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

Abstract

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):

𝐵_𝑛^(𝑀)⋁(𝑓)(𝑥)=^𝑛𝑘=0𝑝_𝑛,𝑘(𝑥)𝑓(𝑘/𝑛)⋁^𝑛_𝑘=0𝑝_𝑛,𝑘(𝑥), (1.1) where 𝑝_𝑛,𝑘(𝑥)=(^𝑛_𝑘)𝑥^𝑘(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 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, ℝ₊, we consider the operations ∨ (maximum) and “⋅”product. Then (ℝ₊,∨,⋅) has a semiring structure and we call it as Max-Product algebra.

Let 𝐼⊂ℝ be a bounded or unbounded interval, and

𝐶𝐵₊(𝐼)={𝑓∶𝐼→ℝ₊;𝑓continuousandboundedon𝐼}. (2.1) The general form of 𝐿_𝑛∶𝐶𝐵₊(𝐼)→𝐶𝐵₊(𝐼), (called here a discrete max-product-type approximation operator) studied in the paper will be

𝐿_𝑛(𝑓)(𝑥)=^𝑛_𝑖=0𝐾_𝑛(𝑥,𝑥_𝑖)⋅𝑓(𝑥_𝑖), (2.2) or

𝐿_𝑛(𝑓)(𝑥)=^∞_𝑖=0𝐾_𝑛(𝑥,𝑥_𝑖)⋅𝑓(𝑥_𝑖), (2.3) where 𝑛∈ℕ,𝑓∈𝐶𝐵₊(𝐼),𝐾_𝑛(⋅,𝑥_𝑖)∈𝐶𝐵₊(𝐼), and 𝑥_𝑖∈𝐼, for all 𝑖. These operators are nonlinear, positive operators and moreover they satisfy a pseudolinearity condition of the form

𝐿_𝑛(𝛼⋅𝑓∨𝛽⋅𝑔)(𝑥)=𝛼⋅𝐿_𝑛(𝑓)(𝑥)∨𝛽⋅𝐿_𝑛(𝑔)(𝑥),∀𝛼,𝛽∈ℝ₊,𝑓,𝑔∶𝐼→ℝ₊. (2.4)

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 [4]). Let 𝐼⊂ℝ be a bounded or unbounded interval: 𝐶𝐵₊(𝐼)={𝑓∶𝐼→ℝ₊;𝑓continuousandboundedon𝐼}, (2.5) and let 𝐿_𝑛∶𝐶𝐵₊(𝐼)→𝐶𝐵₊(𝐼),𝑛∈ℕ, be a sequence of operators satisfying the following properties: (i)if 𝑓,𝑔∈𝐶𝐵₊(𝐼) satisfy 𝑓≤𝑔, then 𝐿_𝑛(𝑓)≤𝐿_𝑛(𝑔) for all 𝑛∈𝑁; (ii)𝐿_𝑛(𝑓+𝑔)≤𝐿_𝑛(𝑓)+𝐿_𝑛(𝑔) for all 𝑓,𝑔∈𝐶𝐵₊(𝐼).
Then for all 𝑓,𝑔∈𝐶𝐵₊(𝐼),𝑛∈𝑁, and 𝑥∈𝐼, we have
|𝐿_𝑛(𝑓)(𝑥)−𝐿_𝑛(𝑔)(𝑥)|≤𝐿_𝑛(|𝑓−𝑔|)(𝑥). (2.6)

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 𝐿_𝑛(𝑓)(𝑥)≤𝐿_𝑛(|𝑓−𝑔|)(𝑥)+𝐿_𝑛(𝑔)(𝑥), that is, 𝐿_𝑛(𝑓)(𝑥)−𝐿_𝑛(𝑔)(𝑥)≤𝐿_𝑛(|𝑓−𝑔|)(𝑥).
Writing now 𝑔=𝑔−𝑓+𝑓≤|𝑓−𝑔|+𝑓 and applying the above reasonings, it follows that 𝐿_𝑛(𝑔)(𝑥)−𝐿_𝑛(𝑓)(𝑥)≤𝐿_𝑛(|𝑓−𝑔|)(𝑥), which, combined with the above inequality, gives |𝐿_𝑛(𝑓)(𝑥)−𝐿_𝑛(𝑔)(𝑥)|≤𝐿_𝑛(|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: 𝐿_𝑛(𝑓∨𝑔)(𝑥)=𝐿_𝑛(𝑓)(𝑥)∨𝐿_𝑛(𝑔)(𝑥),𝑓,𝑔∈𝐶𝐵₊(𝐼). (2.7) Indeed, taking in the above equality 𝑓≤𝑔, 𝑓,𝑔∈𝐶𝐵₊(𝐼), it easily follows that 𝐿_𝑛(𝑓)(𝑥)≤𝐿_𝑛(𝑔)(𝑥).
(2) In addition, it is immediate that the Bernstein max-product operator is positive homogenous, that is, 𝐿_𝑛(𝜆𝑓)=𝜆𝐿_𝑛(𝑓) for all 𝜆≥0.

Corollary 2.3 (see [4]). Let 𝐿_𝑛∶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 |𝑓(𝑥)−𝐿_𝑛1(𝑓)(𝑥)|≤[𝛿𝐿_𝑛(𝜑_𝑥)(𝑥)+𝐿_𝑛(𝑒₀)(𝑥)]𝜔₁(𝑓;𝛿)_𝐼+𝑓(𝑥)⋅|𝐿_𝑛(𝑒₀)(𝑥)−1|, (2.8) where 𝛿>0,𝑒₀(𝑡)=1 for all 𝑡∈𝐼, 𝜑_𝑥(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 𝐿_𝑛(𝜑_𝑥)(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 𝐿_𝑛(𝑓)(𝑥)−𝑓(𝑥)=[𝐿_𝑛(𝑓)(𝑥)−𝑓(𝑥)⋅𝐿_𝑛(𝑒₀)(𝑥)]+𝑓(𝑥)[𝐿_𝑛(𝑒₀)(𝑥)−1], (2.9) it follows (by the positive homogeneity and by Lemma 2.1) that |𝑓(𝑥)−𝐿_𝑛(𝑓)(𝑥)|≤|𝐿_𝑛(𝑓(𝑥))(𝑥)−𝐿_𝑛(𝑓(𝑡))(𝑥)|+|𝑓(𝑥)|⋅|𝐿_𝑛(𝑒₀)(𝑥)−1|≤𝐿_𝑛(|𝑓(𝑡)−𝑓(𝑥)|)(𝑥)+|𝑓(𝑥)|⋅|𝐿_𝑛(𝑒₀)(𝑥)−1|. (2.10) Now, since for all 𝑡,𝑥∈𝐼, we have |𝑓(𝑡)−𝑓(𝑥)|≤𝜔₁(𝑓;|𝑡−𝑥|)_𝐼1≤[𝛿|𝑡−𝑥|+1]𝜔₁(𝑓;𝛿)_𝐼, (2.11) 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 [4]). Suppose that in addition to the conditions in Corollary 2.3, the sequence (𝐿_𝑛)_𝑛 satisfies 𝐿_𝑛(𝑒₀)=𝑒₀, for all 𝑛∈𝑁. Then for all 𝑓∈𝐶𝐵₊(𝐼), 𝑛∈𝑁, and 𝑥∈𝐼, one has |𝑓(𝑥)−𝐿_𝑛1(𝑓)(𝑥)|≤[1+𝛿𝐿_𝑛(𝜑_𝑥)(𝑥)]𝜔₁(𝑓;𝛿)_𝐼. (2.12)

