# Mean Convergence Rate of Derivatives by Lagrange Interpolation on Chebyshev Grids

Authors: Wang Xiulian; Ning Jingrui
Journal: Discrete Dynamics in Nature and Society
 (2011)

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## Abstract

We consider the rate of mean convergence of derivatives by Lagrange interpolation operators based on the Chebyshev nodes. Some estimates of error of the derivatives approximation in terms of the error of best approximation by polynomials are derived. Our results are sharp.


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## Body

## 1. Introduction and Main Results

Mean convergence of Lagrange interpolation based on the zeros of orthogonal polynomials (and possibly some additional points) has been studied for at least 70 years. There is a vast literature on this topic. The authors of [1–3] considered the simultaneous approximation by the Hermite interpolation operators, and we will consider the simultaneous approximation by Lagrange interpolation operators based on the zeros of Chebyshev polynomials. The relevant results can be found in [4–6]. We introduce these results below.Let(1.1)ω(x)=∏k=1N|x-yk|Γk(|x|≤1;-1=y1<y2<⋯<yN=1;Γk>-1;k=1,…,N)
be a so-called generalized Jacobi weight (ω∈GJ), and let(1.2)-1≤x1<x2<⋯<xn≤1
be the zeros of the nth orthogonal polynomial pn(ω) associated with the weight-function ω∈GJ. Let Ln(ω,f) denote the Lagrange interpolating polynomial which interpolates f at the zeros of pn(ω). By using Markov-Bernstein type inequalities in Lp metric, J. Szabados and A. K. Varma [5] reduced the weighted mean convergence of derivatives Ln(r)(ω,f,x) to the weighted mean convergence of Ln(ω,f,x) and obtained the following.  If Lp means functional space equipped with Lp norm and (*)ω(x)∈GJ,ω(x)1/p-1/2(1-x2)1/4∈Lp,
then, for f(r)(x)∈C[-1,1], we have (1.3)∫-11|f(r)(x)-Ln(r)(f,x)|p(1-x2)rp/2ω(x)dx≤CrEn-r-1p(f(r))(n≥r+1).Here and in the following, the constantCr (may be different in the same expression) is independent of n and f but depends on r, and En(·) denotes the error of the best polynomial approximation of degree n of the corresponding function in the L∞ metric.Mastroianni and Nevai [4] get sharper estimates in terms of modulus of continuity instead of the best approximation. It improves some old results. But its proof also needs weighted Markov-Bernstein type inequality in Lp metric and the idea of additional points. For the weight functions not satisfying (*), it is not possible to discuss by their method. To deal with these case, Du and Xu [7] consider the most important special case ω(x)=1/1-x2. Let (1.4)tk=tkn=cos⁡2k-12nπ,k=1,…,n,
be the zeros of Tn(x)=cosnθ,x=cosθ, the nth degree Chebyshev polynomial of the first kind. If f∈C[-1,1], then the well-known Lagrange interpolation polynomial of f based on {tk}k=1n is given by (see [8]) (1.5)Ln(f,x)=∑k=1nf(tk)lk(x),
where (1.6)lk(x)=(-1)k+11-tk2Tn(x)n(x-tk),k=1,…,n.Du and Xu [7] obtained the following.Theorem A.
LetLn(f,x) be as defined as above. Then, for f∈C[-1,1]1, we have
(1.7)(∫-11|f′(x)-Ln′(f,x)|p(1-x2)αdx)1/p≤{CEn-2(f′),α>p2-1,C(ln⁡n)1/pEn-2(f′),α=p2-1,Cn1-((2α+2)/p)En-2(f′),-1<α<p2-1,
and the estimation for -1<α≤(p/2)-1 is sharp.We notice that although the sharp estimate is obtained, the upper bound is notEn-2(f′) for -1≤α≤(p/2)-1. Now we will give a Lagrange interpolation to improve their results.Let(1.8)xk=xkn=cos⁡kπn+1,k=1,…,n,
be the zeros of Un(x)=sin(n+1)θ/sinθ,x=cosθ, the nth degree Chebyshev polynomial of the second kind. If f∈C[-1,1], then the well-known Lagrange interpolation polynomial of f based on {xk}k=1n∪{x0=1,xn+1=-1} is given by (see [9]) (1.9)Qn+2(f,x)=∑k=0n+1f(xk)φk(x),
where (1.10)φ0(x)=(1+x)Un(x)2(n+1),φn+1(x)=(-1)n(x-1)Un(x)2(n+1),φk(x)=(-1)k+1(1-x2)Un(x)(n+1)(x-xk),k=1,…,n.Firstly, we obtain the following.Theorem 1.1.
