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<title>Approximation Theorems for Kantorovich type Lupaş-Stancu Operators based on \(q\)-integers: Approximation Theorems for Kantorovich type Lupaş-Stancu Operators based on \(q\)-integers</title>
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<h1>Approximation Theorems for Kantorovich type Lupaş-Stancu Operators based on \(q\)-integers</h1>
<p class="authors">
<span class="author">Sevilay KIRCI SERENBAY\(^\S \), Özge DALMANOĞLU\(^\ast \)</span>
</p>
<p class="date">January 24, 2017.</p>
</div>
<p>\(^\S \)Baskent University, Department of Mathematics Education, Ankara, Turkey, e-mail: <span class="ttfamily">sevilaykirci@gmail.com.</span> </p>
<p>\(^\ast \)Baskent University, Department of Mathematics Education, Ankara, Turkey, e-mail: <span class="ttfamily">ozgedalmanoglu@gmail.com</span> </p>

<div class="abstract"><p> 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. </p>
<p><b class="bf">MSC.</b> 41A35; 41A36 </p>
<p><b class="bf">Keywords.</b> Lupaş-Kantorovich operators, Modulus of continuity, Peetre’s K-functional, q-integers, rate of convergence, statistical approximation. </p>
</div>
<h1 id="a0000000002">1 Introduction</h1>
<p>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 </p>
<div class="equation" id="1">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p> where</p>
<div class="equation" id="2">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">2</span>
</p>
</div>
<p>is called Lupaş operators <span class="cite">
	[
	<a href="#Lupas" >12</a>
	]
</span>. 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: </p>
<div class="displaymath" id="a0000000003">
  \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}
</div>
<p>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 <span class="cite">
	[
	<a href="#Mahmudov2" >14</a>
	]
</span>  the authors studied Voronovskaja type theorems for the \(q\)-Lupaş operators for fixed \(q{\gt}0\). In <span class="cite">
	[
	<a href="#Ostrovska" >16</a>
	]
</span>, 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 <span class="cite">
	[
	<a href="#Dogru1" >5</a>
	]
</span>, Doğru and Kanat considered a King type modification of Lupaş operators and investigated the statistical approximation properties of the operators. Very recently, Doğru <i class="it">et al.</i> <span class="cite">
	[
	<a href="#Dogru2" >7</a>
	]
</span> introduced a Stancu type generalization of \(q\)-Lupaş operators as </p>
<div class="equation" id="6">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">4</span>
</p>
</div>
<p>where \(b_{n,k}(q;x)\) is given in (<a href="#2">2</a>). They studied the approximation properties and also introduced the \(r\)-th generalization of these operators. </p>
<p>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 <span class="cite">
	[
	<a href="#Mahmudov1" >13</a>
	]
</span> an integral modification, called Kantorovich type generalization of \(q\)-Bernstein operators, have been studied. The authors constructed the operators as </p>
<div class="displaymath" id="a0000000004">
  \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*}
</div>
<p>where \(f\in C[0,1],0{\lt}q{\lt}1\) and </p>
<div class="equation" id="2*">
<p>
  <div class="equation_content">
    \begin{equation}  p_{n,k}(q;{x})= \genfrac {[}{]}{0pt}{1}nk x^{k}(1-x)^{n-k} \label{2*} \end{equation}
  </div>
  <span class="equation_label">5</span>
</p>
</div>
<p>and studied some approximation properties of them. Özarslan and Vedi <span class="cite">
	[
	<a href="#Ozarslan1" >17</a>
	]
</span> introduced \(q\)-Bernstein-Schurer-Kantorovich operators as </p>
<div class="displaymath" id="a0000000005">
  \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*}
</div>
<p>Acu <i class="it">et al.</i> <span class="cite">
	[
	<a href="#Acu" >1</a>
	]
</span> introduced a new q-Stancu-Kantorovich operators as</p>
<div class="displaymath" id="a0000000006">
  \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*}
</div>
<p>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 <span class="cite">
	[
	<a href="#Acu2" >2</a>
	]
</span> where she gave the convergence theorems both in classical and statistical sense and obtained a Voronovskaya type result. </p>
<p>For every \(n\in N\) and \(q\in (0,1)\), Doğru and Kanat <span class="cite">
	[
	<a href="#Dogru3" >6</a>
	]
