Abstract

Let 𝑤𝜆(𝑥)=(1𝑥2)𝜆1/2 and 𝑃𝜆,𝑛(𝑥) be the ultraspherical polynomials with respect to 𝑤𝜆(𝑥). Then, we denote the Stieltjes polynomials with respect to 𝑤𝜆(𝑥) by 𝐸𝜆,𝑛+1(𝑥) satisfying 11𝑤𝜆(𝑥)𝑃𝜆,𝑛(𝑥)𝐸𝜆,𝑛+1(𝑥)𝑥𝑚𝑑𝑥=0, 0𝑚<𝑛+1, 11𝑤𝜆(𝑥)𝑃𝜆,𝑛(𝑥)𝐸𝜆,𝑛+1(𝑥)𝑥𝑚𝑑𝑥0, 𝑚=𝑛+1. In this paper, we investigate asymptotic properties of derivatives of the Stieltjes polynomials 𝐸𝜆,𝑛+1(𝑥) and the product 𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥). Especially, we estimate the even-order derivative values of 𝐸𝜆,𝑛+1(𝑥) and 𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥) at the zeros of 𝐸𝜆,𝑛+1(𝑥) and the product 𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥), respectively. Moreover, we estimate asymptotic representations for the odd derivatives values of 𝐸𝜆,𝑛+1(𝑥) and 𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥) at the zeros of 𝐸𝜆,𝑛+1(𝑥) and 𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥) on a closed subset of (1,1), respectively. These estimates will play important roles in investigating convergence and divergence of the higher-order Hermite-Fejér interpolation polynomials.

1. Introduction

Consider the generalized Stieltjes polynomials 𝐸𝜆,𝑛+1(𝑥) defined (up to a multiplicative constant) by 11𝑤𝜆(𝑥)𝑃𝜆,𝑛(𝑥)𝐸𝜆,𝑛+1(𝑥)𝑥𝑘𝑑𝑥=0,𝑘=0,1,2,,𝑛,𝑛1,(1.1) where 𝑤𝜆(𝑥)=(1𝑥2)𝜆1/2, 𝜆>1/2, and 𝑃𝜆,𝑛(𝑥) is the 𝑛th ultraspherical polynomial for the weight function 𝑤𝜆(𝑥).

The polynomials 𝐸𝜆,𝑛+1(𝑥), introduced by Stieltjes and studied by Szegö, have been used in numerical integration, whereas the polynomials 𝑃𝜆,𝑛(𝑥)𝐸𝜆,𝑛+1(𝑥) have been used in extended Lagrange interpolation. In this paper, we will prove pointwise and asymptotic estimates for the higher-order derivatives of 𝐸𝜆,𝑛+1(𝑥) and 𝑃𝜆,𝑛(𝑥)𝐸𝜆,𝑛+1(𝑥). It is well known that these kind of estimates are useful for studying interpolation processes with multiple nodes.

In 1934, G. Szegö [1] showed that the zeros of the generalized Stieltjes polynomials 𝐸𝜆,𝑛+1(𝑥) are real and inside [1,1] and interlace with the zeros of 𝑃𝜆,𝑛(𝑥) whenever 0𝜆2. Recently, several authors [28] studied further interesting properties for these Stieltjes polynomials. Ehrich and Mastroianni [3, 4] gave accurate pointwise bounds of 𝐸𝜆,𝑛+1(𝑥)(0𝜆1) and the product 𝐹𝜆,2𝑛+1=𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(0𝜆1) on [1,1], and they estimated asymptotic representations for 𝐸𝜆,𝑛+1(𝑥) and 𝐹𝜆,2𝑛+1(𝑥) at the zeros of 𝐸𝜆,𝑛+1(𝑥) and 𝐹𝜆,2𝑛+1(𝑥), respectively. In [6], pointwise upper bounds of 𝐸𝜆,𝑛+1(𝑥), 𝐸𝜆,𝑛+1(𝑥), 𝐹𝜆,2𝑛+1(𝑥), and 𝐹𝜆,2𝑛+1(𝑥) are obtained using the asymptotic differential relations of the first and the second order for the Stieltjes polynomials 𝐸𝜆,𝑛+1(𝑥)(0𝜆1) and 𝐹𝜆,2𝑛+1(𝑥)(0𝜆1). Also the values of 𝐸𝜆,𝑛+1(𝑥) and 𝐹𝜆,2𝑛+1(𝑥) at the zeros of 𝐸𝜆,𝑛+1(𝑥) and 𝐹𝜆,2𝑛+1(𝑥) are estimated in [6]. Moreover, using the results of [6], the Lebesgue constants of Hermite-Fejér interpolatory process are estimated in [7].

In this paper, we find pointwise upper bounds of 𝐸(𝑟)𝜆,𝑛+1(𝑥) and 𝐹(𝑟)𝜆,2𝑛+1(𝑥) for two cases of an odd order and of even order. Using these relations, we investigate asymptotic properties of derivatives of the Stieltjes polynomials 𝐸𝜆,𝑛+1(𝑥) and 𝐹𝜆,2𝑛+1(𝑥) and we also estimate the values of 𝐸(2)𝜆,𝑛+1(𝑥) and 𝐹(2)𝜆,2𝑛+1(𝑥) at the zeros of 𝐸𝜆,𝑛+1(𝑥) and 𝐹𝜆,2𝑛+1(𝑥), respectively. Especially, for the value of 𝐹(2)𝜆,2𝑛+1(𝑥) at the zeros of 𝐹𝜆,2𝑛+1(𝑥), we will estimate 𝑃(𝑟)𝜆,𝑛(𝑥) and 𝐸(𝑟)𝜆,𝑛+1(𝑥) for an odd 𝑟 at the zeros of 𝐸𝜆,𝑛+1(𝑥) and 𝑃𝜆,𝑛(𝑥), respectively. Finally, we investigate asymptotic representations for the values of 𝐸(2+1)𝜆,𝑛+1 and 𝐹(2+1)𝜆,2𝑛+1(𝑥) at the zeros of 𝐸𝜆,𝑛+1(𝑥) and 𝐹𝜆,2𝑛+1(𝑥) on a closed subset of (1,1), respectively. These estimates will play important roles in investigating convergence and divergence of the higher-order Hermite-Fejér interpolation polynomials.

This paper is organized as follows. In Section 2, we will introduce the main results. In Section 3, we will introduce the known results in order to prove the main results. Finally, we will prove the results in Section 4.

2. Main Results

We first introduce some notations, which we use in the following. For the ultraspherical polynomials 𝑃𝜆,𝑛, 𝜆0, we use the normalization 𝑃𝜆,𝑛(1)=𝑛𝑛+2𝜆1 and then we know that 𝑃𝜆,𝑛(1)𝑛2𝜆1. We denote the zeros of 𝑃𝜆,𝑛 by 𝑥(𝜆)𝜈,𝑛, 𝜈=1,,𝑛, and the zeros of Stieltjes polynomials 𝐸𝜆,𝑛+1 by 𝜉(𝜆)𝜇,𝑛+1, 𝜇=1,,𝑛+1. We denote the zeros of 𝐹𝜆,2𝑛+1=𝑃𝜆,𝑛𝐸𝜆,𝑛+1 by 𝑦(𝜆)𝜈,2𝑛+1, 𝜈=1,,2𝑛+1. All nodes are ordered by increasing magnitude. We set 𝜑(𝑥)=1𝑥2, and, for any two sequences {𝑏𝑛}𝑛 and {𝑐𝑛}𝑛 of nonzero real numbers (or functions), we write 𝑏𝑛𝑐𝑛, if there exists a constant 𝐶>0, independent of 𝑛 (and 𝑥) such that 𝑏𝑛𝐶𝑐𝑛 for 𝑛 large enough and write 𝑏𝑛𝑐𝑛 if 𝑏𝑛𝑐𝑛 and 𝑐𝑛𝑏𝑛. We denote by 𝒫𝑛 the space of polynomials of degree at most 𝑛.

For the Chebyshev polynomial 𝑇𝑛(𝑥), note that for 𝜆=0 and 𝜆=1𝐸0,𝑛+1(𝑥)=2𝑛𝜋𝑇𝑛+1(𝑥)𝑇𝑛1𝐸(𝑥)1,𝑛+12(𝑥)=𝜋𝑇𝑛+1(𝑥).(2.1) Therefore, we will consider 𝐸𝜆,𝑛+1(𝑥) for 0<𝜆<1.

