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

${}^{\ast }$Department of Mathematics and Computer Science, University of Oradea, Universităţii 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 [ 15 ] , 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 $\mathrm{m}\mathrm{a}\mathrm{x}\{b_{n,k}(x);k\in \{0,...,n\}\}$ 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 ] , [ 14 ] .

In other two recent papers [ 11 ] and [ 12 ] , 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 [ 13 ] )

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 [ 13 ] , 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-known, we can also have a very pronounced divergence phenomenon in $[a,b]$ (see e.g. Chapter 4 in the book [ 17 ] , see also [ 16 ] , [ 10 ] ).

The goal of the present paper is to determine for $L_n^{(M)}$ the saturation order together with its special class of functions and to obtain a local inverse result.

The plan of the paper goes as follows. In Section 2 the saturation order together with its special class of functions are obtained. Section 3 contains a local inverse approximation result.

2 The Saturation Order

Firstly, we need three simple auxiliary results, Lemmas 2.1-2.3, where $l_{n,k}$ denote the fundamental Lagrange polynomials attached to the knots $x_{n,k}=k/n$, $k\in \{0,1,...,n\}$, $n\in \mathbb{N}$.

Let us denote $J_n(x)=\{k\in \{0,1,...,n\}:l_{n,k}(x)>0\}.$ This implies that

We observe that $\{j,j+1\}\subseteq J_n(x).$ Indeed, for $a=0$ and $b=1$, by using (1.2) we have $sign(l_{n,j}(x))=(-1{)}^{n-j}\cdot (-1{)}^{n-j}=1$ and $sign(l_{n,j+1}(x))=(-1{)}^{n-j-1}\cdot (-1{)}^{n-j-1}=1.$ Then, we denote ${\mathrm{\Omega }}_n(x)=\prod _{i=0}^n(x-x_{n,i})$. The definitions of $l_{n,k}(x)$ and $J_n(x)$ imply $l_{n,k}(x)=\frac{|{\mathrm{\Omega }}_n(x)|}{|x-x_{n,k}|}$ for all $k\in J_n(x).$ We thus obtain that $\underset{k=0}{\overset{n}{\bigvee }}{l}_{n,k}(x)=|{\mathrm{\Omega }}_n(x)|\cdot \underset{k\in J_n(x)}{\bigvee }\frac{1}{|x-x_{n,k}|}.$ Since $\{j,j+1\}\subseteq J_n(x)$ it is immediate that for $x\in [\frac{j}{n},\frac{j+1/2}{n}]$ we have $\underset{k\in J_n(x)}{\bigvee }\frac{1}{|x-x_{n,k}|}=\frac{1}{|x-x_{n,j}|}$ and for $x\in [\frac{j+1/2}{n},\frac{j+1}{n}]$ we have $\underset{k\in J_n(x)}{\bigvee }\frac{1}{|x-x_{n,k}|}=\frac{1}{|x-x_{n,j+1}|}.$ From here we easily get the desired conclusion.

(i) Firstly, by Lemma 2.1 we observe that for $x\in [\frac{(j-1)+1/2}{n},\frac{j}{n}]$ we have $\underset{k=0}{\overset{n}{\bigvee }}{l}_{n,k}(x)=l_{n,j}(x)$. Now, if $j\le n/2$ then it is easy to check that $x:=j/(n+1)\in [\frac{(j-1)+1/2}{n},\frac{j}{n}]$ which implies $\underset{k=0}{\overset{n}{\bigvee }}{l}_{n,k}(j/(n+1))=l_{n,j}(j/(n+1)).$ This implies that

(ii) Since $j\le n/2$, one can easily prove that $j/n\in [\frac{j}{n+1},\frac{j+1/2}{n+1}].$ Therefore, by Lemma 2.1 we obtain $\underset{k=0}{\overset{n+1}{\bigvee }}{l}_{n+1,k}(j/n)=l_{n+1,j}(j/n)$. This implies that

(i) Since $j\ge n/2$ by elementary calculus it is easy to prove that $(j+1)/(n+1)\in [\frac{j}{n},\frac{j+1/2}{n}]$ and by Lemma 2.1 this implies that$\underset{k=0}{\overset{n}{\bigvee }}{l}_{n,k}((j+1)/(n+1))=l_{n,j}((j+1)/(n+1))$. We obtain

(ii) Since $j\ge n/2$, again it is easy to check that $j/n\in [\frac{j+1/2}{n+1},\frac{j+1}{n+1}]$ and by Lemma 2.1 this implies that$\underset{k=0}{\overset{n+1}{\bigvee }}{l}_{n+1,k}(j/n)=l_{n+1,j+1}(j/n)$. We obtain

We are now in position to determine the saturation order and the associated special class of functions for the truncated max-product operator $L_n^{(M)}$.

