Approximation for a generalization of Bernstein operators

Abstract

In this paper, we give a direct approximation theorem, inverse theorem, and equivalent theorem for a generalization of Bernstein operators in the space $L_{p}[0,1]$ ($1\leq p \leq\infty$).

1 Introduction

The Sikkema-Bézier-type generalization of Bernstein-Kantorovich operators is given by

$$\begin{aligned} S_{n,\alpha}(f,x)&=S_{n,\alpha}(f,s_{n},x) \\ &=\sum _{k=0}^{n}\bigl[J_{n,k}^{\alpha }(x)-J_{n,k+1}^{\alpha}(x) \bigr](n+s_{n}+1) \int_{\frac{k}{n+s_{n}+1}}^{\frac {k+1}{n+s_{n}+1}} f(u) \,du, \quad n=1,2,\ldots, \end{aligned}$$

where $J_{n,k}(x)=\sum^{n}_{j=k}p_{n,j}(x)$ are Bézier basic functions, $p_{n,k}(x)={n\choose k}x^{k}(1-x)^{n-k}$, $\alpha\geq1$, and $s_{n}$ is a bounded sequence of natural numbers. If $\alpha=1$ and $s_{n}=0$, then $S_{n,\alpha}(f,x)$ are just the well-known Bernstein-Kantorovich operators [1]

$$B_{n}(f, x)=(n+1)\sum_{k=0}^{n}p_{n,k}(x) \int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}} f(u) \,du,\quad n=1,2,\ldots. $$

Bézier-type operators were introduced by Chang [2], later many results were given in [3–8], and more recent approximation results can be found in [9]. Most of these results are on the rate of convergence of some Bézier-type operators for functions of bounded variation, whereas in the present paper, we give direct, inverse, and equivalent approximation theorems for a generalization of Bézier-type operators in $L_{p}$ spaces. We showed [5] that for Bézier-type operators, the second-order modulus cannot be used, so here we shall use the first-order modulus too. For Sikkema-type operators, we can also see many investigations (see [10]). Next, we state the central approximation theorem for $S_{n,\alpha}(f,x)$ in the spaces $L_{p}[0,1]$ ($1\leq p \leq\infty$), which will be proved in Sections 2 and 3.

Theorem 1.1

For $f\in L_{p}[0, 1]$ ($1\le p\leq\infty$), $\varphi(x)=\sqrt{x(1-x)}$, and $0< \beta<1$, we have

$$ \bigl\Vert S_{n,\alpha}(f, x)-f(x)\bigr\Vert _{p}=O \biggl( \biggl(\frac{1}{\sqrt{n}} \biggr)^{\beta} \biggr) \quad \Leftrightarrow\quad \omega_{\varphi}(f, t)_{p}=O \bigl(t^{\beta} \bigr). $$

(1.1)

In this theorem, we use the first-order modulus defined by

$$\omega_{\varphi}(f, t)_{p}=\sup_{0< h\le t}\biggl\Vert f \biggl(x+\frac{h\varphi (x)}{2} \biggr) -f \biggl(x-\frac{h\varphi(x)}{2} \biggr) \biggr\Vert _{p}, $$

which is equivalent to the K-functionals

$${K}_{\varphi}(f, t)_{p}=\inf_{g\in W_{p}} \bigl\{ \|f-g\|_{p}+t\bigl\Vert \varphi g'\bigr\Vert _{p} \bigr\} $$

and

$$\overline{K}_{\varphi}(f, t)_{p}=\inf_{g\in W_{p}} \bigl\{ \|f-g\|_{p}+t\bigl\Vert \varphi g'\bigr\Vert _{p}+t^{2}\bigl\Vert g'\bigr\Vert _{p} \bigr\} , $$

where $W_{p}=\{f| f\in \mathit{A.C.}_{\mathrm{loc}}, \|f'\|_{p}<\infty\}$. It is well known that [1]

$$ \omega_{\varphi}(f, t)_{p}\sim K_{\varphi}(f, t)_{p}\sim \overline{K}_{\varphi}(f,t)_{p}, $$

(1.2)

where $a\sim b$ means that there exists $C>0$ such that $C^{-1}a\le b\le Ca$.

Throughout this paper, C denotes a constant independent of n and x, but it is not necessarily the same in different cases.

Remark

In [8], we obtained a pointwise approximation for $S_{n,\alpha}(f,x)$. In Theorem 1 of [8], $\lambda=1$, which is the case where $p=\infty$ in (1.1). So in the present paper, by the Riesz-Thorin theorem, we shall only need to prove the case where $p=1$.

