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<title>Some inequalities for a Stancu type operator <i class="it">via</i> (1,1) box convex functions: Some inequalities for a Stancu type operator <i class="it">via</i> (1,1) box convex functions</title>
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<h1>Some inequalities for a Stancu type operator <i class="it">via</i> (1,1) box convex functions</h1>
<p class="authors">
<span class="author">Ioan Gavrea\(^\ast \) Daniel Ianoși\(^\ast \)</span>
</p>
<p class="date">September 20, 2021; accepted: October 16, 2021; published online: November 8, 2021.</p>
</div>
<div class="abstract"><p> In this paper we introduce a Stancu type operator and we prove inequalities of Raşa’s type. </p>
<p><b class="bf">MSC.</b> 41A36, 41A17. </p>
<p><b class="bf">Keywords.</b> Stancu operators, box convex functions. </p>
</div>
<p>\(^\ast \)Department of Mathematics, Technical University of Cluj Napoca, Str. Memorandumului nr. 28, 400114 Cluj-Napoca, Romania, e-mail: <span class="tt">Ioan.Gavrea@math.utcluj.ro</span>, <span class="tt">dan.ianosi@yahoo.com</span>. </p>
<h1 id="a0000000002">1 Introduction</h1>
<p>D.D. Stancu, in <span class="cite">
	[
	<a href="#12" >12</a>
	]
</span> using a probabilistic method, constructs a linear positive polynomial operators \( L_{n,r}^{\alpha ,\beta } \) of Bernstein type, depending on a non-negative integer parameter \(r\) and on two parameters \(\alpha \) and \(\beta \), such that \(0\leq \alpha \leq \beta \). More precisely the operators are defined by </p>
<div class="equation" id="f.1.1">
<p>
  <div class="equation_content">
    \begin{equation}  \left(L_{n,r}^{\alpha ,\beta }f \right)(x)=\sum _{k=0}^{n}w_{n,k,r}(x)f\left( \tfrac {k+\alpha }{n+\beta }\right) \label{f.1.1} \end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p> for \( f\in C[0,1] \), where </p>
<div class="displaymath" id="f.1.2">
  \begin{align} &  w_{n,k,r}(x) = \label{f.1.2} \\ & =\begin{cases}  \binom {n-r}{k} x^k(1-x)^{n-r-k+1}, & \mbox{ if } 0\leq k{\lt}r, \\ \binom {n-r}{k}x^k(1-x)^{n-r-k+1}+\binom {n-r}{k-r}x^{k-r+1}(1-x)^{n-k}, & \mbox{ if } r\leq k\leq n-r, \\ \tbinom {n-r}{k-r} x^{k-r+1}(1-x)^{n-k}, &  \mbox{ if } n-r{\lt} k\leq n . \end{cases} \nonumber \end{align}
</div>
<p>Let \( b_{n,k}(x) \) be the fundamental Bernstein polynomials, \( k,n\in \mathbb {N} \) defined by </p>
<div class="displaymath" id="a0000000003">
  \begin{equation*}  b_{n,k}(x) = \left\{ \begin{array}{l} \binom {n}{k} x^k(1-x)^{n-k}, \mbox{ if } 0\leq k\leq n \\[2mm] 0,\; \: \: \text{otherwise} \end{array} \right. \end{equation*}
</div>
<p>The operators \( L_{n,r}^{\alpha ,\beta } \) can be written in the following forms </p>
<div class="displaymath" id="f.1.3">
  \begin{align}  \left( L_{n,r}^{\alpha ,\beta }f\right) (x)=& \sum _{k=0}^{n-r}b_{n-r,k}(x)\left[(1-x)f\left( \tfrac {k+\alpha }{n+\beta }\right)+xf\left( \tfrac {k+r+\alpha }{n+\beta }\right) \right] \label{f.1.3} \\ \left( L_{n,r}^{\alpha ,\beta }f\right)(x)=& (1-x)\sum _{k=0}^{n-r}b_{n-r,k}(x)f\left( \tfrac {k+\alpha }{n+\beta }\right)+x\sum _{k=r}^{n}b_{n-r,k-r}(x)f\left( \tfrac {k+\alpha }{n+\beta }\right) . \label{f.1.4} \end{align}
</div>
<p><div class="remark_thmwrapper " id="a0000000004">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">1</span>
  </div>
  <div class="remark_thmcontent">
  <p>For \( r=0 \), \( L_{n,0}^{\alpha ,\beta }=L_{n}^{\alpha ,\beta } \) the operator was introduced by D.D. Stancu in <span class="cite">
	[
	<a href="#11" >11</a>
	]
</span>. For \( r=0, \alpha =\beta =0 \), \( L_{n}^{0,0}\) coincides with the Bernstein operator \( B_n \). <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>Inspired by <a href="#f.1.4" class="eqref">7</a> we construct the following operator. </p>
<p>Let \( \varphi \) be an increasing differential function, \( \varphi :[0,1]\rightarrow [0,1], r,n\in \mathbb {N} \) such that \( n{\gt}2r, 0\leq \alpha \leq \beta \leq 1 \). </p>
<p>Then for \( f\in C[0,1] \) we defined \( L_{n,r,\varphi }^{\alpha ,\beta } \) by: </p>
<div class="equation" id="f.1.5">
<p>
  <div class="equation_content">
    \begin{equation}  \left( L_{n,r,\varphi }^{\alpha ,\beta }f\right) (x)=(1-\varphi (x))\sum _{k=0}^{n-r}b_{n-r,k}(x)f\left( \tfrac {k+\alpha }{n+\beta }\right)\\ +\varphi (x)\sum _{k=r}^{n}b_{n-r,k-r}(x)f\left( \tfrac {k+\alpha }{n+\beta }\right), \label{f.1.5} \end{equation}
  </div>
  <span class="equation_label">8</span>
</p>
</div>
<p> \( x\in [0,1] \). </p>
