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Journal of Inequalities and Applications

Erschienen in:

Open Access 01.12.2014 | Research

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

verfasst von: Guofen Liu, Xiuzhong Yang

Erschienen in: Journal of Inequalities and Applications | Ausgabe 1/2014

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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, \infty)

(

1 \leq p \leq \infty

) with Ditzian-Totik modulus.

MSC:41A25, 41A27, 41A36.

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.

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, α} (f, x) = \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) f (t) d t [J_{n, k}^{α} (x) - J_{n, k + 1}^{α} (x)],

(1.1)

where

α \geq 1

f \in L_{p} [0, \infty)

J_{n, k} (x) = \sum_{j = k}^{\infty} s_{n, j} (x)

s_{n, k} = e^{- n x} \frac{{(n x)}^{k}}{k!}

. Obviously

D_{n, α} (f, x)

is bounded and positive in the space

L_{p} [0, \infty)

. When

α = 1

D_{n, α} (f, x)

is the well-known Durrmeyer operator

D_{n, 1} (f, x) = \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) f (t) d t 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, \infty)

(

1 \leq p \leq \infty

φ (x) = \sqrt{x}

\begin{matrix} ω_{φ} {(f, t)}_{p} = sup_{0 < h \leq t} {{∥ f (x + \frac{h φ (x)}{2}) - f (x - \frac{h φ (x)}{2}) ∥}_{p}, x - \frac{h φ (x)}{2} \geq 0}, \\ K_{φ} {(f, t)}_{p} = inf_{g \in W_{p}} {{∥ f - g ∥}_{p} + t {∥ φ g^{'} ∥}_{p}}, \\ {\bar{K}}_{φ} {(f, t)}_{p} = inf_{g \in W_{p}} {{∥ f - g ∥}_{p} + t {∥ φ g^{'} ∥}_{p} + t^{2} {∥ g^{'} ∥}_{p}}, \end{matrix}

where

W_{p} = {f | f \in A . C ._{loc}, {∥ φ f^{'} ∥}_{p} < \infty, {∥ f^{'} ∥}_{p} < \infty}

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

ω_{φ} {(f, t)}_{p} \sim K_{φ} {(f, t)}_{p} \sim {\bar{K}}_{φ} {(f, t)}_{p},

(1.2)

here

a \sim b

means that there exists

c > 0

such that

c^{- 1} a \leq b \leq c a

Now we state our equivalence theorem as follows.

Theorem For

f \in L_{p} [0, \infty)

(

1 \leq p \leq \infty

φ (x) = \sqrt{x}

0 < β < 1

, we have

{∥ D_{n, α} (f, x) - f (x) ∥}_{p} = O ({(\frac{1}{\sqrt{n}})}^{β}) \Leftrightarrow ω_{φ} {(f, t)}_{p} = O (t^{β}) .

(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) > \dots > J_{n, k} (x) > \dots > 0;

(2.1)

(2)

0 < J_{n, k}^{α} (x) - J_{n, k + 1}^{α} (x) \leq α s_{n, k} (x), α \geq 1;

(2.2)

(3)

s_{n, k}^{'} (x) = \frac{n}{φ^{2} (x)} (\frac{k}{n} - x) s_{n, k} (x);

(2.3)

(4)

s_{n, k}^{'} (x) = n s_{n, k - 1} (x) - n s_{n, k} (x), s_{n, - 1} (x) = 0;

(2.4)

(5)

J_{n, 0}^{'} (x) = 0, J_{n, k}^{'} (x) = n s_{n, k - 1} (x) (k = 1, 2, \dots);

(2.5)

(6)

D_{n, 1} ({(t - x)}^{2}, x) \leq n^{- 1} δ_{n}^{2} (x),

(2.6)

where

δ_{n} (x) = φ (x) + \frac{1}{\sqrt{n}}

Now we give the direct theorem.

Theorem 2.1 For

f \in L_{p} [0, \infty)

(

1 \leq p \leq \infty

φ (x) = \sqrt{x}

, we have

{∥ D_{n, α} (f, x) - f (x) ∥}_{p} \leq C ω_{φ} {(f, \frac{1}{\sqrt{n}})}_{p} .

(2.7)

Proof By the definition of

{\bar{K}}_{φ} {(f, t)}_{p}

and the relation (1.2), for fixed n, we can choose

g = g_{n}

such that

{∥ f - g ∥}_{p} + \frac{1}{\sqrt{n}} {∥ φ g^{'} ∥}_{p} + \frac{1}{n} {∥ g^{'} ∥}_{p} \leq C ω_{φ} {(f, \frac{1}{\sqrt{n}})}_{p} .

Since

\begin{aligned} {∥ D_{n, α} f - f ∥}_{p} & \leq {∥ D_{n, α} (f - g) ∥}_{p} + {∥ f - g ∥}_{p} + {∥ D_{n, α} g - g ∥}_{p} \\ \leq C {∥ f - g ∥}_{p} + {∥ D_{n, α} g - g ∥}_{p}, \end{aligned}

(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 = \infty

and

p = 1

p = \infty

. Noting that

g (t) = g (x) + \int_{x}^{t} g^{'} (u) d u

, we write

\begin{array}{rcl} | D_{n, α} (g, x) - g (x) | & = & | D_{n, α} (\int_{x}^{t} g^{'} (u) d u, x) | \\ \leq & {∥ δ_{n} g^{'} ∥}_{\infty} D_{n, α} (| \int_{x}^{t} δ_{n}^{- 1} (u) d u |, x) . \end{array}

Since

| \int_{x}^{t} φ^{- 1} (u) d u | = | \int_{x}^{t} \frac{1}{\sqrt{u}} d u | \leq \frac{2 | t - x |}{φ (x)},

