Chlodowsky-type Szász operators via Boas–Buck-type polynomials and some approximation properties

Abstract

In this paper, we construct the Chlodowsky-type Szász operators defined via Boas–Buck-type polynomials. We prove some approximation properties and obtain the rate of the convergence for these operators. We also study the Voronovskaya-type theorem and weighted approximation.

1 Introduction and preliminaries

The basic sequence of Szász operators is given by

$$ S_{n}(f,\xi )=e^{-n\xi }\sum_{\imath =0}^{\infty }{ \frac{(n\xi )^{\imath }}{\imath !}f \biggl( \frac{\imath }{n} \biggr) } $$

for $x\in{}[ 0,\infty )$. Generalizations of these operators have been studied by many authors. In [21] the authors have obtained a generalization of Szász operators by means of the Appell polynomials defined as follows:

$$ P_{n}(f,\xi )=\frac{e^{-n\xi }}{g(1)}\sum_{\imath =0}^{\infty }{p_{ \imath }(n \xi )f \biggl( \frac{\imath }{n} \biggr) }, $$

where $p_{\imath }(\xi )$, $\imath \geq 0$, are the Appell polynomials defined by

$$ g(t)e^{\xi t}=\sum_{\imath =0}^{\infty }{p_{\imath }( \xi ) \frac{t^{\imath }}{\imath !}}\quad \text{and} \quad g(t)=\sum_{\imath =0}^{\infty }{a_{\imath } \frac{t^{\imath }}{\imath !}}, $$

and $g(t)$ is an analytic function in the disk $\vert t \vert < R$, $R>1$, and $g(1)\neq 0$. A further generalization was given by Ismail [19] by using the Sheffer operators

$$ T_{n}(f,\xi )=\frac{e^{-n\xi H(1)}}{g(1)}\sum_{\imath =0}^{\infty }{s_{\imath }(n \xi )f \biggl( \frac{\imath }{n} \biggr) } $$

for $n\in \mathbb{N}$, where $s_{\imath }(\xi )$, $\imath \geq 0$, are the Sheffer polynomials defined by

$$ g(t)e^{\xi H(t)}=\sum_{\imath =0}^{\infty }{s_{\imath }( \xi ) \frac{t^{\imath }}{\imath !}}, $$

$H(t)=\sum_{\imath =0}^{\infty }{h_{\imath } \frac{t^{\imath }}{\imath !}}$ is an analytic function in the disk $\vert t \vert < R$, $R>1$, $g(1)\neq 0$, and $H^{{\prime }}(1)=1$.

The multiple Sheffer polynomials $\{S_{k_{1},k_{2}}(\xi )\}_{k_{1},k_{2}=0}^{\infty }$ are defined as follows. The generating function is

$$ A(t_{1},t_{2})e^{\xi H(t_{1},t_{2})}=\sum _{k_{1}=0}^{ \infty }\sum_{k_{2}=0}^{ \infty }S_{k_{1},k_{2}}( \xi ) \frac{t_{1}^{k_{1}}t_{2}^{k_{2}}}{k_{1}!k_{2}!}, $$

(1.1)

where $A(t_{1},t_{2})$ and $H(t_{1},t_{2})$ are of the forms

$$ A(t_{1},t_{2})=\sum_{k_{1}=0}^{\infty } \sum_{k_{2}=0}^{ \infty }a_{k_{1},k_{2}} \frac{t_{1}^{k_{1}}t_{2}^{k_{2}}}{k_{1}!k_{2}!} $$

(1.2)

and

$$ H(t_{1},t_{2})=\sum_{k_{1}=0}^{\infty } \sum_{k_{2}=0}^{ \infty }h_{k_{1},k_{2}} \frac{t_{1}^{k_{1}}t_{2}^{k_{2}}}{k_{1}!k_{2}!}, $$

(1.3)

respectively, and satisfy the conditions $A(0,0)=a_{0,0}\neq 0$ and $H(0,0)=h_{0,0}\neq 0$. The positive linear operators involving multiple Sheffer polynomials for $\xi \in{}[ 0,\infty )$ were defined in [3] as follows:

$$ G_{n}(f,\xi )= \frac{e^{-\frac{n\xi }{2}H(1,1)}}{A(1,1)}\sum_{k_{1}=0}^{\infty } \sum_{k_{2}=0}^{\infty } \frac{S_{k_{1},k_{2}} ( \frac{n\xi }{2} ) }{k_{1}!k_{2}!}f \biggl( \frac{k_{1}+k_{2}}{n} \biggr) , $$

(1.4)

provided that the series in the above relations are convergent and the following conditions are satisfied:

• (1) $S_{k_{1},k_{2}}(\xi )\geq 0$, $k_{1},k_{2}\in \mathbb{N}$,

• (2) $A(1,1)\neq 0$, $H_{t_{1}}(1,1)=1$, $H_{t_{2}}(1,1)=1$,

• (3) Series (1.1), (1.2), and (1.3) are convergent for $\vert t_{1} \vert < R$, $\vert t_{2} \vert < R$, and $(R_{1},R_{2})>1$.

In [12] the authors have studied the Kantorovich variant of Szász operators induced by multiple Sheffer polynomials for $\xi \in{}[ 0,\infty )$ as follows:

$$ K_{n}^{(S)}(f,\xi )=\frac{ne^{-\frac{n\xi }{2}H(1,1)}}{A(1,1)}\sum _{k_{1}=0}^{\infty }\sum_{k_{2}=0}^{\infty } \frac{S_{k_{1},k_{2}} ( \frac{n\xi }{2} ) }{k_{1}!k_{2}!} \int _{ \frac{k_{1}+k_{2}}{n}}^{\frac{k_{1}+k_{2}+1}{n}}{f(t)}\,dt, $$

under the condition that the right side of the above relation exists. Szász-type operators involving Charlier polynomials were studied in [2].

We will treat the Chlodowsky variant of the Szász type operators induced by Boas-Buck-type polynomials. The generating functions for the Boas–Buck-type polynomials [20] are

$$ A(t)B \bigl(\xi H(t) \bigr)=\sum_{k=0}^{\infty }{p_{k}( \xi )t^{k}}, $$

(1.5)

where A, B, and H are analytic functions given by the following expressions:

$$\begin{aligned}& A(t)=\sum_{r=0}^{\infty}{a_{r}t^{r}}, \quad a_{0}\neq 0, \end{aligned}$$

(1.6)

$$\begin{aligned}& B(t)=\sum_{r=0}^{\infty}{b_{r}t^{r}}, \quad b_{r}\neq 0, r\geq 0, \end{aligned}$$

(1.7)

$$\begin{aligned}& H(t)=\sum_{r=0}^{\infty}{h_{r}t^{r}}, \quad h_{1}\neq 0. \end{aligned}$$

(1.8)

In what follows, we assume that the above polynomials satisfy the following conditions:

• (1) $A(1)\neq 0$, $H^{{\prime }}(1)=1$, $p_{k}(\xi )\geq 0$, $k=0,1,2,\ldots$ ,

• (2) $B:\mathbb{R}\rightarrow (0,\infty )$,

• (3) The power series (1.5), (1.6), (1.7), and (1.8) are convergent for $\vert t \vert < R$ ($R>1$).

The Chlodowsky variant of the Szász-type operators induced by Boas–Buck-type polynomials given in [26] (see also [1]) is defined as follows:

$$ B_{n}^{\ast }(f;\xi )= \frac{1}{A(1)B ( \frac{n}{b_{n}}\xi H(1) ) }\sum _{k=0}^{\infty }{p_{k} \biggl( \frac{n}{b_{n}}\xi \biggr) f \biggl( \frac{k}{n}b_{n} \biggr) }, $$

(1.9)

where $(b_{n})$ is a numerical positive increasing sequence such that

$$ {b_{n}}\rightarrow \infty , \quad\quad {\frac{b_{n}}{n}}\rightarrow 0 \quad (n\rightarrow \infty ). $$

The sequence $(b_{n})=(\sqrt{n})$ satisfies the above conditions.

We assume the operators $B_{n}^{\ast }$ to be positive. Also, we consider

$$ \lim_{n\rightarrow \infty }{\frac{B^{(k)}(y)}{B(y)}}=1\quad \text{for } k \in \{1,2,3,\ldots,r\}, r\in \mathbb{N}. $$

In the recent years, different classes of operators were studied together with Korovkin- and Voronovskaja-type theorems (see [4–11, 13–18, 23, 27, 28, 30] and [22, 24, 25]).

2 Basic results

Here we calculate the moments and central moments for $B_{n}^{\ast }$ (see [29]).

Lemma 2.1

[26] For all $\xi \in{}[ 0,\infty )$,

$$\begin{aligned}& B_{n}^{\ast }(e_{0};\xi )=1, \\& B_{n}^{\ast }(e_{1};\xi )= \frac{B^{\prime } (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )}x+ \frac{b_{n}}{n} \frac{A^{\prime }(1)}{A(1)}, \\& \begin{aligned} B_{n}^{\ast }(e_{2};\xi )&= \frac{B^{\prime \prime } (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )} \xi ^{2}+\frac{b_{n}}{n} \frac{B^{\prime } (\frac{n}{b_{n}}\xi H(1) ) [ A(1)+2A^{\prime }(1)+H^{\prime \prime }(1)A(1) ] }{A(1)B (\frac{n}{b_{n}}\xi H(1) )}x \\ &\quad {} +\frac{b_{n}^{2}}{n^{2}} \frac{A^{\prime }(1)+A^{\prime \prime }(1)}{A(1)}, \end{aligned} \\& \begin{aligned} B_{n}^{\ast }(e_{3};\xi )&= \frac{B^{\prime \prime \prime } (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )}\xi ^{3}+ \bigl( 3A^{\prime }(1)+3H^{\prime \prime }(1)A(1)+3A(1) \bigr) \frac{B^{\prime \prime } (\frac{n}{b_{n}}\xi H(1) )}{A(1)B (\frac{n}{b_{n}}\xi H(1) )}\frac{b_{n}}{n}\xi ^{2} \\ &\quad {} + \bigl( 3A^{\prime \prime }(1) + 3H^{\prime \prime }(1)A^{ \prime }(1) + H^{\prime \prime \prime }(1)A(1) + 6A^{\prime }(1)+3H^{ \prime \prime }(1)A(1) + A(1) \bigr) \\ &\quad{}\cdot \frac{B^{\prime } (\frac{n}{b_{n}}\xi H(1) )}{A(1)B (\frac{n}{b_{n}}\xi H(1) )}\frac{b_{n}^{2}}{n^{2}} \xi \\ & \quad {} + \bigl( A^{\prime \prime \prime }(1)+3A^{\prime \prime }(1)+A^{ \prime }(1) \bigr) \frac{b_{n}^{3}}{A(1)n^{3}}, \end{aligned} \\& \begin{aligned} B_{n}^{\ast }(e_{4};\xi )&= \frac{B^{(4)} (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )}\xi ^{4}+ \bigl( 4A^{\prime }(1)+6H^{ \prime \prime }(1)A(1)+6A(1) \bigr) \frac{B^{(3)} (\frac{n}{b_{n}}\xi H(1) )}{A(1)B (\frac{n}{b_{n}}\xi H(1) )}\frac{b_{n}}{n}\xi ^{3} \\ &\quad {} + \bigl( 6A^{\prime \prime }(t) + 12H^{\prime \prime }(1) + A^{ \prime }(1) + 4H^{\prime \prime \prime }(1)A(1) + 3H^{\prime \prime }(1)^{2}A(1) + 18A^{\prime }(1) \\ &\quad {} + 18H^{\prime \prime }(1)A(1) + 7A(1) \bigr) \frac{B^{\prime \prime } (\frac{n}{b_{n}}\xi H(1) )}{A(1)B (\frac{n}{b_{n}}\xi H(1) )}\frac{b_{n}^{2}}{n^{2}}\xi ^{2} \\ &\quad {}+ \bigl( 4A^{ \prime \prime \prime }(1)+6A^{\prime \prime }(1)H^{\prime \prime }(1)+4A^{ \prime }(1)H^{\prime \prime \prime }(1)+A(1)H^{(4)}(1)+18A^{\prime \prime }(1) \\ &\quad {} + 18H^{\prime \prime }(1)A^{\prime }(1)+6H^{\prime \prime \prime }(1)A(1)+14A^{\prime }(1)+7H^{\prime \prime }(1)A(1)+A(1) \bigr) \\ &\quad {} \cdot\frac{B^{\prime } (\frac{n}{b_{n}}\xi H(1) )}{A(1)B (\frac{n}{b_{n}}\xi H(1) )} \frac{b_{n}^{3}}{n^{3}}\xi \\ &\quad {} + \bigl( A^{(4)}(1)+6A^{(3)}(1)+7A^{\prime \prime }(1)+A^{\prime }(1) \bigr) \frac{b_{n}^{4}}{A(1)n^{4}}. \end{aligned} \end{aligned}$$

