A Voronovskaja type formula for Soardi’s operators

Abstract

In 1991 Soardi introduced a sequence of positive linear operators $\beta _{n}$ associating to each function $f\in C\left[ 0,1\right]$ a polynomial function which is closely related to the Bernstein polynomials on $\left[ -1,+1\right]$. One of the authors already studied the operators $\beta _{n}$ in several papers. This paper is devoted to other properties of Soardi’s operators. We introduce a version ${\tilde{\beta }}_{n}$ which can be expressed in terms of the classical Bernstein operators and present the relations between $\beta _{n}$ and ${\tilde{\beta }}_{n}$. We derive Voronovskaja-type results for both $\beta _{n}$ and ${\tilde{\beta }}_{n}$. Furthermore, rates of convergence for ${\tilde{\beta }}_{n}$, respectively $\beta _{n}$, are estimated. Finally, we study the first and second moments of $\beta _{n}$.

1 Introduction

In 1991 Soardi [8] introduced the sequence of positive linear operators $\beta _{n}$ associating to each function $f\in C\left[ 0,1\right]$ the polynomial function

$$\begin{aligned} \left( \beta _{n}f\right) \left( x\right) =\sum _{k=0}^{\left\lfloor n/2\right\rfloor }f\left( \frac{n-2k}{n}\right) {\tilde{w}}_{n,k}\left( x\right) , \end{aligned}$$

where

$$\begin{aligned} {\tilde{w}}_{n,\,k}\left( x\right) =\frac{n+1-2k}{\left( n+1\right) 2^{n+1}x} \left( {\begin{array}{c}n+1\\ k\end{array}}\right) \left[ \left( 1-x\right) ^{k}\left( 1+x\right) ^{n+1-k}-\left( 1-x\right) ^{n+1-k}\left( 1+x\right) ^{k}\right] . \end{aligned}$$

Usually, the operators $\beta _{n}$ are given in the form

$$\begin{aligned} \left( \beta _{n}f\right) \left( x\right) =\sum _{k=0}^{m}f\left( \frac{ n-2m+2k}{n}\right) w_{n,k}\left( x\right) , \end{aligned}$$

where $m=\left\lfloor n/2\right\rfloor$ and $w_{n,k}\left( x\right) ={\tilde{w}}_{n,m-k}\left( x\right)$ are the fundamental polynomials. The definition and the proofs in [8] are based on properties of random walks on hypergroups. Soardi proved that, for each $f\in C\left[ 0,1\right]$, the sequence $\left( \beta _{n}f\right)$ is uniformly convergent to f. Furthermore, by an intensive use of probabilistic tools, Soardi [8, Theorem 2] estimated the rate of convergence of $\left( \beta _{n}f\right)$ in terms of the usual modulus of continuity:

$$\begin{aligned} \left\| \beta _{n}f-f\right\| \le \left( 55+\frac{32}{n}\right) \omega \left( f;\frac{1}{\sqrt{n}}\right) ,\text {for}f\in C\left[ 0,1\right] . \end{aligned}$$

Shape preserving properties of the operators $\beta _{n}$ were investigated in [5,6,7]. In particular, if $f\in C \left[ 0,1\right]$ is increasing, then $\beta _{n}f$ is increasing (see [6, Th. 2.1]; this fact will be used in Sect. ). Moreover, if $f\in C\left[ 0,1\right]$ is increasing and convex, then $\beta _{n}f\ge f$ (see [6, Th. 3.1]; this inequality will be instrumental in Sect. ).

For $x\in \left( 0,1\right)$ and bounded functions f on $\left[ 0,1\right]$, Raşa [6, Theorem 4.1] proved the Voronovskaja-type formula

$$\begin{aligned} \left( \beta _{n}f\right) \left( x\right) =f\left( x\right) +\frac{1}{n} \left[ \left( \frac{1}{x}-1\right) f^{\prime }\left( x\right) +\frac{1-x^{2} }{2}f^{\prime \prime }\left( x\right) \right] +o\left( 1/n\right) \end{aligned}$$

as $n\rightarrow \infty$, provided that $f^{\prime \prime }\left( x\right)$ exists.

This paper is devoted to other properties of Soardi’s operators. In Sect.  2 we introduce a version ${\tilde{\beta }}_{n}$ which can be expressed in terms of the classical Bernstein operators. The relations between $\beta _{n}$ and ${\tilde{\beta }}_{n}$ are presented in Sect. . Section 4 contains Voronovskaja-type results for both $\beta _{n}$ and ${\tilde{\beta }}_{n}$. Rates of convergence for ${\tilde{\beta }}_{n}$, respectively $\beta _{n}$, are estimated in Sects. 5 and 2. The last two sections are devoted to the first and second moments of $\beta _{n}$.

2 The variant $\tilde{\beta }_{n}$ and its relation to Bernstein polynomials

In this section we introduce a variant ${\tilde{\beta }}_{n}$ of Soardi’s operator which seems to be more natural. Replacing $f\left( \frac{n-2k}{n} \right)$ with $f\left( \frac{n+1-2k}{n+1}\right)$ leads to the definition

$$\begin{aligned} \left( {\tilde{\beta }}_{n}f\right) \left( x\right) =\sum _{k=0}^{m}f\left( \frac{n+1-2k}{n+1}\right) {\tilde{w}}_{n,k}\left( x\right) , \end{aligned}$$

where $m=\left\lfloor n/2\right\rfloor$. The index manipulation $k\rightarrow n+1-k$ yields

$$\begin{aligned} \left( {\tilde{\beta }}_{n}f\right) \left( x\right) =\sum _{k=n+1-m}^{n+1}f\left( -\frac{n+1-2k}{n+1}\right) {\tilde{w}} _{n,k}\left( x\right) . \end{aligned}$$

For even values of n we have

$$\begin{aligned} 2\left( {\tilde{\beta }}_{n}f\right) \left( x\right) =\sum _{k=0}^{n+1}f\left( \left| \frac{n+1-2k}{n+1}\right| \right) {\tilde{w}}_{n,k}\left( x\right) . \end{aligned}$$

This representation is valid also in the case of odd integers n since the term $f\left( 0\right) {\tilde{w}}_{n,\frac{n+1}{2}}\left( x\right)$ with $k= \frac{n+1}{2}$ is vanishing. Hence, for all $n\ge 0$,

$$\begin{aligned} \left( {\tilde{\beta }}_{n}f\right) \left( x\right) =\frac{1}{2} \sum _{k=0}^{n+1}f\left( \left| \frac{n+1-2k}{n+1}\right| \right) {\tilde{w}}_{n,k}\left( x\right) . \end{aligned}$$

Writing

$$\begin{aligned} \left( {\tilde{\beta }}_{n}f\right) \left( x\right)= & {} \frac{1}{2x} \sum _{k=0}^{n+1}\frac{n+1-2k}{n+1}f\left( \left| 1-2\frac{k}{n+1} \right| \right) \\&\times \left( {\begin{array}{c}n+1\\ k\end{array}}\right) \left[ \left( \frac{1-x}{2}\right) ^{k}\left( \frac{ 1+x}{2}\right) ^{n+1-k}-\left( \frac{1-x}{2}\right) ^{n+1-k}\left( \frac{1+x }{2}\right) ^{k}\right] \end{aligned}$$

we obtain the following relation to the classical Bernstein polynomials.

