Convergence Estimates for Gamma Operator

Abstract

Gamma operator is based on the gamma function and has interesting approximation properties. In the present note, we present some convergence estimates of the Gamma operators based on different modulus of continuities.

1 Introduction

The gamma transform of a function f is given as

$$\begin{aligned} \Gamma _{\alpha ,\beta }f=\frac{\beta ^\alpha }{\Gamma (\alpha )}\int _0^\infty \hbox {e}^{-\beta t}t^{\alpha -1}f(t)\hbox {d}t \end{aligned}$$

where $f\in L_{1,loc}(0,\infty )$ such that $\Gamma _{\alpha ,\beta }|f|<\infty .$ By simple computation, $\Gamma _{\alpha ,\beta }e_r=\frac{(\alpha )_r}{\beta ^r},$ where $e_r(t)=t^r, r=0,1,2,\ldots$ and the Pochhammer symbol $(\alpha )_k=\alpha (\alpha +1)\cdots (\alpha +k-1), (\alpha )_0=1$. In order to preserve the linear functions, Miheşan in [8] considered $\beta =\alpha /x$ and the Gamma operators take the following form:

$$\begin{aligned} G_{\alpha }(f,x)= & {} \left( \frac{\alpha }{x}\right) ^\alpha \frac{1}{\Gamma (\alpha )}\int _0^\infty t^{\alpha -1}\hbox {e}^{-\alpha t/x}f(t)\hbox {d}t. \end{aligned}$$

(1)

As some of the special cases, if

• If $\alpha =nx$, we get Rathore operators [10], defined as

  $$\begin{aligned} R_{n}(f,x)= & {} \frac{n^{nx}}{\Gamma (nx)}\int _0^\infty t^{nx-1}\hbox {e}^{-n t}f(t)\hbox {d}t. \end{aligned}$$

• If $\alpha =n$, we get Post–Widder operators, (see [7]) defined as

  $$\begin{aligned} P_{n}(f,x)= & {} \left( \frac{n}{x}\right) ^n \frac{1}{\Gamma (n)}\int _0^\infty t^{n-1}\hbox {e}^{-n t/x}f(t)\hbox {d}t. \end{aligned}$$

Obviously, the well-known Lupaş operators $L_n$ and the Baskakov operators $V_n$ are composition of Rathore–Szász and Post–Widder–Szász operators, respectively, and one has

$$\begin{aligned} L_n(f,x):=(R_{n}\circ S_n)f=\sum _{k=0}^\infty 2^{-nx}\frac{(nx)_k}{k!.2^k}f\left( \frac{k}{n}\right) \end{aligned}$$

and

$$\begin{aligned} V_n(f,x):=(P_{n}\circ S_n)f=\sum _{k=0}^\infty {{n+k-1}\atopwithdelims ()k}\frac{x^k}{(1+x)^{n+k}}f\left( \frac{k}{n}\right) \end{aligned}$$

where the Szász operators are defined by

$$\begin{aligned} S_n(f,x)=\sum _{k=0}^\infty \hbox {e}^{-nx}\frac{(nx)^k}{k!}f\left( \frac{k}{n}\right) . \end{aligned}$$

In the recent years, several approximation properties concerning convergence behavior of different linear positive operators have been discussed by many researchers. Concerning convergence of linear positive operators, we refer the readers to [2, 4, 5], etc.

Remark 1

The rth-order $r\in \mathbb {N}\cup \{0\}$ moment with $e_r(t)=t^r$ satisfies the relation:

$$\begin{aligned} G_{\alpha }(e_r,x)= & {} \frac{(\alpha )_r.x^r}{\alpha ^r}. \end{aligned}$$

(2)

