Meyer-König and Zeller operators
based on the $q$-integers

Abstract.

By means of the $q$-integers as well as of the Gaussian binomial coefficients we introduce a generalization of the Meyer-König and Zeller operators. For a fixed number $q\in ]0,1],$ the sequence of the generalized Meyer-König and Zeller operators is denoted by ${\left(M_{n,q}\right)}_{n\ge 1}.$ Both a theorem on convergence and a Popoviciu type theorem on the rate of convergence are proved. It is shown that if $f$ is increasing, then $M_{n,q}f$ is also increasing, while if $f$ is convex, then $M_{n,q}f$ is also convex and $M_{n,q}f\ge f,$ generalizing known results when $q=1.$ Likewise, it is shown that if $f$ is convex, then the sequence ${\left(M_{n,q}f\right)}_{n\ge 1}$  is non-increasing as in the case of the classical $M_n$-operators.

1. Introduction

Let $q$ be a number belonging to the interval $]0,1]$ (throughout the paper $q$ will always have this meaning). For each non-negative integer $k,$ the $q$-integer $\left[k\right]$ is defined by

while the $q$-factorial $\left[k\right]!$ is defined by

By means of the $q$-factorial, the Gaussian binomial coefficients are defined by

$$\left[\genfrac{}{}{0.0pt}{}{n}{k}\right]=\frac{\left[n\right]!}{\left[k\right]!\left[n-k\right]!}$$

for all integers $n\ge k\ge 0.$

It is easily seen that the Gaussian binomial coefficients satisfy the recurrence relations

$$\left[\genfrac{}{}{0.0pt}{}{n+1}{k}\right]=\left[\genfrac{}{}{0.0pt}{}{n}{k-1}\right]+q^k\left[\genfrac{}{}{0.0pt}{}{n}{k}\right]$$

and

$$\left[\genfrac{}{}{0.0pt}{}{n+1}{k}\right]=q^{n-k+1}\left[\genfrac{}{}{0.0pt}{}{n}{k-1}\right]+\left[\genfrac{}{}{0.0pt}{}{n}{k}\right]$$

for all integers $n\ge k\ge 1$.

Based on the $q$-integers and on the Gaussian binomial coefficients G.M. Phillips [10] proposed the following generalization of the classical Bernstein operators: for each positive integer $n$ let $B_{n,q}:C[0,1]\to C[0,1]$ be the operator defined by

$$B_{n,q}f\left(x\right)=\sum _{k=0}^{n}f\left(\frac{[k]}{[n]}\right)\left[\genfrac{}{}{0.0pt}{}{n}{k}\right]x^k\prod _{j=0}^{n-k-1}\left(1-q^{j}x\right),$$

where an empty product denotes 1. For $q=1,$ $B_{n,1}$ reduces to the classical Bernstein operator $B_n:C[0,1]\to C[0,1]$

$$B_{n}f\left(x\right)=\sum _{k=0}^{n}f\left(\frac{k}{n}\right)\left(\genfrac{}{}{0pt}{}{n}{k}\right)x^k{\left(1-x\right)}^{n-k}.$$

Approximation properties of these generalized Bernstein operators, which are quite similar with those of the classical Bernstein operators, were investigated in [5], [9], and [10].

Starting from the power series expansion

(1.1)

W. Meyer-König and K. Zeller [8] introduced the sequence ${\left(M_n\right)}_{n\ge 1}$ of linear positive operators $M_n:C[0,1]\to C[0,1]$ defined by

$M_{n}f\left(x\right)$

$M_{n}f\left(1\right)$ $=f\left(1\right).$

For approximation properties of the $M_n$-operators the reader is referred to [3] and [7].

The $q$-generalization of (1.1) is the following power series expansion:

(1.2)

Simple proofs of (1.2) can be found in [1] and [2]. Starting from (1.2) we introduce the sequence ${\left(M_{n,q}\right)}_{n\ge 1}$ of linear positive operators $M_{n,q}:C[0,1]\to C[0,1]$ defined by

(1.3) $M_{n,q}f\left(x\right)$

(1.4) $M_{n,q}f\left(1\right)$ $=f\left(1\right),$

where

$$P_{n,q}\left(x\right)=\prod _{j=0}^n\left(1-q^{j}x\right).$$

It is easy to check (see [8]) that

$$\underset{x\nearrow 1}{lim}{M}_{n,q}f\left(x\right)=f\left(1\right)$$

for all $f\in C[0,1],$so $M_{n,q}$ is well-defined for each positive integer $n$. In fact $M_{n,q}f$ can be defined by (1.3) for each bounded function $f:[0,1]\to ℝ.$ If in addition there exists $\underset{x\nearrow 1}{lim}f\left(x\right)=l,$ then $\underset{x\nearrow 1}{lim}{M}_{n,q}f\left(x\right)=l.$ In this case (1.4) must be replaced by

(1.5) $$M_{n,q}f\left(1\right)=\underset{x\nearrow 1}{lim}f\left(x\right).$$

In analogy with the terminology used in [10], in what follows we will call the $M_{n,q}$-operators generalized Meyer-König and Zeller operators.

