Generalization of quantum calculus and corresponding Hermite–Hadamard inequalities

Abstract

The aim of this paper is first to introduce generalizations of quantum integrals and derivatives which are called $(\phi \,-\,h)$ integrals and $(\phi \,-\,h)$ derivatives, respectively. Then we investigate some implicit integral inequalities for $(\phi \,-\,h)$ integrals. Different classes of convex functions are used to prove these inequalities for symmetric functions. Under certain assumptions, Hermite–Hadamard-type inequalities for q-integrals are deduced. The results presented herein are applicable to convex, m-convex, and $\hbar$-convex functions defined on the non-negative part of the real line.

1 Introduction

Theory of integral inequalities provides useful tools to estimate an integral of functions over intervals. These inequalities give bounds to estimate the cumulative behavior of functions over a specified domain and provide insight into the general behavior of different mathematical objects. On the other hand, Quantum calculus known as q-Calculus is based on the idea of studying the concepts of classical calculus without taking into accounts the limits. In the nineteenth century, Euler [14] investigated several notable results such as Jacobi’s triple product identity and theory of q-hypergeometric functions. The study of quantum integral inequalities has gained much attention of researchers these days. The concept of the definite q-integral was first introduced by Jackson [13] in 1910. Indeed, he was the first who developed q-calculus in a systematic way. In, 2013 and 2014, Tariboon and Ntouyas [23, 24] introduced the concepts of q-derivatives and q-integrals over finite intervals and derived quantum integral inequalities. In 2018, Alp et al. [3] employed the left q-integral operator to develop q-Hermite–Hadamard inequality using convex and quasi-convex functions. Also, Alomari [2] derived the quantum variant of Bernoulli inequality. In 2022, Liu et al. [15] proved the q-Hermite–Hadamard inequality by using $(q-h)$ -derivatives and integrals. Recently, Chen et al. [7] derived some results for $(q-h)$-integrals using $\hbar$-convex and m-convex functions and Farid et al. [9] proved the generalized q-integral inequalities using $(\hbar -m)$ and $(\alpha -m)$ convexities. For some other recent developments in this direction, we refer to [4,5,6, 8, 10, 11, 14, 16, 17, 22] and references mentioned therein.In this paper, a new version of Hermite–Hadamard quantum integral inequalities for convex, m-convex and $\hbar$-convex functions using the left and right $(\phi -h)$-integrals are obtained. The results of this paper can be applied to convex functions defined on non negative part of real line $\mathbb {R}$.

Definition 1

[20] If I is any interval in $\mathbb {R}$. A function $f:{I} \subseteq \mathbb {R}\rightarrow \mathbb {R}$ is said to be convex function if

$$\begin{aligned} f\left( \lambda x+(1-\lambda )y\right) \le \lambda f(x)+(1-\lambda )f(y),\forall x,y\in {I}\text { and }\,\lambda \in [0,1]. \end{aligned}$$

(1.1)

If the inequality is reversed, then f is concave function.

Definition 2

[26] Suppose that ${I}\text { and } {\Omega }$ are any two intervals in $\mathbb {R}$ and $\hbar :(0, 1)\subseteq {\Omega }\rightarrow \mathbb {R}$ is a nonnegative function. A nonnegative function $f:{I}\subseteq \mathbb {R} \rightarrow \mathbb {R}$ is said to be ${\hbar }$-convex if

$$\begin{aligned} f\left( \lambda x+(1-\lambda )y\right) \le \hbar (\lambda ) f(x)+\hbar (1-\lambda )f(y),\forall x,y\in {I}\text { and }\,\lambda \in [0,1] \text { holds}.\nonumber \\ \end{aligned}$$

(1.2)

If the inequality is reversed then f is ${\hbar }$-concave. If ${ \hbar (\lambda )=\lambda }$ in (1.2), then the above inequality reduces to (1.1), and if we take, ${\hbar (\lambda )=\lambda ^{s}}$ in (1.2), we get the definition of s-convex function given in [12].

