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<title>An extension of the Cheney-Sharma operator <br />of the first kind: An extension of the Cheney-Sharma operator <br />of the first kind</title>
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<h1>An extension of the Cheney-Sharma operator <br />of the first kind</h1>
<p class="authors">
<span class="author">Teodora Cătinaş\(^1\) Iulia Buda\(^{1,2}\)</span>
</p>
<p class="date">September 18, 2023; accepted: October 27, 2023; published online: December 27, 2023.</p>
<p class="thanks">\(^1\)Babeş-Bolyai University, Faculty of Mathematics and Computer Science, Str. M. Kogălniceanu Nr. 1, RO-400084 Cluj-Napoca, Romania, e-mail: <span class="tt">teodora.catinas@ubbcluj.ro</span>.<br />\(^2\)Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy, Cluj-Napoca, Romania, e-mail: <span class="tt">iulia.buda@ictp.acad.ro; iulia.mircescu@ubbcluj.ro</span>.</p>
</div>
<div class="abstract"> We extend the Cheney-Sharma operators of the first kind using Stancu type technique and we study some approximation properties of the new operator. We calculate the moments, we study local approximation with respect to a K-functional and the preservation of the Lipschitz constant and order. </div>
<p><b class="bfseries">MSC. </b>41A10, 41A35, 41A36, 47A58. </p>
<p><b class="bfseries">Keywords. </b>Cheney-Sharma operator, Stancu operator, modulus of smoothness, Lipschitz function. </p>
<h1 id="a0000000002">1 Introduction</h1>
<p> In 1964, Cheney and Sharma introduced a generalization of the Bernstein polynomials, </p>
<div class="equation" id="a0000000003">
<p>
  <div class="equation_content">
    \begin{equation}  B_n(f;x):=\sum _{k=0}^n \tbinom {n}{k} f\left(\tfrac {n}{k}\right) x^k (1-x)^{n-k},\end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p> based on Jensen’s generalization of the binomial theorem. In this respect, considering \(\beta {\gt}0\), \(u:=x\), \(v:=1-x\), when \(x\in [0,1]\) and \(m:=n\in \mathbb {N}\), the first identity of the Abel-Jensen formulas, </p>
<div class="displaymath" id="abel1">
  \begin{flalign}  (u+v+m\beta )^m & =\sum _{k=0}^m \tbinom {m}{k} u (u+k\beta )^{k-1} [v+(m-k)\beta ]^{m-k}, \label{abel1}\\ (u+v+m\beta )^m & =\sum _{k=0}^m \tbinom {m}{k} (u+k\beta )^{k} v [v+(m-k)\beta ]^{m-k-1},\label{abel2}\\ (u+v)(u+v+m\beta )^{m-1} & =\sum _{k=0}^m \tbinom {m}{k} u (u+k\beta )^{k-1} v [v+(m-k)\beta ]^{m-k-1},\label{abel3} \end{flalign}
</div>
<p> generates the Cheney-Sharma operator of the first kind, defined, for \(f:[0,1]\to \mathbb {R}\), by (see, <i class="it">e.g.</i>, <span class="cite">
	[
	<a href="index.html#Agr00" >1</a>
	]
</span>, <span class="cite">
	[
	<a href="index.html#cs1" >3</a>
	]
</span>): </p>
<div class="equation" id="Pmare">
<p>
  <div class="equation_content">
    \begin{equation}  P_n^{\beta }(f;x)=\sum _{k=0}^{n} p_{n,k}^{\beta }(x) f\left(\tfrac {k}{n}\right), \label{Pmare} \end{equation}
  </div>
  <span class="equation_label">5</span>
</p>
</div>
<p> where </p>
<div class="equation" id="pmic">
<p>
  <div class="equation_content">
    \begin{equation}  p_{n,k}^{\beta }(x)=\tbinom {n}{k}\tfrac {x(x+k\beta )^{k-1}[1-x+(n-k)\beta ]^{n-k}}{(1+n\beta )^n}. \label{pmic} \end{equation}
  </div>
  <span class="equation_label">6</span>
</p>
</div>
<p>The following properties can be found in <span class="cite">
	[
	<a href="index.html#cs1" >3</a>
	]
</span>. With the usual notations \(e_k(t):=t^k\), \(t\in [0,1]\), \(k=0,1,2,\ldots \), by direct calculation in (<a href="index.html#abel1">2</a>), one easily obtains </p>
<div class="displaymath" id="a0000000004">
  \begin{equation*}  P_{n}^{\beta }(e_0;x)=1. \end{equation*}
</div>
<p>Furthermore, Cheney and Sharma highlighted that </p>
<div class="equation" id="Pe1">
<p>
  <div class="equation_content">
    \begin{equation}  P_{n}^{\beta }(e_1;x)=A_{n}x, \label{Pe1} \end{equation}
  </div>
  <span class="equation_label">7</span>
</p>
</div>
<p> where </p>
<div class="equation" id="Pe1An">
<p>
  <div class="equation_content">
    \begin{equation}  A_n =(1+n\beta )^{-1}\int \limits _{0}^{\infty }e^{-t}\left(1+\tfrac {t\beta }{1+n\beta }\right)^{n-1}dx. \label{Pe1An} \end{equation}