Theorem 2.1. Let 0<𝜆<1 and 𝑟1 be a positive integer. Then, for all 𝑥[𝜉(𝜆)1,𝑛+1,𝜉(𝜆)𝑛+1,𝑛+1], ||𝐸(𝑟)𝜆,𝑛+1||(𝑥)𝑛𝑟+1𝜆𝜑1𝑟𝜆(𝑥).(2.2) Moreover, one has max𝑥[1,1]||𝐸(𝑟)𝜆,𝑛+1||(𝑥)𝑛2𝑟,(2.3) and especially one has, for 𝑥[1,𝜉(𝜆)1,𝑛+1][𝜉(𝜆)𝑛+1,𝑛+1,1], ||𝐸(𝑟)𝜆,𝑛+1||(𝑥)𝑛2𝑟.(2.4)

Theorem 2.2. Let 0<𝜆<1 and 𝑟1 be a positive integer. Then, for all 𝑥[𝜉(𝜆)1,𝑛+1,𝜉(𝜆)𝑛+1,𝑛+1], ||𝐹(𝑟)𝜆,2𝑛+1||(𝑥)𝑛𝑟𝜑12𝜆𝑟(𝑥).(2.5) Moreover, one has max𝑥[1,1]||𝐹(𝑟)𝜆,2𝑛+1||(𝑥)𝑛2𝜆+2𝑟1,(2.6) and especially, for 𝑥[1,𝜉(𝜆)1,𝑛+1][𝜉(𝜆)𝑛+1,𝑛+1,1], ||𝐹(𝑟)𝜆,2𝑛+1||(𝑥)𝑛2𝜆+2𝑟1.(2.7)

In the following, we also estimate the values of 𝐸(2)𝜆,𝑛+1(𝑥) and 𝐹(2)𝜆,2𝑛+1(𝑥), 1 at the zeros {𝜉(𝜆)𝜇,𝑛+1} of 𝐸𝜆,𝑛+1(𝑥) and the zeros {𝑦(𝜆)𝜈,2𝑛+1} of 𝐹𝜆,2𝑛+1(𝑥), respectively.

Theorem 2.3. Let 0<𝜆<1 and 𝑟2 be an even integer. For 1𝜇𝑛+1, one has |||𝐸(𝑟)𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1|||𝑛𝑟𝜑𝑟𝜉(𝜆)𝜇,𝑛+1.(2.8)

Theorem 2.4. Let 0<𝜆<1 and 𝑟2 be an even integer. For 1𝜈2𝑛+1, one has |||𝐹(𝑟)𝜆,2𝑛+1𝑦(𝜆)𝜈,2𝑛+1|||𝑛𝑟1+𝜆𝜑𝑟𝜆𝑦(𝜆)𝜈,2𝑛+1.(2.9)

Finally, we obtain the asymptotic representations for the values of 𝐸(2+1)𝜆,𝑛+1(𝑥) and 𝐹(2+1)𝜆,2𝑛+1(𝑥) at the zeros of 𝐸𝜆,𝑛+1(𝑥) and 𝐹𝜆,2𝑛+1(𝑥) on a closed subset of (1,1), respectively.

Theorem 2.5. Let 0<𝜆<1 and 0<𝜀<1. Suppose |𝜉(𝜆)𝜇,𝑛+1|1𝜀. Then, (a)𝐸(2+1)𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1=(1)(𝑛+1)2𝜑2𝜉(𝜆)𝜇,𝑛+1𝐸𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1𝑛+𝑂2+1,(2.10)(b)𝑃(2)𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1=(1)𝑛(𝑛+2𝜆)𝜑2𝜉(𝜆)𝜇,𝑛+1𝑃𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1𝑛+𝑂2𝜆+23.(2.11) In addition, 𝑃(2)𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1=(1)(𝑛+1)2𝜑2𝜉(𝜆)𝜇,𝑛+1𝑃𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1𝑛+𝑂2+𝜆2.(2.12)

Theorem 2.6. Let 0<𝜆<1 and 0<𝜀<1. Suppose |𝑥(𝜆)𝜈,𝑛|1𝜀. Then, (a)𝐸(2)𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛=(1)(𝑛+1)2𝜑2𝑥(𝜆)𝜈,𝑛𝐸𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛𝑛+𝑂2,(2.13)(b)𝑃(2+1)𝜆,𝑛𝑥(𝜆)𝜈,𝑛=(1)𝑛(𝑛+2𝜆)𝜑2𝑥(𝜆)𝜈,𝑛𝑃𝜆,𝑛𝑥(𝜆)𝜈,𝑛𝑛+𝑂𝜆+22.(2.14) In addition, 𝑃(2+1)𝜆,𝑛𝑥(𝜆)𝜈,𝑛=(1)(𝑛+1)2𝜑2𝑥(𝜆)𝜈,𝑛𝑃𝜆,𝑛𝑥(𝜆)𝜈,𝑛𝑛+𝑂𝜆+21.(2.15)

Theorem 2.7. Let 0<𝜆<1 and 0<𝜀<1. Suppose |𝑦(𝜆)𝜈,2𝑛+1|1𝜀. Then, one has, for a positive integer 1, 𝐹(2+1)𝜆,2𝑛+1𝑦(𝜆)𝜈,2𝑛+1=𝑐(1)(𝑛+1)2𝜑2𝑦(𝜆)𝜈,2𝑛+1𝐹𝜆,2𝑛+1𝑦(𝜆)𝜈,2𝑛+1𝑛+𝑂𝜆+2,(2.16) where 𝑐=421.

3. The Known Results

In this section, we will introduce the known results in [4, 6, 9] to prove main results.

Proposition 3.1. (a) Let 𝜆>1/2. Then, 𝑃𝜆,𝑛(𝑥) satisfies the second-order differential equation as follows: 1𝑥2𝑃𝜆,𝑛(𝑥)(2𝜆+1)𝑥𝑃𝜆,𝑛(𝑥)+𝑛(𝑛+2𝜆)𝑃𝜆,𝑛(𝑥)=0.(3.1)
(b) Let 𝜆>1/2. Then, 𝑃𝜆,𝑛(𝑥)=2𝜆𝑃𝜆+1,𝑛1(𝑥).(3.2)
(c) Let 𝜆>1/2. Then, for 1𝜇𝑛+1, |||𝑃𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1|||𝑛2𝜆1𝜑2𝜉(𝜆)𝜇,𝑛+1.(3.3)
(d) Let 𝜆>1/2. Then, for 1𝜈𝑛, |||𝑃𝜆,𝑛𝑥(𝜆)𝜈,𝑛|||𝑛𝜆𝜑𝜆1𝑥(𝜆)𝜈,𝑛.(3.4)
(e) Let 𝜆>1/2. Then, for 𝑥(1,1) and 𝑟0, ||𝑃(𝑟)𝜆,𝑛||(𝑥)𝑛𝜆+𝑟1𝜑𝜆𝑟(𝑥).(3.5)
(f) Let 0𝜆1. Then, for 1𝜈𝑛, |||𝐸𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛|||𝑛𝜑1𝑥(𝜆)𝜈,𝑛.(3.6)
(g) Let 𝜆>1/2 and 𝑟0. Then, 𝑃𝜆,𝑛(𝑥) satisfies the higher-order differential equation as follows: 1𝑥2𝑃(𝑟+2)𝜆,𝑛(𝑥)(2𝜆+2𝑟+1)𝑥𝑃(𝑟+1)𝜆,𝑛𝑛(𝑥)+2𝑃+2𝜆𝑛𝑟(2𝜆𝑟+2)(𝑟)𝜆,𝑛(𝑥)=0.(3.7)

Proof. (a) It is from [9, (4.2.1)]. (b) It is from [9, (4.7.14)]. (c) It is from [6, Lemma 3.4]. (d) It is from [9, (8.9.7)]. (e) For 𝑟=0, it follows from [9, (7.33.5)], and, for 𝑟1, it comes from (b) and the case of 𝑟=0. (f) It is from [6, Lemma 3.3 (3.23)]. (g) Equation (3.7) comes from (a).

Proposition 3.2 (see [4]). Let 0<𝜆<1. Let 𝜉(𝜆)𝜇,𝑛+1=cos𝜃(𝜆)𝜇,𝑛+1, 𝜇=1,,𝑛+1 and 𝑦(𝜆)𝜈,2𝑛+1=cos𝜓(𝜆)𝜈,2𝑛+1, 𝜈=1,,2𝑛+1. Then, for 𝜇=0,1,,𝑛+2 and 𝜈=0,1,,2𝑛+2, |||𝜃(𝜆)𝜇,𝑛+1𝜃(𝜆)𝜇+1,𝑛+1|||||𝜓(𝜆)𝜈,2𝑛+1𝜓(𝜆)𝜈+1,2𝑛+1||𝑛1,(3.8) where 𝜓(𝜆)0,2𝑛+1=𝜃(𝜆)0,𝑛+1=𝜋 and 𝜓(𝜆)2𝑛+2,2𝑛+1=𝜃(𝜆)𝑛+2,𝑛+1=0.

