Right Quantum Calculus on Finite Intervals with Respect to Another Function and Quantum Hermite–Hadamard Inequalities

Abstract

In this paper, we study right quantum calculus on finite intervals with respect to another function. We present new definitions on the right quantum derivative and right quantum integral of a function with respect to another function and study their basic properties. The new definitions generalize the previous existing results in the literature. We provide applications of the newly defined quantum calculus by obtaining new Hermite–Hadamard-type inequalities for convex, h-convex, and modified h-convex functions.

1. Introduction

Quantum calculus, also known as q-calculus or non-standard calculus, is a mathematical framework that extends traditional calculus to handle discrete and quantum phenomena. It introduces new operators, such as q-derivatives and q-integrals, which capture the discrete nature of the underlying systems. The development of quantum calculus was motivated by the need to model and understand phenomena at the quantum level, where traditional calculus fails to provide accurate descriptions. By incorporating principles from quantum mechanics and discrete mathematics, quantum calculus offers a powerful mathematical framework for solving problems in various fields, including physics, engineering, and computer science.

Historically, in the eighteenth century, Euler obtained the basic formulae in q-calculus. However, Jackson [1] first established what are known as q-derivatives and q-integrals. Currently, there is a significant interest in implementing the q-calculus due to its applications in several areas, such as mathematics, physics, number theory, orthogonal polynomials, hypergeometric functions, and combinatorics; see [2,3]. The q-derivatives and q-integrals were generalized to non-integer orders in [4,5]. For some recent results, we refer the reader to [6,7,8,9,10,11,12,13,14,15] and the references cited therein. The quantum calculus on finite intervals was introduced by Tariboon and Ntouyas [16]. See also [17] for further details on quantum calculus and the recent results.

Recently, in [18], we initiated the study of the quantum calculus on finite intervals with respect to another function. For a function $\Pi :[x,y]\to \mathbb{R}$, the quantum derivative ${ }_x{\mathcal{D}}_{q,\psi }\Pi$ and the quantum integral ${ }_x{\mathcal{I}}_{q,\psi }\Pi$ with respect to the function $\psi$ and quantum number q were defined, and their properties were discussed. The newly defined quantum calculus on finite intervals with respect to another function was applied, and a new Hermite–Hadamard quantum inequality for a convex function was obtained. In this paper, we extend further the quantum calculus on finite intervals with respect to another function by defining the corresponding right quantum derivative ${ }^y{\mathcal{D}}_{q,\psi }\Pi$ and the right quantum integral ${ }^y{\mathcal{I}}_{q,\psi }\Pi$ with respect to the function $\psi$ and quantum number q for a function $\Pi :[x,y]\to \mathbb{R}.$ The basic properties of the right quantum derivative and integral are proved in detail. For the newly defined notions, we prove the corresponding Hermite–Hadamard inequalities for some classes of convex functions. For comprehensive reviews of the Hermite–Hadamard inequality pertaining to fractional integral operators and to quantum calculus, respectively, see [19,20]. For some recent results on Hermite–Hadamard inequality, see [21,22,23,24,25,26] and the references cited therein.

The rest of this paper is organized as follows. In Section 2, we present the new definitions of the right quantum derivative ${ }^y{\mathcal{D}}_{q,\psi }\Pi$ and the right quantum integral ${ }^y{\mathcal{I}}_{q,\psi }\Pi$ with respect to the function $\psi$ and show their basic properties. As an application of the newly defined right quantum notions, we obtain Hermite–Hadamard-type inequalities for convex, h-convex, and modified h-convex functions. Our results are novel and significantly contribute to the literature on this new subject regarding quantum calculus on finite intervals with respect to another function.

2. Right Quantum Calculus on Finite Intervals with Respect to Another Function

Let us recall the new definition of quantum derivative with respect to the function $\psi$ provided recently in [18].

Definition 1. 

Let $\Pi :[x,y]\to \mathbb{R}$ be a continuous function and $q\in (0,1)$ and $\psi :[x,y]\to \mathbb{R},x\ge 0$ be a strictly increasing function. The $q_x$-derivative ${ }_x{\mathcal{D}}_{q,\psi }\Pi$ with respect to the function ψ on $[x,y]$ is defined by

$${ }_x{\mathcal{D}}_{q,\psi }\Pi \left(z\right)=\frac{\Pi \left(z\right)-\Pi (qz+(1-q\left)x\right)}{\psi \left(z\right)-\psi (qz+(1-q\left)x\right)},z\ne x,$$

(1)

and ${ }_x{\mathcal{D}}_{q,\psi }\Pi \left(x\right)={lim}_{z\to x}\{{ }_x{\mathcal{D}}_{q,\psi }\Pi \left(z\right)\}$.

Remark 1. 

$\left(i\right)$ Note that, if $\psi \left(z\right)=z$, then

$${ }_x{\mathcal{D}}_{q,z}\Pi \left(z\right)=\frac{\Pi \left(z\right)-\Pi (qz+(1-q\left)x\right)}{(1-q)(z-x)},$$

which is the q-derivative defined by Tariboon–Ntouyas in [16]. For $x=0$, it provides Jackson’s q-derivative [1]

$${ }_0{\mathcal{D}}_{q,z}\Pi \left(z\right)=\frac{\Pi \left(z\right)-\Pi \left(qz\right)}{(1-q)z}.$$

Now, we define the quantum right $q^y$-derivative of a function $\Pi$ with respect to the function $\psi .$

Definition 2. 

