Ibragimov–Gadjiev operators preserving exponential functions

Abstract

In this paper, a modification of general linear positive operators introduced by Ibragimov and Gadjiev in 1970 is constructed. It is shown that this modification preserves exponential mappings and also contains modified Bernstein-, Szász- and Baskakov-type operators as special cases. The convergence properties of corresponding operators on $[ 0,\infty )$ and in exponentially weighted spaces are investigated. Finally, the quantitative Voronovskaja theorem in terms of modulus of continuity for functions having exponential growth is examined.

1 Introduction

After the construction of a general sequence of positive operators by Ibragimov and Gadjiev [1] in 1970, some authors, inspired by this work, introduced various generalizations of these operators. This is because the results obtained for the generalized operators are also valid for the operators included in them. Thus, Aral and Acar [2] introduced a general class of Durrmeyer operators by modifying the Ibragimov–Gadjiev operators as

$$\begin{aligned} M_{n} ( f;x ) ={}& ( n-m ) \alpha _{n} \mathit{\psi }_{n} ( 0 ) \sum_{\nu =0}^{\infty }K_{n}^{ ( \nu ) } \bigl( x,0,\alpha _{n}\mathit{\psi }_{n} ( 0 ) \bigr) \frac{(-\alpha _{n}\mathit{\psi }_{n} ( 0 ) )^{\nu }}{\nu !} \\ &{}\times \int _{0}^{\infty }f ( y ) K_{n}^{ ( \nu ) } \bigl( y,0,\alpha _{n}\mathit{\psi }_{n} ( 0 ) \bigr) \frac{(-\alpha _{n}\mathit{\psi }_{n} ( 0 ) )^{\nu }}{\nu !}\,dy, \end{aligned}$$

where

$$\begin{aligned} K_{n}^{ ( \nu ) } \bigl( x,0,\alpha _{n}\mathit{\psi }_{n} ( 0 ) \bigr) = \frac{\partial ^{\nu }}{\partial u^{\nu }}K_{n} ( x,t,u ) |_{ \substack{ u=\alpha _{n}\mathit{\psi }_{n} ( t ) \\ t=0}} \end{aligned}$$

$x,t\in \mathbb{R} ^{+}$ and $-\infty < u<\infty$ is a sequence of functions of three variables x, t, u such that $K_{n\text{ }}$ is entire analytic function with respect to variable u for each $x,t\in \mathbb{R} ^{+}$ and for each $n\in \mathbb{N}$. Bozma and Bars [3] defined a Kantorovich-type generalization on a variable bounded interval:

$$\begin{aligned} R_{n} ( f;x ) = ( \beta _{n}+n+2 ) \sum _{\nu =0}^{\infty }K_{n}^{ ( \nu ) } ( x ) \frac{(-\alpha _{n})^{\nu }}{\nu !} \int _{ \frac{\nu +n+1}{\beta _{n}+n+2}}^{\frac{\nu +n+2}{\beta _{n}+n+2}}f ( p ) \,dp, \end{aligned}$$

where $f:L_{1} [ 0,\frac{n+1}{n+2} ] \rightarrow C [ 0, \frac{n+1}{n+2} ]$. Herdem and Büyükyazıcı [4] constructed an extension in q-Calculus of these operators. The q-generalization of Ibragimov–Gadjiev operators have the following form:

$$\begin{aligned} L_{n} ( f;q;x ) =\sum_{\nu =0}^{\infty }q^{ \frac{\nu ( \nu -1 ) }{2}}f \biggl( \frac{ [ \nu ] }{ [ n ] ^{2}\mathit{\psi }_{n} ( 0 ) } \biggr) D_{q,u}K_{n,\nu }^{q} ( x,t,u ) |_{{ \substack{ u=\alpha _{n}\mathit{\psi }_{n} ( t ) \\ t=0}}} \frac{(-\alpha _{n}\mathit{\psi }_{n} ( 0 ) )^{\nu }}{ [ \nu ] !} \end{aligned}$$

for $x\in \mathbb{R} ^{+}$ and any function f defined on $\mathbb{R} ^{+}$. Furthermore, Korovkin-type theorems for continuous and unbounded functions defined on $[ 0,\infty )$ were established, and some representation formulas using q-derivatives were given in [5]. Many other investigations about Ibragimov–Gadjiev operators may also be found in [6–11].

Now, we recall the original construction. Let $\{ \varphi _{n} ( t ) \}$ and $\{ \mathit{\psi }_{n} ( t ) \}$ be the sequence of functions in $C [ 0,A ]$ such that $\varphi _{n} ( 0 ) =0$ and $\mathit{\psi }_{n} ( t ) >0$ for all $t\in [ 0,A ]$, $A>0$. Let also $\{ \alpha _{n} \}$ be a sequence of positive numbers having the following properties $\lim_{n\rightarrow \infty }\frac{\alpha _{n}}{n}=1$, $\lim_{n\rightarrow \infty } \frac{1}{n^{2}\mathit{\psi }_{n} ( 0 ) }=0$.

Assume that a sequence of functions of three variables $\{ K_{n} ( x,t,u ) \}$ ($x,t\in [ 0,A ]$, $-\infty < u<\infty$) satisfies the following conditions:

$1^{o}$ Each function of this family is an entire analytic function with respect to u for fixed $x,t\in [ 0,A ]$;

$2^{o}$ $K_{n} ( x,0,0 ) =1$ for any $x\in [ 0,A ]$ and for all $n\in \mathbb{N}$;

$3^{o}$ $\{ ( -1 ) ^{\nu } [ \frac {\partial ^{\nu }}{\partial u^{\nu }}K_{n} ( x,t,u ) ] _{ \substack{ u=\alpha _{n}\mathit{\psi }_{n} ( t ) \\ t=0}} \} \geq 0$ $( \nu ,n\in \{ 1,2,\ldots \} ;\text{ }x\in [ 0,A ] )$;

$4^{o}$ $-\frac {\partial ^{\nu }}{\partial u^{\nu }}K_{n} ( x,t,u ) |_{ \substack{ u=\alpha _{n}\mathit{\psi }_{n} ( t ) \\ t=0}}=nx [ \frac {\partial ^{\nu -1}}{\partial u^{\nu -1}}K_{n+m} ( x,t,u ) |_{{ \substack{ u=\alpha _{n}\mathit{\psi }_{n} ( t ) \\ t=0}}} ] (\nu ,n\in \{ 1,2,\ldots \} ; x\in [ 0,A ] )$, where m is a number such that $m+n$ is zero or a naturel number.

Under these conditions, Ibragimov–Gadjiev operators are defined as

$$\begin{aligned} G_{n} ( f;x ) =\sum_{\nu =0}^{\infty }f \biggl( \frac{\nu }{n^{2}\mathit{\psi }_{n} ( 0 ) } \biggr) \biggl[ \frac {\partial ^{\nu }}{\partial u^{\nu }}K_{n} ( x,t,u ) |_{ \substack{ u=\alpha _{n}\mathit{\psi }_{n} ( t ) \\ t=0}} \biggr] \frac{(-\alpha _{n}\mathit{\psi }_{n} ( 0 ) )^{\nu }}{\nu !} \end{aligned}$$

(1)

for $x\in \mathbb{R} ^{+}$ and any function f defined on $\mathbb{R} ^{+}$.

