STATISTICAL APPROXIMATION FOR GENUINE MIXED INTEGRAL TYPE OPERATORS

International Journal of Mathematics and Computer Applications Research (IJMCAR) ISSN (P): 2249–6955; ISSN (E): 2249–8060 Vol. 11, Issue 2, Dec 2021, 1–10 © TJPRC Pvt. Ltd.

STATISTICAL APPROXIMATION FOR GENUINE MIXED INTEGRAL TYPE OPERATORS PRERNA SHARMA SVP University of Agriculture and Technology, Meerut, Uttar Pradesh, India ABSTRACT This paper deals with the study the q-analogue of integral type operators namely Szasz-Mirakyan-Baskakov-Stancu operators. For finding the moments, we apply q-derivative and q-Beta functions of these operators. We also examine some estimations e.g. weighted approximation, rate of convergence and point-wise convergence.

AMS Subject Classification: 41A25, 41A30 KEY WORDS: Baskakov Operators, q-Beta Function, q-Analogue & Weighted Approximation

Received: Jun 12, 2021; Accepted: Jul 02, 2021; Published: Jul 24, 2021; Paper Id.: IJMCARDEC20211

Foundation In approximation theory, familyof linear positive operators with the combination of the well known SzaszMirakyan operators and Baskakov operators was established by Gupta and Srivastava [10]. These operators are defined as follows ∞

𝑆𝑛 (𝑓, 𝑥) = (𝑛 − 1) ∑∞ 𝑢=0 𝑙𝑛,𝑢 (𝑥) ∫0 𝑝𝑛,𝑢 (𝑡)𝑓(𝑡)𝑑𝑡, 𝑥 ∈ [0, ∞) where, 𝑙𝑛,𝑢 (𝑥) =

𝑒 −𝑛𝑥(𝑛𝑥)𝑢 𝑢!

Original Article

1. INTRODUCTION

(1.1)

𝑡𝑢 𝑛+𝑢−1 ). , 𝑝𝑛,𝑢 (𝑡) = ( 𝑛+𝑢 .

(1+𝑡) 𝑢

Later on, other approximation properties on such integral type operators were studied by several other researchers such as [7], [9], [12], [18] and [19] etc. In [1], authors have given the rate of convergence for functions including derivatives of bounded variation. Since last few years, application of q-calculus has been a wide area of research for positive linear operators. Manyq-types of generalizations ofintegral type operators and their approximation behaviors were intensively described. q-analogue of various integral operators such as Szasz -Mirakyan operators, Szasz -Mirakyan Beta operators and other operators were proposed by [2], [3], [8], [15], [17], [16]. q-analogue of above operators were proposed and established by [11] as follows 𝑞

∞

𝑞

𝑞

𝑢 𝐴 𝑆𝑛 (𝑓(𝑡), 𝑥) = [𝑛 − 1]𝑞 ∑∞ 𝑢=0 𝑙𝑛,𝑢 (𝑥)𝑞 ∫0 𝑝𝑛,𝑢 (𝑡)𝑓(𝑡)𝑑𝑞 𝑡, 𝑞

where 𝑙𝑛,𝑢 (𝑥) =

𝑢

([𝑛]𝑞 𝑥) [𝑢]𝑞 !

𝑞

𝑢(𝑢−1) 2

(1.2)

𝑢(𝑢−1) 𝑡𝑢 𝑞 (𝑡) = [𝑛 + 𝑢 − 1] 𝑞 2 and 𝑝 . 𝑛,𝑢 𝐸𝑞 ([𝑛]𝑞 𝑥) (1+𝑡)𝑛+𝑢 𝑢 𝑞 𝑞

1

For 𝛼 ∈ [0, 𝛽] and 𝑛 ∈ 𝑁, 0 < 𝑞 < 1, the Stancu- generalization of above operators (1.2) are given below

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editor@tjprc.org

2

Prerna Sharma 𝑞

∞

𝑞

[𝑛]𝑞 𝑡+𝛼

𝑞

𝑢 𝐴 𝑆𝑛,𝛼,𝛽 (𝑓(𝑡), 𝑥) = [𝑛 − 1]𝑞 ∑∞ 𝑢=0 𝑙𝑛,𝑢 (𝑥)𝑞 ∫0 𝑝𝑛,𝑢 (𝑡)𝑓 ([𝑛]

𝑞 𝑡+𝛽

𝑞

) 𝑑𝑞 𝑡,

(1.3)

𝑞

Where 𝑙𝑛,𝑢 (𝑥) and 𝑝𝑛,𝑢 (𝑡) are defined above.

Remark Operators defined in equations (1.1) and (1.2) are linear, when q = 1, α = 0 andβ = 0, then operators (1.3) become the operators discussed in [7].

Here, we specify some notations of q-calculus, these notations can also be studied in [4], [13].

We have, [𝑛]𝑞 =

1−𝑞𝑛 1−𝑞

, for every 𝑛 ∈ 𝑁.

