 with $$m\rightarrow \infty$$ and the inequality
$$2 ( l+1 ) -\frac{r}{\pi }x\leq \biggl\vert \sin \frac{rx}{2} \biggr\vert \quad \text{for }x\in \biggl( \frac{ ( 2l+1 ) \pi }{r}, \frac{2 ( l+1 ) \pi }{r} \biggr) ,$$
we have
$$I_{4}=\sum_{n=r}^{\infty } \int _{\frac{2 ( l+1 ) \pi }{r}-\frac{\pi }{n}}^{\frac{2 ( l+1 ) \pi }{r}-\frac{\pi }{n+1}} \biggl( \frac{2 ( l+1 ) \pi }{r}-x \biggr) ^{-\alpha } \Biggl\vert \sum_{k=n+1} ^{\infty }b_{k}\sin kx \Biggr\vert ^{s}\,dx,$$
and similarly as in the case $$I_{2}$$ we obtain
$$I_{4}\ll \sum_{n=1}^{\infty }n^{\alpha -2-\frac{s}{p}+2s} \vert b _{n} \vert ^{s}.$$
(6)
Finally, combining (3)–(6), we obtain that
$$\int _{-\pi }^{\pi }\omega _{\alpha ,r} ( x ) \bigl\vert g ( x ) \bigr\vert ^{s}\,dx\leqslant C\sum _{n=1}^{ \infty }n^{\alpha -2-\frac{s}{p}+2s} \vert b_{n} \vert ^{s}.$$
The case when $$\phi (x)=\sum_{k=1}^{\infty }b_{k}\cos kx$$ can be proved similarly. □

### 4.5 Proof of Theorem 7

We prove the theorem for the case where $$\phi (x)=\sum_{k=1} ^{\infty }b_{k}\sin kx$$. We follow the method adopted by Tikhonov . Note that if $$1-\theta s<\alpha <1+s$$, then $$\phi \in L^{1}$$. Namely, if $$s>1$$ then using Hölder inequality, we have
\begin{aligned} \int _{0}^{\pi } \bigl\vert \phi ( x ) \bigr\vert \,dx =& \int _{0}^{\pi } \bigl( \omega _{\alpha ,r}(x) \bigr) ^{ \frac{1}{s}} \bigl\vert \phi ( x ) \bigr\vert \biggl( \frac{1}{ \omega _{\alpha ,r}(x)} \biggr) ^{\frac{1}{s}}\,dx \\ \leqslant& \biggl( \int _{0}^{\pi }\omega _{\alpha ,r}(x) \bigl\vert \phi (x) \bigr\vert ^{s}\,dx \biggr) ^{\frac{1}{s}} \biggl( \int _{0}^{\pi } \bigl( \omega _{\alpha ,r}(x)^{- \frac{1}{s}} \bigr) ^{\frac{1}{1-\frac{1}{s}}}\,dx \biggr) ^{1- \frac{1}{s}}. \end{aligned}
We will show that $$\int _{0}^{\pi } ( \omega _{\alpha ,r}(x) ) ^{-\frac{1}{s-1}}\,dx<\infty$$. We can write
$$\int _{0}^{\pi } \bigl( \omega _{\alpha ,r}(x) \bigr) ^{- \frac{1}{s-1}}\,dx=\sum _{l=0}^{ [ \frac{r}{2} ] } \biggl( \int _{\frac{2l\pi }{r}}^{\frac{(2l+1)\pi }{r}} \biggl( x-\frac{2l \pi }{r} \biggr) ^{\frac{\alpha }{s-1}}\,dx+ \int _{\frac{(2l+1)\pi }{r}}^{\frac{2(l+1)\pi }{r}} \biggl( \frac{2(l+1) \pi }{r}-x \biggr) ^{\frac{\alpha }{s-1}}\,dx \biggr) ,$$
when r is an even number, and
\begin{aligned}&\int _{0}^{\pi } \bigl( \omega _{\alpha ,r}(x) \bigr) ^{- \frac{1}{s-1}}\,dx \\ &\quad =\sum _{l=0}^{ [ \frac{r}{2} ] } \int _{\frac{2l\pi }{r}}^{\frac{(2l+1)\pi }{r}} \biggl( x-\frac{2l \pi }{r} \biggr) ^{\frac{\alpha }{s-1}}\,dx+\sum_{l=0}^{ [ \frac{r}{2} ] -1} \int _{\frac{(2l+1)\pi }{r}}^{\frac{2(l+1) \pi }{r}} \biggl( \frac{2(l+1)\pi }{r}-x \biggr) ^{\frac{\alpha }{s-1}}\,dx, \end{aligned}
when r is an odd number.