Proposition 3.3 ([6, Proposition  2.3]). Let 0<𝜆<1. Then, for all 𝑥[1,1], 1𝑥2𝐸𝜆,𝑛+1(𝑥)𝑥𝐸𝜆,𝑛+1(𝑥)+(𝑛+1)2𝐸𝜆,𝑛+1(𝑥)=𝐼𝜆,𝑛(𝑥),(3.9) where 𝐼𝜆,𝑛8(𝑥)=𝛾𝑛(𝜆)[(𝑛+1)/2]𝜈=1(𝑛+1𝜈)𝜈𝛼(𝜆)𝜈,𝑛𝑇𝑛+12𝜈(𝑥).(3.10) Then 𝐼𝜆,𝑛(𝑥) is a polynomial of degree 𝑛1 satisfying max𝑥[1,1]||𝐼𝜆,𝑛||(𝑥)𝑛2.(3.11)

Proposition 3.4 ([4, Theorem  2.1]). Let 0<𝜆<1. Then, for 𝑛0, ||𝐸𝜆,𝑛+1||(𝑥)𝑛1𝜆𝜑1𝜆(𝑥)+11𝑥1.(3.12) Furthermore, 𝐸𝜆,𝑛+1(1)1.

Proposition 3.5 ([6, Theorem  2.5]). Let 0<𝜆<1. (a)For all 𝑥[𝜉(𝜆)1,𝑛+1,𝜉(𝜆)𝑛+1,𝑛+1], ||𝐸𝜆,𝑛+1||(𝑥)𝑛2𝜆𝜑𝜆(𝑥).(3.13) Moreover, one has, for 𝑥[1,𝜉(𝜆)1,𝑛+1][𝜉(𝜆)𝑛+1,𝑛+1,1], ||𝐸𝜆,𝑛+1||(𝑥)𝑛2.(3.14)(b)For all 𝑥[𝜉(𝜆)1,𝑛+1,𝜉(𝜆)𝑛+1,𝑛+1], ||𝐸𝜆,𝑛+1||(𝑥)𝑛3𝜆𝜑1𝜆(𝑥).(3.15) Moreover, one has, for 𝑥[1,𝜉(𝜆)1,𝑛+1][𝜉(𝜆)𝑛+1,𝑛+1,1], ||𝐸𝜆,𝑛+1||(𝑥)𝑛4.(3.16)

Proposition 3.6 ([6, Corollary  2.6]). Let 0<𝜆<1. Then, for all 𝑥[1,1], 1𝑥2𝐹𝜆,2𝑛+1(𝑥)𝑥𝐹𝜆,2𝑛+1(𝑥)+2𝑛2𝐹+2(1+𝜆)𝑛+1𝜆,2𝑛+1(𝑥)=𝐽𝜆,𝑛(𝑥).(3.17) Here, 𝐽𝜆,𝑛(𝑥) is a polynomial of degree of 2𝑛+1 defined in (4.37) such that, for 𝑥[𝜉(𝜆)1,𝑛+1,𝜉(𝜆)𝑛+1,𝑛+1], ||𝐽𝜆,𝑛||(𝑥)𝑛2𝜑12𝜆(𝑥)(3.18) and, for 𝑥[1,𝜉(𝜆)1,𝑛+1][𝜉(𝜆)𝑛+1,𝑛+1,1], ||𝐽𝜆,𝑛||(𝑥)𝑛1+2𝜆.(3.19)

Proposition 3.7 ([6, Corollary  2.7]). Let 0<𝜆<1. (a)For all 𝑥[𝜉(𝜆)1,𝑛+1,𝜉(𝜆)𝑛+1,𝑛+1], ||𝐹𝜆,2𝑛+1||(𝑥)𝑛𝜑2𝜆(𝑥).(3.20) Moreover, one has, for 𝑥[1,𝜉(𝜆)1,𝑛+1][𝜉(𝜆)𝑛+1,𝑛+1,1], ||𝐹𝜆,2𝑛+1||(𝑥)𝑛1+2𝜆.(3.21)(b)For all 𝑥[𝜉(𝜆)1,𝑛+1,𝜉(𝜆)𝑛+1,𝑛+1], ||𝐹𝜆,2𝑛+1||(𝑥)𝑛2𝜑12𝜆(𝑥).(3.22) Moreover, one has, for 𝑥[1,𝜉(𝜆)1,𝑛+1][𝜉(𝜆)𝑛+1,𝑛+1,1], ||𝐹𝜆,2𝑛+1||(𝑥)𝑛3+2𝜆.(3.23)

Proposition 3.8 ([4, Lemma  5.5]). Let 0<𝜆<1. Then, for 𝜇=1,2,,𝑛+1, |||𝐸𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1|||𝑛2𝜆𝜑𝜆𝜉(𝜆)𝜇,𝑛+1(3.24) and, for 𝜈=1,2,,2𝑛+1, |||𝐹𝜆,2𝑛+1𝑦(𝜆)𝜈,2𝑛+1|||𝑛𝜑2𝜆𝑦(𝜆)𝜈,2𝑛+1.(3.25)

We now estimate the second derivatives at the zeros of 𝐸𝜆,𝑛+1 and 𝐹𝜆,2𝑛+1.

Proposition 3.9 ([6, Theorem  2.9]). Let 0<𝜆<1. Then, for 𝜇=1,2,,𝑛+1, |||𝐸𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1|||𝑛2𝜑2𝜉(𝜆)𝜇,𝑛+1(3.26) and, for 𝜈=1,2,,2𝑛+1, |||𝐹𝜆,2𝑛+1𝑦(𝜆)𝜈,2𝑛+1|||𝑛1+𝜆𝜑2𝜆𝑦(𝜆)𝜈,2𝑛+1.(3.27)

4. The Proofs of Main Results

In this section, we let 0<𝜆<1 and 𝑚=(𝑛+1/2). A representation of Stieltjes polynomials 𝐸𝜆,𝑛+1(𝑥) is (cf. [1, 10]) 𝛾𝑛(𝜆)2𝐸𝜆,𝑛+1(cos𝜃)=𝛼(𝜆)0,𝑛cos(𝑛+1)𝜃+𝛼(𝜆)1,𝑛𝛼cos(𝑛1)𝜃++(𝜆)𝑛/2,𝑛1cos𝜃,𝑛even2𝛼(𝜆)𝑛+1/2,𝑛,𝑛odd,(4.1) where 𝛼(𝜆)0,𝑛=𝑓(𝜆)0,𝑛=1,𝜈𝜇=0𝛼(𝜆)𝜇,𝑛𝑓(𝜆)𝜈𝜇,𝑛𝑓=0,𝜈=1,2,,(𝜆)𝜈,𝑛𝜆=1𝜈𝜆1𝛾𝑛+𝜈+𝜆,𝜈=1,2,,𝑛(𝜆)=𝜋Γ(𝑛+2𝜆)Γ(𝑛+𝜆+1)𝜋𝑛𝜆1.(4.2)

In the following, we state the asymptotic differential relation of the higher order of 𝐸𝜆,𝑛+1.

Lemma 4.1. Let 0<𝜆<1. Then, for all 𝑥[1,1] and 𝑟2, 1𝑥2𝐸(𝑟)𝜆,𝑛+1(𝑥)=(2𝑟3)𝑥𝐸(𝑟1)𝜆,𝑛+1(𝑥)+(𝑟2)2(𝑛+1)2𝐸(𝑟2)𝜆,𝑛+1(𝑥)+𝐼(𝑟2)𝜆,𝑛(𝑥)(4.3) and, for 𝑥[𝜉(𝜆)1,𝑛+1,𝜉(𝜆)𝑛+1,𝑛+1], ||𝐼(𝑟2)𝜆,𝑛||(𝑥)𝑛𝑟𝜑2𝑟(𝑥).(4.4) Here, 𝐼𝜆,𝑛(𝑥) is a polynomial of degree 𝑛1 defined in (3.10); 𝐼𝜆,𝑛8(𝑥)=𝛾𝑛𝑚(𝜆)𝜈=1(𝑛+1𝜈)𝜈𝛼(𝜆)𝜈,𝑛𝑇𝑛+12𝜈(𝑥),(4.5) such that max𝑥[1,1]||𝐼(𝑟2)𝜆,𝑛||(𝑥)𝑛2𝑟2.(4.6)

