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

Ioan Raşa

Mircea Ivan

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REVUE D’ANALYSE NUMÉRIQUE ET DE THÉORIE DE L’APPROXIMATIONRev. Anal. Numér. Théor. Approx., vol. 30 (2001) no. 1, pp. 47–54ictp.acad.ro/jnaatA VORONOVSKAYA–TYPE THEOREMMIRCEA IVAN and IOAN RAŞADedicated to the memory of Acad. Tiberiu PopoviciuAbstract. We give an asymptotic estimation for some sequences of divideddifferences. 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 NOTATIONSConsider the points x0 < x1 < . . . < xn on the real axis and let f : [x0, xn]→R be an arbitrary function. Denote by [x0, . . . , xn; f ] the divided difference ofthe function f on the knots x0, . . . , xn, usually defined by[x0, . . . , xn; f ] :=n∑i=0f(xi)(xi−x0)...(xi−xi−1)(xi−xi+1)...(xi−xn) .Consider the polynomial functions ei : R → R, ei(x) = xi, i = 0, 1, . . . . Itis known that [x0, . . . , xn; ei] = 0, i = 0, . . . , n − 1, [x0, . . . , xn; en] = 1. Theproblem was to calculate Ak := [x0, . . . , xn; en+k], k = 1, 2, . . . . In [6] TiberiuPopoviciu uses the identity[x0, . . . , xn;1x−·]= 1(x−x0)...(x−xn) ,to prove the following formulaTechnical University of Cluj–Napoca, Dept. of Mathematics, Str. C. Daicovi-ciu 15, RO-3400, Cluj–Napoca, ROMANIA, e-mail: mircea.ivan@math.utcluj.ro,ioan.rasa@math.utcluj.ro.48 Mircea Ivan, Ioan Raşa 2Ak =∑0≤i0,...,in≤ki0+···+in=kxi00 · · ·xinn .This formula was rediscovered in 1981 by E. Neuman [3]. It does not lookmuch “friendlier” than the initial one,Ak =n∑i=0xn+ki(xi−x0)...(xi−xi−1)(xi−xi+1)...(xi−xn) .Therefore, in [7], it is suggested that a recurrence formula might be moreuseful. We shall use such a formula in order to give an asymptotic estimationfor Ak under some supplementary assumptions on the knots (see Theorem 1).Consider now a triangular matrix of nodes (xn,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 = −xn,i, i = 0, . . . , n, n = 0, 1, . . . .Let a > 0. For n ≥ 1 consider the operator Ln : C[−a−1, a+1]→ C[−a, a],Lnf(x) := n![x+ xn,0, . . . , x+ xn,n;Fn],where f ∈ C[−a− 1, a+ 1], x ∈ [−a, a], Fn ∈ Cn[−a− 1, a+ 1], F (n)n = f.The Ln are positive linear operators of probabilistic type and Bernstein–Schnabl type operators.For particular choices of the matrix (xn,k) various inequalities involving Lnfhave been studied in [4], [5], [8], [12]. If xn,i = −1 + 2in , i = 0, . . . , n, we havealso [10](3) Lnf(x) = 2−n∫ x+1x−1· · ·∫ x+1x−1f(t1+···+tnn)dt1 . . . dtn.Using the Ln operator notation, [7] gives(4)∣∣∣∣Lnf(0)− k−1∑i=0Lne2i(0)(2i)! f(2i)(0)∣∣∣∣ ≤ Lne2k(0)(2k)! ‖f (2k)‖[−1,1].for all f ∈ C2k[−a − 1, a + 1], where ‖ · ‖[−1,1] denotes the uniform norm onC[−1, 1]. As positive operators, Ln have been studied in [9], [10].3 A Voronovskaya–Type Theorem 49They verify:f convex =⇒ Lnf ≥ Ln+1f ≥ f,‖Lnf − f‖ ≤ 2ω(f, 1√3n).We have: Le0 = e0, Le1 = e1, L(e1 − x e0)2(x) = 1(n+1)(n+2)∑ni=0 x2n,i.For equidistant knots xn,i = −1 + 2in , i = 0, . . . , n, we obtainL(e1 − x e0)2(x) = 13n ,hence, using [1, Corollary 4.12], we can prove now that‖Lnf − f‖ ≤ 2.25ω2(f, 1√3n).Our aim is to give a more refined analysis of the convergence behaviour ofthe operators Ln. This is accomplished in Theorem 2.2. MAIN RESULTSTheorem 1. If the triangular matrix (xn,k) satisfies the relations (1), (2)and(5) limn→∞1nn∑i=0x2n,i = 2λ ∈ Rthen, for all k ∈ N, the following equality is fulfilled(6) limn→∞n−k[xn,0, . . . , xn,n; en+2k] =λkk! .The previous relation can be written in the form(7) limn→∞nkLne2k(0) =λkk! (2k)!.Theorem 2. If f ∈ C2k[−a− 1, a+ 1], then, for every matrix (xn,k) satis-fying the conditions (1), (2) and (5), the following Voronovskaya–type relationholds true:(8) limn→∞nk(Lnf(x)−k−1∑i=0Lne2i(0)(2i)! f(2i)(x))= λkk! f(2k)(x),uniformly for x ∈ [−a, a].50 Mircea Ivan, Ioan Raşa 43. PROOF OF THEOREM 1.Consider the polynomial function (e1−xn,0) . . . (e1−xn,n), which we write asen+1−Cn,1en−· · ·−Cn,ne1−Cn,n+1e0. Consider also the sums Sn,p :=n∑i=0xpn,i,p = 1, 2, . . . . Using (2) it is obvious that(9) Sn,p = 0, for odd p.Using (1) it can be easily shown that(10) limn→∞Sn,pnk= 0, p = 1, 2, . . . ; k > 1.We write the relation (5) in the form(11) limn→∞Sn,2n= 2λ.The coefficientsCn,p = (−1)p+1∑0≤i1<...<ip≤nxn,i1 · · ·xn,ip ,can be computed by using Newton’s formulas:Cn,1 = Sn,1Cn,p =1p(Sn,p −p−1∑i=1Sn,iCn,p−i), p = 2, . . . , n+ 1.By considering (9) it follows that:Cn,p = 0, for odd p,and(12)Cn,2 =12Sn,2Cn,2k =12k(Sn,2k −k−1∑i=1Sn,2iCn,2(k−i)), k = 1, . . . , b(n+ 1)/2c.We define γk asγk := limn→∞Cn,2knk, k = 1, 2, . . . .We haveγ1 = limn→∞Cn,2n= limn→∞Sn,22n= λ,and, from (10) and (12), it follows that5 A Voronovskaya–Type Theorem 51γk = −λkγk−1, k ≥ 2hence(13) γk =(−1)k+1λkk!, k ≥ 1.Using the divided difference functional, define the numbers:An,j := [xn,0, . . . , xn,n; en+j ], j = −n,−n+ 1, . . . .It is well known that(14) An,j ={0, if j = −n, . . . ,−1,1, if j = 0.In order to calculate An,j for j ≥ 1, observe that[xn,0, . . . , xn,n; ej−1(e1 − xn,0) . . . (e1 − xn,n)] = 0,that is[xn,0, . . . , xn,n; en+j − Cn,1en+j−1 − · · · − Cn,n+1ej−1] = 0.As a consequence we haveAn,j =n+1∑i=1Cn,iAn,j−i, j = 1, 2, . . .and using (14), we find that(15) An,j =j∑i=1Cn,iAn,j−i, j = 1, . . . , n+ 1.UsingAn,0 = 1An,1 = Cn,1An,0 = 0in (15), it can be deduced that(16) An,p = 0, for odd p,and hence(17) An,2k =k∑i=1Cn,2iAn,2(k−i), 1 ≤ k ≤ n+12 .By defining52 Mircea Ivan, Ioan Raşa 6Bk := limn→∞An,2knk, k ≥ 0,we have B0 = 1, and using (17) we findBk =k∑i=1γiBk−i, k ≥ 1i.e.,(18) Bk =k∑i=1(−1)i+1λii! Bk−i, k ≥ 1.Using (18) we can prove by mathematical induction that(19) Bk =λkk!, k ≥ 0which completes the proof.4. PROOF OF THEOREM 2For arbitrary x ∈ [−a, a] consider the function gx : [−a− 1, a+ 1]→ R,gx := f −2k∑i=0(e1−xe0)ii! f(i)(x).Taylor’s formula implies the existence of a point ξ ∈ (−a− 1, a+ 1), |x− ξ| ≤|x− t|, such thatgx(t) =(t− x)2k(2k)!