$(p,q)$-Generalization of Szász–Mirakjan operators and their approximation properties

Abstract

We introduce a new modification of $(p,q)$-analogue of Szász–Mirakjan operators. Firstly, we give a recurrence relation for the moments of $(p,q)$-analogue of Szász–Mirakjan operators and present some explicit formulae for the moments and central moments up to order 4. Next, we obtain quantitative estimates for the convergence in the polynomial weighted spaces. In addition, we give the Voronovskaya theorem for the new $(p,q)$-Szász–Mirakjan operators.

1 Introduction

In the last two decades, quantum calculus plays a significant role in the approximation of functions by a positive linear operator. In 1987, Lupaş [1] introduced the Bernstein (rational) polynomials based on the q-integers. In 1996, Phillips [2] introduced another generalization of Bernstein polynomials based on q-integers. In [3–23], in the case of $0< q<1$, many operators have been introduced and examined. Among the most important operators there are q-Szász operators. In [18–21] the authors constructed and studied different q-generalizations of Szász–Mirakjan operators in the case $0< q<1$. In 2012, Mahmudov [24] introduced the q-Szász operator in the case $q>1$ and studied quantitative estimates of convergence in polynomial weighted spaces and gave the Voronovskaya theorem.

In recent years, the rapid rise of $(p,q)$-calculus has led to the discovery of new generalizations of Bernstein polynomials containing $(p,q)$-integers. In 2015, Mursaleen [25] introduced $(p,q)$-Bernstein operators and studied approximation properties based on a Korovkin-type approximation theorem of $(p,q)$-Bernstein operators. Also, In 2017, Khan and Lobiyal [26] constructed a $(p,q)$-analogue of Lupaş-Bernstein functions. In [27–39], the authors constructed many operators by using $(p,q)$-integers and studied their approximation properties. Acar [40] introduced $(p,q)$-Szász–Mirakjan operators. In addition, Acar gave a recurrence relation for the moments of these operators. In the same year, H. Sharma and C. Gupta [41] introduced the generalization of the $(p,q)$-Szász–Mirakjan Kantorovich operators and examined their approximation properties. In 2017, Mursaleen, AAH Al-Abied, and Alotaibi [42] constructed new Szász–Mirakjan operators based on $(p,q)$-calculus and studied weighted approximation and a Voronovskaya-type theorem. Also, $(p,q)$-analogues of Szász–Mirakjan–Baskakov operators [18] and Stancu-type Szász–Mirakjan–Baskakov operators [43] were defined, and their approximation properties were investigated. Acar, Agrawal, and Kumar [44] introduced a sequence of $(p,q)$-Szász–Mirakjan operators, and their weighted approximation properties were investigated.

2 Construction of $K_{l,p,q}$ and moment estimations

We give some basic notations and definitions of the $(p,q)$-calculus.

The $(p,q)$-integer and $( p,q )$-factorial are defined by

$$\begin{aligned}& [ l ] _{p,q} :=\left \{ \textstyle\begin{array}{l@{\quad}l}\frac{p^{l}-q^{l}}{p-q}& \text{if }0< q< p\leq1,\\ l &\text{if }p=q=1 \end{array}\displaystyle \right . \quad\text{for }l\in\mathbb{N}, \text{and } [ 0 ] _{p,q}=0, \\& [ l ] _{p,q}! := [ 1 ] _{p,q} [ 2 ] _{p,q}\cdots [ l ] _{p,q} \quad\text{for }l\in\mathbb{N}, \text{and } [ 0 ] _{p,q}!=1. \end{aligned}$$

For integers $0\leq k\leq l$, the $(p,q)$-binomial is defined by

$$\left [ \textstyle\begin{array}{c}l\\ k \end{array}\displaystyle \right ] _{p,q}:= \frac{ [ l ] _{p,q}!}{ [ k ] _{p,q}! [ l-k ] _{p,q}!}. $$

The $( p,q )$-derivative $D_{p,q}g$ of a function $g(z)$ is defined by

$$( D_{p,q}g ) ( z ) :=\frac{g ( pz ) -g ( qz ) }{ ( p-q ) z},\quad z\neq0,\qquad ( D_{p,q}g ) ( 0 ) =g^{{\prime}}(0) $$

The product and quotient formulae for the $(p,q)$-derivative are as follows:

$$\begin{aligned}& D_{p,q} \bigl( f ( z ) g ( z ) \bigr) =g(pz)D_{p,q} \bigl( f ( z ) \bigr) +f(qz)D_{p,q} \bigl( g ( z ) \bigr), \end{aligned}$$

(1)

$$\begin{aligned}& D_{p,q} \biggl( \frac{f ( z ) }{g ( z ) } \biggr) = \frac{g(pz)D_{p,q} ( f ( z ) ) -f(pz)D_{p,q} ( g ( z ) ) }{g(pz)g(qz)} . \end{aligned}$$

(2)

It is known that

$$ D_{p,q}z^{l}= [ l ] _{p,q}z^{l-1}. $$

(3)

The $(p,q)$-analogues of an exponential function, denoted by $e_{p,q} ( z )$ and $E_{p,q} ( z )$, are defined by

$$ e_{p,q} ( z ) :=\sum_{k=0}^{\infty} \frac{p^{\frac {k(k-1)}{2}}z^{k}}{ [ k ] _{p,q}!},\qquad E_{p,q} ( z ) =\sum _{k=0}^{\infty}\frac{q^{\frac{k(k-1)}{2}}z^{k}}{ [ k ] _{p,q}!}, $$

and the $(p,q)$-derivatives of $e_{p,q} ( az )$ and $E_{p,q} ( z )$ are

$$ D_{p,q}e_{p,q} ( az ) =ae_{p,q} ( apz ) ,\qquad D_{p,q}E_{p,q} ( az ) =ae_{p,q} ( aqz ). $$

Further, the $(p,q)$-power is defined by

$$ ( z-y ) _{p,q}^{l}= ( z-y ) ( pz-qy ) \bigl( p^{2}z-q^{2}y \bigr) \cdots \bigl( p^{l-1}z-q^{l-1}y \bigr) . $$

(4)

For any integer l,

$$ D_{p,q} ( z-y ) _{p,q}^{l}= [ l ] _{p,q} ( pz-y ) _{p,q}^{l-1}, $$

(5)

and $D_{p,q} ( z-y ) _{p,q}^{0}=0$.

The formula of the kth $(p,q)$-derivative of the polynomial $( z-y ) _{p,q}^{l}$is

$$ D_{p,q}^{k} ( z-y ) _{p,q}^{l}=p^{\binom{k}{2}} \frac{ [ l ] _{p,q}!}{ [ l-k ] _{p,q}!} \bigl( p^{k}z-y \bigr) _{p,q}^{l-k}, $$

(6)

where $l\in \mathbb{Z} _{+}$ and $0\leq k\leq l$.

The $(p,q)$-analogue of the Taylor formulas for any function $g(z)$ is defined by

$$ g(z)=\sum_{k=0}^{l} ( -1 ) ^{k}q^{-\binom{k}{2}}\frac{ ( D_{p,q}^{k}g ) ( zq^{-k} ) (z-t)_{p,q}^{k}}{ [ k ] _{p,q}!} . $$

(7)

Let $C_{\beta}$ denote the set of all real-valued continuous functions g on $[ 0,\infty )$ such that $w_{\beta}g$ is bounded and uniformly continuous on $[ 0,\infty )$ endowed with the norm

$$\Vert g \Vert _{m}:=\sup_{z\in [ 0,\infty ) }w_{\beta } \bigl\vert g ( z ) \bigr\vert , $$

where $w_{0}(z)=1$, and $w_{\beta}(z)=\frac{1}{1+z^{\beta}}$ for $\beta \in \mathbb{N}$.

