ON CONFIGURATIONS OF POINTS ON THE SPHERE AND APPLICATIONS TO APPROXIMATION OF HOLOMORPHIC FUNCTIONS BY LAGRANGE INTERPOLANTS

. We study certain configurations of points on the unit sphere in R N. As an application, we prove that the sequence of Lagrange interpolation polynomials of holomorphic functions at certain ChungYao lattices converge uniformly to the interpolated functions.

ON CONFIGURATIONS OF POINTS ON THE SPHERE AND APPLICATIONS TO APPROXIMATION OF HOLOMORPHIC FUNCTIONS BY LAGRANGE

INTERPOLANTS

PHUNG VAN MANH

A BSTRACT We study certain configurations of points on the unit sphere in RN As an

ap-plication, we prove that the sequence of Lagrange interpolation polynomials of holomorphic

functions at certain Chung-Yao lattices converge uniformly to the interpolated functions.

1 INTRODUCTION

LetPd(CN) denote the space of polynomials of degree at most d in N complex variables

A subset A of CN that consists of N+dd
distinct points is said to be unisolvent of degree d if, for every function f defined on A, there exists a unique polynomials P ∈Pd(CN) such that P(z) = f (z) for all z ∈ A This polynomial is called the Lagrange polynomial interpolation

of f at A and is denoted by L[A; f ] We are concerned with the problem of approximation of holomorphic functions

Problem 1 LetF be a subclass of entire functions in CN and Ad a unisolvent set of degree d for d= 1, 2, Under what conditions does L[Ad; f ] converge to f uniformly on every compact subset of CN for every f ∈F ?

It is well-known that if N = 1 then a sufficient condition is the boundedness of ∪∞

d=1Ad This

is an immediate consequence of the Hermite Remainder Formula (see [12, p 59]) Moreover, when the interpolation sets are unbounded, there exists a function f ∈ H(C) for which conver-gence does not hold In this case, the problem is valid for a subclassF of H(C) in which the modulus of the interpolation points are controlled by the order of f (see [1] for more details)

It is also proved in [1] that the same results also hold true for Kergin interpolation in CN, a natural generalization of the univariate Lagrange interpolation In contrast to the univariate case, Bloom and Levenberg showed in [5] that the boundedness of the interpolation array (Ad) does not guarantee the uniform convergence of every entire function as soon as N ≥ 2 The trouble here is that the interpolation operator has bad behavior when interpolation points tend

to an algebraic hypersurface of degree d

Problem 2 Let E be a compact subset of CN andF the class of functions which are holomor-phic in a neighborhood of E (which can depend on the functions) Let Ad⊂ CN be a unisolvent set of degree d for d= 1, 2, Under what conditions does L[Ad; f ] converge to f uniformly

on E for every f ∈F ?

We mention that tools from (pluri)potential theory can be used to solve Problem 2 The sufficient conditions are related to the Lebesgue constants, the transfinite diameter and global extremal functions, etc We refer the reader to [2, 4, 18] and the references therein

The aim of this paper is to give answers to the above problems when Ad is a Chung-Yao lat-tice generated by hyperplanes in CN Under natural conditions on (normal) vectors defining the hyperplanes, we prove in Theorem 3.8 that the interpolation polynomials of any holomorphic functions in a sufficiently large domain converge uniformly on a compact subset of CN to the

2000 Mathematics Subject Classification Primary 41A05, 41A63, 52C35.

Key words and phrases Lagrange interpolation, Chung-Yao lattices Configurations on spheres.

1

interpolated functions, provided the boundedness of the interpolation points From Theorem 3.8, we obtain a result regarding the uniform convergence of any entire function on every com-pact set of CN Under the same assumptions on the vectors, we prove in Theorem 3.11 that the convergence in Problem 1 is valid for entire functions of finite order and for the interpolation points that may be unbounded Moreover, the additional conditions regarding the location of interpolation points are similar to those given in [1, Theorem 2.3] In Theorem 3.6, we give explicit Chung-Yao lattices satisfying Theorems 3.8 and 3.11 We mention the equally spaced points in the simplex introduced in [3] and which also satisfy the above two theorems More generally, explicit interpolation points can be obtained by interwining sequences in the complex plane, see for instance [18]

