Introduction - UCSD Mathematicsbli/publications/Li:Lagrange.pdf · LAGRANGE INTERPOLATION AND FINITE ELEMENT SUPERCONVERGENCE 3 element superconvergence estimates [1,5,26,28]. By

LAGRANGE INTERPOLATION AND FINITE ELEMENT

SUPERCONVERGENCE

BO LI

Abstract. We consider the finite element approximation of the Laplacian operatorwith the homogeneous Dirichlet boundary condition, and study the correspondingLagrange interpolation in the context of finite element superconvergence. For d-dimensional Qk-type elements with d ≥ 1 and k ≥ 1, we prove that the interpolationpoints must be the Lobatto points if the Lagrange interpolation and the finiteelement solution are superclose in H1 norm. For d-dimensional Pk-type elements,we consider the standard Lagrange interpolation—the Lagrange interpolation withinterpolation points being the principle lattice points of simplicial elements. Weprove for d ≥ 2 and k ≥ d+1 that such interpolation and the finite element solutionare not superclose in both H1 and L2 norms, and that not all such interpolationpoints are superconvergence points for the finite element approximation.

1. Introduction

Consider the boundary value problemLu = f in Ω,

u = 0 on ∂Ω,(1.1)

where Ω ⊂ Rd is a bounded domain with a Lipschitz continuous boundary ∂Ω, d ≥ 1,

f ∈ L2(Ω), and L : H2(Ω) → L2(Ω) is a second order, linear, self-adjoint, ellipticdifferential operator. Let u ∈ H1

0 (Ω) be its unique weak solution, defined by

A(u, v) = (f, v) ∀v ∈ H10 (Ω),

where A : H10 (Ω) ×H1

0 (Ω) → R is the bilinear, symmetric, continuous, and coerciveform associated with (1.1), and (·, ·) denotes the inner product of L2(Ω). Let τhbe a family of finite element meshes of the domain Ω with the mesh size h → 0. Fixan integer k ≥ 1. For each h, let Sh

k (Ω) ⊂ H1(Ω) ∩ C(Ω) be the corresponding finiteelement space such that Sh

k (Ω)|T ⊇ Pk|T for all T ∈ τh, where Pk is the set of all

polynomials of degree ≤ k. Let

Shk(Ω) = Sh

k (Ω)∩H10 (Ω). Let uh ∈

Shk(Ω) be the finite

element solution, defined by

A(uh, vh) = (f, vh) ∀vh ∈

Shk(Ω).

Date: June 12, 2003.2000 Mathematics Subject Classification. 65N30.Key words and phrases. finite element, Lagrange interpolation, superconvergence.This work was partially supported by the NSF through grant DMS-0072958.

1

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Finally, let Ih : C(Ω) → Shk (Ω) denote the corresponding Lagrange interpolation

operator. The following estimate

h−1‖Ihu− uh‖L2(Ω) + ‖Ihu− uh‖H1(Ω) ≤ Chk (1.2)

is standard, provided that the weak solution u ∈ H10 (Ω) ∩ C(Ω) is smooth enough

and the underlying meshes are quasi-uniform [3,9]. Here and below, we use the letterC to denote a generic, positive constant that is independent of the mesh size h.

The estimate (1.2) is in general optimal. However, in some cases, it can be im-proved. This means that Ihu and uh can be superclose. More precisely, we say thatthe Lagrange interpolation Ihu and the finite element solution uh are superclose inH1 norm, if

‖Ihu− uh‖H1(Ω) = o(

hk)

as h→ 0.

We also say that Ihu and uh are superclose in H1 norm by order (at least) σ > 0, if

‖Ihu− uh‖H1(Ω) ≤ Chk+σ. (1.3)

The following result gives a different expression of the closeness between Ihu and uh

in H1 norm. It is trivially true, and we omit its proof.

Lemma 1.1. If the exact solution u ∈ H10 (Ω) ∩ C(Ω), then

γ‖Ihu− uh‖H1(Ω) ≤ sup

vh∈

Shk(Ω), vh 6=0

|A(u− Ihu, vh)|‖vh‖H1(Ω)

≤M‖Ihu− uh‖H1(Ω),

where γ > 0 and M > 0 are the two constants in the conditions of coercivity andcontinuity, respectively, of the bilinear form A : H1

0 (Ω) ×H10 (Ω) → R,

A(v, v) ≥ γ‖v‖2H1(Ω) ∀v ∈ H1

0 (Ω)

and|A(v, w)| ≤M‖v‖H1(Ω)‖w‖H1(Ω) ∀v, w ∈ H1

0 (Ω).

The supercloseness between the Lagrange interpolation and the finite element so-lution is closely related to the superconvergence of the finite element solution to theexact solution. In fact, if (1.3) holds true, then one can easily obtain the followingestimate of gradient superconvergence

[

1

hd

∑

z∈Zh(Ω)

|∇u(z) − ∇uh(z)|2]1/2

≤ Chk+min(σ,1),

where Zh(Ω) is the set of superconvergence points for the gradient of the Lagrangeinterpolation and ∇ is some kind of average of the gradient [8]. In some cases, onecan obtain a higher order estimate

|A(u− Ihu, vh)| ≤ Chk+σ‖vh‖W 1,p(Ω) ∀vh ∈

Shk(Ω) (1.4)

for some σ > 0 and p ∈ [1,∞). This, together with delicate estimates of a discreteGreen’s function substituting vh in the inequality in (1.4), can lead to pointwise finite

LAGRANGE INTERPOLATION AND FINITE ELEMENT SUPERCONVERGENCE 3

element superconvergence estimates [1, 5, 26, 28]. By Lemma 1.1, the estimate (1.4)is equivalent to the supercloseness estimate (1.3), if p = 2.

In this work, we study the supercloseness between the Lagrange interpolation andthe finite element solution. Our main results are as follows.

1. For d-dimensional Qk-type (tensor product) elements with d ≥ 1 and k ≥ 1, theinterpolation points must be the Lobatto points if the Lagrange interpolationand the finite element solution are superclose in H1 norm, cf. Theorem 2.1.

2. For d-dimensional Pk-type (simplicial) elements with d ≥ 2 and k ≥ d + 1, thestandard Lagrange interpolation—the Lagrange interpolation with its interpo-lation points being the principle lattice points of simplicial elements—and thefinite element solution are not superclose in H1 norm, cf. Theorem 4.1.

3. For d-dimensional Pk-type elements with d ≥ 2 and k ≥ d + 1, not all thestandard Lagrange interpolation points are superconvergence points for the finiteelement solution, cf. Corollary 4.1.

For d-dimensional Qk-type elements with d ≥ 1 and k ≥ 2, the finite elementsolution is superconvergent by one order to the exact solution at the Lobatto points[4,10,16,18,24,25]. This implies that the Lagrange interpolation associated with theLobatto points and the finite element solution are superclose by one order in H1 norm.Here, we prove the converse under the assumption that they are only superclose, butnot necessary superclose by any order, in H1 norm.

For simplicial finite elements, the Lagrange interpolation points can not be ar-bitrarily distributed in general. With good meshes, the standard Lagrange inter-polation and the finite element solution are in fact superclose in H1 norm by oneorder for two-dimensional P1 and P2 elements and for three-dimensional P1 ele-ment [1, 5–7,12–14,17,19,21,22,26–30]. Recently, similar results have been obtainedfor any d-dimensional, linear, simplicial finite elements with a uniform mesh [2]. But,it is still open in general whether or not such supercloseness remains for d-dimensionalPk-type elements with d ≥ 3 and 2 ≤ k ≤ d.