In particular,

$$\begin{aligned} G_{\alpha }(1,x)= & {} 1\\ G_{\alpha }(e_1,x)= & {} x\\ G_{\alpha }(e_2,x)= & {} \left[ 1+\frac{1}{\alpha }\right] x^2\\ G_{\alpha }(e_3,x)= & {} \left[ 1+\frac{3}{\alpha }+\frac{2}{\alpha ^2}\right] x^3\\ G_{\alpha }(e_4,x)= & {} \left[ 1+\frac{6}{\alpha }+\frac{11}{\alpha ^2}+\frac{6}{\alpha ^3}\right] x^4\\ G_{\alpha }(e_5,x)= & {} \left[ 1+\frac{10}{\alpha }+\frac{35}{\alpha ^2}+\frac{50}{\alpha ^3}+\frac{24}{\alpha ^4}\right] x^5\\ G_{\alpha }(e_6,x)= & {} \left[ 1+\frac{15}{\alpha }+\frac{85}{\alpha ^2}+\frac{225}{\alpha ^3}+\frac{274}{\alpha ^4}+\frac{120}{\alpha ^5}\right] x^6.\\ \end{aligned}$$

Also, the moment-generating function of $G_\alpha$ is given by

$$\begin{aligned} G_{\alpha }(\hbox {e}^{\theta t},x)= & {} \left( 1-\frac{\theta x}{\alpha }\right) ^{-\alpha }\\= & {} 1+\theta x+\frac{\alpha (\alpha +1)}{\alpha ^2}\frac{\theta ^2x^2}{2!}+\frac{\alpha (\alpha +1)(\alpha +2)}{\alpha ^3}\frac{\theta ^3x^3}{3!}+\cdots \end{aligned}$$

From this, we observe that the mth-order moment $G_{\alpha }(e_m,x), m=0,1,2,\ldots$ is the coefficient of $\theta ^m/m!$ in the above expansion. By linearity property, the central moments $\mu _{\alpha ,m}(x)=G_\alpha ((t-x)^m,x)$ are given as

$$\begin{aligned} \mu _{\alpha ,0}(x)= & {} 1, \, \, \, \, \, \, \mu _{\alpha ,1}(x)= 0, \, \, \, \, \, \, \mu _{\alpha ,2}(x)= \frac{x^2}{\alpha },\\ \mu _{\alpha ,4}(x)= & {} \left[ \frac{3}{\alpha ^2}+\frac{6}{\alpha ^3}\right] x^4, \, \, \, \, \, \, \mu _{\alpha ,6}(x)= \left[ \frac{15}{\alpha ^3}+\frac{130}{\alpha ^4}+\frac{120}{\alpha ^5}\right] x^6. \end{aligned}$$

The present paper deals with the convergence estimates of the Gamma operators $G_\alpha$. Here we establish some direct results in terms of usual and weighted modulus of continuity. In the end, we estimate a direct result in terms of modulus of continuity of functions of exponential growth.

2 Convergence Estimates

Suppose $C_B[0,\infty )$ be the space of all continuous and bounded functions on $[0,\infty )$ endowed with the norm $\Vert f\Vert =\sup \{|f(x)|:x\in [0,\infty )\}.$ Further, let us consider the following K-functional:

$$\begin{aligned} K_2(f,\delta )=\inf _{g\in C_B^2[0,\infty )}\{\Vert f-g\Vert +\delta \Vert g^{\prime \prime }\Vert \}, \end{aligned}$$

where $\delta >0$ and $C_B^2[0,\infty )=\{g\in C_B[0,\infty ):g^\prime ,g^{\prime \prime }\in C_B[0,\infty )\}.$ Following [2], one has the inequality

$$\begin{aligned} K_2(f,\delta ) \le C\omega _2(f,\sqrt{\delta }), \delta >0.\end{aligned}$$

(3)

Theorem 1

Let $f\in C_B[0,\infty ),$ then we have

$$\begin{aligned} \left| G_{\alpha }(f,x)-f(x)\right| \le C \omega _ 2 \left( f, \frac{x}{\sqrt{\alpha }}\right) . \end{aligned}$$

