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<h1>Caputo Fractional Approximation by Sublinear Operators</h1>
<p class="authors">
<span class="author">George A. Anastassiou\(^\ast \)</span>
</p>
<p class="date">September 25, 2017. Accepted: October 12, 2018. Published online: February 17, 2019.</p>
</div>
<p>\(^\ast \)Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152, U.S.A, e-mail: <span class="tt">ganastss@memphis.edu</span>. </p>

<div class="abstract"><p> Here we consider the approximation of functions by sublinear positive operators with applications to a big variety of Max-Product operators under Caputo fractional differentiability. Our study is based on our general fractional results about positive sublinear operators. We produce Jackson type inequalities under simple initial conditions. So our approach is quantitative by producing inequalities with their right hand sides involving the modulus of continuity of fractional derivative of the function under approximation. </p>
<p><b class="bf">MSC.</b> 26A33, 41A17, 41A25, 41A36. </p>
<p><b class="bf">Keywords.</b> positive sublinear operators, Max-product operators, modulus of continuity, Caputo fractional derivative </p>
</div>
<h1 id="a0000000002">1 Introduction</h1>
<p>The main motivation here is the monograph by B. Bede, L. Coroianu and S. Gal <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	2016
	]
</span>. </p>
<p>Let \(N\in \mathbb {N}\), the well-known Bernstein polynomials <span class="cite">
	[
	<a href="#11" >11</a>
	]
</span> are positive linear operators, defined by the formula </p>
<div class="equation" id="1">
<p>
  <div class="equation_content">
    \begin{equation}  B_{N}\left( f\right) \left( x\right) =\sum _{k=0}^{N} \tbinom Nk x^{k}\left( 1-x\right) ^{N-k}f(\tfrac {k}{N}) ,\text{ \  }x\in \left[ 0,1\right] ,\text{ }f\in C\left( \left[ 0,1\right] \right) .\label{1}\end{equation}
  </div>
  <span class="equation_label">1.1</span>
</p>
</div>
<p>T. Popoviciu in <span class="cite">
	[
	<a href="#12" >12</a>
	]
</span> (1935), proved for \(f\in C\left( \left[ 0,1\right] \right) \) that </p>
<div class="equation" id="2">
<p>
  <div class="equation_content">
    \begin{equation}  \left| B_{N}\left( f\right) \left( x\right) -f\left( x\right) \right| \leq \tfrac {5}{4}\omega _{1}\left( f,\tfrac {1}{\sqrt{N}}\right) ,\text{ \  }\forall \text{ }x\in \left[ 0,1\right] ,\label{2}\end{equation}
  </div>
  <span class="equation_label">1.2</span>
</p>
</div>
<p> where </p>
<div class="equation" id="3">
<p>
  <div class="equation_content">
    \begin{equation}  \omega _{1}\left( f,\delta \right) =\underset {\left| x-y\right| \leq \delta }{\underset {x,y\in \left[ a,b\right] :}{\sup }}\left| f\left( x\right) -f\left( y\right) \right| ,\text{ \  }\delta >0,\label{3}\end{equation}
  </div>
  <span class="equation_label">1.3</span>
</p>
</div>
<p> is the first modulus of continuity, here \(\left[ a,b\right] =\left[ 0,1\right] \). </p>
<p>G.G. Lorentz in <span class="cite">
	[
	<a href="#11" >11</a>
	, 
	p.
	
	21
	]
</span> (1986), proved for \(f\in C^{1}\left( \left[ 0,1\right] \right) \) that </p>
<div class="equation" id="4">
<p>
  <div class="equation_content">
    \begin{equation}  \left| B_{N}\left( f\right) \left( x\right) -f\left( x\right) \right| \leq \tfrac {3}{4\sqrt{N}}\omega _{1}\left( f^{\prime },\tfrac {1}{\sqrt{N}}\right) ,\text{ \  }\forall \text{ }x\in \left[ 0,1\right] ,\label{4}\end{equation}
  </div>
  <span class="equation_label">1.4</span>
</p>
</div>
<p>In <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	1
	]
</span>, the authors introduced the basic Max-product Bernstein operators, </p>
<div class="equation" id="5">
<p>
  <div class="equation_content">
    \begin{equation}  B_{N}^{\left( M\right) }\left( f\right) \left( x\right) =\tfrac {\bigvee \limits _{k=0}^{N}p_{N,k}\left( x\right) f( \frac{k}{N}) }{\bigvee \limits _{k=0}^{N}p_{N,k}\left( x\right) },\text{ \  }N\in \mathbb {N}\text{,}\label{5}\end{equation}
  </div>
  <span class="equation_label">1.5</span>
</p>
</div>
<p> where \(\bigvee \) stands for maximum, and \(p_{N,k}\left( x\right) =\tbinom Nk x^{k}\left( 1-x\right) ^{N-k}\) and \(f:\left[ 0,1\right] \rightarrow \mathbb {R}_{+}=[0,\infty ).\) </p>
<p>These are nonlinear and piecewise rational operators. </p>
<p>The authors in <span class="cite">
	[
	<a href="#5" >5</a>
	]
</span> studied similar such nonlinear operators such as: the Max-product Favard-Szász-Mirakjan operators and their truncated version, the Max-product Baskakov operators and their truncated version, also many other similar specific operators. The study in <span class="cite">
	[
	<a href="#5" >5</a>
	]
</span> is based on presented there general theory of sublinear operators. These Max-product operators tend to converge faster to the on hand function. </p>
<p>So we mention from <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	30
	]
</span>, that for \(f:\left[ 0,1\right] \rightarrow \mathbb {R}_{+}\) continuous, we have the estimate </p>
<div class="equation" id="6">
<p>
  <div class="equation_content">
    \begin{equation}  \left| B_{N}^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq 12\, \omega _{1}\left( f,\tfrac {1}{\sqrt{N+1}}\right) ,\text{ \  for all }N\in \mathbb {N},\text{\  }x\in \left[ 0,1\right] ,\label{6}\end{equation}
  </div>
  <span class="equation_label">1.6</span>
</p>
</div>
<p> Also from <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	36
	]
</span>, we mention that for \(f:\left[ 0,1\right] \rightarrow \mathbb {R}_{+}\) being concave function we get that </p>
<div class="equation" id="7">
<p>
  <div class="equation_content">
    \begin{equation}  \left| B_{N}^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq 2\, \omega _{1}\left( f,\tfrac {1}{N}\right) ,\text{ \  for all }x\in \left[ 0,1\right] ,\label{7}\end{equation}
  </div>
  <span class="equation_label">1.7</span>
</p>
</div>
<p> a much faster convergence. </p>
<p>In this article we expand the study in <span class="cite">
	[
	<a href="#5" >5</a>
	]
</span> by considering Caputo fractional smoothness of functions. So our inequalities are with respect to \(\omega _{1}\left( D^{\alpha }f,\delta \right) \), \(\delta {\gt}0\), where \(D^{\alpha }f\) with \(\alpha {\gt}0\) is the Caputo fractional derivative. </p>
<h1 id="a0000000003">2 Main Results</h1>
<p>We need </p>
<p><div class="definition_thmwrapper " id="d1">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">2.1</span>
  </div>
  <div class="definition_thmcontent">
  <p>Let \(\nu \geq 0\), \(n=\left\lceil \nu \right\rceil \) (\(\left\lceil \cdot \right\rceil \) is the ceiling of the number), \(f\in AC^{n}\left( \left[ a,b\right] \right) \) (space of functions \(f\) with \(f^{\left( n-1\right) }\in AC\left( \left[ a,b\right] \right) \), absolutely continuous functions). We call left Caputo fractional derivative (see <span class="cite">
	[
	<a href="#7" >7</a>
	, 
	p.
	
	49
	]
</span>, <span class="cite">
	[
	<a href="#10" >10</a>
	]
</span>, <span class="cite">
	[
	<a href="#13" >13</a>
	]
</span>) the function </p>
<div class="equation" id="8">
<p>
  <div class="equation_content">
    \begin{equation}  D_{\ast a}^{\nu }f\left( x\right) =\tfrac {1}{\Gamma \left( n-\nu \right) }\int _{a}^{x}\left( x-t\right) ^{n-\nu -1}f^{\left( n\right) }\left( t\right) dt,\text{ \  }\forall \text{ }x\in \left[ a,b\right] , \label{8}\end{equation}
  </div>
  <span class="equation_label">2.8</span>
</p>
</div>
<p> where \(\Gamma \) is the gamma function \(\Gamma \left( v\right) =\int _{0}^{\infty }e^{-t}t^{v-1}dt\), \(v{\gt}0.\) </p>
<p>We set \(D_{\ast a}^{0}f\left( x\right) =f\left( x\right) \), \(\forall \) \(x\in \left[ a,b\right] .\) </p>

  </div>
</div> </p>
<p><div class="lemma_thmwrapper " id="l2">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">2.2</span>
  </div>
  <div class="lemma_thmcontent">
  <p><span class="cite">
	[
	<a href="#2" >2</a>
	]
</span> Let \(\nu {\gt}0\), \(\nu \notin \mathbb {N}\), \(n=\left\lceil \nu \right\rceil \), \(f\in C^{n-1}\left( \left[ a,b\right] \right) \) and \(f^{\left( n\right) }\in L_{\infty }\left( \left[ a,b\right] \right) \). Then \(D_{\ast a}^{\nu }f\left( a\right) =0.\) </p>

  </div>
</div> </p>
<p>We need </p>
<p><div class="definition_thmwrapper " id="d3">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">2.3</span>
  </div>
  <div class="definition_thmcontent">
  <p>(see also <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span>, <span class="cite">
	[
	<a href="#8" >8</a>
	]
</span>, <span class="cite">
	[
	<a href="#10" >10</a>
	]
</span>) Let \(f\in AC^{m}\left( \left[ a,b\right] \right) ,\) \(m=\left\lceil \alpha \right\rceil \), \(\alpha {\gt}0 \). The right Caputo fractional derivative of order \(\alpha {\gt}0\) is given by</p>
<div class="equation" id="9">
<p>
  <div class="equation_content">
    \begin{equation}  D_{b-}^{\alpha }f\left( x\right) =\tfrac {\left( -1\right) ^{m}}{\Gamma \left( m-\alpha \right) }\int _{x}^{b}\left( \zeta -x\right) ^{m-\alpha -1}f^{\left( m\right) }\left( \zeta \right) d\zeta ,\text{ \  }\forall \text{ }x\in \left[ a,b\right] .\label{9}\end{equation}
  </div>
  <span class="equation_label">2.9</span>
</p>
</div>
<p> We set \(D_{b-}^{0}f\left( x\right) =f\left( x\right) \). </p>

  </div>
</div> </p>
<p><div class="lemma_thmwrapper " id="l4">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">2.4</span>
  </div>
  <div class="lemma_thmcontent">
  <p><span class="cite">
	[
	<a href="#2" >2</a>
	]
</span> Let \(f\in C^{m-1}\left( \left[ a,b\right] \right) ,\) \(f^{\left( m\right) }\in L_{\infty }\left( \left[ a,b\right] \right) ,\) \(m=\left\lceil \alpha \right\rceil \), \(\alpha {\gt}0\), \(\alpha \notin \mathbb {N}\). Then \(D_{b-}^{\alpha }f\left( b\right) =0.\) </p>

  </div>
</div> </p>
<p><div class="convention_thmwrapper " id="c5">
  <div class="convention_thmheading">
    <span class="convention_thmcaption">
    Convention
    </span>
    <span class="convention_thmlabel">2.5</span>
  </div>
  <div class="convention_thmcontent">
  <p>We assume that </p>
<div class="equation" id="10">
<p>
  <div class="equation_content">
    \begin{equation}  D_{\ast x_{0}}^{a}f\left( x\right) =0\text{, for }x<x_{0},\label{10}\end{equation}
  </div>
  <span class="equation_label">2.10</span>
</p>
</div>
<p> and </p>
<div class="equation" id="11">
<p>
  <div class="equation_content">
    \begin{equation}  D_{x_{0}-}^{\alpha }f\left( x\right) =0\text{, for }x>x_{0}, \label{11}\end{equation}
  </div>
  <span class="equation_label">2.11</span>
</p>
</div>
<p> for all \(x,x_{0}\in \left[ a,b\right] .\) </p>

  </div>
</div> </p>
<p>We mention </p>
<p><div class="proposition_thmwrapper " id="p6">
  <div class="proposition_thmheading">
    <span class="proposition_thmcaption">
    Proposition
    </span>
    <span class="proposition_thmlabel">2.6</span>
  </div>
  <div class="proposition_thmcontent">
  <p><span class="cite">
	[
	<a href="#2" >2</a>
	]
</span> Let \(f\in C^{n}\left( \left[ a,b\right] \right) \), \(n=\left\lceil \nu \right\rceil \), \(\nu {\gt}0\). Then \(D_{\ast a}^{\nu }f\left( x\right) \) is continuous in \(x\in \left[ a,b\right] .\) </p>

  </div>
</div> </p>
<p><div class="proposition_thmwrapper " id="p7">
  <div class="proposition_thmheading">
    <span class="proposition_thmcaption">
    Proposition
    </span>
    <span class="proposition_thmlabel">2.7</span>
  </div>
  <div class="proposition_thmcontent">
  <p><span class="cite">
	[
	<a href="#2" >2</a>
	]
</span> Let \(f\in C^{m}\left( \left[ a,b\right] \right) \), \(m=\left\lceil \alpha \right\rceil \), \(\alpha {\gt}0\). Then \(D_{b-}^{\alpha }f\left( x\right) \) is continuous in \(x\in \left[ a,b\right] .\) </p>

  </div>
</div> </p>
<p>The modulus of continuity \(\omega _{1}\left( f,\delta \right) \) is defined the same way for bounded functions, see (<a href="#3">1.3</a>), and it is finite. </p>
<p>We make </p>
<p><div class="remark_thmwrapper " id="r8">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">2.8</span>
  </div>
  <div class="remark_thmcontent">
  <p><span class="cite">
	[
	<a href="#2" >2</a>
	]
</span> Let \(f\in C^{n-1}\left( \left[ a,b\right] \right) \), \(f^{\left( n\right) }\in L_{\infty }\left( \left[ a,b\right] \right) \), \(n=\left\lceil \nu \right\rceil \), \(\nu {\gt}0\), \(\nu \notin \mathbb {N}\). Then </p>
<div class="equation" id="12">
<p>
  <div class="equation_content">
    \begin{equation}  \omega _{1}\left( D_{\ast a}^{\nu }f,\delta \right) \leq \tfrac {2\left\|  f^{\left( n\right) }\right\|  _{\infty }}{\Gamma \left( n-\nu +1\right) }\left( b-a\right) ^{n-\nu }.\label{12}\end{equation}
  </div>
  <span class="equation_label">2.12</span>
</p>
</div>
<p> Similarly, let \(f\in C^{m-1}\left( \left[ a,b\right] \right) \), \(f^{\left( m\right) }\in L_{\infty }\left( \left[ a,b\right] \right) \), \(m=\left\lceil \alpha \right\rceil \), \(\alpha {\gt}0\), \(\alpha \notin \mathbb {N}\), then </p>
<div class="equation" id="13.">
<p>
  <div class="equation_content">
    \begin{equation}  \omega _{1}\left( D_{b-}^{\alpha }f,\delta \right) \leq \tfrac {2\left\|  f^{\left( m\right) }\right\|  _{\infty }}{\Gamma \left( m-\alpha +1\right) }\left( b-a\right) ^{m-\alpha }.\label{13.}\end{equation}
  </div>
  <span class="equation_label">2.13</span>
</p>
</div>
<p> That is \(\omega _{1}\left( D_{\ast a}^{\nu }f,\delta \right) \), \(\omega _{1}\left( D_{b-}^{\alpha }f,\delta \right) \) are finite. </p>
<p>Clearly, above \(D_{\ast a}^{\nu }f\) and \(D_{b-}^{\alpha }f\) are bounded, from </p>
<div class="equation" id="14">
<p>
  <div class="equation_content">
    \begin{equation}  \left\vert D_{\ast a}^{\nu }f\left( x\right) \right\vert \leq \tfrac {\Vert f^{\left( n\right) }\Vert _{\infty }}{\Gamma \left( n-\nu +1\right) }\left( b-a\right) ^{n-\nu },\text{ \  }\forall \text{ }x\in \left[ a,b\right] ,\label{14}\end{equation}
  </div>
  <span class="equation_label">2.14</span>
</p>
</div>
<p> see <span class="cite">
	[
	<a href="#2" >2</a>
	]
</span>.<span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>We need </p>
<p><div class="definition_thmwrapper " id="d9">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">2.9</span>
  </div>
  <div class="definition_thmcontent">
  <p>Let \(D_{x_{0}}^{\alpha }f\) denote any of \(D_{x_{0}-}^{\alpha }f\), \(D_{\ast x_{0}}^{\alpha }f\), and \(\delta {\gt}0\). We set </p>
<div class="equation" id="15">
<p>
  <div class="equation_content">
    \begin{equation}  \omega _{1}\left( D_{x_{0}}^{\alpha }f,\delta \right) :=\max \left\{  \omega _{1}\left( D_{x_{0}-}^{\alpha }f,\delta \right) _{\left[ a,x_{0}\right] },\omega _{1}\left( D_{\ast x_{0}}^{\alpha }f,\delta \right) _{\left[ x_{0},b\right] }\right\}  ,\label{15}\end{equation}
  </div>
  <span class="equation_label">2.15</span>
</p>
</div>
<p> where \(x_{0}\in \left[ a,b\right] .\) Here the moduli of continuity are considered over \(\left[ a,x_{0}\right] \) and \(\left[ x_{0},b\right] \), respectively. </p>

  </div>
</div> </p>
<p>We need </p>
<p><div class="theorem_thmwrapper " id="t10">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2.10</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(\alpha {\gt}0\), \(\alpha \notin \mathbb {N}\), \(m=\left\lceil \alpha \right\rceil \), \(x_{0}\in \left[ a,b\right] \subset \mathbb {R}\), \(f\in AC^{m}\left( \left[ a,b\right] ,\mathbb {R}_{+}\right) \) (i.e. \(f^{\left( m-1\right) }\in AC\left( \left[ a,b\right] \right) \), absolutely continuous functions on \(\left[ a,b\right] \)), and \(f^{\left( m\right) }\in L_{\infty }\left( \left[ a,b\right] \right) \). Furthermore we assume that \(f^{\left( k\right) }\left( x_{0}\right) =0\), \(k=1,...,m-1\). Then </p>
<div class="equation" id="16">
<p>
  <div class="equation_content">
    \begin{equation}  \left| f\left( x\right) -f\left( x_{0}\right) \right| \leq \tfrac {\omega _{1}\left( D_{x_{0}}^{\alpha }f,\delta \right) }{\Gamma \left( \alpha +1\right) }\left[ \left| x-x_{0}\right| ^{\alpha }+\tfrac {\left| x-x_{0}\right| ^{\alpha +1}}{\left( \alpha +1\right) \delta }\right] ,\text{ \  }\delta >0,\label{16}\end{equation}
  </div>
  <span class="equation_label">2.16</span>
</p>
</div>
<p> for all \(a\leq x\leq b.\) </p>
<p>If \(0{\lt}\alpha {\lt}1\), then we do not need initial conditions. </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000004">
  <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="#7" >7</a>
	, 
	p.
	
