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Approximation properties of complex q-Balázs-Szabados operators in compact disks

Abstract

This paper deals with approximating properties and convergence results of the complex q-Balázs-Szabados operators attached to analytic functions on compact disks. The order of convergence and the Voronovskaja-type theorem with quantitative estimate of these operators and the exact degree of their approximation are given. Our study extends the approximation properties of the complex q-Balázs-Szabados operators from real intervals to compact disks in the complex plane with quantitative estimate.

MSC:30E10, 41A25.

1 Introduction

In the recent years, applications of q-calculus in the area of approximation theory and number theory have been an active area of research. Details on q-calculus can be found in [13]. Several researchers have purposed the q-analogue of Stancu, Kantorovich and Durrmeyer type operators. Gal [4] studied some approximation properties of the complex q-Bernstein polynomials attached to analytic functions on compact disks.

Also very recently, some authors [57] have studied the approximation properties of some complex operators on complex disks. Balázs [8] defined the Bernstein-type rational functions and gave some convergence theorems for them. In [9], Balázs and Szabados obtained an estimate that had several advantages with respect to that given in [8]. These estimates were obtained by the usual modulus of continuity. The q-form of these operator was given by Doğru. He investigated statistical approximation properties of q-Balázs-Szabados operators [10].

The rational complex Balázs-Szabados operators were defined by Gal [4] as follows:

R n (f;z)= 1 ( 1 + a n z ) n j = 0 n f ( j b n ) ( n j ) ( a n z ) j ,

where D R ={zC:|z|<R} with R> 1 2 , f: D R [R,)C is a function, a n = n β 1 , b n = n β , 0<β 2 3 , nN, zC and z 1 a n .

He obtained the uniform convergence of R n (f;z) to f(z) on compact disks and proved the upper estimate in approximation of these operators. Also, he obtained the Voronovskaja-type result and the exact degree of its approximation.

The goal of this paper is to obtain convergence results for the complex q-Balázs-Szabados operators given by

R n (f;q,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,), 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, .

These operators are obtained simply replacing x by z in the real form of the q-Balázs-Szabados operators introduced in Doğru [10].

The complex q-Balázs-Szabados operators R n (f;q,z) are well defined, linear, and these operators are analytic for all n n 0 and |z|r< [ n 0 ] q 1 β since | 1 a n || 1 q a n || 1 q n 1 a n |.

In this paper, we obtain the following results:

  • the order of convergence for the operators R n (f;q,z),

  • the Voronovskaja-type theorem with quantitative estimate,

  • the exact degree of the approximation for the operators R n (f;q,z).

Throughout the paper, we denote with f r =max{|f(z)|R:z D ¯ r } the norm of f in the space of continuous functions on D ¯ r and with f B [ 0 , ) =sup{|f(x)|R:x[0,)} the norm of f in the space of bounded functions on [0,).

Also, the many results in this study are obtained under the condition that f: D R [R,)C is analytic in D R for r<R, which assures the representation f(z)= k = 0 c k z k for all z D R .

2 Convergence results

The following lemmas will help in the proof of convergence results.

Lemma 1 Let n 0 2, 0<β 2 3 and 1 2 <r<R [ n 0 ] q 1 β 2 . Let us define α k , n , q (z)= R n ( e k ;q,z) for all z D ¯ r , where e k (z)= z k . If f: D R [R,)C is uniformly continuous, bounded on [0,) and analytic in D R , then we have the form

R n (f;q,z)= k = 0 c k α k , n , q (z)

for all z D ¯ r .

Proof For any mN, we define

f m (z)= k = 0 m c k e k (z)if |z|rand f m (z)=f(z)if z(r,).

From the hypothesis on f, it is clear that each f m is bounded on [0,), that is, there exist M( f m )>0 with | f m (z)|M( f m ), which implies that

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

that is all R n ( f m ;q,z) with n n 0 , r< [ n 0 ] q 1 β 2 , mN are well defined for all z D ¯ r .

Defining

f m , k (z)= c k e k (z)if |z|rand f m , k (z)= f ( z ) m + 1 if z(r,),

it is clear that each f m , k is bounded on [0,) and that f m (z)= k = 0 m f m , k (z).

