1 Introduction

Let f be a real function on \(\mathbb {R}\) which is bounded on each finite interval. For \(a,b\in \mathbb {R}\) with \(a<b\), define the function \(f_{a,b}\) on \(\left[ 0,1\right] \) by \(f_{a,b}\left( t\right) =f\left( a+\left( b-a\right) t\right) \). Furthermore, put

$$\begin{aligned} \left\| f\right\| _{a,b}=\sup _{a\le t\le b}\left| f\left( t\right) \right| . \end{aligned}$$

Obviously, \(f_{0,1}\) is the restriction of f to \(\left[ 0,1\right] \) and we have \(\left\| f\right\| _{a,b}=\left\| f_{a,b}\right\| _{0,1}\).

The Bernstein–Chlodovsky operators applied to the function f described above are defined by

$$\begin{aligned} \left( C_{n,a,b}f\right) \left( x\right) =\left( B_{n}f_{a,b}\right) \left( \frac{x-a}{b-a}\right) , \qquad a \le x \le b , \end{aligned}$$

where \(B_{n}\) denote the Bernstein operators defined by

$$\begin{aligned} \left( B_{n}f\right) \left( x\right) =\sum _{\nu =0}^{n}p_{n,\nu }\left( x\right) f\left( \frac{\nu }{n}\right) , \qquad 0 \le x \le 1 , \end{aligned}$$

with Bernstein basis polynomials

$$\begin{aligned} p_{n,\nu }\left( x\right) ={\left( {\begin{array}{c}{n}\\ {\nu }\end{array}}\right) }x^{\nu }\left( 1-x\right) ^{n-\nu },\,\quad 0\le \nu \le n. \end{aligned}$$

In the special case \(\left[ a,b\right] =\left[ 0,1\right] \), we have \( C_{n,0,1}\equiv B_{n}\).

The symmetric version

$$\begin{aligned} \left( C_{n,-c,c}f\right) \left( x\right) =\left( B_{n}f_{-c,c}\right) \left( \frac{x+c}{2c}\right) \end{aligned}$$

with \(-a=b=c>0\) was recently introduced by Kilgore [12, Eq. (7.1)]. Using this Chlodovsky generalization of the Bernstein operators he [12, Theorems 1 and 2 ] gave a constructive proof for the Weierstrass approximation theorem in weighted spaces of continuous functions defined on \(\left[ 0,\infty \right) \) or on \(\left( -\infty ,\infty \right) \). See also [13].

In the following we suppose that the parameters a, b are coupled with n, i.e., \(a=a_{n}\) and \(b=b_{n}\). Because the difference between two nodes of \( C_{n,a,b}\) is at least \(\left( b-a\right) /n\) it is clear that the condition \(b_{n}-a_{n}=o\left( n\right) \) as \(n\rightarrow \infty \) is necessary for having convergence of \(\left( C_{n,a_{n},b_{n}}f\right) \left( x\right) \) to \(f\left( x\right) \).

In the special case \(a_{n}=0<b_{n}\), for \(n\in \mathbb {N}\), these polynomials were introduced by I. Chlodovsky [7] in 1937 in order to approximate functions on infinite intervals. He showed that under the condition (1.3), if a function f satisfies

$$\begin{aligned} \lim _{n\rightarrow \infty }\exp \left( -\frac{\sigma n}{b_{n}}\right) \left\| f\right\| _{b_{n}}=0, \end{aligned}$$
(1.1)

for every \(\sigma >0\), then

$$\begin{aligned} \lim _{n\rightarrow \infty }\left( C_{n,0,b_{n}}f\right) \left( x\right) =f\left( x\right) \end{aligned}$$

at each point x of continuity of f. Moreover, he proved convergence in each continuity point for the large class of functions f satisfying the growth condition \(f\left( t\right) =O\left( \exp \left( t^{p}\right) \right) \) as \(t\rightarrow +\infty \), if the sequence \(\left( b_{n}\right) \) satisfies the condition

$$\begin{aligned} b_{n}=O\left( n^{1/\left( p+1+\eta \right) }\right) \qquad \left( n\rightarrow \infty \right) , \end{aligned}$$
(1.2)

for an arbitrary small \(\eta >0\). For more results on Chlodovsky operators see the survey article [9] by Karsli.

Explicit expressions of the coefficients \(c_{k}^{\left[ b_{n}\right] }\left( f,x\right) \) in terms of Stirling numbers were given by Karsli [10]. He derived the asymptotic expansion if the function f satisfies condition (1.1) for every \(\sigma >0\).

