Approximation Theorems for Kantorovich type Lupaş-Stancu Operators based on \(q\)-integers
Sevilay KIRCI SERENBAY\(^\S \), Özge DALMANOĞLU\(^\ast \)
January 24, 2017.
\(^\S \)Baskent University, Department of Mathematics Education, Ankara, Turkey, e-mail: sevilaykirci@gmail.com.
\(^\ast \)Baskent University, Department of Mathematics Education, Ankara, Turkey, e-mail: ozgedalmanoglu@gmail.com
In this paper, we introduce a Kantorovich generalization of \(q\)-Stancu-Lupaş operators and investigate their approximation properties. The rate of convergence of these operators are obtained by means of modulus of continuity, functions of Lipschitz class and Peetre’s K-functional. We also investigate the convergency of the operators in the statistical sense and give a numerical example in order to estimate the error in the approximation.
MSC. 41A35; 41A36
Keywords. Lupaş-Kantorovich operators, Modulus of continuity, Peetre’s K-functional, q-integers, rate of convergence, statistical approximation.
1 Introduction
For a function \(f(x)\) defined on the interval \([0,1]\), the linear operator \(R_{n,q}:C[0,1]\rightarrow C[0,1]\) defined by
\begin{equation} R_{n,q}(f)=R_{n}(f,q;x)=\sum _{k=0}^{n}f\left( \tfrac {[k]}{[n]}\right) b_{nk}(q;x) \label{1} \end{equation}
1
where
\begin{equation} b_{n,k}(q;x)= \genfrac {[}{]}{0pt}{1}nk \tfrac {q^{k(k-1)/2}x^{k}(1-x)^{n-k}}{\left( 1-x+qx\right) ...(1-x+q^{n-1}x)} \label{2} \end{equation}
2
is called Lupaş operators
[
12
]
. For \(q{\gt}0\), \(R_{n}(f,q;x)\) are linear positive operators on \(C[0,1]\) and for \(q=1\) they turn into the well known Bernstein operators. The following identities hold for the \(R_{n}(f,q;x)\) operators:
\begin{eqnarray} R_{n}(e_{0},q;x) & =& 1 \notag \\ R_{n}(e_{1},q;x) & =& x \label{3} \\ R_{n}(e_{2},q;x) & =& x^{_{2}}+\tfrac {x(1-x)}{[n]}\left( \tfrac {1-x+q^{n}x}{1-x+xq}\right) . \notag \end{eqnarray}
Lupaş investigated the approximation properties of the operators on \(C[0,1]\) and estimated the rate of convergence in terms of modulus of continuity. In
[
14
]
the authors studied Voronovskaja type theorems for the \(q\)-Lupaş operators for fixed \(q{\gt}0\). In
[
16
]
, Ostrovska presented new results for the convergence of the sequence \(R_{n}(f,q_{n};x)\) in \(C[0,1]\). She established approximation theorems for the cases \(q\in (0,1)\) and \(q\in (1,\infty )\), respectively, and studied the convergence of \(\{ R_{n}(f,q_{n};x)\} \), \(q\neq 1\) is fixed, obtaining the limit operator of the Lupaş \(q\)-analogue of the Bernstein operator. In
[
5
]
, Doğru and Kanat considered a King type modification of Lupaş operators and investigated the statistical approximation properties of the operators. Very recently, Doğru et al.
[
7
]
introduced a Stancu type generalization of \(q\)-Lupaş operators as
\begin{equation} R_{n}^{\alpha ,\beta }(f;q,x)=[n+1]\sum \limits _{k=0}^{n}f\left( \tfrac {[k]+[\alpha ]}{[n]+[\beta ]}\right) b_{n,k}(q;x) \label{6} \end{equation}
4
where \(b_{n,k}(q;x)\) is given in (2 ). They studied the approximation properties and also introduced the \(r\)-th generalization of these operators.
Since \(q\)-Bernstein operators has attracted a lot of interest, many generalizations of them have been discovered and studied by several authors. Here we will mention some of them related to our study. For example in
[
13
]
an integral modification, called Kantorovich type generalization of \(q\)-Bernstein operators, have been studied. The authors constructed the operators as
\begin{equation*} B_{n,q}^{\ast }(f;x):=\sum \limits _{k=0}^{n}p_{n,k}(q;{x})\int \limits _{0}^{1}f\left( \tfrac {[k]+q^{k}t}{[n+1]}\right) d_{q}t \end{equation*}
where \(f\in C[0,1],0{\lt}q{\lt}1\) and
\begin{equation} p_{n,k}(q;{x})= \genfrac {[}{]}{0pt}{1}nk x^{k}(1-x)^{n-k} \label{2*} \end{equation}
5
and studied some approximation properties of them. Özarslan and Vedi
[
17
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introduced \(q\)-Bernstein-Schurer-Kantorovich operators as
\begin{equation*} K_{n}^{p}(f;q,x):=\sum \limits _{k=0}^{n+p}p_{n+p,k}(q;{x})\int \limits _{0}^{1}f\left( \tfrac {[k]+q^{k}t}{[n+1]}\right) d_{q}t. \end{equation*}
Acu et al.
[
1
]
introduced a new q-Stancu-Kantorovich operators as
\begin{equation*} S_{n,q}^{\ast (\alpha ,\beta )}(f;x):=\sum \limits _{k=0}^{n}p_{n,k}(q;{x})\int \limits _{0}^{1}f\left( \tfrac {[k]+q^{k}t+\alpha }{[n+1]+\beta }\right) d_{q}t \end{equation*}
where \(0\leq \alpha \leq \beta \), \(f\in C[0,1]\) and \(p_{n,k}(q;{x})\) is given in \((\ref{2*})\). She also established a q-analogue of Stancu-Schurer-Kantorovich operators in
[
2
]
where she gave the convergence theorems both in classical and statistical sense and obtained a Voronovskaya type result.
For every \(n\in N\) and \(q\in (0,1)\), Doğru and Kanat
[
6
]
defined the Kantorovich type modification of Lupaş operators as
\begin{equation} R_{n}(f,q;x)=[n+1]\sum \limits _{k=0}^{n}\left( \int \limits _{[k]/[n+1]}^{[k+1]/[n+1]}f(t)d_{q}t\right) \genfrac {[}{]}{0pt}{1}nk \tfrac {q^{-k}q^{k(k-1)/2}x^{k}(1-x)^{n-k}}{\left( 1-x+qx\right) ...(1-x+q^{n-1}x)}. \label{5} \end{equation}
6
Recently, Agrawal et al.
