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 ℂ. 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 \(C_{r} ( f ) =\min \{ \frac{1}{2}M_{r,1} ( f ) ,\ldots,\frac{1}{2}M_{r,n_{1}-1} ( f ) ,\frac{1}{2}\Vert \frac{(\alpha+1)-z ( \alpha+\beta+2 ) }{ ( \alpha+\beta+2 ) }mz^{m-1}+\frac{z(1-z)}{2}m(m-1)z^{m-2}\Vert _{r} \} \). □