Now the problem turns to find of set of vectors which satisfy certain conditions For sim-plicity, we work with unit vectors of real coordinates They can be viewed as points on the unit sphere SN−1 in RN Let Vd = {n1, , nd} be a set of points on SN−1 We associate Vd

to a number hV d that measures the thickness of Vd (see Section 2 for precise definitions) If each n ∈Vd is regarded as a normal vector of a hyperplane, then the condition on vectors in Theorems 3.8 and 3.11 become lim infd→∞hV d > 0 In Theorem 2.6, we construct a setVdsuch that hV d has this asymptotic behavior We finally note that hV d is well-studied when N = 2 We obtain precise estimates in this case

Notations The product of z = (z1, , zN), w = (w1, , wN) in RNor CN is defined by hz, wi =

∑Nj=1zjwj Let us denote by kzk = (∑Nj=1|zj|2)12 the norm of z The set B(z, R) (resp B(z, R))

is the open ball (resp closed ball) of center z ∈ CN and radius R > 0 Let A be a nonempty set Let An
denote the class of subsets of A containing n elements

2 CONFIGURATIONS OF POINTS ON THE SPHERE

2.1 Distance on the sphere In RN, N ≥ 2, each point can be regarded as a vector Let

e := {ej= (0, 0, 1, 0, , 0) : j = 1, , N} stand for the standard basis for RN We denote

by SN−1 the unit sphere in RN,

SN−1= {x ∈ RN : kxk = 1}

Let V = {n1, , nN−1} be a set of linear independent unit vectors and nk= (nk1, nk2, , nkN),

1 ≤ k ≤ N − 1 We can easily check that the vector nV defined by

nV = det(e, n1, , nN−1) := det









e1 e2 · · · eN

n11 n12 · · · n1N

. .

nN−11 nN−12 · · · nN−1N









(2.1)

is nonzero and orthogonal to V since hn, nVi = det(n, n1, , nN−1) for n ∈ RN Here the determinant in (2.1) is taken pointwisely according to the first row Therefore, it is a normal vector of the hyperplaneHV which passes through V and the origin We thus getHV = {x ∈

RN: hnV, xi = 0} and

dist(n,HV) =|hn, nVi|

Note that dist(n,HV) ≤ 1 for all n ∈ SN−1

Now letVd = {n1, , nd} be a set of d ≥ N points on SN−1 First, we make the assumption thatVdis N-independent, that is, any N vectors inVd are linearly independent, or equivalently det(nk1, nk2, · · · , nkN) 6= 0 for all 1 ≤ k1, k2, , kN≤ d pairwise distinct For each V ∈ Vd

N−1
, a

2

subset consisting of N − 1 vectors ofVd, we write hV,V d for the geometric mean of the distances from all points inVd\V toHV Due to (2.2), we have

hV,V d := ∏

n∈V d \V

dist(n,HV)
1

d−N+1 = ∏n∈V d \V|hn, nVi|
1

d−N+1

Let us define

hV d := min

n

hV,V d : V ∈

 Vd

N− 1

 o

On the other hand, ifVdis not N-independent, then we set hV d = 0

By definition, hV d is independent of the ordering of the vectors inVd In the result below, we fix an order ofVdand identifyVd with (n1, , nd)

Proposition 2.1 The map (n1, , nd) 7→ h{n1, ,nd}is continuous on(SN−1)d

Proof We first prove the continuity of the map at a N-independent tuple Vd = (n1, , nd)

By definition, we can choose a neighborhood Ω ofVd in (SN−1)d such that everyXd∈ Ω is also N-independent Thus formulas (2.3) and (2.4) are well-defined for Xd ∈ Ω and imply the continuity We now turn to the case in which Vd is not N-independent Hence, there exist N dependent vectors, say n1, , nN ∈Vd We can assume that nN = ∑N−1j=1 ajnj where

a1, , aN−1∈ R Since hV d = 0, we only need to verify that

lim

X d →V d

hX d = 0, where Xd is N-independent Write Xd = (x1, , xd) We denote by H the hyperplane

H{x 1 , ,xN−1} Using (2.2) we have

dist(nN,H ) ≤N−1∑

j=1

|aj|dist(nj,H ) ≤N−1∑

j=1

|aj|knj− xjk

It is easily seen that

dist(xN,H ) ≤ knN− xNk + dist(nN,H ) ≤ knN− xNk +

N−1

∑

j=1

|aj|knj− xjk

As xk goes to nkfor k = 1, , N, we have

dist(xN,H ) → 0

Since all perpendicular distances appearing in the definition of hX d are bounded by 1, hX d ≤