The proof of those known results relies on lucky cancellation of inter-element bound-

ary integrals in delicate estimates of the integral form A(u− Ihu, vh) for vh ∈

Shk(Ω).

However, such cancellation seems to be impossible if there exists an element-wise,

bubble-like test function vh ∈

Shk(Ω) that vanishes on the boundary of each element.

Such a function exists if and only if there exists an interior node in each of the sim-plicial elements. This turns out to be true if and only if k ≥ d+ 1 for d-dimensionalPk-type elements. Constructing a bubble-like test function to avoid any possiblecancellation was the original approach in our early work [20] to show the the non-supercloseness for two-dimensional P3 element. Here, we extend such an approachto a general case which is more complicated due to the higher space dimension andhigher polynomial degree.

In proving the non-supercloseness of the Lagrange interpolation to the finite elementsolution for the general d-dimensional Pk-type finite elements with k ≥ d+1, we choosethe underlying domain to be the unit d-dimensional simplex. This allows us to have a

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polynomial of degree exactly k+1 as the solution of the underlying Poisson equationwith the homogeneous Dirichlet boundary condition. In addition, we construct aspecial family of quasi-uniform finite element meshes consisting of enough elementsthat are scaled translations of the unit simplex. Such meshes are uniform for d = 2but non-uniform for d ≥ 3. Calculations based on such meshes are much simplified.With our approach, it is possible to consider a uniform family of finite element meshesof the d-dimensional unit cube, and construct similar but more complicated solutions.Undoubtedly, however, the calculations will be more involved.

In Section 2, we study the optimal Lagrange interpolation points for Qk-type ele-ments. In Section 3, we construct a quasi-uniform family of simplicial finite elementmeshes of a d-dimensional domain for d ≥ 2. With such meshes, we study in Section4 the standard Lagrange interpolation for Pk-type elements. Finally, in Section 5, weprove some auxiliary lemmas.

2. Optimal Lagrange interpolation points for Qk-type finite

elements

Consider the boundary value problem−∆u = f in Ω,

u = 0 on ∂Ω,(2.1)

where f ∈ L2(Ω) and Ω = Πdm=1(am, bm) ⊂ R

d is a d-dimensional rectangular paral-lelepiped with d ≥ 1 and −∞ < am < bm < ∞ for all m = 1, · · · , d. The associatedbilinear form A : H1

0 (Ω) ×H10 (Ω) → R is defined by

A(v, w) = (∇v,∇w) ∀v, w ∈ H10 (Ω).

It is symmetric, continuous, and coercive. The weak solution u ∈ H10 (Ω) of the

boundary value problem (2.1) is defined by

A(u, v) = (f, v) ∀v ∈ H10 (Ω).

Let τh be a family of quasi-uniform rectangular meshes covering Ω with the meshsize h→ 0. We denote a typical mesh by

τh =

d∏

m=1

[xm,jm−1, xm,jm ] : jm = 1, · · · , nm,m = 1, · · · , d

,

where xm,jm = am + jmhm for jm = 0, · · · , nm, hm = (bm − am)/nm, nm ≥ 1 is aninteger for each m = 1, · · · , d, and h = max1≤m≤d hm. For an integer k ≥ 1, letSh

k (Ω) ⊂ H1(Ω) denote the Qk-type finite element space corresponding to the meshτh, i.e., the restriction Sh

k (Ω)|R is exactly Qk|R for each element R ∈ τh, where

Qk = span xα11 · · · xαd

d : α1, · · · , αd are nonnegative integers, α1 + · · · + αd = k .

Let

Shk(Ω) = Sh

k(Ω) ∩H10 (Ω). The finite element solution uh ∈

Shk(Ω) is defined by


Shk(Ω).


For each integer m with 1 ≤ m ≤ d, let ξ(0)m , · · · , ξ(k)

m be k+1 distinct real numberssatisfying

−1 = ξ(0)m < · · · < ξ(k)

m = 1.

We call all the points(

ξ(i1)1 , · · · , ξ(id)

d

)

(im = 0, · · · , k, m = 1, · · · , d) the referenceinterpolation points. We define the Lagrange interpolation points on each element∏d

m=1[xm,jm−1, xm,jm ] ∈ τh (1 ≤ jm ≤ nm, 1 ≤ m ≤ d) by

x(i)m,jm

=hmξ

(i)m + xm,jm−1 + xm,jm

2, i = 0, · · · , k, m = 1, · · · , d.

Finally, we denote by Ih : C(Ω) → Shk (Ω) the Lagrange interpolation operator asso-

ciated with these interpolation points.

Recall that the Jacobi polynomials P(1,1)n (n = 0, 1, · · · ) are orthogonal polynomials

on the interval [−1, 1] with the weight ρ(ξ) = 1− ξ2, normalized by P(1,1)n (1) = n+ 1

[23]. The Rodrigues’ formula for P(1,1)n is

P (1,1)n (ξ) =

(−1)n

2nn! (1 − ξ2)

(

d

dξ

)n[

(1 − ξ2)n+1]

.

For each n ≥ 1, P(1,1)n has exactly n distinct roots in (−1, 1), called Lobatto points

(associated with n). Recall also that the Legendre polynomials are orthogonal poly-nomials on the interval [−1, 1] with the weight ρ(ξ) = 1 [23]. They are given by

Ln(ξ) =1

2nn!

(

d

dξ

)n[

(1 − ξ2)n]

, n = 0, 1, · · · .

It is easy to show that L′n(ξ)∞n=1 is also a sequence of orthogonal polynomials on

[−1, 1] with the weight ρ(ξ) = 1 − ξ2. Consequently, P(1,1)n and L′

n+1 differ only bya nonzero constant. For n ≥ 2, the Lobatto points associated with n − 1 are thusthe roots of L′

n(ξ) in (−1, 1). However, for convenience, we shall call in what followsall the n − 1 distinct roots of L′

n(ξ) in (−1, 1), together with ±1, the Lobatto pointsof order n. (In fact, ±1 are often included in a Lobatto quadrature [11].) We call apoint in R

d a d-dimensional Lobatto point of order n, if each of its d coordinates is aone-dimensional Lobatto point of order n. Obviously, there are (n+1)d d-dimensionalLobatto points of order n.

Together with what is known, the following result implies for Qk-type finite ele-ments with k ≥ 2 that the Lagrange interpolation is superclose to the finite elementsolution in H1 norm if and only if all the interpolation points are the Lobatto points.

Theorem 2.1. Suppose that

‖Ihu− uh‖H1(Ω) = o(

hk)

as h→ 0, (2.2)

whenever the solution u ∈ H10 (Ω) is smooth enough. Then, all the reference interpo-

lation points(

ξ(i1)1 , · · · , ξ(id)

d

)

(0 ≤ im ≤ k, m = 1, · · · , d) must be the d-dimensionalLobatto points of order k.

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Proof. For k = 1, the reference interpolation points are always Lobatto points by ourdefinition. So, we assume that k ≥ 2. We shall show for each m (1 ≤ m ≤ d) that

ξ(0)m , · · · , ξ(k)

m are indeed the k + 1 one-dimensional Lobatto points of order k.Fix an index m with 1 ≤ m ≤ d. Define u ∈ H1

0 (Ω) by

u(x) =

(

xm − am + bm2

)k−1 d∏

l=1

(xl − al)(xl − bl), x = (x1, · · · , xd) ∈ Ω.

Note that u depends on m. Define accordingly f(x) = −∆u(x) for all x ∈ Ω.Obviously, f ∈ L2(Ω), and u ∈ C∞(Ω) solves the boundary value problem (2.1).

