Asymptotic formula in simultaneous approximation for certain Ismail-May-Baskakov operators
Vijay Gupta\(^\ast \) Michael Th. Rassias\(^{\ast \ast }\)
June 30, 2021; accepted: August 29, 2021; published online: February 17, 2022.
The present study deals with a modification of Ismail-May operators with weights of Beta basis functions. We estimate weighted Korovkin’s theorem and difference estimates between two operators and establish an asymptotic formula for the derivative of a function.
MSC. 41A25, 41A30
Keywords. Ismail-May operators, Baskakov basis, asymptotic formula, simultaneous approximation, weighted Korovkin’s theorem.
\(^\ast \)Department of Mathematics, Netaji Subhas University of Technology, Sector 3 Dwarka, New Delhi 110078, India, e-mail: vijaygupta2001@hotmail.com
\({\ast \ast }\)Department of Mathematics and Engineering Sciences, Hellenic Military Academy, 16673 Vari Attikis, Greece,
\({\ast \ast }\)Moscow Institute of Physics and Technology, 141700 Dolgoprudny, Institutskiy per, d. 9, Russia,
\({\ast \ast }\)Institute for Advanced Study, Program in Interdisciplinary Studies, 1 Einstein Dr, Princeton, NJ 08540, USA, e-mail: mthrassias@yahoo.com
1 The Operator
In 1978 Ismail and May
[
13
]
(see also
[
12
]
) studied certain operators of exponential type and introduced some new exponential operators. The following sequence of operators constitutes one of the families of linear positive operator proposed in
[
13
,
(3.14)
]
:
\begin{equation} \label{AE1} (R_nf)(x) = \sum _{\nu =0}^\infty r_{n,\nu }(x)F_{n,\nu }, \end{equation}
1
where \(F_{n,\nu }=f\left(\tfrac {\nu }{n}\right)\) and
\[ r_{n,\nu }(x)=e^{-(n+\nu )x/(1+x)}\tfrac {n(n+\nu )^{\nu -1}}{\nu !}\left(\tfrac {x}{1+x}\right)^\nu . \]
These operators are exponential type operators as
\[ x(1+x)^2[(R_nf)(x)]^\prime =\sum _{\nu =0}^\infty (\nu -nx)r_{n,\nu }(x)f\left(\tfrac {\nu }{n}\right). \]
In recent studies, Gupta and Rassias
[
10
]
presented a collection of the moments of several linear positive operators, Agratini
[
2
]
estimated approximation properties of integral-type operators. The operators \(R_n\) are not suitable in order to approximate integral functions. Additionally, it is observed in
[
6
]
that obtaining the usual Durrmeyer variant of the operators (1) is not possible due to the non-convergence of the integral
\[ \int _0^\infty r_{n,\nu }(u)du. \]
In the recent work
[
6
]
, the hybrid link operators were introduced and the approximation behavior was discussed. We consider here new hybrid operators with weights \(b_{n+2,\nu -1}(t)\) of slightly modified Baskakov basis functions, which for \(x\in [0,\infty )\) are defined as
\begin{eqnarray} (V_nf)(x)& =& \sum _{\nu =0}^\infty r_{n,\nu }(x) G_{n,\nu }(f),\label{AEB1} \end{eqnarray}
where
\[ G_{n,\nu }(f) = \left\{ \begin{array}{lr} (n+1)\int _0^\infty b_{n+2,\nu -1}(u)f(u)du& : 1\le \nu {\lt}\infty \\ f(0) & : \nu =0 \end{array} \right. \]
\[ b_{n,\nu }(u)=\textstyle {{n+\nu -1}\choose {\nu }}\tfrac {u^\nu }{(1+u)^{n+\nu }}. \]
For simplicity, we use the notation
\[ v_{n,\nu }(u)=(n+1) b_{n+2,\nu -1}(u) \]
throughout the paper.
The operators (2) reproduce linear functions. The investigation of convergence of operators in ordinary and simultaneous approximation has been an active area of research over recent years. We mention here some of the related work viz.
[
3
]
,
[
7
]
,
[
8
]
,
[
9
]
,
[
11
]
. In the present paper, we estimate direct result in weighted norm, difference of these operators with discrete operators and prove a formula of asymptotic kind for derivatives of the operators \(V_n.\)
2 Some Lemmas
Lemma
1
If
\[ U_{n,q}(x)=\sum _{\nu =0}^\infty r_{n,\nu }(x)\left(\tfrac {\nu }{n}-x\right)^q,\quad q\in {N}\cup \{ 0\} , \]
then the moments satisfy the relation:
\[ nU_{n,q+1}(x)=x(1+x)^2\left[(U_{n,q}(x))^\prime +qU_{n,q-1}(x)\right]. \]
Also, one can observe that
\[ U_{n,q}(x)=O_x(n^{-[(q+1)/2]}), \]
where \([s]\) stands for the integral part of \(s\).
