IJMMSInternational Journal of Mathematics and Mathematical Sciences1687-04250161-1712Hindawi Publishing Corporation59058910.1155/2009/590589590589Research ArticleApproximation and Shape Preserving Properties of the Bernstein Operator of Max-Product KindBedeBarnabás1CoroianuLucian2GalSorin G.2GovilNarendra Kumar1Department of MathematicsThe University of Texas-Pan American1201 West UniversityEdinburgTX 78539USAutpa.edu2Department of Mathematics and Computer ScienceThe University of OradeaOfficial Postal nr. 1C.P. nr. 114Universitatii 1410087 OradeaRomaniauoradea.ro20099122009200931082009251020092009Copyright © 2009This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Starting from the study of the Shepard nonlinear operator of max-prod type by Bede et al. (2006, 2008), in the book by Gal (2008), Open Problem 5.5.4, pages 324–326, the Bernstein max-prod-type operator is introduced and the question of the approximation order by this operator is raised. In recent paper, Bede and Gal by using a very complicated method to this open question an answer is given by obtaining an upper estimate of the approximation error of the form Cω1(f;1/n) (with an unexplicit absolute constant C>0) and the question of improving the order of approximation ω1(f;1/n) is raised. The first aim of this note is to obtain this order of approximation but by a simpler method, which in addition presents, at least, two advantages: it produces an explicit constant in front of ω1(f;1/n) and it can easily be extended to other max-prod operators of Bernstein type. However, for subclasses of functions f including, for example, that of concave functions, we find the order of approximation ω1(f;1/n), which for many functions f is essentially better than the order of approximation obtained by the linear Bernstein operators. Finally, some shape-preserving properties are obtained.

1. Introduction

Starting from the study of the Shepard nonlinear operator of max-prod-type in [1, 2], by the Open Problem in a recent monograph [3, pages 324–326, 5.5.4], the following nonlinear Bernstein operator of max-prod type is introduced (here means maximum):

Bn(M)(f)(x)=k=0npn,k(x)f(k/n)k=0npn,k(x), where pn,k(x)=(nk)xk(1-x)n-k, for which by a very complicated method in [4, Theorem 6], an upper estimate of the approximation error of the form Cω1(f;1/n) (with C>0 unexplicit absolute constant) is obtained. Also, by Remark 7,2 in the same paper , the question if this order of approximation could be improved is raised.

The first aim of this note is to obtain the same order of approximation but by a simpler method, which in addition presents, at least, two advantages: it produces an explicit constant in front of ω1(f;1/n), and it can easily be extended to other max-prod operators of Bernstein type. Then, one proves by a counterexample that in a sense, for arbitrary f, this order of approximation with respect to ω1(f;·) cannot be improved, giving thus a negative answer to a question raised in [4, Remark 7, 2]. However, for subclasses of functions f including, for example, that of concave functions, we find the order of approximation ω1(f;1/n), which for many functions, f is essentially better than the order of approximation obtained by the linear Bernstein operators. Finally, some shape-preserving properties are presented.

Section 2 presents some general results on nonlinear operators, in Section 3 we prove several auxiliary lemmas, Section 4 contains the approximation results, while in Section 5 we present some shape-preserving properties. The paper ends with Section 6 containing some conclusions concerning the comparisons between the max-product and the linear Bernstein operators.

2. Preliminaries

For the proof of the main results, we need some general considerations on the so-called nonlinear operators of max-prod kind. Over the set of positive reals, +, we consider the operations (maximum) and “·”product. Then (+,,·) has a semiring structure and we call it as Max-Product algebra.

Let I be a bounded or unbounded interval, and

CB+(I)={f:I+;f  continuous  and  bounded  on  I}. The general form of Ln:CB+(I)CB+(I), (called here a discrete max-product-type approximation operator) studied in the paper will be

Ln(f)(x)=i=0nKn(x,xi)·f(xi), or

Ln(f)(x)=i=0Kn(x,xi)·f(xi), where n,  fCB+(I),  Kn(·,xi)CB+(I), and xiI, for all i. These operators are nonlinear, positive operators and moreover they satisfy a pseudolinearity condition of the form

Ln(α·fβ·g)(x)=α·Ln(f)(x)β·Ln(g)(x),α,β+,f,g:I+.

In this section we present some general results on these kinds of operators which will be useful later in the study of the Bernstein max-product-type operator considered in Section 1.

Lemma 2.1 (see [<xref ref-type="bibr" rid="B1">4</xref>]).

Let I be a bounded or unbounded interval: CB+(I)={f:I+;f  continuous  and  bounded  on  I}, and let Ln:CB+(I)CB+(I),  n, be a sequence of operators satisfying the following properties:

if f,gCB+(I) satisfy fg, then Ln(f)Ln(g) for all nN;

Ln(f+g)Ln(f)+Ln(g) for all f,gCB+(I).

Then for all f,gCB+(I),  nN, and xI, we have

|Ln(f)(x)-Ln(g)(x)|Ln(|f-g|)(x).

Proof.

Since it is very simple, we reproduce here the proof in . Let f,gCB+(I). We have f=f-g+g|f-g|+g, which by conditions (i)-(ii) successively implies Ln(f)(x)Ln(|f-g|)(x)+Ln(g)(x), that is, Ln(f)(x)-Ln(g)(x)Ln(|f-g|)(x).

Writing now g=g-f+f|f-g|+f and applying the above reasonings, it follows that Ln(g)(x)-Ln(f)(x)Ln(|f-g|)(x), which, combined with the above inequality, gives |Ln(f)(x)-Ln(g)(x)|Ln(|f-g|)(x).

Remark 2.2.

(1) It is easy to see that the Bernstein max-product operator satisfies the conditions in Lemma 2.1, (i), (ii). In fact, instead of (i), it satisfies the stronger condition: Ln(fg)(x)=Ln(f)(x)Ln(g)(x),f,gCB+(I). Indeed, taking in the above equality fg, f,gCB+(I), it easily follows that Ln(f)(x)Ln(g)(x).

(2) In addition, it is immediate that the Bernstein max-product operator is positive homogenous, that is, Ln(λf)=λLn(f) for all λ0.

Corollary 2.3 (see [<xref ref-type="bibr" rid="B1">4</xref>]).

