# Approximation properties of generalized Baskakov–Schurer–Szasz–Stancu operators preserving $$e^{-2ax}, a>0$$

## Abstract

The current paper deals with a modified form of the Baskakov–Schurer–Szasz–Stancu operators which preserve $$e^{-2ax}$$ for $$a>0$$. The uniform convergence of the modified operators is shown. The rate of convergence is investigated by using the usual modulus of continuity and the exponential modulus of continuity. Then Voronovskaya-type theorem is given for quantitative asymptotic estimation.

## Introduction

Use of linear positive operators has played a crucial role in approximation theory for the last seven decades. In 1950, Szász  defined

$$S_{s}(f;x)= e^{-sx}\sum_{r=0}^{\infty } \frac{(sx)^{r}}{r!}f \biggl( \frac{r}{s} \biggr), \quad s>0$$

for $$x\in [0,\infty )$$. In 1957, Baskakov  proposed

${L}_{s}\left(f;x\right)=\frac{1}{{\left(1+x\right)}^{s}}\sum _{r=0}^{\mathrm{\infty }}\left(\begin{array}{c}s+r-1\\ r\end{array}\right)\frac{{x}^{r}}{{\left(1+x\right)}^{r}}f\left(\frac{r}{s}\right),\phantom{\rule{1em}{0ex}}s\in {\mathbb{N}}^{+},x\in \left[0,\mathrm{\infty }\right).$

In 1962, Schurer  introduced

${B}_{s,p}\left(f;x\right)=\sum _{r=0}^{s+p}\left(\begin{array}{c}s+p\\ r\end{array}\right){x}^{r}{\left(1-x\right)}^{s+p-r}f\left(\frac{r}{s}\right),$

where $$x\in [0,1]$$ and p is a non-negative integer. In 1983, Stancu  studied

${S}_{s}^{\alpha ,\beta }\left(f;x\right)=\sum _{r=0}^{s}\left(\begin{array}{c}s\\ r\end{array}\right){x}^{r}{\left(1-x\right)}^{s-r}f\left(\frac{r+\alpha }{s+\beta }\right),$

satisfying the condition $$0\leq \alpha \leq \beta$$. Various studies related to these operators, such as Baskakov–Szász type operators , Baskakov–Schurer–Szász operators , Baskakov–Szász–Stancu operators , and q-Baskakov–Schurer–Szász–Stancu operators , have been conducted.

In 2010, Aldaz and Render  introduced linear and positive operators preserving 1 and $$e^{x}$$. In 2017, Acar et al.  examined a modified form of the Szász–Mirakyan operators which reproduces constant and $$e^{2ax}, a>0$$. After that, in some studies Szász–Mirakyan operators , Baskakov–Szász–Mirakyan-type operators , Phillips operators , Szász–Mirakyan–Kantorovich operators , and Baskakov operators  preserving constant and exponential function were examined. In addition, Kajla  studied Srivastava–Gupta operators preserving linear functions. On the other hand, Gupta and Tachev  found the general estimation in terms of Pǎltǎneaś modulus of continuity.

In 2018, Bodur et al.  analyzed Baskakov–Szász–Stancu operators preserving exponential functions. Motivated by this paper, we construct a new generalization of the Baskakov–Schurer–Szász–Stancu operators

$\begin{array}{rl}{M}_{s,p}^{\alpha ,\beta }\left(f;x\right)=& \left(s+p\right)\sum _{r=0}^{\mathrm{\infty }}\left(\begin{array}{c}s+p+r-1\\ r\end{array}\right)\frac{{x}^{r}}{{\left(1+x\right)}^{s+p+r}}\\ & ×{\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s+p\right)t}\frac{{\left(s+p\right)}^{r}{t}^{r}}{r!}f\left(\frac{\left(s+p\right)t+\alpha }{s+p+\beta }\right)\phantom{\rule{0.2em}{0ex}}dt,\end{array}$

where s is a positive integer, p is a non-negative integer, and $$0\leq \alpha \leq \beta$$. By taking $$\alpha =0$$ and $$\beta =0$$, we obtain Baskakov–Schurer–Szász operators . In addition, by taking $$s+p=u$$, we get Baskakov–Szász–Stancu operators . Moreover, by taking $$s+p=u$$, $$\alpha =0$$ and $$\beta =0$$, we have Baskakov–Szász operators . We deal with the following modified form:

$\begin{array}{rl}{M}_{s}^{\alpha ,\beta }\left(f;x\right)=& \left(s+p\right)\sum _{r=0}^{\mathrm{\infty }}\left(\begin{array}{c}s+p+r-1\\ r\end{array}\right)\frac{{\left({v}_{s,p}\left(x\right)\right)}^{r}}{{\left(1+{v}_{s,p}\left(x\right)\right)}^{s+p+r}}\\ & ×{\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s+p\right)t}\frac{{\left(s+p\right)}^{r}{t}^{r}}{r!}f\left(\frac{\left(s+p\right)t+\alpha }{s+p+\beta }\right)\phantom{\rule{0.2em}{0ex}}dt.\end{array}$
(1)

Assume that the operators (1) preserve $$e^{-2ax}, a>0$$. In that case, we find the function $$v_{s,p}(x)$$ satisfying $$M_{s}^{\alpha, \beta }(e^{-2at};x)=e^{-2ax}$$ as follows:

$\begin{array}{rl}{e}^{-2ax}& =\left(s+p\right)\sum _{r=0}^{\mathrm{\infty }}\left(\begin{array}{c}s+p+r-1\\ r\end{array}\right)\frac{{\left({v}_{s,p}\left(x\right)\right)}^{r}}{{\left(1+{v}_{s,p}\left(x\right)\right)}^{s+p+r}}{\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s+p\right)t}\frac{{\left(s+p\right)}^{r}{t}^{r}}{r!}{e}^{\frac{-2a\left(\left(s+p\right)t+\alpha \right)}{s+p+\beta }}\phantom{\rule{0.2em}{0ex}}dt\\ & =\frac{s+p+\beta }{s+p+\beta +2a}{e}^{\frac{-2a\alpha }{s+p+\beta }}\sum _{r=0}^{\mathrm{\infty }}\left(\begin{array}{c}s+p+r-1\\ r\end{array}\right){\left(\frac{\left(s+p+\beta \right){v}_{s,p}\left(x\right)}{s+p+\beta +2a}\right)}^{r}\frac{1}{{\left(1+{v}_{s,p}\left(x\right)\right)}^{s+p+r}}.\end{array}$

By a simple computation, we have

$$v_{s,p}(x)=\frac{s+p+\beta +2a}{2a} \biggl\lbrace \biggl( \frac{s+p+ \beta +2a}{s+p+\beta }e^{\frac{2a\alpha }{s+p+\beta }-2ax} \biggr) ^{-1/{s+p}}-1 \biggr\rbrace .$$
(2)

## Some auxiliary results

Here, we present some important equalities and auxiliary lemmas, necessary for the proof of the main theorems.

$$\int _{0}^{\infty }t^{r} e^{-At}\,dt= \frac{r!}{A^{r+1}},\quad A>0.$$
(3)

Negative binomial series is given as follows:

${\left(x+a\right)}^{-s}=\sum _{r=0}^{\mathrm{\infty }}\left(\begin{array}{c}s+r-1\\ r\end{array}\right)\frac{{\left(-x\right)}^{r}}{{a}^{s+r}}.$
(4)

### Lemma 1

Let $$v_{s,p}(x)$$ be given by (2), then we have

$$M_{s}^{\alpha,\beta }\bigl(e^{-At};x\bigr)= \frac{s+p+\beta }{s+p+\beta +A}e ^{\frac{-A\alpha }{s+p+\beta }} \biggl( 1+\frac{Av_{s,p}(x)}{s+p+ \beta +A} \biggr) ^{-(s+p)}.$$
(5)