LetQn(f,x) be as defined as above, 0<p<+∞,α>-1.  Then, for f∈C[-1,1]1, we have
(1.11)(∫-11|f′(x)-Qn′(f,x)|p(1-x2)αdx)1/p≤CEn-2(f′).By Theorem A and Theorem1.1, we know that Qn+2(f,x) have better convergence rate than Ln(f,x) in the case -1≤α≤(p/2)-1. But for continuous function approximation, we noticed that Qn have the same approximation order with Ln, that is, if 0<p<+∞,α>-1, then, for f∈C[-1,1], from Hölder inequality [8, 9], it follows that (1.12)(∫-11|f(x)-Ln(f,x)|p(1-x2)αdx)1/p≤CEn-1(f),(∫-11|f(x)-Qn(f,x)|p(1-x2)αdx)1/p≤CEn-1(f).For high derivatives approximation, how the cases are? Secondly, we will consider second derivative approximation byLn and Qn and obtain the following.Theorem 1.2.
LetQn(f,x) and Ln(f,x) be as defined as above. Then, for f∈C[-1,1]2, we have
(1.13)(∫-11|f′′(x)-Qn′′(f,x)|p(1-x2)αdx)1/p≤{CEn-3(f′′),α>p2-1,C(ln⁡n)1/pEn-3(f′′),α=p2-1,Cn1-(2α+2)/pEn-3(f′′),-1<α<p2-1,(∫-11|f′′(x)-Ln′′(f,x)|p(1-x2)αdx)1/p≤{CEn-3(f′′),α>p-1,C(ln⁡n)1/pEn-3(f′′),α=p-1,Cn2-(2α+2)/pEn-3(f′′),-1<α<p-1,
and the estimation for -1<α≤(p/2)-1 or (-1<α≤p-1) is sharp.From Theorem1.2, we know that for the second derivative approximation, Qn have better approximation orders than Ln in the case -1<α≤p-1.Using the same way as in the proof of Theorem1.2, we can consider the r order derivatives approximation for r≥3, but the computation is more complicated, and we omit the detail.


## 2. Some Lemmas

We introduce some lemmas which are the main tools in our proof.Lemma 2.1 (see [10, p. 519]).
Iff∈C[-1,1]r,  then there exists an algebraic polynomial pn(x) of degree at most n such that
(2.1)|f(j)(x)-pn(j)(x)|≤C[1-x2n]r-jEn-r(f(j)),j=0,1,…,r.In the past, the error estimate depended on the Markov-Bernstein type inequalities inLp metric. In this paper, we will use the inequality in L∞ metric.Lemma 2.2 (see [7, p. 50]).
Letφk(x) be as defined by (1.10), α>-1.  Then, for any fixed p>0,
(2.2)(∫-11|∑k=1nAkφk(x)|p(1-x2)αdx)1/p≤Cmax⁡1≤k≤n|Ak|.To prove our results, we need to build another polynomial integral inequality inL∞ metric. For its proof, we introduce two lemmas.Lemma 2.3 (see [8, p. 914]).
Letv1,v2,…,v2N be distinct integers between 1 and n.  Then, we have
(2.3)∫-11lv1(x)lv2(x)⋯lv2N(x)dx1-x2=0,
and it is well known that
(2.4)∑k=1nlk2(x)≤2.Letx1,…,xn be independent variables, s are positive integers, and (2.5)Vs=(∑k=1nxks)1/s.By the mathematical induction we can obtain the following.Lemma 2.4.
IfN is a positive integer, n>2N, then, the homogeneous symmetrical polynomial of degree 2N:
(2.6)B2N=(∑i=1nxi)2N-(2N)!∑k1<k2<⋯<k2Nxk1⋯xk2N,
can be represented as a homogeneous polynomial of degree 2N about V1,…,V2N:
(2.7)B2N=∑t1≤2N-2,ti≥0Bt1⋯t2NV1t1⋯V2Nt2N.Now we give the inequality inL∞ metric which plays a key role in our paper.Lemma 2.5.
Letℓk(x) be as defined by (2.1), α>-1.  Then, for any fixed p>0,
(2.8)(∫-11|∑k=1nAklk(x)|p(1-x2)αdx)1/p≤Cmax⁡1≤k≤n+1|Ak|.Proof.