  </div>
  <span class="equation_label">8</span>
</p>
</div>
<p>Considering \(\beta =o\left(\tfrac {1}{n}\right)\), \(A_n \leq 1\) and \((A_n)\) tends to \(1\), there is obtained </p>
<div class="displaymath" id="a0000000005">
  \[ P_n^{\beta }(e_1;x)\to x \text{ uniformly on }[0,1]. \]
</div>
<p>Applying a reduction formula, Cheney and Sharma also proved that </p>
<div class="displaymath" id="a0000000006">
  \[ P_n^{\beta }(e_2;x)\to x^2 \text{ uniformly on }[0,1]. \]
</div>
<p>Since \(P_n^{\beta }\) is nonnegative for \(0\leq \beta =o\left(\tfrac {1}{n}\right)\), applying the Korovkin theorem, it is obtained that for all \(f\in C[0,1]\), </p>
<div class="displaymath" id="a0000000007">
  \[ P_n^{\beta }(f;x)\to f \text{ uniformly on }[0,1]. \]
</div>
<p> For \(\beta =0\), one easily obtains the Bernstein operator, </p>
<div class="displaymath" id="a0000000008">
  \[ P_n^0=B_n. \]
</div>
<p>In 1982, Stancu <span class="cite">
	[
	<a href="index.html#stancu2" >6</a>
	]
</span> introduced a new Bernstein type operator, </p>
<div class="displaymath" id="a0000000009">
  \begin{equation*}  L_{n,r}=\sum _{k=0}^{n-r} b_{n-r,k}(x)\left[(1-x)f\left(\tfrac {k}{n}\right)+x f\left(\tfrac {k+r}{n}\right)\right], \end{equation*}
</div>
<p> where \(b_{n,k}\) denote the basis Bernstein polynomials of degree \(n\), </p>
<div class="displaymath" id="a0000000010">
  \begin{equation*}  b_{n,k}=\tbinom {n}{k}x^k(1-x)^{n-k}\text{, } k=0,1,\ldots ,n, \end{equation*}
</div>
<p> for \(f\in C[0,1]\), \(n,r\in \mathbb {N}\) such that \(n{\gt}2r\). </p>
<p>In the present paper we introduce <i class="it">the Stancu type extension of the Cheney and Sharma operator of the first kind</i>, based on an ideea from <span class="cite">
	[
	<a href="index.html#bostanci" >4</a>
	]
</span>, using the Stancu type operator \(L_{n,r}\), and the Cheney and Sharma operator of the first kind, \(P_{n}^{\beta }\), that is given by </p>
<div class="equation" id="SCH1">
<p>
  <div class="equation_content">
    \begin{equation}  {L_P}_{n,r}^{\beta }(f;x):=\sum _{k=0}^{n-r} p_{n-r,k}(x)\left[(1-x)f\left(\tfrac {k}{n}\right)+x f\left(\tfrac {k+r}{n}\right)\right], \label{SCH1} \end{equation}
  </div>
  <span class="equation_label">9</span>
</p>
</div>
<p> where \(p_{n-r,k}\) is given by (<a href="index.html#pmic">6</a>), considering \(n-r\) in places of \(n\), \(f\in C[0,1]\) and \(n,r\in \mathbb {N}\) such that \(n{\gt}2r\). </p>
<p>In order to obtain approximation results, we consider, as in <span class="cite">
	[
	<a href="index.html#cs1" >3</a>
	]
</span>, \(\beta \geq 0 \) such that \(\beta =o\left(\tfrac {1}{n}\right).\) By direct calculation, it is obtained that \({L_P}_{n,r}^{0}\) reduces to the Stancu operator \(L_{n,r}\), while \({L_P}_{n,0}^{\beta }\) represents the Cheney-Sharma operator of the first kind, \(P_n^{\beta }\). </p>
<p>As in <span class="cite">
	[
	<a href="index.html#bostanci" >4</a>
	]
</span>, we are going to calculate the moments of the new operators, using a reduction formula from <span class="cite">
	[
	<a href="index.html#cs1" >3</a>
	]
</span>, and we study local approximation properties with respect to an appropriate K-functional. Moreover, we emphasize the preservation of the Lipschitz constant and the order when applying \({L_P}_{n,r}^{\beta }\) operator to a Lipschitz continuous function, in a similar manner to the one presented in <span class="cite">
	[
	<a href="index.html#bt1" >2</a>
	]
</span> and <span class="cite">
	[
	<a href="index.html#bostanci" >4</a>
	]
</span>. </p>
<h1 id="a0000000011">2 Properties of Cheney-Sharma operator of the first kind</h1>
<p>The purpose of this section is to outline some intermediary results regarding the Cheney and Sharma operators of the first kind that will be used in the sequel in order to prove some properties of the new constructed operator. </p>
<p><div class="lemma_thmwrapper " id="a0000000012">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">1</span>
    <span class="lemma_thmtitle"><span class="cite">
	[
	<a href="index.html#cs1" >3</a>
