${}^{\mathrm{\S }}$Department of Mathematics and Computer Science, University of Oradea, Universitatii str. no. 1, 410087 Oradea, Romania, e-mail :

lcoroianu@uoradea.ro

${}^{\ast }$Department of Mathematics and Computer Science, University of Oradea, Universitatii str. no. 1, 410087 Oradea, Romania, e-mail : galso@uoradea.ro

${}^{\bullet }$The work of both authors has been supported by a grant of the Romanian National Authority for Scientific Research, CNCS–UEFISCDI, project number PN-II-ID-PCE-2011-3-0861.

1 Introduction

Based on the Open Problem 5.5.4, pp. 324-326 in [ 14 ] , in a series of recent papers we have introduced and studied the so-called max-product operators attached to the Bernstein polynomials and to other linear Bernstein-type operators, like those of Favard-Szász-Mirakjan operators (truncated and nontruncated case), see [ 1 ] , [ 3 ] , Meyer-König and Zeller operators, see [ 4 ] , Baskakov operators, see [ 6 ] , [ 7 ] and Bleimann-Butzer-Hahn operators, see [ 5 ] .

For example, in the recent paper [ 2 ] , starting from the linear Bernstein operators $B_n(f)(x)=\sum _{k=0}^n{b}_{n,k}(x)f(k/n)$, where $b_{n,k}(x)=(\genfrac{}{}{0ex}{}{n}{k})x^k(1-x{)}^{n-k}$, written in the equivalent form

and then replacing the sum operator $\mathrm{\Sigma }$ by the maximum operator $\bigvee$, one obtains the nonlinear Bernstein operator of max-product kind

where the notation $\underset{k=0}{\overset{n}{\bigvee }}{b}_{n,k}(x)$ means $max\{b_{n,k}(x);k\in \{0,...,n\}\}$ and similarly for the numerator.

For this max-product operator, nice approximation and shape preserving properties were found in the class of positive valued functions, in e.g. [ 2 ] , [ 12 ] .

In other two recent papers [ 9 ] and [ 10 ] , this idea is applied to the Lagrange interpolation based on the Chebyshev nodes of second kind plus the endpoints, and to the Hermite-Fejér interpolation based on the Chebyshev nodes of first kind respectively, obtaining max-product interpolation operators which, in general, (for example, in the class of positive Lipschitz functions) approximates essentially better than the corresponding Lagrange and Hermite-Fejér interpolation polynomials.

Let $I=[a,b]$, $a<b$ and $f:[a,b]\to \mathbb{R}$. The max-product Lagrange interpolation operator on equidistant knots attached to the function $f$ is given by (see [ 11 ] )

where $x_{n,k}=a+(b-a)k/n$ for all $n\in \mathbb{N}$ and $k\in \{0,1,...,n\}$ and

for all $x\in I$, $n\in \mathbb{N}$ and $k\in \{0,1,...,n\}.$ Note that $L_n^{(M)}(f)$ is a well defined function. Indeed, using the fundamental Lagrange polynomials,

we observe that we can rewrite $l_{n,k}(x)$, $x\in I$, in the form

where

Then, since for any $x\in I$ we have ${\sum }_{i=0}^n{p}_{n,i}(x)=1$ it follows the existence of $i(x)\in \{0,1,...,n\}$ such that $p_{n,i(x)}(x)>0$ and noting that $c_{n,i(x)}>0$ it easily results that $l_{n,i(x)}(x)>0$ and this implies that $\underset{k=0}{\overset{n}{\bigvee }}{l}_{n,k}(x)>0$ for all $x\in I$, which means that indeed $L_n^{(M)}(f)$ is a well defined function on $[a,b]$.

The max-product operator $L_n^{(M)}(f)(x)$ is continuous on $[a,b]$ and has the interpolation properties $L_n^{(M)}(f)(x_{n,j})=f(x_{n,j})$ for all $j\in \{0,1,...,\}$.

Also, according to Corollary 3.2, (i), in [ 11 ] , for positive valued functions, i.e. for $f:[a,b]\to {\mathbb{R}}_{+}$, it satisfies the Jackson-type estimate

where ${\omega }_1{(f;\frac{b-a}{n})}_{[a,b]}$ denotes the modulus of continuity of $f$ on $[a,b]$. This estimate for the Lagrange max-product operator essentially improves for positive valued functions the order of approximation by the classical Lagrange interpolation polynomials on equidistant nodes, when as it is well-know, we can also have a very pronounced divergence phenomenon in $[a,b]$ (see e.g. Chapter 4 in the book [ 17 ] , see also [ 16 ] , [ 8 ] ).

