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Approximation properties of bivariate complex q-Balàzs-Szabados operators of tensor product kind

Abstract

In this study, we consider the bivariate complex q-Balàzs-Szabados operators of the tensor product kind. Approximation properties of these operators attached to analytic functions on compact polydisks are investigated by using the results in the univariate case obtained for q-Balàzs-Szabados operators in (İspir and Yıldız Özkan in J. Inequal. Appl. 2013:361, 2013). In this sense, the upper estimate, the Voronovskaja-type theorem, and the lower estimate are obtained. The exact degree of its approximation is also given.

MSC:30E10, 41A25.

1 Introduction

The approximation properties of the q-analogue operators in compact disks have recently been an active area of the research in the field of the approximation theory [18]. Details of the q-calculus can be found in [911].

Balázs [12] defined the Bernstein-type rational functions. She 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 [13], Balázs and Szabados obtained the best possible estimate under more restrictive conditions, in which both the weight and the order of convergence would be better than [12]. They applied their results to the approximation of certain improper integrals by quadrature sums of positive coefficients based on a finite number of equidistant nodes. The q-form of these operators was given by Doğru. He investigated Korovkin-type statistical approximation properties of these operators for the functions of one and two variables [14]. Atakut and Ispir [15] defined the bivariate real Bernstein-type rational functions of the Bernstein-type rational functions given by Balázs in [12] and proved the approximation theorems for these functions. Ispir and Gupta [16] studied the Bézier variant of generalized Kantrovich-type Balazs operators.

Approximation properties of the rational Balázs-Szabados operators on compact disks in the complex plane were investigated by Gal [17]. He proved the upper estimate in an approximation of these operators. Also, he obtained the exact degree of its approximation by using a Voronovskaja-type result. In [18], the approximation properties given by Gal in the complex plane was extended to the bivariate case.

The complex q-Balázs-Szabados operators was defined in [19] as follows:

R n q (f;z)= 1 s = 0 n 1 ( 1 + q s a n z ) j = 0 n q j ( j 1 ) / 2 f ( [ j ] q b n ) [ n j ] q ( a n z ) j ,

where f: D R [R,)C is uniformly continuous and bounded on [0,) with D R ={zC:|z|<R} for R>0, a n = [ n ] q β 1 , b n = [ n ] q β , q(0,1], 0<β 2 3 , nN, zC, and z 1 q s a n for s=0,1,2, .

We consider the following complex bivariate q-Balázs-Szabados operators of the tensor product kind:

R n , m q 1 , q 2 (f)( z 1 , z 2 )= k = 0 n j = 0 m f ( [ k ] q 1 b n , [ j ] q 2 b m ) p n , k ( z 1 ) p m , j ( z 2 ),
(1)

where f:( D R 1 [ R 1 ,))×( D R 2 [ R 2 ,))C is a uniformly continuous function bounded on [0,)×[0,), a n = [ n ] q 1 β 1 , b n = [ n ] q 1 β , a m = [ m ] q 2 β 1 , b m = [ m ] q 2 β for n,mN, q 1 , q 2 (0,1], 0<β 2 3 .

p n , k ( z 1 )= q 1 k ( k 1 ) / 2 [ n k ] q 1 ( a n z 1 ) k s 1 = 0 n 1 ( 1 + q 1 s 1 a n z 1 ) and p m , j ( z 2 )= q 2 j ( j 1 ) / 2 [ m j ] q 2 ( a m z 2 ) j s 2 = 0 m 1 ( 1 + q 2 s 2 a m z 2 )

for all s 1 =0,1,,n1, s 2 =0,1,,m1 and z 1 , z 2 C with z 1 1 q 1 s 1 a n and z 2 1 q 2 s 2 a m .

The complex bivariate q-Balázs-Szabados operators of the tensor product kind are well defined and linear, and these operators are analytic for all n n 0 , m m 0 , | z 1 | r 1 < [ n 0 ] q 1 1 β and | z 2 | r 2 < [ m 0 ] q 2 1 β .

The aim of this paper is to obtain the exact degree of approximation of the complex bivariate q-Balázs-Szabados operators of the tensor product kind. The Voronovskaja-type theorem in the bivariate case is very different from the univariate case, so the exact degree of approximation of these operators can be obtained for 0<β< 1 2 .

Throughout this paper, we denote by f r 1 , r 2 =max{|f( z 1 , z 2 )|:( z 1 , z 2 ) D ¯ r 1 × D ¯ r 1 } the uniform norm of the function f in the space of continuous functions on D ¯ r 1 × D ¯ r 1 and by f B ( [ 0 , ) × [ 0 , ) ) =sup{|f( z 1 , z 2 )|:( z 1 , z 2 )[0,)×[0,)} the norm of the function f in the space of bounded functions on [0,)×[0,), where D r ={zC:|z|<r} for r>0.