Furthermore, for \(e_i(t)=t^i, i=0,1,2,\ldots \), we have
\[ n(R_ne_{q+1})(x)=x(1+x)^2[(R_ne_q)(x)]^\prime +nx(R_ne_q)(x). \]
Lemma
2
If we denote \(\mu _{n,q}(x)=(V_n(e_1-xe_0)^q)(x)\) for \(q\in {N}\) and \(x\ge 0\), then
\begin{eqnarray*} (n-q) \mu _{n,q+1}(x)& =& x(1+x)^2\left[[\mu _{n,q}(x)]^\prime +q\mu _{n,q-1}(x)\right]\\ & & +q\mu _{n,q}(x)+qx(1+x) \mu _{n,q-1}(x). \end{eqnarray*}
In particular, we have
\[ \mu _{n,0}(x)=1, \mu _{n,1}(x)=0, \mu _{n,2}(x)=\tfrac {x(1+x)(2+x)}{n-1}. \]
In general for all \(x\in [0,\infty )\), we have \(\mu _{n,q}(x)=O_x(n^{-[(q+1)/2]}).\)
Using
\[ x(1+x)^2 r_{n,\nu }^\prime (x)=(\nu -n x)r_{n,\nu }(x) \]
and
\[ u(1+u)v_{n,\nu }^\prime (u)=[(\nu -1)-(n+2)u]v_{n,\nu }(u), \]
we have
\begin{eqnarray*} & & x(1+x)^2[\mu _{n,q}(x)]^\prime =\\ & =& \sum _{\nu =1}^{\infty }x(1+x)^2r_{n,\nu }^\prime (x)\int _0^\infty v_{n,\nu }(u)(e_1-xe_0)^qdu\\ & & +n e^{-n x/(1+x)}(-x)^{q+1}+q (-x)^{q}(1+x)^2e^{-n x/(1+x)}\\ & & -qx(1+x)^2\sum _{\nu =1}^{\infty }r_{n,\nu }(x)\int _0^\infty v_{n,\nu }(u)(e_1-xe_0)^{q-1}du\\ & =& \sum _{\nu =1}^{\infty }(\nu -n x)r_{n,\nu }(x)\int _0^\infty v_{n,\nu }(u)(e_1-xe_0)^qdu\\ & & +n e^{-n x/(1+x)}(-x)^{q+1}-qx(1+x)^2 \mu _{n,q-1}(x)\\ & =& \sum _{\nu =1}^{\infty }r_{n,\nu }(x)\int _0^\infty u(1+u)v_{n,\nu }^\prime (u)(e_1-xe_0)^qdu+(n+2)[ \mu _{n,q+1}(x)\\ & & -(-x)^{q+1}e^{-n x/(1+x)}]+(1+2x)[\mu _{n,q}(x)-(-x)^qe^{-n x/(1+x)}]\\ & & +n e^{-n x/(1+x)}(-x)^{q+1}-qx (1+x)^2 \mu _{n,q-1}(x). \end{eqnarray*}
Thus, we have
\begin{eqnarray*} & & x(1+x)^2\left[[\mu _{n,q}(x)]^\prime +q\mu _{n,q-1}(x)\right]=\\ & =& \sum _{\nu =1}^{\infty }r_{n,\nu }(x)\int _0^\infty [(e_1\! \! -\! \! xe_0)^2\! \! +\! \! (1\! \! +\! \! 2x)(e_1\! \! -\! \! xe_0)\! \! +\! \! x(1\! +\! x)]v_{n,\nu }^\prime (u)(e_1\! \! -\! \! xe_0)^qdu\\ & & +(n+2)[ \mu _{n,q+1}(x)-(-x)^{q+1}e^{-n x/(1+x)}]\\ & & +(1+2x)[\mu _{n,q}(x)-(-x)^qe^{-n x/(1+x)}]+n e^{-n x/(1+x)}(-x)^{q+1}. \end{eqnarray*}
Integrating by parts the last integral, we obtain that
\begin{eqnarray*} & & x(1+x)^2\left[[\mu _{n,q}(x)]^\prime +q\mu _{n,q-1}(x)\right]\\ & =& -(q+2) \mu _{n,q+1}(x)-(q+1)(1+2x)\mu _{n,q}(x)-qx(1+x) \mu _{n,q-1}(x)\\ & & +(n+2) \mu _{n,q+1}(x)+(1+2x)\mu _{n,q}(x). \end{eqnarray*}
This completes the proof of the recurrence relation. The other consequences follow from the recurrence relation.
