Startseite An improvement of the constant in Videnskiĭ’s inequality for Bernstein polynomials
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An improvement of the constant in Videnskiĭ’s inequality for Bernstein polynomials

  • Ulrich Abel ORCID logo EMAIL logo und Hartmut Siebert
Veröffentlicht/Copyright: 7. Februar 2018
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Abstract

In this paper, we deal with improvements on the constant Mn(γ) in the so-called Videnskiĭ inequality

|(Bnf)(x)-f(x)-x(1-x)2nf′′(x)|Mn(γ)x(1-x)nω(f′′;γn)

for a fixed constant γ1 and x[0,1], where Bnf is the Bernstein polynomial, fC2[0,1] and ω is the first order modulus of continuity. Let M(γ)=supnMn(γ). We prove the Videnskiĭ inequality for arbitrary γ1. In particular, we improve the constant M(2)=0.9 (Gonska and Ra şa [8], 2008) to M(2)=0.6875. Finally, we consider Mn(1) for small values of n.

MSC 2010: 41A36; 41A25

1 Introduction

The classical Bernstein polynomials, defined for fC[0,1] by

(Bnf)(x)=ν=0npn,ν(x)f(νn)(x[0,1])

with

pn,ν(x)=(nν)xν(1-x)n-ν(ν=0,,n)

converge uniformly to f on [0,1]. Popoviciu [11] gave the rate of convergence

Bnf-fc0ω(f;n-1/2),

estimated by the ordinary modulus of continuity with the constant c0=32. An elementary proof with constant c0=54 can be found in Lorentz’s book [10]. Sikkema [12] found the best (smallest) value of the constant c0. The asymptotic behaviour was described by Voronovskaja [17] in 1932 (see, e.g., [10, p. 22]). She showed that, for each bounded function f on [0,1] whose second derivative f′′(x) exists at a certain point x[0,1], one has

limnn[(Bnf)(x)-f(x)]=12x(1-x)f′′(x),

the convergence being uniform, for fC2[0,1] (see, e.g., [3, Chapter 10, Theorem 3.1] and cf. [5]). In that same year Bernstein [2] published a generalization. Gonska [6, Theorem 5.1] gave the following quantitative form:

|n[(Bnf)(x)-f(x)]-12x(1-x)f′′(x)|x(1-x)2ω~(f′′;1n2+x(1-x)n),

where ω~ is the least concave majorant of ω satisfying

ω(f;δ)ω~(f;δ)2ω(f;δ)for δ>0.

In 1985, Videnskiĭ [16] proved that, for all x[0,1],

(1.1)|(Bnf)(x)-f(x)-x(1-x)2nf′′(x)|Mx(1-x)nω(f′′;2n)

with the constant M=1 being independent of x. By application of a general theorem due to Sikkema and van der Meer [13], Gonska and Raşa [8, p. 94] recently proved that the inequality is valid with constant M=0.9.

The purpose of the present note is to show by direct methods that the constant M can be diminished to the value

(1.2)M=1116=0.6875.

In 2005, Floater [5] generalized Voronovskaja’s theorem [17] to differentiated Bernstein polynomials. Quantitative results of Floater’s theorem have been given by Gonska and Raşa [9] and Gonska, Heilmann and Raşa [7] in 2009 and 2010, respectively. Further inequalities for the left-hand side of (1.1) with the aid of Ditzian–Totik moduli of first order and average moduli of smoothness were given by Tachev [15] and Finta [4].

2 Main result

We define the sequence of linear operators

(2.1)(Lnf)(x):=n(Bnf)(x)-f(x)x(1-x)-12f′′(x)(x(0,1)),

which allows to write (1.1) as

|(Lnf)(x)|Mω(f′′;2n).

Our main result is the following theorem which states a slightly more general inequality.

Theorem 2.1.

Let γ1 and fC2[0,1]. For all n1 and all x[0,1],

|(Lnf)(x)|M(γ)ω(f′′;γn),

where

M(γ)=12+38γ.

In the particular case n=1, the constant M(γ) can be taken equal to 12.

Remark 2.2.

Notice that for γ=2 the theorem provides inequality (1.1) with the constant M announced in (1.2).

Remark 2.3.

In the convenient case γ=1, we have the constant M(1)=78=0.875.

Proof of Theorem 2.1.

Let γ1 and fC2[0,1]. By taking into account the endpoint interpolation property of the Bernstein polynomials, the assertion is obvious for x{0,1}. Now let x(0,1). By Taylor’s formula we have, for 0νn,

f(νn)=f(x)+f(x)(νn-x)+12f′′(x)(νn-x)2+12(f′′(ξν/n,x)-f′′(x))(νn-x)2

with certain intermediate points ξν/n,x between νn and x. Since Bernstein polynomials preserve linear functions and Bne2=e2+e1n(1-e1), where ek(x)=xk, we conclude that

(Bnf)(x)=f(x)+12f′′(x)x(1-x)n+12ν=0npn,ν(x)(νn-x)2(f′′(ξν/n,x)-f′′(x)).

