Approximation of Jakimovski-Leviatan-Beta type integral operators via q-calculus

1.   Preliminaries and introduction

In 1880, Appell investigated the polynomials named as Appell polynomials (see ^[1]). Jakimovski and Leviatan introduced and modified the Appell polynomials ^[2] in 1969 by the identity defined bellow

where $\beta_{k}(y) = \sum_{i = 0}^{k}\alpha _{i}\frac{y^{n-i}}{ (n-i)!}\; (n\in \mathbb{N})$ and $P(u) = \sum_{k = 0}^{\infty }\alpha _{k}u^{k}, \; P(1)\neq 0$. Let $E[0, \infty)$ denote the set of functions defined on $[0, \infty)$ such that $|f(x)|\leq \kappa e^{\gamma x}$, where $\kappa, \; \gamma$ are positive constants.

We recall some basic notations on $q$-calculus (see ^[3,4,5]). For each non-negative integer $j$, the $q$-integer is defined as

For $|q| < 1,$ the $q$-factorial $\left[j\right] _{q}!$ is defined by

In the standard approach the exponential functions for $q$-calculus:

The improper integral of function $f$ is defined by:

Al-Salam (see ^[6,7]) introduced the family of $q$-Appell polynomials through the generating functions $P_{q}(t) = \sum\limits_{n = 0}^{ \infty }P_{n, q}\frac{t^{n}}{[n]_{q}!}, \; P_{q}(1)\neq 0$. We have

and $q$-differential, $D_{q, x}\big(P_{n, q}(x)\big) = [n]_{q}P_{n-1, q}(x), \; n = 1, 2, \dots$, where $P_{0, q}(x)$ is a non zero constant let say $P_{0, q}$ and $D_{q, x}\big(P_{1, q}(x)\big) = [1]_{q}P_{0, q}(x) = P_{0, q}$. Also $P_{q}(t)e_{q}(tx) = \sum\limits_{n = 0}^{\infty }P_{n, q}(x)\frac{t^{n}}{ [n]_{q}!}, \; \; \; 0 < q < 1$.

Recently in ^[8], authors studied the $q$-analogue of Jakimovski-Levitian operators involving $q$-Appell polynomials; and in ^[9], Stacu type Jakimovski-Levitian-Durmeyer operators have been studied. Recently such $q$-analogues operators have also been studied in ^{[10,11,12,13]}. Our aim is to construct Jakimovski-Leviatan-Beta type $q-$integral operators and show that these positive linear operators are uniformly convergent to a continuous functions. Further, we obtain the Korovkin type results, the rate of convergence as well as some direct theorems.

Let $x\in \lbrack 0, \infty), \; P_{r, q}(x)\geq 0$ and $P_{q}(1)\neq 0$. For every $f\in C_{\zeta }[0, \infty) = \{f\in C[0, \infty):f(t) = O(t^{\zeta }),$ $t\rightarrow \infty$}, and $m\in \mathbb{N}, \; \; 0 < q < 1$, we define

where $\zeta > m$ and

with

2.   Moments of new operators

Lemma 1. Take $e_{i} = t^{i-1}$ for $i = 1, 2, 3, 4, 5$. Then

Proof of Lemma 1.

$\sum\limits_{r = 0}^{\infty }r^{4}\frac{P_{r, q}([m]_{q}x)}{[r]_{q}!}$

Take $f(t) = e_{1}$, then

Take $f(t) = e_{2}$, then

By applying,

Take $f(t) = e_{3}$, then

By applying (2.1) and $[r+2]_{q} = 1+q+q^{2}[r]_{q}$,

Take $f(t) = e_{4}$, then

By a simple calculation we have

$[r+3]_{q}[r+2]_{q}[r+1]_{q}$

If $f(t) = e_{5}$, then

A simple calculation leads to

$[r+4]_{q}[r+3]_{q}[r+2]_{q}[r+1]_{q}$

Lemma 2.

Let $\mu _{j} = (e_{2}-x)^{j}$ for $j = 1, 2$. For all $x\in \lbrack 0, \infty), \; \; \; 0 < q < 1, \; \; \; P_{r, q}(x)\geq 0$ and $P_{q}(1)\neq 0$, we have:

That is,

3.   Korovkin type approximation

We write $C_{B}(\mathbb{R^{+}})$ for the set of all bounded and continuous { functions with

Let

We choose $q = q_{m }$ where $0 < q_{m } < 1$ such that

Theorem 1. For any function $f\in C[0, \infty)\cap E$, we have

uniformly on each compact subset of $[0, \infty)$.