	]
</span></span>
  </div>
  <div class="lemma_thmcontent">
  <p> Let \(x,y\in [0,1]\), \(n\in \mathbb {N}\), \(k=0,1,\ldots ,n\). Then the function </p>
<div class="displaymath" id="a0000000013">
  \begin{equation*}  S(k,n,x,y):=\sum _{\mu =0}^n \tbinom {n}{\mu }(x+\mu \beta )^{\mu +k-1}(y+(n-\mu )\beta )^{n-\mu } \end{equation*}
</div>
<p> satisfies the recurrence relation </p>
<div class="equation" id="recc">
<p>
  <div class="equation_content">
    \begin{equation}  S(k,n,x,y)=S(k-1,n,x,y)+n\beta S(k,n-1,x+\beta ,y). \label{recc} \end{equation}
  </div>
  <span class="equation_label">10</span>
</p>
</div>

  </div>
</div> </p>
<p><div class="corollary_thmwrapper " id="a0000000014">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">2</span>
    <span class="corollary_thmtitle"><span class="cite">
	[
	<a href="index.html#cs1" >3</a>
	]
</span></span>
  </div>
  <div class="corollary_thmcontent">
  <p> By applying recursively the formula (<a href="index.html#recc">10</a>), one obtains </p>
<div class="equation" id="a0000000015">
<p>
  <div class="equation_content">
    \begin{equation}  S(1,n,x,y)=\int \limits _0^{\infty }e^{-t}(x+y+n\beta +t\beta )^n dt, \end{equation}
  </div>
  <span class="equation_label">11</span>
</p>
</div>
<p> and </p>
<div class="displaymath" id="S2">
  \begin{align}  S(2,n,x,y)=& \int \limits _0^{\infty }e^{-t}dt \int \limits _{0}^{\infty }e^{-s}ds [x(x+y+n\beta +t\beta +s\beta )^n \label{S2} \\ & + n\beta ^2 s(x+y+n\beta +t\beta +s\beta )^{n-1}]. \nonumber \end{align}
</div>

  </div>
</div> </p>
<p>The next result refers to the moments of the Cheney-Sharma operator of first kind. </p>
<p><div class="lemma_thmwrapper " id="lemmaMomCS1">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">3</span>
    <span class="lemma_thmtitle">see <span class="cite">
	[
	<a href="index.html#cs1" >3</a>
	]
</span> for a) and b)</span>
  </div>
  <div class="lemma_thmcontent">
  <p> For every \(x\in [0,1]\), \(n\in \mathbb {N}\) one obtains </p>
<div class="displaymath" id="a0000000016">
  \begin{align*}  a) \  P_n^{\beta }(e_0;x)& =1, \\ \nonumber b)\   P_n^{\beta }(e_1;x)& =A_n x,\text{ with }A_n \text{ given by (\ref{Pe1An})}, \\ \nonumber c) \   P_n^{\beta }(e_2;x)& =\tfrac {n-1}{n}[x(x+2\beta )\tilde{A}_n + x(n-2)\beta ^2\tilde{B}_n]+\tfrac {1}{n}A_n x, \end{align*}
</div>
<p> with \(\tilde{A}_n\) and \(\tilde{B}_n\) given by </p>
<div class="displaymath" id="Atilde">
  \begin{align}  \tilde{A}_n& =\tfrac {1}{(1+n\beta )^2}\int \limits _0^{\infty }e^{-t}dt\int \limits _0^{\infty }e^{-s}ds \left(1+\tfrac {t\beta +s\beta }{1+n\beta }\right)^{n-2} \label{Atilde} \\ \tilde{B}_n& =\tfrac {1}{(1+n\beta )^3}\int \limits _0^{\infty }e^{-t}dt\int \limits _0^{\infty }s e^{-s}ds \left(1+\tfrac {t\beta +s\beta }{1+n\beta }\right)^{n-3} \label{Btilde}. \end{align}
</div>
<p>Moreover, considering \(0\leq \beta =o\left(\tfrac {1}{n}\right)\), one determines </p>
<div class="displaymath" id="a0000000017">
  \[ \displaystyle \lim _{n\to \infty }A_n=\displaystyle \lim _{n\to \infty }\tilde{A}_n=\displaystyle \lim _{n\to \infty }\tilde{B}_n=1. \]
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000018">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> a) This result is easily obtained by direct calculation, starting from (<a href="index.html#abel1">2</a>). </p>
<p>b) The second result is presented in <span class="cite">
	[
	<a href="index.html#cs1" >3</a>
	]
</span>. </p>
<p>c) For obtaining \(P_n^{\beta }(e_2;x)\), we consider another method than the one used in <span class="cite">
	[
	<a href="index.html#cs1" >3</a>
	]
</span>, where the authors emphasized some bounds for the terms of \(P_n^{\beta }(e_2;x)\) which are helpful in order to prove that \(P_n^{\beta }(e_2;x)\) converges uniformly to \(x^2\), taking into consideration that \(0\leq \beta =o\left(\tfrac {1}{n}\right)\). But for calculating the moments of the new introduced operator, we need to highlight some coefficients of the monomials \(e_1\) and \(e_2\) that appear in \(P_n^{\beta }(e_2;x)\). </p>
<p>We have <span class="cite">
	[
	<a href="index.html#cs1" >3</a>
	]
