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<h1>On Szász-Mirakyan type operators <br />preserving polynomials</h1>
<p class="authors">
<span class="author">Övgü Gürel Yılmaz\(^\S \), Ali Aral\(^\ast \) Fatma Taşdelen Yeşildal\(^\bullet \)</span>
</p>
<p class="date">June 24, 2016.</p>
</div>
<p>\(^\S \)University of Ankara, Department of Mathematics, 06100, Tandogan, Ankara, Turkey, e-mail: <span class="tt">ogurel@ankara.edu.tr</span>. </p>
<p>\(^\ast \)University of Kırıkkale, Department of Mathematics, Kırıkkale, Turkey, e-mail: <span class="tt">aliaral73@yahoo.com</span>. </p>
<p>\(^\bullet \)University of Ankara, Department of Mathematics, 06100, Tandogan, Ankara, Turkey, e-mail: <span class="tt">tasdelen@science.ankara.edu.tr</span>. </p>

<div class="abstract"><p> In this paper, a modification of Szász-Mirakyan operators is studied <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span> which generalizes the Szász-Mirakyan operators with the property that the linear combination \(e_{2}+\alpha e_{1}\) of the Korovkin’s test functions \(e_{1}\) and \(e_{2}\) are reproduced for \(\alpha \geq 0.\) After providing some computational results, shape preserving properties of mentioned operators are obtained. Moreover, some estimations for the rate of convergence of these operators by using different type modulus of continuity are shown. Furthermore, a Voronovskaya-type formula and an app-roximation result for derivative of operators are calculated. </p>
<p><b class="bf">MSC.</b> 41A25, 41A36 </p>
<p><b class="bf">Keywords.</b> Szász-Mirakyan operators, modified operator, shape preserving properties, Voronovskaya-type theorem. </p>
</div>
<h1 id="a0000000002">1 INTRODUCTION</h1>
<p>Approximation theory is based on finding the best approximation of a function by polynomials or other type of simple functions. For many years, there have been lots of improvements about the approximation theory. In 1853, Russian mathematician Chebyshev focused on this matter. However, the big step was in 1885 when Karl Weierstrass <span class="cite">
	[
	<a href="#Weires" >13</a>
	]
</span> presented the theorem on appro-ximation. </p>
<p>In approximation theory, positive linear operators play an essential role. The study of approximation sequences of linear positive operators was started at the beginning of the 1950s. One of the most important positive linear operators is Bernstein polynomials. Bernstein polynomials on the space \(C\left[ 0,1\right] \) are defined by </p>
<div class="displaymath" id="a0000000003">
  \begin{equation*}  B_{n}f(x)=\sum \limits _{k=0}^{n}f\left( \tfrac {k}{n}\right) \tbinom {n}{k} x^{k}(1-x)^{n-k},\quad n\in \mathbb {N} . \end{equation*}
</div>
<p>Up to now, there have been lots of extensions and modifications of Bernstein polynomials. It was King <span class="cite">
	[
	<a href="#11" >9</a>
	]
</span> who constructed Bernstein type linear positive operator defined on \(C\left[ 0,1\right] \) having an approximation order better then the classical operators such that they reproduce the test function \(e_{0}\) and \(e_{2}.\) This operator has an approximation order better than the classical operators on \(\left[ 0,\frac{1}{3}\right] .\) </p>
<p>Inspiring this fact D. Cardenas-Morales et al. <span class="cite">
	[
	<a href="#5" >4</a>
	]
</span> introduced an operator of King type, which was reproduced \(e_{2}+\alpha e_{1}\) for \(\alpha \geq 0\) and defined by </p>
<div class="equation" id="1">
<p>
  <div class="equation_content">
    \begin{equation}  B_{n,\alpha }f(x)=\sum _{k=0}^{n}\tbinom {n}{k} f\left( \tfrac {k}{n}\right) \left( r_{n,\alpha }^{\ast }(x)\right) ^{k}(1-r_{n,\alpha }^{\ast }(x))^{n-k}, \label{1} \end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p>where </p>
<div class="displaymath" id="a0000000004">
  \begin{equation*}  r_{n,\alpha }^{\ast }(x)=-\tfrac {n\alpha +1}{2(n-1)}+\sqrt{\tfrac {(n\alpha +1)^{2}}{4(n-1)^{2}}+\tfrac {n(\alpha x+x^{2})}{n-1}},\quad n\in \mathbb {N} . \end{equation*}
</div>
<p>They found the shape preserving properties for \(B_{n,\alpha }\) and worked on the comparison with Bernstein polynomials. Furthermore, they also showed that the sequences \(B_{n,\alpha }\) for \(\alpha \geq 0\) are an approximation process. Besides, for different Bernstein Durrmeyer type operators, similar results were given in <span class="cite">
	[
	<a href="#6" >5</a>
	]
</span>. </p>
<p>Szász-Mirakyan operators are the generalizations of Bernstein polynomials on the interval \(\left[ 0,\infty \right) ~ \)which are defined by</p>
<div class="displaymath" id="a0000000005">
  \begin{equation*}  S_{n}f(x)=\sum _{k=0}^{\infty }f\left( \tfrac {k}{n}\right) e^{-nx}\tfrac {\left( nx\right) ^{k}}{k!},\  \  \  n\in \mathbb {N} . \end{equation*}
</div>
<p>Notice that, for all functions \(f:C\left[ 0,\infty \right) \rightarrow \mathbb {R}\) the series at the right hand side convergences absolutely. </p>
<p>There are many papers about different type of generalizations of Szász-Mirakyan operators where the basic properties of approximation are analyzed. In the recent years, the number of the articles related to this fact has increased \(\  (\)see <span class="cite">
	[
	<a href="#3" >2</a>
	]
</span>\(,\) <span class="cite">
	[
	<a href="#8" >7</a>
	]
</span>, <span class="cite">
	[
	<a href="#4" >3</a>
	]
</span> and <span class="cite">
	[
	<a href="#12" >10</a>
	]
</span>\().\) </p>
<p>In order to furnish better error estimation in a certain sense than classical Szász-Mirakyan operators, in <span class="cite">
	[
	<a href="#8" >7</a>
	]
</span>, authors defined the following operators,</p>
<div class="displaymath" id="a0000000006">
  \begin{equation*}  D_{n}^{\ast }f(x)=e^{-nu_{n}^{\ast }(x)}\sum \limits _{k=0}^{\infty }f\left( \tfrac {k}{n}\right) \tfrac {\left( nu_{n}^{\ast }(x)\right) ^{k}}{k!} \end{equation*}
</div>
<p>where </p>
<div class="displaymath" id="a0000000007">
  \begin{equation*}  u_{n}^{\ast }(x)=\tfrac {-1+\sqrt{4n^{2}x^{2}+1}}{2n},\quad n\in \mathbb {N} . \end{equation*}
</div>
<p>Note that, both of the results of <span class="cite">
	[
	<a href="#5" >4</a>
	]
</span> and <span class="cite">
	[
	<a href="#11" >9</a>
	]
</span> were obtained in finite intervals. On the infinite interval, using similar technique, authors introduced Szász-Mirakyan operators King type by reproducing \(e_{1}\) and \(e_{2}\) <span class="cite">
