On the approximation for generalized Szász-Durrmeyer type operators in the space $L_p[0,\mathrm{\infty })$

Abstract

In this paper we give the direct approximation theorem, the inverse theorem, and the equivalence theorem for Szász-Durrmeyer-Bézier operators in the space $L_p[0,\mathrm{\infty })$ ($1\le p\le \mathrm{\infty }$) with Ditzian-Totik modulus.

MSC:41A25, 41A27, 41A36.

1 Introduction

In 1972 Bézier [1] introduced the Bézier basic function and Bézier-type operators are the generalized types of the original operators. The introduction of these operators should have some background. Some properties of the convergence and approximation for some Bézier-type operators have been studied (cf. [2–6]), but there are other aspects that have not yet been considered. For more information as regards the development of the study on this topic or related field, the interested readers can consult the monograph [7] and the paper [8]. In this paper we will consider the direct, inverse and equivalence theorems for the Szász-Durrmeyer-Bézier operator, which is defined by

$$D_{n,\alpha }(f,x)=\sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)f(t)dt[J_{n,k}^{\alpha }(x)-J_{n,k+1}^{\alpha }(x)],$$

(1.1)

where $\alpha \ge 1$, $f\in L_p[0,\mathrm{\infty })$, $J_{n,k}(x)={\sum }_{j=k}^{\mathrm{\infty }}{s}_{n,j}(x)$, $s_{n,k}=e^{-nx}\frac{{(nx)}^k}{k!}$. Obviously $D_{n,\alpha }(f,x)$ is bounded and positive in the space $L_p[0,\mathrm{\infty })$. When $\alpha =1$, $D_{n,\alpha }(f,x)$ is the well-known Durrmeyer operator

$$D_{n,1}(f,x)=\sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)f(t)dt{s}_{n,k}(x).$$

To describe our results, we give the definitions of the first order modulus of smoothness and the K-functional (cf. [9]). For $f\in L_p[0,\mathrm{\infty })$ ($1\le p\le \mathrm{\infty }$), $\phi (x)=\sqrt{x}$,

$$\begin{array}{c}{\omega }_{\phi }{(f,t)}_p=\underset{0<h\le t}{sup}\{{\parallel f(x+\frac{h\phi (x)}{2})-f(x-\frac{h\phi (x)}{2})\parallel }_p,x-\frac{h\phi (x)}{2}\ge 0\},\hfill \\ K_{\phi }{(f,t)}_p=\underset{g\in W_p}{inf}\{{\parallel f-g\parallel }_p+t{\parallel \phi g^{\prime }\parallel }_p\},\hfill \\ {\overline{K}}_{\phi }{(f,t)}_p=\underset{g\in W_p}{inf}\{{\parallel f-g\parallel }_p+t{\parallel \phi g^{\prime }\parallel }_p+t^2{\parallel g^{\prime }\parallel }_p\},\hfill \end{array}$$

where $W_p=\{f|f\in A.C{.}_{\mathrm{loc}},{\parallel \phi f^{\prime }\parallel }_p<\mathrm{\infty },{\parallel f^{\prime }\parallel }_p<\mathrm{\infty }\}$.

It is well known that (cf. [9])

$${\omega }_{\phi }{(f,t)}_p\sim K_{\phi }{(f,t)}_p\sim {\overline{K}}_{\phi }{(f,t)}_p,$$

(1.2)

here $a\sim b$ means that there exists $c>0$ such that $c^{-1}a\le b\le ca$.

Now we state our equivalence theorem as follows.

Theorem For $f\in L_p[0,\mathrm{\infty })$ ($1\le p\le \mathrm{\infty }$), $\phi (x)=\sqrt{x}$, $0<\beta <1$, we have

$${\parallel D_{n,\alpha }(f,x)-f(x)\parallel }_p=O\left({\left(\frac{1}{\sqrt{n}}\right)}^{\beta }\right)\iff {\omega }_{\phi }{(f,t)}_p=O\left(t^{\beta }\right).$$

(1.3)

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

2 Direct theorem

For convenience, we list some basic properties which will be used later and can be found in [9] and [5] or obtained by simple computation:

• (1)
  $$1=J_{n,0}(x)>J_{n,1}(x)>\cdots >J_{n,k}(x)>\cdots >0;$$

  (2.1)
• (2)
  $$0<J_{n,k}^{\alpha }(x)-J_{n,k+1}^{\alpha }(x)\le \alpha s_{n,k}(x),\alpha \ge 1;$$

  (2.2)
• (3)
  $$s_{n,k}^{\prime }(x)=\frac{n}{{\phi }^2(x)}(\frac{k}{n}-x)s_{n,k}(x);$$

  (2.3)
• (4)
  $$s_{n,k}^{\prime }(x)=n{s}_{n,k-1}(x)-n{s}_{n,k}(x),s_{n,-1}(x)=0;$$

  (2.4)
• (5)
  $$J_{n,0}^{\prime }(x)=0,J_{n,k}^{\prime }(x)=n{s}_{n,k-1}(x)(k=1,2,\dots );$$

  (2.5)
• (6)
  $$D_{n,1}({(t-x)}^2,x)\le n^{-1}{\delta }_n^2(x),$$

  (2.6)

where ${\delta }_n(x)=\phi (x)+\frac{1}{\sqrt{n}}$.

Now we give the direct theorem.

Theorem 2.1 For $f\in L_p[0,\mathrm{\infty })$ ($1\le p\le \mathrm{\infty }$), $\phi (x)=\sqrt{x}$, we have

$${\parallel D_{n,\alpha }(f,x)-f(x)\parallel }_p\le C{\omega }_{\phi }{(f,\frac{1}{\sqrt{n}})}_p.$$

(2.7)

Proof By the definition of ${\overline{K}}_{\phi }{(f,t)}_p$ and the relation (1.2), for fixed n, we can choose $g=g_n$ such that

$${\parallel f-g\parallel }_p+\frac{1}{\sqrt{n}}{\parallel \phi g^{\prime }\parallel }_p+\frac{1}{n}{\parallel g^{\prime }\parallel }_p\le C{\omega }_{\phi }{(f,\frac{1}{\sqrt{n}})}_p.$$

Since

$$\begin{array}{rl}{\parallel D_{n,\alpha }f-f\parallel }_p& \le {\parallel D_{n,\alpha }(f-g)\parallel }_p+{\parallel f-g\parallel }_p+{\parallel D_{n,\alpha }g-g\parallel }_p\\ \le C{\parallel f-g\parallel }_p+{\parallel D_{n,\alpha }g-g\parallel }_p,\end{array}$$

(2.8)

we only need to estimate the second term in the above relation. By the Riesz-Thorin theorem (cf. [[10], Theorem 3.6]), we separate the proof of the assertions for $p=\mathrm{\infty }$ and $p=1$.

