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<h1>On Baskakov Operators preserving the exponential function</h1>
<p class="authors">
<span class="author">Övgü Gürel Yılmaz\(^\ast \), Vijay Gupta\(^\bullet \) Ali Aral\(^+\)</span>
</p>
<p class="date">February 2, 2017.</p>
</div>
<p>\(^\ast \) Department of Mathematics, Faculty of Sciences, Ankara University, 06100, Tandogan, Ankara, Turkey, e-mail: <span class="ttfamily">ogurel@ankara.edu.tr</span>. </p>
<p>\(^\bullet \) Department of Mathematics, Netaji Subhas Institute of Technology, Sector 3 Dwarka, New Delhi-110078, India, e-mail: <span class="ttfamily">vijaygupta2001@hotmail.com</span>. </p>
<p>\(^+\) Department of Mathematics, Faculty of Sciences and Arts, Kırıkkale University, Kırıkkale, Turkey, e-mail: <span class="ttfamily">aliaral73@yahoo.com</span>. </p>

<div class="abstract"><p> In this paper, we are concerned about the King-type Baskakov operators defined by means of the preserving functions \(e_{0}\) and \(e^{2ax},\  a{\gt}0\) fixed. Using the modulus of continuity, we show the uniform convergence of the new operators to \(f.\) Also, by analyzing the asymptotic behavior of King-type operators with a Voronovskaya-type theorem, we establish shape preserving properties using the generalized convexity. </p>
<p><b class="bf">MSC.</b> 41A25, 41A36 </p>
<p><b class="bf">Keywords.</b> Baskakov operators, King type operators, Voronovskaya type theorems, modulus of continuity. </p>
</div>
<h1 id="1">1 Introduction</h1>
<p>Approximation theory is one of the important subjects that is frequently used by scientific community. It is divided into many fields one of which is linear positive operators that have an essential role in approximation theory. Some linear positive operators such as Bernstein operators are defined within finite intervals, there are also many linear positive operators that are defined in infinite intervals such as Baskakov operators <span class="cite">
	[
	<a href="#5" >7</a>
	]
</span> defined as</p>
<div class="displaymath" id="a0000000002">
  \begin{equation*}  K_{n}\left( f,x\right) =\tfrac {1}{\left( 1+x\right) ^{n}}\sum \limits _{k=0}^{\infty }f\big( \tfrac {k}{n}\big) \tbinom {n+k-1}k \tfrac {x^{k}}{( 1+x) ^{k}} \label{1} \end{equation*}
</div>
<p> where \(n\in \mathbb {N},\) \(x\in \left[ 0,\infty \right) \) and \(f\in C\left[ 0,\infty \right) \) whenever the above sum converges. </p>
<p>In recent years, several operators were appropriately modified, which preserve the test functions (or the linear combination of them) emerged in this field and many improvements have been made regarding this matter to obtain better approximation. One may see some of the results in the related direction in <span class="cite">
	[
	<a href="#oa1" >2</a>
	]
</span>, <span class="cite">
	[
	<a href="#oa2" >3</a>
	]
</span>, <span class="cite">
	[
	<a href="#2" >4</a>
	]
</span>, <span class="cite">
	[
	<a href="#4" >6</a>
	]
</span>, <span class="cite">
	[
	<a href="#6" >8</a>
	]
</span>, <span class="cite">
	[
	<a href="#10" >12</a>
	]
</span>, <span class="cite">
	[
	<a href="#11" >13</a>
	]
</span>, <span class="cite">
	[
	<a href="#vgrp" >16</a>
	]
</span>, <span class="cite">
	[
	<a href="#vggt" >18</a>
	]
</span>, <span class="cite">
	[
	<a href="#14" >20</a>
	]
</span> and <span class="cite">
	[
	<a href="#16" >22</a>
	]
</span> etc. </p>
<p>In 2003, King <span class="cite">
	[
	<a href="#15" >21</a>
	]
</span> described the modified Bernstein operators which preserve \(e_{0}\) and \(e_{2}\) test functions on \([0,1]\) and examined their approximation properties. Later in 2007, Duman and Özarslan <span class="cite">
	[
	<a href="#12" >15</a>
	]
</span>, introduced Szász-Mirakyan type operators preserving the test function \( e_{0}\) and \(e_{2}\) on the interval \(\left[ 0,\infty \right) \) and achieved better approximation for the generalization of the classical Szász-Mirakyan operators. </p>
<p>In 2006, Morales et al. <span class="cite">
	[
	<a href="#9" >11</a>
	]
</span> have considered the King-type Bernstein polynomials which reproduce the linear combination of test functions for \(\alpha {\gt}0,\) \(e_{2}+\alpha e_{1}\) and analyzed their shape preserving and approximation properties. They gave information about the order of approximation by comparing Bernstein and its modified operators; hence they obtained more general operators than those of King. </p>
<p>Besides preserving the test functions, in 2010 Aldaz and Render <span class="cite">
	[
	<a href="#3" >5</a>
	]
</span> defined linear positive operators which preserve exponential functions and in 2014 Birou <span class="cite">
	[
	<a href="#7" >9</a>
	]
</span> gave some examples about general King-type operators which reproduce polynomials and exponential functions. </p>
<p>Very recently, Acar et al. <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span> proposed the modification of the classical Szász-Mirakyan operators which preserve the functions \(e_{0}\) and \(e^{2ax},\  a{\gt}0.\  \) They investigated the uniform convergence and shape preserving properties of these operators. Using weighted modulus of continuity and first and second order moduli of smoothness, they obtained important approximation properties of the modified operators. A number of results concerning Szász-Mirakyan type operators and generalization of them can be found in <span class="cite">
	[
	<a href="#vgrp" >16</a>
	]
</span>. Also, it was pointed out recently in <span class="cite">
	[
	<a href="#vgnmtr" >17</a>
	]
</span> that by finding the moment generation function, which is common in statistical analysis, one can easily obtain the moments. They obtained m.g.f. and moments of many operators in their recent paper <span class="cite">
	[
	<a href="#vgnmtr" >17</a>
	]
</span>.<br /></p>
<p>The aim of this paper is to consider the modification of the Baskakov operators preserving \(e^{2ax}\). We introduce our operators for \(x\in \left[ 0,\infty \right) ,\) as </p>
<div class="displaymath" id="a0000000003">
  \begin{eqnarray}  K_{n,\beta }(f,x) & =& \sum \limits _{k=0}^{\infty }f\big(\tfrac {k}{n}\big) \tbinom {n+k-1}{k}\tfrac {\left( \beta _{n}\left( x\right) \right) ^{k}}{\left( 1+\beta _{n}\left( x\right) \right) ^{n+k}} \notag \\ & =& \sum \limits _{k=0}^{\infty }f\big( \tfrac {k}{n}\big) \tbinom {-n}{k} \left( -\beta _{n}\left( x\right) \right) ^{k}\left( 1+\beta _{n}\left( x\right) \right) ^{-n-k} \label{3} \\ & =& K_{n}\left( f,\beta _{n}\left( x\right) \right) , \notag \end{eqnarray}
</div>
<p>where </p>
<div class="equation" id="4">
<p>
  <div class="equation_content">
    \begin{equation}  \beta _{n}\left( x\right) =\big( K_{n}(e^{2at},x)\big) ^{-1}\circ e^{2ax}. \label{4} \end{equation}
  </div>
  <span class="equation_label">2</span>
</p>
</div>
<p>In particular, if we take \(\beta _{n}\left( x\right) =x\) in \(\left( \text{\ref{3}}\right) ,\) the King-type Baskakov operators return to classical Baskakov operators. </p>
<p>Let’s find the function \(\beta _{n},\) satisfying the condition,</p>
<div class="displaymath" id="a0000000004">
  \begin{equation*}  K_{n,\beta }(e^{2at},x)=e^{2ax}, \end{equation*}
</div>
<p> which is satisfied for all \(x\in \left[ 0,\infty \right) \) and \(n\in \mathbb {N}.\) </p>
<p>From the definition of King-type Baskakov Operators, we can write</p>
<div class="displaymath" id="a0000000005">
  \begin{eqnarray*}  e^{2ax} & =& K_{n,\beta }(e^{2at},x) \\ e^{2ax} & =& \sum \limits _{k=0}^{\infty }\tbinom {n+k-1}{k} \tfrac {\big( e^{\frac{2a}{n}}\beta _{n}( x) \big) ^{k}}{\left( 1+\beta _{n}\left( x\right) \right) ^{n+k}} \\ e^{\frac{-2ax}{n}} & =& 1+\beta _{n}\left( x\right) \big( 1-e^{\frac{2a}{n}}\big) \end{eqnarray*}
</div>
<p>and finally we get,</p>
<div class="equation" id="A">
<p>
  <div class="equation_content">
    \begin{equation}  \beta _{n}\left( x\right) =\tfrac {e^{\frac{-2ax}{n}}-1}{\big( 1-e^{\frac{2a}{n}}\big) }. \label{A} \end{equation}
  </div>
  <span class="equation_label">3</span>
</p>
</div>
<p>The structure of the paper is as follows. In Section 2, we give some lemmas about new operators which will be used in subsequent sections. In section 3, the uniform convergence of the operators is gained by means of modulus of continuity. The order of approximation is obtained with the help of Voronovskaya-type theorem in Section 4. In the last section, we investigate some results about shape preserving property of King-type Baskakov operators by comparing them to classical operators. </p>
<h1 id="a0000000006">2 Preliminaries</h1>
<p>Here, we mention some lemmas about our operator \(K_{n,\beta }\) which will be used to prove theorems in other sections. </p>
<p><div class="lemma_thmwrapper " id="7">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">1</span>
  </div>
  <div class="lemma_thmcontent">
  <p> Let \(a{\gt}0\) and \(n\in \mathbb {N}.\) For \(f(t)=e^{at},\) the operators get,</p>
<div class="displaymath" id="a0000000007">
  \begin{eqnarray*}  K_{n,\beta }(e^{at},x) & =& \sum \limits _{k=0}^{\infty }\tbinom {n+k-1}{k}\tfrac {\left( e^{\frac{a}{n}}\beta _{n}\left( x\right) \right) ^{k}}{\left( 1+\beta _{n}\left( x\right) \right) ^{n+k}} \notag \\ & =& \big( \beta _{n}\left( x\right) \big( 1-e^{\frac{a}{n}}\big) +1\big) ^{-n} \notag \\ & =& \Big( \tfrac {e^{-\frac{2ax}{n}}-1}{\big( 1-e^{\frac{2a}{n}}\big) }\big( 1-e^{\frac{a}{n}}\big) +1\Big) ^{-n} \notag \\ & =& \Big( \tfrac {e^{\frac{a}{n}}+e^{-\frac{2ax}{n}}}{1+e^{\frac{a}{n}}}\Big) ^{-n}. \end{eqnarray*}
</div>
<p>Moreover, </p>
<div class="displaymath" id="a0000000008">
  \begin{equation*}  \lim _{n\rightarrow \infty }\Big( \tfrac {e^{\frac{a}{n}}+e^{-\frac{2ax}{n}}}{1+e^{\frac{a}{n}}}\Big) ^{-n}=\allowbreak e^{ax}. \label{7} \end{equation*}
</div>

