Modified Stancu operators based on inverse Polya Eggenberger distribution

Abstract

In this paper, we construct a sequence of modified Stancu-Baskakov operators for a real valued function bounded on $[0,\mathrm{\infty })$, based on a function $\tau (x)$. This function $\tau (x)$ is infinite times continuously differentiable on $[0,\mathrm{\infty })$ and satisfy the conditions $\tau (0)=0,{\tau }^{\prime }(x)>0$ and ${\tau }^{″}(x)$ is bounded for all $x\in [0,\mathrm{\infty })$. We study the degree of approximation of these operators by means of the Peetre K-functional and the Ditzian-Totik modulus of smoothness. The quantitative Voronovskaja-type theorems are also established in terms of the first order Ditzian-Totik modulus of smoothness.

Introduction

In 1923, Eggenberger and Pólya [1] were the first to introduce Pólya-Eggenberger distribution. After that, in 1969, Johnson and Kotz [2] gave a short discussion of Pólya-Eggenberger distribution.

The Pólya-Eggenberger distribution X [2] is defined by

$$\begin{array}{rl}& Pr(X=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)\frac{a(a+s)\cdots (a+(u-1)s)b(b+s)\cdots (b+(n-u-1)s)}{(a+b)(a+b+s)\cdots (a+b+(n-1)s)},\\ & k=0,1,2,\dots ,n.\end{array}$$ 1.1

The inverse Pólya-Eggenberger distribution N is defined by

$$\begin{array}{rl}& Pr(N=n+k)=\left(\genfrac{}{}{0ex}{}{(n+k-1)}{k}\right)\frac{a(a+s)\cdots (a+(n-1)s)b(b+s)\cdots (b+(k-1)s)}{(a+b)(a+b+s)\cdots (a+b+(n+k-1)s)},\\ & k=0,1,2,\dots ,n.\end{array}$$ 1.2

In 1970, Stancu [3] introduced a generalization of the Baskakov operators based on inverse Pólya-Eggenberger distribution for a real valued bounded function on $[0,\mathrm{\infty })$, defined by

$$V_n^{[\alpha ]}(f;x)=\sum _{k=0}^{\mathrm{\infty }}{v}_{n,k}(x,\alpha )f\left(\frac{k}{n}\right)=\sum _{k=0}^{\mathrm{\infty }}\left(\genfrac{}{}{0ex}{}{n+k-1}{k}\right)\frac{1^{[n,-\alpha ]}{x}^{[k,-\alpha ]}}{{(1+x)}^{[n+k,-\alpha ]}}f\left(\frac{k}{n}\right),$$ 1.3

where α is a non-negative parameter which may depend only on $n\in \mathbb{N}$ and $a^{[n,h]}=a(a-h)(a-2h)\cdots (a-(n-1)h),a^{[0,h]}=1$ is known as a factorial power of a with increment h. For $\alpha =0$, the operator (1.3) reduces to Baskakov operators [4].

In 1989, Razi [5] studied convergence properties of Stancu-Kantorovich operators based on Pólya-Eggenberger distribution. Very recently, Deo et al. [6] introduced a Stancu-Kantorovich operators based on inverse Pólya-Eggenberger distribution and studied some of its convergence properties. For some other relevant research in this direction we refer the reader to [7–9].

Now, for $\alpha =\frac{1}{n}$, we get a special case of Stancu-Baskakov operators (1.3) defined as

$$V_n^{\langle \frac{1}{n}\rangle }(f;x)=\frac{(2n-1)!}{(n-1)!}\sum _{k=0}^{\mathrm{\infty }}\left(\genfrac{}{}{0ex}{}{n+k-1}{k}\right)\frac{{(nx)}_k}{{(n+nx)}_{n+k}}f\left(\frac{k}{n}\right),$$ 1.4

where ${(a)}_n:=a^{[n,-1]}=a(a+1)\cdots (a+(n-1))$ is called the Pochhammer symbol.

For the Lupas operator, given by

$$L_n(f;x)=\sum _{k=0}^{\mathrm{\infty }}\left(\genfrac{}{}{0ex}{}{n+k-1}{k}\right)\frac{t^k}{{(1+t)}^{n+k}}f\left(\frac{k}{n}\right),$$ 1.5

let ${\mu }_{n,m}(x)=L_n(t^m;x),m\in \mathbb{N}\cup \{0\}$ be the mth order moment.

Lemma 1

For the function ${\mu }_{n,m}(x)$, we have ${\mu }_{n,0}(x)=1$ and we have the recurrence relation

$$x(1+x){\mu }_{n,m}^{\prime }(x)=n{\mu }_{n,m+1}(x)-nx{\mu }_{n,m}(x),m\in \mathbb{N}\cup \{0\},$$ 1.6

where ${\mu }_{n,m}^{\prime }(x)$ is the derivative of ${\mu }_{n,m}(x)$.

Proof

On differentiating ${\mu }_{n,m}(x)$ with respect to x, the proof of the recurrence relation easily follows; hence the details are omitted. □

Remark 1

From Lemma 1, we have

$${\mu }_{n,1}(x)=x,{\mu }_{n,2}(x)=\frac{x+(n+1)x^2}{n},{\mu }_{n,3}(x)=\frac{(n+1)(n+2)x^3+3(n+1)x^2+x}{n^2}.$$

The values of the Stancu-Baskakov operators (1.4) for the test functions $e_i(t)=t^i$, $i=0,1,2$, are given in the following lemma.