(f (2k)(ξ)− f (2k)(x)).For any ε > 0 there exists a number δ > 0 such that|gx(t)| ≤ (t− x)2k εfor all t ∈ [−a− 1, a+ 1], |t− x| < δ.Let C be a constant such that |gx(t)| ≤ C δ2k+2, for all x ∈ [−a, a], t ∈[−a− 1, a+ 1]. Consequently, we obtain|gx(t)| ≤ ε (t− x)2k + C (t− x)2k+2for all x ∈ [−a, a], t ∈ [−a− 1, a+ 1], that is,7 A Voronovskaya–Type Theorem 53|gx| ≤ ε (e1 − xe0)2k + C (e1 − xe0)2k+2,and so,|Lngx(x)| ≤ εLn(e1 − xe0)2k(x) + C Ln(e1 − xe0)2k+2(x).Using the equalityLn(f)(x) = Ln(f ◦ (e1 + xe0))(0),we obtain|Lngx(x)| ≤ εLne2k(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) limn→∞niLne2i(0) =λii!(2i)! i = 1, 2, . . . .Consequently, we obtainlimn→∞nkLngx(x) = 0,uniformly for x ∈ [−a, a], that islimn→∞nk(Lnf(x)−2k∑i=0Lnei(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 let f ∈ C[−a− 1, a+ 1]be 2k–times differentiable at x ∈ [−a, a]. By using the Lemma and [11,Corollary 2] we obtain(22) limn→∞nk(Lnf(x)−k−1∑i=0Lne2i(0)(2i)! f(2i)(x))=λkk!f (2k)(x).(b) If xn,i = −1 + 2i/n, i = 0, . . . , n, then λ = 1/6; in this special case theformula (22) can be found in [2]. In particular, for k = 1 and k = 3,we have(23) limn→∞n(Lnf(x)− f(x)) = 16f′′(x),respectively54 Mircea Ivan, Ioan Raşa 8(24) limn→∞n(n(n(Lnf(x)− f(x))− f′′(x)6)− fIV (x)72)= fV I(x)1296 −fIV (x)180 .(c) In the case of Chebyshev’s knotsxn,k = cos2 2k+12(n+1) π, k = 0, . . . , n,we obtain1n+1n∑k=0cos2 2k+12(n+1) π =12 , ∀n ≥ 1,hence λ = 1.Acknowledgement. The authors gratefully acknowledge Heinz H. Gonskafor his critical remarks on an earlier version.REFERENCES[1] H. H. Gonska and R. K. Kovacheva, The second order modulus revisited: remarks,applications, problems, Conf. Sem. Mat. Univ. Bari, 257, 1994.[2] M. Ivan and I. Raşa, A sequence of positive linear operators, Rev. Anal. Numér.Théor. Approx., 24 (1995), 159–164.[3] E. Neuman, Problem E 2900, Amer. Math. Month., 88 (1981), 538.[4] E. Neuman and J. Pečarić, Inequalities involving multivariate convex functions, J.Math. Anal. Appl., 137 (1989), 541–549.[5] J. Pečarić, An inequality for 3–convex functions, J. Math. Anal. Appl., 90 (1982),213–218.[6] T. Popoviciu, Introduction à la théorie des différences divisées, Bull. Math. de la Soc.Roumaine des Sci., 42 (1940), 65–78.[7] T. Popoviciu, Remarques sur le reste de certaines formules d’approximation d’unedifférence divisée par les dérivées, Buletinul Institutului Politehnic din Iaşi, Serie nouă,13 (17) (1967), 103–109.[8] J. Pečarić and I. Raşa, Inequalities for divided differences of n–convex functions,Studia Univ. Babeş–Bolyai, Math., 33 (1990), 7–10.[9] J. Pečarić and I. Raşa, A linear operator preserving k–convex functions, Bul. Şt.IPCN, 33 (1990), 23–26.[10] I. Raşa, Korovkin approximation and parabolic functions, Conf. Sem. Mat. Univ. Bari,236 (1991).[11] P. C. Sikkema, On some linear positive operators, Indag. Math., 32 (1970), 327–337.[12] D. Zwick, A divided difference inequality for n–convex functions, J. Math. Anal. Appl.,104 (1984), 435–436.Received April 11, 2000.

A Voronovskaya-type theorem

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A Voronovskaya-type theorem

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