# 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 $$L_{p}[0,1]$$ ($$1\leq p \leq\infty$$).

## 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 

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

Bézier-type operators were introduced by Chang , later many results were given in , and more recent approximation results can be found in . 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  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 ). 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 

$$\omega_{\varphi}(f, t)_{p}\sim K_{\varphi}(f, t)_{p}\sim \overline{K}_{\varphi}(f,t)_{p},$$
(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 , we obtained a pointwise approximation for $$S_{n,\alpha}(f,x)$$. In Theorem 1 of , $$\lambda=1$$, 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 $$p=1$$.

## 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 , 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 , (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 , 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 ,

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

The proof is complete. □

## 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 , 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 , 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 , 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. □

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## Acknowledgements

This work was supported by the National Natural Science Foundation of China (grant 11371119), the Key Foundation of Education Department of Hebei Province (grant ZD2016023), and by Natural Science Foundation of Education Department of Hebei Province (grant Z2014031).

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Correspondence to Xiuzhong Yang.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors conceived of the study, participated its design and coordination, drafted the manuscript, participated in the sequence alignment, and read and approved the final manuscript.

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