(2.9)

| \int_{x}^{t} {(\frac{1}{\sqrt{n}})}^{- 1} d u | = \sqrt{n} | t - x |,

(2.10)

and

min {φ^{- 1} (x), \sqrt{n}} \sim δ_{n}^{- 1} (x)

, using the Hölder inequality, we have

\begin{array}{rcl} | D_{n, α} (g, x) - g (x) | & \leq & C δ_{n}^{- 1} (x) {∥ δ_{n} g^{'} ∥}_{\infty} D_{n, α} (| t - x |, x) \\ \leq & C δ_{n}^{- 1} (x) {∥ δ_{n} g^{'} ∥}_{\infty} {(D_{n, α} ({(t - x)}^{2}, x))}^{\frac{1}{2}} . \end{array}

By (2.2) and (2.6) we have

{(D_{n, α} ({(t - x)}^{2}, x))}^{\frac{1}{2}} \leq {(α D_{n, 1} ({(t - x)}^{2}, x))}^{\frac{1}{2}} \leq C n^{- \frac{1}{2}} δ_{n} (x) .

Then

| D_{n, α} (g, x) - g (x) | \leq \frac{C}{\sqrt{n}} {∥ δ_{n} g^{'} ∥}_{\infty} .

(2.11)

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

\begin{aligned} {∥ D_{n, α} (f, x) - f (x) ∥}_{\infty} & \leq C ({∥ f - g ∥}_{\infty} + \frac{1}{\sqrt{n}} {∥ δ_{n} g^{'} ∥}_{\infty}) \\ \leq C ({∥ f - g ∥}_{\infty} + \frac{1}{\sqrt{n}} {∥ φ g^{'} ∥}_{\infty} + \frac{1}{n} {∥ g^{'} ∥}_{\infty}) \\ \leq C ω_{φ} {(f, \frac{1}{\sqrt{n}})}_{\infty} . \end{aligned}

(2.12)

II.

p = 1

. By (2.2) and the Fubini theorem, we have

\begin{aligned} {∥ D_{n, α} (g, x) - g (x) ∥}_{1} \\ \leq α \int_{0}^{\infty} \sum_{k = 0}^{\infty} s_{n, k} (x) n \int_{0}^{\infty} s_{n, k} (t) d t | \int_{x}^{t} g^{'} (u) d u | d x \\ = n α \int_{0}^{\infty} | g^{'} (u) | {\int_{u}^{\infty} \int_{0}^{u} + \int_{0}^{u} \int_{u}^{\infty}} \sum_{k = 0}^{\infty} s_{n, k} (t) s_{n, k} (x) d t d x d u \\ = 2 α \int_{0}^{\infty} | g^{'} (u) | (\sum_{k = 0}^{\infty} n \int_{u}^{\infty} s_{n, k} (t) d t \int_{0}^{u} s_{n, k} (x) d x) d u \\ = : 2 α \int_{0}^{\infty} | g^{'} (u) | H_{n} (u) d u . \end{aligned}

Now we estimate

H_{n} (u)

, by using

\int_{0}^{\infty} s_{n, k} (t) d t = \frac{1}{n}

and

\sum_{k = 0}^{\infty} s_{n, k} (u) = 1

\begin{array}{rcl} H_{n} (u) & = & n \sum_{k = 0}^{\infty} (\int_{0}^{\infty} s_{n, k} (t) d t \int_{0}^{u} s_{n, k} (x) d x - \int_{0}^{u} s_{n, k} (t) d t \int_{0}^{u} s_{n, k} (x) d x) \\ = & u - n \sum_{k = 0}^{\infty} \int_{0}^{u} s_{n, k} (t) d t \int_{0}^{u} s_{n, k} (x) d x \\ = & u - n \sum_{k = 0}^{\infty} (t s_{n, k} (t) |_{0}^{u} - \int_{0}^{u} t s_{n, k}^{'} (t) d t) \int_{0}^{u} s_{n, k} (x) d x \\ = & u - n \sum_{k = 0}^{\infty} u s_{n, k} (u) \int_{0}^{u} s_{n, k} (x) d x + n \sum_{k = 0}^{\infty} \int_{0}^{u} t s_{n, k}^{'} (t) d t \int_{0}^{u} s_{n, k} (x) d x \\ = : & u - I_{1} + I_{2} . \end{array}

Since

\begin{array}{rcl} \int_{0}^{u} s_{n, k} (x) d x & = & \int_{0}^{u} e^{- n x} \frac{{(n x)}^{k}}{k!} d x = - \frac{1}{n} s_{n, k} (u) + \int_{0}^{u} s_{n, k - 1} (x) d x \\ = & - \frac{1}{n} (s_{n, k} (u) + \dots + s_{n, 1} (u)) + \int_{0}^{u} e^{- n x} d x \\ = & - \frac{1}{n} (s_{n, k} (u) + \dots + s_{n, 0} (u)) + \frac{1}{n}, \end{array}

