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<title>Some Properties of the Operators defined by Lupaş: Some Properties of the Operators defined by Lupaş</title>
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<h1>Some Properties of the Operators defined by Lupaş</h1>
<p class="authors">
<span class="author">AYŞEGÜL ERENÇIN\(^\ast \), GÜLEN BAŞCANBAZ-TUNCA\(^\ast \)\(^\ast \) FATMA TAŞDELEN\(^{\ast \ast }\)</span>
</p>
<p class="date">August 14, 2014.<br />Published online: January 23, 2015.</p>
</div>
<p>\(^\ast \) Abant Izzet Baysal University, Faculty of Arts and Science, Department of Mathematics, 14280, Bolu, Turkey, e-mail: <span class="tt">erencina@hotmail.com</span>. </p>
<p>\(^{\ast \ast }\)Ankara University, Faculty of Science, Department of Mathematics, 06100, Tandoǧan, Ankara, Turkey, e-mail: <span class="tt">tunca@science.ankara.edu.tr, tasdelen@science.ankara.edu.tr</span>. </p>

<div class="abstract"><p> In the present paper, we show that a subclass of the operators defined by Lupaş <span class="cite">
	[
	<a href="#lu" >12</a>
	]
</span> preserve properties of the modulus of continuity function and Lipschitz constant and the order of a Lipschitz continuous function. We also concerned with the monotonicity of sequence of such operators for convex functions. </p>
<p><b class="bf">MSC.</b> 41A25, 41A36 </p>
<p><b class="bf">Keywords.</b> Modulus of continuity function, Lipschitz class, monotonicity. </p>
</div>
<h1 id="a0000000002">1 Introduction</h1>
<p>By means of the identity</p>
<div class="displaymath" id="a0000000003">
  \[  \tfrac {1}{(1-a)^{\alpha }}=\sum ^{\infty }_{k=0}\tfrac {(\alpha )_{k}}{k!}a^{k},\quad |a|{\lt}1,  \]
</div>
<p> where \((\alpha )_{0}=1\), \((\alpha )_{k}=\alpha (\alpha +1)\cdots (\alpha +k-1)\), \(k\geq 1\), Lupaş <span class="cite">
	[
	<a href="#lu" >12</a>
	]
</span> proposed the positive linear operators </p>
<div class="displaymath" id="a0000000004">
  \[  {T}_{n}(f;x)=(1-a)^{nx}\sum ^{\infty }_{k=0}\tfrac {(nx)_{k}}{k!}f\left( \tfrac {k}{n}\right) a^{k},\quad x\geq 0  \]
</div>
<p> for the functions \(f:[0,\infty )\rightarrow \mathbb {R}\) and \(n\in \mathbb {N}\). After that Agratini <span class="cite">
	[
	<a href="#ag" >3</a>
	]
</span>, by choosing \(a=\tfrac {1}{2}\) , for the operators </p>
<div class="equation" id="eq:1.1">
<p>
  <div class="equation_content">
    \begin{equation}  \label{eq:1.1}L_{n}(f;x)=2^{-nx}\sum ^{\infty }_{k=0}\tfrac {(nx)_{k}}{2^{k}k!}f\left( \tfrac {k}{n}\right) \end{equation}
  </div>
  <span class="equation_label">1.1</span>
</p>
</div>
<p> obtained some estimates to the order of approximation on a finite interval and proved a Voronovskaya type theorem. Furthermore, he again derived the positive linear operators \(L_{n}\) <i class="it">via</i> a probabilistic approach and presented the Kantorovich and Durrmeyer variants of these operators. In <span class="cite">