2 Direct theorem

To prove the direct theorem, we need the following convergence property of Bernstein-Kantorovich-Bézier operators defined by

$$B_{n,\alpha}(f,x)=\sum_{k=0}^{n} \bigl[J_{n,k}^{\alpha}(x)-J_{n,k+1}^{\alpha }(x) \bigr](n+1) \int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}}f(u)\,du,\quad n=1,2,\ldots. $$

Lemma 2.1

For $f\in L_{p}[0, 1]$ ($1\le p\leq\infty$), we have

$$ \bigl\Vert B_{n,\alpha}(f, x)-f(x)\bigr\Vert _{p}\le C\omega_{\varphi} \biggl(f, \frac{1}{\sqrt{n}} \biggr)_{p}. $$

(2.1)

Proof

For $p=1$, we will have to split estimate (2.1) into estimates on two domains, that is, $x\in E_{n}^{c}=[0, \frac{1}{n}]\cup[1-\frac{1}{n}, 1]$ and $x\in E_{n}=( \frac{1}{n}, 1-\frac{1}{n})$.

First, we choose $g=g_{n}$ such that

$$ \|f-g\|_{1}+\frac{1}{\sqrt{n}}\bigl\Vert \varphi g'\bigr\Vert _{1}\le C\omega_{\varphi}\biggl(f, \frac{1}{\sqrt{n}}\biggr)_{1}. $$

(2.2)

For $x\in E_{n}^{c}$, we have

$$\begin{aligned} \bigl\vert B_{n,\alpha}(g, x)-g(x)\bigr\vert &\le \alpha\sum ^{n}_{k=0}p_{n,k}(x) (n+1) \int_{\frac{k}{n+1}}^{\frac {k+1}{n+1}}\biggl\vert \int_{x}^{t}g'(u)\,du\biggr\vert \,dt \\ &\le \alpha\sum^{n}_{k=0}p_{n,k}(x) (n+1) \int_{\frac{k}{n+1}}^{\frac {k+1}{n+1}} \bigl(\varphi^{-1}(x)+ \varphi^{-1}(t) \bigr)\,dt \int^{1}_{0}\bigl\vert \varphi(u) g'(u)\bigr\vert \,du \\ &\le\alpha\bigl\Vert \varphi g'\bigr\Vert _{1} \Biggl(\varphi^{-1}(x)+\sum^{n}_{k=0}p_{n,k}(x) (n+1) \int_{\frac {k}{n+1}}^{\frac{k+1}{n+1}}\varphi^{-1}(t)\,dt \Biggr). \end{aligned}$$

Noting that

$$\int_{I_{k}}\varphi^{-1}(t)\,dt\le \int_{I_{k}} \biggl(\frac{1}{\sqrt{t}}+\frac {1}{\sqrt{1-t}} \biggr) \,dt\le\frac{4}{\sqrt{n+1}}, $$

we get

$$ \bigl\vert B_{n,\alpha}(g, x)-g(x)\bigr\vert \le C \bigl\Vert \varphi g'\bigr\Vert _{1} \bigl( \varphi^{-1}(x)+\sqrt{n} \bigr). $$

(2.3)

Since

$$\begin{aligned}& \int_{E_{n}^{c}}\varphi^{-1}(x)\,dx\le2 \int_{0}^{\frac{1}{n}}\frac{1}{\sqrt{x}}\,dx +2 \int_{1-\frac{1}{n}}^{1}\frac{1}{\sqrt{1-x}}\,dx\le \frac{8}{\sqrt{n}}, \\& \int_{E_{n}^{c}}\sqrt{n}\,dx= \frac{2}{\sqrt{n}}, \end{aligned}$$

we obtain

$$ \int_{E_{n}^{c}}\bigl\vert B_{n,\alpha}(g, x)-g(x)\bigr\vert \,dx\le C\frac{1}{\sqrt{n}} \bigl\Vert \varphi g'\bigr\Vert _{1}. $$

(2.4)

For $x\in E_{n}$, let $\vert \int^{\frac{k^{*}}{n+1}}_{x}\varphi(u) |g'(u)|\,du\vert =\max_{j=k,k+1}\vert \int^{\frac{j}{n+1}}_{x}\varphi(u) |g'(u)|\,du\vert$, where $k^{*}$ is either k or $k+1$. Then we have

$$\begin{aligned}& \int_{E_{n}}\bigl\vert B_{n,\alpha}(g, x)-g(x)\bigr\vert \,dx \\& \quad \le\alpha \int_{E_{n}}\sum^{n}_{k=0}p_{n,k}(x) (n+1) \int_{I_{k}} \bigl(\varphi^{-1}(x)+\varphi^{-1}(t) \bigr)\,dt \biggl\vert \int^{\frac{k^{*}}{n+1}}_{x}\varphi(u) \bigl\vert g'(u)\bigr\vert \,du\biggr\vert \,dx \\& \quad \le\alpha \int_{E_{n}}\sum^{n}_{k=0}p_{n,k}(x) \biggl(\varphi^{-1}(x) +\sqrt{\frac{n+1}{k+1}} \biggr) \biggl\vert \int^{\frac{k^{*}}{n+1}}_{x}\varphi(u) \bigl\vert g'(u)\bigr\vert \,du\biggr\vert \,dx= : \alpha R_{1}+ \alpha R_{2}. \end{aligned}$$

