Preserving properties
of some Szász-Mirakyan type operators

Jorge Bustamante¹

(Date: March 16, 2024; accepted: April 18, 2024; published online: July 11, 2024.)

Abstract.

For a family of Szász-Mirakyan type operators we prove that they preserve convex-type functions and that a monotonicity property verified by Cheney and Sharma in the case Szász-Mirakyan operators holds for the variation study here. We also verify that several modulus of continuity are preserved.

Key words and phrases:

Szász-Mirakyan type operators, positive linear operators, shape preserving properties

2005 Mathematics Subject Classification:

41A36, 41A99

¹Benemérita Universidad Autónoma de Puebla, Facultad de Ciencias Físico-Matemáticas, Avenida San Claudio y 18 Sur, Colonia San Manuel, Edificio FM1-101B, Ciudad Universitaria, C.P. 72570, Puebla, México. e-mail: jbusta@fcfm.buap.mx.

1. Introduction

Throughout the work $\mathbb{N}$ is the set of all positive integers, $\mathbb{N}_{0}=\mathbb{N}\cup\{0\}$ , and $\mathbb{P}_{n}$ is the family of all algebraic polynomials of degree non greater than $n$ . Moreover, for each $j\in\mathbb{N}_{0}$ , we use the notations

e_{j}(x)=x^{j},\quad\quad x\in\mathbb{R},

and $I=[0,\infty)$ . Let $C(I)$ the family of all continuous functions $f:I\to\mathbb{R}$ .

The Szász-Mirakyan operators are defined by (see [6] and the references therein)

S_{n}(f,x)=e^{-nx}\sum_{k=0}^{\infty}\tfrac{n^{k}}{k!}f\Big{(}\tfrac{k}{n}\Big%{)}x^{k},\quad x\in I.

It is known that $S_{n}(e_{0},x)=1$ and $S_{n}(e_{1},x)=x$ (see [6]).

For a fixed real $p\geq 0$ and $n\in\mathbb{N}$ , Schurer defined ([27] and [28])

(1)

S_{n,p}(f,x)=e^{-(n+p)x}\sum_{k=0}^{\infty}\tfrac{(n+p)^{k}}{k!}f\Big{(}\tfrac%{k}{n}\Big{)}x^{k},\quad x\in I.

Some studies concerning these operators were given by Sikkema in [29] and [30] (see also [26]).

It is known that (see [26, p. 82]), for each $x\geq 0$ and $n\in\mathbb{N}$ , $S_{n,p}(e_{0},x)=1$ and

S_{n,p}(e_{1},x)=x+\tfrac{p\,x}{n}.

Hence one has $S_{n,p}(e_{1},x)=x$ only when $p=0$ .

In this work we study properties of a modification $M_{n,p}$ of Schurer operators satisfying $M_{n,p}(e_{0},x)=1$ and $M_{n,p}(e_{1},x)=x$ .

Let $\{\beta(n)\}$ be an strictly increasing sequence of positive real numbers such that $\lim_{n\to\infty}\beta(n)=\infty$ . For $p\geq 0$ , $n\in\mathbb{N}$ , $x\geq 0$ , and a function $f\in C(I)$ consider the operator

(2)

M_{n,p}(f,x)=e^{-(\beta(n)+p)x}\sum_{k=0}^{\infty}\tfrac{(\beta(n)+p)^{k}}{k!}%f\Big{(}\tfrac{k}{\beta(n)+p}\Big{)}x^{k},

whenever the series converges absolutely. Let $\mathcal{L}(I)$ be the family of all functions $f\in C(I)$ such that, for each $n\in\mathbb{N}$ , the series $M_{n,p}(f)$ converges absolutely.

Notice that $M_{n,p}$ can be considered a more natural extension of Szász-Mirakyan operators. This modification appeared in [8] and [9]. In [8] they were studied in spaces defined by the weight $\varrho_{m}(x)=1/(1+x)^{m}$ , with $m\in\mathbb{N}$ and in [9] some weighted space of bounded functions were considered.

There is a long list of papers devoted to study properties of Szász-Mirakyan operators. Here we recall some of them: [2], [4], [5], [6], [11], [18], [21], [22], [23], [33], [34], [35], [36], and [37]. It is worth asking when the results presented in the cited articles can be extended to the case $M_{n,p}$ operators.

For a fixed $p\geq 0$ , $n\in\mathbb{N}$ , and $x\geq 0$ we use the notations

(3)

g_{n,p}(x)=e^{-(\beta(n)+p)x}\quad\quad{\rm and}\quad a_{n,p}=\beta(n)+p.

For $r\in\mathbb{N}_{0}$ , $C_{r}(I)$ is the family of all $f\in C(I)$ such that

(4)

\|f\|_{r}=\sup\limits_{x\in I}\tfrac{\mid f(x)\mid}{(1+x)^{r}}<\infty.

For $r\in\mathbb{N}_{0}$ , let $C_{r,\infty}(I)$ be the class of all functions $f\in C_{r}(I)$ such that $f(x)/(1+x)^{r}$ has a finite limit as $x\to\infty$ .

In Section 2 we present some general properties of operators $M_{n,p}$ . In Section 3 we show that some known properties related with monotone and convex functions and Szász-Mirakyan operators also holds for the operators $M_{n,p}$ . In Section 4 we prove that several modulus of continuity are preserved (up to a constant) by the operators $M_{n,p}$ .

2. Some basic properties

Since the series

(5)

\sum_{k=0}^{\infty}\tfrac{a_{n,p}^{k}}{k!}x^{k}=e^{(\beta(n)+p)x}=g_{n,p}(x),

converges uniformly on each interval $[0,a]$ , $a>0$ , it can be differentiated term by term. For $i\in\mathbb{N}$ , we will use several times the equations

(6)

g_{n,p}^{(i)}(x)=\sum_{k=i}^{\infty}\tfrac{a_{n,p}^{k}}{(k-i)!}x^{k-i}=\sum_{k%=0}^{\infty}\tfrac{a_{n,p}^{k+i}}{k!}x^{k}=a_{n,p}^{i}g_{n,p}(x).

Theorem 1.

If $i\in\mathbb{N}_{0}$ and

(7)

P_{i+1}(x)=x\,\Big{(}x-\tfrac{1}{a_{n,p}}\Big{)}\cdots\Big{(}x-\tfrac{i}{a_{n,%p}}\Big{)},\quad x\geq 0,

then

(8)

M_{n,p}(P_{i+1},x)=x^{i+1}.

In particular, for each $n\in\mathbb{N}$ and $i\in\mathbb{N}_{0}$ , $e_{i}\in\mathcal{L}[0,\infty)$ and $M_{n,p}(e_{i},x)\in\mathbb{P}_{i}$ .