The corresponding Lipschitz class is given for $0<\alpha\leq2$ by

$$\begin{gathered} \Delta_{j}^{2}g(z):=g(z+2j)-2g(z+j)+g(z), \\ w_{\beta}^{2}(g;\gamma):=\sup_{0< j\leq\gamma} \bigl\Vert \Delta _{j}^{2}g \bigr\Vert ,\qquad\text{Lip}_{\beta}^{2}\alpha:= \bigl\{ g\in C_{\beta}:w_{\beta}^{2}(g;\gamma)=0 \bigl( \gamma^{\alpha} \bigr) ,\gamma\rightarrow0^{+} \bigr\} .\end{gathered} $$

Now we introduce the $( p,q )$-Szász–Mirakjan operator.

Definition 1

Let $0< q< p\leq1$ and $l\in\mathbb{N}$. For $g: [ 0,\infty ) \rightarrow R$, we define the $( p,q )$-Szász–Mirakjan operator as

$$ K_{l,p,q} ( g;z ) :=\sum_{k=0}^{\infty}g \biggl( \frac {p^{l-k} [ k ] _{p,q}}{ [ l ] _{p,q}} \biggr) \frac{p^{k(k-l)}}{q^{k ( k-1 ) /2}}\frac{ [ l ] _{p,q}^{k} z^{k}}{ [ k ] _{p,q}!}e_{p,q} \bigl( - [ l ] _{p,q} p^{k-l+1}q^{-k}z \bigr) . $$

(8)

It is clear that the operator $K_{l,p,q}$ is linear and positive. It is known that the moments $K_{l,p,q} ( t^{m};z )$ play a fundamental role in the approximation theory of positive operators.

Lemma 2

Let$0< q< p\leq1$and$m\in \mathbb{N}$. We have the following recurrence formula:

$$ K_{l,p,q} \bigl( t^{m+1};z \bigr) =\sum _{j=0}^{m}\left ( \textstyle\begin{array} [c]{c}m\\ j \end{array}\displaystyle \right ) \frac{zq^{j}p^{m(l-1)-lj}}{ [ l ] _{p,q}^{m-j}}K_{l,p,q} \bigl( t^{j};pq^{-1}z \bigr) . $$

(9)

Proof

According to the definition of $K_{l,p,q}$ (8), we have

$$\begin{gathered} K_{l,p,q} \bigl( t^{m+1};z \bigr) \\ \quad =\sum_{k=0}^{\infty}\frac{p^{ ( l-k ) (m+1)} [ k ] _{p,q}^{m+1}}{ [ l ] _{p,q}^{m+1}} \frac{p^{k(k-l)}}{q^{k ( k-1 ) /2}}\frac{ [ l ] _{p,q}^{k}z^{k}}{ [ k ] _{p,q}!}e_{p,q} \bigl( - [ l ] _{p,q} p^{k-l+1}q^{-k}z \bigr) \\ \quad =\sum_{k=1}^{\infty}\frac{p^{ ( l-k ) (m+1)} [ k ] _{p,q}^{m}}{ [ l ] _{p,q}^{m}} \frac{p^{k(k-l)}}{q^{k ( k-1 ) /2}}\frac{ [ l ] _{p,q}^{k-1} z^{k}}{ [ k-1 ] _{p,q}!}e_{p,q} \bigl( - [ l ] _{p,q} p^{k-l+1}q^{-k}z \bigr) .\end{gathered} $$

Next, we use the identity $q [ k ] _{p,q}+p^{k}= [ k+1 ] _{p,q}$ to get the desired formula

$$\begin{gathered} K_{l,p,q} \bigl( t^{m+1};z \bigr) \\ \quad =\sum_{k=0}^{\infty}\frac{p^{ ( l-k-1 ) (m+1)} ( q [ k ] _{p,q}+p^{k} ) ^{m}}{ [ l ] _{p,q}^{m}}\frac{p^{(k+1)(k-l+1)}}{q^{k ( k+1 ) /2}}\frac{ [ l ] ^{k}z^{k+1}}{ [ k ] _{p,q}!}e_{p,q} \bigl( - [ l ] _{p,q} p^{k-l+2}q^{-k-1}z \bigr) \\ \quad =\sum_{k=0}^{\infty}\frac{1}{ [ l ] _{p,q}^{m}} \sum_{j=0}^{m}\left ( \textstyle\begin{array} [c]{c}m\\ j \end{array}\displaystyle \right ) q^{j} [ k ] _{p,q}^{j} p^{k ( m-j ) }p^{(l-k-1)(m+1)} \frac{p^{(k+1)(k-l+1)}}{q^{k ( k+1 ) /2}}\frac{ [ l ] _{p,q}^{k}z^{k+1}}{ [ k ] _{p,q}!} \\ \qquad{} \times e_{p,q} \bigl( - [ l ] _{p,q} p^{k-l+2}q^{-k-1}z \bigr) \\ \quad =\sum_{j=0}^{m}\left ( \textstyle\begin{array} [c]{c}m\\ j \end{array}\displaystyle \right ) \frac{zq^{j}p^{m(l-1)-lj}}{ [ l ] _{p,q}^{m-j}}\sum _{k=0}^{\infty}\frac{p^{ ( l-k ) j} [ k ] _{p,q}^{j}}{ [ l ] _{p,q}^{j}} \frac{p^{k(k-l)}}{q^{k ( k-1 ) /2} }\frac{ [ l ] _{p,q}^{k}p^{k}z^{k}}{ [ k ] _{p,q}!q^{k}}e_{q} \bigl( - [ l ] _{p,q}p^{k-l+2}q^{-k-1}z \bigr) \\ \quad =\sum_{j=0}^{m}\left ( \textstyle\begin{array} [c]{c}m\\ j \end{array}\displaystyle \right ) \frac{zq^{j}p^{m(l-1)-lj}}{ [ l ] _{p,q}^{m-j}}K_{l,p,q} \bigl( t^{j};pq^{-1}z \bigr) .\end{gathered} $$

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Lemma 3

Let$0< q< p\leq1$, $z\in [ 0,\infty )$, $l\in l$, and$k\geq0$. We have the following identities related to the$( p,q )$-derivative:

$$\begin{aligned}& zD_{p,q}s_{lk} ( p,q;z ) = [ l ] _{p,q} \biggl( \frac{p^{-k} [ k ] _{p,q}}{ [ l ] _{p,q}}-zp^{-(l-1)} \biggr) s_{lk} ( p,q;pz ) , \\& K_{l,p,q} \bigl( t^{m+1};z \bigr) =\frac{z}{ [ l ] p^{-(l-1)}}D_{p,q}K_{l,p,q} \biggl( t^{m};\frac{z}{p} \biggr) +zK_{l,p,q} \bigl( t^{m};z \bigr) , \end{aligned}$$

(10)

where$s_{lk} ( p,q;z ) =\frac{p^{k(k-l)}}{q^{k ( k-1 ) /2}}\frac{ [ l ] _{p,q}^{k} z^{k}}{ [ k ] _{p,q}!}e_{p,q} ( - [ l ] _{p,q} p^{k-l+1}q^{-k}z )$.

