Open Access

Approximation properties of weighted Kantorovich type operators in compact disks

Journal of Inequalities and Applications20152015:46

https://doi.org/10.1186/s13660-015-0571-1

Received: 13 November 2014

Accepted: 21 January 2015

Published: 5 February 2015

Abstract

In this paper, we discuss the approximation properties of the complex weighted Kantorovich type operators. Quantitative estimates of the convergence, the Voronovskaja type theorem, and saturation of convergence for complex weighted Kantorovich polynomials attached to analytic functions in compact disks will be given. In particular, we show that for functions analytic in \(\{ z\in C:\vert z\vert < R \} \), the rate of approximation by the weighted complex Kantorovich type operators is \(1/n\).

Keywords

weighted Bernstein-Kantorovich operatorsBernstein-Kantorovich operatorcomplex approximation

1 Introduction

The first constructive (and simple) proof of Weierstrass approximation theorem was given by Bernstein [1]. He gave an alternative proof to the Weierstrass approximation theorem and introduced the following polynomial:
$$B_{n}(f) (x)=\sum_{k=0}^{n} p_{n,k}(x)f \biggl( \frac{k}{n} \biggr) ,\quad f: [ 0,1 ] \rightarrow \mathbb{R}, $$
where \(p_{n,k}(x)=\binom{n}{k}x^{k} ( 1-x ) ^{n-k}\), \(x\in [ 0,1 ] \).
For any \(f\in L^{p}([0,1])\), \(1\leq p\leq\infty\), Ditzian and Totik (see [2]) introduced the Kantorovich-Bernstein operator as follows:
$$ K_{n}(f;x)= \sum_{k=0}^{n} p_{n,k}(x) (n+1) \int_{\frac{k+1}{n+1}}^{\frac{k}{n+1}} f(u)\,du, $$
(1)
where \(p_{n,k}(x)=\binom{n}{k}x^{k} ( 1-x ) ^{n-k}\), \(k=0,1,2,\ldots, n\); \(x\in [ 0,1 ]\).
Let
$$ w(x)=x^{\alpha}(1-x)^{\beta},\quad \alpha,\beta>-1, 0\leq x\leq1 , $$
(2)
be the classical Jacobi weights.
Let
$$L_{w}^{p}=\left \{ \begin{array}{@{}l@{\quad}l} \{ f:wf\in L^{p}(0,1)\} , & 1\leq p< \infty,\\ \{ f:f\in ( 0,1 ) , \lim_{x(1-x)\rightarrow0}(wf)(x)=0\} , & p=\infty, \end{array} \right . $$
and equip \(L_{w}^{p}\) with norm
$$\Vert wf\Vert _{p}= \left \{ \begin{array}{@{}l@{\quad}l} ( \int_{0}^{1} \vert (wf)(x)\vert ^{p}\,dx ) ^{1/p}, & 1\leq p< \infty,\\ \sup_{0\leq x\leq1}\vert (wf)(x)\vert ,&p=\infty. \end{array} \right . $$
In [2], Ditzian and Totik studied the case of weighted approximation properties of \(K_{n}(f;x)\) in \(L_{w}^{p}\) under the restrictions \(-\frac{1}{p}<\alpha,\beta<1-\frac{1}{p}\) on the weighted parameters. In [3], Della Vecchia et al. removed the restrictions on α, β and introduced a weighted generalization of the \(K_{n}(f;x)\) as follows:
$$ K_{n}^{\ast}(f;x)=\sum_{k=0}^{n} \frac{\int_{I_{k}}(wf)(t)\,dt}{\int_{I_{k}}w(t)\,dt}p_{n,k}(x),\quad x\in [ 0,1 ] , $$
(3)
where
$$ -\frac{1}{p}< \alpha,\beta,\quad 1\leq p\leq\infty, $$
(4)
and \(f\in L_{w}^{p}\). \(K_{n}^{\ast}(f;x)\) allows a wider class of functions than the operator \(K_{n}(f;x)\). Because, Della Vecchia et al. dropped the restrictions \(\alpha,\beta<1-\frac{1}{p}\) and obtained (see [3]) direct and converse theorems and a Voronovskaya-type relation for the weighted Kantorovich operator (3). Also Della Vecchia et al. solved the saturation problem of the weighted Kantorovich operator (3) (see [4, Theorem 2.1]).
In [5], Yu introduced the following modified operator of \(K_{n} ( f;x ) \):
$$\begin{aligned} K_{n}^{\ast}(f;x) :={}&(1-x)^{n}(n+1) \biggl(2 \int_{I_{1}} f(t)\,dt- \int_{I_{2}} f(t)\,dt\biggr) \\ &{}+ \sum _{k=1}^{n-1} (n+1) \int_{I_{k}} f(u)\,dup_{nk}(x) \\ &{} +x^{n}(n+1) \biggl(2 \int _{I_{n-1}} f(t)\,dt- \int _{I_{n-2}} f(t)\,dt\biggr), \end{aligned}$$
(5)
and direct and converse theorems and a Voronovskaya-type relation were obtained with modification of the classical Kantorovich-Bernstein operators (1) with Jacobi weights \(w(x)=x^{\alpha}(1-x)^{\beta}\), where \(-\frac {1}{p}<\alpha,\beta\).
The goal of the present note is to introduce the complex weighted Kantorovich type operator
$$ K_{n} ( f;z ) = \sum _{k=0}^{n} p_{n,k} ( z ) \frac{ \int_{0}^{1} t^{\alpha}(1-t)^{\beta}f ( \frac{k+t}{n+1} )\,dt}{B(\alpha +1,\beta+1)} , $$
(6)
where \(\frac{-1}{p}<\alpha,\beta\) and \(1\leq p\leq\infty\), \(B(\cdot ,\cdot)\) is a beta function, and we study the convergence properties of \(K_{n} ( f;z ) \). Notice that the approximation properties of complex generalized Kantorovich type operators are studied in [6].

In the recent books of Gal [7, 8] (see references therein), a systematic study of the overconvergence phenomenon in a complex approximation was made for the following important classes of Bernstein-type operators: Bernstein, Bernstein-Faber, Bernstein-Butzer, Bernstein-Stancu, Bernstein-Kantorovich, Favard-Szasz-Mirakjan, Baskakov, Bernstein-Durrmeyer, and Balazs-Szabados.

