Higher order Kantorovich-type Szász–Mirakjan operators

Abstract

In this paper, we define new higher order Kantorovich-type Szász–Mirakjan operators, we give some approximation properties of these operators in terms of various moduli of continuity. We prove a local approximation theorem, a Korovkin-type theorem, and a Voronovskaja-type theorem. We also prove weighted approximation theorems for these new operators.

1 Introduction and auxiliary results

The well-known Bernstein polynomials belonging to a function $f(x)$ defined on the interval $[ 0,1 ]$ are defined as follows:

$$\begin{aligned} B_{n}(f;x)=\sum_{k=0}^{n}f \biggl( \frac{k}{n} \biggr) \begin{pmatrix} n \\ k \end{pmatrix} x^{k}(1-x)^{n-k}\quad (n=1,2,\ldots). \end{aligned}$$

If $f(x)$ is continuous on $[ 0,1 ]$, the polynomials converge uniformly to $f(x)$. These polynomials have an important role in approximation theory and also in other fields of mathematics.

In 1950, for $f\in C[0,\infty )$, Szász [23] defined the operators

$$\begin{aligned} &S_{n} ( f;x ) =\sum_{k=0}^{\infty}p_{n,k}(x)f \biggl( \frac {k}{n} \biggr),\quad x\in {}[ 0,\infty ), n=1,2,\ldots, \\ &\quad\text{where }p_{n,k}(x) =e^{-nx}\frac{(nx)^{k}}{k!}. \end{aligned}$$

In [5], Dubey and Jain proposed the integral modification of the Szász–Mirakjan operators to approximate integrable functions on the interval $[0,\infty ]$, and in [9], Gupta and Sinha studied some direct results on certain Szász–Mirakjan operators. Some related problems were considered by many authors, see for example [1, 2, 5, 10, 13–23] and the references therein.

An operator $L:C[0,1]\rightarrow C[0,1]$ is said to be convex of order $l-1$ if it preserves convexity of order $l-1$, $l\in \mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers. The classical Bernstein operator is an example of a mapping convex of all orders $l-1$, $l\in \mathbb{N}$. For an operator L being convex of order $l-1$, consider

$$\begin{aligned} I_{l}:C[0,1]\rightarrow C[0,1] \end{aligned}$$

given by

$$\begin{aligned} &I_{l}f =f,\quad \text{if }l=0, \\ &(I_{l}f) (x) = \int _{0}^{x}\frac{(x-t)^{l-1}}{(l-1)!}f(t)\,dt,\quad \text{if }l\geq 1. \end{aligned}$$

Suppose that $L(C^{l}[0,1])\subset C^{l}[0,1]$. Let

$$\begin{aligned} Q^{l}:=D^{l}\circ L\circ I_{l},\quad \text{where }D^{l}= \frac{d^{l}}{dx^{l}}. \end{aligned}$$

$Q^{l}$ may be considered as an lth order Kantorovich modification of L. The construction of positive operators $Q^{l}$, $l\geq 0$, is most useful in simultaneous approximation where for appropriate mappings L the difference

$$\begin{aligned} D^{l}Lf-D^{l}f \end{aligned}$$

is considered (see [6, 7, 11]).

On the other hand, we know that Kantorovich-type Szász–Mirakjan operators can be defined as follows:

$$\begin{aligned} K_{n} ( f;x ) =\sum_{k=0}^{\infty}e^{-nx} \frac{(nx)^{k}}{k!} \int _{0}^{1}f \biggl( \frac{k+t}{n} \biggr) \,dt. \end{aligned}$$

By using the lth order integral and the above definition of the Kantorovich-type Szász–Mirakjan operators, we define a new lth order Kantorovich-type Szász–Mirakjan operator as follows:

$$\begin{aligned} K_{n}^{l} ( f;x ) =\sum_{k=0}^{\infty}p_{n,k}(x) \int _{0}^{1}\cdots \int _{0}^{1}f \biggl( \frac{k+t_{1}+\cdots+t_{l}}{n+l} \biggr) \,dt_{1}\cdots dt_{l}, \end{aligned}$$

where $p_{n,k}(x)=e^{-nx}\frac{(nx)^{k}}{k!}$, $n\in \mathbb{N}$, $x\geq 0$, f is a real-valued continuous function defined on $[ 0,\infty )$.

The paper is organized as follows. In the preliminaries section we give some known results and we derive a recurrence formula for the lth order Szász–Mirakjan–Kantorovich operators $K_{n}^{l} ( f;x )$. With the help of the derived recurrence formula, we calculate the moments $K_{n}^{l} ( t^{m};x )$ for $m=0,1,2,3,4$ and we calculate the central moments $K_{n}^{l} ( (t-x)^{m};x )$ for some m. In Sect. 3, we prove a local approximation theorem, a Korovkin-type approximation theorem, and a Voronovskaja-type theorem. We obtain the rate of convergence of these types of operators for Lipschitz-type maximal functions, second order modulus of smoothness and Peetre’s K-functional. In Sect. 4, we investigate weighted approximation properties of the lth order Szász–Mirakjan–Kantorovich operators in terms of the modulus of continuity.

2 Preliminaries

We consider the following class of functions.

Let $C_{B} [ 0,\infty )$ be the space of all real-valued continuous bounded functions f on $[0,\infty )$, endowed with the norm $\Vert f \Vert = \sup_{x\in [ 0,\infty ) } \vert f(x) \vert$.

$$\begin{aligned} C_{B}^{2}[0,\infty ):= \bigl\{ g\in C_{B}[0, \infty ):g^{\prime},g^{ \prime \prime }\in C_{B}[0,\infty ) \bigr\} . \end{aligned}$$

Let $B_{m} [ 0,\infty )$ be the set of all functions f satisfying the condition that $\vert f(x) \vert \leq M_{f}(1+x^{m})$, $x\in [ 0,\infty )$ with some constant $M_{f}$ depending on f. Introduce

$$\begin{aligned} &C_{m} [ 0,\infty ) = \biggl\{ f\in B_{m} [ 0, \infty ) \cap C [ 0,\infty ): \Vert f \Vert _{m}:=\sup_{x\in [ 0,\infty ) } \frac{ \vert f(x) \vert }{1+x^{m}}< \infty \biggr\} , \\ &C_{m}^{\ast} [ 0,\infty ) = \biggl\{ f\in C_{m} [ 0, \infty ):\lim_{x\rightarrow \infty} \frac{ \vert f(x) \vert }{1+x^{m}}< \infty \biggr\} . \end{aligned}$$

In the following lemma we give the moments of the Szász operator up to the fourth order.

Lemma 1

([23])

We have

$$\begin{aligned} &S_{n}(1,x) =1, \\ &S_{n}(t,x) =x, \\ &S_{n}\bigl(t^{2},x\bigr) =x^{2}+ \frac{x}{n}, \\ &S_{n}\bigl(t^{3},x\bigr) =x^{3}+ \frac{3}{n}x^{2}+\frac{1}{n^{2}}x, \\ &S_{n}\bigl(t^{4},x\bigr) =x^{4}+ \frac{6}{n}x^{3}+\frac{7}{n^{2}}x^{2}+ \frac{1}{n^{3}}x. \end{aligned}$$

In the following lemma we derive a recurrence formula for $K_{n}^{l} ( t^{m};x )$ which will be used to calculate moments of the lth order Kantorovich-type Szász–Mirakjan operators.

