Lagrange interpolation and finite element superconvergence

Lagrange Interpolation and Finite ElementSuperconvergenceBo LiDepartment of Mathematics, University of Maryland, College Park, Maryland 20742

Received 30 September 2002; accepted 3 May 2003

DOI 10.1002/num.10078

We consider the finite element approximation of the Laplacian operator with the homogeneous Dirichletboundary condition, and study the corresponding Lagrange interpolation in the context of finite elementsuperconvergence. For d-dimensional Qk-type elements with d � 1 and k � 1, we prove that theinterpolation points must be the Lobatto points if the Lagrange interpolation and the finite element solutionare superclose in H1 norm. For d-dimensional Pk-type elements, we consider the standard Lagrangeinterpolation—the Lagrange interpolation with interpolation points being the principle lattice points ofsimplicial elements. We prove for d � 2 and k � d � 1 that such interpolation and the finite elementsolution are not superclose in both H1 and L2 norms and that not all such interpolation points aresuperconvergence points for the finite element approximation. © 2003 Wiley Periodicals, Inc. Numer MethodsPartial Differential Eq 20: 33–59, 2004

Keywords: finite element; Lagrange interpolation; superconvergence

1. INTRODUCTION

Consider the boundary value problem

��u � f in �,u � 0 on ��, (1.1)

where � � �d is a bounded domain with a Lipschitz continuous boundary ��, d � 1, f � L2(�),and � : H2(�)3 L2(�) is a second-order, linear, self-adjoint, elliptic differential operator. Letu � H0

1(�) be its unique weak solution, defined by

A�u, v� � � f, v� @v � H01��,

Correspondence to: B. Li (e-mail: [email protected])Contract grant sponsor: National Science Foundation; contract grant number: DMS-00729582000 Mathematics Subject Classification: 65N30

© 2003 Wiley Periodicals, Inc.

where A : H01(�) � H0

1(�) 3 � is the bilinear, symmetric, continuous, and coercive formassociated with (1.1), and (�, �) denotes the inner product of L2(�). Let {�h} be a family of finiteelement meshes of the domain � with the mesh size h 3 0. Fix an integer k � 1. For each h,let Sk

h(�) � H1(�) � C(�� ) be the corresponding finite element space such that Skh(�)�T � Pk�T

for all T � �h, where Pk is the set of all polynomials of degree � k. Let Skh(�) � Sk

h(�) � H01(�).

Let uh � Skh(�) be the finite element solution, defined by

A�uh, vh� � � f, vh� @vh � Skh��.

Finally, let Ih : C(�� ) 3 Skh(�) denote the corresponding Lagrange interpolation operator. The

following estimate

h�1�Ihu � uh�L2�� Ihu � uh�H1�� Chk (1.2)

is standard, provided that the weak solution u � H01(�) � C(�� ) is smooth enough and the

underlying meshes are quasi-uniform [1, 2]. Here and below, we use the letter C to denote ageneric, 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 improved. Thismeans that Ihu and uh can be superclose. More precisely, we say that the Lagrange interpolationIhu and the finite element solution uh are superclose in H1 norm, if

�Ihu � uh�H1�� o�hk� as h 3 0.

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

�Ihu � uh�H1�� Chk��. (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 � H01(�) � C(�� ), then

�Ihu � uh�H1�� supvh�S k

h��,vh0

�A�u � Ihu, vh��vh�H1��

� M�Ihu � uh�H1��,

where � 0 and M � 0 are the two constants in the conditions of coercivity and continuity,respectively, of the bilinear form A : H0

1(�) � H01(�) 3 �,

A�v, v� � �v�H1��2 @v � H0

1��

and

�A�v, w�� M�v�H1��w�H1�� @v, w � H01��.

The supercloseness between the Lagrange interpolation and the finite element solution isclosely related to the superconvergence of the finite element solution to the exact solution. In

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fact, if (1.3) holds true, then one can easily obtain the following estimate of gradient super-convergence:

� 1

hd �z��Z h��

�u�z� � �uh�z��2� 1/2

� Chk�min��,1�,

where �Zh(�) is the set of superconvergence points for the gradient of the Lagrange interpolationand � is some kind of average of the gradient [3]. In some cases, one can obtain a higher orderestimate

�A�u � Ihu, vh�� Chk��vh�W1,p�� @vh � Skh�� (1.4)

for some � � 0 and p � [1, �). This, together with delicate estimates of a discrete Green’sfunction substituting vh in the inequality in (1.4), can lead to pointwise finite element super-convergence estimates [4–7]. By Lemma 1.1, the estimate (1.4) is equivalent to the superclose-ness estimate (1.3), if p � 2.

In this work, we study the supercloseness between the Lagrange interpolation and the finiteelement 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 interpolation and the finiteelement 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, the standardLagrange interpolation—the Lagrange interpolation with its interpolation points being theprinciple lattice points of simplicial elements—and the finite element solution are notsuperclose in H1 norm (cf. Theorem 4.1).

3. For d-dimensional Pk-type elements with d � 2 and k � d � 1, not all the standardLagrange interpolation points are superconvergence points for the finite element solution(cf. Corollary 4.1).

For d-dimensional Qk-type elements with d � 1 and k � 2, the finite element solution issuperconvergent by one order to the exact solution at the Lobatto points [8–13]. This impliesthat the Lagrange interpolation associated with the Lobatto points and the finite element solutionare superclose by one order in H1 norm. Here, we prove the converse under the assumption thatthey are only superclose, but not necessary superclose by any order, in H1 norm.