Definition 3

[25] Let $c>0$ and $0\le m\le 1$. The function $f:[0,c]\rightarrow \mathbb {R}$ is called m-convex, if the following inequality holds

$$\begin{aligned} f\left( \lambda x+m(1-\lambda )y\right) \le \lambda f(x)+m(1-\lambda )f(y),\forall x,y\in [0,c]\text { and }\,\lambda \in [0,1]. \nonumber \\ \end{aligned}$$

(1.3)

If $m=1$ in (1.3), then we get (1.1). If we take, $m=0$, then we get a star shaped function. That is, for any point $x\in [0,c]\text { and}\,t\in [0,1]$ we have $f(tx)\le tf(x)$.

Theorem 1

[19] If $f: [a, b] \rightarrow \mathbb {R}$ is a convex function, where $a<b$ and $[a, b]\subset \mathbb {R}$, then, we have the Hermite–Hadamard inequality which is given as follows:

$$\begin{aligned} f\left( \frac{a+b}{2}\right) \le \frac{1}{b-a}\int _{a}^{b}f(x)dx\le \frac{ f(a)+f(b)}{2}. \end{aligned}$$

(1.4)

Definition 4

[20] Let f be a convex function defined on an open interval (a, b), $x_{i}\in (a, b)$ and $\alpha _{i}>0$ with $\sum _{i=1}^{n}\alpha _{i}=1$. Then, the following inequality

$$\begin{aligned} f\left( \sum _{i=1}^{n}\alpha _{i}x_{i}\right) \le \sum _{i=1}^{n} \alpha _{i}f(x_{i}), \end{aligned}$$

(1.5)

is called the Jensen inequality and the function f is known as a J -convex function. The following result is an extension of Hermite–Hadamard inequality to the $\hbar$-convex function [21].

Theorem 2

If $f:I \rightarrow \mathbb {R}$ is a $\hbar$-convex function on the interval $I\subset \mathbb {R}$ and $a, b \in {I}$ where $a<b$. Then, the generalized form of Hermite–Hadamard inequality is given as follows:

$$\begin{aligned} \frac{1}{2\hbar \left( \frac{1}{2}\right) }f\left( \frac{a+b}{2}\right) \le \frac{1 }{b-a}\int _{a}^{b}f(x)dx\le \left( f(a)+f(b)\right) \int _{0}^{1}\hbar (t)dt. \end{aligned}$$

(1.6)

In an ordinary calculus, the classical derivative of a function f is defined as follows:

$$\begin{aligned} \frac{d}{dx}f\left( x \right) = \mathop {\lim }\limits _{x \rightarrow a} \frac{{ f\left( {a} \right) - f\left( x \right) }}{a-x },\ \text {provided that the limit exists}. \end{aligned}$$

(1.7)

Quantum derivatives without involving limit have been defined in various ways. We present two types of derivatives known as q-derivative and the h -derivative for a given function f.Let us recall the notations of a quantum differential given as.

Definition 5

[14] The q-differential is:

$$\begin{aligned} d_{q}f(x)=f(qx)-f(x), \text {whereas the }h\text {-differential is: }\, d_{h}f(x)=f(x+h)-f(x). \end{aligned}$$

(1.8)

In particular, $d_{q}x=(q-1)x$ and $d_{h}x=h$. These two quantum differentials allow us to define the corresponding quantum derivatives.

Definition 6

[14] For a continuous function $f:I\rightarrow \mathbb {R}$,

$$\begin{aligned} D_{q}f(x)=\frac{d_{q}f(x)}{d_{q}x}=\frac{f(qx)-f(x)}{(q-1)x}, and \end{aligned}$$

(1.9)

$$\begin{aligned} D_{h}f(x)=\frac{d_{h}f(x)}{d_{h}x}=\frac{f(x+h)-f(x)}{h}, \end{aligned}$$

(1.10)

are called the q-derivative and h-derivative, respectively. Note that: If f is differentiable, then we have $\lim \limits _{q\rightarrow 1}D_{q}f(x)=\lim \limits _{h\rightarrow 0}D_{h}f(x)=\frac{df(x)}{dx}$.

