1 Introduction

Let \(p=p(x)\) will be a measurable \(2\pi \) - periodic function, \(p_{-}=\inf \left\{ p(x):x\in {\mathbb {R}} \right\} \), \(p^{-}\)\(=\sup \left\{ p(x):x\in {\mathbb {R}} \right\} \), 1 \(\le p_{-}\le p\le \)\(p^{-}\)\(<\infty \) and \(L_{2\pi }^{p} \) will be the space of all measurable \(2\pi \) - periodic functions f such that \(\int _{-\pi }^{\pi }\left| f(x)\right| ^{p(x)}dx<\infty \).

Putting

$$\begin{aligned} \left\| f\right\| _{p}=\left\| f\left( \cdot \right) \right\| _{p\left( \cdot \right) }=\inf \left\{ \alpha >0:\int _{-\pi }^{\pi }\left| \frac{f(x)}{\alpha }\right| ^{p(x)}dx\le 1\right\} \end{aligned}$$

we turn \(L_{2\pi }^{p}\) into a Banach space (see [8]). We write \(\Pi _{2\pi }\) for the set of all \(2\pi \) - periodic variable exponents \(p=p(x)\ge 1\) satisfying the condition

$$\begin{aligned} \left| p(x)-p(y)\right| \ln \frac{2\pi }{\left| x-y\right| } =O\left( 1\right) \text {, }\quad \left( x,y\in \left[ -\pi ,\pi \right] \right) . \end{aligned}$$

For \(f\in L_{2\pi }^{1}\) we will consider the trigonometric Fourier series

$$\begin{aligned} S\left[ f\right] (x):=\frac{a_{0}(f)}{2}+\sum _{\nu =1}^{\infty }(a_{\nu }(f)\cos \nu x+b_{\nu }(f)\sin \nu x) \end{aligned}$$

with the partial sums \(S_{k}\left[ f\right] .\)

We will also need some notations on methods of summability of the series S[f] and modulus of continuity of f in the space \(L_{2\pi }^{p}.\)

If \(A:=\left( a_{n,k}\right) _{0\le n,k\,<\infty }\) be an infinite matrix of real numbers such that

$$\begin{aligned} a_{n,k}\ge 0\text { when }\ k,n=0,1,2,..., \ \lim _{n\rightarrow \infty }a_{n,k}=0\ \text { and }\sum _{k=0}^{\infty }a_{n,k}=1 \end{aligned}$$

or \(A_{0}:=\left( a_{n,k}\right) _{0\le k\le n<\infty },\) where

$$\begin{aligned} a_{n,k}=0\quad \text { when }\ k>n, \end{aligned}$$

then

$$\begin{aligned} T_{n,A}^{\text { }}\left[ f\right] \left( x\right) :=\sum _{k=0}^{\infty }a_{n,k}S_{k}\left[ f\right] \left( x\right) \ \ \ \left( n=0,1,2,...\right) \end{aligned}$$

or

$$\begin{aligned} T_{n,A_{0}}^{\text { }}\left[ f\right] \left( x\right) :=\sum _{k=0}^{n}a_{n,k}S_{k}\left[ f\right] \left( x\right) \ \ \ \left( n=0,1,2,...\right) , \end{aligned}$$

respectively.

Let \(f\in L_{2\pi }^{p}\) with \(p=p(x)\)\(\in \Pi _{2\pi }\). For \(n=0,1,2,...\) we denote the best approximation of f by

$$\begin{aligned} E_{n}(f)_{p}=\inf _{T_{n}}\left\| f-T_{n}\right\| _{p}, \end{aligned}$$

where the infimum is taken over all trigonometric polynomials \( T_{n}(x)=\sum _{k=-n}^{n}c_{k}e^{ikx}\) of the degrees at most n but the role of modulus of continuity of \(f\in L_{2\pi }^{p}\) in the case of a variable exponent \(p=p(x)\) play the function

$$\begin{aligned} \Omega f(\delta )_{p}=\sup _{0<h\le \delta }\left\| f\left( \cdot \right) -\frac{1}{h}\int _{-h}^{h}f(\cdot +t)dt\right\| _{p\left( \cdot \right) }. \end{aligned}$$

It follows from the results of [9] that if \(p=p(x)\)\(\in \Pi _{2\pi }\) , then the function \(\Omega f(\cdot )_{p}\) is continuous on \( [0;\infty )\) and \(\lim _{\delta \rightarrow 0}\Omega f(\delta )_{p}=0.\) It also follows from the above definition that \(\Omega f(\delta )_{p}\) is a non-decreasing function of \(\delta \). We will call \(\Omega f(\cdot )_{p}\) the modulus of continuity of a function \(f\in L_{2\pi }^{p}\). With such modulus of continuity we can define the following class of functions:

$$\begin{aligned} Lip_{p}\left( \omega ,M\right) =\left\{ f\in L_{2\pi }^{p}:\Omega f(\delta )_{p}\le M\omega (\delta ),\text { }\delta >0\right\} , \end{aligned}$$

where \(\omega \) is a function of modulus of continuity type on the interval \( [0,2\pi ],\) i.e. a nondecreasing continuous function having the following properties: \(\omega \left( 0\right) =0,\)\(\omega \left( \delta _{1}+\delta _{2}\right) \le \omega \left( \delta _{1}\right) +\omega \left( \delta _{2}\right) \) for any \(0\le \delta _{1}\le \delta _{2}\le \delta _{1}+\delta _{2}\le 2\pi \) and M is some positive constant.

It was proved in work [10, Theorem 6.1] that if the variable exponent \(p=p(x)\)\(\in \Pi _{2\pi }\) and \(f\in L_{2\pi }^{p}\), then the following Jackson-type inequality holds:

$$\begin{aligned} E_{n}(f)_{p}=O\left( 1\right) \Omega f\left( \frac{1}{n+1}\right) _{p}. \end{aligned}$$

In this paper we generalize and improve the results of A. Guven, D. Israfilov, Xh. Z. Krasniqi and T. N. Shakh-Emirov. We will consider the spaces \(L_{2\pi }^{p(x)}\) with \(p\left( x\right) \ge 1\) and as a measure of approximation we will use the modulus of continuity constructed by the Steklov functions without the modulus of the increments of functions.