Let A be a subset of SN−1 If A contains at least d ≥ N points, then we define

Otherwise, we set hd(A) = 0 The supremum in (2.5) is attained when A is compact by Propo-sition 2.1 Now, let V = (Vd)∞

d=N be an array of points on SN−1 We define the asymptotic behavior of hV d as follows

hV := lim inf

We state a few simple properties which follow immediately from the definitions

Proposition 2.2 (1) 0 ≤ hV d, hd(A) ≤ 1 for allVd andA

(2) hV d and hd(A) are invariant under linear isometries of RN

(3) hV d keeps its value when we replace some elements ofVdby their additive inverses (4) hd(A) = hd(A ∪ (−A)), where −A = {−a : a ∈ A}

3

Proof The first two assertions are trivial Observe that for all V ∈ V d

N−2
, HV coincides with

H(V \{n})∪{−n} for n ∈ X and dist(a,HV) = dist(−a,HV) for a ∈ RN The third assertion follows The last conclusion is an immediate consequence of the third one  2.2 The two-dimensional case We treat the case N = 2 Each point on S1can be regarded as

a complex number LetVd= {n1, , nd}, d ≥ 2 and set nk= eiθkfor 1 ≤ k ≤ d If V = {nk}, thenHV is the straight line passing through nk and the origin Hence

dist(nj,H{nk}) = | sin(θj− θk)| and hnk,V d =

d

∏

j=1, j6=k

| sin(θj− θk)|

1 d−1 (2.7)

We now recall the notation of the d-th diameter and transfinite diameter of a plane compact set

A For details we refer the reader to [16, p 152-158] The d-th diameter of A, denoted by diamd(A), is defined by

diamd(A) := sup{ ∏

1≤ j<k≤d

|xk− xj|
2

d(d−1) : x1, , xd ∈ A} (2.8)

A set {x1, , xd} ⊂ A for which the supremum is attained is called a d-Fekete system for A

It is known that the sequence diamd(A)
∞d=2 is decreasing and its limit is called the transfinite diameter of A,

lim

We have diamd(S1) = dd−11 , diam∞(S1) = 1 More generally, diam∞(A) = sin(α/4) if A ⊂ S1

is an arc that subtends an angle α A d-Fekete system for S1always forms a regular d-polygon These facts can be found in [16, p 135, p 158]

Proposition 2.3 (1) If Γ is a compact subset of S1, then hd(Γ) ≤ (1/2)diamd( ˜Γ) where

˜

Γ = {z2: z ∈ Γ}

(2) If Γ ⊂ S1is an arc that subtends an angle α ≥ π, then there exists a setUd of d points

in Γ such that hU d = (1/2)dd−11 In particular, hd(Γ) = (1/2)dd−11

(3) If Γ ⊂ S1is an arc that subtends an angle α < π, then there exists a setWd of d points

in Γ such that hW d ≥ (1/2)dd−11 sinα

2 In particular, hd(Γ) ≥ (1/2)dd−11 sinα

2 Proof (1) LetVd ⊂ Γ and Vd= {n1, , nd} We write nk= eiθk, 0 ≤ θk< 2π Since |eiφ−

eiθ| = 2| sinφ −θ

2 |, in view of (2.7), we have

hnk,V d = 1

2

d

∏

j=1, j6=k

|e2iθj− e2iθk|
1

Hence

hV d ≤

d

∏

k=1

hnk,V d

1

d = 1 2

d

∏

1≤ j<k≤d

|e2iθj− e2iθk|
2

d(d−1) ≤ 1

2diamd( ˜Γ), (2.11) where ˜Γ = {z2: z ∈ Γ} The first assertion follows

(2) Let Γ be an arc that subtends an angle α ≥ π Since hd(Γ) is invariant under rotation,

we can assume that Γ = {eiθ : 0 ≤ θ ≤ ϕ}, ϕ ≥ π Then ˜Γ = S1 We take Ud = {eiϕk :