It is the main purpose of this paper to investigate the approximation properties of the above introduced operators.

2. The order of Approximation by Generalized Meyer-König and Zeller operators

For each nonnegative integer $i$ let $e_i$ denote the monomial $e_i\left(x\right)=x^i.$

Lemma 2.1.

For all integers $n\ge 3$ and all $x\in [0,1]$ the following relations are valid:

(2.1) $$M_{n,q}{e}_0\left(x\right)=1,$$

(2.2) $$M_{n,q}{e}_1\left(x\right)=x,$$

(2.3) $$M_{n,q}{e}_2\left(x\right)=x^2+\frac{x\left(1-x\right)\left(1-q^{n}x\right)}{\left[n-1\right]}-R_{n,q}\left(x\right),$$

where

(2.4) $$0\le R_{n,q}\left(x\right)\le \frac{[2]q^{n-1}}{\left[n-1\right]\left[n-2\right]}x\left(1-x\right)\left(1-qx\right)\left(1-q^{n}x\right).$$

Proof.

Relation (2.1) is just (1.2), while relation (2.2) follows immediately from (1.2) taking into account that

(2.5) $$\frac{[k]}{\left[n+k\right]}\left[\genfrac{}{}{0.0pt}{}{n+k}{k}\right]=\left[\genfrac{}{}{0.0pt}{}{n+k-1}{k-1}\right]$$

for all positive integers $n$ and $k.$

To prove (2.3)–(2.4) we fix an integer $n\ge 3$ as well as a number $x\in [0,1[$ ((2.3)–(2.4) are trivially true for $x=1$). Taking into account (2.5) we get

$M_{n,q}{e}_2\left(x\right)$ $=P_{n,q}\left(x\right){\displaystyle \sum _{k=1}^{\mathrm{\infty }}}\frac{[k]}{[n+k]}\left[\genfrac{}{}{0.0pt}{}{n+k-1}{k-1}\right]x^k$

$=x{P}_{n,q}\left(x\right){\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\frac{\left[k+1\right]}{\left[n+k+1\right]}\left[\genfrac{}{}{0.0pt}{}{n+k}{k}\right]x^k.$

On the other hand, (1.2) yields

$$e_2\left(x\right)=x{P}_{n,q}\left(x\right)\sum _{k=1}^{\mathrm{\infty }}\left[\genfrac{}{}{0.0pt}{}{n+k-1}{k-1}\right]x^k.$$

Consequently

$M_{n,q}{e}_2\left(x\right)-x^2$ $=\frac{x{P}_{n,q}\left(x\right)}{[n+1]}+x{P}_{n,q}(x){\displaystyle \sum _{k=1}^{\mathrm{\infty }}}\left(\frac{\left[k+1\right]}{\left[n+k+1\right]}\left[\genfrac{}{}{0.0pt}{}{n+k}{k}\right]-\left[\genfrac{}{}{0.0pt}{}{n+k-1}{k-1}\right]\right)x^k.$

By a simple computation

$$\frac{\left[k+1\right]}{\left[n+k+1\right]}\left[\genfrac{}{}{0.0pt}{}{n+k}{k}\right]-\left[\genfrac{}{}{0.0pt}{}{n+k-1}{k-1}\right]=\frac{q^k}{\left[n+k+1\right]}\left[\genfrac{}{}{0.0pt}{}{n+k-1}{k}\right],$$

so

$M_{n,q}{e}_2\left(x\right)$ $=x^2+x{P}_{n,q}\left(x\right){\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\frac{q^k}{\left[n+k+1\right]}\left[\genfrac{}{}{0.0pt}{}{n+k-1}{k}\right]x^k$

$=x^2+\frac{x{P}_{n,q}\left(x\right)}{\left[n-1\right]}{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{q}^k\frac{\left[n+k-1\right]}{\left[n+k+1\right]}\left[\genfrac{}{}{0.0pt}{}{n+k-2}{k}\right]x^k.$