we have

\begin{array}{rcl} I_{1} & = & n \sum_{k = 0}^{\infty} u s_{n, k} (u) [- \frac{1}{n} \sum_{j = 0}^{k} s_{n, j} (u) + \frac{1}{n}] \\ = & \sum_{k = 0}^{\infty} u s_{n, k} (u) - \sum_{k = 0}^{\infty} u s_{n, k} (u) \sum_{j = 0}^{k} s_{n, j} (u) \\ = & u - \sum_{k = 0}^{\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}^{\infty} \int_{0}^{u} t n (s_{n, k - 1} (t) - s_{n, k} (t)) d t \int_{0}^{u} s_{n, k} (x) d x \\ = & n \sum_{k = 0}^{\infty} \int_{0}^{u} t s_{n, k} (t) d t \int_{0}^{u} n (s_{n, k + 1} (x) - s_{n, k} (x)) d x \\ = & n \sum_{k = 0}^{\infty} \int_{0}^{u} t s_{n, k} (t) d t \int_{0}^{u} (- s_{n, k + 1}^{'} (x)) d x \\ = & - n \sum_{k = 0}^{\infty} \int_{0}^{u} \frac{k + 1}{n} s_{n, k + 1} (t) d t s_{n, k + 1} (u) \\ = & - n \sum_{k = 0}^{\infty} \int_{0}^{u} s_{n, k + 1} (t) d t u s_{n, k} (u) \\ = & - n \sum_{k = 0}^{\infty} [- \frac{1}{n} (s_{n, k + 1} (u) + \dots + s_{n, 0} (u)) + \frac{1}{n}] u s_{n, k} (u) \\ = & \sum_{k = 0}^{\infty} u s_{n, k} (u) \sum_{j = 0}^{k + 1} s_{n, j} (u) - \sum_{k = 0}^{\infty} u s_{n, k} (u) \\ = & \sum_{k = 0}^{\infty} u s_{n, k} (u) \sum_{j = 0}^{k} s_{n, j} (u) + u \sum_{k = 0}^{\infty} s_{n, k} (u) s_{n, k + 1} (u) - u . \end{array}

H_{n} (u) = u - u + 2 u \sum_{k = 0}^{\infty} s_{n, k} (u) \sum_{j = 0}^{k} s_{n, j} (u) + u \sum_{k = 0}^{\infty} s_{n, k} (u) s_{n, k + 1} (u) - u .

Since

\begin{array}{rcl} \sum_{k = 0}^{\infty} s_{n, k} (u) \sum_{j = 0}^{k} s_{n, j} (u) & = & \sum_{k = 0}^{\infty} s_{n, k} (u) \sum_{j = k}^{\infty} s_{n, j} (u) \\ = & \sum_{k = 0}^{\infty} s_{n, k} (u) \sum_{j = k + 1}^{\infty} s_{n, j} (u) + \sum_{k = 0}^{\infty} s_{n, k} (u) s_{n, k} (u), \end{array}

we can write

\begin{array}{rcl} H_{n} (u) & = & u (\sum_{k = 0}^{\infty} s_{n, k} (u) \sum_{j = 0}^{k} s_{n, j} (u) + \sum_{k = 0}^{\infty} s_{n, k} (u) \sum_{j = k + 1}^{\infty} s_{n, j} (u)) \\ + u \sum_{k = 0}^{\infty} s_{n, k} (u) (s_{n, k} (u) + s_{n, k + 1} (u)) - u \\ = & u \sum_{k = 0}^{\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) \leq \frac{1}{\sqrt{n u}},

we get

H_{n} (u) \leq 2 \frac{\sqrt{u}}{\sqrt{n}} .

Consequently

\begin{aligned} {∥ D_{n, α} (g, x) - g (x) ∥}_{1} & \leq 4 α \int_{0}^{\infty} | g^{'} (u) | \frac{φ (u)}{\sqrt{n}} d u \\ = \frac{4 α}{\sqrt{n}} {∥ φ^{'} g ∥}_{1} . \end{aligned}

(2.13)

By (2.8) and (2.13) we have

{∥ D_{n, α} (f, x) - f (x) ∥}_{1} \leq C ω_{φ} {(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)

ω_{φ}^{2} {(f, x)}_{p}

cannot be used instead of

ω_{φ} {(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, \infty)

(

1 \leq p \leq \infty

φ (x) = \sqrt{x}

δ_{n} (x) = φ (x) + \frac{1}{\sqrt{n}}

, we have

{∥ δ_{n} D_{n, α}^{'} (f) ∥}_{p} \leq C \sqrt{n} {∥ f ∥}_{p} .

(3.1)

Proof We will show (3.1) for the two cases of

p = \infty

and

p = 1

. Since

\begin{aligned} D_{n, α}^{'} (f, x) & = \sum_{k = 0}^{\infty} α [J_{n, k}^{α - 1} (x) J_{n, k}^{'} (x) - J_{n, k + 1}^{α - 1} (x) J_{n, k + 1}^{'} (x)] n \int_{0}^{\infty} s_{n, k} (t) f (t) d t \\ = α \sum_{k = 0}^{\infty} [(J_{n, k}^{α - 1} (x) - J_{n, k + 1}^{α - 1} (x)) J_{n, k + 1}^{'} (x) + J_{n, k}^{α - 1} (x) s_{n, k}^{'} (x)] n \int_{0}^{\infty} s_{n, k} (t) f (t) d t \end{aligned}

using

\int_{0}^{\infty} s_{n, k} (t) d t = \frac{1}{n}

, we have

\begin{aligned} | D_{n, α}^{'} (f, x) | & \leq α {∥ f ∥}_{\infty} (\sum_{k = 0}^{\infty} [J_{n, k}^{α - 1} (x) - J_{n, k + 1}^{α - 1} (x)] J_{n, k + 1}^{'} (x) + \sum_{k = 0}^{\infty} J_{n, k}^{α - 1} (x) | s_{n, k}^{'} (x) |) \\ = : α {∥ f ∥}_{\infty} (I_{1} + I_{2}) . \end{aligned}

(3.2)

For

x \in E_{n} = (\frac{1}{n}, \infty)

δ_{n} (x) \sim φ (x)

, by (2.1) and (2.3) we get

\begin{aligned} δ_{n} (x) I_{2} & \leq δ_{n} (x) \sum_{k = 0}^{\infty} | s_{n, k}^{'} (x) | \leq \frac{n δ_{n} (x)}{φ^{2} (x)} \sum_{k = 0}^{\infty} | \frac{k}{n} - x | s_{n, k} (x) \\ \leq \frac{n δ_{n} (x)}{φ^{2} (x)} {(\sum_{k = 0}^{\infty} {(\frac{k}{n} - x)}^{2} s_{n, k} (x))}^{\frac{1}{2}} \\ = \frac{n δ_{n} (x)}{φ^{2} (x)} \frac{φ (x)}{\sqrt{n}} \leq 2 \sqrt{n}, \end{aligned}

(3.3)

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

\sum_{k = 0}^{\infty} {(\frac{k}{n} - x)}^{2} s_{n, k} (x) = \frac{φ^{2} (x)}{n} .

