# Approximation and Shape Preserving Properties of the Bernstein Operator of Max-Product Kind

Authors: Barnabás Bede; Lucian Coroianu; Sorin G. Gal
Journal: International Journal of Mathematics and Mathematical Sciences
 (2009)

---

## Abstract

Starting from the study of theShepard 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.


---

## Body

## 1. Introduction

Starting from the study of theShepard 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):(1.1)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 [4], 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 ofω1(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.Section2 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,ℝ+, we consider the operations ∨ (maximum) and “·”product. Then (ℝ+,∨,·) has a semiring structure and we call it as Max-Product algebra.LetI⊂ℝ be a bounded or unbounded interval, and(2.1)CB+(I)={f:I→ℝ+;fcontinuousandboundedonI}.
The general form of Ln:CB+(I)→CB+(I), (called here a discrete max-product-type approximation operator) studied in the paper will be(2.2)Ln(f)(x)=⋁i=0nKn(x,xi)·f(xi),
or(2.3)Ln(f)(x)=⋁i=0∞Kn(x,xi)·f(xi),
where n∈ℕ,f∈CB+(I),Kn(·,xi)∈CB+(I), and xi∈I, for all i. These operators are nonlinear, positive operators and moreover they satisfy a pseudolinearity condition of the form(2.4)Ln(α·f∨β·g)(x)=α·Ln(f)(x)∨β·Ln(g)(x),∀α,β∈ℝ+,f,g:I→ℝ+.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 Section1.Lemma 2.1 (see [4]).
LetI⊂ℝ be a bounded or unbounded interval:
(2.5)CB+(I)={f:I→ℝ+;fcontinuousandboundedonI},
and let Ln:CB+(I)→CB+(I),n∈ℕ, be a sequence of operators satisfying the following properties: (i)
iff,g∈CB+(I) satisfy f≤g, then Ln(f)≤Ln(g) for all n∈N;(ii)
Ln(f+g)≤Ln(f)+Ln(g) for all f,g∈CB+(I).
Then for allf,g∈CB+(I),n∈N, and x∈I, we have
(2.6)|Ln(f)(x)-Ln(g)(x)|≤Ln(|f-g|)(x).Proof.
Since it is very simple, we reproduce here the proof in [4]. Let f,g∈CB+(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 nowg=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:
(2.7)Ln(f∨g)(x)=Ln(f)(x)∨Ln(g)(x),f,g∈CB+(I).
Indeed, taking in the above equality f≤g, f,g∈CB+(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 λ≥0.Corollary 2.3 (see [4]).
LetLn:CB+(I)→CB+(I), n∈N, be a sequence of operators satisfying conditions (i)-(ii) in Lemma 2.1 and in addition being positive homogenous. Then for all f∈CB+(I), n∈N, and x∈I, one has
(2.8)|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 t∈I, φx(t)=|t-x| for all t∈I,x∈I,ω1(f;δ)I=max{|f(x)-f(y)|;x,y∈I,|x-y|≤δ}, and if I is unbounded, then we suppose that there exists Ln(φx)(x)∈ℝ+⋃{+∞},  for any x∈I, n∈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(2.9)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
(2.10)|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,x∈I, we have
(2.11)|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 Corollary2.3 is as follows.Corollary 2.4 (see [4]).
Suppose that in addition to the conditions in Corollary2.3, the sequence (Ln)n satisfies Ln(e0)=e0, for all n∈N. Then for all f∈CB+(I), n∈N, and x∈I, one has
(2.12)|f(x)-Ln(f)(x)|≤[1+1δLn(φx)(x)]ω1(f;δ)I.


