# Upper Bound for Lebesgue Constant of Bivariate Lagrange Interpolation Polynomial on the Second Kind Chebyshev Points

Authors: Juan Liu; Laiyi Zhu
Journal: Journal of Mathematics
 (2022)

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## Abstract

In the paper, we study the upper bound estimation of the Lebesgue constant of the bivariate Lagrange interpolation polynomial based on the common zeros of product Chebyshev polynomials of the second kind on the square−1,12. And, we prove that the growth order of the Lebesgue constant is On+22. This result is different from the Lebesgue constant of Lagrange interpolation polynomial on the unit disk, the growth order of which is On. And, it is different from the Lebesgue constant of the Lagrange interpolation polynomial based on the common zeros of product Chebyshev polynomials of the first kind on the square −1,12, the growth order of which is Olnn2.


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## Body

## 1. Introduction

Chebyshev polynomials play an important role in modern developments, including orthogonal polynomials, polynomial approximation, numerical integration, and spectral methods for partial differential equations (cf. [1]). Especially, the zeros of Chebyshev polynomials are often used in the studies of one-variable Lagrange interpolation polynomials. Many good approximation properties have been obtained over the past decades (cf. [2]). Since multivariate Lagrange interpolation polynomials are difficult to express concretely, many scholars are interested to study them (cf. [3–15]).LetK⊂Rd be a nonempty compact set and V be a subspace of Πnd, where Πnd denotes the space of polynomials with d variables whose degrees do not exceed n and the dimension dimV=N. Then, based on the nodes X≔xkk=1N⊂K, the Lagrange interpolation problem related to V and X can be described as follows: for any function f∈CK, where CK represents the continuous function space on K, we can find a unique polynomial p∈V to satisfy the equation(1)pxk=fxk,k=1,…,N.This polynomial is the so-called Lagrange interpolation polynomial and can be expressed as(2)Lnf,x=∑k=1Nfxklkx,where lkx are the Lagrange interpolation basis functions that satisfy the following formula:(3)lkxj=δkj.The mappingf⟶Lnf can be regarded as an operator from CK to itself, and the norm of the operator is defined as(4)λn=Ln=maxx∈K∑k=1Nlkx,which is called the Lebesgue constant. We know that the uniform convergence of Lnf,x for f∈CK is closely related to the Lebesgue constant.The univariate Lagrange interpolation polynomial and its Lebesgue constant have been extensively studied (cf. [2, 16]). Specially, for K=−1,1 and V=Πn1, the Lebesgue constant Ln≥Clogn and the order of the Lebesgue constant is Ologn when the Chebyshev points are taken as the nodes (cf. [16]).There are relatively few research results on multivariate Lagrange interpolation polynomials. In [3], from Berman’s Theorem, it is shown that for K=Bd, the unit ball in Rd,d≥2, and V=∏nd, the order of the Lebesgue constant is Ond−1/2.It is well known that the Lagrange interpolation polynomial is closely related to cubature formula. Möller (cf. [4]) stated that for centrally symmetric weight functions, the node number of cubature formula satisfies(5)N≥dimΠn−12+n2=n+12+n2,and it is the so-called minimal cubature formula if the number of nodes reaches the lower bound. In [5], Xu studied the relationship between the compact cubature formula and the Lagrange interpolation polynomial. By using this relationship, Xu in [6] established the quadrature formula and the Lagrange interpolation polynomial on K=−1,12, based on the common zeros of the product Chebyshev polynomial of the first kind, which are called minimal cubature formula and Xu-type Lagrange interpolation polynomial on the first kind Chebyshev polynomial. Moreover, for 0<p≤∞, the mean convergence of the interpolation polynomial is also obtained.Bos et al. [7] gave the numerical study of the upper bound of Lebesgue constant of the Xu-type Lagrange interpolation polynomial on the first kind Chebyshev polynomial, the order of which lies in lnn2, and they gave detailed proof of the order in [8]. And, Vecchia et al. [9] gave that the order of the lower bound estimate is lnn2.In [10], for K=−1,12, we gave the compact formulae of the cubature formula and the Lagrange interpolation polynomial based on the common zeros of product Chebyshev polynomials of the second kind, which are called minimal cubature formula and Xu-type Lagrange interpolation polynomial on the second kind Chebyshev polynomial. Furthermore, for 0<p≤2, we studied the mean convergence of the Lagrange interpolation polynomials.In this paper, we study the growth order of the Lebesgue constant and provide a direct elementary proof.Theorem 1.
ForK=−1,12, the upper bound estimate of the Lebesgue constant of Xu-type Lagrange interpolation polynomial on the second kind Chebyshev polynomial in [10] is(6)λn≤1602n+22.Our result gives that the growth order of the Lebesgue constant of Xu-type Lagrange interpolation polynomial on the second kind Chebyshev polynomial on the square−1,12 is On+22. Obviously, it is different from the Lebesgue constant on the disk B2, the growth order of which is On, and is different from the Lebesgue constant of Xu-type Lagrange interpolation polynomial on the first kind Chebyshev polynomial on −1,12, the growth order of which is Olnn2.