[
3
]
studied the approximation properties of Lupaş-
In this paper we present a Kantorovich generalization of the Lupaş-Stancu operators based on the \(q\)-integers. Our purpose is to study the local and global approximation results for these operators. We also investigate statistical approximation properties using Korovkin type statistical approximation theorem.
2 Construction of The Operators
Before proceeding further we recall some basic notations from \(q\)-calculus (see
[
4
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and
[
10
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).
Let \(q{\gt}0\). For each nonnegative integer \(r\), the q-integer \([r]\), the q-factorial \([r]!\) and the q-binomial coefficient \(\genfrac {[}{]}{0pt}{1}rk \), \((r\geq k\geq 0)\) are defined by
\begin{equation*} \lbrack r]:=[r]_{q}:=\left\{ \begin{array}{c} \tfrac {1-q^{r}}{1-q},\hspace{0.5cm}q\neq 1, \\ r,\hspace{1cm}q=1,\end{array}\right. \end{equation*}
\begin{equation*} \lbrack r]!:=\left\{ \begin{array}{c} \lbrack r][r-1]...[1],\ \ \ \hspace{0.5cm}q\geq 1, \\ \ \ \ 1,\ \ \ \ \ \ \ \ \ \ \hspace{0.5cm}q=1,\end{array}\right. \end{equation*}
and
\begin{equation*} \left[ \begin{array}{c} r \\ k\end{array}\right] :=\tfrac {[r]!}{[r-k]![k]!},\quad 0\leq k\leq r, \end{equation*}
respectively. The \(q\)-Jackson integral on the interval \([0,b]\) is defined as
\begin{align} \int \limits _{0}^{b}f(t)d_{q}t=(1-q)b\sum _{j=0}^{\infty }f(q^{j}b)q^{j},\text{ \ \ }0{\lt}q{\lt}1, \label{h} \end{align}
provided that the series is convergent. The Newton’s binomial formula is given by
\begin{equation} (1+x)(1+qx)...(1+q^{n-1}x)=\sum \limits _{k=0}^{n} \genfrac {[}{]}{0pt}{1}nk q^{k(k-1)/2}x^{k}. \label{4} \end{equation}
8
The Euler’s formula is
\begin{equation} \sum _{k=0}^{\infty }\tfrac {q^{k(k-1)/2}x^{k}}{(1-q)^{k}[k]!}=\prod _{k=0}^{\infty }(1+q^{k}x) \end{equation}
9
which can be derived from Newton’s binomial formula. Let \(0{\lt}q{\lt}1.\) We introduce the Kantorovich type \(q\)-Lupaş-Stancu operators as
\begin{equation} R_{n,q}^{(\alpha ,\beta )}(f;x)=\sum \limits _{k=0}^{n} \genfrac {[}{]}{0pt}{1}nk \tfrac {q^{k(k-1)/2}x^{k}(1-x)^{n-k}}{\prod \limits _{s=0}^{n-1}(1-x+q^{s}x)}\left( \int \limits _{0}^{1}f(\tfrac {[k]+q^{k}t+\alpha }{[n+1]+\beta })d_{q}t\right) \label{7} \end{equation}
10
where \(0\leq \alpha \leq \beta \) and \(f\in C[0,1].\)
Lemma
1
For all \(n\in \mathbb {N}\), \(x\in \lbrack 0,1]\) and \(0{\lt}q{\lt}1,\) we have the following equalities:
\begin{align} R_{n,q}^{(\alpha ,\beta )}(1;x) & =1 \notag \\ R_{n,q}^{(\alpha ,\beta )}(t;x) & =\tfrac {[n]}{[n+1]+\beta }\left\{ x+\tfrac {\alpha }{[n]}+\tfrac {1-x+q^{n}x}{[2][n]}\right\} \notag \\ & =\tfrac {1}{[n+1]+\beta }\left( \tfrac {2q}{[2]}[n]x+\alpha +\tfrac {1}{[2]}\right) \label{8} \\ R_{n,q}^{(\alpha ,\beta )}(t^{2};x) & =\tfrac {q}{1-x+qx}\tfrac {[n][n-1]}{\left( [n+1]+\beta \right) ^{2}}x^{2}+\left( 1+\tfrac {2q}{[2]}\tfrac {1-x+q^{n}x}{1-x+qx}+2\alpha \right) \tfrac {[n]}{\left( [n+1]+\beta \right) ^{2}}x \notag \\ & \quad +\left(\textstyle \alpha ^{2}+\textstyle \tfrac {2\alpha }{[2]}\textstyle {(1-x+q^{n}x)}+\tfrac {1}{[3]}\tfrac {\left( 1-x+q^{n}x\right) \left( 1-x+q^{n+1}x\right) }{1-x+qx}\right) \tfrac {1}{\left( [n+1]+\beta \right) ^{2}} \notag \end{align}
Taking \(\tfrac {x}{1-x}\) instead of \(x\) in (
8 ) one gets the first equality of (
11 ). Taking \(\tfrac {qx}{1-x}\) and \(\tfrac {q^{2}x}{1-x}\) instead of \(x\) in \((\ref{4})\) we have