## 3. Auxiliary Results

Since it is easy to check thatBn(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 3.1–3.3, Theorem 4.1, Lemmas 4.2–4.4, 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 eachk,j∈{0,1,2,…n} and x∈[j/(n+1),(j+1)/(n+1)], let us denote(3.1)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 k≥j+1, then(3.2)Mk,n,j(x)=pn,k(x)(k/n-x)pn,j(x),
and if k≤j-1, then(3.3)Mk,n,j(x)=pn,k(x)(x-k/n)pn,j(x).Also, for eachk,j∈{0,1,2,…n}, k≥j+2, and x∈[j/(n+1),(j+1)/(n+1)] let us denote(3.4)M¯k,n,j(x)=pn,k(x)(k/(n+1)-x)pn,j(x),
and for each k,j∈{0,1,2,…n}, k≤j-2, and x∈[j/(n+1),(j+1)/(n+1)] let us denote(3.5)M̲k,n,j(x)=pn,k(x)(x-k/(n+1))pn,j(x).Lemma 3.1.
Letx∈[j/(n+1),(j+1)/(n+1)]: (i)
for allk,j∈{0,1,2,…n}, k≥j+2, one has
(3.6)M¯k,n,j(x)≤Mk,n,j(x)≤3M¯k,n,j(x);(ii)
for allk,j∈{0,1,2,…n}, k≤j-2, one has
(3.7)Mk,n,j(x)≤M̲k,n,j(x)≤6Mk,n,j(x).Proof.
(i) The inequalityM¯k,n,j(x)≤Mk,n,j(x) is immediate.
On the other hand,
(3.8)Mk,n,j(x)M¯k,n,j(x)=k/n-xk/(n+1)-x≤k/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 inequalityMk,n,j(x)≤M̲k,n,j(x) is immediate.
On the other hand,
(3.9)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-1≤2·j+1-kj-k-1=2(1+2j-k-1)≤6,
which proves (ii) and the lemma.Lemma 3.2.
For allk,j∈{0,1,2,…,n} and x∈[j/(n+1),(j+1)/(n+1)], one has
(3.10)mk,n,j(x)≤1.Proof.
We have two cases: 1)k≥j,  2) k≤j.Case 1.
Since clearly the functionh(x)=(1-x)/x is nonincreasing on [j/(n+1),(j+1)/(n+1)], it follows that
(3.11)mk,n,j(x)mk+1,n,j(x)=k+1n-k·1-xx≥k+1n-k·1-(j+1)/(n+1)(j+1)/(n+1)=k+1n-k·n-jj+1≥1,
which implies mj,n,j(x)≥mj+1,n,j(x)≥mj+2,n,j(x)≥⋯≥mn,n,j(x).Case 2.
We get(3.12)mk,n,j(x)mk-1,n,j(x)=n-k+1k·x1-x≥n-k+1k·j/(n+1)1-j/(n+1)=n-k+1k·jn+1-j≥1,
which immediately implies that
(3.13)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.
Letx∈[j/(n+1),(j+1)/(n+1)]. (i)
Ifk∈{j+2,j+3,…,n-1} is such that k-k+1≥j,  then M¯k,n,j(x)≥M¯k+1,n,j(x).(ii)
Ifk∈{1,2,…j-2} is such that k+k≤j, then M̲k,n,j(x)≥M̲k-1,n,j(x).Proof.
(i) We observe that(3.14)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+1≥j implies (k+1)(k-j-1)≥(j+1)(k-j), we obtain
(3.15)M¯k,n,j(x)M¯k+1,n,j(x)≥k+1n-k·n-jj+1·k-j-1k-j≥1.
(ii) We observe that
(3.16)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+k≤j implies j(j-k)≥k(j-k+1), we obtain
(3.17)M̲k,n,j(x)M̲k-1,n,j(x)≥n-k+1k·jn+1-j·j-kj-k+1≥1,
which proves the lemma.Also, a key result in the proof of the main result is the following.Lemma 3.4.
One has(3.18)⋁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 fixedn∈ℕ and 0≤k<k+1≤n, we have
(3.19)0≤pn,k+1(x)≤pn,k(x),iffx∈[0,k+1n+1].
Indeed, the inequality one reduces to
(3.20)0≤(nk+1)xk+1(1-x)n-(k+1)≤(nk)xk(1-x)n-k
after simplifications is equivalent to
(3.21)0≤x[(nk+1)+(nk)]≤(nk).
However, since (nk+1)+(nk)=(n+1k+1), the above inequality immediately becomes equivalent to
(3.22)0≤x≤k+1n+1.
By taking k=0,1,…, in the inequality just proved above, we get
(3.23)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,
(3.24)pn,k+1(x)≤pn,k(x),iffx∈[0,k+1n+1],
and so on,
(3.25)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
(3.26)ifx∈[0,1n+1],thenpn,k(x)≤pn,0(x),∀k=0,1,…,n,ifx∈[1n+1,2n+1],thenpn,k(x)≤pn,1(x),∀k=0,1,…,n,ifx∈[2n+1,3n+1],thenpn,k(x)≤pn,2(x),∀k=0,1,…,n,
and so on, finally
(3.27)ifx∈[nn+1,1],thenpn,k(x)≤pn,n(x),∀k=0,1,…,n,
which proves the lemma.