For simplicial finite elements, the Lagrange interpolation points can not be arbitrarilydistributed in general. With good meshes, the standard Lagrange interpolation and the finiteelement solution are in fact superclose in H1 norm by one order for two-dimensional P1 and P2

elements and for three-dimensional P1 element [4–7, 14–25]. Recently, similar results havebeen obtained for any d-dimensional, linear, simplicial finite elements with a uniform mesh [26].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 boundaryintegrals in delicate estimates of the integral form A(u � Ihu, vh) for vh � Sk

h(�). However, suchcancellation seems to be impossible if there exists an element-wise, bubble-like test functionvh � Sk

h(�) that vanishes on the boundary of each element. Such a function exists if and onlyif there exists an interior node in each of the simplicial elements. This turns out to be true if and

LAGRANGE INTERPOLATION AND FINITE ELEMENT SUPERCONVERGENCE 35

only if k � d � 1 for d-dimensional Pk-type elements. Constructing a bubble-like test functionto avoid any possible cancellation was the original approach in our early work [27] to show thenonsupercloseness for two-dimensional P3 element. Here, we extend such an approach to ageneral case that is more complicated because of the higher space dimension and higherpolynomial degree.

In proving the nonsupercloseness of the Lagrange interpolation to the finite element solutionfor the general d-dimensional Pk-type finite elements with k � d � 1, we choose the underlyingdomain to be the unit d-dimensional simplex. This allows us to have a polynomial of degreeexactly k � 1 as the solution of the underlying Poisson equation with the homogeneous Dirichletboundary condition. In addition, we construct a special family of quasi-uniform finite elementmeshes consisting of enough elements that are scaled translations of the unit simplex. Suchmeshes are uniform for d � 2 but nonuniform for d � 3. Calculations based on such meshes aremuch simplified. With our approach, it is possible to consider a uniform family of finite elementmeshes of 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 elements. InSection 3, we construct a quasi-uniform family of simplicial finite element meshes of ad-dimensional domain for d � 2. With such meshes, we study in Section 4 the standardLagrange interpolation for Pk-type elements. Finally, in Section 5, we prove some auxiliarylemmas.

2. OPTIMAL LAGRANGE INTERPOLATION POINTS FOR QK-TYPE FINITEELEMENTS

Consider the boundary value problem

��u � f in �,u � 0 on ��, (2.1)

where f � L2(�) and � � m�1d (am, bm) � �d is a d-dimensional rectangular parallelpiped with

d � 1 and �� am � bm � � for all m � 1, . . . , d. The associated bilinear form A : H01(�) �

H01(�) 3 � is defined by

A�v, w� � �v, w� @v, w � H01��.

It is symmetric, continuous, and coercive. The weak solution u � H01(�) of the boundary value

problem (2.1) is defined by

A�u, v� � � f, v� @v � H01��.

Let {�h} be a family of quasi-uniform rectangular meshes covering � with the mesh size h3 0. We denote a typical mesh by

�h � � �m�1

d

�xm,jm�1, xm,jm� : jm � 1, . . . , nm, m � 1, . . . , d� ,

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where xm, jm� am � jmhm for jm � 0, . . . , nm, hm � (bm � am)/nm, nm � 1 is an integer for each

m � 1, . . . , d, and h � max1�m�dhm. For an integer k � 1, let Skh(�) � H1(�) denote the

Qk-type finite element space corresponding to the mesh �h, i.e., the restriction Skh(�)�R is exactly

Qk�R for each element R � �h, where Qk � span{x11 . . . xd

d : 1, . . . , d are nonnegativeintegers, 1 � . . . � d � k}. Let Sk

h(�) � Skh(�) � H0

1(�). The finite element solution uh �Sk

h(�) is defined by

A�uh, vh� � � f, vh� @vh � Skh��.

For each integer m with 1 � m � d, let �m(0), . . . , �m

(k) be k � 1 distinct real numbers satisfying

�1 � �m�0� � · · · � �m

�k� � 1.

We call all the points (�1(i1), . . . , �d

(id) (im � 0, . . . , k, m � 1, . . . , d ) the reference interpolationpoints. We define the Lagrange interpolation points on each element m�1

d [xm, jm�1, xm, jm] � �h

(1 � jm � nm, 1 � m � d ) by

xm,jm

�i � �hm�m

�i � � xm,jm�1 � xm,jm

2, i � 0, . . . , k, m � 1, . . . , d.

Finally, we denote by Ih : C(�� ) 3 Skh(�) the Lagrange interpolation operator associated with

these interpolation points.Recall that the Jacobi polynomials Pn

(1,1) (n � 0, 1, . . . ) are orthogonal polynomials on theinterval [�1, 1] with the weight (�) � 1 � �2, normalized by Pn

(1,1)(1) � n � 1 [28]. TheRodrigues’ formula for Pn

(1,1) is

Pn�1,1��

��1�n

2nn!�1 � �2� d

d�n

��1 � �2�n�1�.

For each n � 1, Pn(1,1) has exactly n distinct roots in (�1, 1), called Lobatto points (associated

with n). Recall also that the Legendre polynomials are orthogonal polynomials on the interval[�1, 1] with the weight (�) � 1 [28]. They are given by

Ln�� 1

2nn! d

d�n

��1 � �2�n�, n � 0, 1, . . . .

It is easy to show that {L�n(�)}n�1� is also a sequence of orthogonal polynomials on [�1, 1] with

the weight (�) � 1 � �2. Consequently, Pn(1,1) and L�n�1 differ only by a nonzero constant. For

n � 2, the Lobatto points associated with n � 1 are thus the roots of L�n(�) in (�1, 1). However,for convenience, we shall call in what follows all the n � 1 distinct roots of L�n(�) in (�1, 1),together with �1, the Lobatto points of order n. (In fact, �1 are often included in a Lobattoquadrature [29].) We call a point in �d a d-dimensional Lobatto point of order n, if each of itsd coordinates is a one-dimensional Lobatto point of order n. Obviously, there are (n � 1)d

d-dimensional Lobatto points of order n.