  </div>
</div> </p>
<p><div class="lemma_thmwrapper " id="le2">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">2</span>
  </div>
  <div class="lemma_thmcontent">
  <p> Let \(e_{i}\left( t\right) =t^{i},\) \(i=0,1,2.\  \)For \(a{\gt}0,\) we have,</p>
<div class="displaymath" id="a0000000009">
  \begin{eqnarray*}  K_{n,\beta }(e_{0},x) & =& 1, \\ K_{n,\beta }(e_{1},x) & =& \beta _{n}\left( x\right) , \\ K_{n,\beta }(e_{2},x) & =& \left( \beta _{n}\left( x\right) \right) ^{2}+\tfrac {\beta _{n}\left( x\right) \left( 1+\beta _{n}\left( x\right) \right) }{n}. \end{eqnarray*}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000010">
  <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 \(f(t) = e^{\theta t},\) \(\theta \in \mathbb {R},\) then </p>
<div class="displaymath" id="v-exp">
  \begin{eqnarray}  \label{v-exp} K_{n,\beta }(e^{\theta t},x) & =&  \Big(1+\beta _{n}\left( x\right) -\beta _{n}\left( x\right) e^{\frac{\theta }{n}}\Big)^{-n}. \end{eqnarray}
</div>
<p> Using the software Mathematica, we get the expansion of right side of (<a href="#v-exp">1</a>) in powers of \(\theta \) as follows: </p>
<div class="displaymath" id="a0000000011">
  \begin{eqnarray*} & & K_{n,\beta }(e^{\theta t},x) \\ & =&  1+\beta _{n}\left( x\right) \,  \theta + \left(\tfrac {n (\beta _{n}\left( x\right))^2+(\beta _{n}\left( x\right))^2+\beta _{n}\left( x\right)}{n}\right) \tfrac {\theta ^2}{2!} \\ & &  + \Big(\tfrac {(\beta _{n}\left( x\right))^3+3n(\beta _{n}\left( x\right))^3+2(\beta _{n}\left( x\right))^3+3n(\beta _{n}\left( x\right))^2+3(\beta _{n}\left( x\right)) ^2+\beta _{n}\left( x\right)}{n^2}\Big) \tfrac {\theta ^3}{3!} + \mathcal{O}(\theta ^4). \end{eqnarray*}
</div>
<p> Following <span class="cite">
	[
	<a href="#vgnmtr" >17</a>
	]
</span>, we can say that \(K_{n,\beta }(e^{\theta t},x)\) may be treated as moment ge-nerating function of the operators \(K_{n,\beta },\) which may be utilized to obtain the moments of (<a href="#3">1</a>). Thus using the following equivalence between moments and moment generating function: </p>
<div class="displaymath" id="a0000000012">
  \begin{eqnarray}  K_{n,\beta }(e_r,x) & =&  \Big[ \tfrac {\partial ^r}{\partial \theta ^r} K_{n,\beta }(e^{\theta t},x) \Big]_{\theta =0} \notag \\ & =&  \Big[ \tfrac {\partial ^r}{\partial \theta ^r} \Big\{  \big(1+\beta _{n}\left( x\right)-\beta _{n}\left( x\right)e^{\frac{\theta }{n}}\big)^{-n} \Big\}  \Big]_{\theta =0}, \label{e-mgf} \end{eqnarray}
</div>
<p> we obtain the moments immediately. <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> </p>
<p><div class="lemma_thmwrapper " id="le3">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">3</span>
  </div>
  <div class="lemma_thmcontent">
  <p> Let \(e_{i}^{x}\left( t\right) =\left( t-x\right) ^{i},\) \(i=0,1,2.\) From Lemma <a href="#le2">2</a>, we immediately have </p>
<div class="displaymath" id="a0000000014">
  \begin{eqnarray*}  K_{n,\beta }(e_{0}^{x},x) & =& 1, \\ K_{n,\beta }(e_{1}^{x},x) & =& \beta _{n}\left( x\right) -x, \\ K_{n,\beta }(e_{2}^{x},x) & =& \left( \beta _{n}\left( x\right) -x\right) ^{2}+\tfrac {\beta _{n}\left( x\right) \left( 1+\beta _{n}\left( x\right) \right) }{n}, \end{eqnarray*}
</div>
<p>and also using the software Mathematica, we can compute the following limits,</p>
<div class="displaymath" id="14">
  \begin{eqnarray}  \underset {n\longrightarrow \infty }{\lim }nK_{n,\beta }(e_{1}^{x},x) & =& -ax( x+1) , \label{14} \\ \underset {n\longrightarrow \infty }{\lim }nK_{n,\beta }(e_{2}^{x},x) & =& x(x+1) . \label{15} \end{eqnarray}
</div>