### Proof

Take $$f(t)=e^{-At}$$, then by using (3) and (4) we obtain

$\begin{array}{rl}& {M}_{s}^{\alpha ,\beta }\left({e}^{-At};x\right)\\ & \phantom{\rule{1em}{0ex}}=\left(s+p\right)\sum _{r=0}^{\mathrm{\infty }}\left(\begin{array}{c}s+p+r-1\\ r\end{array}\right)\frac{{\left({v}_{s,p}\left(x\right)\right)}^{r}}{{\left(1+{v}_{s,p}\left(x\right)\right)}^{s+p+r}}\frac{{\left(s+p\right)}^{r}}{r!}{\int }_{0}^{\mathrm{\infty }}{t}^{r}{e}^{-\left(s+p\right)t-\left(\frac{A\left(s+p\right)t+A\alpha }{s+p+\beta }\right)}\phantom{\rule{0.2em}{0ex}}dt\\ & \phantom{\rule{1em}{0ex}}={e}^{\frac{-A\alpha }{s+p+\beta }}\sum _{r=0}^{\mathrm{\infty }}\left(\begin{array}{c}s+p+r-1\\ r\end{array}\right)\frac{{\left({v}_{s,p}\left(x\right)\right)}^{r}}{{\left(1+{v}_{s,p}\left(x\right)\right)}^{s+p+r}}\frac{{\left(s+p\right)}^{r+1}}{r!}{\int }_{0}^{\mathrm{\infty }}{t}^{r}{e}^{-\frac{\left(s+p\right)\left(s+p+\beta +A\right)t}{s+p+\beta }}\phantom{\rule{0.2em}{0ex}}dt\\ & \phantom{\rule{1em}{0ex}}=\frac{s+p+\beta }{s+p+\beta +A}{e}^{\frac{-A\alpha }{s+p+\beta }}\sum _{r=0}^{\mathrm{\infty }}\left(\begin{array}{c}s+p+r-1\\ r\end{array}\right){\left(\frac{\left(s+p+\beta \right){v}_{s,p}\left(x\right)}{s+p+\beta +A}\right)}^{r}\frac{1}{{\left(1+{v}_{s,p}\left(x\right)\right)}^{s+p+r}}\\ & \phantom{\rule{1em}{0ex}}=\frac{s+p+\beta }{s+p+\beta +A}{e}^{\frac{-A\alpha }{s+p+\beta }}{\left(1+\frac{A{v}_{s,p}\left(x\right)}{s+p+\beta +A}\right)}^{-\left(s+p\right)}.\end{array}$

□

### Lemma 2

Let $$e_{k}(t)=t^{k}, k=0,1,2,3,4$$. Then we get the following equalities:

\begin{aligned} &M_{s}^{\alpha,\beta }(e_{0};x) = 1, \\ &M_{s}^{\alpha,\beta }(e_{1};x) = \frac{(s+p)v_{s,p}(x)+\alpha +1}{s+p+ \beta }, \\ &M_{s}^{\alpha,\beta }(e_{2};x) = \frac{(s+p)(s+p+1)v^{2}_{s,p}(x)+(4+2 \alpha )(s+p)v_{s,p}(x)+\alpha ^{2}+2\alpha +2}{(s+p+\beta )^{2}}, \\ &M_{s}^{\alpha,\beta }(e_{3};x) = \frac{(s+p)(s+p+1)(s+p+2)v^{3}_{s,p}(x)+(9+3 \alpha )(s+p)(s+p+1)v^{2}_{s,p}(x)}{(s+p+\beta )^{3}} \\ &\phantom{M_{s}^{\alpha,\beta }(e_{3};x) =}{}+\frac{(3\alpha ^{2}+12\alpha +18)(s+p)v_{s,p}(x)+\alpha ^{3}+3\alpha ^{2}+6\alpha +6}{(s+p+\beta )^{3}}, \\ &M_{s}^{\alpha,\beta }(e_{4};x)\\ &\quad = \frac{(s+p)(s+p+1)(s+p+2)(s+p+3)v ^{4}_{s,p}(x)+(16+4\alpha )(s+p)(s+p+1)(s+p+2)v^{3}_{s,p}(x)}{(s+p+ \beta )^{4}} \\ &\qquad{}+\frac{(72+36\alpha +6\alpha ^{2})(s+p)(s+p+1)v^{2}_{s,p}(x)+(96+72 \alpha +24\alpha ^{2}+4\alpha ^{3})(s+p)v_{s,p}(x)}{(s+p+\beta )^{4}} \\ &\qquad{}+\frac{24+24\alpha +12\alpha ^{2}+4\alpha ^{3}+\alpha ^{4}}{(s+p+ \beta )^{4}}. \end{aligned}

### Proof

Take $$f(t)=e_{1}$$, then by using (3) and (4) we have

$\begin{array}{rl}& {M}_{s}^{\alpha ,\beta }\left({e}_{1};x\right)\\ & \phantom{\rule{1em}{0ex}}=\left(s+p\right)\sum _{r=0}^{\mathrm{\infty }}\left(\begin{array}{c}s+p+r-1\\ r\end{array}\right)\frac{{\left({v}_{s,p}\left(x\right)\right)}^{r}}{{\left(1+{v}_{s,p}\left(x\right)\right)}^{s+p+r}}{\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s+p\right)t}\frac{{\left(s+p\right)}^{r}{t}^{r}}{r!}\left(\frac{\left(s+p\right)t+\alpha }{s+p+\beta }\right)\phantom{\rule{0.2em}{0ex}}dt\\ & \phantom{\rule{1em}{0ex}}=\frac{1}{s+p+\beta }\sum _{r=0}^{\mathrm{\infty }}\left(\begin{array}{c}s+p+r-1\\ r\end{array}\right)\frac{{\left({v}_{s,p}\left(x\right)\right)}^{r}}{{\left(1+{v}_{s,p}\left(x\right)\right)}^{s+p+r}}\frac{{\left(s+p\right)}^{r+2}}{r!}{\int }_{0}^{\mathrm{\infty }}{t}^{r+1}{e}^{-\left(s+p\right)t}\phantom{\rule{0.2em}{0ex}}dt\\ & \phantom{\rule{2em}{0ex}}+\frac{\alpha }{s+p+\beta }\sum _{r=0}^{\mathrm{\infty }}\left(\begin{array}{c}s+p+r-1\\ r\end{array}\right)\frac{{\left({v}_{s,p}\left(x\right)\right)}^{r}}{{\left(1+{v}_{s,p}\left(x\right)\right)}^{s+p+r}}\frac{{\left(s+p\right)}^{r+1}}{r!}{\int }_{0}^{\mathrm{\infty }}{t}^{r}{e}^{-\left(s+p\right)t}\phantom{\rule{0.2em}{0ex}}dt\\ & \phantom{\rule{1em}{0ex}}=\frac{1}{s+p+\beta }\sum _{r=0}^{\mathrm{\infty }}\left(\begin{array}{c}s+p+r-1\\ r\end{array}\right)\frac{{\left({v}_{s,p}\left(x\right)\right)}^{r}}{{\left(1+{v}_{s,p}\left(x\right)\right)}^{s+p+r}}\left(r+1\right)\\ & \phantom{\rule{2em}{0ex}}+\frac{\alpha }{s+p+\beta }\sum _{r=0}^{\mathrm{\infty }}\left(\begin{array}{c}s+p+r-1\\ r\end{array}\right)\frac{{\left({v}_{s,p}\left(x\right)\right)}^{r}}{{\left(1+{v}_{s,p}\left(x\right)\right)}^{s+p+r}}\\ & \phantom{\rule{1em}{0ex}}=\frac{\left(s+p\right){v}_{s,p}\left(x\right)}{s+p+\beta }\sum _{r=0}^{\mathrm{\infty }}\left(\begin{array}{c}s+p+r-1\\ r\end{array}\right)\frac{{\left({v}_{s,p}\left(x\right)\right)}^{r}}{{\left(1+{v}_{s,p}\left(x\right)\right)}^{s+p+r}}\\ & \phantom{\rule{2em}{0ex}}+\frac{\alpha +1}{s+p+\beta }\sum _{r=0}^{\mathrm{\infty }}\left(\begin{array}{c}s+p+r-1\\ r\end{array}\right)\frac{{\left({v}_{s,p}\left(x\right)\right)}^{r}}{{\left(1+{v}_{s,p}\left(x\right)\right)}^{s+p+r}}\\ & \phantom{\rule{1em}{0ex}}=\frac{\left(s+p\right){v}_{s,p}\left(x\right)}{s+p+\beta }{\left(1+{v}_{s,p}\left(x\right)-{v}_{s,p}\left(x\right)\right)}^{-\left(s+p+1\right)}+\frac{\alpha +1}{s+p+\beta }{\left(1+{v}_{s,p}\left(x\right)-{v}_{s,p}\left(x\right)\right)}^{-\left(s+p\right)}\\ & \phantom{\rule{1em}{0ex}}=\frac{\left(s+p\right){v}_{s,p}\left(x\right)}{s+p+\beta }+\frac{\alpha +1}{s+p+\beta }.\end{array}$

