Open Access

Order of approximation by an operator involving biorthogonal polynomials

  • Özlem Öksüzer1Email author,
  • Harun Karsli2 and
  • Fatma Taşdelen Yeşildal1
Journal of Inequalities and Applications20152015:121

https://doi.org/10.1186/s13660-015-0650-3

Received: 23 October 2014

Accepted: 30 March 2015

Published: 4 April 2015

Abstract

The goal of this paper is to estimate the rate of convergence of a linear positive operator involving Konhauser polynomials to bounded variation functions on \([0,1]\). To prove our main result, we have used some methods and techniques of probability theory.

Keywords

Konhauser polynomials order of convergence functions of bounded variation Lebesgue-Stieltjes integration

MSC

41A25 41A35

1 Introduction

In 1965, Konhauser presented the general theory of biorthogonal polynomials [1]. Afterwards, in 1967 [2], he gave the following pair of biorthogonal polynomials: \(Y_{\nu}^{ ( n ) } ( x;k ) \) and \(Z_{\nu}^{ ( n ) } ( x;k ) \) (\(n>-1\), \(k\in \mathbb{Z} ^{+} \)), satisfying
$$ Y_{\nu}^{ ( n ) } ( x;1 ) =Z_{\nu}^{ ( n ) } ( x;1 ) =L_{\nu}^{ ( n ) } ( x ) , $$
where \(L_{\nu}^{ ( n ) } ( x ) \) are classical Laguerre polynomials and \(Y_{\nu}^{ ( n ) } ( x;k ) \) Konhauser polynomials given by
$$ L_{v}^{ ( n ) } ( x ) = \sum_{j=0}^{v} ( -1 ) ^{j}\left ( \begin{array}{@{}c@{}} v+n \\ v-j\end{array} \right ) \frac{x^{j}}{j!}, $$
and
$$ Y_{\nu}^{ ( n ) } ( x;k ) =\frac{1}{\nu!}\sum _{i=0}^{\nu }\frac{x^{i}}{i!}\sum _{j=0}^{i} ( -1 ) ^{j}\left ( \begin{array}{@{}c@{}} i \\ j\end{array} \right ) \biggl( \frac{j+n+1}{k} \biggr) _{\nu}. $$
The classical Meyer-König and Zeller (MKZ) operators are defined in 1960 [3] by
$$\begin{aligned}& M_{n} ( f;x ) = ( 1-x ) ^{n+1}\sum _{k=0}^{\infty }f \biggl( \frac{k}{k+n+1} \biggr) \left ( \begin{array}{@{}c@{}} n+k \\ k\end{array} \right ) x^{k},\quad x\in[0,1), \\& M_{n} ( f;1 ) =f(1). \end{aligned}$$
In order to give the monotonicity properties, Cheney and Sharma introduced the following type modification of the MKZ operators:
$$\begin{aligned} &M_{n}^{\ast} ( f;x ) = ( 1-x ) ^{n+1} \sum_{k=0}^{\infty }f \biggl( \frac{k}{k+n} \biggr) \left ( \begin{array}{@{}c@{}} n+k \\ k\end{array} \right ) x^{k},\quad x \in[0,1) , \\ &M_{n}^{\ast} ( f;1 ) =f(1). \end{aligned}$$
(1)
In [4], they also introduced the operators for \(x\in [ 0,1 ) \) and \(t\in ( -\infty,0 ] \)
$$\begin{aligned} &P_{n} ( f;x ) =\exp \biggl( \frac{tx}{1-x} \biggr) \sum_{k=0}^{\infty}f \biggl( \frac{k}{k+n} \biggr) L_{k}^{(n)} ( t ) x^{k} ( 1-x ) ^{n+1}, \\ &P_{n} ( f;1 ) =f(1), \end{aligned}$$
(2)
where \(L_{k}^{(n)} ( t ) \) denotes the classical Laguerre polynomials.
We consider the sequence of linear positive operators (similarly to [5]), which is another generalization of these operators including \(Y_{\upsilon }^{(n)} ( x;k ) \) Konhauser polynomials. For \(x\in [ 0,1 ) \), \(t\in( -\infty,0 ] \), and \(k< n+1\)
$$ ( L_{n}f ) ( x,t;k ) =\frac{1}{F_{n} ( x,t ) }\sum _{\upsilon=0}^{\infty}f \biggl( \frac{\upsilon k}{k ( \upsilon -1 ) +n+1} \biggr) Y_{\upsilon}^{(n)} ( t;k ) x^{\upsilon}, $$
(3)
where \(\{F_{n} ( x,t ) \}_{n\in \mathbb{N} }\) are the generating functions for the sequence of functions \(\{ Y_{\upsilon }^{(n)} ( x;k ) \}_{v\in \mathbb{N} _{0}}\), namely
$$ F_{n} ( x,t ) =\sum_{\upsilon=0}^{\infty}Y_{\upsilon }^{(n)} ( t;k ) x^{\upsilon},\quad n>0 , $$
(4)
and
$$ F_{n} ( x,t ) = ( 1-x ) ^{-\frac{n+1}{k}}\exp \bigl\{ -t \bigl[ ( 1-x ) ^{-\frac{1}{k}}-1 \bigr] \bigr\} . $$
(5)
\(Y_{\upsilon}^{ ( n ) } ( t;k ) \geq0\) for \(t\in ( -\infty,0 ] \), due to Carlitz [6]. If we choose \(k=1\) in (3), then the operators turn out to be (2). Similarly, if we choose \(k=1\) and \(t=0\) in (3), we get (1), which are called Bernstein power series by Cheney and Sharma in [4].
In view of (4) and (5), one has
$$ 1= ( 1-x ) ^{ ( n+1 ) /k}\exp \bigl\{ t \bigl[ ( 1-x ) ^{-1/k}-1 \bigr] \bigr\} \sum_{\upsilon=0}^{\infty }Y_{\upsilon }^{(n)} ( t;k ) x^{\upsilon},\quad n>0. $$
For simplicity, we set
$$ m_{n}(x,t;k)= ( 1-x ) ^{ ( n+1 ) /k}\exp \bigl\{ t \bigl[ ( 1-x ) ^{-1/k}-1 \bigr] \bigr\} Y_{\upsilon}^{(n)} ( t;k ) x^{\upsilon}. $$
(6)
In view of (6), we can write (3) as
$$ ( L_{n}f ) ( x,t;k ) =\sum_{\upsilon=0}^{\infty }f \biggl( \frac{\upsilon k}{k ( \upsilon-1 ) +n+1} \biggr) m_{n}(x,t;k) , $$
(7)
in which \(x\in [ 0,1 ) \), \(t\in ( -\infty,0 ] \) and \(k< n+1\).