Let $\Pi :[x,y]\to \mathbb{R}$ be a continuous function and $q\in (0,1)$ and $\psi :[x,y]\to \mathbb{R},x\ge 0$ be a strictly increasing function. The right $q^y$-derivative ${ }^y{\mathcal{D}}_{q,\psi }\Pi$ of a function Π with respect to the function ψ, on $[x,y]$, is defined by

$${ }^y{\mathcal{D}}_{q,\psi }\Pi \left(z\right)=\frac{\Pi \left(z\right)-\Pi (qz+(1-q\left)y\right)}{\psi \left(z\right)-\psi (qz+(1-q\left)y\right)}=\frac{\Pi \left(z\right)-\Pi \left({ }^y{Q}_q\left(z\right)\right)}{\psi \left(z\right)-\psi \left({ }^y{Q}_q\left(z\right)\right)},t\ne y,$$

(2)

and ${ }^y{\mathcal{D}}_{q,\psi }\Pi \left(y\right)={lim}_{z\to y}\{{ }^y{\mathcal{D}}_{q,\psi }\Pi \left(z\right)\}$, where

$${ }^y{Q}_q\left(z\right)=qz+(1-q)y$$

(3)

is a q-shifting operator.

Remark 2. 

$\left(i\right)$ If $\psi \left(z\right)=z$, then we obtain the $q^y$-derivative of Π at $z\in [x,y]$ defined in Bermudo et al. [27] as

$${ }^y{\mathcal{D}}_{q,z}\Pi \left(z\right)=\frac{\Pi \left(z\right)-\Pi (qz+(1-q\left)y\right)}{(1-q)(t-y)}.$$

$\left(ii\right)$ If $\psi \left(z\right)=logz$, where $logz={log}_{e}z$ and $z\in [x,y],x>0$, then we have

$${ }^y{\mathcal{D}}_{q,logz}\Pi \left(z\right)=\frac{\Pi \left(z\right)-\Pi (qz+(1-q\left)y\right)}{logz-log(qz+(1-q\left)y\right)},$$

(4)

which is the new definition of quantum derivative in the Hadamard sense. Indeed, it is enough to note that

$$\underset{q\to 1}{lim}\frac{\Pi \left(z\right)-\Pi (qz+(1-q\left)y\right)}{logz-log(qz+(1-q\left)y\right)}=z\frac{d}{dz}\Pi \left(z\right),$$

by L’H$\widehat{o}$pital’s rule, which is the ordinary Hadamard derivative.

Corollary 1. 

If $\Pi \left(z\right)={\psi }^n\left(z\right)$ in (2), $n\in {\mathbb{Z}}^{+},$ then we have

$${ }^y{\mathcal{D}}_{q,\psi }{\psi }^n\left(z\right)=\sum _{i=0}^{n-1}{\psi }^{n-1-i}\left(z\right){\psi }^i(qz+(1-q)y).$$

(5)

Proof. 

Using (2), we obtain

$$\begin{array}{ccc}\hfill { }^y{\mathcal{D}}_{q,\psi }{\psi }^n\left(z\right)& =& \frac{{\psi }^n\left(z\right)-{\psi }^n\left({ }^y{Q}_q\left(z\right)\right)}{\psi \left(z\right)-\psi \left({ }^y{Q}_q\left(z\right)\right)}\hfill \\ & =& \frac{1}{\psi \left(z\right)-\psi \left({ }^y{Q}_q\left(z\right)\right)}[\left(\psi \left(z\right)-\psi \left({ }^y{Q}_q\left(z\right)\right)\right)({\psi }^{n-1}\left(z\right)+{\psi }^{n-2}\left(z\right)\psi \left({ }^y{Q}_q\left(z\right)\right)\hfill \\ & & +\cdots +\psi \left(z\right){\psi }^{n-2}\left({ }^y{Q}_q\left(z\right)\right)+{\psi }^{n-1}\left({ }^y{Q}_q\left(z\right)\right)\left)\right]\hfill \\ & =& \sum _{i=0}^{n-1}{\psi }^{n-1-i}\left(z\right){\psi }^i\left({ }^y{Q}_q\left(z\right)\right),\hfill \end{array}$$

which ends the proof. □

In the next lemma, we summarize the basic properties of the right $q^y$-quantum derivative with respect to another function.

Lemma 1. 

$\left(i\right)$ (Linearity).

$${ }^y{\mathcal{D}}_{q,\psi }(\alpha \Pi \left(z\right)+\beta P\left(z\right))=\alpha { }^y{\mathcal{D}}_{q,\psi }\Pi \left(z\right)+\beta { }^y{\mathcal{D}}_{q,\psi }P\left(z\right),\alpha ,\beta constants.$$

$\left(ii\right)$ (Quantum derivative of product).

$$\begin{array}{ccc}\hfill { }^y{\mathcal{D}}_{q,\psi }(\Pi P)\left(z\right)& =& \Pi \left(z\right){ }^y{\mathcal{D}}_{q,\psi }P\left(z\right)+P(qz+(1-q)y){ }^y{\mathcal{D}}_{q,\psi }\Pi \left(z\right)\hfill \\ & =& P\left(z\right){ }^y{\mathcal{D}}_{q,\psi }\Pi \left(z\right)+\Pi (qz+(1-q)y){ }^y{\mathcal{D}}_{q,\psi }P\left(z\right).\hfill \end{array}$$

$\left(iii\right)$ (Quantum derivative of quotient).

$$\begin{array}{c}\hfill { }^y{\mathcal{D}}_{q,\psi }\left(\frac{\Pi }{P}\right)\left(z\right)=\frac{P\left(z\right){ }^y{\mathcal{D}}_{q,\psi }\Pi \left(z\right)-\Pi \left(z\right){ }^y{\mathcal{D}}_{q,\psi }P\left(z\right)}{P\left(z\right)P(qz+(1-q\left)y\right)},\end{array}$$

where $P\left(z\right)P(qz+(1-q\left)y\right)\ne 0$ for all $z\in [x,y]$.

Proof. 

