Approximation properties for generalized Baskakov-type operators

Abstract

In this paper, we give a generalization of the Baskakov-type operators introduced by Baskakov (Doklady Akademii Nauk SSSR 113:249–251, 1957 (in Russian)) and obtain some direct and inverse results for these new operators.

MSC

41A35, 41A36

Introduction

For a continuous function f on [0,∞) with exponential growth, the Szász operators are given by

$$S_n(f;x)=\sum _{k=0}^{\infty }f(\frac{k}{n})P_{k,n}(x),$$

where $P_{k,n}(x)=e^{-\mathit{\text{nx}}}\frac{{(\mathit{\text{nx}})}^k}{k!}$. In [1], Ditzian proved some important inverse theorems for these operators by using the modulus of continuity defined by

$$\begin{array}{l}{\omega }_2(f;\delta ,A)=\underset{h\le \delta ,x\in \left[0,\infty \right)}{sup}\left|f(x)-2f(x+h)+f(x+2h)\right|e^{-\mathit{\text{Ax}}}\\ =\underset{h\le \delta ,x\in \left[0,\infty \right)}{sup}\left|{\Delta }_h^{2}f(x)\right|e^{-\mathit{\text{Ax}}},\end{array}$$

where supx∈[0,∞)| f(x)e^-Ax| < M.

In 1992, Guo and Zhou [2] gave similar theorems for the following modified Szász operators defined by Mazhar and Totik in [3]:

$$L_n(f;x)=\sum _{k=0}^{\infty }\left(n\underset{0}{\overset{\infty }{\int }}f(t)P_{k,n}(t)\mathit{\text{dt}}\right)P_{k,n}(x).$$

(1)

The authors obtained the following results for these operators:

• (1) Let f ∈ C[0,∞) be a bounded function. Then, for 0 < α < 1,

  $$\left|L_n(f;x)-f(x)\right|\le M{(\frac{x}{n}+\frac{1}{n^2})}^{\alpha /2}$$

holds if and only if

$${\omega }_1(f;\delta )=O({\delta }^{\alpha }),(\delta >0),$$

where

$${\omega }_1(f;\delta )=\underset{0\le h\le \delta }{sup}\underset{x\ge h/2}{sup}\left|f(x+h/2)-f(x-h/2)\right|.$$

(2)

• (2) For f ∈ C[0,∞)∩L _∞ [0,∞) and 0 < α < 1,

  $$\begin{array}{l}{\omega }_1(f;\delta )=O({\delta }^{\alpha })\iff \left|L_n^{\prime }(f;x)\right|\le M{\left(min\left\{n^2,n/x\right\}\right)}^{(1-\alpha )/2}\\ {\omega }_2(f;\delta )=O({\delta }^{\alpha })\iff \left|L_n^{\mathrm{\prime \prime }}(f;x)\right|\le M{\left(min\left\{n^2,n/x\right\}\right)}^{(2-\alpha )/2}\end{array}$$

holds, where ω₁(f;δ) is defined by (2) and

$${\omega }_2(f;\delta )=\underset{0<h\le \delta ,x\in \left[0,\infty \right)}{sup}\left|f(x)-2f(x+h)+f(x+2h)\right|\text{.}$$

(3)

• (3) For f∈C _B [0,∞),

  $${∥L_{n}f-f∥}_{\infty }\le C\left({\omega }_{\phi }^2{(f;\frac{1}{\sqrt{n}})}_{{}_{\infty }}+{\omega }_1(f;\frac{1}{n})+\frac{1}{n}{∥f∥}_{\infty }\right)$$

holds, where C is a constant independent of n, and ${\omega }_{\phi }^2{(f;.)}_{{}_{\infty }}$ is the Ditzian-Totik modulus of smoothness [4] defined by

$$\begin{array}{ll}\hfill {\omega }_{\phi }^2{(f;\delta )}_{{}_{\infty }}& =\underset{0<\mathit{h}\le \delta ,x\in \left[0,\infty \right)}{sup}{∥f(x-\mathit{h}\mathit{\phi }(x))-2f(x)+f(x+\mathit{h}\mathit{\phi }(x))∥}_{\infty },\hfill \\ \mathit{x}\ge \mathit{h},\mathit{\phi }(\mathit{x})=\sqrt{\mathit{x}}\text{.}\end{array}$$

In [5], Baskakov introduced the following sequence of linear operators $\left\{{\mathcal{ℒ}}_n\right\}$ which are generalizations of Bernstein polynomials, Szász operators, and Lupas operators:

$$\begin{array}{c}{\mathcal{ℒ}}_n(f;x)=\sum _{k=0}^{\infty }\frac{{(-x)}^k}{k!}{\phi }_n^{(k)}(x)f(\frac{k}{n})\text{.}\end{array}$$

(4)

Here, $\mathit{x}\in \left[0,b\right)\subset ℝ(b>0$, b can be ∞), $n\in \mathbb{N}$, and ${\left\{{\phi }_n\right\}}_{n\in \mathbb{N}}$ is a sequence of functions defined on [0,b] that have the following properties for all $k,n\in \mathbb{N}:$

• (a) φ _n is analytic on the interval [0,b] including the end points,

• (b) φ _n (0) = 1,

• (c) φ _n is completely monotone, i.e. ${(-1)}^k{\phi }_n^{(k)}(x)\ge 0$,

• (d) there exists a positive integer m ₀ = m ₀(n), such that

  $$-{\phi }_n^{(k)}(x)=n{\phi }_{n+m_0}^{(k-1)}(x)(k=1,2,\dots ),$$
• (e)
  $$\underset{n\to \infty }{\text{lim}}\frac{n}{m_0+n}=1$$

  .