Our paper concerns the rate of pointwise convergence of the operators given by (7). In particular, by means of the techniques of probability theory, we shall estimate the rate of convergence for the operators \(( L_{n}f ) \) for functions of a bounded variation on \([0,1]\) at points x where \(f(x+)\) and \(f(x-)\) exist.

It is worthwhile to say that our present results extend some earlier results. In fact, if we choose \(k=1\) in \(L_{n}f\), then the operators reduce to the operators mentioned [7], and if \(k=1\) and \(t=0\) in \(L_{n}f\), then we get the operator investigated in [8].

For some important papers on different operators related to the present study we refer the readers to Bojanic [9] and Bojanic and Vuilleumier [10] where they estimated the rate of convergence of Fourier series and Fourier-Legendre series of functions of bounded variation, respectively. Cheng [11, 12] gave two results of this type for the Bernstein operators and Szász operators. After the fundamental studies of Bojanic-Vuilleumier and Cheng, their methods have been used widely in many follow up investigations (see, for instance [7, 8, 1315]).

The main theorem of this work reads as follows.

Theorem 1.1

Let f be a function of a bounded variation on \([0,1]\) (\(f\in BV[0,1]\)). Then for each \(x\in ( 0,1 ) \), \(t\in ( -\infty,0 ] \), \(k< n+1\), and n sufficiently large, we have
$$\begin{aligned} &\biggl\vert (L_{n}f) (x,t;k)-\frac{f(x+)+f(x-)}{2}\biggr\vert \\ &\quad\leq\frac{1}{n} \biggl[ \frac{ ( 3x^{2}-4x+2 ) ( k+4nx-t(1-x)^{-1/k} ( k+3x ) ) }{x ( 1-x ) ^{2}}+2 \biggr] \Biggl\{ \sum_{l=1}^{n}\bigvee _{x-x/\sqrt {l}}^{x+(1-x)/\sqrt{l}}(g_{x}) \Biggr\} \\ &\qquad{}+\frac{C^{\ast}}{\sqrt{n}}A(x,t;k)\bigl\{ \bigl\vert f(x+)-f(x-)\bigr\vert +\varepsilon_{n}(x)\bigl\vert f(x)-f(x-)\bigr\vert \bigr\} , \end{aligned}$$
where
$$\begin{aligned} &A(x,t;k) \\ &\quad= \biggl( \frac{1}{k^{3}}e^{t-3t(1-x)^{-1/k}}(1-x)^{-\frac {3(2+k)}{k}}x \bigl( 2e^{2t} \bigl( (1-x)^{\frac{1}{k}}-t \bigr) ^{3}x^{2}\\ &\qquad{}+3e^{t+t(1-x)^{-1/k}} \bigl( t-(1-x)^{\frac{1}{k}} \bigr) (1-x)^{\frac {1}{k}}x \bigl( k \bigl( (1-x)^{\frac{1}{k}}-t \bigr) (1-x)^{\frac{1}{k}} \\ &\qquad{} + \bigl( t^{2}-3t(1-x)^{\frac{1}{k}}+(1-x)^{\frac{2}{k}} \bigr) x \bigr) \\ &\qquad{}+e^{2t(1-x)^{-\frac{1}{k}}}(1-x)^{\frac{2}{k}} \bigl( 3k \bigl( t^{2}-3t(1-x)^{\frac{1}{k}}+(1-x)^{\frac{2}{k}} \bigr) (1-x)^{\frac {1}{k}}x \\ &\qquad{} + \bigl( -t^{3}+6t^{2}(1-x)^{\frac {1}{k}}-7t(1-x)^{\frac{2}{k}}+(1-x)^{\frac{3}{k}} \bigr) x^{2}\\ &\qquad{}+k^{2} \bigl( -t+(1-x)^{\frac{1}{k}} \bigr) (1-x)^{\frac{2}{k}}(1+x) \bigr) \bigr) \biggr) \\ &\qquad{}\Big/ \biggl( \frac{1}{k^{2}}e^{t-2t(1-x)^{-1/k}}(1-x)^{-\frac{2(2+k)}{k}}x \bigl( e^{-t} \bigl( t-(1-x)^{\frac{1}{k}} \bigr) ^{2}x\\ &\qquad{} +e^{t(1-x)^{-\frac{1}{k}}}(1-x)^{\frac{1}{k}} \bigl( k \bigl( (1-x)^{\frac{1}{k}}-t \bigr) (1-x)^{\frac{1}{k}}+ \bigl( t^{2}-3t(1-x)^{\frac{1}{k}}+(1-x)^{\frac{2}{k}} \bigr) x \bigr) \bigr) \biggr) ^{3/2}, \\ &\varepsilon_{n}(x)= \left \{ \begin{array}{@{}l@{\quad}l} 1 ,& x=\frac{\upsilon k^{\prime}}{k^{\prime} ( \upsilon-1 ) +n+1} \textit{ for some }k^{\prime}\in\{0,1,2,\ldots,n\}, \\ 0 ,& x\neq\frac{\upsilon k^{\prime }}{k^{\prime } ( \upsilon-1 ) +n+1}\textit{ for all }k^{\prime }\in \{0,1,2,\ldots,n\}, \end{array} \right . \end{aligned}$$
\(C^{\ast}\) is a certain constant, \(\bigvee_{a}^{b}(g_{x})\) is a total variation of \(g_{x}\) on \([a,b]\) and
$$ g_{x}(s)= \left \{ \begin{array}{@{}l@{\quad}l} f(s)-f(x+), & x< s\leq1, \\ 0, & s=x, \\ f(s)-f(x-), & 0\leq s<x. \end{array} \right . $$

2 Some lemmas

We now mention certain results which are necessary to prove our main theorem.

