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<h1>High order Approximation theory <br />for Banach space valued functions</h1>
<p class="authors">
<span class="author">George A. Anastassiou\(^\ast \)</span>
</p>
<p class="date">February 17, 2017.</p>
</div>
<p>\(^\ast \)Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152, USA, e-mail: <span class="tt">ganastss@memphis.edu</span>. </p>

<div class="abstract"><p> Here we study quantitatively the high degree of approximation of sequences of linear operators acting on Banach space valued differentiable functions to the unit operator. These operators are bounded by real positive linear companion operators. The Banach spaces considered here are general and no positivity assumption is made on the initial linear operators whose we study their approximation properties. We derive pointwise and uniform estimates which imply the approximation of these operators to the unit assuming differentiability of functions. At the end we study the special case where the high order derivative of the on hand function fulfills a convexity condition resulting into sharper estimates. </p>
<p><b class="bf">MSC.</b> 41A17, 41A25, 41A36. </p>
<p><b class="bf">Keywords.</b> Banach space valued differentiable functions, positive linear operator, convexity, modulus of continuity, rate of convergence. </p>
</div>
<h1 id="a0000000002">1 Motivation</h1>
<p>Let \(\left( X,\left\Vert \cdot \right\Vert \right) \) be a Banach space, \(N\in \mathbb {N}\). Consider \(g\in C\left( \left[ 0,1\right] \right) \) and the classic Bernstein polynomials </p>
<div class="equation" id="1">
<p>
  <div class="equation_content">
    \begin{equation}  \big( \widetilde{B}_{N}g\big) \left( t\right) =\sum _{k=0}^{N}g\big( \tfrac {k}{N}\big) \tbinom {N}{k}t^{k}\left( 1-t\right) ^{N-k},\text{ \  }\forall \text{ }t\in \left[ 0,1\right] . \label{1}\end{equation}
  </div>
  <span class="equation_label">1.1</span>
</p>
</div>
<p> Let also \(f\in C\left( \left[ 0,1\right] ,X\right) \) and define the vector valued in \(X\) Bernsein linear operators </p>
<div class="equation" id="2">
<p>
  <div class="equation_content">
    \begin{equation}  \big( B_{N}f\big) \left( t\right) =\sum _{k=0}^{N}f\big(\tfrac {k}{N}\big) \tbinom {N}{k}t^{k}\left( 1-t\right) ^{N-k},\text{ \  }\forall \text{ }t\in \left[ 0,1\right] . \label{2}\end{equation}
  </div>
  <span class="equation_label">1.2</span>
</p>
</div>
<p> That is \(\left( B_{N}f\right) \left( t\right) \in X.\) </p>
<p>Clearly here \(\left\|  f\right\|  \in C\left( \left[ 0,1\right] \right) \). </p>
<p>We notice that </p>
<div class="equation" id="3">
<p>
  <div class="equation_content">
    \begin{equation}  \left\Vert \left( B_{N}f\right) \left( t\right) \right\Vert \leq \sum _{k=0}^{N}\big\Vert f\big( \tfrac {k}{N}\big) \big\Vert \tbinom {N}{k}t^{k}\left( 1-t\right) ^{N-k}=\big( \widetilde{B}_{N}\left( \left\Vert f\right\Vert \right) \big) \left( t\right) ,\forall t\in \left[ 0,1\right] \label{3}\end{equation}
  </div>
  <span class="equation_label">1.3</span>
</p>
</div>
<p>The property </p>
<div class="equation" id="4">
<p>
  <div class="equation_content">
    \begin{equation}  \left\|  \left( B_{N}f\right) \left( t\right) \right\|  \leq \big( \widetilde{B}_{N}\left( \left\|  f\right\|  \right) \big) \left( t\right) ,\text{ \  }\forall t\in \left[ 0,1\right] , \label{4}\end{equation}
  </div>
  <span class="equation_label">1.4</span>
</p>
</div>
<p> is shared by almost all summation/integration similar operators and motivates our work here. </p>
<p>If \(f\left( x\right) =c\in X\) the constant function, then </p>
<div class="equation" id="5">
<p>
  <div class="equation_content">
    \begin{equation}  \left( B_{N}c\right) =c. \label{5}\end{equation}
  </div>
  <span class="equation_label">1.5</span>
</p>
</div>
<p> If \(g\in C\left( \left[ 0,1\right] \right) \) and \(c\in X\), then \(cg\in C\left( \left[ 0,1\right] ,X\right) \) and </p>
<div class="equation" id="6">
<p>
  <div class="equation_content">
    \begin{equation}  \left( B_{N}\left( cg\right) \right) =c\widetilde{B}_{N}\left( g\right) . \label{6}\end{equation}
  </div>
  <span class="equation_label">1.6</span>
</p>
</div>
<p> Again (<a href="#5">1.5</a>), (<a href="#6">1.6</a>) are fulfilled by many summation/integration operators. </p>
<p>In fact here (<a href="#6">1.6</a>) implies (<a href="#5">1.5</a>), when \(g\equiv 1\). </p>
<p>The above can be generalized from \(\left[ 0,1\right] \) to any interval \(\left[ a,b\right] \subset \mathbb {R}\). All this discussion motivates us to consider the following situation. </p>
<p>Let \(L_{N}:C\left( \left[ a,b\right] ,X\right) \hookrightarrow C\left( \left[ a,b\right] ,X\right) \), \(\left( X,\left\|  \cdot \right\|  \right) \) a Banach space, \(L_{N}\) is a linear operator, \(\forall \) \(N\in \mathbb {N}\), \(x_{0}\in \left[ a,b\right] \). Let also \(\widetilde{L}_{N}:C\left( \left[ a,b\right] \right) \hookrightarrow C\left( \left[ a,b\right] \right) \), a sequence of positive linear operators, \(\forall \) \(N\in \mathbb {N}\). </p>
<p>We assume that </p>
<div class="equation" id="7">
<p>
  <div class="equation_content">
    \begin{equation}  \left\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) \right\|  \leq \big( \widetilde{L}_{N}\left( \left\|  f\right\|  \big) \right) \left( x_{0}\right) ,\text{ \  } \label{7}\end{equation}
  </div>
  <span class="equation_label">1.7</span>
</p>
</div>
<p> \(\forall \) \(N\in \mathbb {N}\), \(\forall \) \(x_{0}\in X\), \(\forall \) \(f\in C\left( \left[ a,b\right] ,X\right) .\) </p>
<p>When \(g\in C\left( \left[ a,b\right] \right) \), \(c\in X\), we assume that </p>
<div class="equation" id="8">
<p>
  <div class="equation_content">
    \begin{equation}  \left( L_{N}\left( cg\right) \right) =c\widetilde{L}_{N}\left( g\right) . \label{8}\end{equation}
  </div>
  <span class="equation_label">1.8</span>
</p>
</div>
<p> The special case of </p>
<div class="equation" id="9.">
<p>
  <div class="equation_content">
    \begin{equation}  \widetilde{L}_{N}\left( 1\right) =1, \label{9.}\end{equation}
  </div>
  <span class="equation_label">1.9</span>
</p>
</div>
<p> implies </p>
<div class="equation" id="10">
<p>
  <div class="equation_content">
    \begin{equation}  L_{N}\left( c\right) =c,\text{ \  }\forall \text{ }c\in X. \label{10}\end{equation}
  </div>
  <span class="equation_label">1.10</span>
</p>
</div>
<p> We call \(\widetilde{L}_{N}\) the companion operator of \(L_{N}.\) </p>
<p>Based on the above fundamental properties we study the high order approximation properties of the sequence of linear operators \(\left\{  L_{N}\right\}  _{N\in \mathbb {N}}\), i.e. their high speed convergence to the unit operator. No kind of positivity property of \(\left\{  L_{N}\right\}  _{N\in \mathbb {N}}\) is assumed. Other important motivation comes from <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span>, <span class="cite">
	[
	<a href="#2" >2</a>
	]
</span>, <span class="cite">
	[
	<a href="#3" >3</a>
	]
</span>-<span class="cite">
	[
	<a href="#6" >6</a>
	]
</span>. </p>
<p>Our vector valued differentiation here resembles completely the numerical one, see <span class="cite">
	[
	<a href="#7" >7</a>
	, 
	pp. 83-84
	]
</span>. </p>
<h1 id="a0000000003">2 Main Results</h1>
<p>We need vector Taylor’s formula </p>
<p><div class="theorem_thmwrapper " id="t1.">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2.1</span>
  </div>
  <div class="theorem_thmcontent">
  <p><span class="rm"><p><span class="cite">
	[
	<a href="#7" >7</a>
	, 
	pp. 93
	]
</span>. Let \(n\in \mathbb {N}\) and \(f\in C^{n}\left( \left[ a,b\right] ,X\right) \), where \(\left[ a,b\right] \subset \mathbb {R}\) and \(X\) is a Banach space. Then </p>
<div class="equation" id="11.">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( b\right) =\sum _{i=0}^{n-1}\tfrac {\left( b-a\right) ^{i}}{i!}f^{\left( i\right) }\left( a\right) +\tfrac {1}{\left( n-1\right) !}\int _{a}^{b}\left( b-t\right) ^{n-1}f^{\left( n\right) }\left( t\right) dt. \label{11.}\end{equation}
  </div>
  <span class="equation_label">2.11</span>
</p>
</div>
</span></p>

  </div>
</div> </p>
<p>Above the integral is the usual vector valued Riemann integral, see <span class="cite">
	[
	<a href="#7" >7</a>
	, 
	p. 86
	]
</span>. </p>
<p>We also need </p>
<p><div class="theorem_thmwrapper " id="t2.">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2.2</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(n\in \mathbb {N}\) and \(f\in C^{n}\left( \left[ a,b\right] ,X\right) \), where \(\left[ a,b\right] \subset \mathbb {R}\) and \(X\) is a Banach space. Then </p>
<div class="equation" id="12.">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( a\right) =\sum _{i=0}^{n-1}\tfrac {\left( a-b\right) ^{i}}{i!}f^{\left( i\right) }\left( b\right) +\tfrac {1}{\left( n-1\right) !}\int _{b}^{a}\left( a-t\right) ^{n-1}f^{\left( n\right) }\left( t\right) dt. \label{12.}\end{equation}
  </div>
  <span class="equation_label">2.12</span>
</p>
</div>

  </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> Let </p>
<div class="displaymath" id="a0000000005">
  \[  F\left( x\right) :=\sum _{i=0}^{n-1}\tfrac {\left( a-x\right) ^{i}}{i!}f^{\left( i\right) }\left( x\right) ,\text{ \  }x\in \left[ a,b\right] .  \]
</div>
<p> Here \(F\in C\left( \left[ a,b\right] ,X\right) .\) Notice that \(F\left( a\right) =f\left( a\right) \), and </p>