Lemma 2

For all $n\in \mathbb{N}$, x∈ $[ 0,\infty )$, we have

$$\begin{aligned} K_{n}^{l} \bigl( t^{m};x \bigr) =\sum _{j_{0}+\cdots+j_{l}=m} \begin{pmatrix} m \\ j_{0},\ldots,j_{l}\end{pmatrix} \frac{n^{j_{0}}}{(n+l)^{m}(j_{1}+1)\cdots (j_{l}+1)}S_{n} \bigl( t^{j_{0}},x \bigr), \end{aligned}$$

where $S_{n} ( f,x )$ is the Szász–Mirakjan operator defined in [23].

Proof

We can obtain the recurrence formula with the help of the following equality:

$$\begin{aligned} \biggl( \frac{k+t_{1}+\cdots+t_{l}}{n+l} \biggr) ^{m}={}&\sum _{j_{0}+\cdots+j_{l}=m} \begin{pmatrix} m \\ j_{0},\ldots,j_{l}\end{pmatrix} \frac{k^{j_{0}}t_{1}^{j_{1}}\cdots t_{l}^{j_{l}}}{(n+l)^{m}}\int _{0}^{1}\cdots \int _{0}^{1} \frac{k^{j_{0}}t_{1}^{j_{1}}\cdots t_{l}^{j_{l}}}{(n+l)^{m}} \,dt_{1}\cdots dt_{l}\\ = {}&\frac{k^{j_{0}}}{(n+l)^{m}(j_{1}+1)\cdots (j_{l}+1)}. \end{aligned}$$

Now by direct calculation we write

$$\begin{aligned} K_{n}^{l} \bigl( t^{m};x \bigr) & =\sum _{k=0}^{\infty}p_{n,k} ( x ) \int _{0}^{1}\cdots \int _{0}^{1}f \biggl( \frac {k+t_{1}+\cdots+t_{l}}{n+l} \biggr) ^{m}\,dt_{1}\cdots dt_{l} \\ & =\sum_{k=0}^{\infty}p_{n,k} ( x ) \sum_{j_{0}+\cdots+j_{l}=m} \begin{pmatrix} m \\ j_{0},\ldots,j_{l}\end{pmatrix} \int _{0}^{1}\cdots \int _{0}^{1} \frac{k^{j_{0}}t_{1}^{j_{1}}\cdots t_{l}^{j_{l}}}{(n+l)^{m}}^{m} \,dt_{1}\cdots dt_{l} \\ & =\sum_{k=0}^{\infty}p_{n,k} ( x ) \sum_{j_{0}+\cdots+j_{l}=m} \begin{pmatrix} m \\ j_{0},\ldots,j_{l}\end{pmatrix} \frac{k^{j_{0}}}{(n+l)^{m}(j_{1}+1)\cdots (j_{l}+1)} \,dt_{1}\cdots dt_{l} \\ & =\sum_{j_{0}+\cdots+j_{l}=m} \begin{pmatrix} m \\ j_{0},\ldots,j_{l}\end{pmatrix} \frac{n^{j_{0}}}{(n+l)^{m}(j_{1}+1)\cdots (j_{l}+1)} \sum_{k=0}^{\infty} \frac{k^{j_{0}}}{n^{j_{0}}}p_{n,k} ( x ) \\ & =\sum_{j_{0}+\cdots+j_{l}=m} \begin{pmatrix} m \\ j_{0},\ldots,j_{l}\end{pmatrix} \frac{n^{j_{0}}}{(n+l)^{m}(j_{1}+1)\cdots (j_{l}+1)}S_{n}\bigl(t^{j_{0}},x\bigr), \end{aligned}$$

where $S_{n}(f,x)$ is the Szász–Mirakjan operator. □

Moments and central moments play an important role in approximation theory. In the following lemma we give explicit formulas for the mth ($m=0,1,2,3,4$,) order moments of the lth order Kantorovich-type Szász–Mirakjan operators $K_{n}^{l} ( f;x )$.

Lemma 3

For all $n\in \mathbb{N}$ and $x\in {}[ 0,\infty )$, we have the following equalities:

$$\begin{aligned} &K_{n}^{l} ( 1;x ) =1, \\ &K_{n}^{l} ( t;x ) =\frac{n}{n+l}x+\frac{l}{2(n+l)}, \\ &K_{n}^{l} \bigl( t^{2};x \bigr) = \biggl( \frac{n}{n+l} \biggr) ^{2}x^{2}+ \frac {n(l+1)}{(n+l)^{2}}x+\frac {l(3l+1)}{12(n+l)^{2}}, \\ &K_{n}^{l} \bigl( t^{3};x \bigr) = \biggl( \frac{n}{n+l} \biggr) ^{3}x^{3}+ \frac{3n^{2}l+6n^{2}}{2(n+l)^{3}}x^{2}+ \frac{3nl^{2}+7nl+4n}{4(n+l)^{3}}x+\frac{l^{3}-l^{2}+2l}{8(n+l)^{3}}, \\ &K_{n}^{l} \bigl( t^{4};x \bigr) = \biggl( \frac{n}{n+l} \biggr) ^{4}x^{4}+ \frac{2n^{3}l+6n^{3}}{(n+l)^{4}}x^{3}+ \frac{3n^{2}l^{2}+13n^{2}l+14n^{2}}{2(n+l)^{4}}x^{2} \\ &\phantom{K_{n}^{l} \bigl( t^{4};x \bigr) =}{} +\frac{nl^{3}+2nl^{2}+7nl+2n}{2(n+l)^{4}}x+ \frac{15l^{4}-50l^{3}+185l^{2}-102l}{240(n+l)^{4}}. \end{aligned}$$

Proof

The proof is done by using the recurrence formula given in Lemma 2.

$$\begin{aligned} &K_{n}^{l} ( 1;x ) \quad\text{is obvious. }\\ &K_{n}^{l} ( t;x ) =\sum_{j_{0}+\cdots+j_{l}=1} \begin{pmatrix} 1 \\ j_{0},\ldots,j_{l}\end{pmatrix} \frac{n^{j_{0}}}{(n+l)(j_{1}+1)\cdots (j_{l}+1)}S_{n}\bigl(t^{j_{0}},x \bigr) \\ &\phantom{K_{n}^{l} ( t;x ) } = \begin{pmatrix} l \\ 1 \end{pmatrix} \frac{1}{2(n+l)}+\frac{n}{n+l}x, \\ &K_{n}^{l} \bigl( t^{2};x \bigr) =\sum _{j_{0}+\cdots+j_{l}=2} \begin{pmatrix} 2 \\ j_{0},\ldots,j_{l}\end{pmatrix} \frac{n^{j_{0}}}{(n+l)^{2}(j_{1}+1)\cdots (j_{l}+1)}S_{n} \bigl(t^{j_{0}},x\bigr) \\ &\phantom{K_{n}^{l} \bigl( t^{2};x \bigr) } = \begin{pmatrix} l \\ 2 \end{pmatrix} \frac{2}{4(n+l)^{2}}+ \begin{pmatrix} l \\ 1 \end{pmatrix} \frac{1}{3(n+l)^{2}}+ \begin{pmatrix} l \\ 1 \end{pmatrix} \frac{n}{(n+l)^{2}}x+ \biggl( \frac{n}{n+l} \biggr) ^{2}\biggl(x^{2}+ \frac{x}{n}\biggr) \\ &\phantom{K_{n}^{l} \bigl( t^{2};x \bigr) } = \biggl( \frac{n}{n+l} \biggr) ^{2}x^{2}+ \frac{n(l+1)}{(n+l)^{2}}x+\frac{l(3l+1)}{12(n+l)^{2}}. \end{aligned}$$

$K_{n}^{l} ( t^{3};x )$ and $K_{n}^{l} ( t^{4};x )$ can be done in a similar way. □

In the following lemma we give formulas for the mth order central moments of the lth order Kantorovich-type Szász–Mirakjan operators for $m=1,2,4$.