I. $p=\mathrm{\infty }$. Noting that $g(t)=g(x)+{\int }_x^t{g}^{\prime }(u)du$, we write

$$\begin{array}{rcl}|D_{n,\alpha }(g,x)-g(x)|& =& \left|D_{n,\alpha }({\int }_x^t{g}^{\prime }(u)du,x)\right|\\ \le & {\parallel {\delta }_n{g}^{\prime }\parallel }_{\mathrm{\infty }}{D}_{n,\alpha }\left(\right|{\int }_x^t{\delta }_n^{-1}(u)du|,x).\end{array}$$

Since

$$|{\int }_x^t{\phi }^{-1}(u)du|=\left|{\int }_x^t\frac{1}{\sqrt{u}}du\right|\le \frac{2|t-x|}{\phi (x)},$$

(2.9)

$$\left|{\int }_x^t{\left(\frac{1}{\sqrt{n}}\right)}^{-1}du\right|=\sqrt{n}|t-x|,$$

(2.10)

and $min\{{\phi }^{-1}(x),\sqrt{n}\}\sim {\delta }_n^{-1}(x)$, using the Hölder inequality, we have

$$\begin{array}{rcl}|D_{n,\alpha }(g,x)-g(x)|& \le & C{\delta }_n^{-1}(x){\parallel {\delta }_n{g}^{\prime }\parallel }_{\mathrm{\infty }}{D}_{n,\alpha }(|t-x|,x)\\ \le & C{\delta }_n^{-1}(x){\parallel {\delta }_n{g}^{\prime }\parallel }_{\mathrm{\infty }}{\left(D_{n,\alpha }({(t-x)}^2,x)\right)}^{\frac{1}{2}}.\end{array}$$

By (2.2) and (2.6) we have

$${\left(D_{n,\alpha }({(t-x)}^2,x)\right)}^{\frac{1}{2}}\le {\left(\alpha D_{n,1}({(t-x)}^2,x)\right)}^{\frac{1}{2}}\le C{n}^{-\frac{1}{2}}{\delta }_n(x).$$

Then

$$|D_{n,\alpha }(g,x)-g(x)|\le \frac{C}{\sqrt{n}}{\parallel {\delta }_n{g}^{\prime }\parallel }_{\mathrm{\infty }}.$$

(2.11)

Then, by (2.8) and (2.11), we get

$$\begin{array}{rl}{\parallel D_{n,\alpha }(f,x)-f(x)\parallel }_{\mathrm{\infty }}& \le C({\parallel f-g\parallel }_{\mathrm{\infty }}+\frac{1}{\sqrt{n}}{\parallel {\delta }_n{g}^{\prime }\parallel }_{\mathrm{\infty }})\\ \le C({\parallel f-g\parallel }_{\mathrm{\infty }}+\frac{1}{\sqrt{n}}{\parallel \phi g^{\prime }\parallel }_{\mathrm{\infty }}+\frac{1}{n}{\parallel g^{\prime }\parallel }_{\mathrm{\infty }})\\ \le C{\omega }_{\phi }{(f,\frac{1}{\sqrt{n}})}_{\mathrm{\infty }}.\end{array}$$

(2.12)

II. $p=1$. By (2.2) and the Fubini theorem, we have

$$\begin{array}{r}{\parallel D_{n,\alpha }(g,x)-g(x)\parallel }_1\\ \le \alpha {\int }_0^{\mathrm{\infty }}\sum _{k=0}^{\mathrm{\infty }}{s}_{n,k}(x)n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)dt|{\int }_x^t{g}^{\prime }(u)du|dx\\ =n\alpha {\int }_0^{\mathrm{\infty }}|g^{\prime }(u)|\{{\int }_u^{\mathrm{\infty }}{\int }_0^u+{\int }_0^u{\int }_u^{\mathrm{\infty }}\}\sum _{k=0}^{\mathrm{\infty }}{s}_{n,k}(t)s_{n,k}(x)dtdxdu\\ =2\alpha {\int }_0^{\mathrm{\infty }}|g^{\prime }(u)|(\sum _{k=0}^{\mathrm{\infty }}n{\int }_u^{\mathrm{\infty }}{s}_{n,k}(t)dt{\int }_0^u{s}_{n,k}(x)dx)du\\ =:2\alpha {\int }_0^{\mathrm{\infty }}|g^{\prime }(u)|H_n(u)du.\end{array}$$

Now we estimate $H_n(u)$, by using ${\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)dt=\frac{1}{n}$ and ${\sum }_{k=0}^{\mathrm{\infty }}{s}_{n,k}(u)=1$:

$$\begin{array}{rcl}{H}_n(u)& =& n\sum _{k=0}^{\mathrm{\infty }}({\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)dt{\int }_0^u{s}_{n,k}(x)dx-{\int }_0^u{s}_{n,k}(t)dt{\int }_0^u{s}_{n,k}(x)dx)\\ =& u-n\sum _{k=0}^{\mathrm{\infty }}{\int }_0^u{s}_{n,k}(t)dt{\int }_0^u{s}_{n,k}(x)dx\\ =& u-n\sum _{k=0}^{\mathrm{\infty }}(t{s}_{n,k}(t){|}_0^u-{\int }_0^{u}t{s}_{n,k}^{\prime }(t)dt){\int }_0^u{s}_{n,k}(x)dx\\ =& u-n\sum _{k=0}^{\mathrm{\infty }}u{s}_{n,k}(u){\int }_0^u{s}_{n,k}(x)dx+n\sum _{k=0}^{\mathrm{\infty }}{\int }_0^{u}t{s}_{n,k}^{\prime }(t)dt{\int }_0^u{s}_{n,k}(x)dx\\ =:& u-I_1+I_2.\end{array}$$