<div class="equation" id="13.">
<p>
  <div class="equation_content">
    \begin{equation}  F\left( b\right) =\sum _{i=0}^{n-1}\tfrac {\left( a-b\right) ^{i}}{i!}f^{\left( i\right) }\left( b\right) . \label{13.}\end{equation}
  </div>
  <span class="equation_label">2.13</span>
</p>
</div>
<p> We have </p>
<div class="displaymath" id="a0000000006">
  \[  F^{\prime }\left( x\right) =\tfrac {\left( a-x\right) ^{n-1}}{\left( n-1\right) !}f^{\left( n\right) }\left( x\right) ,\text{ \  }\forall \text{ }x\in \left[ a,b\right] .  \]
</div>
<p> Clearly \(F^{\prime }\in C\left( \left[ a,b\right] ,X\right) .\) </p>
<p>By <span class="cite">
	[
	<a href="#7" >7</a>
	, 
	pp. 92
	]
</span> we get </p>
<div class="equation" id="14.">
<p>
  <div class="equation_content">
    \begin{equation}  F\left( b\right) -F\left( a\right) =\int _{a}^{b}F^{\prime }\left( t\right) dt. \label{14.}\end{equation}
  </div>
  <span class="equation_label">2.14</span>
</p>
</div>
<p> That is we have </p>
<div class="equation" id="15.">
<p>
  <div class="equation_content">
    \begin{equation}  \sum _{i=0}^{n-1}\tfrac {\left( a-b\right) ^{i}}{i!}f^{\left( i\right) }\left( b\right) -f\left( a\right) =\int _{a}^{b}\tfrac {\left( a-t\right) ^{n-1}}{\left( n-1\right) !}f^{\left( n\right) }\left( t\right) dt= -\int _{b}^{a}\tfrac {\left( a-t\right) ^{n-1}}{\left( n-1\right) !}f^{\left( n\right) }\left( t\right) dt, \label{15.}\end{equation}
  </div>
  <span class="equation_label">2.15</span>
</p>
</div>
<p> proving (<a href="#12.">2.12</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>Based on the above Theorems <a href="#t1.">2.1</a>, <a href="#t2.">2.2</a>, we have </p>
<p><div class="corollary_thmwrapper " id="c3.">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">2.3</span>
  </div>
  <div class="corollary_thmcontent">
  <p>Let \(\left( X,\left\Vert \cdot \right\Vert \right) \) be a Banach space and \(f\in C^{n}\left( \left[ a,b\right] ,X\right) \), then we have the vector valued Taylor’s formula </p>
<div class="equation" id="16.">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( y\right) -\sum _{i=0}^{n-1}f^{\left( i\right) }\left( x\right) \tfrac {\left( y-x\right) ^{i}}{i!}=\tfrac {1}{\left( n-1\right) !}\int _{x}^{y}\left( y-t\right) ^{n-1}f^{\left( n\right) }\left( t\right) dt, \label{16.}\end{equation}
  </div>
  <span class="equation_label">2.16</span>
</p>
</div>
<p> \(\forall \) \(x,y\in \left[ a,b\right] \) or </p>
<div class="equation" id="17.">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( y\right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x\right) \tfrac {\left( y-x\right) ^{i}}{i!}=\tfrac {1}{\left( n-1\right) !}\int _{x}^{y}\left( y-t\right) ^{n-1}\big( f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x\right) \big) dt, \label{17.}\end{equation}
  </div>
  <span class="equation_label">2.17</span>
</p>
</div>
<p> \(\forall \) \(x,y\in \left[ a,b\right] .\) </p>

  </div>
</div> </p>
<p>We need </p>
<p><div class="definition_thmwrapper " id="d4.">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">2.4</span>
  </div>
  <div class="definition_thmcontent">
  <p>Let \(f\in C\left( \left[ a,b\right] ,X\right) ,\) where \(\left( X,\left\|  \cdot \right\|  \right) \) is a Banach space. We define </p>
<div class="equation" id="18.">
<p>
  <div class="equation_content">
    \begin{equation}  \omega _{1}\left( f,\delta \right) :=\underset {\left| x-y\right| \leq \delta }{\underset {x,y:}{\sup }}\left\|  f\left( x\right) -f\left( y\right) \right\|  \text{, \  }0<\delta \leq b-a, \label{18.}\end{equation}
  </div>
  <span class="equation_label">2.18</span>
</p>
</div>
<p> the first modulus of continuity of \(f\). </p>

  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="r5.">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">2.5</span>
  </div>
  <div class="remark_thmcontent">
  <p>We study the remainder of (<a href="#17.">2.17</a>): </p>
<div class="equation" id="19.">
<p>
  <div class="equation_content">
    \begin{equation}  R_{n}\left( x,y\right) :=\tfrac {1}{\left( n-1\right) !}\int _{x}^{y}\left( y-t\right) ^{n-1}\big( f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x\right) \big) dt, \label{19.}\end{equation}
  </div>
  <span class="equation_label">2.19</span>
</p>
</div>
<p> \(\forall \) \(x,y\in \left[ a,b\right] .\) </p>
<p>We estimate \(R_{n}\left( x,y\right) \). </p>
<p>Case of \(y\geq x\). We have </p>
<div class="displaymath" id="a0000000008">
  \begin{align}  \left\Vert R_{n}\left( x,y\right) \right\Vert & =\tfrac {1}{\left( n-1\right) !}\Big\Vert \int _{x}^{y}\left( y-t\right) ^{n-1}\big( f^{\left( n\right) }\left( t\right) -f^{\left(n\right)}\left( x\right) \big) dt\Big\Vert \overset {\text{\cite[p. 88]{7}}}{\leq }\nonumber \\ & \leq \tfrac {1}{\left( n-1\right) !}\int _{x}^{y}\left( y-t\right) ^{n-1}\left\Vert f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x\right) \right\Vert dt\label{20.}\\ & \leq \tfrac {1}{\left( n-1\right) !}\int _{x}^{y}\left( y-t\right) ^{n-1}\omega _{1}\big( f^{\left( n\right) },\left\vert t-x\right\vert \big) dt \overset {\text{let}\  h{\gt}0}{=}\nonumber \\ & =\tfrac {1}{\left( n-1\right) !}\int _{x}^{y}\left( y-t\right) ^{n-1}\omega _{1}\big( f^{\left( n\right) },\tfrac {\left\vert t-x\right\vert }{h}h\big) dt\nonumber \\ & \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n-1\right) !}\int _{x}^{y}\left( y-t\right) ^{n-1}\big( 1+\tfrac {\left\vert t-x\right\vert }{h}\big) dt &  \label{21.}\\ & =\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n-1\right) !}\int _{x}^{y}\left( y-t\right) ^{n-1}\big( 1+\tfrac {\left( t-x\right) }{h}\big) dt\nonumber \\ & =\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n-1\right) !}\Big[ \int _{x}^{y}\left( y-t\right) ^{n-1}dt+\tfrac {1}{h}\int _{x}^{y}\left( y-t\right) ^{n-1}\left( t-x\right) ^{2-1}dt\Big] \nonumber \\ & =\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n-1\right) !}\Big[ \tfrac {\left( y-x\right) ^{n}}{n}+\tfrac {1}{h}\tfrac {\Gamma \left( n\right) \Gamma \left( 2\right) }{\Gamma \left( n+2\right) }\left( y-x\right) ^{n+1}\Big] \label{22}\\ & =\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n-1\right) !}\left[ \tfrac {\left( y-x\right) ^{n}}{n}+\tfrac {\left( y-x\right) ^{n+1}}{n\left( n+1\right) h}\right] =\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{n!}\left( y-x\right) ^{n}\left[ 1+\tfrac {\left( y-x\right) }{\left( n+1\right) h}\right] .\nonumber \end{align}
</div>
<p> We have found that </p>
<div class="equation" id="23.">
<p>
  <div class="equation_content">
    \begin{equation}  \left\Vert R_{n}\left( x,y\right) \right\Vert \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{n!}\left( y-x\right) ^{n}\Big[ 1+\tfrac {\left( y-x\right) }{\left( n+1\right) h}\Big] , \label{23.}\end{equation}
  </div>
  <span class="equation_label">2.23</span>
</p>
</div>
<p> for \(y\geq x\), and \(h{\gt}0.\) </p>
<p>Case of \(y\leq x\). </p>
<p>Then </p>
<div class="displaymath" id="a0000000009">
  \begin{align}  \left\|  R_{n}\left( x,y\right) \right\|  & =\tfrac {1}{\left( n-1\right) !}\left\|  \int _{x}^{y}\left( y-t\right) ^{n-1}\big( f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x\right) \big) dt\right\| \nonumber \\ & =\tfrac {1}{\left( n-1\right) !}\left\|  \int _{y}^{x}\left( t-y\right) ^{n-1}\big( f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x\right) \big) dt\right\|  \label{24.}\\ & \leq \tfrac {1}{\left( n-1\right) !}\int _{y}^{x}\left( t-y\right) ^{n-1}\left\|  f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x\right) \right\|  dt\nonumber \\ & \leq \tfrac {1}{\left( n-1\right) !}\int _{y}^{x}\left( t-y\right) ^{n-1}\omega _{1}\big( f^{\left( n\right) },\tfrac {\left| t-x\right| }{h}h\big) dt\nonumber \\ & \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n-1\right) !}\int _{y}^{x}\left( t-y\right) ^{n-1}\big( 1+\tfrac {\left( x-t\right) }{h}\big) dt\nonumber \\ & =\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n-1\right) !}\left[ \int _{y}^{x}\left( t-y\right) ^{n-1}dt+\tfrac {1}{h}\int _{y}^{x}\left( x-t\right) ^{2-1}\left( t-y\right) ^{n-1}dt\right] \label{25.}\\ & =\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n-1\right) !}\left[ \tfrac {\left( x-y\right) ^{n}}{n}+\tfrac {\left( x-y\right) ^{n+1}}{n\left( n+1\right) h}\right] \nonumber \\ & =\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{n!}\left( x-y\right) ^{n}\Big[ 1+\tfrac {\left( x-y\right) }{\left( n+1\right) h}\Big] .\nonumber \end{align}
</div>
<p> Hence </p>
<div class="equation" id="26">
<p>
  <div class="equation_content">
    \begin{equation}  \left\|  R_{n}\left( x,y\right) \right\|  \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{n!}\left( x-y\right) ^{n}\Big[ 1+\tfrac {\left( x-y\right) }{\left( n+1\right) h}\Big] , \label{26}\end{equation}
  </div>
  <span class="equation_label">2.26</span>
</p>
</div>
<p> when \(y\leq x\), \(h{\gt}0\). </p>
<p>We have proved that </p>
<div class="displaymath" id="a0000000010">
  \begin{align}  \left\|  R_{n}\left( x,y\right) \right\|  & =\left\|  \tfrac {1}{\left( n-1\right) !}\int _{x}^{y}\left( y-t\right) ^{n-1}\big( f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x\right) \big) dt\right\|  \leq \nonumber \\ & \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{n!}\left| x-y\right| ^{n}\left[ 1+\tfrac {\left| x-y\right| }{\left( n+1\right) h}\right] , \label{27.}\end{align}
</div>
