Acta Math. Hungar., 125 (4) (2009), 299326. DOI: 10.1007/s10474-009-9023-z First published online July 6, 2009
ON GROWTH OF NORMS OF NEWTON
INTERPOLATING OPERATORS
A. P. GONCHAROV
Department of Mathematics, Bilkent University, 06800 Ankara, Turkey e-mail: goncha@fen.bilkent.edu.tr
(Received January 22, 2009; revised March 12, 2009; accepted March 13, 2009)
Abstract. We consider the problem of growth of the sequence of Lebesgue constants corresponding to the Newton interpolation and estimate the growth of this sequence in the case of a nested family of Chebyshev's points.
1. Introduction Let X =
(
x(n)k)
∞ nn=1, k=1be an in nite triangular array of nodes in [−1, 1]. Let ΛN(X)denote the N-th Lebesgue constant, that is the uniform norm of the Lagrange interpolating operator de ned by the points
(
x(N+1)k)
N+1k=1. It is well-known that the sequence ΛN(X)∞
N=0 has at least logarithmic growth and that the Chebyshev array T is close to the optimal choice. Now suppose that the array X is nested, that is, any row of X consists of the previous row plus one more value. What is the growth of ΛN(X)∞
N=0 in this case? Starting from the classical papers by C. Runge, S. Bernstein, and G. Faber, the problem of approximating properties of Lagrange interpola-tion has attracted atteninterpola-tion of many mathematicians. A variety of results
Key words and phrases: Lebesgue constant, Newton interpolation.
! A. P. GONCHAROV
concerning asymptotic behavior of the Lebesgue constant were obtained for dierent arrays (see e.g. [1], [4], [7]), as well as for dierent metrics. It is rather surprising that the naturally arising corresponding problem for the Newton interpolation was out of consideration. In this respect we can mention only Problem II in [3], where the question of existence of a nested interpolation process for a xed function f ∈ C[0, 1] was posed.
How to choose a sequence X = (xn)∞n=1⊂ [−1, 1] with a moderate growth of ΛN(X)∞N=0 ? One can suppose that such a sequence must at least ap-proximate the equilibrium arcsine distribution dμe(x) = π√dx1−x2. This means
that the nite measures 1
n n
k=1δxk, where δxk stands for the unit mass
lo-cated at xk, converge to dμein the weak∗topology of measures. For example, the Leja sequence L has this approximating property (see e.g. [6], T.V.1.1). R. Taylor and V. Totik [8] proved that the sequenceΛN(L)∞N=0has subex-ponentialgrowth, that is log ΛN(L)
N → 0, as N → ∞. Here and subsequently, log denotes the naturallogarithm.
The Leja sequence is a nested analog of the array F of Fekete points (see e.g. [6] for the de nition of Fekete points). Even for the set [−1, 1], the exact position of the Leja points is not known, whereas the corresponding n-th Fekete points are n-the zeros of n-the Jacobi polynomial Pn−2(1,1) together with the points ±1 (see e.g. [7], T.6.7.1). It should be noted that asymptotically ΛN(F ) N3/2 (see e.g. [4], Ch. 10, Corollary 13). On the other hand, for the Chebyshev array T we have ΛN(T ) π2log N + 1 (see e.g. [5], Theorem 1.2). Thus, a nested analog of the array T can be considered as a possible choice for the desired sequence X.
Here we arrange in a specialmanner the zeros of the Chebyshev polyno-mials (T3s)∞s=0, and show that
ΛN(X) = O exp log2N log 3 + 8 log N
for such nodes. The constant involved in this estimate is not exact, but for a subsequence (Ns)we have the lower bound ΛNs(X) exp
(
log2N
s
log 27 − 5 log Ns
)
. 2. The choice of a sequence of interpolating nodesThe Chebyshev polynomials TN(x) = cos (N arccos x)have the property
TNTM(x) = TNM(x). If M and N are odd, then the polynomial TNM pre-serves all zeros of TM. Therefore one can arrange a nested family of zeros of a subsequence of the Chebyshev polynomials. Here we consider the
polyno-NORMS OF NEWTON INTERPOLATING OPERATORS !
mials T3s for s ∈ N0:={0, 1, 2, . . . }. Let θ(s)j = 2j−13s π2 for j = 1, 2, . . . , 3s.
Then
{
cos θj(s)}
3sj=1 is the set of zeros of T3s. Let Z0 ={0} and for s
∈ N let Zs denote the zeros of T3s that are not zeros of T3s−1. That is
Zs=
{
zj = cos θ(s)j : j = 1, 3, 4, 6, 7, . . . , 3s− 3, 3s− 2, 3s}
. Then Zs can be represented as the union of the set Z−s =
{
cos(
θ(s−1)j +3πs)}
3s−1
j=1 with the set
Zs+ =
{
cos(
θ(s−1)j −3πs)}
3s−1j=1, where the superscripts ± are related to the position of the point from Zs with respect to the corresponding value from
∪s−1 k=0Zk.
We enumerate all points from X := ∪∞
s=0Zs in the following way: let
x1 = 0, x2 = cos5π6 , x3 = cosπ6 and if the points (xk)3 s
k=1with xk= cos ψkare chosen, then let x3s+k= cos (ψk+ π· 3−s−1), x2·3s+k = cos (ψk− π · 3−s−1)for k = 1, 2, . . . , 3s. Thus, we include at rst all points from the set Zs− in the order given by ∪s−1
k=0Zk in the sequence (xk)∞k=1 and afterwards we do so for the corresponding points from Z+
s . For the convenience of the reader, here we give some values of ψk:
ψ1 = π 2, ψ2= π 2 + π 3, ψ3= π 2 − π 3, ψ4 = π 2 + π 9, ψ5 = π 2 + π 3 + π 9, ψ6 = π 2 − π 3 + π 9, ψ7= π 2 − π 9, ψ10= π 2 + π 27, ψ100 = ψ34+19= ψ2·32+1+ π 35 = π 2 − π 33 + π 35.
Let us represent the angle ψn as a combination of δk= π3−k with
k∈ N. By construction of the sequence (ψn)∞n=1, we see that ψ2 = ψ1+ δ1,
ψ3= ψ1− δ1 and for s ∈ N we have ψ3s+k= ψk+ δs+1, ψ2·3s+k = ψk− δs+1
for k = 1, 2, . . . , 3s. For a general formula we x a function κ with the domain {0, 1, 2} such that κ(0) = 0, κ(1) = 1, κ(2) = −1. Then for the ternary expansion n = 1 +s
k=0γk3k with γk ∈ {0, 1, 2} we get ψn= π/2 + s
k=0κ(γk)δk+1. From here it follows that
(1) ψγs3s+···+γq3q+k= ψk+ Δ
for k = 1, 2, . . . , 3q with Δ =s
j=qκ(γj)δj+1. Given the points (xk)N+1k=1, let ωN+1(x) =
N+1
k=1 (x− xk). For k N + 1 we consider the fundamental Lagrange polynomial lk,N(x) = (x−xωkN+1)ω(x)
N+1(xk).