The q-derivative 𝐷𝑞 𝑓 of a function 𝑓 is given by 𝑓(𝑥)−𝑓(𝑞𝑥)

(𝐷𝑞 𝑓)(𝑥) =

(1−𝑞)𝑥

,𝑥 ≠ 0

(1.4)

The q-proper integral are defined as (see [13]) ∞

∞

∫ 𝑓(𝑥)𝑑𝑞 𝑥 = (1 − 𝑞) ∑ 𝑓 ( 𝑛=−∞

0

𝑞𝑛 𝑞𝑛 ) ,𝐴 > 0 𝐴 𝐴

∞

𝑛 𝑛 and ∫0 𝑓(𝑥)𝑑𝑞 𝑥 = (1 − 𝑞)𝑎 ∑∞ 𝑛=−∞ 𝑓(𝑎𝑞 ) 𝑞 , 𝑎 > 0

(1.5)

We have given some auxiliary propositions named as lemmas, which are very helpfulfor the proof of our main results. We obtain approximation behavior of the operators (1.3) in the form of second order modulus of smoothness and classical modulus of continuity. We also present uniform convergence theorems via weighted approximation for the functions belonging to weighted spaces. Lastly, we prove the point-wise estimates for the functions satisfying the Lipschitz conditions.

2. AUXILIARY RESULTS In this section we prove some auxiliary propositions (lemmas) Lemma 1. [11] Moments for these operators are given as 𝑞



𝑆𝑛 (1, 𝑥) = 1



𝑆𝑛 (𝑡, 𝑥) = 𝑞2[𝑛−2] + 𝑞[𝑛−2] , 𝑛 > 2



𝑆𝑛 (𝑡 2 , 𝑥) = 𝑞6 [𝑛−2]

𝑞

[𝑛]𝑞

1

𝑞

𝑞

[𝑛]2 𝑞

𝑞

𝑞 [𝑛−3]𝑞

[𝑛]𝑞 [1+𝑞]2

[2]𝑞

𝑥 2 + 𝑞5[𝑛−2]

𝑞 [𝑛−3]𝑞

=

𝑥 + 𝑞3[𝑛−2]

𝑞 [𝑛−3] 𝑞

,

[𝑛]2𝑞 𝑥 2 + 𝑞[𝑛]𝑞 [1 + 𝑞]2 𝑥 + 𝑞 3 [2]𝑞 , 𝑛 > 3. 𝑞 6 [𝑛 − 2]𝑞 [𝑛 − 3]𝑞

Lemma 2. The central moments are given as 𝛼,𝛽

𝑞

𝑞

∞

𝑞

[𝑛]𝑞 𝑡+𝛼

𝑢 𝐴 𝜇𝑛,𝑚 (𝑥) = 𝑆𝑛,𝛼,𝛽 (𝑡 𝑚 , 𝑥) = [𝑛 − 1]𝑞 ∑∞ 𝑢=0 𝑙𝑛,𝑢 (𝑥)𝑞 ∫0 𝑝𝑛,𝑢 (𝑡)𝑓 ([𝑛]

𝑞

Impact Factor (JCC): 7.3195

𝑚

) 𝑑𝑞 𝑡, then 𝑡+𝛽

NAAS Rating: 3.76

Statistical Approximation for Genuine Mixed Integral Type Operators 𝛼,𝛽

𝑞

𝛼,𝛽

𝑞



𝜇𝑛,0 (𝑥) = 𝑆𝑛,𝛼,𝛽 (1, 𝑥) = 1



𝜇𝑛,1 (𝑥) = 𝑆𝑛,𝛼,𝛽 (𝑡, 𝑥) =



𝜇𝑛,2 (𝑥) = 𝑆𝑛,𝛼,𝛽 (𝑡 2 , 𝑥) = (

𝛼,𝛽

[𝑛]2 𝑞 2 𝑞 [𝑛−2]𝑞 ([𝑛]𝑞 +𝛽) [𝑛]𝑞

𝑞

2

) [

𝑥+

3

[𝑛]𝑞

+

𝑞[𝑛−2]𝑞 ([𝑛]𝑞 +𝛽)

𝛼 ([𝑛]𝑞 +𝛽)

2 2 3 [𝑛]2 𝑞 𝑥 +𝑞[𝑛]𝑞 𝑥[1+𝑞] +[2]𝑞 𝑞

𝑞6 [𝑛−2]𝑞 [𝑛−3]𝑞

([𝑛]𝑞 +𝛽)

,𝑛 > 2

] 2

2𝛼[𝑛]𝑞

[𝑛]𝑞 𝑥 + 𝑞 [𝑛]𝑞 ]+( ) , 𝑛 > 3.

2 [ 2 [𝑛 − 2]𝑞 ([𝑛]𝑞 + 𝛽) ([𝑛]𝑞 + 𝛽) 𝑞

+ 𝛼,𝛽

Proof: Following [11], we have𝜇𝑛,0 (𝑥) = 1, 𝛼,𝛽

∞

𝑞

[𝑛]𝑞 𝑡+𝛼

𝑞

𝑢 𝐴 𝜇𝑛,1 (𝑥) = [𝑛 − 1]𝑞 ∑∞ 𝑢=0 𝑙𝑛,𝑢 (𝑥)𝑞 ∫0 𝑝𝑛,𝑢 (𝑡)𝑓 ([𝑛]

𝑞 𝑡+𝛽

[𝑛]𝑞

=

([𝑛]𝑞 +𝛽)

𝛼

𝑞

𝑆𝑛 (𝑡, 𝑥) +

[𝑛]2 𝑞 2 𝑞 [𝑛−2]𝑞 ([𝑛]𝑞 +𝛽)

𝑞

𝑆𝑛 (1, 𝑥) =

([𝑛]𝑞 +𝛽)

𝛼,𝛽

∞

𝑞

) 𝑑𝑞 𝑡 𝑥+

[𝑛]𝑞 𝑡+𝛼

𝑞

𝑢 𝐴 also, 𝜇𝑛,2 (𝑥) = [𝑛 − 1]𝑞 ∑∞ 𝑢=0 𝑙𝑛,𝑢 (𝑥)𝑞 ∫0 𝑝𝑛,𝑢 (𝑡)𝑓 ([𝑛] 2

[𝑛]𝑞

= (([𝑛]