	[
	<a href="#fa" >5</a>
	]
</span>, a better error estimation and statistical Korovkin type approximation properties of the operators \(L_{n}\) were examined by Dirik. Jain and Pethe <span class="cite">
	[
	<a href="#pf" >9</a>
	]
</span>, as a generalization of Szasz-Mirakjan operators, introduced the operators </p>
<div class="displaymath" id="a0000000005">
  \[  M_{n,\alpha }(f;x)=(1+n\alpha )^{-\tfrac {x}{\alpha }}\sum ^{\infty }_{\nu =0}\left( \alpha +\tfrac {1}{n}\right) ^{-\nu }\tfrac {x^{(\nu ,-\alpha )}}{\nu !} f\left( \tfrac {\nu }{n}\right) ,\quad x\geq 0  \]
</div>
<p> where \(x^{(0,-\alpha )}=1\), \(x^{(\nu ,-\alpha )}=x(x+\alpha )\cdots (x+(\nu -1)\alpha )\), \(0\leq n\alpha \leq 1\) and \(n\in \mathbb {N}\). By setting \(c=c_{n}=\tfrac {1}{n\alpha }\) such that \(c\geq \beta \) for certain constant \(\beta {\gt}0\), Abel and Ivan <span class="cite">
	[
	<a href="#ai" >1</a>
	]
</span> expressed these operators in the equivalent form </p>
<div class="displaymath" id="a0000000006">
  \[  S_{n,c}(f;x)=\sum ^{\infty }_{\nu =0}P^{[c]}_{n,\nu }(x) f\left( \tfrac {\nu }{n}\right) ,\quad x\geq 0  \]
</div>
<p> where \(P^{[c]}_{n,\nu }(x)=\big( \tfrac {c}{1+c}\big) ^{ncx}\binom {ncx+\nu -1}{\nu }(1+c)^{-\nu }\), and studied their local approximation properties and also obtained a complete asymptotic expansion formula. We remark that when \(c=1\) the operators \(S_{n,c}\) reduce to the operators defined by (<a href="#eq:1.1">1.1</a>). In <span class="cite">
	[
	<a href="#af" >6</a>
	]
</span>, Erençin and Taşdelen introduced the following generalization of the operators \(L_{n}\) </p>
<div class="displaymath" id="a0000000007">
  \[  L^{*}_{n}(f;x)=2^{-a_{n}x}\sum ^{\infty }_{k=0}\tfrac {(a_{n}x)_{k}}{2^{k}k!}f\big( \tfrac {k}{b_{n}}\big) ,\qquad x\geq 0  \]
</div>
<p> where \(({a_{n}})\), \(({b_{n}})\) are increasing and unbounded sequences of positive numbers such that </p>
<div class="displaymath" id="a0000000008">
  \[  \tfrac {a_{n}}{b_{n}}=1+{\mathcal O}\big( \tfrac {1}{b_{n}}\big) ,\quad \lim _{n\rightarrow \infty }\tfrac {1}{b_{n}}=0  \]
</div>
<p> and investigated their weighted approximation properties. Later, Erençin and Taşdelen <span class="cite">
	[
	<a href="#af2" >7</a>
	]
</span> estimated the rate of convergence for the Kantorovich type version of the operators \(L^{*}_{n}\) by means of the modulus of continuity, elements of local Lipschitz class and Peetre’s \(K\)-functional. Recently, A-statistical convergence properties of the operators \(L_{n}\) and their Kantorovich type modification were studied by Tarabie in <span class="cite">
	[
	<a href="#ta" >14</a>
	]
</span>. </p>
<p>Note that from Lemma 1 in <span class="cite">