<p> \(\forall \) \(x,y\in \left[ a,b\right] \), \(h{\gt}0\). </p>

  </div>
</div> </p>
<p>We have established </p>
<p><div class="theorem_thmwrapper " id="t6.">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2.6</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(\left( X,\left\|  \cdot \right\|  \right) \) be a Banach space and \(f\in C^{n}\left( \left[ a,b\right] ,X\right) \), \(n\in \mathbb {N}\). Then </p>
<div class="equation" id="28.">
<p>
  <div class="equation_content">
    \begin{equation}  \bigg\|  f\left( y\right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x\right) \tfrac {\left( y-x\right) ^{i}}{i!}\bigg\|  \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{n!}\left| x-y\right| ^{n}\left[ 1+\tfrac {\left| x-y\right| }{\left( n+1\right) h}\right] , \label{28.}\end{equation}
  </div>
  <span class="equation_label">2.28</span>
</p>
</div>
<p> \(\forall \) \(x,y\in \left[ a,b\right] \), \(h{\gt}0\). </p>

  </div>
</div> </p>
<p>It follows our first main result </p>
<p><div class="theorem_thmwrapper " id="t7.">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2.7</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(N\in \mathbb {N}\) and \(L_{N}:C\left( \left[ a,b\right] ,X\right) \rightarrow C\left( \left[ a,b\right] ,X\right) \), where \(\left( X,\left\|  \cdot \right\|  \right) \) is a Banach space and \(L_{N}\) is a linear operator. Let the positive linear operators \(\widetilde{L}_{N}:C\left( \left[ a,b\right] \right) \hookrightarrow C\left( \left[ a,b\right] \right) \), such that </p>
<div class="equation" id="29.">
<p>
  <div class="equation_content">
    \begin{equation}  \left\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) \right\|  \leq \big( \widetilde{L}_{N}\left( \left\|  f\right\|  \right) \big) \left( x_{0}\right) ,\text{ \  } \label{29.}\end{equation}
  </div>
  <span class="equation_label">2.29</span>
</p>
</div>
<p> \(\forall \) \(N\in \mathbb {N},\) \(\forall \) \(f\in C\left( \left[ a,b\right] ,X\right) \), \(\forall \) \(x_{0}\in \left[ a,b\right] .\) </p>
<p>Furthermore assume that </p>
<div class="equation" id="30.">
<p>
  <div class="equation_content">
    \begin{equation}  L_{N}\left( cg\right) =c\widetilde{L}_{N}\left( g\right) ,\text{ \  }\forall \text{ }g\in C\left( \left[ a,b\right] \right) \text{, }\forall \text{ }c\in X. \label{30.}\end{equation}
  </div>
  <span class="equation_label">2.30</span>
</p>
</div>
<p> Let \(n\in \mathbb {N}\), here we deal with \(f\in C^{n}\left( \left[ a,b\right] ,X\right) .\) </p>
<p>Then </p>
<p>1) </p>
<div class="displaymath" id="a0000000011">
  \begin{align} & \bigg\Vert \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -\sum _{i=0}^{n}\tfrac {f^{n\left( i\right) }\left( x_{0}\right) }{i!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{i}\big) \big) \left( x_{0}\right) \bigg\Vert \leq \nonumber \\ & \leq \tfrac {\omega _{1}\big( f^{(n)},\big( \big( \widetilde{L}_{N}\left( \left\vert \cdot -x_{0} \right\vert ^{n+1}\right) \big) \left( x_{0}\right) \big) ^{\frac{1}{n+1}}\big) }{n!} \left( \big( \widetilde{L}_{N}\big( \left\vert \cdot -x_{0}\right\vert ^{n+1}\big) \big) \left( x_{0}\right) \right) ^{\left( \frac{n}{n+1}\right) }\times \nonumber \\ & \quad \times \Big[ \big( \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \big) ^{\frac{1}{n+1}}+\tfrac {1}{n+1}\Big] , \label{31.}\end{align}
</div>
<p>2) </p>
<div class="displaymath" id="a0000000012">
  \begin{align} & \left\Vert \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\Vert \leq \nonumber \\ & \leq \left\Vert f\left( x_{0}\right) \right\Vert \big\vert \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) -1\big\vert +\nonumber \\ & \quad +\sum _{k=0}^{n} \tfrac {\left\Vert f^{(k)}(x_{0}) \right\Vert }{k!} \big( \big( \widetilde{L}_{N}( \vert \cdot -x_{0}\vert ^{k}) \big) (x_{0}) \big) +\nonumber \\ & \quad +\frac{\omega _{1}\big( f^{\left( n\right) },\big( \big( \widetilde{L}_{N}\big( \left\vert \cdot -x_{0}\right\vert ^{n+1}\big) \big) \left( x_{0}\right) \big) ^{\frac{1}{n+1}}\big) }{n!}\big( \big( \widetilde{L}_{N}\big( \big\vert \cdot -x_{0}\big\vert ^{n+1}\big) \big) \left( x_{0}\right) \big) ^{\left( \frac{n}{n+1}\right) }\times \nonumber \\ & \quad \times \Big[ \big( \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \big) ^{\frac{1}{n+1}}+\tfrac {1}{n+1}\Big] , \label{32.}\end{align}
</div>
<p> from <span class="rm">(<a href="#32.">2.32</a>)</span>, and as \(\big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \rightarrow 1\), \(\big( \widetilde{L}_{N}\left( \big\vert \cdot -x_{0}\right\vert ^{n+1}\big) \big) \left( x_{0}\right) \rightarrow 0\), we obtain \(\left( L_{N}\left( f\right) \right) \left( x_{0}\right) \rightarrow f\left( x_{0}\right) \), as \(N\rightarrow \infty ,\) </p>
<p>3) if \(f^{\left( k\right) }\left( x_{0}\right) =0\), \(k=0,1,...,n,\) we get that </p>
<div class="displaymath" id="a0000000013">
  \begin{align} &  \left\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\|  \leq \nonumber \\ & \leq \tfrac {\omega _{1}\big( f^{\left( n\right) },\left( \left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) ^{\frac{1}{n+1}}\big) }{n!}\times \nonumber \\ & \quad \times \big( \big(\widetilde{L}_{N}(| \cdot -x_{0}|^{n+1}) \big) (x_{0}) \big) ^{(\frac{n}{n+1})} \Big[ \big( \big( \widetilde{L}_{N}(1) \big) (x_{0}) \big) ^{\frac{1}{n+1}}+t\tfrac {1}{n+1} \Big] , \label{33.}\end{align}
</div>
<p> an extreme high speed of convergence, </p>
<p>4) one also derives </p>
<div class="displaymath" id="a0000000014">
  \begin{align} & \left\|  \left\|  \left( L_{N}\left( f\right) \right) -f\right\|  \right\|  _{\infty ,\left[ a,b\right] }\leq \nonumber \\ & \leq \left\|  \left\|  f\right\|  \right\|  _{\infty ,\left[ a,b\right] }\left\|  \big( \widetilde{L}_{N}\left( 1\right) \big) -1\right\|  _{\infty ,\left[ a,b\right]}+\nonumber \\ & \quad +\sum _{k=0}^{n}\tfrac {\left\|  \left\|  f^{\left( k\right) }\right\|  \right\|  _{\infty ,\left[ a,b\right] }}{k!}\left\|  \big( \widetilde{L}_{N}\big( | \cdot -x_{0}| ^{k}\big) \big) ( x_{0}) \right\|  _{\infty ,x_{0}\in \left[ a,b\right] }+\nonumber \\ & \quad +\tfrac {\omega _{1}\big( f^{\left( n\right) },\left\|  \left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right\|  _{\infty ,x_{0}\in \left[ a,b\right] }^{\frac{1}{n+1}}\big) }{n!} \times \nonumber \\ & \quad \times \left\|  \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \right\|  _{\infty ,x_{0}\in \left[ a,b\right] }^{\left( \frac{n}{n+1}\right) }\big[ \big\|  \widetilde{L}_{N}\left( 1\right) \big\|  ^{\frac{1}{n+1}}+\tfrac {1}{n+1}\big] , \label{34}\end{align}
</div>
<p> if \(\widetilde{L}_{N}\left( 1\right) \overset {u}{\rightarrow }1\), uniformly, and \(\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \overset {u}{\rightarrow }0\), uniformly in \(x_{0}\in \left[ a,b\right] \), by <span class="rm">(<a href="#34">2.34</a>)</span>, we obtain \(L_{N}\left( f\right) \overset {u}{\rightarrow }f\), uniformly, as \(N\rightarrow \infty .\) </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> 1) One can rewrite (<a href="#28.">2.28</a>) as follows </p>
<div class="equation" id="35.">
<p>
  <div class="equation_content">
    \begin{equation}  \Big\|  f\left( \cdot \right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x_{0}\right) \tfrac {\left( \cdot -x_{0}\right) ^{i}}{i!}\Big\|  \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{n!}\left[ \left| \cdot -x_{0}\right| ^{n}+\tfrac {\left| \cdot -x_{0}\right| ^{n+1}}{\left( n+1\right) h}\right] , \label{35.}\end{equation}
  </div>
  <span class="equation_label">2.35</span>
</p>
</div>
<p> for a fixed \(x_{0}\in \left[ a,b\right] \), \(h{\gt}0\). </p>
<p>We observe that (\(N\in \mathbb {N}\)) </p>
<div class="displaymath" id="a0000000016">
  \begin{align} & \bigg\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -\sum _{i=0}^{n}\tfrac {f^{\left( i\right) }\left( x_{0}\right) }{i!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{i}\big) \big) \left( x_{0}\right) \bigg\|  =\nonumber \\ & =\bigg\|  \big( L_{N}\Big[ f\left( \cdot \right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x_{0}\right) \tfrac {\left( \cdot -x_{0}\right) ^{i}}{i!}\Big] \big) \left( x_{0}\right) \bigg\|  \label{36}\\ & \leq \Big( \widetilde{L}_{N}\Big( \Big\|  f\left( \cdot \right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x_{0}\right) \tfrac {\left( \cdot -x_{0}\right) ^{i}}{i!}\Big\|  \Big) \Big) \left( x_{0}\right)\overset {\text{(by (\ref{35.}))}}{\leq }\nonumber \\ & \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{n!}\Big[ \big( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n}\right) \big) \left( x_{0}\right) +\tfrac {\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) }{\left( n+1\right) h}\Big] =:\left( \xi _{1}\right) . \label{37.}\end{align}
</div>
<p> Above notice that \(\big( f\left( \cdot \right) -\sum _{i=0}^{n}\frac{f^{\left( i\right) }\left( x_{0}\right) }{i!}\left( \cdot -x_{0}\right) ^{i}\big) \in C\left( \left[ a,b\right] ,X\right) \). </p>
<p>By Hölder’s inequality and Riesz representation theorem we obtain </p>
<div class="equation" id="38">
<p>
  <div class="equation_content">
    \begin{equation}  \big( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n}\right) \big) \left( x_{0}\right) \leq \big( \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1} \big) \big) \left( x_{0}\right) \big) ^{\left( \frac{n}{n+1}\right) }\big( \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \big) ^{\frac{1}{\left( n+1\right) }}. \label{38}\end{equation}