Firstly, we will consider the special casep=2N,α=-1/2 by induction on N. For N=1, by (2.3) and (2.4), we obtain
(2.9)∫-11|∑k=1nAklk(x)|2dx1-x2=∑k=1nAk2∫-11lk2(x)dx1-x2+2∑k=1n∑j=1k-1AkAj∫-11lk(x)lj(x)dx1-x2≤max⁡1≤k≤n|Ak|2∫-11∑k=1nlk2(x)dx1-x2≤2πmax⁡1≤k≤n|Ak|2.
Suppose that for0<p≤2(N-1), we have
(2.10)∫-11|∑k=1nAklk(x)|pdx1-x2≤Cpmax⁡1≤k≤n|Ak|p.
Forp=2N, if n≤2N, then, (2.4) gives
(2.11)∫-11|∑k=1nAklk(x)|2Ndx1-x2≤π(4N)2Nmax⁡1≤k≤n|Ak|2N.
If n>2N, then by Lemma 2.4, we know
(2.12)|∑k=1nAklk(x)|2N=(2N)!∑k1<k2<⋯<k2NAk1⋯Ak2Nlk1(x)⋯lk2N(x)+∑t1≤2N-2,ti≥0Bt1⋯t2NV1t1(x)⋯V2Nt2N(x)=I1(x)+I2(x),
where
(2.13)Vs(x)=(∑k=1nAkslks(x))1/s.
From (2.3), it follows that
(2.14)∫-11I1(x)dx1-x2=0.
From (2.4), we know that, for s≥2,
(2.15)|Vs(x)|≤max⁡1≤k≤n|Ak|(∑k=1n|lk(x)|s)1/s≤2max⁡1≤k≤n|Ak|.
By virtue of (2.12) and (2.15), we have
(2.16)|∫-11I2(x)dx1-x2|≤∑t1≤2N-2,ti≥0|Bt1⋯t2N|∫-11|V1t1(x)⋯V2Nt2N(x)|dx1-x2≤∑t1≤2N-2,ti≥02N|Bt1⋯t2N|max⁡1≤k≤n|Ak|2N-t1∫-11|V1t1(x)|dx1-x2≤∑t1≤2N-2,ti≥02N|Bt1⋯t2N|(π+∑i=12N-2Ci)max⁡1≤k≤n|Ak|2N.
From (2.11), (2.12), (2.14), and (2.16), it follows that
(2.17)∫-11|∑k=1nAklk(x)|2Ndx1-x2≤C2Nmax⁡1≤k≤n|Ak|2N.
Now we consider the general case. For arbitraryp>0 and α>-1, it is easy to see that we can choose a positive integer N satisfying p/4N<1 and (α+(p/4N))/(1-(p/2N))>-1. By Hölder inequality and (2.17), we can obtain
(2.18)∫-11|∑k=1nAklk(x)|p(1-x2)αdx≤(∫-11|∑k=1nAklk(x)|2Ndx1-x2)p/2N(∫-11(1-x2)(α+(p/4N))/(1-(p/2N))dx)1-(p/2N)≤Cpmax⁡1≤k≤n|Ak|p.Remark 2.6.
P. Erdös and E. Feldheim [8] give a proof for p=2,4 and α=-1/2. We give a mathematical induction proof for completion.


## 3. Proof of Theorem1.1

We will considerQn+2(f,x) instead of Qn(f,x) for simplicity. For f∈C[-1,1]1, let pn+1(x) be the polynomial of degree at most n+1 satisfying (2.1). It is easily checked that for -1≤x≤1,(3.1)f(x)-Qn+2(f,x)=f(x)-pn+1(x)+Qn+2(pn+1-f,x).
From (3.1), we can conclude that (3.2)f′(x)-Qn+2′(f,x)=f′(x)-pn+1′(x)+Qn+2′(pn+1-f,x)=I1(x)+I2(x).
From (2.1), we can derive (3.3)∫-11|I1(x)|p(1-x2)αdx≤CEnp(f′)∫-11(1-x2)αdx≤CEnp(f′).
It is easy to see that I2(x) is a polynomial of degree at most n. Hence, (3.4)I2(x)=∑k=1n(pn+1(xk)-f(xk))φk′(x)=Ln+1(I2,x)=∑s=1n+1[∑k=1n(pn+1(xk)-f(xk))φk′(ts)]ls(x).By a direct computation, we know(3.5)φk′(ts)=(-1)k+s+11-ts2(n+1)(ts-xk)2+(-1)k+s+1ts(n+1)1-ts2(ts-xk).