Let \(\mathbb{D}_{R}\) be a disc \(\mathbb{D}_{R}:= \{ z\in\mathbb{C}:\vert z\vert < R \} \) in the complex plane . Denote by \(H ( \mathbb{D}_{R} ) \) the space of all analytic functions on \(\mathbb{D}_{R}\). For \(f\in H ( \mathbb{D}_{R} ) \) we assume that \(f ( z ) =\sum_{m=0}^{\infty}a_{m}z^{m}\).

We start with the following quantitative estimates of the convergence for complex weighted Kantorovich type operators attached to an analytic function in a disk of radius \(R>1\) and center 0.

Theorem 1

Let \(f\in H ( \mathbb{D}_{R} ) \). If \(1\leq r< R\), then for all \(\vert z\vert \leq r\) we have
$$ \bigl\vert K_{n} ( f;z ) -f ( z ) \bigr\vert \leq \frac{2}{n}\sum_{m=1}^{\infty} \vert c_{m}\vert m ( m+1 ) r^{m}, $$
(7)
where \(n\in\mathbb{N}\).

The next theorem gives a Voronovskaja type result in compact disks, for complex weighted Kantorovich type operators attached to an analytic function in \(\mathbb{D}_{R}\), where \(R>1\), and with center 0.

Theorem 2

Let \(f\in H ( \mathbb{D}_{R} ) \). If \(1\leq r< R\) then, for all \(\vert z\vert \leq r\), we have
$$\begin{aligned} & \biggl\vert K_{n} ( f;z ) -f ( z ) - \frac {(\alpha+1)-z ( \alpha+\beta+2 ) }{ ( \alpha+\beta+2 ) (n+1)}f^{{\prime}}(z)-\frac{z(1-z)}{2(n+1)}f^{{\prime\prime}}(z)\biggr\vert \\ &\quad \leq\frac{13}{n^{2}}\sum_{m=2}^{\infty} \vert a_{m}\vert m ( m-1 ) ^{2}r^{m}+ \frac{4}{(\alpha+\beta+2)n^{2}}\sum_{m=3}^{\infty }\vert a_{m}\vert m ( m-1 ) r^{m}, \end{aligned}$$
(8)
where \(n\in\mathbb{N}\).

As an application of Theorem 2 we present the order of approximation for complex weighted Kantorovich type operators.

Theorem 3

Let \(f\in H ( \mathbb{D}_{R} ) \). If \(1\leq r< R\) and if f is not a constant function, then the estimate
$$ \bigl\Vert K_{n} ( f ) -f\bigr\Vert _{r}\geq \frac{1}{n}C_{r} ( f ) , \quad n\in\mathbb{N}, $$
(9)
holds, where the constant \(C_{r} ( f ) \) depends on f, α, β, and r but it is independent of n.

2 Auxiliary results

Lemma 4

For all \(n\in\mathbb{N}\), \(m\in\mathbb{N}\cup \{ 0 \} \), \(z\in\mathbb{C}\) we have
$$ K_{n} ( e_{m};z ) =\frac{1}{ ( n+1 ) ^{m}}\sum _{j=0}^{m}\binom{m}{j} \frac{n^{j}B ( \alpha+m-j+1,\beta+1 ) }{B ( \alpha+1,\beta+1 ) }B_{n} ( e_{j};z ) , $$
(10)
where \(e_{m} ( z ) =z^{m}\).

Proof

The recurrence formula can be derived by direct computation:
$$\begin{aligned}[b] K_{n} ( e_{m};z ) & =\frac{1}{ ( n+1 ) ^{m}} \sum_{k=0}^{n} p_{n,k} ( z ) \frac{1}{B ( \alpha+1,\beta+1 ) }\sum_{j=0}^{m} \int_{0}^{1}\binom{m}{j} t^{\alpha}(1-t)^{\beta}{}k^{j}t^{m-j}\,dt \\ & =\frac{1}{ ( n+1 ) ^{m}} \sum _{k=0}^{n} p_{n,k} ( z ) \frac{1}{B ( \alpha+1,\beta+1 ) }\sum_{j=0}^{m} \binom{m}{j} k^{j}\int_{0}^{1}t^{\alpha+m-j}(1-t)^{\beta}\,dt \\ & =\frac{1}{ ( n+1 ) ^{m}}\sum_{j=0}^{m} \binom{m}{j} \frac{n^{j}B(\alpha+m-j+1,\beta+1)}{B ( \alpha+1,\beta +1 ) }B_{n} ( e_{j};z ) . \end{aligned} $$
 □

Lemma 5

We have
$$\begin{aligned}& K_{n} ( e_{0};z ) =1, \\& K_{n} ( e_{1};z ) =\frac{\alpha+1}{ ( \alpha+\beta+2 ) ( n+1 ) }+ \frac{nz}{n+1},\\& \begin{aligned}[b] K_{n} ( e_{2};z ) ={}&\frac{1}{ ( n+1 ) ^{2}}\frac {(\alpha+2)(\alpha+1)}{(\alpha+\beta+3) ( \alpha+\beta+2 ) }+ \frac{2nz}{ ( n+1 ) ^{2}}\frac{ ( \alpha+1 ) }{ ( \alpha+\beta+2 ) }\\ &{}+ \biggl( z^{2}+\frac{z(1-z)}{n} \biggr) \frac{n^{2} }{ ( n+1 ) ^{2}}, \end{aligned} \\& K_{n}\bigl((e_{1}-z);z\bigr)=\frac{(\alpha+1)-z(\alpha+\beta+2)}{(n+1)(\alpha +\beta+2)},\\& \begin{aligned}[b] K_{n}\bigl((e_{1}-z)^{2};z \bigr) ={}&n^{2}\frac{z^{2}+\frac{1}{n}z(1-z)}{(n+1)^{2}}-z^{2}-2z \biggl( \frac{\alpha+1}{(n+1)(\alpha+\beta+2)}+\frac {nz}{n+1} \biggr) \\ &{} +2nz\frac{\alpha+1}{(n+1)^{2}(\alpha+\beta+2)}+\frac{(\alpha+1)(\alpha +2)}{(n+1)^{2}(\alpha+\beta+2)(\alpha+\beta+3)}. \end{aligned} \end{aligned}$$

Lemma 6

For all \(z\in\mathbb{D}_{r}\), \(r\geq1\) we have
$$\bigl\vert K_{n} ( e_{m};z ) \bigr\vert \leq r^{m}, $$
where \(e_{m} ( z ) =z^{m}\).