  </div>
  <span class="equation_label">2.38</span>
</p>
</div>
<p> Therefore </p>
<div class="displaymath" id="a0000000017">
  \begin{align}  \left( \xi _{1}\right) & \leq \tfrac {\omega _{1}\big( f^{\left( n\right) },h\big) }{n!}\left[ \big( \big( \widetilde{L}_{N}\left( \big| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \big) ^{\left( \frac{n}{n+1}\right) }\big( \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \big) ^{\frac{1}{\left( n+1\right) }}\right.\nonumber \\ & \quad \left. +\tfrac {1}{h}\tfrac {\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) }{\left( n+1\right) }\right] =:\left( \xi _{2}\right) . \label{39.}\end{align}
</div>
<p> We choose </p>
<div class="equation" id="40.">
<p>
  <div class="equation_content">
    \begin{equation}  h:=\big( \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \big) ^{\frac{1}{\left( n+1\right) }}, \label{40.}\end{equation}
  </div>
  <span class="equation_label">2.40</span>
</p>
</div>
<p> in case of \(\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) {\gt}0.\) </p>
<p>Then it holds </p>
<div class="displaymath" id="41">
  \begin{align}  \left( \xi _{2}\right) & =\tfrac {\omega _{1}\Big( f^{\left( n\right) },\left( \left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) ^{\frac{1}{\left( n+1\right) }}\Big) }{n!} \label{41}\\ & \quad \times \bigg[ \big( \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \big) ^{\left( \frac{n}{n+1}\right) }\big( \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \big) ^{\frac{1}{\left( n+1\right) }}\nonumber \\ & \quad +\tfrac {\left( \left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) ^{\left( \frac{n}{n+1}\right) }}{\left( n+1\right) }\bigg] =\nonumber \\ & =\tfrac {\omega _{1}\Big( f^{\left( n\right) },\left( \left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) ^{\frac{1}{\left( n+1\right) }}\Big) }{n!}\nonumber \\ & \quad \times \big( \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \big) ^{\left( \frac{n}{n+1}\right) }\Big[ \big( \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \big) ^{\frac{1}{\left( n+1\right) }}+\tfrac {1}{n+1}\Big] . \label{42}\end{align}
</div>
<p> We have proved that </p>
<div class="displaymath" id="a0000000018">
  \begin{align} & \bigg\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -\sum _{i=0}^{n}\tfrac {f^{\left( i\right) }\left( x_{0}\right) }{i!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{i}\big) \big) \left( x_{0}\right) \bigg\|  \leq \nonumber \\ & \leq \tfrac {\omega _{1}\Big( f^{\left( n\right) },\left( \left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) ^{\frac{1}{\left( n+1\right) }}\Big) }{n!}\nonumber \\ & \quad \times \big( \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right)\big) ^{\left( \frac{n}{n+1}\right) }\Big[ \big( \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \big) ^{\frac{1}{\left( n+1\right) }}+\tfrac {1}{n+1}\Big] . \label{43}\end{align}
</div>
<p> By Riesz representation theorem we have </p>
<div class="equation" id="44">
<p>
  <div class="equation_content">
    \begin{equation}  \big( \widetilde{L}_{N}\left( g\right) \big) \left( x_{0}\right) =\int _{\left[ a,b\right] }g\left( t\right) d\mu _{x_{0}}\left( t\right) ,\text{ \  }\forall \text{ }g\in C\left( \left[ a,b\right] \right) , \label{44}\end{equation}
  </div>
  <span class="equation_label">2.44</span>
</p>
</div>
<p> where \(\mu _{x_{0}}\) is a positive finite measure on \(\left[ a,b\right] .\) </p>
<p>That is </p>
<div class="equation" id="45">
<p>
  <div class="equation_content">
    \begin{equation}  \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) =\mu _{x_{0}}\left( \left[ a,b\right] \right) =:M. \label{45}\end{equation}
  </div>
  <span class="equation_label">2.45</span>
</p>
</div>
<p>We have that \(\mu _{x_{0}}\left( \left[ a,b\right] \right) {\gt}0\), because otherwise, if \(\mu _{x_{0}}\left( \left[ a,b\right] \right) =0\), then \(\big( \widetilde{L}_{n}\left( g\right) \big) \left( x_{0}\right) =0\), \(\forall \) \(g\in C\left( \left[ a,b\right] \right) \), and the whole theory here becomes trivial. </p>
<p>Therefore it holds \(\big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) {\gt}0.\) </p>
<p>In case of </p>
<div class="equation" id="46">
<p>
  <div class="equation_content">
    \begin{equation}  \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) =0, \label{46}\end{equation}
  </div>
  <span class="equation_label">2.46</span>
</p>
</div>
<p> we have </p>
<div class="equation" id="47">
<p>
  <div class="equation_content">
    \begin{equation}  \int _{\left[ a,b\right] }\left| t-x_{0}\right| ^{n+1}d\mu _{x_{0}}\left( t\right) =0. \label{47}\end{equation}
  </div>
  <span class="equation_label">2.47</span>
</p>
</div>
<p> The last implies \(\left| t-x_{0}\right| ^{n+1}=0\), a.e, hence \(\left| t-x_{0}\right| =0\), a.e, then \(t-x_{0}=0\) a.e., and \(t=x_{0}\), a.e. on \(\left[ a,b\right] \). Consequently \(\mu _{x_{0}}\left( \left\{  t\in \left[ a,b\right] :t\neq x_{0}\right\}  \right) =0\). That is \(\mu _{x_{0}}=\delta _{x_{0}}M\), where \(\delta _{x_{0}}\) is the Dirac measure at \(\left\{  x_{0}\right\}  \). In that case holds </p>
<div class="equation" id="48">
<p>
  <div class="equation_content">
    \begin{equation}  \big( \widetilde{L}_{N}\left( g\right) \big) \left( x_{0}\right) =g\left( x_{0}\right) M, \qquad \forall g\in C\left( \left[ a,b\right] \right) . \label{48}\end{equation}
  </div>
  <span class="equation_label">2.48</span>
</p>
</div>
<p>Under (<a href="#46">2.46</a>), the right hand side of (<a href="#43">2.43</a>) equals zero. Furthermore it holds </p>
<div class="equation" id="49">
<p>
  <div class="equation_content">
    \begin{equation}  \Big( \widetilde{L}_{N}\Big(\Big\| f\left(\cdot \right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x_{0}\right) \tfrac {\left( \cdot -x_{0}\right) ^{i}}{i!}\Big\|  \Big) \Big) \left( x_{0}\right) \overset {\text{(\ref{48})}}{=}\left\|  f\left( x_{0}\right) -f\left( x_{0}\right) \right\|  M=0. \label{49}\end{equation}
  </div>
  <span class="equation_label">2.49</span>
</p>
</div>
<p> So that by (<a href="#36">2.36</a>) to have </p>
<div class="displaymath" id="a0000000019">
  \begin{align} & \bigg\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -\sum _{i=0}^{n}\tfrac {f^{\left( i\right) }\left( x_{0}\right) }{i!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{i}\big) \big) \left( x_{0}\right) \bigg\|  =\nonumber \\ & =\left\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) M\right\|  =0, \label{50}\end{align}
</div>
<p> also implying </p>
<div class="equation" id="51">
<p>
  <div class="equation_content">
    \begin{equation}  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) =Mf\left( x_{0}\right) . \label{51}\end{equation}
  </div>
  <span class="equation_label">2.51</span>
</p>
</div>
<p> So we have proved that inequality (<a href="#43">2.43</a>) will be always true. </p>
<p>2) Next we see that </p>
<div class="displaymath" id="a0000000020">
  \begin{align} & \big\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \big\|  =\nonumber \\ & =\left\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -\sum _{k=0}^{n}\tfrac {f^{\left( k\right) }\left( x_{0}\right) }{k!}\big( \widetilde{L}_{N}\left( \big( \cdot -x_{0}\right) ^{k}\big) \big) \left( x_{0}\right) +\right.\nonumber \\ & \quad +\left. \sum _{k=0}^{n}\tfrac {f^{\left( k\right) }\left( x_{0}\right) }{k!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{k}\big) \big) \left( x_{0}\right) -f\left( x_{0}\right) \right\|  \leq \label{52}\\ & \leq \bigg\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -\sum _{k=0}^{n}\tfrac {f^{\left( k\right) }\left( x_{0}\right) }{k!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{k}\big) \big) \left( x_{0}\right) \bigg\|  +\nonumber \\ & \quad +\bigg\|  \sum _{k=1}^{n}\tfrac {f^{\left( k\right) }\left( x_{0}\right) }{k!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{k}\big) \big) \left( x_{0}\right) +f\left( x_{0}\right) \big( \widetilde{L}_{N}\left( 1\big) \right) \left( x_{0}\right) -f\big( x_{0}\big) \bigg\|  \overset {\text{(\ref{43})}}{\leq }\nonumber \\ & \leq \left\|  f\left( x_{0}\right) \right\|  \big| \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) -1\big| +\sum _{k=1}^{n}\tfrac {\left\|  f^{\left( k\right) }\left( x_{0}\right) \right\|  }{k!}\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{k}\big) \big) \left( x_{0}\right) +\nonumber \\ & \quad +\tfrac {\omega _{1}\Big( f^{\left( n\right) },\left( \left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) ^{\frac{1}{\left( n+1\right) }}\Big) }{n!}\nonumber \\ & \quad \times \big( \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \big) ^{\left( \frac{n}{n+1}\right) }\Big[ \big( \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \big) ^{\frac{1}{\left( n+1\right) }}+\tfrac {1}{n+1}\Big] . \label{53}\end{align}
</div>
<p> We have proved that </p>
<div class="displaymath" id="a0000000021">