Proof. Notice that $P_{i+1}(x)\in\mathbb{P}_{i+1}$ and, for $k\in\mathbb{N}_{0}$ ,

a_{n,p}^{i+1}P_{i+1}\Big{(}\tfrac{k}{a_{n,p}}\Big{)}=k(k-1)\cdots(k-i).

In particular $P_{i+1}(k/a_{n,p}))=0$ for $0\leq k\leq i$ . Therefore, for each fixed $x>0$ ,

	$\displaystyle a_{n,p}^{i+1}g_{n,p}(x)M_{n,p}(P_{i+1},x)$	$\displaystyle=\sum_{k=i+1}^{\infty}\tfrac{a_{n,p}^{k}x^{k}}{(k-i-1)!}=x^{i+1}%\sum_{k=i+1}^{\infty}\tfrac{a_{n,p}^{k}x^{k-i-1}}{(k-i-1)!}$
		$\displaystyle=x^{i+1}\sum_{k=0}^{\infty}\tfrac{a_{n,p}^{k+i+1}}{k!}x^{k}=x^{i+%1}g_{n}^{(i+1)}(x),$

where we use (6). Therefore $M_{n,p}(P_{i+1},x)=x^{i+1}\in\mathbb{P}_{i+1}$ , for each $i\geq 0$ .

Since, for $i\geq 0$ , $x^{i}$ can be written as a linear combination of the polynomials $P_{1},\ldots,P_{i}$ , we know that $e_{i}\in\mathcal{L}[0,\infty)$ and $M_{n,p}(e_{i},x)\in\mathbb{P}_{i}$ . For $i=0$ it is a simple assertion because $M_{n,p}(e_{0},x)=1$ . $\square$

For the case of Szász-Mirakyan operators the last assertion in Theorem 1 was verified by Becker in [6, Lemma 3].

Proposition 2.

If $r\in\mathbb{N}$ , there exists a constant $C(r)\geq 1$ such that, for every real $a>0$ ,

M_{n,p}((a+e_{1})^{r},x)\leq C(r)(1+a+x)^{r}.

Proof. From Theorem 1 we know that, for each $i\in\mathbb{N}$ , there is an algebraic polynomial $P_{i}\in\mathbb{P}_{n}$ , say $P_{i}(x)=\sum_{k=0}^{i}b_{i,k}x^{k}$ , such that

M_{n,p}(e_{i},x)=\sum_{k=0}^{i}b_{i,k}x^{k}.

If $0\leq x\leq 1$ , then

\Bigg{|}\sum_{k=0}^{i}b_{i,k}x^{k}\Bigg{|}\leq\sum_{k=0}^{i}\mid b_{i,k}\mid%\leq(1+x)^{i}\sum_{k=0}^{i}\mid b_{i,k}\mid.

If $1\leq x$ , then

\Bigg{|}\sum_{k=0}^{i}b_{i,k}x^{k}\Bigg{|}\leq x^{i}\sum_{k=0}^{i}\mid b_{i,k}%\mid\leq(1+x)^{i}\sum_{k=0}^{i}\mid b_{i,k}\mid.

Therefore $0\leq M_{n,p}(e_{i},x)\leq C_{i}(1+x)^{i}$ , where the constant $C_{i}$ depends only on $i$ .

If $a>0$ ,

	$\displaystyle M_{n,p}\Big{(}(a+e_{1})^{r},s\Big{)}$	$\displaystyle=\sum_{j=0}^{r}\tbinom{r}{j}a^{r-j}M_{n,p}(e_{j},s)$
		$\displaystyle\leq C\Big{(}a^{r}+\sum_{j=1}^{r}\tbinom{r}{j}a^{r-j}(1+x)^{j}=C(%1+a+x)^{r}.\square$

Theorem 3 was proved in [9] when $\beta(n)=n$ , but it can be easily extended to the case of a general $\beta(n)$ .

Theorem 3.

The operators $M_{n,p}$ has the following properties:

(i) $M_{n,p}:\mathcal{L}(I)\to C^{1}(I)$ .

(ii) $M_{n,p}(e_{0},x)=1$

(iii) For every $n,m\in\mathbb{N}$ , $f\in\mathcal{L}(I)$ and $x>0$ ,

(9)

\tfrac{1}{a_{n,p}^{m}}M_{n,p}^{(m)}(f,x)=M_{n,p}\Big{(}\Delta^{m}_{1/a_{n,p}}f%(t),x\Big{)},

where $\Delta_{h}^{k}g(u)$ stands the usual $k$ -th forward difference of the function $g$ at $u$ with step $h$ .

The following result can be proved as Theorem 1 in [31] (it is a consequence of the Korovkin theorem).

Theorem 4.

If $f\in\mathcal{L}(I)$ and $a>0$ then $M_{n,p}(f,x)$ converges uniformly to $f(x)$ on $[0,a]$ .

3. Monotonicity and convex functions

For $k\in\mathbb{N}$ , a function $g:I\to\mathbb{R}$ is said to be $k$ -convex, if $\Delta_{h}^{k}g(u)\geq 0$ for each $h>0$ . In particular, $2$ -convexity agrees with the usual notion of convex functions.

For each $k\in\mathbb{N}$ , Szász-Mirakyan operators preserve $k$ -convexity [25]. That is, if $\Delta_{h}^{k}g(u)\geq 0$ and $S_{n}(g,x)$ is well defined, then $\Delta_{h}^{k}S_{n}(g,u)\geq 0$ . If follows from (9) that the operators $M_{n,p}$ share this property Szász-Mirakyan operators. But the assertion must be presented in a more convenient form. Let us explain why we need that. In [39, Th. 1], Zhen proved that, if $f^{\prime}(x)>0$ , then $S_{n}^{\prime}(f,x)>0$ , and if $f^{\prime\prime}(x)>0$ , then $S_{n}^{\prime\prime}(f,x)>0$ . Theorem 5 shows that these types of results are trivial.

Theorem 5.

(i) If $f\in\mathcal{L}(I)$ increases, then $M_{n,p}^{\prime}(f,x)\geq 0$ .

(ii) If $f\in\mathcal{L}(I)$ is convex, then $M_{n,p}^{\prime\prime}(f,x)\geq 0$ .

Proof. It follows directly from (9). $\square$

Cheney and Sharma proved in [10] that, if $f$ is convex, for each $x$ and every $n\in\mathbb{N}$ , $S_{n+1}(f,x)\leq S_{n}(f,x)$ . Horová [15] obtained a converse theorem. In Theorem 6 we verify that a similar result holds for the operators $M_{n,p}$ . A converse result can also be proved (see [15] and [19]). But we do not want to consider that problem here.