(2.5)

To estimate $R_{1}$ and $R_{2}$, we follow [3], pp. 146-147, with a similar method. We now define

$$D(l, n, x)=\biggl\{ k: l\varphi(x)n^{-\frac{1}{2}}\le\biggl\vert \frac{k}{n}-x\biggr\vert < (l +1)\varphi(x)n^{-\frac{1}{2}}\biggr\} . $$

Rewrite $R_{1}$ as follows:

$$R_{1}= \int_{E_{n}}\varphi^{-1}(x)\sum _{l=0}^{\infty}\sum_{k\in D(l,n,x)}p_{n,k}(x) \biggl\vert \int^{\frac{k^{*}}{n+1}}_{x}\varphi(u) \bigl\vert g'(u)\bigr\vert \,du\biggr\vert \,dx. $$

Similarly to [3], (9.6.11), for $x\in E_{n}$, we have

$$\sum_{k\in D(l,n,x)}p_{n,k}(x)\le\frac{C}{(l+1)^{4}}. $$

We now define

$$\begin{aligned}& F(l, x)= \biggl\{ v: v\in[0, 1], |v-x|\le(l+1)\varphi(x)n^{-\frac{1}{2}}+ \frac{1}{n} \biggr\} , \\& G(l, v)= \bigl\{ x: x\in E_{n}, v\in F(l, x) \bigr\} \end{aligned}$$

and by the procedure of [3], p.147, obtain

$$\begin{aligned} R_{1}&\le C\sum_{l=0}^{\infty} \frac{1}{(l+1)^{4}} \int_{E_{n}}\varphi^{-1}(x) \int_{F(l, x)}\varphi(v) \bigl\vert g'(v)\bigr\vert \,dv \,dx \\ &\le C\sum_{l=0}^{\infty}\frac{1}{(l+1)^{4}} \int_{0}^{1}\varphi(v) \bigl\vert g'(v)\bigr\vert \int_{G_{(l, v)}}\varphi^{-1}(x)\,dx \,dv \le C \frac{1}{\sqrt{n}} \bigl\Vert \varphi g'\bigr\Vert _{1}. \end{aligned}$$

(2.6)

On the other hand, for $R_{2}$, we have

$$\begin{aligned} \sum_{k\in D(l,n,x)}p_{n,k}(x)\sqrt{ \frac{n+1}{k+1}}&\le \biggl(\sum_{k\in D(l,n,x)}p_{n,k}(x) \frac{n+1}{k+1} \biggr)^{\frac{1}{2}}= \biggl(\sum_{k\in D(l,n,x)}p_{n+1,k+1}(x) \frac{1}{x} \biggr)^{\frac{1}{2}} \\ & =\varphi^{-1}(x) \biggl( \frac{C}{(1+l)^{3}} \biggr)^{\frac{1}{2}}\le\frac {C}{(1+l)^{4}} \varphi^{-1}(x). \end{aligned}$$

Similarly, we get

$$ R_{2}\le C\frac{1}{\sqrt{n}} \bigl\Vert \varphi g'\bigr\Vert _{1}. $$

(2.7)

Hence, by (2.5)-(2.7) we have

$$ \int_{E_{n}}\bigl\vert B_{n,\alpha}(g, x)-g(x)\bigr\vert \,dx\le\frac{C}{\sqrt{n}}\bigl\Vert \varphi g'\bigr\Vert _{1}. $$

(2.8)

Using (2.4) and (2.8), we complete the proof of Lemma 2.1. □

Theorem 2.2

For $f\in L_{1}[0,1]$, we have

$$ \bigl\Vert S_{n,\alpha}(f, x)-f(x)\bigr\Vert _{1}\le C\omega_{\varphi} \biggl(f, \frac{1}{\sqrt{n}} \biggr)_{1}. $$

(2.9)