Proposition 2.2

[26] We have

$$\begin{aligned}& B_{n}^{\ast } \bigl((e_{1}-\xi );\xi \bigr)= \frac{B^{\prime } (\frac{n}{b_{n}}\xi H(1) )-B (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )}x+\frac{b_{n}}{n}\frac{A^{\prime }(1)}{A(1)}, \\& \begin{aligned} B_{n}^{\ast } \bigl((e_{1}-\xi )^{2}; \xi \bigr)&= \frac{B^{\prime \prime } (\frac{n}{b_{n}}\xi H(1) )-2B^{\prime } (\frac{n}{b_{n}}\xi H(1) )+B (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )} \xi ^{2} \\ &\quad {} +\frac{b_{n}}{n} \frac{(A(1)+2A^{\prime }(1)+A(1)H^{\prime \prime }(1))B^{\prime } (\frac{n}{b_{n}}\xi H(1) )-2A^{\prime }(1)B (\frac{n}{b_{n}}\xi H(1) )}{A(1)B (\frac{n}{b_{n}}\xi H(1) )}\xi \\ & \quad {} +\frac{b_{n}^{2}}{n^{2}} \frac{A^{\prime }(1)+A^{\prime \prime }(1)}{A(1)}, \end{aligned} \\& B_{n}^{\ast } \bigl((e_{1}-\xi )^{4}; \xi \bigr) \\& \quad = \frac{\xi ^{4}}{B (\frac{n}{b_{n}}\xi H(1) )} \biggl[ B^{(4)} \biggl( \frac{n}{b_{n}}\xi H(1) \biggr) -4B^{(3)} \biggl( \frac{n}{b_{n}} \xi H(1) \biggr) +6B^{ \prime \prime } \biggl( \frac{n}{b_{n}}\xi H(1) \biggr) \\& \quad\quad {} - 4B^{\prime } \biggl( \frac{n}{b_{n}}\xi H(1) \biggr) +B \biggl( \frac{n}{b_{n}}\xi H(1) \biggr) \biggr] + \frac{2\xi ^{3}b_{n}}{nA(1)B ( \frac{n}{b_{n}}\xi H(1) ) } \\& \quad\quad {} \cdot \biggl[ \bigl( 2A^{\prime }(1)+3A(1)H^{\prime \prime }(1)+3A(1) \bigr) B^{(3)} \biggl( \frac{n}{b_{n}}\xi H(1) \biggr) -6 \bigl( A^{ \prime }(1)+A(1)H^{\prime \prime }(1)+A(1) \bigr) \\& \quad\quad {} \cdot B^{\prime \prime } \biggl( \frac{n}{b_{n}}\xi H(1) \biggr) +3 \bigl( 2A^{\prime }(1)+A(1)H^{\prime \prime }(1)+A(1) \bigr) B^{\prime } \biggl( \frac{n}{b_{n}}\xi H(1) \biggr) \\& \quad\quad {} -2A^{\prime }(1)B \biggl( \frac{n}{b_{n}}\xi H(1) \biggr) \biggr] +\frac{\xi ^{2}b_{n}^{2}}{n^{2}A(1)B ( \frac{n}{b_{n}}\xi H(1) ) } [ \bigl( 6A^{\prime \prime }(1)+12A^{\prime }(1)H^{\prime \prime }(1) \\& \quad\quad {} +4A(1)H^{\prime \prime \prime }(1)+21A(1)H^{\prime \prime }(1)+18A^{ \prime }(1) +7A(1) \bigr) B^{\prime \prime } \biggl( \frac{n}{b_{n}}\xi H(1) \biggr). \end{aligned}$$

3 Rates of convergence

By $BV[0,\infty )$ we denote the class of all functions of bounded variation on $[0,\infty )$, and by $\bigvee_{a}^{b}{f}$ we denote the total variation of a function f on $[a,b]$, i.e.,

$$ \bigvee_{a}^{b}{f}=V \bigl(f;[a,b] \bigr)=\sup_{P\in \mathbb{P}}{ \Biggl( \sum_{i=1}^{n}{ \bigl\vert f(\xi _{i})-f(\xi _{i-1}) \bigr\vert } \Biggr) }, $$

where $\mathbb{P}$ is the class of all partitions $P:a=\xi _{0}<\xi _{1}<\cdots <\xi _{n}=b$. We denote

$$ C_{2}[0,\infty )= \bigl\{ f\in C[0,\infty ): \bigl\vert f(t) \bigr\vert \leq M_{2} \bigl(1+t^{2} \bigr)\ \forall t\geq 0 \bigr\} , $$

where $M_{2}$ is a constant, and

$$ D_{BV[0,\infty )}= \bigl\{ f\in C_{2}[0,\infty ):f^{{\prime }}\in BV[0, \infty ) \bigr\} . $$

Let

$$ f_{\xi }^{{\prime }}(\theta )=\textstyle\begin{cases} f^{{\prime }}(\theta )-f^{{\prime }}(\xi -) & \text{for } 0\leq \theta < \xi , \\ 0, & \text{for } \theta =\xi , \\ f^{{\prime }}(\theta )-f^{{\prime }}(\xi +) & \text{for }\xi < \theta < \infty .\end{cases} $$

(3.1)

From the construction of operators $B_{n}^{\ast }(f;\xi )$ we obtain the following relation:

$$ B_{n}^{\ast }(f;\xi )= \int _{0}^{\infty }{f(\theta ) \frac{\partial \{K_{n}(\xi ,\theta )\}}{\partial \theta }}\,d \theta , $$

(3.2)

where

$$ K_{n}(\xi ,\theta )=\textstyle\begin{cases} \sum_{k\leq n\theta}{P_{k,n}(\xi )} & \text{for } 0< \theta < \infty , \\ 0 & \text{for } \theta =0,\end{cases} $$

and

$$ P_{k,n}(\xi )=\frac{1}{A(1)B ( \frac{n}{b_{n}}\xi H(1) ) }{p_{k} \biggl( \frac{n}{b_{n}}\xi \biggr) }. $$

Also, let

$$ \beta _{n}(\xi ;\theta )= \int _{0}^{\theta }{ \frac{\partial \{K_{n}(\xi ,u)\}}{\partial u}}\,du. $$

(3.3)

From the above relation it follows that

$$ \beta _{n}(\xi ;\theta )\leq 1. $$

We provide the following result.

Theorem 3.1

Let $f\in D_{BV[0,\infty )}$. Then for sufficiently large n,

$$\begin{aligned}& \bigl\vert B_{n}^{\ast }(f;\xi )-f(\xi ) \bigr\vert \\& \quad \leq \biggl\vert \frac{1}{2} \bigl(f^{{\prime}}(\xi +)+f^{{\prime }}(\xi -) \bigr) \biggr\vert \cdot \bigl\vert B_{n}^{ \ast }(t-\xi ;\xi ) \bigr\vert + \frac{B_{n}^{\ast }((\xi -t)^{2};\xi )}{\xi } \sum_{k=1}^{[\sqrt{n}]}{ \Biggl( \bigvee_{x-\frac{x}{\sqrt{n}}}^{x+\frac{x}{\sqrt{n}}}{f_{x}^{{\prime}}} \Biggr) } \\& \quad\quad {}+\frac{x}{\sqrt{n}} \Biggl( \bigvee_{x- \frac{x}{\sqrt{n}}}^{x+\frac{x}{\sqrt{n}}}{f_{x}^{{\prime }}} \Biggr) + \biggl( \frac{M_{2}}{\xi ^{2}}+4M_{2}+\frac{ \vert f(\xi ) \vert }{\xi ^{2}} \biggr) B_{n}^{\ast } \bigl((t-\xi )^{2};\xi \bigr) \\& \quad\quad {}+ \bigl\vert f^{{\prime }}(\xi +) \bigr\vert \sqrt{B_{n}^{\ast } \bigl((t-\xi )^{2};\xi \bigr)} +\frac{B_{n}^{\ast }((t-\xi )^{2};\xi )}{\xi ^{2}} \bigl\vert f(2\xi )-f(\xi )- \xi f^{{\prime }}(\xi +) \bigr\vert \\& \quad\quad {}+ \biggl\vert \frac{1}{2} \bigl(f^{{\prime }}( \xi +)-f^{{\prime }}(\xi -) \bigr) \biggr\vert \cdot \sqrt{ \bigl\vert B_{n}^{\ast } \bigl((t- \xi )^{2};\xi \bigr) \bigr\vert }. \end{aligned}$$

We need some auxiliary results. We start with the following:

Lemma 3.2

For any $x\in (0,\infty )$ and $n\in \mathbb{N}$, we have

$$\begin{aligned}& \begin{aligned} (1)\quad \beta _{n}(\xi ;t)&= \int _{0}^{t}{ \frac{\partial \{K_{n}(\xi ,u)\}}{\partial u}}\,du \\ &\leq \biggl(\frac{B^{\prime \prime } (\frac{n}{b_{n}}\xi H(1) )-2B^{\prime } (\frac{n}{b_{n}}\xi H(1) )+B (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )}\xi ^{2} \\ &\quad{} +\frac{b_{n}}{n}\frac{(A(1)+2A^{\prime }(1)+A(1)H^{\prime \prime }(1))B^{\prime } (\frac{n}{b_{n}}\xi H(1) )-2A^{\prime }(1)B (\frac{n}{b_{n}}\xi H(1) )}{A(1)B (\frac{n}{b_{n}}\xi H(1) )}\xi \biggr) \\ &\quad {} /{(\xi -t)^{2}}, \end{aligned} \\& \quad \quad \textit{for }0\leq t< \xi ;\\& \begin{aligned} (2)\quad &1-\beta _{n}(\xi ;t) \\ &\quad = \int _{t}^{\infty }{ \frac{\partial \{K_{n}(\xi ,u)\}}{\partial u}}\,du \\ &\quad \leq \biggl(\frac{B^{\prime \prime } (\frac{n}{b_{n}}\xi H(1) )-2B^{\prime } (\frac{n}{b_{n}}\xi H(1) )+B (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )}\xi ^{2} \\ &\quad\quad {} +\frac{b_{n}}{n}\frac{(A(1)+2A^{\prime }(1)+A(1)H^{\prime \prime }(1))B^{\prime } (\frac{n}{b_{n}}\xi H(1) )-2A^{\prime }(1)B (\frac{n}{b_{n}}\xi H(1) )}{A(1)B (\frac{n}{b_{n}}\xi H(1) )}\xi \biggr) \\ &\quad \quad {} /{(t-\xi )^{2}} \end{aligned} \\& \quad \quad \textit{for }\xi < t< \infty . \end{aligned}$$

Proof

(1) Let $0\leq t<\xi$. Then Lemma 2.1 gives

$$ \int _{0}^{t}{\frac{\partial \{K_{n}(x,u)\}}{\partial u}}\,du\leq \int _{0}^{t}{ \biggl( \frac{\xi -u}{\xi -t} \biggr) ^{2} \frac{\partial \{K_{n}(\xi ,u)\}}{\partial u}}\,du\leq \frac{B_{n}^{\ast }((\xi -u)^{2};\xi )}{(\xi -t)^{2}}. $$

(2) In the case $\xi < t<\infty$, in a similar way, we obtain

$$ \int _{t}^{\infty }{\frac{\partial \{K_{n}(\xi ,u)\}}{\partial u}}\,du \leq \int _{0}^{\infty }{ \biggl( \frac{\xi -u}{\xi -t} \biggr) ^{2} \frac{\partial \{K_{n}(\xi ,u)\}}{\partial u}}\,du\leq \frac{B_{n}^{\ast }((\xi -u)^{2};\xi )}{(t-\xi )^{2}}. $$

 □

Lemma 3.3

Let $f\in D_{BV[0,\infty )}$. Then for sufficiently large n,

$$\begin{aligned} \bigl\vert B_{n}^{\ast }(f;\xi )-f(\xi ) \bigr\vert &\leq \biggl\vert \frac{1}{2} \bigl(f^{{\prime}}(\xi +)+f^{{\prime }}(\xi -) \bigr) \biggr\vert \cdot \bigl\vert B_{n}^{ \ast }(t-\xi ;\xi ) \bigr\vert \\ &\quad {} + \vert I_{1} \vert + \vert I_{2} \vert + \biggl\vert \frac{1}{2} \bigl(f^{{\prime }}(\xi +)-f^{{\prime}}(\xi -) \bigr) \biggr\vert \cdot \sqrt{ \bigl\vert B_{n}^{\ast } \bigl((t- \xi )^{2}; \xi \bigr) \bigr\vert }, \end{aligned}$$

where $I_{1}=\int _{0}^{\xi }{ [ \int _{t}^{\xi }{f_{\xi }^{{\prime }}(u)\,du} ] }\frac{\partial K_{n}(\xi ,t)}{\partial t}\,dt$ and $I_{2}=\int _{\xi }^{\infty }{ [ \int _{\xi }^{t}{f_{\xi }^{{\prime}}(u)\,du} ] }\frac{\partial K_{n}(\xi ,t)}{\partial t}\,dt$.

