Some New Anderson Type h and q Integral Inequalities in Quantum Calculus

Abstract

The calculus in the absence of limits is known as quantum calculus. With a difference operator, it substitutes the classical derivative, which permits dealing with sets of functions that are non-differentiations. The theory of integral inequality in quantum calculus is a field of mathematics that has been gaining considerable attention recently. Despite the fact of its application in discrete calculus, it can be applied in fractional calculus as well. In this paper, some new Anderson type q-integral and h-integral inequalities are given using a Feng Qi integral inequality in quantum calculus. These findings are highly beneficial for basic frontier theories, and the techniques offered by technology are extremely useful for those who can stimulate research interest in exploring mathematical applications. Due to the interesting properties in the field of mathematics, integral inequalities have a tied correlation with symmetric convex and convex functions. There exist strong correlations and expansive properties between the different fields of convexity and symmetric function, including probability theory, convex functions, and the geometry of convex functions on convex sets. The main advantage of these essential inequalities is that they can be converted into time-scale calculus. This kind of inevitable inequality can be very helpful in various fields where coordination plays an important role.

1. Introduction

In mathematics, q-calculus is a quantum calculus that calculates without limits. We acquire q-analogue formulas of mathematics in q-calculus that can be captured as a tendency of q toward one. In the same vein of Newton’s efforts on infinite series, Euler first introduced q-calculus. Therefore, the date of q-calculus can be linked back to Euler. Then, F. H. Jackson [1] defined the q-definite integral and introduced a systematic study of q-calculus, which is known as the q-Jackson integral, in 1910. In recent years, due to the high demand of mathematics in the field of quantum calculus, the interest in q-calculus has been increasing. This q-calculus has many applications in different fields of mathematics and other different areas, such as fractals, orthogonal polynomial combinatorics, mechanics, number theory, dynamical systems, special functions, and mechanics for scientific problems in many applied areas.

In numerous branches of mathematics such as differential equations, analysis, geometry, and many other fields, mathematical inequalities have been applied. One such example is Anderson type integral inequalities. Let us recall following lines from [2]. “Consider the following quotient:

$$\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}$$

As x reaches to $x_0$, the limit, gives the well known definition of the differentiation of a function at $x=x_0$, if it exists. However, if we take $x=q{x}_0$ or $x=x_0+h$, where q is a constant different from 1, and h is a constant different from 0 and do not need to take the limit, we enter the fascinating world of Quantum Calculus and will get the definition of q derivative and h derivative using $x=q{x}_0$ and $x=x_0+h$ respectively.”

In [3], Feng Qi studied one of the useful inequalities, which may be stated as follows:

Theorem 1.

Let a positive real number $n\ge 1$, $Q\in C^{\left(n\right)}[m_0,m_1]$ s.t. $Q^{\left(i\right)}\left(m_0\right)\ge 0$ for $i\in \{0,1,\dots n-1\}$ and $Q^{\left(n\right)}\left(m_0\right)\ge n!$. Then, the following is true:

$${\left[{\int }_{m_0}^{m_1}Q\left(\tau \right)d\tau \right]}^{n+1}\le {\int }_{m_0}^{m_1}{\left[Q\left(\tau \right)\right]}^{n+2}d\tau$$

(1)

At the end of this article, Feng Qi proposed an open problem: “What if n is replaced by any positive real number p? In what state is the inequality (1) still true?”

This inequality gained the attention of many mathematicians. Some researchers showed keen interest in this open problem and gave it a try [4,5]. Some of the researchers studied Fenq Qi type inequalities in quantum calculus involving q-integrals and h-integrals [6], and some studied these inequalities in time-scale calculus [7,8], while some researchers further generalized it using the ${\diamond }_{\alpha }$ operator. In this paper, we shall discuss the q-integral and h-integral analogous of a Feng Qi type inequality in quantum calculus by using an Anderson inequality. On the other hand, there is considerable research in the field of q-analysis for achieving quality in quantum computing. Quantum calculus is a series between mathematics and physics and contains a wide range of applications in many fields, such as combinations, mathematical numerical theory, addition theory, and quantum theory [9,10]. Quantum calculus also combines quantum information theory, philosophy, information theory, computer science, and cryptography with several applications [11,12]. Since then, the relevant research has been steadily improving. Specifically, the left quantum integral and the left quantum differential operator were discussed in 2013 [13] by Ntouyas and Tariboon.

Anderson [14] discussed the following fascinating and interesting integral inequality in 1958:

Proposition 1.

If ${\psi }_k$ are convex mappings on the closed interval $[0,1]$, and ${\psi }_k\left(0\right)=0$∀$k\in \{1,\dots ,n\}$, then

$${\int }_0^1{\psi }_1\left(s\right){\psi }_2\left(s\right)\dots {\psi }_n\left(s\right)ds\ge \frac{2^n}{n+1}\left({\int }_0^1{\psi }_1\left(s\right)ds\right)\dots \left({\int }_0^1{\psi }_n\left(s\right)ds\right)$$

(2)

where ${\psi }_1\left(\chi \right)\dots {\psi }_n\left(\chi \right)$ are integrable functions on the closed interval $[0,1]$.

