Quantitative global estimates for generalized double Szasz-Mirakjan operators

Volume 2013, Article ID 613258,8pages http://dx.doi.org/10.1155/2013/613258

Research Article

Quantitative Global Estimates for Generalized Double

Szasz-Mirakjan Operators

Mehmet Ali Özarslan and Hüseyin Aktu

Llu

Eastern Mediterranean University, Gazimagusa, Cyprus, Mersin 10, Turkey

Correspondence should be addressed to Mehmet Ali ¨Ozarslan; mehmetali.ozarslan@emu.edu.tr Received 15 December 2012; Accepted 8 May 2013

Academic Editor: Jingxin Zhang

Copyright © 2013 M. Ali ¨Ozarslan and H. Aktu˘glu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We introduce the generalized double Sz´asz-Mirakjan operators in this paper. We obtain several quantitative estimates for these operators. These estimates help us to determine some function classesS (including some Lipschitz-type spaces) which provide uniform convergence on the whole domain[0, ∞) × [0, ∞).

1. Introduction

The well-known Sz´asz-Mirakjan operators are defined on the spaceA₁as follows: 𝑆_𝑛(𝑓; 𝑥) = 𝑒−𝑛𝑥∑∞ 𝑘=0 𝑓 (𝑘 𝑛) (𝑛𝑥)𝑘 𝑘! , (1)

whereA₁is the set of all real functions on[0, ∞) such that the right-hand side in (1) make sense for all𝑛 > 0 and 𝑥 ∈ [0, ∞). By modifying the Sz´asz-Mirakjan operators as

𝐷_𝑛(𝑓; 𝑥) = 𝑒−𝑛𝑢𝑛(𝑥) ∞ ∑ 𝑘=0 𝑓 (𝑘 𝑛) (𝑛𝑢_𝑛(𝑥))𝑘 𝑘! , (2)

where{𝑢_𝑛(𝑥)} is a sequence of real-valued, continuous func-tions defined on[0, ∞) with 0 ≤ 𝑢_𝑛(𝑥) < ∞, it has been shown in [1] that if one let

𝑢∗_𝑛(𝑥) := −1 + √4𝑛_2𝑛2𝑥2+ 1, 𝑛 ∈ N, (3) then the operators defined by

𝐷∗_𝑛(𝑓; 𝑥) := 𝑆_𝑛(𝑓; 𝑢_𝑛∗(𝑥)) (4) preserve the test function 𝑒₂(𝑥) = 𝑥2 and provide a better error estimation than the operators 𝑆_𝑛(𝑓; 𝑥) for all

𝑓 ∈ 𝐶_𝐵([0, ∞)) and for each 𝑥 ∈ [0, ∞). Note that 𝐶_𝐵([0, ∞)) denotes the space of all bounded and continuous functions on[0, ∞). On the other hand, by letting

V_𝑛(𝑥) := 𝑥 −_2𝑛1; 𝑛 ∈ N, (5) it has been shown in [2] that the operators defined by

𝑉_𝑛∗(𝑓; 𝑥) := 𝑆_𝑛(𝑓; V_𝑛(𝑥)) (6) do not preserve the test functions𝑒₁(𝑥) = 𝑥 and 𝑒₂(𝑥) = 𝑥2 but provide the best error estimation among all the Sz´asz-Mirakjan operators for all 𝑓 ∈ 𝐶_𝐵([0, ∞)) and for each 𝑥 ∈ [1/2, ∞). For the other linear positive operator families which preserve𝑒₂(𝑥) = 𝑥2, we refer [3–9]. On the other hand, in [10,11] the authors considered some operators preserving 𝑒₁(𝑥) = 𝑥.

introduced and investigated different variants of the general double Sz´asz-Mirakjan operators:

𝐷_𝑛(𝑓; 𝑥, 𝑦) : 𝑆_𝑛(𝑓; 𝑢_𝑛(𝑥) , V_𝑛(𝑦)) = 𝑒−𝑛(𝑢𝑛(𝑥)+V𝑛(𝑦)) × ∑∞ 𝑘,𝑙=0 𝑓 (𝑘 𝑛, 𝑙 𝑛) × (𝑛𝑢_𝑛(𝑥))𝑘 𝑘! (𝑛V_𝑛(𝑦))𝑙 𝑙! , 𝑓 ∈ A₂. (8) In [13], they considered the case of operators