</span> </p>
<div class="displaymath" id="a0000000019">
  \[ \tfrac {k^2}{n^2}=\tfrac {n-1}{n}\cdot \tfrac {k}{n}\cdot \tfrac {k-1}{n-1}+\tfrac {k}{n^2},  \]
</div>
<p> that we replace in \(P_n^{\beta }(e_2;x)\) and applying (<a href="index.html#S2">12</a>), we obtain </p>
<div class="displaymath" id="a0000000020">
  \begin{align*}  P_n^{\beta }(e_2;x)& =\tfrac {n-1}{n}(1+n\beta )^{-n} x S(2,n-2,x+2\beta ,1-x) +\tfrac {1}{n} P_n^{\beta }(e_1;x)\\ & =\tfrac {n-1}{n}[x(x+2\beta )\tilde{A}_n + x(n-2)\beta ^2\tilde{B}_n]+\tfrac {1}{n}A_n x, \end{align*}
</div>
<p> where \(\tilde{A}_n\) and \(\tilde{B}_n\) are given in (<a href="index.html#Atilde">11</a>) and (<a href="index.html#Btilde">12</a>), obtained by rewriting the integrals in (<a href="index.html#S2">12</a>). The first limit was evaluated in <span class="cite">
	[
	<a href="index.html#cs1" >3</a>
	]
</span> using the double inequality, </p>
<div class="equation" id="ineqnu">
<p>
  <div class="equation_content">
    \begin{equation}  e^{nu}(1-nu^2)\leq (1+u)^n\leq e^{nu}, \label{ineqnu} \end{equation}
  </div>
  <span class="equation_label">13</span>
</p>
</div>
<p> which will also be helpful for calculating the limits of \(\tilde{A}_n\) and \(\tilde{B}_n\). </p>
<p>Indeed, by applying the right inequality from (<a href="index.html#ineqnu">13</a>), we obtain </p>
<div class="displaymath" id="a0000000021">
  \begin{align*}  \tilde{A}_n\leq &  \tfrac {1}{(1+n\beta )^2}\int \limits _0^{\infty }e^{-t}dt\int \limits _0^{\infty }e^{-s}ds\text{ }e^{\left(\frac{t\beta +s\beta }{1+n\beta }\right)(n-2)}\\ & =\tfrac {1}{(1+n\beta )^2}\left(\int \limits _0^{\infty }e^{-t\left(\tfrac {1+2\beta }{1+n\beta }\right)}dt\right)^2 =\tfrac {1}{(1+2\beta )^2}. \end{align*}
</div>
<p>Similarly, there are obtained the following inequalities: </p>
<div class="displaymath" id="a0000000022">
  \begin{align*}  \frac{1}{(1+2\beta )^2}-\frac{2n\beta ^2}{(1+2\beta )^4}& \leq \tilde{A}_n\leq \frac{1}{(1+2\beta )^2},\\ \frac{1}{(1+3\beta )^3}-\frac{5n\beta ^2}{3(1+3\beta )^5}& \leq \tilde{B}_n\leq \frac{1}{(1+3\beta )^3}. \end{align*}
</div>
<p>Considering \(0\leq \beta =o\left(\tfrac {1}{n}\right)\), it is easily seen that \(\displaystyle \lim _{n\to \infty }\tilde{A}_n=\displaystyle \lim _{n\to \infty }\tilde{B}_n=1,\) hence \(P_n^{\beta }(e_2;x)\) converges uniformly to \(x^2\). <div class="proof_wrapper" id="a0000000023">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<h1 id="a0000000024">3 Properties of the Stancu type extension of the Cheney-Sharma operator of the first kind</h1>
<p>In this section we study some approximation properties for the new operator, \({L_P}_{n,r}^{\beta }\), introduced by us in (<a href="index.html#SCH1">9</a>). </p>
<p>First, we highlight the expressions of the moments of \({L_P}_{n,r}^{\beta }\) in terms of the moments of the Cheney-Sharma operators of the first kind. </p>
<p><div class="lemma_thmwrapper " id="a0000000025">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">4</span>
  </div>
  <div class="lemma_thmcontent">
  <p>For every \(x\in [0,1]\), \(n,r\in \mathbb {N}\) such that \(n{\gt}2r\), we obtain </p>
<div class="displaymath" id="a0000000026">
  \begin{align*}  {L_P}_{n,r}^{\beta }(e_0;x)=& 1,\\ {L_P}_{n,r}^{\beta }(e_1;x)=& \tfrac {n-r}{n}A_{n-r} x +\tfrac {r}{n} x,\\ {L_P}_{n,r}^{\beta }(e_2;x)=& \tfrac {(n-1)(n-r-1)}{n^2}\big[x(x+2\beta )\tilde{A}_{n-r}+x(n-r-2)\beta ^2\tilde{B}_{n-r}\big]\\ & +\tfrac {n-r}{n^2}(1+2xr)A_{n-r}x+\tfrac {r^2}{n^2}x. \end{align*}
</div>

  </div>
</div> <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> Taking into account (<a href="index.html#Pmare">5</a>), (<a href="index.html#pmic">6</a>), (<a href="index.html#SCH1">9</a>), <a href="index.html#lemmaMomCS1">lemma 3</a>, in a similar manner to <span class="cite">
	[
	<a href="index.html#bostanci" >4</a>
	]
</span>, we obtain </p>
<div class="displaymath" id="a0000000028">
