Asymptotic and Non-asymptotic Results in the Approximation by Bernstein Polynomials

Abstract

This paper deals with the approximation of functions by the classical Bernstein polynomials in terms of the Ditzian–Totik modulus of smoothness. Asymptotic and non-asymptotic results are respectively stated for continuous and twice continuously differentiable functions. By using a probabilistic approach, known results are either completed or strengthened.

1 Introduction and Statements of the Main Results

Let ${\mathbb {N}}$ be the set of positive integers and ${\mathbb {N}}_0={\mathbb {N}}\cup \{0\}$. As usual, C[0, 1] denotes the space of all real continuous functions defined on [0, 1], and $C^m[0,1]$, $m\in {\mathbb {N}}_0$, denotes the subspace of all m-times continuously differentiable functions, with the obvious understanding that $C^0[0,1]=C[0,1]$. For $m\in {\mathbb {N}}$, we denote by ${\mathscr {C}}^m[0,1]\supset C^m[0,1]$ the set of functions $f\in C^{m-1}[0,1]$ such that $f^{(m-1)}$ is absolutely continuous, i. e.,

$$\begin{aligned} f^{(m-1)}(y)-f^{(m-1)}(x)=\int _x^yg(u)du,\quad x,y \in [0,1], \end{aligned}$$

for some bounded measurable function g, which can be denoted by $g=f^{(m)}$.

The indicator function of a set A is denoted by $1_A$, and ${\mathbb {E}}$ stands for mathematical expectation.

Let $f\in C[0,1]$. The sup-norm of f is simply denoted by $\Vert f\Vert$, although, more generally, we use the notation $\Vert f\Vert _A=\sup \{|f(x)|:x\in A\}$, $A\subseteq [0,1]$.

The second order central difference of f is defined by

$$\begin{aligned} \Delta _h^2f(x)=f(x+h)-2f(x)+f(x-h),\quad h\ge 0, \end{aligned}$$

whenever $x\pm h\in [0,1]$. The Ditzian–Totik modulus of smoothness of f with weight function $\varphi (x)=\sqrt{x(1-x)}$ is defined by

$$\begin{aligned} \omega ^{\varphi }_2(f;\delta )=\sup \left\{ \left| \Delta _{h\varphi (x)}^2f(x)\right| :\ 0\le h\le \delta ,\ x\pm h\varphi (x)\in [0,1]\right\} ,\quad \delta \ge 0. \end{aligned}$$

The classical first order modulus of continuity is simply denoted by $\omega (f;\delta )$.

In this paper, we will make use of the following important inequality proved by Bustamante [2]:

$$\begin{aligned} \omega _2^{\varphi }(f;\lambda \delta )\le (2+3\lambda ^2)\omega _2^{\varphi }(f;\delta ),\qquad \lambda , \delta \ge 0,\quad \lambda \delta \in [0,1). \end{aligned}$$

(1)

Finally, the nth Bernstein polynomial of f is defined as

$$\begin{aligned} B_nf(x)=\sum _{k=0}^nf\left( \frac{k}{n}\right) p_{n,k}(x),\quad p_{n,k}(x)={n\atopwithdelims ()k}x^k(1-x)^{n-k},\quad k=0,1,\ldots ,n. \end{aligned}$$

We have the probabilistic representation

$$\begin{aligned} B_nf(x)={\mathbb {E}}f\left( \frac{S_n(x)}{n}\right) , \end{aligned}$$

(2)

where $S_n(x)$ is a random variable having the binomial law with parameters n and x, that is to say,

$$\begin{aligned} P(S_n(x)=k)=p_{n,k}(x),\quad k=0,1,\ldots ,n. \end{aligned}$$

(3)

Throughout this paper, whenever we write f, n, x, and y, we are assuming that $f\in C[0,1]$, $n\in {\mathbb {N}}$, and $x,y\in [0,1]$.

