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\title[Max-product operators of Kantorovich type]{Approximation by max-product operators of Kantorovich type
\\ \\
\small{Dedicated to Prof. dr. Heiner Gonska, dr. h.c. of "Babes-Bolyai" University in Cluj-Napoca, on the occasion of his 70th birthday}}
\author{Lucian Coroianu and Sorin G. Gal}
\address{University of Oradea \\ Department of Mathematics and Computer Sciences\\ \\
Universitatii Street, No. 1,\\
410087 Oradea,\\
Romania}
\email{lcoroianu@uoradea.ro, galso@uoradea.ro}
\subjclass{41A35, 41A25, 41A20}
\keywords{max-product operators, max-product operators
of Kantorovich kind, uniform approximation, shape preserving properties, localization results,
max-product Kantorovich-Choquet operators.}
\begin{abstract}
We associate to various linear Kantorovich type approximation operators, nonlinear
max-product operators for which we obtain quantitative approximation results
in the uniform norm, shape preserving properties and localization results.
\end{abstract}
\maketitle
\footnote[1]{Paper presented at the Fourth International Conference on Numerical Analysis and Approximation Theory, Cluj-Napoca, 2018}
\section{Introduction}

The general form of a linear and positive discrete operator attached to $f:I\to [0,
+\infty)$ can be defined by
$$D_{n}(f)(x)=\sum_{k\in I_{n}}p_{n, k}(x)f(x_{n, k}), x\in I, n\in \mathbb{N},$$
where $p_{n, k}(x)$ are various kinds of function basis on $I$ with $\sum_{k\in I_{n}}p_{n, k}(x)=1$, $I_{n}$ are
finite or infinite families of indices and $\{x_{n, k} ; k\in I_{n}\}$
represents a division of $I$.

Based on the Open Problem 5.5.4, pp. 324-326 in \cite{Gal}, to each $D_{n}(f)(x)$, can be attached the max-product type operator defined by
\begin{equation}\label{eeq1}
L_{n}^{(M)}(f)(x)=\frac{\bigvee_{k\in I_{n}}p_{n, k}(x)\cdot f(x_{n, k})}{%
\bigvee_{k\in I_{n}}p_{n, k}(x)}, x\in I, n\in \mathbb{N}.
\end{equation}
Here $\bigvee_{k\in A} a_{k}=\sup_{k\in A}a_{k}$.

Thus, in a series of 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-Sz\'{a}sz-Mirakjan operators (truncated and
nontruncated case), Baskakov operators (truncated and nontruncated case),
Meyer-K\"onig and Zeller operators and Bleimann-Butzer-Hahn operators. All
these results were collected in the very recent research monograph \cite%
{Book-BCG}.

\begin{remark}
The max-product operators can also be naturally called as {\bf possibilistic operators}, since they can be obtained by analogy with the Feller probabilistic scheme used to generate positive and linear operators, by replacing the probability ($\sigma$-additive), with a maxitive set function and the classical integral with the possibilistic integral (see, e.g. \cite{Book-BCG}, Chapter 10, Section 10.2). If, for example, $p_{n, k}(x)$, $n\in \mathbb{N}$, $k=0, ..., n$ is a polynomial basis, then the operators $L_{n}^{(M)}(f)(x)$ become piecewise rational functions.
\end{remark}

Now, to each max-product operator $L_{n}^{(M)}$, we can formally attach its
Kantorovich variant, defined by
\begin{equation}  \label{eeq2}
LK_{n}^{(M)}(f)(x)=\frac{\bigvee_{k\in I_{n}}p_{n, k}(x)\cdot (1/(x_{n,
k+1}-x_{n, k}))\cdot \int_{x_{n, k}}^{x_{n, k+1}}f(t)d t}{\bigvee_{k\in
I_{n}}p_{n, k}(x)},
\end{equation}
with $\{x_{n, k} ; k\in I_{n}\} $ a division of the finite or infinite interval $I$.

The goal of this paper is to study these Kantorovich-type versions for
various max-product operators. Firstly, we prove that these operators are
subadditive, positively homogeneous and monotone. For continuous functions
we prove quantitative estimates, in most of the cases very good Jackson type
estimates, shape preserving properties and localization results.

\section{Uniform and pointwise approximation}

Keeping the notations in the formulas (\ref{eeq1}) and (\ref{eeq2}), let us
denote
\begin{equation*}
C_{+}(I)=\{f:I\to \mathbb{R}_{+} ; f \mbox{ is continuous on } I\},
\end{equation*}
where $I$ is a bounded or unbounded interval and suppose that all $p_{n,
k}(x)$ are continuous functions on $I$, satisfying $p_{n, k}(x)\ge 0$, for
all $x\in I, n\in \mathbb{N}, k\in I_{n}$ and $\sum_{k\in I_{n}}p_{n,
k}(x)=1 $, for all $x\in I, n\in \mathbb{N}$.

In many cases, for the Kantorovich max-product operator $K_{n}^{(M)}$ we
could deduce quantitative estimates in approximation, by using the
elaborated methods we used for the Bernstein-type max-product in the book
\cite{Book-BCG}. However, here we will use a more simple method, which will
be based on the already obtained estimates for the original type max-product
operators denoted by $L_{n}^{(M)}$.

Firstly, we present the following result.

\begin{lemma} (i) For any $f\in C_{+}(I)$, $LK_{n}^{(M)}(f)$%
is continuous on $I$ ;

(ii) If $f\leq g$ then $LK_{n}^{(M)}(f)\leq
LK_{n}^{(M)}(g)$ ;

(iii) $LK_{n}^{(M)}(f+g)\leq LK_{n}^{(M)}(f)+LK_{n}^{(M)}(g)$ ;

(iv) If $f\in C_{+}(I)$ and $\lambda \geq 0$
then $LK_{n}^{(M)}(\lambda f)=\lambda LK_{n}^{(M)}(f)$ ;

(v) If $LK_{n}^{(M)}(e_{0})=e_{0}$, where $e_{0}(x)=1$, for all $%
x\in I$, then for any $f\in C_{+}(I)$, we have 
\begin{equation*}
\left\vert LK_{n}^{(M)}(f)(x)-f(x)\right\vert \leq \left[ 1+\frac{1}{\delta }%
LK_{n}^{(M)}(\varphi _{x})(x)\right] \omega _{1}(f;\delta ),
\end{equation*}%
for any $x\in I$ and $\delta >0$. Here, $\varphi
_{x}(t)=\left\vert t-x\right\vert $, $t\in I$ and $%
\omega_{1}(f; \delta)=\sup\{|f(x)-f(y)|; x, y\in I, |x-y|\le \delta\}$ ;

