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Convergence of the q-Stancu-Szász-Beta type operators

Journal of Inequalities and Applications20142014:354

https://doi.org/10.1186/1029-242X-2014-354

  • Received: 2 April 2014
  • Accepted: 27 August 2014
  • Published:

Abstract

In this paper, we study on q-Stancu-Szász-Beta type operators. We give these operators convergence properties and obtain a weighted approximation theorem in the interval [ 0 , ) .

MSC:41A25, 41A36.

Keywords

  • q-Stancu type operators
  • Szász-Beta type operators
  • weighted approximation

1 Introduction

In [1], Mahmudov constructed q-Szász operators and obtained rate of global convergence in the frame of weighted spaces and a Voronovskaja type theorem for these operators. In [2], Gupta and Mahmudov studied on the q-analog of the Szász-Beta type operators. In [3], Yüksel and Dinlemez gave a Voronovskaja type theorem for q-analog of a certain family Szász-Beta type operators. In [4], Govil and Gupta introduced the q-analog of certain Beta-Szász-Stancu operators. They estimated the moments and established direct results in terms of modulus of continuity and an asymptotic formula for the q-operators. In [514], interesting generalization about q-calculus were given. Our aims are to give approximation properties and a weighted approximation theorem for q-Stancu-Szász-Beta type operators. We use without further explanation the basic notations and formulas, from the theory of q-calculus as set out in [1519]. Let A > 0 and f be a real valued continuous function defined on the interval [ 0 , ) . For 0 < q 1 , q-Stancu-Szász-Beta type operators are defined as
B n , q ( α , β ) ( f , x ) = k = 0 s n , k q ( x ) 0 / A b n , k q ( t ) f ( [ n ] q t + α [ n ] q + β ) d q t ,
(1.1)
where
s n , k q ( x ) = ( [ n ] q x ) k e [ n ] q x [ k ] q !
and
b n , k q ( x ) = q k 2 x k B q ( k + 1 , n ) ( 1 + x ) q n + k + 1 .

If we write q = 1 and α = β = 0 in (1.1), then the operators B n , q ( α , β ) ( f , x ) are reduced to Szász-Beta type operators studied in [2023].

2 Auxiliary results

For the sake of brevity, the notation F s q ( n ) = i = 1 s [ n i ] q and G β q ( n ) = ( [ n ] q + β ) will be used throughout the article. Now we are ready to give the following lemma for the Korovkin test functions.