Together with what is known, the following result implies for Qk-type finite elements withk � 2 that the Lagrange interpolation is superclose to the finite element solution in H1 norm ifand only if all the interpolation points are the Lobatto points.

Theorem 2.1. Suppose that

�Ihu � uh�H1�� o�hk� as h 3 0, (2.2)

whenever the solution u � H01(�) is smooth enough. Then, all the reference interpolation points

(�1(i1), . . . , �d

(id)) (0 � im � k, m � 1, . . . , d ) must be the d-dimensional Lobatto points of or-der k.

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 �m

(0), . . . , �m(k)

are indeed the k � 1 one-dimensional Lobatto points of order k.Fix an index m with 1 � m � d. Define u � H0

1(�) by

u�x� � xm �am � bm

2 k�1 �l�1

d

�xl � al��xl � bl�, x � �x1, . . . , xd� � �� .

Note that u depends on m. Define accordingly f (x) � ��u(x) for all x � �. Obviously, f �L2(�), and u � C�(�� ) solves the boundary value problem (2.1).

For each integer s : 0 � s � k � 2, define vs : [am, bm] 3 � by

vs�xm� � �s2xm � xm,jm�1 � xm,jm

hm , @xm � �xm,jm�1, xm,jm�, jm � 1, . . . , nm,

where the function �s : [�1, 1] 3 � is defined by

�s�� s�2 �1

2�1 � ��

1

2��1�s�1 � ��, � � ��1, 1�.

It is easy to see that

� �s�� s � 1��s � 2��s and �s��1� � �s�1� � 0.

Hence, vs is a continuous piecewise polynomial of degree s � 2 � k, vanishing at all the pointsxm, jm

( jm � 0, . . . , nm). Define vh : �� 3 � for the case d � 1 by vh(x1) � vs(x1) for all x1 �� [a1, b1], and for the case d � 2 by

vh�x� � vs�xm�Wm�x�� @x � �� ,

where

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Wm�x�� l�1,lm

d

�xl � al��xl � bl� @x� � ��,

x� � �x1, . . . , xm�1, xm�1, . . . , xd�,

�� l�1,lm

d

�al, bl�.

Note that vh depends on m and that vh � Skh(�).

Assume that d � 2 temporarily. Let R � l�1d [xl, jl�1, xl, jl

] � �h be an arbitrary element,where 1 � jl � nl and 1 � l � d. The function

u�x� � Wm�x�� i�0

k

�xm � xm,jm

�i � �

is in Qk�R, and agrees with u on all the interpolation points in R. Hence, this function is exactlythe Lagrange interpolation of u on the element R. Thus, we have

�u � Ihu��x� � Wm�x�� i�0

k

�xm � xm,jm

�i � � @x � R.

Let R� � l�1,lmd [xl, jl�1, xl, jl

]. Let � denote the gradient operator with respect to x�. Applyingintegration by parts and using the change of variable

�m �2xm � xm,jm�1 � xm,jm

hm

from xm � [xm, jm�1, xm, jm] to �m � [�1, 1], we obtain that

�R

�u � Ihu��x� � vh�x�dx

��R

��u � Ihu��x� � �vh�x� ��

�xm�u � Ihu��x� �

�

�xmvh�x��dx

��R

��Wm�x��2vs�xm� �i�0

k

�xm � xm,jm�i� �dx � �

R

�Wm�x��2� d

dxm�i�0

k

�xm � xm,jm�i� ��v�s�xm�dx

��R�

��Wm�x��2dx� �xm,jm�1

xm,jm

vs�xm� �i�0

k

�xm � xm,jm�i� �dxm


��R�

�Wm�x��2dx� �xm,jm�1

xm,jm

v�s�xm� �i�0

k

�xm � xm,jm�i� �dxm

�hm

2 k�2

��Wm�L2�R��2 �

�1

1

�s��m� �i�0

k

��m � �m�i��d�m

�hm

2 k

�Wm�L2�R��2 �

�1

1

��s��m� �i�0

k

��m � �m�i��d�m.

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

A�u � Ihu, vh� � �R��h

�R

�u � Ihu��x� � vh�x�dx

��bm � am�

2k�2 hmk�1��Wm�L2��

2 ��1

1

�s��m� �i�0

k

��m � �m�i ��d�m

��bm � am��s � 1��s � 2�

2k hmk�1�Wm�L2��

2 ��1

1

�ms �

i�0

k

��m � �m�i ��d�m.

(2.3)

Similarly, we obtain that

�vh�H1��2 � �

�

��vh�x��2 � �vh�x��2�dx

� �Wm�H1��2 �

jm�1

nm �xm,jm�1

xm,jm

�vs�xm��2dxm � �Wm�L2��2 �

jm�1

nm �xm,jm�1

xm,jm

�v�s�xm��2dxm

��bm � am�

2�Wm�H1��

2 ��1

1

��s��m��2d�m �2�bm � am�

hm2 �Wm�L2��

2 ��1

1

��s��m��2d�m,

leading to

�vh�H1�� bm � am��4 � �bm � am�2�

2��s�H1��1,1��Wm�H1��hm

�1. (2.4)

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

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� 1hmk �

�1

1

�ms �

i�0

k

��m � �m�i ��d�m � 2hm

k�2, (2.5)

where

1 � � 2�bm � am�

4 � �bm � am�2

�s � 1��s � 2��Wm�L2��2

2k��s�H1��1,1��Wm�H1�� 0

and

2 � � 2�bm � am�

4 � �bm � am�2

��Wm�L2��2

2k�2��s�H1��1,1��Wm�H1��

�1

1

�s��m� �i�0

k

��m � �m�i ��d�m � 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�

2k h1k�1 �

�1

1

� 1s �

i�0

k

��1 � �1�i ��d�1

and

�vh�H1��2 �

b1 � a1

2 ��1

1 � ��s��1��2 �4

h12 ��s��1��2�d�1.

