On better approximation order for the max-product Meyer-König and Zeller operator

Sezin Çit; Ogün Doğru

doi:10.1515/dema-2025-0127

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On better approximation order for the max-product Meyer-König and Zeller operator

Published/Copyright: July 30, 2025

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From the journal Demonstratio Mathematica Volume 58 Issue 1

Abstract

Bede et al. (B. Bede, L. Coroianu, and S. G. Gal, Approximation and shape preserving properties of the nonlinear Meyer-König and Zeller operator of max-product kind, Numer. Funct. Anal. Optim. 31 (2010), no. 3, 232–253) defined the max-product Meyer-König and Zeller operator. They examined the approximation and shape preserving properties of this operator, and they found the order of approximation to be y ( 1 − y ) m by the modulus of continuity and claimed that this order of approximation could only be improved in certain subclasses of the functions. In contrast to this claim, we demonstrate that we can obtain a better order of approximation without reducing the function class (by the classical modulus of continuity). We find the degree of approximation to be ( 1 − y ) y 1 α m 1 − 1 α , α = 2 , 3 , … . Since 1 − 1 α tends to 1 for enough big α , we improve this degree of approximation.

Keywords: nonlinear Meyer-König and Zeller operator; max-product kind operators; modulus of continuity

MSC 2010: 41A10; 41A25; 41A36

1 Introduction

As is well known, classical approximation theory approximates a continuous function g : [ a , b ] → R with linear and positive operators when [ a , b ] is a real interval. The approximations achieved by the familiar classes of linear and positive operators (e.g., Bernstein operators) use solely the linear structure over R as underlying algebraic structure.

Bede et al. [1,2] introduced nonlinear operators, which are max-product and max-min operators based on the open problem presented by Gal [3]. These operators showed us that not all approximation operators need to be linear and also that the algebraic structure used need not be the usual algebraic structure of the real numbers. In this current study, we are concerned with max-product operators that are generated by replacing probability by maximum set functions, starting from the idea of obtaining linear positive operators by probabilistic approximation. The construction of nonlinear operators takes into account the possibility of transforming the algebraic structure from the linear one to another more general one [1,4]. Therefore, we generalize the results from the classical approximation theory.

The study of nonlinear max-product operators, starting with the first mention of the max-product Shepard operators [2], has been extensively studied for important classes of linear and positive operators. The approximation properties and shape-preserving properties can be found in previous articles [4–10].

In the literature, there are a lot of studies on the max-product operators. To mention a few of them, in [11], the Kantorovich types of these operators are defined and their approximation degrees are given. When constructing Kantorovich-type max-product operators, the classical integral is replaced by the probabilistic integral [12], which is an important tool in the fuzzy probabilistic approach (probabilistic approximation) for constructing new classes of nonlinear operators. In [13], the statistical approximations of the max-product operators are presented. In addition, q -Bernstein max-product operators based on q -integers, first studied by Duman in [14], are introduced, and their approximation properties are analyzed.

Before max-product Meyer-König and Zeller operator, which is the focus of this article, let us recall the classical Meyer-König and Zeller operator introduced in [15] as follows:

(1) M m ( g ; y ) = ∑ r = 0 ∞ m + r r y r ( 1 − y ) m + 1 g r r + m + 1 ,

where y ∈ [ 0 , 1 ) , m ∈ N . Then, in [16], Cheney and Sharma made a slight modification by replacing g r m + r + 1 by g r m + r . In addition, the classical Meyer-König and Zeller operator it can be written as follows:

M m ( g ; y ) = ∑ r = 0 ∞ m + r r y r ( 1 − y ) m + 1 g r r + m ∑ r = 0 ∞ m + r r y r ( 1 − y ) m + 1 ,

since M m ( e 0 ; y ) = 1 . In [10], Bede at al. replaced the sum operator ∑ by the maximum operator ∨ and gave the definition of the max-product Meyer-König Zeller operator as follows:

(2) Z m ( M ) ( g ) ( y ) ≔ ∨ r = 0 ∞ z m , r ( y ) g r m + r ∨ r = 0 ∞ z m , r ( y ) , y ∈ [ 0 , 1 ) , m ∈ N ,

where z m , r ( y ) = m + r r y r . Also see the degree of approximation and some shape-preserving properties of Z m ( M ) ( g ) ( y ) operators are analyzed in [10].