\begin{align} & \textstyle \prod \limits _{s=1}^{n}(1-x+q^{s}x)=\sum \limits _{k=0}^{n} \genfrac {[}{]}{0pt}{1}nk q^{k(k-1)/2}\left( qx\right) ^{k}(1-x)^{n-k} \label{10} \\ & \textstyle \prod \limits _{s=2}^{n+1}(1-x+q^{s}x)=\sum \limits _{k=0}^{n} \genfrac {[}{]}{0pt}{1}nk q^{k(k-1)/2}\left( q^{2}x\right) ^{k}(1-x)^{n-k} \label{11} \end{align}
respectively. Using the definition of \(q\)-Jackson integral given in \(\left( \ref{h}\right) \) and the first equality of \(\left( \ref{3}\right) \), we can write
\begin{align*} R_{n,q}^{(\alpha ,\beta )}(t;x) =& \sum \limits _{k=0}^{n} \genfrac {[}{]}{0pt}{1}nk \tfrac {q^{k(k-1)/2}x^{k}(1-x)^{n-k}}{\prod \limits _{s=0}^{n-1}(1-x+q^{s}x)}\left(\int \limits _{0}^{1}\tfrac {[k]+q^{k}t+\alpha }{[n+1]+\beta }d_{q}t\right) \\ =& \tfrac {1}{[n+1]+\beta }\sum \limits _{k=0}^{n} \genfrac {[}{]}{0pt}{1}nk \tfrac {q^{k(k-1)/2}x^{k}(1-x)^{n-k}}{\prod \limits _{s=0}^{n-1}(1-x+q^{s}x)}\left( [k]+\tfrac {q^{k}}{[2]}+\alpha \right) \\ =& \tfrac {1}{[n+1]+\beta }\left\{ [n]x\sum \limits _{k=0}^{n-1} \genfrac {[}{]}{0pt}{1}{n-1}k \tfrac {q^{k(k-1)/2}\left( qx\right) ^{k}(1-x)^{n-k-1}}{\prod \limits _{s=0}^{n-1}(1-x+q^{s}x)}\right. \\ & \left. +\tfrac {1}{[2]}\sum \limits _{k=0}^{n} \genfrac {[}{]}{0pt}{1}nk \tfrac {q^{k(k-1)/2}\left( qx\right) ^{k}(1-x)^{n-k}}{\prod \limits _{s=0}^{n-1}(1-x+q^{s}x)}+\alpha \right\} \end{align*}
From the identity (12 ) we get the desired identity for \(R_{n,q}^{(\alpha ,\beta )}(t;x).\)
\begin{align*} & R_{n,q}^{(\alpha ,\beta )}(t^{2};x)= \\ & =\tfrac {1}{\left( [n+1]+\beta \right) ^{2}}\sum \limits _{k=0}^{n} \genfrac {[}{]}{0pt}{1}nk \tfrac {q^{k(k-1)/2}x^{k}(1-x)^{n-k}}{\prod \limits _{s=0}^{n-1}(1-x+q^{s}x)}\left( \int \limits _{0}^{1}\left( [k]+q^{k}t+\alpha \right) ^{2}d_{q}t\right) \\ & =\tfrac {1}{\left( [n+1]+\beta \right) ^{2}}\sum \limits _{k=0}^{n} \genfrac {[}{]}{0pt}{1}nk \tfrac {q^{k(k-1)/2}x^{k}(1-x)^{n-k}}{\prod \limits _{s=0}^{n-1}(1-x+q^{s}x)}\left\{ \textstyle \text{{\small $\left( [k]+\alpha \right) ^{2}$}}+\tfrac {2q^{k}}{[2]}\textstyle \text{{\small $\left( \alpha +[k]\right) $}}+\tfrac {q^{2k}}{[3]}\right\} \\ & =\tfrac {1}{\left( [n+1]+\beta \right) ^{2}}\left\{ qx^{2}[n][n-1]\sum \limits _{k=0}^{n-2} \genfrac {[}{]}{0pt}{1}{n-2}k \tfrac {q^{k(k-1)/2}\left( q^{2}x\right) ^{k}(1-x)^{n-k-2}}{\prod \limits _{s=0}^{n-1}(1-x+q^{s}x)}\right. \\ & \quad \left. +\left( {\small 2\alpha +1}\right) [n]x\sum \limits _{k=0}^{n-1} \genfrac {[}{]}{0pt}{1}{n-1}k \tfrac {q^{k(k-1)/2}\left( qx\right) ^{k}(1-x)^{n-k-1}}{\prod \limits _{s=0}^{n-1}(1-x+q^{s}x)}\right. \\ & \quad \left. +\alpha ^{2}+\tfrac {2\alpha }{[2]}\sum \limits _{k=0}^{n} \genfrac {[}{]}{0pt}{1}nk \tfrac {q^{k(k-1)/2}\left( qx\right) ^{k}(1-x)^{n-k}}{\prod \limits _{s=0}^{n-1}(1-x+q^{s}x)}\right. \\ & \quad \left. +\tfrac {2q}{[2]}[n]x\sum \limits _{k=0}^{n-1} \genfrac {[}{]}{0pt}{1}{n-1}k \tfrac {q^{k(k-1)/2}\left( q^{2}x\right) ^{k}(1-x)^{n-k-1}}{\prod \limits _{s=0}^{n-1}(1-x+q^{s}x)}\right. \\ & \quad \left. +\ \tfrac {1}{[3]}\sum \limits _{k=0}^{n} \genfrac {[}{]}{0pt}{1}nk \tfrac {q^{k(k-1)/2}\left( q^{2}x\right) ^{k}(1-x)^{n-k}}{\prod \limits _{s=0}^{n-1}(1-x+q^{s}x)}\right\} \end{align*}
Using the identities given in (12 ) and (13 ) we get
\begin{align*} R_{n,q}^{(\alpha ,\beta )}(t^{2};x)& =\tfrac {1}{\left( [n+1]+\beta \right) ^{2}}\left\{ \tfrac {qx^{2}[n][n-1]}{1-x+qx}\left( 2\alpha +1\right) [n]x+\alpha ^{2}+\tfrac {2\alpha }{[2]}\left( 1-x+q^{n}x\right) \right. \\ & \quad +\left. \tfrac {2q}{[2]}[n]x\tfrac {1-x+q^{n}x}{1-x+qx}+\tfrac {1}{[3]}\tfrac {\left( 1-x+q^{n}x\right) \left( 1-x+q^{n+1}x\right) }{(1-x+qx)}\right\} . \end{align*}
Arranging the terms we have the desired result.