! A. P. GONCHAROV
quantity ΛN(X) = sup|x|1λN(X; x) is called the Lebesgue constant of or-der N of the sequence X. Clearly, ΛN(X) gives the norm of the Newton interpolating operator
NN : C[−1, 1] −→ ΠN : f → N+1
k=1
f (xk)lk,N(x).
Here and subsequently, ΠN denotes the set of all algebraic polynomials of degree at most N.
The aim of this paper is to estimate the growth of the sequence
ΛN(X)∞N=0.
Given N with 3s< N < 3s+1 we will consider the even trigonometric polynomials
μN(θ) = ωN(cos θ) = N k=1
[cos θ− cos ψk] and ρN(θ) =
3s+1 k=N+1
[cos θ− cos ψk].
Thus, μN(θ)· ρN(θ) = ˜T3s+1(cos θ), where ˜Tnstands for the monic Chebyshev
polynomials 21−nT
n.
Given δ, we de ne for any n ∈ N the polynomial tn(θ, δ) = n k=1 cos θ− cos (θk+ δ) .
Here θk= 2k−1n π2. We will restrict our attention to the values of the
parame-ter δ with |δ| < θ1 = 2nπ . We see that the zeros of tnare de ned by the angles corresponding to the Chebyshev polynomial Tn, but shifted by δ.
For any even trigonometric polynomial t(θ) =n
k=1[cos θ− cos αk] and
δ∈ R, let Sδt denote the polynomial nk=1 cos θ− cos (αk+ δ) . Thus, tn(θ, δ) = SδT˜n(cos θ).
3. On the maximal value of |μN| Let us show that |μN| attains its maximum at 0. Lemma 1. Given m n − 1 and |δ| < π
2n, let 0 α π − mnπ. Then tn(α, δ) tn(α + mπ/n, δ) = (−1) mm k=1 sin (α + θk) + sin δ sin (α + θk)− sin δ.
NORMS OF NEWTON INTERPOLATING OPERATORS ! !
Here for α = θ1+ δ, . . . , θn−m+ δthe fraction above is of the form 0/0, so we consider it to be the corresponding limit.
Proof. Here, tn(α, δ) = n k=1 cos α− cos (θk+ δ) = (−2)n· Π1(α)· Π2(α) with Π1(α) = n k=1 sin1 2 α + 2k− 1 2n π + δ and Π2(α) = n k=1 sin1 2 α−2k− 1 2n π− δ . Similarly, tn(α + mπ/n, δ) = (−2)n· Π1(α + mπ/n)· Π2(α + mπ/n)with Π1(α + mπ/n) = n k=1 sin1 2(α + 2k + 2m− 1 2n π + δ), Π2(α + mπ/n) = n k=1 sin1 2(α + 2m− 2k + 1 2n π− δ).
After cancellation of common terms we get tn(α, δ) tn(α + mπ/n, δ) = sin12
(
α +2nπ + δ)
· · · sin12(
α +2m−12n π + δ)
sin12(
α +2n+12n π + δ)
· · · sin12(
α +2m+2n−12n π + δ)
·sin 1 2(
α−2n−2m+12n π− δ)
· · · sin12(
α−2n−12n π− δ)
sin12(
α +2m−12n π− δ)
· · · sin12(
α + 2nπ − δ)
= m k=1 sin12(
α +2k−12n π + δ)
· sin12(
α−2n−2k+12n π− δ)
sin12(
α +2k−12n π− δ)
· sin12(
α +2n+2k−12n π + δ)
= m k=1 cos(
π2 + δ)
− cos(
π2 − α − θk)
cos(
π2 + δ)
− cos(
π2 + α + θk)
, which gives the desired result.! " A. P. GONCHAROV
Lemma 2. Suppose 0 < α1 α2 · · · αn< π is such that the polyno-mial t(θ) =n
k=1[cos θ− cos αk] attains its maximum modulus at θ = 0. If 0 δ π − αn, then Sδt as well attains its maximum modulus at θ = 0.
Proof. We proceed by induction. The polynomial t1(θ) = cos θ− cos α1
attains its maximum modulus at θ = 0 if and only if π/2 α1 π. In
this case Sδt as well attains its maximum modulus at 0. Suppose the statement is valid for any polynomial tn of the above type. Assume that for tn+1(θ) =
n+1
k=1[cos θ− cos αk] with 0 < α1 · · · αn+1< π we have
tn+1(0)tn+1(θ) for any 0 θ π. Then π/2 αn+1, since otherwise tn+1(π) > tn+1(0). Here, tn+1(θ) = tn(θ)[cos θ− cos αn+1]and Sδtn+1(0) = Sδtn(0)· 1− cos (αn+1+ δ) Sδtn(θ) ·cos θ− cos (αn+1+ δ) for anyθ.
In the next lemma we localize the extrema of tn(θ, δ). For |δ| < 2nπ the function tn(·, δ) has n + 1 local maxima on [0, π] : ϕ0 = 0, ϕj ∈ (θj + δ, θj+1+ δ)for j = 1, 2, . . . , n − 1 and ϕn= π.
Lemma 3. Suppose |δ| < π
2n. For j = 1, 2, . . . , n − 1 we have jπ/n < ϕj <
jπ/n + δ if δ > 0, and jπ/n + δ < ϕj < jπ/n if δ < 0.
Proof. Without loss of generality we can assume δ > 0. Let g(θ) = n
k=1
cos θ− cos (θk+ δ) −1. Then tn(θ, δ) =−tn(θ, δ)g(θ) sin θ. Let us show that
(2) g(jπ/n) < 0, g(jπ/n + δ) > 0 for 1 j n − 1.
This gives a change of sign of the derivative and a local extremal value of tn on the interval (jπ/n, jπ/n + δ).
We will use the representations [cos u− cos v]−1 = 1 2csc u cotv− u 2 − cot v + u 2 = 1 2csc v cotv− u 2 + cot v + u 2 . For 0 a δ we have g(jπ/n + a) = n k=j+1 cos (jπ/n + a)− cos (θk+ δ) −1 − j k=1 cos (θk+ δ)− cos (jπ/n + a) −1.
NORMS OF NEWTON INTERPOLATING OPERATORS ! #
We represent the terms of the rst sum as a dierence of the corresponding cotangents, whereas for the second terms we use the second representation. Then g(jπ/n + a)· 2 sin (jπ/n + a) (3) = n k=j+1 cot 2k− 2j − 1 4n π + δ− a 2 − cot 2k + 2j− 1 4n π + δ + a 2 − j k=1 cot 2k + 2j− 1 4n π + δ + a 2 + cot 2j− 2k + 1 4n π− δ− a 2 = n−j i=1 cot 2i− 1 4n π + δ− a 2 − j i=1 cot 2i− 1 4n π− δ− a 2 − n+j i=j+1 cot 2i− 1 4n π + δ + a 2 .
Choose a = 0. In the case 2j < n we cancel common terms in the rst and the third sums above. Here and in what follows we use the formula
n+j i=n−j+1 cot 2i− 1 4n π + c = j i=1 tan 2i− 1 4n π− c − tan 2i− 1 4n π + c . In this way, g(jπ/n)· 2 sin (jπ/n) = j i=1 cot 2i− 1 4n π + δ 2 − cot 2i− 1 4n π− δ 2 + j i=1 tan 2i− 1 4n π + δ 2 − tan 2i− 1 4n π− δ 2 = j i=1 2 csc 2i− 1 2n π + δ − 2 csc 2i− 1 2n π− δ . The last expression is negative for positive δ.