	[
	<a href="#4" >3</a>
	]
</span>\(.\) </p>
<p>In this paper, as in <span class="cite">
	[
	<a href="#5" >4</a>
	]
</span> for Bernstein polynomials, we consider a similar modification of the Szász–Mirakyan modified operators given in <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span> using the function \(r_{n,\alpha }\) which is defined by \(\left\{  S_{n,\alpha }:C[0,\infty )\rightarrow C[0,\infty )\right\}  _{n{\gt}0}\) for\(\  \alpha \geq 0,~ \)</p>
<div class="displaymath" id="2">
  \begin{align}  S_{n,\alpha }f(x)& :=S_{n,\alpha }(f;x)=\sum _{k=0}^{\infty }f\left( \tfrac {k}{n}\right) \mathcal{P}_{n,k,\alpha }(x) \label{2} \\ \mathcal{P}_{n,k,\alpha }(x)& :=e^{-nr_{n,\alpha }(x)}\tfrac {(nr_{n,\alpha }(x))^{k}}{k!} \nonumber \\ r_{n,\alpha }(x)& =-\tfrac {(\alpha n+1)}{2n}+\sqrt{\tfrac {(\alpha n+1)^{2}}{4n^{2}}+(x^{2}+\alpha x)},\quad n\in \mathbb {N} ,\quad x\geq 0, \nonumber \end{align}
</div>
<p>where \(\left\{  r_{n,\alpha }:[0,\infty )\rightarrow \mathbb {R} \right\}  _{n{\gt}0}~ \)is the sequence of functions. </p>
<p>Noting the fact that when \(n\rightarrow \infty ,\) \(r_{n,\alpha }\rightarrow x \), \(S_{n,\alpha }f\) reduces to the classical Szász-Mirakyan operator. That is, classical Százs-Mirakyan operators turn out to be a limit element of \(S_{n,\alpha }f\) and also if we take \(\alpha =0,\) the sequence \(D_{n}^{\ast }f\) of operators appears which is introduced in <span class="cite">
	[
	<a href="#8" >7</a>
	]
</span>. </p>
<p>In <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span>, authors showed the approximation properties of Szász-Mirakyan mo-dified operator. In the lights of the definition of the operator of (<a href="#2">2</a>), different kinds of results which are related to the mentioned operator are obtained. </p>
<p>The organization of the paper is as follows: </p>
<p>In section 2, shape preserving properties of the Szász-Mirakyan modified operators are investigated. Using the convexity and generalized convexity, relations between the given functions, \(S_{n,\alpha }f\) and \(S_{n}f\) operators are obtained\(.\) Then, the results of Voronovskaya-type theorem are given. Moreover, the rate of convergence properties of this operator for two different modulus of continuities are studied and a theorem which is satisfied by derivative of \(S_{n,\alpha }f~ \)is given. </p>
<p>Throughout the paper, we use following definition and notations. </p>
<p><div class="definition_thmwrapper " id="a0000000008">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">1</span>
  </div>
  <div class="definition_thmcontent">
  <p><span class="cite">
	[
	<a href="#6" >5</a>
	]
</span> A function \(f\in C^{k}[0,\infty )\) (the space of \(k\) times conti-nuously differentiable functions) is said to be \(\tau \) convex of order \(k\in \mathbb {N} \) whenever </p>
<div class="equation" id="3">
<p>
  <div class="equation_content">
    \begin{equation}  D^{k}(f\circ \tau ^{-1})\circ \tau \geq 0. \label{3} \end{equation}
  </div>
  <span class="equation_label">3</span>
</p>
</div>

  </div>
</div> The classical convexity is obtained for \(\tau =e_{1}\) and \(k=2.\) </p>
<p>For \(\tau (x)=e_{2}+\alpha e_{1}\) if <b class="bfseries"> </b>\(f\in C[0,\infty )\) convex functions with respect to \(e_{2}+\alpha e_{1}\) (in classical sense), it fulfills</p>
<div class="displaymath" id="a0000000009">
  \begin{equation*}  \left\vert \begin{tabular}{lll} $1$ 

&  $1$ 

&  $1$ 

\\ $x_{0}^{2}+\alpha x_{0}$ 

&  $x_{1}^{2}+\alpha x_{1}$ 

&  $x_{2}^{2}+\alpha x_{2} $ 

\\ $f(x_{0})$ 

&  $f(x_{1})$ 

&  $f(x_{2})$

\end{tabular}\right\vert \geq 0,\quad 0{\lt}x_{0}{\lt}x_{1}{\lt}x_{2}{\lt}\infty \end{equation*}
</div>
<p>or equivalently to \(\left( \text{\ref{3}}\right) \) for \(f\in C^{2}[0,\infty ),\) \(k=2\) </p>
<div class="displaymath" id="a0000000010">
  \begin{equation*}  f^{\prime \prime }(x)-\tfrac {2}{2x+\alpha }f^{\prime }(x)\geq 0,\quad x{\gt}0 \end{equation*}
</div>
<p>for \(\alpha {\gt}0\). </p>
<p>In this paper, mostly the name of \(\tau \) convexity \( \left( \tau (x)=x^{2}+\alpha x\right) \) is used instead of generalized convexity. </p>
<p>The function space \(C_{2}[0,\infty )\) is defined by\(,\)</p>
<div class="displaymath" id="a0000000011">
  \begin{equation*}  C_{2}[0,\infty )=\left\{  f\in C[0,\infty ):f(x)\leq k_{f}(1+x^{2})\right\}  \end{equation*}
</div>
<p>where \(k_{f}\) is a constant depending on \(f\) and</p>
<div class="displaymath" id="a0000000012">
  \begin{equation*}  C_{2}^{\ast }[0,\infty )=\left\{  f\in C[0,\infty ):\lim _{x\rightarrow \infty }\tfrac {f(x)}{1+x^{2}}=k_{f}\right\}  . \end{equation*}
</div>
<p>The space \(C_{2}[0,\infty )\) is endowed with the norm</p>
<div class="displaymath" id="a0000000013">
  \begin{equation*}  \left\Vert f\right\Vert _{2}=\sup _{x\geq 0}\tfrac {\left\vert f(x)\right\vert }{1+x^{2}}. \end{equation*}
</div>
<h1 id="a0000000014">2 Shape Preserving Properties</h1>
<p>Let \(e_{i}(x)=x^{i},\  i=0,1,2.\) For \(\alpha \geq 0\) and \(n{\gt}0,\) using the results for the Szász-Mirakyan operator (<a href="#2">2</a>), it is found that</p>
<div class="equation" id="M1">
<p>
  <div class="equation_content">
    \begin{equation}  S_{n,\alpha }e_{0}=e_{0},\  \  \  \  \  \  \  \  S_{n,\alpha }e_{1}=r_{n,\alpha }, \  \  \  \  \  \  S_{n,\alpha }e_{2}=\left( r_{n,\alpha }\right) ^{2}+\tfrac {r_{n,\alpha }}{n}. \label{M1} \end{equation}
  </div>
  <span class="equation_label">4</span>
</p>
</div>
<p>In view of the definition of \(~ r_{n,\alpha }\) yields</p>
<div class="equation" id="M2">
<p>
  <div class="equation_content">
    \begin{equation}  S_{n,\alpha }\left( e_{2}+\alpha e_{1}\right) \left( x\right) =x^{2}+\alpha x. \label{M2} \end{equation}
  </div>
  <span class="equation_label">5</span>
</p>
</div>
<p>To obtain the shape preserving properties, we need to find the first and second order derivatives of \(S_{n,\alpha }f.\) For Szász–Mirakyan operators similar results were first established in <span class="cite">
	[
	<a href="#Stancu" >12</a>
	]
</span>. </p>
<p><div class="lemma_thmwrapper " id="Lemma A">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">2</span>
  </div>
  <div class="lemma_thmcontent">
  <p> For any \(f\in C[0,\infty ),\  n\in \mathbb {N} \) and \(x\in \lbrack 0,\infty )\) we have </p>