Proof. For 𝑟2, (4.3) is obtained by 𝑟2 times differentiation of (3.9). Equation (4.6) follows by (3.11) and the use of Markov-Bernstein inequality. Now, we prove (4.4). We know that the Chebyshev polynomial 𝑇𝑛(𝑥) satisfies the second-order differential equation 1𝑥2𝑇𝑛(𝑥)𝑥𝑇𝑛(𝑥)+𝑛2𝑇𝑛(𝑥)=0,(4.7) so we have, by 𝑟2 times differentiation of (4.7), 1𝑥2𝑇𝑛(𝑟)(𝑥)(2𝑟3)𝑥𝑇𝑛(𝑟1)(𝑥)(𝑟2)2𝑛2𝑇𝑛(𝑟2)(𝑥)=0.(4.8) Let for a nonnegative integer 𝑗0, 𝐼𝜆,𝑛,𝑗8(𝑥)=𝛾𝑛𝑚(𝜆)𝜈=1(𝑛+1𝜈)𝜈𝛼(𝜆)𝜈,𝑛||𝑇(𝑗)𝑛+12𝜈||.(𝑥)(4.9) Observe that in the view of Szegö’s result (cf. [1]) 𝛼(𝜆)1,𝑛<𝛼(𝜆)2,𝑛<𝛼(𝜆)3,𝑛<<0,0𝜈=0𝛼(𝜆)𝜈,𝑛<1.(4.10) Then, since |𝐼(𝑗)𝜆,𝑛(𝑥)|𝐼𝜆,𝑛,𝑗(𝑥) (note (4.10)), we will prove that, for 𝑥[𝜉(𝜆)1,𝑛+1,𝜉(𝜆)𝑛+1,𝑛+1] and 𝑗0, 𝐼𝜆,𝑛,𝑗(𝑥)𝑛𝑗+2𝜑𝑗(𝑥)(4.11) instead of (4.4). Since, from the proof of [6, Proposition  2.3], 80<𝛾𝑛𝑚(𝜆)𝜈=1(𝑛+1𝜈)𝜈𝛼(𝜆)𝜈,𝑛𝑛2(4.12) and, for 𝑥=cos𝜃, ||𝑇𝑛+12𝜈(||=|||(𝑥)𝑛+12𝜈)sin(𝑛+12𝜈)𝜃|||sin𝜃𝑛𝜑1(𝑥),(4.13) we obtain that 𝐼𝜆,𝑛,0(𝑥)𝑛2 and 𝐼𝜆,𝑛,1(𝑥)𝑛3𝜑1(𝑥). Using (4.8), we have, for 2𝑗𝑛, 𝐼𝜆,𝑛,𝑗8(𝑥)=𝛾𝑛𝑚(𝜆)𝜈=1(𝑛+1𝜈)𝜈𝛼(𝜆)𝜈,𝑛||𝑇(𝑗)𝑛+12𝜈||8(𝑥)𝛾𝑛𝑚(𝜆)𝜈=1(𝑛+1𝜈)𝜈𝛼(𝜆)𝜈,𝑛×(2𝑗3)|𝑥|1𝑥2||𝑇(𝑗1)𝑛+12𝜈||+||((𝑥)𝑗2)2(𝑛+12𝜈)2||1𝑥2||𝑇(𝑗2)𝑛+12𝜈||(𝑥)(2𝑗3)|𝑥|1𝑥2𝐼𝜆,𝑛,𝑗1(𝑥)+(𝑛+1)21𝑥2𝐼𝜆,𝑛,𝑗2(𝑥).(4.14) Therefore, (4.11) is proved by the mathematical induction on 𝑗. Consequently, we have (4.4).

We obtain pointwise upper bounds of 𝐸(𝑟)𝜆,𝑛+1(𝑥) for two cases of an odd order and an even order in the following.

Lemma 4.2. Let 0<𝜆<1 and 𝑟2. Let 𝑥[𝜉(𝜆)1,𝑛+1,𝜉(𝜆)𝑛+1,𝑛+1]. If 𝑟 is even, then one has ||𝐸(𝑟)𝜆,𝑛+1||(𝑥)𝑛𝑟2𝜑𝑟||𝐸(𝑥)𝜆,𝑛+1||(𝑥)+𝑛𝑟𝜑𝑟||𝐸(𝑥)𝜆,𝑛+1||(𝑥)+𝑛𝑟𝜑𝑟(𝑥),(4.15) and, if 𝑟 is odd, then one has ||𝐸(𝑟)𝜆,𝑛+1||(𝑥)𝑛𝑟1𝜑1𝑟||𝐸(𝑥)𝜆,𝑛+1||(𝑥)+𝑛𝑟1𝜑1𝑟||𝐸(𝑥)𝜆,𝑛+1||(𝑥)+𝑛𝑟𝜑𝑟(𝑥).(4.16)

Proof. Let 𝑟2. From (4.3) and (4.4), we have, for 𝑥[𝜉(𝜆)1,𝑛+1,𝜉(𝜆)𝑛+1,𝑛+1], ||𝐸(𝑟)𝜆,𝑛+1||(𝑥)𝜑2||𝐸(𝑥)(𝑟1)𝜆,𝑛+1||(𝑥)+𝑛2𝜑2||𝐸(𝑥)(𝑟2)𝜆,𝑛+1||(𝑥)+𝑛𝑟𝜑𝑟(𝑥)(4.17) and especially ||𝐸𝜆,𝑛+1||(𝑥)𝜑2||𝐸(𝑥)𝜆,𝑛+1||(𝑥)+𝑛2𝜑2||𝐸(𝑥)𝜆,𝑛+1||(𝑥)+𝑛2𝜑2(𝑥),(4.18) that is, we have (4.15) for 𝑟=2. From Proposition 3.2, we see 1+𝜉(𝜆)1,𝑛+1, 1𝜉(𝜆)𝑛+1,𝑛+11/𝑛, so we have, for 𝑥[𝜉(𝜆)1,𝑛+1,𝜉(𝜆)𝑛+1,𝑛+1], 𝜑1(𝑥)𝑛.(4.19) Then, from (4.17) with 𝑟=3 and (4.18), we know that ||𝐸(3)𝜆,𝑛+1||(𝑥)𝑛2𝜑2||𝐸(𝑥)𝜆,𝑛+1||(𝑥)+𝑛2𝜑4||𝐸(𝑥)𝜆,𝑛+1||(𝑥)+𝑛3𝜑3(𝑥),(4.20) that is, we have (4.16) for 𝑟=3. Assume that (4.15) and (4.16) hold for 3,4,,𝑟1 times differentiation. Let 𝑟 be an even number. Then, we have, from (4.17), (4.19), and the assumptions for 𝑟1 and 𝑟2, ||𝐸(𝑟)𝜆,𝑛+1||(𝑥)𝑛𝑟2𝜑𝑟||𝐸(𝑥)𝜆,𝑛+1||(𝑥)+𝑛𝑟𝜑𝑟||𝐸(𝑥)𝜆,𝑛+1||(𝑥)+𝑛𝑟𝜑𝑟(𝑥),(4.21) that is, we have (4.15). Similarly, we also have (4.16) for an odd 𝑟.

Lemma 4.3. Let 1<𝑥1<𝑥2<<𝑥𝑛<1 and 𝑃(𝑥)=𝑥𝑥1𝑥𝑥2𝑥𝑥𝑛.(4.22) Then, 𝑃(𝑥) is a polynomial of degree of 𝑛1 and has distinct real 𝑛1 zeros in (1,1). Moreover, if one lets 1<𝑦1<𝑦2<<𝑦𝑛1<1 be the zeros of 𝑃(𝑥), then {𝑦𝑖}𝑛𝑖=1 is interlaced with the zeros of 𝑃(𝑥) that is, 𝑥𝑖<𝑦𝑖<𝑥𝑖+1, 𝑖=1,2,,𝑛1.

Proof. Since the sign of 𝑃(𝑥𝑖) is (1)𝑛𝑖, it is proved.

Lemma 4.4. Let 𝑟 be a nonnegative integer. Then, 𝐸(𝑟)𝜆,𝑛+1(𝑥) has distinct 𝑛+1𝑟 real zeros on (1,1). If one lets {𝑥𝑛+1(𝑟,𝑖)}𝑛+1𝑟𝑖=1 be the zeros of the polynomial 𝐸(𝑟)𝜆,𝑛+1(𝑥) with 1<𝑥𝑛+1(𝑟,1)<𝑥𝑛+1(𝑟,2)<<𝑥𝑛+1(𝑟,𝑛+1𝑟)<1,(4.23) then one has, for 1𝑟𝑛 and 𝑘=1,,𝑛+1𝑟, 𝑥𝑛+1(0,𝑘)<𝑥𝑛+1(𝑟,𝑘).(4.24)

Proof. From Lemma 4.3, we know that 𝐸(𝑟)𝜆,𝑛+1 has distinct real 𝑛+1𝑟 zeros on (1,1). By the interlaced zeros property of Lemma 4.3, we see that, for 𝑘=1,,𝑛+1𝑟, 𝑥𝑛+1(𝑟1,𝑘)<𝑥𝑛+1(𝑟,𝑘)<𝑥𝑛+1(𝑟1,𝑘+1).(4.25) Thus, (4.24) is proved.

