Degree of Approximation by Certain Genuine Hybrid Operators

Abstract

This paper is in continuation of our work on certain genuine hybrid operators in (Positivity (Under review)) [3]. First, we discuss some direct results in simultaneous approximation by these operators, e.g. pointwise convergence theorem, Voronovskaja-type theorem and an error estimate in terms of the modulus of continuity. Next, we estimate the rate of convergence for functions having a derivative that coincides a.e. with a function of bounded variation.

1 Introduction

Recently, Gupta and Rassias [5] introduced the Lupaş-Durrmeyer operators based on Polya distribution and discussed some local and global direct results. Also, Gupta [2] studied some other hybrid operators of Durrmeyer type. Păltǎnea [11] (see also [10]) considered a Durrmeyer-type modification of the genuine Szász-Mirakjan operators based on two parameters $\alpha ,\rho >0.$ Inspired by his work, in [3] Gupta et al. introduced certain genuine hybrid operators as follows:

For $c\in \{0,1\}$ and $f\in C_{\gamma } [0,\infty ) := \{f \in C[0,\infty ) : |f(t)|\le M_f\, e^{\gamma t}, \,\mathrm{for\ some}$

$\gamma >0, M_f>0\},$ we define

$$\begin{aligned} B_\alpha ^\rho (f,x)= & {} \sum _{k=1}^{\infty } p_{\alpha ,k}(x,c)\int \limits _0^\infty \theta _{\alpha ,k}^\rho (t)f(t)dt\,+\,p_{\alpha ,0}(x,c)f(0),\end{aligned}$$

(1)

$$\begin{aligned}= & {} \int \limits _0^{\infty } K_{\alpha }^{\rho }(x,t)f(t) dt, \end{aligned}$$

(2)

where

$$\begin{aligned} p_{\alpha ,k}(x,c)= & {} \frac{(-x)^k}{k!}\phi _{\alpha ,c}^{(k)}(x),\theta _{\alpha ,k}^\rho (t)=\frac{\alpha \rho }{\Gamma (k\rho )}e^{-\alpha \rho t}(\alpha \rho t)^{k\rho -1}\\ \mathrm{and}\,\,\,\, K_{\alpha }^{\rho }(x,t)= & {} \displaystyle \sum _{k=1}^{\infty }p_{\alpha ,k}(x,c)\theta _{\alpha ,k}^{\rho }(t)\,+\,p_{\alpha ,0}(x,c)\delta (t);\,\, x\in (0,\infty ). \end{aligned}$$

It is observed that the operators $B_\alpha ^\rho (f,x)$ are well-defined for $\alpha \rho > \gamma .$ We assume that

$$\begin{aligned} \phi _{\alpha ,c}(x)= {\left\{ \begin{array}{ll} \,\,\,\,\,\,e^{-\alpha x}, &{}\mathrm{for}\,\, c=0,\\ (1\,+\,x)^{-\alpha }, &{}\mathrm{for}\,\, c=1. \end{array}\right. } \end{aligned}$$

As shown in paper [3], the operators (1) include several linear positive operators as special cases. Further, we note that the operators (1) preserve the linear functions.

In [3], we studied some direct results, e.g. Voronovskaja-type theorems in ordinary and simultaneous approximation for first-order derivatives as well as results in local and weighted approximation. In this paper, we continue this work by discussing simultaneous approximation for $f^{(r)}(x), r\in \mathbb {N}$ and the rate of convergence of the operators (1) for the functions with derivatives of bounded variation on each finite subinterval of $(0,\infty )$. The paper is organized as follows:

In Sect. 2, we discuss some auxiliary results and then in Sect. 3, we obtain the main results of this paper.

2 Auxiliary Results

For $f : [0,\infty )\rightarrow R,$ we define

$$\begin{aligned} S_{\alpha }(f;x)=\sum _{k=0}^{\infty }p_{\alpha ,k}(x,c)f\bigg (\frac{k}{\alpha }\bigg ) \end{aligned}$$

(3)

such that (3) makes sense for all $x\ge 0.$

For $m\in \mathbb {N}^0=\mathbb {N}\cup \{0\},$ the mth order central moment of the operators $S_{\alpha }$ is given by

$$\begin{aligned} \upsilon _{\alpha ,m}(x) := S_{\alpha }((t-x)^m;x) = \sum _{k=0}^{\infty }p_{\alpha ,k}(x,c)\bigg (\frac{k}{\alpha }-x\bigg )^{m}. \end{aligned}$$

Lemma 1

For the function $\upsilon _{\alpha ,m}(x),$ we have

$$\begin{aligned} \upsilon _{\alpha ,0}(x)=1,\,\,\upsilon _{\alpha ,1}(x)=0 \end{aligned}$$

and

$$\begin{aligned} x(1\,+\,cx)[\upsilon _{\alpha ,m}^{\prime }(x)\,+\,m\upsilon _{\alpha ,m-1}(x)]=\alpha \upsilon _{\alpha ,m\,+\,1}(x). \end{aligned}$$

Thus,

• (i) $\upsilon _{\alpha ,m}(x)$ is a polynomial in x of degree [m / 2]; 

• (ii) for each $x\in [0,\infty ),\upsilon _{\alpha ,m}(x)= O(\alpha ^{-[(m\,+\,1)/2]})$ , where $[\beta ]$ denotes the integral part of $\beta .$

Proof

For the cases $c=0$ and 1,  the proof of this lemma can be found in [8, 12] respectively.

Lemma 2

For the mth order $(m\in \mathbb {N}^0)$ moment of the operators (1) defined as $u_{\alpha ,m}(x):= B_\alpha ^\rho (t^m;x),$ we have

$u_{\alpha ,0}(x)=1, \,\,u_{\alpha ,1}(x)=x,\,\, u_{\alpha ,2}(x)= x^2\,+\,\dfrac{x}{\alpha }\bigg (\dfrac{1}{\rho }\,+\,(1\,+\,cx)\bigg )$

and

$x(1\,+\,cx)u_{\alpha ,m}^{'}(x)=\alpha u_{\alpha ,{m\,+\,1}}(x)-\bigg (\frac{m}{\rho }\,+\,\alpha x\bigg )u_{\alpha ,m}(x),\, m\in \mathbb {N}.$

Consequently, for each $x\in (0,\infty )$ and $m\in \mathbb {N}, u_{\alpha ,m}(x)=x^m\,+\,\alpha ^{-1} (p_m(x,c)\,+\,o(1)),$

where $p_m(x,c)$ is a rational function of x depending on the parameters m and c.

