Some inequalities for a Stancu type operator via (1,1) box convex functions

Ioan Gavrea$^\ast $ Daniel Ianoși$^\ast $

September 20, 2021; accepted: October 16, 2021; published online: November 8, 2021.

In this paper we introduce a Stancu type operator and we prove inequalities of Raşa’s type.

MSC. 41A36, 41A17.

Keywords. Stancu operators, box convex functions.

$^\ast $Department of Mathematics, Technical University of Cluj Napoca, Str. Memorandumului nr. 28, 400114 Cluj-Napoca, Romania, e-mail: Ioan.Gavrea@math.utcluj.ro, dan.ianosi@yahoo.com.

1 Introduction

D.D. Stancu, in [ 12 ] 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

\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}

for $ f\in C[0,1] $, where

\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}

Let $ b_{n,k}(x) $ be the fundamental Bernstein polynomials, $ k,n\in \mathbb {N} $ defined by

\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*}

The operators $ L_{n,r}^{\alpha ,\beta } $ can be written in the following forms

\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}

Remark 1

For $ r=0 $, $ L_{n,0}^{\alpha ,\beta }=L_{n}^{\alpha ,\beta } $ the operator was introduced by D.D. Stancu in [ 11 ] . For $ r=0, \alpha =\beta =0 $, $ L_{n}^{0,0}$ coincides with the Bernstein operator $ B_n $. â–¡

Inspired by 7 we construct the following operator.

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 $.

Then for $ f\in C[0,1] $ we defined $ L_{n,r,\varphi }^{\alpha ,\beta } $ by:

\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}

$ x\in [0,1] $.

The operator defined by 8 can be also written as:

\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}

where

\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}

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} }$.

In [ 10 ] , Ioan Raşa recalled his 25-year-old problem relative to preservation of convexity by the Bernstein-Schnabl operators.

Prove or disprove that

\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}

for each convex function $ f \in C[0,1] $ and for all $ x,y \in [0,1] $.

In [ 8 ] , J. Mrowiec, T. Rajba and S. Wa̧sowicz prove that 11 holds. In the proof of 11 they use the probability theory. As a tool they applied stochastic convex ordering.

In [ 1 ] , U. Abel gave an elementary proof of 11. An extension of 11 was considered in [ 5 ] , 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 [ 5 ] for more details).

In [ 6 ] , Raşa’s conjecture 11 was studied for the case of Baskakov-Mastroianni operators.

Given $ f \in C[0,1] $, denote

\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*}

and for $ q\geq 1 $

\begin{equation*} \Delta _{h}^{q+1}f(x):=\Delta _{h}^{q}\left( \Delta _{h}f(x)\right). \end{equation*}

A function $ f $ defined on $ [0,1] $ is called q-monotone if $\Delta _{h}^{q}f(x)\geq 0 $, for all $ h\geq 0 $ (see [ 2 ] ).

Abel and Leviatan in [ 2 ] proved the following result.

Theorem A

Let $ q,n\in \mathbb {N} $. If $ f\in C[0,1] $ is a q-monotone function, then for all $ x,y \in [0,1] $

\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*}

In [ 7 ] , it was proved that theorem A follows from the fact that the tensorial product of Bernstein polynomials, i.e.

\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*}

preserves $ (q,s) $-convexity (see [ 4 ] ).

The aim of this article is to prove an inequality of type 11 for operators $ L_{n,r,\varphi }^{\alpha ,\beta } $.

2 Main results

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

\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*}

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

\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*}

T. Popoviciu introduced the notion of $(m,n)$ convexity in [ 9 , p. 78 ] . In [ 4 ] , S. Gal and P. Niculescu used box convexity of order $(m,n)$ for $(m,n)$ convexity. We will use the terminology introduced in [ 4 ] and say that the function $ f:I\times J\rightarrow \mathbb {R} $ is box convex of order $(m,n)$ if all divided differences

\begin{equation*} \Bigg[ \begin{matrix} x_1, & \ldots , & x_{m+1} \\ y_1, & \ldots , & y_{n+1} \end{matrix};f\Bigg] \end{equation*}

are non-negative for any choice of distinct points $ x_1,\ldots ,x_{m+1} $ and $ y_1,\ldots ,y_{n+1} $.

For the remaining of this paper, we will consider $ I=J=[0,1] $.

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

\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*}

Lemma 2 [ 9 ]

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.

In this paper we will use lemma 2 for the case $ q=s=1 $.

Definition 3

Let $ U,V $ be two operators $ U,V:C[0,1]\rightarrow C[0,1] $ defined by

\begin{equation*} (Uf)(x)=\sum _{i=0}^n a_i(x)f(x_i) \end{equation*}

\begin{equation*} (Vf)(y)=\sum _{j=0}^m b_j(y)f(y_j). \end{equation*}

for all $ x,y\in I,x_i,y_j\in I, i=0,1,\ldots ,n, j=0,1,\ldots ,m. $

The operator denoted by $ UV, UV:C(I\times I)\rightarrow C(I\times I) $ defined by

\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}

is called the tensor product of the operators $U $ and $ V$.

The following results will be proved in this paper.

Theorem 4

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 9 i.e.

\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}

If $ f $ is $ (1,1) $ box convex function, then $ Lf $ is $ (1,1) $ box convex.

Theorem 5

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:

\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}

Corollary 6

If $ f $ is a $ (1,1) $ box convex function then

\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}

Corollary 7

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:

\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}

3 Proofs

Proof of theorem 4. From 8 we have

\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}

and

\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}

From 17 and 18 we get

\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}

$ f $ being a $ (1,1) $ box convex function, from 19 we obtain

\begin{align*} \frac{\partial ^2L}{\partial x \partial y}(x,y)\geq 0. \end{align*}

From the last inequality and lemma 2 the theorem is proved. â–¡

Proof â–¼

[Proof of theorem 5.] If $ x=x_1 $ or $ y=y_1 $, then the relation 14 is trivial.

Let us suppose that $ x{\gt}x_1 $ and $ y=y_1 $. Then we have

\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*}

From theorem 4 we have $ \frac{\partial ^2L}{\partial u \partial v}(u,v)\geq 0 $, for all $ (u,v)\in I\times I $.

Proof â–¼

[Proof of corollary 6.] Inequality 15 follows from the inequality

\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*}

which follows from 14.

Proof â–¼

[Proof of corollary 7.] 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 16 is inequality 14 for the function $g$.

Remark 8

If $ f $ is a $(1,1) $ box convex function, in 14 the evaluation functional $f\left(\frac{i+\alpha }{n+\beta },\frac{j+\gamma }{m+\delta } \right)$ can be replaced by

\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*}

Remark 9

For $ m=n, r=0, \varphi (x)=\psi (x)=x $ the inequality 16 becomes 11. â–¡

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