Using integration by substitution, we get
\begin{aligned} \int _{0}^{\pi } \bigl( \omega _{\alpha ,r}(x) \bigr) ^{- \frac{1}{s-1}}\,dx =&\sum _{l=0}^{ [ \frac{r}{2} ] } \biggl( \int _{0}^{\frac{\pi }{r}}y^{\frac{\alpha }{s-1}}\,dy+ \int _{0}^{\frac{\pi }{r}}y^{\frac{\alpha }{s-1}}\,dy \biggr) \\ =&2 \biggl( \biggl[ \frac{r}{2} \biggr] +1 \biggr) \frac{s-1}{\alpha +s-1} \biggl( \frac{\pi }{r} \biggr) ^{\frac{\alpha +s-1}{s-1}}, \end{aligned}
when r is an even number, and
\begin{aligned} \int _{0}^{\pi } \bigl( \omega _{\alpha ,r}(x) \bigr) ^{- \frac{1}{s-1}}\,dx =&\sum _{l=0}^{ [ \frac{r}{2} ] } \int _{0}^{\frac{\pi }{r}}y^{\frac{\alpha }{s-1}}\,dy+\sum _{l=0}^{ [ \frac{r}{2} ] -1} \int _{0}^{\frac{ \pi }{r}}y^{\frac{\alpha }{s-1}}\,dx \\ =& \biggl( 2 \biggl[ \frac{r}{2} \biggr] +1 \biggr) \frac{s-1}{\alpha +s-1} \biggl( \frac{\pi }{r} \biggr) ^{\frac{ \alpha +s-1}{s-1}}, \end{aligned}
when r is an odd number.
If $$s=1$$ then $$\alpha >0$$ and
\begin{aligned} \int _{0}^{\pi } \bigl\vert \phi ( x ) \bigr\vert \,dx =& \int _{0}^{\pi }\omega _{\alpha ,r}(x) \bigl\vert \phi (x) \bigr\vert \frac{1}{ \omega _{\alpha ,r}(x)}\,dx \\ \leqslant& \underset{x}{\sup }\frac{1}{\omega _{\alpha ,r}(x)} \int _{0}^{\pi }\omega _{\alpha ,r}(x) \bigl\vert \phi (x) \bigr\vert \,dx= \biggl( \frac{\pi }{r} \biggr) ^{\alpha } \int _{0}^{\pi } \omega _{\alpha ,r}(x) \bigl\vert \phi (x) \bigr\vert \,dx. \end{aligned}
Further, integrating ϕ, we have
$$F ( x ) := \int _{0}^{x}\phi ( t )\,dt= \sum _{n=1}^{\infty }\frac{b_{n}}{n} ( 1-\cos nx ) =2 \sum_{n=1}^{\infty }\frac{b_{n}}{n}\sin ^{2}\frac{nx}{2},$$
and consequently,
$$F \biggl( \frac{\pi }{k} \biggr) \geqslant \sum _{n= [ k/2 ] }^{k}\frac{b_{n}}{n}.$$
(7)
Since $$(b_{n})\in GM ( p,{}_{3}\beta (\theta ),r )$$ and using Lemma 4, we get for $$\theta -\frac{1}{p}<0$$ that
\begin{aligned} b_{v} \leqslant& \sum_{k=v}^{v+r-1}b_{l}= \sum_{d=0}^{ \infty }\sum _{k=2^{d}v}^{2^{d+1}v-1} \vert \triangle _{r}b_{k} \vert \leqslant \sum _{d=0}^{\infty }2^{d}v \Biggl[ \frac{1}{2^{d}v} \sum_{k=2^{d}v}^{2^{d+1}v-1} \vert \triangle _{r}b_{k} \vert ^{p} \Biggr] ^{\frac{1}{p}} \\ \leqslant& C\sum_{d=0}^{\infty } \bigl(2^{d}v\bigr)^{\theta -\frac{1}{p}} \sum _{k= [ \frac{2^{d}v}{c} ] }^{\infty }\frac{b _{k}}{k^{\theta }} \leqslant Cv^{\theta -\frac{1}{p}}\sum_{d=0} ^{\infty } \bigl( 2^{\theta -\frac{1}{p}} \bigr) ^{d}\sum _{k= [ \frac{v}{c} ] }^{\infty }\frac{b_{k}}{k^{ \theta }} \\ \leqslant& \frac{1}{1-2^{\theta -\frac{1}{p}}}Cv^{\theta -\frac{1}{p}} \sum _{k= [ \frac{v}{c} ] }^{\infty }\frac{b_{k}}{k ^{\theta }}\ll v^{\theta -\frac{1}{p}}\sum_{k= [ \frac{v}{c} ] }^{\infty } \frac{b_{k}}{k^{ \theta }} \leqslant Cv^{\theta -\frac{1}{p}}\sum _{d=0}^{\infty } \biggl( 2^{d+1} \biggl[ \frac{v}{c} \biggr] \biggr) ^{1-\theta } \sum _{k=2^{d} [ \frac{v}{c} ] }^{2^{d+1} [ \frac{v}{c} ] }\frac{b_{k}}{k}. \end{aligned}