Therefore,

�vh�H1�� b1 � a1��4 � �b1 � a1�2�

2��s�L2��1,1�h1

�1,

and


� �h1k �

�1

1

�1s �

i�0

k

��1 � �1�i ��d�1 , (2.6)

where

� � � 2�b1 � a1�

4 � �b1 � a1�2

�s � 1��s � 2�

2k��s�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

�ms �

i�0

k

��m � �m�i ��d�m � 0, s � 0, . . . , k � 2.

The polynomial i�1k�1 (� � �m

(i )) of degree k � 1 is thus orthogonal to all the polynomials in Pk�2

on [�1, 1] with the weight (� � �m(0))(� � �m

(k)) � �2 � 1. Hence, it differs from the Jacobipolynomial Pk�1

(1,1) only by a nonzero constant. Consequently, all the points �m(0), . . . , �m

(k) are thek � 1 one-dimensional Lobatto points of order k. y

3. A CONSTRUCTION OF d-DIMENSIONAL SIMPLICIAL FINITE ELEMENTMESHES

We now let d � 2 be an integer and � � �d the open unit simplex

� � � �x1, . . . , xd� � �d : xi � 0, i � 1, . . . , d, �i�1

d

xi � 1� . (3.1)

We shall construct a quasi-uniform family of simplicial finite element meshes {�h} of � suchthat there are O(h�d) elements in �h that are translations of a single d-dimensional simplex�d

�1h�� {�d�1hx : x � �� }, where �d � 0 is a constant depending only on d and h is the mesh

size of �h.For d � 2, the mesh �h 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 a uniform mesh withmesh size h � �2/n. Obviously, there are O(h�2) elements of the mesh that are translations ofthe single two-dimensional simplex �2

�1h�� with �2 � �2.For d � 3, we construct in three steps a simplicial finite element mesh of � with the designed

properties. First, we triangulate the reference unit cube into simplexes. Second, we construct asimplicial finite element mesh of the unit cube by cutting it into many small cubes, triangulatingthem by affine mappings from the triangulated reference unit cube, and gluing them together.Third, we cut the meshed unit cube by the plane ¥i�1

d xi � 1 to define a simplicial finite elementmesh of �.

Step 1. Take the closed unit cube Cd � [0, 1]d � �d as the reference cube and denote by � �(�1, . . . , �d) a generic point in Cd. Define

Bdl � �� Cd : l � 1 � �

i�1

d

�i � l� , l � 1, . . . , d. (3.2)

Obviously,

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� l�1d Bd

l � Cd and int�Bdj � � int�Bd

l � � A if j � l. (3.3)

Notice that Bd1 � Sd, where

Sd � ��1, . . . , �d� � �d : �i � 0, i � 1, . . . , d, �i�1

d

�i � 1� (3.4)

is the reference unit simplex in �d, and both Bd1 and Bd

d are simplexes. But Bdl is not 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 (�1, . . . , �d) (�i � 0 or 1, i � 1, . . . , d ) of Cd lie in the planes¥i�1

d �i � j ( j � 0, . . . , d ), the number of vertices of Cd contained in Bdl is the same as that

contained in the planes ¥i�1d �i � l � 1 and ¥i�1

d �i � l. This number is

dl � 1 � d

l � d � 1l � d � 1,

since 1 � l � d.We now triangulate all the polygons Bd

l (l � 2, . . . , d � 1) into simplexes so that, togetherwith Bd

1 and Bdd, these simplexes can form a simplicial triangulation of Cd. It suffices to

triangulate the boundary of each Bdl into (d � 1)-dimensional simplexes determined by a set of

vertices and then connect the barycenter of Bdl to these vertices. The boundary of Bd

l for each lwith 2 � l � d � 1 is the union of two types of (d � 1)-dimensional polygons

Pd�1m � �� Cd : �

i�1

d

�i � m� , m � l � 1, l,

and

Fd�1l,j,m � Bd

l � �� Cd : �j � m�, j � 1, . . . , d, m � 0, 1.

Consider a first type (d � 1)-dimensional polygon Pd�1m (1 � m � d � 1). If m � 1 or d �

1, then Pd�1m is already a (d � 1)-dimensional simplex. Suppose 2 � m � d � 2. To triangulate

Pd�1m into (d � 1)-dimensional simplexes, we again need only to triangulate the boundary of

Pd�1m into (d � 2)-dimensional simplexes and then connect the barycenter of Pd�1

m to all thevertices in such a (d � 2)-dimensional simplicial triangulation. The boundary of Pd�1

m is theunion of the following sets:

Pd�1m � �� Cd : �j � 0� and Pd�1

m � �� Cd : �j � 1�, j � 1, . . . , d.

Each of these sets is either already a (d � 2)-dimensional simplex (if m � 2 and �j � 1) or stilla first type polygon but of one-dimension lower. For d � 3, both P2

1 and P22 are already

two-dimensional simplexes. Therefore, we conclude by induction that, for d � 3 in general, allthe first type (d � 1)-dimensional polygons Pd�1

m � �d�1 (m � 1, . . . , d � 1) can betriangulated into (d � 1)-dimensional simplexes. Notice that


Pd�1m � �� Cd : �j � 1� � ej � Pd�1

m�1 � �� Cd : �j � 0�,

j � 1, . . . , d, m � 2, . . . , d � 1, (3.5)

where ej � �d is the point with the j-th coordinate 1 and all others 0.Consider now a second type (d � 1)-dimensional polygon Fd�1

l, j,m (2 � l � d � 1, 1 � j �d, m � 0, 1). If l � 2 and m � 1, or l � d � 1 and m � 0, then Fd�1

l, j,m is already a (d �1)-dimensional simplex. Otherwise, Fd�1

l, j,m is a Bdl -type but (d � 1)-dimensional polygon [cf.