<p>The operator defined by <a href="#f.1.5" class="eqref">8</a> can be also written as: </p>
<div class="equation" id="f.1.6">
<p>
  <div class="equation_content">
    \begin{equation}  \left( L_{n,r,\varphi }^{\alpha ,\beta }f\right) (x)=\sum _{k=0}^{n}w_{n,r,k,\varphi }(x)f\left( \tfrac {k+\alpha }{n+\beta }\right), \label{f.1.6} \end{equation}
  </div>
  <span class="equation_label">9</span>
</p>
</div>
<p> where </p>
<div class="equation" id="f.1.7">
<p>
  <div class="equation_content">
    \begin{equation}  w_{n,r,k,\varphi }(x) = \left\{ \begin{array}{l} \displaystyle \left( 1-\varphi (x)\right) b_{n-r,k}(x), \mbox{ if } 0\leq k< r \\ \displaystyle \left(1-\varphi (x)\right)b_{n-r,k}(x)+\varphi (x)b_{n-r,k-r}(x),\mbox{ if } r\leq k\leq n-r \\ \displaystyle \varphi (x)b_{n-r,k-r}(x), \mbox{ if } n-r< k\leq n . \end{array} \right. \label{f.1.7} \end{equation}
  </div>
  <span class="equation_label">10</span>
</p>
</div>
<p>The sequence \( \big( L_{n,r,\varphi }^{\alpha ,\beta }f\big)_{n\in \mathbb {N} } \) is a linear positive sequence and for \( \varphi =x \) it becomes Stancu’s sequence \( \big( L_{n}^{\alpha ,\beta }f\big)_{n\in \mathbb {N} }\). </p>
<p>In <span class="cite">
	[
	<a href="#10" >10</a>
	]
</span>, Ioan Raşa recalled his 25-year-old problem relative to preservation of convexity by the Bernstein-Schnabl operators. </p>
<p>Prove or disprove that </p>
<div class="equation" id="f.1.8">
<p>
  <div class="equation_content">
    \begin{equation}  \sum _{i,j=0}^{n}\left( b_{n,i}(x)b_{n,j}(x)+b_{n,i}(y)b_{n,j}(y)-2b_{n,i}(x)b_{n,j}(y)\right)f\left( \tfrac {i+j}{2n}\right)\geq 0 \label{f.1.8} \end{equation}
  </div>
  <span class="equation_label">11</span>
</p>
</div>
<p> for each convex function \( f \in C[0,1] \) and for all \( x,y \in [0,1] \). </p>
<p>In <span class="cite">
	[
	<a href="#8" >8</a>
	]
</span>, J. Mrowiec, T. Rajba and S. Wa̧sowicz prove that <a href="#f.1.8" class="eqref">11</a> holds. In the proof of <a href="#f.1.8" class="eqref">11</a> they use the probability theory. As a tool they applied stochastic convex ordering. </p>
<p>In <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span>, U. Abel gave an elementary proof of <a href="#f.1.8" class="eqref">11</a>. An extension of <a href="#f.1.8" class="eqref">11</a> was considered in <span class="cite">
	[
	<a href="#5" >5</a>
	]
</span>, where \( b_{n,k}(x) \) were replaced by more general functions and the functional evaluations were replaced by \( A_{\frac{i+j}{2n}}(f) \), where \(\left\lbrace A_{t} \right\rbrace _{t\geq 0} \) is a set of linear positive functionals defined on a linear space of functions satisfying certain assumptions (see <span class="cite">
	[
	<a href="#5" >5</a>
	]
</span> for more details). </p>
<p>In <span class="cite">
	[
	<a href="#6" >6</a>
	]
</span>, Raşa’s conjecture <a href="#f.1.8" class="eqref">11</a> was studied for the case of Baskakov-Mastroianni operators. </p>
<p>Given \( f \in C[0,1] \), denote </p>
<div class="displaymath" id="a0000000005">
  \begin{equation*}  \Delta _{h}^{1}f(x):=\Delta _{h}f(x) := \left\{ \begin{array}{l} \displaystyle f(x+h)-f(x),\:  x,x+h \in [0,1] \\ \displaystyle 0,\:  \: \text{otherwise} \end{array} \right. \end{equation*}
</div>
<p> and for \( q\geq 1 \) </p>
<div class="displaymath" id="a0000000006">
  \begin{equation*}  \Delta _{h}^{q+1}f(x):=\Delta _{h}^{q}\left( \Delta _{h}f(x)\right). \end{equation*}
</div>
<p>A function \( f \) defined on \( [0,1] \) is called <i class="itshape">q</i>-monotone if \(\Delta _{h}^{q}f(x)\geq 0 \), for all \( h\geq 0 \) (see <span class="cite">
	[
	<a href="#2" >2</a>
	]
</span>). </p>
<p>Abel and Leviatan in <span class="cite">
	[
	<a href="#2" >2</a>
	]
</span> proved the following result. </p>
<p><div class="theoremA_thmwrapper " id="th: th.A">
  <div class="theoremA_thmheading">
    <span class="theoremA_thmcaption">
    Theorem
    </span>
    <span class="theoremA_thmlabel">A</span>
  </div>
  <div class="theoremA_thmcontent">
  <p> Let \( q,n\in \mathbb {N} \). If \( f\in C[0,1] \) is a <i class="itshape">q</i>-monotone function, then for all \( x,y \in [0,1] \) </p>
<div class="displaymath" id="a0000000007">
  \begin{align*} & \operatorname {sgn}(x-y)^q \times \\ & \sum _{\partial _1,\ldots ,\partial _q=0}^{n}\sum _{j=0}^{q}(-1)^{q-j}\tbinom {q}{j}\bigg( \prod _{i=1}^{j}b_{n,\partial _i}(x)\bigg) \bigg( \prod _{i=j+1}^{q}b_{n,\partial _i}(y)\bigg) \! \int _0^1 \! f\left(\tfrac {\partial _1+\ldots +\partial _q+\alpha t}{qn+\alpha } \right)dt\geq 0. \end{align*}
</div>