Proof of Theorem 2.1. Let 𝑟1. Equation (2.2) comes from (4.15), (4.16), (3.12), and (3.13). From Propositions 3.4, 3.5, and (4.19), we have max𝑥[1,1]||𝐸(𝑟)𝜆,𝑛+1||(𝑥)𝑛2𝑟,𝑟=1,2.(4.26) Hence, using the Markov-Bernstein inequality, we have (2.3). To prove (2.4), we will use the mathematical induction. We use (3.14). The formula (2.3) holds for 𝑟=1 from (3.14). We suppose that, for 𝑥[1,𝜉(𝜆)1,𝑛+1][𝜉(𝜆)𝑛+1,𝑛+1,1] and 𝑟2, ||𝐸(𝑟1)𝜆,𝑛+1||(𝑥)𝑛2(𝑟1).(4.27) Then, by Lemma 4.4 and (3.8), we have, for 𝑥[𝜉(𝜆)𝑛+1,𝑛+1,1] and 𝑟2, 𝐸(𝑟)𝜆,𝑛+1(𝑥)𝐸(𝑟1)𝜆,𝑛+1=(𝑥)𝑛+2𝑟𝑘=11𝑥𝑥𝑛+11(𝑟1,𝑘)𝜉(𝜆)𝑛+1,𝑛+1𝑥𝑛+11(𝑟1,𝑛+2𝑟)𝜉(𝜆)𝑛+1,𝑛+1𝑥𝑛+1=1(0,𝑛+2𝑟)𝜉(𝜆)𝑛+1,𝑛+1𝜉(𝜆)𝑛+2𝑟,𝑛+11𝜉(𝜆)𝑛+1,𝑛+1𝜉(𝜆)𝑛,𝑛+1𝑛2.(4.28) Here, the last inequality is obtained by Proposition 3.2, that is, 𝜉(𝜆)𝑛+1,𝑛+1𝜉(𝜆)𝑛,𝑛+1=cos𝜃(𝜆)𝑛+1,𝑛+1cos𝜃(𝜆)𝑛,𝑛+1𝑛2.(4.29) Therefore, we have, for 𝑥[𝜉(𝜆)𝑛+1,𝑛+1,1] and 𝑟2, 𝐸(𝑟)𝜆,𝑛+1(𝑥)𝐸(𝑟1)𝜆,𝑛+1(𝑥)𝑛2𝑛2𝑟.(4.30) Hence, from (2.3), we have (2.4). For 𝑥[1,𝜉(𝜆)1,𝑛+1], the proof is similar.

Lemma 4.5. Let be a nonnegative integer and 𝑥[𝜉(𝜆)1,𝑛+1,𝜉(𝜆)𝑛+1,𝑛+1]. Then,

|||𝑥𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)()|||𝑛+1𝜑2𝜆|||𝐼(𝑥),(4.31a)𝜆,𝑛(𝑥)𝑃𝜆,𝑛(𝑥)()|||𝑛+1+𝜆𝜑𝜆||||𝐸(𝑥),(4.31b)𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)()||||𝑛+2𝜑2𝜆1(𝑥).(4.31c)

Proof. (a) When =0, it is obvious from (3.5) and (3.12). Now, suppose 1. From (3.12), (2.2), and (3.5), we have |||𝑥𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)()|||1𝑞+𝑟||𝐸(𝑞)𝜆,𝑛+1(𝑥)𝑃(𝑟+1)𝜆,𝑛||(𝑥)1𝑞+𝑟𝑛𝑞+𝑟+1𝜑(𝑞+𝑟)2𝜆(𝑥)𝑛+1𝜑2𝜆(𝑥).(4.32)
(b) From (4.4) and (3.5), we have |||𝐼𝜆,𝑛(𝑥)𝑃𝜆,𝑛(𝑥)()|||𝑞+𝑟=||𝐼(𝑞)𝜆,𝑛(𝑥)𝑃(𝑟)𝜆,𝑛||(𝑥)𝑛+1+𝜆𝜑𝜆(𝑥).(4.33)
(c) Similarly to the proof of (a), we have, from (2.2) and (3.5), ||||𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)()||||𝑞+𝑟=||𝐸(𝑞+1)𝜆,𝑛+1(𝑥)𝑃(𝑟+1)𝜆,𝑛||(𝑥)𝑛+2𝜑2𝜆1(𝑥).(4.34)

Lemma 4.6. Let 0<𝜆<1. Then, for all 𝑥[1,1] and 𝑟2, 1𝑥2𝐹(𝑟)𝜆,2𝑛+1(𝑥)=(2𝑟5)𝑥𝐹(𝑟1)𝜆,2𝑛+1+(𝑥)(𝑟2)2(𝑛+1)2𝐹𝑛(𝑛+2𝜆)(𝑟2)𝜆,2𝑛+1(𝑥)+𝐽(𝑟2)𝜆,𝑛(𝑥)(4.35) and, for 𝑥[𝜉(𝜆)1,𝑛+1,𝜉(𝜆)𝑛+1,𝑛+1], ||𝐽(𝑟2)𝜆,𝑛||(𝑥)𝑛𝑟𝜑2𝜆𝑟+3(𝑥).(4.36) Here, 𝐽𝜆,𝑛(𝑥) is a polynomial of degree of 2𝑛+1 defined as follows: 𝐽𝜆,𝑛(𝑥)=2𝜆𝑥𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)+21𝑥2𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)+𝐼𝜆,𝑛(𝑥)𝑃𝜆,𝑛(𝑥).(4.37) Furthermore, one has ||𝐽(𝑟2)𝜆,𝑛||(1)𝑛2𝜆+2𝑟3.(4.38)

Proof. Similarly to the proof of Lemma 4.1, (4.35) is obtained by 𝑟2 times differentiation of the second-order differential relation with respect to 𝐹𝜆,2𝑛+1(𝑥), that is, (3.17). So it is sufficient to prove (4.36) and (4.38). From (4.37), we know that 𝐽(𝑟2)𝜆,𝑛(𝑥)=2𝜆𝑥𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(𝑟2)+21𝑥2𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(𝑟2)+𝐼𝜆,𝑛(𝑥)𝑃𝜆,𝑛(𝑥)(𝑟2).(4.39) By Lemma 4.5 (a) and (b), we have, for 𝑥[𝜉(𝜆)1,𝑛+1,𝜉(𝜆)𝑛+1,𝑛+1], |||2𝜆𝑥𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(𝑟2)|||𝑛𝑟1𝜑𝑟+22𝜆|||𝐼(𝑥),𝜆,𝑛(𝑥)𝑃𝜆,𝑛(𝑥)(𝑟2)|||𝑛𝑟1+𝜆𝜑𝑟+2𝜆(𝑥).(4.40) From (4.19) and Lemma 4.5 (c), we have, for 𝑥[𝜉(𝜆)1,𝑛+1,𝜉(𝜆)𝑛+1,𝑛+1] and 𝑟4, ||||21𝑥2𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(𝑟2)||||||||1𝑥2𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(𝑟2)||||+||||𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(𝑟3)||||+||||𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(𝑟4)||||𝑛𝑟𝜑2𝜆𝑟+3(𝑥)+𝑛𝑟1𝜑2𝜆𝑟+2(𝑥)+𝑛𝑟2𝜑2𝜆𝑟+3(𝑥)𝑛𝑟𝜑2𝜆𝑟+3(𝑥).(4.41) When 𝑟=2,3, we can similarly obtain that ||||21𝑥2𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(𝑟2)||||𝑛𝑟𝜑2𝜆𝑟+3(𝑥).(4.42) Therefore, we have (4.36). On the other hand, from (2.4), (4.6), and (3.2), we know that, for a nonnegative integer , 𝐸()𝜆,𝑛+1(1)𝑛2,||𝐼()𝜆,𝑛||(1)𝑛2+2,𝑃()𝜆,𝑛(1)𝑃𝜆+,𝑛(1)𝑛2(𝜆+)1.(4.43) Then, similarly to the proof of Lemma 4.5, we obtain that |||𝑥𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)()|||𝑥=1𝑛2+2𝜆+1,|||𝐼𝜆,𝑛(𝑥)𝑃𝜆,𝑛(𝑥)()|||𝑥=1𝑛2+2𝜆+1,||||𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)()||||𝑥=1𝑛2+2𝜆+3.(4.44) Therefore, we have (4.38).

Lemma 4.7. Let 0<𝜆<1. Then, for 𝑟2, if 𝑟 is even, one has, for 𝑥[𝜉(𝜆)1,𝑛+1,𝜉(𝜆)𝑛+1,𝑛+1], ||𝐹(𝑟)𝜆,2𝑛+1||(𝑥)𝑛𝑟2𝜑𝑟||𝐹(𝑥)𝜆,2𝑛+1||(𝑥)+𝑛𝑟𝜑𝑟||𝐹(𝑥)𝜆,2𝑛+1||(𝑥)+𝑛𝑟𝜑12𝜆𝑟(𝑥)(4.45) and, if 𝑟 is odd, one has ||𝐹(𝑟)𝜆,2𝑛+1||(𝑥)𝑛𝑟1𝜑1𝑟||𝐹(𝑥)𝜆,2𝑛+1||(𝑥)+𝑛𝑟1𝜑1𝑟||𝐹(𝑥)𝜆,2𝑛+1||(𝑥)+𝑛𝑟𝜑12𝜆𝑟(𝑥).(4.46)

Proof. Using (4.35) and (4.36), we obtain the result similarly to the proof of Lemma 4.2.