Proof

Let $g\in C_B^2[0,\infty )$ and $x,t\in [0,\infty ),$ by Taylor’s formula, we have

$$\begin{aligned} g(t)= & {} g(x)+(t-x)g^{\prime } (x)+\int _x^t(t-u)g^{\prime \prime }(u)du. \end{aligned}$$

Then using Remark 1, we have

$$\begin{aligned} \left| G_{\alpha }(g,x)-g(x)\right|= & {} \left| G_{\alpha }\left( \int _x^t (t-u)g^{\prime \prime }(u)du ,x\right) \right| \nonumber \\\le & {} \frac{1}{2} \mu _{\alpha ,2}(x)\ \Vert g^{\prime \prime }\Vert = \frac{x^2}{2\alpha }\Vert g^{\prime \prime }\Vert . \end{aligned}$$

(4)

Also, we have

$$\begin{aligned} |G_{\alpha }(f,x)| \le \Vert f\Vert . \end{aligned}$$

(5)

Therefore, using (4) and (5), we have

$$\begin{aligned} |G_{\alpha }(f,x)-f(x)|\le & {} |G_{\alpha }(f-g,x)-(f-g)(x)|+|G_{\alpha }(g,x)-g(x)|\nonumber \\\le & {} 2 \Vert f-g\Vert +\frac{x^2}{2\alpha } \ \Vert g^{\prime \prime }\Vert . \end{aligned}$$

(6)

Taking infimum over all $g\in C_B^2[0,\infty )$, and using (3), we get the desired result. $\square$

Corollary 1

As special cases, under the same conditions of Theorem 1, we have

• (1) If $\alpha =n$, we get Post–Widder operators and then

  $$\begin{aligned} \left| P_{n}(f,x)-f(x)\right| \le C \omega _ 2 \left( f, \frac{x}{\sqrt{n}}\right) . \end{aligned}$$

• (2) If $\alpha =nx$, we obtain Rathore operators, which satisfy

  $$\begin{aligned} \left| R_{n}(f,x)-f(x)\right| \le C \omega _ 2 \left( f, \sqrt{\frac{x}{n}}\right) . \end{aligned}$$

The weighted modulus $\omega _\varphi (f;h)$, introduced in [9], is defined as

$$\begin{aligned} \omega _\varphi (f;h)=\sup _{\begin{array}{c} |x-y|\le h\varphi \left( \frac{x+y}{2}\right) \\ x\ge 0,y\ge 0,h\ge 0 \end{array}}|f(x)-f(y)|, \end{aligned}$$

where $\varphi (x)=\frac{\sqrt{x}}{1+x^m}, x\in [0,\infty ),m\in \mathbb {N},m\ge 2.$ Aral et al in [1] estimated a Voronovskaja kind asymptotic formula for such weighted modulus of continuity. We consider here those functions which satisfy the property

$$\begin{aligned} \lim _{h\rightarrow 0}\omega _\varphi (f;h)=0, \end{aligned}$$

whenever the function $f\circ e_2$ is uniformly continuous on [0, 1] and the function $f\circ e_r,r=\frac{2}{2m+1}$ is uniformly continuous on $[1,\infty )$, where $e_r(x)=x^r,x\ge 0$. By $W_{\varphi }[0,\infty )$, we mean the subspace of all real-valued functions defined on $[0,\infty )$, for which these conditions hold true. By E, we mean the subspace of $C[0,\infty )$, such that $C_k[0,\infty )\subset E$ with $k=\max \{m+r+1,2r+2,2m\}$ and

$$\begin{aligned} C_k[0,\infty )=\{f\in C[0,\infty ),\,\exists M>0:\,|f(x)|\le M(1+x^k),\forall x\ge 0\},k\in \mathbb {N}. \end{aligned}$$

The following quantitative asymptotic formula was established in [12]:

Theorem A

[12, Th. 2.3] Let $L_n:E \rightarrow C[0,\infty ),\,C_k[0,\infty )\subset E,\,k=\max \{m+3,6,2m\}$ be a sequence of linear positive operators, reproducing the linear functions. If $f\in C^2[0,\infty )\cap E$ and $f^{\prime \prime }\in W_{\varphi }[0,\infty )$, then we have for $x\in (0,\infty )$ that

$$\begin{aligned}&\left| L_n(f,x)-f(x)-\frac{1}{2}f^{\prime \prime }(x)\mu _{n,2}^L(x)\right| \\&\quad \le \frac{1}{2} \left[ \mu _{n,2}^L(x)+\sqrt{2}\sqrt{L_n\left( \left[ 1+\left( x+\frac{|t-x|}{2}\right) ^m\right] ^2;x\right) }\right] \omega _\varphi \left( f^{\prime \prime };\left( \frac{\mu _{n,6}^L}{x}\right) ^{1/2}\right) . \end{aligned}$$

Here $\mu _{n,m}^L(x)=L_n((t-x)^m,x).$

Below we present as the application of Theorem A, the exact estimate for the Gamma operators:

Theorem 2

Under the assumptions of Theorem A, for the family $G_\alpha$ of linear positive operators, we have

$$\begin{aligned}&\left| G_\alpha (f,x)-f(x) -\frac{x^2}{2\alpha }f^{\prime \prime }(x)\right| \\&\quad \le \frac{1}{2} \left[ \frac{x^2}{\alpha }+\sqrt{2}\sqrt{G_\alpha \left( \left[ 1+\left( x+\frac{|t-x|}{2}\right) ^m\right] ^2;x\right) }\right] \\&\qquad \omega _\varphi \biggl (f^{\prime \prime };\biggl (\left[ \frac{15}{\alpha ^3}+\frac{130}{\alpha ^4}+\frac{120}{\alpha ^5}\right] x^5\biggr )^{1/2}\biggr ). \end{aligned}$$

Corollary 2

Under the assumptions of Theorem 2, we have

• (1) If $\alpha =n$, we get Post–Widder operators and then

  $$\begin{aligned}&\left| P_n(f,x)-f(x) -\frac{x^2}{2n}f^{\prime \prime }(x)\right| \\&\quad \le \frac{1}{2} \left[ \frac{x^2}{n}+\sqrt{2}\sqrt{P_n\left( \left[ 1+\left( x+\frac{|t-x|}{2}\right) ^m\right] ^2;x\right) }\right] \\&\qquad \omega _\varphi \biggl (f^{\prime \prime };\biggl (\left[ \frac{15}{n^3}+\frac{130}{n^4}+\frac{120}{n^5}\right] x^5\biggr )^{1/2}\biggr ). \end{aligned}$$

• (2) If $\alpha =nx$, we obtain Rathore operators, which satisfy

  $$\begin{aligned}&\left| R_n(f,x)-f(x) -\frac{x}{2n}f^{\prime \prime }(x)\right| \\&\quad \le \frac{1}{2} \left[ \frac{x}{n}+\sqrt{2}\sqrt{R_n\left( \left[ 1+\left( x+\frac{|t-x|}{2}\right) ^m\right] ^2;x\right) }\right] \\&\qquad \omega _\varphi \biggl (f^{\prime \prime };\biggl (\left[ \frac{15x^2}{n^3}+\frac{130x}{n^4}+\frac{120}{n^5}\right] \biggr )^{1/2}\biggr ). \end{aligned}$$

Also, the following quantitative Voronovskaja type theorem was estimated in [12]:

Theorem B

[12, Theorem 2.3] Let $L_n:E \rightarrow C[0,\infty ),\,C_k[0,\infty )\subset E,\,k=\max \{m+3,4\}$ be sequence of linear positive operators, preserving the linear functions. If $f\in C^2[0,\infty )\cap E$ and $f^{\prime \prime }\in W_{\varphi }[0,\infty )$, then we have for $x\in (0,\infty )$ that

$$\begin{aligned}&\left| L_n(f,x)-f(x)-\frac{1}{2}f^{\prime \prime }(x)\mu _{n,2}^L(x)\right| \\&\quad \le \frac{1}{2} \left[ \mu _{n,2}^L(x)+\frac{\sqrt{2}}{\sqrt{x}}\mu _{n,2}^L(x).C_{n,2,m}(x)\right] \omega _\varphi \left( f''; \sqrt{\frac{\mu _{n,4}^L(x)}{\mu _{n,2}^L(x)}}\right) , \end{aligned}$$

where

$$\begin{aligned} C_{n,2,m}(x)=1+\frac{1}{M_{n,3}^L(x)}.\sum _{k=0}^m{m\atopwithdelims ()k}x^{m-k}\frac{M_{n,k+3}^L(x)}{2^k}. \end{aligned}$$

We suppose for the operators $L_n$ that

$$\begin{aligned} \frac{M^L_{n,k}}{M^L_{n,3}},\,4\le k\le m \end{aligned}$$

is a bounded ratio for fixed x and m, when $n\rightarrow \infty$ and $M^L_{n,r}=L_n(|t-x|^r,x).$

From the estimate $L_n((t-x)^k,x)=O(n^{-[(k+1)/2]}), n\rightarrow \infty$, it is easy to observe that $C_{n,2,m}(x)$ is a bounded term for fixed x and m when $n \rightarrow \infty$. As an application of Theorem B, we have the following result for $G_\alpha$:

Theorem 3

Under the assumptions of Theorem B, for the family $G_\alpha$ of linear positive operators, we have

$$\begin{aligned}&\left| G_\alpha (f,x)-f(x)-\frac{x^2}{2\alpha }f^{\prime \prime }(x)\right| \\&\quad \le \frac{x^2}{2\alpha } \left[ 1+\frac{\sqrt{2}}{\sqrt{x}}.C_{\alpha ,2,m}(x)\right] \omega _\varphi \left( f^{\prime \prime }; x\sqrt{\left( \frac{6}{\alpha ^2}+\frac{3}{\alpha }\right) }\right) , \end{aligned}$$

where

$$\begin{aligned} C_{\alpha ,2,m}(x)=1+\frac{1}{M_{\alpha ,3}^L(x)}.\sum _{k=0}^m{m\atopwithdelims ()k}x^{m-k}\frac{M_{\alpha ,k+3}^L(x)}{2^k}. \end{aligned}$$

We suppose that

$$\begin{aligned} \frac{M^L_{\alpha ,k}}{M^L_{\alpha ,3}},\,4\le k\le m \end{aligned}$$

is a bounded ratio for fixed x and m, when $\alpha \rightarrow \infty$ where $M^L_{\alpha ,r}=G_\alpha (|t-x|^r,x).$

Corollary 3

If $f, f^{\prime \prime }$ satisfy the same conditions as in the assumption of Theorem 3, then we have for $x\in (0,\infty )$ that

$$\begin{aligned}&\lim _{n\rightarrow \infty } \alpha [G_\alpha (f,x)-f(x)]=\frac{x^2}{2}f^{\prime \prime }(x). \end{aligned}$$

For continuous functions on $[0,\infty )$ with exponential growth $\Vert f\Vert _A:=\sup _{x\in [0,\infty )}|f(x)\hbox {e}^{-Ax}|<\infty ,\,A>0,$ the first-order modulus of continuity (see [5, 11]) is defined as

$$\begin{aligned} \omega _1(f,\delta ,A)=\sup _{|h|\le \delta , 0\le x<\infty }|f(x)-f(x+h)|\hbox {e}^{-Ax}. \end{aligned}$$

The second-order modulus of continuity was considered by Ditzian in [3] in similar form.