Lemma
3
If we denote \(T_{n,q}(x)=(V_ne_q)(x)\) for \(q\in {N}\) and \(x\ge 0\), then
\begin{eqnarray*} (n-q) T_{n,q+1}(x)& =& x(1+x)^2[T_{n,q}(x)]^\prime +(q+nx)T_{n,q}(x). \end{eqnarray*}
In general, we have
\begin{eqnarray*} T_{n,q}(x)& =& x^q+\tfrac {q(q-1)}{2}x^{q-1}\tfrac {(1+x)(2+x)}{n-q+1}+O_x(n^{-2}). \end{eqnarray*}
The proof of the lemma follows along the lines of Lemma
1, we thus omit the details.
There exist the polynomials \(\phi _{i,j,q}(x)\) independent of \(n\) and \( \nu \) such that
\[ [x^q(1+x)^{2q}]\tfrac {\partial ^q}{\partial x^q}[r_{n,\nu }(x)]=\sum _{{2i+j\le q}\atop {i,j\ge 0}}n^{i}(\nu -nx)^j\phi _{i,j,q}(x)[r_{n,\nu }(x)]. \]
Lemma
5
If \(\mu _r^{J_{n,\nu }}=\sum _{i=0}^r {r\choose i} (-1)^i J_{n,\nu }(e_{r-i})[ J_{n,\nu }(e_{1})]^i\), then we have
\[ \mu _m^{F_{n,\nu }}=0,m=1,2,3,\ldots , \ \ \ \ \mu _2^{G_{n,\nu }}=\tfrac {\nu ^2+n\nu }{n^2(n-1)}, \ \ \mu _3^{G_{n,\nu }}=\tfrac {4 \nu ^3+6n\nu ^2+2n^2\nu }{n^3(n-1)(n-2)} \]
and
\[ \mu _4^{G_{n,\nu }}=\tfrac {3(6+n) \nu ^4+ 6(6 + n)n \nu ^3+3(8 + n)n^2 \nu ^2+6 n^3 \nu }{n^4(n-1)(n-2)(n-3)}. \]
By definition, we have
\begin{eqnarray*} b^{F_{n,\nu }}& =& F_{n,\nu }(e_1)=\tfrac {\nu }{n}. \end{eqnarray*}
By simple computation, we get
\begin{eqnarray*} \mu _m^{F_{n,\nu }}& =& 0, m=1,2,3,\ldots \end{eqnarray*}
Moreover
\begin{eqnarray*} G_{n,\nu }(e_q)& =& (n+1)\int _0^\infty b_{n+2,\nu -1}(u)u^q du\\ & =& (n+1) \textstyle {n+\nu \choose \nu -1}\displaystyle \int _0^\infty \tfrac {u^{\nu +q-1}}{(1+u)^{n+\nu +1}}du=\tfrac {(\nu +q-1)!(n-q)!}{(\nu -1)!n!} \end{eqnarray*}
Thus
\begin{eqnarray*} b^{G_{n,\nu }}& =& G_{n,\nu }(e_1)=\tfrac {\nu }{n}, \end{eqnarray*}
and
\begin{eqnarray*} \mu _2^{G_{n,\nu }}& =& \sum _{i=0}^2 \textstyle {2\choose i} (-1)^i G_{n,\nu }(e_{2-i})[ G_{n,\nu }(e_{1})]^i\nonumber \\ & =& G_{n,\nu }(e_{2})-[G_{n,\nu }(e_{1})]^2\\ & =& \tfrac {\nu ^2+n\nu }{n^2(n-1)}. \end{eqnarray*}
\begin{align*} \mu _3^{G_{n,\nu }}& =\sum _{i=0}^3 \textstyle {3\choose i} (-1)^i G_{n,\nu }(e_{3-i})[ G_{n,\nu }(e_{1})]^i\nonumber \\ & =G_{n,\nu }(e_{3})-3G_{n,\nu }(e_{2})G_{n,\nu }(e_{1})+3[G_{n,\nu }(e_{1})]^3-[G_{n,\nu }(e_{1})]^3\\ & =\tfrac {1}{n^3(n-1)(n-2)}\left[n^2\nu (\nu \! \! +\! \! 2)(\nu \! \! +\! \! 1)\nu \! \! -\! \! 3n(n\! \! -\! \! 2)(\nu \! \! +\! \! 1)\nu ^2+2(n\! \! -\! \! 1)(n\! \! -\! \! 2)\nu ^3\right]\\ & =\tfrac {4 \nu ^3+6n\nu ^2+2n^2\nu }{n^3(n-1)(n-2)} \end{align*}