Following the works by Ditzian and Ivanov [4] and Totik [9], the rates of uniform convergence for the Bernstein polynomials are characterized by

$$\begin{aligned} K_1\omega _2^{\varphi }\left( f;\frac{1}{\sqrt{n}}\right) \le \Vert B_nf-f\Vert \le K_2\omega _2^{\varphi }\left( f;\frac{1}{\sqrt{n}}\right) , \end{aligned}$$

(4)

for some absolute constants $K_1$ and $K_2$. Whereas no specific values for $K_1$ have been provided yet, different authors completed statement (4) by showing specific values for the constant $K_2$. In this regard, Adell and Sangüesa [1] gave $K_2=4$, Gavrea et al. [5] and Bustamante [2] provided $K_2=3$, and finally, Păltănea [7] proved the validity of $K_2=2.5$, this being the best result up to date and up to our knowledge.

This notwithstanding, if additional smoothness conditions on f are added, then the second inequality in (4) may be valid for values of $K_2$ smaller than 2.5. In this respect, Bustamante and Quesada [3] and Păltănea [8] obtained the following asymptotic result

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\Vert B_nf-f\Vert }{\omega _2^{\varphi }\left( f;1/\sqrt{n}\right) }=\frac{1}{2},\quad f\in C^2[0,1], \end{aligned}$$

(5)

provided that f is not an affine function.

The contribution of this paper is twofold. In first place, we strength statement (5) by giving a non-asymptotic version of it. In fact, we prove the following result.

Theorem 1

If $f\in C^2[0,1]$, then

$$\begin{aligned} \left| \Vert B_nf-f\Vert -\frac{1}{2}\omega _2^{\varphi }\left( f;\frac{1}{\sqrt{n}}\right) \right| \le \frac{1}{4n}\left( \omega \left( f'';\frac{1}{3\sqrt{n}}\right) +\frac{1}{4}\omega _2^{\varphi }\left( f'';\frac{1}{\sqrt{n}}\right) \right) .\nonumber \\ \end{aligned}$$

(6)

As a consequence, statement (5) holds true.

In second place, we complete statement (4) in the following asymptotic form.

Theorem 2

Let $(\tau _n)_{n\ge 1}$ be a sequence of positive real numbers such that

$$\begin{aligned} \tau _n\longrightarrow \infty ,\quad \frac{\tau _n}{n}\longrightarrow 0,\qquad n\rightarrow \infty . \end{aligned}$$

If $f\in C[0,1]$ is not an affine function, then

$$\begin{aligned} \limsup _{n\rightarrow \infty }\frac{1}{\omega _2^{\varphi }\left( f;\frac{1}{\sqrt{n}}\right) }\Vert B_nf-f\Vert _{[\tau _n/n,1-\tau _n/n]}\le \frac{3}{2}. \end{aligned}$$

(7)

Moreover, we have in (4),

$$\begin{aligned} K_2\ge 1. \end{aligned}$$

(8)

This result is based upon Theorem 3 in Sect. 3, which gives estimates of the form

$$\begin{aligned} |B_nf(x)-f(x)|\le K_2(n,x)\omega _2^{\varphi }\left( f;\frac{1}{\sqrt{n}}\right) , \end{aligned}$$

for some explicit constants $K_2(n,x)$ depending on n and x.

The paper is organized as follows. The proof of Theorem 1 is given in Sect. 2. We show Theorem 2 in Sect. 3 with the aid of two kinds of auxiliary results. On the one hand, we define certain smooth approximants $Q_h^af$ of the function $f\in C[0,1]$, by antisymmetrizing in an appropriate way the classical Steklov means of f. On the other hand, we estimate the tail probabilities and the truncated variance of the random variable $S_n(x)$ appearing in the probabilistic representation of $B_nf$ given in (2).

2 Proof of Theorem 1

2.1 Preliminaries

The Taylor’s formula of order $m\in {\mathbb {N}}$ for $f\in {\mathscr {C}}^m[0,1]$, with remainder in integral form can be written as

$$\begin{aligned}&f(y)-\sum _{j=0}^{m-1}\frac{f^{(j)}(x)}{j!}(y-x)^j\nonumber \\&\quad =\frac{(y-x)^m}{(m-1)!}\int _0^1(1-\theta )^{m-1}f^{(m)}(x+(y-x)\theta )d\theta \nonumber \\&\quad =\frac{(y-x)^m}{m!}{\mathbb {E}}f^{(m)}(x+(y-x)\beta _m), \end{aligned}$$

(9)

where $\beta _m$ is a random variable with the beta density $\rho _m(\theta )=m(1-\theta )^{m-1}$, $0\le \theta \le 1$.

