1 Introduction

The applications on the real and complex approximations properties of the operators have been recently an active subject in the area of the approximation theory (see [7, 10–14]).

Balázs [3] defined the Bernstein type rational functions. He gave an estimate for the order of its convergence and proved an asymptotic approximation theorem and a convergence theorem concerning the derivative of these operators.

In [4] , Balázs and Szabados obtained the best possible estimate under the more restrictive conditions, in which both the weight and the order of convergence would be better than [3]. They applied their results to the approximate certain improper integrals by quadrature sums of positive coefficients based on finite number of equidistant nodes.

Atakut and Ispir [2] defined the bivariate real Bernstein type rational functions of the Bernstein type rational functions given by Balázs [3] and proved the approximation theorems for these functions. In [8], Gupta and Ispir studied on the Bezier variant of generalized Kantorovitch type Balázs operators.

The rational complex Balázs-Szabados operators was defined by Gal [6] as follows:

$$\begin{aligned} R_{n}\big ( f;z\big ) =\frac{1}{\big ( 1+a_{n}z\big ) ^{n}}\sum _{j=0}^{n}f\left( \frac{j}{b_{n}}\right) \left( {\begin{array}{c}n\\ j\end{array}}\right) \big ( a_{n}z\big ) ^{j}, \end{aligned}$$

where \(D_{R}=\left\{ z\in :\left| z\right| <R\right\} \) with \(R>0,\) \(f:D_{R}\cup \left[ R,\infty \right) \rightarrow \) is a function, \(a_{n}=n^{\beta -1},\) \(b_{n}=n^{\beta }\) for \(0<\beta \le 2/3,\) \(n\in \mathbb {N} ,\) \(z\in \mathbb {C} \) with \(z\ne -1/{a_{n}}\). He obtained the uniform convergence of \(R_{n}\big ( f;z\big ) \) to \(f\big ( z\big ) \) on compact disk and proved the upper estimate in approximation of these operators. Also, he obtained the Voronovskaja-type formula and the exact degree of its approximation.

Gal extended some complex univariate operators to the case of several complex variables. To illustrate, he studied the approximation properties of the complex Bernstein polynomials, both tensor-product and non-tensor-product types, the complex Favard-Szász-Mirakjan operators of tensor-product kind without exponential growth conditions on f and the complex Baskakov operators of tensor-product kind ([6], see pp. 155–179).

We consider the following real bivariate Balázs-Szabados operators of the univariate Bernstein type rational functions given by Balázs and Szabados[4]

$$\begin{aligned} R_{n,m}\big ( f\big ) \big ( x_{1},x_{2}\big ) =\sum \limits _{k=0}^{n}\sum \limits _{j=0}^{m}f\left( \frac{k}{b_{n}},\frac{j }{b_{m}}\right) p_{n,k}\big ( x_{1}\big ) p_{m,j}\big ( x_{2}\big ) , \end{aligned}$$

where \(f:\left[ 0,\infty \right) \times \left[ 0,\infty \right) \rightarrow \mathbb {R} \) is a function, \(a_{n}=n^{\beta -1},\) \(b_{n}=n^{\beta },\) \( a_{m}=m^{\beta -1},\) \(b_{m}=m^{\beta }\) for n,  \(m\in \mathbb {N} ,\) \(0<\beta \le 2/3,\) and \(x_{1,}\) \(x_{2}\in \mathbb {R} \) with \(x_{1}\ne -1/{a_{n}}\), \(x_{2}\ne -1/{a_{m}}\),

$$\begin{aligned} p_{n,k}\big ( x_{1}\big ) =\frac{\left( {\begin{array}{c}n\\ k\end{array}}\right) \big ( a_{n}x_{1}\big ) ^{k}}{ \big ( 1+a_{n}x_{1}\big ) ^{n}}\quad \text {and}\quad p_{m,j}\big ( x_{2}\big ) = \frac{\left( {\begin{array}{c}m\\ j\end{array}}\right) \big ( a_{m}x_{2}\big ) ^{j}}{\big ( 1+a_{m}x_{2}\big ) ^{m}}. \end{aligned}$$

The real bivariate Balázs-Szabados operators are well defined, linear, and positive. For \(\beta =2/3\), we get the real bivariate Balázs-Szabados operators given by Atakut and Ispir [2].

We define the complex bivariate Balázs-Szabados operators as follows:

$$\begin{aligned} R_{n,m}\big ( f\big ) \big ( z_{1},z_{2}\big ) =\sum \limits _{k=0}^{n}\sum \limits _{j=0}^{m}f\left( \frac{k}{b_{n}},\frac{j }{b_{m}}\right) p_{n,k}\big ( z_{1}\big ) p_{m,j}\big ( z_{2}\big ) , \end{aligned}$$
(1.1)

where \(f:\left( D_{R_{1}}\cup \left[ R_{1},\infty \right) \right) \times \left( D_{R_{2}}\cup \left[ R_{2},\infty \right) \right) \rightarrow \mathbb {C} \) is a function , \(a_{n}=n^{\beta -1},\) \(b_{n}=n^{\beta },\) \( a_{m}=m^{\beta -1},\) \(b_{m}=m^{\beta },\) for n,  \(m\in \mathbb {N} ,\) \(0<\beta \le 2/3,\) and \(z_{1,}\) \(z_{2}\in \mathbb {C} ,\) with \(z_{1}\ne -1/a_{n}\), \(z_{2}\ne -1/{a_{m}}\),

$$\begin{aligned} p_{n,k}\big ( z_{1}\big ) =\frac{\left( {\begin{array}{c}n\\ k\end{array}}\right) \big ( a_{n}z_{1}\big ) ^{k}}{ \big ( 1+a_{n}z_{1}\big ) ^{n}} \quad \text {and}\quad p_{m,j}\big ( z_{2}\big ) = \frac{\left( {\begin{array}{c}m\\ j\end{array}}\right) \big ( a_{m}z_{2}\big ) ^{j}}{\big ( 1+a_{m}z_{2}\big ) ^{m}}. \end{aligned}$$

The complex bivariate Balázs-Szabados operators are well defined, linear, and these operators are analytic for all \(n\ge n_{0},\) \(m\ge m_{0}\) , \(\left| z_{1}\right| \le r_{1}<n_{0}^{1-\beta }\) and \(\left| z_{2}\right| \le r_{2}<m_{0}^{1-\beta }\) since \(z_{1}\ne -1/ {a_{n}}\) and \(z_{2}\ne -1/{a_{m}}.\)

Throughout the paper, we denote with \(\left\| f\right\| _{r_{1},r_{2}}=\max \left\{ \left| f\big ( z_{1},z_{2}\big ) \right| :\big ( z_{1},z_{2}\big ) \right. \) \(\left. \in \bar{D} _{r_{1}}\times \bar{D}_{r_{2}}\right\} \) the uniform norm of f in the space of continuous functions on \(\bar{D}_{r_{1}}\times \bar{D}_{r_{2}}\) and with \(\left\| f\right\| _{B\left( \left[ 0,\infty \right) \times \left[ 0,\infty \right) \right) }=\sup \left\{ \left| f\big ( z_{1},z_{2}\big ) \right| :\big ( z_{1},z_{2}\big ) \in \left[ 0,\infty \right) \times \left[ 0,\infty \right) \right\} \) the uniform norm of f in the space of bounded functions on \(\left[ 0,\infty \right) \times \left[ 0,\infty \right) ,\) where \(D_{r}=\left\{ z\in \mathbb {C} :\left| z\right| <r\right\} \) for \(r>0.\)

2 Auxiliary Results

We need following lemmas and theorems in order to prove the main results for the operators (1.1).

