On Approximation by Linear Combinations of Modified Summation Operators of Integral Type in Orlicz Spaces

Abstract

In this paper, the authors introduce the Orlicz spaces corresponding to the Young function and, by virtue of the equivalent theorem between the modified K-functional and modulus of smoothness, establish the direct, inverse, and equivalent theorems for linear combinations of modified summation operators of integral type in the Orlicz spaces.

1. Introduction and Main Results

Throughout this paper, we use C to denote an absolute constant independent of anything, which may be not necessarily the same in different cases.

There are many types of integral operators (see, for example, [1,2,3,4,5,6,7] and closely related references therein). In the paper [8], Ueki provided a characterization for the boundedness and compactness of the Li-Stević type integral operators

$$T_{\phi }^{g}f\left(z\right)={\int }_0^{z}f\left(\phi \left(\xi \right)\right)g\left(\xi \right)\mathrm{d}\xi$$

from the weighted Bergman space $L_a^p\left(d{A}_{\alpha }\right)$ to the $\beta$-Zygmund space $Z_{\beta }$. Later, Li and Ma [9] investigated the boundedness and compactness of products of composition operators and integral type operators

$$\left(C_{\phi }{I}_{g}f\right)\left(z\right)={\int }_0^{\phi \left(z\right)}{f}^{\prime }\left(\xi \right)g\left(\xi \right)\mathrm{d}\xi$$

from Zygmund-type spaces to $Q_k$ spaces. Recently, the equivalent characterizations for the boundedness and compactness of several integral type operators

$$V_{\Phi }^{h}f\left(z\right)={\int }_0^z{f}^{\prime }(\Phi \left(t\right))h^{\prime }\left(t\right)\mathrm{d}t$$

from $F(p,q,s)$ space to $\alpha$-Bloch-Orlicz and $\beta$-Zygmund-Orlicz spaces were developed in [10]. Gupta and Yadav [11] estimated the approximation by complex summation integral type operator

$$M_n(f,z)=(n+1)\sum _{k=1}^n{p}_{n,k}\left(z\right){\int }_0^{1}f\left(t\right)p_{n,k-1}\left(t\right)\mathrm{d}t+f\left(0\right)p_{n,0}\left(z\right)$$

in compact disks, where $p_{n,k}\left(z\right)=\left(\genfrac{}{}{0pt}{}{n}{k}\right)z^k{(1-z)}^{n-k}$. In [8,9,10,11], some approximating properties of integral type operators in complex spaces were established. Vural, Altin, and Yüksel [12] provided weighted approximation and obtained a rate of convergence of Schurer’s generalization of the q-Hybrid summation operators of integral type

$$\begin{array}{cc}\hfill M_{n,p,q}^{(\alpha ,\beta )}(f,x)& ={[n+p-1]}_q\sum _{k=1}^{\infty }{S}_{n,p,k}^q\left(x\right){\int }_0^{\infty }{P}_{n,p,k-1}^q\left(t\right)f\left(\frac{{[n+p]}_{q}t+\alpha }{{[n+p]}_q+\beta }\right){\mathrm{d}}_q\left(t\right)\hfill \\ & +e^{-{[n+p]}_{q}x}f\left(\frac{\alpha }{{[n+p]}_q+\beta }\right),\hfill \end{array}$$

where

$$S_{n,p,k}^q\left(x\right)={\left({[n+p]}_{q}x\right)}^k\frac{e^{-{[n+p]}_{q}x}}{{\left[k\right]}_q!}\mathrm{and}{P}_{n,p,k}^q={\left(\genfrac{}{}{0pt}{}{n+p+k-1}{k}\right)}_q{q}^{k(k+1)}\frac{x^k}{{(1+x)}_q^{n+p+k}}.$$

In [13], Govil and Gupta considered the simultaneous approximation for the Stancu-type generalization of certain summation operators of integral type

$$G_n^{\alpha ,\beta }(f,x)=\sum _{k=1}^{\infty }{p}_{n,k}\left(x\right){\int }_0^{\infty }{b}_{n,k-1}\left(t\right)f\left(\frac{nt+\alpha }{n+\beta }\right)\mathrm{d}t+{(1+x)}^{-n}f\left(\frac{\alpha }{n+\beta }\right)$$

by hypergeometric series. Srivastava and Gupta [14] introduced and investigated a new sequence of linear positive operators

$$G_{n,c}(f,x)=n\sum _{k=1}^{\infty }{\int }_0^{\infty }{p}_{n,k}(x;c)p_{n+c,k-1}(t;c)f\left(t\right)\mathrm{d}t+{\int }_0^{\infty }{p}_{n,0}(x;c)p_{n,0}(t;c)\delta \left(t\right)f\left(t\right)\mathrm{d}t,$$

which included some well-known operators as its special cases and obtained an estimate on the rate of convergence by means of the decomposition technique for functions of bounded variation. In [15], Gupta, Mohapatra, and Finta studied the mixed summation operators of integral type

$$S_n(f,x)=\sum _{v=1}^{\infty }{s}_{n,v}\left(x\right){\int }_0^{\infty }{b}_{n,v-1}\left(t\right)f\left(t\right)\mathrm{d}t+e^{-nx}f\left(0\right)$$

and obtained the rate of point-wise convergence, where

$$s_{n,v}\left(x\right)=e^{-nx}\frac{{\left(nx\right)}^v}{v!}\mathrm{and}{b}_{n,v}\left(t\right)=\frac{1}{B(n,v+1)}{t}^v{(1+t)}^{-n-v-1}.$$