	54
	]
</span>, we get by left Caputo Taylor formula that </p>
<div class="equation" id="17">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( x\right) =\sum _{k=0}^{m-1}\tfrac {f^{\left( k\right) }\left( x_{0}\right) }{k!}\left( x-x_{0}\right) ^{k}+\tfrac {1}{\Gamma \left( \alpha \right) }\int _{x_{0}}^{x}\left( x-z\right) ^{\alpha -1}D_{\ast x_{0}}^{\alpha }f\left( z\right) dz,\label{17}\end{equation}
  </div>
  <span class="equation_label">2.17</span>
</p>
</div>
<p> for all \(x_{0}\leq x\leq b.\) </p>
<p>Also from <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span>, using the right Caputo fractional Taylor formula we get </p>
<div class="equation" id="18">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( x\right) =\sum _{k=0}^{m-1}\tfrac {f^{\left( k\right) }\left( x_{0}\right) }{k!}\left( x-x_{0}\right) ^{k}+\tfrac {1}{\Gamma \left( \alpha \right) }\int _{x}^{x_{0}}\left( z-x\right) ^{\alpha -1}D_{x_{0}-}^{\alpha }f\left( z\right) dz,\label{18}\end{equation}
  </div>
  <span class="equation_label">2.18</span>
</p>
</div>
<p> for all \(a\leq x\leq x_{0}.\) </p>
<p>By the assumption \(f^{\left( k\right) }\left( x_{0}\right) =0\), \(k=1,...,m-1\), we get </p>
<div class="equation" id="19">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( x\right) -f\left( x_{0}\right) =\tfrac {1}{\Gamma \left( \alpha \right) }\int _{x_{0}}^{x}\left( x-z\right) ^{\alpha -1}D_{\ast x_{0}}^{\alpha }f\left( z\right) dz,\label{19}\end{equation}
  </div>
  <span class="equation_label">2.19</span>
</p>
</div>
<p> for all \(x_{0}\leq x\leq b.\) </p>
<p>And it holds </p>
<div class="equation" id="20">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( x\right) -f\left( x_{0}\right) =\tfrac {1}{\Gamma \left( \alpha \right) }\int _{x}^{x_{0}}\left( z-x\right) ^{\alpha -1}D_{x_{0}-}^{\alpha }f\left( z\right) dz,\label{20}\end{equation}
  </div>
  <span class="equation_label">2.20</span>
</p>
</div>
<p> for all \(a\leq x\leq x_{0}.\) </p>
<p>Notice that when \(0{\lt}\alpha {\lt}1\), then \(m=1\), and (<a href="#19">2.19</a>) and (<a href="#20">2.20</a>) are valid without initial conditions. </p>
<p>Since \(D_{x_{0}-}^{\alpha }f\left( x_{0}\right) =D_{\ast x_{0}}^{\alpha }f\left( x_{0}\right) =0\), we get </p>
<div class="equation" id="21.">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( x\right) -f\left( x_{0}\right) =\tfrac {1}{\Gamma \left( \alpha \right) }\int _{x_{0}}^{x}\left( x-z\right) ^{\alpha -1}\left( \left( D_{\ast x_{0}}^{\alpha }f\right) \left( z\right) -D_{\ast x_{0}}^{\alpha }f\left( x_{0}\right) \right) dz,\label{21.}\end{equation}
  </div>
  <span class="equation_label">2.21</span>
</p>
</div>
<p> \(x_{0}\leq x\leq b,\) and </p>
<div class="equation" id="22">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( x\right) -f\left( x_{0}\right) =\tfrac {1}{\Gamma \left( \alpha \right) }\int _{x}^{x_{0}}\left( z-x\right) ^{\alpha -1}\left( D_{x_{0}-}^{\alpha }f\left( z\right) -D_{x_{0}-}^{\alpha }f\left( x_{0}\right) \right) dz,\label{22}\end{equation}
  </div>
  <span class="equation_label">2.22</span>
</p>
</div>
<p> \(a\leq x\leq x_{0}.\) </p>
<p>We have that (\(x_{0}\leq x\leq b\)) </p>
<div class="displaymath" id="a0000000005">
  \begin{align} & \left| f\left( x\right) -f\left( x_{0}\right) \right| \leq \nonumber \\ &  \leq \tfrac {1}{\Gamma \left( \alpha \right) }\int _{x_{0}}^{x}\left( x-z\right) ^{\alpha -1}\left| \left( D_{\ast x_{0}}^{\alpha }f\right) \left( z\right) -D_{\ast x_{0}}^{\alpha }f\left( x_{0}\right) \right| dz \nonumber \\ & \underset {\text{(}\delta _{1}{\gt}0\text{)}}{\leq }\tfrac {1}{\Gamma \left( \alpha \right) }\int _{x_{0}}^{x}\left( x-z\right) ^{\alpha -1}\omega _{1}\left( D_{\ast x_{0}}^{\alpha }f,\tfrac {\delta _{1}\left| z-x_{0}\right| }{\delta _{1}}\right) _{\left[ x_{0},b\right] }dz \label{23}\\ &  \leq \tfrac {\omega _{1}\left( D_{\ast x_{0}}^{\alpha }f,\delta _{1}\right) _{\left[ x_{0},b\right] }}{\Gamma \left( \alpha \right) }\int _{x_{0}}^{x}\left( x-z\right) ^{\alpha -1}\left( 1+\tfrac {\left( z-x_{0}\right) }{\delta _{1}}\right) dz \nonumber \\ &  = \tfrac {\omega _{1}\left( D_{\ast x_{0}}^{\alpha }f,\delta _{1}\right) _{\left[ x_{0},b\right] }}{\Gamma \left( \alpha \right) }\left[ \tfrac {\left( x-x_{0}\right) ^{\alpha }}{\alpha }+\tfrac {1}{\delta _{1}}\int _{x_{0}}^{x}\left( x-z\right) ^{\alpha -1}\left( z-x_{0}\right) ^{2-1}dz\right] \nonumber \\ &  = \tfrac {\omega _{1}\left( D_{\ast x_{0}}^{\alpha }f,\delta _{1}\right) _{\left[ x_{0},b\right] }}{\Gamma \left( \alpha \right) }\left[ \tfrac {\left( x-x_{0}\right) ^{\alpha }}{\alpha }+\tfrac {1}{\delta _{1}}\tfrac {\Gamma \left( \alpha \right) \Gamma \left( 2\right) }{\Gamma \left( \alpha +2\right) }\left( x-x_{0}\right) ^{\alpha +1}\right] \label{24}\\ &  = \tfrac {\omega _{1}\left( D_{\ast x_{0}}^{\alpha }f,\delta _{1}\right) _{\left[ x_{0},b\right] }}{\Gamma \left( \alpha \right) }\left[ \tfrac {\left( x-x_{0}\right) ^{\alpha }}{\alpha }+\tfrac {1}{\delta _{1}}\tfrac {1}{\left( \alpha +1\right) \alpha }\left( x-x_{0}\right) ^{\alpha +1}\right] \nonumber \\ &  = \tfrac {\omega _{1}\left( D_{\ast x_{0}}^{\alpha }f,\delta _{1}\right) _{\left[ x_{0},b\right] }}{\Gamma \left( \alpha +1\right) }\left[ \left( x-x_{0}\right) ^{\alpha }+\tfrac {\left( x-x_{0}\right) ^{\alpha +1}}{\left( \alpha +1\right) \delta _{1}}\right] . \nonumber \end{align}
</div>
<p> We have proved that </p>
<div class="displaymath" id="25">
  \begin{align}  \left| f\left( x\right) -f\left( x_{0}\right) \right| \leq \tfrac {\omega _{1}\left( D_{\ast x_{0}}^{\alpha }f,\delta _{1}\right) _{\left[ x_{0},b\right] }}{\Gamma \left( \alpha +1\right) }\left[ \left( x-x_{0}\right) ^{\alpha }+\tfrac {\left( x-x_{0}\right) ^{\alpha +1}}{\left( \alpha +1\right) \delta _{1}}\right] ,\label{25}\end{align}
</div>
<p> \(\delta _{1}{\gt}0\), and \(x_{0}\leq x\leq b.\) <br />Similarly acting, we get (\(a\leq x\leq x_{0}\)) </p>
<div class="displaymath" id="a0000000006">
  \begin{align} & \left\vert f\left( x\right) -f\left(x_{0}\right) \right\vert \leq \nonumber \\ &  \leq \tfrac {1}{\Gamma \left( \alpha \right) }\int _{x}^{x_{0}}\left( z-x\right) ^{\alpha -1}\left\vert D_{x_{0}-}^{\alpha }f\left( z\right) -D_{x_{0}-}^{\alpha }f\left( x_{0}\right) \right\vert dz\nonumber \\ & =\tfrac {1}{\Gamma \left( \alpha \right) }\int _{x}^{x_{0}}\left( z-x\right) ^{\alpha -1}\omega _{1}\left( D_{x_{0}-}^{\alpha }f,\tfrac {\delta _{2}(x_{0}-z)}{\delta _{2}}\right) _{\left[ a,x_{0}\right] }dz\label{26}\\ &  \leq \tfrac {\omega _{1}\left( D_{x_{0}-}^{\alpha }f,\delta _{2}\right) _{\left[ a,x_{0}\right] }}{\Gamma \left( \alpha \right) }\left[ \int _{x}^{x_{0}}\left( z-x\right) ^{\alpha -1}\left( 1+\tfrac {x_{0}-z}{\delta _{2}}\right) dz\right] \nonumber \\ & =\tfrac {\omega _{1}\left( D_{x_{0}-}^{\alpha }f,\delta _{2}\right) _{\left[ a,x_{0}\right] }}{\Gamma \left( \alpha \right) }\left[ \tfrac {\left( x_{0}-x\right) ^{\alpha }}{\alpha }+\tfrac {1}{\delta _{2}}\int _{x}^{x_{0}}\left( x_{0}-z\right) ^{2-1}\left( z-x\right) ^{\alpha -1}dz\right]\nonumber \\ & =\tfrac {\omega _{1}\left( D_{x_{0}-}^{\alpha }f,\delta _{2}\right) _{\left[ a,x_{0}\right] }}{\Gamma \left( \alpha \right) }\left[ \tfrac {\left( x_{0}-x\right) ^{\alpha }}{\alpha }+\tfrac {1}{\delta _{2}}\tfrac {\Gamma \left( \alpha \right) \Gamma \left( 2\right) }{\Gamma \left( \alpha +2\right) }\left( x_{0}-x\right) ^{\alpha +1}\right]\nonumber \\ & =\tfrac {\omega _{1}\left( D_{x_{0}-}^{\alpha }f,\delta _{2}\right) _{\left[ a,x_{0}\right] }}{\Gamma \left( \alpha \right) }\left[ \tfrac {\left( x_{0}-x\right) ^{\alpha }}{\alpha }+\tfrac {1}{\delta _{2}}\tfrac {\left( x_{0}-x\right) ^{\alpha +1}}{\left( \alpha +1\right) \alpha }\right] \label{27}\\ & =\tfrac {\omega _{1}\left( D_{x_{0}-}^{\alpha }f,\delta _{2}\right) _{\left[ a,x_{0}\right] }}{\Gamma \left( \alpha +1\right) }\left[ \left( x_{0}-x\right) ^{\alpha }+\tfrac {\left( x_{0}-x\right) ^{\alpha +1}}{\left( \alpha +1\right) \delta _{2}}\right] \nonumber . \end{align}
</div>
<p> We have proved that </p>
<div class="equation" id="28">
<p>
  <div class="equation_content">
    \begin{equation}  \left\vert f\left( x\right) -f\left( x_{0}\right) \right\vert \leq \tfrac {\omega _{1}\left( D_{x_{0}-}^{\alpha }f,\delta _{2}\right) _{\left[ a,x_{0}\right] }}{\Gamma \left( \alpha +1\right) }\left[ \left( x_{0}-x\right) ^{\alpha }+\tfrac {\left( x_{0}-x\right) ^{\alpha +1}}{\left( \alpha +1\right) \delta _{2}}\right] ,\label{28}\end{equation}
  </div>
  <span class="equation_label">2.28</span>
</p>
</div>
<p> \(\delta _{2}{\gt}0\), and (\(a\leq x\leq x_{0}\))\(.\) Choosing \(\delta =\delta _{1}=\delta _{2}{\gt}0\), by (<a href="#25">2.25</a>) and (<a href="#28">2.28</a>), we get (<a href="#16">2.16</a>). <div class="proof_wrapper" id="a0000000007">
  <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>We need </p>
<p><div class="definition_thmwrapper " id="d11">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">2.11</span>
  </div>
  <div class="definition_thmcontent">
  <p>Here \(C_{+}\left( \left[ a,b\right] \right) :=\left\{  f:\left[ a,b\right] \rightarrow \mathbb {R}_{+},\  f\text{ continuous}\right\}  .\) Let \(L_{N}:C_{+}\left( \left[ a,b\right] \right) \rightarrow C_{+}\left( \left[ a,b\right] \right) \), operators, \(\forall \) \(N\in \mathbb {N}\), such that </p>
<p>(i) </p>
<div class="equation" id="29">
<p>
  <div class="equation_content">
    \begin{equation}  L_{N}\left( \alpha f\right) =\alpha L_{N}\left( f\right) ,\   \forall \alpha \geq 0,\forall f\in C_{+}\left( \left[ a,b\right] \right) ,\label{29}\end{equation}
  </div>
  <span class="equation_label">2.29</span>
</p>
</div>
<p>(ii) if \(f,g\in C_{+}\left( \left[ a,b\right] \right) :f\leq g,\) then </p>
<div class="equation" id="30">
<p>
  <div class="equation_content">
    \begin{equation}  L_{N}\left( f\right) \leq L_{N}\left( g\right) \text{, \  }\forall N\in \mathbb {N},\label{30}\end{equation}
  </div>
  <span class="equation_label">2.30</span>
</p>
</div>
<p>(iii) </p>
<div class="equation" id="31">
<p>
  <div class="equation_content">
    \begin{equation}  L_{N}\left( f+g\right) \leq L_{N}\left( f\right) +L_{N}\left( g\right) ,\text{ \  }\forall \text{ }f,g\in C_{+}\left( \left[ a,b\right] \right) .\label{31}\end{equation}
  </div>
  <span class="equation_label">2.31</span>
</p>
</div>
<p>We call \(\left\{  L_{N}\right\}  _{N\in \mathbb {N}}\) positive sublinear operators. </p>

  </div>
</div> </p>
<p>We need a Hölder’s type inequality, see next: </p>
<p><div class="theorem_thmwrapper " id="t12">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2.12</span>
  </div>
  <div class="theorem_thmcontent">
  <p>(see <span class="cite">
	[
	<a href="#3" >3</a>
	]
</span>) Let \(L:C_{+}\left( \left[ a,b\right] \right) \rightarrow C_{+}\left( \left[ a,b\right] \right) \), be a positive sublinear operator and \(f,g\in C_{+}\left( \left[ a,b\right] \right) \), furthermore let \(p,q{\gt}1:\frac{1}{p}+\frac{1}{q}=1\). Assume that \(L\left( \left( f\left( \cdot \right) \right) ^{p}\right) \left( s_{\ast }\right) ,\) \(L\left( \left( g\left( \cdot \right) \right) ^{q}\right) \left( s_{\ast }\right) {\gt}0\) for some \(s_{\ast }\in \left[ a,b\right] \). Then </p>
<div class="equation" id="32.">
<p>
  <div class="equation_content">
    \begin{equation}  L\left( f\left( \cdot \right) g\left( \cdot \right) \right) \left( s_{\ast }\right) \leq \left( L\left( \left( f\left( \cdot \right) \right) ^{p}\right) \left( s_{\ast }\right) \right) ^{\frac{1}{p}}\left( L\left( \left( g\left( \cdot \right) \right) ^{q}\right) \left( s_{\ast }\right) \right) ^{\frac{1}{q}}.\label{32.}\end{equation}
  </div>
  <span class="equation_label">2.32</span>
</p>
</div>