Lemma 1

For a function f on $\left[ 0,1\right]$, we have the relation

$$\begin{aligned} \left( {\tilde{\beta }}_{n}f\right) \left( x\right) =\frac{1}{2x}\left[ \left( B_{n+1}g\right) \left( \frac{1-x}{2}\right) -\left( B_{n+1}g\right) \left( \frac{1+x}{2}\right) \right] , \end{aligned}$$

where

$$\begin{aligned} g\left( t\right) =\left( 1-2t\right) f\left( \left| 1-2t\right| \right) \end{aligned}$$

and $B_{n}g$ denotes the classical Bernstein polynomial on $\left[ 0,1\right]$.

3 Relations among the operators $\beta _{n}$ and $\tilde{ \beta }_{n}$

Consider the operators $\beta _{n}:C\left[ 0,1\right] \rightarrow C\left[ 0,1 \right]$ and ${\tilde{\beta }}_{n}:C\left[ \frac{1}{n+1},1\right] \rightarrow C \left[ 0,1\right]$. Let

$$\begin{aligned} u_{n}&:&\left[ \frac{1}{n+1},1\right] \rightarrow \left[ 0,1\right] ,u_{n}\left( t\right) =\frac{\left( n+1\right) t-1}{n} \\ v_{n}&:&\left[ 0,1\right] \rightarrow \left[ \frac{1}{n+1},1\right] ,v_{n}\left( t\right) =\frac{nt+1}{n+1}. \end{aligned}$$

Then, for $n=1,2,3,\ldots$, $v_{n}=u_{n}^{-1}$. We have $\beta _{n}f={\tilde{\beta }}_{n}\left( f\circ u_{n}\right)$, for $f\in C\left[ 0,1\right]$ and ${\tilde{\beta }}_{n}g=\beta _{n}\left( g\circ v_{n}\right)$, for $g\in C\left[ \frac{1}{n+1},1\right]$. The shape preserving properties of $\beta _{n}$ can be translated to ${\tilde{\beta }}_{n}$. In particular, let $h\in C^{1} \left[ 0,1\right]$. Then, the functions $\left\| h^{\prime }\right\| e_{1}\pm h$ are monotonically increasing, hence $\left\| h^{\prime }\right\| \beta _{n}e_{1}\pm \beta _{n}h$ are monotonically increasing. This implies $\left\| h^{\prime }\right\| \left( \beta _{n}e_{1}\right) ^{\prime }\pm \left( \beta _{n}h\right) ^{\prime }\ge 0$, i.e.,

$$\begin{aligned} -\left\| h^{\prime }\right\| \left( \beta _{n}e_{1}\right) ^{\prime }\le \left( \beta _{n}h\right) ^{\prime }\le \left\| h^{\prime }\right\| \left( \beta _{n}e_{1}\right) ^{\prime }. \end{aligned}$$

Since $0\le \left( \beta _{n}e_{1}\right) ^{\prime }\le \frac{n-1}{n}$ (see [6, Theorem 2.1(i) and Rem. 2.3]) we obtain

$$\begin{aligned} \left\| \left( \beta _{n}h\right) ^{\prime }\right\| \le \frac{n-1}{n} \left\| h^{\prime }\right\| ,\text {for all}h\in C^{1}\left[ 0,1\right] \end{aligned}$$

(1)

(see also [4, Ex. 4.1]).

Now let $g\in C^{1}\left[ \frac{1}{n+1},1\right]$. Then

$$\begin{aligned} \left\| \left( {\tilde{\beta }}_{n}g\right) ^{\prime }\right\|= & {} \left\| \beta _{n}\left( g\circ v_{n}\right) ^{\prime }\right\| \le \frac{n-1}{n}\left\| \left( g\circ v_{n}\right) ^{\prime }\right\| \\= & {} \frac{n-1}{n}\left\| g^{\prime }\left( v_{n}\right) v_{n}^{\prime }\right\| \le \frac{n-1}{n}\left\| g^{\prime }\right\| \cdot \frac{ n}{n+1}, \end{aligned}$$

i.e.,

$$\begin{aligned} \left\| \left( {\tilde{\beta }}_{n}g\right) ^{\prime }\right\| \le \frac{ n-1}{n+1}\left\| g^{\prime }\right\| ,\text { for all}g\in C^{1} \left[ \frac{1}{n+1},1\right] . \end{aligned}$$

(2)

The inequalities (1) and (2) are instrumental in investigating the asymptotic behaviour of the iterates of $\beta _{n}$ and ${\tilde{\beta }}_{n}$; see [4].

Let $f\in C\left[ 0,1\right]$. Then, with $\delta =\sqrt{\frac{3n+1}{\left( n+1\right) ^{2}}\left( 1-x^{2}\right) }$, we obtain from Theorem  below

$$\begin{aligned} \left| \left( \beta _{n}f\right) \left( x\right) -f\left( x\right) \right|= & {} \left| \left( {\tilde{\beta }}_{n}\left( f\circ u_{n}\right) \right) \left( x\right) -f\left( x\right) \right| \\\le & {} \left| \left( {\tilde{\beta }}_{n}\left( f\circ u_{n}\right) \right) \left( x\right) -\left( f\circ u_{n}\right) \left( x\right) \right| +\left| \left( f\circ u_{n}\right) \left( x\right) -f\left( x\right) \right| \\\le & {} 2\omega \left( f\circ u_{n};\delta \right) +\left| f\left( u_{n}\left( x\right) \right) -f\left( x\right) \right| , \end{aligned}$$

where

$$\begin{aligned} \omega \left( f\circ u_{n};\delta \right)= & {} \sup \left\{ \left| \left( f\circ u_{n}\right) \left( t_{1}\right) -\left( f\circ u_{n}\right) \left( t_{2}\right) \right| :\frac{1}{n+1}\le t_{1},t_{2}\le 1,\left| t_{1}-t_{2}\right| \le \delta \right\} \\= & {} \sup \left\{ \left| f\left( u_{n}\left( t_{1}\right) \right) -f\left( u_{n}\left( t_{2}\right) \right) \right| :\frac{1}{n+1}\le t_{1},t_{2}\le 1,\left| t_{1}-t_{2}\right| \le \delta \right\} \\= & {} \sup \left\{ \left| f\left( s_{1}\right) -f\left( s_{2}\right) \right| :0\le s_{1},s_{2}\le 1,\left| s_{1}-s_{2}\right| \le \frac{n+1}{n}\delta \right\} \\= & {} \omega \left( f;\frac{n+1}{n}\delta \right) . \end{aligned}$$