Using (7) yields
\begin{aligned} b_{v} \ll& v^{\theta -\frac{1}{p}}\sum_{d=0}^{\infty } \biggl( 2^{d} \biggl[ \frac{v}{c} \biggr] \biggr) ^{1-\theta }F \biggl( \frac{ \pi }{2^{d+1} [ \frac{v}{c} ] } \biggr) \ll v^{\theta - \frac{1}{p}} \sum_{d=0}^{\infty } \biggl( 2^{d} \biggl[ \frac{v}{c} \biggr] \biggr) ^{-\theta }\sum_{k=2^{d} [ \frac{v}{c} ] }^{2^{d+1} [ \frac{v}{c} ] -1}F \biggl( \frac{\pi }{k} \biggr) \\ \ll& v^{\theta -\frac{1}{p}}\sum_{k= [ \frac{v}{c} ] }^{\infty } \frac{1}{k^{\theta }}F \biggl( \frac{\pi }{k} \biggr) . \end{aligned}
Elementary calculations give
\begin{aligned} \sum_{k=1}^{\infty }k^{\alpha -2+\frac{s}{p}}b_{k}^{s} \ll& \sum_{k=1}^{\infty }k^{\alpha -2+\frac{s}{p}+(\theta - \frac{1}{p})s} \Biggl( \sum_{v= [ \frac{k}{c} ] } ^{\infty } \frac{1}{v^{\theta }}F \biggl( \frac{\pi }{v} \biggr) \Biggr) ^{s} \\ \ll& \sum_{k=1}^{\infty }k^{\alpha -2} \Biggl( \sum_{v= [ \frac{k}{c} ] }^{k}F \biggl( \frac{\pi }{v} \biggr) \Biggr) ^{s}+\sum _{k=1}^{\infty }k^{ \alpha -2+\theta s} \Biggl( \sum _{v=k}^{\infty }\frac{1}{v^{ \theta }}F \biggl( \frac{\pi }{v} \biggr) \Biggr) ^{s} \\ \ll& \sum_{k=1}^{\infty }k^{\alpha -2-s} \Biggl( \sum_{v= [ \frac{k}{c} ] }^{k}vF \biggl( \frac{\pi }{v} \biggr) \Biggr) ^{s}+\sum _{k=1}^{\infty }k^{ \alpha -2+\theta s} \Biggl( \sum _{v=k}^{\infty }\frac{1}{v^{ \theta }}F \biggl( \frac{\pi }{v} \biggr) \Biggr) ^{s}. \end{aligned}
Using Lemma 2, for $$1-\theta s<\alpha <1+s$$, we have
$$\sum_{k=1}^{\infty }k^{\alpha -2-s} \Biggl( \sum_{v= [ \frac{k}{c} ] }^{k}vF \biggl( \frac{\pi }{v} \biggr) \Biggr) ^{s}\ll \sum _{k=1}^{\infty }k ^{(\alpha -2-s)(1-s)} \biggl( kF \biggl( \frac{\pi }{k} \biggr) \biggr) ^{s} \Biggl( \sum _{v=k}^{\infty }v^{\alpha -2-s} \Biggr) ^{s}$$
and
$$\sum_{k=1}^{\infty }k^{\alpha -2+\theta s} \Biggl( \sum_{v=k}^{\infty } \frac{1}{v^{\theta }}F \biggl( \frac{\pi }{v} \biggr) \Biggr) ^{s}\ll \sum_{k=1}^{\infty }k^{(\alpha -2+\theta s)(1-s)} \biggl( \frac{1}{k^{\theta }}F \biggl( \frac{\pi }{k} \biggr) \biggr) ^{s} \Biggl( \sum_{v=1}^{k}v^{\alpha -2+\theta s} \Biggr) ^{s}.$$
Therefore, for $$1-\theta s<\alpha <1+s$$, we get
