View of A Voronovskaya-type theorem

Rev. Anal. Num´er. Th´eor. Approx., vol. 30 (2001) no. 1, pp. 47–54 ictp.acad.ro/jnaat

A VORONOVSKAYA–TYPE THEOREM

MIRCEA IVAN and IOAN RAS¸A

Dedicated to the memory of Acad. Tiberiu Popoviciu

Abstract. We give an asymptotic estimation for some sequences of divided differences. We use this estimation to obtain a Voronovskaya–type formula in- volving linear positive operators.

MSC 2000. 41A36.

Keywords. Divided difference, linear operator, approximation, Voronovskaya- type theorem.

1. INTRODUCTION AND NOTATIONS

Consider the pointsx0 < x1 < . . . < xnon the real axis and letf: [x0, xn]→ Rbe an arbitrary function. Denote by [x₀, . . . , x_n;f] the divided difference of the functionf on the knotsx0, . . . , xn,usually defined by

[x₀, . . . , x_n;f] :=

n

P

i=0

f(xi)

(xi−x0)...(xi−xi−1)(xi−xi+1)...(xi−xn).

Consider the polynomial functions ei:R → R, ei(x) = xⁱ, i= 0,1, . . . . It is known that [x₀, . . . , x_n;e_i] = 0, i = 0, . . . , n−1, [x₀, . . . , x_n;e_n] = 1. The problem was to calculate Ak:= [x0, . . . , xn;en+k], k= 1,2, . . . .In [6] Tiberiu Popoviciu uses the identity

h

x₀, . . . , x_n;_x−·¹ i

= _(x−x ¹

0)...(x−xn), to prove the following formula

Technical University of Cluj–Napoca, Dept. of Mathematics, Str. C. Daicovi- ciu 15, RO-3400, Cluj–Napoca, ROMANIA, e-mail: [email protected], [email protected].

A_k= P

0≤i0,...,in≤k i0+···+in=k

xⁱ₀⁰· · ·xⁱ_nⁿ.

This formula was rediscovered in 1981 by E. Neuman [3]. It does not look much “friendlier” than the initial one,

A_k = Pn

i=0

x^n+k_i

(xi−x₀)...(xi−xi−1)(xi−x_i+1)...(xi−x_n).

Therefore, in [7], it is suggested that a recurrence formula might be more useful. We shall use such a formula in order to give an asymptotic estimation forAk under some supplementary assumptions on the knots (see Theorem 1).

Consider now a triangular matrix of nodes (x_n,k), n= 0,1, . . .;k= 0, . . . , n, (1) −1≤xn,0 < xn,1< . . . < xn,n≤1, n= 0,1, . . .

satisfying the conditions:

(2) xn,n−i =−x_n,i, i= 0, . . . , n, n= 0,1, . . . .

Leta >0. Forn≥1 consider the operatorLn:C[−a−1, a+ 1]→C[−a, a], Lnf(x) :=n![x+xn,0, . . . , x+xn,n;Fn],

wheref ∈C[−a−1, a+ 1], x∈[−a, a], F_n∈Cⁿ[−a−1, a+ 1], Fn⁽ⁿ⁾=f.

The L_n are positive linear operators of probabilistic type and Bernstein–

Schnabl type operators.

For particular choices of the matrix (x_n,k) various inequalities involvingL_nf have been studied in [4], [5], [8], [12]. Ifx_n,i=−1 +²ⁱ_n,i= 0, . . . , n, we have also [10]

(3) Lnf(x) = 2⁻ⁿ Z x+1

x−1

· · · Z x+1

x−1

f ^t1+···+t_n ⁿ

dt1. . . dtn.

Using theL_n operator notation, [7] gives (4)

Lnf(0)−

k−1

P

i=0

Lne2i(0)

(2i)! f⁽²ⁱ⁾(0)

≤ Lne2k(0)

(2k)! kf^(2k)k_[−1,1].

for all f ∈C^2k[−a−1, a+ 1], where k · k_[−1,1] denotes the uniform norm on C[−1,1]. As positive operators,L_n have been studied in [9], [10].

They verify:

f convex =⇒L_nf ≥L_n+1f ≥f, kL_nf−fk ≤2ω

f,^√1

3n

.