We should mention that Ibragimov–Gadjiev operators contain, as a particular case, a series of operators. By choosing $K_{n} ( x,t,u ) = ( 1-\frac{ux}{1+t} ) ^{n},\alpha _{n}\mathit{=n,}$ $\mathit{\psi }_{n} ( 0 ) =\frac{1}{n}$ the operators defined by (1) are transformed into Bernstein polynomials; for $\alpha _{n}\mathit{=n,}$ $\mathit{\psi }_{n} ( 0 ) =\frac{1}{nb_{n}}$ $( \lim_{n\rightarrow \infty }b_{n}=\infty ,\text{ } \lim_{n\rightarrow \infty }\frac{b_{n}}{n}=0 )$, we also get Bernstein–Chlodowsky polynomials. For $K_{n} ( x,t,u ) =e^{-n ( t+ux ) }$, $\alpha _{n}\mathit{=n,}$ $\mathit{\psi }_{n} ( 0 ) =\frac{1}{n}$, we get Szász–Mirakyan operators. Moreover, if we choose $K_{n} ( x,t,u ) =K_{n} ( t+ux )$, $\alpha _{n}\mathit{=n,}$ $\mathit{\psi }_{n} ( 0 ) =\frac{1}{n}$, then we obtain Baskakov operators.

In recent years, by defining the linear positive operators, which preserve the exponential functions by Aldaz and Render [12], several researchers introduced linear positive operators that reproduce the exponential functions by conveniently modifying the well-known operators. In [13], Acar et al. presented a modification of Szász-Mirakyan operators that reproduces the functions 1 and e$^{2ax}$, $a>0$. They discussed approximation properties via a certain weighted modulus of continuity and a quantitative Voronovskaya-type theorem. In [14], recovered a generalization of the Bernstein operators that reproduce the exponential functions e$^{ax}$ and e$^{2ax}$, $a>0$. The authors also showed that this way better approximates functions with modified operators than the classical one. After that, this method was studied and extended in numerous papers. We refer interested readers to [15–32].

In parallel with these developments, this paper aims to construct a new generalization of Ibragimov–Gadjiev operators, $G_{n}^{\ast }$, fixing the function e$^{ax},a>0$. Then, for these operators, we provide some approximation properties and present special cases as examples.

The rest of this work is organized as follows: In Sect. 2, the technique to construct the modified Ibragimov–Gadjiev operators is discussed. In Sect. 3, moments, central moments, and a recurrence formula are calculated. In Sect. 4, convergence properties on $[ 0,\infty )$ and in the light of weighted spaces are investigated. The rate of convergence using the exponential modulus of continuity is also examined. In Sect. 5, it is shown that the new operators, which will be constructed below, contain modified Bernstein-, Szász- and Baskakov-type operators that exist in the literature as a special case. Finally, in the last section, we summarize the main results and give some thoughts that can be applied to expand the scope of this study.

2 Construction of the operators

Let $\{ \varphi _{n} ( t ) \}$ and $\{ \mathit{\psi }_{n} ( t ) \}$ be the sequence of functions in $C [ 0,A ]$ such that $\varphi _{n} ( 0 ) =0$ and $\mathit{\psi }_{n} ( t ) >0$ for all $t\in [ 0,A ]$, $A>0$. Let also $\{ \alpha _{n} \}$ be a sequence of positive numbers such that $\lim_{n\rightarrow \infty }\frac{\alpha _{n}}{n}=1$, $\lim_{n\rightarrow \infty } \frac{1}{n^{2}\mathit{\psi }_{n} ( 0 ) }=0$.

Assume that a sequence of functions of three variables $\{ K_{n} ( \lambda _{n} ( x ) ,t,u ) \}$ ($x,t\in [ 0,A ]$, $-\infty < u<\infty$ and $\lim_{n\rightarrow \infty }\lambda _{n} ( x ) =x$) satisfies the following conditions:

1^∗ Each function of this family is an entire analytic function with respect to u for fixed $x,t\in [ 0,A ]$;

2^∗ $K_{n} ( \lambda _{n} ( x ) ,0,0 ) =1$ for any $x\in [ 0,A ]$ and for all $n\in \mathbb{N}$;

3^∗ $\{ ( -1 ) ^{\nu } [ \frac {\partial ^{\nu }}{\partial u^{\nu }}K_{n} ( \lambda _{n} ( x ) ,t,u ) ] _{ \substack{ u=\alpha _{n}\mathit{\psi }_{n} ( t ) \\ t=0}} \} \geq 0$ $( \nu ,n\in \{ 1,2,\ldots \} ;\text{ }x\in [ 0,A ] )$;

4^∗ $-\frac {\partial ^{\nu }}{\partial u^{\nu }}K_{n} ( \lambda _{n} ( x ) ,t,u ) |_{ \substack{ u=\alpha _{n}\mathit{\psi }_{n} ( t ) \\ t=0}}=n\lambda _{n} ( x ) [ \frac {\partial ^{\nu -1}}{\partial u^{\nu -1}}K_{n+m} ( \lambda _{n} ( x ) ,t,u ) |_{{ \substack{ u=\alpha _{n}\mathit{\psi }_{n} ( t ) \\ t=0}}} ]$ ($\nu ,n\in \{ 1,2,\ldots \}$; $x\in [ 0,A ]$), where m is a number such that $m+n$ is zero or a natural number.

According to these conditions, modified Ibragimov–Gadjiev operators have the following form:

$$\begin{aligned} G_{n}^{\ast } ( f;x ) =\sum_{\nu =0}^{\infty }e^{- \frac{a\nu }{n^{2}\mathit{\psi }_{n} ( 0 ) }}e^{ax}f \biggl( \frac{\nu }{n^{2}\mathit{\psi }_{n} ( 0 ) } \biggr) \biggl[ \frac {\partial ^{\nu }}{\partial u^{\nu }}K_{n} \bigl( \lambda _{n} ( x ) ,t,u \bigr) \big| _{ \substack{ u=\alpha _{n}\mathit{\psi }_{n} ( t ) \\ t=0}} \biggr] \frac{(-\alpha _{n}\mathit{\psi }_{n} ( 0 ) )^{\nu }}{\nu !} \end{aligned}$$

(2)

for $x,a\in \mathbb{R} ^{+}$ and any function f defined on $\mathbb{R} ^{+}$. We also assume that $G_{n}^{\ast }$ satisfies the following condition

$$\begin{aligned} \sum_{\nu =0}^{\infty }e^{- \frac{ \mu \nu }{n^{2}\mathit{\psi }_{n} ( 0 ) }} \biggl[ \frac {\partial ^{\nu }}{\partial u^{\nu }}K_{n} \bigl( \lambda _{n} ( x ) ,t,u \bigr) \big|_{ \substack{ u=\alpha _{n}\mathit{\psi }_{n} ( t ) \\ t=0}} \biggr] \frac{(-\alpha _{n}\mathit{\psi }_{n} ( 0 ) )^{\nu }}{\nu !}=e^{\beta _{n} ( \mu x ) } \end{aligned}$$

(3)

for all $n\in \mathbb{N}$ and $\mu \in \mathbb{Z}$. Here, $\beta _{n}(x)$ is a sequence such that $\lim_{n\rightarrow \infty }\beta _{n}(x)=x$.

It can be easily shown that the operators $G_{n}^{\ast }$ preserve $e^{ax}$, i.e.,

$$\begin{aligned} G_{n}^{\ast } \bigl( e^{at};x \bigr) =e^{ax}. \end{aligned}$$

We can use the Taylor expansion of $\{ K_{n} ( \lambda _{n} ( x ) ,t,u ) \}$ due to condition $( 1^{o} )$. Setting $u=\varphi _{n} ( t )$ and $u_{1}=\alpha _{n}\mathit{\psi }_{n} ( t )$, we have

$$\begin{aligned} K_{n} \bigl( \lambda _{n} ( x ) ,t,\varphi _{n} ( t ) \bigr) =\sum_{\nu =0}^{\infty } \biggl[ \frac {\partial ^{\nu }}{\partial u^{\nu }}K_{n} \bigl( \lambda _{n} ( x ) ,t,u \bigr) \big|_{u=\alpha _{n}\mathit{\psi }_{n} ( t ) } \biggr] \frac{ ( \varphi _{n} ( t ) -\alpha _{n}\mathit{\psi }_{n} ( t ) ) ^{\nu }}{\nu !}. \end{aligned}$$