If we let $Q={\psi }_1={\psi }_2=\cdots ={\psi }_n$ in Equation (2), then we obtain following Feng Qi type integral inequality:

$$\begin{array}{c}\hfill {\int }_0^1{Q}^n\left(\chi \right)d\chi \ge \frac{2^n}{n+1}{\left({\int }_0^{1}Q\left(\chi \right)d\chi \right)}^n.\end{array}$$

(3)

2. q-Integral Inequalities of the Anderson Type

2.1. Notations and Preliminaries

For $q>1$ we recall the notations from [1,7], where $m_0\in \mathbb{C}$:

$$\begin{array}{ccc}\hfill {\left[m_0\right]}_q=\frac{1-q^{m_0}}{1-q},& & {(m_0;q)}_n=\prod _{k=0}^{n-1}(1-m_0{q}^k),n\ge 1\hfill \end{array}$$

$$\begin{array}{ccc}\hfill {\left[0\right]}_q!=1,& & {\left[n\right]}_q!={\left[1\right]}_q{\left[2\right]}_q\dots {\left[n\right]}_q,n\ge 1\hfill \end{array}$$

and

$$\begin{array}{c}\hfill {(\chi -m_0)}_q^n=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}n=0,\chi \in \mathbb{C},\hfill \\ (\chi -m_0)(\chi -q{m}_0)\dots (\chi -q^{n-1}{m}_0),\hfill & \mathrm{if}n\ne 0.\hfill \end{array}\right.\end{array}$$

(4)

The “q-derivative $D_q$Q of a function” Q is defined as

$$\begin{array}{ccc}\hfill \left(D_{q}Q\right)\left(\chi \right)=\frac{Q\left(q\chi \right)-Q\left(\chi \right)}{(q-1)\chi },& & \chi \ne 0,\hfill \end{array}$$

(5)

where $\left(D_{q}Q\right)\left(0\right)=Q^{\prime }\left(0\right)$, provided $Q^{\prime }\left(0\right)$ exists.

The “q-Jackson integral from 0 to $m_0$” is defined by [15]

$$\begin{array}{c}\hfill {\int }_0^{m_0}Q\left(\chi \right)d_q\chi =(q^{-1}-1)m_0\sum _{n=0}^{\infty }Q\left(m_0{q}^{-n}\right)q^{-n},\end{array}$$

(6)

given that the sum converges absolutely.

The “q-Jackson integral in a generic interval $[m_0,m_1]$” is defined as follows (see [15]):

$$\begin{array}{c}\hfill {\int }_{m_0}^{m_1}Q\left(\chi \right)d_q\chi ={\int }_0^{m_0}Q\left(\chi \right)d_q\chi -{\int }_0^{m_1}Q\left(\chi \right)d_q\chi .\end{array}$$

(7)

From [2], we also have that

$$\begin{array}{c}\hfill D_q\left[{\int }_{m_0}^{\chi }Q\left(\tau \right)d_q\tau \right]=Q\left(\chi \right).\end{array}$$

(8)

If $m_1>0$ and $m_0=m_1{q}^n$, where $n\in \mathbb{N}$, we have

$$\begin{array}{c}\hfill {[m_0,m_1]}_q=\{m_1{q}^k:0\le k\le n\}and{(m_0,m_1]}_q={[q^{-1}{m}_0,m_1]}_q.\end{array}$$

2.2. Main Results

Lemma 1.

If we let $p\in \mathbb{R}$ s.t. $p\ge 1$ and λ be a non-negative and monotone function on ${[0,1]}_q$, then

$$\begin{array}{ccc}\hfill {\left[p\right]}_q{\lambda }^{p-1}\left(\chi \right)D_q\lambda \left(\chi \right)\le D_q{\left[\lambda \left(\chi \right)\right]}^p\le {\left[p\right]}_q{\lambda }^{p-1}\left(q\chi \right)D_q\lambda \left(\chi \right),& & \chi \in {(0,1]}_q.\hfill \end{array}$$

Proof. 

We have

$$\begin{array}{c}\hfill D_q\left[{\lambda }^p\right]\left(\chi \right)=\frac{{\lambda }^p\left(q\chi \right)-{\lambda }^p\left(\chi \right)}{(q-1)\chi }=\frac{{\left[p\right]}_q}{(q-1)\chi }{\int }_{\lambda \left(\chi \right)}^{\lambda \left(q\chi \right)}{\tau }^{p-1}{d}_q\tau .\end{array}$$

(9)

Since $\lambda$ is a non-negative and monotonic function, therefore

$$\begin{array}{c}\hfill {\lambda }^{p-1}\left(\chi \right)[\lambda \left(q\chi \right)-\lambda \left(\chi \right)]\le {\int }_{\lambda \left(\chi \right)}^{\lambda \left(q\chi \right)}{\tau }^{p-1}{d}_q\tau \le {\lambda }^{p-1}\left(q\chi \right)[\lambda \left(q\chi \right)-\lambda \left(\chi \right)].\end{array}$$

From Equation (9), we obtain

$$\begin{array}{c}\hfill {\left[p\right]}_q{\lambda }^{p-1}\left(\chi \right)D_q\lambda \left(\chi \right)\le D_q{\left[\lambda \right]}^p\left(\chi \right)\le {\left[p\right]}_q{\lambda }^{p-1}\left(q\chi \right)D_q\lambda \left(\chi \right).\end{array}$$

□

Theorem 2.