(vi) $\left\vert LK_{n}^{(M)}(f)-LK_{n}^{(M)}(g)\right\vert \leq
LK_{n}^{(M)}\left( \left\vert f-g\right\vert \right)$.
\end{lemma}
\begin{proof} The proofs of (i)-(iv) are immediate from the definition of $%
K_{n}^{(M)}$. As for the proof of (v) and (vi), we exactly follow the proof
of e.g., Theorem 1.1.2, pp. 16-17 in \cite{Book-BCG}. 
\end{proof}
\begin{lemma} With the notations in (\ref{eeq1}) and (\ref%
{eeq2}), suppose that, in addition, $|x_{n, k+1}-x_{n, k}|\le \frac{C}{n+1}$
for all $k\in I_{n}$, with $C>0$ an absolute constant. Then, for all $x\in I$
and $n\in \mathbb{N}$, we have
\begin{equation*}
LK_{n}^{(M)}(\varphi_{x})(x)\le L_{n}^{(M)}(\varphi_{x})(x)+\frac{C}{n+1}.
\end{equation*}
\end{lemma}
\begin{proof}
If $f\in C_{+}(I)$, then by the integral mean value
theorem, there exists $\xi_{n, k}\in (x_{n, k}, x_{n, k+1})$, such that $%
\int_{x_{n, k}}^{x_{n, k+1}}f(t)dt=(x_{n, k+1}-x_{n, k})\cdot f(\xi_{n, k})$%
, which immediately leads to
\begin{equation}  \label{meanv}
LK_{n}^{(M)}(f)(x)=\frac{\bigvee_{k\in I_{n}}p_{n, k}(x)\cdot f(\xi_{n, k})}{%
\bigvee_{k\in I_{n}}p_{n, k}(x)}.
\end{equation}
Applying this form for $f(t)=\varphi_{x}(t)$, we get
\begin{equation*}
LK_{n}^{(M)}(\varphi_{x})(x)=\frac{\bigvee_{k\in I_{n}}p_{n, k}(x)\cdot |\xi_{n, k}-x|%
}{\bigvee_{k\in I_{n}}p_{n, k}(x)}
\end{equation*}
\begin{equation*}
\le \frac{\bigvee_{k\in I_{n}}p_{n, k}(x)\cdot |\xi_{n, k}-x_{n, k}|}{%
\bigvee_{k\in I_{n}}p_{n, k}(x)}+L_{n}^{(M)}(\varphi_{x})(x)\le \frac{C}{n+1}%
+L_{n}^{(M)}(\varphi_{x})(x),
\end{equation*}
which proves the lemma. 
\end{proof}
\begin{corollary} With the notations in (\ref{eeq1}) and (\ref%
{eeq2}) and supposing that, in addition, $|x_{n, k+1}-x_{n, k}|\le \frac{C}{%
n+1}$ for all $k\in I_{n}$, for any $f\in C_{+}(I)$, we have 
\begin{equation}
\left\vert LK_{n}^{(M)}(f)(x)-f(x)\right\vert \leq 2\left
[\omega_{1}(f;L_{n}^{(M)}(\varphi _{x})(x))+\omega _{1}(f;C/(n+1) )\right ]\label{eeq4}
\end{equation}%
for any $x\in I$ and $n\in \mathbb{N}$.
\end{corollary}
\begin{proof} By using Lemma 2.2, from the estimate in Lemma 2.1, (v), we
immediately get
\begin{equation*}
\left\vert LK_{n}^{(M)}(f)(x)-f(x)\right\vert \leq
2\omega_{1}(f;L_{n}^{(M)}(\varphi _{x})(x)+C/(n+1))
\end{equation*}
\begin{equation*}
\le 2\left[\omega_{1}(f;L_{n}^{(M)}(\varphi_{x})(x))+\omega _{1}(f;C/(n+1)
)\right ],
\end{equation*}
which proves the corollary. 
\end{proof}
This corollary shows that knowing quantitative estimates in approximation by
a given max-product operator, we can deduce a quantitative estimate for its
Kantorovich variant. Also, this method does not worsen the orders of
approximation of the original operators. Let us exemplify below for several
known max-product operators.

Firstly, let us choose $p_{n, k}(x)={\binom{n}{k}}x^{k}(1-x)^{n-k}$, $I=[0,
1]$, $I_{n}=\{0, ..., n-1\}$ and $x_{n, k}=\frac{k}{n+1}$. In this case, $%
L_{n}^{(M)}$ in (\ref{eeq1}) become the Bernstein max-product operators. Let
us denote by $BK_{n}^{(M)}$ their Kantorovich variant, given by the formula
\begin{equation}  \label{bk}
BK_{n}^{(M)}(f)(x)=\frac{\bigvee_{k=0}^{n}{\binom{n}{k}}x^{k}(1-x)^{n-k}%
\cdot (n+1) \int_{k/(n+1)}^{(k+1)/(n+1)}f(t)d t}{\bigvee_{k=0}^{n}{\binom{n}{%
k}}x^{k}(1-x)^{n-k}}.
\end{equation}
We can state the following result.
\begin{theorem} (i) If $f\in C_{+}([0, 1])$, then we have
\begin{equation*}
|BK_{n}^{(M)}(f)(x)-f(x)|\le 24\omega_{1}(f; 1/\sqrt{n+1})+2\omega_{1}(f;
1/(n+1)), x\in [0, 1], n\in \mathbb{N}.
\end{equation*}
(ii) If $f\in C_{+}([0, 1])$ is concave on $[0, 1]$, then we have
\begin{equation*}
|BK_{n}^{(M)}(f)(x)-f(x)|\le 6\omega_{1}(f; 1/n), x\in [0, 1], n\in \mathbb{N%
}.
\end{equation*}
(iii) If $f\in C_{+}([0, 1])$ is strictly positive on $[0, 1]$, then we have
\begin{equation*}
|BK_{n}^{(M)}(f)(x)-f(x)|\le 2\omega_{1}(f; 1/n)\cdot \left (\frac{%
n\omega_{1}(f; 1/n)}{m_{f}}+4\right )+ 2\omega_{1}(f; 1/n),
\end{equation*}
for all $x\in [0, 1], n\in \mathbb{N}$, where $m_{f}=\min\{f(x); x\in [0,
1]\}$.
\end{theorem}
\begin{proof} (i) is immediate from Corollary 2.3 (with $C=1$) and from
Theorem 2.1.5, p. 30, in \cite{Book-BCG}.

(ii) is immediate from Corollary 2.3 (with $C=1$) and from Corollary 2.1.10,
p. 36 in \cite{Book-BCG}.

(iii) is immediate from Corollary 2.3 (with $C=1$) and from Theorem 2.2.18,
p. 63 in \cite{Book-BCG}. 
\end{proof}
Now, let us choose $p_{n, k}(x)=\frac{(n x)^{k}}{k !}$, $I=[0, +\infty)$, $%
I_{n}=\{0, ..., n, ...,\}$ and $x_{n, k}=\frac{k}{n+1}$. In this case, $%
L_{n}^{(M)}$ in (\ref{eeq1}) become the non-truncated
Favard-Sz\'asz-Mirakjan max-product operators. Let us denote by $%
FK_{n}^{(M)} $ their Kantorovich variant defined by
\begin{equation}  \label{fk}
FK_{n}^{(M)}(f)(x)=\frac{\bigvee_{k=0}^{\infty}\frac{(n x)^{k}}{k !}\cdot
(n+1) \int_{k/(n+1)}^{(k+1)/(n+1)}f(t)d t}{\bigvee_{k=0}^{\infty} \frac{(n
x)^{k}}{k !}}.
\end{equation}
We can state the following result.
\begin{theorem} (i) If $f:[0, +\infty)\to [0, +\infty)$ is
bounded and continuous on $[0, +\infty)$, then we have
\begin{equation*}
|FK_{n}^{(M)}(f)(x)-f(x)|\le 16\omega_{1}(f; \sqrt{x}/\sqrt{n}%
)+2\omega_{1}(f; 1/n), x\in [0, +\infty), n\in \mathbb{N}.
\end{equation*}
(ii) If $f:[0, +\infty)\to [0, +\infty)$ is continuous, bounded,
non-decreasing, concave function on $[0, +\infty)$, then we have
\begin{equation*}
|FK_{n}^{(M)}(f)(x)-f(x)|\le 4\omega_{1}(f; 1/n), x\in [0, +\infty), n\in
\mathbb{N}.
\end{equation*}
\end{theorem}
\begin{proof} (i) is immediate from Corollary 2.3 (with $C=1$) and from
Theorem 3.1.4, p. 162, in \cite{Book-BCG}.

(ii) is immediate from Corollary 2.3 (with $C=1$) and from Corollary 3.1.8,
p. 168 in \cite{Book-BCG}. 
\end{proof}
If we choose $p_{n, k}(x)=\frac{(n x)^{k}}{k !}$, $I=[0, 1]$, $I_{n}=\{0,
..., n\}$ and $x_{n, k}=\frac{k}{n+1}$. In this case, $L_{n}^{(M)}$ in (\ref%
{eeq1}) become the truncated Favard-Sz\'asz-Mirakjan max-product operators.
Let us denote by $TK_{n}^{(M)}$ their Kantorovich variant given by the
formula
\begin{equation} \label{fkt}
TK_{n}^{(M)}(f)(x)=\frac{\bigvee_{k=0}^{n}\frac{(n x)^{k}}{k !}\cdot (n+1)
\int_{k/(n+1)}^{(k+1)/(n+1)}f(t)d t}{\bigvee_{k=0}^{n} \frac{(n x)^{k}}{k !}}.
\end{equation}
We can state the following result.
\begin{theorem} (i) If $f\in C_{+}([0, 1])$, then we have
\begin{equation*}
|TK_{n}^{(M)}(f)(x)-f(x)|\le 12\omega_{1}(f; 1/\sqrt{n})+2\omega_{1}(f;
1/n), x\in [0, 1], n\in \mathbb{N}.
\end{equation*}
(ii) If $f\in C_{+}([0, 1])$ is non-decreasing, concave function on $[0, 1]$%
, then we have
\begin{equation*}
|TK_{n}^{(M)}(f)(x)-f(x)|\le 4\omega_{1}(f; 1/n), x\in [0, 1], n\in \mathbb{N}.
\end{equation*}
\end{theorem}
\begin{proof} (i) is immediate from Corollary 2.3 (with $C=1$) and from
Theorem 3.2.5, p. 178, in \cite{Book-BCG}.