Thus

$$\begin{aligned} \left| \left( \beta _{n}f\right) \left( x\right) -f\left( x\right) \right| \le 2\omega \left( f;\frac{n+1}{n}\delta \right) +\omega \left( f;\frac{\left| 1-x\right| }{n}\right) . \end{aligned}$$

Consequently,

$$\begin{aligned}&\left| \left( \beta _{n}f\right) \left( x\right) -f\left( x\right) \right| \le 2\omega \left( f;\frac{1}{n}\sqrt{\left( 3n+1\right) \left( 1-x^{2}\right) }\right) \nonumber \\&\quad +\omega \left( f;\frac{1-x}{n}\right) ,\text {for}f\in C\left[ 0,1\right] . \end{aligned}$$

(3)

In particular,

$$\begin{aligned} \left| \left( \beta _{n}f\right) \left( x\right) -f\left( x\right) \right| \le 2\omega \left( f;\frac{1}{n}\sqrt{3n+1}\right) +\omega \left( f;\frac{1}{n}\right) ,\text {for}f\in C\left[ 0,1\right] . \end{aligned}$$

See also Soardi’s estimate [8,  Theorem 2]

$$\begin{aligned} \left\| \beta _{n}f-f\right\| \le \left( 55+\frac{32}{n}\right) \omega \left( f;\frac{1}{\sqrt{n}}\right) ,\text {for}f\in C\left[ 0,1\right] . \end{aligned}$$

4 Voronovskaja-type results for the operators $\beta _{n}$ and $\tilde{\beta }_{n}$

In 2000, Raşa [6,  Theorem 4.1] proved the following Voronovskaja-type formula for the operators $\beta _{n}$.

Theorem 1

Let $x\in \left( 0,1\right)$ and f be a bounded function on $\left[ 0,1 \right]$. If $f^{\prime \prime }\left( x\right)$ exists, then

$$\begin{aligned} \left( \beta _{n}f\right) \left( x\right) =f\left( x\right) +\frac{1}{n} \left[ \left( \frac{1}{x}-1\right) f^{\prime }\left( x\right) +\frac{1-x^{2} }{2}f^{\prime \prime }\left( x\right) \right] +o\left( 1/n\right) \end{aligned}$$

as $n\rightarrow \infty$.

If $x\ne 0$, i.e., $t\ne 1/2$, you can insert the well-known asymptotic formulas for $B_{n}$. One obtains

$$\begin{aligned} \left( {\tilde{\beta }}_{n}f\right) \left( x\right) =f\left( x\right) +\frac{ 1-x^{2}}{n}\left[ \frac{1}{x}f^{\prime }\left( x\right) +\frac{1}{2} f^{\prime \prime }\left( x\right) \right] +o\left( 1/n\right) \end{aligned}$$

as $n\rightarrow \infty$. In the special case $x=0$, we can use

$$\begin{aligned} \lim _{x\rightarrow 0}\frac{\left( 1-x\right) ^{k}\left( 1+x\right) ^{n+1-k}-\left( 1-x\right) ^{n+1-k}\left( 1+x\right) ^{k}}{x}=2\left( n+1-2k\right) \end{aligned}$$

in order to obtain

$$\begin{aligned} \left( {\tilde{\beta }}_{n}f\right) \left( 0\right) =\left( n+1\right) \left( B_{n+1}{\hat{g}}\right) \left( \frac{1}{2}\right) , \end{aligned}$$

where

$$\begin{aligned} {\hat{g}}\left( t\right) =\left( 1-2t\right) g\left( t\right) =\left( 1-2t\right) ^{2}f\left( \left| 1-2t\right| \right) . \end{aligned}$$

The asymptotic behaviour can easily be derived if f is an even function which is smooth in $x=0$. If f is not an even function, $\left( B_{n+1} {\hat{g}}\right) \left( \frac{1}{2}\right)$ is an unpleasant expression.

The link to Soardi’s original operator is given by

$$\begin{aligned} \beta _{n}f={\tilde{\beta }}_{n}\left( f\circ u_{n}\right) \end{aligned}$$

(4)

with $u_{n}\left( x\right) =\left( \left( n+1\right) t-1\right) /n$. Therefore,

$$\begin{aligned} \left( \beta _{n}f\right) \left( x\right) =\left( {\tilde{\beta }} _{n}f_{n}\right) \left( x\right) =\left( {\tilde{\beta }}_{n}f\right) \left( x\right) +\frac{x-1}{n}\left( {\tilde{\beta }}_{n}f^{\prime }\right) \left( x\right) +o\left( 1/n\right) \end{aligned}$$

as $n\rightarrow \infty$. A look into the proof of asymptotic formulas for Bernstein polynomials reveals that the latter formula is valid if f is only locally smooth.

We have

$$\begin{aligned} \left( B_{n}f\right) \left( x\right) \sim f\left( x\right) +\sum _{k=1}^{\infty }\frac{c_{k}\left( f,x\right) }{n^{k}} \quad \left( n\rightarrow \infty \right) \end{aligned}$$

with

$$\begin{aligned} c_{k}\left( f,x\right) =\sum _{j=k}^{2k}a_{k,j}\left( x\right) f^{\left( j\right) }\left( x\right) , \end{aligned}$$

where $a_{k,j}\left( x\right)$ are certain polynomials involving Stirling numbers of the first and the second kind. More precisely, we have

$$\begin{aligned} \left( B_{n}f\right) \left( x\right) =f\left( x\right) +\sum _{k=1}^{q}n^{-k}\sum _{s=k}^{2k}\frac{1}{s!}f^{\left( s\right) }\left( x\right) \sum _{\nu =0}^{s}a\left( k,s,\nu \right) \text { }x^{s-\nu }+o \left( n^{-q}\right) \end{aligned}$$

as $n\rightarrow \infty$, where

$$\begin{aligned} a\left( k,s,\nu \right) =\sum _{r\mathbf {=\max }\left\{ \nu ,k\right\} }^{s}\left( -1\right) ^{s-r}\left( {\begin{array}{c}s\\ r\end{array}}\right) S\left( r-\nu ,r-k\right) \text { } \sigma \left( r,r-\nu \right) , \end{aligned}$$

provided that f is bounded on $\left[ 0,1\right]$ and admits a derivative of order 2q at $x\in \left[ 0,1\right]$ (see [1,  Remark 2]).

