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<h1>Some general Kantorovich type operators</h1>
<p class="authors">
<span class="author">Petru I. Braica\(^\ast \) Ovidiu T. Pop\(^\S \)</span>
</p>
<p class="date">August 8, 2012.</p>
</div>
<p>\(^\ast \) Secondary School “Grigore Moisil", 1 Mileniului 1 St. 440037 Satu Mare, Romania, e-mail: <span class="tt">petrubr@yahoo.com</span> </p>
<p>\(^\S \) National College “Mihai Eminescu", 5 Mihai Eminescu St., 440014 Satu Mare, Romania, e-mail: <span class="tt">ovidiutiberiu@yahoo.com</span> </p>

<div class="abstract"><p> A general class of linear and positive operators of Kantorovich-type is constructed. The operators of this type which preserve exactly two test functions from the set \(\{ e_0, e_1, e_2\} \) are determined and their approximation properties and convergence theorems are studied. </p>
<p><b class="bf">MSC.</b> 41A10, 41A35, 41A36. </p>
<p><b class="bf">Keywords.</b> Kantorovich operators, approximation and convergence theorem, Voronovskaja-type theorem. </p>
</div>
<h1 id="a0000000002">1 Introduction</h1>
<p>Let \(\mathbb N\) be the set of positive integers and \(\mathbb N_0=\mathbb N\cup \{ 0\} \). In 1930, L. V. Kantorovich in [3] introduced the linear and positive operators \(K_m:L_1([0,1])\to C([0,1])\), \(m\in \mathbb N_0\), defined for any \(f\in L_1([0,1])\), \(x\in [0,1]\) and \(m\in \mathbb N_0\) by </p>
<div class="equation" id="1.1">
<p>
  <div class="equation_content">
    \begin{equation} \label{1.1} (K_m f)(x)=(m+1)\sum ^m_{k=0}\tbinom {m}{k} x^k(1-x)^{m-k} \int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{m+1}}f(t){\rm d}t. \end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p> The operators from (<a href="#1.1">1</a>) are called Kantorovich operators. We remark that these operators preserve only the test function \(e_0\). </p>
<p>The aim of this paper is to construct and study a general class of linear positive operators which preserve exactly two test functions from the set \(\{ e_0, e_1, e_2\} \). </p>
<p>In [4], J. P. King introduced and studied a Bernstein type operator, which preserves only the test functions \(e_0\) and \(e_2\). Therefore, we say that the operators constructed in this paper are operators of King’s type. </p>
<p>In Section 2, we recall some results from [6], which we use for obtaining the main results of this paper. </p>
<p>In Section 3, respectively 4, we determine the unique operators from Section 2, which preserve exactly the test functions \(e_0\) and \(e_1\), respectively \(e_0\) and \(e_2\). In the Sections 4 and 5, we give approximation, convergence and Voronovskaja’s type theorems for the operators obtained. </p>
<h1 id="a0000000003">2 Preliminaries</h1>
<p>Let \(\mathbb N\) be the set of positive integers and \(\mathbb N_0=\mathbb N\cup \{ 0\} \). In this section we recall some results from [8], which we shall use in the present paper. Let \(I, J\) be real intervals with the property \(I\cap J\neq \emptyset \). For any \(m,k\in \mathbb N_0\), \(m\neq 0\), we consider the functions \(\varphi _{m,k}:J\to \mathbb R\), with the property that \(\varphi _{m,k}(x)\geq 0\), for any \(x\in J\) and the linear positive functionals \(A_{m,k}:E(I)\to \mathbb R\). </p>
<p>For any \(m\in \mathbb N\) we define the operator \(L_m:E(I)\to F(J)\), by </p>
<div class="equation" id="2.1">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.1} (L_m f)(x)=\sum ^\infty _{k=0}\varphi _{m,k}(x) A_{m,k}(f), \end{equation}
  </div>
  <span class="equation_label">2</span>
</p>
</div>
<p> for any \(x\in J\), where \(E(I)\) and \(F(J)\) are linear subspaces of real valued functions defined on \(I\), resp. \(J\), for which the sequence \((L_n)_{n\geq 0}\) defined above is convergent (in the topology of \(F(J)\)). </p>
<p><div class="remark_thmwrapper " id="rem_2.1">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">1</span>
  </div>
  <div class="remark_thmcontent">
  <p> <em>The operators \((L_m)_{m\in \mathbb N}\) are linear and positive on \(E(I\cap J)\).<span class="qed">â–¡</span></em></p>

