# Higher-Order Hermite-Fejér Interpolation for Stieltjes Polynomials

Authors: Hee Sun Jung; Ryozi Sakai
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

Letwλ(x):=(1-x2)λ-1/2 and Pλ,n be the ultraspherical polynomials with respect to wλ(x). Then, we denote the Stieltjes polynomials Eλ,n+1 with respect to wλ(x) satisfying ∫-11‍wλxPλ,nxEλ,n+1xxmdx=0,0≤m<n+1;≠0,m=n+1. In this paper, we consider the higher-order Hermite-Fejér interpolation operator Hn+1,m based on the zeros of Eλ,n+1 and the higher order extended Hermite-Fejér interpolation operator ℋ2n+1,m based on the zeros of Eλ,n+1Pλ,n. When m is even, we show that Lebesgue constants of these interpolation operators are O(nmax{(1-λ)m-2,0})(0<λ<1) and Onmax1-2λm-2,00<λ<1/2, respectively; that is, ℋ2n+1,m=O(nmax{(1-2λ)m-2,0})(0<λ<1) and Hn+1,m=Onmax1-λm-2,00<λ<1/2. In the case of the Hermite-Fejér interpolation polynomials ℋ2n+1,m[·] for 1/2≤λ<1, we can prove the weighted uniform convergence. In addition, when m is odd, we will show that these interpolations diverge for a certain continuous function on [-1,1], proving that Lebesgue constants of these interpolation operators are similar or greater than log n.


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## Body

## 1. Introduction

LetX:={xk,n}⊂[-1,1] and
(1)-1<x1,n<x2,n<⋯<xn-1,n<xn,n<1,ppppppppppppppppppppn=1,2,….
For any real-valued function f on [-1,1] and an integer m≥1, we recall that there exist unique Hermite and Hermite-Fejér interpolatory polynomials of higher order denoted by Hn,m(f,X), and of degree ≤nm-1, defined as follows:
(2)Hn,m(f,X,xk,n)=f(xk,n),1≤k≤n;Hn,m(t)(f,X,xk,n)=0,1≤t≤m-1,ppppppppppppppppppppp1≤k≤n.
We note that, by definition, Hn,1 is the Lagrange, Hn,2 is the Hermite-Fejér, and Hn,4 is the Krylov-Stayermann interpolatory polynomial. By (2), we may write
(3)Hn,m(f,X,x)=∑k=1n‍f(xk,n)hk,n,m(X,x),ppppppppppppppppppn=1,2,….
The polynomials
(4)hk,n,m(X,x)=lk,nm(X,x)∑i=0m-1‍ei,k,n,m(x-xk,n)i,pppppppppppppppppppppppppp1≤k≤n
are unique, of degree exactly nm-1 and satisfy the relations
(5)hk,n,m(t)(X,xl,n)=δ0,tδl,k,1≤k,l≤n,pppppppppppppppppp0≤t≤m-1,
where for nonnegative integers u and v(6)δu,v:={1,u=v;0,u≠v.
Here, lk,n(X,x) are the well-known fundamental Lagrange polynomials of degree n-1 given by
(7)lk,n(X,x):=wn(x)wn′(xk,n)(x-xk,n),wn(x):=∏k=1n‍(x-xk,n),
and the coefficients ei,k may be obtained from the relations
(8)hk,n,m(X,xl,n)=δl,k,1≤k,l≤n;hk,n,m(t)(X,xl,n)=0,1≤t≤m-1,ppppppppppppppppp1≤k,l≤n.
If f∈C(m-1)[-1,1], then the Hermite interpolation polynomial H^n,m(f,X,x) of degree ≤nm-1 with respect to X is defined by
(9)H^n,m(t)(f,X,xk,n):=f(t)(xk,n),pppp1≤k≤n,0≤t≤m-1.
We may express H^n,m(f,X,x) as
(10)H^n,m(f,X,x)=∑t=0m-1∑k=1n‍f(t)(xk,n)ht,k,n,m(X,x),ppppppppppp.ppm=1,2,…,
where for 0≤t≤m-1(11)ht,k,n,m(X,x)=lk,nm(X,x)(x-xk,n)tt!∑i=0m-1-t‍et,i,k,n,m(x-xk,n)i
is the unique polynomial of degree nm-1 satisfying
(12)ht,k,n,m(i)(X,xj,n)=δt,iδk,j,0≤i,t≤m-1,pppppppppppp1≤j,k≤n.
Then, we easily see from the relations (5) and (12) that h0,k,n,m(x)=hk,n,m(x), e0,i,k,n,m=ei,k,n,m, and et,i,k,n,m=e0,i,k,n,m for 1≤k≤n and 0≤i,t≤m-1 (see [1]). Now, we have for any polynomial P of degree ≤nm-1,
(13)P(x)=H^n,m(P,X,x)=Hn,m(P,X,x)+∑t=1m-1∑k=1n‍P(t)(xk,n)ht,k,n,m(X,x).
In what follows, we abbreviate several notations as hk(x):=hk,n,m(x), ei,k:=ei,k,n,m, and et,i,k:=et,i,k,n,m if there is no confusion. Here, we are interested in Hermite-Fejér and Hermite interpolations with respect to X whose elements are the zeros of a sequence of Stieltjes polynomials and the product polynomials of Stieltjes polynomials and the ultraspherical polynomials, respectively. To be precise, we first consider the generalized Stieltjes polynomials Eλ,n+1(x) defined (up to a multiplicative constant) by
(14)∫-11‍wλ(x)Pλ,n(x)Eλ,n+1(x)xkdx=0,pppppppppk=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). In 1935, Szegö [2] 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.For the properties of interpolation operators based at the zeros ofEλ,n+1 and the zeros of Pλ,nEλ,n+1, Ehrich and Mastroianni [3, 4] proved that Lagrange interpolation operators Ln+1 based on the zeros of Eλ,n+1 and extended Lagrange interpolation operators ℒ2n+1 based on the zeros of Eλ,n+1Pλ,n have Lebesgue constants ∥Ln+1∥∞(0<λ<1) and ∥ℒ2n+1∥∞(0<λ≤1/2) of optimal order, that is, O(logn). For the Hermite-Fejér interpolation operator Hn+1 based on the zeros of Eλ,n+1 and the extended Hermite-Fejér interpolation operator ℋ2n+1 based on the zeros of Eλ,n+1Pλ,n, it is proved that Lebesgue constants ∥Hn+1∥∞(0<λ<1) and ∥ℋ2n+1∥∞(0<λ≤1/2) are of optimal order, that is, O(1), in [5]. In this paper, we consider the higher-order Hermite-Fejér interpolation operator Hn+1,m based on the zeros of Eλ,n+1 and the higher-order extended Hermite-Fejér interpolation operator ℋ2n+1,m based on the zeros of Eλ,n+1Pλ,n. When m is even, we show that Lebesgue constants of these interpolation operators are O(nmax{(1-λ)m-2,0}) and O(nmax{(1-2λ)m-2,0}), respectively; that is, ∥Hn+1,m∥=O(nmax{(1-λ)m-2,0})  (0<λ<1) and ∥ℋ2n+1,m∥=O(nmax{(1-2λ)m-2,0})  (0<λ<1/2). In the case of the Hermite-Fejér interpolation polynomials ℋ2n+1,m[·] for 1/2≤λ<1, we can prove the weighted uniform convergence. In addition, when m is odd, we will show that these interpolations diverge for a certain continuous function on [-1,1], proving that Lebesgue constants of these interpolation operators are similar or greater than logn.This paper is organized as follows. In Section2, we will introduce the main results. In Section 3, we will show the auxiliary propositions and estimate the coefficients of Hermite-Fejér interpolation polynomials 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)=O(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 bn≤Ccn and cn≤Cbn. We denote the space of polynomials of degree at most n by 𝒫n.For the Chebyshev polynomialTn(x), note that for λ=0 and λ=1(15)E0,n+1(x)=2nπ(Tn+1(x)-Tn-1(x)),E1,n+1(x)=2πTn+1(x).
In this paper, we let Pn(0)(x):=Tn(x)/n. In these cases, the results are well known or can easily be deduced. Therefore, we will consider the cases for 0<λ<1 and let 0<λ<1 in the following.LetHn+1,m[f] be the Hermite-Fejér interpolation polynomials of f with respect to the zeros of Eλ,n+1(x). Also let ℋ2n+1,m[f] be the Hermite-Fejér interpolation polynomials of f with respect to the zeros of Fλ,2n+1(x). The fundamental Lagrange interpolation polynomials lk,n+1(x) and l*ν,2n+1(x) with respect to Hn+1,m[·] and ℋ2n+1,m[·], respectively, are given by
(16)lμ,n+1(x)=Eλ,n+1(x)Eλ,n+1′(ξμ,n+1(λ))(x-ξμ,n+1(λ)),ppppp.ppμ=1,2,…,n+1,lν,2n+1*(x)=Fλ,2n+1(x)Fλ,2n+1′(yν,2n+1(λ))(x-yν,2n+1(λ)),pppppppppν=1,2,…,2n+1.We let∥Hn+1,m∥ and ∥ℋ2n+1,m∥ be the Lebesgue constants based on the zeros of Eλ,n+1(x) and Fλ,2n+1(x), respectively. That is, the Lebesgue constants ∥Hn+1,m∥ and ∥ℋ2n+1,m∥ are defined as follows:
(17)∥Hn+1,m∥:=supx∈[-1,1]∑μ=1n+1∑i=0m-1‍|ei,μlμ,n+1m(x)(x-ξμ,n+1(λ))i|,(18)∥ℋ2n+1,m∥:=supx∈[-1,1]∑ν=12n+1∑i=0m-1‍|ei,ν*lν,2n+1*m(x)(x-yν,2n+1(λ))i|
and for a nonnegative real function u(x),
(19)∥ℋ2n+1,m∥u:=supx∈[-1,1]u(x)∑ν=12n+1‍∑i=0m-1‍|ei,ν*lν,2n+1*m(x)(x-yν,2n+1(λ))i|,
where ei,μ and ei,ν* are the coefficients of the higher-order Hermite-Fejér interpolation polynomials defined in (4), with respect to Hn+1,m[·] and ℋ2n+1,m[·], respectively.


### 2.1. Uniform Convergence of Hermite-Fejér Interpolation Polynomials of Higher Order

Theorem 1.
Let0<λ<1 and m=2,4,6,…. (a)
(20)∥Hn+1,m∥=O(nmax{(1-λ)m-2,0}).(b)
Suppose that(1-λ)m≤2. Then, for a continuous function f on [-1,1] one has uniformly for x∈[-1,1],
(21)limn→∞|Hn+1,m[f](x)-f(x)|=0.Theorem 2.
Let0<λ<1 and m=2,4,6,….(a)
Then one has for0<λ<1/2(22)∥ℋ2n+1,m∥=O(nmax{(1-2λ)m-2,0})and for1/2≤λ<1(23)∥ℋ2n+1,m∥φ(2λ-1)m=O(1).(b)
For a continuous functionf on [-1,1], if 0<(1-2λ)m≤2, then one has uniformly for x∈[-1,1],
(24)limn→∞|ℋ2n+1,m[f](x)-f(x)|=0and if1/2≤λ<1, then
(25)limn→∞|ℋ2n+1,m[f](x)-f(x)|φ(2λ-1)m(x)=0.


### 2.2. Divergence of Hermite-Fejér Interpolation Polynomials of Higher Order

Theorem 3.
Let0<λ<1 and m=1,3,5,…. Then,
(26)∥Hn+1,m∥~lnn.Theorem 4.
Let0<λ<1 and m=2,4,6,…. Let ɛ>0. Suppose that
(27)|em-1,μ|~nλ+m-2φλ-m(ξμ,n+1(λ)).
If (1-λ)m>2, then
(28)∥Hn+1,m∥~n(1-λ)m-2.Theorem 5.
Let0<λ<1 and m=1,3,5,…. Then,
(29)∥ℋ2n+1,m∥~lnn.Theorem 6.
Let0<λ<1 and m=2,4,6,…. Let ɛ>0. Suppose that
(30)|em-1,ν*|~nλ+m-2φλ-m(yν,2n+1(λ)).
If (1-2λ)m>2, then
(31)∥ℋ2n+1,m∥~nmax{(1-2λ)m-2,0}.If the Lebesgue constant is not bounded, then we know from Helley’s theorem that Hermite-Fejér interpolation does not converge for a certain continuous function on[-1,1].