  </div>
</div> </p>
<p>We make </p>
<p><div class="remark_thmwrapper " id="r13">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">2.13</span>
  </div>
  <div class="remark_thmcontent">
  <p>By <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	17
	]
</span>,we get: let \(f,g\in C_{+}\left( \left[ a,b\right] \right) \), then </p>
<div class="equation" id="33">
<p>
  <div class="equation_content">
    \begin{equation}  \left| L_{N}\left( f\right) \left( x\right) -L_{N}\left( g\right) \left( x\right) \right| \leq L_{N}\left( \left| f-g\right| \right) \left( x\right) ,\text{ \  }\forall \text{ }x\in \left[ a,b\right] .\label{33}\end{equation}
  </div>
  <span class="equation_label">2.33</span>
</p>
</div>
<p> Furthermore, we also have that </p>
<div class="equation" id="34.">
<p>
  <div class="equation_content">
    \begin{equation}  \left| L_{N}\left( f\right) \left( x\right) -f\left( x\right) \right| \leq L_{N}\left( \left| f\left( \cdot \right) -f\left( x\right) \right| \right) \left( x\right) +\left| f\left( x\right) \right| \left| L_{N}\left( e_{0}\right) \left( x\right) -1\right| ,\label{34.}\end{equation}
  </div>
  <span class="equation_label">2.34</span>
</p>
</div>
<p> \(\forall \) \(x\in \left[ a,b\right] \); \(e_{0}\left( t\right) =1.\) </p>
<p>From now on we assume that \(L_{N}\left( 1\right) =1\). Hence it holds </p>
<div class="equation" id="35">
<p>
  <div class="equation_content">
    \begin{equation}  \left| L_{N}\left( f\right) \left( x\right) -f\left( x\right) \right| \leq L_{N}\left( \left| f\left( \cdot \right) -f\left( x\right) \right| \right) \left( x\right) ,\text{ \  }\forall \text{ }x\in \left[ a,b\right] .\label{35}\end{equation}
  </div>
  <span class="equation_label">2.35</span>
</p>
</div>
<p>Using Theorem <a href="#t10">2.10</a> and (<a href="#16">2.16</a>) with (<a href="#35">2.35</a>) we get:</p>
<div class="displaymath" id="a0000000008">
  \begin{align} &  \left| L_{N}\left( f\right) \left( x_{0}\right) -f\left( x_{0}\right) \right| \leq \nonumber \\ &  \leq \tfrac {\omega _{1}\left( D_{x_{0}}^{\alpha }f,\delta \right) }{\Gamma \left( \alpha +1\right) }\label{36}\left[ L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha }\right) \left( x_{0}\right) +\tfrac {L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha +1}\right) \left( x_{0}\right) }{\left( \alpha +1\right) \delta }\right] ,\text{ \  }\delta {\gt}0. \end{align}
</div>
<p> <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>We have proved </p>
<p><div class="theorem_thmwrapper " id="t14">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2.14</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(\alpha {\gt}0\), \(\alpha \notin \mathbb {N}\), \(m=\left\lceil \alpha \right\rceil \), \(x_{0}\in \left[ a,b\right] \subset \mathbb {R}\), \(f\in AC^{m}\left( \left[ a,b\right] ,\mathbb {R}_{+}\right) \), and \(f^{\left( m\right) }\in L_{\infty }\left( \left[ a,b\right] \right) \). Furthermore we assume that <br />\(f^{\left( k\right) }\left( x_{0}\right) =0\), \(k=1,...,m-1\). Let \(L_{N}:C_{+}\left( \left[ a,b\right] \right) \rightarrow C_{+}\left( \left[ a,b\right] \right) \), \(\forall \) \(N\in \mathbb {N},\) be positive sublinear operators, such that \(L_{N}\left( 1\right) =1\), \(\forall \) \(N\in \mathbb {N}\). Then </p>
<div class="displaymath" id="37">
  \begin{align}  \left| L_{N}\left( f\right) \left( x_{0}\right) -f\left( x_{0}\right) \right| & \leq \tfrac {\omega _{1}\left( D_{x_{0}}^{\alpha }f,\delta \right) }{\Gamma \left( \alpha +1\right) } \cdot \left[ L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha }\right) \left( x_{0}\right) +\tfrac {L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha +1}\right) \left( x_{0}\right) }{\left( \alpha +1\right) \delta }\right] \label{37} \end{align}
</div>
<p> \(\delta {\gt}0\), \(\forall \) \(N\in \mathbb {N}\). </p>
<p>In particular <a href="#37" class="eqref">2.37</a> is true for \(\alpha {\gt}1\), \(\alpha \notin \mathbb {N}\). </p>

  </div>
</div> </p>
<p><div class="corollary_thmwrapper " id="c15">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">2.15</span>
  </div>
  <div class="corollary_thmcontent">
  <p>Let \(0{\lt}\alpha {\lt}1\), \(x_{0}\in \left[ a,b\right] \subset \mathbb {R}\), \(f\in AC\left( \left[ a,b\right] ,\mathbb {R}_{+}\right) \), and \(f^{\prime }\in L_{\infty }\left( \left[ a,b\right] \right) \). Let \(L_{N}:C_{+}\left( \left[ a,b\right] \right) \rightarrow C_{+}\left( \left[ a,b\right] \right) \), \(\forall \) \(N\in \mathbb {N},\) be positive sublinear operators, such that \(L_{N}\left( 1\right) =1\), \(\forall \) \(N\in \mathbb {N}\). Then <a href="#37" class="eqref">2.37</a> is valid. </p>

  </div>
</div> </p>
<p>We give </p>
<p><div class="theorem_thmwrapper " id="t16">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2.16</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(0{\lt}\alpha {\lt}1\), \(x_{0}\in \left[ a,b\right] \subset \mathbb {R}\), \(f\in AC\left( \left[ a,b\right] ,\mathbb {R}_{+}\right) \), and \(f^{\prime }\in L_{\infty }\left( \left[ a,b\right] \right) \). Let \(L_{N}\) from \(C_{+}\left( \left[ a,b\right] \right) \) into itself be positive sublinear operators, such that \(L_{N}\left( 1\right) =1\), \(\forall \) \(N\in \mathbb {N}\). Assume that \(L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha +1}\right) \left( x_{0}\right) {\gt}0\), \(\forall \) \(N\in \mathbb {N}\). Then </p>
<div class="displaymath" id="a0000000009">
  \begin{align} & \left| L_{N}\left( f\right) \left( x_{0}\right) -f\left( x_{0}\right) \right| \leq \nonumber \\ & \leq \tfrac {\left( \alpha +2\right) \omega _{1}\left( D_{x_{0}}^{\alpha }f,\left( L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha +1}\right) \left( x_{0}\right) \right) ^{\frac{1}{\alpha +1}}\right) }{\Gamma \left( \alpha +2\right) }\left( L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha +1}\right) \left( x_{0}\right) \right) ^{\frac{\alpha }{\alpha +1}}.\label{38}\end{align}
</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> By Theorem <a href="#t12">2.12</a>, see (<a href="#32.">2.32</a>), we get </p>
<div class="equation" id="39">
<p>
  <div class="equation_content">
    \begin{equation}  L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha }\right) \left( x_{0}\right) \leq \left( L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha +1}\right) \left( x_{0}\right) \right) ^{\frac{\alpha }{\alpha +1}}.\label{39}\end{equation}
  </div>
  <span class="equation_label">2.39</span>
</p>
</div>
<p> Choose </p>
<div class="equation" id="40">
<p>
  <div class="equation_content">
    \begin{equation}  \delta :=\left( L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha +1}\right) \left( x_{0}\right) \right) ^{\frac{1}{\alpha +1}}>0,\label{40}\end{equation}
  </div>
  <span class="equation_label">2.40</span>
</p>
</div>
<p> i.e. \(\delta ^{\alpha +1}=L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha +1}\right) \left( x_{0}\right) .\)<br />By (<a href="#37">2.37</a>) we obtain </p>
<div class="displaymath" id="a0000000011">
  \begin{align} & \left| L_{N}\left( f\right) \left( x_{0}\right) -f\left( x_{0}\right) \right| \leq \nonumber \\ & \leq \tfrac {1}{\Gamma \left( \alpha +1\right) }\omega _{1}\left( D_{x_{0}}^{\alpha }f,\left( L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha +1}\right) \left( x_{0}\right) \right) ^{\frac{1}{\alpha +1}}\right) \cdot \nonumber \\ & \quad \cdot \left[ \left( L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha +1}\right) \left( x_{0}\right) \right) ^{\frac{\alpha }{\alpha +1}}+\tfrac {1}{\left( \alpha +1\right) }\left( L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha +1}\right) \left( x_{0}\right) \right) ^{\frac{\alpha }{\alpha +1}}\right]= \nonumber \end{align}
</div>
<div class="displaymath" id="a0000000012">
  \begin{align} & =\tfrac {\omega _{1}\left( D_{x_{0}}^{\alpha }f,\left( L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha +1}\right) \left( x_{0}\right) \right) ^{\frac{1}{\alpha +1}}\right) }{\Gamma \left( \alpha +1\right) }\cdot \nonumber \\ \label{41} & \quad \cdot \left( L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha +1}\right) \left( x_{0}\right) \right) ^{\frac{\alpha }{\alpha +1}}\left[ 1+\tfrac {1}{\alpha +1}\right] \\ & =\tfrac {\omega _{1}\left( D_{x_{0}}^{\alpha }f,\left( L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha +1}\right) \left( x_{0}\right) \right) ^{\frac{1}{\alpha +1}}\right) }{\Gamma \left( \alpha +1\right) }\left( L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha +1}\right) \left( x_{0}\right) \right) ^{\frac{\alpha }{\alpha +1}}\left( \tfrac {\alpha +2}{\alpha +1}\right)\nonumber \\ & =\tfrac {\left( \alpha +2\right) \omega _{1}\left( D_{x_{0}}^{\alpha }f,\left( L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha +1}\right) \left( x_{0}\right) \right) ^{\frac{1}{\alpha +1}}\right) }{\Gamma \left( \alpha +2\right) }\left( L_{N}\left( \left| \cdot -x_{0}\right| ^{\alpha +1}\right) \left( x_{0}\right) \right) ^{\frac{\alpha }{\alpha +1}},\nonumber \end{align}
</div>
<p> proving (<a href="#38">2.38</a>). <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>
<h1 id="a0000000014">3 Applications</h1>
<p>I) Case \(0{\lt}\alpha {\lt}1.\) </p>
<p>Here we apply Theorem <a href="#t16">2.16</a> to well known Max-product operators. </p>
<p>We make </p>
<p><div class="remark_thmwrapper " id="r17">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">3.1</span>
  </div>
  <div class="remark_thmcontent">
  <p>The Max-product Bernstein operators \(B_{N}^{\left( M\right) }\left( f\right) \left( x\right) \) are defined by (<a href="#5">1.5</a>), see also <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	10
	]
</span>; here \(f:\left[ 0,1\right] \rightarrow \mathbb {R}_{+}\) is a continuous function. </p>
<p>We have \(B_{N}^{\left( M\right) }\left( 1\right) =1\), and </p>
<div class="equation" id="42">
<p>
  <div class="equation_content">
    \begin{equation}  B_{N}^{\left( M\right) }\left( \left| \cdot -x\right| \right) \left( x\right) \leq \tfrac {6}{\sqrt{N+1}}\text{, \  }\forall \text{ }x\in \left[ 0,1\right] \text{, }\forall \text{ }N\in \mathbb {N}\text{,}\label{42}\end{equation}
  </div>
  <span class="equation_label">3.42</span>
</p>
</div>
<p> see <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	31
	]
</span>. </p>
<p>\(B_{N}^{\left( M\right) }\) are positive sublinear operators and thus they possess the monotonicity property, also since \(\left| \cdot -x\right| \leq 1\), then \(\left| \cdot -x\right| ^{\beta }\leq 1\), \(\forall \) \(x\in \left[ 0,1\right] \), \(\forall \) \(\beta {\gt}0.\) </p>
<p>Therefore it holds </p>
<div class="equation" id="43.">
<p>
  <div class="equation_content">
    \begin{equation}  B_{N}^{\left( M\right) }\left( \left| \cdot -x\right| ^{1+\beta }\right) \left( x\right) \leq \tfrac {6}{\sqrt{N+1}}\text{, \  }\forall \text{ }x\in \left[ 0,1\right] \text{, }\forall \text{ }N\in \mathbb {N}\text{, }\forall \text{ }\beta >0.\label{43.}\end{equation}
  </div>
  <span class="equation_label">3.43</span>
</p>
</div>
<p> Furthermore, clearly it holds that </p>
<div class="equation" id="44">
<p>
  <div class="equation_content">
    \begin{equation}  B_{N}^{\left( M\right) }\left( \left| \cdot -x\right| ^{1+\beta }\right) \left( x\right) >0\text{, }\forall \text{ }N\in \mathbb {N}\text{, }\forall \text{ }\beta \geq 0\text{ and any }x\in \left( 0,1\right) .\label{44}\end{equation}
  </div>
  <span class="equation_label">3.44</span>
</p>
</div>
<p> The operator \(B_{N}^{\left( M\right) }\) maps \(C_{+}\left( \left[ 0,1\right] \right) \) into itself. <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>We present </p>
<p><div class="theorem_thmwrapper " id="t18">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.2</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(0{\lt}\alpha {\lt}1\), any \(x\in \left( 0,1\right) \), \(f\in AC\left( \left[ 0,1\right] ,\mathbb {R}_{+}\right) \), and \(f^{\prime }\in L_{\infty }\left( \left[ 0,1\right] \right) \). Then </p>
<div class="equation" id="45">
<p>
  <div class="equation_content">
    \begin{equation}  \left| B_{N}^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq \tfrac {\left( \alpha +2\right) \omega _{1}\left( D_{x}^{\alpha }f,\left( \frac{6}{\sqrt{N+1}}\right) ^{\frac{1}{\alpha +1}}\right) }{\Gamma \left( \alpha +2\right) }\left( \tfrac {6}{\sqrt{N+1}}\right) ^{\frac{\alpha }{\alpha +1}},\label{45}\end{equation}
  </div>
  <span class="equation_label">3.45</span>
</p>
</div>
<p>  \(\forall \) \(N\in \mathbb {N}.\) </p>
<p>As \(N\rightarrow +\infty \), we get \(B_{N}^{\left( M\right) }\left( f\right) \left( x\right) \rightarrow f\left( x\right) \), any \(x\in \left( 0,1\right) . \) </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000015">
  <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 Theorem <a href="#t16">2.16</a> <div class="proof_wrapper" id="a0000000016">
  <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>We continue with </p>
<p><div class="remark_thmwrapper " id="r19">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">3.3</span>
  </div>
  <div class="remark_thmcontent">
  <p>The truncated Favard-Szász-Mirakjan operators are given by </p>
<div class="equation" id="46">
<p>
  <div class="equation_content">
    \begin{equation}  T_{N}^{\left( M\right) }\left( f\right) \left( x\right) =\tfrac {\bigvee \limits _{k=0}^{N}s_{N,k}\left( x\right) f\big( \tfrac {k}{N}\big) }{\bigvee \limits _{k=0}^{N}s_{N,k}\left( x\right) },\text{ \  }x\in \left[ 0,1\right] ,\text{ }N\in \mathbb {N}\text{, }f\in C_{+}\left( \left[ 0,1\right] \right) ,\label{46}\end{equation}
  </div>
  <span class="equation_label">3.46</span>
</p>
</div>
<p> \(s_{N,k}\left( x\right) =\frac{\left( Nx\right) ^{k}}{k!}\), see also <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	11
	]
</span>. </p>
<p>By <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	178-179
	]
</span>,we get that </p>
<div class="equation" id="47">
<p>
  <div class="equation_content">
    \begin{equation}  T_{N}^{\left( M\right) }\left( \left| \cdot -x\right| \right) \left( x\right) \leq \tfrac {3}{\sqrt{N}},\text{ \  }\forall \text{ }x\in \left[ 0,1\right] ,\text{ }\forall \text{ }N\in \mathbb {N}.\label{47}\end{equation}
  </div>
  <span class="equation_label">3.47</span>
</p>
</div>
<p> Clearly it holds </p>
<div class="equation" id="48">
<p>
  <div class="equation_content">
    \begin{equation}  T_{N}^{\left( M\right) }\left( \left| \cdot -x\right| ^{1+\beta }\right) \left( x\right) \leq \tfrac {3}{\sqrt{N}},\text{ \  }\forall \text{ }x\in \left[ 0,1\right] ,\text{ }\forall \text{ }N\in \mathbb {N}\text{, }\forall \text{ }\beta >0.\label{48}\end{equation}
  </div>
  <span class="equation_label">3.48</span>
</p>
</div>
<p> The operators \(T_{N}^{\left( M\right) }\) are positive sublinear operators mapping \(C_{+}\left( \left[ 0,1\right] \right) \) into itself, with \(T_{N}^{\left( M\right) }\left( 1\right) =1\). </p>
<p>Furthermore it holds </p>
<div class="equation" id="49">
<p>
  <div class="equation_content">
    \begin{equation}  T_{N}^{\left( M\right) }\left( \left| \cdot -x\right| ^{\lambda }\right) \left( x\right) =\tfrac {\bigvee \limits _{k=0}^{N}\tfrac {\left( Nx\right) ^{k}}{k!}\left| \tfrac {k}{N}-x\right| ^{\lambda }}{\bigvee \limits _{k=0}^{N}\tfrac {\left( Nx\right) ^{k}}{k!}}>0,\text{ \  }\forall \text{ }x\in (0,1],\text{ }\forall \text{ }\lambda \geq 1\text{, }\forall \text{ }N\in \mathbb {N}.\label{49}\end{equation}
  </div>
  <span class="equation_label">3.49</span>
</p>
</div>
<p> <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>We give </p>
<p><div class="theorem_thmwrapper " id="t20">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.4</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(0{\lt}\alpha {\lt}1\), any \(x\in (0,1]\), \(f\in AC\left( \left[ 0,1\right] ,\mathbb {R}_{+}\right) \), and \(f^{\prime }\in L_{\infty }\left( \left[ 0,1\right] \right) \). Then</p>
<div class="equation" id="50">
<p>
  <div class="equation_content">
    \begin{equation}  \left| T_{N}^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq \tfrac {\left( \alpha +2\right) \omega _{1}\Big( D_{x}^{\alpha }f,\Big( \frac{3}{\sqrt{N}}\big) ^{\frac{1}{\alpha +1}}\Big) }{\Gamma \left( \alpha +2\right) }\left( \tfrac {3}{\sqrt{N}}\right) ^{\frac{\alpha }{\alpha +1}},\text{ \  }\forall \text{ }N\in \mathbb {N}.\label{50}\end{equation}
  </div>
  <span class="equation_label">3.50</span>
</p>
</div>
<p> As \(N\rightarrow +\infty \), we get \(T_{N}^{\left( M\right) }\left( f\right) \left( x\right) \rightarrow f\left( x\right) \), for any \(x\in (0,1].\) </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000017">
  <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 Theorem <a href="#t16">2.16</a>. <div class="proof_wrapper" id="a0000000018">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>We make </p>
<p><div class="remark_thmwrapper " id="r21">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">3.5</span>
  </div>
  <div class="remark_thmcontent">
  <p>Next we study the truncated Max-product Baskakov operators (see <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	11
	]
</span>) </p>
<div class="equation" id="51">
<p>
  <div class="equation_content">
    \begin{equation}  U_{N}^{\left( M\right) }\left( f\right) \left( x\right) =\tfrac {\bigvee \limits _{k=0}^{N}b_{N,k}\left( x\right) f\left( \frac{k}{N}\right) }{\bigvee \limits _{k=0}^{N}b_{N,k}\left( x\right) },\text{ \  }x\in \left[ 0,1\right] ,\text{ }f\in C_{+}\left( \left[ 0,1\right] \right) ,\text{ }N\in \mathbb {N}\text{,}\label{51}\end{equation}
  </div>
  <span class="equation_label">3.51</span>
</p>
</div>
<p> where </p>
<div class="equation" id="52">
<p>
  <div class="equation_content">
    \begin{equation}  b_{N,k}\left( x\right) = \begin{array}[c]{c}\binom {N+k-1}{k} \end{array} \tfrac {x^{k}}{\left( 1+x\right) ^{N+k}}.\label{52}\end{equation}
  </div>
  <span class="equation_label">3.52</span>
</p>
</div>
<p> From <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	pp.
	