We begin with the particular case when $a=0$ and $b=1.$ By hypothesis, there exists $a_n$ $\in \mathbb{R}$, $n\in \mathbb{N}$ with the property $a_n\searrow 0$ as $n\to +\mathrm{\infty }$, such that

Let us choose arbitrary $\epsilon >0.$ Since $a_n\searrow 0$ as $n\to +\mathrm{\infty }$, it follows that there exists $n_0\in \mathbb{N}$ such that $a_n<\epsilon$ for all $n\in \mathbb{N}$, $n\ge n_0$. Noting the above relation we get

Then, from the uniform continuity of $f$ it results the existence of $n_1\in \mathbb{N}$ such that

We will obtain the desired conclusion in two steps: (A) we prove that $f$ is constant on any interval $[a,b]$ with $0<a<b<1/2;$ (B) we prove that $f$ is constant on any interval $[a,b]$ with $1/2<a<b<1.$ Indeed, if (A) holds then thanks to the continuity of $f$ we easily obtain that $f$ is constant on $[0,1/2].$ Similarly, if (B) holds then we obtain that $f$ is constant on $[1/2,1].$ Then, from the continuity of $f$ it easily follows that $f$ is constant on $[0,1].$ So, we start by proving that (A) and (B) hold.

(A) Let us choose arbitrary $a,b\in \mathbb{R}$ such that $0<a<b<1/2.$ Further one, let $x_0$ and $y_0$ be the points where $f$ attaints its minimum and maximum respectively on the interval $[a,b].$ Without any loss of generality we may suppose that $x_0\ne y_0$ (contrariwise it follows that $f$ is constant on $[a,b]$ and there is nothing to prove). We have two subcases: A${}_1)$ $x_0<y_0$ and A${}_2)$ $x_0>y_0.$

  Subcase A${}_1)$ Let $n\in \mathbb{N}$ be with $n>max\{n_0,n_1,2/(y_0-x_0)\}.$ By relation (2.1) it follows that

Moreover, combining the inequality in Lemma 2.2 (i) with the above inequality, we get

Further one, let us choose $j_1\in \{0,1,...,n-1\}$ such that $j_1/n\le y_0\le (j_1+1)/n$ and $x_0\le j_1/n$. Note that there exists such an index $j_1$, because the previous inequalities are equivalent to $n{y}_0-1\le j_1\le n{y}_0$, $n{x}_0\le j_1\le n{y}_0$, while the condition $n>2/(y_0-x_0)$ is equivalent to the condition $n{y}_0-n{x}_0>2$.

Also, from $j_1/n\le y_0\le b<1/2$ it easily follows that $j_1\le n/2$.

As a first consequence, from the relation (2.2) we obtain

Then, since $\underset{l\to \mathrm{\infty }}{lim}\frac{j_1}{n+l}=0$, by $x_0>0$ and $x_0\le j_1/n$ it follows that there exists $l_0\in \mathbb{N}$ such that $\frac{j_1}{n+l_0+1}\le x_0\le \frac{j_1}{n+l_0}$.

It is worth noting here that indeed, the above $l_0$ cannot be equal to $0$, because if we would have $l_0=0$, then we would obtain $j_1/(n+1)\le x_0<y_0\le (j_1+1)/n\le (j_1+2)/(n+1)$, which would imply $y_0-x_0\le 2/(n+1)<2/n$, in contradiction with the supposition that $n>2/(y_0-x_0)$.