(ii) is immediate from Corollary 2.3 (with $C=1$) and from Corollary 3.2.7,
p. 182 in \cite{Book-BCG}.
\end{proof}
Now, let us choose $p_{n, k}(x)={\binom{n+k-1 }{k}}x^{k}/(1+x)^{n+k}$, $%
I=[0, +\infty)$, $I_{n}=\{0, ..., n, ...,\}$ and $x_{n, k}=\frac{k}{n+1}$.
In this case, $L_{n}^{(M)}$ in (\ref{eeq1}) become the non-truncated
Baskakov max-product operators. Let us denote by $VK_{n}^{(M)}$ their
Kantorovich variant defined by
\begin{equation} \label{bask}
VK_{n}^{(M)}(f)(x)=\frac{\bigvee_{k=0}^{\infty}{\binom{n+k-1 }{k}}\frac{x^{k}%
}{(1+x)^{n+k}}\cdot (n+1) \int_{k/(n+1)}^{(k+1)/(n+1)}f(t)d t}{%
\bigvee_{k=0}^{\infty}{\binom{n+k-1 }{k}}\frac{x^{k}}{(1+x)^{n+k}}}.
\end{equation}
We can state the following result.

\begin{theorem} (i) If $f:[0, +\infty)\to [0, +\infty)$ is
bounded and continuous on $[0, +\infty)$, then for all $x\in [0, +\infty)$
and $n\ge 3$, we have
\begin{equation*}
|VK_{n}^{(M)}(f)(x)-f(x)|\le 24\omega_{1}(f; \sqrt{x(x+1)}/\sqrt{n-1}%
)+2\omega_{1}(f; 1/(n+1)).
\end{equation*}
(ii) If $f:[0, +\infty)\to [0, +\infty)$ is continuous, bounded,
non-decreasing, concave function on $[0, +\infty)$, then for $x\in [0,
+\infty)$ and $n\ge 3$ we have
\begin{equation*}
|VK_{n}^{(M)}(f)(x)-f(x)|\le 4\omega_{1}(f; 1/n).
\end{equation*}
\end{theorem}
\begin{proof} (i) is immediate from Corollary 2.3 (with $C=1$) and from
Theorem 4.1.6, p. 196, in \cite{Book-BCG}.

(ii) is immediate from Corollary 2.3 (with $C=1$) and from Corollary 4.1.9,
p. 206 in \cite{Book-BCG}. 
\end{proof}
If we choose $p_{n, k}(x)={\binom{n+k-1 }{k}}x^{k}/(1+x)^{n+k}$, $I=[0, 1]$,
$I_{n}=\{0, ..., n\}$ and $x_{n, k}=\frac{k}{n+1}$, then in this case, $%
L_{n}^{(M)}$ in (\ref{eeq1}) become the truncated Baskakov max-product
operators. Let us denote by $UK_{n}^{(M)}$ their Kantorovich variant defined by
\begin{equation}  \label{baskt}
UK_{n}^{(M)}(f)(x)=\frac{\bigvee_{k=0}^{n}{\binom{n+k-1 }{k}}\frac{x^{k}}{%
(1+x)^{n+k}}\cdot (n+1) \int_{k/(n+1)}^{(k+1)/(n+1)}f(t)d t}{%
\bigvee_{k=0}^{\infty}{\binom{n+k-1 }{k}}\frac{x^{k}}{(1+x)^{n+k}}}.
\end{equation}
We can state the following result.
\begin{theorem} (i) If $f\in C_{+}([0, 1])$, then we have,
\begin{equation*}
|UK_{n}^{(M)}(f)(x)-f(x)|\le 48\omega_{1}(f; 1/\sqrt{n+1})+2\omega_{1}(f;
1/(n+1)), x\in [0, 1], n\ge 2.
\end{equation*}
(ii) If $f\in C_{+}([0, 1])$ is non-decreasing, concave function on $[0, 1]$, then we have
\begin{equation*}
|UK_{n}^{(M)}(f)(x)-f(x)|\le 6\omega_{1}(f; 1/n), x\in [0, 1], n\in \mathbb{N}.
\end{equation*}
\end{theorem}
\begin{proof} (i) is immediate from Corollary 2.3 (with $C=1$) and from
Theorem 4.2.6, p. 217, in \cite{Book-BCG}.

(ii) is immediate from Corollary 2.3 (with $C=1$) and from Corollary 4.2.9,
p. 223 in \cite{Book-BCG}. 
\end{proof}
Now, let us choose $p_{n, k}(x)={\binom{n+k }{k}}x^{k}$, $I=[0, 1]$, $%
I_{n}=\{0, ..., n, ...\}$ and $x_{n, k}=\frac{k}{n+1+k}$. In this case, $%
L_{n}^{(M)}$ in (\ref{eeq1}) become the Meyer-K\"onig and Zeller max-product
operators. Also, it is easy to see that $|x_{n, k+1}-x_{n, k}|\le \frac{1}{%
n+1}$, for all $k\in I_{n}$. Let us denote by $ZK_{n}^{(M)}$ their
Kantorovich variant defined by
\begin{equation}  \label{mkzk}
ZK_{n}^{(M)}(f)(x)=\frac{\bigvee_{k=0}^{\infty}{\binom{n+k }{k}}x^{k}\cdot
\frac{(n+k+1)(n+k+2)}{n+1}\int_{k/(n+1+k)}^{(k+1)/(n+k+2)}f(t)d t}{%
\bigvee_{k=0}^{\infty}{\binom{n+k }{k}}x^{k}}.
\end{equation}
The following result holds.
\begin{theorem} (i) If $f\in C_{+}([0, 1])$, then for $n\ge 4$,
$x\in [0, 1]$, we have
\begin{equation*}
|ZK_{n}^{(M)}(f)(x)-f(x)|\le 36\omega_{1}(f; \sqrt{x}(1-x)/\sqrt{n}%
)+2\omega_{1}(f; 1/n).
\end{equation*}
(ii) If $f\in C_{+}([0, 1])$ is non-decreasing concave function on $[0, 1]$,
then for $x\in [0, 1]$ and $n\ge 2x$ we have
\begin{equation*}
|ZK_{n}^{(M)}(f)(x)-f(x)|\le 4\omega_{1}(f; 1/n).
\end{equation*}
\end{theorem}
\begin{proof} (i) is immediate from Corollary 2.3 (with $C=1$) and from
Theorem 6.1.4, p. 248, in \cite{Book-BCG}.

(ii) is immediate from Corollary 2.3 (with $C=1$) and from Corollary 6.1.7,
p. 256 in \cite{Book-BCG}. 
\end{proof}
In what follows, let us choose $p_{n, k}(x)=h_{n, k}(x)$-the fundamental
Hermite-Fej\'er interpolation polynomials based on the Chebyshev knots of
first kind $x_{n, k}=\cos\left (\frac{2(n-k)+1}{2(n+1)}\pi\right )$, $I=[-1,
1]$, and $I_{n}=\{0, ..., n\}$. In this case, $L_{n}^{(M)}$ in (\ref{eeq1})
become the Hermite-Fej\'er max-product operators. Also, applying he mean
value theorem to $\cos$, it is easy to see that $|x_{n, k+1}-x_{n, k}|\le
\frac{4}{n+1}$, for all $k\in I_{n}$. Let us denote by $HK_{n}^{(M)}$ their
Kantorovich variant defined by
\begin{equation}  \label{hk}
HK_{n}^{(M)}(f)(x)=\frac{\bigvee_{k=0}^{n}h_{n, k}(x)\cdot \frac{1}{x_{n,
k}-x_{n, k+1}}\cdot \int_{x_{n, k}}^{x_{n, k+1}}f(t)d t}{\bigvee_{k=0}^{%
\infty}h_{n, k}(x)},
\end{equation}
where $x_{n, k}=\cos\left (\frac{2(n-k)+1}{2(n+1)}\pi\right )$.