2 Main Results

At the begin we prove the general result.

Theorem 1

Let \(f\in L_{2\pi }^{p}\) with \(p=p(x)\)\(\in \Pi _{2\pi }\). If the conditions

$$\begin{aligned} \sum _{k=0}^{\infty }(k+1)^{\beta }\left| \frac{a_{n,k}}{(k+1)^{\beta }}- \frac{a_{n,k+1}}{(k+2)^{\beta }}\right| =O\left( \frac{1}{n+1}\right) \end{aligned}$$
(1)

for some \(\beta \ge 0\) and

$$\begin{aligned} \sum _{k=0}^{\infty }(k+1)a_{n,k}=O\left( n+1\right) \end{aligned}$$
(2)

hold, then

$$\begin{aligned} \left\| T_{n,A}^{\text { }}\left[ f\right] -f\right\| _{p}=O\left( \Omega f\left( \frac{1}{n+1}\right) _{p} +\sum _{k=0}^{n}a_{n,k}\Omega f\left( \frac{1}{k+1}\right) _{p}\right) . \end{aligned}$$

We observe that all of the lower triangular matrices satisfy (2). In this case we have:

Theorem 2

Let \(f\in L_{2\pi }^{p}\) with \(p=p(x)\)\(\in \Pi _{2\pi }\). If the conditions

$$\begin{aligned} \sum _{k=0}^{n-1}(k+1)^{\beta }\left| \frac{a_{n,k}}{(k+1)^{\beta }}- \frac{a_{n,k+1}}{(k+2)^{\beta }}\right| =O\left( a_{n,n}\right) \end{aligned}$$
(3)

for some \(\beta \ge 0\) and

$$\begin{aligned} (n+1)a_{n,n}=O(1) \end{aligned}$$
(4)

hold, then

$$\begin{aligned} \left\| T_{n,A_{0}}^{\text { }}\left[ f\right] -f\right\| _{p}=O\left( \sum _{k=0}^{n}a_{n,k}\Omega f\left( \frac{1}{k+1}\right) _{p}\right) . \end{aligned}$$

Next we will consider some special cases and we will approximate of \(f\in Lip_{p}\left( \omega ,M\right) \) with \(p=p(x)\)\(\in \Pi _{2\pi }\).

Theorem 3

Let \(f\in Lip_{p}\left( \omega ,M\right) \) with \(p=p(x)\)\(\in \Pi _{2\pi }\). If the conditions (1) for some \(\beta >0\) and (2) hold, then

$$\begin{aligned} \left\| T_{n,A}^{\text { }}\left[ f\right] -f\right\| _{p}=O\left( \omega \left( \frac{1}{n+1}\right) \right) . \end{aligned}$$

In the similar way we obtain, by Theorem 2 the following

Theorem 4

Let \(f\in Lip_{p}\left( \omega ,M\right) \) with \(p=p(x)\)\(\in \Pi _{2\pi }\). If the conditions (3) for some \(\beta >0\) and (4) hold, then

$$\begin{aligned} \left\| T_{n,A_{0}}^{\text { }}\left[ f\right] -f\right\| _{p}=O\left( \omega \left( \frac{1}{n+1}\right) \right) . \end{aligned}$$

Finally, we have some examples and remarks.

Example 1

One can easily verify that \(a_{n,k}=e^{-n}\sum _{j=k}^{\infty }\frac{n^{j}}{ \left( j+1\right) !},\) where \(n,k=0,1,2,...,\) satisfies the conditions (1) for any \(\beta \ge 0\) and (2).

Example 2

We can verify that \(a_{n,k}=\frac{(k+1)^{\beta }-k^{\beta }}{(n+1)^{\beta }}\) for \(k\le n\) and \(a_{n,k}=0\) for \(k>n\), where \(n,k=0,1,2,...,\) satisfies the conditions (3) for any \(\beta >1\) and (4).

Remark 1

If \(((k+1)^{-\beta }a_{n,k})\in HBVS\) for some \(\beta \ge 0\), where

$$\begin{aligned} HBVS:=\left\{ \left( c_{k}\right) \subset [0,\infty ):\sum _{k=0}^{m-1}\left| c_{k}-c_{k+1}\right| =O\left( c_{m}\right) \text { for any }m=1,2,3,...,n\right\} , \end{aligned}$$

then the condition (3) holds.

Remark 2

Let \(f\in Lip_{p}\left( \omega ,M\right) \) with \(\omega \left( \delta \right) =\delta ^{\alpha },\) where \(\alpha \in (0,1]\ \)and \(p=p(x)\)\(\in \Pi _{2\pi }\). From Theorem 4 the results of [1,2,3,4,5, 7] and [11, Theorem 5 (v)], [6, Theorem 5 (vi)] follow at once in the more general and improved forms.More precisely, in the mentioned papers [2,3,4] there is considered the following modulus of continuity

$$\begin{aligned} \Omega _{p}f(\delta )=\sup _{0<h\le \delta }\left\| \frac{1}{h} \int _{-h}^{h}\left| f(\cdot +t)-f\left( \cdot \right) \right| dt\right\| _{p\left( \cdot \right) } \end{aligned}$$

greater than \(\Omega f(\delta )_{p}\) and therefore the class \(Lip_{p}\left( \omega ,M\right) \) constructed by modulus \(\Omega f(\delta )_{p}\) is wider then such one constructed by \(\Omega _{p}f(\delta ).\) Moreover, in these papers and also in [7] there is the assumption \(p\left( x\right) \ge p_{-}>1,\) since for \(p_{-}=1\) the quantity \(\Omega _{p}f(\delta )\) may not tend to 0 when \(\delta \rightarrow 0\) for some \(f\in \)\(L_{2\pi }^{p}.\) In the papers [1, 5, 11, Theorem 5 (v)], [6, Theorem 5 (vi)] where \(p\left( x\right) =const.\) there is used yet bigger the integral modulus of continuity. We can note that in all above cited papers the simultaneous assumptions \(p_{-}=1\) and \(\alpha =1\) are impossible. Furthermore, in Theorem 4 there is considered very general class of matrices defined of the means \(T_{n,A}^{\text { }}\left[ f\right] \) which special cases occur in the above cited papers.