Our aim in this article is to obtain a better approximation degree than [10] with the modulus of continuity without reducing the classes of functions. In particular, the degree of approximation for the Z m ( M ) ( g ) ( y ) operator can be found as ω ( g ; y ( 1 − y ) ⁄ m ) . In addition, Bede et al. claimed that the order of approximation could not be improved except for some subclasses of functions [10]. Contrary to this claim, we will demonstrate that a better degree of approximation can be gained using the classical modulus of continuity. Our motivation in obtaining this improved approximation is that we do not have to use the Cauchy-Schwarz inequality when finding the approximation of the nonlinear operators with the modulus of continuity.

Moreover, with a similar way, the order of approximation can be improved for nonlinear max-product operators whose approximation and rate of convergence properties are examined [17–19].

2 Preliminaries

To determine the significant conclusions, it is essential to provide an overview of the nonlinear operators known as max-product type. In this section, we will review the fundamental definitions and theorems of nonlinear operators as presented in [1,7].

In max-product approximation operators, the algebraic structure is changed from the field of real numbers to an ordered, max-product semiring of positive reals. It has the semiring structure ( R + , ∨ , ⋅ ) with the operations ∨ (maximum) and ⋅ (product) on the set R + and is called max-product algebra.

Let J ⊂ R be bounded or unbounded intervals and

C B + ( J ) = { g : J → R + : g continuous, bounded on J } .

Let L m be the max-product operator, and we consider the general format of L m : C B + ( J ) → C B + ( J ) , as follows:

L m ( g ) ( y ) = ⋁ j = 0 m K m ( y , y j ) g ( y j ) ,

where m ∈ N , g ∈ C B + ( J ) , K m ( ⋅ , y j ) ∈ C B + ( J ) and y j ∈ J , for all j .

These operators are positive, and fulfil the property of pseudo-linearity,

L m ( α g ∨ β h ) ( y ) = α L m ( g ) ( y ) ∨ β L m ( h ) ( y ) , ∀ α , β ∈ R + , g , h ∈ C B + ( J ) ,

which is weaker than linearity. Therefore, they are not linear.

The modulus of continuity is often used in classical approximation theory to evaluate error estimates. In [1], the definition of the modulus of continuity adapted for nonlinear operators is as follows (also its main properties are given in [1]).

Definition 1

[1] Let’s consider the compact metric space ( Y , d ) , and the metric space ( [ 0 , ∞ ) , ∣ . ∣ ) , where ∣ . ∣ is the usual metric of positive real numbers. g : Y → [ 0 , ∞ ) being a bounded function, the modulus of continuity is presented as follows:

ω ( g , δ ) = ∨ { ∣ g ( y ) − g ( q ) ∣ , y , q ∈ Y , d ( y , q ) ≤ δ } .

Also, let us remember the recognized definition of the classical modulus of continuity g ∈ C B + ( J ) and δ > 0 ,

(3) ω ( g , δ ) = max { ∣ g ( y ) − g ( q ) ∣ ; y , q ∈ J , ∣ y − q ∣ ≤ δ } .

Lemma 1

[7] Let’s consider that the interval J ⊂ R is bounded (unbounded) and L m : C B + ( J ) → C B + ( J ) , m ∈ N is a sequence of operators fulfilling the following features for g , h ∈ C B + ( J ) and m ∈ N , which are monotonicity and sublinearity, respectively:

If g ≤ h then L m ( g ) ≤ L m ( h ) ,
L m ( g + h ) ≤ L m ( g ) + L m ( h ) .

Then for every y ∈ J , we have

∣ L m ( g ) ( y ) − L m ( h ) ( y ) ∣ ≤ L m ( ∣ g − h ∣ ) ( y ) .

Corollary 1

[7] Let L m : C B + ( J ) → C B + ( J ) , m ∈ N be positive homogeneous and conditions (i) and (ii) in Lemma 1 above be satisfied. Also let ( L m ) m fulfil the condition L m ( e 0 ) = e 0 , for each m ∈ N . Then for each g ∈ C B + ( J ) , m ∈ N , y ∈ J , we obtain

∣ g ( y ) − L m ( g ) ( y ) ∣ ≤ 1 + 1 δ L m ( φ y ) ( y ) ω ( g , δ ) ,

where δ > 0 , e 0 ( t ) = 1 for all t ∈ J , φ y ( t ) = ∣ t − y ∣ , for all y ∈ J .

3 Auxiliary results

In this section, for the approximation theorem, we present the statements and lemmas given in [10] without proof. We will also give Remark 2, which generalizes Lemma 3 and allows us to improve the degree of approximation of the Z m ( M ) ( g ) ( y ) operator.