Proof of Theorem 1. From the well-known Korovkin's theorem ^[14] (see ^[15,16]), it is enough to show

uniformly on $[0, 1]$.

Using (3.1), $\frac{1}{[m]_{q_{m }}}\rightarrow 0$ and $\frac{ [m]_{q_{m }}}{[m-1]_{q_{m }}}\rightarrow 1\; \; (\nu \rightarrow \infty),$ we have

Which complete the proof.

Let $\varrho (x) = 1+\phi ^{2}(x), \lim_{x\rightarrow \infty }\varrho (x) = \infty$, where $\phi (x)$ is a continuous and strictly increasing function. Let $B_{\varrho }(\mathbb{R}^{+})$ be a set of functions defined on $\mathbb{R}^{+}$ such that there is a constant $M_{f},$

Its subset of continuous functions is denoted by $C_{\varrho }(\mathbb{R} ^{+})$. Note that

is the usual norm on $B_{\varrho }(\mathbb{R}^{+}).$ Let $C_{\varrho }^{0}(\mathbb{R}^{+})$ be a subset of $C_{\varrho }(\mathbb{R}^{+})$ such that

exists and is finite.

Theorem 2. Let $\{\mathcal{J}_{m, q}^{\ast }\}_{m \geq 1}$ be the sequence of linear positive operators (1.5) from $C_{\varrho }(\mathbb{R}^{+})$ into $B_{\varrho }(\mathbb{ R}^{+})$ such that

Then for $f\in C_{\varrho }^{0}(\mathbb{R}^{+})$,

Proof of Theorem 2. Consider $\varphi (x) = x, \; \; \; \varrho (x) = 1+x^{2}$, and

Then for $i = 1, 2, 3$ it is easily proved (by Theorem 1) that

Using Korovkin's theorem, we get

Theorem 3.

Let $x\in \lbrack 0, \infty), \quad f\in C_{\varrho }^{0}(\mathbb{R}^{+})$ with $\varrho (x) = 1+x^{2}$. Then for $P_{r, q}(x)\geq 0, \quad P_{q}(1)\neq 0$, the operators $\mathcal{J}_{m, q}^{\ast }(\, \cdot \,; \, \cdot \, )$ defined by (1.5) satisfying

Proof of Theorem 3. Using Theorem 2, consider

for $i = 1, 2, 3$.

From Lemma 1, for $i = 1$, we have $\mid \mathcal{J}_{m, q}^{\ast }(e_{1};x)-1\mid \rightarrow 0$, and therefore

For $i = 2$

Therefore

For $i = 3$

Hence we have

The Korovkin's theorem completes the proof.

We recall the following spaces:

where $\sigma (x) = 1+x^{2}$. Note that $\parallel f\parallel _{\sigma } = \sup_{x\geq 0}\frac{\mid f(x)\mid }{\sigma (x)}$ is the usual norm on $Q_{\sigma }(\mathbb{R}^{+}).$

Theorem 4. Let $0 < q_{m } < 1$ and $\mathcal{J}_{m, q}^{\ast }(\, \cdot \,; \, \cdot \, )$ be the operators defined by (1.5) with $m > 2$. Then for $f\in Q_{\sigma }^{k}(\mathbb{R}^{+}),$ we have

Proof of Theorem 4. Take $j = 1$, then Lemma 1, yields the first condition. If we take $j = 2, 3$, then from Lemma 1, we get

4.   Order of approximation

The modulus of continuity of $f\in C[0, \infty)$ is defined by

Note that $\lim_{\delta \rightarrow 0+}\varpi (f; \delta) = 0$ for $f\in C[0, \infty)$ and (see (^[17], p. 378)

Theorem 5. For $f\in \tilde{C}[0, \infty), \; \; x\geq 0, \; \; 0 < q < 1$ and $P_{q}(1)\neq 0,$ we have

where $\tilde{C}[0, \infty)$ is the space of uniformly continuous functions on $\mathbb{R}^{+}$ and $\delta _{m, q}^{\ast }$ is defined in Lemma 2.