For each integer s : 0 ≤ s ≤ k − 2, define vs : [am, bm] → R by

vs(xm) = φs

(

2xm − xm,jm−1 − xm,jm

hm

)

,

∀xm ∈ [xm,jm−1, xm,jm ], jm = 1, · · · , nm,

where the function φs : [−1, 1] → R is defined by

φs(ξ) = ξs+2 − 1

2(1 + ξ) − 1

2(−1)s(1 − ξ), ξ ∈ [−1, 1].

It is easy to see that

φ′′s(ξ) = (s+ 1)(s+ 2)ξs and φs(−1) = φs(1) = 0.

Hence, vs is a continuous piecewise polynomial of degree s + 2 ≤ k, vanishing atall the points xm,jm (jm = 0, · · · , nm). Define vh : Ω → R for the case d = 1 byvh(x1) = vs(x1) for all x1 ∈ Ω = [a1, b1], and for the case d ≥ 2 by

vh(x) = vs(xm)Wm(x′) ∀x ∈ Ω,

where

Wm(x′) =d∏

l=1, l 6=m

(xl − al)(xl − bl) ∀x′ ∈ Ω′,

x′ = (x1, · · · , xm−1, xm+1, · · · , xd),

Ω′ =d∏

l=1, l 6=m

(al, bl).

Note that vh depends on m and that vh ∈

Shk(Ω).

Assume that d ≥ 2 temporarily. Let R =∏d

l=1[xl,jl−1, xl,jl] ∈ τh be an arbitrary

element, where 1 ≤ jl ≤ nl and 1 ≤ l ≤ d. The function

u(x) −Wm(x′)k∏

i=0

(

xm − x(i)m,jm

)


is in Qk|R, and agrees with u on all the interpolation points in R. Hence, this functionis exactly the Lagrange interpolation of u on the element R. Thus, we have

(u− Ihu)(x) = Wm(x′)k∏

i=0

(

xm − x(i)m,jm

)

∀x ∈ R.

Let R′ =∏d

l=1, l 6=m[xl,jl−1, xl,jl]. Let ∇′ denote the gradient operator with respect to

x′. Applying integration by parts and using the change of variable

ξm =2xm − xm,jm−1 − xm,jm

hm

from xm ∈ [xm,jm−1, xm,jm ] to ξm ∈ [−1, 1], we obtain that∫

R

∇(u− Ihu)(x) · ∇vh(x) dx

=

∫

R

[

∇′(u− Ihu)(x) · ∇′vh(x) +∂

∂xm

(u− Ihu)(x) ·∂

∂xm

vh(x)

]

dx

=

∫

R

|∇′Wm(x′)|2 vs(xm)k∏

i=0

(

xm − x(i)m,jm

)

dx

+

∫

R

[Wm(x′)]2

[

d

dxm

k∏

i=0

(

xm − x(i)m,jm

)

]

v′s(xm) dx

=

∫

R′

|∇′Wm(x′)|2 dx′∫ xm,jm

xm,jm−1

vs(xm)k∏

i=0

(

xm − x(i)m,jm

)

dxm

−∫

R′

[Wm(x′)]2dx′∫ xm,jm

xm,jm−1

v′′s (xm)k∏

i=0

(

xm − x(i)m,jm

)

dxm

=

(

hm

2

)k+2

‖∇′Wm‖2L2(R′)

∫ 1

−1

φs(ξm)k∏

i=0

(

ξm − ξ(i)m

)

dξm

−(

hm

2

)k

‖Wm‖2L2(R′)

∫ 1

−1

φ′′s(ξm)

k∏

i=0

(

ξm − ξ(i)m

)

dξm.

Consequently, we have by the fact nm = (bm − am)/hm that

A (u− Ihu, vh) =∑

R∈τh

∫

R

∇(u− Ihu)(x) · ∇vh(x) dx

=(bm − am)

2k+2hk+1

m ‖∇′Wm‖2L2(Ω′)

∫ 1

−1

φs(ξm)k∏

i=0

(

ξm − ξ(i)m

)

dξm (2.3)

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−(bm − am)(s+ 1)(s+ 2)

2khk−1

m ‖Wm‖2L2(Ω′)

∫ 1

−1

ξsm

k∏

i=0

(

ξm − ξ(i)m

)

dξm.

Similarly, we obtain that

‖vh‖2H1(Ω) =

∫

Ω

[

|vh(x)|2 + |∇vh(x)|2]

dx

= ‖Wm‖2H1(Ω′)

nm∑

jm=1

∫ xm,jm

xm,jm−1

|vs(xm)|2 dxm

+ ‖Wm‖2L2(Ω′)

nm∑

jm=1

∫ xm,jm

xm,jm−1

|v′s(xm)|2 dxm

=(bm − am)

2‖Wm‖2

H1(Ω′)

∫ 1

−1

|φs(ξm)|2 dξm

+2(bm − am)

h2m

‖Wm‖2L2(Ω′)

∫ 1

−1

|φ′s(ξm)|2 dξm,

leading to

‖vh‖H1(Ω) ≤√

(bm − am)[4 + (bm − am)2]

2‖φs‖H1(−1,1)‖Wm‖H1(Ω′)h

−1m . (2.4)

Therefore, we infer from (2.3) and (2.4) that


≥ α1hkm

∣

∣

∣

∣

∣

∫ 1

−1

ξsm

k∏

i=0

(

ξm − ξ(i)m

)

dξm

∣

∣

∣

∣

∣

− α2hk+2m , (2.5)

where

α1 =

√

2(bm − am)

4 + (bm − am)2

(s+ 1)(s+ 2)‖Wm‖2L2(Ω′)

2k‖φs‖H1(−1,1)‖Wm‖H1(Ω′)

> 0

and

α2 =

√

2(bm − am)

4 + (bm − am)2

‖∇′Wm‖2L2(Ω′)

2k+2‖φs‖H1(−1,1)‖Wm‖H1(Ω′)

·∣

∣

∣

∣

∣

∫ 1

−1

φs(ξm)k∏

i=0

(

ξm − ξ(i)m

)

dξm

∣

∣

∣

∣

∣

≥ 0

are constants independent of h.Assume now d = 1. By a similar but simpler argument, we obtain that

A(u− Ihu, vh) = −(b1 − a1)(s+ 1)(s+ 2)

2khk−1

1

∫ 1

−1

ξs1

k∏

i=0

(

ξ1 − ξ(i)1

)

dξ1


and

‖vh‖2H1(Ω) =

b1 − a1

2

∫ 1

−1

[

|φs(ξ1)|2 +4

h21

∣

∣φ′s(ξ1)

∣

∣

2]

dξ1.

Therefore,

‖vh‖H1(Ω) ≤√

(b1 − a1)[4 + (b1 − a1)2]

2‖φs‖L2(−1,1)h

−11 ,

and


≥ βhk1

∣

∣

∣

∣

∣

∫ 1

−1

ξs1

k∏

i=0

(

ξ1 − ξ(i)1

)

dξ1

∣

∣

∣

∣

∣

, (2.6)

where

β =

√

2(b1 − a1)

4 + (b1 − a1)2

(s+ 1)(s+ 2)

2k‖φs‖L2(−1,1)

> 0

is a constant independent of h.It now follows from Lemma 1.1, (2.2), (2.5), (2.6), and the quasi-uniformity of the

meshes that∫ 1

−1

ξsm

k∏

i=0

(

ξm − ξ(i)m

)

dξm = 0, s = 0, · · · , k − 2.