3 Lemmas

We start with two lemmas from the papers of I. I. Sharapudinov. The first one from [10, Lemma 2.3] will be formulated without a proof but the second one similar to that from the paper [9] but in more general form will be proved.

Lemma 1

(see [10, Lemma 2.3]). Let \(p=p(x)\) be a measurable \(2\pi \)—periodic function, such that 1 \(\le p_{-}\le p\le \)\(p^{-}\)\(<\infty \) and g be a function of two variables, \(2\pi \) - periodic and measurable on \(\left[ -\pi ,\pi \right] \times \left[ -\pi ,\pi \right] .\) Then

$$\begin{aligned} \left\| \int _{-\pi }^{\pi }g(\cdot ,t)dt\right\| _{p}\le 2\int _{-\pi }^{\pi }\left\| g(\cdot ,t)\right\| _{p}dt. \end{aligned}$$

Lemma 2

(cf. [9, Lemma 3.1]). Let \(f\in L_{2\pi }^{p}\) with \(p=p(x)\)\(\in \Pi _{2\pi }\). Then

$$\begin{aligned} \left\| f_{\frac{1}{2\lambda }}\left( \cdot +\tau \right) \right\| _{p}=O\left( 1\right) \left\| f\left( \cdot \right) \right\| _{p} \end{aligned}$$

for every real \(\tau ,\) where \(f_{\frac{1}{2\lambda }}(\tau )=\lambda \int _{- \frac{1}{2\lambda }+\tau }^{\frac{1}{2\lambda }+\tau }f\left( t\right) dt\) with \(\lambda >1\).

Proof

Let \(h=\frac{1}{\left[ \lambda \right] },\)

$$\begin{aligned} x_{k}= & {} \left( kh-1\right) \pi ,{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,}k=0,\pm 1,\pm 2,..., \end{aligned}$$
(5)
$$\begin{aligned} s_{k}= & {} \min \left\{ p\left( x\right) :x_{k-1}\le x\le x_{k+2}\right\} , \text { }k=0,\pm 1,\pm 2,..., \end{aligned}$$
(6)
$$\begin{aligned} p_{t}\left( x\right)= & {} s_{k} \ \ \left( x_{k}-t\le x<x_{k+1}-t\right) , \ \quad k=0,\pm 1,\pm 2,..., \end{aligned}$$
(7)

whence \(p_{t}\left( x\right) =p_{0}\left( x+t\right) \) is a \(2\pi \) - periodic step function such that,

$$\begin{aligned} s_{k-\left( m+1\right) }=p_{t}\left( x\right) \le p\left( x\right) \end{aligned}$$

for \(m\pi h\le t\le \left( m+2\right) \pi h,\) since \(x_{k-\left( m+2\right) }\le x\le x_{k-\left( m-1\right) }\) for such t.

Let \(\left\| f\left( \cdot \right) \right\| _{p}\le 1\) and \(\tau \in {\mathbb {R}} .\) There exists an integer m such that

$$\begin{aligned} m\pi h\le \tau \le \left( m+2\right) \pi h. \end{aligned}$$

It is clear that

$$\begin{aligned} J= & {} \int _{-\pi }^{\pi }\left| f_{\frac{1}{2\lambda }}\left( x+\tau \right) \right| ^{p\left( x\right) }dx=\int _{-\pi }^{\pi }\left| \lambda \int _{-\frac{1}{2\lambda }+x+\tau }^{\frac{1}{2\lambda }+x+\tau }f\left( t\right) dt\right| ^{p\left( x\right) }dx \\= & {} \int _{-\pi -\left( m+1\right) \pi h}^{\pi -\left( m+1\right) \pi h}\left| \lambda \int _{-\frac{1}{2\lambda }+x+\tau }^{\frac{1}{2\lambda } +x+\tau }f\left( t\right) dt\right| ^{p\left( x\right) }dx\\= & {} \sum _{k=0}^{2[\lambda ]-1}\int _{x_{k-\left( m+1\right) }}^{x_{k-m}}\left| \lambda \int _{-\frac{1}{2\lambda }+x+\tau }^{\frac{1}{ 2\lambda }+x+\tau }f\left( t\right) dt\right| ^{p\left( x\right) }dx \\= & {} \sum _{k=0}^{2[\lambda ]-1}\int _{x_{k-\left( m+1\right) }}^{x_{k-m}}\left| \lambda \int _{-\frac{1}{2\lambda }+x+\tau }^{\frac{1}{ 2\lambda }+x+\tau }f\left( t\right) dt\right| ^{p\left( x\right) -s_{k-\left( m+1\right) }+s_{k-\left( m+1\right) }}dx. \end{aligned}$$

