Preserving properties
of some Szász-Mirakyan type operators

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.

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

$$e_j(x)=x^j,x\in ℝ,$$

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

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

$$S_n(f,x)=e^{-nx}\sum _{k=0}^{\mathrm{\infty }}\frac{n^k}{k!}f\left(\frac{k}{n}\right)x^k,x\in I.$$

It is known that $S_n(e_0,x)=1$ and $S_n(e_1,x)=x$ (see [6]).

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

(1) $$S_{n,p}(f,x)=e^{-(n+p)x}\sum _{k=0}^{\mathrm{\infty }}\frac{{(n+p)}^k}{k!}f\left(\frac{k}{n}\right)x^k,x\in I.$$

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 $x\ge 0$ and $n\in \mathbb{N}$, $S_{n,p}(e_0,x)=1$ and

$$S_{n,p}(e_1,x)=x+\frac{px}{n}.$$

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 ${lim}_{n\to \mathrm{\infty }}\beta (n)=\mathrm{\infty }$. For $p\ge 0$, $n\in \mathbb{N}$, $x\ge 0$, and a function $f\in C(I)$ consider the operator

(2) $$M_{n,p}(f,x)=e^{-(\beta (n)+p)x}\sum _{k=0}^{\mathrm{\infty }}\frac{{(\beta (n)+p)}^k}{k!}f\left(\frac{k}{\beta (n)+p}\right)x^k,$$

whenever the series converges absolutely. Let $ℒ(I)$ 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 $p\ge 0$, $n\in \mathbb{N}$, and $x\ge 0$ we use the notations

(3) $$g_{n,p}(x)=e^{-(\beta (n)+p)x}\mathrm{and}{a}_{n,p}=\beta (n)+p.$$

For $r\in {\mathbb{N}}_0$, $C_r(I)$ is the family of all $f\in C(I)$ such that

(4)

For $r\in {\mathbb{N}}_0$, let $C_{r,\mathrm{\infty }}(I)$ be the class of all functions $f\in C_r(I)$ such that $f(x)/{(1+x)}^r$ has a finite limit as $x\to \mathrm{\infty }$.

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) $$\sum _{k=0}^{\mathrm{\infty }}\frac{a_{n,p}^k}{k!}{x}^k=e^{(\beta (n)+p)x}=g_{n,p}(x),$$

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) $$g_{n,p}^{(i)}(x)=\sum _{k=i}^{\mathrm{\infty }}\frac{a_{n,p}^k}{(k-i)!}{x}^{k-i}=\sum _{k=0}^{\mathrm{\infty }}\frac{a_{n,p}^{k+i}}{k!}{x}^k=a_{n,p}^i{g}_{n,p}(x).$$

Theorem 1.

If $i\in {\mathbb{N}}_0$ and

(7) $$P_{i+1}(x)=x\left(x-\frac{1}{a_{n,p}}\right)\mathrm{\cdots }\left(x-\frac{i}{a_{n,p}}\right),x\ge 0,$$

then

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

In particular, for each $n\in \mathbb{N}$ and $i\in {\mathbb{N}}_0$, $e_i\in ℒ[0,\mathrm{\infty })$ 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 $k\in {\mathbb{N}}_0$,

$$a_{n,p}^{i+1}{P}_{i+1}\left(\frac{k}{a_{n,p}}\right)=k(k-1)\mathrm{\cdots }(k-i).$$

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

$a_{n,p}^{i+1}{g}_{n,p}(x)M_{n,p}(P_{i+1},x)$ $={\displaystyle \sum _{k=i+1}^{\mathrm{\infty }}}\frac{a_{n,p}^k{x}^k}{(k-i-1)!}=x^{i+1}{\displaystyle \sum _{k=i+1}^{\mathrm{\infty }}}\frac{a_{n,p}^k{x}^{k-i-1}}{(k-i-1)!}$

$=x^{i+1}{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\frac{a_{n,p}^{k+i+1}}{k!}{x}^k=x^{i+1}{g}_n^{(i+1)}(x),$