	217-218
	]
</span>, we get (\(x\in \left[ 0,1\right] \))</p>
<div class="equation" id="53">
<p>
  <div class="equation_content">
    \begin{equation}  \left( U_{N}^{\left( M\right) }\left( \left| \cdot -x\right| \right) \right) \left( x\right) \leq \tfrac {2\sqrt{3}\left( \sqrt{2}+2\right) }{\sqrt{N+1}}\text{, \  }N\geq 2\text{, }N\in \mathbb {N}.\label{53}\end{equation}
  </div>
  <span class="equation_label">3.53</span>
</p>
</div>
<p> Let \(\lambda \geq 1\), clearly then it holds </p>
<div class="equation" id="54">
<p>
  <div class="equation_content">
    \begin{equation}  \left( U_{N}^{\left( M\right) }\left( \left| \cdot -x\right| ^{\lambda }\right) \right) \left( x\right) \leq \tfrac {2\sqrt{3}\left( \sqrt{2}+2\right) }{\sqrt{N+1}},\text{ \  }\forall \text{ }N\geq 2\text{, }N\in \mathbb {N}.\label{54}\end{equation}
  </div>
  <span class="equation_label">3.54</span>
</p>
</div>
<p> Also it holds \(U_{N}^{\left( M\right) }\left( 1\right) =1\), and \(U_{N}^{\left( M\right) }\) are positive sublinear operators from \(C_{+}\left( \left[ 0,1\right] \right) \) into itself. Furthermore it holds </p>
<div class="equation" id="55">
<p>
  <div class="equation_content">
    \begin{equation}  U_{N}^{\left( M\right) }\left( \left| \cdot -x\right| ^{\lambda }\right) \left( x\right) >0,\text{ \  }\forall \text{ }x\in (0,1],\text{ }\forall \text{ }\lambda \geq 1\text{, }\forall \text{ }N\in \mathbb {N}.\label{55}\end{equation}
  </div>
  <span class="equation_label">3.55</span>
</p>
</div>
<p> <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>We give </p>
<p><div class="theorem_thmwrapper " id="t22">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.6</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(0{\lt}\alpha {\lt}1\), any \(x\in (0,1]\), \(f\in AC\left( \left[ 0,1\right] ,\mathbb {R}_{+}\right) \), and \(f^{\prime }\in L_{\infty }\left( \left[ 0,1\right] \right) \). Then </p>
<div class="displaymath" id="56">
  \begin{align} & \left| U_{N}^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq \tfrac {\left( \alpha +2\right) \omega _{1}\left( D_{x}^{\alpha }f,\left( \frac{2\sqrt{3}\left( \sqrt{2}+2\right) }{\sqrt{N+1}}\right) ^{\frac{1}{\alpha +1}}\right) }{\Gamma \left( \alpha +2\right) }\cdot \left( \tfrac {2\sqrt{3}\left( \sqrt{2}+2\right) }{\sqrt{N+1}}\right) ^{\frac{\alpha }{\alpha +1}},\label{56} \end{align}
</div>
<p> \(\forall \text{ }N\geq 2\text{, }N\in \mathbb {N}\). </p>
<p>As \(N\rightarrow +\infty \), we get \(U_{N}^{\left( M\right) }\left( f\right) \left( x\right) \rightarrow f\left( x\right) \), for any \(x\in (0,1].\) </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000019">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> By Theorem <a href="#t16">2.16</a>. <div class="proof_wrapper" id="a0000000020">
  <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>We continue with </p>
<p><div class="remark_thmwrapper " id="r23">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">3.7</span>
  </div>
  <div class="remark_thmcontent">
  <p>Here we study the Max-product Meyer-Köning and Zeller operators (see <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	11
	]
</span>) defined by </p>
<div class="equation" id="57">
<p>
  <div class="equation_content">
    \begin{equation}  Z_{N}^{\left( M\right) }\left( f\right) \left( x\right) =\frac{\bigvee \limits _{k=0}^{\infty }s_{N,k}\left( x\right) f\big( \frac{k}{N+k}\big) }{\bigvee \limits _{k=0}^{\infty }s_{N,k}\left( x\right) }, \quad \forall N\in \mathbb {N}\text{, }f\in C_{+}\left( \left[ 0,1\right] \right) ,\label{57}\end{equation}
  </div>
  <span class="equation_label">3.57</span>
</p>
</div>
<p> \(s_{N,k}\left( x\right) = \begin{array}[c]{c}\binom {N+k} {k} \end{array} x^{k}\), \(x\in \left[ 0,1\right] \). </p>
<p>By <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	253
	]
</span>, we get that </p>
<div class="equation" id="58">
<p>
  <div class="equation_content">
    \begin{equation}  Z_{N}^{\left( M\right) }\left( \left| \cdot -x\right| \right) \left( x\right) \leq \tfrac {8\left( 1+\sqrt{5}\right) }{3}\tfrac {\sqrt{x}\left( 1-x\right) }{\sqrt{N}}\text{, \  }\forall \text{ }x\in \left[ 0,1\right] ,\text{ }\forall \text{ }N\geq 4\text{, }N\in \mathbb {N}.\label{58}\end{equation}
  </div>
  <span class="equation_label">3.58</span>
</p>
</div>
<p> As before we get that (for \(\lambda \geq 1\))</p>
<div class="equation" id="59">
<p>
  <div class="equation_content">
    \begin{equation}  Z_{N}^{\left( M\right) }\left( \left| \cdot -x\right| ^{\lambda }\right) \left( x\right) \leq \tfrac {8\left( 1+\sqrt{5}\right) }{3}\tfrac {\sqrt{x}\left( 1-x\right) }{\sqrt{N}}:=\rho \left( x\right) ,\text{ \  }\label{59}\end{equation}
  </div>
  <span class="equation_label">3.59</span>
</p>
</div>
<p> \(\forall \) \(x\in \left[ 0,1\right] \), \(N\geq 4\), \(N\in \mathbb {N}.\) </p>
<p>Also it holds \(Z_{N}^{\left( M\right) }\left( 1\right) =1\), and \(Z_{N}^{\left( M\right) }\) are positive sublinear operators from \(C_{+}\left( \left[ 0,1\right] \right) \) into itself. Also it holds </p>
<div class="equation" id="60">
<p>
  <div class="equation_content">
    \begin{equation}  Z_{N}^{\left( M\right) }\left( \left| \cdot -x\right| ^{\lambda }\right) \left( x\right) >0,\quad \forall \text{ }x\in (0,1),\text{ }\forall \text{ }\lambda \geq 1\text{, }\forall \text{ }N\in \mathbb {N}.\label{60}\end{equation}
  </div>
  <span class="equation_label">3.60</span>
</p>
</div>

  </div>
</div> </p>
<p>We give </p>
<p><div class="theorem_thmwrapper " id="t24">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.8</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(0{\lt}\alpha {\lt}1\), any \(x\in (0,1)\), \(f\in AC\left( \left[ 0,1\right] ,\mathbb {R}_{+}\right) \), and \(f^{\prime }\in L_{\infty }\left( \left[ 0,1\right] \right) \). Then </p>
<div class="equation" id="61">
<p>
  <div class="equation_content">
    \begin{equation}  \left| Z_{N}^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq \tfrac {\left( \alpha +2\right) \omega _{1}\left( D_{x}^{\alpha }f,\left( \rho \left( x\right) \right) ^{\frac{1}{\alpha +1}}\right) }{\Gamma \left( \alpha +2\right) }\left( \rho \left( x\right) \right) ^{\frac{\alpha }{\alpha +1}}\label{61}\end{equation}
  </div>
  <span class="equation_label">3.61</span>
</p>
</div>
<p> \(\forall \) \(N\geq 4\), \(N\in \mathbb {N}.\) </p>
<p>As \(N\rightarrow +\infty \), we get \(Z_{N}^{\left( M\right) }\left( f\right) \left( x\right) \rightarrow f\left( x\right) \), for any \(x\in (0,1).\) </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> By Theorem <a href="#t16">2.16</a>. <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>We continue with </p>
<p><div class="remark_thmwrapper " id="r25">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">3.9</span>
  </div>
  <div class="remark_thmcontent">
  <p>Here we deal with the Max-product truncated sampling operators (see <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	13
	]
</span>) defined by </p>
<div class="equation" id="62">
<p>
  <div class="equation_content">
    \begin{equation}  W_{N}^{\left( M\right) }\left( f\right) \left( x\right) =\frac{\bigvee \limits _{k=0}^{N}\frac{\sin \left( Nx-k\pi \right) }{Nx-k\pi }f\left( \frac{k\pi }{N}\right) }{\bigvee \limits _{k=0}^{N}\frac{\sin \left( Nx-k\pi \right) }{Nx-k\pi }},\text{ }\label{62}\end{equation}
  </div>
  <span class="equation_label">3.62</span>
</p>
</div>
<p> and </p>
<div class="equation" id="63">
<p>
  <div class="equation_content">
    \begin{equation}  K_{N}^{\left( M\right) }\left( f\right) \left( x\right) =\frac{\bigvee \limits _{k=0}^{N}\frac{\sin ^{2}\left( Nx-k\pi \right) }{\left( Nx-k\pi \right) ^{2}}f\left( \frac{k\pi }{N}\right) }{\bigvee \limits _{k=0}^{N}\frac{\sin ^{2}\left( Nx-k\pi \right) }{\left( Nx-k\pi \right) ^{2}}},\text{ }\label{63}\end{equation}
  </div>
  <span class="equation_label">3.63</span>
</p>
</div>
<p> \(\forall \) \(x\in \left[ 0,\pi \right] \), \(f:\left[ 0,\pi \right] \rightarrow \mathbb {R}_{+}\) a continuous function. </p>
<p>Following <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	343
	]
</span>, and making the convention \(\frac{\sin \left( 0\right) }{0}=1\) and denoting \(s_{N,k}\left( x\right) =\frac{\sin \left( Nx-k\pi \right) }{Nx-k\pi }\), we get that \(s_{N,k}\left( \frac{k\pi }{N}\right) =1\), and \(s_{N,k}\left( \frac{j\pi }{N}\right) =0\), if \(k\neq j\), furthermore \(W_{N}^{\left( M\right) }\left( f\right) ( \frac{j\pi }{N}) =f( \frac{j\pi }{N}) \), for all \(j\in \left\{  0,...,N\right\}  .\) </p>
<p>Clearly \(W_{N}^{\left( M\right) }\left( f\right) \) is a well-defined function for all \(x\in \left[ 0,\pi \right] \), and it is continuous on \(\left[ 0,\pi \right] \), also \(W_{N}^{\left( M\right) }\left( 1\right) =1.\) </p>
<p>By <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	344
	]
</span>, \(W_{N}^{\left( M\right) }\) are positive sublinear operators. </p>
<p>Call \(I_{N}^{+}\left( x\right) =\left\{  k\in \left\{  0,1,...,N\right\}  ;s_{N,k}\left( x\right) {\gt}0\right\}  \), and set \(x_{N,k}:=\frac{k\pi }{N}\), \(k\in \left\{  0,1,...,N\right\}  \). </p>
<p>We see that </p>
<div class="equation" id="64">
<p>
  <div class="equation_content">
    \begin{equation}  W_{N}^{\left( M\right) }\left( f\right) \left( x\right) =\frac{\bigvee \limits _{k\in I_{N}^{+}\left( x\right) }s_{N,k}\left( x\right) f\left( x_{N,k}\right) }{\bigvee \limits _{k\in I_{N}^{+}\left( x\right) }s_{N,k}\left( x\right) }.\label{64}\end{equation}
  </div>
  <span class="equation_label">3.64</span>
</p>
</div>
<p> By <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	346
	]
</span>,we have</p>
<div class="equation" id="65">
<p>
  <div class="equation_content">
    \begin{equation}  W_{N}^{\left( M\right) }\left( \left| \cdot -x\right| \right) \left( x\right) \leq \tfrac {\pi }{2N},\text{ \  }\forall \text{ }N\in \mathbb {N},\text{ }\forall \text{ }x\in \left[ 0,\pi \right] .\label{65}\end{equation}
  </div>
  <span class="equation_label">3.65</span>
</p>
</div>
<p> Notice also \(\left| x_{N,k}-x\right| \leq \pi \), \(\forall \) \(x\in \left[ 0,\pi \right] .\) </p>
<p>Therefore (\(\lambda \geq 1\)) it holds </p>
<div class="equation" id="66.">
<p>
  <div class="equation_content">
    \begin{equation}  W_{N}^{\left( M\right) }\left( \left| \cdot -x\right| ^{\lambda }\right) \left( x\right) \leq \tfrac {\pi ^{\lambda -1}\pi }{2N}=\tfrac {\pi ^{\lambda }}{2N},\text{ \  }\forall \text{ }x\in \left[ 0,\pi \right] \text{, }\forall \text{ }N\in \mathbb {N}.\label{66.}\end{equation}
  </div>
  <span class="equation_label">3.66</span>
</p>
</div>
<p> If \(x\in \left( \tfrac {j\pi }{N},\tfrac {\left( j+1\right) \pi }{N}\right) \), with \(j\in \left\{  0,1,...,N\right\}  \), we obtain \(nx-j\pi \in \left( 0,\pi \right) \) and thus \(s_{N,j}\left( x\right) =\tfrac {\sin \left( Nx-j\pi \right) }{Nx-j\pi }{\gt}0\), see <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	pp.
	
