LAGRANGE INTERPOLATION AND FINITE ELEMENT SUPERCONVERGENCE BO LI Abstract. We consider the finite element approximation of the Laplacian operator with the homogeneous Dirichlet boundary condition, and study the corresponding Lagrange interpolation in the context of finite element superconvergence. For ddimensional Qk -type elements with d ≥ 1 and k ≥ 1, we prove that the interpolation points must be the Lobatto points if the Lagrange interpolation and the finite element solution are superclose in H 1 norm. For d-dimensional Pk -type elements, we consider the standard Lagrange interpolation—the Lagrange interpolation with interpolation points being the principle lattice points of simplicial elements. We prove for d ≥ 2 and k ≥ d + 1 that such interpolation and the finite element solution are not superclose in both H 1 and L2 norms, and that not all such interpolation points are superconvergence points for the finite element approximation.

1. Introduction Consider the boundary value problem  Lu = f u=0

in Ω, on ∂Ω,

(1.1)

where Ω ⊂ Rd is a bounded domain with a Lipschitz continuous boundary ∂Ω, d ≥ 1, f ∈ L2 (Ω), and L : H 2 (Ω) → L2 (Ω) is a second order, linear, self-adjoint, elliptic differential operator. Let u ∈ H01 (Ω) be its unique weak solution, defined by ∀v ∈ H01 (Ω),

A(u, v) = (f, v)

where A : H01 (Ω) × H01 (Ω) → R is the bilinear, symmetric, continuous, and coercive form associated with (1.1), and (·, ·) denotes the inner product of L2 (Ω). Let {τh } be a family of finite element meshes of the domain Ω with the mesh size h → 0. Fix ¯ be the corresponding finite an integer k ≥ 1. For each h, let Skh (Ω) ⊂ H 1 (Ω) ∩ C(Ω) h element space such that Sk (Ω)|T ⊇ Pk |T for all T ∈ τh , where Pk is the set of all ◦

◦

polynomials of degree ≤ k. Let S hk (Ω) = Skh (Ω) ∩ H01 (Ω). Let uh ∈ S hk (Ω) be the finite element solution, defined by A(uh , vh ) = (f, vh )

◦

∀vh ∈ S hk (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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¯ → S h (Ω) denote the corresponding Lagrange interpolation Finally, let Ih : C(Ω) k operator. The following estimate h−1 kIh u − uh kL2 (Ω) + kIh u − uh kH 1 (Ω) ≤ Chk (1.2) ¯ is smooth enough is standard, provided that the weak solution u ∈ H01 (Ω) ∩ C(Ω) and the underlying meshes are quasi-uniform [3,9]. Here and below, we use the letter C 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 improved. This means that Ih u and uh can be superclose. More precisely, we say that the Lagrange interpolation Ih u and the finite element solution uh are superclose in H 1 norm, if
as h → 0. kIh u − uh kH 1 (Ω) = o hk

We also say that Ih u and uh are superclose in H 1 norm by order (at least) σ > 0, if kIh u − uh kH 1 (Ω) ≤ Chk+σ .

(1.3)

The following result gives a different expression of the closeness between Ih u and uh in H 1 norm. It is trivially true, and we omit its proof. ¯ then Lemma 1.1. If the exact solution u ∈ H01 (Ω) ∩ C(Ω), γkIh u − uh kH 1 (Ω) ≤

sup ◦

v h ∈S h k (Ω), vh 6=0

|A(u − Ih u, vh )| ≤ M kIh u − uh kH 1 (Ω) , kvh kH 1 (Ω)

where γ > 0 and M > 0 are the two constants in the conditions of coercivity and continuity, respectively, of the bilinear form A : H01 (Ω) × H01 (Ω) → R, and

A(v, v) ≥ γkvk2H 1 (Ω)

∀v ∈ H01 (Ω)

|A(v, w)| ≤ M kvkH 1 (Ω) kwkH 1 (Ω)

∀v, w ∈ H01 (Ω).

The supercloseness between the Lagrange interpolation and the finite element solution is closely related to the superconvergence of the finite element solution to the exact solution. In fact, if (1.3) holds true, then one can easily obtain the following estimate of gradient superconvergence  1/2 1 X 2 ¯ h (z)| ≤ Chk+min(σ,1) , |∇u(z) − ∇u d h z∈Zh (Ω)

where Zh (Ω) is the set of superconvergence points for the gradient of the Lagrange ¯ is some kind of average of the gradient [8]. In some cases, one interpolation and ∇ can obtain a higher order estimate |A(u − Ih u, vh )| ≤ Chk+σ kvh kW 1,p (Ω)

◦

∀vh ∈ S hk (Ω)

(1.4)

for some σ > 0 and p ∈ [1, ∞). This, together with delicate estimates of a discrete Green’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 and the finite element solution. Our main results are as follows. 1. For d-dimensional Qk -type (tensor product) elements with d ≥ 1 and k ≥ 1, the interpolation points must be the Lobatto points if the Lagrange interpolation and the finite element solution are superclose in H 1 norm, cf. Theorem 2.1. 2. For d-dimensional Pk -type (simplicial) elements with d ≥ 2 and k ≥ d + 1, the standard Lagrange interpolation—the Lagrange interpolation with its interpolation points being the principle lattice points of simplicial elements—and the finite element solution are not superclose in H 1 norm, cf. Theorem 4.1. 3. For d-dimensional Pk -type elements with d ≥ 2 and k ≥ d + 1, not all the standard Lagrange 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 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 the Lobatto points and the finite element solution are superclose by one order in H 1 norm. Here, we prove the converse under the assumption that they are only superclose, but not necessary superclose by any order, in H 1 norm. For simplicial finite elements, the Lagrange interpolation points can not be arbitrarily distributed in general. With good meshes, the standard Lagrange interpolation and the finite element solution are in fact superclose in H 1 norm by one order for two-dimensional P1 and P2 elements and for three-dimensional P1 element [1, 5–7, 12–14, 17, 19, 21, 22, 26–30]. Recently, similar results have been obtained for 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-dimensional Pk -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 − Ih u, vh ) for vh ∈ S hk (Ω). However, such cancellation seems to be impossible if there exists an element-wise, ◦