Proof

Indeed, using the inequality \(\vert B_{n} ( e_{j};z ) \vert \leq r^{j}\) (see [7]) we get
$$\begin{aligned} \bigl\vert K_{n} ( e_{m};z ) \bigr\vert & \leq\sum _{j=0}^{m}\binom{m}{j} \frac{n^{j}B ( \alpha+m-j+1,\beta+1 ) }{ ( n+1 ) ^{m}B ( \alpha+1,\beta+1 ) }\bigl\vert B_{n} ( e_{j};z ) \bigr\vert \\ & \leq\frac{1}{ ( n+1 ) ^{m}}\sum_{j=0}^{m} \binom{m}{j} n^{j}r^{m}= \biggl( \frac{1+n}{n+1} \biggr) ^{m}r^{m}=r^{m}. \end{aligned}$$
 □

Lemma 7

For all \(z\in\mathbb{C}\), \(z\in\mathbb{N\cup\{}0\mathbb{\}}\) we have
$$ \begin{aligned}[b] K_{n} ( e_{m+1};z ) ={}&\frac{z ( 1-z ) }{n}K_{n} ( e_{m};z ) +zK_{n} ( e_{m};z )\\ &{} +\frac{1}{ ( n+1 ) ^{m}}\sum_{j=0}^{m+1} \binom{m+1}{j} n^{j-1}\frac{B ( \alpha+m-j+2,\beta+1 ) }{B ( \alpha+1,\beta+1 ) } \\ &{}\times\biggl( \frac{n}{n+1}- \frac{j}{m+1} \biggr) B_{n} ( e_{j};z ) . \end{aligned} $$
(11)

Proof

We know that (see [7])
$$\frac{z ( 1-z ) }{n}B_{n}^{{\prime}} ( e_{j};z ) =B_{n} ( e_{j+1};z ) -zB_{n} ( e_{j};z ) . $$
Taking the derivative of (10) and using the above formula we have
$$\begin{aligned} \frac{z ( 1-z ) }{n}K_{n}^{{\prime}} ( e_{m};z ) & = \frac{z ( 1-z ) }{n}\sum_{j=0}^{m} \binom{m}{j} \frac{n^{j}B ( \alpha+m-j+1,\beta+1 ) }{ ( n+1 ) ^{m}B ( \alpha+1,\beta+1 ) }B_{n}^{{\prime}} ( e_{j};z ) \\ & =\sum_{j=0}^{m}\binom{m}{j} \frac{n^{j}B ( \alpha+m-j+1,\beta+1 ) }{ ( n+1 ) ^{m}B ( \alpha+1,\beta+1 ) } \bigl( B_{n} ( e_{j+1};z ) -zB_{n} ( e_{j};z ) \bigr) \\ & =\sum_{j=1}^{m+1} \binom{m}{j-1} \frac{n^{j-1}B ( \alpha+m-j+2,\beta+1 ) }{ ( n+1 ) ^{m}B ( \alpha+1,\beta+1 ) }B_{n} ( e_{j};z ) -zK_{n} ( e_{m};z ) . \end{aligned}$$
It follows that
$$\begin{aligned} K_{n} ( e_{m+1};z ) ={}&\frac{z ( 1-z ) }{n}K_{n}^{{\prime}} ( e_{m};z ) +zK_{n} ( e_{m};z ) \\ &{} +\sum_{j=0}^{m+1} \binom{m+1}{j} \frac{n^{j}B ( \alpha+m-j+2,\beta+1 ) }{ ( n+1 ) ^{m+1}B ( \alpha+1,\beta+1 ) }B_{n} ( e_{j};z ) \\ &{} -\sum_{j=1}^{m+1} \binom{m}{j-1} \frac{n^{j-1}B ( \alpha+m-j+2,\beta+1 ) }{ ( n+1 ) ^{m+1}B ( \alpha+1,\beta+1 ) }B_{n} ( e_{j};z ) \\ ={}&\frac{z ( 1-z ) }{n}K_{n}^{{\prime}} ( e_{m};z ) +zK_{n} ( e_{m};z ) +\frac{B ( \alpha+m+2,\beta+1 ) }{ ( n+1 ) ^{m+1}B ( \alpha+1,\beta+1 ) } \\ &{} +\sum_{j=1}^{m+1} \binom{m+1}{j} \frac{n^{j-1}B ( \alpha+m-j+2,\beta+1 ) }{ ( n+1 ) ^{m}B ( \alpha+1,\beta+1 ) } \biggl( \frac{-j}{m+1}+\frac{n}{n+1} \biggr) B_{n} ( e_{j};z ) \\ ={}&\frac{z ( 1-z ) }{n}K_{n}^{{\prime}} ( e_{m};z ) +zK_{n} ( e_{m};z ) \\ & {}+\sum_{j=0}^{m+1} \binom{m+1}{j} \frac{n^{j-1}B ( \alpha+m-j+2,\beta+1 ) }{ ( n+1 ) ^{m}B ( \alpha+1,\beta+1 ) } \biggl( \frac{-j}{m+1}+\frac{n}{n+1} \biggr) B_{n} ( e_{j};z ) . \end{aligned}$$
Here we used the identity
$$\binom{m}{j-1} =\binom{m+1}{j} \frac{j}{ ( m+1 ) }. $$
 □
Define
$$\begin{aligned} E_{n,m} ( z ) :={}&K_{n} ( e_{m};z ) -e_{m} ( z ) -\frac{(\alpha+1)-z ( \alpha+\beta+2 ) }{ ( \alpha+\beta+2 ) (n+1)}mz^{m-1} - \frac{z(1-z)}{2(n+1)}m(m-1)z^{m-2}. \end{aligned}$$

Lemma 8

Let \(n,m\in\mathbb{N}\), we have the following recurrence formula:
$$\begin{aligned} E_{n,m} ( z ) ={}&\frac{z ( 1-z ) }{n} \bigl( K_{n,q} ( e_{m-1};z ) -e_{m-1} ( z ) \bigr) ^{{\prime}}\\ &{}+zE_{n,m-1} ( z ) -\frac{m-1}{n(n+1)}z^{m-1} ( 1-z )\\ &{} -\frac{(\alpha+1)-z(\alpha+\beta+2)}{(\alpha+\beta +2)(n+1)}z^{m-1}\\ &{}+\frac {1}{ ( n+1 ) ^{m}}\sum _{j=0}^{m}\binom{m}{j} \frac{n^{j}B ( \alpha+m-j+1,\beta+1 ) }{B ( \alpha+1,\beta+1 ) } \frac{mn-j(n+1)}{mn}B_{n} ( e_{j};z ) . \end{aligned}$$

Proof

It is immediate that \(E_{n,m} ( z ) \) is a polynomial of degree less than or equal to m and that \(E_{n,0} ( z ) =E_{n,1} ( z ) =0\).