Throughout the paper we assume that the sequences \(\left( a_{n}\right) \) and \(\left( b_{n}\right) \) satisfy

$$\begin{aligned} b_{n}-a_{n}>0,\,\quad \lim _{n\rightarrow \infty }\left( -a_{n}\right) =\lim _{n\rightarrow \infty }b_{n}=+\infty ,\text { }\quad \text {and }\quad \lim _{n\rightarrow \infty }\frac{-a_{n}b_{n}}{n}=0. \end{aligned}$$
(1.3)

The purpose of this note is a pointwise complete asymptotic expansion for the sequence of Bernstein–Chlodovsky operators in the form:

$$\begin{aligned} \left( C_{n,a_{n},b_{n}}f\right) \left( x\right) \sim f\left( x\right) +\sum _{k=1}^{\infty }c_{k}^{\left[ a_{n},b_{n}\right] }\left( f,x\right) \left( \frac{-a_{n}b_{n}}{n}\right) ^{k}+o\left( \left( \frac{-a_{n}b_{n}}{n} \right) ^{q}\right) \end{aligned}$$
(1.4)

as \(n\rightarrow \infty \), for sufficiently smooth functions f satisfying \( f\left( t\right) =O\left( \exp \left( \alpha t^{p}\right) \right) \) as \( t\rightarrow +\infty \), provided that the sequences \(\left( a_{n}\right) ,\) \( \left( b_{n}\right) \) satisfy \(\left( -a_{n}b_{n}\right) =o\left( n^{1/\left( p+1\right) }\right) \) as \(n\rightarrow \infty \). The coefficients \(c_{k}^{\left[ a_{n},b_{n}\right] }\left( f,x\right) \), which depend on f and \(a_{n},b_{n}\), are bounded with respect to n.

The latter formula means that, for each fixed \(x>0\) and for all positive integers q:

$$\begin{aligned} \left( C_{n,a_{n},b_{n}}f\right) \left( x\right) =f\left( x\right) +\sum _{k=1}^{q}c_{k}^{\left[ a_{n},b_{n}\right] }\left( f,x\right) \left( \frac{-a_{n}b_{n}}{n}\right) ^{k}+o\left( \left( \frac{-a_{n}b_{n}}{n} \right) ^{q}\right) \end{aligned}$$

as \(n\rightarrow \infty \).

2 Main Result

For real constants \(\alpha \ge 0\) and \(p\ge 0\), let \(W_{\alpha ,p}\) denote the class of functions \(f\in C\left( \mathbb {R}\right) \) satisfying the growth condition:

$$\begin{aligned} f\left( t\right) =O\left( \exp \left( \alpha \left| t\right| ^{p}\right) \right) \qquad \left( \left| t\right| \rightarrow +\infty \right) . \end{aligned}$$

Note that in the special instance \(p=0\) the class \(W_{\alpha ,0}\) consists of the bounded continuous functions on \(\mathbb {R}\). Since \(W_{0,p}\) and \( W_{\alpha ,0}\) coincide we consider only the case \(\alpha >0\).

Recall that the Stirling numbers \(s\left( n,k\right) \) and \(S\left( n,k\right) \) of first and second kind, respectively, are defined by the relations:

$$\begin{aligned} z^{\underline{n}}=\sum _{k=0}^{n}s\left( n,k\right) z^{k} \qquad \text {and } \qquad z^{n}=\sum _{k=0}^{n}S\left( n,k\right) z^{\underline{k}}\qquad \left( z\in \mathbb {C}\right) , \end{aligned}$$

where \(z^{\underline{0}}=1\) and \(z^{\underline{n}}=z\left( z-1\right) \cdots \left( z-n+1\right) \), for \(n\in \mathbb {N}\), denote the falling factorials.

The following theorem is the main result.

Theorem 2.1

Let \(\alpha ,p\ge 0\). Suppose that the function \(f\in W_{\alpha ,p}\) is 2q times differentiable in the point \( x>0\). Let \(\left( -a_{n}\right) \) and \(\left( b_{n}\right) \) be sequences of reals tending to infinity and satisfying the growth condition:

$$\begin{aligned} -a_{n}b_{n}=o\left( n^{1/\left( p+1\right) }\right) \text { }\qquad \left( n\rightarrow \infty \right) . \end{aligned}$$
(2.1)

Then, for any positive integer q, the Bernstein–Chlodovsky operators \( C_{n,a_{n},b_{n}}\) possess the asymptotic expansion:

$$\begin{aligned} \left( C_{n,a_{n},b_{n}}f\right) \left( x\right) =f\left( x\right) +\sum _{k=1}^{q}c_{k}^{\left[ a_{n},b_{n}\right] }\left( f,x\right) \left( \frac{-a_{n}b_{n}}{n}\right) ^{k}+o\left( \left( \frac{-a_{n}b_{n}}{n} \right) ^{q}\right) \end{aligned}$$

as \(n\rightarrow \infty \), where

$$\begin{aligned} c_{k}^{\left[ a_{n},b_{n}\right] }\left( f,x\right) =O\left( 1\right) \text { }\qquad \left( n\rightarrow \infty \right) . \end{aligned}$$
(2.2)

The coefficients \(c_{k}^{\left[ a,b\right] }\left( f,x\right) \) have the explicit representation:

$$\begin{aligned} c_{k}^{\left[ a,b\right] }\left( f,x\right) =\sum _{s=k}^{2k}\frac{f^{\left( s\right) }\left( x\right) }{s!}\sum _{j=0}^{s}A\left( k,s,j\right) \frac{ \left( b-a\right) ^{j}\left( x-a\right) ^{s-j}}{\left( -ab\right) ^{k}}\text { }\qquad \left( a<b\right) \end{aligned}$$

with numbers

$$\begin{aligned} A\left( k,s,j\right) =\sum _{r=\max \left\{ j,k\right\} }^{s}\left( -1\right) ^{s-r}\left( {\begin{array}{c}s\\ r\end{array}}\right) s\left( r-j,r-k\right) S\left( r,r-j\right) . \end{aligned}$$
(2.3)

Remark 2.2

Note that the coefficients \(c_{k}^{\left[ a_{n},b_{n}\right] }\left( f,x\right) \) depend on n but are bounded with respect to n.