Proof of Theorem 2.2. Equation (2.5) comes from Lemma 4.7 and Proposition 3.7. We will show (2.6) and (2.7). From Proposition 3.7, we see max𝑥[1,1]||𝐹𝜆,2𝑛+1||(𝑥)𝑛1+2𝜆.(4.47) Hence, using Markov-Bernstein inequality for 𝐹𝜆,2𝑛+1𝒫2𝑛, we have (2.6). Now, we show (2.7). By Proposition 3.7 (a), it is true for 𝑟=1. We suppose that, for 𝑟2, ||𝐹(𝑟1)𝜆,2𝑛+1||(𝑥)𝑛2𝑟+2𝜆3,𝑥1,𝜉(𝜆)1,𝑛+1𝜉(𝜆)𝑛+1,𝑛+1,1.(4.48) As the proof of Theorem 2.1, we have, for 𝑟2 and for 𝑥[𝜉(𝜆)𝑛+1,𝑛+1,1], 𝐹(𝑟)𝜆,2𝑛+1(𝑥)𝐹(𝑟1)𝜆,2𝑛+1(𝑥)𝑛2.(4.49) Therefore, we see that by induction with Proposition 3.7, (2.7) holds for every 𝑟=1,2,3,.

Corollary 4.8. Let 0<𝜆<1 and 𝑟2. Then, for 𝑥[1,𝜉(𝜆)1,𝑛+1][𝜉(𝜆)𝑛+1,𝑛+1,1], ||𝐽(𝑟2)𝜆,𝑛||(𝑥)𝑛2𝑟+2𝜆3.(4.50)

Proof. Corollary 4.8 comes from (4.35), (2.7), and (3.8).

Proof of Theorem 2.3. Equation (2.8) comes from (4.15) and (3.24).

Lemma 4.9. For 1𝜇𝑛+1 and 1𝜈2𝑛+1, |||𝑃𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1|||𝑛𝜆1𝜑𝜆𝜉(𝜆)𝜇,𝑛+1|||𝐸,(4.51)𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛|||𝑛1𝜆𝜑1𝜆𝑥(𝜆)𝜈,𝑛.(4.52)

Proof. Since we know from (3.24) and (3.25) that 𝑛𝜑2𝜆𝜉(𝜆)𝜇,𝑛+1|||𝐹𝜆,2𝑛+1𝜉(𝜆)𝜇,𝑛+1|||=|||𝐸𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1𝑃𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1|||𝑛2𝜆𝜑𝜆𝜉(𝜆)𝜇,𝑛+1|||𝑃𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1|||,(4.53) (4.51) is obviously proved. Similarly, since we have, from (3.4) and (3.25), 𝑛𝜑2𝜆𝑥(𝜆)𝜈,𝑛|||𝐹𝜆,2𝑛+1𝑥(𝜆)𝜈,𝑛|||=|||𝐸𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛𝑃𝜆,𝑛𝑥(𝜆)𝜈,𝑛|||𝑛𝜆𝜑𝜆1𝑥(𝜆)𝜈,𝑛|||𝐸𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛|||,(4.54) (4.52) is obtained.

Lemma 4.10. Let 0<𝜆<1 and 𝑟1. Let 𝑟 be an odd integer. (a)For 1𝜇𝑛+1, |||𝑃(𝑟)𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1|||𝑛2𝜆+𝑟2𝜑𝑟1𝜉(𝜆)𝜇,𝑛+1.(4.55)(b)For 1𝜈𝑛, |||𝐸(𝑟)𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛|||𝑛𝑟𝜑𝑟𝑥(𝜆)𝜈,𝑛.(4.56)

Proof. (a) We know, from (3.3), |||𝑃𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1|||𝑛2𝜆1𝜑2𝜉(𝜆)𝜇,𝑛+1.(4.57) So (4.55) holds for 𝑘=1. Assume that, for 𝑘=1,2,,, |||𝑃(2𝑘1)𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1|||𝑛2𝜆+2𝑘3𝜑2𝑘𝜉(𝜆)𝜇,𝑛+1.(4.58) Then, we have from (3.5), (3.7), and (4.58) that |||𝑃(2+1)𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1|||𝜑2𝜉(𝜆)𝜇,𝑛+1|||𝑃(2)𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1|||+𝑛2𝜑2𝜉(𝜆)𝜇,𝑛+1|||𝑃(21)𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1|||𝑛2𝜆+21𝜑22𝜉(𝜆)𝜇,𝑛+1.(4.59) Therefore, we have the result using the mathematical induction.
(b) For an odd integer 𝑟1, we have from (3.6), (4.16), and (4.52) |||𝐸(𝑟)𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛|||𝑛𝑟1𝜑1𝑟|||𝐸(𝑥)𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛|||+|||𝐸𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛|||+𝑛𝑟𝜑𝑟𝑥(𝜆)𝜈,𝑛𝑛𝑟𝜑𝑟𝑥(𝜆)𝜈,𝑛.(4.60)

Lemma 4.11. Let 𝑟2. If 𝑟 is even, then |||𝑃(𝑟)𝜆,𝑛𝑥(𝜆)𝜈,𝑛|||𝑛𝜆+𝑟2𝜑𝜆𝑟1𝑥(𝜆)𝜈,𝑛.(4.61)

Proof. It is easily proved from (3.4) and (3.7).

Lemma 4.12. Let 𝑘 be a positive integer. Then, one has, for 1𝜈2𝑛+1, |||𝐽(2𝑘)𝜆,𝑛𝑦(𝜆)𝜈,2𝑛+1|||𝑛2𝑘+1+𝜆𝜑2𝑘𝜆𝑦(𝜆)𝜈,2𝑛+1.(4.62)