Lemma 2

[10]

The Stancu-Baskakov operators (1.4) verify:

• (i) $V_n^{\langle \frac{1}{n}\rangle }(1;x)=1$,
• (ii) $V_n^{\langle \frac{1}{n}\rangle }(t;x)=\frac{nx}{n-1}$,
• (iii) $V_n^{\langle \frac{1}{n}\rangle }(t^2;x)=\frac{n^2}{(n-1)(n-2)}[x^2+\frac{x(x+1)}{n}+\frac{1}{n}(1-\frac{1}{n})x]$.
• (iv) $V_n^{\langle \frac{1}{n}\rangle }(t^3;x)=\frac{n^3}{(n-1)(n-2)(n-3)}[\frac{(n+1)(n+2)}{n^2}{x}^3+\frac{3(2n^2+n-1)}{n^3}{x}^2+\frac{(2n-1)(3n-1)}{n^4}x]$
• (v) $V_n^{\langle \frac{1}{n}\rangle }(t^4;x)=\frac{n^4}{(n-1)(n-2)(n-3)(n-4)}[\frac{(n+1)(n+2)(n+3)}{n^3}{x}^4+\frac{6(n+1)(n+2)(2n-1)}{n^4}{x}^3+\frac{6(6n^3+n^2-4n+1)}{n^5}{x}^2+\frac{26n^2-27n+7}{n^5}x]$.

Proof

The identities (i)-(iii) are proved in [10], hence we give the proof of the identity (iv). The identity (v) can be proved in a similar manner.

We have

$$\begin{array}{rcl}{V}_n^{\langle \frac{1}{n}\rangle }(t^3;x)& =& \frac{(2n-1!)}{(n-1)!}\sum _{k=0}^{\mathrm{\infty }}\left(\genfrac{}{}{0ex}{}{(n+k-1)}{k}\right)\frac{{(nx)}_k}{{(n+nx)}_{n+k}}{\left(\frac{k}{n}\right)}^3\\ & =& \frac{1}{B(nx,n)}{\int }_0^{\mathrm{\infty }}\frac{t^{nx-1}}{{(1+t)}^{nx+n}}{\mu }_{n,3}(t)dt,\end{array}$$

where $B(nx,n)$ is the Beta function.

Therefore using Remark 1, we get

$$V_n^{\langle \frac{1}{n}\rangle }(t^3;x)=\frac{1}{B(nx,n)}{\int }_0^{\mathrm{\infty }}\frac{t^{nx-1}}{{(1+t)}^{nx+n}}\left[\frac{(n+1)(n+2)t^3+3(n+1)t^2+t}{n^2}\right]dt.$$

Now, by a simple calculation, we get the required result. □

As a consequence of Lemma 2, we obtain the following.

Lemma 3

For the Stancu-Baskakov operator (1.4), the following equalities hold:

• (i) $V_n^{\langle \frac{1}{n}\rangle }((t-x);x)=\frac{x}{n-1}$,
• (ii) $V_n^{\langle \frac{1}{n}\rangle }({(t-x)}^2;x)=\frac{2nx(x+1)+(2x-1)x}{(n-1)(n-2)}$,
• (iii) $V_n^{\langle \frac{1}{n}\rangle }({(t-x)}^4;x)=\frac{1}{n(n-1)(n-2)(n-3)(n-4)}[12n(n^2-13n+2)x^3(x+1)+12n(n^2+8n-13)x^2(x+1)+(26n^2+48n-22)x(x+1)+(29-75n)x]$.

Let $0\le r_n(x)\le 1$ be a sequence of continuous functions for each $x\in [0,1]$ and $n\in \mathbb{N}$. Using this sequence $r_n(x)$, for any $f\in C[0,1]$, King [11] proposed the following modification of the Bernstein polynomial for a better approximation:

$$((B_{n}f)\circ r_n)(x)=\sum _{k=0}^{n}f\left(\frac{k}{n}\right)\left(\genfrac{}{}{0ex}{}{n}{k}\right){(r_n(x))}^k{(1-r_n(x))}^{n-k}.$$

Gonska et al. [12] introduced a sequence of King-type operators $D_n^{\tau }:C[0,1]\to C[0,1]$ defined as

$$D_n^{\tau }f=(B_{n}f)\circ {(B_n\tau )}^{-1}\circ \tau ,$$

where $\tau \in C[0,1]$ such that $\tau (0)=0,\tau (1)=1$ and ${\tau }^{\prime }(x)>0$ for each $x\in [0,1]$ and studied global smoothness preservation, the approximation of decreasing and convex functions, the validity of a Voronovskaja-type theorem and a recursion formula generalizing a corresponding result for the classical Bernstein operators.

Motivated by the above work, in the present paper we introduce modified Stancu-Baskakov operators based on a function $\tau (x)$ and obtain the rate of approximation of these operators with the help of Peetre’s K-functional and the Ditzian-Totik modulus of smoothness. Also, we prove a quantitative Voronovskaja-type theorem by using the first order Ditzian-Totik modulus of smoothness.

Throughout this paper, we assume that C denotes a constant not necessarily the same at each occurence.

Modified Stancu-Baskakov operators

Let $\tau (x)$ be continuously differentiable ∞ times on $[0,\mathrm{\infty })$, such that $\tau (0)=0,{\tau }^{\prime }(x)>0$ and ${\tau }^{″}(x)$ is bounded for all $x\in [0,\mathrm{\infty })$. We introduce a sequence of Stancu-Baskakov operators for $f\in C_B[0,\mathrm{\infty })$, the space of all continuous and bounded functions on $[0,\mathrm{\infty })$, endowed with the norm $\parallel f\parallel ={sup}_{x\in [0,\mathrm{\infty })}|f(x)|$, by

$$V_n^{\langle \frac{1}{n},\tau \rangle }(f;x)=\sum _{k=0}^{\mathrm{\infty }}{p}_{n,k}^{\langle \frac{1}{n},\tau \rangle }(x)(f\circ {\tau }^{-1})\left(\frac{k}{n}\right),x\in [0,\mathrm{\infty }),$$ 2.1

where

$$p_{n,k}^{\langle \frac{1}{n},\tau \rangle }(x)=\frac{(2n-1)!}{(n-1)!}\left(\genfrac{}{}{0ex}{}{n+k-1}{k}\right)\frac{{(n\tau (x))}_k}{{(n(1+\tau (x)))}_{n+k}}.$$