	343-344
	]
</span>. </p>
<p>Consequently it holds (\(\lambda \geq 1\)) </p>
<div class="equation" id="67">
<p>
  <div class="equation_content">
    \begin{equation}  W_{N}^{\left( M\right) }\left( \left| \cdot -x\right| ^{\lambda }\right) \left( x\right) =\tfrac {\bigvee \limits _{k\in I_{N}^{+}\left( x\right) }s_{N,k}\left( x\right) \left| x_{N,k}-x\right| ^{\lambda }}{\bigvee \limits _{k\in I_{N}^{+}\left( x\right) }s_{N,k}\left( x\right) }>0,\text{ \  }\forall \text{ }x\in \left[ 0,\pi \right] \text{,}\label{67}\end{equation}
  </div>
  <span class="equation_label">3.67</span>
</p>
</div>
<p> such that \(x\neq x_{N,k}\), for any \(k\in \left\{  0,1,...,N\right\}  .\) <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>We give </p>
<p><div class="theorem_thmwrapper " id="t26">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.10</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(0{\lt}\alpha {\lt}1\), any \(x\in \left[ 0,\pi \right] \) be such that \(x\neq \tfrac {k\pi }{N}\), \(k\in \left\{  0,1,...,N\right\}  \), \(\forall \) \(N\in \mathbb {N}\); \(f\in AC\left( \left[ 0,\pi \right] ,\mathbb {R}_{+}\right) \), and \(f^{\prime }\in L_{\infty }\left( \left[ 0,\pi \right] \right) \). Then </p>
<div class="equation" id="68">
<p>
  <div class="equation_content">
    \begin{equation}  \left| W_{N}^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq \tfrac {\left( \alpha +2\right) \omega _{1}\left( D_{x}^{\alpha }f,\left( \frac{\pi ^{\alpha +1}}{2N}\right) ^{\frac{1}{\alpha +1}}\right) }{\Gamma \left( \alpha +2\right) }\left( \tfrac {\pi ^{\alpha +1}}{2N}\right) ^{\frac{\alpha }{\alpha +1}},\text{ \  }\forall \text{ }N\in \mathbb {N}.\label{68}\end{equation}
  </div>
  <span class="equation_label">3.68</span>
</p>
</div>
<p> As \(N\rightarrow +\infty \), we get \(W_{N}^{\left( M\right) }\left( f\right) \left( x\right) \rightarrow f\left( x\right) \). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000023">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> By Theorem <a href="#t16">2.16</a>. <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> </p>
<p>We make </p>
<p><div class="remark_thmwrapper " id="r27">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">3.11</span>
  </div>
  <div class="remark_thmcontent">
  <p>Here we continue with the Max-product truncated sampling operators (see <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	13
	]
</span>) defined by </p>
<div class="equation" id="69">
<p>
  <div class="equation_content">
    \begin{equation}  K_{N}^{\left( M\right) }\left( f\right) \left( x\right) =\frac{\bigvee \limits _{k=0}^{N}\frac{\sin ^{2}\left( Nx-k\pi \right) }{\left( Nx-k\pi \right) ^{2}}f\big( \frac{k\pi }{N}\big) }{\bigvee \limits _{k=0}^{N}\frac{\sin ^{2}\left( Nx-k\pi \right) }{\left( Nx-k\pi \right) ^{2}}},\text{ }\label{69}\end{equation}
  </div>
  <span class="equation_label">3.69</span>
</p>
</div>
<p> \(\forall \) \(x\in \left[ 0,\pi \right] \), \(f:\left[ 0,\pi \right] \rightarrow \mathbb {R}_{+}\) a continuous function. </p>
<p>Following <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	350
	]
</span>, and making the convention \(\tfrac {\sin \left( 0\right) }{0}=1\) and denoting \(s_{N,k}\left( x\right) =\tfrac {\sin ^{2}\left( Nx-k\pi \right) }{\left( Nx-k\pi \right) ^{2}}\), we get that \(s_{N,k}\left( \tfrac {k\pi }{N}\right) =1\), and \(s_{N,k}\left( \tfrac {j\pi }{N}\right) =0\), if \(k\neq j\), furthermore \(K_{N}^{\left( M\right) }\left( f\right) \left( \tfrac {j\pi }{N}\right) =f\left( \tfrac {j\pi }{N}\right) \), for all \(j\in \left\{  0,...,N\right\}  .\) </p>
<p>Since \(s_{N,j}\left( \tfrac {j\pi }{N}\right) =1\) it follows that \(\bigvee _{k=0}^{N}s_{N,k}\left( \tfrac {j\pi }{N}\right) \geq 1{\gt}0,\) for all \(j\in \left\{  0,1,...,N\right\}  \). Hence \(K_{N}^{\left( M\right) }\left( f\right) \) is well-defined function for all \(x\in \left[ 0,\pi \right] \), and it is continuous on \(\left[ 0,\pi \right] \), also \(K_{N}^{\left( M\right) }\left( 1\right) =1.\) By <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	350
	]
</span>, \(K_{N}^{\left( M\right) } \) are positive sublinear operators. </p>
<p>Denote \(x_{N,k}:=\frac{k\pi }{N}\), \(k\in \left\{  0,1,...,N\right\}  \). </p>
<p>By <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	352
	]
</span>, we have</p>
<div class="equation" id="70">
<p>
  <div class="equation_content">
    \begin{equation}  K_{N}^{\left( M\right) }\left( \left| \cdot -x\right| \right) \left( x\right) \leq \tfrac {\pi }{2N},\text{ \  }\forall \text{ }N\in \mathbb {N},\text{ }\forall \text{ }x\in \left[ 0,\pi \right] .\label{70}\end{equation}
  </div>
  <span class="equation_label">3.70</span>
</p>
</div>
<p> Notice also \(\left| x_{N,k}-x\right| \leq \pi \), \(\forall \) \(x\in \left[ 0,\pi \right] .\) </p>
<p>Therefore (\(\lambda \geq 1\)) it holds </p>
<div class="equation" id="71">
<p>
  <div class="equation_content">
    \begin{equation}  K_{N}^{\left( M\right) }\left( \left| \cdot -x\right| ^{\lambda }\right) \left( x\right) \leq \tfrac {\pi ^{\lambda -1}\pi }{2N}=\tfrac {\pi ^{\lambda }}{2N},\text{ \  }\forall \text{ }x\in \left[ 0,\pi \right] \text{, }\forall \text{ }N\in \mathbb {N}.\label{71}\end{equation}
  </div>
  <span class="equation_label">3.71</span>
</p>
</div>
<p> If \(x\in \left( \tfrac {j\pi }{N},\tfrac {\left( j+1\right) \pi }{N}\right) \), with \(j\in \left\{  0,1,...,N\right\}  \), we obtain \(nx-j\pi \in \left( 0,\pi \right) \) and thus \(s_{N,j}\left( x\right) =\tfrac {\sin ^{2}\left( Nx-j\pi \right) }{\left( Nx-j\pi \right) ^{2}}{\gt}0\), see <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	pp.
	
	350
	]
</span>. </p>
<p>Consequently it holds (\(\lambda \geq 1\)) </p>
<div class="equation" id="72">
<p>
  <div class="equation_content">
    \begin{equation}  K_{N}^{\left( M\right) }\left( \left| \cdot -x\right| ^{\lambda }\right) \left( x\right) =\tfrac {\bigvee \limits _{k=0}^{N}s_{N,k}\left( x\right) \left| x_{N,k}-x\right| ^{\lambda }}{\bigvee \limits _{k=0}^{N}s_{N,k}\left( x\right) }>0,\text{ \  }\forall \text{ }x\in \left[ 0,\pi \right] \text{,} \label{72}\end{equation}
  </div>
  <span class="equation_label">3.72</span>
</p>
</div>
<p> such that \(x\neq x_{N,k}\), for any \(k\in \left\{  0,1,...,N\right\}  .\) <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>We give </p>
<p><div class="theorem_thmwrapper " id="t28">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.12</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(0{\lt}\alpha {\lt}1\), \(x\in \left[ 0,\pi \right] \) be such that \(x\neq \frac{k\pi }{N}\), \(k\in \left\{  0,1,...,N\right\}  \), \(\forall \) \(N\in \mathbb {N}\); \(f\in AC\left( \left[ 0,\pi \right] ,\mathbb {R}_{+}\right) \), and \(f^{\prime }\in L_{\infty }\left( \left[ 0,\pi \right] \right) \). Then </p>
<div class="equation" id="73">
<p>
  <div class="equation_content">
    \begin{equation}  \left| K_{N}^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq \tfrac {\left( \alpha +2\right) \omega _{1}\left( D_{x}^{\alpha }f,\left( \tfrac {\pi ^{\alpha +1}}{2N}\right) ^{\frac{1}{\alpha +1}}\right) }{\Gamma \left( \alpha +2\right) }\left( \tfrac {\pi ^{\alpha +1}}{2N}\right) ^{\frac{\alpha }{\alpha +1}},\text{ \  }\forall \text{ }N\in \mathbb {N}.\label{73}\end{equation}
  </div>
  <span class="equation_label">3.73</span>
</p>
</div>
<p> As \(N\rightarrow +\infty \), we get \(K_{N}^{\left( M\right) }\left( f\right) \left( x\right) \rightarrow f\left( x\right) \). </p>

  </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> By Theorem <a href="#t16">2.16</a>. <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> </p>
<p>When \(\alpha =\frac{1}{2}\) we get: </p>
<p><div class="corollary_thmwrapper " id="a0000000027">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">3.13</span>
  </div>
  <div class="corollary_thmcontent">
  <p>Let \(f\in AC\left( \left[ 0,1\right] ,\mathbb {R}_{+}\right) \), \(f^{\prime }\in L_{\infty }\left( \left[ 0,1\right] \right) \). Then </p>
<div class="equation" id="74">
<p>
  <div class="equation_content">
    \begin{equation}  \left| B_{N}^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq \tfrac {10\sqrt[3]{6}\omega _{1}\left( D_{x}^{\frac{1}{2}}f,\frac{\sqrt[3]{36}}{\sqrt[3]{N+1}}\right) }{3\sqrt{\pi }\sqrt[6]{N+1}},\text{ \  }\forall \text{ }x\in \left( 0,1\right) \text{, }\forall \text{ }N\in \mathbb {N}.\label{74}\end{equation}
  </div>
  <span class="equation_label">3.74</span>
</p>
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000028">
  <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 Theorem <a href="#t18">3.2</a>. <div class="proof_wrapper" id="a0000000029">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>Due to lack of space we avoid to give other applications when \(\alpha =\frac{1}{2}\) from the other Max-product operators. </p>
<p>II) Case \(\alpha {\gt}1\), \(\alpha \notin \mathbb {N}\). </p>
<p>Here we apply Theorem <a href="#t14">2.14</a> to well known Max-product operators. </p>
<p>We present </p>
<p><div class="theorem_thmwrapper " id="t30">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.14</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(\alpha {\gt}1,\) \(\alpha \notin \mathbb {N}\) \(m=\left\lceil \alpha \right\rceil \), \(x\in \left[ 0,1\right] \), \(f\in AC^{m}\left( \left[ 0,1\right] ,\mathbb {R}_{+}\right) \), and \(f^{\left( m\right) }\in L_{\infty }\left( \left[ 0,1\right] \right) \). Furthermore we assume that \(f^{\left( k\right) }\left( x\right) =0\), \(k=1,...,m-1\). Then </p>
<div class="displaymath" id="75">
  \begin{align} & \left| B_{N}^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq \tfrac {\omega _{1}\left( D_{x}^{\alpha }f,\big( \frac{6}{\sqrt{N+1}}\big) ^{\frac{1}{\alpha +1}}\right) }{\Gamma \left( \alpha +1\right) }\cdot \label{75}\left[ \tfrac {6}{\sqrt{N+1}}+\tfrac {1}{\left( \alpha +1\right) }\left( \tfrac {6}{\sqrt{N+1}}\right) ^{\frac{\alpha }{\alpha +1}}\right], \\ & \text{\  }\forall \text{ }N\in \mathbb {N}. \nonumber \end{align}
</div>
<p>We get \(\underset {N\rightarrow +\infty }{\lim }B_{N}^{\left( M\right) }\left( f\right) \left( x\right) =f\left( x\right) .\) </p>

  </div>
</div> </p>
<p><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> Applying (<a href="#37">2.37</a>) for \(B_{N}^{\left( M\right) }\) and using (<a href="#43.">3.43</a>), we get </p>
<div class="equation" id="76">
<p>
  <div class="equation_content">
    \begin{equation}  \left| B_{N}^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq \tfrac {\omega _{1}\left( D_{x}^{\alpha }f,\delta \right) }{\Gamma \left( \alpha +1\right) }\left[ \tfrac {6}{\sqrt{N+1}}+\tfrac {\frac{6}{\sqrt{N+1}}}{\left( \alpha +1\right) \delta }\right] .\label{76}\end{equation}
  </div>
  <span class="equation_label">3.76</span>
</p>
</div>
<p> Choose \(\delta =\left( \tfrac {6}{\sqrt{N+1}}\right) ^{\frac{1}{\alpha +1}}\), then \(\delta ^{\alpha +1}=\frac{6}{\sqrt{N+1}}\), and apply it to (<a href="#76">3.76</a>). Clearly we derive (<a href="#75">3.75</a>). <div class="proof_wrapper" id="a0000000031">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>We continue with </p>
<p><div class="theorem_thmwrapper " id="t31">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.15</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Same assumptions as in Theorem <a href="#t30">3.14</a>. Then </p>
<div class="displaymath" id="77">
  \begin{align} & \left| T_{N}^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq \tfrac {\omega _{1}\left( D_{x}^{\alpha }f,\left( \frac{3}{\sqrt{N}}\right) ^{\frac{1}{\alpha +1}}\right) }{\Gamma \left( \alpha +1\right) } \label{77}\left[ \tfrac {3}{\sqrt{N}}+\tfrac {1}{\left( \alpha +1\right) }\left( \tfrac {3}{\sqrt{N}}\right) ^{\frac{\alpha }{\alpha +1}}\right] , \end{align}
</div>
<p> \(\forall \text{ }N\in \mathbb {N}.\) </p>
<p>We get \(\underset {N\rightarrow +\infty }{\lim }T_{N}^{\left( M\right) }\left( f\right) \left( x\right) =f\left( x\right) .\) </p>

  </div>
</div> </p>
<p><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> Use of Theorem <a href="#t14">2.14</a>, similar to the proof of Theorem <a href="#t30">3.14</a>. <div class="proof_wrapper" id="a0000000033">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>We give </p>
<p><div class="theorem_thmwrapper " id="t32">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.16</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Same assumptions as in Theorem <a href="#t30">3.14</a>. Then </p>
<div class="displaymath" id="a0000000034">
  \begin{align} & \left| U_{N}^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq \nonumber \\ & \leq \tfrac {\omega _{1}\left( D_{x}^{\alpha }f,\left( \frac{2\sqrt{3}\left( \sqrt{2}+2\right) }{\sqrt{N+1}}\right) ^{\frac{1}{\alpha +1}}\right) }{\Gamma \left( \alpha +1\right) }\cdot \label{78}\left[ \tfrac {2\sqrt{3}\left( \sqrt{2}+2\right) }{\sqrt{N+1}}+\tfrac {1}{\left( \alpha +1\right) }\left( \tfrac {2\sqrt{3}\left( \sqrt{2}+2\right) }{\sqrt{N+1}}\right) ^{\frac{\alpha }{\alpha +1}}\right], \end{align}
</div>
<p> \(\forall \text{ }N\in \mathbb {N},\text{ }N\geq 2.\) </p>
<p>We get \(\underset {N\rightarrow +\infty }{\lim }U_{N}^{\left( M\right) }\left( f\right) \left( x\right) =f\left( x\right) .\) </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> Use of Theorem <a href="#t14">2.14</a>, similar to the proof of Theorem <a href="#t30">3.14</a>. <div class="proof_wrapper" id="a0000000036">
  <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>We give </p>
<p><div class="theorem_thmwrapper " id="t33">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.17</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Same assumptions as in Theorem <a href="#t30">3.14</a>. Then </p>
<div class="displaymath" id="79">
  \begin{align} & \left| Z_{N}^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq \tfrac {\omega _{1}\left( D_{x}^{\alpha }f,\left( \rho \left( x\right) \right) ^{\frac{1}{\alpha +1}}\right) }{\Gamma \left( \alpha +1\right) }\cdot \label{79}\left[ \rho \left( x\right) +\tfrac {1}{\left( \alpha +1\right) }\left( \rho \left( x\right) \right) ^{\frac{\alpha }{\alpha +1}}\right], \end{align}
</div>
<p> \(\forall \text{ }N\in \mathbb {N}, \text{ }N\geq 4.\) </p>
<p>We get \(\underset {N\rightarrow +\infty }{\lim }Z_{N}^{\left( M\right) }\left( f\right) \left( x\right) =f\left( x\right) ,\) where \(\rho \left( x\right) \) is as in <a href="#59" class="eqref">3.59</a>. </p>