	[
	<a href="#ag" >3</a>
	]
</span> we have </p>
<div class="displaymath" id="a0000000009">
  \[ \begin{split} &  L_{n}(1;x)=1,\\ &  L_{n}(t;x)=x. \end{split}  \]
</div>
<p>In this paper, for the operators \(L_{n}\) defined by (<a href="#eq:1.1">1.1</a>), we firstly show that when \(f\) is a general function of modulus of continuity, then \(L_{n}(f;x):=L_{n}(f)\) is also a function of modulus of continuity with the help of the same technique of Li <span class="cite">
	[
	<a href="#liz" >11</a>
	]
</span>. Later, we also show that the operators \(L_{n}\) preserve the Lipschitz constant and the order of a Lipschitz continuous function. Furthermore, we discuss the monotonicity of the operators \(L_{n}\) for \(n\) under the convexity of \(f\). We note that in the literature there are a number of papers containing preservation properties of positive linear operators. Some of them are <span class="cite">
	[
	<a href="#aa" >2</a>
	]
</span>, <span class="cite">
	[
	<a href="#bep" >4</a>
	]
</span>, <span class="cite">
	[
	<a href="#agf" >8</a>
	]
</span>, <span class="cite">
	[
	<a href="#kp" >10</a>
	]
</span> and <span class="cite">
	[
	<a href="#ti" >15</a>
	]
</span>. </p>
<h1 id="a0000000010">2 Some properties of the operators \(L_{\lowercase {n}}\)</h1>
<p>In order to give some properties of the operators defined by (<a href="#eq:1.1">1.1</a>) let us recall some definitions. </p>
<p>Let \(f\) be a real valued continuous function defined on \([0,\infty )\). Then \(f\) is said to be Lipschitz continuous of order \(\gamma \) \((0{\lt}\gamma \leq 1)\) on \([0,\infty )\), if there exists \(M{\gt} 0\) such that </p>
<div class="displaymath" id="a0000000011">
  \[  \left| f(x)-f(y)\right| \leq M\left| x-y\right| ^{\gamma } \]
</div>
<p> for all \(x,y\in [0,\infty )\). The set of Lipschitz continuous functions of order \(\gamma \) with Lipschitz constant \(M\) is denoted by \(\operatorname {Lip}_{M}(\gamma )\). </p>
<p>A real valued continuous function \(f\) is said to be convex on \([0,\infty )\), if </p>
<div class="displaymath" id="a0000000012">
  \[  f\Big( \sum ^{n}_{i=1}\alpha _{i}t_{i}\Big) \leq \sum ^{n}_{i=1}\alpha _{i}f(t_{i})  \]
</div>
<p> for all \(t_{1},t_{2},\cdots ,t_{n}\in [0,\infty )\) and for all non-negative numbers \(\alpha _{1},\alpha _{2},\cdots ,\alpha _{n}\) such that \(\alpha _{1}+\alpha _{2}+\cdots +\alpha _{n}=1\). </p>
<p>Also, a continuous and non-negative function \(\omega \) defined on \([0,\infty )\) is called the modulus of continuity function, if each of the following conditions is satisfied: </p>
<ul class="itemize">
  <li><p>\(\omega (u+v)\leq \omega (u)+\omega (v)\) for \(u,v\in [0,\infty )\), i.e., \(\omega \) is subadditive, </p>
</li>
  <li><p>\(\omega (u)\geq \omega (v)\) for \(u\geq v\), i.e., \(\omega \) is non-decreasing, </p>