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

Since, for $i\ge 0$, $x^i$ can be written as a linear combination of the polynomials $P_1,\mathrm{\dots },P_i$, we know that $e_i\in ℒ[0,\mathrm{\infty })$ and $M_{n,p}(e_i,x)\in {\mathbb{P}}_i$. For $i=0$ it is a simple assertion because $M_{n,p}(e_0,x)=1$. $\mathrm{\square }$

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 $C(r)\ge 1$ such that, for every real $a>0$,

$$M_{n,p}({(a+e_1)}^r,x)\le C(r){(1+a+x)}^r.$$

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 $0\le x\le 1$, then

$$\left|\sum _{k=0}^i{b}_{i,k}{x}^k\right|\le \sum _{k=0}^i\mid b_{i,k}\mid \le {(1+x)}^i\sum _{k=0}^i\mid b_{i,k}\mid .$$

If $1\le x$, then

$$\left|\sum _{k=0}^i{b}_{i,k}{x}^k\right|\le x^i\sum _{k=0}^i\mid b_{i,k}\mid \le {(1+x)}^i\sum _{k=0}^i\mid b_{i,k}\mid .$$

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

If $a>0$,

$M_{n,p}({(a+e_1)}^r,s)$ $={\displaystyle \sum _{j=0}^r}\left(\genfrac{}{}{0pt}{}{r}{j}\right)a^{r-j}{M}_{n,p}(e_j,s)$

$\le C(a^r+{\displaystyle \sum _{j=1}^r}\left(\genfrac{}{}{0pt}{}{r}{j}\right)a^{r-j}{(1+x)}^j=C{(1+a+x)}^r.\mathrm{\square }$

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) $M_{n,p}:ℒ(I)\to C^1(I)$.

(ii) $M_{n,p}(e_0,x)=1$

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

(9) $$\frac{1}{a_{n,p}^m}{M}_{n,p}^{(m)}(f,x)=M_{n,p}({\mathrm{\Delta }}_{1/a_{n,p}}^{m}f(t),x),$$

where ${\mathrm{\Delta }}_h^{k}g(u)$ 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 $f\in ℒ(I)$ 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 $g:I\to ℝ$ is said to be $k$-convex, if ${\mathrm{\Delta }}_h^{k}g(u)\ge 0$ 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 ${\mathrm{\Delta }}_h^{k}g(u)\ge 0$ and $S_n(g,x)$ is well defined, then ${\mathrm{\Delta }}_h^k{S}_n(g,u)\ge 0$. 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 $f\in ℒ(I)$ increases, then $M_{n,p}^{\prime }(f,x)\ge 0$.

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

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

Cheney and Sharma proved in [10] that, if $f$ is convex, for each $x$ and every $n\in \mathbb{N}$, $S_{n+1}(f,x)\le S_n(f,x)$. 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 $f\in ℒ(I)$ is convex then, for each $x\ge 0$ and $n\in \mathbb{N}$,

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

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

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

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

$$c_{n,k}={(\beta (n+1)-\beta (n))}^k\mathrm{and}{b}_{n,p}=\frac{a_{n,p}}{\beta (n+1)-\beta (n)},$$

then

$$c_{n,k}\sum _{r=0}^k\left(\genfrac{}{}{0pt}{}{k}{r}\right)b_{n,p}^r={(\beta (n+1)-\beta (n))}^k{\left(\frac{\beta (n)+p}{(\beta (n+1)-\beta (n))}+1\right)}^k=a_{n+1,p}^k.$$

That is

$$\frac{c_{n,k}}{a_{n+1,p}^k}k!\sum _{r=0}^k\frac{1}{r!}\frac{b_{n,p}^r}{(k-r)!}=1.$$