Lemma 4

The operator defined by (2.1) satisfies the following equalities:

• (i) $V_n^{\langle \frac{1}{n},\tau \rangle }(1;x)=1$,
• (ii) $V_n^{\langle \frac{1}{n},\tau \rangle }(\tau (t);x)=\frac{n\tau (x)}{n-1}$,
• (iii) $V_n^{\langle \frac{1}{n},\tau \rangle }({\tau }^2(t);x)=\frac{n^2}{(n-1)(n-2)}[{\tau }^2(x)+\frac{\tau (x)(\tau (x)+1)}{n}+\frac{1}{n}(1-\frac{1}{n})\tau (x)]$,
• (iv) $V_n^{\langle \frac{1}{n}\rangle }({\tau }^3(t);x)=\frac{n^3}{(n-1)(n-2)(n-3)}[\frac{(n+1)(n+2)}{n^2}{\tau }^3(x)+\frac{3(2n^2+n-1)}{n^3}{\tau }^2(x)+\frac{(2n-1)(3n-1)}{n^4}\tau (x)]$,
• (v) $V_n^{\langle \frac{1}{n}\rangle }({\tau }^4(t);x)=\frac{n^4}{(n-1)(n-2)(n-3)(n-4)}[\frac{(n+1)(n+2)(n+3)}{n^3}{\tau }^4(x)+\frac{6(n+1)(n+2)(2n-1)}{n^4}{\tau }^3(x)+\frac{6(6n^3+n^2-4n+1)}{n^5}{\tau }^2(x)+\frac{26n^2-27n+7}{n^5}\tau (x)]$.

Proof

The proof of lemma is straightforward on using Lemma 2. Hence we omit the details. □

Let the mth order central moment for the operators given by (2.1) be defined as

$${\mu }_{n,m}^{\tau }(x)=V_n^{\langle \frac{1}{n},\tau \rangle }({(\tau (t)-\tau (x))}^m;x).$$

Lemma 5

For the central moment operator ${\mu }_{n,m}^{\tau }(x)$, the following equalities hold:

• (i) ${\mu }_{n,1}^{\tau }(x)=\frac{\tau (x)}{n-1}$,
• (ii) ${\mu }_{n,2}^{\tau }(x)=\frac{2n{\varphi }_{\tau }^2(x)+(2\tau (x)-1)\tau (x)}{(n-1)(n-2)}$,
• (iii) ${\mu }_{n,4}^{\tau }(x)=\frac{1}{n(n-1)(n-2)(n-3)(n-4)}[12n(n^2-13n+2){\tau }^2(x){\varphi }_{\tau }^2(x)+12n(n^2+8n-13)\tau (x){\varphi }_{\tau }^2(x)+(26n^2+48n-22){\varphi }_{\tau }^2(x)+(29-75n)\tau (x)]$,

where ${\varphi }_{\tau (x)}^2(x)=\tau (x)(\tau (x)+1)$.

Proof

Using the definition (2.1) of the modified Stancu-Baskakov operators and Lemma 4, the proof of the lemma easily follows. Hence, the details are omitted. □

Let

$$W^2=\{g\in C_B[0,\mathrm{\infty }):g^{\prime },g^{″}\in C_B[0,\mathrm{\infty })\}.$$

For $f\in C_B[0,\mathrm{\infty })$ and $\delta >0$, the Peetre K-functional [13] is defined by

$$K(f;\delta )=\underset{g\in W^2}{inf}\{\parallel f-g\parallel +\delta {\parallel g\parallel }_{W^2}\},$$

where

$${\parallel g\parallel }_{W^2}=\parallel g\parallel +\parallel g^{\prime }\parallel +\parallel g^{″}\parallel .$$

From [14], Proposition 3.4.1, there exists a constant $C>0$ independent of f and δ such that

$$K(f;\delta )\le C({\omega }_2(f;\sqrt{\delta })+min\{1,\delta \}\parallel f\parallel ),$$ 2.2

where ${\omega }_2$ is the second order modulus of smoothness of $f\in C_B[0,\mathrm{\infty })$ and is defined as

$${\omega }_2(f;\delta )=\underset{0<|h|\le \delta }{sup}\underset{x,x+2h\in [0,\mathrm{\infty })}{sup}|f(x+2h)-2f(x+h)+f(x)|.$$

In the following, we assume that ${inf}_{x\in [0,\mathrm{\infty })}{\tau }^{\prime }(x)\ge a,a\in {\mathbb{R}}^{+}:=(0,\mathrm{\infty })$.

Next, we recall the definitions of the Ditzian-Totik first order modulus of smoothness and the K-functional [15]. Let ${\varphi }_{\tau }(x):=\sqrt{\tau (x)(1+\tau (x))}$ and $f\in C_B[0,\mathrm{\infty })$. The first order modulus of smoothness is given by

$${\omega }_{{\varphi }_{\tau }}(f;t)=\underset{0<h\le t}{sup}\left\{\right|f(x+\frac{h{\varphi }_{\tau }(x)}{2})-f(x-\frac{h{\varphi }_{\tau }(x)}{2})|,x\pm \frac{h{\varphi }_{\tau }(x)}{2}\in [0,\mathrm{\infty })\}.$$