  </div>
</div> </p>
<p>In <span class="cite">
	[
	<a href="#7" >7</a>
	]
</span>, it was proved that <a href="#th: th.A">theorem A</a> follows from the fact that the tensorial product of Bernstein polynomials, <i class="it">i.e.</i> </p>
<div class="displaymath" id="a0000000008">
  \begin{equation*}  \left( B_{n,m}f\right)(x,y) =\sum _{i=0}^{n}\sum _{j=0}^{m}b_{n,i}(x)b_{m,j}(y)f\big(\tfrac {i}{n},\tfrac {j}{m} \big) \end{equation*}
</div>
<p> preserves \( (q,s) \)-convexity (see <span class="cite">
	[
	<a href="#4" >4</a>
	]
</span>). </p>
<p>The aim of this article is to prove an inequality of type <a href="#f.1.8" class="eqref">11</a> for operators \( L_{n,r,\varphi }^{\alpha ,\beta } \). </p>
<h1 id="a0000000009">2 Main results</h1>
<p>If \(f(x,y)\) is a function defined on the rectangle \( I\times J, I=[a,b], J=[c,d], x_1,\ldots ,x_m \) are distinct points from \( I \) and \(y_1,\ldots ,y_n \) are distinct points from \( J \), then the double divided difference is defined by </p>
<div class="displaymath" id="a0000000010">
  \begin{align*}  \Bigg[ \begin{matrix}  x_1, \ldots ,x_m 