</span> defined the Kantorovich type modification of Lupaş operators as</p>
<div class="equation" id="5">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">6</span>
</p>
</div>
<p>Recently, Agrawal <i class="it">et al.</i> <span class="cite">
	[
	<a href="#Agrawal" >3</a>
	]
</span> studied the approximation properties of Lupaş-<br />Kantorovich operators based on Pólya distribution. </p>
<p>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. </p>
<h1 id="a0000000007">2 Construction of The Operators</h1>
<p>Before proceeding further we recall some basic notations from \(q\)-calculus (see <span class="cite">
	[
	<a href="#ARAL2" >4</a>
	]
</span> and <span class="cite">
	[
	<a href="#KAC" >10</a>
	]
</span>). </p>
<p>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 </p>
<div class="displaymath" id="a0000000008">
  \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*}
</div>
<div class="displaymath" id="a0000000009">
  \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*}
</div>
<p>and </p>
<div class="displaymath" id="a0000000010">
  \begin{equation*}  \left[ \begin{array}{c} r \\ k\end{array}\right] :=\tfrac {[r]!}{[r-k]![k]!},\quad 0\leq k\leq r, \end{equation*}
</div>
<p>respectively. The \(q\)-Jackson integral on the interval \([0,b]\) is defined as </p>
<div class="displaymath" id="h">
  \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}
</div>
<p>provided that the series is convergent. The Newton’s binomial formula is given by </p>
<div class="equation" id="4">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">8</span>
</p>
</div>
<p>The Euler’s formula is </p>
<div class="equation" id="a0000000011">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">9</span>
</p>
</div>
<p>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</p>
<div class="equation" id="7">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">10</span>
</p>
</div>
<p>where \(0\leq \alpha \leq \beta \) and \(f\in C[0,1].\) </p>
<p><div class="lemma_thmwrapper " id="Lemma1">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">1</span>
  </div>
  <div class="lemma_thmcontent">
  <p> For all \(n\in \mathbb {N}\), \(x\in \lbrack 0,1]\) and \(0{\lt}q{\lt}1,\) we have the following equalities:</p>
<div class="displaymath" id="a0000000012">
  \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}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000013">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Taking \(\tfrac {x}{1-x}\) instead of \(x\) in (<a href="#4">8</a>) one gets the first equality of (<a href="#8">11</a>). Taking \(\tfrac {qx}{1-x}\) and \(\tfrac {q^{2}x}{1-x}\) instead of \(x\) in \((\ref{4})\) we have</p>
<div class="displaymath" id="10">
  \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}
</div>
<p>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 </p>
<div class="displaymath" id="a0000000014">
  \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*}
</div>
<p>From the identity (<a href="#10">12</a>) we get the desired identity for \(R_{n,q}^{(\alpha ,\beta )}(t;x).\) </p>
<div class="displaymath" id="a0000000015">
  \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*}
</div>
<p>Using the identities given in (<a href="#10">12</a>) and (<a href="#11">13</a>) we get</p>
<div class="displaymath" id="a0000000016">
  \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*}
</div>
<p>Arranging the terms we have the desired result. </p>
<p><div class="remark_thmwrapper " id="a0000000017">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">2</span>
  </div>
  <div class="remark_thmcontent">
  <p>From Lemma 1 we have,</p>
<div class="equation" id="12">
<p>
  <div class="equation_content">
    \begin{equation}  R_{n,q}^{(\alpha ,\beta )}(t-x;x)=\tfrac {-\left( q^{n}+\beta \right) }{[n+1]+\beta }x+\tfrac {1}{[n+1]+\beta }\left\{  \alpha +\tfrac {1-x+q^{n}x}{[2]}\right\}  \label{12} \end{equation}
  </div>
  <span class="equation_label">14</span>
</p>
</div>
<div class="displaymath" id="a0000000018">
  \begin{align}  R_{n,q}^{(\alpha ,\beta )}(\left( t-x\right) ^{2};x)\leq & \left( \tfrac {\lbrack n]}{\left( [n+1]+\beta \right) }-1\right) ^{2}+\left( 3+2\alpha \right) \tfrac {[n]}{\left( [n+1]+\beta \right) ^{2}} \notag \\ &  +\tfrac {\left( \alpha +1\right) ^{2}}{\left( [n+1]+\beta \right) ^{2}}+\tfrac {2\left( \alpha +1\right) }{\left( [n+1]+\beta \right) } \label{13} \end{align}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000019">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>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 \(R_{n,q}^{(\alpha ,\beta )}(f;x).\)</p>