Proof

By Lemma 2.1 we have

$$\begin{aligned}& \bigl\Vert S_{n,\alpha}(f,x)-f(x)\bigr\Vert _{1} \\& \quad \leq \bigl\Vert S_{n,\alpha }(f,x)-B_{n,\alpha}(f,x)\bigr\Vert _{1}+ \bigl\Vert B_{n,\alpha}(f,x)-f(x)\bigr\Vert _{1} \\& \quad \leq \int_{0}^{1}\sum^{n}_{k=0} \bigl[J_{n,k}^{\alpha}(x)-J_{n,k+1}^{\alpha }(x) \bigr](n+1) \int_{\frac{k}{n+1}}^{\frac{k+1}{n+1}} \biggl\vert f\biggl( \frac {n+1}{n+s_{n}+1}u\biggr)-f(u)\biggr\vert \, du\, dx \\& \qquad {}+C\omega_{\varphi} \biggl(f,\frac{1}{\sqrt{n}} \biggr)_{1} \\& \quad \leq C \biggl(\omega_{1}\biggl(f,\frac{1}{n} \biggr)_{1}+\omega_{\varphi} \biggl(f,\frac {1}{\sqrt{n}} \biggr)_{1} \biggr), \end{aligned}$$

and, by (3.1.5) in [1],

$$\omega_{1}\biggl(f,\frac{1}{n}\biggr)_{1}\leq C \omega_{\varphi} \biggl(f,\frac{1}{\sqrt {n}} \biggr)_{1}. $$

The proof is complete. □

3 Inverse theorem

Lemma 3.1

For $f\in L_{1}[0, 1]$, $\varphi(x)=\sqrt {x(1-x)}$, and $\delta_{n}(x)=\varphi(x)+\frac{1}{\sqrt{n}}$, we have

$$ \bigl\Vert \delta_{n}(x)S_{n,\alpha}'(f,x) \bigr\Vert _{1}\le C\sqrt{n}\|f\|_{1}, $$

(3.1)

and, furthermore, for $f\in W_{1}$,

$$ \bigl\Vert \delta_{n}(x)S_{n,\alpha}'(f,x) \bigr\Vert _{1}\le C\bigl\Vert \delta_{n}f' \bigr\Vert _{1}. $$

(3.2)

Proof

First, we prove (3.1), that is,

$$ \bigl\Vert \delta_{n}(x)S_{n,\alpha}'(f,x) \bigr\Vert _{1}\le C\sqrt{n}\|f\|_{1}. $$

(3.3)

Write $a_{k}(f)=(n+s_{n}+1)\int_{I_{k}} f(t)\,dt$, where $I_{k}=[\frac {k}{n+s_{n}+1},\frac{k+1}{n+s_{n}+1}]$. Noting that $J_{n, n+1}(x)=0$, we have

$$\begin{aligned} \bigl\vert S_{n,\alpha}'(f,x)\bigr\vert \le& \alpha\sum^{n-1}_{k=0} \bigl\vert a_{k}(f)\bigr\vert \bigl[J^{\alpha-1}_{n,k}(x)- J^{\alpha-1}_{n,k+1}(x) \bigr]J'_{n,k+1}(x) \\ &{}+\alpha\sum^{n}_{k=0}\bigl\vert a_{k}(f)\bigr\vert J^{\alpha-1}_{n,k}(x)\bigl\vert p'_{n,k}(x)\bigr\vert \\ =:&\alpha (J_{1}+J_{2} ). \end{aligned}$$

(3.4)

Then

$$\begin{aligned} \int_{0}^{1}\bigl\vert \delta_{n}(x) S_{n,\alpha}'(f, x)\bigr\vert \,dx \leq&\alpha \int_{0}^{1}\delta_{n}(x) (J_{1}+J_{2})\,dx \\ =&\alpha \biggl( \int_{E_{n}^{c}}+ \int_{E_{n}} \biggr)\delta_{n}(x) (J_{1}+J_{2} )\,dx. \end{aligned}$$

(3.5)

Next, we estimate the four parts in (3.5):

$$\begin{aligned} \int_{E_{n}^{c}}\delta_{n}(x)J_{2}\,dx \le& \int_{E_{n}^{c}}\delta_{n}(x)\sum ^{n}_{k=1}\bigl\vert a_{k}(f)\bigr\vert n \bigl(p_{n-1, k-1}(x)+p_{n-1, k}(x) \bigr)\,dx \\ &{}+ \int_{E_{n}^{c}}\delta_{n}(x)\bigl\vert a_{0}(f)\bigr\vert np_{n-1,0}(x) \,dx . \end{aligned}$$

For $x\in E_{n}^{c}$, $\delta_{n}(x)\le \frac{2}{\sqrt{n}}$, and since $\int_{0}^{1}p_{n-1, k}(x) \,dx=\frac{1}{n}$, we get

$$\begin{aligned} \int_{E_{n}^{c}}\delta_{n}(x)J_{2}\,dx \le& \frac{4}{\sqrt{n}}\sum^{n}_{k=1} (n+s_{n}+1) \int_{I_{k}}\bigl\vert f(t)\bigr\vert \,dt+\frac{2(n+s_{n}+1)}{\sqrt{n}} \int_{0}^{\frac{1}{n+s_{n}+1}}\bigl\vert f(t)\bigr\vert \,dt \\ \le& C\sqrt{n}\|f\|_{1} . \end{aligned}$$

(3.6)

