Weighted Approximation of Functions by the Szász–Mirakjan–Kantorovich Operator

Abstract

We investigate the weighted approximation of functions in $L_p$-norm by Kantorovich modifications of the classical Szász–Mirakjan operator, with weights of type $(1+x)^{\alpha }$, $\alpha \in \mathbb {R}$. By defining an appropriate K-functional we prove direct inequality for them.

1 Introduction

The classical Szász–Mirakjan operator (see [9, 10]) is defined for bounded functions f(x) in $[0,\infty )$ by the formula

$$\begin{aligned} S_n f(x)=S_n(f;x) = \sum _{k=0}^{\infty } f\left( \frac{k}{n}\right) s_{n,k}(x),\quad x\ge 0, \end{aligned}$$

(1.1)

where

$$\begin{aligned} s_{n,k}(x) = e^{-nx}\frac{(nx)^k}{k!} \end{aligned}$$

and the Kantorovich modification of $S_n$ is defined (see, for instance, [3, Chapter 9]) by

$$\begin{aligned} {{\tilde{S}}}_nf(x)= {{\tilde{S}}}_n(f;x)= \sum _{k=0}^{\infty } s_{n,k}(x)\,n\int _{\frac{k}{n}}^{\frac{k+1}{n}}f(u)du,\quad x\ge 0. \end{aligned}$$

(1.2)

This operator is well-defined for every function f(x), which is summable on any finite closed subinterval of $[0,\infty )$.

There are many papers about weighted approximation of functions by $S_n$ in uniform norm—see, for instance, the bibliography of [6]. That is not the case about weighted approximation of functions by Kantorovich modifications of $S_n$. The best results, to our knowledge, are the next inequalities of weak type in terminology of [2], proved in [3, p.159, Theorem 10.1.3.].

Let $w^*(x)=x^{\gamma (0)}(1+x)^{\gamma (\infty )}$ where $\gamma (\infty )$ is arbitrary and $-1/p<\gamma (0)<1-1/p$ for $1 \le p \le \infty$, for $p=1$ and $p=\infty$ $\gamma (0)$ may also be equal to zero.

Theorem 1.1

Suppose $w^*f \in L_p[0,\infty )$ and either $1\le p \le \infty$ and $\alpha <1$, or $1<p<\infty$ and $\alpha \le 1$. Then for $\tilde{S}_n^*$ the next equivalency is true.

$$\begin{aligned} \left\| w^*\left( {{\tilde{S}}}_nf-f\right) \right\| _{L_p[0,\infty )} =O\Big (n^{-\alpha }\Big )\Leftrightarrow \left\| w^*\Delta _{h\sqrt{\varphi }}^2f \right\| _{L_p[2h^2,\infty )}=O\Big (h^{2\alpha }\Big ). \end{aligned}$$

Here $\Vert \circ \Vert _{p(J)}$ stands for the usual $L_p$-norm on the interval J, $\ \varphi (x)=x$ and

$$\begin{aligned} \Delta _{h\sqrt{\varphi (x)}}^2(f,x)=f\left( x-h\sqrt{\varphi (x)}\right) -2f(x)+f\left( x+h\sqrt{\varphi (x)}\right) . \end{aligned}$$

Our goal in this paper is to investigate the approximation of functions in the $L_p$-norm by the Szász–Mirakjan–Kantorovich operator. We prove Jackson-type inequality for the weighted error of approximation and by defining an appropriate K-functional we prove direct inequality for it.

Before stating our main result, let us introduce the needed notations. The weights under consideration in our survey are

$$\begin{aligned} w(x)=(1+x)^{\alpha }, \quad \alpha \in {\mathbb {R}}. \end{aligned}$$

(1.3)

By $\varphi (x)=x$ we denote the weight which is naturally connected with the second moment of the Szász–Mirakjan operator. The first derivative operator is denoted by $D=\frac{d}{dx}$. Thus, $Dg(x)=g'(x)$ and $D^k g(x)=g^{(k)}(x)$ for every natural k. We define the second order differential operator ${{\tilde{D}}}$ by the formula

$$\begin{aligned} {{\tilde{D}}} g(x)=D(\varphi Dg)(x)=xg''(x)+g'(x). \end{aligned}$$

The space $AC_{loc}(0,\infty )$ consists of the functions which are absolutely continuous in [a, b] for every $[a,b] \subset (0,\infty )$. We also set

$$\begin{aligned}&L_p(w) = \{f : f,Df \in AC_{loc}(0,\infty ), \ w(x)f(x) \in L_p[0,\infty ) \}, \\&W_p(w) =\left\{ \begin{array}{cc} \displaystyle \{f : f,Df \in AC_{loc}(0,\infty ), w(x){{\tilde{D}}}f \in L_p[0,\infty ), \lim _{x\rightarrow {0_+}}\varphi (x) Df(x) = 0\} , \alpha \le 0\\ \displaystyle \{f : f,Df \in AC_{loc}(0,\infty ), w(x){{\tilde{D}}}f \in L_p[0,\infty ), \lim _{x\rightarrow {0_+,\infty }}\varphi (x) Df(x) = 0\} , \alpha >0 \end{array} \right. , \\&L_p(w)+W_p(w) = \big \{f : f=f_1+f_2, \ f_1\in L_p(w), \ f_2\in W_p(w)\big \}. \end{aligned}$$

Also, we define a K-functional $K_w(f,t)_p$ for $t>0$, by

$$\begin{aligned} K_w(f,t)_p = \inf \left\{ \Vert w(f-g)\Vert _p+t\left\| w{{\tilde{D}}}g\right\| _p : f-g \in L_p(w), \ g\in W_p(w) \right\} .\nonumber \\ \end{aligned}$$

(1.4)

The relation “$\theta _1(f,t)$ is equivalent to $\theta _2(f,t)$”, in notation: $\theta _1(f,t) \sim \theta _2(f,t)$, means that there exists a positive constant C independent of f and t such that

$$\begin{aligned} C^{-1}\theta _1(f,t) \le \theta _2(f,t) \le C\theta _1(f,t). \end{aligned}$$

Above and throughout C denotes a positive constant, not necessarily the same at each occurrence, which is independent of the function f(x) (or g(x)), and the parameter n (or t) in the specified range.

Our main results are the following theorems. The first one is a Jackson-type inequality. It shows that the rate of convergence of ${{\tilde{S}}}_n$ is at least $n^{-1}$ if the approximated function is smooth enough.