  </div>
</div> </p>
<h1 id="a0000000015">3 Main Result</h1>
<p>In this section we deal with the uniform convergence of \(K_{n,\beta }(f,x)\) to \(f\) on \(\left[ 0,\infty \right) \) by means of the modulus of continuity. First we will give the theorem about uniform convergence for linear positive operators in terms of exponential functions and second theorem is about the rate of the uniform convergence. </p>
<p>Let us denote by \(C^{\ast }\left[ 0,\infty \right) ,\) the Banach space of all real-valued continuous functions on \(\left[ 0,\infty \right) \) with the property \(\underset {x\longrightarrow \infty }{\lim }f(x)\) exists and finite, given with the uniform norm. </p>
<p><div class="theorem_thmwrapper " id="19">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">4</span>
  </div>
  <div class="theorem_thmcontent">
  <p> <span class="cite">
	[
	<a href="#8" >10</a>
	]
</span> If the sequence \(A_{n}:\) \(C^{\ast }\left[ 0,\infty \right) \rightharpoonup C^{\ast }\left[ 0,\infty \right) \) of positive linear operators satisfies the conditions</p>
<div class="displaymath" id="a0000000016">
  \begin{equation*}  \underset {n\longrightarrow \infty }{\lim }A_{n}\big( e^{-kt},x\big) =e^{-kx},\   \   \  k=0,1,2. \label{16} \end{equation*}
</div>
<p>uniformly in \(\left[ 0,\infty \right) \) then for \(f\in \left[ 0,\infty \right) ,\)</p>
<div class="displaymath" id="a0000000017">
  \begin{equation*}  \underset {n\longrightarrow \infty }{\lim }A_{n}\left( f,x\right) =f\left( x\right) \label{17} \end{equation*}
</div>
<p>uniformly in \(\left[ 0,\infty \right) .\) </p>

  </div>
</div> </p>
<p>Now we mention the theorem which is about the estimate for the rate of the uniform convergence. In this theorem, we will use the definition of the modulus of continuity of \(f\) in <span class="cite">
	[
	<a href="#13" >19</a>
	]
</span>, for any \(\delta {\gt}0\) and each \(x\in \) \(\left[ 0,\infty \right) \  \)is defined by,</p>
<div class="displaymath" id="a0000000018">
  \begin{equation*}  \omega ^{\ast }\left( f,\delta \right) =\sup _{\substack { x,t\geq 0 \\ \begin{bgroup} \vert e^{-x}-e^{-t} \vert \leq \delta 