If we have the function Q on ${[0,1]}_q$, then it satisfies

$$\begin{array}{c}\hfill Q\left(0\right)=0,D_{q}Q\left(\chi \right)\ge \frac{2^p}{p+1}{\left[p\right]}_q{\chi }^{p-2}for\chi \in {(0,1]}_{q}andp\ge 2,(yes,accordingtopower)\end{array}$$

(10)

Then, it is true that

$$\begin{array}{c}\hfill {\int }_0^1{Q}^p\left(q\chi \right)d_q\chi \ge \frac{2^p}{p+1}{\left({\int }_0^{1}Q\left(\chi \right)d_q\chi \right)}^p.\end{array}$$

Proof. 

Include ${\displaystyle \lambda \left(\chi \right)={\int }_0^{\chi }Q\left(u\right)d_{q}u}$ and

$$\begin{array}{c}\hfill F\left(\chi \right)={\int }_0^{\chi }{Q}^p\left(qu\right)d_{q}u-\frac{2^p}{p+1}{\left({\int }_0^{\chi }Q\left(u\right)d_{q}u\right)}^p.\end{array}$$

We have

$$\begin{array}{c}\hfill D_{q}F\left(\chi \right)=Q^p\left(q\chi \right)-\frac{2^p}{p+1}{D}_q\left[{\lambda }^p\right]\left(\chi \right).\end{array}$$

Since Q and F both are increasing on ${[0,1]}_q$, we obtain the following from Lemma 1:

$$\begin{array}{ccc}\hfill D_{q}F\left(\chi \right)& \ge & Q^p\left(q\chi \right)-\frac{{\left[p\right]}_q}{p+1}{2}^p{\lambda }^{p-1}\left(\chi \right)Q\left(\chi \right)\hfill \\ \hfill & \ge & Q^p\left(q\chi \right)-\frac{{\left[p\right]}_q}{p+1}{2}^p{\lambda }^{p-1}\left(\chi \right)Q\left(q\chi \right)=Q\left(q\chi \right)h\left(\chi \right),\hfill \end{array}$$

(11)

where $h\left(\chi \right)=Q^{p-1}\left(q\chi \right)-\frac{{\left[p\right]}_q}{p+1}{2}^p{\lambda }^{p-1}\left(\chi \right)$.

We also have

$$\begin{array}{c}\hfill D_{q}h\left(\chi \right)=D_q{Q}^{p-1}\left(q\chi \right)-\frac{{\left[p\right]}_q}{p+1}{2}^p{D}_q{\lambda }^{p-1}\left(\chi \right).\end{array}$$

By using Lemma 1 again, we obtained

$$\begin{array}{ccc}\hfill D_{q}h\left(\chi \right)& \ge & (p-1)Q^{p-2}\left(\chi \right)D_{q}Q\left(q\chi \right)-\frac{{\left[p\right]}_q(p-1)}{p+1}{2}^p{\lambda }^{p-2}\left(\chi \right)D_q\lambda \left(q\chi \right)\hfill \end{array}$$

(12)

$$\begin{array}{ccc}\hfill & \ge & (p-1)Q\left(\chi \right)[Q^{p-3}\left(\chi \right)D_{q}Q\left(q\chi \right)-\frac{{\left[p\right]}_q}{p+1}{2}^p{\lambda }^{p-2}\left(q\chi \right)].\hfill \end{array}$$

(13)

Since Q is increasing, we have

$$\begin{array}{c}\hfill {\int }_0^{\chi }Q\left(u\right)d_{q}u\le Q\left(\chi \right)(q\chi -0)=q\chi Q\left(\chi \right).\end{array}$$

Now, using the assumptions of this theorem and the inequalities in Equations (12) and (13), we obtain

$$\begin{array}{c}\hfill D_{q}h\left(\chi \right)\ge (p-1)Q^{p-1}\left(\chi \right)[D_{q}Q\left(q\chi \right)-\frac{{\left[p\right]}_q}{p+1}{2}^p{\left(q\chi \right)}^{p-2}]\ge 0,\end{array}$$

In addition, from the fact that $h\left(0\right)=Q^p\left(0\right)=0$, we obtain $h\left(\chi \right)\ge 0$, $\chi \in {[0,1]}_q$.

From $F\left(0\right)=0$ and $D_{q}F\left(\chi \right)=Q\left(\chi \right)h\left(\chi \right)\ge 0$, it is given that $F\left(\chi \right)\ge 0$ ∀$\chi \in {[0,1]}_q$, particularly

$$\begin{array}{c}\hfill F\left(1\right)={\int }_0^1{Q}^p\left(qu\right)d_{q}u-\frac{2^p}{p+1}{\left({\int }_0^{1}Q\left(u\right)d_{q}u\right)}^p\ge 0.\end{array}$$

□

Corollary 1.