Lemma 2.1

For every \(x\in[0,1)\), \(t\in( -\infty ,0 ] \), and \(k< n+1\), we have
$$\begin{aligned} & ( L_{n}1 ) ( x,t;k ) =1, \\ & ( L_{n}s ) ( x,t;k ) \leq x-\frac{tx}{n(1-x)^{1/k}}, \\ &\bigl(L_{n}s^{2}\bigr) (x,t;k) < x^{2}- \frac{tx^{2}}{n(1-x)^{1/k}}-\frac{ktx}{n(1-x)^{1/k}}+\frac{kx}{n} , \end{aligned}$$
(8)
and
$$ \bigl(L_{n}(s-x)^{2}\bigr) (x,t;k)\leq\frac{x ( k+4nx-t(1-x)^{-1/k} ( k+3x ) ) }{n}. $$
(9)

Proof

For the proof of (8) see [16].

To prove (9), we can use the following equality:
$$\begin{aligned} \bigl(L_{n}(s-x)^{2}\bigr) (x,t;k) =&\sum _{v=0}^{\infty} \biggl( \frac {\upsilon k}{k ( \upsilon-1 ) +n+1}-x \biggr) ^{2}m_{n}(x,t;k) \\ \leq&\sum_{v=0}^{\infty} \biggl( \frac{\upsilon k}{k ( \upsilon -1 ) +n+1} \biggr) ^{2}m_{n}(x,t;k)\\ &{}+2x\sum _{v=0}^{\infty } \biggl( \frac{\upsilon k}{k ( \upsilon-1 ) +n+1} \biggr) m_{n}(x,t;k) +x^{2}\sum_{v=0}^{\infty}m_{n}(x,t;k). \end{aligned}$$
From (8), one has (9). So, this completes the proof of Lemma 2.1. □

Lemma 2.2

For \(x\in(0,1)\), \(t\leq0\),
$$\begin{aligned} \lambda_{n}(x,t,y;k) =&\int_{0}^{y}K_{n}(x,t,s;k)\,ds, \quad 0\leq s\leq x \\ \leq&\frac{x ( k+4nx-t(1-x)^{-1/k} ( k+3x ) ) }{n ( x-y ) ^{2}}, \end{aligned}$$
where
$$ K_{n}(x,t,y;k)= \left \{ \begin{array}{@{}l@{\quad}l} \sum_{k\leq\frac{(n+1)y}{v(1-y)+y}}m_{n}(x,t;k), & 0< y<1, \\ 0, & y=0 . \end{array} \right . $$

Proof

For \(s< y<x\),
$$\begin{aligned} \lambda_{n}(x,t,y;k) =&\int_{0}^{y}K_{n} ( x,t,s;k )\,ds \\ \leq&\int_{0}^{y}K_{n} ( x,t,s;k ) \biggl( \frac {x-s}{x-y} \biggr) ^{2}\,ds \\ =&\frac{1}{ ( x-y ) ^{2}}\int_{0}^{y}K_{n} ( x,t,s;k ) ( x-s ) ^{2}\,ds \\ \leq&\frac{1}{ ( x-y ) ^{2}}\bigl(L_{n}(s-x)^{2}\bigr) (x,t;k) \\ \leq&\frac{x ( k+4nx-t(1-x)^{-1/k} ( k+3x ) ) }{n ( x-y ) ^{2}}. \end{aligned}$$
 □

Lemma 2.3

For \(x\in [ 0,1 ) \), \(t\in( -\infty ,0 ] \), and \(k< n+1\),
$$\begin{aligned}& \sum_{\upsilon=0}^{\infty}vY_{\upsilon}^{(0)} ( t;k ) x^{\upsilon}=\frac{e^{t-t(1-x)^{-1/k}} ( -t+(1-x)^{\frac {1}{k}} ) (1-x)^{-1-\frac{2}{k}}x}{k}, \end{aligned}$$
(10)
$$\begin{aligned}& \begin{aligned}[b] \sum_{\upsilon=0}^{\infty}v^{2}Y_{\upsilon}^{(0)} ( t;k ) x^{\upsilon} ={}&\frac{1}{k^{2}}e^{t-t(1-x)^{-1/k}}(1-x)^{-2-\frac{3}{k} }x \bigl( k \bigl( (1-x)^{\frac{1}{k}}-t \bigr) (1-x)^{\frac{1}{k}} \\ &{} + \bigl( t^{2}-3t(1-x)^{\frac{1}{k}}+(1-x)^{\frac{2}{k}} \bigr) x \bigr) , \end{aligned} \end{aligned}$$
(11)
$$\begin{aligned}& \sum_{\upsilon=0}^{\infty}v^{3}Y_{\upsilon}^{(0)} ( t;k ) x^{\upsilon} =\frac{1}{k^{3}}e^{t-t(1-x)^{-1/k}}(1-x)^{-3-\frac{4}{k} }x \bigl( 3k \bigl( t^{2}-3t(1-x)^{\frac{1}{k}}(1-x)^{\frac{2}{k}} \bigr) (1-x)^{\frac{1}{k}}x \\& \hphantom{\sum_{\upsilon=0}^{\infty}v^{3}Y_{\upsilon}^{(0)} ( t;k ) x^{\upsilon} =}{}+ \bigl( -t^{3}+6t^{2}(1-x)^{\frac{1}{k}}-7t(1-x)^{\frac{2}{k}}+(1-x)^{ \frac{3}{k}} \bigr) x^{2} \\& \hphantom{\sum_{\upsilon=0}^{\infty}v^{3}Y_{\upsilon}^{(0)} ( t;k ) x^{\upsilon} =}{}+ k^{2} \bigl( -t+(1-x)^{\frac{1}{k}} \bigr) (1-x)^{\frac{2}{k}}(1+x) \bigr) . \end{aligned}$$
(12)