For $\left(i\right)$, we have

$$\begin{array}{ccc}\hfill { }^y{\mathcal{D}}_{q,\psi }(\alpha \Pi \left(z\right)+\beta P\left(z\right))& =& \frac{\alpha \Pi \left(z\right)+\beta P\left(z\right)-\alpha \Pi \left({ }^y{Q}_q\left(z\right)\right)-\beta P\left({ }^y{Q}_q\left(z\right)\right)}{\psi \left(z\right)-\psi \left({ }^y{Q}_q\left(z\right)\right)}\hfill \\ & =& \frac{\alpha \left[\Pi \left(z\right)-\Pi \left({ }^y{Q}_q\left(z\right)\right)\right]+\beta \left[P\left(z\right)-P\left({ }^y{Q}_q\left(z\right)\right)\right]}{\psi \left(z\right)-\psi \left({ }^y{Q}_q\left(z\right)\right)}\hfill \\ & =& \alpha { }^y{\mathcal{D}}_{q,\psi }\Pi \left(z\right)+\beta { }^y{\mathcal{D}}_{q,\psi }P\left(z\right).\hfill \end{array}$$

Next, for the first part of $\left(ii\right)$, we have

$$\begin{array}{ccc}\hfill { }^y{\mathcal{D}}_{q,\psi }(\Pi P)\left(z\right)& =& \frac{(\Pi P)\left(z\right)-(\Pi P)\left({ }^y{Q}_q\left(z\right)\right)}{\psi \left(z\right)-\psi \left({ }^y{Q}_q\left(z\right)\right)}\hfill \\ & =& \frac{\Pi \left(z\right)P\left(z\right)-P\left({ }^y{Q}_q\left(z\right)\right)\Pi \left(z\right)-\Pi \left({ }^y{Q}_q\left(z\right)\right)P\left({ }^y{Q}_q\left(z\right)\right)+P\left({ }^y{Q}_q\left(z\right)\right)\Pi \left(z\right)}{\psi \left(z\right)-\psi \left({ }^y{Q}_q\left(z\right)\right)}\hfill \\ & =& \Pi \left(z\right){ }^y{\mathcal{D}}_{q,\psi }P\left(z\right)+P\left({ }^y{Q}_q\left(z\right)\right){ }^y{\mathcal{D}}_{q,\psi }\Pi \left(z\right).\hfill \end{array}$$

We can prove the second equality in a similar way. To show $\left(iii\right)$, we have

$$\begin{array}{ccc}\hfill { }^y{\mathcal{D}}_{q,\psi }\left(\frac{\Pi }{P}\right)\left(z\right)& =& \frac{\left(\frac{\Pi }{P}\right)\left(z\right)-\left(\frac{\Pi }{P}\right)\left({ }^y{Q}_q\left(z\right)\right)}{\psi \left(z\right)-\psi \left({ }^y{Q}_q\left(z\right)\right)}\hfill \\ & =& \frac{\Pi \left(z\right)P\left({ }^y{Q}_q\left(z\right)\right)+\Pi \left(z\right)P\left(z\right)-\Pi \left({ }^y{Q}_q\left(z\right)\right)P\left(z\right)-\Pi \left(z\right)P\left(z\right)}{P\left(z\right)P\left({ }^y{Q}_q\left(z\right)\right)\left(\psi \left(z\right)-\psi \left({ }^y{Q}_q\left(z\right)\right)\right)}\hfill \\ & =& \frac{P\left(z\right){ }^y{\mathcal{D}}_{q,\psi }\Pi \left(z\right)-\Pi \left(z\right){ }^y{\mathcal{D}}_{q,\psi }P\left(z\right)}{P\left(z\right)P\left({ }^y{Q}_q\left(z\right)\right)}.\hfill \end{array}$$

 □

To derive the right quantum integral with respect to another function, we define a q-shifting operator by

$${ }^y{Q}_q\Pi \left(z\right)=\Pi \left({ }^y{Q}_q\left(z\right)\right)=\Pi (qz+(1-q)y),\mathrm{with}{ }^y{Q}_q^0\Pi \left(z\right)=\Pi \left(z\right).$$

Then, we can see that ${ }^y{Q}_q^n\Pi \left(z\right)=\Pi \left({ }^y{Q}_q^n\left(z\right)\right)=\Pi (q^{n}z+(1-q^n)y)$ for all $n\in {\mathbb{Z}}_0$. To show this, for $n=0$, we have ${ }^{y}Q{ }_q^0\Pi \left(z\right)=\Pi (q^{0}z+(1-q^0)y)=\Pi \left(z\right)$. Assume that ${ }^y{Q}_q^k\Pi \left(z\right)=\Pi (q^{k}z+(1-q^k)y)$ holds for $n=k.$ Then, we obtain

$$\begin{array}{ccc}\hfill { }^y{Q}_q^{k+1}\Pi \left(z\right)& =& { }^y{Q}_q\left({ }^y{Q}_q^k\Pi \left(z\right)\right)\hfill \\ & =& { }^y{Q}_q\left(\Pi (q^{k}z+(1-q^k)y)\right)\hfill \\ & =& \Pi (q^k(qz+(1-q)y)+(1-q^k)y)\hfill \\ & =& \Pi (q^{k+1}z+(1-q^{k+1})y).\hfill \end{array}$$

Thus, it holds that $n=k+1,$ and, hence, for every $n,$ by mathematical induction.