## 2.1. Uniform Convergence of Hermite-Fejér Interpolation Polynomials of Higher Order

Theorem 1.
Let0<λ<1 and m=2,4,6,…. (a)
(20)∥Hn+1,m∥=O(nmax{(1-λ)m-2,0}).(b)
Suppose that(1-λ)m≤2. Then, for a continuous function f on [-1,1] one has uniformly for x∈[-1,1],
(21)limn→∞|Hn+1,m[f](x)-f(x)|=0.Theorem 2.
Let0<λ<1 and m=2,4,6,….(a)
Then one has for0<λ<1/2(22)∥ℋ2n+1,m∥=O(nmax{(1-2λ)m-2,0})and for1/2≤λ<1(23)∥ℋ2n+1,m∥φ(2λ-1)m=O(1).(b)
For a continuous functionf on [-1,1], if 0<(1-2λ)m≤2, then one has uniformly for x∈[-1,1],
(24)limn→∞|ℋ2n+1,m[f](x)-f(x)|=0and if1/2≤λ<1, then
(25)limn→∞|ℋ2n+1,m[f](x)-f(x)|φ(2λ-1)m(x)=0.


## 2.2. Divergence of Hermite-Fejér Interpolation Polynomials of Higher Order

Theorem 3.
Let0<λ<1 and m=1,3,5,…. Then,
(26)∥Hn+1,m∥~lnn.Theorem 4.
Let0<λ<1 and m=2,4,6,…. Let ɛ>0. Suppose that
(27)|em-1,μ|~nλ+m-2φλ-m(ξμ,n+1(λ)).
If (1-λ)m>2, then
(28)∥Hn+1,m∥~n(1-λ)m-2.Theorem 5.
Let0<λ<1 and m=1,3,5,…. Then,
(29)∥ℋ2n+1,m∥~lnn.Theorem 6.
Let0<λ<1 and m=2,4,6,…. Let ɛ>0. Suppose that
(30)|em-1,ν*|~nλ+m-2φλ-m(yν,2n+1(λ)).
If (1-2λ)m>2, then
(31)∥ℋ2n+1,m∥~nmax{(1-2λ)m-2,0}.If the Lebesgue constant is not bounded, then we know from Helley’s theorem that Hermite-Fejér interpolation does not converge for a certain continuous function on[-1,1].


## 3. Estimation of the Coefficients of Higher-Order Hermite-Fejér Interpolation Polynomials

Proposition 7.
Let0<λ<1.(1)
See [4, Theorem2.1],forn≥0,
(32)|Eλ,n+1(x)|≤C(n1-λφ1-λ(x)+1)-1≤x≤1.Furthermore,Eλ,n+1(1)≥C.(2)
See [4, Theorem  2.1] and[6,  7.33.5], forn≥0,
(33)|Fλ,2n+1(x)|≤Cφ1-2λ(x),-1+Cn2≤x≤1-Cn2.Proposition 8 (see [4, Lemma  5.5]).
Let0<λ<1. Then, for μ=1,2,…,n+1,
(34)|Eλ,n+1′(ξμ,n+1(λ))|~n2-λφ-λ(ξμ,n+1(λ))
and for ν=1,2,…,2n+1,
(35)|Fλ,2n+1′(yν,2n+1(λ))|~nφ-2λ(yν,2n+1(λ)).Proposition 9 (see [7, Theorem  2.1]).
Let0<λ<1 and r≥1 a positive integer. Then, for all x∈[ξ1,n+1(λ),ξn+1,n+1(λ)],
(36)|Eλ,n+1(r)(x)|≤Cnr+1-λφ1-r-λ(x).
Moreover, one has for x∈[-1,ξ1,n+1(λ)]∪[ξn+1,n+1(λ),1],
(37)|Eλ,n+1(r)(x)|~n2r.Proposition 10 (see [7, Theorem  2.2]).
Let0<λ<1 and r≥1 a positive integer. Then, for all x∈[ξ1,n+1(λ),ξn+1,n+1(λ)],
(38)|Fλ,2n+1(r)(x)|≤Cnrφ1-2λ-r(x).
Moreover, one has forx∈[-1,ξ1,n+1(λ)]∪[ξn+1,n+1(λ),1],
(39)|Fλ,2n+1(r)(x)|~n2λ+2r-1.Proposition 11 (see [7, Theorems  2.3, 2.4]).
Let0<λ<1 and r≥2 an even integer.(a)
For1≤μ≤n+1,
(40)|Eλ,n+1(r)(ξμ,n+1(λ))|≤Cnrφ-r(ξμ,n+1(λ)).(b)
For1≤ν≤2n+1,
(41)|Fλ,2n+1(r)(yν,2n+1(λ))|≤Cnr-1+λφ-r-λ(yν,2n+1(λ)).Proposition 12 (see [7, Lemma  4.9]).
For1≤ν≤2n+1, one has
(42)|Eλ,n+1(xν,n(λ))|~n1-λφ1-λ(xν,n(λ)).Proposition 13 (see [7, Theorems  2.6, 2.7]).
Let0<λ<1 and 0<ɛ<1.(a)
For|ξμ,n+1(λ)|≤1-ɛand a positive integerℓ≥0, one has(43)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)
For|yν,2n+1(λ)|≤1-ɛand a positive integerℓ≥0, one has(44)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),wherecℓ=2ℓ-1+4ℓ/2.Theorem 14.
Let0<λ<1, r≥0 and n,m≥1.(a)
Uniformly for1≤μ≤n+1(45)|[lμ,n+1m](r)(ξμ,n+1(λ))|≤Cnrφ-r(ξμ,n+1(λ))
and if r is odd,
(46)|[lμ,n+1m](r)(ξμ,n+1(λ))|≤Cnλ+r-1φ-r-1+λ(ξμ,n+1(λ)).(b)
Uniformly for0≤s≤m-1,
(47)|es,μ|≤Cnsφ-s(ξμ,n+1(λ))
and if s is odd,
(48)|es,μ|≤Cnλ+s-1φ-s-1+λ(ξμ,n+1(λ)).Theorem 15.
Let0<λ<1, r≥0 and n,m≥1.(a)
Uniformly for1≤ν≤2n+1,
(49)|[lν,2n+1*m](r)(yν,2n+1λ)|≤Cnrφ-r(yν,2n+1λ)
and if r is odd, one has
(50)|[lν,2n+1*m](r)(yν,2n+1λ)|≤Cnλ+r-1φ-r-1+λ(yν,2n+1λ).(b)
Uniformly for0≤s≤m-1,
(51)|es,ν*|≤Cnsφ-s(yν,2n+1λ)
and if s is odd,
(52)|es,ν*|≤Cnλ+s-1φ-s-1+λ(yν,2n+1λ).Theorem 16.
Let0<λ<1 and 0<ɛ<1. Suppose that |ξμ,n+1(λ)|≤1-ɛ. Then, there exists a constant ϕℓ(m)>0 depending only on m and ℓ such that
(53)(lμ,n+1m)(2ℓ)(ξμ,n+1(λ))=(-1)ℓϕℓ(m)(n+1)2ℓ×φ-2ℓ(ξμ,n+1(λ))(1+O(nλ-1)).Theorem 17.
Let0<λ<1 and 0<ɛ<1. Suppose that |ξμ,n+1(λ)|≤1-ɛ. Then, there exists a constant Ψℓ(-m) with Ψℓ(-m)>0 depending only on m and ℓ such that
(54)e2ℓ,μ=(-1)ℓ1(2ℓ)!Ψℓ(-m)(n+1)2ℓφ-2ℓ(ξμ,n+1(λ))×(1+O(nλ-1)).Theorem 18.
Let0<λ<1 and 0<ɛ<1. Suppose that |yν,2n+1(λ)|≤1-ɛ. Then, there exists a constant αℓ(m)>0 depending only on m and ℓ such that
(55)(lν,2n+1*m)(2ℓ)(yν,2n+1(λ))=(-1)ℓαℓ(m)(n+1)2ℓ×φ-2ℓ(yν,2n+1(λ))(1+O(nλ-1)).Theorem 19.
Let0<λ<1 and 0<ɛ<1. Suppose that |yν,2n+1(λ)|≤1-ɛ. Then, there exists a constant Φℓ(-m) with Φℓ(-m)>0 depending only on m and ℓ such that
(56)e2ℓ,ν*=(-1)ℓ1(2ℓ)!Φℓ(-m)(n+1)2ℓφ-2ℓ(yν,2n+1(λ))×(1+O(nλ-1)).Proof of Theorem14.
We prove by induction onm. Since
(57)[Eλ,n+1′(ξμ,n+1(λ))(x-ξμ,n+1(λ))lμ,n+1(x)]x=ξμ,n+1(λ)(r+1)=Eλ,n+1(r+1)(ξμ,n+1(λ)),
we know that
(58)lμ,n+1(r)(ξμ,n+1(λ))=Eλ,n+1(r+1)(ξμ,n+1(λ))(r+1)Eλ,n+1′(ξμ,n+1(λ)).
So, it holds for m=1 by (34) and (36). Now, assume that it holds for 1,2,…,m-1. Then, using Leibnitz’s rule for differentiation, we obtain
(59)|[lμ,n+1m](r)(ξμ,n+1(λ))|≤C∑i=0r(ri)|lμ,n+1(i)(ξμ,n+1(λ))|×|[lμ,n+1m-1](r-i)(ξμ,n+1(λ))|≤Cnrφ-r(ξμ,n+1(λ)).
Suppose that r is odd. Then,
(60)|[lμ,n+1m](r)(ξμ,n+1(λ))|≤C∑i=0r‍(ri)|lμ,n+1(i)(ξμ,n+1(λ))|×|[lμ,n+1m-1](r-i)(ξμ,n+1(λ))|:=(∑i=even‍+∑i=odd‍).
Since r-i is odd for an even i and r-i is even for an odd i, we have by the mathematical induction on m, (34), (36), (40), and (58),
(61)∑i=evenorodd‍|lμ,n+1(i)(ξμ,n+1(λ))||[lμ,n+1m-1](r-i)(ξμ,n+1(λ))|≤Cnλ+r-1φ-r+λ-1(ξμ,n+1(λ)).
These complete the proofs of (45) and (46). To prove (47) and (48), we proceed by induction on s. Firstly, for s=0, (47) is trivial since e0,μ=1. For s≥1, we have by [8, (3.3)] and [8, (3.4)]
(62)0=hμ(s)(ξμ,n+1(λ))=∑i=0s‍ei,μ(si)i![lμ,n+1m](s-i)(ξμ,n+1(λ)),
so that
(63)es,μ=-1s!∑i=0s-1‍ei,μ(si)i![lμ,n+1m](s-i)(ξμ,n+1(λ)).
Thus, if we assume that (47) holds for s=0,1,…,t-1, t≥1, then by (45), we have
(64)|et,μ|≤C∑i=0t-1‍|ei,μ|[lμ,n+1m](t-i)(ξμ,n+1(λ))≤Cntφt(ξμ,n+1(λ)).
Suppose that t is odd. Then,
(65)|et,μ|≤C(∑i=event-1‍+∑i=oddt-1‍)|ei,μ|[lμ,n+1m](t-i)(ξμ,n+1(λ)).
Since t-i is odd for an even i and t-i is even for an odd i, we have by the mathematical induction, (45), (46), and (47)
(66)∑i=evenoroddt-1‍|ei,μ|[lμ,n+1m](t-i)(ξμ,n+1(λ))≤Cnλ+t-1φλ-t-1(ξμ,n+1(λ)).
These complete the proofs of (47) and (48).Proof of Theorem15.
Using (35), (38), and (41), this is proved by the same method as the proof of Theorem 14.Forj=0,1,… define ϕj(1):=(2j+1)-1 and for k≥2(67)ϕj(k):=∑r=0j‍12j-2r+1(2j2r)ϕr(k-1).
We rewrite the relation (67) in the form for ν=1,2,3…,
(68)ϕ0(ν):=1
and for j=1,2,3,…,  ν=2,3,4,…,
(69)ϕj(ν)-ϕj(ν-1)=12j+1∑r=0j-1‍(2j+12r)ϕr(ν-1).
Now, for every j, we will introduce an auxiliary polynomial determined by {Ψj(y)}j=1∞ as the following lemma.Lemma 20 (see [9, Lemma  11]).
(i)
Forj=0,1,2,…, there exists a unique polynomial Ψj(y) of degree j such that
(70)Ψj(ν)=ϕj(ν),ν=1,2,3,….
(ii)
Ψ
0
(
y
)
=
1 and Ψj(0)=0, j=1,2,….Lemma 21 (see [9, Lemma  13]).
Ify<0, then for j=0,1,2,…,
(71)(-1)jΨj(y)>0.Lemma 22 (see [9, Proof of Lemma  14]).
For positive integerss and m,
(72)∑r=0s‍(2s2r)Ψr(-m)ϕs-r(m)=0.Proof of Theorem16.
Similarly to Theorem14, we use mathematical induction with respect to m. From (58), (43), and (34), we know that
(73)lμ,n+1(2ℓ)(ξμ,n+1)=(-1)ℓϕℓ(1)(n+1)2ℓφ-2ℓ(ξμ,n+1(λ))×(1+O(nλ-1)),|lμ,n+1(2ℓ-1)(ξμ,n+1)|≤O(n2ℓ-2+λ).
Then, from the following relations:
(74)(lμ,n+1m)(2ℓ)(ξμ,n+1)=∑0≤2r≤2ℓ‍(2ℓ2r)(lμ,n+1m-1)(2r)(ξμ,n+1)lμ,n+1(2ℓ-2r)(ξμ,n+1)+∑1≤2r-1≤2ℓ‍(2ℓ2r-1)(lμ,n+1m-1)(2r-1)ppppppppp×(ξμ,n+1)lμ,n+1(2ℓ-2r+1)(ξμ,n+1)=(-1)ℓϕℓ(m)(n+1)2ℓφ-2ℓ(ξμ,n+1(λ))(1+O(nλ-1)),(lμ,n+1m)(2ℓ-1)(ξμ,n+1)=∑0≤2r≤2ℓ-1‍(2ℓ-12r)(lμ,n+1m-1)(2r)(ξμ,n+1)lμ,n+1(2ℓ-2r-1)(ξμ,n+1)+∑1≤2r-1≤2ℓ-1‍(2ℓ-12r-1)(lμ,n+1m-1)(2r-1)ppppppppp×(ξμ,n+1)lμ,n+1(2ℓ-2r)(ξμ,n+1)=O(n2ℓ-2+λ),
we have the results by induction with respect to m.Proof of Theorem17.
We prove (54) by induction on s. Since e0,μ=1 and Ψ0(y)=1, (54) holds for s=0. From (63), we write e2s,μ in the form of
(75)e2s,μ=-∑r=0s-1‍1(2s-2r)!e2r,μ(lμ,n+1m)(2s-2r)(ξμ,n+1(λ))-∑r=1s‍1(2s-2r+1)!e2r-1,μ(lμ,n+1m)(2s-2r+1)(ξμ,n+1(λ))=:I+II.
Then, by (46) and (48), |II| is O(n2s+2λ-2). For 0≤r≤s-1, we suppose (54). Then, since we know from (53)
(76)(lμ,n+1m)(2s-2r)(ξμ,n+1(λ))=(-1)s-rϕs-r(m)(n+1)2(s-r)×φ-2(s-r)(ξμ,n+1(λ))(1+O(nλ-1)),e2r,μ=(-1)r1(2r)!Ψr(-m)(n+1)2r×φ-2r(ξμ,n+1(λ))(1+O(nλ-1)),
we have for I(77)∑r=0s-1‍=(-1)s+1(2s)!(n+1)2sφ-2s(ξμ,n+1(λ))×∑r=0s-1‍(2s2r)Ψr(-m)ϕs-r(m)(1+O(nλ-1)).
Then, using Lemma 22 and ϕ0(m)=1, we have the following form:
(78)e2s,μ=(-1)s(2s)!Ψs(-m)(n+1)2sφ-2s(ξμ,n+1(λ))×(1+O(nλ-1)).
Therefore, we proved the result.Proof of Theorems18 and 19.
These theorems are proved by the same method as the above theorems.