Using (11) we get
$$\begin{aligned} E_{n,m} ( z ) ={}&\frac{z ( 1-z ) }{n} \bigl( K_{n} ( e_{m-1};z ) -e_{m-1} ( z ) \bigr) ^{{\prime} }+ \frac{ ( m-1 ) }{n}z^{m-1} ( 1-z ) \\ &{} +z \biggl( E_{n,m-1} ( z ) +\frac{(\alpha+1)-z(\alpha+\beta +2)}{(\alpha+\beta+2)(n+1)} ( m-1 ) z^{m-2}\\ &{}+\frac{z(1-z)}{2(n+1)}(m-1) (m-2)z^{m-3} \biggr) \\ &{} -\frac{(\alpha+1)-z(\alpha+\beta+2)}{(\alpha+\beta+2)(n+1)}mz^{m-1}-\frac{z(1-z)}{2(n+1)}m(m-1)z^{m-2} \\ &{} +\frac{1}{ ( n+1 ) ^{m}}\sum_{j=0}^{m} \binom{m}{j} \frac{n^{j}B ( \alpha+m-j+1,\beta+1 ) }{B ( \alpha+1,\beta+1 ) }\frac{mn-j(n+1)}{mn}B_{n} ( e_{j};z ) , \\ E_{n,m} ( z ) ={}&\frac{z ( 1-z ) }{n} \bigl( K_{n} ( e_{m-1};z ) -e_{m-1} ( z ) \bigr) ^{{\prime} }+z^{m-1} \frac{(\alpha+1)-z(\alpha+\beta+2)}{(\alpha+\beta+2)(n+1)} ( m-1-m ) \\ &{} +z^{m-1} ( 1-z ) \frac{ ( m-1 ) (m-2-m)}{2(n+1)}+z^{m-1} ( 1-z ) \frac{ ( m-1 ) }{n} \\ &{} +\frac{1}{ ( n+1 ) ^{m}}\sum_{j=0}^{m} \binom{m}{j} \frac{n^{j}B ( \alpha+m-j+1,\beta+1 ) }{B ( \alpha+1,\beta+1 ) }\frac{mn-j(n+1)}{mn}B_{n} ( e_{j};z ) , \\ E_{n,m} ( z ) ={}&\frac{z ( 1-z ) }{n} \bigl( K_{n,q} ( e_{m-1};z ) -e_{m-1} ( z ) \bigr) ^{{\prime}}\\ &{}+zE_{n,m-1} ( z ) +\frac{(m-1)}{n(n+1)}z^{m-1} ( 1-z ) -\frac{(\alpha+1)-z(\alpha+\beta+2)}{(\alpha+\beta +2)(n+1)}z^{m-1}\\ &{}+\frac {1}{ ( n+1 ) ^{m}}\sum _{j=0}^{m}\binom{m}{j} \frac{n^{j}B ( \alpha+m-j+1,\beta+1 ) }{B ( \alpha+1,\beta+1 ) } \frac{mn-j(n+1)}{mn}B_{n} ( e_{j};z ) , \end{aligned}$$
which is the desired recurrence formula. □