Proof

For \(n=0\), let \(Y_{\upsilon}^{(n)} ( x;k ) \) be the Konhauser polynomials’ generating function defined by
$$ \sum_{\upsilon=0}^{\infty}Y_{\upsilon}^{(0)} ( t;k ) x^{\upsilon }= ( 1-x ) ^{-\frac{1}{k}}\exp \bigl\{ -t \bigl[ ( 1-x ) ^{-\frac{1}{k}}-1 \bigr] \bigr\} . $$
Taking derivatives of both sides with respect to x yields
$$ \sum_{\upsilon=0}^{\infty}vY_{\upsilon}^{(0)} ( t;k ) x^{\upsilon-1}=\frac{e^{t-t(1-x)^{-1/k}} ( -t+(1-x)^{\frac{1}{k}} ) (1-x)^{-1-\frac{2}{k}}}{k}. $$
Editing the equation, we have
$$ \sum_{\upsilon=0}^{\infty}vY_{\upsilon}^{(0)} ( t;k ) x^{\upsilon}=\frac{e^{t-t(1-x)^{-1/k}} ( -t+(1-x)^{\frac {1}{k}} ) (1-x)^{-1-\frac{2}{k}}x}{k}. $$
The proofs of (11), (12) are similar. □

Lemma 2.4

Let \(\zeta_{1}\) be the random variable with
$$ P(\zeta_{1}=k)= ( 1-x ) ^{1/k}\exp \bigl\{ t \bigl[ ( 1-x ) ^{-1/k}-1 \bigr] \bigr\} Y_{\upsilon}^{(0)} ( t;k ) x^{\upsilon}. $$
Then letting \(a_{1}=E\vert \zeta_{1}\vert \), \(b_{1}^{2}=E ( \zeta_{1}-a_{1} ) ^{2}\),
$$\begin{aligned} E ( \zeta_{1}-a_{1} ) ^{2} =& \frac{1}{k^{2}}e^{t-2t(1-x)^{-1/k}}(1-x)^{-\frac{2(2+k)}{k}}x \bigl( e^{-t} \bigl( t-(1-x)^{\frac{1}{k}} \bigr) ^{2}x \\ &{} +e^{t(1-x)^{-\frac{1}{k}}}(1-x)^{\frac{1}{k}} \bigl( k \bigl( (1-x)^{\frac{1}{k}}-t \bigr) (1-x)^{\frac{1}{k}}\\ &{}+ \bigl( t^{2}-3t(1-x)^{\frac {1}{k}}+(1-x)^{\frac{2}{k}} \bigr) x \bigr) \bigr) \end{aligned}$$
and
$$\begin{aligned} E\vert \zeta_{1}-a_{1}\vert ^{3} =& \frac{1}{k^{3}}e^{t-3t(1-x)^{-1/k}}(1-x)^{-\frac{3(2+k)}{k}}x \bigl( 2e^{2t} \bigl( (1-x)^{\frac{1}{k}}-t \bigr) ^{3}x^{2} \\ &{}+3e^{t+t(1-x)^{-1/k}} \bigl( t-(1-x)^{\frac{1}{k}} \bigr) (1-x)^{\frac {1}{k}}x \bigl( k \bigl( (1-x)^{\frac{1}{k}}-t \bigr) (1-x)^{\frac{1}{k}} \\ &{} + \bigl( t^{2}-3t(1-x)^{\frac{1}{k}}+(1-x)^{\frac{2}{k}} \bigr) x \bigr) \\ &{}+e^{2t(1-x)^{-\frac{1}{k}}}(1-x)^{\frac{2}{k}} \bigl( 3k \bigl( t^{2}-3t(1-x)^{\frac{1}{k}}+(1-x)^{\frac{2}{k}} \bigr) (1-x)^{\frac {1}{k}}x \\ &{}+ \bigl( -t^{3}+6t^{2}(1-x)^{\frac{1}{k}}-7t(1-x)^{\frac{2}{k}}+(1-x)^{ \frac{3}{k}} \bigr) x^{2} \\ &{} +k^{2} \bigl( -t+(1-x)^{\frac{1}{k}} \bigr) (1-x)^{\frac {2}{k}}(1+x) \bigr) \bigr) . \end{aligned}$$

Proof

For \(n=0\), we have
$$\begin{aligned}& a_{1}=E(\zeta_{1})=\sum_{\upsilon=0}^{\infty}vY_{\upsilon }^{(0)} ( t;k ) x^{\upsilon}=\frac{e^{t-t(1-x)^{-1/k}} ( -t+(1-x)^{\frac {1}{k}} ) (1-x)^{-1-\frac{2}{k}}x}{k}, \\& b_{1}^{2} =E ( \zeta_{1}-a_{1} ) ^{2}=E\bigl(\zeta _{1}^{2} \bigr)-2a_{1}E(\zeta_{1})+a_{1}^{2} =E\bigl(\zeta_{1}^{2}\bigr)-\bigl(E(\zeta_{1}) \bigr)^{2}, \end{aligned}$$
and
$$\begin{aligned} E\vert \zeta_{1}-a_{1}\vert ^{3} =E\bigl( \zeta _{1}^{3}\bigr)-3a_{1}E\bigl( \zeta_{1}^{2}\bigr)+3a_{1}^{2}E( \zeta_{1})-a_{1}^{3} =E\bigl(\zeta_{1}^{3}\bigr)-3E(\zeta_{1})E \bigl(\zeta_{1}^{2}\bigr)+2 \bigl( E(\zeta _{1}) \bigr) ^{3}. \end{aligned}$$
If we use (10)-(12), the proof of the Lemma 2.4 is completed. □

Next, we recall the well-known Berry-Esséen bound for the classical central limit theorem of probability theory.