Since $\varphi _{n} ( 0 ) =0$ and $K_{n} ( \lambda _{n} ( x ) ,0,0 ) =1$, by taking $t=0$ the above equation turns into

$$\begin{aligned} \sum_{\nu =0}^{\infty } \biggl[ \frac {\partial ^{\nu }}{\partial u^{\nu }}K_{n} \bigl( \lambda _{n} ( x ) ,t,u \bigr) \big|_{ \substack{ u=\alpha _{n}\mathit{\psi }_{n} ( t ) \\ t=0}} \biggr] \frac{ ( -\alpha _{n}\mathit{\psi }_{n} ( 0 ) ) ^{\nu }}{\nu !}=1. \end{aligned}$$

(4)

Using Eq. (4), we get

$$\begin{aligned} G_{n}^{\ast } \bigl( e^{at};x \bigr) &=\sum _{\nu =0}^{ \infty }e^{-\frac{a\nu }{n^{2}\mathit{\psi }_{n} ( 0 ) }}e^{ax}e^{ \frac{a\nu }{n^{2}\mathit{\psi }_{n} ( 0 ) }} \biggl[ \frac {\partial ^{\nu }}{\partial u^{\nu }}K_{n} \bigl( \lambda _{n} ( x ) ,t,u \bigr) \big| _{ \substack{ u=\alpha _{n}\mathit{\psi }_{n} ( t ) \\ t=0}} \biggr] \frac{(-\alpha _{n}\mathit{\psi }_{n} ( 0 ) )^{\nu }}{\nu !} \\ &=e^{ax}\sum_{\nu =0}^{\infty } \biggl[ \frac {\partial ^{\nu }}{\partial u^{\nu }}K_{n} \bigl( \lambda _{n} ( x ) ,t,u \bigr) \big| _{ \substack{ u=\alpha _{n}\mathit{\psi }_{n} ( t ) \\ t=0}} \biggr] \frac{(-\alpha _{n}\mathit{\psi }_{n} ( 0 ) )^{\nu }}{\nu !} \\ &=e^{ax}. \end{aligned}$$

3 Auxilary results

In this part, we mention some obvious properties of the modified Ibragimov–Gadjiev operators.

Lemma 1

Let $f ( t ) =e^{\theta at}$, $\theta \in \mathbb{Z}$. Then, for the operators defined by (2), we have

$$\begin{aligned} G_{n}^{\ast } \bigl( e^{\theta at};x \bigr) =e^{ax+\beta _{n}( ( \theta -1 ) ax)} \end{aligned}$$

here $\beta _{n}$ is the same as (3).

We give the following lemma without proof since it similarly follows from the work by the authors dealing with moments for the q-Ibragimov–Gadjiev operators [4].

Lemma 2

Let $e_{i} ( t ) :=t^{i}$, $i=0,1,2$. Then the operators $G_{n}^{\ast }$ satisfies

$$\begin{aligned} &G_{n}^{\ast } ( e_{0};x ) =e^{ax+\beta _{n} ( -ax ) }, \\ &G_{n}^{\ast } ( e_{1};x ) =e^{ ( ax+\beta _{n+m} ( -ax ) - \frac{a}{n^{2}\mathit{\psi }_{n} ( 0 ) } ) } \frac{\alpha _{n}}{n}\lambda _{n} (x), \\ &G_{n}^{\ast } ( e_{2};x ) =e^{ ( ^{ax+e^{\beta _{n+2m}(-ax)}- \frac{2a}{n^{2}\mathit{\psi }_{n} ( 0 ) }} ) } \biggl( \frac{\alpha _{n}}{n} \biggr) ^{2}\frac{n+m}{n}\lambda _{n} ^{2}(x) \\ &\phantom{G_{n}^{\ast } ( e_{2};x ) =}{}+e^{ ( ^{ax+\beta _{n+m}(-ax)- \frac{2a}{n^{2}\mathit{\psi }_{n} ( 0 ) }} ) } \frac{\alpha _{n}}{n}\frac{1}{n^{2}\mathit{\psi }_{n} ( 0 ) }\lambda _{n} (x), \\ &G_{n}^{\ast } ( e_{3};x ) =e^{ ( ^{ax+\beta _{n+3m}(-ax)- \frac{3a}{n^{2}\mathit{\psi }_{n} ( 0 ) }} ) } \biggl( \frac{\alpha _{n}}{n} \biggr) ^{3} \frac{ ( n+m ) ( n+2m ) }{n^{2}}\lambda _{n} ^{3}(x) \\ &\phantom{G_{n}^{\ast } ( e_{3};x ) =}{}+e^{ ( ax+\beta _{n+2m}(-ax)- \frac{2a}{n^{2}\mathit{\psi }_{n} ( 0 ) } ) } \biggl( \frac{\alpha _{n}}{n} \biggr) ^{2} \frac{3 ( n+m ) }{n} \frac{1}{n^{2}\mathit{\psi }_{n} ( 0 ) }\lambda _{n} ^{2}(x) \\ &\phantom{G_{n}^{\ast } ( e_{3};x ) =}{}+e^{ ( ax+\beta _{n+m}(-ax)- \frac{a}{n^{2}\mathit{\psi }_{n} ( 0 ) } ) } \frac{\alpha _{n}}{n} \biggl( \frac{1}{n^{2}\mathit{\psi }_{n} ( 0 ) } \biggr) ^{2} \lambda _{n} (x), \\ &G_{n}^{\ast } ( e_{4};x ) =e^{ ( ^{ax+\beta _{n+4m}(-ax)- \frac{4a}{n^{2}\mathit{\psi }_{n} ( 0 ) }} ) } \biggl( \frac{\alpha _{n}}{n} \biggr) ^{4} \frac{ ( n+m ) ( n+2m ) ( n+3m ) }{n^{3}}\lambda _{n} ^{4}(x) \\ &\phantom{G_{n}^{\ast } ( e_{4};x ) =}{}+e^{ ( ax+\beta _{n+3m}(-ax)- \frac{3a}{n^{2}\mathit{\psi }_{n} ( 0 ) } ) } \biggl( \frac{\alpha _{n}}{n} \biggr) ^{3} \frac{6 ( n+m ) ( n+2m ) }{n^{2}} \frac{1}{n^{2}\mathit{\psi }_{n} ( 0 ) }\lambda _{n} ^{3}(x) \\ &\phantom{G_{n}^{\ast } ( e_{4};x ) =}{}+e^{ ( ax+\beta _{n+2m}(-ax)-\frac{2a}{n^{2}\mathit{\psi }_{n} ( 0 ) } ) } \biggl( \frac{\alpha _{n}}{n} \biggr) ^{2} \frac{7 ( n+m ) }{n} \biggl( \frac{1}{n^{2}\mathit{\psi }_{n} ( 0 ) } \biggr) ^{2}\lambda _{n} ^{2}(x) \\ &\phantom{G_{n}^{\ast } ( e_{4};x ) =}{}+e^{ ( ^{ax+\beta _{n+m}(-ax)-\frac{a}{n^{2}\mathit{\psi }_{n} ( 0 ) }} ) } \frac{\alpha _{n}}{n} \biggl( \frac{1}{n^{2}\mathit{\psi }_{n} ( 0 ) } \biggr) ^{3}\lambda _{n} (x). \end{aligned}$$