The convergence results will be obtained under the condition that f:( D R 1 [ R 1 ,))×( D R 2 [ R 2 ,))C is analytic in D R 1 × D R 2 for r 1 < R 1 and r 2 < R 2 , which ensures the representation f( z 1 , z 2 )= k = 0 f k ( z 2 ) z 1 k , where f k ( z 2 )= j = 0 c k , j z 2 j for all ( z 1 , z 2 ) D R 1 × D R 2 .

2 Auxiliary results

Let q=( q n ) be a sequence satisfying

lim n q n =1and lim n q n n =c(0c<1).
(2)

We need the following lemmas in order to prove the main results for the operators (1).

Lemma 1 Let n 0 , m 0 2, 0<β 2 3 , 1 2 < r 1 < R 1 [ n 0 ] q 1 1 β 2 and 1 2 < r 2 < R 2 [ m 0 ] q 1 1 β 2 . If f:( D R 1 [ R 1 ,))×( D R 2 [ R 2 ,))C is a uniformly continuous function bounded on [0,)×[0,) and analytic in D R 1 × D R 2 then we have the form

R n , m q 1 , q 2 (f)( z 1 , z 2 )= k = 0 j = 0 c k , j R n , m q 1 , q 2 ( e k , j )( z 1 , z 2 )

for all ( z 1 , z 2 ) D r 1 × D r 2 , where ( e k , j )( z 1 , z 2 )= e 1 k ( z 1 ) e 2 j ( z 2 ) with e 1 k ( z 1 )= z 1 k , e 2 j ( z 2 )= z 2 j for k,jN.

Proof For any s,rN, we define

f s , r ( z 1 , z 2 ) = k = 0 s j = 0 r c k , j e k , j ( z 1 , z 2 ) if  | z 1 | r 1 , | z 2 | r 2 and  f s , r ( z 1 , z 2 ) = f ( z 1 , z 2 ) if  ( z 1 , z 2 ) ( r 1 , ) × ( r 2 , ) .

From the hypothesis on f, it is clear that each f s , r is bounded on [0,)×[0,), which implies that

| R n , m q 1 , q 2 ( f s , r )( z 1 , z 2 )| k = 0 n j = 0 m | p n , k ( z 1 )|| p m , j ( z 2 )| M f s , r <,

where M f s , r is a constant depending on f s , r , so all R n , m q 1 , q 2 ( f s , r ) are well defined for all n,mN, n n 0 , m m 0 , r 1 < [ n 0 ] q 1 β 2 , r 2 < [ m 0 ] q 1 β 2 and ( z 1 , z 2 ) D r 1 × D r 2 .

Defining

f s , r , k , j ( z 1 , z 2 ) = c k , j e k , j ( z 1 , z 2 ) if  | z 1 | r 1 , | z 2 | r 2 and f s , r , k , j ( z 1 , z 2 ) = f ( z 1 , z 2 ) ( s + 1 ) ( r + 1 ) if  ( z 1 , z 2 ) ( r 1 , ) × ( r 2 , ) .

It is clear that each f s , r , k , j is bounded on [0,)×[0,) and

f s , r ( z 1 , z 2 )= k = 0 s j = 0 r f s , r , k , j ( z 1 , z 2 ).

From the linearity of R n , m q 1 , q 2 , we have

R n , m q 1 , q 2 ( f s , r )( z 1 , z 2 )= k = 0 s j = 0 r c k , j R n , m q 1 , q 2 ( e k , j )( z 1 , z 2 ).

It suffices to prove that

lim s , r R n , m q 1 , q 2 ( f s , r )( z 1 , z 2 )= R n , m q 1 , q 2 (f)( z 1 , z 2 )

for any fixed n,mN, n n 0 , m m 0 , | z 1 | r 1 and | z 2 | r 2 . Since

f s , r f B ( [ 0 , ) × [ 0 , ) ) f s , r f r 1 , r 2 ,

we can write

| R n , m q 1 , q 2 ( f s , r ) ( z 1 , z 2 ) R n , m q 1 , q 2 ( f ) ( z 1 , z 2 ) | M r 1 , r 2 , m , n q 1 , q 2 f s , r f B ( [ 0 , ) × [ 0 , ) ) M r 1 , r 2 , m , n q 1 , q 2 f s , r f r 1 , r 2
(3)

for | z 1 | r 1 and | z 2 | r 2 .