Combining (3.4) and (3.5), we derive (3.6)I2(x)=∑s=1n+1[∑k=1n(pn+1(xk)-f(xk))(-1)k+s+11-ts2(n+1)(ts-xk)2]ls(x)+∑s=1n+1[∑k=1n(pn+1(xk)-f(xk))(-1)k+s+1ts(n+1)1-ts2(ts-xk)]ls(x)=J1(x)+J2(x).
We consider J1(x) first. For an arbitrary 1≤s≤n+1, (3.7)|∑k=1n(pn+1(xk)-f(xk))(-1)k+s+11-ts2(n+1)(ts-xk)2|≤CEn(f′)n2∑k=1n1-xk21-ts2(ts-xk)2≤CEn(f′)n2(∑k=1n1-ts2(ts-xk)2+∑k=1n1-xk2(ts-xk)2).
Similar to [9, p. 71], we have (3.8)∑k=1n(1-x2)Un2(x)(x-xk)2=(1-x2)[(Un′(x))2-Un(x)Un′′(x)],∑k=1n(1-xk2)Un2(x)(x-xk)2=∑k=1n(1-x2)Un2(x)(x-xk)2+2xUn(x)Un′(x)-nUn2(x).
By [9, p. 71], we know (3.9)(1-x2)Un′(x)=xUn(x)-(n+1)Tn+1(x),(3.10)(1-x2)Un′′(x)=3xUn′(x)-n(n+2)Un(x).
Let x=ts, then by (3.8), (3.9), and (3.10), we obtain (3.11)∑k=1n1-ts2(ts-xk)2=n(n+2)-2ts21-ts2≤n(n+2),(3.12)∑k=1n1-xk2(ts-xk)2=n2+n.
From (3.7), (3.11), and (3.12), we obtain that for an arbitrary 1≤s≤n+1, (3.13)|∑k=1n(pn+1(xk)-f(xk))(-1)k+s+11-ts2(n+1)(ts-xk)2|≤CEn(f′).
From (2.8) and (3.13), we can obtain (3.14)(∫-11|J1(x)|p(1-x2)αdx)1/p≤CEn(f′).
Now we consider J2(x). Exchanging the summation order, we have (3.15)J2(x)=∑k=1n(-1)k(pn+1(xk)-f(xk))n+1[∑s=1n+1(-1)s+1ts1-ts2(ts-xk)ls(x)]=∑k=1n(-1)k(pn+1(xk)-f(xk))(n+1)2[∑s=1n+1tsTn+1(x)(ts-xk)(x-ts)].
It is easy to know (3.16)∑s=1n+1Tn+1(x)x-ts=Tn+1′(x)=(n+1)Un(x).
Let x=xk, then, we have (3.17)∑s=1n+11xk-ts=0.
By (3.16), (3.17), and the identity (3.18)tsTn+1(x)(ts-xk)(x-ts)=xTn+1(x)(x-xk)(x-ts)-xkTn+1(x)(x-xk)(xk-ts),
we conclude that (3.19)∑s=1n+1tsTn+1(x)(ts-xk)(x-ts)=(n+1)xUn(x)x-xk.
From (3.15) and (3.19), it follows that (3.20)J2(x)=∑k=1n(-1)k(pn+1(xk)-f(xk))n+1⋅xUn(x)x-xk.
For an arbitrary 1≤k≤n, by (3.20), (2.1), |Un′(xk)|=(n+1)/(1-xk2),k=1,2,…,n, and a simple computation, we can obtain (3.21)|J2(xk)|=|(-1)k(pn+1(xk)-f(xk))n+1⋅xkUn′(xk)|≤CEn(f′).
For k=0, by (2.1), Un(1)=n+1 and a simple computation we obtain (3.22)|J2(1)|=|∑k=1n(-1)k(pn+1(xk)-f(xk))1-xk|≤CEn(f′)n∑k=1n11-xk.
From 2x/π≤sinx≤x,forallx∈[0,π/2], we derive (3.23)∑k=1n11-xk=∑k=1n12sin⁡kπ/2(n+1)≤∑k=1nn+1k≤Cnln⁡n.
Hence, (3.24)|J2(1)|≤Cln⁡nEn(f′).
Similarly, (3.25)|J2(-1)|≤Cln⁡nEn(f′).
The fact that J2(x) is an algebraic polynomial of degree at most n implies (3.26)J2(x)=Qn+2(J2,x)=J2(1)φ0(x)+J2(-1)φn+1(x)+∑k=1nJ2(xk)φk(x).