  \begin{align*} & {L_P}_{n,r}^{\beta }(e_0;x)=P_{n-r}^{\beta }(e_0;x)=1, \\ & {L_P}_{n,r}^{\beta }(e_1;x)=\tfrac {n-r}{n}P_{n-r}^{\beta }(e_1;x) +\tfrac {r}{n} x P_{n-r}^{\beta }(e_0;x) =\tfrac {n-r}{n}A_{n-r} x +\tfrac {r}{n} x, \\ & {L_P}_{n,r}^{\beta }(e_2;x)=\tfrac {(n-r)^2}{n^2}P_{n-r}^{\beta }(e_2;x)+2xr\tfrac {n-r}{n^2}P_{n-r}^{\beta }(e_1;x)+\tfrac {r^2}{n^2}x P_{n-r}^{\beta }(e_0;x) \\ & =\tfrac {(n-r)^2}{n^2}\left\{ \tfrac {n-r-1}{n-r}[x(x+2\beta )\tilde{A}_{n-r}+x(n-r-2)\beta ^2\tilde{B}_{n-r}]+\tfrac {1}{n-r}A_{n-r}x\right\}  \\ & \quad +2xr\tfrac {n-r}{n^2}A_{n-r}x+\tfrac {r^2}{n^2}x \\ & =\tfrac {(n-1)(n-r-1)}{n^2}[x(x+2\beta )\tilde{A}_{n-r}+x(n-r-2)\beta ^2\tilde{B}_{n-r}] \\ & \quad +\tfrac {n-r}{n^2}(1+2xr)A_{n-r}x+\tfrac {r^2}{n^2}x. \end{align*}
</div>
<p> <div class="proof_wrapper" id="a0000000029">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="lemma_thmwrapper " id="lemmaIneq1">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">5</span>
  </div>
  <div class="lemma_thmcontent">
  <p> For every \(f\in C[0,1]\), we have </p>
<div class="displaymath" id="a0000000030">
  \[ \| {L_P}_{n,r}^{\beta }\| \leq \| f\| . \]
</div>

  </div>
</div> <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> Considering the expression of \({L_P}_{n,r}^{\beta }(f;x)\) and Lemma <a href="index.html#lemmaMomCS1">3</a>, we get </p>
<div class="displaymath" id="a0000000032">
  \begin{align*}  |{L_P}_{n,r}^{\beta }(f;x)|& =\left|\sum _{k=0}^{n-r} p_{n-r,k}(x)\left[(1-x)f\left(\tfrac {k}{n}\right)+x f\left(\tfrac {k+r}{n}\right)\right]\right|\\ & \leq \sum _{k=0}^{n-r} p_{n-r,k}(x) \left|(1-x)f\left(\tfrac {k}{n}\right)+x f\left(\tfrac {k+r}{n}\right)\right|\\ & \leq \sum _{k=0}^{n-r} p_{n-r,k}(x) \left[(1-x)\left|f\left(\tfrac {k}{n}\right)\right|+x \left|f\left(\tfrac {k+r}{n}\right)\right|\right]\\ & \leq \| f\|  \sum _{k=0}^{n-r} p_{n-r,k}(x) (1-x+x) \\ & =\| f\|  P_{n-r}^{\beta }(e_0;x)\\ & =\| f\| . \end{align*}
</div>
<p> <div class="proof_wrapper" id="a0000000033">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>The next approximation result is based on the equivalence between the first order modulus of smoothness and the following K-functional, </p>
<div class="displaymath" id="a0000000034">
  \[ K_1 (f,\delta )=\inf _{g\in W^1}\left\{ \| f-g\| +\delta \| g'\| \right\} , \]
</div>
<p> where \(\delta {\gt}0\) and \(W^1:=\left\{ g\in C[0,1]: g'\in C[0,1]\right\} \). </p>
<p>Considering the first order modulus of smoothness associated to a function \(f\in C[0,1]\), given by \(\omega (f,\delta )=\sup \limits _{\substack {0\leq h\leq \delta ; x,x+h\in [0,1]}}|f(x+h)-f(x)|,\) in <span class="cite">
	[
	<a href="index.html#agratini" >7</a>
	]
</span>, it is asserted that there exists a positive constant \(C{\gt}0\) such that </p>
<div class="equation" id="equiv">
<p>
  <div class="equation_content">
    \begin{equation}  C^{-1} \cdot \omega (f,\delta )\leq K_1(f,\delta )\leq C\cdot \omega (f,\delta ). \label{equiv} \end{equation}
  </div>
  <span class="equation_label">14</span>
</p>
</div>
<p><div class="theorem_thmwrapper " id="a0000000035">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">6</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(f\in C[0,1]\), \(x\in [0,1]\) \(n,r\in \mathbb {N}\) such that \(n\geq 2r\), and \(0\leq \beta =o\left(\frac{1}{n}\right).\) Denoting </p>
<div class="displaymath" id="a0000000036">
  \begin{align*}  \delta _{n,r}(x):& =\left|{L_P}_{n,r}^{\beta }((e_1-x);x)\right|=\tfrac {n-r}{n}x(1-A_{n-r}) \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000037">
  \begin{align*}  \tilde{\delta }_{n,r}(x)=\frac{\delta _{n,r}(x)}{2}, \end{align*}
</div>
<p> we obtain </p>
<div class="displaymath" id="a0000000038">
  \[ |{L_P}_{n,r}^{\beta }(f;x)-f(x)|\leq C\cdot \omega (f,\tilde{\delta }_{n,r}(x)). \]
</div>

  </div>
</div> <div class="proof_wrapper" id="a0000000039">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> For \(g\in W^1\) and \(x,t\in [0,1]\), by the Lagrange theorem, it results that there exists \(c_{x,t}\) between \(x\) and \(t\) such that </p>
<div class="displaymath" id="a0000000040">
  \[ g(t)-g(x)=g'(c_{x,t})(t-x). \]
</div>
<p>Applying the operator \({L_P}_{n,r}^{\beta }\) and taking into account the linearity of the operator, we get </p>
<div class="displaymath" id="a0000000041">