</li>
  <li><p>\(\lim _{u\rightarrow 0^{+}}\omega (u)=\omega (0)=0\) </p>
</li>
</ul>
<p> (see p. 106 in <span class="cite">
	[
	<a href="#mp" >13</a>
	]
</span>). </p>
<p><div class="thm_thmwrapper " id="a0000000013">
  <div class="thm_thmheading">
    <span class="thm_thmcaption">
    Theorem
    </span>
    <span class="thm_thmlabel">1</span>
  </div>
  <div class="thm_thmcontent">
  <p>If \(\omega \) is a modulus of continuity function, then \(L_{n}(\omega )\) is also a modulus of continuity function. </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000014">
  <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 \(x,y\in [0,\infty )\) and \(x\leq y\). Then we have </p>
<div class="displaymath" id="a0000000015">
  \[ L_{n}(f;y)= 2^{-ny}\sum ^{\infty }_{k=0}\tfrac {(ny)_{k}}{2^{k}k!}f\left( \tfrac {k}{n}\right) = 2^{-ny}\sum ^{\infty }_{k=0}\tfrac {(n(x+(y-x)))_{k}}{2^{k}k!}f\left( \tfrac {k}{n}\right) .  \]
</div>
<p> Since </p>
<div class="displaymath" id="a0000000016">
  \[  (n(x+(y-x)))_{k}=\sum ^{k}_{i=0}\tbinom {k}{i}(nx)_{i}(n(y-x))_{k-i} \]
</div>
<p> one may write </p>
<div class="displaymath" id="a0000000017">
  \[  L_{n}(f;y)=2^{-ny}\sum ^{\infty }_{k=0}\sum ^{k}_{i=0}\tfrac {1}{2^{k}k!}\tbinom {k}{i}(nx)_{i}(n(y-x))_{k-i}f\left( \tfrac {k}{n}\right) .  \]
</div>
<p> Changing the order of the above summations and then taking \(k-i=j\), we reach to </p>
<div class="equation" id="eq:2.1">
<p>
  <div class="equation_content">
    \begin{equation}  \label{eq:2.1}L_{n}(f;y)= 2^{-ny}\sum ^{\infty }_{i=0}\sum ^{\infty }_{k=i}\tfrac {1}{2^{k}k!}\tbinom {k}{i}(nx)_{i}(n(y-x))_{k-i}f\left( \tfrac {k}{n}\right) = \end{equation}
  </div>
  <span class="equation_label">2.1</span>
</p>
</div>
<div class="displaymath" id="a0000000018">
  \begin{equation*}  = 2^{-ny}\sum ^{\infty }_{i=0}\sum ^{\infty }_{j=0}\tfrac {(nx)_{i}}{2^{i}i!}\tfrac {(n(y-x))_{j}}{2^{j}j!}f\big( \tfrac {i+j}{n}\big) . \end{equation*}
</div>
<p> On the other hand using the identity \(2^{n(y-x)}=\sum ^{\infty }_{j=0}\tfrac {(n(y-x))_{j}}{2^{j}j!}\), we get </p>
<div class="equation" id="eq:2.2">
<p>
  <div class="equation_content">
    \begin{equation}  \label{eq:2.2}\begin{split}  L_{n}(f;x)= &  2^{-nx}\sum ^{\infty }_{i=0}\tfrac {(nx)_{i}}{2^{i}i!}f\left( \tfrac {i}{n}\right) \\ = &  2^{-ny}2^{n(y-x)}\sum ^{\infty }_{i=0}\tfrac {(nx)_{i}}{2^{i}i!}f\left( \tfrac {i}{n}\right) \\ = &  2^{-ny}\sum ^{\infty }_{i=0}\sum ^{\infty }_{j=0}\tfrac {(nx)_{i}}{2^{i}i!}\tfrac {(n(y-x))_{j}}{2^{j}j!}f\left( \tfrac {i}{n}\right) . \end{split} \end{equation}
  </div>
  <span class="equation_label">2.2</span>
</p>
</div>
<p> Hence from (<a href="#eq:2.1">2.1</a>) and (<a href="#eq:2.2">2.2</a>) it follows that </p>
<div class="equation" id="eq:2.3">