From (2), we set

$$\begin{array}{c}\hfill H\left(z\right)=\frac{\Pi \left(z\right)-\Pi (qz+(1-q\left)y\right)}{\psi \left(z\right)-\psi (qz+(1-q\left)y\right)}=\frac{(1-{ }^y{Q}_q)\Pi \left(z\right)}{\psi \left(z\right)-\psi (qz+(1-q\left)y\right)},\end{array}$$

and we obtain

$$\begin{array}{ccc}\hfill \Pi \left(z\right)& =& \frac{1}{1-{ }^y{Q}_q}\left[\psi \left(z\right)-\psi (qz+(1-q\left)y\right)\right]H\left(z\right)\hfill \\ & =& \sum _{i=0}^{\infty }{ }^y{Q}_q^i\left[\psi \left(z\right)-\psi (qz+(1-q\left)y\right)\right]H\left(z\right)\hfill \\ & =& \sum _{i=0}^{\infty }\left[\psi (q^{i}z+(1-q^i)y)-\psi (q^{i+1}z+(1-q^{i+1})y)\right]H(q^{i}z+(1-q^i)y),\hfill \end{array}$$

provided that the right-hand side is convergent. From this concept of quantum antiderivative, we are in a position to define the right quantum integral on finite intervals with respect to another function.

Definition 3. 

Let $\Pi :[x,y]\to \mathbb{R},$ $q\in (0,1)$, and $\psi :[x,y]\to \mathbb{R}, x\ge 0$ be a strictly increasing function. The $q^y$-integral ${ }^y{\mathcal{I}}_{q,\psi }\Pi$ with respect to the function ψ on $[x,y]$, is defined by

$$\begin{array}{ccc}\hfill { }^y{\mathcal{I}}_{q,\psi }\Pi \left(z\right)& =& {\int }_z^y\Pi \left(s\right){ }^y{d}_q^{\psi }s\hfill \\ & =& \sum _{i=0}^{\infty }\left[\psi (q^{i}z+(1-q^i)y)-\psi (q^{i+1}z+(1-q^{i+1})y)\right]\Pi (q^{i}z+(1-q^i)y)\hfill \\ & =& \sum _{i=0}^{\infty }\left[\psi \left({ }^y{Q}_q^i\left(z\right)\right)-\psi \left({ }^y{Q}_q^{i+1}\left(z\right)\right)\right]\Pi \left({ }^y{Q}_q^i\left(z\right)\right),\hfill \end{array}$$

(6)

which is well-defined if the right-hand side exists. Moreover, for $c\in (x,y)$, the right quantum integral can be written as

$$\begin{array}{ccc}\hfill {\int }_z^c\Pi \left(s\right){ }^y{d}_q^{\psi }s& =& {\int }_z^y\Pi \left(s\right){ }^y{d}_q^{\psi }s-{\int }_c^y\Pi \left(s\right){ }^y{d}_q^{\psi }s\hfill \\ & =& \sum _{i=0}^{\infty }\left[\psi (q^{i}z+(1-q^i)y)-\psi (q^{i+1}z+(1-q^{i+1})y)\right]\Pi (q^{i}z+(1-q^i)y)\hfill \\ & & -\sum _{i=0}^{\infty }\left[\psi (q^{i}c+(1-q^i)y)-\psi (q^{i+1}c+(1-q^{i+1})y)\right]\Pi (q^{i}c+(1-q^i)y).\hfill \end{array}$$

Remark 3. 

If $\psi \left(z\right)=-z$, then we obtain $q^y$-definite quantum integral defined by Bermudo et al. [27] by

$${ }^y{\mathcal{I}}_{q,z}\Pi \left(z\right)=(1-q)(y-z)\sum _{i=0}^{\infty }{q}^i\Pi (q^{i}z+(1-q^i)y).$$

Remark 4. 

If $z=0$ and $y=1$ in Remark 3, then

$${\int }_0^1\Pi \left(s\right){ }^1{d}_{q}s=(1-q)\sum _{i=0}^{\infty }{q}^i\Pi (1-q^i).$$

Example 1. 

Let $\psi \left(z\right)={(y-z)}^m$, $\Pi \left(z\right)={(y-z)}^n$, $m,n>0.$ Then,

$$\begin{array}{ccc}\hfill {\int }_z^y{(y-s)}^n{ }^y{d}_q^{{(y-z)}^m}s& =& \sum _{i=0}^{\infty }\left[{\left(q^i(y-z)\right)}^m-{\left(q^{i+1}(y-z)\right)}^m\right]{\left(q^i(y-z)\right)}^n\hfill \\ & =& {(y-z)}^{m+n}\left[\sum _{i=0}^{\infty }{\left(q^{m+n}\right)}^i-q^m\sum _{i=0}^{\infty }{\left(q^{m+n}\right)}^i\right]\hfill \\ & =& {(y-z)}^{m+n}\left(\frac{1-q^m}{1-q^{m+n}}\right)\hfill \end{array}$$

since $y-{ }^y{Q}_q^i\left(z\right)=q^i(y-z)$.

Now, some basic relations of right quantum calculus with respect to another function will be proved.

Theorem 1. 

The following relations hold:

$\left(i\right)$${ }^y{\mathcal{D}}_{q,\psi }\left({ }^y{\mathcal{I}}_{q,\psi }\Pi \right)\left(z\right)=\Pi \left(z\right)$.

$\left(ii\right)$${ }^y{\mathcal{I}}_{q,\psi }\left({ }^y{\mathcal{D}}_{q,\psi }\Pi \right)\left(z\right)=\Pi \left(z\right)-\Pi \left(y\right)$, where $z\in (x,y]$.

Proof. 

