About this Journal Submit a Manuscript Table of Contents
Journal of Applied Mathematics
Volume 2012 (2012), Article ID 546784, 28 pages
http://dx.doi.org/10.1155/2012/546784
Research Article

Approximation by Lupas-Type Operators and Szász-Mirakyan-Type Operators

1Department of Mathematics Education, Sungkyunkwan University, Seoul 110-745, Republic of Korea
2Department of Mathematics, Meijo University, Nagoya 468-8502, Japan

Received 28 July 2011; Accepted 5 January 2012

Academic Editor: Yuantong Gu

Copyright © 2012 Hee Sun Jung and Ryozi Sakai. 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.

Abstract

Lupas-type operators and Szász-Mirakyan-type operators are the modifications of Bernstein polynomials to infinite intervals. In this paper, we investigate the convergence of Lupas-type operators and Szász-Mirakyan-type operators on [0,).

1. Introduction and Main Results

For 𝑓𝐶([0,1]), Bernstein operator 𝐵𝑛𝑓(𝑥) is defined as follows: Let𝑝𝑛,𝑘𝑛𝑘𝑥(𝑥)=𝑘(1𝑥)𝑛𝑘,0𝑘𝑛,(1.1) and then we define𝐵𝑛𝑓(𝑥)=𝑛𝑘=0𝑝𝑛,𝑘𝑘(𝑥)𝑓𝑛.(1.2) Derriennic [1] gave a modified operator of 𝐵𝑛𝑓 such as𝑀𝑛𝑓(𝑥)=(𝑛+1)𝑛𝑘=0𝑝𝑛,𝑘(𝑥)10𝑝𝑛,𝑘(𝑡)𝑓(𝑡)𝑑𝑡,(1.3) and obtained the result that for 𝑓𝐶(2)([0,1]),lim𝑛𝑀𝑛𝑓=(𝑥)𝑓(𝑥)(12𝑥)𝑓(𝑥)+𝑥(1𝑥)𝑓(𝑥).(1.4) Lupas investigated a family of linear positive operators which mapped the class of all bounded and continuous functions on [0,) into 𝐶[0,) such that𝐿𝑛𝑓(𝑥)=𝑘=0𝑘𝑥𝑛+𝑘1𝑘(1+𝑥)𝑛+𝑘𝑓𝑘𝑛[,𝑥0,).(1.5) Moreover, Sahai and Prasad [2] modified Lupas operators as follows: Let 𝑓 be integrable on [0,) and let 𝑛 be a positive integer. Then we define𝑀𝑛[𝑓](𝑥)=(𝑛1)𝑘=0𝑃𝑛,𝑘(𝑥)0𝑃𝑛,𝑘([𝑦)𝑓(𝑦)𝑑𝑦,𝑥0,),(1.6) where𝑃𝑛,𝑘𝑘𝑥(𝑥)=𝑛+𝑘1𝑘(1+𝑥)𝑛+𝑘.(1.7) In this paper, we assume that 𝑛 is a positive integer. Then they obtained the following;

Theorem 1.1 (see [2], Theorem 1). If 𝑓 is integrable on [0,) and admits its (𝑟+1)th and (𝑟+2)th derivatives, which are bounded at a point 𝑥[0,), and 𝑓(𝑟)(𝑥)=𝑂(𝑥𝛼) (𝛼 is a positive integer 2) as 𝑥, then lim𝑛𝑛𝑀𝑛[𝑓](𝑟)(𝑥)𝑓(𝑟)(𝑥)=(𝑟+1)(12𝑥)𝑓(𝑟+1)(𝑥)+𝑥(1𝑥)𝑓(𝑟+2)(𝑥).(1.8)

Theorem 1.1 holds only for bounded 𝑥𝐾, so it does not mean the norm convergence on [0,). In this paper, we improve Theorem 1.1 with respect to the norm convergence on [0,).

Let 0<𝑝 and let 𝑤 be a positive weight, that is, 𝑤(𝑥)0 for 𝑥. For a function 𝑔 on [0,), we define the norm by 𝑔𝐿𝑝([0,))=[0,)||||𝑔(𝑡)𝑝𝑑𝑡1/𝑝,0<𝑝<sup[0,)||||𝑔(𝑡),𝑝=.(1.9) For convenience, for nonnegative integers 𝑛2, 𝑟, and 𝑛𝑟20, we let 𝐴𝑛,𝑟=(𝑛1)!(𝑛2)!(.𝑛𝑟2)!(𝑛+𝑟1)!(1.10) Then we have the following results:

Theorem 1.2. Let 0<𝑝. Let 𝛼 and 𝑟 be nonnegative integers and 𝑛𝑟20. Let 𝑓𝐶(𝑟+1)([0,)) satisfy ||𝑓(𝑟)||(𝑥)𝑂(1)(𝑥+1)𝛼,||𝑓(𝑟+1)||(𝑥)𝑂(1)(𝑥+1)𝛼+2.(1.11) Then we have uniformly for 𝑓 and 𝑛, |||𝐴𝑛,𝑟𝑀𝑛[𝑓](𝑟)(𝑥)𝑓(𝑟)|||1(𝑥)=𝑂𝑛1/3(𝑥+1)𝛼+2.(1.12) In particular, if (𝑥+1)𝛼+2𝑤(𝑥)𝐿𝑝([0,))<, then we have uniformly for 𝑛, 𝐴𝑛,𝑟𝑀𝑛[𝑓](𝑟)(𝑥)𝑓(𝑟)(𝑥)𝑤(𝑥)𝐿𝑝([0,))1=𝑂𝑛1/3.(1.13)

Remark 1.3. (a) We see that for nonnegative integers 𝑛2, 𝑟, and 𝑛𝑟20, (𝑛1)!(𝑛2)!(𝑛𝑟2)!(𝑛+𝑟1)!1as𝑛.(1.14)
(b) The following weight is useful. 𝑤𝜆1(𝑥)=1+𝑥𝜆1𝜆>𝑝+𝛼+2,0<𝑝<,𝜆𝛼+2,𝑝=.(1.15)

Let 1𝜓(𝑥)=.1+𝑥(1.16)

Theorem 1.4. Let 𝑟 and 𝛽 be nonnegative integers and 𝑛𝑟20. Let 𝑓𝐶(𝑟+2)([0,)) satisfy 𝑓(𝑟+1)(𝑥)𝜓2𝛽+1(𝑥)𝐿([0,))𝑓<,(𝑟+2)(𝑥)𝜓2𝛽(𝑥)𝐿([0,))<.(1.17) Then we have uniformly for 𝑓 and 𝑛, |||𝐴𝑛,𝑟𝑀𝑛[𝑓](𝑟)(𝑥)𝑓(𝑟)|||𝜓(𝑥)2𝛽+21(𝑥)𝑂𝑛𝑓(𝑟+1)(𝑥)𝜓2𝛽+1(𝑥)𝐿([0,))+𝑓(𝑟+2)(𝑥)𝜓2𝛽(𝑥)𝐿([0,)).(1.18)