<p>
  <div class="equation_content">
    \begin{equation}  \label{eq:2.3}L_{n}(f;y)-L_{n}(f;x)=2^{-ny}\sum ^{\infty }_{i=0}\sum ^{\infty }_{j=0}\tfrac {(nx)_{i}}{2^{i}i!}\tfrac {(n(y-x))_{j}}{2^{j}j!} \left[ f\big( \tfrac {i+j}{n}\big) -f\left( \tfrac {i}{n}\right) \right] . \end{equation}
  </div>
  <span class="equation_label">2.6</span>
</p>
</div>
<p> Thus by means of the equality (<a href="#eq:2.3">2.6</a>) and the subadditivity of \(\omega \), we can write </p>
<div class="displaymath" id="a0000000019">
  \[ \begin{split}  L_{n}(\omega ;y)-L_{n}(\omega ;x)= &  2^{-ny}\sum ^{\infty }_{i=0}\sum ^{\infty }_{j=0}\tfrac {(nx)_{i}}{2^{i}i!}\tfrac {(n(y-x))_{j}}{2^{j}j!} \left[ \omega \big( \tfrac {i+j}{n}\big) -\omega \left( \tfrac {i}{n}\right) \right] \\ \leq &  2^{-ny}\sum ^{\infty }_{i=0}\sum ^{\infty }_{j=0}\tfrac {(nx)_{i}}{2^{i}i!}\tfrac {(n(y-x))_{j}}{2^{j}j!}\omega \big( \tfrac {j}{n}\big) \\ = &  2^{-nx}\sum ^{\infty }_{i=0}\tfrac {(nx)_{i}}{2^{i}i!}2^{-n(y-x)}\sum ^{\infty }_{j=0}\tfrac {(n(y-x))_{j}}{2^{j}j!}\omega \big( \tfrac {j}{n}\big) \\ = &  L_{n}(1;x)2^{-n(y-x)}\sum ^{\infty }_{j=0}\tfrac {(n(y-x))_{j}}{2^{j}j!}\omega \big( \tfrac {j}{n}\big) \\ = &  2^{-n(y-x)}\sum ^{\infty }_{j=0}\tfrac {(n(y-x))_{j}}{2^{j}j!}\omega \big( \tfrac {j}{n}\big) \\ = &  L_{n}(\omega ;y-x). \end{split}  \]
</div>
<p> This shows the subadditivity of \(L_{n}(\omega )\). We also infer from (<a href="#eq:2.3">2.6</a>) that \(L_{n}(\omega ;y)\geq L_{n}(\omega ;x)\) for \(y\geq x\) which means that \(L_{n}(\omega )\) is non-decreasing. Finally the property \(L_{n}(\omega ;0)=\omega (0)=0\) is clear. Thus we may conclude that \(L_{n}(\omega )\) is a modulus of continuity function. </p>
<p>Now we introduce the second result of this section with the following theorem. </p>
<p><div class="thm_thmwrapper " id="a0000000020">
  <div class="thm_thmheading">
    <span class="thm_thmcaption">
    Theorem
    </span>
    <span class="thm_thmlabel">2</span>
  </div>
  <div class="thm_thmcontent">
  <p>If \(f \in \operatorname {Lip}_{M}(\gamma )\), then \(L_{n}(f)\in \operatorname {Lip}_{M}(\gamma )\). </p>

  </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>Suppose that \(x\leq y\). By using the facts \(f \in \operatorname {Lip}_{M}(\gamma )\) and \(L_{n}(1,x)=1\) from (<a href="#eq:2.3">2.6</a>) we can write </p>
<div class="displaymath" id="a0000000022">
  \begin{align*}  \left| L_{n}(f;y)-L_{n}(f;x)\right| &  \leq M 2^{-ny}\sum ^{\infty }_{i=0}\sum ^{\infty }_{j=0}\tfrac {(nx)_{i}}{2^{i}i!}\tfrac {(n(y-x))_{j}}{2^{j}j!}\big( \tfrac {j}{n}\big) ^{\gamma }\\ &  =M2^{-nx}\sum ^{\infty }_{i=0}\tfrac {(nx)_{i}}{2^{i}i!}2^{-n(y-x)}\sum ^{\infty }_{j=0}\tfrac {(n(y-x))_{j}}{2^{j}j!}\big( \tfrac {j}{n}\big) ^{\gamma }\\ &  =ML_{n}(1,x)2^{-n(y-x)}\sum ^{\infty }_{j=0}\tfrac {(n(y-x))_{j}}{2^{j}j!}\big( \tfrac {j}{n}\big) ^{\gamma }\\ &  =M2^{-n(y-x)}\sum ^{\infty }_{j=0}\tfrac {(n(y-x))_{j}}{2^{j}j!}\big( \tfrac {j}{n}\big) ^{\gamma }. \end{align*}