𝑞

[𝑛]𝑞

= (([𝑛]

2𝛼[𝑛]𝑞

) 𝑆𝑛𝑞 (𝑡 2 , 𝑥) + +𝛽) ([𝑛]

𝑞

2

) [

𝑞 +𝛽)

+𝛽)2

𝑞6 [𝑛−2]𝑞 [𝑛−3]𝑞

𝑞

]+

2𝛼[𝑛]𝑞 2

([𝑛]𝑞 +𝛽)

[

𝑞

+

𝛼

,

([𝑛]𝑞 +𝛽)

2

) 𝑑𝑞 𝑡 𝑡+𝛽

2

𝛼

𝑞

𝑆𝑛 (𝑡, 𝑥) + (([𝑛]

2 2 3 [𝑛]2 𝑞 𝑥 +𝑞[𝑛]𝑞 𝑥[2] 𝑞 +[2]𝑞 𝑞

[𝑛]𝑞 𝑞[𝑛−2]𝑞 ([𝑛]𝑞 +𝛽)

) 𝑆𝑛𝑞 (1, 𝑥) +𝛽) [𝑛]𝑞 𝑥+𝑞

𝑞2 [𝑛−2]𝑞

]+(

𝛼 ([𝑛]𝑞

2

) . +𝛽)

Corollary 1. Using Lemma 2, we set 𝑞

𝑆𝑛,𝛼,𝛽 ((𝑡 − 𝑥)2 , 𝑥) = ( +( +

[𝑛]4 𝑞 2 6 𝑞 [𝑛−2]𝑞 [𝑛−3]𝑞 ([𝑛]𝑞 +𝛽)

2 [𝑛]3 𝑞 [2]𝑞 2 5 𝑞 [𝑛−2]𝑞 [𝑛−3]𝑞 ([𝑛]𝑞 +𝛽)

[𝑛]2 𝑞 [2]𝑞 2 3 [𝑛−2] [𝑛−3] 𝑞 𝑞 𝑞 ([𝑛]𝑞 +𝛽)

and𝛾𝑛 (𝑞, 𝑥) =

+

+

− 𝑞2[𝑛−2]

2[𝑛]2 𝑞𝛼 2 2 𝑞 [𝑛−2]𝑞 [𝑛−3]𝑞 ([𝑛]𝑞 +𝛽) 2𝛼[𝑛]𝑞 2

𝑞[𝑛−2]𝑞 ([𝑛]𝑞 +𝛽)

+

2[𝑛]𝑞

− 𝑞[𝑛−2]

𝑞 ([𝑛]𝑞 +𝛽)

𝛼2 2

([𝑛]𝑞 +𝛽)

([2]𝑞 [𝑛]𝑞 −𝑞2 𝛽[𝑛−2]𝑞 )𝑥+𝑞[𝑛]𝑞 +𝛼𝑞2 [𝑛−2]𝑞 𝑞2 [𝑛−2]𝑞 ([𝑛]𝑞 +𝛽)

2[𝑛]2 𝑞

𝑞 [𝑛−3]𝑞 ([𝑛]𝑞 +𝛽)

+ 1) 𝑥 2 2𝛼

− ([𝑛]

)𝑥

𝑞 +𝛽)

= 𝛿𝑛 (𝑞, 𝑥)

.

3. MAIN RESULTS Definition: Let 𝐶𝐵 [0, ∞)denotes the space of all functions f which are continuous and bounded on the interval[0, ∞) endowed with the norm ‖𝑓‖ = sup |𝑓(𝑥)|. 𝑥∈[0,∞)

We define the Peetre’s K-functional as, 𝐾2 (𝑓, 𝛿) =

inf

{‖𝑓 − 𝑗‖ + 𝛿 ‖𝑗′′‖},

2 [0,∞) 𝑗∈𝐶𝐵

where, 𝐶𝐵2 [0, ∞) = {𝑗 ∈ 𝐶𝐵 [0, ∞): 𝑗 ′ , 𝑗′′ ∈ 𝐶𝐵 [0, ∞)}.By [5], ∃ an absolute positive constant “C” such that 𝐾2 (𝑓, 𝛿) ≤ 𝐶𝜔2 (𝑓, √𝛿), www.tjprc.org

(3.1) editor@tjprc.org

4

Prerna Sharma

where 𝛿 > 0, we write the second order modulus of smoothness as below 𝜔2 (𝑓, 𝛿) = sup

sup |𝑓(𝑥 + 2ℎ) − 2𝑓(𝑥 + ℎ) + 𝑓(𝑥)|.

0≤ℎ≤𝛿 𝑥∈[0,∞)

In this continuation the usual modulus of continuity of 𝑓 ∈ 𝐶𝐵 [0, ∞) is given as 𝜔(𝑓, 𝛿) = sup sup |𝑓(𝑥 + ℎ) − 𝑓(𝑥)|.