The inequality $\frac{j_1}{n+l_0+1}\le x_0\le \frac{j_1}{n+l_0}$ and (2.1) also implies that

Since $j_1\le n/2$, applying successively relation (2.3) we obtain

Taking the sum of all these inequalities we get

Then, by relations (2.4)–(2.5) we obtain

and since $0\le f(y_0)-f(x_0)$, we obtain

On the other hand, since $0<x_0\le j_1/(n+l_0),$ after some simple calculations we get (note that $j_1\le n/2)$

Using this information in relation (2.6) we obtain

where $\epsilon >0$ was chosen arbitrary. Therefore, passing in the previous inequality with $\epsilon \searrow 0$, we obtain $f(x_0)=f(y_0)$ (here, it is important that $x_0>0$ ). Since on the interval $[a,b]$ the maximum value and the minimum value of the function $f$ coincide, we obtain that $f$ is a constant function on the interval $[a,b]$ and hence we obtained the desired conclusion for this case.

  Subcase A${}_2)$ Let us choose arbitrary $n\in \mathbb{N}$, $n>max\{n_0,n_1,2/(x_0-y_0)\}.$ By relation (2.1) it follows that

Moreover, combining the inequality in Lemma 2.2 (ii) with the above inequality, we get

Let $j_1$ and $l_0$ be chosen as in the previous case, with the difference that now we have $j_1/(n+l_0+1)\le y_0\le j_1/(n+l_0)$ and $j_1/n\le x_0\le (j_1+1)/n$. Applying successively the above inequality (2.7) we get

Taking the sum of all these inequalities and then reasoning as in the previous case we obtain that

Now, reasoning again as in the previous case we obtain

Again, we easily obtain that $f(x_0)=f(y_0)$ which implies that $f$ is constant on $[a,b].$ Summarizing, we obtain that (A) holds.

(B) Let us choose arbitrary $a,b\in \mathbb{R}$ such that $1/2<a<b<1.$ Further one, let $x_0$ and $y_0$ be the points where $f$ attaints its minimum and maximum respectively on the interval $[a,b].$ Without any loss of generality we may suppose that $x_0\ne y_0$ (contrariwise it follows that $f$ is constant on $[a,b]$ and there is nothing to prove). We have two subcases: B${}_1)$ $x_0<y_0$ and B${}_2)$ $x_0>y_0.$

  Subcase B${}_1)$ Let us choose arbitrary $n\in \mathbb{N}$, $n>max\{n_0,n_1,2/(y_0-x_0)\}.$ By relation (2.1) it follows that

Moreover, combining the inequality in Lemma 2.3 (ii) with the above inequality, we get

Further one, let us choose $j_1\in \{1,2,...,n\}$ such that $(j_1-1)/n\le x_0\le j_1/n$ and $j_1/n\le y_0$. Note that there exists such an index $j_1$, because the previous inequalities are equivalent to $n{x}_0\le j_1\le n{x}_0+1$, $n{x}_0\le j_1\le n{y}_0$, while the condition $n>2/(y_0-x_0)$ is equivalent to the condition $n{y}_0-n{x}_0>2$.

Also, from $1/2<x_0\le \frac{j_1}{n}$, it easily follows that $j_1\ge n/2$.

As a first consequence, from relation (2.2) we obtain

Then, since $\underset{l\to \mathrm{\infty }}{lim}\frac{j_1+l}{n+l}=1$, by $y_0<1$ and $j_1/n\le y_0$ it follows that there exists $l_0\in \mathbb{N}$ such that $\frac{j_1+l_0}{n+l_0}\le y_0\le \frac{j_1+l_0+1}{n+l_0+1}$.

It is worth noting here that the above $l_0$ cannot be equal to $0$, because if we would have $l_0=0$ then we would obtain $(j_1-1)/n\le x_0\le j_1/n\le y_0\le (j_1+1)/(n+1)\le (j_1+1)/n$, which would imply $y_0-x_0\le 2/n$, in contradiction with the supposition that $n>2/(y_0-x_0)$.