In [12,13,14,15], some approximating properties of integral type operators were discussed in $C[0,\infty )$, which is a special case of the Orlicz space. For $f\in L_{\Phi }^{\ast }[0,\infty )$, the modified summation operators of integral type $B_n(f,x)$ are defined in [16] as

$$B_n(f,x)=\frac{1}{n+1}\sum _{k=1}^{\infty }{b}_{n,k}\left(x\right){\int }_0^{\infty }{b}_{n,k}\left(t\right)f\left(t\right)\mathrm{d}t,x\in [0,\infty ),$$

(1)

where $b_{n,k}\left(x\right)=\frac{(n+k)!}{(k-1)!n!}\frac{x^{k-1}}{{(1+x)}^{n+k+1}}$ for $k,n\ge 1$. Recently, Han and Wu [16] obtained the following direct, inverse, and equivalent theorems of modified summation operators of integral type in Orlicz spaces.

Theorem 1

(Direct theorem [16]). Let $f\in L_{\Phi }^{\ast }[0,\infty )$, ${\phi }^2\left(x\right)=x(1+x)$, and $\Psi \in {\Delta }_2$. Then

$$\parallel B_n{\left(f\right)-f\parallel }_{\Phi }\le C\left[{\omega }_{2,\phi }{\left(f,\frac{1}{\sqrt{n}}\right)}_{\Phi }+{\omega }_1{\left(f,\frac{1}{n}\right)}_{\Phi }+\frac{{\parallel f\parallel }_{\Phi }}{n}\right].$$

Theorem 2

(Inverse theorem [16]). Let $f\in L_{\Phi }^{\ast }[0,\infty )$, $0\le \alpha <2$, and $\parallel B_n{\left(f\right)-f\parallel }_{\Phi }=O\left(n^{-\alpha /2}\right)$. Then

$${\omega }_{2,\phi }{(f,t)}_{\Phi }=O\left(t^{\alpha }\right)and{\omega }_1{(f,t)}_{\Phi }=O\left(t^{\alpha /2}\right).$$

Theorem 3

(Equivalence theorem [16]). Let $f\in L_{\Phi }^{\ast }[0,\infty )$ and $0\le \alpha <2$. Then

$$\parallel B_n{\left(f\right)-f\parallel }_{\Phi }=O\left(n^{-\alpha /2}\right)ifandonlyif{\omega }_{2,\phi }{(f,t)}_{\Phi }=O\left(t^{\alpha }\right)and{\omega }_1{(f,t)}_{\Phi }=O\left(t^{\alpha /2}\right).$$

In recent years, since the Orlicz spaces are more general than the classical $L_p$ spaces, which is composed of measurable functions $f\left(x\right)$ such that ${\left|f\left(x\right)\right|}^p$ are integrable, there is growing interest in problems of approximation in Orlicz spaces.

For smoothly proceeding, we recall from [17] some definitions and related results.

A continuous convex function $\Phi \left(t\right)$ on $[0,\infty )$ is called a Young function if it satisfies

$$\underset{t\to 0^{+}}{lim}\frac{\Phi \left(t\right)}{t}=0\mathrm{and}\underset{t\to \infty }{lim}\frac{\Phi \left(t\right)}{t}=\infty .$$

For a given Young function $\Phi \left(t\right)$, its complementary Young function is denoted by $\Psi \left(t\right)$.

It is clear that the convexity of $\Phi \left(t\right)$ can lead to $\Phi \left(\alpha t\right)\le \alpha \Phi \left(t\right)$ for $\alpha \in [0,1]$. In particular, one has $\Phi \left(\alpha t\right)<\alpha \Phi \left(t\right)$ for $\alpha \in (0,1).$

A Young function $\Phi \left(t\right)$ is said to satisfy the ${\Delta }_2$-condition, denoted by $\Phi \in {\Delta }_2$, if there exist $t_0>0$ and $C\ge 1$ such that $\Phi \left(2t\right)\le C\Phi \left(t\right)$ for $t\ge t_0$.

Let $\Phi \left(t\right)$ be a Young function. We define the Orlicz class $L_{\Phi }[0,\infty )$ as the collection of all Lebesgue measurable functions $u\left(x\right)$ on $[0,\infty )$. Since the integral

$$\rho (u,\Phi )={\int }_0^{\infty }\Phi \left(\right|u\left(x\right)\left|\right)\mathrm{d}x$$

is finite, we define the Orlicz space $L_{\Phi }^{\ast }[0,\infty )$ as the linear hull of $L_{\Phi }[0,\infty )$ under the Luxemburg norm

$${\parallel u\parallel }_{(\Phi )}=\underset{\lambda >0}{inf}\left\{\lambda :\rho \left(\frac{u}{\lambda },\Phi \right)\le 1\right\}.$$

The Orlicz norm, which is equivalent to the Luxemburg norm on $L_{\Phi }^{\ast }[0,\infty )$, is given by

$${\parallel u\parallel }_{\Phi }=\underset{\rho (v,\Psi )\le 1}{sup}\left|{\int }_0^{\infty }u\left(x\right)v\left(x\right)\mathrm{d}x\right|$$

and satisfies

$${\parallel u\parallel }_{(\Phi )}\le {\parallel u\parallel }_{\Phi }\le 2{\parallel u\parallel }_{(\Phi )}.$$