  \begin{align} & \left\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\| \nonumber \\ & \leq \left\|  f\left( x_{0}\right) \right\|  \left| \left( \widetilde{L}_{N}\left( 1\right) \right) \left( x_{0}\right) -1\right| +\nonumber \\ & \quad +\sum _{k=1}^{n}\tfrac {\left\|  f^{\left( k\right) }\left( x_{0}\right) \right\|  }{k!}\big( \widetilde{L}_{N}\big( \big| \cdot -x_{0}\big| ^{k}\big) \big) \left( x_{0}\right) +\nonumber \\ & \quad +\tfrac {\omega _{1}\big( f^{\left( n\right) },\left( \left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) ^{\frac{1}{\left( n+1\right) }}\big) }{n!} \label{54}\\ & \quad \times \big( \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \big) ^{\left( \frac{n}{n+1}\right) }\Big[ \big( \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \big) ^{\frac{1}{\left( n+1\right) }}+\tfrac {1}{n+1}\Big].\nonumber \end{align}
</div>
<p> By Hölder’s inequality for \(k=1,...,n,\) we obtain </p>
<div class="equation" id="55">
<p>
  <div class="equation_content">
    \begin{equation}  \big( \widetilde{L}_{N}\big( \big| \cdot -x_{0}\big| ^{k}\big) \big) \left( x_{0}\right) \leq \big( \big( \widetilde{L}_{N}\big( \big| \cdot -x_{0}\big| ^{n+1}\big) \big) \big( x_{0}\big) \big) ^{\left( \frac{k}{n+1}\right) }\big( \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \big) ^{\left( \frac{n+1-k}{n+1}\right) }. \label{55}\end{equation}
  </div>
  <span class="equation_label">2.55</span>
</p>
</div>
<p> Clealry by (<a href="#54">2.54</a>) and (<a href="#55">2.55</a>), when \(\big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \rightarrow 1\) and \(\big( \widetilde{L}_{N}\big( \big| \cdot -x_{0}\big| ^{n+1}\big) \big) \left( x_{0}\right) \rightarrow 0\), we obtain \(\left( L_{N}\left( f\right) \right) \left( x_{0}\right) \rightarrow f\left( x_{0}\right) \), as \(N\rightarrow \infty \). Notice that \(\big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \) will be bounded. </p>
<p>3) If \(f^{\left( k\right) }\left( x_{0}\right) =0\), \(k=0,1,...,n,\) we get that </p>
<div class="displaymath" id="a0000000022">
  \begin{align} & \left\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\|  \overset {\text{(54)}}{\leq }\nonumber \\ & \leq \tfrac {\omega _{1}\Big( f^{\left( n\right) },\left( \left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) ^{\frac{1}{\left( n+1\right) }}\Big) }{n!}\nonumber \\ & \quad \times \Big( \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \Big) ^{\left( \frac{n}{n+1}\right) }\Big[ \big( \big( \widetilde{L}_{N}\left( 1\big) \right) \left( x_{0}\right) \big) ^{\frac{1}{\left( n+1\right) }}+\tfrac {1}{n+1}\Big] , \label{56}\end{align}
</div>
<p> an extreme high speed of convergence. </p>
<p>4) One also derives from (<a href="#54">2.54</a>) that </p>
<div class="displaymath" id="a0000000023">
  \begin{align} & \left\|  \left\|  \left( L_{N}\left( f\right) \right) -f\right\|  \right\|  _{\infty ,\left[ a,b\right] }\leq \nonumber \\ & \leq \left\|  \left\|  f\right\|  \right\|  _{\infty ,\left[ a,b\right] }\left\|  \widetilde{L}_{N}\left( 1\right) -1\right\|  _{\infty ,\left[ a,b\right] }+\nonumber \\ & \quad +\sum _{k=1}^{n}\tfrac {\left\|  \left\|  f^{\left( k\right) }\right\|  \right\|  _{\infty ,\left[ a,b\right] }}{k!}\left\|  \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{k}\big) \big) \left( x_{0}\right) \right\|  _{\infty ,x_{0}\in \left[ a,b\right] }+\nonumber \\ & \quad +\tfrac {\omega _{1}\Big( f^{\left( n\right) },\left\|  \left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right\|  _{\infty ,x_{0}\in \left[ a,b\right] }^{\frac{1}{\left( n+1\right) }}\Big) }{n!}\nonumber \\ & \quad \times \left\|  \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \right\|  _{\infty ,x_{0}\in \left[ a,b\right] }^{\left( \frac{n}{n+1}\right) }\big[ \|  \widetilde{L}_{N}\left( 1\right) \|  ^{\frac{1}{\left( n+1\right) }}+\tfrac {1}{n+1}\big] . \label{57}\end{align}
</div>
<p> Inequality (<a href="#55">2.55</a>), for \(k=1,...,n\), implies </p>
<div class="displaymath" id="a0000000024">
  \begin{align} & \left\|  \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{k}\big) \big) \left( x_{0}\right) \right\|  _{\infty ,x_{0}\in \left[ a,b\right] }\nonumber \\ & \leq \left\|  \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \right\|  _{\infty ,x_{0}\in \left[ a,b\right] }^{\left( \frac{k}{n+1}\right) }\left\|  \widetilde{L}_{N}\left( 1\right) \right\|  _{\infty ,x_{0}\in \left[ a,b\right] }^{\left( \frac{n+1-k}{n+1}\right) }. \label{58}\end{align}
</div>
<p> Consequently, if \(\widetilde{L}_{N}\left( 1\right) \overset {u}{\rightarrow }1 \), uniformly, and \(\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \overset {u}{\rightarrow }0\), uniformly in \(x_{0}\in \left[ a,b\right] \), by (<a href="#57">2.57</a>) and (<a href="#58">2.58</a>), we obtain \(L_{N}\left( f\right) \overset {u}{\rightarrow }f\), uniformly, as \(N\rightarrow \infty .\) </p>
<p>Here the assumption \(\widetilde{L}_{N}\left( 1\right) \overset {u}{\rightarrow }1\), uniformly, as \(N\rightarrow \infty \), implies that \(\big\|  \widetilde{L}_{N}\left( 1\right) \big\|  \) is bounded. </p>
<p>The proof of the theorem now is complete. <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> </p>
<p>We make </p>
<p><div class="remark_thmwrapper " id="r9">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">2.8</span>
  </div>
  <div class="remark_thmcontent">
  <p>Let \(\left( X,\left\|  \cdot \right\|  \right) \) be a Banach space and \(f\in C^{n}\left( \left[ a,b\right] ,X\right) \), \(n\in \mathbb {N}\), and \(x_{0}\in \left( a,b\right) \) be fixed. Then </p>
<div class="displaymath" id="a0000000026">
  \begin{align}  f\left( y\right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x_{0}\right) \tfrac {\left( y-x_{0}\right) ^{i}}{i!}& =\tfrac {1}{\left( n-1\right) !}\int _{x_{0}}^{y}\left( y-t\right) ^{n-1}\big( f^{\left( n\right)} \left( t\right) -f^{\left( n\right) }\left( x_{0}\right) \big) dt\nonumber \\ & =:R_{n}\left( x_{0},y\right) \text{, \  }\forall \text{ }y\in \left[ a,b\right] \label{59}\end{align}
</div>
<p> We assume that \(g\left( t\right) :=\big\|  f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x_{0}\right) \big\|  \) is convex in \(t\in \left[ a,b\right] .\) </p>
<p>We consider \(0{\lt}h\leq \min \left( x_{0}-a,b-x_{0}\right) .\) Obviously \(g\left( x_{0}\right) =0\). Then by Lemma 8.1.1, p. 243 of <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span>, we obtain </p>
<div class="equation" id="60">
<p>
  <div class="equation_content">
    \begin{equation}  g\left( t\right) \leq \tfrac {\omega _{1}\left( g,h\right) }{h}\left| t-x_{0}\right| \text{, \  }\forall \text{ }t\in \left[ a,b\right] . \label{60}\end{equation}
  </div>
  <span class="equation_label">2.60</span>
</p>
</div>
<p> For any \(t_{1},t_{2}\in \left[ a,b\right] :\left| t_{1}-t_{2}\right| \leq h\) we get </p>
<div class="displaymath" id="a0000000027">
  \begin{align} & \Big| \big\|  f^{\left( n\right) }\left( t_{1}\right) -f^{\left( n\right) }\left( x_{0}\right) \big\|  -\big\|  f^{\left( n\right) }\left( t_{2}\right) -f^{\left( n\right) }\left( x_{0}\right) \big\|  \Big|\nonumber \\ & \leq \big\|  f^{\left( n\right) }\left( t_{1}\right) -f^{\left( n\right) }\left( t_{2}\right) \big\|  \leq \omega _{1}\left( f^{n},h\right) . \label{61}\end{align}
</div>
<p> That is </p>
<div class="equation" id="62">
<p>
  <div class="equation_content">
    \begin{equation}  \omega _{1}\left( g,h\right) \leq \omega _{1}\big( f^{\left( n\right) },h\big) . \label{62}\end{equation}
  </div>
  <span class="equation_label">2.62</span>
</p>
</div>
<p> The last implies </p>
<div class="equation" id="63">
<p>
  <div class="equation_content">
    \begin{equation}  \big\|  f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x_{0}\right) \big\|  \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{h}\left| t-x_{0}\right| \text{, \  \  }\forall \text{ }t\in \left[ a,b\right] . \label{63}\end{equation}
  </div>
  <span class="equation_label">2.63</span>
</p>
</div>
<p> We estimate \(R_{n}\left( x_{0},y\right) .\) </p>
<p>Case of \(y\geq x_{0}.\) We have </p>
<div class="displaymath" id="a0000000028">
  \begin{align}  \left\|  R_{n}\left( x_{0},y\right) \right\|  & =\tfrac {1}{\left( n-1\right) !}\left\|  \int _{x_{0}}^{y}\left( y-t\right) ^{n-1}\big( f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x_{0}\right) \big) dt\right\|  \overset {\text{(\cite[pp. 88]{7})}}{\leq }\nonumber \\ & \leq \tfrac {1}{\left( n-1\right) !}\int _{x_{0}}^{y}\left( y-t\right) ^{n-1}\left\|  f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x_{0}\right) \right\|  dt\overset {\text{(\ref{63})}}{\leq }\nonumber \\ & \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n-1\right) !h}\int _{x_{0}}^{y}\left( y-t\right) ^{n-1}\left( t-x_{0}\right) ^{2-1}dt=\nonumber \\ & =\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n-1\right) !h}\tfrac {\Gamma \left( n\right) \Gamma \left( 2\right) }{\Gamma \left( n+2\right) }\left( y-x_{0}\right) ^{n+1}=\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n+1\right) !h}\left( y-x_{0}\right) ^{n+1}. \label{64}\end{align}
</div>
<p> We proved that </p>
<div class="equation" id="65">
<p>
  <div class="equation_content">
    \begin{equation}  \left\|  R_{n}\left( x_{0},y\right) \right\|  \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{\left( n+1\right) !h}\left( y-x_{0}\right) ^{n+1},\text{ \  }\forall \text{ }y\geq x_{0}. \label{65}\end{equation}
  </div>
  <span class="equation_label">2.65</span>
</p>
</div>
<p>Case of \(y\leq x_{0}\). Then </p>
<div class="displaymath" id="a0000000029">