By the same manner, other results can be obtained. □

### Lemma 3

Let us briefly denote $$\phi _{x}^{k}(t)=(t-x)^{k}$$ for $$k=0,1,2,4$$. Then we obtain the following equalities for the central moments:

\begin{aligned} &M_{s}^{\alpha,\beta }\bigl(\phi _{x}^{0};x\bigr) = 1, \\ &M_{s}^{\alpha,\beta }\bigl(\phi _{x}^{1};x\bigr) = \frac{(s+p)v_{s,p}(x)+\alpha +1}{s+p+\beta }-x, \\ &M_{s}^{\alpha,\beta }\bigl(\phi _{x}^{2};x\bigr) \\ &\quad= \frac{(s+p)(s+p+1)v^{2}_{s,p}(x)+(4+2 \alpha )(s+p)v_{s,p}(x)+\alpha ^{2}+2\alpha +2}{(s+p+\beta )^{2}} \\ &\qquad{}-\frac{2x ( (s+p)v_{s,p}(x)+\alpha +1 ) }{s+p+\beta }+x ^{2}, \\ &M_{s}^{\alpha,\beta }\bigl(\phi _{x}^{4};x\bigr)\\ &\quad= \frac{(s+p)(s+p+1)(s+p+2)(s+p+3)v ^{4}_{s,p}(x)+(16+4\alpha )(s+p)(s+p+1)(s+p+2)v^{3}_{s,p}(x)}{(s+p+ \beta )^{4}} \\ &\qquad{}+\frac{(72+36\alpha +6\alpha ^{2})(s+p)(s+p+1)v^{2}_{s,p}(x)+(96+72 \alpha +24\alpha ^{2}+4\alpha ^{3})(s+p)v_{s,p}(x)}{(s+p+\beta )^{4}} \\ &\qquad{}+\frac{24+24\alpha +12\alpha ^{2}+4\alpha ^{3}+\alpha ^{4}}{(s+p+ \beta )^{4}}-4x \biggl(\frac{(s+p)(s+p+1)(s+p+2)v^{3}_{s,p}(x)}{(s+p+ \beta )^{3}} \\ &\qquad{}+\frac{(9+3\alpha )(s+p)(s+p+1)v^{2}_{s,p}(x)+(3\alpha ^{2}+12 \alpha +18)(s+p)v_{s,p}(x)+\alpha ^{3}+3\alpha ^{2}+6\alpha +6}{(s+p+ \beta )^{3}} \biggr) \\ &\qquad{}+6x^{2} \biggl( \frac{(s+p)(s+p+1)v^{2}_{s,p}(x)+(4+2\alpha )(s+p)v _{s,p}(x)+\alpha ^{2}+2\alpha +2}{(s+p+\beta )^{2}} \biggr) \\ &\qquad{}-4x^{3} \biggl(\frac{(s+p)v_{s,p}(x)+\alpha +1}{s+p+\beta } \biggr) +x^{4}. \end{aligned}

### Proof

We use the linearity of the $$M_{s}^{\alpha,\beta }$$ operators and Lemma 2 $$M_{s}^{\alpha,\beta }(\phi _{x}^{0};x)=M_{s}^{\alpha,\beta }(e_{0};x)$$, $$M_{s}^{\alpha,\beta }(\phi _{x}^{1};x)= M_{s}^{\alpha,\beta }(e_{1};x)-x M_{s}^{\alpha,\beta }(e_{0};x)$$, $$M_{s}^{\alpha,\beta }(\phi _{x}^{2};x)= M_{s}^{\alpha,\beta }(e_{2};x)-2x M_{s}^{\alpha,\beta }(e_{1};x)+ x ^{2}M_{s}^{\alpha,\beta }(e_{0};x)$$, $$M_{s}^{\alpha,\beta }(\phi _{x}^{4};x)= M_{s}^{\alpha,\beta }(e_{4};x)-4x M_{s}^{\alpha,\beta }(e_{3};x)+6x ^{2}M_{s}^{\alpha,\beta }(e_{2};x)-4x^{3}M_{s}^{\alpha,\beta }(e _{1};x)+x^{4}M_{s}^{\alpha,\beta }(e_{0};x)$$. □

### Remark 1

Considering the definition of $$v_{s,p}(x)$$, we obtain the following limits for every $$x\in [0,\infty )$$ and $$0\leq \alpha \leq \beta$$:

$$\lim_{s\rightarrow \infty } sM_{s}^{\alpha,\beta }\bigl( \phi _{x} ^{1};x\bigr)=2ax+ax^{2}$$
(6)

and

$$\lim_{s\rightarrow \infty } sM_{s}^{\alpha,\beta } \bigl(\phi _{x} ^{2};x\bigr)= 2x+x^{2}.$$
(7)

## Main results

Let $${C^{*}[0,\infty )}$$ denote the subspace of all real-valued continuous functions on $${[0,\infty )}$$ with the condition that $$\lim_{m\rightarrow \infty }f(x)$$ exists and is finite, equipped with the uniform norm. The uniform convergence of a sequence of linear positive operators is demonstrated by Boyanov and Veselinov . We present the following theorem according to  for the newly constructed operators (1).

### Theorem 1

If the linear positive operators (1) satisfy

$$\lim_{s\rightarrow \infty } M_{s}^{\alpha,\beta } \bigl(e^{-mt};x\bigr)=e ^{-mx},\quad m=0,1,2,$$
(8)

uniformly in $$[0,\infty )$$, then for each $$f\in {C^{*}[0,\infty )}$$

$$\lim_{s\rightarrow \infty } M_{s}^{\alpha,\beta } (f;x)=f(x)$$
(9)

uniformly in $$[0,\infty )$$.

### Proof

We have already known that $$\lim_{s\rightarrow \infty } M _{s}^{\alpha,\beta } (1;x)=1$$. Considering equality (5) with $$v_{s,p}(x)$$ given in (2), we have

\begin{aligned} M_{s}^{\alpha,\beta }\bigl(e^{-t};x\bigr)& = \frac{s+p+\beta }{s+p+\beta +1}e ^{\frac{-\alpha }{s+p+\beta }} \biggl( 1+ \frac{v_{s,p}(x)}{s+p+\beta +1} \biggr) ^{-(s+p)} \\ &= e^{-x}+\frac{(1-2a)(2+x)xe^{-x}}{2(s+p)}+\mathcal{O} \bigl( {(s+p)^{-2}} \bigr) \end{aligned}
(10)

and

\begin{aligned} M_{s}^{\alpha,\beta }\bigl(e^{-2t};x\bigr) &= \frac{s+p+\beta }{s+p+\beta +2}e ^{\frac{-2\alpha }{s+p+\beta }} \biggl( 1+\frac{2v_{s,p}(x)}{s+p+ \beta +2} \biggr) ^{-(s+p)} \\ &= e^{-2x}+\frac{(1-a)(4+2x)xe^{-2x}}{(s+p)}+\mathcal{O} \bigl( {(s+p)^{-2}} \bigr). \end{aligned}
(11)

Hence, we prove that

$$\lim_{s\rightarrow \infty } M_{s}^{\alpha,\beta } \bigl(e^{-mt};x\bigr)=e ^{-mx},\quad m=0,1,2,$$

uniformly in $$[0,\infty )$$. This means that, for any $$f\in {C^{*}[0, \infty )}$$, $$\lim_{s\rightarrow \infty } M_{s}^{\alpha, \beta } (f;x)=f(x)$$ uniformly in $$[0,\infty )$$. □

After about four decades later than Boyanov and Veselinov , Holhoş  studied the uniform convergence of a sequence of linear positive operators. He obtained the following theorem for an effective estimation of the linear positive operators.