If 2j n then arguing as above we get for g(jπ/n) · 2 sin (jπ/n) the sim-ilar negative form
n−j i=1 2 csc 2i− 1 2n π + δ − 2 csc 2i− 1 2n π− δ .
! $ A. P. GONCHAROV
This proves the rst inequality in (2).
We now turn to the case a = δ in (3). If 2j < n then g(jπ/n + δ)· 2 sin (jπ/n + δ) = n−j i=j+1 cot2i− 1 4n π− cot 2i− 1 4n π + δ + j i=1 tan 2i− 1 4n π + δ − tan 2i− 1 4n π− δ , which is positive for δ > 0.
For 2j n we get g(jπ/n + δ)· 2 sin (jπ/n + δ) = j i=n−j+1 cot 2i− 1 4n π− δ − cot 2i− 1 4n π + n−j i=1 tan 2i− 1 4n π + δ − tan 2i− 1 4n π− δ > 0,
since for j n − 1 all arguments of the cotangents above are positive. This gives (2) andcompletes the proof.
Lemma 4. Suppose |δ| < π
2n. If δ > 0 then the values
(
tn(ϕj, δ))
nj=0 decrease with respect to j. For δ < 0 we get an increasing set of values.Proof. Suppose δ > 0. Let us x j with 1 j n andcompare the values
|
tn(
ϕj−πn, δ)|
and tn(ϕj, δ). By Lemma 1,|
tn(
ϕj− πn, δ)|
tn(ϕj, δ) =|
sin(
ϕj −πn+ θ1)
+ sin δ|
|
sin(
ϕj −πn+ θ1)
− sin δ|
.By Lemma 3, we can consider the expression on the right without modu-lus. Clearly, the fraction derived is larger than 1 for δ > 0. Thentn(ϕj−1, δ)
|
tn(
ϕj−πn, δ)|
, since ϕj−πn ∈ (θj−1+ δ, θj+ δ)and ϕj−1provides a maxi-mal value oftn(·, δ) on this interval. From this, tn(ϕj−1, δ) tn(ϕj, δ).Similar arguments apply to the case δ < 0.
The next lemma deals with the product of polynomials of type t3k.
Lemma 5. Let t(θ) = t3q(θ, δq+1+ λ)· t3p(θ,−δq+1+ λ) with 0 p < q,
|λ| < 1
NORMS OF NEWTON INTERPOLATING OPERATORS ! %
Proof. The proof will be divided into four steps. The rst three of them correspond to the case p 1.
1) We prove the inequalityt(α) t(α + δp) for0 α π − δp. This means that the function t(·) attains its maximum on the interval[0, δp].
2) For 0 θ β := δp− δq+1+ λwe show t(0) t(θ). Thus it remains to consider only the values β < θ < δp.
3) We show that t> 0 on (β, δ p),so max
[β,δp]
t(θ) = max
{
t(β),t(δp)}
. Then t(0) t(β),by 2),and t(0)t(δp),by 1).4) We consider the case p = 0 separately. To prove 1),let us x 0 α π − δp. Since
sin u− sin v sin u + sin v =
tanu−v2 tanu+v2 , we use Lemma 1 in the form
t(α) t(α + δp) =
|
tan 1 2(
α + θ(p)1 − δq+1+ λ)|
|
tan12(
α + θ(p)1 + δq+1− λ)|
· 3q−p k=1|
tan12(
α + θ(q)k + δq+1+ λ)|
|
tan12(
α + θ(q)k − δq+1− λ)|
. Here θ(q)k =(
k−12)
δq. Therefore, θ(p)1 = 12δp= 123q−pδq, θ3(q)q−p = δp−12δq and,as is easy to check,arguments of all tangents above belong to the interval (0, π/2), so we can remove the signs of modulus. Thus we want to prove the inequality (4) 3q−p k=1 tan12(
α + θ(q)k + δq+1+ λ)
tan12(
α + θ(q)k − δq+1− λ)
tan12(
α + θ1(p)+ δq+1− λ)
tan12(
α + θ1(p)− δq+1+ λ)
.Let us consider the function f(x) = tan (A+x)tan (A−x) for (5) 0 < x < A with x + A < π/2.
! & A. P. GONCHAROV
The values A = 1
2
(
α + θ(q)k)
, x = 12(δq+1+ λ) satisfy (5) for k = 1, 2, . . . , 3q−p, therefore the left fraction in (4) attains its minimal value when λ is minimal. Similarly, we can take A = 12
(
α + θ(p)1)
and x = 12(δq+1− λ), also satisfying (5). Then the right fraction in (4) attains its maximal value when x is maximal, that is λ is minimal. Therefore it is enough to prove (4) for λ =−12δq+1. Set σ := 12δq+1. We want to show3q−p k=1 tan12
(
α + θ(q)k + σ)
tan12(
α + θ(q)k − σ)
tan12(
α + θ(p)1 + 3σ)
tan12(
α + θ(p)1 − 3σ)
, which is equivalent to 3q−p k=1 1 + 2 sin σ sin(
α + θ(q)k)
− sin σ 1 + 2 sin 3σ sin(
α + θ1(p))
− sin 3σ .We can restrict ourselves in the product on the left only to the values k = 1, k = 12(3q−p+ 1)and k = 3q−p, since all terms in this product exceed 1. Here θ(3(q)q−p+1)/2= θ1(p)= 12δp. We use the following notations:
b1 = sin α +1 2δq − sin σ, b2= sin α + 1 2δp − sin σ, b3 = sin α + δp−1 2δq − sin σ, a = sin α +1 2δp − sin 3σ. Then the desired inequality takes the form
3 k=1 1 +2 sin σ bk 1 +2 sin 3σ a .
We multiply all terms on the left, neglect the term containing sin3σ, and use
the inequality sin 3σ < 3 sin σ in order to replace the desired inequality by 1 b1 + 1 b2 + 1 b3 + 2 sin σ 1 b1b2 + 1 b1b3 + 1 b2b3 3 a. The last inequality is equivalent to
NORMS OF NEWTON INTERPOLATING OPERATORS ! '
We express both parts of the inequality above as functions of the variable y := sin
(
α +12δp)
. Clearly, y > 0. Also for brevity, let μ := 12(δp− δq). Then b2 = y− sin σ, a = y − sin 3σ, b1+ b3 = 2y cos μ− 2 sin σ. Therefore, b1+ b2+ b3 = y(2 cos μ + 1)− 3 sin σ. Also,
b1b3 = y2− 2y sin σ cos μ + sin2σ− sin2μ
and
b2(b1+ b3) + b1b3 = y2(2 cos μ + 1)− 2y sin σ(2 cos μ + 1) + 3 sin2σ− sin2μ. The left part of (6) can be written in new terms as A3y3+ A2y2+ A1y
+ A0with A3= 2 cos μ + 1, A2=− sin 3σ(2 cos μ+1), A1=−3 sin2σ−sin2μ,
and A0= sin 3σ(3 sin2σ + sin2μ). On the other hand, on the right we
have B3y3+ B2y2+ B1y + B0 with B3= 3, B2=−3 sin σ(2 cos μ + 1), B1 =
3 sin2σ(2 cos μ + 1)− sin2μ , B0 =−3 sin σ (sin2σ− sin2μ). We transfer
A3y3 to the right, whereas B2y2+ B1y + B0 to the left side of (6), so we
reduce the inequality to the following: (A2− B2)y2+ (A1− B1)y + A0− B0
(B3− A3)y3, that is
4 sin3σ(2 cos μ + 1)· y2+
2 sin2μ− 12 sin2σ cos2μ 2
· y + 4 sin3σ(3 cos2σ− sin2μ) 4 sin2 μ
2 · y
3.