In equation (3), taking the limit as s,r and using lim s , r f s , r f r 1 , r 2 =0, we get the result. □

Lemma 2 Let n 0 , m 0 2, 0<β 2 3 , 1 2 < r 1 < R 1 [ n 0 ] q 1 1 β 2 and 1 2 < r 2 < R 2 [ m 0 ] q 2 1 β 2 . For all n n 0 , m m 0 , | z 1 | r 1 , | z 2 | r 2 and k=0,1,2, the following inequality holds:

| R n , m q 1 , q 2 ( e k , j )( z 1 , z 2 )|k!j! ( 20 r 1 ) k ( 20 r 2 ) j .

Proof Using Lemma 4 in [19], the lemma is easily proved, so we omit the proof of the lemma. □

3 Main results

Let us denote by A C the space of all uniformly continuous complex valued functions defined on ( D R 1 [ R 1 ,))×( D R 2 [ R 2 ,)), bounded on [0,)×[0,) and analytic in D R 1 × D R 2 and for which there exist M>0, 0< A 1 < 1 20 r 1 and 0< A 2 < 1 20 r 2 with | c k , j |M A 1 k A 2 j k ! j ! for all k,j=0,1,2, (which implies |f( z 1 , z 2 )|M e A 1 | z 1 | + A 2 | z 2 | for all ( z 1 , z 2 ) D R 1 × D R 2 ).

We have the following upper estimate.

Theorem 1 Let q 1 =( q 1 , n ) and q 2 =( q 2 , m ) be sequences satisfying the conditions given in equation (2) and let n 0 , m 0 2, 0<β 2 3 , 1 2 < r 1 < R 1 [ n 0 ] q 1 1 β 2 and 1 2 < r 2 < R 2 [ m 0 ] q 2 1 β 2 . If f A C , then for all n n 0 , m m 0 , | z 1 | r 1 and | z 2 | r 2 the following inequality holds:

| R n , m q 1 , q 2 (f)( z 1 , z 2 )f( z 1 , z 2 )| ( a n + 1 b n ) C 3 (f)+ ( a m + 1 b m ) C 4 (f),

where

C 3 ( f ) = max { M r 1 r 2 e 2 r 1 A 1 + r 2 A 2 , 9 M e r 2 A 2 k = 1 ( k 1 ) ( 20 r 1 A 1 ) k 1 } , C 4 ( f ) = max { 2 M ( r 2 ) 2 e 2 r 2 A 2 k = 0 ( 20 r 1 A 1 ) k , 9 M k = 0 ( 20 r 1 A 1 ) k j = 1 ( j 1 ) ( 20 r 2 A 2 ) j 1 } ,

and also the series k = 0 ( 20 r 1 A 1 ) k , k = 1 (k1) ( 20 r 1 A 1 ) k 1 and j = 1 (j1) ( 20 r 2 A 2 ) j 1 are convergent.

Proof Using Lemma 1, we can write

| R n , m q 1 , q 2 (f)( z 1 , z 2 )f( z 1 , z 2 )| k = 0 j = 0 | c k , j || R n , m q 1 , q 2 ( e k , j )( z 1 , z 2 ) e k , j ( z 1 , z 2 )|.
(4)

Taking into account Lemma 4 in [19] and the estimate given in the proof of Theorem 2 in [19], for all | z 1 | r 1 and | z 2 | r 2 , we obtain

| R n , m q 1 , q 2 ( e k , j ) ( z 1 , z 2 ) e k , j ( z 1 , z 2 ) | = | R n q 1 ( e 1 k ) ( z 1 ) . R m q 2 ( e 2 j ) ( z 2 ) z 1 k z 2 j | | R n q 1 ( e 1 k ) ( z 1 ) | | R m q 2 ( e 2 j ) ( z 2 ) z 2 j | + | z 2 j | | R n q 1 ( e 1 k ) ( z 1 ) z 1 k | ( k ! ) ( 20 r 1 ) k { 2 a m ( r 2 ) 2 j ( 2 r 2 ) j 1 + 9 b m ( j 1 ) ( j ! ) ( 20 r 2 ) j 1 } + ( r 2 ) j { 2 a n ( r 1 ) 2 k ( 2 r 1 ) k 1 + 9 b n ( k 1 ) ( k ! ) ( 20 r 1 ) k 1 } = 2 a n ( r 1 ) 2 ( 2 r 1 ) k 1 j ( r 2 ) j + 2 a m ( r 2 ) 2 ( k ! ) ( 20 r 1 ) k ( 2 r 2 ) j 1 + 9 b n ( k 1 ) ( k ! ) ( 20 r 1 ) k 1 ( r 2 ) j + 9 b m ( k ! ) ( 20 r 1 ) k ( j 1 ) ( j ! ) ( 20 r 2 ) j 1 .
(5)

Applying equation (5) in equation (4), we get

| R n , m q 1 , q 2 ( f ) ( z 1 , z 2 ) f ( z 1 , z 2 ) | a n M r 1 r 2 e 2 r 1 A 1 + r 2 A 2 + 2 a m M ( r 2 ) 2 e 2 r 2 A 2 k = 0 ( 20 r 1 A 1 ) k + 9 M b n e r 2 A 2 k = 1 ( k 1 ) ( 20 r 1 A 1 ) k 1 + 9 M b m k = 0 ( 20 r 1 A 1 ) k j = 1 ( j 1 ) ( 20 r 2 A 2 ) j 1 .