Theorem 6.

(i) If $f\in\mathcal{L}(I)$ is convex then, for each $x\geq 0$ and $n\in\mathbb{N}$ ,

f(x)\leq M_{n+1,p}(f,x)\leq M_{n,p}(f,x).

(ii) If $f\in\mathcal{L}(I)$ is concave then, for each $x\geq 0$ and $n\in\mathbb{N}$ ,

M_{n,p}(f,x)\leq M_{n+1,p}(f,x)\leq f(x).

Proof. Assume $f$ is convex. If we set

c_{n,k}=(\beta(n+1)-\beta(n))^{k}\quad{\rm and}\quad b_{n,p}=\tfrac{a_{n,p}}{%\beta(n+1)-\beta(n)},

then

c_{n,k}\sum_{r=0}^{k}\tbinom{k}{r}b_{n,p}^{r}=(\beta(n+1)-\beta(n))^{k}\Big{(}%\tfrac{\beta(n)+p}{(\beta(n+1)-\beta(n))}+1\Big{)}^{k}=a_{n+1,p}^{k}.

That is

\tfrac{c_{n,k}}{a_{n+1,p}^{k}}\,k!\sum_{r=0}^{k}\tfrac{1}{r!}\tfrac{b_{n,p}^{r%}}{(k-r)!}=1.

Therefore

	$\displaystyle\tfrac{c_{n,k}}{a_{n+1,p}^{k}}k!\sum_{r=0}^{k}\tfrac{r}{a_{n,p}}%\tfrac{1}{r!}\tfrac{b_{n,p}^{r}}{(k-r)!}$	$\displaystyle=\tfrac{k}{a_{n+1,p}^{k}}\tfrac{c_{n,k}}{(\beta(n+1)-\beta(n))}%\sum_{r=1}^{k}\tbinom{k-1}{r}b_{n,p}^{r-1}$
		$\displaystyle=\tfrac{k}{a_{n+1,p}^{k}}\tfrac{c_{n,k}}{(\beta(n+1)-\beta(n))}%\Big{(}\tfrac{\beta(n)+p}{(\beta(n+1)-\beta(n))}+1\Big{)}^{k-1}$
		$\displaystyle=\tfrac{k}{a_{n+1,p}^{k}}\Big{(}\beta(n+1)+p\Big{)}^{k-1}=\tfrac{%k}{a_{n+1,p}}.$

This proves that $k/a_{n+1,p}$ is a convex combination of the points $\{r/a_{n,p}:\,0\leq r\leq k\}$ .

If $f$ is convex, then

\displaystyle f\Big{(}\tfrac{k}{a_{n+1,p}}\Big{)}\leq\tfrac{c_{n,k}}{a_{n+1,p}%^{k}}k!\sum_{r=0}^{k}f\Big{(}\tfrac{r}{a_{n,p}}\Big{)}\tfrac{1}{r!}\tfrac{b_{n%,p}^{r}}{(k-r)!}=\tfrac{k!}{a_{n+1,p}^{k}}\sum_{r=0}^{k}c_{n,k}^{k-r}f\Big{(}%\tfrac{r}{a_{n,p}}\Big{)}\tfrac{1}{r!}\tfrac{a_{n,p}^{r}}{(k-r)!}.

By the Cauchy multiplication rule for product of series,

	$\displaystyle e^{(\beta(n+1)-\beta(n))x}\sum_{k=0}^{\infty}\tfrac{a_{n,p}^{k}}%{k!}f\Big{(}\tfrac{k}{a_{n,p}}\Big{)}x^{k}$	$\displaystyle=\sum_{k=0}^{\infty}\Big{\{}\sum_{m+r=k}\tfrac{((\beta(n+1)-\beta%(n))x)^{m}}{m!}\tfrac{a_{n,p}^{r}}{r!}f\Big{(}\tfrac{r}{a_{n,p}}\Big{)}x^{r}%\Big{\}}$
		$\displaystyle=\sum_{k=0}^{\infty}\Big{\{}\sum_{r=0}^{k}\tfrac{(\beta(n+1)-%\beta(n))^{k-r}}{(k-r)!}\tfrac{a_{n,p}^{r}}{r!}f\Big{(}\tfrac{r}{a_{n,p}}\Big{%)}\Big{\}}x^{k}.$

Therefore

	$\displaystyle e^{a_{n+1,p}x}\Big{(}M_{n,p}(f,x)-M_{n+1,p}(f,x)\Big{)}=$
	$\displaystyle=e^{(\beta(n+1)-\beta(n))x}\sum_{k=0}^{\infty}\tfrac{a_{n,p}^{k}}%{k!}f\Big{(}\tfrac{k}{a_{n,p}}\Big{)}x^{k}-\sum_{k=0}^{\infty}\tfrac{a_{n+1,p}%^{k}}{k!}f\Big{(}\tfrac{k}{a_{n+1,p}}\Big{)}x^{k}$
	$\displaystyle=\sum_{k=0}^{\infty}\Big{\{}\sum_{r=0}^{k}\tfrac{(\beta(n+1)-%\beta(n))^{k-r}}{(k-r)!}\tfrac{a_{n,p}^{r}}{r!}f\Big{(}\tfrac{r}{a_{n,p}}\Big{%)}-\tfrac{a_{n+1,p}^{k}}{k!}f\Big{(}\tfrac{k}{a_{n+1,p}}\Big{)}\Big{\}}x^{k}%\geq 0.$

This proves that $M_{n,p}(f,x)\geq M_{n+1,p}(f,x)$ . From Theorem 4 we know that $M_{n,p}(f,x)\to f(x)$ as $n\to\infty$ (pointwise convergence). Thus $M_{n+1,p}(f,x)\geq f(x)$ .

The concave functions follows by changing $f$ by $-f$ . $\square$

Fix $n\in\mathbb{N}$ and let $f\in C_{r}(I)$ be a non-negative function (see (4)).

For a non-negative function $f\in C_{r}(I)$ , in [38], Zhao proved that if $f(x)/x$ is non-increasing on $(0,\infty)$ , then for each $n\geq 1$ , $S_{n}(f,x)/x$ is non-increasing. A similar result can be proved for the operators $M_{n,p}$ by modifying the arguments of Zhao. Since the work [38] is not well known, we include the complete proof. Notice that the condition $f\in C_{r}(I)$ (assumed by Zhao) will be replaced by the more general $f\in\mathcal{L}(I)$ .

Theorem 7.

Let $f\in\mathcal{L}(I)$ be a non-negative function. If $f(x)/x$ is non-increasing on $(0,\infty)$ , then for each $n\in\mathbb{N}$ , $M_{n,p}(f,x)/x$ is non-increasing.