Lemma 1

If $f\in C^2[0,1]$ and $\delta \ge 0$, then

$$\begin{aligned} \left| \omega _2^{\varphi }(f;\delta )-\delta ^2\left\| \varphi ^2 f''\right\| \right| \le \frac{\delta ^2}{8}\omega _2^{\varphi }\left( f'';\delta \right) . \end{aligned}$$

Proof

Let $h\ge 0$ with $x\pm h\in [0,1]$. Using (9) with $m=2$, we get

$$\begin{aligned} f(x-h)= & {} f(x)-f'(x)h+\frac{f''(x)}{2}h^2\\&+\frac{h^2}{2}{\mathbb {E}}\left( f''(x-h\beta _2)-f''(x)\right) , \end{aligned}$$

as well as

$$\begin{aligned} f(x+h)= & {} f(x)+f'(x)h+\frac{f''(x)}{2}h^2\\&+\frac{h^2}{2}{\mathbb {E}}\left( f''(x+h\beta _2)-f''(x)\right) . \end{aligned}$$

Adding these two identities, we obtain

$$\begin{aligned} \Delta _h^2f(x)= & {} f''(x)h^2\nonumber \\&+\frac{h^2}{2}{\mathbb {E}}\left( f''(x+h\beta _2)-2f''(x)+f''(x-h\beta _2)\right) . \end{aligned}$$

(10)

Replacing in (10) h by $h\varphi (x)$ and applying the reverse triangular inequality, we have

$$\begin{aligned}&\left| \omega _2^{\varphi }(f;\delta )-\delta ^2\left\| \varphi ^2 f''\right\| \right| \le \frac{\delta ^2}{2}\left\| \varphi ^2\right\| \omega _2^{\varphi }\left( f'';\delta \right) \\&\quad = \frac{\delta ^2}{8}\omega _2^{\varphi }(f'';\delta ), \end{aligned}$$

thus completing the proof. $\square$

Gonska et al. [6] showed that

$$\begin{aligned} \left\| B_nf-f-\frac{1}{2n}\varphi ^2f''\right\| \le \frac{1}{4n}\omega \left( f'';\frac{1}{3\sqrt{n}}\right) . \end{aligned}$$

(11)

2.2 Proof of Theorem 1

Statement (6) is an inmediate consequence of (11), Lemma 1 with $\delta =1/\sqrt{n}$, and the reverse and direct triangular inequalities. On the other hand, we have from Lemma 1

$$\begin{aligned}&\omega _2^{\varphi }\left( f;\frac{1}{\sqrt{n}}\right) =\frac{1}{n}\left\| \varphi ^2f''\right\| +o\left( \frac{1}{n}\right) , \end{aligned}$$

since $f\in C^2[0,1]$. Thus, statement (5) readily follows from (6), and completes the proof.

3 Proof of Theorem 2

3.1 Auxiliary Results

Let $0<h\le 1/3$. We consider the Steklov means of f defined as

$$\begin{aligned}&P_hf(y)=\int _{-1}^1\int _{-1}^1f\left( y+\frac{h}{2}(v_1+v_2)\right) dv_1dv_2\\&=\int _{-1}^1f(y+hv)\rho (v)dv,\quad h\le y\le 1-h,\end{aligned}$$

where

$$\begin{aligned} \rho (v)= & {} (1+v)1_{[-1,0]}+(1-v)1_{(0,1]},\quad -1\le v\le 1. \end{aligned}$$

In probabilistic terms, the Steklov means of f can be written as follows. Let $V_1$ and $V_2$ be independent identically distributed random variables having the uniform distribution on $[-1,1]$ and set $V=(V_1+V_2)/2$. Since $\rho (v)$ is the probability density of V, we can write

$$\begin{aligned} P_hf(y)= & {} {\mathbb {E}}f(y+hV),\quad h\le y\le 1-h. \end{aligned}$$

(12)

Lemma 2

Let $0<h\le 1/3$ and let $P_hf(y)$ be as in (12). Then,

• (a)
  $$\begin{aligned} \left| P_hf(y)-f(y)\right| \le \frac{1}{2}\omega _2^{\varphi }\left( f;\frac{h}{\varphi (y)}\right) . \end{aligned}$$
• (b)
  $$\begin{aligned} \left| (P_hf)''(y)\right| \le \frac{1}{h^2}\omega _2^{\varphi }\left( f;\frac{h}{\varphi (y)}\right) . \end{aligned}$$