Choosing C 3 (f) and C 4 (f) as given in the theorem, we reach the desired result. □

For f( z 1 , z 2 ), we define the parametric extensions of the Voronovskaja formula by

z 1 L n , q 1 ( f ) ( z 1 , z 2 ) : = R n q 1 ( f ( , z 2 ) ) ( z 1 ) f ( z 1 , z 2 ) ψ n , q 1 1 ( z 1 ) f z 1 ( z 1 , z 2 ) 1 2 ψ n , q 1 2 ( z 1 ) 2 f z 1 2 ( z 1 , z 2 )

and

z 2 L m , q 2 ( f ) ( z 1 , z 2 ) : = R m q 2 ( f ( z 1 , ) ) ( z 2 ) f ( z 1 , z 2 ) ψ m , q 2 1 ( z 2 ) f z 2 ( z 1 , z 2 ) 1 2 ψ m , q 2 2 ( z 2 ) 2 f z 2 2 ( z 1 , z 2 ) ,

where ψ k , q i (z)= R k q ( ( t z ) i ;z) for i=1,2 given in Lemma 6 in [19].

Their product (composition) gives

z 2 L m , q 2 ( f ) ( z 1 , z 2 ) z 1 L n , q 1 ( f ) ( z 1 , z 2 ) = R m q 2 ( R n q 1 ( f ( , ) ) ( z 1 ) f ( z 1 , ) ψ n , q 1 1 ( z 1 ) f z 1 ( z 1 , ) ψ n , q 1 2 ( z 1 ) 2 f z 1 2 ( z 1 , ) ) ( z 2 ) [ R n q 1 ( f ( , z 2 ) ) ( z 1 ) f ( z 1 , z 2 ) ψ n , q 1 1 ( z 1 ) f z 1 ( z 1 , z 2 ) ψ n , q 1 2 ( z 1 ) 2 f z 1 2 ( z 1 , z 2 ) ] ψ m , q 2 1 ( z 2 ) [ R n q 1 ( f z 2 ( , z 2 ) ) ( z 1 ) f z 2 ( z 1 , z 2 ) ψ n , q 1 1 ( z 1 ) 2 f z 2 z 1 ( z 1 , z 2 ) ψ n , q 1 2 ( z 1 ) 3 f z 2 z 1 2 ( z 1 , z 2 ) ] ψ m , q 2 2 ( z 2 ) [ R n q 1 ( 2 f z 2 2 ( , z 2 ) ) ( z 1 ) 2 f z 2 2 ( z 1 , z 2 ) ψ n , q 1 1 ( z 1 ) 3 f z 2 2 z 1 ( z 1 , z 2 ) ψ n , q 1 2 ( z 1 ) 4 f z 2 2 z 1 2 ( z 1 , z 2 ) ] : = E 1 E 2 + E 3 E 4 .
(6)

After a simple calculation, we obtain the commutativity property,

z 2 L m , q 2 (f)( z 1 , z 2 ) z 1 L n , q 1 (f)( z 1 , z 2 )= z 1 L n , q 1 (f)( z 1 , z 2 ) z 2 L m , q 2 (f)( z 1 , z 2 ).

In the following a Voronovskaja-type result for the operators (1) is presented. It will be the product of the parametric extensions generated by Voronovskaja’s formula in the univariate case.

Theorem 2 Let q 1 =( q 1 , n ) and q 2 =( q 2 , m ) be sequences satisfying the conditions given in equation (2) and let n 0 , m 0 2, 0<β 2 3 , 1 2 < r 1 < R 1 [ n 0 ] q 1 1 β 2 and 1 2 < r 2 < R 2 [ m 0 ] q 2 1 β 2 . If f A C , then for all n n 0 , m m 0 , | z 1 | r 1 and | z 2 | r 2 the following inequality holds:

| z 2 L m , q 2 (f)( z 1 , z 2 ) z 1 L n , q 1 (f)( z 1 , z 2 )| C 5 (f) [ ( a n + 1 b n ) 2 + ( a m + 1 b m ) 2 ] ,

where C 5 (f)= 1 2 max{ C r 1 , r 2 1 (f), C r 1 , r 2 2 (f)}, C , and C are fixed constants,