## 4. The Proofs of Theorems

### 4.1. Proof of Theorems1 and 2

From now on, we assume that0<λ<1. We first state some known results for the Stieltjes polynomials. Let xη,n(λ):=cosϕη,n(λ), η=1,…,n, ξμ,n+1(λ):=cosθμ,n+1(λ), μ=1,…,n+1, and yν,2n+1(λ):=cosψν,2n+1(λ), ν=1,2,…,2n+1. Ehrich and Mastroianni [4] proved that for μ=0,1,…,n+1 and ν=0,1,…,2n+1(79)|ϕη,n(λ)-ϕη+1,n(λ)|~|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):=ϕ0,n(λ)=π and ψ2n+2,2n+1(λ):=θn+2,n+1(λ):=ϕn+1,n(λ):=0. It implies that for μ=0,1,…,n+2 and ν=0,1,…,2n+2,
(80)ξμ+1,n+1(λ)-ξμ,n+1(λ)~1nφ(ξμ,n+1(λ));yν+1,2n+1(λ)-yν,2n+1(λ)~1nφ(yν,2n+1(λ)),φ(ξμ+1,n+1(λ))~φ(ξμ,n+1(λ));φ(yν+1,2n+1(λ))~φ(yν,2n+1(λ)),
where y0,2n+1(λ):=ξ0,n+1(λ):=-1 and y2n+2,2n+1(λ):=ξn+2,n+1(λ):=1.Lemma 23.
Letk be a positive integer and 0<x<1-C/n2. Then,(a)
forα>0,
(81)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-α(t)φk-1(t)|x-t|kdt≤Cφ-α(x)nk-1O(nmax{α-2k,0})×{1,k⩾2;lnn,k=1;(b)
forα1>0 and ɛ1>0,
(82)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-α1-ɛ1(t)φk-1(t)|x-t|kdt≤Cφ-α1(x)nk-1+ɛ1O(nmax{α1-2k,0});(c)
forα>0,
(83)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φα(t)φk-1(t)|x-t|kdt≤Cnk-1{1,k⩾2;lnn,k=1;(d)
forα1>0 and ɛ1>0,
(84)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φα1-ɛ1(t)φk-1(t)|x-t|kdt≤Cnk-1+ɛ1.Proof.
(a)(85)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-α+k-1(t)|x-t|kdt~∫0x*‍+∫x*(1+x*)/2‍+∫(1+x*)/21-C/n2‍φ-α+k-1(t)|x-t|kdt=:I1+I2+I3,
where x*:=x-φ(x)/n and x*:=x+φ(x)/n.
I
1: Suppose that 0≤t≤x*. Then, since φ-α(t)≤φ-α(x), we have
(86)∫0x*‍φ-α+k-1(t)|x-t|kdt≤Cφ-α(x)∫0x*‍φk-1(t)|x-t|kdt.
Since we know for 0≤t≤x*,
(87)φ(t)~(x-t)1/2+φ(x),
we have
(88)∫0x*‍φk-1(t)|x-t|kdt~∫0x*‍1|x-t|(k-1)/2|x-t|dt+φk-1(x)×∫0x*‍1|x-t|k-1|x-t|dt≤C{|x-x*|-(k-1)/2+φk-1(x)|x-x*|-k+1,k≥2;lnn,k=1≤C{nk-1,k≥2;lnn,k=1.
Therefore, we have for I1(89)I1≤Cφ-α(x){nk-1,k≥2;lnn,k=1.
I
2: Since φ(t)~φ(x) for x*≤t≤(1+x*)/2,
(90)∫x*(1+x*)/2‍φ-α+k-1(t)|x-t|kdt~φ-α+k-1(x)∫x*(1+x*)/2‍1|x-t|kdt≤Cφ-α(x){nk-1,k≥2;lnn,k=1.
Therefore, we have
(91)I2≤Cφ-α(x){nk-1,k≥2;lnn,k=1.
I
3: Since |x-t|≥C|1-t| for (1+x*)/2<t<1-C/n2, we have for I3(92)∫(1+x*)/21-C/n2‍φ-α+k-1(t)|x-t|kdt=∫(1+x*)/21-C/n2‍φ-α+k-1(t)|x-t|k-α/2|x-t|α/2dt≤C1|(1+x*)/2-x|α/2×∫(1+x*)/21-C/n2‍1(1-t)(k+1)/2φ-α+2k(t)|x-t|-α/2+kdt.
Since for (1+x*)/2<t<1-C/n2,
(93)φ-α+2k(t)|x-t|-α/2+k≤C{1,if-α+2k≥0;nα-2k,if-α+2k<0=O(nmax{α-2k,0}),
with |(1+x*)/2-x|~1-x~φ2(x), we have
(94)I3=∫(1+x*)/21-C/n2‍φ-α+k-1(t)|x-t|kdt≤O(nmax{α-2k,0})φ-α(x)×∫(1+x*)/21-C/n2‍1(1-t)(k+1)/2dt~nk-1φ-α(x)O(nmax{α-2k,0}){1,k⩾2;lnn,k=1.
Therefore, we have the result (a).
(b) Similarly to (a), we let(95)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-α1-ɛ1+k-1(t)|x-t|kdt~∫0x*‍+∫x*(1+x*)/2‍+∫(1+x*)/21-C/n2‍φ-α1+k-1-ϵ1(t)|x-t|kdt=:I1+I2+I3.
Then, for I1, we have using (87)
(96)∫0x*‍φ-α1+k-1-ϵ1(t)|x-t|kdt≤Cφ-α1(x)∫0x*‍(φ-ϵ1(t)|x-t|(k-1)/2|x-t|ppppppppppppp+φk-1(x)φ-ϵ1(t)|x-t|k-1|x-t|)dt≤Cφ-α1(x)nk-1∫0x*‍φ-ϵ1(t)x-tdt,(97)∫0x*‍φ-ϵ1(t)x-tdt≤C∫0x*‍1(x-t)ϵ1/2+1dt≤C(nφ(x))ϵ1/2≤Cnɛ1
by the use ofφ(t)≥C|x-t|1/2. Therefore, we have for I1(98)I1≤Cφ-α1(x)nk-1+ɛ1.
I
2: Since 1-t≥C(t-x) and φ(t)~φ(x) for x*≤t≤(1+x*)/2,
(99)∫x*(1+x*)/2‍φ-α1+k-1-ϵ1(t)|x-t|kdt~φ-α1(x)nk-1∫x*(1+x*)/2‍φ-ϵ1(t)(t-x)dt≤Cφ-α1(x)nk-1∫x*(1+x*)/2‍1(t-x)ϵ1/2+1dt≤Cφ-α1(x)nk-1+ɛ1.
Therefore,
(100)I2≤Cφ-α1(x)nk-1+ɛ1.
I
3: Since |x-t|≥C|(1+x*)/2-x| for (1+x*)/2<t<1-C/n2, we have using (93)
(101)φ-α1+k-1-ϵ1(t)|x-t|k=φ-α1+k-1-ϵ1(t)|x-t|k-α1/2|x-t|α1/2≤C1|(1+x*)/2-x|α1/2φ-α1+k-1-ϵ1(t)|x-t|k-α1/2=Cφ-α1(x)1(1-t)(k-1+ɛ1)/2+1φ-α1+2k(t)|x-t|k-α1/2=φ-α1(x)O(nmax{α-2k,0})1(1-t)(k-1+ɛ1)/2+1,
and we know that
(102)∫(1+x*)/21-C/n2‍1(1-t)(k-1+ɛ1)/2+1dt≤Cnk-1+ɛ1.
Therefore, we see that
(103)∫(1+x*)/21-C/n2‍φ-α1+k-1-ϵ1(t)|x-t|kdt≤Cφ-α1(x)nk-1+ɛ1O(nmax{α1-2k,0}).
Consequently, we have the result (b).
(c) and (d): Since|1-t2|≤C(|1-x2|+|t-x|), we know φ(t)≤C(φ(x)+|x-t|1/2). Then, we have
(104)∫0≤t≤1-C/n2,|x-t|>φ(x)/nφα+k-1(t)|x-t|kdt≤C(φα+k-1(x)∫1|x-t|kdt+∫‍1|x-t|(k+1)/2dt)≤Cnk-1{1,k⩾2;lnnk=1,
because we see if k=1, then
(105)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍1|x-t|kdt≤Clnn,∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍1|x-t|(k+1)/2dt≤Clnn
and if k≥2, then
(106)φα+k-1(x)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍1|x-t|kdt≤Cφα+k-1(x)|x-t|k-1≤Cφα+k-1(x)(nφ(x))k-1≤Cnk-1,∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍1|x-t|(k+1)/2dt≤Cnk-1;
that is, we have (c). Similarly,
(107)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φα1+k-1-ϵ1(t)|x-t|kdt≤C(φα1+k-1(x)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-ϵ1(t)|x-t|kdt+∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-ϵ1(t)|x-t|(k+1)/2dt):=C(A+B).
Then, for A(108)A=φα1+k-1(x)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-ϵ1(t)|x-t|kdt≤Cφα1+k-1(x)(nφ(x))k-1∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-ϵ1(t)|x-t|dt≤Cnk-1∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-ϵ1(t)|x-t|dt
and for B(109)B=∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-ϵ1(t)|x-t|(k+1)/2dt≤Cnk-1∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-ϵ1(t)|x-t|dt.
On the other hand, by (97) and (99), we know that
(110)∫0x*‍φ-ϵ1(t)x-tdt≤Cnɛ1,∫x*(1+x*)/2‍φ-ϵ1(t)(t-x)dt≤Cnɛ1.
Since |x-t|≥C|1-t| for (1+x*)/2<t<1-C/n2, we have
(111)∫(1+x*)/21-C/n2‍φ-ϵ1(t)(t-x)dt≤C∫(1+x*)/21-C/n2‍1(1-t)ϵ1/2+1dt≤Cnɛ1.
Therefore, we have
(112)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-ϵ1(t)|x-t|dt≤Cnk-1+ɛ1.
So, by (108) and (109) we have (d).Lemma 24.
Let0≤i≤m-1. Let a>0 be a fixed, sufficiently small constant. Let x∈[ξk-1,n+1(λ),ξk+1,n+1(λ)]∩[0,1-a/n2] for some 1≤k≤n+1. Then, one has uniformly for n(113)∑μ∉[k-2,k+2],1≤μ≤n+1φ(1-λ)m(x)nm-iφλm-i(ξμ,n+1(λ))|x-ξμ,n+1(λ)|m-i≤O(nmax{(1-λ)m-2(m-i),0}){1,0≤i≤m-2;lnn,i=m-1.Proof.
For the simplicity, we denote(114)∑μ∉[k-2,k+2]‍:=∑μ∉[k-2,k+2],1≤μ≤n+1φ(1-λ)m(x)nm-iφλm-i(ξμ,n+1(λ))|x-ξμ,n+1(λ)|m-i.
Then, we know that(115)∑μ∉[k-2,k+2]‍=φ(1-λ)m(x)nm-i-1×∑μ∉[k-2,k+2]φ(ξμ,n+1(λ))nφ(λ-1)m+m-i-1(ξμ,n+1(λ))|x-ξμ,n+1(λ)|m-i.‍
Then, for someC>0 with 1-C/n2>(1+ξn+1,n+1(λ))/2, we have by (80)
(116)∑μ∉[k-2,k+2]φ(ξμ,n+1(λ))nφ(λ-1)m+m-i-1(ξμ,n+1(λ))|x-ξμ,n+1(λ)|m-i~∫[-1+C/n2,1-C/n2]∖[ξk-2,n+1(λ),ξk+2,n+1(λ)]φ(λ-1)m+m-i-1(t)|x-t|m-idt.‍
Therefore, we have the result from Lemma23 (a) with α=(1-λ)m, k=m-i.For convenience, we let(117)Hn+1,m,i[f](x)∶=∑μ=1n+1‍ei,μlμ,n+1m(x)(x-ξμ,n+1(λ))if(ξμ,n+1(λ)),
then
(118)Hn+1,m[f](x):=∑i=0m-1‍Hn+1,m,i[f](x).Let(119)∥Hn+1,m,i∥:=supx∈[-1,1]∑μ=1n+1‍|ei,μlμ,n+1m(x)(x-ξμ,n+1(λ))i|.Lemma 25.
Letm≥2 be a positive integer. Then, one has for 0≤i≤m-1(120)∥Hn+1,m,i∥=O(nmax{(1-λ)m-2(m-i),0})×{1,0≤i≤m-2;lnn,i=m-1.Proof.