3. Auxiliary Results

Since it is easy to check that 𝐵_𝑛^(𝑀)(𝑓)(0)−𝑓(0)=𝐵_𝑛^(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 each 𝑘,𝑗∈{0,1,2,…𝑛} and 𝑥∈[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)], let us denote

𝑀_{𝑘,𝑛,𝑗}𝑝(𝑥)=^𝑛,𝑘(𝑥)|𝑘/𝑛−𝑥|𝑝_𝑛,𝑗(𝑥),𝑚_{𝑘,𝑛,𝑗}𝑝(𝑥)=^𝑛,𝑘(𝑥)𝑝_𝑛,𝑗.(𝑥) (3.1) It is clear that if 𝑘≥𝑗+1, then

𝑀_{𝑘,𝑛,𝑗}𝑝(𝑥)=^𝑛,𝑘(𝑥)(𝑘/𝑛−𝑥)𝑝_𝑛,𝑗,(𝑥) (3.2) and if 𝑘≤𝑗−1, then

𝑀_{𝑘,𝑛,𝑗}𝑝(𝑥)=^𝑛,𝑘(𝑥)(𝑥−𝑘/𝑛)𝑝_𝑛,𝑗.(𝑥) (3.3)

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

𝑀_{𝑘,𝑛,𝑗}𝑝(𝑥)=^𝑛,𝑘(𝑥)(𝑘/(𝑛+1)−𝑥)𝑝_𝑛,𝑗(𝑥), (3.4) and for each 𝑘,𝑗∈{0,1,2,…𝑛}, 𝑘≤𝑗−2, and 𝑥∈[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)] let us denote

𝑀_{𝑘,𝑛,𝑗}𝑝(𝑥)=^𝑛,𝑘(𝑥)(𝑥−𝑘/(𝑛+1))𝑝_𝑛,𝑗.(𝑥) (3.5)

Lemma 3.1. Let 𝑥∈[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)]: (i)for all 𝑘,𝑗∈{0,1,2,…𝑛}, 𝑘≥𝑗+2, one has 𝑀_{𝑘,𝑛,𝑗}(𝑥)≤𝑀_{𝑘,𝑛,𝑗}(𝑥)≤3𝑀_{𝑘,𝑛,𝑗}(𝑥); (3.6)(ii)for all 𝑘,𝑗∈{0,1,2,…𝑛}, 𝑘≤𝑗−2, one has 𝑀_{𝑘,𝑛,𝑗}(𝑥)≤𝑀_{𝑘,𝑛,𝑗}(𝑥)≤6𝑀_{𝑘,𝑛,𝑗}(𝑥). (3.7)

Proof. (i) The inequality 𝑀_{𝑘,𝑛,𝑗}(𝑥)≤𝑀_{𝑘,𝑛,𝑗}(𝑥) is immediate.
On the other hand,
𝑀_{𝑘,𝑛,𝑗}(𝑥)𝑀_{𝑘,𝑛,𝑗}=(𝑥)𝑘/𝑛−𝑥≤𝑘/(𝑛+1)−𝑥𝑘/𝑛−𝑗/(𝑛+1)≤𝑘/(𝑛+1)−(𝑗+1)/(𝑛+1)𝑘𝑛+𝑘−𝑛𝑗=𝑛(𝑘−𝑗−1)𝑘−𝑗+𝑘𝑘−𝑗−1𝑛(𝑘−𝑗−1)≤3, (3.8) which proves (i).
(ii) The inequality 𝑀_{𝑘,𝑛,𝑗}(𝑥)≤𝑀_{𝑘,𝑛,𝑗}(𝑥) is immediate.
On the other hand,
𝑀_{𝑘,𝑛,𝑗}(𝑥)𝑀_{𝑘,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−𝑛𝑛+1𝑛⋅𝑗+1−𝑘𝑗−𝑘−1≤2⋅𝑗+1−𝑘2𝑗−𝑘−1=2(1+𝑗−𝑘−1)≤6, (3.9) 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 𝑚_{𝑘,𝑛,𝑗}(𝑥)≤1. (3.10)

Proof. We have two cases: 1) 𝑘≥𝑗, 2) 𝑘≤𝑗.Case 1. Since clearly the function ℎ(𝑥)=(1−𝑥)/𝑥 is nonincreasing on [𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)], it follows that 𝑚_{𝑘,𝑛,𝑗}(𝑥)𝑚_{𝑘+1,𝑛,𝑗}=(𝑥)𝑘+1⋅𝑛−𝑘1−𝑥𝑥≥𝑘+1⋅𝑛−𝑘1−(𝑗+1)/(𝑛+1)=(𝑗+1)/(𝑛+1)𝑘+1⋅𝑛−𝑘𝑛−𝑗𝑗+1≥1, (3.11) which implies 𝑚_{𝑗,𝑛,𝑗}(𝑥)≥𝑚_{𝑗+1,𝑛,𝑗}(𝑥)≥𝑚_{𝑗+2,𝑛,𝑗}(𝑥)≥⋯≥𝑚_{𝑛,𝑛,𝑗}(𝑥).Case 2. We get 𝑚_{𝑘,𝑛,𝑗}(𝑥)𝑚_{𝑘−1,𝑛,𝑗}=(𝑥)𝑛−𝑘+1𝑘⋅𝑥≥1−𝑥𝑛−𝑘+1𝑘⋅𝑗/(𝑛+1)=1−𝑗/(𝑛+1)𝑛−𝑘+1𝑘⋅𝑗𝑛+1−𝑗≥1, (3.12) which immediately implies that 𝑚_{𝑗,𝑛,𝑗}(𝑥)≥𝑚_{𝑗−1,𝑛,𝑗}(𝑥)≥𝑚_{𝑗−2,𝑛,𝑗}(𝑥)≥⋯≥𝑚_0,𝑛,𝑗(𝑥). (3.13) Since 𝑚_{𝑗,𝑛,𝑗}(𝑥)=1, the conclusion of the lemma is immediate.

Lemma 3.3. Let 𝑥∈[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)]. (i)If 𝑘∈{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)If 𝑘∈{1,2,…𝑗−2} is such that 𝑘+√𝑘≤𝑗, then 𝑀_{𝑘,𝑛,𝑗}(𝑥)≥𝑀_{𝑘−1,𝑛,𝑗}(𝑥).

Proof. (i) We observe that 𝑀_{𝑘,𝑛,𝑗}(𝑥)𝑀_{𝑘+1,𝑛,𝑗}=(𝑥)𝑘+1⋅𝑛−𝑘1−𝑥𝑥⋅𝑘/(𝑛+1)−𝑥(𝑘+1)/(𝑛+1)−𝑥. (3.14) 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 𝑀_{𝑘,𝑛,𝑗}(𝑥)𝑀_{𝑘+1,𝑛,𝑗}≥(𝑥)𝑘+1⋅𝑛−𝑘𝑛−𝑗⋅𝑗+1𝑘−𝑗−1𝑘−𝑗≥1. (3.15)
 (ii) We observe that
𝑀_{𝑘,𝑛,𝑗}(𝑥)𝑀_{𝑘−1,𝑛,𝑗}=(𝑥)𝑛−𝑘+1𝑘⋅𝑥⋅1−𝑥𝑥−𝑘/(𝑛+1).𝑥−(𝑘−1)/(𝑛+1) (3.16) 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 𝑀_{𝑘,𝑛,𝑗}(𝑥)𝑀_{𝑘−1,𝑛,𝑗}≥(𝑥)𝑛−𝑘+1𝑘⋅𝑗⋅𝑛+1−𝑗𝑗−𝑘𝑗−𝑘+1≥1, (3.17) which proves the lemma.