Remark 2.3

Our assumption (2.1) on the sequence \(\left( -a_{n}b_{n}\right) \) corresponds to Chlodovsky’s condition (1.2). Furthermore, it is related to the assumption in the case \(a_{n}=0\) (see, [3, Theorem 1, Eq. (4)]).

In the special case \(q=1\) Theorem 2.1 implies the following Voronovskaja-type result.

Corollary 2.4

Let \(\alpha ,p\ge 0\). Suppose that the function \(f\in W_{\alpha ,p}\) admits a second derivative at the point \(x>0\). Let \(\left( -a_{n}\right) \) and \(\left( b_{n}\right) \) be sequences of reals tending to infinity and satisfying the growth condition (2.1) . Then, the Bernstein–Chlodovsky operators \(C_{n,a_{n},b_{n}}\) satisfy the asymptotic relation:

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{n}{-a_{n}b_{n}}\left( \left( C_{n,a_{n},b_{n}}f\right) \left( x\right) -f\left( x\right) \right) =\frac{1 }{2}f^{\left( 2\right) }\left( x\right) \end{aligned}$$
(2.4)

Remark 2.5

The expansion in Theorem 2.1 is completely different to the (pointwise) complete asymptotic expansion:

$$\begin{aligned} \left( B_{n}f\right) \left( x\right) \sim f\left( x\right) +\sum _{k=1}^{\infty }c_{k}^{\left[ 0,1\right] }\left( f,x\right) n^{-k}\text { }\qquad \left( n\rightarrow \infty \right) , \end{aligned}$$

for the classical Bernstein polynomials \(B_{n}\), which is valid for all bounded functions \(f:\left[ 0,1\right] \rightarrow \mathbb {R}\) being sufficiently smooth in \(x\in \left[ 0,1\right] \). The Voronovskaja formula states that

$$\begin{aligned} \lim _{n\rightarrow \infty }n\left( \left( B_{n}f\right) \left( x\right) -f\left( x\right) \right) =\frac{1}{2}x\left( 1-x\right) f^{\left( 2\right) }\left( x\right) . \end{aligned}$$

The same is true in the case of the classical Bernstein–Chlodovsky operators \(C_{n,0,b_{n}}\). Their Voronovskaja-type formula:

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{n}{b_{n}}\left( \left( C_{n,0,b_{n}}f\right) \left( x\right) -f\left( x\right) \right) =\frac{1}{2} xf^{\left( 2\right) }\left( x\right) \end{aligned}$$

was derived in 1960 by Albrycht and Radecki [5]. For further history consult the survey article [9].

3 Auxiliary Results and Proof of the Main Theorem

Our starting-point is an explicit representation of the central moments of the Bernstein polynomials in terms of Stirling numbers of the first and second kind. In the following we write \(e_{m}\left( x\right) =x^{m}\), \(m\in \mathbb {N}_{0}\), for the m-th monomial and \(\psi _{x}\left( t\right) =t-x\) for \(x\in \mathbb {R}\).

Lemma 3.1

The central moments of the Bernstein polynomials possess the representation

$$\begin{aligned} \left( B_{n}\psi _{x}^{s}\right) \left( x\right) =\sum _{k=\left\lfloor \frac{ s+1}{2}\right\rfloor }^{s}n^{-k}\sum _{j=0}^{s}A\left( k,s,j\right) x^{s-j} \end{aligned}$$

\(\left( s=0,1,2,\ldots \right) \), where the coefficients are given by Eq. (2.3).

For a proof see, e.g., [2].

Lemma 3.2

The central moments of the Bernstein–Chlodovsky operators possess the representation:

$$\begin{aligned} \left( C_{n,a,b}\psi _{x}^{s}\right) \left( x\right) =\sum _{k=\left\lfloor \frac{s+1}{2}\right\rfloor }^{s}n^{-k}\sum _{j=0}^{s}A\left( k,s,j\right) \left( b-a\right) ^{j}\left( x-a\right) ^{s-j}, \end{aligned}$$
(3.1)

where the coefficients \(A\left( k,s,j\right) \) are given by Eq. (2.3) .

Proof

We have

$$\begin{aligned} \left( C_{n,a,b}\psi _{x}^{s}\right) \left( x\right)= & {} \sum _{\nu =0}^{n}p_{n,\nu }\left( \frac{x-a}{b-a}\right) \left( a+\left( b-a\right) \frac{\nu }{n}-x\right) ^{s} \\= & {} \left( b-a\right) ^{s}\sum _{\nu =0}^{n}p_{n,\nu }\left( \frac{x-a}{b-a} \right) \left( \frac{\nu }{n}-\frac{x-a}{b-a}\right) ^{s} \\= & {} \left( b-a\right) ^{s}\left( B_{n}\psi _{\frac{x-a}{b-a}}^{s}\right) \left( \frac{x-a}{b-a}\right) \end{aligned}$$

and the lemma follows by Lemma 3.1. \(\square \)