Proof of Theorem 5. Using Cauchy-Schwarz inequality and (4.1), (4.2), we get

$\mid \mathcal{J}_{m, q}^{\ast }(f; x)-f(x)\mid$

Choosing $\delta = \delta _{m, q}^{\ast } = \sqrt{\mathcal{J}_{m, q}^{\ast }((t-x)^{2};x)}$, then we get our result.

5.   Rate of convergence

For $f\in C[0, \infty)$, $M > 0$ and $0 < \nu \leq 1$, we define

Theorem 6. For each $f\in Lip_{M}(\nu), \; \; M > 0, \; \; \; 0 < \nu \leq 1$ and $0 < q < 1,$ we have

where $\lambda _{m, q}(x) = \mathcal{J}_{m, q}^{\ast }\left(\mu _{2};x\right)$.

Proof of Theorem 6. Hölder inequality and (5.1) imply that

So that

$\mid \mathcal{J}_{m, q}^{\ast }(f; x)-f(x)\mid$

Which complete the proof.

Denote

with

also

Theorem 7. For any $\psi \in C_{B}^{2}(\mathbb{R} ^{+}),$ we have

where $\lambda _{m, q}(x)$ is given in Theorem 6.

Proof of Theorem 7. We applying the generalized mean value theorem in the Taylor series expansion, then

By linearity, we have

which gives

$\mid \mathcal{J}_{m, q}^{\ast }(\psi; x)-\psi (x)\mid$

Hence

$\mid \mathcal{J}_{m, q}^{\ast }(\psi; x)-\psi (x)\mid$

From (5.3), we have $\parallel \psi ^{\prime }\parallel _{C_{B}[0, \infty)}\leq \parallel \psi \parallel _{C_{B}^{2}[0, \infty)}$.

$\mid \mathcal{J}_{m, q}^{\ast }(\psi; x)-\psi (x)\mid$

This completes the proof.

6.   Direct theorem

The Peetre's $K$-functional ^[18] is defined by

where

Note that $K_{2}(f, \delta)\leq C\varpi _{2}(f, \delta ^{\frac{1}{2} }), \; \delta > 0$, $C > 0,$ where

is the second order modulus of continuity.

Theorem 8. For $f\in C_{B}(\mathbb{R}^{+}),$ we have

Proof of Theorem 8. Applying Theorem 7, we get

From (5.3) clearly we have $\parallel \psi \parallel _{C_{B}[0, \infty)}\leq \parallel \psi \parallel _{C_{B}^{2}[0, \infty)}$.

Therefore

$\mid \mathcal{J}_{m, q}^{\ast }(f; x)-f(x)\mid$

Taking infimum over all $\psi \in C_{B}^{2}(\mathbb{R}^{+})$ and using (6.1), we get

For an absolute constant $\mathcal{D} > 0$, we use the relation ^[19]

This complete the proof.

For an arbitrary $f\in Q_{\sigma }^{k}(\mathbb{R}^{+}),$ we define (see ^[20])

Note that $\lim_{\delta \rightarrow 0}\Omega (f, \delta)\rightarrow 0$ and

where $f\in Q_{\sigma }^{k}(\mathbb{R}^{+})$ and $t, x\in \lbrack 0, \infty)$.

Theorem 9. Let $q = q_m$, then for $f\in Q_{\sigma }^{k}(\mathbb{R}^{+})$, we have

where $\mathcal{A}_1, \mathcal{A}_2 > 0$ and

Proof of Theorem 9. To prove this theorem our aim is to use the results (6.4), (6.5) and then apply Cauchy-Schwarz inequality, thus we see

and

In the light of Lemma 2, we easily see that

where

For a constant $\mathcal{A}_1 > 0$ we have

Similarly a small calculation lead us

where

and

As $\frac{1}{[m-i]_{q_m} } = 0$ for all $i = 1, 2$ when $m \to \infty$, imply that for a constant $\mathcal{A}_{2} > 0$, easily we have

(6.8), imply that

Hence, by combining (6.7)–(6.11) and (6.6), and finally if we choosing $\delta = \sqrt{\Psi_{m, q_m}(m)}$ after taking supremum $y\in \lbrack 0, \Psi_{m, q_m}(m))$, we are denumerable to get the result.

Acknowledgments

This project was funded by the Deanship of Scientific Research (DSR) at King Abdulaziz University, Jeddah, under grant no. G: 1434-130-1440. The authors, therefore, acknowledge with thanks DSR for technical and financial support.

Conflict of interest

The authors declare no conflict of interest in this paper.