Lemma 1

[11] (see also [5]) For every positive number $h>0$ and $k\in \mathbb {N}$, the following holds true

$$\begin{aligned} \omega _1(f,kh,A)\le k\cdot \hbox {e}^{A(k-1)h}\cdot \omega _1(f,h,A). \end{aligned}$$

The main result in terms of exponential modulus of continuity is the following theorem:

Theorem 4

For the family of linear positive operators $G_\alpha$, if $f\in C^2[0,\infty )\cap E$ and $f''\in \hbox {Lip}(\beta ,A),\,0<\beta \le 1$, then for $\alpha >2Ax$, we have for $x\in [0,\infty )$

$$\begin{aligned}&\left| G_\alpha \left( f,x\right) -f\left( x\right) - \frac{x^2}{2\alpha }f^{\prime \prime }\left( x\right) \right| \\&\quad \le \left[ \hbox {e}^{2Ax}+\frac{C(A,x)}{2}+\frac{\sqrt{C(2A,x)}}{2}\right] \cdot \mu _{\alpha ,2}(x)\cdot \omega _1\left( f^{\prime \prime },\sqrt{\frac{\mu _{\alpha ,4}(x)}{\mu _{\alpha ,2}(x)}},A\right) , \end{aligned}$$

where $\mu _{\alpha ,2}(x)$ and $\mu _{\alpha ,4}(x)$ are given in Remark 1, $C(A,x)= 2^{2(Ax+1)}\left( 1+\frac{Ax}{2}\right)$ and the spaces $\hbox {Lip}(\beta ,A),0<\beta \le 1$ consist of all functions such that $\omega _1(f,\delta ,A)\le M\delta ^\beta$ for all $\delta <1.$

Proof

For the function $f \in C^2[0,\infty )$, Taylor’s expansion at the point $x\in [0,\infty )$ is given by

$$\begin{aligned} f(t)=f(x)+(t-x)f^{\prime }(x)+\frac{(t-x)^2}{2!}f^{\prime \prime }(x)+\varepsilon _2(t,x), \end{aligned}$$

(7)

where

$$\begin{aligned} \varepsilon _2(t,x)=\frac{f^{\prime \prime }(\xi )-f^{\prime \prime }(x)}{2} (t-x)^2 \end{aligned}$$

and $\xi$ lies between x and t, also applying the operator $G_\alpha$ on (7) and using Remark 1, we have

$$\begin{aligned} \left| G_\alpha (f,x)-f(x)-\frac{x^2}{2\alpha }f^{\prime \prime }(x)\right| \le G_\alpha \left( \left| \varepsilon _2(t,x)\right| , x \right) \end{aligned}$$

(8)

In order to estimate the proof of theorem, we estimate $G_\alpha \left( \left| \varepsilon _2(t,x)\right| ,x \right) .$ By simple computations (see [5, p. 101]), we have

$$\begin{aligned} \left| \varepsilon _2(t,x)\right| \le \frac{1}{2} \left( \hbox {e}^{2Ax}+\hbox {e}^{At}\right) \left( 1+\frac{|t-x|}{h}\right) \omega _1\left( f^{\prime \prime },h,A\right) |t-x|^2. \end{aligned}$$

Consequently,

$$\begin{aligned}&G_\alpha \left( \left| \varepsilon _2(t,x)\right| , x \right) \le \frac{1}{2} \left[ G_\alpha \left( \left( \hbox {e}^{2Ax}+\hbox {e}^{At}\right) .\left( |t-x|^2+\frac{|t-x|^3}{h} \right) ;x \right) \right] \nonumber \\&\qquad \omega _1\left( f^{\prime \prime },h,A\right) . \end{aligned}$$

(9)

With operators (1), we have

$$\begin{aligned} G_\alpha \left( t^m\hbox {e}^{At},x\right) = \left( \frac{\alpha }{x}\right) ^\alpha \frac{\Gamma (m+\alpha )}{\Gamma (\alpha )}\left( \frac{\alpha }{x}-A\right) ^{-m-\alpha }, Ax<\alpha \end{aligned}$$

(10)

