# Asymptotic Properties of Derivatives of the Stieltjes Polynomials

Authors: Hee Sun Jung; Ryozi Sakai
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

Letwλ(x):=(1−x2)λ−1/2 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, 0≤m<n+1, ∫−11wλ(x)Pλ,n(x)Eλ,n+1(x)xmdx≠0, 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.


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## Body

## 1. Introduction

Consider the generalized Stieltjes polynomialsEλ,n+1(x) defined (up to a multiplicative constant) by
(1.1)∫-11wλ(x)Pλ,n(x)Eλ,n+1(x)xkdx=0,k=0,1,2,…,n,n⩾1,
where wλ(x)=(1-x2)λ-1/2, λ>-1/2, and Pλ,n(x) is the nth ultraspherical polynomial for the weight function wλ(x).The polynomialsEλ,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 0⩽λ⩽2. Recently, several authors [2–8] studied further interesting properties for these Stieltjes polynomials. Ehrich and Mastroianni [3, 4] gave accurate pointwise bounds of Eλ,n+1(x)(0⩽λ⩽1) and the product Fλ,2n+1∶=Eλ,n+1(x)Pλ,n(x)(0⩽λ⩽1) 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)(0⩽λ⩽1) and Fλ,2n+1(x)(0⩽λ⩽1). 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 ofEλ,n+1(r)(x) and Fλ,2n+1(r)(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 Eλ,n+1(2ℓ)(x) and Fλ,2n+1(2ℓ)(x) at the zeros of Eλ,n+1(x) and Fλ,2n+1(x), respectively. Especially, for the value of Fλ,2n+1(2ℓ)(x) at the zeros of Fλ,2n+1(x), we will estimate Pλ,n(r)(x) and Eλ,n+1(r)(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 Eλ,n+1(2ℓ+1) and Fλ,2n+1(2ℓ+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 Section2, 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 polynomialsPλ,n, λ≠0, we use the normalization Pλ,n(1)=(n+2λ-1n) and then we know that Pλ,n(1)~n2λ-1. We denote the zeros of Pλ,n by xν,n(λ), ν=1,…,n, and the zeros of Stieltjes polynomials Eλ,n+1 by ξμ,n+1(λ), μ=1,…,n+1. We denote the zeros of Fλ,2n+1∶=Pλ,nEλ,n+1 by yν,2n+1(λ), ν=1,…,2n+1. All nodes are ordered by increasing magnitude. We set φ(x)∶=1-x2, and, for any two sequences {bn}n and {cn}n of nonzero real numbers (or functions), we write bn≲cn, if there exists a constant C>0, independent of n (and x) such that bn⩽Ccn for n large enough and write bn~cn if bn≲cn and cn≲bn. We denote by 𝒫n the space of polynomials of degree at most n.For the Chebyshev polynomialTn(x), note that for λ=0 and λ=1(2.1)E0,n+1(x)=2nπ(Tn+1(x)-Tn-1(x))E1,n+1(x)=2πTn+1(x).
Therefore, we will consider Eλ,n+1(x) for 0<λ<1.Theorem 2.1.
Let0<λ<1 and r⩾1 be a positive integer. Then, for all x∈[ξ1,n+1(λ),ξn+1,n+1(λ)],
(2.2)|Eλ,n+1(r)(x)|≲nr+1-λφ1-r-λ(x).
Moreover, one has
(2.3)max⁡x∈[-1,1]|Eλ,n+1(r)(x)|~n2r,
and especially one has, for x∈[-1,ξ1,n+1(λ)]∪[ξn+1,n+1(λ),1],
(2.4)|Eλ,n+1(r)(x)|~n2r.Theorem 2.2.
Let0<λ<1 and r⩾1 be a positive integer. Then, for all x∈[ξ1,n+1(λ),ξn+1,n+1(λ)],
(2.5)|Fλ,2n+1(r)(x)|≲nrφ1-2λ-r(x).
Moreover, one has
(2.6)max⁡x∈[-1,1]|Fλ,2n+1(r)(x)|≲n2λ+2r-1,
and especially, for x∈[-1,ξ1,n+1(λ)]∪[ξn+1,n+1(λ),1],
(2.7)|Fλ,2n+1(r)(x)|~n2λ+2r-1.In the following, we also estimate the values ofEλ,n+1(2ℓ)(x) and Fλ,2n+1(2ℓ)(x), ℓ⩾1 at the zeros {ξμ,n+1(λ)} of Eλ,n+1(x) and the zeros {yν,2n+1(λ)} of Fλ,2n+1(x), respectively.Theorem 2.3.
Let0<λ<1 and r⩾2 be an even integer. For 1⩽μ⩽n+1, one has
(2.8)|Eλ,n+1(r)(ξμ,n+1(λ))|≲nrφ-r(ξμ,n+1(λ)).Theorem 2.4.
Let0<λ<1 and r⩾2 be an even integer. For 1⩽ν⩽2n+1, one has
(2.9)|Fλ,2n+1(r)(yν,2n+1(λ))|≲nr-1+λφ-r-λ(yν,2n+1(λ)).Finally, we obtain the asymptotic representations for the values ofEλ,n+1(2ℓ+1)(x) and Fλ,2n+1(2ℓ+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.
Let0<λ<1 and 0<ɛ<1. Suppose |ξμ,n+1(λ)|⩽1-ɛ. Then, (a)
(2.10)Eλ,n+1(2l+1)(ξμ,n+1(λ))=(-1)l(n+1)2lφ-2l(ξμ,n+1(λ))Eλ,n+1′(ξμ,n+1(λ))+O(n2l+1),(b)
(2.11)Pλ,n(2l)(ξμ,n+1(λ))=(-1)lnl(n+2λ)lφ-2l(ξμ,n+1(λ))Pλ,n(ξμ,n+1(λ))+O(n2λ+2l-3).
In addition,
(2.12)Pλ,n(2l)(ξμ,n+1(λ))=(-1)l(n+1)2lφ-2l(ξμ,n+1(λ))Pλ,n(ξμ,n+1(λ))+O(n2l+λ-2).Theorem 2.6.
Let0<λ<1 and 0<ɛ<1. Suppose |xν,n(λ)|⩽1-ɛ. Then, (a)
(2.13)Eλ,n+1(2l)(xν,n(λ))=(-1)l(n+1)2lφ-2l(xν,n(λ))Eλ,n+1(xν,n(λ))+O(n2l),(b)
(2.14)Pλ,n(2l+1)(xν,n(λ))=(-1)lnl(n+2λ)lφ-2l(xν,n(λ))Pλ,n′(xν,n(λ))+O(nλ+2l-2).