The polynomial∏k−1

i=1

(

ξ−ξ(i)m

)

of degree k−1 is thus orthogonal to all the polynomials

in Pk−2 on [−1, 1] with the weight(

ξ− ξ(0)m

)(

ξ− ξ(k)m

)

= ξ2 − 1. Hence, it differs from

the Jacobi polynomial P(1,1)k−1 only by a nonzero constant. Consequently, all the points

ξ(0)m , · · · , ξ(k)

m are the k + 1 one-dimensional Lobatto points of order k.

3. A construction of d-dimensional simplicial finite element meshes

We now let d ≥ 2 be an integer and Ω ⊂ Rd the open unit simplex

Ω =

(x1, · · · , xd) ∈ Rd : xi > 0, i = 1, · · · , d,

d∑

i=1

xi < 1

. (3.1)

We shall construct a quasi-uniform family of simplicial finite element meshes τhof Ω such that there are O(h−d) elements in τh that are translations of a single d-dimensional simplex σ−1

d hΩ = σ−1d hx : x ∈ Ω, where σd > 0 is a constant depending

only on d and h is the mesh size of τh.For d = 2, the mesh τh can be defined by three families of parallel lines x1 = i/n,

x2 = j/n, and x1 +x2 = l/n, where n ≥ 1 is an integer and i, j, l = 0, · · · , n. This is auniform mesh with mesh size h =

√2/n. Obviously, there are O(h−2) elements of the

mesh that are translations of the single 2-dimensional simplex σ−12 hΩ with σ2 =

√2.

For d ≥ 3, we construct in three steps a simplicial finite element mesh of Ω withthe designed properties. First, we triangulate the reference unit cube into simplexes.Second, we construct a simplicial finite element mesh of the unit cube by cutting itinto many small cubes, triangulating them by affine mappings from the triangulated

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reference unit cube, and gluing them together. Third, we cut the meshed unit cubeby the plane

∑di=1 xi = 1 to define a simplicial finite element mesh of Ω.

Step 1. Take the closed unit cube Cd = [0, 1]d ⊂ Rd as the reference cube and

denote by ξ = (ξ1, · · · , ξd) a generic point in Cd. Define

Bld =

ξ ∈ Cd : l − 1 ≤d∑

i=1

ξi ≤ l

, l = 1, · · · , d. (3.2)

Obviously,

∪dl=1B

ld = Cd and int(Bj

d) ∩ int(Bld) = ∅ if j 6= l. (3.3)

Notice that B1d = Sd, where

Sd =

(ξ1, · · · , ξd) ∈ Rd : ξi ≥ 0, i = 1, · · · , d,

d∑

i=1

ξi ≤ 1

(3.4)

is the reference unit simplex in Rd, and both B1

d and Bdd are simplexes. But Bl

d isnot a simplex if 1 < l < d. This is because that the number of vertices in a d-dimensional simplex is d+ 1. However, since all the vertices (ξ1, · · · , ξd) (ξi = 0 or 1,

i = 1, · · · , d) of Cd lie in the planes∑d

i=1 ξi = j (j = 0, · · · , d), the number of vertices

of Cd contained in Bld is the same as that contained in the planes

∑di=1 ξi = l− 1 and

∑di=1 ξi = l. This number is

(

dl − 1

)

+

(

dl

)

=

(

d+ 1l

)

> d+ 1,

since 1 < l < d.We now triangulate all the polygons B l

d (l = 2, · · · d − 1) into simplexes so that,together with B1

d and Bdd , these simplexes can form a simplicial triangulation of Cd.

It suffices to triangulate the boundary of each B ld into (d− 1)-dimensional simplexes

determined by a set of vertices, and then connect the barycenter of B ld to these

vertices. The boundary of Bld for each l with 2 ≤ l ≤ d− 1 is the union of two types

of (d− 1)-dimensional polygons

Pmd−1 = ξ ∈ Cd :

d∑

i=1

ξi = m, m = l − 1, l,

and

F l,j,md−1 = Bl

d ∩ ξ ∈ Cd : ξj = m, j = 1, · · · , d, m = 0, 1.

Consider a first type (d− 1)-dimensional polygon Pmd−1 (1 ≤ m ≤ d− 1). If m = 1

or d − 1, then Pmd−1 is already a (d − 1)-dimensional simplex. Suppose 2 ≤ m ≤

d− 2. To triangulate Pmd−1 into (d− 1)-dimensional simplexes, we again need only to

triangulate the boundary of Pmd−1 into (d−2)-dimensional simplexes and then connect


the barycenter of Pmd−1 to all the vertices in such a (d − 2)-dimensional simplicial

triangulation. The boundary of Pmd−1 is the union of the following sets:

Pmd−1 ∩ ξ ∈ Cd : ξj = 0 and Pm

d−1 ∩ ξ ∈ Cd : ξj = 1, j = 1, · · · , d.Each of these sets is either already a (d − 2)-dimensional simplex (if m = 2 andξj = 1) or still a first type polygon but of one-dimension lower. For d = 3, both P 1

2

and P 22 are already 2-dimensional simplexes. Therefore, we conclude by induction

that, for d ≥ 3 in general, all the first type (d−1)-dimensional polygons Pmd−1 ⊂ R

d−1

(m = 1, · · · , d − 1) can be triangulated into (d − 1)-dimensional simplexes. Noticethat

Pmd−1 ∩ ξ ∈ Cd : ξj = 1 = ej + Pm−1

d−1 ∩ ξ ∈ Cd : ξj = 0,j = 1, · · · , d, m = 2, · · · , d− 1, (3.5)

where ej ∈ Rd is the point with the j-th coordinate 1 and all others 0.

Consider now a second type (d− 1)-dimensional polygon F l,j,md−1 (2 ≤ l ≤ d− 1, 1 ≤

j ≤ d, m = 0, 1). If l = 2 and m = 1, or l = d− 1 and m = 0, then F l,j,md−1 is already a

(d − 1)-dimensional simplex. Otherwise, F l,j,md−1 is a Bl

d-type but (d − 1)-dimensionalpolygon, cf. (3.2). For d = 3, there are altogether six of such 2-dimensional polygonsF 2,j,m

2 (j = 1, 2, 3, m = 0, 1). All of them are 2-dimensional simplexes. So, by

induction, the second type (d− 1)-dimensional polygons F l,j,md−1 with d ≥ 3 can all be

triangulated into (d− 1)-dimensional simplexes. Notice that

F l,j,1d−1 = ej + F l−1,j,0

d−1 , l = 2, · · · , d− 1, j = 1, · · · , d. (3.6)

Finally, for each l ∈ 2, · · · , d − 1, we connect the barycenter of polygon B ld to

all the vertices in the constructed (d− 1)-dimensional triangulation of the boundaryof Bl

d. This results in a triangulation of B ld into d-dimensional simplexes. All these

simplexes in the triangulation of B ld for l = 2, · · · , d− 1, together with the simplexes

B1d = Sd and Bd

d , form a simplicial triangulation of the unit cube Cd.By the construction, cf. (3.2), (3.5), and (3.6), the simplicial triangulation of the

reference unit cube Cd satisfies the following properties.

1. The unit simplex Sd is a simplicial element of the triangulation.2. Triangulation symmetry: for each integer i with 1 ≤ i ≤ d, the restriction of

the simplicial triangulation of the reference unit cube Cd on the two faces ξi = 0and ξi = 1 results in the same (d−1)-dimensional simplicial triangulation of the(d− 1)-dimensional unit cube

C id−1 =

(ξ1, · · · , ξi−1, ξi+1, · · · , ξd) ∈ Rd−1 :

0 ≤ ξj ≤ 1, j = 1, · · · , i− 1, i+ 1, · · · , d

.