The following result holds.
\begin{theorem} If $f\in C_{+}([-1, 1])$, then for $n\in
\mathbb{N}$, $x\in [-1, 1]$, we have
\begin{equation*}
|HK_{n}^{(M)}(f)(x)-f(x)|\le 30\omega_{1}(f; 1/n).
\end{equation*}
\end{theorem}
\begin{proof} 
It is immediate from Corollary 2.3 (with $C=4$) and from
Theorem 7.1.5, p. 286, in \cite{Book-BCG}. 
\end{proof}
Now, let us consider choose $p_{n, k}(x)=e^{-|x-k/(n+1)|}$, $I=(-\infty,
+\infty)$, $I_{n}=\mathbb{Z}$-the set of integers and $x_{n, k}=\frac{k}{n+1}
$. In this case, $L_{n}^{(M)}$ in (\ref{eeq1}) become the Picard max-product
operators. Let us denote by ${\mathcal{P}}K_{n}^{(M)}$ their Kantorovich
variant defined by
\begin{equation}  \label{pick}
{\mathcal{P}}K_{n}^{(M)}(f)(x)=\frac{\bigvee_{k=0}^{\infty}e^{-|x-k/(n+1)|}%
\cdot (n+1) \int_{k/(n+1)}^{(k+1)/(n+1)}f(t)d t}{\bigvee_{k=0}^{%
\infty}e^{-|x-k/(n+1)|}}.
\end{equation}
We can state the following result.
\begin{theorem} If $f:\mathbb{R}\to [0, +\infty)$ is bounded
and uniformly continuous on $\mathbb{R}$, then we have
\begin{equation*}
|{\mathcal{P}}K_{n}^{(M)}(f)(x)-f(x)|\le 6\omega_{1}(f; 1/n), x\in \mathbb{R}%
, n\in \mathbb{N}.
\end{equation*}
\end{theorem}
\begin{proof}  It is immediate from Corollary 2.3 (with $C=1$) and from
Theorem 10.3.1, p. 423, in \cite{Book-BCG}. 
\end{proof}
In what follows, let us choose $p_{n, k}(x)=e^{-(x-k/(n+1))^{2}}$, $%
I=(-\infty, +\infty)$, $I_{n}=\mathbb{Z}$-the set of integers and $x_{n, k}=%
\frac{k}{n+1}$. In this case, $L_{n}^{(M)}$ in (\ref{eeq1}) become the
Weierstrass max-product operators. Let us denote by ${\mathcal{W}}%
K_{n}^{(M)} $ their Kantorovich variant defined by
\begin{equation}  \label{weik}
{\mathcal{W}}K_{n}^{(M)}(f)(x)=\frac{\bigvee_{k=0}^{%
\infty}e^{-(x-k/(n+1))^{2}}\cdot (n+1) \int_{k/(n+1)}^{(k+1)/(n+1)}f(t)d t}{%
\bigvee_{k=0}^{\infty}e^{-(x-k/(n+1))^{2}}}.
\end{equation}
We can state the following result.
\begin{theorem} If $f:\mathbb{R}\to [0, +\infty)$ is bounded
and uniformly continuous on $\mathbb{R}$, then we have
\begin{equation*}
|{\mathcal{W}}K_{n}^{(M)}(f)(x)-f(x)|\le 4\omega_{1}(f; 1/\sqrt{n}%
)+2\omega_{1}(f; 1/n), x\in \mathbb{R}, n\in \mathbb{N}.
\end{equation*}
\end{theorem}
\begin{proof} It is immediate from Corollary 2.3 (with $C=1$) and from
Theorem 10.3.3, p. 425, in \cite{Book-BCG}. 
\end{proof}
At the end of this section, let us choose $p_{n, k}(x)=\frac{1}{%
n^{2}(x-k/n)^{2}+1}$, $I=(-\infty, +\infty)$, $I_{n}=\mathbb{Z}$-the set of
integers and $x_{n, k}=\frac{k}{n+1}$. In this case, $L_{n}^{(M)}$ in (\ref%
{eeq1}) become the Poisson-Cauchy max-product operators. Let us denote by ${%
\mathcal{C}}K_{n}^{(M)}$ their Kantorovich variant
\begin{equation}  \label{pck}
{\mathcal{C}}K_{n}^{(M)}(f)(x)=\frac{\bigvee_{k=0}^{\infty}\frac{1}{%
n^{2}(x-k/(n+1))^{2}+1}\cdot (n+1) \int_{k/(n+1)}^{(k+1)/(n+1)}f(t)d t}{%
\bigvee_{k=0}^{\infty}\frac{1}{n^{2}(x-k/(n+1))^{2}+1}}.
\end{equation}
We can state the following result.
\begin{theorem} If $f:\mathbb{R}\to [0, +\infty)$ is bounded
and uniformly continuous on $\mathbb{R}$, then we have
\begin{equation*}
|{\mathcal{C}}K_{n}^{(M)}(f)(x)-f(x)|\le 6\omega_{1}(f; 1/n), x\in \mathbb{R}%
, n\in \mathbb{N}.
\end{equation*}
\end{theorem}
\begin{proof}  It is immediate from Corollary 2.3 (with $C=1$) and from
Theorem 10.3.5, p. 426, in \cite{Book-BCG}. 
\end{proof}
\begin{remark} All the Kantorovich kind max-product operators $%
LK_{n}^{(M)}$ given by (\ref{eeq2}) are defined and used for approximation
of positive valued functions. But, they can be used for approximation of
lower bounded functions of variable sign too, by introducing the new
operators
\begin{equation*}
N_{n}^{(M)}(f)(x)=LK_{n}^{(M)}(f+c)(x)-c,
\end{equation*}
where $c>0$ is such that $f(x)+c>0$, for all $x$ in the domain of definition
of $f$.

It is easy to see that the operators $N_{n}^{(M)}$ give the same
approximation orders as $LK_{n}^{(M)}$.
\end{remark}

\section{Shape preserving properties for the Bernstein-Kantorovich max-product operators}

In this section we deal with the shape preserving properties of the
Bernstein-Kantorovich max-product operators $BK_{n}^{(M)}$ given by (\ref{bk}).

We can prove the following.
\begin{theorem} Let $f\in C_{+}([0, 1])$. 

(i) If $f$ is non-decreasing (non-increasing) on $[0, 1]$, then for
all $n\in \mathbb{N}$, $BK_{n}^{(M)}(f)$ is non-decreasing (non-increasing,
respectively) on $[0, 1]$. 

(ii) If $f$ is quasi-convex on $[0,1]$ then for all $n\in \mathbb{N}$%
, $BK_{n}^{(M)}(f)$ is quasi-convex on $[0, 1]$. Here quasi-convexity on $%
[0, 1]$ means that $f(\lambda x+(1-\lambda)y)\le \max\{f(x), f(y)\}$, for
all $x, y, \lambda\in [0, 1]$.
\end{theorem}
\begin{proof} (i) By using the formula (\ref{meanv}) for $LK_{n}^{(M)}$,
we can write $BK_{n}^{(M)}(f)$ under the form
\begin{equation*}
BK_{n}^{(M)}(f)(x)=\frac{\bigvee_{k=0}^{n}{\binom{n}{k}}x^{k}(1-x)^{n-k}%
\cdot f(\xi_{n, k})}{\bigvee_{k=0}^{n}{\binom{n}{k}}x^{k}(1-x)^{n-k}},
\end{equation*}
where $\xi_{n, k}\in (x_{n, k}, x_{n, k+1})$, for all $k=0, ..., n$.

Then, by analogy with the proofs for the Bernstein max-product operators
(see \cite{Book-BCG}, pp. 39-41, the proofs for the Bernstein-Kantorovich
max-product operators, will be based on the properties of the functions
\begin{equation*}
f_{k, n, j}(x)=\frac{{\binom{n}{k}}}{{\binom{n}{j}}}\cdot \left (\frac{x}{1-x%
}\right)^{k-j}\cdot f(\xi_{n, k}).
\end{equation*}
Now, analyzing the proofs of Lemma 2.1.13, Corollary 2.1.14, Theorem 2.1.15
and Corollary 2.1.16 in \cite{Book-BCG}, pp. 39-41, it is easy to see that
they work identically for the above $f_{k, n, j}$ too and we immediately
obtain the required conclusions.