Lemma 1 Let e m ( t ) = t m , m = 0 , 1 , 2 , we get
(i) B n , q ( α , β ) ( e 0 , x ) = 1 , (ii) B n , q ( α , β ) ( e 1 , x ) = [ n ] q 2 x q 2 G β q ( n ) F 1 q ( n ) + [ n ] q q G β q ( n ) F 1 q ( n ) + α G β q ( n ) , (iii) B n , q ( α , β ) ( e 2 , x ) = [ n ] q 4 x 2 q 6 G β q ( n ) 2 F 2 q ( n ) + { [ n ] q 3 q 5 G β q ( n ) 2 F 2 q ( n ) B n , q ( α , β ) ( e 2 , x ) = + ( 1 + [ 2 ] q ) [ n ] q 3 q 4 G β q ( n ) 2 F 2 q ( n ) + 2 α [ n ] q 2 q 2 G β q ( n ) 2 F 1 q ( n ) } x B n , q ( α , β ) ( e 2 , x ) = + [ 2 ] q [ n ] q 2 q 3 G β q ( n ) 2 F 2 q ( n ) + 2 α [ n ] q q G β q ( n ) 2 F 1 q ( n ) + α 2 G β q ( n ) 2 .
Proof Using the q-Gamma and q-Beta functions in [15, 24], we obtain the following equality:
q k 2 0 / A 1 B ( k + 1 , n ) t k + m ( 1 + t ) q n + k + 1 d q t = [ m + k ] q ! [ n m 1 ] q ! q { 2 k 2 ( k + m ) ( k + m + 1 ) } / 2 [ k ] q ! [ n 1 ] q ! .
(2.1)
Then, using (2.1), for m = 0 , we get
B n , q ( α , β ) ( e 0 , x ) = e [ n ] q x k = 0 ( [ n ] q x ) k [ k ] q ! q k ( k 1 ) / 2 = e [ n ] q x E q [ n ] q x = 1 ,
and the proof of (i) is finished. With a direct computation, we obtain (ii) as follows:
B n , q ( α , β ) ( e 1 , x ) = [ n ] q G β q ( n ) F 1 q ( n ) k = 1 ( [ n ] q x ) k [ k 1 ] q ! q k ( k 3 ) 2 / 2 e [ n ] q x + [ n ] q G β q ( n ) F 1 q ( n ) k = 0 ( [ n ] q x ) k [ k ] q ! q k ( k 1 ) 2 / 2 e [ n ] q x + α G β q ( n ) k = 0 ( [ n ] q x ) k [ k ] q ! q k ( k 1 ) / 2 e [ n ] q x = [ n ] q 2 x q 2 G β q ( n ) F 1 q ( n ) E q [ n ] q x e [ n ] q x + [ n ] q q G β q ( n ) F 1 q ( n ) E q [ n ] q x e [ n ] q x + α G β q ( n ) E q [ n ] q x e [ n ] q x = [ n ] q 2 x q 2 G β q ( n ) F 1 q ( n ) + [ n ] q q G β q ( n ) F 1 q ( n ) + α G β q ( n ) .
Using the equality
[ n ] q = [ s ] q + q s [ n s ] q , 0 s n ,
(2.2)
we get
B n , q ( α , β ) ( e 2 , x ) = [ n ] q 4 x 2 q 6 G β q ( n ) 2 F 2 q ( n ) + { [ n ] q 3 q 5 G β q ( n ) 2 F 2 q ( n ) + ( 1 + [ 2 ] q ) [ n ] q 3 q 4 G β q ( n ) 2 F 2 q ( n ) + 2 α [ n ] q 2 q 2 G β q ( n ) 2 F 1 q ( n ) } x + [ 2 ] q [ n ] q 2 q 3 G β q ( n ) 2 F 2 q ( n ) + 2 α [ n ] q q G β q ( n ) 2 F 1 q ( n ) + α 2 G β q ( n ) 2 ,

and so we have the proof of (iii). □

To obtain our main results we need to compute the second moment.