Proof

Since V takes values in $[-1,1]$ and is symmetric (i. e., V and $-V$ have the same law), we see that

$$\begin{aligned} \left| P_hf(y)-f(y)\right| =\frac{1}{2}\left| {\mathbb {E}}(f(y+hV) +f(y-hV)-2f(y))\right| \le \frac{1}{2}\omega _2^{\varphi }\left( f;\frac{h}{\varphi (y)}\right) , \end{aligned}$$

thus showing (a). On the other hand, it can be checked that

$$\begin{aligned} P_hf(y)= & {} \frac{1}{h^2}\left( f_{(2)}(y+h)+f_{(2)}(y-h)-2f_{(2)}(y)\right) , \end{aligned}$$

where $f_{(2)}$ is a second antiderivative of f. This readily implies part (b) and completes the proof. $\square$

We will make use of the approximant $P_hf$, whose domain is the interval $[h,1-h]$, to define a further one whose domain is the whole interval [0, 1], keeping at the same time analogous properties to those given in Lemma 2. To this end, we assume that

$$\begin{aligned} n\ge 3,\qquad 0<a<\frac{\varphi (a/2)}{\sqrt{n}}+a\le 1. \end{aligned}$$

(13)

and take

$$\begin{aligned} h= & {} \frac{\varphi (ax)}{\sqrt{n}},\qquad \frac{1}{a(n+1)}\le x\le \frac{1}{2}. \end{aligned}$$

(14)

It turns out that

$$\begin{aligned} h\le \min (ax,1/3). \end{aligned}$$

(15)

Now, we define the approximant $Q_h^af(y)$ by antisymmetrizing $P_hf(y)$ around the axes $y=ax$ and $y=1-ax$ as follows

$$\begin{aligned} Q_h^af(y)=\left\{ \begin{array}{ll} 2P_hf(ax)-P_hf(2ax-y), &{}\quad y\in [0,ax); \\ P_hf(y), &{} \quad y\in [ax,1-ax]; \\ 2P_hf(1-ax)-P_hf(2(1-ax)-y), &{}\quad y\in (1-ax,1]. \end{array} \right. \end{aligned}$$

(16)

The fact that $Q_h^af$ is well defined readily follows from (13) and (14). Also, note that $Q_h^af$ is twice differentiable except at the points ax and $1-ax$. In these two points, $Q_h^af$ only has sided second derivatives. This implies that $Q_h^af\in {\mathscr {C}}^2[0,1]$.

Lemma 3

Let $R_a=[ax,1-ax]$. Under assumptions (13) and (14), we have

• (a) If $y\in R_a$, then

  $$\begin{aligned} \left| Q_h^af(y)-f(y)\right| \le \frac{1}{2}\omega _2^{\varphi }\left( f;\frac{1}{\sqrt{n}}\right) ,\quad \left| (Q_h^af)''(y)\right| \le \frac{1}{h^2}\omega _2^{\varphi }\left( f;\frac{1}{\sqrt{n}}\right) . \end{aligned}$$

• (b) If $y\notin R_a$, then

  $$\begin{aligned} \left| Q_h^af(y)-f(y)\right| \le \left( \frac{7}{2}+\frac{3\sqrt{anx}}{(1-a)^{3/2}}\right) \omega _2^{\varphi }\left( f;\frac{h}{\varphi (h)}\right) , \end{aligned}$$

  and

  $$\begin{aligned} \left| (Q_h^af)''(y)\right| \le \frac{1}{h^2}\omega _2^{\varphi }\left( f;\frac{h}{\varphi (h)}\right) . \end{aligned}$$

Proof

(a) If $y\in R_a$, then

$$\begin{aligned} \frac{h}{\varphi (y)}= & {} \frac{\varphi (ax)}{\varphi (y)}\frac{1}{\sqrt{n}}\le \frac{1}{\sqrt{n}}. \end{aligned}$$

(17)

Thus, the first inequality in part (a) follows from Lemma 2(a) and definition (16), whereas the second one follows from Lemma 2(b).