C r 1 , r 2 1 ( f ) = M r 1 3 e r 2 A 2 k = 2 ( k 2 ) ( k 1 ) k ( k + 1 ) ( 20 r 1 A 1 ) k 3 × max { C e r 2 A 2 j = 0 ( 20 r 2 A 2 ) j , C , C r 2 A 2 , C ( 1 + r 2 + r 2 2 ) r 2 A 2 2 }

and

C r 1 , r 2 2 ( f ) = M r 2 3 e r 1 A 1 j = 2 ( j 2 ) ( j 1 ) j ( j + 1 ) ( 20 r 2 A 2 ) j 3 × max { C e r 1 A 1 k = 0 ( 20 r 1 A 1 ) k , C , C r 1 A 1 , C ( 1 + r 1 + r 1 2 ) r 1 A 1 2 } .

Proof From the analyticity of f in D R 1 × D R 2 , since all partial derivatives of f are analytic in D R 1 × D R 2 , using Lemma 1, we can write

R n q 1 ( f ( , z 2 ) ) ( z 1 ) f ( z 1 , z 2 ) ψ n , q 1 1 ( z 1 ) f z 1 ( z 1 , z 2 ) ψ n , q 1 2 ( z 1 ) 2 f z 1 2 ( z 1 , z 2 ) = k = 2 f k ( z 2 ) [ R n q 1 ( e 1 k ) ( z 1 ) e 1 k ( z 1 ) ψ n , q 1 1 ( z 1 ) k z 1 k 1 ψ n , q 1 2 ( z 1 ) k ( k 1 ) z 1 k 2 ] .
(7)

Applying now R m q 2 to equation (7) with respect to z 2 and Lemma 1 in [19], we obtain

E 1 = k = 2 R m q 2 ( f k ) ( z 2 ) [ R n q 1 ( e 1 k ) ( z 1 ) e 1 k ( z 1 ) ψ n , q 1 1 ( z 1 ) k z 1 k 1 ψ n , q 1 2 ( z 1 ) k ( k 1 ) z 1 k 2 ] = k = 2 j = 0 c k , j R m q 2 ( e 2 j ) ( z 2 ) [ R n q 1 ( e 1 k ) ( z 1 ) e 1 k ( z 1 ) ψ n , q 1 1 ( z 1 ) k z 1 k 1 ψ n , q 1 2 ( z 1 ) k ( k 1 ) z 1 k 2 ] .
(8)

In equation (8), passing now to absolute value for | z 1 | r 1 and | z 2 | r 2 and taking into account the Lemma 4 in [19] and the estimate given in the proof of Theorem 3 in [19], it follows that

| E 1 | ( a n + 1 b n ) 2 j = 0 | c k , j | j ! ( 20 r 2 ) j k = 2 C ( k 2 ) ( k 1 ) k ( k + 1 ) k ! ( 20 r 1 ) k + 3 ( a n + 1 b n ) 2 M C r 1 3 j = 0 ( 20 r 2 A 2 ) j k = 2 ( k 2 ) ( k 1 ) k ( k + 1 ) ( 20 r 1 A 1 ) k 3
(9)

for | z 1 | r 1 and | z 2 | r 2 .

Similarly, using the estimate given in the proof of Theorem 3 in [19] for | z 1 | r 1 and | z 2 | r 2 we have

| E 2 | k = 2 | f k ( z 2 ) | | R n q 1 ( e 1 k ) ( z 1 ) e 1 k ( z 1 ) ψ n , q 1 1 ( z 1 ) k z 1 k 1 ψ n , q 1 2 ( z 1 ) k ( k 1 ) z 1 k 2 | ( a n + 1 b n ) 2 j = 0 | c k , j | r 2 j k = 2 C ( k 2 ) ( k 1 ) k ( k + 1 ) k ! ( 20 r 1 ) k + 3 ( a n + 1 b n ) 2 M C r 1 3 e r 2 A 2 k = 2 ( k 2 ) ( k 1 ) k ( k + 1 ) ( 20 r 1 A 1 ) k 3 .
(10)

Using

R n q 1 ( f z 2 ( , z 2 ) ) ( z 1 ) = k = 0 f k z 2 ( z 2 ) R n q 1 ( e 1 k ) ( z 1 ) = k = 0 j = 1 c k , j j z 2 j 1 R n q 1 ( e 1 k ) ( z 1 ) ,

we can write

E 3 = ψ m , q 2 1 ( z 2 ) [ R n q 1 ( f z 2 ( , z 2 ) ) ( z 1 ) f z 2 ( z 1 , z 2 ) ψ n , q 1 1 ( z 1 ) 2 f z 2 z 1 ( z 1 , z 2 ) ψ n , q 1 2 ( z 1 ) 3 f z 2 z 1 2 ( z 1 , z 2 ) ] = ψ m , q 2 1 ( z 2 ) k = 2 j = 1 c k , j j z 2 j 1 [ R n q 1 ( e 1 k ) ( z 1 ) e 1 k ( z 1 ) ψ n , q 1 1 ( z 1 ) k z 1 k 1 ψ n , q 1 2 ( z 1 ) k ( k 1 ) z 1 k 2 ] .