ϕk = kπ/d, k = 0, , d − 1} Evidently, e2iϕk, k = 0, , d − 1, are the vertices of a regular d-polygon Therefore they form a d-Fekete system for S1 Moreover, in view of (2.10), we realize that heiϕk,U d does not depend on k It follows that the two inequalities in (2.11) are thus equalities whenVdis replaced byUd This gives

hd(Γ) ≥ hUd = 1

2diamd( ˜Γ) =

1

2d

1

4

The reverse inequality is a consequence of (1) Note that the proof of (2) also gives the follow-ing relation that we will use in the proof of (3),

d−1

∏

j=0, j6=k

| sin( j − k)π

d | = hd−1

e iϕk ,U d

= hU d

d−1

2d−1, 0 ≤ k ≤ d − 1. (2.13) (3) Assume that Γ is an arc that subtends an angle at most π and Γ = {eiθ : −ω ≤ θ ≤ ω} with 0 < ω < π

2 To prove the desired estimates, we will construct a set of points in Γ, say

Wd = {eiθj : j = 1, , d}, such that hW d ≥ (1/2)dd−11 sin ω Our construction is motivated by [8] Let us set

φj= (2 j − 1)π

2d , ξj= cos φj and θj= arcsin(ξjsin ω), j= 1, , d (2.14) Obviously, θj∈ (−ω, ω) and ξj= −ξd+1− j for all 1 ≤ j ≤ d For 1 ≤ k ≤ d we have

hd−1

e iθk,W d

=

d

∏

j=1, j6=k

| sin(θj− θk)| =

d

∏

j=1, j6=k

| sin arcsin(ξjsin ω) − arcsin(ξksin ω)
| (2.15)

Using the angle sum identity for the sin and the identity cos arcsin(ξjsin ω)
=q1 − ξ2j sin2ω

we get

hd−1

e iθk,W d = (sin ω)d−1

d

∏

j=1, j6=k

ξj

q

1 − ξk2sin2ω − ξk

q

1 − ξj2sin2ω (2.16)

We denote by Pk the product of the d − 1 factors at the right hand side of (2.16) and by qj the j-factor of Pk In other words, Pk= ∏dj=1, j6=kqj It remains to prove that

Pk≥ d

Since ξj= −ξd+1− j for all 1 ≤ j ≤ d, we have

qjqd+1− j= |ξ2j − ξk2| = | cos2φj− cos2φk| = | sin(φk− φk) sin(φj+ φk)| (2.18) The proof of (2.17) will be divided into two cases

Case 1 d+ 1 = 2k (of course d must be odd ) By (2.18) we can write

Pk2=

d

∏

j=1, j6=k

| sin(φj− φk) sin(φj+ φk)| (2.19) Looking at the values of the φj’s given by (2.14) and using (2.13) we have

d

∏

j=1, j6=k

| sin(φj− φk)| = d

On the other hand, since φj+ φd+1− j = π, we get

| sin(φj+ φk)| = | sin(φd+1− j− φk)| = | sin(φd+1− j− φd+1−k)| (2.21) Consequently,

d

∏

j=1, j6=k

| sin(φj+ φk)| =

d

∏

j=1, j6=k

| sin(φd+1− j− φd+1−k)| = d

2d−1. (2.22)

5

Combining the above calculations we finally obtain Pk= d

2d−1 Case 2 d+ 1 6= 2k Applying (2.18) again, we get

Pk2= q2d+1−k

d

∏

j=1, j6=k, j6=d+1−k

| sin(φj− φk) sin(φj+ φk)| (2.23)

On the other hand, since ξk+ ξd+1−k= 0 and φk+ φd+1−k= π,

qd+1−k= |2ξk

q

1 − ξk2sin2ω | ≥ | sin(2φk)| = | sin(φd+1−k− φk)| (2.24) Combining (2.24) with (2.23) and using (2.21) we obtain

Pk2≥

d

∏

j=1, j6=k

| sin(φj− φk)|

d

∏

j=1, j6=d+1−k

| sin(φj− φd+1−k)| = d

2d−1

2

, (2.25)

where we use (2.13) in the last equation This gives Pk≥ d

2 d−1, and the proof is complete  Corollary 2.4 If Γ ⊂ S1is an arc that subtends an angle α, then diamd(Γ) ≥ dd−11 sinα

4 Proof Suppose that Γ = {eiθ : 0 ≤ θ ≤ α} and γ = {eiθ : 0 ≤ θ ≤ α/2} Proposition 2.3 now yields

(1/2)diamd(Γ) ≥ hd(γ) ≥ (1/2)dd−11 sin(α/4), and the corollary follows Note that the properties of transfinite diameter give the weaker estimate,

diamd(Γ) ≥ diam∞(Γ) = sin(α/4)