Let us define the weighted modulus of smoothness by 𝜔𝑘(𝑓;𝜂;𝑡)=sup0𝑡Δ𝑘𝑓()𝜂()𝐿([0,)),𝑡0,𝑘=1,2,(1.19) where Δ1𝑓(𝑥)=𝑓(𝑥+)𝑓(𝑥),Δ2𝑓(𝑥)=𝑓(𝑥)2𝑓(𝑥+)+𝑓(𝑥+2).(1.20)

Theorem 1.5. Let 𝛽 and 𝑟 be nonnegative integers and 𝑛𝑟20. Let 𝑓𝐶𝑟([0,)). Then we have uniformly for 𝑓 and 𝑛, 𝐴𝑛,𝑟𝑀𝑛[𝑓](𝑟)(𝑥)𝑓(𝑟)𝜓(𝑥)2𝛽+2(𝑥)𝐿([0,))1𝐶𝑛𝜔1𝑓(𝑟);𝜓2𝛽+1;1𝑛+𝜔2𝑓(𝑟);𝜓2𝛽;1𝑛.(1.21)

The Szász-Mirakyan operators are also generalizations of Bernstein polynomials on infinite intervals. They are defined by: 𝑆𝑛(𝑓)(𝑥)=𝑘=0𝑆𝑛,𝑘(𝑘𝑥)𝑓𝑛,(1.22) where 𝑆𝑛,𝑘𝑒(𝑥)=𝑛𝑥(𝑛𝑥)𝑘.𝑘!(1.23)

In [3], the class of Szász-Mirakyan operators 𝑆𝑛;𝑟,𝑞(𝑓;𝑥) was defined as follows: 𝑆𝑛;𝑟,𝑞(1𝑓;𝑥)=𝐴𝑟(𝑛𝑥)𝑘=0(𝑛𝑥)𝑟𝑘𝑓(𝑟𝑘)!𝑟𝑘[𝑛+𝑞,𝑥0,),(1.24) where 𝑞>0 and 𝐴𝑟(𝑡)=𝑘=0𝑡𝑟𝑘[(𝑟𝑘)!,𝑡0,).(1.25)

Theorem 1.6 (see [3]). Let 𝑞>0 and 𝑟 be fixed numbers. Then there exists 𝑀𝑞,𝑟=const. >0 depending only on 𝑞 and 𝑟 such that, for every uniformly continuous and bounded function 𝑓(𝑥)𝑒𝑞𝑥 on [0,), the following inequalities hold;(a)𝑆𝑛;𝑞,𝑟𝜑(𝑓;𝑥)𝑓(𝑥)(𝑥)𝑒𝑞𝑥𝐿([0,))𝑀𝑞,𝑟1𝑓𝑛+𝑞(𝑥)𝑒𝑞𝑥𝐿([0,))+𝑓(𝑥)𝑒𝑞𝑥𝐿([0,));(1.26)(b)𝑆𝑛;𝑞,𝑟𝜑(𝑓;𝑥)𝑓(𝑥)(𝑥)𝑒𝑞𝑥𝐿([0,))𝑀𝑞,𝑟1𝛿𝑛+𝑞𝑛,𝑞𝜔1𝑓;𝑒𝑞𝑥;𝛿𝑛,𝑞+𝜔2𝑓;𝑒𝑞𝑥;𝛿𝑛,𝑞,(1.27) where 𝛿𝑛,𝑞=(𝑛+𝑞)1/2. (c) for every fixed 𝑥[0,), we have for every continuous 𝑓 with 𝑓(𝑗)(𝑥)𝑒𝑞𝑥, 𝑗=0,1,2, bounded on [0,),lim𝑛𝑛𝑆𝑛;𝑞,𝑟(𝑓;𝑥)𝑓(𝑥)=𝑞𝑥𝑓𝑥(𝑥)+2𝑓(𝑥).(1.28)

Now, we modify the Szász-Mirakyan operators as follows: let 𝑓 be integrable on [0,), then we define𝑄𝑛,𝛽[𝑓](𝑥)=(𝑛+𝛽)𝑘=0𝑆𝑛,𝑘(𝑥)0𝑆𝑛+𝛽,𝑘([𝑦)𝑓(𝑦)𝑑𝑦,𝑥0,),(1.29) where 𝛽 is a nonnegative integer. Then we have the following results:

Theorem 1.7. Let 𝛼, 𝛽 and 𝑟 be nonnegative integers. Let 𝑓𝐶(𝑟+1)([0,)) satisfies ||𝑓(𝑟)||(𝑥)𝑂(1)𝑒𝛽𝑥(𝑥+1)𝛼,||𝑓(𝑟+1)||(𝑥)𝑂(1)𝑒𝛽𝑥(𝑥+1)𝛼+2.(1.30) Then one has uniformly for 𝑓 and 𝑛, ||||𝑛+𝛽𝑛𝑟𝑄𝑛,𝛽[𝑓](𝑟)(𝑥)𝑓(𝑟)||||1(𝑥)=𝑂𝑛1/3𝑒𝛽𝑥(𝑥+1)𝛼+2.(1.31) In particular, let 0<𝑝. If one supposes 𝑒𝛽𝑥(𝑥+1)𝛼+2𝑤(𝑥)𝐿𝑝([0,))<, then one has uniformly for 𝑓 and 𝑛, 𝑛+𝛽𝑛𝑟𝑄𝑛,𝛽[𝑓](𝑟)(𝑥)𝑓(𝑟)(𝑥)𝑤(𝑥)𝐿𝑝([0,))1=𝑂𝑛1/3.(1.32)

Remark 1.8. (a) We note that for nonnegative integers 𝛽 and 𝑟, 𝑛+𝛽𝑛𝑟1as𝑛.(1.33)
(b) The following weight is useful. 𝑤𝜆,𝛽(𝑥)=𝑒𝛽𝑥𝑤𝜆(𝑥),(1.34) where 𝑤𝜆(𝑥) is defined in Remark 1.3.

Theorem 1.9. Let 𝛽, 𝛾, and 𝑟 be nonnegative integers. Let 𝑓𝐶(𝑟+2)([0,)) satisfies 𝑓(𝑟+1)(𝑥)𝑒𝛽𝑥𝜓2𝛾+1(𝑥)𝐿([0,))𝑓<,(𝑟+2)(𝑥)𝑒𝛽𝑥𝜓2𝛾(𝑥)𝐿([0,))<.(1.35) Then one has uniformly for 𝑓 and 𝑛, ||||𝑛+𝛽𝑛𝑟𝑄𝑛,𝛽[𝑓](𝑟)(𝑥)𝑓(𝑟)𝑒(𝑥)𝛽x𝜓2𝛾+2||||1(𝑥)𝑂𝑛𝑓(𝑟+1)(𝑥)𝑒𝛽𝑥𝜓2𝛾+1(𝑥)𝐿([0,))+𝑓(𝑟+2)(𝑥)𝑒𝛽𝑥𝜓2𝛾(𝑥)𝐿([0,)).(1.36)

Theorem 1.10. Let 𝛽, 𝛾, and 𝑟 be nonnegative integers. Then one has for 𝑓𝐶𝑟([0,)), 𝑛+𝛽𝑛𝑟𝑄𝑛,𝛽[𝑓](𝑟)(𝑥)𝑓(𝑟)𝑒(𝑥)𝛽𝑥𝜓2𝛾+2(𝑥)𝐿([0,))1𝐶𝑛𝜔1𝑓;𝑒𝛽𝑥𝜓2𝛾+11(𝑥);𝑛+𝜔2𝑓;𝑒𝛽𝑥𝜓2𝛾1(𝑥);𝑛.(1.37)

2. Proofs of Results

First, we will prove results for Lupas-type operators such as Theorems 1.2, 1.4, and 1.5. To prove theorems, we need some lemmas.