bubble-like test function vh ∈ S hk (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 simplicial elements. This turns out to be true if and only if k ≥ d + 1 for d-dimensional Pk -type elements. Constructing a bubble-like test function to avoid any possible cancellation was the original approach in our early work [20] to show the the nonsupercloseness for two-dimensional P3 element. Here, we extend such an approach to a general case which is more complicated due to the higher space dimension and higher polynomial degree. In proving the non-supercloseness of the Lagrange interpolation to the finite element solution for the general d-dimensional Pk -type finite elements with k ≥ d+1, we choose the underlying domain to be the unit d-dimensional simplex. This allows us to have a

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BO LI

polynomial of degree exactly k + 1 as the solution of the underlying Poisson equation with the homogeneous Dirichlet boundary condition. In addition, we construct a special family of quasi-uniform finite element meshes consisting of enough elements that are scaled translations of the unit simplex. Such meshes are uniform for d = 2 but 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 meshes 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. In Section 3, we construct a quasi-uniform family of simplicial finite element meshes of a d-dimensional domain for d ≥ 2. With such meshes, we study in Section 4 the standard Lagrange interpolation for Pk -type elements. Finally, in Section 5, we prove some auxiliary lemmas. 2. Optimal Lagrange interpolation points for Qk -type finite elements Consider the boundary value problem  −∆u = f u=0

in Ω, on ∂Ω,

(2.1)

where f ∈ L2 (Ω) and Ω = Πdm=1 (am , bm ) ⊂ Rd is a d-dimensional rectangular parallelepiped with d ≥ 1 and −∞ < am < bm < ∞ for all m = 1, · · · , d. The associated bilinear form A : H01 (Ω) × H01 (Ω) → R is defined by A(v, w) = (∇v, ∇w)

∀v, w ∈ H01 (Ω).

It is symmetric, continuous, and coercive. The weak solution u ∈ H01 (Ω) of the boundary value problem (2.1) is defined by A(u, v) = (f, v)

∀v ∈ H01 (Ω).

Let {τh } be a family of quasi-uniform rectangular meshes covering Ω with the mesh size h → 0. We denote a typical mesh by Y  d τh = [xm,jm −1 , xm,jm ] : jm = 1, · · · , nm , m = 1, · · · , d , m=1

where xm,jm = am + jm hm for jm = 0, · · · , nm , hm = (bm − am )/nm , nm ≥ 1 is an integer for each m = 1, · · · , d, and h = max1≤m≤d hm . For an integer k ≥ 1, let Skh (Ω) ⊂ H 1 (Ω) denote the Qk -type finite element space corresponding to the mesh τh , i.e., the restriction Skh (Ω)|R is exactly Qk |R for each element R ∈ τh , where Qk = span {xα1 1 · · · xαd d : α1 , · · · , αd are nonnegative integers, α1 + · · · + αd = k} . ◦

◦

Let S hk (Ω) = S hk (Ω) ∩ H01 (Ω). The finite element solution uh ∈ S hk (Ω) is defined by A(uh , vh ) = (f, vh )

◦

∀vh ∈ S hk (Ω).

LAGRANGE INTERPOLATION AND FINITE ELEMENT SUPERCONVERGENCE

5

(k)

(0)

For each integer m with 1 ≤ m ≤ d, let ξm , · · · , ξm be k + 1 distinct real numbers satisfying (0) (k) −1 = ξm < · · · < ξm = 1.
(i ) (i ) We call all the points ξ1 1 , · · · , ξd d (im = 0, · · · , k, m = 1, · · · , d) the reference interpolation points. We define the Lagrange interpolation points on each element Q d m=1 [xm,jm −1 , xm,jm ] ∈ τh (1 ≤ jm ≤ nm , 1 ≤ m ≤ d) by (i)

hm ξm + xm,jm −1 + xm,jm , i = 0, · · · , k, m = 1, · · · , d. 2 ¯ → S h (Ω) the Lagrange interpolation operator assoFinally, we denote by Ih : C(Ω) k ciated with these interpolation points. (1,1) Recall that the Jacobi polynomials Pn (n = 0, 1, · · · ) are orthogonal polynomials (1,1) on the interval [−1, 1] with the weight ρ(ξ) = 1 − ξ 2 , normalized by Pn (1) = n + 1 (1,1) [23]. The Rodrigues’ formula for Pn is  n   (−1)n d 2 n+1 (1,1) . (1 − ξ ) Pn (ξ) = n 2 n! (1 − ξ 2 ) dξ (i)

xm,jm =

(1,1)

For each n ≥ 1, Pn 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 ρ(ξ) = 1 [23]. They are given by  n   d 1 (1 − ξ 2 )n , n = 0, 1, · · · . Ln (ξ) = n 2 n! dξ