Remark 1

[10]

In all expressions and in the approximation theorem, we will suppose that g : [ 0 , 1 ] → R + is continuous on [ 0 , 1 ] . Also, we will take 0 < y < 1 . Because of Z m ( M ) ( g ) ( 0 ) − g ( 0 ) = Z m ( M ) ( g ) ( 1 ) − g ( 1 ) = 0 for all m .
The operator Z m ( M ) ( g ) ( y ) provides all the conditions of Lemma 1 and Corollary 1.

For each r , s ∈ { 0 , 1 , 2 , … } and y ∈ s m + s , s + 1 m + s + 1 , let us determine the following expressions:

m r , m , s ( y ) ≔ z m , r ( y ) z m , s ( y ) , M r , m , s ( y ) ≔ z m , r ( y ) z m , s ( y ) r m + r − y

similar to [10]. We obtain the following states:

when r ≥ s + 1

M r , m , s ( y ) = m r , m , s ( y ) r m + r − y ,

when r ≤ s

M r , m , s ( y ) = m r , m , s ( y ) y − r m + r ,

where z m , r ( y ) = m + r r y r .

Lemma 2

[10] For all r , s ∈ { 0 , 1 , 2 , … } and y ∈ s m + s , s + 1 m + s + 1 , we obtain

m r , m , s ( y ) ≤ 1 .

Lemma 3

[10] Let y ∈ s m + s , s + 1 m + s + 1 .

If r ∈ { s + 1 , s + 2 , … } is such that
(4) s ≤ r − r + 1 + ( s + 1 ) 2 m ,
then we have
M r , m , s ( y ) ≥ M r + 1 , m , s ( y ) .
If r ∈ { 0 , 1 , … , s } is such that
(5) s ≥ r + r + s 2 m ,
then we have
M r , m , s ( y ) ≥ M r − 1 , m , s ( y ) .

Remark 2

(i) Let us write the inequality (4), hypothesis (i) of Lemma 3, as follows:

r + 1 + ( s + 1 ) 2 m ≤ ( r − s ) 2 r + 1 + ( s + 1 ) 2 m ( r − s ) α − 2 ≤ ( r − s ) α ,

where α = 2 , 3 , … . In this case, since r ≥ s + 1 , i.e., r − s ≥ 1 , we find 1 ≤ ( r − s ) α − 2 . If we apply this to the aforementioned equation, we obtain

r + 1 + ( s + 1 ) 2 m ≤ ( r − s ) α

(6) s ≤ r − r + 1 + ( s + 1 ) 2 m 1 α .

Therefore, for every α = 2 , 3 , … , if r ∈ { s + 1 , s + 2 , … } and inequality (6) is satisfied, then we obtain M r , m , s ( y ) ≥ M r + 1 , m , s ( y ) .

(ii) Similarly, if we revise inequality (5) in hypothesis (ii) of Lemma 3

( s − r ) 2 ≥ r + s 2 m ( s − r ) α ≥ r + s 2 m ( s − r ) α − 2 ,

where α = 2 , 3 , … and s ≠ 0 . In this case, r ≤ s but since inequality (5) is not satisfied for r = s , we should take r < s or r ≤ s − 1 , i.e., 1 ≤ s − r , so, we find 1 ≤ ( s − r ) α − 2 . If we apply this to the above equation, we obtain

( s − r ) α ≥ r + s 2 m

(7) s ≥ r + r + s 2 m 1 α .

Therefore, for every α = 2 , 3 , … , if r ∈ { 0 , 1 , … , s } is such that inequality (7), we obtain M r , m , s ( y ) ≥ M r − 1 , m , s ( y ) .

Lemma 4

[10] We obtain

⋁ r = 0 ∞ z m , r ( y ) = z m , s ( y ) , for e a c h y ∈ s m + s , s + 1 m + s + 1 , s = 0 , 1 , 2 , …

where z m , r ( y ) = m + r r y r .

4 Approximation results

The main goal is to improve the estimation of the operators Z m ( M ) ( g ) ( y ) for the function g by using the modulus of continuity. Due to the theorem provided, it can be observed that the degree of approximation may be enhanced when α reaches a sufficiently big value. Furthermore, using α = 2 gives approximation results that are consistent with the findings in [10].

Theorem 1

Let’s consider the continuous function g : [ 0 , 1 ] → R + . For the operators in (2), we have

∣ Z m ( M ) ( g ) ( y ) − g ( y ) ∣ ≤ ( 1 + 8 ( 1 − y ) y 1 α ) ω g ; 1 m 1 − 1 α , y ∈ [ 0 , 1 ] , m ≥ 4 ,

where α = 2 , 3 , … and ω ( g ; δ ) is the classical modulus of continuity defined in (3).