The identity \(\left( \ref{12}\right) \) is obvious. For the inequality \(\left( \ref{13}\right) \) we use the following second central moment of the operator \(R_{n,q}^{(\alpha ,\beta )}(f;x).\)
\begin{align} & R_{n,q}^{(\alpha ,\beta )} (\left( t-x\right) ^{2};x)= \nonumber \\ =& \tfrac {q}{1-x+qx}\tfrac {[n][n-1]}{\left( [n+1]+\beta \right) ^{2}}x^{2} \notag \\ & +\left( 1+\tfrac {2q}{[2]}\tfrac {1-x+q^{n}x}{1-x+qx}+2\alpha \right) \tfrac {[n]}{\left( [n+1]+\beta \right) ^{2}}x \notag \\ & +\left\{ \left( \alpha ^{2}+\tfrac {2\alpha }{[2]}\left( 1-x+q^{n}x\right) \right) +\tfrac {1}{[3]}\tfrac {\left( 1-x+q^{n}x\right) \left( 1-x+q^{n+1}x\right) }{(1-x+qx)}\right\} \tfrac {1}{\left( [n+1]+\beta \right) ^{2}} \label{14} \\ & -2x\tfrac {[n]}{[n+1]+\beta }\left\{ x+\tfrac {\alpha }{[n]}+\tfrac {1-x+q^{n}x}{[2][n]}\right\} +x^{2} \notag \end{align}
For \(0{\lt}q{\lt}1\) and \(0\leq x\leq 1,\) we have \(\tfrac {q}{1-x+qx}\leq 1.\) Also using the inequality \([n-1]{\lt}[n]\) we can write
\begin{equation*} \left( \tfrac {q}{1-x+qx}\tfrac {[n][n-1]}{\left( [n+1]+\beta \right) ^{2}}-2\tfrac {[n]}{[n+1]+\beta }+1\right) x^{2}\leq \left( \tfrac {\lbrack n]}{[n+1]+\beta }-1\right) ^{2}x^{2}. \end{equation*}
Since \(\max \limits _{0\leq x\leq 1}\tfrac {\left( 1-x+q^{n}x\right) }{(1-x+qx)}=1\) and \(1-x+q^{n}x\leq 1\) \(,\) we have,
\begin{equation*} \left( 1+\tfrac {2q}{[2]}\tfrac {1-x+q^{n}x}{1-x+qx}+2\alpha \right) \leq 3+2\alpha \end{equation*}
and
\begin{align*} \left( \alpha ^{2}+\tfrac {2\alpha }{[2]}\left( 1-x+q^{n}x\right) \right) +\tfrac {1}{[3]}\tfrac {\left( 1-x+q^{n}x\right) \left( 1-x+q^{n+1}x\right) }{(1-x+qx)}& \leq \alpha ^{2}+2\alpha +1\\ & =\left( \alpha +1\right) ^{2}. \end{align*}
Using the above inequalities in \(\left( \ref{14}\right) \) and keeping in mind that \(0\leq x\leq 1,\) we finally get the desired result.
3 Direct Estimates
In this section, we give some direct theorems for the operators \(R_{n,q}^{(\alpha ,\beta )}(f;x).\) In what follows we denote by \(\left\Vert .\right\Vert =\left\Vert .\right\Vert _{C[0,1]}\) the uniform norm on \(C[0,1]. \)
Theorem
3
Let \(f\in C[0,1]\) and \(q:=\left( q_{n}\right) ,0{\lt}q_{n}{\lt}1\) be a sequence satisfying the condition
\begin{equation} \lim _{n\rightarrow \infty }q_{n}=1\text{.} \label{A} \end{equation}
17
Then we have
\begin{equation*} \lim _{n\rightarrow \infty }\left\Vert R_{n,q_{n}}^{(\alpha ,\beta )}(f;.)-f(.)\right\Vert =0. \end{equation*}
From Lemma 1 and Korovkin’s theorem, the proof is obvious because \([n]_{q_{n}}\rightarrow \infty \) as \(n\rightarrow \infty .\) Let \(f\in C[0,1].\) The modulus of continuity of \(f\) is defined by
\begin{equation*} w(f;\delta )=\sup \limits _{\substack { {t,x\in }\left[ 0,1\right] \\ \begin{bgroup} \left\vert t-x\right\vert {\leq \delta }
\end{bgroup}}}\left\vert f(t)-f(x)\right\vert . \end{equation*}
It is well known that for any \(\delta {\gt}0\) and each \(t\in \lbrack 0,1]\)
\begin{equation} \left\vert f(t)-f(x)\right\vert \leq w(f;\delta )\left( 1+\tfrac {\left\vert t-x\right\vert }{\delta }\right) \text{.} \label{15} \end{equation}
18
The next theorem gives us the rate of convergence of the operators \(R_{n,q}^{(\alpha ,\beta )}(f;x)\) in terms of modulus of continuity.
Theorem
4
If \(0{\lt}q{\lt}1\) , then for any \(f\in C[0,1]\), we have
\begin{equation*} \left\Vert R_{n,q}^{(\alpha ,\beta )}(f;x)-f(x)\right\Vert \leq 2w(f;\sqrt{\delta _{n,q}}) \end{equation*}
where \(\delta _{n,q}=\left( \tfrac {[n]}{\left( [n+1]+\beta \right) }-1\right) ^{2}+\left( 3+2\alpha \right) \tfrac {[n]}{\left( [n+1]+\beta \right) ^{2}}+\tfrac {\left( \alpha +1\right) ^{2}}{\left( [n+1]+\beta \right) ^{2}}+\tfrac {2\left( \alpha +1\right) }{\left( [n+1]+\beta \right) }. \)
We have
\begin{eqnarray} |R_{n,q}^{(\alpha ,\beta )}(f;x)-f(x)| & =& \left\vert \sum \limits _{k=0}^{n}b_{n,k}\left( q;x\right)\displaystyle \int \limits _{0}^{1}\left( f\left( \tfrac {[k]+q^{k}t+\alpha }{[n+1]+\beta }\right) -f(x)\right) d_{q}t\right\vert \notag \\ & \leq & \sum \limits _{k=0}^{n}b_{n,k}\left( q;x\right) \displaystyle \int \limits _{0}^{1}\left( \tfrac {\left\vert \frac{[k]+q^{k}t+\alpha }{[n+1]+\beta }-x\right\vert }{\delta }+1\right) w(f;\delta )d_{q}t \notag \end{eqnarray}
Using Cauchy-Schwarz inequality we have,
\begin{equation*} \int \limits _{0}^{1}\left\vert \tfrac {[k]+q^{k}t+\alpha }{[n+1]+\beta }-x\right\vert d_{q}t\leq \left\{ \int \limits _{0}^{1}\left( \tfrac {[k]+q^{k}t+\alpha }{[n+1]+\beta }-x\right) ^{2}d_{q}t\right\} ^{1/2} \end{equation*}
from which we can write
\begin{equation*} \sum \limits _{k=0}^{n}b_{n,k}\left( q;x\right) \displaystyle \int \limits _{0}^{1}\left( \left\vert \tfrac {[k]+q^{k}t+\alpha }{[n+1]+\beta }-x\right\vert d_{q}t\right) \leq \sum \limits _{k=0}^{n}b_{n,k}\left( q;x\right) \left\{ \displaystyle \int \limits _{0}^{1}\left( \tfrac {[k]+q^{k}t+\alpha }{[n+1]+\beta }-x\right) ^{2}d_{q}t\right\} ^{1/2} \end{equation*}