Therefore

$\frac{c_{n,k}}{a_{n+1,p}^k}k!{\displaystyle \sum _{r=0}^k}\frac{r}{a_{n,p}}\frac{1}{r!}\frac{b_{n,p}^r}{(k-r)!}$ $=\frac{k}{a_{n+1,p}^k}\frac{c_{n,k}}{(\beta (n+1)-\beta (n))}{\displaystyle \sum _{r=1}^k}\left(\genfrac{}{}{0pt}{}{k-1}{r}\right)b_{n,p}^{r-1}$

$=\frac{k}{a_{n+1,p}^k}\frac{c_{n,k}}{(\beta (n+1)-\beta (n))}{\left(\frac{\beta (n)+p}{(\beta (n+1)-\beta (n))}+1\right)}^{k-1}$

$=\frac{k}{a_{n+1,p}^k}{\left(\beta (n+1)+p\right)}^{k-1}=\frac{k}{a_{n+1,p}}.$

This proves that $k/a_{n+1,p}$ is a convex combination of the points $\{r/a_{n,p}:\mathrm{\hspace{0.17em}0}\le r\le k\}$.

If $f$ is convex, then

$f\left(\frac{k}{a_{n+1,p}}\right)\le \frac{c_{n,k}}{a_{n+1,p}^k}k!{\displaystyle \sum _{r=0}^k}f\left(\frac{r}{a_{n,p}}\right)\frac{1}{r!}\frac{b_{n,p}^r}{(k-r)!}=\frac{k!}{a_{n+1,p}^k}{\displaystyle \sum _{r=0}^k}{c}_{n,k}^{k-r}f\left(\frac{r}{a_{n,p}}\right)\frac{1}{r!}\frac{a_{n,p}^r}{(k-r)!}.$

By the Cauchy multiplication rule for product of series,

$e^{(\beta (n+1)-\beta (n))x}{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\frac{a_{n,p}^k}{k!}f\left(\frac{k}{a_{n,p}}\right)x^k$ $={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\left\{{\displaystyle \sum _{m+r=k}}\frac{{((\beta (n+1)-\beta (n))x)}^m}{m!}\frac{a_{n,p}^r}{r!}f\left(\frac{r}{a_{n,p}}\right)x^r\right\}$

$={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\left\{{\displaystyle \sum _{r=0}^k}\frac{{(\beta (n+1)-\beta (n))}^{k-r}}{(k-r)!}\frac{a_{n,p}^r}{r!}f\left(\frac{r}{a_{n,p}}\right)\right\}{x}^k.$

Therefore

$e^{a_{n+1,p}x}\left(M_{n,p}(f,x)-M_{n+1,p}(f,x)\right)=$

$=e^{(\beta (n+1)-\beta (n))x}{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\frac{a_{n,p}^k}{k!}f\left(\frac{k}{a_{n,p}}\right)x^k-{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\frac{a_{n+1,p}^k}{k!}f\left(\frac{k}{a_{n+1,p}}\right)x^k$

$={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\left\{{\displaystyle \sum _{r=0}^k}\frac{{(\beta (n+1)-\beta (n))}^{k-r}}{(k-r)!}\frac{a_{n,p}^r}{r!}f\left(\frac{r}{a_{n,p}}\right)-\frac{a_{n+1,p}^k}{k!}f\left(\frac{k}{a_{n+1,p}}\right)\right\}{x}^k\ge 0.$

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

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

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 $(0,\mathrm{\infty })$, then for each $n\ge 1$, $S_n(f,x)/x$ 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 $f\in ℒ(I)$.

Theorem 7.