Let Ln:CB+(I)CB+(I), nN, be a sequence of operators satisfying conditions (i)-(ii) in Lemma 2.1 and in addition being positive homogenous. Then for all fCB+(I), nN, and xI, one has |f(x)-Ln(f)(x)|[1δLn(φx)(x)+Ln(e0)(x)]ω1(f;δ)I+f(x)·|Ln(e0)(x)-1|, where δ>0,  e0(t)=1 for all tI, φx(t)=|t-x| for all tI,  xI,  ω1(f;δ)I=max{|f(x)-f(y)|;x,yI,|x-y|δ}, and if I is unbounded, then we suppose that there exists Ln(φx)(x)+{+}, for any xI, n.

Proof.

The proof is identical with that for positive linear operators and because of its simplicity, we reproduce it what follows. Indeed, from the identity Ln(f)(x)-f(x)=[Ln(f)(x)-f(x)·Ln(e0)(x)]+f(x)[Ln(e0)(x)-1], it follows (by the positive homogeneity and by Lemma 2.1) that |f(x)-Ln(f)(x)||Ln(f(x))(x)-Ln(f(t))(x)|+|f(x)|·|Ln(e0)(x)-1|Ln(|f(t)-f(x)|)(x)+|f(x)|·|Ln(e0)(x)-1|. Now, since for all t,xI, we have |f(t)-f(x)|ω1(f;|t-x|)I[1δ|t-x|+1]ω1(f;δ)I, replacing the above, we immediately obtain the estimate in the statement.

An immediate consequence of Corollary 2.3 is as follows.

Corollary 2.4 (see [<xref ref-type="bibr" rid="B1">4</xref>]).

Suppose that in addition to the conditions in Corollary 2.3, the sequence (Ln)n satisfies Ln(e0)=e0, for all nN. Then for all fCB+(I), nN, and xI, one has |f(x)-Ln(f)(x)|[1+1δLn(φx)(x)]ω1(f;δ)I.

3. Auxiliary Results

Since it is easy to check that Bn(M)(f)(0)-f(0)=Bn(M)(f)(1)-f(1)=0 for all n, notice that in the notations, proofs and statements of the all approximation results, that is, in Lemmas 3.13.3, Theorem 4.1, Lemmas 4.24.4, Corollaries 4.6, 4.7, in fact we always may suppose that 0<x<1. For the proofs of the main results, we need some notations and auxiliary results, as follows.

For each k,j{0,1,2,n} and x[j/(n+1),(j+1)/(n+1)], let us denote

Mk,n,j(x)=pn,k(x)|k/n-x|pn,j(x),mk,n,j(x)=pn,k(x)pn,j(x). It is clear that if kj+1, then

Mk,n,j(x)=pn,k(x)(k/n-x)pn,j(x), and if kj-1, then

Mk,n,j(x)=pn,k(x)(x-k/n)pn,j(x).

Also, for each k,j{0,1,2,n},   kj+2, and x[j/(n+1),(j+1)/(n+1)] let us denote

M¯k,n,j(x)=pn,k(x)(k/(n+1)-x)pn,j(x), and for each k,j{0,1,2,n}, kj-2, and x[j/(n+1),(j+1)/(n+1)] let us denote

M̲k,n,j(x)=pn,k(x)(x-k/(n+1))pn,j(x).

Lemma 3.1.

Let x[j/(n+1),(j+1)/(n+1)]:

for all k,j{0,1,2,n},   kj+2, one has M¯k,n,j(x)Mk,n,j(x)3M¯k,n,j(x);

for all k,j{0,1,2,n},   kj-2, one has Mk,n,j(x)M̲k,n,j(x)6Mk,n,j(x).

Proof.

(i) The inequality M¯k,n,j(x)Mk,n,j(x) is immediate.

On the other hand,

Mk,n,j(x)M¯k,n,j(x)=k/n-xk/(n+1)-xk/n-j/(n+1)k/(n+1)-(j+1)/(n+1)kn+k-njn(k-j-1)=k-jk-j-1+kn(k-j-1)3, which proves (i).

(ii) The inequality Mk,n,j(x)M̲k,n,j(x) is immediate.

On the other hand,

M̲k,n,j(x)Mk,n,j(x)=x-k/(n+1)x-k/n(j+1)/(n+1)-k/(n+1)j/(n+1)-k/n=(n+1)(j+1-k)nj-nk-k(n+1)(j+1-k)nj-nk-n=n+1n·j+1-kj-k-12·j+1-kj-k-1=2(1+2j-k-1)6, which proves (ii) and the lemma.

Lemma 3.2.

For all k,j{0,1,2,,n} and x[j/(n+1),(j+1)/(n+1)], one has mk,n,j(x)1.

Proof.

We have two cases: 1) kj, 2) kj.Case 1.

Since clearly the function h(x)=(1-x)/x is nonincreasing on [j/(n+1),(j+1)/(n+1)], it follows that mk,n,j(x)mk+1,n,j(x)=k+1n-k·1-xxk+1n-k·1-(j+1)/(n+1)(j+1)/(n+1)=k+1n-k·n-jj+11, which implies mj,n,j(x)mj+1,n,j(x)mj+2,n,j(x)mn,n,j(x).

Case 2.

We get mk,n,j(x)mk-1,n,j(x)=n-k+1k·x1-xn-k+1k·j/(n+1)1-j/(n+1)=n-k+1k·jn+1-j1, which immediately implies that mj,n,j(x)mj-1,n,j(x)mj-2,n,j(x)m0,n,j(x). Since mj,n,j(x)=1, the conclusion of the lemma is immediate.

Lemma 3.3.

Let x[j/(n+1),(j+1)/(n+1)].

If k{j+2,j+3,,n-1} is such that k-k+1j, then M¯k,n,j(x)M¯k+1,n,j(x).

If k{1,2,j-2} is such that k+kj, then M̲k,n,j(x)M̲k-1,n,j(x).

Proof.

(i) We observe that M¯k,n,j(x)M¯k+1,n,j(x)=k+1n-k·1-xx·k/(n+1)-x(k+1)/(n+1)-x. Since the function g(x)=((1-x)/x)(k/(n+1)-x)/((k+1)/(n+1)-x) clearly is nonincreasing, it follows that g(x)g((j+1)/(n+1))=((n-j)/(j+1))·((k-j-1)/(k-j)) for all x[j/(n+1),(j+1)/(n+1)]. Then, since the condition k-k+1j implies (k+1)(k-j-1)(j+1)(k-j), we obtain M¯k,n,j(x)M¯k+1,n,j(x)k+1n-k·n-jj+1·k-j-1k-j1.