<p>\(a)\)</p>
<div class="displaymath" id="a0000000015">
  \begin{equation*}  S_{n,\alpha }^{\prime }f(x)=nr_{n,\alpha }^{\prime }(x)\sum \limits _{k=0}^{\infty }\left\{  f\left( \tfrac {k+1}{n}\right) -f\left( \tfrac {k}{n}\right) \right\}  \mathcal{P}_{n,k,\alpha }(x), \end{equation*}
</div>
<p>\(b)\) </p>
<div class="displaymath" id="a0000000016">
  \begin{eqnarray*}  S_{n,\alpha }^{\prime \prime }f(x) & =& nr_{n,\alpha }^{\prime \prime }(x)\sum \limits _{k=0}^{\infty }\left\{  f\left( \tfrac {k+1}{n}\right) -f\left( \tfrac {k}{n}\right) \right\}  \mathcal{P}_{n,k,\alpha }(x) \\ & & +n^{2}\left( r_{n,\alpha }^{\prime }(x)\right) ^{2}\sum \limits _{k=0}^{\infty }\left\{  f\left( \tfrac {k+2}{n}\right) -2f\left( \tfrac {k+1}{n}\right) +f\left( \tfrac {k}{n}\right) \right\}  \mathcal{P}_{n,k,\alpha }(x). \end{eqnarray*}
</div>
<p>Calculating the first and the second order derivatives of  \(r_{n,\alpha },\) it is directly seen that</p>
<div class="equation" id="K3">
<p>
  <div class="equation_content">
    \begin{equation}  r_{n,\alpha }^{\prime }(x)=\tfrac {1}{2}\tfrac {(2x+\alpha )}{\sqrt{\frac{(\alpha n+1)^{2}}{4n^{2}}+(x^{2}+\alpha x)}} \label{K3} \end{equation}
  </div>
  <span class="equation_label">6</span>
</p>
</div>
<p>and</p>
<div class="displaymath" id="a0000000017">
  \begin{equation*}  r_{n,\alpha }^{\prime \prime }(x)=\tfrac {1}{2}\left( \tfrac {\left( \alpha n+1\right) ^{2}}{4n^{2}}+\left( x^{2}+\alpha x\right) \right) ^{-\frac{3}{2}}\left( \tfrac {\alpha }{n}+\tfrac {1}{2n^{2}}\right) . \end{equation*}
</div>
<p>\(c)\) For \(r_{n,\alpha }(x)\in \lbrack 0,\infty )/\left\{  \frac{k}{n}, k=0,1,...\right\}  ,\) we have</p>
<div class="displaymath" id="a0000000018">
  \begin{equation*}  S_{n,\alpha }f(x)-f(r_{n,\alpha }(x))=\tfrac {r_{n,\alpha }(x)}{n}\sum \limits _{k=0}^{\infty }f\big[ r_{n,\alpha }(x),\tfrac {k}{n},\tfrac {k+1}{n}\big] \mathcal{P}_{n,k,\alpha }(x), \end{equation*}
</div>
<p>where \(f\left[ x_{0},x_{1},x_{2}\right] \) is the divided differences of \(f\) with respect to \(x_{0},x_{1}\) and \(x_{2}\) such that \(0\leq x_{0}{\lt}x_{1}{\lt}\) \(x_{2}{\lt}\infty .\) </p>

  </div>
</div> </p>
<p>By applying first and second order derivatives of the operators, it leads to the following theorem: </p>
<p><div class="theorem_thmwrapper " id="a0000000019">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Suppose that \(f\in C[0,\infty ),\  n\in \mathbb {N} \) and \(x\in \lbrack 0,\infty ).\) Then we have </p>
<p>\(i)\) If \(f\in C[0,\infty )\) is increasing, then \(S_{n,\alpha }f\) is increasing, </p>
<p>\(ii)\  \)If \(f\in C[0,\infty )\) is increasing and convex, then \(S_{n,\alpha }f\) is convex. </p>

  </div>
</div> </p>
<p><div class="theorem_thmwrapper " id="a0000000020">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">4</span>
  </div>
  <div class="theorem_thmcontent">
  <p>If \(f\) is convex and decreasing, we have </p>
<div class="equation" id="G1">
<p>
  <div class="equation_content">
    \begin{equation}  S_{n,\alpha }f\left( x\right) \geq f\left( x\right) ,\  x\in \lbrack 0,\infty ). \label{G1} \end{equation}
  </div>
  <span class="equation_label">7</span>
</p>
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000021">
  <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 know that a function \(f\) is convex if and only if all second order divided differences of \(f\) are nonnegative, (see <span class="cite">
	[
	<a href="#Phillps" >11</a>
	, 
	p. 259
	]
</span>). Thus, using Lemma <a href="#Lemma A">2</a>, \(c)\) we have \(S_{n,\alpha }f\left( x\right) \geq f\left( r_{n,\alpha }(x)\right) ,\) \(x\in \lbrack 0,\infty ).\) Also we know that <span class="cite">
	[
	<a href="#1" >1</a>
	, 
	Lemma 2.2
	]
</span> the inequality </p>
<div class="displaymath" id="A1">
  \begin{align}  0{\lt}r_{n,\alpha }(x){\lt}x{\lt}\infty \label{A1} \end{align}
</div>
<p>holds true. Considering \(f\) is decreasing function, we have the desired result. <div class="proof_wrapper" id="a0000000022">
  <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>As an immediate consequence of the above result, one can stated the following theorem. </p>
<p><div class="theorem_thmwrapper " id="a0000000023">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">5</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(n\in \mathbb {N} ,\  \alpha \in \lbrack 0,\infty )\) and \(f\in C[0,\infty )\) be convex with respect to \(e_{2}+\alpha e_{1},\) \(\alpha {\gt}0\). Then we have </p>
<div class="displaymath" id="a0000000024">
  \begin{equation*}  f(x)\leq S_{n,\alpha }f(x)\leq S_{n}f(x),\quad x{\gt}0. \end{equation*}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000025">
  <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 the remark of <span class="cite">
	[
	<a href="#13" >14</a>
	, 
	Remark, p.426
	]
</span>, we know that </p>
<div class="displaymath" id="a0000000026">
  \begin{equation*}  S_{n,\alpha }f(x)\geq f(x),\quad x\geq 0 \end{equation*}
</div>
<p>because of the \(f\) is convex with respect to \(\tau =\tfrac {e_{2}+\alpha e_{1}}{1+\alpha },\  \) \(\alpha {\gt}0.\  \)It is known that the Szász-Mirakyan operators, \(n\in \mathbb {N} \), of a convex function \(f\), satisfy </p>
<div class="displaymath" id="a0000000027">
  \begin{equation*}  S_{n}f(x)\geq f(x) \end{equation*}
</div>
<p>for all \(n\in \mathbb {N} \) and \(\  \alpha \in \lbrack 0,\infty )\) (see <span class="cite">
	[
	<a href="#13" >14</a>
	, 
	Remark, p. 438
	]
</span>). Thus, since \(\tau \) is convex we have </p>
<div class="displaymath" id="a0000000028">
  \begin{equation*}  S_{n}(\tau )\geq \tau . \end{equation*}
</div>
<p>Since \(\left( S_{n}(\tau )\right) ^{-1}\) is increasing, so we get</p>
<div class="displaymath" id="a0000000029">
  \begin{eqnarray*}  \left( S_{n}(\tau )\right) ^{-1}\circ \left( S_{n}(\tau )\right) & \geq & \left( S_{n}(\tau )\right) ^{-1}\circ \tau . \end{eqnarray*}
</div>
<p>Thus, we have </p>
<div class="displaymath" id="a0000000030">
  \begin{equation*}  x\geq \left( \left( S_{n}(\tau )\right) ^{-1}\circ \tau \right) (x). \end{equation*}
</div>
<p>Applying the operator \(S_{n}f\) on both sides of the above inequality (\(S_{n}\) is monotone operator), then we obtain</p>
<div class="displaymath" id="a0000000031">
  \begin{equation*}  S_{n}f(x)\geq S_{n}(\left( S_{n}(\tau )\right) ^{-1}\circ \tau )(x)=S_{n,\alpha }f(x), \end{equation*}