### Theorem 2

()

For a sequence of linear positive operators $$A_{s}: C ^{*}[0,\infty )\longrightarrow C^{*}[0,\infty )$$, we have

$$\Vert A_{s}f-f \Vert _{[0,\infty )}\leq \Vert f \Vert _{[0,\infty )}\delta _{s}+(2+\delta _{s})\omega ^{*}(f,\sqrt{\delta _{s}+2\sigma _{s}+\rho _{s}})$$

for every function $$f\in C^{*}[0,\infty )$$, where

\begin{aligned} &\bigl\Vert A_{s}(e_{0})-1 \bigr\Vert _{[0,\infty )} = \delta _{s}, \\ &\bigl\Vert A_{s}\bigl(e^{-t}\bigr)-e^{-x} \bigr\Vert _{[0,\infty )} = \sigma _{s}, \\ &\bigl\Vert A_{s}\bigl(e^{-2t}\bigr)-e^{-2x} \bigr\Vert _{[0,\infty )} = \rho _{s} \end{aligned}

and $$\omega ^{*}(f,\eta )=\sup_{|e^{-x}-e^{-t}| \leq \eta, x,t>0} |f(t)-f(x)|$$ denotes the modulus of continuity. Here, $$\delta _{s}, \sigma _{s}$$, and $$\rho _{s}$$ tend to zero as $$s\rightarrow \infty$$.

In the same manner with the above theorem, we present a quantitative estimation of the Baskakov–Schurer–Szasz–Stancu operators which preserve $$e^{-2ax}, a>0$$, as follows.

### Theorem 3

For $$f\in C^{*}[0,\infty )$$, we have the following inequality:

$$\bigl\Vert M_{s}^{\alpha,\beta }f-f \bigr\Vert _{[0,\infty )}\leq 2\omega ^{*}(f,\sqrt{2\sigma _{s,p}+\rho _{s,p}}),$$
(12)

where

\begin{aligned} &\bigl\Vert M_{s}^{\alpha,\beta }\bigl(e^{-t} \bigr)-e^{-x} \bigr\Vert _{[0,\infty )} = \sigma _{s,p}, \\ &\bigl\Vert M_{s}^{\alpha,\beta }\bigl(e^{-2t} \bigr)-e^{-2x} \bigr\Vert _{[0,\infty )} = \rho _{s,p}. \end{aligned}

Here, $$\sigma _{s,p}$$ and $$\rho _{s,p}$$ tend to zero as $$s\rightarrow \infty$$. So, $$M_{s}^{\alpha,\beta }f$$ converges to f uniformly.

### Proof

The Baskakov–Schurer–Szasz–Stancu operators $$M_{s}^{\alpha,\beta }$$ preserve constant. Thus, $$\delta _{s,p}= \Vert M_{s}^{\alpha,\beta }(e_{0})-1 \Vert _{[0,\infty )}=0$$. In order to calculate $$\sigma _{s,p}$$, we take into consideration equality (10). So, we obtain

$$M_{s}^{\alpha,\beta }\bigl(e^{-t};x\bigr)- e^{-x}= \frac{(1-2a)(2+x)xe^{-x}}{2(s+p)}+\mathcal{O} \bigl( {(s+p)^{-2}} \bigr).$$

Since

$$\sup_{x\in [0,\infty )} xe^{-x}=\frac{1}{e},\qquad \sup_{x\in [0,\infty )} x^{2}e^{-x}= \frac{4}{e^{2}},$$

we achieve

\begin{aligned} \sigma _{s,p} = \bigl\Vert M_{s}^{\alpha,\beta } \bigl(e^{-t}\bigr)-e^{-x} \bigr\Vert _{[0,\infty )}= \frac{(1-2a)}{e(s+p)}+\frac{2(1-2a)}{e^{2}(s+p)}+ \mathcal{O} \bigl( {(s+p)^{-2}} \bigr). \end{aligned}

In the same way, with the help of equality (11), we have

$$M_{s}^{\alpha,\beta }\bigl(e^{-2t};x\bigr)- e^{-2x}= \frac{(1-a)(4+2x)xe^{-2x}}{(s+p)}+\mathcal{O} \bigl( {(s+p)^{-2}} \bigr).$$

By using

$$\sup_{x\in [0,\infty )} xe^{-2x}=\frac{1}{2e},\qquad \sup_{x\in [0,\infty )} x^{2}e^{-2x}= \frac{1}{e^{2}},$$

we get

$$\rho _{s,p}= \bigl\Vert M_{s}^{\alpha,\beta } \bigl(e^{-2t}\bigr)-e^{-2x} \bigr\Vert _{[0,\infty )}= \frac{2(1-a)}{e(s+p)}+\frac{2(1-a)}{e^{2}(s+p)}+ \mathcal{O} \bigl( {(s+p)^{-2}} \bigr).$$

Consequently, $$\sigma _{s,p}$$ and $$\rho _{s,p}$$ tend to zero as $$s\rightarrow \infty$$. □

In Sect. 4, we investigate the rate of convergence by using the usual modulus of continuity.

## The usual modulus of continuity

The class of all bounded and uniform continuous functions f on $$[0,\infty )$$ is denoted by $$C_{B}[0,\infty )$$ endowed with the norm $$\Vert f \Vert _{C_{B}}=\sup_{x\geq 0} |f(x)|$$. For $$f\in C_{B}[0,\infty )$$, the modulus of continuity is given by

$$\omega (f,\delta ):=\sup_{0< h< \delta } \sup_{ x,x+h \in {[0,\infty )}} \bigl\vert f(x+h)-f(x) \bigr\vert .$$

The second order modulus of continuity of the function $$f\in C_{B}[0, \infty )$$ is defined by

$$\omega _{2}(f,\delta ):=\sup_{0< h< \sqrt{\delta } }\sup _{ x,x+h\in {[0,\infty )}} \bigl\vert f(x+2h)-2f(x+h)+f(x) \bigr\vert ,$$

where $$\delta >0$$. Peetre’s K-functionals are described as

$$K_{2}(f,\delta ):=\inf_{g \in C_{B}^{2}[0,\infty ) }\bigl\{ \Vert f-g \Vert _{C_{B}[0,\infty )}+\delta \Vert g \Vert _{C_{B}^{2}[0,\infty )}\bigr\} .$$

Here, $$C_{B}^{2}[0,\infty )$$ denotes the space of the functions f, for which $$f'$$ and $$f''$$ belong to $$C_{B}[0,\infty )$$. The relation between the second order modulus of continuity and Peetre’s K-functional is given by 

$$K_{2}(f,\delta )\leq M\omega _{2}(f,\sqrt{\delta }),$$

where $$M>0$$.

### Lemma 4

For $$f\in C_{B}[0,\infty )$$, we have $$|M_{s}^{\alpha,\beta } (f;x)| \leq \Vert f \Vert$$.

### Theorem 4

Let $$f\in C_{B}[0,\infty )$$. Then, for all $$x \in {[0,\infty )}$$, there exists a positive constant M such that

\begin{aligned} \bigl\vert M_{s}^{\alpha,\beta }(f;x)-f(x) \bigr\vert \leq M\omega _{2}(f,\sqrt{\mu _{s,p}})+\omega \biggl(f, \biggl\vert \frac{(s+p)v_{s,p}(x)+\alpha +1}{s+p+ \beta } \biggr\vert \biggr), \end{aligned}
(13)

where

\begin{aligned} \mu _{s,p} ={}& \frac{(s+p)(2s+2p+1)}{(s+p+\beta )^{2}}v^{2}_{s,p}(x)+ \biggl(\frac{(4\alpha +6)(s+p)}{(s+p+\beta )^{2}}-\frac{4x(s+p)}{s+p+ \beta } \biggr) v_{s,p}(x) \\ &{}+\frac{2\alpha ^{2}+4\alpha +3}{(s+p+\beta )^{2}}-\frac{4x(\alpha +1)}{s+p+ \beta } +2x^{2}. \end{aligned}
(14)

Here, $$v_{s,p}(x)$$ is the same as in (2).