For

x \in E_{n}^{c} = [0, \frac{1}{n}]

δ_{n} (x) \sim \frac{1}{\sqrt{n}}

, by (2.4) we have

δ_{n} (x) I_{2} \leq δ_{n} (x) \sum_{k = 0}^{\infty} n (s_{n, k - 1} (x) + s_{n, k} (x)) \leq 4 \sqrt{n} .

(3.4)

By (3.3) and (3.4), we get

δ_{n} (x) I_{2} \leq 6 \sqrt{n} .

(3.5)

Noting

J_{n, 0}^{'} (x) = 0

, we have

\begin{aligned} I_{1} & = \sum_{k = 0}^{\infty} (J_{n, k}^{α - 1} (x) (J_{n, k}^{'} (x) - s_{n, k}^{'} (x)) - J_{n, k + 1}^{α - 1} (x) J_{n, k + 1}^{'} (x)) \\ \leq \sum_{k = 0}^{\infty} J_{n, k}^{α - 1} (x) J_{n, k}^{'} (x) - \sum_{k = 0}^{\infty} J_{n, k + 1}^{α - 1} (x) J_{n, k + 1}^{'} (x) + \sum_{k = 0}^{\infty} J_{n, k}^{α - 1} (x) | s_{n, k}^{'} (x) | \\ = I_{2}, \end{aligned}

then

δ_{n} (x) I_{1} \leq 6 \sqrt{n} .

(3.6)

Combining (3.2), (3.5), and (3.6) we get for

p = \infty

{∥ δ_{n} D_{n, α}^{'} (f) ∥}_{\infty} \leq C \sqrt{n} {∥ f ∥}_{\infty} .

(3.7)

For

p = 1

, we have

\begin{aligned} | D_{n, α}^{'} (f) | & \leq α \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | f (t) | d t [(J_{n, k}^{α - 1} (x) - J_{n, k + 1}^{α - 1} (x)) J_{n, k + 1}^{'} (x) + J_{n, k}^{α - 1} (x) | s_{n, k}^{'} (x) |] \\ = : α ({\tilde{J}}_{1} + {\tilde{J}}_{2}) . \end{aligned}

(3.8)

Hence we can write

\int_{0}^{\infty} | δ_{n} (x) D_{n, α}^{'} (f, x) | d x \leq α \int_{0}^{\infty} δ_{n} (x) ({\tilde{J}}_{1} + {\tilde{J}}_{2}) d x = α (\int_{E_{n}^{c}} + \int_{E_{n}}) δ_{n} (x) ({\tilde{J}}_{1} + {\tilde{J}}_{2}) d x .

(3.9)

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

\begin{array}{rcl} \int_{E_{n}^{c}} δ_{n} (x) {\tilde{J}}_{2} d x & \leq & \int_{E_{n}^{c}} δ_{n} (x) \sum_{k = 1}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | f (t) | d t n (s_{n, k - 1} (x) + s_{n, k} (x)) d x \\ + \int_{E_{n}^{c}} δ_{n} (x) n \int_{0}^{\infty} s_{n, 0} (t) | f (t) | d t n s_{n, 0} (x) d x . \end{array}

For

x \in E_{n}^{c}

δ_{n} (x) \leq \frac{2}{\sqrt{n}}

s_{n, 0} (t) \leq 1

, noting

\int_{0}^{\infty} s_{n, k} (x) d x = \frac{1}{n}

, we have

\int_{E_{n}^{c}} δ_{n} (x) {\tilde{J}}_{2} d x \leq \frac{4 n}{\sqrt{n}} {∥ f ∥}_{1} + \frac{2 n}{\sqrt{n}} {∥ f ∥}_{1} \leq 6 \sqrt{n} {∥ f ∥}_{1} .

(3.10)

Since

J_{n, k}^{α - 1} (x) - J_{n, k + 1}^{α - 1} (x) \leq 1

and

J_{n, k + 1}^{'} (x) = n s_{n, k} (x)

, we have

\int_{E_{n}^{c}} δ_{n} (x) {\tilde{J}}_{1} d x \leq \int_{E_{n}^{c}} δ_{n} (x) \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | f (t) | d t n s_{n, k} (x) d x \leq 2 \sqrt{n} {∥ f ∥}_{1} .

(3.11)

To estimate

\int_{E_{n}} δ_{n} (x) {\tilde{J}}_{2} d x

, we will need the relation [[9], p.129, (9.4.15)]

\int_{E_{n}} \frac{{(\frac{k}{n} - x)}^{2}}{φ^{2} (x)} s_{n, k} (x) d x \leq C n^{- 2} .