By [10, (26)], it is sufficient to prove (120) for 0≤x≤1-a/n2, where a>0 is a fixed, sufficiently small constant. In the following case, let x∈[ξk-1,n+1(λ),ξk+1,n+1(λ)]∩[0,1-a/n2] for some 1≤k≤n+1. Then, we have from (32), (34), and (47)
(121)∑μ∉[k-2,k+2]|ei,μlμ,n+1m(x)(x-ξμ,n+1(λ))i|=∑μ∉[k-2,k+2]‍|ei,μ||Eλ,n+1(x)Eλ,n+1′(ξμ,n+1(λ))|m1|x-ξμ,n+1(λ)|m-i≤C∑μ∉[k-2,k+2]‍niφi(ξμ,n+1(λ))|n1-λφ1-λ(x)n2-λφ-λ(ξμ,n+1(λ))|m1|x-ξμ,n+1(λ)|m-i≤Cφ(1-λ)m(x)nm-i∑μ∉[k-2,k+2]‍φλm-i(ξμ,n+1(λ))|x-ξμ,n+1(λ)|m-i=O(nmax{(1-λ)m-2(m-i),0}){1,0≤i≤m-2;lnn,i=m-1.
The last inequality follows from Lemma24. On the other hand, if μ∈[k-2,k+2], there exists θμ between x and ξμ,n+1(λ) such that we see from (34) and (36) with r=1,
(122)|lμ,n+1(x)|=|Eλ,n+1′(θμ)Eλ,n+1′(ξμ,n+1(λ))|≤C.
Hence, from (47) and (80), we conclude that
(123)∑μ∈[k-2,k+2]1≤μ≤n+1|ei,μlμ,n+1m(x)(x-ξμ,n+1(λ))i|≤C.
Therefore, we have the result.Lemma 26.
Letm≥2 be an even integer. Then,
(124)∥Hn+1,m,m-1∥=O(nmax{(1-λ)m-2,0}).Proof.
First, we note thatm-1 is odd. Let x∈[ξk-1,n+1(λ),ξk+1,n+1(λ)]∩[0,1-a/n2] (ξ0,n+1(λ):=-1, ξn+2,n+1(λ):=1) for some 1≤μ≤n+1. Then, considering α1, ɛ1 as (1-λ)m, 1-λ, we have from (32), (34), (48), and Lemma 23(b)
(125)∑μ∉[k-2,k+2]|em-1,μlμ,n+1m(x)(x-ξμ,n+1(λ))m-1|=∑μ∉[k-2,k+2]‍|em-1,μ||Eλ,n+1(x)Eλ,n+1′(ξμ,n+1(λ))|m1|x-ξμ,n+1(λ)|≤Cφ(1-λ)m(x)n2-λ∑μ∉[k-2,k+2]‍φλm-m+1-(1-λ)(ξμ,n+1(λ))|x-ξμ,n+1(λ)|=φ(1-λ)m(x)n1-λ∑μ∉[k-2,k+2]‍φ(ξμ,n+1(λ))nφ(λ-1)m-(1-λ)(ξμ,n+1(λ))|x-ξμ,n+1(λ)|~φ(1-λ)m(x)n1-λ∫‍φ(λ-1)m-(1-λ)(t)|x-t|dt=O(nmax{(1-λ)m-2,0}),
where it is integrated under [-1+C/n2,1-C/n2]∖[ξk-2,n+1(λ),ξk+2,n+1(λ)]. Besides, we have, similarly to (123),
(126)∑μ∈[k-2,k+2]‍|em-1,μlμ,n+1m(x)(x-ξμ,n+1(λ))i|≤C.
Therefore, we have the result.Lemma 27.
For any polynomialR∈𝒫nm+m-1, one has
(127)|R(x)-Hn+1,m[R](x)|≤C∑t=1m-1‍∑i=0m-1-t‍∥R(t)φt∥∞ntO(nmax{(1-λ)m-2(m-t-i),0})lnn.Proof.
Letx∈[ξk-1,n+1(λ),ξk+1,n+1(λ)]∩[0,1-a/n2]  (ξ0,n+1(λ):=-1,ξn+2,n+1(λ):=1) for some 1≤μ≤n+1. 									(128)R(x)-Hn+1,m[R](x)=H^n+1,m[R](x)-Hn+1,m[R](x)=∑t=1m-1‍∑μ=1n+1‍R(t)(ξμ,n+1(λ))htμ(x)=∑μ=1n+1‍∑t=1m-1‍∑i=0m-1-t‍et,i,μt!R(t)(ξμ,n+1(λ))lμ,n+1m(x)(x-ξμ,n+1(λ))t+i=∑μ∉[k-2,k+2]‍+∑μ∈[k-2,k+2]‍.
Here,et,i,k is the coefficient of the higher-order Hermite interpolation polynomial H^n+1,m[R] based on the zeros of Eλ,n+1, defined in (11). Since et,i,k=e0,i,k=ei,k, we can see from (47) that uniformly for 0≤i≤m-1,
(129)|et,i,μ|≤Cniφ-i(ξμ,n+1(λ)).
Hence, using (32) and (34), we see for μ∉[k-2,k+2](130)|et,i,μt!R(t)(ξμ,n+1(λ))lμ,n+1m(x)(x-ξμ,n+1(λ))t+i|≤C∥R(t)φt∥∞ntφ(1-λ)m(x)nm-(t+i)φλm-t-i(ξμ,n+1(λ))|x-ξμ,n+1(λ)|m-(t+i).
Here, using Lemma24 with t+i as i for μ∉[k-2,k+2], we have the right formula in the lemma. We also see that for μ∈[k-2,k+2](131)|et,i,μt!R(t)(ξμ,n+1(λ))lμ,n+1m(x)(x-ξμ,n+1(λ))t+i|≤C∥R(t)φt∥∞nt.
Consequently, we have(132)|R(x)-Hn+1,m[R](x)|≤C∑t=1m-1‍∑i=0m-1-t‍∥R(t)φt∥∞ntO(nmax{(1-λ)m-2(m-t-i),0})lnn.Proof of Theorem1.
(a) From Lemma26, the result is trivially proved.
(b) Sincef is continuous on [-1,1], for given ɛ>0, there exists a polynomial R such that for x∈[-1,1](133)|f(x)-R(x)|<ɛ.
Then, one has from Lemmas 26 and 27(134)limn→∞|Hn+1,m[f](x)-f(x)|≤|f(x)-R(x)|+limn→∞|Hn+1,m[f-R](x)|+limn→∞|Hn+1,m[R](x)-R(x)|≤C(1+∥Hn+1,m∥)∥f-R∥∞+limn→∞∑t=1m-1‍∑i=0m-1-t‍∥R(t)φt∥∞ntlnn≤Cɛ.
This implies (21).Lemma 28.
Let0≤i≤m-1. Let a>0 be a fixed, sufficiently small constant. x∈[yk-1,2n+1(λ),yk+1,2n+1(λ)]∩[0,1-a/n2]  (y0,2n+1(λ):=-1,y2n+2,2n+1(λ):=1) for some 1≤k≤2n+1. Then we have uniformly for n(135)∑ν∉[k-2,k+2],1≤ν≤2n+1φ(1-2λ)m(x)nm-iφ2λm-i(yν,2n+1(λ))|x-yν,2n+1(λ)|m-i≤C{An,i(x)if0<λ<12;Bn,i(x)if12≤λ<1,
where
(136)An,i(x)=nmax{(1-2λ)m-2(m-i),0}×{1,0≤i≤m-2;lnn,i=m-1,(137)Bn,i(x)=φ(1-2λ)m(x)×{1,0≤i≤m-2;lnn,i=m-1.Proof.
When0<λ<1/2, we can show the lemma as the proof of Lemma 24 (see (116)). If 1/2≤λ<1, we have from Lemma 23(c),
(138)∑ν∉[k-2,k+2],1≤ν≤2n+1≤Cφ(1-2λ)m(x)nm-i-1×∫[-1+a/n2,1-a/n2]∖[yk-1,2n+1(λ),yk+1,2n+1(λ)]φ(2λ-1)m+m-i-1(t)|x-t|m-idt‍≤C{An,i(x)if0<λ<12;Bn,i(x)if12≤λ<1.Let(139)ℋ2n+1,m,i[f](x)∶=∑ν=12n+1‍ei,ν*lν,2n+1*m(x)×(x-yν,2n+1(λ))if(yν,2n+1(λ)),
then
(140)ℋ2n+1,m[f](x):=∑i=0m-1‍ℋ2n+1,m,i[f](x).Let(141)∥ℋ2n+1,m,i∥:=supx∈[-1,1]∑ν=12n+1‍|ei,ν*lν,2n+1*m(x)(x-yν,2n+1(λ))i|
and for a nonnegative real function u(x),
(142)∥ℋ2n+1,m,i∥u∶=supx∈[-1,1]u(x)×∑ν=12n+1‍|ei,ν*lν,2n+1*m(x)(x-yν,2n+1(λ))i|.Lemma 29.
Letm≥2 be a positive integer. If 0<λ<1/2, then one has for 0≤i≤m-1(143)∥ℋ2n+1,m,i∥=O(nmax{(1-2λ)m-2(m-i),0})×{1,0≤i≤m-2;lnn,i=m-1.
If1/2≤λ<1, then one has for 0≤i≤m-1(144)∥ℋ2n+1,m,i∥φ(2λ-1)m=O(1)×{1,0≤i≤m-2;lnn,i=m-1.Proof.
Similarly to the proof of Lemma25, using (51), we have from Proposition 7(2), Proposition 8 (35), and Lemma 28(145)∑ν∉[k-2,k+2],1≤ν≤2n+1|ei,ν*lν,2n+1*m(x)(x-yν,2n+1(λ))i|≤C∑ν∉[k-2,k+2],1≤ν≤2n+1φ(1-2λ)m(x)nm-iφ2λm-i(yν,2n+1(λ))|x-yν,2n+1(λ)|m-i≤C{An,i(x)if0<λ<12;Bn,i(x)if12≤λ<1,
where An,i(x) and Bn,i(x) are defined by (136) or (137). Besides, we easily know from (51), (35), (38), and (80)
(146)∑ν∈[k-2,k+2],1≤ν≤2n+1‍|ei,ν*lν,2n+1*m(x)(x-yν,2n+1(λ))i|≤C.
Therefore, we have the result.Let(147)ℋ2n+1,m,i[f](x):=∑ν=12n+1‍ei,ν*lν,2n+1*m(x)×(x-yν,2n+1(λ))if(yν,2n+1(λ)),
then
(148)ℋ2n+1,m[f](x):=∑i=0m-1‍ℋ2n+1,m,i[f](x).Let(149)∥ℋ2n+1,m,i∥:=supx∈[-1,1]∑ν=12n+1‍|ei,ν*lν,2n+1*m(x)(x-yν,2n+1(λ))i|.Lemma 30.
Letm≥2 be an even integer. Then, one has for 0<λ<1/2(150)∥ℋ2n+1,m,m-1∥=O(nmax{(1-2λ)m-2,0})
and for 1/2≤λ<1(151)∥ℋ2n+1,m,m-1∥φ(2λ-1)m=O(1).Proof.
First, we note thatm-1 is odd. Similarly to the proof of Lemma 26, using Proposition 7(2), Proposition 8 (35), and (51), we have
(152)∑:=∑ν∉[k-2,k+2],1≤ν≤2n+1|em-1,ν*lν,2n+1*m(x)(x-yν,2n+1(λ))m-1|≤Cφ(1-2λ)m(x)n2-λ×∑ν∉[k-2,k+2],1≤ν≤2n+1‍φ2λm-m+1-(1-λ)(yν,2n+1(λ))|x-yν,2n+1(λ)|≤Cφ(1-2λ)m(x)n1-λ∫A‍φ(2λ-1)m-(1-λ)(t)|x-t|dt,
where A:=[-1+a/n2,1-a/n2]∖[yk-1,2n+1(λ),yk+1,2n+1(λ)]. If 0<λ<1/2, considering α1 and ɛ1 as (1-2λ)m and (1-λ), respectively, we have from Lemma 23(b)
(153)∫A‍φ(2λ-1)m-(1-λ)(t)|x-t|dt≤Cφ(2λ-1)m(x)n1-λ×O(nmax{(1-2λ)m-2,0})
and if 1/2≤λ<1, considering α1 and ɛ1 as (2λ-1)m and (1-λ), respectively, we have from Lemma 23 (d)
(154)∫A‍φ(2λ-1)m-(1-λ)(t)|x-t|dt≤Cn1-λ.
Hence, we conclude that(155)∑‍≤C{nmax{(1-2λ)m-2,0},0<λ<12;φ(1-2λ)m(x),12≤λ<1.
Besides, we easily know that(156)∑ν∈[k-2,k+2],1≤ν≤2n+1|em-1,ν*lν,2n+1*m(x)(x-yν,2n+1(λ))m-1|≤C.
Therefore, we have the result.Lemma 31.
For any polynomialR∈𝒫2nm+m-1, one has
(157)|R(x)-ℋ2n+1,m[R](x)|≤C∑t=1m-1‍∑i=0m-1-t‍∥R(t)φt∥∞nt×{An,t+i(x)if0<λ<12;Bn,t+i(x)if12≤λ<1.Proof.
We have using Lemma28(158)|R(x)-ℋ2n+1,m[R](x)|≤C∑t=1m-1‍∑i=0m-1-t‍∑ν=12n+1|R(t)(yν,2n+1(λ))φt(yν,2n+1(λ))|ntoooooooooooooooo×φ(1-2λ)m(x)nm-(t+i)φ2λm-t-i(yν,2n+1(λ))|x-yν,2n+1(λ)|m-(t+i)≤C∑t=1m-1‍∑i=0m-1-t‍∥R(t)φt∥∞nt{An,t+i(x),if0<λ<12;Bn,t+i(x),if12≤λ<1.Proof of Theorem2.
Using Lemmas29, 30, and 31, it is similar to the proof of Theorem 1.