Lemma 3

[3] For $m\in \mathbb {N}^0,$ if the mth order central moment $\mu _{\alpha ,m}(x)$ for the operators $B_{\alpha }^{\rho }$ is defined as

$$\begin{aligned} \mu _{\alpha ,m}(x):=B_\alpha ^\rho ((t-x)^{m},x)=\sum _{k=1}^{\infty }p_{\alpha ,k}(x,c)\int \limits _0^\infty \theta _{\alpha ,k}^\rho (t)(t-x)^mdt\,+\,p_{\alpha ,0}(x,c)(-x)^m, \end{aligned}$$

then we have the following recurrence relation:

$$\begin{aligned} \alpha \mu _{\alpha ,m\,+\,1}(x)= & {} x(1\,+\,cx)\mu _{\alpha ,m}^\prime (x)\,+\,mx\left[ \frac{1}{\rho }\,+\,(1\,+\,cx)\right] \mu _{\alpha ,m-1}(x)\,+\,\frac{m}{\rho }\mu _{\alpha ,m}(x). \end{aligned}$$

Consequently,

• (i) $\mu _{\alpha ,0}(x)=1,\,\, \mu _{\alpha ,1}(x)=0,\,\, \mu _{\alpha ,2}(x)=\dfrac{\{1\,+\,\rho (1\,+\,cx)\}x}{\alpha \rho };$

• (ii) $\mu _{\alpha ,m}(x)$ is a polynomial in x of degree atmost m; 

• (iii) for every $x\in (0,\infty )$, $\mu _{\alpha ,m}(x)=O\bigg (\alpha ^{-[(m\,+\,1)/2]}\bigg );$

• (iv) the coefficients of $\alpha ^{-m}$ in $\mu _{\alpha ,2m}(x)$ and $\mu _{\alpha ,2m-1}(x)$ are $(2m-1)!! \bigg \{x\bigg (\dfrac{1}{\rho }\,+\,(1\,+\,cx)\bigg )\bigg \}^{m}$ and $\dfrac{(2m-1)!! (m-1)}{3} x^{m-1} \bigg (\dfrac{1}{\rho }\,+\,(1\,+\,cx)\bigg )^{m-2} \bigg \{(1\,+\,cx)\bigg (\dfrac{1}{\rho }\,+\,(1\,+\,2cx)\bigg )\,+\,\dfrac{2}{\rho }\bigg (\dfrac{1}{\rho }\,+\,(1\,+\,cx)\bigg )\bigg \}$ respectively.

Corollary 1

For $x\in [0,\infty )$ and $\alpha >0,$ it is observed that

$$\mu _{\alpha ,2}(x)\le \dfrac{\lambda x(1\,+\,cx)}{\alpha },\,\,{\text {where}} \,\,\lambda =1\,+\,\frac{1}{\rho }>1.$$

Corollary 2

[3] Let $\gamma$ and $\delta$ be any two positive real numbers and $[a,b]\subset (0,\infty )$ be any bounded interval. Then, for any $m>0$ there exists a constant $M^\prime$ independent of $\alpha$ such that

$$\begin{aligned} \bigg \Vert \displaystyle \sum _{k=1}^{\infty }p_{\alpha ,k}(x,c)\int \limits _{|t-x|\ge \delta } \theta _{\alpha ,k}^\rho (t)e^{\gamma t}dt\bigg \Vert \le M^\prime \alpha ^{-m}, \end{aligned}$$

where $\Vert .\Vert$ is the sup-norm over [a, b].

Lemma 4

For every $x\in (0,\infty )$ and $r\in \mathbb {N}^0,$ there exist polynomials $q_{i,j,r}(x)$ in x independent of $\alpha$ and k such that

$$\begin{aligned} \frac{d^r}{dx^r}p_{\alpha ,k}(x,c)=p_{\alpha ,k}(x,c)\displaystyle \sum _{\begin{array}{c} 2i\,+\,j\le r\\ i,j\ge 0 \end{array}} \alpha ^i (k-\alpha x)^j\frac{(q_{i,j,r}(x,c))}{(p(x,c))^r}, \end{aligned}$$

where $p(x,c)=x(1\,+\,cx).$

Proof

For the cases $c=0,1,$ the proof of this lemma can be seen in [8, 12] respectively.

3 Main Results

3.1 Simultaneous Approximation

Throughout this section, we assume that $0<a<b<\infty .$

In the following theorem, we show that the derivative $B_\alpha ^{\rho (r)}(f;.)$ is also an approximation process for $f^{(r)}.$

Theorem 1

(Basic convergence theorem) Let $f\in C_{\gamma }[0,\infty ).$ If $f^{(r)}$ exists at a point $x\in (0,\infty ),$ then we have

$$\begin{aligned} \lim _{\alpha \rightarrow \infty }\bigg (\frac{d^r}{d\omega ^r}B_\alpha ^\rho (f;\omega ) \bigg )_{\omega =x}=f^{(r)}(x). \end{aligned}$$

(4)

Further, if $f^{(r)}$ is continuous on $(a-\eta , b\,+\,\eta ), \eta >0,$ then the limit in (4) holds uniformly in [a, b].

Proof

By our hypothesis, we have

$$\begin{aligned} f(t)= \sum _{\nu =0}^{r}\frac{f^{(\nu )}(x)}{\nu {!}}(t-x)^\nu \,+\,\psi (t,x) (t-x)^r, \,\, t\in [0,\infty ), \end{aligned}$$

where the function $\psi (t,x)\rightarrow 0$ as $t\rightarrow x.$ From the above equation, we may write

$$\begin{aligned} \bigg (\frac{d^r}{d\omega ^r}B_\alpha ^\rho (f(t);\omega )\bigg )_{\omega =x}= & {} \sum _{\nu =0}^{r}\frac{f^{(\nu )}(x)}{\nu {!}}\bigg (\frac{d^r}{d\omega ^r} B_\alpha ^\rho (t-x)^{\nu };\omega )\bigg )_{\omega =x} \\&+\,\bigg (\frac{d^r}{d\omega ^r}B_\alpha ^\rho (\psi (t,x)(t-x)^r;\omega )\bigg )_{\omega =x}\\= & {} : I_1\,+\,I_2, \mathrm{say}. \end{aligned}$$

First, we estimate $I_1.$

$$\begin{aligned} I_1= & {} \sum _{\nu =0}^{r}\frac{f^{(\nu )}(x)}{\nu !}\bigg \{\frac{d^r}{d\omega ^r} \bigg (\sum _{j=0}^{\nu }\begin{pmatrix}\nu \\ j\end{pmatrix}(-x)^{\nu -j}B_\alpha ^\rho (t^j;\omega )\bigg ) _{\omega =x}\bigg \}\\ {}= & {} \sum _{\nu =0}^{r}\frac{f^{(\nu )}(x)}{\nu {!}}\sum _{j=0} ^{\nu }\begin{pmatrix}\nu \\ j\end{pmatrix}(-x)^{\nu -j}\bigg (\frac{d^r}{d\omega ^r}B_\alpha ^\rho (t^j;\omega ) \bigg )_{\omega =x}\\ {}= & {} \sum _{\nu =0}^{r-1}\frac{f^{(\nu )}(x)}{\nu {!}}\sum _{j=0} ^{\nu }\begin{pmatrix}\nu \\ j\end{pmatrix}(-x)^{\nu -j}\bigg (\frac{d^r}{d\omega ^r}B_\alpha ^\rho (t^j;\omega ) \bigg )_{\omega =x}\,\\&+\,\frac{f^{(r)}(x)}{r{!}}\sum _{j=0}^{r}\begin{pmatrix}r \\ j\end{pmatrix}(-x)^{r-j} \bigg (\frac{d^r}{d\omega ^r}B_\alpha ^\rho (t^j;\omega )\bigg )_{\omega =x}\\&\, :=&I_3\,+\,I_4, \mathrm{say.} \end{aligned}$$