𝑢_𝑛(1)(𝑥) := −1 + √4𝑛_2𝑛2𝑥2+ 1, V(1) 𝑛 (𝑦) := −1 + √4𝑛2_𝑦2_{+ 1} 2𝑛 , 𝑛 ∈ N, (9)

which preserve the test function𝑒_2,0(𝑥, 𝑦) + 𝑒_0,2(𝑥, 𝑦) := 𝑥2+ 𝑦2and provide a better error estimation than the operators 𝑆_𝑛(𝑓; 𝑥, 𝑦) for all 𝑓 ∈ 𝐶_𝐵([0, ∞) × [0, ∞)) and for each 𝑥, 𝑦 ∈ [0, ∞). On the other hand, in [14], they considered the case

𝑢(2)_𝑛 (𝑥, 𝛼) := − (𝑛𝛼 + 1) + √4𝑛2(𝑥2+ 𝛼𝑥) + (𝑛𝛼 + 1) 2 2𝑛 , V(2)_𝑛 (𝑦, 𝛽) := − (𝑛𝛽 + 1) + √4𝑛2(𝑦2+ 𝛽𝑦) + (𝑛𝛽 + 1) 2 2𝑛 , 𝑛 ∈ N, 𝛼, 𝛽 ∈ R. (10) Note that for this case, the operators 𝐷_𝑛(𝑓; 𝑥, 𝑦) do not preserve any test function (i.e.,𝑒_0,0(𝑥, 𝑦) = 1, 𝑒_1,0(𝑥, 𝑦) = 𝑥, 𝑒_0,1(𝑥, 𝑦) = 𝑦, and 𝑒_2,0(𝑥, 𝑦)+𝑒_0,2(𝑥, 𝑦) = 𝑥2+𝑦2) but provide a better error estimation than the operators𝑆_𝑛(𝑓; 𝑥, 𝑦) for all 𝑓 ∈ 𝐶_𝐵([0, ∞) × [0, ∞)) and 𝑥, 𝑦 ∈ [0, 1].

Finally, we should note that, following the similar argu-ments as used in [2], the best error estimation among all the general double Sz´asz-Mirakjan operators can be obtained from the case:

𝑢(3)_𝑛 (𝑥) := 𝑥 −_2𝑛1 , V(3)_𝑛 (𝑦) := 𝑦 − 1

2𝑛, 𝑛 ∈ N, (11) for all𝑓 ∈ 𝐶_𝐵([0, ∞) × [0, ∞)) and 𝑥, 𝑦 ∈ [1/2, ∞).

For the operators 𝐷_𝑛(𝑓; 𝑥, 𝑦) the following Lemma is straightforward.

Lemma 1. Let x = (𝑥, 𝑦), t = (𝑡, 𝑠), 𝑒𝑖,𝑗(x) = 𝑥𝑖𝑦𝑗, 𝑖, 𝑗 =

0, 1, 2, and 𝜓2

x(t) = ‖t − x‖2. Then, for each𝑥, 𝑦 ≥0 and 𝑛 >

1, one has (a)𝐷_𝑛(𝑒_0,0; 𝑥, 𝑦) = 1, (b)𝐷_𝑛(𝑒_1,0; 𝑥, 𝑦) = 𝑢_𝑛(𝑥), 𝐷_𝑛(𝑒_0,1; 𝑥, 𝑦) = V_𝑛(𝑦), (c)𝐷_𝑛(𝑒_2,0+𝑒_0,2; 𝑥, 𝑦) = 𝑢_𝑛2(𝑥)+V2_𝑛(𝑦)+((𝑢_𝑛(𝑥)+V_𝑛(𝑦))/𝑛), (d)𝐷_𝑛(𝜓2_x(t); 𝑥, 𝑦) = (𝑢_𝑛(𝑥) − 𝑥)2 + (V_𝑛(𝑦) − 𝑦)2 + ((𝑢𝑛(𝑥) + V𝑛(𝑦))/𝑛).

2. Global Results

In this section we first introduce the following Lipschitz-type space:

Lip∗_𝑀(𝛼) := {𝑓 ∈ 𝐶 ([0, ∞) × [0, ∞)) :

󵄨󵄨󵄨󵄨𝑓(t) − 𝑓(x)󵄨󵄨󵄨󵄨 ≤ 𝑀 ‖t − x‖_{(‖t‖ + 𝑥 + 𝑦)}𝛼 𝛼/2;

𝑡, 𝑠; 𝑥, 𝑦 ∈ (0, ∞) } , (12)

wheret = (𝑡, 𝑠), x = (𝑥, 𝑦), 𝑀 is any positive constant, and 0 < 𝛼 ≤ 1.

We should note that this space is the bivariate extension of Lipschitz-type space considered earlier by Szasz [15]. For the space Lip∗_𝑀(𝛼) with 0 < 𝛼 ≤ 1, we have the following approximation result.