Let x=cosθ. By (3.24) and a simple computation similar to [11, p. 204], we obtain that, for p>0 and α>-1, (3.27)∫-11|J2(1)φ0(x)|p(1-x2)αdx≤Cln⁡pnEnp(f′)(n+1)p∫0π|sin⁡nθ|psin⁡p-2α-1θdθ≤CEnp(f′).
Similarly, (3.28)∫-11|J2(-1)φn+1(x)|p(1-x2)αdx≤CEnp(f′).
By virtue of (2.2) and (3.21), we have (3.29)∫-11|∑k=1nJ2(xk)φk(x)|p(1-x2)αdx≤CEnp(f′).
From (3.26), (3.27), (3.28), and (3.29), it follows that (3.30)(∫-11|J2(x)|p(1-x2)αdx)1/p≤CEn(f′).
By (3.2), (3.3), (3.6), (3.14), and (3.30), we obtain the upper estimate.


## 4. Proof of Theorem1.2

We considerQn first. We will consider Qn+2(f,x) instead of Qn(f,x) for simplicity. For f∈C[-1,1]2, let pn+1(x) be the polynomial of degree at most n+1 satisfying (2.1). From (3.1), it follows that (4.1)f′′(x)-Qn+2′′(f,x)=f′′(x)-pn+1′′(x)+Qn+2′′(pn+1-f,x)=M1(x)+M2(x).From (2.1), we can derive (4.2)∫-11|M1(x)|p(1-x2)αdx≤CEn-1p(f′′).Similar to (3.4), (4.3)M2(x)=∑s=1n+1[∑k=1n(pn+1(xk)-f(xk))φk′′(ts)]ls(x).By a direct computation, we get(4.4)φk′′(ts)=2(-1)k+s1-ts2(n+1)(ts-xk)3+2(-1)k+sts(n+1)1-ts2(ts-xk)2+(n+1)(-1)k+s+11-ts2(ts-xk)+(-1)k+s+1(n+1)(1-ts2)3/2(ts-xk).Equations (4.3) and (4.4) yield (4.5)M2(x)=∑s=1n+1[∑k=1n(pn+1(xk)-f(xk))2(-1)k+s1-ts2(n+1)(ts-xk)3]ls(x)+∑s=1n+1[∑k=1n(pn+1(xk)-f(xk))2(-1)k+sts(n+1)1-ts2(ts-xk)2]ls(x)+∑s=1n+1[∑k=1n(pn+1(xk)-f(xk))(n+1)(-1)k+s+11-ts2(ts-xk)]ls(x)+∑s=1n+1[∑k=1n(pn+1(xk)-f(xk))(-1)k+s+1(n+1)(1-ts2)3/2(ts-xk)]ls(x)=N1(x)+N2(x)+N3(x)+N4(x).
We consider N1(x) now. For an arbitrary 1≤s≤n+1, from (2.1), (3.12), and ∑k=1n|φk(x)|2≤2 (see [9]), it follows that (4.6)|∑k=1n(pn+1(xk)-f(xk))2(-1)k+s1-ts2(n+1)(ts-xk)3|≤CEn-1(f′′)(n+1)3∑k=1n(1-xk2)1-ts2|ts-xk|3=CEn-1(f′′)(n+1)2∑k=1n(1-xk2)|φk(ts)||ts-xk|2≤CEn-1(f′′)(n+1)2∑k=1n(1-xk2)|ts-xk|2≤CEn-1(f′′).
From (2.8) and (4.6), we can obtain (4.7)(∫-11|N1(x)|p(1-x2)αdx)1/p≤CEn-1(f′′).
Now we consider N2(x). From 2x/π≤sinx≤x,forallx∈[0,π/2], it follows that 1-ts2≥sin(π/2(n+1))≥1/(n+1). By (2.1) and (3.12), we have that, for an arbitrary 1≤s≤n+1, (4.8)|∑k=1n(pn+1(xk)-f(xk))2(-1)k+sts(n+1)1-ts2(ts-xk)2|≤CEn-1(f′′)(n+1)31-ts2∑k=1n1-xk2(ts-xk)2≤CEn-1(f′′).
From (2.8) and (4.8), we can obtain (4.9)(∫-11|N2(x)|p(1-x2)αdx)1/p≤CEn-1(f′′).
For the N3(x), similar to (3.15), we have (4.10)N3(x)=∑k=1n(-1)k(pn+1(xk)-f(xk))[∑s=1n+1Tn+1(x)(ts-xk)(x-ts)]=(n+1)∑k=1n(-1)k(pn+1(xk)-f(xk))Un(x)x-xk.