By using Definitions 2 and 3, we have

$$\begin{array}{ccc}\hfill { }^y{\mathcal{D}}_{q,\psi }\left({ }^y{\mathcal{I}}_{q,\psi }\Pi \right)\left(z\right)& =& { }^y{\mathcal{D}}_{q,\psi }\left\{\sum _{i=0}^{\infty }\left[\psi \left({ }^y{Q}_q^i\left(z\right)\right)-\psi \left({ }^y{Q}_q^{i+1}\left(z\right)\right)\right]\Pi \left({ }^y{Q}_q^i\left(z\right)\right)\right\}\hfill \\ & =& \frac{1}{(\psi \left(z\right)-\psi \left({ }^y{Q}_q\left(z\right)\right))}\{\sum _{i=0}^{\infty }\left[\psi \left({ }^y{Q}_q^i\left(z\right)\right)-\psi \left({ }^y{Q}_q^{i+1}\left(z\right)\right)\right]\Pi \left({ }^y{Q}_q^i\left(z\right)\right)\hfill \\ & & -\sum _{i=0}^{\infty }\left[\psi \left({ }^y{Q}_q^{i+1}\left(z\right)\right)-\psi \left({ }^y{Q}_q^{i+2}\left(z\right)\right)\right]\Pi \left({ }^y{Q}_q^{i+1}\left(z\right)\right)\}\hfill \\ & =& \frac{1}{(\psi \left(z\right)-\psi \left({ }^y{Q}_q\left(z\right)\right))}\hfill \\ & & \times [\left(\psi \left(z\right)-\psi \left({ }^y{Q}_q\left(z\right)\right)\right)\Pi \left(z\right)+\left(\psi \left({ }^y{Q}_q\left(z\right)\right)-\psi \left({ }^y{Q}_q^2\left(z\right)\right)\right)\Pi \left({ }^y{Q}_q\left(z\right)\right)\hfill \\ & & +\left(\psi \left({ }^y{Q}_q^2\left(z\right)\right)-\psi \left({ }^y{Q}_q^3\left(z\right)\right)\right)\Pi \left({ }^y{Q}_q^2\left(z\right)\right)+\cdots \hfill \\ & & -\{\left(\psi \left({ }^y{Q}_q\left(z\right)\right)-\psi \left({ }^y{Q}_q^2\left(z\right)\right)\right)\Pi \left({ }^y{Q}_q\left(z\right)\right)\hfill \\ & & +\left(\psi \left({ }^y{Q}_q^2\left(z\right)\right)-\psi \left({ }^y{Q}_q^3\left(z\right)\right)\right)\Pi \left({ }^y{Q}_q^2\left(z\right)\right)+\cdots \left\}\right]\hfill \\ & =& \Pi \left(z\right),\hfill \end{array}$$

which proves $\left(i\right).$

Also, we have

$$\begin{array}{ccc}\hfill { }^y{\mathcal{I}}_{q,\psi }\left({ }^y{\mathcal{D}}_{q,\psi }\Pi \right)\left(z\right)& =& { }^y{\mathcal{I}}_{q,\psi }\left\{\frac{\Pi \left(z\right)-\Pi \left({ }^y{Q}_q\left(z\right)\right)}{\psi \left(z\right)-\psi \left({ }^y{Q}_q\left(z\right)\right)}\right\}\hfill \\ & =& \sum _{i=0}^{\infty }\left[\psi \left({ }^y{Q}_q^i\left(z\right)\right)-\psi \left({ }^y{Q}_q^{i+1}\left(z\right)\right)\right]\frac{\Pi \left({ }^y{Q}_q^i\left(z\right)\right)-\Pi \left({ }^y{Q}_q^{i+1}\left(z\right)\right)}{\psi \left({ }^y{Q}_q^i\left(z\right)\right)-\psi \left({ }^y{Q}_q^{i+1}\left(z\right)\right)}\hfill \\ & =& \Pi \left(z\right)-\Pi \left({ }^y{Q}_q\left(z\right)\right)+\left(\Pi \left({ }^y{Q}_q\left(z\right)\right)-\Pi \left({ }^y{Q}_q^2\left(z\right)\right)\right)\hfill \\ & & +\left(\Pi \left({ }^y{Q}_q^2\left(z\right)\right)-\Pi \left({ }^y{Q}_q^3\left(z\right)\right)\right)+\cdots \hfill \\ & =& \Pi \left(z\right)-\Pi \left(y\right),\hfill \end{array}$$

with ${lim}_{u\to \infty }\Pi (q^{u}z+(1-q^u)y)=\Pi \left(y\right),$ and thus $\left(ii\right)$ is proved. □

3. Right Quantum Hermite–Hadamard Inequalities

In this section, we apply the newly defined right quantum integral with respect to another function to establish new Hermite–Hadamard-type inequalities for convex functions on $[x,y].$ We recall that $\Pi$ is called convex if

$$\Pi (\lambda x+(1-\lambda \left)y\right)\le \lambda \Pi \left(x\right)+(1-\lambda )\Pi \left(y\right)$$

for all $\lambda \in [0,1]$.

Theorem 2 

([18]). Assume that $\Pi :[x,y]\to \mathbb{R}$ is a convex differentiable function on $(x,y)$. Then, the left quantum integral of Π with respect to $\psi \left(z\right)={(z-x)}^m$, $m>0$ satisfies

$$\begin{array}{ccc}\hfill \Pi \left(\frac{y(1-q^m)+x{q}^m(1-q)}{1-q^{m+1}}\right)& \le & \frac{1}{{(y-x)}^m}{\int }_x^y\Pi \left(s\right){ }_x{d}_q^{{(z-x)}^m}s\hfill \\ & \le & \frac{(1-q^m)\Pi \left(y\right)+q^m(1-q)\Pi \left(x\right)}{1-q^{m+1}}.\hfill \end{array}$$

(7)

Using the idea of proving Theorem 2, we prove a new right quantum Hermite–Hadamard inequality.

Theorem 3. 

Assume that $\Pi :[x,y]\to \mathbb{R}$ is a convex differentiable function on $(x,y)$. Then, the right quantum integral of Π with respect to $\psi \left(z\right)={(y-z)}^m$, $m>0$ satisfies

$$\begin{array}{ccc}\hfill \Pi \left(\frac{x(1-q^m)+y{q}^m(1-q)}{1-q^{m+1}}\right)& \le & \frac{1}{{(y-x)}^m}{\int }_x^y\Pi \left(s\right){ }^y{d}_q^{{(y-z)}^m}s\hfill \\ & \le & \frac{(1-q^m)\Pi \left(x\right)+q^m(1-q)\Pi \left(y\right)}{1-q^{m+1}}.\hfill \end{array}$$

(8)

Proof. 