and
\begin{align*} \mu _4^{G_{n,\nu }}& =\sum _{i=0}^4 \textstyle {4\choose i} (-1)^i G_{n,\nu }(e_{4-i})[ G_{n,\nu }(e_{1})]^i\nonumber \\ & =G_{n,\nu }(e_{4})-4G_{n,\nu }(e_{3})G_{n,\nu }(e_{1})+6G_{n,\nu }(e_{2})[G_{n,\nu }(e_{1})]^2\\ \nonumber & -4G_{n,\nu }(e_{1})[G_{n,\nu }(e_{1})]^3+[G_{n,\nu }(e_{1})]^4\nonumber \\ & =\tfrac {1}{n^4(n-1)(n-2)(n-3)}\biggl[n^3(\nu \! +\! 3)(\nu \! +\! 2)(\nu \! +\! 1)\nu \! -\! 4n^2(n\! -\! 3)(\nu \! +\! 2)(\nu \! +\! 1)\nu ^2\\ & \quad +6n(n-2)(n-3)(\nu +1)\nu ^3-3\nu ^4(n-1)(n-2)(n-3)\biggr]\\ & =\tfrac {1}{n^4(n-1)(n-2)(n-3)}\biggl[6 n^3 \nu + 3(8 + n) n^2\nu ^2 + 6( 6 + n)n \nu ^3 + 3(6 + n) \nu ^4\biggr]. \end{align*}
The other estimates follow similarly, we thus omit the details.
3 Direct Estimates and Asymptotic Formula
Let
\[ B_{m}[0,\infty )=\{ f:|f(x)|\le c_f(1+x^m), \forall x\in [0,\infty )\} ,\quad m{\gt}3 \]
where \(c_f\) is an absolute constant dependent on \(f,\) but independent of \(x.\) Let
\[ C_{m}[0,\infty )=C[0,\infty )\cap B_{m}[0,\infty ). \]
For each \(f\in C_{m}[0,\infty )\) we consider modulus of continuity (see
[
14
]
) as
\[ \Omega (f,\delta )=\sup _{|\triangle x|{\lt}\delta ,x\in R^+}\tfrac {|f(x+\triangle x)-f(x)|}{(1+h^m)(1+x^m)}. \]
Additionally, \(C^*_{m}[0,\infty )\) denotes the subspace of continuous functions \(f \in B_{m}[0,\infty )\), for which
\[ \lim _{x\to \infty } |f(x)|(1+x^m)^{-1}{\lt}\infty . \]
We consider the norm by
\[ \Vert f\Vert _m=\sup _{0\le x{\lt}\infty } \tfrac {|f(x)|}{(1+x^m)}. \]
Theorem
6
If \(f\in C^*_m\left[ 0,\infty \right)\), then
\[ \displaystyle \lim _{n\to \infty }\Vert (V_nf)-f\Vert _m=0. \]
By the weighted Korovkin theorem due to Gadjiev
[
4
]
, we know that if \(f\in C^*_m\left[ 0,\infty \right)\), satisfying
\[ \lim _{n\to \infty }\Vert (V_ne_i)-e_i\Vert _m=0, \ \ i=0,1,2, \]
then we have
\[ \lim _{n\to \infty }\Vert (V_nf)(x)-f(x)\Vert _m=0. \]
In order to prove the result, we use lemma 3. Clearly, the result holds for \(i=0,1\). Thus, in order to complete the proof we proceed for
\begin{eqnarray*} \lim _{n\to \infty }\Vert (V_ne_2)(x)-e_2\Vert _2& =& \lim _{n\to \infty } \tfrac {1}{(1+x^m)}\biggl[x^2+\tfrac {x(1+x)(2+x)}{n-1}-x^2\biggr]\\ & =& \lim _{n\to \infty } \tfrac {1}{(1+x^m)}\biggl[\tfrac {x(1+x)(2+x)}{n-1}\biggr]=0. \end{eqnarray*}
This completes the proof of the theorem.