We have: Le₀ =e₀, Le₁ =e₁, L(e₁−x e₀)²(x) = _(n+1)(n+2)¹ Pn i=0x²_n,i. For equidistant knotsx_n,i=−1 +²ⁱ_n,i= 0, . . . , n, we obtain

L(e₁−x e₀)²(x) = _3n¹ , hence, using [1, Corollary 4.12], we can prove now that

kL_nf−fk ≤2.25ω₂ f,^√1

3n

.

Our aim is to give a more refined analysis of the convergence behaviour of the operators Ln.This is accomplished in Theorem 2.

2. MAIN RESULTS

Theorem 1. If the triangular matrix (x_n,k) satisfies the relations (1), (2) and

(5) lim

n→∞

1 n

n

P

i=0

x²_n,i= 2λ∈R then, for all k∈N,the following equality is fulfilled

(6) lim

n→∞n^−k[x_n,0, . . . , x_n,n;e_n+2k] = ^λ_k!^k. The previous relation can be written in the form

(7) lim

n→∞ n^kL_ne_2k(0) = ^λ_k!^k(2k)!.

Theorem 2. If f ∈C^2k[−a−1, a+ 1], then, for every matrix (x_n,k) satis- fying the conditions(1),(2)and(5), the following Voronovskaya–type relation holds true:

(8) lim

n→∞ n^k

Lnf(x)−

k−1

P

i=0

Lne2i(0)

(2i)! f⁽²ⁱ⁾(x)

= ^λ_k!^kf^(2k)(x), uniformly for x∈[−a, a].

3. PROOF OF THEOREM 1.

Consider the polynomial function (e₁−x_n,0). . .(e₁−x_n,n),which we write as en+1−C_n,1en− · · · −C_n,ne1−C_n,n+1e0.Consider also the sumsSn,p:=

n

P

i=0

x^p_n,i, p= 1,2, . . . .Using (2) it is obvious that

(9) S_n,p= 0, for oddp.

Using (1) it can be easily shown that

(10) lim

n→∞

Sn,p

n^k = 0, p= 1,2, . . .; k >1.

We write the relation (5) in the form

(11) lim

n→∞

Sn,2

n = 2λ.

The coefficients

C_n,p= (−1)^p+1 P

0≤i₁<...<ip≤n

x_n,i1· · ·x_n,ip, can be computed by using Newton’s formulas:

C_n,1 = S_n,1 C_n,p = ¹_p

S_n,p−

p−1

P

i=1

S_n,iCn,p−i

, p= 2, . . . , n+ 1.

By considering (9) it follows that:

C_n,p= 0, for odd p, and

(12)

Cn,2 = ¹₂Sn,2

C_n,2k = _2k¹ S_n,2k−

k−1

X

i=1

S_n,2iC_n,2(k−i)

!

, k= 1, . . . ,b(n+ 1)/2c.

We define γ_k as

γ_k:= lim

n→∞

C_n,2k

n^k , k= 1,2, . . . . We have

γ1= lim

n→∞

Cn,2

n = lim

n→∞

Sn,2

2n =λ, and, from (10) and (12), it follows that

γ_k=−^λ_kγk−1, k≥2 hence

(13) γ_k= (−1)^k+1λ^k

k! , k≥1.

Using the divided difference functional, define the numbers:

A_n,j := [x_n,0, . . . , x_n,n;e_n+j], j=−n,−n+ 1, . . . . It is well known that

(14) A_n,j =

0, ifj=−n, . . . ,−1, 1, ifj= 0.

In order to calculate An,j forj≥1,observe that

[xn,0, . . . , xn,n;ej−1(e1−xn,0). . .(e1−xn,n)] = 0, that is

[x_n,0, . . . , x_n,n;e_n+j−C_n,1en+j−1− · · · −C_n,n+1ej−1] = 0.

As a consequence we have A_n,j =

n+1P

i=1

C_n,iAn,j−i, j= 1,2, . . . and using (14),we find that

(15) A_n,j =

j

P

i=1

C_n,iAn,j−i, j= 1, . . . , n+ 1.