Lemma 3

Let $\mu _{n,r} ( x ) =G_{n}^{\ast } ( ( t-x ) ^{r};x )$, $r=0,1,2,4$. Then by considering above Lemma, we have

$$\begin{aligned} &\mu _{n,0} ( x ) =e^{ax+\beta _{n} ( -ax ) }, \\ &\mu _{n,1} ( x ) =\frac{\alpha _{n}}{n}\lambda _{n} (x)e^{ ( ax+\beta _{n+m} ( -ax ) - \frac{a}{n^{2}\mathit{\psi }_{n} ( 0 ) } ) }-xe^{ax+\beta _{n} ( -ax ) }, \\ &\mu _{n,2} ( x ) = \biggl( \frac{\alpha _{n}}{n} \biggr) ^{2} \frac{n+m}{n}\lambda _{n} ^{2}(x)e^{ ( ^{ax+e^{\beta _{n+2m}(-ax)}- \frac{2a}{n^{2}\mathit{\psi }_{n} ( 0 ) }} ) } \\ &\phantom{\mu _{n,2} ( x ) =}{}-\frac{\alpha _{n}}{n} \frac{1}{n^{2}\mathit{\psi }_{n} ( 0 ) }\lambda _{n} (x)e^{ ( ^{ax+\beta _{n+m}(-ax)- \frac{2a}{n^{2}\mathit{\psi }_{n} ( 0 ) }} ) } \\ &\phantom{\mu _{n,2} ( x ) =}{}-2x\frac{\alpha _{n}}{n}\lambda _{n} (x)e^{ ( ax+\beta _{n+m} ( -ax ) - \frac{a}{n^{2}\mathit{\psi }_{n} ( 0 ) } ) }+x^{2}e^{ax+\beta _{n} ( -ax ) }, \\ &\mu _{n,4} ( x ) =e^{ ( ax+\beta _{n+4m}(-ax)- \frac{4a}{n^{2}\mathit{\psi }_{n} ( 0 ) } ) } \biggl( \frac{\alpha _{n}}{n} \biggr) ^{4} \frac{ ( n+m ) ( n+2m ) ( n+3m ) }{n^{3}}\lambda _{n} ^{4}(x) \\ &\phantom{\mu _{n,4} ( x ) =}{}+e^{ ( ^{ax+\beta _{n+3m}(-ax)- \frac{3a}{n^{2}\mathit{\psi }_{n} ( 0 ) }} ) } \biggl( \frac{\alpha _{n}}{n} \biggr) ^{3} \frac{6 ( n+m ) ( n+2m ) }{n^{2}} \frac{1}{n^{2}\mathit{\psi }_{n} ( 0 ) }\lambda _{n} ^{3}(x) \\ &\phantom{\mu _{n,4} ( x ) =}{}+e^{ ( ^{ax+\beta _{n+2m}(-ax)- \frac{2a}{n^{2}\mathit{\psi }_{n} ( 0 ) }} ) } \biggl( \frac{\alpha _{n}}{n} \biggr) ^{2} \frac{7 ( n+m ) }{n} \biggl( \frac{1}{n^{2}\mathit{\psi }_{n} ( 0 ) } \biggr) ^{2}\lambda _{n} ^{2}(x) \\ &\phantom{\mu _{n,4} ( x ) =}{}+e^{ ( ^{ax+\beta _{n+m}(-ax)-\frac{a}{n^{2}\mathit{\psi }_{n} ( 0 ) }} ) } \frac{\alpha _{n}}{n} \biggl( \frac{1}{n^{2}\mathit{\psi }_{n} ( 0 ) } \biggr) ^{3}\lambda _{n} (x) \\ &\phantom{\mu _{n,4} ( x ) =}{}-4x \biggl\{ e^{ ( ^{ax+\beta _{n+3m}(-ax)- \frac{3a}{n^{2}\mathit{\psi }_{n} ( 0 ) }} ) } \biggl( \frac{\alpha _{n}}{n} \biggr) ^{3} \frac{ ( n+m ) ( n+2m ) }{n^{2}}\lambda _{n} ^{3}(x) \\ &\phantom{\mu _{n,4} ( x ) =}{}+e^{ ( ax+\beta _{n+2m}(-ax)- \frac{2a}{n^{2}\mathit{\psi }_{n} ( 0 ) } ) } \biggl( \frac{\alpha _{n}}{n} \biggr) ^{2} \frac{3 ( n+m ) }{n} \frac{1}{n^{2}\mathit{\psi }_{n} ( 0 ) }\lambda _{n} ^{2}(x) \\ &\phantom{\mu _{n,4} ( x ) =}{} +e^{ ( ax+\beta _{n+m}(-ax)-\frac{a}{n^{2}\mathit{\psi }_{n} ( 0 ) } ) } \frac{\alpha _{n}}{n} \biggl( \frac{1}{n^{2}\mathit{\psi }_{n} ( 0 ) } \biggr) ^{2} \lambda _{n} (x) \biggr\} \\ & \phantom{\mu _{n,4} ( x ) =}{}+6x^{2} \biggl\{ e^{ ( ^{ax+e^{ \beta _{n+2m}(-ax)}- \frac{2a}{n^{2}\mathit{\psi }_{n} ( 0 ) }} ) } \biggl( \frac{\alpha _{n}}{n} \biggr) ^{2}\frac{n+m}{n}\lambda _{n} ^{2}(x) \textit{ }\\ &\phantom{\mu _{n,4} ( x ) =}{} +e^{ ( ^{ax+\beta _{n+m}(-ax)- \frac{2a}{n^{2}\mathit{\psi }_{n} ( 0 ) }} ) } \frac{\alpha _{n}}{n}\frac{1}{n^{2}\mathit{\psi }_{n} ( 0 ) }\lambda _{n} (x) \biggr\} \\ &\phantom{\mu _{n,4} ( x ) =}{}-4x^{3}e^{ ( ax+\beta _{n+m} ( -ax ) -\frac{a}{n^{2}\mathit{\psi }_{n} ( 0 ) } ) } \frac{\alpha _{n}}{n}\lambda _{n} (x)+x^{4}. \end{aligned}$$

4 Main results

In this section, first, we analyze the uniform convergence of $G_{n}^{\ast }$ on $[ 0,\infty )$ by means of the modulus of continuity. In 1970, Boyanov and Vaselinov [33] gave approximation properties of a function in an infinite interval.

Now, suppose that $C^{\ast } [ 0,\infty )$ denotes the Banach space of all real-valued continuous functions on $[ 0,\infty )$ with the property $\lim_{x\rightarrow \infty }f ( x )$ existing and finite, given with uniform norm $\Vert . \Vert _{C^{\ast } [ 0,\infty ) }$.

Theorem A

[33] If the sequence $A_{n}:C^{\ast } [ 0,\infty ) \rightarrow C^{\ast } [ 0,\infty )$ of linear positive operators satisfies the conditions

$$\begin{aligned} \lim_{n\rightarrow \infty }A_{n} \bigl( e^{-kt};x \bigr) =e^{-kx},\quad k=0,1,2, \end{aligned}$$

uniformly in $[ 0,\infty )$ then for $f\in [ 0,\infty )$,

$$\begin{aligned} \lim_{n\rightarrow \infty }A_{n} ( f;x ) =f ( x ) . \end{aligned}$$

Later, Holhoş [34] expanded Theorem A to find the rate of uniform convergence.

Theorem B

[34] Let $A_{n}:C^{\ast } [ 0,\infty ) \rightarrow C^{\ast } [ 0,\infty )$ be a sequence of positive linear operators satisfying that

$$\begin{aligned} &\bigl\Vert A_{n} ( 1;x ) -1 \bigr\Vert _{ [ 0, \infty ) } =a_{n}, \\ &\bigl\Vert A_{n} \bigl( e^{-t};x \bigr) -e^{-x} \bigr\Vert _{ [ 0,\infty ) } =b_{n}, \\ &\bigl\Vert A_{n} \bigl( e^{-2t};x \bigr) -e^{-2x} \bigr\Vert _{ [ 0,\infty ) } =c_{n}. \end{aligned}$$

Then, for every function of $f\in C^{\ast } [ 0,\infty )$,

$$\begin{aligned} \bigl\Vert A_{n} ( f;x ) -f ( x ) \bigr\Vert _{ [ 0,\infty ) }\leq \Vert f \Vert _{ [ 0,\infty ) }a_{n}+ ( 2+a_{n} ) \omega ^{ \ast } ( f,\sqrt{a_{n}+2b_{n}+c_{n}} ) \end{aligned}$$

where

$$\begin{aligned} \omega ^{\ast } ( f,\delta ) =\max_{ \substack{ \vert e^{-x}-e^{-t} \vert \leq \delta \\ x,t>0}} \bigl\vert f ( t ) -f ( x ) \bigr\vert , \quad \delta >0 \end{aligned}$$

is the modulus of continuity. Further,

$$\begin{aligned} \bigl\vert f ( t ) -f ( x ) \bigr\vert \leq \biggl( 1+\frac{ ( e^{-x}-e^{-t} ) ^{2}}{\delta ^{2}} \biggr) \omega ^{\ast } ( f,\delta ) . \end{aligned}$$

Now, we are ready to prove our main theorem about uniform convergence of $G_{n}^{\ast }$.