## 2. The Lebesgue Constant of Xu-Type Lagrange Interpolation Polynomial on the Second Kind Chebyshev Polynomial

In order to prove Theorem1, by using reproducing kernel, we give the expression of the Lebesgue constant λn in this section.First, we briefly introduce the Xu-type Lagrange interpolation polynomial on the second kind Chebyshev polynomial in [10].Letℕ0 denote the set of nonnegative integers. For n∈ℕ0, Chebyshev polynomial of the second kind Unx (cf. [17]) is defined by(7)Unx=sinn+1θsinθ,x=cosθ,θ∈0,π,and they are orthogonal polynomials with respect to the second kind Chebyshev weight w1x=2/π1−x2.The product Chebyshev polynomial of the second kind of degreen on −1,12 (cf. [5]) is defined by(8)Pknx,y=Un−kxUky,x,y∈−1,12,k=0,1,…,n,n∈ℕ0,and correspondingly, the product Chebyshev weight function of the second kind is(9)W1x,y=w1xw1y=4π21−x21−y2,x,y∈−1,12.Forx,y∈−1,12, the reproducing kernel of the product Chebyshev polynomials is defined by(10)Knx,y=∑k=0n−1∑j=0kPjkxPjky,n=1,2,….LetX=xk,l|k,l=2i,2j+1,i=1,2,…,n+1/2,j=0,1,…,n/2∪xk,l|k,l=2i+1,2j,i=0,1,…,n/2,j=1,2,…,n+1/2, where xk,l=zk,zl, zk=cosθk,θk=kπ/n+2,k=0,1,…,n+2, be nodes; the Xu-type Lagrange interpolation polynomial on the second kind Chebyshev polynomial is(11)Lnf,x=∑k=1Nfxklkx=∑k=1NfxkKn∗x,xkKn∗xk,xk=∑i=1n+1/2∑j=0n/2λ2i,2j+1fx2i,2j+1Kn∗x,x2i,2j+1+∑i=0n/2∑j=1n+1/2λ2i+1,2jfx2i+1,2jKn∗x,x2i+1,2j,where x=x,y∈−1,12, Kn∗x,y=1/2Kn+1x,y+Knx,y,(12)λ2i,2j+1=Kn∗x2i,2j+1,x2i,2j+1−1=8sinθ2isinθ2j+12n+22,λ2i+1,2j=Kn∗x2i+1,2j,x2i+1,2j−1=8sinθ2i+1sinθ2j2n+22,i,j=0,1,…,n+12.Obviously, the node numberN of formula (11) is 2n+1/2n/2+1. When n=2m, N=nn+2/2, which reaches the lower bound dimΠn−12+n/2. When n=2m−1, N=n+12/2, which is one more than the lower bound (cf. [4]).(13)λn=maxx∈−1,12∑k=1NKn∗x,xkKn∗xk,xkis called Lebesgue constant of Xu-type Lagrange interpolation polynomial on the second kind Chebyshev polynomial. Writing(14)λn=maxx∈−1,12∑i=1n+1/2∑j=0n/2Kn∗x,x2i,2j+1Kn∗x2i,2j+1,x2i,2j+1+∑i=0n/2∑j=1n+1/2Kn∗x,x2i+1,2jKn∗x2i+1,2j,x2i+1,2j,(15)Λn1x≔∑i=1n+1/2∑j=0n/2Kn∗x,x2i,2j+1Kn∗x2i,2j+1,x2i,2j+1,Λn2x≔∑i=0n/2∑j=1n+1/2Kn∗x,x2i+1,2jKn∗x2i+1,2j,x2i+1,2j,Λnx≔Λn1x+Λn2x,then(16)Λn=maxx∈−1,12Λnx=maxx∈−1,12Λn1x+Λn2x.The expression ofKn∗x,y given in [10] is(17)Kn∗x,y=18sinφ1sinφ2sinϕ1sinϕ2∑k=12∑l=12εiεjFnφ1+εkϕ1,φ2+εlϕ2+Fn+1φ1+εkϕ1,φ2+εlϕ2,where εk=−1k,k=1,2,x=cosφ1,cosφ2,andy=cosϕ1,cosϕ2. Then, (18)Fnτ1,τ2=cosn+2τ1−cosn+2τ2+cosn+1τ1−cosn+1τ24cosτ1−cosτ2.Lemma 1.
Ifϕ1=θ2iandϕ2=θ2j+1,i=1,2,…,n+1/2andj=0,1,…,n/2, then(19)sinn+2φ1+εkϕ1+φ2+εlϕ22=−1iεk+jεlεlcosn+2φ1+φ22,(20)sinn+2φ1+εkϕ1−φ2−εlϕ22=−1iεk−jεl−εlcosn+2φ1−φ22.
Ifϕ1=θ2i+1andϕ2=θ2j,i=0,1,2,…,n/2andj=1,2,…,n+1/2, then(21)sinn+2φ1+εkϕ1+φ2+εlϕ22=−1iεk+jεlεkcosn+2φ1+φ22,sinn+2φ1+εkϕ1−φ2−εlϕ22=−1iεk−jεlεkcosn+2φ1−φ22,where εk=−1k,k=1,2,andεl=−1l,l=1,2.
We only prove formula (19); other formulae can be proved similarly. For ϕ1=θ2i,ϕ2=θ2j+1,εk=−1k,andεl=−1l, we have(22)cosn+2εkϕ1+εlϕ22=0.
Hence,(23)sinn+2φ1+εkϕ1+φ2+εlϕ22=cosn+2φ1+φ22siniεk+jεlπ+π2εl=−1iεk+jεlεlcosn+2φ1+φ22.Lemma 2.
Ifϕ1=θ2iandϕ2=θ2j+1,i=1,2,…,n+1/2andj=0,1,…,n/2, then(24)∑k=12∑l=12εkεlFnφ1+εkϕ1,φ2+εlϕ2+Fn+1φ1+εkϕ1,φ2+εlϕ2=∑k=12∑l=12εkεlcos2φ1+εkϕ12Un+1φ1+εkϕ1+φ2+εlϕ22×Un+1φ1+εkϕ1−φ2−εlϕ22=∑k=12∑l=12εkεl2cos2φ1+εkϕ1/2cosn+2φ1+φ2/2cosn+2φ1−φ2/2cosφ1+εkϕ1−cosφ2+εlϕ2.
Ifϕ1=θ2i+1andϕ2=θ2j,i=0,1,2,…,n/2andj=1,2,…,n+1/2, then(25)∑k=12∑l=12εkεlFnφ1+εkϕ1,φ2+εlϕ2+Fn+1φ1+εkϕ1,φ2+εlϕ2=∑k=12∑l=12εkεlcos2φ1+εkϕ12Un+1φ1+εkϕ1+φ2+εlϕ22×Un+1φ1+εkϕ1−φ2−εlϕ22=−∑k=12∑l=12εkεl2cos2φ1+εkϕ1/2cosn+2φ1+φ2/2cosn+2φ1−φ2/2cosφ1+εkϕ1−cosφ2+εlϕ2.where εk=−1k,k=1,2,andεl=−1l,l=1,2.