Definition 7

[10] For a continuous function $f:I\rightarrow \mathbb {R}$, the $(q-h)$-derivative is expressed as:

$$\begin{aligned} D_{q}^{h}f(x)=\frac{d_{q}^{h}f(x)}{d_{q}^{h}x}=\frac{f(q(x+h))-f(x)}{ (q-1)x+qh},\ x_{0}:=\frac{qh}{1-q}\ne x \end{aligned}$$

(1.11)

where $q\in (0, 1)$ and $h\in \mathbb {R }.$ If $x=x_{0}$, then we have $D_{q}^{h}f(x_{0})=\lim \limits _{x\rightarrow x_{0}}D_{q}^{h}f(x)$.

When $h=0$ and $q=1$ in (7), we get quantum derivative and plank‘s derivative, respectively defined in (6).The Jackson integral or q-integral according to [13], is expressed as:

Definition 8

For a continuous function f, $0<q<1$ the q-integral on [0, a] is expressed as:

$$\begin{aligned} \int _{0}^{a}f(x)d_{q}x=\sum _{n=0}^{\infty }a(1-q)q^{n}f(q^{n}a). \end{aligned}$$

(1.12)

Note that,

$$\begin{aligned} \lim \limits _{q\rightarrow 1}\int _{0}^{a}f(x)d_{q}x=\int _{0}^{a}f(x)dx. \end{aligned}$$

(1.13)

If $c\in (0, b)$, then q-integral of f is defined as:

$$\begin{aligned} \int _{c}^{b}f(x)d_{q}x=\int _{0}^{b}f(x)d_{q}x-\int _{0}^{c}f(x)d_{q}x. \end{aligned}$$

(1.14)

Definition 9

[24] Assume that $f:[a,b]\rightarrow \mathbb {R}$ is a continuous function, $x\in [a,b]$, and $0<q<1$. Then $q_{a}$-definite integral on [a, b] is defined as:

$$\begin{aligned} \int _{a}^{x}f(t)_{a}dt_{q}=(1-q)(x-a)\sum _{n=0}^{\infty }q^{n}f(q^{n}a+(1-q^{n})x),x>a. \end{aligned}$$

Also, ${q^{b}}$-definite integral on [a, b] is defined as:

$$\begin{aligned} \int _{x}^{b}f(t)\,^{b}dt_{q}=(1-q)(b-x)\sum _{n=0}^{\infty }q^{n}f(q^{n}x+(1-q^{n})b),x<b. \end{aligned}$$

When $a=0$ in (9), we get Jackson integral defined in (8).

Definition 10

[15] Let $f:[a,b]\rightarrow \mathbb {R}$ be a continuous function, $x\in [a,b]$, $0<q<1$ and $h\in \mathbb {R}$. Then $(q_{a}-h)$-integral on [a, b] is defined as:

$$\begin{aligned} \int _{a}^{x}f(t)_{h}d_{q}^{a}t=((1-q)(x-a)+qh)\sum _{n=0}^{\infty }q^{n}f(q^{n}a+(1-q^{n})x+nq^{n}h),x>a, \end{aligned}$$

whereas, $({q^{b}-h)}$-integral on [a, b] is defined as:

$$\begin{aligned} \int _{x}^{b}f(t)_{h}d_{q}^{b}t=((1-q)(b-x)+qh)\sum _{n=0}^{\infty }q^{n}f(q^{n}x+(1-q^{n})b+nq^{n}h),x<b. \end{aligned}$$

When $h=0$ in Definition (10), we get integrals defined in (9). Let us now recall the results on quantum Hermite–Hadamard integral inequalities:

Theorem 3

[3] Let $0<q<1$, and $f:[a,b]\rightarrow \mathbb {R}$ a differentiable convex function on an interval (a, b). Then, the q -Hermite–Hadamard inequality on $q_{a}$-integral is given as:

$$\begin{aligned} f\left( \frac{aq+b}{1+q}\right) \le \frac{1}{b-a}\int _{a}^{b}f(x)d_{q}^{a}x \le \frac{qf(a)+f(b)}{1+q}. \end{aligned}$$

(1.15)