From the linearity of R n (f;q,z), we have

R n ( f m ;q,z)= k = 0 m c k α k , n , q (z)for all |z|r.

It suffices to prove that

lim m R n ( f m ;q,z)= R n (f;q,z)

for any fixed nN, n n 0 and |z|r.

We have the following inequality for all |z|r:

| R n ( f m ;q,z) R n (f;q,z)| M r , n , q f m f r ,
(1)

where M r , n , q = s = 0 n 1 ( 1 + q s a n r ) ( 1 q s a n r ) .

Using (1), lim m f m f r =0 and f m f B [ 0 , ) f m f r , the proof of the lemma is finished. □

Lemma 2 If we denote ( β + z ) q n = s = 0 n 1 (β+ q s z), then the following formula holds:

D q [ 1 ( β + z ) q n ] = [ n ] q ( β + z ) q n + 1 ,

where β is a fixed real number and zC.

Proof We can write ( β + z ) q n as follows:

( β + z ) q n = q n ( n 1 ) / 2 ( z + q n + 1 β ) q n .
(2)

In [3] (see p.10, Proposition 3.3), we already have the following formula:

D q [ ( β + z ) q n ] = [ n ] q ( β + z ) q n 1 .
(3)

Using (2) and (3), we get

D q [ ( β + z ) q n ] = q n ( n 1 ) / 2 [ n ] q ( z + q n + 1 β ) q n 1 = [ n ] q q n 1 q ( n 1 ) ( n 2 ) / 2 ( z + q n + 2 ( q 1 β ) ) q n 1 = [ n ] q q n 1 ( q 1 β + z ) q n 1 = [ n ] q ( β + q z ) q n 1 .
(4)

From (4), we obtain the result. □

Lemma 3 We have the following recurrence formula for the complex q-Balázs-Szabados operators R n (f;q,z):

α k + 1 , n , q (z)= ( 1 + q n a n z ) z ( 1 + a n z ) b n D q [ α k , n , q ( z ) ] + z 1 + a n z α k , n , q (z),

where α k , n , q (z)= R n ( e k ;q,z) for all nN, zC and k=0,1,2, .

Proof Firstly, we calculate D q [ α k , n , q (z)] as follows:

D q [ α k , n , q ( z ) ] = D q [ 1 s = 0 n 1 ( 1 + q s a n z ) ] j = 0 n q j ( j 1 ) / 2 ( [ j ] q b n ) k [ n j ] q ( a n z ) j + 1 s = 0 n 1 ( 1 + q s + 1 a n z ) j = 0 n q j ( j 1 ) / 2 ( [ j ] q b n ) k [ n j ] q ( a n ) j D q [ z j ] .
(5)

Considering Lemma 2 and using D q [ z j ]= [ j ] q z j 1 in (5), we get

D q [ α k , n , q ( z ) ] = b n 1 + q n a n z 1 s = 0 n 1 ( 1 + q s a n z ) α k , n , q ( z ) + b n ( 1 + a n z ) z ( 1 + q n a n z ) α k + 1 , n , q ( z ) .
(6)

From (6), the proof of the lemma is finished. □

Corollary 1 ([11], p.143, Corollary 1.10.4)

Let f(z)= p k ( z ) j = 1 k ( z a j ) , where p k (z) is a polynomial of degreek, and we suppose that | a j |R>1 for all j=1,2,,k. If 1r<R, then for all |z|r we have

| f (z)| R + r R r k r f r .

Under hypothesis of the corollary above, by the mean value theorem [12] in complex analysis, we have

| D q [ f ( z ) ] | R + r R r k r f r .
(7)

Lemma 4 Let n 0 2, 0<β 2 3 and 1 2 <r<R [ n 0 ] q 1 β 2 . For all n n 0 , |z|r and k=0,1,2, , we have

| α k , n , q (z)|k! ( 20 r ) k .

Proof Taking the absolute value of the recurrence formula in Lemma 3 and using the triangle inequality, we get

| α k + 1 , n , q (z)| ( 1 + q n a n | z | ) | z | | 1 a n | z | | b n | D q [ α k , n , q ( z ) ] |+ | z | | 1 a n | z | | | α k , n , q (z)|.
(8)

In order to get an upper estimate for | D q [ α k , n , q (z)]|, by using (7), we obtain

| D q [ α k , n , q ( z ) ] | [ n 0 ] q 1 β + r [ n 0 ] q 1 β r k r α k , n , q r .