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

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

Since

$$\frac{M_{n,p}(f,x)}{x}=f(0)\frac{g_{n,p}(x)}{x}+g_{n,p}(x)\sum _{k=1}^{\mathrm{\infty }}\frac{a_{n,p}^k}{k!}f\left(\frac{k}{a_n}\right)x^{k-1}$$

and

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

${\displaystyle \sum _{k=1}^{\mathrm{\infty }}}\frac{a_{n,p}^k}{k!}f\left(\frac{k}{a_{n,p}}\right)\frac{d}{dx}\left(g_{n,p}(x)x^{k-1}\right)=$

$=g_{n,p}(x){\displaystyle \sum _{k=2}^{\mathrm{\infty }}}\frac{a_{n,p}^k(k-1)}{k!}f\left(\frac{k}{a_{n,p}}\right)x^{k-2}-g_{n,p}(x){\displaystyle \sum _{k=1}^{\mathrm{\infty }}}\frac{a_{n,p}^{k+1}}{k!}f\left(\frac{k}{a_n}\right)x^{k-1}$

$=g_{n,p}(x){\displaystyle \sum _{k=1}^{\mathrm{\infty }}}\frac{a_{n,p}^{k+1}k}{(k+1)!}f\left(\frac{k+1}{a_{n,p}}\right)x^{k-1}-g_{n,p}(x){\displaystyle \sum _{k=1}^{\mathrm{\infty }}}\frac{a_{n,p}^{k+1}}{k!}f\left(\frac{k}{a_{n,p}}\right)x^{k-1}$

$=g_{n,p}(x){\displaystyle \sum _{k=1}^{\mathrm{\infty }}}\left\{\frac{a_{n,p}}{k+1}f\left(\frac{k+1}{a_{n,p}}\right)-\frac{a_n}{k}f\left(\frac{k}{a_n}\right)\right\}\frac{a_{n,p}^k{x}^{k-1}}{(k-1)!}\le 0.$

The result is proved. $\mathrm{\square }$

4. Preservation of modulus of continuity

Definition 8.

A function $\omega :I\to {ℝ}^{+}$ 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 ${ℝ}^{+}$.

Definition 9.

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

$$\omega (s+t)\le \omega (s)+\omega (t).$$

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

(10) $$\omega (\lambda t)\le (1+\lambda )\omega (f,t).$$

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) $\tilde{\omega }$ such that

(11) $$\omega (t)\le \tilde{\omega }(t)\le 2\omega (t).$$

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

$$\mathrm{\Lambda }(\omega ,A)=\{f\in C(I):\omega (f,t)\le A\omega (t)\},$$

it is asserted in [16] that $f\in \mathrm{\Lambda }(\omega ,A)$ if and only if $S_n(f)\in \mathrm{\Lambda }(\omega ,A)$, for each each $n\in \mathbb{N}$. On the other hand, in [4] the authors considered functions $f$ such that , where $\omega (f,t)$ is the usual modulus of continuity. Of course the condition 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

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 $C_0^{\ast }=\{f\in C_0:\omega (f,t)>0\}$. In [14] Hermann proved that

$$\underset{f\in C_0^{\ast }}{sup}\frac{{\Vert S_n(f)-f\Vert }_C}{\omega (f,1/n)}=\mathrm{\infty }.$$

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

$M_{n,p}(f,x+s)-M_{n,p}(f,x)=e^{-a_{n,p}(x+s)}{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\frac{a_{n,p}^k{x}^k}{k!}{\displaystyle \sum _{i=1}^{\mathrm{\infty }}}\frac{a_{n,p}^i}{i!}{s}^i\left(f\left(\frac{k+i}{a_{n,p}}\right)-f\left(\frac{k}{a_{n,p}}\right)\right).$

Proof. Notice that

$e^{a_{n,p}(x+s)}{M}_{n,p}(f,x+s)=$

$={\displaystyle \sum _{j=0}^{\mathrm{\infty }}}f\left(\frac{j}{a_{n,p}}\right)\frac{a_{n,p}^j}{j!}{(x+s)}^j={\displaystyle \sum _{j=0}^{\mathrm{\infty }}}f\left(\frac{j}{a_{n,p}}\right)\frac{a_{n,p}^j}{j!}{\displaystyle \sum _{k=0}^j}\left(\genfrac{}{}{0pt}{}{j}{k}\right)x^k{s}^{j-k}=$