This implies that

(2.2)x(1-x)n|(Lnf)(x)|12ν=0npn,ν(x)(νn-x)2ω(f′′;|ξν/n,x-x|).

The case n=1 will be considered separately. In this case the right-hand side of the latter inequality becomes

12x2(1-x)ω(f′′;|ξ0,x-x|)+12x(1-x)2ω(f′′;|ξ1,x-x|)

and we obtain

x(1-x)|(L1f)(x)|12x(1-x)ω(f′′;1).

Taking advantage of the well-known obvious inequality

(2.3)ω(g;λh)λω(g;h),

which is valid for any function g and λ,h>0, we obtain, for δ>0,

x(1-x)|(L1f)(x)|12x(1-x)δ-1ω(f′′;δ).

In particular, for δ=γ1, it follows that

x(1-x)|(L1f)(x)|12x(1-x)ω(f′′;γ),

which proves the assertion for n=1.

We continue with the case n2. A consequence of (2.3) is

ω(f′′;|ξν/n,x-x|)ω(f′′;|νn-x|)δ-1|νn-x|ω(f′′;δ)(1+δ-1|νn-x|)ω(f′′;δ).

Hence, by inequality (2.2),

2x(1-x)n|(Lnf)(x)|0νn,|νn-x|<δpn,ν(x)(νn-x)2ω(f′′;δ)+0νn,|νn-x|δpn,ν(x)(νn-x)2(1+δ-1|νn-x|)ω(f′′;δ).

If |νn-x|δ, then δ-1|νn-x|δ-2|νn-x|2, so we conclude that

2x(1-x)n|(Lnf)(x)|ν=0npn,ν(x)(νn-x)2ω(f′′;δ)+δ-20νn,|νn-x|δpn,ν(x)(νn-x)4ω(f′′;δ)
[(Bnψx2)(x)+δ-2(Bnψx4)(x)]ω(f′′;δ),

where the function ψx is defined by ψx(t)=t-x. Using (Bnψx2)(x)=xn(1-x) and

(Bnψx4)(x)=x(1-x)n3(1-6x(1-x)+3nx(1-x))

(see, e.g., [10, p. 14]), we have

|(Lnf)(x)|(12+1-6x(1-x)+3nx(1-x)2δ2n2)ω(f′′;δ).

For n2, the nominator 1+(3n-6)x(1-x) can be estimated by 14(3n-2)<34n, since 3n-2 is nonnegative and 0x(1-x)14 on [0,1]. Hence, we obtain, for each n2,

|(Lnf)(x)|(12+38δ2n)ω(f′′;δ).

Inserting the special value δ=γn proves Theorem 2.1. ∎

3 Best constants in the case δ=1n for some small n

Let us now consider the special choice δ=1n in the inequality

supx(0,1)|(Lnf)(x)|ω(f′′;δ)Mn(n).

We are interested in the best (smallest) value Mn[opt] of the constant Mn, i.e.,

Mn[opt]:=supfXnsupx(0,1)|(Lnf)(x)|,

where Xn consists of all functions fC2[0,1] with ω(f′′;1n)=1. By the interpolation property of Bn we have to consider only the case x(0,1).

3.1 The case n=1

We prove that M1[opt]=12. For n=1, by Theorem 2.1 we know that M1[opt]12. Put bk=1-1k (k) and define a sequence of functions fk(x)=k6(x-bk)+3, where u+=u if u0 and u+=0 if u<0. Obviously, fkC2[0,1] and fk′′(x)=k(x-bk)+ and we see that ω(fk′′;1)=fk′′(1)-fk′′(bk)=1. We show that

limksupx(0,1)|(L1fk)(x)|=12.

Since

x(1-x)(L1f)(x)=(1-x)f(0)+xf(1)-f(x)-12x(1-x)f′′(x)

and fk(x)=k6(x-bk)+3, we have

(L1fk)(x)=fk(1)1-x=16(1-x)k216k(0<x<bk).

Now let x(bk,1]. We obtain

(L1fk)(x)=f(1)-f(x)x(1-x)-fk(1)x-12fk′′(x).

By the mean value theorem we conclude that

|f(1)-f(x)x(1-x)|=x-1|fk(ξ)|bk-1k2(1-bk)2=12(k-1).

Furthermore,

0fk(1)x16kbk=16(k-1)

and

012fk′′(x)=12k(x-bk)12.

Hence, limk(L1fk)(1)=-12 and the claim follows by continuity.

3.2 The cases n2

For the considerations we use the following auxiliary results.

Lemma 3.1 (Stancu [14, equation (11.5)]).

For each function f:[0,1]R there holds

(3.1)(Bnf)(x)-f(x)=x(1-x)nν=0n-1pn-1,ν(x)[x,νn,ν+1n;f]

provided that x[0,1]{νnν=0,,n}.