(2)

For $f\in L_{\Phi }^{\ast }[0,\infty )$, the weighted K-functional $K_{r,\phi }(f,t^r)$, the modified weighted K-functional ${\overline{K}}_{r,\phi }(f,t^r)$, and the weighted modulus of smoothness ${\omega }_{{}_{r,\phi }}{(f,t)}_{\Phi }$ are given, respectively, by

$$\begin{array}{c}{K}_{r,\phi }{(f,t^r)}_{\Phi }=\underset{g}{inf}\left\{{\parallel f-g\parallel }_{\Phi }+t^r{∥{\phi }^r{g}^{\left(r\right)}∥}_{\Phi }:g^{(r-1)}\in A{C}_{loc}\right\},\\ {\overline{K}}_{r,\phi }{(f,t^r)}_{\Phi }=\underset{g}{inf}\left\{{\parallel f-g\parallel }_{\Phi }+t^r{∥{\phi }^r{g}^{\left(r\right)}∥}_{\Phi }+t^{2r}{∥g^{\left(r\right)}∥}_{\Phi }:g^{(r-1)}\in A{C}_{loc}\right\},\end{array}$$

and

$${\omega }_{{}_{r,\phi }}{(f,t)}_{\Phi }=\underset{0<h\le t}{sup}{\parallel {\Delta }_{h\phi }^{r}f\parallel }_{\Phi },$$

where $\phi \left(x\right)=\sqrt{x}$, $\phi \left(x\right)=\sqrt{x(1+x)}$, or $\phi \left(x\right)=x$, and $g^{(r-1)}\in A{C}_{loc}$ means that g is $r-1$ times differentiable and $g^{(r-1)}$ is absolutely continuous in every closed finite interval $[c,d]\subseteq [0,\infty )$.

Between the weighted modulus of smoothness and the modified weighted K-functional, there exists the following equivalent theorem.

Theorem 4

([16]). Let $f\in L_{\Phi }^{\ast }[0,\infty )$. Then there exist some constants C and $t_0$ such that

$$\frac{{\omega }_{{}_{r,\phi }}{(f,t)}_{\Phi }}{C}\le {\overline{K}}_{r,\phi }{(f,t^r)}_{\Phi }\le C{\omega }_{{}_{r,\phi }}{(f,t)}_{\Phi },0<t\le t_0.$$

(3)

Between the weighted modulus of smoothness and the weighted K-functional, there exists the following equivalent theorem.

Theorem 5

([18]). Let $f\in L_{\Phi }^{\ast }[0,\infty )$. Then there are some constants C and $t_0$ such that

$$\frac{{\omega }_{r,\phi }{(f,t)}_{\Phi }}{C}\le K_{r,\phi }{(f,t^r)}_{\Phi }\le C{\omega }_{r,\phi }{(f,t)}_{\Phi },0<t\le t_0.$$

(4)

Currently, there are few results about linear combinations of modified summation operators of integral type $B_n(f,x)$. In this article, we investigate the approximation of linear combinations of modified summation operators of integral type $B_n(f,x)$ in Orlicz spaces $L_{\Phi }^{\ast }[0,\infty )$. The linear combinations of modified summation operators of integral type $B_n(f,x)$ are defined as

$$L_{n,r}(f,x)=\sum _{i=0}^{2r-1}{c}_i\left(n\right)B_{n_i}(f,x),$$

where

$$\begin{array}{c}n\le n_0\le n_1<\dots <n_{2r-1}\le C_n,{\displaystyle \sum _{i=0}^{2r-1}}{c}_i\left(n\right)=1,{\displaystyle \sum _{i=0}^{2r-1}}\left|c_i\left(n\right)\right|<C,\\ {\displaystyle \sum _{i=0}^{2r-1}}{c}_i\left(n\right)B_{n_i}\left({(t-x)}^k,x\right)=0,k=0,1,2,\dots ,2r-1.\end{array}$$

(5)

Our main results in this paper can be stated as the following three theorems.

Theorem 6

(Direct theorem). Let $f\in L_{\Phi }^{\ast }[0,\infty )$, $n\in \mathbb{N}$, $\Psi \in {\Delta }_2$, and $\phi \left(x\right)=\sqrt{x(1+x)}$. Then

$$\parallel L_{n,r}{\left(f\right)-f\parallel }_{\Phi }\le C{\omega }_{2r,\phi }{\left(f,\frac{1}{\sqrt{n}}\right)}_{\Phi }.$$

Theorem 7

(Inverse theorem). Let $f\in L_{\Phi }^{\ast }[0,\infty )$, $n\ge 2r$, and ${\phi }^2\left(x\right)=x(1+x)$. Then

$${\omega }_{2r,\phi }{\left(f,\frac{1}{n^{r/2}}\right)}_{\Phi }\le \frac{C}{n^r}\sum _{k=1}^n{k}^{r-1}{\parallel L_{n,r}\left(f\right)-f\parallel }_{\Phi }.$$

Theorem 8

(Equivalent theorem). Let $f\in L_{\Phi }^{\ast }[0,\infty )$, $n\ge 2r$, ${\phi }^2\left(x\right)=x(1+x)$, and $\Psi \in {\Delta }_2$. Then

$$\parallel L_{n,r}{\left(f\right)-f\parallel }_{\Phi }=O\left(\psi \left(\frac{1}{n^{1/2}}\right)\right),n\to \infty ifandonlyif{\omega }_{2r,\phi }{(f,t)}_{\Phi }=O\left(\psi \left(t\right)\right),t\to 0^{+}.$$

These main results improve some conclusions in [19] and increase the approximating speed of corresponding operators.