3. For any integer j with 1 ≤ j ≤ d, the plane∑d

i=1 ξi = j does not intersect theinterior of any simplicial element of the triangulation.

12 BO LI

Step 2. Fix an integer n ≥ 1 and use planes xi = j/n (i = 1, · · · , d, j = 0, · · · , n)

to cut the unit cube [0, 1]d into nd small cubes. Let cd =∏d

i=1[x0i , x

0i + 1/n] denote

a typical such small cube. Define G : Cd → cd by G(ξ) = (1/n)ξ + x0 for allξ ∈ Cd, where x0 =

(

x01, · · · , x0

d

)

. Obviously, it is a one-to-one and onto, orientationpreserving, and affine mapping from the reference unit cube Cd to the small cubecd. Therefore, together with the constructed simplicial triangulation of the referenceunit cube Cd, the mapping G : Cd → cd defines a simplicial triangulation of the smallcube cd. By the arbitrariness of cd and the property of triangulation symmetry ofthe simplicial triangulation of the reference unit cube, we have in fact constructed asimplicial finite element mesh of the unit cube [0, 1]d. The mesh size is h = σd/n,where σd is the maximum of diameters of simplexes in the constructed triangulationof the reference unit cube Cd.

The constructed simplicial finite element mesh of the unit cube [0, 1]d satisfies thefollowing properties.

1. Each small cube cd =∏d

i=1[x0i , x

0i + 1/n] contains one simplicial element

sd =

(x1, · · · , xd) ∈ cd :d∑

i=1

(

xi − x0i

)

≤ 1

n

,

which is a translation of the simplex (1/n)Sd = σ−1d hSd. Thus, there are nd =

(σd/h)d simplicial elements in the mesh that are translations of the single simplex

σ−1d hSd.

2. The plane∑d

i=1 xi = 1 does not intersect the interior of any simplicial element.

3. If the plane∑d

i=1 xi = 1 intersects the interior of a small cube cd =∏d

i=1[x0i , x

0i +

1/n], then the simplex sd must be in Ω.

The first property follows from the first property in Step 1 and our construction ofthe finite element mesh of the unit cube [0, 1]d. To show the other two properties, we

consider a typical small cube cd =∏d

i=1[x0i , x

0i +1/n]. After the change of variables ξ =

n(x−x0), where x0 = (x01, · · · , x0

d), the cube cd is transformed into the reference unit

cube Cd and the plane∑d

i=1 xi = 1 into∑d

i=1 ξi = j0, where j0 = n(

1 −∑di=1 x

0i

)

.

Notice that j0 is an integer, since all nx0i (i = 1, . . . , d) are integers. If j0 /∈ 1, · · · , d−

1, then the plane∑d

i=1 xi = 1 does not cut the interior of the small cube cd, byour triangulation of the reference unit cube Cd, cf. (3.2) and (3.3). Otherwise,

j0 ∈ 1, · · · , d− 1. In this case, the plane∑d

i=1 xi = 1 cuts the interior of the smallcube cd but not the interior of any simplicial element, by the last property stated inStep 1. Moreover,

∑di=1(x

0i + 1/n) ≤ 1, since j0 ≥ 1. Hence, the small simplex sd is

contained in Ω.Step 3. Cut the constructed simplicial finite element mesh of the unit cube by

the plane∑d

i=1 xi = 1. By the second property in Step 2, we have constructed asimplicial finite element mesh τh of the domain Ω. Since the d-dimensional volumeof Ω is 1/d! and that of each small cube is 1/nd, it follows from the first property in


Step 2, this mesh τh contains O(h−d) simplicial elements which are translations of thesingle simplex σ−1

d hSd = σ−1d hΩ.

Letting n = 1, · · · , we then obtain a family of simplicial finite element meshes τh.Since we only use a single reference triangulation to construct each mesh, the familyof meshes τh are quasi-uniform.

We summarize our results in the following theorem.

Theorem 3.1. Let d ≥ 2 be an integer and Ω ⊂ Rd the d-dimensional open unit

simplex. Then, there exist a quasi-uniform family of simplicial finite element meshesτh of Ω such that each of the meshes τh contains O(h−d) simplicial elements whichare translations of the single simplex σ−1

d hSd, where h is the mesh size of τh andσd > 0 a constant depending only on the dimension d.

4. On the standard Lagrange interpolation for d-dimensional Pk-type

finite elements

Let d ≥ 2 be an integer and Ω ⊂ Rd the open unit simplex defined in (3.1). Let k

be an integer such that k ≥ d+ 1. Define u : Ω → R by

u(x) =

(

1 −d∑

i=1

xi

)k+1−d d∏

i=1

xi ∀x = (x1, · · · , xd) ∈ Ω.

Define also f(x) = −∆u(x) for x ∈ Ω. Obviously, u ∈ H10 (Ω) ∩ C∞(Ω) solves the

boundary value problem−∆u = f in Ω,

u = 0 on ∂Ω.(4.1)

Equivalently, u ∈ H10 (Ω) is the weak solution, defined by

A(u, v) = (f, v) ∀v ∈ H10 (Ω),

where A : H10 (Ω) ×H1

0 (Ω) → R, defined by

A(v, w) = (∇v,∇w) ∀v, w ∈ H10 (Ω),

is the bilinear form associated with the boundary value problem (4.1), It is symmetric,continuous, and coercive.

Let n ≥ 1 be an integer and τh the corresponding simplicial finite element meshof Ω constructed in Section 3. Let Sh

k (Ω) ⊂ H1(Ω) denote the Pk-type finite elementspace corresponding to the mesh τh, i.e., the restriction Sh

k (Ω)|T is exactly Pk|T for

each element T ∈ τh. Let

Shk(Ω) = Sh

k (Ω) ∩ H10 (Ω). The finite element solution

uh ∈

Shk(Ω) is defined by


Shk(Ω).

Finally, denote by Ih : C(Ω) → Shk (Ω) the standard Lagrange interpolation whose

interpolation points are the principle lattice points of all simplex elements of τh [3,9].

14 BO LI

Theorem 4.1. Let d and k be integers such that d ≥ 2 and k ≥ d + 1. With thequasi-uniform family of simplicial finite element meshes constructed in Section 3, wehave that

‖Ihu− uh‖H1(Ω) ≥ ζd,khk, (4.2)

where ζd,k > 0 is a constant depending only on d and k.

Proof. We shall call T ∈ τh a corner simplicial element if

T =

(x1, · · · , xd) ∈ Rd : xi −jin

≥ 0, i = 1, · · · , d,d∑

i=1

(

xi −jin

)

≤ 1

n

for some integers ji with 0 ≤ ji ≤ n− 1, i = 1, . . . , d. For such an element, we denoteits d+ 1 vertices by

x(0) =

(

j1n, · · · , jd

n

)

,

x(i) =

(

j1n, · · · , ji−1

n,ji + 1

n,ji+1

n, · · · , jd

n

)

, i = 1, · · · , d.

For each x ∈ T , let λi(x) (i = 0, · · · , d) be the barycentric coordinates of x definedby λi ∈ P1|T , and λi

(

x(j))

= 1 if i = j and 0 if i 6= j. Explicitly,

λi(x) = n

(

xi −jin

)

, i = 1, · · · , d, x = (x1, · · · , xd) ∈ T,

λ0(x) = 1 −d∑

i=1

λi(x), x ∈ T.

Define ψT : T → R by

ψT (x) =

[

d∏

i=1

λi(x)

]

k−d−1∏

j=0

[

λ0(x) −j

k

]

∀x ∈ T.