It is easy to show that {L0n (ξ)}∞ n=1 is also a sequence of orthogonal polynomials on (1,1) [−1, 1] with the weight ρ(ξ) = 1 − ξ 2 . Consequently, Pn and L0n+1 differ only by a nonzero constant. For n ≥ 2, the Lobatto points associated with n − 1 are thus the roots of L0n (ξ) in (−1, 1). However, for convenience, we shall call in what follows all the n − 1 distinct roots of L0n (ξ) in (−1, 1), together with ±1, the Lobatto points of order n. (In fact, ±1 are often included in a Lobatto quadrature [11].) We call a point in Rd a d-dimensional Lobatto point of order n, if each of its d 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 with k ≥ 2 that the Lagrange interpolation is superclose to the finite element solution in H 1 norm if and only if all the interpolation points are the Lobatto points. Theorem 2.1. Suppose that kIh u − uh kH 1 (Ω) = o hk




as h → 0,

(2.2)

whenever the solution u ∈ H01 (Ω) is smooth enough. Then, all the reference interpo(i ) (i )
lation points ξ1 1 , · · · , ξd d (0 ≤ im ≤ k, m = 1, · · · , d) must be the d-dimensional Lobatto points of order k.

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BO LI

Proof. For k = 1, the reference interpolation points are always Lobatto points by our definition. So, we assume that k ≥ 2. We shall show for each m (1 ≤ m ≤ d) that (0) (k) ξm , · · · , ξm are indeed the k + 1 one-dimensional Lobatto points of order k. Fix an index m with 1 ≤ m ≤ d. Define u ∈ H01 (Ω) by  k−1 Y d am + b m ¯ (xl − al )(xl − bl ), x = (x1 , · · · , xd ) ∈ Ω. u(x) = xm − 2 l=1

Note that u depends on m. Define accordingly f (x) = −∆u(x) for all x ∈ Ω. ¯ solves the boundary value problem (2.1). Obviously, f ∈ L2 (Ω), and u ∈ C ∞ (Ω) For each integer s : 0 ≤ s ≤ k − 2, define vs : [am , bm ] → R by   2xm − xm,jm −1 − xm,jm vs (xm ) = φs , hm ∀xm ∈ [xm,jm −1 , xm,jm ], jm = 1, · · · , nm , where the function φs : [−1, 1] → R is defined by

1 1 φs (ξ) = ξ s+2 − (1 + ξ) − (−1)s (1 − ξ), 2 2 It is easy to see that φ00s (ξ) = (s + 1)(s + 2)ξ s

and

ξ ∈ [−1, 1].

φs (−1) = φs (1) = 0.

Hence, vs is a continuous piecewise polynomial of degree s + 2 ≤ k, vanishing at ¯ → R for the case d = 1 by all the points xm,jm (jm = 0, · · · , nm ). Define vh : Ω ¯ vh (x1 ) = vs (x1 ) for all x1 ∈ Ω = [a1 , b1 ], and for the case d ≥ 2 by vh (x) = vs (xm )Wm (x0 )

¯ ∀x ∈ Ω,

where 0

Wm (x ) =

d Y

l=1, l6=m 0

∀x0 ∈ Ω0 ,

(xl − al )(xl − bl )

x = (x1 , · · · , xm−1 , xm+1 , · · · , xd ), 0

Ω =

d Y

(al , bl ).

l=1, l6=m ◦

Note that vh depends on m and that vh ∈ S hk (Ω). Q Assume that d ≥ 2 temporarily. Let R = dl=1 [xl,jl −1 , xl,jl ] ∈ τh be an arbitrary element, where 1 ≤ jl ≤ nl and 1 ≤ l ≤ d. The function 0

u(x) − Wm (x )

k Y i=0

(i)

xm − xm,jm




LAGRANGE INTERPOLATION AND FINITE ELEMENT SUPERCONVERGENCE

7

is in Qk |R , and agrees with u on all the interpolation points in R. Hence, this function is exactly the Lagrange interpolation of u on the element R. Thus, we have 0

(u − Ih u)(x) = Wm (x )

k  Y i=0

(i)

xm − xm,jm



∀x ∈ R.

Q Let R0 = dl=1, l6=m [xl,jl −1 , xl,jl ]. Let ∇0 denote the gradient operator with respect to x0 . 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 Z ∇(u − Ih u)(x) · ∇vh (x) dx R  Z  ∂ ∂ 0 0 ∇ (u − Ih u)(x) · ∇ vh (x) + = (u − Ih u)(x) · vh (x) dx ∂xm ∂xm R Z k Y (i)
0 0 2 = |∇ Wm (x )| vs (xm ) xm − xm,jm dx R

i=0

k Y

"

#
d 2 (i) + [Wm (x0 )] xm − xm,jm vs0 (xm ) dx dx m i=0 R Z xm,jm Z k Y (i)
0 0 2 0 vs (xm ) xm − xm,jm dxm = |∇ Wm (x )| dx Z

xm,jm −1

R0

− =

Z

0

2

[Wm (x )] dx R0

0

Z

xm,jm

xm,jm −1

i=0

vs00 (xm )

k Y i=0

(i)
xm − xm,jm dxm

k 1 Y
hm (i) dξm k∇0 Wm k2L2 (R0 ) φs (ξm ) ξm − ξ m 2 −1 i=0  k Z 1 k Y
hm (i) φ00s (ξm ) dξm . − ξm − ξ m kWm k2L2 (R0 ) 2 −1 i=0



k+2

Z

Consequently, we have by the fact nm = (bm − am )/hm that XZ A (u − Ih u, vh ) = ∇(u − Ih u)(x) · ∇vh (x) dx R∈τh