Theorem 2. For any 𝑓 ∈ 𝐿𝑖𝑝∗_𝑀(𝛼), 𝛼 ∈ (0, 1] and for each

𝑥, 𝑦 ∈ (0, ∞), 𝑛 ∈ N, one has 󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ 𝑀 (𝑥 + 𝑦)𝛼/2[(𝑢𝑛(𝑥) − 𝑥) 2_{+ (V} 𝑛(𝑦) − 𝑦)2 +𝑢𝑛(𝑥) + V𝑛(𝑦) 𝑛 ] 𝛼/2 . (13)

Proof. Take𝛼 = 1. Then, for 𝑓 ∈ Lip∗_𝑀(1) and for each 𝑥, 𝑦 ∈

Using the Cauchy-Schwarz inequality, we obtain 󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ 𝑀 (𝑥 + 𝑦)1/2√𝐷𝑛(𝜓 2 x(t) ; 𝑥, 𝑦) = 𝑀 (𝑥 + 𝑦)1/2 × √(𝑢_𝑛(𝑥) − 𝑥)2+ (V_𝑛(𝑦) − 𝑦)2+𝑢𝑛(𝑥) + V𝑛(𝑦) 𝑛 . (15) Secondly let0 < 𝛼 < 1. Then, for 𝑓 ∈ Lip∗_𝑀(𝛼) and for each 𝑥, 𝑦 ∈ (0, ∞), we have 󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ 𝐷𝑛(󵄨󵄨󵄨󵄨𝑓 (𝑡, 𝑠) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨;𝑥,𝑦) ≤ 𝑀𝐷_𝑛( ‖t − x‖𝛼 (‖t‖ + 𝑥 + 𝑦)𝛼/2; 𝑥, 𝑦) ≤ 𝑀 (𝑥 + 𝑦)𝛼/2𝐷𝑛(‖t − x‖ 𝛼_{; 𝑥, 𝑦) .} (16) Applying the H¨older inequality with𝑝 = 2/𝛼 and 𝑞 = 2/(2 − 𝛼), we have, for any 𝑓 ∈ Lip∗_𝑀(𝛼),

󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ 𝑀 (𝑥 + 𝑦)𝛼/2[𝐷𝑛(𝜓 2 x(t) ; 𝑥, 𝑦)] 𝛼/2 = 𝑀 (𝑥 + 𝑦)𝛼/2 × [ (𝑢_𝑛(𝑥) − 𝑥)2+ (V_𝑛(𝑦) − 𝑦)2 +𝑢𝑛(𝑥) + V_𝑛 𝑛(𝑦)]𝛼/2, (17) Hence, the result.

The following lemma will be used in the rest of the paper.

Lemma 3. One has, for each 𝑥, 𝑦 > 0,

𝐷_𝑛(√(√𝑡 − √𝑥)2+ (√𝑠 − √𝑦)2; 𝑥, 𝑦) ≤ 1 √𝑥√(𝑢𝑛(𝑥) − 𝑥) 2₊𝑢𝑛(𝑥) 𝑛 + 1 √𝑦√(V𝑛(𝑦) − 𝑦) 2₊V𝑛(𝑦) 𝑛 . (18)

Proof. Using the fact that √𝑎 + 𝑏 ≤ √𝑎 + √𝑏 (𝑎, 𝑏 ≥ 0), we

get 𝐷_𝑛(√(√𝑡 − √𝑥)2+ (√𝑠 − √𝑦)2; 𝑥, 𝑦) = 𝑒−𝑛(𝑢𝑛(𝑥)+V𝑛(𝑦)) ∞ ∑ 𝑘,𝑙=0 √(√ 𝑘_𝑛− √𝑥) 2 + (√_𝑛𝑙 − √𝑦) 2 ×(𝑛𝑢𝑛(𝑥)) 𝑘 𝑘! (𝑛V_𝑛(𝑦))𝑙 𝑙! ≤ 𝑒−𝑛𝑢𝑛(𝑥) ∞ ∑ 𝑘=0 󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨 󵄨󵄨󵄨√ 𝑘𝑛− √𝑥󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨_󵄨󵄨󵄨 (𝑛𝑢_𝑛(𝑥))𝑘 𝑘! + 𝑒−𝑛V𝑛(𝑦) ∞ ∑ 𝑙=0 󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨 󵄨󵄨󵄨√ 𝑙𝑛− √𝑦 󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨 󵄨󵄨󵄨 (𝑛V_𝑛(𝑦))𝑙 𝑙! = 𝑒−𝑛𝑢𝑛(𝑥) ∞ ∑ 𝑘=0 |𝑘/𝑛 − 𝑥| √𝑘/𝑛 + √𝑥 (𝑛𝑢𝑛(𝑥))𝑘 𝑘! + 𝑒−𝑛V𝑛(𝑦) ∞ ∑ 𝑙=0 󵄨󵄨󵄨󵄨𝑙/𝑛 − 𝑦󵄨󵄨󵄨󵄨 √𝑙/𝑛 + √𝑦 (𝑛V_𝑛(𝑦))𝑙 𝑙! ≤𝑒−𝑛𝑢𝑛(𝑥) √𝑥 ∞ ∑ 𝑘=0 󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨𝑘𝑛− 𝑥󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨 (𝑛𝑢_𝑛(𝑥))𝑘 𝑘! +𝑒−𝑛V𝑛(𝑦) √𝑦 ∞ ∑ 𝑙=0 󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨𝑛𝑙 − 𝑦󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨 (𝑛V_𝑛(𝑦))𝑙 𝑙! . (19) Finally, applying the Cauchy-Schwarz inequality, we write