We claim that ψT differs only by a nonzero constant from the local shape functionassociated with the nodal point

x =

(

j1n

+1

nk, · · · , jd

n+

1

nk

)

∈ T

whose barycentric coordinate is

(λ0(x), λ1(x), · · · , λd(x)) =

(

k − d

k,1

k, · · · , 1

k

)

∈ Rd+1.

In fact, ψT ∈ Pk|T . Moreover, any nodal point x ∈ T has the barycentric coordinatesλi(x) = mi/k for some integer mi with 0 ≤ mi ≤ k, i = 0, · · · , d, and m0 =

k −∑di=1mi. If 0 ≤ m0 ≤ k − d − 1, then ψT (x) = 0. If k − d + 1 ≤ m0 ≤ k,

then at least one mi = 0 (1 ≤ i ≤ d), implying that ψT (x) = 0. If m0 = k − d,


then∑d

i=1mi = d. In this case, if for some i (1 ≤ i ≤ d) mi = 0, then ψT (x) = 0.Otherwise, all mi = 1 (i = 1, · · · , d), and x = x. But, ψT (x) = (k − d)!/kk > 0.

Denoting by IT : C(T ) → Pk|T the local Lagrange interpolation operator on T—therestriction of Ih : C(Ω) → Sk

h(Ω) onto C(T ), we then have

(IT (λ0ψT ))(x) = λ0(x)ψT (x) =k − d

kψT (x) ∀x ∈ T.

Consequently, since on T , u(x)−n−k−1λ0(x)ψT (x) is a polynomial of degree ≤ k, andIT : C(T ) → Pk|T is a projection on Pk|T , we have that

u(x) − (ITu)(x) = n−k−1 [(λ0ψT )(x) − (IT (λ0ψT ))(x)]

= n−k−1

[

λ0(x) −k − d

k

]

ψT (x) (4.3)

= n−k−1

[

d∏

i=1

λi(x)

]

k−d∏

j=0

[

λ0(x) −j

k

]

∀x ∈ T.

We now define vT ∈ Pd+1|T ⊆ Pk|T by vT (x) =∏d

i=0 λi(x) for all x ∈ T . By asimple calculation, we have that

∆vT (x) = −2n2

d∑

i=1

d∏

j=1, j 6=i

λj(x) ∀x ∈ T. (4.4)

Moreover, using the change of variables ξi = λi(x) (i = 1, · · · , d) from x ∈ T toξ ∈ Sd, we obtain that

‖vT‖2H1(T ) = n−d

∫

Sd

d∏

i=0

ξ2i dξ + n2−d

∫

Sd

∣

∣

∣

∣

∇ξ

( d∏

i=0

ξi

)∣

∣

∣

∣

2

dξ, (4.5)

where ξ0 = 1 −∑di=1 ξi and ∇ξ is the gradient with respect to ξ.

By (4.3), u − ITu vanishes on the boundary of T . Therefore, by integration byparts, (4.3), (4.4), and the change of variables ξi = λi(x) (i = 1, · · · , d) from x ∈ Tto ξ ∈ Sd, we get that

∫

T

∇(u− ITu)(x) · ∇vT (x) dx

= −∫

T

(u− ITu)(x) ∆vT (x) dx (4.6)

= 2n1−k

∫

T

[

d∏

i=1

λi(x)

]

k−d∏

j=0

[

λ0(x) −j

k

]

d∑

i=1

d∏

j=1, j 6=i

λj(x) dx

= 2n1−k−d

∫

Sd

(

d∏

i=1

ξi

)[

k−d∏

j=0

(

ξ0 −j

k

)

]

d∑

i=1

d∏

j=1, j 6=i

ξj dξ.

16 BO LI

Denote by τ ′h the collection of all the corner simplicial elements in τh. Definevh : Ω → R by vh = 0 on all elements in τh\τ ′h and vh = vT on any element T ∈ τ ′h.

We have that vh ∈

Shk(Ω), since for each T ∈ τ ′h, vT ∈ Pk|T vanishes on the boundary

of T . Moreover, we have by (4.5) that

‖vh‖2H1(Ω) = n2−d|τ ′h|

[

n−2

∫

Sd

d∏

i=0

ξ2i dξ +

∫

Sd

∣

∣

∣

∣

∇ξ

( d∏

i=1

ξi

)∣

∣

∣

∣

2

dξ

]

,

where |τ ′h| is the number of elements in τ ′h, and by (4.6) that

A(u− Ihu, vh) =∑

T∈τ ′

h

∫

T

∇(u− ITu)(x) · ∇vT (x) dx = 2µd,k|τ ′h|n1−k−d,

where

µd,k =

∫

Sd

(

d∏

i=1

ξi

)[

k−d∏

j=0

(

ξ0 −j

k

)

]

d∑

i=1

d∏

j=1, j 6=i

ξj dξ

is a constant depending only on d and k. Consequently,

|A(u− Ihu, vh)|‖vh‖H(Ω)

≥(

2|µd,k|√νd

)

|τ ′h|1/2n−k−d/2, (4.7)

where

νd =

∫

Sd

[

d∏

i=0

ξ2i +

∣

∣

∣

∣

∇ξ

( d∏

i=1

ξi

)∣

∣

∣

∣

2]

dξ > 0

is a constant depending only on d.It follows from the construction of the mesh τh in Section 3 that h = σd/n and

|τ ′h| ≥ κdh−d for some constants σd > 0 and κd > 0 that depend only on d. Moreover,

µd,k 6= 0 by Lemma 5.1 below. Therefore, the desired inequality (4.2) follows from(4.7) and Lemma 1.1 with

ζd,k =

√

4κdµ2d,k

σ2k+dd νd,k

> 0,

where we use the fact that the constant M in the continuity condition in Lemma 1.1can be taken as 1 in the present case.

Corollary 4.1. Let d and k be integers such that d ≥ 2 and k ≥ d + 1. With thequasi-uniform family of simplicial finite element meshes constructed in Section 3, wehave that

maxz∈Nh

|u(z) − uh(z)| ≥ θd,khk+1, (4.8)

where Nh is the set of all the standard Lagrange interpolation points and θd,k > 0 isa constant depending only on d and k.


Proof. Notice that (Ihu)(z) = u(z) for all z ∈ Nh. Thus, if (4.8) were not true, thenwe would have

‖Ihu− uh‖L∞(Ω) = o(

hk+1)

as h→ 0.

This would lead to

‖Ihu− uh‖L2(Ω) = o(

hk+1)

as h→ 0,

and further to‖Ihu− uh‖H1(Ω) = o

(

hk)

as h→ 0

by an inverse estimate, contradicting the assertion of Theorem 4.1.

5. Auxiliary Lemmas

Lemma 5.1. We have for any integers d and k satisfying d ≥ 2 and k ≥ d+ 1 that

µd,k :=

∫

Sd

(

d∏

i=1

ξi

)[

k−d∏

j=0

(

ξ0 −j

k

)

]

d∑

i=1

d∏

j=1, j 6=i

ξj dξ 6= 0,

where Sd is the d-dimensional unit simplex defined in (3.4) and ξ0 = 1 −∑di=1 ξi.

Proof. By the symmetry about the variables ξ1, · · · , ξd, and the change of variablesηi = ξi (i = 1, · · · , d− 1) and ηd = 1 −∑d

i=1 ξi, we have

µd,k = d

∫

Sd

(

d∏

i=1

ξi

)[

k−d∏

j=0

(

ξ0 −j

k

)

]

d−1∏

i=1

ξi dξ

= d

∫

Sd

(

d−1∏

i=1

η2i

)(

1 −d∑

i=1

ηi

)

k−d∏

j=0

(

ηd −j

k

)

dη (5.1)

= d

∫ 1

0

dηd

∫ 1−ηd

0

dηd−1 · · ·∫ 1−

Pdi=j+1 ηi

0

dηj · · ·∫ 1−

Pdi=2 ηi

0

dη1

(

d−1∏

i=1

η2i

)(

1 −d∑

i=1

ηi

)

k−d∏

j=0

(

ηd −j

k

)

.