In addition,
(2.15)Pλ,n(2l+1)(xν,n(λ))=(-1)l(n+1)2lφ-2l(xν,n(λ))Pλ,n′(xν,n(λ))+O(nλ+2l-1).Theorem 2.7.
Let0<λ<1 and 0<ɛ<1. Suppose |yν,2n+1(λ)|⩽1-ɛ. Then, one has, for a positive integer ℓ⩾1,
(2.16)Fλ,2n+1(2l+1)(yν,2n+1(λ))=cl(-1)l(n+1)2lφ-2l(yν,2n+1(λ))Fλ,2n+1′(yν,2n+1(λ))+O(nλ+2l),
where cℓ=4ℓ-2ℓ-1.


## 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) Letλ>-1/2. Then, Pλ,n(x) satisfies the second-order differential equation as follows:
(3.1)(1-x2)Pλ,n″(x)-(2λ+1)xPλ,n′(x)+n(n+2λ)Pλ,n(x)=0.
(b) Letλ>-1/2. Then,
(3.2)Pλ,n′(x)=2λPλ+1,n-1(x).
(c) Letλ>-1/2. Then, for 1⩽μ⩽n+1,
(3.3)|Pλ,n′(ξμ,n+1(λ))|≲n2λ-1φ-2(ξμ,n+1(λ)).
(d) Letλ>-1/2. Then, for 1⩽ν⩽n,
(3.4)|Pλ,n′(xν,n(λ))|~nλφ-λ-1(xν,n(λ)).
(e) Letλ>-1/2. Then, for x∈(-1,1) and r⩾0,
(3.5)|Pλ,n(r)(x)|≲nλ+r-1φ-λ-r(x).
(f) Let0⩽λ⩽1. Then, for 1⩽ν⩽n,
(3.6)|Eλ,n+1′(xν,n(λ))|≲nφ-1(xν,n(λ)).
(g) Letλ>-1/2 and r⩾0. Then, Pλ,n(x) satisfies the higher-order differential equation as follows:
(3.7)(1-x2)Pλ,n(r+2)(x)-(2λ+2r+1)xPλ,n(r+1)(x)+(n2+2λn-r(2λ-r+2))Pλ,n(r)(x)=0.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 r⩾1, 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]).
Let0<λ<1. Let ξμ,n+1(λ)∶=cosθμ,n+1(λ), μ=1,…,n+1 and yν,2n+1(λ)∶=cosψν,2n+1(λ), ν=1,…,2n+1. Then, for μ=0,1,…,n+2 and ν=0,1,…,2n+2,
(3.8)|θμ,n+1(λ)-θμ+1,n+1(λ)|~|ψν,2n+1(λ)-ψν+1,2n+1(λ)|~n-1,
where ψ0,2n+1(λ)∶=θ0,n+1(λ)∶=π and ψ2n+2,2n+1(λ)∶=θn+2,n+1(λ)∶=0.Proposition 3.3 ([6, Proposition  2.3]).
Let0<λ<1. Then, for all x∈[-1,1],
(3.9)(1-x2)Eλ,n+1″(x)-xEλ,n+1′(x)+(n+1)2Eλ,n+1(x)=Iλ,n(x),
where
(3.10)Iλ,n(x)=8γn(λ)∑ν=1[(n+1)/2]‍(n+1-ν)ναν,n(λ)Tn+1-2ν(x).
Then Iλ,n(x) is a polynomial of degree n-1 satisfying
(3.11)max⁡x∈[-1,1]|Iλ,n(x)|≲n2.Proposition 3.4 ([4, Theorem  2.1]).
Let0<λ<1. Then, for n⩾0,
(3.12)|Eλ,n+1(x)|≲n1-λφ1-λ(x)+1-1⩽x⩽1.
Furthermore, Eλ,n+1(1)≳1.Proposition 3.5 ([6, Theorem  2.5]).
Let0<λ<1. (a)
For allx∈[ξ1,n+1(λ),ξn+1,n+1(λ)],
(3.13)|Eλ,n+1′(x)|≲n2-λφ-λ(x).
Moreover, one has, for x∈[-1,ξ1,n+1(λ)]∪[ξn+1,n+1(λ),1],
(3.14)|Eλ,n+1′(x)|~n2.(b)
For allx∈[ξ1,n+1(λ),ξn+1,n+1(λ)],
(3.15)|Eλ,n+1″(x)|≲n3-λφ-1-λ(x).
Moreover, one has, for x∈[-1,ξ1,n+1(λ)]∪[ξn+1,n+1(λ),1],
(3.16)|Eλ,n+1″(x)|~n4.Proposition 3.6 ([6, Corollary  2.6]).
Let0<λ<1. Then, for all x∈[-1,1],
(3.17)(1-x2)Fλ,2n+1″(x)-xFλ,2n+1′(x)+(2n2+2(1+λ)n+1)Fλ,2n+1(x)=Jλ,n(x).
Here, Jλ,n(x) is a polynomial of degree of 2n+1 defined in (4.37) such that, for x∈[ξ1,n+1(λ),ξn+1,n+1(λ)],
(3.18)|Jλ,n(x)|≲n2φ1-2λ(x)
and, for x∈[-1,ξ1,n+1(λ)]∪[ξn+1,n+1(λ),1],
(3.19)|Jλ,n(x)|≲n1+2λ.Proposition 3.7 ([6, Corollary  2.7]).
Let0<λ<1. (a)
For allx∈[ξ1,n+1(λ),ξn+1,n+1(λ)],
(3.20)|Fλ,2n+1′(x)|≲nφ-2λ(x).
Moreover, one has, for x∈[-1,ξ1,n+1(λ)]∪[ξn+1,n+1(λ),1],
(3.21)|Fλ,2n+1′(x)|~n1+2λ.(b)
For allx∈[ξ1,n+1(λ),ξn+1,n+1(λ)],
(3.22)|Fλ,2n+1″(x)|≲n2φ-1-2λ(x).
Moreover, one has, for x∈[-1,ξ1,n+1(λ)]∪[ξn+1,n+1(λ),1],
(3.23)|Fλ,2n+1″(x)|~n3+2λ.Proposition 3.8 ([4, Lemma  5.5]).