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

Lemma 3.4. One has ^𝑛_𝑘=0𝑝_𝑛,𝑘(𝑥)=𝑝_𝑛,𝑗𝑗(𝑥),∀𝑥∈[,𝑛+1𝑗+1𝑛+1],𝑗=0,1,…,𝑛, (3.18) where 𝑝_𝑛,𝑘(𝑥)=(^𝑛_𝑘)𝑥^𝑘(1−𝑥)^𝑛−𝑘.

Proof. First we show that for fixed 𝑛∈ℕ and 0≤𝑘<𝑘+1≤𝑛, we have 0≤𝑝_𝑛,𝑘+1(𝑥)≤𝑝_𝑛,𝑘(𝑥),iff𝑥∈[0,𝑘+1𝑛+1]. (3.19) Indeed, the inequality one reduces to 𝑛0≤𝑥𝑘+1^𝑘+1(1−𝑥)^{𝑛−(𝑘+1)}≤𝑛𝑘𝑥^𝑘(1−𝑥)^𝑛−𝑘 (3.20) after simplifications is equivalent to 𝑛+𝑛𝑘𝑛𝑘0≤𝑥[𝑘+1]≤. (3.21) However, since (^𝑛_𝑘+1)+(^𝑛_𝑘)=(^𝑛+1_𝑘+1), the above inequality immediately becomes equivalent to 0≤𝑥≤𝑘+1.𝑛+1 (3.22) By taking 𝑘=0,1,…, in the inequality just proved above, we get 𝑝_𝑛,1(𝑥)≤𝑝_𝑛,01(𝑥),iff𝑥∈[0,𝑝𝑛+1],_𝑛,2(𝑥)≤𝑝_𝑛,12(𝑥),iff𝑥∈[0,𝑝𝑛+1],_𝑛,3(𝑥)≤𝑝_𝑛,23(𝑥),iff𝑥∈[0,𝑛+1], (3.23) and so on, 𝑝_𝑛,𝑘+1(𝑥)≤𝑝_𝑛,𝑘(𝑥),iff𝑥∈[0,𝑘+1𝑛+1], (3.24) and so on, 𝑝_{𝑛,𝑛−2}(𝑥)≤𝑝_{𝑛,𝑛−3}(𝑥),iff𝑥∈[0,𝑛−2𝑝𝑛+1],_{𝑛,𝑛−1}(𝑥)≤𝑝_{𝑛,𝑛−2}(𝑥),iff𝑥∈[0,𝑛−1𝑝𝑛+1],_𝑛,𝑛(𝑥)≤𝑝_{𝑛,𝑛−1}𝑛(𝑥),iff𝑥∈[0,𝑛+1]. (3.25) From all these inequalities, reasoning by recurrence we easily obtain 1if𝑥∈[0,𝑛+1],then𝑝_𝑛,𝑘(𝑥)≤𝑝_𝑛,01(𝑥),∀𝑘=0,1,…,𝑛,if𝑥∈[,2𝑛+1𝑛+1],then𝑝_𝑛,𝑘(𝑥)≤𝑝_𝑛,12(𝑥),∀𝑘=0,1,…,𝑛,if𝑥∈[,3𝑛+1𝑛+1],then𝑝_𝑛,𝑘(𝑥)≤𝑝_𝑛,2(𝑥),∀𝑘=0,1,…,𝑛, (3.26) and so on, finally 𝑛if𝑥∈[𝑛+1,1],then𝑝_𝑛,𝑘(𝑥)≤𝑝_𝑛,𝑛(𝑥),∀𝑘=0,1,…,𝑛, (3.27) which proves the lemma.

4. Approximation Results

If 𝐵_𝑛^(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]→ℝ₊ is continuous, then one has the estimate |𝐵_𝑛^(𝑀)(𝑓)(𝑥)−𝑓(𝑥)|≤12𝜔₁1(𝑓;√𝑛+1),∀𝑛∈ℕ,𝑥∈[0,1], (4.1) where 𝜔₁(𝑓;𝛿)=sup{|𝑓(𝑥)−𝑓(𝑦)|;𝑥,𝑦∈[0,1],|𝑥−𝑦|≤𝛿}. (4.2)