The inequality $\frac{j_1+l_0}{n+l_0}\le y_0\le \frac{j_1+l_0+1}{n+l_0+1}$ and (2.1) also implies that

Since by $j_1\ge n/2$ it is very easy to verify that for $l\in \{0,1,...,l_0\}$ we have $j_1+l\ge (n+l)/2$, applying successively relation (2.8) we obtain

Taking the sum of all these inequalities and then reasoning as in the previous cases we obtain that

and then

On the other hand, by $\frac{j_1+l_0}{n+l_0}\le y_0$ it follows (note that $y_0<1)$

Using the above inequality in relation (2.11) we easily obtain $0\le f(y_0)-f(x_0)\le \epsilon (y_0/(1-y_0)+2).$ Now reasoning as in the subcase A${}_1)$ we obtain $f(x_0)=f(y_0)$ and we immediately conclude that $f$ is constant on $[a,b].$

Subcase B${}_2)$ Let us choose arbitrary $n\in \mathbb{N}$, $n>max\{n_0,n_1,2/(x_0-y_0)\}.$ By relation (2.1) it follows that

Moreover, combining the inequality in Lemma 2.3 (i) with the above inequality, we get

Let $j_1$ and $l_0$ be chosen as in the previous case, with the difference that now we have $(j_1-1)/n\le y_0\le j_1/n$ and $\frac{j_1+l_0}{n+l_0}\le x_0\le \frac{j_1+l_0+1}{n+l_0+1}$. Applying successively the above inequality (2.12) we get

Taking the sum of all these inequalities and then reasoning as in the previous case we obtain that

Now, reasoning again as in the previous case we obtain

and since by the same method like in the previous case we have $l_0\le \frac{n{x}_0}{1-x_0}$, we easily obtain $0\le f(y_0)-f(x_0)\le \epsilon (x_0/(1-x_0)+2).$ This easily implies that $f(x_0)=f(y_0)$, which means that $f$ is constant on $[a,b].$ Summarizing, we obtain that (B) holds.

Now, by the discussion just before the beginning of the case (A), we conclude that $f$ is constant on the whole interval $[0,1].$

At the end, we discuss now the general case when the Lagrange max-prod operator is attached to functions defined on an interval $[a,b]$ with $a<b.$ To make distinction between the general case and the particular case of the interval $[0,1]$ in what follows 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 a function $f\in C([a,b])$ we have $\Vert L_n^{(M)}(f)-f\Vert =o(1/n).$ Let us define the function $g:[0,1]\to \mathbb{[}a,b]$, $g(y)=a+(b-a)y.$ It is immediate that for any $x\in [a,b]$ there exists an unique $y(x)\in [0,1]$ such that $f(x)=(f\circ g)(y(x)).$ Then we observe that for any $x\in [a,b]$ we have

The above equalities imply

for all $x\in [a,b].$ This last formula together with the previous relation

easily implies that

for all $x\in [a,b]$ which now easily implies that $\Vert {\bar{L}}_n^{(M)}(f\circ g)-(f\circ g)\Vert =o(1/n).$ Consequently, we can apply the conclusion of the particular case considered at the beginning of the proof and we thus conclude that $f\circ g$ is a constant function. This easily implies that $f$ is a constant function and now the proof is complete.

3 Local Inverse Result

According to Corollary 3.2, (i) in [ 13 ] , the saturation order $\frac{1}{n}$ in the above Theorem 2.4 is attained for positive Lipschitz functions on $[a,b]$.

Conversely, we can present the following local inverse result.

The proof of Theorem 3.1 requires the following three lemmas.

We prove only the direct implication since the converse one is immediate. Since $f$ is continuous on the interval $[\alpha ,\beta ]$, it easily follows that for each $n\in \mathbb{N}$, $n\ge 2$, $1/n\le \beta -\alpha ,$ there exist $x_n,y_n\in [\alpha ,\beta ]$ satisfying $|x_n-y_n|\le 1/n$ and ${\omega }_1(f,1/n{)}_{[\alpha ,\beta ]}=|f(x_n)-f(y_n)|.$ Clearly that by hypothesis and without any loss of generality, we may suppose that $x_n\ne y_n$ and $x_n<y_n,$ for all $n\in \mathbb{N}$.