(3.2)]. For d � 3, there are altogether six of such two-dimensional polygons F22, j,m ( j � 1, 2, 3,

m � 0, 1). All of them are two-dimensional simplexes. So, by induction, the second type (d �1)-dimensional polygons Fd�1

l, j,m with d � 3 can all be triangulated into (d � 1)-dimensionalsimplexes. Notice that

Fd�1l,j,1 � ej � Fd�1

l�1,j,0, l � 2, . . . , d � 1, j � 1, . . . , d. (3.6)

Finally, for each l � {2, . . . , d � 1}, we connect the barycenter of polygon Bdl to all the

vertices in the constructed (d � 1)-dimensional triangulation of the boundary of Bdl . This results

in a triangulation of Bdl into d-dimensional simplexes. All these simplexes in the triangulation

of Bdl for l � 2, . . . , d � 1, together with the simplexes Bd

1 � Sd and Bdd, 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 �i � 0 and �i � 1 results inthe same (d � 1)-dimensional simplicial triangulation of the (d � 1)-dimensional unitcubeCd�1

i � ��1, . . . , �i�1, �i�1, . . . , �d� � �d�1 :

0 � �j � 1, j � 1, . . . , i � 1, i � 1, . . . , d�.

3. For any integer j with 1 � j � d, the plane ¥i�1d �i � j does not intersect the interior of

any simplicial element of the triangulation.

Step 2. Fix an integer n � 1 and use planes xi � j/n (i � 1, . . . , d, j � 0, . . . , n) to cut theunit cube [0, 1]d into nd small cubes. Let cd � i�1

d [xi0, xi

0 � 1/n] denote a typical such smallcube. Define G : Cd 3 cd by G(�) � (1/n)� � x0 for all � � Cd, where x0 � (x1

0, . . . , xd0).

Obviously, it is a one-to-one and onto, orientation preserving, and affine mapping from thereference unit cube Cd to the small cube cd. Therefore, together with the constructed simplicialtriangulation of the reference unit cube Cd, the mapping G : Cd 3 cd defines a simplicialtriangulation of the small cube cd. By the arbitrariness of cd and the property of triangulationsymmetry of the simplicial triangulation of the reference unit cube, we have in fact constructeda simplicial finite element mesh of the unit cube [0, 1]d. The mesh size is h � �d /n, where �d

is the maximum of diameters of simplexes in the constructed triangulation of the reference unitcube Cd.

The constructed simplicial finite element mesh of the unit cube [0, 1]d satisfies the followingproperties:

1. Each small cube cd � i�1d [xi

0, xi0 � 1/n] contains one simplicial element

44 LI

sd � � � x1, . . . , xd� � cd : �i�1

d

� xi � xi0� �

1

n� ,

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

simplicial elements in the mesh that are translations of the single simplex �d�1hSd.

2. The plane ¥i�1d xi � 1 does not intersect the interior of any simplicial element.

3. If the plane ¥i�1d xi � 1 intersects the interior of a small cube cd � i�1

d [xi0, xi

0 � 1/n],then the simplex sd must be in �� .

The first property follows from the first property in Step 1 and our construction of the finiteelement mesh of the unit cube [0, 1]d. To show the other two properties, we consider a typicalsmall cube cd � i�1

d [xi0, xi

0 � 1/n]. After the change of variables � � n(x � x0), where x0 �(x1

0, . . . , xd0), the cube cd is transformed into the reference unit cube Cd and the plane ¥i�1

d xi �1 into ¥i�1

d �i � j0, where j0 � n(1 � ¥i�1d xi

0). Notice that j0 is an integer, since all nxi0 (i �

1, . . . , d ) are integers. If j0 � {1, . . . , d � 1}, then the plane ¥i�1d xi � 1 does not cut the

interior of the small cube cd, by our triangulation of the reference unit cube Cd [cf. (3.2) and(3.3)]. Otherwise, j0 � {1, . . . , d � 1}. In this case, the plane ¥i�1

d xi � 1 cuts the interior ofthe small cube cd but not the interior of any simplicial element, by the last property stated in Step1. Moreover, ¥i�1

d (xi0 � 1/n) � 1, since j0 � 1. Hence, the small simplex sd is contained in �� .

Step 3. Cut the constructed simplicial finite element mesh of the unit cube by the plane ¥i�1d

xi � 1. By the second property in Step 2, we have constructed a simplicial finite element mesh�h of the domain �. Since the d-dimensional volume of � is 1/d! and that of each small cubeis 1/nd, it follows from the first property in Step 2, this mesh �h contains O(h�d) simplicialelements that are translations of the single simplex �d

�1hSd � �d�1h�� .

Letting n � 1, . . . , we then obtain a family of simplicial finite element meshes {�h}. Sincewe only use a single reference triangulation to construct each mesh, the family of meshes {�h}are quasi-uniform.

We summarize our results in the following theorem.

Theorem 3.1. Let d � 2 be an integer and � � �d the d-dimensional open unit simplex. Then,there exist a quasi-uniform family of simplicial finite element meshes {�h} of � such that eachof the meshes �h contains O(h�d) simplicial elements which are translations of the singlesimplex �d

�1hSd, where h is the mesh size of �h and �d � 0 a constant depending only on thedimension d.

4. ON THE STANDARD LAGRANGE INTERPOLATION FOR d-DIMENSIONAL PK-TYPE FINITE ELEMENTS

Let d � 2 be an integer and � � �d the open unit simplex defined in (3.1). Let k be an integersuch that k � d � 1. Define u : �� 3 � by

u�x� � 1 � �i�1

d

xi k�1�d

�i�1

d

xi @x � �x1, . . . , xd� � �� .