Since

$$\begin{array}{rcl}{\int }_0^u{s}_{n,k}(x)dx& =& {\int }_0^u{e}^{-nx}\frac{{(nx)}^k}{k!}dx=-\frac{1}{n}{s}_{n,k}(u)+{\int }_0^u{s}_{n,k-1}(x)dx\\ =& -\frac{1}{n}(s_{n,k}(u)+\cdots +s_{n,1}(u))+{\int }_0^u{e}^{-nx}dx\\ =& -\frac{1}{n}(s_{n,k}(u)+\cdots +s_{n,0}(u))+\frac{1}{n},\end{array}$$

we have

$$\begin{array}{rcl}{I}_1& =& n\sum _{k=0}^{\mathrm{\infty }}u{s}_{n,k}(u)[-\frac{1}{n}\sum _{j=0}^k{s}_{n,j}(u)+\frac{1}{n}]\\ =& \sum _{k=0}^{\mathrm{\infty }}u{s}_{n,k}(u)-\sum _{k=0}^{\mathrm{\infty }}u{s}_{n,k}(u)\sum _{j=0}^k{s}_{n,j}(u)\\ =& u-\sum _{k=0}^{\mathrm{\infty }}u{s}_{n,k}(u)\sum _{j=0}^k{s}_{n,j}(u).\end{array}$$

Using the equation $\frac{k+1}{n}{s}_{n,k+1}(u)=u{s}_{n,k}(u)$ and (2.4), we have

$$\begin{array}{rcl}{I}_2& =& n\sum _{k=0}^{\mathrm{\infty }}{\int }_0^{u}tn(s_{n,k-1}(t)-s_{n,k}(t))dt{\int }_0^u{s}_{n,k}(x)dx\\ =& n\sum _{k=0}^{\mathrm{\infty }}{\int }_0^{u}t{s}_{n,k}(t)dt{\int }_0^{u}n(s_{n,k+1}(x)-s_{n,k}(x))dx\\ =& n\sum _{k=0}^{\mathrm{\infty }}{\int }_0^{u}t{s}_{n,k}(t)dt{\int }_0^u(-s_{n,k+1}^{\prime }(x))dx\\ =& -n\sum _{k=0}^{\mathrm{\infty }}{\int }_0^u\frac{k+1}{n}{s}_{n,k+1}(t)dt{s}_{n,k+1}(u)\\ =& -n\sum _{k=0}^{\mathrm{\infty }}{\int }_0^u{s}_{n,k+1}(t)dtu{s}_{n,k}(u)\\ =& -n\sum _{k=0}^{\mathrm{\infty }}[-\frac{1}{n}(s_{n,k+1}(u)+\cdots +s_{n,0}(u))+\frac{1}{n}]u{s}_{n,k}(u)\\ =& \sum _{k=0}^{\mathrm{\infty }}u{s}_{n,k}(u)\sum _{j=0}^{k+1}{s}_{n,j}(u)-\sum _{k=0}^{\mathrm{\infty }}u{s}_{n,k}(u)\\ =& \sum _{k=0}^{\mathrm{\infty }}u{s}_{n,k}(u)\sum _{j=0}^k{s}_{n,j}(u)+u\sum _{k=0}^{\mathrm{\infty }}{s}_{n,k}(u)s_{n,k+1}(u)-u.\end{array}$$

So

$$H_n(u)=u-u+2u\sum _{k=0}^{\mathrm{\infty }}{s}_{n,k}(u)\sum _{j=0}^k{s}_{n,j}(u)+u\sum _{k=0}^{\mathrm{\infty }}{s}_{n,k}(u)s_{n,k+1}(u)-u.$$

Since

$$\begin{array}{rcl}\sum _{k=0}^{\mathrm{\infty }}{s}_{n,k}(u)\sum _{j=0}^k{s}_{n,j}(u)& =& \sum _{k=0}^{\mathrm{\infty }}{s}_{n,k}(u)\sum _{j=k}^{\mathrm{\infty }}{s}_{n,j}(u)\\ =& \sum _{k=0}^{\mathrm{\infty }}{s}_{n,k}(u)\sum _{j=k+1}^{\mathrm{\infty }}{s}_{n,j}(u)+\sum _{k=0}^{\mathrm{\infty }}{s}_{n,k}(u)s_{n,k}(u),\end{array}$$

we can write

$$\begin{array}{rcl}{H}_n(u)& =& u(\sum _{k=0}^{\mathrm{\infty }}{s}_{n,k}(u)\sum _{j=0}^k{s}_{n,j}(u)+\sum _{k=0}^{\mathrm{\infty }}{s}_{n,k}(u)\sum _{j=k+1}^{\mathrm{\infty }}{s}_{n,j}(u))\\ +u\sum _{k=0}^{\mathrm{\infty }}{s}_{n,k}(u)(s_{n,k}(u)+s_{n,k+1}(u))-u\\ =& u\sum _{k=0}^{\mathrm{\infty }}{s}_{n,k}(u)(s_{n,k}(u)+s_{n,k+1}(u)).\end{array}$$

Using the result of [[5], Lemma 3]

$$s_{n,k}(u)\le \frac{1}{\sqrt{nu}},$$

we get

$$H_n(u)\le 2\frac{\sqrt{u}}{\sqrt{n}}.$$

Consequently

$$\begin{array}{rl}{\parallel D_{n,\alpha }(g,x)-g(x)\parallel }_1& \le 4\alpha {\int }_0^{\mathrm{\infty }}|g^{\prime }(u)|\frac{\phi (u)}{\sqrt{n}}du\\ =\frac{4\alpha }{\sqrt{n}}{\parallel {\phi }^{\prime }g\parallel }_1.\end{array}$$

(2.13)

By (2.8) and (2.13) we have

$${\parallel D_{n,\alpha }(f,x)-f(x)\parallel }_1\le C{\omega }_{\phi }{(f,\frac{1}{\sqrt{n}})}_1.$$

(2.14)

From (2.12) and (2.14), (2.7) is obtained. □

Remark 1 In [11] we show that the second order modulus cannot be used for the Baskakov-Bézier operators. Similarly in (2.7) ${\omega }_{\phi }^2{(f,x)}_p$ cannot be used instead of ${\omega }_{\phi }{(f,x)}_p$.

3 Inverse theorem

To prove the inverse theorem, we need the following lemmas.