(ii) Since as in the case of the max-product Bernstein operators in
Corollary 2.1.18, p. 41 in \cite{Book-BCG}, this point is based on the
properties from the above point (i) and on the properties in the above Lemma
2.1, (i)-(iv), we easily get the required conclusion for this point too.
\end{proof}

In what follows, we will prove that $BK_{n}^{(M)}$ preserves quasi-concavity
too. This property holds in the case of the operator $B_{n}^{(M)}$ (By
Theorem 5.1 in \cite{Co-GAL-AA}). However, it is difficult to adapt the
proof to our case. Instead, we can prove this property by finding a direct
correspondence between the operators $B_{n}^{(M)}$ and $BK_{n}^{(M)}$.

Let us notice that the operator $BK_{n}^{(M)}$ can be obtained from the
operator $B_{n}^{(M)}$. Suppose that $f$ is arbitrary in $C_{+}\left(
[0,1]\right) $. Let us consider
\begin{equation}
f_{n}(x)=(n+1)\int_{nx/(n+1)}^{(nx+1)/(n+1)}f(t)dt  \label{fn}
\end{equation}%
It is readily seen that $B_{n}^{(M)}(f_{n})(x)=BK_{n}^{(M)}(f)(x)$, for all $%
x\in \lbrack 0,1]$. We also notice that $f_{n}\in C_{+}\left( [0,1]\right) $%
. What is more, if $f$ is strictly positive then so is $f_{n}$.

A function $f:[a,b]\rightarrow \mathbb{R}$ is quasi-concave if $-f$ is
quasi-convex. If $f$ is continuous, quasi-concavity equivalently means that
there exists $c\in \lbrack a,b]$ such that $f$ is nondecreasing on $[a,c]$
and nonincreasing on $[c,b].$

We are now in position to prove that $BK_{n}^{(M)}$ preserves
quasi-concavity too.
\begin{theorem} Let $f\in C_{+}([0,1])$. If $f$%
is quasi-concave on $\left[ 0,1\right] $ then $%
BK_{n}^{(M)}(f)$ is quasi-concave on $\left[ 0,1\right] $.
\end{theorem}
\begin{proof} For some arbitrary $n\geq 1$ let us consider the function $%
f_{n}$ given by (\ref{fn}). Moreover, let $c\in \lbrack 0,1]$ such that $f$
is nondecreasing on $[0,c]$ and nonincreasing on $[c,1]$. Then, let $j(c)\in
\{0,....,n\}$ such that
\begin{equation*}
\frac{j(c)}{n+1}\leq c\leq \frac{j(c)+1}{n+1}.
\end{equation*}%
Next, we consider the function $g_{n}$ which interpolates $f_{n}$ at all the
knots $\frac{k}{n}$, $k=0,1,...,n$, and which is continuous on $[0,1]$ and
affine on any interval $\left[ \frac{k}{n},\frac{k+1}{n}\right] $, $%
k=0,1,...,n-1$. It means that $g_{n}$ is the continuous polygonal line which
interpolates $f_{n}$ at all the knots $\frac{k}{n}$, $k=0,1,...,n$. This
easily implies that $B_{n}^{(M)}(f_{n})(x)=B_{n}^{(M)}(g_{n})(x)$, $x\in
\lbrack 0,1]$, hence, $BK_{n}^{(M)}(f)(x)=B_{n}^{(M)}(g_{n})(x)$, $x\in
\lbrack 0,1]$. Let us now choose arbitrary $0\leq k_{1}<k_{2}\leq j(c)-1$.
We have $g_{n}\left( \frac{k_{1}}{n}\right)
=(n+1)\int_{k_{1}/(n+1)}^{\left( k_{1}+1\right) /(n+1)}f(t)dt$ and $%
g_{n}\left( \frac{k_{2}}{n}\right) =(n+1)\int_{k_{2}/(n+1)}^{\left(
k_{2}+1\right) /(n+1)}f(t)dt$. As $\frac{k_{1}+1}{n+1}\leq \frac{k_{2}}{n+1}$
and $f$ is nondecreasing on $[0,\frac{k_{2}+1}{n+1}]$, we easily obtain
(after applying the mean value theorem) that $g_{n}\left( \frac{k_{1}}{n}%
\right) \leq g_{n}\left( \frac{k_{2}}{n}\right) $. The construction of $%
g_{n} $ easily implies that $g_{n}$ is nondecreasing on $\left[ 0,\frac{%
j(c)-1}{n}\right] $. By similar reasoning we get that $g_{n}$ is
nonincreasing on $\left[ \frac{j(c)+1}{n},1\right] $. Now, suppose that $%
f\left( \frac{j(c)}{n+1}\right) \geq f\left( \frac{j(c)+1}{n+1}\right) $.
The quasi-concavity of $f$ implies that $f(x)\geq f\left( \frac{j(c)+1}{n+1}%
\right) $ for any $x\in \left[ \frac{j(c)}{n+1},\frac{j(c)+1}{n+1}\right] $.
Since there exists $x_{0}\in \left[ \frac{j(c)}{n+1},\frac{j(c)+1}{n+1}%
\right] $ such that $(n+1)\int_{j(c)/(n+1)}^{\left( j(c)+1\right)
/(n+1)}f(t)dt=f(x_{0})=g_{n}\left( \frac{j(c)}{n}\right) $, and since $%
f\left( \frac{j(c)+1}{n+1}\right) $ $\geq g_{n}\left( \frac{j(c)+1}{n}%
\right) $ (this is true indeed as $f$ is nondecreasing on $\left[ \frac{%
j(c)+1}{n+1},1\right] $), we get that $g_{n}\left( \frac{j(c)}{n}\right)
\geq g_{n}\left( \frac{j(c)+1}{n}\right) $. Therefore, $g_{n}$ is
nonincreasing on $\left[ \frac{j(c)}{n},\frac{j(c)+1}{n}\right] $. This
implies that $g_{n}$ is nondecreasing on $\left[ 0,\frac{j(c)-1}{n}\right] $
and nonincreasing on $\left[ \frac{j(c)}{n},1\right] $. But $f$ is affine on
$\left[ \frac{j(c)-1}{n},\frac{j(c)}{n}\right] $ which means that it is
monotone on this interval. Clearly this implies that $g_{n}$ is either
nondecreasing on $\left[ 0,\frac{j(c)-1}{n}\right] $ and nonincreasing on $%
\left[ \frac{j(c)-1}{n},1\right] $ or, it is nondecreasing on $\left[ 0,%
\frac{j(c)}{n}\right] $ and nonincreasing on $\left[ \frac{j(c)}{n},1\right]
$. It means that $g_{n}$ is quasi-concave on $[0,1]$. By similar reasonings
we get to the same conclusion if $f\left( \frac{j(c)}{n+1}\right) \leq
f\left( \frac{j(c)+1}{n+1}\right) $. The only difference is that now $g_{n}$
is either nondecreasing on $\left[ 0,\frac{j(c)}{n}\right] $ and
nonincreasing on $\left[ \frac{j(c)}{n},1\right] $ or, it is nondecreasing
on $\left[ 0,\frac{j(c)+1}{n}\right] $ and nonincreasing on $\left[ \frac{%
j(c)+1}{n},1\right] $. Thus, we just proved that $g_{n}$ is quasi-concave on
$[0,1]$. By Theorem 5.1 in \cite{Co-GAL-AA} (see also Theorem 2.2.22 in the
book , it follows that $B_{n}^{(M)}(g_{n})$ is quasi-concave on $[0,1]$. As $%
B_{n}^{(M)}(g_{n})=BK_{n}^{(M)}(f)$, it follows that $BK_{n}^{(M)}(f)$ is
quasi-concave on $[0,1]$.
\end{proof}
As an important side remark, let us note that in Theorem 5.1 of paper \cite%
{Co-GAL-AA}(see also the book \cite{Book-BCG}), it is proved that if $f$ is
quasi-concave and $c$ is a maximum point of $f$ then there exists a maximum
point of $B_{n}^{(M)}(f)$\ such that $\left\vert c-c\prime \right\vert $\ $%
\leq \frac{1}{n+1}$. By the construction\ of $g_{n}$ it \ follows that one
maximum point of $g_{n}$ is between the values $\frac{j(c)-1}{n}$, $\frac{%
j(c)}{n}$ or $\frac{j(c)+1}{n}$. If we denote this value with $c_{n}$ then
one can easily check that $\left\vert c_{n}-c\right\vert \leq \frac{2}{n}$.
Now, applying the afore mentioned property obtained in \cite{Co-GAL-AA}, let
$c\prime $ be a maximum point of \ $B_{n}^{(M)}(g_{n})=BK_{n}^{(M)}(f)$,
such that $\left\vert c\prime -c_{n}\right\vert \leq \frac{1}{n+1}$. This
easily implies that \ $\left\vert c\prime -c\right\vert \leq \frac{3}{n}$.\
\ So, we obtained a quite similar result for the operator $BK_{n}^{(M)}$ in
comparison with the operator $B_{n}^{(M)}$.