</div>
<p> Now applying Hölder’s inequality one gets </p>
<div class="displaymath" id="a0000000023">
  \[ \left| L_{n}(f;y)-L_{n}(f;x)\right| \leq M \Big( 2^{-n(y-x)}\sum ^{\infty }_{j=0}\tfrac {(n(y-x))_{j}}{2^{j}j!}\tfrac {j}{n}\Big) ^{\gamma } =M\left( L_{n}(t;y-x)\right) ^{\gamma }.  \]
</div>
<p> Since \(L_{n}(t;x)=x\) the above inequality implies that </p>
<div class="displaymath" id="a0000000024">
  \[  \left| L_{n}(f;y)-L_{n}(f;x)\right| \leq M(y-x)^{\gamma }.  \]
</div>
<p> Similarly, it can be shown that when \(x{\gt}y\) our claim is valid. </p>
<p>Now, we will study the monotonicity of the sequence of the operators \(L_{n}(f;x)\) defined by (<a href="#eq:1.1">1.1</a>) when the function \(f\) is convex. </p>
<p><div class="thm_thmwrapper " id="a0000000025">
  <div class="thm_thmheading">
    <span class="thm_thmcaption">
    Theorem
    </span>
    <span class="thm_thmlabel">3</span>
  </div>
  <div class="thm_thmcontent">
  <p>If \(f\) is a convex function defined on \([0,\infty )\), then \(L_{n}(f;x)\) is strictly monotonically decreasing, unless \(f\) is the linear function (in which case \(L_{n}(f;x)=L_{n+1}(f;x)\) for all \(n\)). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000026">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>We have </p>
<div class="displaymath" id="a0000000027">
  \[ \begin{split}  L_{n} &  (f;x)-L_{n+1}(f;x)=\\ &  = 2^{-nx}\sum ^{\infty }_{k=0}\tfrac {(nx)_{k}}{2^{k}k!} f\left( \tfrac {k}{n}\right) -2^{-(n+1)x}\sum ^{\infty }_{k=0}\tfrac {((n+1)x)_{k}}{2^{k}k!} f\big( \tfrac {k}{n+1}\big) \\ &  =2^{-(n+1)x}\left\{  2^{x}\sum ^{\infty }_{k=0}\tfrac {(nx)_{k}}{2^{k}k!} f\left( \tfrac {k}{n}\right) -\sum ^{\infty }_{k=0}\tfrac {((n+1)x)_{k}}{2^{k}k!} f\big( \tfrac {k}{n+1}\big) \right\}  . \end{split}  \]
</div>
<p> Using the identity \(2^{x}=\sum ^{\infty }_{l=0}\tfrac {(x)_{l}}{2^{l}l !}\), one may write </p>
<div class="displaymath" id="eq:2.4">
  \begin{align}  \label{eq:2.4}& L_{n} (f;x)-L_{n+1}(f;x)= \\ &  =2^{-(n+1)x}\left\{  \sum ^{\infty }_{l=0}\tfrac {(x)_{l}}{2^{l}l!}\sum ^{\infty }_{k=0}\tfrac {(nx)_{k}}{2^{k}k!} f\left( \tfrac {k}{n}\right) -\sum ^{\infty }_{k=0}\tfrac {((n+1)x)_{k}}{2^{k}k!} f\big( \tfrac {k}{n+1}\big) \right\} = \nonumber \end{align}
</div>
<div class="displaymath" id="a0000000028">
  \begin{align*} &  =2^{-(n+1)x}\left\{  \sum ^{\infty }_{l=0}\sum ^{\infty }_{k=l}\tfrac {(nx)_{k-l}(x)_{l}}{2^{k}l!(k-l)!}f\big( \tfrac {k-l}{n}\big) -\sum ^{\infty }_{k=0}\tfrac {((n+1)x)_{k}}{2^{k}k!} f\big( \tfrac {k}{n+1}\big) \right\}  \\ &  =2^{-(n+1)x}\left\{  \sum ^{\infty }_{k=0}\left[ \sum ^{k}_{l=0}\tfrac {(nx)_{l}(x)_{k-l}}{2^{k}l!(k-l)!}f\left( \tfrac {l}{n}\right) -\tfrac {((n+1)x)_{k}}{2^{k}k!} f\big( \tfrac {k}{n+1}\big) \right] \right\}  . \end{align*}