By (18), we have(26)Fnτ1,τ2+Fn+1τ1,τ2=−cosτ1/2cos2n+3/2τ1+cos2n+5/2τ1−cosτ2/2cos2n+3/2τ2+cos2n+5/2τ24sinτ1+τ2/2sinτ1−τ2/2=−cosn+2τ1cos2τ1/2−cosn+2τ2cos2τ2/22sinτ1+τ2/2sinτ1−τ2/2=−cosn+2τ1−cosn+2τ2cos2τ1/2+cosn+2τ2cos2τ1/2−cos2τ2/22sinτ1+τ2/2sinτ1−τ2/2=cos2τ12Un+1τ1+τ22Un+1τ1−τ22+12cosn+2τ2.
From (19) and (20), we can obtain (24). And, we can similarly prove (25).Lemma 3.
The following relation holds:(27)I=∑k=12∑l=12εkεlcos2φ1+εkϕ1/2cosφ1+εkϕ1−cosφ2+εlϕ2=4sinφ1sinϕ1sinφ2sinϕ2,−cos2φ1−ϕ1/2cosφ1+ϕ1−cosφ2−ϕ2−cos2φ2−ϕ2/2cosφ1−ϕ1−cosφ2+ϕ2∏k=12∏l=12cosφ1+εkϕ1−cosφ2+εlϕ2,where εk=−1k,k=1,2,andεl=−1l,l=1,2. We have(28)I=∑k=12∑l=12εkεlcos2φ1+εkϕ1/2cosφ1+εkϕ1−cosφ2+εlϕ2=cos2φ1+ϕ12⋅−cosφ2−ϕ2+cosφ2+ϕ2cosφ1+ϕ1−cosφ2+ϕ2cosφ1+ϕ1−cosφ2−ϕ2−cos2φ1−ϕ12⋅−cosφ2−ϕ2+cosφ2+ϕ2cosφ1−ϕ1−cosφ2+ϕ2cosφ1−ϕ1−cosφ2−ϕ2=−2sinφ2sinϕ2cos2φ1+ϕ1/2cosφ1+ϕ1−cosφ2+ϕ2cosφ1+ϕ1−cosφ2−ϕ2−cos2φ1−ϕ1/2cosφ1−ϕ1−cosφ2+ϕ2cosφ1−ϕ1−cosφ2−ϕ2=2sinφ1sinϕ1sinφ2sinϕ21cosφ1+ϕ1−cosφ2+ϕ2cosφ1+ϕ1−cosφ2−ϕ2−4cos2φ1−ϕ1/2cosφ1cosϕ1−cosφ2cosϕ2∏k=12∏l=12cosφ1+εkϕ1−cosφ2+εlϕ2=4sinφ1sinϕ1sinφ2sinϕ2,−cos2φ1−ϕ1/2cosφ1+ϕ1−cosφ2−ϕ2−cos2φ2−ϕ2/2cosφ1−ϕ1−cosφ2+ϕ2∏k=12∏l=12cosφ1+εkϕ1−cosφ2+εlϕ2.
By Lemmas2 and 3, the following result can be obtained.Lemma 4.
Ifx=cosφ1,cosφ2,y=cosϕ1,cosϕ2,ϕ1=θ2i,andϕ2=θ2j+1,i=1,2,…,n+1/2andj=0,1,…,n/2, then(29)Kn∗x,y=cosn+2φ1+φ2/2cosn+2φ1−φ2/24sinφ1sinφ2sinϕ1sinϕ2I=cosn+2φ1+φ22cosn+2φ1−φ22,−cos2φ1−ϕ1/2cosφ1+ϕ1−cosφ2−ϕ2−cos2φ2−ϕ2/2cosφ1−ϕ1−cosφ2+ϕ2∏k=12∏l=12cosφ1+εkϕ1−cosφ2+εlϕ2.
Ifϕ1=θ2i+1andϕ2=θ2j,i=0,1,2,…,n/2andj=1,2,…,n+1/2, then(30)Kn∗x,y=−cosn+2φ1+φ2/2cosn+2φ1−φ2/24sinφ1sinφ2sinϕ1sinϕ2I=cosn+2φ1+φ22cosn+2φ1−φ22⋅cos2φ1−ϕ1/2cosφ1+ϕ1−cosφ2−ϕ2+cos2φ2−ϕ2/2cosφ1−ϕ1−cosφ2+ϕ2∏k=12∏l=12cosφ1+εkϕ1−cosφ2+εlϕ2.
Furthermore, we can obtain the following lemma.Lemma 5.
Letx=cosφ1,cosφ2andy=cosϕ1,cosϕ2. If ϕ1=θ2iandϕ2=θ2j+1,i=1,2,…,n+1/2andj=0,1,…,n/2, then(31)Λn1x=∑i=1n+1/2∑j=0n/2sinϕ1sinϕ22n+22cosn+2φ1+φ22cosn+2φ1−φ22cos2φ1−ϕ1/2sinφ1+ϕ1+φ2+ϕ2/2sinφ1+ϕ1−φ2−ϕ2/2sinφ1−ϕ1+φ2+ϕ2/2sinφ1−ϕ1−φ2−ϕ2/2×1sinφ1−ϕ1+φ2−ϕ2/2sinφ1−ϕ1−φ2+ϕ2/2+cos2φ2−ϕ2/2sinφ1+ϕ1+φ2+ϕ2/2sinφ1+ϕ1−φ2−ϕ2/2×1sinφ1+ϕ1+φ2−ϕ2/2sinφ1+ϕ1−φ2+ϕ2/2sinφ1−ϕ1+φ2−ϕ2/2sinφ1−ϕ1−φ2+ϕ2/2.
Ifϕ1=θ2i+1andϕ2=θ2j,i=0,1,…,n/2andj=1,2…,n+1/2, then(32)Λn2x=∑i=0n/2∑j=1n+1/2sinϕ1sinϕ22n+22cosn+2φ1+φ22cosn+2φ1−φ22cos2φ1−ϕ1/2sinφ1+ϕ1+φ2+ϕ2/2sinφ1+ϕ1−φ2−ϕ2/2sinφ1−ϕ1+φ2+ϕ2/2sinφ1−ϕ1−φ2−ϕ2/2×1sinφ1−ϕ1+φ2−ϕ2/2sinφ1−ϕ1−φ2+ϕ2/2+cos2φ2−ϕ2/2sinφ1+ϕ1+φ2+ϕ2/2sinφ1+ϕ1−φ2−ϕ2/2×1sinφ1+ϕ1+φ2−ϕ2/2sinφ1+ϕ1−φ2+ϕ2/2sinφ1−ϕ1+φ2−ϕ2/2sinφ1−ϕ1−φ2+ϕ2/2.