</div>
<p>which completes the proof. <div class="proof_wrapper" id="a0000000032">
  <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="theorem_thmwrapper " id="a0000000033">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">6</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let<b class="bfseries"> </b>\(n\in \mathbb {N} ,\  \alpha \in \lbrack 0,\infty )\)<b class="bfseries"> </b>and<b class="bfseries"> </b>\(f\in C[0,\infty )\) then</p>
<div class="displaymath" id="a0000000034">
  \begin{equation*}  \lim _{\alpha \rightarrow \infty }S_{n,\alpha }f(x)=S_{n}f(x) \end{equation*}
</div>
<p>uniformly for any closed interval \(\left[ a,b\right] \subset \lbrack 0,\infty ).\) </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000035">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> <b class="bfseries"> </b>It is easily checked that</p>
<div class="displaymath" id="a0000000036">
  \begin{equation*}  \lim _{\alpha \rightarrow \infty }r_{n,\alpha }(x)=x \end{equation*}
</div>
<p>is uniform on \(\left[ a,b\right] .\) This completes the proof. <div class="proof_wrapper" id="a0000000037">
  <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="a0000000038">3 Asymptotic Expression</h1>
<p>We begin by the following Voronovskaya-type theorem: </p>
<p><div class="theorem_thmwrapper " id="a0000000039">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">7</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(f\in C_{2}^{\ast }[0,\infty )\) such that \(f^{\prime },\) \(f^{\prime \prime }\in C_{2}^{\ast }[0,\infty ).\) Then we have</p>
<div class="displaymath" id="a0000000040">
  \begin{equation*}  \lim _{n\rightarrow \infty }2n\left( S_{n,\alpha }f(x)-f(x)\right) =x\left( f^{\prime \prime }(x)-\tfrac {2}{2x+\alpha }f^{\prime }(x)\right) , \end{equation*}
</div>
<p>for every \(x\in \lbrack 0,\infty ).\) </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> Let \(f,\) \(f^{\prime }\) and \(f^{\prime \prime }\in C_{2}^{\ast }[0,\infty )\) and \(x\in \lbrack 0,\infty )\) be fixed. By the Taylor formula, we can write</p>
<div class="equation" id="T1">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( t\right) =f\left( x\right) +f^{\prime }\left( x\right) \left( t-x\right) +\tfrac {1}{2}f^{\prime \prime }\left( x\right) \left( t-x\right) ^{2}+\lambda _{x}\left( t\right) \left( t-x\right) ^{2}, \label{T1} \end{equation}
  </div>
  <span class="equation_label">9</span>
</p>
</div>
<p> where \(\lambda _{x}\left( t\right) \in C_{2}^{\ast }[0,\infty )\) and \(\lim _{t\rightarrow x}\lambda _{x}\left( t\right) =0.\) From (<a href="#T1">9</a>) we have</p>
<div class="displaymath" id="a0000000042">
  \begin{eqnarray*}  2n\left( S_{n,\alpha }f(x)-f(x)\right) & =& f^{\prime }\left( x\right) 2nS_{n,\alpha }\left( e_{1}-xe_{0}\right) (x) \\ & & +\tfrac {1}{2}f^{\prime \prime }\left( x\right) 2nS_{n,\alpha }\left( e_{1}-xe_{0}\right) ^{2}(x)\\ & & +2nS_{n,\alpha }\lambda _{x}\left( \cdot \right) \left( e_{1}-xe_{0}\right) ^{2}(x). \end{eqnarray*}
</div>
<p>By the Cauchy-Schwarz inequality, we have </p>
<div class="equation" id="T2">
<p>
  <div class="equation_content">
    \begin{equation}  S_{n,\alpha }\lambda _{x}\left( \cdot \right) \left( e_{1}-xe_{0}\right) ^{2}(x)\leq \sqrt{S_{n,\alpha }\lambda _{x}^{2}\left( \cdot \right) \left( x\right) }\sqrt{S_{n,\alpha }\left( e_{1}-xe_{0}\right) ^{4}(x)}. \label{T2} \end{equation}
  </div>
  <span class="equation_label">10</span>
</p>
</div>
<p> To prove Voronovskaya-type theorem, we must compute \(S_{n,\alpha }e_{i}^{x},\  i=1,2,3,4\). </p>
<p>Here \(e_{i}^{x}(t):=(t-x)^{i}.\) Employing the definition of \(r_{n,\alpha },\) we can compute following limits. </p>
<div class="displaymath" id="a0000000043">
  \begin{eqnarray*}  \lim _{n\rightarrow \infty }2nS_{n,\alpha }e_{1}^{x}(x)& =& \lim _{n\rightarrow \infty }2nS_{n,\alpha }\left( e_{1}-xe_{0}\right) (x)\\ & =& \lim _{n\rightarrow \infty }2n\left( r_{n,\alpha }(x)-x\right) \\ & =& -\tfrac {2x}{\left( \alpha +2x\right) } \end{eqnarray*}
</div>
<p>and using the last limit, we get</p>
<div class="displaymath" id="a0000000044">
  \begin{eqnarray*}  \lim _{n\rightarrow \infty }2nS_{n,\alpha }e_{2}^{x}(x) & =& \lim _{n\rightarrow \infty }2n(S_{n,\alpha }e_{2}(x)-2xS_{n,\alpha }e_{1}(x)+x^{2}) \\ & =& \lim _{n\rightarrow \infty }2n(S_{n,\alpha }\left( e_{2}+\alpha e_{1}\right) (x)-\left( \alpha +2x\right) r_{n,\alpha }(x)+x^{2}) \\ & =& \lim _{n\rightarrow \infty }2n\left( x^{2}+\alpha x-\left( \alpha +2x\right) r_{n,\alpha }(x)+x^{2}\right) \\ & =& \lim _{n\rightarrow \infty }2n\left( \alpha +2x\right) \left( x-r_{n,\alpha }(x)\right) \\ & =& 2x. \end{eqnarray*}
</div>
<p>Moreover, we compute the following via Mathematica</p>
<div class="displaymath" id="a0000000045">
  \begin{align*}  nS_{n,\alpha }e_{4}(x) =& \tfrac {n^{4}(2x^{4}+4x^{3}\alpha +6x^{2}\alpha ^{2}+4x\alpha ^{3}+\alpha ^{4})\! \! -\! \! n^{3}(10x^{2}\alpha +10x\alpha ^{2}+2\alpha ^{3})}{2n^{3}} \\ & \quad -\tfrac {n^{3}(2x^{2}\alpha +2x\alpha ^{2}-\alpha ^{3})\sqrt{1+2n\alpha +n^{2}(2x+\alpha )^{2}}}{2n^{3}} \\ & \quad +\tfrac {n^{2}(4x^{2}+4x\alpha +3\alpha ^{2})\sqrt{1+2n\alpha +n^{2}(2x+\alpha )^{2}}}{2n^{3}}+o(1) \\ nS_{n,\alpha }e_{3}(x) =& \tfrac {-n^{3}(3x^{2}\alpha +3x\alpha ^{2}+\alpha ^{3})+n^{2}(3x^{2}+3x\alpha )}{2n^{2}} \\ & \quad +\tfrac {n^{2}(x^{2}+\alpha x+\alpha ^{2})\sqrt{1+2n\alpha +n^{2}(2x+\alpha )^{2}}}{2n^{2}} \\ & \quad -\tfrac {n\alpha \sqrt{1+2n\alpha +n^{2}(2x+\alpha )^{2}}}{2n^{2}}+o(1) \\ nS_{n,\alpha }e_{2}(x) =& \tfrac {1}{2}\left( \alpha +n(2x^{2}+2x\alpha +\alpha ^{2})-\alpha \sqrt{1+2n\alpha +n^{2}(2x+\alpha )^{2}}\right) \\ nS_{n,\alpha }e_{1}(x) =& -\tfrac {1}{2}\left( 1+n\alpha \right) +\tfrac {1}{2}\sqrt{1+2n\alpha +n^{2}(2x+\alpha )^{2}}. \end{align*}
</div>
<p>Then if we use equalities which are mentioned above, we obtain </p>
<div class="displaymath" id="a0000000046">
  \begin{align*}  nS_{n,\alpha }e_{4}^{x}(x) =& n\left( S_{n,\alpha }e_{4}-4xS_{n,\alpha }e_{3}+6x^{2}S_{n,\alpha }e_{2}-4x^{3}S_{n,\alpha }e_{1}+x^{4}S_{n,\alpha }e_{0}\right) (x) \\ =& \tfrac {n^{4}(16x^{4}+32x^{3}\alpha +24x^{2}\alpha ^{2}+8x\alpha ^{3}+\alpha ^{4})}{2n^{3}} \\ & \quad -\tfrac {n^{3}(8x^{3}+12x^{2}\alpha +6x\alpha ^{2}+\alpha ^{3})\sqrt{1+2n\alpha +n^{2}(2x+\alpha )^{2}}}{2n^{3}} \\ & \quad -\tfrac {n^{3}(8x^{3}+16x^{2}\alpha +10x\alpha ^{2}+2\alpha ^{3})}{2n^{3}} \\ & \quad +\tfrac {n^{2}(4x^{2}+8x\alpha +3\alpha ^{2})\sqrt{1+2n\alpha +n^{2}(2x+\alpha )^{2}}}{2n^{3}}+o(1). \end{align*}