By the Hölder inequality and (2.3), we get

\begin{aligned} \int_{E_{n}} δ_{n} (x) {\tilde{J}}_{2} d x \\ \leq 2 \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | f (t) | d t \int_{E_{n}} φ (x) \frac{n}{φ^{2} (x)} | \frac{k}{n} - x | s_{n, k} (x) d x \\ \leq 2 \sum_{k = 0}^{\infty} n^{2} \int_{0}^{\infty} s_{n, k} (t) | f (t) | d t {(\int_{E_{n}} \frac{{(\frac{k}{n} - x)}^{2}}{φ^{2} (x)} s_{n, k} (x) d x)}^{\frac{1}{2}} {(\int_{E_{n}} s_{n, k} (x) d x)}^{\frac{1}{2}} \\ \leq C \sqrt{n} \sum_{k = 0}^{\infty} \int_{0}^{\infty} s_{n, k} (t) | f (t) | d t \\ = C \sqrt{n} {∥ f ∥}_{1} . \end{aligned}

(3.12)

In order to estimate

\int_{E_{n}} δ_{n} (x) {\tilde{J}}_{1} d x

, we consider the two cases of

α \geq 2

and

1 < α < 2

(when

α = 1

{\tilde{J}}_{1} = 0

For

α \geq 2

J_{n, k}^{α - 1} (x) - J_{n, k + 1}^{α - 1} (x) \leq (α - 1) s_{n, k} (x)

. Using integration by parts, we can deduce

\begin{array}{rcl} \int_{E_{n}} δ_{n} (x) {\tilde{J}}_{1} d x & \leq & C \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | f (t) | d t \int_{E_{n}} φ (x) s_{n, k} (x) J_{n, k + 1}^{'} (x) d x \\ = & C \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | f (t) | d t \\ \times (φ (x) s_{n, k} (x) J_{n, k + 1} (x) |_{\frac{1}{n}}^{\infty} - \int_{\frac{1}{n}}^{\infty} J_{n, k + 1} (x) d (φ (x) s_{n, k} (x))) \\ = & C \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | f (t) | d t φ (x) s_{n, k} (x) J_{n, k + 1} (x) |_{\frac{1}{n}}^{\infty} \\ - C \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | f (t) | d t \\ \times (\int_{\frac{1}{n}}^{\infty} J_{n, k + 1} (x) \frac{1}{2 \sqrt{x}} s_{n, k} (x) d x + \int_{\frac{1}{n}}^{\infty} φ (x) s_{n, k}^{'} (x) J_{n, k + 1} (x) d x) . \end{array}

Noting that

φ (x) s_{n, k} (x) J_{n, k + 1} (x) |_{\frac{1}{n}}^{\infty} < 0

and

\int_{\frac{1}{n}}^{\infty} J_{n, k + 1} (x) \frac{1}{2 \sqrt{x}} s_{n, k} (x) d x > 0

, and from (2.3), we have

\begin{aligned} \int_{E_{n}} δ_{n} (x) {\tilde{J}}_{1} d x \\ \leq C \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | f (t) | d t \int_{\frac{1}{n}}^{\infty} | φ (x) s_{n, k}^{'} (x) J_{n, k + 1} (x) | d x \\ \leq C \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | f (t) | d t n {(\int_{E_{n}} \frac{{(\frac{k}{n} - x)}^{2}}{φ^{2} (x)} s_{n, k} (x) d x)}^{\frac{1}{2}} {(\int_{E_{n}} s_{n, k} (x) d x)}^{\frac{1}{2}} \\ \leq C \sqrt{n} {∥ f ∥}_{1} . \end{aligned}

(3.13)

For

1 < α < 2

, using the mean value theorem, we know

J_{n, k}^{α - 1} (x) - J_{n, k + 1}^{α - 1} (x) = (α - 1) {(ξ_{k} (x))}^{α - 2} s_{n, k} (x),

where

J_{n, k + 1} (x) < ξ_{k} (x) < J_{n, k} (x)

and

α - 2 < 0

, then

J_{n, k}^{α - 1} (x) - J_{n, k + 1}^{α - 1} (x) \leq (α - 1) J_{n, k + 1}^{α - 2} (x) s_{n, k} (x) .

For

1 < α < 2

, we get from the procedure of (3.13)

\begin{aligned} \int_{E_{n}} δ_{n} (x) {\tilde{J}}_{1} d x & \leq \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | f (t) | d t \int_{E_{n}} (α - 1) φ (x) s_{n, k} (x) J_{n, k + 1}^{α - 2} (x) J_{n, k + 1}^{'} (x) d x \\ = \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | f (t) | d t \int_{E_{n}} φ (x) s_{n, k} (x) d J_{n, k + 1}^{α - 1} (x) \\ \leq \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | f (t) | d t \int_{E_{n}} φ (x) | s_{n, k}^{'} (x) | d x \\ \leq C \sqrt{n} {∥ f ∥}_{1} . \end{aligned}

(3.14)

Combining (3.13) and (3.14), we get for

α \geq 1

\int_{E_{n}} δ_{n} (x) {\tilde{J}}_{1} d x \leq C \sqrt{n} {∥ f ∥}_{1} .

(3.15)

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

{∥ δ_{n} D_{n, α}^{'} (f) ∥}_{1} \leq C \sqrt{n} {∥ f ∥}_{1} .

(3.16)

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

Lemma 3.2 For

f \in W_{p}

φ (x) = \sqrt{x}

δ_{n} (x) = φ (x) + \frac{1}{\sqrt{n}}

, we have

{∥ δ_{n} D_{n, α}^{'} (f) ∥}_{p} \leq C {∥ δ_{n} f^{'} ∥}_{p} .

(3.17)

Proof By the Riesz-Thorin theorem, we shall prove Lemma 3.2 for

p = \infty

and

p = 1

. For

f \in W_{p}

and noting that

D_{n, α}^{'} (1, x) = 0

, we have

\sum_{k = 0}^{\infty} n \int_{0}^{\infty} f (x) s_{n, k} (t) d t {(J_{n, k}^{α} (x) - J_{n, k + 1}^{α} (x))}^{'} = 0 .