## 3. Proof of Theorem 1

The proof of Theorem1 is given in this section. And, since the estimates of Λn1x and Λn2x are similar, we need to only estimate Λn1x.Settingτ1=φ1+φ2/2∈0,π,τ2=φ1−φ2/2∈−π/2,π/2, we have(33)Kn∗x,x2i,2j+1Kn∗x2i,2j+1,x2i,2j+1≔Ai,j1+Bi,j1≔Ai,j11Ai,j12+Bi,j11Bi,j12,where(34)Ai,j11=sinθ2isinθ2j+1n+2⋅cosn+2τ2cosφ1−θ2i/2sinτ2+θ2i−θ2j+1/2sinτ2−θ2i+θ2j+1/2sinτ2−θ2i−θ2j+1/2,Ai,j12=sinθ2isinθ2j+1n+2⋅cosn+2τ1cosφ1−θ2i/2sinτ1+θ2i+θ2j+1/2sinτ1−θ2i−θ2j+1/2sinτ1−θ2i+θ2j+1/2,Bi,j11=sinθ2isinθ2j+1n+2⋅cosn+2τ2cosφ2−θ2j+1/2sinτ2+θ2i−θ2j+1/2sinτ2+θ2i+θ2j+1/2sinτ2−θ2i−θ2j+1/2,Bi,j12=sinθ2isinθ2j+1n+2⋅cosn+2τ1cosφ2−θ2j+1/2sinτ1+θ2i+θ2j+1/2sinτ1+θ2i−θ2j+1/2sinτ1−θ2i+θ2j+1/2.To prove Theorem1, we first prove some lemmas.Lemma 6.
Ifτ2∈0,θ1/2 and τ1∈0,θ1/2∪π−θ1/2,π, we have the following:(1)
For0≤j≤i−2 or i+1≤j≤n/2,(35)Ai,j11≤4sinθ2isinθ2j+1cosn+2τ2n+2sin2θ2i−θ2j+1/2.Forj=i or j=i−1,(36)Ai,j11≤2n+2.(2)
For0≤j≤i−2,(37)Ai,j12≤8cosn+2τ1n+2sinθ2i−θ2j+1/2.Fori+1≤j≤n/2,(38)Ai,j12≤4cosn+2τ1n+2sinθ2i−θ2j+1/2.Forj=i or j=i−1,(39)Ai,j12≤4.(1)
We first consider the case ofτ2∈0,θ1/2.For0≤j≤i−2, since θ2i−θ2j+1/2≤θ2i−θ2j+1/2+τ2≤π/2, we obtain(40)sinτ2+θ2i−θ2j+12≥sinθ2i−θ2j+12.Noticing thatθ2i+θ2j+1/2−θ1/2≤θ2i+θ2j+1/2−τ2≤θ2i+θ2j+1/2≤π and sinx is a convex function on 0,π, we have(41)sinθ2i+θ2j+12−τ2≥minsinθ2i+θ2j+12,sinθ2i+θ2j+12−θ12≥12sinθ2i+θ2j+12≥12sinθ2isinθ2j+1.Forθ2i−θ2j+1/2−θ1/2≤θ2i−θ2j+1/2−τ2≤θ2i−θ2j+1/2≤π/2, then(42)sinθ2i−θ2j+12−τ2≥sinθ2i−θ2j+12−θ12≥12sinθ2i−θ2j+12.By (40)–(42), we can obtain (35).Fori+1≤j≤n/2, since θ3/2≤θ2j+1−θ2i/2−θ1/2≤θ2j+1−θ2i/2−τ2<π/2, then(43)sinθ2j+1−θ2i2−τ2≥sinθ2j+1−θ2i2−θ12≥12sinθ2j+1−θ2i2.Forθ2j+1−θ2i/2≤θ2j+1−θ2i/2+τ2≤θ2j+1−θ2i/2+θ1/2≤θj≤π/2, we have(44)sinθ2j+1−θ2i2+τ2≥sinθ2j+1−θ2i2,and combining (41), we can obtain (35).Forj=i or j=i−1, it is easy to prove that Ai,j11≤2n+2.Whenτ2∈−θ1/2,0, by setting τ2′=−τ2∈0,θ1/2, we can similarly prove (35) and (36).(2)
Ifτ1∈0,θ1/2, for 0≤j≤i−2, since θ2i+θ2j+1/2≤θ2i+θ2j+1/2+τ1≤θ2i+θ2j+1/2+θ1/2≤π−θ2, then(45)sinθ2i+θ2j+12+τ1≥minsinθ2i+θ2j+12,sinθ2i+θ2j+12+θ12≥12sinθ2i+θ2j+12≥12sinθ2isinθ2j+1.And, we have(46)sinθ2i+θ2j+12−τ1≥12sinθ2isinθ2j+1,sinθ2i−θ2j+12−τ1≥12sinθ2i−θ2j+12,so we obtain (37).Fori+1≤j≤n/2, considering(47)sinθ2i+θ2j+12−τ1≥12sinθ2isinθ2j+1,sinθ2j+1−θ2i2+τ1≥sinθ2j+1−θ2i2,we have (38).Forj=i or j=i−1, it is easy to prove that Ai,j12≤4.
Whenτ1∈π−θ1/2,π, setting τ1′=π−τ1∈0,θ1/2, similar to the case of τ1∈0,θ1/2, the estimation of Ai,j12 can be obtained.
In the same way, we can obtain the following estimates ofBi,j11 and Bi,j12.Lemma 7.
