Inverse and Saturation Theorems for Some Summation Integral Linear Positive Operators

Anshul Srivastava¹, V. K. Singh²

¹Dr. Akhilesh Das Gupta Institute of Technology and Management, GGSIPU, New Delhi, India.
²Deen Dayal Upadhyay Gorakhpur University, Gorakhpur, Uttar Pradesh, India.

DOI: 10.4236/jamp.2019.710170                

Abstract

Inverse and saturation theorems have been studied for linear combination of some summation integral hybrid linear positive operators. Linear approximating technique and Steklov means have been used to obtain better approximation results.

Keywords

Linear Combination, Hybrid Operators, Steklov Means, Order of Approximation

1. Introduction

Development of linear positive operators brings major contribution in the field of approximation theory. Several mathematicians [1] [2] [3] [4] [5] have worked on hybrid linear positive operators. They improved rate of convergence by taking their linear combination. Here in this paper we consider a sequence of hybrid operators, combination of Beta and Baskakov basis functions,

$\left(B_{\left(n\right)}f\right)\left(x\right)=\frac{\left(n-1\right)}{n}{\displaystyle {\sum }_{i=0}^{\infty }{b}_{\left(n,i\right)}\left(x\right){\displaystyle {\int }_0^{\infty }{p}_{\left(n,i\right)}\left(t\right)f\left(t\right)\text{d}t}}$, (1)

for every, $x\in \left[0,\infty \right)$ and $f\in L_p\left[0,\infty \right),p\ge 1$

Here,

$b_{\left(n,i\right)}\left(x\right)=\frac{1}{B\left(i+1,n\right)}\frac{x^i}{{\left(1+x\right)}^{n+i+1}}$

and

$p_{\left(n,i\right)}\left(t\right)=\left(\begin{array}{c}n+i-1\\ i\end{array}\right)\frac{t^i}{{\left(1+t\right)}^{n+i}}$ (2)

where, $B\left(i+1,n\right)=\frac{i!\left(n-1\right)!}{\left(n+1\right)!}$ is Beta function.

Clearly, the above operators are linear positive operators and reproduce only constant functions.

Also, $B_{\left(n\right)}\left(1,x\right)=1.$

These operators can be used to approximate Lebesgue integrable functions and can be $L_p$ approximation methods.

The order of approximation for these operators is at its best at $O\left(n^{-1}\right)$. We can improve the order of approximation by taking linear combination of these operators.

Let $d_0,d_1,d_2,\cdots ,d_n$ be $k+1$ arbitrary but fixed distinct positive integers. Then linear combination $B_{\left(n\right)}\left(f,k,x\right)$ of $B_{\left(d_{j}n\right)}\left(f,x\right)$, $j=0,1,\cdots ,n$ is defined by,

$B_{\left(n\right)}\left(f,k,x\right)=\frac{1}{\Delta }\left|\begin{array}{ccccc}{B}_{\left(d_{0}n\right)}\left(f,x\right)& d_0^{-1}& d_0^{-2}& \cdots & d_0^{-n}\\ B_{\left(d_{1}n\right)}\left(f,x\right)& d_1^{-1}& d_1^{-2}& \cdots & d_1^{-n}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ B_{\left(d_{n}n\right)}\left(f,x\right)& d_n^{-1}& d_n^{-2}& \cdots & d_n^{-n}\end{array}\right|$ (3)

where $\Delta$ is Vandermonde determinant obtained by replacing the operator column of above determinant by entries 1 given by

$\Delta =\left|\begin{array}{ccccc}1& d_0^{-1}& d_0^{-2}& \cdots & d_0^{-n}\\ 1& d_1^{-1}& d_1^{-2}& \cdots & d_1^{-n}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 1& d_n^{-1}& d_n^{-2}& \cdots & d_n^{-n}\end{array}\right|$

Simplification of (3) leads to,

$B_{\left(n\right)}\left(f,k,x\right)={\displaystyle {\sum }_{j=0}^{k}C\left(j,k\right)B_{\left(d_{j}n\right)}\left(f,x\right)}$ (4)

where, $C\left(j,k\right)={\displaystyle {\prod }_{i=0,i\ne j}^n\frac{d_j}{d_j-d_i}}$ $k\ne 0$ and $C\left(0,0\right)=1$

Let $0<a_1<a_3<a_2<b_2<b_3<b_1<\infty$ and $I_i=\left[a_i,b_i\right]$, $i=1,2,3$. Also let $\left[\alpha \right]$ denote integral part of $\alpha$.