Lemma 2.1. Let 𝑚 and 𝑟 be nonnegative integers and 𝑛>𝑚+𝑟+1. Let 𝑇𝑛,𝑚,𝑟(𝑥)=(𝑛𝑟1)𝑘=0𝑃𝑛+r,𝑘(𝑥)0𝑃𝑛𝑟,𝑘+𝑟(𝑦)(𝑦𝑥)𝑚𝑑𝑦.(2.1) Then(i)𝑇𝑛,0,𝑟(𝑥)=1, (ii)𝑇𝑛,1,𝑟(𝑥)=(𝑟+1)(1+2𝑥),𝑇(𝑛𝑟+2)𝑛,2,𝑟(𝑥)=2(𝑛1)𝑥(1+𝑥)+(𝑛𝑟2)(𝑛𝑟3)(𝑟+1)(𝑟+2)(1+2𝑥)2;(𝑛𝑟2)(𝑛𝑟3)(2.2)(iii)for 𝑚1, (𝑛𝑚𝑟2)𝑇𝑛,𝑚+1,𝑟𝑇(𝑥)=𝑥(1+𝑥)𝑛,𝑚,𝑟(𝑥)+2𝑚𝑇𝑛,𝑚1,𝑟(𝑥)+(𝑚+𝑟+1)(1+2𝑥)𝑇𝑛,𝑚,𝑟,(2.3) where 𝑛>𝑚+𝑟+2; (iv)for 𝑚0, 𝑇𝑛,𝑚,𝑟1(𝑥)=𝑂𝑛[(𝑚+1)/2]𝑞𝑛,𝑚,𝑟(𝑥),(2.4) where 𝑞𝑛,𝑚,𝑟(𝑥) is a polynomial of degree 𝑚 such that the coefficients are bounded independently of 𝑛 and they are positive for 𝑛>𝑚+𝑟+1.

Proof. (i), (ii), and (iii) have been proved in [2, Lemma 1]. So we may show only the part of (2.4). For 𝑚=1,2, (2.4) holds. Let us assume (2.4) for 𝑚(2). We note 𝑇𝑛,𝑚,𝑟1(𝑥)=𝑂𝑛[(𝑚+1)/2]𝑞𝑚,𝑟(𝑥),𝑞𝑚,𝑟(𝑥)𝒫𝑚1.(2.5) So, we have by the assumption of induction, (𝑛𝑚𝑟2)𝑇𝑛,𝑚+1,𝑟𝑇(𝑥)=𝑥(1+𝑥)𝑛,𝑚,𝑟(𝑥)+2𝑚𝑇𝑛,𝑚1,𝑟(𝑥)+(𝑚+𝑟+1)(1+2𝑥)𝑇𝑛,𝑚,𝑟𝐶1𝑥(1+𝑥)𝑛[(𝑚+1)/2]𝑞𝑚,𝑟1(𝑥)+2𝑚𝑛[𝑚/2]𝑞𝑚1,𝑟+1(𝑥)(𝑚+𝑟+1)(1+2𝑥)𝑛[(𝑚+1)/2]𝑞𝑚,𝑟(𝑥).(2.6) Here, if 𝑚 is even, then 𝑚+12𝑚+1=2+1=𝑚+22=𝑚+22,𝑚2𝑚+1=2+1=𝑚+22=𝑚+22,(2.7) and if 𝑚 is odd, then 𝑚+12+1=𝑚+12+1=𝑚+22,𝑚2+1=𝑚12+1=𝑚+12=𝑚+22.(2.8) Hence, we have 𝑇𝑛,𝑚+1,𝑟1(𝑥)=𝑂𝑛[(𝑚+2)/2]𝑞𝑛,𝑚+1,𝑟(𝑥),(2.9) and here we see that 𝑞𝑛,𝑚+1,𝑟(𝑥) is a polynomial of degree 𝑚+1 such that the coefficients of 𝑞𝑛,𝑚+1,𝑟(𝑥) are bounded independently of 𝑛. Moreover, we see from (2.6) that the coefficients of 𝑞𝑛,𝑚+1,𝑟(𝑥) are positive for 𝑛>𝑚+𝑟+2.

Lemma 2.2 (see [2, Lemma 2]). Let 𝑟 be a nonnegative integer and 𝑛𝑟20. Then one has for 𝑓𝐶𝑟([0,)): 𝑀𝑛[𝑓](𝑟)(𝑥)=(𝑛𝑟1)!(𝑛+𝑟1)!(𝑛1)!(𝑛2)!𝑘=0𝑃𝑛+𝑟,𝑘(𝑥)0𝑃𝑛𝑟,𝑘+𝑟(𝑦)𝑓(𝑟)(𝑦)𝑑𝑦.(2.10)

Let𝑀𝑛[𝑓](𝑟)(𝑥)=(𝑛𝑟1)𝑘=0𝑃𝑛+𝑟,𝑘(𝑥)0𝑃𝑛𝑟,𝑘+𝑟(𝑦)𝑓(𝑟)(𝑦)𝑑𝑦.(2.11) Then we have𝑀𝑛[𝑓](𝑟)(𝑥)=𝐴𝑛,𝑟𝑀𝑛[𝑓](𝑟)(𝑥),(2.12) where 𝐴𝑛,𝑟 is defined by (1.10).

