Quantitative Global Estimates for Generalized Double Szasz-Mirakjan Operators

Abstract

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

1. Introduction

The well-known Szász-Mirakjan operators are defined on the space 𝒜₁ as follows: 𝑆_𝑛𝑓;𝑥=𝑒^{∞−𝑛𝑥}_𝑘=0𝑓𝑘𝑛(𝑛𝑥)^𝑘,𝑘! (1) where 𝒜₁ is the set of all real functions on [0,∞) such that the right-hand side in (1) make sense for all 𝑛>0 and 𝑥∈[0,∞). By modifying the Szász-Mirakjan operators as 𝐷_𝑛𝑓;𝑥=𝑒^{−𝑛𝑢𝑛∞(𝑥)}_𝑘=0𝑓𝑘𝑛(𝑛𝑢_𝑛(𝑥))^𝑘,𝑘! (2) where {𝑢_𝑛(𝑥)} is a sequence of real-valued, continuous functions defined on [0,∞) with 0≤𝑢_𝑛(𝑥)<∞, it has been shown in [1] that if one let 𝑢^∗_𝑛(𝑥)√∶=−1+4𝑛²𝑥²+12𝑛,𝑛∈ℕ, (3) then the operators defined by 𝐷^∗_𝑛𝑓;𝑥∶=𝑆_𝑛𝑓;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)1∶=𝑥−2𝑛;𝑛∈ℕ, (5) it has been shown in [2] that the operators defined by 𝑉^∗_𝑛𝑓;𝑥∶=𝑆_𝑛𝑓;𝑣_𝑛(𝑥) (6) do not preserve the test functions 𝑒₁(𝑥)=𝑥 and 𝑒₂(𝑥)=𝑥² but provide the best error estimation among all the Szász-Mirakjan operators for all 𝑓∈𝐶_𝐵([0,∞)) and for each 𝑥∈[1/2,∞). For the other linear positive operator families which preserve 𝑒₂(𝑥)=𝑥², we refer [3–9]. On the other hand, in [10, 11] the authors considered some operators preserving 𝑒₁(𝑥)=𝑥.

Favard was the first to introduce the double Szász-Mirakjan operators [12]: 𝑆_𝑛𝑓;𝑥,𝑦=𝑒^{∞−𝑛(𝑥+𝑦)}_𝑘,𝑙=0𝑓𝑘𝑛,𝑙𝑛(𝑛𝑥)^𝑘𝑘!(𝑛𝑦)^𝑙𝑙!,𝑓∈𝒜₂, (7) where 𝒜₂ is the set of all real functions on [0,∞)×[0,∞) such that the right-hand side in (7) has a meaning for all 𝑛>0 and 𝑥,𝑦∈[0,∞). Recently, Dirik and Demirci have introduced and investigated different variants of the general double Szász-Mirakjan operators: 𝐷_𝑛𝑓;𝑥,𝑦∶𝑆_𝑛𝑓;𝑢_𝑛(𝑥),𝑣_𝑛𝑦=𝑒^{−𝑛(𝑢𝑛(𝑥)+𝑣𝑛(𝑦))}×^∞_𝑘,𝑙=0𝑓𝑘𝑛,𝑙𝑛×𝑛𝑢_𝑛(𝑥)^𝑘𝑘!𝑛𝑣_𝑛𝑦^𝑙,𝑙!𝑓∈𝒜₂. (8) In [13], they considered the case of operators 𝑢_𝑛⁽¹⁾(𝑥)√∶=−1+4𝑛²𝑥²+1,𝑣2𝑛_𝑛(1)𝑦∶=−1+4𝑛²𝑦²+1,2𝑛𝑛∈ℕ, (9) which preserve the test function 𝑒_2,0(𝑥,𝑦)+𝑒_0,2(𝑥,𝑦)∶=𝑥²+𝑦² and 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_𝑛⁽²⁾()−()+𝑥,𝛼∶=𝑛𝛼+14𝑛²𝑥²+𝛼𝑥+(𝑛𝛼+1)²,𝑣2𝑛_𝑛(2)−+𝑦,𝛽∶=𝑛𝛽+14𝑛²𝑦²+𝛽𝑦+𝑛𝛽+1²,2𝑛𝑛∈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 operators 𝑆_𝑛(𝑓;𝑥,𝑦) for all 𝑓∈𝐶_𝐵([0,∞)×[0,∞)) and 𝑥,𝑦∈[0,1].