The coe cient of y2 is positive. Also, σ = π
2·3q+1 54π, as q 2.
There-fore 3 cos2σ > sin2μ. Since 0 < y < 1, it remains to show that
2 sin2μ− 12 sin2σ cos2 μ
2 4 sin
2μ
2, which can be reduced to tan2 μ
2cos μ 3 sin2σ.
Here, μ 1
2(δq−1− δq) = 6σ. It follows that tan2 μ2 9σ2. On the other
hand, μ < 1
2δp π/6, which gives the desired inequality and completes the proof of the rst step.
To prove 2), let us denote by ϕ(p)1 the point on the interval
(
θ1(p)− δq+1+ λ, θ(p)2 − δq+1+ λ)
where t3p attains its minimal value. By Lemma 3, β = δp− δq+1+ λ < ϕ(p)1 < δp. Therefore, the polynomial t3p decreases on the interval [0, β]. Also, by
! A. P. GONCHAROV
Lemma 4, the values
|
t3q(
ϕ(q)j , δq+1+ λ)|
of the corresponding local maxima decrease as j increases. Therefore, t(0) t(θ) for0 θ β.3) We turn to the values θ ∈ I := (β, δp). We cannot apply the arguments of Lemma 1 to this case. For example, if λ −σ and α −3σ = −θ(q)1 ,
then α + δp ∈ I. One can showthat t(α) <t(α + δp) here, since the value
t3q(α, δq+1+ λ) is rather small.
Let us analyze the behavior of the polynomial t on I in detail. Clearly, the polynomial t3p is negative here and, what is more, the values of t3p on I
are near to its local minimum. On the other hand, the polynomial t3q has a
root at the point z := δp−12δq+1+ λ∈ I, furthermore this is a root with odd multiplicity 3q−p. Therefore, t3q changes its sign from plus to minus when θ passes this point. Thus, t < 0 on (β, z) and t > 0 on (z, δp). Let us showthat
t > 0 on I. We have t3p(θ) =−t3p(θ)gp(θ) sin θ, t3q(θ) =−t3q(θ)gq(θ) sin θ,
where gp(θ) = 3p k=1
[
cos θ− cos(
θ(p)k − δq+1+ λ)]
−1, gq(θ) = 3q k=1[
cos θ− cos(
θ(q)k + δq+1+ λ)]
−1. Hence, t(θ) =−t(θ)g(θ) sin θ with g = gp+ gq. Note that the functions gq and g have a vertical asymptote at z. We see that the inequalities g > 0 on (β, z) and g < 0 on (z, δp) will give the positivity of t on I.
The function f(x) =n
k=1x−x1 k decreases on any interval of its
continu-ity. Therefore the function f(cos θ) increases on any such interval and it is enough to showthat
(7) g(δp) < 0 and g(β) > 0.
Let us rst examine the value δp. Arguing as in Lemma 3 (for the case
j = 1, a = 0), we get gp(δp)· sin δp = csc 1 2δp− δq+1+ λ − csc 1 2δp+ δq+1− λ . Similarly, gq(δp)· sin δp = 3 q−p k=1
[
csc(
θk(q)+ δq+1+ λ)
− csc(
θk(q)− δq+1− λ)]
.NORMS OF NEWTON INTERPOLATING OPERATORS !
In order to get the rst inequality in (7), we need to show that csc
(
θ(p)1 − δq+1+ λ)
+ 3 q−p k=1 csc(
θ(q)k + δq+1+ λ)
< csc(
θ(p)1 + δq+1− λ)
+ 3 q−p k=1 csc(
θk(q)− δq+1− λ)
.Since the left part of this inequality increases, whereas the right part decreases, when λ becomes smaller, it is enough to consider the value λ = −1 2δq+1 =−σ. We want to show (8) 3 q−p k=1
[
csc(
θ(q)k + σ)
− csc(
θk(q)− σ)]
< csc(
θ1(p)+ 3σ)
− csc(
θ1(p)− 3σ)
. The function f(x) = csc (x + 3σ) − csc (x − 3σ) increases as x increases with 3σ < x < π − 3σ. Therefore it su ces to consider the smallest value of θ(p)1 that is the case q = p + 1, when θ1(p)= 9σ. On the other hand, wecan take only the rst term in the sum in (8), since all expressions in the square brackets there are negative. Thus, for θ(q)1 = 3σ, we reduce (8) to the
inequality
csc (4σ)− csc (2σ) < csc (12σ) − csc (6σ), which is valid for 0 < σ π
54, as is easy to check. Thus, g(δp) < 0.
Our next goal is to show the second inequality in (7). In the same manner we get the representation
gp(β)· 2 sin β = 3p−2 k=1 cot 2k + 1 4 δp − cot 2k + 1 4 δp− δq+1+ λ + tan δp 4 − δq+1+ λ − tan δp 4 + δq+1− λ .
This value is negative and, as a function of λ with |λ| < σ, it attains its minimum when λ = −σ. Therefore,
gp(β)· 2 sin β − sin 6σ cos
(
δ4p − 3σ)
· cos(
δ4p + 3σ)
(9) + 3 p−2 k=1 − sin 3σ sin2k+14 δp· sin(
2k+14 δp− 3σ)
.! A. P. GONCHAROV
The term gq(β)can be handled in the same way. We get
gq(β) = 3q k=3q−p cos (3q−pδq− 2σ + λ) − cos k−1 2 δq+ 2σ + λ −1 − 3q−p −1 k=1 cos k−1 2 δq+ 2σ + λ − cos (3q−pδ q− 2σ + λ) −1 . As in Lemma 3, the terms of the rst (second) sum can be represented as a dierence (a sum) of the corresponding cotangents. Hence, gq(β)· 2 sin β = A1− A2− B1− B2, where A1= 3q−3 q−p+1 k=1 cot1 2(6k− 5)σ, A2 = 3q−3 q−p+1 k=1 cot1 2(2δp+ 6kσ− 9σ + 2λ), B1 = 3q−p −1 k=1 cot1 2(6k− 1)σ, B2= 3q−p −1 k=1 cot1 2(δp+ 6kσ− 3σ + 2λ). We decompose A1 in the sum
A11+ A12+ A13= 3q−p −1 k=1 + 2·3 q−p−1 k=3q−p + 3q−3 q−p+1 k=2·3q−p . Also, A2= A21+ A22= 3q−3·3 q−p+2 k=1 + 3q−3 q−p+1 k=3q−3·3q−p+3 . Here the dierence
A13− A21= 3q−3·3 q−p+2 k=1 cot1 2(2δp+ 6kσ− 11σ) − cot1 2(2δp+ 6kσ− 9σ + 2λ)
is positive, since |λ| < σ. We want to estimate gq(β)from below, so we can neglect this dierence. Also,
A12− B2= cot1 2(δp− 5σ) + 3q−p −1 k=1 cot1 2 δp+ (6k− 5)σ − cot12δp+ (6k− 3)σ + 2λ> cot1 2(δp− 5σ).