Let $n\in \mathbb{N}$ and the function Q on ${[0,1]}_q$ satisfy

$$\begin{array}{c}\hfill Q\left(0\right)=0,D_{q}Q\left(\chi \right)\ge \frac{n+1}{n+2}{2}^{n+1}{\chi }^n,\chi \in {(0,1]}_q.\end{array}$$

Then, we have

$$\begin{array}{c}\hfill {\int }_0^1{Q}^{n+1}\left(q\chi \right)d_q\chi \ge \frac{2^{n+1}}{n+2}{\left({\int }_0^{1}Q\left(\chi \right)d_q\chi \right)}^{n+1}.\end{array}$$

Proof. 

The results follow directly by substituting $p=n+1$ into Theorem 2. □

Corollary 2.

Let $n\in \mathbb{N}$ and the function Q on ${[0,1]}_q$ satisfy

$$\begin{array}{c}\hfill D_q^{i}Q\left(0\right)\ge 0,0\le i\le n-1,D_q^{n}Q\left(\chi \right)\ge \frac{n+1}{n+2}{2}^{n+1}{\left[n\right]}_q!,\chi \in {(0,1]}_q.\end{array}$$

Then, we have

$$\begin{array}{c}\hfill {\int }_0^1{\left(Q\left(q\chi \right)\right)}^{n+2}{d}_q\chi \ge \frac{2^{n+2}}{n+3}{\left({\int }_0^{1}Q\left(\chi \right)d_q\chi \right)}^{n+2}.\end{array}$$

Proof. 

Since ${\displaystyle D_q^{n}Q\left(\chi \right)\ge \frac{n+1}{n+2}{2}^{n+1}{\left[n\right]}_q!}$, therefore, by q-integrating $(n-1)$ times on $[0,\chi ]$, we obtain

$$\begin{array}{c}\hfill D_{q}Q\left(\chi \right)\ge \frac{n+1}{n+2}{2}^{n+1}{\chi }^n.\end{array}$$

We obtain our required result by using Corollary 1. □

Theorem 3.

Let $p\in \mathbb{R}$ s.t. $p\ge 1$ and the function Q on ${[0,1]}_q$ satisfy

$$\begin{array}{c}\hfill Q\left(0\right)=0,p\le D_{q}Q\left(q\chi \right),\forall \chi \in {(0,1]}_q.\end{array}$$

(14)

Then, we have

$$\begin{array}{c}\hfill {\int }_0^1{Q}^{p+2}\left(q\chi \right)d_q\chi \ge \frac{2^{p+2}}{p+3}{\left({\int }_0^{1}Q\left(\chi \right)d_q\chi \right)}^{p+2}.\end{array}$$

(15)

Proof. 

Include ${\displaystyle \lambda \left(\chi \right)={\int }_0^{x}Q\left(u\right)d_{q}u}$ and

$$\begin{array}{c}\hfill F\left(\chi \right)={\int }_0^{\chi }{Q}^{p+2}\left(qu\right)d_{q}u-\frac{2^{p+2}}{p+3}{\left({\int }_0^{\chi }Q\left(u\right)d_{q}u\right)}^{p+2}\chi \in {[0,1]}_q.\end{array}$$

(16)

We have

$$\begin{array}{c}\hfill D_{q}F\left(\chi \right)=Q^{p+2}\left(q\chi \right)-\frac{2^{p+2}}{p+3}{D}_q\left[{\lambda }^{p+2}\right]\left(\chi \right)\chi \in {[0,1]}_q.\end{array}$$

(17)

Since Q and $\lambda$ both are increasing on ${[0,1]}_q$, we obtain the following from Lemma 1 for $\chi \in {(0,1]}_q$:

$$\begin{array}{ccc}\hfill D_{q}F\left(\chi \right)& \ge & Q^{p+2}\left(q\chi \right)-\frac{p+2}{p+3}{2}^{p+2}{\lambda }^{p+1}\left(\chi \right)Q\left(\chi \right)\hfill \\ \hfill & \ge & Q^{p+2}\left(q\chi \right)-\frac{p+2}{p+3}{2}^{p+2}{\lambda }^{p+2}\left(\chi \right)Q\left(q\chi \right)=Q\left(q\chi \right)h\left(\chi \right),\hfill \end{array}$$

(18)

Here, ${\displaystyle h\left(\chi \right)=Q^{p+1}\left(q\chi \right)-\frac{p+2}{p+3}{2}^{p+2}{\lambda }^{p+1}\left(\chi \right)}$.