Proof. We will prove that $(d/dx)(M_{n,p}(f,x)/x)\leq 0$ . We use the notations in (3).

Since

\tfrac{M_{n,p}(f,x)}{x}=f(0)\tfrac{g_{n,p}(x)}{x}+g_{n,p}(x)\sum_{k=1}^{\infty%}\tfrac{a_{n,p}^{k}}{k!}f\Big{(}\tfrac{k}{a_{n}}\Big{)}x^{k-1}

and

\tfrac{d}{dx}\tfrac{g_{n,p}(x)}{x}=\tfrac{g_{n,p}(x)}{x^{2}}\Big{(}-a_{n,p}x-1%\Big{)}<0,

we should consider the derivative of the previous series. Note that

	$\displaystyle\sum_{k=1}^{\infty}\tfrac{a_{n,p}^{k}}{k!}f\Big{(}\tfrac{k}{a_{n,%p}}\Big{)}\tfrac{d}{dx}\Big{(}g_{n,p}(x)x^{k-1}\Big{)}=$
	$\displaystyle=g_{n,p}(x)\sum_{k=2}^{\infty}\tfrac{a_{n,p}^{k}(k-1)}{k!}f\Big{(%}\tfrac{k}{a_{n,p}}\Big{)}x^{k-2}-g_{n,p}(x)\sum_{k=1}^{\infty}\tfrac{a_{n,p}^%{k+1}}{k!}f\Big{(}\tfrac{k}{a_{n}}\Big{)}x^{k-1}$
	$\displaystyle=g_{n,p}(x)\sum_{k=1}^{\infty}\tfrac{a_{n,p}^{k+1}\,k}{(k+1)!}f%\Big{(}\tfrac{k+1}{a_{n,p}}\Big{)}x^{k-1}-g_{n,p}(x)\sum_{k=1}^{\infty}\tfrac{%a_{n,p}^{k+1}}{k!}f\Big{(}\tfrac{k}{a_{n,p}}\Big{)}x^{k-1}$
	$\displaystyle=g_{n,p}(x)\sum_{k=1}^{\infty}\Big{\{}\tfrac{a_{n,p}}{k+1}f\Big{(%}\tfrac{k+1}{a_{n,p}}\Big{)}-\tfrac{a_{n}}{k}f\Big{(}\tfrac{k}{a_{n}}\Big{)}%\Big{\}}\tfrac{a_{n,p}^{k}\,\,x^{k-1}}{(k-1)!}\leq 0.$

The result is proved. $\square$

4. Preservation of modulus of continuity

Definition 8.

A function $\omega:I\to\mathbb{R}^{+}$ is called a modulus of continuity if $\omega(0)=0$ , $\lim_{t\to 0}\omega(t)=0$ , $\omega$ is non-negative and non-decreasing in $I$ and $\omega(t)$ is continuous in $\mathbb{R}^{+}$ .

Definition 9.

A function $\omega:I\to\mathbb{R}^{+}$ is called subadditive if for any $s,t\geq 0$

\omega(s+t)\leq\omega(s)+\omega(t).

If a subadditive function $\omega:I\to\mathbb{R}^{+}$ is continuous at zero and $\omega(0)=0$ , then it is continuous. If $\omega$ is subadditive, then $\omega(2t)\leq 2\omega(t)$ and it follows from standard arguments that, if $t,\lambda>0$ , then

(10)

\omega(\lambda t)\leq(1+\lambda)\,\omega(f,t).

It is known that (see [12, p. 43]), for any modulus of continuity $\omega$ on $I$ , there exists a concave modulus of continuity (the least concave majorant) $\widetilde{\omega}$ such that

(11)

\omega(t)\leq\widetilde{\omega}(t)\leq 2\omega(t).

For Szász-Mirakyan operators preservation of the usual modulus of continuity has been considered in [32], [16] and [4]. For instance, if $\omega(t)$ is a concave modulus of continuity and

\Lambda(\omega,A)=\Big{\{}f\in C(I):\,\,\omega(f,t)\leq A\omega(t)\Big{\}},

it is asserted in [16] that $f\in\Lambda(\omega,A)$ if and only if $S_{n}(f)\in\Lambda(\omega,A)$ , for each each $n\in\mathbb{N}$ . On the other hand, in [4] the authors considered functions $f$ such that $0<\omega(f,1)<\infty$ , where $\omega(f,t)$ is the usual modulus of continuity. Of course the condition $0<\omega(f,1)$ holds whenever $f$ is not a constant function.

Of course, since the usual modulus of continuity is not well defined for all $f\in C(I)$ , such a result must be handled with care. In fact in [14] Hermann presented a negative result. Let

C_{0}=\Big{\{}f\in C(I):\,\sup_{x\in I}\mid f(x+\delta)-f(x)\mid<\infty\,\,\,%\,{\rm for\,\,\,any}\,\,\delta>0\Big{\}}.

Notice that for any $f\in C_{0}$ the usual modulus of continuity is well defined, but the conditions $f\in C_{0}$ and $\delta\to 0$ does not necessarily imply $\omega(f,\delta)\to 0$ .

Set $C_{0}^{*}=\{f\in C_{0}:\,\omega(f,t)>0\}$ . In [14] Hermann proved that

\sup_{f\in C_{0}^{*}}\tfrac{\|S_{n}(f)-f\|_{C}}{\omega(f,1/n)}=\infty.

In this section we prove some results related with preservation of some modulus of continuity by the operators $M_{n,p}$ .

Although Theorem 3 is sufficient to prove the preservation of convexity of different order by the operators $M_{n,p}$ , we need other kind of representations for studying modulus of continuity.

The ideas for the proof of Proposition 10 have been used for different authors in the case of Szász-Mirakyan operators (see [32] and [16]).

Proposition 10.

If $f\in\mathcal{L}(I)$ , $n\in\mathbb{N}$ , $x\in I$ and $s>0$ , then

\displaystyle M_{n,p}(f,x+s)\!-\!M_{n,p}(f,x)\!=\!e^{-a_{n,p}(x+s)}\sum_{k=0}^%{\infty}\tfrac{a_{n,p}^{k}x^{k}}{k!}\sum_{i=1}^{\infty}\tfrac{a_{n,p}^{i}}{i!}%s^{i}\Big{(}f\Big{(}\tfrac{k+i}{a_{n,p}}\Big{)}\!-\!f\Big{(}\tfrac{k}{a_{n,p}}%\Big{)}\Big{)}.