For the operators ${\mathcal{ℒ}}_n(f;x)$ given by (4), we have (see [5]):

$${\mathcal{ℒ}}_n(1;x)=1,$$

(5)

$${\mathcal{ℒ}}_n(t;x)=x,$$

(6)

$${\mathcal{ℒ}}_n(t^2;x)=\frac{n(m_0+n)}{n^2}{x}^2+\frac{x}{n},$$

(7)

and

$${\mathcal{ℒ}}_n({(t-\mathit{x})}^2;\mathit{x})=\frac{m_0}{n}{\mathit{x}}^2+\frac{1}{n}\mathit{x}\text{.}$$

In the present paper, inspired by the operators (1) and (4), we introduce a generalization of the Baskakov operators as follows:

$${\mathcal{ℒ}}_n^{\ast }(f;x)=\sum _{k=0}^{\infty }\left(\frac{1}{{\gamma }_n}\underset{0}{\overset{\infty }{\int }}f\left(t\right)\frac{{\left(-t\right)}^k}{k!}{\phi }_n^{\left(k\right)}\left(t\right)\mathit{\text{dt}}\right)\frac{{\left(-x\right)}^k}{k!}{\phi }_n^{\left(k\right)}\left(x\right),$$

(8)

where ${\gamma }_n=\underset{0}{\overset{\infty }{\int }}{\phi }_n\left(t\right)\mathit{\text{dt}}<\infty$ for all $n\in \mathbb{N}$ and φ _n also satisfies the following condition:

$$\underset{x\to \infty }{\text{lim}}{\mathit{x}}^k{\phi }_n^{\left(\mathit{k}-1\right)}\left(\mathit{x}\right)=0,k,n\in \mathbb{N}$$

and

$$\underset{n\to \infty }{\text{lim}}\frac{{\gamma }_{n-\upsilon m_0}}{{\gamma }_n}=1,\upsilon =1,2,3\text{.}$$

In this study, we shall give some direct and inverse results for the new operators defined by (8).

Note that if φ _n (x) = e^-nxin (8), we get the operators L _n (f;x) defined by (1). Also, very important results were obtained by Mazhar and Totik in [3]. Recently, integral-type modification of some operators based on q-integers have been studied by Gupta et al. [6], Gupta and Kim [7], and Kim [8].

Main results

Now, we give the following lemma which will be used for the proof of theorems:

Lemma 1. The following equalities hold:

$$\begin{array}{ll}\hfill {\mathcal{ℒ}}_n^{\ast }\left(1,x\right)& =1,\hfill \\ \hfill {\mathcal{ℒ}}_n^{\ast }\left(t,x\right)& =\frac{{\gamma }_{n-m_0}}{{\gamma }_n}\left(\frac{\mathit{\text{nx}}}{n-m_0}+\frac{1}{n-m_0}\right),\hfill \\ \hfill {\mathcal{ℒ}}_n^{\ast }\left(t^2,x\right)& =\frac{{\gamma }_{n-2m_0}}{{\gamma }_n}\left[\frac{n\left(m_0+n\right)}{\left(n-2m_0\right)\left(n-m_0\right)}{x}^2\right.\hfill \\ +\left.\frac{4\mathit{\text{nx}}}{\left(n-2m_0\right)\left(n-m_0\right)}+\frac{2}{\left(n-2m_0\right)\left(n-m_0\right)}\right]\text{.}\hfill \end{array}$$

From the definition of operators ${\mathcal{ℒ}}_n^{\ast }$ and Lemma 1, we have

$$\begin{array}{ll}\hfill {\mathcal{ℒ}}_n^{\ast }\left(\left(t-x\right){}^2,x\right)& =\left(\frac{{\gamma }_{n-2m_0}}{{\gamma }_n}\frac{n\left(m_0+n\right)}{\left(n-2m_0\right)\left(n-m_0\right)}-\frac{2n}{n-m_0}\frac{{\gamma }_{n-m_0}}{{\gamma }_n}+1\right)x^2\\ +\left(\frac{4n}{\left(n-2m_0\right)\left(n-m_0\right)}\frac{{\gamma }_{n-2m_0}}{{\gamma }_n}-\frac{2}{n-m_0}\frac{{\gamma }_{n-m_0}}{{\gamma }_n}\right)x\hfill \\ +\frac{2}{\left(n-2m_0\right)\left(n-m_0\right)}\frac{{\gamma }_{n-2m_0}}{{\gamma }_n}\text{.}\hfill \end{array}$$

(9)