Proof

We take the $(p,q)$-derivative of $s_{lk} ( p,q;z )$:

$$\begin{aligned} D_{p,q}s_{lk} ( p,q;z ) & =\frac{p^{k ( k-l ) }}{q^{k ( k-1 ) /2} [ k ] _{p,q}!}D_{p,q} \bigl( z^{k}e_{p,q} \bigl( - [ l ] _{p,q}p^{k-l+1}q^{-k}z \bigr) \bigr) \\ & =\frac{p^{k ( k-l ) }}{q^{k ( k-1 ) /2} [ k ] _{p,q}!} \bigl( [ k ] _{p,q}z^{k-1}e_{p,q} \bigl( - [ l ] _{p,q}p^{k-l+1}q^{-k} ( pz ) \bigr) \bigr) \\ & =\frac{p^{k ( k-l ) }}{q^{k ( k-1 ) /2} [ k ] _{p,q}!} \bigl( (qz)^{k} [ l ] _{p,q} p^{k-l+1}q^{-k} e_{p,q} \bigl( - [ l ] _{p,q}p^{k-l+1}q^{-k} ( pz ) \bigr) \bigr) . \end{aligned}$$

Then

$$\begin{aligned} zD_{p,q}s_{lk} ( p,q;z ) & =p^{-k} [ k ] _{p,q}\frac{ p^{k ( k-l ) }}{q^{k ( k-1 ) /2}}\frac{ ( pz ) ^{k}}{ [ k ] _{p,q}!}e_{p,q} \bigl( - [ l ] _{p,q}p^{k-l+1}q^{-k} ( pz ) \bigr) \\ &\quad -z [ l ] _{p,q}p^{-(l-1)}\frac{ p^{k ( k-l ) }}{q^{k ( k-1 ) /2}} \frac{ ( pz ) ^{k}}{ [ k ] _{p,q}!}e_{p,q} \bigl( - [ l ] _{p,q}p^{k-l+1}q^{-k} ( pz ) \bigr) \\ & =p^{-k} [ k ] _{p,q}s_{lk} ( p,q;pz ) -z [ l ] _{p,q}p^{-(l-1)}s_{lk} ( p,q;pz ) \\ & = [ l ] _{p,q} \biggl( \frac{p^{-k} [ k ] _{p,q}}{ [ l ] _{p,q}}-z p^{-(l-1)} \biggr) s_{lk} ( p,q;pz ) . \end{aligned}$$

Using the obtained formula and the definition of the operator $K_{l,p,q}$, we get the second desired formula:

$$\begin{aligned} zD_{p,q}K_{l,p,q} \bigl( t^{m};z \bigr) & = [ l ] \sum_{k=0}^{\infty} \biggl( \frac{p^{ ( l-k ) } [ k ] _{p,q} }{ [ l ] _{p,q}} \biggr) ^{m}p^{-(l-1)} \biggl( \frac{p^{-k} [ k ] _{p,q}}{p^{-(l-1)} [ l ] _{p,q}}-z \biggr) s_{lk} ( p,q;pz ) \\ & = [ l ] \sum_{k=0}^{\infty} \biggl( \frac{ [ k ] _{p,q}}{ [ l ] _{p,q}} \biggr) ^{m+1}p^{(l-k) ( m+1 ) } \frac{p^{-(l-1)}}{p}s_{lk} ( p,q;pz ) \\ & \quad- [ l ] _{p,q} p^{-(l-1)} z\sum _{k=0}^{\infty} \biggl( \frac{ [ k ] _{p,q}}{ [ l ] _{p,q}} \biggr) ^{m}p^{(l-k)m}s_{lk} ( p,q;pz ) \\ & =p^{-l} [ l ] _{p,q}K_{l,p,q} \bigl( t^{m+1};pz \bigr) - [ l ] _{p,q} p^{-(l-1)}zK_{l,p,q} \bigl( t^{m};pz \bigr) . \end{aligned}$$

 □

Lemma 4

For$0< q< p\leq1$and$l\in \mathbb{N}$, we have

$$\begin{aligned}& K_{l,p,q} ( 1;z ) =1,\quad\quad K_{l,p,q} ( t;z ) =z,\qquad K_{l,p,q} \bigl( t^{2};z \bigr) =z^{2}+ \frac{1}{p^{-(l-1)} [ l ] }z, \\& K_{l,p,q} \bigl( t^{3};z \bigr) =z^{3}+ \frac{2p+q}{p^{-(l-1)}p [ l ] _{p,q}}z^{2}+\frac{1}{p^{-2(l-1)} [ l ] _{p,q}^{2}}z, \\& \begin{aligned}K_{l,p,q} \bigl( t^{4};z \bigr) & =z^{4}+ \biggl( \frac{3p^{2}+2qp+q^{2}}{p^{2}} \biggr) \frac{z^{3}}{p^{-(l-1)} [ l ] _{p,q}} \\ &\quad + \biggl( \frac{3p^{2}+3pq+q^{2}}{p^{2}} \biggr) \frac{z^{2}}{p^{-2(l-1)} [ l ] _{p,q}^{2}}+ \frac{1}{p^{-3(l-1)} [ l ] _{p,q}^{3}}z.\end{aligned} \end{aligned}$$

Proof

From the $(p,q)$-Taylor theorem [45] we have

$$\psi_{l} ( t ) =\sum_{k=0}^{\infty} \frac{ ( -1 ) ^{k}q^{-\binom{k}{2}}}{ [ k ] _{p,q}!} \bigl( D_{p,q}^{k}\psi _{l} \bigr) \bigl( zq^{-k} \bigr) ( z\ominus t ) _{p,q}^{k}.$$

For $t=0$, having in mind the equalities

$$\begin{aligned} ( z ) _{p,q}^{k} & =z^{k}p^{k ( k-1 ) /2}, D_{p,q}^{k}e_{p,q} \bigl( - [ l ] _{p,q} p^{-(l-1)}z \bigr) \\ & = ( -1 ) ^{k}p^{k ( k-1 ) /2}p^{-(l-1)k} [ l ] _{p,q}^{k}e_{q} \bigl( - [ l ] _{p,q} p^{-(l-1)}p^{k}z \bigr) \end{aligned}$$

for $\psi_{l} ( z ) =e_{p,q} ( - [ l ] _{p,q} p^{-(l-1)}z )$, we get the formula

$$\begin{aligned} 1 & =\psi_{l} ( 0 ) =\sum_{k=0}^{\infty} \frac{ ( -1 ) ^{k} q^{-\binom{k}{2}} ( z ) _{p,q}^{k}}{ [ k ] _{p,q}!} \bigl( D_{p,q}^{k} \varphi_{l} \bigr) \bigl( zq^{-k} \bigr) \\ & =\sum_{k=0}^{\infty}\frac{ ( -1 ) ^{k} q^{-\binom{k}{2} }z^{k}p^{k ( k-l ) }, }{ [ k ] _{q}!q^{k ( k-1 ) /2}} ( -1 ) ^{k} [ l ] _{p,q}^{k}e_{p,q} \bigl( - [ l ] _{p,q} p^{-(l-1-k)}q^{-k}z \bigr) \\ & =\sum_{k=0}^{\infty}\frac{ [ l ] _{p,q}^{k} z^{k}}{ [ k ] _{p,q}!p^{-k ( k-1 ) /2}} \biggl( \frac{p}{q} \biggr) ^{k ( k-1 ) /2}p^{-(l-1)k}e_{p,q} \biggl( - [ l ] _{p,q} p^{-(l-1)} \biggl( \frac{p}{q} \biggr) ^{k}z \biggr) , \end{aligned}$$

that is, $K_{l,p,q} ( 1;z ) =1$.

For $i=2,3,4$, recurrence formula (10) gives us the following results:

$$\begin{aligned}& \begin{aligned} K_{l,p,q} \bigl( t^{2};z \bigr) & =\frac{z}{p^{-(l-1)} [ l ] _{p,q}} \biggl\{ D_{p,q}K_{l,p,q} \biggl( t;\frac{z}{p} \biggr) +\frac{ [ l ] _{p,q}}{p^{(l-1)}}K_{l,p,q} ( t;z ) \biggr\} \\ & =\frac{z}{p^{-(l-1)} [ l ] _{p,q}} \biggl\{ 1+\frac{ [ l ] _{p,q}}{p^{(l-1)}}z \biggr\} = \frac{z}{p^{-(l-1)} [ l ] _{p,q}}+z^{2}, \end{aligned} \\& \begin{aligned} K_{l,p,q} \bigl( t^{3};z \bigr) & =\frac{z}{p^{-(l-1)} [ l ] _{p,q}}D_{p,q}K_{l,p,q} \biggl( t^{2};\frac{z}{p} \biggr) +zK_{l,p,q} \bigl( t^{2};z \bigr) \\ & =\frac{z}{p^{-(l-1)} [ l ] _{p,q}}\frac{1}{ [ l ] _{p,q}p^{-(l-1)}}+ [ 2 ] _{p,q} \frac{z}{p}+z \biggl\{ z^{2}+\frac {z}{p^{-(l-1)} [ l ] _{p,q}} \biggr\} \\ & =z^{3}+\frac{ ( 2p+q ) }{p^{-(l-2)} [ l ] _{p,q}}z^{2}+ \frac{z}{ [ l ] _{p,q}^{2}p^{-2(l-1)}}, \end{aligned} \\& \begin{aligned} K_{l,p,q} \bigl( t^{4};z \bigr) & =\frac{z}{p^{-(l-1)} [ l ] _{p,q}}D_{p,q}K_{l,p,q} \biggl( t^{3};\frac{z}{p} \biggr) +zK_{l,p,q} \bigl( t^{3};z \bigr) \\ & =\frac{z}{p^{-(l-1)} [ l ] _{p,q}} \biggl\{ \frac{ [ 3 ] _{p,q}z^{2}}{p^{2}}+\frac{ [ 2 ] _{p,q} ( 2p+q ) }{pp^{-(l-2)} [ l ] _{p,q}}z+ \frac{1}{ [ l ] _{p,q}^{2}p^{-2(l-1)}} \biggr\} \\ &\quad +z \biggl\{ z^{3}+\frac{ ( 2p+q ) }{p^{-(l-2)} [ l ] _{p,q}}z^{2}+ \frac{z}{ [ l ] _{p,q}^{2}p^{-2(l-1)}} \biggr\} \\ & =z^{4}+ \biggl( +\frac{3p^{2}+2qp+q^{2}}{p^{2}} \biggr) \frac{z^{3}}{p^{-(l-1)} [ l ] _{p,q}} \\ & \quad+ \biggl( \frac{3p^{2}+3pq+q^{2}}{p^{2}} \biggr) \frac{z^{2}}{p^{-2(l-1)} [ l ] ^{2}}+ \frac{1}{p^{-3(l-1)} [ l ] ^{3}}z. \end{aligned} \end{aligned}$$

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Lemma 5

For every$z\in [ 0,\infty )$, we have

$$\begin{aligned}& K_{l,p,q} \bigl( ( t-z ) ;z \bigr) =0, \end{aligned}$$

(11)

$$\begin{aligned}& K_{l,p,q} \bigl( ( t-z ) ^{2};z \bigr) = \frac{z}{p^{-(l-1)} [ l ] _{p,q}}, \\& K_{l,p,q} \bigl( ( t-z ) ^{3};z \bigr) = \frac{1}{ [ l ] _{p,q}^{2} p^{-2(l-1)}}z+\frac{z^{2}}{p^{-(l-1)} [ l ] _{p,q}} \biggl( \frac{q}{p}-1 \biggr) , \\& K_{l,p,q} \bigl( ( t-z ) ^{4};z \bigr) = \frac{1}{ [ l ] _{p,q}^{3} p^{-3(l-1)}}z + \biggl( \frac{3pq+q^{2}-p^{2}}{p^{2}} \biggr) \frac{z^{2}}{ [ l ] _{p,q}^{2} p^{-2(l-1)}}+ \frac{ ( p-q ) ^{2}z^{3}}{p^{2}p^{-(l-1)} [ l ] _{p,q}}. \end{aligned}$$

(12)

Proof

In fact, we may easily calculate third- and fourth-order central moments as follows:

$$\begin{aligned}& \begin{aligned} K_{l,p,q} \bigl( ( t-z ) ^{3};z \bigr) & =K_{l,p,q} \bigl( t^{3};z \bigr) -3zK_{l,p,q} \bigl( t^{2};z \bigr) +3z^{2}K_{l,p,q} ( t;z ) -z^{3} \\ & = \biggl( z^{3}+\frac{2p+q}{ [ l ] _{p,q} pp^{-(l-1)}}z^{2}+ \frac{1}{ [ l ] _{p,q}^{2}p^{-2(l-1)}}z \biggr) \\ & \quad-3z \biggl( z^{2}+\frac{z}{ [ l ] _{p,q} p^{-(l-1)}} \biggr) +3z^{3}-z^{3} \\ & =\frac{1}{ [ l ] _{p,q}^{2} p^{-2(l-1)}}z+\frac{z^{2} ( q-p ) }{pp^{-(l-1)} [ l ] _{p,q}}, \end{aligned} \\& \begin{aligned} & K_{l,p,q} \bigl( ( t-z ) ^{4};z \bigr) \\ &\quad =K_{l,p,q} \bigl( t^{4};z \bigr) -4zK_{l,p,q} \bigl( t^{3};z \bigr) +6z^{2}K_{l,p,q} \bigl( t^{2};z \bigr) -4z^{3}K_{l,p,q} ( t;z ) +z^{4} \\ &\quad =z^{4}+ \biggl( \frac{3p^{2}+2qp+q^{2}}{p^{2}} \biggr) \frac{z^{3}}{p^{-(l-1)} [ l ] _{p,q}}+ \biggl( \frac{3p^{2}+3pq+q^{2}}{p^{2} } \biggr) \frac{z^{2}}{p^{-2(l-1)} [ l ] ^{2}}+ \frac{1}{p^{-3(l-1)} [ l ] ^{3}}z \\ &\qquad -4z \biggl( z^{3}+\frac{ ( 2p+q ) }{p^{-(l-2)} [ l ] _{p,q}}z^{2}+ \frac{z}{ [ l ] _{p,q}^{2}p^{-2(l-1)}} \biggr) +6z^{2} \biggl( \frac{z}{p^{-(l-1)} [ l ] _{p,q}}+z^{2} \biggr) -4z^{4}+z^{4} \\ &\quad =\frac{1}{ [ l ] _{p,q}^{3} p^{-3(l-1)}}z+ \biggl( \frac {3pq+q^{2}-p^{2}}{p^{2}} \biggr) \frac{z^{2}}{ [ l ] _{p,q}^{2} p^{-2(l-1)}}+\frac{ ( p-q ) ^{2}z^{3}}{p^{2}p^{-(l-1)} [ l ] _{p,q}}.\end{aligned} \end{aligned}$$

 □

Remark 6

For $0< q< p \leq1$,

$$\lim_{l\rightarrow\infty} [ l ] _{p,q}=0\quad\text{or}\quad \frac{1}{p-q}.$$

In our study, we assume that $q=q_{l}\in ( 0,1 )$ and $p=p_{l}\in ( q,1 ]$ are such that

$$\lim_{l\rightarrow\infty}q_{l}=1,\qquad\lim_{l\rightarrow \infty } p_{l}=1, $$

and

$$\lim_{l\rightarrow\infty}q_{l}^{l}=1,\qquad \lim_{l\rightarrow \infty }p_{l}^{l}=1. $$

Therefore

$$\lim_{l\rightarrow\infty} [ l ] _{p_{l},q_{l}}=\infty. $$

For all $0< q< p\leq1$ and $j\geq0$, the $( p,q )$-difference operators are defined as

$$\Delta_{p,q}^{0}g(z_{j})=0,\qquad \Delta_{p,q}^{1}g(z_{j})= \Delta_{p,q}g(z_{j}), $$

and

$$\Delta_{p,q}^{k+1}g(z_{j})=p^{k} \Delta_{p,q}^{k}g(z_{j+1})-q^{k} \Delta _{p,q}^{k}g(z_{j+1}), $$

where $z_{j}=\frac{p^{l-j} [ j ] _{p,q}}{ [ l ] _{p,q}}$. Using this definition, we can prove the following lemmas.