𝐷_𝑛(√(√𝑡 − √𝑥)2+ (√𝑠 − √𝑦)2; 𝑥, 𝑦) ≤ 1 √𝑥√𝑒−𝑛𝑢𝑛(𝑥) ∞ ∑ 𝑘=0 (𝑘 𝑛− 𝑥) 2_(𝑛𝑢 𝑛(𝑥))𝑘 𝑘! + 1 √𝑦√𝑒−𝑛V𝑛(𝑦) ∞ ∑ 𝑙=0 (_𝑛𝑙 − 𝑦)2(𝑛V𝑛(𝑦)) 𝑙 𝑙! = 1 √𝑥√(𝑢𝑛(𝑥) − 𝑥) 2₊𝑢𝑛(𝑥) 𝑛 + 1 √𝑦√(V𝑛(𝑦) − 𝑦) 2₊V𝑛(𝑦) 𝑛 . (20)

Recall that, for all𝑓 ∈ 𝐶_𝐵([0, ∞) × [0, ∞)), the modulus of𝑓 denoted by 𝜔 (𝑓; 𝛿) is defined as 𝜔 (𝑓; 𝛿) := sup { 󵄨󵄨󵄨󵄨𝑓 (𝑡, 𝑠) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 : √(𝑡 − 𝑥)2_{+ (𝑠 − 𝑦)}2_{< 𝛿,} (𝑡, 𝑠) , (𝑥, 𝑦) ∈ [0, ∞) × [0, ∞) } . (21)

Theorem 4. Let 𝑓∗(𝑥, 𝑦) = 𝑓(𝑥2, 𝑦2). Then one has, for each

𝑥, 𝑦 > 0, 󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ 2𝜔 (𝑓∗; 𝛿𝑛(𝑥, 𝑦)) , (22) where 𝛿_𝑛(𝑥, 𝑦) := 1 √𝑥√(𝑢𝑛(𝑥) − 𝑥)2+𝑢𝑛_𝑛(𝑥) + 1 √𝑦√(V𝑛(𝑦) − 𝑦) 2₊V𝑛(𝑦) 𝑛 . (23)

Proof. We directly have

󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ 𝐷𝑛(󵄨󵄨󵄨󵄨𝑓 (𝑡, 𝑠) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨;𝑥,𝑦) = 𝐷_{𝑛(󵄨󵄨󵄨󵄨󵄨𝑓}∗(√𝑡, √𝑠) − 𝑓∗(√𝑥, √𝑦)󵄨󵄨󵄨󵄨_{󵄨 ; 𝑥, 𝑦)} ≤ 𝐷_𝑛(𝜔 (𝑓∗; √(√𝑡 − √𝑥)2+ (√𝑠 − √𝑦)2) ; 𝑥, 𝑦) = 𝑒−𝑛(𝑢𝑛(𝑥)+V𝑛(𝑦)) × ∑∞ 𝑘,𝑙=0 𝜔 (𝑓∗; √(√𝑘 𝑛− √𝑥) 2 + (√𝑙 𝑛− √𝑦) 2 ; 𝑥, 𝑦) ×(𝑛𝑢𝑛(𝑥))𝑘 𝑘! (𝑛V𝑛(𝑦))𝑙 𝑙! . (24) Therefore, 󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 = 𝑒−𝑛(𝑢𝑛(𝑥)+V𝑛(𝑦)) ∞ ∑ 𝑘,𝑙=0 (𝑛𝑢_𝑛(𝑥))𝑘 𝑘! (𝑛V_𝑛(𝑦))𝑙 𝑙! × 𝜔 (𝑓∗; √(√𝑘/𝑛 − √𝑥) 2 + (√𝑙/𝑛 − √𝑦)2 𝐷_𝑛(√(√𝑡 − √𝑥)2+ (√𝑠 − √𝑦)2; 𝑥, 𝑦) × 𝐷𝑛( ((√𝑡 − √𝑥)2 +(√𝑠 − √𝑦)2)1/2; 𝑥, 𝑦) ; 𝑥, 𝑦) . (25) Because of the fact that