$={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}{\displaystyle \sum _{j=k}^{\mathrm{\infty }}}\left(\genfrac{}{}{0pt}{}{j}{k}\right)f\left(\frac{j}{a_{n,p}}\right)\frac{a_{n,p}^j}{j!}{x}^k{s}^{j-k}={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\frac{x^k}{k!}{\displaystyle \sum _{j=k}^{\mathrm{\infty }}}f\left(\frac{j}{a_{n,p}}\right)\frac{a_{n,p}^j}{(j-k)!}{s}^{j-k}$

$={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\frac{x^k}{k!}{\displaystyle \sum _{i=0}^{\mathrm{\infty }}}f\left(\frac{i+k}{a_{n,p}}\right)\frac{a_{n,p}^{i+k}}{i!}{s}^i.$

On the other hand,

$e^{a_{n,p}(x+s)}{M}_{n,p}(f,x)=e^{a_{n,p}s}{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}f\left(\frac{k}{a_{n,p}}\right)\frac{a_{n,p}^k}{k!}{x}^k={\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\frac{a_{n,p}^k{x}^k}{k!}\left({\displaystyle \sum _{i=0}^{\mathrm{\infty }}}\frac{a_{n,p}^i}{i!}{s}^{i}f\left(\frac{k}{a_{n,p}}\right)\right).$

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

Let $U{C}_b(I)$ the class of all bounded uniformly continuous functions $f:I\to ℝ$. For $f\in U{C}_b(I)$ and $t\ge 0$, define

(12) $$\omega (f,t)=\underset{0\le h\le t}{sup}\underset{x\ge 0}{sup}\mid f(x+h)-f(x)\mid .$$

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

Theorem 11.

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

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

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

If $f\in U{C}_b(I)$, then $f\in ℒ(I)$. From Proposition 10 one has

$\mid M_{n,p}(f,x+s)-M_{n,p}(f,x)\mid \le$

$\le e^{-a_{n,p}(x+s)}{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\frac{a_{n,p}^k{x}^k}{k!}{\displaystyle \sum _{i=0}^{\mathrm{\infty }}}\frac{a_{n,p}^i}{i!}{s}^i\mid f\left(\frac{k+i}{a_{n,p}}\right)-f\left(\frac{k}{a_{n,p}}\right)\mid$

$\le e^{-a_{n,p}(x+s)}{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\frac{a_{n,p}^k{x}^k}{k!}{\displaystyle \sum _{i=0}^{\mathrm{\infty }}}\frac{a_{n,p}^i}{i!}{s}^i\omega (f,\frac{i}{a_{n,p}})$

$=M_{n,p}(\omega (f,e_1),s)\le M_{n,p}(\tilde{\omega }(f,e_1),s).$

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

$$M_{n,p}(\tilde{\omega }(f),s)\le \tilde{\omega }(f,s)\le 2\omega (f,s).$$

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

$$\mid M_{n,p}(f,x)-M_{n,p}(f,y)\mid =\mid M_{n,p}(f,y+t)-M_{n,p}(f,y)\mid \le \epsilon .$$

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

For $f\in U{C}_b(I)$, , and $t>0$ define

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

$$K^{\alpha }(f)=\underset{0\le t}{sup}{\theta }_{\alpha }(f,t).$$

For , let us set ${\mathrm{Lip}}^{\alpha }(I)$ for the family of all $f\in U{C}_b(I)$ such that

For , we also we consider the subspace

(13) $${\mathrm{lip}}^{\alpha }(I)=\{f\in {\mathrm{Lip}}^{\alpha }(I):\underset{t\to 0}{lim}{\theta }_{\alpha }(f,t)=0\}.$$

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 ${\mathrm{lip}}^{\alpha }(I)$ by the operators $M_{n,p}$.

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

Theorem 12.