\\ y_1, \ldots ,y_n 

\end{matrix};f\Bigg]=& \big[x_1,\ldots , x_m;\left[ y_1,\ldots ,y_n;f(x,\cdot )\right] \big] \\ =& \big[y_1,\ldots ,y_n;\left[ x_1,\ldots ,x_m;f(\cdot ,y)\right] \big]. \end{align*}
</div>
<p>Here, for the distinct points \( z_k\), \(k=\overline{1,n} \) and a function \( g \) defined on a set that contains these points, \( \left[z_1,\ldots ,z_n;g \right] \) denote the classical divided difference given by </p>
<div class="displaymath" id="a0000000011">
  \begin{align*}  \left[z_1;g \right]=& g(z_1) \\ \left[z_1,z_2;g \right]=& \frac{g(z_1)-g(z_2)}{z_1-z_2} \\ \left[z_1,z_2,\ldots ,z_{n-1},z_n;g \right]=& \frac{\left[z_1,\ldots ,z_{n-1};g \right]-\left[z_2,\ldots ,z_{n};g \right]}{z_1-z_n}. \end{align*}
</div>
<p>T. Popoviciu introduced the notion of \((m,n)\) convexity in <span class="cite">
	[
	<a href="#9" >9</a>
	, 
	p.
	
	78
	]
</span>. In <span class="cite">
	[
	<a href="#4" >4</a>
	]
</span>, S. Gal and P. Niculescu used box convexity of order \((m,n)\) for \((m,n)\) convexity. We will use the terminology introduced in <span class="cite">
	[
	<a href="#4" >4</a>
	]
</span> and say that the function \( f:I\times J\rightarrow \mathbb {R} \) is box convex of order \((m,n)\) if all divided differences </p>
<div class="displaymath" id="a0000000012">
  \begin{equation*}  \Bigg[ \begin{matrix}  x_1, 

&  \ldots ,

& x_{m+1} 

\\ y_1, 

&  \ldots ,

& y_{n+1} 

\end{matrix};f\Bigg] \end{equation*}
</div>
<p> are non-negative for any choice of distinct points \( x_1,\ldots ,x_{m+1} \) and \( y_1,\ldots ,y_{n+1} \). </p>
<p>For the remaining of this paper, we will consider \( I=J=[0,1] \). </p>
<p>In particular, let us observe that when \( m=n=1 \) for \( f:I\times I \rightarrow \mathbb {R}\), box convexity of order \( (1,1) \) is equivalent to </p>
<div class="displaymath" id="a0000000013">
  \begin{align*} &  \Bigg[ \begin{matrix}  x_1, 

&  x_{2} 

\\ y_1, 

&  y_{2} 

\end{matrix};f\Bigg] =\tfrac {1}{(x_1-x_2)(y_1-y_2)}\Big[ f(x_1,y_1)-f(x_1,y_2)-f(x_2,y_1)+f(x_2,y_2)\Big]\geq 0. \end{align*}
</div>
<p><div class="lemma_thmwrapper " id="lm: lm.2.1">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">2</span>
    <span class="lemma_thmtitle"><span class="cite">
	[
	<a href="#9" >9</a>
	]
</span></span>
  </div>
  <div class="lemma_thmcontent">
  <p>  If \( f\in C^{q+s}(I\times I)\) and \( \frac{\partial ^{q+s}f}{\partial x^{q} \partial y^{s}}\geq 0 \) on \( I\times I \), then \( f \) is \( (q,s) \) box convex. </p>