Proof

For $f\in D_{BV[0,\infty )}$, we may write

$$ \begin{aligned} f^{{\prime }}(t)={}&\frac{1}{2} \bigl[f^{{\prime }}( \xi +)+f^{{\prime }}( \xi -) \bigr]+f_{\xi }^{{\prime }}(t)+ \frac{1}{2} \bigl[f^{{\prime }}(\xi +)-f^{{\prime}}(\xi -) \bigr] \operatorname{sgn}(t-\xi ) \\ &{}+\delta _{\xi }(t) \biggl[ f^{{\prime }}(t)- \frac{1}{2} \bigl[f^{{\prime }}(\xi +)+f^{{\prime }}(\xi -) \bigr] \biggr] , \end{aligned} $$

where

$$ \delta _{\xi }(t)=\textstyle\begin{cases} 1, & t=\xi , \\ 0, & t\neq \xi .\end{cases} $$

From the above facts we get

$$\begin{aligned}& \bigl\vert B_{n}^{\ast }(f;\xi )-f(\xi ) \bigr\vert \\& \quad = \int _{0}^{\infty }{ \bigl(f(t)-f(\xi ) \bigr) \frac{\partial \{K_{n}(\xi ,t)\}}{\partial t}}\,dt= \int _{0}^{\infty }{ \biggl[ \int _{\xi }^{t}{f^{{\prime }}(u)\,du} \biggr] \frac{\partial \{K_{n}(\xi ,t)\}}{\partial t}}\,dt\\& \quad = \int _{0}^{\infty }{ \biggl[ \int _{\xi }^{t} \biggl\{ \frac{1}{2} \bigl[f^{{\prime}}(\xi +)+f^{{\prime }}(\xi -) \bigr]+f_{\xi }^{{\prime }}(u)+ \frac{1}{2} \bigl[f^{{\prime }}(\xi +)-f^{{\prime }}(\xi -) \bigr]\operatorname{sgn}(u-\xi )}\\& \quad\quad{} +\delta _{\xi }(u) \biggl[ f^{{\prime }}(u)-\frac{1}{2} \bigl[f^{{\prime }}( \xi +)+f^{{\prime }}(\xi -) \bigr] \biggr] \,du \biggr\} \frac{\partial \{K_{n}(\xi ,t)\}}{\partial t} \biggr]\,dt. \end{aligned}$$

Since $\int _{\xi }^{t}{\delta _{\xi }(u)}\,du=0$, we obtain

$$\begin{aligned} B_{n}^{\ast }(f;\xi )-f(\xi )&=\frac{1}{2} \bigl[f^{{\prime }}(\xi +)+f^{{\prime}}(\xi -) \bigr] \int _{0}^{\infty }{(t-\xi ) \frac{\partial \{K_{n}(\xi ,t)\}}{\partial t}} \,dt\\ &\quad {} + \int _{0}^{\infty }{ \biggl[ \int _{\xi }^{t}{f_{\xi }^{{\prime }}(u) \,du} \biggr] \frac{\partial \{K_{n}(\xi ,t)\}}{\partial t}}\,dt \\ &\quad\quad{} +\frac{1}{2} \bigl[f^{{\prime}}( \xi +)-f^{{\prime }}(\xi -) \bigr] \int _{0}^{\infty }{(t-\xi ) \frac{\partial \{K_{n}(\xi ,t)\}}{\partial t}} \,dt. \end{aligned}$$

Let us now break the second term in the above relation into two parts:

$$\begin{aligned} \int _{0}^{\infty }{ \biggl[ \int _{\xi }^{t}{f^{{\prime }}(u)\,du} \biggr] \frac{\partial \{K_{n}(\xi ,t)\}}{\partial t}}\,dt&=- \int _{0}^{\xi }{ \biggl[ \int _{t}^{\xi }{f_{\xi }^{{\prime }}(u) \,du} \biggr] \frac{\partial \{K_{n}(\xi ,t)\}}{\partial t}}\,dt \\ &\quad{} + \int _{\xi }^{\infty }{ \biggl[ \int _{\xi }^{t}{f_{\xi }^{{\prime }}(u) \,du} \biggr] \frac{\partial \{K_{n}(\xi ,t)\}}{\partial t}}\,dt=-I_{1}+I_{2}. \end{aligned}$$

Now we have the following estimate:

$$\begin{aligned}& \begin{aligned} \bigl\vert B_{n}^{\ast }(f;\xi )-f(\xi ) \bigr\vert \leq{}& \biggl\vert \frac{1}{2} \bigl[f^{{\prime}}(\xi +)+f^{{\prime }}(\xi -) \bigr] \biggr\vert \cdot \bigl\vert B_{n}^{ \ast } \bigl((t-\xi ),\xi \bigr) \bigr\vert + \vert I_{1} \vert + \vert I_{2} \vert \\ &{}+ \biggl\vert \frac{1}{2} \bigl[f^{{\prime }}(\xi +)-f^{{\prime }}( \xi -) \bigr] \biggr\vert \cdot B_{n}^{\ast } \bigl( \vert t- \xi \vert , \xi \bigr).\end{aligned} \end{aligned}$$

Applying the Cauchy–Schwarz inequality to the above relation, we get

$$\begin{aligned}& \begin{aligned} \bigl\vert B_{n}^{\ast }(f;\xi )-f(\xi ) \bigr\vert \leq{}& \biggl\vert \frac{1}{2} \bigl[f^{{\prime}}(\xi +)+f^{{\prime }}(\xi -) \bigr] \biggr\vert \cdot \bigl\vert B_{n}^{ \ast } \bigl((t-\xi ),\xi \bigr) \bigr\vert + \vert I_{1} \vert + \vert I_{2} \vert \\ &{} + \biggl\vert \frac{1}{2} \bigl[f^{{\prime }}(\xi +)-f^{{\prime }}( \xi -) \bigr] \biggr\vert \cdot \sqrt{B_{n}^{\ast } \bigl((t-\xi )^{2},\xi \bigr)}.\end{aligned} \end{aligned}$$

 □

Lemma 3.4

Let $f\in D_{BV[0,\infty )}$, and let n be sufficiently large. Then

$$ \vert I_{1} \vert \leq \frac{B_{n}^{\ast }((\xi -u)^{2};\xi )}{\xi }\sum _{k=1}^{[ \sqrt{n}]}{ \Biggl( \bigvee _{\xi -\frac{\xi }{\sqrt{n}}}^{\xi }{f_{\xi }^{{\prime}}} \Biggr) }+\frac{\xi }{\sqrt{n}} \Biggl( \bigvee_{\xi - \frac{\xi }{\sqrt{n}}}^{\xi }{f_{\xi }^{{\prime }}} \Biggr) , $$

where $I_{1}=\int _{0}^{\xi }{ [ \int _{t}^{\xi }{f_{\xi }^{{\prime }}(u)\,du} ] }\frac{\partial K_{n}(\xi ,t)}{\partial t}\,dt$.

Proof

By integration by parts we have

$$\begin{aligned} \vert I_{1} \vert &= \biggl\vert \int _{0}^{\xi }{ \biggl[ \int _{\xi }^{t}{f_{\xi }^{{\prime}}(u) \,du} \biggr] }\frac{\partial K_{n}(\xi ,t)}{\partial t}\,dt \biggr\vert \\ & = \biggl\vert \int _{0}^{\xi }{f_{\xi }(t)} \frac{\partial K_{n}(\xi ,t)}{\partial t}\,dt \biggr\vert + \biggl\vert \int _{0}^{\xi }{ \biggl[ \int _{0}^{t}{ \frac{\partial K_{n}(\xi ,u)}{\partial t}\,du} \biggr] f_{\xi }^{{\prime }}(t)}\,dt \biggr\vert \\ &\leq \int _{0}^{\xi -\frac{\xi }{\sqrt{n}}}{ \biggl\vert \int _{0}^{t}{ \frac{\partial K_{n}(\xi ,u)}{\partial t}\,du} \biggr\vert \bigl\vert f_{\xi }^{{\prime}}(t) \bigr\vert } \,dt+ \int _{\xi -\frac{\xi }{\sqrt{n}}}^{\xi }{ \biggl\vert \int _{0}^{t}{ \frac{\partial K_{n}(\xi ,u)}{\partial t}\,du} \biggr\vert \bigl\vert f_{\xi }^{{\prime}}(t) \bigr\vert }\,dt. \end{aligned}$$

Using Lemma 3.2, we obtain

$$\begin{aligned} \vert I_{1} \vert &\leq B_{n}^{\ast } \bigl(( \xi -u)^{2};\xi \bigr) \int _{0}^{\xi - \frac{\xi }{\sqrt{n}}}{ \Biggl( \bigvee _{t}^{\xi }{f_{\xi }^{{\prime }}} \Biggr) \frac{1}{(\xi -t)^{2}}}\,dt+ \int _{\xi -\frac{\xi }{\sqrt{n}}}^{\xi }{ \Biggl( \bigvee _{\xi -\frac{\xi }{\sqrt{n}}}^{\xi }{f_{\xi }^{{\prime }}} \Biggr) } \,dt\\ &\leq B_{n}^{\ast } \bigl((\xi -u)^{2};\xi \bigr) \int _{0}^{\xi - \frac{\xi }{\sqrt{n}}}{ \Biggl( \bigvee _{t}^{\xi }{f_{\xi }^{{\prime}}} \Biggr) \frac{1}{(\xi -t)^{2}}}\,dt+\frac{\xi }{\sqrt{n}} \Biggl( \bigvee _{\xi -\frac{\xi }{\sqrt{n}}}^{ \xi }{f_{\xi }^{{\prime }}} \Biggr) . \end{aligned}$$

Substituting $u=\frac{\xi }{\xi -t}$, we get

$$ \int _{0}^{\xi -\frac{\xi }{\sqrt{n}}}{ \Biggl( \bigvee _{t}^{\xi }{f_{ \xi }^{{\prime }}} \Biggr) \frac{1}{(\xi -t)^{2}}}\,dt=\frac{1}{\xi } \int _{1}^{\sqrt{n}}{ \Biggl( \bigvee _{\xi -\frac{\xi }{\sqrt{n}}}^{\xi }{f_{ \xi }^{{\prime }}} \Biggr) } \,du\leq \frac{1}{\xi }\sum_{k=1}^{[\sqrt{n}]}{ \Biggl( \bigvee_{\xi -\frac{\xi }{\sqrt{n}}}^{\xi }{f_{\xi }^{{\prime}}} \Biggr) }, $$

which yields that

$$ \vert I_{1} \vert \leq \frac{B_{n}^{\ast }((\xi -u)^{2};\xi )}{\xi }\sum _{k=1}^{[ \sqrt{n}]}{ \Biggl( \bigvee _{\xi -\frac{\xi }{\sqrt{n}}}^{\xi }{f_{x}^{{\prime}}} \Biggr) }+\frac{\xi }{\sqrt{n}} \Biggl( \bigvee_{\xi - \frac{\xi }{\sqrt{n}}}^{\xi }{f_{\xi }^{{\prime }}} \Biggr) . $$

 □

Lemma 3.5

Let $f\in D_{BV[0,\infty )}$, and let n be sufficiently large. Then

$$\begin{aligned} \vert I_{2} \vert &\leq \biggl( \frac{M_{2}}{\xi ^{2}}+4M_{2}+ \frac{ \vert f(\xi ) \vert }{\xi ^{2}} \biggr) B_{n}^{\ast } \bigl((t- \xi )^{2};\xi \bigr)+ \bigl\vert f^{{\prime }}(\xi +) \bigr\vert \sqrt {B_{n}^{\ast } \bigl((t-\xi )^{2}; \xi \bigr)}\\ &\quad {} +\frac{B_{n}^{\ast }((t-\xi )^{2};\xi )}{(t-\xi )^{2}} \bigl\vert f(2\xi )-f( \xi )-\xi f^{{\prime }}(\xi +) \bigr\vert + \frac{\xi }{\sqrt{n}} \Biggl( \bigvee_{\xi }^{\xi +\frac{\xi }{\sqrt{n}}}{f_{\xi }^{{\prime }}} \Biggr) \\ &\quad {} +B_{n}^{\ast } \bigl((t- \xi )^{2};\xi \bigr) \sum_{k=1}^{[\sqrt{n}]}{ \Biggl( \bigvee _{\xi }^{ \xi +\frac{\xi }{k}}{f_{\xi }^{{\prime }}} \Biggr) }, \end{aligned}$$

where $I_{2}=\int _{\xi }^{\infty }{ [ \int _{\xi }^{t}{f_{\xi }^{{\prime}}(u)\,du} ] }\frac{\partial K_{n}(\xi ,t)}{\partial t}\,dt$.

Proof

By the properties of integrals we have

$$\begin{aligned}& \begin{gathered} \biggl\vert \int _{\xi }^{\infty }{ \biggl[ \int _{\xi }^{t}{f_{\xi }^{{\prime}}(u) \,du} \biggr] }\frac{\partial K_{n}(\xi ,t)}{\partial t}\,dt \biggr\vert \\ \quad \leq \biggl\vert \int _{2\xi }^{\infty }{ \biggl[ \int _{ \xi }^{t}{f_{\xi }^{{\prime }}(u) \,du} \biggr] } \frac{\partial K_{n}(\xi ,t)}{\partial t}\,dt \biggr\vert + \biggl\vert \int _{\xi }^{2\xi }{ \biggl[ \int _{\xi }^{t}{f_{\xi }^{{\prime }}(u) \,du} \biggr] }\frac{\partial K_{n}(\xi ,t)}{\partial t}\,dt \biggr\vert \\ \quad =I_{2}^{{\prime }}+I_{2}^{{\prime \prime }},\end{gathered} \\& \begin{aligned} I_{2}^{{\prime }}&= \biggl\vert \int _{2\xi }^{\infty }{ \biggl[ \int _{ \xi }^{t}{ \bigl(f_{\xi }^{{\prime }}(u)-f_{\xi }^{{\prime }}( \xi +) \bigr)\,du} \biggr] } \frac{\partial K_{n}(\xi ,t)}{\partial t}\,dt \biggr\vert \\ &\leq \biggl\vert \int _{2\xi }^{\infty }{{ \bigl(f(t)-f(\xi ) \bigr)}} \frac{\partial K_{n}(\xi ,t)}{\partial t}\,dt \biggr\vert + \bigl\vert f^{{\prime }}(\xi +) \bigr\vert \biggl\vert \int _{2\xi }^{\infty }{(t-\xi ) \frac{\partial K_{n}(\xi ,t)}{\partial t}}\,dt \biggr\vert \\ &\leq \int _{2\xi }^{ \infty }{ \bigl\vert f(t) \bigr\vert \frac{\partial K_{n}(\xi ,t)}{\partial t}}\,dt+ \bigl\vert f(\xi ) \bigr\vert \int _{2 \xi }^{\infty }{\frac{\partial K_{n}(\xi ,t)}{\partial t}}\,dt+ \bigl\vert f^{{\prime }}(\xi +) \bigr\vert \int _{2\xi }^{\infty }{ \vert t-\xi \vert \frac{\partial K_{n}(\xi ,t)}{\partial t}}\,dt \\ &=A_{1}+A_{2}+A_{3}.\end{aligned} \end{aligned}$$