2. Proof of the Direct Theorem

In order to prove the direct theorem, we need several lemmas below.

Lemma 1.

The modified summation operator of integral type $B_n(f,x)$ defined in Equation (1) satisfies

$$B_n(1,x)=1and{B}_n\left({(t-x)}^{2r},x\right)\le C{\left[\frac{{\delta }_n^2\left(x\right)}{n}\right]}^r,$$

where ${\delta }_n^2\left(x\right)=max\left\{{\phi }^2\left(x\right),\frac{1}{n}\right\}$, $\phi \left(x\right)=\sqrt{x(1+x)}$, $r\in \mathbb{N}$, and C is a positive constant.

Proof. 

This follows from simple calculation. □

Lemma 2

([19]). If u locates between x and t, then

$$\frac{{(t-u)}^{2r-1}}{{\phi }^{2r}\left(u\right)}\le \frac{{|t-x|}^{2r-1}}{{\phi }^{2(r-1)}\left(x\right)}\frac{1}{x}\left(\frac{1}{1+x}+\frac{1}{1+t}\right).$$

Lemma 3.

Let $f\in L_{\Phi }^{\ast }[0,\infty )$. Then

$$\parallel L_{n,r}{\left(f\right)\parallel }_{\Phi }\le C{\parallel f\parallel }_{\Phi }.$$

Proof. 

By Lemma 3.2 in [16], we have

$$\parallel B_n{\left(f\right)\parallel }_{\Phi }\le 2{\parallel f\parallel }_{\Phi }.$$

Using Equation (5), we obtain

$$\parallel L_{n,r}{\left(f\right)\parallel }_{\Phi }={∥\sum _{i=0}^{2r-1}{c}_i\left(n\right)B_{n_i}\left(f\right)∥}_{\Phi }\le \sum _{i=0}^{2r-1}|c_i\left(n\right)|\parallel B_{n_i}{\left(f\right)\parallel }_{\Phi }\le C{\parallel f\parallel }_{\Phi }.$$

The proof of Lemma 3 is complete. □

Lemma 4

([18]). For $f\in L_{\Phi }^{\ast }[0,\infty )$ and $\Psi \in {\Delta }_2$, we have

$${\parallel \theta \left(f\right)\parallel }_{\Phi }\le C{\parallel f\parallel }_{\Phi },$$

where

$$\theta (f,x)=\underset{\begin{array}{c}0\le t<\infty \\ t\ne x\end{array}}{sup}\frac{1}{t-x}{\int }_x^{t}f\left(u\right)\mathrm{d}u$$

is the Hardy-Littlewood function of $f\left(x\right)$, and C is a positive constant.

We are now in a position to prove Theorem 6.

Proof of Theorem 6.

Let

$$U=W_{\Phi }^{2r}\left\{g:g^{(2r-1)}\in A{C}_{loc},{\phi }^{2r}{g}^{\left(2r\right)}\in L_{\Phi }^{\ast }[0,\infty )\right\}.$$

Taylor’s formula with integral remainder of $g\in U$ reads

$$g\left(t\right)=\sum _{i=0}^{2r-1}\frac{g^{\left(i\right)}\left(x\right)}{i!}{(t-x)}^i+R_{2r}(g,t,x),$$

where

$$R_{2r}(g,t,x)=\frac{1}{(2r-1)!}{\int }_x^t{(t-u)}^{2r-1}{g}^{\left(2r\right)}\left(u\right)\mathrm{d}u,x\in [0,\infty ).$$

From Equation (5), it follows that $L_{n,r}(g,x)-g\left(x\right)=L_{n,r}(R_{2r}(g,t,x),x)$ and

$$\parallel L_{n,r}{\left(g\right)-g\parallel }_{\Phi }={\parallel L_{n,r}(R_{2r}(g,·,x),x)\parallel }_{\Phi }.$$

(6)

Now we estimate $|R_{2r}(g,t,x)|$. As $x\in [\frac{1}{n},\infty )$, we have ${\delta }_n^2\left(x\right)={\phi }^2\left(x\right)$. Applying Lemma 2 leads to

$$\begin{array}{cc}\hfill |R_{2r}(g,t,x)|& \le \frac{1}{(2r-1)!}\left|{\int }_x^t\frac{{(t-u)}^{2r-1}}{{\phi }^{2r}\left(u\right)}{\delta }_n^{2r}\left(u\right)g^{\left(2r\right)}\left(u\right)\mathrm{d}u\right|\hfill \\ & \le \frac{1}{(2r-1)!}\frac{{|t-x|}^{2r-1}}{{\phi }^{2r-2}\left(x\right)}\frac{1}{x}\left(\frac{1}{1+x}+\frac{1}{1+t}\right)\left|{\int }_x^t{\delta }_n^{2r}\left(u\right)g^{\left(2r\right)}\left(u\right)\mathrm{d}u\right|\hfill \\ & \le \frac{1}{(2r-1)!}\frac{{(t-x)}^{2r}}{{\phi }^{2r-2}\left(x\right)}\left[\frac{1}{x(1+x)}+\frac{1}{x(1+t)}\right]\left|\theta ({\delta }_n^{2r}{g}^{\left(2r\right)},x)\right|\hfill \\ & \triangleq I_1+I_2.\hfill \end{array}$$