(ii) We observe that

M̲k,n,j(x)M̲k-1,n,j(x)=n-k+1k·x1-x·x-k/(n+1)x-(k-1)/(n+1). Since the function h(x)=(x/(1-x))·(x-k/(n+1))/(x-(k-1)/(n+1)) is nondecreasing, it follows that h(x)h(j/(n+1))=(j/(n+1-j))·((j-k)/(j-k+1)) for all x[j/(n+1),(j+1)/(n+1)]. Then, since the condition k+kj implies j(j-k)k(j-k+1), we obtain M̲k,n,j(x)M̲k-1,n,j(x)n-k+1k·jn+1-j·j-kj-k+11, which proves the lemma.

Also, a key result in the proof of the main result is the following.

Lemma 3.4.

One has k=0npn,k(x)=pn,j(x),x[jn+1,j+1n+1],j=0,1,,n, where pn,k(x)=(nk)xk(1-x)n-k.

Proof.

First we show that for fixed n and 0k<k+1n, we have 0pn,k+1(x)pn,k(x),iffx[0,k+1n+1]. Indeed, the inequality one reduces to 0(nk+1)xk+1(1-x)n-(k+1)(nk)xk(1-x)n-k after simplifications is equivalent to 0x[(nk+1)+(nk)](nk). However, since (nk+1)+(nk)=(n+1k+1), the above inequality immediately becomes equivalent to 0xk+1n+1. By taking k=0,1,, in the inequality just proved above, we get pn,1(x)pn,0(x),iffx[0,1n+1],pn,2(x)pn,1(x),iffx[0,2n+1],pn,3(x)pn,2(x),iffx[0,3n+1], and so on, pn,k+1(x)pn,k(x),iffx[0,k+1n+1], and so on, pn,n-2(x)pn,n-3(x),iffx[0,n-2n+1],pn,n-1(x)pn,n-2(x),iffx[0,n-1n+1],pn,n(x)pn,n-1(x),iffx[0,nn+1]. From all these inequalities, reasoning by recurrence we easily obtain ifx[0,1n+1],then  pn,k(x)pn,0(x),k=0,1,,n,ifx[1n+1,2n+1],then  pn,k(x)pn,1(x),k=0,1,,n,ifx[2n+1,3n+1],then  pn,k(x)pn,2(x),k=0,1,,n, and so on, finally ifx[nn+1,1],then  pn,k(x)pn,n(x),k=0,1,,n, which proves the lemma.

4. Approximation Results

If Bn(M)(f)(x) represents the nonlinear Bernstein operator of max-product type defined in Section 1, then the first main result of this section is the following.

Theorem 4.1.

If f:[0,1]+ is continuous, then one has the estimate |Bn(M)(f)(x)-f(x)|12ω1(f;1n+1),n,x[0,1], where ω1(f;δ)=sup{|f(x)-f(y)|;x,y[0,1],  |x-y|δ}.

Proof.

It is easy to check that the max-product Bernstein operators fulfill the conditions in Corollary 2.4 and we have |Bn(M)(f)(x)-f(x)|(1+1δnBn(M)(φx)(x))ω1(f;δn), where φx(t)=|t-x|. So, it is enough to estimate En(x):=Bn(M)(φx)(x)=k=0npn,k(x)|k/n-x|k=0npn,k(x).

Let x[j/(n+1),(j+1)/(n+1)], where j{0,,n} is fixed, arbitrary. By Lemma 3.4 we easily obtain

En(x)=maxk=0,,n{Mk,n,j(x)},x[jn+1,j+1n+1]. In all what follows we may suppose that j{1,,n}, because for j=0, simple calculation shows that in this case, we get En(x)1/n, for all x[0,1/(n+1)]. So it remains to obtain an upper estimate for each Mk,n,j(x) when j=1,,n is fixed, x[j/(n+1),(j+1)/(n+1)] and k=0,,n. In fact, we will prove that Mk,n,j(x)6n+1,x[jn+1,j+1n+1],k=0,,n, which immediately will implies that En(x)6n+1,x[0,1],n, and taking δn=6/n+1 in (4.3), we immediately obtain the estimate in the statement.

In order to prove (4.6) we distinguish the following cases: 1) k{j-1,j,j+1}, 2) kj+2 and, 3) kj-2.

Case 1.

If k=j, then Mj,n,j(x)=|j/n-x|. Since x[j/(n+1),(j+1)/(n+1)], it easily follows that Mj,n,j(x)1/(n+1).

If k=j+1, then Mj+1,n,j(x)=mj+1,n,j(x)((j+1)/n-x). Since by Lemma 3.2 we have mj+1,n,j(x)1, we obtain Mj+1,n,j(x)(j+1)/n-x(j+1)/n-j/(n+1)=(n+j+1)/n(n+1)3/(n+1).

If k=j-1, then Mj-1,n,j(x)=mj-1,n,j(x)(x-(j-1)/n)(j+1)/(n+1)-(j-1)/n=(2n-(j+1))/n(n+1)2/(n+1).

Case 2.

Subcase a

Suppose first that k-k+1<j. We get M¯k,n,j(x)=mk,n,j(x)(kn+1-x)kn+1-xkn+1-jn+1kn+1-k-k+1n+1=k+1n+11n+1.

Subcase b

Suppose now that k-k+1j. Since the function g(x)=x-x+1 is nondecreasing on the interval [0,), it follows that there exists k¯{0,1,2,n}, of maximum value, such that k¯-k¯+1<j. Then, for k1=k¯+1, we get k1-k1+1j and M¯k¯+1,n,j(x)=mk¯+1,n,j(x)(k¯+1n+1-x)k¯+1n+1-xk¯+1n+1-jn+1k¯+1n+1-k¯-k¯+1n+1=k¯+1+1n+12n+1. Also, we have k1j+2. Indeed, this is a consequence of the fact that g is nondecreasing on the interval [0,) and because it is easy to see that g(j+1)<j. By Lemma 3.3, (i) it follows that M¯k¯+1,n,j(x)M¯k¯+2,n,j(x)M¯n,n,j(x). We thus obtain M¯k,n,j(x)2/n+1 for any k{k¯+1,k¯+2,n}.