Now we will estimate

$$ A_{1}\leq M_{2} \int _{2\xi }^{\infty }{ \bigl(1+t^{2} \bigr) \frac{\partial K_{n}(\xi ,t)}{\partial t}}\,dt. $$

Since $t\geq 2\xi$, that is, $t-\xi \geq \xi$, we get

$$\begin{aligned} A_{1}&\leq M_{2} \int _{2\xi }^{\infty }{\frac{(t-\xi )^{2}}{\xi ^{2}} \frac{\partial K_{n}(\xi ,t)}{\partial t}} \,dt+4M_{2} \int _{2\xi }^{\infty }{(t- \xi )^{2} \frac{\partial K_{n}(\xi ,t)}{\partial t}}\,dt\\ &\leq \biggl( \frac{M_{2}}{\xi ^{2}}+4M_{2} \biggr) B_{n}^{\ast } \bigl((t- \xi )^{2};\xi \bigr). \end{aligned}$$

Now we estimate

$$ A_{2}\leq \bigl\vert f(\xi ) \bigr\vert \int _{2\xi }^{\infty }{ \frac{(t-\xi )^{2}}{\xi ^{2}} \frac{\partial K_{n}(\xi ,t)}{\partial t}}\,dt\leq \frac{ \vert f(\xi ) \vert }{\xi ^{2}}B_{n}^{\ast } \bigl((t-\xi )^{2};\xi \bigr) $$

and

$$ A_{3}\leq \bigl\vert f^{{\prime }}(\xi +) \bigr\vert \int _{2x}^{\infty }{ \vert t-\xi \vert \frac{\partial K_{n}(\xi ,t)}{\partial t}}\,dt\leq \bigl\vert f^{{\prime }}(\xi +) \bigr\vert \sqrt {B_{n}^{\ast } \bigl((t-\xi )^{2};\xi \bigr)}. $$

From last three relations we get the upper bound

$$ \bigl\vert I_{2}^{{\prime }} \bigr\vert \leq \biggl( \frac{M_{2}}{\xi ^{2}}+4M_{2}+ \frac{ \vert f(\xi ) \vert }{\xi ^{2}} \biggr) B_{n}^{\ast } \bigl((t-\xi )^{2};\xi \bigr)+ \bigl\vert f^{{\prime }}( \xi +) \bigr\vert \sqrt {B_{n}^{\ast } \bigl((t-\xi )^{2};\xi \bigr)}. $$

Now we estimate

$$\begin{aligned} \bigl\vert I_{2}^{{\prime \prime }} \bigr\vert = \biggl\vert \int _{\xi }^{2\xi }{ \biggl[ \int _{\xi }^{t}{f_{\xi }^{{\prime }}(u) \,du} \biggr] } \frac{\partial K_{n}(\xi ,t)}{\partial t}\,dt \biggr\vert &\leq \bigl\vert 1-\beta _{n}( \xi ,2\xi ) \bigr\vert \biggl\vert \int _{\xi }^{2\xi }{f_{\xi }^{{\prime }}(u) \,du} \biggr\vert \\ &\quad {} + \biggl\vert \int _{\xi }^{2\xi }f_{\xi }^{{\prime }}(t) \bigl(1-\beta _{n}( \xi ,t) \bigr)\,dt \biggr\vert . \end{aligned}$$

From Lemma 3.2 we have

$$\begin{aligned}& \begin{aligned} \bigl\vert I_{2}^{{\prime \prime }} \bigr\vert &\leq \frac{B_{n}^{\ast }((t-\xi )^{2};\xi )}{\xi ^{2}} \biggl\vert \int _{ \xi }^{2\xi }{ \bigl(f{^{\prime }}(u)-f^{{\prime }}( \xi +) \bigr)}\,du \biggr\vert + \biggl\vert \int _{\xi }^{\xi +\frac{\xi }{\sqrt{n}}}{f_{ \xi }{^{\prime }}(t) \bigl(1-\beta _{n}(\xi ,t) \bigr)}\,dt \biggr\vert \\ &\quad {} + \biggl\vert \int _{\xi +\frac{\xi }{\sqrt{n}}}^{2\xi }{f_{\xi }{^{ \prime }}(t) \bigl(1-\beta _{n}(\xi ,t) \bigr)}\,dt \biggr\vert \\ & = \frac{B_{n}^{\ast }((t-\xi )^{2};\xi )}{\xi ^{2}} \bigl\vert f(2\xi )-f(\xi )- \xi f^{{\prime }}(\xi +) \bigr\vert \biggl\vert \int _{\xi }^{\xi +\frac{\xi }{\sqrt{n}}}{f_{\xi }{^{ \prime }}(t) \bigl(1-\beta _{n}(\xi ,t) \bigr)}\,dt \biggr\vert \\ &\quad{} + \biggl\vert \int _{\xi + \frac{\xi }{\sqrt{n}}}^{2\xi }{f_{\xi }{^{\prime }}(t) \bigl(1-\beta _{n}(\xi ,t) \bigr)}\,dt \biggr\vert ,\end{aligned} \\& \biggl\vert \int _{\xi }^{\xi +\frac{\xi }{\sqrt{n}}}{f_{\xi }{^{ \prime }}(t) \bigl(1-\beta _{n}(\xi ,t) \bigr)}\,dt \biggr\vert \leq \int _{\xi }^{\xi + \frac{\xi }{\sqrt{n}}}{ \Biggl( \bigvee _{\xi }^{t}{f_{\xi }^{{\prime }}} \Biggr) } \,dt\leq \frac{\xi }{\sqrt{n}} \Biggl( \bigvee_{\xi }^{\xi + \frac{\xi }{\sqrt{n}}}{f_{\xi }^{{\prime }}} \Biggr) , \end{aligned}$$

and

$$ \biggl\vert \int _{\xi +\frac{\xi }{\sqrt{n}}}^{2\xi }{f_{\xi }{^{ \prime }}(t) \bigl(1-\beta _{n}(\xi ,t) \bigr)}\,dt \biggr\vert \leq B_{n}^{\ast } \bigl((t-\xi )^{2}; \xi \bigr) \int _{\xi +\frac{\xi }{\sqrt{n}}}^{2\xi }{ \Biggl( \bigvee _{ \xi }^{t}{f_{\xi }^{{\prime }}} \Biggr) }\frac{1}{(\xi -t)^{2}}\,dt. $$

For $u=\frac{\xi }{t-\xi }$, we get

$$\begin{aligned} \biggl\vert \int _{\xi +\frac{\xi }{\sqrt{n}}}^{2\xi }{f_{\xi }{^{ \prime }}(t) \bigl(1-\beta _{n}(\xi ,t) \bigr)}\,dt \biggr\vert &\leq \frac{B_{n}^{\ast }((t-\xi )^{2};\xi )}{\xi } \int _{1}^{\sqrt{n}}{ \Biggl( \bigvee _{\xi }^{\xi +\frac{\xi }{u}}{f_{\xi }^{{\prime }}} \Biggr) } \,du\\ &\leq \frac{B_{n}^{\ast }((t-\xi )^{2};\xi )}{\xi }\sum_{k=1}^{[\sqrt{n}]}{ \Biggl( \bigvee_{\xi }^{\xi +\frac{\xi }{k}}{f_{\xi }^{{\prime }}} \Biggr) }. \end{aligned}$$

From the last relations we obtain that

$$ \begin{aligned} \bigl\vert I_{2}^{{\prime \prime }} \bigr\vert \leq{}& \frac{B_{n}^{\ast }((t-\xi )^{2};\xi )}{\xi ^{2}} \bigl\vert f(2\xi )-f(\xi )- \xi f^{{\prime }}(\xi +) \bigr\vert +\frac{\xi }{\sqrt{n}} \Biggl( \bigvee _{\xi }^{\xi +\frac{\xi }{\sqrt{n}}}{f_{\xi }^{{\prime}}} \Biggr) \\ &{}+\frac{B_{n}^{\ast }((t-\xi )^{2};\xi )}{\xi }\sum_{k=1}^{[ \sqrt{n}]}{ \Biggl( \bigvee_{\xi }^{\xi +\frac{\xi }{k}}{f_{\xi }^{{\prime }}} \Biggr) }. \end{aligned} $$

Hence

$$\begin{aligned}& \begin{aligned} \vert I_{2} \vert \leq \bigl\vert I_{2}^{{\prime }} \bigr\vert + \bigl\vert I_{2}^{{\prime \prime }} \bigr\vert &\leq \biggl( \frac{M_{2}}{\xi ^{2}}+4M_{2}+ \frac{ \vert f(\xi ) \vert }{\xi ^{2}} \biggr) B_{n}^{\ast } \bigl((t-\xi )^{2};\xi \bigr)+ \bigl\vert f^{{\prime }}(\xi +) \bigr\vert \sqrt {B_{n}^{\ast } \bigl((t-\xi )^{2};\xi \bigr)}\\ &\quad {} +\frac{B_{n}^{\ast }((t-\xi )^{2};\xi )}{\xi ^{2}} \bigl\vert f(2\xi )-f(\xi )- \xi f^{{\prime }}(\xi +) \bigr\vert + \frac{\xi }{\sqrt{n}} \Biggl( \bigvee_{ \xi }^{\xi +\frac{\xi }{\sqrt{n}}}{f_{\xi }^{{\prime }}} \Biggr) \\ &\quad{} + \frac{B_{n}^{\ast }((t-\xi )^{2};\xi )}{\xi }\sum_{k=1}^{[\sqrt{n}]}{ \Biggl( \bigvee_{\xi }^{\xi +\frac{\xi }{k}}{f_{\xi }^{{\prime }}} \Biggr) }.\end{aligned} \end{aligned}$$

 □

Proof of Theorem 3.1

Based on Lemmas 3.2, 3.3, 3.4 and 3.5, we get the following estimate:

$$\begin{aligned}& \bigl\vert B_{n}^{\ast }(f;\xi )-f(\xi ) \bigr\vert \\ & \quad \leq \biggl\vert \frac{1}{2} \bigl(f^{{\prime}}(\xi +)+f^{{\prime }}(\xi -) \bigr) \biggr\vert \cdot \bigl\vert B_{n}^{ \ast }(t-\xi ;\xi ) \bigr\vert \\ & \quad\quad {} + \vert I_{1} \vert + \vert I_{2} \vert + \biggl\vert \frac{1}{2} \bigl(f^{{\prime }}(\xi +)-f^{{\prime}}(\xi -) \bigr) \biggr\vert \cdot \sqrt{ \bigl\vert B_{n}^{\ast } \bigl((t- \xi )^{2}; \xi \bigr) \bigr\vert }\\ & \quad \leq \biggl\vert \frac{1}{2} \bigl(f^{{\prime }}(\xi +)+f^{{\prime }}( \xi -) \bigr) \biggr\vert \cdot \bigl\vert B_{n}^{\ast }(t- \xi ;\xi ) \bigr\vert + \frac{B_{n}^{\ast }((\xi -t)^{2};\xi )}{\xi } \sum_{k=1}^{[\sqrt{n}]}{ \Biggl( \bigvee _{\xi -\frac{\xi }{\sqrt{n}}}^{\xi }{f_{\xi }^{{\prime}}} \Biggr) } \\ & \quad\quad {} +\frac{\xi }{\sqrt{n}} \Biggl( \bigvee_{\xi -\frac{\xi }{\sqrt{n}}}^{\xi }{f_{\xi }^{{\prime}}} \Biggr) + \biggl( \frac{M_{f}}{\xi ^{2}}+4M_{f}+\frac{ \vert f(\xi ) \vert }{\xi ^{2}} \biggr) B_{n}^{\ast } \bigl((t-\xi )^{2};\xi \bigr) \\ & \quad\quad {}+ \bigl\vert f^{{\prime }}(\xi +) \bigr\vert \sqrt{B_{n}^{\ast } \bigl((t-\xi )^{2};\xi \bigr)}+\frac{B_{n}^{\ast }((t-\xi )^{2};x)}{\xi ^{2}} \bigl\vert f(2\xi )-f(\xi )-\xi f^{{\prime}}(\xi +) \bigr\vert \\ & \quad\quad {} + \frac{\xi }{\sqrt{n}} \Biggl( \bigvee_{\xi }^{\xi+\frac{\xi }{\sqrt{n}}}{f_{\xi }^{{\prime }}} \Biggr) + \frac{B_{n}^{\ast }((t-\xi )^{2};\xi )}{\xi }\sum_{k=1}^{[\sqrt{n}]}{ \Biggl( \bigvee_{\xi }^{\xi +\frac{\xi }{k}}{f_{\xi }^{{\prime }}} \Biggr) }\\ & \quad\quad {} + \biggl\vert \frac{1}{2} \bigl(f^{{\prime }}(\xi +)-f^{{\prime }}( \xi -) \bigr) \biggr\vert \cdot \sqrt{ \bigl\vert B_{n}^{\ast } \bigl((t-\xi )^{2};\xi \bigr) \bigr\vert }. \end{aligned}$$