It is worth noting that saturation and local inverse results for $L_n^{(M)}(f)(x)$ were obtained in [ 13 ] .

The plan of the paper goes as follows. In Section 2 an interesting strong localization result for the Lagrange max-product operator $L_n^{(M)}$ is obtained. At the end of the section and as consequences of this localization result, a local direct result and an interesting local shape preserving property are proved. Section 3 contains comparisons with some linear interpolation operators of rational type.

It is worth noting in Section 2 the strong localization result expressed by Theorem 2.1, that shows that if the continuous strictly positive functions $f$ and $g$ coincide on a subinterval $[\alpha ,\beta ]\subset [0,1]$, then for sufficiently large values of $n$, $L_n^{(M)}(f)$ and $L_n^{(M)}(g)$ coincide on subintervals sufficiently close to $[\alpha ,\beta ]$. Clearly, Corollary 2.4 shows that $L_n^{(M)}(f)$ is very suitable to approximate continuous functions which are constant on some subintervals. Namely, if $f$ is a continuous strictly positive function which is constant on some subintervals $[{\alpha }_i,{\beta }_i]$, $i=1,...,p$, of $[a,b]$, then for sufficiently large $n$, $L_n^{(M)}(f)$ takes the same constant values on subintervals sufficiently close to each $[{\alpha }_i,{\beta }_i]$, $i=1,...,p$.

2 Localization Results

Let $l_{n,k}$ denote the fundamental Lagrange polynomials attached to the knots $x_{n,k}=k/n$, $k\in \{0,1,...,n\}$, $n\in \mathbb{N}$.

The main result of this section is the following localization result.

Let us choose arbitrary $x\in [c,d]$ and for each $n\in \mathbb{N}$ let $j_x\in \{0,1,...,n\}$ ( $j_x$ depends on $n$ too, but there is no need at all to complicate on the notations) be such that $x\in [j_x/n,(j_x+1)/n].$ Then we know that

where $J_n(x)=\{k\in \{0,1,...,n\}:l_{n,k}(x)>0\}$ and $l_{n,k}$, $k\in \{0,1,...,n\}$ are the Lagrange fundamental polynomials attached to the knots $x_{n,k}=k/n$, $k\in \{0,1,...,n\}.$ Since $x\in [c,d]\cap [j_x/n,(j_x+1)/n]$ and since $a<c<d<b$ it is immediate that for $n\ge n_0$ where $n_0$ is chosen such that $1/n_0<min\{c-a,d-b\}$, we obtain $a<j_x/n<b$ which gives $na<j_x<nb$ for all $n\ge n_0$ (indeed, if we would suppose that there exists $n>n_0$ which does not satisfy the previous double inequalities, then we would easily get a contradiction).

It is important to notice here that $n_0$ does not depend on $x.$ From the inequalities $na<j_x<b$ it follows that if $n\ge n_0$ then for any $x\in [c,d]$ there exists ${\alpha }_x\in [a,b]$ such that $j_x=n{\alpha }_x$.

In what follows, it will serve to our purpose to use the sequence $(a_n{)}_{n\ge 1}$, $a_n=\sqrt{n}$. For this sequence there exists $n_1\in \mathbb{N}$ such that $na-a_n>0$ for all $n\ge n_1.$

Our intention is to prove as an intermediate result, that there exists an absolute constant $N_0\in \mathbb{N}$ which does not depend of $x\in [c,d]$ such that for any $n\ge N_0$ and $x\in [c,d]$ we have $\underset{k=0}{\overset{n}{\bigvee }}{l}_{n,k}(x)f(\frac{k}{n})=\underset{k\in I_{n,x}}{\bigvee }{l}_{n,k}(x)f(\frac{k}{n})$ where $I_{n,x}=\{k\in J_n(x):j_x-a_n\le k\le j_x+a_n\}.$ In order to obtain this conclusion, for $n\ge max\{n_0,n_1\}$ let us choose $k\in J_n(x)\setminus I_{n,x}$. We have two cases: i) $k+a_n<j_x$ and ii) $j_x+a_n<k.$