Under the condition r< [ n 0 ] q 1 β 2 , it holds [ n 0 ] q 1 β + r [ n 0 ] q 1 β r <3, which implies

| D q [ α k , n , q ( z ) ] | 3 k r α k , n , q r .
(9)

Applying (9) to (8) and passing to norm, we get

α k + 1 , n , q r ( 1 + q n a n r ) 3 k ( 1 a n r ) b n α k , n , q r + r 1 a n r α k , n , q r .

From the hypothesis of the lemma, we have 1 1 a n r <2, 1+ q n a n r< 3 2 , and 1 b n <1, which implies

α k + 1 , n , q r 20r(k+1) α k , n , q r .

Taking step by step k=0,1,2, , we obtain

α k + 1 , n , q r ( 20 r ) k + 1 (k+1)!.

Using | α k + 1 , n , q | α k + 1 , n , q r and replacing k+1 with k, the proof of the lemma is finished. □

Let q={ q n } be a sequence satisfying the following conditions:

lim n q n =1and lim n q n n =c(0c<1).
(10)

Now we are in a position to prove the following convergence result.

Theorem 1 Let { q n } be a sequence satisfying the conditions (10) with q n (0,1] for all nN, and let n 0 2, 0<β 2 3 and 1 2 <r<R [ n 0 ] q n 1 β 2 . If f: D R [R,)C is uniformly continuous, bounded on [0,) and analytic in D R and there exist M>0, 0<A< 1 20 r with | c k |M A k k ! (which implies |f(z)|M e A | z | for all z D R ), then the sequence { R n ( f ; q n , z ) } n n 0 is uniformly convergent to f in D ¯ r .

Proof From Lemma 2 and Lemma 6, for all n n 0 and |z|r, we have

| R n (f; q n ,z)| k = 0 | c k || α k , n , q n (z)| k = 0 M A k k ! k! ( 20 r ) k =M k = 0 ( 20 A r ) k ,

where the series k = 0 ( 20 A r ) k is convergent for 0<A< 1 20 r .

Since lim n R n (f; q n ,x)=f(x) for all x[0,r] (see [10]), by Vitali’s theorem (see [13], p.112, Theorem 3.2.10), it follows that { R n (f; q n ,z)} uniformly converges to f(z) in D ¯ r . □

We can give the following upper estimate in the approximation of R n (f; q n ,z).

Theorem 2 Let { q n } be a sequence satisfying the conditions (10) with q n (0,1] for all nN, and let n 0 2, 0<β 2 3 and 1 2 <r<R [ n 0 ] q n 1 β 2 . If f: D R [R,)C is uniformly continuous, bounded on [0,) and analytic in D R and there exist M>0, 0<A< 1 20 r with | c k |M A k k ! (which implies |f(z)|M e A | z | for all z D R ), then the following upper estimate holds:

| R n (f; q n ,z)f(z)| C r 1 (f) ( a n + 1 b n ) ,

where C r 1 (f)=max{9MA k = 1 (k1) ( 20 A r ) k 1 ,2 r 2 MA e 2 A r } and k = 1 (k1) ( 20 A r ) k 1 <.

Proof Using the recurrence formula in Lemma 4, we have

| α k + 1 , n , q n ( z ) z k + 1 | ( 1 + q n n a n | z | ) | z | | 1 a n | z | | b n | D q n [ α k , n , q n ( z ) z k ] | + | z | | 1 a n | z | | | α k , n , q n ( z ) z k | + 1 b n ( 1 + q n n a n | z | ) | 1 a n | z | | [ k ] q n | z | k + a n | 1 a n | z | | | z | k + 2 .

For |z|r, we get

| α k + 1 , n , q n ( z ) z k + 1 | ( 1 + q n n a n r ) r ( 1 a n r ) b n | D q n [ α k , n , q n ( z ) ] | + r 1 a n r | α k , n , q n ( z ) z k | + 2 b n ( 1 + q n n a n r ) ( 1 a n r ) [ k ] q n r k + a n 1 a n r r k + 2 .