As we have seen in the proof of Lemma 3.2 the central moments of the Bernstein–Chlodovsky operators can be expressed in terms of Bernstein polynomials:

$$\begin{aligned} \left( C_{n,a,b}\psi _{x}^{s}\right) \left( x\right) =\left( b-a\right) ^{s}\left( B_{n}\psi _{\frac{x-a}{b-a}}^{s}\right) \left( \frac{x-a}{b-a} \right) . \end{aligned}$$

As a consequence from well-known properties of the Bernstein polynomials we obtain the following result:

Lemma 3.3

For \(a\le x\le b\), it holds

$$\begin{aligned} \left( C_{n,a,b}\psi _{x}^{2s}\right) \left( x\right)= & {} \sum _{k=s}^{2s}\left( \frac{\left( x-a\right) \left( b-x\right) }{n} \right) ^{k} \\&\times \sum _{i=2s-k}^{s}d_{2s,i}\left( \frac{\left( x-a\right) \left( b-x\right) }{\left( b-a\right) ^{2}}\right) ^{i-k}\left( b-a\right) ^{2s-2k}, \\ \left( C_{n,a,b}\psi _{x}^{2s+1}\right) \left( x\right)= & {} \left( b+a-2x\right) \sum _{k=s+1}^{2s+1}\left( \frac{\left( x-a\right) \left( b-x\right) }{n}\right) ^{k} \\&\times \sum _{i=2s+1-k}^{s}d_{2s+1,i}\left( \frac{\left( x-a\right) \left( b-x\right) }{\left( b-a\right) ^{2}}\right) ^{i-k}\left( b-a\right) ^{2s-2k}, \end{aligned}$$

where \(d_{s,i}\) are certain real numbers.

Proof

Taking advantage of the well-known formulas for the central moments of the Bernstein polynomials (see [8, Chapt. 10, Theorem 1.1]):

$$\begin{aligned} \left( B_{n}\psi _{x}^{2s}\right) \left( x\right)= & {} \sum _{j=0}^{s}n^{j-2s}\sum _{i=j}^{s}d_{2s,i}\left( x\left( 1-x\right) \right) ^{i}, \\ \left( B_{n}\psi _{x}^{2s+1}\right) \left( x\right)= & {} \left( 1-2x\right) \sum _{j=0}^{s}n^{j-\left( 2s+1\right) }\sum _{i=j}^{s}d_{2s+1,i}\left( x\left( 1-x\right) \right) ^{i}, \end{aligned}$$

where \(d_{s,i}\) are certain real numbers, we obtain

$$\begin{aligned} \left( C_{n,a,b}\psi _{x}^{2s}\right) \left( x\right)= & {} \left( b-a\right) ^{2s}\sum _{k=s}^{2s}\frac{1}{n^{k}}\sum _{i=2s-k}^{s}d_{2s,i}\left( \frac{ \left( x-a\right) \left( b-x\right) }{\left( b-a\right) ^{2}}\right) ^{i}, \\ \left( C_{n,a,b}\psi _{x}^{2s+1}\right) \left( x\right)= & {} \left( b-a\right) ^{2s}\left( b+a-2x\right) \sum _{k=s+1}^{2s+1}\frac{1}{n^{k}} \\&\times \sum _{i=2s+1-k}^{s}d_{2s+1,i}\left( \frac{\left( x-a\right) \left( b-x\right) }{\left( b-a\right) ^{2}}\right) ^{i}. \end{aligned}$$

This can be rewritten in the form as stated in the lemma. \(\square \)

For the sake of brevity, in the following, we write

$$\begin{aligned} \left( C_{n,a,b}\psi _{x}^{s}\right) \left( x\right) =\sum _{k=\left\lfloor \frac{s+1}{2}\right\rfloor }^{s}\left( \frac{-ab}{n}\right) ^{k}Q_{k,s}\left( a,b,s;x\right) . \end{aligned}$$
(3.2)

Note that, for \(a\le x\le b\), with \(a<0<b\), we have

$$\begin{aligned} \left| \frac{\left( x-a\right) \left( b-x\right) }{n}\right|\le & {} \left| 1-\frac{x}{a}\right| \left| 1-\frac{x}{b}\right| \cdot \frac{\left| ab\right| }{n}, \\ \left| \frac{\left( x-a\right) \left( b-x\right) }{\left( b-a\right) ^{2} }\right|\le & {} \frac{1}{4}, \\ \left| \frac{b+a-2x}{b-a}\right|\le & {} 1, \\ \left( b-a\right) ^{m}\le & {} 1, \quad \text {for }m\le 0, \quad \text {if } b-a\ge 1. \end{aligned}$$

This immediately implies the following estimate for the central moment of the Bernstein–Chlodovsky operators.

Lemma 3.4

Let \(\left( -a_{n}\right) \) and \(\left( b_{n}\right) \) be sequences of reals tending to infinity and satisfying the condition \(-a_{n}b_{n}=o\left( n\right) \) as \(n\rightarrow \infty \). Then, for \(s=0,1,2,\ldots \), the quantities \(Q_{k,s}\left( a_{n},b_{n},s;x\right) \) \(\left( \left\lfloor \frac{s+1}{2}\right\rfloor \le k\le s\right) \) are bounded with respect to n, and

$$\begin{aligned} \left( C_{n,a_{n},b_{n}}\psi _{x}^{s}\right) \left( x\right) =O\left( \left( \frac{-a_{n}b_{n}}{n}\right) ^{\left\lfloor \frac{s+1}{2}\right\rfloor }\right) \text { }\qquad \left( n\rightarrow \infty \right) . \end{aligned}$$

A crucial tool is the following estimate due to Bernstein (see [14, Theorem  1.5.3, p. 18f]).