Let us consider the sequences $(a_n{)}_{n\ge 1}$ and $(b_n{)}_{n\ge 1},$ $a_n=n{\omega }_1(f,1/n{)}_{[\alpha ,\beta ]}=n|f(x_n)-f(y_n)|$ and $b_n=n\cdot M_n(\alpha ,\beta ).$

Let us fix $n\in \mathbb{N}$. Since $f$ is uniformly continuous on $[\alpha ,\beta ]$, it follows that there exists $m\in \mathbb{N}$ such that for all $x,y\in [0,1]$ satisfying $|x-y|\le 1/m$ we have $|f(x)-f(y)|\le 1/n$. In addition, we may choose sufficiently large $m\in \mathbb{N}$ such that $y_n-x_n>2/m$, that is $m>2/(y_n-x_n)$.

Since $0<\alpha <y_n\le \beta <1/2$, clearly there exists $j\in \{1,...,m-1\}$ (depending on $m$ and $n$) such that $j/m\le y_n\le (j+1)/m.$

Since $\underset{l\to \mathrm{\infty }}{lim}j/(m+l)=0$ and since $x_n\ge \alpha >0$, it results the existence of $l_0\in \mathbb{N}$ (depending on $j$ and $m$) such that $j/(m+l_0+1)\le x_n\le j/(m+l_0).$

By the inequalities $x_n\le j/(m+l_0)<j/m\le y_n$, we get

where $p\in \{0,1,...,l_0\}$ is such that

On the other hand, we observe that $max\{|j/(m+l_0)-x_n|,|j/m-y_n|\}\le 1/m$, which implies $|f(x_n)-f(j/(m+l_0))|\le 1/n$ and $|f(j/m)-f(y_n)|\le 1/n.$ We thus obtain that

By the inequalities $x_n\le j/(m+l_0)\le j/m\le y_n$ we get $j/m-j/(m+l_0)\le y_n-x_n\le 1/n$ and this implies $j{l}_0/(m(m+l_0))\le 1/n$ and then $l_0\le m/j\cdot (m+l_0)/n\le 1/\alpha \cdot (m+l_0)/n$. (Here we used that $\alpha \le x_n<j/m$).

Then, by the inequalities $0<\alpha \le x_n\le A:=j/(m+l_0)\le B:=j/m\le y_n\le \beta$ we easily get $B/A\le \beta /\alpha$, which immediately implies $j/(m+l_0)\ge j/m\cdot \alpha /\beta$. From here we get $m+l_0\le m\beta /\alpha$, that is $l_0\le m(\beta /\alpha -1).$ Replacing this last inequality in the inequality $l_0\le 1/\alpha \cdot (m+l_0)/n$ just proved above, we conclude that $l_0\le \beta /{\alpha }^2\cdot m/n.$

Replacing now in relation (3.1) and then multiplying with $n,$ we get

and clearly this implies that $a_n\le 2+\beta /{\alpha }^2\cdot M_{m+p}(\alpha ,\beta ).$ Summarizing, for any $n\in \mathbb{N}$ there exist $m+p\in \mathbb{N}$ such that $a_n\le \beta /{\alpha }^2\cdot b_{m+p}+2$. Since $m>2/(y_n-x_n)$ and $y_n-x_n<1/n$, we get $m>2n$. Therefore, by $\underset{n\to \mathrm{\infty }}{lim sup}{a}_n=\mathrm{\infty }$, it easily follows that $\underset{n\to \mathrm{\infty }}{lim sup}{b}_n=\mathrm{\infty }$ and the lemma is proved.

In an absolutely similar manner we obtain the following.

Also, we can prove:

If $\alpha <1/2<\beta$ then by the hypothesis it is elementary to prove that either $\underset{n\to \mathrm{\infty }}{lim sup}n\cdot {\omega }_1(f,1/n{)}_{[\alpha ,1/2]}=\mathrm{\infty }$ or $\underset{n\to \mathrm{\infty }}{lim sup}n\cdot {\omega }_1(f,1/n{)}_{[1/2,\beta ]}=\mathrm{\infty }.$ Therefore, without any loss of generality we may suppose that we have only two cases: (i) $0<\alpha <\beta \le 1/2$ and (ii) $1/2\le \alpha <\beta <1.$

Case (i) For fixed $n\in \mathbb{N}$ with $n\ge 1/(\beta -\alpha )$, let us choose $k(n)\in \{1,...,n\}$ such that $\alpha \le \frac{k(n)}{n+1}\le \frac{k(n)}{n}\le \beta$ and

Note that such an index $k(n)$ exists, because the inequalities $\alpha \le k(n)/(n+1)\le k(n)/n\le \beta$ imply $\alpha (n+1)\le k(n)\le \beta n$, where $\beta n-\alpha (n+1)\ge 1$.