R

(bm − am ) k+1 0 = hm k∇ Wm k2L2 (Ω0 ) k+2 2

Z

1

φs (ξm ) −1

k Y i=0


(i) ξm − ξ m dξm

(2.3)

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BO LI

(bm − am )(s + 1)(s + 2) k−1 − hm kWm k2L2 (Ω0 ) k 2

Z

1 −1

Similarly, we obtain that kvh k2H 1 (Ω)

= =

Z

Ω



k Y i=0


(i) dξm . ξm − ξ m

 |vh (x)|2 + |∇vh (x)|2 dx

kWm k2H 1 (Ω0 ) +

s ξm

nm Z X

jm =1

kWm k2L2 (Ω0 )

xm,jm

xm,jm −1

nm X

jm =1

Z

|vs (xm )|2 dxm

xm,jm xm,jm −1

(bm − am ) = kWm k2H 1 (Ω0 ) 2

Z

|vs0 (xm )|2 dxm

1 −1

|φs (ξm )|2 dξm

2(bm − am ) kWm k2L2 (Ω0 ) + 2 hm

Z

1 −1

|φ0s (ξm )|2 dξm ,

leading to kvh kH 1 (Ω) ≤

r

(bm − am )[4 + (bm − am )2 ] kφs kH 1 (−1,1) kWm kH 1 (Ω0 ) h−1 m . 2

Therefore, we infer from (2.3) and (2.4) that Z k 1 Y
|A(u − Ih u, vh )| (i) k s ξm − ξm dξm − α2 hk+2 ≥ α 1 hm ξm m , kvh kH 1 (Ω) −1 i=0 where

α1 =

s

and α2 =

2 2(bm − am ) (s + 1)(s + 2)kWm kL2 (Ω0 ) >0 4 + (bm − am )2 2k kφs kH 1 (−1,1) kWm kH 1 (Ω0 )

s

k∇0 Wm k2L2 (Ω0 ) 2(bm − am ) 4 + (bm − am )2 2k+2 kφs kH 1 (−1,1) kWm kH 1 (Ω0 ) Z k 1 Y
(i) · φs (ξm ) ξm − ξ m dξm ≥ 0 −1 i=0

are constants independent of h. Assume now d = 1. By a similar but simpler argument, we obtain that Z k (b1 − a1 )(s + 1)(s + 2) k−1 1 s Y (i)
h ξ ξ − ξ dξ1 A(u − Ih u, vh ) = − 1 1 1 1 2k −1 i=0

(2.4)

(2.5)

LAGRANGE INTERPOLATION AND FINITE ELEMENT SUPERCONVERGENCE

and kvh k2H 1 (Ω)

b1 − a 1 = 2

Therefore, r

1 −1



 2 4 0 |φs (ξ1 )| + 2 φs (ξ1 ) dξ1 . h1 2

(b1 − a1 )[4 + (b1 − a1 )2 ] kφs kL2 (−1,1) h−1 1 , 2 Z k 1 Y
|A(u − Ih u, vh )| (i) k s ≥ βh1 ξ1 ξ1 − ξ1 dξ1 , −1 kvh kH 1 (Ω)

kvh kH 1 (Ω) ≤ and

Z

9

(2.6)

i=0

where

β=

s

2(b1 − a1 ) (s + 1)(s + 2) >0 4 + (b1 − a1 )2 2k kφs kL2 (−1,1)

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 Z 1 k Y
s (i) dξm = 0, s = 0, · · · , k − 2. ξm ξm − ξ m −1

i=0

(i)
The polynomial i=1 ξ −ξm of degree k−1 is thus orthogonal to all the polynomials (0)
(k)
in Pk−2 on [−1, 1] with the weight ξ − ξm ξ − ξm = ξ 2 − 1. Hence, it differs from (1,1) the Jacobi polynomial Pk−1 only by a nonzero constant. Consequently, all the points (0) (k) ξm , · · · , ξm are the k + 1 one-dimensional Lobatto points of order k.

Qk−1

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 

d

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

d X i=1

xi < 1 .

(3.1)

We shall construct a quasi-uniform family of simplicial finite element meshes {τ h } of Ω such that there are O(h−d ) elements in τh that are translations of a single d¯ = {σ −1 hx : x ∈ Ω}, ¯ where σd > 0 is a constant depending dimensional simplex σd−1 hΩ d 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 a the uniform mesh with mesh size h = 2/n. Obviously, there are O(h−2 ) elements of √ ¯ with σ2 = 2. mesh that are translations of the single 2-dimensional simplex σ2−1 hΩ For d ≥ 3, we construct in three steps a simplicial finite element mesh of Ω with the 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 it into many small cubes, triangulating them by affine mappings from the triangulated

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BO LI

reference unitPcube, and gluing them together. Third, we cut the meshed unit cube by 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 Bdl

d X



= ξ ∈ Cd : l − 1 ≤

Obviously,

∪dl=1 Bdl = Cd

i=1

ξi ≤ l ,

l = 1, · · · , d.