Theorem 1

For $f\in C^{\ast } [ 0,\infty )$, we have

$$\begin{aligned} \bigl\Vert G_{n}^{\ast }f-f \bigr\Vert _{ [ 0,\infty ) } \leq \Vert f \Vert _{ [ 0,\infty ) }a_{n}+ ( 2+a_{n} ) \omega ^{\ast } ( f,\sqrt{a_{n}+2b_{n}+c_{n}} ) \end{aligned}$$

where

$$\begin{aligned} &a_{n} = \bigl\Vert G_{n}^{\ast } ( 1;x ) -1 \bigr\Vert _{ [ 0,\infty ) } \\ &b_{n} = \bigl\Vert G_{n}^{\ast } \bigl( e^{-t};x \bigr) -e^{-x} \bigr\Vert _{ [ 0,\infty ) } \\ &c_{n} = \bigl\Vert G_{n}^{\ast } \bigl( e^{-2t};x \bigr) -e^{-2x} \bigr\Vert _{ [ 0,\infty ) }. \end{aligned}$$

Moreover, $a_{n}$, $b_{n}$ and $c_{n}$ tend to zero as n goes to infinity so that $G_{n}^{\ast }$ converges uniformly to f.

Proof

Considering Lemma 1 and the definition of the operators $G_{n}^{\ast }$, we can write

$$\begin{aligned} &G_{n}^{\ast } ( 1;x ) =e^{ax+\beta _{n} ( -ax ) }, \\ &G_{n} \bigl( e^{-t};x \bigr) =\sum _{\nu =0}^{\infty }e^{- \frac{a\nu }{n^{2}\mathit{\psi }_{n} ( 0 ) }}e^{ax}e^{- \frac{\nu }{n^{2}\mathit{\psi }_{n} ( 0 ) }} \biggl[ \frac {\partial ^{\nu }}{\partial u^{\nu }}K_{n} \bigl( \lambda ( x ) ,t,u \bigr) \big|_{ \substack{ u=\alpha _{n}\mathit{\psi }_{n} ( t ) \\ t=0}} \biggr] \frac{(-\alpha _{n}\mathit{\psi }_{n} ( 0 ) )^{\nu }}{\nu !} \\ &\phantom{G_{n} \bigl( e^{-t};x \bigr) }=e^{ax}\sum_{\nu =0}^{\infty } \bigl( e^{- \frac{\nu ( a+1 ) }{n^{2}\mathit{\psi }_{n} ( 0 ) }} \bigr) \biggl[ \frac {\partial ^{\nu }}{\partial u^{\nu }}K_{n} \bigl( \lambda ( x ) ,t,u \bigr) \big|_{ \substack{ u=\alpha _{n}\mathit{\psi }_{n} ( t ) \\ t=0}} \biggr] \frac{(-\alpha _{n}\mathit{\psi }_{n} ( 0 ) )^{\nu }}{\nu !}\text{ } \\ &\phantom{G_{n} \bigl( e^{-t};x \bigr) }=e^{ax}e^{\beta _{n}(-(a+1)x)} \\ &G_{n} \bigl( e^{-2t};x \bigr) =\sum _{\nu =0}^{\infty }e^{- \frac{a\nu }{n^{2}\mathit{\psi }_{n} ( 0 ) }}e^{ax}e^{- \frac{2\nu }{n^{2}\mathit{\psi }_{n} ( 0 ) }} \biggl[ \frac {\partial ^{\nu }}{\partial u^{\nu }}K_{n} \bigl( \lambda ( x ) ,t,u \bigr) \big|_{ \substack{ u=\alpha _{n}\mathit{\psi }_{n} ( t ) \\ t=0}} \biggr] \frac{(-\alpha _{n}\mathit{\psi }_{n} ( 0 ) )^{\nu }}{\nu !} \\ &\phantom{G_{n} \bigl( e^{-2t};x \bigr)}=e^{ax}\sum_{\nu =0}^{\infty }e^{- \frac{(a+2)\nu }{n^{2}\mathit{\psi }_{n} ( 0 ) }} \biggl[ \frac {\partial ^{\nu }}{\partial u^{\nu }}K_{n} \bigl( \lambda ( x ) ,t,u \bigr) \big|_{ \substack{ u=\alpha _{n}\mathit{\psi }_{n} ( t ) \\ t=0}} \biggr] \frac{(-\alpha _{n}\mathit{\psi }_{n} ( 0 ) )^{\nu }}{\nu !} \\ &\phantom{G_{n} \bigl( e^{-2t};x \bigr)}=e^{ax}e^{\beta _{n}(-(a+2)x)}. \end{aligned}$$

Thus, the result follows immediately from (4) and Theorem B. □

Now, we examine the behavior of the operators $G_{n}^{\ast }$ on some weighted spaces and then prove a quantitative Voronovskaja theorem in terms of modulus of continuity for functions having exponential growth.

Set $\varphi ( x ) =1+e^{2ax}$, $x\in \mathbb{R} ^{+}$, and consider the following weighted spaces:

$$\begin{aligned} &B_{\varphi } \bigl( \mathbb{R} ^{+} \bigr) = \bigl\{ f:\mathbb{R} ^{+}\rightarrow \mathbb{R} :\text{ } \bigl\vert f ( x ) \bigr\vert \text{ }\leq M_{f}\varphi ( x ) \bigr\} , \\ &C_{\varphi } \bigl( \mathbb{R} ^{+} \bigr) =C\bigl(\mathbb{R} ^{+}\bigr)\cap B_{\varphi } \bigl( \mathbb{R} ^{+} \bigr) , \\ &C_{\varphi }^{k} \bigl( \mathbb{R} ^{+} \bigr) = \biggl\{ f\in C\bigl(\mathbb{R} ^{+}\bigr):\lim _{x\rightarrow \infty } \frac{f ( x ) }{\varphi ( x ) }=k_{f} \biggr\} , \end{aligned}$$

where $M_{f}$ and $k_{f}$ are constants depending on f. All three spaces are normed spaces with the norm

$$\begin{aligned} \Vert f \Vert _{\varphi }=\sup_{x\in \mathbb{R} ^{+}} \frac{ \vert f ( x ) \vert }{\varphi ( x ) }. \end{aligned}$$

It is obvious that for any $f\in C_{\varphi }^{k} ( \mathbb{R} ^{+} )$, the inequality

$$\begin{aligned} \bigl\Vert G_{n}^{\ast } ( f ) \bigr\Vert _{\varphi } \leq \Vert f \Vert _{\varphi } \end{aligned}$$

holds, and we conclude that $G_{n}^{\ast }$ maps $C_{\varphi }^{k} ( \mathbb{R} ^{+} )$ to $C_{\varphi }^{k} ( \mathbb{R} ^{+} )$ [25].