Lemma 2 Let q ( 0 , 1 ) and n > 2 . Then we have the following inequality:
B n , q ( α , β ) ( ( t x ) 2 , x ) ( 2 ( 1 q 4 ) q 6 + 164 ( α + β + 1 ) 2 [ n ] q q 6 F 2 q ( n ) ) x ( x + 1 ) + 6 ( α + 1 ) 2 q 3 G β q ( n ) .
Proof From the linearity of the B n , q ( α , β ) operators and Lemma 1, we write the second moment as
B n , q ( α , β ) ( ( t x ) 2 , x ) = { [ n ] q 4 q 6 G β q ( n ) 2 F 2 q ( n ) 2 [ n ] q 2 q 2 G β q ( n ) F 1 q ( n ) + 1 } x 2 + { { 1 + ( 1 + [ 2 ] q ) q } [ n ] q 3 q 5 G β q ( n ) 2 F 2 q ( n ) + 2 α [ n ] q 2 q 2 G β q ( n ) 2 F 1 q ( n ) 2 [ n ] q q G β q ( n ) F 1 q ( n ) 2 α G β q ( n ) } x + [ 2 ] q [ n ] q 2 q 3 G β q ( n ) 2 F 2 q ( n ) + 2 α [ n ] q q G β q ( n ) 2 F 1 q ( n ) + α 2 G β q ( n ) 2 { [ n ] q 4 q 6 G β q ( n ) 2 F 2 q ( n ) 2 [ n ] q 2 q 2 G β q ( n ) F 1 q ( n ) + 1 + { 1 + ( 1 + [ 2 ] q ) q } [ n ] q 3 q 5 G β q ( n ) 2 F 2 q ( n ) + 2 α [ n ] q 2 q 2 G β q ( n ) 2 F 1 q ( n ) } x ( x + 1 ) + [ 2 ] q [ n ] q 2 q 3 G β q ( n ) 2 F 2 q ( n ) + 2 α [ n ] q q G β q ( n ) 2 F 1 q ( n ) + α 2 G β q ( n ) 2 { [ n ] q 4 ( 1 + q 6 ) 2 q 4 [ n 2 ] q 4 + 2 β q 6 [ n ] q [ n 1 ] q [ n 2 ] q q 6 G β q ( n ) 2 F 2 q ( n ) + ( q + q 2 + [ 2 ] q q 2 ) [ n ] q 3 q 6 G β q ( n ) 2 F 2 q ( n ) + q 6 β 2 [ n 1 ] q [ n 2 ] q q 6 G β q ( n ) 2 F 2 q ( n ) + 2 α q 4 [ n ] q 2 [ n 2 ] q q 6 G β q ( n ) 2 F 2 q ( n ) } x ( x + 1 ) + { [ 2 ] q + 2 α q 2 + α 2 q 3 } [ n ] q q 3 G β q ( n ) F 2 q ( n ) .
From (2.2), we have
B n , q ( α , β ) ( ( t x ) 2 , x ) { [ n 2 ] q 4 ( q 14 + q 8 2 q 4 ) q 6 G β q ( n ) 2 F 2 q ( n ) + ( 1 + q 6 ) { 4 [ 2 ] q q 6 [ n 2 ] q 3 + 6 [ 2 ] q 2 q 4 [ n 2 ] q 2 + 4 [ 2 ] q 3 q 2 [ n 2 ] q + [ 2 ] q 4 } q 6 G β q ( n ) 2 F 2 q ( n ) + ( q + q 2 + [ 2 ] q q 2 + 2 β q 6 + 2 α q 4 ) [ n ] q 3 + β 2 q 6 [ n ] q 2 q 6 G β q ( n ) 2 F 2 q ( n ) } x ( x + 1 ) + ( [ 2 ] q + q 2 ) ( [ 2 ] q + 2 α q 2 + α 2 q 3 ) q 3 G β q ( n ) F 1 q ( n ) ( 2 ( 1 q 4 ) q 6 + 164 ( α + β + 1 ) 2 [ n ] q q 6 F 2 q ( n ) ) x ( x + 1 ) + 6 ( α + 1 ) 2 q 3 G β q ( n ) .

And the proof of Lemma 2 is now finished. □

3 Direct estimates

Now in our considerations, C B [ 0 , ) denotes the set of all bounded-continuous functions from [ 0 , ) to . C B [ 0 , ) is a normed space with the norm f B = sup { | f ( x ) | : x [ 0 , ) } . We denote the first modulus of continuity on the finite interval [ 0 , b ] , b > 0 ,
ω [ 0 , b ] ( f , δ ) = sup 0 < h δ , x [ 0 , b ] | f ( x + h ) f ( x ) | .
(3.1)
The Peetre K-functional is defined by
K 2 ( f , δ ) = inf { f g B + δ g B : g W 2 } , δ > 0 ,
where W 2 = { g C B [ 0 , ) : g , g C B [ 0 , ) } . By Theorem 2.4 in [25], p.177, there exists a positive constant C such that
K 2 ( f , δ ) C ω 2 ( f , δ ) ,
(3.2)
where
ω 2 ( f , δ ) = sup 0 < h δ sup x [ 0 , ) | f ( x + 2 h ) 2 f ( x + h ) f ( x ) | .