First, we estimate $I_4.$

$$\begin{aligned} I_4= & {} \frac{f^{(r)}(x)}{r!}\sum _{j=0}^{r-1}\begin{pmatrix}r \\ j\end{pmatrix}(-x)^{r-j} \bigg (\frac{d^r}{d\omega ^r}B_\alpha ^\rho (t^j;\omega )\bigg )_{\omega =x} \,+\,\frac{f^{(r)}(x)}{r!}\bigg (\frac{d^r}{d\omega ^r}B_\alpha ^\rho (t^r;\omega )\bigg ) _{\omega =x}\\ {}:= & {} I_5\,+\,I_6, \mathrm{say}. \end{aligned}$$

Using Lemma 2, we get

$I_6= f^{(r)}(x)\,+\,O\bigg (\dfrac{1}{\alpha }\bigg ), I_3=O\bigg (\dfrac{1}{\alpha }\bigg )$ and $I_5= O\bigg (\dfrac{1}{\alpha }\bigg ),$ as $\alpha \rightarrow \infty .$

Combining the above estimates, for each $x\in (0,\infty )$ we obtain $I_1 \rightarrow f^{(r)}(x)$ as $\alpha \rightarrow \infty .$

Next, we estimate $I_2.$ By making use of Lemma 4, we have

$$\begin{aligned} |I_2|\le & {} \sum _{k=1}^{\infty }\frac{p_{\alpha ,k}(x,c)}{(p(x,c))^r} \sum _{\begin{array}{c} 2i\,+\,j\le r\\ i,j\ge 0 \end{array}}\alpha ^i|k-\alpha x|^j|q_{i,j,r}(x,c)| \int \limits _{0}^{\infty }\theta _{\alpha ,k}^{\rho }(t)|\psi (t,x)||(t-x)^r| dt\\&\,+\,\bigg |\bigg (\frac{d^r}{d\omega ^r}p_{\alpha ,0}(\omega ,c)\bigg )_{\omega =x}\bigg ||\psi (0,x)(-x)^r| \\:= & {} I_7\,+\,I_8, \,\,\mathrm{say}. \end{aligned}$$

Since $\psi (t,x)\rightarrow 0$ as $t\rightarrow x,$ for a given $\varepsilon >0$ there exists a $\delta >0$ such that

$|\psi (t,x)|<\varepsilon$ whenever $|t-x|<\delta .$ For $|t-x|\ge \delta ,\, |(t-x)^r\psi (t,x)|\le M e^{\gamma t},$ for some constant $M>0.$

Again, using Lemma 4, we have

$$\begin{aligned} |I_7|\le & {} \sum _{k=1}^{\infty }\sum _{\begin{array}{c} 2i\,+\,j\le r\\ i,j\ge 0 \end{array}} \alpha ^i |k-\alpha x|^j\frac{|q_{i,j,r}(x,c)|}{(p(x,c))^r}p_{\alpha ,k}(x,c)\bigg (\varepsilon \int \limits _{|t-x|<\delta } \theta _{\alpha ,k}^{\rho }(t)|t-x|^r dt \\&+\,M\int \limits _{|t-x|\ge \delta }\theta _{\alpha ,k}^{\rho }(t)e^{\gamma t}dt\bigg ) := I_9\,+\,I_{10}, \,\,\mathrm{say}. \end{aligned}$$

Let $K=\displaystyle \sup _{\begin{array}{c} 2i\,+\,j\le r\\ i,j\ge 0 \end{array}}\frac{|q_{i,j,r}(x,c)|}{(p(x,c))^r}.$ By applying the Schwarz inequality, Lemmas 1 and 3, we get

$$\begin{aligned} |I_9|\le & {} \varepsilon K\sum _{k=1}^{\infty }\sum _{\begin{array}{c} 2i\,+\,j\le r\\ i,j\ge 0 \end{array}} \alpha ^i|k-\alpha x|^jp_{\alpha ,k}(x,c)\bigg (\int \limits _{0}^{\infty } \theta _{\alpha ,k}^{\rho }(t)(t-x)^{2r} dt\bigg )^{\frac{1}{2}}\\\le & {} \varepsilon K\sum _{\begin{array}{c} 2i\,+\,j\le r\\ i,j\ge 0 \end{array}}\alpha ^{i\,+\,j} \bigg (\sum _{k=1}^{\infty }\bigg (\frac{k}{\alpha }-x\bigg )^{2j}p_{\alpha ,k}(x,c)\bigg )^{\frac{1}{2}} \\&\bigg (\sum _{k=1}^{\infty }p_{\alpha ,k}(x,c)\int \limits _{0}^{\infty }\theta _{\alpha ,k} ^{\rho }(t)(t-x)^{2r} dt\bigg )^{\frac{1}{2}}\\ {}\le & {} \varepsilon K\sum _{\begin{array}{c} 2i\,+\,j\le r\\ i,j\ge 0 \end{array}}\alpha ^{i\,+\,j}\bigg (\upsilon _{\alpha ,2j}(x)-x^{2j}\phi _{\alpha ,c}(x)\bigg )^{\frac{1}{2}} \\&\bigg (B_{\alpha }^{\rho }((t-x)^{2r};x))-x^{2r}\phi _{\alpha ,c}(x)\bigg ) ^{\frac{1}{2}}\\ {}= & {} \varepsilon \sum _{\begin{array}{c} 2i\,+\,j\le r\\ i,j\ge 0 \end{array}}\alpha ^{i\,+\,j}\{O(\alpha ^{-j})\,+\,O(\alpha ^{-s_1})\}^{1/2}\\&\times \{O(\alpha ^{-r})\,+\,O(\alpha ^{-s_2})\}^{1/2},\,\,\mathrm{for\ any}\,\,s_1,s_2>0. \end{aligned}$$

$$\begin{aligned}&\mathrm{Choosing}\,\, s_1,\, s_2\,\, \mathrm{such \ that}\,\, s_1>j\,\, \mathrm{and} \,\,s_2>r,\,\,\mathrm{we\ have}\,\, |I_9|= \varepsilon \\&\quad \sum _{\begin{array}{c} 2i\,+\,j\le r\\ i,j\ge 0 \end{array}}\alpha ^{i\,+\,j}O(\alpha ^{-j/2})O(\alpha ^{-r/2}) = \varepsilon . O(1). \end{aligned}$$

Since $\varepsilon >0$ is arbitrary, $I_9\rightarrow 0$ as $\alpha \rightarrow \infty .$