Applying Cauchy-Schwarz inequality once more, the right hand side of the above inequality becomes
\begin{align} \sum \limits _{k=0}^{n}b_{n,k}& \left( q;x\right) \left\{ \displaystyle \int \limits _{0}^{1}\left( \tfrac {[k]+q^{k}t+\alpha }{[n+1]+\beta }-x\right) ^{2}d_{q}t\right\} ^{1/2} \notag \\ & \leq \left\{ \sum \limits _{k=0}^{n}b_{n,k}\left( q;x\right) \displaystyle \int \limits _{0}^{1}\left( \tfrac {[k]+q^{k}t+\alpha }{[n+1]+\beta }-x\right) ^{2}d_{q}t\right\} ^{1/2}\left\{ \textstyle \sum \limits _{k=0}^{n}b_{n,k}\left( q;x\right) \right\} ^{1/2}. \notag \end{align}
Hence, by the first equality of \(\left( \ref{3}\right) ,\) we have
\begin{align*} |R_{n,q}^{(\alpha ,\beta )}(f;x)-f(x)| & \leq w(f;\delta )\left\{ 1+\tfrac {1}{\delta }\left[ \sum \limits _{k=0}^{n}b_{n,k}\left( q;x\right) \displaystyle \int \limits _{0}^{1}\left( \tfrac {[k]+q^{k}t+\alpha }{[n+1]+\beta }-x\right) ^{2}d_{q}t\right] ^{1/2}\right\} \\ & =w(f;\delta )\left\{ 1+\tfrac {1}{\delta }\left( R_{n,q}^{(\alpha ,\beta )}(\left( t-x\right) ^{2};x)\right) ^{1/2}\right\} \end{align*}
Taking maximum of both sides over the interval \([0,1],\) we have
\begin{equation*} \left\Vert R_{n,q}^{(\alpha ,\beta )}(f;.)-f(.)\right\Vert \leq w(f;\delta )\left\{ 1+\tfrac {1}{\delta }\left( \delta _{n,q}\right) ^{1/2}\right\} \end{equation*}
Choosing \(\delta =\left( \delta _{n,q}\right) ^{1/2}\) we get the result.
For \(0{\lt}\alpha \leq 1\), a function \(f\in C[0,1]\) belongs to \(\operatorname {Lip}_{M}\left( \alpha \right) \) if
\begin{equation*} \left\vert f(t)-f(x)\right\vert \leq M\left\vert t-x\right\vert ^{\alpha } \end{equation*}
is satisfied for some \(M{\gt}0\) and for all \(t,x\in \lbrack 0,1]\). The following theorem gives us the rate of convergence of the operators in terms of the functions of Lipschitz class.
Theorem
5
Let \(f\in \operatorname {Lip}_{M}\left( \alpha \right) \) and \(q:=(q_{n}),0{\lt}q_{n}{\lt}1\), be a sequence satisfying the conditions given in \(\left( \ref{A}\right) .\) Then
\begin{equation*} \left\Vert R_{n,q}^{(\alpha ,\beta )}(f;.)-f(.)\right\Vert \leq M\left( \delta _{n,q}\right) ^{\alpha /2} \end{equation*}
where \(\left( \delta _{n,q}\right) \) is given in Theorem 4.
By linearity and positivity of the operator and using the condition \(f\in \operatorname {Lip}_{M}\left( \alpha \right) ,\) we have
\begin{equation} |R_{n,q}^{(\alpha ,\beta )}(f;x)-f(x)|\leq M\sum \limits _{k=0}^{n}b_{n,k}\left( q;x\right) \int \limits _{0}^{1}\left\vert \tfrac {[k]+q^{k}t+\alpha }{[n+1]+\beta }-x\right\vert ^{\alpha }d_{q}t. \end{equation}
19
The Hölder’s inequality with \(p=\tfrac {2}{\alpha }\) and \(q=\tfrac {2}{2-\alpha }\) gives us
\begin{equation*} |R_{n,q}^{(\alpha ,\beta )}(f;x)-f(x)|\leq M\sum \limits _{k=0}^{n}b_{n,k}\left( q;x\right) \left( \int \limits _{0}^{1}\left( \tfrac {[k]+q^{k}t+\alpha }{[n+1]+\beta }-x\right) ^{2}d_{q}t\right) ^{\alpha /2}. \end{equation*}
Applying the Hölder’s inequality once more for the sum term, we obtain
\begin{align*} & |R_{n,q}^{(\alpha ,\beta )}(f;x)-f(x)| \leq \\ \leq & M\left( \sum \limits _{k=0}^{n}b_{n,k}\left( q;x\right) \displaystyle \int \limits _{0}^{1}\left( \tfrac {[k]+q^{k}t+\alpha }{[n+1]+\beta }-x\right) ^{2}d_{q}t\right) ^{\alpha /2} \left( \textstyle \sum \limits _{k=0}^{n}b_{n,k}\left( q;x\right) \right) ^{\left( 2-\alpha \right) /2} \\ =& M\left( R_{n,q}^{(\alpha ,\beta )}(\left( t-x\right) ^{2};x)\right) ^{\alpha /2} \end{align*}
Taking maximum of both sides of the above inequality over \([0,1]\), we get the desired result.
Lastly, we will study the rate of convergence of the operators \(R_{n,q}^{(\alpha ,\beta )}(f;x)\) by means of Peetre’s K-functionals. Remember that the Peetre’s K-functional is defined by
\begin{equation} K_{2}(f;\delta )=\inf _{g\in C^{2}[0,1]}\left\{ \left\Vert f-g\right\Vert +\delta \left\Vert g^{\prime \prime }\right\Vert \right\} . \label{K} \end{equation}
20
Recall that the second modulus of a function is defined by
\begin{equation*} w_{2}(f;\delta )=\sup _{0\leq h\leq \delta }\sup _{x\in \lbrack 0,1]}\left\vert f(x+2h)-2f(x+h)+f(x)\right\vert . \end{equation*}
It is known
[
11
,
p. 177, Th. 2.4
]
that there exists a positive constant \(C{\gt}0\) such that
\begin{equation} K_{2}(f;\delta )\leq Cw_{2}(f;\sqrt{\delta }). \label{L} \end{equation}
21
We need the following Lemma for the proof of the theorem on Peetre’s K-functional.
Lemma
6
For \(f\in C[0,1]\) and \(x\in \lbrack 0,1]\) one has
\begin{equation*} \left\vert R_{n;q}^{(\alpha ,\beta )}(f,x)\right\vert \leq \left\Vert f\right\Vert \end{equation*}
The proof follows from the linearity of the operator \(R_{n,q}^{(\alpha ,\beta )}(f,x)\) and from the first identity of Lemma 1.