Theorem 2. Let f ∈ C[0,∞)be a bounded function, and 0 < α ≤ 1.If the usual modulus of smoothness ω₁(f,t)defined by (2) satisfies

$${\omega }_1\left(f,t\right)=O\left(t^{\alpha }\right)\left(t>0\right),$$

(10)

then

$$\begin{array}{ll}\hfill \left|{\mathcal{ℒ}}_n^{\ast }\left(f,x\right)-f\left(x\right)\right|\le K& \left(\left(\frac{{\gamma }_{n-2m_0}}{{\gamma }_n}\frac{n\left(m_0+n\right)}{\left(n-2m_0\right)\left(n-m_0\right)}-\frac{2n}{n-m_0}\frac{{\gamma }_{n-m_0}}{{\gamma }_n}+1\right)x^2\right.\hfill \\ +\left(\frac{4n}{\left(n-2m_0\right)\left(n-m_0\right)}\frac{{\gamma }_{n-2m_0}}{{\gamma }_n}-\frac{2}{n-m_0}\frac{{\gamma }_{n-m_0}}{{\gamma }_n}\right)x\hfill \\ {\left.+\frac{2}{\left(n-2m_0\right)\left(n-m_0\right)}\frac{{\gamma }_{n-2m_0}}{{\gamma }_n}\right)}^{\alpha /2}\hfill \end{array}$$

holds.

Proof. Using the definition of the operators ${\mathcal{ℒ}}_n^{\ast }$ and the equality (9), we obtain the following inequality:

$$\begin{array}{l}\left|{\mathcal{ℒ}}_n^{\ast }\left(f,x\right)-f(x)\right|\le \sum _{k=0}^{\infty }\left(\frac{1}{\underset{0}{\overset{\infty }{\int }}{\phi }_n(t)\mathit{\text{dt}}}\underset{0}{\overset{\infty }{\int }}\left|f(t)-f(x)\right|\frac{{\left(-t\right)}^k}{k!}\right.\\ \left.{\phi }_n^{(k)}(t)\mathit{\text{dt}}\right)\frac{{\left(-x\right)}^k}{k!}{\phi }_n^{(k)}\left(x\right)\\ \le \sum _{k=0}^{\infty }\left(\frac{1}{\underset{0}{\overset{\infty }{\int }}{\phi }_n\left(t\right)\mathit{\text{dt}}}\underset{0}{\overset{\infty }{\int }}{\omega }_1\left(\left|t-x\right|\right)\frac{{\left(-t\right)}^k}{k!}\right.\\ \left.{\phi }_n^{(k)}(t)\mathit{\text{dt}}\right)\frac{{\left(-x\right)}^k}{k!}{\phi }_n^{(k)}(x)\\ \le {\omega }_1\left(f,{\delta }_n\right)\left\{1+\frac{1}{{\delta }_n}\left(\sum _{k=0}^{\infty }\frac{1}{\underset{0}{\overset{\infty }{\int }}{\phi }_n(t)\mathit{\text{dt}}}\underset{0}{\overset{\infty }{\int }}\left|t\right.\right.\right.\\ \left.\left.\left.-x\right|p_{n,k}(t)\mathit{\text{dt}}\right)p_{n,k}(x)\right\}\\ \le {\omega }_1\left(f,{\delta }_n\right)\left\{1+\frac{1}{{\delta }_n}{\left({\mathcal{ℒ}}_n^{\ast }\left({\left(t-x\right)}^2,x\right)\right)}^{1/2}\right\}\\ ={\omega }_1\left(f,{\delta }_n\right)\left\{1+\frac{1}{{\delta }_n}\right.\\ \left(\left(\frac{{\gamma }_{n-2m_0}}{{\gamma }_n}\frac{n\left(m_0+n\right)}{\left(n-2m_0\right)\left(n-m_0\right)}\right.\right.\\ \left.-\frac{2n}{n-m_0}\frac{{\gamma }_{n-m_0}}{{\gamma }_n}+1\right)x^2\\ +\left(\frac{4n}{\left(n-2m_0\right)\left(n-m_0\right)}\frac{{\gamma }_{n-2m_0}}{{\gamma }_n}\right.\\ \left.-\frac{2}{n-m_0}\frac{{\gamma }_{n-m_0}}{{\gamma }_n}\right)x\\ \left.{\left.+\frac{2}{n-m_0}\frac{{\gamma }_{n-2m_0}}{{\gamma }_n}\right)}^{1/2}\right\}\text{.}\end{array}$$

If we choose δ _n as

$$\begin{array}{l}{\delta }_n=\left[\left(\frac{{\gamma }_{n-2m_0}}{{\gamma }_n}\frac{n\left(m_0+n\right)}{\left(n-2m_0\right)\left(n-m_0\right)}-\frac{2n}{n-m_0}\frac{{\gamma }_{n-m_0}}{{\gamma }_n}+1\right)x^2\right.\\ +\left(\frac{4n}{\left(n-2m_0\right)\left(n-m_0\right)}\frac{{\gamma }_{n-2m_0}}{{\gamma }_n}-\frac{2}{n-m_0}\frac{{\gamma }_{n-m_0}}{{\gamma }_n}\right)x\\ {\left.+\frac{2}{\left(n-2m_0\right)\left(n-m_0\right)}\frac{{\gamma }_{n-2m_0}}{{\gamma }_n}\right]}^{1/2},\end{array}$$

then we get

$$\begin{array}{ll}\hfill \left|{\mathcal{ℒ}}_n^{\ast }\left(f,x\right)-f(x)\right|& \le 2{\omega }_1\left(f,\left[\left(\frac{{\gamma }_{n-2m_0}}{{\gamma }_n}\frac{n\left(m_0+n\right)}{\left(n-2m_0\right)\left(n-m_0\right)}\right.\right.\right.\hfill \\ \left.-\frac{2n}{n-m_0}\frac{{\gamma }_{n-m_0}}{{\gamma }_n}+1\right)x^2\hfill \\ +\left(\frac{4n}{\left(n-2m_0\right)\left(n-m_0\right)}\frac{{\gamma }_{n-2m_0}}{{\gamma }_n}\right.\hfill \\ \left.-\frac{2}{n-m_0}\frac{{\gamma }_{n-m_0}}{{\gamma }_n}\right)x\hfill \\ \left.{\left.+\frac{2}{\left(n-2m_0\right)\left(n-m_0\right)}\frac{{\gamma }_{n-2m_0}}{{\gamma }_n}\right]}^{1/2}\right)\text{.}\hfill \end{array}$$