</div>
<p>Finally by letting \(n\rightarrow \infty ,~ \)we get</p>
<div class="displaymath" id="a0000000047">
  \begin{align*}  \lim _{n\rightarrow \infty }nS_{n,\alpha }e_{4}^{x}(x) =& -\tfrac {(8x^{3}+12x^{2}\alpha +6x\alpha ^{2}+\alpha ^{3})\alpha }{2(2x+\alpha )}-(4x^{3}+8x^{2}\alpha +5x\alpha ^{2}+\alpha ^{3}) \\ & \quad +\tfrac {(2x+\alpha )(4x^{2}+8x\alpha +3\alpha ^{2})}{2} \\ =& 0. \end{align*}
</div>
<p>Putting this results in (<a href="#T2">10</a>), we have the desired result. <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> </p>
<h1 id="a0000000049">4 Rate of Convergence of \(S_{ {\lowercase {n}} ,\alpha }{\lowercase {f}} \)</h1>
<p>In this Section, the rate of convergence of \(S_{n,\alpha }\) operators in terms of both the weighted modulus of continuity and classical one is obtained. </p>
<p>Examining relations (<a href="#M1">4</a>) and using the fact that \(r_{n,\alpha }\rightarrow x\) as \(n\rightarrow \infty ,\) then, on the basis of Korovkin’s first theorem, we observe that \(S_{n,\alpha }\) is an approximation process on compact subsets included in \([0,\infty ).\) Now, we want to give sufficient conditions which ensure both uniform convergence of the sequence \(S_{n,\alpha }\) to the identity operator on the whole interval \([0,\infty )\) and the rate of convergence. For Bernstein type operators, a similar result was first established in <span class="cite">
	[
	<a href="#7" >6</a>
	]
</span>. This problem was further studied by de la Cal and Carcamo in <span class="cite">
	[
	<a href="#7" >6</a>
	]
</span>. </p>
<p><div class="theorem_thmwrapper " id="a0000000050">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">8</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Assume that \(f:[0,\infty )\rightarrow \mathbb {R}\) is bounded. Let </p>
<div class="displaymath" id="a0000000051">
  \begin{equation*}  f^{\ast }(z)=f(z^{2}),\quad z\in \lbrack 0,\infty ). \end{equation*}
</div>
<p>We have for all \(x\geq 0\) and \(\alpha {\gt}0\)</p>
<div class="equation" id="S">
<p>
  <div class="equation_content">
    \begin{equation}  \left\vert S_{n,\alpha }f(x)-f(x)\right\vert \leq 6w\left( f^{\ast };\tfrac {1}{\sqrt{n}}\right) . \label{S} \end{equation}
  </div>
  <span class="equation_label">11</span>
</p>
</div>
<p>Therefore, \(S_{n,\alpha }f\) converges to \(f\) uniformly on \([0,\infty )\) as \(n\rightarrow \infty \), whenever \(f^{\ast }\  \)is uniformly continuous. </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000052">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Let \(x{\gt}0\) be arbitrary fixed. Since</p>
<div class="displaymath" id="a0000000053">
  \begin{equation*}  S_{n,\alpha }f(x)=S_{n,\alpha }(f^{\ast }(\sqrt{e_{1}});x) \end{equation*}
</div>
<p>using the the definition of modulus of smoothness \(w\left( f^{\ast };\cdot \right) ,\) we have </p>
<div class="displaymath" id="a0000000054">
  \begin{align*}  \left\vert S_{n,\alpha }f(x)-f(x)\right\vert & =\left\vert S_{n,\alpha }(f^{\ast }(\sqrt{e_{1}});x)-f^{\ast }(\sqrt{x})\right\vert \\ & \leq S_{n,\alpha }\left( \left\vert f^{\ast }(\sqrt{e_{1}})-f^{\ast }(\sqrt{x})\right\vert ;x\right) \\ & \leq S_{n,\alpha }\left( w\left( f^{\ast };\vert \sqrt{t}-\sqrt{x}\vert \right) ;x\right) \\ & =S_{n,\alpha }\! \! \left( w\left( f^{\ast };\tfrac {\left\vert \sqrt{t}\! -\! \sqrt{x}\right\vert }{S_{n,\alpha }\! \left( \vert \sqrt{e_{1}}\! -\! \sqrt{x}e_{0}\vert ;x\right) }S_{n,\alpha }\! \left( \vert \sqrt{e_{1}}\! -\! \sqrt{x}e_{0}\vert ;x\right) \right) \right) . \end{align*}
</div>
<p>Further on, using the property \(w\left( f^{\ast };\right) \), we have</p>
<div class="displaymath" id="a0000000055">
  \begin{align}  \left\vert S_{n,\alpha }f(x)-f(x)\right\vert & \leq w\left( f^{\ast };S_{n,\alpha }\left( \left\vert \sqrt{e_{1}}-\sqrt{x}e_{0}\right\vert ;x\right) \right) \notag \\ & \quad \times \left( 1+\tfrac {1}{S_{n,\alpha }\left( \vert \sqrt{e_{1}}-\sqrt{x}e_{0}\vert ;x\right) }S_{n,\alpha }\left( \left\vert \sqrt{e_{1}}-\sqrt{x}e_{0}\right\vert ;x\right) \right) \notag \\ & =2w\left( f^{\ast };S_{n,\alpha }\left( \left\vert \sqrt{e_{1}}-\sqrt{x}e_{0}\right\vert ;x\right) \right) . \label{S2} \end{align}
</div>
<p>Using Cauchy- Schwarz inequality, we have </p>
<div class="displaymath" id="a0000000056">
  \begin{eqnarray}  S_{n,\alpha }\left( \left\vert \sqrt{e_{1}}-\sqrt{x}e_{0}\right\vert ;x\right) & =& S_{n,\alpha }\left( \tfrac {\left\vert e_{1}-xe_{0}\right\vert }{\sqrt{e_{1}}+\sqrt{x}};x\right) \notag \\ & \leq & \tfrac {1}{\sqrt{x}}S_{n,\alpha }\left( (e_{1}-xe_{0})^{2};x\right) ^{1/2}. \label{S1} \end{eqnarray}
</div>
<p>From (<a href="#M1">4</a>) and (<a href="#M2">5</a>), we deduce</p>
<div class="displaymath" id="a0000000057">
  \begin{eqnarray*}  S_{n,\alpha }\left( (e_{1}-xe_{0})^{2};x\right) & =& S_{n,\alpha }\left( e_{2}+\alpha e_{1}\right) \left( x\right) -\left( 2x+\alpha \right) S_{n,\alpha }\left( e_{1}\right) \left( x\right) +x^{2} \\ & =& \left( 2x+\alpha \right) \left( x-r_{n,\alpha }(x)\right) . \end{eqnarray*}
</div>
<p>Clearly</p>
<div class="displaymath" id="a0000000058">
  \begin{align}  x-r_{n,\alpha }(x) =& x+\tfrac {(\alpha n+1)}{2n}-\sqrt{\tfrac {(\alpha n+1)^{2}}{4n^{2}}+(x^{2}+\alpha x)} \notag \\ =& \tfrac {x}{n\left( x+\frac{(\alpha n+1)}{2n}+\sqrt{\frac{(\alpha n+1)^{2}}{4n^{2}}+(x^{2}+\alpha x)}\right) } \notag \\ \leq & \tfrac {x}{nx+\alpha n+1}. \label{K2} \end{align}
</div>
<p> According to (<a href="#S1">13</a>) we have</p>
<div class="displaymath" id="a0000000059">
  \begin{equation*}  S_{n,\alpha }\left( \left\vert \sqrt{e_{1}}-\sqrt{x}e_{0}\right\vert ;x\right) \leq \tfrac {\sqrt{3}}{\sqrt{n}}. \end{equation*}
</div>
<p>Thus by (<a href="#S2">12</a>), we have the inequality (<a href="#S">11</a>). </p>
<p>Under the hypothesis of our theorem, \(f^{\ast }\) is uniformly continous on \([0,\infty ),\) we know that \(\lim _{\delta \rightarrow 0}w\left( f^{\ast };\delta \right) =0.\) Since the inequality (<a href="#S">11</a>) valid for all \(x\in \lbrack 0,\infty )\) leads us to the conclusion of our theorem. <div class="proof_wrapper" id="a0000000060">