Lemma 1

For all \(n,m\in \mathbb {N}, \) we have

$$\begin{aligned}&R_{n,m}\left( e_{0,0}\right) \big ( x_{1},x_{2}\big ) =1,\\&R_{n,m}\left( e_{1,0}\right) \big ( x_{1},x_{2}\big ) =\frac{x_{1}}{ 1+a_{n}x_{1}},\\&R_{n,m}\left( e_{0,1}\right) \big ( x_{1},x_{2}\big ) =\frac{x_{2}}{ 1+a_{m}x_{2}},\\&R_{n,m}\left( e_{2,0}\right) \big ( x_{1},x_{2}\big ) =\frac{x_{1}^{2}}{ \big ( 1+a_{n}x_{1}\big ) ^{2}}+\frac{x_{1}}{b_{n}\big ( 1+a_{n}x_{1}\big ) ^{2}},\\&R_{n,m}\left( e_{0,2}\right) \big ( x_{1},x_{2}\big ) =\frac{x_{2}^{2}}{ \big ( 1+a_{m}x_{2}\big ) ^{2}}+\frac{x_{2}}{b_{m}\big ( 1+a_{m}x_{2}\big ) ^{2}}, \end{aligned}$$

where \(\left( e_{i,j}\right) \big ( x_{1},x_{2}\big ) =e_{1}^{i}\left( x_{1}\right) e_{2}^{j}\left( x_{2}\right) \) with \(e_{1}^{i}\left( x_{1}\right) =x_{1}^{i}\) and \(e_{2}^{j}\left( x_{2}\right) =x_{2}^{j}\) for \(i,j=0,1,2.\)

Proof

Using Barbosu technique in [5] and Lemma 2.1 in [3], the lemma can be easily proved, so we will omit the proof of the lemma. \(\square \)

Lemma 2

Let \(f:\left[ 0,\infty \right) \times \left[ 0,\infty \right) \rightarrow \mathbb {R} \) be a continuous function. If

$$\begin{aligned}&\lim _{n,m\rightarrow \infty }\left\| R_{n,m}\left( e_{0,0}\right) -e_{0,0}\right\| _{r_{1},r_{2}}=0,\\&\lim _{n,m\rightarrow \infty }\left\| R_{n,m}\left( e_{1,0}\right) -e_{1,0}\right\| _{r_{1},r_{2}}=0,\\&\lim _{n,m\rightarrow \infty }\left\| R_{n,m}\left( e_{0,1}\right) -e_{0,1}\right\| _{r_{1},r_{2}}=0,\\&\lim _{n,m\rightarrow \infty }\left\| R_{n,m}\left( e_{2,0}+e_{0,2}\right) -\left( e_{2,0}+e_{0,2}\right) \right\| _{r_{1},r_{2}}=0, \end{aligned}$$

then \(\left\{ R_{n,m}\left( f\right) \right\} \) converges uniformly to f on \(\left[ 0,r_{1}\right] \times \left[ 0,r_{2}\right] \) for \(r_{1},r_{2}>0.\)

Proof

From Lemma 2, taking into account Volkov’s theorem in [15] (also see in [1], p.245),the lemma can be easily proved, so we will omit the proof of the lemma. \(\square \)

Lemma 3

Let \(n_{0},m_{0}\ge 2,\) \(0<\beta \le 2/3\), \(1/2 <r_{1}<R_{1}\le n_{0}^{1-\beta }/2\) and \(1/2 <r_{2}<R_{2}\le m_{0}^{1-\beta }/2\). If \(f:\left( D_{R_{1}}\cup \left[ R_{1},\infty \right) \right) \times \left( D_{R_{2}}\cup \left[ R_{2},\infty \right) \right) \rightarrow \mathbb {C} \) is a uniformly continuous bounded on \(\left[ 0,\infty \right) \times \left[ 0,\infty \right) \) and analytic in \(D_{R_{1}}\times D_{R_{2}}\), then we have the form

$$\begin{aligned} R_{n,m}\big ( f\big ) \big ( z_{1},z_{2}\big ) =\sum \limits _{k=0}^{\infty }\sum \limits _{j=0}^{\infty }c_{k,j}R_{n,m}\big ( e_{k,j}\big ) \big ( z_{1},z_{2}\big ) \end{aligned}$$

for all \(\big ( z_{1},z_{2}\big ) \in D_{r_{1}}\times D_{r_{2}},\) where \( \left( e_{k,j}\right) \big ( z_{1},z_{2}\big ) =e_{1}^{k}\left( z_{1}\right) e_{2}^{j}\left( z_{2}\right) \) with \(e_{1}^{k}\left( z_{1}\right) =z_{1}^{k}\) and \(e_{2}^{j}\left( z_{2}\right) =z_{2}^{j}\) for \( k,j\in \mathbb {N}.\)

Proof

For any \(s,r\in \mathbb {N} ,\) we define

$$\begin{aligned} f_{s,r}\left( z_{1},z_{2}\right)= & {} \sum \limits _{k=0}^{s}\sum \limits _{j=0}^{r}c_{k,j}e_{k,j}\left( z_{1},z_{2}\right) \,\text {if}\,\left| z_{1}\right| \le r_{1},\left| z_{2}\right| \le r_{2} \end{aligned}$$

and

$$\begin{aligned} f_{s,r}\left( z_{1},z_{2}\right)= & {} f\left( z_{1},z_{2}\right) \,\text {if}\, \left( z_{1},z_{2}\right) \in \left( r_{1},\infty \right) \times \left( r_{2},\infty \right) . \end{aligned}$$

From the hypothesis on f, it is clear that each \(f_{s,r}\) is bounded on \( \left[ 0,\infty \right) \times \left[ 0,\infty \right) ,\) which implies that

$$\begin{aligned} \left| R_{n,m}\left( f_{s,r}\right) \big ( z_{1},z_{2}\big ) \right| \le \sum \limits _{k=0}^{n}\sum \limits _{j=0}^{m}\left| p_{n,k}\left( z_{1}\right) \right| \left| p_{m,j}\left( z_{2}\right) \right| M_{f_{s,r}}<\infty , \end{aligned}$$

where \(M_{f_{s,r}}\) is a constant depending on \(f_{s,r},\) so all \( R_{n,m}\left( f_{s,r}\right) \) are well defined for all n,  \(m\in \mathbb {N} ,\) \(n\ge n_{0},\) \(m\ge m_{0},\) \(r_{1}<n_{0}^{1-\beta }/2,\) \(r_{2}< m_{0}^{1-\beta }/2\) and \(\big ( z_{1},z_{2}\big ) \in D_{r_{1}}\times D_{r_{2}}\).

Defining

$$\begin{aligned} f_{s,r,k,j}\left( z_{1},z_{2}\right)= & {} c_{k,j}e_{k,j}\left( z_{1},z_{2}\right) \,\text {if}\,\left| z_{1}\right| \le r_{1},\, \left| z_{2}\right| \le r_{2} \end{aligned}$$

and

$$\begin{aligned} f_{s,r,k,j}\left( z_{1},z_{2}\right)= & {} \frac{f\left( z_{1},z_{2}\right) }{ \left( s+1\right) \left( r+1\right) }\,\text {if}\,\left( z_{1},z_{2}\right) \in \left( r_{1},\infty \right) \times \left( r_{2},\infty \right) . \end{aligned}$$

It is clear that each \(f_{s,r,k,j}\) is bounded on \(\left[ 0,\infty \right) \times \left[ 0,\infty \right) \) and

$$\begin{aligned} f_{s,r}\left( z_{1},z_{2}\right) =\sum \limits _{k=0}^{s}\sum \limits _{j=0}^{r}f_{s,r,k,j}\left( z_{1},z_{2}\right) . \end{aligned}$$

From the linearity of \(R_{n,m},\) we have

$$\begin{aligned} R_{n,m}\left( f_{s,r}\right) \left( z_{1},z_{2}\right) =\sum \limits _{k=0}^{s}\sum \limits _{j=0}^{r}c_{k,j}R_{n,m}\left( e_{k,j}\right) \left( z_{1},z_{2}\right) . \end{aligned}$$

It suffices to prove that

$$\begin{aligned} \lim _{s,r\rightarrow \infty }R_{n,m}\left( f_{s,r}\right) \left( z_{1},z_{2}\right) =R_{n,m}\left( f\right) \left( z_{1},z_{2}\right) \end{aligned}$$

for any fixed n,  \(m\in \mathbb {N} ,\) \(n\ge n_{0},\) \(m\ge m_{0}\), \(\left| z_{1}\right| \le r_{1}\) and \(\left| z_{2}\right| \le r_{2}.\) Since

$$\begin{aligned} \left\| f_{s,r}-f\right\| _{B\left( \left[ 0,\infty \right) \times \left[ 0,\infty \right) \right) }\le \left\| f_{s,r}-f\right\| _{r_{1},r_{2}}, \end{aligned}$$

we can write

$$\begin{aligned}&\left| R_{n,m}\left( f_{s,r}\right) \big ( z_{1},z_{2}\big ) -R_{n,m}\left( f\right) \big ( z_{1},z_{2}\big ) \right| \nonumber \\&\quad \le M_{r_{1},r_{2},m,n}\left\| f_{s,r}-f\right\| _{B\left( \left[ 0,\infty \right) \times \left[ 0,\infty \right) \right) } \nonumber \\&\quad \le M_{r_{1},r_{2},m,n}\left\| f_{s,r}-f\right\| _{r_{1},r_{2}} \end{aligned}$$
(2.1)

for \(\left| z_{1}\right| \le r_{1}\) and \(\left| z_{2}\right| \le r_{2}.\)

In (2.1), taking limit as \(s,r\rightarrow \infty \) and using \( \lim \limits _{s,r\rightarrow \infty }\left\| f_{s,r}-f\right\| _{r_{1},r_{2}}=0,\) we get the result. \(\square \)

Next lemma and theorem will help to get the convergence result of the operators (1.1).