Lemma 3.5

(Bernstein)For \(0\le t\le 1\), the inequality

$$\begin{aligned} 0\le z\le \frac{3}{2}\sqrt{nt\left( 1-t\right) } \end{aligned}$$

implies

$$\begin{aligned} \sum _{\begin{array}{c} \nu \\ \left| \nu -nt\right| \ge 2z\sqrt{nt\left( 1-t\right) } \end{array}}p_{n\nu }\left( t\right) \le 2\exp \left( -z^{2}\right) . \end{aligned}$$

The next lemma presents a form of Lemma 3.5 which is more useful for application to Chlodovsky operators on the real line. It follows the idea of Albrycht and Radecki [5] who proved a similar result for the classical Chlodovsky operators.

Lemma 3.6

Let \(a<x<b\). If \(0<\delta \le 3\left( x-a\right) \left( b-x\right) /\left( b-a\right) \) it holds

$$\begin{aligned} \sum _{\begin{array}{c} \nu \\ \left| a+\left( b-a\right) \frac{\nu }{n} \right| \ge \delta \end{array}}p_{n\nu }\left( \frac{x-a}{b-a}\right) \le 2\exp \left( -\frac{n\delta ^{2}}{4\left( x-a\right) \left( b-x\right) }\right) \end{aligned}$$

Proof of Lemma 3.6

Putting \(t=\left( x-a\right) /\left( b-a\right) \) in Lemma 3.5, we have

$$\begin{aligned}&\sum _{\begin{array}{c} \nu \\ \left| \nu -n\frac{x-a}{b-a}\right| \ge 2z\sqrt{n\frac{x-a}{b-a}\frac{b-x}{b-a}} \end{array}}p_{n\nu }\left( \frac{x-a}{b-a} \right) \\&\quad =\sum _{\begin{array}{c} \nu \\ \left| a-x+\left( b-a\right) \frac{\nu }{n} \right| \ge 2z\sqrt{n^{-1}\left( x-a\right) \left( b-x\right) } \end{array}} p_{n\nu }\left( \frac{x-a}{b-a}\right) \le 2\exp \left( -z^{2}\right) \end{aligned}$$

if \(0\le z\le \frac{3}{2}\sqrt{n\frac{x-a}{b-a}\frac{b-x}{b-a}}\). Choose \( \delta =2z\sqrt{n^{-1}\left( x-a\right) \left( b-x\right) }\). Then

$$\begin{aligned} \sum _{\begin{array}{c} \nu \\ \left| a-x+\left( b-a\right) \frac{\nu }{n} \right| \ge \delta \end{array}}p_{n\nu }\left( \frac{x-a}{b-a}\right) \le 2\exp \left( -\frac{n\delta ^{2}}{4\left( x-a\right) \left( b-x\right) }\right) \end{aligned}$$

if \(0\le \delta /\left( 2\sqrt{n^{-1}\left( x-a\right) \left( b-x\right) } \right) \le \frac{3}{2}\sqrt{n\frac{x-a}{b-a}\frac{b-x}{b-a}}\). The latter inequality is equivalent to

$$\begin{aligned} \delta \le 3\frac{\left( x-a\right) \left( b-x\right) }{b-a} \end{aligned}$$

which is a condition of the lemma. \(\square \)

Lemma 3.7

Let \(a<x<b\) and \(0<\delta \le 3\left( x-a\right) \left( b-x\right) /\left( b-a\right) \). If a bounded function \(f:\left[ a,b\right] \rightarrow \mathbb { R}\) satisfies \(f\left( t\right) =0\), for all \(t\in \left( x-\delta ,x+\delta \right) \cap \left[ a,b\right] \), it follows the estimate

$$\begin{aligned} \left| \left( C_{n,a,b}f\right) \left( x\right) \right| \le 2\exp \left( -\frac{n\delta ^{2}}{4\left( x-a\right) \left( b-x\right) }\right) \left\| f\right\| _{a,b}. \end{aligned}$$

Proof

Because of \(f\left( a+\left( b-a\right) \frac{\nu }{n}\right) =0\) for all \( \nu \in \left\{ 0,\ldots ,n\right\} \) with \(\left| a+\left( b-a\right) \frac{\nu }{n}-x\right| <\delta \) we have

$$\begin{aligned} \left| \left( C_{n,a,b}f\right) \left( x\right) \right|= & {} \left| \underset{\left| a+\left( b-a\right) \frac{\nu }{n} \right| \ge \delta }{\sum \limits _{\nu }}p_{n,\nu }\left( \frac{x-a}{b-a} \right) f\left( a+\left( b-a\right) \frac{\nu }{n}\right) \right| \\\le & {} \left\| f\right\| _{a,b}\underset{\left| a+\left( b-a\right) \frac{\nu }{n}\right| \ge \delta }{\sum \limits _{\nu }} p_{n,\nu }\left( \frac{x-a}{b-a}\right) \end{aligned}$$

and the assertion follows by an application of Lemma 3.6. \(\square \)

A direct consequence is the following localization result for Bernstein–Chlodovsky polynomials which is interesting in itself.