 Corollary 2.5 Let Γ ⊂ S1be an arc that subtends an angle α Then there exists an array (Vd)

of points in Γ such that

lim

d→∞hV d = lim

d→∞hd(Γ) =

(

(1/2) sin(α/2) if α < π Proof If α ≥ π, the conclusion is trivial by Proposition 2.3(2) In case 0 < α < π, we assume that Γ = {eiθ : 0 ≤ θ ≤ α} Then ˜Γ = {eiθ: 0 ≤ θ ≤ 2α} and diam∞( ˜Γ) = sin(α /2) From the first and third part of Proposition 2.3 we have

(1/2)dd−11 sin(α/2) ≤ hW d ≤ hd(Γ) ≤ (1/2)diamd( ˜Γ)

2.3 A lower bound for the general case In higher dimensions, we will construct a precise array V = (Vd)∞

d=N on SN−1 and get a lower bound for hV We need to compute the

vec-tor nV and the scalar product hnj, nVi for V ∈ Vd

N−1
and nj ∈Vd It can be done when the computations reduce to univariate Vandermonde determinants which are recalled now Let

F = (p1, , pm) be a tuple of m univariate polynomials and T = {t1, ,tm} a set of m real numbers The Vandermonde determinant corresponding toF and T is defined as follows

Theorem 2.6 There exists an arrayV = (Vd)∞

d=N of points onSN−1 such that

hV := lim infd→∞ hV d ≥√ 1

6

Corollary 2.7 We have

lim inf

d→∞ hd(SN−1) ≥ √ 1

N(2e)N−1.

We divide the proof into a sequence of lemmas The first lemma gives the formula for generalized Vandermonde determinants For convenience, we give its simple proof

Lemma 2.8 Let m ≥ 1 and Fk = (1, ,tk−1,tk+1, ,tm), resulting by removing tk from (1,t, ,tm), be tuples of monomials for k = 0, , m Then, for Tm= {t1, ,tm}, we have

VDM(Fk; Tm) = σm−k(t1, ,tm) ∏

1≤i< j≤m

(tj− ti), (2.28) where σ0= 1 and σj is the elementary symmetric polynomial of degree j for1 ≤ j ≤ m,

σj(t1, ,tm) = ∑

1≤k1<···<k j ≤m

tk1· · ·tkj

Proof By continuity, we only prove the lemma for ti6= tj, i 6= j Let us recall the formula for the univariate Vandermonde determinant,

VDM(Fm; Tm) = ∏

1≤i< j≤m

We consider a polynomial of degree m defined as follows

p(t) = det









1 t · · · tm

1 t1 · · · tm

1

.

1 tm · · · tm

m









Expanding the determinant along the first row we obtain

p(t) =

m

∑

k=0

(−1)kVDM(Fk; Tm)tk (2.31)

It is clear that p(tj) = 0 for all 1 ≤ j ≤ m Hence

p(t) = (−1)mVDM(Fm; Tm)

m

∏

k=1

(t − tk) = (−1)mVDM(Fm; Tm)

m

∑

k=0

(−1)m−kσm−ktk (2.32) Comparing the coefficients of tk in (2.31) and (2.32) we get the desired equalities  Lemma 2.9 Let b > 0 and Td = {t1, ,td} ⊂ R satisfy |tj− tk| ≥ b

d for j6= k Then

d

∏

j=1, j6=k

|tj− tk| ≥ 2

b

b 2e

d

, ∀1 ≤ k ≤ d

Proof Without loss of generality we assume that t1< t2< · · · < td Then |tj− tk| ≥ |( j−k)bd | for all 1 ≤ j, k ≤ d Thus

d

∏

j=1, j6=k

|tj− tk| ≥

d

∏

j=1, j6=k

|( j − k)b

d

d−1

(k − 1)!(d − k)! (2.33) The proof reduces to showing that

(k − 1)!(d − k)! ≥ 2

d

d 2e

d

7

If d is even, say d = 2m, then (k − 1)!(d − k)! ≥ (m!)m ≥ 1

m

m e

2m

, here we use the inequality m! ≥ me
min the second estimate, and (2.34) follows

If d is odd, say d = 2m + 1, then

(k − 1)!(d − k)! ≥ (m!)2≥ m

e

2m

= 2 d

d 2e

(1 +2m1 )2m ≥ 2

d

d 2e

d

Lemma 2.10 Let d ≥ N ≥ 2 and 2a ≥ b > 0 Let Td = {t1, ,td} be a set of real numbers in [−a, a] such that |tj− tk| ≥ b/d for j 6= k Consider the set of unit vectors