Therefore, in both subcases, by Lemma 3.1, (i) too, we get Mk,n,j(x)6/n+1.

Case 3.

Subcase a

Suppose first that k+kj. Then we obtain M̲k,n,j(x)=mk,n,j(x)(x-kn+1)j+1n+1-kn+1k+k+1n+1-kn+1=k+1n+1n+1n+12n+1.

Subcase b

Suppose now that k+k<j. Let k̃{0,1,2,n} be the minimum value such that k̃+k̃j. Then k2=k̃-1 satisfies k2+k2<j and M̲k-1,n,j(x)=mk-1,n,j(x)(x-k̃-1n+1)j+1n+1-k̃-1n+1k̃+k̃+1n+1-k̃-1n+1=k̃+2n+13n+1. Also, because in this case we have j2, it is immediate that k2j-2. By Lemma 3.3, (ii), it follows that M̲k-1,n,j(x)M̲k-2,n,j(x)M̲0,n,j(x). We obtain M̲k,n,j(x)3/n+1 for any kj-2 and x[j/(n+1),(j+1)/(n+1)].

In both subcases, by Lemma 3.1, (ii) too, we get Mk,n,j(x)3/n+1.

In conclusion, collecting all the estimates in the above cases and subcases we easily get the relationship (4.6), which completes the proof.

Remarks.

(1) The order of approximation in terms of ω1(f;n) in Theorem 4.1 cannot be improved, in the sense that the order of maxx[0,1]{En(x)} is exactly 1/n (here En(x) is defined in the proof of Theorem 4.1). Indeed, for n, let us take jn=[n/2],  kn=jn+[n],   xn=(jn+1)/(n+1) and denote ñ=n-[n/2]. Then, we can write M¯kn,n,jn(xn)=(nkn)xnkn(1-xn)n-kn(njn)xnjn(1-xn)n-jn(knn+1-xn)=(ñ-[n]+1)(ñ-[n]+2)ñ([n/2]+1)([n/2]+2)([n/2]+[n])([n/2]+1ñ)[n]·[n]-1n+1.

Since 2[n/2]n-1, we easily get [n/2]+1ñ, which implies (([n/2]+1)/ñ  )[n]1 for all n. On the other hand,

(ñ-[n]+1)(ñ-[n]+2)ñ([n/2]+1)([n/2]+2)([n/2]+[n])(ñ-[n]+1[n/2]+[n])[n](n/2-n+1n/2+n)n.

Because limn((n/2-n+1)/(n/2+n))n=e-4, there exists n0 such that

(ñ-[n]+1)(ñ-[n]+2)ñ([n/2]+1)([n/2]+2)([n/2]+[n])e-5 for all nn0. It follows M¯kn,n,jn(xn)e-5([n]-1)n+1e-56n, for all nmax{n0,4}. Taking into account Lemma 3.1, (i) too, it follows that for all nmax{n0,4}, we have Mkn,n,jn(xn)e-5/6n, which implies the desired conclusion.

(2) With respect to the method of the proof in , the method in this paper presents, at least, two advantages: it produces the explicit constant 12 in front of ω1(f;1/n+1) and its ideas can be easily used for other max-prod Bernstein operators too, which will be done in several forthcoming papers.

In what follows, we will prove that for large subclasses of functions f, the order of approximation ω1(f;1/n+1) in Theorem 4.1 can essentially be improved to ω1(f;1/n).

For this purpose, for any k,j{0,1,,n}, let us define the functions fk,n,j:[j/(n+1),(j+1)/(n+1)]:

fk,n,j(x)=mk,n,j(x)f(kn)=(nk)(nj)(x1-x)k-jf(kn). Then it is clear that for any j{0,1,,n} and x[j/(n+1),(j+1)/(n+1)], we can write

Bn(M)(f)(x)=k=0nfk,n,j(x). Also we need the following four auxiliary lemmas.

Lemma 4.2.

Let f:[0,1][0,) be such that Bn(M)(f)(x)=max{fj,n,j(x),fj+1,n,j(x)}x[jn+1,j+1n+1]. Then |Bn(M)(f)(x)-f(x)|2ω1(f;1n),x[jn+1,j+1n+1], where ω1(f;δ)=max{|f(x)-f(y)|;x,y[0,1],  |x-y|δ}.

Proof.

We distinguish the two following cases.

Let x[j/(n+1),(j+1)/(n+1)] be fixed such that Bn(M)(f)(x)=fj,n,j(x). Because by simple calculation we have -1/(n+1)x-j/n1/(n+1) and fj,n,j(x)=f(j/n), it follows that |Bn(M)(f)(x)-f(x)|ω1(f;1n+1).

Let x[j/(n+1),(j+1)/(n+1)] be such that Bn(M)(f)(x)=fj+1,n,j(x). We have two subcases:

Bn(M)(f)(x)f(x), when evidently fj,n,j(x)fj+1,n,j(x)f(x) and we immediately get |Bn(M)(f)(x)-f(x)|=|fj+1,n,j(x)-f(x)|=f(x)-fj+1,n,j(x)f(x)-f(jn)ω1(f;1n+1);

Bn(M)(f)(x)>f(x), when |Bn(M)(f)(x)-f(x)|=fj+1,n,j(x)-f(x)=mj+1,n,j(x)f(j+1n)-f(x)f(j+1n)-f(x). Because 0(j+1)/n-x(j+1)/n-j/(n+1)=j/n(n+1)+1/n<2/n, it follows f((j+1)/n)-f(x)2ω1(f;1/n), which proves the lemma.

Lemma 4.3.

Let f:[0,1][0,) be such that Bn(M)(f)(x)=max{fj,n,j(x),fj-1,n,j(x)}x[jn+1,j+1n+1]. Then |Bn(M)(f)(x)-f(x)|2ω1(f;1n),x[jn+1,j+1n+1].

Proof.

We distinguish the two following cases:

Bn(M)(f)(x)=fj,n,j(x), when as in Lemma 4.2 we get |Bn(M)(f)(x)-f(x)|ω1(f;1n+1),

Bn(M)(f)(x)=fj-1,n,j(x), when we have two subcases:

Bn(M)(f)(x)f(x), when as in the case of Lemma 4.2 we obtain |Bn(M)(f)(x)-f(x)|ω1(f;1n+1).