  </div>
</div> For \(m\in \mathbb N\) and \(i\in \mathbb N_0\), we define \((T_{m,i})\) by </p>
<div class="equation" id="2.2">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.2} (T_{m,i}L_m)(x)=m^i(L_m\psi ^i_x)(x)=m^i\sum ^\infty _{k=0} \varphi _{m,k}(x)A_{m,k}(\psi ^i_x) \end{equation}
  </div>
  <span class="equation_label">3</span>
</p>
</div>
<p> for any \(x\in I\cap J\), \(\psi ^i_x(t)=(t-x)^i\), \(t\in I\). </p>
<p>In that follows \(s\in \mathbb N_0\) is even and we assume that the next two conditions: </p>
<p>\(\bullet \) there exist the smallest \(\alpha _s, \alpha _{s+2}\in [0,+\infty )\), so that </p>
<div class="equation" id="2.3">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.3} \lim _{m\to \infty }\tfrac {(T_{m,j}L_m)(x)}{m^{\alpha _j}}= B_j(x)\in \mathbb R\end{equation}
  </div>
  <span class="equation_label">4</span>
</p>
</div>
<p> for any \(x\in I\cap J\) and \(j\in \{ s, s+2\} \) </p>
<div class="equation" id="2.4">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.4} \alpha _{s+2}<\alpha _s+2. \end{equation}
  </div>
  <span class="equation_label">5</span>
</p>
</div>
<p>\(\bullet \) \(I\cap J\) is an interval. </p>
<p><div class="theorem_thmwrapper " id="th_2.1">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2</span>
  </div>
  <div class="theorem_thmcontent">
  <p> <em>(see [8])</em> Let \(f\in E(I)\) be a function. If \(x\in I\cap J\) and \(f\) is \(s\) times differentiable in a neighborhood of \(x\), \(f^{(s)}\) is continuous in \(x\), then </p>
<div class="equation" id="2.5">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.5} \lim _{m\to \infty }m^{s-\alpha _s}\left((L_m f)(x)-\sum ^s_{i=0} \tfrac {f^{(i)}(x)}{m^i i!}\,  (T_{m,j} L_m)(x)\right)=0. \end{equation}
  </div>
  <span class="equation_label">6</span>
</p>
</div>
<p> Assume that \(f\) is \(s\) times differentiable on \(I\), \(f^{(s)}\) is continuous on \(I\) and there exists a compact interval \(K\subset I\cap J\), such that there exists \(m(s)\in \mathbb N\) and constant \(k_j\in \mathbb R\) depending on \(K\), so for \(m\geq m(s)\) and \(x\in K\) the following inequalities </p>
<div class="equation" id="2.6">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.6} \tfrac {(T_{m,j}L_m)(x)}{m^{\alpha _j}}\leq k_j, \end{equation}
  </div>
  <span class="equation_label">7</span>
</p>
</div>
<p> hold for \(j\in \{ s,s+2\} \).<br />Then the convergence expressed by \((\ref{2.5})\) is uniform on \(K\) and </p>
<div class="displaymath" id="2.7">
  \begin{gather} \label{2.7} m^{s-\alpha _s}\left|(L_m f)(x)-\sum ^s_{i=0} \tfrac {f^{(i)}(x)}{m^i i!}\, (T_{m,i} L_m)(x)\right|\\ \leq \tfrac {1}{s!}\, (k_s+k_{s+2})\omega \left(f^{(s)}; \tfrac {1}{\sqrt{m^{2+\alpha _s-\alpha _{s+2}}}}\right),\nonumber \end{gather}
</div>
<p> for any \(x\in K\), \(m\geq m(s)\), where \(\omega (f;\cdot )\) denotes the modulus of continuity of the function \(f\). </p>

  </div>
</div> </p>
<p><div class="corollary_thmwrapper " id="cor_2.1">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">3</span>
  </div>
  <div class="corollary_thmcontent">
  <p> Let \(f:I\to \mathbb R\) be a \(s\) times differentiable function on \(I\cap J\) with \(f^{(s)}\) continuous on \(I\cap J\). Then </p>
<div class="equation" id="2.8">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.8} \lim _{m\to \infty } (L_m f)(x)=B_0(x) f(x) \end{equation}
  </div>
  <span class="equation_label">9</span>
</p>
</div>
<p> if \(s=0\) and \(\alpha _0=0\), where \(B_0\) is defined by \((\ref{2.3})\). If \(s\geq 2\), then </p>
<div class="equation" id="2.9">
<p>
  <div class="equation_content">
    \begin{equation} \label{2.9} \lim _{m\to \infty }m^{s-\alpha _s} \left[\! (L_m f)(x)-\! \sum ^{s-1}_{i=0}\tfrac {1}{m^i i!} \, (T_{m,i}L_m)(x)f^{(i)}(x)\right]\! \! =\! \tfrac {1}{s!}\, B_s(x)f^{(s)}(x), \end{equation}
  </div>
  <span class="equation_label">10</span>
</p>
</div>
<p> where \(B_s\) are defined by \((\ref{2.3})\). </p>
<p>If \(f\) is a \(s\) times differentiable function on \(I\cap J\), with \(f^{(s)}\) continuous and bounded on \(I\cap J\) and <em>(<a href="#2.6">7</a>)</em> takes place for an interval \(K\subset I\cap J\), then the convergence in <em>(<a href="#2.8">9</a>)</em> and <em>(<a href="#2.9">10</a>)</em> are uniform on \(K\). </p>

  </div>
</div> </p>
<h1 id="a0000000004">3 The construction of a general linear and positive operators</h1>
<p>Let \(J\subset R\) be an interval, \(m_0\in \mathbb N_0\), \(m_0\geq 2\) given, \(\mathbb N_1=\{ m\in \mathbb N|m\geq m_0\} \), the function \(\alpha _m, \beta _m:J\to \mathbb R\), \(\alpha _m(x)\geq 0\), \(\beta _m(x)\geq 0\) for any \(x\in J\) and \(m\in \mathbb N_1\). <div class="definition_thmwrapper " id="def_3.1">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">4</span>
  </div>
  <div class="definition_thmcontent">
  <p><em>For \(m\in \mathbb N_1\), we define the operator of the following form <div class="equation" id="3.1">
<p>
  <div class="equation_content">
    \begin{equation} \label{3.1} (K^*_m f)(x)=(m+1)\sum ^m_{k=0}\tbinom {m}{k} \alpha ^k_m(x)\beta ^{m-k}_m(x)\int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{m+1}}f(t){\rm d}t \end{equation}
  </div>
  <span class="equation_label">11</span>
</p>
</div> for any \(f\in L_1([0,1])\) and \(x\in J\). </em></p>