𝜔 (𝑓; 𝜆𝛿) ≤ (1 + 𝜆) 𝜔 (𝑓; 𝛿) , (26) we have 󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ 𝜔 (𝑓∗; 𝐷_𝑛(√(√𝑡 − √𝑥)2+ (√𝑠 − √𝑦)2; 𝑥, 𝑦)) × 𝑒−𝑛(𝑢𝑛(𝑥)+V𝑛(𝑦)) × ∑∞ 𝑘,𝑙=0 [ [ [ [ [ 1 + √(√𝑘/𝑛 − √𝑥) 2 + (√𝑙/𝑛 − √𝑦)2 𝐷𝑛(√(√𝑡 − √𝑥)2+ (√𝑠 − √𝑦)2; 𝑥, 𝑦) ] ] ] ] ] ×(𝑛𝑢𝑛(𝑥)) 𝑘 𝑘! (𝑛V_𝑛(𝑦))𝑙 𝑙! , (27) and hence 󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ 2𝜔 (𝑓∗_{; 𝐷} 𝑛(√(√𝑡 − √𝑥)2+ (√𝑠 − √𝑦)2; 𝑥, 𝑦)) . (28) Finally, usingLemma 3, the proof is completed.

Theorem 5. Let 𝑓∗_{(𝑥, 𝑦) = 𝑓(𝑥}2_{, 𝑦}2_{). Let}

𝑓∗∈ 𝐿𝑖𝑝_𝑀(𝛼) := {𝑓∗∈ 𝐶𝐵([0, ∞) × [0, ∞)) :

󵄨󵄨󵄨󵄨𝑓∗_{(t) − 𝑓}∗

(x)󵄨󵄨󵄨󵄨 ≤ 𝑀‖t − x‖𝛼; 𝑡, 𝑠; 𝑥, 𝑦 ∈ (0, ∞)} ,

(29)

wheret = (𝑡, 𝑠), x = (𝑥, 𝑦), 𝑀 is any positive constant, and

0 < 𝛼 ≤ 1. Then

󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ 𝑀𝛿𝛼𝑛(𝑥, 𝑦) , (30)

Proof. We directly have 󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ 𝐷𝑛(󵄨󵄨󵄨󵄨𝑓 (𝑡, 𝑠) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨;𝑥,𝑦) = 𝐷_{𝑛(󵄨󵄨󵄨󵄨󵄨𝑓}∗(√𝑡, √𝑠) − 𝑓∗(√𝑥, √𝑦)󵄨󵄨󵄨󵄨_{󵄨 ; 𝑥, 𝑦)} ≤ 𝑀𝐷_𝑛(((√𝑡 − √𝑥)2+ (√𝑠 − √𝑦)2)𝛼/2; 𝑥, 𝑦) = 𝑀𝑒−𝑛(𝑢𝑛(𝑥)+V𝑛(𝑦)) × ∑∞ 𝑘,𝑙=0 ((√𝑘 𝑛− √𝑥) 2 + (√𝑙 𝑛− √𝑦) 2 ) 𝛼/2 ×(𝑛𝑢𝑛(𝑥)) 𝑘 𝑘! (𝑛V_𝑛(𝑦))𝑙 𝑙! . (31)

Applying the H¨older inequality with𝑝 = 1/𝛼 and 𝑞 = 1/(1 − 𝛼), we have 󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ 𝑀[𝐷𝑛(√(√𝑡 − √𝑥)2+ (√𝑠 − √𝑦)2; 𝑥, 𝑦)] 𝛼 . (32) UsingLemma 3, we get the result.

3. Concluding Remarks

In this section we show that taking𝑢_𝑛(𝑥) = 𝑥 and V_𝑛(𝑦) = 𝑦 or𝑢_𝑛(𝑥) = 𝑢(𝑖)_𝑛 (𝑥) and V_𝑛(𝑦) = V(𝑖)_𝑛 (𝑦), 𝑖 = 1, 3, in Theorems

2,4, and5gives global results. Also we present the results obtained by Theorems2,4, and5for𝑢_𝑛(𝑥) = 𝑢(2)_𝑛 (𝑥) and V_𝑛(𝑦) = V(2)

𝑛 (𝑦).