Theorem
7
Let \(f\in C[0,1],x\in \lbrack 0,1]\) and \(0{\lt}q{\lt}1.\) Then there exist a positive constant \(C\) such that
\begin{equation*} \left\vert R_{n;q}^{(\alpha ,\beta )}(f,x)-f(x)\right\vert \leq Cw_{2}(f;\sqrt{\alpha _{n,q}})+w(f;\beta _{n,q}(x)) \end{equation*}
where \(\alpha _{n,q}=\delta _{n,q}+\tfrac {2}{\left( [n+1]+\beta \right) ^{2}}\left\{ 3\left( [2]\beta +1\right) ^{2}+2q^{2n+2}\right\} \) and
\(\beta _{n,q}(x)=\left\vert \tfrac {1}{[n+1]+\beta }\left( \tfrac {2q}{[2]}[n]x+\alpha +\tfrac {1}{[2]}\right) -x\right\vert \).
Consider the following auxiliary operators \(\widetilde{R}_{n;q}^{(\alpha ,\beta )}(f,x)\) defined by
\begin{equation} \widetilde{R}_{n;q}^{(\alpha ,\beta )}(f,x)=R_{n;q}^{(\alpha ,\beta )}(f,x)+f(x)-f\left( \tfrac {1}{[n+1]+\beta }\left( \tfrac {2q}{[2]}[n]x+\alpha +\tfrac {1}{[2]}\right) \right) \label{M} \end{equation}
22
Since \(R_{n;q}^{(\alpha ,\beta )}\) is linear, from Lemma 1,
\begin{equation*} \widetilde{R}_{n;q}^{(\alpha ,\beta )}(t-x;q,x)=0. \end{equation*}
By Taylor’s theorem we have
\begin{equation*} g(t)=g(x)+g^{\prime }(x)(t-x)+\textstyle \int \limits _{x}^{t}(t-u)g^{\prime \prime }(u)du. \end{equation*}
Applying \(\widetilde{R}_{n;q}^{(\alpha ,\beta )}\) to the both side of the above equality, we get
\begin{align*} & \widetilde{R}_{n;q}^{(\alpha ,\beta )}\left( g;x\right) -g(x) = \\ & =\widetilde{R}_{n;q}^{(\alpha ,\beta )}\left( \displaystyle \int \limits _{x}^{t}(t-u)g^{\prime \prime }(u)du;q,x\right) \\ & =R_{n;q}^{(\alpha ,\beta )}\left( \displaystyle \int \limits _{x}^{t}(t-u)g^{\prime \prime }(u)du\right) - \end{align*}
\begin{align*} & -\int \limits _{x}^{{\frac{1}{[n+1]+\beta }{\left( \frac{2q}{[2]}[n]x+\alpha +\frac{1}{[2]}\right)}}}\left( \tfrac {1}{[n+1]+\beta }\left( \tfrac {2q}{[2]}[n]x+\alpha +\tfrac {1}{[2]}\right) -u\right) g^{\prime \prime }(u)du \end{align*}
Hence we have
\begin{align} & \left\vert \widetilde{R}_{n;q}^{(\alpha ,\beta )}\left( g;x\right) -g(x)\right\vert \leq \left\Vert g^{\prime \prime }\right\Vert \left\{ R_{n;q}^{(\alpha ,\beta )}\left( \left\vert \displaystyle \int \limits _{x}^{t}(t-u)du\right\vert ;x\right) \right. \notag \\ & \left. +\left\vert \displaystyle \int \limits _{x}^{\frac{1}{[n+1]+\beta }\left( \frac{2q}{[2]}[n]x+\alpha +\frac{1}{[2]}\right) }\left( \tfrac {1}{[n+1]+\beta }\left( \tfrac {2q}{[2]}[n]x+\alpha +\tfrac {1}{[2]}\right) -u\right) du\right\vert \right\} \notag \\ & \leq \left\Vert g^{\prime \prime }\right\Vert \left\{ R_{n;q}^{(\alpha ,\beta )}\left( (t-x)^{2};x\right) +\left( \tfrac {1}{[n+1]+\beta }\left( \tfrac {2q}{[2]}[n]x+\alpha +\tfrac {1}{[2]}\right) -x\right) ^{2}\right\} \label{B} \end{align}
For the last term of the above inequality we can write
\begin{align*} & \left( \tfrac {1}{[n+1]+\beta }\left( \tfrac {2q}{[2]}[n]x+\alpha +\tfrac {1}{[2]}\right) -x\right) ^{2} \leq \\ & \leq 2\left\{ \left( \tfrac {2q[n]}{[2]\left( [n+1]+\beta \right) }-1\right) ^{2}x^{2}+\left( \tfrac {\alpha +\frac{1}{[2]}}{[n+1]+\beta }\right) ^{2}\right\} \\ & =\tfrac {2}{[2]^{2}\left( [n+1]+\beta \right) ^{2}}\left\{ \left( 1+q^{n+1}+[2]\beta \right) ^{2}x^{2}+\left( [2]\alpha +1\right) ^{2}\right\} \end{align*}
Since \(0\leq x\leq 1\) and \(\alpha \leq \beta ,\) we have
\begin{align*} & \left( \tfrac {1}{[n+1]+\beta }\left( \tfrac {2q}{[2]}[n]x+\alpha +\tfrac {1}{[2]}\right) -x\right) ^{2} \leq \\ & \leq \tfrac {2}{[2]^{2}\left( [n+1]+\beta \right) ^{2}}\left\{ 2\left( \left( [2]\beta +1\right) ^{2}+q^{2n+2}\right) +\left( [2]\beta +1\right) ^{2}\right\} \end{align*}
from \(\ \)which we get
\begin{equation} \left\vert \widetilde{R}_{n;q}^{(\alpha ,\beta )}\left( g;x\right) -g(x)\right\vert \leq \left\Vert g^{\prime \prime }\right\Vert \left\{ \delta _{n,q}+\tfrac {2}{\left( [n+1]+\beta \right) ^{2}}\left\{ 3\left( [2]\beta +1\right) ^{2}+2q^{2n+2}\right\} \right\} \label{17} \end{equation}