  </div>
</div> <div class="lemma_thmwrapper " id="lem_3.1">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">5</span>
  </div>
  <div class="lemma_thmcontent">
  <p> The following identities </p>
<div class="equation" id="3.2">
<p>
  <div class="equation_content">
    \begin{equation} \label{3.2} (K^*_m e_0)(x)=(\alpha _m (x)+\beta _m(x))^m, \end{equation}
  </div>
  <span class="equation_label">12</span>
</p>
</div>
<div class="equation" id="3.3">
<p>
  <div class="equation_content">
    \begin{equation} \label{3.3} (K^*_m e_1)(x)=\tfrac {(\alpha _m (x)+\beta _m(x))^{m-1}}{2(m+1)}\,  ((2m+1)\alpha _m(x)+\beta _m(x)), \end{equation}
  </div>
  <span class="equation_label">13</span>
</p>
</div>
<p> and </p>
<div class="displaymath" id="3.4">
  \begin{align} \label{3.4} (K^*_m e_2)(x) & =\tfrac {(\alpha _m(x)+\beta _m(x))^{m-2}}{3(m+1)^2}\,  \big(3m(m-1)\alpha ^2_m(x)\\ & \quad +6m\alpha _m(x)(\alpha _m(x)+\beta _m(x))+ (\alpha _m(x)+\beta _m(x))^2\big),\nonumber \end{align}
</div>
<p> hold, for any \(x\in J\) and any \(m\in \mathbb N_1\). </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000005">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> For \(m\in \mathbb N_1\) and \(k\in \{ 0,1,\dots , m\} \) we have </p>
<div class="displaymath" id="a0000000006">
  \begin{align*} &  \int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{m+1}}e_0(t){\rm d}t=\tfrac {1}{m+1}\, ,\quad \int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{m+1}}e_1(t){\rm d}t=\tfrac {2k+1}{(m+1)^2}\\ &  \text{and}\; \;  \int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{m+1}}e_2(t){\rm d}t=\tfrac {3k^2 +3k+1}{(m+1)^3}\, . \end{align*}
</div>
<p> Then </p>
<div class="displaymath" id="a0000000007">
  \begin{align*}  (K^*_m e_0)(x) &  =(m+1)\sum ^m_{k=0}\tbinom {m}{k}\alpha ^k_m(x)\beta ^{m-k}_m(x) \int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{k+1}}e_0(t){\rm d}t\\ &  =\sum ^m_{k=0}\tbinom {m}{k}\alpha _m^k(x)\beta ^{m-k}_m(x), \end{align*}
</div>
<p> so (<a href="#3.2">12</a>) holds; </p>
<div class="displaymath" id="a0000000008">
  \begin{align*}  (K^*_m e_1)(x)&  =(m+1)\sum ^m_{k=0}\tbinom {m}{k}\alpha ^k_m(x)\beta ^{m-k}_m(x) \int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{m+1}}e_1(t){\rm d}t\\ &  =\tfrac {1}{2(m+1)}\bigg(2m\alpha _m(x)\sum ^m_{k=1}\tbinom {m-1}{k-1}\alpha ^{k-1}_m(x)\beta ^{m-k}_m(x)\\ &  \qquad +\sum ^m_{k=0}\tbinom {m}{k}\alpha _m^k(x)\beta ^{m-1}_m(x)\bigg), \end{align*}
</div>
<p> so (<a href="#3.3">13</a>) holds and </p>
<div class="displaymath" id="a0000000009">
  \begin{gather*}  (K^*_m e_2)(x)=(m+1)\sum ^m_{k=0}\tbinom {m}{k}\alpha ^k_m(x)\beta ^{m-k}_m(x)\int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{m+1}} e_2(t){\rm d}t\\ =\tfrac {1}{3(m+1)^2}\bigg(3m(m+1)\alpha ^2_m(x)\sum ^m_{k=2}\tbinom {m-2}{k-2}\alpha ^{k-2}_m(x) \beta ^{m-k}_m(x)\\ +6m\alpha _m(x)\sum ^m_{k=1}\tbinom {m-1}{k-1}\alpha ^{k-1}_m(x)\beta ^{m-k}_m(x) +\sum ^m_{k=0} \tbinom {m}{k}\alpha ^k_m(x)\beta ^{m-k}_m(x)\bigg), \end{gather*}
</div>
<p> from where (<a href="#3.4">14</a>) follows. <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> <div class="remark_thmwrapper " id="rem_3.1">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">6</span>
  </div>
  <div class="remark_thmcontent">
  <p><em>In the following, we will use Theorem <a href="#th_2.1">2</a>, where \(I=[0,1]\), <div class="equation" id="3.5">
<p>
  <div class="equation_content">
    \begin{equation} \label{3.5} \varphi _{m,k}(x)=(m+1)\tbinom {m}{k}\alpha ^k_m(x)\beta ^{m-k}_m(x) \end{equation}
  </div>
  <span class="equation_label">15</span>
</p>
</div> and <div class="equation" id="3.6">
<p>
  <div class="equation_content">
    \begin{equation} \label{3.6} A_{m,k}(x)=\int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{m+1}}f(t){\rm d}t \end{equation}
  </div>
  <span class="equation_label">16</span>
</p>
</div> for any \(x\in J\), \(f\in L_1([0,1])\), \(m\in \mathbb N_1\) and \(k\in \{ 0,1,\dots ,m\} \).<span class="qed">â–¡</span></em></p>

  </div>
</div> </p>
<h1 id="a0000000011">4 Kantorovich-type operators preserving the test <br />functions \(e_0\) and \(e_1\)</h1>
<p>In this case, we impose the conditions \((K^*_m e_0)(x)=e_0(x)\) and \((K^*_m e_1)(x)=e_1(x)\), for any \(x\in J\) and \(m\in \mathbb N_1\). From the conditions above, taking (<a href="#3.2">12</a>) and (<a href="#3.3">13</a>) into account, we have </p>
<div class="displaymath" id="a0000000012">
  \begin{align*} &  (\alpha _m (x)+\beta (x))^m=1\; \;  \text{and}\\ &  \tfrac {(\alpha _m(x)+\beta _m(x))^{m-1}}{2(m+1)}\, ((2m+1)\alpha _m(x)+\beta _m(x))=x, \end{align*}
</div>
<p> from where </p>
<div class="equation" id="4.1">
<p>
  <div class="equation_content">
    \begin{equation} \label{4.1} \alpha _m(x)=\tfrac {2(m+1)x-1}{2m}\, , \end{equation}
  </div>
  <span class="equation_label">17</span>
</p>
</div>
<div class="equation" id="4.2">
<p>
  <div class="equation_content">
    \begin{equation} \label{4.2} \beta _m(x)=\tfrac {2m+1-2(m+1)x}{2m}\, , \end{equation}
  </div>
  <span class="equation_label">18</span>
</p>
</div>
<p> for any \(x\in [0,1]\) and \(m\in \mathbb N_1\).<br />From \(\alpha _m(x)\geq 0\) and \(\beta _m(x)\geq 0\), for any \(m\in \mathbb N_1\), we have </p>
<div class="equation" id="4.3">
<p>
  <div class="equation_content">
    \begin{equation} \label{4.3} \tfrac {1}{2(m+1)}\leq x\leq \tfrac {2m+1}{2(m+1)}\, . \end{equation}
  </div>
  <span class="equation_label">19</span>
</p>
</div>
<p><div class="lemma_thmwrapper " id="lem_4.1">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">7</span>
  </div>
  <div class="lemma_thmcontent">
  <p> The following </p>
<div class="equation" id="4.4">
<p>
  <div class="equation_content">
    \begin{equation} \label{4.4} \left[\tfrac {1}{2(m_0+1)}\, ; \tfrac {2m_0+1}{2(m_0+1)}\right]\subset \left[\tfrac {1}{2(m+1)}\, ;\tfrac {2m+1}{2(m+1)}\right]\subset [0,1] \end{equation}
  </div>
  <span class="equation_label">20</span>
</p>
</div>
<p> hold for any \(m\in \mathbb N_1\). </p>