Further, the appropriate K-functional is defined by

$$K_{{\varphi }_{\tau }}(f;t)=\underset{g\in W_{{\varphi }_{\tau }}[0,\mathrm{\infty })}{inf}\{\parallel f-g\parallel +t\parallel {\varphi }_{\tau }{g}^{\prime }\parallel \}(t>0),$$

where $W_{{\varphi }_{\tau }}[0,\mathrm{\infty })=\{g:g\in A{C}_{\mathrm{loc}}[0,\mathrm{\infty }),\parallel {\varphi }_{\tau }{g}^{\prime }\parallel <\mathrm{\infty }\}$ and $g\in A{C}_{\mathrm{loc}}[0,\mathrm{\infty })$ means that g is absolutely continuous on every interval $[a,b]\subset [0,\mathrm{\infty })$. It is well known [15], p.11, that there exists a constant $C>0$ such that

$$K_{{\varphi }_{\tau }}(f;t)\le C{\omega }_{{\varphi }_{\tau }}(f;t).$$ 2.3

Theorem 1

If $f\in C_B[0,\mathrm{\infty })$, then

$$\parallel V_n^{\langle \frac{1}{n},\tau \rangle }f\parallel \le \parallel f\parallel .$$

Proof

By the definition of the modified Stancu-Baskakov operators (2.1) and using Lemma 4 we have

$$|V_n^{\langle \frac{1}{n},\tau \rangle }(f;x)|\le \sum _{k=0}^n{p}_{n,k}^{\langle \frac{1}{n},\tau \rangle }(x)\left|(f\circ {\tau }^{-1})\left(\frac{k}{n}\right)\right|\le \parallel f\circ {\tau }^{-1}\parallel V_n^{\langle \frac{1}{n},\tau \rangle }(1;x)=\parallel f\parallel ,$$

for every $x\in [0,\mathrm{\infty })$. Hence the required result is immediate. □

Theorem 2

Let $f\in C_B[0,\mathrm{\infty })$. Then, for $n\ge 3$, there exists a constant $C>0$ such that

$$|V_n^{\langle \frac{1}{n},\tau \rangle }(f;x)-f(x)|\le C\{{\omega }_2(f;\frac{{\varphi }_{\tau }(x)}{\sqrt{n-2}})+\frac{{\varphi }_{\tau }^2(x)}{n-2}\parallel f\parallel \}+\omega (f\circ {\tau }^{-1};\left(\frac{\tau (x)}{n-1}\right)),$$

on each compact subset of $[0,\mathrm{\infty })$.

Proof

Let U be a compact subset of $[0,\mathrm{\infty })$. For each $x\in U$, first we define an auxiliary operator as

$$V_n^{\ast \langle \frac{1}{n},\tau \rangle }(f;x)=V_n^{\langle \frac{1}{n},\tau \rangle }(f;x)-f\circ {\tau }^{-1}\left(\frac{n\tau (x)}{n-1}\right)+f(x).$$ 2.4

Now, using Lemma 4, we have

$$V_n^{\ast \langle \frac{1}{n},\tau \rangle }(1;x)=1\text{and}{V}_n^{\ast \langle \frac{1}{n},\tau \rangle }(\tau (t);x)=\tau (x)\text{hence }{V}_n^{\ast \langle \frac{1}{n},\tau \rangle }(\tau (t)-\tau (x);x)=0.$$

Let $g\in W^2$, $x\in U$ and $t\in [0,\mathrm{\infty })$. Then by Taylor’s expansion, and using results in [16], p.32, we get

$$\begin{array}{rcl}g(t)& =& (g\circ {\tau }^{-1})(\tau (t))\\ & =& (g\circ {\tau }^{-1})(\tau (x))+{(g\circ {\tau }^{-1})}^{\prime }(\tau (x))(\tau (t)-\tau (x))+{\int }_{\tau (x)}^{\tau (t)}(\tau (t)-u){(g\circ {\tau }^{-1})}^{″}(u)du\\ & =& g(x)+{(g\circ {\tau }^{-1})}^{\prime }(\tau (x))(\tau (t)-\tau (x))+{\int }_{\tau (x)}^{\tau (t)}(\tau (t)-u)\frac{g^{″}({\tau }^{-1}(u))}{{[{\tau }^{\prime }({\tau }^{-1}(u))]}^2}du\\ & & -{\int }_{\tau (x)}^{\tau (t)}(\tau (t)-u)\frac{g^{\prime }({\tau }^{-1}(u)){\tau }^{″}({\tau }^{-1}(u))}{{[{\tau }^{\prime }({\tau }^{-1}(u))]}^3}du.\end{array}$$ 2.5

Now, applying the operator $V_n^{\ast \langle \frac{1}{n},\tau \rangle }(\cdot ;x)$ to both sides of the above equality, we get

$$\begin{array}{rl}& V_n^{\ast \langle \frac{1}{n},\tau \rangle }(g;x)-g(x)\\ & ={(g\circ {\tau }^{-1})}^{\prime }(\tau (x))V_n^{\ast \langle \frac{1}{n},\tau \rangle }((\tau (t)-\tau (x));x)\\ & +V_n^{\ast \langle \frac{1}{n},\tau \rangle }({\int }_{\tau (x)}^{\tau (t)}(\tau (t)-u)\frac{g^{″}({\tau }^{-1}(u))}{{[{\tau }^{\prime }({\tau }^{-1}(u))]}^2}du;x)\\ & -V_n^{\ast \langle \frac{1}{n},\tau \rangle }({\int }_{\tau (x)}^{\tau (t)}(\tau (t)-u)\frac{g^{\prime }({\tau }^{-1}(u)){\tau }^{″}({\tau }^{-1}(u))}{{[{\tau }^{\prime }({\tau }^{-1}(u))]}^3}du;x)\\ & =V_n^{\langle \frac{1}{n},\tau \rangle }({\int }_{\tau (x)}^{\tau (t)}(\tau (t)-u)\frac{g^{″}({\tau }^{-1}(u))}{{[{\tau }^{\prime }({\tau }^{-1}(u))]}^2}du;x)\\ & -{\int }_{\tau (x)}^{\frac{n\tau (x)}{n-1}}(\frac{n\tau (x)}{n-1}-u)\frac{g^{″}({\tau }^{-1}(u))}{{[{\tau }^{\prime }({\tau }^{-1}(u))]}^2}du\\ & -V_n^{\langle \frac{1}{n},\tau \rangle }({\int }_{\tau (x)}^{\tau (t)}(\tau (t)-u)\frac{g^{\prime }({\tau }^{-1}(u)){\tau }^{″}({\tau }^{-1}(u))}{{[{\tau }^{\prime }({\tau }^{-1}(u))]}^3}du;x)\\ & +{\int }_{\tau (x)}^{\frac{n\tau (x)}{n-1}}(\frac{n\tau (x)}{n-1}-u)\frac{g^{\prime }({\tau }^{-1}(u)){\tau }^{″}({\tau }^{-1}(u))}{{[{\tau }^{\prime }({\tau }^{-1}(u))]}^3}du.\end{array}$$ 2.6