Proof. Notice that

	$\displaystyle e^{a_{n,p}(x+s)}M_{n,p}(f,x+s)=$
	$\displaystyle=\sum_{j=0}^{\infty}f\Big{(}\tfrac{j}{a_{n,p}}\Big{)}\tfrac{a_{n,%p}^{j}}{j!}(x+s)^{j}=\sum_{j=0}^{\infty}f\Big{(}\tfrac{j}{a_{n,p}}\Big{)}%\tfrac{a_{n,p}^{j}}{j!}\sum_{k=0}^{j}\tbinom{j}{k}x^{k}s^{j-k}=$
	$\displaystyle=\sum_{k=0}^{\infty}\sum_{j=k}^{\infty}\tbinom{j}{k}f\Big{(}%\tfrac{j}{a_{n,p}}\Big{)}\tfrac{a_{n,p}^{j}}{j!}x^{k}s^{j-k}=\sum_{k=0}^{%\infty}\tfrac{x^{k}}{k!}\sum_{j=k}^{\infty}f\Big{(}\tfrac{j}{a_{n,p}}\Big{)}%\tfrac{a_{n,p}^{j}}{(j-k)!}s^{j-k}$
	$\displaystyle=\sum_{k=0}^{\infty}\tfrac{x^{k}}{k!}\sum_{i=0}^{\infty}f\Big{(}%\tfrac{i+k}{a_{n,p}}\Big{)}\tfrac{a_{n,p}^{i+k}}{i!}s^{i}.$

On the other hand,

\displaystyle e^{a_{n,p}(x+s)}M_{n,p}(f,x)\!=\!e^{a_{n,p}s}\sum_{k=0}^{\infty}%f\Big{(}\tfrac{k}{a_{n,p}}\Big{)}\tfrac{a_{n,p}^{k}}{k!}x^{k}\!=\!\sum_{k=0}^{%\infty}\tfrac{a_{n,p}^{k}x^{k}}{k!}\Big{(}\sum_{i=0}^{\infty}\tfrac{a_{n,p}^{i%}}{i!}s^{i}f\Big{(}\tfrac{k}{a_{n,p}}\Big{)}\Big{)}.

It follows from the equation given above the announced result. $\square$

Let $UC_{b}(I)$ the class of all bounded uniformly continuous functions $f:I\to\mathbb{R}$ . For $f\in UC_{b}(I)$ and $t\geq 0$ , define

(12)

\omega(f,t)=\sup_{0\leq h\leq t}\,\sup_{x\geq 0}\,\mid f(x+h)-f(x)\mid.

It can be proved that $\omega(f,t)$ is subadditive modulus of continuity in the sense of Definition 8.

Theorem 11.

If $f\in UC_{b}(I)$ , $n\in\mathbb{N}$ , and $s>0$ , then $M_{n,p}(f,x)$ is uniformly continuous and

\omega(M_{n,p}(f),s)\leq 2\,\omega(f,s).

Proof. Let $\widetilde{\omega}(f,t)$ be the least concave majorant of $\omega(f,t)$ .

If $f\in UC_{b}(I)$ , then $f\in\mathcal{L}(I)$ . From Proposition 10 one has

	$\displaystyle\mid M_{n,p}(f,x+s)-M_{n,p}(f,x)\mid\leq$
	$\displaystyle\leq e^{-a_{n,p}(x+s)}\sum_{k=0}^{\infty}\tfrac{a_{n,p}^{k}x^{k}}%{k!}\sum_{i=0}^{\infty}\tfrac{a_{n,p}^{i}}{i!}s^{i}\mid f\Big{(}\tfrac{k+i}{a_%{n,p}}\Big{)}-f\Big{(}\tfrac{k}{a_{n,p}}\Big{)}\mid$
	$\displaystyle\leq e^{-a_{n,p}(x+s)}\sum_{k=0}^{\infty}\tfrac{a_{n,p}^{k}x^{k}}%{k!}\sum_{i=0}^{\infty}\tfrac{a_{n,p}^{i}}{i!}s^{i}\omega\Big{(}f,\tfrac{i}{a_%{n,p}}\Big{)}$
	$\displaystyle=M_{n,p}(\omega(f,e_{1}),s)\leq M_{n,p}(\widetilde{\omega}(f,e_{1%}),s).$

Since $\widetilde{\omega}(f,t)$ is a concave function, it follows from Theorem 6 that

\hskip 73.97716ptM_{n,p}(\widetilde{\omega}(f),s)\leq\widetilde{\omega}(f,s)%\leq 2\,\omega(f,s).

In particular if $\varepsilon>0$ , $\omega(f,s)\leq\varepsilon/2$ , $0\leq y<x$ , $x-y\leq s$ and we set $x=y+t$

\mid M_{n,p}(f,x)-M_{n,p}(f,y)\mid=\mid M_{n,p}(f,y+t)-M_{n,p}(f,y)\mid\leq\varepsilon.

This proves that $M_{n,p}(f)$ is uniformly continuous. $\square$

For $f\in UC_{b}(I)$ , $0<\alpha\leq 1$ , and $t>0$ define

\theta_{\alpha}(f,t)=\sup_{0<s\leq t}\,\,\sup_{x\in I,\,0<h\leq s}\tfrac{\mid f%(x+h)-f(x)\mid}{h^{\alpha}},

$\theta_{\alpha}(f,0)=0$ , and

K^{\alpha}(f)=\sup_{0\leq t}\,\theta_{\alpha}(f,t).

For $0<\alpha\leq 1$ , let us set $\operatorname{Lip}^{\alpha}(I)$ for the family of all $f\in UC_{b}(I)$ such that

K^{\alpha}(f)<\infty.

For $0<\alpha<1$ , we also we consider the subspace

(13)

\operatorname{lip}^{\alpha}(I)=\left\{f\in\operatorname{Lip}^{\alpha}(I):\,\,%\lim_{t\rightarrow 0}\theta_{\alpha}(f,t)=0~{}~{}\right\}.

This type of spaces appears when we study the approximation in Hölder type norms (see [7]).

We will analyze the problem of the preservation of the constants $K^{\alpha}(f)$ and the class $\operatorname{lip}^{\alpha}(I)$ by the operators $M_{n,p}$ .

For an analogous of Theorem 12 for Szász-Mirakyan operators see [16] and [13].

Theorem 12.

(i) If $0<\alpha\leq 1$ and $f\in\operatorname{Lip}^{\alpha}(I)$ , then $M_{n,p}(f)\in\operatorname{Lip}^{\alpha}(I)$ , and

(14)

K^{\alpha}(M_{n,p}(f))\leq K^{\alpha}(f),

for each $n\in\mathbb{N}$ .

(ii) If $0<\alpha\leq 1$ , $f\in\mathcal{L}(I)$ , $M_{n,p}(f)\in\operatorname{Lip}^{\alpha}(I)$ , for each $n\in\mathbb{N}$ , and

K:=\sup_{n\in\mathbb{N}}K^{\alpha}(M_{n,p}(f))<\infty,

then $f\in\operatorname{Lip}^{\alpha}(I)$ .