(3.2)

0≤ℎ≤𝛿 𝑥∈[0,∞)

Theorem 1. Let 𝑓 ∈ 𝐶𝐵 [0, ∞), then for all 𝑗 ∈ 𝐶𝐵2 [0, ∞), we have ̅ 𝑞 (𝑗, 𝑥) − 𝑗(𝑥)| ≤ (𝛿𝑛 (𝑞, 𝑥) + 𝛾𝑛2 (𝑞, 𝑥))‖𝑗′′‖, |𝑆𝑛,𝛼,𝛽

(3.3)

2

[𝑛]𝑞 𝑥 ̅ 𝑞 (𝑓, 𝑥) = 𝑆 𝑞 ( where,𝑆𝑛,𝛼,𝛽 𝑛,,𝛼,𝛽 + 𝑓(𝑥) − 𝑓 𝑞2 [𝑛−2] ([𝑛]

[𝑛]𝑞

𝑞 +𝛽)

𝑞

𝑥 + 𝑞[𝑛−2]

𝑞 ([𝑛]𝑞 +𝛽)

𝛼

+ ([𝑛]

).

(3.4)

𝑞 +𝛽)

Proof: From (3.4) and Lemma 2, we have ̅𝑆 𝑞 (𝑓, 𝑥)(𝑡 − 𝑥, 𝑥) = 0 𝑛,𝛼,𝛽

(3.5)

Suppose 𝑥 ∈ [0, ∞) and 𝑗 ∈ 𝐶𝐵2 [0, ∞). Using Taylor’s expansion 𝑡

𝑗(𝑡) − 𝑗(𝑥) = (𝑡 − 𝑥)𝑗 ′ (𝑥) + ∫𝑥 (𝑡 − 𝑦)𝑗 ′′ (𝑦)𝑑𝑦. ̅𝑞 Applying the operators 𝑆𝑛,𝛼,𝛽 to both the sides of above equality and taken in to account (3.5), we obtain ̅ 𝑞 (𝑗, 𝑥) − 𝑗(𝑥) = 𝑆̅ 𝑞 ((𝑡 − 𝑥)𝑗 ′ (𝑥), 𝑥) + 𝑆̅ 𝑞 (∫𝑡(𝑡 − 𝑦)𝑗 ′′ (𝑦)𝑑𝑦, 𝑥) 𝑆𝑛,𝛼,𝛽 𝑛,𝛼,𝛽 𝑛,𝛼,𝛽 𝑥 𝑞

𝑡

𝑞

̅ ((𝑡 − 𝑥)𝑗 ′ (𝑥), 𝑥) + 𝑆𝑛,𝛼,𝛽 (∫𝑥 (𝑡 − 𝑦)𝑗 ′′ (𝑦)𝑑𝑦, 𝑥) = 𝑗 ′ (𝑥)𝑆𝑛,𝛼,𝛽 [𝑛]2 𝛼 𝑞 𝑥+𝑞[𝑛]𝑞 𝑥+ ([𝑛]𝑞 +𝛽) 𝑞2 [𝑛−2]𝑞 ([𝑛]𝑞 +𝛽)

− ∫𝑥

=

𝑞 𝑆𝑛,𝛼,𝛽

𝑡 (∫𝑥 (𝑡

− 𝑦)𝑗

(

[𝑛]2 𝑞 𝑥+𝑞[𝑛]𝑞 𝑞2 [𝑛−2]𝑞 ([𝑛]𝑞 +𝛽)

′′ (𝑦)𝑑𝑦,

+

𝛼 ([𝑛]𝑞 +𝛽)

− 𝑦) 𝑗 ′′ (𝑦)𝑑𝑦

[𝑛]2 𝛼 𝑞 𝑥+𝑞[𝑛]𝑞 𝑥+ ([𝑛]𝑞 +𝛽) 𝑞2 [𝑛−2]𝑞 ([𝑛]𝑞 +𝛽)

𝑥 ) − ∫𝑥

𝑡

(

[𝑛]2 𝑞 𝑥+𝑞[𝑛]𝑞 𝑞2 [𝑛−2]𝑞 ([𝑛]𝑞 +𝛽)

𝑡

𝛼

+ ([𝑛]

𝑞 +𝛽)

− 𝑦) 𝑗 ′′ (𝑦)𝑑𝑦.

𝑡

We get, |∫𝑥 (𝑡 − 𝑦)𝑗 ′′ (𝑦)𝑑𝑦 | ≤ |∫𝑥 (𝑡 − 𝑦)|𝑗 ′′ (𝑦)|𝑑𝑦 | ≤ ‖𝑗′′‖ |∫𝑥 (𝑡 − 𝑦)𝑑𝑦| ≤ (𝑡 − 𝑥)2 |𝑗′′|, and [𝑛]2 𝛼 𝑞 𝑥+𝑞[𝑛]𝑞 2 [𝑛−2] ([𝑛] +𝛽)𝑥+([𝑛]𝑞 +𝛽) 𝑞 𝑞

|∫𝑥𝑞

[𝑛]2 𝑞 𝑥+𝑞[𝑛]𝑞

≤ (𝑞2[𝑛−2]

𝑞 ([𝑛]𝑞

(

[𝑛]2 𝑞 𝑥+𝑞[𝑛]𝑞 2 𝑞 [𝑛−2]𝑞 ([𝑛]𝑞 +𝛽)

𝑞

𝛼 ([𝑛]𝑞 +𝛽)

− 𝑦) 𝑗 ′′ (𝑦)𝑑𝑦|

2

𝛼

+ ([𝑛] +𝛽)

+

− 𝑦) ‖𝑗′′‖ = 𝛾𝑛2 (𝑞, 𝑥)‖𝑗 ′′ ‖. +𝛽)