Lemma 4

Let \(n_{0},m_{0}\ge 2,\) \(0<\beta \le 2/3\), \(1/2 <r_{1}<R_{1}\le n_{0}^{1-\beta }/2\) and \(1/2 <r_{2}<R_{2}\le m_{0}^{1-\beta }/2\). For all \(n\ge n_{0},\) \(m\ge m_{0}\), \(\left| z_{1}\right| \le r_{1}\) and \(\left| z_{2}\right| \le r_{2}, \)the following inequality holds

$$\begin{aligned} \left| R_{n,m}\left( e_{k,j}\right) \big ( z_{1},z_{2}\big ) \right| \le k!j!\left( 2r_{1}\right) ^{k}\left( 2r_{2}\right) ^{j}. \end{aligned}$$

Proof

Using Lemma 1.10.2 in [6] (see p. 141),  the lemma is easily proved, so we will omit the proof of the theorem. \(\square \)

3 Main Results

Let us denote with \(A_{C}\left( f\right) \) the space of the all complex valued functions, which are uniformly continuous on \(\left( D_{R_{1}}\cup \left[ R_{1},\infty \right) \right) \times \left( D_{R_{2}}\cup \left[ R_{2},\infty \right) \right) ,\) bounded on \(\left[ 0,\infty \right) \times \left[ 0,\infty \right) \) and analytic in \(D_{R_{1}}\times D_{R_{2}}\), that is \(f\left( z_{1},z_{2}\right) =\sum \nolimits _{k=0}^{\infty }f_{k}\left( z_{2}\right) z_{1}^{k}\)for all \(\left( z_{1},z_{2}\right) \in D_{R_{1}}\times D_{R_{2}}\) with \(f_{k}\left( z_{2}\right) =\sum \nolimits _{j=0}^{\infty }c_{k,j}z_{2}^{j},\) and for which there exist \( M>0,\) \(0<A_{1}<1/2r_{1}\) and \(0<A_{2}<1/2r_{2}\) with \( \left| c_{k,j}\right| \le MA_{1}^{k}A_{2}^{j}/k!j!\) (which implies \(\left| f\left( z_{1},z_{2}\right) \right| \le Me^{A_{1}\left| z_{1}\right| +A_{2}\left| z_{2}\right| }\) for all \(\left( z_{1},z_{2}\right) \in D_{R_{1}}\times D_{R_{2}}).\)

Now, we can give the following convergence result.

Theorem 1

Let \(n_{0},m_{0}\ge 2,\) \(0<\beta \le 2/3\), \(1/2 <r_{1}<R_{1}\le n_{0}^{1-\beta }/2\) and \(1/2 <r_{2}<R_{2}\le m_{0}^{1-\beta }/2\). If \(f\in A_{C}\left( f\right) \), then the sequence of operators \(\left\{ R_{n,m}\left( f\right) \left( z_{1},z_{2}\right) \right\} \) is uniformly convergent to f on \(\bar{D} _{r_{1}}\times \bar{D}_{r_{2}}\) for all \(n\ge n_{0}\) and \(m\ge m_{0}.\)

Proof

From Lemma 3 and Lemma 4, we can write

$$\begin{aligned} \left| R_{n,m}\left( f\right) \big ( z_{1},z_{2}\big ) \right|\le & {} \sum \limits _{k=0}^{\infty }\sum \limits _{j=0}^{\infty }\left| c_{k,j}\right| \left| R_{n,m}\left( e_{k,j}\right) \big ( z_{1},z_{2}\big ) \right| \nonumber \\\le & {} M\sum \limits _{k=0}^{\infty }\sum \limits _{j=0}^{\infty }\big ( 2r_{1}A_{1}\big ) ^{k}\big ( 2r_{2}A_{2}\big ) ^{j}<\infty , \end{aligned}$$
(3.1)

where the series \(\sum \nolimits _{k=0}^{\infty }\sum \nolimits _{j=0}^{\infty }\big ( 2r_{1}A_{1}\big ) ^{k}\big ( 2r_{2}A_{2}\big ) ^{j}\) is convergent since \(0<A_{1}<1/2r_{1}\) and \(0<A_{2}<1/2r_{2}.\)

On the other hand, from Lemma 2, we know that

$$\begin{aligned} \lim _{n,m\rightarrow \infty }R_{n,m}\left( f\right) \big ( x_{1},x_{2}\big ) =f\big ( x_{1},x_{2}\big ) \end{aligned}$$
(3.2)

for all \(\big ( x_{1},x_{2}\big ) \in \left[ 0,r_{1}\right] \times \left[ 0,r_{2}\right] .\)

Using (3.1) and (3.2) and taking into account the Vitali’s theorem [9] (see p.112, Theorem 3.2.10), it follows that \(\left\{ R_{n,m}\left( f\right) \big ( z_{1},z_{2}\big ) \right\} \) uniformly converges to f on \(\bar{D}_{r_{1}}\times \bar{D}_{r_{2}}\) for all \(n\ge n_{0}\) and \(m\ge m_{0}.\)

We have the following upper estimate.

Theorem 2

Let \(n_{0},m_{0}\ge 2,\) \(0<\beta \le 2/3\), \(1/2 <r_{1}<R_{1}\le n_{0}^{1-\beta }/2\) and \(1/2 <r_{2}<R_{2}\le m_{0}^{1-\beta }/2\). If \(f\in A_{C}\left( f\right) \), then for all \(n\ge \max \left\{ n_{0},1/r_{1}^{1/\beta }\right\} ,\) \(m\ge \max \left\{ m_{0},1/r_{2}^{1/\beta }\right\} ,\) \(\left| z_{1}\right| \le r_{1}\) and \(\left| z_{2}\right| \le r_{2}\) the following inequality holds

$$\begin{aligned} \left| R_{n,m}\left( f\right) \big ( z_{1},z_{2}\big ) -f\big ( z_{1},z_{2}\big ) \right| \le \left( a_{n}+\frac{1}{b_{n}}\right) C^{1}\left( f\right) +\left( a_{m}+\frac{1}{b_{m}}\right) C^{2}\left( f\right) , \end{aligned}$$

where

$$\begin{aligned} C^{1}\left( f\right)= & {} \max \left\{ 2r_{1}Me^{2r_{1}A_{1}+r_{2}A_{2}},\frac{ 4M}{r_{1}}e^{r_{2}A_{2}}\sum \limits _{k=1}^{\infty }k\big ( 2r_{1}A_{1}\big ) ^{k-1}\right\} , \\ C^{2}\left( f\right)= & {} \max \left\{ 2r_{2}Me^{2r_{2}A_{2}}\sum \limits _{k=0}^{\infty }\big ( 2r_{1}A_{1}\big ) ^{k},\frac{4M}{r_{2}}\sum \limits _{k=0}^{\infty }\big ( 2r_{1}A_{1}\big ) ^{k}\sum \limits _{j=1}^{\infty }j\big ( 2r_{2}A_{2}\big ) ^{j-1}\right\} , \end{aligned}$$

and also the series \(\sum \nolimits _{k=0}^{\infty }\big ( 2r_{1}A_{1}\big ) ^{k},\) \(\sum \nolimits _{k=1}^{\infty }k\big ( 2r_{1}A_{1}\big ) ^{k-1}\) and \( \sum \nolimits _{j=1}^{\infty }j\big ( 2r_{2}A_{2}\big ) ^{j-1}\) are convergent.