### Proof

We define the auxiliary operators $$\tilde{M}_{s}^{\alpha,\beta }:C _{B}[0,\infty )\rightarrow C_{B}[0,\infty )$$

$$\tilde{M}_{s}^{\alpha,\beta }(g;x)=M_{s}^{\alpha,\beta }(g;x)+g(x)-g \biggl( \frac{(s+p)v_{s,p}(x)+\alpha +1}{s+p+\beta } \biggr),$$
(15)

where $$v_{s,p}(x)$$ is as given by (2). Note that the operators (15) are positive and linear. By using the Taylor expansion for $$g \in C_{B}^{2}[0,\infty )$$, we have

$$g(t)=g(x)+(t-x)g'(x)+ \int _{x}^{t}(t-u)g''(u) \,du,\quad x,t\in {[0,\infty )}.$$
(16)

Applying $$\tilde{M}_{s}^{\alpha,\beta }$$ operators to the both sides of equation (16) and using Lemma 3, we obtain

\begin{aligned} &\bigl\vert \tilde{M}_{s}^{\alpha,\beta }(g;x)-g(x) \bigr\vert \\ &\quad = \biggl\vert \tilde{M}_{s}^{\alpha,\beta } \biggl( \int _{x}^{t}(t-u)g''(u) \,du;x \biggr) \biggr\vert \\ &\quad\leq \biggl\vert M_{s}^{\alpha,\beta } \biggl( \int _{x}^{t}(t-u)g''(u) \,du;x \biggr) \biggr\vert \\ &\qquad{}+ \biggl\vert \int _{x}^{ \frac{(s+p)v_{s,p}(x)+\alpha +1}{s+p+\beta } } \biggl(\frac{(s+p)v_{s,p}(x)+\alpha +1}{s+p+\beta } -u \biggr)g''(u)\,du \biggr\vert . \end{aligned}
(17)

Further,

\begin{aligned} &\biggl\vert M_{s}^{\alpha,\beta } \biggl( \int _{x}^{t}(t-u)g''(u) \,du;x \biggr) \biggr\vert \\ &\quad\leq M_{s}^{\alpha,\beta } \biggl( \int _{x}^{t} \vert t-u \vert \bigl\vert g''(u) \bigr\vert \,du;x \biggr) \leq \bigl\Vert g'' \bigr\Vert M_{s}^{\alpha,\beta }\bigl( \phi _{x}^{2};x\bigr) \end{aligned}
(18)

and

\begin{aligned} & \biggl\vert \int _{x}^{\frac{(s+p)v_{s,p}(x)+\alpha +1}{s+p+\beta }} \biggl( \frac{(s+p)v_{s,p}(x)+\alpha +1}{s+p+\beta } -u \biggr)g''(u)\,du \biggr\vert \\ &\quad \leq \bigl\Vert g'' \bigr\Vert \biggl(\frac{(s+p)v_{s,p}(x)+\alpha +1}{s+p+\beta }-x \biggr) ^{2}. \end{aligned}
(19)

Rewrite (18) and (19) in (17), then we have

\begin{aligned} &\bigl\vert \tilde{M}_{s}^{\alpha,\beta }(g;x)-g(x) \bigr\vert \\ &\quad \leq \bigl\Vert g'' \bigr\Vert \biggl(M_{s}^{\alpha,\beta } \bigl(\phi _{x}^{2};x\bigr)+ \biggl(\frac{(s+p)v _{s,p}(x)+\alpha +1}{s+p+\beta } -x \biggr)^{2} \biggr) \\ &\quad = \bigl\Vert g'' \bigr\Vert \biggl( \frac{(s+p)(s+p+1)v^{2}_{s,p}(x)+(4+2\alpha )(s+p)v _{s,p}(x)+\alpha ^{2}+2\alpha +2}{(s+p+\beta )^{2}} \\ &\qquad{} -\frac{2x ( (s+p)v_{s,p}(x)+\alpha +1 ) }{s+p+ \beta }+x^{2} + \biggl(\frac{(s+p)v_{s,p}(x)+\alpha +1}{s+p+\beta } -x \biggr) ^{2} \biggr) \\ &\quad= \bigl\Vert g'' \bigr\Vert \biggl( \frac{(s+p)(2s+2p+1)}{(s+p+\beta )^{2}}v^{2}_{s,p}(x)+ \biggl(\frac{(4\alpha +6)(s+p)}{(s+p+\beta )^{2}}- \frac{4x(s+p)}{s+p+ \beta } \biggr) v_{s,p}(x) \\ &\qquad{}+\frac{2\alpha ^{2}+4\alpha +3}{(s+p+\beta )^{2}}-\frac{4x( \alpha +1)}{s+p+\beta } +2x^{2} \biggr) \\ &\quad:= \bigl\Vert g'' \bigr\Vert \mu _{s,p}, \end{aligned}
(20)

where

\begin{aligned} \mu _{s,p} ={}& \frac{(s+p)(2s+2p+1)}{(s+p+\beta )^{2}}v^{2}_{s,p}(x)+ \biggl(\frac{(4\alpha +6)(s+p)}{(s+p+\beta )^{2}}-\frac{4x(s+p)}{s+p+ \beta } \biggr) v_{s,p}(x) \\ &{}+\frac{2\alpha ^{2}+4\alpha +3}{(s+p+\beta )^{2}}-\frac{4x(\alpha +1)}{s+p+ \beta } +2x^{2}. \end{aligned}
(21)

By using the auxiliary operators (15) and Lemma 4, we get

$$\bigl\Vert \tilde{M}_{s}^{\alpha,\beta }(f;x) \bigr\Vert \leq \bigl\Vert M_{s}^{\alpha,\beta }(f;x) \bigr\Vert +2 \Vert f \Vert \leq 3 \Vert f \Vert .$$
(22)

From (15), (20), and (22), for every $$g\in {C_{B} ^{2}[0,\infty )}$$, we obtain

\begin{aligned} &\bigl\vert M_{s}^{\alpha,\beta }(f;x)-f(x) \bigr\vert \\ &\quad = \biggl\vert \tilde{M}_{s}^{\alpha, \beta }(f;x)-f(x)+f \biggl( \frac{(s+p)v_{s,p}(x)+\alpha +1}{s+p+ \beta } \biggr) -f(x) \\ &\qquad{}+ \tilde{M}_{s}^{\alpha,\beta }(g;x)- \tilde{M}_{s}^{\alpha ,\beta }(g;x)+ g(x)-g(x) \biggr\vert \\ &\quad\leq \bigl\vert \tilde{M}_{s}^{\alpha,\beta }(f-g;x)-(f-g) (x) \bigr\vert + \biggl\vert f \biggl( \frac{(s+p)v_{s,p}(x)+\alpha +1}{s+p+\beta } \biggr) -f(x) \biggr\vert \\ &\qquad{}+ \bigl\vert \tilde{M}_{s}^{\alpha,\beta }(g;x)-g(x) \bigr\vert \\ &\quad\leq 4 \Vert f-g \Vert + \bigl\Vert g'' \bigr\Vert \mu _{s,p}+ \biggl\vert f \biggl( \frac{(s+p)v_{s,p}(x)+ \alpha +1}{s+p+\beta } \biggr) -f(x) \biggr\vert \\ &\quad \leq K_{2}(f,\mu _{s,p})+\omega \biggl(f, \biggl\vert \frac{(s+p)v_{s,p}(x)+ \alpha +1}{s+p+\beta } -x \biggr\vert \biggr) \\ &\quad \leq M\omega _{2}(f,\sqrt{\mu _{s,p}})+\omega \biggl(f, \biggl\vert \frac{(s+p)v _{s,p}(x)+\alpha +1}{s+p+\beta }-x \biggr\vert \biggr). \end{aligned}
(23)

□

### Remark 2

We see that $$\mu _{s,p}=\frac{x(2+x)}{s+p}+O((s+p)^{-2})\rightarrow 0$$, when $$s\rightarrow \infty$$. This result guarantees the convergence of Theorem 4.