Using

An,0 = 1

An,1 = Cn,1An,0 = 0 in (15), it can be deduced that

(16) A_n,p= 0, for odd p,

and hence

(17) A_n,2k=

k

P

i=1

C_n,2iA_n,2(k−i), 1≤k≤ ⁿ⁺¹₂ . By defining

Bk:= lim

n→∞

A_n,2k

n^k , k≥0, we have B0 = 1,and using (17) we find

B_k=

k

P

i=1

γ_iBk−i, k≥1 i.e.,

(18) B_k=

k

P

i=1

(−1)ⁱ⁺¹λⁱ

i! Bk−i, k≥1.

Using (18) we can prove by mathematical induction that

(19) B_k= λ^k

k!, k≥0 which completes the proof.

4. PROOF OF THEOREM 2

For arbitraryx∈[−a, a] consider the functiong_x : [−a−1, a+ 1]→R, gx:=f −

2k

P

i=0

(e1−xe₀)ⁱ

i! f⁽ⁱ⁾(x).

Taylor’s formula implies the existence of a point ξ∈(−a−1, a+ 1),|x−ξ| ≤

|x−t|,such that

g_x(t) =(t−x)^2k

(2k)! (f^(2k)(ξ)−f^(2k)(x)).

For anyε >0 there exists a number δ >0 such that

|g_x(t)| ≤(t−x)^2kε for all t∈[−a−1, a+ 1],|t−x|< δ.

Let C be a constant such that |g_x(t)| ≤ C δ^2k+2, for all x ∈ [−a, a], t ∈ [−a−1, a+ 1]. Consequently, we obtain

|g_x(t)| ≤ε(t−x)^2k+C(t−x)^2k+2 for all x∈[−a, a], t∈[−a−1, a+ 1], that is,

|g_x| ≤ε(e₁−xe₀)^2k+C(e₁−xe₀)^2k+2, and so,

|L_ngx(x)| ≤ε Ln(e1−xe0)^2k(x) +C Ln(e1−xe0)^2k+2(x).

Using the equality

Ln(f)(x) =Ln(f ◦(e1+xe0))(0), we obtain

|L_ngx(x)| ≤ε Lne2k(0) +C Lne2k+2(0).

Taking into account the fact that (20) Lnei(0) = n!i!

(n+i)!An,i, i= 1,2, . . . it follows

(21) lim

n→∞nⁱL_ne_2i(0) = λⁱ

i! (2i)! i= 1,2, . . . . Consequently, we obtain

n→∞lim n^kL_ng_x(x) = 0, uniformly for x∈[−a, a],that is

n→∞lim n^k

Lnf(x)−

2k

P

i=0 Lnei(0)

i! f⁽ⁱ⁾(x)

= 0.

Finally, using (16) the relation (8) is proved.

5. REMARKS

(a) Suppose that (1), (2) and (5) are satisfied and letf ∈C[−a−1, a+ 1]

be 2k–times differentiable atx∈[−a, a].By using the Lemma and [11, Corollary 2] we obtain

(22) lim

n→∞n^k

L_nf(x)−

k−1

P

i=0

Lne2i(0)

(2i)! f⁽²ⁱ⁾(x)

= λ^k

k!f^(2k)(x).

(b) If xn,i=−1 + 2i/n,i= 0, . . . , n, thenλ= 1/6; in this special case the formula (22) can be found in [2]. In particular, for k = 1 andk = 3, we have

(23) lim

n→∞n(Lnf(x)−f(x)) = ¹₆f⁰⁰(x), respectively

(24) lim

n→∞n

n

n Lnf(x)−f(x)

− ^f00₆^(x)

−^fIV₇₂^(x)

= ^{fV I}₁₂₉₆^(x) −^fIV₁₈₀^(x). (c) In the case of Chebyshev’s knots

x_n,k= cos^{2 2k+1}_2(n+1)π, k= 0, . . . , n, we obtain

1 n+1

n

P

k=0

cos^{2 2k+1}_2(n+1)π= ¹₂, ∀n≥1, henceλ= 1.

Acknowledgement. The authors gratefully acknowledge Heinz H. Gonska for his critical remarks on an earlier version.

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Received April 11, 2000.