Proof

Using Lemma 3, we can write

$$\begin{aligned}&\left| R_{n,m}\left( f\right) \big ( z_{1},z_{2}\big ) -f\big ( z_{1},z_{2}\big ) \right| \le \sum \limits _{k=0}^{\infty }\sum \limits _{j=0}^{\infty }\left| c_{k,j}\right| \left| R_{n,m}\left( e_{k,j}\right) \big ( z_{1},z_{2}\big )\right. \nonumber \\&\quad \quad \left. -e_{k,j}\big ( z_{1},z_{2}\big ) \right| . \end{aligned}$$
(3.3)

Taking into account Lemma 1.10.2 and the estimate given in the proof of Theorem 1.10.5 in [6] (see p. 141 pp 144-145), for all \(\left| z_{1}\right| \le r_{1}\) and \(\left| z_{2}\right| \le r_{2},\) we obtain

$$\begin{aligned}&\left| R_{n,m}\left( e_{k,j}\right) \big ( z_{1},z_{2}\big ) -e_{k,j}\big ( z_{1},z_{2}\big ) \right| \nonumber \\&\quad =\left| R_{n}\left( e_{1}^{k}\right) \left( z_{1}\right) .R_{m}\left( e_{2}^{j}\right) \left( z_{2}\right) -z_{1}^{k}z_{2}^{j}\right| \nonumber \\&\quad \le \left| R_{n}\left( e_{1}^{k}\right) \left( z_{1}\right) \right| \left| R_{m}\left( e_{2}^{j}\right) \left( z_{2}\right) -z_{2}^{j}\right| +\left| z_{2}^{j}\right| \left| R_{n}\left( e_{1}^{k}\right) \left( z_{1}\right) -z_{1}^{k}\right| \nonumber \\&\quad \le \left( k!\right) \left( 2r_{1}\right) ^{k}\left\{ \left( 2a_{m}r_{2}\right) \left( 2r_{2}\right) ^{j}+\frac{8\left( 2r_{2}\right) ^{j-1}}{b_{m}}j\left( j!\right) \right\} \nonumber \\&\qquad +r_{2}^{j}\left\{ 2a_{n}r_{1}\left( 2r_{1}\right) ^{k}+\frac{8\left( 2r_{1}\right) ^{k-1}}{b_{n}}k\left( k!\right) \right\} \nonumber \\&=\left( 2a_{n}r_{1}\right) \left( 2r_{1}\right) ^{k}r_{2}^{j}+\left( 2a_{m}r_{2}\right) \left( k!\right) \left( 2r_{1}\right) ^{k}\left( 2r_{2}\right) ^{j} \nonumber \\&\qquad +\frac{8k\left( k!\right) \left( 2r_{1}\right) ^{k-1}r_{2}^{j}}{b_{n}}+ \frac{8\left( k!\right) \left( 2r_{1}\right) ^{k}j\left( j!\right) \left( 2r_{2}\right) ^{j-1}}{b_{m}}. \end{aligned}$$
(3.4)

Applying (3.4) in(3.3), we get

$$\begin{aligned} \left| R_{n,m}\left( f\right) \big ( z_{1},z_{2}\big ) -f\big ( z_{1},z_{2}\big ) \right|\le & {} 2a_{n}r_{1}M\sum \limits _{k=0}^{\infty }\sum \limits _{j=0}^{\infty }\frac{\big ( 2r_{1}A_{1}\big ) ^{k}\left( r_{2}A_{2}\right) ^{j}}{k!j!} \\&+2a_{m}r_{2}M\sum \limits _{k=0}^{\infty }\sum \limits _{j=0}^{\infty }\frac{ \big ( 2r_{1}A_{1}\big ) ^{k}\left( 2r_{2}A_{2}\right) ^{j}}{j!} \\&+\frac{8MA_{1}}{b_{n}}\sum \limits _{k=1}^{\infty }\sum \limits _{j=0}^{\infty }\frac{k\big ( 2r_{1}A_{1}\big ) ^{k-1}\left( r_{2}A_{2}\right) ^{j}}{j!} \\&+\frac{8MA_{2}}{b_{m}}\sum \limits _{k=0}^{\infty }\sum \limits _{j=1}^{\infty }\big ( 2r_{1}A_{1}\big ) ^{k}j\left( 2r_{2}A_{2}\right) ^{j-1} \end{aligned}$$

which complete the prove. \(\square \)

In what follows a Voronovskaja-type result for the operators (1.1) is presented. It will be the product of the parametric extensions generated by Voronovskaja’s formula in univariate case in Theorem 1.10.6 in [6] (see p. 145).

Theorem 3

Let \(n_{0},m_{0}\ge 2,\) \(0<\beta \le 2/3\), \(1/2 <r_{1}<R_{1}\le n_{0}^{1-\beta }/2\) and \(1/2 <r_{2}<R_{2}\le m_{0}^{1-\beta }/2\). If \(f\in A_{C}\left( f\right) \), then for all \(n\ge \max \left\{ n_{0},1/r_{1}^{1/\beta }\right\} ,\) \(m\ge \max \left\{ m_{0},1/r_{2}^{1/\beta }\right\} ,\) \(\left| z_{1}\right| \le r_{1}\) and \(\left| z_{2}\right| \le r_{2}\) the following inequality holds

$$\begin{aligned}&\left| z_{2}T_{m}\left( f\right) \big ( z_{1},z_{2}\big ) \circ z_{1}T_{n}\left( f\right) \big ( z_{1},z_{2}\big ) \right| \le C^{3}\left( f\right) \left[ \left( a_{n}+\frac{1}{b_{n}}\right) ^{2}+\left( a_{m}+\frac{1}{b_{m}}\right) ^{2}\right] , \end{aligned}$$

where \(C^{3}\left( f\right) =1/2\max \left\{ C_{r_{1},r_{2}}^{r_{1}^{*}}\left( f\right) ,C_{r_{1},r_{2}}^{r_{2}^{*}}\left( f\right) \right\} \) with

$$\begin{aligned} C_{r_{1},r_{2}}^{r_{1}^{*}}\left( f\right)= & {} MC_{r_{1,}r_{1}^{ *}}e^{A_{2}r_{2}}\sum \limits _{k=2}^{\infty }\left( k-1\right) k\left( k+1\right) \left( k+2\right) \left( 2A_{1}r_{1}\right) ^{k} \\&\times \max \left\{ 1,e^{-A_{2}r_{2}}\sum \limits _{j=0}^{\infty }\left( 2A_{2}r_{2}\right) ^{j},\frac{a_{m}r_{2}^{2}}{1-a_{m}r_{2}},\frac{ a_{m}^{2}b_{m}r_{2}^{4}+r_{2}}{2b_{m}\left( 1-a_{m}r_{2}\right) ^{2}} \right\} \\ C_{r_{1},r_{2}}^{r_{2}^{*}}\left( f\right)= & {} MC_{r_{_{2},}r_{2}^{ *}}e^{A_{1}r_{1}}\sum \limits _{j=2}^{\infty }\left( j-1\right) j\left( j+1\right) \left( j+2\right) \left( 2A_{2}r_{2}\right) ^{j} \\&\times \max \left\{ 1,e^{-A_{1}r_{1}}\sum \limits _{k=0}^{\infty }\left( 2A_{1}r_{1}\right) ^{k},\frac{a_{n}r_{1}^{2}}{1-a_{n}r_{1}},\frac{ a_{n}^{2}b_{n}r_{1}^{4}+r_{1}}{2b_{n}\left( 1-a_{n}r_{1}\right) ^{2}} \right\} , \end{aligned}$$

for \(1/2<r_{1}<r_{1}^{*}<n_{0}^{1-\beta }/2\) and \( 1/2<r_{2}<r_{2}^{*}\le m_{0}^{1-\beta }/2.\)

Proof

For \(f\left( z_{1},z_{2}\right) ,\) we define the parametric extensions of the Voronovskaja’s formula by

$$\begin{aligned} z_{1}T_{n}\left( f\right) \left( z_{1},z_{2}\right):= & {} R_{n}\left( f\left( .,z_{2}\right) \right) \left( z_{1}\right) -f\left( z_{1},z_{2}\right) + \frac{a_{n}z_{1}^{2}}{1+a_{n}z_{1}}\frac{\partial f}{\partial z_{1}}\left( z_{1},z_{2}\right) \\&-\frac{a_{n}^{2}b_{n}z_{1}^{4}+z_{1}}{2b_{n}\left( 1+a_{n}z_{1}\right) ^{2} }\frac{\partial ^{2}f}{\partial z_{1}^{2}}\left( z_{1},z_{2}\right) \end{aligned}$$

and

$$\begin{aligned} z_{2}T_{m}\left( f\right) \left( z_{1},z_{2}\right):= & {} R_{m}\left( f\left( z_{1},.\right) \right) \left( z_{2}\right) -f\left( z_{1},z_{2}\right) + \frac{a_{m}z_{2}^{2}}{1+a_{m}z_{2}}\frac{\partial f}{\partial z_{2}}\left( z_{1},z_{2}\right) \\&-\frac{a_{m}^{2}b_{m}z_{2}^{4}+z_{2}}{2b_{m}\left( 1+a_{m}z_{2}\right) ^{2} }\frac{\partial ^{2}f}{\partial z_{2}^{2}}\left( z_{1},z_{2}\right) . \end{aligned}$$