Considering Lemma 6 in [19] and the estimate given in the proof of Theorem 3 in [19], for | z 1 | r 1 and | z 2 | r 2 , we obtain

| E 3 | ( a n + 1 b n ) 2 | ψ m , q 2 1 ( z 2 ) | k = 2 j = 1 | c k , j | j r 2 j 1 C ( k 2 ) ( k 1 ) k ( k + 1 ) k ! ( 20 r 1 ) k + 3 ( a n + 1 b n ) 2 M C r 1 3 r 2 A 2 e r 2 A 2 k = 2 ( k 2 ) ( k 1 ) k ( k + 1 ) ( 20 r 1 A 1 ) k 3
(11)

and also, using

R n ( 2 f z 2 2 ( , z 2 ) ) ( z 1 ) = k = 0 2 f k z 2 2 ( z 2 ) R n ( e 1 k ) ( z 1 ) = k = 0 j = 2 c k , j j ( j 1 ) z 2 j 2 R n ( e 1 k ) ( z 1 ) ,

we can write

E 4 = ψ m , q 2 2 ( z 2 ) [ R n ( 2 f z 2 2 ( , z 2 ) ) ( z 1 ) 2 f z 2 2 ( z 1 , z 2 ) ψ n , q 1 1 ( z 1 ) 3 f z 1 z 2 2 ( z 1 , z 2 ) ψ n , q 1 2 ( z 1 ) 4 f z 1 2 z 2 2 ( z 1 , z 2 ) ] = ψ m , q 2 2 ( z 2 ) k = 2 j = 2 c k , j j ( j 1 ) z 2 j 2 [ R n q 1 ( e 1 k ) ( z 1 ) e 1 k ( z 1 ) ψ n , q 1 1 ( z 1 ) k z 1 k 1 ψ n , q 1 2 ( z 1 ) k ( k 1 ) z 1 k 2 ] .

Taking into account Lemma 6 in [19] and the estimate given in the proof of Theorem 3 in [19], for | z 1 | r 1 and | z 2 | r 2 we get

| E 4 | ( a n + 1 b n ) 2 | ψ m , q 2 2 ( z 2 ) | j = 2 j ( j 1 ) r 2 j 2 × k = 2 | c k , j | C ( k 2 ) ( k 1 ) k ( k + 1 ) k ! ( 20 r 1 ) k + 3 ( a n + 1 b n ) 2 M C r 1 3 ( 1 + r 2 + r 2 2 ) r 2 A 2 2 e A 2 r 2 × k = 2 ( k 2 ) ( k 1 ) k ( k + 1 ) ( 20 A 1 r 1 ) k 3
(12)

for | z 1 | r 1 and | z 2 | r 2 . Using equations (9)-(12), we get

| z 2 L m , q 2 ( f ) ( z 1 , z 2 ) z 1 L n , q 1 ( f ) ( z 1 , z 2 ) | | E 1 | + | E 2 | + | E 3 | + | E 4 | C r 1 , r 2 1 ( f ) ( a n + 1 b n ) 2 .

If we estimate | z 1 L n , q 1 (f)( z 1 , z 2 ) z 2 L m , q 2 (f)( z 1 , z 2 )|, then by reason of the 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, we reach the result. □

Let us denote by A C ( 2 ) the space of all complex valued functions where they and their first and second partial derivatives are uniformly continuous on ( D R 1 [ R 1 ,))×( D R 2 [ R 2 ,)), bounded on [0,)×[0,) and analytic in D R 1 × D R 2 , and there exist M>0, 0< A 1 < 1 20 r 1 , 0< A 2 < 1 20 r 2 with | c k , j |M A 1 k A 2 j k ! j ! (which implies |f( z 1 , z 2 )|M e A 1 | z 1 | + A 2 | z 2 | for all ( z 1 , z 2 ) D R 1 × D R 2 ).

Theorems 1 and 2 will be used to find the exact degree in the approximation of R n , n q 1 , q 2 (f). In this sense, we have the following lower estimate.