Theorem 4

[3] Let $0<q<1$, and $f:[a,b]\rightarrow \mathbb {R}$ a differentiable convex function on an interval (a, b), the q-Hermite–Hadamard inequality on $q_{a}$-integral is given by:

$$\begin{aligned} f\left( \frac{a+bq}{1+q}\right) +f^{\prime }\left( \frac{a+bq}{1+q}\right) \frac{(b-a)(1-q)}{1+q}\le \frac{1}{b-a}\int _{a}^{b}f(x)d_{q}^{a}x \le \frac{ qf(a)+f(b)}{1+q}. \end{aligned}$$

(1.16)

Theorem 5

[3] Let $0<q<1$, and $f:[a,b]\rightarrow \mathbb {R}$ a differentiable convex function on (a, b), the q-Hermite–Hadamard inequality on $q_{a}$ -integral is given as:

$$\begin{aligned} f\left( \frac{a+b}{1+q}\right) +f^{\prime }\left( \frac{a+b}{1+q}\right) \frac{(b-a)(1-q)}{2(1+q)}\le \frac{1}{b-a}\int _{a}^{b}\!f(x)d_{q}^{a}x \le \frac{qf(a)+f(b)}{1+q}. \end{aligned}$$

(1.17)

Furthermore, the q-Hadamrd inequality for convex functions was established for $q_{b}$-integral.

Theorem 6

[4] Let $f:[a,b]\rightarrow \mathbb {R}$ be a differentiable convex function on an interval (a, b). Then, the q-Hadamard inequality on $q_{b}$ -integral is given as:

$$\begin{aligned} f\left( \frac{a+bq}{1+q}\right) \le \frac{1}{b-a}\int _{a}^{b}f(x)d_{q}^{b}x \le \frac{f(a)+qf(b)}{1+q}. \end{aligned}$$

(1.18)

The aforementioned quantum integral inequalities have further generalized forms (see [1, 15, 18, 24]).

2 A genaralization of qauntum calculus

In this section, we define $(\phi -h)$-derivative and $(\phi -h)$-integral.

Definition 11

Let $f:{I}\rightarrow \mathbb {R}$ be a continuous function, $\Omega$ be an interval containing (0, 1) and $\phi :\Omega \rightarrow \Omega$ be a function such that $\phi (x)(x+h)\in {I}$ for all $x\in \Omega$ and $h\in \mathbb {R}$. Then $(\phi -h)$-derivative is defined as follows:

$$\begin{aligned} _{h}D_{\phi }(f(x))=\frac{_{h}d_{\phi }f(x)}{_{h}d_{\phi }(x)}=\frac{f(\phi (x)(x+h))-f(x)}{(\phi (x)-1)x+\phi (x)h},x\ne \frac{\phi (x)h}{1-\phi (x)}:=u, \nonumber \\ \end{aligned}$$

(2.1)

where $\phi (x)\in (0,1)$. If $x=u$, then we have $_{h}D_{\phi }(f(u))=\lim \limits _{x\rightarrow u}(_{h}D_{\phi }(f(x)))$.When $h=0$ in (2.1), then we have the concept of $\phi$-derivative given as follows:

$$\begin{aligned} D_{\phi }(f(x))=\frac{f(x\phi (x))-f(x)}{(\phi (x)-1)x},x\ne 0,\text {where } \,\phi (x)\in (0,1). \end{aligned}$$

(2.2)

When $\phi (x)=q$ and $h=0$ in Definition 11, then we have q-derivative given in Definition 6, When $\phi (x)=q$ in he Definition 11, then we have $(q-h)$-derivative presented in Definition 7.

Example 1

Let $f(x)=x^{n}$, where n is positive integer. Take, $\phi (x)=x^{2}$, $\forall x\in (0,1)$. Then

$$\begin{aligned} \begin{aligned} _{h}D_{\phi }(x)^{n}&=\frac{(x^{2}(x+h))^{n}-x^{n}}{(x^{2}-1)x+x^{2}h}\\ {}&= \frac{((x^{2})^{n}-1)x^{n}}{(x^{2}-1)x+x^{2}h}+\frac{(x^{2})^{n} \left[ nx^{n-1}h+\frac{n(n-1)}{2!}x^{n-2}h^{2}+\cdots +h^{n} \right] }{(x^{2}-1)x+x^{2}h}. \end{aligned}\nonumber \\ \end{aligned}$$