### 4.2.  Proof of Theorems3 and 4

Proof of Theorems3 and 4.
Suppose thatm=1,3,5,…, and let
(159)Λn+1,m(x):=∑μ=1n+1‍∑i=0m-1‍|ei,μlμ,n+1m(x)(x-ξμ,n+1(λ))i|.
From (17), we know that
(160)∥Hn+1,m∥=supx∈[-1,1]Λn+1,m(x).
Letx0 be the least positive zero of Pλ,n(x). Then, we have
(161)Λn+1,m(x0)≥∑-1/2≤ξμ,n+1(λ)≤0‍|Eλ,n+1(x0)Eλ,n+1′(ξμ,n+1(λ))|mppppppp×(|em-1,μ|x0-ξμ,n+1(λ)-∑i=0m-2‍|ei,μ|(x0-ξμ,n+1(λ))m-i).
Since we know from (42), (34), and (47)
(162)∑-1/2≤ξμ,n+1(λ)≤0‍|Eλ,n+1(x0)Eλ,n+1′(ξμ,n+1(λ))|m|em-1,μ|x0-ξμ,n+1(λ)~∑-1/2≤ξμ,n+1(λ)≤0‍1n1x0-ξμ,n+1(λ)~∫-1/20‍1x0-tdt~lnn,∑-1/2≤ξμ,n+1(λ)≤0‍|Eλ,n+1(x0)Eλ,n+1′(ξμ,n+1(λ))|m∑i=0m-2‍|ei,μ|(x0-ξμ,n+1(λ))m-i~∑i=0m-2‍1nm-i-1∑-1/2≤ξμ,n+1(λ)≤0‍1n1(x0-ξμ,n+1(λ))m-i~∑i=0m-2‍1nm-i-1∫-1/20‍1(x0-t)m-idt~1.
Thus, we have(163)Λn+1,m(x0)≥Clnn.
Therefore, we have the result from Lemma29.Remark 32.
Similarly, if we letxT be the least zero of Pλ,n(x) with xT>c and if we consider Λn+1,m(x) for (-1+c)/2≤ξμ,n+1(λ)≤c, then we have for -1<c<d<1(164)max[c,d]Λn+1,m(x)≥Clnn.Proof of Theorem4.
Suppose thatm=2,4,…. Let x0 be the least positive zero of Pλ,n(x). Then, similarly to the proof of Theorem 3, we have using the assumption (27)
(165)Λn+1,m(x0)≥∑-1≤ξμ,n+1(λ)≤-1/2‍|Eλ,n+1(x0)Eλ,n+1′(ξμ,n+1(λ))|m(|em-1,μ|-∑i=0m-2‍|ei,μ|)≥C∑-1≤ξμ,n+1(λ)≤-1/2‍φ(ξμ,n+1(λ))nnλ-1φ(λ-1)(m+1)(ξμ,n+1(λ))~nλ-1∫-1+1/n2-1/2‍φ(λ-1)(m+1)(t)dt~n(1-λ)m-2.
Here, we used the followings:(166)nλ+m-2φ-λ+m(ξμ,n+1(λ))-C1∑i=0m-2‍niφi(ξμ,n+1(λ))≥C2nλ+m-2φ-λ+m(ξμ,n+1(λ)),∫-1+1/n2-1/2‍φ(λ-1)(m+1)(t)dt~n(1-λ)m-1-λ
because ((λ-1)m+λ-1)/2<(-3+λ)/2<-1. Thus, we have ∥Hn+1,m∥≥Cn(1-λ)m-2, and it implies (28) from (20).Proposition 33 (see [6, (8.21.10)]).
Forɛ≤θ≤π-ɛ,
(167)Pλ,n(cosθ)=Γ(λ+1/2)Γ(2λ)nλ-1π-1/22λsin-λθ×cos{(n+λ)θ-λπ2}+O(nλ-2).Proposition 34 (see [11, Theorem]).
Forɛ≤θ≤π-ɛ,
(168)Eλ,n+1(cosθ)=n1-λπ-1/222-λsin1-λθ×cos{(n+λ)θ-(λ-1)π2}+o(n1-λ).Lemma 35.
Forɛ≤θ≤π-ɛ,
(169)Fλ,2n+1(cosθ)=-2Γ(λ+1/2)Γ(2λ)×π-1sin1-2λθsin{2(n+λ)θ-λπ}+o(1).Proof.
It is proved from (167) and (168).Proof of Theorem5.
Suppose thatm=1,3,5,…, and let
(170)Λ~2n+1,m(x):=∑ν=12n+1‍∑i=0m-1‍|ei,ν*lν,2n+1*m(x)(x-yν,2n+1(λ))i|.
From (18), we know that
(171)∥ℋ2n+1,m∥=supx∈[-1,1]Λ~2n+1,m(x).
Letx0:=cos((λ+n+1/2)π/2(n+λ)). Then, we know Fν,2n+1(x0)~1 from (169). Then, similarly to the proof of Theorem 3, we have from (35), (56), and (51), for x>0,
(172)Λ~2n+1,m(x0)≥∑-1/2≤yν,2n+1(λ)≤0‍|Fν,2n+1(x0)Fν,2n+1′(yν,2n+1(λ))|mppppppppp×(|em-1,ν*|x0-yν,2n+1(λ)-∑i=0m-2‍|ei,ν*|(x0-yν,2n+1(λ))m-i)≥Clnn.
Therefore, we have the result from Lemma29.Proof of Theorem6.
Suppose thatm=2,4,…. Let x0:=cos((λ+n+1/2)π/2(n+λ)). Then, since we know that Fν,2n+1(x0)~1 as we see above, similarly to the proof of Theorem 4, we have using the assumption (30)
(173)Λ~2n+1,m(x0)≥∑-1≤yν,2n+1(λ)≤-1/2‍|Fν,2n+1(x0)Fν,2n+1′(yν,2n+1(λ))|moooooooooo×(|em-1,ν*|-∑i=0m-2‍|ei,ν*|)≥C∑-1≤yν,2n+1(λ)≤-1/2‍(φ2λ(yν,2n+1(λ))n)mnλ+m-2φ-λ+m(yν,2n+1(λ))~nλ-1∫-1+1/n2-1/2‍φ(2λ-1)m+(λ-1)(t)dt~n(1-2λ)m-2.
Here, we used the following:(174)(φ2λ(yν,2n+1(λ))n)m(nλ+m-2φ-λ+m(yν,2n+1(λ))-C1nm-2φm-2(yν,2n+1(λ)))≥C2(φ2λ(yν,2n+1(λ))n)mnλ+m-2φ-λ+m(yν,2n+1(λ)),∫-1+1/n2-1/2‍φ(2λ-1)m+(λ-1)(t)dt~n(1-2λ)m-1-λ
because ((2λ-1)m+λ-1)/2<(-3+λ)/2<-1. Therefore, we have the result.