</div>
<p> Now we only need to show that for all \(k=0,1,\ldots \), </p>
<div class="equation" id="eq:2.5">
<p>
  <div class="equation_content">
    \begin{equation}  \label{eq:2.5}f\big( \tfrac {k}{n+1}\big) \leq \tfrac {1}{((n+1)x)_{k}}\sum ^{k}_{l=0}\tbinom {k}{l}(nx)_{l}(x)_{k-l}f\left( \tfrac {l}{n}\right) , \end{equation}
  </div>
  <span class="equation_label">2.17</span>
</p>
</div>
<p> which is a direct result of convexity. In fact, set </p>
<div class="displaymath" id="a0000000029">
  \[  \alpha _{l}=\tbinom {k}{l}\tfrac {(nx)_{l}(x)_{k-l}}{((n+1)x)_{k}}\geq 0 \qquad {\rm and\quad } t_{l}=\tfrac {l}{n}.  \]
</div>
<p> Therefore with the help of the identity \(((n+1)x)_{k}=\sum ^{k}_{l=0}\tbinom {k}{l}(nx)_{l}(x)_{k-l}\), it is clear that </p>
<div class="displaymath" id="a0000000030">
  \[  \sum ^{k}_{l=0}\alpha _{l}=\tfrac {1}{((n+1)x)_{k}}\sum ^{k}_{l=0}\tbinom {k}{l}(nx)_{l}(x)_{k-l}=1.  \]
</div>
<p> On the other hand, we have </p>
<div class="displaymath" id="a0000000031">
  \[ \begin{split}  \sum ^{k}_{l=0}\alpha _{l}t_{l}= &  \tfrac {1}{((n+1)x)_{k}}\sum ^{k}_{l=0}\tbinom {k}{l}(nx)_{l}(x)_{k-l}\tfrac {l}{n} \\ = &  \tfrac {1}{n((n+1)x)_{k}}\sum ^{k}_{l=1}\tfrac {k!}{(l-1)!(k-l)!}(nx)_{l}(x)_{k-l}\\ = &  \tfrac {k}{n((n+1)x)_{k}}\sum ^{k-1}_{l=0}\tbinom {k-1}{l}(nx)_{l+1}(x)_{k-l-1}. \end{split}  \]
</div>
<p> Since </p>
<div class="displaymath" id="a0000000032">
  \[  (nx)_{l+1}=nx(nx+1)_{l},\quad ((n+1)x)_{k}=(n+1)x((n+1)x+1)_{k-1} \]
</div>
<p> and </p>
<div class="displaymath" id="a0000000033">
  \[  ((n+1)x+1)_{k-1}=\sum ^{k-1}_{l=0}\tbinom {k-1}{l}(nx+1)_{l}(x)_{k-l-1} \]
</div>
<p> one may write </p>
<div class="displaymath" id="a0000000034">
  \[ \sum ^{k}_{l=0}\alpha _{l}t_{l}= \tfrac {k}{(n+1)((n+1)x+1)_{k-1}}\sum ^{k-1}_{l=0}\tbinom {k-1}{l}(nx+1)_{l}(x)_{k-l-1} = \tfrac {k}{n+1} \]
</div>
<p> which, making use of the convexity of \(f\), gives the inequality (<a href="#eq:2.5">2.17</a>). Hence from (<a href="#eq:2.4">2.16</a>) we arrive at the desired result. Clearly \(L_{n}(f;x)=L_{n+1}(f;x)\) occurs only if </p>
<div class="displaymath" id="a0000000035">
  \[ \begin{split}  f\Big( \tfrac {1}{((n+1)x)_{k}}\sum ^{k}_{l=0}\tbinom {k}{l}(nx)_{l}(x)_{k-l}\tfrac {l}{n}\Big) = &  f\big( \tfrac {k}{n+1}\big) \\ = &  \tfrac {1}{((n+1)x)_{k}}\sum ^{k}_{l=0}\tbinom {k}{l}(nx)_{l}(x)_{k-l}f\left( \tfrac {l}{n}\right) \\ \end{split}  \]
</div>
<p> for all \(k=0,1,\ldots \) This implies that \(f\) is linear in \([0,\infty )\). Thus the proof is completed. </p>
<p><small class="footnotesize">  </small></p>
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