Now, we estimate $I_{10}.$ By applying Cauchy–Schwarz inequality, Lemma 1 and Corollary 2, we obtain

$$\begin{aligned} |I_{10}|\le & {} M K \sum _{k=1}^{\infty }\sum _{\begin{array}{c} 2i\,+\,j\le r\\ i,j\ge 0 \end{array}} \alpha ^i|k-\alpha x|^j p_{\alpha ,k}(x,c)\int \limits _{|t-x|\ge \delta } \theta _{\alpha ,k}^{\rho }(t)e^{\gamma t} dt\\\le & {} M_1 \sum _{\begin{array}{c} 2i\,+\,j\le r\\ i,j\ge 0 \end{array}}\alpha ^{i\,+\,j}\bigg (\sum _{k=1}^{\infty } \bigg (\frac{k}{\alpha }- x\bigg )^{2j} p_{\alpha ,k}(x,c)\bigg )^{1/2}\\&\times \,\bigg (\sum _{k=1}^{\infty } p_{\alpha ,k}(x,c)\int \limits _{|t-x|\ge \delta }\theta _{\alpha ,k}^{\rho }(t)e^{2\gamma t} dt\bigg )^{1/2}, \mathrm{where}\, M_1=M K\\\le & {} M_1 \sum _{\begin{array}{c} 2i\,+\,j\le r\\ i,j\ge 0 \end{array}}\alpha ^{i\,+\,j}\bigg (\upsilon _{\alpha ,2j}(x)-x^{2j}\phi _{\alpha ,c}(x)\bigg )^{1/2}\\&\times \,\bigg (\sum _{k=1}^{\infty } p_{\alpha ,k}(x,c)\int \limits _{|t-x|\ge \delta }\theta _{\alpha ,k}^{\rho }(t)e^{2\gamma t} dt\bigg )^{1/2}\\= & {} \sum _{\begin{array}{c} 2i\,+\,j\le r\\ i,j\ge 0 \end{array}}\alpha ^{i\,+\,j}\{O(\alpha ^{-j})\,+\,O(\alpha ^{-m_1})\}^{1/2} \\&\quad \times \,\{O(\alpha ^{-m_2})\}^{1/2},\,\,\mathrm{for\ any}\,\,m_1, m_2>0.\\ \mathrm{Choosing}\,\, m_1>j,\,\, \mathrm{we\ get} \\ |I_{10}|= & {} \sum _{\begin{array}{c} 2i\,+\,j\le r\\ i,j\ge 0 \end{array}} \alpha ^{i\,+\,j} O(\alpha ^{-j/2}) O(\alpha ^{-m_2/2})=\,O(\alpha ^{(r-m_2)/2}), \end{aligned}$$

which implies that $I_{10}=o(1),$ as $\alpha \rightarrow \infty ,$ on choosing $m_2>r.$ Next, we estimate $I_8.$ We may write

$$\begin{aligned} |I_8|= & {} \bigg |\bigg (\frac{d^r}{d\omega ^r}p_{\alpha ,0}(\omega ,c)\bigg )_{\omega =x}\bigg ||\psi (0,x)|x^r\\= & {} |\phi _{\alpha ,c}^{(r)}(x)|\,\,|\psi (0,x)|x^r. \end{aligned}$$

Now, we observe that $\phi _{\alpha ,0}^{(r)}(x)= e^{-\alpha x}(-\alpha )^{r}$ and $\phi _{\alpha ,1}^{(r)}(x)=\dfrac{(-1)^r(\alpha )_r}{(1\,+\,x)^{\alpha \,+\,r}},$ which implies that $I_8=O(\alpha ^{-p})$ for any $p>0,$ in view of the fact that $|\psi (0,x)x^r|\le N_1,$ for some $N_1>0.$

By combining the estimates $I_7-I_{10},$ we obtain $I_2\rightarrow 0$ as $\alpha \rightarrow \infty .$

To prove the uniformity assertion, it is sufficient to remark that $\delta (\varepsilon )$ in the above proof can be chosen to be independent of $x\in [a,b]$ and also that the other estimates hold uniformity in $x\in [a,b]$. This completes the proof of the theorem.

Next, we establish an asymptotic formula.

Theorem 2

(Voronovskaja type result) Let $f\in C_{\gamma }[0,\infty ).$ If f admits a derivative of order $(r\,+\,2)$ at a fixed point $x\in (0,\infty ),$ then we have

$$\begin{aligned} \lim _{\alpha \rightarrow \infty }\alpha \bigg (\bigg (\frac{d^r}{d\omega ^r}B_\alpha ^\rho (f;\omega )\bigg ) _{\omega =x}-f^{(r)}(x)\bigg )=\displaystyle \sum _{\nu =1}^{r\,+\,2}Q(\nu ,r,c,a,x)f^{(\nu )}(x), \end{aligned}$$

(5)

where $Q(\nu ,r,c,a,x)$ are certain rational functions of x independent of $\alpha .$

Further, if $f^{(r\,+\,2)}$ is continuous on $(a-\eta , b\,+\,\eta ), \eta >0,$ then the limit in (5) holds uniformly in [a, b].

Proof

From the Taylor’s theorem, for $t\in [0,\infty )$ we may write

$$\begin{aligned} f(t)=\sum _{\nu =0}^{r\,+\,2}\frac{f^{(\nu )}(x)}{\nu !}(t-x)^\nu \,+\,\psi (t,x)(t-x)^{r\,+\,2}, \end{aligned}$$

(6)

where the function $\psi (t,x)\rightarrow 0$ as $t\rightarrow x.$

Now, from Eq. (6), we have

$$\begin{aligned} \bigg (\frac{d^r}{d\omega ^r}B_\alpha ^\rho (f(t);\omega )\bigg )_{\omega =x}= & {} \sum _{\nu =0}^{r\,+\,2} \frac{f^{(\nu )}(x)}{\nu !}\bigg (\frac{d^r}{dw^{r}}(B_\alpha ^\rho ((t-x)^{\nu };\omega )\bigg ) _{\omega =x} \\&+\,\bigg (\frac{d^r}{d\omega ^r}B_\alpha ^\rho (\psi (t,x)(t-x)^{r\,+\,2};\omega ) \bigg )_{\omega =x}\\ {}= & {} \sum _{\nu =0}^{r\,+\,2}\frac{f^{(\nu )}(x)}{\nu !} \sum _{j=0}^{\nu }\begin{pmatrix}\nu \\ j\end{pmatrix}(-x)^{\nu -j}\bigg (\frac{d^r}{d\omega ^r}B_\alpha ^\rho (t^j;\omega ) \bigg )_{\omega =x}\\&+\,\bigg (\frac{d^r}{d\omega ^r}B_\alpha ^\rho (\psi (t,x))(t-x)^{r\,+\,2};\omega \bigg ) _{\omega =x}\\ {}:= & {} J_1\,+\,J_2, \,\, {say}. \end{aligned}$$

Proceeding in a manner similar to the estimate of $I_2$ in Theorem 1, for each $x\in (0,\infty )$ we get $\alpha J_2 \rightarrow 0$ as $\alpha \rightarrow \infty .$