Theorem
7
For \(f^{(s)}\in C_B[0,\infty )\) , \(0\le s\le 4\), \(s\in \mathbf{N}\), \(x\in [0,\infty )\) and \(n\in \mathbf{N}\), we have
\begin{eqnarray*} & & |((V_nf)-(R_nf))(x)|\le \\ & \le & \tfrac {\| f^{iv}\| }{24(n-1)(n-2)(n-3)}\biggl[6 x + 3(8 + n)\left( x^2+\tfrac {x(1+x)^2}{n}\right)\\ & & + 6( 6 + n)\left(x^3+\tfrac {3x^2(1+x)^2}{n}+\tfrac {x(1+x)^3(1+3x)}{n^2}\right)\\ & & + 3(6 + n) \left(x^4+\tfrac {6x^3(1+x)^2}{n}+\tfrac {x^2(1+x)^3(7+15x)}{n^2}+\tfrac {x(1+x)^4(1+10x+15x^2)}{n^3}\right)\biggr]\\ & & +\tfrac {\| f^{\prime \prime \prime }\| }{3(n-1)(n-2)}\left[2x^3+\tfrac {6x^2(1+x)^2}{n}+\tfrac {2x(1+x)^3(1+3x)}{n^2}+3x^2+\tfrac {3x(1+x)^2}{n}+x\right]\\ & & + \tfrac {\| f^{\prime \prime }\| }{2(n-1)}\left[x^2+\tfrac {x(1+x)^2}{n}+x\right]. \end{eqnarray*}
Using
lemma 5 and
lemma 1, we have
\begin{eqnarray*} \alpha (x)& =& \sum _{\nu =0}^{\infty }r_{n,\nu }(x) (\mu _{4}^{F_{n,\nu }}+\mu _4^{G_{n,\nu }})\\ & =& \tfrac {1}{24n^4(n-1)(n-2)(n-3)}\sum _{\nu =0}^{\infty }r_{n,\nu }(x)\biggl[6 n^3 \nu + 3(8 + n) n^2\nu ^2 \\ & & + 6( 6 + n)n \nu ^3 + 3(6 + n) \nu ^4\biggr]\\ & =& \tfrac {1}{24(n-1)(n-2)(n-3)}\biggl[6 (R_ne_1)(x) + 3(8 + n)(R_ne_2)(x)\\ & & + 6( 6 + n)(R_ne_3)(x)+ 3(6 + n) (R_ne_4)(x)\biggr]\\ & =& \tfrac {1}{24(n-1)(n-2)(n-3)}\biggl[6 x + 3(8 + n)\left( x^2+\tfrac {x(1+x)^2}{n}\right)\\ & & + 6( 6 + n)\left(x^3+\tfrac {3x^2(1+x)^2}{n}+\tfrac {x(1+x)^3(1+3x)}{n^2}\right)\\ & & + 3(6\! \! +\! \! n) \left(x^4\! +\! \tfrac {6x^3(1+x)^2}{n}\! +\! \tfrac {x^2(1+x)^3(7+15x)}{n^2}\! +\! \tfrac {x(1+x)^4(1+10x+15x^2)}{n^3}\right)\biggr]. \end{eqnarray*}
Additionally,
\begin{eqnarray*} \beta (x)& =& \tfrac {1}{3!}\sum _{\nu =0}^{\infty }r_{n,\nu }(x)\left| \mu _{3}^{F_{n,\nu }}-\mu _{3}^{G_{n,\nu }}\right|\\ & =& \tfrac {1}{3!}\sum _{\nu =0}^{\infty }r_{n,\nu }(x)\tfrac {4 \nu ^3+6n\nu ^2+2n^2\nu }{n^3(n-1)(n-2)}\\ & =& \tfrac {1}{3(n-1)(n-2)}\left[2(R_ne_3)(x)+3(R_ne_2)(x)+(R_ne_1)(x)\right]\\ & =& \tfrac {1}{3(n-1)(n-2)}\left[2x^3+\tfrac {6x^2(1+x)^2}{n}+\tfrac {2x(1+x)^3(1+3x)}{n^2}+3x^2+\tfrac {3x(1+x)^2}{n}+x\right]. \end{eqnarray*}
Furthermore, we have
\begin{eqnarray*} \gamma (x)& =& \tfrac {1}{2!}\sum _{\nu =0}^{\infty }r_{n,\nu }(x)\left| \mu _{2}^{F_{n,\nu }}-\mu _2^{G_{n,\nu }} \right|\\ & =& \tfrac {1}{2!}\sum _{\nu =0}^{\infty }r_{n,\nu }(x)\tfrac {\nu ^2+n\nu }{n^2(n-1)}\\ & =& \tfrac {1}{2(n-1)}[(R_ne_2)(x)+(R_ne_1)(x)]=\tfrac {1}{2(n-1)}\left[x^2+\tfrac {x(1+x)^2}{n}+x\right]. \end{eqnarray*}
Finally, we get
\begin{eqnarray*} \delta ^2(x)& =& \sum _{\nu =0}^{\infty }r_{n,\nu }(x)(b^{F_{n,\nu }}-b^{G_{n,\nu }})^2=0. \end{eqnarray*}
Following
[
5
,
Theorem 3
]
and
[
1
,
Theorem 4
]
, we derive that
\begin{eqnarray*} |((V_nf)-(R_nf))(x)| & \le & \| f^{iv}\| \alpha (x)+\| f^{\prime \prime \prime }\| \beta (x)+\| f^{\prime \prime }\| \gamma {(x)}\\ & & +2\omega (f,\delta (x)). \end{eqnarray*}
Substituting the values of the above estimates, we get the desired result.