By simple computations using (10), we get

$$\begin{aligned} G_\alpha \left( (t-x)^2\hbox {e}^{At},x\right)= & {} \left( \frac{\alpha }{x}\right) ^\alpha \left( \frac{\alpha }{x}-A\right) ^{-\alpha -2}(A^2x^2+\alpha )\nonumber \\= & {} \left( \frac{\alpha }{x}\right) ^{\alpha +2}\left( \frac{\alpha }{x}-A\right) ^{-\alpha -2}\left( 1+\frac{A^2x^2}{\alpha }\right) \mu _{\alpha ,2}(x)\nonumber \\= & {} \left( 1-\frac{Ax}{\alpha }\right) ^{-\alpha -2}\left( 1+\frac{A^2x^2}{\alpha }\right) \mu _{\alpha ,2}(x). \end{aligned}$$

(11)

Let $\alpha >2Ax,$ thus (11) becomes

$$\begin{aligned} G_\alpha \left( (t-x)^2\hbox {e}^{At},x\right)\le & {} 2^{2(Ax+1)}\left( 1+\frac{Ax}{2}\right) \mu _{\alpha ,2}(x):=C(A,x)\mu _{\alpha ,2}(x). \end{aligned}$$

(12)

Also, by Cauchy–Schwarz inequality we get

$$\begin{aligned} G_\alpha \left( |t-x|^3\hbox {e}^{At},x\right)\le & {} \sqrt{ G_\alpha \left( (t-x)^2\hbox {e}^{2At},x\right) }.\sqrt{\mu _{\alpha ,4}(x)}\nonumber \\\le & {} \sqrt{C(2A,x)\mu _{\alpha ,2}(x)}.\sqrt{\mu _{\alpha ,4}(x)}. \end{aligned}$$

(13)

Substituting $\displaystyle h:=\sqrt{\frac{\mu _{\alpha ,4}(x)}{\mu _{\alpha ,2}(x)}}$ in (9) and combining (8), (12) and (13), we obtain the desired result. $\square$

Remark 2

For fixed $x\in [0,\infty )$, when $\alpha \rightarrow \infty$, we observe that

$$\begin{aligned} \frac{\mu _{\alpha ,4}(x)}{\mu _{\alpha ,2}(x)}=\left( \frac{6}{\alpha ^2}+\frac{3}{\alpha }\right) x^2\rightarrow 0 ,\ \alpha \rightarrow \infty , \end{aligned}$$

which guarantees the convergence of Theorems 3 and 4.

In the next convergence estimate, we denote by $C^{*}[0,\infty )$, the linear space of real-valued continuous functions on $[0,\infty )$ with the property that $\displaystyle \lim _{x \rightarrow \infty } f(x)$ exists and is finite, endowed with the uniform norm.

Also, for every $\delta \ge 0$ the modulus of continuity (see [6]) is given by

$$\begin{aligned} \omega ^{*}(f,\delta ) = \sup _{\begin{array}{c} x,t \ge 0 \\ |\hbox {e}^{-x}-\hbox {e}^{-t}|\le \delta \end{array}} |f(x) -f(t)|. \end{aligned}$$

Theorem 5

Let $f, f^{\prime \prime } \in C^{*}[0,\infty ),$ then, for $x\in [0,\infty ),$ the following inequality holds:

$$\begin{aligned}&\left| \alpha \ [G_\alpha (f,x)-f(x)] -\frac{x^2}{2} f^{\prime \prime }(x) \right| \\&\quad \le \frac{\omega ^{*}(f^{\prime \prime },\alpha ^{-1/2})}{2}\left[ x^2 + x^2\sqrt{\frac{3(\alpha +2)}{\alpha }} \left[ \alpha ^2G_\alpha \left( \left( \hbox {e}^{-x}-\hbox {e}^{-t}\right) ^4,x \right) \right] ^{1/2} \right] . \end{aligned}$$

Proof

By the Taylor’s formula, we have

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

where $\xi$ lies between x and t and

$$\begin{aligned} h\left( \xi ,x\right) :=\frac{f^{\prime \prime }\left( \xi \right) -f^{\prime \prime }\left( x\right) }{2} \end{aligned}$$