Lemma 4

For all $n\in \mathbb{N}$, we have the following central moments:

$$\begin{aligned} &K_{n}^{l} \bigl( (t-x);x \bigr) =\frac{l(1-2x)}{2(n+l)}, \\ &K_{n}^{l} \bigl( (t-x)^{2};x \bigr) = \frac{l^{2}}{(n+l)^{2}}x^{2}+\frac{n-l^{2}}{(n+l)^{2}}x+ \frac{3l^{2}+l}{12(n+l)^{2}}, \\ &K_{n}^{l} \bigl( (t-x)^{4};x \bigr) = \biggl( \frac{l}{n+l} \biggr) ^{4}x^{4}+ \frac{6nl^{2}+2l^{4}}{(n+l)^{4}}x^{3}+ \frac{-12nl^{2}+6n^{2}-8nl+3l^{4}+l^{3}}{2(n+l)^{4}}x^{2} \\ &\phantom{K_{n}^{l} \bigl( (t-x)^{4};x \bigr)=}{} +\frac{3nl^{2}+5nl+2n-l^{4}+l^{3}-2l^{2}}{2(n+l)^{4}}x+ \frac{15l^{4}-50l^{3}+185l^{2}-102l}{240(n+l)^{4}}. \end{aligned}$$

Proof

The proof is done by using Lemma 3 and the linearity of the operators.

$$\begin{aligned} &K_{n}^{l} \bigl( (t-x);x \bigr) =K_{n}^{l} ( t;x ) -x= \frac {n}{n+l}x+\frac{l}{2(n+l)}-x=\frac{l(1-2x)}{2(n+l)}, \\ &K_{n}^{l} \bigl( (t-x)^{2};x \bigr) =K_{n}^{l} \bigl( t^{2};x \bigr) -2xK_{n}^{l} ( t;x ) +x^{2} \\ &\phantom{K_{n}^{l} \bigl( (t-x)^{2};x \bigr) } = \biggl( \frac{n}{n+l} \biggr) ^{2}x^{2}+ \frac {n(l+1)}{(n+l)^{2}}x+\frac {l(3l+1)}{12(n+l)^{2}}-2x \biggl( \frac{n}{n+l}x+ \frac{l}{2(n+l)} \biggr) +x^{2} \\ &\phantom{K_{n}^{l} \bigl( (t-x)^{2};x \bigr) } =\frac{l^{2}}{(n+l)^{2}}x^{2}+\frac{n-l^{2}}{(n+l)^{2}}x+ \frac{3l^{2}+l}{12(n+l)^{2}}, \\ &K_{n}^{l} \bigl( (t-x)^{4};x \bigr) \\ &\quad=K_{n}^{l} \bigl( t^{4};x \bigr) -4xK_{n}^{l} \bigl( t^{3};x \bigr) +6x^{2}K_{n}^{l} \bigl( t^{2};x \bigr) -4x^{3}K_{n}^{l} ( t;x ) +x^{4} \\ &\quad = \biggl( \frac{n}{n+l} \biggr) ^{4}x^{4}+ \frac{2n^{3}l+6n^{3}}{(n+l)^{4}}x^{3}+\frac{3n^{2}l^{2}+13n^{2}l+14n^{2}}{2(n+l)^{4}}x^{2} \\ &\qquad{} +\frac{nl^{3}+2nl^{2}+7nl+2n}{2(n+l)^{4}}x+ \frac{15l^{4}-50l^{3}+185l^{2}-102l}{240(n+l)^{4}} \\ &\qquad{} -4x \biggl( \biggl( \frac{n}{n+l} \biggr) ^{3}x^{3}+ \frac{3n^{2}l+6n^{2}}{2(n+l)^{3}}x^{2}+\frac{3nl^{2}+7nl+4n}{4(n+l)^{3}}x+ \frac{l^{3}-l^{2}+2l}{8(n+l)^{3}} \biggr) \\ &\qquad{}+6x^{2} \biggl( \biggl( \frac{n}{n+l} \biggr) ^{2}x^{2}+ \frac {n(l+1)}{(n+l)^{2}}x+\frac {l(3l+1)}{12(n+l)^{2}} \biggr) -4x^{3} \biggl( \frac{n}{n+l}x+\frac{l}{2(n+l)} \biggr) +x^{4} \\ &\quad = \biggl( \frac{l}{n+l} \biggr) ^{4}x^{4}+ \frac{6nl^{2}+2l^{4}}{(n+l)^{4}}x^{3}+\frac{-12nl^{2}+6n^{2}-8nl+3l^{4}+l^{3}}{2(n+l)^{4}}x^{2} \\ &\qquad{} +\frac{3nl^{2}+5nl+2n-l^{4}+l^{3}-2l^{2}}{2(n+l)^{4}}x+ \frac{15l^{4}-50l^{3}+185l^{2}-102l}{240(n+l)^{4}}. \end{aligned}$$

 □

One of the main problems in approximation theory is to estimate the rate of convergence for sequences of positive linear operators. Voronovskaja-type formulas are one of the most important tools for studying their asymptotic behavior. In the following lemma we give two limits that later will be used to prove Voronovskaja-type theorem for the lth order Kantorovich-type Szász–Mirakjan operators.

Lemma 5

For $x\in {}[ 0,\infty )$ and $n\rightarrow \infty$, we have the following limits:

$$\begin{aligned} &\mathit{(i)}\quad\lim_{n\rightarrow \infty}nK_{n}^{l} ( t-x;x ) = \frac {l(1-2x)}{2}, \\ &\mathit{(ii)}\quad\lim_{n\rightarrow \infty}nK_{n}^{l} \bigl( ( t-x ) ^{2};x \bigr) =x. \end{aligned}$$

Proof

The proof is trivial with the use of the formulas $K_{n}^{l} ( t-x;x )$ and $K_{n}^{l} ( (t-x)^{2};x )$ given in Lemma 3,

$$\begin{aligned} &\mathit{(i)}\quad\lim_{n\rightarrow \infty}nK_{n}^{l} ( t-x;x ) = \lim _{n\rightarrow \infty}\frac{nl(1-2x)}{2(n+l)}= \frac {l(1-2x)}{2}, \\ &\mathit{(ii)}\quad\lim_{n\rightarrow \infty}nK_{n}^{l} \bigl( ( t-x ) ^{2};x \bigr) =\lim_{n\rightarrow \infty} \biggl\{ \frac{nl^{2}}{(n+l)^{2}}x^{2}+ \frac{n^{2}-nl^{2}}{(n+l)^{2}}x+\frac{3nl^{2}+nl}{12(n+l)^{2}}\biggr\} \\ &\phantom{\mathit{(ii)}\quad\lim_{n\rightarrow \infty}nK_{n}^{l} \bigl( ( t-x ) ^{2};x \bigr) } =x. \end{aligned}$$