Theorem
8
Let \(f\in C[0,\infty )\) with \(|f(t)|\leq C (1+t)^{\gamma }\) for some \(\gamma {\gt}0, t\ge 0.\) If \(f^{(q+2)}\) exists at a point \(x\in (0,\infty ),\) then we have
\begin{eqnarray*} \lim _{n\rightarrow \infty }n(V_n^{(q)}f)(x) -f^{(q)}(x))& =& \tfrac {q(q-1)(q-2)}{2}f^{(q-1)}(x)+\tfrac {3q(q-1)(x+1)}{2}f^{(q)}(x)\\ & & +\tfrac {q[(q^2+2)x^2+3(q+1)x+2]}{2}f^{(q+1)}(x)\\ & & +\tfrac {x(1+x)(2+x)}{2}f^{(q+2)}(x). \end{eqnarray*}
Applying Taylor’s expansion, we have From Taylor’s theorem, we have
\begin{eqnarray} (V_n^{(q)}f)(x)& =& \sum _{v=0}^{q+2}\tfrac {f^{(v)}(x)}{v!}\bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}(V_n(e_1-xe_0)^{v})(\kappa )\bigg)_{\kappa =x} \nonumber \\ & & + \bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}(V_n\psi (t,x)(e_1-xe_0)^{q+2})(\kappa )\bigg)_{\kappa =x}\nonumber \\ & =& \sum _{v=0}^{q+2}\tfrac {f^{(v)}(x)}{v!}\sum _{j=0}^{v}\textstyle {v\choose j}(-x)^{v-j}\bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,j}(\kappa )\bigg)_{\kappa =x}\nonumber \\ & & + \bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}(V_n\psi (u,x)(e_1-xe_0)^{q+2})(\kappa )\bigg)_{\kappa =x}\nonumber \\ & :=& I_1+I_2\nonumber \end{eqnarray}
where the function \(\psi (u,x)\rightarrow 0\) as \(u\rightarrow x.\) Obviously \(I_1\) is estimated as
\begin{eqnarray*} I_1& =& \tfrac {f^{(q-1)}(x)}{(q-1)!}\bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,q-1}(\kappa )\bigg)_{\kappa =x}\nonumber \\ & & +\tfrac {f^{(q)}(x)}{q!}\left[q(-x)\bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,q-1}(\kappa )\bigg)_{\kappa =x} +\bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,q}(\kappa )\bigg)_{\kappa =x}\right]\nonumber \\ & & +\tfrac {f^{(q+1)}(x)}{(q+1)!}\bigg[\tfrac {(q+1)q}{2}x^2\bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,q-1}(\kappa )\bigg)_{\kappa =x} \! +\! (q+1)(\! -\! x)\bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,q}(\kappa )\bigg)_{\kappa =x}\nonumber \\ & & +\bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,q+1}(\kappa )\bigg)_{\kappa =x}\biggr]\nonumber \\ & & +\tfrac {f^{(q+2)}(x)}{(q+2)!}\biggl[\tfrac {(q+2)(q+1)q}{3!}(-x)^3 \bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,q-1}(\kappa )\bigg)_{\kappa =x} \\ & & +\tfrac {(q+2)(q+1)}{2}x^2 \bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,q}(\kappa )\bigg)_{\kappa =x}\nonumber \\ & & +(q+2)(-x)\bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,q+1}(\kappa )\bigg)_{\kappa =x}+ \bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}T_{n,q+2}(\kappa )\bigg)_{\kappa =x}\biggr].\nonumber \end{eqnarray*}
Applying lemma 3, we have