From Lemma 1, we conclude that

$$\begin{array}{c}{B}_n(I_1,x)=\frac{1}{n+1}{\displaystyle \sum _{k=1}^{\infty }}{b}_{n,k}\left(x\right){\int }_0^{\infty }{b}_{n,k}\left(t\right)\frac{1}{(2r-1)!}\frac{{(t-x)}^{2r}}{{\phi }^{2r}\left(x\right)}\left|\theta ({\delta }_n^{2r}{g}^{\left(2r\right)},x)\right|\mathrm{d}t\\ =\frac{\left|\theta ({\delta }_n^{2r}{g}^{\left(2r\right)},x)\right|{\phi }^{-2r}\left(x\right)}{(2r-1)!}{B}_n\left({(t-x)}^{2r},x\right)\le \frac{C}{n^r}\left|\theta ({\delta }_n^{2r}{g}^{\left(2r\right)},x)\right|\end{array}$$

(7)

and

$$\begin{array}{cc}\hfill B_n(I_2,x)& =\frac{\left|\theta ({\delta }_n^{2r}{g}^{\left(2r\right)},x)\right|}{n+1}\sum _{k=1}^{\infty }{b}_{n,k}\left(x\right){\int }_0^{\infty }{b}_{n,k}\left(t\right)\frac{1}{(2r-1)!}\frac{{(t-x)}^{2r}}{{\phi }^{2r-2}\left(x\right)}\frac{1}{x(1+t)}\mathrm{d}t\hfill \\ & =\frac{\left|\theta ({\delta }_n^{2r}{g}^{\left(2r\right)},x)\right|{\phi }^{-2r}\left(x\right)}{(2r-1)!(n+1)}\sum _{k=1}^{\infty }{\left(\frac{n+1}{n+k+1}\right)}^2{b}_{n+1,k}\left(x\right){\int }_0^{\infty }{b}_{n+1,k}\left(t\right){(t-x)}^{2r}\mathrm{d}t\hfill \\ & \le \frac{C}{n^r}\left|\theta ({\delta }_n^{2r}{g}^{\left(2r\right)},x)\right|.\hfill \end{array}$$

(8)

Hence, by Inequalities (7) and (8) and Lemma 4, it follows that

$$\begin{array}{c}\hfill \parallel B_n(R_{2r}(g,·,x),x){\parallel }_{\Phi [\frac{1}{n},\infty )}\le \frac{C}{n^r}{∥\theta ({\delta }_n^{2r}{g}^{\left(2r\right)},x)∥}_{\Phi [\frac{1}{n},\infty )}\le \frac{C}{n^r}{∥{\delta }_n^{2r}{g}^{\left(2r\right)}∥}_{\Phi [\frac{1}{n},\infty )}.\end{array}$$

(9)

For $x\in [0,\frac{1}{n})$ and ${\delta }_n^2\left(x\right)=\frac{1}{n}$, we have

$$\begin{array}{c}|R_{2r}(g,t,x)|=\frac{1}{(2r-1)!}\left|{\int }_x^t{(t-u)}^{2r-1}{g}^{\left(2r\right)}\left(u\right)\mathrm{d}u\right|\\ =\frac{1}{(2r-1)!}\left|{\int }_x^t\frac{{(t-u)}^{2r-1}}{1/n^r}{\delta }_n^{2r}\left(u\right)g^{\left(2r\right)}\left(u\right)\mathrm{d}u\right|\le \frac{1}{(2r-1)!}{(x-t)}^{2r}{n}^r\left|\theta ({\delta }_n^{2r}{g}^{\left(2r\right)},x)\right|.\end{array}$$

Using Lemmas 1 and 4 arrives at

$$B_n\left(\right|R_{2r}(g,t,x)|,x)\le \frac{n^r}{(2r-1)!}\frac{C}{n^r}{\delta }_n^{2r}\left(x\right)\left|\theta ({\delta }_n^{2r}{g}^{\left(2r\right)},x)\right|\le \frac{C}{n^r}\left|\theta ({\delta }_n^{2r}{g}^{\left(2r\right)},x)\right|$$

and

$$\parallel B_n(R_{2r}(g,·,x),x){\parallel }_{\Phi [0,\frac{1}{n})}\le \frac{C}{n^r}{∥\theta \left({\delta }_n^{2r}{g}^{\left(2r\right)}\right)∥}_{\Phi [0,\frac{1}{n})}\le \frac{C}{n^r}{∥{\delta }_n^{2r}{g}^{\left(2r\right)}∥}_{\Phi [0,\frac{1}{n})}.$$