\end{bgroup}}}\left\vert f\left( x\right) -f\left( t\right) \right\vert . \label{18} \end{equation*}
</div>
<p>This modulus of continuity has the property</p>
<div class="displaymath" id="a0000000019">
  \begin{equation*}  \left\vert f\left( t\right) -f\left( x\right) \right\vert \leq \Big( 1+\tfrac {\left( e^{-x}-e^{-t}\right) ^{2}}{\delta ^{2}}\Big) \omega ^{\ast }\left( f,\delta \right) ,\   \   \   \delta {\gt}0. \label{19} \end{equation*}
</div>
<p><div class="theorem_thmwrapper " id="20">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">5</span>
  </div>
  <div class="theorem_thmcontent">
  <p> <span class="cite">
	[
	<a href="#13" >19</a>
	]
</span> Let \(A_{n}:\) \(C^{\ast }\left[ 0,\infty \right) \rightharpoonup C^{\ast }\left[ 0,\infty \right) \) be a sequence of linear positive operators with</p>
<div class="displaymath" id="a0000000020">
  \begin{eqnarray*}  \left\Vert A_{n}\left( 1,x\right) -1\right\Vert _{\left[ 0,\infty \right) } & =& a_{n}, \\ \big\Vert A_{n}\big( e^{-t},x\big) -e^{-x}\big\Vert _{\left[ 0,\infty \right) } & =& b_{n}, \\ \big\Vert A_{n}\big( e^{-2t},x\big) -e^{-2x}\big\Vert _{\left[ 0,\infty \right) } & =& c_{n}. \end{eqnarray*}
</div>
<p>Then </p>
<div class="displaymath" id="a0000000021">
  \begin{equation*}  \left\Vert A_{n}\left( f,x\right) -f(x)\right\Vert _{\left[ 0,\infty \right) }\leq \left\Vert f\right\Vert _{\left[ 0,\infty \right) }a_{n}+\left( 2+a_{n}\right) \omega ^{\ast }\big( f,\sqrt{a_{n}+2b_{n}+c_{n}}\big) \label{20} \end{equation*}
</div>
<p>for every function of \(f\in C^{\ast }\left[ 0,\infty \right) .\) </p>