## 4. Approximation Results

IfBn(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.
Iff:[0,1]→ℝ+ is continuous, then one has the estimate
(4.1)|Bn(M)(f)(x)-f(x)|≤12ω1(f;1n+1),∀n∈ℕ,x∈[0,1],
where
(4.2)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 Corollary2.4 and we have
(4.3)|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
(4.4)En(x):=Bn(M)(φx)(x)=⋁k=0npn,k(x)|k/n-x|⋁k=0npn,k(x).
Letx∈[j/(n+1),(j+1)/(n+1)], where j∈{0,…,n} is fixed, arbitrary. By Lemma 3.4 we easily obtain
(4.5)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
(4.6)Mk,n,j(x)≤6n+1,∀x∈[jn+1,j+1n+1],k=0,…,n,
which immediately will implies that
(4.7)En(x)≤6n+1,∀x∈[0,1],n∈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) k≥j+2 and, 3) k≤j-2.
Case 1.
Ifk=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).
Ifk=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).
Ifk=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 thatk-k+1<j. We get
(4.8)M¯k,n,j(x)=mk,n,j(x)(kn+1-x)≤kn+1-x≤kn+1-jn+1≤kn+1-k-k+1n+1=k+1n+1≤1n+1.Subcase b
Suppose now thatk-k+1≥j. 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+1≥j and
(4.9)M¯k¯+1,n,j(x)=mk¯+1,n,j(x)(k¯+1n+1-x)≤k¯+1n+1-x≤k¯+1n+1-jn+1≤k¯+1n+1-k¯-k¯+1n+1=k¯+1+1n+1≤2n+1.
Also, we have k1≥j+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 Lemma3.1, (i) too, we get Mk,n,j(x)≤6/n+1.
Case 3.
Subcase a
Suppose first thatk+k≥j. Then we obtain
(4.10)M̲k,n,j(x)=mk,n,j(x)(x-kn+1)≤j+1n+1-kn+1≤k+k+1n+1-kn+1=k+1n+1≤n+1n+1≤2n+1.Subcase b
Suppose now thatk+k<j. Let k̃∈{0,1,2,…n} be the minimum value such that k̃+k̃≥j. Then k2=k̃-1 satisfies k2+k2<j and
(4.11)M̲k-1,n,j(x)=mk-1,n,j(x)(x-k̃-1n+1)≤j+1n+1-k̃-1n+1≤k̃+k̃+1n+1-k̃-1n+1=k̃+2n+1≤3n+1.
Also, because in this case we have j≥2, it is immediate that k2≤j-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 k≤j-2 and x∈[j/(n+1),(j+1)/(n+1)].
In both subcases, by Lemma3.1, (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∈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
(4.12)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.
Since2[n/2]≥n-1, we easily get [n/2]+1≥ñ, which implies (([n/2]+1)/ñ)[n]≥1 for all n∈ℕ. On the other hand,
(4.13)(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.
Becauselimn→∞((n/2-n+1)/(n/2+n))n=e-4, there exists n0∈ℕ such that
(4.14)(ñ-[n]+1)(ñ-[n]+2)⋯ñ([n/2]+1)([n/2]+2)⋯([n/2]+[n])≥e-5
for all n≥n0. It follows
(4.15)M¯kn,n,jn(xn)≥e-5([n]-1)n+1≥e-56n,
for all n≥max{n0,4}. Taking into account Lemma 3.1, (i) too, it follows that for all n≥max{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 [4], 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 functionsf, 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 anyk,j∈{0,1,…,n}, let us define the functions fk,n,j:[j/(n+1),(j+1)/(n+1)]→ℝ:(4.16)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(4.17)Bn(M)(f)(x)=⋁k=0nfk,n,j(x).
Also we need the following four auxiliary lemmas.Lemma 4.2.
Letf:[0,1]→[0,∞) be such that
(4.18)Bn(M)(f)(x)=max{fj,n,j(x),fj+1,n,j(x)}∀x∈[jn+1,j+1n+1].
Then
(4.19)|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.Case  (i).
Letx∈[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/n≤1/(n+1) and fj,n,j(x)=f(j/n), it follows that
(4.20)|Bn(M)(f)(x)-f(x)|≤ω1(f;1n+1).Case  (ii).
Letx∈[j/(n+1),(j+1)/(n+1)] be such that Bn(M)(f)(x)=fj+1,n,j(x). We have two subcases:(a)
Bn(M)(f)(x)≤f(x), when evidently fj,n,j(x)≤fj+1,n,j(x)≤f(x) and we immediately get