Bn(M)(f)(x)>f(x), when by using the same idea as in the subcase (b) of Lemma 4.2 and taking into account that 0x-j-1nj+1n+1-j-1n=-jn(n+1)+1n+1+1n<2n, we obtain |Bn(M)(f)(x)-f(x)|2ω1(f;1n), which proves the lemma.

Lemma 4.4.

Let f:[0,1][0,) be such that Bn(M)(f)(x)=max{fj-1,n,j(x),fj,n,j(x),fj+1,n,j(x)}, for all x[j/(n+1),(j+1)/(n+1)]. Then |Bn(M)(f)(x)-f(x)|2ω1(f;1n),x[j(n+1),(j+1)(n+1)].

Proof.

Let x[j/(n+1),(j+1)/(n+1)]. If Bn(M)(f)(x)=fj,n,j(x) or Bn(M)(f)(x)=fj+1,n,j(x), then Bn(M)(f)(x)=max{fj,n,j(x),fj+1,n,j(x)} and from Lemma 4.2, it follows that |Bn(M)(f)(x)-f(x)|2ω1(f;1n). If Bn(M)(f)(x)=fj-1,n,j(x), then Bn(M)(f)(x)=max{fj,n,j(x),fj-1,n,j(x)} and from Lemma 4.3, we get |Bn(M)(f)(x)-f(x)|2ω1(f;1n), which ends the proof.

Lemma 4.5.

Let f:[0,1][0,) be concave. Then the following two properties hold:

the function g:(0,1][0,),  g(x)=f(x)/x is nonincreasing;

the function h:[0,1)[0,),  h(x)=f(x)/(1-x) is nondecreasing.

Proof.

(i) Let x,y(0,1] be with xy. Then f(x)=f(xyy+y-xy0)xyf(y)+y-xyf(0)xyf(y), which implies that f(x)/xf(y)/y.

(ii) Let x,y[0,1) be with xy. Then

f(x)=f(1-x1-yy+x-y1-y1)1-x1-yf(y)+x-y1-yf(1)1-x1-yf(y), which implies f(x)/(1-x)f(y)/(1-y).

Corollary 4.6.

Let f:[0,1][0,) be a concave function. Then |Bn(M)(f)(x)-f(x)|2ω1(f;1n),x[0,1].

Proof.

Let x[0,1] and j{0,1,n} such that x[j/(n+1),(j+1)/(n+1)]. Let k{0,1,n} be with kj. Then fk+1,n,j(x)=(nk+1)(nj)(x1-x)k+1-jf(k+1n)=(nk)(nj)n-kk+1(x1-x)k-jx1-xf(k+1n). From Lemma 4.5, (i), we get f((k+1)/n)/(k+1)/nf(k/n)/k/n, that is, f((k+1)/n)((k+1)/k)(f(k/n)). Since x/(1-x)(j+1)/(n-j), we get fk+1,n,j(x)(nk)(nj)n-kk+1(x1-x)k-jj+1n-j·k+1kf(kn)=fk,n,j(x)j+1k·n-kn-j. It is immediate that for kj+1, it follows that fk,n,j(x)fk+1,n,j(x). Thus we obtain fj+1,n,j(x)fj+2,n,j(x)fn,j,n(x).

Now let k{0,1,n} be with kj. Then

fk-1,n,j(x)=(nk-1)(nj)(x1-x)k-1-jf(k-1n)=(nk)(nj)·kn-k+1(x1-x)k-j1-xxf(k-1n). From Lemma 4.5, (ii), we get f(k/n)/(1-k/n)f((k-1)/n)/(1-(k-1)/n), that is, f(k/n)((n-k)/(n-k+1))(f((k-1)/n)). Because (1-x)/x(n+1-j)/j, we get fk-1,n,j(x)(nk)(nj)kn-k+1(x1-x)k-jn+1-jj·n-k+1n-kf(kn)=fk,n,j(x)kj·n+1-jn-k. For kj-1 it is immediate that fk,n,j(x)fk-1,n,j(x), which implies fj-1,n.j(x)fj-2,n.j(x)f0,n,j(x). From (4.38) and (4.42), we obtain Bn(M)(f)(x)=max{fj-1,n,j(x),fj,n,j(x),fj+1,n,j(x)}, which combined with Lemma 4.4 implies |Bn(M)(f)(x)-f(x)|2ω1(f;1n), and proves the corollary.

Corollary 4.7.

(i) If f:[0,1][0,) is nondecreasing and such that the function g:(0,1][0,),  g(x)=f(x)/x is nonincreasing, then |Bn(M)(f)(x)-f(x)|2ω1(f;1n),x[0,1].

(ii) If f:[0,1][0,) is nonincreasing and such that the function h:[0,1)[0,),  h(x)=f(x)/(1-x) is nondecreasing, then

|Bn(M)(f)(x)-f(x)|2ω1(f;1n),x[0,1].

Proof.

(i) Since f is nondecreasing it follows (see the proof of Theorem 5.5 in Section 5) that Bn(M)(f)(x)=kjnfk,n,j(x),x[jn+1,j+1n+1]. Following the proof of Corollary 4.6, we get Bn(M)(f)(x)=max{fj,n,j(x),fj+1,n,j(x)},x[jn+1,j+1n+1], and from Lemma 4.2, we obtain |Bn(M)(f)(x)-f(x)|2ω1(f;1n).

(ii) Since f is nonincreasing, it follows (see the proof of Corollary 5.6 in Section 5) that

Bn(M)(f)(x)=k0jfk,n,j(x),x[jn+1,j+1n+1]. Following the proof of Corollary 4.6, we get Bn(M)(f)(x)=max{fj-1,n,j(x),fj,n,j(x)}, and from Lemma 4.3 we obtain |Bn(M)(f)(x)-f(x)|2ω1(f;1n).

Remark 4.8.