Combining this with Equation (9) leads to

$$\parallel B_n(R_{2r}(g,·,x),x){\parallel }_{\Phi [0,\infty )}\le \frac{C}{n^r}{∥{\delta }_n^{2r}{g}^{\left(2r\right)}∥}_{\Phi [0,\infty )}$$

and, consequently,

$$\begin{array}{c}\parallel L_{n,r}(R_{2r}(g,·,x),x){\parallel }_{\Phi }\le \sum _{i=0}^{2r-1}|c_i|\parallel B_{n_i}(R_{2r}(g,·,x),x){\parallel }_{\Phi }\hfill \\ \hfill \le \sum _{i=0}^{2r-1}|c_i|\frac{C}{n_i^r}{∥{\delta }_{n_i}^{2r}{g}^{\left(2r\right)}∥}_{\Phi }\le \sum _{i=0}^{2r-1}\left|c_i\right|\frac{C}{n^r}{∥{\delta }_n^{2r}{g}^{\left(2r\right)}∥}_{\Phi }\le \frac{C}{n^r}{∥{\delta }_n^{2r}{g}^{\left(2r\right)}∥}_{\Phi }.\end{array}$$

Then, applying the above inequality, Inequalities (3) and (6), and Lemma 3, we obtain

$$\begin{array}{c}\parallel L_{n,r}{\left(f\right)-f\parallel }_{\Phi }\le \parallel L_{n,r}{(f-g)-(f-g)\parallel }_{\Phi }+{\parallel L_{n,r}\left(g\right)-g\parallel }_{\Phi }\hfill \\ \hfill \le {C\parallel f-g\parallel }_{\Phi }+\frac{C}{n^r}{∥{\delta }_n^{2r}{g}^{\left(2r\right)}∥}_{\Phi }\le C{\omega }_{2r,\phi }{\left(f,\frac{1}{\sqrt{n}}\right)}_{\Phi }.\end{array}$$

The proof of the direct theorem is complete. □

3. Proofs of the Inverse and Equivalent Theorems

For proving Theorems 7 and 8, we need the following lemmas.

Lemma 5.

If $f\in L_{\Phi }^{\ast }[0,\infty )$ and $n\ge 2r$, then

$${∥{\phi }^{2r}{L}_{n,r}^{\left(2r\right)}\left(f\right)∥}_{\Phi }\le C{n}^r{\parallel f\parallel }_{\Phi }.$$

Proof. 

Since

$$\begin{array}{cc}\hfill B_n^{\left(2r\right)}(f,x)& =\frac{1}{n+1}\sum _{k=1}^{\infty }{b}_{n,k}^{\left(2r\right)}\left(x\right){\int }_0^{\infty }{b}_{n,k}\left(t\right)f\left(t\right)\mathrm{d}t\hfill \\ & =\frac{1}{n+1}\sum _{k=1}^{\infty }\frac{(n+k)!}{(k-1)!n!}\sum _{i=0}^{2r}\left(\genfrac{}{}{0pt}{}{2r}{i}\right){(-1)}^i{D}^{2r-i}\left(x^{k-1}\right)D^i\left(\frac{1}{{(1+x)}^{n+k}}\right){\int }_0^{\infty }{b}_{n,k}\left(t\right)f\left(t\right)\mathrm{d}t\hfill \\ & =\prod _{j=2}^{2r}(n+j)\sum _{k=1}^{\infty }{b}_{n+2r,k}\left(x\right){\int }_0^{\infty }\sum _{i=0}^{2r}\left(\genfrac{}{}{0pt}{}{2r}{i}\right){(-1)}^i{b}_{n,k+2r-i}\left(t\right)f\left(t\right)\mathrm{d}t\hfill \\ & \le \prod _{j=2}^{2r}(n+j)\sum _{k=1}^{\infty }{b}_{n+2r,k}\left(x\right){\int }_0^{\infty }\sum _{i=0}^{2r}\left(\genfrac{}{}{0pt}{}{2r}{i}\right)b_{n,k+2r-i}\left(t\right)f\left(t\right)\mathrm{d}t,\hfill \end{array}$$

(10)

we have

$$\begin{array}{c}{\phi }^{2r}\left(x\right)B_n^{\left(2r\right)}(f,x)\le \prod _{j=2}^{2r}(n+j)\sum _{k=1}^{\infty }{b}_{n+2r,k}{\phi }^{2r}\left(x\right){\int }_0^{\infty }\sum _{i=0}^{2r}\left(\genfrac{}{}{0pt}{}{2r}{i}\right)b_{n,k+2r-i}\left(t\right)f\left(t\right)\mathrm{d}t\hfill \\ \hfill \begin{array}{c}\hspace{1em}\hspace{1em}\hspace{1em}\le \prod _{j=2}^{2r}(n+j)\sum _{k=1}^{\infty }{b}_{n,k+r}\left(x\right)\frac{(k+r-1)!n!}{(n+r+k)!}\frac{(n+2r+k)!}{(k-1)!(n+2r)!}{\int }_0^{\infty }\sum _{i=0}^{2r}\left(\genfrac{}{}{0pt}{}{2r}{i}\right)b_{n,k+2r-i}\left(t\right)f\left(t\right)\mathrm{d}t\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\le \frac{C{n}^r}{n+1}\sum _{i=0}^{2r}\left(\genfrac{}{}{0pt}{}{2r}{i}\right)\sum _{k=1}^{\infty }{b}_{n,k+r}\left(x\right){\int }_0^{\infty }{b}_{n,k+2r-i}\left(t\right)f\left(t\right)\mathrm{d}t.\hfill \end{array}\end{array}$$