In Sect. 5, we obtain the rate of convergence by using the exponential modulus of continuity.

## The exponential modulus of continuity

For $$f\in C[0,\infty )$$, the exponential growth of order $$B>0$$ is given by

$$\Vert f \Vert _{B}:=\sup_{x\in {[0,\infty )}} \bigl\vert f(x)e^{-Bx} \bigr\vert < \infty.$$
(24)

The first order modulus of continuity of functions with exponential growth is defined as

$$\omega _{1}(f,\delta,B)=\mathop{\sup_{h\leq \delta }}_{x\in {[0, \infty )}} \bigl\vert f(x)-f(x+h) \bigr\vert e^{-Bx}.$$
(25)

Let $$f\in \operatorname{Lip}(c,B)$$ for some $$0< c\leq 1$$. Then, for each $$\delta <1$$,

$$\omega _{1}(f,\delta,B)\leq M\delta ^{c}.$$
(26)

Let K be a subspace of $$C[0,\infty )$$ which contains functions f with exponential growth, $$\Vert f \Vert _{B}<\infty$$.

### Theorem 5

Let $${M}_{s}^{\alpha,\beta }:K\rightarrow C[0,\infty )$$ be the sequence of linear positive operators preserving $$e^{-2ax}$$, $$a>0$$. We assume that $${M}_{s}^{\alpha,\beta }$$ satisfy

$${M}_{s}^{\alpha,\beta }\bigl((t-x)^{2}e^{Bt};x \bigr)\leq C_{a}(B,x) {M}_{s} ^{\alpha,\beta }\bigl(\phi _{x}^{2};x\bigr)$$
(27)

for fixed $$x\in {[0,\infty )}$$ and for $$B>0$$. Additionally, if $$f\in C^{2}[0,\infty )\cap K$$, $$0< c\leq 1$$, and $$f''\in \operatorname{Lip}(c,B)$$, then for fixed $$x\in {[0,\infty )}$$, we have

\begin{aligned} &\biggl\vert {M}_{s}^{\alpha,\beta }(f;x)-f(x)-f'(x){M}_{s}^{\alpha, \beta } \bigl(\phi _{x}^{1};x\bigr)-\frac{f''(x)}{2}{M}_{s}^{\alpha,\beta } \bigl( \phi _{x}^{2};x \bigr) \biggr\vert \\ &\quad\leq {M}_{s}^{\alpha,\beta } \bigl( \phi _{x}^{2};x \bigr) \biggl(\frac{ \sqrt{C_{a}(2B,x)}}{2}+\frac{C_{a}(B,x)}{2}+e^{2Bx} \biggr) \omega _{1} \biggl( f'',\sqrt{ \frac{{M}_{s}^{\alpha,\beta } ( \phi _{x}^{4};x )}{ {M}_{s}^{\alpha,\beta } ( \phi _{x}^{2};x )}},B \biggr). \end{aligned}

### Proof

We begin with the Taylor expansion of the function $$f\in C^{2}[0,\infty )$$ at $$x\in {[0,\infty )}$$.

$$f(t)=f(x)+(t-x)f'(x)+\frac{(t-x)^{2}}{2!}f''(x)+H_{2}(f;t,x),$$
(28)

where $$H_{2}(f;t,x)= \frac{ ( f''(\eta )-f''(x) ) (t-x)^{2}}{2}$$ is the remainder term. Here, η is between t and x. Applying the operators $${M}_{s}^{\alpha,\beta }$$ to equality (28), we obtain

\begin{aligned} &\biggl\vert {M}_{s}^{\alpha,\beta }(f;x)-f(x)-f'(x){M}_{s}^{\alpha, \beta } \bigl(\phi _{x}^{1};x\bigr)-\frac{f''(x)}{2}{M}_{s}^{\alpha,\beta } \bigl( \phi _{x}^{2};x \bigr) \biggr\vert \\ &\quad = \bigl\vert {M}_{s}^{\alpha,\beta }\bigl(H _{2}(f;t,x);x\bigr) \bigr\vert \\ &\quad\leq {M}_{s}^{\alpha,\beta }\bigl( \bigl\vert H_{2}(f;t,x) \bigr\vert ;x \bigr). \end{aligned}
(29)

Here,

\begin{aligned} H_{2}(f;t,x)=\frac{ ( f''(\eta )-f''(x) ) (t-x)^{2}}{2} \leq \frac{ (t-x)^{2}}{2} \textstyle\begin{cases} e^{Bx} \omega _{1}(f'',h,B),& \vert t-x \vert \leq h, \\ e^{Bx} \omega _{1}(f'',kh,B),& h\leq \vert t-x \vert \leq kh. \end{cases}\displaystyle \end{aligned}

Tachev et al.  proved that, for each $$h>0$$ and $$k\in \mathbb{N}$$,

$$\omega _{1}(f,kh,B)\leq ke^{B(k-1)h} \omega _{1}(f,h,B).$$
(30)

By using inequality (30), we get

\begin{aligned} &\frac{ e^{Bx}(t-x)^{2}}{2}\omega _{1}\bigl(f'',kh,B \bigr) \\ &\quad\leq \frac{ e^{Bx}(t-x)^{2}}{2} ke^{B(k-1)h} \omega _{1} \bigl(f'',h,B\bigr) \\ &\quad\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'',h,B \bigr) \\ &\quad \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'',h,B\bigr). \end{aligned}

Therefore,

\begin{aligned} \bigl\vert H_{2}(f;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'',h,B\bigr). \end{aligned}
(31)

Applying the operators $${M}_{s}^{\alpha,\beta }$$ to inequality (31), we write

\begin{aligned} &{M}_{s}^{\alpha,\beta }\bigl( \bigl\vert H_{2}(f;t,x) \bigr\vert ;x \bigr)\\ &\quad \leq \frac{1}{2} {M}_{s}^{\alpha,\beta } \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'',h,B\bigr) \\ &\quad= \biggl( \frac{1}{2h} {M}_{s}^{\alpha,\beta } \bigl( \vert t-x \vert ^{3}e ^{Bt};x \bigr)+ \frac{1}{2} {M}_{s}^{\alpha,\beta } \bigl( \vert t-x \vert ^{2}e ^{Bt};x \bigr) \\ &\qquad{}+ \frac{e^{2Bx}}{2h} {M}_{s}^{\alpha,\beta } \bigl( \vert t-x \vert ^{3};x \bigr)+ \frac{e^{2Bx}}{2} {M}_{s}^{\alpha,\beta } \bigl( \vert t-x \vert ^{2};x \bigr) \biggr) \omega _{1}\bigl(f'',h,B\bigr). \end{aligned}

By some computations we obtain

\begin{aligned} &{M}_{s}^{\alpha,\beta } \bigl( \vert t-x \vert ^{2}e^{Bt};x \bigr) \\ &\quad = {M}_{s} ^{\alpha,\beta } \bigl( t^{2}e^{Bt};x \bigr)-2x {M}_{s}^{\alpha, \beta } \bigl( te^{Bt};x \bigr)+x^{2} {M}_{s}^{\alpha,\beta } \bigl( e^{Bt};x \bigr) \\ &\quad = e^{\frac{B\alpha }{s+p+\beta }}\frac{(s+p)(s+p+1)(s+p+\beta )^{3}v _{s,p}(x)^{2}}{(s+p+\beta -B)^{5}} \biggl(1-\frac{Bv_{s,p}(x)}{s+p+ \beta -B} \biggr)^{-(s+p+2)} \\ &\qquad{}+e^{\frac{B\alpha }{s+p+\beta }} \biggl( \frac{4(s+p)(s+p+\beta )^{2}v _{s,p}(x)}{(s+p+\beta -B)^{4}}+\frac{2\alpha (s+p)(s+p+\beta )v_{s,p}(x)}{(s+p+ \beta -B)^{3}} \\ &\qquad{}- \frac{2x(s+p)(s+p+\beta )^{2}v_{s,p}(x)}{(s+p+\beta -B)^{3}} \biggr) \biggl(1-\frac{Bv_{s,p}(x)}{s+p+\beta -B} \biggr)^{-(s+p+1)} \\ &\qquad{}+e^{\frac{B\alpha }{s+p+\beta }} \biggl(\frac{2(s+p+\beta )}{(s+p+ \beta -B)^{3}}+\frac{2\alpha }{(s+p+\beta -B)^{2}}+ \frac{\alpha ^{2}}{(s+p+ \beta )(s+p+\beta -B)} \\ &\qquad{}-\frac{2x(s+p+\beta )}{(s+p+\beta -B)^{2}}-\frac{2x\alpha }{(s+p+ \beta -B)}+\frac{x^{2}(s+p+\beta )}{(s+p+\beta -B)} \biggr) \biggl(1-\frac{Bv _{s,p}(x)}{s+p+\beta -B} \biggr)^{-(s+p)} \\ &\quad = e^{Bx} \biggl(1+ \frac{B(12+12(1+2a+B)x+4(1+6a+3B)x^{2}+3(2a+B)x^{3})}{2(2+x) (s+p)}\\ &\qquad{}+O\bigl((s+p)^{-2} \bigr) \biggr) {M}_{s}^{\alpha,\beta } \bigl( \phi _{x}^{2};x \bigr). \end{aligned}