Lemma 7

For all$0< q< p\leq1$and$j,k\in \mathbb{N} \cup \{ 0 \}$, we have

$$g [ z_{j,}z_{j+1},\ldots,z_{j+k} ] = \frac{p^{-(l-1)k} [ l ] _{p,q}^{k} p^{k(k-1)/2}\Delta _{p,q}^{k}g(z_{j})}{p^{-k(2j+k-1)/2}q^{k(2j+k-1)/2} [ k ] _{p,q}!}. $$

Lemma 8

For all$0< q< p\leq1$, we have

$$\Delta_{p,q}^{k}g(z_{0})=g \biggl[ 0, \frac{1}{p^{-(l-1)} [ l ] _{p,q}},\ldots,\frac{ [ k ] _{p,q}}{p^{-(l-k)} [ l ] _{p,q}} \biggr] \frac{ q^{k(k-1)/2} [ k ] _{p,q}!}{p^{k(k-l)} [ l ] _{p,q}^{k}}. $$

Lemma 9

We have

$$\Delta_{p,q}^{k}g(z_{j})=\sum _{i=0}^{k}\frac{ ( -1 ) ^{i}q^{\frac{i(i-1)}{2}}p^{-i(k-i)}}{p^{\frac{i(i-1)}{2}}}\left [ \textstyle\begin{array}{c}k\\ i \end{array}\displaystyle \right ] _{p,q}g(z_{j+k-i}). $$

Lemma 10

The$(p,q)$-Szász–Mirakjan operator can be represented as

$$K_{l,p,q} ( g;z ) =\sum_{k=0}^{\infty}g \biggl[ 0,\frac {1}{p^{-(l-1)} [ l ] _{p,q}},\ldots,\frac{ [ k ] _{p,q}}{p^{-(l-k)} [ l ] _{p,q}} \biggr] z^{k}. $$

Proof

Indeed,

$$\begin{aligned} K_{l,p,q} ( g;z ) & =\sum_{k=0}^{\infty}g \biggl( \frac {p^{l-k} [ k ] _{p,q}}{ [ l ] _{p,q}} \biggr) \frac{p^{k(k-l)}}{q^{k ( k-1 ) /2}}\frac{ [ l ] _{p,q}^{k} z^{k}}{ [ k ] _{p,q}!}e_{p,q} \bigl( - [ l ] _{p,q} p^{k-l+1}q^{-k}z \bigr) \\ & =\sum_{k=0}^{\infty} g \biggl( \frac{p^{l-k} [ k ] _{p,q}}{ [ l ] _{p,q}} \biggr) \frac{p^{k(k-l)}}{q^{k ( k-1 ) /2}}\frac{ [ l ] _{p,q}^{k} z^{k}}{ [ k ] _{p,q}!}\sum_{j=0}^{\infty} \frac{ ( -1 ) ^{j} [ l ] _{p,q}^{j} p^{ ( k-l+1 ) j} z^{j}}{p^{-j(j-1)/2} q^{kj} [ j ] _{p,q}!} \\ & =\sum_{k=0}^{\infty}\sum _{j=0}^{k} g \biggl( \frac{p^{l-k+j} [ k-j ] _{p,q}}{ [ l ] _{p,q}} \biggr) \frac{p^{ ( k-j ) (k-j-l)}}{q^{ ( k-j ) (k-j-1)/2}}\\ &\quad\times \frac{ [ l ] _{p,q}^{k-j} z^{k-j}}{ [ k-j ] _{p,q}!}\frac{ ( -1 ) ^{j} [ l ] _{p,q}^{j} p^{ ( k-j-l+1 ) j} z^{j}}{p^{-j(j-1)/2} q^{ ( k-j ) j} [ j ] _{p,q}!} \\ & =\sum_{k=0}^{\infty}\Delta_{p,q}^{k}g(z_{0}) \frac{p^{k(k-l)} [ l ] _{p,q}^{k} z^{k}}{q^{\frac{k(k-1)}{2}} [ k ] _{p,q}!}. \end{aligned}$$

 □

The next result gives an explicit formula for the moments $K_{l,p,q} ( t^{m};z )$ in terms of Stirling numbers, which is a $(p,q)$-analogue of Becker’s formula; see [46].

Lemma 11

For$0< q< p\leq1$and$m\in l$, we have

$$ K_{l,p,q} \bigl( t^{m};z \bigr) =\sum _{k=1}^{m}\mathbb{S}_{p,q} ( m,k ) \frac{z^{k}}{p^{-(l-1)(m-k)} [ l ] _{p,q}^{m-k}}, $$

(13)

where

$$\mathbb{S}_{p,q} ( m,k ) =\frac{1}{q^{\frac{k(k-1)}{2}}p^{\frac{-k(k-1)}{2}} [ k ] _{p,q}!}\sum _{j=0}^{k}\frac{ ( -1 ) ^{j}q^{\frac{j(j-1)}{2}}}{p^{{\frac{- ( k-j ) (k-j-1)}{2}}}}\left [ \textstyle\begin{array}{c}k\\ j \end{array}\displaystyle \right ] _{p,q}p^{-(k-j-1)m} [ k-j ] _{p,q}^{m}$$

are the second-type Stirling polynomials satisfying the equalities

$$\begin{aligned}& \mathbb{S}_{p,q} ( m+1,j ) =p^{-(j-1)} [ j ] _{p,q}\mathbb{S}_{p,q} ( m,j ) +\mathbb{S}_{p,q} ( m,j-1 ) ,\quad m\geq0, j\geq1, \\& \mathbb{S}_{p,q} ( 0,0 ) =1,\qquad \mathbb{S}_{p,q} ( m,0 ) =0,\quad m>0,\qquad \mathbb{S}_{p,q} ( m,j ) =0,\quad m< j. \end{aligned}$$

(14)

Clearly, $K_{l,p,q} ( t^{m};z )$are polynomials of degreemwithout a constant term.

Proof

Because of $K_{l,p,q} ( t;z ) =z$ and $K_{l,p,q} ( t^{2};z ) =z^{2}+\frac{z}{p^{-(l-1)} [ l ] _{p,q}}$, representation (13) holds for $m=1,2$ with $\mathbb{S}_{p,q} ( 2,1 ) =1$, $\mathbb{S}_{p,q} ( 1,1 ) =1$.

Using mathematical induction, assume (13) to be valued for m. Then from Lemma 3 we get

$$\begin{gathered} K_{l,p,q} \bigl( t^{m};z \bigr) =\sum _{k=1}^{m}\mathbb{S}_{p,q} ( m,k ) \frac{z^{k}}{p^{-(l-1)(m-k)} [ l ] _{p,q}^{m-k}}, \\ K_{l,p,q} \bigl( t^{m+1};pz \bigr) \\ \quad =\frac{zp^{l}}{ [ l ] _{p,q }}D_{p,q}K_{l,p,q} \bigl( t^{m};z \bigr) +zpK_{l,p,q} \bigl( t^{m};pz \bigr) \\ \quad =\frac{zp^{l}}{ [ l ] _{p,q}}\sum_{j=1}^{m} [ j ] _{p,q}\mathbb{S}_{p,q} ( m,j ) \frac{z^{j-1}}{p^{-(l-1)(m-j)} [ l ] _{p,q}^{m-j}}+zp\sum_{j=1}^{m} \mathbb{S}_{p,q} ( m,j ) \frac{ ( pz ) ^{j}}{p^{-(l-1)(m-j)} [ l ] _{p,q}^{m-j}} \\ \quad =\frac{1}{p^{-l}}\sum_{j=1}^{m} [ j ] _{p,q}\mathbb{S}_{p,q} ( m,j ) \frac{z^{j}}{p^{-(l-1)(m-j)} [ l ] _{p,q}^{m-j+1}}+\sum_{j=1}^{m} \mathbb{S}_{p,q} ( m,j ) \frac{ ( pz ) ^{j+1}}{p^{-(l-1)(m-j)} [ l ] _{p,q}^{m-j}} \\ \quad =\sum_{j=1}^{m}p^{-(j-1)} [ j ] _{p,q}\mathbb{S}_{p,q} ( m,j ) \frac{ ( zp ) ^{j}}{ p^{-(l-1)(m-j+1)} [ l ] _{p,q}^{m-j+1}}+ \sum_{j=1}^{m}\mathbb{S}_{p,q} ( m,j ) \frac{ ( pz ) ^{j+1}}{p^{-(l-1)(m-j)} [ l ] _{p,q}^{m-j}} \\ \quad =\frac{zp}{p^{-(l-1)m} [ l ] _{p,q}^{m}}\mathbb{S}_{p,q} ( m,1 ) + ( pz ) ^{m+1}\mathbb{S}_{p,q} ( m,m ) \\ \qquad +\sum_{j=2}^{m} \bigl( p^{-(j-1)} [ j ] _{p,q}\mathbb{S}_{p,q} ( m,j ) +\mathbb{S}_{p,q} ( m,j-1 ) \bigr) \frac{ ( zp ) ^{j}}{ p^{-(l-1)(m-j+1)} [ l ] _{p,q}^{m-j+1}}.\end{gathered}$$

 □

Remark 12

For $p=q=1$, formulae (14) become recurrence formulas satisfied by the second-type Stirling numbers from [8].