Let $f\in C\left[ 0,1\right]$. We define f on $\left[ -1,+1\right]$ such that f becomes an even function, i.e., $f\left( -x\right) =f\left( x\right)$. Put $\varphi \left( t\right) =1-2t$. If $x\ne 0$, i.e., $t\ne 1/2$, we have

$$\begin{aligned} g\left( t\right) =\varphi \left( t\right) f\left( \varphi \left( t\right) \right) \end{aligned}$$

and

$$\begin{aligned} g^{\left( j\right) }\left( t\right)= & {} \left( -2\right) ^{j}\left[ \left( 1-2t\right) f^{\left( j\right) }\left( \varphi \left( t\right) \right) +jf^{\left( j-1\right) }\left( \varphi \left( t\right) \right) \right] , \\ g^{\left( j\right) }\left( \frac{1-x}{2}\right)= & {} \left( -2\right) ^{j} \left[ xf^{\left( j\right) }\left( x\right) +jf^{\left( j-1\right) }\left( x\right) \right] , \\ g^{\left( j\right) }\left( \frac{1+x}{2}\right)= & {} \left( -2\right) ^{j} \left[ -xf^{\left( j\right) }\left( -x\right) +jf^{\left( j-1\right) }\left( -x\right) \right] \\= & {} -2^{j}\left[ xf^{\left( j\right) }\left( x\right) +jf^{\left( j-1\right) }\left( x\right) \right] . \end{aligned}$$

Then

$$\begin{aligned} \left( {\tilde{\beta }}_{n-1}f\right) \left( x\right) =\frac{1}{x}\left( B_{n}g\right) \left( \frac{1-x}{2}\right) \sim f\left( x\right) +\sum _{k=1}^{\infty }\frac{x^{-1}c_{k}\left( g,\frac{1-x}{2}\right) }{n^{k}}\quad \left( n\rightarrow \infty \right) . \end{aligned}$$

5 An estimate of the rate of convergence for the operators $\tilde{ \beta }_{n}$

In this section we derive an estimate for the rate of convergence for the operators ${\tilde{\beta }}_{n}$ in terms of the ordinary modulus of continuity $\omega \left( f,\delta \right)$.

Put $\tilde{g}\left( x\right) =g\left( 1-x\right)$. Then $\left( B_{n}g\right) \left( \frac{1+x}{2}\right) =\left( B_{n}\mathbf {g}\right) \left( \frac{1-x}{2}\right)$ and

$$\begin{aligned} \left( {\tilde{\beta }}_{n-1}f\right) \left( x\right) =\frac{1}{2x}\left( B_{n}\left( g-\tilde{g}\right) \right) \left( \frac{1-x}{2}\right) . \end{aligned}$$

For functions of the form

$$\begin{aligned} g\left( t\right) =\left( 1-2t\right) f\left( \left| 1-2t\right| \right) , \end{aligned}$$

we have $\tilde{g}=-g$. Hence,

$$\begin{aligned} \left( {\tilde{\beta }}_{n-1}f\right) \left( x\right) =\frac{1}{x}\left( B_{n}g\right) \left( \frac{1-x}{2}\right) . \end{aligned}$$

(5)

Lemma 2

For all $n\in {\mathbb {N}}$,

$$\begin{aligned} \left( {\tilde{\beta }}_{n}e_{1}\right) \left( x\right) \ge x \left( x\in \left[ 0,1\right] \right) . \end{aligned}$$

Proof

With the notations of Sect. 3 we have

$$\begin{aligned}&{\tilde{\beta }}_{n}e_{1}=\beta _{n}\left( e_{1}\circ v_{n}\right) =\beta _{n}v_{n}=\beta _{n}\left( \frac{n}{n+1}e_{1}+\frac{1}{n+1}e_{0}\right) \\&= \frac{n}{n+1}\beta _{n}e_{1}+\frac{1}{n+1}\beta _{n}e_{0}. \end{aligned}$$

Since $\beta _{n}$ preserves constant functions and $\beta _{n}f\ge f$, for all increasing and convex functions $f\in C\left[ 0,1\right]$, we obtain

$$\begin{aligned} {\tilde{\beta }}_{n}e_{1}\ge \frac{n}{n+1}e_{1}+\frac{1}{n+1}e_{0}=e_{1}+\frac{ e_{0}-e_{1}}{n+1}\ge e_{1}. \end{aligned}$$

$\square$

For reals t, x, put $\psi _{x}\left( t\right) =t-x$.

Lemma 3

For all $n\in {\mathbb {N}}$, the second central moment of ${\tilde{\beta }}_{n}$ satisfies the estimate

$$\begin{aligned} \left( {\tilde{\beta }}_{n}\psi _{x}^{2}\right) \left( x\right) \le \frac{3n+1 }{\left( n+1\right) ^{2}}\left( 1-x^{2}\right) \quad \left( x\in \left[ 0,1\right] \right) . \end{aligned}$$

Remark 1

The constant on the right-hand side is best possible on $\left[ 0,1\right]$ because, for $x=0$, we have $\left( {\tilde{\beta }}_{n}\psi _{0}^{2}\right) \left( 0\right) =\left( {\tilde{\beta }}_{n}e_{2}\right) \left( 0\right) =\left( 3n+1\right) /\left( n+1\right) ^{2}$.