  \begin{align}  |{L_P}_{n,r}^{\beta }(g;x)-g(x)|& =|g’(c_{x,t})|\cdot |{L_P}_{n,r}^{\beta }(e_1-x;x)|\nonumber \\ &  \leq \| g’\|  \cdot \delta _{n,r}(x). \nonumber \end{align}
</div>
<p>Hence, by <a href="index.html#lemmaIneq1">lemma 5</a>, we have </p>
<div class="displaymath" id="a0000000042">
  \begin{align}  |{L_P}_{n,r}^{\beta }(f;x)-f(x)|& =|{L_P}_{n,r}^{\beta }(f-g;x)-(f-g)(x)+{L_P}_{n,r}^{\beta }(g;x)-g(x)|\nonumber \\ & \leq |{L_P}_{n,r}^{\beta }(f-g;x)-(f-g)(x)| +|{L_P}_{n,r}^{\beta }(g;x)-g(x)|\nonumber \\ & \leq 2\| f-g\|  + \| g’\| \cdot \delta _{n,r}(x)\nonumber \\ & =2\left(\| f-g\| +\tilde\delta _{n,r}(x)\cdot \| g’\| \right).\nonumber \end{align}
</div>
<p>Applying inequality (<a href="index.html#equiv">14</a>), we obtain </p>
<div class="displaymath" id="a0000000043">
  \begin{align*}  |{L_P}_{n,r}^{\beta }(f;x)-f(x)|& \leq 2\cdot K_1(f,\tilde{\delta }_{n,r}(x))\leq C \cdot \omega (f,\tilde{\delta }_{n,r}(x)). \end{align*}
</div>
<p> <div class="proof_wrapper" id="a0000000044">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>Next, we study the preservation of the Lipschitz constant and the order of a Lipschitz continuous function by applying the operator \({L_P}_{n,r}^{\beta }\). </p>
<p><div class="theorem_thmwrapper " id="a0000000045">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">7</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(f\in \mathrm{Lip}_M(\alpha ,[0,1])\). Then </p>
<div class="displaymath" id="a0000000046">
  \[ {L_P}_{n,r}^{\beta }f\in \mathrm{Lip}_M(\alpha ,[0,1]), \]
</div>
<p> for \(n\in \mathbb {N}\), where \(\mathrm{Lip}_M(\alpha ,[0,1])\) denotes the class of Lipschitz continuous functions on \([0,1]\) of order \(\alpha \in (0,1]\) and constant \(M\), defined by </p>
<div class="displaymath" id="a0000000047">
  \[ \mathrm{Lip}_M(\alpha ,[0,1])=\left\{ f\in C[0,1]: |f(x)-f(y)|\leq M|x-y|^{\alpha }\right\} . \]
</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 a similar manner to the one applied in <span class="cite">
	[
	<a href="index.html#bt1" >2</a>
	]
</span> and <span class="cite">
	[
	<a href="index.html#bostanci" >4</a>
	]
</span>. </p>
<p>Considering \(x,y\in [0,1]\), \(x\leq y\), we have </p>
<div class="displaymath" id="a0000000049">
  \begin{align}  {L_P}_{n,r}^{\beta }(f;y)=& \sum _{k=0}^{n-r} p_{n-r,k}(y)\left[(1-y)f\left(\tfrac {k}{n}\right)+y f\left(\tfrac {k+r}{n}\right)\right]\nonumber \\ =& \tfrac {1}{\left(1+(n-r)\beta \right)^{n-r}} \sum _{k=0}^{n-r} \tbinom {n-r}{k} y (y+k\beta )^{k-1} \label{u1}\\ & \cdot \left[1-y+(n-r-k)\beta \right]^{n-r-k}\left[(1-y)f\left(\tfrac {k}{n}\right)+y f\left(\tfrac {k+r}{n}\right)\right].\nonumber \end{align}
</div>
<p>Using the third Abel-Jensen identity (<a href="index.html#abel3">4</a>), with \(u:=x\), \(v:=y-x\), \(m:=k\), we obtain </p>
<div class="displaymath" id="a0000000050">
  \[ y(y+k\beta )^{k-1}=\sum _{j=0}^k \tbinom {k}{j} x (x+j\beta )^{j-1} (y-x)[y-x+(k-j)\beta ]^{k-j-1}. \]
</div>
<p>Replacing this in (<a href="index.html#u1">15</a>), we get </p>
<div class="displaymath" id="a0000000051">
  \begin{align*} {L_P}_{n,r}^{\beta }(f;y)=& \tfrac {1}{\left(1+(n-r)\beta \right)^{n-r}} \sum _{j=0}^{n-r} \tbinom {n-r}{j}\sum _{k=0}^j \tbinom {j}{k} x (x+k\beta )^{k-1} (y-x)\\ & \cdot \left[y-x+(j-k)\beta \right]^{j-k-1} \left[1-y+(n-r-j)\beta \right]^{n-r-j}\\ & \cdot \left[(1-y)f\left(\tfrac {j}{n}\right)+y f\left(\tfrac {j+r}{n}\right)\right].\end{align*}
</div>
<p> Denoting \(l:=j-k\), we have </p>
<div class="displaymath" id="a0000000052">
  \begin{align*} {L_P}_{n,r}^{\beta }(f;y)=& \tfrac {1}{\left(1+(n-r)\beta \right)^{n-r}} \sum _{k=0}^{n-r} \sum _{l=0}^{n-r-k} \tbinom {n-r}{k}\tbinom {n-r-k}{l} x (x+k\beta )^{k-1} (y-x)\\ & \cdot \left(y-x+l\beta \right)^{l-1} \left[1-y+(n-r-k-l)\beta \right]^{n-r-k-l}\\ & \cdot \left[(1-y)f\left(\tfrac {k+l}{n}\right)+y f\left(\tfrac {k+l+r}{n}\right)\right].\end{align*}
</div>
<p>Now considering </p>
<div class="displaymath" id="lpfx">