Since

$$ \Biggl( \bigvee_{a}^{b}{f} \Biggr) + \Biggl( \bigvee_{b}^{c}{f} \Biggr) = \Biggl( \bigvee_{a}^{c}{f} \Biggr) , $$

we obtain

$$\begin{aligned}& \bigl\vert B_{n}^{\ast }(f;\xi )-f(\xi ) \bigr\vert \\ & \quad \leq \biggl\vert \frac{1}{2} \bigl(f^{{\prime}}(\xi +)+f^{{\prime }}(\xi -) \bigr) \biggr\vert \cdot \bigl\vert B_{n}^{ \ast }(t-\xi ;\xi ) \bigr\vert \\ & \quad\quad {} + \vert I_{1} \vert + \vert I_{2} \vert + \biggl\vert \frac{1}{2} \bigl(f^{{\prime }}(\xi +)-f^{{\prime}}(\xi -) \bigr) \biggr\vert \cdot \sqrt{ \bigl\vert B_{n}^{\ast } \bigl((t- \xi )^{2}; \xi \bigr) \bigr\vert }\\ & \quad \leq \biggl\vert \frac{1}{2} \bigl(f^{{\prime }}(\xi +)+f^{{\prime }}( \xi -) \bigr) \biggr\vert \cdot \bigl\vert B_{n}^{\ast }(t- \xi ;\xi ) \bigr\vert + \frac{B_{n}^{\ast }((\xi -t)^{2};\xi )}{\xi } \sum_{k=1}^{[\sqrt{n}]}{ \Biggl( \bigvee _{\xi -\frac{\xi }{\sqrt{n}}}^{\xi + \frac{\xi }{\sqrt{n}}}{f_{\xi }^{{\prime }}} \Biggr) } \\ & \quad\quad {}+\frac{\xi }{\sqrt{n}} \Biggl( \bigvee_{\xi -\frac{\xi }{\sqrt{n}}}^{ \xi +\frac{\xi }{\sqrt{n}}}{f_{\xi }^{{\prime }}} \Biggr) + \biggl( \frac{M_{f}}{\xi ^{2}}+4M_{f}+\frac{ \vert f(\xi ) \vert }{\xi ^{2}} \biggr) B_{n}^{\ast } \bigl((t-\xi )^{2};\xi \bigr) \\ & \quad\quad {}+ \bigl\vert f^{{\prime }}(\xi +) \bigr\vert \sqrt{B_{n}^{\ast } \bigl((t-\xi )^{2};\xi \bigr)} +\frac{B_{n}^{\ast }((t-\xi )^{2};\xi )}{\xi ^{2}} \bigl\vert f(2\xi )-f(\xi )- \xi f^{{\prime }}(\xi +) \bigr\vert \\ & \quad\quad {} + \biggl\vert \frac{1}{2} \bigl(f^{{\prime }}( \xi +)-f^{{\prime }}(\xi -) \bigr) \biggr\vert \cdot \sqrt{ \bigl\vert B_{n}^{\ast } \bigl((t- \xi )^{2};\xi \bigr) \bigr\vert }. \end{aligned}$$

 □

4 Voronovskaya-type theorems

The Voronovskaya-type theorem for the Chlodowsky-type Szász operators based on Boas–Buck-type polynomials under certain conditions is known. First, we introduce following assumptions [26]:

$$\begin{aligned} & \lim_{n\rightarrow \infty } \frac{n}{b_{n}} \frac{B^{\prime } (\frac{n}{b_{n}}\xi H(1) )-B (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )}=l_{1}( \xi ); \end{aligned}$$

(4.1)

$$\begin{aligned} & \lim_{n\rightarrow \infty } \frac{n}{b_{n}} \frac{B^{\prime \prime } (\frac{n}{b_{n}}\xi H(1) )-2B^{\prime } (\frac{n}{b_{n}}\xi H(1) )+B (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )}=l_{2}( \xi ); \end{aligned}$$

(4.2)

$$\begin{aligned} & \lim_{n\rightarrow \infty } \biggl( \frac{n}{b_{n}} \biggr) ^{2}\frac{1}{B (\frac{n}{b_{n}}\xi H(1) )} \biggl[ B^{(4)} \biggl( \frac{n}{b_{n}}\xi H(1) \biggr) -4B^{(3)} \biggl( \frac{n}{b_{n}} \xi H(1) \biggr) +6B^{\prime \prime } \biggl( \frac{n}{b_{n}}\xi H(1) \biggr) \\ & \quad {} - 4B^{\prime } \biggl( \frac{n}{b_{n}}\xi H(1) \biggr) +B \biggl( \frac{n}{b_{n}}\xi H(1) \biggr) \biggr] =l_{3}(\xi ); \end{aligned}$$

(4.3)

$$\begin{aligned} & \lim_{n\rightarrow \infty } \frac{n}{b_{n}} \frac{1}{B ( \frac{n}{b_{n}}\xi H(1) ) A(1)} \biggl[ \bigl( 2A^{\prime }(1)+3A(1)H^{ \prime \prime }(1)+3A(1) \bigr) B^{(3)} \biggl( \frac{n}{b_{n}}\xi H(1) \biggr) -6 \bigl( A^{\prime }(1) \\ & \quad {} + A(1)H^{\prime \prime }(1)+A(1) \bigr) B^{\prime \prime } \biggl( \frac{n}{b_{n}}\xi H(1) \biggr) +3 \bigl( 2A^{\prime }(1)+A(1)H^{ \prime \prime }(1)+A(1) \bigr) B^{\prime } \biggl( \frac{n}{b_{n}} \xi H(1) \biggr) \\ & \quad {} -2A^{\prime }(1)B \biggl( \frac{n}{b_{n}}\xi H(1) \biggr) \biggr] =l_{4}(\xi ). \end{aligned}$$

(4.4)

Remark 4.1

[26] As a consequence of the above assumption, we obtain

• i) $\lim_{n\rightarrow \infty } \frac{n}{b_{n}}B_{n}^{\ast } ( e_{1}-\xi ;\xi ) =\eta _{1}(\xi )$,

• ii) $\lim_{n\rightarrow \infty } \frac{n}{b_{n}}B_{n}^{\ast } ( (e_{1}-\xi )^{2};\xi ) =\eta _{2}( \xi )$,

• iii) $\lim_{n\rightarrow \infty } ( \frac{n}{b_{n}} ) ^{2}B_{n}^{\ast } ( (e_{1}-\xi )^{4}; \xi ) =\eta _{3}(\xi )$,

where

$$\begin{aligned} & \eta _{1}(\xi )=\xi l_{1}(\xi )+\frac{A^{\prime }(1)}{A(1)}, \qquad \eta _{2}(\xi )=\xi ^{2}l_{2}(\xi )+\xi \bigl( 1+H^{\prime \prime }(1) \bigr) , \\ & \eta _{3}(\xi )=\xi ^{4}l_{3}(\xi )+2\xi ^{3}l_{4}(\xi )+3\xi ^{2} \bigl( H^{\prime \prime }(1)^{2}+2H^{\prime \prime }(1)+1 \bigr) . \end{aligned}$$

Theorem 4.2

[26] (Voronovskaya-type theorem) For every $f\in C_{E}({\mathbb{R}}_{0}^{+})$ such that $f^{\prime },f^{\prime \prime }\in C_{E}({\mathbb{R}}_{0}^{+})$, we have

$$ \lim_{n\rightarrow \infty }\frac{n}{b_{n}} \bigl[B_{n}^{\ast }(f; \xi )-f( \xi ) \bigr]= \biggl( \xi l_{1}(\xi )+\frac{A^{\prime }(1)}{A(1)} \biggr) f^{ \prime }(\xi )+\frac{1}{2} ( \xi ^{2}l_{2}( \xi )+\xi \bigl(1+H^{ \prime \prime }(1) \bigr) f^{\prime \prime }(\xi ), $$

uniformly with respect to $\xi \in{}[ 0,a]$, $a>0$, where $l_{i}(\xi )$, $i=1,2$, are defined in (4.1) and (4.2).

Example 4.3

Write

$$ NB_{n}^{\ast }(h,\xi )=(1+u_{n})B_{n}^{\ast }(h, \xi ), $$

where

$$ u_{n}=\textstyle\begin{cases} \frac{b_{m}^{2}}{m^{2}}, & m^{2}-m\leq n\leq m^{2}-1, \\ \frac{b_{m}^{3}}{m^{3}}, & n=m^{2}, m\in \mathbb{N}\setminus \{1\}, \\ 0 & \text{otherwise}.\end{cases} $$

Lemma 4.4

For the fourth-order central moment, we have the following estimate:

$$ \biggl( \frac{n}{b_{n}} \biggr) ^{2}NB_{n}^{\ast } \bigl((y-\xi )^{4};\xi \bigr) \rightarrow \eta _{3}(\xi ) \quad \textit{on } {}[ 0,M] \textit{ as } n\rightarrow \infty . $$

Proof

From Proposition 2.2 we have

$$ \biggl( \frac{n}{b_{n}} \biggr) ^{2}NB_{n}^{\ast } \bigl((y-\xi )^{4};\xi \bigr)= \biggl( \frac{n}{b_{n}} \biggr) ^{2}(1+u_{n})B_{n}^{\ast } \bigl((y-\xi )^{4}; \xi \bigr), $$

from which we obtain that

$$ \lim_{n\rightarrow \infty }{ \biggl( \frac{n}{b_{n}} \biggr) ^{2}(1+u_{n})B_{n}^{ \ast } \bigl((y-\xi )^{4};\xi \bigr)}=\eta _{3}(\xi )\quad \text{on }[0,M]. $$

 □

Theorem 4.5

Let $f\in C^{B}[0,\infty )$, the space of bounded and continuous functions in $[0,\infty )$, and suppose that $f^{{\prime }},f^{{\prime \prime }}\in C^{B}[0,\infty )$. Then

$$ \begin{aligned}& \biggl( \frac{n}{b_{n}} \biggr) \bigl[NB_{n}^{\ast }(f;\xi )-f( \xi ) \bigr] \\ &\quad \sim f^{{\prime }}(\xi ) \biggl( l_{1}(\xi )\xi + \frac{A^{\prime }(1)}{A(1)} \biggr) \\ &\quad\quad {} +\frac{f^{{\prime \prime }}(\xi )}{2} \biggl(l_{2}(\xi )\xi ^{2}+ \frac{(A(1)+2A^{\prime }(1)+A(1)H^{\prime \prime }(1))-2A^{\prime }(1)}{A(1)B} \xi \biggr) (st_{T}) \end{aligned}$$

for each $x\in{}[ 0,M]$ and any finite M.

Proof

Taylor’s formula gives

$$ f(y)=f(\xi )+(y-\xi )f^{{\prime }}(\xi )+ \frac{1}{2}(y-\xi )^{2}f^{{ \prime \prime }}(\xi )+(y-\xi )^{2}\psi (y- \xi ), $$

(4.5)

where $\psi (y-\xi )\rightarrow 0$ as $y-\xi \rightarrow 0$. Applying $NB_{n}^{\ast }$ to both sides of relation (4.5), we get

$$\begin{aligned} NB_{n}^{\ast }(f)={}&(1+u_{n})f(\xi )+(1+u_{n})f^{{\prime }}(\xi ) \biggl( \frac{B^{\prime } (\frac{n}{b_{n}}\xi H(1) )-B (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )}x+ \frac{b_{n}}{n}\frac{A^{\prime }(1)}{A(1)} \biggr) \\ & {} +(1+u_{n})\frac{f^{{\prime \prime }}(\xi )}{2} \biggl( \frac{B^{\prime \prime } (\frac{n}{b_{n}}\xi H(1) )-2B^{\prime } (\frac{n}{b_{n}}\xi H(1) )+B (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )} \xi ^{2} \\ & {} +\frac{b_{n}}{n} \frac{(A(1)+2A^{\prime }(1)+A(1)H^{\prime \prime }(1))B^{\prime } (\frac{n}{b_{n}}\xi H(1) )-2A^{\prime }(1)B (\frac{n}{b_{n}}\xi H(1) )}{A(1)B (\frac{n}{b_{n}}\xi H(1) )}x \\ & {} +\frac{b_{n}^{2}}{n^{2}} \frac{A^{\prime }(1)+A^{\prime \prime }(1)}{A(1)} \biggr)+(1+u_{n})NB_{n}^{\ast } \bigl(\Phi ^{2}\psi (y-\xi );\xi \bigr). \end{aligned}$$

This yields

$$\begin{aligned} \biggl( \frac{n}{b_{n}} \biggr) NB_{n}^{\ast }(f)={}& \biggl( \frac{n}{b_{n}} \biggr) (1+u_{n})f(\xi ) \\ &{}+ \biggl( \frac{n}{b_{n}} \biggr) (1+u_{n})f^{{\prime}}(\xi ) \biggl( \frac{B^{\prime } (\frac{n}{b_{n}}\xi H(1) )-B (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )}x+ \frac{b_{n}}{n}\frac{A^{\prime }(1)}{A(1)} \biggr) \\ & {} + \biggl( \frac{n}{b_{n}} \biggr) (1+u_{n}) \frac{f^{{\prime \prime }}(\xi )}{2} \biggl( \frac{B^{\prime \prime } (\frac{n}{b_{n}}\xi H(1) )-2B^{\prime } (\frac{n}{b_{n}}\xi H(1) )+B (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )}\xi ^{2} \\ & {} +\frac{b_{n}}{n} \frac{(A(1)+2A^{\prime }(1)+A(1)H^{\prime \prime }(1))B^{\prime } (\frac{n}{b_{n}}\xi H(1) )-2A^{\prime }(1)B (\frac{n}{b_{n}}\xi H(1) )}{A(1)B (\frac{n}{b_{n}}\xi H(1) )}\xi \\ & {} +\frac{b_{n}^{2}}{n^{2}} \frac{A^{\prime }(1)+A^{\prime \prime }(1)}{A(1)} \biggr)+ \biggl( \frac{n}{b_{n}} \biggr) (1+u_{n})NB_{n}^{\ast } \bigl(\Phi ^{2} \psi (y-\xi );\xi \bigr). \end{aligned}$$