Proof. From (4.37), we know that 𝐽𝜆,𝑛(𝑥)=2𝜆𝑥𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)+21𝑥2𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)+𝐼𝜆,𝑛(𝑥)𝑃𝜆,𝑛(𝑥).(4.63) From Lemma 4.5 (a) and (b), we know that for 𝑥[𝑦(𝜆)1,2𝑛+1,𝑦(𝜆)2𝑛+1,2𝑛+1] (=[𝜉(𝜆)1,𝑛+1,𝜉(𝜆)𝑛+1,𝑛+1]) |||𝑥𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(2𝑘)|||𝑛2𝑘+1𝜑2𝑘2𝜆|||𝐼(𝑥)𝜆,𝑛(𝑥)𝑃𝜆,𝑛(𝑥)(2𝑘)|||𝑛2𝑘+1+𝜆𝜑2𝑘𝜆(𝑥).(4.64) On the other hand, we estimate |((1𝑥2)𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥))(2𝑘)| splitting into three terms as follows: ||||1𝑥2𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(2𝑘)||||𝑞+𝑟=2𝑘||1𝑥2𝐸(𝑞+1)𝜆,𝑛+1(𝑥)𝑃(𝑟+1)𝜆,𝑛||||||𝐸(𝑥)+𝑂(1)𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(2𝑘1)||||+||||𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(2𝑘2)||||.(4.65) Here, from Lemma 4.5 (c), we have for 𝑥[𝑦(𝜆)1,2𝑛+1,𝑦(𝜆)2𝑛+1,2𝑛+1]||||𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(2𝑘1)||||𝑛2𝑘+1𝜑2𝜆2𝑘||||𝐸(𝑥),𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(2𝑘2)||||𝑛2𝑘𝜑2𝜆2𝑘+1(𝑥)𝑛2𝑘+1𝜑2𝜆2𝑘(𝑥).(4.66) For the first term, we also split into two terms as follows: 𝑞+𝑟=2𝑘||1𝑥2𝐸(𝑞+1)𝜆,𝑛+1(𝑥)𝑃(𝑟+1)𝜆,𝑛(||=𝑥)𝑞+𝑟=2𝑘,𝑞even,𝑟even||1𝑥2𝐸(𝑞+1)𝜆,𝑛+1(𝑥)𝑃(𝑟+1)𝜆,𝑛||+(𝑥)𝑞+𝑟=2𝑘,𝑞odd,𝑟odd||1𝑥2𝐸(𝑞+1)𝜆,𝑛+1(𝑥)𝑃(𝑟+1)𝜆,𝑛(||𝑥)=𝐴1(𝑥)+𝐴2(𝑥).(4.67) From (2.2) and (4.55), we know that for even 𝑞 and 𝑟|||𝐸(𝑞+1)𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1|||𝑛𝑞+2𝜆𝜑𝑞𝜆𝜉(𝜆)𝜇,𝑛+1,|||𝑃(𝑟+1)𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1|||𝑛2𝜆+𝑟1𝜑𝑟2𝜉(𝜆)𝜇,𝑛+1.(4.68) Also, we know from (4.56) and (3.5) that for even 𝑞 and 𝑟|||𝐸(𝑞+1)𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛|||𝑛𝑞+1𝜑𝑞1𝑥(𝜆)𝜈,𝑛,|||𝑃(𝑟+1)𝜆,𝑛𝑥(𝜆)𝜈,𝑛|||𝑛𝜆+𝑟𝜑𝜆𝑟1𝑥(𝜆)𝜈,𝑛.(4.69) Then, we have |||𝐴1𝜉(𝜆)𝜇,𝑛+1|||𝜑2𝜉(𝜆)𝜇,𝑛+1𝑞+𝑟=2𝑘,𝑞even,𝑟even|||𝐸(𝑞+1)𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1𝑃(𝑟+1)𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1|||𝑛2𝑘+1+𝜆𝜑2𝑘𝜆𝜉(𝜆)𝜇,𝑛+1,|||𝐴1𝑥(𝜆)𝜈,𝑛|||𝜑2𝑥(𝜆)𝜈,𝑛𝑞+𝑟=2𝑘,𝑞even,𝑟even|||𝐸(𝑞+1)𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛𝑃(𝑟+1)𝜆,𝑛𝑥(𝜆)𝜈,𝑛|||𝑛2𝑘+1+𝜆𝜑𝜆2𝑘𝑥(𝜆)𝜈,𝑛.(4.70) Similarly, for odd 𝑞 and 𝑟, we have, by (2.8) and (3.5), |||𝐸(𝑞+1)𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1𝑃(𝑟+1)𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1|||𝑛2𝑘+1+𝜆𝜑2𝑘𝜆2𝜉(𝜆)𝜇,𝑛+1(4.71) and, by (2.2) and (4.61), |||𝐸(𝑞+1)𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛𝑃(𝑟+1)𝜆,𝑛𝑥(𝜆)𝜈,𝑛|||𝑛2𝑘+1+𝜆𝜑𝜆2𝑘2𝑥(𝜆)𝜈,𝑛.(4.72) Thus, we have |||𝐴2𝜉(𝜆)𝜇,𝑛+1|||𝜑2𝜉(𝜆)𝜇,𝑛+1𝑞+𝑟=2𝑘,𝑞odd,𝑟odd|||𝐸(𝑞+1)𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1𝑃(𝑟+1)𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1|||𝑛2𝑘+1+𝜆𝜑2𝑘𝜆𝜉(𝜆)𝜇,𝑛+1,|||𝐴2𝑥(𝜆)𝜈,𝑛|||𝜑2𝑥(𝜆)𝜈,𝑛𝑞+𝑟=2𝑘,𝑞odd,𝑟odd|||𝐸(𝑞+1)𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛𝑃(𝑟+1)𝜆,𝑛𝑥(𝜆)𝜈,𝑛|||𝑛2𝑘+1+𝜆𝜑𝜆2𝑘𝑥(𝜆)𝜈,𝑛.(4.73) Therefore, we have the result.

Proof of Theorem 2.4. When 𝑟=2, (2.9) holds from (3.27). Let even 𝑟>2, and suppose that (2.9) holds for 𝑟2. Since we know by (4.35), (2.5), (4.62), and (4.19) 𝜑2𝑦(𝜆)𝜈,2𝑛+1|||𝐹(𝑟)𝜆,2𝑛+1𝑦(𝜆)𝜈,2𝑛+1||||||𝐹(𝑟1)𝜆,2𝑛+1𝑦(𝜆)𝜈,2𝑛+1|||+𝑛2|||𝐹(𝑟2)𝜆,2𝑛+1𝑦(𝜆)𝜈,2𝑛+1|||+|||𝐽(𝑟2)𝜆,𝑛𝑦(𝜆)𝜈,2𝑛+1|||𝑛2|||𝐹(𝑟2)𝜆,2𝑛+1𝑦(𝜆)𝜈,2𝑛+1|||+𝑛𝑟1+𝜆𝜑𝑟+2𝜆𝑦(𝜆)𝜈,2𝑛+1,(4.74) we obtain, using mathematical induction, 𝜑2𝑦(𝜆)𝜈,2𝑛+1|||𝐹(𝑟)𝜆,2𝑛+1𝑦(𝜆)𝜈,2𝑛+1|||𝑛𝑟1+𝜆𝜑𝑟+2𝜆𝑦(𝜆)𝜈,2𝑛+1.(4.75) Therefore, (2.9) is proved.

Proof of Theorem 2.5. (a) From (4.3), we know that 𝜑2𝜉(𝜆)𝜇,𝑛+1𝐸(3)𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1=(𝑛+1)2𝐸𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1𝐸+𝑂𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1𝐸+𝑂𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1𝐼+𝑂𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1.(4.76) Therefore, we have, by (2.2), (3.24), and (3.26), 𝐸(3)𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1=(𝑛+1)2𝜑2𝜉(𝜆)𝜇,𝑛+1𝐸𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1𝑛+𝑂3.(4.77) Suppose that, for an integer 2, 𝐸(21)𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1=(1)1(𝑛+1)2(1)𝜑2(1)𝜉(𝜆)𝜇,𝑛+1𝐸𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1𝑛+𝑂21.(4.78) Then, from (4.3), we obtain 𝜑2𝜉(𝜆)𝜇,𝑛+1𝐸(2+1)𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1=(𝑛+1)2𝐸(21)𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1𝐸+𝑂(2)𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1𝐸+𝑂(21)𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1𝐼+𝑂(21)𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1.(4.79) Therefore, by (2.2), (2.8), and (4.4), we have 𝐸(2+1)𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1=(𝑛+1)2𝜑2𝜉(𝜆)𝜇,𝑛+1𝐸(21)𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1𝑛+𝑂2+1=(1)(𝑛+1)2𝜑2𝜉(𝜆)𝜇,𝑛+1𝐸𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1𝑛+𝑂2+1.(4.80)
(b) Similarly to the proof of (a), by (4.51), (3.3), and (3.7), we can obtain 𝑃(2)𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1=(1)𝑛(𝑛+2𝜆)𝜑2𝜉(𝜆)𝜇,𝑛+1𝑃𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1𝑛+𝑂2𝜆+23.(4.81) In addition, we see that, from (4.51), 𝑃(2)𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1=(1)(𝑛+1)2𝜑2𝜉(𝜆)𝜇,𝑛+1𝑃𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1𝑛+𝑂21𝑃𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1𝑛+𝑂2𝜆+23=(1)(𝑛+1)2𝜑2𝜉(𝜆)𝜇,𝑛+1𝑃𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1𝑛+𝑂2+𝜆2.(4.82)

Proof of Theorem 2.6. (a) From (3.9), we know that 𝜑2𝑥(𝜆)𝜈,𝑛𝐸𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛&=(𝑛+1)2𝐸𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛𝐸+𝑂𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛𝐼+𝑂𝜆,𝑛𝑥(𝜆)𝜈,𝑛.(4.83) Therefore, by (4.4) and (4.56), we have 𝐸𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛=(𝑛+1)2𝜑2𝑥(𝜆)𝜈,𝑛𝐸𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛𝑛+𝑂2.(4.84) Then, we obtain from (4.3), (2.2), and (4.56) that 𝜑2𝑥(𝜆)𝜈,𝑛𝐸(2)𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛=(𝑛+1)2𝐸(22)𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛𝑛+𝑂2.(4.85) Therefore, we have the result inductively.
(b) From (3.7), we know that 𝜑2𝑥(𝜆)𝜈,𝑛𝑃(3)𝜆,𝑛𝑥(𝜆)𝜈,𝑛=𝑛(𝑛+2𝜆)𝑃𝜆,𝑛𝑥(𝜆)𝜈,𝑛𝑃+𝑂𝜆,𝑛𝑥(𝜆)𝜈,𝑛𝑃+𝑂𝜆,𝑛𝑥(𝜆)𝜈,𝑛.(4.86) Therefore, by (3.4) and (4.61), we have 𝑃(3)𝜆,𝑛𝑥(𝜆)𝜈,𝑛=𝑛(𝑛+2𝜆)𝜑2𝑥(𝜆)𝜈,𝑛𝑃𝜆,𝑛𝑥(𝜆)𝜈,𝑛𝑛+𝑂𝜆.(4.87) Suppose that, for an integer 2, 𝑃(21)𝜆,𝑛𝑥(𝜆)𝜈,𝑛=(1)1𝑛1(𝑛+2𝜆)1𝜑2(1)𝑥(𝜆)𝜈,𝑛𝑃𝜆,𝑛𝑥(𝜆)𝜈,𝑛𝑛+𝑂𝜆+24.(4.88) Then from (3.5), (3.7), and (4.61) 𝑃(2+1)𝜆,𝑛𝑥(𝜆)𝜈,𝑛=(1)𝑛(𝑛+2𝜆)𝜑2𝑥(𝜆)𝜈,𝑛𝑃𝜆,𝑛𝑥(𝜆)𝜈,𝑛𝑛+𝑂𝜆+22.(4.89) Here, we see that for 2{𝑛(𝑛+2𝜆)}=(𝑛+1)2𝑛+𝑂21.(4.90) Hence, we obtain, from (3.4), 𝑃(2+1)𝜆,𝑛𝑥(𝜆)𝜈,𝑛=(1)(𝑛+1)2𝜑2𝑥(𝜆)𝜈,𝑛𝑃𝜆,𝑛𝑥(𝜆)𝜈,𝑛𝑛+𝑂21𝑃𝜆,𝑛𝑥(𝜆)𝜈,𝑛𝑛+𝑂𝜆+22=(1)(𝑛+1)2𝜑2𝑥(𝜆)𝜈,𝑛𝑃𝜆,𝑛𝑥(𝜆)𝜈,𝑛𝑛+𝑂𝜆+21.(4.91)