3 $( p,q )$-Szász–Mirakjan operators in a polynomial weighted space

Lemma 13

For given any fixed$\beta\in\mathbb{N}\cup \{ 0 \}$and$0< q< p\leq1$, we have

$$ \bigl\Vert K_{l,p,q} ( 1/w_{\beta};z ) \bigr\Vert _{\beta}\leq K_{1} ( p,q,\beta ) ,\quad l\in l, $$

(15)

where$K_{1} ( p,q,\beta )$are positive constants. Moreover, for every$g\in C_{\beta}$, we have

$$ \bigl\Vert K_{l,p,q} ( g ) \bigr\Vert _{\beta}\leq K_{1} ( p,q,\beta ) \Vert g \Vert _{\beta},\quad l\in l. $$

(16)

Thus$K_{l,p,q}$is a linear positive operator from$C_{\beta}$into$C_{\beta}$.

Proof

Inequality (15) is obvious for $\beta=0$. Let $\beta\geq1$. Then by (13) we have

$$\begin{aligned} w_{\beta} ( z ) M_{l,q} ( 1/w_{\beta};z ) & =w_{\beta } ( z ) M_{l,q} \bigl( 1+z^{\beta};z \bigr) \\ & =w_{\beta} ( z ) M_{l,q} ( 1;z ) +w_{\beta} ( z ) M_{l,q} \bigl( z^{\beta};z \bigr) \\ & =w_{\beta} ( z ) +w_{\beta} ( z ) \sum _{j=1}^{\beta }\mathbb{S}_{p,q} ( \beta,j ) \frac{z^{j}}{p^{-(l-1)(\beta -j)} [ l ] _{p,q}^{\beta-j}}\leq K_{1} ( p,q,\beta ) , \end{aligned}$$

where $K_{1} ( p,q,\beta ) >0$ is a constant depending on β, p, and q. From this (15) follows. Moreover, for every $g\in C_{\beta}$,

$$\bigl\Vert K_{l,p,q} ( g ) \bigr\Vert _{\beta}\leq \Vert g \Vert _{\beta} \bigl\Vert K_{l,p,q} ( 1/w_{\beta} ) \bigr\Vert _{\beta}.$$

By applying (15) we obtain

$$\bigl\Vert K_{l,p,q} ( g ) \bigr\Vert _{\beta}\leq K_{1} ( p,q,\beta ) \Vert g \Vert . $$

 □

Lemma 14

For given any fixed$\beta\in\mathbb{N}\cup \{ 0 \}$and$0< q< p\leq1$, we have

$$ \biggl\Vert K_{l,p,q} \biggl( \frac{ ( t-\cdot ) ^{2}}{w_{\beta } ( t ) };\cdot \biggr) \biggr\Vert _{\beta}\leq\frac{K_{2} ( p,q,\beta ) }{p^{-(l-1)} [ l ] _{p,q}},\quad l\in l, $$

(17)

where$K_{2} ( p,q,\beta )$are positive constants.

Proof

Formula (11) imply (17) for $\beta=0$. We have

$$K_{l,p,q} \biggl( \frac{ ( t-z ) ^{2}}{w_{\beta} ( t ) };z \biggr) =K_{l,p,q} \bigl( ( t-z ) ^{2};z \bigr) +K_{l,p,q} \bigl( ( t-z ) ^{2}t^{\beta};z \bigr) $$

for $\beta,l\in\mathbb{N}$. If $\beta=1$, then we get

$$\begin{aligned} K_{l,p,q} \bigl( ( t-z ) ^{2} ( 1+t ) ;z \bigr) & =K_{l,p,q} \bigl( ( t-z ) ^{2};z \bigr) +K_{l,p,q} \bigl( ( t-z ) ^{2}t;z \bigr) \\ & =K_{l,p,q} \bigl( ( t-z ) ^{3};z \bigr) + ( 1+z ) K_{l,p,q} \bigl( ( t-z ) ^{2};z \bigr) , \end{aligned}$$

which by Lemma 5 yields (17) for $\beta=1$.

Let $\beta\geq2$. By applying (13) we get

$$\begin{aligned} & w_{\beta} ( z ) K_{l,p,q} \bigl( ( t-z ) ^{2}t^{\beta};z \bigr) \\ & \quad=w_{\beta} ( z ) \bigl( K_{l,p,q} \bigl( t^{\beta+2};z \bigr) -2zK_{l,p,q} \bigl( t^{\beta+1};z \bigr) +z^{2}K_{l,p,q} \bigl( t^{\beta };z \bigr) \bigr) \\ &\quad =w_{\beta} ( z ) \Biggl( z^{\beta+2}+\sum _{j=1}^{\beta +1}\mathbb{S}_{p,q} ( \beta+2,j ) \frac{z^{j}}{p^{-(l-1) ( \beta+2-j ) } [ l ] _{p,q}^{\beta+2-j}} \\ &\qquad -2z^{\beta+2}-2\sum_{j=1}^{\beta} \mathbb{S}_{p,q} ( \beta +1,j ) \frac{z^{j+1}}{p^{-(l-1) ( \beta+1-j ) } [ l ] _{p,q}^{\beta+1-j}} \\ & \qquad+z^{\beta+2}+\sum_{j=1}^{\beta-1} \mathbb{S}_{p,q} ( \beta,j ) \frac{z^{j+2}}{p^{-(l-1) ( \beta-j ) } [ l ] _{p,q}^{\beta-j}}\Biggr) \\ &\quad =w_{\beta} ( z ) \Biggl( \sum_{j=2}^{\beta} \mathbb{S}_{p,q} ( \beta+2,j ) -2\mathbb{S}_{p,q} ( \beta+1,j ) +\mathbb{S}_{p,q} ( \beta,j-1 ) \Biggr) \frac{z^{j+1}}{p^{-(l-1)(\beta+1-j)} [ l ] _{p,q}^{\beta+1-j}} \\ &\qquad +\mathbb{S}_{p,q} ( \beta+2,1 ) \frac{z}{p^{-(l-1)(\beta +1)} [ l ] _{p,q}^{\beta+1}}+ \bigl( \mathbb{S}_{p,q} ( \beta+2,2 ) -2\mathbb{S}_{p,q} ( p+2,1 ) \bigr) \frac {z^{2}}{p^{-(l-1)\beta} [ l ] _{p,q}^{\beta}} \\ & \quad=w_{\beta} ( z ) \frac{z}{p^{-(l-1)} [ l ] _{p,q}}\mathcal{\wp}_{\beta} ( p,q;z ) , \end{aligned}$$

where $\mathcal{\wp}_{\beta} ( p,q;z )$ is a polynomial of degree β. Therefore we have

$$w_{\beta} ( z ) K_{l,p,q} \bigl( ( t-z ) ^{2}t^{\beta };z \bigr) \leq K_{2} ( p,q,\beta ) \frac{z}{p^{-(l-1)} [ l ] _{p,q}}. $$

 □

In the next theorem, we give an approximation property of $K_{l,p,q}$.