Theorem 2

For each function $f\in C_{\varphi }^{k} ( \mathbb{R} ^{+} )$,

$$\begin{aligned} \lim_{n\rightarrow \infty } \bigl\Vert G_{n}^{\ast } ( f ) -f \bigr\Vert _{\varphi }=0. \end{aligned}$$

Proof

Using the general result established in [25], it is sufficient to verify that following three conditions

$$\begin{aligned} \lim_{n\rightarrow \infty } \bigl\Vert G_{n}^{\ast } \bigl( e^{ \upsilon at} \bigr) -e^{\upsilon at} \bigr\Vert _{\varphi }=0, \quad\upsilon =0,1,2. \end{aligned}$$

(5)

For $\upsilon =0$, from Lemma 3, one has

$$\begin{aligned} \bigl\Vert G_{n}^{\ast } ( 1 ) -1 \bigr\Vert _{\varphi }= \sup_{x\in \mathbb{R} ^{+}} \frac{ \vert e^{ax+\beta _{n} ( -ax ) }-1 \vert }{1+e^{2ax}}. \end{aligned}$$

By passing to limit condition, using equality (4), we have

$$\begin{aligned} \lim_{n\rightarrow \infty } \bigl\Vert G_{n}^{\ast } ( 1 ) -1 \bigr\Vert _{\varphi }=0. \end{aligned}$$

We now prove for $\upsilon =2$, similarly from Lemma 3 and equality (4), we get

$$\begin{aligned} \bigl\Vert G_{n}^{\ast } \bigl( e^{2at} \bigr) -e^{2ax} \bigr\Vert _{\varphi } &=\sup_{x\in \mathbb{R} ^{+}} \frac{ \vert e^{ax+\beta _{n}(ax)}-e^{2ax} \vert }{1+e^{2ax}} \\ &\leq \frac{e^{ax}}{1+e^{2ax}} \bigl\vert e^{\beta _{n}(ax)}-e^{ax} \bigr\vert \\ &\leq \bigl\vert e^{\beta _{n}(ax)}-e^{ax} \bigr\vert , \end{aligned}$$

which leads to

$$\begin{aligned} \lim_{n\rightarrow \infty } \bigl\Vert G_{n}^{\ast } \bigl( e^{2at} \bigr) -e^{2ax} \bigr\Vert _{\varphi }=0. \end{aligned}$$

Since $G_{n}^{\ast } ( e^{at};x ) =e^{ax}$, condition (5) is implemented for $\upsilon =1$. Hence, the proof is completed. □

Theorem 3

Let $G_{n}^{\ast }$: $K\rightarrow C [ 0,\infty )$ be the sequence of linear positive operators preserving $e^{ax}$, $a>0$. We suppose that for each constant $B>0$ and fixed $x\in [ 0,\infty )$, $G_{n}^{\ast }$ satisfy

$$\begin{aligned} G_{n}^{\ast } \bigl( ( t-x ) ^{2}e^{Bt};x \bigr) \leq C_{a} ( B,x ) \mu _{n,2} ( x ) . \end{aligned}$$

Additionally, if $f\in C^{2} [ 0,\infty )$ ∩K and $f^{{\prime \prime }}\in Lip ( c,B )$, $0< c\leq 1$, then for $x\in [ 0,\infty )$,

$$\begin{aligned} & \biggl\vert G_{n}^{\ast } ( f;x ) -f ( x ) -f^{ \prime } ( x ) \mu _{n,1}-\frac{1}{2}f^{\prime \prime } ( x ) \mu _{n,2} ( x ) \biggr\vert \\ &\quad\leq \mu _{n,2} ( x ) \biggl( \frac{\sqrt{C_{a} ( 2B,x ) }}{2}+ \frac{C_{a} ( B,x ) }{2}+e^{2Bx} \biggr) \omega _{1} \biggl( f^{\prime \prime },\sqrt{ \frac{\mu _{n,4} ( x ) }{\mu _{n,2} ( x ) }},B \biggr) . \end{aligned}$$

Proof

By considering the Taylor expansion of the function $f\in C^{2} [ 0,\infty )$ at $x\in [ 0,\infty )$, we obtain

$$\begin{aligned} f ( t ) =f ( x ) +f^{\prime } ( x ) ( t-x ) +f^{\prime \prime } ( x ) \frac{ ( t-x ) ^{2}}{2}+h ( t,x ), \end{aligned}$$

(6)

where

$$\begin{aligned} h ( t,x ) =\frac{ ( t-x ) ^{2}}{2} \bigl( f^{ \prime \prime } ( \xi ) -f^{\prime \prime } ( x ) \bigr) , \quad x< \xi < t. \end{aligned}$$

Applying the operators $G_{n}^{\ast }$ to equality (6), we have

$$\begin{aligned} \biggl\vert G_{n}^{\ast } ( f;x ) -f ( x ) -f^{ \prime } ( x ) \mu _{n,1}-\frac{1}{2}f^{\prime \prime } ( x ) \mu _{n,2} ( x ) \biggr\vert &= \bigl\vert G_{n}^{\ast } \bigl( h ( t,x ) ;x \bigr) \bigr\vert \\ &\leq G_{n}^{\ast } \bigl( \bigl\vert h ( t,x ) \bigr\vert ;x \bigr) . \end{aligned}$$

(7)

Additionally,

$$\begin{aligned} h ( t,x ) =\frac{ ( t-x ) ^{2}}{2} \bigl( f^{ \prime \prime } ( \xi ) -f^{\prime \prime } ( x ) \bigr) \leq \frac{ ( t-x ) ^{2}}{2}\textstyle\begin{cases} e^{Bx}\omega _{1} ( f^{\prime \prime },h,B ) , & \vert t-x \vert \leq h, \\ e^{Bx}\omega _{1} ( f^{\prime \prime },kh,B ) , & h\leq \vert t-x \vert \leq kh,\end{cases}\displaystyle \end{aligned}$$

It was proved by Tachev et al. [35] that, for each $h>0$ and $k\in \mathbb{N}$,

$$\begin{aligned} \omega _{1} ( f,kh,B ) \leq ke^{B ( k-1 ) h} \omega _{1} ( f,h,B ) . \end{aligned}$$

With the help of the above inequality, we obtain

$$\begin{aligned} \frac{ ( t-x ) ^{2}e^{Bx}}{2}\omega _{1} \bigl( f^{\prime \prime },kh,B \bigr) & \leq \frac{ ( t-x ) ^{2}e^{Bx}}{2}ke^{B ( k-1 ) h}\omega _{1} \bigl( f^{\prime \prime },h,B \bigr) \\ &\leq \frac{ ( t-x ) ^{2}}{2} \biggl( \frac{ \vert t-x \vert }{h}+1 \biggr) e^{Bx}e^{B \vert t-x \vert }\omega _{1} \bigl( f^{\prime \prime },h,B \bigr) \\ &\leq \frac{ ( t-x ) ^{2}}{2} \biggl( \frac{ \vert t-x \vert }{h}+1 \biggr) \bigl( e^{Bt}+e^{2Bx} \bigr) \omega _{1} \bigl( f^{\prime \prime },h,B \bigr) . \end{aligned}$$

Thus,

$$\begin{aligned} \bigl\vert h ( t,x ) \bigr\vert \leq \frac{ ( t-x ) ^{2}}{2} \biggl( \frac{ \vert t-x \vert }{h}+1 \biggr) \bigl( e^{Bt}+e^{2Bx} \bigr) \omega _{1} \bigl( f^{\prime \prime },h,B \bigr) . \end{aligned}$$

Applying the operators $G_{n}^{\ast }$ to both sides of the above inequality, we have

$$\begin{aligned} G_{n}^{\ast } \bigl( \bigl\vert h ( t,x ) \bigr\vert ;x \bigr) \leq {}&\frac{1}{2}G_{n}^{\ast } \biggl( \biggl( \frac{ \vert t-x \vert ^{3}}{h}+ \vert t-x \vert ^{2} \biggr) \bigl( e^{Bt}+e^{2Bx} \bigr) ;x \biggr) \omega _{1} \bigl( f^{\prime \prime },h,B \bigr) \\ ={}& \biggl( \frac{1}{2h}G_{n}^{\ast } \bigl( \vert t-x \vert ^{3}e^{Bt};x \bigr) +\frac{1}{2}G_{n}^{\ast } \bigl( \vert t-x \vert ^{2}e^{Bt};x \bigr) \biggr) \\ &{}+\frac{e^{2Bx}}{2h}G_{n}^{\ast } \bigl( \vert t-x \vert ^{3};x \bigr) +\frac{e^{2Bx}}{2}G_{n}^{\ast } \bigl( \vert t-x \vert ^{2};x \bigr) \omega _{1} \bigl( f^{\prime \prime },h,B \bigr) . \end{aligned}$$