  \begin{align}  \left\|  R_{n}\left( x_{0},y\right) \right\|  & =\tfrac {1}{\left( n-1\right) !}\left\|  \int _{x_{0}}^{y}\left( y-t\right) ^{n-1}\big( f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x_{0}\right) \big) dt\right\|  \nonumber \\ & =\tfrac {1}{\left( n-1\right) !}\left\|  \int _{y}^{x_{0}}\left( t-y\right) ^{n-1}\big( f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x_{0}\right) \big) dt\right\|  \nonumber \\ & \leq \tfrac {1}{\left( n-1\right) !}\int _{y}^{x_{0}}\left( t-y\right) ^{n-1}\left\|  f^{\left( n\right) }\left( t\right) -f^{\left( n\right) }\left( x_{0}\right) \right\|  dt\overset {\text{(\ref{63})}}{\leq }\nonumber \\ & \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{h\left( n-1\right) !}\int _{y}^{x_{0}}\left( x_{0}-t\right) ^{2-1}\left( t-y\right) ^{n-1}dt \label{66} \\ & =\tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{h\left( n+1\right) !}\left( x_{0}-y\right) ^{n+1}.\nonumber \end{align}
</div>
<p> That is proving </p>
<div class="equation" id="67">
<p>
  <div class="equation_content">
    \begin{equation}  \left\|  R_{n}\left( x_{0},y\right) \right\|  \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{h\left( n+1\right) !}\left( x_{0}-y\right) ^{n+1},\text{ \  }\forall \text{ }y\in \left[ a,b\right] :y\leq x_{0}. \label{67}\end{equation}
  </div>
  <span class="equation_label">2.67</span>
</p>
</div>
<p>We have established that </p>
<div class="equation" id="68">
<p>
  <div class="equation_content">
    \begin{equation}  \left\|  R_{n}\left( x_{0},y\right) \right\|  \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{h\left( n+1\right) !}\left| y-x_{0}\right| ^{n+1}\text{, \  }\forall \text{ }y\in \left[ a,b\right] , \label{68}\end{equation}
  </div>
  <span class="equation_label">2.68</span>
</p>
</div>
<p> where \(0{\lt}h\leq \min \left( x_{0}-a,b-x_{0}\right) \), \(x_{0}\in \left( a,b\right) \), and \(\left\|  f\left( \cdot \right) -f\left( x_{0}\right) \right\|  \) is convex over \(\left[ a,b\right] .\) </p>

  </div>
</div> </p>
<p>We have proved </p>
<p><div class="theorem_thmwrapper " id="t9">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2.9</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(\left( X,\left\|  \cdot \right\|  \right) \) be a Banach space and \(f\in C^{n}\left( \left[ a,b\right] ,X\right) \), \(n\in \mathbb {N}\), and \(x_{0}\in \left( a,b\right) \) be fixed. Let \(0{\lt}h\leq \min \left( x_{0}-a,b-x_{0}\right) \), and assume that \(\left\|  f^{\left( n\right) }\left( \cdot \right) -f^{\left( n\right) }\left( x_{0}\right) \right\|  \) is convex over \(\left[ a,b\right] \). Then </p>
<div class="equation" id="69">
<p>
  <div class="equation_content">
    \begin{equation}  \bigg\|  f\left( y\right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x_{0}\right) \tfrac {\left( y-x_{0}\right) ^{i}}{i!}\bigg\|  \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{h\left( n+1\right) !}\left| y-x_{0}\right| ^{n+1}\text{, \  }\forall \text{ }y\in \left[ a,b\right] . \label{69}\end{equation}
  </div>
  <span class="equation_label">2.69</span>
</p>
</div>

  </div>
</div> </p>
<p>We give our second main result under convexity. </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 \(N\in \mathbb {N}\) and \(L_{N}:C\left( \left[ a,b\right] ,X\right) \rightarrow C\left( \left[ a,b\right] ,X\right) \), where \(\left( X,\left\|  \cdot \right\|  \right) \) is a Banach space and \(L_{N}\) is a linear operator. Let the positive linear operators \(\widetilde{L}_{N}:C\left( \left[ a,b\right] \right) \hookrightarrow C\left( \left[ a,b\right] \right) \), such that </p>
<div class="equation" id="70">
<p>
  <div class="equation_content">
    \begin{equation}  \left\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) \right\|  \leq \big( \widetilde{L}_{N}\left( \left\|  f\right\|  \right) \big) \left( x_{0}\right) ,\text{ \  } \label{70}\end{equation}
  </div>
  <span class="equation_label">2.70</span>
</p>
</div>
<p> \(\forall \) \(N\in \mathbb {N},\) \(\forall \) \(f\in C\left( \left[ a,b\right] ,X\right) \), where \(x_{0}\in \left( a,b\right) \) is fixed\(.\) </p>
<p>Furthermore assume that </p>
<div class="equation" id="71">
<p>
  <div class="equation_content">
    \begin{equation}  L_{N}\left( cg\right) =c\widetilde{L}_{N}\left( g\right) ,\text{ \  }\forall \text{ }g\in C\left( \left[ a,b\right] \right) \text{, }\forall \text{ }c\in X. \label{71}\end{equation}
  </div>
  <span class="equation_label">2.71</span>
</p>
</div>
<p> Let \(n\in \mathbb {N}\), here we deal with \(f\in C^{n}\left( \left[ a,b\right] ,X\right) .\) </p>
<p>We further assume that \(\big\|  f^{\left( n\right) }\left( \cdot \right) -f^{\left( n\right) }\left( x_{0}\right) \big\|  \) is convex over \(\left[ a,b\right] \), and </p>
<div class="equation" id="72">
<p>
  <div class="equation_content">
    \begin{equation}  0\leq \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \leq \min \left( x_{0}-a,b-x_{0}\right) . \label{72}\end{equation}
  </div>
  <span class="equation_label">2.72</span>
</p>
</div>
<p>Then </p>
<p>1) </p>
<div class="displaymath" id="73">
  \begin{align} & \bigg\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -\sum _{i=0}^{n}\tfrac {f^{\left( i\right) }\left( x_{0}\right) }{i!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{i}\big) \big)\bigg\|  \left( x_{0}\right) \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) }{\left( n+1\right) !}, \label{73}\end{align}
</div>
<p>2) </p>
<div class="displaymath" id="a0000000030">
  \begin{align}  \left\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\|  & \leq \left\|  f\left( x_{0}\right) \right\|  \left| \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) -1\right| +\nonumber \\ & \quad +\sum _{k=1}^{n}\tfrac {\left\|  f^{\left( k\right) }\left( x_{0}\right) \right\|  }{k!}\big( \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{k}\big) \big) \left( x_{0}\right) \big) +\nonumber \\ & \quad +\tfrac {\omega _{1}\left( f^{\left( n\right) },\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) }{\left( n+1\right)!}, \label{74}\end{align}
</div>
<p> as \(\big( \widetilde{L}_{N}\left(1\right) \big) \left( x_{0}\right) \rightarrow 1\), and \(\big( \widetilde{L}_{N} \big( \left| \cdot -x_{0}\right|^{n+1}\big) \big) \left( x_{0}\right) \rightarrow 0\), we obtain \(\left( L_{N}\left( f\right) \right) \left( x_{0}\right) \rightarrow f\left( x_{0}\right) \), as \(N\rightarrow \infty ,\) </p>
<p>3) if \(f^{\left( k\right) }\left( x_{0}\right) =0\), \(k=0,1,...,n,\) we get that </p>
<div class="equation" id="75">
<p>
  <div class="equation_content">
    \begin{equation}  \left\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\|  \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) }{\left( n+1\right) !}, \label{75}\end{equation}
  </div>
  <span class="equation_label">2.75</span>
</p>
</div>
<p> a high speed of convergence. </p>

  </div>
</div> </p>
<p><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> 1) One can rewrite (<a href="#69">2.69</a>) as follows </p>
<div class="equation" id="76">
<p>
  <div class="equation_content">
    \begin{equation}  \bigg\|  f\left( \cdot \right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x_{0}\right) \tfrac {\left( \cdot -x_{0}\right) ^{i}}{i!}\bigg\|  \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{h\left( n+1\right) !}\left| \cdot -x_{0}\right| ^{n+1}\text{, \  }\forall \text{ }y\in \left[ a,b\right] . \label{76}\end{equation}
  </div>
  <span class="equation_label">2.76</span>
</p>
</div>
<p> We observe that </p>
<div class="displaymath" id="77">
  \begin{align} & \bigg\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -\sum _{i=0}^{n}\tfrac {f^{\left( i\right) }\left( x_{0}\right) }{i!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{i}\big) \big) \left( x_{0}\right) \bigg\|  = \label{77}\\ & =\bigg\|  \Big( L_{N}\Big[ f\left( \cdot \right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x_{0}\right) \tfrac {\left( \cdot -x_{0}\right) ^{i}}{i!}\Big] \Big) \left( x_{0}\right) \bigg\|  \leq \nonumber \\ & \leq \Big(\widetilde{L}_{N}\Big( \Big\|  f\left( \cdot \right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x_{0}\right) \tfrac {\left( \cdot -x_{0}\right) ^{i}}{i!}\Big\|  \Big) \Big) \left( x_{0}\right) \overset {\text{(by (\ref{76}))}}{\leq } \label{78}\\ & \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },h\right) }{h\left( n+1\right) !}\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right)=\nonumber \\ & =\tfrac {\omega _{1}\left( f^{\left( n\right) },\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) }{\left( n+1\right) !}, \label{79}\end{align}
</div>
<p> by choosing </p>
<div class="equation" id="80">
<p>
  <div class="equation_content">
    \begin{equation}  h:=\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) , \label{80}\end{equation}
  </div>
  <span class="equation_label">2.80</span>
</p>
</div>
<p> if \(\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) {\gt}0.\) </p>
<p>We have proved that </p>
<div class="displaymath" id="a0000000032">
  \begin{align} & \bigg\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -\sum _{i=0}^{n}\tfrac {f^{\left( i\right) }\left( x_{0}\right) }{i!}\big( \widetilde{L}_{N}\left( \left( \cdot -x_{0}\right) ^{i}\right) \big) \left( x_{0}\right) \bigg\|  \nonumber \\ & \leq \tfrac {\omega _{1}\left( f^{\left( n\right) },\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) }{\left( n+1\right) !}. \label{81}\end{align}
</div>
<p> By Riesz representation theorem we have </p>