  </div>
</div> </p>
<p>Now we will give the main result about uniform convergence of \(K_{n,\beta }\) by means of the modulus of continuity which is defined above. </p>
<p><div class="theorem_thmwrapper " id="21">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">6</span>
  </div>
  <div class="theorem_thmcontent">
  <p> For \(f\in C^{\ast }\left[ 0,\infty \right) ,\) we have </p>
<div class="displaymath" id="a0000000022">
  \begin{equation*}  \left\Vert K_{n,\beta }f-f\right\Vert _{\left[ 0,\infty \right) }\leq 2\omega ^{\ast }\big( f,\sqrt{2b_{n}+c_{n}}\big) , \label{21} \end{equation*}
</div>
<p> where </p>
<div class="displaymath" id="a0000000023">
  \begin{eqnarray*}  \big\Vert K_{n,\beta }\big( e^{-t},x\big) -e^{-x}\big\Vert _{\left[ 0,\infty \right) } & =& b_{n}, \\ \big\Vert K_{n,\beta }\big( e^{-2t},x\big) -e^{-2x}\big\Vert _{\left[ 0,\infty \right) } & =& c_{n}. \end{eqnarray*}
</div>
<p>Here when \(n\longrightarrow \infty ,\) \(b_{n}\) and \(c_{n}\) tend to zero and \(\left\{  K_{n,\beta }f\right\}  \) converges uniformly to the function f. </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000024">
  <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 show that the conditions of the Theorem <a href="#19">4</a>. From Lemma <a href="#le2">2</a>, we can write that</p>
<div class="displaymath" id="a0000000025">
  \begin{equation*}  \left\Vert K_{n,\beta }\left( 1,x\right) -1\right\Vert _{\left[ 0,\infty \right) }=a_{n}=0. \end{equation*}
</div>
<p>Now we can examine that for the images of the \(e^{-kt},\  k=1,2.\) under the operators \(K_{n,\beta }f.\) To show it, we will use the \(\left( \text{\ref{3}}\right) \) and \(\left( \text{\ref{A}}\right) .\) </p>
<p>For \(f(t)=e^{-t},\)</p>
<div class="displaymath" id="a0000000026">
  \begin{eqnarray*}  K_{n,\beta }\big( e^{-t},x\big) & =& \sum \limits _{k=0}^{\infty }\tbinom {n+k-1}{k}\tfrac {\big( e^{\frac{-1}{n}}\beta _{n}\left( x\right) \big) ^{k}}{\left( 1+\beta _{n}\left( x\right) \right) ^{n+k}} \\ & =& \sum \limits _{k=0}^{\infty }\tbinom {-n}{k}\Big( -e_{n}^{\frac{-1}{n}}\beta _{n}\left( x\right) \Big) ^{k}\left( 1+\beta _{n}\left( x\right) \right) ^{-n-k} \\ & =& \big( \beta _{n}\left( x\right) \big( 1-e^{-\frac{1}{n}}\big) +1\big) ^{-n} \\ & =& \Big( \tfrac {e^{-\frac{2ax}{n}}-1}{\big( 1-e^{\frac{2a}{n}}\big) }\big( 1-e^{-\frac{1}{n}}\big) +1\Big) ^{-n}. \end{eqnarray*}
</div>
<p> Expanding the right hand side of above expression using Mathematica, we find that </p>
<div class="displaymath" id="a0000000027">
  \begin{align*}  K_{n,\beta }\big( e^{-t},x\big)=& e^{-x}+\tfrac {e^{-x}}{2n}x(1+x)(1+2a) \\ & +\tfrac {e^{-x}}{24n^2}(1+2a)e^{-x}x(1+x)[-4-4a-5x-2ax+3x^2+6ax^2] \\ & +{\mathcal O}(n^{-3}). \end{align*}
</div>
<p> Next, for \(f(t)=e^{-2t},\) after simple computation, we have </p>
<div class="displaymath" id="a0000000028">
  \begin{eqnarray*}  K_{n,\beta }\big( e^{-2t},x\big) & =& \Big( \tfrac {e^{-\frac{2ax}{n}}-1}{\big( 1-e^{\frac{2a}{n}}\big) }\big( 1-e^{-\frac{2}{n}}\big) +1\Big) ^{-n}. \end{eqnarray*}
</div>
<p> Finally, expanding the right hand side of above expression using Mathematica, we get </p>
<div class="displaymath" id="a0000000029">
  \begin{align*}  K_{n,\beta }\big( e^{-2t},x\big) =& e^{-2x}+\tfrac {e^{-2x}}{n}2(1+a)x(1+x) \\ & +\tfrac {2e^{-2x}}{3n^2}x(1+x)(1+a)(-2-a-x+ax+3x^2+3ax^2) \\ & +{\mathcal O}(n^{-3}). \end{align*}
</div>
<p> The result follows immediately from the above identities. <div class="proof_wrapper" id="a0000000030">
  <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="a0000000031">4 Aymptotic Expression</h1>
<p>Voronovskaya-type theorem is the main tool to talk about the order of the approximation in approximation theory. Now, we will analyse the asymptotic behavior of our operators \(K_{n,\beta }\) in view of the Voronovskaya-type theorem. </p>
<p><div class="theorem_thmwrapper " id="t4">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">7</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(f,\  f^{\prime \prime }\in C^{\ast }[0,\infty )\). Then we have, </p>
<div class="displaymath" id="a0000000032">
  \begin{align*} & \big\vert n\left[ K_{n,\beta }(f,x)-f(x)\right] +ax(x+1)f^{\prime }(x)-\tfrac {x(x+1)}{2}f^{\prime \prime }(x)\big\vert \leq \\ & \leq \left\vert p_{n}(x)\right\vert \left\vert f^{\prime }(x)\right\vert +\left\vert q_{n}(x)\right\vert \left\vert f^{\prime \prime }(x)\right\vert +2\left( 2q_{n}+x(x+1)+r_{n}(x)\right) \omega ^{\ast }\big( f^{\prime \prime },\tfrac {1}{\sqrt{n}}\big) , \end{align*}
</div>
<p>where</p>
<div class="displaymath" id="a0000000033">
  \begin{eqnarray*}  p_{n}(x) & =& nK_{n,\beta }\left( e_{1}^{x}(t),x\right) +ax(x+1), \\ q_{n}(x) & =& \tfrac {1}{2}\left( nK_{n,\beta }\left( e_{2}^{x}(t),x\right) -x(x+1)\right) , \\ r_{n}(x) & =& n^{2}\sqrt{K_{n,\beta }\big( \left( e^{-x}-e^{-t}\right) ^{4},x\big) }\sqrt{K_{n,\beta }(\left( t-x)^{4},x\right) .} \end{eqnarray*}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000034">
  <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 Taylor’s expansion of \(f\) at the point \(x,\) we have </p>
<div class="displaymath" id="a0000000035">
  \begin{equation*}  f(t)=f(x)+f^{\prime }(x)(t-x)+\tfrac {f^{\prime \prime }(x)}{2}(t-x)^{2}+h(t,x)(t-x)^{2}, \end{equation*}
</div>
<p>where, </p>
<div class="displaymath" id="a0000000036">
  \begin{equation*}  h(t,x)=\tfrac {f^{\prime \prime }(\eta )-f^{\prime \prime }(x)}{2}, \end{equation*}
</div>
<p>and \(\eta \) is a number lying between \(x\) and \(t\). </p>
<p>If we apply the operators \(K_{n,\beta }\  \) on both sides of the above Taylor’s expansion and use the linearity property of \(K_{n,\beta },\) we can write </p>
<div class="displaymath" id="a0000000037">
  \begin{align} & K_{n,\beta }\left( f,x\right) -f(x)-f^{\prime }(x)K_{n,\beta }\left( e_{1}^{x}(t),x\right) -\tfrac {f^{\prime \prime }(x)}{2}K_{n,\beta }\left( e_{2}^{x}(t),x\right)\notag =\\ & =K_{n,\beta }\left( h(t,x)e_{2}^{x}(t),x\right) . \label{22} \end{align}
</div>
<p>We multiply both sides of the \(\left( \text{\ref{22}}\right) \) with \(n\) and applying Lemma <a href="#le3">3</a>, we have </p>
<div class="displaymath" id="a0000000038">
  \begin{align*} & \big\vert n\left[ K_{n,\beta }\left( f,x\right) -f(x)\right] +ax(x+1)f^{\prime }(x)-\tfrac {x(x+1)}{2}f^{\prime \prime }(x)\big\vert \leq \\ & \leq \big\vert \left( nK_{n,\beta }\left( e_{1}^{x}(t),x\right) +ax(x+1)\right) \big\vert \big\vert f^{\prime }(x)\big\vert \\ &  \quad +\tfrac {1}{2}\big\vert nK_{n,\beta }\left( e_{2}^{x}(t),x\right) -x(x+1)\big\vert \left\vert f^{\prime \prime }(x)\right\vert +\left\vert nK_{n,\beta }\left( h(t,x)e_{2}^{x}(t),x\right) \right\vert . \end{align*}
</div>
<p>If we define \(p_{n}\) and \(q_{n}\) with following equations, </p>
<div class="displaymath" id="a0000000039">
  \begin{eqnarray*}  p_{n}(x) & =& nK_{n,\beta }\left( e_{1}^{x}(t),x\right) +ax(x+1), \\ q_{n}(x) & =& \tfrac {1}{2}\left( nK_{n,\beta }\left( e_{2}^{x}(t),x\right) -x(x+1)\right) . \end{eqnarray*}
</div>
<p>we can write </p>
<div class="displaymath" id="a0000000040">
  \begin{align} & \big\vert n\left[ K_{n,\beta }\left( f,x\right) -f(x)\right] +ax(x+1)f^{\prime }(x)-\tfrac {x(x+1)}{2}f^{\prime \prime }(x)\big\vert \notag \leq \\ & \leq \left\vert p_{n}(x)\right\vert \left\vert f^{\prime }(x)\right\vert +\tfrac {1}{2}\left\vert q_{n}(x)\right\vert \left\vert f^{\prime \prime }(x)\right\vert +\left\vert nK_{n,\beta }\left( h(t,x)e_{2}^{x}(t),x\right) \right\vert . \label{25} \end{align}
