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

  • Bogdan SzalEmail author
  • Maciej Kubiak
Open Access
Research
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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 [2]. Later, Theorem 1 was showed in [7] 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 [12] 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 [13].
In [3] 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 [3] and Leindler [7], 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 [9]):

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 [9] 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 [9]. 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 [8].

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 [9] (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 [3] 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

([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 [9]. 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.

Notes

Acknowledgements

The publication costs of this article were covered by University of Zielona Góra.

Availability of data and materials

Not applicable.

Authors’ contributions

The study was carried out in collaboration with equal responsibility. All authors read and approved the final manuscript.

Funding

Not applicable.

Competing interests

The authors declare that they have no competing interests.

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Authors and Affiliations

  1. 1.Faculty of Mathematics, Computer Science and EconometricsUniversity of Zielona GóraZielona GóraPoland

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