Lemma 2.5

(Berry-Esséen)

Let \(\{\zeta _{k}\}_{k=1}^{\infty}\) be a sequence of independent and identically distributed random variables with finite variance such that the expectation \(E(\zeta_{1})=a_{1}\in \mathbb{R} \), the variance \(\operatorname{Var}(\zeta_{1})=E ( \zeta_{1}-a_{1} ) ^{2}=b_{1}^{2}>0\). Assume \(E\vert \zeta_{1}-a_{1}\vert ^{3}<\infty\), then there exists a constant C, \(1/\sqrt{2\pi}\leq C\leq 0.82\), such that for all \(n=1,2,\ldots\) and all t,
$$ \Biggl\vert P \Biggl( \frac{1}{b_{1}\sqrt{n}}\sum_{k=1}^{n}( \zeta _{k}-a_{1})\leq t \Biggr) -\frac{1}{\sqrt{2\pi}}\int _{-\infty }^{t}e^{-u^{2}/2}\,du\Biggr\vert < C \frac{E\vert \zeta _{1}-a_{1}\vert ^{3}}{b_{1}^{3}\sqrt{n}}. $$

Its proof can be found in Shiryayev [17].

Lemma 2.6

For \(x\in(0,1)\), \(t\in ( -\infty,0 ] \) and \(k< n+1\), we have
$$ \biggl\vert \sum_{\frac{(n+1)x}{v(1-x)+x}< k}m_{n}(x,t;k)- \frac {1}{2}\biggr\vert \leq\frac{C}{\sqrt{n}}A(x,t;k); $$
\(A(x,t;k)\) is given in Theorem  1.1.

Proof

By direct calculation, one has from Lemmas 2.4-2.5 the desired result. □

Lemma 2.7

For all \(x\in(0,1)\) define the function \(\operatorname{sgn}(s-x)\) by
$$ \operatorname{sgn}(s-x)= \left \{ \begin{array}{@{}l@{\quad}l} 1, & s>x, \\ 0, & s=x, \\ -1, & s< x. \end{array} \right . $$
We have
$$ \bigl(L_{n}\operatorname{sgn}(s-x)\bigr) (x,t;k)=2\sum _{\frac{(n+1)x}{v(1-x)+x}< k}m_{n}(x,t;k)-1+\varepsilon_{n}(x)m_{n} \bigl(x,t;k^{\prime}\bigr), $$
where \(\varepsilon_{n}(x)\) is as in Theorem  1.1 and \(k^{\prime}=\frac {vk}{k(v-1)+n+1}\).

Proof

If we apply the operator \(L_{n}\) to the function of \(\operatorname{sgn}(s-x)\), we have
$$\begin{aligned} \bigl(L_{n}\operatorname{sgn}(s-x)\bigr) (x,t) =&\sum _{v=0}^{\infty} \operatorname{sgn} \biggl( \frac {\upsilon k}{k ( \upsilon-1 ) +n+1}-x \biggr) m_{n}(x,t;k) \\ =&\sum_{\frac{(n+1)x}{v(1-x)+x}< k}m_{n}(x,t;k)+\sum _{k<\frac{(n+1)x}{v(1-x)+x}}(-1)m_{n}(x,t;k), \end{aligned}$$
and we can write
$$\begin{aligned}& (L_{n}1) (x,t) =\sum_{\frac {(n+1)x}{v(1-x)+x}< k}m_{n}(x,t;k)+ \sum _{k<\frac{(n+1)x}{v(1-x)+x}}m_{n}(x,t;k)+\varepsilon _{n}(x)m_{n}\bigl(x,t;k^{\prime}\bigr), \\& \sum_{k<\frac{(n+1)x}{v(1-x)+x}}m_{n}(x,t;k) =1- \biggl[ \sum _{\frac{(n+1)x}{v(1-x)+x}<k}m_{n}(x,t;k)-\varepsilon _{n}(x)m_{n}\bigl(x,t;k^{\prime}\bigr) \biggr] . \end{aligned}$$
Thus
$$\begin{aligned}[b] &\bigl(L_{n}\operatorname{sgn}(s-x)\bigr) (x,t;k) \\ &\quad=\sum _{\frac{(n+1)x}{v(1-x)+x}< k}m_{n}(x,t;k)- \biggl[ 1-\sum _{\frac{(n+1)x}{v(1-x)+x}<k}m_{n}(x,t;k)-\varepsilon_{n}(x)m_{n} \bigl(x,t;k^{\prime}\bigr) \biggr]\\ &\quad=2\sum_{\frac{(n+1)x}{v(1-x)+x}<k}m_{n}(x,t;k)-1+\varepsilon _{n}(x)m_{n}\bigl(x,t;k^{\prime}\bigr). \end{aligned} $$
This completes the proof of Lemma 2.7. □

Lemma 2.8

If the conditions of Theorem  1.1 hold, we have for all \(x\in(0,1)\)
$$\begin{aligned} &\biggl\vert \frac{f(x+)-f(x-)}{2}\bigl(L_{n} \operatorname{sgn}(s-x)\bigr) (x,t;k)+ \biggl[ f(x)-\frac{f(x+)+f(x-)}{2} \biggr] (L_{n}\delta) (x,t;k)\biggr\vert \\ &\quad\leq\frac{C^{\ast}}{\sqrt{n}}A(x,t;k) \bigl\{ \bigl\vert f(x+)-f(x-)\bigr\vert +\varepsilon_{n}(x)\bigl\vert f(x)-f(x-)\bigr\vert \bigr\} , \end{aligned}$$
(13)
where \(\delta_{x}(s)\) is the Dirac delta function, \(C^{\ast}\) is a certain constant, and \(A(x,t;k)\) is given in Theorem  1.1.