  <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>Now, we focus on weighted space \(C_{2}[0,\infty )\). Using (<a href="#M1">4</a>) and (<a href="#A1">8</a>), then we obtain</p>
<div class="displaymath" id="a0000000061">
  \begin{align*}  \dfrac {S_{n,\alpha }\left( e_{2}+e_{0}\right) \left( x\right) }{1+x^{2}}=& \dfrac {1+r_{n,\alpha }^{2}\left( x\right) +\frac{r_{n,\alpha }\left( x\right) }{n}}{1+x^{2}} \\ \leq & \dfrac {1+x^{2}+\frac{x}{n}}{1+x^{2}} \\ \leq & 3.\end{align*}
</div>
<p> Therefore, we can say that \(S_{n,\alpha }\) acts from \(C_{2}[0,\infty )\) to \(C_{2}[0,\infty ).\) Also, we give an estimation in terms of following weighted modulus of continuity. It is known that, if \(f\) is not uniformly continuous on the interval \([0,\infty )\), then the usual first modulus of continuity \(w(f,\delta )\) does not tend to zero, as \(\delta \rightarrow 0\).<b class="bfseries"> </b>Here, we use the following weighted modulus of continuity to gain this property. For \(f\in C_{2}[0,\infty )\) and for every \(\delta {\gt}0,\) the weighted modulus of continuity considered in <span class="cite">
	[
	<a href="#10" >8</a>
	]
</span> is defined as follows: </p>
<div class="equation" id="U1">
<p>
  <div class="equation_content">
    \begin{equation}  \Omega (f,\delta )=\sup _{\substack { x\geq 0 \\ \left\vert h\right\vert <\delta }}\frac{\left\vert f(x+h)-f(x)\right\vert }{(1+x^{2})(1+h^{2})}. \label{U1} \end{equation}
  </div>
  <span class="equation_label">15</span>
</p>
</div>
<p>It is known that for every \(f\in C_{2}^{\ast }[0,\infty ),\) \(\Omega (f,\delta ),\) \(\delta {\gt}0,\) the following properties hold true</p>
<div class="displaymath" id="a0000000062">
  \begin{equation*}  \lim _{\delta \rightarrow 0}\Omega (f,\delta )=0 \end{equation*}
</div>
<p>and </p>
<div class="equation" id="U2">
<p>
  <div class="equation_content">
    \begin{equation}  \Omega (f,\lambda \delta )\leq 2\left( 1+\lambda \right) \left( 1+\delta ^{2}\right) \Omega (f,\delta ),\qquad \text{ }\lambda >0. \label{U2} \end{equation}
  </div>
  <span class="equation_label">16</span>
</p>
</div>
<p>For \(f\in C_{2}[0,\infty ),\) by (<a href="#U1">15</a>) and (<a href="#U2">16</a>), we can write </p>
<div class="equation" id="U3">
<p>
  <div class="equation_content">
    \begin{equation}  \left\vert f\left( t\right) -f(x)\right\vert \leq 2(1+x^{2})(1+\delta ^{2})\left( 1+\tfrac {\left\vert t-x\right\vert }{\delta }\right) \left( 1+\left( t-x\right) ^{2}\right) \Omega (f,\delta ),\  \  t,x\geq 0. \label{U3} \end{equation}
  </div>
  <span class="equation_label">17</span>
</p>
</div>
<p><div class="theorem_thmwrapper " id="a0000000063">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">9</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(f\in C_{2}^{\ast }[0,\infty ),\) then for all \(x\in \lbrack 0,\infty )\) we have the following inequality </p>
<div class="displaymath" id="a0000000064">
  \begin{equation*}  \frac{\left\vert S_{n,\alpha }f(x)-f(x)\right\vert }{\left( 1+x^{2}\right) ^{5/2}}\leq C_{\alpha }\Omega (f,\tfrac {1}{\sqrt{n}}), \end{equation*}
</div>
<p>where \(C_{\alpha }\) is a positive constant depending only on \(\alpha .\) </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000065">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Let \(n\in \mathbb {N}\) and \(f\in C_{2}^{\ast }[0,\infty ).\) Considering (<a href="#U3">17</a>) with \(\lambda =\left\vert t-x\right\vert \delta ^{-1},\) from Cauchy-Schwarz inequality we can write </p>
<div class="displaymath" id="a0000000066">
  \begin{align*}  \left\vert S_{n,\alpha }f(x)-f(x)\right\vert & =\left\vert \sum \limits _{k=0}^{\infty }\left( f\left( \tfrac {k}{n}\right) -f(x)\right) e^{-nr_{n,\alpha }(x)}\frac{(nr_{n,\alpha }(x))^{k}}{k!}\right\vert \\ & \leq \sum \limits _{k=0}^{\infty }\left\vert f\left( \tfrac {k}{n}\right) -f(x)\right\vert e^{-nr_{n,\alpha }(x)}\frac{(nr_{n,\alpha }(x))^{k}}{k!} \\ & \leq 2(1+x^{2})(1+\delta ^{2})\Omega (f,\delta )\times \\ & \quad \times \sum \limits _{k=0}^{\infty }\left( 1+\tfrac {1}{\delta }\left\vert \tfrac {k}{n}-x\right\vert \right) \left( 1+\left( \tfrac {k}{n}-x\right) ^{2}\right) e^{-nr_{n,\alpha }(x)}\tfrac {(nr_{n,\alpha }(x))^{k}}{k!} \\ & =2(1+x^{2})(1+\delta ^{2})\Omega (f,\delta )\times \\ & \quad \times \Big\{  1\! +\! S_{n,\alpha }e_{2}^{x}+\tfrac {1}{\delta }\left( S_{n,\alpha }e_{2}^{x}\right) ^{1/2} +\tfrac {1}{\delta }\left( S_{n,\alpha }e_{2}^{x}\right) ^{1/2}\left( S_{n,\alpha }e_{4}^{x}\right) ^{1/2}\Big\}  \end{align*}
</div>
<p> and choosing \(\delta =\tfrac {1}{\sqrt{n}},\) then there is a constant \(C_{\alpha }\) depending on \(\alpha \) such that we have</p>
<div class="displaymath" id="a0000000067">
  \begin{equation*}  \frac{\left\vert S_{n,\alpha }f(x)-f(x)\right\vert }{\left( 1+x^{2}\right) ^{5/2}}\leq C_{\alpha }\Omega \Big(f,\tfrac {1}{\sqrt{n}}\Big) \end{equation*}
</div>
<p>and the proof is completed. <div class="proof_wrapper" id="a0000000068">
  <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="a0000000069">5 Convergence of derivative of \(S_{\lowercase {n},\alpha } \lowercase {f}\)</h1>
<p>Before considering the main results of this section, we state the derivative of the operator (<a href="#2">2</a>) in following lemma. </p>
<p><div class="lemma_thmwrapper " id="a0000000070">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">10</span>
  </div>
  <div class="lemma_thmcontent">
  <p>Let \(f\) be a continuously differentiable on \([0,\infty ).\) Then, we have</p>
<div class="equation" id="K1">
<p>
  <div class="equation_content">
    \begin{equation}  S_{n,\alpha }^{\prime }f(x)=e^{-nr_{n,\alpha }(x)}r_{n,\alpha }^{\prime }(x)\sum \limits _{k=0}^{\infty }f^{\prime }\left( \tfrac {k+\phi _{k}}{n}\right) \frac{(nr_{n,\alpha }(x))^{k}}{k!},\quad 0<\phi _{k}<1. \label{K1} \end{equation}
  </div>
  <span class="equation_label">18</span>
</p>
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000071">
  <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 <a href="#Lemma A">2</a>, we know that</p>
<div class="displaymath" id="a0000000072">
  \begin{equation*}  S_{n,\alpha }^{\prime }f(x)=nr_{n,\alpha }^{\prime }(x)\sum \limits _{k=0}^{\infty }\left\{  f\left( \tfrac {k+1}{n}\right) -f\left( \tfrac {k}{n}\right) \right\}  \mathcal{P}_{n,k,\alpha }(x). \end{equation*}
</div>
<p>It is well known that</p>
<div class="displaymath" id="a0000000073">