<div class="equation" id="82">
<p>
  <div class="equation_content">
    \begin{equation}  \big( \widetilde{L}_{N}\left( g\right) \big) \left( x_{0}\right) =\int _{\left[ a,b\right] }g\left( t\right) d\mu _{x_{0}}\left( t\right) ,\text{ \  }\forall \text{ }g\in C\left( \left[ a,b\right] \right) , \label{82}\end{equation}
  </div>
  <span class="equation_label">2.82</span>
</p>
</div>
<p> where \(\mu _{x_{0}}\) is a positive finite measure on \(\left[ a,b\right] .\) </p>
<p>That is </p>
<div class="equation" id="83">
<p>
  <div class="equation_content">
    \begin{equation}  \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) =\mu _{x_{0}}\left( \left[ a,b\right] \right) =:M. \label{83}\end{equation}
  </div>
  <span class="equation_label">2.83</span>
</p>
</div>
<p> Without loss of generality we assume that \(M{\gt}0\), if \(M=0\), then our theory is trivial. </p>
<p>In case of </p>
<div class="equation" id="84">
<p>
  <div class="equation_content">
    \begin{equation}  \big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) =0, \label{84}\end{equation}
  </div>
  <span class="equation_label">2.84</span>
</p>
</div>
<p> we have </p>
<div class="equation" id="85">
<p>
  <div class="equation_content">
    \begin{equation}  \int _{\left[ a,b\right] }\left| t-x_{0}\right| ^{n+1}d\mu _{x_{0}}\left( t\right) =0. \label{85}\end{equation}
  </div>
  <span class="equation_label">2.85</span>
</p>
</div>
<p> The last implies \(\left| t-x_{0}\right| ^{n+1}=0\), a.e, hence \(\left| t-x_{0}\right| =0\), a.e, then \(t-x_{0}=0\) a.e., and \(t=x_{0}\), a.e. on \(\left[ a,b\right] \). Consequently \(\mu _{x_{0}}\left( \left\{  t\in \left[ a,b\right] :t\neq x_{0}\right\}  \right) =0\). That is \(\mu _{x_{0}}=\delta _{x_{0}}M\), where \(\delta _{x_{0}}\) is the Dirac measure at \(\left\{  x_{0}\right\}  \). In that case holds </p>
<div class="equation" id="86">
<p>
  <div class="equation_content">
    \begin{equation}  \big( \widetilde{L}_{N}\left( g\right) \big) \left( x_{0}\right) =g\left( x_{0}\right) M, \label{86}\end{equation}
  </div>
  <span class="equation_label">2.86</span>
</p>
</div>
<p> \(\forall \) \(g\in C\left( \left[ a,b\right] \right) .\) </p>
<p>Under (<a href="#84">2.84</a>), the right hand side of (<a href="#81">2.81</a>) equals zero. Furthermore it holds </p>
<div class="displaymath" id="87">
  \begin{align}  \Big( \widetilde{L}_{N}\Big( \Big\|  f\left( \cdot \right) -\sum _{i=0}^{n}f^{\left( i\right) }\left( x_{0}\right) \tfrac {\left( \cdot -x_{0}\right) ^{i}}{i!}\Big\|  \Big) \Big) \left( x_{0}\right) \overset {\text{(\ref{86})}}{=}\left\|  f\left( x_{0}\right) -f\left( x_{0}\right) \right\|  M=0. \label{87}\end{align}
</div>
<p> So that by (<a href="#77">2.77</a>) to have </p>
<div class="equation" id="88">
<p>
  <div class="equation_content">
    \begin{equation}  \bigg\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -\sum _{i=0}^{n}\tfrac {f^{\left( i\right) }\left( x_{0}\right) }{i!}\left( \widetilde{L}_{N}\left( \left( \cdot -x_{0}\right) ^{i}\right) \right) \left( x_{0}\right) \bigg\|  =0. \label{88}\end{equation}
  </div>
  <span class="equation_label">2.88</span>
</p>
</div>
<p> Therefore inequality (<a href="#81">2.81</a>) is true again and always. </p>
<p>2) Next again we see that </p>
<div class="displaymath" id="a0000000033">
  \begin{align} & \left\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\|  =\nonumber \\ & =\left\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -\sum _{k=0}^{n}\tfrac {f^{\left( k\right) }\left( x_{0}\right) }{k!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{k}\big) \big) \left( x_{0}\right) +\right.\nonumber \\ & \quad +\left. \sum _{k=0}^{n}\tfrac {f^{\left( k\right) }\left( x_{0}\right) }{k!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{k}\big) \big) \left( x_{0}\right) -f\left( x_{0}\right) \right\|  \leq \label{89}\\ & \leq \bigg\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -\sum _{k=0}^{n}\tfrac {f^{\left( k\right) }\left( x_{0}\right) }{k!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{k}\big) \big) \left( x_{0}\right) \bigg\|  +\nonumber \\ & \quad +\bigg\|  \sum _{k=1}^{n}\tfrac {f^{\left( k\right) }\left( x_{0}\right) }{k!}\big( \widetilde{L}_{N}\big( \left( \cdot -x_{0}\right) ^{k}\big) \big) \left( x_{0}\right) +f\left( x_{0}\right) \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) -f\left( x_{0}\right) \bigg\|  \overset {\text{(\ref{81})}}{\leq }\nonumber \\ & \leq \big\|  f\left( x_{0}\right) \big\|  \left| \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) -1\right| +\sum _{k=1}^{n}\tfrac {\left\|  f^{\left( k\right) }\left( x_{0}\right) \right\|  }{k!}\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{k}\big) \big) \left( x_{0}\right) +\nonumber \\ & \quad +\tfrac {\omega _{1}\left( f^{\left( n\right) },\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) }{\left( n+1\right) !}. \label{90}\end{align}
</div>
<p> We have proved that </p>
<div class="displaymath" id="a0000000034">
  \begin{align}  \left\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\|  \leq &  \left\|  f\left( x_{0}\right) \right\|  \left| \big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) -1\right| +\nonumber \\ & \quad +\sum _{k=1}^{n}\tfrac {\left\|  f^{\left( k\right) }\left( x_{0}\right) \right\|  }{k!}\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{k}\big) \big) \left( x_{0}\right) +\nonumber \\ & \quad +\tfrac {\omega _{1}\left( f^{\left( n\right) },\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) }{\left( n+1\right) !}. \label{91}\end{align}
</div>
<p> Clearly by (<a href="#55">2.55</a>) and (<a href="#91">2.91</a>), when \(\big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \rightarrow 1\) and \(\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \rightarrow 0\), we obtain \(\left( L_{N}\left( f\right) \right) \left( x_{0}\right) \rightarrow f\left( x_{0}\right) \), as \(N\rightarrow \infty \). Notice that \(\big( \widetilde{L}_{N}\left( 1\right) \big) \left( x_{0}\right) \) will be bounded. </p>
<p>3) If \(f^{\left( k\right) }\left( x_{0}\right) =0\), \(k=0,1,...,n,\) we get that </p>
<div class="displaymath" id="92">
  \begin{align}  \left\|  \left( L_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\|  \overset {\text{(\ref{91})}}{\leq } \tfrac {\omega _{1}\left( f^{\left( n\right) },\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) }{\left( n+1\right) !}, \label{92}\end{align}
</div>
<p> a high speed of convergence. </p>
<p>The theorem is proved. <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> </p>
<p><div class="theorem_thmwrapper " id="t11">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2.11</span>
  </div>
  <div class="theorem_thmcontent">
  <p>All as in Theorem <span class="rm"><a href="#t10">2.10</a></span>. Consider \(n\in \mathbb {N}\) an odd number. Then inequalities <span class="rm">(<a href="#73">2.73</a>)</span>–<span class="rm">(<a href="#75">2.75</a>)</span> are sharp, in fact they are attained by \(f_{\ast }\left( t\right) =\vec{i}\left| t-x_{0}\right| ^{n+1}\), where \(\vec{i}\in X\), \(\big\|  \vec{i}\big\|  =1\), \(\forall \) \(t\in \left[ a,b\right] .\) </p>

  </div>
</div> </p>
<p><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> Let \(n\) an odd natural number, then \(n+1\) is even. We consider \(f_{\ast }\left( t\right) =\vec{i}\left| t-x_{0}\right| ^{n+1}=\vec{i}\left( t-x_{0}\right) ^{n+1}\in X\), where \(\vec{i}\in X\), \(\big\|  \vec{i}\big\|  =1\). We have that \(f_{\ast }\in C^{n}\left( \left[ a,b\right] ,X\right) \) and </p>
<div class="equation" id="93">
<p>
  <div class="equation_content">
    \begin{equation}  f_{\ast }^{\left( n\right) }\left( t\right) =\vec{i}\left( n+1\right) !\left( t-x_{0}\right) \text{, \  }\forall t\in \left[ a,b\right] , \label{93}\end{equation}
  </div>
  <span class="equation_label">2.93</span>
</p>
</div>
<p> along with \(f_{\ast }^{\left( k\right) }\left( x_{0}\right) =0\), \(k=0,1,...,n.\) </p>
<p>Furthermore it holds </p>
<div class="equation" id="94">
<p>
  <div class="equation_content">
    \begin{equation}  \left\|  f_{\ast }^{\left( n\right) }\left( t\right) -f_{\ast }^{\left( n\right) }\left( x_{0}\right) \right\|  =\left( n+1\right) !\left| t-x_{0}\right| , \label{94}\end{equation}
  </div>
  <span class="equation_label">2.94</span>
</p>
</div>
<p> which is a convex function in \(t\in \left[ a,b\right] \). So we apply \(f_{\ast }\) to inequalities (<a href="#73">2.73</a>)-(<a href="#75">2.75</a>). </p>
<p>1) The left hand side of (<a href="#73">2.73</a>) equals \(\left( \widetilde{L}_{N}( \left| t-x_{0}\right| ^{n+1}) \right) \left( x_{0}\right) \). The right hand side of (<a href="#73">2.73</a>) is </p>
<div class="displaymath" id="95">
  \begin{align} &  \tfrac {\omega _{1}\big( f_{\ast }^{\left( n\right) },\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \big) }{\left( n+1\right) !}= \label{95}\\[2mm]& =\tfrac {\omega _{1}\left( \vec{i}\left( n+1\right) !\left( t-x_{0}\right) ,\left( \widetilde{L}_{N}\left( \left| \cdot -x_{0}\right| ^{n+1}\right) \right) \left( x_{0}\right) \right) }{\left( n+1\right) !} \nonumber \\[2mm]& =\end{align}
</div>
<p>=<br /> =c t_1,t_2 a,b :<br /> t_1-t_2 ≤( L_N(  ·-x_0 ^n+1) )  x_0   t_1-t_2 = L_N  ·-x_0 ^n+1   x_0 . </p>
<div class="displaymath" id="a0000000037">
  \end{align}
</div>