Proof

We have

$$\begin{aligned} \left( {\tilde{\beta }}_{n}\psi _{x}^{2}\right) \left( x\right) =\left( {\tilde{\beta }}_{n}e_{2}\right) \left( x\right) -2x\left( {\tilde{\beta }} _{n}e_{1}\right) \left( x\right) +x^{2}\left( {\tilde{\beta }}_{n}e_{0}\right) \left( x\right) \le \left( {\tilde{\beta }}_{n}e_{2}\right) \left( x\right) -x^{2} \end{aligned}$$

on $\left[ 0,1\right]$, where we used the inequality of Lemma 2. The desired estimate now follows from

$$\begin{aligned} \left( {\tilde{\beta }}_{n-1}e_{2}\right) \left( x\right) =x^{2}+\left( 1-x^{2}\right) \left( 3n-2\right) /n^{2}. \end{aligned}$$

$\square$

Theorem 2

Let $f:C\left[ 0,1\right]$. For all $n\in {\mathbb {N}}$, and $\delta >0$,

$$\begin{aligned} \left| \left( {\tilde{\beta }}_{n}f\right) \left( x\right) -f\left( x\right) \right| \le \left( 1+\frac{1}{\delta }\sqrt{\frac{3n+1}{\left( n+1\right) ^{2}}\left( 1-x^{2}\right) }\right) \omega \left( f,\delta \right) \quad \left( x\in \left[ 0,1\right] \right) . \end{aligned}$$

Proof of Theorem 2

The estimate follows from Lemma 3 by standard arguments (see, e.g., [2, Theorem5.1.2]). $\square$

Putting $\delta =\sqrt{3/\left( n+1\right) }$ immediately yields the following consequence.

Corollary 1

For all $n\in {\mathbb {N}}$,

$$\begin{aligned} \left| \left( {\tilde{\beta }}_{n}f\right) \left( x\right) -f\left( x\right) \right| \le \left( 1+\sqrt{1-x^{2}}\right) \omega \left( f, \sqrt{\frac{3}{n+1}}\right) \quad \left( x\in \left[ 0,1\right] \right) . \end{aligned}$$

6 An estimate of rate of convergence for the Soardi operator

As already mentioned in the introduction Soardi [8, Theorem 2] estimated the rate of convergence of the operators $\beta _{n}$ in terms of the ordinary modulus of continuity:

$$\begin{aligned} \left\| \beta _{n}f-f\right\| \le \left( 55+\frac{32}{n}\right) \omega \left( f;\frac{1}{\sqrt{n}}\right) ,\text {for}f\in C\left[ 0,1\right] . \end{aligned}$$

In this section we improve this estimate considerably by diminishing the absolute constant.

Theorem 3

Let $f:C\left[ 0,1\right]$. For all $n\in {\mathbb {N}}$, Soardi’s operator $\beta _{n}$ satisfies the estimate

$$\begin{aligned} \left\| \beta _{n}f-f\right\| \le \left( 1+\sqrt{3+\frac{2}{n}} \right) \omega \left( f;\frac{1}{\sqrt{n}}\right) \quad \left( n\in {\mathbb {N}}\right) . \end{aligned}$$

Remark 2

In particular, we have

$$\begin{aligned} \left\| \beta _{n}f-f\right\| \le c\cdot \omega \left( f;\frac{1}{ \sqrt{n}}\right) \quad \left( n\in {\mathbb {N}}\right) , \end{aligned}$$

where $c=\left( 1+\sqrt{5}\right) \approx 3.236$.

The essential ingredient of the proof is the following estimate of the second central moment of the operators $\beta _{n}$.

Lemma 4

For all $n\in {\mathbb {N}}$, the second central moment of $\beta _{n}$ satisfies the estimate

$$\begin{aligned} \left( \beta _{n}\psi _{x}^{2}\right) \left( x\right) \le \frac{3}{n}\left( 1-x^{2}\right) +\frac{2}{n^{2}}\left( 1-x\right) \quad \left( x\in \left[ 0,1\right] \right) . \end{aligned}$$

Remark 3

Since $1-x\le 1-x^{2}$ on $\left[ 0,1\right]$, we have

$$\begin{aligned} \left( \beta _{n}\psi _{x}^{2}\right) \left( x\right) \le \frac{5}{n}\left( 1-x^{2}\right) \quad \left( x\in \left[ 0,1\right] \right) . \end{aligned}$$

Furthermore, for each $\varepsilon >0$, there is an index $n_{0}$ such that for each $n>n_{0}$,

$$\begin{aligned} \left( \beta _{n}\psi _{x}^{2}\right) \left( x\right) \le \frac{ 3+\varepsilon }{n}\left( 1-x^{2}\right) \quad \left( x\in \left[ 0,1 \right] \right) . \end{aligned}$$

Proof of Lemma 4

Using the relation $\beta _{n}f={\tilde{\beta }}_{n}\left( f\circ u_{n}\right)$ from Sect. 3 with $u_{n}\left( x\right) =\left( \left( n+1\right) t-1\right) /n$ we obtain

$$\begin{aligned} \beta _{n}\psi _{x}^{2}= & {} \left( \frac{n+1}{n}\right) ^{2}{\tilde{\beta }} _{n}\left( e_{1}-\frac{nx+1}{n+1}e_{0}\right) ^{2}\\= & {} \left( \frac{n+1}{n} \right) ^{2}\left[ {\tilde{\beta }}_{n}e_{2}-2\frac{nx+1}{n+1}{\tilde{\beta }} _{n}e_{1}+\left( \frac{nx+1}{n+1}\right) ^{2}e_{0}\right] . \end{aligned}$$

By Lemma 2,

$$\begin{aligned} \left( \beta _{n}\psi _{x}^{2}\right) \left( x\right)\le & {} \left( \frac{n+1 }{n}\right) ^{2}\left[ x^{2}+\left( 1-x^{2}\right) \frac{3n+1}{\left( n+1\right) ^{2}}-2\frac{nx+1}{n+1}x+\left( \frac{nx+1}{n+1}\right) ^{2} \right] \\= & {} \frac{3n+1}{n^{2}}\left( 1-x^{2}\right) +\left( \frac{n+1}{n}\right) ^{2}\left( x-\frac{nx+1}{n+1}\right) ^{2} \\= & {} \frac{3n+1}{n^{2}}\left( 1-x^{2}\right) +\left( \frac{x-1}{n}\right) ^{2} \\= & {} \frac{3}{n}\left( 1-x^{2}\right) +\frac{1-x^{2}+\left( x-1\right) ^{2}}{ n^{2}} \\= & {} \frac{3}{n}\left( 1-x^{2}\right) +\frac{2}{n^{2}}\left( 1-x\right) , \end{aligned}$$

which is the desired estimate. $\square$

Proof of Theorem 3

By Lemma 4, it holds

$$\begin{aligned} \left( \beta _{n}\psi _{x}^{2}\right) \left( x\right) \le \frac{3}{n}+\frac{ 2}{n^{2}} \quad \left( x\in \left[ 0,1\right] \right) . \end{aligned}$$