Then

\begin{aligned} | D_{n, α}^{'} (f, x) | = & | \sum_{k = 0}^{\infty} n \int_{0}^{\infty} (f (t) - f (x)) s_{n, k} (t) d t {(J_{n, k}^{α} (x) - J_{n, k + 1}^{α} (x))}^{'} | \\ = & | \sum_{k = 0}^{\infty} n \int_{0}^{\infty} \int_{x}^{t} f^{'} (u) d u s_{n, k} (t) d t {(J_{n, k}^{α} (x) - J_{n, k + 1}^{α} (x))}^{'} | \\ \leq & α \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | \int_{x}^{t} f^{'} (u) d u | d t \\ \times {[J_{n, k}^{α - 1} (x) - J_{n, k + 1}^{α - 1} (x)] J_{n, k + 1}^{'} (x) + J_{n, k}^{α - 1} (x) | s_{n, k}^{'} (x) |} . \end{aligned}

(3.18)

By (2.9) and (2.10) we have

| \int_{x}^{t} δ_{n} (u) d u | \leq C δ_{n}^{- 1} (x) | t - x |,

hence

\begin{aligned} | δ_{n} (x) D_{n, α}^{'} (f, x) | \leq & C {∥ δ_{n} f^{'} ∥}_{\infty} \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | t - x | d t \\ \times {[J_{n, k}^{α - 1} (x) - J_{n, k + 1}^{α - 1} (x)] J_{n, k + 1}^{'} (x) + J_{n, k}^{α - 1} (x) | s_{n, k}^{'} (x) |} \\ = : & C {∥ δ_{n} f^{'} ∥}_{\infty} (J_{1} + J_{2}) . \end{aligned}

(3.19)

For

x \in E_{n}^{c}

δ_{n} (x) \sim \frac{1}{\sqrt{n}}

and by (2.1) and (2.4) we have

\begin{array}{rcl} J_{2} & = & \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | t - x | d t J_{n, k}^{α - 1} (x) | s_{n, k}^{'} (x) | \\ \leq & n \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | t - x | d t (s_{n, k - 1} (x) + s_{n, k} (x)) \\ = & n \sum_{k = 1}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | t - x | d t s_{n, k - 1} (x) + n \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | t - x | d t s_{n, k} (x) \\ = : & K_{1} + K_{2} . \end{array}

By (2.6) we get

K_{2} = n D_{n, 1} (| t - x |, x) \leq n {(D_{n, 1} ({(t - x)}^{2}, x))}^{\frac{1}{2}} \leq \sqrt{n} δ_{n} (x) \leq 2 .

For

K_{1}

, we write

K_{1} = n \sum_{k = 2}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | t - x | d t s_{n, k - 1} (x) + n^{2} \int_{0}^{\infty} s_{n, 1} (t) | t - x | d t s_{n, 0} (x) .

First, using (2.6),

\int_{0}^{\infty} s_{n, k} (t) d t = \frac{1}{n}

, and the Hölder inequality, we have

\begin{aligned} n \sum_{k = 2}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | t - x | d t s_{n, k - 1} (x) \\ = n \sum_{k = 2}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | t - x | d t s_{n, k} (x) \frac{k}{n x} \\ \leq n {(\sum_{k = 2}^{\infty} n^{2} {(\int_{0}^{\infty} s_{n, k} (t) | t - x | d t)}^{2} s_{n, k} (x))}^{\frac{1}{2}} {(\sum_{k = 2}^{\infty} s_{n, k} (x) \frac{k^{2}}{{(n x)}^{2}})}^{\frac{1}{2}} \\ \leq n {(\sum_{k = 2}^{\infty} n^{2} \int_{0}^{\infty} s_{n, k} (t) d t \cdot \int_{0}^{\infty} s_{n, k} (t) {(t - x)}^{2} d t s_{n, k} (x))}^{\frac{1}{2}} {(\sum_{k = 2}^{\infty} s_{n, k} (x) \frac{k^{2}}{{(n x)}^{2}})}^{\frac{1}{2}} \\ \leq C n {(\sum_{k = 2}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) {(t - x)}^{2} d t s_{n, k} (x))}^{\frac{1}{2}} {(\sum_{k = 2}^{\infty} s_{n, k - 2} (x))}^{\frac{1}{2}} \\ \leq C n \cdot \frac{δ_{n} (x)}{\sqrt{n}} \\ \leq C . \end{aligned}

Next, for

x \in E_{n}^{c}

e^{- n x} \leq 1

, and

\int_{0}^{\infty} s_{n, k} (t) d t = \frac{1}{n}

, we have

\begin{array}{rcl} n^{2} \int_{0}^{\infty} s_{n, 1} (t) | t - x | d t s_{n, 0} (x) & \leq & n^{2} \int_{0}^{\infty} s_{n, 1} (t) (t + x) d t e^{- n x} \\ \leq & n^{2} (\frac{2}{n} \int_{0}^{\infty} s_{n, 2} (t) d t + x \int_{0}^{\infty} s_{n, 1} (t) d t) \\ = & n^{2} (\frac{2}{n^{2}} + \frac{x}{n}) \\ \leq & 3 . \end{array}

Then we get

J_{2} \leq C, x \in E_{n}^{c} .

(3.20)

Noting that

J_{n, k}^{α - 1} (x) - J_{n, k + 1}^{α - 1} (x) \leq 1

, by (2.5) we have

J_{1} \leq K_{2} \leq 2 .