We also have

$$\begin{array}{c}\hfill D_{q}h\left(\chi \right)=D_q{Q}^{p+1}\left(q\chi \right)-\frac{p+2}{p+3}{2}^{p+2}{D}_q{\lambda }^{p+1}\left(\chi \right).\end{array}$$

By using Lemma 1 again, we obtain

$$\begin{array}{ccc}\hfill D_{q}h\left(\chi \right)& \ge & (p+1)Q^p\left(\chi \right)D_{q}Q\left(q\chi \right)-\frac{(p+1)(p+2)}{p+3}{2}^{p+2}{\lambda }^p\left(\chi \right)Q\left(\chi \right)\hfill \end{array}$$

(19)

$$\begin{array}{ccc}\hfill & \ge & (p+1)Q\left(\chi \right)[Q^{p-1}\left(\chi \right)D_{q}Q\left(q\chi \right)-\frac{p+2}{p+3}{2}^{p+3}{\lambda }^p\left(q\chi \right)].\hfill \end{array}$$

(20)

Since the function Q is increasing, for $\chi \in {[0,1]}_q$, we have

$$\begin{array}{c}\hfill {\int }_0^{q\chi }Q\left(u\right)d_{q}u\le Q\left(q\chi \right)(1-0)=Q\left(q\chi \right).\end{array}$$

Therefore, we have

$$\begin{array}{c}\hfill D_{q}h\left(\chi \right)\ge (p+1)Q^p\left(\chi \right)[D_{q}Q\left(q\chi \right)-\frac{p+2}{p+3}{2}^{p+3}]\ge 0,\end{array}$$

We deduce from the conditions given in Equation (14) that h is increasing on ${[0,1]}_q$.

Finally, since $h\left(0\right)=Q^{p+1}\left(0\right)=0$, it follows that F increases and $F\left(1\right)\ge F\left(0\right)=0$, which completes the proof. □

Theorem 4.

Let Q be an increasing function such that $Q\left(0\right)=0$. Then, for all $p\ge 0$, we have

$$\begin{array}{c}\hfill {\int }_0^1{\left[Q\left(q\chi \right)\right]}^{2p+1}{d}_q\chi \ge \frac{2^{2p+1}}{2p+2}{\left[{\int }_0^{1}Q\left(\chi \right)d_q\chi \right]}^{2p+1},\end{array}$$

(21)

provided that $D_{q}f\ge \frac{2p+1}{2p+2}{2}^{2p+1}$.

Proof. 

For $\tau \in {[0,1]}_q$, we put forth

$$\begin{array}{c}\hfill F\left(\tau \right)={\int }_0^{\tau }{\left[Q\left(q\chi \right)\right]}^{2p+1}{d}_q\chi -\frac{2^{2p+1}}{2p+2}{\left[{\int }_0^{t}Q\left(\chi \right)d_q\chi \right]}^{2p+1}and\lambda \left(\tau \right)={\int }_0^{\tau }Q\left(q\chi \right)d_q\chi .\end{array}$$

Then, for $\tau \in {[0,1]}_q$, we have

$$\begin{array}{ccc}\hfill D_{q}F\left(\tau \right)& =& {\left[Q\left(q\tau \right)\right]}^{2p+1}-\frac{2^{2p+1}}{2p+2}{D}_q{\left[\lambda \left(\tau \right)\right]}^{2p+1}\hfill \\ \hfill & \ge & {\left[Q\left(\tau \right)\right]}^{2p+1}-\frac{2p+1}{2p+2}{2}^{2p+1}{\lambda }^{2p}\left(\tau \right)Q\left(qt\right)\hfill \\ & \ge & {\left[Q\left(q\tau \right)\right]}^{2p+1}-\frac{2p+1}{2p+2}{2}^{2p+1}{\lambda }^{2p}\left(\tau \right)Q\left(q\tau \right)\hfill \end{array}$$

(22)

$$\begin{array}{ccc}& =& Q\left(q\tau \right)\left({\left[Q\left(q\tau \right)\right]}^{2p}-\frac{2p+1}{2p+2}{2}^{2p+1}{\lambda }^{2p}\left(\tau \right)\right)\hfill \\ & =& Q\left(q\tau \right)G\left(\tau \right),\hfill \end{array}$$

(23)

where ${\displaystyle G\left(\tau \right)={\left[Q\left(q\tau \right)\right]}^{2p}-\frac{2p+1}{2p+2}{2}^{2p+1}{\lambda }^{2p}\left(\tau \right)}$.

Furthermore, we have

$$\begin{array}{ccc}\hfill D_{q}G\left(\tau \right)& \ge & 2p{Q}^{2p-1}\left(qt\right)D_{q}Q\left(\tau \right)-\frac{2p(2p+1)}{2p+2}{2}^{2p+1}{\lambda }^{2p-1}\left(\tau \right)Q\left(qt\right)\hfill \\ \hfill & \ge & 2pQ\left(qt\right)\left(Q^{2p}\left(qt\right)D_{q}Q\left(\tau \right)-\frac{2p+1}{2p+2}{2}^{2p+1}{\lambda }^{2p-1}\left(\tau \right)\right).\hfill \end{array}$$

(24)

Since the function Q is increasing, we have the following for $\chi \in {[0,1]}_q$:

$$\begin{array}{c}\hfill {\int }_0^{q\chi }Q\left(qu\right)d_{q}u\le Q\left(q\chi \right)(1-0)=Q\left(q\chi \right).\end{array}$$