Theorem 1.2

Let ${{\tilde{S}}}_n$ be defined by (1.2), w(x) by (1.3) and $1 \le p \le \infty$. Then there exist absolute constant $C>0$ such that for all $f\in W_p(w)$ and all $n\in {\mathbb {N}}$ there hold

$$\begin{aligned} \left\| w\left( {{\tilde{S}}}_n f - f\right) \right\| _p \le \frac{C(\alpha )}{n}\left\| w{{\tilde{D}}}f\right\| _p. \end{aligned}$$

Theorem 1.3

Let ${{\tilde{S}}}_n$ be defined by (1.2), the K-functional be given by (1.4), w(x) by (1.3) and $1 \le p \le \infty$. Then there exist absolute constant $C>0$ such that for all $f\in L_p(w)+ W_p(w)$ and all $n\in {\mathbb {N}}$ there hold

$$\begin{aligned} \Vert w({{\tilde{S}}}_nf-f) \Vert _p \le C K_w\left( f,\frac{1}{n}\right) _p. \end{aligned}$$

Remark 1.4

The inequalities in Theorems 1.2 and 1.3 are stronger than the results mentioned above. They are stronger even for $w(x)=1$ - see [8, p. 4]

Remark 1.5

Very important question is how to characterize the K-functionals $K_w\left( f,t\right) _p$ by appropriate moduli of smoothness. To our knowledge it is completely open even for $w(x)=1$. In series of papers [4, 5, 7] the authors introduced new moduli of smoothness and characterized the next weighted K-functionals

$$\begin{aligned} K_w^*(f,t)_p = \inf \left\{ \Vert w(f-g)\Vert _p+t\left\| w D^2g\right\| _p : f-g, D^2g \in L_p(w) \right\} , \end{aligned}$$

which are different from $K_w(f,t)_p$. But under some additional restrictions on the functions f, for $p>1$, they could be used in order to characterize the K-functionals $K_w(f,t)_p$. For $p=1$ probably new moduli are needed, even in the unweighted case, i.e. $w(x)=1$.

2 Auxiliary Results

In this section we collect some properties of $S_n$, ${{\tilde{S}}}_n$ and $s_{n,k}$, which can be found in [3, 11, 12], or verified by direct computation. Here we also prove all the lemmas we need to establish the main result.

We begin with the relations:

$$\begin{aligned}&\sum _{k=0}^\infty s_{n,k}(x)=1,\quad x\ge 0, \end{aligned}$$

(2.1)

$$\begin{aligned}&\sum _{k=0}^\infty k\,s_{n,k}(x)=nx,\quad x\ge 0,\end{aligned}$$

(2.2)

$$\begin{aligned}&\int _0^\infty s_{n,k}(x)\,dx =\frac{1}{n}. \end{aligned}$$

(2.3)

The first several moments of the operators $S_n$ and ${{\tilde{S}}}_n$ are

$$\begin{aligned}&S_n(1,x) = 1, \quad S_n(\circ -x,x) = 0, \quad S_n\left( (\circ -x)^2,x\right) = \frac{\varphi (x)}{n}; \end{aligned}$$

(2.4)

$$\begin{aligned}&{{\tilde{S}}}_n(1,x) = 1, \quad {{\tilde{S}}}_n(\circ -x,x) = \frac{1}{2n},\quad {{\tilde{S}}}_n((\circ -x)^2,x) =\frac{\varphi (x)}{n}+\frac{1}{2n^2}. \end{aligned}$$

(2.5)

Generally, it was shown in [3, (9.4.14)]

$$\begin{aligned} S_n\left( (\circ -x)^{2m},x\right) \le C(m)\left( \frac{\varphi (x)}{n}\right) ^m \quad \text{ for } \quad x \ge \frac{1}{n}, \quad m\in {\mathbb {N}} . \end{aligned}$$

(2.6)

Also, in [3] the next inequalities are proved. In [3, p.161, section 10.2] the next inequality about the boundedness of ${{\tilde{S}}}_n$ in weighted norm is proved, i.e. for every function $f\in L_p(w)$ the next inequality is true

$$\begin{aligned} \left\| w{{\tilde{S}}}_nf \right\| _p \le C(\alpha ) \left\| wf \right\| _p, \end{aligned}$$

(2.7)

and in [3, p.163]

$$\begin{aligned} \sum _{k=0}^\infty s_{n,k}(x)\left( 1+\frac{k}{n} \right) ^m \le C(1+x)^m \quad \text{ where }\quad m\in {\mathbb {Z}}. \end{aligned}$$

(2.8)

We need to prove some additional lemmas. The first one is a simple generalization of inequality (2.8).

Lemma 2.1

For $\alpha \in {\mathbb {R}}$ there exists a constant $C(\alpha )$ such that for every natural $n \ge |\alpha |$ and every $x \in [0,\infty )$

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

(2.9)

Proof

Let $m\in {\mathbb {N}}$ be the smallest integer such that $m>|\alpha |$. By Holder’s inequality we have

$$\begin{aligned} \sum _{k=0}^\infty s_{n,k}(x)\left( 1+\frac{k}{n}\right) ^\alpha \le \left\{ \sum _{k=0}^\infty s_{n,k}(x)\left( 1+\frac{k}{n}\right) ^{sign(\alpha )m} \right\} ^{\frac{|\alpha |}{m}} \left\{ \sum _{k=0}^\infty s_{n,k}(x) \right\} ^{1-\frac{|\alpha |}{m}}. \end{aligned}$$

and the lemma follows from  (2.8) and  (2.4). $\square$

We need the next very important technical result.

Lemma 2.2

For every integer m there exists a constant C(m) such that for every naturals n and k, $n>|m|$ the next inequalities are true

$$\begin{aligned}&1-\frac{C(m)}{n\left( 1+\frac{k}{n} \right) ^2} \le \frac{n^{m+1}(n+k)!}{(n+k+m)!}\int _0^\infty s_{n,k}(x)(1+x)^mdx \le 1. \end{aligned}$$

(2.10)

Proof

We consider two cases.

• 1. $m \ge 0$.

  Obviously the inequalities are true for $m=0$. And for $m>0$ we have

  $$\begin{aligned} \int _0^\infty s_{n,k}(x)(1+x)^{m+1}dx=&\int _0^\infty s_{n,k}(x)(1+x)^mdx\\&+ \frac{k+1}{n}\int _0^\infty s_{n,k+1}(x)(1+x)^mdx \end{aligned}$$

  and consequently

  $$\begin{aligned}&\frac{n^{m+2}(n+k)!}{(n+k+m+1)!}\int _0^\infty s_{n,k}(x)(1+x)^{m+1}dx\\&\quad = \frac{n}{n+k+m+1}\frac{n^{m+1}(n+k)!}{(n+k+m)!}\int _0^\infty s_{n,k}(x)(1+x)^mdx\\&\qquad +\frac{k+1}{n+k+1} \frac{n^{m+1}(n+k+1)!}{(n+k+m+1)!}\int _0^\infty s_{n,k+1}(x)(1+x)^mdx. \end{aligned}$$

  Now, inductively we have

  $$\begin{aligned} \frac{n^{m+2}(n+k)!}{(n+k+m+1)!}&\int _0^\infty s_{n,k}(x)(1+x)^{m+1}dx< \frac{n}{n+k+m+1}+\frac{k+1}{n+k+1}<1 \end{aligned}$$

  and

  $$\begin{aligned}&\frac{n^{m+2}(n+k)!}{(n+k+m+1)!} \int _0^\infty s_{n,k}(x)(1+x)^{m+1}dx\\&\quad > \frac{n}{n+k+m+1}\left[ 1-\frac{C_1(m)}{n\left( 1+\frac{k}{n} \right) ^2} \right] +\frac{k+1}{n+k+1}\left[ 1-\frac{C_2(m)}{n\left( 1+\frac{k+1}{n} \right) ^2} \right] \\&\quad = \frac{n}{n+k+m+1}+\frac{k+1}{n+k+1}-\frac{C_3(m)}{n\left( 1+\frac{k}{n} \right) ^2} \\&\quad =1-\frac{m}{n\left( 1+\frac{k+1}{n} \right) \left( 1+\frac{k+m+1}{n} \right) }-\frac{C_3(m)}{n\left( 1+\frac{k}{n} \right) ^2} =1-\frac{C(m)}{n\left( 1+\frac{k}{n} \right) ^2}. \end{aligned}$$

• 2. $m< 0$.