For an arbitrary 1≤k≤n, by (2.1) and a simple computation, we can obtain (4.11)|N3(xk)|=(n+1)|(pn+1(xk)-f(xk))⋅Un′(xk)|≤CEn-1(f′′).
For k=0, (2.1) leads to (4.12)|N3(1)|=(n+1)2|∑k=1n(-1)k(pn+1(xk)-f(xk))1-xk|≤CnEn-1(f′′).
Similarly, (4.13)N3(-1)≤CnEn-1(f′′).
Similar to (3.30), from (4.10), (4.11), (4.12), and (4.13), it follows that (4.14)(∫-11|N3(x)|p(1-x2)αdx)1/p≤{CEn-1(f′′),α>p2-1,C(ln⁡n)1/pEn-1(f′′),α=p2-1,Cn1-(2α+2)/pEn-1(f′′),-1<α<p2-1.
For the N4(x), similar to (3.15), we have (4.15)N4(x)=∑k=1n(-1)k(pn+1(xk)-f(xk))(n+1)2[∑s=1n+1Tn+1(x)(1-ts2)(ts-xk)(x-ts)].
It is easy to verify (4.16)Tn+1(x)(1-ts2)(ts-xk)(x-ts)=1x-xk[Tn+1(x)(1-ts2)(x-ts)+Tn+1(x)(1-ts2)(ts-xk)].
For a≠±1, it is easy to verify that (4.17)1(1-x2)(x-a)=-12(1+a)(1+x)+12(1-a)(1-x)+1(1-a2)(x-a).
From (4.17), (3.16) and (4.18)∑s=1n+111+ts=∑s=1n+111-ts=(n+1)2,we obtain (4.19)∑s=1n+1Tn+1(x)(1-ts2)(x-ts)=Tn+1(x)2(1+x)∑s=1n+111+ts-Tn+1(x)2(1-x)∑s=1n+111-ts+11-x2∑s=1n+1Tn+1(x)x-ts=-(n+1)2xTn+1(x)1-x2+(n+1)Un(x)1-x2,∑s=1n+1Tn+1(x)(1-ts2)(ts-xk)=(n+1)2xkTn+1(x)1-xk2.
From (4.16), (4.19), (61), (3.9), and a direct computation, we get (4.20)∑s=1n+1Tn+1(x)(1-ts2)(ts-xk)(x-ts)=-(n+1)2(1+xxk)Tn+1(x)(1-x2)(1-xk2)+(n+1)Un(x)(1-x2)(x-xk)=(n+1)(1+xxk)Un′(x)1-xk2+(n+1)xkUn(x)1-xk2+(n+1)Un(x)(1-xk2)(x-xk).
From (4.15) and (4.20), we obtain (4.21)N4(x)=∑k=1n(-1)k(pn+1(xk)-f(xk))(n+1)[(1+xxk)Un′(x)1-xk2+xkUn(x)1-xk2]+∑k=1n(-1)k(pn+1(xk)-f(xk))(n+1)Un(x)(1-xk2)(x-xk)=N41(x)+N42(x).
For N41(x), from (2.1), we can obtain (4.22)|N41(x)|≤|Un′(x)|+|Un(x)|(n+1)2En-1(f′′).
By (3.9), Markov inequality, and ∥Un(x)∥∞=n+1, we obtain (4.23)|Un′(x)|≤2(n+1)1-x2,|Un′(x)|≤n2(n+1).
So for an arbitrary 0≤A≤1, (4.24)|Un′(x)|≤21-A(n+1)1+2A(1-x2)1-A≤8n1+2A(1-x2)1-A.
Let x=cosθ. From 2α+1>-1, we can choose A such that 0<A<1 and 2α+1-2p+2pA>-1. Then by (4.24) and 2x/π≤sinx≤x,forallx∈[0,π/2], we can obtain (4.25)∫-11|Un′(x)|p(1-x2)αdx=2∫-10|Un′(x)|p(1-x2)αdx≤23p+1(∫0π/2(n+1)np(1+2A)sin⁡2α+1-2p+2pAθdθ+∫π/2(n+1)π/2npsin⁡1+2α-2pθdθ)≤{Cnp,α>p-1,Cnpln⁡⁡n,α=p-1,Cn3p-2α-2,-1<α<p-1.
From ∥Un(x)∥∞=n+1, it follows that (4.26)∫-11|Un(x)|p(1-x2)αdx≤Cnp.