𝑞

𝑡

𝑞

̅ ̅ (𝑗, 𝑥) − 𝑗(𝑥)| = |𝑆𝑛,𝛼,𝛽 (∫𝑥 (𝑡 − 𝑦)𝑗 ′′ (𝑦)𝑑𝑦, 𝑥 ) We follow that, |𝑆𝑛,𝛼,𝛽 [𝑛]2 𝛼 𝑞 𝑥+𝑞[𝑛]𝑞 𝑥+ ([𝑛]𝑞 +𝛽) 𝑞2 [𝑛−2]𝑞 ([𝑛]𝑞 +𝛽)

− ∫𝑥

𝑞

(

[𝑛]2 𝑞 𝑥+𝑞[𝑛]𝑞 𝑞2 [𝑛−2]𝑞 ([𝑛]𝑞 +𝛽)

≤ 𝑆𝑛,,𝛼,𝛽 ((𝑡 − 𝑥)2 ‖𝑗 ′′ ‖, 𝑥) + (

Impact Factor (JCC): 7.3195

𝛼

+ ([𝑛]

[𝑛]2 𝑞 𝑥+𝑞[𝑛]𝑞 𝑞2 [𝑛−2]𝑞 ([𝑛]𝑞 +𝛽)

𝑞 +𝛽)

+

− 𝑦) 𝑗 ′′ (𝑦)𝑑𝑦|

𝛼 ([𝑛]𝑞 +𝛽)

2

− 𝑥) ‖𝑗′′‖ = 𝛿𝑛 (𝑞, 𝑥) + 𝛾𝑛2 (𝑞, 𝑥)‖𝑗 ′′ ‖.

NAAS Rating: 3.76

Statistical Approximation for Genuine Mixed Integral Type Operators

5

Theorem 2.Suppose𝑓 ∈ 𝐶𝐵 [0, ∞), then for all∈ [0, ∞), ∃ a constant 𝐾 > 0, such that 𝑞 (𝑓, 𝑥) − 𝑓(𝑥)| ≤ 𝐾𝜔2 (𝑓, √𝛿𝑛 (𝑞, 𝑥) + 𝛾𝑛2 (𝑞, 𝑥))𝜔(𝑓, 𝛾𝑛 (𝑞, 𝑥)) |𝑆𝑛,𝛼,𝛽

Proof. From (3.4), for 𝑗 ∈ 𝐶𝐵2 [0, ∞), we can get 𝑞 ̅ 𝑞 (𝑓, 𝑥) − 𝑓(𝑥)| + |𝑓(𝑥) − 𝑓 ( (𝑓, 𝑥) − 𝑓(𝑥)| ≤ |𝑆𝑛,𝛼,𝛽 |𝑆𝑛,𝛼,𝛽

𝑞

̅ (𝑓, −𝑗, 𝑥) − (𝑓 − 𝑗)(𝑥)| + |𝑓(𝑥) − 𝑓 ( ≤ |𝑆𝑛,𝛼,𝛽

[𝑛]2 𝑞 𝑥+𝑞[𝑛]𝑞

𝛼

𝑞2 [𝑛−2]𝑞 ([𝑛]𝑞 +𝛽)

[𝑛]2 𝑞 𝑥+𝑞[𝑛]𝑞 2 𝑞 [𝑛−2]𝑞 ([𝑛]𝑞 +𝛽)

+

𝛼

)|

+ ([𝑛]

𝑞 +𝛽)

̅ 𝑞 (𝑗, 𝑥) − 𝑗(𝑥)| )|+|𝑆𝑛,𝛼,𝛽

([𝑛]𝑞 +𝛽)

[𝑛]2

̅ 𝑞 (𝑓, −𝑗, 𝑥)| + |(𝑓 − 𝑗)(𝑥)| + |𝑓(𝑥) − 𝑓 ( 2 𝑞𝑥+𝑞[𝑛]𝑞 + 𝛼 )|+|𝑆̅ 𝑞 (𝑗, 𝑥) − 𝑗(𝑥)|. ≤ |𝑆𝑛,𝛼,𝛽 𝑛,𝛼,𝛽 𝑞 [𝑛−2] ([𝑛] +𝛽) ([𝑛] +𝛽) 𝑞

𝑞

𝑞

𝑞

̅ Taking boundedness of the operators 𝑆𝑛,𝛼,𝛽 and using (3.3), we get [𝑛]2 𝑞 𝑥+𝑞[𝑛]𝑞

𝑞

|𝑆𝑛,𝛼,𝛽 (𝑓, 𝑥) − 𝑓(𝑥)| ≤ 4‖𝑓 − 𝑗‖ + |𝑓(𝑥) − 𝑓 (

𝑞2 [𝑛−2]𝑞 ([𝑛]𝑞 +𝛽)

+

𝛼

)|

([𝑛]𝑞 +𝛽)

+(𝛿𝑛 (𝑞, 𝑥) + 𝛾𝑛2 (𝑞, 𝑥))‖𝑗 ′′ ‖ ≤ 4‖𝑓 − 𝑗‖ + (𝛿𝑛 (𝑞, 𝑥) + 𝛾𝑛2 (𝑞, 𝑥))‖𝑗 ′′ ‖ + 𝜔(𝑓, 𝛾𝑛 (𝑞, 𝑥)).