\section{Approximation of Lipschitz functions by Bern\-ste\-in\--Kan\-to\-ro\-vich max-product operators}

Let us return to the functions $f_{n}$ given in (\ref{fn}) and let us find
now an upper bound for the approximation of $f$ by $f_{n}$ in terms of the
uniform norm. For some $x\in \lbrack 0,1]$, using the mean value theorem,
there exists $\xi _{x}\in \left[ \frac{nx}{n+1},\frac{nx+1}{n+1}\right] $
such that $f_{n}(x)=f(\xi _{x})$. We also easily notice that $\left\vert \xi
_{x}-x\right\vert \leq \frac{1}{n+1}$. It means that
\begin{equation}
\left\vert f(x)-f_{n}(x)\right\vert \leq \omega _{1}(f;1/(n+1)),x\in \mathbb{%
R},n\in \mathbb{N}.  \label{estim f-fn}
\end{equation}%
In particular, if $f$ is Lipschitz with constant $C$ then $f_{n}$ is
Lipschitz continuous with constant $3C$. These estimation are useful to
prove some inverse results in the case of the operator $BK_{n}^{(M)}$ by
using analogue results already obtained for the operator $B_{n}^{(M)}$.

Below we present a result which gives for the class of Lipschitz
function the order of approximation $1/n$ \ in the approximation by the
operator $BK_{n}^{(M)}$, hence an analogue result which holds in the case of
the operator $B_{n}^{(M)}$.

\begin{theorem} Suppose that $f$ is Lipschitz on $%
[0,1]$ with Lipschitz constant $C$ and suppose that the
lower bound of $f$ is $m_{f}>0$. Then we have
\begin{equation*}
\left\Vert BK_{n}^{(M)}(f)-f\right\Vert \leq 2C\left( \frac{C}{m_{f}}%
+5\right) \cdot \frac{1}{n}\text{, }n\geq 1.
\end{equation*}
\end{theorem}
\begin{proof} The estimation is immediate using the estimation from
Corollary 2.4, (iii), taking into account that $\omega _{1}(f;1/n)\leq C/n$.
\end{proof}

\section{Localization results for Bern\-ste\-in\--Kan\-torovich max-product operators}

We firstly prove a very strong localization property of the operator $%
BK_{n}^{(M)}$.

\begin{theorem} Let $f,g:[0,1]\rightarrow \lbrack 0,\infty )$%
be both bounded on $[0,1]$  with strictly positive lower
bounds and suppose that there exist $a,b\in \lbrack 0,1]$, $%
0<a<b<1$ such that $f(x)=g(x)$ for all $x\in \lbrack
a,b].$ Then for all $c,d\in \lbrack a,b]$ satisfying $%
a<c<d<b$ there exists $\widetilde{n}\in \mathbb{N}$
depending only on $f,g,a,b,c,d$ such that $%
BK_{n}^{(M)}(f)(x)=BK_{n}^{(M)}(g)(x)$ for all $x\in \lbrack c,d]$%
and $n\in \mathbb{N}$ with $n\geq \widetilde{n}.$
\end{theorem}
\begin{proof} Let us choose arbitrary $x\in \lbrack c,d]$ and for each $%
n\in \mathbb{N}$ let $j_{x}\in \{0,1,...,n\}$ be such that $x\in \lbrack
j_{x}/(n+1),(j_{x}+1)/(n+1)].$ Then by relation (4.17) in \cite%
{Be-Co-Ga} we have
\begin{equation}
BK_{n}^{(M)}(f)(x)=B_{n}^{(M)}(f_{n})(x)=\bigvee\limits_{k=0}^{n}\left(
f_{n}\right) _{k,n,j_{x}}(x),  \label{Bk-BN}
\end{equation}%
where for $k\in \{0,1,...,n\}$ we have
\begin{equation}
\left( f_{n}\right) _{k,n,j_{x}}=\frac{{\binom{n}{k}}}{{\binom{n}{j_{x}}}}%
\left( \frac{x}{1-x}\right) ^{k-j_{x}}f_{n}\left( \frac{k}{n}\right) .
\label{gn}
\end{equation}%
and each $f_{n}$ is given by (\ref{fn}). Let us denote with $m_{f},M_{f}$
and $m_{f_{n}},M_{f_{n}}$ respectively, the minimums and maximum values of
the functions $f$ and $f_{n}$, respectively. By the mean value theorem, one
can easily notice that for any $x\in \lbrack 0,1]$ there exists $\xi
_{n,x}\in \lbrack 0,1]$ such that $f_{n}(x)=f(\xi _{n,x})$. It means that $%
0<m_{f}\leq m_{f_{n}}\leq M_{f_{n}}\leq M_{f}$. In what follows, the proof
is very similar to the proof of Theorem 2.1 in \cite{Co-Gal-Localization}
(see also Theorem 2.4.1 in \cite{Book-BCG}). However, as often we will use $%
f_{n}$ instead of $f$, especially since the constants obtained in the proof
of Theorem 2.1 in \cite{Co-Gal-Localization} depend on $f$, in our setting
these constants would depend on $f_{n}$, hence, they would depend on $n$, if
we would apply directly the results in \cite{Co-Gal-Localization}.
Therefore, there are some differences in the two proofs as our intention is
to obtain constants that do not depend on $f_{n}$.