Therefore, we have

$$\begin{array}{c}\hfill D_{q}G\left(\tau \right)\ge 2p{Q}^{2p}\left(qt\right)[D_{q}Q\left(\tau \right)-\frac{2p+1}{2p+2}{2}^{2p+1}]\ge 0,\end{array}$$

Hence, G is increasing on ${[0,1]}_q$. Moreover, we have

$$\begin{array}{c}\hfill G\left(0\right)={\left[Q\left(0\right)\right]}^{2p}-\frac{2p+1}{2p+2}{2}^{2p+1}{\lambda }^{2p}\left(0\right)=0,\end{array}$$

for all $t\in {(0,1]}_q$, $G\left(\tau \right)>G\left(0\right)=0$, implying that $D_{q}F\left(\tau \right)>0$, ∀$\tau \in {(0,1]}_q$. Thus, F is increasing on ${[0,1]}_q$. In particular, $F\left(1\right)>F\left(0\right)=0$, which proves our claim. □

3. h-Integral Inequalities of the Anderson Type

3.1. Notations and Preliminaries

Let us recall some definitions from [15].

Let $h\ne 0$. A “quantum derivative of a function” Q, denoted by $D_{h}Q$, is given by

$$\begin{array}{c}\hfill D_{h}Q\left(\chi \right)=\frac{Q(\chi +h)-Q\left(\chi \right)}{h}.\end{array}$$

(25)

One can easily verify that

$$\begin{array}{ccc}\hfill D_h\left(Q\left(\chi \right)\lambda \left(\chi \right)\right)& =& Q\left(\chi \right)D_h\lambda \left(\chi \right)+\lambda (\chi +h)D_{h}Q\left(\chi \right)\hfill \end{array}$$

(26)

$$\begin{array}{ccc}\hfill D_h\left(\frac{Q\left(\chi \right)}{\lambda \left(\chi \right)}\right)& =& \frac{\lambda \left(\chi \right)D_{h}Q\left(\chi \right)-f\left(\chi \right)D_h\lambda \left(\chi \right)}{\lambda \left(\chi \right)\lambda (\chi +h)}\hfill \end{array}$$

(27)

If $Q^{\prime }\left(0\right)$ exists, then $D_{h}Q\left(0\right)=Q^{\prime }\left(0\right)$. As $h\to 0$, we find an ordinary derivative.

If $m_1-m_0\in h\mathbb{Z}$, the definite h-integral is defined by [4]

$$\begin{array}{c}\hfill {\int }_{m_0}^{m_1}Q\left(\chi \right)d_{h}x=\left\{\begin{array}{cc}h\left(Q\left(m_0\right)+Q(m_0+h)+\cdots +Q(m_1-h)\right),\hfill & \mathrm{if}{m}_0<m_1,\hfill \\ 0,\hfill & \mathrm{if}{m}_0=m_1\hfill \\ -h\left(Q\left(m_1\right)+Q(m_1+h)+\cdots +Q(m_0-h)\right),\hfill & \mathrm{if}{m}_0>m_1.\hfill \end{array}\right.\end{array}$$

(28)

The following theorem, whose proof can be found in [4], justifies Equation (28):

Theorem 5.

If F is an h-antiderivative of Q and $m_1-m_0\in h\mathbb{Z}$, then

$${\int }_{m_0}^{m_1}Q\left(\chi \right)d_h\chi =F\left(m_1\right)-F\left(m_0\right).$$

(29)

By applying Theorem 5 to $D_h\left(Q\left(\chi \right)\lambda \left(\chi \right)\right)$ and using Equation (26), one can find the following:

$$\begin{array}{c}\hfill {\int }_{m_0}^{m_1}Q\left(\chi \right)d_h\left(\chi \right)=Q\left(m_1\right)\lambda \left(m_1\right)-Q\left(m_0\right)\lambda \left(m_0\right)-{\int }_0^1\lambda (\chi +h)d_{h}Q\left(\chi \right).\end{array}$$

(30)

For any function Q, one can easily verify

$$\begin{array}{c}\hfill D_h\left[{\int }_a^{x}Q\left(\tau \right)d_h\tau \right]=Q\left(\chi \right).\end{array}$$

(31)

3.2. Main Results

Before we proceed further, we need a lemma:

Lemma 2.

If we let $n\in \mathbb{R}$ s.t. $n\ge 1$ and the function λ be non-negative increasing on $[0,1]$, then

$$\begin{array}{c}\hfill n{\lambda }^n\left(\chi \right)D_h\lambda \left(\chi \right)\le D_h{\left[\lambda \left(\chi \right)\right]}^n\le n{\lambda }^n\left(\chi \right)D_h\lambda \left(\chi \right),\end{array}$$

(32)

Proof. 