Theorem 3 Let q 1 =( q 1 , n ) and q 2 =( q 2 , n ) be sequences satisfying the conditions given in equation (2) and let n 0 2, 0<β< 1 2 , 1 2 < r 1 < R 1 [ n 0 ] q 1 1 β 2 and 1 2 < r 2 < R 2 [ n 0 ] q 2 1 β 2 . If f A C ( 2 ) and f is not a solution of the complex partial differential equation

K(f)( z 1 , z 2 )= z 1 2 f z 1 2 ( z 1 , z 2 )+ z 2 2 f z 2 2 ( z 1 , z 2 )=0,

then for all n n 0 we have

R n , n q 1 , q 2 ( f ) f r 1 , r 2 1 36 ( 1 + a n b n ) ( a n + 1 b n ) K ( f ) r 1 , r 2 .

Proof From equation (6), we can write

R n , n q 1 , q 2 ( f ) ( z 1 , z 2 ) f ( z 1 , z 2 ) = 2 ( a n + 1 b n ) { K n ( f ) ( z 1 , z 2 ) + 2 ( a n + 1 b n ) [ D n ( f ) ( z 1 , z 2 ) 4 ( a n + 1 b n ) 2 ] + E n ( f ) ( z 1 , z 2 ) + F n ( f ) ( z 1 , z 2 ) + G n ( f ) ( z 1 , z 2 ) } ,
(13)

where

D n ( f ) ( z 1 , z 2 ) = z 2 L n , q 2 ( f ) ( z 1 , z 2 ) z 1 L n , q 1 ( f ) ( z 1 , z 2 ) , E n ( f ) ( z 1 , z 2 ) = z 1 L n , q 1 ( f ) ( z 1 , z 2 ) + z 2 L n , q 2 ( f ) ( z 1 , z 2 ) 2 ( a n + 1 b n ) , F n ( f ) ( z 1 , z 2 ) = h = 1 4 F n h ( f ) ( z 1 , z 2 )

with

F n 1 ( f ) ( z 1 , z 2 ) = b n ψ n , q 1 1 ( z 1 ) 2 ( 1 + a n b n ) [ R n q 2 ( f z 1 ( z 1 , ) ) ( z 2 ) f z 1 ( z 1 , z 2 ) ] , F n 2 ( f ) ( z 1 , z 2 ) = b n ψ n , q 2 1 ( z 2 ) 2 ( 1 + a n b n ) [ R n q 1 ( f z 2 ( , z 2 ) ) ( z 1 ) f z 2 ( z 1 , z 2 ) ] , F n 3 ( f ) ( z 1 , z 2 ) = b n ψ n , q 1 2 ( z 1 ) 4 ( 1 + a n b n ) [ R n q 2 ( 2 f z 1 2 ( z 1 , ) ) ( z 2 ) 2 f z 1 2 ( z 1 , z 2 ) ] , F n 4 ( f ) ( z 1 , z 2 ) = b n ψ n , q 2 2 ( z 2 ) 4 ( 1 + a n b n ) [ R n q 1 ( 2 f z 2 2 ( , z 2 ) ) ( z 1 ) 2 f z 2 2 ( z 1 , z 2 ) ] , G n ( f ) ( z 1 , z 2 ) = b n ψ n , q 1 1 ( z 1 ) 2 ( 1 + a n b n ) f z 1 ( z 1 , z 2 ) + b n ψ n , q 2 1 ( z 2 ) 2 ( 1 + a n b n ) f z 2 ( z 1 , z 2 ) G n ( f ) ( z 1 , z 2 ) = b n ψ n , q 2 1 ( z 2 ) ψ n , q 1 1 ( z 1 ) 2 ( 1 + a n b n ) 2 f z 2 z 1 ( z 1 , z 2 ) G n ( f ) ( z 1 , z 2 ) = b n ψ n , q 2 1 ( z 2 ) ψ n , q 1 2 ( z 1 ) 4 ( 1 + a n b n ) 3 f z 2 z 1 2 ( z 1 , z 2 ) G n ( f ) ( z 1 , z 2 ) = b n ψ n , q 2 2 ( z 2 ) ψ n , q 1 1 ( z 1 ) 4 ( 1 + a n b n ) 3 f z 2 2 z 1 ( z 1 , z 2 ) G n ( f ) ( z 1 , z 2 ) = b n ψ n , q 2 2 ( z 2 ) ψ n , q 1 2 ( z 1 ) 8 ( 1 + a n b n ) 4 f z 2 2 z 1 2 ( z 1 , z 2 ) ,

and

K n (f)( z 1 , z 2 )= b n 4 ( 1 + a n b n ) { ψ n , q 1 2 ( z 1 ) 2 f z 1 2 ( z 1 , z 2 ) + ψ n , q 2 2 ( z 2 ) 2 f z 2 2 ( z 1 , z 2 ) } .