Let $1\le p\le \infty$, $f\in L_p\left[0,\infty \right)$. Then for sufficiently small $\eta >0$, the steklov mean $f_{\eta ,m}$ of m-th order corresponding to f is defined as,

$f_{\eta ,m}\left(t\right)={\eta }^{-m}{\displaystyle {\int }_{-\pi /2}^{\pi /2}{\displaystyle {\int }_{-\pi /2}^{\pi /2}\cdots {\displaystyle {\int }_{-\pi /2}^{\pi /2}\left[f\left(t\right)+{\left(-1\right)}^{m-1}{\Delta }^m{\displaystyle {\sum }_{i=1}^{m{t}_i}f\left(t\right)}\right]\text{d}{t}_1\text{d}{t}_2\cdots \text{d}{t}_m}}}$ (5)

We will use the following results:

a) $f_{\eta ,m}$ has derivatives up to order m, $f_{\eta ,m}^{\left(m-1\right)}\in AC\left[a_1,b_1\right]$ and $f_{\eta ,m}^{\left(m\right)}$ exists a.e and belongs to $L_p\left(a_1,b_1\right)$.

b) ${\Vert f_{\eta ,m}^{\left(r\right)}\Vert }_{L_{p\left[a_2,b_2\right]}}\le G_m{\eta }^{-r}{\omega }_r,r=1,2,3,\cdots ,m$

c) ${\Vert f-f_{\eta ,m}\Vert }_{L_{p\left[a_2,b_2\right]}}\le G_{m+1}{\omega }_m\left(f,\eta ,p,\left[a_1,b_1\right]\right)$

d) ${\Vert f_{\eta ,m}\Vert }_{L_{p\left[a_2,b_2\right]}}\le G_{m+2}{\Vert f\Vert }_{L_p\left(a_1,b_1\right)}$

${\Vert f_{\eta ,m}^{\left(m\right)}\Vert }_{L_{p\left[a_2,b_2\right]}}\le G_{m+3}{\eta }^{-m}{\Vert f\Vert }_{L_p\left(a_1,b_1\right)}$ (6)

where $G_i$ ’s are certain constants independent of f and η.

The present chapter deals with inverse and saturation results for these operators using linear approximating methods.

Operators (1) can be written as,

$\left(B_{\left(n\right)}f\right)\left(x\right)={\displaystyle {\int }_0^{\infty }{P}_{\left(n\right)}\left(x,t\right)f\left(t\right)\text{d}t}$ (7)

where the kernel, $P_{\left(n\right)}\left(x,t\right)=\frac{\left(n-1\right)}{n}{\displaystyle {\sum }_{i=0}^{\infty }{b}_{\left(n,i\right)}\left(x\right)p_{\left(n,i\right)}(t)}$

2. Some Auxiliary Results

Here, we will present some definitions, results and lemmas which we will be needing in our main theorems.

Definition 2.1.

Jensen’s Inequality. It generalizes the statement that the secant line of a convex function lies above the graph of the function. The secant line consists of weighted means of the convex function (for $t\in \left[0,1\right]$ ).

$f\left(t{x}_1+\left(1-t\right)x_2\right)\le tf\left(x_1\right)+\left(1-t\right)f(x2)$

Definition 2.2.

Fubini’s Theorem. It gives conditions under which it is possible to compute a double integral by usng iterated integral. Order of integration may be switched if the double integral yields a finite answer when the integrand is replaced by its absolute value.

Lemma 2.3. [3] There exists polynomials $s_{\left(k,j,r\right)}\left(x\right)$ independent of $i$ and $n$ such that,

$\frac{{\text{d}}^r}{\text{d}{x}^r}\left(\frac{x^i}{{\left(1+x\right)}^{n+i}}\right)={\displaystyle {\sum }_{\begin{array}{c}2j+k\le r\\ j,k\ge 0\end{array}}{n}^i{\left(i-nx\right)}^j{s}_{\left(k,j,r\right)}\left(x\right)\frac{x^{i-r}}{{\left(1+x\right)}^{n+i+r}}}$

Also,

${\left(x^2+x\right)}^r\frac{{\text{d}}^r}{\text{d}{x}^r}\left(b_{\left(n,i\right)}\left(x\right)\right)={\displaystyle {\sum }_{\begin{array}{c}2j+k\le r\\ j,k\ge 0\end{array}}{\left(n+1\right)}^k{\left(i-\left(n+1\right)x\right)}^j{s}_{\left(k,j,r\right)}\left(x\right)b_{\left(n,i\right)}(x)}$

Lemma 2.4. For $m\in N_0,n\in N,x\in \left[0,\infty \right)$, the mth order central moment is defined by,