Proof

For every y ∈ s m + s , s + 1 m + s + 1 , the operator Z m ( M ) ( g ) ( y ) fulfils the situations in Corollary 1. Thus, we obtain the next inequality with φ y ( t ) = ∣ t − y ∣

(8) ∣ Z m ( M ) ( g ) ( y ) − g ( y ) ∣ ≤ 1 + 1 δ m Z m ( M ) ( φ y ) ( y ) ω ( g , δ m ) ,

where δ m is a sequence of positive real numbers. Let’s define the expression

E m ( y ) ≔ Z m ( M ) ( φ y ) ( y )

= ⋁ r = 0 ∞ m + r r y r r m + r − y ⋁ r = 0 ∞ m + r r y r , y ∈ [ 0 , 1 ] .

Let y ∈ s m + s , s + 1 m + s + 1 , where s ∈ { 0 , 1 , … } is constant and arbitrary. By using Lemma 4, we may write that

E m ( y ) = max r = 0 , 1 , … { M r , m , s ( y ) } .

Initially, let we examine the s = 0 case, M r , m , 0 ( y ) = m + r r y r r m + r − y , where r = 0 , 1 , 2 , … and y ∈ 0 , 1 m + 1 and α = 2 , 3 , … .

When r = 0 , we obtain

M 0 , m , 0 ( y ) = y = y 1 α y 1 − 1 α ≤ y 1 α ( m + 1 ) 1 − 1 α ≤ y 1 α m 1 − 1 α = ( 1 − y ) y 1 α m 1 − 1 α 1 ( 1 − y ) ≤ ( 1 − y ) y 1 α m 1 − 1 α m + 1 m ≤ 2 ( 1 − y ) y 1 α m 1 − 1 α .

In the aforementioned inequality, we used the following information. 0 ≤ y ≤ 1 m + 1 or 1 − 1 m + 1 ≤ 1 − y ≤ 1 , then we have 1 1 − y ≤ m + 1 m ≤ 2 .

And when r = 1 , we obtain

M 1 , m , 0 ( y ) = m + 1 1 y 1 m + 1 − y ≤ y ≤ 2 ( 1 − y ) y 1 α m 1 − 1 α ,

when r = 2 , we obtain

M 2 , m , 0 ( y ) = m + 2 2 y 2 2 m + 2 − y ≤ ( m + 1 ) ( m + 2 ) 2 y 2 2 m + 2 = ( m + 1 ) y 2 ≤ ( m + 1 ) y 1 m + 1 = y ≤ 2 ( 1 − y ) y 1 α m 1 − 1 α .

So, we have

E m ( y ) = max r ∈ { 0 , 1 , 2 } { M r , m , 0 ( y ) } , y ∈ 0 , 1 m + 1 .

In fact, when r ≥ 2 , this becomes

E m ( y ) = max r ∈ { 0 , 1 , … } { M r , m , 0 ( y ) } , y ∈ 0 , 1 m + 1 .

Moreover, by Remark 2 ( i ) , we know that M r , m , 0 ( y ) ≥ M r + 1 , m , 0 ( y ) . Thus, we obtain an upper estimate for E m ( y ) ≤ 2 ( 1 − y ) y 1 α m 1 − 1 α , for each y ∈ 0 , 1 m + 1 , when s = 0 .

At the moment, it is time to look for an upper estimate for each M r , m , s ( y ) for s = 1 , 2 , … , y ∈ s m + s , s + 1 m + s + 1 and r ∈ { 0 , 1 , … } .

In fact, if we may show the next one inequality

(9) M r , m , s ( y ) ≤ 8 ( 1 − y ) y 1 α m 1 − 1 α ,

where α = 2 , 3 , … and for each y ∈ s m + s , s + 1 m + s + 1 , r = 0 , 1 , 2 , … this is exactly equivalent to

E m ( y ) ≤ 8 ( 1 − y ) y 1 α m 1 − 1 α , for all y ∈ [ 0 , 1 ] and m ≥ 4 .

This allows us to complete the proof by taking δ m = 1 m 1 − 1 α in (8).

In order to show the inequality (9), we have just written earlier, let us examine the following situations:

(1) r ≥ s + 1 ; (2) r ≤ s .