Using (9), 1 1 a n r <2, and 1+ q n n a n r< 3 2 , we obtain

| α k + 1 , n , q n (z) z k + 1 | 9 k k ! b n ( 20 r ) k +2r| α k , n , q n (z) z k |+ 6 b n [ k ] q n r k +2 a n r k + 2 .

Since 6 [ k ] q n r k 9kk! ( 20 r ) k for all k=0,1,2, , we can write

| α k + 1 , n , q n (z) z k + 1 | 18 k k ! b n ( 20 r ) k +2r| α k , n , q n (z) z k |+2 a n r k + 2 .

Taking k=0,1,2, step by step, finally we arrive at

| α k , n , q n (z) z k | 9 b n (k1)k! ( 20 r ) k 1 +2 a n r 2 k ( 2 r ) k 1 ,
(11)

which implies

| R n ( f ; q n , z ) f ( z ) | k = 1 | c k | | α k , n , q n ( z ) z k | k = 1 M A k k ! { 9 b n ( k 1 ) k ! ( 20 r ) k 1 + 2 a n r 2 k ( 2 r ) k 1 } = 9 M A b n k = 1 ( k 1 ) ( 20 A r ) k 1 + 2 a n r 2 M A k = 1 ( 20 A r ) k 1 ( k 1 ) ! = 9 M A b n k = 1 ( k 1 ) ( 20 A r ) k 1 + 2 a n r 2 M A e 2 A r .

Choosing C r 1 (f)=max{9MA k = 1 (k1) ( 20 A r ) k 1 ,2 r 2 MA e 2 A r }, we obtain the desired result.

Here the series k = 0 ( 20 A r ) k is convergent for 0<A< 1 20 r and the series is absolutely convergent in D ¯ r , it easily follows that k = 1 (k1) ( 20 A r ) k 1 <. □

The following lemmas will help in the proof of the next theorem.

Lemma 5 For all nN, we have

R n ( e 0 ;q,z)=1,
(12)
R n ( e 1 ;q,z)= z 1 + a n z ,
(13)
R n ( e 2 ;q,z)= ( 1 a n b n ) q z 2 ( 1 + a n z ) ( 1 + a n q z ) + z b n ( 1 + a n z ) ,
(14)

where e k (z)= z k for k=0,1,2.

Proof (12) and (13) are obtained simply replacing x by z in Lemma 3.1 and Lemma 3.2 in [10]. Also, using [ n ] q =1+q [ n 1 ] q and a n b n = 1 [ n ] q and replacing x by z in Lemma 3.3 in [10], (14) is obtained. □

Lemma 6 For all nN, the following equalities for the operators R n (f;q,z) hold:

ψ n , q 1 (z)= a n z 2 1 + a n z ,
(15)
ψ n , q 2 ( z ) = z b n ( 1 + a n z ) ( 1 + a n q z ) ( 1 q ) z 2 ( 1 + a n z ) ( 1 + a n q z ) ψ n , q 2 ( z ) = a n ( 1 q ) z 3 ( 1 + a n z ) ( 1 + a n q z ) + a n 2 q z 4 ( 1 + a n z ) ( 1 + a n q z ) ,
(16)

where ψ n , q i (z)= R n ( ( t e 1 ) i ;q,z) for i=1,2.

Proof From Lemma 5, the proof can be easily got, so we omit it. □

Now, we present a quantitative Voronovskaja-type formula.

Let us define

A k , n , q n (z)= R n (f; q n ,z)f(z) ψ n , q 1 (z) f (z) 1 2 ψ n , q 2 (z) f (z).
(17)

Theorem 3 Let { q n } be a sequence satisfying the conditions (10) with q n (0,1] for all nN, n 0 2, 0<β 2 3 and 1 2 <r<R [ n 0 ] q n 1 β 2 . If f: D R [R,)C is uniformly continuous, bounded on [0,) and analytic in D R and there exist M>0, 0<A< 1 20 r with | c k |M A k k ! (which implies |f(z)|M e A | z | for all z D R ), then for all n n 0 and |z|r, we have

| A k , n , q n (z)| C r 2 (f) ( a n + 1 b n ) 2 ,

where C r 2 (f)= C M r 3 k = 3 (k2)(k1)k(k+1) ( 20 r A ) k 3 < and C is a fixed real number.