Their product (composition) gives

$$\begin{aligned}&\!\!\!z_{2}T_{m}\left( f\right) \left( z_{1},z_{2}\right) \circ z_{1}T_{n}\left( f\right) \left( z_{1},z_{2}\right) \\&\!\!\!\quad =R_{m}\left[ R_{n}\big ( f\left( .,.\right) \big ) \left( z_{1}\right) -f\left( z_{1},.\right) +\frac{a_{n}z_{1}^{2}}{1+a_{n}z_{1}}\frac{\partial f }{\partial z_{1}}\left( z_{1},.\right) \right. \\&\!\!\!\qquad \left. -\frac{a_{n}^{2}b_{n}z_{1}^{4}+z_{1}}{2b_{n}\left( 1+a_{n}z_{1}\right) ^{2}}\frac{\partial ^{2}f}{\partial z_{1}^{2}}\left( z_{1},.\right) \right] \left( z_{2}\right) -\left[ R_{n}\big ( f\left( .,z_{2}\right) \big ) \left( z_{1}\right) -f\left( z_{1},z_{2}\right) \right. \\&\!\!\!\qquad \left. +\frac{a_{n}z_{1}^{2}}{1+a_{n}z_{1}}\frac{ \partial f}{\partial z_{1}}\left( z_{1},z_{2}\right) -\frac{a_{n}^{2}b_{n}z_{1}^{4}+z_{1}}{2b_{n}\left( 1+a_{n}z_{1}\right) ^{2}}\frac{\partial ^{2}f}{\partial z_{1}^{2}}\left( z_{1},z_{2}\right) \right] +\frac{a_{m}z_{2}^{2}}{1+a_{m}z_{2}}\frac{\partial }{\partial z_{2}} \\&\!\!\!\qquad \times \left[ R_{n}\big ( f\left( .,z_{2}\right) \big ) \left( z_{1}\right) -f\left( z_{1},z_{2}\right) +\frac{a_{n}z_{1}^{2}}{1+a_{n}z_{1}}\frac{\partial f}{ \partial z_{1}}\left( z_{1},z_{2}\right) \right. \\&\!\!\!\qquad \left. -\frac{a_{n}^{2}b_{n}z_{1}^{4}+z_{1}}{2b_{n}\left( 1+a_{n}z_{1}\right) ^{2}}\frac{\partial ^{2}f}{\partial z_{1}^{2}}\left( z_{1},z_{2}\right) \right] -\frac{a_{m}^{2}b_{m}z_{2}^{4}+z_{2}}{2b_{m}\left( 1+a_{m}z_{2}\right) ^{2} }\frac{\partial ^{2}}{\partial z_{2}^{2}} \\&\!\!\!\qquad \times \left[ R_{n}\big ( f\left( .,z_{2}\right) \big ) \left( z_{1}\right) -f\left( z_{1},z_{2}\right) + \frac{a_{n}z_{1}^{2}}{1+a_{n}z_{1}}\frac{\partial f}{\partial z_{1}}\left( z_{1},z_{2}\right) \right. \\&\!\!\!\qquad \left. -\frac{a_{n}^{2}b_{n}z_{1}^{4}+z_{1}}{2b_{n}\left( 1+a_{n}z_{1}\right) ^{2}}\frac{\partial ^{2}f}{\partial z_{1}^{2}}\left( z_{1},z_{2}\right) \right] :=E_{1}-E_{2}+E_{3}-E_{4}. \end{aligned}$$

After simple calculation, it is obtained the commutativity property

$$\begin{aligned} z_{2}T_{m}\left( f\right) \left( z_{1},z_{2}\right) \circ z_{1}T_{n}\left( f\right) \left( z_{1},z_{2}\right) =z_{1}T_{n}\left( f\right) \left( z_{1},z_{2}\right) \circ z_{2}T_{m}\left( f\right) \left( z_{1},z_{2}\right) .\qquad \end{aligned}$$
(3.5)

From the analyticity of f, since the all partial derivatives of f are analytic in \(D_{R_{1}}\times D_{R_{2}},\) using Lemma 3, we can write

$$\begin{aligned}&R_{n}\big ( f\left( .,z_{2}\right) \big ) \left( z_{1}\right) -f\left( z_{1},z_{2}\right) +\frac{a_{n}z_{1}^{2}}{1+a_{n}z_{1}}\frac{\partial f}{ \partial z_{1}}\left( z_{1},z_{2}\right) -\frac{a_{n}^{2}b_{n}z_{1}^{4}+z_{1}}{2b_{n}\left( 1+a_{n}z_{1}\right) ^{2} }\frac{\partial ^{2}f}{\partial z_{1}^{2}}\left( z_{1},z_{2}\right) \nonumber \\&\quad =\sum \limits _{k=2}^{\infty }f_{k}\left( z_{2}\right) \left[ R_{n}\left( e_{1}^{k}\right) \left( z_{1}\right) -e_{1}^{k}\left( z_{1}\right) +\frac{ \left( a_{n}z_{1}^{2}\right) kz_{1}^{k-1}}{1+a_{n}z_{1}} -\frac{\left( a_{n}^{2}b_{n}z_{1}^{4}+z_{1}\right) k\left( k-1\right) z_{1}^{k-2}}{2b_{n}\left( 1+a_{n}z_{1}\right) ^{2}}\right] .\nonumber \\ \end{aligned}$$
(3.6)

Applying now \(R_{m}\) to (3.6) with respect to \(z_{2}\), we obtain

$$\begin{aligned} E_{1}= & {} \sum \limits _{k=2}^{\infty }R_{m}\left( f_{k}\right) \left( z_{2}\right) \left[ R_{n}\left( e_{1}^{k}\right) \left( z_{1}\right) -e_{1}^{k}\left( z_{1}\right) \right. \nonumber \\&\left. +\frac{\left( a_{n}z_{1}^{2}\right) kz_{1}^{k-1}}{1+a_{n}z_{1}}- \frac{\left( a_{n}^{2}b_{n}z_{1}^{4}+z_{1}\right) k\left( k-1\right) z_{1}^{k-2}}{2b_{n}\left( 1+a_{n}z_{1}\right) ^{2}}\right] \nonumber \\= & {} \sum \limits _{k=2}^{\infty }\sum \limits _{j=0}^{\infty }c_{k,j}R_{m}\left( e_{2}^{j}\right) \left( z_{2}\right) \left[ R_{n}\left( e_{1}^{k}\right) \left( z_{1}\right) -e_{1}^{k}\left( z_{1}\right) \right. \nonumber \\&\left. +\frac{\left( a_{n}z_{1}^{2}\right) kz_{1}^{k-1}}{1+a_{n}z_{1}}- \frac{\left( a_{n}^{2}b_{n}z_{1}^{4}+z_{1}\right) k\left( k-1\right) z_{1}^{k-2}}{2b_{n}\left( 1+a_{n}z_{1}\right) ^{2}}\right] . \end{aligned}$$
(3.7)

In (3.7), passing now to absolute value for \(\left| z_{1}\right| \le r_{1}\) and \(\left| z_{2}\right| \le r_{2}\) and taking into account the Lemma 1.10.2 and the estimate given in the proof of Theorem 1.10.6 in [6] (see p. 141 and pp. 145–146), it follows

$$\begin{aligned} \left| E_{1}\right|\le & {} \left( a_{n}+\frac{1}{b_{n}}\right) ^{2}C_{r_{1,}r_{1}^{*}} \nonumber \\&\times \sum \limits _{k=2}^{\infty }\sum \limits _{j=0}^{\infty }\left| c_{k,j}\right| j!\left( 2r_{2}\right) ^{j}\left( k-1\right) k\left( k+1\right) \left( k+2\right) k!\left( 2r_{1}\right) ^{k} \nonumber \\\le & {} \left( a_{n}+\frac{1}{b_{n}}\right) ^{2}MC_{r_{1,}r_{1}^{*}}e^{A_{2}r_{2}} \nonumber \\&\times \sum \limits _{k=2}^{\infty }\left( k-1\right) k\left( k+1\right) \left( k+2\right) \left( 2A_{1}r_{1}\right) ^{k}\sum \limits _{j=0}^{\infty }\left( 2A_{2}r_{2}\right) ^{j}, \end{aligned}$$
(3.8)

where \(r_{1}<r_{1}^{*}<R<n_{0}^{1-\beta }/2\) with \(r_{1}^{*}< 1/2A_{1}\) and \(C_{r_{1,}r_{1}^{*}}\) is a constant depending on \( r_{1}\) and \(r_{1}^{*}.\)