(3.2)

int(Bdj ) ∩ int(Bdl ) = ∅ if j 6= l.

and

(3.3)

Notice that Bd1 = Sd , where 

d

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

d X i=1

ξi ≤ 1



(3.4)

is the reference unit simplex in Rd , and both Bd1 and Bdd are simplexes. But Bdl is not a simplex if 1 < l < d. This is because that the number of vertices in a ddimensional simplex is d + 1. However, since all the vertices (ξ1 , · · · , ξd ) (ξi = 0 or 1, P i = 1, · · · , d) of Cd lie in the planes di=1 ξi = j (j = 0, · · · , d), the number of vertices P of Cd contained in Bdl is the same as that contained in the planes di=1 ξi = l − 1 and Pd i=1 ξi = l. This number is       d d d+1 + = > d + 1, l−1 l l since 1 < l < d. We now triangulate all the polygons Bdl (l = 2, · · · d − 1) into simplexes so that, together with Bd1 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 Bdl for each l with 2 ≤ l ≤ d − 1 is the union of two types of (d − 1)-dimensional polygons m = {ξ ∈ Cd : Pd−1

and

d X

ξi = m},

i=1

l,j,m Fd−1 = Bdl ∩ {ξ ∈ Cd : ξj = m},

m = l − 1, l, j = 1, · · · , d, m = 0, 1.

m Consider a first type (d − 1)-dimensional polygon Pd−1 (1 ≤ m ≤ d − 1). If m = 1 m or d − 1, then Pd−1 is already a (d − 1)-dimensional simplex. Suppose 2 ≤ m ≤ m d − 2. To triangulate Pd−1 into (d − 1)-dimensional simplexes, we again need only to m triangulate the boundary of Pd−1 into (d − 2)-dimensional simplexes and then connect

LAGRANGE INTERPOLATION AND FINITE ELEMENT SUPERCONVERGENCE

11

m the barycenter of Pd−1 to all the vertices in such a (d − 2)-dimensional simplicial m triangulation. The boundary of Pd−1 is the union of the following sets: m m Pd−1 ∩ {ξ ∈ Cd : ξj = 0} and Pd−1 ∩ {ξ ∈ Cd : ξj = 1},

j = 1, · · · , d.

Each of these sets is either already a (d − 2)-dimensional simplex (if m = 2 and ξj = 1) or still a first type polygon but of one-dimension lower. For d = 3, both P 21 and P22 are already 2-dimensional simplexes. Therefore, we conclude by induction m that, for d ≥ 3 in general, all the first type (d − 1)-dimensional polygons P d−1 ⊂ Rd−1 (m = 1, · · · , d − 1) can be triangulated into (d − 1)-dimensional simplexes. Notice that m−1 m Pd−1 ∩ {ξ ∈ Cd : ξj = 1} = ej + Pd−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. l,j,m Consider now a second type (d − 1)-dimensional polygon Fd−1 (2 ≤ l ≤ d − 1, 1 ≤ l,j,m j ≤ d, m = 0, 1). If l = 2 and m = 1, or l = d − 1 and m = 0, then Fd−1 is already a l,j,m l (d − 1)-dimensional simplex. Otherwise, Fd−1 is a Bd -type but (d − 1)-dimensional polygon, cf. (3.2). For d = 3, there are altogether six of such 2-dimensional polygons F22,j,m (j = 1, 2, 3, m = 0, 1). All of them are 2-dimensional simplexes. So, by l,j,m induction, the second type (d − 1)-dimensional polygons Fd−1 with d ≥ 3 can all be triangulated into (d − 1)-dimensional simplexes. Notice that l,j,1 l−1,j,0 Fd−1 = ej + Fd−1 ,

l = 2, · · · , d − 1, j = 1, · · · , d.

(3.6)

Finally, for each l ∈ {2, · · · , d − 1}, we connect the barycenter of polygon B dl 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 Bd1 = 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 ξi = 0 and ξi = 1 results in the same (d − 1)-dimensional simplicial triangulation of the (d − 1)-dimensional unit cube  i Cd−1 = (ξ1 , · · · , ξi−1 , ξi+1 , · · · , ξd ) ∈ Rd−1 : 0 ≤ ξj ≤ 1, j = 1, · · · , i − 1, i + 1, · · · , d . P 3. For any integer j with 1 ≤ j ≤ d, the plane di=1 ξi = j does not intersect the interior 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) Q to cut the unit cube [0, 1]d into nd small cubes. Let cd = di=1 [x0i , x0i + 1/n] denote a typical such small cube. Define G : Cd → cd by G(ξ) = (1/n)ξ + x0 for all
0 0 0 ξ ∈ Cd , where x = x1 , · · · , xd . Obviously, it is a one-to-one and onto, orientation preserving, and affine mapping from the reference unit cube Cd to the small cube cd . Therefore, together with the constructed simplicial triangulation of the reference unit cube Cd , the mapping G : Cd → cd defines a simplicial triangulation of the small cube cd . By the arbitrariness of cd and the property of triangulation symmetry of the simplicial triangulation of the reference unit cube, we have in fact constructed a simplicial 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 triangulation of the reference unit cube Cd . The constructed simplicial finite element mesh of the unit cube [0, 1]d satisfies the following properties. Q 1. Each small cube cd = di=1 [x0i , x0i + 1/n] contains one simplicial element sd =



(x1 , · · · , xd ) ∈ cd :

d X i=1

xi −

x0i




 1 , ≤ n

which is a translation of the simplex (1/n)Sd = σd−1 hSd . Thus, there are nd = (σd /h)d simplicial elements in the mesh that are translations of the single simplex σd−1 hSd . P 2. The plane di=1 xi = 1 does not intersect the interior of any simplicial element. P Q 3. If the plane di=1 xi = 1 intersects the interior of a small cube cd = di=1 [x0i , x0i + ¯ 1/n], then the simplex sd must be in Ω.