  Let us denote

  $$\begin{aligned} I_{k,m}=\int _0^\infty s_{n,k}(x)(1+x)^{m}dx. \end{aligned}$$

  We have

  $$\begin{aligned} I_{k,m}=\frac{n}{k}I_{k-1,m+1}-\frac{n}{k}I_{k-1,m} \end{aligned}$$

  (2.11)

  and after integrating by parts

  $$\begin{aligned} I_{k,m}=I_{k-1,m}+\frac{m}{n}I_{k,m-1}. \end{aligned}$$

  (2.12)

  Multiplying (2.12) by $\frac{n}{k}$ and summing with (2.11) we obtain

  $$\begin{aligned} I_{k,m}=\frac{n}{n+k}I_{k-1,m+1}+\frac{m}{n+k}I_{k,m-1} \end{aligned}$$

  (2.13)

  and consequently

  $$\begin{aligned} I_{k,m} \le \frac{n}{n+k}I_{k-1,m+1}. \end{aligned}$$

  (2.14)

• 2a. $k\ge |m|$.

  Applying (2.14) m times and using (2.3) we have

  $$\begin{aligned} I_{k,m} \le \frac{n}{n+k}...\frac{n}{n+k+m+1}I_{k+m,0}= \frac{(n+k+m)!}{n^{m+1}(n+k)!} \end{aligned}$$

  (2.15)

  i.e

  $$\begin{aligned} \frac{n^{m+1}(n+k)!}{(n+k+m)!}\int _0^\infty s_{n,k}(x)(1+x)^mdx \le 1. \end{aligned}$$

  From (2.15) we have

  $$\begin{aligned} I_{k,m-1} \le \frac{(n+k+m-1)!}{n^{m}(n+k)!} \end{aligned}$$

  and than from (2.13) it follows

  $$\begin{aligned} I_{k,m} \ge \frac{n}{n+k}I_{k-1,m+1}+\frac{m}{n+k} \frac{(n+k+m-1)!}{n^{m}(n+k)!} \end{aligned}$$

  or

  $$\begin{aligned} \frac{n^{m+1}(n+k)!}{(n+k+m)!}I_{k,m}\ge & {} \frac{n^{m+2}(n+k-1)!}{(n+k+m)!}I_{k-1,m+1}\\&+\frac{mn}{(n+k)(n+k+m)} \frac{(n+k+m-1)!}{n^{m}(n+k)!} \end{aligned}$$

  i.e.

  $$\begin{aligned} \frac{n^{m+1}(n+k)!}{(n+k+m)!}I_{k,m} \ge \frac{n^{m+2}(n+k-1)!}{(n+k+m)!}I_{k-1,m+1} +\frac{C(m)}{n\left( 1+\frac{k}{n} \right) ^2}. \end{aligned}$$

  Applying this inequality m times we obtain (2.10).

• 2b. $k< |m|$.

  Applying (2.14) k times

  $$\begin{aligned} I_{k,m} \le \frac{n^kn!}{(n+k)!}I_{0,m+k}. \end{aligned}$$

  But

  $$\begin{aligned} I_{0,m+k}=\int _0^\infty s_{n,0}(x)(1+x)^{m+k}dx<\int _0^\infty s_{n,0}(x)dx=\frac{1}{n} \end{aligned}$$

  and consequently

  $$\begin{aligned} I_{k,m} \le \frac{n^{k-1}n!}{(n+k)!} \le \frac{(n+k+m)!}{n^{m+1}(n+k)!}, \end{aligned}$$

  i.e.

  $$\begin{aligned} \frac{n^{m+1}(n+k)!}{(n+k+m)!}\int _0^\infty s_{n,k}(x)(1+x)^mdx \le 1. \end{aligned}$$

  We have

  $$\begin{aligned} n\int _0^\infty s_{n,k}(x)(1+x)^mdx&> n\int _0^\infty \frac{e^{-(n-m)}(nx)^k}{k!}dx\\&=\left( \frac{n}{n-m}\right) ^{k} n\int _0^\infty s_{n-m,k}(x)dx\\&=\left( \frac{n}{n-m}\right) ^{k}>1+\frac{km}{n-m}>1-\frac{C(m)}{n}. \end{aligned}$$

  Then

  $$\begin{aligned} \frac{n^{m+1}(n+k)!}{(n+k+m)!}\int _0^\infty s_{n,k}(x)(1+x)^mdx&>\prod _{k=0}^{|m|-1}\left( 1+\frac{k-i}{n} \right) \left( 1-\frac{C(m)}{n} \right) \\&>1-\frac{C(m)}{n}. \end{aligned}$$

  The lemma is proved.

$\square$

The next lemma is an elementary consequence of this lemma.

Lemma 2.3

For w(x) defined by (1.3) the next inequality is true

$$\begin{aligned} \int _0^\infty w(x)s_{n,k}(x)dx\le \frac{C(\alpha )}{n}w\left( \frac{k}{n} \right) . \end{aligned}$$

(2.16)

Proof

By Holder’s inequality applied for the smallest integer m such that $m>|\alpha |$ we have

$$\begin{aligned} \int _0^\infty w(x)s_{n,k}(x)dx \le \left\{ \int _0^\infty s_{n,k}(x)(1+x)^{sign(\alpha )m}dx \right\} ^{\frac{|\alpha |}{m}} \left\{ \int _0^\infty s_{n,k}(x)dx \right\} ^{1-\frac{|\alpha |}{m}} \end{aligned}$$

and the lemma follows from (2.10) and (2.3). $\square$

We need more technical results. Let us denote by $\phi$ the function $\phi (z)=\ln z$. In [8, p.7, 2.20 and p.11, 2.26] the next two estimations for ${{\tilde{S}}}$ are proved

$$\begin{aligned} \left\| {{\tilde{S}}}_n\phi -\phi \right\| _1 \le \frac{C}{n} \end{aligned}$$

(2.17)

and

$$\begin{aligned} \left\| x\left( {{\tilde{S}}}_n\phi -\phi \right) \right\| _\infty \le \frac{C}{n}. \end{aligned}$$

(2.18)

Lemma 2.4

Let $1\le p\le \infty$. Then for all $f \in W_p(w)$ and $n\in {\mathbb {N}}$ there holds

$$\begin{aligned} \left\| \varphi w\left( {{\tilde{S}}}_n\phi -\phi \right) Df\right\| _p \le \frac{C(\alpha )}{n}\left\| w{{\tilde{D}}}f\right\| _p. \end{aligned}$$

Proof

We consider the cases $\alpha \le 0$ and $\alpha > 0$ separately.

• 1. $\alpha \le 0$.