Next, we estimate $J_1.$

$$\begin{aligned} J_1= & {} \sum _{\nu =0}^{r-1}\frac{f^{(\nu )}(x)}{\nu !}\sum _{j=0}^{\nu }\begin{pmatrix}\nu \\ j\end{pmatrix} (-x)^{\nu -j}\bigg (\frac{d^r}{d\omega ^r}B_\alpha ^\rho (t^j;\omega )\bigg )_{\omega =x}\\&\,+\,\frac{f^{(r)}(x)}{r!}\sum _{j=0}^{r}\begin{pmatrix}r \\ j\end{pmatrix}(-x)^{r-j}\bigg (\frac{d^r}{d\omega ^r}B_\alpha ^\rho (t^j;\omega )\bigg )_{\omega =x}\\&\,+\,\frac{f^{(r\,+\,1)}(x)}{(r\,+\,1)!}\sum _{j=0}^{r\,+\,1}\begin{pmatrix}r\,+\,1 \\ j\end{pmatrix}(-x)^{r\,+\,1-j}\bigg (\frac{d^r}{d\omega ^r}B_\alpha ^\rho (t^j;\omega )\bigg )_{\omega =x}\\&\,+\,\frac{f^{(r\,+\,2)}(x)}{(r\,+\,2)!}\sum _{j=0}^{r\,+\,2}\begin{pmatrix}r\,+\,2 \\ j\end{pmatrix}(-x)^{r\,+\,2-j}\bigg (\frac{d^r}{d\omega ^r}B_\alpha ^\rho (t^j;\omega )\bigg )_{\omega =x}. \end{aligned}$$

Making use of Lemma 2, we have

$$\begin{aligned} J_1 = f^{(r)}(x)\,+\,\alpha ^{-1}\bigg (\displaystyle \sum _{\nu =1}^{r\,+\,2}Q(\nu ,r,c,a,x)f^{(\nu )} (x)\,+\,o(1)\bigg ). \end{aligned}$$

Thus, from the estimates of $J_1$ and $J_2,$ the required result follows.

The uniformity assertion follows as in the proof of Theorem 1. This completes the proof.

The next result provides an estimate of the degree of approximation in $B_\alpha ^{\rho (r)}(f;x)\rightarrow f^{(r)}(x), r\in \mathbb {N}.$

Theorem 3

(Degree of approximation) Let $r\le q \le r\,+\,2, f\in C_{\gamma }[0,\infty )$ and $f^{(q)}$ exist and be continuous on $(a-\eta , b\,+\,\eta )$ where $\eta >0$ is sufficiently small. Then, for sufficiently large $\alpha$

$$\begin{aligned}&\bigg \Vert \bigg (\frac{d^r}{d\omega ^r} B_\alpha ^\rho (f;\omega )\bigg )_{\omega =x}\\&\quad -f^{(r)}(x) \bigg \Vert _{C[a,b]}\le \max \{C_1\alpha ^{-(q-r)/2}\omega _{f^{(q)}}(\alpha ^{-1/2},(a-\eta ,b\,+\,\eta )), C_2\,\,\alpha ^{-1}\}, \end{aligned}$$

where $C_1=C_1(r,c)$ and $C_2=C_2(r,f,c).$

Proof

By our hypothesis we have,

$$\begin{aligned} f(t)=\sum _{i=0}^{q}\frac{f^{(i)}(x)}{i!}(t-x)^i \,+\,\frac{f^{(q)}(\xi )-f^{(q)}(x)}{q!}(t-x)^{q} \chi (t)\,+\,\phi (t,x)(1-\chi (t)), \end{aligned}$$

(7)

where $\xi$ lies between t and x and $\chi (t)$ is the characteristic function of $(a-\eta , b\,+\,\eta ).$ The function $\phi (t,x)$ for $t\in [a,b]$ is bounded by $M e^{\gamma t}$ for some constant $M >0.$

We operate $\dfrac{d^r}{d\omega ^r} B_\alpha ^\rho (.;\omega )$ on the equality (7) and break the right-hand side into three parts $E_1, E_2$ and $E_3,$ say, corresponding to the three terms on the right-hand side of Eq. (7).

Now, treating $E_1$ in a manner similar to the treatment of $J_1$ of Theorem 2, we get $E_1=f^{(r)}(x)\,+\,O(\alpha ^{-1}),$ uniformly in $x\in [a,b].$

Making use of the inequality

$$\begin{aligned} |f^{(q)}(\xi )-f^{(q)}(x)|\le \bigg (1\,+\,\frac{|t-x|}{\delta }\bigg )\omega _{f^{(q)}} (\delta ),\, \delta >0, \end{aligned}$$

and Lemma 4, we get

$$\begin{aligned} |E_2|\le & {} \frac{\omega _{f^{(q)}}(\delta )}{q!}\bigg \{\sum _{k=1}^{\infty }\sum _{\begin{array}{c} 2i\,+\,j\le r\\ i,j\ge 0 \end{array}} \frac{\alpha ^i|k-\alpha x|^j |q_{i,j,r}(x,c)|}{(p(x,c))^r}p_{\alpha ,k}(x,c)\\&\times \int \limits _{0} ^{\infty }\theta _{\alpha ,k}^{\rho }(t)\bigg (1\,+\,\frac{|t-x|}{\delta }\bigg )|t-x|^q \chi (t) dt\\&\,+\,\bigg (x^q\,+\,\frac{x^{q\,+\,1}}{\delta }\bigg )\phi _{\alpha ,c}^{(r)}(x)\bigg \}\\= & {} E_4 \,+\, E_5. \end{aligned}$$

Finally, let

$$\begin{aligned} S_1=\displaystyle \sup _{x\in [a,b]}\sup _{\begin{array}{c} 2i\,+\,j\le r\\ i,j\ge 0 \end{array}} \frac{|q_{i,j,r}(x,c)|}{(p(x,c))^r}, \end{aligned}$$

then by applying Schwarz inequality, Lemmas 1 and 3, we obtain

$$\begin{aligned} E_4\le & {} \frac{\omega _{f^{(q)}}(\delta ) S_1}{q!} \sum _{\begin{array}{c} 2i\,+\,j\le r\\ i,j\ge 0 \end{array}} \alpha ^{i\,+\,j}\bigg (\sum _{k=1}^{\infty }\bigg (\frac{k}{\alpha }- x\bigg )^{2j}p_{\alpha ,k}(x,c)\bigg )^{1/2} \\&\times \bigg \{\bigg (\sum _{k=1}^{\infty }p_{\alpha ,k}(x,c)\int \limits _{0}^{\infty }\theta _{\alpha ,k} ^{\rho }(t)(t-x)^{2q} dt\bigg )^{1/2}\\&\,+\,\frac{1}{\delta }\bigg (\sum _{k=1}^{\infty }p_{\alpha ,k}(x,c)\int \limits _{0} ^{\infty }\theta _{\alpha ,k}^{\rho }(t)(t-x)^{2q\,+\,2} dt\bigg )^{1/2} \bigg \}\\\le & {} \omega _{f^{(q)}}(\delta ) S_1 \sum _{\begin{array}{c} 2i\,+\,j\le r\\ i,j\ge 0 \end{array}} \alpha ^{i\,+\,j}\bigg (\upsilon _{\alpha ,2j}(x)-x^{2j} \phi _{\alpha ,c}(x)\bigg )^{1/2}\\&\,\times \bigg \{\bigg (B_{\alpha }^{\rho }((t-x)^{2q};x)-x^{2q} \phi _{\alpha ,c}(x)\bigg )^{1/2} \\&\,+\,\frac{1}{\delta }\bigg (B_{\alpha }^{\rho }((t-x)^{2q\,+\,2};x)-x^{2q\,+\,2} \phi _{\alpha ,c}(x)\bigg )^{1/2} \bigg \}\\ {}= & {} \omega _{f^{(q)}}(\delta )\sum _{\begin{array}{c} 2i\,+\,j\le r\\ i,j\ge 0 \end{array}}\alpha ^{i\,+\,j}\{O(\alpha ^{-j}) \,+\,O(\alpha ^{-s_1})\}^{1/2}\\&\times \{(O(\alpha ^{-q})\,+\,O(\alpha ^{-s_2})\}^{1/2} \,+\,\frac{1}{\delta }\{(O(\alpha ^{-(q\,+\,1)})\\&\,+\,O(\alpha ^{-s_3}))\}^{1/2},\,\, \,{for \ any}\,\, s_1,s_2,s_3>0. \end{aligned}$$