The first property follows from the first property in Step 1 and our construction of the finite element mesh of the unit cube [0, 1]d . To show the other two properties, we Q consider a typical small cube cd = di=1 [x0i , x0i +1/n]. After the change of variables ξ = n(x − x0 ), where x0 = (x01 , · · · , x0d ), the cube cd is transformed into the reference unit  P P P cube Cd and the plane di=1 xi = 1 into di=1 ξi = j0 , where j0 = n 1 − di=1 x0i .

Notice that j0 is an integer, since all nx0i (i = 1, . . . , d) are integers. If j0 ∈ / {1, · · · , d− Pd 1}, then the plane i=1 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, P j0 ∈ {1, · · · , d − 1}. In this case, the plane di=1 xi = 1 cuts the interior of the small cube cd but not thePinterior of any simplicial element, by the last property stated in Step 1. Moreover, di=1 (x0i + 1/n) ≤ 1, since j0 ≥ 1. Hence, the small simplex sd is ¯ contained in Ω. Step 3. P Cut the constructed simplicial finite element mesh of the unit cube by the plane di=1 xi = 1. By the second property in Step 2, we have constructed a simplicial finite element mesh τh of the domain Ω. Since the d-dimensional volume of Ω is 1/d! and that of each small cube is 1/nd , it follows from the first property in

LAGRANGE INTERPOLATION AND FINITE ELEMENT SUPERCONVERGENCE

13

Step 2, this mesh τh contains O(h−d ) simplicial elements which are translations of the ¯ single simplex σd−1 hSd = σd−1 hΩ. Letting n = 1, · · · , we then obtain a family of simplicial finite element meshes {τ h }. Since we only use a single reference triangulation to construct each mesh, the family of 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 meshes {τh } of Ω such that each of the meshes τh contains O(h−d ) simplicial elements which are translations of the single simplex σd−1 hSd , where h is the mesh size of τh and σd > 0 a constant depending only on the dimension d. 4. On the standard Lagrange interpolation for d-dimensional P k -type finite elements Let d ≥ 2 be an integer and Ω ⊂ Rd the open unit simplex defined in (3.1). Let k ¯ → R by be an integer such that k ≥ d + 1. Define u : Ω  k+1−d Y d d X ¯ u(x) = 1 − xi xi ∀x = (x1 , · · · , xd ) ∈ Ω. i=1

i=1

¯ solves the Define also f (x) = −∆u(x) for x ∈ Ω. Obviously, u ∈ H01 (Ω) ∩ C ∞ (Ω) boundary value problem  −∆u = f in Ω, (4.1) u=0 on ∂Ω.

Equivalently, u ∈ H01 (Ω) is the weak solution, defined by A(u, v) = (f, v)

where A : H01 (Ω) × H01 (Ω) → R, defined by A(v, w) = (∇v, ∇w)

∀v ∈ H01 (Ω), ∀v, w ∈ H01 (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 mesh of Ω constructed in Section 3. Let Skh (Ω) ⊂ H 1 (Ω) denote the Pk -type finite element space corresponding to the mesh τh , i.e., the restriction Skh (Ω)|T is exactly Pk |T for ◦

each element T ∈ τh . Let S hk (Ω) = Skh (Ω) ∩ H01 (Ω). The finite element solution ◦

uh ∈ S hk (Ω) is defined by

◦

A(uh , vh ) = (f, vh ) ∀vh ∈ S hk (Ω). ¯ → S h (Ω) the standard Lagrange interpolation whose Finally, denote by Ih : C(Ω) k interpolation points are the principle lattice points of all simplex elements of τ h [3,9].

14

BO LI

Theorem 4.1. Let d and k be integers such that d ≥ 2 and k ≥ d + 1. With the quasi-uniform family of simplicial finite element meshes constructed in Section 3, we have that kIh u − uh kH 1 (Ω) ≥ ζd,k hk , (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    d  X ji 1 ji d T = (x1 , · · · , xd ) ∈ R : xi − ≥ 0, i = 1, · · · , d, ≤ xi − n n n i=1

for some integers ji with 0 ≤ ji ≤ n − 1, i = 1, . . . , d. For such an element, we denote its d + 1 vertices by   jd j1 (0) , ,··· , x = n n   j1 ji−1 ji + 1 ji+1 jd (i) x = ,··· , , , ,··· , , i = 1, · · · , d. n n n n n

For each x ∈ T , let λi (x)
(i = 0, · · · , d) be the barycentric coordinates of x defined by λi ∈ P1 |T , and λi x(j) = 1 if i = j and 0 if i 6= j. Explicitly,   ji λi (x) = n xi − , i = 1, · · · , d, x = (x1 , · · · , xd ) ∈ T, n λ0 (x) = 1 −

d X

λi (x),

i=1

Define ψT : T → R by

ψT (x) =

"

d Y

λi (x)

i=1

x ∈ T. # k−d−1 Y j=0

j λ0 (x) − k



∀x ∈ T.

We claim that ψT differs only by a nonzero constant from the local shape function associated with the nodal point   j1 1 jd 1 x˜ = + ,··· , + ∈T n nk n nk whose barycentric coordinate is

(λ0 (˜ x), λ1 (˜ x), · · · , λd (˜ x)) =



1 k−d 1 , ,··· , k k k



∈ Rd+1 .