Proof of Theorem 1.2. Let |𝑦𝑥|1. By the second inequality in (1.11), ||𝑓(𝑟)(𝑦)𝑓(𝑟)||=||||||𝑓(𝑥)𝑦𝑥(𝑟+1)||||||||𝜉(𝜉)𝑂(1)𝑦𝑥𝛼+2||||||(𝑂(1)𝑦𝑥𝑥+1)𝛼+2.(2.13) Let 𝜀=𝑛𝛾,0<𝛾<1, ||||𝑀𝑛[𝑓](𝑟)(𝑥)𝑓(𝑟)||||=|||||(𝑥)(𝑛𝑟1)𝑘=0𝑃𝑛+𝑟,𝑘×(𝑥)|𝑦𝑥|<𝜀𝑃𝑛𝑟,𝑘+𝑟||𝑓(𝑦)(𝑟)(𝑦)𝑓(𝑟)||(𝑥)𝑑𝑦+|𝑦𝑥|𝜀𝑃𝑛𝑟,𝑘+𝑟||𝑓(𝑦)(𝑟)(𝑦)𝑓(𝑟)|||||||(𝑥)𝑑𝑦=𝐴+𝐵.(2.14) First, we see by (2.13) and Lemma 2.1, |||||𝐴=𝑂(1)(𝑛𝑟1)𝑘=0𝑃𝑛+𝑟,𝑘(𝑥)|𝑦𝑥|<𝜀𝑃𝑛𝑟,𝑘+𝑟||||(𝑦)𝑦𝑥(𝑥+1)𝛼+2|||||||𝑇𝑑𝑦𝑂(1)𝜀𝑛,0,𝑟||(𝑥)(𝑥+1)𝛼+2=𝑂(1)𝜀(𝑥+1)𝛼+2.(2.15) Next, we estimate 𝐵. By the first inequality in (1.11), |||||𝐵𝐶(𝑛𝑟1)𝑘=0𝑃𝑛+𝑟,𝑘(𝑥)|𝑦𝑥|>𝜀𝑃𝑛𝑟,𝑘+𝑟||𝑓(𝑦)(𝑟)||+||𝑓(𝑦)(𝑟)||||||||||||(𝑥)𝑑𝑦𝐶(𝑛𝑟1)𝑘=0𝑃𝑛+𝑟,𝑘(𝑥)|𝑦𝑥|>𝜀𝑃𝑛𝑟,𝑘+𝑟(𝑦)((𝑦+1)𝛼+(𝑥+1)𝛼|||||.)𝑑𝑦(2.16) Here, using (𝑦+1)𝛼=((𝑦𝑥)+𝑥+1)𝛼=𝛼𝑖=0𝛼𝑖(𝑦𝑥)𝑖(𝑥+1)𝛼𝑖(2.17) and the notation: 𝑖=1,(𝑖odd)0,(𝑖even),(2.18) we have ||||𝐵𝐶(𝑛𝑟1)𝑘=0𝑃𝑛+𝑟,𝑘(𝑥)|𝑦𝑥|>𝜀𝑃𝑛𝑟,𝑘+𝑟(𝑦)×𝛼𝑖=1𝛼𝑖(𝑦𝑥)𝑖(𝑥+1)𝛼𝑖+2(𝑥+1)𝛼||(||𝑑𝑦𝐶𝑛𝑟1)𝑘=0𝑃𝑛+𝑟,𝑘(𝑥)|𝑦𝑥|>𝜀𝑃𝑛𝑟,𝑘+𝑟(𝑦)×𝛼𝑖=1𝛼𝑖(𝑦𝑥)𝑖|||𝑦𝑥𝜀|||𝑖(𝑥+1)𝛼𝑖||||𝑑𝑦+𝐶(𝑛𝑟1)𝑘=0𝑃𝑛+𝑟,𝑘(𝑥)|𝑦𝑥|>𝜀𝑃𝑛𝑟,𝑘+𝑟(𝑦)𝑦𝑥𝜀2(𝑥+1)𝛼𝑑𝑦=𝐵1+𝐵2.(2.19) Then, we obtain 𝐵1𝐶𝛼𝑖=1𝛼𝑖||𝑇𝑛,𝑖+𝑖,𝑟||1𝜀𝑖(𝑥+1)𝛼𝑖𝐶𝛼𝑖=1𝛼𝑖𝑂𝑛𝛾𝑖𝑛[(𝑖+𝑖+1)/2]||𝑞𝑛,𝑖+𝑖,𝑟(||(𝑥)𝑥+1)𝛼𝑖1𝑂𝑛[(𝑖+𝑖+1)/2]𝛾𝑖(𝑥+1)𝛼+𝑖1𝑂𝑛1𝛾(𝑥+1)𝛼+1.(2.20) Here, we used the following that for 𝑖1, 𝑖+𝑖+12𝛾𝑖1𝛾,(2.21) because 𝑖+𝑖+12𝛾𝑖=𝑖+12𝑖𝛾,𝑖odd,2,𝑖even.(2.22) And we know that 𝐵2||𝑇𝐶𝑛,2,𝑟||1(𝑥)𝜀2𝑥𝛼1𝑂𝑛[3/2]||𝑞𝑛,2,𝑟||1(𝑥)𝜀2𝑥𝛼𝑛𝑂2𝛾𝑛[3/2]||𝑞𝑛,2,𝑟||𝑥(𝑥)𝛼1𝑂𝑛12𝛾(𝑥+1)𝛼+2.(2.23) Thus, we obtain 1𝐵𝑂𝑛12𝛾(𝑥+1)𝛼+2.(2.24) Therefore, we have uniformly on 𝑛, ||||𝑀𝑛[𝑓](𝑟)(𝑥)𝑓(𝑟)||||1(𝑥)𝑂𝑛𝛾(𝑥+1)𝛼+21+𝑂𝑛12𝛾(𝑥+1)𝛼+2.(2.25) Here, if we let 𝛾=1/3, then we have ||||𝑀𝑛[𝑓](𝑟)(𝑥)𝑓(𝑟)||||1(𝑥)=𝑂𝑛1/3(𝑥+1)𝛼+2,(2.26) that is, (1.12) is proved. So, we also have a norm convergence (1.13).