Vd= {nj= vj/kvjk : vj= (1,tj, ,tN−1j ), j = 1, , d}

Then there exists a positive constant C1depending only on a, b and N such that

hV d ≥ C

1 d−N+1

where C=



b 2e(a+1)

N−1

1

√ 1+a 2 +···+a 2N−2 Proof Note that Vd is N-independent This property is a consequence of (2.40) below It suffices to show that

hV,V d ≥ C

1 d−N+1

1 C for all V ∈

 Vd

N− 1



Without loss of generality we can assume that V = {n1, n2, , nN−1} Looking at the formula

of det(e, n1, , nN−1) in (2.1) and expanding the determinant along the first row we can write

nV = det(e, n1, , nN−1) =

N−1

∑

k=0

(−1)kDkek+1, (2.35)

where Dk is the determinant of (N − 1) × (N − 1) matrix obtained by deleting the first row and the (k + 1)-th column of the corresponding matrix Set δ = ∏N−1s=1 kv1sk In view of (2.1) and the definition of nk, we conclude from Lemma 2.8 that

Dk= δ VDM(Fk; TN−1) = δ σN−1−k(TN−1) ∏

1≤n<m≤N−1

(tm− tn), ∀0 ≤ k ≤ N − 1, (2.36)

where Fk = (1, ,tk−1,tk+1, ,tN−1) and TN−1 = {t1, ,tN−1} Hence (2.35) and (2.36) give

knVk = δ ∏

1≤n<m≤N−1

|tm− tn|

N−1

∑

k=0

σk(TN−1)2

1

Since σ0= 1 and tj6= tk for j 6= k, (2.37) shows that nV is a nonzero vector To get an upper bound for knVk, we note that

N−1

∑

k=0

σk(TN−1)2≤

N−1

∑

k=0

N − 1 k

2

a2k< (a + 1)2(N−1) (2.38) since TN−1⊂ [−a, a] Hence

knVk ≤ δ (a + 1)N−1 ∏

1≤n<m≤N−1

8

By the definition of nV in (2.1) and the formula for a univariate Vandermonde determinant, we have

|hnj, nVi| = | det(nj, n1, , nN−1)|

kvjk

N−1

∏

k=1

|tk− tj| ∏

1≤n<m≤N−1

|tm− tn|, ∀N ≤ j ≤ d (2.40) Combining (2.39) with (2.40) we obtain

|hnj, nVi|

knVk ≥

∏N−1k=1 |tk− tj| (a + 1)N−1kvjk ≥

1 (a + 1)N−1M

N−1

∏

k=1

|tk− tj|, ∀N ≤ j ≤ d, (2.41) where M :=√

1 + a2+ · · · + a2N−2≥ kvjk Since |tk− tj| ≤ 2a for every 1 ≤ j, k ≤ d, (2.41) the formula of hV,V d shows that

hd−N+1V,V d ≥  1

(a + 1)N−1M

d−N+1 d

∏

j=N

N−1

∏

k=1

|tk− tj|

(2a)(N−1)(N−2)

(a + 1)N−1M

d−N+1 N−1

∏

k=1

d

∏

j=1, j6=k

|tk− tj|

≥ C1Cd−N+1 Here we use Lemma 2.9 in the last estimate The constant C1is given by

(2a)(N−1)(N−2)(

2

b)N−1( b

2e)

(N−1)2

We finally remark that when N = 2, the empty product ∏1≤n<m≤N−1|tm− tn| is taken to be 1

Proof of Thereom 2.6 For each d ≥ N, let Vd be the tuple of unit vectors defined in Lemma 2.10 SetV = (Vd)∞

d=N By the estimate in Lemma 2.10, we have

hV = lim inf

2e(a + 1)

√

1 + a2+ · · · + a2N−2 (2.42)

We will find the maximal value of the right hand side of (2.42) The AM-GM inequality gives

2e(a + 1)