Choosing $s_1,s_2,s_3$ such that $s_1>j, s_2>q, s_3>q\,+\,1,$ we have

$$\begin{aligned} |E_4|= & {} \omega _{f^{(q)}}(\delta )\sum _{\begin{array}{c} 2i\,+\,j\le r\\ i,j\ge 0 \end{array}} \alpha ^{i\,+\,j}O\bigg (\frac{1}{\alpha ^{j/2}}\bigg ) \bigg \{O\bigg (\frac{1}{\alpha ^{q/2}}\bigg )\,+\,\frac{1}{\delta }O\bigg (\frac{1}{\alpha ^{(q\,+\,1)/2}}\bigg )\bigg \}. \end{aligned}$$

Now, on choosing $\delta =\alpha ^{-1/2},$ we get

$$\begin{aligned} |E_4|\le & {} C_1 \alpha ^{-(q-r)/2} \omega _{f^{(q)}}(\alpha ^{-1/2},(a-\eta ,b\,+\,\eta )). \end{aligned}$$

Next, proceeding in a manner similar to the estimate of $I_8$ in Theorem 1, we have $E_5=O(\alpha ^{-p}),$ for any $p>0.$ Choosing $p>1,$ we have $E_5=O(\alpha ^{-1}),$ as $\alpha \rightarrow \infty .$

Finally, proceeding along the lines of the estimate of $I_{10}$ of Theorem 2, we obtain $E_3=o(\alpha ^{-1})$ as $\alpha \rightarrow \infty .$

On combining the estimates of $E_1-E_5,$ we get the required result.

3.2 Rate of Convergence

In this section, we shall estimate the rate of convergence for the generalized hybrid operators $B_{\alpha }^{\rho }$ for functions with derivatives of bounded variation. In recent years, several researchers have obtained results in this direction for different sequences of linear positive operators. We refer the reader to some of the related papers (cf. [1, 4, 6, 7, 9], etc.).

Let $f\in DBV_{\gamma }[0,\infty ),$ $\gamma \ge 0$ be the class of all functions defined on $[0,\infty ),$ having a derivative that coincides, a.e. with a function of bounded variation on every finite subinterval of $[0,\infty )$ and $|f(t)|\le Mt^{\gamma },$ $\forall \,\,\ t>0.$

It turns out that for $f\in DBV_{\gamma }[0,\infty ),$ we may write

$$\begin{aligned} f(x)=\int \limits _{0}^{x}g(t)dt\,+\,f(0), \end{aligned}$$

where g(t) is a function of bounded variation on each finite subinterval of $[0,\infty ).$

Lemma 5

For all $x\in (0,\infty ),\, \lambda >1$ and $\alpha$ sufficiently large, we have

• (i) $\lambda _{\alpha }^{\rho }(x,t) = \displaystyle \int \limits _{0}^t K_{\alpha }^{\rho }(x,u)du \le \dfrac{1}{(x-t)^2}\dfrac{\lambda x(1\,+\,cx)}{\alpha }, \,\,0\le t<x;$

• (ii) $1-\lambda _{\alpha }^{\rho }(x,z) = \displaystyle \int \limits _{z}^{\infty } K_{\alpha }^{\rho }(x,u)du \le \dfrac{1}{(z-x)^2}\dfrac{\lambda x(1\,+\,cx)}{\alpha },\,\,x<z<\infty .$

Proof

First we prove (i).

$$\begin{aligned} \lambda _{\alpha }^{\rho }(x,t)= & {} \int \limits _{0}^t K_{\alpha }^{\rho }(x,u)du\le \int \limits _{0}^t\bigg (\frac{x-u}{x-t}\bigg )^2 K_{\alpha }^{\rho }(x,u)du\\ {}\le & {} \frac{1}{(x-t)^2}B_{\alpha }^{\rho }((u-x)^2;x)\\\le & {} \dfrac{1}{(x-t)^2}\dfrac{\lambda x(1\,+\,cx)}{\alpha }. \end{aligned}$$

The proof of (ii) is similar.

Theorem 4

Let $f\in DBV_{\gamma }[0,\infty ),\gamma \ge 0.$ Then for every $x\in (0,\infty ),\, r (\in \mathbb {N})>2\gamma$ and sufficiently large $\alpha$, we have

$$\begin{aligned} |B_{\alpha }^{\rho }(f;x)-f(x)|\le & {} \bigg |\frac{f^\prime (x\,+\,)-f^\prime (x-)}{2}\bigg |\bigg \{\frac{\lambda x(1\,+\,cx)}{\alpha }\bigg \}^{1/2} \\&\,+\,\frac{x}{\sqrt{\alpha }}\bigvee _{x-\frac{x}{\sqrt{\alpha }}}^{x\,+\,\frac{x}{\sqrt{\alpha }}}(f^\prime _x) \,+\,\frac{\lambda (1\,+\,cx)}{\alpha }\sum _{m=1}^{\sqrt{[\alpha ]}}\bigvee _{x-\frac{x}{m}}^{x\,+\,\frac{x}{m}}(f^\prime _x)\\&\,+\,|f^{\prime }(x\,+\,)|\bigg \{\frac{\lambda x(1\,+\,cx)}{\alpha }\bigg \}^{1/2} \\&\,+\,|f(2x)-f(x)-xf^{\prime }(x\,+\,)|\frac{\lambda (1\,+\,cx)}{\alpha x}\\&\,+\,M^\prime \frac{A(r,x)}{\alpha ^{\gamma /2}}\,+\,|f(x)|\frac{\lambda (1\,+\,cx)}{\alpha x}, \end{aligned}$$

where

$$\begin{aligned} f^\prime _x(t)= {\left\{ \begin{array}{ll} f^\prime (t)-f^\prime (x\,+\,), &{}x < t < \infty \\ 0 &{}\,\,\,\, t=x\\ f^\prime (t)-f^\prime (x-), &{} 0\le t < x \end{array}\right. }, \end{aligned}$$

$\bigvee _a^b(f^\prime (x))$ is the total variation of $f^{\prime }_x$ on [a, b],  A(r, x) is a constant depending on r and x and $M^\prime$ is a constant depending on f and $\gamma .$