Consequently, using (10) in the above inequality, we finally get

$$\begin{array}{c}\left|{\mathcal{ℒ}}_n^{\ast }\left(f,x\right)-f(x)\right|\le 2K\left(\left(\frac{{\gamma }_{n-2m_0}}{{\gamma }_n}\frac{n\left(m_0+n\right)}{\left(n-2m_0\right)\left(n-m_0\right)}\right.\right.\hfill \\ \left.-\frac{2n}{n-m_0}\frac{{\gamma }_{n-m_0}}{{\gamma }_n}+1\right)x^2\hfill \\ +\left(\frac{4n}{\left(n-2m_0\right)\left(n-m_0\right)}\frac{{\gamma }_{n-2m_0}}{{\gamma }_n}\right.\hfill \\ \left.-\frac{2}{n-m_0}\frac{{\gamma }_{n-m_0}}{{\gamma }_n}\right)x\hfill \\ {\left.+\frac{2}{\left(n-2m_0\right)\left(n-m_0\right)}\frac{{\gamma }_{n-2m_0}}{{\gamma }_n}\right)}^{\alpha /2}\text{.}\hfill \\ \square \hfill \end{array}$$

Theorem 3. Let f ∈ C[0,∞)be a bounded function, and 0 < α < 1.If

$$\left|{\mathcal{ℒ}}_n^{\ast }\left(f,x\right)-f(x)\right|\le K{\left(\frac{1}{n^2}+\frac{1}{n^2(1+m_{0}x)}\right)}^{\alpha /2}$$

for some positive constant K,then

$${\omega }_1\left(f,t\right)=O\left(t^{\alpha }\right),\left(t>0\right),$$

where ω₁(f,t) is the usual modulus of smoothness of f defined by (2).

Proof. For δ > 0, let

$$f_{\delta }(x)=\frac{1}{\delta }\underset{0}{\overset{\infty }{\int }}f\left(x+s\right)\mathrm{ds.}$$

For the function f ∈ C[0,∞) ∩ L _∞ [0,∞), the following inequalities hold (see [9]).

$${∥f_{\delta }-f∥}_{\infty }\le {\omega }_1\left(f,\delta \right),$$

(11)

$${∥f_{\delta }^{\prime }∥}_{\infty }\le \frac{1}{\delta }{\omega }_1\left(f,\delta \right)\text{.}$$

(12)

Now, we find the derivative of ${\mathcal{ℒ}}_n^{\ast }\left(f,x\right)$ with respect to x. From the definition of the operators ${\mathcal{ℒ}}_n^{\ast }$, we can write

$$\begin{array}{c}\frac{d}{\mathit{\text{dx}}}{\mathcal{ℒ}}_n^{\ast }\left(f,x\right)=\frac{n}{x}\sum _{k=0}^{\infty }\left(\frac{1}{\underset{0}{\overset{\infty }{\int }}{\phi }_n(t)\mathit{\text{dt}}}\underset{0}{\overset{\infty }{\int }}f(t)\frac{{\left(-t\right)}^k}{k!}{\phi }_n^{(k)}\left(t\right)\mathit{\text{dt}}\right)\hfill \\ \frac{{\left(-x\right)}^k}{k!}{\phi }_n^{(k)}\left(x\right)\left\{\left(\frac{k}{n}-x\right)+x\left(1-\frac{{\phi }_{n+m_0}^{(k)}(x)}{{\phi }_n^{(k)}(x)}\right)\right\}\hfill \\ =n\sum _{k=0}^{\infty }\left(\frac{1}{\underset{0}{\overset{\infty }{\int }}{\phi }_n(t)\mathit{\text{dt}}}\underset{0}{\overset{\infty }{\int }}f(t)\left(p_{n,k+1}(t)\right.\right.\hfill \\ \left.\left.-p_{n,k}(t)\right)\mathit{\text{dt}}\right)\frac{{\left(-x\right)}^k}{k!}{\phi }_{n+m_0}^{(k)}\left(x\right)\text{.}\hfill \end{array}$$