Finally, we should note that, following the similar arguments as used in [2], the best error estimation among all the general double Szász-Mirakjan operators can be obtained from the case: 𝑢_𝑛⁽³⁾(𝑥)1∶=𝑥−2𝑛,𝑣_𝑛⁽³⁾𝑦1∶=𝑦−2𝑛,𝑛∈ℕ, (11) for all 𝑓∈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), 𝑒_𝑖,𝑗(𝐱)=𝑥^𝑖𝑦^𝑗, 𝑖,𝑗=0,1,2, and𝜓²_𝐱(𝐭)=‖𝐭−𝐱‖². Then, for each 𝑥,𝑦≥0 and𝑛>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)=𝑢²_𝑛(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+((𝑢_𝑛(𝑥)+𝑣_𝑛(𝑦))/𝑛).

2. Global Results

In this section we first introduce the following Lipschitz-type space: Lip^∗_𝑀(𝛼))×[∶𝑓(𝐭)(𝐱‖‖∶=𝑓∈𝐶([0,∞0,∞))−𝑓)≤𝑀𝐭−𝐱^𝛼‖𝐭‖+𝑥+𝑦^𝛼/2;𝑡,𝑠;𝑥,𝑦∈(0,∞), (12) where 𝐭=(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 𝐷_𝑛≤𝑀𝑓;𝑥,𝑦−𝑓𝑥,𝑦𝑥+𝑦^𝛼/2𝑢_𝑛(𝑥)−𝑥²+𝑣_𝑛𝑦−𝑦²+𝑢_𝑛(𝑥)+𝑣_𝑛𝑦𝑛^𝛼/2. (13)

Proof. Take 𝛼=1. Then, for 𝑓∈Lip^∗_𝑀(1) and for each 𝑥,𝑦∈(0,∞), we get 𝐷_𝑛𝑓;𝑥,𝑦−𝑓𝑥,𝑦≤𝐷_𝑛𝑓(𝑡,𝑠)−𝑓𝑥,𝑦;𝑥,𝑦≤𝑀𝐷_𝑛‖𝐭−𝐱‖‖𝐭‖+𝑥+𝑦^1/2≤𝑀;𝑥,𝑦𝑥+𝑦^1/2𝐷_𝑛‖𝐭−𝐱‖;𝑥,𝑦. (14) Using the Cauchy-Schwarz inequality, we obtain 𝐷_𝑛≤𝑀𝑓;𝑥,𝑦−𝑓𝑥,𝑦𝑥+𝑦^1/2𝐷_𝑛𝜓²_𝐱(𝐭)=𝑀;𝑥,𝑦𝑥+𝑦^1/2×𝑢_𝑛(𝑥)−𝑥²+𝑣_𝑛𝑦−𝑦²+𝑢_𝑛(𝑥)+𝑣_𝑛𝑦𝑛. (15) Secondly let 0<𝛼<1. Then, for 𝑓∈Lip^∗_𝑀(𝛼) and for each 𝑥,𝑦∈(0,∞), we have 𝐷_𝑛𝑓;𝑥,𝑦−𝑓𝑥,𝑦≤𝐷_𝑛𝑓(𝑡,𝑠)−𝑓𝑥,𝑦;𝑥,𝑦≤𝑀𝐷_𝑛‖𝐭−𝐱‖^𝛼‖𝐭‖+𝑥+𝑦^𝛼/2≤𝑀;𝑥,𝑦𝑥+𝑦^𝛼/2𝐷_𝑛‖𝐭−𝐱‖^𝛼;𝑥,𝑦. (16) Applying the Hölder inequality with 𝑝=2/𝛼 and 𝑞=2/(2−𝛼), we have, for any 𝑓∈Lip^∗_𝑀(𝛼),  𝐷_𝑛≤𝑀𝑓;𝑥,𝑦−𝑓𝑥,𝑦𝑥+𝑦^𝛼/2𝐷_𝑛𝜓²_𝐱(𝐭);𝑥,𝑦^𝛼/2=𝑀𝑥+𝑦^𝛼/2×𝑢_𝑛(𝑥)−𝑥²+𝑣_𝑛(𝑦)−𝑦²+𝑢_𝑛(𝑥)+𝑣_𝑛(𝑦)𝑛^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−√𝑥²+√𝑠−√𝑦²≤1;𝑥,𝑦√𝑥𝑢_𝑛(𝑥)−𝑥²+𝑢_𝑛(𝑥)𝑛+1√𝑦𝑣_𝑛𝑦−𝑦²+𝑣_𝑛𝑦𝑛. (18)