  We have

  $$\begin{aligned}&\left| \varphi (x) w(x)Df(x)\right| =\left| xw(x)Df(x)\right| \\&\quad =\left| w(x)\int _0^x {{\tilde{D}}}f(t)dt \right| \le \int _0^x \left| w(t){{\tilde{D}}}f(t) \right| dt, \end{aligned}$$

  and consequently

  $$\begin{aligned} \left| \varphi (x)w(x)Df(x)\right| \le \left| \int _0^\infty w(t)\tilde{D}f(t)dt \right| =\left\| w{{\tilde{D}}}f \right\| _1 \end{aligned}$$

  and

  $$\begin{aligned} \left| \varphi (x)w(x)Df(x)\right| \le x\left\| w{{\tilde{D}}}f \right\| _\infty . \end{aligned}$$

  Then for $p=1$

  $$\begin{aligned} \left\| w\varphi \left( {{\tilde{S}}}_n\phi -\phi \right) Df\right\| _1 \le \left\| w{{\tilde{D}}}f\right\| _1 \left\| {{\tilde{S}}}_n\phi -\phi \right\| _1 \le \frac{C}{n}\left\| w{{\tilde{D}}}f\right\| _1 \end{aligned}$$

  where for the last inequality we used the estimation (2.17). For $p=\infty$

  $$\begin{aligned} \left\| w\varphi \left( {{\tilde{S}}}_n\phi -\phi \right) Df\right\| _\infty \le \left\| w{{\tilde{D}}}f\right\| _1 \left\| \varphi \left( \tilde{S}_n\phi -\phi \right) \right\| _\infty \le \frac{C}{n} \left\| w\tilde{D}f\right\| _\infty \end{aligned}$$

  where for the last inequality we used the estimation (2.18).

• 2. $\alpha > 0$.

  For $L_1$ we use the representation

  $$\begin{aligned} \left| \varphi (x)w(x)Df(x)\right| =\left| w(x)\int _x^\infty \tilde{D}f(t)dt \right| \le \int _v^\infty \left| w(t){{\tilde{D}}}f(t) \right| dt. \end{aligned}$$

  Then as above

  $$\begin{aligned} \left\| w\varphi \left( {{\tilde{S}}}_n\phi -\phi \right) Df\right\| _1 \le \left\| w{{\tilde{D}}}f\right\| _1 \left\| {{\tilde{S}}}_n\phi -\phi \right\| _1 \le \frac{C}{n} \left\| w{{\tilde{D}}}f\right\| _1. \end{aligned}$$

  For $L_\infty$ we consider two cases.

• 2a. $x \le 1$.

  In this case $(1+x)^\alpha \le 2^\alpha (1+t)^\alpha$ for $0 \le t \le 1$. Consequently,

  $$\begin{aligned} \left| \varphi (x)w(x) \left( {{\tilde{S}}}_n(\phi ;x)-\phi (x) \right) Df(x)\right|&=\left| w(x)\int _0^x {{\tilde{D}}}f(t)dt \left[ {{\tilde{S}}}_n(\phi ;x)-\phi (x) \right] \right| \\&\le 2^\alpha \int _0^x \left| w(t){{\tilde{D}}}f(t)\right| dt \left| {{\tilde{S}}}_n(\phi ;x)-\phi (x) \right| \\&\le C(\alpha )\left\| w{{\tilde{D}}}f\right\| _\infty \left| x\left( {{\tilde{S}}}_n(\phi ;x)-\phi (x) \right) \right| \\&\le C(\alpha )\left\| w{{\tilde{D}}}f\right\| _\infty \left\| \varphi \left( {{\tilde{S}}}_n\phi -\phi \right) \right\| _\infty \\&\le \frac{C(\alpha )}{n} \end{aligned}$$

  where we used again the estimation (2.18).

• 2b. $x>1$.

  In this case $(1+x)^\alpha \sim x^\alpha$. And for $0<\alpha <1$ by using the Hardy’s inequality and the estimation (2.18) we obtain

  $$\begin{aligned}&\Big |w(x)\varphi (x) \left( {{\tilde{S}}}_n(\phi ;x)-\phi (x) \right) Df(x)\Big |\\&\quad \le \left| \frac{w(x)}{x}\int _0^x {{\tilde{D}}}f(t)dt\right| \left| x\left( {{\tilde{S}}}_n(\phi ;x)-\phi (x)\right) \right| \\&\quad \le C\left( \frac{1}{x^{1-\alpha }}\int _0^x \left| t\tilde{D}f(t)\right| \frac{dt}{t}\right) \left\| \varphi \left( {{\tilde{S}}}_n\phi -\phi \right) \right\| _\infty \\&\quad \le \frac{C(\alpha )}{n}\left\| w{{\tilde{D}}}f\right\| _\infty . \end{aligned}$$

  For $\alpha \ge 1$ again, using the Hardy’s inequality and the estimation (2.18), we obtain

  $$\begin{aligned}&\Big |w(x)\varphi (x) \left( {{\tilde{S}}}_n(\phi ,x)-\phi (x) \right) Df(x)\Big |\\&\quad \le \left| \frac{w(x)}{x}\int _x^\infty {{\tilde{D}}}f(t)dt\right| \left| x\left( {{\tilde{S}}}_n(\phi ;x)-\phi (x)\right) \right| \\&\quad \le C\left( x^{\alpha -1}\int _x^\infty \left| t\tilde{D}f(t)\right| \frac{dt}{t}\right) \left\| \varphi \left( {{\tilde{S}}}_n\phi -\phi \right) \right\| _\infty \\&\quad \le \frac{C(\alpha )}{n}\left\| w{{\tilde{D}}}f\right\| _\infty . \end{aligned}$$

$\square$

Lemma 2.5

For every $x\in [0,\infty )$ and $n\in {\mathbb {N}}$ there holds

$$\begin{aligned} \left| w(x){{\tilde{S}}}_n\left( \frac{ (\cdot )-x}{w(\cdot )};x\right) \right| \le \frac{C}{n}. \end{aligned}$$

Proof

We have by using the Lagrange’s formula

$$\begin{aligned} \left| w(x){{\tilde{S}}}_n\left( \frac{ (\cdot )-x}{w(\cdot )};x\right) \right|&=\left| w(x)\sum _{k=0}^{\infty }s_{n,k}(x)n\int _{\frac{k}{n}}^{\frac{k+1}{n}}\frac{t-x}{w(t)} dt\right| \\&\le \left| \sum _{k=0}^{\infty }s_{n,k}(x)\frac{w(x)}{w\left( \frac{k}{n} \right) } n\int _{\frac{k}{n}}^{\frac{k+1}{n}}\left( t-x\right) dt\right| \\&\quad +w(x)\sum _{k=0}^{\infty }s_{n,k}(x) n\int _{\frac{k}{n}}^{\frac{k+1}{n}}\left| t-x\right| \left| t-\frac{k}{n}\right| \frac{|w'(\xi )|}{w^2(\xi )}dt \end{aligned}$$

where

$$\begin{aligned} \xi \in \left[ \frac{k}{n}, \frac{k+1}{n}\right] . \end{aligned}$$

Now

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

From (2.9)

$$\begin{aligned} \frac{1}{2n}\sum _{k=0}^{\infty }s_{n,k}(x)\frac{w(x)}{w\left( \frac{k}{n} \right) }\le \frac{C}{n}. \end{aligned}$$

(2.19)

Since

$$\begin{aligned} s_{n,k}(x)\frac{k}{n}=xs_{n,k-1}(x) \end{aligned}$$

it follows again from (2.9) that

$$\begin{aligned} \left| \sum _{k=0}^{\infty }s_{n,k}(x)\frac{w(x)}{w\left( \frac{k}{n} \right) } \left( \frac{k}{n}-x \right) \right|&\le xw(x)\sum _{k=0}^{\infty }s_{n,k}(x)\left| w^{-1}\left( \frac{k+1}{n} \right) - w^{-1}\left( \frac{k}{n} \right) \right| \nonumber \\&\le \frac{Cxw(x)}{n}\sum _{k=0}^{\infty }\frac{s_{n,k}(x)}{\left( 1+\frac{k}{n} \right) ^{\alpha +1}} \le \frac{Cx}{n(1+x)} \le \frac{C}{n}. \end{aligned}$$