 □

3 Local approximation

In this section, we establish local approximation theorem for the lth order Kantorovich-type Szász–Mirakjan operators. We consider the Peetre’s K-functional

$$\begin{aligned} K_{2}(f,\delta ):=\inf \bigl\{ \Vert f-g \Vert +\delta \bigl\Vert g^{\prime \prime} \bigr\Vert :g\in C_{B}^{2}[0, \infty ) \bigr\} ,\quad \delta \geq 0. \end{aligned}$$

Then from the known result in [4], there exists an absolute constant $C>0$ such that

$$\begin{aligned} K_{2}(f,\delta )\leq C\omega _{2}(f,\sqrt{\delta}), \end{aligned}$$

(1)

where

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

is the second modulus of smoothness of $f\in C_{B}[0,\infty )$.

In the following theorem we state the first main result for the local approximation of our operators $K_{n}^{l}(f;x)$.

Theorem 6

There exists an absolute constant $C>0$ such that

$$\begin{aligned} \bigl\vert K_{n}^{l}(f;x)-f(x) \bigr\vert \leq C\omega _{2}\bigl(f,\sqrt{ \delta _{n}(x)}\bigr)+\omega \bigl( f, \theta _{n}(x) \bigr), \end{aligned}$$

where

$$\begin{aligned} &f \in C_{B}[0,\infty ], \quad\delta _{n}(x)=K_{n}^{l} \bigl((t-x)^{2};x\bigr)+ \bigl( K_{n}^{l} \bigl((t-x);x\bigr) \bigr) ^{2}\\ &\phantom{f \in C_{B}[0,\infty ], \quad\delta _{n}(x)}=\frac{6l^{2}+l}{12(n+l)^{2}}+ \frac{n-2l^{2}}{(n+l)^{2}}x+\frac{2l^{2}}{(n+l)^{2}}x^{2}, \\ &\theta _{n}(x) = \bigl\vert K_{n}^{l} \bigl((t-x);x\bigr) \bigr\vert = \biggl\vert \frac{l(1-2x)}{2(n+l)} \biggr\vert ,\quad 0\leq x< \infty. \end{aligned}$$

Proof

Let

$$\begin{aligned} \widetilde{K}_{n}^{l}(f;x)=K_{n}^{l}(f;x)+f(x)-f \bigl( \mu _{n}(x) \bigr), \end{aligned}$$

where $f\in C_{B}[0,\infty ]$, $\mu _{n}(x)=K_{n}^{l}((t-x);x)+x=\frac {l+2nx}{2(n+l)}$. Note that $\widetilde{K}_{n}^{l}((t-x);x)=0$. By using Taylor’s formula, we have

$$\begin{aligned} g(t)=g(x)+g^{\prime}(x) (t-x)+ \int _{x}^{t}(t-s)g^{\prime \prime}(s)\,ds,\quad g\in C_{B}^{2}[0,\infty ). \end{aligned}$$

Applying $\widetilde{K}_{n}^{l}$ to both sides of the above equation, we have

$$\begin{aligned} &\widetilde{K}_{n}^{l}(g;x)-g(x) \\ &\quad =\widetilde{K}_{n}^{l} \bigl((t-x)g^{ \prime }(x);x\bigr)+\widetilde{K}_{n}^{l} \biggl( \int _{x}^{t}(t-s)g^{ \prime \prime }(s)\,ds;x \biggr) \\ &\quad =g^{\prime}(x)\widetilde{K}_{n}^{l}\bigl((t-x);x \bigr)+K_{n}^{l} \biggl( \int _{x}^{t}(t-s)g^{\prime \prime}(s) \,ds;x \biggr) - \int _{x}^{\mu _{n}(x)} \bigl( \mu _{n}(x)-s \bigr) g^{\prime \prime}(s)\,ds \\ &\quad =K_{n}^{l} \biggl( \int _{x}^{t}(t-s)g^{\prime \prime}(s)\,ds;x \biggr) - \int _{x}^{\mu _{n}(x)} \bigl( \mu _{n}(x)-s \bigr) g^{\prime \prime}(s)\,ds. \end{aligned}$$

On the other hand,

$$\begin{aligned} \biggl\vert \int _{x}^{t}(t-s)g^{\prime \prime}(s)\,ds \biggr\vert \leq \int _{x}^{t}(t-s) \bigl\vert g^{\prime \prime}(s) \bigr\vert \,ds\leq \bigl\Vert g^{\prime \prime} \bigr\Vert \int _{x}^{t}(t-s)\,ds\leq \bigl\Vert g^{\prime \prime } \bigr\Vert (t-s)^{2} \end{aligned}$$

and

$$\begin{aligned} \biggl\vert \int _{x}^{\mu _{n}(x)} \bigl( \mu _{n}(x)-s \bigr) g^{ \prime \prime }(s)\,ds \biggr\vert \leq \bigl\Vert g^{\prime \prime} \bigr\Vert \bigl(\mu _{n}(x)-x\bigr)^{2}= \bigl\Vert g^{\prime \prime} \bigr\Vert \bigl( K_{n}^{l}(t-x;x) \bigr) ^{2}, \end{aligned}$$

which implies

$$\begin{aligned} \bigl\vert \widetilde{K}_{n}^{l}(g;x)-g(x) \bigr\vert & \leq \biggl\vert K_{n}^{l} \biggl( \int _{x}^{t}(t-s)g^{\prime \prime}(s)\,ds;x \biggr) \biggr\vert + \biggl\vert \int _{x}^{\mu _{n}(x)} \bigl( \mu _{n}(x)-s \bigr) g^{\prime \prime }(s)\,ds \biggr\vert \\ & \leq \bigl\Vert g^{\prime \prime} \bigr\Vert \bigl\{ K_{n}^{l}\bigl((t-x)^{2};x\bigr)+\bigl(K_{n}^{l} ( t-x );x \bigr)^{2} \bigr\} \\ & = \bigl\Vert g^{\prime \prime} \bigr\Vert \delta _{n}(x). \end{aligned}$$

(2)

We also have

$$\begin{aligned} \bigl\vert \widetilde{K}_{n}^{l}(f;x) \bigr\vert \leq \bigl\vert K_{n}^{l}(f;x) \bigr\vert + \bigl\vert f(x) \bigr\vert + \bigl\vert f\bigl( \mu _{n}(x)\bigr) \bigr\vert \leq K_{n}^{l} \bigl( \vert f \vert ;x \bigr) +2 \Vert f \Vert \leq 3 \Vert f \Vert . \end{aligned}$$

Using (2) and the uniform boundedness of $\widetilde{K}_{n}^{l}$, we get

$$\begin{aligned} \bigl\vert K_{n}^{l}(f;x)-f(x) \bigr\vert & \leq \bigl\vert \widetilde{K}_{n}^{l}(f-g;x) \bigr\vert + \bigl\vert \widetilde{K}_{n}^{l}(g;x)-g(x) \bigr\vert + \bigl\vert f(x)-g(x) \bigr\vert + \bigl\vert f \bigl( \mu _{n}(x) \bigr) -f(x) \bigr\vert \\ & \leq 4 \Vert f-g \Vert + \bigl\Vert g^{\prime \prime} \bigr\Vert \delta _{n}(x)+\omega \bigl( f,\theta _{n}(x) \bigr) \text{.} \end{aligned}$$

If we take the infimum on the right hand side over all $g\in C_{B}^{2}[0,\infty )$, we obtain

$$\begin{aligned} \bigl\vert K_{n}^{l}(f;x)-f(x) \bigr\vert \leq 4K_{2} \bigl( f; \delta _{n}(x) \bigr) +\omega \bigl( f, \theta _{n}(x) \bigr), \end{aligned}$$

which together with (1) gives the proof of the theorem. □

Corollary 7

Let $A>0$. Then, for each $f\in C[0,\infty )$, the sequence of operators $K_{n}^{l}(f;x)$ converges to f uniformly on $[ 0,A ]$.