Case i) Firstly, note that $j_x\in J_n(x)$, because $sign(l_{n,j}(x))=(-1{)}^{n-j}\cdot (-1{)}^{n-j}=1$. Noting that $k/n<(j_x-a_n)/n$ and $nx\ge j_x,$ we get   

Then, denoting the infimum and the supremum of $f$ on $[a,b]$ with $m_f$ and $M_f$ respectively (according to the hypotheses these values are strictly positive), we get that

Since $\underset{n\to \mathrm{\infty }}{lim}\sqrt{n}\cdot \frac{m_f}{M_f}=\mathrm{\infty },$ it follows that there exists $n_2\in \mathbb{N}$, $n_2\ge max\{n_0,n_1\}$ such that $\frac{l_{n,j_x}(x)f(\frac{j_x}{n})}{l_{n,k}(x)f(\frac{k}{n})}>1$ for all $x\in [c,d]$, $n\ge n_2$ and $k\in \{0,1,...,n\}$, $k<j_x-a_n$ (as $k\notin I_{n,x}$). In addition, it is important to notice that $n_2$ does not depend on $x\in [c,d]$, but of course it depends on $f.$

Case ii) The proof is identical with the proof of the above Case i) and therefore we conclude that there exists an absolute constant $n_3\in \mathbb{N}$ which depends only on $a,b,c,d,f$ such that

for all $x\in [c,d]$, $n\ge n_3$ and $k\in \{0,1,...,n\}$, $k>j_x+a_n$.

Analyzing the results obtained in cases i)-ii), it results that for all $x\in [c,d]$, $n\ge N_0$,$N_0=max\{n_2,n_3\}$ and $k\in \{0,1,...,n\}$, with $k<j_x-a_n$ or $k>j_x+a_n$, we have

Since from the Case i) we know that $j_x\in J_n(x)$ and since this easily implies that actually $j_x\in I_{n,x}$, we obtain our preliminary result, that is

where $I_{n,x}=\{k\in J_n(x):j_x-a_n\le k\le j_x+a_n\}.$

Next, let us choose arbitrary $x\in [c,d]$ and $n\in \mathbb{N}$ so that $n\ge N_0.$ If there exists $k\in$ $I_{n,x}$ such that $k/n\notin [c,d]$ then we distinguish two cases. Either $k/n<c$ or $k/n>d.$ In the first case we observe that

Since $\underset{n\to \mathrm{\infty }}{lim}\frac{a_n+1}{n}=0,$ it results that for sufficiently large $n$ we necessarily have $\frac{a_n+1}{n}<c-a$ which clearly implies that $k/n\in [a,c].$ In the same manner, when $k/n>d$, for sufficiently large $n$ we necessarily have $k/n\in [d,b].$

Summarizing, there exists ${\tilde{N}}_1\in \mathbb{N}$ independent of any $x\in [c,d]$, such that

and in addition for any $x\in [c,d]$, $n\ge {\tilde{N}}_1$ and $k\in I_{n,x}$, we have $k/n\in [a,b].$ Also, it is easy to check that ${\tilde{N}}_1$ depends only on $a,b,c,d,f.$ We thus obtain that

and in addition for any $x\in [c,d]$, $n\ge {\tilde{N}}_1$ and $k\in I_{n,x}$, we have $k/n\in [a,b].$

Reasoning for the function $g$ exactly as in the case of the function $f$, it follows that there exists ${\tilde{N}}_2\in \mathbb{N}$ which depends only on $a,b,c,d,g$ such that

and in addition for any $x\in [c,d]$, $n\ge {\tilde{N}}_2$ and $k\in I_{n,x}$, we have $k/n\in [a,b].$ Taking $\tilde{n}=max\{{\tilde{N}}_1,{\tilde{N}}_2\}$, we easily obtain the desired conclusion.

We can easily extend the above result to arbitrary intervals, as follows.