(2.20)

The second inequality above follows from the easily proved inequality

$$\begin{aligned} \left| \left( 1+\frac{k+1}{n} \right) ^{\beta }-\left( 1+\frac{k}{n} \right) ^{\beta } \right| \le \frac{C}{n}\left( 1+\frac{k}{n} \right) ^{\beta -1} \quad \text{ where }\quad \beta \in {\mathbb {R}}. \end{aligned}$$

From (2.19) and (2.20) we obtain

$$\begin{aligned} \left| \sum _{k=0}^{\infty }s_{n,k}(x)\frac{w(x)}{w\left( \frac{k}{n} \right) } n\int _{\frac{k}{n}}^{\frac{k+1}{n}}\left( t-x\right) dt\right| \le \frac{C}{2n}. \end{aligned}$$

(2.21)

Now, for $t \in \left[ \frac{k}{n}, \frac{k+1}{n}\right]$

$$\begin{aligned} \left| t-\frac{k}{n}\right| \frac{|w'(\xi )|}{w^2(\xi )}\le \frac{C}{n\left( 1+\frac{k}{n} \right) w\left( \frac{k}{n}\right) } \end{aligned}$$

and

$$\begin{aligned} |t-x| \le \left| \frac{k}{n}-x \right| +\left| \frac{k+1}{n}-x \right| . \end{aligned}$$

So

$$\begin{aligned}&\left| w(x)\sum _{k=0}^{\infty }s_{n,k}(x) n\int _{\frac{k}{n}}^{\frac{k+1}{n}}\left| t-x\right| \left| t-\frac{k}{n}\right| \frac{|w'(\xi )|}{w^2(\xi )}dt\right| dt\\&\quad \le \frac{Cw(x)}{n}\sum _{k=0}^{\infty }\frac{s_{n,k}(x)}{\left( 1+\frac{k}{n} \right) ^{\alpha +1}} \left| \frac{k}{n}-x \right| +\frac{Cw(x)}{n}\sum _{k=0}^{\infty }\frac{s_{n,k}(x)}{\left( 1+\frac{k}{n} \right) ^{\alpha +1}} \left| \frac{k+1}{n}-x \right| . \end{aligned}$$

We will estimate the first term. The estimation of the second is similar.

By applying Cauchy’s inequality we get

$$\begin{aligned} \sum _{k=0}^{\infty }\frac{s_{n,k}(x)}{\left( 1+\frac{k}{n} \right) ^{\alpha +1}} \left| \frac{k}{n}-x \right| \le \left[ \sum _{k=0}^{\infty }\frac{s_{n,k}(x)}{\left( 1+\frac{k}{n} \right) ^{2(\alpha +1)}} \right] ^{1/2} \left[ \sum _{k=0}^{\infty } s_{n,k}(x)\left( \frac{k}{n}-x \right) ^2\right] ^{1/2}. \end{aligned}$$

From (2.9)

$$\begin{aligned} \sum _{k=0}^{\infty }\frac{s_{n,k}(x)}{\left( 1+\frac{k}{n} \right) ^{2(\alpha +1)}} \le \frac{C}{(1+x)^{2(\alpha +1)}}, \end{aligned}$$

from (2.4)

$$\begin{aligned} \sum _{k=0}^{\infty } s_{n,k}(x)\left( \frac{k}{n}-x \right) ^2=\frac{x}{n}, \end{aligned}$$

and consequently

$$\begin{aligned} \frac{Cw(x)}{n}\sum _{k=0}^{\infty }\frac{s_{n,k}(x)}{\left( 1+\frac{k}{n} \right) ^{\alpha +1}} \left| \frac{k}{n}-x \right| \le \frac{C\sqrt{x}}{(1+x)n^{3/2}}\le \frac{C}{n}. \end{aligned}$$

$\square$

Lemma 2.6

For every $x\in [0,\infty )$ and $n\in {\mathbb {N}}$ there holds

$$\begin{aligned} \left| xw(x){{\tilde{S}}}_n\left( \frac{\ln x- \ln (\cdot )}{w(\cdot )};x\right) \right| \le \frac{C}{n}. \end{aligned}$$

Proof

Again, by using the Lagrange’s formula we have

$$\begin{aligned}&\left| xw(x){{\tilde{S}}}_n\left( \frac{ (\cdot )-x}{w(\cdot )};x\right) \right| \\&\quad \le x\left| \sum _{k=0}^{\infty }s_{n,k}(x)\frac{w(x)}{w\left( \frac{k}{n} \right) } n\int _{\frac{k}{n}}^{\frac{k+1}{n}}\left( \ln x-\ln t\right) dt\right| \\&\qquad +xw(x)\sum _{k=0}^{\infty }s_{n,k}(x) n\int _{\frac{k}{n}}^{\frac{k+1}{n}}\left| \ln x-\ln t\right| \left| t-\frac{k}{n}\right| \frac{|w'(\xi )|}{w^2(\xi )}dt. \end{aligned}$$

For the last term in the RHS using

$$\begin{aligned} x|\ln x-\ln t|\le |t-x| \end{aligned}$$

we have

$$\begin{aligned}&xw(x)\sum _{k=0}^{\infty }s_{n,k}(x) n\int _{\frac{k}{n}}^{\frac{k+1}{n}}\left| \ln x-\ln t\right| \left| t-\frac{k}{n}\right| \frac{|w'(\xi )|}{w^2(\xi )}dt\\&\quad \le w(x)\sum _{k=0}^{\infty }s_{n,k}(x) n\int _{\frac{k}{n}}^{\frac{k+1}{n}}\left| t-x\right| \left| t-\frac{k}{n}\right| \frac{|w'(\xi )|}{w^2(\xi )}dt \end{aligned}$$

and we already estimated it in the previous lemma.

For the first term we have

$$\begin{aligned}&x\left| \sum _{k=0}^{\infty }s_{n,k}(x)\frac{w(x)}{w\left( \frac{k}{n} \right) } n\int _{\frac{k}{n}}^{\frac{k+1}{n}}\left( \ln x-\ln t \right) dt \right| \\&\quad \le x\left| \sum _{k=0}^{\infty }s_{n,k}(x)\frac{w(x)}{w\left( \frac{k}{n} \right) } \left( \ln \frac{k+1}{n}-\ln x \right) \right| +Cx\sum _{k=0}^{\infty }s_{n,k}(x) \frac{w(x)}{w\left( \frac{k}{n} \right) } \frac{1}{k}. \end{aligned}$$

The last inequality follows from

$$\begin{aligned} n\int _{\frac{k}{n}}^{\frac{k+1}{n}}\left( \ln x-\ln t \right) dt =\ln x -\ln \frac{k+1}{n}+O\left( \frac{1}{k} \right) . \end{aligned}$$

Since

$$\begin{aligned} x\sum _{k=0}^{\infty }s_{n,k}(x)\frac{w(x)}{w\left( \frac{k}{n} \right) }\frac{1}{k}\le \frac{C}{n} \end{aligned}$$

it follows that

$$\begin{aligned} xw(x)\left| {{\tilde{S}}}_n\left( \frac{\ln x- \ln (\cdot )}{w(\cdot )}:x\right) \right| \le x\left| \sum _{k=0}^{\infty }s_{n,k}(x)\frac{w(x)}{w\left( \frac{k}{n} \right) } \left( \ln \frac{k+1}{n}-\ln x \right) \right| + \frac{C}{n}. \end{aligned}$$

Now we consider two cases.