Since

$$\frac{\left[n+k-1\right]}{\left[n+k+1\right]}=1-\frac{[2]q^{n+k-1}}{\left[n+k+1\right]},$$

we get

$M_{n,q}{e}_2\left(x\right)$ $=x^2+\frac{x{P}_{n,q}\left(x\right)}{\left[n-1\right]}{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\left[\genfrac{}{}{0.0pt}{}{n+k-2}{k}\right]{\left(qx\right)}^k$

$-\frac{\left[2\right]q^{n-1}x{P}_{n,q}\left(x\right)}{\left[n-1\right]}{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\frac{1}{\left[n+k+1\right]}\left[\genfrac{}{}{0.0pt}{}{n+k-2}{k}\right]{\left(q^{2}x\right)}^k.$

Relation (1.2) ensures that

$$\sum _{k=0}^{\mathrm{\infty }}\left[\genfrac{}{}{0.0pt}{}{n+k-2}{k}\right]{\left(qx\right)}^k=\frac{1}{\prod _{j=0}^{n-2}\left(1-q^j\cdot qx\right)}=\frac{\left(1-x\right)\left(1-q^{n}x\right)}{P_{n,q}\left(x\right)},$$

hence

$$M_{n,q}{e}_2\left(x\right)=x^2+\frac{x\left(1-x\right)\left(1-q^{n}x\right)}{\left[n-1\right]}-R_{n,q}\left(x\right),$$

where

$0$ $\le R_{n,q}\left(x\right)=\frac{\left[2\right]q^{n-1}x{P}_{n,q}\left(x\right)}{\left[n-1\right]}{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\frac{1}{\left[n+k+1\right]}\left[\genfrac{}{}{0.0pt}{}{n+k-2}{k}\right]{\left(q^{2}x\right)}^k$

$\le \frac{\left[2\right]q^{n-1}x{P}_{n,q}\left(x\right)}{\left[n-1\right]}{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\frac{1}{\left[n+k-2\right]}\left[\genfrac{}{}{0.0pt}{}{n+k-2}{k}\right]{\left(q^{2}x\right)}^k$

$=\frac{\left[2\right]q^{n-1}x{P}_{n,q}\left(x\right)}{\left[n-1\right]\left[n-2\right]}{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\left[\genfrac{}{}{0.0pt}{}{n+k-3}{k}\right]{\left(q^{2}x\right)}^k$

$=\frac{\left[2\right]q^{n-1}x{P}_{n,q}\left(x\right)}{\left[n-1\right]\left[n-2\right]}\cdot \frac{1}{\prod _{j=0}^{n-3}\left(1-q^j\cdot q^{2}x\right)}$

$=\frac{\left[2\right]q^{n-1}}{\left[n-1\right]\left[n-2\right]}x\left(1-x\right)\left(1-qx\right)\left(1-q^{n}x\right).$

The completes the proof of (2.3)–(2.4). ∎

From (2.4) it follows that $R_{n,q}\left(x\right)\to 0$ as $n\to \mathrm{\infty }$ for all $x\in [0,1].$ On the order hand, for we have $\left[n-1\right]\to \frac{1}{1-q}$ as $n\to \mathrm{\infty },$ so

$$M_{n,q}{e}_2\left(x\right)\to x^2+x\left(1-x\right)(1-q),\text{as }n\to \mathrm{\infty }.$$

Hence ${\left(M_{n,q}{e}_2\right)}_{n\ge 1}$ does not converge to $e_2.$ However, the following estimation turns out to be very useful in what follows:

$\left|M_{n,q}{e}_2\left(x\right)-e_2\left(x\right)\right|$ $\le \frac{x\left(1-x\right)\left(1-q^{n}x\right)}{\left[n-1\right]}+\frac{\left[2\right]q^{n-1}x\left(1-x\right)\left(1-qx\right)\left(1-q^{n}x\right)}{\left[n-1\right]\left[n-2\right]}$

$\le \frac{1}{4\left[n-1\right]}+\frac{[2]}{4\left[n-1\right]}=\frac{2+q}{4\left[n\right]}\cdot \frac{\left[n\right]}{\left[n-1\right]},$

and since

$$\frac{\left[n\right]}{\left[n-1\right]}=\frac{1-q^n}{1-q^{n-1}}\le 1+q,$$

we finally get

(2.6) $$\left|M_{n,q}{e}_2\left(x\right)-e_2\left(x\right)\right|\le \frac{\left(2+q\right)\left(1+q\right)}{4\left[n\right]}\le \frac{3}{2\left[n\right]}$$

for all integers $n\ge 3$ and all $x\in [0,1].$

By proceeding like G. M. Phillips [10], in order to obtain a convergent sequence of generalized Meyer-König and Zeller operators, we let $q=q_n$ depend on $n.$ More precisely, we choose a sequence ${\left(q_n\right)}_{n\ge 1}$ of real numbers satisfying