Applying the operator $G_\alpha$ to above equality and using Remark 1, we can write that

$$\begin{aligned}&\left| G_\alpha \left( f,x\right) -\mu _{\alpha ,0}(x) f\left( x\right) - \mu _{\alpha ,1}(x) \ f^{\prime }(x) -\frac{1}{2} \ \mu _{\alpha ,2}(x) f^{\prime \prime }\left( x\right) \right| \nonumber \\&\quad = \left| G_\alpha \left( h\left( \xi ,x\right) \ (t-x)^2,x\right) \right| . \end{aligned}$$

Thus, we get

$$\begin{aligned} \left| \alpha \ [G_\alpha (f,x)-f(x)] -\frac{x^2}{2} f^{\prime \prime }(x) \right|= & {} \left| \alpha \ G_\alpha \left( h\left( \xi ,x\right) \ (t-x)^2,x\right) \right| . \end{aligned}$$

Using the property of $\omega ^{*}(.,\delta )$ and applying Cauchy–Schwarz inequality, we obtain

$$\begin{aligned}&\alpha \ G_\alpha \left( |h\left( \xi ,x\right) | \ (t-x)^2,x\right) \\&\quad \le \frac{1}{2} \ \alpha \ \omega ^{*}(f^{\prime \prime },\delta ) \ \mu _{\alpha ,2}(x)+ \frac{\alpha }{2 \delta ^2} \ \omega ^{*}(f^{\prime \prime },\delta ) \ \\&\qquad \left[ G_\alpha \left( \left( \hbox {e}^{-x}-\hbox {e}^{-t}\right) ^4,x \right) \, \cdot \, \mu _{\alpha ,4}(x)\right] ^{1/2}\\&\quad \le \frac{\omega ^{*}(f^{\prime \prime },\delta )}{2}\left[ x^2 + \frac{\alpha }{ \delta ^2} \left[ G_\alpha \left( \left( \hbox {e}^{-x}-\hbox {e}^{-t}\right) ^4,x \right) \, \cdot \, \left( \frac{3(\alpha +2)x^4}{\alpha ^3}\right) \right] ^{1/2} \right] . \end{aligned}$$

Choosing $\delta =\alpha ^{-1/2},$ we finally get the desired result. $\square$

Remark 3

The convergence of the Gamma operators $G_\alpha$ in the above theorem takes place for $\alpha$ sufficiently large. Using (10) and the software Mathematica, we find that

$$\begin{aligned}&\lim \limits _{\alpha \rightarrow \infty } \alpha ^2 \ G_\alpha \left( \left( \hbox {e}^{-x}-\hbox {e}^{-t}\right) ^4,x \right) \\&\quad = \lim \limits _{\alpha \rightarrow \infty } \alpha ^2 \ \left( G_\alpha (\hbox {e}^{-4t},x)-4\hbox {e}^{-x}G_\alpha (\hbox {e}^{-3t},x)+6\hbox {e}^{-2x}G_\alpha (\hbox {e}^{ -2t},x)\right. \\&\qquad \left. -4\hbox {e}^{-3x}G_\alpha (\hbox {e}^{-t},x)+\hbox {e}^{-4x} \right) \\&\quad = \lim _{\alpha \rightarrow \infty } \alpha ^2 \ \biggl [ \left( \frac{\alpha }{\alpha +4x}\right) ^{\alpha }-4\hbox {e}^{-x} \left( \frac{\alpha }{\alpha +3x}\right) ^{\alpha }+6\hbox {e}^{-2x}\left( \frac{\alpha }{\alpha +2x}\right) ^{\alpha }\\&\qquad -4\hbox {e}^{-3x}\left( \frac{\alpha }{\alpha +x}\right) ^{\alpha } +\hbox {e}^{-4x}\biggr ]\\&\quad = 3\hbox {e}^{-4x}x^4. \end{aligned}$$