Proof. Using the fact that √𝑎+𝑏≤√𝑎+√𝑏(𝑎,𝑏≥0), we get 𝐷_𝑛√𝑡−√𝑥²+√𝑠−√𝑦²;𝑥,𝑦=𝑒^{−𝑛(𝑢𝑛(𝑥)+𝑣𝑛∞(𝑦))}_𝑘,𝑙=0𝑘𝑛−√𝑥²+𝑙𝑛−√𝑦²×𝑛𝑢_𝑛(𝑥)^𝑘𝑘!𝑛𝑣_𝑛𝑦^𝑙𝑙!≤𝑒^{−𝑛𝑢𝑛∞(𝑥)}_𝑘=0𝑘𝑛−√𝑥𝑛𝑢_𝑛(𝑥)^𝑘𝑘!+𝑒^{−𝑛𝑣𝑛∞(𝑦)}_𝑙=0𝑙𝑛−√𝑦𝑛𝑣_𝑛𝑦^𝑙𝑙!=𝑒^{−𝑛𝑢𝑛∞(𝑥)}_𝑘=0|𝑘/𝑛−𝑥|𝑘/𝑛+√𝑥𝑛𝑢_𝑛(𝑥)^𝑘𝑘!+𝑒^{−𝑛𝑣𝑛∞(𝑦)}_𝑙=0𝑙/𝑛−𝑦𝑙/𝑛+√𝑦𝑛𝑣_𝑛𝑦^𝑙≤𝑒𝑙!_{−𝑛𝑢𝑛(𝑥)}√𝑥^∞_𝑘=0𝑘𝑛−𝑥𝑛𝑢_𝑛(𝑥)^𝑘+𝑒𝑘!_{−𝑛𝑣𝑛(𝑦)}√𝑦^∞_𝑙=0𝑙𝑛−𝑦𝑛𝑣_𝑛𝑦^𝑙.𝑙! (19) Finally, applying the Cauchy-Schwarz inequality, we write 𝐷_𝑛√𝑡−√𝑥²+√𝑠−√𝑦²≤1;𝑥,𝑦√𝑥𝑒^{−𝑛𝑢𝑛∞(𝑥)}_𝑘=0𝑘𝑛−𝑥²𝑛𝑢_𝑛(𝑥)^𝑘+1𝑘!√𝑦𝑒^{−𝑛𝑣𝑛∞(𝑦)}_𝑙=0𝑙𝑛−𝑦²𝑛𝑣_𝑛𝑦^𝑙=1𝑙!√𝑥𝑢_𝑛(𝑥)−𝑥²+𝑢_𝑛(𝑥)𝑛+1√𝑦𝑣_𝑛𝑦−𝑦²+𝑣_𝑛𝑦𝑛. (20) Using Lemma 1, we get the result.

Recall that, for all 𝑓∈𝐶_𝐵([0,∞)×[0,∞)), the modulus of 𝑓 denoted by 𝜔(𝑓;𝛿) is defined as 𝜔𝑓;𝛿∶=sup𝑓(𝑡,𝑠)−𝑓𝑥,𝑦∶(𝑡−𝑥)²+𝑠−𝑦²(),∈[)×[.<𝛿,𝑡,𝑠𝑥,𝑦0,∞0,∞) (21)

Theorem 4. Let 𝑓^∗(𝑥,𝑦)=𝑓(𝑥²,𝑦²). Then one has, for each 𝑥,𝑦>0, 𝐷_𝑛𝑓;𝑥,𝑦−𝑓𝑥,𝑦≤2𝜔𝑓^∗;𝛿_𝑛𝑥,𝑦, (22) where 𝛿_𝑛1𝑥,𝑦∶=√𝑥𝑢_𝑛(𝑥)−𝑥²+𝑢_𝑛(𝑥)𝑛+1√𝑦𝑣_𝑛𝑦−𝑦²+𝑣_𝑛𝑦𝑛. (23)