  </div>
</div> <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> Because the function \(\tfrac {1}{2(m+1)}\) is decreasing and the function <br />\(\tfrac {2m+1}{2(m+1)}\)  is increasing, relation (<a href="#4.4">20</a>) follows. <div class="proof_wrapper" id="a0000000014">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Taking the remarks above, we construct the sequence of operators \((K^*_{1,m})_{m\geq m_0}\). <div class="definition_thmwrapper " id="def_4.1">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">8</span>
  </div>
  <div class="definition_thmcontent">
  <p><em>If \(m\in \mathbb N_1\), we define the operator <div class="displaymath" id="4.5">
  \begin{gather} \label{4.5} (K^*_{1,m}f)(x)\\ =\tfrac {m+1}{(2m)^m}\sum ^m_{k=0}\tbinom {m}{k} (2(m+1)x-1)^k(2m+1-2(m+1)x)^{m-k}\int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{m+1}}f(t){\rm d}t\nonumber \end{gather}
</div> for any \(f\in L_1([0,1])\) and any \(x\in \bigg[\tfrac {1}{2(m_0+1)}\, ;\tfrac {2m_0+1}{2(m_0+1)}\bigg]\). </em></p>

  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="rem_4.1">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">9</span>
  </div>
  <div class="remark_thmcontent">
  <p><em>In this case, we note \(J=\bigg[\tfrac {1}{2(m_0+1)}\, ;\tfrac {2m_0+1}{2(m_0+1)}\bigg]{=}I^{(1)}_{(m_0)}\).<span class="qed">â–¡</span></em></p>

  </div>
</div> <div class="lemma_thmwrapper " id="lem_4.2">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">10</span>
  </div>
  <div class="lemma_thmcontent">
  <p> We have </p>
<div class="equation" id="4.6">
<p>
  <div class="equation_content">
    \begin{equation} \label{4.6} (K^*_{1,m}e_0)(x)=1, \end{equation}
  </div>
  <span class="equation_label">22</span>
</p>
</div>
<div class="equation" id="4.7">
<p>
  <div class="equation_content">
    \begin{equation} \label{4.7} (K^*_{1,m}e_1)(x)=x \end{equation}
  </div>
  <span class="equation_label">23</span>
</p>
</div>
<p> and </p>
<div class="equation" id="4.8">
<p>
  <div class="equation_content">
    \begin{equation} \label{4.8} (K^*_{1,m}e_2)(x)=\tfrac {m-1}{m}\, x^2+\tfrac {1}{m}\, x-\tfrac {5m+3}{12m(m+1)^2} \end{equation}
  </div>
  <span class="equation_label">24</span>
</p>
</div>
<p> for any \(x\in I^{(1)}_{(m_0)}\) and \(m\in \mathbb N_1\). </p>

  </div>
</div> <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> Results immediately from the condition above and (<a href="#3.4">14</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> <div class="lemma_thmwrapper " id="lem_4.3">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">11</span>
  </div>
  <div class="lemma_thmcontent">
  <p> The following identities </p>
<div class="equation" id="4.9">
<p>
  <div class="equation_content">
    \begin{equation} \label{4.9} (T_{m,0} K^*_{1,m})(x)=1, \end{equation}
  </div>
  <span class="equation_label">25</span>
</p>
</div>
<div class="equation" id="4.10">
<p>
  <div class="equation_content">
    \begin{equation} \label{4.10} (T_{m,1}K^*_{1,m})(x)=0 \end{equation}
  </div>
  <span class="equation_label">26</span>
</p>
</div>
<p> and </p>
<div class="equation" id="4.11">
<p>
  <div class="equation_content">
    \begin{equation} \label{4.11} (T_{m,2}K^*_{1,m})(x)=mx(1-x)-\tfrac {m(5m+3)}{12(m+1)^2} \end{equation}
  </div>
  <span class="equation_label">27</span>
</p>
</div>
<p> hold, for any \(x\in I^{(1)}_{(m_0)}\) and \(m\in \mathbb N_1\) </p>