Considering Theorems 2 and 3 in [19], we get

lim n E n (f)( z 1 , z 2 )=0and lim n F n (f)( z 1 , z 2 )=0.

Under the conditions of the theorem, since lim n a n =0, lim n 1 b n =0, lim n a n b n =0 for 0<β< 1 2 , it is also clear that

lim n G n (f)( z 1 , z 2 )=0.

From Theorem 2, we obtain

lim n 2 ( a n + 1 b n ) [ D n ( f ) 4 ( a n + 1 b n ) 2 ] + E n ( f ) + F n ( f ) + G n ( f ) r 1 , r 2 =0.

Using lim n a n b n =0 for 0<β< 1 2 and 1 1 + a n | z 1 | 2 3 , we get

K n ( f ) r 1 , r 2 1 18 ( 1 + a n b n ) | z 1 || 2 f z 1 2 ( z 1 , z 2 )|.
(14)

Similarly, it follows that

K n ( f ) r 1 , r 2 1 18 ( 1 + a n b n ) | z 2 || 2 f z 2 2 ( z 1 , z 2 )|.
(15)

From equations (14) and (15), we can write

K n ( f ) r 1 , r 2 1 36 ( 1 + a n b n ) K ( f ) r 1 , r 2 .
(16)

In equation (13), taking into account the inequalities

H + T r 1 , r 2 | H r 1 , r 2 T r 1 , r 2 | H r 1 , r 2 T r 1 , r 2 ,

and equation (16), it follows that

R n , n q 1 , q 2 ( f ) f r 1 , r 2 2 ( a n + 1 b n ) { K n ( f ) r 1 , r 2 2 ( a n + 1 b n ) [ D n ( f ) 4 ( a n + 1 b n ) 2 ] + E n ( f ) + F n ( f ) + G n ( f ) r 1 , r 2 } ( a n + 1 b n ) K n ( f ) r 1 , r 2 ( a n + 1 b n ) 1 36 ( 1 + a n b n ) K ( f ) r 1 , r 2

for all n n 0 with n 0 depending only f, r 1 and r 2 . We used that by hypothesis we have K ( f ) r 1 , r 2 >0. □

Combining Theorem 2 with Theorem 3, we immediately obtain the following result giving the exact degree of the operators (1).

Corollary 1 Suppose that the hypothesis in the statement of Theorem  3 holds. If the Taylor series of f contains at least one term of the form c k , 0 z 1 k with c k , 0 0 and k2 or of the form c 0 , j z 2 j with c 0 , j 0 and j2, then for all n n 0 we have

R n , n q 1 , q 2 ( f ) f r 1 , r 2 ( a n + 1 b n ) .

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

z 1 2 f z 1 2 ( z 1 , z 2 )+ z 2 2 f z 2 2 ( z 1 , z 2 )=0,| z 1 |< R 1 ,| z 2 |< R 2 .

Indeed, suppose the contrary. Since a simple calculation gives

z 1 2 f z 1 2 ( z 1 , z 2 ) + z 2 2 f z 2 2 ( z 1 , z 2 ) = k = 1 c k + 1 , 0 k ( k + 1 ) z 1 k + k = 1 c k + 1 , 1 k ( k + 1 ) z 1 k z 2 + 2 j = 2 c 2 , j z 1 z 2 j + k = 2 j = 2 c k + 1 , j k ( k + 1 ) z 1 k z 2 j , + j = 1 c 0 , j + 1 j ( j + 1 ) z 2 j + j = 1 c 1 , j + 1 j ( j + 1 ) z 1 z 2 j + 2 k = 2 c k , 2 z 1 k z 2 + k = 2 j = 2 c k , j + 1 j ( j + 1 ) z 1 k z 2 j ,

by setting equal to zero and by the identification of the coefficients, from the terms under the first and fifth sign ∑, we immediately get c k + 1 , 0 = c 0 , j + 1 =0, for all k=1,2, and j=1,2, , which contradicts the hypothesis on f. Therefore the hypothesis and the lower estimate in Theorem 3 are satisfied, which completes the proof. □

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The author is grateful to the editor and the reviewers for making valuable suggestions, leading to a better presentation of the work.

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Yıldız Özkan, E. Approximation properties of bivariate complex q-Balàzs-Szabados operators of tensor product kind. J Inequal Appl 2014, 20 (2014). https://doi.org/10.1186/1029-242X-2014-20

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Keywords

  • complex approximation
  • q-Balàzs-Szabados operators
  • order of convergence
  • Voronovskaja-type theorem
  • exact degree of approximation