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].