$T_{\left(n,m\right)}\left(x\right)=B_{\left(n\right)}\left({\left(t-x\right)}^m,x\right)=\frac{\left(n-1\right)}{n}{\displaystyle {\sum }_{i=0}^{\infty }{b}_{\left(n,i\right)}\left(x\right){\displaystyle {\int }_0^{\infty }{p}_{\left(n,i\right)}\left(t\right){\left(t-x\right)}^m\text{d}t}}$

then,

1) For each $x\in \left[0,\infty \right)$, we have, $T_{\left(n,m\right)}\left(x\right)=O\left(n^{-\left[\left(m+1\right)/2\right]}\right)$

2) $T_{\left(n,0\right)}=1$

3) $T_{\left(n,1\right)}=\left(3x+1\right)/\left(n-2\right)$

4) $T_{\left(n,2\right)}=\frac{2\left\{nx\left(1+x\right)+7x^2+5x+1\right\}}{n^2-5n+6}$

5) For n sufficiently large and $c>2$, $B_{\left(n\right)}\left(\left|t-x\right|,x\right)\le {\left[T_{\left(n,2\right)}\right]}^{1/2}\le \sqrt{\frac{C{x}^2+Cx}{n}}$

We have the following recurrence relation,

$\begin{array}{c}\left(n-m-2\right)T_{\left(n,m+1\right)}\left(x\right)=\left(x^2+x\right)T_{\left(n,m\right)}^{\left(1\right)}\left(x\right)+2m\left(x^2+x\right)T_{\left(n,m-1\right)}\left(x\right)\\ +\left[2x+m+1+x\left(2m+1\right)\right]T_{\left(n,m\right)}\left(x\right)\end{array}$ (8)

Proof. $\left(x^2+x\right){b^{\prime }}_{\left(n,i\right)}\left(x\right)=\left(i-nx-x\right)b_{\left(n,i\right)}(x)$

$\left(t^2+t\right){p^{\prime }}_{\left(n,i\right)}\left(t\right)=\left(i-nt\right)p_{\left(n,i\right)}(t)$

Also,

$\begin{array}{c}\left(x^2+x\right)T_{\left(n,m\right)}^{\left(1\right)}\left(x\right)=\frac{\left(n-1\right)}{n}{\displaystyle {\sum }_{i=0}^{\infty }\left(x^2+x\right){b^{\prime }}_{\left(n,i\right)}\left(x\right)}\\ \times {\displaystyle {\int }_0^{\infty }{p}_{\left(n,i\right)}\left(t\right){\left(t-x\right)}^m\text{d}t}-m\left(x^2+x\right)T_{\left(n,m-1\right)}(x)\end{array}$

This implies,

$\begin{array}{l}\left(x^2+x\right)T_{\left(n,m\right)}^{\left(1\right)}\left(x\right)+m\left(x^2+x\right)T_{\left(n,m-1\right)}\left(x\right)\\ =\frac{\left(n-1\right)}{n}{\displaystyle {\sum }_{i=0}^{\infty }\left(i-nx-x\right)b_{\left(n,i\right)}\left(x\right){\displaystyle {\int }_0^{\infty }{p}_{\left(n,i\right)}\left(t\right){\left(t-x\right)}^m\text{d}t}}\\ -\frac{\left(n-1\right)}{n}{\displaystyle {\sum }_{i=0}^{\infty }{b}_{\left(n,i\right)}\left(x\right){\displaystyle {\int }_0^{\infty }\left\{\left(i-nt\right)+\left(nt-nx\right)-x\right\}{p}_{\left(n,i\right)}\left(t\right){\left(t-x\right)}^m\text{d}t}}\\ =\frac{\left(n-1\right)}{n}{\displaystyle {\sum }_{i=0}^{\infty }{b}_{\left(n,i\right)}\left(x\right){\displaystyle {\int }_0^{\infty }\left(t^2+t\right){p^{\prime }}_{\left(n,i\right)}\left(t\right){\left(t-x\right)}^m\text{d}t}}+n{T}_{\left(n,m+1\right)}\left(x\right)-x{T}_{\left(n,m\right)}(x)\end{array}$

$\begin{array}{l}=\frac{\left(n-1\right)}{n}{\displaystyle {\sum }_{i=0}^{\infty }{b}_{\left(n,i\right)}\left(x\right){\displaystyle {\int }_0^{\infty }[\left(t-x\right)+2x\left(t-x\right)+{\left(t-x\right)}^2}}\\ +\left(x^2+x\right)]p_{\left(n,i\right)}\left(t\right){\left(t-x\right)}^m\text{d}t+n{T}_{\left(n,m+1\right)}\left(x\right)-x{T}_{\left(n,m\right)}\left(x\right)\\ =-\left(m+1\right)\left(2x+1\right)T_{\left(n,m\right)}\left(x\right)-\left(m+2\right)T_{\left(n,m+1\right)}\left(x\right)\\ -m\left(x^2+x\right)T_{\left(n,m-1\right)}\left(x\right)+n{T}_{\left(n,m+1\right)}\left(x\right)-x{T}_{\left(n,m\right)}(x)\end{array}$

Hence the result (8).