Ifτ2∈0,θ1/2 and τ1∈0,θ1/2∪π−θ1/2,π, then we have the following:(1)
For0≤j≤i−2 or i+1≤j≤n/2,(48)Bi,j11≤4sinθ2isinθ2j+1cosn+2τ2n+2sin2θ2i−θ2j+1/2.Forj=i or j=i−1, Bi,j11≤2n+2.(2)
For0≤j≤i−2,(49)Bi,j12≤4cosn+2τ1n+2sinθ2i−θ2j+1/2.Fori+1≤j≤n/2,(50)Bi,j12≤8cosn+2τ1n+2sinθ2i−θ2j+1/2.Forj=i or j=i−1, Bi,j12≤4.Lemma 8.
Ifτ1∈θ1/2,π/4∪3π/4,π−θ1/2, then Ai,j12≤42,Bi,j12≤42.(1)
We first prove the case ofτ1∈θ1/2,π/4. If 0<θ2i+θ2j+1/2<π/2, we have(51)sinθ2i+θ2j+12+τ1≥22sinθ2i+θ2j+12≥24sinθ2i,(52)cosn+2τ1sinθ2j+1sinτ1−θ2i−θ2j+1/2sinτ1−θ2i+θ2j+1/2≤cosn+2τ1cosτ1−θ2i+θ2j+1/2sinτ1−θ2i+θ2j+1/2+cosn+2τ1cosτ1−θ2i−θ2j+1/2sinτ1−θ2i−θ2j+1/2≤2n+2.Ifπ/2≤θ2i+θ2j+1/2≤π, we obtain(53)sinθ2i+θ2j+12−τ1≥22sinθ2i+θ2j+12≥24sinθ2j+1,(54)cosn+2τ1sinθ2isinτ1+θ2i+θ2j+1/2sinτ1−θ2i−θ2j+1/2≤2n+2.In summary, we can obtainAi,j12≤42.(2)
Ifτ1∈3π/4,π−θ1/2, by setting τ1′=π−τ1∈θ1/2,π/2 and the same as the case of τ1∈θ1/2,π/4, the conclusion of the lemma can be proved.
Similar to Lemma8, we can obtain the following conclusion.Lemma 9.
Ifτ1∈θ1/2,π/3∪2π/3,π−θ1/2, then Ai,j12≤8,Bi,j12≤8.Lemma 10.
Ifτ2∈θ1/2,π/4, then Ai,j11≤4,Bi,j11≤4.
The estimates ofAi,j11 and Bi,j11 are similar, so we only take Ai,j11 as an example.
We first prove the case ofτ2∈θ1/2,π/4. For every i, there is j0 so that τ2−θ2i+θ2j0+1/2≤θ1/2 holds, that is, θ2i+θ2j0+1/2−θ1/2≤τ2≤θ2i+θ2j0+1/2+θ1/2.(i)
Forj=j0, we have θ2i−θ1/2≤τ2+θ2i−θ2j0+1/2≤θ2i+θ1/2; then,(55)sinτ2+θ2i−θ2j0+12≥minsinθ2i+θ12,sinθ2i−θ12≥12sinθ2i.Because ofθ2j0+1−θ1/2≤τ2−θ2i−θ2j0+1/2≤θ2j0+1+θ1/2, we have(56)sinτ2−θ2i−θ2j0+12≥minsinθ2j0+1+θ12,sinθ2j0+1−θ12≥12sinθ2j0+1,and on account ofcosn+2τ2/sinτ2−θ2i+θ2j+1/2≤n+2, we obtain Ai,j11≤4.(ii)
The remaining part will be discussed in two situations:j0≤i−1 and j0≥i.(a)
The casej0≤i−1.Forj≤j0−1≤i−2, since π/2≥τ2−θ2i−θ2j+1/2≥θ2j+1, we have sinτ2−θ2i−θ2j+1/2≥sinθ2j+1. And, because of(57)sinθ2icosn+2τ2sinτ2+θ2i−θ2j+1/2sinτ2−θ2i+θ2j+1/2≤2n+2,we obtainAi,j11≤2.Forj0+1≤j≤i−1≤n+1/2−1, considering that θi≤τ2+θ2i−θ2j+1/2<θ2i, we have sinτ2+θ2i−θ2j+1/2≥minsinθ2i,sinθi, so(58)sinθ2j+1cosn+2τ2sinτ2−θ2j+1−θ2i/2sinτ2−θ2i−θ2j+1/2≤2n+2.And, using the following (61), similar to (57), we obtain Ai,j11≤4.Forj≥i≥j0+1, since θ2j+1/2≤τ2+θ2j+1−θ2i/2≤θ2j+1−θ1/2, then(59)sinτ2+θ2j+1−θ2i2≥minsinθ2j+12,sinθ2j+1.Therefore,(60)sinθ2j+1sinτ2+θ2j+1−θ2i/2≤2.And, on account of (57), we have Ai,j11≤4.(b)
The casej0≥i.For0≤j≤i−1≤j0−1, on account of θ2i≤τ2+θ2i−θ2j+1/2≤π/2, we obtain sinτ2+θ2i−θ2j+1/2≥sinθ2i. And, considering that(61)sinθ2j+1cosn+2τ2sinτ2−θ2j+1+θ2i/2sinτ2−θ2i−θ2j+1/2≤2n+2,we obtainAi,j11≤2.Fori≤j<j0−1, on account of π/4≥τ2−θ2j+1−θ2i/2>θ2i, we have sinτ2−θ2j+1−θ2i/2≥sinθ2i. Since θ2j+1≤τ2+θ2j+1−θ2i/2<π/2, we obtain sinτ2+θ2j+1−θ2i/2≥sinθ2j+1. And, considering that cosn+2τ2/sinτ2−θ2i+θ2j+1/2≤n+2, we obtain Ai,j11≤1.Fori+1≤j0+1≤j≤n/2, since θ2j+1/2≤τ2+θ2j+1−θ2i/2<θ2j+1, we obtain(62)sinτ2+θ2j+1−θ2i2≥minsinθ2j+1,sinθ2j+12.Thus,(63)sinθ2j+1sinτ2+θ2i−θ2j+1/2≤maxsinθ2j+1sinθ2j+1,sinθ2j+1sinθ2j+1/2≤2.And on account of (57), we have Ai,j11≤4.To sum up, ifτ2∈θ1/2,π/4, we obtain Ai,j11≤4.