Corollary 6. For any 𝑓 ∈ 𝐿𝑖𝑝∗

𝑀(𝛼), 𝛼 ∈ (0, 1] and for all

𝑥, 𝑦 ∈ (0, ∞), 𝑛 ∈ N, one has

󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ _𝑛𝑀_𝛼/2, (33)

uniformly as𝑛 → ∞, for the following pairs of 𝑢_𝑛(𝑥) and

V_𝑛(𝑥):

(i)𝑢_𝑛(𝑥) = 𝑥 and V_𝑛(𝑦) = 𝑦,

(ii)𝑢_𝑛(𝑥) = (−1 + √4𝑛2𝑥2+ 1)/2𝑛 and V_𝑛(𝑦) = (−1 + √4𝑛2_𝑦2_{+ 1)/2𝑛,}

(iii)𝑢_𝑛(𝑥) = 𝑥 − 1/2𝑛 and V_𝑛(𝑦) = 𝑦 − 1/2𝑛.

Proof. (i) Taking𝑢_𝑛(𝑥) = 𝑥 and V_𝑛(𝑦) = 𝑦 in (13), we directly

have

󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ _𝑛𝑀_𝛼/2. (34)

(ii) Taking𝑢_𝑛(𝑥) = (−1 + √4𝑛2𝑥2+ 1)/2𝑛 and V_𝑛(𝑦) = (−1 + √4𝑛2_𝑦2_{+ 1)/2𝑛 in (13) gives} 󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 = 𝑀 (𝑥 + 𝑦)𝛼/2[ 1 𝑛(𝑥 + 𝑦 − 𝑥√4𝑛2𝑥2+ 1 − 𝑦√4𝑛2_𝑦2_{+ 1 + 2𝑛𝑥}2_{+ 2𝑛𝑦}2_)]𝛼/2 ≤ 𝑀 (𝑥 + 𝑦)𝛼/2[ 1 𝑛(𝑥 + 𝑦 − 𝑥√4𝑛2𝑥2 − 𝑦√4𝑛2_𝑦2_{+ 2𝑛𝑥}2_{+ 2𝑛𝑦}2_)]𝛼/2 = 𝑀 𝑛𝛼/2. (35) (iii) Taking𝑢_𝑛(𝑥) = 𝑥 − 1/2𝑛 and V_𝑛(𝑦) = 𝑦 − 1/2𝑛 in (13), we have 󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 = 𝑀 (𝑥 + 𝑦)𝛼/2[(− 1 2𝑛) 2 + (− 1 2𝑛) 2 +𝑥 + 𝑦 − 1/𝑛 𝑛 ] 𝛼/2 ≤ 𝑀 (𝑥 + 𝑦)𝛼/2[ 1 2𝑛2(2𝑛𝑥 + 2𝑛𝑦 − 1)] 𝛼/2 = 𝑀 (𝑥 + 𝑦)𝛼/2[ 1 𝑛(𝑥 + 𝑦)] 𝛼/2 = 𝑀 𝑛𝛼/2. (36)

Corollary 7. Let 𝑓∗_{(𝑥, 𝑦) = 𝑓(𝑥}2_{, 𝑦}2_{). Then one has}

󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ 2𝜔 (𝑓∗;_√𝑛2 ) , (37)

uniformly as𝑛 → ∞, for the following pairs of 𝑢_𝑛(𝑥) and

V_𝑛(𝑥):

(i)𝑢_𝑛(𝑥) = 𝑥 and V_𝑛(𝑦) = 𝑦,

(ii)𝑢_𝑛(𝑥) = (−1 + √4𝑛2𝑥2+ 1)/2𝑛 and V_𝑛(𝑦) = (−1 + √4𝑛2_𝑦2_{+ 1)/2𝑛,}

(iii)𝑢_𝑛(𝑥) = 𝑥 − 1/2𝑛 and V_𝑛(𝑦) = 𝑦 − 1/2𝑛.