Using [2, (5.1.5)], we obtain

$$\begin{aligned} \left| \left( \beta _{n}f\right) \left( x\right) -f\left( x\right) \right|\le & {} \left( 1+\sqrt{n\left( \beta _{n}\psi _{x}^{2}\right) \left( x\right) }\right) \omega \left( f;\frac{1}{\sqrt{n}}\right) \\\le & {} \left( 1+\sqrt{3+\frac{2}{n}}\right) \omega \left( f;\frac{1}{\sqrt{n}} \right) . \end{aligned}$$

This completes the proof. $\square$

7 The second moment of $\beta _{n}$

We have

$$\begin{aligned} \left( \beta _{n}e_{2}\right) \left( x\right)= & {} \left( {\tilde{\beta }} _{n}u_{n}^{2}\right) \left( x\right) =\left( {\tilde{\beta }}_{n}\left( \frac{ n+1}{n}e_{1}-\frac{1}{n}e_{0}\right) ^{2}\right) \left( x\right) \\= & {} \left( \frac{n+1}{n}\right) ^{2}\left( {\tilde{\beta }}e_{2}\right) \left( x\right) +\frac{1}{n^{2}}-2\frac{n+1}{n^{2}}\left( {\tilde{\beta }} _{n}e_{1}\right) \left( x\right) . \end{aligned}$$

Since

$$\begin{aligned} \left( {\tilde{\beta }}e_{2}\right) \left( x\right) =x^{2}+\frac{3n+1}{\left( n+1\right) ^{2}}\left( 1-x^{2}\right) \text {and }\left( {\tilde{\beta }}_{n}e_{1}\right) \left( x\right) \ge x \end{aligned}$$

we obtain

$$\begin{aligned} \left( \beta _{n}e_{2}\right) \left( x\right)\le & {} \left( \frac{n+1}{n} \right) ^{2}\left( x^{2}+\frac{3n+1}{\left( n+1\right) ^{2}}\left( 1-x^{2}\right) \right) +\frac{1}{n^{2}}-2\frac{n+1}{n^{2}}x \\\le & {} \left( \frac{n+1}{n}\right) ^{2}x^{2}+\frac{3n+1}{n^{2}}-\frac{3n+1}{ n^{2}}x^{2}+\frac{1}{n^{2}}-\frac{2n+2}{n^{2}}x \\= & {} x^{2}+\frac{1}{n^{2}}\left( -nx^{2}-\left( 2n+2\right) x+3n+2\right) \\= & {} x^{2}+\frac{1-x}{n^{2}}\left( nx+3n+2\right) . \end{aligned}$$

It follows

$$\begin{aligned} 0\le \left( \beta _{n}e_{2}\right) \left( x\right) -x^{2}\le \frac{nx+3n+2 }{n^{2}}\left( 1-x\right) . \end{aligned}$$

8 The value $\left( \beta _{n}e_{1}\right) \left( 0\right)$ of the first moment

The operator $\beta _{n}$ does not reproduce the function $e_{1}\left( x\right) =x$, $x\in \left[ 0,1\right]$. But $\beta _{n}e_{1}$ is increasing and convex ([6, Th. 2.1]), $\beta _{n}e_{1}\ge e_{1}$ ([6, Th. 3.1]), and $\beta _{n}e_{1}\left( 1\right) =1$. Consequently,

$$\begin{aligned} 0\le \beta _{n}e_{1}\left( x\right) -x\le \beta _{n}e_{1}\left( 0\right) \left( 1-x\right) ,\text {for}x\in \left[ 0,1\right] . \end{aligned}$$

So, we need a good control on $\beta _{n}e_{1}(0)$. This is our aim in what follows.

By Eq. ( 3) , we infer that

$$\begin{aligned} x\le \left( \beta _{n}e_{1}\right) \left( x\right) \le x+\frac{2}{n}\sqrt{ \left( 3n+1\right) \left( 1-x^{2}\right) }+\frac{1-x}{n},\text {for} \quad x\in \left[ 0,1\right] . \end{aligned}$$

In particular, it follows that

$$\begin{aligned} 0\le \left( \beta _{n}e_{1}\right) \left( 0\right) \le \frac{1+2\sqrt{3n+1} }{n}\sim 2\sqrt{\frac{3}{n}} \quad \left( n\rightarrow \infty \right) . \end{aligned}$$

In the next section we derive closed expressions for $\left( \beta _{n}e_{1}\right) \left( 0\right)$ and study its asymptotic behaviour as n tends to infinity. We prove that the exact asymptotic rate of convergence is

$$\begin{aligned} \left( \beta _{n}e_{1}\right) \left( 0\right) \sim \frac{2\sqrt{2}}{\sqrt{ \pi n}} \quad \left( n\rightarrow \infty \right) . \end{aligned}$$

Note that $2\sqrt{3}\approx 3.4641$ and $2\sqrt{2/\pi }\approx 1.59577$.

Theorem 4

At $x=0$, the first moment of Soardi’s operator has the explicit representation

$$\begin{aligned} \left( \beta _{2n}e_{1}\right) \left( 0\right)= & {} \frac{1}{2^{2n}}\left( 2+ \frac{1}{2n}\right) \left( {\begin{array}{c}2n\\ n\end{array}}\right) -\frac{1}{2n}, \\ \left( \beta _{2n-1}e_{1}\right) \left( 0\right)= & {} \frac{1}{2^{2n-1}}\left( 1+\frac{1}{2n-1}\right) \left( {\begin{array}{c}2n\\ n\end{array}}\right) -\frac{1}{2n-1} \end{aligned}$$

and satisfies the asymptotic relation

$$\begin{aligned} \left( \beta _{n}e_{1}\right) \left( 0\right) =\frac{2\sqrt{2}}{\sqrt{\pi n}} -\frac{1}{n}+O\left( n^{-3/2}\right) \quad \left( n\rightarrow \infty \right) . \end{aligned}$$

Proof

Since

$$\begin{aligned} \lim _{x\rightarrow 0}\frac{1}{x}\left[ \left( 1-x\right) ^{k}\left( 1+x\right) ^{n+1-k}-\left( 1-x\right) ^{n+1-k}\left( 1+x\right) ^{k}\right] =2\left( n+1-2k\right) , \end{aligned}$$

we have

$$\begin{aligned} {\tilde{w}}_{n,k}\left( 0\right) =\left( {\begin{array}{c}n+1\\ k\end{array}}\right) \frac{2\left( n+1-2k\right) ^{2}}{\left( n+1\right) 2^{n+1}} \end{aligned}$$

and

$$\begin{aligned} \left( \beta _{n}e_{r}\right) \left( 0\right) =\frac{1}{2^{n}\left( n+1\right) n^{r}}\sum _{k=0}^{m}\left( {\begin{array}{c}n+1\\ k\end{array}}\right) \left( n+1-2k\right) ^{2}\left( n-2k\right) ^{r}. \end{aligned}$$