From (4.22), (4.25), and (4.26), it follows that (4.27)(∫-11|N41(x)|p(1-x2)αdx)1/p≤{CEn-1(f′′),α≥p2-1,Cn1-(2α+2)/pEn-1(f′′),-1<α<p2-1.
For N42(x), from (2.1) and a simple computation, we can obtain that, for 1≤k≤n, (4.28)|N42(xk)|=1(n+1)(1-xk2)|(pn+1(xk)-f(xk))⋅Un′(xk)|≤CEn-1(f′′).
Let x=1. Then, from (3.10) and (4.29)∑s=1n-1Un(x)x-xs=Un′(x),
we obtain (4.30)∑k=1n11-xk=n(n+2)3.
From (2.1), it follows that (4.31)|N42(1)|=|∑k=1n(-1)k(pn+1(xk)-f(xk))(1-xk)(1-xk2)|≤CEn-1(f′′)(n+1)2∑k=1n11-xk≤CEn-1(f′′).
Similarly, (4.32)|N42(-1)|≤CEn-1(f′′).
Similar to (3.30), from (4.28), (4.31), and (4.32), we can obtain (4.33)(∫-11|N42(x)|p(1-x2)αdx)1/p≤CEn-1(f′′).
From (4.1), (4.2), (4.5), (4.7), (4.9), (4.14), (4.21), (4.27), and (4.33), we obtain the upper estimate.On the other hand, forp≥2α+2, let f(x)=(1-x2)Un(x). Then,(4.34)f′′(x)=-2(n+2)(n+1)Tn(x)+qn-1(x),
here, qn-1(x) is a polynomial of degree at most n-1. Hence, (4.35)En-1(f′′)=2(n+2)(n+1).
It is easy to verify that (4.36)Qn+2′′(f,x)=0,f′′(x)=-(n+1)2Un(x)+-Un(x)+(n+1)xTn+1(x)1-x2.Letx=cosθ, then, (2kπ+π)/2(n+1)≤θ≤(2kπ+2π)/2(n+1) implies that Tn+1(x)Un(x)≤0. Therefore, (4.37)∫-11|f′′(x)-Qn′′(f,x)|p(1-x2)αdx≥∫01|f′′(x)-Qn′′(f,x)|p(1-x2)αdx≥n2p∑k=0[(n+1)/2]∫2kπ/2(n+1)(2kπ+π)/2(n+1)|sin⁡(n+1)θ|psin⁡p-2α-1θdθ≥{Cn2pln⁡⁡n,α=p2-1,Cn3p-2α-2,-1<α<p2-1,≥{Cln⁡⁡nEn-1p(f′′),α=p2-1,Cnp-2α-2En-1p(f′′),-1<α<p2-1,We considerLn in the following. For f∈C[-1,1]2, let pn-1(x) be the polynomial of degree at most n-1 satisfying (2.1). Then, (4.38)f′′(x)-Ln′′(f,x)=f′′(x)-pn-1′′(x)+Ln′′(pn-1-f,x)=K1(x)+K2(x).
From (2.1), we can derive (4.39)∫-11|K1(x)|p(1-x2)αdx≤CEn-3p(f′′).
If f∈C[-1,1], then the well-known Lagrange interpolation polynomial of f based on {xk}k=1n is given by (4.40)Rn(f,x)=∑k=1nf(xk)ϕk(x),
where (4.41)ϕk(x)=(-1)k+1(1-xk2)Un(x)(n+1)(x-xk),k=1,…,n.
Similar to (3.4), we have (4.42)K2(x)=Rn-1(K2,x)=∑s=1n-1[∑k=1n(pn-1(tk)-f(tk))lk′′(xs)]ϕs(x).
By a direct computation, we obtain (4.43)lk′′(xs)=n(-1)k+s1-tk2(1-xs2)(xs-tk)+2(-1)k+s+11-tk2n(xs-tk)3,s=1,…,n-1.
From (4.42) and (4.43), it follows that (4.44)K2(x)=∑s=1n-1[∑k=1n(pn-1(tk)-f(tk))n(-1)k+s1-tk2(1-xs2)(xs-tk)]ϕs(x)+∑s=1n-1[∑k=1n(pn-1(tk)-f(tk))2(-1)k+s+11-tk2n(xs-tk)3]ϕs(x)=A1(x)+A2(x).
Exchanging the summation order, we have (4.45)A1(x)=∑k=1n(-1)k+1(pn-1(tk)-f(tk))1-tk2[∑s=1n-1Un-1(x)(xs-tk)(x-xs)].
For an arbitrary 1≤s≤n-1, (4.46)Un-1(x)(xs-tk)(x-xs)=1x-tk(Un-1(x)xs-tk+Un-1(x)x-xs).