Theorem 8

Let $f\in C_{2}^{\ast} [ 0,\infty )$. Then $\lim_{n\rightarrow \infty}K_{n}^{l}(f;x)=f(x)$, uniformly on $[0,A]$.

Proof

Since

$$\begin{aligned} K_{n}^{l}(1;x)\rightarrow 1,\qquad K_{n}^{l}(t;x) \rightarrow x,\qquad K_{n}^{l}\bigl(t^{2};x \bigr)\rightarrow x^{2}\quad\text{as }n\rightarrow \infty {,} \end{aligned}$$

uniformly in $[ 0,\infty )$. By the Korovkin theorem, $K_{n}^{l}(f;x)$ converges to $f(x)$ uniformly on $[0,A]$. □

Theorem 9

Let $n\geq l^{2}$, $f\in C_{2} [ 0,\infty )$ and $\omega _{A+1}(f,\delta )=\sup_{ \vert t-x \vert \leq \delta}\sup_{x,t\in {}[ 0,A+1]} \vert f(t)-f(x) \vert$ be the modulus of continuity on the interval $[ 0,A+1 ] \subset [ 0,\infty )$, where $A>0$. Then we have

$$\begin{aligned} \bigl\Vert K_{n}^{l}(f;x)-f(x) \bigr\Vert _{C [ 0,A ] } \leq 4M_{f} \bigl( 1+A^{2} \bigr) \alpha _{n}(A)+2\omega _{A+1} \bigl( f,\sqrt{\alpha _{n}(A)} \bigr), \end{aligned}$$

where $\alpha _{n}(A)=K_{n}^{l}((t-x)^{2};A)$.

Proof

For $x\in [ 0,A ]$ and $t\geq 0$, we can get (see [8], Eq. 3.3)

$$\begin{aligned} \bigl\vert f(t)-f(x) \bigr\vert \leq 4M_{f} \bigl( 1+A^{2} \bigr) (t-x)^{2}+ \biggl( 1+\frac{ \vert t-x \vert }{\delta} \biggr) \omega _{A+1} ( f,\delta ). \end{aligned}$$

Now, by the Cauchy–Schwarz inequality, we have

$$\begin{aligned} \bigl\vert K_{n}^{l}(f;x)-f(x) \bigr\vert & \leq K_{n}^{l} \bigl( \bigl\vert f(t)-f(x) \bigr\vert ;x \bigr) \\ & \leq 4M_{f} \bigl( 1+A^{2} \bigr) K_{n}^{l} \bigl( (t-x)^{2};x \bigr) + \biggl( 1+K_{n}^{l} \biggl( \frac{ \vert t-x \vert }{\delta };x \biggr) \biggr) \omega _{A+1} ( f,\delta ) \\ & \leq 4M_{f} \bigl( 1+A^{2} \bigr) K_{n}^{l} \bigl( (t-x)^{2};x \bigr) +\omega _{A+1} ( f,\delta ) \biggl( 1+ \frac{1}{\delta} \bigl( K_{n}^{l} \bigl( (t-x)^{2};x \bigr) \bigr) ^{ \frac{1}{2}} \biggr). \end{aligned}$$

For $x\in [ 0,A ]$, using Lemma 4,

$$\begin{aligned} K_{n}^{l} \bigl( (t-x)^{2};x \bigr) = \frac{x^{2}l^{2}}{(n+l)^{2}}+\frac{x(n-l^{2})}{(n+l)^{2}}+\frac{3l^{2}+l}{12(n+l)^{2}}\leq \alpha _{n}(A). \end{aligned}$$

Thus we get

$$\begin{aligned} \bigl\vert K_{n}^{l}(f;x)-f(x) \bigr\vert \leq 4M_{f} \bigl( 1+A^{2} \bigr) \alpha _{n}(A)+ \omega _{A+1} ( f,\delta ) \biggl( 1+\frac{1}{\delta } \bigl( \alpha _{n}(A) \bigr) ^{\frac{1}{2}} \biggr). \end{aligned}$$

By taking $\delta =\sqrt{\alpha _{n}(A)}$, we get the desired result. □

In the following theorem we give a Voronovskaja-type result for the lth order Kantorovich-type Szász–Mirakjan operators.

Theorem 10

For any $f\in C_{B}^{2}[0,\infty )$, the following asymptotic equality holds:

$$\begin{aligned} \lim_{n\rightarrow \infty}n\bigl(K_{n}^{l}(f;x)-f(x) \bigr)=\frac {l(1-2x)}{2}f^{ \prime }(x)+\frac{1}{2}xf^{\prime \prime}(x) \end{aligned}$$

uniformly on $[0,A]$.

Proof

Let $f\in C_{B}^{2}[0,\infty )$ and $x\in {}[ 0,\infty )$ be fixed. By using Taylor’s formula, we write

$$\begin{aligned} f(t)=f(x)+f^{\prime}(x) (t-x)+\frac{1}{2}f^{\prime \prime}(x) (t-x)^{2}+r(t,x) (t-x)^{2} {,} \end{aligned}$$

(3)

where the function $r(t,x)$ is the Peano form of the remainder, $r(t,x)\in C_{B}[0,\infty )$ and $\lim_{t\rightarrow x}r(t,x)=0$. Applying $K_{n}^{l}$ to (3), we obtain

$$\begin{aligned} & n \bigl( K_{n}^{l}(f;x)-f(x) \bigr) \\ &\quad =nf^{\prime}(x)K_{n}^{l}(t-x;x)+ \frac{n}{2}f^{\prime \prime}(x)K_{n}^{l}\bigl( ( t-x ) ^{2};x\bigr)+nK_{n}^{l}\bigl(r(t,x) ( t-x ) ^{2};x\bigr). \end{aligned}$$

By using the Cauchy–Schwarz inequality, we get

$$\begin{aligned} K_{n}^{l}\bigl(r(t,x) ( t-x ) ^{2};x\bigr)\leq \sqrt{K_{n}^{l}\bigl(r^{2}(t,x);x \bigr)}\sqrt{K_{n}^{l}\bigl((t-x)^{4};x \bigr).} \end{aligned}$$

(4)