Now over all𝑗 ∈ 𝐶𝐵2 [0, ∞), considering infimum on right hand side and applying equation (3.1), we follow 𝑞 (𝑓, 𝑥) − 𝑓(𝑥)| ≤ 4𝐾2 (𝑓, 𝛿𝑛 (𝑞, 𝑥) + 𝛾𝑛2 (𝑞, 𝑥)) + 𝜔(𝑓, 𝛾𝑛 (𝑞, 𝑥)) |𝑆𝑛,𝛼,𝛽

≤ 4𝑀𝜔2 (𝑓, √𝛿𝑛 (𝑞, 𝑥) + 𝛾𝑛2 (𝑞, 𝑥)) + 𝜔(𝑓, 𝛾𝑛 (𝑞, 𝑥)) ≤ 𝐾𝜔2 (𝑓, √𝛿𝑛 (𝑞, 𝑥) + 𝛾𝑛2 (𝑞, 𝑥)) + 𝜔(𝑓, 𝛾𝑛 (𝑞, 𝑥)), where K=4M>0.

Theorem 3. Let f be a function which is bounded and integrable on the interval [0, ∞). For 𝑞 = 𝑞𝑛 ∈ (0, 1),suppose second order derivative offunction f exists at a fixed point 𝑥 ∈ [0, ∞), such that 𝑞𝑛 → 1 𝑎𝑠 𝑛 → ∞, than 𝑥2

𝑞

𝑛 (𝑓, 𝑥) − 𝑓(𝑥)] = [(2 − 𝛽)𝑥 + 1 − 𝛼]𝑓 ′ (𝑥) + ( + 𝑥) 𝑓 ′′ (𝑥). lim [𝑛]𝑞𝑛 [𝑆𝑛,𝛼,𝛽 2

𝑛→∞

Proof: For the proof of above theorem, we use Taylor’s expansion 1

𝑓(𝑡) − 𝑓(𝑥) = (𝑡 − 𝑥)𝑓 ′ (𝑥) + (𝑡 − 𝑥)2 (2 𝑓 ′′ (𝑥) + 𝜀(𝑡 − 𝑥)), 𝑞

Where 𝜀 is bounded andlim 𝜀(𝑡) = 0. Now applying the operators 𝑆𝑛,𝛼,𝛽 (𝑓, 𝑥) to the above relation, we have 𝑡→0

𝑞

1

𝑞

𝑞

𝑞

𝑛 𝑛 𝑛 ((𝑡 − 𝑥)2 , 𝑥) + 𝑆𝑛,𝛼,𝛽 (𝜀(𝑡 − 𝑥)(𝑡 − 𝑥)2 , 𝑥) ((𝑡 − 𝑥), 𝑥) + 𝑓 ′′ (𝑥)𝑆𝑛,𝛼,𝛽 𝑆𝑛,𝛼,𝛽 (𝑓, 𝑥) − 𝑓(𝑥) = 𝑓 ′ (𝑥)𝑆𝑛,𝛼,𝛽

2

1

𝑞

𝑛 (𝜀(𝑡 − 𝑥)(𝑡 − 𝑥)2 , 𝑥), = 𝑓 ′ (𝑥)𝛾𝑛 (𝑞𝑛 , 𝑥) + 2 𝑓 ′′ (𝑥)𝛿𝑛 (𝑞𝑛 , 𝑥) + 𝑆𝑛,𝛼,𝛽

where 𝛾𝑛 (𝑞𝑛 , 𝑥) 𝑎𝑛𝑑 𝛿𝑛 (𝑞𝑛 , 𝑥) can be seen in equation (3.2).

Now using inequality of Cauchy-Schwarz, we have 𝑞

𝑞

1/2

𝑛 𝑛 (𝜀(𝑡 − 𝑥)(𝑡 − 𝑥)2 , 𝑥) ≤ (𝑆𝑛,𝛼,𝛽 [𝑛]𝑞 𝑆𝑛,𝛼,𝛽 (𝜀 2 (𝑡 − 𝑥)))

1/2

𝑞𝑛 ([𝑛]2𝑞𝑛 𝑆𝑛,𝛼,𝛽 ((𝑡 − 𝑥)4 , 𝑥))

𝑞

𝑛 ((𝑡 − 𝑥)4 , 𝑥) = 0, Using Lemma 2, we have lim [𝑛]2𝑞𝑛 𝑆𝑛,𝛼,𝛽

𝑛→∞

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Prerna Sharma

since lim 𝛾𝑛 (𝑞𝑛 , 𝑥) = (2 − 𝛽)𝑥 + 1 − 𝛼 and lim 𝛼𝑛 (𝑞𝑛 , 𝑥) = 𝑥 2 + 2𝑥. 𝑛→∞

𝑛→∞

This is the required result.

Definition 2: Let 𝐻𝑥 2 [0, ∞)represents the set of all available functions fwhich are defined on the interval [0, ∞), satisfying the condition |𝑓(𝑥)| ≤ 𝑀𝑓 (1 + 𝑥 2 ), where 𝑀𝑓 is a constant depending only on f and 𝐶𝑥 2 [0, ∞) is the subspace of all functions which are continuous and belongs to 𝐻𝑥 2 [0, ∞). Let 𝐶𝑥∗2 [0, ∞) is the subspace of all functionsf belonging to𝐶𝑥 2 [0, ∞), for which lim

𝑓(𝑥)

𝑥→∞ 1+𝑥 2

is finite.We define norm on 𝐶𝑥∗2 [0, ∞) as below

|𝑓(𝑥)|

‖𝑓‖𝑥 2 = sup

2 𝑥∈[0,∞) 1+𝑥

.