Case (1). Subcase (a). Let r ≥ s + 1 and assume first that

s ≥ r − ( r + 1 ) + ( s + 1 ) 2 m 1 α

or equivalently,

r ≤ s + ( r + 1 ) + ( s + 1 ) 2 m 1 α .

Then, by applying Lemma 2, we obtain

M r , m , s ( y ) = m r , m , s ( y ) r m + r − y ≤ r m + r − y ≤ r m + r − s m + s ≤ s + ( r + 1 ) + ( s + 1 ) 2 m 1 α m + s + ( r + 1 ) + ( s + 1 ) 2 m 1 α − s m + s = m ( r + 1 ) + ( s + 1 ) 2 m 1 α ( m + s ) m + s + ( r + 1 ) + ( s + 1 ) 2 m 1 α .

Now we observe that r + 1 ≤ 2 ( s + 1 ) .

Let’s define the following function ϕ :

(10) ϕ ( r ) ≔ r − ( r + 1 ) + ( s + 1 ) 2 m 1 α .

Suppose that, r + 1 > 2 s + 2 , which implies r ≥ 2 s + 2 . Since α = 2 , 3 , … , we have 1 α − 1 < 0 , and clearly ( r + 1 ) + ( s + 1 ) 2 m > 0 . Therefore, we obtain the following result:

ϕ ′ ( r ) = 1 − 1 α ( r + 1 ) + ( s + 1 ) 2 m 1 α − 1 = 1 − 1 α ( r + 1 ) + ( s + 1 ) 2 m 1 − 1 α > 0 .

Since the function ϕ is nondecreasing on [ 0 , ∞ ) , we obtain ϕ ( r ) ≥ ϕ ( 2 s + 2 ) . This implies that

s ≥ r − ( r + 1 ) + ( s + 1 ) 2 m 1 α ≥ 2 s + 2 − 2 s + 3 + ( s + 1 ) 2 m 1 α

that is,

2 s + 3 + ( s + 1 ) 2 m 1 α ≥ s + 2

2 s + 3 + ( s + 1 ) 2 m ≥ ( s + 2 ) α .

This inequality leads to a contradiction for all α ≥ 2 . In particular, when α = 2 , the condition

2 s + 3 + ( s + 1 ) 2 m ≥ s 2 + 4 s + 4 = 2 s + 3 + ( s + 1 ) 2

implies

( s + 1 ) 2 m ≥ ( s + 1 ) 2 ,

which is clearly not possible. Therefore, the inequality cannot hold for any α ≥ 2 .

As a consequence,

( r + 1 ) + ( s + 1 ) 2 m 1 α ≤ 2 ( s + 1 ) + ( s + 1 ) 2 m 1 α ,

and also we obtain

M r , m , s ( y ) ≤ m ( s + 1 ) 2 + ( s + 1 ) m 1 α ( m + s ) m + s + ( s + 1 ) 2 + ( s + 1 ) m 1 α .

Since y ∈ s m + s , s + 1 m + s + 1 , then y ≤ s + 1 m + s + 1 after simple calculations it follows m + y − 1 ≤ ( 1 − y ) ( m + s ) , i.e., m + y − 1 1 − y ≤ m + s . Also, since s m + s ≤ y , then we have s ( 1 − y ) ≤ m y , i.e., s ≤ m y 1 − y . So we obtain s + 1 ≤ 2 s ≤ 2 m y 1 − y .

Now, let’s define the following function ϑ :

ϑ ( ξ , s ) ≔ m ξ ( m + s ) ( m + s + ξ ) .

The function is increasing with respect to ξ , and decreasing with respect to s , when ξ = ( s + 1 ) 2 + ( s + 1 ) m 1 α . So, we obtain

M r , m , s ( y ) ≤ m 4 m y 1 − y 1 + m y m ( 1 − y ) 1 α m + y − 1 1 − y m + y − 1 1 − y + 4 m y 1 − y 1 + m y m ( 1 − y ) 1 α = m 4 1 α ( m y ) 1 α ( 1 − y ) 2 α m + y − 1 1 − y 2 + ( m + y − 1 ) 4 1 α ( m y ) 1 α ( 1 − y ) 2 α + 1 = m 4 1 α ( m y ) 1 α ( 1 − y ) 2 α ( m + y − 1 ) 2 ( 1 − y ) 2 α − 1 + ( m + y − 1 ) 4 1 α ( m y ) 1 α ( 1 − y ) 2 α + 1 = 4 1 α m 1 + 1 α ( 1 − y ) y 1 α ( m + y − 1 ) 2 ( 1 − y ) 2 α − 1 + ( m + y − 1 ) 4 1 α ( m y ) 1 α .