In fact, ψT ∈ Pk |T . Moreover, any nodal point x ∈ T has the barycentric coordinates λi (x) = mi /k for some integer mi with 0 ≤ mi ≤ k, i = 0, · · · , d, and m0 = P k − di=1 mi . 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,

LAGRANGE INTERPOLATION AND FINITE ELEMENT SUPERCONVERGENCE

15

P then di=1 mi = 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)!/k k > 0. Denoting by IT : C(T ) → Pk |T the local Lagrange interpolation operator on T —the ¯ → S k (Ω) onto C(T ), we then have restriction of Ih : C(Ω) h k−d ψT (x) ∀x ∈ T. k Consequently, since on T , u(x) − n−k−1 λ0 (x)ψT (x) is a polynomial of degree ≤ k, and IT : C(T ) → Pk |T is a projection on Pk |T , we have that (IT (λ0 ψT ))(x) = λ0 (˜ x)ψT (x) =

u(x) − (IT u)(x) = n−k−1 [(λ0 ψT )(x) − (IT (λ0 ψT ))(x)]   k−d −k−1 ψT (x) =n λ0 (x) − k # k−d " d  Y Y j −k−1 λ0 (x) − =n λi (x) k j=0 i=1 Qd

We now define vT ∈ Pd+1 |T ⊆ Pk |T by vT (x) = simple calculation, we have that ∆vT (x) = −2n

2

d d X Y

i=0

λj (x)

i=1 j=1, j6=i

Moreover, using the change of variables ξi = λi (x) ξ ∈ Sd , we obtain that Z Y Z d 2 −d 2 2−d kvT kH 1 (T ) = n ξi dξ + n Sd i=0

Sd

(4.3) ∀x ∈ T.

λi (x) for all x ∈ T . By a

∀x ∈ T.

(4.4)

(i = 1, · · · , d) from x ∈ T to Y  2 d ∇ξ dξ, ξ i

(4.5)

i=0

P where ξ0 = 1 − di=1 ξi and ∇ξ is the gradient with respect to ξ. By (4.3), u − IT u vanishes on the boundary of T . Therefore, by integration by parts, (4.3), (4.4), and the change of variables ξi = λi (x) (i = 1, · · · , d) from x ∈ T to ξ ∈ Sd , we get that Z ∇(u − IT u)(x) · ∇vT (x) dx T Z = − (u − IT u)(x) ∆vT (x) dx (4.6) T # (k−d ) X Z "Y d d d Y Y j λj (x) dx = 2n1−k λi (x) λ0 (x) − k T i=1 i=1 j=1, j6=i j=0 # d ! "k−d  Z d d X Y Y Y j 1−k−d ξj dξ. = 2n ξi ξ0 − k Sd i=1 j=1, j6=i i=1 j=0

16

BO LI

Denote by τh0 the collection of all the corner simplicial elements in τh . Define ¯ → R by vh = 0 on all elements in τh \τ 0 and vh = vT on any element T ∈ τ 0 . vh : Ω h h ◦

We have that vh ∈ S hk (Ω), since for each T ∈ τh0 , vT ∈ Pk |T vanishes on the boundary of T . Moreover, we have by (4.5) that "  2 # Z Y Z Y d d ∇ξ dξ , ξ ξi2 dξ + kvh k2H 1 (Ω) = n2−d |τh0 | n−2 i Sd i=0

Sd

i=1

where |τh0 | is the number of elements in τh0 , and by (4.6) that XZ ∇(u − IT u)(x) · ∇vT (x) dx = 2µd,k |τh0 |n1−k−d , A(u − Ih u, vh ) = T

T ∈τh0

where µd,k =

Z

d Y

! "k−d # X d d Y Y j ξi ξ0 − ξj dξ k i=1 j=0 i=1 j=1, j6=i

Sd

is a constant depending only on d and k. Consequently,   |A(u − Ih u, vh )| 2|µd,k | ≥ |τh0 |1/2 n−k−d/2 , √ kvh kH ( Ω) νd

(4.7)

where νd =

Z

Sd

"

d Y i=0

ξi2

 d  2 # Y + ∇ξ ξi dξ > 0 i=1

is a constant depending only on d. It follows from the construction of the mesh τh in Section 3 that h = σd /n and |τh0 | ≥ κd h−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 s 4κd µ2d,k > 0, ζd,k = σd2k+d νd,k where we use the fact that the constant M in the continuity condition in Lemma 1.1 can 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 the quasi-uniform family of simplicial finite element meshes constructed in Section 3, we have that (4.8) max |u(z) − uh (z)| ≥ θd,k hk+1 , z∈Nh

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

LAGRANGE INTERPOLATION AND FINITE ELEMENT SUPERCONVERGENCE

17

Proof. Notice that (Ih u)(z) = u(z) for all z ∈ Nh . Thus, if (4.8) were not true, then we would have
as h → 0. kIh u − uh kL∞ (Ω) = o hk+1 This would lead to
as h → 0, kIh u − uh kL2 (Ω) = o hk+1

and further to


kIh u − uh kH 1 (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 ! "k−d # X Z d d d Y Y Y j ξ0 − ξj dξ 6= 0, µd,k := ξi k Sd i=1 j=1, j6=i i=1 j=0 P 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 Pd ξ1 , · · · , ξd , and the change of variables ηi = ξi (i = 1, · · · , d − 1) and ηd = 1 − i=1 ξi , we have ! "k−d # Y Z d d−1 Y Y j ξ0 − µd,k = d ξi ξi dξ k Sd j=0 i=1 i=1 ! ! k−d  Z d−1 d Y X Y j 2 =d ηi 1− dη (5.1) ηi ηd − k Sd i=1 i=1 j=0 Z 1 Z 1−Pdi=2 ηi Z 1−ηd Z 1−Pdi=j+1 ηi =d dηd dη1 dηj · · · dηd−1 · · · 0 0 0 0 ! ! k−d  d−1 d Y X Y j 2 . ηi 1− ηi ηd − k i=1 i=1 j=0 Set

E1 = and Ej =

Z

1− 0

Pd

Z

1−

Pd

i=2

ηi

0

i=j+1

ηi

η12

ηj2 Ej−1 dηj

1−

d X i=1

ηi

!

dη1

for j = 2, · · · , d − 1.