Lemma 2.5. For sufficiently large n and certain polynomials $Q\left(p,k,x\right)$ in x of degree p/2 there holds,

$B_{\left(n\right)}\left({\left(t-x\right)}^p,k,x\right)=\frac{1}{n^{\left(k+1\right)}}\left\{Q\left(p,k,x\right)+O\left(1\right)\right\}$

for every $p\in N$.

Proof. From the above lemma 2.4 we have,

$B_{\left(n\right)}\left({\left(t-x\right)}^p,x\right)={\displaystyle {\sum }_{i=0}^{p/2}{P}_{\left(i\right)}\left(x\right)\left\{\frac{1}{n^{\frac{\left[p+1\right]}{2}+i}}\right\}}$

Here, $P_{\left(i\right)}$ ’s are certain polynomials in x of degree at most i.

Now using lemma 2.4 we will have the required result.

Lemma 2.6. [4] [5] Let for $q\left(t\right)\in L_1\left[0,\infty \right)$ and $n\in N$, we have,

${\Vert \frac{\left(n-1\right)}{n}{\displaystyle {\int }_0^{\infty }{b}_{\left(n,i\right)}\left(x\right)p_{\left(n,i\right)}\left(t\right)\frac{{\left(i-nt\right)}^m}{n^m}q\left(t\right)\text{d}t}\Vert }_{L_1\left[0,\infty \right)}\le K\frac{1}{n^{m/2}}{\Vert q\left(t\right)\Vert }_{L_1\left[0,\infty \right)}$ (9)

where, $q\left(t\right)$ has a compact support and K is some constant independent of n and q.

Lemma 2.7. [4] [5] Let for $q\left(w\right)\in L_p\left[0,\infty \right),p>1$ and $j\in N\cup \left\{0\right\}$, $m>0$, we have

$\begin{array}{l}{\Vert \frac{\left(n-1\right)}{n}{\displaystyle {\int }_0^{\infty }{\displaystyle {\sum }_{i=0}^{\infty }{b}_{\left(n,i\right)}\left(x\right)p_{\left(n,i\right)}\left(t\right)\frac{{\left(i-nx\right)}^k}{n^k}{\displaystyle {\int }_x^t{\left(t-w\right)}^{j}q\left(w\right)\text{d}w\text{d}t}}}\Vert }_{L_p\left(a_2,b_2\right)}\\ \le H\left\{\frac{1}{n^{\left(k+j+1\right)/2}}{\Vert h\Vert }_{L_p\left(a_1,b_1\right)}+\frac{1}{n^m}{\Vert h\Vert }_{L_p\left[0,\infty \right)}\right\}\end{array}$ (10)

Here, H is some constant independent of n and q.

Lemma 2.8. [2] [4] Let $q\in L_p\left[0,\infty \right),p\ge 1$ and $supph\subset \left[a_2,b_2\right]$, then,

${\Vert B_{\left(n\right)}^{\left(2k+2\right)}\left(q,.\right)\Vert }_{L_p\left(a_2,b_2\right)}\le I{n}^{k+1}{\Vert q\Vert }_{L_p\left(a_2,b_2\right)}$ (11)

Here, I is some constant independent of n and q.

Lemma 2.9. [1] [2] [4] Let $q^{\left(2k+1\right)}\in AC\left(a_2,b_2\right)$ and $q^{\left(2k+2\right)}\in L_p\left(a_2,b_2\right)$, then,

${\Vert B_{\left(n\right)}^{\left(2k+2\right)}\left(q,.\right)\Vert }_{L_p\left(a_2,b_2\right)}\le J{n}^{k+1}{\Vert q^{\left(2k+2\right)}\Vert }_{L_p\left(a_2,b_2\right)}$ (12)

Here, J is some constant independent of n and q.