  </div>
</div> </p>
<p><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> Use of Theorem <a href="#t14">2.14</a>, similar to the proof of Theorem <a href="#t30">3.14</a>. <div class="proof_wrapper" id="a0000000038">
  <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>We continue with </p>
<p><div class="theorem_thmwrapper " id="t34">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.18</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(\alpha {\gt}1,\) \(\alpha \notin \mathbb {N}\), \(m=\left\lceil \alpha \right\rceil \), \(x\in \left[ 0,\pi \right] \subset \mathbb {R}\), \(f\in AC^{m}\left( \left[ 0,\pi \right] ,\mathbb {R}_{+}\right) \), and \(f^{\left( m\right) }\in L_{\infty }\left( \left[ 0,\pi \right] \right) \). Furthermore we assume that \(f^{\left( k\right) }\left( x\right)\) \( =0\), \(k=1,...,m-1\). Then </p>
<div class="displaymath" id="80">
  \begin{align} & \left| W_{N}^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq \tfrac {\omega _{1}\Big( D_{x}^{\alpha }f,\big( \frac{\pi ^{\alpha +1}}{2N}\big) ^{\frac{1}{\alpha +1}}\Big) }{\Gamma \left( \alpha +1\right) }\cdot \label{80}\left[ \tfrac {\pi ^{\alpha }}{2N}+\tfrac {1}{\left( \alpha +1\right) }\left( \tfrac {\pi ^{\alpha +1}}{2N}\right) ^{\frac{\alpha }{\alpha +1}}\right], \end{align}
</div>
<p> \(\forall \text{ }N\in \mathbb {N}.\) </p>
<p>We have that \(\underset {N\rightarrow +\infty }{\lim }W_{N}^{\left( M\right) }\left( f\right) \left( x\right) =f\left( x\right) .\) </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000039">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Applying (<a href="#37">2.37</a>) for \(W_{N}^{\left( M\right) }\) and using (<a href="#66.">3.66</a>), we get </p>
<div class="equation" id="81">
<p>
  <div class="equation_content">
    \begin{equation}  \left| W_{N}^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq \tfrac {\omega _{1}\left( D_{x}^{\alpha }f,\delta \right) }{\Gamma \left( \alpha +1\right) }\left[ \tfrac {\pi ^{\alpha }}{2N}+\tfrac {\frac{\pi ^{\alpha +1}}{2N}}{\left( \alpha +1\right) \delta }\right] .\label{81}\end{equation}
  </div>
  <span class="equation_label">3.81</span>
</p>
</div>
<p> Choose \(\delta =\left( \tfrac {\pi ^{\alpha +1}}{2N}\right) ^{\frac{1}{\alpha +1}}\), i.e. \(\delta ^{\alpha +1}=\tfrac {\pi ^{\alpha +1}}{2N}\), and \(\delta ^{\alpha }=\left( \tfrac {\pi ^{\alpha +1}}{2N}\right) ^{\frac{\alpha }{\alpha +1}}\). We use the last into (<a href="#81">3.81</a>) and we obtain (<a href="#80">3.80</a>). <div class="proof_wrapper" id="a0000000040">
  <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>We also have </p>
<p><div class="theorem_thmwrapper " id="t35">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.19</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(\alpha {\gt}1,\) \(\alpha \notin \mathbb {N}\), \(m=\left\lceil \alpha \right\rceil \), \(x\in \left[ 0,\pi \right] \subset \mathbb {R}\), \(f\in AC^{m}\left( \left[ 0,\pi \right] ,\mathbb {R}_{+}\right) \), and \(f^{\left( m\right) }\in L_{\infty }\left( \left[ 0,\pi \right] \right) \). Furthermore we assume that <br />\(f^{\left( k\right) }\left( x\right) =0\), \(k=1,...,m-1\). Then </p>
<div class="displaymath" id="82">
  \begin{align} & \left| K_{N}^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq \tfrac {\omega _{1}\left( D_{x}^{\alpha }f,\left( \frac{\pi ^{\alpha +1}}{2N}\right) ^{\frac{1}{\alpha +1}}\right) }{\Gamma \left( \alpha +1\right) }\cdot \label{82}\left[ \tfrac {\pi ^{\alpha }}{2N}+\tfrac {1}{\left( \alpha +1\right) }\left( \tfrac {\pi ^{\alpha +1}}{2N}\right) ^{\frac{\alpha }{\alpha +1}}\right] , \end{align}
</div>
<p> \( \forall \text{ }N\in \mathbb {N}.\) </p>
<p>We have that \(\underset {N\rightarrow +\infty }{\lim }K_{N}^{\left( M\right) }\left( f\right) \left( x\right) =f\left( x\right) .\) </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> As in Theorem <a href="#t34">3.18</a>. <div class="proof_wrapper" id="a0000000042">
  <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>We make </p>
<p><div class="remark_thmwrapper " id="r36">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">3.20</span>
  </div>
  <div class="remark_thmcontent">
  <p>We mention the interpolation Hermite-Fejer polynomials on Chebyshev knots of the first kind (see <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	4
	]
</span>): Let \(f:\left[ -1,1\right] \rightarrow \mathbb {R}\) and based on the knots \(x_{N,k}=\cos \left( \tfrac {\left( 2\left( N-k\right) +1\right) }{2\left( N+1\right) }\pi \right) \in \left( -1,1\right) \), \(k\in \left\{  0,...,N\right\}  \), \(-1{\lt}x_{N,0}{\lt}x_{N,1}{\lt}...{\lt}x_{N,N}{\lt}1\), which are the roots of the first kind Chebyshev polynomial \(T_{N+1}\left( x\right) =\cos \left( \left( N+1\right) \arccos x\right) \), we define (see Fejér <span class="cite">
	[
	<a href="#9" >9</a>
	]
</span>) </p>
<div class="equation" id="83">
<p>
  <div class="equation_content">
    \begin{equation}  H_{2N+1}\left( f\right) \left( x\right) =\sum _{k=0}^{N}h_{N,k}\left( x\right) f\left( x_{N,k}\right) ,\label{83}\end{equation}
  </div>
  <span class="equation_label">3.83</span>
</p>
</div>
<p> where </p>
<div class="equation" id="84">
<p>
  <div class="equation_content">
    \begin{equation}  h_{N,k}\left( x\right) =\left( 1-x\cdot x_{N,k}\right) \left( \tfrac {T_{N+1}\left( x\right) }{\left( N+1\right) \left( x-x_{N,k}\right) }\right) ^{2},\label{84}\end{equation}
  </div>
  <span class="equation_label">3.84</span>
</p>
</div>
<p> the fundamental interpolation polynomials. </p>
<p>The Max-product interpolation Hermite-Fejér operators on Chebyshev <br />knots of the first kind (see <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	12
	]
</span>) are defined by </p>
<div class="equation" id="85">
<p>
  <div class="equation_content">
    \begin{equation}  H_{2N+1}^{\left( M\right) }\left( f\right) \left( x\right) =\tfrac {\bigvee \limits _{k=0}^{N}h_{N,k}\left( x\right) f\left( x_{N,k}\right) }{\bigvee \limits _{k=0}^{N}h_{N,k}\left( x\right) },\text{ \  }\forall \text{ }N\in \mathbb {N}\text{,}\label{85}\end{equation}
  </div>
  <span class="equation_label">3.85</span>
</p>
</div>
<p> where \(f:\left[ -1,1\right] \rightarrow \mathbb {R}_{+}\) is continuous. </p>
<p>Call </p>
<div class="equation" id="86">
<p>
  <div class="equation_content">
    \begin{equation}  E_{N}\left( x\right) :=H_{2N+1}^{\left( M\right) }\left( \left| \cdot -x\right| \right) \left( x\right) =\tfrac {\bigvee \limits _{k=0}^{N}h_{N,k}\left( x\right) \left| x_{N,k}-x\right| }{\bigvee \limits _{k=0}^{N}h_{N,k}\left( x\right) }, \quad x\in \left[ -1,1\right] . \label{86}\end{equation}
  </div>
  <span class="equation_label">3.86</span>
</p>
</div>
<p> Then by <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	287
	]
</span>, we obtain that </p>
<div class="equation" id="87">
<p>
  <div class="equation_content">
    \begin{equation}  E_{N}\left( x\right) \leq \tfrac {2\pi }{N+1}\text{, \  }\forall \text{ }x\in \left[ -1,1\right] ,\text{ }N\in \mathbb {N}\text{.}\label{87}\end{equation}
  </div>
  <span class="equation_label">3.87</span>
</p>
</div>
<p> For \(m{\gt}1\), we get </p>
<div class="displaymath" id="a0000000043">
  \begin{align}  H_{2N+1}^{\left( M\right) }\left( \left| \cdot -x\right| ^{m}\right) \left( x\right) & =\tfrac {\bigvee \limits _{k=0}^{N}h_{N,k}\left( x\right) \left| x_{N,k}-x\right| ^{m}}{\bigvee \limits _{k=0}^{N}h_{N,k}\left( x\right) }= \nonumber \\ & =\tfrac {\bigvee \limits _{k=0}^{N}h_{N,k}\left( x\right) \left| x_{N,k}-x\right| \left| x_{N,k}-x\right| ^{m-1}}{\bigvee \limits _{k=0}^{N}h_{N,k}\left( x\right) }\leq 2^{m-1}\tfrac {\bigvee \limits _{k=0}^{N}h_{N,k}\left( x\right) \left| x_{N,k}-x\right| }{\bigvee \limits _{k=0}^{N}h_{N,k}\left( x\right) } \nonumber \\ & \leq \tfrac {2^{m}\pi }{N+1}, \quad \forall x\in \left[ -1,1\right] ,\text{ }N\in \mathbb {N}\text{.} \label{88}\end{align}
</div>
<p>Hence it holds </p>
<div class="equation" id="89">
<p>
  <div class="equation_content">
    \begin{equation}  H_{2N+1}^{\left( M\right) }\left( \left| \cdot -x\right| ^{m}\right) \left( x\right) \leq \tfrac {2^{m}\pi }{N+1}, \quad \forall x\in \left[ -1,1\right] ,\text{ }m>1\text{, }\forall \text{ }N\in \mathbb {N}\text{.}\label{89}\end{equation}
  </div>
  <span class="equation_label">3.89</span>
</p>
</div>
<p> Furthermore we have </p>
<div class="equation" id="90">
<p>
  <div class="equation_content">
    \begin{equation}  H_{2N+1}^{\left( M\right) }\left( 1\right) \left( x\right) =1\text{, \  }\forall \text{ }x\in \left[ -1,1\right] ,\label{90}\end{equation}
  </div>
  <span class="equation_label">3.90</span>
</p>
</div>
<p> and \(H_{2N+1}^{\left( M\right) }\) maps continuous functions to continuous functions over \(\left[ -1,1\right] \) and for any \(x\in \mathbb {R}\) we have \(\bigvee _{k=0}^{N}h_{N,k}\left( x\right) {\gt}0\). </p>
<p>We also have \(h_{N,k}\left( x_{N,k}\right) =1\), and \(h_{N,k}\left( x_{N,j}\right) =0\), if \(k\neq j\), furthermore it holds \(H_{2N+1}^{\left( M\right) }\left( f\right) \left( x_{N,j}\right) =f\left( x_{N,j}\right) \), for all \(j\in \left\{  0,...,N\right\}  \), see <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	282
	]
</span>. </p>
<p>\(H_{2N+1}^{\left( M\right) }\) are positive sublinear operators, <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	282
	]
</span>. <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>We give </p>
<p><div class="theorem_thmwrapper " id="t37">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.21</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(\alpha \! \! {\gt}\! \! 1,\) \(\alpha \notin \mathbb {N}\), \(m\! \! =\! \! \left\lceil \alpha \right\rceil \), \(x\in \left[ \! -\! 1,1\right] \), \(f\in AC^{m}\left( \left[ \! -\! 1,1\right] ,\mathbb {R}_{+}\right) \), and \(f^{\left( m\right) }\in L_{\infty }\left( \left[ -1,1\right] \right) \). Furthermore we assume that \(f^{\left( k\right) }\left( x\right) =0\), \(k=1,...,m-1\). Then </p>
<div class="displaymath" id="91">
  \begin{align} & \left| H_{2N+1}^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq \tfrac {\omega _{1}\left( D_{x}^{\alpha }f,\big( \frac{2^{\alpha +1}\pi }{N+1}\big) ^{\frac{1}{\alpha +1}}\right) }{\Gamma \left( \alpha +1\right) }\cdot \label{91}\left[ \tfrac {2^{\alpha }\pi }{N+1}+\tfrac {1}{\left( \alpha +1\right) }\left( \tfrac {2^{\alpha +1}\pi }{N+1}\right) ^{\frac{\alpha }{\alpha +1}}\right] , \end{align}
</div>
<p> \(\forall \text{ }N\in \mathbb {N}.\) </p>
<p>Furthermore it holds \(\underset {N\rightarrow +\infty }{\lim }H_{2N+1}^{\left( M\right) }\left( f\right) \left( x\right) =f\left( x\right) .\) </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000044">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> By Theorem <a href="#t14">2.14</a>, choose \(\delta :=\left( \tfrac {2^{\alpha +1}\pi }{N+1}\right) ^{\frac{1}{\alpha +1}}\), use (<a href="#37">2.37</a>), (<a href="#89">3.89</a>). <div class="proof_wrapper" id="a0000000045">
  <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>We continue with </p>
<p><div class="remark_thmwrapper " id="r38">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">3.22</span>
  </div>
  <div class="remark_thmcontent">
  <p>Here we deal with Lagrange interpolation polynomials on Chebyshev knots of second kind plus the endpoints \(\pm 1\) (see <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	5
	]
</span>). These polynomials are linear operators attached to \(f:\left[ -1,1\right] \rightarrow \mathbb {R}\) and to the knots \(x_{N,k}=\cos \left( \left( \tfrac {N-k}{N-1}\right) \pi \right) \in \left[ -1,1\right] \), \(k=1,...,N,\) \(N\in \mathbb {N}\), which are the roots of \(\omega _{N}\left( x\right) =\sin \left( N-1\right) t\sin t\), \(x=\cos t\). Notice that \(x_{N,1}=-1\) and \(x_{N,N}=1.\) Their formula is given by <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	377
	]
</span> </p>
<div class="equation" id="92">
<p>
  <div class="equation_content">
    \begin{equation}  L_{N}\left( f\right) \left( x\right) =\sum _{k=1}^{N}l_{N,k}\left( x\right) f\left( x_{N,k}\right) ,\label{92}\end{equation}
  </div>
  <span class="equation_label">3.92</span>
</p>
</div>
<p> where </p>
<div class="equation" id="93">
<p>
  <div class="equation_content">
    \begin{equation}  l_{N,k}\left( x\right) =\tfrac {\left( -1\right) ^{k-1}\omega _{N}\left( x\right) }{\left( 1+\delta _{k,1}+\delta _{k,N}\right) \left( N-1\right) \left( x-x_{N,k}\right) },\label{93}\end{equation}
  </div>
  <span class="equation_label">3.93</span>
</p>
</div>
<p> \(N\geq 2\), \(k=1,...,N\), and \(\omega _{N}\left( x\right) =\prod _{k=1}^{N}\left( x-x_{N,k}\right) \) and \(\delta _{i,j}\) denotes the Kronecher’s symbol, that is \(\delta _{i,j}=1\), if \(i=j\), and \(\delta _{i,j}=0 \), if \(i\neq j\). </p>
<p>The Max-product Lagrange interpolation operators on Chebyshev knots of second kind, plus the endpoints \(\pm 1\), are defined by <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	12
	]
</span> </p>
<div class="equation" id="94">
<p>
  <div class="equation_content">
    \begin{equation}  L_{N}^{\left( M\right) }\left( f\right) \left( x\right) =\tfrac {\bigvee \limits _{k=1}^{N}l_{N,k}\left( x\right) f\left( x_{N,k}\right) }{\bigvee \limits _{k=1}^{N}l_{N,k}\left( x\right) },\quad x\in \left[ -1,1\right] ,\label{94}\end{equation}
  </div>
  <span class="equation_label">3.94</span>
</p>
</div>
<p> where \(f:\left[ -1,1\right] \rightarrow \mathbb {R}_{+}\) continuous. </p>
<p>First we see that \(L_{N}^{\left( M\right) }\left( f\right) \left( x\right) \) is well defined and continuous for any \(x\in \left[ -1,1\right] \). Following <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	289
	]
</span>, because \(\sum _{k=1}^{N}l_{N,k}\left( x\right) =1\), \(\forall \) \(x\in \mathbb {R}\), for any \(x\) there exists \(k\in \left\{  1,...,N\right\}  :l_{N,k}\left( x\right) {\gt}0\), hence \(\bigvee _{k=1}^{N}l_{N,k}\left( x\right) {\gt}0\). We have that \(l_{N,k}\left( x_{N,k}\right) =1\), and \(l_{N,k}\left( x_{N,j}\right) =0\), if \(k\neq j\). Furthermore it holds \(L_{N}^{\left( M\right) }\left( f\right) \left( x_{N,j}\right) =f\left( x_{N,j}\right) \), all \(j\in \left\{  1,...,N\right\}  ,\) and \(L_{N}^{\left( M\right) }\left( 1\right) =1\). </p>
<p>Call \(I_{N}^{+}\left( x\right) =\left\{  k\in \left\{  1,...,N\right\}  ;l_{N,k}\left( x\right) {\gt}0\right\}  \), then \(I_{N}^{+}\left( x\right) \neq \emptyset \). </p>
<p>So for \(f\in C_{+}\left( \left[ -1,1\right] \right) \) we get </p>
<div class="equation" id="95">
<p>
  <div class="equation_content">
    \begin{equation}  L_{N}^{\left( M\right) }\left( f\right) \left( x\right) =\tfrac {\bigvee \limits _{k\in I_{N}^{+}\left( x\right) }l_{N,k}\left( x\right) f\left( x_{N,k}\right) }{\bigvee \limits _{k\in I_{N}^{+}\left( x\right) }l_{N,k}\left( x\right) }\geq 0.\label{95}\end{equation}
  </div>
  <span class="equation_label">3.95</span>
</p>
</div>
<p> Notice here that \(\left| x_{N,k}-x\right| \leq 2\), \(\forall \) \(x\in \left[ -1,1\right] .\) </p>
<p>By <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	297
	]
</span>, we get that </p>
<div class="displaymath" id="96">
  \begin{align}  L_{N}^{\left( M\right) }\left( \left| \cdot -x\right| \right) \left( x\right) =\tfrac {\bigvee \limits _{k=1}^{N}l_{N,k}\left( x\right) \left| x_{N,k}-x\right| }{\bigvee \limits _{k=1}^{N}l_{N,k}\left( x\right) }= \tfrac {\bigvee \limits _{k\in I_{N}^{+}\left( x\right) }l_{N,k}\left( x\right) \left| x_{N,k}-x\right| }{\bigvee \limits _{k\in I_{N}^{+}\left( x\right) }l_{N,k}\left( x\right) }\leq \tfrac {\pi ^{2}}{6\left( N-1\right) },\label{96}\end{align}
</div>
<p> \(N\geq 3\), \(\forall \) \(x\in \left( -1,1\right) \), \(N\) is odd. </p>
<p>We get that (\(m{\gt}1\)) </p>
<div class="equation" id="97">
<p>
  <div class="equation_content">
    \begin{equation}  L_{N}^{\left( M\right) }\left( \left| \cdot -x\right| ^{m}\right) \left( x\right) =\tfrac {\bigvee \limits _{k\in I_{N}^{+}\left( x\right) }l_{N,k}\left( x\right) \left| x_{N,k}-x\right| ^{m}}{\bigvee \limits _{k\in I_{N}^{+}\left( x\right) }l_{N,k}\left( x\right) }\leq \tfrac {2^{m-1}\pi ^{2}}{6\left( N-1\right) },\label{97}\end{equation}
  </div>
  <span class="equation_label">3.97</span>
</p>
</div>
<p> \(N\geq 3\) odd, \(\forall \) \(x\in \left( -1,1\right) .\) </p>
<p>\(L_{N}^{\left( M\right) }\) are positive sublinear operators, <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	290
	]
</span>. <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>We give </p>
<p><div class="theorem_thmwrapper " id="t39">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.23</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Same assumptions as in Theorem <span class="rm"><a href="#t37">3.21</a></span>. Then </p>
<div class="displaymath" id="98">
  \begin{align} & \left| L_{N}^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq \tfrac {\omega _{1}\left( D_{x}^{\alpha }f,\left( \frac{2^{\alpha }\pi ^{2}}{6\left( N-1\right) }\right) ^{\frac{1}{\alpha +1}}\right) }{\Gamma \left( \alpha +1\right) } \label{98}\left[ \tfrac {2^{\alpha -1}\pi ^{2}}{6\left( N-1\right) }+\tfrac {1}{\left( \alpha +1\right) }\left( \tfrac {2^{\alpha }\pi ^{2}}{6\left( N-1\right) }\right) ^{\frac{\alpha }{\alpha +1}}\right], \end{align}
</div>
<p> \(\forall \text{ }N\in \mathbb {N}:N\geq 3\text{, odd.}\) </p>
<p>It holds \(\underset {N\rightarrow +\infty }{\lim }L_{N}^{\left( M\right) }\left( f\right) \left( x\right) =f\left( x\right) .\) </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000046">
  <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 Theorem <a href="#t14">2.14</a>, choose \(\delta :=\left( \tfrac {2^{\alpha }\pi ^{2}}{6\left( N-1\right) }\right) ^{\frac{1}{\alpha +1}}\), use of (<a href="#37">2.37</a>) and (<a href="#97">3.97</a>). At \(\pm 1\) the left hand side of (<a href="#98">3.98</a>) is zero, thus (<a href="#98">3.98</a>) is trivially true. <div class="proof_wrapper" id="a0000000047">
  <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>We make </p>
<p><div class="remark_thmwrapper " id="r40">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">3.24</span>
  </div>
  <div class="remark_thmcontent">
  <p>Let \(f\in C_{+}\left( \left[ -1,1\right] \right) \), \(N\geq 4\), \(N\in \mathbb {N}\), \(N\) even. </p>
<p>By <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	p.
	