Proof. We directly have 𝐷_𝑛𝑓;𝑥,𝑦−𝑓𝑥,𝑦≤𝐷_𝑛𝑓(𝑡,𝑠)−𝑓𝑥,𝑦;𝑥,𝑦=𝐷_𝑛𝑓^∗√𝑡,√𝑠−𝑓^∗√𝑥,√𝑦;𝑥,𝑦≤𝐷_𝑛𝜔𝑓^∗;√𝑡−√𝑥²+√𝑠−√𝑦²;𝑥,𝑦=𝑒^{−𝑛(𝑢𝑛(𝑥)+𝑣𝑛(𝑦))}×^∞_𝑘,𝑙=0𝜔𝑓^∗;𝑘𝑛−√𝑥²+𝑙𝑛−√𝑦²×;𝑥,𝑦𝑛𝑢_𝑛(𝑥)^𝑘𝑘!𝑛𝑣_𝑛𝑦^𝑙.𝑙! (24) Therefore, 𝐷_𝑛𝑓;𝑥,𝑦−𝑓𝑥,𝑦=𝑒^{−𝑛(𝑢𝑛(𝑥)+𝑣𝑛∞(𝑦))}_𝑘,𝑙=0𝑛𝑢_𝑛(𝑥)^𝑘𝑘!𝑛𝑣_𝑛𝑦^𝑙𝑓𝑙!×𝜔^∗;𝑘/𝑛−√𝑥²+𝑙/𝑛−√𝑦²𝐷_𝑛√𝑡−√𝑥²+√𝑠−√𝑦²;𝑥,𝑦×𝐷_𝑛√𝑡−√𝑥²+√𝑠−√𝑦^21/2;𝑥,𝑦;𝑥,𝑦. (25) Because of the fact that 𝜔𝑓;𝜆𝛿≤(1+𝜆)𝜔𝑓;𝛿, (26) we have 𝐷_𝑛𝑓𝑓;𝑥,𝑦−𝑓𝑥,𝑦≤𝜔^∗;𝐷_𝑛√𝑡−√𝑥²+√𝑠−√𝑦²;𝑥,𝑦×𝑒^{−𝑛(𝑢𝑛(𝑥)+𝑣𝑛(𝑦))}×^∞_𝑘,𝑙=01+𝑘/𝑛−√𝑥²+𝑙/𝑛−√𝑦²𝐷_𝑛√𝑡−√𝑥²+√𝑠−√𝑦²×;𝑥,𝑦𝑛𝑢_𝑛(𝑥)^𝑘𝑘!𝑛𝑣_𝑛𝑦^𝑙,𝑙! (27) and hence 𝐷_𝑛𝑓𝑓;𝑥,𝑦−𝑓𝑥,𝑦≤2𝜔^∗;𝐷_𝑛√𝑡−√𝑥²+√𝑠−√𝑦²;𝑥,𝑦. (28) Finally, using Lemma 3, the proof is completed.

Theorem 5. Let 𝑓^∗(𝑥,𝑦)=𝑓(𝑥²,𝑦²). Let 𝑓^∗∈𝐿𝑖𝑝_𝑀(𝛼)∶=𝑓^∗∈𝐶_𝐵)×[∶𝑓([0,∞0,∞))_∗(𝐭)−𝑓^∗(𝐱)≤𝑀‖𝐭−𝐱‖^𝛼;𝑡,𝑠;𝑥,𝑦∈(0,∞), (29) where 𝐭=(𝑡,𝑠),𝐱=(𝑥,𝑦),𝑀 is any positive constant, and 0<𝛼≤1. Then 𝐷_𝑛𝑓;𝑥,𝑦−𝑓𝑥,𝑦≤𝑀𝛿^𝛼_𝑛𝑥,𝑦, (30) where 𝛿_𝑛(𝑥,𝑦) is the same as in Theorem 4.

Proof. We directly have 𝐷_𝑛𝑓;𝑥,𝑦−𝑓𝑥,𝑦≤𝐷_𝑛𝑓(𝑡,𝑠)−𝑓𝑥,𝑦;𝑥,𝑦=𝐷_𝑛𝑓^∗√𝑡,√𝑠−𝑓^∗√𝑥,√𝑦;𝑥,𝑦≤𝑀𝐷_𝑛√𝑡−√𝑥²+√𝑠−√𝑦^2𝛼/2;𝑥,𝑦=𝑀𝑒^{−𝑛(𝑢𝑛(𝑥)+𝑣𝑛(𝑦))}×^∞_𝑘,𝑙=0𝑘𝑛−√𝑥²+𝑙𝑛−√𝑦^2𝛼/2×𝑛𝑢_𝑛(𝑥)^𝑘𝑘!𝑛𝑣_𝑛𝑦^𝑙.𝑙! (31) Applying the Hölder inequality with 𝑝=1/𝛼 and 𝑞=1/(1−𝛼),we have 𝐷_𝑛𝐷𝑓;𝑥,𝑦−𝑓𝑥,𝑦≤𝑀_𝑛√𝑡−√𝑥²+√𝑠−√𝑦²;𝑥,𝑦^𝛼. (32) Using Lemma 3, we get the result.