Proposition 3.8

(Localization theorem). Let \(\alpha ,p\ge 0\) be fixed constants and propose that \(f\in W_{\alpha ,p} \) satisfies the estimate

$$\begin{aligned} \left| f\left( t\right) \right| \le K\exp \left( \alpha \left| t\right| ^{p}\right) \text { }\qquad \left( t\in \mathbb {R}\right) . \end{aligned}$$

Furthermore, fix the real number \(x\in \left( a,b\right) \) and let \(\delta >0 \). Then \(f\left( t\right) =0\), for all \(t\in \left( x-\delta ,x+\delta \right) \), implies

$$\begin{aligned} \left| \left( C_{n,a,b}f\right) \left( x\right) \right| \le 2K\exp \left( -\frac{n\delta ^{2}}{4\left( x-a\right) \left( b-x\right) }\right) \exp \left( \alpha \max \left\{ \left| a\right| ,\left| b\right| \right\} ^{p}\right) . \end{aligned}$$

Suppose that \(\lim _{n\rightarrow \infty }\left( -a_{n}\right) =\lim _{n\rightarrow \infty }b_{n}=+\infty \). Then, for sufficiently large values of n, we can assume that \(a_{n}<-1<1<b_{n}\), such that \(\max \left\{ \left| a_{n}\right| ,\left| b_{n}\right| \right\} \le -a_{n}b_{n}\). Since \(\left( x-a_{n}\right) \left( b_{n}-x\right) =O\left( -a_{n}b_{n}\right) \) as \(n\rightarrow \infty \), there is as positive constant \(M\left( x\right) \) (independent of n), such that

$$\begin{aligned} \frac{1}{4\left( x-a_{n}\right) \left( b_{n}-x\right) }\ge \frac{M\left( x\right) }{-a_{n}b_{n}}, \end{aligned}$$

for sufficiently large values of n. Hence, we have

$$\begin{aligned} \exp \left( -\frac{n\delta ^{2}}{4\left( x-a\right) \left( b-x\right) } \right) \le \exp \left( -M\left( x\right) \frac{n\delta ^{2}}{-a_{n}b_{n}} \right) . \end{aligned}$$

We conclude that, for large n,

$$\begin{aligned} \left| \left( C_{n,a_{n},b_{n}}f\right) \left( x\right) \right|\le & {} 2K\exp \left( -M\left( x\right) \frac{n\delta ^{2}}{-a_{n}b_{n}} +\alpha \left( -a_{n}b_{n}\right) ^{p}\right) \\= & {} 2K\exp \left( -M\left( x\right) \frac{n}{-a_{n}b_{n}}\left( \delta ^{2}- \frac{\alpha }{M\left( x\right) }\cdot \frac{\left( -a_{n}b_{n}\right) ^{p+1} }{n}\right) \right) . \end{aligned}$$

Proof of Theorem 2.1

Suppose that f is continuous on \(\mathbb {R}\) being 2q times differentiable at the point \(x\in \mathbb {R}\). Define the function \(h_{x}\) by

$$\begin{aligned} f=\sum _{s=0}^{2q}\frac{f^{\left( s\right) }\left( x\right) }{s!}\psi _{x}^{s}+h_{x}\psi _{x}^{2q} \end{aligned}$$
(3.3)

and \(h_{x}\left( x\right) =0\). It is a consequence of Taylor’s theorem that \( h_{x}\) is continuous at x. Hence, \(h_{x}\in C\left( \mathbb {R}\right) \). Applying the operator \(C_{n,a,b}\) to both sides of Eq. (3.3) we obtain

$$\begin{aligned} \left( C_{n,a,b}f\right) \left( x\right) =\sum _{s=0}^{2q}\frac{f^{\left( s\right) }\left( x\right) }{s!}\left( C_{n,a,b}\psi _{x}^{s}\right) \left( x\right) +\left( C_{n,a,b}\left( h_{x}\psi _{x}^{2q}\right) \right) \left( x\right) . \end{aligned}$$

The first sum is equal to

$$\begin{aligned} \sum _{s=0}^{2q}\frac{f^{\left( s\right) }\left( x\right) }{s!}\left( C_{n,a,b}\psi _{x}^{s}\right) \left( x\right)= & {} \sum _{s=0}^{2q}\frac{ f^{\left( s\right) }\left( x\right) }{s!}\sum _{k=\left\lfloor \frac{s+1}{2} \right\rfloor }^{s}\left( \frac{-ab}{n}\right) ^{k}Q_{k,s}\left( a,b,s;x\right) \\= & {} \sum _{k=0}^{2q}\left( \frac{-ab}{n}\right) ^{k}\sum _{s=k}^{2k}\frac{ f^{\left( s\right) }\left( x\right) }{s!}Q_{k,s}\left( a,b,s;x\right) \\= & {} \sum _{k=0}^{2q}c_{k}^{\left[ a,b\right] }\left( f,x\right) \left( \frac{ -ab}{n}\right) ^{k}. \end{aligned}$$

