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