(3.21)

For

x \in E_{n}

δ_{n} (x) \sim φ (x)

, using (2.3),

\int_{0}^{\infty} s_{n, k} (t) d t = \frac{1}{n}

and the Hölder inequality, we have

\begin{aligned} J_{2} & \leq \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | t - x | d t \frac{n}{φ^{2} (x)} | \frac{k}{n} - x | s_{n, k} (x) \\ \leq {(\sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) {(t - x)}^{2} d t s_{n, k} (x))}^{\frac{1}{2}} {(\sum_{k = 0}^{\infty} {(\frac{k}{n} - x)}^{2} s_{n, k} (x))}^{\frac{1}{2}} \cdot \frac{n}{φ^{2} (x)} \\ \leq \frac{δ_{n} (x)}{\sqrt{n}} \cdot \frac{φ (x)}{\sqrt{n}} \cdot \frac{n}{φ^{2} (x)} \\ \leq 2 . \end{aligned}

Noting that

J_{n, 0}^{'} (x) = 0

, one has

\begin{array}{rcl} J_{1} & = & \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | t - x | d t [J_{n, k}^{α - 1} (x) - J_{n, k + 1}^{α - 1} (x)] J_{n, k + 1}^{'} (x) \\ \leq & \sum_{k = 1}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | t - x | d t J_{n, k}^{α - 1} (x) J_{n, k}^{'} (x) - \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | t - x | d t J_{n, k + 1}^{α - 1} (x) J_{n, k + 1}^{'} (x) \\ + \sum_{k = 0}^{\infty} n \int_{0}^{\infty} s_{n, k} (t) | t - x | d t J_{n, k}^{α - 1} (x) | s_{n, k}^{'} (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} & \leq & \sum_{k = 1}^{\infty} n \int_{0}^{\infty} (s_{n, k} (t) - s_{n, k - 1} (t)) | t - x | d t J_{n, k}^{α - 1} (x) J_{n, k}^{'} (x) \\ \leq & \sum_{k = 1}^{\infty} n | \int_{0}^{\infty} \frac{1}{n} s_{n, k}^{'} (t) | t - x | d t | n s_{n, k - 1} (x) \\ \leq & \sum_{k = 1}^{\infty} (| \int_{0}^{x} s_{n, k}^{'} (t) (x - t) d t | + | \int_{x}^{\infty} s_{n, k}^{'} (t) (t - x) d t |) n s_{n, k - 1} (x) \\ = & \sum_{k = 1}^{\infty} (| \int_{0}^{x} (x - t) d s_{n, k} (t) | + | \int_{x}^{\infty} (t - x) d s_{n, k} (t) |) n s_{n, k - 1} (x) \\ \leq & \sum_{k = 1}^{\infty} \int_{0}^{\infty} s_{n, k} (t) d t n s_{n, k - 1} (x) \\ = & 1 . \end{array}

Thus

J_{1} \leq J_{2} + J_{3} \leq 3 .

(3.22)

So we get

{∥ δ_{n} D_{n, α}^{'} (f) ∥}_{\infty} \leq C {∥ δ_{n} f^{'} ∥}_{\infty} .

(3.23)

For

p = 1

, using (2.1), (2.4), (2.5), and integration by parts, we have

\begin{array}{rcl} D_{n, α}^{'} (f, x) & = & α \sum_{k = 1}^{\infty} n \int_{0}^{\infty} f (t) s_{n, k} (t) d t J_{n, k}^{α - 1} (x) J_{n, k}^{'} (x) \\ - α \sum_{k = 0}^{\infty} n \int_{0}^{\infty} f (t) s_{n, k} (t) d t J_{n, k + 1}^{α - 1} (x) J_{n, k + 1}^{'} (x) \\ = & α \sum_{k = 1}^{\infty} n \int_{0}^{\infty} f (t) (s_{n, k} (t) - s_{n, k - 1} (t)) d t J_{n, k}^{α - 1} (x) J_{n, k}^{'} (x) \\ = & - α \sum_{k = 1}^{\infty} \int_{0}^{\infty} f (t) s_{n, k}^{'} (t) d t J_{n, k}^{α - 1} (x) J_{n, k}^{'} (x) \\ = & α \sum_{k = 1}^{\infty} \int_{0}^{\infty} f^{'} (t) s_{n, k} (t) d t J_{n, k}^{α - 1} (x) J_{n, k}^{'} (x) \\ \leq & α \sum_{k = 1}^{\infty} \int_{0}^{\infty} f^{'} (t) s_{n, k} (t) d t n s_{n, k - 1} (x) . \end{array}

Let

\begin{aligned} {∥ δ_{n} D_{n, α}^{'} (f) ∥}_{1} & = {∥ δ_{n} D_{n, α}^{'} (f) ∥}_{1}^{E_{n}^{c}} + {∥ δ_{n} D_{n, α}^{'} (f) ∥}_{1}^{E_{n}} \\ = : \tilde{K_{1}} + \tilde{K_{2}} . \end{aligned}

(3.24)

For

x \in E_{n}^{c}

, noting that

\sqrt{n} δ_{n} (t) \geq 1

and

\sqrt{n} δ_{n} (x) \leq 2

, we have

\begin{aligned} \tilde{K_{1}} & \leq α \int_{E_{n}^{c}} δ_{n} (x) \sum_{k = 1}^{\infty} \int_{0}^{\infty} | f^{'} (t) | s_{n, k} (t) d t n s_{n, k - 1} (x) d x \\ \leq α \int_{E_{n}^{c}} \sqrt{n} δ_{n} (x) \sum_{k = 1}^{\infty} \int_{0}^{\infty} | δ_{n} (t) f^{'} (t) | s_{n, k} (t) d t n s_{n, k - 1} (x) d x \\ \leq 2 α \sum_{k = 1}^{\infty} \int_{0}^{\infty} | δ_{n} (t) f^{'} (t) | s_{n, k} (t) d t n \int_{E_{n}^{c}} s_{n, k - 1} (x) d x \\ \leq 2 α {∥ δ_{n} f^{'} ∥}_{1} . \end{aligned}

(3.25)

For

x \in E_{n}

, we estimate

\tilde{K_{2}}

\begin{array}{rcl} \tilde{K_{2}} & \leq & 2 α \int_{E_{n}} φ (x) \sum_{k = 1}^{\infty} \int_{0}^{\infty} \frac{| δ_{n} (t) f^{'} (t) |}{φ (t)} s_{n, k} (t) d t n s_{n, k - 1} (x) d x \\ = & 2 n α \sum_{k = 1}^{\infty} \int_{0}^{\infty} \frac{| δ_{n} (t) f^{'} (t) |}{φ (t)} s_{n, k} (t) d t \int_{E_{n}} φ (x) s_{n, k - 1} (x) d x . \end{array}

