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. □