We need the set $I_{n,x}=\{k\in \{0,1,...,n\}:j_{x}-a_{n}\leq k\leq
j_{x}+a_{n}\}$, where $a_{n}=\left[ \sqrt[3]{n^{2}}\right] $ (here $\left[ a%
\right] $ denotes the integer part of $a$). Now, suppose that $k\notin
I_{n,x}$ , and let us discuss first the case when $k<j_{x}-a_{n}$. If we
look over the proof of Theorem 2.1 in \cite{Co-Gal-Localization}, we observe
that this proof is split in cases i) and ii). Case i) corresponds to the
case when $k<j_{x}-a_{n}$. Furthermore this case is divided in two subcases i%
$_{a}$) and i$_{b}$). In subcase i$_{a}$) the inequality $\frac{%
f_{j_{x},n,,j_{x}}(x)}{f_{k,n,j_{x}}(x)}\geq \left( 1+\frac{a_{n}}{nb-a_{n}}%
\right) ^{a_{n}}\cdot \frac{f(j_{x}/n)}{f(k/n)}$ is obtained which then
gives $\frac{f_{j_{x},n,,j_{x}}(x)}{f_{k,n,j_{x}}(x)}\geq \left( 1+\frac{%
a_{n}}{nb-a_{n}}\right) ^{a_{n}}\cdot \frac{m_{f}}{M_{f}}$. Applying this
reasoning but considering $f_{n}$ instead of $f$, we get $\frac{\left(
f_{n}\right) _{j_{x},n,j_{x}}(x)}{\left( f_{n}\right) _{k,n,j_{x}}(x)}\geq
\left( 1+\frac{a_{n}}{nb-a_{n}}\right) ^{a_{n}}\cdot \frac{f_{n}(j_{x}/n)}{%
f_{n}(k/n)}$. But since $m_{f}\leq m_{f_{n}}\leq M_{f_{n}}\leq M_{f}$, we
get $\frac{\left( f_{n}\right) _{j_{x},n,j_{x}}(x)}{\left( f_{n}\right)
_{k,n,j_{x}}(x)}\geq \left( 1+\frac{a_{n}}{nb-a_{n}}\right) ^{a_{n}}\cdot
\frac{m_{f}}{M_{f}}$. We get the same conclusion all cases and subcases,
that is, any lower bound for $\frac{f_{j_{x},n,,j_{x}}(x)}{f_{k,n,j_{x}}(x)}$
is also a lower bound for $\frac{\left( f_{n}\right) _{j_{x},n,j_{x}}(x)}{%
\left( f_{n}\right) _{k,n,j_{x}}(x)}$, for any $k$ outside of $I_{n,x}$.
Since in..., we proved that there exists $N_{0}\in \mathbb{N}$ which may
depend only on $f,a,b,c,d$, such that for any $n\geq N_{0}$, $k\in
\{0,1,...,n\}$, with $k<j_{x}-a_{n}$ or $k>j_{x}+a_{n}$, we have $\frac{%
f_{j_{x},n,,j_{x}}(x)}{f_{k,n,j_{x}}(x)}\geq 1$, it follows that $\frac{%
\left( f_{n}\right) _{j_{x},n,j_{x}}(x)}{\left( f_{n}\right) _{k,n,j_{x}}(x)}%
\geq 1$, for any $n\geq N_{0}$, $k\in \{0,1,...,n\}$, with $k<j_{x}-a_{n}$
or $k>j_{x}+a_{n}$. Combining this fact with relations (\ref{Bk-BN})-(\ref%
{gn}), we get that
\begin{equation*}
BK_{n}^{(M)}(f)(x)=\bigvee\limits_{k\in I_{n,x}}\left( f_{n}\right)
_{k,n,j_{x}}(x),x\in \lbrack c,d],\text{ }n\geq N_{0}.
\end{equation*}%
Using a similar reasoning as in the proof of Theorem 2.1 in \cite%
{Co-Gal-Localization}, in what follows, we will prove that $N_{0}$ can be
replaced if necessary with a larger value $\widetilde{N}_{1}$ such that $[%
\frac{k}{n+1},\frac{k+1}{n+1}]\subseteq \lbrack a,b]$ for any $k\in I_{n,x}$%
. Let us choose arbitrary $x\in \lbrack c,d]$ and $n\in \mathbb{N}$ so that $%
n\geq N_{0}.$ If there exists $k\in $ $I_{n,x}$ such that $k/\left(
n+1\right) \notin \lbrack c,d]$ then we distinguish two cases. Either $\frac{%
k}{n+1}<c$ or $\frac{k}{n+1}>d.$ In the first case we observe that
\begin{equation*}
0<c-\frac{k}{n+1}\leq x-\frac{k}{n+1}\leq \frac{j_{x}+1}{n+1}-\frac{k}{n+1}%
\leq \frac{j_{x}+1}{n+1}-\frac{k}{n+1}\leq \frac{a_{n}+1}{n+1}.
\end{equation*}%
Since $\lim\limits_{n\rightarrow \infty }\frac{a_{n}+1}{n+1}=0,$ it results
that for sufficiently large $n$ we necessarily have $\frac{a_{n}+1}{n+1}<c-a$
which \ clearly implies that $\frac{k}{n+1}\in \lbrack a,c].$ In the same
manner, when $\frac{k}{n+1}>d$, for sufficiently large $n$ we necessarily
have $\frac{k}{n+1}\in \lbrack d,b].$ By similar reasoning it results that
for sufficiently large $n$ we necessarily have $\frac{k}{n+1}\in \lbrack
a,b].$Summarizing, there exists a constant $\widetilde{N}_{1}\in \mathbb{N}$
independent of any $x\in \lbrack c,d]$ such that
\begin{equation*}
BK_{n}^{(M)}(f)(x)=\bigvee\limits_{k\in I_{n,x}}\left( f_{n}\right)
_{k,n,j_{x}}(x)\text{, }x\in \lbrack c,d],\text{ }n\geq \widetilde{N}_{1}
\end{equation*}%
and in addition for any $x\in \lbrack c,d]$, $n\geq \widetilde{N}_{1}$ and $%
k\in I_{n,x}$, we have $[\frac{k}{n+1},\frac{k+1}{n+1}]\subseteq \lbrack
a,b].$ Also, it is easy to check that $\widetilde{N}_{1}$ depends only on $%
a,b,c,d,f.$

Now, for $k\in \{0,1,...,n\}$ taking
\begin{equation*}
\left( g_{n}\right) _{k,n,j_{x}}=\frac{{\binom{n}{k}}}{{\binom{n}{j_{x}}}}%
\left( \frac{x}{1-x}\right) ^{k-j_{x}}g_{n}\left( \frac{k}{n}\right) ,
\end{equation*}%
applying the same reasoning, there exists $\widetilde{N}_{2}\in \mathbb{N}$
which may depend only on $a,b,c,d,g$, such that
\begin{equation*}
BK_{n}^{(M)}(g)(x)=\bigvee\limits_{k\in I_{n,x}}\left( g_{n}\right)
_{k,n,j_{x}}(x)\text{, }x\in \lbrack c,d],\text{ }n\geq \widetilde{N}_{2}
\end{equation*}%
and in addition for any $x\in \lbrack c,d]$, $n\geq \widetilde{N}_{2}$ and $%
k\in I_{n,x}$, we have $[\frac{k}{n+1},\frac{k+1}{n+1}]\subseteq \lbrack
a,b] $ $.$ Since $f(x)=g(x)$, $x\in \lbrack a,b]$, we get that for any $%
n\geq \widetilde{n}=\max \{\widetilde{N}_{1},\widetilde{N}_{2}\}$, $k\in
I_{n.x}$ and $x\in \lbrack c,d]$, it holds that $\left( f_{n}\right)
_{k,n,j_{x}}(x)=\left( g_{n}\right) _{k,n,j_{x}}(x)$. Thus, for any $n\geq
\widetilde{n}$ and $x\in \lbrack c,d]$, we have $%
BK_{n}^{(M)}(f)(x)=BK_{n}^{(M)}(g)(x)$. The proof is complete now. 
\end{proof}
As in the case of the Bernstein max-product operator, we can present a local
direct approximation result as an immediate consequence of the localization
result in Theorem 5.1.
\begin{corollary} Let $f:[0,1]\rightarrow \lbrack 0,\infty )$%
be bounded on $[0,1]$ with the lower bound strictly positive and $%
0<a<b<1$ be such that $f|_{[a,b]}\in Lip\,[a,b]$ with Lipschitz
constant $\overline{C}.$ Then, for any $c,d\in \lbrack 0,1]$%
satisfying $a<c<d<b$, we have 
\begin{equation*}
\left\vert BK_{n}^{(M)}(f)(x)-f(x)\right\vert \leq \frac{C}{n}\text{
for all }n\in \mathbb{N}\text{ and }x\in \lbrack c,d],
\end{equation*}%
where the constant $C$ depends only on $f$ and  $%
a,b,c,d.$
\end{corollary}
\begin{proof} Let us define the function $F:[0,1]\rightarrow \mathbb{R}$,%
\begin{equation*}
F(x)=\left\{
\begin{array}{ccc}
f(a) & \text{if} & x\in \lbrack 0,a], \\
f(x) & \text{if} & x\in \lbrack a,b], \\
f(b) & \text{if} & x\in \lbrack b,1].%
\end{array}%
\right.
\end{equation*}
The hypothesis immediately imply that $F$ is a strictly positive Lipschitz
function on $[0,1]$. Then, according to Theorem 4.1 and noting that the
minimum of $F$ is above the minimum of $f$, $m_{f}$, it results that
\begin{equation*}
\left\vert BK_{n}^{(M)}(F)(x)-F(x)\right\vert \leq 2\overline{C}\left( \frac{%
\overline{C}}{m_{f}}+5\right) \cdot \frac{1}{n},\text{for all }x\in \lbrack
0,1],n\in \mathbb{N}.
\end{equation*}

Now, let us choose arbitrary $c,d\in \lbrack a,b]$ such that $a<c<d<b.$
Then, by Theorem 5.1 it results the existence of $\widetilde{n}\in \mathbb{N}
$ which depends only on $a,b,c,d,f,F$ such that $%
BK_{n}^{(M)}(F)(x)=BK_{n}^{(M)}(f)(x)$ for all $x\in \lbrack c,d].$ But
since actually the function $F$ depends on the function $f$, by simple
reasonings we get that in fact $\widetilde{n}$ depends only on $a,b,c,d$ and
$f$.