Proof From Lemma 1 and the analyticity of f, we can write

| A k , n , q n (z)| k = 2 | c k || E k , n , q n (z)|,
(18)

where

E k , n , q n ( z ) = α k , n , q n ( z ) z k + a n k z k + 1 1 + a n z ( k 1 ) k z k 1 2 b n ( 1 + a n z ) ( 1 + a n q n z ) + ( 1 q n ) ( k 1 ) k z k 2 ( 1 + a n z ) ( 1 + a n q n z ) + a n ( 1 q n ) ( k 1 ) k z k + 1 2 ( 1 + a n z ) ( 1 + a n q n z ) a n 2 q n ( k 1 ) k z k + 2 2 ( 1 + a n z ) ( 1 + a n q n z ) .
(19)

Using Lemma 5, we easily obtain that E 0 , n , q (z)= E 1 , n , q (z)= E 2 , n , q (z)=0.

Combining (19) with the recurrence formula in Lemma 3, a simple calculation leads us to the following recurrence formula:

E k + 1 , n , q n (z)= ( 1 + q n n a n z ) z b n ( 1 + a n z ) D q n [ E k , n , q n ( z ) ] + z 1 + a n z E k , n , q n (z)+ F k , n , q n (z),
(20)

where

F k , n , q n ( z ) = ( k [ k ] q n ) z k b n ( 1 + a n z ) 2 ( 1 + a n q n z ) + a n 2 k z k + 3 ( 1 + a n z ) 2 ( 1 q n ) k z k + 1 ( 1 + a n z ) 2 ( 1 + a n q n z ) + a n ( 1 q n ) k z k + 2 ( 1 + a n z ) 2 ( 1 + a n q n z ) a n 2 q n k z k + 3 ( 1 + a n z ) 2 ( 1 + a n q n z ) a n k ( k + 1 ) z k + 1 2 b n ( 1 + a n z ) 2 ( 1 + a n q n z ) + a n ( 1 q n ) k ( k + 1 ) z k + 2 2 ( 1 + a n z ) 2 ( 1 + a n q n z ) + a n 2 ( 1 q n ) k ( k + 1 ) z k + 3 2 ( 1 + a n z ) 2 ( 1 + a n q n z ) a n 3 q n k ( k + 1 ) z k + 4 2 ( 1 + a n z ) 2 ( 1 + a n q n z ) a n ( 1 + q n n a n z ) ( ( k 1 ) [ k + 1 ] q n q n [ k 1 ] q n ) z k + 1 b n ( 1 + a n z ) 2 ( 1 + a n q n z ) a n 2 ( 1 + q n n a n z ) ( k 1 ) q n [ k ] q n z k + 2 b n ( 1 + a n z ) 2 ( 1 + a n q n z ) + a n q n n [ k ] q n z k + 1 b n ( 1 + a n z ) 2 ( 1 + a n q n z ) ( 1 + q n n a n z ) [ k 1 ] q n ( k 1 ) k z k 1 2 b n 2 ( 1 + a n z ) ( 1 + a n q n z ) ( 1 + a n q n 2 z ) + ( 1 q n ) ( 1 + q n n a n z ) [ k ] q n ( k 1 ) k z k 2 b n ( 1 + a n z ) ( 1 + a n q n z ) ( 1 + a n q n 2 z ) + a n ( 1 q n ) ( 1 + q n n a n z ) [ k + 1 ] q n ( k 1 ) k z k + 1 2 b n ( 1 + a n z ) ( 1 + a n q n z ) ( 1 + a n q n 2 z ) a n 2 q n ( 1 + q n n a n z ) [ k + 2 ] q n ( k 1 ) k z k + 2 2 b n ( 1 + a n z ) ( 1 + a n q n z ) ( 1 + a n q n 2 z ) + a n ( 1 + q n n a n z ) ( 1 + q n ) ( k 1 ) k z k 2 b n 2 ( 1 + a n z ) 2 ( 1 + a n q n z ) ( 1 + a n q n 2 z ) a n ( 1 q n ) ( 1 + q n ) ( 1 + q n n a n z ) ( k 1 ) k z k + 1 2 b n ( 1 + a n z ) 2 ( 1 + a n q n z ) ( 1 + a n q n 2 z ) a n 2 ( 1 q n ) ( 1 + q n ) ( 1 + q n n a n z ) ( k 1 ) k z k + 2 2 b n ( 1 + a n z ) 2 ( 1 + a n q n z ) ( 1 + a n q n 2 z ) a n 3 q n ( 1 + q n ) ( 1 + q n n a n z ) ( k 1 ) k z k + 3 2 b n ( 1 + a n z ) 2 ( 1 + a n q n z ) ( 1 + a n q n 2 z ) .