Consider the point ${\displaystyle c:=\frac{x(1-q^m)+y{q}^m(1-q)}{1-q^{m+1}}.}$ Then, $c\in [x,y]$ since ${lim}_{q\to 0}c=x$ and

$$\underset{q\to 1}{lim}\frac{x(1-q^m)+y{q}^m(1-q)}{1-q^{m+1}}=\frac{mx+y}{m+1}\in (x,y)$$

for all $m>0$, by L’H$\widehat{\mathrm{o}}$pital’s rule with respect to q.

Note that there exists a tangent line $\varphi \left(z\right)$ under the curve of $\Pi \left(z\right)$, i.e., for $z\in (x,y)$,

$$\varphi \left(z\right)=\Pi \left(c\right)+{\Pi }^{\prime }\left(c\right)(z-c)\le \Pi \left(z\right),$$

by the fact that $\Pi$ is differentiable convex function on $(x,y).$ In order to prove the left side of (8), we take quantum integration with respect to a function $\psi \left(z\right)={(y-z)}^m$, $m>0$ and apply the formula in Example 1 for $n=0,1$, and we have

$$\begin{array}{ccc}& & {\int }_x^y\varphi \left(s\right){ }^y{d}_q^{{(y-z)}^m}ds\hfill \\ & =& {\int }_x^y\left[\Pi \left(c\right)+{\Pi }^{\prime }\left(c\right)(s-c)\right]{ }^y{d}_q^{{(y-z)}^m}s\hfill \\ & =& {(y-x)}^m\Pi \left(c\right)+{\Pi }^{\prime }\left(c\right){\int }_a^y(y-c-(y-s)){ }^y{d}_q^{{(y-z)}^m}s\hfill \\ & =& {(y-x)}^m\Pi \left(c\right)+{\Pi }^{\prime }\left(c\right)\left[\frac{{(y-x)}^{m+1}(1-q^m)}{1-q^{m+1}}-\frac{{(y-x)}^{m+1}(1-q^m)}{1-q^{m+1}}\right]\hfill \\ & =& {(y-x)}^m\Pi \left(\frac{x(1-q^m)+y{q}^m(1-q)}{1-q^{m+1}}\right)\le {\int }_x^y\Pi \left(s\right){ }^y{d}_q^{{(y-z)}^m}s.\hfill \end{array}$$

To prove the right-hand inequality, we consider the line connecting the points $(x,\Pi (x\left)\right)$ and $(y,\Pi (y\left)\right)$

$$\chi \left(z\right)=\Pi \left(y\right)+\left(\frac{\Pi \left(y\right)-\Pi \left(x\right)}{y-x}\right)(z-y),$$

which, from convexity, implies that $\Pi \left(z\right)\le \chi \left(z\right)$ for all $z\in [x,y].$ Hence,

$$\begin{array}{ccc}\hfill {\int }_x^y\chi \left(s\right){ }^y{d}_q^{{(y-z)}^m}s& =& {\int }_x^y\left[\Pi \left(y\right)+\left(\frac{\Pi \left(y\right)-\Pi \left(x\right)}{y-x}\right)(s-y)\right]{ }^y{d}_q^{{(y-z)}^m}s\hfill \\ & =& {(y-x)}^m\left[\frac{(1-q^m)\Pi \left(x\right)+q^m(1-q)\Pi \left(y\right)}{1-q^{m+1}}\right]\hfill \\ & \ge & {\int }_x^y\Pi \left(s\right){ }^y{d}_q^{{(y-z)}^m}s.\hfill \end{array}$$

The proof is completed by combining both cases. □

Remark 5. 

If $m=1$, then (8) is reduced to

$$\Pi \left(\frac{x+qy}{1+q}\right)\le \frac{1}{(y-x)}{\int }_x^y\Pi \left(s\right){ }^y{d}_{q}s\le \frac{\Pi \left(x\right)+q\Pi \left(y\right)}{1+q},$$

which appears in [27].

From Theorems 2 and 3, we yield the next corollary.

Corollary 2. 

Let $\Pi :[x,y]\to \mathbb{R}$ be a convex differentiable function on $(x,y)$ and $q\in (0,1)$. Then, we have

$$\begin{array}{c}\hfill \Pi \left(\frac{y(1-q^m)+x{q}^m(1-q)}{1-q^{m+1}}\right)+\Pi \left(\frac{x(1-q^m)+y{q}^m(1-q)}{1-q^{m+1}}\right)\\ \hfill \le \frac{1}{{(y-x)}^m}\left[{\int }_x^y\Pi \left(s\right){ }_x{d}_q^{{(z-x)}^m}s+{\int }_x^y\Pi \left(s\right){ }^y{d}_q^{{(y-z)}^m}s\right]\le \Pi \left(x\right)+\Pi \left(y\right).\end{array}$$

(9)

Corollary 3. 

Let $\Pi :[x,y]\to \mathbb{R}$ be a convex differentiable function on $(x,y)$ and $q\in (0,1)$. Then, we have

$$\begin{array}{c}\hfill 2\Pi \left(\frac{x+y}{2}\right)\le \frac{1}{{(y-x)}^m}\left[{\int }_x^y\Pi \left(s\right){ }_x{d}_q^{{(z-x)}^m}s+{\int }_x^y\Pi \left(s\right){ }^y{d}_q^{{(y-z)}^m}s\right]\le \Pi \left(x\right)+\Pi \left(y\right).\end{array}$$

(10)

Proof. 