Proof. It is easy to check that the max-product Bernstein operators fulfill the conditions in Corollary 2.4 and we have |𝐵_𝑛^(𝑀)1(𝑓)(𝑥)−𝑓(𝑥)|≤(1+𝛿_𝑛𝐵_𝑛^(𝑀)(𝜑_𝑥)(𝑥))𝜔₁(𝑓;𝛿_𝑛), (4.3) where 𝜑_𝑥(𝑡)=|𝑡−𝑥|. So, it is enough to estimate 𝐸_𝑛(𝑥)∶=𝐵_𝑛^(𝑀)(𝜑_𝑥⋁)(𝑥)=^𝑛𝑘=0𝑝_𝑛,𝑘(𝑥)|𝑘/𝑛−𝑥|⋁^𝑛_𝑘=0𝑝_𝑛,𝑘(𝑥). (4.4)
Let 𝑥∈[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)], where 𝑗∈{0,…,𝑛} is fixed, arbitrary. By Lemma 3.4 we easily obtain
𝐸_𝑛(𝑥)=max_{𝑘=0,…,𝑛}{𝑀_{𝑘,𝑛,𝑗}𝑗(𝑥)},𝑥∈[,𝑛+1𝑗+1𝑛+1]. (4.5) In all what follows we may suppose that 𝑗∈{1,…,𝑛}, because for 𝑗=0, simple calculation shows that in this case, we get 𝐸_𝑛(𝑥)≤1/𝑛, for all 𝑥∈[0,1/(𝑛+1)]. So it remains to obtain an upper estimate for each 𝑀_{𝑘,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 𝑀_{𝑘,𝑛,𝑗}6(𝑥)≤√𝑗𝑛+1,∀𝑥∈[,𝑛+1𝑗+1𝑛+1],𝑘=0,…,𝑛, (4.6) which immediately will implies that 𝐸_𝑛6(𝑥)≤√𝑛+1,∀𝑥∈[0,1],𝑛∈ℕ, (4.7) and taking 𝛿_𝑛=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. If 𝑘=𝑗, then 𝑀_{𝑗,𝑛,𝑗}(𝑥)=|𝑗/𝑛−𝑥|. Since 𝑥∈[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)], it easily follows that 𝑀_{𝑗,𝑛,𝑗}(𝑥)≤1/(𝑛+1).
If 𝑘=𝑗+1, then 𝑀_{𝑗+1,𝑛,𝑗}(𝑥)=𝑚_{𝑗+1,𝑛,𝑗}(𝑥)((𝑗+1)/𝑛−𝑥). Since by Lemma 3.2 we have 𝑚_{𝑗+1,𝑛,𝑗}(𝑥)≤1, we obtain 𝑀_{𝑗+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 𝑀_{𝑗−1,𝑛,𝑗}(𝑥)=𝑚_{𝑗−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)=𝑚_{𝑘,𝑛,𝑗}𝑘(𝑥)(𝑘𝑛+1−𝑥)≤𝑘𝑛+1−𝑥≤−𝑗𝑛+1≤𝑘𝑛+1−√𝑛+1𝑘−𝑘+1=√𝑛+1𝑘+1≤1𝑛+1√.𝑛+1 (4.8)Subcase b
Suppose now that 𝑘−√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 𝑀_{𝑘+1,𝑛,𝑗}(𝑥)=𝑚_{𝑘+1,𝑛,𝑗}(𝑥)(𝑘+1𝑛+1−𝑥)≤𝑘+1≤𝑛+1−𝑥𝑘+1−𝑗𝑛+1≤𝑛+1𝑘+1−𝑛+1𝑘−𝑘+1=𝑛+1𝑘+1+1≤2𝑛+1√.𝑛+1 (4.9) 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 Lemma 3.1, (i) too, we get 𝑀_{𝑘,𝑛,𝑗}(𝑥)≤6/√𝑛+1.
Case 3. Subcase a
Suppose first that 𝑘+√𝑘≥𝑗. Then we obtain 𝑀_{𝑘,𝑛,𝑗}(𝑥)=𝑚_{𝑘,𝑛,𝑗}𝑘(𝑥)(𝑥−𝑛+1)≤𝑗+1−𝑘𝑛+1≤√𝑛+1𝑘+𝑘+1−𝑘𝑛+1=√𝑛+1𝑘+1≤√𝑛+1𝑛+1≤2𝑛+1√.𝑛+1 (4.10)Subcase b
Suppose now that 𝑘+√𝑘<𝑗. Let 𝑘∈{0,1,2,…𝑛} be the minimum value such that 𝑘+𝑘≥𝑗. Then 𝑘₂=𝑘−1 satisfies 𝑘₂+𝑘₂<𝑗 and 𝑀_{𝑘−1,𝑛,𝑗}(𝑥)=𝑚_{𝑘−1,𝑛,𝑗}(𝑥)(𝑥−𝑘−1𝑛+1)≤𝑗+1−𝑛+1𝑘−1≤𝑛+1𝑘+𝑘+1−𝑛+1𝑘−1=𝑛+1𝑘+2≤3𝑛+1√.𝑛+1 (4.11) Also, because in this case we have 𝑗≥2, it is immediate that 𝑘₂≤𝑗−2. By Lemma 3.3, (ii), it follows that 𝑀_{𝑘−1,𝑛,𝑗}(𝑥)≥𝑀_{𝑘−2,𝑛,𝑗}(𝑥)≥⋯≥𝑀_0,𝑛,𝑗(𝑥). We obtain 𝑀_{𝑘,𝑛,𝑗}(𝑥)≤3/√𝑛+1 for any 𝑘≤𝑗−2 and 𝑥∈[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)].
In both subcases, by Lemma 3.1, (ii) too, we get 𝑀_{𝑘,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 max_𝑥∈[0,1]{𝐸_𝑛(𝑥)} is exactly 1/√𝑛 (here 𝐸_𝑛(𝑥) is defined in the proof of Theorem 4.1). Indeed, for 𝑛∈ℕ, let us take 𝑗_𝑛=[𝑛/2],𝑘_𝑛=𝑗_𝑛+[√𝑛], 𝑥_𝑛=(𝑗_𝑛+1)/(𝑛+1) and denote 𝑛=𝑛−[𝑛/2]. Then, we can write 𝑀_{𝑘𝑛,𝑛,𝑗𝑛}(𝑥_𝑛()=^𝑛𝑘𝑛)𝑥^𝑘𝑛_𝑛(1−𝑥_𝑛)^{𝑛−𝑘𝑛}(^𝑛_𝑗𝑛)𝑥^𝑗𝑛_𝑛(1−𝑥_𝑛)^{𝑛−𝑗𝑛}(𝑘_𝑛𝑛+1−𝑥_𝑛)=(𝑛−[√𝑛]+1)(𝑛−[√𝑛]+2)⋯𝑛([𝑛/2]+1)([𝑛/2]+2)⋯([𝑛/2]+[√(𝑛])[𝑛/2]+1𝑛)^[√𝑛]⋅[√𝑛]−1.𝑛+1 (4.12)
Since 2[𝑛/2]≥𝑛−1, we easily get [𝑛/2]+1≥𝑛, which implies (([𝑛/2]+1)/𝑛)^[√𝑛]≥1 for all 𝑛∈ℕ. On the other hand,
(𝑛−[√𝑛]+1)(𝑛−[√𝑛]+2)⋯𝑛([𝑛/2]+1)([𝑛/2]+2)⋯([𝑛/2]+[√√𝑛])≥(𝑛−[𝑛]+1[𝑛/2]+[√)𝑛]^[√𝑛]√≥(𝑛/2−𝑛+1𝑛/2+√𝑛)^√𝑛. (4.13)
Because lim_𝑛→∞((𝑛/2−√𝑛+1)/(𝑛/2+√𝑛))^√𝑛=𝑒⁻⁴, there exists 𝑛₀∈ℕ such that
(𝑛−[√𝑛]+1)(𝑛−[√𝑛]+2)⋯𝑛([𝑛/2]+1)([𝑛/2]+2)⋯([𝑛/2]+[√𝑛])≥𝑒⁻⁵ (4.14) for all 𝑛≥𝑛₀. It follows 𝑀_{𝑘𝑛,𝑛,𝑗𝑛}(𝑥_𝑛𝑒)≥⁻⁵([√𝑛]−1)≥𝑒𝑛+16√⁻⁵𝑛, (4.15) 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 𝑀_{𝑘𝑛,𝑛,𝑗𝑛}(𝑥_𝑛)≥𝑒⁻⁵/6√𝑛, 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 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 𝑓_{𝑘,𝑛,𝑗}∶[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)]→ℝ∶

𝑓_{𝑘,𝑛,𝑗}(𝑥)=𝑚_{𝑘,𝑛,𝑗}𝑘(𝑥)𝑓(𝑛()=^𝑛𝑘)(^𝑛_𝑗)(𝑥)1−𝑥^𝑘−𝑗𝑘𝑓(𝑛). (4.16) Then it is clear that for any 𝑗∈{0,1,…,𝑛} and 𝑥∈[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)], we can write

𝐵_𝑛^(𝑀)(𝑓)(𝑥)=^𝑛_𝑘=0𝑓_{𝑘,𝑛,𝑗}(𝑥). (4.17) Also we need the following four auxiliary lemmas.

Lemma 4.2. Let 𝑓∶[0,1]→[0,∞) be such that 𝐵_𝑛^(𝑀)(𝑓)(𝑥)=max{𝑓_{𝑗,𝑛,𝑗}(𝑥),𝑓_{𝑗+1,𝑛,𝑗}𝑗(𝑥)}∀𝑥∈[,𝑛+1𝑗+1𝑛+1]. (4.18) Then |𝐵_𝑛^(𝑀)(𝑓)(𝑥)−𝑓(𝑥)|≤2𝜔₁1(𝑓;𝑛𝑗),∀𝑥∈[,𝑛+1𝑗+1𝑛+1], (4.19) 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).Let 𝑥∈[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)] be fixed such that 𝐵_𝑛^(𝑀)(𝑓)(𝑥)=𝑓_{𝑗,𝑛,𝑗}(𝑥). Because by simple calculation we have −1/(𝑛+1)≤𝑥−𝑗/𝑛≤1/(𝑛+1) and 𝑓_{𝑗,𝑛,𝑗}(𝑥)=𝑓(𝑗/𝑛), it follows that |𝐵_𝑛^(𝑀)(𝑓)(𝑥)−𝑓(𝑥)|≤𝜔₁1(𝑓;𝑛+1). (4.20)Case  (ii).Let 𝑥∈[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)] be such that 𝐵_𝑛^(𝑀)(𝑓)(𝑥)=𝑓_{𝑗+1,𝑛,𝑗}(𝑥). We have two subcases: (a)𝐵_𝑛^(𝑀)(𝑓)(𝑥)≤𝑓(𝑥), when evidently 𝑓_{𝑗,𝑛,𝑗}(𝑥)≤𝑓_{𝑗+1,𝑛,𝑗}(𝑥)≤𝑓(𝑥) and we immediately get |𝐵_𝑛^(𝑀)(𝑓)(𝑥)−𝑓(𝑥)|=|𝑓_{𝑗+1,𝑛,𝑗}(𝑥)−𝑓(𝑥)|=𝑓(𝑥)−𝑓_{𝑗+1,𝑛,𝑗}𝑗(𝑥)≤𝑓(𝑥)−𝑓(𝑛)≤𝜔₁1(𝑓;𝑛+1); (4.21)(b)𝐵_𝑛^(𝑀)(𝑓)(𝑥)>𝑓(𝑥), when |𝐵_𝑛^(𝑀)(𝑓)(𝑥)−𝑓(𝑥)|=𝑓_{𝑗+1,𝑛,𝑗}(𝑥)−𝑓(𝑥)=𝑚_{𝑗+1,𝑛,𝑗}(𝑥)𝑓(𝑗+1𝑛)−𝑓(𝑥)≤𝑓(𝑗+1𝑛)−𝑓(𝑥). (4.22) 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 𝐵_𝑛^(𝑀)(𝑓)(𝑥)=max{𝑓_{𝑗,𝑛,𝑗}(𝑥),𝑓_{𝑗−1,𝑛,𝑗}𝑗(𝑥)}∀𝑥∈[,𝑛+1𝑗+1𝑛+1]. (4.23) Then |𝐵_𝑛^(𝑀)(𝑓)(𝑥)−𝑓(𝑥)|≤2𝜔₁1(𝑓;𝑛𝑗),∀𝑥∈[,𝑛+1𝑗+1𝑛+1]. (4.24)