  \begin{equation*}  f\left( \tfrac {k+1}{n}\right) -f\left( \tfrac {k}{n}\right) =\tfrac {1}{n}f\left[ \tfrac {k}{n},\tfrac {k+1}{n}\right] , \end{equation*}
</div>
<p>where \(f\left[ \frac{k}{n},\frac{k+1}{n}\right] \) is the divided difference of the points \(\frac{k}{n}\) and \(\frac{k+1}{n}.\) </p>
<p>Differentiating (<a href="#2">2</a>) with respect to \(x\) and applying the definition of the divided difference in first derivative of \(S_{n,\alpha },\) we obtain</p>
<div class="displaymath" id="a0000000074">
  \begin{equation*}  S_{n,\alpha }^{\prime }f(x)=e^{-nr_{n,\alpha }(x)}r_{n,\alpha }^{\prime }(x)\sum \limits _{k=0}^{\infty }f\left[ \tfrac {k}{n},\tfrac {k+1}{n}\right] \tfrac {(nr_{n,\alpha }(x))^{k}}{k!}. \end{equation*}
</div>
<p>Moreover, since the divided difference of \(f\) satisfies the equality</p>
<div class="displaymath" id="a0000000075">
  \begin{equation*}  f\left[ \tfrac {k}{n},\tfrac {k+1}{n}\right] =f^{\prime }(\xi ),\quad \tfrac {k}{n}{\lt}\xi {\lt}\tfrac {k+1}{n}, \end{equation*}
</div>
<p>taking \(\xi =\tfrac {k+\phi _{k}}{n},\  0{\lt}\phi _{k}{\lt}1,\) we get</p>
<div class="displaymath" id="a0000000076">
  \begin{equation*}  S_{n,\alpha }^{\prime }f(x)=e^{-nr_{n,\alpha }(x)}r_{n,\alpha }^{\prime }(x)\sum \limits _{k=0}^{\infty }f^{\prime }\left( \tfrac {k+\phi _{k}}{n}\right) \tfrac {(nr_{n,\alpha }(x))^{k}}{k!},\  \  \  \  0{\lt}\phi _{k}{\lt}1 \end{equation*}
</div>
<p>which completes the proof. <div class="proof_wrapper" id="a0000000077">
  <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>First, we shall prove weighted convergence of first derivative of operator (<a href="#2">2</a>) for \(f\in C_{2}^{\ast }[0,\infty )\) in Lipschitz norm. Let \(f\) be continuously differentiable function, which belongs to \(C_{2}[0,\infty )\) and also \(f^{\prime }\) satisfies the Lipschitz condition that is</p>
<div class="displaymath" id="a0000000078">
  \begin{equation*}  \left\vert f^{\prime }(x)-f^{\prime }(t)\right\vert \leq M\left\vert x-t\right\vert ^{\beta }\  ,\  0{\lt}\beta \leq 1\  ,\  \text{for any }x,t\geq 0. \end{equation*}
</div>
<p>In this case, we write \(f^{\prime }\in Lip_{M}\beta .\) </p>
<p><div class="theorem_thmwrapper " id="a0000000079">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">11</span>
  </div>
  <div class="theorem_thmcontent">
  <p>If the function \(f\) is continuously differentiable on \([0,\infty )\) and \(f^{\prime }\) belongs to \(Lip_{M}\beta ,\) then we obtain</p>
<div class="displaymath" id="a0000000080">
  \begin{equation*}  \lim _{n\rightarrow \infty }\sup _{x\in \lbrack 0,\infty )}\frac{\left\vert S_{n,\alpha }^{\prime }f(x)-f^{\prime }(x)\right\vert }{1+x^{\beta +1}}=0. \end{equation*}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000081">
  <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 (<a href="#K1">18</a>), we can write</p>
<div class="displaymath" id="a0000000082">
  \begin{eqnarray*}  \left\vert S_{n,\alpha }^{\prime }f(x)-f^{\prime }(x)\right\vert & \leq &  \left\vert S _{n,\alpha }^{\prime }f(x)-f^{\prime }(r_{n,\alpha }(x))r_{n,\alpha }^{\prime }(x)\right\vert \\ & & +\left\vert f^{\prime }(r_{n,\alpha }(x))r_{n,\alpha }^{\prime }(x)-f^{^{\prime }}(x)\right\vert \\ & \leq & r_{n,\alpha }^{\prime }(x)\sum \limits _{k=0}^{\infty }\left\vert f^{\prime }\left( \tfrac {k+\phi _{k}}{n}\right) -f^{\prime }(r_{n,\alpha }(x))\right\vert \mathcal{P}_{n,k,\alpha }(x) \\ & & +\left\vert f^{\prime }(r_{n,\alpha }(x))r_{n,\alpha }^{\prime }(x)-f^{\prime }(x)\right\vert . \end{eqnarray*}
</div>
<p>By reason of \(f^{\prime }\in Lip_{M}^{\beta }\), we attain</p>
<div class="displaymath" id="a0000000083">
  \begin{eqnarray*}  \left\vert S_{n,\alpha }^{\prime }f(x)-f^{\prime }(x)\right\vert & \leq & Mr_{n,\alpha }^{\prime }(x)\sum \limits _{k=0}^{\infty }\left\vert \left( \tfrac {k+\phi _{k}}{n}\right) -r_{n,\alpha }(x)\right\vert ^{\beta }\mathcal{P}_{n,k,\alpha }(x) \\ & & +\left\vert f^{\prime }(r_{n,\alpha }(x))r_{n,\alpha }^{\prime }(x)-f^{\prime }(x)\right\vert . \end{eqnarray*}
</div>
<p>Since \(\phi _{k}{\lt}1\), we deduce</p>
<div class="displaymath" id="a0000000084">
  \begin{eqnarray*}  \left\vert S_{n,\alpha }^{\prime }f(x)-f^{\prime }(x)\right\vert & \leq & Mr_{n,\alpha }^{\prime }(x)\sum \limits _{k=0}^{\infty }\left\vert \left( \tfrac {k+1}{n}\right) -r_{n,\alpha }(x)\right\vert ^{\beta }\mathcal{P}_{n,k,\alpha }(x) \\ & & +\left\vert f^{\prime }(r_{n,\alpha }(x))r_{n,\alpha }^{\prime }(x)-f^{\prime }(x)\right\vert . \end{eqnarray*}
</div>
<p>Applying Holder inequality, we have</p>
<div class="displaymath" id="a0000000085">
  \begin{align*} & \left\vert S_{n,\alpha }^{\prime }f(x)-f^{\prime }(x)\right\vert \leq \\ & \leq Mr_{n,\alpha }^{\prime }(x)S_{n,\alpha }\left( \left( e_{1}\left( t\right) +\tfrac {1}{n}-r_{n,\alpha }(x)\right) ^{2};x\right) ^{\frac{\beta }{2}} +\left\vert f^{\prime }(r_{n,\alpha }(x))r_{n,\alpha }^{\prime }(x)-f^{\prime }(x)\right\vert \\ & =Mr_{n,\alpha }^{\prime }(x)\left( S_{n,\alpha }\left( \left( e_{1}\left( t\right) \! -\! r_{n,\alpha }(x)\right) ^{2};x\right)+2\tfrac {1}{n}S_{n,\alpha }\left( e_{1}\left( t\right) \! -\! r_{n,\alpha }(x);x\right) +\tfrac {1}{n^{2}}\right) ^{\frac{\beta }{2}} \\ & \quad +\left\vert f^{\prime }(r_{n,\alpha }(x))r_{n,\alpha }^{\prime }(x)-f^{\prime }(x)\right\vert . \end{align*}
</div>
<p> Let us remark that </p>
<div class="displaymath" id="a0000000086">
  \begin{align}  \sup _{x\in \lbrack 0,\infty )}& \frac{r_{n,\alpha }^{\prime }(x)S_{n,\alpha }\left( \left( e_{1}\left( t\right) -r_{n,\alpha }(x)\right) ^{2};x\right) ^{\frac{\beta }{2}}}{1+x^{\beta +1}} \nonumber \\ & =\sup _{x\in \lbrack 0,\infty )}\frac{r_{n,\alpha }(x)^{\frac{\beta }{2}}}{\left( 1+x^{\beta +1}\right) n}\rightarrow 0, \quad \text{ as}\  n\rightarrow \infty , \label{L1} \end{align}
</div>
<p>\(S_{n,\alpha }\left( e_{1}\left( t\right) -r_{n,\alpha }(x);x\right) =0\) and </p>
<div class="equation" id="L2">
<p>
  <div class="equation_content">
    \begin{equation}  \sup _{x\in \lbrack 0,\infty )}\frac{r_{n,\alpha }^{\prime }(x)}{1+x^{\beta +1}}\rightarrow 0, \quad \text{ as\  }n\rightarrow \infty . \label{L2} \end{equation}
  </div>
  <span class="equation_label">20</span>
</p>
</div>
<p>On the other hand,</p>
<div class="displaymath" id="a0000000087">