Again, for each $x\in U$, we have

$$\begin{array}{rl}& |V_n^{\ast \langle \frac{1}{n},\tau \rangle }(g;x)-g(x)|\\ & \le \frac{1}{2}\frac{\parallel g^{″}\parallel }{a^2}{V}_n^{\langle \frac{1}{n},\tau \rangle }({(\tau (t)-\tau (x))}^2;x)+\frac{1}{2}\frac{\parallel g^{″}\parallel }{a^2}{(\frac{n\tau (x)}{n-1}-\tau (x))}^2\\ & +\frac{1}{2}\frac{\parallel g^{\prime }\parallel \parallel {\tau }^{″}\parallel }{a^3}{V}_n^{\langle \frac{1}{n},\tau \rangle }({(\tau (t)-\tau (x))}^2;x)+\frac{1}{2}\frac{\parallel g^{\prime }\parallel \parallel {\tau }^{″}\parallel }{a^3}{(\frac{n\tau (x)}{n-1}-\tau (x))}^2\\ & =\frac{1}{2}(\frac{\parallel g^{″}\parallel }{a^2}+\frac{\parallel g^{\prime }\parallel \parallel {\tau }^{″}\parallel }{a^3})[V_n^{\langle \frac{1}{n},\tau \rangle }({(\tau (t)-\tau (x))}^2;x)+{\left(\frac{\tau (x)}{n-1}\right)}^2].\end{array}$$ 2.7

Now, using the definition of the auxiliary operators, Theorem 1 and inequality (2.7), for each $x\in U$ we have

$$\begin{array}{rl}& |V_n^{\langle \frac{1}{n},\tau \rangle }(f;x)-f(x)|\\ & \le |V_n^{\ast \langle \frac{1}{n},\tau \rangle }(f-g;x)|+|V_n^{\ast \langle \frac{1}{n},\tau \rangle }(g;x)-g(x)|\\ & +|g(x)-f(x)|+|f\circ {\tau }^{-1}\left(\frac{n\tau (x)}{n-1}\right)-f\circ {\tau }^{-1}(\tau (x))|\\ & \le 4\parallel f-g\parallel +\frac{1}{2}(\frac{\parallel g^{″}\parallel }{a^2}+\frac{\parallel g^{\prime }\parallel \parallel {\tau }^{″}\parallel }{a^3})[V_n^{\langle \frac{1}{n},\tau \rangle }({(\tau (t)-\tau (x))}^2;x)+{\left(\frac{\tau (x)}{n-1}\right)}^2]\\ & +\omega (f\circ {\tau }^{-1};\left(\frac{\tau (x)}{n-1}\right)).\end{array}$$ 2.8

Let $C=max(4,\frac{4}{a^2},\frac{4}{a^3}\parallel {\tau }^{″}\parallel )$, we get

$$|V_n^{\langle \frac{1}{n},\tau \rangle }(f;x)-f(x)|\le C(\parallel f-g\parallel +{\parallel g\parallel }_{W^2}\frac{{\varphi }_{\tau }^2(x)}{n-2})+\omega (f\circ {\tau }^{-1};\left(\frac{\tau (x)}{n-1}\right)).$$ 2.9

Taking the infimum on the right side of the above inequality over all $g\in W^2$ and for all $x\in U$, we have

$$|V_n^{\langle \frac{1}{n},\tau \rangle }(f;x)-f(x)|\le CK(f;\frac{{\varphi }_{\tau }^2(x)}{n-2})+\omega (f\circ {\tau }^{-1};\left(\frac{\tau (x)}{n-1}\right)),$$ 2.10

using equation (2.2), we get the required result. □

Theorem 3

Let $f\in C_B[0,\mathrm{\infty })$. Then for every $x\in [0,\mathrm{\infty })$, and $n\ge 3$ we have

$$|V_n^{\langle \frac{1}{n},\tau \rangle }(f;x)-f(x)|\le C{\omega }_{{\varphi }_{\tau }}(f;\frac{\sqrt{6}c(x)}{a\sqrt{(n-2)}}).$$

Proof

For any $g\in W_{{\varphi }_{\tau }}[0,\mathrm{\infty })$, by Taylor’s expansion, we have

$$g(t)=(g\circ {\tau }^{-1})(\tau (t))=(g\circ {\tau }^{-1})(\tau (x))+{\int }_{\tau (x)}^{\tau (t)}{(g\circ {\tau }^{-1})}^{\prime }(u)du.$$

Applying the operator $V_n^{\langle \frac{1}{n},\tau \rangle }(\cdot ;x)$ on both sides of the above equality, we get

$$|V_n^{\langle \frac{1}{n},\tau \rangle }(g;x)-g(x)|=\left|V_n^{\langle \frac{1}{n},\tau \rangle }({\int }_{\tau (x)}^{\tau (t)}{(g\circ {\tau }^{-1})}^{\prime }(u)du)\right|.$$ 2.11