Proof

By direct calculation, we get
$$ (L_{n}\delta) (x,t;k)=\varepsilon_{n}(x)m_{n} \bigl(x,t;k^{\prime}\bigr). $$
One has
$$\begin{aligned} &\biggl\vert \frac{f(x+)-f(x-)}{2}\bigl(L_{n}\operatorname{sgn}(s-x) \bigr) (x,t;k)+ \biggl[ f(x)-\frac{f(x+)+f(x-)}{2} \biggr] (L_{n}\delta) (x,t;k)\biggr\vert \\ &\quad\leq\biggl\vert \frac{f(x+)-f(x-)}{2}\bigl(L_{n} \operatorname{sgn}(s-x)\bigr) (x,t;k)\biggr\vert +\biggl\vert \biggl[ f(x)- \frac{f(x+)+f(x-)}{2} \biggr] (L_{n}\delta ) (x,t;k)\biggr\vert \\ &\quad\leq\biggl\vert \frac{f(x+)-f(x-)}{2} \biggl[ 2\sum _{\frac {(n+1)x}{v(1-x)+x}< k}m_{n}(x,t;k)-1+\varepsilon_{n}(x)m_{n} \bigl(x,t;k^{\prime }\bigr) \biggr] \biggr\vert \\ &\qquad{}+\biggl\vert \biggl[ f(x)-\frac{f(x+)+f(x-)}{2} \biggr] \varepsilon _{n}(x)m_{n}\bigl(x,t;k^{\prime}\bigr)\biggr\vert \\ &\quad\leq\bigl\vert f(x+)-f(x-)\bigr\vert \biggl\vert \sum _{\frac {(n+1)x}{v(1-x)+x}<k}m_{n}(x,t;k)-\frac{1}{2}\biggr\vert +\bigl\vert f(x)-f(x-)\bigr\vert \varepsilon_{n}(x)m_{n} \bigl(x,t;k^{\prime}\bigr). \end{aligned}$$
According to Lemma 2.6, one has
$$ \biggl\vert \sum_{\frac{(n+1)x}{v(1-x)+x}< k}m_{n}(x,t;k)- \frac {1}{2}\biggr\vert \leq\frac{C}{\sqrt{n}}A(x,t;k), $$
(14)
and using the method of proof of Lemma 2.6 and evaluations which are similar to the work in [13], we have
$$ m_{n}\bigl(x,t;k^{\prime}\bigr)=:\frac{C_{1}}{\sqrt{n}}A ( x,t;k ) . $$
(15)
Set \(C^{\ast}=\max \{ C,C_{1} \} \). Consequently from (14) and (15) we get (13).

This completes the proof of Lemma 2.8. □

Lemma 2.9

For n sufficiently large, we have
$$\begin{aligned} &\bigl\vert (L_{n}g_{x}) (x,t;k)\bigr\vert \\ &\quad\leq \frac{1}{n} \biggl[ \frac{ ( 3x^{2}-4x+2 ) ( k+4nx-t(1-x)^{-1/k} ( k+3x ) ) }{x ( 1-x ) ^{2}}+2 \biggr] \Biggl\{ \sum _{l=1}^{n}\bigvee_{x-x/\sqrt{l}}^{x+(1-x)/\sqrt{l}}(g_{x}) \Biggr\} , \end{aligned}$$
\(g_{x} ( s ) \) is given in Theorem  1.1.

Proof

We recall the Lebesgue-Stieltjes integral representations
$$ (L_{n}f) (x,t;k)=\int_{0}^{1}f(y)\,d_{y} \lambda_{n}(x,t,y;k), $$
(16)
where \(\lambda_{n}(x,t,y;k)=\int_{0}^{y}K_{n}(x,t,s;k)\,ds\), \(0\leq s\leq x\). From (16), we can rewrite \((L_{n}g_{x})(x,t;k)\) as follows:
$$\begin{aligned} \bigl\vert (L_{n}g_{x}) (x,t;k)\bigr\vert =&\Biggl\vert \sum_{v=0}^{\infty}g_{x} \biggl( \frac{\upsilon k}{k ( \upsilon -1 ) +n+1} \biggr) m_{n}(x,t;k)\Biggr\vert \\ =&\biggl\vert \int_{0}^{1}g_{x}(y)\,d_{y} \lambda _{n}(x,t,y;k)\biggr\vert . \end{aligned}$$
(17)
To estimate (17), we decompose it into three parts as follows:
$$ \int_{0}^{1}g_{x}(y)\,d_{y} \lambda _{n}(x,t,y;k)=(I_{1}g_{x}) (x,t;k)+(I_{2}g_{x}) (x,t;k)+(I_{3}g_{x}) (x,t;k), $$
(18)
where
$$\begin{aligned}& (I_{1}g_{x}) (x,t)=\int_{0}^{x-x/\sqrt{n}}g_{x}(y)\,d_{y} \lambda _{n}(x,t,y;k), \\& (I_{2}g_{x}) (x,t)=\int_{x-x/\sqrt{n}}^{x+(1-x)/\sqrt{n}}g_{x}(y)\,d_{y} \lambda_{n}(x,t,y;k), \\& (I_{3}g_{x}) (x,t)=\int_{x+(1-x)/\sqrt{n}}^{1}g_{x}(y)\,d_{y} \lambda _{n}(x,t,y;k). \end{aligned}$$
We shall evaluate \((I_{1}g_{x})(x,t)\), \((I_{2}g_{x})(x,t)\), and \((I_{3}g_{x})(x,t)\).

First we estimate \((I_{2}g_{x})(x,t)\), for \(y\in [ x-x/\sqrt {n},x+(1-x)/\sqrt{n} ]\).