\begin{eqnarray*} I_1& =& \tfrac {f^{(q-1)}(x)}{(q-1)!}\left[\tfrac {q!(q-1)(q-2)}{2(n-q+2)}\right]+f^{(q)}(x)\left[1+\tfrac {3nq(q-1)(x+1)}{2(n-q+2)(n-q+1)}+O(n^{-2})\right]\\ & & +f^{(q+1)}(x)\left[\tfrac {nq[(q^2+2)x^2+3(q+1)x+2]}{2(n-q)(n-q+2)}+O(n^{-2})\right]\nonumber \\ & & +\tfrac {f^{(q+2)}(x)}{(q+2)!}\biggl[\tfrac {x^3}{2(n-q+1)}+\tfrac {3x^2}{2(n-q-1)}+\tfrac {nx}{(n-q)(n-q-1)}\biggr].\nonumber \end{eqnarray*}
Thus
\begin{eqnarray} \lim _{n\rightarrow \infty }n((V_n^{(q)}f)(x)-f^{(q)}(x))& =& \tfrac {q(q-1)(q-2)}{2}f^{(q-1)}(x)+\tfrac {3q(q-1)(x+1)}{2}f^{(q)}(x)\nonumber \\ & & +\tfrac {q[(q^2+2)x^2+3(q+1)x+2]}{2}f^{(q+1)}(x)\nonumber \\ & & +\tfrac {x(1+x)(2+x)}{2}f^{(q+2)}(x)+\lim _{n\rightarrow \infty }n I_2.\nonumber \end{eqnarray}
Next, in view of lemma 4, we have
\begin{align*} & |I_2|\leq \\ & \leq \left(\sum _{\nu =1}^{\infty }\sum _{{2i+j\leq q}\atop { i,j\geq 0}}n^{i}|\nu \! \! -\! \! n \kappa |^{j}\tfrac {|\phi _{i,j,q}(\kappa )|}{\kappa ^q(1+\kappa )^{2q}} r_{n,\nu }(\kappa )\int _0^{\infty }v_{n,\nu }(u)|\psi (u,x)|.|u\! \! -\! \! x|^{q+2} du\right)_{\kappa =x}\nonumber \\ & \quad +|\psi (0,x)(-x)^{q+2}| \bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}r_{n,0}(\kappa )\bigg)_{\kappa =x}\nonumber \\ & := E_1+E_2.\nonumber \end{align*}
Because of the fact \(\psi (u,x)\rightarrow 0\) as \(u\rightarrow x,\) for a given \(\epsilon {\gt}0\) there will exist a \(\delta {\gt}0\) such that \(|\psi (u,x)|{\lt}\epsilon \) whenever \(|u-x|{\lt}\delta .\)
Furthermore, if \(m \ge \{ \gamma ,q+2\} \), and \(m\) a positive integer
\[ |(u-x)^{q+2}\psi (u,x)|\leq M |u-x|^m \]
for \(|u-x|\geq \delta .\) Thus
\begin{align} |E_1|& \leq \sum _{\nu =1}^{\infty }\sum _{{2i+j\leq q}\atop {i,j\geq 0}}n^{i}|\nu -n x|^{j}\tfrac {|\phi _{i,j,q}(x)|}{x^q(1+x)^{2q}} {r_{n,\nu }(x)}\bigg(\epsilon \int _{|u-x|{\lt}\delta } v_{n,\nu }(u)|u-x|^{q+2}du\nonumber \\ & \quad +M\int _{|u-x|\geq \delta } v_{n,\nu }(u) |u-x|^m du\bigg)\nonumber \\ & :=J_1+J_2.\nonumber \end{align}
Let us consider
\[ K=\displaystyle \sup _{{2i+j\leq q}\atop { i,j\geq 0}}\tfrac {|\phi _{i,j,q}(x)|}{x^q(1+x)^{2q}}. \]
Using Schwarz’s inequality, lemma 1, lemma 2 and lemma 3, we deduce that
\begin{eqnarray*} J_1& =& \epsilon \, \ K\sum _{\nu =1}^{\infty }\sum _{ {2i+j\leq q}\atop {i,j\geq 0}}n^{i}|\nu -n x|^{j}r_{n,\nu }(x) \bigg(\int _{0}^{\infty }v_{n,\nu }(u)du\bigg)^{1/2}\\ & & \bigg(\int _{0}^{\infty }v_{n,\nu }(u)(e_1-xe_0)^{2q+4}du\bigg)^{1/2}\nonumber \\ & \leq & \epsilon \, \ K\sum _{ {2i+j\leq q}\atop { i,j\geq 0}} n^{i+j}\bigg(\sum _{\nu =0}^{\infty }r_{n,\nu } (x)\bigg(\tfrac {\nu }{n}-x\bigg)^{2j}-x^{2j}r_{n,0}(x)\bigg)^{1/2}\nonumber \\ & & \bigg(V_n((e_1-xe_0)^{2q+4},x)-x^{2q+4}r_{n,0}(x)\bigg)^{1/2}\nonumber \\ & =& \epsilon \sum _{{2i+j\leq q}\atop {i,j\geq 0}}n^{i+j}\bigg\{ O_x\bigg(\tfrac {1}{n^j}\bigg)+O_x\bigg(\tfrac {1}{n^s}\bigg)\bigg\} ^{1/2} \bigg\{ O_x\bigg(\tfrac {1}{n^{q+2}}\bigg)+O_x\bigg(\tfrac {1}{n^p}\bigg)\bigg\} ^{1/2}, \end{eqnarray*}