Now, using the properties of the operators (4), we obtain

$$\begin{array}{c}\left|\frac{d}{\mathit{\text{dx}}}{\mathcal{ℒ}}_n^{\ast }\left(f_{\delta }-f,x\right)\right|=\frac{n}{x}\sum _{k=0}^{\infty }\left(\frac{1}{\underset{0}{\overset{\infty }{\int }}{\phi }_n(t)\mathit{\text{dt}}}\underset{0}{\overset{\infty }{\int }}\left|f_{\delta }(t)-f(t)\right|\frac{{\left(-t\right)}^k}{k!}{\phi }_n^{(k)}\left(t\right)\mathit{\text{dt}}\right)\hfill \\ \times \frac{{\left(-x\right)}^k}{k!}{\phi }_n^{(k)}(x)\left[\left(\frac{k}{n}-x\right)+x\left(1-\frac{{\phi }_{n+m_0}^{(k)}(x)}{{\phi }_n^{(k)}(x)}\right)\right]\hfill \\ \le \frac{n}{x}{∥f_{\delta }-f∥}_{\infty }\left\{{\mathcal{ℒ}}_n\left(\left|t-x\right|,x\right)\right.\hfill \\ \left.+x\sum _{k=0}^{\infty }\frac{{\left(-x\right)}^k}{k!}{\phi }_n^{(k)}(x)\left(1-\frac{{\phi }_{n+m_0}^{(k)}(x)}{{\phi }_n^{(k)}(x)}\right)\right\}\hfill \\ =\frac{n}{x}{∥f_{\delta }-f∥}_{\infty }{\left(\frac{m_0}{n}{x}^2+\frac{x}{n}\right)}^{1/2}\hfill \\ \le \frac{n}{x}{\omega }_1\left(f;\delta \right){\left(\frac{m_0}{n}{x}^2+\frac{x}{n}\right)}^{1/2},\hfill \end{array}$$

(13)

where we have used the inequality,

$${\mathcal{ℒ}}_n\left(\left|t-x\right|,x\right)\le {\left({\mathcal{ℒ}}_n{\left(t-x\right)}^2,x\right)}^{1/2}={\left(\frac{m_0}{n}{x}^2+\frac{x}{n}\right)}^{1/2}\text{.}$$

On the other hand, we also have

$$\begin{array}{c}\left|\frac{d}{\mathit{\text{dx}}}{\mathcal{ℒ}}_n^{\ast }\left(f_{\delta }-f,x\right)\right|\hfill \\ =\left|n\sum _{k=0}^{\infty }\left(\frac{1}{\underset{0}{\overset{\infty }{\int }}{\phi }_n(t)\mathit{\text{dt}}}\underset{0}{\overset{\infty }{\int }}\left|f_{\delta }(t)-f(t)\right|\right.\right.\hfill \\ \times \left.\left.\left(p_{n,k+1}(t)-p_{n,k}(t)\right)\mathit{\text{dt}}\right)\frac{{\left(-x\right)}^k}{k!}{\phi }_{n+m_0}^{(k)}(x)\right|\hfill \\ \le 2n{∥f_{\delta }-f∥}_{\infty }{\phi }_{n+m_0}(0)\hfill \\ \le 2n{\omega }_1\left(f;\delta \right)\text{.}\hfill \end{array}$$

Using the two estimates of $\left|\frac{d}{\mathit{\text{dx}}}{\mathcal{ℒ}}_n^{\ast }\left(f_{\delta }-f,x\right)\right|$ obtained above, we get

$$\left|\frac{d}{\mathit{\text{dx}}}{\mathcal{ℒ}}_n^{\ast }\left(f_{\delta }-f,x\right)\right|\le 2{\omega }_1\left(f;\delta \right)min\left\{{\left(n{m}_0+\frac{n}{x}\right)}^{1/2},n\right\}\text{.}$$

Also, one can easily obtain that

$$\begin{array}{c}\left|\frac{d}{\mathit{\text{dx}}}{\mathcal{ℒ}}_n^{\ast }\left(f_{\delta },x\right)\right|=\left|\frac{n}{m_0-n}\sum _{k=0}^{\infty }\left(\frac{1}{\underset{0}{\overset{\infty }{\int }}{\phi }_n(t)\mathit{\text{dt}}}\underset{0}{\overset{\infty }{\int }}{f}_{\delta }(t)\left(p_{n-m_0,k+1}^{\prime }(t)\right)\mathit{\text{dt}}\right)\right.\hfill \\ \times \left.\frac{{\left(-x\right)}^k}{k!}{\phi }_{n+m_0}^{(k)}(x)\right|\hfill \\ \le \frac{1}{\delta }\frac{{\gamma }_{n-m_0}}{{\gamma }_n}\frac{n}{n-m_0}{\omega }_1\left(f;\delta \right)\text{.}\hfill \end{array}$$