NORMS OF NEWTON INTERPOLATING OPERATORS ! ! Therefore, gq(β)· 2 sin β > A11− A22− B1+ cot1 2(δp− 5σ) = 3q−p −1 k=1 cot1 2(6k− 5)σ − cot 1 2(6k− 1)σ + 3q−p −1 k=1 tan1 2(6kσ− 3σ + 2λ) − tan 1 2(6kσ− 3σ − 2λ) + cot1 2(δp− 5σ) + tan 1 2(δp− 3σ + 2λ). The value λ = −σ only reduces this expression. From here,
gq(β)· 2 sin β 3q−p −1 k=1 4 cos (6k− 3)σ sin 2σ sin (6k− 5)σ sin (6k − 1)σ + 2 sin (δp− 5σ). Combining this with (9) we get the desired inequality g(β) > 0 provided
2 sin (δp− 5σ) > sin 6σ cos (δp/4− 3σ) · cos (δp/4 + 3σ) and 3q−p −1 k=1 4 cos (6k− 3)σ sin 2σ sin (6k− 5)σ sin (6k − 1)σ > 3 p−2 k=1 sin 3σ sin2k+14 δp· sin
(
2k+14 δp− 3σ)
. It is a simple matter to check the rst inequality for 0 < σ π54 and δp with 18σ = 3δq δp π/3. On the other hand, increase of p will enlarge the right part of the second inequality and decrease the left one. Therefore we can assume p = q − 1, that is δp= 18σ. Moreover, we consider only the rst term on the left. It remains to show that
4 cos 3σ sin 2σ sin σ sin 5σ > 3 p−2 k=1 sin 3σ sin
(
9k +92)
σ sin(
9k +32)
σ. For the sum on the right we have3 p−2 k=1 < 3 p−2 k=1 3σ sin29kσ < π2 4 3 p−2 k=1 3σ 81k2σ2 < π4 648σ,
! " A. P. GONCHAROV since 9kσ < π/2. Eventually, 4 cos 3σ sin 2σ sin σ sin 5σ > 8 cos 3σ cos σ 5σ > 2 5σ, which exceeds π4
648σ for the given values of σ.
4) To deal with p = 0, we note that
t1(θ,−δq+1+ λ) = cos θ− cos (ψ1− δq+1+ λ) = cos θ− sin (δq+1− λ). If θ π/2 thent1(θ) t1(0). Indeed, this is evident for θ π/2 − δq+1+ λ. If π/2 − δq+1+ λ < θ π/2 then
t1(θ) t1(π/2) = sin (δq+1− λ) t1(0), since 2 sin (δq+1− λ) 2 sin 3σ 1 for σ = 2·3πq 18π.
Furthermore, by Lemma 4, the values of the local maxima of |t3q|
de-crease, so t(0) t(θ) for 0 θ π/2. Let us x θ > π/2. For t1 we get
(10) t1(0) t1(θ) > 1− sin 3σ − cos θ + sin 3σ 1− sin 3σ 1 + sin 3σ > 1− 3σ 1 + 3σ. Given θ, we choose m such that
θm(q)+ δq+1+ λ θ < θ(q)m+1+ δq+1+ λ.
Thent3q(θ) t3q(ϕm), where by Lemma 3,mδq < ϕm < mδq+ δq+1+ λ.
Since θ > π/2, we have m 2. We compare, by means of Lemma 1, the values of t3q(·) at the points ϕm and α := ϕm− mδq. Clearly, 0 < α <
δq+1+ λ < θ1(q) and t3q(0) > t3q(α) > 0. Thus, t3q(0) t3q(θ) t3q(α) t3q(θ) = m k=1 sin
(
α + θk(q))
+ sin (δq+1+ λ) sin(
α + θk(q))
− sin (δq+1+ λ).All fractions in the product exceed 1, so we can consider only the rst two terms. Moreover, arguing as above, we see that the expression takes its minimal value when λ is minimal, whereas α is maximal. Therefore,
t3q(0) t3q(θ) > 2 k=1 1 + 2 sin σ sin
(
θ1(q)+ θk(q))
− sin σ .NORMS OF NEWTON INTERPOLATING OPERATORS ! #
Since θ1(q)= 3σ, θ2(q)= 9σ, and sin u + sin v sin (u + v) for 0 u, v π,
we get t3q(0) t3q(θ) > 1 +2 sin σ sin 5σ 1 + 2 sin σ sin 11σ > 1 + 2 sin σ· 16 55σ. Combining this with (10) we see that t(0) >t(θ), provided
16 55 sin σ σ 3σ 1− 3σ.
The left part of this inequality decreases, whereas the right part is an increasing function, so we have to check the inequality only for maximal σ.
Suppose q 2. Then 0 < σ π
54 and 16·5455·π sin54π 0.29. On the other
hand, π
18−π < 0.22, which proves the inequality.
We cannot apply the above mentioned arguments for q = 1. However one can use in this case a direct computation. Here
t1(θ,−δ2+ λ) = cos θ− sin π 9 − λ , t3(θ, δ2+ λ) = cos θ− cos π 6 + π 9 + λ ·cos θ− cos π 2 + π 9 + λ · cos θ− cos 5π 6 + π 9 + λ
and we consider the values |λ| < π
18 and θ ∈ I :=
[
π2, π]
. We see thatdiminu-tion of λ will decrease the value of t = t1t3at θ = 0 and will increase the values
of |t| at its extremal points on I. Therefore, in the worst case λ = −π
18, we get t(θ) = cos θ−1 2 cos θ− cos2π 9 cos θ− cos5π 9 cos θ− cos8π 9 . As before, t(θ) =−t(θ)g(θ) sin θ, where
g(θ) = cos θ− cos2π 9 −1 + cos θ− cos5π 9 −1 + cos θ− cos8π 9 −1 + cos θ−1 2 −1 .
! $ A. P. GONCHAROV
We see that t
(
π2
)
> 0. Since g(
π2)
≈ 3.51 > 0 and t has only two zeros on I,the polynomial t decreases on
[
π2, ϕ
]
and then increases on [ϕ, π], where ϕis a point of minimum of t with 5π
9 < ϕ < 8π9 . Thus, it is enough to check
only three inequalities: t(0) t
(
π2
)
, t(0) t(π), t(0) t(ϕ). Directcom-putation gives t(0) ≈ 0.2663, t
(
π2
)
≈ 0.0625, t(π) ≈ 0.1320. In order to getthe third inequality, let us nd a small enough interval (a, b) ⊂
(
5π9 ,8π9
)
suchthat g(a) · g(b) < 0, so ϕ ∈ (a, b). I f a < θ < b then, clearly, t(θ) < c := 1 2 − cos b cos2π 9 − cos b · cos5π 9 − cos b cos a− cos8π 9 . Now the values a = 52π
72, b = 53π72 can be taken. Indeed, g(a) ≈ −0.3483, g(b)
≈ 0.2498 and c ≈ 0.2526 < t(0).