Define also f (x) � ��u(x) for x � �. Obviously, u � H01(�) � C�(�� ) solves the boundary

value problem

��u � f in �,u � 0 on ��. (4.1)

Equivalently, u � H01(�) is the weak solution, defined by

A�u, v� � � f, v� @v � H01��,

where A : H01(�) � H0

1(�) 3 �, defined by

A�v, w� � �v, w� @v, w � H01��,

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 �h the corresponding simplicial finite element mesh of �constructed in Section 3. Let Sk

h(�) � H1(�) denote the Pk-type finite element space corre-sponding to the mesh �h, i.e., the restriction Sk

h(�)�T is exactly Pk�T for each element T � �h. LetSk

h(�) � Skh(�) � H0

1(�). The finite element solution uh � Skh(�) is defined by

A�uh, vh� � � f, vh� @vh � Skh��.

Finally, denote by Ih : C(�� ) 3 Skh(�) the standard Lagrange interpolation whose interpolation

points are the principle lattice points of all simplex elements of �h [1, 2].

Theorem 4.1. Let d and k be integers such that d � 2 and k � d � 1. With the quasi-uniformfamily of simplicial finite element meshes constructed in Section 3, we have that

�Ihu � uh�H1�� d,khk, (4.2)

where �d,k � 0 is a constant depending only on d and k.Proof. We shall call T � �h a corner simplicial element if

T � � �x1, . . . , xd� � Rd : xi �ji

n� 0, i � 1, . . . , d, �

i�1

d xi �ji

n �1

n�for some integers ji with 0 � ji � n � 1, i � 1, . . . , d. For such an element, we denote its d �1 vertices by

x�0� � j1

n, . . . ,

jd

n ,

x�i � � j1

n, . . . ,

ji�1

n,

ji � 1

n,

ji�1

n, . . . ,

jd

n , i � 1, . . . , d.

46 LI

For each x � T, let �i(x) (i � 0, . . . , d ) be the barycentric coordinates of x defined by �i � P1�T,and �i(x( j)) � 1 if i � j and 0 if i j. Explicitly,

�i�x� � nxi �ji

n , i � 1, . . . , d, x � �x1, . . . , xd� � T,

�0�x� � 1 � �i�1

d

�i�x�, x � T.

Define �T : T 3 � by

�T�x� � ��i�1

d

�i�x�� j�0

k�d�1 ��0�x� �j

k� @x � T.

We claim that �T differs only by a nonzero constant from the local shape function associatedwith the nodal point

x � j1

n�

1

nk, . . . ,

jd

n�

1

nk � T

whose barycentric coordinate is

��0�x�, �1�x�, . . . , �d�x�� k � d

k,

1

k, . . . ,

1

k � �d�1.

In fact, �T � Pk�T. Moreover, any nodal point x � T has the barycentric coordinates �i(x) � mi /kfor some integer mi with 0 � mi � k, i � 0, . . . , d, and m0 � k � ¥i�1

d mi. If 0 � m0 � k �d � 1, then �T(x) � 0. If k � d � 1 � m0 � k, then at least one mi � 0 (1 � i � d ), implyingthat �T(x) � 0. If m0 � k � d, then ¥i�1

d mi � d. In this case, if for some i (1 � i � d) mi � 0,then �T(x) � 0. Otherwise, all mi � 1 (i � 1, . . . , d), and x � x. But, �T( x) � (k � d)!/kk � 0.

Denoting by IT : C(T )3 Pk�T the local Lagrange interpolation operator on T—the restrictionof Ih : C(�� ) 3 Sh

k(�) onto C(T ), we then have

�IT��0�T��x� � �0�x��T�x� �k � d

k�T�x� @x � T.

Consequently, since on T, u(x) � n�k�1�0(x)�T(x) is a polynomial of degree � k, and IT : C(T )3 Pk�T is a projection on Pk�T, we have that

u�x� � �ITu��x� � n�k�1��0�T��x� � �IT��0�T��x��

� n�k�1��0�x� �k � d

k ��T�x�

� n�k�1 ��i�1

d

�i�x�� j�0

k�d ��0�x� �j

k� � x � T. (4.3)


We now define vT � Pd�1�T � Pk�T by vT(x) � i�0d �i(x) for all x � T. By a simple

calculation, we have that

�vT�x� � �2n2 �i�1

d �j�1,ji

d

�j�x� @x � T. (4.4)

Moreover, using the change of variables �i � �i(x) (i � 1, . . . , d ) from x � T to � � Sd, weobtain that

�vT�H1�T �2 � n�d �

Sd

�i�0

d

�i2d� � n2�d �

Sd

��i�0

d

�i 2

d�, (4.5)

where �0 � 1 � ¥i�1d �i and � is the gradient with respect to �.

By (4.3), u � ITu vanishes on the boundary of T. Therefore, by integration by parts, (4.3),(4.4), and the change of variables �i � �i(x) (i � 1, . . . , d ) from x � T to � � Sd, we get that

�T

�u � ITu��x� � vT�x�dx � � �T

�u � ITu��x��vT�x�dx

� 2n1�k �T

��i�1

d

�i�x��j�0

k�d ��0�x� �j

k�� i�1

d �j�1,ji

d

�j�x�dx

� 2n1�k�d �Sd

�i�1

d

�i��j�0

k�d �0 �j

k� �i�1

d �j�1,ji

d

�jd�. (4.6)

Denote by ��h the collection of all the corner simplicial elements in �h. Define vh : �� 3 � byvh � 0 on all elements in �h��h and vh � vT on any element T � ��h. We have that vh � Sk

h(�),since for each T � ��h, vT � Pk�T vanishes on the boundary of T. Moreover, we have by (4.5) that