Lemma 4.13. Let 0<𝜀<1 and |𝑦(𝜆)𝜈,2𝑛+1|1𝜀. Then, for a nonnegative integer 0, 𝐽(2+1)𝜆,𝑛𝑦(𝜆)𝜈,2𝑛+1=22+1(1)+1(𝑛+1)2+2𝜑2𝑦(𝜆)𝜈,2𝑛+1𝐹𝑦(𝜆)𝜈,2𝑛+1𝑛+𝑂𝜆+2+2.(4.92)

Proof. Let |𝑦(𝜆)𝜈,2𝑛+1|1𝜀. From (4.37) and Lemma 4.5, we see that 𝐽(2+1)𝜆,𝑛𝑦(𝜆)𝜈,2𝑛+1=21𝑥2𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(2+1)𝐸4𝑥(2+1)𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(2)𝐸2(2+1)𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(21)+2𝜆𝑥𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(2+1)+𝐼𝜆,𝑛(𝑥)𝑃𝜆,𝑛(𝑥)(2+1)|||𝑥=𝑦(𝜆)𝜈,2𝑛+1=21𝑥2𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(2+1)||||𝑥=𝑦(𝜆)𝜈,2𝑛+1𝑛+𝑂2+2+𝜆.(4.93) Here, we let 2(2+1)(𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥))(21)=0 when =0. To estimate the first term, we split it into two terms as follows: 𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(2+1)=0𝑖2+1,𝑖even+0𝑖2+1,𝑖odd𝑖𝐸2+1(𝑖+1)𝜆,𝑛+1(𝑥)𝑃(2+2i)𝜆,𝑛(𝑥).(4.94) Then, using |𝑥(𝜆)𝜈,𝑛|1𝜀, from (2.13) and (2.15), we obtain 0𝑖2+1,𝑖odd𝑖𝐸2+1(𝑖+1)𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛𝑃(2+2𝑖)𝜆,𝑛𝑥(𝜆)𝜈,𝑛=(1)+1(𝑛+1)2+2𝜑22𝑥(𝜆)𝜈,𝑛𝐹𝜆,2𝑛+1𝑥(𝜆)𝜈,𝑛0𝑖2+1,𝑖odd𝑖𝑛2+1+𝑂𝜆+2+2=22(1)+1(𝑛+1)2+2𝜑22𝑥(𝜆)𝜈,𝑛𝐹𝜆,2𝑛+1𝑥(𝜆)𝜈,𝑛𝑛+𝑂𝜆+2+2(4.95) and, from (4.56) and from (4.61), 0𝑖2+1,𝑖even𝑖𝐸2+1(𝑖+1)𝜆,𝑛+1𝑥(𝜆)𝜈,𝑛𝑃(2+2𝑖)𝜆,𝑛𝑥(𝜆)𝜈,𝑛𝑛=𝑂𝜆+2+1.(4.96) Thus, we have, for |𝑥(𝜆)𝜈,𝑛|1𝜀, 𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(2+1)||||𝑥=𝑥(𝜆)𝜈,𝑛=22(1)+1(𝑛+1)2+2𝜑22𝑥(𝜆)𝜈,𝑛𝐹𝜆,2𝑛+1𝑥(𝜆)𝜈,𝑛𝑛+𝑂𝜆+2+2.(4.97) Similarly, noting |𝜉(𝜆)𝜇,𝑛|1𝜀, from (2.10) and (2.12) 0𝑖2+1,𝑖even𝑖𝐸2+1(𝑖+1)𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1𝑃(2+2𝑖)𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1=22(1)+1(𝑛+1)2+2𝜑22𝜉(𝜆)𝜇,𝑛+1𝐹𝜆,2𝑛+1𝜉(𝜆)𝜇,𝑛+1𝑛+𝑂𝜆+2+2(4.98) and, from (2.8) and (4.55), 0𝑖2+1,𝑖odd𝑖𝐸2+1(𝑖+1)𝜆,𝑛+1𝜉(𝜆)𝜇,𝑛+1𝑃(2+2𝑖)𝜆,𝑛𝜉(𝜆)𝜇,𝑛+1𝑛=𝑂2𝜆+2+1.(4.99) Then, we have, for |𝜉(𝜆)𝜇,𝑛+1|1𝜀, 𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(2+1)||||𝑥=𝜉(𝜆)𝜇,𝑛+1=22(1)+1(𝑛+1)2+2𝜑22𝜉(𝜆)𝜇,𝑛+1𝐹𝜆,2𝑛+1𝜉(𝜆)𝜇,𝑛+1𝑛+𝑂𝜆+2+2.(4.100) Therefore, we have, for |𝑦(𝜆)𝜈,2𝑛+1|1𝜀, 𝐸𝜆,𝑛+1(𝑥)𝑃𝜆,𝑛(𝑥)(2+1)||||𝑥=𝑦(𝜆)𝜈,2𝑛+1=22(1)+1(𝑛+1)2+2𝜑22𝑦(𝜆)𝜈,2𝑛+1𝐹𝜆,2𝑛+1𝑦(𝜆)𝜈,2𝑛+1𝑛+𝑂𝜆+2+2.(4.101) Thus, we have, for |𝑦(𝜆)𝜈,2𝑛+1|1𝜀, 𝐽(2+1)𝜆,𝑛𝑦(𝜆)𝜈,2𝑛+1=22+1(1)+1(𝑛+1)2+2𝜑2𝑦(𝜆)𝜈,2𝑛+1𝐹𝜆,2𝑛+1𝑦(𝜆)𝜈,2𝑛+1𝑛+𝑂𝜆+2+2.(4.102)

Proof of Theorem 2.7. From (4.35), (3.25), (3.27), and Lemma 4.13, we have 𝜑2𝑦(𝜆)𝜈,2𝑛+1𝐹(3)𝜆,2𝑛+1𝑦(𝜆)𝜈,2𝑛+1=2(𝑛+1)2𝐹𝜆,2𝑛+1𝑦(𝜆)𝜈,2𝑛+1+𝐽𝜆,𝑛𝑦(𝜆)𝜈,2𝑛+1+𝑂𝑛𝐹𝜆,2𝑛+1𝑦(𝜆)𝜈,2𝑛+1𝐹+𝑂𝜆,2𝑛+1𝑦(𝜆)𝜈,2𝑛+1=3(𝑛+1)2𝐹𝜆,2𝑛+1𝑦(𝜆)𝜈,2𝑛+1𝑛+𝑂𝜆+2.(4.103) Suppose that 𝐹(21)𝜆,2𝑛+1𝑦(𝜆)𝜈,2𝑛+1=𝑐1(1)1(𝑛+1)22𝜑2+2𝑦(𝜆)𝜈,2𝑛+1𝐹𝜆,2𝑛+1𝑦(𝜆)𝜈,2𝑛+1𝑛+𝑂𝜆+22.(4.104) Then, we obtain from (4.35) 𝜑2𝑦(𝜆)𝜈,2𝑛+1𝐹(2+1)𝜆,2𝑛+1𝑦(𝜆)𝜈,2𝑛+1=2𝑐1+221(1)(𝑛+1)2𝜑2+2𝑦(𝜆)𝜈,2𝑛+1𝐹𝜆,2𝑛+1𝑦(𝜆)𝜈,2𝑛+1𝑛+𝑂𝜆+2=𝑐(1)(𝑛+1)2𝜑2+2𝑦(𝜆)𝜈,2𝑛+1𝐹𝜆,2𝑛+1𝑦(𝜆)𝜈,2n+1𝑛+𝑂𝜆+2.(4.105) Therefore, (2.16) is proved.

Acknowledgment

The authors thank the referees for many kind suggestions and comments.