Lemma 2.10. [1] [2] [4] [5] Let $f\in C^{2k+2}\left(a_1,b_1\right)$ have a compact support, there holds for $n\to \infty$

$B_{\left(n\right)}\left(f,k,x\right)-f\left(x\right)=\frac{1}{n^{k+1}}\left\{{\displaystyle {\sum }_{j=0}^{2\left(k+1\right)}{P}_{\left(j\right)}\left(k,x\right)f^{\left(j\right)}\left(x\right)}+O\left(1\right)\right\}$ (13)

uniformly in $x\in \left(a_1,b_1\right)$, where $P_{\left(j\right)}\left(k,x\right)$ is a polynomial in x of degree j and does not vanish for all $i=1,2,3,\cdots ,2\left(k+1\right)$.

3. Inverse Theorem

Theorem 3.1. Let $0<\alpha <2\left(k+1\right)$, $f\in L_p\left[0,\infty \right),p\ge 1$, and

${\Vert B_{\left(n\right)}\left(f,k,x\right)-f\left(x\right)\Vert }_{L_p\left(a_1,b_1\right)}=O\left(\frac{1}{n^{\alpha /2}}\right)$

then, ${\omega }_{2\left(k+1\right)}\left(f,\tau ,p,\left[a_2,b_2\right]\right)=O\left({\tau }^{\alpha }\right)$ for $n\to \infty ,\tau \to 0$

Proof. Let $fh=\bar{f}$ for all $\gamma \le \tau$ and $a_1<x_1<x_2<x_3<a_2<b_2<y_3<y_2<y_1<b_1$, where, $supph\subset \left[x_2,y_2\right]$ for any function $g\in C_0^{2\left(k+1\right)}$ and $h\left(t\right)=1$ on $\left[x_3,y_3\right]$

${\Delta }_{\gamma }^{2\left(k+1\right)}\bar{f}\left(x\right)={\Delta }_{\gamma }^{2\left(k+1\right)}\left(\bar{f}\left(t\right)-B_{\left(n\right)}\left(\bar{f},k,x\right)\right)+{\Delta }_{\gamma }^{2\left(k+1\right)}\left(B_{\left(n\right)}\left(\bar{f},k,x\right)\right)$

Now solving 2^nd term using Jensen’s inequality and Fubini’s theorem we have

$\begin{array}{c}\Vert {\Delta }_{\gamma }^{2\left(k+1\right)}\bar{f}{\left(x\right)}_{L_p\left[x_2,y_2\right]}\Vert \le {\Vert {\Delta }_{\gamma }^{2\left(k+1\right)}\left(\bar{f}\left(t\right)-B_{\left(n\right)}\left(\bar{f},k,x\right)\right)\Vert }_{L_p\left[x_2,y_2\right]}\\ +{\gamma }^{2\left(k+1\right)}{\Vert B_{\left(n\right)}^{2\left(k+1\right)}\left(\bar{f},k,x\right)\Vert }_{L_p\left[x_2,y_2+2\gamma \left(k+1\right)\right]}\\ \le {\Vert {\Delta }_{\gamma }^{2\left(k+1\right)}\left(\bar{f}\left(t\right)-B_{\left(n\right)}\left(\bar{f},k,x\right)\right)\Vert }_{L_p\left[x_2,y_2\right]}\\ +{\gamma }^{2\left(k+1\right)}{\Vert B_{\left(n\right)}^{2\left(k+1\right)}\left(\bar{f}-{\bar{f}}_{\eta ,2\left(k+1\right)},k,x\right)\Vert }_{L_p\left[x_2,y_2+2\gamma \left(k+1\right)\right]}\\ +{\Vert B_{\left(n\right)}^{2\left(k+1\right)}\left({\bar{f}}_{\eta ,2\left(k+1\right)},k,x\right)\Vert }_{L_p\left[x_2,y_2+2\gamma \left(k+1\right)\right]}\end{array}$

Applying (6), (11), (12) we have

$\begin{array}{l}{\Vert {\Delta }_{\gamma }^{2\left(k+1\right)}\bar{f}\left(x\right)\Vert }_{L_p\left[x_2,y_2\right]}\\ \le {\Vert {\Delta }_{\gamma }^{2\left(k+1\right)}\left(\bar{f}\left(t\right)-B_{\left(n\right)}\left(\bar{f},k,x\right)\right)\Vert }_{L_p\left[x_2,y_2\right]}\\ +L{\gamma }^{2\left(k+1\right)}\left(n^{\left(k+1\right)}+\frac{1}{{\eta }^{2\left(k+1\right)}}\right){\omega }_{2\left(k+1\right)}\left(\bar{f},\eta ,p,\left[x_2,y_2\right]\right)\end{array}$

For proving our theorem, we will be using mathematical induction:

Step 1: For $\alpha \le 1$

Now we will prove for $n\to \infty$,

${\Vert {\Delta }_{\gamma }^{2\left(k+1\right)}\left(\bar{f}\left(t\right)-B_{\left(n\right)}\left(\bar{f},k,x\right)\right)\Vert }_{L_p\left[x_2,y_2\right]}=O\left(\frac{1}{n^{\alpha /2}}\right)$

therefore,

${\omega }_{2\left(k+1\right)}\left(\bar{f},\tau ,p,\left[x_2,y_2\right]\right)=O,\tau \to 0$

$\bar{f}\left(t\right)=f\left(t\right)$ for $t\in \left[x_3,y_3\right]$

$\begin{array}{l}{\Vert B_{\left(n\right)}\left(\bar{f},k,x\right)-\bar{f}\Vert }_{L_p\left[x_2,y_2\right]}\\ \le {\Vert B_{\left(n\right)}\left(h\left(x\right)\left(f\left(t\right)-f\left(x\right)\right),k,x\right)\Vert }_{L_p\left[x_2,y_2\right]}\\ +{\Vert B_{\left(n\right)}\left(f\left(t\right)\left(h\left(t\right)-h\left(x\right)\right),k,x\right)\Vert }_{L_p\left[x_2,y_2\right]}\end{array}$

${\Vert B_{\left(n\right)}\left(\bar{f},k,x\right)-\bar{f}\Vert }_{L_p\left[x_2,y_2\right]}=O\left(\frac{1}{n^{1/2}}\right)+O\left(\frac{1}{n^{\alpha /2}}\right)$

Step 2: For $\left(m-1\right)<\alpha <m$

Let $\left[c,d\right]\subset \left(a_1,b_1\right)$

${\omega }_{2\left(k+1\right)}\left(f,\tau ,p,\left[c,d\right]\right)=O\left({\tau }^{\left(m-1\right)+\beta }\right),\tau \to 0,0<\beta <1$

Here, we have, $\phi \left(t\right)$ as a characteristic function of $\left[x_1,y_1\right]$

Due to smoothness of f we have,

$\begin{array}{l}{\Vert B_{\left(n\right)}\left(\bar{f},k,x\right)-\bar{f}\Vert }_{L_p\left[x_2,y_2\right]}\\ \le {\displaystyle {\sum }_{i=0}^{r-2}\frac{1}{i!}{\Vert f^{\left(i\right)}\left(x\right)B_{\left(n\right)}\left({\left(t-x\right)}^i\left(h\left(t\right)-h\left(x\right)\right),k,x\right)\Vert }_{L_p\left[x_2,y_2\right]}}\\ +\frac{1}{\left(r-2\right)!}{\Vert B_{\left(n\right)}\left(\phi \left(t\right)\left(h\left(t\right)-h\left(x\right)\left({\displaystyle {\int }_x^t\frac{{\left(t-w\right)}^r}{{\left(t-w\right)}^2}\left(f^{\left(r-1\right)}\left(w\right)-f^{\left(r-1\right)}\left(x\right)\right)\text{d}w}\right),k,x\right)\right)\Vert }_{L_p\left[x_2,y_2\right]}\\ +{\Vert B_{\left(n\right)}\left(\left(f\left(x\right)-{\displaystyle {\sum }_{i=0}^{r-2}\frac{{\left(t-x\right)}^i}{i!}{f}^{\left(i\right)}\left(x\right)}\right)\left(1-\phi \left(t\right)\right)\left(h\left(t\right)-h\left(x\right)\right),k,x\right)\Vert }_{L_p\left[x_2,y_2\right]}\\ =H_1+H_2+H_3\end{array}$

where, for $t\in \left[0,\infty \right)$ and $x\in \left[x_2,y_2\right]$

Using lemma 2.4, we have for $n\to \infty$

$H_1{H}_3=O\left(\frac{1}{n^{\left(k+1\right)}}\right)$

Now, using direct theorem in [6]