Gadzhiev proved the weighted Korovkin-type theorems in [26]. We give the Gadzhiev results in weighted spaces. Let ρ ( x ) = 1 + x 2 and the weighted spaces C ρ [ 0 , ) denote the space of all continuous functions f, satisfying | f ( x ) | M f ρ ( x ) , where M f is a constant depending only on f. C ρ [ 0 , ) is a normed space with the norm f ρ = sup { | f ( x ) | ρ ( x ) : x R + { 0 } } and C ρ [ 0 , ) denotes the subspace of all functions f C ρ [ 0 , ) for which lim | x | | f ( x ) | ρ ( x ) exists finitely.

Thus we are ready to give direct results. The following lemma is routine and its proof is omitted.

Lemma 3 Let
B ¯ n , q ( α , β ) ( f , x ) = B n , q ( α , β ) ( f , x ) f ( D n , q ( α , β ) ( x ) ) + f ( x ) .
(3.3)
Then the following assertions hold for the operators (3.3):
(i) B ¯ n , q ( α , β ) ( 1 , x ) = 1 , (ii) B ¯ n , q ( α , β ) ( t , x ) = x , (iii) B ¯ n , q ( α , β ) ( t x , x ) = 0 ,

where D n , q ( α , β ) ( x ) = [ n ] q 2 x q 2 G β q ( n ) F 1 q ( n ) + [ n ] q q G β q ( n ) F 1 q ( n ) + α G β q ( n ) .

Lemma 4 Let q ( 0 , 1 ) and n > 2 . Then for every x [ 0 , ) and f C B [ 0 , ) , we have the inequality
| B ¯ n , q ( α , β ) ( f , x ) f ( x ) | δ n , q ( α , β ) ( x ) f B ,

where δ n , q ( α , β ) ( x ) = ( 2 ( 1 q 4 ) q 6 + 263 ( α + β + 1 ) 2 q 6 F 1 q ( n ) ) x ( x + 1 ) + 5 ( α + 1 ) 2 q 3 G β q ( n ) .

Proof Using Taylor’s expansion
f ( t ) = f ( x ) + ( t x ) f ( x ) + x t ( t u ) f ( u ) d u
and Lemma 3, we obtain
B ¯ n , q ( α , β ) ( f , x ) f ( x ) = B ¯ n , q ( α , β ) ( x t ( t u ) f ( u ) d u , x ) .
Then, using Lemma 1 and the inequality
| x t ( t u ) f ( u ) d u | f B ( t x ) 2 2 ,
we get
| B ¯ n , q ( α , β ) ( f , x ) f ( x ) | | B n , q ( α , β ) ( x t ( t u ) f ( u ) d u , x ) x D n , q ( α , β ) ( x ) { D n , q ( α , β ) ( x ) u } f ( u ) d u | f B 2 { ( 2 ( 1 q 4 ) q 6 + 164 ( α + β + 1 ) 2 [ n ] q q 6 F 2 q ( n ) + ( [ n ] q 2 q 2 G β q ( n ) F 1 q ( n ) 1 ) 2 + 2 [ n ] q 3 q 3 G β q ( n ) 2 F 1 q ( n ) 2 + 2 [ n ] q 2 α q 2 G β q ( n ) 2 F 1 q ( n ) ) x ( x + 1 ) + ( [ n ] q + α q [ n 1 ] q q G β q ( n ) F 1 q ( n ) ) 2 + 6 ( α + 1 ) 2 q 3 G β q ( n ) F 1 q ( n ) } f B 2 { ( 4 ( 1 q 4 ) q 6 + 526 ( α + β + 1 ) 2 q 6 F 1 q ( n ) ) x ( x + 1 ) + 10 ( α + 1 ) 2 q 3 G β q ( n ) } .