For any t > 0 and 0 < h ≤ t, t∈(0,∞), we can write

$$\begin{array}{c}\left|f\left(x+h\right)-f(x)\right|\le \left|f\left(x+h\right)-{\mathcal{ℒ}}_n^{\ast }\left(f,x+h\right)\right|+\left|f(x)-{\mathcal{ℒ}}_n^{\ast }\left(f,x\right)\right|\hfill \\ +\left|\underset{0}{\overset{h}{\int }}\frac{d}{\mathit{\text{dx}}}{\mathcal{ℒ}}_n^{\ast }\left(f_{\delta },x+u\right)\mathit{\text{du}}\right|+\left|\underset{0}{\overset{h}{\int }}\frac{d}{\mathit{\text{dx}}}{\mathcal{ℒ}}_n^{\ast }\left(f-f_{\delta },x+u\right)\mathit{\text{du}}\right|\hfill \\ \le 2K{\left(\delta \left(n,m_0,x,h\right)\right)}^{\alpha }+\underset{0}{\overset{h}{\int }}\frac{1}{\delta }\frac{2n}{{}_{n-m_0}}\frac{{\gamma }_{n-m_0}}{{\gamma }_n}{\omega }_1\left(f;\delta \right)\mathit{\text{du}}\hfill \\ +\underset{0}{\overset{h}{\int }}2{\omega }_1\left(f;\delta \right)min\left\{{\left(n{m}_0+\frac{n}{x+u}\right)}^{1/2},n\right\}\mathit{\text{du}}\hfill \\ =2K{\left(\delta \left(n,m_0,x,h\right)\right)}^{\alpha }+\frac{h}{\delta }\frac{2n}{{}_{n-m_0}}\frac{{\gamma }_{n-m_0}}{{\gamma }_n}{\omega }_1\left(f;\delta \right)\hfill \\ +2{\omega }_1\left(f;\delta \right)\underset{0}{\overset{h}{\int }}min\left\{{\left(n{m}_0+\frac{n}{x+u}\right)}^{1/2},n\right\}\mathit{\text{du}}\hfill \\ \le 2K{\left(\delta \left(n,m_0,x,h\right)\right)}^{\alpha }+\frac{h}{\delta }\frac{2n}{{}_{n-m_0}}\frac{{\gamma }_{n-m_0}}{{\gamma }_n}{\omega }_1\left(f;\delta \right)\hfill \\ +6{\omega }_1\left(f;\delta \right)\frac{h}{\sqrt{\frac{1}{n^2}+\frac{1}{n^2(1+m_0(x+h))}}}\hfill \\ \le 2K{\left(\delta \left(n,m_0,x,h\right)\right)}^{\alpha }+6h{\omega }_1\left(f;\delta \right)\left[\frac{n}{{}_{n-m_0}}\frac{1}{\delta }+\frac{1}{\delta \left(n,m_0,x,h\right)}\right],\hfill \end{array}$$

where

$$\delta \left(n,m_0,x,h\right)={\left(\frac{1}{n^2}+\frac{1}{n^2(1+m_0(x+h))}\right)}^{1/2}\text{.}$$

Note that for any $n\in \mathbb{N}$, we have

$$\frac{1}{2}\delta \left(n,m_0,x,h\right)\le \delta \left(n+1,m_0,x,h\right)\le \delta \left(n,m_0,x,h\right)\text{.}$$

Hence, for $0<\delta <\frac{1}{2}$, we can choose $n\in \mathbb{N}$ such that

$$\delta \left(n,m_0,x,h\right)<\delta \le 2\delta \left(n,m_0,x,h\right)\text{.}$$

For sufficiently large n, we get

$$\begin{array}{l}\left|f\left(x+h\right)-f(x)\right|\le 2K{\left(\delta \left(n,m_0,x,h\right)\right)}^{\alpha }+6h{\omega }_1\left(f;\delta \right)\left[\frac{2}{\delta }+\frac{1}{\delta \left(n,m_0,x,h\right)}\right]\hfill \\ \le 2K{\delta }^{\alpha }+24\frac{h}{\delta }{\omega }_1\left(f;\delta \right)\hfill \\ \le max\left\{2K,24\right\}\left({\delta }^{\alpha }+\frac{h}{\delta }{\omega }_1\left(f;\delta \right)\right)\text{.}\hfill \end{array}$$

From last inequality, for 0 < h ≤ t, we get

$${\omega }_1\left(f;t\right)\le K^{\prime }\left({\delta }^{\alpha }+\frac{t}{\delta }{\omega }_1\left(f;\delta \right)\right)$$

which implies ω₁(f;t) = O(t^α), as desired. □

Theorem 4. For f ∈ C[0,∞) ∩ L _∞ [0,∞), 0 < α < 2, we have

$$\begin{array}{c}{\omega }_2\left(f,t\right)=O\left(t^{\alpha }\right)\iff \left|\frac{d^2}{d{x}^2}{\mathcal{ℒ}}_n^{\ast }\left(f,x\right)\right|\hfill \\ \le Mmin{\left\{\frac{2n}{x}+4n^2+6n{m}_0,4n\left(m_0+n\right)\right\}}^{(2-\alpha )/2},(n>2m_0),\hfill \end{array}$$

where ω₂(f,.)is the modulus of smoothness of f defined by (3).

Proof. (:⇒) We assume that ω₂(f,t) ≤ M t^α. For g ∈ C _B [0,∞), we get the second-order derivative of the operator ${\mathcal{ℒ}}_n^{\ast }\left(g,x\right)$ with respect to x as

$$\begin{array}{c}\frac{d^2}{d{x}^2}{\mathcal{ℒ}}_n^{\ast }\left(g,x\right)=n\left(m_0+n\right)\sum _{k=0}^{\infty }\left(\frac{1}{\underset{0}{\overset{\infty }{\int }}{\phi }_n(t)\mathit{\text{dt}}}\underset{0}{\overset{\infty }{\int }}g(t)\left(p_{n,k}(t)\right.\right.\hfill \\ \left.\left.-2p_{n,k+1}\left(t\right)+p_{n,k+2}(t)\right)\mathit{\text{dt}}\right)\frac{{\left(-x\right)}^k}{k!}{\phi }_{n+m_0}^{(k)}(x)\hfill \end{array}$$

$$=n\left(m_0+n\right)x^{-2}\sum _{k=0}^{\infty }\left(\frac{1}{\underset{0}{\overset{\infty }{\int }}{\phi }_n(t)\mathit{\text{dt}}}\underset{0}{\overset{\infty }{\int }}f(t)p_{n,k}(t)\mathit{\text{dt}}\right)\text{.}$$