Let0<λ<1. Then, for μ=1,2,…,n+1,
(3.24)|Eλ,n+1′(ξμ,n+1(λ))|~n2-λφ-λ(ξμ,n+1(λ))
and, for ν=1,2,…,2n+1,
(3.25)|Fλ,2n+1′(yν,2n+1(λ))|~nφ-2λ(yν,2n+1(λ)).We now estimate the second derivatives at the zeros ofEλ,n+1 and Fλ,2n+1.Proposition 3.9 ([6, Theorem  2.9]).
Let0<λ<1. Then, for μ=1,2,…,n+1,
(3.26)|Eλ,n+1″(ξμ,n+1(λ))|≲n2φ-2(ξμ,n+1(λ))
and, for ν=1,2,…,2n+1,
(3.27)|Fλ,2n+1″(yν,2n+1(λ))|≲n1+λφ-2-λ(yν,2n+1(λ)).


## 4.  The Proofs of Main Results

In this section, we let0<λ<1 and m=⌊(n+1/2)⌋. A representation of Stieltjes polynomials Eλ,n+1(x) is (cf. [1, 10])
(4.1)γn(λ)2Eλ,n+1(cos⁡θ)=α0,n(λ)cos⁡(n+1)θ+α1,n(λ)cos⁡(n-1)θ+⋯+{αn/2,n(λ)cos⁡θ,neven12αn+1/2,n(λ),nodd,
where
(4.2)α0,n(λ)=f0,n(λ)=1,∑μ=0ν‍αμ,n(λ)fν-μ,n(λ)=0,ν=1,2,…,fν,n(λ)∶=(1-λν)(1-λn+ν+λ),ν=1,2,…,γn(λ)=πΓ(n+2λ)Γ(n+λ+1)~πnλ-1.In the following, we state the asymptotic differential relation of the higher order ofEλ,n+1.Lemma 4.1.
Let0<λ<1. Then, for all x∈[-1,1] and r⩾2,
(4.3)(1-x2)Eλ,n+1(r)(x)=(2r-3)xEλ,n+1(r-1)(x)+((r-2)2-(n+1)2)Eλ,n+1(r-2)(x)+Iλ,n(r-2)(x)
and, for x∈[ξ1,n+1(λ),ξn+1,n+1(λ)],
(4.4)|Iλ,n(r-2)(x)|≲nrφ2-r(x).
Here, Iλ,n(x) is a polynomial of degree n-1 defined in (3.10);
(4.5)Iλ,n(x)=8γn(λ)∑ν=1m‍(n+1-ν)ναν,n(λ)Tn+1-2ν(x),
such that
(4.6)max⁡x∈[-1,1]|Iλ,n(r-2)(x)|≲n2r-2.Proof.
Forr⩾2, (4.3) is obtained by r-2 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
(4.7)(1-x2)Tn″(x)-xTn′(x)+n2Tn(x)=0,
so we have, by r-2 times differentiation of (4.7),
(4.8)(1-x2)Tn(r)(x)-(2r-3)xTn(r-1)(x)-((r-2)2-n2)Tn(r-2)(x)=0.
Let for a nonnegative integer j⩾0,
(4.9)Iλ,n,j(x)∶=-8γn(λ)∑ν=1m‍(n+1-ν)ναν,n(λ)|Tn+1-2ν(j)(x)|.
Observe that in the view of Szegö’s result (cf. [1])
(4.10)α1,n(λ)<α2,n(λ)<α3,n(λ)<⋯<0,0⩽∑ν=0∞‍αν,n(λ)<1.
Then, since |Iλ,n(j)(x)|⩽Iλ,n,j(x) (note (4.10)), we will prove that, for x∈[ξ1,n+1(λ),ξn+1,n+1(λ)] and j⩾0,
(4.11)Iλ,n,j(x)≲nj+2φ-j(x)
instead of (4.4). Since, from the proof of [6, Proposition  2.3],
(4.12)0<-8γn(λ)∑ν=1m‍(n+1-ν)ναν,n(λ)≲n2
and, for x=cosθ,
(4.13)|Tn+1-2ν′(x)|=|(n+1-2ν)sin⁡(n+1-2ν)θsin⁡θ|≲nφ-1(x),
we obtain that Iλ,n,0(x)≲n2 and Iλ,n,1(x)≲n3φ-1(x). Using (4.8), we have, for 2⩽j⩽n,
(4.14)Iλ,n,j(x)=-8γn(λ)∑ν=1m‍(n+1-ν)ναν,n(λ)|Tn+1-2ν(j)(x)|Iλ,n,j(x)≤-8γn(λ)∑ν=1m‍(n+1-ν)ναν,n(λ)Iλ,n,j(x)≤-×((2j-3)|x|1-x2|Tn+1-2ν(j-1)(x)|+|(j-2)2-(n+1-2ν)2|1-x2|Tn+1-2ν(j-2)(x)|)Iλ,n,j(x)≲(2j-3)|x|1-x2Iλ,n,j-1(x)+(n+1)21-x2Iλ,n,j-2(x).
Therefore, (4.11) is proved by the mathematical induction on j. Consequently, we have (4.4).We obtain pointwise upper bounds ofEλ,n+1(r)(x) for two cases of an odd order and an even order in the following.Lemma 4.2.
Let0<λ<1 and r⩾2. Let x∈[ξ1,n+1(λ),ξn+1,n+1(λ)]. If r is even, then one has
(4.15)|Eλ,n+1(r)(x)|≲nr-2φ-r(x)|Eλ,n+1′(x)|+nrφ-r(x)|Eλ,n+1(x)|+nrφ-r(x),
and, if r is odd, then one has
(4.16)|Eλ,n+1(r)(x)|≲nr-1φ1-r(x)|Eλ,n+1′(x)|+nr-1φ-1-r(x)|Eλ,n+1(x)|+nrφ-r(x).Proof.
Letr⩾2. From (4.3) and (4.4), we have, for x∈[ξ1,n+1(λ),ξn+1,n+1(λ)],
(4.17)|Eλ,n+1(r)(x)|≲φ-2(x)|Eλ,n+1(r-1)(x)|+n2φ-2(x)|Eλ,n+1(r-2)(x)|+nrφ-r(x)
and especially
(4.18)|Eλ,n+1″(x)|≲φ-2(x)|Eλ,n+1′(x)|+n2φ-2(x)|Eλ,n+1(x)|+n2φ-2(x),
that is, we have (4.15) for r=2. From Proposition 3.2, we see 1+ξ1,n+1(λ), 1-ξn+1,n+1(λ)≳1/n, so we have, for x∈[ξ1,n+1(λ),ξn+1,n+1(λ)],
(4.19)φ-1(x)≲n.