Further, similarly as in [9, Theorem 2.1],

$$\begin{aligned} \left| \int _{-\frac{1}{2\lambda }+x+\tau }^{\frac{1}{2\lambda }+x+\tau }f\left( t\right) dt\right|\le & {} \int _{-\frac{1}{2\lambda }+x+\tau }^{ \frac{1}{2\lambda }+x+\tau }\left| f\left( t\right) \right| dt\le \int _{-\pi +x+\tau }^{\pi +x+\tau }\left| f\left( t\right) \right| dt=\int _{-\pi }^{\pi }\left| f\left( t\right) \right| dt \\= & {} \left\| f\left( \cdot \right) \right\| _{1}\le (2\pi +1)\left\| f\left( \cdot \right) \right\| _{p}\le (2\pi +1) \end{aligned}$$

and for \(x_{k-\left( m+1\right) }\le x\le x_{k-m}\)

$$\begin{aligned} p(x)-s_{k-\left( m+1\right) }=O\left( \frac{1}{\ln \lambda }\right) , \end{aligned}$$

whence

$$\begin{aligned} \lambda ^{p\left( x\right) -s_{k-\left( m+1\right) }}=O\left( \lambda ^{1/\ln \lambda }\right) =O\left( 1\right) . \end{aligned}$$

Next, by the Jensen inequality,

$$\begin{aligned} J= & {} O\left( 1\right) \sum _{k=0}^{2[\lambda ]-1}\int _{x_{k-\left( m+1\right) }}^{x_{k-m}}\left| \lambda \int _{-\frac{1}{2\lambda }+x+\tau }^{\frac{1}{ 2\lambda }+x+\tau }f\left( t\right) dt\right| ^{s_{k-\left( m+1\right) }}dx \\= & {} O\left( 1\right) \sum _{k=0}^{2[\lambda ]-1}\int _{x_{k-\left( m+1\right) }}^{x_{k-m}}\left[ \lambda \int _{-\frac{1}{2\lambda }+x+\tau }^{\frac{1}{ 2\lambda }+x+\tau }\left| f\left( t\right) \right| ^{s_{k-\left( m+1\right) }}dt\right] dx \\= & {} O\left( 1\right) \sum _{k=0}^{2[\lambda ]-1}\int _{x_{k-\left( m+1\right) }}^{x_{k-m}}\left[ \lambda \int _{-\frac{1}{2\lambda }+\tau }^{\frac{1}{ 2\lambda }+\tau }\left| f\left( t+x\right) \right| ^{s_{k-\left( m+1\right) }}dt\right] dx \\= & {} O\left( 1\right) \lambda \int _{-\frac{1}{2\lambda }+\tau }^{\frac{1}{ 2\lambda }+\tau }\left[ \sum _{k=0}^{2[\lambda ]-1}\int _{x_{k-\left( m+1\right) }}^{x_{k-m}}\left| f\left( t+x\right) \right| ^{s_{k-\left( m+1\right) }}dx\right] dt \\= & {} O\left( 1\right) \lambda \int _{-\frac{1}{2\lambda }-\tau }^{\frac{1}{ 2\lambda }-\tau }\left[ \sum _{k=0}^{2[\lambda ]-1}\int _{x_{k-\left( m+1\right) }}^{x_{k-m}}\left| f\left( x-t\right) \right| ^{s_{k-\left( m+1\right) }}dx\right] dt \\= & {} O\left( 1\right) \lambda \int _{-\frac{1}{2\lambda }-\tau }^{\frac{1}{ 2\lambda }-\tau }\left[ \sum _{k=0}^{2[\lambda ]-1}\int _{x_{k-\left( m+1\right) }-t}^{x_{k-m}-t}\left| f\left( x\right) \right| ^{s_{k-\left( m+1\right) }}dx\right] dt \\= & {} O\left( 1\right) \lambda \int _{-\frac{1}{2\lambda }-\tau }^{\frac{1}{ 2\lambda }-\tau }\left[ \sum _{k=0}^{2[\lambda ]-1}\int _{x_{k-\left( m+1\right) }-t}^{x_{k-m}-t}\left| f\left( x\right) \right| ^{p_{t}\left( x\right) }dx\right] dt \\= & {} O\left( 1\right) \lambda \int _{-\frac{1}{2\lambda }-\tau }^{\frac{1}{ 2\lambda }-\tau }\left[ \int _{-\pi -\left( m+1\right) \pi h-t}^{\pi -\left( m+1\right) \pi h-t}\left| f\left( x\right) \right| ^{p_{t}\left( x\right) }dx\right] dt \\= & {} O\left( 1\right) \lambda \int _{-\frac{1}{2\lambda }-\tau }^{\frac{1}{ 2\lambda }-\tau }\left[ \int _{-\pi }^{\pi }\left| f\left( x\right) \right| ^{p_{t}\left( x\right) }dx\right] dt=O\left( 1\right) , \end{aligned}$$

since similarly as in [9, p. 143] \(\int _{-\pi }^{\pi }\left| f\left( x\right) \right| ^{p_{t}\left( x\right) }dx=O\left( 1\right) .\) Thus our result follows. \(\square \)

Next we present some estimate of the kernel.

Lemma 3

If \(\beta \ge 0\) and \(0<t\le \pi \), then

$$\begin{aligned} \left| \sum _{k=0}^{n}(k+1)^{\beta }\frac{\sin \frac{(2l+1)t}{2}}{2\sin \frac{t}{2}}\right| \le \pi ^{2}\frac{(n+1)^{\beta }}{t^{2}}. \end{aligned}$$

Proof

Using the Abel transformation

$$\begin{aligned}&\left| \sum _{k=0}^{n}(k+1)^{\beta }\frac{\sin \frac{(2l+1)t}{2}}{2\sin \frac{t}{2}}\right| \\&\quad \le \left| \sum _{k=0}^{n-1}\left[ (k+1)^{\beta }-(k+2)^{\beta }\right] \sum _{l=0}^{k}\frac{\sin \frac{(2l+1)t}{2}}{2\sin \frac{t}{2}}+(n+1)^{\beta }\sum _{l=0}^{n}\frac{\sin \frac{(2l+1)t}{2}}{2\sin \frac{t}{2}}\right| . \end{aligned}$$