Proof. We distinguish the two following cases:Case  (i).𝐵_𝑛^(𝑀)(𝑓)(𝑥)=𝑓_{𝑗,𝑛,𝑗}(𝑥), when as in Lemma 4.2 we get |𝐵_𝑛^(𝑀)(𝑓)(𝑥)−𝑓(𝑥)|≤𝜔₁1(𝑓;𝑛+1), (4.25)Case  (ii).𝐵_𝑛^(𝑀)(𝑓)(𝑥)=𝑓_{𝑗−1,𝑛,𝑗}(𝑥), when we have two subcases:(a)𝐵_𝑛^(𝑀)(𝑓)(𝑥)≤𝑓(𝑥), when as in the case of Lemma 4.2 we obtain |𝐵_𝑛^(𝑀)(𝑓)(𝑥)−𝑓(𝑥)|≤𝜔₁1(𝑓;𝑛+1). (4.26)(b)𝐵_𝑛^(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 0≤𝑥−𝑗−1𝑛≤𝑗+1−𝑛+1𝑗−1𝑛=−𝑗+1𝑛(𝑛+1)+1𝑛+1𝑛<2𝑛, (4.27) we obtain |𝐵_𝑛^(𝑀)(𝑓)(𝑥)−𝑓(𝑥)|≤2𝜔₁1(𝑓;𝑛), (4.28) which proves the lemma.

Lemma 4.4. Let 𝑓∶[0,1]→[0,∞) be such that 𝐵_𝑛^(𝑀)(𝑓)(𝑥)=max{𝑓_{𝑗−1,𝑛,𝑗}(𝑥),𝑓_{𝑗,𝑛,𝑗}(𝑥),𝑓_{𝑗+1,𝑛,𝑗}(𝑥)}, (4.29) for all 𝑥∈[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)]. Then |𝐵_𝑛^(𝑀)(𝑓)(𝑥)−𝑓(𝑥)|≤2𝜔₁1(𝑓;𝑛𝑗),∀𝑥∈[,(𝑛+1)(𝑗+1)(𝑛+1)]. (4.30)

Proof. Let 𝑥∈[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)]. If 𝐵_𝑛^(𝑀)(𝑓)(𝑥)=𝑓_{𝑗,𝑛,𝑗}(𝑥) or 𝐵_𝑛^(𝑀)(𝑓)(𝑥)=𝑓_{𝑗+1,𝑛,𝑗}(𝑥), then 𝐵_𝑛^(𝑀)(𝑓)(𝑥)=max{𝑓_{𝑗,𝑛,𝑗}(𝑥),𝑓_{𝑗+1,𝑛,𝑗}(𝑥)} and from Lemma 4.2, it follows that |𝐵_𝑛^(𝑀)(𝑓)(𝑥)−𝑓(𝑥)|≤2𝜔₁1(𝑓;𝑛). (4.31) If 𝐵_𝑛^(𝑀)(𝑓)(𝑥)=𝑓_{𝑗−1,𝑛,𝑗}(𝑥), then 𝐵_𝑛^(𝑀)(𝑓)(𝑥)=max{𝑓_{𝑗,𝑛,𝑗}(𝑥),𝑓_{𝑗−1,𝑛,𝑗}(𝑥)} and from Lemma 4.3, we get |𝐵_𝑛^(𝑀)(𝑓)(𝑥)−𝑓(𝑥)|≤2𝜔₁1(𝑓;𝑛), (4.32) which ends the proof.

Lemma 4.5. Let 𝑓∶[0,1]→[0,∞) be concave. Then the following two properties hold: (i)the function 𝑔∶(0,1]→[0,∞),𝑔(𝑥)=𝑓(𝑥)/𝑥 is nonincreasing; (ii)the function ℎ∶[0,1)→[0,∞),ℎ(𝑥)=𝑓(𝑥)/(1−𝑥) is nondecreasing.

Proof. (i) Let 𝑥,𝑦∈(0,1] be with 𝑥≤𝑦. Then 𝑥𝑓(𝑥)=𝑓(𝑦𝑦+𝑦−𝑥𝑦𝑥0)≥𝑦𝑓(𝑦)+𝑦−𝑥𝑦𝑥𝑓(0)≥𝑦𝑓(𝑦), (4.33) which implies that 𝑓(𝑥)/𝑥≥𝑓(𝑦)/𝑦.
(ii) Let 𝑥,𝑦∈[0,1) be with 𝑥≥𝑦. Then
𝑓(𝑥)=𝑓(1−𝑥1−𝑦𝑦+𝑥−𝑦1−𝑦1)≥1−𝑥1−𝑦𝑓(𝑦)+𝑥−𝑦1−𝑦𝑓(1)≥1−𝑥1−𝑦𝑓(𝑦), (4.34) which implies 𝑓(𝑥)/(1−𝑥)≥𝑓(𝑦)/(1−𝑦).

Corollary 4.6. Let 𝑓∶[0,1]→[0,∞) be a concave function. Then |𝐵_𝑛^(𝑀)(𝑓)(𝑥)−𝑓(𝑥)|≤2𝜔₁1(𝑓;𝑛),∀𝑥∈[0,1]. (4.35)