Proof of Theorem 1.4. We know that for 𝑓𝐶(𝑟+2)([0,)), 𝑓(𝑟)(𝑡)=𝑓(𝑟)(𝑥)+𝑓(𝑟+1)(𝑥)(𝑡𝑥)+𝑡𝑥(𝑡𝑢)𝑓(𝑟+2)(||||𝑢)𝑑𝑢,(2.27)𝑡𝑥(𝑡𝑢)𝑓(𝑟+2)||||𝑓(𝑢)𝑑𝑢𝐶(𝑟+2)(𝑥)𝜓2𝛽(𝑥)𝐿([0,))(1+𝑥)2𝛽+(1+𝑡)2𝛽(𝑡𝑥)2,(2.28) where 𝜓(𝑡)=1/(1+𝑥). Then we obtain from (2.10) and (2.27), 𝑀𝑛[𝑓](𝑟)(𝑥)=𝑓(𝑟)(𝑥)+𝑓(𝑟+1)(𝑥)𝑇𝑛,1,𝑟(𝑥)+(𝑛𝑟1)𝑘=0𝑃𝑛+𝑟,𝑘(𝑥)0𝑃𝑛𝑟,𝑘+𝑟(𝑦)𝑦𝑥(𝑦𝑢)𝑓(𝑟+2)(𝑢)𝑑𝑢𝑑𝑦(2.29) and from (2.28), |||||(𝑛𝑟1)𝑘=0𝑃𝑛+𝑟,𝑘(𝑥)0𝑃𝑛𝑟,𝑘+𝑟(𝑦)𝑦𝑥(𝑦𝑢)𝑓(𝑟+2)|||||𝑓(𝑢)𝑑𝑢𝑑𝑦(𝑟+2)(𝑥)𝜓2𝛽(𝑥)𝐿([0,))×(1+𝑥)2𝛽||𝑇𝑛,2,𝑟||+|||||(𝑥)(𝑛𝑟1)𝑘=0𝑃𝑛+𝑟,𝑘(𝑥)0𝑃𝑛𝑟,𝑘+𝑟(𝑦)(1+𝑦)2𝛽(𝑦𝑥)2|||||.𝑑𝑦(2.30) Using (1+𝑦)2𝛽𝐶((𝑦𝑥)2𝛽+(1+𝑥)2𝛽), we have |||||(𝑛𝑟1)𝑘=0𝑃𝑛+𝑟,𝑘(𝑥)0𝑃𝑛𝑟,𝑘+𝑟(𝑦)(1+𝑦)2𝛽(𝑦𝑥)2|||||||𝑇𝑑𝑦𝐶𝑛,2𝛽+2,𝑟||(𝑥)+(1+𝑥)2𝛽||𝑇𝑛,2,𝑟||.(𝑥)(2.31) Therefore, we have ||||𝑀𝑛[𝑓](𝑟)(𝑥)𝑓(𝑟)||||𝜓(𝑥)2𝛽+2||𝑓(𝑥)(𝑟+1)(𝑥)𝜓2𝛽+1||||𝑇(𝑥)𝑛,1,𝑟||𝑓(𝑥)𝜓(𝑥)+𝐶(𝑟+2)(𝑥)𝜓2𝛽(𝑥)𝐿([0,))(1+𝑥)2𝛽||𝑇𝑛,2,𝑟(||𝜓𝑥)2𝛽+2(𝑓𝑥)+𝐶(𝑟+2)(𝑥)𝜓2𝛽(𝑥)𝐿([0,))||𝑇𝑛,2𝛽+2,𝑟||𝜓(𝑥)2𝛽+21(𝑥)𝑂𝑛||𝑓(𝑟+1)(𝑥)𝜓2𝛽+1||||𝑔(𝑥)𝑛,1,𝑟||𝜓1(𝑥)(𝑥)+𝑂𝑛𝑓(𝑟+2)(𝑥)𝜓2𝛽(𝑥)𝐿([0,))(1+𝑥)2𝛽||𝑔𝑛,2,𝑟||𝜓(𝑥)2𝛽+21(𝑥)+𝑂𝑛𝛽+1𝑓(𝑟+2)(𝑥)𝜓2𝛽(𝑥)𝐿([0,))||𝑔𝑛,2𝛽+2,𝑟(||𝜓𝑥)2𝛽+2(𝑥).(2.32) Since we know that for 𝑥[0,), ||𝑔𝑛,1,𝑟||(𝑥)𝜓(𝑥)𝐶,(1+𝑥)2𝛽||𝑔𝑛,2,𝑟||𝜓(𝑥)2𝛽+2||𝑔(𝑥)𝐶,𝑛,2𝛽+2,𝑟||𝜓(𝑥)2𝛽+2(𝑥)𝐶,(2.33) we have ||||𝑀𝑛[𝑓](𝑟)(𝑥)𝑓(𝑟)||||𝜓(𝑥)2𝛽+21(𝑥)𝑂𝑛𝑓(𝑟+1)(𝑥)𝜓2𝛽+1(𝑥)𝐿([0,))+𝑓(𝑟+2)(𝑥)𝜓2𝛽(𝑥)𝐿([0,)).(2.34)

Lemma 2.3. Let 𝑟 and 𝛽 be nonnegative integers and 𝑛𝑟20. Let 𝑓𝐶(𝑟)([0,)) satisfies 𝑓(𝑟)𝜓2𝛽𝐿([0,))<.(2.35) Then one has uniformly for 𝑛, 𝑓 and 𝑥[0,), ||||𝑀𝑛[𝑓](𝑟)||||𝜓(𝑥)2𝛽𝑓(𝑥)𝐶(𝑟)𝜓2𝛽𝐿([0,)).(2.36)

Proof. Using (1+𝑦)2𝛽𝐶((𝑦𝑥)2𝛽+(1+𝑥)2𝛽), we have |||||(𝑛𝑟1)𝑘=0𝑃𝑛+𝑟,𝑘(𝑥)0𝑃𝑛𝑟,𝑘+𝑟(𝑦)(1+𝑦)2𝛽|||||𝜓𝑑𝑦𝐶2𝛽||𝑇(𝑥)+𝑛,2𝛽,𝑟||(𝑥).(2.37) The assumption (2.35) means ||𝑓(𝑟)||(𝑦)𝐶(1+𝑦)2𝛽.(2.38) Then we can obtain by (2.10), ||||𝑀𝑛[𝑓](𝑟)|||||||||(𝑥)𝐶(𝑛𝑟1)𝑘=0𝑃𝑛+𝑟,𝑘(𝑥)0𝑃𝑛𝑟,𝑘+𝑟(𝑦)(1+𝑦)2𝛽|||||𝑓𝑑𝑦(𝑟)𝜓2𝛽𝐿([0,))𝜓𝐶2𝛽||𝑇(𝑥)+𝑛,2𝛽,𝑟||𝑓(𝑥)(𝑟)𝜓2𝛽𝐿([0,))𝜓𝐶2𝛽1(𝑥)+𝑂𝑛𝛽||𝑞𝑛,2𝛽,𝑟||𝑓(𝑥)(𝑟)𝜓2𝛽𝐿([0,)).(2.39) Consequently, since |𝑞𝑛,2𝛽,𝑟(𝑥)|𝜓2𝛽(𝑥) is uniformly bounded on [0,), we have the result.

The Steklov function [𝑓](𝑥) for 𝑓𝐶([0,)) is defined as follows:[𝑓]4(𝑥)=20/2[]2𝑓(𝑥+𝑠+𝑡)𝑓(𝑥+2(𝑠+𝑡))𝑑𝑠𝑑𝑡,𝑥0,>0.(2.40)

Then for the Steklov function [𝑓](𝑥) with respect to 𝑓𝐶([0,)), we have the following properties.

Lemma 2.4 (cf.[4]). Let 𝑓(𝑥)𝐶([0,)) and 𝜂(𝑥) be a positive and nonincreasing function on [0,). Then (i) [𝑓](𝑥)𝐶2([0,));
(ii) [𝑓](𝑥)𝑓(𝑥)𝜂(𝑥)𝐿([0,))𝜔2𝑓;𝜂;2;(2.41)
(iii) [𝑓](𝑥)𝜂(𝑥)𝐿([0,))4𝜔1𝑓;𝜂;2𝜂(𝑥)+1𝜂(𝑥+(/2))𝜔1(𝑓;𝜂;)𝜂(𝑥);𝜂(𝑥+)(2.42)
(iv) [𝑓](𝑥)𝜂(𝑥)𝐿([0,))422𝜔2𝑓;𝜂;2+14𝜔2.(𝑓;𝜂;)(2.43)