3. Concluding Remarks

In this section we show that taking 𝑢_𝑛(𝑥)=𝑥 and 𝑣_𝑛(𝑦)=𝑦 or 𝑢_𝑛(𝑥)=𝑢_𝑛^(𝑖)(𝑥) and 𝑣_𝑛(𝑦)=𝑣_𝑛^(𝑖)(𝑦),𝑖=1,3, in Theorems 2, 4, and 5 gives global results. Also we present the results obtained by Theorems 2, 4, and 5 for 𝑢_𝑛(𝑥)=𝑢_𝑛⁽²⁾(𝑥) and 𝑣_𝑛(𝑦)=𝑣_𝑛⁽²⁾(𝑦).

Corollary 6. For any 𝑓∈𝐿𝑖𝑝^∗_𝑀(𝛼),𝛼∈(0,1] and for all 𝑥,𝑦∈(0,∞),𝑛∈ℕ, one has 𝐷_𝑛≤𝑀𝑓;𝑥,𝑦−𝑓𝑥,𝑦𝑛^𝛼/2, (33) uniformly as 𝑛→∞, 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+4𝑛²𝑦²+1)/2𝑛,(iii)𝑢_𝑛(𝑥)=𝑥−1/2𝑛 and 𝑣_𝑛(𝑦)=𝑦−1/2𝑛.

Proof. (i) Taking 𝑢_𝑛(𝑥)=𝑥 and 𝑣_𝑛(𝑦)=𝑦 in (13), we directly have 𝐷_𝑛≤𝑀𝑓;𝑥,𝑦−𝑓𝑥,𝑦𝑛^𝛼/2. (34) (ii) Taking 𝑢_𝑛(𝑥)=(−1+√4𝑛²𝑥²+1)/2𝑛 and 𝑣_𝑛(𝑦)=(−1+4𝑛²𝑦²+1)/2𝑛 in (13) gives 𝐷_𝑛=𝑀𝑓;𝑥,𝑦−𝑓𝑥,𝑦𝑥+𝑦^𝛼/21𝑛𝑥+𝑦−𝑥√4𝑛²𝑥²+1−𝑦4𝑛²𝑦²+1+2𝑛𝑥²+2𝑛𝑦^2𝛼/2≤𝑀𝑥+𝑦^𝛼/21𝑛𝑥+𝑦−𝑥√4𝑛²𝑥²−𝑦4𝑛²𝑦²+2𝑛𝑥²+2𝑛𝑦^2𝛼/2=𝑀𝑛^𝛼/2. (35) (iii) Taking 𝑢_𝑛(𝑥)=𝑥−1/2𝑛 and 𝑣_𝑛(𝑦)=𝑦−1/2𝑛 in (13), we have 𝐷_𝑛=𝑀𝑓;𝑥,𝑦−𝑓𝑥,𝑦𝑥+𝑦^𝛼/2−12𝑛²+−12𝑛²+𝑥+𝑦−1/𝑛𝑛^𝛼/2≤𝑀𝑥+𝑦^𝛼/212𝑛²2𝑛𝑥+2𝑛𝑦−1^𝛼/2=𝑀𝑥+𝑦^𝛼/21𝑛𝑥+𝑦^𝛼/2=𝑀𝑛^𝛼/2. (36)

Corollary 7. Let 𝑓^∗(𝑥,𝑦)=𝑓(𝑥²,𝑦²). Then one has 𝐷_𝑛𝑓;𝑥,𝑦−𝑓𝑥,𝑦≤2𝜔𝑓^∗;2√𝑛, (37) uniformly as 𝑛→∞, 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+4𝑛²𝑦²+1)/2𝑛,(iii)𝑢_𝑛(𝑥)=𝑥−1/2𝑛 and 𝑣_𝑛(𝑦)=𝑦−1/2𝑛.