  \begin{equation*}  f^{\prime }(r_{n,\alpha }(x))r_{n,\alpha }^{\prime }(x)-f^{\prime }(x)=\left( f^{\prime }(r_{n,\alpha }(x))-f^{\prime }(x)\right) r_{n,\alpha }^{\prime }(x)-f^{\prime }(x)\left( 1-r_{n,\alpha }^{\prime }(x)\right) . \end{equation*}
</div>
<p>Since \(f^{\prime }\) \(\in \) \(Lip_{M}\beta ,\) we can write</p>
<div class="displaymath" id="a0000000088">
  \begin{equation*}  \left\vert f^{\prime }(r_{n,\alpha }(x))-f^{\prime }(x)\right\vert \leq M\left\vert r_{n,\alpha }(x)-x\right\vert ^{\beta } \end{equation*}
</div>
<p>and</p>
<div class="displaymath" id="a0000000089">
  \begin{equation*}  \left\vert f^{\prime }(x)\right\vert \leq \left\vert f^{\prime }(0)\right\vert +Mx^{\beta }\leq M_{f}\left( 1+x^{\beta }\right) , \end{equation*}
</div>
<p>where \(M_{f}=\max \left\{  \left\vert f^{\prime }(0)\right\vert ,M\right\}  .\) </p>
<p>Now by (<a href="#K3">6</a>) and (<a href="#K2">14</a>) , we get</p>
<div class="displaymath" id="a0000000090">
  \begin{eqnarray}  \frac{\left\vert f^{\prime }(r_{n,\alpha }(x))r_{n,\alpha }^{\prime }(x)-f^{\prime }(x)\right\vert }{1+x^{\beta +1}} & \leq & \frac{\left\vert f^{\prime }(r_{n,\alpha }(x))-f^{\prime }(x)\right\vert }{1+x^{\beta +1}}r_{n,\alpha }^{\prime }(x) \nonumber \\ & & +\frac{\left\vert f^{\prime }(x)\right\vert }{1+x^{\beta +1}}\left\vert 1-r_{n,\alpha }^{\prime }(x)\right\vert \notag \\ & \leq & M\frac{\left\vert r_{n,\alpha }(x)-x\right\vert ^{\beta }}{1+x^{\beta +1}}+\frac{\left\vert f^{\prime }(x)\right\vert }{1+x^{\beta +1}}\left\vert 1-r_{n,\alpha }^{\prime }(x)\right\vert \notag \\ & \leq & M\frac{x^{\beta }}{\left( 1+x^{\beta +1}\right) \left( \alpha n+1\right) ^{\beta }} \nonumber \\ & & +M_{f}\frac{2}{\alpha ^{2}}\frac{1+x^{\beta }}{1+x^{\beta +1}}\left( \frac{\alpha }{2n}+\frac{1}{4n^{2}}\right) . \label{L3} \end{eqnarray}
</div>
<p>Combining (<a href="#L1">19</a>), (<a href="#L2">20</a>) and (<a href="#L3">21</a>), we have the the desired result. <div class="proof_wrapper" id="a0000000091">
  <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="a0000000092">6 Comparison with \(S_{\lowercase {n}}({\lowercase {f}})\)</h1>
<p>Let us denote \(w(f,\delta )\) be the first order modulus of continuity of \(f\in C_{B}[0,\infty )\) (the space of all bounded and continuous functions on\([0,\infty )\))\(,\) where </p>
<div class="displaymath" id="a0000000093">
  \begin{equation*}  w(f;\delta )=\sup _{0{\lt}\left\vert x-y\right\vert {\lt}\delta }\left\vert f(x)-f(y)\right\vert . \end{equation*}
</div>
<p>We have the following estimates for \(S_{n}f\), \(D_{n}^{\ast }f\) and \(S_{n,\alpha }f\) in terms of the modulus of continuity \(w(f,\delta )\), </p>
<div class="displaymath" id="a0000000094">
  \begin{eqnarray*}  \left\vert S_{n}f(x)-f(x)\right\vert & \leq & 2w(f;\delta _{1}(x)) \\ \left\vert S_{n,\alpha }f(x)-f(x)\right\vert & \leq & 2w(f,\delta _{2}(x)) \end{eqnarray*}
</div>
<p>where \(\delta _{1}^{2}(x)=\frac{x}{n}\) and \(\delta _{2}^{2}(x)=(\alpha +2x)(x-r_{n,\alpha }(x)).\) </p>
<p>In the following theorem, we present analogues theorem for \(S_{n,\alpha }f\) to show a better order of approximation. </p>
<p><div class="theorem_thmwrapper " id="a0000000095">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">12</span>
  </div>
  <div class="theorem_thmcontent">
  <p>For every \(f\in C_{B}[0,\infty ),\) \(x\geq 0\) and \(n\in \mathbb {N} ,\) we have </p>
<div class="displaymath" id="a0000000096">
  \begin{equation*}  \delta _{2}(x)\leq \delta _{1}(x) \end{equation*}
</div>
<p>and one can get the best approximation using \(S_{n,\alpha }f.\) </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000097">
  <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 order of approximation to a function \(f\in C_{B}[0,\infty ),\) given by the sequence \(S_{n,\alpha }f\) will be at least as good as of \(S_{n}f\) whenever</p>
<div class="displaymath" id="a0000000098">
  \begin{equation*}  2x^{2}+\alpha x-r_{n,\alpha }(x)(\alpha +2x)\leq \tfrac {x}{n}. \end{equation*}
</div>
<p>Let</p>
<div class="displaymath" id="a0000000099">
  \begin{equation*}  K_{n,\alpha }=2x^{2}+\alpha x-r_{n,\alpha }(x)(\alpha +2x)-\tfrac {x}{n}. \end{equation*}
</div>
<p>For \(x\in \lbrack 0,\infty ),~ \alpha \in \lbrack 0,\infty )\) and \(n{\gt}0,~ \)the only root is \(x=0\) and one can see that \(K_{n,\alpha }\) never changes the sign in \((0,\infty ).~ \)To analyze the sign of the \(K_{n,\alpha }\), we can use the first derivative of \(K_{n,\alpha }\)</p>
<div class="displaymath" id="a0000000100">
  \begin{equation*}  K_{n,\alpha }(0)=0,\  \  \  \  \  K_{n,\alpha }^{\prime }(x){\lt}0 \end{equation*}
</div>
<p>and so that for \(x\in \lbrack 0,\infty ),\) we get \(2x^{2}+\alpha x-r_{n,\alpha }(x)(\alpha +2x)\leq \frac{x}{n}\) that is \(\delta _{2}\leq \delta _{1}.\) <div class="proof_wrapper" id="a0000000101">
  <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><small class="footnotesize">  </small></p>
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</dd>
  <dt><a name="3">2</a></dt>
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  <dd><p><a href ="http://dx.doi.org/10.1016/j.amc.2014.01.046"> <i class="sc">D. Cardenas-Morales, P. Garrancho </i> and <i class="sc">I. Raşa</i>, <i class="it">Approximation properties of Bernstein-Durrmeyer type operators</i>, Appl. Math. Comput., <b class="bf">232</b> (2014), pp.&#160;1–8. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
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</dd>
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</dd>
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  <dd><p><a href ="https://doi.org/10.2298/FIL1301173O"> <i class="sc">O. Duman</i> and <i class="sc">M. Ali Özarslan</i>, <i class="it">Local approximation properties for certain King type operators</i>, Filomat, <b class="bf">27</b> (2013), pp.&#160;173–181. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
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  <dd><p><i class="sc">D. D. Stancu</i>, <i class="it">Approximation of functions by a new class of linear polynomial operators</i>, Rev. Roumanine Math. Pures Appl., <b class="bf">13</b> (1968), pp.&#160;1173–1194. </p>
</dd>
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</dd>
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</a> </p>
</dd>
</dl>


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