We observe that $r^{2}(x,x)=0$ and $r^{2}(.,x)\in C_{B}[0,\infty )$. Now from Corollary 7 it follows that

$$\begin{aligned} \lim_{n\rightarrow \infty}K_{n}^{l}\bigl(r^{2}(t,x);x \bigr)=r^{2}(x,x)=0 \end{aligned}$$

(5)

uniformly with respect to $x\in {}[ 0,A]$. Finally, from (4), (5), and Lemma 5, we get immediately

$$\begin{aligned} \lim_{n\rightarrow \infty}nK_{n}^{l}\bigl(r(t,x) ( t-x ) ^{2};x\bigr)=0, \end{aligned}$$

which completes the proof. □

Theorem 11

Let $\alpha \in (0,1]$ and S be any subset of the interval $[0,\infty )$. Then, if $f\in C_{B}[0,\infty )$ is locally $Lip(\alpha )$, i.e., the condition

$$\begin{aligned} \bigl\vert f(y)-f(x) \bigr\vert \leq L \vert y-x \vert ^{ \alpha },\quad y\in S\textit{ and }x\in {}[ 0,\infty ) \end{aligned}$$

(6)

holds, then, for each $x\in {}[ 0,\infty )$, we have

$$\begin{aligned} \bigl\vert K_{n}^{l}(f;x)-f(x) \bigr\vert \leq L \bigl\{ \lambda _{n}^{\frac{\alpha}{2}}(x)+2\bigl(d(x,S) \bigr)^{\alpha} \bigr\} , \end{aligned}$$

where $\lambda _{n}(x)=\frac{3l^{2}+l}{12(n+l)^{2}}+ \frac{n-l^{2}}{(n+l)^{2}}x+\frac{l^{2}}{(n+l)^{2}}x^{2}$, L is a constant depending on α and f, and $d(x,S)$ is the distance between x and S defined as

$$\begin{aligned} d(x,S)=\inf \bigl\{ \vert t-x \vert :t\in S \bigr\} . \end{aligned}$$

Proof

Let S̄ be the closure of S in $[0,\infty )$. Then there exists a point $x_{0}\in \bar{S}$ such that $\vert x-x_{0} \vert =d(x,S)$. By the triangle inequality

$$\begin{aligned} \bigl\vert f(t)-f(x) \bigr\vert \leq \bigl\vert f(t)-f(x_{0}) \bigr\vert + \bigl\vert f(x)-f(x_{0}) \bigr\vert \end{aligned}$$

and by (6), we get

$$\begin{aligned} \bigl\vert K_{n}^{l}(f;x)-f(x) \bigr\vert & \leq K_{n}^{l} \bigl( \bigl\vert f(t)-f(x_{0}) \bigr\vert ;x \bigr) +K_{n}^{l} \bigl( \bigl\vert f(x)-f(x_{0}) \bigr\vert ;x \bigr) \\ & \leq L \bigl\{ K_{n}^{l} \bigl( \vert t-x_{0} \vert ^{ \alpha };x \bigr) + \vert x-x_{0} \vert ^{\alpha} \bigr\} \\ & \leq L \bigl\{ K_{n}^{l} \bigl( \vert t-x \vert ^{ \alpha }+ \vert x-x_{0} \vert ^{\alpha};x \bigr) + \vert x-x_{0} \vert ^{\alpha} \bigr\} \\ & \leq L \bigl\{ K_{n}^{l} \bigl( \vert t-x \vert ^{ \alpha };x \bigr) +2 \vert x-x_{0} \vert ^{\alpha} \bigr\} . \end{aligned}$$

Now, by using the Hölder inequality with $p=\frac{2}{\alpha}$ and $q=\frac{2}{2-\alpha}$, we get

$$\begin{aligned} \bigl\vert K_{n}^{l}(f;x)-f(x) \bigr\vert & \leq L \bigl\{ \bigl[ K_{n}^{l} \bigl( \vert t-x \vert ^{\alpha p};x \bigr) \bigr] ^{\frac{1}{p}} \bigl[ K_{n}^{l} \bigl( 1^{q};x \bigr) \bigr] ^{ \frac{1}{q}}+2 \bigl( d(x,S) \bigr) ^{\alpha} \bigr\} \\ & =L \bigl\{ \bigl[ K_{n}^{l} \bigl( \vert t-x \vert ^{2};x \bigr) \bigr] ^{\frac{\alpha}{2}}+2 \bigl( d(x,S) \bigr) ^{ \alpha } \bigr\} \\ & =L \biggl\{ \biggl[ \frac{3l^{2}+l}{12(n+l)^{2}}+ \frac{n-l^{2}}{(n+l)^{2}}x+ \frac{l^{2}}{(n+l)^{2}}x^{2} \biggr] ^{\frac{\alpha}{2}}+2 \bigl( d(x,S) \bigr) ^{\alpha} \biggr\} \\ & =L \bigl\{ \bigl(\lambda _{n}(x)\bigr)^{\frac{\alpha}{2}}+2 \bigl( d(x,S) \bigr) ^{\alpha} \bigr\} , \end{aligned}$$

and the proof is completed. □

4 Weighted approximation

In this section, we give weighted approximation theorems for the lth order Kantorovich-type Szász–Mirakjan operators. We will use the following two lemmas which can be found in [3] and [12].

Lemma 12

For $m\in \mathbb{N}$, we have

$$\begin{aligned} S_{n}\bigl(t^{j_{0}};x\bigr)=\sum_{j=1}^{j_{0}}a_{j_{0},j} \frac {x^{j}}{n^{j_{0}-j}}, \end{aligned}$$

(7)

where

$$\begin{aligned} &a_{j_{0}+1,j} =ja_{j_{0},j}+a_{j_{0},j-1},\quad j_{0}\geq 0, j\geq 1, \\ &a_{0,0} =1, \qquad a_{j_{0},0}=0,\quad j_{0}>0,\qquad a_{j_{0},j}=0,\quad j_{0}< j. \end{aligned}$$

Lemma 13

Let $m\in \mathbb{N} \cup \{ 0 \}$ and $l\in \mathbb{Z} ^{+}$ be fixed. Then there exists a positive constant $C_{m}(l)$ such that

$$\begin{aligned} \bigl\Vert K_{n}^{l}\bigl(1+t^{m};x\bigr) \bigr\Vert _{m}\leq C_{m}(l),\quad n\in \mathbb{N}. \end{aligned}$$

(8)

Moreover, for every $f\in C_{2}^{\ast} [ 0,\infty )$, we have

$$\begin{aligned} \bigl\Vert K_{n}^{l}(f;x) \bigr\Vert _{m}\leq C_{m}(l) \Vert f \Vert _{m},\quad n\in \mathbb{N}. \end{aligned}$$

(9)

Thus $K_{n}^{l}$ is a linear positive operator from $C_{m}^{\ast} [ 0,\infty )$ into $C_{m}^{\ast} [ 0,\infty )$ for any $m\in \mathbb{N} \cup \{ 0 \}$.