Since \(\sum _{l=0}^{n}\frac{\sin \frac{(2l+1)t}{2}}{2\sin \frac{t}{2}}=\frac{ \sin ^{2}\frac{(n+1)t}{2}}{2\sin ^{2}\frac{t}{2}},\) the above expression does not exceed

$$\begin{aligned}&\sum _{k=0}^{n-1}\left| (k+1)^{\beta }-(k+2)^{\beta }\right| \frac{ \sin ^{2}\frac{(k+1)t}{2}}{2\sin ^{2}\frac{t}{2}}+(n+1)^{\beta }\frac{\sin ^{2}\frac{(n+1)t}{2}}{2\sin ^{2}\frac{t}{2}} \\&\quad \le \frac{\pi ^{2}}{2t^{2}}\left[ \sum _{k=0}^{n-1}\left| (k+1)^{\beta }-(k+2)^{\beta }\right| +(n+1)^{\beta }\right] \\&\quad \le \frac{\pi ^{2}}{2t^{2}}\left[ 2(n+1)^{\beta }-1\right] \le \frac{\pi ^{2}}{t^{2}}(n+1)^{\beta }, \end{aligned}$$

whence the desired estimate follows. \(\square \)

Application of Lemma 3 gives next more general estimate.

Lemma 4

If (1) for some \(\beta \ge 0\) and (2) hold, then

$$\begin{aligned} \frac{1}{2\pi }\int _{-\pi }^{\pi }\left| \sum _{k=0}^{\infty }a_{n,k}\sum _{\nu =0}^{k}l_{\nu }\cos \nu t\right| dt=O\left( 1\right) , \text { where }l_{\nu }=\left\{ \begin{array}{c} 1\text {, when }\nu =0, \\ \frac{\pi }{4\sin \frac{\pi }{8}},\text { when }\nu >0. \end{array} \right. \end{aligned}$$

Proof

Since

$$\begin{aligned} 1+\frac{\pi }{4\sin \frac{\pi }{8}}\sum _{\nu =1}^{k}\cos \nu t=1+\frac{\pi }{ 4\sin \frac{\pi }{8}}\left( \frac{\sin \frac{(2k+1)t}{2}}{2\sin \frac{t}{2}}- \frac{1}{2}\right) \end{aligned}$$

therefore

$$\begin{aligned}&\frac{1}{2\pi }\int _{-\pi }^{\pi }\left| \sum _{k=0}^{\infty }a_{n,k}\left\{ 1-\frac{\pi }{8\sin \frac{\pi }{8}}+\frac{\pi }{4\sin \frac{ \pi }{8}}\frac{\sin \frac{(2k+1)t}{2}}{2\sin \frac{t}{2}}\right\} \right| dt \\&\quad \le \left| 1-\frac{\pi }{8\sin \frac{\pi }{8}}\right| +\frac{1}{ 8\sin \frac{\pi }{8}}\int _{-\pi }^{\pi }\left| \sum _{k=0}^{\infty }a_{n,k}\frac{\sin \frac{(2k+1)t}{2}}{2\sin \frac{t}{2}}\right| dt. \end{aligned}$$

Using the Abel transformation, Lemma 3 and (1) we get

$$\begin{aligned}&\left| \sum _{k=0}^{\infty }\frac{a_{n,k}}{(k+1)^{\beta }}(k+1)^{\beta } \frac{\sin \frac{(2k+1)t}{2}}{2\sin \frac{t}{2}}\right| \\&\quad =\left| \sum _{k=0}^{\infty }\left[ \frac{a_{n,k}}{(k+1)^{\beta }}- \frac{a_{n,k+1}}{(k+2)^{\beta }}\right] \sum _{l=0}^{k}(l+1)^{\beta }\frac{ \sin \frac{(2l+1)t}{2}}{2\sin \frac{t}{2}}\right| \\&\quad \le \frac{\pi ^{2}}{t^{2}}\sum _{k=0}^{\infty }(k+1)^{\beta }\left| \frac{a_{n,k}}{(k+1)^{\beta }}-\frac{a_{n,k+1}}{(k+2)^{\beta }}\right| = \frac{O\left( 1\right) }{t^{2}(n+1)}. \end{aligned}$$

Thus

$$\begin{aligned}&\int _{-\pi }^{\pi }\left| \sum _{k=0}^{\infty }a_{n,k}\frac{\sin \frac{ (2k+1)t}{2}}{2\sin \frac{t}{2}}\right| dt \\&\quad =\left( \int _{0}^{\frac{\pi }{n+1}}+\int _{\frac{\pi }{n+1}}^{\pi }\right) \left| \sum _{k=0}^{\infty }a_{n,k}\frac{\sin \frac{(2k+1)t}{2}}{2\sin \frac{t}{2}}\right| dt \\&\quad \le \int _{0}^{\frac{\pi }{n+1}}\sum _{k=0}^{\infty }a_{n,k}\frac{2k+1}{2} dt+\int _{\frac{\pi }{n+1}}^{\pi }\frac{O\left( 1\right) }{t^{2}(n+1)}dt \end{aligned}$$

and by the assumption (2)

$$\begin{aligned}&\int _{-\pi }^{\pi }\left| \sum _{k=0}^{n}a_{n,k}\frac{\sin \frac{(2k+1)t }{2}}{2\sin \frac{t}{2}}\right| dt \\&\quad \le \frac{\pi }{n+1}O\left( n+1\right) +\frac{O\left( 1\right) }{n+1}\left( \frac{\pi }{ n+1}\right) ^{-1}=O\left( 1\right) , \end{aligned}$$

whence our estimate follows. \(\square \)

We will yet to need the following approximation result.