A new interesting proof of Stancu’s formula is given in [5, p. 132]. In case when x coincides with νn or 1n(ν+1), an extension of Stancu’s formula to the whole interval [0,1] can be obtained using a limit process if f possesses a derivative on [0,1]. For instance, the divided difference on multiple knots [νn,νn,ν+1n;f] is equal to n2(f(ν+1n)-f(νn))-nf(νn).

Lemma 3.2 (see, e.g., [10, Chapter 4, Section 7, equation (7.16), p. 123]).

Let mN and fCm[a,b]. If all points x0,,xm[a,b] are distinct, there holds

[x0,,xm;f]=1(m-1)!abf(m)(t)[x0,,xm;(-t)+m-1]dt.

Because of ν=0n-1pn-1,ν(x)=1, Stancu’s formula (3.1) implies

(Bnf)(x)-f(x)-x(1-x)2nf′′(x)=x(1-x)nν=0n-1pn-1,ν(x)([x,νn,ν+1n;f]-12f′′(x)).

By Lemma 3.2,

[x,νn,ν+1n;f]=01f′′(t)[x,νn,ν+1n;(-t)+]dt(x(0,1)).

In the particular case f(x)=x2, we see that

(3.2)1=[x,νn,ν+1n;f]=201[x,νn,ν+1n;(-t)+]dt

and therefore, by (2.1),

(3.3)(Lnf)(x)=ν=0n-1pn-1,ν(x)01(f′′(t)-f′′(x))[x,νn,ν+1n;(-t)+]dt.

Note that [x0,x1,x2;(-t)+]0 for all pairwise different points x0,x1,x2 with 0x0<x1<x21, and that [x0,x1,x2;(-t)+]=0 if t<x0 or t>x2.

Taking advantage of inequality (2.3), we obtain, for δ>0,

|f′′(t)-f′′(x)|ω(f′′;|t-x|)δ-1|t-x|ω(f′′;δ).

By (3.2) and (3.3), we conclude that, for δ>0,

|(Lnf)(x)|cn(δ,x)ω(f′′;δ),

where

(3.4)cn(δ,x)=ν=0n-1pn-1,ν(x)01δ-1|t-x|[x,νn,ν+1n;(-t)+]dt.

We define

Cn:=supx(0,1)cn(1n,x).

A list giving the exact values of Cn for small n can be found in [1, p. 31]:

C1=12=0.500000,
C2=18(97+1402-102+1)0.504142,
C3=1162(45340+66783-1963-99)0.503828,
C4=132(783-119)0.503124.

We conjecture that Mn[opt]=Cn (n). Since the proofs are rather long and tedious, we omit the calculations for n=3 and n=4. In order to illustrate the method, we restrict ourselves in proving that

C2=18(97+1402-102+1)

by estimating c2(12,x).

3.3 The case n=2

Put δ=δ2=120.707. By (3.4), we have c2(δ2,x)=(1-x)I1+xI2 with

I1=01δ-1|t-x|[x,0,12;(-t)+]dt,
I2=01δ-1|t-x|[x,12,1;(-t)+]dt.

If 1-δ2xδ2, we have |t-x|δ2 for all t[0,1], which implies I1=I2=12 such that c2(δ2,x)=12. Now we consider the case 0<x<1-δ20.293. Since [x,0,12;(-t)+]=0 for t[0,12],

I1=012[x,0,12;(-t)+]dt=012[x,0,12;(-t)+]dt=12.

Since [x,12,1;(-t)+]=0 for t[x,1],

I2=xx+δ2[x,12,1;(-t)+]dt+2x+δ21[x,12,1;(-t)+]dt
=x1[x,12,1;(-t)+]dt+x+δ21[x,12,1;(-t)+]dt.

By (3.2), the first integral is equal to 12. Furthermore, [x,12,1;(-t)+]=(1-t)+(1-x)(1-12) if x+δ2t1. Hence,

I2=12+11-x(1-x-δ2)2.

Finally,

c2(δ2,x)=x3-(2-2)x2-(2-1)x+121-x.

The critical points of c2(δ2,x) are given by

x1=14(3-1+42)0.104978,
x2=12(2-2)0.292893,
x3=14(3+1+42)1.39502.

Note that c2(δ2,0)=c2(δ2,1-δ2)=12. Since x2,x3(0,1-δ2), we conclude that

c2(δ2,x)c2(δ2,x1)=18(97+1402-102+1)0.504142

for all x(0,1-δ2). The symmetry c2(δ2,x)=c2(δ2,1-x) implies this estimate also for x(δ2,1).

This completes the proof of

C2=18(97+1402-102+1).

Acknowledgements

The authors are grateful to the anonymous reviewer for valuable suggestions which improved the final outcome of the manuscript. In particular, he encouraged us to a deeper study of the inequality in the case of small n.

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Received: 2016-08-18
Revised: 2016-12-19
Accepted: 2016-12-29
Published Online: 2018-02-07
Published in Print: 2020-03-01

© 2020 Walter de Gruyter GmbH, Berlin/Boston

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