$$\sum_{k=1}^{\infty }k^{\alpha -2+\frac{s}{p}}b_{k}^{s} \ll \sum_{k=1}^{\infty }k^{\alpha -2+s} \Biggl(F \biggl( \frac{\pi }{k} \biggr) \Biggr) ^{s}.$$
Denoting by $$d_{v}:=\int _{\frac{\pi }{v+1}}^{\frac{\pi }{v}} \vert \phi ( x ) \vert \,dx$$, we get
$$\sum_{k=1}^{\infty }k^{\alpha -2+\frac{s}{p}}b_{k}^{s} \ll \sum_{k=1}^{\infty }k^{\alpha -2+s} \Biggl( \sum_{v=k} ^{\infty }d_{v} \Biggr) ^{s}.$$
By Lemma 2, for $$\alpha >1-s$$, we obtain
\begin{aligned} \sum_{k=1}^{\infty }k^{\alpha -2+s} \Biggl( \sum_{v=k} ^{\infty }d_{v} \Biggr) ^{s} \ll& \sum_{k=1}^{\infty }k^{(\alpha -2+s)(1-s)}d_{k}^{s} \Biggl( \sum_{v=1}^{k}v^{\alpha -2+s} \Biggr) ^{s} \\ \ll& \sum_{k=1}^{\infty }k^{(\alpha -2+s)(1-s)}k^{(\alpha -2+s+1)s}d _{k}^{s}=\sum_{k=1}^{\infty }k^{\alpha -2+2s}d_{k}^{s}. \end{aligned}
Applying Hölder inequality when $$s>1$$, we have
$$d_{k}^{s}\ll \frac{1}{k^{2(s-1)}} \int _{\frac{\pi }{(k+1)}}^{\frac{ \pi }{k}} \bigl\vert \phi (x) \bigr\vert ^{s}\,dx.$$
Finally, using the latter estimate, we get
\begin{aligned} \sum_{k=1}^{\infty }k^{\alpha -2+\frac{s}{p}}b_{k}^{s} \ll& \sum_{k=1}^{\infty }k^{\alpha -2+2s}d_{k}^{s} \\ \leqslant& \sum_{k=1}^{r}k^{\alpha -2+2s} \biggl( \int _{\frac{\pi }{(k+1)}}^{\frac{\pi }{k}} \bigl\vert \phi (x) \bigr\vert \,dx \biggr) ^{s}+\sum_{k=r}^{\infty }k^{\alpha } \int _{\frac{\pi }{(k+1)}}^{\frac{\pi }{k}} \bigl\vert \phi (x) \bigr\vert ^{s}\,dx \\ \ll& \biggl( \int _{0}^{\pi } \bigl\vert \phi ( x ) \bigr\vert \,dx \biggr) ^{s}+\sum_{k=r}^{\infty } \int _{\frac{\pi }{k+1}}^{\frac{\pi }{k}}x^{-\alpha } \bigl\vert \phi ( x ) \bigr\vert ^{s}\,dx \\ \leqslant& \biggl( \int _{0}^{\pi } \bigl\vert \phi ( x ) \bigr\vert \,dx \biggr) ^{s}+ \int _{0}^{\pi }\omega _{\alpha ,r} ( x ) \bigl\vert \phi ( x ) \bigr\vert ^{p}\,dx< \infty . \end{aligned}

The case when $$\phi (x)=\sum_{k=1}^{\infty }b_{k}\cos kx$$ can by proved similarly. □

## 5 Conclusions

We have introduced two new classes of p-bounded variation sequences, $$\overline{GM}(p,\beta ,r)$$ and $$GM(p,\beta ,r)$$, where $$\beta := ( \beta _{n} )$$ is a nonnegative sequence, p a positive real number, $$r\in \mathbb{N}$$, $$\theta \in (0,1]$$. Moreover, we have studied properties of such classes and obtained a sufficient and necessary condition for weighted integrability of functions defined by trigonometric series with coefficients belonging to these classes. In particular, from our theorems we derive all related earlier results.

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