We see that |t| attains its maximum value at 0, just as in all previous cases.
The main result of this section is
Lemma 6. For any N ∈ N, 0 θ π we have μN(0)μN(θ).
Proof. If N = 3s with s ∈ N0, then the function μ3s(θ) = 21−3scos 3sθ attains its maximum modulus at 0. Also, μ2(θ) = cos θ(cos θ− cos 5π/6) has
this property. Therefore we have the desired inequality for 1 N 3. By induction, assume it is valid for N 3s. Suppose 3s< N < 3s+1. I f N 2· 3s, that is N = 3s+ kfor 1 k 3s, then ψN = ψk+ δs+1. By induction, μk(0)μk(θ) for allθ, so, by Lemma 2, μN(0)μN(θ) for allθ.
Let us consider the remaining case N = 2 · 3s+ γ
r· 3r+· · · + γq· 3q+
γp· 3p. Here γk = 0, that is γk = 1 or γk= 2, and s > r > · · · > q > p 0. The rst 3s angles (ψ
k)3 s
k=1 give us the polynomial μ3s, the next (ψk)2·3 s k=3s+1
are zeros of t3s(·, δs+1). For the numbers (2 · 3s+ k)3k=1r we have the angles
ψ2·3s+k= ψk− δs+1 with k = 1, . . . , 3r, and the corresponding polynomial
t3r(·, −δs+1). I f γr= 2 then for the next 3r indices (2 · 3s+ 3r+ k)3k=1r we get the angles ψ2·3s+3r+k=−δs+1+ δr+1+ ψkfor k = 1, . . . , 3r, and the
cor-responding polynomial t3r(·, −δs+1+ δr+1). Therefore,
μN = μ3s· t3s(·, δs+1)· t3r(·, −δs+1)· · · for γr = 1, and
NORMS OF NEWTON INTERPOLATING OPERATORS ! %
Here · · · denotes the product of certain polynomials t3k with k < r. Both
functions t3s(·, δs+1)· t3r(·, −δs+1) and t3r(·, −δs+1+ δr+1) have maximum modulus at 0, the rst by Lemma 5, and the second by Lemma 2. Continu-ing in this way, from (1) we get in . . . the last term
t3p·, −δs+1+κ(γr)δr+1+· · · + κ(γq)δq+1
for γp = 1, and this polynomial together with
t3p·, −δs+1+κ(γr)δr+1+· · · + κ(γq)δq+1+ δp+1
for γp= 2. The last polynomial has the form t3p(·, δ) with δ > 0, so we can apply Lemma 2 for it. In the case γq= 1 the previous polynomial is of the same type. If γq= 2then we combine the polynomial t3p(·, −δs+1+· · ·−δq+1) with t3q(·, −δs+1+· · · + δq+1), which is the second in q-th pair. Thus we get
the product t = t3q(·, λ + δq+1)· t3p(·, λ − δq+1) with λ =sk=q+1κ(γk)δk+1, where κ(γs) =−1. It is easy to check that
−1 2δq+1+ 1 2δs+1 λ 1 2δq+1− 5 2δs+1. Therefore, |λ| < 1
2δq+1 and Lemma 5 can be applied to the polynomial t. By that we decompose μN into a product of terms every one of which has maximum modulus at 0, so μN also has this property.
4. On the growth of ΛN
Values of the polynomial tn(·, δ) can be expressed in terms of the product
πn(b) := sinπ 4n + b sin 3π 4n + b · · · sin 2n− 1 4n π + b . For example, as is easy to see,
tn(0, δ) = 2nπ2n(δ/2), tn(π, δ) = (−2)nπn2(−δ/2).
For some values of the parameter b, the exact magnitude of πn(b)can be found.
Lemma 7. For any n ∈ N we have πn(0) = 2−n+
1
! & A. P. GONCHAROV
Proof. Let us show rst that for any m ∈ N
(11) 2m m k=1 cos kπ 2m + 1 = 1.
The Chebyshev polynomial T2m+1 has local extrema at the points cos2m+1kπ
for k = 1, . . . , 2m. Hence, T2m+1 (x) = 22m(2m + 1) 2m k=1 x− cos kπ 2m + 1 .
On the other hand,
T2m+1 (x) = (2m + 1)sin (2m + 1)θ
sin θ for x = cos θ with 0 θ π. Taking x = 0 in both expressions yields
T2m+1 (0) = 22m(2m + 1) 2m k=1 cos kπ 2m + 1 = (2m + 1)(−1) m. Since 2m k=m+1 cos kπ 2m + 1 = (−1) mm k=1 cos kπ 2m + 1, (11) follows.
By means of (11) we can express πn(0) = n
k=1sin2k−14n π.
Let us rst examine the odd value n = 2m + 1. Here, πn(0) = m k=1 sin2k− 1 4n π· 1 √ 2 · m k=1 cos2k− 1 4n π = 2 −m−1 2 · m k=1 sin2k− 1 2n π. The last product is equal to m
k=1coskπn, which is 2−m, by (11). Thus,
πn(0) = 2−m−12 2−m = 2−n+12.
Similarly, π2m(0) = 2−mπm(0), andby that, π2qk(0) = 2−k(2q−1)πk(0)for
any k, q ∈ N. For the general case, we write n in the form n = 2q(2m + 1) with m, q ∈ N0. Combining the previous representations we get
NORMS OF NEWTON INTERPOLATING OPERATORS ! '
To express πn:= πn
(
4nπ)
= n−1k=1sinkπ2n we apply similar arguments.
In-stead of (11) we use (12) 2m m k=1 cos 2k− 1 4m + 2π = √ 2m + 1 for m ∈ N. Indeed, T2m+1(x) = 22mxmk=1
(
x2− cos2 2k−14m+2π)
. Thus,T2m+1 (0) = lim x→0 T2m+1(x) x = 2 2m(−1)mm k=1 cos2 2k− 1 4m + 2π. But as above, T
2m+1(0) = (−1)m(2m + 1). Comparing both forms of
T2m+1 (0)we obtain (12).
Arguing as above, we see that π2m= 2−m+12πm and π2qk= 2κπk with κ = (−2q+ 1)k + q/2. Also, π2m+1= 2−m m k=1 sin kπ 2m + 1 = 2 −mm k=1 cos 2k− 1 4m + 2π = 4 −m√2m + 1, by (12). Thus for any n = 2q(2m + 1) we have
πn= 2(2m+1)(−2q+1)+q/2· 4−m√2m + 1 = 2−n+1√n, which is the desired conclusion.
Corollary 1. If 0 δ π
2n then 21−n max0θπtn(θ, δ) 22−nn. If −π
2n + ε σ 0 for some 0 < ε < 2nπ then tn(π, σ)/tn(0, σ) (π/ε)2. Proof. Indeed, if 0 δ π
2n then by Lemma 4 we have
max 0θπtn(θ, δ) = tn(0, δ) = 2 nπ2 n(δ/2) 2nπn2 π 4n = 22−nn. Also, tn(0, δ) ˜Tn(0) = 21−n.