Lemma 3.1 For $f\in L_p[0,\mathrm{\infty })$ ($1\le p\le \mathrm{\infty }$), $\phi (x)=\sqrt{x}$, ${\delta }_n(x)=\phi (x)+\frac{1}{\sqrt{n}}$, we have

$${\parallel {\delta }_n{D}_{n,\alpha }^{\prime }(f)\parallel }_p\le C\sqrt{n}{\parallel f\parallel }_p.$$

(3.1)

Proof We will show (3.1) for the two cases of $p=\mathrm{\infty }$ and $p=1$. Since

$$\begin{array}{rl}{D}_{n,\alpha }^{\prime }(f,x)& =\sum _{k=0}^{\mathrm{\infty }}\alpha [J_{n,k}^{\alpha -1}(x)J_{n,k}^{\prime }(x)-J_{n,k+1}^{\alpha -1}(x)J_{n,k+1}^{\prime }(x)]n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)f(t)dt\\ =\alpha \sum _{k=0}^{\mathrm{\infty }}[(J_{n,k}^{\alpha -1}(x)-J_{n,k+1}^{\alpha -1}(x))J_{n,k+1}^{\prime }(x)+J_{n,k}^{\alpha -1}(x)s_{n,k}^{\prime }(x)]n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)f(t)dt\end{array}$$

using ${\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)dt=\frac{1}{n}$, we have

$$\begin{array}{rl}|D_{n,\alpha }^{\prime }(f,x)|& \le \alpha {\parallel f\parallel }_{\mathrm{\infty }}(\sum _{k=0}^{\mathrm{\infty }}[J_{n,k}^{\alpha -1}(x)-J_{n,k+1}^{\alpha -1}(x)]J_{n,k+1}^{\prime }(x)+\sum _{k=0}^{\mathrm{\infty }}{J}_{n,k}^{\alpha -1}(x)|s_{n,k}^{\prime }(x)|)\\ =:\alpha {\parallel f\parallel }_{\mathrm{\infty }}(I_1+I_2).\end{array}$$

(3.2)

For $x\in E_n=(\frac{1}{n},\mathrm{\infty })$, ${\delta }_n(x)\sim \phi (x)$, by (2.1) and (2.3) we get

$$\begin{array}{rl}{\delta }_n(x)I_2& \le {\delta }_n(x)\sum _{k=0}^{\mathrm{\infty }}|s_{n,k}^{\prime }(x)|\le \frac{n{\delta }_n(x)}{{\phi }^2(x)}\sum _{k=0}^{\mathrm{\infty }}|\frac{k}{n}-x|s_{n,k}(x)\\ \le \frac{n{\delta }_n(x)}{{\phi }^2(x)}{(\sum _{k=0}^{\mathrm{\infty }}{(\frac{k}{n}-x)}^2{s}_{n,k}(x))}^{\frac{1}{2}}\\ =\frac{n{\delta }_n(x)}{{\phi }^2(x)}\frac{\phi (x)}{\sqrt{n}}\le 2\sqrt{n},\end{array}$$

(3.3)

here we used (cf. [[9], p.128, Lemma 9.4.3])

$$\sum _{k=0}^{\mathrm{\infty }}{(\frac{k}{n}-x)}^2{s}_{n,k}(x)=\frac{{\phi }^2(x)}{n}.$$

For $x\in E_n^c=[0,\frac{1}{n}]$, ${\delta }_n(x)\sim \frac{1}{\sqrt{n}}$, by (2.4) we have

$${\delta }_n(x)I_2\le {\delta }_n(x)\sum _{k=0}^{\mathrm{\infty }}n(s_{n,k-1}(x)+s_{n,k}(x))\le 4\sqrt{n}.$$

(3.4)

By (3.3) and (3.4), we get

$${\delta }_n(x)I_2\le 6\sqrt{n}.$$

(3.5)

Noting $J_{n,0}^{\prime }(x)=0$, we have

$$\begin{array}{rl}{I}_1& =\sum _{k=0}^{\mathrm{\infty }}(J_{n,k}^{\alpha -1}(x)(J_{n,k}^{\prime }(x)-s_{n,k}^{\prime }(x))-J_{n,k+1}^{\alpha -1}(x)J_{n,k+1}^{\prime }(x))\\ \le \sum _{k=0}^{\mathrm{\infty }}{J}_{n,k}^{\alpha -1}(x)J_{n,k}^{\prime }(x)-\sum _{k=0}^{\mathrm{\infty }}{J}_{n,k+1}^{\alpha -1}(x)J_{n,k+1}^{\prime }(x)+\sum _{k=0}^{\mathrm{\infty }}{J}_{n,k}^{\alpha -1}(x)|s_{n,k}^{\prime }(x)|\\ =I_2,\end{array}$$

then

$${\delta }_n(x)I_1\le 6\sqrt{n}.$$

(3.6)

Combining (3.2), (3.5), and (3.6) we get for $p=\mathrm{\infty }$

$${\parallel {\delta }_n{D}_{n,\alpha }^{\prime }(f)\parallel }_{\mathrm{\infty }}\le C\sqrt{n}{\parallel f\parallel }_{\mathrm{\infty }}.$$

(3.7)

For $p=1$, we have

$$\begin{array}{rl}|D_{n,\alpha }^{\prime }(f)|& \le \alpha \sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|f(t)|dt[(J_{n,k}^{\alpha -1}(x)-J_{n,k+1}^{\alpha -1}(x))J_{n,k+1}^{\prime }(x)+J_{n,k}^{\alpha -1}(x)|s_{n,k}^{\prime }(x)|]\\ =:\alpha ({\tilde{J}}_1+{\tilde{J}}_2).\end{array}$$

(3.8)

Hence we can write

$${\int }_0^{\mathrm{\infty }}|{\delta }_n(x)D_{n,\alpha }^{\prime }(f,x)|dx\le \alpha {\int }_0^{\mathrm{\infty }}{\delta }_n(x)({\tilde{J}}_1+{\tilde{J}}_2)dx=\alpha ({\int }_{E_n^c}+{\int }_{E_n}){\delta }_n(x)({\tilde{J}}_1+{\tilde{J}}_2)dx.$$

(3.9)

Now we estimate the last part of (3.9) in four phases:

$$\begin{array}{rcl}{\int }_{E_n^c}{\delta }_n(x){\tilde{J}}_{2}dx& \le & {\int }_{E_n^c}{\delta }_n(x)\sum _{k=1}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|f(t)|dtn(s_{n,k-1}(x)+s_{n,k}(x))dx\\ +{\int }_{E_n^c}{\delta }_n(x)n{\int }_0^{\mathrm{\infty }}{s}_{n,0}(t)|f(t)|dtn{s}_{n,0}(x)dx.\end{array}$$