(i) If and $f\in {\mathrm{Lip}}^{\alpha }(I)$, then $M_{n,p}(f)\in {\mathrm{Lip}}^{\alpha }(I)$, and

(14) $$K^{\alpha }(M_{n,p}(f))\le K^{\alpha }(f),$$

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

(ii) If , $f\in ℒ(I)$, $M_{n,p}(f)\in {\mathrm{Lip}}^{\alpha }(I)$, for each $n\in \mathbb{N}$, and

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

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 $M_{n,p}(g,x)\le g(x)$.

For any $k\in {\mathbb{N}}_0$ and $i\in \mathbb{N}$,

$$\mid f\left(\frac{k+i}{a_{n,p}}\right)-f\left(\frac{k}{a_{n,p}}\right)\mid \le \omega (f,\frac{i}{a_{n,p}})\le {\theta }_{\alpha }(f,\frac{i}{a_{n,p}}){\left(\frac{i}{a_{n,p}}\right)}^{\alpha }\le K^{\alpha }(f)g\left(\frac{i}{a_{n,p}}\right).$$

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

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

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

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

$\mid f(x+h)-f(x)\mid$ $\le \mid f(x+h)-M_{n,p}(f,x+h)\mid$

$+\mid M_{n,p}(f,x+h)-M_{n,p}(f,x)\mid +\mid M_{n,p}(f,x)-f(x)\mid$

$\le \mid f(x+h)-M_{n,p}(f,x+h)\mid +K{h}^{\alpha }+\mid M_{n,p}(f,x)-f(x)\mid .$

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

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

Proposition 13.

For and each $f\in {\mathrm{lip}}^{\alpha }(I)$,

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

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

On the other hand, given $\epsilon >0$, for any , there exists $x_h\in I$ such that

$$\omega (f,h)\le \epsilon h^{\alpha }+\mid f(x_h+h)-f(x_h)\mid .$$

Therefore

Proposition 14.

If , for each $f\in {\mathrm{lip}}^{\alpha }(I)$, 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 and fix any $s$ and $h$ such that and .

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

We still have to consider the case . Since and , one has

$\frac{\mid f(x+h)-f(x)\mid }{h^{\alpha }}$ $\le \frac{\mid f(x+h-t+t)-f(x+h-t)\mid }{t^{\alpha }}+\frac{\mid f(x+h-t)-f(x)\mid }{{(h-t)}^{\alpha }}$

$\le {\theta }_{\alpha }(f,t)+{\theta }_{\alpha }(f,h-t)\le {\theta }_{\alpha }(f,t)+{\theta }_{\alpha }(f,v),$

because ${\theta }_{\alpha }(f,t)$ increases and $h-t\le s-t\le v$. Therefore

$${\theta }_{\alpha }(f,t+v)\le {\theta }_{\alpha }(f,t)+{\theta }_{\alpha }(f,v).$$

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

Theorem 15.

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

$${\theta }_{\alpha }(M_{n,p}(f),t)\le 2{\theta }_{\alpha }(f,t).$$

Proof. If , taking into account Theorem 11, one has

$$\mid M_{n,p}(f,x+s)-M_{n,p}(f,x)\mid \le 2\omega (f,s)=2\frac{\omega (f,s)}{s^{\alpha }}{s}^{\alpha }\le 2{\theta }_{\alpha }(f,s)s^{\alpha }\le 2{\theta }_{\alpha }(f,t)s^{\alpha }.$$

This is sufficient to prove the result. $\mathrm{\square }$

For each $r\ge 0$, $f\in C_r(I)$ (see (4)), and $t\ge 0$, define

$${\mathrm{\Omega }}_r(f,t)=\underset{0\le s\le t}{sup}\underset{x\ge 0}{sup}\frac{\mid f(x+s)-f(x)\mid }{{(1+x+s)}^r}.$$

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

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

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

$$\frac{1}{2^r}{\mathrm{\Omega }}_r(f,t)\le {\tilde{\mathrm{\Omega }}}_r(f,t)\le {\mathrm{\Omega }}_r(f,t).$$