Theorem 15

Let$g\in C_{p}^{2}$, $0< q< p\leq1$, and$z\in{}[ 0,\infty)$. There exist positive constants$K_{3} ( p,q,\beta ) >0$such that

$$w_{\beta} ( z ) \bigl\vert K_{l,p,q} ( g;z ) -g ( z ) \bigr\vert \leq K_{3} ( p,q,\beta ) \bigl\Vert g^{\prime\prime} \bigr\Vert \frac{z}{p^{-(l-1)} [ l ] _{p,q}}. $$

Proof

By the Taylor formula

$$g ( t ) =g ( z ) +g^{\prime} ( z ) ( t-z ) + \int_{z}^{t} \int_{z}^{s}g^{\prime\prime} ( u ) \,du \,ds,\quad g\in C_{p}^{2}, $$

we obtain that

$$\begin{aligned} & w_{\beta} ( z ) \bigl\vert K_{l,p,q} ( g;z ) -g ( z ) \bigr\vert \\ &\quad =w_{\beta} ( z ) \biggl\vert K_{l,p,q} \biggl( \int_{z}^{t}\int_{z}^{s}g^{\prime\prime} ( u ) \,du \,ds;z \biggr) \biggr\vert \\ &\quad \leq w_{\beta} ( z ) K_{l,p,q} \biggl( \biggl\vert \int_{z}^{t} \int_{z}^{s}g^{\prime\prime} ( u ) \,du \,ds \biggr\vert ;z \biggr) \\ &\quad \leq w_{\beta} ( z ) K_{l,p,q} \biggl( \bigl\Vert g^{\prime \prime} \bigr\Vert _{\beta} \biggl\vert \int_{z}^{t} \int_{z}^{s} \bigl( 1+u^{m} \bigr) \,du \,ds \biggr\vert ;z \biggr) \\ &\quad \leq w_{\beta} ( z ) \frac{1}{2} \bigl\Vert g^{\prime\prime } \bigr\Vert _{\beta}K_{l,p,q} \bigl( ( t-z ) ^{2} \bigl( 1/w_{\beta} ( z ) +1/w_{\beta} ( t ) \bigr) ;z \bigr) \\ &\quad \leq\frac{1}{2} \bigl\Vert g^{\prime\prime} \bigr\Vert _{\beta} \bigl( K_{l,p,q} \bigl( ( t-z ) ^{2};z \bigr) +w_{\beta} ( z ) M_{l,q} \bigl( ( t-z ) ^{2}w_{\beta} ( t ) ;z \bigr) \bigr) \\ &\quad \leq K_{3} ( p,q,z ) \bigl\Vert g^{\prime\prime} \bigr\Vert _{\beta}\frac{z}{p^{-(l-1)} [ l ] _{p,q}}. \end{aligned}$$

 □

We consider the modified Steklov means

$$g_{h}(z):=\frac{4}{h^{2}} \int_{0}^{\frac{h}{2}} \int_{0}^{\frac{h}{2}} \bigl[ 2g(z+s+t)-g \bigl(z+2(s+t)\bigr) \bigr] \,ds\,dt, $$

which have the following properties:

$$\begin{gathered} g(z)-g_{h}(z) =\frac{4}{h^{2}} \int_{0}^{\frac{h}{2}} \int _{0}^{\frac{h}{2}}\Delta_{s+t}^{2}g(z) \,ds\,dt, \\ g_{h}^{\prime\prime}(z) =h^{-2} \bigl( 8 \Delta_{\frac{h}{2}}^{2}g(z)-\Delta_{h}^{2}g(z) \bigr),\end{gathered} $$

and therefore

$$\begin{gathered} \Vert g-g_{h} \Vert _{\beta} \leq \omega_{\beta}^{2}(g;h), \\ \bigl\Vert g_{h}^{\prime\prime} \bigr\Vert _{\beta} \leq\frac {1}{9h^{2}}\omega_{\beta}^{2}(g;h).\end{gathered} $$

We may prove the following so-called direct approximation theorem.

Theorem 16

For given any$\beta\in\mathbb{N\cup} \{ 0 \} ,g\in C_{\beta}$, $z\in{}[0,\infty)$, and$0< q< p\leq1$, we have

$$w_{\beta}(z) \bigl\vert K_{l,p,q} ( g;z ) -g(z) \bigr\vert \leq M_{\beta}\omega_{\beta}^{2} \biggl( g; \sqrt{\frac{z}{p^{-(l-1)}[l]_{p,q}}} \biggr) =M_{p} \omega_{p}^{2} \biggl( g;\sqrt{\frac{p^{l-1} ( q-p ) z}{ ( q^{l}-p^{l} ) }} \biggr) . $$

Particularly, if$\mathrm{Lip}_{\beta}^{2}\alpha$for some$\alpha\in(0,2]$, then

$$w_{p}(z) \bigl\vert K_{l,p,q} ( g;z ) -g(z) \bigr\vert \leq M_{\beta} \biggl( \frac{z}{p^{-(l-1)}[l]_{p,q}} \biggr) ^{\frac{\alpha}{2}}. $$

Proof

For $g\in C_{\beta}$ and $h>0$,

$$\bigl\vert K_{l,p,q}(g;z)-g(z) \bigr\vert \leq \bigl\vert K_{l,p,q} \bigl( ( g-g_{h} ) ;z \bigr) -(g-g_{h}) (z) \bigr\vert + \bigl\vert K_{l,p,q} ( g_{h};z ) -g_{h}(z) \bigr\vert , $$

and therefore

$$\begin{aligned} w_{\beta}(z) \bigl\vert K_{l,p,q} ( g;z ) -g(z) \bigr\vert & \leq \Vert g-g_{h} \Vert _{\beta} \biggl( w_{\beta }(z)K_{l,p,q} \biggl( \frac{1}{w_{\beta}(t)};z \biggr) +1 \biggr) \\ & \quad+K_{3} ( p,q,\beta ) \bigl\Vert g_{h}^{\prime\prime} \bigr\Vert _{\beta}\frac{z}{p^{-(l-1)}[l]_{p,q}}. \end{aligned}$$

Since $w_{\beta}(z)K_{l,p,q} ( \frac{1}{w_{\beta}(t)};z ) \leq K_{1} ( p,q,\beta )$,we get that

$$w_{\beta}(z) \bigl\vert K_{l,p,q} ( g;z ) -g(z) \bigr\vert \leq l(p,q,\beta)w_{\beta}^{2}(g,h) \biggl( 1+ \frac {z}{h^{2}p^{-(l-1)}[l]_{p,q}.} \biggr) $$

Thus choosing $h=\sqrt{\frac{z}{p^{-(l-1)} [ l ] _{p,q}}}$, we complete the proof. □

Corollary 17

If$\beta\in \mathbb{N} \cup \{ 0 \}$, $g\in C_{\beta}$, $0< q< p\leq1$, and$z\in {}[0,\infty)$, then

$$\lim_{l\rightarrow\infty}K_{l,p,q} ( g;z ) =g(z) $$

uniformly on every$[ c,d ]$, $0\leq c< d$.

4 Convergence of $( p,q )$-Szász–Mirakjan operators

In [47, Theorem 1] and [48, Theorem1], Totik and de la Cal investigated the class problem of all continuous functions g such that $K_{l,p,q} ( g )$ converges to g uniformly on the whole interval $[0,\infty)$ as $l\rightarrow\infty$. The following thorem is a $( p,q )$-analogue of Theorem 1 in [48].

Theorem 18

Assume that$g: [ 0,\infty ) \rightarrow \mathbb{R}$is either bounded or uniformly continuous. Let

$$g^{\ast} ( z ) =g \bigl( z^{2} \bigr) ,\quad z\in [ 0,\infty ) . $$

Then, for all$t>0$and$z\geq0$,

$$ \bigl\vert K_{l,p,q} ( g;z ) -g ( z ) \bigr\vert \leq2\omega \biggl( g^{\ast};\sqrt{\frac{1}{p^{-(l-1)} [ l ] _{p,q}}} \biggr) . $$

(18)

Therefore$K_{l,p,q} ( g;z )$converges toguniformly on$[ 0,\infty )$as$l\rightarrow\infty$whenever$g^{\ast}$is uniformly continuous.