Note that \(c_{0}^{\left[ a,b\right] }\left( f,x\right) =1\). Eq. (2.2) is a consequence of Lemma 3.4. We conclude that

$$\begin{aligned} \sum _{s=0}^{2q}\frac{f^{\left( s\right) }\left( x\right) }{s!}\left( C_{n,a_{n},b_{n}}\psi _{x}^{s}\right) \left( x\right) =\sum _{k=0}^{q}c_{k}^{ \left[ a_{n},b_{n}\right] }\left( f,x\right) \left( \frac{-a_{n}b_{n}}{n} \right) ^{k}+o\left( \left( \frac{-a_{n}b_{n}}{n}\right) ^{q}\right) \end{aligned}$$

as \(n\rightarrow \infty \). In order to complete the proof we have to show that the remainder can be estimated by

$$\begin{aligned} \left( C_{n,a_{n},b_{n}}\left( h_{x}\psi _{x}^{2q}\right) \right) \left( x\right) =o\left( \left( \frac{-a_{n}b_{n}}{n}\right) ^{q}\right) \text { } \qquad \left( n\rightarrow \infty \right) . \end{aligned}$$

To this end let \(\left( \delta _{n}\right) \) be a sequence of positive numbers such that

$$\begin{aligned} \delta _{n}^{2}=\frac{-a_{n}b_{n}}{nM\left( x\right) }\left( \alpha \left( -a_{n}b_{n}\right) ^{p}-q\log \frac{-a_{n}b_{n}}{n}+\sqrt{\frac{n}{ -a_{n}b_{n}}}\right) \text { }\qquad \left( n\in \mathbb {N}\right) . \end{aligned}$$
(3.4)

Note that the conditions (1.3) and (2.1) imply that \(\delta _{n}=o\left( 1\right) \) as \( n\rightarrow \infty \). Define

$$\begin{aligned} \varepsilon _{n}=\sup \left\{ \left| h_{x}\left( t\right) \right| :t\in \left( x-\delta _{n},x+\delta _{n}\right) \right\} . \end{aligned}$$

Because \(h_{x}\) is continuous with \(h_{x}\left( x\right) =0\) we have \( \varepsilon _{n}=o\left( 1\right) \) as \(n\rightarrow \infty \). We split the remainder into two parts

$$\begin{aligned}&\left( C_{n,a_{n},b_{n}}\left( h_{x}\psi _{x}^{2q}\right) \right) \left( x\right) \\&\quad =\underset{\left| a_{n}+\left( b_{n}-a_{n}\right) \frac{\nu }{n} -x\right| <\delta _{n}}{\sum \limits _{\nu }}p_{n,\nu }\left( \frac{x-a_{n} }{b_{n}-a_{n}}\right) \left( h_{x}\psi _{x}^{2q}\right) \left( a_{n}+\left( b_{n}-a_{n}\right) \frac{\nu }{n}\right) \\&\qquad +\underset{\left| a_{n}+\left( b_{n}-a_{n}\right) \frac{\nu }{n} -x\right| \ge \delta _{n}}{\sum \limits _{\nu }}p_{n,\nu }\left( \frac{ x-a_{n}}{b_{n}-a_{n}}\right) \left( h_{x}\psi _{x}^{2q}\right) \left( a_{n}+\left( b_{n}-a_{n}\right) \frac{\nu }{n}\right) \\&\quad =\sum \nolimits _{1}+\sum \nolimits _{2}, \end{aligned}$$

say. Let us start with the estimate of the first sum:

$$\begin{aligned} \left| \sum \nolimits _{1}\right|\le & {} \varepsilon _{n}\underset{ \left| a_{n}+\left( b_{n}-a_{n}\right) \frac{\nu }{n}-x\right| <\delta _{n}}{\sum \limits _{\nu }}p_{n,\nu }\left( \frac{x-a_{n}}{b_{n}-a_{n}} \right) \psi _{x}^{2q}\left( a_{n}+\left( b_{n}-a_{n}\right) \frac{\nu }{n} \right) \\\le & {} \varepsilon _{n}\left( C_{n,a_{n},b_{n}}\psi _{x}^{2q}\right) \left( x\right) =\varepsilon _{n}O\left( \left( \frac{-a_{n}b_{n}}{n}\right) ^{q}\right) =o\left( \left( \frac{-a_{n}b_{n}}{n}\right) ^{q}\right) \end{aligned}$$

as \(n\rightarrow \infty \), where we used Lemma 3.4. By the Taylor formula (3.3), the second sum can be rewritten as

$$\begin{aligned} \sum \nolimits _{2}= & {} \underset{\left| a_{n}+\left( b_{n}-a_{n}\right) \frac{\nu }{n}-x\right| \ge \delta _{n}}{\sum \limits _{\nu }}p_{n,\nu }\left( \frac{x-a_{n}}{b_{n}-a_{n}}\right) \\&\times \left( f\left( a_{n}+\left( b_{n}-a_{n}\right) \frac{\nu }{n} \right) -\sum _{s=0}^{2q}\frac{f^{\left( s\right) }\left( x\right) }{s!}\psi _{x}^{s}\left( a_{n}+\left( b_{n}-a_{n}\right) \frac{\nu }{n}\right) \right) \end{aligned}$$