By simple reasonings, it follows that if f:[0,1][0,) is a convex, nondecreasing function satisfying f(x)/xf(1) for all x[0,1], then the function g:(0,1][0,),  g(x)=f(x)/x is nonincreasing and as a consequence for f is valid the conclusion of Corollary 4.7, (i). Indeed, for simplicity, let us suppose that fC1[0,1] and denote F(x)=xf(x)-f(x), x[0,1]. Then g(x)=F(x)/x2, for all x(0,1]. Since the inequality f(x)/xf(1) can be written as (f(1)-f(x))/(1-x)f(1), for all x[0,1), passing to limit with x1, it follows f(1)f(1), which implies (since f is nondecreasing) F(x)xf(1)-f(x)xf(1)-xf(1)=x[f(1)-f(1)]0,x(0,1], which means that g(x) is nonincreasing.

An example of function satisfying the above conditions is f(x)=ex,   x[0,1].

Analogously, if f:[0,1][0,) is a convex, nonincreasing function satisfying f(x)/(1-x)f(0), then for f is valid the conclusion of Corollary 4.7, (ii). An example of function satisfying these conditions is f(x)=e-x, x[0,1].

5. Shape-Preserving Properties

In this section, we will present some shape preserving properties, by proving that the max-product Bernstein operator preserves the monotonicity and the quasiconvexity. First, we have the following simple result.

Lemma 5.1.

For any arbitrary function f:[0,1]+, Bn(M)(f)(x) is positive, continuous on [0,1], and satisfies Bn(M)(f)(0)=f(0),  Bn(M)(f)(1)=f(1).

Proof.

Since pn,k(x)>0 for all x(0,1), n, k{0,,n}, it follows that the denominator k=0npn,k(x)>0 for all x(0,1) and n. However, the numerator is a maximum of continuous functions on [0,1], so it is a continuous function on [0,1], and this implies that Bn(M)(f)(x) is continuous on (0,1). To prove now the continuity of Bn(M)(f)(x) at x=0 and x=1, we observe that pn,k(0)=0 for all k{1,2,,n},   pn,k(0)=1 for k=0 and pn,k(1)=0 for all k{0,1,,n-1},   pn,k(1)=1 for k=n, which implies that k=0npn,k(x)=1 in the case of x=0 and x=1. The fact that Bn(M)(f)(x) coincides with f(x) at x=0 and x=1 immediately follows from the above considerations, which proves the theorem.

Remark 5.2.

Note that because of the continuity of Bn(M)(f)(x) on [0,1], it will suffice to prove the shape properties of Bn(M)(f)(x) on (0,1) only. As a consequence, in the notations and proofs below, we always may suppose that 0<x<1.

As in Section 4, for any k,j{0,1,,n}, let us consider the functions fk,n,j:[j/(n+1),(j+1)/(n+1)],

fk,n,j(x)=mk,n,j(x)f(kn)=(nk)(nj)(x1-x)k-jf(kn). For any j{0,1,,n} and x[j/(n+1),(j+1)/(n+1)], we can write Bn(M)(f)(x)=k=0nfk,n,j(x).

Lemma 5.3.

If f:[0,1]+ is a nondecreasing function, then for any k,j{0,1,n},  kj and x[j/(n+1),(j+1)/(n+1)], one has fk,n,j(x)fk-1,n,j(x).

Proof.

Because kj, by the proof of Lemma 3.2, Case 2, it follows that mk,n,j(x)mk-1,n,j(x). From the monotonicity of f, we get f(k/n)f((k-1)/n). Thus, we obtain mk,n,j(x)f(kn)mk-1,n,j(x)f(k-1n), which proves the lemma.

Corollary 5.4.

If f:[0,1]+ is nonincreasing, then fk,n,j(x)fk+1,n,j(x) for any k,j{0,1,n},  kj, and x[j/(n+1),(j+1)/(n+1)].

Proof.

Because kj, by the proof of Lemma 3.2, Case 1, it follows that mk,n,j(x)mk+1,n,j(x). From the monotonicity of f, we get f(k/n)f((k+1)/n). Thus we obtain mk,n,j(x)f(kn)mk+1,n,j(x)f(k+1n), which proves the corollary.

Theorem 5.5.

If f:[0,1]+ is nondecreasing, then Bn(M)(f) is nondecreasing.

Proof.

Because Bn(M)(f) is continuous on [0,1], it suffices to prove that on each subinterval of the form [j/(n+1),(j+1)/(n+1)], with j{0,1,n}, Bn(M)(f) is nondecreasing.

So let j{0,1,n} and x[j/(n+1),(j+1)/(n+1)]. Because f is nondecreasing, from Lemma 5.3 it follows that

fj,n,j(x)fj-1,n,j(x)fj-2,n,j(x)f0,n,j(x), but then it is immediate that Bn(M)(f)(x)=kjnfk,n,j(x), for all x[j/(n+1),(j+1)/(n+1)]. Clearly that for kj, the function fk,n,j is nondecreasing and since Bn(M)(f) is defined as the maximum of nondecreasing functions, it follows that it is nondecreasing.

Corollary 5.6.

If f:[0,1]+ is nonincreasing, then Bn(M)(f) is nonincreasing.

Proof.

Because Bn(M)(f) is continuous on [0,1], it suffices to prove that on each subinterval of the form [j/(n+1),(j+1)/(n+1)], with j{0,1,n}, Bn(M)(f) is nonincreasing.

So let j{0,1,n} and x[j/(n+1),(j+1)/(n+1)]. Because f is nonincreasing, from Corollary 5.4, it follows that

fj,n,j(x)fj+1,n,j(x)fj+2,n,j(x)fn,n,j(x), but then it is immediate that Bn(M)(f)(x)=k0jfk,n,j(x), for all x[j/(n+1),(j+1)/(n+1)]. Clearly that for kj the function fk,n,j is nonincreasing and since Bn(M)(f) is defined as the maximum of nonincreasing functions, it follows that it is nonincreasing.

In what follows, let us consider the following concept generalizing the monotonicity and convexity.

Definition 5.7.

Let f:[0,1] be continuous on [0,1]. One says that the function f:[0,1] is quasiconvex on [0,1] if it satisfies the inequality f(λx+(1-λ)y)max{f(x),f(y)},x,y,λ[0,1], (see, e.g., [3, page 4, (iv)]).

Remark 5.8.

By , the continuous function f is quasiconvex on [0,1] equivalently means that there exists a point c[0,1] such that f is nonincreasing on [0,c] and nondecreasing on [c,1]. The class of quasiconvex functions includes the class of nondecreasing functions and the class of nonincreasing functions. Also, it obviously includes the class of convex functions on [0,1].