√

1 + a2+ · · · + a2N−2 ≤ b

4e√ a

N−1 1

√

NaN−1 ≤ √ 1

N(2e)N−1. Note that the maximum is attained when a = 1 and b = 2, and the proof is complete  Remark 2.11 Examining the proof of Lemma 2.10 and looking at (2.38), we have the better estimate

hV d ≥ C

1 d−N+1 1

 b 2e

q

∑N−1k=0 N−1k

2

a2k· ∑N−1k=0 a2k

Taking a = 1 and b = 2, we obtain

lim inf

eN−1

q

N 2N−2N−1

Open question Find the asymptotic behavior of hd(SN−1) as d → ∞ and an array of unit vectorsV = (Vd)∞

d=N such that hV attains the asymptotic behavior (if it exists)

9

3 APPLICATIONS TO APPROXIMATION OF HOLOMORPHIC FUNCTIONS

In this section, we essentially follow the notational conventions presented in [7, 10]

3.1 Chung-Yao lattices A hyperplane ` in CN is defined by an equation ` = {z ∈ CN :

hn, zi + c = 0}, |nk = 1 From now on, we assume that the (normal) vector in the definition of the hyperplane is a unit vector For convenience, we write `(z) = hn, zi + c and ˜`(z) = hn, zi

A set H of N hyperplanes in CN is said to be in general position if their intersection is a sin-gleton, that is ∩Nj=1`j= {ϑH} If `j(z) = hnj, zi + cj, then H is in general position if and only

if det(n1, , nN) 6= 0 Here and subsequently, we identify {ϑH} with ϑH More generally, a family H = {`1, , `d} of d ≥ N hyperplanes in CN is said to be in general position if

(1) Every H ∈ H

N
, a subset of N hyperplanes of H, is in general position;

(2) The map H ∈ H

N
7→ ϑH = ∩`∈H` is one-to-one

The set ΘH = {ϑH: H ∈ H

N
}, consisting of d

N
points, is called a Chung-Yao lattice of order

d It is well-known that ΘH is a unisolvent set of degree d − N We can now state a theorem due to Chung and Yao

Theorem 3.1 Let H = {`1, , `d} be a family of d ≥ N hyperplanes in general position in

CN If`j is given by`j= {z ∈ CN: hnj, zi + cj= 0} and f is a function defined on ΘH, then

L[ΘH; f ](z) = ∑

H∈(H

N)

f(ϑH)l(ΘH, ϑH; z), (3.1)

where the fundamental Lagrange interpolation polynomial (FLIP) is given by

l(ΘH, ϑH; z) = ∏

`∈H\H

`(z)

`(ϑH), H∈

 H N



Now if K = {`k1, , `kN−1} ⊂ H, k1< k2< · · · < kN−1, then ∩K = ∩`∈K` is a complex line

in CN It passes through a point a and is parallel to the vector nV = det(e, nk1, , nkN−1), where

V = {nk1, , nkN−1} is the set of nomal vectors of hyperplanes in K Hence we can write

∩K = {a + nVt: t ∈ C}

From now on, we write nK instead of nV With this notation, we have

˜

`j(nK) = hnj, nKi = hnj, nVi (3.3) Note that the last term is actually introduced in Section 2 in which it plays an important role 3.2 The de Boor error formula Let Ω be a convex open set in CN We will denote by H(Ω) the set of all holomorphic functions in Ω To every A = {a0, , ak} ⊂ Ω and f ∈ H(Ω), we associate a symetric k-linear form denoted by [A, ·]( f ) and defined by

[a0, , ak|u1, , uk]( f ) =

Z

[A]

Du1 Dukf =

Z

[A]

Dkf(·)(u1, , uk), uj∈ CN, (3.4)

where

Z

[A]

h= Z

∆k

h(a0+

k

∑

j=1

ξj(aj− a0))dξ1 dξk (3.5)

and ∆k= {ξ = (ξ1, , ξk) : ξj≥ 0, ∑k

j=1ξj≤ 1} This k-linear form is called the multivariate divided difference of f at A Using the above notations, we can state a beautiful remainder formula for Lagrange polynomial at Chung-Yao lattices due to de Boor, (see [7, Theorem 3.1])

10

We denote by Pk the product of the d − factors at the right hand side of (2.16) and by qj the j-factor of Pk In other words, Pk= ∏dj=1,... class="page_container" data-page="10">

3 APPLICATIONS TO APPROXIMATION OF HOLOMORPHIC FUNCTIONS< /small>

In this section, we essentially follow the notational conventions presented... divide the proof into a sequence of lemmas The first lemma gives the formula for generalized Vandermonde determinants For convenience, we give its simple proof

Lemma 2.8 Let m ≥ and Fk