�vh�H1��2 � n2�d��h��n�2 �

Sd

�i�0

d

�i2d� � �

Sd

��i�1

d

�i 2

d�� ,

where ��h� is the number of elements in ��h, and by (4.6) that

A�u � Ihu, vh� � �T��h

�T

�u � ITu��x� � vT�x�dx � 2�d,k��h�n1�k�d,

where

48 LI

�d,k � �Sd

�i�1

d

�i� �j�0

k�d �0 �j

k� �i�1

d �j�1,ji

d

�jd�

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

�A�u � Ihu, vh��vh�H��

� 2��d,k��d

��h�1/2n�k�d/2, (4.7)

where

�d � �Sd

��i�0

d

�i2 � ��

i�1

d

�i 2�d� � 0

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

for some constants �d � 0 and �d � 0 that depend only on d. Moreover, �d,k 0 by Lemma5.1 below. Therefore, the desired inequality (4.2) follows from (4.7) and Lemma 1.1 with

�d,k � � 4�d�d,k2

�d2k�d�d,k

� 0,

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

Corollary 4.1. Let d and k be integers such that d � 2 and k � d � 1. With the quasi-uniformfamily of simplicial finite element meshes constructed in Section 3, we have that

maxz��h

�u�z� � uh�z�� d,khk�1, (4.8)

where �h is the set of all the standard Lagrange interpolation points and �d,k � 0 is a constantdepending only on d and k.

Proof. Notice that (Ihu)(z) � u(z) for all z � �h. Thus, if (4.8) were not true, then wewould have

�Ihu � uh�L�� o�hk�1� as h 3 0.

This would lead to

�Ihu � uh�L2�� o�hk�1� as h 3 0,

and further to

�Ihu � uh�H1�� o�hk� as h 3 0


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

5. AUXILIARY LEMMAS

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

�d,k :� �Sd

�i�1

d

�i��j�0

k�d �0 �j

k� �i�1

d �j�1,ji

d

�jd� � 0,

where Sd is the d-dimensional unit simplex defined in (3.4) and �0 � 1 � ¥i�1d �i.

Proof. By the symmetry about the variables �1, . . . , �d, and the change of variables �i ��i (i � 1, . . . , d � 1) and �d � 1 � ¥i�1

d �i, we have

�d,k � d �Sd

�i�1

d

�i� �j�0

k�d �0 �j

k� �i�1

d�1

�id�

� d �Sd

�i�1

d�1

�i21 � �

i�1

d

�i �j�0

k�d �d �j

kd�

� d �0

1

d�d �0

1��d

d�d�1 · · · �0

1�¥i�j�1d �i

d�j · · · �0

1�¥i�2d �i

d�1 �i�1

d�1

�i2

� 1 � �i�1

d

�i �j�0

k�d �d �j

k . (5.1)

Set

E1 � �0

1�¥i�2d �i

�12 1 � �

i�1

d

�id�1

and

Ej � �0

1�¥i�j�1d �i

�j2Ej�1d�j for j � 2, . . . , d � 1.

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

�j2 � �1 � �

i�j

d

�i � 1 � �i�j�1

d

�i� 2

� 1 � �i�j

d

�i 2

� 21 � �i�j

d

�i1 � �i�j�1

d

�i � 1 � �i�j�1

d

�i 2

,

50 LI

we obtain that

Ej � ��i�1

j 2

�3i � 1��3i ��3i � 1��1 � �i�j�1

d

�i 3j�1

, j � 1, . . . , d � 1.

It then follows from (5.1) that

�d,k � d �0

1

Ed�1 �j�0

k�d �d �j

kd�d

� d��i�1

d�1 2

�3i � 1��3i��3i � 1�� 0

1

�1 � �d�3d�2 �

j�0

k�d �d �j

kd�d.

This is a nonzero constant by Lemma 5.2 below. y

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

Jd,k :� �0

1

�1 � t�3d�2 �i�0

k�d t �i

kdt��0 if k � d is even,�0 if k � d is odd. (5.2)

Proof. Denote q � k � d � 1 and �l(t) � i�0l (t � ti) for any integer l � 0, where tr �

r/k for any real r.

Case 1. q � k � d is even. Let �l(t) � �0t �l(s)ds. By [30; Lemma 4, p. 309], we have �q(t) �

0 for all t � (0, tq) and �q(tq) � 0. Moreover, for tq � t � 1, we have

�q�t� � �q�tq� � �tq

t

�q�s�ds � �tq

t

�q�s�ds � 0,

since �q(s) � 0 for all s � tq. Therefore, we obtain by integration by parts that

Jd,k � �0

1

�1 � t�3d�2��q�t�dt � �3d � 2� �0

1

�1 � t�3d�3�q�t�dt

� �3d � 2� �0

tq

�1 � t�3d�3�q�t�dt � �3d � 2� �tq

1

�1 � t�3d�3�q�t�dt � 0.


Case 2. q � k � d is odd. Direct calculations lead to

Jd,d�q � ��d � 1�q�d �

�d � q�q �i��1

q

�3d � i ��1 � 0, q � 1, 3, 5, 7, (5.3)

where

1�d � � 1 � 0,

3�d � � 12�4d2 � d � 3� � 0,

5�d � � 6�1257d4 � 513d3 � 3031d2 � 225d � 2400� � 0,

7�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 � �0

tq

�1 � t�3d�2�q�t�dt � �tq

1

�1 � t�3d�2�q�t�dt � Id,k � Md,k. (5.4)

By straight forward calculations, we obtain that

Md,k :� �j�1

d �tq�j�1

tq�j

�1 � t�3d�2 �i�0

q

�t � ti�dt � �j�1

d ��i�0

q

�tq�j � ti�� tq�j�1

tq�j

�1 � t�3d�2dt

�1

�3d � 1�k2d�k �j�1

d ��i�0

q

�i � j ��d � j � 1�3d�1 � �d � j �3d�1�

�1

�3d � 1�k2d�k ��i�0

q

�1 � i ��d3d�1 � �j�1

d�1 ��i�0

q

�i � j � 1� � �i�0

q

�i � j ��d � j �3d�1��

q � 1

�3d � 1�k2d�k �j�0

d�1 ��i�1

q

�i � j ��d � j �3d�1, (5.5)

where in the third step we use the Abel summation identity

�j�1

d

uj�vj�1 � vj� � u1v0 � udvd � �j�1

d�1

�uj�1 � uj�vj

with uj � i�0q (i � j ) and vj � (d � j )3d�1.