	298
	]
</span>, we get </p>
<div class="equation" id="99">
<p>
  <div class="equation_content">
    \begin{equation}  L_{N}^{\left( M\right) }\left( \left| \cdot -x\right| \right) \left( x\right) \leq \tfrac {4\pi ^{2}}{3\left( N-1\right) }=\tfrac {2^{2}\pi ^{2}}{3\left( N-1\right) },\text{ \  }\forall x\in \left( -1,1\right) .\label{99}\end{equation}
  </div>
  <span class="equation_label">3.99</span>
</p>
</div>
<p> Hence (\(m{\gt}1\)) </p>
<div class="equation" id="100">
<p>
  <div class="equation_content">
    \begin{equation}  L_{N}^{\left( M\right) }\left( \left| \cdot -x\right| ^{m}\right) \left( x\right) \leq \tfrac {2^{m+1}\pi ^{2}}{3\left( N-1\right) },\text{ \  }\forall \text{ }x\in \left( -1,1\right) .\label{100}\end{equation}
  </div>
  <span class="equation_label">3.100</span>
</p>
</div>
<p> <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>We present </p>
<p><div class="theorem_thmwrapper " id="t41">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.25</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Same assumptions as in Theorem <span class="rm"><a href="#t37">3.21</a></span>. Then </p>
<div class="displaymath" id="101">
  \begin{align} & \left| L_{N}^{\left( M\right) }\left( f\right) \left( x\right) \! -\! f\left( x\right) \right| \leq \tfrac {\omega _{1}\left( D_{x}^{\alpha }f,\left( \frac{2^{\alpha +2}\pi ^{2}}{3\left( N-1\right) }\right) ^{\frac{1}{\alpha +1}}\right) }{\Gamma \left( \alpha +1\right) } \label{101}\left[ \tfrac {2^{\alpha +1}\pi ^{2}}{3\left( N-1\right) }+\tfrac {1}{\left( \alpha +1\right) }\left( \tfrac {2^{\alpha +2}\pi ^{2}}{3\left( N-1\right) }\right) ^{\frac{\alpha }{\alpha +1}}\right] , \end{align}
</div>
<p> \(\forall \text{ }N\in \mathbb {N},\text{ }N\geq 4\text{, }N\) is even. </p>
<p>It holds \(\underset {N\rightarrow +\infty }{\lim }L_{N}^{\left( M\right) }\left( f\right) \left( x\right) =f\left( x\right) .\) </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000048">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> By Theorem <a href="#t14">2.14</a>, choose \(\delta :=\left( \tfrac {2^{\alpha +2}\pi ^{2}}{3\left( N-1\right) }\right) ^{\frac{1}{\alpha +1}}\), use of (<a href="#37">2.37</a>) and (<a href="#100">3.100</a>). At \(\pm 1\), (<a href="#101">3.101</a>) is trivially true. <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>We need </p>
<p><div class="definition_thmwrapper " id="d42">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">3.26</span>
  </div>
  <div class="definition_thmcontent">
  <p>(<span class="cite">
	[
	<a href="#6" >6</a>
	, 
	p.
	
	41
	]
</span>). Let \(I\subset \mathbb {R}\) be an interval of finite or infinite length, and \(f:I\rightarrow \mathbb {R}\) a bounded or uniformly continuous function. We define the first modulus of continuity </p>
<div class="equation" id="102">
<p>
  <div class="equation_content">
    \begin{equation}  \omega _{1}\left( f,\delta \right) _{I}=\underset {\left| x-y\right| \leq \delta }{\underset {x,y\in I}{\sup }}\left| f\left( x\right) -f\left( y\right) \right| ,\text{ \  }\delta >0.\label{102}\end{equation}
  </div>
  <span class="equation_label">3.102</span>
</p>
</div>
<p> Clearly, it holds \(\omega _{1}\left( f,\delta \right) _{I}{\lt}+\infty .\) </p>
<p>We also have </p>
<div class="equation" id="103">
<p>
  <div class="equation_content">
    \begin{equation}  \omega _{1}\left( f,r\delta \right) _{I}\leq \left( r+1\right) \omega _{1}\left( f,\delta \right) _{I},\text{ \  any }r\geq 0.\label{103}\end{equation}
  </div>
  <span class="equation_label">3.103</span>
</p>
</div>

  </div>
</div> </p>
<p><div class="convention_thmwrapper " id="c43">
  <div class="convention_thmheading">
    <span class="convention_thmcaption">
    Convention
    </span>
    <span class="convention_thmlabel">3.27</span>
  </div>
  <div class="convention_thmcontent">
  <p>Let a real number \(m{\gt}1\), from now on we assume that \(D_{x_{0}-}^{m}f\) is either bounded or uniformly continuous function on \((-\infty ,x_{0}]\), similarly from now on we assume that \(D_{\ast x_{0}}^{m}f\) is either bounded or uniformly continuous function on \([x_{0},+\infty ).\) </p>

  </div>
</div> </p>
<p>We need </p>
<p><div class="definition_thmwrapper " id="d44">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">3.28</span>
  </div>
  <div class="definition_thmcontent">
  <p>Let \(D_{x_{0}}^{m}f\) (real number \(m{\gt}1\)) denote any of \(D_{x_{0}-}^{m}f\), \(D_{\ast x_{0}}^{m}f\) and \(\delta {\gt}0\). We set </p>
<div class="equation" id="104">
<p>
  <div class="equation_content">
    \begin{equation}  \omega _{1}\left( D_{x_{0}}^{m}f,\delta \right) _{\mathbb {R}}:=\max \left\{  \omega _{1}\left( D_{x_{0}-}^{m}f,\delta \right) _{(-\infty ,x_{0}]},\omega _{1}\left( D_{\ast x_{0}}^{m}f,\delta \right) _{[x_{0},+\infty )}\right\}  ,\label{104}\end{equation}
  </div>
  <span class="equation_label">3.104</span>
</p>
</div>
<p> where \(x_{0}\in \mathbb {R}\). Notice that \(\omega _{1}\left( D_{x_{0}}^{m}f,\delta \right) _{\mathbb {R}}{\lt}+\infty .\) </p>

  </div>
</div> </p>
<p>We will use </p>
<p><div class="theorem_thmwrapper " id="t45">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.29</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let the real number \(m{\gt}0\), \(m\notin \mathbb {N}\), \(\lambda =\left\lceil m\right\rceil \), \(x_{0}\in \mathbb {R}\), \(f\in AC^{\lambda }\left( \left[ a,b\right] ,\mathbb {R}_{+}\right) \) (i.e. \(f^{\left( \lambda -1\right) }\in AC\left[ a,b\right] ,\) absolutely continuous functions on \(\left[ a,b\right] \)), \(\forall \) \(\left[ a,b\right] \subset \mathbb {R}\), and \(f^{\left( \lambda \right) }\in L_{\infty }\left( \mathbb {R}\right) \). Furthermore we assume that \(f^{\left( k\right) }\left( x_{0}\right) =0\), \(k=1,...,\lambda -1\). The Convention <span class="rm"><a href="#c43">3.27</a></span> is imposed here. Then </p>
<div class="equation" id="105">
<p>
  <div class="equation_content">
    \begin{equation}  \left| f\left( x\right) -f\left( x_{0}\right) \right| \leq \tfrac {\omega _{1}\left( D_{x_{0}}^{m}f,\delta \right) _{\mathbb {R}}}{\Gamma \left( m+1\right) }\left[ \left| x-x_{0}\right| ^{m}+\tfrac {\left| x-x_{0}\right| ^{m+1}}{\left( m+1\right) \delta }\right] ,\text{ \  }\delta >0,\label{105}\end{equation}
  </div>
  <span class="equation_label">3.105</span>
</p>
</div>
<p> for all \(x\in \mathbb {R}\). </p>
<p>If \(0{\lt}m{\lt}1\), then we do not need initial conditions. </p>

  </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> Similar to Theorem <a href="#t10">2.10</a>. <div class="proof_wrapper" id="a0000000051">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>We continue with </p>
<p><div class="remark_thmwrapper " id="r46">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">3.30</span>
  </div>
  <div class="remark_thmcontent">
  <p>Let \(b:\mathbb {R}\rightarrow \mathbb {R}_{+}\) be a centered (it takes a global maximum at \(0\)) bell-shaped function, with compact support \(\left[ -T,T\right] \), \(T{\gt}0\) (that is \(b\left( x\right) {\gt}0\) for all \(x\in \left( -T,T\right) \)) and \(I=\int _{-T}^{T}b\left( x\right) dx{\gt}0.\) </p>
<p>The Cardaliaguet-Euvrard neural network operators are defined by (see <span class="cite">
	[
	<a href="#4" >4</a>
	]
</span>) </p>
<div class="equation" id="106">
<p>
  <div class="equation_content">
    \begin{equation}  C_{N,\alpha }\left( f\right) \left( x\right) =\sum _{k=-N^{2}}^{N^{2}}\tfrac {f\left( \frac{k}{n}\right) }{IN^{1-\alpha }}b\left( N^{1-\alpha }\left( x-\tfrac {k}{N}\right) \right) ,\label{106}\end{equation}
  </div>
  <span class="equation_label">3.106</span>
</p>
</div>
<p> \(0{\lt}\alpha {\lt}1\), \(N\in \mathbb {N}\) and typically here \(f:\mathbb {R}\rightarrow \mathbb {R}\) is continuous and bounded or uniformly continuous on \(\mathbb {R}\). </p>
<p>\(CB\left( \mathbb {R}\right) \) denotes the continuous and bounded function on \(\mathbb {R}\), and </p>
<div class="displaymath" id="a0000000052">
  \[  CB_{+}\left( \mathbb {R}\right) =\left\{  f:\mathbb {R}\rightarrow \lbrack 0,\infty );\text{ }f\in CB\left( \mathbb {R}\right) \right\}  .  \]
</div>
<p> The corresponding max-product Cardaliaguet-Euvrard neural network operators will be given by </p>
<div class="equation" id="107">
<p>
  <div class="equation_content">
    \begin{equation}  C_{N,\alpha }^{\left( M\right) }\left( f\right) \left( x\right) =\tfrac {\bigvee \limits _{k=-N^{2}}^{N^{2}}b\left( N^{1-\alpha }\left( x-\frac{k}{N}\right) \right) f\left( \frac{k}{N}\right) }{\bigvee \limits _{k=-N^{2}}^{N^{2}}b\left( N^{1-\alpha }\left( x-\frac{k}{N}\right) \right) }, \label{107}\end{equation}
  </div>
  <span class="equation_label">3.107</span>
</p>
</div>
<p> \(x\in \mathbb {R}\), typically here \(f\in CB_{+}\left( \mathbb {R}\right) \), see also <span class="cite">
	[
	<a href="#4" >4</a>
	]
</span>. </p>
<p>Next we follow <span class="cite">
	[
	<a href="#4" >4</a>
	]
</span>. </p>
<p>For any \(x\in \mathbb {R}\), denoting </p>
<div class="displaymath" id="a0000000053">
  \[  J_{T,N}\left( x\right) =\left\{  k\in \mathbb {Z}\text{; }-N^{2}\leq k\leq N^{2}\text{, }N^{1-\alpha }\left( x-\tfrac {k}{N}\right) \in \left( -T,T\right) \right\}  ,  \]
</div>
<p> we can write </p>
<div class="equation" id="108">
<p>
  <div class="equation_content">
    \begin{equation}  C_{N,\alpha }^{\left( M\right) }\left( f\right) \left( x\right) =\tfrac {\bigvee \limits _{k\in J_{T,N}\left( x\right) }b\left( N^{1-\alpha }\left( x-\frac{k}{N}\right) \right) f\left( \frac{k}{N}\right) }{\bigvee \limits _{k\in J_{T,N}\left( x\right) }b\left( N^{1-\alpha }\left( x-\frac{k}{N}\right) \right) },\label{108}\end{equation}
  </div>
  <span class="equation_label">3.108</span>
</p>
</div>
<p> \(x\in \mathbb {R}\), \(N{\gt}\max \left\{  T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\}  \), where \(J_{T,N}\left( x\right) \neq \emptyset \). Indeed, we have \(\bigvee _{k\in J_{T,N}\left( x\right) }b\left( N^{1-\alpha }\left( x-\frac{k}{N}\right) \right) {\gt}0\), \(\forall \) \(x\in \mathbb {R}\) and \(N{\gt}\max \left\{  T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\}  .\) </p>
<p>We have that \(C_{N,\alpha }^{\left( M\right) }\left( 1\right) \left( x\right) =1\), \(\forall \) \(x\in \mathbb {R}\) and \(N{\gt}\max \left\{  T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\}  \). <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>See in <span class="cite">
	[
	<a href="#4" >4</a>
	, 
	Lemma 2.1, Corollary 2.2 and Remarks
	]
</span>. </p>
<p>We need </p>
<p><div class="theorem_thmwrapper " id="t47">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.31</span>
  </div>
  <div class="theorem_thmcontent">
  <p><span class="cite">
	[
	<a href="#4" >4</a>
	]
</span>. Let \(b\left( x\right) \) be a centered bell-shaped function, continuous and with compact support \(\left[ -T,T\right] \), \(T{\gt}0\), \(0{\lt}\alpha {\lt}1\) and \(C_{N,\alpha }^{\left( M\right) }\) be defined as in <a href="#107" class="eqref">3.107</a>. </p>
<p>(i) If \(\left| f\left( x\right) \right| \leq c\) for all \(x\in \mathbb {R}\) then \(\left| C_{N,\alpha }^{\left( M\right) }\left( f\right) \left( x\right) \right| \leq c\), for all \(x\in \mathbb {R}\) and \(N{\gt}\max \left\{  T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\}  \) and \(C_{N,\alpha }^{\left( M\right) }\left( f\right) \left( x\right) \) is continuous at any point \(x\in \mathbb {R}\), for all \(N{\gt}\max \left\{  T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\}  ;\) </p>
<p>(ii) If \(f,g\in CB_{+}\left( \mathbb {R}\right) \) satisfy \(f\left( x\right) \leq g\left( x\right) \) for all \(x\in \mathbb {R}\), then \(C_{N,\alpha }^{\left( M\right) }\left( f\right) \left( x\right) \leq C_{N,\alpha }^{\left( M\right) }\left( g\right) \left( x\right) \) for all \(x\in \mathbb {R}\) and \(N{\gt}\max \left\{  T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\}  ;\) </p>
<p>(iii) \(C_{N,\alpha }^{\left( M\right) }\left( f+g\right) \left( x\right) \leq C_{N,\alpha }^{\left( M\right) }\left( f\right) \left( x\right) +C_{N,\alpha }^{\left( M\right) }\left( g\right) \left( x\right) \) for all \(f,g\in CB_{+}\left( \mathbb {R}\right) \), \(x\in \mathbb {R}\) and \(N{\gt}\max \left\{  T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\}  ;\) </p>
<p>(iv) For all \(f,g\in CB_{+}\left( \mathbb {R}\right) \), \(x\in \mathbb {R}\) and \(N{\gt}\max \left\{  T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\}  \), we have </p>
<div class="displaymath" id="a0000000054">
  \[  \left| C_{N,\alpha }^{\left( M\right) }\left( f\right) \left( x\right) -C_{N,\alpha }^{\left( M\right) }\left( g\right) \left( x\right) \right| \leq C_{N,\alpha }^{\left( M\right) }\left( \left| f-g\right| \right) \left( x\right) ;  \]
</div>
<p>(v) \(C_{N,\alpha }^{\left( M\right) }\) is positive homogeneous, that is \(C_{N,\alpha }^{\left( M\right) }\left( \lambda f\right) \left( x\right) =\lambda C_{N,\alpha }^{\left( M\right) }\left( f\right) \left( x\right) \) for all \(\lambda \geq 0\), \(x\in \mathbb {R}\), \(N{\gt}\max \left\{  T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\}  \) and \(f\in CB_{+}\left( \mathbb {R}\right) .\) </p>