<div class="displaymath" id="a0000000020">
  \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}
</div>
<p>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 </p>
<div class="displaymath" id="a0000000021">
  \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*}
</div>
<p>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, </p>
<div class="displaymath" id="a0000000022">
  \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*}
</div>
<p>and </p>
<div class="displaymath" id="a0000000023">
  \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*}
</div>
<p>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. </p>
<h1 id="a0000000024">3 Direct Estimates</h1>
<p>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 \(C[0,1]. \) </p>
<p><div class="theorem_thmwrapper " id="a0000000025">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(f\in C[0,1]\) and \(q:=\left( q_{n}\right) ,0{\lt}q_{n}{\lt}1\) be a sequence satisfying the condition </p>
<div class="equation" id="A">
<p>
  <div class="equation_content">
    \begin{equation}  \lim _{n\rightarrow \infty }q_{n}=1\text{.} \label{A} \end{equation}
  </div>
  <span class="equation_label">17</span>
</p>
</div>
<p>Then we have </p>
<div class="displaymath" id="a0000000026">
  \begin{equation*}  \lim _{n\rightarrow \infty }\left\Vert R_{n,q_{n}}^{(\alpha ,\beta )}(f;.)-f(.)\right\Vert =0. \end{equation*}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000027">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>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 </p>
<div class="displaymath" id="a0000000028">
  \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*}
</div>
<p>It is well known that for any \(\delta {\gt}0\) and each \(t\in \lbrack 0,1]\)</p>
<div class="equation" id="15">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">18</span>
</p>
</div>
<p>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. </p>
<p><div class="theorem_thmwrapper " id="a0000000029">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">4</span>
  </div>
  <div class="theorem_thmcontent">
  <p>If \(0{\lt}q{\lt}1\) , then for any \(f\in C[0,1]\), we have</p>
<div class="displaymath" id="a0000000030">
  \begin{equation*}  \left\Vert R_{n,q}^{(\alpha ,\beta )}(f;x)-f(x)\right\Vert \leq 2w(f;\sqrt{\delta _{n,q}}) \end{equation*}
</div>
<p>where \(\delta _{n,q}=\left( \tfrac {[n]}{\left( [n+1]+\beta \right) }-1\right) ^{2}+\left( 3+2\alpha \right) \tfrac {[n]}{\left( [n+1]+\beta \right) ^{2}}+\tfrac {\left( \alpha +1\right) ^{2}}{\left( [n+1]+\beta \right) ^{2}}+\tfrac {2\left( \alpha +1\right) }{\left( [n+1]+\beta \right) }. \) </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000031">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>We have </p>
<div class="displaymath" id="a0000000032">
  \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}
</div>
<p>Using Cauchy-Schwarz inequality we have, </p>
<div class="displaymath" id="a0000000033">
  \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*}
</div>
<p>from which we can write </p>
<div class="displaymath" id="a0000000034">
  \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*}
</div>
<p>Applying Cauchy-Schwarz inequality once more, the right hand side of the above inequality becomes</p>
<div class="displaymath" id="a0000000035">
  \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}
</div>
<p>Hence, by the first equality of \(\left( \ref{3}\right) ,\) we have </p>
<div class="displaymath" id="a0000000036">
  \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*}
</div>
<p>Taking maximum of both sides over the interval  \([0,1],\) we have</p>
<div class="displaymath" id="a0000000037">
  \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*}
</div>
<p>Choosing \(\delta =\left( \delta _{n,q}\right) ^{1/2}\)  we get the result. </p>
<p>For \(0{\lt}\alpha \leq 1\), a function \(f\in C[0,1]\) belongs to \(\operatorname {Lip}_{M}\left( \alpha \right) \) if</p>
<div class="displaymath" id="a0000000038">
  \begin{equation*}  \left\vert f(t)-f(x)\right\vert \leq M\left\vert t-x\right\vert ^{\alpha } \end{equation*}
</div>