52 LI

Since q � 9 is odd, we have by [30; Lemma 2, p. 309] that �q(t) � �q(tq � t) for all t �[tq/2, tq]. Thus, by the change of variable tq � t 3 s from t � [tq/2, tq] to s � [0, tq/2], we get

Id,k :��0

tq/2

�1 � t�3d�2�q�t�dt � �tq/2

tq

�1 � t�3d�2�q�tq � t�dt

� �0

tq/2

��1 � t�3d�2 � �1 � tq � t�3d�2��q�t�dt

� �j�1

q0 �t2j�2

t2j

��1 � t�3d�2 � �1 � tq � t�3d�2��q�t�dt

� �t2q0

tq/2

��1 � t�3d�2 � �1 � tq � t�3d�2��q�t�dt

� �j�1

q0

Hd,kj� Gd,k, (5.6)

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 :� �t2q0

tq/2

��1 � t�3d�2 � �1 � tq � t�3d�2��q�t�dt � 0. (5.7)

If (q � 1)/2 is even, then for any t � (t2q0, tq/2), �q(t) has 2q0 � 1 negative factors. Hence, it

is negative. Thus (5.7) holds true. If (q � 1)/2 is odd, then for any t � (t2q0�1/2, t2q0�1), �q(t) �0, since it has 2q0 � 3 negative factors. Hence, by the change of variable t 3 t � 1/k from[t2q0�1, t2q0�3/2] to [t2q0

, t2q0�1/2], we obtain that

Gd,k � �t2q0

t2q0�1/2

� �t2q0�1/2

t2q0�1

� �t2q0�1

t2q0�3/2 ��1 � t�3d�2 � �1 � tq � t�3d�2��q�t�dt

� �t2q0

t2q0�1/2

��1 � t�3d�2 � �1 � tq � t�3d�2��q�t�dt

� �t2q0�1

t2q0�3/2

��1 � t�3d�2 � �1 � tq � t�3d�2��q�t�dt


� �t2q0

t2q0�1/2

��1 � t�3d�2 � �1 � tq � t�3d�2��q�t�dt

� �t2q0

t2q0�1/2 �1 �1

k� t 3d�2

� 1 � tq �1

k� t 3d�2��q1

k� tdt

� � �t2q0

t2q0�1/2

gd,k�t��q�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 �q�1(t). So, �q�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 3 t � 1/k from [t2j�1, t2j] to [t2j�2,

t2j�1], we get that

Hd,k,j :� �t2j�2

t2j

��1 � t�3d�2 � �1 � tq � t�3d�2��q�t�dt

� �t2j�2

t2j�1

� �t2j�1

t2j ��1 � t�3d�2 � �1 � tq � t�3d�2��q�t�dt

� � �t2j�2

t2j�1

gd,k�t��q�1�t�dt, (5.10)

where gd,k is defined in (5.8). For each t � (t2j�2, t2j�1), �q�1(t) has 4q0 � 2j � 4 negativefactors. So, �q�1(t) � 0. Using the fact that (�1)sfd,k

(s) (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

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and

g�d,k�t� � ��fd,k�t� � fd,kt �1

k� � �tq � t�f�d,k�t� � t �1

kf�d,kt �1

k� tq � 2t �

1

kf�d,kt �1

k� 0 (5.11)

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 �q�1(t) � 0 for t � (t0, t1), weconclude that

Id,k � Hd,k,1

� �gd,k�t1� �t0

t1

�q�1�t�dt

� �tq�3fd,k�t1� �i�2

q�1

�ti � t1� �t0

t1

t�t1 � t�dt

� ��q � 3��q � 2�!

6k2d�k �q � d � 1�3d�2. (5.13)

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� �j�0

d�1 ��i�1

q

�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 this inequality holdstrue for all q � 9. Therefore, since

6�q � 1�

�3d � 1��q � 3��

2

d 1 �1

3d � 11 �4

q � 3 �4

d, d � 3, q � 9,

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


�j�0

d�1 ��i�1

q

�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�! �m�j

3d�2 3d � 2m �d � j �3d�2�m�q � j � 1�m

� ��i�1

q

�i � j�� d � j�3d�2 �m�j

3d�2 ��3d � 2� · · · �3d � 1 � m��q � j � 1�m

��j � 1� · · · m��q � 1� · · · �q � j��d � j�m

� ��i�1

q

�i � j�� d � j�3d�2 �m�j

3d�2 ��3d � 2� · · · �3d � 1 � m��q � j � 1�m�j�1

��j � 1� · · · m��q � j��d � j�m

� ��i�1

q

�i � j�� d � j�3d�2Sj,

that is,

�q � 2�!�q � d � 1�3d�2Sj�1 � ��

i�1

q

�i � j �� d � j �3d�2, j � 0, . . . , d � 1, (5.15)

where

Sj � �m�j

3d�2 ��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 summation S0 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

� 9 (5.16)

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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

� 21. (5.17)

For 2 � j � d � 1, we get by keeping only the term with m � min(3j � 1, q � j � 1) in thesum 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

�j�2

d�1

Sj�1 � 31�q �

j�2

d�1 � 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

�j�0

d�1

Sj�1 � 1/4.

Consequently, by summing (5.15) over j � 0, . . . , d � 1, we obtain the desired inequality(5.14), since d � 1. The proof is complete. y

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Lagrange interpolation and finite element superconvergence

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