  </div>
</div> <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 using Lemma <a href="#lem_4.2">10</a> and relation (<a href="#2.2">3</a>), we have </p>
<div class="displaymath" id="a0000000018">
  \begin{align*}  (T_{m,0}K^*_{1,m})(x) & =(K^*_{1,m}e_0)(x)=1,\\ (T_{m,1}K^*_{1,m})(x) & =m(K^*_{1,m}\psi _x)(x) =m((K^*_{1,m}e_1)(x)-x(K^*_{1,m}e_0)(x))=0 \end{align*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000019">
  \[  (T_{m,2}K^*_{1,m}(x)=m^2(K^*_{1,m}\psi ^2_x)(x)=m^2\big((K^*_{1,m}e_1)(x)+x^2 (K^*_{1,m}e_0)(x)\big),  \]
</div>
<p> from where (<a href="#4.11">27</a>) follows. <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> <div class="lemma_thmwrapper " id="lem_4.4">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">12</span>
  </div>
  <div class="lemma_thmcontent">
  <p> We have that </p>
<div class="equation" id="4.12">
<p>
  <div class="equation_content">
    \begin{equation} \label{4.12} \lim _{m\to \infty }(T_{m,0}K^*_{1,m})(x)=1, \end{equation}
  </div>
  <span class="equation_label">28</span>
</p>
</div>
<div class="equation" id="4.13">
<p>
  <div class="equation_content">
    \begin{equation} \label{4.13} \lim _{m\to \infty }\tfrac {(T_{m,2}K^*_{1,m})(x)}{m}=x(1-x) \end{equation}
  </div>
  <span class="equation_label">29</span>
</p>
</div>
<p> for any \(x\in I^{(1)}_{(m_0)}\) and \(m(0)\in \mathbb N\) exists such that </p>
<div class="equation" id="4.14">
<p>
  <div class="equation_content">
    \begin{equation} \label{4.14} \tfrac {(T_{m,2}K^*_{1,m})(x)}{m}\leq \tfrac {5}{4} \end{equation}
  </div>
  <span class="equation_label">30</span>
</p>
</div>
<p> for any \(x\in I^{(1)}_{(m_0)}\) and \(m\in \mathbb N_1\), \(m\geq m(0)\). </p>

  </div>
</div> <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> The relation (<a href="#4.12">28</a>) and (<a href="#4.13">29</a>) results taking (<a href="#4.9">25</a>) and (<a href="#4.12">28</a>) into account. By using the definition of limit a function and because \(x(1-x)\leq \tfrac {1}{4}\)  for any \(x\in [0,1]\), from (<a href="#4.13">29</a>) the inequality (<a href="#4.14">30</a>) is obtained. <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> <div class="theorem_thmwrapper " id="th_4.1">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">13</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(f:[0,1]\to \mathbb R\) be a continuous function on \([0,1]\). Then </p>
<div class="equation" id="4.15">
<p>
  <div class="equation_content">
    \begin{equation} \label{4.15} \lim _{m\to \infty }K^*_{1,m} f=f \end{equation}
  </div>
  <span class="equation_label">31</span>
</p>
</div>
<p> uniformly on \(I^{(1)}_{(m_0)}\) and there exists \(m(0)\in \mathbb N_1\) such that </p>
<div class="equation" id="4.16">
<p>
  <div class="equation_content">
    \begin{equation} \label{4.16} \left|(K^*_{1,m}f)(x)-f(x)\right|\leq \tfrac {9}{4}\, \omega \left(f; \tfrac {1}{\sqrt{m}}\right) \end{equation}
  </div>
  <span class="equation_label">32</span>
</p>
</div>
<p> for any \(x\in I^{(1)}_{(m_0)}\) and \(m\in \mathbb N_1\), \(m\geq m_0\). </p>

  </div>
</div> <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> We apply Theorem <a href="#th_2.1">2</a> and Corollary <a href="#cor_2.1">3</a> for \(s=0\), \(\alpha _0=0\), \(\alpha _2=1\), \(k_0=1\) and \(k_2=\tfrac {9}{4}\) . <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> <div class="theorem_thmwrapper " id="th_4.2">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">14</span>
  </div>
  <div class="theorem_thmcontent">
  <p> If \(f\in C([0,1])\), \(x\in I^{(1)}_{(m_0)}\), \(f\) is two times differentiable in neighborhood of \(x\) and \(f^{(2)}\) is continuous on \(x\), then </p>
<div class="equation" id="4.17">
<p>
  <div class="equation_content">
    \begin{equation} \label{4.17} \lim _{m\to \infty }m\left((K^*_{1,m}f)(x)-f(x)\right)= \tfrac {1}{2}\, x(1-x)f^{(2)}(x). \end{equation}
  </div>
  <span class="equation_label">33</span>
</p>
</div>

  </div>
</div> <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> We use the results from Corollary <a href="#cor_2.1">3</a> for \(s=2\). <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>
<h1 id="a0000000027">5 Kantorovich-type operators preserving the test <br />functions \(e_0\) and \(e_2\)</h1>
<p>In this section, we impose the conditions \((K^*_m e_0)(x)=e_0(x)\) and <br />\((K^*_m e_2)(x)=e_2(x)\), for any \(x\in J\) and \(m\in \mathbb N_1\). Then, taking (<a href="#3.2">12</a>) and (<a href="#3.4">14</a>) into account, we have \((\alpha _m(x)+\beta _m(x))^m=1\) and </p>
<div class="displaymath" id="a0000000028">
  \begin{align*} &  \tfrac {(\alpha _m(x)+\beta _m(x))^{m-2}}{3(m+1)^2}\,  \big(3m(m-1)\alpha ^2_m(x)+6m\alpha _m(x)(\alpha _m(x)+\beta _m(x))\\ & \quad + (\alpha _m(x)+\beta _m(x))^2\big)=x^2, \end{align*}
</div>
<p> from where </p>
<div class="equation" id="5.1">
<p>
  <div class="equation_content">
    \begin{equation} \label{5.1} \alpha _m(x)+\beta _m(x)=1 \end{equation}
  </div>
  <span class="equation_label">34</span>
</p>
</div>
<p> and </p>
<div class="equation" id="5.2">
<p>
  <div class="equation_content">
    \begin{equation} \label{5.2} 3m(m-1)\alpha ^2_m(x)+6m\alpha _m(x)+1-3(m+1)^2 x^2=0. \end{equation}
  </div>
  <span class="equation_label">35</span>
</p>
</div>
<p> The discriminant of the equation (<a href="#5.2">35</a>) is </p>
<div class="displaymath" id="a0000000029">
  \[  \Delta _m=12 m(2m+1+3(m-1)(m+1)^2 x^2)\geq 0,  \]
</div>
<p> for any \(x\in J\) and any \(m\in \mathbb N_1\), \(m\geq 2\) and we note </p>
<div class="equation" id="5.3">
<p>
  <div class="equation_content">
    \begin{equation} \label{5.3} \delta _m(x)=3m(2m+1+3(m-1)(m+1)^2 x^2), \end{equation}
  </div>
  <span class="equation_label">36</span>
</p>
</div>
<p> \(x\in J\), \(m\in \mathbb N_1\), \(m\geq 2\). </p>
<p>If \(m\in \mathbb N_1\), \(m\geq 2\), then for </p>
<div class="equation" id="5.4">
<p>
  <div class="equation_content">
    \begin{equation} \label{5.4} x\geq \tfrac {1}{(m+1)\sqrt{3}} \end{equation}
  </div>
  <span class="equation_label">37</span>
</p>
</div>
<p> the inequality \(\tfrac {1-3(m+1)^2 x^2}{3m(m-1)}\leq 0\) is true, so the equation from (<a href="#5.2">35</a>) has exactly one positive solution. This is </p>
<div class="equation" id="5.5">
<p>
  <div class="equation_content">
    \begin{equation} \label{5.5} \alpha _m(x)=\tfrac {-3m+\sqrt{\delta _m(x)}}{3m(m-1)} \end{equation}
  </div>
  <span class="equation_label">38</span>
</p>
</div>
<p> and then </p>
<div class="equation" id="5.6">
<p>
  <div class="equation_content">
    \begin{equation} \label{5.6} \beta _m(x)=\tfrac {3m^2-\sqrt{\delta _m(x)}}{3m(m-1)} \end{equation}
  </div>
  <span class="equation_label">39</span>
</p>
</div>
<p> where \(x\in J\), and to satisfy (<a href="#5.4">37</a>), \(m\in \mathbb N_1\), \(m\geq 2\). </p>
<p><div class="lemma_thmwrapper " id="lem_5.1">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">15</span>
  </div>
  <div class="lemma_thmcontent">
  <p> Let \(m\in \mathbb N_1\), \(m\geq 2\). Then \(\beta _m(x)\geq 0\), \(x\geq 0\) if and only if </p>
<div class="equation" id="5.7">
<p>
  <div class="equation_content">
    \begin{equation} \label{5.7} 0\leq x\leq \tfrac {\sqrt{3m^2+3m+1}}{(m+1)\sqrt{3}}\, . \end{equation}
  </div>
  <span class="equation_label">40</span>
</p>
</div>