Then, from (4.17) with r=3 and (4.18), we know that
(4.20)|Eλ,n+1(3)(x)|≲n2φ-2(x)|Eλ,n+1′(x)|+n2φ-4(x)|Eλ,n+1(x)|+n3φ-3(x),
that is, we have (4.16) for r=3. Assume that (4.15) and (4.16) hold for 3,4,…,r-1 times differentiation. Let r be an even number. Then, we have, from (4.17), (4.19), and the assumptions for r-1 and r-2,
(4.21)|Eλ,n+1(r)(x)|≲nr-2φ-r(x)|Eλ,n+1′(x)|+nrφ-r(x)|Eλ,n+1(x)|+nrφ-r(x),
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
(4.22)P(x)∶=(x-x1)(x-x2)⋯(x-xn).
Then, P′(x) is a polynomial of degree of n-1 and has distinct real n-1 zeros in (-1,1). Moreover, if one lets -1<y1<y2<⋯<yn-1<1 be the zeros of P′(x), then {yi}i=1n is interlaced with the zeros of P(x) that is, xi<yi<xi+1, i=1,2,…,n-1.Proof.
Since the sign ofP′(xi) is (-1)n-i, it is proved.Lemma 4.4.
Letr be a nonnegative integer. Then, Eλ,n+1(r)(x) has distinct n+1-r real zeros on (-1,1). If one lets {xn+1(r,i)}i=1n+1-r be the zeros of the polynomial Eλ,n+1(r)(x) with
(4.23)-1<xn+1(r,1)<xn+1(r,2)<⋯<xn+1(r,n+1-r)<1,
then one has, for 1⩽r⩽n and k=1,…,n+1-r,
(4.24)xn+1(0,k)<xn+1(r,k).Proof.
From Lemma4.3, we know that Eλ,n+1(r) has distinct real n+1-r zeros on (-1,1). By the interlaced zeros property of Lemma 4.3, we see that, for k=1,…,n+1-r,
(4.25)xn+1(r-1,k)<xn+1(r,k)<xn+1(r-1,k+1).
Thus, (4.24) is proved.Proof of Theorem2.1.
Letr⩾1. 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
(4.26)max⁡x∈[-1,1]|Eλ,n+1(r)(x)|~n2r,r=1,2.
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,ξ1,n+1(λ)]∪[ξn+1,n+1(λ),1] and r⩾2,
(4.27)|Eλ,n+1(r-1)(x)|~n2(r-1).
Then, by Lemma 4.4 and (3.8), we have, for x∈[ξn+1,n+1(λ),1] and r⩾2,
(4.28)Eλ,n+1(r)(x)Eλ,n+1(r-1)(x)∑k=1n+2-r‍=∑k=1n+2-r‍1x-xn+1(r-1,k)⩾1ξn+1,n+1(λ)-xn+1(r-1,n+2-r)∑k=1n+2-r‍⩾1ξn+1,n+1(λ)-xn+1(0,n+2-r)=1ξn+1,n+1(λ)-ξn+2-r,n+1(λ)∑k=1n+2-r‍≳1ξn+1,n+1(λ)-ξn,n+1(λ)≳n2.
Here, the last inequality is obtained by Proposition 3.2, that is,
(4.29)ξn+1,n+1(λ)-ξn,n+1(λ)=cos⁡θn+1,n+1(λ)-cos⁡θn,n+1(λ)~n-2.
Therefore, we have, for x∈[ξn+1,n+1(λ),1] and r⩾2,
(4.30)Eλ,n+1(r)(x)≳Eλ,n+1(r-1)(x)n2~n2r.
Hence, from (2.3), we have (2.4). For x∈[-1,ξ1,n+1(λ)], the proof is similar.Lemma 4.5.
Letℓ be a nonnegative integer and x∈[ξ1,n+1(λ),ξn+1,n+1(λ)]. Then,(4.31a)
|
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λ
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λ
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(4.31b)
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(4.31c)
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.Proof.
(a) Whenℓ=0, it is obvious from (3.5) and (3.12). Now, suppose ℓ⩾1. From (3.12), (2.2), and (3.5), we have
(4.32)|(xEλ,n+1(x)Pλ,n′(x))(l)|≲∑l-1⩽q+r⩽l‍|Eλ,n+1(q)(x)Pλ,n(r+1)(x)||(xEλ,n+1(x)Pλ,n′(x))(l)|≲∑l-1⩽q+r⩽l‍nq+r+1φ-(q+r)-2λ(x)≲nl+1φ-l-2λ(x).
(b) From (4.4) and (3.5), we have
(4.33)|(Iλ,n(x)Pλ,n(x))(l)|≲∑q+r=l‍|Iλ,n(q)(x)Pλ,n(r)(x)||(Iλ,n(x)Pλ,n(x))(l)|≲nl+1+λφ-l-λ(x).
(c) Similarly to the proof of (a), we have, from (2.2) and (3.5),
(4.34)|(Eλ,n+1′(x)Pλ,n′(x))(l)|≲∑q+r=l‍|Eλ,n+1(q+1)(x)Pλ,n(r+1)(x)|≲nl+2φ-2λ-l-1(x).Lemma 4.6.
Let0<λ<1. Then, for all x∈[-1,1] and r⩾2,
(4.35)(1-x2)Fλ,2n+1(r)(x)=(2r-5)xFλ,2n+1(r-1)(x)(1-x2)Fλ,2n+1(r)(x)=2+((r-2)2-(n+1)2-n(n+2λ))Fλ,2n+1(r-2)(x)+Jλ,n(r-2)(x)
and, for x∈[ξ1,n+1(λ),ξn+1,n+1(λ)],
(4.36)|Jλ,n(r-2)(x)|≲nrφ-2λ-r+3(x).
Here, Jλ,n(x) is a polynomial of degree of 2n+1 defined as follows:
(4.37)Jλ,n(x)=2λxEλ,n+1(x)Pλ,n′(x)Jλ,n(x)=2+2(1-x2)Eλ,n+1′(x)Pλ,n′(x)+Iλ,n(x)Pλ,n(x).
Furthermore, one has
(4.38)|Jλ,n(r-2)(1)|≲n2λ+2r-3.Proof.