In the following results, C i will denote fixed real numbers for i=1,2,3.

Under the hypothesis of Theorem 3, we have

| 1 1 + q n s a n z | 1 1 q n s a n r <2for s=0,1,2,
(21)
a n r< 1 2 and1+ q n n a n r< 3 2 ,
(22)
1 q n a n b n andk [ k ] q n a n b n ( k 1 ) k 2 ,
(23)
[ k ] q n k, a n <1and 1 b n <1.
(24)

Using (21)-(24), for |z|r, we get

| F k , n , q n ( z ) | C 1 ( a n 2 + a n b n + 1 b n 2 ) k ( k + 1 ) ( k + 2 ) × max { r k 1 , r k , r k + 1 , r k + 2 , r k + 3 , r k + 4 } C 1 ( a n + 1 b n ) 2 k ( k + 1 ) ( k + 2 ) ( 2 r ) k + 4 .
(25)

On the other hand, for |z|r, we have

| ( 1 + q n n a n z ) z b n ( 1 + a n z ) D q n [ E k , n , q n ( z ) ] | ( 1 + q n n a n r ) r b n ( 1 a n r ) 3 k r E k , n , q n r 3 k ( 1 + q n n a n r ) b n ( 1 a n r ) { α k , n , q n e k r + a n k r k + 1 1 a n r + ( k 1 ) k r k 1 2 b n ( 1 a n r ) ( 1 a n q n r ) + ( 1 q n ) ( k 1 ) k r k 2 ( 1 a n r ) ( 1 a n q n r ) + a n ( 1 q n ) ( k 1 ) k r k + 1 2 ( 1 a n r ) ( 1 a n q n r ) + a n 2 q n ( k 1 ) k r k + 2 2 ( 1 a n r ) ( 1 a n q n r ) } .

Taking into account (11) in the proof of Theorem 2, we obtain

| ( 1 + q n n a n z ) z b n ( 1 + a n z ) D q n [ E k , n , q n ( z ) ] | C 2 1 b n ( a n + 1 b n ) ( k 1 ) k ( k + 1 ) × ( k ! ) ( 20 r ) k + 2 C 2 ( a n + 1 b n ) 2 ( k 1 ) k ( k + 1 ) ( k ! ) ( 20 r ) k + 2 .
(26)

Considering (25) and (26) in (20), we get

| E k + 1 , n , q n (z)|2r| E k , n , q n (z)|+ C 3 ( a n + 1 b n ) 2 k(k+1)(k+2)(k+1)! ( 20 r ) k + 4 .

Since E 0 , n , q (z)= E 1 , n , q (z)= E 2 , n , q (z)=0, taking k=2,3,4, in the last inequality step by step, finally we arrive at

| E k , n , q n (z)| C 3 ( a n + 1 b n ) 2 (k2)(k1)k(k+1)(k!) ( 20 r ) k + 3 .
(27)

Finally, considering (27) in (18) and using 20rA<1, the proof of the theorem is complete. □

Remark 1 For 0<q1, since 1 [ n ] q 1q as n, therefore a n = ( 1 [ n ] q ) 1 β ( 1 q ) 1 β and 1 b n = ( 1 [ n ] q ) β ( 1 q ) β as n. If a sequence { q n } satisfies the conditions (10), then 1 [ n ] q 0 as n; therefore a n = ( 1 [ n ] q ) 1 β 0 and 1 b n = ( 1 [ n ] q ) β 0 as n.

Under the conditions (10), Theorem 2 and Theorem 3 show that { R n ( f ; q n , z ) } n n 0 uniformly converges to f(z) in D ¯ r .