<p>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. </p>
<p><div class="theorem_thmwrapper " id="a0000000039">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">5</span>
  </div>
  <div class="theorem_thmcontent">
  <p>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</p>
<div class="displaymath" id="a0000000040">
  \begin{equation*}  \left\Vert R_{n,q}^{(\alpha ,\beta )}(f;.)-f(.)\right\Vert \leq M\left( \delta _{n,q}\right) ^{\alpha /2} \end{equation*}
</div>
<p>where \(\left( \delta _{n,q}\right) \) is given  in Theorem 4. </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000041">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>By linearity and positivity of the operator and using the condition \(f\in \operatorname {Lip}_{M}\left( \alpha \right) ,\)  we have</p>
<div class="equation" id="a0000000042">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">19</span>
</p>
</div>
<p>The Hölder’s inequality with \(p=\tfrac {2}{\alpha }\) and \(q=\tfrac {2}{2-\alpha }\) gives us </p>
<div class="displaymath" id="a0000000043">
  \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*}
</div>
<p>Applying the Hölder’s inequality once more for the sum term, we obtain</p>
<div class="displaymath" id="a0000000044">
  \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*}
</div>
<p>Taking maximum of both sides of the above inequality over \([0,1]\), we get the desired result. </p>
<p>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</p>
<div class="equation" id="K">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">20</span>
</p>
</div>
<p>Recall that the second modulus of a function is defined by </p>
<div class="displaymath" id="a0000000045">
  \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*}
</div>
<p>It is known <span class="cite">
	[
	<a href="#Lorentz" >11</a>
	, 
	p. 177, Th. 2.4
	]
</span> that there exists a positive constant \(C{\gt}0\) such that</p>
<div class="equation" id="L">
<p>
  <div class="equation_content">
    \begin{equation}  K_{2}(f;\delta )\leq Cw_{2}(f;\sqrt{\delta }). \label{L} \end{equation}
  </div>
  <span class="equation_label">21</span>
</p>
</div>
<p>We need the following Lemma for the proof of the theorem on Peetre’s K-functional. </p>
<p><div class="lemma_thmwrapper " id="a0000000046">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">6</span>
  </div>
  <div class="lemma_thmcontent">
  <p>For \(f\in C[0,1]\) and \(x\in \lbrack 0,1]\) one has</p>
<div class="displaymath" id="a0000000047">
  \begin{equation*}  \left\vert R_{n;q}^{(\alpha ,\beta )}(f,x)\right\vert \leq \left\Vert f\right\Vert \end{equation*}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000048">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>The proof follows from the linearity of the operator \(R_{n,q}^{(\alpha ,\beta )}(f,x)\) and from the first identity of Lemma 1. </p>
<p><div class="theorem_thmwrapper " id="a0000000049">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">7</span>
  </div>
  <div class="theorem_thmcontent">
  <p>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 </p>
<div class="displaymath" id="a0000000050">
  \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*}
</div>
<p>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 </p>
<p>\(\beta _{n,q}(x)=\left\vert \tfrac {1}{[n+1]+\beta }\left( \tfrac {2q}{[2]}[n]x+\alpha +\tfrac {1}{[2]}\right) -x\right\vert \). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000051">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Consider the following auxiliary operators \(\widetilde{R}_{n;q}^{(\alpha ,\beta )}(f,x)\) defined by</p>
<div class="equation" id="M">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">22</span>
</p>
</div>
<p>Since \(R_{n;q}^{(\alpha ,\beta )}\) is linear, from Lemma 1,</p>
<div class="displaymath" id="a0000000052">
  \begin{equation*}  \widetilde{R}_{n;q}^{(\alpha ,\beta )}(t-x;q,x)=0. \end{equation*}
</div>
<p>By Taylor’s theorem we have</p>
<div class="displaymath" id="a0000000053">
  \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*}
</div>
<p>Applying \(\widetilde{R}_{n;q}^{(\alpha ,\beta )}\) to the both side of the above equality, we get </p>
<div class="displaymath" id="a0000000054">
  \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*}
</div>
<div class="displaymath" id="a0000000055">
  \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*}