For $x\in E_n^c$, ${\delta }_n(x)\le \frac{2}{\sqrt{n}}$, $s_{n,0}(t)\le 1$, noting ${\int }_0^{\mathrm{\infty }}{s}_{n,k}(x)dx=\frac{1}{n}$, we have

$${\int }_{E_n^c}{\delta }_n(x){\tilde{J}}_{2}dx\le \frac{4n}{\sqrt{n}}{\parallel f\parallel }_1+\frac{2n}{\sqrt{n}}{\parallel f\parallel }_1\le 6\sqrt{n}{\parallel f\parallel }_1.$$

(3.10)

Since $J_{n,k}^{\alpha -1}(x)-J_{n,k+1}^{\alpha -1}(x)\le 1$ and $J_{n,k+1}^{\prime }(x)=n{s}_{n,k}(x)$, we have

$${\int }_{E_n^c}{\delta }_n(x){\tilde{J}}_{1}dx\le {\int }_{E_n^c}{\delta }_n(x)\sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|f(t)|dtn{s}_{n,k}(x)dx\le 2\sqrt{n}{\parallel f\parallel }_1.$$

(3.11)

To estimate ${\int }_{E_n}{\delta }_n(x){\tilde{J}}_{2}dx$, we will need the relation [[9], p.129, (9.4.15)]

$${\int }_{E_n}\frac{{(\frac{k}{n}-x)}^2}{{\phi }^2(x)}{s}_{n,k}(x)dx\le C{n}^{-2}.$$

By the Hölder inequality and (2.3), we get

$$\begin{array}{r}{\int }_{E_n}{\delta }_n(x){\tilde{J}}_{2}dx\\ \le 2\sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|f(t)|dt{\int }_{E_n}\phi (x)\frac{n}{{\phi }^2(x)}|\frac{k}{n}-x|s_{n,k}(x)dx\\ \le 2\sum _{k=0}^{\mathrm{\infty }}{n}^2{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|f(t)|dt{({\int }_{E_n}\frac{{(\frac{k}{n}-x)}^2}{{\phi }^2(x)}{s}_{n,k}(x)dx)}^{\frac{1}{2}}{({\int }_{E_n}{s}_{n,k}(x)dx)}^{\frac{1}{2}}\\ \le C\sqrt{n}\sum _{k=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|f(t)|dt\\ =C\sqrt{n}{\parallel f\parallel }_1.\end{array}$$

(3.12)

In order to estimate ${\int }_{E_n}{\delta }_n(x){\tilde{J}}_{1}dx$, we consider the two cases of $\alpha \ge 2$ and $1<\alpha <2$ (when $\alpha =1$, ${\tilde{J}}_1=0$).

For $\alpha \ge 2$, $J_{n,k}^{\alpha -1}(x)-J_{n,k+1}^{\alpha -1}(x)\le (\alpha -1)s_{n,k}(x)$. Using integration by parts, we can deduce

$$\begin{array}{rcl}{\int }_{E_n}{\delta }_n(x){\tilde{J}}_{1}dx& \le & C\sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|f(t)|dt{\int }_{E_n}\phi (x)s_{n,k}(x)J_{n,k+1}^{\prime }(x)dx\\ =& C\sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|f(t)|dt\\ \times (\phi (x)s_{n,k}(x)J_{n,k+1}(x){|}_{\frac{1}{n}}^{\mathrm{\infty }}-{\int }_{\frac{1}{n}}^{\mathrm{\infty }}{J}_{n,k+1}(x)d(\phi (x)s_{n,k}(x)))\\ =& C\sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|f(t)|dt\phi (x)s_{n,k}(x)J_{n,k+1}(x){|}_{\frac{1}{n}}^{\mathrm{\infty }}\\ -C\sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|f(t)|dt\\ \times ({\int }_{\frac{1}{n}}^{\mathrm{\infty }}{J}_{n,k+1}(x)\frac{1}{2\sqrt{x}}{s}_{n,k}(x)dx+{\int }_{\frac{1}{n}}^{\mathrm{\infty }}\phi (x)s_{n,k}^{\prime }(x)J_{n,k+1}(x)dx).\end{array}$$

Noting that $\phi (x)s_{n,k}(x)J_{n,k+1}(x){|}_{\frac{1}{n}}^{\mathrm{\infty }}<0$ and ${\int }_{\frac{1}{n}}^{\mathrm{\infty }}{J}_{n,k+1}(x)\frac{1}{2\sqrt{x}}{s}_{n,k}(x)dx>0$, and from (2.3), we have

$$\begin{array}{r}{\int }_{E_n}{\delta }_n(x){\tilde{J}}_{1}dx\\ \le C\sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|f(t)|dt{\int }_{\frac{1}{n}}^{\mathrm{\infty }}|\phi (x)s_{n,k}^{\prime }(x)J_{n,k+1}(x)|dx\\ \le C\sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|f(t)|dtn{({\int }_{E_n}\frac{{(\frac{k}{n}-x)}^2}{{\phi }^2(x)}{s}_{n,k}(x)dx)}^{\frac{1}{2}}{({\int }_{E_n}{s}_{n,k}(x)dx)}^{\frac{1}{2}}\\ \le C\sqrt{n}{\parallel f\parallel }_1.\end{array}$$

(3.13)

For $1<\alpha <2$, using the mean value theorem, we know

$$J_{n,k}^{\alpha -1}(x)-J_{n,k+1}^{\alpha -1}(x)=(\alpha -1){({\xi }_k(x))}^{\alpha -2}{s}_{n,k}(x),$$

where $J_{n,k+1}(x)<{\xi }_k(x)<J_{n,k}(x)$ and $\alpha -2<0$, then

$$J_{n,k}^{\alpha -1}(x)-J_{n,k+1}^{\alpha -1}(x)\le (\alpha -1)J_{n,k+1}^{\alpha -2}(x)s_{n,k}(x).$$

For $1<\alpha <2$, we get from the procedure of (3.13)

$$\begin{array}{rl}{\int }_{E_n}{\delta }_n(x){\tilde{J}}_{1}dx& \le \sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|f(t)|dt{\int }_{E_n}(\alpha -1)\phi (x)s_{n,k}(x)J_{n,k+1}^{\alpha -2}(x)J_{n,k+1}^{\prime }(x)dx\\ =\sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|f(t)|dt{\int }_{E_n}\phi (x)s_{n,k}(x)d{J}_{n,k+1}^{\alpha -1}(x)\\ \le \sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|f(t)|dt{\int }_{E_n}\phi (x)|s_{n,k}^{\prime }(x)|dx\\ \le C\sqrt{n}{\parallel f\parallel }_1.\end{array}$$