3 Proofs of the main results

Proof of Theorem 1

By use of the above recurrence we obtain the following relationship:
$$\begin{aligned} K_{n} ( e_{m};z ) -e_{m} ( z ) ={}& \frac{z ( 1-z ) }{n}K_{n}^{{\prime}} ( e_{m-1};z ) +z \bigl( K_{n} ( e_{m-1};z ) -e_{m-1} ( z ) \bigr) \\ &{} +\frac{1}{ ( n+1 ) ^{m}}\sum_{j=0}^{m} \binom{m}{j} \frac{n^{j}B ( \alpha+m-j+1,\beta+1 ) }{B ( \alpha+1,\beta+1 ) } \\ &{}\times\frac{mn-j ( n+1 ) }{mn}B_{n} ( e_{j};z ) . \end{aligned}$$
(12)
For \(\vert z\vert \leq r\) we can easily estimate the sum in the above formula as follows:
$$\begin{aligned} & \Biggl\vert \frac{1}{ ( n+1 ) ^{m}}\sum_{j=0}^{m} \binom{m}{j} n^{j}\frac{B ( \alpha+m-j+1,\beta+1 ) }{B ( \alpha+1,\beta+1 ) } \biggl( 1-\frac{j}{m}- \frac{j}{mn} \biggr) B_{n} ( e_{j};z ) \Biggr\vert \\ &\quad \leq\frac{1}{ ( n+1 ) ^{m}} \Biggl( \sum_{j=0}^{m-1} \binom{m-1}{j} \frac{m}{m-j}\frac{n^{j}B ( \alpha+m-j+1,\beta+1 ) }{B ( \alpha+1,\beta+1 ) }\biggl\vert 1- \frac{j}{m}-\frac{j}{mn}\biggr\vert \Biggr) \bigl\vert B_{n} ( e_{j};z ) \bigr\vert \\ &\qquad{} +\frac{B ( \alpha+1,\beta+1 ) }{B ( \alpha+1,\beta +1 ) }\frac{n^{m-1}}{ ( n+1 ) ^{m}}r^{m} \\ &\quad \leq\frac{1}{ ( n+1 ) ^{m}} \Biggl( \sum_{j=0}^{m-1} \binom{m-1}{j} \frac{mn^{j}}{m-j}\biggl\vert 1-\frac{j}{m}- \frac{j}{mn}\biggr\vert \Biggr) \bigl\vert B_{n} ( e_{j};z ) \bigr\vert +\frac{n^{m-1} }{ ( n+1 ) ^{m}}r^{m} \\ &\quad \leq\frac{2m ( n+1 ) ^{m-1}+(n+1)^{m-1}}{ ( n+1 ) ^{m}}r^{m} \\ &\quad =\frac{ ( n+1 ) ^{m-1}(2m+1)}{ ( n+1 ) ^{m}}r^{m}=\frac{(2m+1)}{(n+1)}r^{m}. \end{aligned}$$
It is well known that, by a linear transformation, the Bernstein inequality in the closed unit disk becomes
$$\bigl\vert P_{m}^{\prime} ( z ) \bigr\vert \leq \frac{m}{r}\Vert P_{m}\Vert _{r}, \quad\mbox{for all }\vert z\vert \leq r, r\geq1, $$
for all \(\vert z\vert \leq r\), where \(P_{m} ( z ) \) is a complex polynomial of degree ≤m. From the above recurrence formula (12) we get
$$\begin{aligned} \bigl\vert K_{n} ( e_{m};z ) -e_{m} ( z ) \bigr\vert & \leq\frac{\vert z\vert \vert 1-z\vert }{n}\bigl\vert K_{n}^{{\prime}} ( e_{m-1};z ) \bigr\vert +\vert z\vert \bigl\vert K_{n} ( e_{m-1};z ) -e_{m-1} ( z ) \bigr\vert + \frac{(2m+1)}{(n+1)}r^{m} \\ & \leq\frac{r ( 1+r ) }{n}\frac{m-1}{r}\bigl\Vert K_{n} ( e_{m-1} ) \bigr\Vert _{r}+r\bigl\vert K_{n} ( e_{m-1};z ) -e_{m-1} ( z ) \bigr\vert +\!\frac{(2m+1)}{(n+1)}r^{m} \\ & \leq r\bigl\vert K_{n} ( e_{m-1};z ) -e_{m-1} ( z ) \bigr\vert +\frac{2 ( m-1 ) }{n}r^{m}+\frac{(2m+1)}{ ( n+1 ) }r^{m} \\ & \leq r\bigl\vert K_{n} ( e_{m-1};z ) -e_{m-1} ( z ) \bigr\vert +\frac{4m}{n}r^{m}. \end{aligned}$$
By writing the last inequality for \(m=1,2,\ldots\) , we can easily obtain, step by step, the following:
$$\begin{aligned} & \bigl\vert K_{n} ( e_{m};z ) -e_{m} ( z ) \bigr\vert \\ &\quad\leq\frac{4m}{n}r^{m}+r\frac{4(m-1)}{n}r^{m-1}+r^{2} \frac{4 ( m-2 ) }{n}r^{m-2}+\cdots+r^{m-1}\frac{4}{n}r \\ &\quad =\frac{4}{n}r^{m} ( m+m-1+\cdots+1 ) \leq \frac{2m ( m+1 ) }{n}r^{m}. \end{aligned}$$
(13)
Since \(K_{n,q} ( f;z ) \) is analytic in \(\mathbb{D}_{R}\) (see [7, p.6]), we can write
$$K_{n} ( f;z ) =\sum_{m=0}^{\infty}a_{m}K_{n} ( e_{m};z ), \quad z\in\mathbb{D}_{R}, $$
which together with (13) immediately implies for all \(\vert z\vert \leq r\)
$$\bigl\vert K_{n} ( f;z ) -f ( z ) \bigr\vert \leq \sum _{m=0}^{\infty} \vert a_{m}\vert \bigl\vert K_{n} ( e_{m};z ) -e_{m} ( z ) \bigr\vert \leq\frac{2}{n}\sum_{m=1}^{\infty} \vert c_{m}\vert m ( m+1 ) r^{m}. $$
 □