(b) Suppose that $y\in [0,ax)$. By (16), we can write

$$\begin{aligned} Q_h^af(y)-f(y)&=2\left( P_hf(ax)-f(ax)\right) \nonumber \\&\quad -\left( P_hf(2ax-y)-f(2ax-y)\right) \nonumber \\&\quad -\left( f(2ax-y)+f(y)-2f(ax)\right) . \end{aligned}$$

(18)

Since $h\le ax\le 2ax-y\le 1-h$, we see that

$$\begin{aligned} \varphi (ax)\ge \varphi (h),\quad \varphi (2ax-y)\ge \varphi (h). \end{aligned}$$

(19)

We therefore have from Lemma 2(a)

$$\begin{aligned}&\left| Q_h^af(y)-f(y)\right| \le \frac{3}{2}\omega _2^{\varphi }\left( f;\frac{h}{\varphi (h)}\right) \nonumber \\&\quad +\omega _2^{\varphi }\left( f;\frac{ax}{\varphi (ax)}\right) . \end{aligned}$$

(20)

Applying (1) with $\lambda =ax\varphi (h)/(h\varphi (ax))$ and $\delta =h/\varphi (h)$, we obtain

$$\begin{aligned} \omega _2^{\varphi }\left( f;\frac{ax}{\varphi (ax)}\right)&\le \left( 2+\frac{3(ax)^2\varphi ^2(h)}{\varphi ^2(ax)h^2}\right) \omega _2^{\varphi }\left( f;\frac{h}{\varphi (h)}\right) \nonumber \\&\le \left( 2+\frac{3\sqrt{anx}}{(1-a)^{3/2}}\right) \omega _2^{\varphi }\left( f;\frac{h}{\varphi (h)}\right) , \end{aligned}$$

(21)

as follows from (14) and some simple computations. Hence, the first inequality follows from (20) and (21).

On the other hand, we have from (16), (19), and Lemma 2(b)

$$\begin{aligned} \left| (Q_h^af)''(y)\right|= & {} \left| (P_hf)''(2ax-y)\right| \le \frac{1}{h^2}\omega _2^{\varphi }\left( f;\frac{h}{\varphi (2ax-y)}\right) \\\le & {} \frac{1}{h^2}\omega _2^{\varphi }\left( f;\frac{h}{\varphi (h)}\right) . \end{aligned}$$

If $y\in (1-ax,1]$, the proof es similar. $\square$

The following estimates concerning the random variable $S_n(x)/n$ will be needed.

Lemma 4

In the setting of Lemma 3, denote by $r=1-a$. Then,

• (a)
  $$\begin{aligned} P\left( \frac{S_n(x)}{n}\notin R_a\right) \le e^{-nxr^2/2}+3e^{-nxr^2/(2e)}=:\epsilon _n(x). \end{aligned}$$
• (b)
  $$\begin{aligned}&\frac{1}{h^2}{\mathbb {E}}\left( \frac{S_n(x)}{n}-x\right) ^2 1_{\left\{ \frac{S_n(x)}{n}\notin R_a\right\} } \\&\quad \le \frac{nx}{a(1-ax)}\left( e^{-nxr^2/2}+6e^{-(n-2)xr^2/(2e)}\right) =:\delta _n(x). \end{aligned}$$

Proof

(a) As follows from (3), we have

$$\begin{aligned} {\mathbb {E}}e^{\theta S_n(x)}=\left( 1+x(e^{\theta }-1)\right) ^n,\quad \theta \in {\mathbb {R}}. \end{aligned}$$

(22)

Let $\theta \ge 0$. By (22) and Chebyshev’s inequality, we have

$$\begin{aligned}&P\left( S_n(x)<anx\right) =P\left( e^{-\theta S_n(x)}>e^{-\theta anx}\right) \le {\mathbb {E}}e^{-\theta S_n(x)+\theta anx}\nonumber \\&\quad =e^{-n\left( -\log \left( 1-x\left( 1-e^{-\theta }\right) \right) -\theta ax\right) }\le e^{-nx\left( \left( 1-e^{-\theta }\right) -a\theta \right) }\le e^{-nx\left( r\theta -\theta ^2/2\right) },\qquad \quad \end{aligned}$$