  </div>
</div> </p>
<p>In this paper we will use <a href="#lm: lm.2.1">lemma 2</a> for the case \( q=s=1 \). </p>
<p><div class="definition_thmwrapper " id="def: def.2.2">
  <div class="definition_thmheading">
    <span class="definition_thmcaption">
    Definition
    </span>
    <span class="definition_thmlabel">3</span>
  </div>
  <div class="definition_thmcontent">
  <p> Let \( U,V \) be two operators \( U,V:C[0,1]\rightarrow C[0,1] \) defined by </p>
<div class="displaymath" id="a0000000014">
  \begin{equation*}  (Uf)(x)=\sum _{i=0}^n a_i(x)f(x_i) \end{equation*}
</div>
<div class="displaymath" id="a0000000015">
  \begin{equation*}  (Vf)(y)=\sum _{j=0}^m b_j(y)f(y_j). \end{equation*}
</div>
<p> for all \( x,y\in I,x_i,y_j\in I, i=0,1,\ldots ,n, j=0,1,\ldots ,m. \) </p>
<p>The operator denoted by \( UV, UV:C(I\times I)\rightarrow C(I\times I) \) defined by </p>
<div class="equation" id="f.2.1">
<p>
  <div class="equation_content">
    \begin{equation}  UVf(x,y)=\sum _{i=0}^{n}\sum _{j=0}^{m}f(x_i,y_j)a_i(x)b_j(y) \label{f.2.1} \end{equation}
  </div>
  <span class="equation_label">12</span>
</p>
</div>
<p> is called the tensor product of the operators \(U \) and \( V\). </p>

  </div>
</div> </p>
<p>The following results will be proved in this paper. </p>
<p><div class="theorem_thmwrapper " id="th: th.2.3">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">4</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \( \varphi ,\psi :[0,1]\rightarrow [0,1] \) be two increasing differentiable functions and let \( L=L_{n,r_1,\varphi }^{\alpha ,\beta } L_{m,r_2,\psi }^{\gamma ,\delta }\) the tensor product of operators \( L_{n,r_1,\varphi }^{\alpha ,\beta }, L_{m,r_2,\psi }^{\gamma ,\delta }\) defined as in <a href="#f.1.6" class="eqref">9</a> <i class="it">i.e.</i> </p>
<div class="equation" id="f.2.2">
<p>
  <div class="equation_content">
    \begin{equation}  (Lf)(x,y)=\sum _{i=0}^{n}\sum _{j=0}^{m}w_{n,r_1,i,\varphi }(x)w_{m,r_2,j,\psi }(y)f\left(\tfrac {i+\alpha }{n+\beta },\tfrac {j+\gamma }{m+\delta } \right). \label{f.2.2} \end{equation}
  </div>
  <span class="equation_label">13</span>
</p>
</div>
<p>If \( f \) is \( (1,1) \) box convex function, then \( Lf \) is \( (1,1) \) box convex. </p>

  </div>
</div> </p>
<p><div class="theorem_thmwrapper " id="th: th.2.4">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">5</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \( f \) be a \( (1,1) \) box convex function on \( I\times I \) and \( x,x_1,y,y_1 \in [0,1] \). The following inequality holds: </p>
<div class="displaymath" id="f.2.3">
  \begin{align}  \label{f.2.3} & \operatorname {sgn}\left[(x-x_1)(y-y_1) \right]\cdot \\ & \cdot \sum _{i=0}^{n}\sum _{j=0}^{m}\left( w_{n,r_1,\varphi }(x)-w_{n,r_1,\varphi }(x_1)\right)\left( w_{m,r_2,\psi }(y)-w_{m,r_2,\psi }(y_1)\right) f\! \left(\tfrac {i+\alpha }{n+\beta },\tfrac {j+\gamma }{m+\delta } \right)\geq 0. \nonumber \end{align}
</div>