• 1. $x\le \frac{1}{n}$.

  In this case:

  $$\begin{aligned}&x\left| \sum _{k=0}^{\infty }s_{n,k}(x)\frac{w(x)}{w\left( \frac{k}{n} \right) } \left( \ln \frac{k+1}{n}-\ln x \right) \right| \le x\sum _{k=0}^{\infty }s_{n,k}(x)\frac{w(x)}{w\left( \frac{k}{n} \right) } \left( \ln (k+1)-\ln (nx) \right) \\&\quad \le x\sum _{k=0}^{\infty }s_{n,k}(x)\frac{w(x)}{w\left( \frac{k}{n} \right) }\ln (k+1) + x |\ln (nx) |\sum _{k=0}^{\infty }s_{n,k}(x)\frac{w(x)}{w\left( \frac{k}{n} \right) }\\&\quad \le x\sum _{k=0}^{\infty }s_{n,k}(x)\frac{w(x)}{w\left( \frac{k}{n} \right) } (k+1) +C x |\ln (nx) | \le C\left( nx^2+x+\frac{1}{n}\right) \le \frac{C}{n} \end{aligned}$$

  because of (2.9), (2.2) and the simple inequalities

  $$\begin{aligned} nx^2\le \frac{1}{n} \quad \text{ and }\quad x |\ln (nx) |\le \frac{1}{n}. \end{aligned}$$

• 2. $x> \frac{1}{n}$.

  By Taylor’s formula

  $$\begin{aligned} \ln \frac{k+1}{n}=\ln x +\frac{1}{x}\left( \frac{k+1}{n}-x \right) -\int _{x}^{\frac{k+1}{n}}\left( \frac{k+1}{n}-t \right) \frac{dt}{t^2} \end{aligned}$$

  and consequently

  $$\begin{aligned}&x\left| \sum _{k=0}^{\infty }s_{n,k}(x)\frac{w(x)}{w\left( \frac{k}{n} \right) } \left( \ln \frac{k+1}{n}-\ln x \right) \right| \nonumber \\&\quad \le \left| \sum _{k=0}^{\infty }s_{n,k}(x)\frac{w(x)}{w\left( \frac{k}{n} \right) } \left( \frac{k+1}{n}-x \right) \right| \nonumber \\&\qquad +x\left| \sum _{k=0}^{\infty }s_{n,k}(x)\frac{w(x)}{w\left( \frac{k}{n} \right) }\int _{x}^{\frac{k+1}{n}}\left( \frac{k+1}{n}-t \right) \frac{dt}{t^2}\right| . \end{aligned}$$

  (2.22)

  The estimation of the first term in RHS of (2.22) is analogous to the estimation of

  $$\begin{aligned} \left| \sum _{k=0}^{\infty }s_{n,k}(x)\frac{w(x)}{w\left( \frac{k}{n} \right) } \left( \frac{k}{n}-x \right) \right| \end{aligned}$$

  in the previous lemma.

  For the second term in RHS of (2.22) since

  $$\begin{aligned} \int _{x}^{\frac{k+1}{n}}\left( \frac{k+1}{n}-t \right) \frac{dt}{t^2} \le \left| \frac{k+1}{n}-x \right| \int _{x}^{\frac{k+1}{n}}\frac{dt}{t^2} =\frac{n}{(k+1)x}\left( \frac{k+1}{n}-x \right) ^2 \end{aligned}$$

  we have by applying the Cauchy’s inequality, (2.9) and (2.6)

  $$\begin{aligned}&x\left| \sum _{k=0}^{\infty }s_{n,k}(x)\frac{w(x)}{w\left( \frac{k}{n} \right) }\int _{x}^{\frac{k+1}{n}}\left( \frac{k+1}{n}-t \right) \frac{dt}{t^2}\right| \\&\quad \le \sum _{k=0}^{\infty }s_{n,k}(x)\frac{w(x)}{w\left( \frac{k}{n} \right) }\frac{n}{k+1}\left( \frac{k+1}{n}-x \right) ^2\\&\quad =\frac{w(x)}{x}\sum _{k=0}^{\infty }\frac{s_{n,k+1}(x)}{w\left( \frac{k}{n} \right) }\left( \frac{k+1}{n}-x \right) ^2\\&\quad \le \frac{w(x)}{x} \left[ \sum _{k=0}^{\infty }\frac{s_{n,k+1}(x)}{w\left( \frac{k}{n} \right) }\right] ^{1/2} \left[ \sum _{k=0}^{\infty }s_{n,k+1}(x)\left( \frac{k+1}{n}-x \right) ^4\right] ^{1/2}\\&\quad \le \frac{w(x)}{x} \left[ \sum _{k=0}^{\infty }\frac{s_{n,k}(x)}{w\left( \frac{k}{n} \right) }\right] ^{1/2} \left[ \sum _{k=0}^{\infty }s_{n,k+1}(x)\left( \frac{k+1}{n}-x \right) ^4\right] ^{1/2}\\&\quad \le \frac{w(x)}{x}\left[ \frac{C}{w(x)}\right] ^{1/2}\left[ C\left( \frac{x}{n}\right) ^2\right] ^{1/2}=\frac{C}{n}. \end{aligned}$$

$\square$

Lemma 2.7

Let $1\le p\le \infty$. Then for all $f \in W_p(w)$ and $n\in {\mathbb {N}}$ there holds

$$\begin{aligned} \left\| w(x){{\tilde{S}}}_n\left( \int _x^{(\cdot )}\left[ \phi (\cdot )-\phi (u) \right] {{\tilde{D}}}f(u)du\right) \right\| _p \le \frac{C(\alpha )}{n}\left\| w{{\tilde{D}}}f\right\| _p. \end{aligned}$$

Proof

We have

$$\begin{aligned}&\left| w(x){{\tilde{S}}}_n\left( \int _x^{(\cdot )}\left[ \phi (\cdot )-\phi (u) \right] {{\tilde{D}}}f(u)du\right) \right| \\&\quad =w(x)\sum _{k=0}^{\infty }s_{n,k}(x)n \int _{\frac{k}{n}}^{\frac{k+1}{n}}\left( \int _x^{t}\left[ \ln t-\ln u \right] \frac{w(u){{\tilde{D}}}f(u)}{w(u)}du\right) dt\\&\quad \le C\sum _{k=0}^{\infty }s_{n,k}(x)\frac{w(x)}{w(k/n)}n \int _{\frac{k}{n}}^{\frac{k+1}{n}}\left( \int _x^{t}\left[ \ln t-\ln u \right] \left| w(u){{\tilde{D}}}f(u) \right| du\right) dt\\&\quad =C\sum _{k=0}^{\infty }s_{n,k}(x)\left[ \frac{n(1+x)}{n+k}\right] ^\alpha b_{k,1} \end{aligned}$$

where

$$\begin{aligned} b_{k,1}=n \int _{\frac{k}{n}}^{\frac{k+1}{n}}\left( \int _x^{t}\left[ \ln t-\ln u \right] \left| w(u){{\tilde{D}}}f(u) \right| du\right) dt. \end{aligned}$$