Note that

$$\begin{array}{cc}\hfill \Pi \left(\frac{x+y}{2}\right)& =\Pi \left(\frac{y(1-q^m)+x{q}^m(1-q)}{2(1-q^{m+1})}+\frac{x(1-q^m)+y{q}^m(1-q)}{2(1-q^{m+1})}\right)\hfill \\ & \le \frac{1}{2}\Pi \left(\frac{y(1-q^m)+x{q}^m(1-q)}{(1-q^{m+1})}\right)+\frac{1}{2}\Pi \left(\frac{x(1-q^m)+y{q}^m(1-q)}{(1-q^{m+1})}\right).\hfill \end{array}$$

From Corollary 2, we have the result. □

Now, we prove new inequalities for $q_x$- and $q^y$-integrals with respect to other functions for h convex functions.

Definition 4 

([28]). Let $h:[0,1]\to {\mathbb{R}}_0^{+}$ and $\Pi :[x,y]\subset \mathbb{R}\to {\mathbb{R}}_0^{+}$. Then, the function Π is called h-convex if we have

$$\Pi (\lambda u+(1-\lambda \left)v\right)\le h\left(\lambda \right)\Pi \left(u\right)+h(1-\lambda )\Pi \left(v\right)$$

for all $u,v\in [x,y]$ and $\lambda \in (0,1).$

Theorem 4. 

Let $\Pi :[x,y]\to {\mathbb{R}}_0^{+}$ be an h-convex function such that $h\left(\frac{1}{2}\right)\ne 0$ and $q\in (0,1)$. Then, we have

$$\begin{array}{cc}\hfill \frac{1}{h\left(\frac{1}{2}\right)}\Pi \left(\frac{x+y}{2}\right)& \le \frac{1}{(y-x)}\left[{\int }_x^y\Pi \left(t\right){ }_x{d}_q^{(z-x)}t+{\int }_x^y\Pi \left(t\right){ }^y{d}_q^{(y-z)}t\right]\hfill \\ & \le \left[{\int }_0^{1}h\left(z\right)d_{q}z+{\int }_0^{1}h(1-z)d_{q}z\right]\left[\Pi \left(x\right)+\Pi \left(y\right)\right].\hfill \end{array}$$

(11)

Proof. 

We have for all $u,v\in [x,y]$ that

$$\Pi \left(\frac{u+v}{2}\right)\le h\left(\frac{1}{2}\right)\left(\Pi \left(u\right)+\Pi \left(v\right)\right)$$

by the h-convexity of $\Pi .$ By putting $u=zy+(1-z)x$ and $v=zx+(1-z)y$, we obtain

$$\frac{1}{h\left(\frac{1}{2}\right)}\Pi \left(\frac{x+y}{2}\right)\le \Pi (zy+(1-z)x)+\Pi (zx+(1-z)y).$$

(12)

q-integrating both sides of (12), according to Jackson integral, we have

$${\int }_x^y\frac{1}{h\left(\frac{1}{2}\right)}\Pi \left(\frac{x+y}{2}\right)d_{q}z=\frac{1}{h\left(\frac{1}{2}\right)}\Pi \left(\frac{x+y}{2}\right)(y-x),$$

(13)

$$\begin{array}{ccc}\hfill {\int }_x^y\Pi (zy+(1-z)x)d_{q}z& =& (y-x)(1-q)\sum _{i=0}^{\infty }{q}^i\Pi (q^{i}y+(1-q^i)x)\hfill \\ & =& \sum _{i=0}^{\infty }\left[q^i(y-x)-q^{i+1}(y-x)\right]\Pi (q^{i}y+(1-q^i)x)\hfill \\ & =& {\int }_x^y\Pi {\left(t\right)}_x{d}_q^{(z-x)}t,\hfill \end{array}$$

(14)

$$\begin{array}{ccc}\hfill {\int }_x^y\Pi (zx+(1-z)y)d_{q}z& =& (y-x)(1-q)\sum _{i=0}^{\infty }{q}^i\Pi (q^{i}x+(1-q^i)y)\hfill \\ & =& \sum _{i=0}^{\infty }\left[q^i(y-x)-q^{i+1}(y-x)\right]\Pi (q^{i}x+(1-q^i)y)\hfill \\ & =& {\int }_x^y\Pi {\left(t\right)}^y{d}_q^{(y-z)}t.\hfill \end{array}$$

(15)

From (13)–(15), we have the first inequality of (11).

From the definitions of $q_x$ and $q^y$-integrals, we have

$$\begin{array}{cc}\hfill \frac{1}{y-x}{\int }_x^y\Pi \left(t\right){ }_x{d}_q^{(z-x)}t& =(1-q)\sum _{i=0}^{\infty }{q}^i\Pi (q^{i}y+(1-q^i)x)\hfill \\ \hfill & =(1-q)\sum _{i=0}^{\infty }{q}^i\Pi (x+q^i(y-x))\hfill \\ \hfill & ={\int }_0^1\Pi (x+z(y-x))d_{q}z.\hfill \end{array}$$

By h-convexity of $\Pi$, we have

$${\int }_0^1\Pi (x+z(y-x))d_{q}z={\int }_0^1\Pi (zy+(1-z)x)d_{q}z\le \Pi \left(y\right){\int }_0^{1}h\left(z\right)d_{q}z+\Pi \left(x\right){\int }_0^{1}h(1-z)d_{q}z.$$

Similarly,

$$\begin{array}{cc}\hfill \frac{1}{y-x}{\int }_x^y\Pi \left(t\right){ }^y{d}_q^{(y-z)}t& =(1-q)\sum _{i=0}^{\infty }{q}^i\Pi (q^{i}x+(1-q^i)y)\hfill \\ \hfill & =(1-q)\sum _{i=0}^{\infty }{q}^i\Pi (y-q^i(y-x))\hfill \\ \hfill & ={\int }_0^1\Pi (y-z(y-x))d_{q}z\hfill \end{array}$$

and

$${\int }_0^1\Pi (y-z(y-x))d_{q}z\le \Pi \left(x\right){\int }_0^{1}h\left(z\right)d_{q}z+\Pi \left(y\right){\int }_0^{1}h(1-z)d_{q}z.$$

So, we obtain the right-hand side of inequality (11). □

Example 2. 