Whenτ2∈−π/4,−θ1/2, by setting τ2′=−τ2∈θ1/2,π/2, we can similarly prove the conclusion.Lemma 11.
Ifτ2∈π/4,π/2, then Ai,j11≤42,Bi,j11≤42.
The estimates ofAi,j11 and Bi,j11 are similar, so we only take Ai,j11 as an example.
Forj≤i−1, we (64)sinτ2+θ2i−θ2j+12≥22cosθ2i−θ2j+12≥24sinθ2i.
And, using (61), we obtain Ai,j11≤42.
Fori≤j, we have sinτ2+θ2j+1−θ2i/2≥2/4sinθ2j+1 and (57). So, we obtain Ai,j11≤42.
Ifτ2∈−π/2,−π/4, by setting τ2′=−τ2∈π/4,π/2, we can similarly prove the conclusion.
To sum up, whenτ2∈π/4,π/2, we have Ai,j11≤42.Proof.
of Theorem 1. Sinceτ1∈0,π,τ2∈−π/2,π/2 and τ2≤τ1, it will be convenient to divide the argument into several cases as follows:(1)
Ifτ2∈0,θ1/2, then τ1 may belong to the following intervals 0,θ1/2∪π−θ1/2,π, θ1/2,π/4∪3π/4,π−θ1/2, and π/4,3π/4(2)
Ifτ2∈θ1/2,π/4, then τ1 may belong to the following intervals θ1/2,π/3∪2π/3,π−θ1/2, π/3,2/3π, and π−θ1/2,π(3)
Ifτ2∈π/4,π/2, then τ1 may belong to the following intervals π/4,π/2−θ1/2∪π/2+θ1/2,3/4π, π/2−θ1/2,π/2, π/2,π/2+θ1/2, 3/4π,π−θ1/2, and π−θ1/2,π
Next, we will discuss each case separately.Case 1.
Ifτ2∈0,θ1/2,τ1∈0,θ1/2∪π−θ1/2,π, then(65)Λn1x≤28n+22.
By Lemma6, we can obtain(66)∑i=1n+1/2∑j=0i−2Ai,j1≤∑i=1n+1/2∑j=0i−232sinθ2isinθ2j+1cosn+2τ1cosn+2τ2n+22sin3θ2i−θ2j+1/2≤∑i=2n+1/2∑j=0i−232n+22sin3θ2i−θ2j+1/2≤∑i=2n+1/2∑j=0i−232n+22i−2j−13≤∑i=2n+1/232n+2∫1i−212x−13dx=∑i=2n+1/28n+21−12i−32≤4n+22,(67)∑i=1n+1/2∑j=i+1n/2Ai,j1≤∑i=1n/2−1∑j=i+1n/216sinθ2isinθ2j+1cosn+2τ1cosn+2τ2n+22sin3θ2j+1−θ2i/2≤∑i=1n/2−1∑j=i+1n/216n+22sin3θ2j+1−θ2i/2≤∑i=1n/2−1∑j=i+1n/216n+22j−2i+13≤2n+22.
Forj=i or j=i−1, by Lemma 6, we obtain(68)∑i=1n/2Ai,j1≤∑i=1n/28n+2≤4n+22.
From (66)–(68), we can obtain(69)∑i=1n+1/2∑j=0n/2Ai,j1≤14n+22,τ2∈0,θ12,τ1∈0,θ12∪π−θ12,π.
Similarly, by Lemma7, we have(70)∑i=1n+1/2∑j=0n/2Bi,j1≤14n+22,τ2∈0,θ12,τ1∈0,θ12∪π−θ12,π.
Therefore,(71)Λn1x≤28n+22,τ2∈0,θ12,τ1∈0,θ12∪π−θ12,π.Case 2.
Ifτ2∈0,θ1/2andτ1∈θ1/2,π/4∪3π/4,π−θ1/2, then(72)Λn1x≤322n+22.
By Lemmas6 and 8, it can be proved that(73)∑i=1n+1/2∑j=0i−2Ai,j1≤∑i=1n+1/2∑j=0i−2162sinθ2isinθ2j+1cosn+2τ2n+2sin2θ2i−θ2j+1/2≤∑i=2n+1/2∑j=0i−2162n+2sin2θ2i−θ2j+1/2≤∑i=2n+1/2∑j=0i−2162n+22i−2j−12≤42n+22,(74)∑i=1n+1/2∑j=i+1n/2Ai,j1≤∑i=1n/2−1∑j=i+1n/2162sinθ2isinθ2j+1cosn+2τ2n+2sin2θ2j+1−θ2i/2≤∑i=1n/2−1∑j=i+1n/2162n+2sin2θ2j+1−θ2i/2≤∑i=1n/2−1∑j=i+1n/2162n+22j−2i+12≤42n+22.
Forj=i or j=i−1, we have(75)∑i=1n/2Ai,j1≤∑i=1n/282n+2≤42n+22.
Combining (73)–(75), we obtain(76)∑i=1n+1/2∑j=0n/2Ai,j1≤162n+22,τ2∈0,θ12,τ1∈θ12,π4∪3π4,π−θ12.
In the same way, we have(77)∑i=1n+1/2∑j=0n/2Bi,j1≤162n+22,τ2∈0,θ12,τ1∈θ12,π4∪3π4,π−θ12.
In conclusion, it can be seen that(78)Λn1x≤322n+22,τ2∈0,θ12,τ1∈θ12,π4∪3π4,π−θ12.Case 3.
Ifτ2∈0,θ1/2,τ1∈π/4,3π/4, then(79)Λn1x≤4n+22.
Sinceτ2≤θ1/2≤π/6, we have(80)sinφ1sinφ2=sin2τ1−sin2τ2≥14.
Furthermore, we can obtain(81)Λn1x≤∑i=1n+1/2∑j=0n/2sinθ2isinθ2j+1n+22sinφ1sinφ2×∑k=12∑l=12Un+1τ1+εkθ2i+εlθ2j+12Un+1τ2+εkθ2i+εlθ2j+12≤4n+22.Case 4.
Ifτ2∈θ1/2,π/4andτ1∈θ1/2,π/3∪2π/3,π−θ1/2, then(82)Λn1x≤16n+22.