We have

$$\begin{array}{c}\hfill D_h\left[{\lambda }^n\right]\left(\chi \right)=\frac{{\lambda }^n(\chi +h)-{\lambda }^n\left(\chi \right)}{h}=\frac{n}{h}{\int }_{\lambda \left(\chi \right)}^{\lambda (\chi +h)}{\tau }^{n-1}{d}_h\tau .\end{array}$$

(33)

Since $\lambda$ is a non-negative increasing function, we have

$$\begin{array}{c}\hfill {\lambda }^n\left(\chi \right)[\lambda (\chi +h)-\lambda \left(\chi \right)]\le {\int }_{\lambda \left(\chi \right)}^{\lambda (\chi +h)}{\tau }^{n-1}{d}_h\tau \le {\lambda }^n(\chi +h)[\lambda (\chi +h)-\lambda \left(\chi \right)].\end{array}$$

(34)

Therefore, through Equations (33) and (34), we obtain our required result. □

Theorem 6.

Let function Q be non-negative and increasing on $[0,1]$ and satisfy

$$\begin{array}{c}\hfill Q^{\xi -2}\left(\chi \right)D_{h}Q\left(\chi \right)\ge \frac{\xi }{\xi +1}{2}^{\xi }{\left(\chi +2h\right)}^{\xi -2}{Q}^{\xi -1}\left(\chi +2h\right)for\chi \ge 2and(1-0)\in hz.\end{array}$$

Then, we have

$$\begin{array}{c}\hfill {\int }_0^1{\left[Q\left(\chi \right)\right]}^{\xi }{d}_{h}x\ge \frac{2^{\xi }}{\xi +1}{\left[\left({\int }_0^{1}Q\left(q\chi \right)d_{h}x\right)\right]}^{\xi }.\end{array}$$

(35)

Proof. 

For $\chi \in [0,1]$, let

$$\begin{array}{c}\hfill F\left(\chi \right)={\int }_0^{\chi }{\left[Q\left(u\right)\right]}^{\xi }{d}_{h}u-\frac{2^{\xi }}{\xi +1}{\left[\left({\int }_0^{\chi }Q\left(u\right)d_{h}u\right)\right]}^{\xi }.\end{array}$$

and $\lambda \left(\chi \right)={\int }_0^{\chi }Q\left(u\right)d_{h}u$. By virtue of Lemma 2, it follows that

$$\begin{array}{c}\hfill D_{h}F\left(\chi \right)=F^{\xi }\left(\chi \right)-\frac{2^{\xi }}{\xi +1}{D}_h\left[{\lambda }^{\xi }\right]\left(\chi \right).\\ \hfill \ge Q^{\xi }\left(\chi \right)-\frac{2^{\xi }}{\xi +1}{\lambda }^{\xi -1}(\chi +h)Q\left(\chi \right)\\ \hfill =Q\left(\chi \right)\left[Q^{\xi -1}\left(\chi \right)-\frac{2^{\xi }}{\xi +1}{\lambda }^{\xi -1}(\chi +h)\right],\end{array}$$

(36)

where

$$F_1\left(\chi \right)=Q^{\xi }\left(\chi \right)-\frac{2^{\xi }}{\xi +1}{\lambda }^{\xi -1}(\chi +h)Q\left(\chi \right).$$

With Lemma 2, we have

$$\begin{array}{ccc}\hfill D_h{F}_1& =& D_h\left(Q^{\xi -1}\left(\chi \right)-\frac{2^{\xi }}{\xi +1}{\lambda }^{\xi -1}(\chi +h)Q\left(\chi \right)\right)\hfill \end{array}$$

(37)

$$\begin{array}{ccc}& \ge & \left(\xi -1\right)Q^{\xi -2}\left(\chi \right)D_{h}Q\left(\chi \right)-\frac{\xi (\xi -1)}{\xi +1}{2}^{\xi }{\lambda }^{\xi -2}(\chi +2h)D_h\lambda (\chi +h)\hfill \\ \hfill & =& \left(n-1\right)Q^{\xi -2}\left(\chi \right)D_{h}Q\left(\chi \right)-\frac{n(\xi -1)}{\xi +1}{2}^{\xi }{\lambda }^{\xi -2}(\chi +2h)Q(\chi +h)\hfill \end{array}$$

(38)

Since the function $\lambda$ is increasing and non-negative, therefore

$$\begin{array}{c}\hfill \lambda (\chi +2h)={\int }_0^{\chi +2h}Q\left(u\right)d_{h}u\le (\chi +2h-0)Q(\chi +2h)\end{array}$$

(39)

Hence, we have

$$\begin{array}{c}\hfill D_h{F}_1\left(\chi \right)\ge (\xi -1)Q^{\xi -2}\left(\chi \right)D_{h}Q\left(\chi \right)-\frac{\xi (\xi -1)}{\xi +1}{2}^{\xi }{\lambda }^{\xi -2}(\chi +2h)Q(\chi +2h).\end{array}$$

$$\begin{array}{c}\hfill =(\xi -1)Q^{\xi -2}\left(\chi \right)D_{h}Q\left(\chi \right)-\frac{\xi (\xi -1)}{\xi +1}{2}^{\xi }{\lambda }^{\xi -2}(\chi +2h-0)Q^{\xi -2}(\chi +2h)\end{array}$$

(40)

$$\begin{array}{c}\hfill =(\xi -1)\left(Q^{\xi -2}\left(\chi \right)D_{h}Q\left(\chi \right)-\frac{\xi }{\xi +1}{2}^{\xi }{(\chi +2h)}^{\xi -2}{Q}^{\xi -2}(\chi +2h)\right)\ge 0.\end{array}$$

(41)

which assures that $F_1$ is increasing. Hence, $F_1\left(\chi \right)\ge F_1\left(0\right)\ge 0$ and $D_{h}F\left(\chi \right)\ge 0$, so F is increasing since $F\left(\chi \right)\ge F\left(0\right)=0.$ □

Theorem 7.