From [16], we have

$$\begin{array}{rl}|{\int }_{\tau (x)}^{\tau (t)}{(g\circ {\tau }^{-1})}^{\prime }(u)du|& =|{\int }_x^t\frac{g^{\prime }(y)}{{\tau }^{\prime }(y)}{\tau }^{\prime }(y)dy|=|{\int }_x^t\frac{{\varphi }_{\tau }(y)}{{\varphi }_{\tau }(y)}\frac{g^{\prime }(y)}{{\tau }^{\prime }(y)}{\tau }^{\prime }(y)dy|\\ & \le \frac{\parallel {\varphi }_{\tau }{g}^{\prime }\parallel }{a}\left|{\int }_x^t\frac{{\tau }^{\prime }(y)}{{\varphi }_{\tau }(y)}dy\right|\end{array}$$ 2.12

and

$$\begin{array}{rcl}\left|{\int }_x^t\frac{{\tau }^{\prime }(y)}{{\varphi }_{\tau }(y)}dy\right|& \le & |{\int }_x^t(\frac{1}{\sqrt{\tau (y)}}+\frac{1}{\sqrt{1+\tau (y)}}){\tau }^{\prime }(y)dy|\\ & \le & |{\int }_x^t\frac{1}{\sqrt{\tau (y)}}{\tau }^{\prime }(y)dy|+|{\int }_x^t\frac{1}{\sqrt{1+\tau (y)}}{\tau }^{\prime }(y)dy|\\ & =& 2\{|\sqrt{\tau (t)}-\sqrt{\tau (x)}|+|\sqrt{1+\tau (t)}-\sqrt{1+\tau (x)}|\}\\ & <& 2|\tau (t)-\tau (x)|(\frac{1}{\sqrt{\tau (x)}}+\frac{1}{\sqrt{1+\tau (x)}})\\ & =& \frac{2|\tau (t)-\tau (x)|}{\sqrt{\tau (x)(1+\tau (x))}}(\sqrt{1+\tau (x)}+\sqrt{\tau (x)})\\ & =& \frac{2|\tau (t)-\tau (x)|}{\sqrt{\tau (x)(1+\tau (x))}}c(x)\\ & =& \frac{2c(x)|\tau (t)-\tau (x)|}{{\varphi }_{\tau }(x)}.\end{array}$$ 2.13

Now, from equations (2.12)-(2.13) and using the Cauchy-Schwarz inequality, we obtain

$$\begin{array}{rcl}|V_n^{\langle \frac{1}{n},\tau \rangle }(g;x)-g(x)|& \le & \frac{2c(x)\parallel {\varphi }_{\tau }{g}^{\prime }\parallel }{a{\varphi }_{\tau }(x)}{V}_n^{\langle \frac{1}{n},\tau \rangle }\left(\right|\tau (t)-\tau (x)|;x)\\ & \le & \frac{2c(x)\parallel {\varphi }_{\tau }{g}^{\prime }\parallel }{a{\varphi }_{\tau }(x)}{V}_n^{\langle \frac{1}{n},\tau \rangle }{({(\tau (t)-\tau (x))}^2;x)}^{\frac{1}{2}}\\ & =& \frac{2c(x)\parallel {\varphi }_{\tau }{g}^{\prime }\parallel }{a{\varphi }_{\tau }(x)}{\left[\frac{2n{\varphi }_{\tau }^2(x)+(2\tau (x)-1)\tau (x)}{(n-1)(n-2)}\right]}^{\frac{1}{2}}.\end{array}$$ 2.14

Thus, for $f\in C_B[0,\mathrm{\infty })$ and any $g\in W_{{\varphi }_{\tau }}[0,\mathrm{\infty })$, we have

$$\begin{array}{rcl}|V_n^{\langle \frac{1}{n},\tau \rangle }(f;x)-f(x)|& \le & |V_n^{\langle \frac{1}{n},\tau \rangle }(f-g;x)|+|f(x)-g(x)|+|V_n^{\langle \frac{1}{n},\tau \rangle }(g;x)-g(x)|\\ & \le & 2\parallel f-g\parallel +\frac{2c(x)\parallel {\varphi }_{\tau }{g}^{\prime }\parallel }{a{\varphi }_{\tau }(x)}{\left[\frac{2n{\varphi }_{\tau }^2(x)+(2\tau (x)-1)\tau (x)}{(n-1)(n-2)}\right]}^{\frac{1}{2}}\\ & =& \frac{2c(x)\parallel {\varphi }_{\tau }{g}^{\prime }\parallel }{a}{\left[\frac{2(n+1)}{(n-1)(n-2)}\right]}^{\frac{1}{2}}+2\parallel f-g\parallel \\ & \le & 2\{\parallel f-g\parallel +\frac{\sqrt{6}c(x)}{a\sqrt{(n-2)}}\parallel {\varphi }_{\tau }{g}^{\prime }\parallel \}.\end{array}$$ 2.15

Taking the infimum on the right side of the above inequality over all $g\in W_{{\varphi }_{\tau }}[0,\mathrm{\infty })$, we get

$$|V_n^{\langle \frac{1}{n},\tau \rangle }(f;x)-f(x)|\le 2K_{{\varphi }_{\tau }}(f;\frac{\sqrt{6}c(x)}{a\sqrt{(n-2)}}).$$