Since $J^{\alpha-1}_{n,k}(x)-J^{\alpha-1}_{n,k+1}(x)\le 1$ and $J'_{n,k+1}(x)=np_{n-1, k}(x)$, it is easy to see that

$$ \int_{E_{n}^{c}}\delta_{n}(x)J_{1}\,dx\le C \sqrt{n}\|f\|_{1} . $$

(3.7)

To estimate $\int_{E_{n}}\delta_{n}(x)J_{2}\,dx$, we recall that by [3], p.129, (9.4.15),

$$\int_{E_{n}}\frac{(\frac{k}{n}-x)^{2}}{\varphi^{2}(x)}p_{n,k}(x)\,dx\le Cn^{-2} $$

with $p'_{n,k}(x)=\frac{n}{\varphi^{2}(x)} (\frac{k}{n}-x )p_{n,k}(x)$, $x\in (0, 1)$. By the Hölder inequality we have

$$\begin{aligned} \int_{E_{n}}\delta_{n}(x)J_{2}\,dx&\le 2\sum ^{n}_{k=0}\bigl\vert a_{k}(f) \bigr\vert \int_{E_{n}}\varphi(x)\cdot\frac{n}{\varphi ^{2}(x)}\biggl\vert \frac{k}{n}-x\biggr\vert p_{n,k}(x)\,dx \\ &\le 2n\sum^{n}_{k=0}\bigl\vert a_{k}(f)\bigr\vert (n+1)^{-\frac{1}{2}} \biggl( \int_{E_{n}} \frac{ (\frac{k}{n}-x )^{2}}{\varphi^{2}(x)}p_{n,k}(x)\,dx \biggr)^{\frac{1}{2}} \\ &\le C\sqrt{n}\sum^{n}_{k=0} \int_{I_{k}}\bigl\vert f(t)\bigr\vert \,dt=C\sqrt{n}\|f \|_{1}. \end{aligned}$$

(3.8)

To estimate $\int_{E_{n}}\delta_{n}(x)J_{1}\,dx$, we will consider two cases, $\alpha\ge2$ and $1<\alpha<2$ ($J_{1}=0$ when $\alpha=1$).

For $\alpha\ge2$, we have $J^{\alpha-1}_{n,k}(x)- J^{\alpha-1}_{n,k+1}(x)\le(\alpha-1)p_{n,k}(x)$, and we need a result of [4], p.375,

$$ p_{n,k}(x)\le \frac{1}{\sqrt{2e}}\frac{1}{\sqrt{nx(1-x)}} \quad \mbox{for } 0\le k\le n . $$

(3.9)

Since $J'_{n,k}(x)=np_{n-1, k-1}(x) \ge0$, we have

$$\begin{aligned} \int_{E_{n}}\delta_{n}(x)J_{1}\,dx&\le C\sum ^{n-1}_{k=0}\bigl\vert a_{k}(f) \bigr\vert \int_{E_{n}}\varphi(x)p_{n,k}(x) n p_{n-1,k}(x) \,dx \\ &\le C\sum^{n}_{k=0}\bigl\vert a_{k}(f)\bigr\vert \frac{1}{\sqrt{n}}\le C\sqrt{n}\|f\|_{1}. \end{aligned}$$

(3.10)

For $1<\alpha<2$, applying $u(a)-u(b)=u'(\xi)(a-b)$ ($a<\xi<b$), we get that there exists $\xi_{k}(x)$, $J_{n,k+1}(x)<\xi_{k}(x)<J_{n,k}(x)$ such that

$$J^{\alpha-1}_{n,k}(x)- J^{\alpha-1}_{n,k+1}(x)=( \alpha-1) \bigl(\xi_{k}(x)\bigr)^{\alpha-2}p_{n,k}(x)\le ( \alpha-1)J^{\alpha-2}_{n,k+1}(x)p_{n,k}(x). $$

Hence, we have

$$\begin{aligned} \int_{E_{n}}\delta_{n}(x)J_{1}\,dx&\le C \int_{E_{n}}\varphi(x)\sum^{n-1}_{k=0} \bigl\vert a_{k}(f)\bigr\vert p_{n,k}(x) (\alpha -1)J^{\alpha-2}_{n,k+1}(x) J'_{n,k+1}(x)\,dx \\ &=: C\sum^{n-1}_{k=0}\bigl\vert a_{k}(f)\bigr\vert J. \end{aligned}$$

(3.11)