24
by \(\left( \ref{B}\right) .\) On the other hand from \(\left( \ref{M}\right) \) and Lemma 6, we have
\begin{eqnarray*} \left\vert \widetilde{R}_{n;q}^{(\alpha ,\beta )}\left( f;x\right) \right\vert & \leq & \left\vert R_{n;q}^{(\alpha ,\beta )}(f;x)\right\vert +2\left\Vert f\right\Vert \\ & \leq & 3\left\Vert f\right\Vert . \end{eqnarray*}
Thus, from \(\left( \ref{M}\right) \) and \(\left( \ref{17}\right) \) we can write
\begin{align*} & \textstyle {\left\vert R_{n;q}^{(\alpha ,\beta )}(f;x)-f(x)\right\vert }\leq \\ & \leq \left\vert \widetilde{R}_{n;q}^{(\alpha ,\beta )}\left( f;x\right) -f\left( x\right) \right\vert +\left\vert f\left( x\right) -f\left( \tfrac {1}{[n+1]+\beta }\left( \tfrac {2q}{[2]}[n]x+\alpha +\tfrac {1}{[2]}\right) \right) \right\vert \\ & \leq \left\vert \widetilde{R}_{n;q}^{(\alpha ,\beta )}\left( f-g;x\right) \right\vert +\left\vert \left( f-g\right) \left( x\right) \right\vert +\left\vert \widetilde{R}_{n;q}^{(\alpha ,\beta )}\left( g;x\right) -g\left( x\right) \right\vert \\ & \quad +\left\vert f\left( x\right) -f\left( \tfrac {1}{[n+1]+\beta }\left( \tfrac {2q}{[2]}[n]x+\alpha +\tfrac {1}{[2]}\right) \right) \right\vert \\ & \leq 4\left\Vert f-g\right\Vert +\left\Vert g^{\prime \prime }\right\Vert \alpha _{n,q}+\left\vert f\left( x\right) -f\left( \tfrac {1}{[n+1]+\beta }\left( \tfrac {2q}{[2]}[n]x+\alpha +\tfrac {1}{[2]}\right) \right) \right\vert \end{align*}
Taking the infimum on the right hand side over all \(g\in C^{2}[0,1]\) and using (20 ) and (21 ) we get
\begin{eqnarray*} \left\vert R_{n;q}^{(\alpha ,\beta )}(f;x)-f(x)\right\vert & \leq & 4K_{2}(f;\alpha _{n,q})+w(f;\beta _{n,q}(x)) \\ & \leq & Cw_{2}(f;\sqrt{\alpha _{n,q}})+w(f;\beta _{n,q}(x)). \end{eqnarray*}
which completes the proof.
4 Statistical convergence properties
Before proceeding further let us recall the concept of statistical convergence, which was first introduced by H. Fast
[
8
]
in 1951 and has been studied frequently in approximation theory for the last two decades.
The natural density of a set \(K\subseteq N\) is defined by
\begin{equation*} \delta (K)=\lim _{n}\tfrac {1}{n}\left\vert \{ k:\; k\leq n,\text{ }k\in K\} \right\vert \end{equation*}
provided the limit exists (see
[
15
]
); here \(\left\vert A\right\vert \) denotes the cardinality of the set \(A.\) A sequence \(x=(x_{k})\) is called statistically convergent to a number L if, for every \(\epsilon {\gt}0\)
\begin{equation*} \delta \{ k:\; |x_{k}-L|\geq \epsilon \} =0 \end{equation*}
and it is denoted as \(st-\displaystyle \lim _{k}x_{k}=L\).
A.D. Gadjiev and C. Orhan
[
9
]
proved the following Bohman-Korovkin type approximation theorem using the concept of the statistical convergence.
Theorem A
[
9
]
If the sequence of linear positive operators \(A_{n}:C[a,b]\rightarrow C[a,b]\) satisfies the conditions,
\begin{equation*} st-\lim _{n}\Vert A_{n}(e_{\nu };.)-e_{\nu }(.)\Vert _{C[a,b]}=0\; \; \; ,\; \; \; e_{\nu }(t)=t^{\nu } \end{equation*}
for \(\nu =0,1,2\), then for any function \(f\in C[a,b]\),
\begin{equation*} st-\lim _{n}\Vert A_{n}(f;.)-f(.)\Vert _{C[a,b]}=0. \end{equation*}
Theorem
9
Let \((q_{n})\), \(0{\lt}q_{n}{\lt}1\), be a sequence satisfying
\begin{equation} st-\lim _{n}q_{n}=1\; \; \; \; \text{and}\; \; \; \; st-\lim _{n}q_{n}^{n}=c\in (0,1). \label{18} \end{equation}
25
Then for all \(f\in C[0,1]\), the operator \(R_{n;q_{n}}^{(\alpha ,\beta )}\) satisfies
\begin{equation*} st-\lim _{n}\Vert R_{n;q_{n}}^{(\alpha ,\beta )}(f,.)-f(.)\Vert =0. \end{equation*}
It is enough to prove that
\begin{equation*} st-\lim _{n}\Vert R_{n;q_{n}}^{(\alpha ,\beta )}(e_{i};.)-e_{i}(.)\Vert =0, \end{equation*}
for \(e_{i}(t)=t^{i},i=0,1,2\), then the proof follows from Theorem A.