</div>
<p> Hence we have </p>
<div class="displaymath" id="a0000000056">
  \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}
</div>
<p>For the last term of the above inequality we can write</p>
<div class="displaymath" id="a0000000057">
  \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*}
</div>
<p>Since \(0\leq x\leq 1\) and \(\alpha \leq \beta ,\) we have</p>
<div class="displaymath" id="a0000000058">
  \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*}
</div>
<p>from \(\  \)which we get</p>
<div class="equation" id="17">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">24</span>
</p>
</div>
<p>by \(\left( \ref{B}\right) .\) On the other hand from \(\left( \ref{M}\right) \) and Lemma 6, we have </p>
<div class="displaymath" id="a0000000059">
  \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*}
</div>
<p>Thus, from \(\left( \ref{M}\right) \) and  \(\left( \ref{17}\right) \) we can write<small class="small"><div class="displaymath" id="a0000000060">
  \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*}
</div> </small>Taking the infimum on the right hand side over all \(g\in C^{2}[0,1]\) and using (<a href="#K">20</a>) and (<a href="#L">21</a>) we get </p>
<div class="displaymath" id="a0000000061">
  \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*}
</div>
<p>which completes the proof. </p>
<p><div class="remark_thmwrapper " id="a0000000062">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">8</span>
  </div>
  <div class="remark_thmcontent">
  <p>For \(q_{n}\rightarrow 1\) as \(n\rightarrow \infty ,\) we have \(\alpha _{n,q_{n}}\rightarrow 0\) and \(\beta _{n,q_{n}}(x)\rightarrow 0.\) </p>

  </div>
</div> </p>
<h1 id="a0000000063">4 Statistical convergence properties</h1>
<p>Before proceeding further let us recall the concept of statistical convergence, which was first introduced by H. Fast <span class="cite">
	[
	<a href="#FAST" >8</a>
	]
</span> in 1951 and has been studied frequently in approximation theory for the last two decades. </p>
<p>The natural density of a set \(K\subseteq N\) is defined by </p>
<div class="displaymath" id="a0000000064">
  \begin{equation*}  \delta (K)=\lim _{n}\tfrac {1}{n}\left\vert \{ k:\; k\leq n,\text{ }k\in K\}  \right\vert \end{equation*}
</div>
<p>provided the limit exists (see <span class="cite">
	[
	<a href="#NIV" >15</a>
	]
</span>); 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 \(\epsilon {\gt}0\) </p>
<div class="displaymath" id="a0000000065">
  \begin{equation*}  \delta \{ k:\; |x_{k}-L|\geq \epsilon \} =0 \end{equation*}
</div>
<p>and it is denoted as \(st-\displaystyle \lim _{k}x_{k}=L\).<br /></p>
<p>A.D. Gadjiev and C. Orhan <span class="cite">
	[
	<a href="#GAD" >9</a>
	]
</span> proved the following Bohman-Korovkin type approximation theorem using the concept of the statistical convergence. </p>
<p><div class="theorema_thmwrapper " id="a0000000066">
  <div class="theorema_thmheading">
    <span class="theorema_thmcaption">
    Theorem A
    </span>
  </div>
  <div class="theorema_thmcontent">
  <p><span class="cite">
	[
	<a href="#GAD" >9</a>
	]
</span> If the sequence of linear positive operators \(A_{n}:C[a,b]\rightarrow C[a,b]\) satisfies the conditions, </p>
<div class="displaymath" id="a0000000067">
  \begin{equation*}  st-\lim _{n}\Vert A_{n}(e_{\nu };.)-e_{\nu }(.)\Vert _{C[a,b]}=0\;  \;  \; ,\;  \;  \; e_{\nu }(t)=t^{\nu } \end{equation*}
</div>
<p>for \(\nu =0,1,2\), then for any function \(f\in C[a,b]\), </p>
<div class="displaymath" id="a0000000068">
  \begin{equation*}  st-\lim _{n}\Vert A_{n}(f;.)-f(.)\Vert _{C[a,b]}=0. \end{equation*}
</div>

  </div>
</div> </p>
<p><div class="theorem_thmwrapper " id="T1">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">9</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \((q_{n})\), \(0{\lt}q_{n}{\lt}1\), be a sequence satisfying </p>
<div class="equation" id="18">
<p>
  <div class="equation_content">
    \begin{equation}  st-\lim _{n}q_{n}=1\;  \;  \;  \;  \text{and}\;  \;  \;  \; st-\lim _{n}q_{n}^{n}=c\in (0,1). \label{18} \end{equation}
  </div>
  <span class="equation_label">25</span>
</p>
</div>
<p>Then for all \(f\in C[0,1]\), the operator \(R_{n;q_{n}}^{(\alpha ,\beta )}\) satisfies </p>
<div class="displaymath" id="a0000000069">