Proof

By the hypothesis, we may write

$$\begin{aligned} f^{\prime }(t)= & {} \frac{1}{2}\bigg (f^\prime (x\,+\,)\,+\,f^\prime (x-)\bigg )\,+\,f_x^\prime (t)\nonumber \\&\,+\,\frac{1}{2}\bigg (f^\prime (x\,+\,)-f^\prime (x-)\bigg )sgn(t-x) \nonumber \\&\,+\,\delta _x(t)\bigg (f^\prime (t)-\frac{1}{2}\bigg (f^\prime (x\,+\,)\,+\,f^\prime (x-)\bigg )\bigg ), \end{aligned}$$

(8)

where

$$\begin{aligned} \delta _x(t)= {\left\{ \begin{array}{ll} 1 &{} t=x\\ 0 &{} t\ne x. \end{array}\right. } \end{aligned}$$

From Eqs. (2) and (8), we have

$$\begin{aligned} B_{\alpha }^{\rho }(f;x)-f(x)= & {} \int \limits _0^{\infty }K_{\alpha }^{\rho }(x,t)f(t)dt-f(x)=\int \limits _0^{\infty }(f(t)-f(x))K_{\alpha }^{\rho }(x,t)dt\\= & {} \int \limits _0^{x}(f(t)-f(x))K_{\alpha }^{\rho }(x,t)dt\,+\,\int \limits _x^{\infty }(f(t)-f(x))K_{\alpha }^{\rho }(x,t)dt\\= & {} -\int \limits _0^{x}\bigg (\int \limits _t^{x}f^{\prime }(u)du\bigg )K_{\alpha }^{\rho }(x,t)dt\,+\,\int \limits _x^{\infty }\bigg (\int \limits _x^{t}f^{\prime }(u)du\bigg )K_{\alpha }^{\rho }(x,t)dt\\= & {} I_1(x)\,+\,I_2(x),\,\, say. \end{aligned}$$

Using Eq. (8), we get

$$\begin{aligned} I_1(x)= & {} \int \limits _0^{x}\bigg \{\int \limits _t^x\frac{1}{2}\bigg (f^\prime (x\,+\,)\,+\,f^\prime (x-)\bigg )\,+\,f_x^\prime (u)\,+\,\frac{1}{2}\bigg (f^\prime (x\,+\,)-f^\prime (x-)\bigg )sgn(u-x)\\&\,+\,\delta _x(u)\bigg (f^\prime (u)-\frac{1}{2}\bigg (f^\prime (x\,+\,)\,+\,f^\prime (x-)\bigg )\bigg )du\bigg \}K_{\alpha }^{\rho }(x,t)dt. \end{aligned}$$

Since $\int \limits _x^t\delta _x(u)du =0,$ we have

$$\begin{aligned} I_1(x)= & {} \frac{1}{2}\bigg (f^\prime (x\,+\,)\,+\,f^\prime (x-)\bigg )\int \limits _0^{x}(x-t)K_{\alpha }^{\rho }(x,t)dt\,+\,\int \limits _0^x \bigg (\int \limits _x^t f_x^\prime (u)du\bigg )K_{\alpha }^{\rho }(x,t)dt\nonumber \\&-\frac{1}{2}\bigg (f^\prime (x\,+\,)-f^\prime (x-)\bigg )\int \limits _0^{x}|x-t|K_{\alpha }^{\rho }(x,t)dt. \end{aligned}$$

(9)

Proceeding similarly, we find that

$$\begin{aligned} I_2(x)= & {} \frac{1}{2}\bigg (f^\prime (x\,+\,)\,+\,f^\prime (x-)\bigg )\int \limits _x^{\infty }(t-x)K_{\alpha }^{\rho }(x,t)dt\,+\,\int \limits _x^\infty \bigg (\int \limits _x^t f_x^\prime (u)du\bigg )K_{\alpha }^{\rho }(x,t)dt\nonumber \\&\,+\,\frac{1}{2}\bigg (f^\prime (x\,+\,)-f^\prime (x-)\bigg )\int \limits _x^{\infty }|t-x|K_{\alpha }^{\rho }(x,t)dt. \end{aligned}$$

(10)

By combining (9) and (10), we get

$$\begin{aligned} B_{\alpha }^{\rho }(f;x)-f(x)= & {} \frac{1}{2}\bigg (f^\prime (x\,+\,)\,+\,f^\prime (x-)\bigg )\int \limits _0^{\infty }(t-x)K_{\alpha }^{\rho }(x,t)dt \\&\,+\,\frac{1}{2}\bigg (f^\prime (x\,+\,)-f^\prime (x-)\bigg )\int \limits _0^{\infty }|t-x|K_{\alpha }^{\rho }(x,t)dt \nonumber \\&-\int \limits _0^x \bigg (\int \limits _t^x f_x^\prime (u)du\bigg )K_{\alpha }^{\rho }(x,t)dt\,+\,\int \limits _x^\infty \bigg (\int \limits _x^t f_x^\prime (u)du\bigg )K_{\alpha }^{\rho }(x,t)dt. \end{aligned}$$

Hence

$$\begin{aligned} |B_{\alpha }^{\rho }(f;x)-f(x)|\le & {} \bigg |\frac{f^\prime (x\,+\,)\,+\,f^\prime (x-)}{2}\bigg ||B_{\alpha }^{\rho }(t-x;x)|\,+\,\bigg |\frac{f^\prime (x\,+\,)-f^\prime (x-)}{2}\bigg | B_{\alpha }^{\rho }(|t-x|;x) \nonumber \\&\,+\,\bigg |\int \limits _0^x \bigg (\int \limits _t^x f_x^\prime (u)du\bigg )K_{\alpha }^{\rho }(x,t)dt\bigg |\,+\,\bigg |\int \limits _x^\infty \bigg (\int \limits _x^t f_x^\prime (u)du\bigg )K_{\alpha }^{\rho }(x,t)dt\bigg |.\nonumber \\ \end{aligned}$$

(11)

On application of Lemma 5 and integration by parts, we obtain

$$\begin{aligned} \int \limits _0^x \bigg (\int \limits _t^x f_x^\prime (u)du\bigg )K_{\alpha }^{\rho }(x,t)dt=\int \limits _0^x \bigg (\int \limits _t^x f_x^\prime (u)du\bigg )\frac{\partial }{\partial t}\lambda _{\alpha }^{\rho }(x,t)dt= \int \limits _0^x f_x^{\prime }(t)\lambda _{\alpha }^{\rho }(x,t)dt. \end{aligned}$$

Thus,

$$\begin{aligned} \bigg |\int \limits _0^x \bigg (\int \limits _t^x f_x^\prime (u)du\bigg )K_{\alpha }^{\rho }(x,t)dt\bigg |\le & {} \int \limits _0^x |f_x^{\prime }(t)|\lambda _{\alpha }^{\rho }(x,t)dt\nonumber \\\le & {} \int \limits _0^{x-\frac{x}{\sqrt{\alpha }}} |f_x^{\prime }(t)|\lambda _{\alpha }^{\rho }(x,t)dt\,+\,\int \limits _{x-\frac{x}{\sqrt{\alpha }}}^x |f_x^{\prime }(t)|\lambda _{\alpha }^{\rho }(x,t)dt. \end{aligned}$$