Proof. (i) For 𝑓𝐶([0,)), we have the Steklov functions [𝑓](𝑥) and [𝑓](𝑥) as follows. We note [𝑓]4(𝑥)=20/2𝑥𝑥+/22𝑓(𝑢+𝑡)𝑑𝑢𝑥𝑥+12𝑓(𝑢+2𝑡)𝑑𝑢𝑑𝑡,𝑥0,>0.(2.44) Then, we can see from (2.44), [𝑓]4(𝑥)=20/22𝑓𝑥+21+𝑡𝑓(𝑥+𝑡)2=4(𝑓(𝑥++2𝑡)𝑓(𝑥+2𝑡))𝑑𝑡20/22Δ1/21𝑓(𝑥+𝑡)2Δ1𝑓(𝑥+2𝑡)𝑑𝑡.(2.45) Similarly to (2.44), we know [𝑓]4(𝑥)=22𝑥𝑥+/2𝑓𝑢+21𝑓(𝑢)𝑑𝑢4𝑥𝑥+.(𝑓(𝑢+)𝑓(𝑢))𝑑𝑢(2.46) Therefore, we have from (2.46), [𝑓]4(𝑥)=22𝑓(𝑥+)2𝑓𝑥+21+𝑓(𝑥)4=4(𝑓(𝑥+2)2𝑓(𝑥+)+𝑓(𝑥))22Δ2/21𝑓(𝑥)4Δ2.𝑓(𝑥)(2.47) Therefore, (i) is proved.
(ii) We easily see from (2.44) that ||𝑓[𝑓](𝑥)𝜂||=||||4(𝑥)(𝑥)20/2Δ2𝑠+𝑡||||𝑓(𝑥)𝜂(𝑥)𝑑𝑠𝑑𝑡𝜔2𝑓;𝜂;2.(2.48)
(iii) From (2.46), we have ||[𝑓]||||||4(𝑥)𝜂(𝑥)20/22Δ1/2𝑓(𝑥+𝑡)𝜂(𝑥+𝑡)𝜂(𝑥)||||+||||4𝜂(𝑥+𝑡)𝑑𝑡20/212Δ1𝑓(𝑥+2𝑡)𝜂(𝑥+2𝑡)𝜂(x)||||4𝜂(𝑥+2𝑡)𝑑𝑡𝜔1𝑓;𝜂;2𝜂(𝑥)+1𝜂(𝑥+/2)𝜔1(𝑓;𝜂;)𝜂(𝑥).𝜂(𝑥+)(2.49)
(iv) From (2.47), we have ||[𝑓]||4(𝑥)𝜂(𝑥)22𝜔2𝑓;𝜂;2+14𝜔2.(𝑓;𝜂;)(2.50)

Proof of Theorem 1.5. We know that for 𝑓(𝑥)𝐶𝑟([0,)), [𝑓](𝑟)𝑓(𝑥)=(𝑟)[𝑓](𝑥),(𝑟+1)𝑓(𝑥)=(𝑟)[𝑓](𝑥),(𝑟+2)𝑓(𝑥)=(𝑟)(𝑥).(2.51) Then, we have 𝑀𝑛[𝑓](𝑟)(𝑥)𝑓(𝑟)𝜓(𝑥)2𝛽+2(𝑥)𝐿([0,))𝑀𝑛[𝑓]𝑓(𝑟)(𝑥)𝜓2𝛽+2(𝑥)𝐿([0,))+𝑀𝑛[𝑓](𝑟)(𝑥)[𝑓(𝑟)](𝜓𝑥)2𝛽+2(𝑥)𝐿([0,))+[𝑓(𝑟)](𝑥)𝑓(𝑟)𝜓(𝑥)2𝛽+2(𝑥)𝐿([0,)).(2.52) From (2.51) and (2.41) of Lemma 2.4, 𝑀𝑛[𝑓]𝑓(𝑟)(𝑥)𝜓2𝛽+2(𝑥)𝐿([0,))𝑓(𝑟)[𝑓](𝑥)(𝑟)𝜓(𝑥)2𝛽+2(𝑥)𝐿([0,))=𝑓(𝑟)𝑓(𝑥)(𝑟)𝜓(𝑥)2𝛽+2(𝑥)𝐿([0,))𝜔2𝑓(𝑟);𝜓2𝛽+2.;(2.53) Here, we suppose 0<1 and then we know that 𝜓(𝑥)𝜓(𝑥+)2,𝜓(𝑥)𝜓(𝑥+/2)2.(2.54) From Theorem 1.4, (2.51), (2.42), and (2.43) of Lemma 2.4, we have 𝑀𝑛[𝑓](𝑟)(𝑥)[𝑓](𝑟)𝜓(𝑥)2𝛽+2(𝑥)𝐿([0,))1𝑂𝑛[𝑓](𝑟+1)(𝑥)𝜓2𝛽+1(𝑥)𝐿([0,))+[𝑓](𝑟+2)(𝑥)𝜓2𝛽(𝑥)𝐿([0,))1𝑂𝑛1𝜔1𝑓(𝑟);𝜓2𝛽+1+1;2𝜔2𝑓(𝑟);𝜓2𝛽.;(2.55) Therefore, we have (𝑀𝑛[𝑓])(𝑟)(𝑥)𝑓(𝑟)𝜓(𝑥)2𝛽+2(𝑥)𝐿([0,))1𝑂𝑛1𝜔1𝑓(𝑟);𝜓2𝛽+1+1;2𝜔2𝑓(𝑟);𝜓2𝛽;+𝜔2𝑓(𝑟);𝜓2𝛽+2.;(2.56) If we let =1/𝑛, then 𝑀𝑛[𝑓](𝑟)(𝑥)𝑓(𝑟)𝜓(𝑥)2𝛽+2(𝑥)𝐿([0,))1𝐶𝑛𝜔1𝑓(𝑟);𝜓2𝛽+1;1𝑛+𝜔2𝑓(𝑟);𝜓2𝛽;1𝑛,(2.57) because 𝜔2(𝑓(𝑟);𝜓2𝛽+2;1/𝑛)𝜔2(𝑓(𝑟);𝜓2𝛽;1/𝑛).

From now on, we will prove Theorems 1.7, 1.9, and 1.10, which are the results for the Szász-Mirakyan operators, analogously to the case of Lupas-type operators.

Lemma 2.5. Let 𝑟 be a nonnegative integer. Then one has for 𝑓𝐶𝑟([0,)), 𝑄𝑛,𝛽[𝑓](𝑟)(𝑛𝑥)=(𝑛+𝛽)𝑛+𝛽𝑟𝑘=0𝑆𝑛,𝑘(𝑥)0𝑆𝑛+𝛽,𝑘+𝑟(𝑦)𝑓(𝑟)([𝑦)𝑑𝑦,𝑥0,).(2.58)