Proof of Theorem 2

A simple calculation and the use of the recurrence formula (10) lead us to the following relationship:
$$\begin{aligned} E_{n,m} ( z ) ={}&\frac{z ( 1-z ) }{n} \bigl( K_{n,q} ( e_{m-1};z ) -e_{m-1} ( z ) \bigr) ^{{\prime}}+zE_{n,m-1} ( z ) +\frac{m-1}{n ( n+1 ) }z^{m-1} ( 1-z ) \\ &{} +\frac{1}{n+1} \bigl( z^{m}-B_{n} ( e_{m};z ) \bigr) +\frac {1}{n+1} \biggl( 1-\frac{n^{m-1}}{ ( n+1 ) ^{m-1}} \biggr) B_{n} ( e_{m};z ) \\ &{} +\frac{ ( \alpha+1 ) }{(\alpha+\beta+2) ( n+1 ) } \biggl( \frac{n^{m-1}}{ ( n+1 ) ^{m-1}}-1 \biggr) B_{n} ( e_{m-1};z ) \\ &{}+\frac{\alpha+1}{(\alpha+\beta+2) ( n+1 ) } \bigl( B_{n} ( e_{m-1};z ) -z^{m-1} \bigr) -\frac{ ( m-1 ) n^{m-2}}{(\alpha+\beta+2) ( n+1 ) ^{m}}B_{n} ( e_{m-1};z ) \\ &{} +\frac{1}{ ( n+1 ) ^{m}}\sum_{j=0}^{m-2} \binom{m}{j} \frac{n^{j}B ( \alpha+m-j+1,\beta+1 ) }{B ( \alpha+1,\beta+1 ) }\frac{mn-j ( n+1 ) }{mn}B_{n} ( e_{j};z ) \\ :={}&\sum_{k=1}^{9}I_{k}. \end{aligned}$$
(14)
Firstly we estimate \(I_{3}\), \(I_{8}\). It is clear that
$$ \begin{aligned} &\vert I_{3}\vert \leq \frac{m-1}{n ( n+1 ) }r^{m-1} ( 1+r ) , \\ &\vert I_{8}\vert \leq\frac{ ( m-1 ) }{(\alpha +\beta+2) ( n+1 ) ^{2}}\bigl\vert B_{n,q} ( e_{m-1};z ) \bigr\vert \leq\frac{ ( m-1 ) }{(\alpha+\beta+2) ( n+1 ) ^{2}}r^{m-1}. \end{aligned} $$
(15)
Secondly using the known inequality
$$1-\prod_{k=1}^{m}x_{k}\leq\sum _{k=1}^{m} ( 1-x_{k} ) , \quad 0 \leq x_{k}\leq1, $$
to estimate \(I_{5}\), \(I_{6}\), \(I_{9}\), we have
$$\begin{aligned} & \vert I_{5}\vert \leq \frac{1}{n+1} \biggl( 1-\frac{n^{m-1}}{ ( n+1 ) ^{m-1}} \biggr) \bigl\vert B_{n} ( e_{m};z ) \bigr\vert \leq\frac{m-1}{ ( n+1 ) ^{2}}r^{m}, \\ &\vert I_{6}\vert \leq\frac{\alpha+1}{(\alpha+\beta+2) ( n+1 ) } \biggl( 1- \frac{n^{m-1}}{ ( n+1 ) ^{m-1}} \biggr) \bigl\vert B_{n} ( e_{m-1};z ) \bigr\vert \leq\frac{ ( m-1 ) }{ ( n+1 ) ^{2}}r^{m-1}, \\ &| I_{9}| \leq\frac{1}{ ( n+1 ) ^{m}}\sum _{j=0}^{m-2} \binom{m-2}{j} \frac{m ( m-1 ) }{ ( m-j ) ( m-j-1 ) }\\ &\hphantom{| I_{9}|=}{}\times \frac{n^{j}B ( \alpha+m-j+1,\beta+1 ) }{B ( \alpha+1,\beta+1 ) } \biggl( 1-\frac{j}{m}-\frac{j}{mn} \biggr) r^{j} \\ &\hphantom{| I_{9}|} \leq\frac{2m ( m-1 ) ( \alpha+1 ) ( \alpha+2 ) ( n+1 ) ^{m-2}}{(\alpha+\beta+2)(\alpha +\beta+3) ( n+1 ) ^{m}}r^{m}\leq\frac{2m ( m-1 ) }{ ( n+1 ) ^{2}}r^{m}. \end{aligned}$$
(16)
Finally we estimate \(I_{4}\), \(I_{7}\). We use [7, Theorem 1.5.1]:
$$\begin{aligned} \vert I_{4}\vert +\vert I_{7}\vert & \leq\frac{1}{n+1}\bigl\vert z^{m}-B_{n} ( e_{m};z ) \bigr\vert +\frac {1}{(\alpha+\beta+2) ( n+1 ) }\bigl\vert B_{n} ( e_{m-1};z ) -z^{m-1}\bigr\vert \\ & \leq\frac{3m ( m-1 ) }{2n(n+1)}(1+r)r^{m-1}+\frac{3(m-1) ( m-2 ) }{2(\alpha+\beta+2)n(n+1)}(1+r)r^{m-2}. \end{aligned}$$
(17)
Using (13), (15), (16), and (17) in (14) finally we have (\(m\geq3\))