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

$${\tilde{\mathrm{\Omega }}}_r(S_n(f),t)\le C{\tilde{\mathrm{\Omega }}}_r(f,t).$$

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

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

$$\mathrm{\Omega }(f,t)=\underset{0\le s\le t}{sup}\underset{x\in I}{sup}\frac{\mid f(x+s)-f(x)\mid }{{(1+s)}^2{(1+x)}^2}.$$

For $0\le t\le 1$, $\mathrm{\Omega }(f,t)$ and ${\mathrm{\Omega }}_2(f,t)$ are equivalent. In fact, suppose that $s\le 1$. First one has

$$(1+s^2)(1+x^2)=1+s^2+x^2+s^2{x}^2\le 2(1+s^2+x^2)\le 2{(1+x+s)}^2.$$

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

$${(1+s+x)}^2=1+2x+2s+x^2+2xs+s^2\le 7(1+s^2+x^2)\le 7(1+s^2)(1+x^2).$$

and, if $x>1$,

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

Therefore

$$\frac{1}{2}{\mathrm{\Omega }}_2(f,t)\le \mathrm{\Omega }(f,t)\le 7{\mathrm{\Omega }}_2(f,t).$$

Proposition 16.

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

Proof. It is clear that ${\mathrm{\Omega }}_r(f,0)=0$ and ${\mathrm{\Omega }}_r(f,t)$ 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 ${\mathrm{\Omega }}_r(f,t)$ is a subadditive. In order to prove continuity, it is sufficient to verify continuity a zero, but if follows from the condition ${lim}_{x\to \mathrm{\infty }}f(x)=0$.

(b) Assume $r>0$. Denote $A={lim}_{x\to \mathrm{\infty }}f(x)/{(1+x)}^r$. Given $\epsilon >0$, there exists $x_0$ such that

If $t>0$ and , then

$\underset{x\ge 0}{sup}\frac{\mid f(x+s)-f(x)\mid }{{(1+x+s)}^r}$ $\le \underset{0\le x\le x_0}{sup}\frac{\mid f(x+s)-f(x)\mid }{{(1+x+s)}^r}+\underset{x\ge x_0}{sup}\frac{\mid f(x+s)-f(x)\mid }{{(1+x+s)}^r}$

$\le \underset{0\le x\le x_0}{sup}\mid f(x+s)-f(x)\mid +\underset{x\ge x_0}{sup}\frac{\mid f(x+s)-A\mid }{{(1+x+s)}^r}+\underset{x\ge x_0}{sup}\frac{\mid f(x)-A\mid }{{(1+x)}^r}$

$\le {\omega }_1{(f,t)}_{[0,x_0+1]}+\epsilon ,$

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 ${lim}_{t\to 0}{\mathrm{\Omega }}_r(f,t)\to 0=0$.

(c) Let us verify that ${\mathrm{\Omega }}_r(f,t)$ is subadditive: ${\mathrm{\Omega }}_r(f,v+t)\le {\mathrm{\Omega }}_r(f,v)+{\mathrm{\Omega }}_r(f,t)$. Without losing generality we assume that .

Fijemos $x\ge 0$ and .

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

Let us consider the case . Since , one has . Therefore

$\frac{\mid f(x+s)-f(x)\mid }{{(1+x+s)}^r}$ $\le \frac{\mid f(x+s-t+t)-f(x+s-t)\mid }{{(1+(x+s-t)+t)}^r}+\frac{\mid f(x+s-t)-f(x)\mid }{{(1+x+s-t)}^r}$

$\le {\mathrm{\Omega }}_r(f,t)+{\mathrm{\Omega }}_r(f,s-t)\le {\mathrm{\Omega }}_r(f,t)+{\mathrm{\Omega }}_r(f,v).$

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

Theorem 17.