  \begin{equation*}  st-\lim _{n}\Vert R_{n;q_{n}}^{(\alpha ,\beta )}(f,.)-f(.)\Vert =0. \end{equation*}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000070">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>It is enough to prove that </p>
<div class="displaymath" id="a0000000071">
  \begin{equation*}  st-\lim _{n}\Vert R_{n;q_{n}}^{(\alpha ,\beta )}(e_{i};.)-e_{i}(.)\Vert =0, \end{equation*}
</div>
<p>for \(e_{i}(t)=t^{i},i=0,1,2\), then the proof follows from Theorem A. </p>
<p>For \(i=0\), it is clear from the first identity of Lemma 1 that </p>
<div class="equation" id="21">
<p>
  <div class="equation_content">
    \begin{equation}  st-\displaystyle \lim _{n}\Vert R_{n;q_{n}}^{(\alpha ,\beta )}(e_{0};.)-e_{0}(.)\Vert =0. \label{21} \end{equation}
  </div>
  <span class="equation_label">26</span>
</p>
</div>
<p>For \(i=1\), again Lemma 1 implies, </p>
<div class="displaymath" id="a0000000072">
  \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*}
</div>
<p>from which we can write </p>
<div class="equation" id="23">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">27</span>
</p>
</div>
<p>Now for a given \(\epsilon {\gt}0\), let us define the following sets: </p>
<div class="displaymath" id="a0000000073">
  \begin{equation*}  T:=\{ n:\Vert R_{n,q_{n}}^{(\alpha ,\beta )}(e_{1},.)-e_{1}(.)\Vert \geq \epsilon \} , \end{equation*}
</div>
<div class="displaymath" id="a0000000074">
  \begin{equation*}  T_{1}:=\{ n:\tfrac {1+q^{n+1}+[2]\beta }{[2]([n+1]+\beta )}\geq \tfrac {\epsilon }{2}\} . \end{equation*}
</div>
<div class="displaymath" id="a0000000075">
  \begin{equation*}  T_{2}:=\{ n:\tfrac {\alpha +\frac{1}{[2]}}{[n+1]+\beta }\geq \tfrac {\epsilon }{2}\} . \end{equation*}
</div>
<p>From (<a href="#23">27</a>) it is clear that \(T\subseteq T_{1}\cup T_{2}\). So we can write, </p>
<div class="equation" id="22***">
<p>
  <div class="equation_content">
    \begin{equation}  \delta \left( T\right) \leq \delta \left( T_{1}\right) +\delta \left( T_{2}\right) \label{22***} \end{equation}
  </div>
  <span class="equation_label">28</span>
</p>
</div>
<p>From the conditions (<a href="#18">25</a>), we have </p>
<div class="displaymath" id="a0000000076">
  \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*}
</div>
<p>which implies that the right hand side of the inequality (<a href="#22***">28</a>) is zero. Therefore we have, </p>
<div class="displaymath" id="a0000000077">
  \begin{equation*}  \delta \{ n:\Vert R_{n,q_{n}}^{(\alpha ,\beta )}(e_{1},.)-e_{1}(.)\Vert \geq \epsilon \} =0 \end{equation*}
</div>
<p>which implies </p>
<div class="equation" id="22**">
<p>
  <div class="equation_content">
    \begin{equation}  st-\displaystyle \lim _{n}\Vert R_{n,q_{n}}^{(\alpha ,\beta )}(e_{1},.)-e_{1}(.)\Vert =0. \label{22**} \end{equation}
  </div>
  <span class="equation_label">29</span>
</p>
</div>
<p>Lastly for \(i=2\) we can write </p>
<div class="displaymath" id="a0000000078">
  \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}
</div>
<p>from which we have </p>
<div class="displaymath" id="a0000000079">
  \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}
</div>
<p>Now, for a given \(\epsilon {\gt}0\), let us define the following sets: </p>
<div class="displaymath" id="a0000000080">
  \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*}
</div>
<p>From (<a href="#30">30</a>) it is clear that \(K\subseteq K_{1}\cup K_{2}\cup K_{3}\). Therefore we have, </p>
<div class="equation" id="24">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">31</span>
</p>
</div>
<p>Taking the conditions given in (<a href="#18">25</a>) into account, one has </p>
<div class="displaymath" id="24*">
  \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}
</div>
<p>From (<a href="#24*">32</a>) the right hand side of (<a href="#24">31</a>) becomes zero and hence we get </p>
<div class="displaymath" id="a0000000081">
  \begin{equation*}  \delta \{ n:\Vert R_{n,q_{n}}^{(\alpha ,\beta )}(e_{2},.)-e_{2}(.)\Vert \geq \epsilon \} =0 \end{equation*}
</div>
<p>i.e., </p>
<div class="equation" id="24**">
<p>
  <div class="equation_content">
    \begin{equation}  st-\lim _{n}\Vert R_{n,q_{n}}^{(\alpha ,\beta )}(e_{2},.)-e_{2}(.)\Vert =0. \label{24**} \end{equation}
  </div>
  <span class="equation_label">33</span>
</p>
</div>
<p>Now by (<a href="#21">26</a>), (<a href="#22**">29</a>) and (<a href="#24**">33</a>) we conclude from Theorem A that for all \(f\in C[0,1]\) </p>