</div>
<p>Using \(\left( \text{\ref{14}}\right) \) and \(\left( \text{\ref{15}}\right) ,\) we can see that if we let the \(n\) goes the infinity, \(p_{n}(x)\) and \(q_{n}(x) \) tend to zero at any point \(x\). </p>
<p>To finish the proof, we have to calculate the term \(\left\vert nK_{n,\beta }\left( h(t,x)e_{2}^{x}(t),x\right) \right\vert \). We know that,</p>
<div class="displaymath" id="a0000000041">
  \begin{equation*}  \left\vert f(t)-f(x)\right\vert \leq \Big( 1+\tfrac {\left( e^{-x}-e^{-t}\right) ^{2}}{\delta ^{2}}\Big) \omega ^{\ast }\left( f,\delta \right) . \end{equation*}
</div>
<p>From this equality, we get</p>
<div class="displaymath" id="a0000000042">
  \begin{equation*}  \left\vert h(t,x)\right\vert \leq \Big( 1+\tfrac {\left( e^{-x}-e^{-t}\right) ^{2}}{\delta ^{2}}\Big) \omega ^{\ast }\left( f^{\prime \prime },\delta \right) . \end{equation*}
</div>
<p>Here for \(\delta {\gt}0,\) if \(\left\vert e^{-x}-e^{-t}\right\vert \leq \delta \) then \(\left\vert h(t,x)\right\vert \leq 2\omega ^{\ast }( f^{\prime \prime },\delta ) \) and if </p>
<div class="displaymath" id="a0000000043">
  \[ \big\vert e^{-x}-e^{-t}\big\vert {\gt}\delta  \]
</div>
<p> then \(\left\vert h(t,x)\right\vert \leq 2\big( \tfrac {( e^{-x}-e^{-t}) ^{2}}{\delta ^{2}}\big) \omega ^{\ast }\left( f^{\prime \prime },\delta \right) \) so we can write, </p>
<div class="displaymath" id="a0000000044">
  \begin{eqnarray*}  \left\vert h(t,x)\right\vert \leq 2\Big( 1+\tfrac {( e^{-x}-e^{-t}) ^{2}}{\delta ^{2}}\Big) \omega ^{\ast }\left( f^{\prime \prime },\delta \right). \end{eqnarray*}
</div>
<p>If we use it \(\left( \text{\ref{25}}\right) ,\) we obtain </p>
<div class="displaymath" id="a0000000045">
  \begin{eqnarray*}  nK_{n,\beta }\left( h(t,x)e_{2}^{x}(t),x\right) & \leq & 2n\omega ^{\ast }\left( f^{\prime \prime },\delta \right) K_{n,\beta }\left( e_{2}^{x}(t),x\right) \\ & & +\tfrac {2n}{\delta ^{2}}\omega ^{\ast }\left( f^{\prime \prime },\delta \right) K_{n,\beta }\big( \big( e^{-x}-e^{-t}\big) ^{2}e_{2}^{x}(t),x\big) . \end{eqnarray*}
</div>
<p>By applying Cauchy-Schwarz inequality, we obtain </p>
<div class="displaymath" id="a0000000046">
  \begin{align*} & nK_{n,\beta }\left( h(t,x)e_{2}^{x}(t),x\right) \leq \\ & \leq 2n\omega ^{\ast }\left( f^{\prime \prime },\delta \right) K_{n,\beta }\left( e_{2}^{x}(t),x\right) \\ & \quad +\tfrac {2n}{\delta ^{2}}\omega ^{\ast }\left( f^{\prime \prime },\delta \right) \sqrt{K_{n,\beta }\big( \left( e^{-x}-e^{-t}\right) ^{4},x\big) }\sqrt{K_{n,\beta }\left( e_{4}^{x}(t),x\right) }. \end{align*}
</div>
<p>If we choose \(\delta =\tfrac {1}{\sqrt{n}}\) and define </p>
<div class="displaymath" id="a0000000047">
  \begin{eqnarray*}  r_{n}(x)=\sqrt{n^{2}K_{n,\beta }\big( \left( e^{-x}-e^{-t}\right) ^{4},x\big) }\sqrt{n^{2}(K_{n,\beta }\left( t-x)^{4},x\right) }, \end{eqnarray*}
</div>
<p> we achieve </p>
<div class="displaymath" id="a0000000048">
  \begin{align*} & \left\vert n\left[ K_{n,\beta }(f,x)-f(x)\right] +ax(x+1)f^{\prime }(x)-\tfrac {x(x+1)}{2}f^{\prime \prime }(x)\right\vert \leq \\ & \leq \left\vert p_{n}(x)\right\vert \left\vert f^{\prime }(x)\right\vert +\left\vert q_{n}(x)\right\vert \left\vert f^{\prime \prime }(x)\right\vert +2\left( 2q_{n}+x(x+1)+r_{n}(x)\right) \omega ^{\ast }\big( f^{\prime \prime },\tfrac {1}{\sqrt{n}}\big) . \end{align*}
</div>
<p>Hence we have the desired result. <div class="proof_wrapper" id="a0000000049">
  <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="t5">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">8</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(x\in \lbrack 0,\infty )\) and \(f,\  f^{\prime \prime }\in C^{\ast }[0,\infty ),\) we have</p>
<div class="equation" id="26">
<p>
  <div class="equation_content">
    \begin{equation}  \lim _{n\longrightarrow \infty }n\left[ K_{n,\beta }(f,x)-f(x)\right] =-ax(x+1)f^{\prime }(x)+\tfrac {x(x+1)}{2}f^{\prime \prime }(x). \label{26} \end{equation}
  </div>
  <span class="equation_label">7</span>
</p>
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000050">
  <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 the Taylor’s expansion of \(f\), we have </p>
<div class="equation" id="e22">
<p>
  <div class="equation_content">
    \begin{equation}  \label{e22} f(t) = f(x) + f^{\prime }(x)(t-x) + \tfrac {1}{2} f^{\prime \prime }(x)(t-x)^{2} + r(t,x)(t-x)^{2}, \end{equation}
  </div>
  <span class="equation_label">8</span>
</p>
</div>
<p> where \(\displaystyle \lim _{t\rightarrow x}r(t,x)=0.\) Operating \(K_{n,\beta }\) to the identity (<a href="#e22">8</a>), we obtain </p>
<div class="displaymath" id="a0000000051">
  \begin{align*}  K_{n,\beta }(f,x)- f(x) \! = \!  & K_{n,\beta }(t-x,x) f^{\prime }(x) \! + \!  K_{n,\beta }\big(\left(t \! - \! x\right)^{2},x\big) \tfrac {f^{\prime \prime }(x)}{2} \\ & \! + \! K_{n,\beta }\big( r\left( t,x\right) \left( t \! - \! x\right)^{2},x\big). \end{align*}
</div>
<p> Using the Cauchy-Schwarz inequality, we have </p>
<div class="equation" id="e23">
<p>
  <div class="equation_content">
    \begin{equation}  \label{e23} K_{n,\beta }\big(r\left(t,x\right)\left(t-x\right)^{2},x\big)\leq \sqrt{K_{n,\beta }\left(r^{2}\left(t,x\right) ,x\right)} \,  \sqrt{K_{n,\beta }\left(\big(t-x\right)^{4},x\big)}. \end{equation}
  </div>
  <span class="equation_label">9</span>
</p>
</div>
<p> Also, we have </p>
<div class="equation" id="e24">
<p>
  <div class="equation_content">
    \begin{equation}  \label{e24} \lim _{n\rightarrow \infty }K_{n,\beta }\big(r^{2}\left(t,x\right),x\big)=r^{2}\left(x,x\right) = 0. \end{equation}
  </div>
  <span class="equation_label">10</span>
</p>
</div>
<p> Now from (<a href="#e23">9</a>), (<a href="#e24">10</a>) and (<a href="#e-mgf">2</a>), we get </p>
<div class="displaymath" id="a0000000052">
  \begin{align*}  \lim _{n\rightarrow \infty } n K_{n,\beta }\big(r\left(t,x\right) \left(t-x\right)^{2}, x\big) = 0. \end{align*}
</div>
<p> Thus we have </p>
<div class="displaymath" id="a0000000053">
  \begin{align*} & \lim _{n\rightarrow \infty } n \left(K_{n,\beta }(f,x)-f(x)\right)=\\ & =\lim _{n\rightarrow \infty } n \big[K_{n,\beta }(t\! -\! x,x) f^{\prime }(x)\! +\!  \tfrac {1}{2}f^{\prime \prime }(x) K_{n,\beta }(\left(t\! -\! x\right)^{2},x)\! +\!  K_{n,\beta }(r\left(t,x\right)\left(t\! -\! x\right)^{2},x)\big] . \end{align*}
</div>
<p> The result follows immediately by (<a href="#14">3</a>) and (<a href="#15">4</a>). <div class="proof_wrapper" id="a0000000054">
  <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="a0000000055">5 Better Estimation by means of Generalized Convexity</h1>
<p>Now we deal with the comparison with the operators \(K_{n,\beta }\) with the classical Baskakov operators by way of generalized convexity. To do this, first of all we give the definition of generalized convexity. We can see these definitions <span class="cite">
	[
	<a href="#7" >9</a>
	]
</span>, <span class="cite">
	[
	<a href="#17" >23</a>
	]
</span>. </p>
<p><div class="definition_thmwrapper " id="27">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">9</span>
  </div>
  <div class="definition_thmcontent">
  <p>A function \(f\in C[0,\infty )\) is said to be convex with respect to the function \(\tau \) if, </p>
<div class="displaymath" id="a0000000056">
  \begin{equation*}  \left\vert \begin{tabular}{lll} $1$ 