  </div>
</div> </p>
<p><div class="corollary_thmwrapper " id="cor: cor.2.5">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">6</span>
  </div>
  <div class="corollary_thmcontent">
  <p> If \( f \) is a \( (1,1) \) box convex function then </p>
<div class="displaymath" id="f.2.4">
  \begin{align}  \label{f.2.4} \sum _{i=0}^{n}\sum _{j=0}^{m}\Big[ & w_{n,r_1,\varphi }(x)w_{m,r_2,\psi }(x)+w_{n,r_1,\varphi }(y)w_{m,r_2,\psi }(y) \\ &  -w_{n,r_1,\varphi }(x)w_{m,r_2,\psi }(y)-w_{n,r_1,\varphi }(y)w_{m,r_2,\psi }(x)\Big] f \left(\tfrac {i+\alpha }{n+\beta },\tfrac {j+\gamma }{m+\delta } \right)\geq 0. \nonumber \end{align}
</div>

  </div>
</div> </p>
<p><div class="corollary_thmwrapper " id="cor: cor.2.6">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">7</span>
  </div>
  <div class="corollary_thmcontent">
  <p> If \( f:[0,1]\rightarrow \mathbb {R} \) is a convex function, then for all \( x,x_1,y,y_1 \in [0,1] \) the following inequality holds: </p>
<div class="displaymath" id="f.2.5">
  \begin{align}  \label{f.2.5} & \operatorname {sgn}\left[(x-x_1)(y-y_1) \right]\cdot \\ & \cdot \sum _{i=0}^{n}\sum _{j=0}^{m}\! \big[ w_{n,r_1,\varphi }(x)\! -\! w_{n,r_1,\varphi }(x_1)\big]\big[ w_{m,r_2,\psi }(y)\! -\! w_{m,r_2,\psi }(y_1)\big] \! f\! \left(\tfrac {i+\alpha }{2(n+\beta )}\! +\! \tfrac {j+\gamma }{2(m+\delta )} \right) \! \! \geq \!  0. \nonumber \end{align}
</div>