Proof. (i) Set $g(x)=x^{\alpha}$ . Since the function $g(x)$ is concave function and $M_{n,p}(g,x)\to g(x)$ (Theorem 4) and it follows Theorem 6 that $M_{n,p}(g,x)\leq g(x)$ .

For any $k\in\mathbb{N}_{0}$ and $i\in\mathbb{N}$ ,

\mid f\Big{(}\tfrac{k+i}{a_{n,p}}\Big{)}-f\Big{(}\tfrac{k}{a_{n,p}}\Big{)}\mid%\leq\omega\Big{(}f,\tfrac{i}{a_{n,p}}\Big{)}\leq\theta_{\alpha}\Big{(}f,\tfrac%{i}{a_{n,p}}\Big{)}\Big{(}\tfrac{i}{a_{n,p}}\Big{)}^{\alpha}\leq K^{\alpha}(f)%g\Big{(}\tfrac{i}{a_{n,p}}\Big{)}.

From Proposition 10 we know that, for $x\in I$ and $h>0$ ,

\mid M_{n,p}(f,x+h)-M_{n,p}(f,x)\mid\leq K^{\alpha}(f)M_{n,p}(g,h)\leq K^{%\alpha}(f)h^{\alpha}.

(ii) From Theorem 4 we know that, for each fixed $x\in I$ , $M_{n,p}(f,x)\to f(x)$ , as $n\to\infty$ .

For $x\geq 0$ , $h>0$ fixed, and each $n\in\mathbb{N}$ , one has

	$\displaystyle\mid f(x+h)-f(x)\mid$	$\displaystyle\leq\mid f(x+h)-M_{n,p}(f,x+h)\mid$
		$\displaystyle\quad+\mid M_{n,p}(f,x+h)-M_{n,p}(f,x)\mid+\mid M_{n,p}(f,x)-f(x)\mid$
		$\displaystyle\leq\mid f(x+h)-M_{n,p}(f,x+h)\mid+Kh^{\alpha}+\mid M_{n,p}(f,x)-%f(x)\mid.$

The result follows by taking $n\to\infty$ . $\square$

For the preservation of the class $\operatorname{lip}^{\alpha}(I)$ we need some previous results.

Proposition 13.

For $0<\alpha<1$ and each $f\in\operatorname{lip}^{\alpha}(I)$ ,

\theta_{\alpha}(f,t)=\sup_{0<s\leq t}\,\,\sup_{0<h\leq s}\tfrac{\omega(f,h)}{h%^{\alpha}},

where $\omega(f,t)$ is defined by (12).

Proof. By definition, if $f\in UC_{b}(I)$ , then $\omega(f,s)$ is well defined. It is clear that

\sup_{x\in I,\,0<h\leq s}\tfrac{\mid f(x+h)-f(x)\mid}{h^{\alpha}}\leq\sup_{0<h%\leq s}\tfrac{\omega(f,h)}{h^{\alpha}}.

On the other hand, given $\varepsilon>0$ , for any $0<h\leq s$ , there exists $x_{h}\in I$ such that

\omega(f,h)\leq\varepsilon h^{\alpha}+\mid f(x_{h}+h)-f(x_{h})\mid.

Therefore

\tfrac{\omega(f,h)}{h^{\alpha}}\leq\varepsilon+\tfrac{\mid f(x_{h}+h)-f(x_{h})%\mid}{h^{\alpha}}\leq\varepsilon+\sup_{x\in I,\,0<h\leq s}\tfrac{\mid f(x+h)-f%(x)\mid}{h^{\alpha}}.\square

Proposition 14.

If $0<\alpha<1$ , for each $f\in\operatorname{lip}^{\alpha}(I)$ , the functional $\theta_{\alpha}(f,t)$ is a subadditive modulus of continuity.

Proof. (a) By definition $\theta_{\alpha}(f,0)=0$ and $\theta_{\alpha}(f,t)\to 0$ as $t\to 0$ . Moreover it is clear that $\theta_{\alpha}(f,t)$ is non-negative and non-decreasing in $I$

(b) Let us verify that $\theta_{\alpha}(f,t)$ is subadditive. Assume $0<v\leq t$ and fix any $s$ and $h$ such that $0<s\leq v+t$ and $0<h\leq s$ .

If $x\in I$ and $h\leq t$ it is clear that

\tfrac{\mid f(x+h)-f(x)\mid}{h^{\alpha}}\leq\sup_{0<u\leq t}\sup_{y\in I,\,0<w%\leq u}\tfrac{\mid f(y+w)-f(y)\mid}{w^{\alpha}}=\theta_{\alpha}(f,t).

We still have to consider the case $v\leq t<h$ . Since $t<h$ and $0<h-t<h$ , one has

	$\displaystyle\tfrac{\mid f(x+h)-f(x)\mid}{h^{\alpha}}$	$\displaystyle\leq\tfrac{\mid f(x+h-t+t)-f(x+h-t)\mid}{t^{\alpha}}+\tfrac{\mid f%(x+h-t)-f(x)\mid}{(h-t)^{\alpha}}$
		$\displaystyle\leq\theta_{\alpha}(f,t)+\theta_{\alpha}(f,h-t)\leq\theta_{\alpha%}(f,t)+\theta_{\alpha}(f,v),$

because $\theta_{\alpha}(f,t)$ increases and $h-t\leq s-t\leq v$ . Therefore

\theta_{\alpha}(f,t+v)\leq\theta_{\alpha}(f,t)+\theta_{\alpha}(f,v).

(c) Taking into account that $\theta_{\alpha}(f,0)=0$ and $\theta(f,t)$ is subadditive, it is a continuous function. $\square$

Theorem 15.

If $0<\alpha<1$ , $f\in\operatorname{lip}^{\alpha}(I)$ , $n\in\mathbb{N}$ , and $t>0$ , then

\theta_{\alpha}(M_{n,p}(f),t)\leq 2\theta_{\alpha}(f,t).

Proof. If $0<s\leq t$ , taking into account Theorem 11, one has

\mid M_{n,p}(f,x+s)-M_{n,p}(f,x)\mid\leq 2\,\omega(f,s)=2\,\tfrac{\omega(f,s)}%{s^{\alpha}}s^{\alpha}\leq 2\,\theta_{\alpha}(f,s)s^{\alpha}\leq 2\,\theta_{%\alpha}(f,t)s^{\alpha}.