Proof. We know that 𝑆(𝑟)𝑛,𝑘(𝑥)=𝑟𝑖=0𝑟𝑖(𝑒𝑛𝑦)(𝑟𝑖)(𝑛𝑦)𝑘(𝑖)=𝑘!𝑟𝑖=0𝑟𝑖(1)𝑟(1)𝑖𝑛𝑟𝑆𝑛,𝑘𝑖𝑆(𝑥),(𝑟)𝑛,𝑘(𝑥)=𝑟𝑖=0𝑟𝑖(𝑒𝑛𝑦)(𝑖)(𝑛𝑦)𝑘(𝑟𝑖)=𝑘!𝑟𝑖=0𝑟𝑖(1)𝑖𝑛𝑟𝑆𝑛,𝑘𝑟+𝑖(𝑥).(2.59) Therefore, we have 𝑄𝑛,𝛽[𝑓](𝑟)(𝑥)=(𝑛+𝛽)𝑘=0𝑆(𝑟)𝑛,𝑘(𝑥)0𝑆𝑛+𝛽,𝑘(𝑦)𝑓(𝑦)𝑑𝑦=(𝑛+𝛽)𝑟𝑖=0𝑘=𝑖𝑟𝑖(1)𝑟(1)𝑖𝑛𝑟𝑆𝑛,𝑘𝑖(𝑥)0𝑆𝑛+𝛽,𝑘(𝑛𝑦)𝑓(𝑦)𝑑𝑦=(𝑛+𝛽)𝑛+𝛽𝑟𝑘=0𝑆𝑛,𝑘(𝑥)0𝑟𝑖=0𝑟𝑖(1)𝑟(1)𝑖(𝑛+𝛽)𝑟𝑆𝑛+𝛽,𝑘+𝑖𝑛(𝑦)𝑓(𝑦)𝑑𝑦=(𝑛+𝛽)𝑛+𝛽𝑟𝑘=0𝑆𝑛,𝑘(𝑥)0(1)𝑟𝑆(𝑟)𝑛+𝛽,𝑘+𝑟𝑛(𝑦)𝑓(𝑦)𝑑𝑦=(𝑛+𝛽)𝑛+𝛽𝑟𝑘=0𝑆𝑛,𝑘(𝑥)0𝑆𝑛+𝛽,𝑘+𝑟(𝑦)𝑓(𝑟)(𝑦)𝑑𝑦.(2.60)

Lemma 2.6. Let 𝑎, 𝑏, and 𝑚 be nonnegative integers. 𝑅𝑛,𝑚,𝑟(𝑎,𝑏;𝑥)=(𝑛+𝑏)𝑘=0𝑆𝑛+𝑎,𝑘(𝑥)0𝑆𝑛+𝑏,𝑘+𝑟(𝑦)(𝑦𝑥)𝑚𝑑𝑦.(2.61) Then one has(i)𝑅𝑛,0,𝑟(𝑎,𝑏;𝑥)=1 and 𝑅𝑛,1,𝑟(𝑎,𝑏;𝑥)=((𝑎𝑏)𝑥+𝑟+1)/(𝑛+𝑏); (ii)For 𝑚1(𝑛+𝑏)R𝑛,𝑚+1,𝑟(𝑎,𝑏;𝑥)=𝑥𝑅(1)𝑛,𝑚,𝑟(𝑎,𝑏;𝑥)+((𝑎𝑏)𝑥+𝑚+𝑟+1)𝑅𝑛,𝑚,𝑟(𝑎,𝑏;𝑥)+2𝑥𝑚𝑅𝑛,𝑚1,𝑟(𝑎,𝑏;𝑥);(2.62)(iii)𝑅𝑛,𝑚,𝑟1(𝑎,𝑏;𝑥)=𝑂𝑛[(𝑚+1)/2]𝑔𝑛,𝑚,𝑟(𝑎,𝑏;𝑥),(2.63) where 𝑔𝑛,𝑚,𝑟(𝑎,𝑏;𝑥) is a polynomial of degree 𝑚 such that the coefficients of 𝑔𝑛,𝑚,𝑟(𝑎,𝑏;𝑥) are bounded independently of 𝑛.

Proof. Let 𝑅𝑛,𝑚,𝑟(𝑥)=𝑅𝑛,𝑚,𝑟(𝑎,𝑏;𝑥). Then (i) 𝑅𝑛,0,𝑟(𝑥)=(𝑛+𝑏)𝑘=0𝑆𝑛+𝑎,𝑘(𝑥)0𝑆𝑛+𝑏,𝑘+𝑟(𝑅𝑦)𝑑𝑦=1,𝑛,1,𝑟(𝑥)=(𝑛+𝑏)𝑘=0𝑆𝑛+𝑎,𝑘(𝑥)0𝑆𝑛+𝑏,𝑘+𝑟=(𝑦)(𝑦𝑥)𝑑𝑦𝑘=0𝑆𝑛+𝑎,𝑘(𝑥)𝑘+𝑟+1=𝑛+𝑏𝑥𝑘=0𝑆𝑛+𝑎,𝑘𝑘(𝑥)+𝑛+𝑏𝑟+1=𝑛+𝑏𝑥(𝑛+𝑎)𝑥+𝑛+𝑏𝑟+1𝑛+𝑏𝑥=(𝑎𝑏)𝑥+𝑟+1.𝑛+𝑏(2.64)
(ii) Using 𝑥𝑆(1)𝑛,𝑘(𝑥)=(𝑘𝑛𝑥)𝑆𝑛,𝑘(𝑥), we obtain 𝑥𝑅(1)𝑛,𝑚,𝑟(𝑥)+𝑚𝑅𝑛,𝑚1,𝑟(𝑥)=(𝑛+𝑏)𝑘=0𝑥𝑆(1)𝑛+𝑎,𝑘(𝑥)0𝑆𝑛+𝑏,𝑘+𝑟(𝑦)(𝑦𝑥)𝑚𝑑𝑦=(𝑛+𝑏)𝑘=0𝑆𝑛,𝑘(𝑥)0(𝑘(𝑛+𝑎)𝑥)𝑆𝑛+𝑏,𝑘+𝑟(𝑦)(𝑦𝑥)𝑚𝑑𝑦.(2.65) Here, we see (𝑘(𝑛+𝑎)𝑥)𝑆𝑛+𝑏,𝑘+𝑟(𝑦)=((𝑘+𝑟(𝑛+𝑏)𝑦)(𝑟+(𝑎𝑏)𝑥)+(𝑛+𝑏)(𝑦𝑥))𝑆𝑛+𝑏,𝑘+𝑟(𝑦)=𝑦𝑆(1)𝑛+𝑏,𝑘+𝑟(𝑦)(𝑟+(𝑎𝑏)𝑥)𝑆𝑛+𝑏,𝑘+𝑟(𝑦)+(𝑛+𝑏)𝑆𝑛+𝑏,𝑘+𝑟(𝑦)(𝑦𝑥).(2.66) Then substituting (2.66) for (2.65), we consider the following; 𝑥𝑅(1)𝑛,𝑚,𝑟(𝑥)+𝑚𝑅𝑛,𝑚1,𝑟(𝑥)=(𝑛+𝑏)𝑘=0𝑆𝑛,𝑘(𝑥)0(𝑦𝑆(1)𝑛+𝑏,𝑘+𝑟(𝑦)(𝑟+(𝑎𝑏)𝑥)𝑆𝑛+𝑏,𝑘+𝑟+(𝑦)(𝑛+𝑏)𝑆𝑛+𝑏,𝑘+𝑟(𝑦)(𝑦𝑥))(𝑦𝑥)𝑚𝑑𝑦=1+2+3.(2.67) Then, we have 1=(𝑛+𝑏)𝑘=0𝑆𝑛,𝑘(𝑥)0𝑦𝑆(1)𝑛+𝑏,𝑘+𝑟(𝑦)(𝑦𝑥)𝑚=𝑑𝑦(𝑛+𝑏)𝑘=0𝑆𝑛+𝑎,𝑘(𝑥)0𝑆(1)𝑛+𝑏,𝑘+𝑟(𝑦)(𝑦𝑥)𝑚+1𝑑𝑦+𝑥(𝑛+𝑏)𝑘=0𝑆𝑛+𝑎,𝑘(𝑥)0𝑆(1)𝑛+𝑏,𝑘+𝑟(𝑦)(𝑦𝑥)𝑚𝑑𝑦=(𝑚+1)𝑅𝑛,𝑚,𝑟(𝑥)𝑥𝑚𝑅𝑛,𝑚1,𝑟(𝑥).(2.68) Here the last equation follows by parts of integration. Furthermore, we have 1+2=(𝑟+(𝑎𝑏)𝑥)𝑅𝑛,𝑚,𝑟(𝑥)+(𝑛+𝑏)𝑅𝑛,𝑚+1,𝑟(𝑥).(2.69) Therefore, we have (𝑛+𝑏)𝑅𝑛,𝑚+1,𝑟(𝑥)=𝑥𝑅(1)𝑛,𝑚,𝑟(𝑥)+((𝑎𝑏)𝑥+𝑚+𝑟+1)𝑅𝑛,𝑚,𝑟(𝑥)+2𝑥𝑚𝑅𝑛,𝑚1,𝑟(𝑥).(2.70)
(iii) It is proved by the same method as the proof of Lemma 2.1 (iv).