  \begin{align} {L_P}_{n,r}^{\beta }(f;x)=& \tfrac {1}{\left(1+(n-r)\beta \right)^{n-r}}\sum _{k=0}^{n-r} \tbinom {n-r}{k}x(x+k\beta )^{k-1}\label{lpfx}\\ & \cdot [1-x+(n-r-k)\beta ]^{n-r-k}\left[(1-x)f\left(\tfrac {k}{n}\right)+x f\left(\tfrac {k+r}{n}\right)\right]\nonumber , \end{align}
</div>
<p> by taking \(u:=y-x\), \(v:=1-y\) and \(m:=n-r-k\) in the first Abel-Jensen identity (<a href="index.html#abel1">2</a>), <i class="it">i.e.</i>, </p>
<div class="displaymath" id="a0000000053">
  \begin{align*} \left[1-x+(n-r-k)\beta \right]^{n-r-k}=& \sum _{j=0}^{n-r-k}\tbinom {n-r-k}{j}(y-x)(y-x+j\beta )^{j-1}\\ & \cdot \left[1-y+(n-r-k-j)\beta \right]^{n-r-k-j}\end{align*}
</div>
<p> and replacing in (<a href="index.html#lpfx">16</a>), it results </p>
<div class="displaymath" id="a0000000054">
  \begin{align*} & {L_P}_{n,r}^{\beta }(f;x)=\tfrac {1}{\left(1+(n-r)\beta \right)^{n-r}}\sum _{k=0}^{n-r} \sum _{l=0}^{n-r-k}\tbinom {n-r}{k}\tbinom {n-r-k}{l}x \\ & \cdot (x+k\beta )^{k-1}(y-x)(y-x+l\beta )^{l-1}[1-y+(n-r-k-l)\beta ]^{n-r-k-l}\nonumber \\ & \cdot \left[(1-x)f\left(\tfrac {k}{n}\right)+x f\left(\tfrac {k+r}{n}\right)\right]\nonumber . \end{align*}
</div>
<p>We obtain </p>
<div class="displaymath" id="a0000000055">
  \begin{align} & {L_P}_{n,r}^{\beta }(f;y)-{L_P}_{n,r}^{\beta }(f;x)=\tfrac {1}{\left(1+(n-r)\beta \right)^{n-r}}\sum _{k=0}^{n-r} \sum _{l=0}^{n-r-k}\tbinom {n-r}{k}\tbinom {n-r-k}{l}\nonumber \\ & \cdot x(x+k\beta )^{k-1}(y-x)(y-x+l\beta )^{l-1}\left[1-y+(n-r-k-l)\beta \right]^{n-r-k-l}\nonumber \\ & \cdot \left[(1-y)\left(f\left(\tfrac {k+l}{n}\right)-f\left(\tfrac {k}{n}\right)\right)+x\left(f\left(\tfrac {k+l+r}{n}\right)-f\left(\tfrac {k+r}{n}\right)\right)\right.\nonumber \\ & \left. +(y-x)\left(f\left(\tfrac {k+l+r}{n}\right)-f\left(\tfrac {k}{n}\right)\right)\right].\nonumber \end{align}
</div>
<p>Using the hypothesis that \(f\in \text{Lip}_M(\alpha ,[0,1])\), it results </p>
<div class="displaymath" id="a0000000056">
  \begin{align*} & \left| {L_P}_{n,r}^{\beta }(f;y)\right. \left. -{L_P}_{n,r}^{\beta }(f;x) \right| \leq \tfrac {M}{\left(1+(n-r)\beta \right)^{n-r}}\sum _{k=0}^{n-r} \sum _{l=0}^{n-r-k}\tbinom {n-r}{k}\tbinom {n-r-k}{l}\\ & \cdot x(x+k\beta )^{k-1}(y-x)(y-x+l\beta )^{l-1}\left[1-y+(n-r-k-l)\beta \right]^{n-r-k-l} \\ & \cdot \left[\left(1-(y-x)\right)\left(\tfrac {l}{n}\right)^{\alpha }+(y-x)\left(\tfrac {l+r}{n}\right)^{\alpha }\right], \end{align*}
</div>
<p> and furthermore, </p>
<div class="displaymath" id="a0000000057">
  \begin{align*} & \left| {L_P}_{n,r}^{\beta }(f;y)\right.\left.-{L_P}_{n,r}^{\beta }(f;x) \right| \leq \tfrac {M}{\left(1+(n-r)\beta \right)^{n-r}}\sum _{l=0}^{n-r} \tbinom {n-r}{l} (y-x)(y-x+l\beta )^{l-1}\\ & \cdot \sum _{k=0}^{n-r-l} \tbinom {n-r-l}{k} x(x+k\beta )^{k-1}\left[1-y+(n-r-k-l)\beta \right]^{n-r-k-l} \\ & \cdot \left[\left(1-(y-x)\right)\left(\tfrac {l}{n}\right)^{\alpha }+(y-x)\left(\tfrac {l+r}{n}\right)^{\alpha }\right]. \end{align*}
</div>
<p>Using (<a href="index.html#abel1">2</a>) with \(u:=x\), \(v:=1-y\) and \(m:=n-r-l\), one obtains </p>
<div class="displaymath" id="a0000000058">
  \begin{align*}  \left[1-(y-x)+(n-r-l)\beta \right]^{n-r-l} =& \sum _{k=0}^{n-r-l}\tbinom {n-r-l}{k}x(x+k\beta )^{k-1}\\ & \cdot \left[1-y+(n-r-l-k)\beta \right]^{n-r-l-k}. \end{align*}
</div>
<p>So we get, </p>
<div class="displaymath" id="a0000000059">
  \begin{align*} &  \left| {L_P}_{n,r}^{\beta }(f;y)\right.-\left.{L_P}_{n,r}^{\beta }(f;x) \right| \leq \tfrac {M}{\left(1+(n-r)\beta \right)^{n-r}}\sum _{l=0}^{n-r} \tbinom {n-r}{l} (y-x)(y-x+l\beta )^{l-1}\\ & \cdot \left[1-(y-x)+(n-r-l)\beta \right]^{n-r-l} \left[\left(1-(y-x)\right)\left(\tfrac {l}{n}\right)^{\alpha }+(y-x)\left(\tfrac {l+r}{n}\right)^{\alpha }\right]. \end{align*}
</div>
<p>Recall that \(g(t)=t^{\alpha }\) is concave for \(\alpha \in (0,1]\), meaning that </p>
<div class="displaymath" id="a0000000060">
  \[ \alpha _1 \cdot x_1^{\alpha }+\alpha _2 \cdot x_2^{\alpha }\leq (\alpha _1 \cdot x_1 +\alpha _2 \cdot x_2)^{\alpha }\text{, for }\alpha _1+\alpha _2=1. \]
</div>
<p>In our case, \(x_1:=\frac{l}{n}\), \(x_2:=\frac{l+r}{n}\), \(\alpha _1:=1-(y-x)\) and \(\alpha _2:=y-x\), so it results </p>
<div class="displaymath" id="a0000000061">