&  $1$ 

&  $1$ 

\\ $\tau \left( x_{0}\right) $ 

&  $\tau \left( x_{1}\right) $ 

&  $\tau \left( x_{2}\right) $ 

\\ $f(x_{0})$ 

&  $f(x_{1})$ 

&  $f(x_{2})$

\end{tabular}\right\vert \geq 0\   \  ,\   \   \  0{\lt}x_{0}{\lt}x_{1}{\lt}x_{2}{\lt}\infty . \label{27} \end{equation*}
</div>

  </div>
</div> </p>
<p>We take account of the convexity in the classical sense is obtained for \(\tau \) = \(e_{1}\). </p>
<p><div class="definition_thmwrapper " id="28">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">10</span>
  </div>
  <div class="definition_thmcontent">
  <p>A function \(f\in C^{k}[0,\infty )\) (the space of \(k\) times continuously differentiable functions) is said to be \(\tau \) convex of order \(k\in \mathbb {N}\) whenever </p>
<div class="displaymath" id="a0000000057">
  \begin{equation*}  D^{k}(f\circ \tau ^{-1})\circ \tau \geq 0. \label{28} \end{equation*}
</div>

  </div>
</div> </p>
<p>The classical convexity is obtained for \(\tau =e_{1}\) and \(k=2.\) </p>
<p>For the operators \(K_{n,\beta },\) we are talking about generalized convexity with respect to the function \(\tau =e^{2ax}.\) </p>
<p>Next we will give the following remark about \(\tau \) convexity which is given in <span class="cite">
	[
	<a href="#7" >9</a>
	]
</span>. </p>
<p><div class="remark_thmwrapper " id="29">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">11</span>
  </div>
  <div class="remark_thmcontent">
  <p>The function \(f\in C^{2}[0,\infty )\) is convex with respect to the function \(\tau \) if and only if</p>
<div class="displaymath" id="a0000000058">
  \begin{equation*}  f^{\prime \prime }(x)\geq \tfrac {\tau ^{\prime \prime }(x)}{\tau ^{\prime }(x)}f^{\prime }(x),\   \   \  x\in \lbrack 0,\infty ). \label{29} \end{equation*}
</div>

  </div>
</div> </p>
<p>According to the remark which is above, we can write that the function \(f\in C^{2}[0,\infty )\) is convex with respect to the function \(\tau =e^{2ax}\) if and only if</p>
<div class="equation" id="30">
<p>
  <div class="equation_content">
    \begin{equation}  f^{\prime \prime }(x)\geq 2af^{\prime }(x),\   \  x\in \lbrack 0,\infty ).\    \label{30} \end{equation}
  </div>
  <span class="equation_label">11</span>
</p>
</div>
<p><div class="theorem_thmwrapper " id="t6">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">12</span>
  </div>
  <div class="theorem_thmcontent">
  <p> If the function \(f\in C^{2}\left[ 0,\infty \right) \) is strictly \(\varphi \)-convex with respect to \(\tau \left( x\right) =e^{2ax},\) \(a{\gt}0,\) then for all \(x\geq 0\) there exists \(n_{0}=n_{0}(x)\in \mathbb {N}\) such that for \(n\geq n_{0}\) there holds</p>
<div class="equation" id="31">
<p>
  <div class="equation_content">
    \begin{equation}  f(x)\leq K_{n,\beta }(f,x) \label{31} \end{equation}
  </div>
  <span class="equation_label">12</span>
</p>
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000059">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Using (<a href="#26">7</a>) and (<a href="#30">11</a>), the proof is obtained immediately. <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><div class="theorem_thmwrapper " id="t7">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">13</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(f\in C[0,\infty )\) and \(K_{n}\) be a classical Baskakov operator which is defined on \(C[0,\infty ),\) we can write<br />i)\(\  \)If\(\  f\  \)is convex, then\(\  f(x)\leq K_{n}\left( f,x\right) .\)<br />ii) If\(\  f\  i\)s convex, then\(\  K_{n+1}\left( f,x\right) \leq K_{n}\left( f,x\right) .\)<br />iii) If \(f\) is increasing (decreasing), then \(K_{n}\left( f,x\right) \) is increasing (decreasing). </p>