By an argument of induction on j (1 ≤ j ≤ d − 1) using the expression !#2 ! " d d X X ηi ηi − 1 − ηj2 = 1− i=j

i=j+1

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BO LI

=

1−

we obtain that " j Y Ej = i=1

d X

ηi

i=j

!2

−2 1−

2 (3i − 1)(3i)(3i + 1)

d X i=j

#

ηi

!

1−

1− d X

ηi

i=j+1

ηi

i=j+1

d X

!3j+1

!

,

+

1−

d X

i=j+1

ηi

!2

,

j = 1, · · · , d − 1.

It then follows from (5.1) that  Z 1 k−d Y j Ed−1 ηd − µd,k = d dηd k 0 j=0 #Z "d−1  k−d 1 Y Y j 2 3d−2 dηd . ηd − (1 − ηd ) =d (3i − 1)(3i)(3i + 1) 0 k j=0 i=1

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   Z 1 k−d Y i >0 if k − d is even, 3d−2 Jd,k := dt (1 − t) t− (5.2) < 0 if k − d is odd. k 0 i=0 Q Proof. Denote q = k − d ≥ 1 and ωl (t) = li=0 (t − ti ) for any integer l ≥ 0, where tr = r/k for any real r. Rt Case 1: q = k − d is even. Let Ωl (t) = 0 ωl (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 Z t Z t Ωq (t) = Ωq (tq ) + ωq (s) ds = ωq (s) ds > 0, tq

tq

since ωq (s) > 0 for all s ≥ tq . Therefore, we obtain by integration by parts that Z 1 Z 1 3d−2 0 Jd,k = (1 − t) Ωq (t) dt = (3d − 2) (1 − t)3d−3 Ωq (t) dt 0 0 Z tq Z 1 3d−3 (1 − t)3d−3 Ωq (t) dt > 0. = (3d − 2) (1 − t) Ωq (t) dt + (3d − 2) 0

tq

Case 2: q = k − d is odd. Direct calculations lead to q (d − 1)γq (d) Y (3d + i)−1 < 0, Jd,d+q = − (d + q)q i=−1

where

γ1 (d) = 1 > 0, γ3 (d) = 12(4d2 − d + 3) > 0,

q = 1, 3, 5, 7,

(5.3)

LAGRANGE INTERPOLATION AND FINITE ELEMENT SUPERCONVERGENCE

19

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 Z tq Z 1 3d−2 Jd,k = (1 − t) ωq (t) dt + (1 − t)3d−2 ωq (t) dt = Id,k + Md,k . 0

(5.4)

tq

By straight forward calculations, we obtain that q d Z tq+j Y X (1 − t)3d−2 (t − ti ) dt Md,k := tq+j−1

j=1


0. 0 Moreover, fd,k (t) > 0 and fd,k (t) < 0 for all t ∈ (0, tq/2 ). Thus,   1 1 gd,k (t) ≥ (tq − t)fd,k (t) − t + fd,k (t) ≥ fd,k (t) > 0 ∀t ∈ (t2q0 , t2q0 +1/2 ). k k 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 Z t2j   (1 − t)3d−2 + (1 − tq + t)3d−2 ωq (t) dt Hd,k,j := t2j−2

Z

= =−

t2j−1

+ t2j−2

Z

t2j−1

Z

t2j

t2j−1

!



 (1 − t)3d−2 + (1 − tq + t)3d−2 ωq (t) dt

(5.10)

gd,k (t)ωq−1 (t) dt, t2j−2

where gd,k is defined in (5.8). For each t ∈ (t2j−2 , t2j−1 ), ωq−1 (t) has 4q0 − 2j + 4 (s) negative factors. So, ωq−1 (t) > 0. Using the fact that (−1)s fd,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       1 1 1 4 gd,k (t) ≥ tq − 2t − fd,k t + ≥ fd,k t + >0 k k k k and 0 (t) gd,k

      1 1 1 0 0 t+ = − fd,k (t) + fd,k t + + (tq − t)fd,k (t) − t + f k k d,k k     1 1 0 < tq − 2t − fd,k t + (5.11) k k tq−3 fd,k (t1 ) > 0,

t0 < t < t1 .

Consequently, by (5.6), (5.7), (5.10), (5.12), and the fact that ω q−1 (t) > 0 for t ∈ (t0 , t1 ), we conclude that Id,k < Hd,k,1 < −gd,k (t1 )

Z

t1

ωq−1 (t) dt t0

(5.13)

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BO LI

< −tq−3 fd,k (t1 )

q−1 Y i=2

(ti − t1 )

Z

t1 t0

t(t1 − t) dt

(q − 3)(q − 2)! (q + d − 1)3d−2 . 2d+k 6k It follows now from (5.4), (5.5), and (5.13) that we need only to show that " q # d−1 Y X 6(q + 1) (i + j) (d − j)3d−2 ≤ (q − 2)!(q + d − 1)3d−2 (3d − 1)(q − 3) j=0 i=1