Finally, using equation (2.3), the theorem is immediate. □

Theorem 4

For any $f\in C^2[0,\mathrm{\infty })$ and $x\in [0,\mathrm{\infty })$, the following inequality hold:

$$\begin{array}{rl}& |V_n^{\langle \frac{1}{n},\tau \rangle }(f;x)-f(x)-\frac{f^{\prime }(x)}{{\tau }^{\prime }(x)}{\mu }_{n,1}^{\tau }(x)-\frac{1}{2}[\frac{f^{″}(x)}{{({\tau }^{\prime }(x))}^2}-f^{\prime }(x)\frac{{\tau }^{″}(x)}{{({\tau }^{\prime }(x))}^3}]{\mu }_{n,2}^{\tau }(x)|\\ & \le {({\mu }_{n,2}^{\tau }(x))}^{\frac{1}{2}}[\parallel {(f\circ {\tau }^{-1})}^{″}-g\parallel {({\mu }_{n,2}^{\tau }(x))}^{\frac{1}{2}}+\frac{2\parallel {\varphi }_{\tau }{g}^{\prime }\parallel }{a}{\varphi }_{\tau }^{-1}(x){({\mu }_{n,4}^{\tau }(x))}^{1/2}].\end{array}$$

Proof

Let $f\in C^2[0,\mathrm{\infty })$ and $x,t\in [0,\mathrm{\infty })$. Then by Taylor’s expansion, we have

$$\begin{array}{rl}f(t)& =(f\circ {\tau }^{-1})(\tau (t))\\ & =(f\circ {\tau }^{-1})(\tau (x))+{(f\circ {\tau }^{-1})}^{\prime }(\tau (x))(\tau (t)-\tau (x))+{\int }_{\tau (x)}^{\tau (t)}(\tau (t)-u){(f\circ {\tau }^{-1})}^{″}(u)du.\end{array}$$

Hence,

$$\begin{array}{rl}& f(t)-f(x)-{(f\circ {\tau }^{-1})}^{\prime }(\tau (x))(\tau (t)-\tau (x))-\frac{1}{2}{(f\circ {\tau }^{-1})}^{″}(\tau (x)){(\tau (t)-\tau (x))}^2\\ & ={\int }_{\tau (x)}^{\tau (t)}(\tau (t)-u){(f\circ {\tau }^{-1})}^{″}(u)du-{\int }_{\tau (x)}^{\tau (t)}(\tau (t)-u){(f\circ {\tau }^{-1})}^{″}(\tau (x))du\\ & ={\int }_{\tau (x)}^{\tau (t)}(\tau (t)-u)[{(f\circ {\tau }^{-1})}^{″}(u)-{(f\circ {\tau }^{-1})}^{″}(\tau (x))]du.\end{array}$$

Applying $V_n^{\langle \frac{1}{n},\tau \rangle }$ to both sides of the above relation, we get

$$\begin{array}{rl}& |V_n^{\langle \frac{1}{n},\tau \rangle }(f;x)-f(x)-\frac{f^{\prime }(x)}{{\tau }^{\prime }(x)}{\mu }_{n,1}^{\tau }(x)-\frac{1}{2}[\frac{f^{″}(x)}{{({\tau }^{\prime }(x))}^2}-f^{\prime }(x)\frac{{\tau }^{″}(x)}{{({\tau }^{\prime }(x))}^3}]{\mu }_{n,2}^{\tau }(x)|\\ & =\left|V_n^{\langle \frac{1}{n},\tau \rangle }({\int }_{\tau (x)}^{\tau (t)}(\tau (t)-u)[{(f\circ {\tau }^{-1})}^{″}(u)-{(f\circ {\tau }^{-1})}^{″}(\tau (x))]du;x)\right|\\ & \le V_n^{\langle \frac{1}{n},\tau \rangle }\left(\right|{\int }_{\tau (x)}^{\tau (t)}|\tau (t)-u||{(f\circ {\tau }^{-1})}^{″}(u)-{(f\circ {\tau }^{-1})}^{″}(\tau (x))|du|;x).\end{array}$$ 2.16

For $g\in W_{{\varphi }_{\tau }[0,\mathrm{\infty })}$, we have

$$\begin{array}{rl}& \left|{\int }_{\tau (x)}^{\tau (t)}\right|\tau (t)-u\left|\right|{(f\circ {\tau }^{-1})}^{″}(u)-{(f\circ {\tau }^{-1})}^{″}(\tau (x))\left|du\right|\\ & \le \left|{\int }_{\tau (x)}^{\tau (t)}\right|\tau (t)-u\left|\right|{(f\circ {\tau }^{-1})}^{″}(u)-(g\circ {\tau }^{-1})(u)\left|du\right|\\ & +\left|{\int }_{\tau (x)}^{\tau (t)}\right|\tau (t)-u\left|\right|(g\circ {\tau }^{-1})(u)-(g\circ {\tau }^{-1})(\tau (x))\left|du\right|\\ & +\left|{\int }_{\tau (x)}^{\tau (t)}\right|\tau (t)-u\left|\right|(g\circ {\tau }^{-1})(\tau (x))-{(f\circ {\tau }^{-1})}^{″}(\tau (x))\left|du\right|\\ & =\left|{\int }_x^t\right|{(f\circ {\tau }^{-1})}^{″}(\tau (y))-g(y)\left|\right|\tau (t)-\tau (y)|{\tau }^{\prime }(y)dy|\\ & +\left|{\int }_x^t\right|g(y)-g(x)\left|\right|\tau (t)-\tau (y)|{\tau }^{\prime }(y)dy|\\ & +\left|{\int }_x^t\right|g(x)-{(f\circ {\tau }^{-1})}^{″}(\tau (x))\left|\right|\tau (t)-\tau (y)|{\tau }^{\prime }(y)dy|\\ & \le 2\parallel {(f\circ {\tau }^{-1})}^{″}-g\parallel |{\int }_x^t|\tau (t)-\tau (y)|{\tau }^{\prime }(y)dy|\\ & +\left|{\int }_x^t\right|{\int }_x^y|g^{\prime }(v)|dv||\tau (t)-\tau (y)|{\tau }^{\prime }(y)dy|\\ & \le \parallel {(f\circ {\tau }^{-1})}^{″}-g\parallel {(\tau (t)-\tau (x))}^2+\parallel {\varphi }_{\tau }{g}^{\prime }\parallel \left|{\int }_x^t\right|{\int }_x^y\frac{dv}{{\varphi }_{\tau }(v)}||\tau (t)-\tau (y)|{\tau }^{\prime }(y)dy|.\end{array}$$