Corollary 5.9.

If f:[0,1]+ is continuous and quasiconvex on [0,1], then for all n, Bn(M)(f) is quasiconvex on [0,1].

Proof.

If f is nonincreasing (or nondecreasing) on [0,1] (i.e., the point c=1 (or c=0) in Remark 5.8), then by the Corollary 5.6 (or Theorem 5.5, resp.), it follows that for all n, Bn(M)(f) is nonincreasing (or nondecreasing) on [0,1].

Suppose now that there exists c(0,1), such that f is nonincreasing on [0,c] and nondecreasing on [c,1]. Define the functions F,G:[0,1]+ by F(x)=f(x) for all x[0,c], F(x)=f(c) for all x[c,1] and G(x)=f(c) for all x[0,c], G(x)=f(x) for all x[c,1].

It is clear that F is nonincreasing and continuous on [0,1], G is nondecreasing and continuous on [0,1], and f(x)=max{F(x),G(x)}, for all x[0,1].

However, it is easy to show (see also Remark 2.2 after the proof of Lemma 2.1) that

Bn(M)(f)(x)=max{Bn(M)(F)(x),Bn(M)(G)(x)},x[0,1], where by Corollary 5.6 and Theorem 5.5, Bn(M)(F)(x) is nonincreasing and continuous on [0,1] and Bn(M)(G)(x) is nondecreasing and continuous on [0,1]. We have two cases: 1) Bn(M)(F)(x) and Bn(M)(G)(x) do not intersect each other; 2) Bn(M)(F)(x) and Bn(M)(G)(x) intersect each other.Case 1.

We have max{Bn(M)(F)(x),Bn(M)(G)(x)}=Bn(M)(F)(x) for all x[0,1] or max{Bn(M)(F)(x),Bn(M)(G)(x)}=Bn(M)(G)(x) for all x[0,1], which obviously proves that Bn(M)(f)(x) is quasiconvex on [0,1].

Case 2.

In this case, it is clear that there exists a point c[0,1] such that Bn(M)(f)(x) is nonincreasing on [0,c] and nondecreasing on [c,1], which by the result in  implies that Bn(M)(f)(x) is quasiconvex on [0,1] and proves the corollary.

Remark 5.10.

The preservation of the quasiconvexity by the linear Bernstein operators was proved in .

It is of interest to exactly calculate Bn(M)(f)(x) for f(x)=e0(x)=1 and for f(x)=e1(x)=x. In this sense, we can state the following.

Lemma 5.11.

For all x[0,1] and n, one has Bn(M)(e0)(x)=1 and Bn(M)(e1)(x)=x·pn-1,0(x)pn,0(x)=x1-x,if  x[0,1n+1],Bn(M)(e1)(x)=x·pn-1,0(x)pn,1(x)=1n,if  x[1n+1,1n],Bn(M)(e1)(x)=x·pn-1,1(x)pn,1(x)=x1-x·n-1n,if  x[1n,2n+1],Bn(M)(e1)(x)=x·pn-1,1(x)pn,2(x)=2n,if  x[2n+1,2n],Bn(M)(e1)(x)=x·pn-1,2(x)pn,2(x)=x1-x·n-2n,if  x[2n,3n+1],Bn(M)(e1)(x)=x·pn-1,2(x)pn,3(x)=3n,if  x[3n+1,3n], and so on, in general one has Bn(M)(e1)(x)=x1-x·n-jn,if  x[jn,j+1n+1],Bn(M)(e1)(x)=j+1n,if  x[j+1n+1,(j+1)n], for j{0,1,,n-1}.

Proof.

The formula Bn(M)(e0)(x)=1 is immediate by the definition of Bn(M)(f)(x).

To find the formula for Bn(M)(e1)(x), we will use the explicit formula in Lemma 3.4 which says that

k=0npn,k(x)=pn,j(x),x[jn+1,j+1n+1],j=0,1,,n, where pn,k(x)=(nk)xk(1-x)n-k.

Indeed, since

maxk=0,,n{pn,k(x)kn}=maxk=1,,n{pn,k(x)kn}=x·maxk=0,,n-1{pn-1,k(x)}, this follows by applying Lemma 3.4 to both expressions maxk=0,,n{pn,k(x)}, maxk=0,,n-1{pn-1,k(x)}, taking into account that we get the following division of the interval [0,1]0<1n+11n2n+12n3n+13n4n+14n.

Remarks.

(1) The convexity of f on [0,1] is not preserved by Bn(M)(f) as can be seen from Lemma 5.11. Indeed, while f(x)=e1(x)=x is obviously convex on [0,1], it is easy to see that Bn(M)(e1) is not convex on [0,1].

(2) Also, if f is supposed to be starshaped on [0,1] (i.e., f(λx)λf(x) for all x,λ[0,1]), then again by Lemma 5.11, it follows that Bn(M)(f) for f(x)=e1(x) is not starshaped on [0,1], although e1(x) obviously is starshaped on [0,1].

Despite of the absence of the preservation of the convexity, we can prove the interesting property that for any arbitrary function f, the max-product Bernstein operator Bn(M)(f) is piecewise convex on [0,1]. We present the following.

Theorem 5.12.

For any function f:[0,1][0,), Bn(M)(f) is convex on any interval of the form [j/(n+1),(j+1)/(n+1)], j=0,1,,n.

Proof.

For any k,j{0,1,,n}, let us consider the functions fk,n,j:[j/(n+1),(j+1)/(n+1)],fk,n,j(x)=mk,n,j(x)f(kn)=(nk)(nj)(x1-x)k-jf(kn). Clearly, we have Bn(M)(f)(x)=k=0nfk,n,j(x), for any j{0,1,,n} and x[j/(n+1),(j+1)/(n+1)].

We will prove that for any fixed j, each function fk,n,j(x) is convex on [j/(n+1),(j+1)/(n+1)], which will imply that Bn(M)(f) can be written as a maximum of some convex functions on [j/(n+1),(j+1)/(n+1)].

Since f0, it suffices to prove that the functions gk,j:[0,1], gk,j(x)=(x/(1-x))k-j are convex on [j/(n+1),(j+1)/(n+1)].

For k=j, gj,j is constant so is convex.