(2.7)

Then we have

hence

$$\left[n\right]=1+q_n+q_n^2+\mathrm{\dots }+q_n^{n-1}\ge n-\frac{n\left(n-1\right)}{2n}=\frac{n+1}{2}.$$

Taking into account (2.6) we deduce that

$$\Vert M_{n,q_n}{e}_2-e_2\Vert \le \frac{3}{n+1}$$

for all $n\ge 3.$ Consequently, the sequence ${\left(M_{n,q_n}{e}_2\right)}_{n\ge 1}$ converges uniformly to $e_2$ on $[0,1].$ By applying the well-known Bohman-Korovkin theorem, we can conclude that the following theorem hods:

Theorem 2.2.

If ${\left(q_n\right)}_{n\ge 1}$ is a sequence of real numbers satisfying (2.7), then for each $f\in C[0,1]$ the sequence ${\left(M_{n,q_n}f\right)}_{n\ge 1}$ converges uniformly to $f$ on$[0,1].$

Given a function $f:[0,1]\to ℝ$ as well as a positive real number $\delta ,$ let

$$\omega (f,\delta )=sup\left\{\right|f\left(x_1\right)-f\left(x_2\right)|:x_1,x_2\in [0,1],|x_1-x_2|\le \delta \}$$

denote the usual modulus of continuity of $f.$ From (2.6), by means of Theorem 2.2 in [7], we deduce the following Popoviciu-type theorem for the $M_{n,q}$-operators.

Theorem 2.3.

If $f\in C[0,1],$ then for each integer $n\ge 3$ we have

$$\Vert M_{n,q}f-f\Vert \le \frac{5}{2}\omega (f,\frac{1}{\sqrt{\left[n\right]}}).$$

Likewise, from (2.6), by means of Theorem 2.4 in [7], we deduce the following Lorentz-type theorem for the $M_{n,q}$-operators.

Theorem 2.4.

If $f\in C^1[0,1],$ then for each integer $n\ge 3$ we have

$$\Vert M_{n,q}f-f\Vert \le \frac{3+\sqrt{6}}{2\sqrt{\left[n\right]}}\omega (f^{\prime },\frac{1}{\sqrt{\left[n\right]}}).$$

Since the proofs of Theorem 2.3 and Theorem 2.4 are quite similar with those of Corollary 2.3 and Corollary 2.5, respectively, in [7], we omit them.

3. Convexity and generalized Meyer-König and Zeller operators

For any real sequence $a,$ finite of infinite, we denote by $v\left(a\right)$ the number of strict sign changes in $a.$ Given a function $f:[0,1]\to ℝ$, let $V\left(f\right)$ be the number of sign changes of $f$ in $[0,1],$ i.e.

$$V\left(f\right)=supv(f\left(x_1\right),\mathrm{\dots },f\left(x_m\right))$$

where the supremum is taken over all increasing sequences for all positive integers $m.$

An operator $L$ assigning to each function $f:[0,1]\to ℝ$ the function $Lf:[0,1]\to ℝ$ is said to be a variation diminishing operator (cf. [11]) if

$$V\left(Lf\right)\le V\left(f\right),\text{for all functions }f:[0,1]\to ℝ.$$

Theorem 3.1.

For each positive integer $n,$ the generalized Meyer-König and Zeller operator $M_{n,q}$ is a variation diminishing operator.

Proof.

By means of the well-known Descartes’ rule of signs it is easy to prove that if $a={\left(a_k\right)}_{k\ge 0}$ is a sequence of real numbers such that the power series ${\sum }_{k\ge 0}{a}_k{x}^k$ converges uniformly on $[0,1]$ to a function $g,$ then

$$V\left(g\right)\le v\left(a\right).$$

Taking this into account we have

$V\left(M_{n,q}f\right)$ $=V\left({\displaystyle \sum _{k=0}^{\mathrm{\infty }}}f\left(\frac{\left[k\right]}{\left[n+k\right]}\right)\left[\genfrac{}{}{0.0pt}{}{n+k}{k}\right]x^k\right)$

$\le V\left({\left(f\left(\frac{\left[k\right]}{\left[n+k\right]}\right)\left[\genfrac{}{}{0.0pt}{}{n+k}{k}\right]\right)}_{k\ge 0}\right)$

$\le V\left(f\right)$

for all continuous functions $f:[0,1]\to ℝ$. This proves the assertion of the theorem. ∎

Remark 3.1.