## 4.1. Proof of Theorems1 and 2

From now on, we assume that0<λ<1. We first state some known results for the Stieltjes polynomials. Let xη,n(λ):=cosϕη,n(λ), η=1,…,n, ξμ,n+1(λ):=cosθμ,n+1(λ), μ=1,…,n+1, and yν,2n+1(λ):=cosψν,2n+1(λ), ν=1,2,…,2n+1. Ehrich and Mastroianni [4] proved that for μ=0,1,…,n+1 and ν=0,1,…,2n+1(79)|ϕη,n(λ)-ϕη+1,n(λ)|~|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):=ϕ0,n(λ)=π and ψ2n+2,2n+1(λ):=θn+2,n+1(λ):=ϕn+1,n(λ):=0. It implies that for μ=0,1,…,n+2 and ν=0,1,…,2n+2,
(80)ξμ+1,n+1(λ)-ξμ,n+1(λ)~1nφ(ξμ,n+1(λ));yν+1,2n+1(λ)-yν,2n+1(λ)~1nφ(yν,2n+1(λ)),φ(ξμ+1,n+1(λ))~φ(ξμ,n+1(λ));φ(yν+1,2n+1(λ))~φ(yν,2n+1(λ)),
where y0,2n+1(λ):=ξ0,n+1(λ):=-1 and y2n+2,2n+1(λ):=ξn+2,n+1(λ):=1.Lemma 23.
Letk be a positive integer and 0<x<1-C/n2. Then,(a)
forα>0,
(81)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-α(t)φk-1(t)|x-t|kdt≤Cφ-α(x)nk-1O(nmax{α-2k,0})×{1,k⩾2;lnn,k=1;(b)
forα1>0 and ɛ1>0,
(82)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-α1-ɛ1(t)φk-1(t)|x-t|kdt≤Cφ-α1(x)nk-1+ɛ1O(nmax{α1-2k,0});(c)
forα>0,
(83)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φα(t)φk-1(t)|x-t|kdt≤Cnk-1{1,k⩾2;lnn,k=1;(d)
forα1>0 and ɛ1>0,
(84)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φα1-ɛ1(t)φk-1(t)|x-t|kdt≤Cnk-1+ɛ1.Proof.
(a)(85)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-α+k-1(t)|x-t|kdt~∫0x*‍+∫x*(1+x*)/2‍+∫(1+x*)/21-C/n2‍φ-α+k-1(t)|x-t|kdt=:I1+I2+I3,
where x*:=x-φ(x)/n and x*:=x+φ(x)/n.
I
1: Suppose that 0≤t≤x*. Then, since φ-α(t)≤φ-α(x), we have
(86)∫0x*‍φ-α+k-1(t)|x-t|kdt≤Cφ-α(x)∫0x*‍φk-1(t)|x-t|kdt.
Since we know for 0≤t≤x*,
(87)φ(t)~(x-t)1/2+φ(x),
we have
(88)∫0x*‍φk-1(t)|x-t|kdt~∫0x*‍1|x-t|(k-1)/2|x-t|dt+φk-1(x)×∫0x*‍1|x-t|k-1|x-t|dt≤C{|x-x*|-(k-1)/2+φk-1(x)|x-x*|-k+1,k≥2;lnn,k=1≤C{nk-1,k≥2;lnn,k=1.
Therefore, we have for I1(89)I1≤Cφ-α(x){nk-1,k≥2;lnn,k=1.
I
2: Since φ(t)~φ(x) for x*≤t≤(1+x*)/2,
(90)∫x*(1+x*)/2‍φ-α+k-1(t)|x-t|kdt~φ-α+k-1(x)∫x*(1+x*)/2‍1|x-t|kdt≤Cφ-α(x){nk-1,k≥2;lnn,k=1.
Therefore, we have
(91)I2≤Cφ-α(x){nk-1,k≥2;lnn,k=1.
I
3: Since |x-t|≥C|1-t| for (1+x*)/2<t<1-C/n2, we have for I3(92)∫(1+x*)/21-C/n2‍φ-α+k-1(t)|x-t|kdt=∫(1+x*)/21-C/n2‍φ-α+k-1(t)|x-t|k-α/2|x-t|α/2dt≤C1|(1+x*)/2-x|α/2×∫(1+x*)/21-C/n2‍1(1-t)(k+1)/2φ-α+2k(t)|x-t|-α/2+kdt.
Since for (1+x*)/2<t<1-C/n2,
(93)φ-α+2k(t)|x-t|-α/2+k≤C{1,if-α+2k≥0;nα-2k,if-α+2k<0=O(nmax{α-2k,0}),
with |(1+x*)/2-x|~1-x~φ2(x), we have
(94)I3=∫(1+x*)/21-C/n2‍φ-α+k-1(t)|x-t|kdt≤O(nmax{α-2k,0})φ-α(x)×∫(1+x*)/21-C/n2‍1(1-t)(k+1)/2dt~nk-1φ-α(x)O(nmax{α-2k,0}){1,k⩾2;lnn,k=1.
Therefore, we have the result (a).
(b) Similarly to (a), we let(95)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-α1-ɛ1+k-1(t)|x-t|kdt~∫0x*‍+∫x*(1+x*)/2‍+∫(1+x*)/21-C/n2‍φ-α1+k-1-ϵ1(t)|x-t|kdt=:I1+I2+I3.
Then, for I1, we have using (87)
(96)∫0x*‍φ-α1+k-1-ϵ1(t)|x-t|kdt≤Cφ-α1(x)∫0x*‍(φ-ϵ1(t)|x-t|(k-1)/2|x-t|ppppppppppppp+φk-1(x)φ-ϵ1(t)|x-t|k-1|x-t|)dt≤Cφ-α1(x)nk-1∫0x*‍φ-ϵ1(t)x-tdt,(97)∫0x*‍φ-ϵ1(t)x-tdt≤C∫0x*‍1(x-t)ϵ1/2+1dt≤C(nφ(x))ϵ1/2≤Cnɛ1
by the use ofφ(t)≥C|x-t|1/2. Therefore, we have for I1(98)I1≤Cφ-α1(x)nk-1+ɛ1.
I
2: Since 1-t≥C(t-x) and φ(t)~φ(x) for x*≤t≤(1+x*)/2,
(99)∫x*(1+x*)/2‍φ-α1+k-1-ϵ1(t)|x-t|kdt~φ-α1(x)nk-1∫x*(1+x*)/2‍φ-ϵ1(t)(t-x)dt≤Cφ-α1(x)nk-1∫x*(1+x*)/2‍1(t-x)ϵ1/2+1dt≤Cφ-α1(x)nk-1+ɛ1.
Therefore,
(100)I2≤Cφ-α1(x)nk-1+ɛ1.
I
3: Since |x-t|≥C|(1+x*)/2-x| for (1+x*)/2<t<1-C/n2, we have using (93)
(101)φ-α1+k-1-ϵ1(t)|x-t|k=φ-α1+k-1-ϵ1(t)|x-t|k-α1/2|x-t|α1/2≤C1|(1+x*)/2-x|α1/2φ-α1+k-1-ϵ1(t)|x-t|k-α1/2=Cφ-α1(x)1(1-t)(k-1+ɛ1)/2+1φ-α1+2k(t)|x-t|k-α1/2=φ-α1(x)O(nmax{α-2k,0})1(1-t)(k-1+ɛ1)/2+1,
and we know that
(102)∫(1+x*)/21-C/n2‍1(1-t)(k-1+ɛ1)/2+1dt≤Cnk-1+ɛ1.
Therefore, we see that
(103)∫(1+x*)/21-C/n2‍φ-α1+k-1-ϵ1(t)|x-t|kdt≤Cφ-α1(x)nk-1+ɛ1O(nmax{α1-2k,0}).
Consequently, we have the result (b).
(c) and (d): Since|1-t2|≤C(|1-x2|+|t-x|), we know φ(t)≤C(φ(x)+|x-t|1/2). Then, we have
(104)∫0≤t≤1-C/n2,|x-t|>φ(x)/nφα+k-1(t)|x-t|kdt≤C(φα+k-1(x)∫1|x-t|kdt+∫‍1|x-t|(k+1)/2dt)≤Cnk-1{1,k⩾2;lnnk=1,
because we see if k=1, then
(105)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍1|x-t|kdt≤Clnn,∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍1|x-t|(k+1)/2dt≤Clnn
and if k≥2, then
(106)φα+k-1(x)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍1|x-t|kdt≤Cφα+k-1(x)|x-t|k-1≤Cφα+k-1(x)(nφ(x))k-1≤Cnk-1,∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍1|x-t|(k+1)/2dt≤Cnk-1;
that is, we have (c). Similarly,
(107)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φα1+k-1-ϵ1(t)|x-t|kdt≤C(φα1+k-1(x)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-ϵ1(t)|x-t|kdt+∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-ϵ1(t)|x-t|(k+1)/2dt):=C(A+B).
Then, for A(108)A=φα1+k-1(x)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-ϵ1(t)|x-t|kdt≤Cφα1+k-1(x)(nφ(x))k-1∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-ϵ1(t)|x-t|dt≤Cnk-1∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-ϵ1(t)|x-t|dt
and for B(109)B=∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-ϵ1(t)|x-t|(k+1)/2dt≤Cnk-1∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-ϵ1(t)|x-t|dt.
On the other hand, by (97) and (99), we know that
(110)∫0x*‍φ-ϵ1(t)x-tdt≤Cnɛ1,∫x*(1+x*)/2‍φ-ϵ1(t)(t-x)dt≤Cnɛ1.
Since |x-t|≥C|1-t| for (1+x*)/2<t<1-C/n2, we have
(111)∫(1+x*)/21-C/n2‍φ-ϵ1(t)(t-x)dt≤C∫(1+x*)/21-C/n2‍1(1-t)ϵ1/2+1dt≤Cnɛ1.
Therefore, we have
(112)∫0≤t≤1-C/n2,|x-t|>φ(x)/n‍φ-ϵ1(t)|x-t|dt≤Cnk-1+ɛ1.
So, by (108) and (109) we have (d).Lemma 24.
Let0≤i≤m-1. Let a>0 be a fixed, sufficiently small constant. Let x∈[ξk-1,n+1(λ),ξk+1,n+1(λ)]∩[0,1-a/n2] for some 1≤k≤n+1. Then, one has uniformly for n(113)∑μ∉[k-2,k+2],1≤μ≤n+1φ(1-λ)m(x)nm-iφλm-i(ξμ,n+1(λ))|x-ξμ,n+1(λ)|m-i≤O(nmax{(1-λ)m-2(m-i),0}){1,0≤i≤m-2;lnn,i=m-1.Proof.
For the simplicity, we denote(114)∑μ∉[k-2,k+2]‍:=∑μ∉[k-2,k+2],1≤μ≤n+1φ(1-λ)m(x)nm-iφλm-i(ξμ,n+1(λ))|x-ξμ,n+1(λ)|m-i.
Then, we know that(115)∑μ∉[k-2,k+2]‍=φ(1-λ)m(x)nm-i-1×∑μ∉[k-2,k+2]φ(ξμ,n+1(λ))nφ(λ-1)m+m-i-1(ξμ,n+1(λ))|x-ξμ,n+1(λ)|m-i.‍
Then, for someC>0 with 1-C/n2>(1+ξn+1,n+1(λ))/2, we have by (80)
(116)∑μ∉[k-2,k+2]φ(ξμ,n+1(λ))nφ(λ-1)m+m-i-1(ξμ,n+1(λ))|x-ξμ,n+1(λ)|m-i~∫[-1+C/n2,1-C/n2]∖[ξk-2,n+1(λ),ξk+2,n+1(λ)]φ(λ-1)m+m-i-1(t)|x-t|m-idt.‍
Therefore, we have the result from Lemma23 (a) with α=(1-λ)m, k=m-i.For convenience, we let(117)Hn+1,m,i[f](x)∶=∑μ=1n+1‍ei,μlμ,n+1m(x)(x-ξμ,n+1(λ))if(ξμ,n+1(λ)),
then
(118)Hn+1,m[f](x):=∑i=0m-1‍Hn+1,m,i[f](x).Let(119)∥Hn+1,m,i∥:=supx∈[-1,1]∑μ=1n+1‍|ei,μlμ,n+1m(x)(x-ξμ,n+1(λ))i|.Lemma 25.
Letm≥2 be a positive integer. Then, one has for 0≤i≤m-1(120)∥Hn+1,m,i∥=O(nmax{(1-λ)m-2(m-i),0})×{1,0≤i≤m-2;lnn,i=m-1.Proof.