Therefore, by Jensen’s inequality [20] and the inequality (2), we obtain

$$\begin{array}{c}{∥{\phi }^{2r}\left(x\right)B_n^{\left(2r\right)}\left(f\right)∥}_{\Phi }\hfill \\ \hfill \begin{array}{c}\hspace{1em}\hspace{1em}\le 2\underset{\lambda >0}{inf}\left\{\lambda :{\int }_0^{\infty }\Phi \left(\frac{1}{n+1}\sum _{i=0}^{2r}\left(\genfrac{}{}{0pt}{}{2r}{i}\right)\sum _{k=1}^{\infty }{b}_{n,k+r}\left(x\right){\int }_0^{\infty }{b}_{n,k+2r-i}\left(t\right)\frac{C{n}^r\left|f\left(t\right)\right|}{\lambda }\mathrm{d}t\right)\mathrm{d}x\le 1\right\}\hfill \\ \hspace{1em}\hspace{1em}\le 2\underset{\lambda >0}{inf}\left\{\lambda :{\int }_0^{\infty }\frac{1}{2^{2r}}\sum _{i=0}^{2r}\left(\genfrac{}{}{0pt}{}{2r}{i}\right)\Phi \left(\frac{1}{n+1}\sum _{k=1}^{\infty }{b}_{n,k+r}\left(x\right){\int }_0^{\infty }{b}_{n,k+2r-i}\left(t\right)\frac{C{n}^r\left|f\left(t\right)\right|}{\lambda }\mathrm{d}t\right)\mathrm{d}x\le 1\right\}\hfill \\ \hspace{1em}\hspace{1em}\le 2\underset{\lambda >0}{inf}\left\{\lambda :{\int }_0^{\infty }\frac{1}{2^{2r}}\sum _{i=0}^{2r}\left(\genfrac{}{}{0pt}{}{2r}{i}\right)\Phi \left(\frac{1}{n+1}\sum _{k=1}^{\infty }{b}_{n,k}\left(x\right){\int }_0^{\infty }{b}_{n,k+r-i}\left(t\right)\frac{C{n}^r\left|f\left(t\right)\right|}{\lambda }\mathrm{d}t\right)\mathrm{d}x\le 1\right\}\hfill \\ \hspace{1em}\hspace{1em}\le 2\underset{\lambda >0}{inf}\left\{\lambda :{\int }_0^{\infty }\frac{1}{2^{2r}}\sum _{i=0}^{2r}\left(\genfrac{}{}{0pt}{}{2r}{i}\right)\frac{1}{n+1}\sum _{k=1}^{\infty }{b}_{n,k}\left(x\right){\int }_0^{\infty }{b}_{n,k+r-i}\left(t\right)\Phi \left(\frac{C{n}^r\left|f\left(t\right)\right|}{\lambda }\right)\mathrm{d}t\mathrm{d}x\le 1\right\}\hfill \\ \hspace{1em}\hspace{1em}\le C{n}^r{\parallel f\parallel }_{\Phi },\hfill \end{array}\end{array}$$

where $b_{n,k+r-i}\left(t\right)=0$ for $n+r-i\le 0$. Combining this with Equation (5) leads to

$${∥{\phi }^{2r}\left(x\right)L_{n,r}^{\left(2r\right)}\left(f\right)∥}_{\Phi }\le \sum _{i=0}^{2r-1}|c_i\left(n\right)|{∥{\phi }^{2r}\left(x\right)B_{n_i}^{\left(2r\right)}\left(f\right)∥}_{\Phi }\le \sum _{i=0}^{2r-1}|c_i\left(n\right)|C{n}_i^r{\parallel f\parallel }_{\Phi }\le C{n}^r{\parallel f\parallel }_{\Phi }.$$

Lemma 5 is thus proved. □

Lemma 6.

Let $f\in L_{\Phi }^{\ast }[0,\infty )$ and $n\ge 2r$. Then

$${∥{\phi }^{2r}{L}_{n,r}^{\left(2r\right)}\left(f\right)∥}_{\Phi }\le C{∥{\phi }^{2r}{f}^{\left(2r\right)}∥}_{\Phi }.$$

Proof. 

Integrating by parts $2r$ times in Equation (10) gives

$$\begin{array}{cc}\hfill B_n^{\left(2r\right)}(f,x)& =\frac{1}{n+1}\prod _{j=1}^{2r}(n+j)\sum _{k=1}^{\infty }{b}_{n+2r,k}\left(x\right){\int }_0^{\infty }\sum _{i=0}^{2r}\left(\genfrac{}{}{0pt}{}{2r}{i}\right){(-1)}^i{b}_{n,k+2r-i}\left(t\right)f\left(t\right)\mathrm{d}t\hfill \\ & =\frac{(n+1)(n+2)\dots (n+2r)}{n(n-1)\dots (n-2r+1)}\frac{1}{n+1}\sum _{k=1}^{\infty }{b}_{n+2r,k}\left(x\right){\int }_0^{\infty }{b}_{n-2r,k+2r}^{\left(2r\right)}\left(t\right)f\left(t\right)\mathrm{d}t\hfill \\ & =\frac{(n+2r)!(n-2r)!}{{(n!)}^2}\frac{1}{n+1}\sum _{k=1}^{\infty }{b}_{n+2r,k}\left(x\right){\int }_0^{\infty }{b}_{n-2r,k+2r}\left(t\right)f^{\left(2r\right)}\left(t\right)\mathrm{d}t.\hfill \end{array}$$