Set

E1 =

∫ 1−Pd

i=2 ηi

0

η21

(

1 −d∑

i=1

ηi

)

dη1

and

Ej =

∫ 1−Pd

i=j+1 ηi

0

η2jEj−1 dηj for j = 2, · · · , d− 1.

By an argument of induction on j (1 ≤ j ≤ d− 1) using the expression

η2j =

[(

1 −d∑

i=j

ηi

)

−(

1 −d∑

i=j+1

ηi

)]2

18 BO LI

=

(

1 −d∑

i=j

ηi

)2

− 2

(

1 −d∑

i=j

ηi

)(

1 −d∑

i=j+1

ηi

)

+

(

1 −d∑

i=j+1

ηi

)2

,

we obtain that

Ej =

[

j∏

i=1

2

(3i− 1)(3i)(3i+ 1)

](

1 −d∑

i=j+1

ηi

)3j+1

, j = 1, · · · , d− 1.

It then follows from (5.1) that

µd,k = d

∫ 1

0

Ed−1

k−d∏

j=0

(

ηd −j

k

)

dηd

= d

[

d−1∏

i=1

2

(3i− 1)(3i)(3i+ 1)

]

∫ 1

0

(1 − ηd)3d−2

k−d∏

j=0

(

ηd −j

k

)

dηd.

This is a nonzero constant by Lemma 5.2 below.

Lemma 5.2. We have for any integers d and k satisfying d ≥ 2 and k ≥ d+ 1 that

Jd,k :=

∫ 1

0

(1 − t)3d−2

k−d∏

i=0

(

t− i

k

)

dt

> 0 if k − d is even,< 0 if k − d is odd.

(5.2)

Proof. Denote q = k − d ≥ 1 and ωl(t) =∏l

i=0(t − ti) for any integer l ≥ 0, wheretr = r/k for any real r.

Case 1: q = k−d is even. Let Ωl(t) =∫ t

0ωl(s) ds. By [15] (Lemma 4 on page 309),

we have Ωq(t) > 0 for all t ∈ (0, tq) and Ωq(tq) = 0. Moreover, for tq < t ≤ 1, we have

Ωq(t) = Ωq(tq) +

∫ t

tq

ωq(s) ds =

∫ t

tq

ωq(s) ds > 0,

since ωq(s) > 0 for all s ≥ tq. Therefore, we obtain by integration by parts that

Jd,k =

∫ 1

0

(1 − t)3d−2Ω′q(t) dt = (3d− 2)

∫ 1

0

(1 − t)3d−3Ωq(t) dt

= (3d− 2)

∫ tq

0

(1 − t)3d−3Ωq(t) dt+ (3d− 2)

∫ 1

tq

(1 − t)3d−3Ωq(t) dt > 0.

Case 2: q = k − d is odd. Direct calculations lead to

Jd,d+q = −(d− 1)γq(d)

(d+ q)q

q∏

i=−1

(3d+ i)−1 < 0, q = 1, 3, 5, 7, (5.3)

where

γ1(d) = 1 > 0,

γ3(d) = 12(4d2 − d+ 3) > 0,


γ5(d) = 6(1257d4 + 513d3 + 3031d2 − 225d+ 2400) > 0,

γ7(d) = 72(36163d6 + 83793d5 + 250663d4 + 144355d3

+ 410070d2 + 53452d+ 352800) > 0.

Therefore, we may and shall assume that q = k − d ≥ 9. We have

Jd,k =

∫ tq

0

(1 − t)3d−2ωq(t) dt+

∫ 1

tq

(1 − t)3d−2ωq(t) dt = Id,k +Md,k. (5.4)

By straight forward calculations, we obtain that

Md,k :=d∑

j=1

∫ tq+j

tq+j−1

(1 − t)3d−2

q∏

i=0

(t− ti) dt

<

d∑

j=1

[

q∏

i=0

(tq+j − ti)

]

∫ tq+j

tq+j−1

(1 − t)3d−2dt

=1

(3d− 1)k2d+k

d∑

j=1

[

q∏

i=0

(i+ j)

]

[

(d− j + 1)3d−1 − (d− j)3d−1]

(5.5)

=1

(3d− 1)k2d+k

[

q∏

i=0

(1 + i)

]

d3d−1

+d−1∑

j=1

[

q∏

i=0

(i+ j + 1) −q∏

i=0

(i+ j)

]

(d− j)3d−1

=q + 1

(3d− 1)k2d+k

d−1∑

j=0

[

q∏

i=1

(i+ j)

]

(d− j)3d−1,

where in the third step we use the Abel summation identity

d∑

j=1

uj(vj−1 − vj) = u1v0 − udvd +d−1∑

j=1

(uj+1 − uj)vj

with uj =∏q

i=0(i+ j) and vj = (d− j)3d−1.Since q ≥ 9 is odd, we have by [15] (Lemma 2 on page 309) that ωq(t) = ωq(tq − t)

for all t ∈ [tq/2, tq]. Thus, by the change of variable tq − t → s from t ∈ [tq/2, tq] tos ∈ [0, tq/2], we get

Id,k :=

∫ tq/2

0

(1 − t)3d−2ωq(t) dt+

∫ tq

tq/2

(1 − t)3d−2ωq(tq − t) dt

=

∫ tq/2

0

[

(1 − t)3d−2 + (1 − tq + t)3d−2]

ωq(t) dt

20 BO LI

=

q0∑

j=1

∫ t2j

t2j−2

[

(1 − t)3d−2 + (1 − tq + t)3d−2]

ωq(t) dt (5.6)

+

∫ tq/2

t2q0

[

(1 − t)3d−2 + (1 − tq + t)3d−2]

ωq(t) dt

=

q0∑

j=1

Hd,k,j +Gd,k,

where

q0 =

(q − 3)/4 if (q − 1)/2 is odd,(q − 1)/4 if (q − 1)/2 is even.

We show now that

Gd,k :=

∫ tq/2

t2q0

[

(1 − t)3d−2 + (1 − tq + t)3d−2]

ωq(t) dt < 0. (5.7)

If (q − 1)/2 is even, then for any t ∈ (t2q0 , tq/2), ωq(t) has 2q0 − 1 negative factors.Hence, it is negative. Thus (5.7) holds true. If (q − 1)/2 is odd, then for anyt ∈ (t2q0+1/2, t2q0+1), ωq(t) < 0, since it has 2q0 + 3 negative factors. Hence, bythe change of variable t → t − 1/k from [t2q0+1, t2q0+3/2] to [t2q0 , t2q0+1/2], we obtainthat

Gd,k =

(

∫ t2q0+1/2

t2q0

+

∫ t2q0+1

t2q0+1/2

+

∫ t2q0+3/2

t2q0+1

)

[

(1 − t)3d−2 + (1 − tq + t)3d−2]

ωq(t) dt

<

∫ t2q0+1/2

t2q0

[

(1 − t)3d−2 + (1 − tq + t)3d−2]

ωq(t) dt

+

∫ t2q0+3/2

t2q0+1

[

(1 − t)3d−2 + (1 − tq + t)3d−2]

ωq(t) dt

=

∫ t2q0+1/2

t2q0

[

(1 − t)3d−2 + (1 − tq + t)3d−2]

ωq(t) dt

+

∫ t2q0+1/2

t2q0

[

(

1 − 1

k− t

)3d−2

+

(

1 − tq +1

k+ t

)3d−2]

ωq

(

1

k+ t

)

dt

= −∫ t2q0+1/2

t2q0

gd,k(t)ωq−1(t) dt,

where

gd,k(t) = (tq − t)fk,d(t) −(

t+1

k

)

fd,k

(

t+1

k

)

, (5.8)

and

fd,k(t) = (1 − t)3d−2 + (1 − tq + t)3d−2. (5.9)


For t ∈ (t2q0 , t2q0+1/2), there are 2q0 + 2 negative factors in ωq−1(t). So, ωq−1(t) > 0.Moreover, fd,k(t) > 0 and f ′

d,k(t) < 0 for all t ∈ (0, tq/2). Thus,

gd,k(t) ≥ (tq − t)fd,k(t) −(

t+1

k

)

fd,k(t) ≥1

kfd,k(t) > 0 ∀t ∈ (t2q0 , t2q0+1/2).