It is well known that (4.47)∑s=1n-1Un-1(x)x-xs=Un-1′(x).
Let x=tk. Then, from (4.47) and (3.9), it follows that (4.48)∑s=1n-11tk-xs=Un-1′(tk)Un-1(tk)=tk1-tk2.
Combining (4.47), (4.48), and (3.9), we obtain (4.49)∑s=1n-11x-tk(Un-1(x)xs-tk+Un-1(x)x-xs)=1x-tk-tkUn-1(x)+(1-tk2)Un-1′(x)1-tk2=Un-1(x)+(x+tk)Un-1′(x)1-tk2+1x-tk-xUn-1(x)+(1-x2)Un-1′(x)1-tk2=Un-1(x)+(x+tk)Un-1′(x)1-tk2-nTn(x)(x-tk)(1-tk2).
From (4.45) and (4.49), it follows that (4.50)A1(x)=∑k=1n(-1)k+1(pn-1(tk)-f(tk))Un-1(x)+(x+tk)Un-1′(x)1-tk2+∑k=1n(-1)k(pn-1(tk)-f(tk))nTn(x)(x-tk)1-tk2=A11(x)+A12(x).
By (2.1), we have (4.51)|A11(x)|≤CEn-3(f′′)(|Un-1′(x)|+|Un(x)|)n.
From (4.51), (4.25), and (4.26), it follows that (4.52)(∫-11|A11(x)|p(1-x2)αdx)1/p≤{CEn-3(f′′),α>p-1,C(ln⁡n)1/pEn-3(f′′),α=p-1,Cn2-(2α+2)/pEn-3(f′′),-1<α<p-1.
From (2.1) and |Tn′(tk)|=n/1-tk2, it follows that, for 1≤k≤n, (4.53)|A12(tk)|=|(-1)k+1(pn-1(tk)-f(tk))nTn′(tk)1-tk2|≤CEn-3(f′′).
From (4.53), (2.8), and (4.54)A12(x)=∑k=1nA12(tk)lk(x),
we know (4.55)(∫-11|A12(x)|p(1-x2)αdx)1/p≤CEn-3(f′′).We considerA2(x) now. For an arbitrary 1≤s≤n-1, by (2.1) and (2.4), we obtain (4.56)|∑k=1n(pn-1(tk)-f(tk))2(-1)k+s+11-tk2n(xs-tk)3|≤CEn-3(f′′)∑k=1n(1-tk2)3/2n3|xs-tk|3=CEn-3(f′′)∑k=1n|lk3(xs)|≤CEn-3(f′′).
From (4.56), (4.44), and (4.41), we can obtain (4.57)|A2(xs)|≤CEn-3(f′′),1≤s≤n-1.|A2(1)|≤CnEn-3(f′′),|A2(-1)|≤CnEn-3(f′′).
Similar to the proof of (3.30), by (4.57), (95), and (4.58)A2(x)=∑s=0nA2(xs)φs(x),
we can obtain (4.59)(∫-11|A2(x)|p(1-x2)αdx)1/p≤{CEn-1(f′′),α>p2-1,C(ln⁡n)1/pEn-1(f′′),α=p2-1,Cn1-(2α+2)/pEn-1(f′′),-1<α<p2-1.
From (4.38), (4.39), (4.44), (4.50), (4.52), (4.55), and (4.59), we can obtain the upper estimate.On the other hand, letf(x)=Tn(x). Then, it is easy to see (4.60)Ln(f,x)=0,(4.61)f′′(x)=4n(n-1)Tn-2(x)+qn-3(x),
here, qn-3(x) is a polynomial of degree at most n-3. Consequently, due to (4.61), we get (4.62)En-3(f′′)=4n(n-1).
Let x=cosθ. From (4.60), (4.62), (3.9), Tn′′(x)=nUn-1′(x), and the odevity of Un-1′(x), it follows that (4.63)∫-11|f′′(x)-Ln′′(f,x)|p(1-x2)αdx=2En-3p(f′′)4p(n-1)p∫01|Un-1′(x)|p(1-x2)αdx≥En-3p(f′′)8p∑k=1n/2∫(kπ-π)/4/nkπ/n1|sin⁡θ|2p-2α-1dθ≥{CEn-3p(f′′),α>p-1,Cln⁡⁡nEn-3p(f′′),α=p-1,Cn2p-2α-2En-3p(f′′),-1<α<p-1.



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*Source: 503561-2011-11-14.xml*