(3.14)

Combining (3.13) and (3.14), we get for $\alpha \ge 1$

$${\int }_{E_n}{\delta }_n(x){\tilde{J}}_{1}dx\le C\sqrt{n}{\parallel f\parallel }_1.$$

(3.15)

From (3.8)-(3.12) and (3.15), we obtain

$${\parallel {\delta }_n{D}_{n,\alpha }^{\prime }(f)\parallel }_1\le C\sqrt{n}{\parallel f\parallel }_1.$$

(3.16)

By (3.7) and (3.16), Lemma 3.1 holds. □

Lemma 3.2 For $f\in W_p$, $\phi (x)=\sqrt{x}$, ${\delta }_n(x)=\phi (x)+\frac{1}{\sqrt{n}}$, we have

$${\parallel {\delta }_n{D}_{n,\alpha }^{\prime }(f)\parallel }_p\le C{\parallel {\delta }_n{f}^{\prime }\parallel }_p.$$

(3.17)

Proof By the Riesz-Thorin theorem, we shall prove Lemma 3.2 for $p=\mathrm{\infty }$ and $p=1$. For $f\in W_p$ and noting that $D_{n,\alpha }^{\prime }(1,x)=0$, we have

$$\sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}f(x)s_{n,k}(t)dt{(J_{n,k}^{\alpha }(x)-J_{n,k+1}^{\alpha }(x))}^{\prime }=0.$$

Then

$$\begin{array}{rl}|D_{n,\alpha }^{\prime }(f,x)|=& |\sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}(f(t)-f(x))s_{n,k}(t)dt{(J_{n,k}^{\alpha }(x)-J_{n,k+1}^{\alpha }(x))}^{\prime }|\\ =& |\sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{\int }_x^t{f}^{\prime }(u)du{s}_{n,k}(t)dt{(J_{n,k}^{\alpha }(x)-J_{n,k+1}^{\alpha }(x))}^{\prime }|\\ \le & \alpha \sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|{\int }_x^t{f}^{\prime }(u)du|dt\\ \times \{[J_{n,k}^{\alpha -1}(x)-J_{n,k+1}^{\alpha -1}(x)]J_{n,k+1}^{\prime }(x)+J_{n,k}^{\alpha -1}(x)|s_{n,k}^{\prime }(x)|\}.\end{array}$$

(3.18)

By (2.9) and (2.10) we have

$$|{\int }_x^t{\delta }_n(u)du|\le C{\delta }_n^{-1}(x)|t-x|,$$

hence

$$\begin{array}{rl}|{\delta }_n(x)D_{n,\alpha }^{\prime }(f,x)|\le & C{\parallel {\delta }_n{f}^{\prime }\parallel }_{\mathrm{\infty }}\sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|t-x|dt\\ \times \{[J_{n,k}^{\alpha -1}(x)-J_{n,k+1}^{\alpha -1}(x)]J_{n,k+1}^{\prime }(x)+J_{n,k}^{\alpha -1}(x)|s_{n,k}^{\prime }(x)|\}\\ =:& C{\parallel {\delta }_n{f}^{\prime }\parallel }_{\mathrm{\infty }}(J_1+J_2).\end{array}$$

(3.19)

For $x\in E_n^c$, ${\delta }_n(x)\sim \frac{1}{\sqrt{n}}$ and by (2.1) and (2.4) we have

$$\begin{array}{rcl}{J}_2& =& \sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|t-x|dt{J}_{n,k}^{\alpha -1}(x)|s_{n,k}^{\prime }(x)|\\ \le & n\sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|t-x|dt(s_{n,k-1}(x)+s_{n,k}(x))\\ =& n\sum _{k=1}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|t-x|dt{s}_{n,k-1}(x)+n\sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|t-x|dt{s}_{n,k}(x)\\ =:& K_1+K_2.\end{array}$$

By (2.6) we get

$$K_2=n{D}_{n,1}(|t-x|,x)\le n{\left(D_{n,1}({(t-x)}^2,x)\right)}^{\frac{1}{2}}\le \sqrt{n}{\delta }_n(x)\le 2.$$

For $K_1$, we write

$$K_1=n\sum _{k=2}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|t-x|dt{s}_{n,k-1}(x)+n^2{\int }_0^{\mathrm{\infty }}{s}_{n,1}(t)|t-x|dt{s}_{n,0}(x).$$

First, using (2.6), ${\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)dt=\frac{1}{n}$, and the Hölder inequality, we have

$$\begin{array}{r}n\sum _{k=2}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|t-x|dt{s}_{n,k-1}(x)\\ =n\sum _{k=2}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|t-x|dt{s}_{n,k}(x)\frac{k}{nx}\\ \le n{(\sum _{k=2}^{\mathrm{\infty }}{n}^2{({\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|t-x|dt)}^2{s}_{n,k}(x))}^{\frac{1}{2}}{(\sum _{k=2}^{\mathrm{\infty }}{s}_{n,k}(x)\frac{k^2}{{(nx)}^2})}^{\frac{1}{2}}\\ \le n{(\sum _{k=2}^{\mathrm{\infty }}{n}^2{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)dt\cdot {\int }_0^{\mathrm{\infty }}{s}_{n,k}(t){(t-x)}^{2}dt{s}_{n,k}(x))}^{\frac{1}{2}}{(\sum _{k=2}^{\mathrm{\infty }}{s}_{n,k}(x)\frac{k^2}{{(nx)}^2})}^{\frac{1}{2}}\\ \le Cn{(\sum _{k=2}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t){(t-x)}^{2}dt{s}_{n,k}(x))}^{\frac{1}{2}}{(\sum _{k=2}^{\mathrm{\infty }}{s}_{n,k-2}(x))}^{\frac{1}{2}}\\ \le Cn\cdot \frac{{\delta }_n(x)}{\sqrt{n}}\\ \le C.\end{array}$$

Next, for $x\in E_n^c$, $e^{-nx}\le 1$, and ${\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)dt=\frac{1}{n}$, we have

$$\begin{array}{rcl}{n}^2{\int }_0^{\mathrm{\infty }}{s}_{n,1}(t)|t-x|dt{s}_{n,0}(x)& \le & n^2{\int }_0^{\mathrm{\infty }}{s}_{n,1}(t)(t+x)dt{e}^{-nx}\\ \le & n^2(\frac{2}{n}{\int }_0^{\mathrm{\infty }}{s}_{n,2}(t)dt+x{\int }_0^{\mathrm{\infty }}{s}_{n,1}(t)dt)\\ =& n^2(\frac{2}{n^2}+\frac{x}{n})\\ \le & 3.\end{array}$$