  </div>
</div> <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> From \(\beta _m(x)\geq 0\) we have \(3m^2\geq \sqrt{\delta _m(x)}\), equivalent after calculus to \(x^2\leq \tfrac {3m^2+3m+1}{3(m+1)^2}\) , from where (<a href="#5.7">40</a>) follows. <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> <div class="lemma_thmwrapper " id="lem_5.2">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">16</span>
  </div>
  <div class="lemma_thmcontent">
  <p> Let \(m\in \mathbb N_1\), \(m\geq 2\). If \(x\in \bigg[\tfrac {1}{(m+1)\sqrt{3}}\, ;\tfrac {\sqrt{3m^2+3m+1}}{(m+1)\sqrt{3}}\bigg]\), then \(\alpha _m(x)\geq 0\) and \(\beta _m(x)\geq 0\). </p>

  </div>
</div> <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> Results immediately from (<a href="#5.4">37</a>) and (<a href="#5.7">40</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><div class="lemma_thmwrapper " id="lem_5.3">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">17</span>
  </div>
  <div class="lemma_thmcontent">
  <p> The following </p>
<div class="equation" id="5.8">
<p>
  <div class="equation_content">
    \begin{equation} \label{5.8} \left[\tfrac {1}{(m_0\! +\! 1)\sqrt{3}}\, ; \tfrac {\sqrt{3m^2_0\! +\! 3m_0\! +\! 1}}{(m_0+1)\sqrt{3}}\right] \! \subset \!  \left[\tfrac {1}{(m\! +\! 1)\sqrt{3}}\, ;\tfrac {\sqrt{3m^2\! +\! 3m\! +\! 1}}{(m+1)\sqrt{3}}\right] \subset [0,1] \end{equation}
  </div>
  <span class="equation_label">41</span>
</p>
</div>
<p> hold, for any \(m\in \mathbb N_1\). </p>

  </div>
</div> <div class="proof_wrapper" id="a0000000034">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> By using that the functions \(\tfrac {1}{(m+1)\sqrt{3}}\)  and \(\tfrac {\sqrt{3m^2+3m+1}}{(m+1)\sqrt{3}}\) are decreasing, relations from (<a href="#5.8">41</a>) follows. <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> <div class="definition_thmwrapper " id="def_5.1">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">18</span>
  </div>
  <div class="definition_thmcontent">
  <p><em>If \(m\in \mathbb N_1\), we define the operator \(K^*_{2,m}\) by <div class="displaymath" id="5.9">
  \begin{align} \label{5.9} (K^*_{2,m}f)(x) &  =\tfrac {m+1}{3m(m-1))^m}\sum ^m_{k=0}\tbinom {m}{k} \left(-3m+\sqrt{\delta _m(x)}\right)^k\\ & \quad \times \left(3m^2-\sqrt{\delta _m(x)}\right)^{m-k} \int ^{\tfrac {k+1}{m+1}}_{\tfrac {k}{m+1}}f(t)dt\nonumber \end{align}
</div> for any \(f\in L_1([0,1])\) and any \(x\in \bigg[\tfrac {1}{(m_0+1)\sqrt{3}}\, ;\tfrac {\sqrt{3m_0^2+3m_0+1}}{(m_0+1)\sqrt{3}}\bigg]\). </em></p>

  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="rem_5.1">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">19</span>
  </div>
  <div class="remark_thmcontent">
  <p><em>In this section, we note <div class="displaymath" id="a0000000036">
  \[  J=\bigg[\tfrac {1}{(m_0+1)\sqrt{3}}\, ;\tfrac {\sqrt{3m_0^2+3m_0+1}}{(m_0+1)\sqrt{3}}\bigg] {=}I^{(2)}_{(m_0)}.\hfil \qed  \]
</div> </em></p>