Although one can calculate it for arbitrary $r\in {\mathbb {N}}$, we restrict ourselves to $r=1$. Let us first consider the case of even parameters 2n:

$$\begin{aligned} \left( \beta _{2n}e_{1}\right) \left( 0\right) =\frac{1}{2^{2n}\left( 2n+1\right) n}\sum _{k=0}^{n}\left( {\begin{array}{c}2n+1\\ k\end{array}}\right) \left( 2n+1-2k\right) ^{2}\left( n-k\right) . \end{aligned}$$

Writing

$$\begin{aligned}&\left( 2n+1-2k\right) ^{2}\left( n-k\right) =-4k^{{\underline{3}}}+4\left( 3n-2\right) k^{{\underline{2}}}\\&\quad +\left( -12n^{2}+4n-1\right) k+n\left( 2n+1\right) ^{2} \end{aligned}$$

we obtain

$$\begin{aligned}&\sum _{k=0}^{n}\left( {\begin{array}{c}2n+1\\ k\end{array}}\right) \left( 2n+1-2k\right) ^{2}\left( n-k\right) \\&\quad =-4\left( 2n+1\right) ^{{\underline{3}}}\sum _{k=3}^{n}\left( {\begin{array}{c}2n-2\\ k-3\end{array}}\right) +4\left( 3n-2\right) \left( 2n+1\right) ^{{\underline{2}}}\sum _{k=2}^{n}\left( {\begin{array}{c} 2n-1\\ k-2\end{array}}\right) \\&\qquad +\left( -12n^{2}+4n-1\right) \left( 2n+1\right) \sum _{k=1}^{n}\left( {\begin{array}{c}2n\\ k-1\end{array}}\right) +n\left( 2n+1\right) ^{2}\sum _{k=0}^{n}\left( {\begin{array}{c}2n+1\\ k\end{array}}\right) \\&\quad =-4\left( 2n+1\right) ^{{\underline{3}}}\sum _{k=0}^{n-3}\left( {\begin{array}{c}2n-2\\ k\end{array}}\right) +4\left( 3n-2\right) \left( 2n+1\right) ^{{\underline{2}}}\sum _{k=0}^{n-2} \left( {\begin{array}{c}2n-1\\ k\end{array}}\right) \\&\qquad +\left( -12n^{2}+4n-1\right) \left( 2n+1\right) \sum _{k=0}^{n-1}\left( {\begin{array}{c}2n\\ k\end{array}}\right) +n\left( 2n+1\right) ^{2}\sum _{k=0}^{n}\left( {\begin{array}{c}2n+1\\ k\end{array}}\right) \\&\quad =:A+B+C+D. \end{aligned}$$

Now

$$\begin{aligned} A= & {} -4\left( 2n+1\right) ^{{\underline{3}}}\frac{2^{2n-2}-\left( {\begin{array}{c}2n-2\\ n-2\end{array}}\right) - \left( {\begin{array}{c}2n-2\\ n-1\end{array}}\right) -\left( {\begin{array}{c}2n-2\\ n\end{array}}\right) }{2} \\= & {} -2^{2n-1}\left( 2n+1\right) ^{{\underline{3}}}+4\left( 2n+1\right) \left[ n\left( n-1\right) \left( {\begin{array}{c}2n\\ n\end{array}}\right) +\frac{n^{2}}{2}\left( {\begin{array}{c}2n\\ n\end{array}}\right) \right] \\= & {} -2^{2n-1}\left( 2n+1\right) ^{{\underline{3}}}+2n\left( 2n+1\right) \left( 3n-2\right) \left( {\begin{array}{c}2n\\ n\end{array}}\right) , \\ B= & {} 4\left( 3n-2\right) \left( 2n+1\right) ^{{\underline{2}}}\frac{2^{2n-1}- \left( {\begin{array}{c}2n-1\\ n-1\end{array}}\right) -\left( {\begin{array}{c}2n-1\\ n\end{array}}\right) }{2} \\= & {} 2^{2n}\left( 3n-2\right) \left( 2n+1\right) ^{{\underline{2}}}-2\left( 3n-2\right) \left( 2n+1\right) ^{{\underline{2}}}\left( {\begin{array}{c}2n\\ n\end{array}}\right) , \\ C= & {} \left( -12n^{2}+4n-1\right) \left( 2n+1\right) \frac{2^{2n}-\left( {\begin{array}{c}2n\\ n \end{array}}\right) }{2}, \\ D= & {} 2^{2n}n\left( 2n+1\right) ^{2}. \end{aligned}$$

Finally,

$$\begin{aligned} \left( \beta _{2n}e_{1}\right) \left( 0\right)= & {} \frac{1}{2^{2n}\left( 2n+1\right) n}\left( A+B+C+D\right) \\= & {} -\left( 2n-1\right) +\left( 6n-4\right) +\left( -6n+2-\frac{1}{2n}\right) +\left( 2n+1\right) \\&+\frac{1}{2^{2n}}\left( {\begin{array}{c}2n\\ n\end{array}}\right) \left[ 2\left( 3n-2\right) -4\left( 3n-2\right) +\left( 6n-2+\frac{1}{2n}\right) \right] \\= & {} -\frac{1}{2n}+\frac{1}{2^{2n}}\left( {\begin{array}{c}2n\\ n\end{array}}\right) \left( 2+\frac{1}{2n}\right) . \end{aligned}$$

The well-known asymptotic behaviour of the central binomial coefficient (cf. Catalan constant $\frac{1}{n+1}\left( {\begin{array}{c}2n\\ n\end{array}}\right)$)

$$\begin{aligned} \left( {\begin{array}{c}2n\\ n\end{array}}\right) =\frac{4^{n}}{\sqrt{\pi n}}\left( 1+O\left( n^{-1}\right) \right) \quad \left( n\rightarrow \infty \right) \end{aligned}$$

leads to the asymptotic formula

$$\begin{aligned} \left( \beta _{2n}e_{1}\right) \left( 0\right) =\frac{2}{\sqrt{\pi n}}-\frac{ 1}{2n}+O\left( n^{-3/2}\right) \quad \left( n\rightarrow \infty \right) . \end{aligned}$$