By [10, (26)], it is sufficient to prove (120) for 0≤x≤1-a/n2, where a>0 is a fixed, sufficiently small constant. In the following case, let x∈[ξk-1,n+1(λ),ξk+1,n+1(λ)]∩[0,1-a/n2] for some 1≤k≤n+1. Then, we have from (32), (34), and (47)
(121)∑μ∉[k-2,k+2]|ei,μlμ,n+1m(x)(x-ξμ,n+1(λ))i|=∑μ∉[k-2,k+2]‍|ei,μ||Eλ,n+1(x)Eλ,n+1′(ξμ,n+1(λ))|m1|x-ξμ,n+1(λ)|m-i≤C∑μ∉[k-2,k+2]‍niφi(ξμ,n+1(λ))|n1-λφ1-λ(x)n2-λφ-λ(ξμ,n+1(λ))|m1|x-ξμ,n+1(λ)|m-i≤Cφ(1-λ)m(x)nm-i∑μ∉[k-2,k+2]‍φλm-i(ξμ,n+1(λ))|x-ξμ,n+1(λ)|m-i=O(nmax{(1-λ)m-2(m-i),0}){1,0≤i≤m-2;lnn,i=m-1.
The last inequality follows from Lemma24. On the other hand, if μ∈[k-2,k+2], there exists θμ between x and ξμ,n+1(λ) such that we see from (34) and (36) with r=1,
(122)|lμ,n+1(x)|=|Eλ,n+1′(θμ)Eλ,n+1′(ξμ,n+1(λ))|≤C.
Hence, from (47) and (80), we conclude that
(123)∑μ∈[k-2,k+2]1≤μ≤n+1|ei,μlμ,n+1m(x)(x-ξμ,n+1(λ))i|≤C.
Therefore, we have the result.Lemma 26.
Letm≥2 be an even integer. Then,
(124)∥Hn+1,m,m-1∥=O(nmax{(1-λ)m-2,0}).Proof.
First, we note thatm-1 is odd. Let x∈[ξk-1,n+1(λ),ξk+1,n+1(λ)]∩[0,1-a/n2] (ξ0,n+1(λ):=-1, ξn+2,n+1(λ):=1) for some 1≤μ≤n+1. Then, considering α1, ɛ1 as (1-λ)m, 1-λ, we have from (32), (34), (48), and Lemma 23(b)
(125)∑μ∉[k-2,k+2]|em-1,μlμ,n+1m(x)(x-ξμ,n+1(λ))m-1|=∑μ∉[k-2,k+2]‍|em-1,μ||Eλ,n+1(x)Eλ,n+1′(ξμ,n+1(λ))|m1|x-ξμ,n+1(λ)|≤Cφ(1-λ)m(x)n2-λ∑μ∉[k-2,k+2]‍φλm-m+1-(1-λ)(ξμ,n+1(λ))|x-ξμ,n+1(λ)|=φ(1-λ)m(x)n1-λ∑μ∉[k-2,k+2]‍φ(ξμ,n+1(λ))nφ(λ-1)m-(1-λ)(ξμ,n+1(λ))|x-ξμ,n+1(λ)|~φ(1-λ)m(x)n1-λ∫‍φ(λ-1)m-(1-λ)(t)|x-t|dt=O(nmax{(1-λ)m-2,0}),
where it is integrated under [-1+C/n2,1-C/n2]∖[ξk-2,n+1(λ),ξk+2,n+1(λ)]. Besides, we have, similarly to (123),
(126)∑μ∈[k-2,k+2]‍|em-1,μlμ,n+1m(x)(x-ξμ,n+1(λ))i|≤C.
Therefore, we have the result.Lemma 27.
For any polynomialR∈𝒫nm+m-1, one has
(127)|R(x)-Hn+1,m[R](x)|≤C∑t=1m-1‍∑i=0m-1-t‍∥R(t)φt∥∞ntO(nmax{(1-λ)m-2(m-t-i),0})lnn.Proof.
Letx∈[ξk-1,n+1(λ),ξk+1,n+1(λ)]∩[0,1-a/n2]  (ξ0,n+1(λ):=-1,ξn+2,n+1(λ):=1) for some 1≤μ≤n+1. 									(128)R(x)-Hn+1,m[R](x)=H^n+1,m[R](x)-Hn+1,m[R](x)=∑t=1m-1‍∑μ=1n+1‍R(t)(ξμ,n+1(λ))htμ(x)=∑μ=1n+1‍∑t=1m-1‍∑i=0m-1-t‍et,i,μt!R(t)(ξμ,n+1(λ))lμ,n+1m(x)(x-ξμ,n+1(λ))t+i=∑μ∉[k-2,k+2]‍+∑μ∈[k-2,k+2]‍.
Here,et,i,k is the coefficient of the higher-order Hermite interpolation polynomial H^n+1,m[R] based on the zeros of Eλ,n+1, defined in (11). Since et,i,k=e0,i,k=ei,k, we can see from (47) that uniformly for 0≤i≤m-1,
(129)|et,i,μ|≤Cniφ-i(ξμ,n+1(λ)).
Hence, using (32) and (34), we see for μ∉[k-2,k+2](130)|et,i,μt!R(t)(ξμ,n+1(λ))lμ,n+1m(x)(x-ξμ,n+1(λ))t+i|≤C∥R(t)φt∥∞ntφ(1-λ)m(x)nm-(t+i)φλm-t-i(ξμ,n+1(λ))|x-ξμ,n+1(λ)|m-(t+i).
Here, using Lemma24 with t+i as i for μ∉[k-2,k+2], we have the right formula in the lemma. We also see that for μ∈[k-2,k+2](131)|et,i,μt!R(t)(ξμ,n+1(λ))lμ,n+1m(x)(x-ξμ,n+1(λ))t+i|≤C∥R(t)φt∥∞nt.
Consequently, we have(132)|R(x)-Hn+1,m[R](x)|≤C∑t=1m-1‍∑i=0m-1-t‍∥R(t)φt∥∞ntO(nmax{(1-λ)m-2(m-t-i),0})lnn.Proof of Theorem1.
(a) From Lemma26, the result is trivially proved.
(b) Sincef is continuous on [-1,1], for given ɛ>0, there exists a polynomial R such that for x∈[-1,1](133)|f(x)-R(x)|<ɛ.
Then, one has from Lemmas 26 and 27(134)limn→∞|Hn+1,m[f](x)-f(x)|≤|f(x)-R(x)|+limn→∞|Hn+1,m[f-R](x)|+limn→∞|Hn+1,m[R](x)-R(x)|≤C(1+∥Hn+1,m∥)∥f-R∥∞+limn→∞∑t=1m-1‍∑i=0m-1-t‍∥R(t)φt∥∞ntlnn≤Cɛ.
This implies (21).Lemma 28.
Let0≤i≤m-1. Let a>0 be a fixed, sufficiently small constant. x∈[yk-1,2n+1(λ),yk+1,2n+1(λ)]∩[0,1-a/n2]  (y0,2n+1(λ):=-1,y2n+2,2n+1(λ):=1) for some 1≤k≤2n+1. Then we have uniformly for n(135)∑ν∉[k-2,k+2],1≤ν≤2n+1φ(1-2λ)m(x)nm-iφ2λm-i(yν,2n+1(λ))|x-yν,2n+1(λ)|m-i≤C{An,i(x)if0<λ<12;Bn,i(x)if12≤λ<1,
where
(136)An,i(x)=nmax{(1-2λ)m-2(m-i),0}×{1,0≤i≤m-2;lnn,i=m-1,(137)Bn,i(x)=φ(1-2λ)m(x)×{1,0≤i≤m-2;lnn,i=m-1.Proof.
When0<λ<1/2, we can show the lemma as the proof of Lemma 24 (see (116)). If 1/2≤λ<1, we have from Lemma 23(c),
(138)∑ν∉[k-2,k+2],1≤ν≤2n+1≤Cφ(1-2λ)m(x)nm-i-1×∫[-1+a/n2,1-a/n2]∖[yk-1,2n+1(λ),yk+1,2n+1(λ)]φ(2λ-1)m+m-i-1(t)|x-t|m-idt‍≤C{An,i(x)if0<λ<12;Bn,i(x)if12≤λ<1.Let(139)ℋ2n+1,m,i[f](x)∶=∑ν=12n+1‍ei,ν*lν,2n+1*m(x)×(x-yν,2n+1(λ))if(yν,2n+1(λ)),
then
(140)ℋ2n+1,m[f](x):=∑i=0m-1‍ℋ2n+1,m,i[f](x).Let(141)∥ℋ2n+1,m,i∥:=supx∈[-1,1]∑ν=12n+1‍|ei,ν*lν,2n+1*m(x)(x-yν,2n+1(λ))i|
and for a nonnegative real function u(x),
(142)∥ℋ2n+1,m,i∥u∶=supx∈[-1,1]u(x)×∑ν=12n+1‍|ei,ν*lν,2n+1*m(x)(x-yν,2n+1(λ))i|.Lemma 29.
Letm≥2 be a positive integer. If 0<λ<1/2, then one has for 0≤i≤m-1(143)∥ℋ2n+1,m,i∥=O(nmax{(1-2λ)m-2(m-i),0})×{1,0≤i≤m-2;lnn,i=m-1.
If1/2≤λ<1, then one has for 0≤i≤m-1(144)∥ℋ2n+1,m,i∥φ(2λ-1)m=O(1)×{1,0≤i≤m-2;lnn,i=m-1.Proof.
Similarly to the proof of Lemma25, using (51), we have from Proposition 7(2), Proposition 8 (35), and Lemma 28(145)∑ν∉[k-2,k+2],1≤ν≤2n+1|ei,ν*lν,2n+1*m(x)(x-yν,2n+1(λ))i|≤C∑ν∉[k-2,k+2],1≤ν≤2n+1φ(1-2λ)m(x)nm-iφ2λm-i(yν,2n+1(λ))|x-yν,2n+1(λ)|m-i≤C{An,i(x)if0<λ<12;Bn,i(x)if12≤λ<1,
where An,i(x) and Bn,i(x) are defined by (136) or (137). Besides, we easily know from (51), (35), (38), and (80)
(146)∑ν∈[k-2,k+2],1≤ν≤2n+1‍|ei,ν*lν,2n+1*m(x)(x-yν,2n+1(λ))i|≤C.
Therefore, we have the result.Let(147)ℋ2n+1,m,i[f](x):=∑ν=12n+1‍ei,ν*lν,2n+1*m(x)×(x-yν,2n+1(λ))if(yν,2n+1(λ)),
then
(148)ℋ2n+1,m[f](x):=∑i=0m-1‍ℋ2n+1,m,i[f](x).Let(149)∥ℋ2n+1,m,i∥:=supx∈[-1,1]∑ν=12n+1‍|ei,ν*lν,2n+1*m(x)(x-yν,2n+1(λ))i|.Lemma 30.
Letm≥2 be an even integer. Then, one has for 0<λ<1/2(150)∥ℋ2n+1,m,m-1∥=O(nmax{(1-2λ)m-2,0})
and for 1/2≤λ<1(151)∥ℋ2n+1,m,m-1∥φ(2λ-1)m=O(1).Proof.
First, we note thatm-1 is odd. Similarly to the proof of Lemma 26, using Proposition 7(2), Proposition 8 (35), and (51), we have
(152)∑:=∑ν∉[k-2,k+2],1≤ν≤2n+1|em-1,ν*lν,2n+1*m(x)(x-yν,2n+1(λ))m-1|≤Cφ(1-2λ)m(x)n2-λ×∑ν∉[k-2,k+2],1≤ν≤2n+1‍φ2λm-m+1-(1-λ)(yν,2n+1(λ))|x-yν,2n+1(λ)|≤Cφ(1-2λ)m(x)n1-λ∫A‍φ(2λ-1)m-(1-λ)(t)|x-t|dt,
where A:=[-1+a/n2,1-a/n2]∖[yk-1,2n+1(λ),yk+1,2n+1(λ)]. If 0<λ<1/2, considering α1 and ɛ1 as (1-2λ)m and (1-λ), respectively, we have from Lemma 23(b)
(153)∫A‍φ(2λ-1)m-(1-λ)(t)|x-t|dt≤Cφ(2λ-1)m(x)n1-λ×O(nmax{(1-2λ)m-2,0})
and if 1/2≤λ<1, considering α1 and ɛ1 as (2λ-1)m and (1-λ), respectively, we have from Lemma 23 (d)
(154)∫A‍φ(2λ-1)m-(1-λ)(t)|x-t|dt≤Cn1-λ.
Hence, we conclude that(155)∑‍≤C{nmax{(1-2λ)m-2,0},0<λ<12;φ(1-2λ)m(x),12≤λ<1.
Besides, we easily know that(156)∑ν∈[k-2,k+2],1≤ν≤2n+1|em-1,ν*lν,2n+1*m(x)(x-yν,2n+1(λ))m-1|≤C.
Therefore, we have the result.Lemma 31.
For any polynomialR∈𝒫2nm+m-1, one has
(157)|R(x)-ℋ2n+1,m[R](x)|≤C∑t=1m-1‍∑i=0m-1-t‍∥R(t)φt∥∞nt×{An,t+i(x)if0<λ<12;Bn,t+i(x)if12≤λ<1.Proof.
We have using Lemma28(158)|R(x)-ℋ2n+1,m[R](x)|≤C∑t=1m-1‍∑i=0m-1-t‍∑ν=12n+1|R(t)(yν,2n+1(λ))φt(yν,2n+1(λ))|ntoooooooooooooooo×φ(1-2λ)m(x)nm-(t+i)φ2λm-t-i(yν,2n+1(λ))|x-yν,2n+1(λ)|m-(t+i)≤C∑t=1m-1‍∑i=0m-1-t‍∥R(t)φt∥∞nt{An,t+i(x),if0<λ<12;Bn,t+i(x),if12≤λ<1.Proof of Theorem2.
Using Lemmas29, 30, and 31, it is similar to the proof of Theorem 1.