(4.21)|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);(b)
Bn(M)(f)(x)>f(x), when
(4.22)|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.
Letf:[0,1]→[0,∞) be such that
(4.23)Bn(M)(f)(x)=max{fj,n,j(x),fj-1,n,j(x)}∀x∈[jn+1,j+1n+1].
Then
(4.24)|Bn(M)(f)(x)-f(x)|≤2ω1(f;1n),∀x∈[jn+1,j+1n+1].Proof.
We distinguish the two following cases:Case  (i).
Bn(M)(f)(x)=fj,n,j(x), when as in Lemma 4.2 we get
(4.25)|Bn(M)(f)(x)-f(x)|≤ω1(f;1n+1),Case  (ii).
Bn(M)(f)(x)=fj-1,n,j(x), when we have two subcases:(a)
Bn(M)(f)(x)≤f(x), when as in the case of Lemma 4.2 we obtain
(4.26)|Bn(M)(f)(x)-f(x)|≤ω1(f;1n+1).(b)
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
(4.27)0≤x-j-1n≤j+1n+1-j-1n=-jn(n+1)+1n+1+1n<2n,
we obtain
(4.28)|Bn(M)(f)(x)-f(x)|≤2ω1(f;1n),
which proves the lemma.Lemma 4.4.
Letf:[0,1]→[0,∞) be such that
(4.29)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
(4.30)|Bn(M)(f)(x)-f(x)|≤2ω1(f;1n),∀x∈[j(n+1),(j+1)(n+1)].Proof.
Letx∈[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
(4.31)|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
(4.32)|Bn(M)(f)(x)-f(x)|≤2ω1(f;1n),
which ends the proof.Lemma 4.5.
Letf:[0,1]→[0,∞) be concave. Then the following two properties hold: (i)
the functiong:(0,1]→[0,∞),g(x)=f(x)/x is nonincreasing;(ii)
the functionh:[0,1)→[0,∞),h(x)=f(x)/(1-x) is nondecreasing.Proof.
(i) Letx,y∈(0,1] be with x≤y. Then
(4.33)f(x)=f(xyy+y-xy0)≥xyf(y)+y-xyf(0)≥xyf(y),
which implies that f(x)/x≥f(y)/y.
(ii) Letx,y∈[0,1) be with x≥y. Then
(4.34)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.
Letf:[0,1]→[0,∞) be a concave function. Then
(4.35)|Bn(M)(f)(x)-f(x)|≤2ω1(f;1n),∀x∈[0,1].Proof.
Letx∈[0,1] and j∈{0,1,…n} such that x∈[j/(n+1),(j+1)/(n+1)]. Let k∈{0,1,…n} be with k≥j. Then
(4.36)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)/n≤f(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
(4.37)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 k≥j+1, it follows that fk,n,j(x)≥fk+1,n,j(x). Thus we obtain
(4.38)fj+1,n,j(x)≥fj+2,n,j(x)≥⋯≥fn,j,n(x).
Now letk∈{0,1,…n} be with k≤j. Then
(4.39)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
(4.40)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 k≤j-1 it is immediate that fk,n,j(x)≥fk-1,n,j(x), which implies
(4.42)fj-1,n.j(x)≥fj-2,n.j(x)≥⋯≥f0,n,j(x).
From (4.38) and (4.42), we obtain
(4.41)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
(4.43)|Bn(M)(f)(x)-f(x)|≤2ω1(f;1n),
and proves the corollary.Corollary 4.7.
(i) Iff:[0,1]→[0,∞) is nondecreasing and such that the function g:(0,1]→[0,∞),g(x)=f(x)/x is nonincreasing, then
(4.44)|Bn(M)(f)(x)-f(x)|≤2ω1(f;1n),∀x∈[0,1].
(ii) Iff:[0,1]→[0,∞) is nonincreasing and such that the function h:[0,1)→[0,∞),h(x)=f(x)/(1-x) is nondecreasing, then
(4.45)|Bn(M)(f)(x)-f(x)|≤2ω1(f;1n),∀x∈[0,1].Proof.
(i) Sincef is nondecreasing it follows (see the proof of Theorem 5.5 in Section 5) that
(4.46)Bn(M)(f)(x)=⋁k≥jnfk,n,j(x),∀x∈[jn+1,j+1n+1].
Following the proof of Corollary 4.6, we get
(4.47)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
(4.48)|Bn(M)(f)(x)-f(x)|≤2ω1(f;1n).
(ii) Sincef is nonincreasing, it follows (see the proof of Corollary 5.6 in Section 5) that
(4.49)Bn(M)(f)(x)=⋁k≥0jfk,n,j(x),∀x∈[jn+1,j+1n+1].
Following the proof of Corollary 4.6, we get
(4.50)Bn(M)(f)(x)=max{fj-1,n,j(x),fj,n,j(x)},
and from Lemma 4.3 we obtain
(4.51)|Bn(M)(f)(x)-f(x)|≤2ω1(f;1n).Remark 4.8.