Using the inequality

$$\frac{|y-v|}{v(1+v)}\le \frac{|y-x|}{x(1+x)},x<v<y,$$

we can write

$$\frac{|\tau (y)-\tau (v)|}{\tau (v)(1+\tau (v))}\le \frac{|\tau (y)-\tau (x)|}{\tau (x)(1+\tau (x))}.$$

Therefore,

$$\begin{array}{rl}& \left|{\int }_{\tau (x)}^{\tau (t)}\right|\tau (t)-u\left|\right|{(f\circ {\tau }^{-1})}^{″}(u)-{(f\circ {\tau }^{-1})}^{″}(\tau (x))\left|du\right|\\ & \le \parallel {(f\circ {\tau }^{-1})}^{″}-g\parallel {(\tau (t)-\tau (x))}^2\\ & +\parallel {\varphi }_{\tau }{g}^{\prime }\parallel \left|{\int }_x^t\right|{\int }_x^y\frac{{|\tau (y)-\tau (x)|}^{1/2}}{{\tau }^{\prime }(v){\varphi }_{\tau }(x)}\cdot \frac{{\tau }^{\prime }(v)}{{|\tau (y)-\tau (v)|}^{1/2}}dv||\tau (t)-\tau (y)|{\tau }^{\prime }(y)dy|\\ & \le \parallel {(f\circ {\tau }^{-1})}^{″}-g\parallel {(\tau (t)-\tau (x))}^2\\ & +2\frac{\parallel {\varphi }_{\tau }{g}^{\prime }\parallel }{a}{\varphi }_{\tau }^{-1}(x)|{\int }_x^t|\tau (y)-\tau (x)||\tau (t)-\tau (y)|{\tau }^{\prime }(y)dy|\\ & \le \parallel {(f\circ {\tau }^{-1})}^{″}-g\parallel {(\tau (t)-\tau (x))}^2+2\frac{\parallel {\varphi }_{\tau }{g}^{\prime }\parallel }{a}{\varphi }_{\tau }^{-1}(x)|{\int }_x^t{(\tau (t)-\tau (x))}^2{\tau }^{\prime }(y)dy|\\ & \le \parallel {(f\circ {\tau }^{-1})}^{″}-g\parallel {(\tau (t)-\tau (x))}^2+2\frac{\parallel {\varphi }_{\tau }{g}^{\prime }\parallel }{a}{\varphi }_{\tau }^{-1}(x){|\tau (t)-\tau (x)|}^3.\end{array}$$ 2.17

Now combining equations (2.16)-(2.17), applying Lemma 3 and the Cauchy-Schwarz inequality, we get

$$\begin{array}{rl}& |V_n^{\langle \frac{1}{n},\tau \rangle }(f;x)-f(x)-\frac{f^{\prime }(x)}{{\tau }^{\prime }(x)}{\mu }_{n,1}^{\tau }(x)-\frac{1}{2}[\frac{f^{″}(x)}{{({\tau }^{\prime }(x))}^2}-f^{\prime }(x)\frac{{\tau }^{″}(x)}{{({\tau }^{\prime }(x))}^3}]{\mu }_{n,2}^{\tau }(x)|\\ & \le \parallel {(f\circ {\tau }^{-1})}^{″}-g\parallel V_n^{\langle \frac{1}{n},\tau \rangle }({(\tau (t)-\tau (x))}^2;x)+2\frac{\parallel {\varphi }_{\tau }{g}^{\prime }\parallel }{a}{\varphi }_{\tau }^{-1}(x)V_n^{\langle \frac{1}{n},\tau \rangle }({|\tau (t)-\tau (x)|}^3;x)\\ & \le \parallel {(f\circ {\tau }^{-1})}^{″}-g\parallel V_n^{\langle \frac{1}{n},\tau \rangle }({(\tau (t)-\tau (x))}^2;x)\\ & +\frac{2\parallel {\varphi }_{\tau }{g}^{\prime }\parallel }{a}{\varphi }_{\tau }^{-1}(x){\left[V_n^{\langle \frac{1}{n},\tau \rangle }({(\tau (t)-\tau (x))}^2;x)\right]}^{1/2}{\left[V_n^{\langle \frac{1}{n},\tau \rangle }({(\tau (t)-\tau (x))}^4;x)\right]}^{1/2}\\ & =\parallel {(f\circ {\tau }^{-1})}^{″}-g\parallel {\mu }_{n,2}^{\tau }(x)+\frac{2\parallel {\varphi }_{\tau }{g}^{\prime }\parallel }{a}{\varphi }_{\tau }^{-1}(x){({\mu }_{n,2}^{\tau }(x))}^{1/2}{({\mu }_{n,4}^{\tau }(x))}^{1/2}\\ & ={({\mu }_{n,2}^{\tau }(x))}^{\frac{1}{2}}[\parallel {(f\circ {\tau }^{-1})}^{″}-g\parallel {({\mu }_{n,2}^{\tau }(x))}^{\frac{1}{2}}+\frac{2\parallel {\varphi }_{\tau }{g}^{\prime }\parallel }{a}{\varphi }_{\tau }^{-1}(x){({\mu }_{n,4}^{\tau }(x))}^{1/2}].\end{array}$$

This completes the proof of the theorem. □