Similarly, using Lemma 3, we have

$$\begin{aligned} \left| E_{2}\right|\le & {} \sum \limits _{k=2}^{\infty }\left| f_{k}\left( z_{2}\right) \right| \left| R_{n}\left( e_{1}^{k}\right) \left( z_{1}\right) -e_{1}^{k}\left( z_{1}\right) +\frac{\left( a_{n}z_{1}^{2}\right) kz_{1}^{k-1}}{1+a_{n}z_{1}}\right. \nonumber \\&\left. -\frac{\left( a_{n}^{2}b_{n}z_{1}^{4}+z_{1}\right) k\left( k-1\right) z_{1}^{k-2}}{2b_{n}\left( 1+a_{n}z_{1}\right) ^{2}}\right| \nonumber \\\le & {} \left( a_{n}+\frac{1}{b_{n}}\right) ^{2}C_{r_{1,}r_{1}^{*}} \sum \limits _{k=2}^{\infty }\sum \limits _{j=0}^{\infty }\left| c_{k,j}\right| r_{2}^{j}\left( k-1\right) k\left( k+1\right) \left( k+2\right) \left( k!\right) \left( 2r_{1}\right) ^{k} \nonumber \\\le & {} \left( a_{n}+\frac{1}{b_{n}}\right) ^{2}MC_{r_{1,}r_{1}^{*}}e^{A_{2}r_{2}} \sum \limits _{k=2}^{\infty }\left( k-1\right) k\left( k+1\right) \left( k+2\right) \left( 2A_{1}r_{1}\right) ^{k}. \end{aligned}$$
(3.9)

Using

$$\begin{aligned} R_{n}\left( \frac{\partial f}{\partial z_{2}}\left( .,z_{2}\right) \right) \left( z_{1}\right)= & {} \sum \limits _{k=0}^{\infty }\frac{\partial f_{k}}{ \partial z_{2}}\left( z_{2}\right) R_{n}\left( e_{1}^{k}\right) \left( z_{1}\right) \\= & {} \sum \limits _{k=0}^{\infty }\sum \limits _{j=1}^{\infty }c_{k,j}jz_{2}^{j-1}R_{n}\left( e_{1}^{k}\right) \left( z_{1}\right) , \end{aligned}$$

we can write

$$\begin{aligned} E_{3}= & {} \frac{a_{m}z_{2}^{2}}{1+a_{m}z_{2}}\left[ R_{n}\left( \frac{ \partial f}{\partial z_{2}}\left( .,z_{2}\right) \right) \left( z_{1}\right) -\frac{\partial f}{\partial z_{2}}\left( z_{1},z_{2}\right) \right. \\&\left. +\frac{a_{n}z_{1}^{2}}{1+a_{n}z_{1}}\frac{\partial ^{2}f}{\partial z_{1}\partial z_{2}}\left( z_{1},z_{2}\right) -\frac{ a_{n}^{2}b_{n}z_{1}^{4}+z_{1}}{2b_{n}\left( 1+a_{n}z_{1}\right) ^{2}}\frac{ \partial ^{3}f}{\partial z_{1}^{2}\partial z_{2}}\left( z_{1},z_{2}\right) \right] \\= & {} \frac{a_{m}z_{2}^{2}}{1+a_{m}z_{2}}\sum \limits _{k=2}^{\infty }\sum \limits _{j=1}^{\infty }c_{k,j}jz_{2}^{j-1}\left[ R_{n}\left( e_{1}^{k}\right) \left( z_{1}\right) -e_{1}^{k}\left( z_{1}\right) \right. \\&\left. +\frac{\left( a_{n}z_{1}^{2}\right) kz_{1}^{k-1}}{1+a_{n}z_{1}}- \frac{\left( a_{n}^{2}b_{n}z_{1}^{4}+z_{1}\right) k\left( k-1\right) z_{1}^{k-2}}{2b_{n}\left( 1+a_{n}z_{1}\right) ^{2}}\right] , \end{aligned}$$

which implies

$$\begin{aligned} \left| E_{3}\right|\le & {} \left( a_{n}+\frac{1}{b_{n}}\right) ^{2} \frac{a_{m}r_{2}^{2}}{1-a_{m}r_{2}}C_{r_{1,}r_{1}^{*}} \nonumber \\&\times \sum \limits _{k=2}^{\infty }\sum \limits _{j=1}^{\infty }\left| c_{k,j}\right| jr_{2}^{j-1}\left( k-1\right) k\left( k+1\right) \left( k+2\right) \left( k!\right) \left( 2r_{1}\right) ^{k}, \nonumber \\\le & {} \left( a_{n}+\frac{1}{b_{n}}\right) ^{2}\frac{a_{m}r_{2}^{2}}{ 1-a_{m}r_{2}} \nonumber \\&\times MC_{r_{1,}r_{1}^{*}}e^{A_{2}r_{2}}\sum \limits _{k=2}^{\infty }\left( k-1\right) k\left( k+1\right) \left( k+2\right) \left( 2A_{1}r_{1}\right) ^{k}. \end{aligned}$$
(3.10)

And also, using

$$\begin{aligned} R_{n}\left( \frac{\partial ^{2}f}{\partial z_{2}^{2}}\left( .,z_{2}\right) \right) \left( z_{1}\right)= & {} \sum \limits _{k=0}^{\infty }\frac{\partial ^{2}f_{k}}{\partial z_{2}^{2}}\left( z_{2}\right) R_{n}\left( e_{1}^{k}\right) \left( z_{1}\right) \\= & {} \sum \limits _{k=0}^{\infty }\sum \limits _{j=2}^{\infty }c_{k,j}j\left( j-1\right) z_{2}^{j-2}R_{n}\left( e_{1}^{k}\right) \left( z_{1}\right) , \end{aligned}$$

we can write

$$\begin{aligned} E_{4}= & {} \frac{a_{m}^{2}b_{m}z_{2}^{4}+z_{2}}{2b_{m}\left( 1+a_{m}z_{2}\right) ^{2}}\left[ R_{n}\left( \frac{\partial ^{2}f}{\partial z_{2}^{2}}\left( .,z_{2}\right) \right) \left( z_{1}\right) -\frac{\partial ^{2}f}{\partial z_{2}^{2}}\left( z_{1},z_{2}\right) \right. \\&\left. +\frac{a_{n}z_{1}^{2}}{1+a_{n}z_{1}}\frac{\partial ^{3}f}{\partial z_{1}\partial z_{2}^{2}}\left( z_{1},z_{2}\right) -\frac{ a_{n}^{2}b_{n}z_{1}^{4}+z_{1}}{2b_{n}\left( 1+a_{n}z_{1}\right) ^{2}}\frac{ \partial ^{4}f}{\partial z_{1}^{2}\partial z_{2}^{2}}\left( z_{1},z_{2}\right) \right] \\= & {} \frac{a_{m}^{2}b_{m}z_{2}^{4}+z_{2}}{2b_{m}\left( 1+a_{m}z_{2}\right) ^{2} }\sum \limits _{k=2}^{\infty }\sum \limits _{j=2}^{\infty }c_{k,j}j\left( j-1\right) z_{2}^{j-2}\left[ R_{n}\left( e_{1}^{k}\right) \left( z_{1}\right) -e_{1}^{k}\left( z_{1}\right) \right. \\&\left. +\frac{\left( a_{n}z_{1}^{2}\right) kz_{1}^{k-1}}{1+a_{n}z_{1}}- \frac{\left( a_{n}^{2}b_{n}z_{1}^{4}+z_{1}\right) k\left( k-1\right) z_{1}^{k-2}}{2b_{n}\left( 1+a_{n}z_{1}\right) ^{2}}\right] , \end{aligned}$$

which implies

$$\begin{aligned} \left| E_{4}\right|\le & {} \left( a_{n}+\frac{1}{b_{n}}\right) ^{2} \frac{a_{m}^{2}b_{m}r_{2}^{4}+r_{2}}{2b_{m}\left( 1-a_{m}r_{2}\right) ^{2}} C_{r_{1,}r_{1}^{*}}\sum \limits _{j=2}^{\infty }j\left( j-1\right) r_{2}^{j-2} \nonumber \\&\times \sum \limits _{k=2}^{\infty }\left| c_{k,j}\right| \left( k-1\right) k\left( k+1\right) \left( k+2\right) \left( k!\right) \left( 2r_{1}\right) ^{k}, \nonumber \\\le & {} \left( a_{n}+\frac{1}{b_{n}}\right) ^{2}\frac{ a_{m}^{2}b_{m}r_{2}^{4}+r_{2}}{2b_{m}\left( 1-a_{m}r_{2}\right) ^{2}} \nonumber \\&\times MC_{r_{1,}r_{1}^{*}}e^{A_{2}r_{2}}\sum \limits _{k=2}^{\infty }\left( k-1\right) k\left( k+1\right) \left( k+2\right) \left( 2A_{1}r_{1}\right) ^{k}. \end{aligned}$$
(3.11)

Using (3.7)–(3.11), we get

$$\begin{aligned} \left| z_{2}T_{m}\left( f\right) \left( z_{1},z_{2}\right) \circ z_{1}T_{n}\left( f\right) \left( z_{1},z_{2}\right) \right|\le & {} \left| E_{1}\right| +\left| E_{2}\right| +\left| E_{3}\right| +\left| E_{4}\right| \\\le & {} C_{r_{1},r_{2}}^{r_{1}^{*}}\left( f\right) \left( a_{n}+\frac{1}{ b_{n}}\right) ^{2}. \end{aligned}$$