(2.3)

By taking $h=0$, we have

$$\begin{aligned} \begin{aligned} D_{x^{2}}(x)^{n}&=\frac{(x^{2}.x)^{n}- x^{n}}{(x^{2}-1)x}= \frac{((x^{2})^{n}-1)x^{n}}{(x^{2}-1)x}=x^{n-1} \frac{((x^{2})^{n}-1)}{(x^{2}-1)}\\&=x^{n-1}((x^{2})^{n-1}+(x^{2})^{n-2}+\cdots +x^{2}+1). \end{aligned} \end{aligned}$$

(2.4)

If $\phi (x)=q$, then we have $D_{q}(x)^{n}=[n]_{q}x^{n-1}$, $q\in (0,1)$ where $[n]_{q}=\frac{1-q^{n}}{1-q}.$

Definition 12

Let $f:[a,b]\rightarrow \mathbb {R}$ be a continuous function, $\Omega$ an interval containing (0, 1) and $\phi :\Omega \rightarrow \Omega$ be a function such that $\phi (x)(x+h)\in {I}$, $\forall x\in \Omega$, $h\in \mathbb {R}$, and $\phi (x)\in (0,1)$. The left $(\phi -h)$-integral is defined as:

$$\begin{aligned}{} & {} \int _{a}^{x}f(t)_{h}d_{\phi }t=((1-\phi (x))(x-a)\nonumber \\{} & {} \quad +\phi (x)h)\sum _{n=0}^{\infty }\phi ^{n}(x)f \left( \phi ^{n}(x)a+(1-\phi ^{n}(x))x+nh\phi ^{n}(x)\right) ,x>a, \end{aligned}$$

(2.5)

and the right $(\phi -h)$-integral is defined as:

$$\begin{aligned}{} & {} \int _{x}^{b}f(t)_{h}d_{\phi }t=((1-\phi (x))(b-x)\nonumber \\{} & {} \quad +\phi (x)h)\sum _{n=0}^{\infty }\phi ^{n}(x)f \left( \phi ^{n}(x).x+(1-\phi ^{n}(x))b++nh\phi ^{n}(x)\right) ,b>x. \end{aligned}$$

(2.6)

When $\phi (x)=q$ and $h=0$ in Definition 12, then we have Definition 9, When $\phi (x)=q$ in Definition 12, then we have Definition 10.

Example 2

If $0<\phi (x)<1$, and $f(x)=x$ for all $x\in [a,b]$, then we have

$$\begin{aligned} \begin{aligned}&\int ^{x}_{a}f(x)_{h}d^{a}_{\phi }x \\&\quad =((1-\phi (x))(x-a)+\phi (x)h)\sum _{n=0}^{\infty }\phi ^{n}(x)f( \phi ^{n}(x).x+(1-\phi ^{n}(x))a+n\phi ^{n}(x)h)\\&\quad =((1-\phi (x))(x-a)+\phi (x)h)\times \left( \sum _{n=0}^{\infty }\phi ^{2n}(x).x+ \sum _{n=0}^{\infty }(\phi ^{n}(x)-\phi ^{2n}(x))a+h\sum _{n=0}^{\infty }n \phi ^{2n}(x)\right) \\&\quad =((1-\phi (x))(x-a)+\phi (x)h)\left( \frac{x+a\phi (x)}{1-\phi ^{2}(x)}+nh\frac{\phi ^{2}(x)}{(1-\phi ^{2}(x))^{2}} \right) , \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&\int ^{b}_{x}f(x)_{h}d^{b}_{\phi }x \\&\quad =((1-\phi (x))(b-x)+\phi (x)h)\sum _{n=0}^{\infty }\phi ^{n}(x)f( \phi ^{n}(x).x+(1-\phi ^{n}(x))b+n\phi ^{n}(x)h)\\&\quad =((1-\phi (x))(b-x)+\phi (x)h)\\&\qquad \times \left( \sum _{n=0}^{\infty }\phi ^{2n}(x).x+\sum _{n=0}^{\infty }(\phi ^{n}(x)- \phi ^{2n}(x))b+h\sum _{n=0}^{\infty }n\phi ^{2n}(x)\right) \\&\quad =((1- \phi (x))(b-x)+\phi (x)h)\left( \frac{x+b\phi (x)}{1-\phi ^{2}(x)}+nh\frac{ \phi ^{2}(x)}{(1-\phi ^{2}(x))^{2}}\right) .&\end{aligned} \end{aligned}$$