Note that \(g_{x}(x)=0\) and \(\vert \int_{x-x/\sqrt {n}}^{x+(1-x)/\sqrt{n}}\,d_{y}\lambda_{n}(x,t,y;k)\vert \leq1\), so we have
$$\begin{aligned} \bigl\vert (I_{2}g_{x}) (x,t;k)\bigr\vert =&\biggl\vert \int_{x-x/\sqrt{n}}^{x+(1-x)/\sqrt{n}} \bigl(g_{x}(y)-g_{x}(x)\bigr)\,d_{y}\lambda _{n}(x,t,y;k)\biggr\vert \\ \leq&\int_{x-x/\sqrt{n}}^{x+(1-x)/\sqrt{n}}\bigl\vert g_{x}(y)-g_{x}(x)\bigr\vert \bigl\vert \,d_{y} \lambda _{n}(x,t,y;k)\bigr\vert \\ \leq&\bigvee_{x-x/\sqrt{n}}^{x+(1-x)/\sqrt{n}}(g_{x}) \leq\frac {1}{n-1}\sum_{l=2}^{n}\bigvee _{x-x/\sqrt{l}}^{x+(1-x)/\sqrt{l} }(g_{x}). \end{aligned}$$
(19)
Next we estimate \((I_{1}g_{x})(x,t)\). Using partial Lebesgue-Stieltjes integrations, we obtain
$$\begin{aligned} (I_{1}g_{x}) (x,t;k) =&\int_{0}^{x-x/\sqrt {n}}g_{x}(y)\,d_{y} \lambda _{n}(x,t,y;k) \\ =&g_{x}(x-x/\sqrt{n})\lambda_{n}(x,t,x-x/ \sqrt{n};k)-g_{x}(0)\lambda _{n}(x,t,0;k) \\ &{}-\int_{0}^{x-x/\sqrt{n}}\lambda_{n}(x,t,y;k)\,d_{y}g_{x}(y). \end{aligned}$$
Since \(\vert g_{x}(x-x/\sqrt{n})\vert =\vert g_{x}(x-x/\sqrt{n})-g_{x}(x)\vert \leq\bigvee_{x-x/\sqrt{n}}^{x}(g_{x})\), it follows that
$$ \bigl\vert (I_{1}g_{x}) (x,t;k)\bigr\vert \leq\bigvee _{x-x/\sqrt {n}}^{x}(g_{x}) \lambda_{n}(x,t,x-x/\sqrt{n};k)+\int_{0}^{x-x/\sqrt {n}} \lambda_{n}(x,t,y;k)\,d_{y} \Biggl( -\bigvee _{y}^{x}(g_{x}) \Biggr) . $$
From Lemma 2.2, it is clear that
$$ \lambda_{n}(x,t,x-x/\sqrt{n};k)\leq\frac{x ( k+4nx-t(1-x)^{-1/k} ( k+3x ) ) }{n ( \frac{x}{\sqrt{n}} ) ^{2}}. $$
It follows that
$$\begin{aligned} \bigl\vert (I_{1}g_{x}) (x,t;k)\bigr\vert \leq& \bigvee_{x-x/\sqrt{n}}^{x}(g_{x}) \frac{x ( k+4nx-t(1-x)^{-1/k} ( k+3x ) ) }{n ( \frac{x}{\sqrt{n}} ) ^{2}} \\ &{}+\int_{0}^{x-x/\sqrt{n}}\frac{x ( k+4nx-t(1-x)^{-1/k} ( k+3x ) ) }{n(x-y)^{2}}\,d_{y} \Biggl( -\bigvee_{y}^{x}(g_{x}) \Biggr) \\ =&\bigvee_{x-x/\sqrt{n}}^{x}(g_{x}) \frac{x ( k+4nx-t(1-x)^{-1/k} ( k+3x ) ) }{n ( \frac{x}{\sqrt {n}} ) ^{2}} \\ &{}+\frac{x ( k+4nx-t(1-x)^{-1/k} ( k+3x ) ) }{n}\int_{0}^{x-x/\sqrt{n}} \frac{1}{(x-y)^{2}}\,d_{y} \Biggl( -\bigvee_{y}^{x}(g_{x}) \Biggr) . \end{aligned}$$
Furthermore, since
$$\begin{aligned} &\int_{0}^{x-x/\sqrt{n}}\frac{1}{(x-y)^{2}}\,d_{y} \Biggl( -\bigvee_{y}^{x}(g_{x}) \Biggr) \\ &\quad=-\frac{1}{(x-y)^{2}}\bigvee_{y}^{x}(g_{x}) \Big|_{0}^{x-x/\sqrt{n}}+\int_{0}^{x-x/\sqrt{n}} \frac{2}{(x-y)^{3}}\bigvee_{y}^{x}(g_{x})\,dy \\ &\quad=-\frac{1}{ ( \frac{x}{\sqrt{n}} ) ^{2}}\bigvee_{x-x/\sqrt{n}}^{x}(g_{x})+ \frac{1}{x^{2}}\bigvee_{0}^{x}(g_{x})+ \int _{0}^{x-x/\sqrt{n}}\frac{2}{(x-y)^{3}}\bigvee _{y}^{x}(g_{x})\,dy, \end{aligned}$$
and putting \(y=x-x/\sqrt{u}\) in the last integral, we get
$$ \int_{0}^{x-x/\sqrt{n}}\frac{2}{(x-y)^{3}} \bigvee _{y}^{x}(g_{x})\,dy= \frac{1}{x^{2}}\int_{1}^{n}\bigvee _{x-x/\sqrt{u}}^{x}(g_{x})\,du=\frac{1}{x^{2}}\sum _{l=1}^{n} \bigvee _{x-x/\sqrt{l}}^{x}(g_{x}). $$
Consequently,
$$\begin{aligned} \bigl\vert (I_{1}g_{x}) (x,t)\bigr\vert \leq&\bigvee _{x-x/\sqrt{n}}^{x}(g_{x}) \frac{x ( k+4nx-t(1-x)^{-1/k} ( k+3x ) ) }{n ( \frac{x}{\sqrt{n}} ) ^{2}} \\ &{}+\frac{x ( k+4nx-t(1-x)^{-1/k} ( k+3x ) ) }{n} \Biggl\{ -\frac{1}{ ( \frac{x}{\sqrt{n}} ) ^{2}}\bigvee _{x-x/\sqrt {n}}^{x}(g_{x}) \\ &{} +\frac{1}{x^{2}}\bigvee_{0}^{x}(g_{x})+ \frac{1}{x^{2}}\sum_{l=1}^{n} \bigvee_{x-x/\sqrt{l}}^{x}(g_{x}) \Biggr\} \\ =&\frac{x ( k+4nx-t(1-x)^{-1/k} ( k+3x ) ) }{n} \Biggl\{ \frac{1}{x^{2}}\bigvee _{0}^{x}(g_{x})+\frac{1}{x^{2}}\sum_{l=1}^{n}\bigvee _{x-x/\sqrt{l}}^{x}(g_{x}) \Biggr\} \\ =&\frac{k+4nx-t(1-x)^{-1/k} ( k+3x ) }{nx} \Biggl\{ \bigvee_{0}^{x}(g_{x})+ \sum_{l=1}^{n}\bigvee _{x-x/\sqrt{l}}^{x}(g_{x}) \Biggr\} . \end{aligned}$$
Moreover,
$$ \bigvee_{0}^{x}(g_{x})\leq \sum_{l=1}^{n}\bigvee _{x-x/\sqrt{l}}^{x}(g_{x})\leq\sum _{l=1}^{n}\bigvee_{x-x/\sqrt {l}}^{x+(1-x)/\sqrt{l}}(g_{x}) $$
yields
$$ \bigl\vert (I_{1}g_{x}) (x,t)\bigr\vert \leq \frac{2 ( k+4nx-t(1-x)^{-1/k} ( k+3x ) ) }{nx} \Biggl\{ \sum_{l=1}^{n} \bigvee_{x-x/\sqrt{l}}^{x+(1-x)/\sqrt{l}}(g_{x}) \Biggr\} . $$
(20)
Using a similar method for estimating \(\vert (I_{3}g_{x})(x,t)\vert \), we get
$$ \bigl\vert (I_{3}g_{x}) (x,t)\bigr\vert \leq \frac{x ( k+4nx-t(1-x)^{-1/k} ( k+3x ) ) }{n(1-x)^{2}} \Biggl\{ \sum_{l=1}^{n} \bigvee_{x}^{x+(1-x)/\sqrt{l}}(g_{x}) \Biggr\} . $$
(21)
Putting (19)-(21) in (18), we get
$$\begin{aligned} \bigl\vert (L_{n}g_{x}) (x,t)\bigr\vert \leq&\bigl\vert (I_{1}g_{x}) (x,t)\bigr\vert +\bigl\vert (I_{2}g_{x}) (x,t)\bigr\vert +\bigl\vert (I_{3}g_{x}) (x,t)\bigr\vert \\ \leq&\frac{2 ( k+4nx-t(1-x)^{-1/k} ( k+3x ) ) }{nx} \Biggl\{ \sum_{l=1}^{n} \bigvee_{x-x/\sqrt {l}}^{x+(1-x)/\sqrt{l}}(g_{x}) \Biggr\} \\ &{}+\frac{1}{n-1}\sum_{l=2}^{n} \bigvee_{x-x/\sqrt {l}}^{x+(1-x)/\sqrt{l}}(g_{x}) \\ &{}+\frac{x ( k+4nx-t(1-x)^{-1/k} ( k+3x ) ) }{n(1-x)^{2}} \Biggl\{ \sum_{l=1}^{n} \bigvee_{x}^{x+(1-x)/\sqrt{l}}(g_{x}) \Biggr\} . \end{aligned}$$
(22)
Obviously,
$$ \frac{2}{x}+\frac{x}{(1-x)^{2}}=\frac{3x^{2}-4x+2}{x(1-x)^{2}} $$
and
$$ \bigvee_{x}^{x+(1-x)/\sqrt{l}}(g_{x})\leq \bigvee_{x-x/\sqrt{l}}^{x+(1-x)/\sqrt{l}}(g_{x}). $$
Hence, we obtain from (22)
$$\begin{aligned} \bigl\vert (L_{n}g_{x}) (x,t)\bigr\vert \leq& \biggl( \frac{2}{x}+\frac {x}{(1-x)^{2}} \biggr) \biggl( \frac{k+4nx-t(1-x)^{-1/k} ( k+3x ) }{n} \biggr) \Biggl\{ \sum_{l=1}^{n}\bigvee _{x-x/\sqrt {l}}^{x+(1-x)/\sqrt{l}}(g_{x}) \Biggr\} \\ &{}+\frac{1}{n-1}\sum_{l=2}^{n} \bigvee_{x-x/\sqrt {l}}^{x+(1-x)/\sqrt{l}}(g_{x}) \\ \leq&\frac{1}{n} \biggl[ \frac{ ( 3x^{2}-4x+2 ) ( k+4nx-t(1-x)^{-1/k} ( k+3x ) ) }{x(1-x)^{2}}+2 \biggr] \\ &{}\times\Biggl\{ \sum _{l=1}^{n}\bigvee _{x-x/\sqrt{l}}^{x+(1-x)/\sqrt{l}}(g_{x}) \Biggr\} . \end{aligned}$$
This inequality is equivalent to the one to be proved. □