Therefore

$$\begin{aligned}& \biggl\vert \biggl( \frac{n}{b_{n}} \biggr) \biggl[NB_{n}^{\ast } (f;\xi )-f(\xi )-f^{{\prime }} (\xi ) \biggl( \frac{B^{\prime } (\frac{n}{b_{n}}\xi H(1) )-B (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )}x+ \frac{A^{\prime }(1)}{A(1)} \biggr) \\& \quad\quad{} -\frac{f^{{\prime \prime }}(\xi )}{2} \biggl(\frac{B^{\prime \prime } (\frac{n}{b_{n}}\xi H(1) )-2B^{\prime } (\frac{n}{b_{n}}\xi H(1) )+B (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )}{ \xi }^{2} \\& \quad\quad{} + \frac{(A(1)+2A^{\prime }(1)+A(1)H^{\prime \prime }(1))B^{\prime } (\frac{n}{b_{n}}\xi H(1) )-2A^{\prime }(1)B (\frac{n}{b_{n}}\xi H(1) )}{A(1)B (\frac{n}{b_{n}}\xi H(1) )} \xi \biggr) \biggr] \biggr\vert \\& \quad \leq \biggl( \frac{n}{b_{n}} \biggr) { Ku}_{n} + \biggl( \frac{n}{b_{n}} \biggr) { K}_{1}{ u}_{n} \biggl\vert \biggl( \frac{B^{\prime } (\frac{n}{b_{n}}\xi H(1) )-B (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )} \xi + \frac{A^{\prime }(1)}{A(1)} \biggr) \biggr\vert \\& \quad\quad{} + \biggl( \frac{b_{n}}{n} \biggr) \frac{K_{2}}{2} \biggl\vert \frac{A^{\prime }(1)+A^{\prime \prime }(1)}{A(1)} \biggr\vert \\& \quad\quad{} + \biggl( \frac{n}{b_{n}} \biggr) { u}_{n} \frac{K_{2}}{2} \biggl\vert \frac{B^{\prime \prime } (\frac{n}{b_{n}}\xi H(1) )-2B^{\prime } (\frac{n}{b_{n}}\xi H(1) )+B (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )}{ \xi }^{2} + \frac{b_{n}}{n} \frac{A^{\prime }(1)+A^{\prime \prime }(1)}{A(1)} \biggr\vert \\& \quad\quad{} + \biggl( \frac{n}{b_{n}} \biggr) \bigl\vert NB_{n}^{\ast } \bigl((y-\xi )^{2} \psi (y-\xi ); \xi \bigr) \bigr\vert +u_{n} \biggl( \frac{n}{b_{n}} \biggr) \bigl\vert NB_{n}^{ \ast } \bigl((y-\xi )^{2} \psi (y-\xi );\xi \bigr) \bigr\vert , \end{aligned}$$

where $K=\sup_{\xi \in{}[ 0,M]}{ \vert f(\xi ) \vert }$, $K_{1}=\sup_{\xi \in{}[ 0,M]}{ \vert f^{{\prime }}(\xi ) \vert }$, and $K_{2}=\sup_{\xi \in{}[ 0,M]}{ \vert f^{{\prime \prime }}(\xi ) \vert }$.

Now we will prove that

$$ \lim_{n\rightarrow \infty }{ \biggl( \frac{n}{b_{n}} \biggr) \bigl\vert NB_{n}^{ \ast } \bigl((y-\xi )^{2}\psi (y-\xi );\xi \bigr) \bigr\vert }=0. $$

Applying the Cauchy–Schwartz inequality, we get

$$ \biggl( \frac{n}{b_{n}} \biggr) \bigl\vert NB_{n}^{\ast } \bigl((y- \xi )^{2}\psi (y- \xi );\xi \bigr) \bigr\vert \leq \biggl[ \biggl( \frac{n}{b_{n}} \biggr) ^{2}NB_{n}^{ \ast } \bigl((y-\xi )^{4};\xi \bigr) \biggr] ^{\frac{1}{2}} \cdot {} \bigl[ NB_{n}^{ \ast } \bigl(\psi ^{2};\xi \bigr) \bigr]^{\frac{1}{2}}. $$

(4.6)

Also, by setting $\eta _{\xi }(y)=(\psi (y-\xi ))^{2}$ we have that $\eta _{\xi }(\xi )=0$ and $\eta _{\xi }(\cdot )\in C[0,M]$. So

$$ NB_{n}^{\ast }(\eta _{\xi })\rightarrow 0(st_{ \mathfrak{T}})\quad \text{on }{}[ 0,M]. $$

(4.7)

Now from the last relation, (4.6), (4.7), and Lemma 4.4 we obtain that

$$ \biggl( \frac{n}{b_{n}} \biggr) ^{2}NB_{n}^{\ast } \bigl((y- \xi )^{2}\psi (y- \xi );\xi \bigr)\rightarrow 0(st_{\mathfrak{T}})\quad \text{on } {}[ 0,M]. $$

(4.8)

From the definition of the sequence $(u_{n})$ we obtain $( \frac{n}{b_{n}} ) u_{n}\rightarrow 0(st_{\mathfrak{T}})$ on $[0,M]$.

Let $\epsilon >0$. Define the following sets:

$$\begin{aligned}& \begin{aligned} A={}& \biggl\vert \{n: \vert \biggl( \frac{n}{b_{n}} \biggr) \biggl[NB_{n}^{ \ast }(f; \xi )-f(\xi )-f^{{\prime }}(\xi ) \biggl( \frac{B^{\prime } (\frac{n}{b_{n}}\xi H(1) )-B (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )}\xi + \frac{A^{\prime }(1)}{A(1)} \biggr) \\ & {} -\frac{f^{{\prime \prime }}(\xi )}{2} \biggl( \frac{B^{\prime \prime } (\frac{n}{b_{n}}\xi H(1) )-2B^{\prime } (\frac{n}{b_{n}}\xi H(1) )+B (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )} \xi ^{2} \\ & {} + \frac{(A(1)+2A^{\prime }(1)+A(1)H^{\prime \prime }(1))B^{\prime } (\frac{n}{b_{n}}\xi H(1) )-2A^{\prime }(1)B (\frac{n}{b_{n}}\xi H(1) )}{A(1)B (\frac{n}{b_{n}}\xi H(1) )}\xi \biggr) \biggr] \biggr\vert , \end{aligned} \\& A_{1} = \biggl\vert \biggl\{ n: \biggl\vert \biggl( \frac{n}{b_{n}} \biggr) u_{n} \biggr\vert \geq \frac{\epsilon }{3K} \biggr\} \biggr\vert , \\& A_{2} = \biggl\vert \biggl\{ n: \biggl\vert \biggl( \frac{n}{b_{n}} \biggr) NB_{n}^{\ast } \bigl((y-\xi )^{2}\psi (y-\xi );\xi \bigr) \biggr\vert \geq \frac{\epsilon }{3} \biggr\} \biggr\vert , \\& A_{3} = \biggl\vert \biggl\{ n: \biggl\vert \biggl( \frac{n}{b_{n}} \biggr) u_{n}NB_{n}^{\ast } \bigl((y-\xi )^{2}\psi (y-\xi );\xi \bigr) \biggr\vert \geq \frac{\epsilon }{3} \biggr\} \biggr\vert . \end{aligned}$$

From last relations we obtain that $A\leq A_{1}+A_{2}+A_{3}$. Hence the result follows. □

Theorem 4.6

Let $f,f^{{\prime }},f^{{\prime \prime }}\in C^{B}[0,\infty )$ and $\lim_{n\rightarrow \infty }{ ( \frac{n}{b_{n}} ) ^{3}B_{n}^{ \ast }((e_{1}-\xi )^{6},\xi )}=\eta _{4}(\xi )$. Then

$$\begin{aligned} & \biggl\vert \biggl( \frac{n}{b_{n}} \biggr) \bigl( B_{n}^{\ast }(f, \xi )-f( \xi ) \bigr) -f^{{\prime }}(\xi ) \biggl( \frac{n}{b_{n}} \biggr) \biggl( \frac{B^{\prime } (\frac{n}{b_{n}}\xi H(1) )-B (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )}x+\frac{b_{n}}{n} \frac{A^{\prime }(1)}{A(1)} \biggr) \\ &\quad {} -\frac{f^{{\prime \prime }}(\xi )}{2}\cdot \biggl( \frac{n}{b_{n}} \biggr) \biggl[ \frac{B^{\prime \prime } (\frac{n}{b_{n}}\xi H(1) )-2B^{\prime } (\frac{n}{b_{n}}\xi H(1) )+B (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )}\xi ^{2} \\ &\quad {} +\frac{b_{n}}{n} \frac{(A(1)+2A^{\prime }(1)+A(1)H^{\prime \prime }(1))B^{\prime } (\frac{n}{b_{n}}\xi H(1) )-2A^{\prime }(1)B (\frac{n}{b_{n}}\xi H(1) )}{A(1)B (\frac{n}{b_{n}}\xi H(1) )}x \\ &\quad {} +\frac{b_{n}^{2}}{n^{2}} \frac{A^{\prime }(1)+A^{\prime \prime }(1)}{A(1)} \biggr] \biggr\vert =O(1) \omega \biggl( f^{{\prime \prime }}, \biggl( \frac{b_{n}}{{n}} \biggr) ^{-\frac{1}{2}} \biggr) \end{aligned}$$

as $n\rightarrow \infty$ for every $\xi \in{}[ 0,\infty )$.

Proof

By Taylor’s theorem we get

$$ f(u)=f(\xi )+f^{{\prime }}(\xi ) (u-\xi )+ \frac{f^{{\prime \prime }}(\xi )}{2}(u-\xi )^{2}+R(u,\xi ), $$

where $R(u,\xi )= \frac{f^{{\prime \prime }}(\theta )-f^{{\prime \prime }}(\xi )}{2}(u- \xi )^{2}$ for $\theta \in (u,\xi )$. From this we have

$$ \biggl\vert B_{n}^{\ast }(f,\xi )-f(\xi )-f^{{\prime }}( \xi )B_{n}^{ \ast } \bigl((u-\xi );\xi \bigr)- \frac{f^{{\prime \prime }}(\xi )}{2}B_{n}^{ \ast } \bigl((u-\xi )^{2}; \xi \bigr) \biggr\vert \leq B_{n}^{\ast } \bigl( \bigl\vert R(u,\xi ) \bigr\vert , \xi \bigr), $$

from which we get that

$$\begin{aligned} & \biggl\vert \biggl( \frac{n}{b_{n}} \biggr) \bigl( B_{n}^{\ast }(f, \xi )-f( \xi ) \bigr) -f^{{\prime }}(\xi ) \biggl( \frac{n}{b_{n}} \biggr) \biggl( \frac{B^{\prime } (\frac{n}{b_{n}}\xi H(1) )-B (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )}\xi +\frac{b_{n}}{n} \frac{A^{\prime }(1)}{A(1)} \biggr) \\ & \quad\quad {} - \frac{f^{{\prime \prime }}(\xi )}{2}\cdot \biggl( \frac{n}{b_{n}} \biggr) \biggl[ \frac{B^{\prime \prime } (\frac{n}{b_{n}}\xi H(1) )-2B^{\prime } (\frac{n}{b_{n}}\xi H(1) )+B (\frac{n}{b_{n}}\xi H(1) )}{B (\frac{n}{b_{n}}\xi H(1) )}\xi ^{2} \\ &\quad\quad {} +\frac{b_{n}}{n} \frac{(A(1)+2A^{\prime }(1)+A(1)H^{\prime \prime }(1))B^{\prime } (\frac{n}{b_{n}}\xi H(1) )-2A^{\prime }(1)B (\frac{n}{b_{n}}\xi H(1) )}{A(1)B (\frac{n}{b_{n}}\xi H(1) )}\xi \\ &\quad\quad {} +\frac{b_{n}^{2}}{n^{2}} \frac{A^{\prime }(1)+A^{\prime \prime }(1)}{A(1)} \biggr] \biggr\vert \\ &\quad \leq \biggl( \frac{n}{b_{n}} \biggr) \cdot B_{n}^{\ast } \bigl( \bigl\vert R(u,\xi ) \bigr\vert , \xi \bigr). \end{aligned}$$

From the properties of modulus of continuity we obtain

$$ \biggl\vert \frac{f^{{\prime \prime }}(\theta )-f^{{\prime \prime }}(\xi )}{2!} \biggr\vert \leq \frac{1}{2!} \biggl( 1+ \frac{ \vert \theta -\xi \vert }{\delta } \biggr) \omega \bigl(f^{{\prime \prime }},\delta \bigr). $$

We know that

$$ \biggl\vert \frac{f^{{\prime \prime }}(\theta )-f^{{\prime \prime }}(\xi )}{2!} \biggr\vert \leq \textstyle\begin{cases} \omega (f^{{\prime \prime },\delta }), & \vert u-\xi \vert \leq \delta , \\ \frac{(t-\xi )^{4}}{\delta ^{4}}\omega (f^{{\prime \prime }},\delta ), & \vert u-\xi \vert \geq \delta .\end{cases} $$

For $0<\delta <1$, we obtain that

$$ \biggl\vert \frac{f^{{\prime \prime }}(\theta )-f^{{\prime \prime }}(\xi )}{2!} \biggr\vert \leq \omega \bigl(f^{{\prime \prime }},\delta \bigr) \biggl( 1+ \frac{(u-\xi )^{4}}{\delta ^{4}} \biggr) , $$

which implies that

$$ \bigl\vert R(u,\xi ) \bigr\vert \leq \omega \bigl(f^{{\prime \prime }},\delta \bigr) \biggl( 1+ \frac{(u-\xi )^{4}}{\delta ^{4}} \biggr) (u-\xi )^{2}=\omega \bigl(f^{{ \prime \prime }}, \delta \bigr) \biggl( (u-\xi )^{2}+ \frac{(u-\xi )^{6}}{\delta ^{4}} \biggr) . $$

By the linearity of $B_{n}^{\ast }$ and the above relation we obtain

$$ B_{n}^{\ast } \bigl( \bigl\vert R(u,\xi ) \bigr\vert ,\xi \bigr)\leq \omega \bigl(f^{{\prime \prime }}, \delta \bigr) \biggl( B_{n}^{\ast } \bigl((u-\xi )^{2},\xi \bigr)+ \frac{1}{\delta ^{4}}B_{n}^{\ast } \bigl((u-\xi )^{6}, \xi \bigr) \biggr) . $$

By Remark 4.1, for any $x\in{}[ 0,\infty )$, we obtain

$$ B_{n}^{\ast } \bigl( \bigl\vert R(u,\xi ) \bigr\vert ,\xi \bigr)\leq \omega \bigl(f^{{\prime \prime }}, \delta \bigr) \biggl( O \biggl( \frac{b_{n}}{n} \biggr) + \frac{1}{\delta ^{4}}O \biggl( \frac{b_{n}}{n} \biggr) ^{3} \biggr) =O \biggl( \frac{b_{n}}{n} \biggr) \omega \bigl(f^{{\prime \prime }}, \delta _{n} \bigr). $$

We complete the proof by taking $\delta _{n}= ( \frac{b_{n}}{n} ) ^{-\frac{1}{2}}$. □

We prove the following results under the conditions given in the assumptions.