  \begin{align*}  \left| {L_P}_{n,r}^{\beta }(f;y)\right.& \left.-{L_P}_{n,r}^{\beta }(f;x) \right| \leq \tfrac {M}{\left(1+(n-r)\beta \right)^{n-r}}\sum _{l=0}^{n-r} \tbinom {n-r}{l} (y-x)(y-x+l\beta )^{l-1}\\ & \cdot \left[1-(y-x)+(n-r-l)\beta \right]^{n-r-l} \left[\left(1-(y-x)\right)\tfrac {l}{n}+(y-x)\tfrac {l+r}{n}\right]^{\alpha }\\ & =M\sum _{l=0}^{n-r} p_{n-r,l}(y-x)\left[\left(1-(y-x)\right)\tfrac {l}{n}+(y-x)\tfrac {l+r}{n}\right]^{\alpha }. \end{align*}
</div>
<p><b class="bfseries">Case \(\alpha =1\).</b> The last inequality becomes </p>
<div class="displaymath" id="a0000000062">
  \begin{align*}  \left| {L_P}_{n,r}^{\beta }(f;y)-{L_P}_{n,r}^{\beta }(f;x) \right| & \leq {L_P}_{n,r}^{\beta }(e_1,y-x)\\ & =M(y-x)\left(\tfrac {n-r}{n}A_{n-r}+\tfrac {r}{n}\right)\\ & \leq M(y-x), \end{align*}
</div>
<p> since \((A_n) \to 1\) (see <span class="cite">
	[
	<a href="index.html#cs1" >3</a>
	]
</span>), as it has been mentioned in the first section. </p>
<p><b class="bfseries">Case \(0{\lt}\alpha {\lt}1\).</b> We will apply the Hölder inequality for \(p:=\frac{1}{\alpha }\) and \(q:=\frac{1}{1-\alpha }\) s.t. \(\frac{1}{p}+\frac{1}{q}=1\). Taking into account that \({L_P}_{n,r}^{\beta }(e_0;x)=1\), it results </p>
<div class="displaymath" id="a0000000063">
  \begin{align*} & \left| {L_P}_{n,r}^{\beta }(f;y)-{L_P}_{n,r}^{\beta }(f;x) \right| \leq \\ & \leq M\left\{ \sum _{l=0}^{n-r} p_{n-r,l}(y\! -\! x)\left[\left(1-(y\! -\! x)\right)\tfrac {l}{n}+(y\! -\! x)\tfrac {l+r}{n}\right]\right\} ^{\alpha }\! \cdot \!  \left\{ \sum _{l=0}^{n-r} p_{n-r,l}(y\! -\! x)\right\} ^{1-\alpha }\\ & =M\left({L_P}_{n,r}^{\beta }(e_1;y-x)\right)^{\alpha }\cdot \left({L_P}_{n,r}^{\beta }(e_0;y-x)\right)^{1-\alpha }\\ & =M\cdot (y-x)^{\alpha }\cdot \left(\tfrac {n-r}{n}A_{n-r}+\tfrac {r}{n}\right)^{\alpha }\\ & \leq M\cdot (y-x)^{\alpha }. \end{align*}
</div>
<p> <div class="proof_wrapper" id="a0000000064">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<dl class="bibliography">
  <dt><a name="Agr00">1</a></dt>
  <dd><p><i class="sc">0. Agratini</i>, <i class="it">Approximation by linear operators</i>, Cluj University Press, 2000. </p>
</dd>
  <dt><a name="bt1">2</a></dt>
  <dd><p><i class="sc">G. Başcanbaz-Tunca, A. Erençin, F. Taşdelen</i>, <i class="it">Some properties of Bernstein type Cheney and Sharma Operators</i>, General Mathematics, <b class="bf">24</b> (2016), pp. 17–25. </p>
</dd>
  <dt><a name="cs1">3</a></dt>
  <dd><p><i class="sc">E.W. Cheney, A. Sharma</i>, <i class="it">On a generalization of Bernstein polynomials</i>, Riv. Mat. Univ. Parma, <b class="bf">2</b> (1964), pp. 77–84. </p>
</dd>
  <dt><a name="bostanci">4</a></dt>
  <dd><p><a href ="https://doi.org/10.33993/jnaat472-1133"> <i class="sc">T. Bostanci, G. Başcanbaz-Tunca</i>, <i class="it">A Stancu type extension of Cheney and Sharma operator</i>, J. Numer. Anal. Approx. Theory, <b class="bf">47</b> (2018), pp. 124–134, <a href="https://doi.org/10.33993/jnaat472-1133">https://doi.org/10.33993/jnaat472-1133</a>. </a> </p>
</dd>
  <dt><a name="stancu1">5</a></dt>
  <dd><p><i class="sc">D.D. Stancu, C. Cismaşiu</i>, <i class="it">On an approximating linear positive operator of Cheney-Sharma</i>, Rev. Anal. Numér. Théor. Approx., <b class="bf">26</b> (1997), pp. 221–227. </p>
</dd>
  <dt><a name="stancu2">6</a></dt>
  <dd><p><a href ="https://doi.org/10.1007/978-3-0348-6308-7_23"> <i class="sc">D.D. Stancu</i>, <i class="it">Quadrature formulas constructed by using certain linear positive operators</i>, Numerical Integration (Proc. Conf., Oberwolfach, 1981), ISNM 57 (1982), pp. 241–251, <a href="https://doi.org/10.1007/978-3-0348-6308-7_23">https://doi.org/10.1007/978-3-0348-6308-7_23</a>. </a> </p>
</dd>
  <dt><a name="agratini">7</a></dt>
  <dd><p><i class="sc">D.D. Stancu, G. Coman, O. Agratini, R.T. Trîmbițaș, P. Blaga, I. Chiorean</i>, <i class="it">Analiză numerică și teoria aproximării</i>, Presa Universitară Clujeană, 2001 (in Romanian). </p>
</dd>
  <dt><a name="a0000000065"></a></dt>
  <dd></dd>
  <dt><a name=""></a></dt>
  <dd> </dd>
</dl>


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