Using some computations, we get

$$\begin{aligned} G_{n}^{\ast } \bigl( \vert t-x \vert ^{2}e^{Bt};x \bigr) ={}&G_{n}^{ \ast } \bigl( t^{2}e^{Bt};x \bigr) -2xG_{n}^{\ast } \bigl( te^{Bt};x \bigr) +x^{2}G_{n}^{\ast } \bigl( e^{Bt};x \bigr) \\ ={}&e^{ ( \beta _{n+2m} ( ( B-a ) x ) +ax+ \frac{ ( B-a ) 2}{n^{2}\mathit{\psi }_{n} ( 0 ) } ) } \biggl( \frac{\alpha _{n}}{n} \biggr) ^{2} \frac{ ( n+m ) }{n}\lambda ^{2}(x) \\ &{}+e^{ ( \beta _{n+m} ( ( B-a ) x ) +ax+ \frac{ ( B-a ) }{n^{2}\mathit{\psi }_{n} ( 0 ) } ) } \frac{\alpha _{n}}{n} \frac{1}{n^{2}\mathit{\psi }_{n} ( 0 ) }\lambda (x) \\ &{}-2xe^{ ( ^{\beta _{n+m} ( ( B-a ) x ) +ax+ \frac{ ( B-a ) }{n^{2}\mathit{\psi }_{n} ( 0 ) }} ) }\frac{\alpha _{n}}{n}\lambda (x)+x^{2}e^{ax+\beta _{n} ( ( B-a ) x ) }, \end{aligned}$$

For sufficiently large n, it is obvious that

$$\begin{aligned} G_{n}^{\ast } \bigl( \vert t-x \vert ^{2}e^{Bt};x \bigr) \leq C_{a} ( B,x ) \mu _{n,2} ( x ) . \end{aligned}$$

(8)

Making use of the Cauchy–Schwarz inequality, we have the following inequalities

$$\begin{aligned} &G_{n}^{\ast } \bigl( \vert t-x \vert ^{3}e^{Bt};x \bigr) \leq \sqrt{G_{n}^{\ast } \bigl( \vert t-x \vert ^{2}e^{2Bt};x \bigr) }\sqrt {G_{n}^{\ast } \bigl( \vert t-x \vert ^{4};x \bigr) } \\ &\phantom{G_{n}^{\ast } \bigl( \vert t-x \vert ^{3}e^{Bt};x \bigr) }\leq \sqrt{C_{a} ( 2B,x ) \mu _{n,2} ( x ) }\sqrt{\mu _{n,4} ( x ) }, \end{aligned}$$

(9)

$$\begin{aligned} & G_{n}^{\ast } \bigl( \vert t-x \vert ^{3};x \bigr) \leq \sqrt{G_{n}^{\ast } \bigl( \vert t-x \vert ^{4};x \bigr) } \sqrt{G_{n}^{\ast } \bigl( \vert t-x \vert ^{2};x \bigr) } \\ &\phantom{G_{n}^{\ast } \bigl( \vert t-x \vert ^{3};x \bigr) }\leq \sqrt{\mu _{n,4} ( x ) }\sqrt{\mu _{n,2} ( x ) }. \end{aligned}$$

(10)

Thus, using inequalities (8), (9), and (10) in (7), we obtain

$$\begin{aligned} & \biggl\vert G_{n}^{\ast } ( f;x ) -f ( x ) -f^{ \prime } ( x ) \mu _{n,1}-\frac{1}{2}f^{\prime \prime } ( x ) \mu _{n,2} ( x ) \biggr\vert \\ &\quad\leq \biggl( \frac{1}{2h}\sqrt{C_{a} ( 2B,x ) \mu _{n,2} ( x ) }\sqrt{\mu _{n,4} ( x ) }+\frac{1}{2}C_{a} ( B,x ) \mu _{n,2} ( x ) \\ &\qquad{} +\frac{e^{2Bx}}{2h}\sqrt{\mu _{n,4} ( x ) }\sqrt{ \mu _{n,2} ( x ) }+\frac{e^{2Bx}}{2}\mu _{n,2} ( x ) \biggr) \omega _{1} \bigl( f^{\prime \prime },h,B \bigr) . \end{aligned}$$

Finally, when $h=\sqrt{ \frac{\mu _{n,4} ( x ) }{\mu _{n,2} ( x ) }}$ is chosen and substituted in the above equality, we get

$$\begin{aligned} & \biggl\vert G_{n}^{\ast } ( f;x ) -f ( x ) -f^{ \prime } ( x ) \mu _{n,1}-\frac{1}{2}f^{\prime \prime } ( x ) \mu _{n,2} ( x ) \biggr\vert \\ &\quad\leq \mu _{n,2} ( x ) \biggl( \frac{\sqrt{C_{a} ( 2B,x ) }}{2}+ \frac{C_{a} ( B,x ) }{2}+e^{2Bx} \biggr) \omega _{1} \biggl( f^{\prime \prime },\sqrt{ \frac{\mu _{n,4} ( x ) }{\mu _{n,2} ( x ) }},B \biggr) . \end{aligned}$$

Note that, for fixed $x\in [ 0,\infty )$, $\frac{\mu _{n,4} ( x ) }{\mu _{n,2} ( x ) } \rightarrow 0$ as $n\rightarrow \infty$, guarantees the convergence of Theorem 2. □

5 An application of modified Ibragimov–Gadjiev operators

Just like classical Ibragimov–Gadjiev operators, modified Ibragimov–Gadjiev operators also contain some modified operators preserving exponential functions under appropriate selection of $K_{n} ( \lambda _{n} ( x ) ,t,u ) ,\alpha _{n}$ and $\mathit{\psi }_{n} ( 0 )$.

If property 4^∗ is applied ν-times to the $K_{n} ( \lambda _{n} ( x ) ,t,u )$, the operators defined by (1) can be reduced to the form

$$\begin{aligned} G_{n}^{\ast } ( f;x ) ={}&\sum_{\nu =0}^{\infty }e^{- \frac{a\nu }{n^{2}\mathit{\psi }_{n} ( 0 ) }}e^{ax}f \biggl( \frac{\nu }{n^{2}\mathit{\psi }_{n} ( 0 ) } \biggr) \frac{n ( n+m ) ... ( n+ ( \nu -1 ) m ) }{\nu !} \\ &{}\times \bigl(\lambda _{n} ( x ) \alpha _{n}\mathit{\psi }_{n} ( 0 ) \bigr)^{\nu }K_{n+\nu m} \bigl( \lambda _{n} ( x ) ,0,\alpha _{n}\mathit{\psi }_{n} ( 0 ) \bigr) . \end{aligned}$$

(11)

1. In case

$$\begin{aligned} K_{n} \bigl( \lambda _{n} ( x ) ,t,u \bigr) = \biggl( 1- \frac{u\lambda _{n} ( x ) }{1+t} \biggr) ^{n} , \end{aligned}$$

the operator (11) turns into the form

$$\begin{aligned} G_{n}^{\ast } ( f;x ) ={}&\sum_{\nu =0}^{\infty }e^{- \frac{a\nu }{n^{2}\mathit{\psi }_{n} ( 0 ) }}e^{ax}f \biggl( \frac{\nu }{n^{2}\mathit{\psi }_{n} ( 0 ) } \biggr) \binom{n}{\nu } \\ &{}\times\bigl( \lambda _{n} ( x ) \alpha _{n}\mathit{\psi }_{n} ( 0 ) \bigr)^{\nu } \bigl( 1-\alpha _{n}\mathit{\psi }_{n} ( 0 ) \lambda _{n} ( x ) \bigr) ^{n-\nu }. \end{aligned}$$