Theorem 4.7

Let $f\in C^{B}[0,\infty )$ and $f^{{\prime }},f^{{\prime \prime }}\in C[0,\infty )$. Then

$$ \lim_{n\rightarrow \infty }\frac{n}{b_{n}} \bigl[B_{n}^{\ast }(fg, \xi )-B_{n}^{ \ast }(f,\xi )B_{n}^{\ast }(g, \xi ) \bigr]=\frac{1}{2} ( \xi ^{2}l_{2}( \xi )+\xi \bigl(1+H^{\prime \prime }(1) \bigr) f^{{\prime }}(\xi )g^{{\prime}}(\xi ) $$

for any $x\in{}[ 0,M]$, where $M>0$.

Proof

After some calculations, we obtain

$$\begin{aligned}& \frac{n}{b_{n}} \bigl[B_{n}^{\ast }(fg,\xi )-B_{n}^{\ast }(f,\xi )B_{n}^{ \ast }(g,\xi ) \bigr] \\& \quad = \biggl[\frac{n}{b_{n}} \bigl( B_{n}^{\ast }(fg,\xi )-fg \bigr) ) - \biggl( \xi l_{1}(\xi )+\frac{A^{\prime }(1)}{A(1)} \biggr) (fg)^{{\prime}}(\xi ) \\& \quad\quad {}- \frac{1}{2} ( \xi ^{2}l_{2}(\xi )+\xi \bigl(1+H^{\prime \prime }(1) \bigr) \frac{(fg)^{{\prime \prime }}(\xi )}{2} \biggr] \\& \quad\quad {} -g(\xi ) \biggl[ \frac{n}{b_{n}} \bigl( B_{n}^{\ast }(f,\xi )-f(\xi ) \bigr) - \biggl( \xi l_{1}(\xi )+ \frac{A^{\prime }(1)}{A(1)} \biggr) f^{{\prime }}(\xi ) \\& \quad\quad {}- \frac{1}{2} ( \xi ^{2}l_{2}(\xi )+\xi \bigl(1+H^{\prime \prime }(1) \bigr) \frac{f^{{\prime \prime }}(\xi )}{2} \biggr] \\& \quad\quad {}-B_{n}^{\ast }(f, \xi ) \biggl[\frac{n}{b_{n}} \bigl( B_{n}^{\ast }(g,\xi )-g(\xi ) \bigr) - \biggl( \xi l_{1}( \xi )+\frac{A^{\prime }(1)}{A(1)} \biggr) g^{{\prime }}(\xi ) \\& \quad\quad {} -\frac{1}{2} ( \xi ^{2}l_{2}(\xi )+\xi \bigl(1+H^{\prime \prime }(1) \bigr) \frac{g^{{\prime \prime }}(\xi )}{2} \biggr]+\frac{1}{2} ( \xi ^{2}l_{2}(\xi )+\xi \bigl(1+H^{\prime \prime }(1) \bigr) f^{{\prime}}(\xi )g^{{\prime }}(\xi ) \\& \quad\quad {} +\frac{1}{2} ( \xi ^{2}l_{2}(\xi )+\xi \bigl(1+H^{\prime \prime }(1) \bigr) \frac{g^{{\prime \prime }}(\xi )}{2} \bigl[f(\xi )-B_{n}^{\ast }(f, \xi ) \bigr] \\& \quad\quad {}+ \biggl( \xi l_{1}( \xi )+\frac{A^{\prime }(1)}{A(1)} \biggr) g^{{\prime}}(\xi ) \bigl[f(\xi )-B_{n}^{\ast }(f,\xi ) \bigr]. \end{aligned}$$

Now the proof follows from Theorem 4.2 and Proposition 2.2. □

5 Weighted approximation

Now we will study some properties of $B_{n}^{\ast }$ in weighted spaces. Also, we will suppose that

$$ \lim_{n\rightarrow \infty }{\frac{B^{(k)}(y)}{B(y)}}=1\quad \text{for every } k=1,2,\ldots,r; r\in \mathbb{N}. $$

Let $\rho (x)=x^{2}+1$ be the weight function, and let $M_{f}$ be a positive constant. We write

• (i) $B_{\rho }[0,\infty )$ for the space of bounded functions $\vert f(x) \vert \leq M_{f}\rho (x)$ with $\Vert f \Vert _{\rho }=\sup_{x\geq 0}\frac{ \vert f(x) \vert }{\rho (x)}$.

• (ii) $C_{\rho }[0,\infty )$ for the subspace of continuous functions in $B_{\rho }[0,\infty )$.

• (iii) $C_{\rho }^{\ast }[0,\infty )$ for the space of functions $f\in C_{\rho }[0,\infty )$ with fn ite $\lim_{x\rightarrow \infty }\frac{f(x)}{\rho (x)}$.

The weighted modulus of continuity $\Omega (f;\delta )$ is defined by

$$ \Omega (f;\delta )=\sup_{x\geq 0, 0< \vert h \vert \leq \delta } \frac{ \vert f(x+h)-f(x) \vert }{(1+h^{2})\rho (x)}\quad \text{for all } f\in C_{\rho }^{\ast }[0, \infty ). $$

For any $\mu \in{}[ 0,\infty )$,

$$ \Omega (f;\mu \delta )\leq 2(1+\mu ) \bigl(1+\delta ^{2} \bigr)\Omega (f;\delta ), $$

and

$$ \bigl\vert f(t)-f(x) \bigr\vert \leq 2 \biggl( \frac{ \vert t-x \vert }{\delta }+1 \biggr) \bigl(1+\delta ^{2} \bigr) \Omega (f;\delta ) \bigl(1+x^{2} \bigr) \bigl(1+(t-x)^{2} \bigr), \quad f\in C_{\rho }^{\ast }[0, \infty ). $$

Theorem 5.1

For $f\in C_{\rho }^{\ast }[0,\infty )$, we have

$$ \lim_{n\rightarrow \infty } \bigl\Vert B_{n}^{\ast }(f;x)-f(x) \bigr\Vert _{ \rho }=0. $$

Proof

It suffices to check that $B_{n}^{\ast }(e_{i};x)$ uniformly converges to $e_{i}$ as $n\rightarrow \infty$, where $e_{i}(x)=x^{i}$, $i=0,1,2$, and apply the weighted Korovkin-type theorem. Using Lemma 2.1, the case $i=0$ is trivial. Now

$$ \bigl\Vert B_{n}^{\ast }e_{1}-e_{1} \bigr\Vert _{\rho }=\sup_{x \geq 0} \biggl\{ \frac{ \vert B_{n}^{\ast }e_{1}-e_{1} \vert }{\rho (x)} \biggr\} \leq \sup_{x\geq 0} \frac{ \vert \alpha _{1}(n,x) \vert }{\rho (x)}, $$

and by a similar consideration, we have

$$ \bigl\Vert B_{n}^{\ast }e_{2}-e_{2} \bigr\Vert _{\rho }=\sup_{x \geq 0} \biggl\{ \frac{ \vert B_{n}^{\ast }e_{2}-e_{2} \vert }{\rho (x)} \biggr\} \leq \sup_{x\geq 0} \biggl\{ \frac{ \vert \alpha _{2}(n,x) \vert }{\rho (x)} \biggr\} , $$

where

$$\begin{aligned}& \alpha _{1}(n,x)= \biggl( \frac{B^{{\prime }} ( \frac{n}{b_{n}}xH(1) ) }{B ( \frac{n}{b_{n}}xH(1) ) }-1 \biggr) x+ \frac{b_{n}}{n}\cdot \frac{A^{{\prime }}(1)}{A(1)}, \\& \begin{aligned} \alpha _{2}(n,x)&= \biggl( \frac{B^{\prime \prime } (\frac{n}{b_{n}}xH(1) )}{B (\frac{n}{b_{n}}xH(1) )}-1 \biggr) x^{2}+ \frac{b_{n}}{n} \frac{B^{\prime } (\frac{n}{b_{n}}xH(1) ) [ A(1)+2A^{\prime }(1)+H^{\prime \prime }(1)A(1) ] }{A(1)B (\frac{n}{b_{n}}xH(1) )}x \\ &\quad {} +\frac{b_{n}^{2}}{n^{2}} \frac{A^{\prime }(1)+A^{\prime \prime }(1)}{A(1)}. \end{aligned} \end{aligned}$$

We conclude that

$$ \lim_{n} \bigl\Vert B_{n}^{\ast }e_{i}-e_{i} \bigr\Vert _{\rho }=\lim_{n \rightarrow \infty } \bigl\Vert B_{n}^{\ast }e_{i}-e_{i} \bigr\Vert _{\rho }=0\quad (i=0,1,2), $$

which finishes the proof. □

Theorem 5.2

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

$$\begin{aligned} \sup_{x\in{} [0,\infty )}{ \frac{ \vert B_{n}^{*}(f;x)-f(x) \vert }{(1+x^{2})(A(n,x)+B(n,x)x+C(n,x)x^{2}+D(n,x)x^{3}+E(n,x)x^{4})}}\leq K \Omega \bigl(f;n^{-\frac{1}{4}} \bigr) \end{aligned}$$

for sufficiently large n, $A(n,x)$, $B(n,x)$, $C(n,x)$, $D(n,x)$, and $E(n,x)$ depend on n and x, and K is a positive constant.

Proof

For $x \in {}[0, \infty )$, we have

$$ B_{n}^{*}(f;x)-f(x) = \frac{1}{A(1)B (\frac{n}{b_{n}}xH(1) )}\sum _{k=0}^{\infty}{p_{k} \biggl( \frac{n}{b_{n}}x \biggr) \biggl[f \biggl( \frac{k}{n}b_{n} \biggr)-f(x) \biggr]}. $$

Using the properties of the weighted modulus, we obtain

$$\begin{aligned} & \bigl\vert B_{n}^{\ast }(f;x)-f(x) \bigr\vert \\ &\quad \leq \frac{1}{A(1)B ( \frac{n}{b_{n}}xH(1) ) }\sum_{k=0}^{ \infty }p_{k} \biggl( \frac{n}{b_{n}}x \biggr) 2 \bigl(1+\delta _{n}^{2} \bigr) \Omega (f;\delta _{n}) \bigl(1+x^{2} \bigr) \\ &\quad\quad{}\cdot { \biggl( \frac{ \vert ( \frac{k}{n}b_{n} ) -x \vert }{\delta _{n}}+1 \biggr) \bigl(1+(t-x)^{2} \bigr)}. \end{aligned}$$

Let us denote by $S(t,x)= ( \frac{ \vert ( \frac{k}{n}b_{n} ) -x \vert }{\delta _{n}}+1 ) (1+(t-x)^{2})$. Then

$$ S(t,x)\leq \textstyle\begin{cases} 2(1+\delta _{n}^{2})&\text{if } \vert \frac{k}{n}b_{n}-x \vert \leq \delta _{n}, \\ 2(1+\delta _{n}^{2}) \frac{ ( \frac{k}{n}b_{n}-x ) ^{4}}{\delta _{n}^{4}}& \text{if } \vert \frac{k}{n}b_{n}-x \vert \geq \delta _{n}.\end{cases} $$

From last relation we get that

$$ S(x,t)\leq 2 \bigl(1+\delta _{n}^{2} \bigr) \biggl(1+ \frac{ ( \frac{k}{n}b_{n} -x )^{4}}{\delta _{n}^{4}} \biggr). $$

So

$$\begin{aligned} \begin{aligned} & \bigl\vert B_{n}^{*}(f;x)-f(x) \bigr\vert \\ &\quad \leq 4 \frac{1}{A(1)B (\frac{n}{b_{n}}xH(1) )}\sum_{k=0}^{\infty}p_{k} \biggl( \frac{n}{b_{n}}x \biggr) \bigl(1+\delta _{n}^{2} \bigr) \Omega (f;\delta _{n}) \bigl(1+x^{2} \bigr) \\ &\quad\quad{}\cdot { \bigl(1+ \delta _{n}^{2} \bigr) \biggl(1+ \frac{ ( \frac{k}{n}b_{n} -x )^{4}}{\delta _{n}^{4}} \biggr)}. \end{aligned} \end{aligned}$$