$\begin{array}{l}{\displaystyle {\int }_{x_2}^{y_2}{\left|B_{\left(n\right)}\left(\phi \left(t\right)\left(h\left(t\right)-h\left(x\right)\left({\displaystyle {\int }_x^t\frac{{\left(t-w\right)}^r}{{\left(t-w\right)}^2}\left(f^{\left(r-1\right)}\left(w\right)-f^{\left(r-1\right)}\left(x\right)\right)\text{d}w}\right),x\right)\right)\right|}^p}\\ \le R{\displaystyle {\int }_{x_2}^{y_2}{\displaystyle {\int }_{x_1}^{y_1}{V}_{\left(n\right)}\left(x,t\right)\frac{{\left|t-x\right|}^{rp}}{\left|t-x\right|}}}{\displaystyle {\int }_x^t\phi \left(w\right){\left|f^{\left(r-1\right)}\left(w\right)-f^{\left(r-1\right)}\left(x\right)\right|}^p\text{d}w\text{d}t\text{d}x}\\ \le R{\displaystyle {\sum }_{l=1}^r{\displaystyle {\int }_{x_2}^{y_2}\{{\displaystyle {\int }_{x+\frac{l}{\sqrt{n}}}^{x+\frac{\left(l+1\right)}{\sqrt{n}}}{V}_{\left(n\right)}\left(x,t\right)\frac{n^{2p}}{l^{4p}}}\frac{{\left|t-x\right|}^{\left(r+4\right)p}}{\left|t-x\right|}}}\\ \times {\displaystyle {\int }_x^{x+\frac{\left(l+1\right)}{\sqrt{n}}}\phi \left(w\right){\left|f^{\left(r-1\right)}\left(w\right)-f^{\left(r-1\right)}\left(x\right)\right|}^p\text{d}w\text{d}t}\\ +{\displaystyle {\int }_{x-\frac{\left(l+1\right)}{\sqrt{n}}}^{x-\frac{l}{\sqrt{n}}}{V}_{\left(n\right)}\left(x,t\right)\frac{n^{2p}}{l^{4p}}\frac{{\left|t-x\right|}^{\left(r+4\right)p}}{\left|t-x\right|}}\end{array}$

$\begin{array}{l}\times {\displaystyle {\int }_{x-\frac{\left(l+1\right)}{\sqrt{n}}}^x\phi \left(w\right){\left|f^{\left(r-1\right)}\left(w\right)-f^{\left(r-1\right)}\left(x\right)\right|}^p\text{d}w\text{d}t}\}\text{d}x\\ +{\displaystyle {\int }_{x_2}^{y_2}{\displaystyle {\int }_{x_2-\left(\frac{1}{\sqrt{n}}\right)}^{y_2+\left(\frac{1}{\sqrt{n}}\right)}{V}_{\left(n\right)}\left(x,t\right)\frac{{\left|t-x\right|}^{rp}}{\left|t-x\right|}}}{\displaystyle {\int }_{x-\left(\frac{1}{\sqrt{n}}\right)}^{x+\left(\frac{1}{\sqrt{n}}\right)}\phi \left(w\right){\left|f^{\left(r-1\right)}\left(w\right)-f^{\left(r-1\right)}\left(t\right)\right|}^p\text{d}w\text{d}t\text{d}x}\\ \le R\{{\displaystyle {\sum }_{l=1}^r\frac{n^{\left(-rp+1\right)/2}}{l^{4p}}{\displaystyle {\int }_0^{\frac{\left(l+1\right)}{\sqrt{n}}}{\left(\omega \left(f^{\left(r-1\right)},w,p,\left[x_1,y_1\right]\right)\right)}^p\text{d}w}}\\ +n^{\left(-rp+1\right)/2}{\displaystyle {\int }_0^{\frac{1}{\sqrt{n}}}{\left(\omega \left(f^{\left(r-1\right)},w,p,\left[x_1,y_1\right]\right)\right)}^p\text{d}w}\}\end{array}$

Using lemma 2.4 and interchanging integration we have

$\le R\left\{{\displaystyle {\sum }_{l=1}^r\frac{n^{\left(-rp+1\right)/2}}{l^{4p}}}{\displaystyle {\int }_0^{\frac{\left(l+1\right)}{\sqrt{n}}}{\left(O\left({\omega }^{\beta }\right)\right)}^p\text{d}w}+n^{\left(-rp+1\right)/2}{\displaystyle {\int }_0^{\frac{1}{\sqrt{n}}}{\left(O\left({\omega }^{\beta }\right)\right)}^p\text{d}w}\right\}$

We have for $n\to \infty$

$H_2=O\left(\frac{1}{n^{\left(r+\beta \right)/2}}\right)$

Combining all the results

$H_1{H}_2{H}_3=O\left(\frac{1}{n^{\left(k+1\right)}}\right)O\left(\frac{1}{n^{\left(r+\beta \right)/2}}\right)$

Thus we have the theorem.