To estimate the other term, we note that tn(π, σ) 2nπ2n
π 4n
= 22−nn. On the other hand,
πn(σ/2) sin ε/2 · n−1 k=1 sin kπ 2n + ε 2 ε ππn π 4n , so tn(0, σ) 22−nn ε π 2 .
! A. P. GONCHAROV
Our next objective is to analyze the remaining polynomial ρN =
3s+1 k=N+1
cos (·) − cos ψk .
This polynomial can be handled in the same way as μN in Lemma 6. Lemma 8. Let 3s< N < 3s+1 for some s ∈ N0, 0 θ π. Thenρ
N(θ) 182· 3s2+5sρ
N(0).
Proof. Let M = 3s+1− N, so deg ρ
N = M and 1 M 2 · 3s− 1. We use the ternary decomposition
M = γs· 3s+ γr· 3r+ γq· 3q+· · · + γp· 3p with s > r > q > · · · > p 0. Here γs∈ {0, 1} and γr, . . . , γp ∈ {1, 2}. In order to decompose ρN into a product of polynomials of type t3k, we sort out all angles (ψk)3k=N+1s+1 in the
inverse order, that is from ψ3s+1 to ψN+1.
Suppose rst γs= 1. Then the angles (ψ2·3s+k)3 s
k=1 are zeros of the poly-nomial t3s(·, −δs+1). The next 3r angles have indices
(3s+ 2· 3s−1+· · · + 2 · 3r+ k)3k=1r and correspond to t3r(·, δs+1− δs− · · · − δr+1).
If γs= 0then the polynomial t3r(·, −δs+1−δs−· · ·−δr+1)has zeros at the points ψj with indices j = 2 · 3s+ 2· 3s−1+· · · + 2 · 3r+ kwhere k = 1, . . . , 3r. Since γs−1=· · · = γr+1= 0, so κ(2 − γk) =−1 for r + 1 k s − 1, the rst terms in the decomposition of ρN are
tγ3ss(·, −δs+1)· t3r
·, κ(2 − γs)δs+1+· · · + κ(2 − γr+1)δr+2− δr+1
. If γr= 2 then the next indices
(2− γs)3s+ 2· 3s−1+· · · + 3r+ k3k=1r give t3r·, κ(2 − γs)δs+1+· · · + κ(2 − γr+1)δr+2+ δr+1.
Continuing in this way, we get
ρN = tγ3ss(·, −δs+1)· t3r·, κ(2 − γs)δs+1+· · · + κ(2 − γr+1)δr+2− δr+1 · tγr−1 3r ·, κ(2 − γs)δs+1+· · · + κ(2 − γr+1)δr+2+ δr+1 · · · · · t3p·, κ(2 − γs)δs+1+· · · + κ(2 − γp+1)δp+2− δp+1 · tγ3pp−1(·, · · · + δp+1).
NORMS OF NEWTON INTERPOLATING OPERATORS !
For every term of type tγk−1
3k (·, δ) we have δ > 0. Therefore,
|
tγ3kk−1(θ, δ)|
tγk−1
3k (0, δ)for any 0 θ π, by Lemma 4, and we can neglect these terms
while estimating ρN(θ)/ρN(0) from above. All other terms (with δ < 0) attain the maximal modulus value at θ = π, by Lemma 4. It follows that (13) ρN(θ) ρN(0) tγ3ss(π,−δs+1) tγ3ss(0,−δs+1) · k=p,...,q,r t3k(π, σk) t3k(0, σk) , where σk =κ(2 − γs)δs+1+· · · + κ(2 − γk+1)δk+2− δk+1.
Suppose γs= 0. Then the rst term on the right in (13) is absent. Let us estimate all σk from above. At rst, σr=−δs+1− δs− · · · − δr+1 =−12δr+12δs+1. From Corollary 1 we conclude that t3r(π, σr)/t3r(0, σr) (2π/δs+1)2 = 4· 32(s+1). Secondly,
σq =−δs+1− δs− · · · − δr+2+κ(2 − γr)δr+1− δr− · · · − δq+2− δq+1. But here γr = 0, so κ(2 − γr) 0. Therefore,
σq −1 2δq+ 1 2δr− 1 2δr+1+ 1 2δs+1>− 1 2δq+ δr+1. In the worst case, when r = s − 1, we get
σq >−1 2δq+ δs and t3q(π, σq)/t3q(0, σq) 3 2s. In general, σk=−δs+1− δs− · · · − δr+2+κ(2 − γr)δr+1− δr− · · · − δm+2 +κ(2 − γm)δm+1− δm− · · · − δk+2− δk+1,
where m = min {j > k : γj = 0}. Thus, σk>−12δk+ δm+1 and the product on the right in (13) can be estimated from above by
4· 32(s+1)32s· · · 32 = 4· 3(s+2)(s+1)= 36· 3s2+3s. For the case γs= 1 we apply similar arguments. Here,
−δs+1=− 1 2δs+ 1 2δs+1, so t3s(π,−δs+1)/t3s(0,−δs+1) 4 · 3 2(s+1).
! A. P. GONCHAROV Also, σr = δs+1− δs− · · · − δr+1 =−1 2δr+ 5 6δs>− 1 2δr+ δs+1.
For any other value of k the lower bound of σk is the same as for the case
γs = 0. Therefore,
ρN(θ)/ρN(0) 4·32(s+1)·32(s+1+s+···+1)= 4·3(s+1)(s+4)= 182·3s2+5s. We are now able to estimate the fundamental Lagrange polynomials cor-responding to the given sequence.
Lemma 9. For any N ∈ N, 1 k N + 1, and |x| 1, we havelk,N(x) 4πN2l
k,N(1).
Proof. If N = 1 then for both polynomials l1,1(x) =−√23x and l2,1(x)
= 2x+√√3
3 we have triviallylk,1(x) lk,1(1). If N = 3s− 1 for some s ∈ N, then l
k,N(x) = (x−xTk3s)T(x)
3s(xk) andlk,N(x)
4/π (see e.g. [5], p. 58). On the other hand,
lk,N(1) = (1− xk)T3s(xk) −1 3−2s,
by the Markov inequality (see e.g. [5], p. 123), which gives the result for this case.
Suppose 3s N 3s+1− 2 for some s ∈ N. Fix k N + 1 and 0 θ π. Here,
lk,N(cos θ) = μN+1(θ)
(cos θ− cos ψk)ωN+1(xk) and the desired inequality has the form
(14) 1− cos θ
| cos θ − cos ψk|·
μN+1(θ)
μN+1(0) 4πN
2.
The angle ψk is equal to
(
i−12)
δs+1 with some 1 < i 3s+1. The valuei = 1is excluded since in this case we have ψk= 12δs+1, that is k = 3s+1, but
k N + 1 3s+1− 1. Thus, 32δs+1 ψk π −12δs+1.
Let us rst consider the case |θ − ψk| > 12δs+1. Here, 32δs+1 θ + ψk 2π−32δs+1. Indeed, if ψk= π−12δs+1 then θ < π − δs+1. Otherwise, ψk π−32δs+1 with any θ π.