Theorem 4. Let 𝑞 = 𝑞𝑛 satisfies 𝑞𝑛 ∈ (0, 1) and let 𝑞𝑛 → 1 𝑎𝑠 𝑛 → ∞. Then for each 𝑓 ∈ 𝐶𝑥 2 [0, ∞), we have 𝑞𝑛 (𝑓) − 𝑓‖ lim ‖𝑆𝑛,𝛼,𝛽

𝑛→∞

Proof:

= 0.

𝑥2

Following the weighted Korokvin theorem in [6], we see it 𝑞

𝑛 (𝑡 𝑘 , 𝑥) − 𝑥 𝑘 ‖ followinglim ‖𝑆𝑛,𝛼,𝛽

𝑛→∞

𝑥2

will be enough to verify the

= 0, where k is a positve integer and𝑘 ∈ [0, 2].

(3.6)

𝑞

𝑛 (1, 𝑥) = 1, so theorem holds for 𝑘 = 0, to get the proof of the theorem, it is suffices to show that Since 𝑆𝑛,𝛼,𝛽

𝑞

𝑛 (𝑡 𝑘 , 𝑥) = 𝑥 𝑘 , 𝑘 = 1, 2. 𝑆𝑛,𝛼,𝛽

𝑞

𝑛 (𝑡, 𝑥) − 𝑥‖ Now ‖𝑆𝑛,𝛼,𝛽

≤

≤ sup

𝑥2

2 2 ([𝑛]2 𝑞 −𝑞 [𝑛−2]𝑞 ([𝑛]𝑞 +𝛽)𝑥+𝑞[𝑛]𝑞 +𝛼𝑞 [𝑛−2]𝑞

1

𝑞2 [𝑛−2]𝑞 ([𝑛]𝑞 +𝛽)

1+𝑥 2

𝑥∈[0,∞)

([2]𝑞 [𝑛]𝑞 [−𝑞2 [𝑛−2]𝑞 𝛽)+𝑞[𝑛]𝑞 +𝛼𝑞2 [𝑛−2]𝑞 𝑞2 [𝑛−2]𝑞 ([𝑛]𝑞 +𝛽)

2 𝑥∈[0,∞) 1+𝑥

𝑞𝑛 (𝑡, 𝑥) − 𝑥‖ which implies lim ‖𝑆𝑛,𝛼,𝛽 𝑛→∞

𝑞

𝑛 (𝑡 2 , 𝑥) − 𝑥 2 ‖ Also ‖𝑆𝑛,𝛼,𝛽

1

+ sup

1+𝑥 2

𝑥∈[0,∞)

=(

+[

[

2

𝛼

2 [𝑛]𝑞 𝑥[2]2 𝑞 +𝑞 [2]𝑞

𝑞5 [𝑛−2]𝑞 [𝑛−3]𝑞 [𝑛]2 𝑞 6 [𝑛−2] 𝑞 𝑞 [𝑛−3]𝑞

) (

([𝑛]𝑞 +𝛽)

[𝑛]𝑞 (1+𝑞)2 5 [𝑛−2] 𝑞 𝑞 [𝑛−3]𝑞

𝛼

+ [(([𝑛]

𝑞

𝑥2

+

2

2[𝑛]𝑞 𝛼

𝑥+1

sup

= 0.

𝑥2

≤ sup

,

𝑥2

𝑥∈[0,∞) 1+𝑥

( 2

[𝑛]𝑞

2

) (

𝑞6 [𝑛−2]𝑞 [𝑛−3]𝑞

([𝑛]𝑞 +𝛽)

2[𝑛]𝑞 𝛼 [𝑛]𝑞 𝑥+𝑞 2 ([𝑛] +𝛽) 𝑞2 [𝑛−2] 1+𝑥 𝑞 𝑞 𝑥∈[0,∞) 1

] + sup

−(

([𝑛]𝑞 +𝛽)

[𝑛]𝑞

[𝑛]𝑞

[

2

) ) sup

[𝑛]2 𝑞

𝑥2

2 𝑥∈[0,∞) 1+𝑥

−(

] + sup 𝑥∈[0,∞)

([𝑛]𝑞 +𝛽) [𝑛]𝑞

1 1+𝑥 2

(

2

) ) 𝛼

2

)

([𝑛]𝑞 +𝛽)

,

𝑥

] sup

2 2 2 ([𝑛]𝑞 +𝛽) 𝑞 [𝑛−2]𝑞 𝑥∈[0,∞) 1+𝑥

[2]𝑞

) + 3[𝑛−2] +𝛽) 𝑞

𝑞 [𝑛−3] 𝑞

+ 2[𝑛]𝑞

𝛼 ([𝑛]𝑞 +𝛽)

𝑞𝑛 (𝑡 2 , 𝑥) − 𝑥 2 ‖ which implies that lim ‖𝑆𝑛,𝛼,𝛽 𝑛→∞

1

2

𝑥2

. 𝑞[𝑛−2] ] sup

1

2 𝑞 𝑥∈[0,∞) 1+𝑥

,

= 0.

Hence the proof.

Impact Factor (JCC): 7.3195

NAAS Rating: 3.76

Statistical Approximation for Genuine Mixed Integral Type Operators

7

Theorem 5. Suppose 𝛼 ∈ (0, 1] and let Sdenotes any bounded subset of[0, ∞). If 𝑓 ∈ 𝐶𝐵 (0, ∞) ∩ 𝐿𝑖𝑝𝑀(𝛼), then we 𝛼

𝑞

know,|𝑆𝑛,𝛼,𝛽 (𝑓, 𝑥) − 𝑓(𝑥)| ≤ 𝐾 {𝛿𝑛2 (𝑞, 𝑥) + 2(𝑑(𝑥, 𝑆))𝛼 }, where K is an arbitrary constant which depends on α and d(x, S) being the distance between x and subset S, which can be defined as 𝑑(𝑥, 𝑆) = 𝑖𝑛𝑓{(𝑡 − 𝑥): 𝑡 ∈ 𝑆𝑎𝑛𝑑𝑥 ∈ [0, ∞)} and 𝛿𝑛 (𝑞, 𝑥) are defined in Corollary 1.