Proof. (i) Taking𝑢_𝑛(𝑥) = 𝑥 and V_𝑛(𝑦) = 𝑦 in (23), we directly

have

(ii) Taking𝑢_𝑛(𝑥) = (−1 + √4𝑛2𝑥2+ 1)/2𝑛 and V_𝑛(𝑦) = (−1 + √4𝑛2_𝑦2_{+ 1)/2𝑛 in (23) gives} 󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ 2𝜔 (𝑓∗; 𝛿𝑛(𝑥, 𝑦)) , (39) where 𝛿_𝑛(𝑥, 𝑦) := 1 √𝑥√𝑦(√𝑦√ 1 𝑛(𝑥 − 𝑥√4𝑛2𝑥2+ 1 + 2𝑛𝑥2) + √𝑥√1 𝑛(𝑦 − 𝑦√4𝑛2𝑦2+ 1 + 2𝑛𝑦2)) ≤ 1 √𝑥√𝑦(√𝑦√ 1 𝑛𝑥 + √𝑥√ 1 𝑛𝑦) = 2 √𝑛. (40) (iii) Taking𝑢_𝑛(𝑥) = 𝑥 − 1/2𝑛 and V_𝑛(𝑦) = 𝑦 − 1/2𝑛 in (23), we have 󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ 2𝜔 (𝑓∗; 𝛿𝑛(𝑥, 𝑦)) , (41) where 𝛿𝑛(𝑥, 𝑦) := _√𝑥1 √( 1_2𝑛) 2 +𝑥_𝑛−_2𝑛1₂ + 1 √𝑦√( 12𝑛) 2 +𝑦_𝑛−_2𝑛1₂ = 1 2√𝑥√𝑦(√𝑥√ 1 𝑛2(4𝑛𝑦 − 1) + √𝑦√_𝑛1₂(4𝑛𝑥 − 1)) ≤ 1 √𝑥√𝑦(√𝑥√ 1 𝑛𝑦 + √𝑦√ 1 𝑛𝑥) = 2 √𝑛. (42)

Corollary 8. Let 𝑓∗_{(𝑥, 𝑦) = 𝑓(𝑥}2_{, 𝑦}2_{), and let}

𝑓∗ ∈ 𝐿𝑖𝑝_𝑀(𝛼) := {𝑓∗∈ 𝐶_𝐵([0, ∞) × [0, ∞)) : 󵄨󵄨󵄨󵄨𝑓∗_{(t) − 𝑓}∗

(x)󵄨󵄨󵄨󵄨 ≤ 𝑀‖t − x‖𝛼; 𝑡, 𝑠; 𝑥, 𝑦 ∈ (0, ∞) } ,

(43)

wheret = (𝑡, 𝑠), x = (𝑥, 𝑦), 𝑀 is any positive constant, and

0 < 𝛼 ≤ 1. Then

󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ 𝑀(4_𝑛) 𝛼/2

, (44)

uniformly as𝑛 → ∞, for the following pairs of 𝑢_𝑛(𝑥) and

V_𝑛(𝑦) :

(i)𝑢_𝑛(𝑥) = 𝑥 and V_𝑛(𝑦) = 𝑦,

(ii)𝑢_𝑛(𝑥) = (−1 + √4𝑛2𝑥2+ 1)/2𝑛 and V_𝑛(𝑦) = (−1 + √4𝑛2_𝑦2_{+ 1)/2𝑛,}

(iii)𝑢_𝑛(𝑥) = 𝑥 − 1/2𝑛 and V_𝑛(𝑦) = 𝑦 − 1/2𝑛.

Proof. (i) Taking𝑢_𝑛(𝑥) = 𝑥 and V_𝑛(𝑦) = 𝑦 in (23), we directly

have

󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ 𝑀(4_𝑛) 𝛼/2

. (45)

(ii) Taking𝑢_𝑛(𝑥) = (−1 + √4𝑛2𝑥2+ 1)/2𝑛 and V_𝑛(𝑦) = (−1 + √4𝑛2_𝑦2_{+ 1)/2𝑛 in (23) gives} 󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ 𝑀𝛿𝛼/2𝑛 (𝑥, 𝑦) , (46) where 𝛿_𝑛(𝑥, 𝑦) := 𝛿_𝑛(𝑥, 𝑦) := 1 √𝑥√𝑦 × (√𝑦√_𝑛1(𝑥 − 𝑥√4𝑛2_𝑥2_{+ 1 + 2𝑛𝑥}2₎ + √𝑥√1 𝑛(𝑦 − 𝑦√4𝑛2𝑦2+ 1 + 2𝑛𝑦2)) ≤ 1 √𝑥√𝑦(√𝑦√ 1 𝑛𝑥 + √𝑥√ 1 𝑛𝑦) = 2 √𝑛. (47) (iii) Taking𝑢_𝑛(𝑥) = 𝑥 − 1/2𝑛 and V_𝑛(𝑦) = 𝑦 − 1/2𝑛 in (23), we have

where 𝛿_𝑛(𝑥, 𝑦) := 1 √𝑥√( 12𝑛) 2 +𝑥 𝑛− 1 2𝑛2 + 1 √𝑦√( 12𝑛) 2 +𝑦_𝑛−_2𝑛1₂ = 1 2√𝑥√𝑦(√𝑥√ 1 𝑛2(4𝑛𝑦 − 1) + √𝑦√_𝑛1₂ (4𝑛𝑥 − 1)) ≤ 1 √𝑥√𝑦(√𝑥√ 1 𝑛𝑦 + √𝑦√ 1 𝑛𝑥) = 2 √𝑛. (49)