For \(i=0\), it is clear from the first identity of Lemma 1 that
\begin{equation} st-\displaystyle \lim _{n}\Vert R_{n;q_{n}}^{(\alpha ,\beta )}(e_{0};.)-e_{0}(.)\Vert =0. \label{21} \end{equation}
26
For \(i=1\), again Lemma 1 implies,
\begin{equation*} R_{n,q}^{(\alpha ,\beta )}(e_{1},x)-e_{1}(x)=\left( \tfrac {2q}{[2]}\tfrac {[n]}{[n+1]+\beta }-1\right) x+\left( \tfrac {\alpha +\frac{1}{[2]}}{[n+1]+\beta }\right) \end{equation*}
from which we can write
\begin{equation} |R_{n,q}^{(\alpha ,\beta )}(e_{1},x)-e_{1}(x)|\leq \tfrac {1+q^{n+1}+[2]\beta }{[2]([n+1]+\beta )}x+\tfrac {\alpha +\frac{1}{[2]}}{[n+1]+\beta } \label{23} \end{equation}
27
Now for a given \(\epsilon {\gt}0\), let us define the following sets:
\begin{equation*} T:=\{ n:\Vert R_{n,q_{n}}^{(\alpha ,\beta )}(e_{1},.)-e_{1}(.)\Vert \geq \epsilon \} , \end{equation*}
\begin{equation*} T_{1}:=\{ n:\tfrac {1+q^{n+1}+[2]\beta }{[2]([n+1]+\beta )}\geq \tfrac {\epsilon }{2}\} . \end{equation*}
\begin{equation*} T_{2}:=\{ n:\tfrac {\alpha +\frac{1}{[2]}}{[n+1]+\beta }\geq \tfrac {\epsilon }{2}\} . \end{equation*}
From (27 ) it is clear that \(T\subseteq T_{1}\cup T_{2}\). So we can write,
\begin{equation} \delta \left( T\right) \leq \delta \left( T_{1}\right) +\delta \left( T_{2}\right) \label{22***} \end{equation}
28
From the conditions (25 ), we have
\begin{equation*} st-\lim _{n}\tfrac {1+q^{n+1}+[2]\beta }{[2]([n+1]+\beta )}=0 \\ \ \text{and} \\ \ st-\lim _{n}\tfrac {\alpha +\frac{1}{[2]}}{[n+1]+\beta }=0 \end{equation*}
which implies that the right hand side of the inequality (28 ) is zero. Therefore we have,
\begin{equation*} \delta \{ n:\Vert R_{n,q_{n}}^{(\alpha ,\beta )}(e_{1},.)-e_{1}(.)\Vert \geq \epsilon \} =0 \end{equation*}
which implies
\begin{equation} st-\displaystyle \lim _{n}\Vert R_{n,q_{n}}^{(\alpha ,\beta )}(e_{1},.)-e_{1}(.)\Vert =0. \label{22**} \end{equation}
29
Lastly for \(i=2\) we can write
\begin{align} |R_{n,q}^{(\alpha ,\beta )}(e_{2},x)& -e_{2}(x)| =\left( 1-\tfrac {q}{1-x+qx}\tfrac {[n][n-1]}{\left( [n+1]+\beta \right) ^{2}}\right) x^{2} \notag \\ & +\left( 1+\tfrac {2q}{[2]}\tfrac {1-x+q^{n}x}{1-x+qx}+2\alpha \right) \tfrac {[n]}{\left( [n+1]+\beta \right) ^{2}}x \notag \\ & +\left( \alpha ^{2}+\tfrac {2\alpha }{[2]}(1-x+q^{n}x)+\tfrac {1}{[3]}\tfrac {\left( 1-x+q^{n}x\right) \left( 1-x+q^{n+1}x\right) }{1-x+qx}\right) \tfrac {1}{\left( [n+1]+\beta \right) ^{2}} \notag \end{align}
from which we have
\begin{align} \Vert R_{n,q}^{(\alpha ,\beta )}(e_{2},.)-e_{2}(.)\Vert \leq & \left( 1-q\tfrac {[n][n-1]}{\left( [n+1]+\beta \right) ^{2}}\right) \notag \\ & +\left( 3+2\alpha \right) \tfrac {[n]}{\left( [n+1]+\beta \right) ^{2}}+\left( \alpha ^{2}+\tfrac {2\alpha }{[2]}+\tfrac {1}{[3]}\right) \tfrac {1}{\left( [n+1]+\beta \right) ^{2}} \label{30} \end{align}
Now, for a given \(\epsilon {\gt}0\), let us define the following sets:
\begin{align*} & K:=\{ n:\Vert R_{n,q_{n}}^{(\alpha ,\beta )}(e_{2},.)-e_{2}(.)\Vert \geq \epsilon \} , \\ & K_{1}:=\{ n:\left( 1-q\tfrac {[n][n-1]}{\left( [n+1]+\beta \right) ^{2}}\right) \geq \tfrac {\epsilon }{3}\} \\ & K_{2}:=\{ n:\left( 3+2\alpha \right) \tfrac {[n]}{\left( [n+1]+\beta \right) ^{2}}\geq \tfrac {\epsilon }{3}\} \\ & K_{3}:=\{ n:\left( \alpha ^{2}+\tfrac {2\alpha }{[2]}+\tfrac {1}{[3]}\right) \tfrac {1}{\left( [n+1]+\beta \right) ^{2}}\geq \tfrac {\epsilon }{3}\} \end{align*}
From (30 ) it is clear that \(K\subseteq K_{1}\cup K_{2}\cup K_{3}\). Therefore we have,
\begin{equation} \delta \left( K\right) \leq \delta \left( K_{1}\right) +\delta \left( K_{2}\right) +\delta \left( K_{3}\right) . \label{24} \end{equation}
31
Taking the conditions given in (25 ) into account, one has
\begin{eqnarray} st-\lim _{n}\left( 1-q\tfrac {[n][n-1]}{\left( [n+1]+\beta \right) ^{2}}\right) & =& 0 \notag \label{24*} \\ st-\lim _{n}\left( 3+2\alpha \right) \tfrac {[n]}{\left( [n+1]+\beta \right) ^{2}} & =& 0 \\ st-\lim _{n}\left( \alpha ^{2}+\tfrac {2\alpha }{[2]}+\tfrac {1}{[3]}\right) \tfrac {1}{\left( [n+1]+\beta \right) ^{2}} & =& 0 \notag \end{eqnarray}
From (32 ) the right hand side of (31 ) becomes zero and hence we get
\begin{equation*} \delta \{ n:\Vert R_{n,q_{n}}^{(\alpha ,\beta )}(e_{2},.)-e_{2}(.)\Vert \geq \epsilon \} =0 \end{equation*}
i.e.,
\begin{equation} st-\lim _{n}\Vert R_{n,q_{n}}^{(\alpha ,\beta )}(e_{2},.)-e_{2}(.)\Vert =0. \label{24**} \end{equation}
33
Now by (26 ), (29 ) and (33 ) we conclude from Theorem A that for all \(f\in C[0,1]\)
\begin{equation*} st-\lim _{n}\Vert R_{n,q_{n}}^{(\alpha ,\beta )}(f,.)-f(.)\Vert =0. \end{equation*}
Example
10
Taking \(f(x)=x^{2},\) (curve 4), we compute the error estimation of \(q-\)Lupaş Kantorovich operators given by \(\left( \ref{7}\right) \) for \(q=0.5\) (curve 3), \(q=0.7\) (curve 2) and \(q=0.85\) (curve 1).
Table
1
Error estimates of \(R_{n,q_{n}}^{(\protect \alpha ,\protect \beta )}(f,x)\) for different values of \(q.\) (\(n=30\), \(\protect \alpha =1\) and \(\protect \beta =4.)\)
Figure
1
Estimation of the \(q\)-Lupaş-Kantorovich operators to the function \(f(x) = x^{2}\) for \(q = 0.5; 0.7\) and \(1\).
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