Lemma 5

Let \(f\in L_{2\pi }^{p}\) with \(p=p(x)\)\(\in \Pi _{2\pi }\) and let \(t_{n}\) be a trigonometric polynomial of the degree at most n, such that \( \left\| f-t_{n}\right\| _{p}=O\left( 1\right) \Omega f(\frac{1}{n+1} )_{p}\). If (1) for some \(\beta \ge 0\) and (2) hold, then

$$\begin{aligned} \left\| \sum _{k=0}^{\infty }a_{n,k}S_{k}\left[ f-t_{n}\right] \right\| _{p}=O\left( \Omega f\left( \frac{1}{n+1}\right) _{p}\right) . \end{aligned}$$

Proof

First of all, under the notation \(f_{h}\left( t\right) =\frac{1}{2h} \int _{-h}^{h}f\left( y+t\right) dy,\) we will prove that

$$\begin{aligned} a_{0}(f)=a_{0}(f_{h}),\text { }a_{\nu }(f)=\frac{\nu h}{\sin \nu h}a_{\nu }(f_{h}),\,\ \left( \nu =1,2,3,...\right) \end{aligned}$$

and

$$\begin{aligned} b_{\nu }(f)=\frac{\nu h}{\sin \nu h}b_{\nu }(f_{h}),\,\ \left( \nu =1,2,3,...\right) . \end{aligned}$$

Really, it is clear that \(a_{0}(f_{h})=a_{0}(f)\) and for \(\nu =1,2,3,...,\)

$$\begin{aligned} \begin{array}{c} a_{\nu } \\ b_{\nu } \end{array} (f_{h})= & {} \frac{1}{2\pi }\int _{-\pi }^{\pi }\frac{1}{2h}\int _{-h}^{h}f \left( y+t\right) dy \begin{array}{c} \cos \\ \sin \end{array} \nu tdt \\= & {} \frac{1}{2\pi }\frac{1}{2h}\int _{-h}^{h}\left( \int _{-\pi }^{\pi }f\left( y+t\right) \begin{array}{c} \cos \\ \sin \end{array} \nu tdt\right) dy \\= & {} \frac{1}{2\pi }\frac{1}{2h}\int _{-h}^{h}\left( \int _{-\pi }^{\pi }f\left( t\right) \begin{array}{c} \cos \\ \sin \end{array} \nu \left( t-y\right) dt\right) dy \\= & {} \frac{1}{2\pi }\frac{1}{2h}\int _{-h}^{h}\left( \int _{-\pi }^{\pi }f\left( t\right) \left( \begin{array}{c} \cos \\ \sin \end{array} \nu t\cos \nu y\pm \begin{array}{c} \sin \\ \cos \end{array} \nu t\sin \nu y\right) dt\right) dy \\= & {} \frac{1}{2\pi }\frac{1}{2h}\int _{-h}^{h}\left( \cos \nu y\int _{-\pi }^{\pi }f\left( t\right) \begin{array}{c} \cos \\ \sin \end{array} \nu tdt\pm \sin \nu y\int _{-\pi }^{\pi }f\left( t\right) \begin{array}{c} \sin \\ \cos \end{array} \nu tdt\right) dy \\= & {} \begin{array}{c} a_{\nu } \\ b_{\nu } \end{array} (f)\frac{1}{2h}\int _{-h}^{h}\cos \nu tdt\pm \begin{array}{c} b_{\nu } \\ a_{\nu } \end{array} (f)\frac{1}{2h}\int _{-h}^{h}\sin \nu tdt= \begin{array}{c} a_{\nu } \\ b_{\nu } \end{array} (f)\frac{\sin \nu h}{\nu h}. \end{aligned}$$

Hence

$$\begin{aligned} S_{k}\left[ f\right] \left( x\right)= & {} \frac{a_{0}(f)}{2}+\sum _{\nu =1}^{k}(a_{\nu }(f)\cos \nu x+b_{\nu }(f)\sin \nu x) \\= & {} \frac{a_{0}(f_{h})}{2}+\sum _{\nu =1}^{k}\frac{\nu h}{\sin \nu h}a_{\nu }(f_{h})\cos \nu x+\sum _{\nu =1}^{k}\frac{\nu h}{\sin \nu h}b_{\nu }(f_{h})\sin \nu x \\= & {} \frac{1}{2\pi }\int _{-\pi }^{\pi }f_{h}\left( t\right) dt+\sum _{\nu =1}^{k}\frac{\nu h}{\sin \nu h}\frac{1}{\pi }\int _{-\pi }^{\pi }f_{h}\left( t\right) \cos \nu tdt\cos \nu x \\&+\sum _{\nu =1}^{k}\frac{\nu h}{\sin \nu h}\frac{1}{\pi }\int _{-\pi }^{\pi }f_{h}\left( t\right) \sin \nu tdt\sin \nu x \\= & {} \frac{1}{2\pi }\int _{-\pi }^{\pi }f_{h}\left( t\right) dt+\sum _{\nu =1}^{k}\frac{\nu h}{\sin \nu h}\frac{1}{\pi }\int _{-\pi }^{\pi }f_{h}\left( t\right) \cos \nu \left( t-x\right) dt \\= & {} \frac{1}{2\pi }\int _{-\pi }^{\pi }f_{h}\left( t\right) dt+\frac{1}{\pi } \sum _{\nu =1}^{k}\frac{\nu h}{\sin \nu h}\int _{-\pi }^{\pi }f_{h}\left( t+x\right) \cos \nu tdt \\= & {} \frac{1}{\pi }\int _{-\pi }^{\pi }f_{h}\left( t+x\right) \left( \frac{1}{2} +\sum _{\nu =1}^{k}\frac{\nu h}{\sin \nu h}\cos \nu t\right) dt. \end{aligned}$$