Accordingly,

$$\begin{array}{c}{\phi }^{2r}\left(x\right)B_n^{\left(2r\right)}(f,x)\hfill \\ \hfill \begin{array}{c}\hspace{1em}\hspace{1em}\hspace{1em}=\frac{(n+2r)!(n-2r)!}{{(n!)}^2}\frac{1}{n+1}\sum _{k=1}^{\infty }{b}_{n+2r,k}\left(x\right){\phi }^{2r}\left(x\right){\int }_0^{\infty }{b}_{n-2r,k+2r}\left(t\right){\phi }^{-2r}\left(t\right){\phi }^{2r}\left(t\right)f^{\left(2r\right)}\left(t\right)\mathrm{d}t\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\le \frac{C}{n+1}\sum _{k=1}^{\infty }{b}_{n-2r+2,k+r}\left(x\right){\int }_0^{\infty }{b}_{n,k+r}\left(t\right){\phi }^{2r}\left(t\right)f^{\left(2r\right)}\left(t\right)\mathrm{d}t.\hfill \end{array}\end{array}$$

Employing Inequality (2) and Jensen’s inequality [20] reveals

$$\begin{array}{c}{∥{\phi }^{2r}\left(x\right)B_n^{\left(2r\right)}\left(f\right)∥}_{\Phi }\hfill \\ \hfill \begin{array}{c}\hspace{1em}\hspace{1em}\hspace{1em}\le 2\underset{\lambda >0}{inf}\left\{\lambda :{\int }_0^{\infty }\Phi \left(\frac{C}{n+1}\sum _{k=1}^{\infty }{b}_{n-2r+2,k+r}\left(x\right){\int }_0^{\infty }{b}_{n,k+r}\left(t\right)\frac{{\phi }^{2r}\left(t\right)\left|f^{\left(2r\right)}\left(t\right)\right|}{\lambda }\mathrm{d}t\right)\mathrm{d}x\le 1\right\}\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\le 2\underset{\lambda >0}{inf}\left\{\lambda :{\int }_0^{\infty }\Phi \left(\frac{C}{n+1}\sum _{k=1}^{\infty }{b}_{n-2r+2,k}\left(x\right){\int }_0^{\infty }{b}_{n,k}\left(t\right)\frac{{\phi }^{2r}\left(t\right)\left|f^{\left(2r\right)}\left(t\right)\right|}{\lambda }\mathrm{d}t\right)\mathrm{d}x\le 1\right\}\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\le 2\underset{\lambda >0}{inf}\left\{\lambda :{\int }_0^{\infty }\frac{{\sum }_{k=1}^{\infty }{b}_{n-2r+2,k}\left(x\right)}{n-2r+3}{\int }_0^{\infty }{b}_{n,k}\left(t\right)\Phi \left(\frac{\frac{C(n-2r+3)}{n+1}{\phi }^{2r}\left(t\right)\left|f^{\left(2r\right)}\left(t\right)\right|}{\lambda }\right)\mathrm{d}t\mathrm{d}x\le 1\right\}\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\le 2\underset{\lambda >0}{inf}\left\{\lambda :\frac{n+1}{n-2r+3}{\int }_0^{\infty }\Phi \left(\frac{\frac{C(n-2r+3)}{n+1}{\phi }^{2r}\left(t\right)\left|f^{\left(2r\right)}\left(t\right)\right|}{\lambda }\right)\mathrm{d}t\le 1\right\}\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\le C{∥{\phi }^{2r}{f}^{\left(2r\right)}∥}_{\Phi }.\hfill \end{array}\end{array}$$

Applying the above inequality and Inequality (5) results in

$$\begin{array}{c}{∥{\phi }^{2r}{L}_{n,r}^{\left(2r\right)}\left(f\right)∥}_{\Phi }={∥\sum _{i=0}^{2r-1}{c}_i\left(n\right){\phi }^{2r}{B}_{n_i}^{\left(2r\right)}\left(f\right)∥}_{\Phi }\le \sum _{i=0}^{2r-1}\left|c_i\left(n\right)\right|{∥{\phi }^{2r}{B}_{n_i}^{\left(2r\right)}\left(f\right)∥}_{\Phi }\\ \le C\sum _{i=0}^{2r-1}\left|c_i\left(n\right)\right|{∥{\phi }^{2r}{f}^{\left(2r\right)}∥}_{\Phi }\le C{∥{\phi }^{2r}{f}^{\left(2r\right)}∥}_{\Phi }.\end{array}$$

The lemma is proved. □

Proof of Theorem 7.

From Lemmas 5 and 6 and [21] (Theorem 2.2), we obtain

$$K_{2r,\phi }{\left(f,\frac{1}{n^{r/2}}\right)}_{\Phi }\le \frac{C}{n^r}\sum _{k=1}^n{k}^{r-1}{\parallel L_{n,r}\left(f\right)-f\parallel }_{\Phi }.$$

Utilizing Inequality (4) concludes the inverse theorem. □

Proof of Theorem 8.

Using the so-called order function $\psi \left(t\right)=t^{\alpha }{|lnt|}^{\beta }{e}^{{|lnt|}^{\gamma }}$ for $0<\alpha <1$, $\beta \in \mathbb{R}$, and $\gamma <1$ and combining Theorems 7 and 8 conclude the equivalent theorem. □