(23)

where we have used the inequalities

$$\begin{aligned} -\log (1-u)\ge u,\quad u\ge 0,\qquad 1-e^{-\theta }\ge \theta -\frac{\theta ^2}{2},\quad \theta \ge 0. \end{aligned}$$

Choosing $\theta =r$ in (23) (the value minimizing the exponent), we get

$$\begin{aligned} P\left( S_n(x)<anx\right) \le e^{-nxr^2/2}. \end{aligned}$$

(24)

On the other hand, we claim that

$$\begin{aligned} P\left( S_n(x)>n(1-ax)\right) \le P\left( S_n(x)>n(1-ax)-1\right) \le 3e^{-nxr^2/(2e)}. \end{aligned}$$

(25)

Indeed, let $0\le \theta \le 1$. Using the inequalities

$$\begin{aligned} \log (1+u)\le u,\quad u\ge 0,\qquad e^{\theta }-1\le \theta +\frac{e\theta ^2}{2},\quad 0\le \theta \le 1, \end{aligned}$$

we have, as in the proof of (24),

$$\begin{aligned}&P\left( S_n(x)>n(1-ax)-1\right) =P\left( e^{\theta S_n(x)}>e^{\theta n(1-ax)-\theta }\right) \nonumber \\&\quad \le {\mathbb {E}}e^{\theta S_n(x)-n\theta (1-ax)+\theta }\le 3{\mathbb {E}}e^{\theta S_n(x)-n\theta (1-ax)}\nonumber \\&\quad =3e^{n\left( \log \left( 1+x\left( e^{\theta }-1\right) \right) -\theta (1-ax)\right) } \le 3e^{n\left( x\theta +ex\theta ^2/2-\theta (1-ax)\right) }\nonumber \\&\quad =3e^{n\theta (2x-1)}e^{nx(e\theta ^2/2-r\theta )} \le 3e^{nx\left( e\theta ^2/2-r\theta \right) }, \end{aligned}$$

(26)

since $x\le 1/2$. Thus, claim (25) follows by choosing $\theta =r/e$ in (26). Hence, part (a) follows from (24) and (25).

(b) From (24), we see that

$$\begin{aligned}&{\mathbb {E}}\left( \frac{S_n(x)}{n}-x\right) ^2 1_{\{S_n(x)<anx\}}\nonumber \\&\quad \le x^2P\left( S_n(x)<anx \right) \le x^2e^{-nxr^2/2}. \end{aligned}$$

(27)

On the other hand, since $1-ax\ge 1/2$, we have

$$\begin{aligned} \frac{k}{k-1}\le \frac{n/2}{n/2-1}=\frac{n}{n-2},\quad k>n(1-ax). \end{aligned}$$

We therefore have

$$\begin{aligned}&{\mathbb {E}}\left( \frac{S_n(x)}{n}-x\right) ^2 1_{\{S_n(x)>n(1-ax)\}}\le \frac{1}{n^2}{\mathbb {E}}S_n(x)^21_{\{S_n(x)>n(1-ax)\}} \nonumber \\&\quad =\frac{1}{n^2}\sum _{k>n(1-ax)}{n\atopwithdelims ()k}k^2x^k(1-x)^{n-k}\nonumber \\&\quad =\frac{n-1}{n}x^2\sum _{k>n(1-ax)}{n-2\atopwithdelims ()k-2}\frac{k}{k-1}x^{k-2}(1-x)^{n-k}\nonumber \\&\quad \le \frac{n-1}{n-2}x^2P\left( S_{n-2}(x)>n(1-ax)-2\right) \nonumber \\&\quad \le 2x^2P\left( S_{n-2}(x)>n(1-ax)-2\right) , \end{aligned}$$

(28)

since $n\ge 3$. Observe that

$$\begin{aligned} n(1-ax)-2=(n-2)(1-ax)-2ax\ge (n-2)(1-ax)-1, \end{aligned}$$

as follows from assumptions (13) and (14). By (25), the right-hand side in (28) can be bounded above by

$$\begin{aligned} 2x^2P\left( S_{n-2}(x)>(n-2)(1-ax)-1\right) \le 6x^2e^{-(n-2)xr^2/(2e)}. \end{aligned}$$

This, together with (27) and (28), shows part (b) and completes the proof.

$\square$

We are in a position to give the following local estimate.