Since $p_{n,k}(x)=\frac{n(1-x)}{n-k}p_{n-1,k}(x)$ for $k< n$, from (3.9) we can deduce that

$$\begin{aligned} J&\le \int_{E_{n}}\frac{\alpha-1}{\sqrt{n}}\cdot\frac{n(1-x)}{n-k}J^{\alpha -2}_{n,k+1}(x) J'_{n,k+1}(x)\,dx \\ &\le\frac{\sqrt{n}}{n-k} \int_{0}^{1}(1-x)\, dJ^{\alpha-1}_{n,k+1}(x) =\frac{\sqrt{n}}{n-k} \int_{0}^{1}J^{\alpha-1}_{n,k+1}(x)\,dx \\ &=\frac{\sqrt{n}}{n-k} \biggl( \int_{0}^{\frac{k}{n}}J^{\alpha-1}_{n,k+1}(x)\,dx + \int^{1}_{\frac{k}{n}}J^{\alpha-1}_{n,k+1}(x)\,dx \biggr) \\ &=:\frac{\sqrt{n}}{n-k} (L_{1}+L_{2} ) \end{aligned}$$

(3.12)

and

$$ \frac{\sqrt{n}}{n-k}L_{2}\le\frac{\sqrt{n}}{n-k}\biggl(1- \frac{k}{n}\biggr)=\frac {1}{\sqrt{n}}. $$

(3.13)

In order to estimate $\frac{\sqrt{n}}{n-k}L_{1}$, choose $l\in \mathbb{N}$ such that $l(\alpha-1)>1$. Then, for $k< n$, we have

$$\begin{aligned} J_{n,k+1}&=\sum_{j=k+1}^{n} \frac{(n+l-j)\cdots (n-j+1)}{(n+l)\cdots(n+1)}p_{n+l,j}(x) (1-x)^{-l} \\ &\le\frac{(n+l-k)^{l}}{(n+1)^{l}} (1-x)^{-l}. \end{aligned}$$

Therefore, for $k\le n-1$, we get

$$\begin{aligned} \frac{\sqrt{n}}{n-k}L_{1} \le& \frac{\sqrt{n}}{n-k} \biggl( \frac{n+l-k}{n+1} \biggr)^{l(\alpha-1)} \int _{0}^{\frac{k}{n}}(1-x)^{-l(\alpha-1)}\,dx \\ \le& 2^{l(\alpha-1)}\frac{\sqrt{n}}{n-k} \biggl[ \biggl(\frac{l}{n+1} \biggr)^{l(\alpha-1)} + \biggl(\frac{n-k}{n+1} \biggr)^{l(\alpha-1)} \biggr] \\ &{}\cdot\frac{1}{l(\alpha-1)-1} \biggl[\biggl(1-\frac{k}{n}\biggr)^{1-l(\alpha-1)}-1 \biggr] \\ \le& C_{\alpha}\frac{\sqrt{n}}{n-k} \biggl(\frac{n-k}{n+1} \biggr)^{l(\alpha -1)} \biggl(\frac{n-k}{n} \biggr)^{1-l(\alpha-1)} \le C \frac{1}{\sqrt{n}}. \end{aligned}$$

(3.14)

By (3.11)-(3.14), for $1<\alpha<2$, we obtain

$$ \int_{E_{n}}\delta_{n}(x)J_{1}\,dx\le C\sum ^{n-1}_{k=0}\bigl\vert a_{k}(f) \bigr\vert \frac{1}{\sqrt{n}}\le C\sqrt{n}\|f\|_{1}. $$

(3.15)

Estimates (3.5)-(3.8) and (3.15) imply (3.1).

Now we prove (3.2). For $f\in W_{p}$, noting that $J'_{n,0}(x)=0$, we have

$$\begin{aligned}& S_{n,\alpha}'(f,x) \\& \quad =\alpha \Biggl[\sum^{n}_{k=1}a_{k}(f)J^{\alpha-1}_{n,k}(x)J'_{n,k}(x)- \sum^{n}_{k=1}a_{k-1}(f)J^{\alpha-1}_{n,k}(x)J'_{n,k}(x) \Biggr] \\& \quad =\alpha\sum^{n}_{k=1}(n+s_{n}+1) \biggl[ \int_{\frac{k}{n+s_{n}+1}}^{\frac {k+1}{n+s_{n}+1}}f(u)\,du- \int_{\frac{k-1}{n+s_{n}+1}}^{\frac{k}{n+s_{n}+1}}f(u)\,du \biggr]J^{\alpha -1}_{n,k}(x)J'_{n,k}(x) \\& \quad =\alpha\sum^{n}_{k=1}(n+s_{n}+1) \int_{0}^{\frac{1}{n+s_{n}+1}} \biggl[f\biggl(\frac{k}{n+s_{n}+1}+u \biggr)- f\biggl(\frac{k-1}{n+s_{n}+1}+u\biggr) \biggr]\, duJ^{\alpha-1}_{n,k}(x)J'_{n,k}(x) \\& \quad =\alpha\sum^{n}_{k=1}(n+s_{n}+1) \int_{0}^{\frac{1}{n+s_{n}+1}} \int_{0}^{\frac {1}{n+s_{n}+1}} f'\biggl( \frac{k-1}{n+s_{n}+1}+u+v\biggr)\, dv\, duJ^{\alpha-1}_{n,k}(x)J'_{n,k}(x). \end{aligned}$$