Here, let’s denote the function ψ as follows:

ψ ( y ) ≔ ( m + y − 1 ) ( m + y − 1 ) ( 1 − y ) 2 α − 1 + 4 1 α ( m y ) 1 α .

Then ψ is nondecreasing when 0 < y < 1 . In fact, since,

ψ ′ ( y ) = ( m + y − 1 ) ( 1 − y ) 2 α − 1 + 4 1 α ( m y ) 1 α + ( m + y − 1 ) ( 1 − y ) 2 α − 1 + ( m + y − 1 ) 2 α − 1 ( − 1 ) ( 1 − y ) 2 α − 2 + 4 1 α m α ( m y ) 1 α − 1

because of α ≥ 2 or 2 α − 1 ≤ 0 , then ψ ′ ( y ) > 0 when 0 < y < 1 . Thus, we have ψ ( y ) ≥ ψ ( 0 ) = ( m − 1 ) 2 . Since m ≥ 4 , then ( m − 1 ) 2 ≥ m 2 2 , so ψ ( y ) ≥ m 2 2 .

Finally, for this case, by using our last result, we obtain

M r , m , s ( y ) ≤ 2 4 1 α m 1 + 1 α ( 1 − y ) y 1 α m 2 = 2 4 1 α ( 1 − y ) y 1 α m 1 − 1 α ≤ 4 ( 1 − y ) y 1 α m 1 − 1 α .

Subcase (b). Let r ≥ s + 1 and assume now that s < r − ( r + 1 ) + ( s + 1 ) 2 m 1 α . In subcase (a), the function ϕ ( r ) = r − ( r + 1 ) + ( s + 1 ) 2 m 1 α defined by equation (10) is non-decreasing with respect to r ≥ 0 . Hence, it could be concluded that there is a maximum value r ¯ ∈ { 1 , 2 , … } , fulfilling the condition r ¯ − ( r ¯ + 1 ) + ( s + 1 ) 2 m 1 α < s . So, for r 1 = r ¯ + 1 , we obtain r 1 − ( r 1 + 1 ) + ( s + 1 ) 2 m 1 α ≥ s . Also, by the way we chose r ¯ , it implies that r 1 ≤ 2 ( s + 1 ) . Thus, similar to subcase (a), we have

M r ¯ + 1 , m , s ( y ) = m r ¯ + 1 , m , s ( y ) r ¯ + 1 m + r ¯ + 1 − y ≤ 4 1 α ( 1 − y ) y 1 α m 1 − 1 α .

Since

r 1 > r 1 − ( r 1 + 1 ) + ( s + 1 ) 2 m 1 α ≥ s ,

it follows that r 1 ≥ s + 1 . Then, by Remark 2 ( i ) we obtain M r ¯ + 1 , m , s ( y ) ≥ M r ¯ + 2 , m , s ( y ) ≥ … . As a result, for all

M r , m , s ( y ) < 4 1 α ( 1 − y ) y 1 α m 1 − 1 α

r ∈ { r ¯ + 1 , r ¯ + 2 , … } , we obtain

Case (2). Subcase (a). Let r ≤ s and assume first that s ≤ r + r + s 2 m 1 α or s − r + s 2 m 1 α ≤ r . We have

M r , m , s ( y ) = m r , m , s ( y ) y − r m + r ≤ s + 1 m + s + 1 − r m + r ≤ s + 1 m + s + 1 − s − r + s 2 m 1 α m + s − r + s 2 m 1 α = m r + s 2 m 1 α + 1 ( m + s + 1 ) m + s − r + s 2 m 1 α .

By taking into account that r ≤ s and s ≤ m y 1 − y , we have

M r , m , s ( y ) ≤ m s + s 2 m 1 α + 1 ( m + s + 1 ) m + s − s + s 2 m 1 α ≤ m m y 1 − y + m y 1 − y 2 m 1 α + 1 ( m + s + 1 ) m + s − m y 1 − y + m y 1 − y 2 m 1 α = m m y 1 − y 1 1 − y 1 α + 1 ( m + s + 1 ) m + s − m y 1 − y 1 1 − y 1 α = m ( m y ) 1 α + ( 1 − y ) 2 α ( m + s + 1 ) ( m + s ) ( 1 − y ) 2 α − ( m y ) 1 α .