Proof

Inequality (8) is obvious for $m=0$. Let $m\geq 1$. Then, by Lemma 12, we have

$$\begin{aligned} &\frac{1}{1+x^{m}}K_{n}^{l}\bigl(1+t^{m};x\bigr)\\ &\quad= \frac{1}{1+x^{m}}+ \frac{1}{1+x^{m}}\sum_{j_{0}+\cdots+j_{l}=m} \binom{m}{j_{0,}\cdots j_{l}} \frac{n^{j_{0}}}{(n+l)^{m}(j_{1}+1)\cdots (j_{l}+1)} \sum_{j=1}^{j_{0}}a_{j_{0},j} \frac{x^{j}}{n^{j_{0}-j}}. \end{aligned}$$

Thus

$$\begin{aligned} \frac{1}{1+x^{m}}K_{n}^{l}\bigl(1+t^{m};x\bigr) \leq 1+k_{m}(l)=C_{m}(l), \end{aligned}$$

where $C_{m}(l)$ is a positive constant depending on m and l. On the other hand,

$$\begin{aligned} \bigl\Vert K_{n}^{l}(f;x) \bigr\Vert _{m}\leq \Vert f \Vert _{m} \bigl\Vert K_{n}^{l} \bigl(1+t^{m};x\bigr) \bigr\Vert _{m} \end{aligned}$$

for every $f\in C_{m}^{\ast} [ 0,\infty )$. By applying (8), we obtain (9). □

Theorem 14

For each $f\in C_{2}^{\ast} [ 0,\infty )$, one has

$$\begin{aligned} \lim_{n\rightarrow \infty} \bigl\Vert K_{n}^{l}(f;x)-f(x) \bigr\Vert _{2}=0. \end{aligned}$$

Proof

To prove this theorem, we need to use a Korovkin-type theorem on weighted approximation. That is, it is sufficient to verify the following three conditions:

$$\begin{aligned} \lim_{n\rightarrow \infty} \bigl\Vert K_{n}^{l} \bigl(t^{m};x\bigr)-x^{m} \bigr\Vert _{2}=0,\quad m=0,1,2. \end{aligned}$$

For $m=0$, it is obvious. For $m=1$, we have

$$\begin{aligned} \lim_{n\rightarrow \infty} \bigl\Vert K_{n}^{l}(t;x)-x \bigr\Vert _{2} & =\sup_{x\geq 0} \frac{ \vert K_{n}^{l}(t;x)-x \vert }{1+x^{2}} \\ & =\sup_{x\geq 0}\frac{1}{1+x^{2}} \biggl\vert \frac{l}{2(n+l)}+\frac{n}{ ( n+l ) }x-x \biggr\vert \\ & \leq \frac{l}{2(n+l)}\sup_{x\geq 0}\frac{1}{1+x^{2}}+ \frac {l}{ ( n+l ) }\sup_{x\geq 0} \frac{x}{1+x^{2}} \\ & \leq \frac{l}{2(n+l)}+\frac{l}{ ( n+l ) }= \frac{3l}{2(n+l)}, \end{aligned}$$

and by a similar way, we can write

$$\begin{aligned} &\lim_{n\rightarrow \infty} \bigl\Vert K_{n}^{l} \bigl(t^{2};x\bigr)-x \bigr\Vert _{2}\\ & \quad=\sup _{x\geq 0} \frac{ \vert K_{n}^{l}(t^{2};x)-x^{2} \vert }{1+x^{2}} \\ &\quad =\sup_{x\geq 0}\frac{1}{1+x^{2}} \biggl\vert \frac{n^{2}}{(n+l)^{2}}x^{2}+\frac{n(l+1)}{(n+l)^{2}}x+\frac{3l^{2}+l}{12(n+l)^{2}}-x^{2} \biggr\vert \\ & \quad\leq \biggl\vert \frac{-l^{2}-2nl}{(n+l)^{2}} \biggr\vert \sup_{x\geq 0} \frac{x^{2}}{1+x^{2}}+ \frac{n(l+1)}{(n+l)^{2}}\sup_{x\geq 0} \frac{x}{1+x^{2}}+ \frac{3l^{2}+l}{12(n+l)^{2}}\sup_{x\geq 0} \frac {1}{1+x^{2}} \\ & \quad\leq \frac{l^{2}+2nl}{(n+l)^{2}}+\frac{n(l+1)}{(n+l)^{2}}+ \frac{3l^{2}+l}{12(n+l)^{2}}, \end{aligned}$$

which implies that

$$\begin{aligned} \lim_{n\rightarrow \infty} \bigl\Vert K_{n}^{l} \bigl(t^{m};x\bigr)-x^{m} \bigr\Vert _{2}=0,\quad m=0,1,2. \end{aligned}$$

 □

Theorem 15

For each $f\in C_{2}^{\ast} [ 0,\infty )$ and all $\beta >0$, one has

$$\begin{aligned} \lim_{n\rightarrow \infty}\sup_{x\geq 0}\frac{ \vert K_{n}^{l}(f;x)-f(x) \vert }{(1+x^{2})^{1+\beta}}=0. \end{aligned}$$

Proof

For any fixed $0< A<\infty$ and by Lemma 13, we have

$$\begin{aligned} \sup_{x\geq 0} \frac{ \vert K_{n}^{l}(f;x)-f(x) \vert }{ ( 1+x^{2} ) ^{1+\beta}} ={}&\sup_{x\leq A}\frac{ \vert K_{n}^{l}(f;x)-f(x) \vert }{ ( 1+x^{2} ) ^{1+\beta}}+\sup_{x\geq A} \frac{ \vert K_{n}^{l}(f;x)-f(x) \vert }{ ( 1+x^{2} ) ^{1+\beta}} \\ \leq{}& \sup_{x\leq A} \bigl\vert K_{n}^{l}(f;x)-f(x) \bigr\vert +\sup_{x\geq A} \frac{ \vert K_{n}^{l}(f;x) \vert + \vert f(x) \vert }{ ( 1+x^{2} ) ^{1+\beta}} \\ \leq{}& \bigl\Vert K_{n}^{l}(f)-f \bigr\Vert _{C [ 0,A ] }+ \Vert f \Vert _{2}\sup_{x\geq A} \frac{\vert K_{n}^{l}(1+t^{2};x\vert }{ ( 1+x^{2} ) ^{1+\beta}} \\ &{} +\sup_{x\geq A} \frac{ \vert f(x) \vert }{ ( 1+x^{2} ) ^{1+\beta}} \\ ={}&J_{1}+J_{2}+J_{3}. \end{aligned}$$

(10)

Using Theorem 9, we can see that $J_{1}$ goes to zero as $n\rightarrow \infty$.