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 𝑘≥𝑗. Then 𝑓_{𝑘+1,𝑛,𝑗}((𝑥)=^𝑛𝑘+1)(^𝑛_𝑗)(𝑥)1−𝑥^{𝑘+1−𝑗}𝑓(𝑘+1𝑛)=(^𝑛_𝑘)(^𝑛_𝑗)𝑛−𝑘(𝑥𝑘+1)1−𝑥^𝑘−𝑗𝑥1−𝑥𝑓(𝑘+1𝑛). (4.36) 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 𝑓_{𝑘+1,𝑛,𝑗}((𝑥)≤^𝑛𝑘)(^𝑛_𝑗)𝑛−𝑘(𝑥𝑘+1)1−𝑥^𝑘−𝑗𝑗+1⋅𝑛−𝑗𝑘+1𝑘𝑘𝑓(𝑛)=𝑓_{𝑘,𝑛,𝑗}(𝑥)𝑗+1𝑘⋅𝑛−𝑘.𝑛−𝑗 (4.37) It is immediate that for 𝑘≥𝑗+1, it follows that 𝑓_{𝑘,𝑛,𝑗}(𝑥)≥𝑓_{𝑘+1,𝑛,𝑗}(𝑥). Thus we obtain 𝑓_{𝑗+1,𝑛,𝑗}(𝑥)≥𝑓_{𝑗+2,𝑛,𝑗}(𝑥)≥⋯≥𝑓_{𝑛,𝑗,𝑛}(𝑥). (4.38)
Now let 𝑘∈{0,1,…𝑛} be with 𝑘≤𝑗. Then
𝑓_{𝑘−1,𝑛,𝑗}((𝑥)=^𝑛𝑘−1)(^𝑛_𝑗)(𝑥)1−𝑥^{𝑘−1−𝑗}𝑓(𝑘−1𝑛)=(^𝑛_𝑘)(^𝑛_𝑗)⋅𝑘(𝑥𝑛−𝑘+1)1−𝑥^𝑘−𝑗1−𝑥𝑥𝑓(𝑘−1𝑛). (4.39) 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 𝑓_{𝑘−1,𝑛,𝑗}((𝑥)≤^𝑛𝑘)(^𝑛_𝑗)𝑘(𝑥𝑛−𝑘+1)1−𝑥^𝑘−𝑗𝑛+1−𝑗𝑗⋅𝑛−𝑘+1𝑘𝑛−𝑘𝑓(𝑛)=𝑓_{𝑘,𝑛,𝑗}𝑘(𝑥)𝑗⋅𝑛+1−𝑗.𝑛−𝑘 (4.40) For 𝑘≤𝑗−1 it is immediate that 𝑓_{𝑘,𝑛,𝑗}(𝑥)≥𝑓_{𝑘−1,𝑛,𝑗}(𝑥), which implies 𝑓_{𝑗−1,𝑛.𝑗}(𝑥)≥𝑓_{𝑗−2,𝑛.𝑗}(𝑥)≥⋯≥𝑓_0,𝑛,𝑗(𝑥). (4.42) From (4.38) and (4.42), we obtain 𝐵_𝑛^(𝑀)(𝑓)(𝑥)=max{𝑓_{𝑗−1,𝑛,𝑗}(𝑥),𝑓_{𝑗,𝑛,𝑗}(𝑥),𝑓_{𝑗+1,𝑛,𝑗}(𝑥)}, (4.41) which combined with Lemma 4.4 implies |𝐵_𝑛^(𝑀)(𝑓)(𝑥)−𝑓(𝑥)|≤2𝜔₁1(𝑓;𝑛), (4.43) 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 |𝐵_𝑛^(𝑀)(𝑓)(𝑥)−𝑓(𝑥)|≤2𝜔₁1(𝑓;𝑛),∀𝑥∈[0,1]. (4.44)
 (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
|𝐵_𝑛^(𝑀)(𝑓)(𝑥)−𝑓(𝑥)|≤2𝜔₁1(𝑓;𝑛),∀𝑥∈[0,1]. (4.45)

Proof. (i) Since 𝑓 is nondecreasing it follows (see the proof of Theorem 5.5 in Section 5) that 𝐵_𝑛^(𝑀)(𝑓)(𝑥)=^𝑛_𝑘≥𝑗𝑓_{𝑘,𝑛,𝑗}𝑗(𝑥),∀𝑥∈[,𝑛+1𝑗+1𝑛+1]. (4.46) Following the proof of Corollary 4.6, we get 𝐵_𝑛^(𝑀)(𝑓)(𝑥)=max{𝑓_{𝑗,𝑛,𝑗}(𝑥),𝑓_{𝑗+1,𝑛,𝑗}𝑗(𝑥)},∀𝑥∈[,𝑛+1𝑗+1𝑛+1], (4.47) and from Lemma 4.2, we obtain |𝐵_𝑛^(𝑀)(𝑓)(𝑥)−𝑓(𝑥)|≤2𝜔₁1(𝑓;𝑛). (4.48)
  (ii) Since 𝑓 is nonincreasing, it follows (see the proof of Corollary 5.6 in Section 5) that
𝐵_𝑛^(𝑀)(𝑓)(𝑥)=^𝑗_𝑘≥0𝑓_{𝑘,𝑛,𝑗}𝑗(𝑥),∀𝑥∈[,𝑛+1𝑗+1𝑛+1]. (4.49) Following the proof of Corollary 4.6, we get 𝐵_𝑛^(𝑀)(𝑓)(𝑥)=max{𝑓_{𝑗−1,𝑛,𝑗}(𝑥),𝑓_{𝑗,𝑛,𝑗}(𝑥)}, (4.50) and from Lemma 4.3 we obtain |𝐵_𝑛^(𝑀)(𝑓)(𝑥)−𝑓(𝑥)|≤2𝜔₁1(𝑓;𝑛). (4.51)

Remark 4.8. By simple reasonings, it follows that if 𝑓∶[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) 𝐹(𝑥)≤𝑥𝑓^′(1)−𝑓(𝑥)≤𝑥𝑓^′(1)−𝑥𝑓(1)=𝑥[𝑓^′(1)−𝑓(1)]≤0,∀𝑥∈(0,1], (4.52) which means that 𝑔(𝑥) is nonincreasing.

An example of function satisfying the above conditions is 𝑓(𝑥)=𝑒^𝑥, 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]→ℝ₊, 𝐵_𝑛^(𝑀)(𝑓)(𝑥) is positive, continuous on [0,1], and satisfies 𝐵_𝑛^(𝑀)(𝑓)(0)=𝑓(0),𝐵_𝑛^(𝑀)(𝑓)(1)=𝑓(1).

Proof. Since 𝑝_𝑛,𝑘(𝑥)>0 for all 𝑥∈(0,1), 𝑛∈ℕ, 𝑘∈{0,…,𝑛}, it follows that the denominator ⋁^𝑛_𝑘=0𝑝_𝑛,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 𝐵_𝑛^(𝑀)(𝑓)(𝑥) is continuous on (0,1). To prove now the continuity of 𝐵_𝑛^(𝑀)(𝑓)(𝑥) at 𝑥=0 and 𝑥=1, we observe that 𝑝_𝑛,𝑘(0)=0 for all 𝑘∈{1,2,…,𝑛}, 𝑝_𝑛,𝑘(0)=1 for 𝑘=0 and 𝑝_𝑛,𝑘(1)=0 for all 𝑘∈{0,1,…,𝑛−1}, 𝑝_𝑛,𝑘(1)=1 for 𝑘=𝑛, which implies that ⋁^𝑛_𝑘=0𝑝_𝑛,𝑘(𝑥)=1 in the case of 𝑥=0 and 𝑥=1. The fact that 𝐵_𝑛^(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 𝐵_𝑛^(𝑀)(𝑓)(𝑥) on [0,1], it will suffice to prove the shape properties of 𝐵_𝑛^(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 𝑓_{𝑘,𝑛,𝑗}∶[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)]→ℝ,
𝑓_{𝑘,𝑛,𝑗}(𝑥)=𝑚_{𝑘,𝑛,𝑗}𝑘(𝑥)𝑓(𝑛()=^𝑛𝑘)(^𝑛_𝑗)(𝑥)1−𝑥^𝑘−𝑗𝑘𝑓(𝑛). (5.1) For any 𝑗∈{0,1,…,𝑛} and 𝑥∈[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)], we can write 𝐵_𝑛^(𝑀)(𝑓)(𝑥)=^𝑛_𝑘=0𝑓_{𝑘,𝑛,𝑗}(𝑥). (5.2)

Lemma 5.3. If 𝑓∶[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 𝑓_{𝑘,𝑛,𝑗}(𝑥)≥𝑓_{𝑘−1,𝑛,𝑗}(𝑥).