<p>Hence the right hand side of (<a href="#73">2.73</a>) equals also \(\big( \widetilde{L}_{N}\big( \left| \cdot -x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \). That is (<a href="#73">2.73</a>) is an attained inequality by \(f_{\ast }.\) </p>
<p>2) The left hand side of (<a href="#74">2.74</a>) equals \(\big( \widetilde{L}_{N}\big( \left| t-x_{0}\right| ^{n+1}\big) \big) \left( x_{0}\right) \), which equals its right hand side. That is (<a href="#74">2.74</a>) is an attained inequality by \(f_{\ast }.\) </p>
<p>3) Same as above, inequality (<a href="#75">2.75</a>) is attained by \(f_{\ast }.\) </p>
<p>We have proved that (<a href="#73">2.73</a>)-(<a href="#75">2.75</a>) are sharp inequalities. <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><div class="corollary_thmwrapper " id="c12">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">2.12</span>
  </div>
  <div class="corollary_thmcontent">
  <p>(to Theorem <span class="rm"><a href="#t7.">2.7</a>)</span> Let \(\left( X,\left\|  \cdot \right\|  \right) \) be a Banach space and any \(f\in C^{1}\left( \left[ 0,1\right] ,X\right) \). Then </p>
<p>1) </p>
<div class="displaymath" id="97">
  \begin{align}  \left\|  \left( B_{N}\left( f\right) \right) \left( x\right) -f\left( x\right) \right\|  \leq \tfrac {3}{2}\omega _{1}\left( f^{\prime },\sqrt{\tfrac {x\left( 1-x\right) }{N}}\right) \sqrt{\tfrac {x\left( 1-x\right) }{N}} & \leq \tfrac {0.75}{\sqrt{N}}\omega _{1}\big( f^{\prime },\tfrac {1}{2\sqrt{N}}\big),\label{97} \end{align}
</div>
<p> \(\forall x\in \left[ 0,1\right] , \forall N\in \mathbb {N}\), and </p>
<p>2) </p>
<div class="equation" id="98">
<p>
  <div class="equation_content">
    \begin{equation}  \left\|  \left\|  B_{N}\left( f\right) -f\right\|  \right\|  _{\infty ,\left[ 0,1\right] }\leq \tfrac {0.75}{\sqrt{N}}\omega _{1}\big( f^{\prime },\tfrac {1}{2\sqrt{N}}\big) . \label{98}\end{equation}
  </div>
  <span class="equation_label">2.98</span>
</p>
</div>
<p> Hence \(B_{n}\left( f\right) \overset {u}{\rightarrow }f\), uniformly, as \(N\rightarrow \infty .\) </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> The operators \(B_{N}\), \(\widetilde{B}_{N}\) fulfill (<a href="#29.">2.29</a>), (<a href="#30.">2.30</a>). We have that \(\widetilde{B}_{N}\left( 1\right) =1\), \(\big( \widetilde{B}_{N}\left( id\right) \big) \left( x\right) =x\), </p>
<div class="equation" id="99">
<p>
  <div class="equation_content">
    \begin{equation}  \big( \widetilde{B}_{N}\left( \left( \cdot -x\right) \right) \big) \left( x\right) =0\text{,} \label{99}\end{equation}
  </div>
  <span class="equation_label">2.99</span>
</p>
</div>
<p> and </p>
<div class="displaymath" id="a0000000040">
  \[  \big( \widetilde{B}_{N}\big( \left( \cdot -x\right) ^{2}\big) \big) \left( x\right) =\tfrac {x\left( 1-x\right) }{N}\leq \tfrac {1}{4N},\text{ \  }\forall \text{ }x\in \left[ 0,1\right] .  \]
</div>
<p> We use (<a href="#31.">2.31</a>) for \(n=1\). We have (by use of (<a href="#99">2.99</a>)) that</p>
<div class="displaymath" id="a0000000041">
  \begin{align} & \left\|  \left( B_{N}\left( f\right) \right) \left( x\right) -f\left( x\right) \right\|  \leq \nonumber \\ & \leq \tfrac {3}{2}\omega _{1}\big( f^{\prime },\big( \big( \widetilde{B}_{N}\big( \left( \cdot -x\right) ^{2}\big) \big) \left( x\right) \big) ^{\frac{1}{2}}\big) \big( \big( \widetilde{B}_{N}\left( \big( \cdot -x\right) ^{2}\big) \big) \left( x\right) \big) ^{\frac{1}{2}}= \label{100}\\ & =\tfrac {3}{2}\omega _{1}\left( f^{\prime },\sqrt{\tfrac {x\left( 1-x\right) }{N}}\right) \sqrt{\tfrac {x\left( 1-x\right) }{N}}\leq \tfrac {3}{2}\omega _{1}\left( f^{\prime },\tfrac {1}{2\sqrt{N}}\right) \tfrac {1}{2\sqrt{N}}=\nonumber \\ & =\tfrac {3}{4\sqrt{N}}\omega _{1}\left( f^{\prime },\tfrac {1}{2\sqrt{N}}\right) =\tfrac {0.75}{\sqrt{N}}\omega _{1}\left( f^{\prime },\tfrac {1}{2\sqrt{N}}\right) . \label{101}\end{align}
</div>
<p><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 finish with </p>
<p><div class="corollary_thmwrapper " id="c13">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">2.13</span>
  </div>
  <div class="corollary_thmcontent">
  <p><span class="rm">(to</span> Theorem <span class="rm"><a href="#t10">2.10</a>)</span> Let \(\left( X,\left\|  \cdot \right\|  \right) \) be a Banach space and any \(f\in C^{1}\left( \left[ 0,1\right] ,X\right) \) such that \(\left\|  f^{\prime }\left( t\right) -f^{\prime }\left( x_{0}\right) \right\|  \) is convex function in \(t\in \left[ 0,1\right] \), where \(x_{0}\in \left( 0,1\right) \) is a fixed number. Then </p>
<div class="displaymath" id="102">
  \begin{align}  \left\|  \left( B_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\|  & \leq \tfrac {1}{2}\omega _{1}\big( f^{\prime },\tfrac {x_{0}\left( 1-x_{0}\right) }{N}\big) \leq \tfrac {1}{2}\omega _{1}\left( f^{\prime },\tfrac {1}{4N}\right) ,\text{ \  }\forall \text{ }N\in \mathbb {N}. \label{102}\end{align}
</div>
<p> Above notice the high speed of convergence \(\frac{1}{N}\) under the convexity assumption. </p>
<p>Inequalities <span class="rm">(<a href="#102">2.102</a>)</span> are sharp. The first part of <span class="rm">(<a href="#102">2.102</a>)</span> is attained by \(\vec{i}\left( t-x_{0}\right) ^{2}\), \(\vec{i}\in X\), \(\big\|  \vec{i}\big\|  =1\), \(\forall \) \(t\in \left[ 0,1\right] \). The second part of <span class="rm">(<a href="#102">2.102</a>)</span> is equality at \(x_{0}=\frac{1}{2}.\) </p>
<p>As \(N\rightarrow \infty \), we have that \(\left( B_{N}\left( f\right) \right) \left( x_{0}\right) \rightarrow f\left( x_{0}\right) \). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000043">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Let \(x_{0}\in \left( 0,1\right) \), then \(x_{0}\left( 1-x_{0}\right) \leq x_{0}\) and \(x_{0}\left( 1-x_{0}\right) \leq 1-x_{0}\), hence \(x_{0}\left( 1-x_{0}\right) \leq \min \left( x_{0},1-x_{0}\right) \) and </p>
<div class="equation" id="103">
<p>
  <div class="equation_content">
    \begin{equation}  \tfrac {x_{0}\left( 1-x_{0}\right) }{N}\leq \min \left( x_{0},1-x_{0}\right) \text{, \  }\forall \text{ }N\in \mathbb {N}. \label{103}\end{equation}
  </div>
  <span class="equation_label">2.103</span>
</p>
</div>
<p> The last shows that (see (<a href="#99">2.99</a>)) </p>
<div class="equation" id="104">
<p>
  <div class="equation_content">
    \begin{equation}  0<\big( \widetilde{B}_{N}\big( \left( \cdot -x_{0}\right) ^{2}\big) \big) \left( x_{0}\right) \leq \min \left( x_{0},1-x_{0}\right) \text{, \  }\forall \text{ }N\in \mathbb {N}. \label{104}\end{equation}
  </div>
  <span class="equation_label">2.104</span>
</p>
</div>
<p> Let here \(f\in C^{1}\left( \left[ 0,1\right] ,X\right) \) such that \(\left\Vert f^{\prime }\left( \cdot \right) -f^{\prime }\left( x_{0}\right) \right\Vert \) is convex over \(\left[ 0,1\right] .\) Then, by (<a href="#73">2.73</a>), we get </p>
<div class="displaymath" id="105">
  \begin{align}  \left\Vert \left( B_{N}\left( f\right) \right) \left( x_{0}\right) -f\left( x_{0}\right) \right\Vert & \leq \tfrac {\omega _{1}\left( f^{\prime },\left( \widetilde{B}_{N}\left( \left( \cdot -x_{0}\right) ^{2}\right) \right) \left( x_{0}\right) \right) }{2} \label{105}\\ & =\tfrac {\omega _{1}\big( f^{\prime },\frac{x_{0}\left( 1-x_{0}\right) }{N}\big) }{2}\leq \tfrac {1}{2}\omega _{1}\big( f^{\prime },\tfrac {1} {4N}\nonumber \big) . \end{align}
</div>
<p><div class="proof_wrapper" id="a0000000044">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
<dl class="bibliography">
  <dt><a name="1">1</a></dt>
  <dd><p><i class="sc">G.A. Anastassiou</i>, <i class="itshape">Moments in Probability and Approximation Theory</i>, Pitman Research Notes in Math., <b class="bf">287</b>, Longman Sci. &amp; Tech., Harlow, U.K., 1993. </p>
</dd>
  <dt><a name="2">2</a></dt>
  <dd><p><a href ="https://doi.org/10.14492/hokmj/1351257969"> <i class="sc">G.A. Anastassiou</i>, <i class="itshape">Lattice homomorphism - Korovkin type inequalities for vector valued functions</i>, Hokkaido Mathematical Journal, Vol. 26 (1997), pp.&#160;337–364. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="3">3</a></dt>
  <dd><p><i class="sc">T. Nishishiraho</i>, <i class="itshape">Bernstein-type approximation processes for vector-valued functions</i>, Acta Math. Hungar. <b class="bf">84</b> (1999), no. 1-2, pp.&#160;149–158. </p>
</dd>
  <dt><a name="4">4</a></dt>
  <dd><p><i class="sc">R. Paltanea</i>, <i class="itshape">Vector variants of some approximation theorems of Korovkin and of Sendov and Popov. Constructive theory of functions</i>, pp.&#160;366–373, DARBA, Sofia, 2003. </p>
</dd>
  <dt><a name="5">5</a></dt>
  <dd><p><i class="sc">R. Paltanea</i>, <i class="itshape">Approximation of functions in Banach spaces using linear and positive operators</i>, in Proceedings of RoGer seminar 2004, Mediamira Science Publisher, Cluj-Napoca, 2004, pp.&#160;5–20. </p>
</dd>
  <dt><a name="6">6</a></dt>
  <dd><p><a href ="https://ictp.acad.ro/approximation-higher-order-convex-functions-i/"> <i class="sc">T. Popoviciu</i>, <i class="itshape">Sur l’approximation des fonctions convexes d’ordre superieur</i>, Mathematica, <b class="bf">10</b> (1935), pp.&#160;49–54, Cluj, Romania. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="7">7</a></dt>
  <dd><p><i class="sc">G.E. Shilov</i>, <i class="itshape">Elementary Functional Analysis</i>, Dover Publications, Inc., New York, 1996. </p>
</dd>
</dl>


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