Theorem 3

In the setting of Lemma 4, we have

$$\begin{aligned} \left| B_nf(x)-f(x)\right| \le \left( 1+\frac{1}{2}\frac{\varphi ^2(x)}{\varphi ^2(ax)}\right) \omega _2^{\varphi }\left( f;\frac{1}{\sqrt{n}}\right) +\nu _n(x)\omega _2^{\varphi }\left( f;\frac{h}{\varphi (h)}\right) , \end{aligned}$$

where

$$\begin{aligned}&\nu _n(x)=\left( \frac{7}{2}+\frac{3\sqrt{anx}}{(1-a)^{3/2}}\right) \epsilon _n(x)+\frac{1}{2}\delta _n(x). \end{aligned}$$

(29)

Proof

We use the notation $Qf(y)=Q_h^af(y)$ and write

$$\begin{aligned} B_nf(x)-f(x)&=(Qf(x)-f(x))+(B_nf(x)-B_n(Qf)(x)\nonumber \\&\quad +(B_n(Qf)(x)-Qf(x))\nonumber \\&=:I+II+III. \end{aligned}$$

(30)

By Lemma 3(a), we have

$$\begin{aligned} |I|\le \frac{1}{2}\omega _2^{\varphi }\left( f;\frac{1}{\sqrt{n}}\right) . \end{aligned}$$

(31)

By (2) and Lemma 3(a) and (b), we see that

$$\begin{aligned} |II|&= \left| {\mathbb {E}}Qf\left( \frac{S_n(x)}{n}\right) -{\mathbb {E}}f\left( \frac{S_n(x)}{n}\right) \right| \nonumber \\&\le \frac{1}{2}\omega _2^{\varphi }\left( f;\frac{1}{\sqrt{n}}\right) \nonumber \\&\quad +\left( \frac{7}{2}+\frac{3\sqrt{anx}}{(1-a)^{3/2}}\right) P\left( \frac{S_n(x)}{n}\notin R_a\right) \omega _2^{\varphi }\left( f;\frac{h}{\varphi (h)}\right) . \end{aligned}$$

(32)

Finally, denote by $\xi _n(x)=x+(S_n(x)/n-x)\beta _2$. Applying (9) with $m=2$ and Lemma 3, we get

$$\begin{aligned} |III|&=\frac{1}{2}\left| {\mathbb {E}}(Qf)''(\xi _n(x))\left( \frac{S_n(x)}{n}-x\right) ^2\right| \nonumber \\&\le \frac{1}{2h^2}\omega _2^{\varphi }\left( f;\frac{1}{\sqrt{n}}\right) {\mathbb {E}}\left( \frac{S_n(x)}{n}-x\right) ^2 1_{\{S_n(x)/n\in R_a\}}\nonumber \\&\quad +\frac{1}{2h^2}\omega _2^{\varphi }\left( f;\frac{h}{\varphi (h)}\right) {\mathbb {E}}\left( \frac{S_n(x)}{n}-x\right) ^2 1_{\{S_n(x)/n\notin R_a\}}\nonumber \\&\le \frac{1}{2}\frac{\varphi ^2(x)}{\varphi ^2(ax)}\omega _2^{\varphi }\left( f;\frac{1}{\sqrt{n}}\right) \nonumber \\&\quad +\frac{1}{2h^2}\omega _2^{\varphi }\left( f;\frac{h}{\varphi (h)}\right) {\mathbb {E}}\left( \frac{S_n(x)}{n}-x\right) ^2 1_{\{S_n(x)/n\notin R_a\}}, \end{aligned}$$

(33)

where we have used (14), the inequality $1/\sqrt{n}\le h/\varphi (h)$, and the well known fact that

$$\begin{aligned} {\mathbb {E}}\left( \frac{S_n(x)}{n}-x\right) ^2= & {} \frac{\varphi ^2(x)}{n}. \end{aligned}$$

The result follows from (30)–(33) and Lemma 4. $\square$

3.2 Proof of Theorem 2

Since the random variables $S_n(x)$ and $n-S_n(1-x)$ have the same law, we have

$$\begin{aligned} B_nf(1-x)-f(1-x)= & {} {\mathbb {E}}f\left( 1-\frac{S_n(x)}{n}\right) -f(1-x). \end{aligned}$$

On the other hand, if $g(y)=f(1-y)$, we obviously have

$$\begin{aligned} \omega _2^{\varphi }(g;\delta )= & {} \omega _2^{\varphi }(f;\delta ),\quad \delta \ge 0. \end{aligned}$$

Thus, without loss of generality, we can assume that $0<x\le 1/2$.