Let $\mu$ be the smallest integer such that $\mu >|\alpha |$. By Holder’s inequality we have

$$\begin{aligned}&\sum _{k=0}^{\infty }s_{n,k}(x) \left[ \frac{n(1+x)}{n+k}\right] ^\alpha b_{k,1}\\&\quad \le \left\{ \sum _{k=0}^{\infty }s_{n,k}(x) \left[ \frac{n(1+x)}{n+k}\right] ^{sign(\alpha )\mu } b_{k,1} \right\} ^{\frac{|\alpha |}{\mu }} \left\{ \sum _{k=0}^{\infty }s_{n,k}(x)b_{k,1} \right\} ^{1-\frac{|\alpha |}{\mu }}. \end{aligned}$$

We define a new operator ${{\tilde{S}}}_{n,\alpha }$ by the formula

$$\begin{aligned} {{\tilde{S}}}_{n,\alpha }f(x)=\int _0^\infty K_n(x,t)f(t)dt =\sum _{k=0}^{\infty }s_{n,k}(x)(1+x)^m\frac{n^{m+1}(n+k)!}{(n+k+m)!} \int _{\frac{k}{n}}^{\frac{k+1}{n}}f(t)dt \end{aligned}$$

where $m=sign(\alpha )\mu$.

For the estimation of the $L_1$-norm by applying the Holder’s inequality again we have

$$\begin{aligned}&\left\| \sum _{k=0}^{\infty }s_{n,k}(x)\left[ \frac{n(1+x)}{n+k}\right] ^\alpha b_{k,1} \right\| _1 \\&\quad \le \left\| \sum _{k=0}^{\infty }s_{n,k}(x)\left[ \frac{n(1+x)}{n+k}\right] ^m b_{k,1} \right\| _1^{\frac{|\alpha |}{\mu }} \left\| \sum _{k=0}^{\infty }s_{n,k}(x) b_{k,1} \right\| _1^{1-\frac{|\alpha |}{\mu }}. \end{aligned}$$

For the second term in RHS we can use the estimation (2.17) - simply replacing ${{\tilde{D}}}f$ by $w{{\tilde{D}}}f$. For the first term since

$$\begin{aligned} \left( \frac{n}{n+k} \right) ^{m}\le C\frac{n^{m}(n+k)!}{(n+k+m)!} \end{aligned}$$

we have

$$\begin{aligned}&\left\| \sum _{k=0}^{\infty }s_{n,k}(x)\left[ \frac{n(1+x)}{n+k}\right] ^m b_{k,1} \right\| _1\\&\quad \le C \left\| {{\tilde{S}}}_{n,\alpha } \left( \int _x^{(\cdot )}\left[ \ln {(\cdot )}-\ln u \right] \left| w(u){{\tilde{D}}}f(u) \right| du\right) \right\| _1\\&\quad =C\left( \int _0^x+\int _x^\infty \right) =C\left( I_1(x)+I_2(x) \right) . \end{aligned}$$

Changing the order of integration twice in both of the integrals we obtain

$$\begin{aligned} \int _0^\infty I_1(x)dx=\int _0^\infty \left| w(u){{\tilde{D}}}f(u) \right| \left( \int _u^\infty {{\tilde{S}}}_{n,\alpha } \left( [\ln u-\ln (\cdot )]_+;x \right) dx \right) du \end{aligned}$$

and

$$\begin{aligned} \int _0^\infty I_2(x)dx=\int _0^\infty \left| w(u){{\tilde{D}}}f(u) \right| \left( \int _0^u {{\tilde{S}}}_{n,\alpha } \left( [\ln (\cdot )-\ln u]_+;x \right) dx \right) du \end{aligned}$$

where

$$\begin{aligned} \left[ f(z) \right] _+=\frac{1}{2}\left( \left| f(z) \right| +f(z) \right) . \end{aligned}$$

Consequently,

$$\begin{aligned} \int _0^\infty I_1(x)dx+\int _0^\infty I_2(x)dx=\int _0^\infty \left| w(u){{\tilde{D}}}f(u) \right| I(u) du \end{aligned}$$

where

$$\begin{aligned} I(u)=\int _u^\infty {{\tilde{S}}}_{n,\alpha } \left( [\ln u-\ln (\cdot )]_+;x \right) dx +\int _0^u {{\tilde{S}}}_{n,\alpha } \left( [\ln (\cdot )-\ln u]_+;x \right) dx. \end{aligned}$$

We have by equation (2.10) of Lemma  2.2 for every function $h(x)\in {\mathbb {L}}_1[0,\infty )$

$$\begin{aligned} \left\| {{\tilde{S}}}_{n,\alpha }h \right\| _1 =\sum _{k=0}^{\infty } \frac{n^{m}(n+k)!}{(n+k+m)!} \int _0^\infty s_{n,k}(x)(1+x)^mdx \int _{\frac{k}{n}}^{\frac{k+1}{n}}h(t)dt \le \Vert h\Vert _1 \end{aligned}$$

and

$$\begin{aligned} \left\| {{\tilde{S}}}_{n,\alpha }h \right\| _1&\ge \sum _{k=0}^{\infty } \left( 1-\frac{C}{n\left( 1+\frac{k}{n} \right) ^2}\right) \int _{\frac{k}{n}}^{\frac{k+1}{n}}h(t)dt\\&= \Vert h\Vert _1-\frac{C}{n}\sum _{k=0}^{\infty }\frac{1}{\left( 1+\frac{k}{n} \right) ^2} \int _{\frac{k}{n}}^{\frac{k+1}{n}}|h(t)|dt. \end{aligned}$$

Also,

$$\begin{aligned} \sum _{k=0}^{\infty }\frac{1}{\left( 1+\frac{k}{n} \right) ^2} \int _{\frac{k}{n}}^{\frac{k+1}{n}}|h(t)|dt&\le \sum _{k=0}^{\infty }\frac{1}{\left( 1+\frac{k}{n} \right) ^2} \int _{\frac{k}{n}}^{\frac{k+1}{n}} |\ln t|dt\\&\le \frac{1}{n}\sum _{k=0}^{\infty }\frac{\left| \ln {\frac{k+1}{n}}\right| }{\left( 1+\frac{k}{n} \right) ^2}\le \frac{C}{n} \end{aligned}$$

i.e.

$$\begin{aligned} \int _0^\infty {{\tilde{S}}}_{n,\alpha } \left( [\ln u-\ln (\cdot )]_+;x \right) dx =\int _0^u \left( \ln u-\ln x\right) dx +O\left( \frac{1}{n} \right) . \end{aligned}$$

Since

$$\begin{aligned} \int _0^\infty {{\tilde{S}}}_{n,\alpha }dx=\int _0^u \tilde{S}_{n,\alpha }dx+\int _u^\infty {{\tilde{S}}}_{n,\alpha }dx \end{aligned}$$

it follows that

$$\begin{aligned}&\int _u^\infty {{\tilde{S}}}_{n,\alpha } \left( [\ln u-\ln (\cdot )]_+;x \right) dx\\&\quad =\int _0^u \left( \ln u-\ln x\right) dx- \int _0^u \tilde{S}_{n,\alpha } \left( [\ln u-\ln (\cdot )]_+;x \right) dx +O\left( \frac{1}{n} \right) . \end{aligned}$$