3 Proof of the main theorem

Now we are ready to establish our main theorem.

Proof

For any \(f\in BV[0,1]\), we decompose f into four parts on \([0,1]\) for sufficiently large n,
$$\begin{aligned} f(s) =&\frac{f(x+)+f(x-)}{2}+g_{x}(s)+\frac{f(x+)-f(x-)}{2} \operatorname{sgn}(s-x) \\ &{}+\delta_{x}(s) \biggl[ f(x)-\frac{f(x+)+f(x-)}{2} \biggr], \end{aligned}$$
(23)
where
$$ \delta_{x}(s)= \left \{ \begin{array}{@{}l@{\quad}l} 1, & x=s, \\ 0, & x\neq s. \end{array} \right . $$
If we apply the operator \(L_{n}\) to the both sides of (23), then we have
$$\begin{aligned} (L_{n}f) (x,t;k) =&\frac{f(x+)+f(x-)}{2}(L_{n}1) (x,t;k)+(L_{n}g_{x}) (x,t;k) \\ &{}+\frac{f(x+)-f(x-)}{2}\bigl(L_{n}\operatorname{sgn}(s-x)\bigr) (x,t;k) \\ &{}+ \biggl[ f(x)-\frac{f(x+)+f(x-)}{2} \biggr] (L_{n} \delta_{x}) (x,t;k). \end{aligned}$$
It follows that
$$\begin{aligned} &\biggl\vert (L_{n}f) (x,t;k)-\frac{f(x+)+f(x-)}{2}\biggr\vert \\ &\quad\leq\bigl\vert (L_{n}g_{x}) (x,t;k)\bigr\vert + \biggl\vert \frac{f(x+)-f(x-)}{2}\bigl(L_{n}\operatorname{sgn}(s-x) \bigr) (x,t;k) \\ &\qquad{} + \biggl[ f(x)-\frac{f(x+)+f(x-)}{2} \biggr] (L_{n} \delta _{x}) (x,t;k)\biggr\vert . \end{aligned}$$
By Lemmas 2.8-2.9, we get the required result. Thus the proof is completed. □

Declarations

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, University of Ankara
(2)
Department of Mathematics, Faculty of Science and Arts, Abant Izzet Baysal University

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