Consequently, denoting \(t_{n,h}\left( t\right) =\frac{1}{2h} \int _{-h}^{h}t_{n}\left( y+t\right) dy,\) we obtain

$$\begin{aligned}&S_{k}\left[ f-t_{n}\right] \left( x\right) \\&\quad =\frac{1}{\pi }\int _{-\pi }^{\pi }\left[ f-t_{n}\right] _{h}\left( t+x\right) \left( \frac{1}{2}+\sum _{\nu =1}^{k}\frac{\nu h}{\sin \nu h}\cos \nu t\right) dt \\&\quad =\frac{1}{\pi }\int _{-\pi }^{\pi }\left( f_{h}\left( t+x\right) -t_{n,h}\left( t+x\right) \right) \left( \frac{1}{2}+\sum _{\nu =1}^{k}\frac{ \nu h}{\sin \nu h}\cos \nu t\right) dt \end{aligned}$$

and therefore

$$\begin{aligned}&\sum _{k=0}^{\infty }a_{n,k}S_{k}\left[ f-t_{n}\right] \left( x\right) \\&\quad =\sum _{k=0}^{\infty }a_{n,k}\frac{1}{\pi }\int _{-\pi }^{\pi }\left( f_{h}\left( t+x\right) -t_{n,h}\left( t+x\right) \right) \left( \frac{1}{2} +\sum _{\nu =1}^{k}\frac{\nu h}{\sin \nu h}\cos \nu t\right) dt \\&\quad =\frac{1}{\pi }\int _{-\pi }^{\pi }\left( f_{h}\left( t+x\right) -t_{n,h}\left( t+x\right) \right) \sum _{k=0}^{\infty }a_{n,k}\left( \frac{1}{ 2}+\sum _{\nu =1}^{k}\frac{\nu h}{\sin \nu h}\cos \nu t\right) dt. \end{aligned}$$

Hence, by Lemma 1 and Lemma 2 with \(0<h<\frac{1}{2}\) and \(\left| t\right| \le \pi \)

$$\begin{aligned}&\left\| \sum _{k=0}^{\infty }a_{n,k}S_{k}\left[ f-t_{n}\right] \right\| _{p} \\&\quad \le \frac{2}{\pi }\int _{-\pi }^{\pi }\left\| f_{h}\left( t+\cdot \right) -t_{n,h}\left( t+\cdot \right) \right\| _{p}\left| \sum _{k=0}^{\infty }a_{n,k}\left( \frac{1}{2}+\sum _{\nu =1}^{k}\frac{\nu h}{ \sin \nu h}\cos \nu t\right) \right| dt \\&\quad =O\left( 1\right) \frac{1}{\pi }\int _{-\pi }^{\pi }\left\| f-t_{n}\right\| _{p}\left| \sum _{k=0}^{\infty }a_{n,k}\left( \frac{1}{ 2}+\sum _{\nu =1}^{k}\frac{\nu h}{\sin \nu h}\cos \nu t\right) \right| dt \end{aligned}$$

and further, taking \(h=\frac{\pi }{8\nu }<\frac{1}{2}\) for \(v=1,2,...\), we have

$$\begin{aligned}&\left\| \sum _{k=0}^{\infty }a_{n,k}S_{k}\left[ f-t_{n}\right] \right\| _{p} \\&\quad =O\left( 1\right) \frac{1}{2\pi }\int _{-\pi }^{\pi }\left| \sum _{k=0}^{\infty }a_{n,k}\left( 1+\frac{\pi }{4\sin \frac{\pi }{8}} \sum _{\nu =1}^{k}\cos \nu t\right) dt\right| \left\| f-t_{n}\right\| _{p}. \end{aligned}$$

Since, by Lemma 4

$$\begin{aligned} \frac{1}{2\pi }\int _{-\pi }^{\pi }\left| \sum _{k=0}^{\infty }a_{n,k}\left( 1+\frac{\pi }{4\sin \frac{\pi }{8}}\sum _{\nu =1}^{k}\cos \nu t\right) \right| dt=O\left( 1\right) \end{aligned}$$

therefore, by the assumptions,

$$\begin{aligned} \left\| \sum _{k=0}^{\infty }a_{n,k}S_{k}\left[ f-t_{n}\right] \right\| _{p}=O\left( 1\right) \left\| f-t_{n}\right\| _{p}=O\left( 1\right) \Omega f\left( \frac{1}{n+1}\right) _{p} \end{aligned}$$

and our proof is ended. \(\square \)

4 Proofs of Main Results

4.1 Proof of Theorem 1

If \(t_{n}\) is a such polynomial that \(\left\| f-t_{n}\right\| _{p}=O\left( \Omega f(\frac{1}{n+1})_{p}\right) ,\) then

$$\begin{aligned}&\left\| T_{n,A}^{\text { }}\left[ f\right] -f\right\| _{p} \\&\quad =\left\| T_{n,A}^{\text { }}\left[ f\right] -\sum _{k=0}^{n}a_{n,k}t_{k}-\sum _{k=n+1}^{ \infty }a_{n,k}t_{n}+\sum _{k=0}^{n}a_{n,k}t_{k}+\sum _{k=n+1}^{\infty }a_{n,k}t_{n}-f\right\| _{p}\\&\quad \le \left\| T_{n,A}^{\text { }}\left[ f\right] - \sum _{k=0}^{n}a_{n,k}t_{k}-\sum _{k=n+1}^{\infty }a_{n,k}t_{n}\right\| _{p}+\left\| \sum _{k=0}^{n}a_{n,k}t_{k}+\sum _{k=n+1}^{\infty }a_{n,k}t_{n}-f\right\| _{p} \\&\quad =\left\| \sum _{k=0}^{n}a_{n,k}\left\{ S_{k}\left[ f\right] -t_{k}\right\} +\sum _{k=n+1}^{\infty }a_{n,k}\left\{ S_{k}\left[ f\right] -t_{n}\right\} \right\| _{p} \\&\quad \quad +\left\| \sum _{k=0}^{n}a_{n,k}\left( f-t_{k}\right) +\sum _{k=n+1}^{\infty }a_{n,k}\left( f-t_{n}\right) \right\| _{p} \\&\quad \le \left\| \sum _{k=0}^{\infty }a_{n,k}S_{k}\left[ f-t_{n}\right] \right\| _{p}+\left\| \sum _{k=0}^{n}a_{n,k}\left( f-t_{k}\right) \right\| _{p}+\left\| \sum _{k=n+1}^{\infty }a_{n,k}\left( f-t_{n}\right) \right\| _{p} \\&\quad =\left\| \sum _{k=0}^{\infty }a_{n,k}S_{k}\left[ f-t_{n}\right] \right\| _{p}+O\left( 1\right) \sum _{k=0}^{n}a_{n,k}\Omega f\left( \frac{1}{k+1} \right) _{p} \\&\quad \quad +\sum _{k=n+1}^{\infty }a_{n,k}O\left( \Omega f\left( \frac{1}{n+1} \right) _{p}\right) , \end{aligned}$$