  </div>
</div> </p>
<h1 id="a0000000016">3 Proofs</h1>
<p><i class="itshape">Proof of <a href="#th: th.2.3">theorem 4</a>.</i> From <a href="#f.1.5" class="eqref">8</a> we have </p>
<div class="displaymath" id="f.2.6">
  \begin{align}  \label{f.2.6} & L’_{n,r_1,\varphi }(f)(x)= \\ & =\varphi ’(x)\sum _{i=0}^{n-r_1}b_{n-r_1,i}(x)\left(f\left( \tfrac {i+\alpha +r}{n+\beta }\right)-f\left( \tfrac {i+\alpha }{n+\beta }\right) \right) \nonumber \\ & \quad +\left( 1-\varphi (x)\right)(n-r_1) \sum _{i=0}^{n-r_1-1}b_{n-r_1-1,i}(x)\left(f\left( \tfrac {i+\alpha +1}{n+\beta }\right)-f\left( \tfrac {i+\alpha }{n+\beta }\right) \right) \nonumber \\ & \quad +\varphi (x)(n-r_1) \sum _{i=0}^{n-r_1-1}b_{n-r_1-1,i}(x)\left(f\left( \tfrac {i+r_1+\alpha +1}{n+\beta }\right)-f\left( \tfrac {i+r_1+\alpha }{n+\beta }\right) \right) \nonumber \\ & =\varphi ’(x)\tfrac {r_1}{n+\beta }\sum _{i=0}^{n-r_1}b_{n-r_1,i}(x)\left[ \tfrac {i+\alpha }{n+\beta },\tfrac {i+\alpha +r_1}{n+\beta };f\right] \nonumber \\ & \quad +\left( 1-\varphi (x)\right)\tfrac {n-r_1}{n+\beta }\sum _{i=0}^{n-r_1-1}b_{n-r_1-1,i}(x) \left[ \tfrac {i+\alpha }{n+\beta },\tfrac {i+\alpha +1}{n+\beta };f\right] \nonumber \\ & \quad +\varphi (x)\tfrac {n-r_1}{n+\beta } \sum _{i=0}^{n-r_1-1}b_{n-r_1-1,i}(x) \left[ \tfrac {i+r_1+\alpha }{n+\beta },\tfrac {i+r_1+\alpha +1}{n+\beta };f\right] \nonumber \end{align}
</div>
<p> and </p>
<div class="displaymath" id="f.2.7">
  \begin{align}  \label{f.2.7} L’_{m,r_2,\psi }(f)(y) =& \psi ’(y)\tfrac {r_2}{m+\delta }\sum _{j=0}^{m-r_2}b_{m-r_2,j}(y)\left[ \tfrac {j+\gamma }{m+\delta },\tfrac {j+\gamma +r_2}{m+\delta };f\right] \\ & +\left( 1-\psi (y)\right)\tfrac {m-r_2}{m+\delta }\sum _{j=0}^{m-r_2-1}b_{m-r_2-1,j}(y) \left[ \tfrac {j+\gamma }{m+\delta },\tfrac {j+\gamma +1}{m+\delta };f\right] \nonumber \\ & +\psi (y)\tfrac {m-r_2}{m+\delta } \sum _{j=0}^{m-r_2-1}b_{m-r_2-1,j}(y) \left[ \tfrac {j+r_2+\gamma }{m+\delta },\tfrac {j+r_2+\gamma +1}{m+\delta };f\right] \nonumber \end{align}
</div>
<p>From <a href="#f.2.6" class="eqref">17</a> and <a href="#f.2.7" class="eqref">18</a> we get </p>
<div class="displaymath" id="f.2.8">
  \begin{align}  \label{f.2.8} & \frac{\partial ^2L}{\partial x \partial y}(x,y)= \\ & =\varphi ’(x)\psi ’(y)\tfrac {r_1}{n+\beta } \tfrac {r_2}{m+\delta } \sum _{i=0}^{n-r_1}\sum _{j=0}^{m-r_2}b_{n-r_1,i}(x)b_{m-r_2,j}(y)\Bigg[ \begin{matrix}  \frac{i+\alpha }{n+\beta }, 

&  \frac{i+\alpha +r_1}{n+\beta } \nonumber 

\\ \frac{j+\gamma }{m+\delta }, 

&  \frac{j+\gamma +r_2}{m+\delta } 

\end{matrix};f\Bigg] \nonumber \\ & \quad +\left(1-\varphi (x) \right) \left( 1-\psi (y)\right)\tfrac {(n-r_1)(m-r_2)}{(n+\beta )(m+\delta )} \nonumber \\ & \quad \times \sum _{i=0}^{n-r_1-1}\sum _{j=0}^{m-r_2-1}b_{n-r_1-1,i}(x)b_{m-r_2-1,j}(y)\Bigg[ \begin{matrix}  \frac{i+\alpha }{n+\beta }, 

&  \frac{i+\alpha +1}{n+\beta } \nonumber 

\\ \frac{j+\gamma }{m+\delta }, 

&  \frac{j+\gamma +1}{m+\delta } 

\end{matrix};f\Bigg] \nonumber \\ & \quad +\varphi (x) \psi (y)\tfrac {(n-r_1)(m-r_2)}{(n+\beta )(m+\delta )} \nonumber \\ & \times \sum _{i=0}^{n-r_1-1}\sum _{j=0}^{m-r_2-1}b_{n-r_1-1,i}(x)b_{m-r_2-1,j}(y)\Bigg[ \begin{matrix}  \frac{i+r_1+\alpha }{n+\beta }, 