Similarly to the proof of Lemma4.1, (4.35) is obtained by r-2 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
(4.39)Jλ,n(r-2)(x)=(2λxEλ,n+1(x)Pλ,n′(x))(r-2)Jλ,n(r-2)(x)=+(2(1-x2)Eλ,n+1′(x)Pλ,n′(x))(r-2)+(Iλ,n(x)Pλ,n(x))(r-2).
By Lemma 4.5 (a) and (b), we have, for x∈[ξ1,n+1(λ),ξn+1,n+1(λ)],
(4.40)|(2λxEλ,n+1(x)Pλ,n′(x))(r-2)|≲nr-1φ-r+2-2λ(x),|(Iλ,n(x)Pλ,n(x))(r-2)|≲nr-1+λφ-r+2-λ(x).
From (4.19) and Lemma 4.5 (c), we have, for x∈[ξ1,n+1(λ),ξn+1,n+1(λ)] and r⩾4,
(4.41)|(2(1-x2)Eλ,n+1′(x)Pλ,n′(x))(r-2)|2(1-x2)≲|(1-x2)(Eλ,n+1′(x)Pλ,n′(x))(r-2)|+|(Eλ,n+1′(x)Pλ,n′(x))(r-3)|2(1-x2)+|(Eλ,n+1′(x)Pλ,n′(x))(r-4)|2(1-x2)≲nrφ-2λ-r+3(x)+nr-1φ-2λ-r+2(x)+nr-2φ-2λ-r+3(x)2(1-x2)≲nrφ-2λ-r+3(x).
When r=2,3, we can similarly obtain that
(4.42)|(2(1-x2)Eλ,n+1′(x)Pλ,n′(x))(r-2)|≲nrφ-2λ-r+3(x).
Therefore, we have (4.36). On the other hand, from (2.4), (4.6), and (3.2), we know that, for a nonnegative integer ℓ,
(4.43)Eλ,n+1(l)(1)~n2l,|Iλ,n(l)(1)|≲n2l+2,Pλ,n(l)(1)~Pλ+l,n-l(1)~n2(λ+l)-1.
Then, similarly to the proof of Lemma 4.5, we obtain that
(4.44)|(xEλ,n+1(x)Pλ,n′(x))(l)|x=1≲n2l+2λ+1,|(Iλ,n(x)Pλ,n(x))(l)|x=1≲n2l+2λ+1,|(Eλ,n+1′(x)Pλ,n′(x))(l)|x=1≲n2l+2λ+3.
Therefore, we have (4.38).Lemma 4.7.
Let0<λ<1. Then, for r⩾2, if r is even, one has, for x∈[ξ1,n+1(λ),ξn+1,n+1(λ)],
(4.45)|Fλ,2n+1(r)(x)|≲nr-2φ-r(x)|Fλ,2n+1′(x)|+nrφ-r(x)|Fλ,2n+1(x)|+nrφ1-2λ-r(x)
and, if r is odd, one has
(4.46)|Fλ,2n+1(r)(x)|≲nr-1φ1-r(x)|Fλ,2n+1′(x)|+nr-1φ-1-r(x)|Fλ,2n+1(x)|+nrφ1-2λ-r(x).Proof.
Using (4.35) and (4.36), we obtain the result similarly to the proof of Lemma 4.2.Proof of Theorem2.2.
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
(4.47)max⁡x∈[-1,1]|Fλ,2n+1′(x)|~n1+2λ.
Hence, using Markov-Bernstein inequality for Fλ,2n+1′∈𝒫2n, we have (2.6). Now, we show (2.7). By Proposition 3.7 (a), it is true for r=1. We suppose that, for r⩾2,
(4.48)|Fλ,2n+1(r-1)(x)|~n2r+2λ-3,x∈[-1,ξ1,n+1(λ)]∪[ξn+1,n+1(λ),1].
As the proof of Theorem 2.1, we have, for r⩾2 and for x∈[ξn+1,n+1(λ),1],
(4.49)Fλ,2n+1(r)(x)Fλ,2n+1(r-1)(x)≳n2.
Therefore, we see that by induction with Proposition 3.7, (2.7) holds for every r=1,2,3,….Corollary 4.8.
Let0<λ<1 and r⩾2. Then, for x∈[-1,ξ1,n+1(λ)]∪[ξn+1,n+1(λ),1],
(4.50)|Jλ,n(r-2)(x)|≲n2r+2λ-3.Proof.
Corollary4.8 comes from (4.35), (2.7), and (3.8).Proof of Theorem2.3.
Equation (2.8) comes from (4.15) and (3.24).Lemma 4.9.
For1⩽μ⩽n+1 and 1⩽ν⩽2n+1,
(4.51)|Pλ,n(ξμ,n+1(λ))|~nλ-1φ-λ(ξμ,n+1(λ)),(4.52)|Eλ,n+1(xν,n(λ))|~n1-λφ1-λ(xν,n(λ)).Proof.
Since we know from (3.24) and (3.25) that
(4.53)nφ-2λ(ξμ,n+1(λ))~|Fλ,2n+1′(ξμ,n+1(λ))|=|Eλ,n+1′(ξμ,n+1(λ))Pλ,n(ξμ,n+1(λ))|~n2-λφ-λ(ξμ,n+1(λ))|Pλ,n(ξμ,n+1(λ))|,
(4.51) is obviously proved. Similarly, since we have, from (3.4) and (3.25),
(4.54)nφ-2λ(xν,n(λ))~|Fλ,2n+1′(xν,n(λ))|=|Eλ,n+1(xν,n(λ))Pλ,n′(xν,n(λ))|~nλφ-λ-1(xν,n(λ))|Eλ,n+1(xν,n(λ))|,
(4.52) is obtained.Lemma 4.10.
Let0<λ<1 and r⩾1. Let r be an odd integer. (a)
For1⩽μ⩽n+1,
(4.55)|Pλ,n(r)(ξμ,n+1(λ))|≲n2λ+r-2φ-r-1(ξμ,n+1(λ)).(b)
For1⩽ν⩽n,
(4.56)|Eλ,n+1(r)(xν,n(λ))|≲nrφ-r(xν,n(λ)).Proof.
(a) We know, from (3.3),
(4.57)|Pλ,n′(ξμ,n+1(λ))|≲n2λ-1φ-2(ξμ,n+1(λ)).
So (4.55) holds for k=1. Assume that, for k=1,2,…,ℓ,
(4.58)|Pλ,n(2k-1)(ξμ,n+1(λ))|≲n2λ+2k-3φ-2k(ξμ,n+1(λ)).