Then we get

$$J_2\le C,x\in E_n^c.$$

(3.20)

Noting that $J_{n,k}^{\alpha -1}(x)-J_{n,k+1}^{\alpha -1}(x)\le 1$, by (2.5) we have

$$J_1\le K_2\le 2.$$

(3.21)

For $x\in E_n$, ${\delta }_n(x)\sim \phi (x)$, using (2.3), ${\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)dt=\frac{1}{n}$ and the Hölder inequality, we have

$$\begin{array}{rl}{J}_2& \le \sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|t-x|dt\frac{n}{{\phi }^2(x)}|\frac{k}{n}-x|s_{n,k}(x)\\ \le {(\sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t){(t-x)}^{2}dt{s}_{n,k}(x))}^{\frac{1}{2}}{(\sum _{k=0}^{\mathrm{\infty }}{(\frac{k}{n}-x)}^2{s}_{n,k}(x))}^{\frac{1}{2}}\cdot \frac{n}{{\phi }^2(x)}\\ \le \frac{{\delta }_n(x)}{\sqrt{n}}\cdot \frac{\phi (x)}{\sqrt{n}}\cdot \frac{n}{{\phi }^2(x)}\\ \le 2.\end{array}$$

Noting that $J_{n,0}^{\prime }(x)=0$, one has

$$\begin{array}{rcl}{J}_1& =& \sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|t-x|dt[J_{n,k}^{\alpha -1}(x)-J_{n,k+1}^{\alpha -1}(x)]J_{n,k+1}^{\prime }(x)\\ \le & \sum _{k=1}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|t-x|dt{J}_{n,k}^{\alpha -1}(x)J_{n,k}^{\prime }(x)-\sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|t-x|dt{J}_{n,k+1}^{\alpha -1}(x)J_{n,k+1}^{\prime }(x)\\ +\sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)|t-x|dt{J}_{n,k}^{\alpha -1}(x)|s_{n,k}^{\prime }(x)|.\end{array}$$

The third term of the above is $J_2$, $J_3$ denotes the difference of the front two terms, and we need only to consider $J_3$. By (2.1), (2.4), (2.5), and integration by parts, we have

$$\begin{array}{rcl}{J}_3& \le & \sum _{k=1}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}(s_{n,k}(t)-s_{n,k-1}(t))|t-x|dt{J}_{n,k}^{\alpha -1}(x)J_{n,k}^{\prime }(x)\\ \le & \sum _{k=1}^{\mathrm{\infty }}n|{\int }_0^{\mathrm{\infty }}\frac{1}{n}{s}_{n,k}^{\prime }(t)|t-x|dt|n{s}_{n,k-1}(x)\\ \le & \sum _{k=1}^{\mathrm{\infty }}\left(\right|{\int }_0^x{s}_{n,k}^{\prime }(t)(x-t)dt|+|{\int }_x^{\mathrm{\infty }}{s}_{n,k}^{\prime }(t)(t-x)dt\left|\right)n{s}_{n,k-1}(x)\\ =& \sum _{k=1}^{\mathrm{\infty }}\left(\right|{\int }_0^x(x-t)d{s}_{n,k}(t)|+|{\int }_x^{\mathrm{\infty }}(t-x)d{s}_{n,k}(t)\left|\right)n{s}_{n,k-1}(x)\\ \le & \sum _{k=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{s}_{n,k}(t)dtn{s}_{n,k-1}(x)\\ =& 1.\end{array}$$

Thus

$$J_1\le J_2+J_3\le 3.$$

(3.22)

So we get

$${\parallel {\delta }_n{D}_{n,\alpha }^{\prime }(f)\parallel }_{\mathrm{\infty }}\le C{\parallel {\delta }_n{f}^{\prime }\parallel }_{\mathrm{\infty }}.$$

(3.23)

For $p=1$, using (2.1), (2.4), (2.5), and integration by parts, we have

$$\begin{array}{rcl}{D}_{n,\alpha }^{\prime }(f,x)& =& \alpha \sum _{k=1}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}f(t)s_{n,k}(t)dt{J}_{n,k}^{\alpha -1}(x)J_{n,k}^{\prime }(x)\\ -\alpha \sum _{k=0}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}f(t)s_{n,k}(t)dt{J}_{n,k+1}^{\alpha -1}(x)J_{n,k+1}^{\prime }(x)\\ =& \alpha \sum _{k=1}^{\mathrm{\infty }}n{\int }_0^{\mathrm{\infty }}f(t)(s_{n,k}(t)-s_{n,k-1}(t))dt{J}_{n,k}^{\alpha -1}(x)J_{n,k}^{\prime }(x)\\ =& -\alpha \sum _{k=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}f(t)s_{n,k}^{\prime }(t)dt{J}_{n,k}^{\alpha -1}(x)J_{n,k}^{\prime }(x)\\ =& \alpha \sum _{k=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{f}^{\prime }(t)s_{n,k}(t)dt{J}_{n,k}^{\alpha -1}(x)J_{n,k}^{\prime }(x)\\ \le & \alpha \sum _{k=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{f}^{\prime }(t)s_{n,k}(t)dtn{s}_{n,k-1}(x).\end{array}$$

Let

$$\begin{array}{rl}{\parallel {\delta }_n{D}_{n,\alpha }^{\prime }(f)\parallel }_1& ={\parallel {\delta }_n{D}_{n,\alpha }^{\prime }(f)\parallel }_1^{E_n^c}+{\parallel {\delta }_n{D}_{n,\alpha }^{\prime }(f)\parallel }_1^{E_n}\\ =:\widetilde{K_1}+\widetilde{K_2}.\end{array}$$

(3.24)