&  \frac{i+r_1+\alpha +1}{n+\beta } \nonumber 

\\ \quad \frac{j+r_2+\gamma }{m+\delta }, 

&  \frac{j+r_2+\gamma +1}{m+\delta } 

\end{matrix};f\Bigg], \end{align}
</div>
<p>\( f \) being a \( (1,1) \) box convex function, from <a href="#f.2.8" class="eqref">19</a> we obtain </p>
<div class="displaymath" id="a0000000017">
  \begin{align*}  \frac{\partial ^2L}{\partial x \partial y}(x,y)\geq 0. \end{align*}
</div>
<p> From the last inequality and <a href="#lm: lm.2.1">lemma 2</a> the theorem is proved. <span class="qed">â–¡</span></p>
<p><div class="proof_wrapper" id="a0000000018">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>[Proof of <a href="#th: th.2.4">theorem 5</a>.] If \( x=x_1 \) or \( y=y_1 \), then the relation <a href="#f.2.3" class="eqref">14</a> is trivial. </p>
<p>Let us suppose that \( x{\gt}x_1 \) and \( y=y_1 \). Then we have </p>
<div class="displaymath" id="a0000000019">
  \begin{align*} & \sum _{i=0}^{n}\sum _{j=0}^{m}[ w_{n,r_1,\varphi }(x)-w_{n,r_1,\varphi }(x_1)][ w_{m,r_2,\psi }(y)-w_{m,r_2,\psi }(y_1)] f\left(\tfrac {i+\alpha }{n+\beta },\tfrac {j+\gamma }{m+\delta } \right)= \\ & =\int _{x_1}^{x} \int _{y_1}^{y}\left( \sum _{i=0}^{n}\sum _{j=0}^{m} w’_{n,r_1,\varphi }(u)w’_{m,r_2,\psi }(v)f\left( \tfrac {i+\alpha }{n+\beta },\tfrac {j+\gamma }{m+\delta }\right)\right)dudv \\ & =\int _{x_1}^{x} \int _{y_1}^{y} \tfrac {\partial ^2L}{\partial x \partial y}(u,v)dudv. \end{align*}
</div>
<p>From <a href="#th: th.2.3">theorem 4</a> we have \( \frac{\partial ^2L}{\partial u \partial v}(u,v)\geq 0 \), for all \( (u,v)\in I\times I \). </p>
<p><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>[Proof of <a href="#cor: cor.2.5">corollary 6</a>.] Inequality <a href="#f.2.4" class="eqref">15</a> follows from the inequality </p>
<div class="displaymath" id="a0000000021">
  \begin{align*} & \sum _{i=0}^{n}\sum _{j=0}^{m}\left( w_{n,r_1,\varphi }(x)-w_{n,r_1,\varphi }(y)\right)\left( w_{m,r_2,\psi }(x)-w_{m,r_2,\psi }(y)\right) f\left(\tfrac {i+\alpha }{n+\beta },\tfrac {j+\gamma }{m+\delta } \right)\geq 0, \end{align*}
</div>
<p> which follows from <a href="#f.2.3" class="eqref">14</a>. </p>
<p><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>[Proof of <a href="#cor: cor.2.6">corollary 7</a>.] If the function \(f\) is a convex function then the function \(g:I\times I\rightarrow \mathbb {R}, g(x,y)=f\left(\frac{x+y}{2} \right) \) is a \((1,1) \) box convex function and so inequality <a href="#f.2.5" class="eqref">16</a> is inequality <a href="#f.2.3" class="eqref">14</a> for the function \(g\). </p>
<p><div class="remark_thmwrapper " id="a0000000023">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">8</span>
  </div>
  <div class="remark_thmcontent">
  <p>If \( f \) is a \((1,1) \) box convex function, in <a href="#f.2.3" class="eqref">14</a> the evaluation functional \(f\left(\frac{i+\alpha }{n+\beta },\frac{j+\gamma }{m+\delta } \right)\) can be replaced by </p>
<div class="displaymath" id="a0000000024">
  \begin{equation*}  \int _{0}^{1} \int _{0}^{1}f\left(\tfrac {i+\alpha u}{n+\beta },\tfrac {j+\gamma v}{m+\delta } \right) dudv, \qquad i=0,1,\ldots ,n, j=0,1,\ldots ,m. \end{equation*}
</div>

  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="a0000000025">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">9</span>
  </div>
  <div class="remark_thmcontent">
  <p>For \( m=n, r=0, \varphi (x)=\psi (x)=x \) the inequality <a href="#f.2.5" class="eqref">16</a> becomes <a href="#f.1.8" class="eqref">11</a>. <span class="qed">â–¡</span></p>

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


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