  </div>
</div> <div class="lemma_thmwrapper " id="lem_5.4">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">20</span>
  </div>
  <div class="lemma_thmcontent">
  <p> We have </p>
<div class="equation" id="5.10">
<p>
  <div class="equation_content">
    \begin{equation} \label{5.10} (K^*_{2,m}e_0)(x)=1, \end{equation}
  </div>
  <span class="equation_label">43</span>
</p>
</div>
<div class="equation" id="5.11">
<p>
  <div class="equation_content">
    \begin{equation} \label{5.11} (K^*_{2,m}e_1)(x)=\tfrac {2\sqrt{\delta _m(x)}-3m-3}{6(m-1)(m+1)} \end{equation}
  </div>
  <span class="equation_label">44</span>
</p>
</div>
<p> and </p>
<div class="equation" id="5.12">
<p>
  <div class="equation_content">
    \begin{equation} \label{5.12} (K^*_{1,m} e_2)(x)=x^2 \end{equation}
  </div>
  <span class="equation_label">45</span>
</p>
</div>
<p> for any \(x\in I^{(2)}_{(m_0)}\) and \(m\in \mathbb N_1\). </p>

  </div>
</div> <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> It is inferred from the conditions above and (<a href="#3.3">13</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> <div class="lemma_thmwrapper " id="lem_5.5">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">21</span>
  </div>
  <div class="lemma_thmcontent">
  <p> The following identities </p>
<div class="equation" id="5.13">
<p>
  <div class="equation_content">
    \begin{equation} \label{5.13} (T_{m,0}K^*_{1,m})(x)=1, \end{equation}
  </div>
  <span class="equation_label">46</span>
</p>
</div>
<div class="equation" id="5.14">
<p>
  <div class="equation_content">
    \begin{equation} \label{5.14} (T_{m,1}K^*_{2,m})(x)=m\left(\tfrac {2\sqrt{\delta _m(x)}-3m-3}{6(m-1)(m+1)}-x\right) \end{equation}
  </div>
  <span class="equation_label">47</span>
</p>
</div>
<p> and </p>
<div class="equation" id="5.15">
<p>
  <div class="equation_content">
    \begin{equation} \label{5.15} (T_{m,2}K^*_{2,m})(x)=2m^2 x\left(x-\tfrac {2\sqrt{\delta _m(x)}-3m-3}{6(m-1)(m+1)}\right). \end{equation}
  </div>
  <span class="equation_label">48</span>
</p>
</div>
<p> hold, for any \(x\in I^{(2)}_{(m_0)}\) and \(m\in \mathbb N_1\) </p>

  </div>
</div> <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> By using Lemma <a href="#lem_4.4">12</a> and (<a href="#2.2">3</a>), we have that </p>
<div class="displaymath" id="a0000000040">
  \begin{align*}  (T_{m,0} K^*_{2,m})(x) & =(K^*_{2,m} e_0)(x)=1,\\ (T_{m,1}K^*_{2,m})(x) & =m(K^*_{2,m}\psi _x)(x)=m\left((K^*_{2,m}e_1)(x)-x(K^*_{2,m}e_0)(x)\right), \end{align*}
</div>
<p> so (<a href="#5.14">47</a>) holds and </p>
<div class="displaymath" id="a0000000041">
  \begin{align*}  (T_{m,2}K^*_{2,m})(x) &  =m^2(K^*_{2,m}\psi ^2_x)(x)\\ &  =m^2\left((K^*_{2,m}e_2)(x)-2x (K^*_{2,m}e_1)(x)+x^2(K_{2,m}e_0)(x)\right), \end{align*}
</div>
<p> from where (<a href="#5.15">48</a>) is obtained. <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> <div class="lemma_thmwrapper " id="lem_5.6">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">22</span>
  </div>
  <div class="lemma_thmcontent">
  <p> The following identity </p>
<div class="equation" id="5.16">
<p>
  <div class="equation_content">
    \begin{equation} \label{5.16} \lim _{m\to \infty }m\left(\tfrac {2\sqrt{\delta _m(x)}-3m-3}{6(m-1)(m+1)}-x\right)= \tfrac {x-1}{2} \end{equation}
  </div>
  <span class="equation_label">49</span>
</p>
</div>
<p> holds for any \(x\in I^{(2)}_{(m_0)}\). </p>