From (1.5) it follows that $V\left(M_{n,q}f\right)\le V\left(f\right)$ for every bounded function $f:[0,1]\to ℝ$ for which there exists ${lim}_{x\nearrow 1}f\left(x\right).$ Taking account of (2.1) and (2.2), by the above theorem we deduce that

(3.1) $$V\left(M_{n,q}f-p\right)=V\left(M_{n,q}\left(f-p\right)\right)\le V\left(f-p\right)$$

for every bounded function $f:[0,1]\to ℝ$ for which there exists ${lim}_{x\nearrow 1}f\left(x\right)$ and every linear polynomial $p.$ A standard reasoning based on (3.1) (see, for instance, [5], [11], [6]) yields the following theorem. ∎

Theorem 3.2.

For each positive integer $n$ the following assertions are true:

• If $f:[0,1]\to ℝ$ is an increasing (decreasing) function, then $M_{n,q}f$ is also increasing (decreasing).

• If $f:[0,1]\to ℝ$ is a convex function, then $M_{n,q}f$ is also convex and $M_{n,q}f\left(x\right)\ge f\left(x\right)$ for all $x\in [0,1].$

Like in the case of the classical Meyer-König and Zeller operators, we have the following.

Theorem 3.3.

If $f:[0,1]\to ℝ$ is a convex function, then for each $x\in [0,1]$ the sequence ${\left(M_{n,q}f\left(x\right)\right)}_{n\ge 1}$ is non-increasing.

Proof.

The assertion of the theorem being trivially true for $x=1,$ we may assume that Let $n$ be any positive integer. We have

${\displaystyle \frac{M_{n,q}f\left(x\right)-M_{n+1,q}f\left(x\right)}{P_{n,q}\left(x\right)}}$ $=q^{n+1}x{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}f\left(\frac{\left[k\right]}{\left[n+k+1\right]}\right)\left[\genfrac{}{}{0.0pt}{}{n+k+1}{k}\right]x^k$

$-{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}f\left(\frac{\left[k\right]}{\left[n+k+1\right]}\right)\left[\genfrac{}{}{0.0pt}{}{n+k+1}{k}\right]x^k$

$+{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}f\left(\frac{\left[k\right]}{[n+k]}\right)\left[\genfrac{}{}{0.0pt}{}{n+k}{k}\right]x^k.$

Using the recurrence formula

$$\left[\genfrac{}{}{0.0pt}{}{n+k+1}{k}\right]=q^{n+1}\left[\genfrac{}{}{0.0pt}{}{n+k}{k-1}\right]+\left[\genfrac{}{}{0.0pt}{}{n+k}{k}\right]$$

in the second sum we get

${\displaystyle \frac{M_{n,q}f\left(x\right)-M_{n+1,q}f\left(x\right)}{P_{n,q}\left(x\right)}}=$

$=x{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\left[\genfrac{}{}{0.0pt}{}{n+k}{k}\right]x^k\{q^{n+1}\frac{\left[n+k+1\right]}{\left[n+1\right]}f\left(\frac{\left[k\right]}{\left[n+k+1\right]}\right)-q^{n+1}\frac{\left[n+k+1\right]}{\left[n+1\right]}f\left(\frac{\left[k+1\right]}{\left[n+k+2\right]}\right)$

$-\frac{\left[n+k+1\right]}{\left[n+1\right]}f\left(\frac{\left[k+1\right]}{\left[n+k+2\right]}\right)+\frac{\left[n+k+1\right]}{\left[k+1\right]}f\left(\frac{\left[k+1\right]}{\left[n+k+1\right]}\right)\}.$

As a simple computation shows, the expression between the braces equals to

$$\frac{q^{n+2k+1}}{\left[n+k+1\right]\left[n+k+2\right]}[\frac{\left[k\right]}{\left[n+k+1\right]},\frac{\left[k+1\right]}{\left[n+k+2\right]},\frac{\left[k+1\right]}{\left[n+k+1\right]};f]\ge 0$$

because $f$ is convex. Consequently, $M_{n,q}f\left(x\right)\ge M_{n+1,q}f\left(x\right).$ ∎

Received September 2, 1999.

“Babeş Bolyai” University

Faculty of Mathematics and Computer Science

1, M. Kogălniceanu St.

3400 Cluj-Napoca, Romania

E-mail: ttrif@math.ubbcluj.ro

Added in proof. After the paper had been sent to the typography, the author found out that the sequence of generalized Meyer-König and Zeller operators considered here, had already been introduced and investigated by Luciana Lupaş, A $q$-analogue of the Meyer-König and Zeller operator, Anal. Univ. Oradea, 2, pp. 62-66, 1992. Thus, part of the results in the present paper (Theorem 2.2 and Theorem 3.3) were established for the first time in the previously quoted article by Luciana Lupaş.