Hence,

$$\begin{aligned} \bigl\vert S_{n,\alpha}'(f,x)\bigr\vert \le& \alpha\sum^{n-1}_{k=0} \int _{0}^{\frac{2}{n+s_{n}+1}} \biggl\vert f'\biggl( \frac{k}{n+s_{n}+1}+u\biggr)\biggr\vert \, duJ'_{n,k+1}(x) \\ =&\alpha\Biggl( \int_{0}^{\frac{2}{n+s_{n}+1}}\bigl\vert f'(u)\bigr\vert \, duJ'_{n,1}(x) + \int_{0}^{\frac{2}{n+s_{n}+1}}\biggl\vert f'\biggl( \frac{n-1}{n+s_{n}+1}+u\biggr) \biggr\vert \, duJ'_{n,n}(x) \\ &{}+\sum^{n-2}_{k=1} \int_{0}^{\frac{2}{n+s_{n}+1}}\biggl\vert f'\biggl( \frac {k}{n+s_{n}+1}+u\biggr)\biggr\vert \, duJ'_{n,k+1}(x) \Biggr) \\ =&\alpha (Q_{1}+Q_{2}+Q_{3} ). \end{aligned}$$

(3.16)

First, we estimate $\int_{0}^{1}\delta_{n}(x)Q_{3}\,dx$. For $1\le k\le n-2$ and $0< u<\frac{2}{n+s_{n}+1}$, we have $\frac{k}{n}(1-\frac{k}{n})\le C(\frac{k}{n+s_{n}+1}+u)(1-\frac{k}{n+s_{n}+1}-u)$ and (similarly to [1], p.155)

$$\begin{aligned}& \int_{0}^{\frac{2}{n+s_{n}+1}}\biggl\vert f'\biggl( \frac{k}{n+s_{n}+1}+u\biggr)\biggr\vert \,du \\& \quad \le C \int_{0}^{\frac{2}{n+s_{n}+1}}\biggl\vert \varphi\biggl( \frac{k}{n+s_{n}+1}+u\biggr)f'\biggl(\frac {k}{n+s_{n}+1}+u\biggr)\biggr\vert \,du\varphi^{-1}\biggl(\frac{k}{n}\biggr) \\& \quad \le C\varphi^{-1}\biggl(\frac{k}{n}\biggr) \int_{\frac{k}{n+s_{n}+1}}^{\frac {k+2}{n+s_{n}+1}}\bigl\vert \varphi(u)f'(u) \bigr\vert \,du. \end{aligned}$$

Therefore,

$$\int_{0}^{1}\delta_{n}(x)Q_{3} \,dx\le Cn\sum^{n-2}_{k=1} \int_{\frac{k}{n+s_{n}+1}}^{\frac{k+2}{n+s_{n}+1}}\bigl\vert \varphi (u)f'(u)\bigr\vert \,du \int_{0}^{1}\delta_{n}(x) \varphi^{-1}\biggl(\frac{k}{n}\biggr)p_{n-1,k}(x)\,dx. $$

Noting that $\varphi^{-1}(\frac{k}{n})<\sqrt{n}$ for $0< k< n-1$, we get

$$\begin{aligned}& \int_{0}^{1}\biggl(\varphi(x)+\frac{1}{\sqrt{n}} \biggr)\varphi^{-1}\biggl(\frac {k}{n}\biggr)p_{n-1,k}(x) \,dx \\& \quad \le\frac{2}{\sqrt{n}} \biggl( \int_{0}^{1} \biggl(\varphi^{2}(x)+ \frac{1}{n} \biggr)\varphi^{-2}\biggl(\frac{k}{n} \biggr)p_{n-1,k}(x)\,dx \biggr)^{\frac{1}{2}} \\& \quad \le\frac{C}{\sqrt{n}} \biggl( \int_{0}^{1}\varphi^{2}(x) \frac{n}{k}\frac {n}{n-k}p_{n-1,k}(x)\,dx + \int_{0}^{1}p_{n-1,k}(x)\,dx \biggr)^{\frac{1}{2}} \\& \quad \le\frac{C}{\sqrt{n}} \biggl( \int_{0}^{1}p_{n+1,k+1}(x)\,dx+ \frac{1}{n} \biggr)^{\frac{1}{2}}\le C\frac{1}{n}. \end{aligned}$$

Hence, we have

$$ \|\delta_{n}Q_{3}\|_{1}\le C\bigl\Vert \varphi f'\bigr\Vert _{1}\le C\bigl\Vert \delta_{n}f'\bigr\Vert _{1}. $$

(3.17)