Since $f_x^{\prime }(x)=0$ and $\lambda _{\alpha }^{\rho }(x,t)\le 1,$ we get

$$\begin{aligned} \int \limits _{x-\frac{x}{\sqrt{\alpha }}}^x |f_x^{\prime }(t)|\lambda _{\alpha }^{\rho }(x,t)dt= & {} \int \limits _{x-\frac{x}{\sqrt{\alpha }}}^x |f_x^{\prime }(t)-f_x^{\prime }(x)|\lambda _{\alpha }^{\rho }(x,t)dt\le \int \limits _{x-\frac{x}{\sqrt{\alpha }}}^x \bigvee _t^x(f_x^{\prime })dt\\\le & {} \bigvee _{x-\frac{x}{\sqrt{\alpha }}}^x(f_x^{\prime })\int \limits _{x-\frac{x}{\sqrt{\alpha }}}^x dt=\frac{x}{\sqrt{\alpha }}\bigvee _{x-\frac{x}{\sqrt{\alpha }}}^x(f_x^{\prime }). \end{aligned}$$

Similarly, using Lemma 5 and putting $t=x-\frac{x}{u},$ we get

$$\begin{aligned} \int \limits _0^{x-\frac{x}{\sqrt{\alpha }}} |f_x^{\prime }(t)|\lambda _{\alpha }^{\rho }(x,t)dt\le & {} \frac{\lambda x(1\,+\,cx)}{\alpha }\int \limits _0^{x-\frac{x}{\sqrt{\alpha }}} |f_x^{\prime }(t)|\frac{dt}{(x-t)^2}\\ {}\le & {} \frac{\lambda x(1\,+\,cx)}{\alpha }\int \limits _0^{x-\frac{x}{\sqrt{\alpha }}}\bigvee _t^x(f_x^{\prime })\frac{dt}{(x-t)^2}\\= & {} \frac{\lambda (1\,+\,cx)}{\alpha }\int \limits _1^{\sqrt{\alpha }}\bigvee _{x-\frac{x}{u}}^x(f_x^{\prime })du\le \frac{\lambda (1\,+\,cx)}{\alpha }\sum _{m=1}^{[\sqrt{\alpha }]}\bigvee _{x-\frac{x}{m}}^x(f_x^{\prime }). \end{aligned}$$

Consequently,

$$\begin{aligned} \bigg |\int \limits _0^x\bigg (\int \limits _t^xf_x^\prime (u)du\bigg )K_{\alpha }^{\rho }(x,t)dt\bigg |\le \frac{x}{\sqrt{\alpha }}\bigvee _{x-\frac{x}{\sqrt{\alpha }}}^x(f_x^{\prime }) \,+\,\frac{\lambda (1\,+\,cx)}{\alpha }\sum _{m=1}^{[\sqrt{\alpha }]}\bigvee _{x-\frac{x}{m}}^x(f_x^{\prime }). \end{aligned}$$

(12)

Also, we have

$$\begin{aligned} \bigg |\int \limits _x^\infty \bigg (\int \limits _x^t f_x^\prime (u)du\bigg )K_{\alpha }^{\rho }(x,t)dt\bigg |\le & {} \bigg |\int \limits _x^{2x}\bigg (\int \limits _x^t f_x^\prime (u)du\bigg ) \frac{\partial }{\partial t}(1-\lambda _{\alpha }^{\rho }(x,t))dt\bigg |\\&\,+\,\bigg |\int \limits _{2x}^\infty \bigg (\int \limits _x^t f_x^\prime (u)du\bigg )K_{\alpha }^{\rho }(x,t)dt\bigg |\\\le & {} \bigg |\int \limits _{2x}^\infty (f(t)-f(x))K_{\alpha }^{\rho }(x,t)dt\bigg | \\&\,+\,|f^{\prime }(x\,+\,)|\bigg |\int \limits _{2x}^\infty (t-x)K_{\alpha }^{\rho }(x,t)dt\bigg |\\&\,+\,\bigg |\int \limits _x^{2x}f_x^{\prime }(u)du\bigg ||1-\lambda _{\alpha }^{\rho }(x,2x)| \\&\,+\,\int \limits _x^{2x}|f_x^{\prime }(t)|(1-\lambda _{\alpha }^{\rho }(x,t))dt. \end{aligned}$$

Applying Lemma 5, we get

$$\begin{aligned} \bigg |\int \limits _x^\infty \bigg (\int \limits _x^t f_x^\prime (u)du\bigg )K_{\alpha }^{\rho }(x,t)dt\bigg |\le & {} M \int \limits _{2x}^\infty t^{\gamma }K_{\alpha }^{\rho }(x,t)dt \,+\,|f(x)|\int \limits _{2x}^\infty K_{\alpha }^{\rho }(x,t)dt\nonumber \\&\,+\,|f^{\prime }(x\,+\,)|\bigg \{\frac{\lambda x(1\,+\,cx)}{\alpha }\bigg \}^{1/2} \nonumber \\&\,+\,\frac{\lambda (1\,+\,cx)}{\alpha x}|f(2x)-f(x)-xf^{\prime }(x\,+\,)|\nonumber \\&\,+\,\frac{x}{\sqrt{\alpha }}\bigvee _x^{x\,+\,\frac{x}{\sqrt{\alpha }}}(f_x^{\prime })\,+\,\frac{\lambda (1\,+\,cx)}{\alpha }\sum _{m=1}^{[\sqrt{\alpha }]}\bigvee _x^{x\,+\,\frac{x}{m}}(f_x^{\prime }).\nonumber \\ \end{aligned}$$

(13)

We note that we can choose $r\in \mathbb {N}$ such that $2r> \gamma .$

Since $t\le 2(t-x)$ and $x\le t-x$ when $t\ge 2x,$ using H$\ddot{\text {o}}$lder’s inequality and Lemma 3, we obtain

$M \displaystyle \int \limits _{2x}^\infty t^{\gamma }K_{\alpha }^{\rho }(x,t)dt\,+\,|f(x)|\int \limits _{2x}^\infty K_{\alpha }^{\rho }(x,t)dt$

$$\begin{aligned}\le & {} 2^{\gamma } M \int \limits _{2x}^\infty (t-x)^{\gamma }K_{\alpha }^{\rho }(x,t)dt\,+\,\frac{|f(x)|}{x^2}\int \limits _{2x}^\infty (t-x)^2 K_{\alpha }^{\rho }(x,t)dt\nonumber \\\le & {} 2^{\gamma } M \bigg (\int \limits _{0}^\infty (t-x)^{2r}K_{\alpha }^{\rho }(x,t)dt\bigg )^{\gamma /2r}\,+\, |f(x)|\frac{\lambda (1\,+\,cx)}{\alpha x}\nonumber \\\le & {} M^\prime \frac{A(r,x)}{\alpha ^{\gamma /2}}\,+\,|f(x)|\frac{\lambda (1\,+\,cx)}{\alpha x},\,\, {\text {where}}\,\, M^\prime = 2^{\gamma } M. \end{aligned}$$

(14)

Using Lemma 3 and combining (11), (12), (13) and (14), we get the required result.