Let $n\ge 1$. If the function Q is non-negative and increasing on $[0,1]$ and satisfies

$$\begin{array}{c}\hfill Q^{n-1}\left(\chi \right)D_{h}Q\left(\chi \right)\ge 2^n{(\chi +2h)}^{n-1}{Q}^n(\chi +2h)\end{array}$$

(42)

then

$$\begin{array}{c}\hfill {\int }_0^1{Q}^{n+1}\left(\chi \right)d_h\left(\chi \right)\ge \frac{2^n}{n+1}{\left({\int }_0^{1}Q\left(\chi \right)d_h\left(\chi \right)\right)}^{n+1}\end{array}$$

(43)

Proof. 

For $\chi \in [0,1]$, let

$$\begin{array}{c}\hfill F\left(\chi \right)={\int }_0^{\chi }{Q}^{n+1}\left(u\right)d_h\left(\tau \right)-\frac{2^n}{n+1}{\left({\int }_0^{\chi }Q\left(u\right)d_h\left(u\right)\right)}^{n+1}\end{array}$$

and

$$\begin{array}{c}\hfill \lambda \left(\chi \right)={\int }_0^{\chi }Q\left(u\right)d_h\left(u\right).\end{array}$$

Utilizing Lemma 2 gives

$$\begin{array}{ccc}\hfill D_{h}F\left(\chi \right)& =& Q^{n+1}\left(\chi \right)-\frac{2^n}{n+1}{D}_h\left({\lambda }^{n+1}\left(\chi \right)\right)\hfill \\ & \ge & Q^{n+1}\left(\chi \right)-\frac{2^n}{n+1}(n+1){\lambda }^n(\chi +h)Q\left(\chi \right)\hfill \\ & =& Q^{n+1}\left(\chi \right)-2^n{\lambda }^n(\chi +h)Q\left(\chi \right)\hfill \\ & =& Q\left(\chi \right)\left[Q^n\left(\chi \right)-2^n{\lambda }^n(\chi +h)\right]\hfill \end{array}$$

where

$$\begin{array}{c}\hfill F_1\left(\chi \right)=Q^n\left(\chi \right)-2^n{\lambda }^n(\chi +h).\end{array}$$

From Lemma 2, it follows that

$$\begin{array}{ccc}\hfill D_h{F}_1\left(\chi \right)& =& D_h\left(Q^n\left(\chi \right)\right)-2^n{D}_h\left({\lambda }^n(\chi +h)\right)\hfill \\ & \ge & n{Q}^{n-1}\left(\chi \right)D_{h}Q\left(\chi \right)-2^n.n{\lambda }^{n-1}(\chi +2h)Q(\chi +h).\hfill \end{array}$$

Since Q is increasing and non-negative, then

$$\begin{array}{c}\hfill \lambda (\chi +2h)={\int }_0^{\chi +2h}Q\left(u\right)d_h\left(u\right)\le (\chi +2h)Q(\chi +2h)\end{array}$$

(44)

Hence, we have

$$\begin{array}{ccc}\hfill D_h{F}_1\left(\chi \right)& \ge & n{Q}^{n-1}\left(\chi \right)D_{h}Q\left(\chi \right)-n.2^n{(\chi +2h)}^{n-1}{Q}^{n-1}(\chi +2h)Q(\chi +h)\hfill \\ & \ge & n{Q}^{n-1}\left(\chi \right)D_{h}Q\left(\chi \right)-n.2^n{(\chi +2h)}^{n-1}{Q}^n(\chi +2h)\hfill \end{array}$$

$$\begin{array}{c}\hfill D_h{F}_1\left(\chi \right)=n\left(Q^{n-1}\left(\chi \right)D_{n}Q\left(\chi \right)-2^n{(\chi +2h)}^{n-1}{Q}^n(\chi +2h)\right)\ge 0\end{array}$$

(45)

We conclude that $F_1$ is an increasing function. Hence, $F_1\left(\chi \right)\ge F_1\left(0\right)=0$ and $D_{h}F\left(\chi \right)\ge 0$. Therefore, F increases since $F\left(\chi \right)\ge F\left(0\right)=0.$ □

4. Conclusions

• In the current study, we analyzed a Feng Qi type q-integral inequality and h-integral inequality and then discussed an Anderson type integral inequality in quantum calculus;
• We considered how to translate some Anderson type integral inequalities into quantum calculus (i.e., q-integral and h integral inequalities);
• By transforming these Anderson type integral inequalities into the equivalent q-integral inequalities and h-integral inequalities, we established a solution method for the corresponding fractional integral inequalities. Our work further develops the solution of time-scale calculus.