This is sufficient to prove the result. $\square$

For each $r\geq 0$ , $f\in C_{r}(I)$ (see (4)), and $t\geq 0$ , define

\Omega_{r}(f,t)=\,\sup_{0\leq s\leq t}\,\,\sup_{x\geq 0}\tfrac{\mid f(x+s)-f(x%)\mid}{(1+x+s)^{r}}.

We will use this modulus only in the case $f\in C_{r,\infty}(I)$ .

Before proving some properties of this modulus, let us compare them with others that have been used previously.

The following functional was considered by Kratz and Stadtmüller in [20]. For $r\in\mathbb{N}$ and a function $f\in C_{r}(I)$ set

\widetilde{\Omega}_{r}(f,t)=\sup_{s,v\in I,\mid s-v\mid\leq t}\tfrac{\mid f(s)%-f(v)\mid}{(1+s+v)^{r}}=\sup_{x\geq 0}\,\,\sup_{0<s\leq t}\tfrac{\mid f(x+s)-f%(x)\mid}{(1+2x+s)^{r}}.

Taking into account that $1+x+s\leq 1+2x+s\leq 2(1+x+s)$ , we know that

\tfrac{1}{2^{r}}\Omega_{r}(f,t)\leq\widetilde{\Omega}_{r}(f,t)\leq\Omega_{r}(f%,t).

Kratz and Stadtmüller proved that, for Szász-Mirakyan operators, there exists a constant $C$ such that, for all $f\in C_{r}(I)$ , every $t\geq 0$ and each $n\in\mathbb{N}$ ,

\widetilde{\Omega}_{r}(S_{n}(f),t)\leq C\widetilde{\Omega}_{r}(f,t).

They did not proved that $\lim_{t\to 0^{+}}\widetilde{\Omega}_{r}(f,t)=0$ . We will verify that, if $f\in C_{r,\infty}(I)$ , then $\lim_{t\to 0^{+}}\Omega_{r}(f,t)=0$ .

For $f\in C_{2,\infty}(I)$ , another modulus was considered in [3] by setting

\Omega(f,t)=\,\sup_{0\leq s\leq t}\,\,\sup_{x\in I}\tfrac{\mid f(x+s)-f(x)\mid%}{(1+s)^{2}(1+x)^{2}}.

For $0\leq t\leq 1$ , $\Omega(f,t)$ and $\Omega_{2}(f,t)$ are equivalent. In fact, suppose that $s\leq 1$ . First one has

(1+s^{2})(1+x^{2})=1+s^{2}+x^{2}+s^{2}x^{2}\leq 2(1+s^{2}+x^{2})\leq 2(1+x+s)^%{2}.

On the other hand, if $x\leq 1$ ,

(1+s+x)^{2}=1+2x+2s+x^{2}+2xs+s^{2}\leq 7(1+s^{2}+x^{2})\leq 7(1+s^{2})(1+x^{2%}).

and, if $x>1$ ,

(1+s+x)^{2}\leq 3+5x^{2}+s^{2}\leq 5(1+s^{2}+x^{2})\leq 5(1+s^{2})(1+x^{2}).

Therefore

\tfrac{1}{2}\Omega_{2}(f,t)\leq\Omega(f,t)\leq 7\Omega_{2}(f,t).

Proposition 16.

If $r$ is a non negative real and $f\in C_{r,\infty}(I)$ , then $\Omega_{r}(f,t)$ is a subadditive modulus of continuity in the sense of Definition 8.

Proof. It is clear that $\Omega_{r}(f,0)=0$ and $\Omega_{r}(f,t)$ is non-negative and non-decreasing in $I$

(a) We consider first the case $r=0$ . As in the case of the classical modulus of continuity, it is easy to prove that the functional $\Omega_{r}(f,t)$ is a subadditive. In order to prove continuity, it is sufficient to verify continuity a zero, but if follows from the condition $\lim_{x\to\infty}f(x)=0$ .

(b) Assume $r>0$ . Denote $A=\lim_{x\to\infty}f(x)/(1+x)^{r}$ . Given $\varepsilon>0$ , there exists $x_{0}$ such that

\left|\tfrac{f(x)}{(1+x)^{r}}-A\right|<\tfrac{\varepsilon}{2},\quad\quad x\geqx%_{0}.

If $t>0$ and $0<s\leq t\leq 1$ , then

	$\displaystyle\sup_{x\geq 0}\tfrac{\mid f(x+s)-f(x)\mid}{(1+x+s)^{r}}$	$\displaystyle\leq\sup_{0\leq x\leq x_{0}}\tfrac{\mid f(x+s)-f(x)\mid}{(1+x+s)^%{r}}+\sup_{x\geq x_{0}}\tfrac{\mid f(x+s)-f(x)\mid}{(1+x+s)^{r}}$
		$\displaystyle\leq\sup_{0\leq x\leq x_{0}}\mid f(x+s)-f(x)\mid+\sup_{x\geq x_{0%}}\tfrac{\mid f(x+s)-A\mid}{(1+x+s)^{r}}+\sup_{x\geq x_{0}}\tfrac{\mid f(x)-A%\mid}{(1+x)^{r}}$
		$\displaystyle\leq\omega_{1}(f,t)_{[0,x_{0}+1]}+\varepsilon,$

where $\omega_{1}(f,t)_{[0,x_{0}+1]}$ is the usual modulus of continuity in the interval $[0,x_{0}+1]$ .

This is sufficient to prove that $\lim_{t\to 0}\Omega_{r}(f,t)\to 0=0$ .

(c) Let us verify that $\Omega_{r}(f,t)$ is subadditive: $\Omega_{r}(f,v+t)\leq\Omega_{r}(f,v)+\Omega_{r}(f,t)$ . Without losing generality we assume that $0<v\leq t$ .

Fijemos $x\geq 0$ and $0<s\leq t+v$ .

If $s\leq t$ , it is clear that

\tfrac{\mid f(x+s)-f(x)\mid}{1+(x+s)^{r}}\leq\sup_{0<s\leq t}\tfrac{\mid f(x+s%)-f(x)\mid}{1+(x+s)^{r}}=\Omega_{r}(f,t).

Let us consider the case $v\leq t<s$ . Since $0<s-t$ , one has $(1+x+s-t)^{r}<(1+x+s)^{r}$ . Therefore

	$\displaystyle\tfrac{\mid f(x+s)-f(x)\mid}{(1+x+s)^{r}}$	$\displaystyle\leq\tfrac{\mid f(x+s-t+t)-f(x+s-t)\mid}{(1+(x+s-t)+t)^{r}}+%\tfrac{\mid f(x+s-t)-f(x)\mid}{(1+x+s-t)^{r}}$
		$\displaystyle\leq\Omega_{r}(f,t)+\Omega_{r}(f,s-t)\leq\Omega_{r}(f,t)+\Omega_{%r}(f,v).$

It is sufficient to prove that $\Omega_{r}(f,t)$ is a modulus of continuity. $\square$

Theorem 17.