By simple reasonings, it follows that iff:[0,1]→[0,∞) is a convex, nondecreasing function satisfying f(x)/x≥f(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 f∈C1[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)/x≥f(1) can be written as (f(1)-f(x))/(1-x)≤f(1), for all x∈[0,1), passing to limit with x→1, it follows f′(1)≤f(1), which implies (since f′ is nondecreasing)
(4.52)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 isf(x)=ex, x∈[0,1].Analogously, iff:[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 functionf:[0,1]→ℝ+, 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.
Sincepn,k(x)>0 for all x∈(0,1), n∈ℕ, k∈{0,…,n}, it follows that the denominator ⋁k=0npn,k(x)>0 for all x∈(0,1) and n∈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 ofBn(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 Section4, for any k,j∈{0,1,…,n}, let us consider the functions fk,n,j:[j/(n+1),(j+1)/(n+1)]→ℝ,
(5.1)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
(5.2)Bn(M)(f)(x)=⋁k=0nfk,n,j(x).Lemma 5.3.
Iff:[0,1]→ℝ+ is a nondecreasing function, then for any k,j∈{0,1,…n},k≤j and x∈[j/(n+1),(j+1)/(n+1)], one has fk,n,j(x)≥fk-1,n,j(x).Proof.
Becausek≤j, 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
(5.3)mk,n,j(x)f(kn)≥mk-1,n,j(x)f(k-1n),
which proves the lemma.Corollary 5.4.
Iff:[0,1]→ℝ+ is nonincreasing, then fk,n,j(x)≥fk+1,n,j(x) for any k,j∈{0,1,…n},k≥j, and x∈[j/(n+1),(j+1)/(n+1)].Proof.
Becausek≥j, 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
(5.4)mk,n,j(x)f(kn)≥mk+1,n,j(x)f(k+1n),
which proves the corollary.Theorem 5.5.
Iff:[0,1]→ℝ+ is nondecreasing, then Bn(M)(f) is nondecreasing.Proof.
BecauseBn(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 letj∈{0,1,…n} and x∈[j/(n+1),(j+1)/(n+1)]. Because f is nondecreasing, from Lemma 5.3 it follows that
(5.5)fj,n,j(x)≥fj-1,n,j(x)≥fj-2,n,j(x)≥⋯≥f0,n,j(x),
but then it is immediate that
(5.6)Bn(M)(f)(x)=⋁k≥jnfk,n,j(x),
for all x∈[j/(n+1),(j+1)/(n+1)]. Clearly that for k≥j, 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.
Iff:[0,1]→ℝ+ is nonincreasing, then Bn(M)(f) is nonincreasing.Proof.
BecauseBn(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 letj∈{0,1,…n} and x∈[j/(n+1),(j+1)/(n+1)]. Because f is nonincreasing, from Corollary 5.4, it follows that
(5.7)fj,n,j(x)≥fj+1,n,j(x)≥fj+2,n,j(x)≥⋯≥fn,n,j(x),
but then it is immediate that
(5.8)Bn(M)(f)(x)=⋁k≥0jfk,n,j(x),
for all x∈[j/(n+1),(j+1)/(n+1)]. Clearly that for k≤j 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.
Letf:[0,1]→ℝ be continuous on [0,1]. One says that the function f:[0,1]→ℝ is quasiconvex on [0,1] if it satisfies the inequality
(5.9)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 [5], 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.
Iff:[0,1]→ℝ+ is continuous and quasiconvex on [0,1], then for all n∈ℕ, Bn(M)(f) is quasiconvex on [0,1].Proof.
Iff 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∈ℕ, Bn(M)(f) is nonincreasing (or nondecreasing) on [0,1].
Suppose now that there existsc∈(0,1), such that f is nonincreasing on [0,c] and nondecreasing on [c,1]. Define the functions F,G:[0,1]→ℝ+ 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 thatF 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 Remark2.2 after the proof of Lemma 2.1) that