By Lemmas9 and 10, it is easy to prove this conclusion.Case 5.
Ifτ2∈θ1/2,π/4andτ1∈π/3,2/3π, then(83)Λn1x≤4n+22.
Ifτ2∈θ1/2,π/4andτ1∈π/3,2π/3, since sinφ1sinφ2=sin2τ1−sin2τ2≥1/4, then similar to (81), we obtain Λn1x≤4n+22.Case 6.
Ifτ2∈θ1/2,π/4andτ1∈π−θ1/2,π, then(84)Λn1x≤8n+22.
Ifτ1∈π−θ1/2,π, it is easy to prove that Ai,j12≤4. Combining the results of Lemma 10, we can obtain(85)∑i=1n+1/2∑j=0n/2Ai,j1≤∑i=1n+1/2∑j=0n/216≤4n+22.
Similarly, we can obtain(86)∑i=1n+1/2∑j=0n/2Bi,j1≤∑i=1n+1/2∑j=0n/216≤4n+22.
Thus,Λn1x≤8n+22,τ2∈θ1/2,π/4,andτ1∈π−θ1/2,π.Case 7.
Ifτ2∈π/4,π/2forτ1∈π/4,π/2−θ1/2∪π/2+θ1/2,3/4π, then(87)Λn1x≤802n+22.
By Lemma11, we know Ai,j11≤42andτ2∈π/4,π/2. Next, let us estimate Ai,j12, for τ1∈π/4,π/2−θ1/2. For every j, there is i0 such that τ1−θ2i0+θ2j+1/2≤θ1/2 holds; that is, θ2i0+θ2j+1/2−θ1/2≤τ1≤θ2i0+θ2j+1/2+θ1/2.(i)
Fori=i0, since π/2−θ1/2≤τ1+θ2i0+θ2j+1/2≤π−θ1/2, then(88)sinτ1+θ2i0+θ2j+12≥sinθ12≥1n+2.And, because ofθ2j+1−θ1/2≤τ1−θ2i0−θ2j+1/2≤θ2j+1+θ1/2, we can obtain(89)sinτ1−θ2i0−θ2j+12≥minsinθ2j+1+θ12,sinθ2j+1−θ12≥12sinθ2j+1.Furthermore, considering thatcosn+2τ1/sinτ1−θ2i0+θ2j+1/2≤n+2, we have Ai0,j12≤2n+2. Thus,(90)∑j=0n/2Ai0,j1≤∑j=0n/282n+2≤42n+22.(ii)
For0≤i≤i0−1, the two cases j≤i−1 and j≥i are discussed separately.(a)
Ifj≤i−1, for θ2j+1≤τ1−θ2i−θ2j+1/2≤θ2i+θ2i0+1/2, we have(91)sinτ1−θ2i−θ2j+12≥minsinθ2i+θ2i0+12,sinθ2j+1≥min12sinθ2i,sinθ2j+1,sosinθ2isinθ2j+1/sinτ1−θ2i−θ2j+1/2≤2. And, considering that(92)sinτ1+θ2i+θ2j+12sinτ1−θ2i+θ2j+12=sin2τ1−sin2θ2i+θ2j+12=sinτ1+sinθ2i+θ2j+12sinτ1−sinθ2i+θ2j+12≥sin2τ12−θ2i+θ2j+14,we obtain(93)Ai,j12≤2n+2sin2τ1/2−θ2i+θ2j+1/4.(b)
Ifj≥i, we have(94)sinτ1+θ2j+1−θ2i2≥sinπ4cosθ2j+1−θ2i2≥22sinθ2j+1sinθ2i.And, considering that(95)sinτ1+θ2i+θ2j+12sinτ1−θ2i+θ2j+12=sinτ1+sinθ2i+θ2j+12sinτ1−sinθ2i+θ2j+12≥2sinθ2isinθ2j+1sin2τ12−θ2i+θ2j+14,we obtain(96)Ai,j12≤1n+2sin2τ1/2−θ2i+θ2j+1/4.In summary, we obtain(97)∑j=0n/2∑i=1i0−1Ai,j1≤∑j=0n/2∑i=1i0−142n+2⋅2sin2τ1/2−θ2i+θ2j+1/2≤∑j=0n/2∑i=1i0−1322n+22i0−i−12≤322n+2∑j=0n/21+∫1i0−112x−12dx≤242n+22.(iii)
Fori0+1≤i≤n+1/2, the following two cases are discussed separately.(a)
Ifj≥i, from (96), we have(98)∑j=0n/2∑i=i0+1n/2Ai,j1≤∑j=0n/2∑i=i0+1n/242n+2⋅1sin2θ2i+θ2j+1/4−τ1/2≤∑j=0n/2∑i=i0+1n/2162n+222i0−i−12≤122n+22.(b)
Ifj≤i−1, for every j, there is i1 such that τ1−θ2i1−θ2j+1/2≤θ1/2 holds, that is,(99)θ2i1−θ2j+12−θ12≤τ1≤θ2i1−θ2j+12+θ12.Fori=i1, since θ2i1−θ1/2≤τ1+θ2i1+θ2j+1/2≤θ2i1+θ1/2, then(100)sinτ1+θ2i+θ2j+12≥minsinθ2i1+θ12,sinθ2i1−θ12≥12sinθ2i1.On account of−θ2j+1−θ1/2≤τ1−θ2i1+θ2j+1/2≤−θ2j+1+θ1/2, we obtain(101)sinτ1−θ2i+θ2j+12≥minsinθ2j+1−θ12,sinθ2j+1+θ12≥12sinθ2j+1.And because ofcosn+2τ1/sinτ1−θ2i−θ2j+1/2≤n+2, we obtain Ai,j12≤4.Fori≠i1, since θ2i1−1+θ2i/2≤τ1+θ2i+θ2j+1/2≤θ2i1+1+θ2i/2, then(102)sinτ1+θ2i+θ2j+12≥minsinθ2i1+1+θ2i2,sinθ2i1−1+θ2i2≥12sinθ2i.And, because of(103)cosn+2τ1sinθ2j+1sinτ1−θ2i−θ2j+1/2sinτ1−θ2i+θ2j+1/2≤2n+2,we obtainAi,j12≤4.Thus, forj≤i−1, we have(104)∑j=0n/2∑i=i0+1n+1/2Ai,j1≤42n+22.Based on the above results, ifτ2∈π/4,π/2andτ1∈π/4,π/2−θ1/2, we can obtain(105)∑j=0n/2∑i=1n+1/2Ai,j1≤402n+22.Similarly, ifτ2∈π/4,π/2andτ1∈π/4,π/2−θ1/2, we can obtain(106)∑j=0n/2∑i=1n+1/2Bi,j1≤402n+22.Therefore, we obtainΛn1x≤802n+22,τ2∈π/4,π/2,andτ1∈π/4,π/2−θ1/2.