If we estimate \(\left| z_{1}T_{n}\left( f\right) \left( z_{1},z_{2}\right) \circ z_{2}T_{m}\left( f\right) \left( z_{1},z_{2}\right) \right| ,\) then by reason of symmetry we get a similar order of approximation, simply interchanging above the places of n with m and \( r_{1}\) with \(r_{2}.\)

In conclusion, using the commutativity property given in (3.5), we reach the result. \(\square \)

Let us denote with \(A_{C}^{\left( 2\right) }\left( f\right) \) the space of the all complex valued functions that they and their first and second partial derivatives are uniformly continuous on \(\left( D_{R_{1}}\cup \left[ R_{1},\infty \right) \right) \times \left( D_{R_{2}}\cup \left[ R_{2},\infty \right) \right) ,\) bounded on \(\left[ 0,\infty \right) \times \left[ 0,\infty \right) \) and analytic in \(D_{R_{1}}\times D_{R_{2}}\) and there exist \(M>0,\) \(0<A_{1}<1/2r_{1},\) \(0<A_{2}<1/2r_{2}\) with \( \left| c_{k,j}\right| \le MA_{1}^{k}A_{2}^{j}/k!j! (\) which implies \(\left| f\left( z_{1},z_{2}\right) \right| \le Me^{A_{1}\left| z_{1}\right| +A_{2}\left| z_{2}\right| }\) for all \(\left( z_{1},z_{2}\right) \in D_{R_{1}}\times D_{R_{2}}).\)

Theorem 2 and Theorem 3 will be used to find the exact degree in approximation of \(R_{n,n}\left( f\right) .\) In this sense we have the following lower estimate.

Theorem 4

Let \(n_{0}\ge 2,\) \(0<\beta <1/2\), \(1/2<r_{1}<R_{1}\le n_{0}^{1-\beta }/2\), and \(1/2<r_{2}<R_{2}\le n_{0}^{1-\beta }/2\). If \(f\in A_{C}^{\left( 2\right) }\left( f\right) \ \) and f is not a solution of the complex partial differential equation

$$\begin{aligned} K\left( f\right) \big ( z_{1},z_{2}\big ) =z_{1}\frac{\partial ^{2}f}{ \partial z_{1}^{2}}\big ( z_{1},z_{2}\big ) +z_{2}\frac{\partial ^{2}f}{ \partial z_{2}^{2}}\big ( z_{1},z_{2}\big ) =0 , \end{aligned}$$
(3.12)

then for all \(n\ge n_{0}\) we have

$$\begin{aligned} \left\| R_{n,n}\left( f\right) -f\right\| _{r_{1},r_{2}}\ge \frac{1}{ 36\left( 1+a_{n}b_{n}\right) }\left( a_{n}+\frac{1}{b_{n}}\right) \left\| K\left( f\right) \right\| _{r_{1},r_{2}}. \end{aligned}$$

Proof

We can write

$$\begin{aligned}&R_{n,n}\left( f\right) \big ( z_{1},z_{2}\big ) -f\big ( z_{1},z_{2}\big )\nonumber \\&\quad =2\left( a_{n}+\frac{1}{b_{n}}\right) \left\{ K_{n}\left( f\right) \left( z_{1},z_{2}\right) +2\left( a_{n}+\frac{1}{b_{n}}\right) \left[ \frac{ D_{n}\left( f\right) \left( z_{1},z_{2}\right) }{4\left( a_{n}+\frac{1}{b_{n} }\right) ^{2}}\right] \right. \nonumber \\&\qquad \left. +E_{n}\left( f\right) \left( z_{1},z_{2}\right) +F_{n}\left( f\right) \left( z_{1},z_{2}\right) \right\} , \end{aligned}$$
(3.13)

where

$$\begin{aligned} D_{n}\left( f\right) \left( z_{1},z_{2}\right)= & {} z_{2}T_{n}\left( f\right) \left( z_{1},z_{2}\right) \circ z_{1}T_{n}\left( f\right) \left( z_{1},z_{2}\right) ,\\ E_{n}\left( f\right) \left( z_{1},z_{2}\right)= & {} \sum \limits _{h=1}^{4}E_{n}^{h}\left( f\right) \left( z_{1},z_{2}\right) \end{aligned}$$

with

$$\begin{aligned} E_{n}^{1}\left( f\right) \left( z_{1},z_{2}\right)= & {} \frac{ -a_{n}b_{n}z_{1}^{2}}{2\left( 1+a_{n}b_{n}\right) \left( 1+a_{n}z_{1}\right) }\left[ R_{n}\left( \frac{\partial f}{\partial z_{1}}\left( z_{1},. \right) \right) \left( z_{2}\right) -\frac{\partial f}{\partial z_{1}}\left( z_{1},z_{2}\right) \right] ,\\ E_{n}^{2}\left( f\right) \left( z_{1},z_{2}\right)= & {} \frac{ -a_{n}b_{n}z_{2}^{2}}{2\left( 1+a_{n}b_{n}\right) \left( 1+a_{n}z_{2}\right) }\left[ R_{n}\left( \frac{\partial f}{\partial z_{2}}\left( .,z_{2}\right) \right) \left( z_{2}\right) -\frac{\partial f}{\partial z_{2}}\left( z_{1},z_{2}\right) \right] ,\\ E_{n}^{3}\left( f\right) \left( z_{1},z_{2}\right)= & {} \frac{ a_{n}^{2}b_{n}z_{1}^{4}+z_{1}}{4\left( 1+a_{n}b_{n}\right) \left( 1+a_{n}z_{1}\right) ^{2}}\left[ R_{n}\left( \frac{\partial ^{2}f}{\partial z_{1}^{2}}\left( z_{1},.\right) \right) \left( z_{2}\right) -\frac{ \partial ^{2}f}{\partial z_{1}^{2}}\left( z_{1},z_{2}\right) \right] ,\\ E_{n}^{4}\left( f\right) \left( z_{1},z_{2}\right)= & {} \frac{ a_{n}^{2}b_{n}z_{2}^{4}+z_{2}}{4\left( 1+a_{n}b_{n}\right) \left( 1+a_{n}z_{2}\right) ^{2}}\left[ R_{n}\left( \frac{\partial ^{2}f}{\partial z_{2}^{2}}\left( .,z_{2}\right) \right) \left( z_{1}\right) -\frac{\partial ^{2}f}{\partial z_{2}^{2}}\left( z_{1},z_{2}\right) \right] ,\\ F_{n}\left( f\right) \left( z_{1},z_{2}\right)= & {} \frac{z_{1}T_{n}\left( f\right) \left( z_{1},z_{2}\right) +z_{2}T_{n}\left( f\right) \left( z_{1},z_{2}\right) }{2\left( a_{n}+\frac{1}{b_{n}}\right) } \\&-\frac{a_{n}b_{n}z_{1}^{2}}{2\left( 1+a_{n}z_{1}\right) } \frac{\partial f}{\partial z_{1}}\left( z_{1},z_{2}\right) -\frac{a_{n}b_{n}z_{2}^{2}}{2\left( 1+a_{n}z_{2}\right) } \frac{\partial f}{\partial z_{2}}\left( z_{1},z_{2}\right) \\&-\frac{a_{n}^{2}b_{n}z_{1}^{2}z_{2}^{2}}{2\left( 1+a_{n}b_{n}\right) \left( 1+a_{n}z_{1}\right) \left( 1+a_{n}z_{2}\right) } \frac{\partial ^{2}f}{\partial z_{1}\partial z_{2}}\left( z_{1},z_{2}\right) \\&+\frac{\left( a_{n}^{2}b_{n}z_{1}^{4}+z_{1}\right) \left( a_{n}z_{2}^{2}\right) }{4\left( 1+a_{n}b_{n}\right) \left( 1+a_{n}z_{1}\right) ^{2}\left( 1+a_{n}z_{2}\right) }\frac{\partial ^{3}f}{\partial z_{1}^{2}\partial z_{2}}\left( z_{1},z_{2}\right) \\&+\frac{\left( a_{n}^{2}b_{n}z_{2}^{4}+z_{2}\right) \left( a_{n}z_{1}^{2}\right) }{4\left( 1+a_{n}b_{n}\right) \left( 1+a_{n}z_{1}\right) \left( 1+a_{n}z_{2}\right) ^{2}}\frac{\partial ^{3}f}{ \partial z_{1}\partial z_{2}^{2}}\left( z_{1},z_{2}\right) \\&-\frac{\left( a_{n}^{2}b_{n}z_{1}^{4}+z_{1}\right) \left( a_{n}^{2}b_{n}z_{2}^{4}+z_{2}\right) }{8b_{n}\left( 1+a_{n}b_{n}\right) \left( 1+a_{n}z_{1}\right) ^{2}\left( 1+a_{n}z_{2}\right) ^{2}}\frac{ \partial ^{4}f}{\partial z_{1}^{2}\partial z_{2}^{2}}\left( z_{1},z_{2}\right) , \end{aligned}$$

and

$$\begin{aligned} K_{n}\left( f\right) \left( z_{1},z_{2}\right)= & {} \frac{1}{1+a_{n}b_{n}} \left\{ \frac{ a_{n}^{2}b_{n}z_{1}^{4}+z_{1}}{4\left( 1+a_{n}z_{1}\right) ^{2}}\frac{ \partial ^{2}f}{\partial z_{1}^{2}}\left( z_{1},z_{2}\right) \right. \left. \!+\frac{ a_{n}^{2}b_{n}z_{2}^{4}+z_{2}}{4\left( 1+a_{n}z_{2}\right) ^{2}}\frac{ \partial ^{2}f}{\partial z_{2}^{2}}\left( z_{1},z_{2}\right) \right\} . \end{aligned}$$