Proof. Because 𝑘≤𝑗, by the proof of Lemma 3.2, Case 2, it follows that 𝑚_{𝑘,𝑛,𝑗}(𝑥)≥𝑚_{𝑘−1,𝑛,𝑗}(𝑥). From the monotonicity of 𝑓, we get 𝑓(𝑘/𝑛)≥𝑓((𝑘−1)/𝑛). Thus, we obtain 𝑚_{𝑘,𝑛,𝑗}𝑘(𝑥)𝑓(𝑛)≥𝑚_{𝑘−1,𝑛,𝑗}(𝑥)𝑓(𝑘−1𝑛), (5.3) which proves the lemma.

Corollary 5.4. If 𝑓∶[0,1]→ℝ₊ is nonincreasing, then 𝑓_{𝑘,𝑛,𝑗}(𝑥)≥𝑓_{𝑘+1,𝑛,𝑗}(𝑥) for any 𝑘,𝑗∈{0,1,…𝑛},𝑘≥𝑗, and 𝑥∈[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)].

Proof. Because 𝑘≥𝑗, by the proof of Lemma 3.2, Case 1, it follows that 𝑚_{𝑘,𝑛,𝑗}(𝑥)≥𝑚_{𝑘+1,𝑛,𝑗}(𝑥). From the monotonicity of 𝑓, we get 𝑓(𝑘/𝑛)≥𝑓((𝑘+1)/𝑛). Thus we obtain 𝑚_{𝑘,𝑛,𝑗}𝑘(𝑥)𝑓(𝑛)≥𝑚_{𝑘+1,𝑛,𝑗}(𝑥)𝑓(𝑘+1𝑛), (5.4) which proves the corollary.

Theorem 5.5. If 𝑓∶[0,1]→ℝ₊ is nondecreasing, then 𝐵_𝑛^(𝑀)(𝑓) is nondecreasing.

Proof. Because 𝐵_𝑛^(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}, 𝐵_𝑛^(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
𝑓_{𝑗,𝑛,𝑗}(𝑥)≥𝑓_{𝑗−1,𝑛,𝑗}(𝑥)≥𝑓_{𝑗−2,𝑛,𝑗}(𝑥)≥⋯≥𝑓_0,𝑛,𝑗(𝑥), (5.5) but then it is immediate that 𝐵_𝑛^(𝑀)(𝑓)(𝑥)=^𝑛_𝑘≥𝑗𝑓_{𝑘,𝑛,𝑗}(𝑥), (5.6) for all 𝑥∈[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)]. Clearly that for 𝑘≥𝑗, the function 𝑓_{𝑘,𝑛,𝑗} is nondecreasing and since 𝐵_𝑛^(𝑀)(𝑓) is defined as the maximum of nondecreasing functions, it follows that it is nondecreasing.

Corollary 5.6. If 𝑓∶[0,1]→ℝ₊ is nonincreasing, then 𝐵_𝑛^(𝑀)(𝑓) is nonincreasing.

Proof. Because 𝐵_𝑛^(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}, 𝐵_𝑛^(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
𝑓_{𝑗,𝑛,𝑗}(𝑥)≥𝑓_{𝑗+1,𝑛,𝑗}(𝑥)≥𝑓_{𝑗+2,𝑛,𝑗}(𝑥)≥⋯≥𝑓_{𝑛,𝑛,𝑗}(𝑥), (5.7) but then it is immediate that 𝐵_𝑛^(𝑀)(𝑓)(𝑥)=^𝑗_𝑘≥0𝑓_{𝑘,𝑛,𝑗}(𝑥), (5.8) for all 𝑥∈[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)]. Clearly that for 𝑘≤𝑗 the function 𝑓_{𝑘,𝑛,𝑗} is nonincreasing and since 𝐵_𝑛^(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]→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 𝑓(𝜆𝑥+(1−𝜆)𝑦)≤max{𝑓(𝑥),𝑓(𝑦)},∀𝑥,𝑦,𝜆∈[0,1], (5.9) (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. If 𝑓∶[0,1]→ℝ₊ is continuous and quasiconvex on [0,1], then for all 𝑛∈ℕ, 𝐵_𝑛^(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∈ℕ, 𝐵_𝑛^(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]→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 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
𝐵_𝑛^(𝑀)(𝑓)(𝑥)=max{𝐵_𝑛^(𝑀)(𝐹)(𝑥),𝐵_𝑛^(𝑀)(𝐺)(𝑥)},∀𝑥∈[0,1], (5.10) where by Corollary 5.6 and Theorem 5.5, 𝐵_𝑛^(𝑀)(𝐹)(𝑥) is nonincreasing and continuous on [0,1] and 𝐵_𝑛^(𝑀)(𝐺)(𝑥) is nondecreasing and continuous on [0,1]. We have two cases: 1) 𝐵_𝑛^(𝑀)(𝐹)(𝑥) and 𝐵_𝑛^(𝑀)(𝐺)(𝑥) do not intersect each other; 2) 𝐵_𝑛^(𝑀)(𝐹)(𝑥) and 𝐵_𝑛^(𝑀)(𝐺)(𝑥) intersect each other.Case 1. We have max{𝐵_𝑛^(𝑀)(𝐹)(𝑥),𝐵_𝑛^(𝑀)(𝐺)(𝑥)}=𝐵_𝑛^(𝑀)(𝐹)(𝑥) for all 𝑥∈[0,1] or max{𝐵_𝑛^(𝑀)(𝐹)(𝑥),𝐵_𝑛^(𝑀)(𝐺)(𝑥)}=𝐵_𝑛^(𝑀)(𝐺)(𝑥) for all 𝑥∈[0,1], which obviously proves that 𝐵_𝑛^(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 𝐵_𝑛^(𝑀)(𝑓)(𝑥) is nonincreasing on [0,𝑐^′] and nondecreasing on [𝑐^′,1], which by the result in [5] implies that 𝐵_𝑛^(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 calculate 𝐵_𝑛^(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∈ℕ, one has 𝐵_𝑛^(𝑀)(𝑒₀)(𝑥)=1 and 𝐵_𝑛^(𝑀)(𝑒₁𝑝)(𝑥)=𝑥⋅^𝑛−1,0(𝑥)𝑝_𝑛,0=𝑥(𝑥)11−𝑥,if𝑥∈[0,𝐵𝑛+1],_𝑛(𝑀)(𝑒₁𝑝)(𝑥)=𝑥⋅^𝑛−1,0(𝑥)𝑝_𝑛,1=1(𝑥)𝑛1,if𝑥∈[,1𝑛+1𝑛𝐵],_𝑛(𝑀)(𝑒₁𝑝)(𝑥)=𝑥⋅^𝑛−1,1(𝑥)𝑝_𝑛,1=𝑥(𝑥)⋅1−𝑥𝑛−1𝑛1,if𝑥∈[𝑛,2𝐵𝑛+1],_𝑛(𝑀)(𝑒₁𝑝)(𝑥)=𝑥⋅^𝑛−1,1(𝑥)𝑝_𝑛,2=2(𝑥)𝑛2,if𝑥∈[,2𝑛+1𝑛𝐵],_𝑛(𝑀)(𝑒₁𝑝)(𝑥)=𝑥⋅^𝑛−1,2(𝑥)𝑝_𝑛,2=𝑥(𝑥)⋅1−𝑥𝑛−2𝑛2,if𝑥∈[𝑛,3𝐵𝑛+1],_𝑛(𝑀)(𝑒₁𝑝)(𝑥)=𝑥⋅^𝑛−1,2(𝑥)𝑝_𝑛,3=3(𝑥)𝑛3,if𝑥∈[,3𝑛+1𝑛], (5.11) and so on, in general one has 𝐵_𝑛^(𝑀)(𝑒₁𝑥)(𝑥)=⋅1−𝑥𝑛−𝑗𝑛𝑗,if𝑥∈[𝑛,𝑗+1𝐵𝑛+1],_𝑛(𝑀)(𝑒₁)(𝑥)=𝑗+1𝑛,if𝑥∈[𝑗+1,𝑛+1(𝑗+1)𝑛], (5.12) for 𝑗∈{0,1,…,𝑛−1}.