(5.10)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 havemax{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 pointc′∈[0,1] such that Bn(M)(f)(x) is nonincreasing on [0,c′] and nondecreasing on [c′,1], which by the result in [5] 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 [6].It is of interest to exactly calculateBn(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 allx∈[0,1] and n∈ℕ, one has Bn(M)(e0)(x)=1 and
(5.11)Bn(M)(e1)(x)=x·pn-1,0(x)pn,0(x)=x1-x,ifx∈[0,1n+1],Bn(M)(e1)(x)=x·pn-1,0(x)pn,1(x)=1n,ifx∈[1n+1,1n],Bn(M)(e1)(x)=x·pn-1,1(x)pn,1(x)=x1-x·n-1n,ifx∈[1n,2n+1],Bn(M)(e1)(x)=x·pn-1,1(x)pn,2(x)=2n,ifx∈[2n+1,2n],Bn(M)(e1)(x)=x·pn-1,2(x)pn,2(x)=x1-x·n-2n,ifx∈[2n,3n+1],Bn(M)(e1)(x)=x·pn-1,2(x)pn,3(x)=3n,ifx∈[3n+1,3n],
and so on, in general one has
(5.12)Bn(M)(e1)(x)=x1-x·n-jn,ifx∈[jn,j+1n+1],Bn(M)(e1)(x)=j+1n,ifx∈[j+1n+1,(j+1)n],
for j∈{0,1,…,n-1}.Proof.
The formulaBn(M)(e0)(x)=1 is immediate by the definition of Bn(M)(f)(x).
To find the formula forBn(M)(e1)(x), we will use the explicit formula in Lemma 3.4 which says that
(5.13)⋁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
(5.14)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](5.15)0<1n+1≤1n≤2n+1≤2n≤3n+1≤3n≤4n+1≤4n⋯.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 functionf, the max-product Bernstein operator Bn(M)(f) is piecewise convex on [0,1]. We present the following.Theorem 5.12.
For any functionf:[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 anyk,j∈{0,1,…,n}, let us consider the functions fk,n,j:[j/(n+1),(j+1)/(n+1)]→ℝ,(5.16)fk,n,j(x)=mk,n,j(x)f(kn)=(nk)(nj)(x1-x)k-jf(kn).
Clearly, we have
(5.17)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 fixedj, 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)].
Sincef≥0, it suffices to prove that the functions gk,j:[0,1]→ℝ, gk,j(x)=(x/(1-x))k-j are convex on [j/(n+1),(j+1)/(n+1)].
Fork=j, gj,j is constant so is convex.
Fork=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)].
Fork=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)].
Ifk≥j+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)].
Ifk≤j-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 functionsgk,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 althoughBn(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.
Forf belonging to the set
(5.18)S[0,1]={f:[0,1]→ℝ;f∈C1[0,1],f(0)=0,fisnondecreasingon[0,1]},
let us define the following subadditive and positive homogenous operators (as function of f):
(5.19)Ln(f)(x)=∫0xBn(M)(f′)(t)dt,x∈[0,1],n∈ℕ.
If f∈S[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, sincef 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 f′≥0 on [0,1] and by the relationship Ln′(f)(x)=Bn(M)(f′)(x)≥0 for all x∈[0,1].
Also, writingf(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)=sinx≥0, f′′(x)=cosx≥0 and f′′′(x)=-sinx≤0, 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 operatorBn(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])(6.1)∥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(6.2)ω2φ(f;δ)=sup{sup{|f(x+hφ(x))-2f(x)+f(x-hφ(x))|;x∈Ih},h∈[0,δ]},
with φ(x)=x(1-x), δ≤1 and Ih=[h2/(1+h2),1/(1+h2)].Now, iff is, for example, a nondecreasing concave polygonal line on [0,1], then by simple reasonings we get that ω2φ(f;δ)~δ for δ≤1, 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 forBn(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 Section5 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.



---

*Source: 590589-2009-12-09.xml*