Since $\beta \le 1/2$, it results that $k(n)\le n/2$ and hence we can use the conclusion of Lemma 2.2. This means that we have

and

If $f(k(n)/n)\ge f(k(n)/(n+1))$ then

and this implies

If $f(k(n)/n)<f(k(n)/(n+1))$then

and this implies

In conclusion, for any $n\in \mathbb{N}$ with $n\ge 1/(\beta -\alpha )$, we have

Since by Lemma 3.2 we have $\underset{n\to \mathrm{\infty }}{lim sup}n\cdot M_n(\alpha ,\beta )=\mathrm{\infty }$, it easily follows now that $\underset{n\to \mathrm{\infty }}{lim sup}n\cdot {\Vert L_n^{(M)}(f)-f\Vert }_{[\alpha ,\beta ]}=\mathrm{\infty }.$

Case (ii) The proof is similar with that of the Case (i), which proves the lemma.

Now we are in position to prove Theorem 3.1.

Proof of Theorem 3.1. Using the same type of reasoning as in the proof of Theorem 2.4 it suffices to deal only with the particular case when $a=0$ and $b=1.$ Firstly we prove that $f$ is a Lipschitz function on $[\alpha ,\beta ]$ if and only if $\underset{n\to \mathrm{\infty }}{lim sup}n\cdot {\omega }_1(f,1/n{)}_{[\alpha ,\beta ]}<\mathrm{\infty }$. Indeed, if $f$ is a Lipschitz function on $[\alpha ,\beta ]$ then evidently that there exists $M>0$ such that we have $n\cdot {\omega }_1(f,1/n{)}_{[\alpha ,\beta ]}\le M$, which implies $\underset{n\to \mathrm{\infty }}{lim sup}n\cdot {\omega }_1(f,1/n{)}_{[\alpha ,\beta ]}\le M<\mathrm{\infty }$.

Conversely, $\underset{n\to \mathrm{\infty }}{lim sup}n\cdot {\omega }_1(f,1/n{)}_{[\alpha ,\beta ]}\le M<\mathrm{\infty }$ implies ${\omega }_1(f,1/n{)}_{[\alpha ,\beta ]}\le \frac{M}{n}$, for all $n\in \mathbb{N}$. For arbitrary $h\in (0,1)$, let $n\in \mathbb{N}$ be such that $\frac{1}{n+1}\le h\le \frac{1}{n}$. It follows ${\omega }_1(f;h{)}_{[\alpha ,\beta ]}\le {\omega }_1(f,1/n{)}_{[\alpha ,\beta ]}\le \frac{M}{n}\le \frac{2M}{n+1}\le 2Mh$, that is ${\omega }_1(f;h{)}_{[\alpha ,\beta ]}\le 2Mh$, for all $h\in [0,1]$, which obviously is equivalent with the fact that $f$ is a Lipschitz function on $[\alpha ,\beta ]$ (indeed, for fixed $x,y\in [\alpha ,\beta ]$ we have $|f(x)-f(y)|\le {\omega }_1(f;|x-y|{)}_{[\alpha ,\beta ]}\le 2M|x-y|$).

Now, by the hypothesis it follows $n\cdot \Vert L_n^{(M)}(f)-f{\Vert }_{[\alpha ,\beta ]}\le M$, for all $n\in \mathbb{N}$. Supposing that $f$ is not a Lipschitz function on $[\alpha ,\beta ]$, by the above considerations it follows that $\underset{n\to \mathrm{\infty }}{lim sup}n\cdot {\omega }_1(f,1/n{)}_{[\alpha ,\beta ]}=\mathrm{\infty }$. But then by Lemma 3.4 we get

which is a contradiction. The theorem is proved. $\text{Unsupported use of hfill}$