Consequently,

$$\begin{aligned} I(u)&=\int _0^u \left( \ln u-\ln x\right) dx + \int _0^u \tilde{S}_{n,\alpha } \left( [\ln (\cdot )-\ln u]_+ - [\ln u-\ln (\cdot )]_+;x \right) dx\\&\quad +O\left( \frac{1}{n} \right) \\&=\int _0^u \left( \ln u-\ln x\right) dx + \int _0^u \tilde{S}_{n,\alpha } \left( \ln (\cdot )-\ln u;x \right) dx +O\left( \frac{1}{n} \right) \\&= \int _0^u\left( {{\tilde{S}}}_{n,\alpha } \left( \ln (\cdot );x\right) -\ln x \right) dx+ +O\left( \frac{1}{n} \right) \\&\le \int _0^\infty \left( {{\tilde{S}}}_{n,\alpha } \left( \ln (\cdot );x\right) -\ln x \right) dx+ +O\left( \frac{1}{n} \right) \\&=\left\| {{\tilde{S}}}_{n,\alpha } \ln (\cdot )-\ln (\cdot ) \right\| _1+O\left( \frac{1}{n} \right) \\&\le \left\| {{\tilde{S}}}_{n,\alpha } \ln (\cdot )- {{\tilde{S}}}_{n} \ln (\cdot ) \right\| _1 + \left\| {{\tilde{S}}}_{n} \ln (\cdot )- \ln (\cdot )\right\| _1 +O\left( \frac{1}{n} \right) \\ \end{aligned}$$

and by using the estimate (2.17) and Lemma 2.2 we complete the proof of lemma for $p=1$.

Now we estimate the $L_\infty$-norm. We have

$$\begin{aligned}&\left| w(x){{\tilde{S}}}_n\left( \int _x^{(\cdot )}(\ln (\cdot )-\ln x) {{\tilde{D}}}f(u)du;x\right) \right| \\&\quad \le \left\| w{{\tilde{D}}}f \right\| _\infty \left\| w(x){{\tilde{S}}}_n\left( \int _x^{(\cdot )} \frac{\ln (\cdot )-\ln u}{w(u)}du;x\right) \right\| _\infty . \end{aligned}$$

From the obvious $w^{-1}(u)\le w^{-1}(\cdot )+w^{-1}(x)$ for u between $(\cdot )$ and x it follows that

$$\begin{aligned}&\left\| w(x){{\tilde{S}}}_n\left( \int _x^{(\cdot )} \frac{\ln (\cdot )-\ln u}{w(u)}du;x\right) \right\| _\infty \nonumber \\&\quad \le \left| {{\tilde{S}}}_n\left( \int _x^{(\cdot )}(\ln (\cdot )-\ln u)du;x\right) \right| +\left\| w(x){{\tilde{S}}}_n\left( \int _x^{(\cdot )} \frac{\ln (\cdot )-\ln u}{w(\cdot )}du;x\right) \right\| _\infty . \end{aligned}$$

(2.23)

For the first term in the RHS of (2.23) we have

$$\begin{aligned} {{\tilde{S}}}_n\left( \int _x^{(\cdot )}(\ln (\cdot )-\ln u)du;x\right) =\frac{1}{2n}-x\left( {{\tilde{S}}}_n(\ln (\cdot );x)-\ln x \right) \end{aligned}$$

and by using the estimation (2.18) we obtain the needed estimate.

For the second term in the RHS of (2.23) we have

$$\begin{aligned}&w(x){{\tilde{S}}}_n\left( \int _x^{(\cdot )} \frac{\ln (\cdot )-\ln u}{w(\cdot )}du;x\right) \\&\quad =w(x) \left[ x{{\tilde{S}}}_n\left( \frac{\ln x- \ln (\cdot )}{w(\cdot )};x\right) +{{\tilde{S}}}_n\left( \frac{ (\cdot )-x}{w(\cdot )};x\right) \right] \\&\quad =xw(x){{\tilde{S}}}_n\left( \frac{\ln x- \ln (\cdot )}{w(\cdot )};x\right) +w(x){{\tilde{S}}}_n\left( \frac{ (\cdot )-x}{w(\cdot )};x\right) . \end{aligned}$$

By using Lemma 2.5 and Lemma 2.6 we complete the proof of lemma for $p=\infty$. $\square$

3 Proofs of Theorems 1.2 and 1.3

Proof of Theorem 1.2

We follow the argument in [1, pp. 41–42].

We have for $x,t>0$

$$\begin{aligned} f(t)=f(x)+\varphi (x)\left( \phi (t)-\phi (x) \right) Df(x)+\int _x^t\left( \phi (t)-\phi (u) \right) {{\tilde{D}}}f(u)du. \end{aligned}$$

Applying ${{\tilde{S}}}_n$ to both sides with regard to t and using  (2.5), we obtain

$$\begin{aligned}&{{\tilde{S}}}_n(f;x)-f(x)\\&\quad =\varphi (x)\left[ \tilde{S}_n(\phi (\cdot );x)-\phi (x) \right] Df(x) +\tilde{S}_n\left( \int _x^{(\cdot )}\left[ \phi (\cdot )-\phi (u) \right] \tilde{D}f(u)du\right) \end{aligned}$$

and

$$\begin{aligned} w(x)\left[ {{\tilde{S}}}_n(f,x)-f(x)\right]= & {} w(x)\varphi (x)\left[ \tilde{S}_n(\phi (\cdot );x)-\phi (x) \right] Df(x)\\&+w(x)\tilde{S}_n\left( \int _x^{(\cdot )}\left[ \phi (\cdot )-\phi (u) \right] \tilde{D}f(u)du\right) . \end{aligned}$$

Consequently

$$\begin{aligned} \left\| w\left( {{\tilde{S}}}_nf-f\right) \right\| _p\\ \le \left\| w\varphi \left( {{\tilde{S}}}_n\phi -\phi \right) Df\right\| _p +\left\| w(x)\tilde{S}_n\left( \int _x^{(\cdot )}\left[ \phi (\cdot )-\phi (u) \right] \tilde{D}f(u)du\right) \right\| _p. \end{aligned}$$

By using Lemmas 2.4 and 2.7 we complete the proof. $\square$

Proof of Theorem 1.3

To recall, we denote by C positive constants, not necessarily the same at each occurrence, which are independent of f, g, n, and l. We prove the theorem by means of a standard argument.

Let $1\le p\le \infty$. For any $g \in {{\tilde{W}}}_p(w)$ such that $f-g\in L_p(w)$ we have in virtue of (2.7) and Theorem 1.2

$$\begin{aligned} \left\| w(f-{{\tilde{S}}}_n f)\right\| _p&\le \Vert w(f-g)\Vert _p +\left\| w{{\tilde{S}}}_n (f-g)\right\| _p + \frac{C}{n}\left\| w{{\tilde{D}}} g\right\| _p\\&\le C\left( \Vert w(f-g)\Vert _p + \frac{1}{n}\left\| w{{\tilde{D}}} g\right\| _p\right) . \end{aligned}$$

Taking the infimum on g we arrive at the left-hand side inequality in the theorem. $\square$