Proof. (i) Taking 𝑢_𝑛(𝑥)=𝑥 and 𝑣_𝑛(𝑦)=𝑦 in (23), we directly have 𝐷_𝑛𝑓;𝑥,𝑦−𝑓𝑥,𝑦≤2𝜔𝑓^∗;2√𝑛. (38) (ii) Taking 𝑢_𝑛(𝑥)=(−1+√4𝑛²𝑥²+1)/2𝑛 and 𝑣_𝑛(𝑦)=(−1+4𝑛²𝑦²+1)/2𝑛 in (23) gives 𝐷_𝑛𝑓;𝑥,𝑦−𝑓𝑥,𝑦≤2𝜔𝑓^∗;𝛿_𝑛𝑥,𝑦, (39) where 𝛿_𝑛1𝑥,𝑦∶=√𝑥√𝑦√𝑦1𝑛𝑥−𝑥√4𝑛²𝑥²+1+2𝑛𝑥²+√𝑥1𝑛𝑦−𝑦4𝑛²𝑦²+1+2𝑛𝑦²≤1√𝑥√𝑦√𝑦1𝑛𝑥+√𝑥1𝑛𝑦=2√𝑛. (40) (iii) Taking 𝑢_𝑛(𝑥)=𝑥−1/2𝑛 and 𝑣_𝑛(𝑦)=𝑦−1/2𝑛 in (23), we have 𝐷_𝑛𝑓;𝑥,𝑦−𝑓𝑥,𝑦≤2𝜔𝑓^∗;𝛿_𝑛𝑥,𝑦, (41) where 𝛿_𝑛1𝑥,𝑦∶=√𝑥12𝑛²+𝑥𝑛−12𝑛²+1√𝑦12𝑛²+𝑦𝑛−12𝑛²=12√𝑥√𝑦√𝑥1𝑛²+√4𝑛𝑦−1𝑦1𝑛²()≤14𝑛𝑥−1√𝑥√𝑦√𝑥1𝑛𝑦+√𝑦1𝑛𝑥=2√𝑛. (42)

Corollary 8. Let 𝑓^∗(𝑥,𝑦)=𝑓(𝑥²,𝑦²), and let 𝑓^∗∈𝐿𝑖𝑝_𝑀(𝛼)∶=𝑓^∗∈𝐶_𝐵)×[∶𝑓([0,∞0,∞))_∗(𝐭)−𝑓^∗(𝐱)≤𝑀‖𝐭−𝐱‖^𝛼;𝑡,𝑠;𝑥,𝑦∈(0,∞), (43) where 𝐭=(𝑡,𝑠),𝐱=(𝑥,𝑦),𝑀 is any positive constant, and 0<𝛼≤1. Then 𝐷_𝑛4𝑓;𝑥,𝑦−𝑓𝑥,𝑦≤𝑀𝑛^𝛼/2, (44) uniformly as 𝑛→∞, 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+4𝑛²𝑦²+1)/2𝑛,(iii)𝑢_𝑛(𝑥)=𝑥−1/2𝑛 and 𝑣_𝑛(𝑦)=𝑦−1/2𝑛.

Proof. (i) Taking 𝑢_𝑛(𝑥)=𝑥 and 𝑣_𝑛(𝑦)=𝑦 in (23), we directly have 𝐷_𝑛4𝑓;𝑥,𝑦−𝑓𝑥,𝑦≤𝑀𝑛^𝛼/2. (45) (ii) Taking 𝑢_𝑛(𝑥)=(−1+√4𝑛²𝑥²+1)/2𝑛 and 𝑣_𝑛(𝑦)=(−1+4𝑛²𝑦²+1)/2𝑛 in (23) gives 𝐷_𝑛𝑓;𝑥,𝑦−𝑓𝑥,𝑦≤𝑀𝛿_𝑛^𝛼/2𝑥,𝑦, (46)  where 𝛿_𝑛𝑥,𝑦∶=𝛿_𝑛1𝑥,𝑦∶=√𝑥√𝑦×√𝑦1𝑛𝑥−𝑥√4𝑛²𝑥²+1+2𝑛𝑥²+√𝑥1𝑛𝑦−𝑦4𝑛²𝑦²+1+2𝑛𝑦²≤1√𝑥√𝑦√𝑦1𝑛𝑥+√𝑥1𝑛𝑦=2√𝑛. (47) (iii) Taking 𝑢_𝑛(𝑥)=𝑥−1/2𝑛 and 𝑣_𝑛(𝑦)=𝑦−1/2𝑛 in (23), we have 𝐷_𝑛𝑓;𝑥,𝑦−𝑓𝑥,𝑦≤𝑀𝛿_𝑛^𝛼/2𝑥,𝑦, (48)  where 𝛿_𝑛1𝑥,𝑦∶=√𝑥12𝑛²+𝑥𝑛−12𝑛²+1√𝑦12𝑛²+𝑦𝑛−12𝑛²=12√𝑥√𝑦√𝑥1𝑛²+√4𝑛𝑦−1𝑦1𝑛²()≤14𝑛𝑥−1√𝑥√𝑦√𝑥1𝑛𝑦+√𝑦1𝑛𝑥=2√𝑛. (49)