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

$${\mathrm{\Omega }}_r(M_{n,p}(f),t)\le C{\mathrm{\Omega }}_r(f,t).$$

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

$\left|f\left(\frac{k+i}{a_{n,p}}\right)-f\left(\frac{k}{a_{n,p}}\right)\right|$ $\le {\mathrm{\Omega }}_r\left(\frac{i}{a_{n,p}}\right){\left(1+\frac{k}{a_{n,p}}+\frac{i}{a_{n,p}}\right)}^r$

$\le {\mathrm{\Omega }}_r(f,s)\left(1+\frac{i}{a_{n,p}s}\right){\left(1+\frac{k}{a_{n,p}}+\frac{i}{a_{n,p}}\right)}^r,$

because ${\mathrm{\Omega }}_r(f,s)$ is a subadditive modulus.

Therefore (see Proposition 10)

$\mid M_{n,p}(f,x+s)-M_{n,p}(f,x)\mid =$

$=\left|e^{-a_{n,p}(x+s)}{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\frac{a_{n,p}^k{x}^k}{k!}{\displaystyle \sum _{i=0}^{\mathrm{\infty }}}\frac{a_{n,p}^i}{i!}{s}^i\left(f\left(\frac{k+i}{a_{n,p}}\right)-f\left(\frac{k}{a_{n,p}}\right)\right)\right|$

$\le \frac{{\mathrm{\Omega }}_r(f,s)}{e^{a_{n,p}(x+s)}}{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\frac{a_{n,p}^k{x}^k}{k!}{\displaystyle \sum _{i=0}^{\mathrm{\infty }}}\frac{a_{n,p}^i}{i!}{s}^i\left(1+\frac{i}{a_{n,p}s}\right){\left(1+\frac{k}{a_{n,p}}+\frac{i}{a_{n,p}}\right)}^r.$

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

$e^{-a_{n,p}s}{\displaystyle \sum _{i=0}^{\mathrm{\infty }}}\frac{a_{n,p}^i}{i!}{s}^i{\left(1+\frac{k}{a_{n,p}}+\frac{i}{a_{n,p}}\right)}^r=$

$=M_{n,p}({\left(1+\frac{k}{a_{n,p}}+e_1\right)}^r,s)\le C(r){\left(2+\frac{k}{a_{n,p}}+s\right)}^r\le 2^{r}C(r){\left(1+\frac{k}{a_{n,p}}+s\right)}^r.$

On the other hand

$\frac{e^{-a_{n,p}s}}{s}{\displaystyle \sum _{i=1}^{\mathrm{\infty }}}\frac{a_{n,p}^i}{i!}{s}^i\frac{i}{a_{n,p}}{\left(1+\frac{k}{a_{n,p}}+\frac{i}{a_{n,p}}\right)}^r=$

$=e^{-a_{n,p}s}{\displaystyle \sum _{i=1}^{\mathrm{\infty }}}\frac{a_{n,p}^{i-1}}{(i-1)!}{s}^{i-1}{\left(1+\frac{k}{a_{n,p}}+\frac{i}{a_{n,p}}\right)}^r$

$=M_{n,p}({\left(1+\frac{1}{a_{n,p}}+\frac{k}{a_{n,p}}+e_1\right)}^r,s)\le M_{n,p}({\left(2+\frac{k}{a_{n,p}}+e_1\right)}^r,s)$

$\le C(r){\left(3+\frac{k}{a_{n,p}}+s\right)}^r\le 3^{r}C(r){\left(1+\frac{k}{a_{n,p}}+s\right)}^r.$

From the estimates given above one has

$\mid M_{n,p}(f,x+s)-M_{n,p}(f,x)\mid$ $\le 3^{r}C(r)\frac{{\mathrm{\Omega }}_r(f,s)}{e^{a_{n,p}x}}{\displaystyle \sum _{k=0}^{\mathrm{\infty }}}\frac{a_{n,p}^k{x}^k}{k!}{\left(1+\frac{k}{a_{n,p}}+s\right)}^r$

$\le 6^{r}C(r){\mathrm{\Omega }}_r(f,s){(1+x+s)}^r,$

where we use again Proposition 2. $\mathrm{\square }$