Hence,

$$\left|\frac{d^2}{d{x}^2}{\mathcal{ℒ}}_n^{\ast }\left(g,x\right)\right|\le 4n\left(m_0+n\right){∥g∥}_{{}_{\infty }}$$

and

$$\left|\frac{d^2}{d{x}^2}{\mathcal{ℒ}}_n^{\ast }\left(g,x\right)\right|\le \left(\frac{2n}{x}+4n^2+6n{m}_0\right){∥g∥}_{{}_{\infty }}\text{.}$$

Now for f ∈ C[0,∞)∩L _∞ [0,∞), let us define the Steklov function as

$$f_d(x)=\frac{4}{d^2}\int \underset{0}{\overset{d/2}{\int }}\left(2f\left(x+u+v\right)-f\left(x+2u+2v\right)\right)\mathrm{dudv.}$$

Then,

$$∥f-f_d∥\le {\omega }_2\left(f,d\right)$$

and

$$∥f_d^{\parallel }∥\le \frac{9}{d^2}{\omega }_2\left(f,d\right)$$

For f _d , one can verify

$$\begin{array}{c}\left|\frac{d^2}{d{x}^2}{\mathcal{ℒ}}_n^{\ast }\left(f_d,x\right)\right|=\left|\sum _{k=0}^{\infty }\left(\frac{1}{\underset{0}{\overset{\infty }{\int }}{\phi }_n(t)\mathit{\text{dt}}}\underset{0}{\overset{\infty }{\int }}{f}_d^{\parallel }(t)p_{n+2m_0,k+2}\left(t\right)\mathit{\text{dt}}\right)\right|\hfill \\ \le \frac{1}{n\left(m_0+n\right)}\frac{9}{d^2}{\omega }_2\left(f,d\right)\text{.}\hfill \end{array}$$

Choosing $d=min{\left\{\frac{2n}{x}+4n^2+6n{m}_0,4n\left(m_0+n\right)\right\}}^{-1/2}$, we get

$$\begin{array}{c}\left|\frac{d^2}{d{x}^2}{\mathcal{ℒ}}_n^{\ast }\left(f,x\right)\right|\le \left|\frac{d^2}{d{x}^2}{\mathcal{ℒ}}_n^{\ast }\left(f_d,x\right)\right|+\left|\frac{d^2}{d{x}^2}{\mathcal{ℒ}}_n^{\ast }\left(f-f_d,x\right)\right|\hfill \\ \le \frac{1}{n\left(m_0+n\right)}\frac{9}{d^2}{\omega }_2\left(f,d\right)+min\left\{\frac{2n}{x}\right.\hfill \\ \left.+4n^2+6n{m}_0,4n\left(m_0+n\right)\right\}{\omega }_2\left(f,d\right),\hfill \end{array}$$

which proves the necessity part of the theorem. (⇐:) Now, in order to prove the sufficiency part of the theorem, we define the combination of ${\mathcal{ℒ}}_{n,1}^{\ast }$ as follows

$${\mathcal{ℒ}}_{n,1}^{\ast }\left(f,x\right)=a_0(n){\mathcal{ℒ}}_{n_0}\left(f,x\right)+a_1(n){\mathcal{ℒ}}_{n_1}\left(f,x\right),$$

where |a₀(n)| + |a₁(n)| ≤ B, n = n₀ < n₁ ≤ A n with A and B as absolute constants having the property

$${\mathcal{ℒ}}_{n,1}^{\ast }\left(t^i,x\right)=x^i,i=0,1\text{.}$$

Using the methods in [9, 10] for f ∈ C[0,∞)∩L _∞ [0,∞), we have

$$\begin{array}{c}\left|{\mathcal{ℒ}}_{n,1}^{\ast }\left(f,x\right)-f(x)\right|\le M{\omega }_2\left(f,\sqrt{{\delta }_n(x)}\right),{\delta }_n(x)\hfill \\ =\frac{2\left(m_{0}x+1\right)\left(\left(n+m_0\right)x+1\right)}{\left(n-2m_0\right)\left(n-m_0\right)}\text{.}\hfill \end{array}$$

For m,n ∈ N, x ∈ (0, ∞ ),0 < h ≤ t, we have

$$\begin{array}{c}\left|{\mathcal{ℒ}}_m^{\ast }\left(f,x\right)-2{\mathcal{ℒ}}_m^{\ast }\left(f,x+h\right)+{\mathcal{ℒ}}_m^{\ast }\left(f,x+2h\right)\right|\hfill \\ \le 4M{\omega }_2\left({\mathcal{ℒ}}_m^{\ast }f,\sqrt{{\delta }_n\left(x+2h\right)}\right)\hfill \\ +\int \underset{0}{\overset{h}{\int }}\left|\frac{d^2}{d{x}^2}\underset{n}{\overset{\ast }{\mathcal{ℒ}}}\left(\underset{n,1}{\overset{\ast }{\mathcal{ℒ}}}f,x+u+v\right)\right|\mathrm{dudv}\text{.}\hfill \end{array}$$

(14)

Now, we shall estimate the second term on the right-hand side of the above inequality. Firstly, we have

$$\left|\frac{d^2}{d{x}^2}{\mathcal{ℒ}}_{n,1}^{\ast }\left(f,x\right)\right|\le 4B{(\mathit{\text{An}})}^{2-\alpha }\text{.}$$