since

$$\begin{aligned} S_{k}\left[ f-t_{n}\right] \left( x\right) =\left\{ \begin{array}{c} S_{k}f\left( x\right) -t_{k}\left( x\right) ,\text { for }k\le n. \\ S_{k}f\left( x\right) -t_{n}\left( x\right) ,\text { for }k\ge n. \end{array} \right. \end{aligned}$$

We note that by Lemma 5

$$\begin{aligned} \left\| \sum _{k=0}^{\infty }a_{n,k}S_{k}\left[ f-t_{n}\right] \right\| _{p}=O\left( \Omega f\left( \frac{1}{n+1}\right) _{p}\right) . \end{aligned}$$

Thus our result follows. \(\square \)

4.2 Proof of Theorem 2

We can note that the assumptions on entries of \(A_{0}\) yield

$$\begin{aligned}&\sum _{k=0}^{\infty }(k+1)^{\beta }\left| \frac{a_{n,k}}{(k+1)^{\beta }} -\frac{a_{n,k+1}}{(k+2)^{\beta }}\right| \\&\quad =\sum _{k=0}^{n-1}(k+1)^{\beta }\left| \frac{a_{n,k}}{(k+1)^{\beta }}- \frac{a_{n,k+1}}{(k+2)^{\beta }}\right| +a_{n,n} \\&\quad =O\left( a_{n,n}\right) +a_{n,n}=O\left( a_{n,n}\right) =O\left( \frac{1}{n+1}\right) , \end{aligned}$$

and

$$\begin{aligned} \sum _{k=0}^{\infty }(k+1)a_{n,k}=\sum _{k=0}^{n}(k+1)a_{n,k}\le \left( n+1\right) \sum _{k=0}^{n}a_{n,k}=n+1. \end{aligned}$$

Thus the result follows by Theorem 1, since using the monotonicity of \( \Omega f(\delta )_{p}\) with respect to \(\delta >0\)

$$\begin{aligned} \sum _{k=0}^{n}a_{n,k}\Omega f\left( \frac{1}{k+1}\right) _{p}\ge \Omega f\left( \frac{1}{n+1} \right) _{p}\sum _{k=0}^{n}a_{n,k}=\Omega f\left( \frac{1}{n+1}\right) _{p}. \end{aligned}$$

This ends the proof of Theorem 2. \(\square \)

4.3 Proof of Theorem 3

The subadditivity of \(\omega \) implies \(\omega \left( n\delta \right) \le n\omega \left( \delta \right) ,\) whence \(\omega \left( \lambda \delta \right) \le \left( \lambda +1\right) \omega \left( \delta \right) \) and therefore \(\frac{\omega \left( \delta _{2}\right) }{\delta _{2}}\le 2\frac{ \omega \left( \delta _{1}\right) }{\delta _{1}}\) since \(\omega \left( \delta _{2}\right) =\omega \left( \frac{\delta _{1}}{\delta _{1}}\delta _{2}\right) \le \left( \frac{\delta _{2}}{\delta _{1}}+1\right) \omega \left( \delta _{1}\right) =\left( \frac{\delta _{2}}{\delta _{1}}+\frac{\delta _{1}}{ \delta _{1}}\right) \omega \left( \delta _{1}\right) \le \left( \frac{ \delta _{2}}{\delta _{1}}+\frac{\delta _{2}}{\delta _{1}}\right) \omega \left( \delta _{1}\right) =2\frac{\delta _{2}}{\delta _{1}}\omega \left( \delta _{1}\right) ,\) where \(n\in {\mathbb {N}} _{0},\)\(\lambda \ge 0\) and \(0< \delta _{1}\le \delta _{2}.\) Hence, by ( 1) with \(\beta >0,\)

$$\begin{aligned}&\sum _{k=0}^{n}a_{n,k}\omega \left( \frac{1}{k+1}\right) =\sum _{k=0}^{n}\frac{a_{n,k}}{k+1} \frac{\omega \left( \frac{1}{k+1}\right) }{\frac{1}{k+1}}\\&\quad \le 2\left( n+1\right) \omega \left( \frac{1}{n+1}\right) \sum _{k=0}^{n}\frac{a_{n,k}}{ k+1} \\&\quad = 2\left( n+1\right) \omega \left( \frac{1}{n+1}\right) \sum _{k=0}^{\infty }(k+1)^{\beta -1}\frac{a_{n,k}}{(k+1)^{\beta }} \\&\quad \le 2\left( n+1\right) \omega \left( \frac{1}{n+1}\right) \sum _{k=0}^{\infty }\left| \frac{a_{n,k}}{(k+1)^{\beta }}-\frac{a_{n,k+1}}{(k+2)^{\beta }} \right| \sum _{l=0}^{k}(l+1)^{\beta -1} \\&\quad =O\left( n+1\right) \omega \left( \frac{1}{n+1}\right) \sum _{k=0}^{\infty }(k+1)^{\beta }\left| \frac{a_{n,k}}{(k+1)^{\beta }}-\frac{a_{n,k+1}}{(k+2)^{\beta }} \right| \\&\quad =\left( n+1\right) \omega \left( \frac{1}{n+1}\right) O\left( \frac{1}{n+1}\right) =O\left( \omega \left( \frac{1}{n+1}\right) \right) . \end{aligned}$$

Thus, by Theorem 1, our result follows. \(\square \)