$$\begin{aligned} &\bigl\vert E_{n,m} ( z ) \bigr\vert \\ &\quad \leq\frac{r ( 1+r ) }{n}\bigl\vert \bigl( K_{n} ( e_{m-1};z ) -e_{m-1} ( z ) \bigr) ^{{\prime}}\bigr\vert +r\bigl\vert E_{n,m-1} ( z ) \bigr\vert +\frac{m-1}{n ( n+1 ) }r^{m-1} ( 1+r ) \\ &\qquad{} +\frac{3m ( m-1 ) }{2n(n+1)}(1+r)r^{m-1}+\frac{m-1}{ ( n+1 ) ^{2}}r^{m}+ \frac{ ( m-1 ) }{ ( n+1 ) ^{2}}r^{m-1}\\ &\qquad{}+\frac{3(m-1) ( m-2 ) }{2(\alpha+\beta+2)n(n+1)}(1+r)r^{m-2} +\frac{ ( m-1 ) }{(\alpha+\beta+2) ( n+1 ) ^{2}}r^{m-1}+\frac{2m ( m-1 ) }{ ( n+1 ) ^{2}}r^{m} \\ &\quad \leq\frac{r ( 1+r ) }{n}\bigl\vert \bigl( K_{n} ( e_{m-1};z ) -e_{m-1} ( z ) \bigr) ^{{\prime}}\bigr\vert +r\bigl\vert E_{n,m-1} ( z ) \bigr\vert \\ &\qquad{} +\frac{2 ( m-1 ) }{n^{2}}r^{m}+\frac{3m ( m-1 ) }{n^{2}}r^{m}+ \frac{m-1}{n^{2}}r^{m}+\frac{ ( m-1 ) }{n^{2}}r^{m}+ \frac{3(m-1) ( m-2 ) }{(\alpha+\beta+2)n^{2}}r^{m} \\ &\qquad{} +\frac{ ( m-1 ) }{(\alpha+\beta+2)n^{2}}r^{m}+\frac{2m ( m-1 ) }{n^{2}}r^{m} \\ &\quad \leq\frac{r ( 1+r ) }{n}\bigl\vert \bigl( K_{n} ( e_{m-1};z ) -e_{m-1} ( z ) \bigr) ^{{\prime}}\bigr\vert +r\bigl\vert E_{n,m-1} ( z ) \bigr\vert \\ &\qquad{} +\frac{2m ( m-1 ) }{n^{2}}r^{m}+\frac{3m ( m-1 ) }{n^{2}}r^{m}+ \frac{m ( m-1 ) }{n^{2}}r^{m}+\frac{m ( m-1 ) }{n^{2}}r^{m}\\ &\qquad{}+ \frac{3(m-1) ( m-2 ) }{(\alpha +\beta+2)n^{2}}r^{m} +\frac{m ( m-1 ) }{(\alpha+\beta+2)n^{2}}r^{m}+\frac{2m ( m-1 ) }{n^{2}}r^{m} \\ &\quad \leq\frac{r ( 1+r ) }{n}\bigl\vert \bigl( K_{n} ( e_{m-1};z ) -e_{m-1} ( z ) \bigr) ^{{\prime}}\bigr\vert +r\bigl\vert E_{n,m-1} ( z ) \bigr\vert \\ &\qquad{} +\frac{9m ( m-1 ) }{n^{2}}r^{m}+\frac{4m(m-1)}{(\alpha +\beta+2)n^{2}}r^{m} \\ &\quad \leq\frac{r ( 1+r ) }{n}\frac{m-1}{r}\bigl\Vert K_{n} ( e_{m-1};z ) -e_{m-1} ( z ) \bigr\Vert _{r}+r\bigl\vert E_{n,m-1} ( z ) \bigr\vert \\ &\qquad{} +\frac{9m ( m-1 ) }{n^{2}}r^{m}+ \frac{4m(m-1)}{(\alpha+\beta+2)n^{2}}r^{m} \\ &\quad \leq\frac{ ( m-1 ) ( 1+r ) }{n}\frac{2 ( m-1 ) m}{n}r^{m-1}+r\bigl\vert E_{n,m-1} ( z ) \bigr\vert +\frac{9m ( m-1 ) }{n^{2}}r^{m}+ \frac{4m(m-1)}{(\alpha +\beta+2)n^{2}}r^{m} \\ &\quad \leq r\bigl\vert E_{n,m-1} ( z ) \bigr\vert + \frac{4m ( m-1 ) ^{2}}{n^{2}}r^{m}+\frac{9m ( m-1 ) }{n^{2}}r^{m}+ \frac{4m(m-1)}{(\alpha+\beta+2)n^{2}}r^{m} \\ &\quad \leq r\bigl\vert E_{n,m-1} ( z ) \bigr\vert + \frac{13m ( m-1 ) ^{2}}{n^{2}}r^{m}+\frac{4m(m-1)}{(\alpha+\beta+2)n^{2}}r^{m}. \end{aligned}$$
As a consequence, we get
$$\bigl\vert E_{n,m} ( z ) \bigr\vert \leq\frac{13m ( m-1 ) ^{2}}{n^{2}}r^{m}+ \frac{4m(m-1)}{(\alpha+\beta+2)n^{2}}r^{m}. $$
The result follows from the fact that
$$\begin{aligned} & \biggl\vert K_{n} ( f;z ) -f ( z ) -\frac {(\alpha+1)-z ( \alpha+\beta+2 ) }{ ( \alpha+\beta+2 ) (n+1)}f^{{\prime}}(z)- \frac{z(1-z)}{2(n+1)}f^{{\prime\prime}}(z)\biggr\vert \\ &\quad \leq\sum_{m=2}^{\infty} \vert a_{m}\vert \bigl\vert E_{n,m} ( z ) \bigr\vert \\ &\quad \leq\frac{13}{n^{2}}\sum_{m=2}^{\infty} \vert a_{m}\vert m ( m-1 ) ^{2}r^{m}+ \frac{4}{(\alpha+\beta+2)n^{2}}\sum_{m=3}^{\infty }\vert a_{m}\vert m ( m-1 ) r^{m}. \end{aligned}$$
(18)