Since $\sqrt{n}\delta_{n}(u)\ge1$, for $Q_{1}$, we write

$$\begin{aligned} \delta_{n}(x)Q_{1}&=\delta_{n}(x) \int_{0}^{\frac{2}{n+s_{n}+1}}\bigl\vert f'(u) \bigr\vert \, duJ'_{n,1}(x) \\ &\le\delta_{n}(x)n \int_{0}^{\frac{2}{n+s_{n}+1}}\sqrt{n}\bigl\vert \delta _{n}(u)f'(u)\bigr\vert \,du p_{n-1, 0}(x) \\ &\le\bigl\Vert \delta_{n}f'\bigr\Vert _{1} \delta_{n}(x)n^{\frac{3}{2}}p_{n-1, 0}(x). \end{aligned}$$

Therefore, we have

$$\begin{aligned} \int_{0}^{1} \delta_{n}(x)Q_{1} \,dx&\le C\bigl\Vert \delta_{n}f'\bigr\Vert _{1} \biggl( \int _{E_{n}^{c}}np_{n-1,0}(x)\,dx+ \int_{E_{n}}\varphi(x)n^{\frac{3}{2}}p_{n-1,0}(x)\,dx \biggr) \\ &\le C\bigl\Vert \delta_{n}f'\bigr\Vert _{1} \biggl[1+n \biggl( \int_{E_{n}}\varphi ^{2}(x)p_{n-1,0}(x)\,dx \biggr)^{\frac{1}{2}} \biggr] \\ &=C\bigl\Vert \delta_{n}f'\bigr\Vert _{1} \biggl[1+n \biggl( \int_{E_{n}}\frac {1}{n+1}p_{n+1,1}(x)\,dx \biggr)^{\frac{1}{2}} \biggr] \le C\bigl\Vert \delta_{n}f' \bigr\Vert _{1}. \end{aligned}$$

(3.18)

Similarly, we have

$$ \int_{0}^{1} \delta_{n}(x)Q_{2} \,dx\le C\bigl\Vert \delta_{n}f'\bigr\Vert _{1}. $$

(3.19)

By (3.16)-(3.19) we obtain

$$\bigl\Vert \delta_{n}(x)L_{n\alpha}'(f, x)\bigr\Vert _{1}\le C\bigl\Vert \delta_{n}f'\bigr\Vert _{1}. $$

This is (3.2). The proof of Lemma 3.1 is complete. □

Theorem 3.2

Let $f\in L_{p}[0, 1]$ ($1\le p\le\infty$), $\varphi(x)=\sqrt{x(1-x)}$, and $0<\beta<1$. Then

$$\bigl\Vert S_{n,\alpha}(f, x)-f(x)\bigr\Vert _{p}=O \bigl(n^{- \frac{\beta}{2}} \bigr) $$

implies $\omega_{\varphi}(f, t)_{p}=O (t^{\beta} )$.

Proof

By Lemma 3.1, for appropriate g, we have

$$\begin{aligned} K_{\varphi}(f, t)_{p}&\le\bigl\Vert f-L_{n\alpha}(f)\bigr\Vert _{p}+t\bigl\Vert \delta_{n}L'_{n\alpha}(f) \bigr\Vert _{p} \\ &\le Cn^{-\frac{\beta}{2}}+t \bigl(\bigl\Vert \delta_{n}L'_{n\alpha}(f-g) \bigr\Vert _{p}+\bigl\Vert \delta_{n}L'_{n\alpha}(g) \bigr\Vert _{p} \bigr) \\ &\le Cn^{-\frac{\beta}{2}}+Ct \bigl(\sqrt{n} \Vert f-g\Vert _{p}+\bigl\Vert \delta_{n}g'\bigr\Vert _{p} \bigr) \\ &\le Cn^{-\frac{\beta}{2}}+Ct\sqrt{n} \biggl(\Vert f-g\Vert _{p}+ \frac{1}{\sqrt{n}}\bigl\Vert \varphi g'\bigr\Vert _{p}+ \frac{1}{n}\bigl\Vert g'\bigr\Vert _{p} \biggr) \\ &\le C \biggl(n^{-\frac{\beta}{2}}+\frac{t}{n^{-\frac{1}{2}}}\overline{K}_{\varphi} \bigl(f, n^{-\frac{1}{2}}\bigr)_{p} \biggr) \\ &\le C \biggl(n^{-\frac{\beta}{2}}+\frac{t}{n^{-\frac{1}{2}}}K_{\varphi}\bigl(f, n^{-\frac{1}{2}}\bigr)_{p} \biggr), \end{aligned}$$

which by the Berens-Lorentz lemma implies that

$$ K_{\varphi}(f, t)_{p}=O\bigl(t^{\beta} \bigr). $$

(3.20)

From relation (1.2) and (3.20) we see that the proof of Theorem 3.2 is complete. □