Then, we have from (3.5), (3.7), and (4.58) that
(4.59)|Pλ,n(2l+1)(ξμ,n+1(λ))|Pλ,n(2l+1)≲φ-2(ξμ,n+1(λ))|Pλ,n(2l)(ξμ,n+1(λ))|+n2φ-2(ξμ,n+1(λ))|Pλ,n(2l-1)(ξμ,n+1(λ))|Pλ,n(2l+1)≲n2λ+2l-1φ-2l-2(ξμ,n+1(λ)).
Therefore, we have the result using the mathematical induction.
(b) For an odd integerr⩾1, we have from (3.6), (4.16), and (4.52)
(4.60)|Eλ,n+1(r)(xν,n(λ))|Eλ,n+1≲nr-1φ1-r(x)(|Eλ,n+1′(xν,n(λ))|+|Eλ,n+1(xν,n(λ))|)+nrφ-r(xν,n(λ))Eλ,n+1≲nrφ-r(xν,n(λ)).Lemma 4.11.
Letr⩾2. If r is even, then
(4.61)|Pλ,n(r)(xν,n(λ))|≲nλ+r-2φ-λ-r-1(xν,n(λ)).Proof.
It is easily proved from (3.4) and (3.7).Lemma 4.12.
Letk be a positive integer. Then, one has, for 1⩽ν⩽2n+1,
(4.62)|Jλ,n(2k)(yν,2n+1(λ))|≲n2k+1+λφ-2k-λ(yν,2n+1(λ)).Proof.
From (4.37), we know that
(4.63)Jλ,n(x)=2λxEλ,n+1(x)Pλ,n′(x)Jλ,n(x)=2+2(1-x2)Eλ,n+1′(x)Pλ,n′(x)+Iλ,n(x)Pλ,n(x).
From Lemma 4.5 (a) and (b), we know that for x∈[y1,2n+1(λ),y2n+1,2n+1(λ)] (=[ξ1,n+1(λ),ξn+1,n+1(λ)])
(4.64)|(xEλ,n+1(x)Pλ,n′(x))(2k)|≲n2k+1φ-2k-2λ(x)|(Iλ,n(x)Pλ,n(x))(2k)|≲n2k+1+λφ-2k-λ(x).
On the other hand, we estimate |((1-x2)Eλ,n+1′(x)Pλ,n′(x))(2k)| splitting into three terms as follows:
(4.65)|((1-x2)Eλ,n+1′(x)Pλ,n′(x))(2k)|1-x2≲∑q+r=2k‍|(1-x2)Eλ,n+1(q+1)(x)Pλ,n(r+1)(x)|1-x21-+O(1)[|(Eλ,n+1′(x)Pλ,n′(x))(2k-1)|+|(Eλ,n+1′(x)Pλ,n′(x))(2k-2)|].
Here, from Lemma 4.5 (c), we have for x∈[y1,2n+1(λ),y2n+1,2n+1(λ)](4.66)|(Eλ,n+1′(x)Pλ,n′(x))(2k-1)|≲n2k+1φ-2λ-2k(x),|(Eλ,n+1′(x)Pλ,n′(x))(2k-2)|≲n2kφ-2λ-2k+1(x)≲n2k+1φ-2λ-2k(x).
For the first term, we also split into two terms as follows:
(4.67)∑q+r=2k|(1-x2)Eλ,n+1(q+1)(x)Pλ,n(r+1)(x)|1-x21-x2=∑q+r=2k,q:even,r:even|(1-x2)Eλ,n+1(q+1)(x)Pλ,n(r+1)(x)|1-x21-x21-x2x2+∑q+r=2k,q:odd,r:odd|(1-x2)Eλ,n+1(q+1)(x)Pλ,n(r+1)(x)|1-x21-x2:=A1(x)+A2(x).
From (2.2) and (4.55), we know that for even q and r(4.68)|Eλ,n+1(q+1)(ξμ,n+1(λ))|≲nq+2-λφ-q-λ(ξμ,n+1(λ)),|Pλ,n(r+1)(ξμ,n+1(λ))|≲n2λ+r-1φ-r-2(ξμ,n+1(λ)).
Also, we know from (4.56) and (3.5) that for even q and r(4.69)|Eλ,n+1(q+1)(xν,n(λ))|≲nq+1φ-q-1(xν,n(λ)),|Pλ,n(r+1)(xν,n(λ))|≲nλ+rφ-λ-r-1(xν,n(λ)).
Then, we have
(4.70)|A1(ξμ,n+1(λ))|ξμ,n+1(λ)≲φ2(ξμ,n+1(λ))∑q+r=2k,q:even,r:even‍|Eλ,n+1(q+1)(ξμ,n+1(λ))Pλ,n(r+1)(ξμ,n+1(λ))|ξμ,n+1(λ)≲n2k+1+λφ-2k-λ(ξμ,n+1(λ)),|A1(xν,n(λ))|≲φ2(xν,n(λ))∑q+r=2k,q:even,r:even‍|Eλ,n+1(q+1)(xν,n(λ))Pλ,n(r+1)(xν,n(λ))||A1(xν,n(λ))|≲n2k+1+λφ-λ-2k(xν,n(λ)).
Similarly, for odd q and r, we have, by (2.8) and (3.5),
(4.71)|Eλ,n+1(q+1)(ξμ,n+1(λ))Pλ,n(r+1)(ξμ,n+1(λ))|≲n2k+1+λφ-2k-λ-2(ξμ,n+1(λ))
and, by (2.2) and (4.61),
(4.72)|Eλ,n+1(q+1)(xν,n(λ))Pλ,n(r+1)(xν,n(λ))|≲n2k+1+λφ-λ-2k-2(xν,n(λ)).
Thus, we have
(4.73)|A2(ξμ,n+1(λ))|A2≲φ2(ξμ,n+1(λ))∑q+r=2k,q:odd,r:odd‍|Eλ,n+1(q+1)(ξμ,n+1(λ))Pλ,n(r+1)(ξμ,n+1(λ))|A2≲n2k+1+λφ-2k-λ(ξμ,n+1(λ)),|A2(xν,n(λ))|≲φ2(xν,n(λ))∑q+r=2k,q:odd,r:odd‍|Eλ,n+1(q+1)(xν,n(λ))Pλ,n(r+1)(xν,n(λ))||A2(xν,n(λ))|≲n2k+1+λφ-λ-2k(xν,n(λ)).
Therefore, we have the result.Proof of Theorem2.4.