Note that since \(f^{ ( 3 ) }=\sum_{m=4}^{\infty}a_{m}m ( m-1 ) ( m-2 ) z^{m-3}\) and the series is absolutely convergent for all \(\vert z\vert < R\), the finiteness of the involved constants in the statement easily follows. □

Proof of Theorem 3

For all \(z\in\mathbb{D}_{R}\) and \(n\in\mathbb{N}\) we get
$$\begin{aligned} &K_{n,q} ( f;z ) -f ( z ) \\ &\quad=\frac{1}{n+1} \biggl\{ \frac{(\alpha+1)-z ( \alpha+\beta+2 ) }{ ( \alpha +\beta+2 ) }f^{{\prime}}(z)+\frac{z(1-z)}{2}f^{{^{\prime\prime }}}(z) \\ &\qquad{}+ ( n+1 ) \biggl( K_{n,q} ( f;z ) -f ( z ) -\frac{(\alpha+1)-z ( \alpha+\beta+2 ) }{ ( \alpha +\beta+2 ) }f^{{\prime}}(z)mz^{m-1}- \frac{z(1-z)}{2(n+1)}f^{{{\prime\prime}}}(z) \biggr) \biggr\} . \end{aligned}$$
We apply
$$\Vert F+G\Vert _{r}\geq\bigl\vert \Vert F\Vert _{r}-\Vert G\Vert _{r}\bigr\vert \geq \Vert F\Vert _{r}-\Vert G\Vert _{r}$$
to get
$$\begin{aligned} &\bigl\Vert K_{n,q} ( f ) -f\bigr\Vert _{r}\\ &\quad\geq \frac{1}{n+1} \biggl\{ \biggl\Vert \frac{(\alpha+1)-z ( \alpha+\beta+2 ) }{ ( \alpha+\beta+2 ) }f^{{\prime}}(z)+ \frac{z(1-z)}{2}f^{{^{\prime \prime}}}(z)\biggr\Vert _{r} \\ &\qquad{}- ( n+1 ) \biggl\Vert K_{n} ( f;z ) -f ( z ) -\frac{(\alpha+1)-z ( \alpha+\beta+2 ) }{ ( \alpha +\beta+2 ) }f^{{\prime}}(z)- \frac{z(1-z)}{2(n+1)}f^{{{\prime \prime}}}(z)\biggr\Vert _{r} \biggr\} . \end{aligned}$$
Because by hypothesis f is not a constant in \(\mathbb{D}_{R}\), it follows
$$\biggl\Vert \frac{(\alpha+1)-z ( \alpha+\beta+2 ) }{ ( \alpha+\beta+2 ) }mz^{m-1}+\frac{z(1-z)}{2}m(m-1)z^{m-2} \biggr\Vert _{r}>0. $$
Indeed, assuming the contrary it follows that \(\frac{(\alpha+1)-z ( \alpha+\beta+2 ) }{ ( \alpha+\beta+2 ) }mz^{m-1}+\frac{z(1-z)}{2}m(m-1)z^{m-2}=0\), for all \(z\in\overline{\mathbb {D}}_{R}\), that is,
$$\begin{aligned} &\sum_{m=1}^{\infty}a_{m} \biggl( \frac{(\alpha+1)}{ ( \alpha +\beta+2 ) }-z \biggr) mz^{m-1}+\sum_{m=2}^{\infty}a_{m} \frac {z(1-z)}{2}m(m-1)z^{m-2} =0, \\ &\sum_{m=1}^{\infty}a_{m} \frac{(\alpha+1)}{ ( \alpha+\beta+2 ) }mz^{m-1}-\sum_{m=1}^{\infty}a_{m}mz^{m}+ \frac{1}{2}\sum_{m=2}^{\infty}a_{m}m(m-1)z^{m-1}\\ &\quad{}- \frac{1}{2}\sum_{m=2}^{\infty}a_{m}m(m-1)z^{m} =0, \\ &0 =\frac{(\alpha+1)}{ ( \alpha+\beta+2 ) }a_{1}+\sum_{m=1}^{\infty}a_{m+1} \frac{(\alpha+1)}{ ( \alpha+\beta+2 ) } ( m+1 ) z^{m}-\sum_{m=1}^{\infty}a_{m}mz^{m}\\ &\hphantom{0 =}{}+ \frac{1}{2}\sum_{m=1}^{\infty}a_{m+1} ( m+1 ) (m)z^{m}-\frac {1}{2}\sum_{m=2}^{\infty}a_{m}m(m-1)z^{m},\\ &\frac{(\alpha+1)}{ ( \alpha+\beta+2 ) }a_{1}+\sum_{m=1}^{\infty } \biggl( a_{m+1}\frac{(\alpha+1)}{ ( \alpha+\beta+2 ) } ( m+1 ) -a_{m}m\\ &\quad{}+ \frac{1}{2}a_{m+1} ( m+1 ) (m)-\frac{1}{2}a_{m}m(m-1) \biggr) z^{m}=0, \\ &a_{1}=0, \\ &a_{m+1}\frac{(\alpha+1)}{ ( \alpha+\beta+2 ) } ( m+1 ) -a_{m}m+ \frac{1}{2}a_{m+1} ( m+1 ) (m)-\frac{1}{2}a_{m}m(m-1)=0, \\ &a_{m+1} \biggl( \frac{(\alpha+1)}{ ( \alpha+\beta+2 ) } ( m+1 ) +\frac{1}{2} ( m+1 ) (m) \biggr) =a_{m} \biggl( m+\frac{1}{2}m(m-1) \biggr), \\ &a_{m+1}=\frac{a_{m} ( m+\frac{1}{2}m(m-1) ) }{\frac{(\alpha +1)}{ ( \alpha+\beta+2 ) } ( m+1 ) +\frac{1}{2} ( m+1 ) (m)}, \end{aligned}$$
for all \(z\in\overline{\mathbb{D}}_{R}\backslash \{ 0 \} \). Thus \(a_{m}=0\), \(m=1,2,3,\ldots \) . Thus, f is constant, which is in contradiction with the hypothesis.
Now, by Theorem 2 we have
$$\begin{aligned} & ( n+1 ) \biggl\vert K_{n,q} ( f;z ) -f ( z ) -\frac{(\alpha+1)-z ( \alpha+\beta+2 ) }{ ( \alpha+\beta+2 ) }mz^{m-1}- \frac{z(1-z)}{2}m(m-1)z^{m-2}\biggr\vert \\ &\quad \leq\frac{n+1}{n} \Biggl( \frac{13}{n}\sum _{m=2}^{\infty} \vert a_{m}\vert m^{2} ( m-1 ) ^{2}r^{m}+\frac{4}{(\alpha +\beta+2)n}\sum _{m=2}^{\infty} \vert a_{m}\vert (m-1) ( m-2 ) r^{m} \Biggr) \rightarrow0 \\ &\qquad \mbox{as }n\rightarrow \infty. \end{aligned}$$
Consequently, there exists \(n_{1}\) (depending only on f and r) such that, for all \(n\geq n_{1}\), we have
$$\begin{aligned} & \biggl\Vert \frac{(\alpha+1)-z ( \alpha+\beta+2 ) }{ ( \alpha+\beta+2 ) }mz^{m-1}+\frac{z(1-z)}{2}m(m-1)z^{m-2} \biggr\Vert _{r} \\ &\qquad{} - ( n+1 ) \biggl\Vert K_{n} ( f;z ) -f ( z ) - \frac{1}{ ( n+1 ) } \biggl( \frac{(\alpha+1)-z ( \alpha+\beta+2 ) }{ ( \alpha+\beta+2 ) }mz^{m-1}\\ &\qquad{}+ \frac{z(1-z)}{2}m(m-1)z^{m-2} \biggr) \biggr\Vert _{r} \\ & \quad\geq\frac{1}{2}\biggl\Vert \frac{(\alpha+1)-z ( \alpha+\beta +2 ) }{ ( \alpha+\beta+2 ) }mz^{m-1}+ \frac{z(1-z)}{2}m(m-1)z^{m-2}\biggr\Vert _{r}, \end{aligned}$$
which implies
$$\begin{aligned}[b] &\bigl\Vert K_{n} ( f ) -f\bigr\Vert _{r}\geq \frac{1}{n+1}\frac {1}{2}\biggl\Vert \frac{(\alpha+1)-z ( \alpha+\beta+2 ) }{ ( \alpha+\beta+2 ) }mz^{m-1}+ \frac{z(1-z)}{2}m(m-1)z^{m-2}\biggr\Vert _{r},\\ &\quad \mbox{for all }n\geq n_{1}. \end{aligned} $$
For \(1\leq n\leq n_{1}-1\) we have
$$\bigl\Vert K_{n} ( f ) -f\bigr\Vert _{r}\geq \frac{1}{2(n+1)} \bigl( ( n+1 ) \bigl\Vert K_{n} ( f ) -f\bigr\Vert _{r} \bigr) =\frac{1}{2(n+1)}M_{r,n} ( f ) >0, $$
which finally implies that
$$\bigl\Vert K_{n} ( f ) -f\bigr\Vert _{r}\geq \frac{1}{n+1}C_{r} ( f ) , $$
for all n, with \(C_{r} ( f ) =\min \{ \frac{1}{2}M_{r,1} ( f ) ,\ldots,\frac{1}{2}M_{r,n_{1}-1} ( f ) ,\frac{1}{2}\Vert \frac{(\alpha+1)-z ( \alpha+\beta+2 ) }{ ( \alpha+\beta+2 ) }mz^{m-1}+\frac{z(1-z)}{2}m(m-1)z^{m-2}\Vert _{r} \} \). □

Declarations

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Authors’ Affiliations

(1)
Eastern Mediterranean University

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© Mahmudov and Kara; licensee Springer. 2015