On the other hand, we also have

$$\begin{array}{c}\left|\frac{d^2}{d{x}^2}{\mathcal{ℒ}}_{n,1}^{\ast }\left(f,x\right)\right|\le B{\left(\frac{\mathit{\text{An}}}{x}+6{(\mathit{\text{An}})}^2\right)}^{(2-\alpha )/2}\hfill \\ \le \mathit{\text{MBA}}{n}^{(2-\alpha )/2}{\left(\frac{1}{x}\right)}^{(2-\alpha )/2}\text{.}\hfill \end{array}$$

From which, we can write

$${\left(\frac{1}{x}\right)}^{\alpha /2-1}\left|\frac{d^2}{d{x}^2}{\mathcal{ℒ}}_{n,1}^{\ast }\left(f,x\right)\right|\le \mathit{\text{MBA}}{n}^{(2-\alpha )/2}\text{.}$$

Hence, using the above inequality, we get

$$\begin{array}{c}\left|{\left(\frac{1}{x}\right)}^{\alpha /2-1}\frac{d^2}{d{x}^2}{\mathcal{ℒ}}_m^{\ast }\left({\mathcal{ℒ}}_{n,1}^{\ast }f,x\right)\right|\hfill \\ =\left|{\left(\frac{1}{x}\right)}^{\alpha /2-1}\sum _{k=0}^{\infty }\left(\frac{1}{\underset{0}{\overset{\infty }{\int }}{\phi }_n(t)\mathit{\text{dt}}}\underset{0}{\overset{\infty }{\int }}\frac{d^2}{d{x}^2}{\mathcal{ℒ}}_{n,1}^{\ast }\left(f,t\right)p_{n+2m_0,k+2}(t)\mathit{\text{dt}}\right)\right.\hfill \\ \left.\times \frac{{\left(-x\right)}^k}{k!}{\phi }_{n+m_0}^{\left(k\right)}(x)\right|\hfill \\ \le \mathit{\text{MBA}}{n}^{(2-\alpha )/2}\text{.}\hfill \end{array}$$

After making some arrangements and then taking the integral of both sides of the above inequality, we get

$$\begin{array}{c}\int \underset{0}{\overset{h}{\int }}\left|\frac{d^2}{d{x}^2}\underset{n}{\overset{\ast }{\mathcal{ℒ}}}\left(\underset{n,1}{\overset{\ast }{\mathcal{ℒ}}}f,x+u+v\right)\right|\hfill \\ \mathit{\text{dudv}}\le \mathit{\text{MBA}}{n}^{(2-\alpha )/2}\int \underset{0}{\overset{h}{\int }}{\left|\frac{1}{x+u+v}\right|}^{\frac{\alpha -1}{2}}\mathit{\text{dudv}}\hfill \\ \le \mathit{\text{MBA}}{n}^{(2-\alpha )/2}{h}^2{\left(\frac{n}{x+2h}\right)}^{(2-\alpha )/2}\text{.}\hfill \end{array}$$

(15)

Now, substituting (15) into (14), we finally obtain

$$\begin{array}{c}\left|{\mathcal{ℒ}}_m^{\ast }\left(f,x\right)-2\mathcal{ℒ}\right.\left.m_{\ast }\left(f,x+h\right)+{\mathcal{ℒ}}_m^{\ast }\left(f,x+2h\right)\right|\hfill \\ \le 4M{\omega }_2\left({\mathcal{ℒ}}_m^{\ast }f,\sqrt{{\delta }_n\left(x+2h\right)}\right)+M_1{h}^2{\left(M_n\left(x+2h\right)\right)}^{(2-\alpha )},\hfill \end{array}$$

(16)

where $M_n(x)=\sqrt{\frac{x}{n}}$. Choosing $n\in \mathbb{N}$ such that

$$\frac{t}{2C}\le max\left\{\sqrt{{\delta }_n\left(x+2h\right)},M_n\left(x+2h\right)\right\}\le \frac{t}{C},$$

we obtain from (16) by induction

$$\begin{array}{c}{\omega }_2\left({\mathcal{ℒ}}_m^{\ast }f,t\right)\le 4M{\omega }_2\left({\mathcal{ℒ}}_m^{\ast }f,\frac{t}{C}\right)+{\left(2C\right)}^{2-\alpha }{M}_2{t}^{\alpha }\hfill \\ .\hfill \\ .\hfill \\ .\hfill \\ \le t^2{(4M)}^k{C}^{-2k}∥\frac{d^2}{d{x}^2}{\mathcal{ℒ}}_m^{\ast }f∥+{\left(2C\right)}^{2-\alpha }{M}_2{t}^{\alpha }\frac{C^{\alpha }}{C^{\alpha }-4M}\text{.}\hfill \end{array}$$

(17)

If we take C = (1 + 4M)^1/α and let k → ∞, we obtain

$${\omega }_2\left({\mathcal{ℒ}}_m^{\ast }f,t\right)\le \frac{1}{C^{\alpha }-4M}4C^2{M}_2{t}^{\alpha }$$

which implies that ω₂(f,t) = O(t^α), where $\frac{1}{C^{\alpha }-4M}4C^2{M}_2$ is independent of m. So, the proof is completed. □