Since $$s+p\geq 1$$,

$${M}_{s}^{\alpha,\beta } \bigl( \vert t-x \vert ^{2}e^{Bt};x \bigr)\leq C_{a}(B,x) {M}_{s}^{\alpha,\beta } \bigl( \phi _{x}^{2};x \bigr).$$
(32)

We have the following inequalities with the help of Cauchy–Schwarz inequality:

\begin{aligned} &{M}_{s}^{\alpha,\beta } \bigl( \vert t-x \vert ^{3}e^{Bt};x \bigr) \\ &\quad \leq \sqrt{ {M}_{s}^{\alpha,\beta } \bigl( \vert t-x \vert ^{2}e^{2Bt};x \bigr)} \sqrt{ {M}_{s}^{\alpha,\beta } \bigl( \vert t-x \vert ^{4};x \bigr)} \\ &\quad \leq \sqrt{C_{a}(2B,x){M}_{s}^{\alpha,\beta } \bigl( \phi _{x} ^{2};x \bigr) } \sqrt{ {M}_{s}^{\alpha,\beta } \bigl( \phi _{x}^{4};x \bigr)}, \end{aligned}
(33)
\begin{aligned} &{M}_{s}^{\alpha,\beta } \bigl( \vert t-x \vert ^{3};x \bigr) \\ &\quad \leq \sqrt{ {M}_{s}^{\alpha,\beta } \bigl( \vert t-x \vert ^{4};x \bigr)}\sqrt{ {M}_{s} ^{\alpha,\beta } \bigl( \vert t-x \vert ^{2};x \bigr)} \\ &\quad \leq \sqrt{{M}_{s}^{\alpha,\beta } \bigl( \phi _{x}^{4};x \bigr) }\sqrt{{M}_{s}^{\alpha,\beta } \bigl( \phi _{x}^{2};x \bigr)}. \end{aligned}
(34)

Thus, by using inequalities (32), (33), and (34) in (29), we write

\begin{aligned} &\biggl\vert {M}_{s}^{\alpha,\beta }(f;x)-f(x)-f'(x){M}_{s}^{\alpha, \beta } \bigl(\phi _{x}^{1};x\bigr)-\frac{f''(x)}{2}{M}_{s}^{\alpha,\beta } \bigl( \phi _{x}^{2};x \bigr) \biggr\vert \\ &\quad\leq \biggl( \frac{1}{2h}\sqrt{C_{a}(2B,x) {M}_{s}^{\alpha, \beta } \bigl( \phi _{x}^{2};x \bigr) }\sqrt{ {M}_{s}^{\alpha, \beta } \bigl( \phi _{x}^{4};x \bigr)} \\ &\qquad{} + \frac{1}{2} C_{a}(B,x) {M}_{s}^{\alpha,\beta } \bigl( \phi _{x}^{2};x \bigr)+ \frac{e^{2Bx}}{2h}\sqrt {{M}_{s}^{\alpha, \beta } \bigl( \phi _{x}^{4};x \bigr) }\sqrt{{M}_{s}^{\alpha, \beta } \bigl( \phi _{x}^{2};x \bigr)} \\ &\qquad{} + \frac{e^{2Bx}}{2}{M}_{s}^{\alpha,\beta } \bigl( \phi _{x} ^{2};x \bigr) \biggr) \omega _{1} \bigl(f'',h,B\bigr). \end{aligned}
(35)

Finally, when we choose $$h=\sqrt{\frac{{M}_{s}^{\alpha,\beta } ( \phi _{x}^{4};x )}{ {M}_{s}^{\alpha,\beta } ( \phi _{x}^{2};x )}}$$ and substitute it in (35), we obtain

\begin{aligned} &\biggl\vert {M}_{s}^{\alpha,\beta }(f;x)-f(x)-f'(x){M}_{s}^{\alpha, \beta } \bigl(\phi _{x}^{1};x\bigr)-\frac{f''(x)}{2}{M}_{s}^{\alpha,\beta } \bigl( \phi _{x}^{2};x \bigr) \biggr\vert \\ &\quad \leq {M}_{s}^{\alpha,\beta } \bigl( \phi _{x}^{2};x \bigr) \biggl(\frac{ \sqrt{C_{a}(2B,x)}}{2}+\frac{C_{a}(B,x)}{2}+e^{2Bx} \biggr) \omega _{1} \biggl( f'',\sqrt{ \frac{ {M}_{s}^{\alpha,\beta } ( \phi _{x}^{4};x )}{{M}_{s}^{\alpha,\beta } ( \phi _{x}^{2};x )}},B \biggr). \end{aligned}

Note that, for fixed $$x\in {[0,\infty )}$$, $$\frac{{M}_{s}^{\alpha, \beta } ( \phi _{x}^{4};x )}{ {M}_{s}^{\alpha,\beta } ( \phi _{x}^{2};x )}=\frac{5x(2+x)}{s+p}+O((s+p)^{-2})\rightarrow 0$$ as $$s\rightarrow \infty$$. This result guarantees the convergence of Theorem 5. □

In Sect. 6, we give the Voronovskaya-type theorem to examine the asymptotic behavior of the constructed operators (1). For the quantitative Voronovskaya-type theorems, we refer to the pioneering works  and .

## Voronovskaya-type theorem

### Theorem 6

For $$f,f''\in C^{*}[0,\infty )$$ and $$x\in {[0,\infty )}$$, we have the inequality

\begin{aligned} &\biggl\vert s \bigl( {M}_{s}^{\alpha,\beta }(f;x)-f(x) \bigr)- \bigl(2ax+ax ^{2}\bigr)f'(x)-\biggl(x+\frac{x^{2}}{2} \biggr)f''(x) \biggr\vert \\ &\quad \leq \bigl\vert r_{s,p}(x) \bigr\vert \bigl\vert f'(x) \bigr\vert \\ &\qquad{}+ \bigl\vert t_{s,p}(x) \bigr\vert \bigl\vert f''(x) \bigr\vert +2\bigl(2t_{s,p}(x)+2x+x^{2}+z_{s,p}(x) \bigr)\omega ^{*}\bigl(f'',s ^{-1/2} \bigr), \end{aligned}

where

\begin{aligned} &r_{s,p}(x) = s {M}_{s}^{\alpha,\beta }\bigl(\phi _{x}^{1};x\bigr)-\bigl(2ax+ax^{2}\bigr), \\ &t_{s,p}(x) = \frac{s}{2} {M}_{s}^{\alpha,\beta }\bigl( \phi _{x}^{2};x\bigr)- \biggl(x+\frac{x^{2}}{2} \biggr), \\ &z_{s,p}(x) = s^{2} \sqrt{ {M}_{s}^{\alpha,\beta } \bigl(\bigl(e^{-x}-e^{-t}\bigr)^{4};x\bigr)}\sqrt { {M} _{s}^{\alpha,\beta }\bigl(\phi _{x}^{4};x \bigr)}. \end{aligned}