Therefore, for arbitrary $x\in \lbrack c,d]$ and $n\in \mathbb{N}$ with $%
n\geq \widetilde{n}$ we obtain
\begin{equation*}
\left\vert BK_{n}^{(M)}(f)(x)-f(x)\right\vert =\left\vert
BK_{n}^{(M)}(F)(x)-F(x)\right\vert \leq 2\overline{C}\left( \frac{\overline{C%
}}{m_{f}}+5\right) \cdot \frac{1}{n},
\end{equation*}%
where $C_{1}$ and $\widetilde{n}$ depend only on $a,b,c,d$ and $f$.

Now, denoting
\begin{equation*}
C_{2}=\max_{1\leq n<\widetilde{n}}\{n\cdot \Vert BK_{n}^{(M)}(f)-f\Vert
_{\lbrack c,d]}\},
\end{equation*}%
we finally obtain
\begin{equation*}
|BK_{n}^{(M)}(f)(x)-f(x)|\leq \frac{C}{n},\text{for all }n\in \mathbb{N}%
,\,x\in \lbrack c,d],
\end{equation*}%
with $C=\max \{2\overline{C}\left( \frac{\overline{C}}{m_{f}}+5\right)
,C_{2}\}$ depending only on $a,b,c,d$ and $f$. 
\end{proof}
In a previous section we proved that $BK_{n}^{(M)}$ preserves monotonicity
and more generally quasi-convexity. By the localization result in Theorem
5.1 and then applying a very similar reasoning to the one used in the proof
of Corollary 5.2, we obtain local versions for these shape preserving
properties. Indeed, in all cases it will suffice to consider the same $F$ as
in the proof of Corollary 5.2 as this function will be monotone or
quasi-convex/quasi-concave, respectively, whenever $f$ will be monotone or
quasi-convex/quasi-concave, respectively. For this reason we omit the proofs
of the following corollaries (see also the corresponding local shape
preserving properties proved for the operator $B_{n}^{(M)}$ in \cite%
{Co-Gal-Localization}).
\begin{corollary} Let $f:[0,1]\rightarrow \lbrack 0,\infty )$%
be bounded on $[0,1]$ with strictly positive lower bound
and suppose that there exists $a,b\in \lbrack 0,1]$, $0<a<b<1$, such that $f$ is nondecreasing (nonincreasing)
on $[a,b]$. Then for any $c,d\in \lbrack a,b]$ with $
a<c<d<b,$ there exists $\widetilde{n}\in \mathbb{N}$
depending only on $a,b,c,d$ and $f$, such that $
B_{n}^{(M)}(f)$ is nondecreasing (nonincreasing) on $%
[c,d]$ for all $n\in \mathbb{N}$ with $n\geq \widetilde{n%
}.$
\end{corollary}
\begin{corollary} Let $f:[0,1]\rightarrow \lbrack 0,\infty )$%
be a continuous and strictly positive function and suppose that
there exists $a,b\in \lbrack 0,1]$\textit{, }$0<a<b<1$, such that $%
f$ is quasi-convex on $[a,b]$. Then for any $c,d\in
\lbrack a,b]$ with $a<c<d<b,$ there exists $\widetilde{n}%
\in \mathbb{N}$ depending only on $a,b,c,d$ and $f$%, 
such that $B_{n}^{(M)}(f)$ is quasi-convex
on $[c,d]$ for all $n\in \mathbb{N}$ with $n\geq
\widetilde{n}.$
\end{corollary}
\begin{corollary} Let $f:[0,1]\rightarrow \lbrack 0,\infty )$%
be a continuous and strictly positive function and suppose that
there exists $a,b\in \lbrack 0,1]$, $0<a<b<1$, such that $%
f$ is quasi-concave on $[a,b]$. Then for any $c,d\in
\lbrack a,b]$ with $a<c<d<b,$ there exists $\widetilde{n}%
\in \mathbb{N}$ depending only on $a,b,c,d$ and $f$, such that $B_{n}^{(M)}(f)$ is quasi-concave
on $[c,d]$ for all $n\in \mathbb{N}$ with $n\geq
\widetilde{n}.$
\end{corollary}
\begin{remark} As in the cases of Bernstein-type max-product operators studied in the research monograph \cite{Book-BCG},
for the the max-product Kantorovich
type operators we can find natural interpretation as possibilistic operators, which can be deduced from the Feller scheme written in terms of the
possibilistic integral. These approaches also offer new proofs for the
uniform convergence, based on a Chebyshev type inequality in the theory of
possibility.
\end{remark}
\begin{remark} In the recently submitted paper \cite{CG-sub}, we have introduced the more general Kantorovich
max-product operators based on a generalized $(\varphi, \psi)$-kernel, by the formula
\begin{equation}\label{Kantor}
K_{n}^{(M)}(f;\varphi ,\psi )(x)=\frac{1}{b}\cdot \frac{\bigvee_{k=0}^{n}%
\frac{\varphi (nx-kb)}{\psi (nx-kb)}\cdot \left[ (n+1)%
\int_{kb/(n+1)}^{(k+1)b/(n+1)}f\left( v\right) dv\right] }{\bigvee_{k=0}^{n}%
\frac{\varphi (nx-kb)}{\psi (nx-kb)}},
\end{equation}%
where $b>0$, $f:[0,b]\rightarrow \mathbb{R}_{+}$, $f\in L^{p}[0,b]$, $1\leq
p\leq \infty $ and  $\varphi $ and $\psi $ satisfy some properties specific to max-product operators and proved pointwise, uniform or $L^{p}$ convergence
quantitative approximation results. For particular choices of $(\varphi, \psi)$, we have obtained approximation results for many other max-product Kantorovich operators, including for example the sampling operators based on sinc-type kernels.
\end{remark}
\begin{remark} In another recently in preparation paper \cite{CG-sub2}, we have generalized the max-product Kantorovich operators from the above Remark 2),
by replacing the classical linear integral $\int d v$, by the nonlinear Choquet integral $(C)\int d\mu(v)$ with respect to a monotone and submodular set function $\mu$ obtaining and studying  the max-product Kantorovich-Choquet operators given by the formula
$$K_{n}^{(M)}(f;\varphi ,\psi )(x)$$
\begin{equation}
=\frac{1}{b}\cdot \frac{\bigvee_{k=0}^{n}%
\frac{\varphi (nx-kb)}{\psi (nx-kb)}\cdot \left[
(C)\int_{kb/(n+1)}^{(k+1)b/(n+1)}f\left( v\right) d\mu(v)/\mu\left (\left [\frac{k b}{n+1}, \frac{(k+1)b}{n+1}\right ]\right )\right] }{\bigvee_{k=0}^{n}%
\frac{\varphi (nx-kb)}{\psi (nx-kb)}},
\end{equation}
It is worth noting that the max-product Kantorovich-Choquet operators are {\bf doubly nonlinear operators} : firstly due to $\max$ and secondly, due to the Choquet integral.
\end{remark}

{\bf Acknowledgement.} The work of both authors was supported by a grant of the Romanian National Authority for Scientific Research, CNCS-UEFISCDI,
project number PN-III-P1-1.1-PD-2016-1416.



\begin{thebibliography}{99}

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\bibitem{Book-BCG} B. Bede, L. Coroianu, S. G. Gal, \emph{Approximation
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\bibitem{CG-sub} L. Coroianu, S. G. Gal, \emph{Approximation by truncated max-product operators of
Kantorovich-type based on generalized $(\varphi, \psi)$-kernels}, Math. Meth. Appl. Sci., online access,
2018 : 1-14, DOI :10.1002/mma.5262


\bibitem{CG-sub2} L. Coroianu, S. G. Gal, \emph{Approximation by max-product operators of
Kantorovich-Choquet type based on generalized $(\varphi, \psi)$-kernels}, in preparation.



\bibitem{Co-GAL-AA} L. Coroianu, S. G. Gal, \emph{Classes of functions
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\bibitem{Co-Gal-Localization} L. Coroianu, S. G. Gal, \emph{%
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\bibitem{Gal} S. G. Gal, \emph{Shape-Preserving Approximation by Real and
Complex Polynomials}, XIV + 352 pp., Birkh\"{a}user, Boston, Basel, Berlin, 2008, ISBN 978-0-8176-4702-5.



\end{thebibliography}

\end{document}