And the proof of the Lemma 4 is now completed. □

Theorem 1 Let ( q n ) ( 0 , 1 ) a sequence such that q n 1 as n . Then for every n > 2 , x [ 0 , ) and f C B [ 0 , ) , we have the inequality
| B n , q n ( α , β ) ( f , x ) f ( x ) | 2 M ω 2 ( f , δ n , q n ( α , β ) ( x ) ) + w ( f , η n , q n ( α , β ) ( x ) ) ,

where η n , q n ( α , β ) ( x ) = ( [ n ] q n 2 q n 2 G β q n ( n ) F 1 q n ( n ) 1 ) x + [ n ] q n q n G β q n ( n ) F 1 q n ( n ) + α G β q n ( n ) .

Proof Using (3.3) for any g W 2 , we obtain the following inequality:
| B n , q n ( α , β ) ( f , x ) f ( x ) | | B ¯ n , q n ( α , β ) ( f g , x ) ( f g ) ( x ) + B ¯ n , q n ( α , β ) ( g , x ) g ( x ) | + | f ( [ n ] q n 2 q n 2 G β q n ( n ) F 1 q n ( n ) x + [ n ] q n q n G β q n ( n ) F 1 q n ( n ) + α G β q n ( n ) ) f ( x ) | .
From Lemma 4, we get
| B n , q n ( α , β ) ( f , x ) f ( x ) | 2 f g B + δ n , q n ( α , β ) ( x ) g + | f ( [ n ] q n 2 q n 2 G β q n ( n ) F 1 q n ( n ) x + [ n ] q n q n G β q n ( n ) F 1 q n ( n ) + α G β q n ( n ) ) f ( x ) | .
By using equality (3.1) we have
| B n , q n ( α , β ) ( f , x ) f ( x ) | 2 f g B + δ n , q n ( α , β ) ( x ) g B + w ( f , η n , q n ( α , β ) ( x ) ) .

Taking the infimum over g W 2 on the right-hand side of the above inequality and using the inequality (3.2), we get the desired result. □

Theorem 2 Let ( q n ) ( 0 , 1 ) a sequence such that q n 1 as n . Then f C ρ [ 0 , ) , and we have
lim n B n , q n ( α , β ) ( f ) f ρ = 0 .
Proof From Lemma 1, it is obvious that B n , q n ( α , β ) ( e 0 ) e 0 ρ = 0 . Since | [ n ] q n 2 q n 2 G β q n ( n ) F 1 q n ( n ) x + [ n ] q n q n G β q n ( n ) F 1 q n ( n ) + α G β q n ( n ) x | ( x + 1 ) o ( 1 ) and x + 1 1 + x 2 is positive and bounded from above for each x 0 , we obtain
B n , q n ( α , β ) ( e 1 ) e 1 ρ x + 1 1 + x 2 o ( 1 ) .

And then lim n B n , q n ( α , β ) ( e 1 ) e 1 ρ = 0 .

Similarly for every n > 2 , we write
B n , q n ( α , β ) ( e 2 ) e 2 ρ = sup x [ 0 , ) { | ( [ n ] q n 4 q n 6 G β q n ( n ) 2 F 2 q n ( n ) 1 ) x 2 1 + x 2 + { ( 1 + ( 1 + [ 2 ] q n ) q n ) [ n ] q n 3 + 2 α q 2 [ n ] q n 2 [ n 1 ] q n q n 5 G β q n ( n ) 2 F 2 q n ( n ) } x + [ 2 ] q n [ n ] q n 2 q n 3 G β q n ( n ) 2 F 2 q n ( n ) 1 + x 2 + + 2 α q n 2 [ n ] q n [ n 2 ] q n q n 3 G β q n ( n ) 2 F 2 q n ( n ) + α 2 G β q n ( n ) 2 | 1 + x 2 } sup x [ 0 , ) 1 + x + x 2 1 + x 2 o ( 1 ) ,

we get lim n B n , q n ( α , β ) ( e 2 ) e 2 ρ = 0 . Thus, from AD Gadzhiev’s theorem in [26], we obtain the desired result of Theorem 2. □

Declarations

Acknowledgements

The author would like thank the referee for many helpful comments.

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, Gazi University, Teknikokullar, Ankara, 06500, Turkey

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