From Theorem 2 and Theorem 3, we get the following consequence.

Theorem 4 Let { q n } be a sequence satisfying the conditions (10) with q n (0,1] for all nN, n 0 2, 0<β 2 3 , β 1 2 and 1 2 <r<R [ n 0 ] q n 1 β 2 . Suppose that f: D R [R,)C is uniformly continuous, bounded on [0,) and analytic in D R and there exist M>0, 0<A< 1 20 r with | c k |M A k k ! (which implies |f(z)|M e A | z | for all z D R ). If f is not a polynomial of degree ≤1, then for all n n 0 we have

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

Proof We can write

R n (f; q n ,z)f(z)= ( a n + 1 b n ) { G ( z ) + H n ( z ) } ,
(28)

where

G ( z ) = a n a n + 1 / b n z 2 f ( z ) 1 + a n z + 1 a n b n + 1 z f ( z ) 2 ( 1 + a n z ) ( 1 + a n q n z ) 1 q n a n + 1 / b n z 2 f ( z ) 2 ( 1 + a n z ) ( 1 + a n q n z ) a n ( 1 q n ) a n + 1 / b n z 3 f ( z ) 2 ( 1 + a n z ) ( 1 + a n q n z ) + a n 2 a n + 1 / b n q n z 4 f ( z ) 2 ( 1 + a n z ) ( 1 + a n q n z )
(29)

and

H n (z)= ( a n + 1 b n ) [ 1 ( a n + 1 b n ) 2 A k , n , q n ( z ) ] ,
(30)

and also ( H n ( z ) ) n N is a sequence of analytic functions uniformly convergent to zero for all |z|r.

Since a n + 1 b n 0 as n, and taking into account Theorem 3, it remains only to show that for sufficiently large n and for all |z|r, we have |G(z)|>ρ>0, where ρ is independent of n.

If 2β1<0, then the term 1 a n b n + 1 1 as n, while the other terms converge to zero, so there exists a natural number n 1 N with n 1 n 0 so that for all n n 1 and |z|r, we have

|G(z)| 1 2 | z f ( z ) 2 ( 1 + a n z ) ( 1 + a n q n z ) | 1 4 | z f ( z ) | ( 1 + r ) 2 .
(31)

If 2β1>0, then the term a n a n + 1 / b n 1 as n, while the other terms converge to zero. So, there exists a natural number n 2 N with n 2 n 0 so that for all n n 2 and |z|r, we have

|G(z)| 1 2 | z 2 f ( z ) 1 + a n z | 1 2 | z 2 f ( z ) | 1 + r .
(32)

In the case of 2β1=0, that is, β= 1 2 , we obtain a n 2 a n + 1 / b n = [ n ] q n 1 / 2 , as n, so that the case β= 1 2 remains unsettled.

Choosing n 3 =max{ n 1 , n 2 }, considering (31) and (32), for all n n 3 , we get

R n ( f ; q n , ) f r ( a n + 1 b n ) | G r H n r | ( a n + 1 b n ) 1 2 G r .

For all n{ n 0 ,, n 3 1}, we get

R n ( f ; q n , ) f r ( a n + 1 b n ) M r , n , q n (z)

with M r , n , q n (z)= 1 a n + 1 / b n R n ( f ; q n , ) f r >0, which finally implies

R n ( f ; q n , ) f r ( a n + 1 b n ) C r (f)
(33)

for all n n 0 , with C r (f)=min{ M r , n 0 , q n (z),, M r , n 3 1 , q n (z), 1 2 G r }.

From (33) and Theorem 3, the proof is complete. □

Remark 2 Recently, it is much more interesting to study these operators in the case q>1. Authors continue to study that case.

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

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The main idea of this paper is proposed by NI. All authors contributed equally in writing this article and read and approved the final manuscript.

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İspir, N., Yıldız Özkan, E. Approximation properties of complex q-Balázs-Szabados operators in compact disks. J Inequal Appl 2013, 361 (2013). https://doi.org/10.1186/1029-242X-2013-361

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Keywords

  • complex q-Balázs-Szabados operators
  • order of convergence
  • Voronovskaja-type theorem
  • approximation in compact disks