Proof. The formula 𝐵_𝑛^(𝑀)(𝑒₀)(𝑥)=1 is immediate by the definition of 𝐵_𝑛^(𝑀)(𝑓)(𝑥).
To find the formula for 𝐵_𝑛^(𝑀)(𝑒₁)(𝑥), we will use the explicit formula in Lemma 3.4 which says that
^𝑛_𝑘=0𝑝_𝑛,𝑘(𝑥)=𝑝_𝑛,𝑗𝑗(𝑥),∀𝑥∈[,𝑛+1𝑗+1𝑛+1],𝑗=0,1,…,𝑛, (5.13) where 𝑝_𝑛,𝑘(𝑥)=(^𝑛_𝑘)𝑥^𝑘(1−𝑥)^𝑛−𝑘.
Indeed, since
max_{𝑘=0,…,𝑛}{𝑝_𝑛,𝑘𝑘(𝑥)𝑛}=max_{𝑘=1,…,𝑛}{𝑝_𝑛,𝑘𝑘(𝑥)𝑛}=𝑥⋅max_{𝑘=0,…,𝑛−1}{𝑝_{𝑛−1,𝑘}(𝑥)}, (5.14) this follows by applying Lemma 3.4 to both expressions max_{𝑘=0,…,𝑛}{𝑝_𝑛,𝑘(𝑥)}, max_{𝑘=0,…,𝑛−1}{𝑝_{𝑛−1,𝑘}(𝑥)}, taking into account that we get the following division of the interval [0,1]10<≤1𝑛+1𝑛≤2≤2𝑛+1𝑛≤3≤3𝑛+1𝑛≤4≤4𝑛+1𝑛⋯. (5.15)

Remarks. (1) The convexity of 𝑓 on [0,1] is not preserved by 𝐵_𝑛^(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 𝐵_𝑛^(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 𝐵_𝑛^(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 𝐵_𝑛^(𝑀)(𝑓) is piecewise convex on [0,1]. We present the following.

Theorem 5.12. For any function 𝑓∶[0,1]→[0,∞), 𝐵_𝑛^(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 𝑓_{𝑘,𝑛,𝑗}∶[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)]→ℝ,𝑓_{𝑘,𝑛,𝑗}(𝑥)=𝑚_{𝑘,𝑛,𝑗}𝑘(𝑥)𝑓(𝑛()=^𝑛𝑘)(^𝑛_𝑗)(𝑥)1−𝑥^𝑘−𝑗𝑘𝑓(𝑛). (5.16) Clearly, we have 𝐵_𝑛^(𝑀)(𝑓)(𝑥)=^𝑛_𝑘=0𝑓_{𝑘,𝑛,𝑗}(𝑥), (5.17) 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 𝑓_{𝑘,𝑛,𝑗}(𝑥) is convex on [𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)], which will imply that 𝐵_𝑛^(𝑀)(𝑓) can be written as a maximum of some convex functions on [𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)].
Since 𝑓≥0, it suffices to prove that the functions 𝑔_𝑘,𝑗∶[0,1]→ℝ, 𝑔_𝑘,𝑗(𝑥)=(𝑥/(1−𝑥))^𝑘−𝑗 are convex on [𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)].
For 𝑘=𝑗, 𝑔_𝑗,𝑗 is constant so is convex.
For 𝑘=𝑗+1, we get 𝑔_𝑗+1,𝑗(𝑥)=𝑥/(1−𝑥) for any 𝑥∈[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)]. Then 𝑔^′′_𝑗+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 𝑔_{𝑗−1,𝑗}(𝑥)=(1−𝑥)/𝑥 for any 𝑥∈[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)]. Then 𝑔^′′_{𝑗−1,𝑗}(𝑥)=2/𝑥³>0 for any 𝑥∈[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)].
If 𝑘≥𝑗+2, then 𝑔^′′_𝑘,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 𝑘≤𝑗−2, then 𝑔^′′_𝑘,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 𝑔^′′_𝑘,𝑗(𝑥)>0 for any 𝑥∈[𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)].
Since all the functions 𝑔_𝑘,𝑗 are convex on [𝑗/(𝑛+1),(𝑗+1)/(𝑛+1)], we get that 𝐵_𝑛^(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 𝐵_𝑛^(𝑀)(𝑓) does not preserve the convexity too, by using 𝐵_𝑛^(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]→ℝ;𝑓∈𝐶¹[0,1],𝑓(0)=0,𝑓isnondecreasingon[0,1]}, (5.18) let us define the following subadditive and positive homogenous operators (as function of 𝑓): 𝐿_𝑛(𝑓)(𝑥)=^𝑥₀𝐵_𝑛^(𝑀)(𝑓^′)(𝑡)𝑑𝑡,𝑥∈[0,1],𝑛∈ℕ. (5.19) If 𝑓∈𝑆[0,1] is convex, then 𝐿_𝑛(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 𝐿_𝑛(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 𝐵_𝑛^(𝑀)(𝑓^′)(𝑥) is nondecreasing and, therefore, we get the convexity of 𝐿_𝑛(𝑓)(𝑥) on [0,1]. The monotonicity of 𝐿_𝑛(𝑓)(𝑥) is immediate by 𝑓^′≥0 on [0,1] and by the relationship 𝐿^′_𝑛(𝑓)(𝑥)=𝐵_𝑛^(𝑀)(𝑓^′)(𝑥)≥0 for all 𝑥∈[0,1].
Also, writing 𝑓(𝑥)=∫^𝑥₀𝑓^′(𝑡)𝑑𝑡 and supposing that 𝑓^′ is concave, by Corollary 4.6, we get that the order of approximation of 𝑓 by 𝐿_𝑛(𝑓) is 𝜔₁(𝑓^′;1/𝑛). In addition, 𝐿_𝑛(𝑓)(𝑥) obviously is of 𝐶¹-class (which is not the case of original operator 𝐵_𝑛^(𝑀)(𝑓)(𝑥)) and 𝐿^′_𝑛(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 𝐿_𝑛(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 𝐵_𝑛^(𝑀)(𝑓) with the linear Bernstein operator 𝐵_𝑛(𝑓)(𝑥)=∑^𝑛_𝑘=0𝑝_𝑛,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])

‖𝐵_𝑛(𝑓)−𝑓‖∼𝜔^𝜑₂1(𝑓;√𝑛), (6.1) where ‖𝑓‖=sup{|𝑓(𝑥)|;𝑥∈[0,1]} and 𝜔^𝜑₂(𝑓;𝛿) is the Ditzian-Totik second-order modulus of smoothness given by

𝜔^𝜑₂(𝑓;𝛿)=sup{sup{|𝑓(𝑥+ℎ𝜑(𝑥))−2𝑓(𝑥)+𝑓(𝑥−ℎ𝜑(𝑥))|;𝑥∈𝐼_ℎ},ℎ∈[0,𝛿]}, (6.2) with 𝜑(𝑥)=𝑥(1−𝑥), 𝛿≤1 and 𝐼_ℎ=[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 𝜔^𝜑₂(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 for 𝐵_𝑛^(𝑀)(𝑒₁)(𝑥) in Lemma 5.11. Indeed, by direct calculation can be easily proved that ‖𝐵_𝑛^(𝑀)(𝑒₁)−𝑒₁‖∼1/𝑛, while it is well known that ‖𝐵_𝑛(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.