(12)

Conditions $( 1^{\ast } )$–$( 4^{\ast } )$ are fulfilled, and $m=-1$. For $\alpha _{n}=n$, $\mathit{\psi }_{n} ( 0 ) =\frac{1}{n}$, we have modified Bernstein operators

$$\begin{aligned} G_{n}^{\ast } ( f;x ) =\sum_{\nu =0}^{\infty }e^{- \frac{a\nu }{n}}e^{ax}f \biggl( \frac{\nu }{n^{2}\mathit{\psi }_{n} ( 0 ) } \biggr) \binom{n}{\nu }\bigl(\lambda _{n} ( x ) \bigr)^{\nu } \bigl( 1- \lambda _{n} ( x ) \bigr) ^{n-\nu }, \end{aligned}$$

where

$$\begin{aligned} \lambda _{n} ( x ) =\frac{e^{-ax/n}-1}{e^{a/n}-1} \end{aligned}$$

defined by Aral et al. [14]. Use of L’Hospital’s rule gives

$$\begin{aligned} \lim_{n\rightarrow \infty }\lambda _{n} ( x ) =x \end{aligned}$$

as claimed.

2. In (12), for $\alpha _{n}=n$, $\mathit{\psi }_{n} ( 0 ) =\frac{1}{nb_{n}}$ $( \lim b_{n}= \infty ,\text{ lim}\frac{b_{n}}{n}=0 )$, the operator (11) becomes modified Bernstein–Chlodowsky operators

$$\begin{aligned} G_{n}^{\ast } ( f;x ) =\sum_{\nu =0}^{\infty }e^{ \frac{-a\nu b_{n}}{n}}e^{ax}f \biggl( \frac{\nu b_{n}}{n} \biggr) \binom{n}{\nu }\biggl( \frac{\lambda _{n} ( x ) }{b_{n}} \biggr)^{\nu } \biggl( 1- \frac{\lambda _{n} ( x ) }{b_{n}} \biggr) ^{n-\nu } \end{aligned}$$

with

$$\begin{aligned} \lambda _{n} ( x ) =b_{n}\frac{e^{ax/n}-1}{e^{ab_{n}/n}-1},\qquad \lim_{n\rightarrow \infty }\lambda _{n} ( x ) =x \end{aligned}$$

defined by Özsaraç et al. [26].

3. By choosing

$$\begin{aligned} K_{n} \bigl( \lambda _{n} ( x ) ,t,u \bigr) =e^{-n ( t+u\lambda _{n} ( x ) ) }, \end{aligned}$$

conditions $( 1^{\ast } )$–$( 4^{\ast } )$ are fulfilled, and $m=0$. The operator (11) becomes

$$\begin{aligned} G_{n}^{\ast } ( f;x ) =\sum_{\nu =0}^{\infty }e^{- \frac{a\nu }{n^{2}\mathit{\psi }_{n} ( 0 ) }}e^{ax}f \biggl( \frac{\nu }{n^{2}\mathit{\psi }_{n} ( 0 ) } \biggr) \frac{ ( n\lambda _{n} ( x ) ) ^{\nu }}{\nu !}\bigl(\alpha _{n}\mathit{\psi }_{n} ( 0 ) \bigr)^{\nu }e^{-n ( \alpha _{n}\mathit{\psi }_{n} ( 0 ) \lambda _{n} ( x ) ) }. \end{aligned}$$

(13)

If we choose $\alpha _{n}=n$, $\mathit{\psi }_{n} ( 0 ) = \frac{1}{n}$, we get modified Szász–Mirakjan operators

$$\begin{aligned} G_{n}^{\ast } ( f;x ) =e^{-n\lambda _{n} ( x ) } \sum _{\nu =0}^{\infty }e^{-\frac{a\nu }{n}}e^{ax}f \biggl( \frac{\nu }{n} \biggr) \frac{ ( n\lambda _{n} ( x ) ) ^{\nu }}{\nu !}, \end{aligned}$$

where

$$\begin{aligned} \lambda _{n} ( x ) = \frac{-ax}{n ( e^{-a/n}-1 ) } \end{aligned}$$

introduced by Acu et al. [15].

Besides, with the choice of $\lambda _{n} ( x )$ in (13) as

$$\begin{aligned} \lambda _{n} ( x ) =\frac{ax}{n ( e^{a/n}-1 ) } \end{aligned}$$

another variant of Szász–Mirakjan operators is obtained, which was presented by Goyal [24]. As can be seen easily, using L’Hospital’s rule, the limit of both $\lambda _{n} ( x )$ yields

$$\begin{aligned} \lim_{n\rightarrow \infty }\lambda _{n} ( x ) =x. \end{aligned}$$

It must be noted that the new variants of Szász–Mirakjan operators obtained by different selection of $\lambda _{n} ( x )$ differ in terms of the functions they preserve as well as their structural features.

4. In addition, if we choose

$$\begin{aligned} K_{n} \bigl( \lambda _{n} ( x ) ,t,u \bigr) = \bigl( 1+t+u \lambda _{n} ( x ) \bigr) ^{-n} , \end{aligned}$$

all conditions are fulfilled and $m=1$. Thus, the operator defined by (11) turns into

$$\begin{aligned} G_{n} ( f;x ) ={}&\sum_{\nu =0}^{\infty }e^{- \frac{a\nu }{n^{2}\mathit{\psi }_{n} ( 0 ) }}e^{ax}f \biggl( \frac{\nu }{n^{2}\mathit{\psi }_{n} ( 0 ) } \biggr) \binom{n+\nu -1}{\nu } \\ &{}\times \bigl(\lambda _{n} ( x ) \alpha _{n}\mathit{\psi }_{n} ( 0 ) \bigr)^{\nu }K_{n+\nu } \bigl( \lambda _{n} ( x ) ,0,\alpha _{n}\mathit{\psi }_{n} ( 0 ) \bigr) . \end{aligned}$$

Choosing $\alpha _{n}=n$, $\mathit{\psi }_{n} ( 0 ) = \frac{1}{n}$, we get Baskakov operators

$$\begin{aligned} G_{n} ( f;x ) =\sum_{\nu =0}^{\infty }e^{- \frac{a\nu }{n}}e^{ax}f \biggl( \frac{\nu }{n} \biggr) \binom{n+\nu -1}{\nu } \frac{\lambda _{n}^{\nu } ( x ) }{ ( 1+\lambda _{n} ( x ) ) ^{n+\nu }}, \end{aligned}$$

where

$$\begin{aligned} \lambda _{n} ( x ) = \frac{e^{-\frac{ax}{n}}-1}{1-e^{\frac{a}{n}}},\qquad \lim _{n\rightarrow \infty }\lambda _{n} ( x ) =x \end{aligned}$$

defined by Özsaraç and Acar [25].

By choosing the appropriate sequences of $K_{n} ( \lambda _{n} ( x ) ,t,u )$, $\alpha _{n}$ and $\mathit{\psi }_{n} ( 0 )$, one can obtain other new operators, and we leave it to readers.

6 Conclusions

Through this work, a new generalization of Ibragimov–Gadjiev operators, which fixes the function e$^{ax},a>0$, has been constructed. Then, for these operators, some approximation properties have been provided, and it has been shown that the newly defined operators contain modified Bernstein-, Szász-, and Baskakov-type operators, which were studied by several authors, as special cases. The relationship between these operators obtained by different choices of $\lambda _{n} ( x )$ has also been revealed.

It is worth noting to readers that one can obtain new operators by taking different sequences of $K_{n} ( \lambda _{n} ( x ) ,t,u )$, $\alpha _{n}$ and $\mathit{\psi }_{n} ( 0 )$. Moreover, the other approximation properties not covered in this study may also be investigated.