Whenr=2, (2.9) holds from (3.27). Let even r>2, and suppose that (2.9) holds for r-2. Since we know by (4.35), (2.5), (4.62), and (4.19)
(4.74)φ2(yν,2n+1(λ))|Fλ,2n+1(r)(yν,2n+1(λ))|Fλ,2n+1(r)Fλ,2n+1(r)≲|Fλ,2n+1(r-1)(yν,2n+1(λ))|+n2|Fλ,2n+1(r-2)(yν,2n+1(λ))|+|Jλ,n(r-2)(yν,2n+1(λ))|Fλ,2n+1(r)Fλ,2n+1(r)≲n2|Fλ,2n+1(r-2)(yν,2n+1(λ))|+nr-1+λφ-r+2-λ(yν,2n+1(λ)),
we obtain, using mathematical induction,
(4.75)φ2(yν,2n+1(λ))|Fλ,2n+1(r)(yν,2n+1(λ))|≲nr-1+λφ-r+2-λ(yν,2n+1(λ)).
Therefore, (2.9) is proved.Proof of Theorem2.5.
(a) From (4.3), we know that
(4.76)φ2(ξμ,n+1(λ))Eλ,n+1(3)(ξμ,n+1(λ))=-(n+1)2Eλ,n+1′(ξμ,n+1(λ))+O(Eλ,n+1′(ξμ,n+1(λ)))φ2(ξμ,n+1(λ))Eλ,n+1(3)(ξμ,n+1(λ))+O(Eλ,n+1″(ξμ,n+1(λ)))+O(Iλ,n′(ξμ,n+1(λ))).
Therefore, we have, by (2.2), (3.24), and (3.26),
(4.77)Eλ,n+1(3)(ξμ,n+1(λ))=-(n+1)2φ-2(ξμ,n+1(λ))Eλ,n+1′(ξμ,n+1(λ))+O(n3).
Suppose that, for an integer ℓ⩾2,
(4.78)Eλ,n+1(2l-1)(ξμ,n+1(λ))=(-1)l-1(n+1)2(l-1)φ-2(l-1)(ξμ,n+1(λ))Eλ,n+1′(ξμ,n+1(λ))+O(n2l-1).
Then, from (4.3), we obtain
(4.79)φ2(ξμ,n+1(λ))Eλ,n+1(2l+1)(ξμ,n+1(λ))=-(n+1)2Eλ,n+1(2l-1)(ξμ,n+1(λ))+O(Eλ,n+1(2l)(ξμ,n+1(λ)))+O(Eλ,n+1(2l-1)(ξμ,n+1(λ)))+O(Iλ,n(2l-1)(ξμ,n+1(λ))).
Therefore, by (2.2), (2.8), and (4.4), we have
(4.80)Eλ,n+1(2l+1)(ξμ,n+1(λ))=-(n+1)2φ-2(ξμ,n+1(λ))Eλ,n+1(2l-1)(ξμ,n+1(λ))+O(n2l+1)=(-1)l(n+1)2lφ-2l((ξμ,n+1(λ)))Eλ,n+1′(ξμ,n+1(λ))+O(n2l+1).
(b) Similarly to the proof of (a), by (4.51), (3.3), and (3.7), we can obtain
(4.81)Pλ,n(2l)(ξμ,n+1(λ))=(-1)lnl(n+2λ)lφ-2l(ξμ,n+1(λ))Pλ,n(ξμ,n+1(λ))+O(n2λ+2l-3).
In addition, we see that, from (4.51),
(4.82)Pλ,n(2l)(ξμ,n+1(λ))=(-1)l(n+1)2lφ-2l(ξμ,n+1(λ))Pλ,n(ξμ,n+1(λ))+O(n2l-1Pλ,n(ξμ,n+1(λ)))+O(n2λ+2l-3)=(-1)l(n+1)2lφ-2l(ξμ,n+1(λ))Pλ,n(ξμ,n+1(λ))+O(n2l+λ-2).Proof of Theorem2.6.
(a) From (3.9), we know that
(4.83)φ2(xν,n(λ))Eλ,n+1″(xν,n(λ))&=-(n+1)2Eλ,n+1(xν,n(λ))+O(Eλ,n+1′(xν,n(λ)))+O(Iλ,n(xν,n(λ))).
Therefore, by (4.4) and (4.56), we have
(4.84)Eλ,n+1″(xν,n(λ))=-(n+1)2φ-2(xν,n(λ))Eλ,n+1(xν,n(λ))+O(n2).
Then, we obtain from (4.3), (2.2), and (4.56) that
(4.85)φ2(xν,n(λ))Eλ,n+1(2l)(xν,n(λ))=-(n+1)2Eλ,n+1(2l-2)(xν,n(λ))+O(n2l).
Therefore, we have the result inductively.
(b) From (3.7), we know that
(4.86)φ2(xν,n(λ))Pλ,n(3)(xν,n(λ))=-n(n+2λ)Pλ,n′(xν,n(λ))+O(Pλ,n′′(xν,n(λ)))+O(Pλ,n′(xν,n(λ))).
Therefore, by (3.4) and (4.61), we have
(4.87)Pλ,n(3)(xν,n(λ))=-n(n+2λ)φ-2(xν,n(λ))Pλ,n′(xν,n(λ))+O(nλ).
Suppose that, for an integer ℓ⩾2,
(4.88)Pλ,n(2l-1)(xν,n(λ))=(-1)l-1nl-1(n+2λ)l-1φ-2(l-1)(xν,n(λ))Pλ,n′(xν,n(λ))+O(nλ+2l-4).
Then from (3.5), (3.7), and (4.61)
(4.89)Pλ,n(2l+1)(xν,n(λ))=(-1)lnl(n+2λ)lφ-2l(xν,n(λ))Pλ,n′(xν,n(λ))+O(nλ+2l-2).
Here, we see that for ℓ⩾2(4.90){n(n+2λ)}l=(n+1)2l+O(n2l-1).
Hence, we obtain, from (3.4),
(4.91)Pλ,n(2l+1)(xν,n(λ))=(-1)l(n+1)2lφ-2l(xν,n(λ))Pλ,n′(xν,n(λ))+O(n2l-1Pλ,n′(xν,n(λ)))+O(nλ+2l-2)=(-1)l(n+1)2lφ-2l(xν,n(λ))Pλ,n′(xν,n(λ))+O(nλ+2l-1).Lemma 4.13.