  </div>
</div> <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> We have </p>
<div class="displaymath" id="a0000000044">
  \begin{gather*}  \lim _{m\to \infty }\left(\tfrac {m^2}{(m-1)(m+1)}\cdot \tfrac {\sqrt{\delta _m(x)}-3(m-1)(m+1)x}{3m} -\tfrac {m}{2(m-1)}\right)\\ =-\tfrac {1}{2}+\lim _{m\to \infty }\tfrac {\sqrt{\delta _m(x)}-3(m-1)(m+1)x}{3m}\\ =-\tfrac {1}{2}+\lim _{m\to \infty }\tfrac {\delta _m(x)-9(m-1)^2(m+1)^2 x^2} {3m\left(\sqrt{\delta _m(x)}+3(m-1)(m+1)x\right)} \end{gather*}
</div>
<p> and after a few calculations, identity (<a href="#5.16">49</a>) follows. <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> <div class="lemma_thmwrapper " id="lem_5.7">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">23</span>
  </div>
  <div class="lemma_thmcontent">
  <p> We have that </p>
<div class="equation" id="5.17">
<p>
  <div class="equation_content">
    \begin{equation} \label{5.17} \lim _{m\to \infty }(T_{m,0}K^*_{2,m})(x)=1, \end{equation}
  </div>
  <span class="equation_label">50</span>
</p>
</div>
<div class="equation" id="5.18">
<p>
  <div class="equation_content">
    \begin{equation} \label{5.18} \lim _{m\to \infty }\tfrac {(T_{m,2}K^*_{2,m})(x)}{m}=x(1-x) \end{equation}
  </div>
  <span class="equation_label">51</span>
</p>
</div>
<p> for any \(x\in I^{(2)}_{(m_0)}\) and \(m(0)\in \mathbb N\) exists such that </p>
<div class="equation" id="5.19">
<p>
  <div class="equation_content">
    \begin{equation} \label{5.19} \tfrac {(T_{m,2}K^*_{2,m})(x)}{m}\leq \tfrac {5}{4} \end{equation}
  </div>
  <span class="equation_label">52</span>
</p>
</div>
<p> for any \(x\in I^{(2)}_{(m_0)}\) and \(m\in \mathbb N_1\), \(m\geq m(0)\). </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> The relations (<a href="#5.17">50</a>) and (<a href="#5.18">51</a>) imply (<a href="#5.13">46</a>), (<a href="#5.15">48</a>) and (<a href="#5.16">49</a>). By using the definition of the limit of a function and because \(x(1-x)\leq \tfrac {1}{4}\)  for any \(x\in [0,1]\), from (<a href="#5.18">51</a>) the relation (<a href="#5.19">52</a>) is obtained. <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> <div class="theorem_thmwrapper " id="th_5.1">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">24</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(f:[0,1]\to \mathbb R\) be a continuous function on \([0,1]\). Then </p>
<div class="equation" id="5.20">
<p>
  <div class="equation_content">
    \begin{equation} \label{5.20} \lim _{m\to \infty }K^*_{2,m} f=f \end{equation}
  </div>
  <span class="equation_label">53</span>
</p>
</div>
<p> uniformly on \(x\in I^{(2)}_{(m_0)}\) and there exists \(m(0)\in \mathbb N\) such that </p>
<div class="equation" id="5.21">
<p>
  <div class="equation_content">
    \begin{equation} \label{5.21} \left|(K_{2,m}^* f)(x)-f(x)\right|\leq \tfrac {9}{4}\, \omega \left(f; \tfrac {1}{\sqrt{m}}\right) \end{equation}
  </div>
  <span class="equation_label">54</span>
</p>
</div>
<p> for any \(x\in I^{(2)}_{(m_0)}\) and any \(m\in \mathbb N_1\), \(m\geq m(0)\). </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> Theorem <a href="#th_5.1">24</a> is a results from Theorem <a href="#th_2.1">2</a> and Corollary <a href="#cor_2.1">3</a> for \(s=0\), \(\alpha _0=0\), \(\alpha _2=1\), \(k_0=1\) and \(k_2=\tfrac {5}{4}\) . <div class="proof_wrapper" id="a0000000049">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="theorem_thmwrapper " id="th_5.2">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">25</span>
  </div>
  <div class="theorem_thmcontent">
  <p> If \(f\in C([0,1])\), \(x\in I^{(2)}_{(m_0)}\), \(f\) is two times differentiable in a neighborhood of \(x\), \(f^{(2)}\) is continuous in \(x\), then </p>
<div class="equation" id="5.22">
<p>
  <div class="equation_content">
    \begin{equation} \label{5.22} \lim _{m\to \infty }m\left((K^*_{2,m}f)(x)-f(x)\right)= \tfrac {x-1}{2}\,  f^{(1)}(x)+\tfrac {x(1-x)}{2}\,  f^{(2)}(x). \end{equation}
  </div>
  <span class="equation_label">55</span>
</p>
</div>

  </div>
</div> <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> Taking Lemma <a href="#lem_5.6">22</a> into account and applying Theorem <a href="#th_2.1">2</a> for \(s=2\), we obtain the relation (<a href="#5.22">55</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><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">O. Agratini</i>, <em>An asymptotic formula for a class of approximation processes of King’s type</em>, Studia Sci. Math. Hungar, <b class="bfseries">47</b> (2010) No. 4, pp.&#160;435–444. </p>
</dd>
  <dt><a name="2">2</a></dt>
  <dd><p><i class="sc">P.I. Braica, O. T.Pop</i> and <i class="sc">A. D. Indrea</i>, <em>About a King-type operator</em>, Appl. Math. Inf. Sci., <b class="bfseries">6</b> (2012) No. 1, pp.&#160;145–148. </p>
</dd>
  <dt><a name="3">3</a></dt>
  <dd><p><i class="sc">L.V. Kantorovich</i>, <em>Sur certain développements suivant les polinômes de la forme de S. Bernstein</em>, C. R. Acad. URSS, <b class="bfseries">I</b>, <b class="bfseries">II</b> (1930), pp.&#160;563–568, pp.&#160;595–600. </p>
</dd>
  <dt><a name="4">4</a></dt>
  <dd><p><i class="sc">J.P. King</i>, <em>Positive linear operators which preserve \(x^2\)</em>, Acta Math. Hungar., <b class="bfseries">99</b> (3) (2003), pp.&#160;203–208. </p>
</dd>
  <dt><a name="5">5</a></dt>
  <dd><p><i class="sc">H. Gonska</i> and <i class="sc">P. Pitul</i>, <em>Remarks on an article of J. P. King</em>, Comment. Math. Univ. Carolina, <b class="bfseries">46</b> (2005) No. 4, pp.&#160;645–665. </p>
</dd>
  <dt><a name="6">6</a></dt>
  <dd><p><i class="sc">O.T. Pop</i>, <em>About some linear and positive operators defined by infinite sum</em>, Dem. Math., <b class="bfseries">XXXIX</b> (2006) No. 2, pp.&#160;377–388. </p>
</dd>
  <dt><a name="7">7</a></dt>
  <dd><p><i class="sc">O.T. Pop, D. Bărbosu</i> and <i class="sc">P.I. Braica</i>, <em>Some general linears and positive operators</em> (to appear) </p>
</dd>
  <dt><a name="8">8</a></dt>
  <dd><p><a href ="http://ictp.acad.ro/jnaat/journal/article/view/2005-vol34-no1-art9"> <i class="sc">O.T. Pop</i>, <em>The generalization of Voronovskaja’s theorem for a class of linear and positive operators</em>, Rev. Anal. Numér. Théor. Approx., <b class="bfseries">34</b> (2005), pp.&#160;79–91. <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="9">9</a></dt>
  <dd><p><i class="sc">E. Voronovskaja</i>, <em>Détermination de la forme asymptotique d’approximation des fonctions par les polynômes de S. N. Bernstein</em>, C. R. Acad. Sci. URSS, 1932, pp.&#160;79–85. </p>
</dd>
</dl>


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