<div class="displaymath" id="a0000000082">
  \begin{equation*}  st-\lim _{n}\Vert R_{n,q_{n}}^{(\alpha ,\beta )}(f,.)-f(.)\Vert =0. \end{equation*}
</div>

<p><div class="example_thmwrapper " id="a0000000083">
  <div class="example_thmheading">
    <span class="example_thmcaption">
    Example
    </span>
    <span class="example_thmlabel">10</span>
  </div>
  <div class="example_thmcontent">
  <p>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). </p>

  </div>
</div> </p>
<div class="table"  id="fig.h=0.001">
   <small class="footnotesize"><table class="tabular">
  <tr>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> x </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> Error bound for \(q=0.5\) </p>

    </td>
    <td  style="text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> Error bound for \(q=0.7\) </p>

    </td>
    <td  style="text-align:left" 
        rowspan=""
        colspan="">
      <p> Error bound for \(q=0.85\) </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>0 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.6574686544 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.6508325077 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> 0.4855162530 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>0.3 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.0569703701 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.3717804595 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> 0.2747085315 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>0.5 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.1057573294 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.0748125294 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> 0.0569703701 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p>0.8 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.0476530351 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black" 
        rowspan=""
        colspan="">
      <p> 0.0418758717 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left" 
        rowspan=""
        colspan="">
      <p> 0.0211787652 </p>

    </td>
  </tr>
  <tr>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p>1 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0.1345946106 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-right:1px solid black; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0.1266259064 </p>

    </td>
    <td  style="border-top-style:solid; border-top-color:black; border-top-width:1px; text-align:left; border-bottom-style:solid; border-bottom-color:black; border-bottom-width:1px" 
        rowspan=""
        colspan="">
      <p> 0.05804583687 </p>

    </td>
  </tr>
</table> </small> <figcaption>
  <span class="caption_title">Table</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">Error estimates of \(R_{n,q_{n}}^{(\protect \alpha ,\protect \beta )}(f,x)\) for different values of \(q.\) (\(n=30\), \(\protect \alpha =1\) and \(\protect \beta =4.)\) </span> 
</figcaption> 
</div>
<figure id="a0000000084">
  
<div class="centered"> <img src="img-0001.png" alt="\includegraphics[height=3.3572in, width=4.0776in]{mapleportal_2.png}" style="height:3.3572in; width:4.0776in" />
 </div>
<figcaption>
  <span class="caption_title">Figure</span> 
  <span class="caption_ref">1</span> 
  <span class="caption_text">Estimation of the \(q\)-Lupaş-Kantorovich operators to the function \(f(x) = x^{2}\) for \(q = 0.5; 0.7\) and \(1\).</span> 
</figcaption>

</figure>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
<dl class="bibliography">
  <dt><a name="Acu">1</a></dt>
  <dd><p><span class="scshape">A.M. Acu, D. Bărbosu</span> and <span class="scshape">D.F. Sofonea</span>, <i class="itshape">Note on a q-analogue of Stancu-Kantorovich operators</i>, Miskolc Mathematical Notes, <b class="bfseries">16</b> (2015) 1, pp.&#160;3–15. </p>
</dd>
  <dt><a name="Acu2">2</a></dt>
  <dd><p><span class="scshape">A.M. Acu</span>, <i class="itshape">Stancu-Schurer-Kantorovich operators based on q-integers</i>, Applied Mathematics and Computation, <b class="bfseries">259</b> (2015), pp.&#160;896–907. </p>
</dd>
  <dt><a name="Agrawal">3</a></dt>