Proof

By the definition of $g^{\ast}$ we have

$$K_{l,p,q} ( g;z ) =K_{l,p,q} \bigl( g^{\ast} ( \sqrt{\cdot } ) ;z \bigr) . $$

Thus we can write

$$\begin{aligned} \bigl\vert K_{l,p,q} ( g;z ) -g ( z ) \bigr\vert & = \bigl\vert K_{l,p,q} \bigl( g^{\ast} ( \sqrt{\cdot} ) ;z \bigr) -g^{\ast} ( \sqrt{z} ) \bigr\vert \\ & = \Biggl\vert \sum_{k=0}^{\infty} \biggl( g^{\ast} \biggl( \sqrt{\frac{ [ k ] _{p,q}}{p^{-(l-k)} [ l ] _{p,q}}} \biggr) -g^{\ast } ( \sqrt{z} ) \biggr) s_{l,k} ( p,q;z ) \Biggr\vert \\ & \leq\sum_{k=0}^{\infty} \biggl\vert \biggl( g^{\ast} \biggl( \sqrt {\frac{ [ k ] _{p,q}}{p^{-(l-k)} [ l ] _{p,q}}} \biggr) -g^{\ast} ( \sqrt{z} ) \biggr) \biggr\vert s_{l,k} ( p,q;z ) \\ & \leq\sum_{k=0}^{\infty}\omega \biggl( g^{\ast}; \biggl\vert \sqrt {\frac{ [ k ] _{p,q}}{p^{-(l-k)} [ l ] _{p,q}}}-\sqrt {z} \biggr\vert \biggr) s_{l,k} ( q;z ) \\ & \leq\sum_{k=0}^{\infty}\omega \biggl( g^{\ast};\frac{ \vert \sqrt {\frac{ [ k ] _{p,q}}{p^{-(l-k)} [ l ] _{p,q}}}-\sqrt {z} \vert }{K_{l,p,q} ( \vert \sqrt{\cdot}-\sqrt{z} \vert ;z ) }K_{l,p,q} \bigl( \vert \sqrt{\cdot}-\sqrt{z} \vert ;z \bigr) \biggr) s_{l,k} ( p,q;z ) . \end{aligned}$$

Finally, from the inequality

$$\omega \bigl( g^{\ast};\alpha\delta \bigr) \leq ( 1+\alpha ) \omega \bigl( g^{\ast};\delta \bigr) ,\quad \alpha,\delta\geq0, $$

we obtain

$$\begin{aligned} \bigl\vert K_{l,p,q} ( g;z ) -g ( z ) \bigr\vert & \leq\omega \bigl( g^{\ast};K_{l,p,q} \bigl( \vert \sqrt{\cdot}-\sqrt {z} \vert ;z \bigr) \bigr) \sum_{k=0}^{\infty} \biggl( 1+\frac { \vert \sqrt{\frac{ [ k ] _{p,q}}{p^{-(l-k)} [ l ] _{p,q}}}-\sqrt{z} \vert }{K_{l,p,q} ( \vert \sqrt{\cdot}-\sqrt{z} \vert ;z ) } \biggr) s_{l,k} ( p,q;z ) \\ & =2\omega \bigl( g^{\ast};K_{l,p,q} \bigl( \vert \sqrt{ \cdot}-\sqrt {z} \vert ;z \bigr) \bigr) . \end{aligned}$$

To complete the proof, we need to show that for all $t>0$ and $z>0$, we have

$$M_{l,q} \bigl( \vert \sqrt{\cdot}-\sqrt{z} \vert ;z \bigr) \leq \sqrt{\frac{1}{p^{-(l-1)} [ l ] }}. $$

Indeed, from the Cauchy–Schwarz inequality it follows that

$$\begin{aligned} & K_{l,p,q} \bigl( \vert \sqrt{\cdot}-\sqrt{z} \vert ;z \bigr) \\ &\quad =\sum_{k=0}^{\infty} \biggl\vert \sqrt{ \frac{ [ k ] _{p,q}}{p^{-(l-k)} [ l ] _{p,q}}}-\sqrt{z} \biggr\vert s_{l,k} ( p,q;z ) \\ & \quad=\sum_{k=0}^{\infty}\frac{ \vert \frac{ [ k ] _{p,q}}{p^{-(l-k)} [ l ] _{p,q}}-z \vert }{\sqrt{\frac{ [ k ] _{p,q}}{p^{-(l-k)} [ l ] _{p,q}}}+\sqrt {z}}s_{l,k} ( p,q;z ) \leq\frac{1}{\sqrt{z}}\sum_{k=0}^{\infty} \biggl\vert \frac{ [ k ] _{p,q}}{p^{-(l-k)} [ l ] _{p,q}}-z \biggr\vert s_{l,k} ( p,q;z ) \\ & \quad\leq\frac{1}{\sqrt{z}}\sqrt{\sum_{k=0}^{\infty} \biggl\vert \frac{ [ k ] _{p,q}}{p^{-(l-k)} [ l ] _{p,q}}-z \biggr\vert ^{2}s_{l,k} ( q;z ) }=\frac{1}{\sqrt{z}}\sqrt{K_{l,p,q} \bigl( ( \cdot-z ) ^{2};z \bigr) } \\ &\quad =\frac{1}{\sqrt{z}}\sqrt{\frac{1}{p^{-(l-1)} [ l ] _{p,q}}z}=\sqrt{ \frac{1}{p^{-(l-1)} [ l ] _{p,q}}}. \end{aligned}$$

 □

Our next results is a Voronovskaya-type theorem for $( p,q )$-Szász–Mirakjan operators.

Theorem 19

Let$0< q< p\leq1$. For any$g\in C_{\beta}^{2} [ 0,\infty )$, we have the equality

$$\lim_{l\rightarrow\infty} [ l ] _{q_{l}} \bigl( K_{l,p,q} ( g;z ) -g ( z ) \bigr) =\frac{z}{2} g^{\prime\prime} ( z ) $$

for every$z\in [ 0,\infty )$.

Proof

Let $z\in [ 0,\infty )$ be fixed. By the Taylor formula we may write

$$ g ( t ) =g ( z ) +g^{\prime} ( z ) ( t-z ) +\frac{1}{2}g^{\prime\prime} ( z ) ( t-z ) ^{2}+r ( t;z ) ( t-z ) ^{2}, $$

(19)

where $r ( t;z )$ is the Peano form of the remainder, $r ( \cdot;z ) \in C_{\beta}$, and $\lim_{t\rightarrow z}r ( t;z ) =0$. Applying $K_{l,p,q}$ to (19), we obtain

$$\begin{aligned} [ l ] _{p_{,}q} \bigl( K_{l,p_{,}q} ( g;z ) -g ( z ) \bigr) &=g^{\prime} ( z ) [ l ] _{p_{,},q}K_{l,p,q} ( t-z;z ) \\ &\quad+\frac{1}{2}g^{\prime\prime} ( z ) [ l ] _{p_{,},q}K_{l,p,q} \bigl( ( t-z ) ^{2};z \bigr) + [ l ] _{p_{,},q}K_{l,p,q} \bigl( r ( t;z ) ( t-z ) ^{2};z \bigr) .\end{aligned} $$

Applying the Cauchy–Schwarz inequality, we have

$$ K_{l,p,q} \bigl( r ( t;z ) ( t-z ) ^{2};z \bigr) \leq \sqrt{K_{l,p,q} \bigl( r^{2} ( t;z ) ;z \bigr) }\sqrt{K_{l,p,q} \bigl( ( t-z ) ^{4};z \bigr) }. $$

(20)

Obviously, $r^{2} ( z;z ) =0$. Then it follows from Corollary 17 that

$$ \lim_{l\rightarrow\infty}K_{l,p,q} \bigl( r^{2} ( t;z ) ;z \bigr) =r^{2} ( z;z ) =0. $$

(21)

Now from (20), (21), and Lemma 5 we immediately get

$$\lim_{l\rightarrow\infty} [ l ] _{p,q}K_{l,p,q} \bigl( r ( t;z ) ( t-z ) ^{2};z \bigr) =0. $$

The proof is completed. □