Therefore, (5.7) also holds true.Fix now j ∈ 1, · · · , q0. By the change of variable t → t− 1/k from [t2j−1, t2j] to

[t2j−2, t2j−1], we get that

Hd,k,j :=

∫ t2j

t2j−2

[

(1 − t)3d−2 + (1 − tq + t)3d−2]

ωq(t) dt

=

(

∫ t2j−1

t2j−2

+

∫ t2j

t2j−1

)

[

(1 − t)3d−2 + (1 − tq + t)3d−2]

ωq(t) dt (5.10)

= −∫ t2j−1

t2j−2

gd,k(t)ωq−1(t) dt,

where gd,k is defined in (5.8). For each t ∈ (t2j−2, t2j−1), ωq−1(t) has 4q0 − 2j + 4

negative factors. So, ωq−1(t) > 0. Using the fact that (−1)sf(s)d,k(t) > 0 for all

t ∈ (0, tq/2) and s = 0, 1, 2, where fd,k is defined in (5.9), we easily obtain that

gd,k(t) ≥(

tq − 2t− 1

k

)

fd,k

(

t+1

k

)

≥ 4

kfd,k

(

t+1

k

)

> 0

and

g′d,k(t) = −[

fd,k(t) + fd,k

(

t+1

k

)]

+ (tq − t)f ′d,k(t) −

(

t+1

k

)

f ′d,k

(

t+1

k

)

<

(

tq − 2t− 1

k

)

f ′d,k

(

t+1

k

)

(5.11)

< 0

for all t ∈ (t2j−2, t2j−1). Therefore,

Hd,k,j < 0, j = 1, · · · , q0. (5.12)

By (5.11), we have that

gd,k(t) ≥ gd,k(t1) = tq−1fd,k(t1) − t2f(t2) > tq−3fd,k(t1) > 0, t0 < t < t1.

Consequently, by (5.6), (5.7), (5.10), (5.12), and the fact that ωq−1(t) > 0 for t ∈(t0, t1), we conclude that

Id,k < Hd,k,1

< −gd,k(t1)

∫ t1

t0

ωq−1(t) dt (5.13)

22 BO LI

< −tq−3fd,k(t1)

q−1∏

i=2

(ti − t1)

∫ t1

t0

t(t1 − t) dt

< −(q − 3)(q − 2)!

6k2d+k(q + d− 1)3d−2.

It follows now from (5.4), (5.5), and (5.13) that we need only to show that

6(q + 1)

(3d− 1)(q − 3)

d−1∑

j=0

[

q∏

i=1

(i+ j)

]

(d− j)3d−2 ≤ (q − 2)!(q + d− 1)3d−2

for all the integers d ≥ 2 and q ≥ 9. For d = 2, one can easily verify that thisinequality holds true for all q ≥ 9. Therefore, since

6(q + 1)

(3d− 1)(q − 3)=

2

d

(

1 +1

3d− 1

)(

1 +4

q − 3

)

≤ 4

d, d ≥ 3, q ≥ 9,

to complete the proof of the lemma, we need only to show that

d−1∑

j=0

[

q∏

i=1

(i+ j)

]

(d− j)3d−2 ≤ d

4(q − 2)!(q + d− 1)3d−2, d ≥ 3, q ≥ 9. (5.14)

For each index j with 0 ≤ j ≤ d− 1, we have by the binomial formula that

(q − 2)!(q + d− 1)3d−2

= (q − 2)! [(d− j) + (q + j − 1)]3d−2

≥ (q − 2)!3d−2∑

m=j

(

3d− 2m

)

(d− j)3d−2−m(q + j − 1)m

=

[

q∏

i=1

(i+ j)

]

(d− j)3d−2

3d−2∑

m=j

[(3d− 2) · · · (3d− 1 −m)](q + j − 1)m

[(j + 1) · · ·m] [(q − 1) · · · (q + j)](d− j)m

≥[

q∏

i=1

(i+ j)

]

(d− j)3d−2

3d−2∑

m=j

[(3d− 2) · · · (3d− 1 −m)](q + j − 1)m−j−1

[(j + 1) · · ·m](q + j)(d− j)m

=

[

q∏

i=1

(i+ j)

]

(d− j)3d−2Sj,

that is,

(q − 2)!(q + d− 1)3d−2S−1j ≥

[

q∏

i=1

(i+ j)

]

(d− j)3d−2, j = 0, · · · , d− 1, (5.15)

where

Sj =3d−2∑

m=j

[(3d− 2) · · · (3d− 1 −m)](q + j − 1)m−j−1

[(j + 1) · · ·m](q + j)(d− j)m, j = 0, · · · , d− 1.


For j = 0 and j = 1, keeping only the term with m = 4 and m = 5 in the summationS0 and S1, respectively, and using the fact that q ≥ 9, we get that

S0 ≥[(3d− 2)(3d− 3)(3d− 4)(3d− 5)](q − 1)3

24qd4

≥ [(2d) · (2d) · d · d](q − 1)3

24qd4(5.16)

≥ 9

and that

S1 ≥[(3d− 2)(3d− 3)(3d− 4)(3d− 5)(3d− 6)]q3

120(q + 1)(d− 1)5

≥ [3(d− 1) · 3(d− 1) · 2(d− 1) · 2(d− 1) · (d− 1)]q3

120(q + 1)(d− 1)5(5.17)

≥ 21.

For 2 ≤ j ≤ d− 1, we get by keeping only the term with m = min(3j − 1, q + j − 1)in the sum Sj that

Sj ≥[(3d− 2) · · · (3d− 1 −m)](q + j − 1)m−j−1

[(j + 1) · · ·m](q + j)(d− j)m

≥ [3(d− j)]m(q + j − 1)m−j−1

[(j + 1) · · ·m](q + j)(d− j)m

≥ 3m

(j + 1)(j + 2)

≥ 3q+j−1

(j + 1)(j + 2),

leading to

d−1∑

j=2

S−1j ≤ 31−q

d−1∑

j=2

(j + 1)(j + 2)

3j=

11 · 3d − 6d2 − 24d− 27

4 · 3d+q−1≤ 1

37. (5.18)

It follows from (5.16)–(5.18) that

d−1∑

j=0

S−1j ≤ 1/4.

Consequently, by summing (5.15) over j = 0, · · · , d − 1, we obtain the desired in-equality (5.14), since d ≥ 1. The proof is complete.

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24 BO LI

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Department of Mathematics, University of Maryland, College Park, MD 20742,

U.S.A.

E-mail address: [email protected]

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