Under the conditions of theorem, since \(\lim \limits _{n\rightarrow \infty }a_{n}=0\), \(\lim \limits _{n\rightarrow \infty }1/b_{n}= 0\) and \(\lim \limits _{n\rightarrow \infty }a_{n}b_{n}= 0\) for \(0<\beta <1/2\), considering Theorem 1.10.5 and Theorem 1.10.6 in [6] (see p. 145), it is clear that

$$\begin{aligned} \lim _{n\rightarrow \infty }E_{n}\left( f\right) \left( z_{1},z_{2}\right) = 0\,\text {and}\, \lim _{n\rightarrow \infty }F_{n}\left( f\right) \left( z_{1},z_{2}\right) =0. \end{aligned}$$
(3.14)

From Theorem 3, we obtain

$$\begin{aligned} \lim _{n\rightarrow \infty }\left\| 2\left( a_{n}+\frac{1}{b_{n}}\right) \left[ \frac{D_{n}\left( f\right) }{4\left( a_{n}+\frac{1}{b_{n}}\right) ^{2}}\right] +E_{n}\left( f\right) +F_{n}\left( f\right) \right\| _{r_{1},r_{2}}= 0. \end{aligned}$$

Using \(\lim \limits _{n\rightarrow \infty }a_{n}b_{n}= 0 \) for \(0<\beta <1/2\) and \(1 /1+a_{n}\left| z_{1}\right| \ge 2/3,\) we get

$$\begin{aligned} \left\| K_{n}\left( f\right) \right\| _{r_{1},r_{2}}\ge \frac{1}{ 18\left( 1+a_{n}b_{n}\right) }\left| z_{1}\right| \left| \frac{ \partial ^{2}f}{\partial z_{1}^{2}}\left( z_{1},z_{2}\right) \right| . \end{aligned}$$
(3.15)

Similarly, it follows

$$\begin{aligned} \left\| K_{n}\left( f\right) \right\| _{r_{1},r_{2}}\ge \frac{1}{ 18\left( 1+a_{n}b_{n}\right) }\left| z_{2}\right| \left| \frac{ \partial ^{2}f}{\partial z_{2}^{2}}\left( z_{1},z_{2}\right) \right| . \end{aligned}$$
(3.16)

From (3.15) and (3.16), we can write

$$\begin{aligned} \left\| K_{n}\left( f\right) \right\| _{r_{1},r_{2}}\ge \frac{1}{ 36\left( 1+a_{n}b_{n}\right) }\left\| K\left( f\right) \right\| _{r_{1},r_{2}}. \end{aligned}$$
(3.17)

In (3.13), taking into account the inequalities

$$\begin{aligned} \left\| H+G\right\| _{r_{1},r_{2}}\ge \left| \left\| H\right\| _{r_{1},r_{2}}-\left\| G\right\| _{r_{1},r_{2}}\right| \ge \left\| H\right\| _{r_{1},r_{2}}-\left\| G\right\| _{r_{1},r_{2}}, \end{aligned}$$

and (3.17), it follows

$$\begin{aligned}&\left\| R_{n,n}\left( f\right) -f\right\| _{r_{1},r_{2}} \\&\quad \ge 2\left( a_{n}+\frac{1}{b_{n}}\right) \left\{ \left\| K_{n}\left( f\right) \right\| _{r_{1},r_{2}}\right. \\&\qquad \left. -\left\| 2\left( a_{n}+\frac{1}{b_{n}}\right) \left[ \frac{ D_{n}\left( f\right) }{4\left( a_{n}+\frac{1}{b_{n}}\right) ^{2}}\right] +E_{n}\left( f\right) +F_{n}\left( f\right) \right\| _{_{r_{1},r_{2}}}\right\} \\&\quad \ge \left( a_{n}+\frac{1}{b_{n}}\right) \left\| K_{n}\left( f\right) \right\| _{r_{1},r_{2}} \\&\quad \ge \left( a_{n}+\frac{1}{b_{n}}\right) \frac{1}{36\left( 1+a_{n}b_{n}\right) }\left\| K\left( f\right) \right\| _{r_{1},r_{2}} \end{aligned}$$

for all \(n\ge n_{0}\) with \(n_{0}\) depending only \(f,r_{1}\) and \(r_{2.}\) We used here that by hypothesis we have \(\left\| K\left( f\right) \right\| _{r_{1},r_{2}}>0.\) \(\square \)

Combining Theorem 3 with Theorem 4, we immediately obtain the following result giving the exact degree of the operators (1.1).

Corollary 1

Suppose that the hypothesis in the statement of Theorem 6 holds. If Taylor series of f contains at least one term of the form \(c_{k+1,0}z_{1}^{k}\) with \(c_{k+1,0}\ne 0\) and \(k=1,2,\ldots \) or of the form \(c_{0,j+1}z_{2}^{j}\) with \(c_{0,j+1}\ne 0\) and \(j=1,2,\ldots \), then for all \(n\ge n_{0}\) we have

$$\begin{aligned} \left\| R_{n,n}\left( f\right) -f\right\| _{r_{1},r_{2}}\sim \left( a_{n}+\frac{1}{b_{n}}\right) . \end{aligned}$$

Proof

It suffices to prove that under the hypothesis on f, it cannot be a solution of the complex partial differential equation

$$\begin{aligned} z_{1}\frac{\partial ^{2}f}{\partial z_{1}^{2}}\left( z_{1},z_{2}\right) +z_{2}\frac{\partial ^{2}f}{\partial z_{2}^{2}}\left( z_{1},z_{2}\right) =0, \,\left| z_{1}\right| <R_{1},\left| z_{2}\right| <R_{2.} \end{aligned}$$

Indeed, suppose the contrary. Since simple calculation gives

$$\begin{aligned} z_{1}\frac{\partial ^{2}f}{\partial z_{1}^{2}}\left( z_{1},z_{2}\right) +z_{2}\frac{\partial ^{2}f}{\partial z_{2}^{2}}\left( z_{1},z_{2}\right)= & {} \sum \limits _{k=1}^{\infty }c_{k+1,0}k\left( k+1\right) z_{1}^{k}+\sum \limits _{k=1}^{\infty }c_{k+1,1}k\left( k+1\right) z_{1}^{k}z_{2} \\&+2\sum \limits _{j=2}^{\infty }c_{2,j}z_{1}z_{2}^{j}+\sum \limits _{k=2}^{\infty }\sum \limits _{j=2}^{\infty }c_{k+1,j}k\left( k+1\right) z_{1}^{k}z_{2}^{j},\\&+\,\sum \limits _{j=1}^{\infty }c_{0,j+1}j\left( j+1\right) z_{2}^{j}+\sum \limits _{j=1}^{\infty }c_{1,j+1}j\left( j+1\right) z_{1}z_{2}^{j} \\&+2\sum \limits _{k=2}^{\infty }c_{k,2}z_{1}^{k}z_{2}+\sum \limits _{k=2}^{\infty }\sum \limits _{j=2}^{\infty }c_{k,j+1}j\left( j+1\right) z_{1}^{k}z_{2}^{j}, \end{aligned}$$

by equaling with zero and by the identification of coefficients, from the terms under the first and fifth sign \(\sum ,\) we immediately get that \( c_{k+1,0}=c_{0,j+1}=0,\) for all \(k=1,2,\ldots \) and \(j=1,2,\ldots ,\) which contradicts the hypothesis on f. Therefore the hypothesis and the lower estimate in Theorem 4 satisfy, which complete the proof. \(\square \)