If $r$ is a non negative real, there exists a constant $C$ such that, for $f\in C_{r,\infty}(I)$ , $n\in\mathbb{N}$ , and $t>0$ ,

\Omega_{r}(M_{n,p}(f),t)\leq C\Omega_{r}(f,t).

Proof. Notice that, for $s>0$ , taking into account (10), with $t=s$ and $\lambda=i/(sa_{n,p})$ ,

	$\displaystyle\left\|f\Big{(}\tfrac{k+i}{a_{n,p}}\Big{)}-f\Big{(}\tfrac{k}{a_{n,%p}}\Big{)}\right\|$	$\displaystyle\leq\Omega_{r}\Big{(}\tfrac{i}{a_{n,p}}\Big{)}\Big{(}1+\tfrac{k}{%a_{n,p}}+\tfrac{i}{a_{n,p}}\Big{)}^{r}$
		$\displaystyle\leq\Omega_{r}(f,s)\Big{(}1+\tfrac{i}{a_{n,p}s}\Big{)}\Big{(}1+%\tfrac{k}{a_{n,p}}+\tfrac{i}{a_{n,p}}\Big{)}^{r},$

because $\Omega_{r}(f,s)$ is a subadditive modulus.

Therefore (see Proposition 10)

	$\displaystyle\mid M_{n,p}(f,x+s)-M_{n,p}(f,x)\mid=$
	$\displaystyle=\left\|e^{-a_{n,p}(x+s)}\sum_{k=0}^{\infty}\tfrac{a_{n,p}^{k}x^{k%}}{k!}\sum_{i=0}^{\infty}\tfrac{a_{n,p}^{i}}{i!}s^{i}\Big{(}f\Big{(}\tfrac{k+i%}{a_{n,p}}\Big{)}-f\Big{(}\tfrac{k}{a_{n,p}}\Big{)}\Big{)}\right\|$
	$\displaystyle\leq\tfrac{\Omega_{r}(f,s)}{e^{a_{n,p}(x+s)}}\sum_{k=0}^{\infty}%\tfrac{a_{n,p}^{k}x^{k}}{k!}\sum_{i=0}^{\infty}\tfrac{a_{n,p}^{i}}{i!}s^{i}%\Big{(}1+\tfrac{i}{a_{n,p}s}\Big{)}\Big{(}1+\tfrac{k}{a_{n,p}}+\tfrac{i}{a_{n,%p}}\Big{)}^{r}.$

Taking into account Proposition 2 (with $a=1+k/a_{n,p}$ ), we obtain

	$\displaystyle e^{-a_{n,p}s}\sum_{i=0}^{\infty}\tfrac{a_{n,p}^{i}}{i!}s^{i}\Big%{(}1+\tfrac{k}{a_{n,p}}+\tfrac{i}{a_{n,p}}\Big{)}^{r}=$
	$\displaystyle=M_{n,p}\Big{(}\Big{(}1+\tfrac{k}{a_{n,p}}+e_{1}\Big{)}^{r},s\Big%{)}\leq C(r)\Big{(}2+\tfrac{k}{a_{n,p}}+s\Big{)}^{r}\leq 2^{r}C(r)\Big{(}1+%\tfrac{k}{a_{n,p}}+s\Big{)}^{r}.$

On the other hand

	$\displaystyle\tfrac{e^{-a_{n,p}s}}{s}\sum_{i=1}^{\infty}\tfrac{a_{n,p}^{i}}{i!%}s^{i}\tfrac{i}{a_{n,p}}\Big{(}1+\tfrac{k}{a_{n,p}}+\tfrac{i}{a_{n,p}}\Big{)}^%{r}=$
	$\displaystyle=e^{-a_{n,p}s}\sum_{i=1}^{\infty}\tfrac{a_{n,p}^{i-1}}{(i-1)!}s^{%i-1}\Big{(}1+\tfrac{k}{a_{n,p}}+\tfrac{i}{a_{n,p}}\Big{)}^{r}$
	$\displaystyle=M_{n,p}\Big{(}\Big{(}1+\tfrac{1}{a_{n,p}}+\tfrac{k}{a_{n,p}}+e_{%1}\Big{)}^{r},s\Big{)}\leq M_{n,p}\Big{(}\Big{(}2+\tfrac{k}{a_{n,p}}+e_{1}\Big%{)}^{r},s\Big{)}$
	$\displaystyle\leq C(r)\Big{(}3+\tfrac{k}{a_{n,p}}+s\Big{)}^{r}\leq 3^{r}C(r)%\Big{(}1+\tfrac{k}{a_{n,p}}+s\Big{)}^{r}.$

From the estimates given above one has

	$\displaystyle\mid M_{n,p}(f,x+s)-M_{n,p}(f,x)\mid$	$\displaystyle\leq 3^{r}C(r)\tfrac{\Omega_{r}(f,s)}{e^{a_{n,p}x}}\sum_{k=0}^{%\infty}\tfrac{a_{n,p}^{k}x^{k}}{k!}\Big{(}1+\tfrac{k}{a_{n,p}}+s\Big{)}^{r}$
		$\displaystyle\leq 6^{r}C(r)\Omega_{r}(f,s)(1+x+s)^{r},$

where we use again Proposition 2. $\square$

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Preserving properties
of some Szász-Mirakyan type operators

Abstract.

Key words and phrases:

2005 Mathematics Subject Classification:

1. Introduction

2. Some basic properties

Theorem 1.

Proposition 2.

Theorem 3.

Theorem 4.

3. Monotonicity and convex functions

Theorem 5.

Theorem 6.

Theorem 7.

4. Preservation of modulus of continuity

Definition 8.

Definition 9.

Proposition 10.

Theorem 11.

Theorem 12.

Proposition 13.

Proposition 14.

Theorem 15.

Proposition 16.

Theorem 17.

References

Information

Indexing and abstracting

Publisher

Preserving properties of some Szász-Mirakyan type operators

Abstract.

Key words and phrases:

2005 Mathematics Subject Classification:

1. Introduction

2. Some basic properties

Theorem 1.

Proposition 2.

Theorem 3.

Theorem 4.

3. Monotonicity and convex functions

Theorem 5.

Theorem 6.

Theorem 7.

4. Preservation of modulus of continuity

Definition 8.

Definition 9.

Proposition 10.

Theorem 11.

Theorem 12.

Proposition 13.

Proposition 14.

Theorem 15.

Proposition 16.

Theorem 17.

References

Information

Indexing and abstracting

Publisher

Preserving properties
of some Szász-Mirakyan type operators