Using the Hölder inequality and

2 \sqrt{a b} \leq a + b

(

a, b > 0

), we have

\begin{aligned} \int_{0}^{\infty} \frac{| δ_{n} (t) f^{'} (t) |}{φ (t)} s_{n, k} (t) d t \\ \leq {(\int_{0}^{\infty} | δ_{n} (t) f^{'} (t) | s_{n, k} (t) d t \int_{0}^{\infty} \frac{| δ_{n} (t) f^{'} (t) |}{φ^{2} (t)} s_{n, k} (t) d t)}^{\frac{1}{2}} \\ = {(\int_{0}^{\infty} | δ_{n} (t) f^{'} (t) | s_{n, k} (t) d t \int_{0}^{\infty} | δ_{n} (t) f^{'} (t) | \frac{n}{k} s_{n, k - 1} (t) d t)}^{\frac{1}{2}} \\ \leq {(\frac{n}{k})}^{\frac{1}{2}} (\int_{0}^{\infty} | δ_{n} (t) f^{'} (t) | s_{n, k} (t) d t + \int_{0}^{\infty} | δ_{n} (t) f^{'} (t) | s_{n, k - 1} (t) d t) \end{aligned}

and

\begin{array}{rcl} \int_{E_{n}} φ (x) s_{n, k - 1} (x) d x & \leq & \frac{1}{\sqrt{n}} {(\int_{E_{n}} φ^{2} (x) s_{n, k - 1} (x) d x)}^{\frac{1}{2}} \\ = & \frac{1}{\sqrt{n}} {(\frac{k}{n})}^{\frac{1}{2}} {(\int_{E_{n}} s_{n, k} (x) d x)}^{\frac{1}{2}} \\ \leq & \frac{1}{n} {(\frac{k}{n})}^{\frac{1}{2}} . \end{array}

Therefore we have

\begin{aligned} \tilde{K_{2}} & \leq 2 α \sum_{k = 1}^{\infty} (\int_{0}^{\infty} | δ_{n} (t) f^{'} (t) | s_{n, k} (t) d t + \int_{0}^{\infty} | δ_{n} (t) f^{'} (t) | s_{n, k - 1} (t) d t) \\ \leq 2 α {∥ δ_{n} f^{'} ∥}_{1} . \end{aligned}

(3.26)

By (3.24)-(3.26) we have

{∥ δ_{n} D_{n, α}^{'} (f) ∥}_{1} \leq C {∥ δ_{n} f^{'} ∥}_{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, \infty)

(

1 \leq p \leq \infty

φ (x) = \sqrt{x}

0 < β < 1

{∥ D_{n, α} (f, x) - f (x) ∥}_{p} = O (n^{- \frac{β}{2}})

implies

ω_{φ} {(f, t)}_{p} = O (t^{β})

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

\begin{aligned} K_{φ} {(f, t)}_{p} & \leq {∥ f - D_{n, α} (f) ∥}_{p} + t {∥ δ_{n} D_{n, α}^{'} (f) ∥}_{p} \\ \leq C n^{- \frac{β}{2}} + t ({∥ δ_{n} D_{n, α}^{'} (f - g) ∥}_{p} + {∥ δ_{n} D_{n, α}^{'} (g) ∥}_{p}) \\ \leq C n^{- \frac{β}{2}} + C t (\sqrt{n} {∥ f - g ∥}_{p} + {∥ δ_{n} g^{'} ∥}_{p}) \\ \leq C n^{- \frac{β}{2}} + C t \sqrt{n} ({∥ f - g ∥}_{p} + \frac{1}{\sqrt{n}} {∥ φ g^{'} ∥}_{p} + \frac{1}{n} {∥ g^{'} ∥}_{p}) \\ \leq C (n^{- \frac{β}{2}} + \frac{t}{n^{- \frac{1}{2}}} {\bar{K}}_{φ} {(f, n^{- \frac{1}{2}})}_{p}) \\ \leq C (n^{- \frac{β}{2}} + \frac{t}{n^{- \frac{1}{2}}} K_{φ} {(f, n^{- \frac{1}{2}})}_{p}) \end{aligned}

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

K_{φ} {(f, t)}_{p} = O (t^{β}) .

(3.28)

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

Acknowledgements

Xiuzhong Yang was supported by the National Natural Science Foundation of China (grant No. 11371119) and both authors were supported by the Natural Science Foundation of Education Department of Hebei Province (grant No. Z2014031). The authors express their thanks to the referees for their constructive suggestions in improving the quality of the paper.

Open AccessThis article is distributed under the terms of the Creative Commons Attribution 4.0 International License (https://creativecommons.org/licenses/by/4.0), which permits use, duplication, adaptation, distribution, and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

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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Titel: On the approximation for generalized Szász-Durrmeyer type operators in the space
verfasst von: Guofen Liu
Xiuzhong Yang
Publikationsdatum: 01.12.2014
Verlag: Springer International Publishing
Erschienen in: Journal of Inequalities and Applications / Ausgabe 1/2014
Elektronische ISSN: 1029-242X
DOI: https://doi.org/10.1186/1029-242X-2014-447

Springer Professional

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

Abstract

Competing interests

Authors’ contributions

1 Introduction

2 Direct theorem

3 Inverse theorem

Acknowledgements

Competing interests

Authors’ contributions

Premium Partner

Springer Professional

Abstract

Competing interests

Authors’ contributions

1 Introduction

2 Direct theorem

3 Inverse theorem

Acknowledgements

Competing interests

Authors’ contributions

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