  </div>
</div> </p>
<p>We make </p>
<p><div class="remark_thmwrapper " id="r48">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">3.32</span>
  </div>
  <div class="remark_thmcontent">
  <p>We have that </p>
<div class="equation" id="109">
<p>
  <div class="equation_content">
    \begin{equation}  E_{N,\alpha }\left( x\right) :=C_{N,\alpha }^{\left( M\right) }\left( \left| \cdot -x\right| \right) \left( x\right) =\tfrac {\bigvee \limits _{k\in J_{T,N}\left( x\right) }b\left( N^{1-\alpha }\left( x-\frac{k}{N}\right) \right) \left| x-\frac{k}{N}\right| }{\bigvee \limits _{k\in J_{T,N}\left( x\right) }b\left( N^{1-\alpha }\left( x-\frac{k}{N}\right) \right) },\label{109}\end{equation}
  </div>
  <span class="equation_label">3.109</span>
</p>
</div>
<p> \(\forall \) \(x\in \mathbb {R}\), and \(N{\gt}\max \left\{  T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\}  .\) <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>We mention from <span class="cite">
	[
	<a href="#4" >4</a>
	]
</span> the following: </p>
<p><div class="theorem_thmwrapper " id="t49">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.33</span>
  </div>
  <div class="theorem_thmcontent">
  <p><span class="cite">
	[
	<a href="#4" >4</a>
	]
</span>. Let \(b\left( x\right) \) be a centered bell-shaped function, continuous and with compact support \(\left[ -T,T\right] \), \(T{\gt}0\) and \(0{\lt}\alpha {\lt}1\). In addition, suppose that the following requirements are fulfilled: </p>
<p>(i) There exist \(0{\lt}m_{1}\leq M_{1}{\lt}\infty \) such that \(m_{1}\left( T-x\right) \leq b\left( x\right) \leq M_{1}\left( T-x\right) \), \(\forall \) \(x\in \left[ 0,T\right] ;\) </p>
<p>(ii) There exist \(0{\lt}m_{2}\leq M_{2}{\lt}\infty \) such that \(m_{2}\left( x+T\right) \leq b\left( x\right) \leq M_{2}\left( x+T\right) \), \(\forall \) \(x\in \left[ -T,0\right] \). </p>
<p>Then for all \(f\in CB_{+}\left( \mathbb {R}\right) \), \(x\in \mathbb {R}\) and for all \(N\in \mathbb {N}\) satisying \(N{\gt}\max \left\{  T+\left| x\right| ,\left( \frac{2}{T}\right) ^{\frac{1}{\alpha }}\right\}  \), we have the estimate </p>
<div class="equation" id="110">
<p>
  <div class="equation_content">
    \begin{equation}  \left| C_{N,\alpha }^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq c\,  \omega _{1}\left( f,N^{\alpha -1}\right) _{\mathbb {R}},\label{110}\end{equation}
  </div>
  <span class="equation_label">3.110</span>
</p>
</div>
<p> where </p>
<div class="displaymath" id="a0000000055">
  \[  c:=2\left( \max \left\{  \tfrac {TM_{2}}{2m_{2}},\tfrac {TM_{1}}{2m_{1}}\right\}  +1\right) ,  \]
</div>
<p> and </p>
<div class="displaymath" id="a0000000056">
  \[  \omega _{1}\left( f,\delta \right) _{\mathbb {R}}:=\underset {\left| x-y\right| \leq \delta }{\underset {x,y\in \mathbb {R}:}{\sup }}\left| f\left( x\right) -f\left( y\right) \right| .  \]
</div>

  </div>
</div> </p>
<p>We make </p>
<p><div class="remark_thmwrapper " id="r50">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">3.34</span>
  </div>
  <div class="remark_thmcontent">
  <p>In <span class="cite">
	[
	<a href="#4" >4</a>
	]
</span>, was proved that </p>
<div class="equation" id="111">
<p>
  <div class="equation_content">
    \begin{equation}  E_{N,\alpha }\left( x\right) \leq \max \left\{  \tfrac {TM_{2}}{2m_{2}},\tfrac {TM_{1}}{2m_{1}}\right\}  N^{\alpha -1},\text{ \  }\forall \text{ }N>\max \left\{  T+\left| x\right| ,\big(\tfrac {2}{T}\big) ^{\frac{1}{\alpha }}\right\}  .\label{111}\end{equation}
  </div>
  <span class="equation_label">3.111</span>
</p>
</div>
<p> That is </p>
<div class="equation" id="112">
<p>
  <div class="equation_content">
    \begin{equation}  C_{N,\alpha }^{\left( M\right) }\left( \left| \cdot -x\right| \right) \left( x\right) \leq \max \left\{  \tfrac {TM_{2}}{2m_{2}},\tfrac {TM_{1}}{2m_{1}}\right\}  N^{\alpha -1},\text{ \  }\forall \text{ }N>\max \left\{  T+\left| x\right| ,\big( \tfrac {2}{T}\big) ^{\frac{1}{\alpha }}\right\}  .\label{112}\end{equation}
  </div>
  <span class="equation_label">3.112</span>
</p>
</div>
<p> From (<a href="#109">3.109</a>) we have that \(\left| x-\frac{k}{N}\right| \leq \frac{T}{N^{1-\alpha }}.\) </p>
<p>Hence (\(m{\gt}1\)) (\(\forall \) \(x\in \mathbb {R}\) and \(N{\gt}\max \Big\{  T+\left| x\right| ,\big( \frac{2}{T}\big) ^{\frac{1}{\alpha }}\Big\}  \)) </p>
<div class="displaymath" id="113">
  \begin{align}  C_{N,\alpha }^{\left( M\right) }\left( \left| \cdot -x\right| ^{m}\right) \left( x\right) & =\tfrac {\bigvee \limits _{k\in J_{T,N}\left( x\right) }b\left( N^{1-\alpha }\left( x-\frac{k}{N}\right) \right) \left| x-\frac{k}{N}\right| ^{m}}{\bigvee \limits _{k\in J_{T,N}\left( x\right) }b\left( N^{1-\alpha }\left( x-\frac{k}{N}\right) \right) } \nonumber \label{113}\\ & \leq \left( \tfrac {T}{N^{1-\alpha }}\right) ^{m-1}\max \left\{  \tfrac {TM_{2}}{2m_{2}},\tfrac {TM_{1}}{2m_{1}}\right\}  N^{\alpha -1},\text{ \  } \nonumber \\ \forall \text{ }N{\gt}\max \left\{  T\! \! +\! \! \left| x\right| ,\left( \tfrac {2}{T}\right) ^{\frac{1}{\alpha }}\right\}  . \end{align}
</div>
<p> Then (\(m{\gt}1\)) it holds </p>
<div class="displaymath" id="a0000000057">
  \begin{align}  C_{N,\alpha }^{\left( M\right) }\left( \left| \cdot -x\right| ^{m}\right) \left( x\right) & \leq T^{m-1}\max \left\{  \tfrac {TM_{2}}{2m_{2}},\tfrac {TM_{1}}{2m_{1}}\right\}  \tfrac {1}{N^{m\left( 1-\alpha \right) }},\text{ \  } \nonumber \\ \forall \text{ }N{\gt}\max \left\{  T+\left| x\right| ,\left( \tfrac {2}{T}\right) ^{\frac{1}{\alpha }}\right\}  .\label{114}\end{align}
</div>
<p> Call </p>
<div class="equation" id="115">
<p>
  <div class="equation_content">
    \begin{equation}  \theta :=\max \left\{  \tfrac {TM_{2}}{2m_{2}},\tfrac {TM_{1}}{2m_{1}}\right\}  >0.\label{115}\end{equation}
  </div>
  <span class="equation_label">3.115</span>
</p>
</div>
<p> Consequently (\(m{\gt}1\)) we derive </p>
<div class="equation" id="116">
<p>
  <div class="equation_content">
    \begin{equation}  C_{N,\alpha }^{\left( M\right) }\left( \left| \cdot -x\right| ^{m}\right) \left( x\right) \leq \tfrac {\theta T^{m-1}}{N^{m\left( 1-\alpha \right) }},\text{ \  }\forall \text{ }N>\max \left\{  T+\left| x\right| ,\left( \tfrac {2}{T}\right) ^{\frac{1}{\alpha }}\right\}  .\label{116}\end{equation}
  </div>
  <span class="equation_label">3.116</span>
</p>
</div>
<p> <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>We need </p>
<p><div class="theorem_thmwrapper " id="t51">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.35</span>
  </div>
  <div class="theorem_thmcontent">
  <p>All here as in Theorem <a href="#t45">3.29</a>, where \(x=x_{0}\in \mathbb {R}\) is fixed. Let \(b\) be a centered bell-shaped function, continuous and with compact support \(\left[ -T,T\right] \), \(T{\gt}0\), \(0{\lt}\alpha {\lt}1\) and \(C_{N,\alpha }^{\left( M\right) }\) be defined as in <a href="#107" class="eqref">3.107</a>. Then </p>
<div class="displaymath" id="117">
  \begin{align}  \left| C_{N,\alpha }^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq \nonumber \tfrac {\omega _{1}\left( D_{x}^{m}f,\delta \right) _{\mathbb {R}}}{\Gamma \left( m+1\right) }\left[ C_{N,\alpha }^{\left( M\right) }\left( \left| \cdot -x\right| ^{m}\right) \left( x\right) +\tfrac {C_{N,\alpha }^{\left( M\right) }\left( \left| \cdot -x\right| ^{m+1}\right) \left( x\right) }{\left( m+1\right) \delta }\right] ,\end{align}
</div>
<p> <br />\(\forall \) \(N\in \mathbb {N}:N{\gt}\max \left\{  T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\}  .\) </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000058">
  <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 Theorem <a href="#t45">3.29</a> and (<a href="#105">3.105</a>) we get </p>
<div class="equation" id="118">
<p>
  <div class="equation_content">
    \begin{equation}  \left| f\left( \cdot \right) -f\left( x\right) \right| \leq \tfrac {\omega _{1}\left( D_{x}^{m}f,\delta \right) _{\mathbb {R}}}{\Gamma \left( m+1\right) }\left[ \left| \cdot -x\right| ^{m}+\tfrac {\left| \cdot -x\right| ^{m+1}}{\left( m+1\right) \delta }\right] ,\text{ \  }\delta >0,\label{118}\end{equation}
  </div>
  <span class="equation_label">3.117</span>
</p>
</div>
<p> true over \(\mathbb {R}\). </p>
<p>As in Theorem <a href="#t47">3.31</a> and using similar reasoning and \(C_{N,\alpha }^{\left( M\right) }\left( 1\right) =1\), we get </p>
<div class="displaymath" id="a0000000059">
  \begin{align} & \left| C_{N,\alpha }^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq C_{N,\alpha }^{\left( M\right) }\left( \left| f\left( \cdot \right) -f\left( x\right) \right| \right) \left( x\right) \overset {\text{(\ref{118})}}{\leq } \nonumber \\ & \overset {\text{(\ref{118})}}{\leq } \tfrac {\omega _{1}\left( D_{x}^{m}f,\delta \right) _{\mathbb {R}}}{\Gamma \left( m+1\right) }\left[ C_{N,\alpha }^{\left( M\right) }\left( \left| \cdot -x\right| ^{m}\right) \left( x\right) +\tfrac {C_{N,\alpha }^{\left( M\right) }\left( \left| \cdot -x\right| ^{m+1}\right) \left( x\right) }{\left( m+1\right) \delta }\right] ,\label{119}\end{align}
</div>
<p> <br />\(\forall \) \(N\in \mathbb {N}:N{\gt}\max \left\{  T+\left| x\right| ,T^{-\frac{1}{\alpha }}\right\}  .\) <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>We continue with </p>
<p><div class="theorem_thmwrapper " id="t52">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.36</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Here all as in Theorem <a href="#t45">3.29</a>, where \(x=x_{0}\in \mathbb {R}\) is fixed and \(m{\gt}1\). Also the same assumptions as in Theorem <a href="#t49">3.33</a>. Then </p>
<div class="displaymath" id="a0000000061">
  \begin{align}  \left| C_{N,\alpha }^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| & \leq \tfrac {1}{\Gamma \left( m+1\right) }\omega _{1}\left( D_{x}^{m}f,\big( \tfrac {\theta T^{m}}{N^{\left( m+1\right) \left( 1-\alpha \right) }}\big) ^{\frac{1}{m+1}}\right) _{\mathbb {R}}\cdot \nonumber \\ & \quad \cdot \left[ \tfrac {\theta T^{m-1}}{N^{m\left( 1-\alpha \right) }}+\tfrac {1}{\left( m+1\right) }\left( \tfrac {\theta T^{m}}{N^{\left( m+1\right) \left( 1-\alpha \right) }}\right) ^{\frac{m}{m+1}}\right] ,\label{120}\end{align}
</div>
<p> \(\forall \) \(N\in \mathbb {N}:N{\gt}\max \left\{  T+\left| x\right| ,\big( \frac{2}{T}\big) ^{\frac{1}{\alpha }}\right\}  .\) </p>
<p>We have that \(\underset {N\rightarrow +\infty }{\lim }C_{N,\alpha }^{\left( M\right) }\left( f\right) \left( x\right) =f\left( x\right) .\) </p>

  </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> We apply Theorem <a href="#t51">3.35</a>. In (<a href="#117">3.117</a>) we choose </p>
<div class="displaymath" id="a0000000063">
  \[  \delta :=\left( \tfrac {\theta T^{m}}{N^{\left( m+1\right) \left( 1-\alpha \right) }}\right) ^{\frac{1}{m+1}},  \]
</div>
<p> thus \(\delta ^{m+1}=\frac{\theta T^{m}}{N^{\left( m+1\right) \left( 1-\alpha \right) }}\), and </p>
<div class="equation" id="121">
<p>
  <div class="equation_content">
    \begin{equation}  \delta ^{m}=\left( \tfrac {\theta T^{m}}{N^{\left( m+1\right) \left( 1-\alpha \right) }}\right) ^{\frac{m}{m+1}}.\label{121}\end{equation}
  </div>
  <span class="equation_label">3.120</span>
</p>
</div>
<p>Therefore we have</p>
<div class="displaymath" id="a0000000064">
  \begin{align} & \left\vert C_{N,\alpha }^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right\vert \overset {\text{(\ref{116})}}{\leq }\nonumber \\ & \overset {\text{(\ref{116})}}{\leq }\tfrac {1}{\Gamma \left( m+1\right) }\omega _{1}\left( D_{x}^{m}f,\left( \tfrac {\theta T^{m}}{N^{\left( m+1\right) \left( 1-\alpha \right) }}\right) ^{\frac{1}{m+1}}\right) _{\mathbb {R}}\cdot \label{122}\left[ \tfrac {\theta T^{m-1}}{N^{m\left( 1-\alpha \right) }}\! +\! \tfrac {1}{\left( m+1\right) \delta }\tfrac {\theta T^{m}}{N^{\left( m+1\right) \left( 1-\alpha \right) }}\right] \nonumber \\ & =\tfrac {1}{\Gamma \left( m+1\right) }\omega _{1}\left( D_{x}^{m}f,\left( \tfrac {\theta T^{m}}{N^{\left( m+1\right) \left( 1-\alpha \right) }}\right) ^{\frac{1}{m+1}}\right) _{_{\mathbb {R}}}\left[ \tfrac {\theta T^{m-1}}{N^{m\left( 1-\alpha \right) }}+\tfrac {1}{\left( m+1\right) \delta }\delta ^{m+1}\right] \nonumber  \\ &  \overset {\text{(\ref{121})}}{\! \! =\! \! }\tfrac {1}{\Gamma \left( m+1\right) }\omega _{1}\left( D_{x}^{m}f,\big( \tfrac {\theta T^{m}}{N^{\left( m+1\right) \left( 1-\alpha \right) }}\big) ^{\frac{1}{m+1}}\right)_{\mathbb {R}} \left[ \tfrac {\theta T^{m-1}}{N^{m\left( 1-\alpha \right) }}\! +\! \tfrac {1}{\left( m+1\right) }\big( \tfrac {\theta T^{m}}{N^{\left( m+1\right) \left( 1-\alpha \right) }}\big) ^{\frac{m}{m+1}}\right] \end{align}
</div>
<p> \(\forall \) \(N\in \mathbb {N}:N{\gt}\max \left\{  T+\left\vert x\right\vert ,\left( \frac{2}{T}\right) ^{\frac{1}{\alpha }}\right\} , \) proving the inequality (<a href="#120">3.119</a>). <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> </p>
<p>We finish with (case of \(\alpha =1.5\)) </p>
<p><div class="corollary_thmwrapper " id="c53">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">3.37</span>
  </div>
  <div class="corollary_thmcontent">
  <p>Let \(x\in \left[ 0,1\right] \), \(f\in AC^{2}\left( \left[ 0,1\right] ,\mathbb {R}_{+}\right) \) and \(f^{\left( 2\right) }\in L_{\infty }\left( \left[ 0,1\right] \right) \). Assume that \(f^{\prime }\left( x\right) =0\). Then </p>
<div class="displaymath" id="124">
  \begin{align}  \left| B_{N}^{\left( M\right) }\left( f\right) \left( x\right) -f\left( x\right) \right| \leq \tfrac {4\omega _{1}\Big( D_{x}^{1.5}f,\big( \frac{6}{\sqrt{N+1}}\big) ^{\frac{2}{5}}\Big) }{3\sqrt{\pi }} \left[ \tfrac {6}{\sqrt{N+1}}+\tfrac {2}{5}\Big( \tfrac {6}{\sqrt{N+1}}\Big) ^{\frac{3}{5}}\right] , \label{124}\end{align}
</div>
<p> \(\forall \text{ }N\in \mathbb {N}\text{.}\) </p>

  </div>
</div> </p>
<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> By Theorem <a href="#t30">3.14</a>, apply (<a href="#75">3.75</a>). <div class="proof_wrapper" id="a0000000067">
  <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>Due to lack of space we do not give other example applications. </p>
<p><small class="footnotesize">  </small></p>
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</a> </p>
</dd>
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</dd>
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</dd>
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</dd>
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</dd>
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</dd>
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</dd>
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</dd>
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</dd>
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  <dd><p><i class="sc">G.S. Frederico, D.F.M. Torres</i>, <i class="it">Fractional Optimal Control in the sense of Caputo and the fractional Noether’s theorem</i>, International Mathematical Forum, vol. <b class="bf">3</b> (2008), no. 10, pp.&#160;479–493. </p>
</dd>
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</dd>
  <dt><a name="12">12</a></dt>
  <dd><p><i class="sc">T. Popoviciu</i>, <i class="it">Sur l’approximation de fonctions convexes d’order superieur</i>, Mathematica (Cluj), <b class="bf">10</b> (1935), pp.&#160;49–54. </p>
</dd>
  <dt><a name="13">13</a></dt>
  <dd><p><i class="sc">S.G. Samko, A.A. Kilbas, O.I. Marichev</i>, <i class="it">Fractional Integrals and Derivatives, Theory and Applications</i>, (Gordon and Breach, Amsterdam, 1993) [English translation from the Russian, Integrals and Derivatives of Fractional Order and Some of Their Applications (Nauka i Tekhnika, Minsk, 1987)]. </p>
</dd>
</dl>


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