4. Saturation Result

Theorem 4.1. For $f\in L_p\left[0,\infty \right)$, $1\le p<\infty$ and ${\Vert B_{\left(n\right)}\left(f,k,x\right)-f\Vert }_{L_p\left(a_1,b_1\right)}$, then, f coincides almost everywhere with a function F on $\left[a_2,b_2\right]$ having $2\left(k+1\right)$ th derivative such that,

a) $F^{\left(2k+1\right)}\in A.C\left(a_2,b_2\right),\forall p>1$

b) $F^{2\left(k+1\right)}\in L_p\left(a_2,b_2\right),\forall p>1$

c) $F^{\left(2k+1\right)}\in L_p\left(a_2,b_2\right),\forall p=1$

d) $F^{\left(2k\right)}\in A.C\left(a_2,b_2\right),\forall p=1$

for $a_1<x_1<x_2<a_2<b_2<y_2<y_1<b_1$

Proof. Let $q\in C_0^{2\left(k+1\right)},suppq\subset \left(a_2,b_2\right)$ such that $q=1$ on $\left[x_1,y_1\right]$ and $fq=\bar{f}$, then for $0<\alpha <2\left(k+1\right)$, we have,

${\Vert B_{\left(n\right)}\left(f,k,x\right)-f\left(x\right)\Vert }_{L_p\left(x_1,y_1\right)}=O\left(\frac{1}{n^{\alpha /2}}\right)$

${\omega }_{2\left(k+1\right)}\left(f,\tau ,p,\left[x_2,y_2\right]\right)=O(\tau \alpha )$

Now,

$\begin{array}{l}{\Vert B_{\left(n\right)}\left(\bar{f},k,x\right)-\bar{f}\left(x\right)\Vert }_{L_{p\left[x_2,y_2\right]}}\\ \le {\Vert B_{\left(n\right)}\left(f,k,x\right)-f\left(x\right)\Vert }_{L_{p\left[x_2,y_2\right]}}+{\Vert B_{\left(n\right)}\left(\bar{f}-f,k,x\right)\Vert }_{L_{p\left[x_2,y_2\right]}}\end{array}$

Here, ${\Vert B_{\left(n\right)}\left(\bar{f}-f,k,x\right)\Vert }_{L_{p\left[x_2,y_2\right]}}$ is arbitrarily small, so we have,

${\Vert B_{\left(n\right)}\left(\bar{f},k,x\right)-\bar{f}\left(x\right)\Vert }_{L_{p\left[x_2,y_2\right]}}=O\left(\frac{1}{n^{\left(k+1\right)}}\right)$, $n\to \infty$

Case 1: for $p>1$

Consider a sequence $\left\{n_j\right\}$ and a function $h\left(x\right)\in L_{p\left[x_2,y_2\right]}$ such that for every $g\in C_0^{2\left(k+1\right)}$ and $suppg\subset \left(a_1,b_1\right)$, we have from Alaoglu’s theorem,

${\lim }_{n_j\to \infty }{n}_j^{k+1}\langle B_{\left(n\right)}\left(\bar{f},k,x\right)-\bar{f},g\rangle =\langle h,g\rangle$ (14)

By lemma 2.10, we have

$\begin{array}{l}{\lim }_{n_j\to \infty }{n}_j^{k+1}\langle B_{\left(n\right)}\left(\bar{f},k,x\right)-\bar{f},g\rangle \\ ={\lim }_{n_j\to \infty }{n}_j^{k+1}\left({\lim }_{n_j\to \infty }{n}_j^{k+1}\langle B_{\left(n\right)}\left(g,k,x\right)-g,\bar{f}\rangle \right)\\ =\langle {\displaystyle {\sum }_{i=1}^{2\left(k+1\right)}{P}_{\left(i\right)}\left(k,x\right)\frac{{\partial }^i}{\partial x^i}\bar{f}},g\rangle \\ =\langle g,{\displaystyle {\sum }_{i=1}^{2\left(k+1\right)}{P}_{\left(i\right)}^{*}\left(k_i,x\right)\frac{{\partial }^i}{\partial x^i}\bar{f}}\rangle \end{array}$ (15)

Comparing (14) and (15) we have

$h={\displaystyle {\sum }_{i=1}^{2\left(k+1\right)}{P}_{\left(j\right)}\left(k,x\right)\frac{{\partial }^i}{\partial x^i}\bar{f}}$

Here, ${\bar{f}}^{\left(2k+1\right)}\in A.C\left(x_2,y_2\right)$

So, $F^{\left(2k+1\right)}\in A.C\left(a_2,b_2\right)$

Similarly, ${\bar{f}}^{\left(2k+1\right)}\in A.C\left(x_2,y_2\right)$

So, $F^{\left(2k+1\right)}\in A.C\left(a_2,b_2\right)$

We have (i) and (ii)

Case 2: for $p=1$

Proceeding in the similar manner as above, we get (iii) and (iv).

Hence the theorem.

5. Conclusion

We have improved order of approximation by taking suitable linear combinations. Inverse and saturation results have been developed for our hybrid operators.

Acknowledgements

The author is grateful to the referee for many suggestions that have improved this paper a lot.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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