and we obtain

$$\begin{aligned} \left| \sum \nolimits _{2}\right|\le & {} 2\exp \left( -M\left( x\right) \frac{n\delta _{n}^{2}}{-a_{n}b_{n}}\right) \\&\times \left( \left\| f\right\| _{a_{n},b_{n}}+\sum _{s=0}^{2q}\frac{\left| f^{\left( s\right) }\left( x\right) \right| }{s!}\max \left\{ \left| x-a_{n}\right| ^{s},\left| b_{n}-x\right| ^{s}\right\} \right) , \end{aligned}$$

where in the last step Lemma 3.6 was applied. Note that

$$\begin{aligned} \sum _{s=0}^{2q}\frac{\left| f^{\left( s\right) }\left( x\right) \right| }{s!}\max \left\{ \left| x-a_{n}\right| ^{s},\left| b_{n}-x\right| ^{s}\right\} =O\left( \left( -a_{n}b_{n}\right) ^{2q}\right) \text { }\qquad \left( n\rightarrow \infty \right) . \end{aligned}$$

Hence,

$$\begin{aligned} \sum \nolimits _{2}= & {} O\left( \exp \left( \alpha \left( -a_{n}b_{n}\right) ^{p}-M\left( x\right) \frac{n\delta _{n}^{2}}{-a_{n}b_{n}}\right) \right) \\&+ \ O\left( \exp \left( 2q\log \left( -a_{n}b_{n}\right) -M\left( x\right) \frac{n\delta _{n}^{2}}{-a_{n}b_{n}}\right) \right) \end{aligned}$$

as \(n\rightarrow \infty \). In the case \(p=0\), i.e., f is bounded on \(\mathbb {R}\), we have

$$\begin{aligned} \sum \nolimits _{2}=O\left( \exp \left( 2q\log \left( -a_{n}b_{n}\right) -M\left( x\right) \frac{n\delta _{n}^{2}}{-a_{n}b_{n}}\right) \right) \text { }\qquad \left( n\rightarrow \infty \right) . \end{aligned}$$

We can assume that \(\alpha >0\). Therefore, in the case \(p>0\), we have

$$\begin{aligned} \sum \nolimits _{2}=O\left( \exp \left( \alpha \left( -a_{n}b_{n}\right) ^{p}-M\left( x\right) \frac{n\delta _{n}^{2}}{-a_{n}b_{n}}\right) \right) \text { }\qquad \left( n\rightarrow \infty \right) . \end{aligned}$$

Obviously, it is sufficient to estimate the latter relation. By Eq. (3.4), we infer that

$$\begin{aligned} \sum \nolimits _{2}= & {} O\left( \exp \left( q\log \frac{-a_{n}b_{n}}{n}-\sqrt{ \frac{n}{-a_{n}b_{n}}}\right) \right) \\= & {} O\left( \left( \frac{-a_{n}b_{n}}{n}\right) ^{q}e^{-\sqrt{n/\left( -a_{n}b_{n}\right) }}\right) \text { }\qquad \left( n\rightarrow \infty \right) . \end{aligned}$$

Finally, we conclude that the remainder can be estimated by

$$\begin{aligned} \left( C_{n,a_{n},b_{n}}\left( h_{x}\psi _{x}^{2q}\right) \right) \left( x\right) =o\left( \left( \frac{-a_{n}b_{n}}{n}\right) ^{q}\right) \text { } \qquad \left( n\rightarrow \infty \right) \end{aligned}$$

which completes the proof of the theorem. \(\square \)

Proof of Corollary 2.4

In the special case \(q=1\), Theorem 2.1 states that

$$\begin{aligned} \left( C_{n,a_{n},b_{n}}f\right) \left( x\right) =f\left( x\right) +c_{1}^{ \left[ a_{n},b_{n}\right] }\left( f,x\right) \frac{-a_{n}b_{n}}{n}+o\left( \frac{-a_{n}b_{n}}{n}\right) \text { }\qquad \left( n\rightarrow \infty \right) \end{aligned}$$

which can be rewritten in the form

$$\begin{aligned} \frac{n}{-a_{n}b_{n}}\left( \left( C_{n,a_{n},b_{n}}f\right) \left( x\right) -f\left( x\right) \right) =c_{1}^{\left[ a_{n},b_{n}\right] }\left( f,x\right) +o\left( 1\right) \text { }\qquad \left( n\rightarrow \infty \right) . \end{aligned}$$

We have

$$\begin{aligned} c_{1}^{\left[ a,b\right] }\left( f,x\right) =\frac{n}{-ab}\cdot \frac{ f^{\left( 2\right) }\left( x\right) }{2!}\left( C_{n,a,b}\psi _{x}^{2}\right) \left( x\right) =\frac{1}{2}f^{\left( 2\right) }\left( x\right) \frac{\left( x-a\right) \left( b-x\right) }{-ab} \end{aligned}$$

and the desired formula follows because

$$\begin{aligned} \lim _{n\rightarrow \infty }c_{1}^{\left[ a_{n},b_{n}\right] }\left( f,x\right) =\frac{1}{2}f^{\left( 2\right) }\left( x\right) \lim _{n\rightarrow \infty }\frac{\left( x-a_{n}\right) \left( b_{n}-x\right) }{-a_{n}b_{n}}=\frac{1}{2}f^{\left( 2\right) }\left( x\right) . \end{aligned}$$

This completes the proof. \(\square \)