# 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.

## 1 Introduction

Use of linear positive operators has played a crucial role in approximation theory for the last seven decades. In 1950, Szász [22] 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 [6] 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 [20] 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 [21] 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 [12], Baskakov–Schurer–Szász operators [18], Baskakov–Szász–Stancu operators [17], and q-Baskakov–Schurer–Szász–Stancu operators [19], have been conducted.

In 2010, Aldaz and Render [4] introduced linear and positive operators preserving 1 and $$e^{x}$$. In 2017, Acar et al. [2] 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 [5], Baskakov–Szász–Mirakyan-type operators [10], Phillips operators [13], Szász–Mirakyan–Kantorovich operators [11], and Baskakov operators [24] preserving constant and exponential function were examined. In addition, Kajla [16] studied Srivastava–Gupta operators preserving linear functions. On the other hand, Gupta and Tachev [14] found the general estimation in terms of Pǎltǎneaś modulus of continuity.

In 2018, Bodur et al. [7] 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 [18]. In addition, by taking $$s+p=u$$, we get Baskakov–Szász–Stancu operators [17]. Moreover, by taking $$s+p=u$$, $$\alpha =0$$ and $$\beta =0$$, we have Baskakov–Szász operators [12]. 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)

## 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)

## 3 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 [8]. We present the following theorem according to [8] 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 [8], Holhoş [15] 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

([15])

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.

## 4 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 [9]

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

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.

## 5 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. [23] 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 [1] and [3].

## 6 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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Sofyalıoğlu, M., Kanat, K. Approximation properties of generalized Baskakov–Schurer–Szasz–Stancu operators preserving $$e^{-2ax}, a>0$$. J Inequal Appl 2019, 112 (2019). https://doi.org/10.1186/s13660-019-2062-2

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### Keywords

• Exponential functions
• Modulus of continuity
• Voronovskaya-type theorem