Let0<ɛ<1 and |yν,2n+1(λ)|⩽1-ɛ. Then, for a nonnegative integer ℓ⩾0,
(4.92)Jλ,n(2l+1)(yν,2n+1(λ))=22l+1(-1)l+1(n+1)2l+2φ-2l(yν,2n+1(λ))F′(yν,2n+1(λ))+O(nλ+2l+2).Proof.
Let|yν,2n+1(λ)|⩽1-ɛ. From (4.37) and Lemma 4.5, we see that
(4.93)Jλ,n(2l+1)(yν,2n+1(λ))Jλ,n(2l+1)=2(1-x2)(Eλ,n+1′(x)Pλ,n′(x))(2l+1)-4x(2l+1)(Eλ,n+1′(x)Pλ,n′(x))(2l)Jλ,n(2l+1)=2-2l(2l+1)(Eλ,n+1′(x)Pλ,n′(x))(2l-1)Jλ,n(2l+1)=2+2λ(xEλ,n+1(x)Pλ,n′(x))(2l+1)+(Iλ,n(x)Pλ,n(x))(2l+1)|x=yν,2n+1(λ)Jλ,n(2l+1)=2(1-x2)(Eλ,n+1′(x)Pλ,n′(x))(2l+1)|x=yν,2n+1(λ)+O(n2l+2+λ).
Here, we let 2ℓ(2ℓ+1)(Eλ,n+1′(x)Pλ,n′(x))(2ℓ-1)=0 when ℓ=0. To estimate the first term, we split it into two terms as follows:
(4.94)(Eλ,n+1′(x)Pλ,n′(x))(2l+1)Eλ,n+1′=∑0⩽i⩽2l+1,i:even‍+∑0⩽i⩽2l+1,i:odd‍(2l+1i)Eλ,n+1(i+1)(x)Pλ,n(2l+2-i)(x).
Then, using |xν,n(λ)|⩽1-ɛ, from (2.13) and (2.15), we obtain
(4.95)∑0⩽i⩽2l+1,i:odd‍(2l+1i)Eλ,n+1(i+1)(xν,n(λ))Pλ,n(2l+2-i)(xν,n(λ))22l=(-1)l+1(n+1)2l+2φ-2l-2(xν,n(λ))Fλ,2n+1′(xν,n(λ))∑0⩽i⩽2l+1,i:odd‍(2l+1i)+O(nλ+2l+2)22l=22l(-1)l+1(n+1)2l+2φ-2l-2(xν,n(λ))Fλ,2n+1′(xν,n(λ))+O(nλ+2l+2)
and, from (4.56) and from (4.61),
(4.96)∑0⩽i⩽2l+1,i:even‍(2l+1i)Eλ,n+1(i+1)(xν,n(λ))Pλ,n(2l+2-i)(xν,n(λ))=O(nλ+2l+1).
Thus, we have, for |xν,n(λ)|⩽1-ɛ,
(4.97)(Eλ,n+1′(x)Pλ,n′(x))(2l+1)|x=xν,n(λ)Eλ,n+1′=22l(-1)l+1(n+1)2l+2φ-2l-2(xν,n(λ))Fλ,2n+1′(xν,n(λ))+O(nλ+2l+2).
Similarly, noting |ξμ,n(λ)|⩽1-ɛ, from (2.10) and (2.12)
(4.98)∑0⩽i⩽2l+1,i:even‍(2l+1i)Eλ,n+1(i+1)(ξμ,n+1(λ))Pλ,n(2l+2-i)(ξμ,n+1(λ))22l2=22l(-1)l+1(n+1)2l+2φ-2l-2(ξμ,n+1(λ))Fλ,2n+1′(ξμ,n+1(λ))+O(nλ+2l+2)
and, from (2.8) and (4.55),
(4.99)∑0⩽i⩽2l+1,i:odd‍(2l+1i)Eλ,n+1(i+1)(ξμ,n+1(λ))Pλ,n(2l+2-i)(ξμ,n+1(λ))=O(n2λ+2l+1).
Then, we have, for |ξμ,n+1(λ)|⩽1-ɛ,
(4.100)(Eλ,n+1′(x)Pλ,n′(x))(2l+1)|x=ξμ,n+1(λ)Eλ,n+1′=22l(-1)l+1(n+1)2l+2φ-2l-2(ξμ,n+1(λ))Fλ,2n+1′(ξμ,n+1(λ))+O(nλ+2l+2).
Therefore, we have, for |yν,2n+1(λ)|⩽1-ɛ,
(4.101)(Eλ,n+1′(x)Pλ,n′(x))(2l+1)|x=yν,2n+1(λ)Eλ,n+1′=22l(-1)l+1(n+1)2l+2φ-2l-2(yν,2n+1(λ))Fλ,2n+1′(yν,2n+1(λ))+O(nλ+2l+2).
Thus, we have, for |yν,2n+1(λ)|⩽1-ɛ,
(4.102)Jλ,n(2l+1)(yν,2n+1(λ))=22l+1(-1)l+1(n+1)2l+2φ-2l(yν,2n+1(λ))Fλ,2n+1′(yν,2n+1(λ))+O(nλ+2l+2).Proof of Theorem2.7.
From (4.35), (3.25), (3.27), and Lemma 4.13, we have
(4.103)φ2(yν,2n+1(λ))Fλ,2n+1(3)(yν,2n+1(λ))φ2=-2(n+1)2Fλ,2n+1′(yν,2n+1(λ))+Jλ,n′(yν,2n+1(λ))φ2φ2φ2+O(nFλ,2n+1′(yν,2n+1(λ)))+O(Fλ,2n+1′′(yν,2n+1(λ)))φ2=-3(n+1)2Fλ,2n+1′(yν,2n+1(λ))+O(nλ+2).
Suppose that
(4.104)Fλ,2n+1(2l-1)(yν,2n+1(λ))=cl-1(-1)l-1(n+1)2l-2φ-2l+2(yν,2n+1(λ))Fλ,2n+1′(yν,2n+1(λ))+O(nλ+2l-2).
Then, we obtain from (4.35)
(4.105)φ2(yν,2n+1(λ))Fλ,2n+1(2l+1)(yν,2n+1(λ))φ2=(2cl-1+22l-1)(-1)l(n+1)2lφ-2l+2(yν,2n+1(λ))Fλ,2n+1′(yν,2n+1(λ))+O(nλ+2l)φ2=cl(-1)l(n+1)2lφ-2l+2(yν,2n+1(λ))Fλ,2n+1′(yν,2n+1(λ))+O(nλ+2l).
Therefore, (2.16) is proved.



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*Source: 482935-2012-08-09.xml*