Also because of m + y − 1 1 − y = m 1 − y − 1 ≤ s + m and we have

M r , m , s ( y ) ≤ m ( m y ) 1 α + ( 1 − y ) 2 α m 1 − y − 1 + 1 m 1 − y − 1 ( 1 − y ) 2 α − ( m y ) 1 α = ( 1 − y ) ( m y ) 1 α + ( 1 − y ) 2 α m ( 1 − y ) 2 α − 1 − ( m y ) 1 α − ( 1 − y ) 2 α ≤ ( 1 − y ) ( m y ) 1 α + 1 m ( 1 − y ) 2 α − 1 − ( m y ) 1 α − ( 1 − y ) 2 α .

In the last inequality, we used ( 1 − y ) 2 α ≤ 1 .

Since s ≥ 1 and m ≥ 4 also y ∈ s m + s , s + 1 m + s + 1 , then 4 5 ≤ m y it follows 4 5 1 α ≤ ( m y ) 1 α , so 1 ≤ 5 4 1 α ( m y ) 1 α . We obtain ( m y ) 1 α + 1 ≤ ( m y ) 1 α + 5 4 1 α ( m y ) 1 α = 1 + 5 4 1 α ( m y ) 1 α ≤ 5 2 ( m y ) 1 α .

Also, let’s denote the function λ as follows:

λ ( y ) ≔ m ( 1 − y ) 2 α − 1 − ( m y ) 1 α − ( 1 − y ) 2 α .

We have

λ ′ ( y ) = m 2 α − 1 ( − 1 ) ( 1 − y ) 2 α − 2 − m α ( m y ) 1 α − 1 − 2 α ( − 1 ) ( 1 − y ) 2 α − 1 = m ( α − 2 ) α ( 1 − y ) 2 − 2 α − m α ( m y ) 1 − 1 α + 2 α ( 1 − y ) 1 − 2 α = m ( α − 2 ) ( m y ) 1 − 1 α − m ( 1 − y ) 2 − 2 α + 2 ( 1 − y ) α ( 1 − y ) 2 − 2 α ( m y ) 1 − 1 α .

So, because of m ≥ 4 , the function λ is nonincreasing for α = 2 , and for α ≥ 3 , the function λ is nondecreasing when 0 < y < 1 .

Let α = 2 . Then we have λ ( y ) ≥ λ ( 1 ) ,

lim y → 1 m ( 1 − y ) 1 − 2 α − ( m y ) 1 α + ( 1 − y ) 2 α ≥ lim y → 1 m 2 .

Now, let α ≥ 3 . Then we have λ ( y ) ≥ λ ( 0 ) = m − 1 and since m ≥ 4 , then clearly it follows λ ( 0 ) ≥ m 2 .

Therefore, in both case, we have λ ( y ) ≥ m 2 or 1 λ ( y ) ≤ 2 m .

Using what we found earlier, we have

M r , m , s ( y ) ≤ 5 2 ( 1 − y ) ( m y ) 1 α m 2 = 5 ( 1 − y ) y 1 α m 1 − 1 α .

Subcase (b). Now let r ≤ s and assume that

s > r + r + s 2 m 1 α .

Let r ˜ ∈ { 1 , 2 , … , s } be the minimum value such that

r ˜ + r ˜ + s 2 m 1 α > s

or equivalently,

r ˜ > s − r ˜ + s 2 m 1 α

is satisfied. Define r 2 = r ˜ − 1 , then it follows that r 2 + r 2 + s 2 m 1 α ≤ s . Following the same reasoning as in subcase (a), we obtain

M r ˜ − 1 , m , s ( y ) = m r ˜ − 1 , m , s ( y ) y − r ˜ − 1 m + r ˜ − 1 ≤ s + 1 m + s + 1 − r ˜ − 1 m + r ˜ − 1 ≤ s + 1 m + s + 1 − s − r ˜ + s 2 m 1 α − 1 m + s − r ˜ + s 2 m 1 α − 1 = m r ˜ + s 2 m 1 α + 2 ( m + s + 1 ) m + s − r ˜ + s 2 m 1 α − 1 .

And since r ˜ ≤ s and s ≤ m y 1 − y , then similar to the aforementioned subcase, we have

M r ˜ − 1 , m , s ( y ) ≤ m s + s 2 m 1 α + 2 ( m + s + 1 ) m + s − s + s 2 m 1 α − 1

≤ m m y 1 − y 1 1 − y 1 α + 2 ( m + s + 1 ) m + s − m y 1 − y 1 1 − y 1 α − 1 = m ( m y ) 1 α + 2 ( 1 − y ) 2 α ( m + s + 1 ) ( m + s − 1 ) ( 1 − y ) 2 α − ( m y ) 1 α .