Let $[x,y]=[0,1]$, $\Pi \left(z\right)=z^2$, $h\left(z\right)=z$ and $q=\frac{1}{2}.$ Then, Π is convex on $[0,1].$ We have $h\left(\frac{1}{2}\right)=\frac{1}{2}$, $\Pi \left(\frac{x+y}{2}\right)=\frac{1}{4}$, $\Pi \left(x\right)=0,$ $\Pi \left(y\right)=1,$ and we find

$${\int }_x^y\Pi \left(t\right){ }_x{d}_q^{(z-x)}t=\frac{4}{7},{\int }_x^y\Pi \left(t\right){ }^y{d}_q^{(y-z)}x=\frac{5}{21},{\int }_0^{1}h\left(z\right)d_{q}z=\frac{2}{3},{\int }_0^{1}h(1-z)d_{q}z=\frac{1}{3}.$$

Hence,

$$\begin{array}{ccc}\hfill \frac{1}{2}=\frac{1}{h\left(\frac{1}{2}\right)}\Pi \left(\frac{x+y}{2}\right)& \le & \frac{1}{(y-x)}\left[{\int }_x^y\Pi \left(t\right){ }_x{d}_q^{(z-x)}t+{\int }_x^y\Pi \left(t\right){ }^y{d}_q^{(y-z)}t\right]=\frac{4}{7}+\frac{5}{21}\hfill \\ & \le & \left[{\int }_0^{1}h\left(z\right)d_{q}z+{\int }_0^{1}h(1-z)d_{q}z\right]\left[\Pi \left(x\right)+\Pi \left(y\right)\right]=\frac{2}{3}+\frac{1}{3},\hfill \end{array}$$

which means that inequality (11) holds.

Then, we obtain another inequality concerning modified h-convex functions.

Definition 5 

([29]). Let $h:[0,1]\to {\mathbb{R}}_0^{+}$ and $\Pi :[x,y]\subset \mathbb{R}\to {\mathbb{R}}_0^{+}$. If

$$\Pi (\lambda u+(1-\lambda \left)v\right)\le h\left(\lambda \right)\Pi \left(u\right)+(1-h(\lambda \left)\right)\Pi \left(v\right)$$

holds for any $u,v\in [x,y]$ and $\lambda \in [0,1],$ then Π is called modified h-convex function.

Theorem 5. 

Let $\Pi :[x,y]\to {\mathbb{R}}_0^{+}$ be a modified h-convex function and $q\in (0,1)$. Then, we have

$$2\Pi \left(\frac{x+y}{2}\right)\le \frac{1}{(y-x)}\left[{\int }_x^y\Pi \left(t\right){ }_x{d}_q^{(z-x)}t+{\int }_x^y\Pi \left(t\right){ }^y{d}_q^{(y-z)}t\right]\le \Pi \left(x\right)+\Pi \left(y\right).$$

(16)

Proof. 

Because $\Pi$ is modified h-convex, then for all $u,v\in [x,y]$ we have

$$\Pi \left(\frac{u+v}{2}\right)\le h\left(\frac{1}{2}\right)\Pi \left(u\right)+\left(1-h\left(\frac{1}{2}\right)\right)\Pi \left(v\right)$$

and

$$\Pi \left(\frac{u+v}{2}\right)\le h\left(\frac{1}{2}\right)\Pi \left(v\right)+\left(1-h\left(\frac{1}{2}\right)\right)\Pi \left(u\right).$$

Adding these two inequalities, we obtain $2\Pi \left(\frac{u+v}{2}\right)\le \Pi \left(u\right)+\Pi \left(v\right).$ By replacing $u=zy+(1-z)x$ and $v=zx+(1-z)y$ and q-integrating both sides according to Jackson integral, we have the first inequality of (16).

The second inequality is proved by combining the following two inequalities.

$${\int }_x^y\Pi \left(t\right){ }_x{d}_q^{(z-x)}t={\int }_0^1\Pi (x+z(y-x))d_{q}z\le \Pi \left(y\right){\int }_0^{1}h\left(z\right)d_{q}z+\Pi \left(x\right){\int }_0^1[1-h\left(z\right)]d_{q}z,$$

$${\int }_x^y\Pi \left(t\right){ }^y{d}_q^{(y-z)}t={\int }_0^1\Pi (y-z(y-x))d_{q}z\le \Pi \left(x\right){\int }_0^{1}h\left(z\right)d_{q}z+\Pi \left(y\right){\int }_0^1[1-h\left(z\right)]d_{q}z.$$

 □

4. Conclusions

In the present research, we have further discussed the quantum calculus on finite intervals with respect to another function, which was initiated by the authors in [18], where the derivative ${ }_x{\mathcal{D}}_{q,\psi }$ and the integral ${ }_x{\mathcal{I}}_{q,\psi }$ were defined and their properties were studied. Here, we defined the corresponding quantum derivative ${ }^y{\mathcal{D}}_{q,\psi }$ and quantum integral ${ }^y{\mathcal{I}}_{q,\psi }$ with respect to another function and studied their basic properties. As applications of the newly defined right quantum calculus with respect to another function, we obtained new Hermite–Hadamard-type inequalities for convex, h-convex, and modified h-convex functions.

For future study, we plan to investigate if we can apply the new definition of quantum calculus in finite intervals with respect to another function to other types of quantum inequalities, such as Fejer, Ostrowski, Grüss, etc.