For k=j+1, we get gj+1,j(x)=x/(1-x) for any x[j/(n+1),(j+1)/(n+1)]. Then gj+1,j′′(x)=2/(1-x)3>0 for any x[j/(n+1),(j+1)/(n+1)].

For k=j-1, it follows that gj-1,j(x)=(1-x)/x for any x[j/(n+1),(j+1)/(n+1)]. Then gj-1,j′′(x)=2/x3>0 for any x[j/(n+1),(j+1)/(n+1)].

If kj+2, then gk,j′′(x)=((k-j)/(1-x)4)(x/(1-x))k-j-2(k-j-1+2x)>0 for any x[j/(n+1),(j+1)/(n+1)].

If kj-2, then gk,j′′(x)=((k-j)/(1-x)4)(x/(1-x))k-j-2(k-j-1+2x). Sice (k-j-1+2x)k-j+1-1 for any x[j/(n+1),(j+1)/(n+1)], it follows that (k-j)(k-j-1+2x)>0, which implies gk,j′′(x)>0 for any x[j/(n+1),(j+1)/(n+1)].

Since all the functions gk,j are convex on [j/(n+1),(j+1)/(n+1)], we get that Bn(M)(f) is convex on [j/(n+1),(j+1)/(n+1)] as maximum of these functions, which proves the theorem.

At the end of this section, let us note that although Bn(M)(f) does not preserve the convexity too, by using Bn(M)(f) it easily can be constructed new nonlinear operators which converge to the function and preserve the convexity too.

Indeed, in this sense, for example, we present the following.

Theorem 5.13.

For f belonging to the set S[0,1]={f:[0,1];fC1[0,1],  f(0)=0,  f  is  nondecreasing  on  [0,1]}, let us define the following subadditive and positive homogenous operators (as function of f): Ln(f)(x)=0xBn(M)(f)(t)dt,x[0,1],n. If fS[0,1] is convex, then Ln(f)(x) is nondecreasing and convex on [0,1]. In addition, if f is concave on [0,1], then the order of approximation of f through Ln(f) is ω1(f;1/n).

Proof.

Indeed, since f is convex, it follows that f(x) is nondecreasing on [0,1], which by Theorem 5.5 implies that Bn(M)(f)(x) is nondecreasing and, therefore, we get the convexity of Ln(f)(x) on [0,1]. The monotonicity of Ln(f)(x) is immediate by f0 on [0,1] and by the relationship Ln(f)(x)=Bn(M)(f)(x)0 for all x[0,1].

Also, writing f(x)=0xf(t)dt and supposing that f is concave, by Corollary 4.6, we get that the order of approximation of f by Ln(f) is ω1(f;1/n). In addition, Ln(f)(x) obviously is of C1-class (which is not the case of original operator Bn(M)(f)(x)) and Ln(f)(x) converges uniformly to f on [0,1] with the same order of approximation ω1(f;1/n).

Remarks.

(1) A simple example of function f verifying the statement of Theorem 5.13 is f(x)=1-cosx, because in this case, we easily get that f(0)=0, f(x)=sinx0, f′′(x)=cosx0 and f′′′(x)=-sinx0, for all x[0,1].

(2) In the definition of Ln(f)(x) in the above Theorem 5.13, obviously that the values f(k/n) are involved. To involve values of f only but without to loose the properties mentioned in Theorem 5.13, we can replace there f(k/n) by, for example, (f((k+1)/n)-f(k/n))/((k+1)/n-k/n)=n[f((k+1)/n)-f(k/n)] or by (f((k+1)/(n+1))-f(k/n))/((k+1)/(n+1)-k/n).

6. Comparisons with the Linear Bernstein Operator

In this section, we compare the max-product Bernstein operator Bn(M)(f) with the linear Bernstein operator Bn(f)(x)=k=0npn,k(x)f(k/n). First, it is known that for the linear Bernstein operator, the best possible uniform approximation result is given by the equivalence: (see [7, 8])

Bn(f)-f~ω2φ(f;1n), where f=sup{|f(x)|;x[0,1]} and ω2φ(f;δ) is the Ditzian-Totik second-order modulus of smoothness given by

ω2φ(f;δ)=sup{sup{|f(x+hφ(x))-2f(x)+f(x-hφ(x))|;xIh},  h[0,δ]}, with φ(x)=x(1-x),   δ1 and Ih=[h2/(1+h2),1/(1+h2)].

Now, if f is, for example, a nondecreasing concave polygonal line on [0,1], then by simple reasonings we get that ω2φ(f;δ)~δ for δ1, which shows that the order of approximation obtained in this case by the linear Bernstein operator is exactly 1/n. On the other hand, since such of function f obviously is a Lipschitz function on [0,1] (as having bounded all the derivative numbers) by Corollary 4.6, we get that the order of approximation by the max-product Bernstein operator is less than 1/n, which is essentially better than 1/n. In a similar manner, by Corollary 4.7 and by the Remark 4.8 after this corollary, we can produce many subclasses of functions for which the order of approximation given by the max-product Bernstein operator is essentially better than the order of approximation given by the linear Bernstein operator. In fact, the Corollaries 4.6 and 4.7 have no corespondent in the case of linear Bernstein operator. All these prove the advantages we may have in some cases, by using the max-product Bernstein operator. Intuitively, the max-product Bernstein operator has better approximation properties than its linear counterpart, for nondifferentiable functions in a finite number of points (with the graphs having some "corners"), as an example for functions defined as a maximum of a finite number of continuous functions on [0,1].

On the other hand, in other cases (e.g., for differentiable functions), the linear Bernstein operator has better approximation properties than the max-product Bernstein operator, as can be seen from the formula for Bn(M)(e1)(x) in Lemma 5.11. Indeed, by direct calculation can be easily proved that Bn(M)(e1)-e1~1/n, while it is well known that Bn(e1)-e1=0.

Concerning now the shape-preserving properties, it is clear from Section 5 that the linear Bernstein operator has better properties. However, for some particular classes of functions, the type of construction in Theorem 5.13, combined with Corollaries 4.6 and 4.7, can produce max-product Bernstein-type operators with good preservation properties (e.g., preserving monotonicity and convexity) and giving in some cases (supposing, e.g., that f is a concave polygonal line), the same order of approximation as the linear Bernstein operator.

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