Also because of m 1 − y − 1 ≤ m + s , then we obtain

M r ˜ − 1 , m , s ( y ) ≤ m ( m y ) 1 α + 2 ( 1 − y ) 2 α m 1 − y − 1 + 1 m 1 − y − 2 ( 1 − y ) 2 α − ( m y ) 1 α = ( 1 − y ) ( m y ) 1 α + 2 ( 1 − y ) 2 α m ( 1 − y ) 2 α − 1 − ( m y ) 1 α − 2 ( 1 − y ) 2 α ≤ ( 1 − y ) ( m y ) 1 α + 2 m ( 1 − y ) 2 α − 1 − ( m y ) 1 α − 2 ( 1 − y ) 2 α .

Since 1 ≤ 5 4 1 α ( m y ) 1 α then, we obtain ( m y ) 1 α + 2 ≤ ( m y ) 1 α + 2 5 4 1 α ( m y ) 1 α = 1 + 2 5 4 1 α ( m y ) 1 α . Also, let’s denote the function μ as follows:

μ ( y ) ≔ m ( 1 − y ) 2 α − 1 − ( m y ) 1 α − 2 ( 1 − y ) 2 α .

We have

μ ′ ( y ) = m 2 α − 1 ( − 1 ) ( 1 − y ) 2 α − 2 − m α ( m y ) 1 α − 1 − 2 2 α ( − 1 ) ( 1 − y ) 2 α − 1 = m ( α − 2 ) α ( 1 − y ) 2 − 2 α − m α ( m y ) 1 − 1 α + 4 α ( 1 − y ) 1 − 2 α = m ( α − 2 ) ( m y ) 1 − 1 α − m ( 1 − y ) 2 − 2 α + 4 ( 1 − y ) α ( 1 − y ) 2 − 2 α ( m y ) 1 − 1 α .

So, because of m ≥ 4 , the function μ ( y ) for α = 2 , is nonincreasing and for α ≥ 3 , is nondecreasing for 0 < y < 1 .

Let α = 2 , we have μ ( y ) ≥ μ ( 1 ) ,

lim y → 1 m ( 1 − y ) 1 − 2 α − ( m y ) 1 α + 2 ( 1 − y ) 2 α ≥ lim y → 1 m 2 .

Now, let α ≥ 3 , we have μ ( y ) ≥ μ ( 0 ) = m − 2 and since m ≥ 4 , then μ ( 0 ) ≥ m 2 .

In both case, we have μ ( y ) ≥ m 2 or 1 μ ( y ) ≤ 2 m .

Therefore, we obtain

M r ˜ − 1 , m , s ( y ) ≤ 1 + 2 5 4 1 α ( 1 − y ) ( m y ) 1 α m 2 ≤ 2 1 + 2 5 4 1 α ( 1 − y ) y 1 α m 1 − 1 α ≤ 8 ( 1 − y ) y 1 α m 1 − 1 α .

In the light of Remark 2, (ii), following M r ˜ − 1 , m , s ( y ) ≥ M r ˜ − 2 , m , s ( y ) ≥ … ≥ M 0 , m , s ( y ) . Thus, we obtain

□ M r , m , s ( y ) ≤ 8 ( 1 − y ) y 1 α m 1 − 1 α .

Corollary 2

The study [10] showed that the degree of approximation for Z m ( M ) ( g ) ( y ) operators is ( 1 − y ) y m , using the modulus of continuity. Nevertheless, thanks to Theorem 1, we have shown that the degree of approximation is ( 1 − y ) y 1 α m 1 − 1 α . For sufficiently big α , the expression 1 ⁄ m 1 − 1 α tends to 1 ⁄ m . Consequently, this choice of α improves the order of approximation, since 1 − 1 α ≥ 1 2 for α = 2 , 3 , … .

Acknowledgments

The authors sincerely thank the referees for their valuable comments and helpful suggestions and the editors for their efforts in managing the review process.

Funding information: The authors state no funding involved.
Author contributions: The authors contributed equally to this article.
Conflict of interest: The authors state no conflict of interest.
Ethical approval: Not applicable.
Data availability statement: Data sharing not applicable to this article as no datasets were generated or analyzed during the current study.

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Received: 2024-10-15

Revised: 2025-02-18

Accepted: 2025-02-28

Published Online: 2025-07-30

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/dema-2025-0127

Keywords for this article

nonlinear Meyer-König and Zeller operator; max-product kind operators; modulus of continuity

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