Now we consider the case of odd parameters $2n-1$:

$$\begin{aligned} \left( \beta _{2n-1}e_{1}\right) \left( 0\right) =\frac{1}{2^{2n-2}n\left( 2n-1\right) }\sum _{k=0}^{n-1}\left( {\begin{array}{c}2n\\ k\end{array}}\right) \left( n-k\right) ^{2}\left( 2n-1-2k\right) . \end{aligned}$$

Writing

$$\begin{aligned} \left( n-k\right) ^{2}\left( 2n-1-2k\right) =-2k^{{\underline{3}}}+\left( 6n-7\right) k^{{\underline{2}}}+\left( -6n^{2}+8n-3\right) k+n^{2}\left( 2n-1\right) \end{aligned}$$

we obtain

$$\begin{aligned}&\sum _{k=0}^{n-1}\left( {\begin{array}{c}2n\\ k\end{array}}\right) \left( n-k\right) ^{2}\left( 2n-1-2k\right) \\&\quad =-2\left( 2n\right) ^{{\underline{3}}}\sum _{k=3}^{n-1}\left( {\begin{array}{c}2n-3\\ k-3\end{array}}\right) +\left( 6n-7\right) \left( 2n\right) ^{{\underline{2}}}\sum _{k=2}^{n-1}\left( {\begin{array}{c} 2n-2\\ k-2\end{array}}\right) \\&\qquad +\left( -6n^{2}+8n-3\right) \left( 2n\right) \sum _{k=1}^{n-1}\left( {\begin{array}{c}2n-1\\ k-1\end{array}}\right) +n^{2}\left( 2n-1\right) \sum _{k=0}^{n-1}\left( {\begin{array}{c}2n\\ k\end{array}}\right) \\&\quad =:A+B+C+D. \end{aligned}$$

Now

$$\begin{aligned} A= & {} -2\left( 2n\right) ^{{\underline{3}}}\sum _{k=0}^{n-4}\left( {\begin{array}{c}2n-3\\ k\end{array}}\right) =-2\left( 2n\right) ^{{\underline{3}}}\frac{2^{2n-3}-2\left( {\begin{array}{c}2n-3\\ n-3\end{array}}\right) -2 \left( {\begin{array}{c}2n-3\\ n-2\end{array}}\right) }{2} \\= & {} -\left( 2n\right) ^{{\underline{3}}}2^{2n-3}+2\left( {\begin{array}{c}2n\\ n\end{array}}\right) n\left( n-1\right) \left( n-2\right) +2\left( {\begin{array}{c}2n\\ n\end{array}}\right) n^{2}\left( n-1\right) \\= & {} -\left( 2n\right) ^{{\underline{3}}}2^{2n-3}+4n\left( n-1\right) ^{2}\left( {\begin{array}{c} 2n\\ n\end{array}}\right) , \\ B= & {} \left( 6n-7\right) \left( 2n\right) ^{{\underline{2}}}\sum _{k=0}^{n-3} \left( {\begin{array}{c}2n-2\\ k\end{array}}\right) =\left( 6n-7\right) \left( 2n\right) ^{{\underline{2}}}\frac{ 2^{2n-2}-2\left( {\begin{array}{c}2n-2\\ n-2\end{array}}\right) -\left( {\begin{array}{c}2n-2\\ n-1\end{array}}\right) }{2} \\= & {} \left( 6n-7\right) \left[ \left( 2n\right) ^{{\underline{2}}}2^{2n-3}- \left( {\begin{array}{c}2n\\ n\end{array}}\right) n\left( n-1\right) -\frac{1}{2}\left( {\begin{array}{c}2n\\ n\end{array}}\right) n^{2}\right] \\= & {} \left( 6n-7\right) \left( 2n\right) ^{{\underline{2}}}2^{2n-3}-\frac{1}{2} \left( 6n-7\right) n\left( 3n-2\right) \left( {\begin{array}{c}2n\\ n\end{array}}\right) , \\ C= & {} \left( -6n^{2}+8n-3\right) \left( 2n\right) \sum _{k=0}^{n-2}\left( {\begin{array}{c}2n-1 \\ k\end{array}}\right) \\= & {} \left( -6n^{2}+8n-3\right) \left( 2n\right) \frac{2^{2n-1}-2\left( {\begin{array}{c}2n-1 \\ n-1\end{array}}\right) }{2} \\= & {} \left( -6n^{2}+8n-3\right) n\left( 2^{2n-1}-\left( {\begin{array}{c}2n\\ n\end{array}}\right) \right) , \\ D= & {} n^{2}\left( 2n-1\right) \frac{1}{2}\left( 2^{2n}-\left( {\begin{array}{c}2n\\ n\end{array}}\right) \right) . \end{aligned}$$

Finally,

$$\begin{aligned}&\left( \beta _{2n-1}e_{1}\right) \left( 0\right) \\&\quad =\frac{1}{2^{2n-2}n\left( 2n-1\right) }\left( A+B+C+D\right) \\&\quad =-\left( 2n-2\right) +\left( 6n-7\right) +\frac{2\left( -6n^{2}+8n-3\right) }{2n-1}+\frac{2n\left( 2n-1\right) }{2n-1} \\&\qquad +\frac{1}{2^{2n-2}\left( 2n-1\right) }\left( {\begin{array}{c}2n\\ n\end{array}}\right) \\&\qquad \left[ 4\left( n-1\right) ^{2}-\frac{1}{2}\left( 6n-7\right) \left( 3n-2\right) +\left( 6n^{2}-8n+3\right) -\frac{1}{2}n\left( 2n-1\right) \right] \\&\quad =4n-5+\frac{-8n^{2}+14n-6}{2n-1}+\frac{n}{2^{2n-2}\left( 2n-1\right) } \left( {\begin{array}{c}2n\\ n\end{array}}\right) \\&\quad =\frac{-1}{2n-1}+\frac{2\left( 2n-1\right) +2}{2^{2n}\left( 2n-1\right) } \left( {\begin{array}{c}2n\\ n\end{array}}\right) =\frac{-1}{2n-1}+\frac{1}{2^{2n}}\left( {\begin{array}{c}2n\\ n\end{array}}\right) \left( 2+\frac{2}{ 2n-1}\right) . \end{aligned}$$

This proves the explicit representation for odd values of the parameter. As above we obtain the asymptotic formula

$$\begin{aligned} \left( \beta _{2n-1}e_{1}\right) \left( 0\right) =\frac{2}{\sqrt{\pi n}}- \frac{1}{2n-1}+O\left( n^{-3/2}\right) \quad \left( n\rightarrow \infty \right) . \end{aligned}$$

This completes the proof. $\square$