Remark 9. Corollaries 7 and 8 conclude that 𝑓 is a real continuous and bounded function on [0,∞)×[0,∞)and if 𝑓^∗(𝑥,𝑦)=𝑓(𝑥²,𝑦²) is uniformly continuous on [0,∞)×[0,∞), then 𝐷_𝑛(𝑓) converges uniformly to 𝑓 as 𝑛→∞. Note that the one variable version of Corollary 7 was given in [16].

Corollary 10. Take 𝑢_𝑛(𝑥)=𝑢_𝑛⁽²⁾=−+𝑥,𝛾𝑛𝛾+14𝑛²𝑥²+𝛾𝑥+𝑛𝛾+1²,𝑣2𝑛_𝑛𝑦=𝑣_𝑛⁽²⁾=−+𝑦,𝛽𝑛𝛽+14𝑛²𝑦²+𝛽𝑦+𝑛𝛽+1²,2𝑛 (50) where 𝛼,𝛽∈ℝ.Then  (i) for any 𝑓∈𝐿𝑖𝑝^∗_𝑀(𝛼), 𝛼∈(0,1] and for each 𝑥,𝑦∈(0,∞), 𝑛∈ℕ, one has 𝐷_𝑛≤𝑀𝑓;𝑥,𝑦−𝑓𝑥,𝑦2𝑛𝑥+𝑦^𝛼/2𝛿𝑥,𝛾+𝛿𝑦,𝛽^𝛼/2, (51) where 𝛿=×𝑥,𝛾2𝑥+𝛾𝑛𝛾−𝑛²2𝑥+𝛾²+2𝑛𝛾+1+2𝑛𝑥+1, (52)  (ii) let 𝑓^∗(𝑥,𝑦)=𝑓(𝑥²,𝑦²). Then one has for each 𝑥,𝑦>0𝐷_𝑛𝑓;𝑥,𝑦−𝑓𝑥,𝑦≤2𝜔𝑓^∗;𝛿𝑥,𝛾+𝛿𝑦,𝛽, (53) where 𝛿=1𝑥,𝛾√×12𝑥𝑛2𝑥+𝛾𝑛𝛾−𝑛²2𝑥+𝛾²+2𝑛𝛾+1+2𝑛𝑥+1^1/2, (54)  (iii) let 𝑓^∗(𝑥,𝑦)=𝑓(𝑥²,𝑦²), and let 𝑓^∗∈𝐿𝑖𝑝_𝑀(𝛼)∶=𝑓^∗∈𝐶_𝐵)×[∶𝑓([0,∞0,∞))_∗(𝐭)−𝑓^∗(𝐱)≤𝑀‖𝐭−𝐱‖^𝛼;𝑡,𝑠;𝑥,𝑦∈(0,∞), (55) where 𝐭=(𝑡,𝑠), 𝐱=(𝑥,𝑦), 𝑀 is any positive constant, and 0<𝛼≤1. Then 𝐷_𝑛𝑓;𝑥,𝑦−𝑓𝑥,𝑦≤𝑀𝛿(𝑥,𝛾)+𝛿(𝑦,𝛽)^𝛼/2, (56) where 𝛿(𝑥,𝛾) is the same given in Corollary 10(𝑖𝑖).

It should be mentioned that, for 𝛼=0 and 𝛽=0, 𝑢_𝑛⁽²⁾(𝑥,0)=(−1+√4𝑛²𝑥²+1)/2𝑛 and 𝑣_𝑛⁽²⁾(𝑦,0)=(−1+4𝑛²𝑦²+1)/2𝑛. Therefore, Corollary 10(i), Corollary 10(ii), and Corollary 10(iii) reduce to Corollary 6(ii), Corollary 7(ii), and Corollary 8(ii), respectively.