In the setting of Lemma 4, we claim that

$$\begin{aligned}&\omega _2^{\varphi }\left( f;\frac{h}{\varphi (h)}\right) \nonumber \\&\quad \le \left( 2+3\sqrt{\frac{anx}{1-a}}\right) \omega _2^{\varphi }\left( f;\frac{1}{\sqrt{n}}\right) . \end{aligned}$$

(34)

Actually, choose $\lambda =h\sqrt{n}/\varphi (h)$ and $\delta =1/\sqrt{n}$. By definition (14) and the fact that $h\le ax$, we see that

$$\begin{aligned} \lambda ^2=\frac{h^2n}{\varphi ^2(h)}=\frac{\varphi ^2(ax)}{\varphi ^2(h)}=\frac{ax(1-ax)}{h(1-h)} \le \frac{ax}{h}=\frac{ax\sqrt{n}}{\varphi (ax)}=\frac{\sqrt{anx}}{\sqrt{1-ax}}\le \sqrt{\frac{anx}{1-a}}. \end{aligned}$$

This, in conjunction with (1), shows claim (34).

From Theorem 3 and (34), we have

$$\begin{aligned} \frac{|B_nf(x)-f(x)|}{\omega _2^{\varphi }(f;1/\sqrt{n})}\le 1+\frac{1}{2} \frac{\varphi ^2(x)}{\varphi ^2(ax)}+\left( 2+3\sqrt{\frac{anx}{1-a}}\right) \nu _n(x), \end{aligned}$$

(35)

where $\nu _n(x)$ is defined in (29). Recalling the definitions of $\epsilon _n(x)$ and $\delta _n(x)$ given in Lemma 4, we see that

$$\begin{aligned} \left( 2+3\sqrt{\frac{anx}{1-a}}\right) \nu _n(x)\le P_3(\sqrt{nx})e^{-cnx}, \end{aligned}$$

(36)

where $P_3(\cdot )$ is a polynomial of degree three and c is a positive constant not depending on n and x. Observe that, whenever $t_n\rightarrow \infty$, as $n\rightarrow \infty$, we have

$$\begin{aligned} \lim _{n\rightarrow \infty } \sup _{u\ge t_n} P_3\left( \sqrt{u}\right) e^{-cu}=0. \end{aligned}$$

(37)

Let $\tau _n$ be as in Theorem 2. From (35) and (36), we get

$$\begin{aligned} \frac{1}{\omega _2^{\varphi }\left( f;1/\sqrt{n}\right) } \Vert B_nf-f\Vert _{[\tau _n/n,1/2]}\le 1+\frac{1}{2a}+\sup _{u\ge \tau _n} P_3\left( \sqrt{u}\right) e^{-cu}. \end{aligned}$$

By (37) and the fact that $\tau _n\rightarrow \infty$, as $n\rightarrow \infty$, this implies that

$$\begin{aligned} \limsup _{n\rightarrow \infty } \frac{1}{\omega _2^{\varphi }\left( f;1/\sqrt{n}\right) } \Vert B_nf-f\Vert _{[\tau _n/n,1/2]}\le 1+\frac{1}{2a}, \end{aligned}$$

which shows (7), since $0<a<1$ is arbitrary.

On the other hand, let $x\in (0,1/n)$. Consider the function

$$\begin{aligned} f_x(y)=\left( 1-\frac{y}{x}\right) 1_{[0,x]}(y). \end{aligned}$$

Observe that $\omega _2^{\varphi }(f_x;1/\sqrt{n})=1$, as well as

$$\begin{aligned} B_nf_x(x)-f_x(x)= & {} {\mathbb {E}}f_x\left( \frac{S_n(x)}{n}\right) =P(S_n(x)=0)=(1-x)^n, \end{aligned}$$

thus implying that $K_2\ge (1-x)^n$. Therefore, letting $x\rightarrow 0$, we see that $K_2\ge 1$. This shows (8) and completes the proof.