Proof: Following the properties related to infimum, there exists at least one point y in the closure of S, y∈ S: 𝑑(𝑥, 𝑆) = |𝑦 − 𝑥|.

Taking in view the triangular inequality, we get |𝑓(𝑡) − 𝑓(𝑥)| ≤ |𝑓(𝑡) − 𝑓(𝑦) + |𝑓(𝑦) − 𝑓(𝑥)|| 𝑞

𝑞

Hence, |𝑆𝑛,𝛼,𝛽 (𝑡, 𝑥) − 𝑓(𝑥)| ≤ 𝑆𝑛,𝛼,𝛽 (|𝑓(𝑡) − 𝑓(𝑥)|, 𝑥) 𝑞

𝑞

≤ 𝑆𝑛,𝛼,𝛽 (𝑡, 𝑥)(|𝑓(𝑡) − 𝑓(𝑦)|, 𝑥) + 𝑆𝑛,𝛼,𝛽 (|𝑓(𝑦) − 𝑓(𝑥)|, 𝑥) 𝑞

𝑞

≤ 𝐾{𝑆𝑛,𝛼,𝛽 (|𝑡 − 𝑦 |𝛼 , 𝑥) + |𝑦 − 𝑥|𝛼 } ≤ 𝐾{𝑆𝑛,𝛼,𝛽 (|𝑡 − 𝑥|𝛼 + |𝑥 − 𝑦 |𝛼 , 𝑥) + |𝑦 − 𝑥|𝛼 } 𝑞

≤ 𝐾{𝑆𝑛,𝛼,𝛽 (|𝑡 − 𝑥|𝛼 , 𝑥) + 2|𝑦 − 𝑥|𝛼 }.

2

2

1

1

1

2

We opt𝑙1 = 𝛼 , 𝑙2 = 2−𝛼 𝑎𝑛𝑑 𝑤𝑒 𝑔𝑒𝑡 𝑙 + 𝑙 = 1, and then using Holder’s inequality, we have 1/𝑙1

𝑞 𝑞 (𝑡, 𝑥) − 𝑓(𝑥)| ≤ 𝐾 {[𝑆𝑛,𝛼,𝛽 (|𝑡 − 𝑥|𝛼𝑙1 , 𝑥)] |𝑆𝑛,𝛼,𝛽 𝛼/2

𝑞

= 𝐾 {[𝑆𝑛,𝛼,𝛽 (|𝑡 − 𝑥|2 , 𝑥)]

𝑞

1/𝑙2

× [𝑆𝑛,𝛼,𝛽 (1𝑙2 , 𝑥)]

+ 2|𝑦 − 𝑥|𝛼 }

𝛼

+ 2|𝑦 − 𝑥|2 } = 𝐾 {𝛿𝑛2 (𝑞, 𝑥) + 2(𝑑(𝑥, 𝑆))𝛼 }.

Hence, we get the proof of the theorem.

Now, to find local direct estimates of the operators defined in (1.3), following [14] and using Lipcshitz type maximal function of order α. 𝜔 ̅𝛼 (𝑓, 𝑥) =

sup 𝑡≠𝑥,𝑡∈[0,∞)

|𝑓(𝑡)−𝑓(𝑥)| |𝑡−𝑥| 𝛼

, 𝑥 ∈ (0, ∞] 𝑎𝑛𝑑 𝛼 ∈ (0, 1].

(3.7)

Theorem 6. Let 𝑓 ∈ 𝐶𝐵 [0, ∞), and 0 ≤ 𝛼 ≤ 1, then for every 𝑥 ∈ (0, ∞], we have 𝛼

𝑞

|𝑆𝑛,𝛼,𝛽 (𝑓, 𝑥) − 𝑓(𝑥)| ≤ 𝜔 ̅𝛼 (𝑓, 𝑥)𝛿𝑛2 (𝑥).

Proof: From the equation (3.7), we have 𝑞 𝑞 (𝑓, 𝑥) − 𝑓(𝑥)| ≤ 𝜔 |𝑆𝑛,𝛼,𝛽 ̅𝛼 (𝑓, 𝑥)𝑆𝑛,𝛼,𝛽 (𝑡, 𝑥)(|𝑡 − 𝑥|𝛼 , 𝑥).

2

2

𝛼

2−𝛼

Applying the Holder inequality, with𝑙1 = , 𝑙2 =

, we get 𝛼/2

𝑞 𝑞 (𝑓, 𝑥) − 𝑓(𝑥)| ≤ 𝜔 |𝑆𝑛,𝛼,𝛽 ̅𝛼 (𝑓, 𝑥)[𝑆𝑛,𝛼,𝛽 (|𝑡 − 𝑥|𝛼 , 𝑥)]

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𝛼

≤𝜔 ̅𝛼 (𝑓, 𝑥)𝛿𝑛2 (𝑥).

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Prerna Sharma

REFERENCES 1.

T. Acar, V. Gupta, A. Aral, Rate of convergence for generalized Szasz operators, Bull. Math. Sci. 1(2011), 99-113.

2.

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