### Proof

By the Taylor expansion for a function f, we write

$$f(t)=f(x)+(t-x)f'(x)+\frac{(t-x)^{2}}{2}f''(x)+k(t,x) (t-x)^{2},$$
(36)

where

$$k(t,x):=\frac{f''(\xi )-f''(x)}{2}.$$

Here, $$k(t,x)$$ is the remainder term and ξ is a number between x and t. Applying the $${M}_{s}^{\alpha,\beta }$$ operators to (36), we obtain

$${M}_{s}^{\alpha,\beta }(f;x)-f(x)=f'(x){M}_{s}^{\alpha,\beta } \bigl(\phi _{x}^{1};x\bigr)+\frac{f''(x)}{2}{M}_{s}^{\alpha,\beta } \bigl(\phi _{x}^{2};x\bigr)+ {M}_{s}^{\alpha,\beta } \bigl(k(t,x)\phi _{x}^{2};x\bigr).$$

Then

\begin{aligned} &\biggl\vert s \bigl[ {M}_{s}^{\alpha,\beta }(f;x)-f(x) \bigr]- \bigl(2ax+ax ^{2}\bigr)f'(x)-\biggl(x+\frac{x^{2}}{2} \biggr)f''(x) \biggr\vert \\ & \quad\leq \bigl\vert s{M}_{s}^{\alpha,\beta }\bigl(\phi _{x}^{1};x\bigr)-\bigl(2ax+ax^{2}\bigr) \bigr\vert \bigl\vert f'(x) \bigr\vert +\frac{1}{2} \bigl\vert s{M}_{s}^{\alpha,\beta }\bigl(\phi _{x} ^{2};x\bigr)-\bigl(2x+x^{2}\bigr) \bigr\vert \bigl\vert f''(x) \bigr\vert \\ &\qquad{}+ \bigl\vert s{M}_{s}^{\alpha,\beta }\bigl(k(t,x)\phi _{x}^{2};x\bigr) \bigr\vert . \end{aligned}

We briefly denote that $$r_{s,p}(x):=s{M}_{s}^{\alpha,\beta }(\phi _{x}^{1};x)-(2ax+ax^{2})$$ and $$t_{s,p}(x):=\frac{s}{2} {M}_{s}^{\alpha,\beta }(\phi _{x}^{2};x)- (x+\frac{x^{2}}{2} )$$. Thus,

\begin{aligned} &\biggl\vert s \bigl[{M}_{s}^{\alpha,\beta }(f;x)-f(x) \bigr]- \bigl(2ax+ax^{2}\bigr)f'(x)-\biggl(x+\frac{x ^{2}}{2} \biggr)f''(x) \biggr\vert \\ &\quad \leq \bigl\vert r_{s,p}(x) \bigr\vert \bigl\vert f'(x) \bigr\vert + \vert t_{s,p} \vert \bigl\vert f''(x) \bigr\vert \\ &\qquad{}+ \bigl\vert s {M}_{s}^{\alpha,\beta }\bigl(k(t,x)\phi _{x}^{2};x\bigr) \bigr\vert . \end{aligned}

Note that by using equalities (6) and (7), $$r_{s,p}(x)$$ and $$t_{s,p}(x)$$ go to zero as $$s\rightarrow \infty$$. Now, we deal with the term $$\vert s{M}_{s}^{\alpha,\beta }(k(t,x)\phi _{x}^{2};x) \vert$$.

$$\bigl\vert f(t)-f(x) \bigr\vert \leq \biggl(1+\frac{(e^{-x}-e^{-t})^{2}}{ \eta ^{2}} \biggr) \omega ^{*}(f,\eta ).$$

Using this inequality, we have

$$\bigl\vert k(t,x) \bigr\vert \leq \biggl(1+\frac{(e^{-x}-e^{-t})^{2}}{\eta ^{2}} \biggr) \omega ^{*}\bigl(f'',\eta \bigr).$$

For $$\eta >0$$, if $$|e^{-x}-e^{-t}|>\eta$$, then $$\vert k(t,x) \vert \leq \frac{2(e^{-x}-e^{-t})^{2}}{\eta ^{2}} \omega ^{*}(f'',\eta )$$ and if $$|e^{-x}-e^{-t}|\leq \eta$$, then $$\vert k(t,x) \vert \leq 2 \omega ^{*}(f'',\eta )$$. Thus, we write $$\vert k(t,x) \vert \leq 2 (1+\frac{(e^{-x}-e^{-t})^{2}}{\eta ^{2}} ) \omega ^{*}(f'',\eta )$$. Therefore,

\begin{aligned} &\bigl\vert s{M}_{s}^{\alpha,\beta }\bigl(k(t,x)\phi _{x}^{2};x\bigr) \bigr\vert \\ &\quad \leq s{M}_{s}^{\alpha,\beta } \bigl( \bigl\vert k(t,x) \bigr\vert \phi _{x}^{2};x \bigr) \\ &\quad\leq 2s\omega ^{*}\bigl(f'',\eta \bigr){M}_{s}^{\alpha,\beta }\bigl(\phi _{x}^{2};x \bigr)+\frac{2s}{ \eta ^{2}}\omega ^{*}\bigl(f'', \eta \bigr){M}_{s}^{\alpha,\beta } \bigl( \bigl(e^{-x}-e ^{-t}\bigr)^{2}\phi _{x}^{2};x \bigr) \\ &\quad\leq 2s\omega ^{*}\bigl(f'',\eta \bigr){M}_{s}^{\alpha,\beta }\bigl(\phi _{x}^{2};x \bigr) \\ &\qquad{}+\frac{2s}{\eta ^{2}}\omega ^{*}\bigl(f'', \eta \bigr)\sqrt{{M}_{s}^{\alpha, \beta } \bigl( \bigl(e^{-x}-e^{-t}\bigr)^{4};x \bigr) }\sqrt {{M}_{s}^{\alpha ,\beta } \bigl( \phi _{x}^{4};x \bigr) }. \end{aligned}

If we choose $$\eta =1/\sqrt{s}$$ and $$z_{s,p}:=\sqrt{s^{2}{M}_{s} ^{\alpha,\beta } ( (e^{-x}-e^{-t})^{4};x ) }\sqrt{s^{2} {M}_{s}^{\alpha,\beta } ( \phi _{x}^{4};x ) }$$, we get

\begin{aligned} &\biggl\vert s \bigl({M}_{s}^{\alpha,\beta }(f;x)-f(x) \bigr)- \bigl(2ax+ax ^{2}\bigr)f'(x)-\biggl(x+\frac{x^{2}}{2} \biggr)f''(x) \biggr\vert \\ &\quad \leq \bigl\vert r_{s,p}(x) \bigr\vert \bigl\vert f'(x) \bigr\vert \\ &\qquad{}+ \bigl\vert t_{s,p}(x) \bigr\vert \bigl\vert f''(x) \bigr\vert +\bigl(4t_{s,p}(x)+ 4x+2x^{2}+2z_{s,p}(x)\bigr)\omega ^{*} \bigl(f'',s^{-1/2}\bigr). \end{aligned}

□

### Remark 3

We obtain the following result by some calculations:

$$\lim_{s\rightarrow \infty } s^{2}{M}_{s}^{\alpha,\beta } \bigl( \phi _{x}^{4};x\bigr)=3x^{2}(2+x)^{2}.$$
(37)

Additionally, we get the following result:

$$\lim_{s\rightarrow \infty }s^{2}{M}_{s}^{\alpha,\beta } \bigl(\bigl(e ^{-t}-e^{-x}\bigr)^{4};x \bigr)=3x^{2}(2+x)^{2}e^{-4x}.$$
(38)

We give the following corollary as a result of Theorem 6 and Remark 3.

### Corollary 1

Suppose that $$f,f''\in C^{*}[0,\infty )$$ and $$x\in {[0,\infty )}$$. Then the equality

$$\lim_{s\rightarrow \infty } s \bigl( {M}_{s}^{\alpha,\beta }(f;x)-f(x) \bigr) =\bigl(2ax+ax^{2}\bigr)f'(x)+\biggl(x+ \frac{x^{2}}{2}\biggr)f''(x)$$
(39)

holds.

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