By Theorem 14, we can get

$$\begin{aligned} J_{2} & = \Vert f \Vert _{2} \lim_{n\rightarrow \infty} \sup_{x\geq A} \frac{\vert K_{n}^{l}(1+t^{2};x\vert }{ ( 1+x^{2} ) ^{1+\beta}} \\ & =\sup_{x\geq A} \frac{ \Vert f \Vert _{2}}{ ( 1+x^{2} ) ^{\beta}}\leq \frac{ \Vert f \Vert _{2}}{ ( 1+A^{2} ) ^{\beta}}. \end{aligned}$$

Since $\vert f(x) \vert \leq M_{f}(1+x^{2})$,

$$\begin{aligned} J_{3}=\sup_{x\geq A} \frac{ \vert f(x) \vert }{ ( 1+x^{2} ) ^{1+\beta}}\leq \sup _{x\geq A} \frac{M_{f}}{ ( 1+x^{2} ) ^{\beta}}\leq \frac{M_{f}}{ ( 1+A^{2} ) ^{\beta}}. \end{aligned}$$

If we choose A large enough, we get

$$\begin{aligned} J_{2}\rightarrow 0\quad\text{and}\quad J_{3}\rightarrow 0\quad\text{as }n \rightarrow \infty. \end{aligned}$$

Hence by (10) we obtain the desired result

$$\begin{aligned} \lim_{n\rightarrow \infty} \sup_{x\geq 0}\frac{ \vert K_{n}^{l}(f;x)-f(x) \vert }{(1+x^{2})^{1+\beta}}=0. \end{aligned}$$

 □

For $f\in C_{2}^{\ast} [ 0,\infty )$, the weighted modulus of continuity is defined as

$$\begin{aligned} \Omega _{m}(f,\delta )= \sup_{x\geq 0, 0< h\leq \delta}\frac{ \vert f(x+h)-f(x) \vert }{1+(x+h)^{m}}. \end{aligned}$$

Lemma 16

If $f\in C_{m}^{\ast} [ 0,\infty ),m\in \mathbb{N}$, then

• (i) $\Omega _{m}(f,\delta )$ is a monotone increasing function of δ,

• (ii) $\lim_{\delta \rightarrow \infty}\Omega _{m}(f,\delta )=0$,

• (iii) for any $\rho \in [ 0,\infty ),\Omega _{m}(f,\rho \delta )\leq (1+\rho )\Omega _{m}(f,\delta )$.

Theorem 17

If $f\in C_{m}^{\ast} [ 0,\infty )$, then

$$\begin{aligned} \bigl\Vert K_{n}^{l}(f)-f \bigr\Vert _{m+1}\leq k\Omega _{m} \biggl( f, \frac {1}{\sqrt{n+l}} \biggr), \end{aligned}$$

where k is a constant independent of f and n.

Proof

From the definition of $\Omega _{m}(f,\delta )$ and Lemma 16, we may write

$$\begin{aligned} \bigl\vert f(t)-f(x) \bigr\vert & \leq \bigl( 1+ \bigl( x+ \vert t-x \vert \bigr) ^{m} \bigr) \biggl( \frac{ \vert t-x \vert }{\delta}+1 \biggr) \Omega _{m}(f, \delta ) \\ & \leq \bigl( 1+ ( 2x+t ) ^{m} \bigr) \biggl( \frac{ \vert t-x \vert }{\delta}+1 \biggr) \Omega _{m}(f, \delta ). \end{aligned}$$

Then we have

$$\begin{aligned} \bigl\vert K_{n}^{l}(f;x)-f(x) \bigr\vert & \leq K_{n}^{l}\vert \bigl(f(t)-f(x)\vert;x\bigr) \\ & \leq \Omega _{m}(f,\delta )K_{n}^{l} \bigl( 1+(2x+t)^{m};x \bigr) +K_{n}^{l} \bigl( \bigl( 1+(2x+t)^{m} \bigr);x \bigr) \\ & =\Omega _{m}(f,\delta )K_{n}^{l} \bigl( 1+(2x+t)^{m};x \bigr) +I_{1}. \end{aligned}$$

Applying the Cauchy–Schwarz inequality to $I_{1}$, we get

$$\begin{aligned} I_{1}\leq K_{n}^{l} \bigl( \bigl(1+(2x+t)^{m} \bigr) ^{2};x\bigr))^{1/2} \biggl( K_{n}^{l} \biggl( \frac{ \vert t-x \vert ^{2}}{\delta ^{2}};x \biggr) \biggr) ^{1/2}. \end{aligned}$$

Therefore,

$$\begin{aligned} \bigl\vert K_{n}^{l}(f;x)-f(x) \bigr\vert \leq {}&\Omega _{m}(f, \delta )K_{n}^{l} \bigl(1+(2x+t)^{m};x\bigr)\\ &{}+K_{n}^{l} \bigl( \bigl(1+(2x+t)^{m} \bigr) ^{2};x \bigr))^{1/2} \biggl( K_{n}^{l} \biggl( \frac{ \vert t-x \vert ^{2}}{\delta ^{2}};x \biggr) \biggr) ^{1/2}. \end{aligned}$$

From Lemmas 13 and 12, we have

$$\begin{aligned} &K_{n}^{l}\bigl(1+(2x+t)^{m};x\bigr)\leq C_{m}(l) \bigl( 1+x^{m} \bigr),\\ &K_{n}^{l} \bigl( \bigl(1+(2x+t)^{m} \bigr) ^{2};x\bigr))^{1/2} \biggl( K_{n}^{l} \biggl( \frac{ \vert t-x \vert ^{2}}{\delta ^{2}};x \biggr) \biggr) ^{1/2}\leq C_{m}^{1}(l) \bigl( 1+x^{m} \bigr). \end{aligned}$$

Also, from Lemma 4, we have

$$\begin{aligned} \biggl( K_{n}^{l} \biggl( \frac{ \vert t-x \vert ^{2}}{\delta ^{2}};x \biggr) \biggr) ^{1/2} & \leq \frac{1}{\delta}\sqrt{ \frac{x^{2}l^{2}}{(n+l)^{2}}+ \frac{x(n-l^{2})}{(n+l)^{2}}+ \frac{3l^{2}+l}{12(n+l)^{2}}} \\ & \leq \frac{l(1+x)}{\delta \sqrt{ ( n+l ) }}. \end{aligned}$$

So, if we combine all these results, we get

$$\begin{aligned} \bigl\vert K_{n}^{l}(f;x)-f(x) \bigr\vert & \leq \Omega _{m}(f, \delta ) \biggl( C_{m}(l) \bigl( 1+x^{m} \bigr) +C_{m}^{1}(l) \frac{ ( 1+x^{m} ) ( 1+x ) l}{\delta \sqrt{ ( n+l ) }} \biggr) \\ & =\Omega _{m}(f,\delta ) \biggl( C_{m}(l) \bigl( 1+x^{m} \bigr) +C_{m}^{1}(l)C_{1} \frac{l ( 1+x^{m+1} ) }{\delta \sqrt{ ( n+l ) }} \biggr), \end{aligned}$$

where

$$\begin{aligned} C_{1}=\sup_{x\geq 0}\frac{1+x^{m}+x+x^{m+1}}{1+x^{m+1}}. \end{aligned}$$

In the above inequality, if we substitute $\frac{1}{\sqrt{n+l}}$ instead of δ, we obtain the desired result. □

5 Conclusion

In this paper, by using the lth order integration and the definition of the Kantorovich type Szász–Mirakjan operators, we defined a new lth order Kantorovich-type Szász–Mirakjan operator. We derived a recurrence formula, and with the help of this formula we calculated the moments $K_{n}^{l} ( t^{m};x )$ for $m=0,1,2,3,4$ and we calculated the central moments $K_{n}^{l} ( (t-x)^{m};x )$ for $m=1,2,4$. We established a local approximation theorem, a Korovkin-type approximation theorem, and a Voronovskaja-type theorem. We obtained the rate of convergence of these types of operators for Lipschitz-type maximal functions, second order modulus of smoothness, and Peetre’s K-functional. At last we investigated weighted approximation properties of the lth order Szász–Mirakjan–Kantorovich operators in terms of the modulus of continuity.