  <dd><p><a href ="https://doi.org/10.1007/s1221"> <span class="scshape">P.N. Agrawal</span>, <span class="scshape">N. Ispir</span> and <span class="scshape">A. Kajla</span> <i class="itshape">Approximation properties of Lupaş-Kantorovich operators based on Pólya distribution</i>, Rend. Circ. Mat. Palermo (2016) 65: pp.&#160;185–208. <img src="img-0002.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="ARAL2">4</a></dt>
  <dd><p><span class="scshape">A. Aral</span>, <span class="scshape">V. Gupta</span>, <span class="scshape">R. P. Agarwal</span>, <i class="itshape">Applications of q-Calculus in Operator Theory</i>, Springer, Berlin, 2013. </p>
</dd>
  <dt><a name="Dogru1">5</a></dt>
  <dd><p><a href ="https://doi.org/10.1016/j.camwa.2011.12.033"> <span class="scshape">O. Doğru</span> and <span class="scshape">K. Kanat</span>, <i class="itshape">Statistical approximation properties of  King-type modification of Lupaş operators</i>, Comput. Math. Appl., 64 (2012) pp.&#160;511–517. <img src="img-0002.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="Dogru3">6</a></dt>
  <dd><p><a href ="https://doi.org/10.1016/j.mcm.2011.10.059"> <span class="scshape">O. Doğru</span> and <span class="scshape">K. Kanat</span>, <i class="itshape">On statistical approximation properties of the Kantorovich type Lupaş operators</i>, Mathematical and Computer Modelling, 55, 3–4, (2012), pp.&#160;1610–1621. <img src="img-0002.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="Dogru2">7</a></dt>
  <dd><p><a href ="http://www.jstor.org/stable/24898690"> <span class="scshape">O. Doğru</span>, <span class="scshape">G.  Içöz</span>  and <span class="scshape">K. Kanat</span>, <i class="itshape">On the rates of convergence of the q-Lupaş-Stancu operators</i>, Filomat 30:5 (2016), pp.&#160;1151–1160. <img src="img-0002.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="FAST">8</a></dt>
  <dd><p><span class="scshape">H. Fast</span>, <i class="itshape">Sur la convergence statistique</i>, Colloq. Math Studia Mathematica, <b class="bfseries">2</b> (1951), pp.&#160;241–244. </p>
</dd>
  <dt><a name="GAD">9</a></dt>
  <dd><p><span class="scshape">A.D. Gadjiev</span> and <span class="scshape">C. Orhan</span>, <i class="itshape">Some approximation properties via statistical convergence</i>, Rocky Mountain J. Math., <b class="bfseries">32</b> (2002), pp.&#160;129–138. </p>
</dd>
  <dt><a name="KAC">10</a></dt>
  <dd><p><span class="scshape">V.&#160;Kac</span> and <span class="scshape">P.&#160;Cheung</span>, <i class="itshape">Quantum Calculus</i>, Springer–Verlag, New York–Berlin–Heidelberg, 1953. </p>
</dd>
  <dt><a name="Lorentz">11</a></dt>
  <dd><p><span class="scshape">R.A. DeVore</span> and <span class="scshape">G.G. Lorentz</span>, <i class="itshape">Constructive Approximation</i>, Springer-Verlag, Berlin, 1993.</p>
</dd>
  <dt><a name="Lupas">12</a></dt>
  <dd><p><span class="scshape">A. Lupaş</span>, <i class="itshape">A q-analogue of the Bernstein operator</i>, University of Cluj-Napoca, Seminar on Numerical and Statistical Calculus, 9 (1987), pp.&#160;85–92. </p>
</dd>
  <dt><a name="Mahmudov1">13</a></dt>
  <dd><p><span class="scshape">N.I. Mahmudov</span> and <span class="scshape">P. Sabancıgil</span>, <i class="itshape">Approximation theorems for q-Bernstein-Kantorovich operators</i>, Filomat 27:4 (2013) pp.&#160;721–730. </p>
</dd>
  <dt><a name="Mahmudov2">14</a></dt>
  <dd><p><span class="scshape">N.I. Mahmudov</span> and <span class="scshape">P. Sabancıgil</span>, <i class="itshape">Voronovskaja type theorem for the Lupaş \(q\)-analogue of the Bernstein operators</i>, Math. Commun. 17, (2012) pp.&#160;83–91. </p>
</dd>
  <dt><a name="NIV">15</a></dt>
  <dd><p><span class="scshape">Niven I</span>. , <span class="scshape">Zuckerman H.S.</span> and <span class="scshape">Montgomery H.</span>, <i class="itshape">An Introduction to the Theory of Numbers</i>, \(5^{th}\) edition, Wiley, New York, 1991. </p>
</dd>
  <dt><a name="Ostrovska">16</a></dt>
  <dd><p><span class="scshape">S. Ostrovska</span>, <i class="itshape">On the Lupaş q-analogue of the Bernstein operator</i>, Rocky Mountain J. Math., 36 5 (2006), pp.&#160;1615–1629. </p>
</dd>
  <dt><a name="Ozarslan1">17</a></dt>
  <dd><p><span class="scshape">M.A. Özarslan</span> and <span class="scshape">T. Vedi</span>, <i class="itshape">q-Bernstein-Schurer-Kantorovich operators</i>, J. Ineq. Appl.,(2013) p.&#160;444. </p>
</dd>
</dl>


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