Remark 9. Corollaries 7 and 8 conclude that 𝑓 is a real

continuous and bounded function on[0, ∞) × [0, ∞) and if𝑓∗(𝑥, 𝑦) = 𝑓(𝑥2, 𝑦2) is uniformly continuous on [0, ∞) × [0, ∞), then 𝐷𝑛(𝑓) converges uniformly to 𝑓 as 𝑛 → ∞.

Note that the one variable version ofCorollary 7was given in [16]. Corollary 10. Take 𝑢_𝑛(𝑥) = 𝑢(2)_𝑛 (𝑥, 𝛾) = − (𝑛𝛾 + 1) + √4𝑛2(𝑥2+ 𝛾𝑥) + (𝑛𝛾 + 1) 2 2𝑛 , V_𝑛(𝑦) = V(2)_𝑛 (𝑦, 𝛽) = − (𝑛𝛽 + 1) + √4𝑛2(𝑦2+ 𝛽𝑦) + (𝑛𝛽 + 1) 2 2𝑛 , (50) where𝛼, 𝛽 ∈ R. Then

(i) for any𝑓 ∈ 𝐿𝑖𝑝∗_𝑀(𝛼), 𝛼 ∈ (0, 1] and for each 𝑥, 𝑦 ∈ (0, ∞), 𝑛 ∈ N, one has 󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ 𝑀 [2𝑛 (𝑥 + 𝑦)]𝛼/2[𝛿 (𝑥, 𝛾) + 𝛿 (𝑦, 𝛽)] 𝛼/2_, (51) where 𝛿 (𝑥, 𝛾) = [ (2𝑥 + 𝛾) × (𝑛𝛾 − √𝑛2_{(2𝑥 + 𝛾)}2_{+ 2𝑛𝛾 + 1 + 2𝑛𝑥 + 1)] ,} (52)

(ii) let𝑓∗(𝑥, 𝑦) = 𝑓(𝑥2, 𝑦2). Then one has for each 𝑥, 𝑦 > 0 󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ 2𝜔 (𝑓∗; 𝛿 (𝑥, 𝛾) + 𝛿 (𝑦, 𝛽)) , (53) where 𝛿 (𝑥, 𝛾) = 1 √2𝑥 × (_𝑛1(2𝑥 + 𝛾) (𝑛𝛾 − √𝑛2_{(2𝑥 + 𝛾)}2_{+ 2𝑛𝛾 + 1} +2𝑛𝑥 + 1))1/2, (54) (iii) let𝑓∗(𝑥, 𝑦) = 𝑓(𝑥2, 𝑦2), and let

𝑓∗_{∈ Lip} 𝑀(𝛼) := {𝑓∗∈ 𝐶𝐵([0, ∞) × [0, ∞)) : 󵄨󵄨󵄨󵄨𝑓∗_{(t) − 𝑓}∗ (x)󵄨󵄨󵄨󵄨 ≤ 𝑀‖t − x‖𝛼; 𝑡, 𝑠; 𝑥, 𝑦 ∈ (0, ∞) } , (55)

wheret = (𝑡, 𝑠), x = (𝑥, 𝑦), 𝑀 is any positive constant,

and0 < 𝛼 ≤ 1. Then

󵄨󵄨󵄨󵄨𝐷𝑛(𝑓; 𝑥, 𝑦) − 𝑓 (𝑥, 𝑦)󵄨󵄨󵄨󵄨 ≤ 𝑀[𝛿(𝑥, 𝛾) + 𝛿(𝑦, 𝛽)]𝛼/2, (56)

where𝛿(𝑥, 𝛾) is the same given inCorollary 10(𝑖𝑖).

It should be mentioned that, for 𝛼 = 0 and 𝛽 = 0, 𝑢_𝑛(2)(𝑥, 0) = (−1 + √4𝑛2_𝑥2_{+ 1)/2𝑛 and}

V(2)_𝑛 (𝑦, 0) = (−1 + √4𝑛2_𝑦2_{+ 1)/2𝑛. Therefore,}

Corollary 10(i), Corollary 10(ii), and Corollary 10(iii)

reduce toCorollary 6(ii),Corollary 7(ii), andCorollary 8(ii),

respectively.

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d’Ankara A1, vol. 58, no. 1, pp. 17–22, 2009.

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