for any \(s,p{\gt}0\), choosing \(s\) and \(p\) such that \(s{\gt}j\) and \(p{\gt}q+2\), we obtain
\begin{eqnarray} J_1& \leq & \epsilon \sum _{{2i+j\leq q}\atop { i,j\geq 0}} n^{i+j}O_x\bigg(\tfrac {1}{n^{j/2}}\bigg)O_x\bigg(\tfrac {1}{n^{q/2+1}}\bigg)=\epsilon . O_x(n^{-1}).\nonumber \end{eqnarray}
Since \(\epsilon {\gt}0\) is arbitrary, \(n J_1\rightarrow 0\) as \(n\rightarrow \infty .\)
Applying again Schwarz’s inequality, lemma 1 and lemma 3, we obtain
\begin{eqnarray*} J_2& \leq & M_1\sum _{{2i+j\leq q}\atop { i,j\geq 0}}n^{i+j} \bigg(\sum _{\nu =0}^{\infty }\bigg(\tfrac {\nu }{n}-x\bigg)^{2j}r_{n,\nu }(x)-x^{2j}r_{n,0}(x)\bigg)^{1/2}\\ & & \bigg(\sum _{\nu =1}^{\infty }r_{n,\nu } (x)\int _{|u-x|\geq \delta }v_{n,\nu }(u)(e_1-xe_0)^{2m}du\bigg)^{1/2}\\ & \leq & M_1\sum _{{2i+j\leq q}\atop {i,j\geq 0}}n^{i+j}\bigg\{ O_x\bigg(\tfrac {1}{n^j}\bigg)+O_x\bigg(\tfrac {1}{n^s}\bigg)\bigg\} ^{1/2} \bigg\{ O_x\bigg(\tfrac {1}{n^m}\bigg)+O_x\bigg(\tfrac {1}{n^l}\bigg)\bigg\} ^{1/2}, \end{eqnarray*}
for any \(s,l{\gt}0\). We now choose \(s,l\) such that \( s{\gt} j, l{\gt}m\). Then, we derive that
\begin{eqnarray*} J_2& \leq & M_1 \sum _{{2i+j\leq q}\atop { i,j\geq 0}}n^{i+j}O_x\bigg(\tfrac {1}{n^{j/2}}\bigg)O_x\bigg(\tfrac {1}{n^{m/2}}\bigg)= O_x\bigg(\tfrac {1}{n^{(m-q)/2}}\bigg), \end{eqnarray*}
which implies that \(n J_2\rightarrow 0\), as \(n\rightarrow \infty \), on choosing \(m {\gt}q+2.\)
From the above estimates of \(J_1\) and \(J_2,\) \(n E_1\rightarrow 0\), as \(n\rightarrow \infty .\) Finally, we have
\[ |E_2|=|\psi (0,x)(-x)^q| \bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}r_{n,0}(\kappa )\bigg)_{\kappa =x}. \]
Since \(|\psi (0,x)(-x)^q|{\lt} N_1\) for some \(N_1{\gt}0,\) also
\[ \bigg(\tfrac {\partial ^q}{\partial \kappa ^{q}}r_{n,0}(\kappa )\bigg)_{\kappa =x}=\bigg[\tfrac {\partial ^q}{\partial \kappa ^{q}} \bigg(e^{-n \kappa /(1+\kappa )}\bigg)\bigg]_{\kappa =x} \]
which yields that \(nE_2\rightarrow 0\) as \(n\rightarrow \infty .\) By collecting the estimates of \(E_1\) and \(E_2,\) we obtain \(n I_2\rightarrow 0\) as \(n\rightarrow \infty .\) Therefore, combining \(I_1\) and \(I_2,\) the required result follows.
Corollary
9
If \(f\in C[0,\infty )\) and \(|f(t)|\leq C (1+t)^{\gamma }\) for some \(\gamma {\gt}0, t\ge 0\) with the existence of \(f^{\prime \prime }\) in \(x\in (0,\infty ),\) then we have
\begin{eqnarray*} \lim _{n\rightarrow \infty }n((V_nf)(x) -f(x))& =& \tfrac {x(x^2+3x+2)}{2}f^{\prime \prime }(x). \end{eqnarray*}
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