After some calculations, we get

$$\begin{aligned} & \sum_{k=0}^{\infty }p_{k} \biggl( \frac{n}{b_{n}}x \biggr) \biggl( 1+ \frac{ ( \frac{k}{n}b_{n}-x ) ^{4}}{\delta _{n}^{4}} \biggr) \\ &\quad =\sum_{k=0}^{\infty }p_{k} \biggl( \frac{n}{b_{n}}x \biggr) + \frac{1}{\delta _{n}^{4}}\sum _{k=0}^{\infty }p_{k} \biggl( \frac{n}{b_{n}}x \biggr) \biggl[ \biggl( \frac{k}{n} \biggr) ^{4}b_{n}^{4}-4 \biggl( \frac{k}{n} \biggr) ^{3}b_{n}^{3}x+6 \biggl( \frac{k}{n} \biggr) ^{2}b_{n}^{2}x^{2} \\ &\quad\quad{} -4 \biggl( \frac{k}{n} \biggr) b_{n}x^{3}+x^{4} \biggr] \\ &\quad = \biggl( 1+\frac{x^{4}}{\delta _{n}^{4}} \biggr) A(1)B \biggl( \frac{n}{b_{n}}xH(1) \biggr)+\frac{b_{n}^{4}}{n^{4}\delta _{n}^{4}}\sum _{k=0}^{\infty }k^{4}p_{k} \biggl( \frac{n}{b_{n}}x \biggr) -4 \frac{xb_{n}^{3}}{n^{3}\delta _{n}^{4}}\sum _{k=0}^{\infty }k^{3}p_{k} \biggl( \frac{n}{b_{n}}x \biggr) \\ &\quad\quad {} +6\frac{x^{2}b_{n}^{2}}{n^{2}\delta _{n}^{4}}\sum _{k=0}^{\infty }k^{2}p_{k} \biggl( \frac{n}{b_{n}}x \biggr) -4\frac{x^{3}b_{n}}{n\delta _{n}^{4}}\sum_{k=0}^{\infty }kp_{k} \biggl( \frac{n}{b_{n}}x \biggr) . \end{aligned}$$

From these relations and Lemma 2.1 of [26]) we get

$$\begin{aligned}& \sum_{k=0}^{\infty}p_{k} \biggl( \frac{n}{b_{n}}x \biggr) \biggl(1+ \frac{ ( \frac{k}{n}b_{n} -x )^{4}}{\delta _{n}^{4}} \biggr) \\& \quad = \biggl(1+ \frac{x^{4}}{\delta _{n}^{4}} \biggr) A(1)B \biggl(\frac{n}{b_{n}}x H(1) \biggr) \\& \quad \quad {} -4\frac{x^{3}b_{n}}{n\delta _{n}^{4}} \biggl[ A^{\prime}(1)B \biggl( \frac{n}{b_{n} }x H(1) \biggr) +\frac{n}{b_{n}}x A(1)B^{\prime} \biggl( \frac{n}{b_{n}}x H(1) \biggr) \biggr] \\& \quad \quad {} +6 \frac{x^{2}b_{n}^{2}}{n^{2}\delta _{n}^{4}} \biggl[ \frac{n^{2}}{b_{n}^{2}}x^{2} A(1){B^{\prime \prime} \biggl(\frac{n}{b_{n}}x H(1) \biggr)} \\& \quad \quad {}+ \frac{n}{b_{n}}x{ \bigl( A(1)+2A^{\prime}(1)+H^{\prime \prime}(1)A(1) \bigr)B^{\prime} \biggl(\frac{n}{b_{n}}x H(1) \biggr)} \\& \quad \quad {} + \bigl(A^{\prime}(1)+A^{ \prime \prime}(1) \bigr)B \biggl( \frac{n}{b_{n}}x H(1) \biggr) \biggr] -4\frac{xb_{n}^{3}}{n^{3}\delta _{n}^{4}} \biggl[ \frac{n^{3}}{b_{n}^{3}}x^{3} A(1)B^{\prime \prime \prime} \biggl( \frac{n}{b_{n}}x H(1) \biggr) \\& \quad \quad {} + \frac{n^{2}}{b_{n}^{2}}x^{2} \bigl( 3A^{\prime}(1)+3H^{\prime \prime}(1)A(1)+3A(1) \bigr)B^{\prime \prime} \biggl(\frac{n}{b_{n}}x H(1) \biggr) \\& \quad \quad {} + \frac{n}{b_{n}}x \bigl(3A^{\prime \prime}(1) + 3H^{\prime \prime}(1)A^{\prime}(1) + H^{\prime \prime \prime}(1)A(1) + 6A^{\prime}(1) \\& \quad \quad {}+ 3H^{\prime \prime}(1)A(1) + A(1) \bigr)B^{\prime } \biggl(\frac{n}{b_{n}}x H(1) \biggr) + \bigl(A^{\prime \prime \prime}(1)+3A^{\prime \prime}(1)+A^{\prime}(1) \bigr)B \biggl( \frac{n}{b_{n}}x H(1) \biggr) \biggr] \\& \quad \quad {} +\frac{b_{n}^{4}}{n^{4}\delta _{n}^{4}} \biggl[ \frac{n^{4}}{b_{n}^{4}}x^{4} A(1)B^{(4)} \biggl(\frac{n}{b_{n}}x H(1) \biggr) \\& \quad \quad {}+ \frac{n^{3}}{b_{n}^{3}}x^{3} \bigl( 4A^{ \prime}(1)+6H^{\prime \prime}(1)A(1)+6A(1) \bigr)B^{\prime \prime \prime} \biggl( \frac{n}{b_{n}}x H(1) \biggr) \\& \quad \quad {} + \frac{n^{2}}{b_{n}^{2}}x^{2} \bigl(6A^{\prime \prime}(t) + 12H^{ \prime \prime}(1) + A^{\prime}(1) + 4H^{\prime \prime \prime}(1)A(1) + 3H^{ \prime \prime}(1)^{2}A(1) + 18A^{\prime}(1) \\& \quad \quad {} + 18H^{\prime \prime}(1)A(1) + 7A(1) \bigr) B^{\prime \prime} \biggl(\frac{n}{b_{n}}x H(1) \biggr) + \bigl(4A^{\prime \prime \prime}(1) + 6A^{\prime \prime}(1)H^{ \prime \prime}(1) \bigr] \\& \quad \quad {}+ 4A^{\prime}(1)H^{\prime \prime \prime}(1) + A(1)H^{(4)}(1) + 18A^{\prime \prime}(1) \\& \quad \quad {} + 18H^{\prime \prime}(1)A^{\prime}(1)+6H^{\prime \prime \prime}(1)A(1)+14A^{\prime}(1)+7H^{\prime \prime}(1)A(1)+A(1) \bigr)\frac{n}{b_{n}}xB^{ \prime} \biggl(\frac{n}{b_{n}}x H(1) \biggr) \\& \quad \quad {} + \bigl(A^{(4)}(1)+6A^{(3)}(1)+7A^{\prime \prime}(1)+A^{\prime}(1) \bigr)B \biggl(\frac{n}{b_{n}}x H(1) \biggr) \biggr]. \end{aligned}$$

From last two relations we get

$$\begin{aligned}& \bigl\vert B_{n}^{*}(f;x)-f(x) \bigr\vert \\& \quad \leq 4 \frac{(1+\delta _{n}^{2})^{2} \Omega (f;\delta _{n})(1+x^{2})}{A(1)B (\frac{n}{b_{n}}xH(1) )}\sum_{k=0}^{\infty}p_{k} \biggl( \frac{n}{b_{n}}x \biggr) \biggl(1+ \frac{ ( \frac{k}{n}b_{n} -x )^{4}}{\delta _{n}^{4}} \biggr) \\& \quad \leq 4 \frac{(1+\delta _{n}^{2})^{2} \Omega (f;\delta _{n})(1+x^{2})}{A(1)B (\frac{n}{b_{n}}xH(1) )} \biggl\{ \biggl(1+\frac{x^{4}}{\delta _{n}^{4}} \biggr) A(1)B \biggl(\frac{n}{b_{n}}x H(1) \biggr) \\& \quad \quad {} -4\frac{x^{3}b_{n}}{n\delta _{n}^{4}} \biggl[ A^{\prime}(1)B \biggl( \frac{n}{b_{n} }x H(1) \biggr)+\frac{n}{b_{n}}x A(1)B^{\prime} \biggl( \frac{n}{b_{n}}x H(1) \biggr) \biggr] \\& \quad \quad {} +6 \frac{x^{2}b_{n}^{2}}{n^{2}\delta _{n}^{4}} \biggl[ \frac{n^{2}}{b_{n}^{2}}x^{2} A(1){B^{\prime \prime} \biggl(\frac{n}{b_{n}}x H(1) \biggr)} \\& \quad \quad {} + \frac{n}{b_{n}}x{ \bigl( A(1)+2A^{\prime}(1)+H^{\prime \prime}(1)A(1) \bigr)B^{\prime} \biggl(\frac{n}{b_{n}}x H(1) \biggr)} \\& \quad \quad {} + \bigl(A^{\prime}(1)+A^{ \prime \prime}(1) \bigr)B \biggl( \frac{n}{b_{n}}x H(1) \biggr) \biggr] \\& \quad \quad {} -4\frac{xb_{n}^{3}}{n^{3}\delta _{n}^{4}} \biggl[ \frac{n^{3}}{b_{n}^{3}}x^{3} A(1)B^{\prime \prime \prime} \biggl( \frac{n}{b_{n}}x H(1) \biggr) \\& \quad \quad {}+ \frac{n^{2}}{b_{n}^{2}}x^{2} \bigl( 3A^{\prime}(1)+3H^{\prime \prime}(1)A(1)+3A(1) \bigr)B^{\prime \prime} \biggl(\frac{n}{b_{n}}x H(1) \biggr) \\& \quad \quad {} + \frac{n}{b_{n}}x \bigl(3A^{\prime \prime}(1) + 3H^{\prime \prime}(1)A^{\prime}(1) + H^{\prime \prime \prime}(1)A(1) + 6A^{\prime}(1) + 3H^{\prime \prime}(1)A(1) \\& \quad \quad {} + A(1) \bigr)B^{\prime } \biggl(\frac{n}{b_{n}}x H(1) \biggr) + \bigl(A^{\prime \prime \prime}(1)+3A^{\prime \prime}(1)+A^{\prime}(1) \bigr)B \biggl( \frac{n}{b_{n}}x H(1) \biggr) \biggr] \\& \quad \quad {} +\frac{b_{n}^{4}}{n^{4}\delta _{n}^{4}} \biggl[ \frac{n^{4}}{b_{n}^{4}}x^{4} A(1)B^{(4)} \biggl(\frac{n}{b_{n}}x H(1) \biggr)+ \frac{n^{3}}{b_{n}^{3}}x^{3} \bigl( 4A^{ \prime}(1)+6H^{\prime \prime}(1)A(1) \\& \quad \quad {} +6A(1) \bigr)B^{\prime \prime \prime} \biggl( \frac{n}{b_{n}}x H(1) \biggr) + \frac{n^{2}}{b_{n}^{2}}x^{2} \bigl(6A^{\prime \prime}(t) + 12H^{ \prime \prime}(1) + A^{\prime}(1) + 4H^{\prime \prime \prime}(1)A(1) \\& \quad \quad {} + 3H^{ \prime \prime}(1)^{2}A(1) + 18A^{\prime}(1) + 18H^{\prime \prime}(1)A(1) + 7A(1) \bigr) B^{\prime \prime} \biggl(\frac{n}{b_{n}}x H(1) \biggr) \\& \quad \quad {}+ \bigl(4A^{\prime \prime \prime}(1) + 6A^{\prime \prime}(1)H^{ \prime \prime}(1) \bigr] + 4A^{\prime}(1)H^{\prime \prime \prime}(1) + A(1)H^{(4)}(1) + 18A^{\prime \prime}(1) \\& \quad \quad {} + 18H^{\prime \prime}(1)A^{\prime}(1)+6H^{\prime \prime \prime}(1)A(1)+14A^{\prime}(1)+7H^{\prime \prime}(1)A(1)+A(1) \bigr)\frac{n}{b_{n}}xB^{ \prime} \biggl(\frac{n}{b_{n}}x H(1) \biggr) \\& \quad \quad {} + \bigl(A^{(4)}(1)+6A^{(3)}(1)+7A^{\prime \prime}(1)+A^{\prime}(1) \bigr)B \biggl(\frac{n}{b_{n}}x H(1) \biggr) \biggr] \biggr\} . \end{aligned}$$

For $\delta _{n}=n^{-\frac{1}{4}}$, we have

$$\begin{aligned}& \bigl\vert B_{n}^{\ast }(f;x)-f(x) \bigr\vert \\& \quad \leq 16 \Omega (f;\delta _{n}) \bigl(1+x^{2} \bigr) \bigl( A(n,x)+B(n,x)x+C(n,x)x^{2}+D(n,x)x^{3}+E(n,x)x^{4} \bigr) , \end{aligned}$$

where $A(n,x)$, $B(n,x)$, $C(n,x)$, $D(n,x)$, and $E(n,x)$ depend on n and x.

Now from last relation we obtain

$$ \sup_{x\in{}[ 0,\infty )}{ \frac{ \vert B_{n}^{\ast }(f;x)-f(x) \vert }{(1+x^{2})(A(n,x)+B(n,x)x+C(n,x)x^{2}+D(n,x)x^{3}+E(n,x)x^{4})}}\leq K \Omega \bigl( f;n^{-\frac{1}{4}} \bigr) . $$

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