For $x\in E_n^c$, noting that $\sqrt{n}{\delta }_n(t)\ge 1$ and $\sqrt{n}{\delta }_n(x)\le 2$, we have

$$\begin{array}{rl}\widetilde{K_1}& \le \alpha {\int }_{E_n^c}{\delta }_n(x)\sum _{k=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}|f^{\prime }(t)|s_{n,k}(t)dtn{s}_{n,k-1}(x)dx\\ \le \alpha {\int }_{E_n^c}\sqrt{n}{\delta }_n(x)\sum _{k=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}|{\delta }_n(t)f^{\prime }(t)|s_{n,k}(t)dtn{s}_{n,k-1}(x)dx\\ \le 2\alpha \sum _{k=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}|{\delta }_n(t)f^{\prime }(t)|s_{n,k}(t)dtn{\int }_{E_n^c}{s}_{n,k-1}(x)dx\\ \le 2\alpha {\parallel {\delta }_n{f}^{\prime }\parallel }_1.\end{array}$$

(3.25)

For $x\in E_n$, we estimate $\widetilde{K_2}$

$$\begin{array}{rcl}\widetilde{K_2}& \le & 2\alpha {\int }_{E_n}\phi (x)\sum _{k=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{|{\delta }_n(t)f^{\prime }(t)|}{\phi (t)}{s}_{n,k}(t)dtn{s}_{n,k-1}(x)dx\\ =& 2n\alpha \sum _{k=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{|{\delta }_n(t)f^{\prime }(t)|}{\phi (t)}{s}_{n,k}(t)dt{\int }_{E_n}\phi (x)s_{n,k-1}(x)dx.\end{array}$$

Using the Hölder inequality and $2\sqrt{ab}\le a+b$ ($a,b>0$), we have

$$\begin{array}{r}{\int }_0^{\mathrm{\infty }}\frac{|{\delta }_n(t)f^{\prime }(t)|}{\phi (t)}{s}_{n,k}(t)dt\\ \le {({\int }_0^{\mathrm{\infty }}|{\delta }_n(t)f^{\prime }(t)|s_{n,k}(t)dt{\int }_0^{\mathrm{\infty }}\frac{|{\delta }_n(t)f^{\prime }(t)|}{{\phi }^2(t)}{s}_{n,k}(t)dt)}^{\frac{1}{2}}\\ ={({\int }_0^{\mathrm{\infty }}|{\delta }_n(t)f^{\prime }(t)|s_{n,k}(t)dt{\int }_0^{\mathrm{\infty }}|{\delta }_n(t)f^{\prime }(t)|\frac{n}{k}{s}_{n,k-1}(t)dt)}^{\frac{1}{2}}\\ \le {\left(\frac{n}{k}\right)}^{\frac{1}{2}}({\int }_0^{\mathrm{\infty }}|{\delta }_n(t)f^{\prime }(t)|s_{n,k}(t)dt+{\int }_0^{\mathrm{\infty }}|{\delta }_n(t)f^{\prime }(t)|s_{n,k-1}(t)dt)\end{array}$$

and

$$\begin{array}{rcl}{\int }_{E_n}\phi (x)s_{n,k-1}(x)dx& \le & \frac{1}{\sqrt{n}}{({\int }_{E_n}{\phi }^2(x)s_{n,k-1}(x)dx)}^{\frac{1}{2}}\\ =& \frac{1}{\sqrt{n}}{\left(\frac{k}{n}\right)}^{\frac{1}{2}}{({\int }_{E_n}{s}_{n,k}(x)dx)}^{\frac{1}{2}}\\ \le & \frac{1}{n}{\left(\frac{k}{n}\right)}^{\frac{1}{2}}.\end{array}$$

Therefore we have

$$\begin{array}{rl}\widetilde{K_2}& \le 2\alpha \sum _{k=1}^{\mathrm{\infty }}({\int }_0^{\mathrm{\infty }}|{\delta }_n(t)f^{\prime }(t)|s_{n,k}(t)dt+{\int }_0^{\mathrm{\infty }}|{\delta }_n(t)f^{\prime }(t)|s_{n,k-1}(t)dt)\\ \le 2\alpha {\parallel {\delta }_n{f}^{\prime }\parallel }_1.\end{array}$$

(3.26)

By (3.24)-(3.26) we have

$${\parallel {\delta }_n{D}_{n,\alpha }^{\prime }(f)\parallel }_1\le C{\parallel {\delta }_n{f}^{\prime }\parallel }_1.$$

(3.27)

By (3.23) and (3.27), Lemma 3.2 holds. □

Using Lemmas 3.1 and 3.2, we can prove the inverse theorem.

Theorem 3.3 For $f\in L_p[0,\mathrm{\infty })$ ($1\le p\le \mathrm{\infty }$), $\phi (x)=\sqrt{x}$, $0<\beta <1$,

$${\parallel D_{n,\alpha }(f,x)-f(x)\parallel }_p=O\left(n^{-\frac{\beta }{2}}\right)$$

implies ${\omega }_{\phi }{(f,t)}_p=O(t^{\beta })$.

Proof Using Lemmas 3.1 and 3.2, for a suitable function g, we have

$$\begin{array}{rl}{K}_{\phi }{(f,t)}_p& \le {\parallel f-D_{n,\alpha }(f)\parallel }_p+t{\parallel {\delta }_n{D}_{n,\alpha }^{\prime }(f)\parallel }_p\\ \le C{n}^{-\frac{\beta }{2}}+t({\parallel {\delta }_n{D}_{n,\alpha }^{\prime }(f-g)\parallel }_p+{\parallel {\delta }_n{D}_{n,\alpha }^{\prime }(g)\parallel }_p)\\ \le C{n}^{-\frac{\beta }{2}}+Ct(\sqrt{n}{\parallel f-g\parallel }_p+{\parallel {\delta }_n{g}^{\prime }\parallel }_p)\\ \le C{n}^{-\frac{\beta }{2}}+Ct\sqrt{n}({\parallel f-g\parallel }_p+\frac{1}{\sqrt{n}}{\parallel \phi g^{\prime }\parallel }_p+\frac{1}{n}{\parallel g^{\prime }\parallel }_p)\\ \le C(n^{-\frac{\beta }{2}}+\frac{t}{n^{-\frac{1}{2}}}{\overline{K}}_{\phi }{(f,n^{-\frac{1}{2}})}_p)\\ \le C(n^{-\frac{\beta }{2}}+\frac{t}{n^{-\frac{1}{2}}}{K}_{\phi }{(f,n^{-\frac{1}{2}})}_p)\end{array}$$

which by the Berens-Lorentz lemma (cf. [[9], Lemma 9.3.4]) implies that

$$K_{\phi }{(f,t)}_p=O\left(t^{\beta }\right).$$

(3.28)

From (1.2) and (3.28), we see that the proof of Theorem 3.3 is completed. □