  </div>
</div> </p>
<p><div class="theorem_thmwrapper " id="32">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">14</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(f\in C[0,\infty )\) be a decreasing and convex function. Then we have</p>
<div class="displaymath" id="a0000000061">
  \begin{equation*}  f(x)\leq K_{n+1,\beta }\left( f,x\right) \leq K_{n,\beta }(f,x). \label{32} \end{equation*}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000062">
  <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 Theorem <a href="#t7">13</a>, we can write \(\beta _{n}(x)\leq \beta _{n+1}(x).\) Using this inequality,</p>
<div class="displaymath" id="a0000000063">
  \begin{eqnarray*}  K_{n,\beta }(f,x)-K_{n+1,\beta }\left( f,x\right) & =& K_{n}\left( f,\beta _{n}(x)\right) -K_{n+1}\left( f,\beta _{n+1}(x)\right) \\ & =& K_{n}\left( f,\beta _{n}(x)\right) -K_{n+1}\left( f,\beta _{n}(x)\right) \\ & & +K_{n+1}\left( f,\beta _{n}(x)\right) -K_{n+1}\left( f,\beta _{n+1}(x)\right) . \end{eqnarray*}
</div>
<p>Since \(f\) is decreasing and convex, we get</p>
<div class="displaymath" id="a0000000064">
  \begin{eqnarray*}  K_{n,\beta }(f,x)-K_{n+1,\beta }\left( f,x\right) & \geq & 0 \\ K_{n,\beta }(f,x) & \geq & K_{n+1,\beta }\left( f,x\right) . \end{eqnarray*}
</div>
<p>Also from Theorem <a href="#t5">8</a>, there exists \(n_{1}=n_{1}(x)\in \mathbb {N} ,\) such that for \(n\geq n_{1}\), we have</p>
<div class="displaymath" id="a0000000065">
  \begin{equation*}  f(x)\leq K_{n+1,\beta }\left( f,x\right) . \end{equation*}
</div>
<p> <div class="proof_wrapper" id="a0000000066">
  <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="33">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">15</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(f\in C[0,\infty )\) be a increasing and convex with respect to the function \(\tau =e^{2ax}.\) Then </p>
<div class="displaymath" id="a0000000067">
  \begin{equation*}  f(x)\leq K_{n,\beta }( f,x) \leq K_{n}(f,x). \label{33} \end{equation*}
</div>

  </div>
</div> </p>
<p><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> From <span class="cite">
	[
	<a href="#17" >23</a>
	]
</span>, we can write </p>
<div class="displaymath" id="a0000000069">
  \begin{equation*}  f(x)\leq K_{n,\beta }(f,x). \end{equation*}
</div>
<p>Since \(f\) is convex with respect to the function \(\tau =e^{2ax}\), we have from Theorem <a href="#t7">13</a>, that </p>
<div class="displaymath" id="a0000000070">
  \begin{equation*}  f(x)\leq K_{n}( f,x) . \end{equation*}
</div>
<p>Also \(\tau =e^{2ax}\) is a convex function, so that we can write \(\tau \) instead of \(f\)</p>
<div class="displaymath" id="a0000000071">
  \begin{equation*}  e^{2ax}\leq K_{n}\big( e^{2at},x\big) . \end{equation*}
</div>
<p>Using the \(\left( K_{n}\left( \tau \right) \right) ^{-1}\) as an increasing function,</p>
<div class="displaymath" id="a0000000072">
  \begin{equation*}  K_{n}\big( e^{2at},x\big) ^{-1}\circ e^{2ax}\leq K_{n}\big( e^{2at},x\big) ^{-1}\circ K_{n}\big( e^{2at},x\big) . \end{equation*}
</div>
<p>Hence from the \(\left( \text{\ref{4}}\right) ,\) we have</p>
<div class="equation" id="34">
<p>
  <div class="equation_content">
    \begin{equation}  \beta _{n}\left( x\right) \leq x. \label{34} \end{equation}
  </div>
  <span class="equation_label">13</span>
</p>
</div>
<p>Applying the operators \(K_{n}\) both sides of \(\left( \text{\ref{34}}\right) , \)</p>
<div class="equation" id="35">
<p>
  <div class="equation_content">
    \begin{equation}  K_{n}(f,\beta _{n}\left( x\right) )=K_{n,\beta }(f,x)\leq K_{n}\left( f,x\right) \label{35} \end{equation}
  </div>
  <span class="equation_label">14</span>
</p>
</div>
<p>Combining \(\left( \text{\ref{31}}\right) \) and \(\left( \text{\ref{35}}\right) ,\  \)we get desired result. <div class="proof_wrapper" id="a0000000073">
  <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="remark_thmwrapper " id="a0000000074">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">16</span>
  </div>
  <div class="remark_thmcontent">
  <p>We can also use the definition of Baskakov operators, considered by Chen <span class="cite">
	[
	<a href="#chen" >14</a>
	]
</span>: </p>
<div class="displaymath" id="a0000000075">
  \begin{equation*}  \overset {\thicksim }{V}_{n}f(x)=\sum \limits _{k=0}^{\infty }\tfrac {n\left( n+\alpha \right) \left( n+2\alpha \right) ...\left[ n+\left( k-1\right) \alpha \right] }{k!}\tfrac {x^{k}}{\left( 1+\alpha x\right) ^{\frac{n}{\alpha }+k}}f\big( \tfrac {k}{n}\big) .\end{equation*}
</div>
<p>Using this definition, we can define another King type of Baskakov operators,</p>
<div class="displaymath" id="a0000000076">
  \begin{equation*}  V_{n}f(x)=\sum \limits _{k=0}^{\infty }\tfrac {n\left( n+\alpha \right) \left( n+2\alpha \right) ...\left[ n+\left( k-1\right) \alpha \right] }{k!}\tfrac {\left( b_{n}(x)\right) ^{k}}{\left( 1+\alpha b_{n}(x)\right) ^{\frac{n}{\alpha }+k}}f\big( \tfrac {k}{n}\big) , \end{equation*}
</div>
<p>\(x\geq 0,\) \(n\in \mathbb {N} .\)<br />In the last expression by setting \(\alpha =1,\) we get the definition given in \(\left( \text{\ref{3}}\right) .\) </p>

  </div>
</div> </p>
<p><div class="acknowledgement_thmwrapper " id="a0000000077">
  <div class="acknowledgement_thmheading">
    <span class="acknowledgement_thmcaption">
    Acknowledgements
    </span>
  </div>
  <div class="acknowledgement_thmcontent">
  <p>Authors are thankful to the reviewer for valuable suggestions leading to overall improvements in the paper. </p>

  </div>
</div> </p>
<p><small class="footnotesize">  </small></p>
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</dl>


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