Proof of Theorem 1.7. Let |𝑦𝑥|1. By the second inequality in (1.30), ||𝑓(𝑟)(𝑦)𝑓(𝑟)||=||||||𝑓(𝑥)𝑦𝑥(𝑟+1)||||||||𝑒(𝜉)𝐶𝑦𝑥𝛽𝜉𝜉𝛼+2||||||𝑒𝐶𝑦𝑥𝛽𝑥(𝑥+1)𝛼+2.(2.71) Let 𝜀=𝑛𝛾, 0<𝛾<1, ||||𝑛+𝛽𝑛𝑟𝑄𝑛,𝛽[𝑓](𝑟)(𝑥)𝑓(𝑟)||||=|||||(𝑥)(𝑛+𝛽)𝑘=0𝑆𝑛,𝑘(𝑥)|𝑦𝑥|<𝜀𝑆𝑛+𝛽,𝑘+𝑟||𝑓(𝑦)(𝑟)(𝑦)𝑓(𝑟)||+(𝑥)𝑑𝑦|𝑦𝑥|𝜀𝑆𝑛+𝛽,𝑘+𝑟||𝑓(𝑦)(𝑟)(𝑦)𝑓(𝑟)|||||||(𝑥)𝑑𝑦=𝐴+𝐵.(2.72) First, we see that by (2.71) and Lemma 2.6(i), 𝐴𝐶(𝑛+𝛽)𝑘=0𝑆𝑛,𝑘(𝑥)|𝑦𝑥|<𝜀𝑆𝑛+𝛽,𝑘+𝑟(||||𝑦)𝑦𝑥𝑑𝑦𝑒𝛽𝑥(𝑥+1)𝛼+2𝐶𝜀𝑒𝛽𝑥(𝑥+1)𝛼+21𝑂𝑛𝛾𝑒𝛽𝑥(𝑥+1)𝛼+2.(2.73) Next, to estimate 𝐵, we split it into two parts: |||||𝐵=(𝑛+𝛽)𝑘=0𝑆𝑛,𝑘(𝑥)|𝑦𝑥|𝜀𝑆𝑛+𝛽,𝑘+𝑟||𝑓(𝑦)(𝑟)(𝑦)𝑓(𝑟)|||||||(𝑥)𝑑𝑦(𝑛+𝛽)𝑘=0𝑆𝑛,𝑘(𝑥)|𝑦𝑥|𝜀𝑆𝑛+𝛽,𝑘+𝑟(||𝑓𝑦)(𝑟)(||+||𝑓𝑦)(𝑟)(||𝑥)𝑑𝑦(𝑛+𝛽)𝑘=0𝑆𝑛,𝑘(𝑥)|𝑦𝑥|𝜀𝑆𝑛+𝛽,𝑘+𝑟𝑒(𝑦)𝛽𝑦(𝑦+1)𝛼+𝑒𝛽𝑥(𝑥+1)𝛼𝑑𝑦=𝐵1+𝐵2.(2.74) First, we estimate 𝐵1=(𝑛+𝛽)𝑘=0𝑆𝑛,𝑘(𝑥)|𝑦𝑥|>𝜀𝑆𝑛+𝛽,𝑘+𝑟(𝑦)𝑒𝛽𝑦(𝑦+1)𝛼𝑑𝑦.(2.75) Then, using the following facts: 𝑆𝑛+𝛽,𝑘+𝑟(𝑦)𝑒𝛽𝑦=𝑆𝑛,𝑘+𝑟(𝑦)𝑛+𝛽𝑛𝑘+𝑟,(2.76)𝑛+𝛽𝑛𝑘+𝑟𝑆𝑛,𝑘(𝑥)=𝑛+𝛽𝑛𝑟𝑆𝑛+𝛽,𝑘(𝑥)𝑒𝛽𝑥,(2.77)(𝑦+1)𝛼=((𝑦𝑥)+𝑥+1)𝛼=𝛼𝑖=0𝛼𝑖(𝑦𝑥)𝑖(𝑥+1)𝛼𝑖,(2.78) we have 𝐵1𝐶𝑒𝛽𝑥(𝑛+𝛽)𝑛+𝛽𝑛𝑟𝑘=0𝑆𝑛+𝛽,𝑘(𝑥)|𝑦𝑥|>𝜀𝑆𝑛,𝑘+𝑟(𝑦)(𝑦+1)𝛼𝑑𝑦=𝐶𝑒𝛽𝑥𝑛𝑛+𝛽𝑛𝑟+1𝑘=0𝑆𝑛+𝛽,𝑘×(𝑥)|𝑦𝑥|>𝜀𝑆𝑛,𝑘+𝑟(𝑦)𝛼𝑖=1𝛼𝑖(𝑦x)𝑖(𝑥+1)𝛼𝑖𝑑𝑦+𝐶𝑒𝛽𝑥𝑛𝑛+𝛽𝑛𝑟𝑘=0𝑆𝑛+𝛽,𝑘(𝑥)|𝑦𝑥|>𝜀𝑆𝑛,𝑘+𝑟(𝑦)(𝑥+1)𝛼𝑑𝑦=𝐶𝑒𝛽𝑥𝐵11+𝐵12.(2.79) Then, using (2.18) and Lemma 2.6, we have 𝐵11=𝑛𝑘=0𝑆𝑛+𝛽,𝑘(𝑥)|𝑦𝑥|>𝜀𝑆𝑛,k+𝑟(𝑦)𝛼𝑖=1𝛼𝑖(𝑦𝑥)𝑖(𝑥+1)𝛼𝑖𝑑𝑦𝑛𝑘=0𝑆𝑛+𝛽,𝑘(𝑥)|𝑦𝑥|>𝜀