We obtain the desired conclusion as a direct consequence of the previous theorem. Indeed, firstly to make a distinction, we denote with ${\bar{L}}_n^{(M)}$ the Lagrange max-product operator attached to functions defined on the interval $[0,1].$ In addition, in what follows, for all all $n\in \mathbb{N}$ and $k\in \{0,1,...,n\}$ we denote with $l_{n,k}^1$ the fundamental Lagrange polynomials defined on the interval $[0,1].$

Suppose now that for the two functions $f,g\in C([a,b])$ we have $f(x)=g(x)$, for all $x\in [a^{\prime },b^{\prime }]$. Let us define the function $h:[0,1]\to \mathbb{[}a,b]$, $h(y)=a+(b-a)y.$ It is immediate that for any $x\in [a,b]$ there exists an unique $y(x)=h^{-1}(x)\in [0,1]$ such that $f(x)=(f\circ h)(y(x))$ and $g(x)=(g\circ h)(y(x))$.

Then we observe that for any $x\in [a,b]$ we have

The above equalities imply

and analogously $L_n^{(M)}(g)(x)={\bar{L}}_n^{(M)}(g\circ h)(y(x))$, for all $x\in [a,b].$

Then, our result is immediate by applying Theorem 2.1 to ${\bar{L}}_n^{(M)}(g\circ h)(y(x))$ and ${\bar{L}}_n^{(M)}(g\circ h)(y(x))$, where recall that $f\circ h,g\circ h:[0,1]\to [0,+\mathrm{\infty })$, $y(x)=h^{-1}(x)$ and $h:[0,1]\to [a,b]$, $h(x)=a+(b-a)x$.

The next direct approximation result is now an immediate consequence of the localization result in Theorem 2.2, as follows.

Let us define the function $F:[a,b]\to \mathbb{R}$,

The hypothesis imply that $F$ is continuous and strictly positive on $[a,b]$ and according to Corollary 3.2 in [ 11 ] it results that

Since by the definition of $F$ we have ${\omega }_1(F,\frac{b-a}{n}{)}_{[a,b]}\le {\omega }_1(f,\frac{b-a}{n}{)}_{[a,b]}$ and since by the relation (2.1) it easily follows ${\omega }_1(f,\frac{b-a}{n}{)}_{[a,b]}\le C_0(b-a)/n$, we get

Now, let us choose arbitrary $c,d\in [a^{\prime },b^{\prime }]$ such that $a^{\prime }<c<d<b^{\prime }.$ Then, by Theorem 2.2 (applicable to $f$ and $F$) it results the existence of $\tilde{n}\in \mathbb{N}$ which depends only on $a,b,a^{\prime },b^{\prime }c,d,f,F$ such that $L_n^{(M)}(F)(x)=L_n^{(M)}(f)(x)$ for all $x\in [c,d].$ But since actually the function $F$ depends on the function $f$, it is clear that in fact $\tilde{n}$ depends only on $a,b,a^{\prime },b^{\prime }c,d$ and $f$.

Therefore, for arbitrary $x\in [c,d]$ and $n\in \mathbb{N}$ with $n\ge \tilde{n}$ we obtain

where $C_0$ and $\tilde{n}$ depend only on $a,b,a^{\prime },b^{\prime }c,d$ and $f$.

Now, denoting

we finally obtain

with $C=max\{2C_0(b-a),C_1\}$ depending only on $a,b,c,d$ and $f$. This proves the corollary. At the end of this section, as a consequence of the localization result in Theorem 2.2 we present a locally constant preserving property.

Let $g:[a,b]\to {\mathbb{R}}_{+}$ be given by $g(x)=\alpha >0$ for all $x\in [a,b]$. Since $f(x)=g(x)$ for all $x\in [a^{\prime },b^{\prime }]$ and since obviously $L_n^{(M)}(g)(x)=\alpha$ for all $x\in [a,b]$, by Theorem 2.2 we easily obtain the desired conclusion.

3 Final Remarks

Let us note that in Hermann-Vértesi [ 15 ] , starting from a Lagrange interpolatory process (convergent or not)

with

new linear interpolatory (rational) operators of the form

are constructed, for which in the case when $r>2$ and the knots $x_{n,k}$ satisfy some special requirements (e.g. some Jacobi knots), the Jackson-type order of approximation

is obtained (see e.g. Theorem 3.2 in Hermann-Vértesi [ 15 ] ).

In other words, for the linear rational construction $R_n(f)(x)$, we get the same order of approximation as for the interpolatory max-product operator (which is piecewise rational)

But clearly that with respect to $R_n(f)(x)$, the max-product rational operator $L_n^{(M)}(f)(x)$ presents several advantages, pointed out by the next remarks.