## 4.2.  Proof of Theorems3 and 4

Proof of Theorems3 and 4.
Suppose thatm=1,3,5,…, and let
(159)Λn+1,m(x):=∑μ=1n+1‍∑i=0m-1‍|ei,μlμ,n+1m(x)(x-ξμ,n+1(λ))i|.
From (17), we know that
(160)∥Hn+1,m∥=supx∈[-1,1]Λn+1,m(x).
Letx0 be the least positive zero of Pλ,n(x). Then, we have
(161)Λn+1,m(x0)≥∑-1/2≤ξμ,n+1(λ)≤0‍|Eλ,n+1(x0)Eλ,n+1′(ξμ,n+1(λ))|mppppppp×(|em-1,μ|x0-ξμ,n+1(λ)-∑i=0m-2‍|ei,μ|(x0-ξμ,n+1(λ))m-i).
Since we know from (42), (34), and (47)
(162)∑-1/2≤ξμ,n+1(λ)≤0‍|Eλ,n+1(x0)Eλ,n+1′(ξμ,n+1(λ))|m|em-1,μ|x0-ξμ,n+1(λ)~∑-1/2≤ξμ,n+1(λ)≤0‍1n1x0-ξμ,n+1(λ)~∫-1/20‍1x0-tdt~lnn,∑-1/2≤ξμ,n+1(λ)≤0‍|Eλ,n+1(x0)Eλ,n+1′(ξμ,n+1(λ))|m∑i=0m-2‍|ei,μ|(x0-ξμ,n+1(λ))m-i~∑i=0m-2‍1nm-i-1∑-1/2≤ξμ,n+1(λ)≤0‍1n1(x0-ξμ,n+1(λ))m-i~∑i=0m-2‍1nm-i-1∫-1/20‍1(x0-t)m-idt~1.
Thus, we have(163)Λn+1,m(x0)≥Clnn.
Therefore, we have the result from Lemma29.Remark 32.
Similarly, if we letxT be the least zero of Pλ,n(x) with xT>c and if we consider Λn+1,m(x) for (-1+c)/2≤ξμ,n+1(λ)≤c, then we have for -1<c<d<1(164)max[c,d]Λn+1,m(x)≥Clnn.Proof of Theorem4.
Suppose thatm=2,4,…. Let x0 be the least positive zero of Pλ,n(x). Then, similarly to the proof of Theorem 3, we have using the assumption (27)
(165)Λn+1,m(x0)≥∑-1≤ξμ,n+1(λ)≤-1/2‍|Eλ,n+1(x0)Eλ,n+1′(ξμ,n+1(λ))|m(|em-1,μ|-∑i=0m-2‍|ei,μ|)≥C∑-1≤ξμ,n+1(λ)≤-1/2‍φ(ξμ,n+1(λ))nnλ-1φ(λ-1)(m+1)(ξμ,n+1(λ))~nλ-1∫-1+1/n2-1/2‍φ(λ-1)(m+1)(t)dt~n(1-λ)m-2.
Here, we used the followings:(166)nλ+m-2φ-λ+m(ξμ,n+1(λ))-C1∑i=0m-2‍niφi(ξμ,n+1(λ))≥C2nλ+m-2φ-λ+m(ξμ,n+1(λ)),∫-1+1/n2-1/2‍φ(λ-1)(m+1)(t)dt~n(1-λ)m-1-λ
because ((λ-1)m+λ-1)/2<(-3+λ)/2<-1. Thus, we have ∥Hn+1,m∥≥Cn(1-λ)m-2, and it implies (28) from (20).Proposition 33 (see [6, (8.21.10)]).
Forɛ≤θ≤π-ɛ,
(167)Pλ,n(cosθ)=Γ(λ+1/2)Γ(2λ)nλ-1π-1/22λsin-λθ×cos{(n+λ)θ-λπ2}+O(nλ-2).Proposition 34 (see [11, Theorem]).
Forɛ≤θ≤π-ɛ,
(168)Eλ,n+1(cosθ)=n1-λπ-1/222-λsin1-λθ×cos{(n+λ)θ-(λ-1)π2}+o(n1-λ).Lemma 35.
Forɛ≤θ≤π-ɛ,
(169)Fλ,2n+1(cosθ)=-2Γ(λ+1/2)Γ(2λ)×π-1sin1-2λθsin{2(n+λ)θ-λπ}+o(1).Proof.
It is proved from (167) and (168).Proof of Theorem5.
Suppose thatm=1,3,5,…, and let
(170)Λ~2n+1,m(x):=∑ν=12n+1‍∑i=0m-1‍|ei,ν*lν,2n+1*m(x)(x-yν,2n+1(λ))i|.
From (18), we know that
(171)∥ℋ2n+1,m∥=supx∈[-1,1]Λ~2n+1,m(x).
Letx0:=cos((λ+n+1/2)π/2(n+λ)). Then, we know Fν,2n+1(x0)~1 from (169). Then, similarly to the proof of Theorem 3, we have from (35), (56), and (51), for x>0,
(172)Λ~2n+1,m(x0)≥∑-1/2≤yν,2n+1(λ)≤0‍|Fν,2n+1(x0)Fν,2n+1′(yν,2n+1(λ))|mppppppppp×(|em-1,ν*|x0-yν,2n+1(λ)-∑i=0m-2‍|ei,ν*|(x0-yν,2n+1(λ))m-i)≥Clnn.
Therefore, we have the result from Lemma29.Proof of Theorem6.
Suppose thatm=2,4,…. Let x0:=cos((λ+n+1/2)π/2(n+λ)). Then, since we know that Fν,2n+1(x0)~1 as we see above, similarly to the proof of Theorem 4, we have using the assumption (30)
(173)Λ~2n+1,m(x0)≥∑-1≤yν,2n+1(λ)≤-1/2‍|Fν,2n+1(x0)Fν,2n+1′(yν,2n+1(λ))|moooooooooo×(|em-1,ν*|-∑i=0m-2‍|ei,ν*|)≥C∑-1≤yν,2n+1(λ)≤-1/2‍(φ2λ(yν,2n+1(λ))n)mnλ+m-2φ-λ+m(yν,2n+1(λ))~nλ-1∫-1+1/n2-1/2‍φ(2λ-1)m+(λ-1)(t)dt~n(1-2λ)m-2.
Here, we used the following:(174)(φ2λ(yν,2n+1(λ))n)m(nλ+m-2φ-λ+m(yν,2n+1(λ))-C1nm-2φm-2(yν,2n+1(λ)))≥C2(φ2λ(yν,2n+1(λ))n)mnλ+m-2φ-λ+m(yν,2n+1(λ)),∫-1+1/n2-1/2‍φ(2λ-1)m+(λ-1)(t)dt~n(1-2λ)m-1-λ
because ((2λ-1)m+λ-1)/2<(-3+λ)/2<-1. Therefore, we have the result.



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*Source: 542653-2013-11-26.xml*