And, we can similarly prove the case ofτ2∈π/4,π/2andτ1∈π/2+θ1/2,3/4π. Then, the conclusion of Case 7 is obtained.Case 8.
Ifτ2∈π/4,π/2andτ1∈π/2−θ1/2,π/2, then(107)Λn1x≤662n+22.
Ifτ1∈π/2−θ1/2,π/2, for every j, there is i0=n+1/2−j such that τ1−θ2i0+θ2j+1/2≤θ1/2 holds.(i)
For1≤i≤i0−1, the following two cases are discussed separately:(a)
Ifj≤i−1, since θ2j+1≤τ1−θ2i−θ2j+1/2<π/2, then sinτ1−θ2i−θ2j+1/2≥sinθ2j+1. And, because of π/2+θ1≤τ1+θ2i+θ2j+1/2≤π−θ1/2, we have(108)sinτ1+θ2i+θ2j+12≥sinπ2+θ2i+θ2j+12≥1−2i+2j+1n+2.And, considering thatθ1/2≤τ1−θ2i+θ2j+1/2<π/2, we obtain(109)sinτ1−θ2i+θ2j+12≥sinπ2−θ12−θ2i+θ2j+12≥1−2i+2j+2n+2.Hence,(110)Ai,j12≤n+2n−2i−2j2.(b)
Ifj≥i, we have(111)sinτ1+θ2j+1−θ2i2≥sinπ6cosθ2j+1−θ2i2≥12sinθ2j+1sinθ2i.And. by (108) and (109), we obtain(112)Ai,j12≤2n+2n−2i−2j2.Therefore,(113)∑j=0n/2∑i=1i0−1Ai,j1≤82n+2∑j=0n+1/2−2∑i=1n+1/2−j−11n−2i−2j2≤82n+2∑j=0n+1/2−2∫1n+1/2−j−11n−2j−2x2dx+1≤62n+22.(ii)
Fori0+2≤i≤n+1/2, the following two cases are discussed separately:(a)
Ifj≤i−1, since π/2−θ1≤τ1−θ2i−θ2j+1/2≤π/2−θ1/2, then(114)sinθ2isinθ2j+1sinτ1−θ2i−θ2j+1/2≤sinθ2isinθ2j+1sinθi+θ1/2sinθj+θ1/2≤4.Because of0≤θi−i0−θ1/2≤θ2i+θ2j+1/2−τ1≤θi−i0+θ1/2<π/2, we can obtain(115)sinθ2i+θ2j+12−τ1≥sinθ2i+θ2j+12−π2≥2i+2j+1n+2−1.Owing toθ1/2≤τ1+θ2i+θ2j+1/2−π≤π/2−3θ1/2, we have(116)sinτ1+θ2i+θ2j+12−π≥sinθ2i+θ2j+12−θ12−π2≥2i+2jn+2−1.Then,(117)Ai,j11≤4n+22i+2j−n−12i+j−n+2.(b)
Ifj≥i, from (111), (115), (116), we know that(118)Ai,j11≤2n+22i+2j−n−12i+j−n+2.Hence,(119)∑j=0n/2∑i=i0+2n+1/2Ai,j1≤∑j=0n/2∑i=i0+2n+1/242n+2⋅4n+22i+2j−n−12i+j−n+2≤∑j=0n+1/2−1∑i=i0+2n+1/2162n+22i+j−n+22≤∑j=0n+1/2−1162n+2∫i0+2n+1/212x+2j−n−22dx+1≤122n+22.(iii)
Fori=i0 or i=i0+1, it is easy to prove that Ai,j12≤2n+2. Then, we have(120)∑j=0n/2Ai,j≤∑j=0n/282n+2≤42n+22.Based on the above conclusions, we can obtain(121)∑i=1n+1/2∑j=0n/2Ai,j≤262n+22.Similarly, we can prove that(122)∑i=1n+1∑j=0n/2Bi,j≤402n+22.Then,Λn1x≤662n+22,τ2∈π/4,π/2,andτ1∈π/2−θ1/2,π/2.Case 9.
Ifτ2∈π/4,π/2andτ1∈π/2,π/2+θ1/2, then(123)Λn1x≤382n+22.
Ifτ2∈π/4,π/2andτ1∈π/2,π/2+θ1/2, similar to the estimates of (121) and (122), we can obtain(124)∑i=1n+1/2∑j=0n/2Ai,j≤1422+n2,∑i=1n+1/2∑j=0n/2Bi,j≤242n+22.
Therefore,Λn1x≤382n+22,τ2∈π/4,π/2,andτ1∈π/2,π/2+θ1/2.Case 10.
Ifτ2∈π/4,π/2andτ1∈3π/4,π−θ1/2, then(125)Λn1x≤16n+22.
By Lemma8 and 11, we can easily obtain the conclusion.Case 11.
Ifτ2∈π/4,π/2andτ1∈π−θ1/2,π, then(126)Λn1x≤82n+22.Proof.
Ifτ1∈π−θ1/2,π, it is easy to prove that Ai,j12≤4,Bi,j12≤4. And, by Lemma 11, we can obtain the conclusion.
Based on the conclusion of Case1–Case 11, we obtain Λn1x≤802n+22,τ2∈0,π/2,τ1∈0,π.
And similarly, we haveΛn2x≤802n+22,τ2∈0,π/2,andτ1∈0,π.
Then, the proof of Theorem1 is completed.



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