NORMS OF NEWTON INTERPOLATING OPERATORS ! !
Therefore,
| cos θ − cos ψk| = 2 sin 1 2|θ − ψk| · sin 1 2(θ + ψk) > 2 sin 1 4δs+1· sin 3 4δs+1 > 2 4 π2 1 4δs+1 3 4δs+1= 1 6 · 3 −2s 1 6· N −2. This inequality together with Lemma 6 imply (14).
For the case |θ − ψk| 12δs+1 we use the representation
μN+1(θ) = μN+1(ψk) + μN+1(ξ)· (θ − ψk) and the Bernstein inequality (see e.g. [5], p. 118):
μN+1(ξ) (N + 1)· max
0θπμN+1(θ).
Since μN+1(ψk) = 0, we get by Lemma 6,
μN+1(θ) (N + 1) · |θ − ψk| · μN+1(0). To prove (14) it is enough to show
(15) (N + 1)· |θ − ψk| 4πN2| cos θ − cos ψk|. Here, 3
2δs+1 θ + ψk 2π − 12δs+1. Therefore, sin12(θ + ψk) sin14δs+1
> 6N1 . On the other hand, sin12|θ − ψk| > π1|θ − ψk|. Hence, | cos θ − cos ψk| > |θ − ψk|(3πN)−1, which gives (15), since N + 1 43N for N 3.
Theorem 1. For the Lebesgue constants ΛN(X) corresponding to the Newton interpolation at the points X = (xn)∞n=1 we have
ΛN(X) C · N8· Nlog N/ log 3 for any N 1. Here C = 2 · 722.
Proof. Let Ns= 3s− 1. If N = Ns then (xk)N+1k=1 are exactly the zeros of T3s and (see e.g. [5], p. 19) ΛN(X) 2πlog N + 1. Therefore we can make
the assumption: 3s N 3s+1− 2 for some s ∈ N0. For such N we have lk,N(1) =lk,Ns+1(1) · ρN+1(ψk)
! " A. P. GONCHAROV
For the Lagrange fundamental polynomials at Chebyshev's points we have lk,Ns+1(1) 4/π (see for instance [5], p. 58).
Lemma 9 and Lemma 8now show that for any |x| 1 we have lk,N(x) 16· 182· N2· 3s2+5s 722· N7· 3s2.
This gives the desired result, since ΛN(X)is a sum of N + 1 terms with bound presented above and N + 1 2N.
Example. Let us estimate from below the value lj,N(1) for N =
1
2(3s+1− 1) and j = 12(3s+ 1). Here, ψj = max1kNψk = π−12δs. There-fore, xj =− cos12δs. Since N = 3s+ 3s−1+· · · + 3 + 1, the polynomial μN has the decomposition μN = ˜T3ss−1q=0t3q(·, δq+2+· · · + δs+1). Hence,
(16) lj,N(1) = lj,Ns(1)· s−1 q=0 t3q(0, σq) t3q(ψj, σq) with σq = 12(δq+1− δs+1).
Here the value of lj,Ns(1) is rather small. Indeed,
lj,Ns(1) = T3s(1) T3s(xj)(1− xj) = 3−s· tan1 4δs> π 4· 9s.
But because of the positivity of all σq, the product on the right side of (16) has a growth of order at least 3s2/3, as we will see now. Since all terms in the product are larger than 1, we can exclude the value q = 0. For 1 q s − 1, let K = 1
2(3q− 1). It is easily seen that t3q(0, σq)/t3q(ψj, σq) > A2q, where Aq= 3q k=1 tan 3k− 1 6 δq− 1 4δs+1 = cot 3K + 1 6 δq+ 1 4δs+1 ·K k=1 tan 3k− 1 6 δq− 1 4δs+1 tan 3k− 2 6 δq+ 1 4δs+1 .
The rst cotangent is larger than 1 since its argument does not exceed π/4. Also we can neglect the term in the product corresponding the value k = 1. For other terms we use the formula
tan v
NORMS OF NEWTON INTERPOLATING OPERATORS ! # From here, log Aq > K k=2 log(1 + ak) > K k=2 ak−1 2 K k=2 a2k, where ak = tan 1 6δq− 1 2δs+1 · cot 3k− 2 6 δq+ 1 4δs+1 + tan 3k− 1 6 δq− 1 4δs+1 . The expression in square brackets can be written as
cot (b− ε) + tan (b + ε) = 2 cos 2ε sin 2b− sin 2ε with b = 1
2
(
k−12)
δq, ε = 121 δq− 14δs+1. Here, b > 9ε, so [· · · ] < (2k−1)δ8 q. Therefore,Kk=2a2k< 12, as is easy to check.
In order to estimate akfrom below, we use the bound cot t > 1t−2t, which
is valid for 0 < t < π. It follows that ak> 1 6δq− 1 2δs+1 · cot 1 2kδq > 1 6δq− 1 2δs+1 · 2 kδq − kδq 4 . Straightforward estimation givesK
k=2ak>κ · log K − 0.193, where κ = 3−1 − 3q−s−1. Therefore, A q> 3q·κ· e−0.43 and s−1 q=1 t3q(0, σq)/t3q(ψj, σq) > e−0.86·s· 3κ1 with κ1 = 1 3s(s− 1) − 2 · 3 −s−1 s−1 q=1 q3q. Since the last sum is equal to 1
43s(2s− 3) +34, we have κ1 > 13s2−23s. Even-tually, lj,N(1) > π 4 e0.86· 38/3−s· 3s2/3> 3s2/3−4s. The latter exceeds exp
(
log2Ns! $ A. P. GONCHAROV: NORMS OF NEWTON INTERPOLATING OPERATORS
Question. One can suggest that more symmetric in ll of the levels Zs will decrease the size of the Lebesgue constants corresponding to a new se-quence. In this connection we can reformulate the question from [2]: Is it possible to reach a polynomial growth of the sequenceΛN(X)∞
N=0for some another rearrangement of points from the sequence (xk)∞k=1?
References
[1] L. Brutman, Lebesgue functions for polynomial interpolation a survey, Ann. Numer. Math., " (1997), 111127.
[2] A. Goncharov, What is the size of the Lebesgue constant for Newton interpolation?, in: Constructive Theory of Functions (Varna, 2005), edited by B. D. Bojanov, p. 144.
[3] E. Levin and B. Shekhtman, Two problems on interpolation, Constr. Approx., (1995), 513515.
[4] Paul G. Nevai, Orthogonal Polynomials, Memoirs of AMS, Vol.18, 123 (Providence, 1979).
[5] T. J. Rivlin, The Chebyshev Polynomials, second edition, Pure and Applied Mathe-matics (New York, 1990).
[6] E. B. Sa and V. Totik, Logarithmic Potentials with External Fields, Springer-Verlag (1997).
[7] G. Szegö, Orthogonal Polynomials, fourth edition, AMS Coll. Publ., Vol. 23 (Provi-dence, 1975).
[8] R. Taylor and V. Totik, Lebesgue constants for Leja points, IMA Journal of Num. Anal. (to appear).