If we take $f(x)=1$ in Example 2, we get:

$$\begin{aligned} \int _{a}^{x} \, _{h}d^{a}_{\phi }x= \frac{(1-\phi (x))(x-a)+\phi (x)h}{1-\phi (x) }. \end{aligned}$$

(2.7)

$$\begin{aligned} \int _{x}^{b} \, _{h}d^{b}_{\phi }x=\frac{(1-\phi (x))(b-x)+\phi (x)h}{1-\phi (x)}. \end{aligned}$$

(2.8)

Example 3

Let $k\in \mathbb {N}$, $\phi (x)=x$ and $f(x)=x^{k}$, for all $x\in [a,b]$, then

$$\begin{aligned} \begin{aligned} \int ^{x}_{a}x^{k}\,_{h}d_{\phi }x&=((1-\phi (x))(x-a)+\phi (x)h)\sum _{n=0}^{\infty }\phi ^{n}(x)( \phi ^{n}(x)(x+nh)\\&\quad +(1-\phi ^{n}(x))a)^{k} \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned}&(\phi ^{n}(x)(x+nh)+(1-\phi ^{n}(x))a)^{k}\\&\quad =\begin{pmatrix} k\\ 0 \end{pmatrix}(\phi ^{n}(x)(x+nh))^{k}((1-\phi ^{n}(x))a)^{0} +\begin{pmatrix} k\\ 1 \end{pmatrix}(\phi ^{n}(x)(x+nh))^{k-1}((1-\phi ^{n}(x))a)^{1}\\&\qquad +\begin{pmatrix} k\\ 2 \end{pmatrix}(\phi ^{n}(x)(x+nh))^{k-2}((1-\phi ^{n}(x))a)^{2}+\cdots \\&\qquad + \begin{pmatrix} k\\ k \end{pmatrix}(\phi ^{n}(x)(x+nh))^{0}((1-\phi ^{n}(x))a)^{k}. \end{aligned} \end{aligned}$$

If we take $\phi (x)=x$, $x\in (0,1)$ and $k=3$, then we obtain that

$$\begin{aligned} \begin{aligned}&\int ^{x}_{a}x^{3}\,_{h}d_{x}x =((1-x)(x-a)+xh)\sum _{n=0}^{\infty }x^{n}( x^{n}(x+nh)+(1-x^{n})a)^{3}\end{aligned} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned}&(x^{n}(x+nh)+(1-x^{n})a)^{3}\\&\quad =\begin{pmatrix} 3\\ 0 \end{pmatrix}(x^{n}(x+nh))^{3}((1-x^{n})a)^{0}+\begin{pmatrix} 3\\ 1 \end{pmatrix}(x^{n}(x+nh))^{2}((1-x^{n})a)^{1}\\&\qquad +\begin{pmatrix} 3\\ 2 \end{pmatrix}(x^{n}(x+nh))^{1}((1-x^{n})a)^{2}++\begin{pmatrix} 3\\ 3 \end{pmatrix}(x^{n}(x+nh))^{0}((1-x^{n})a)^{3}. \end{aligned} \end{aligned}$$

Hence, we have

$$\begin{aligned} \int ^{x}_{a}x^{3} \,_{h}d_{x}x&=((1-x)(x-a)+xh)[\sum _{n=0}^{\infty }x^{n}(x^{n}(x+nh))^{3}\\ {}&\quad +3 \sum _{n=0}^{\infty }x^{n}(x^{n}(x+nh))^{2}((1-x^{n})a)\\&\quad +3\sum _{n=0}^{ \infty }x^{n}(x^{n}(x+nh))((1-x^{n})a)^{2}\\ {}&\quad +\sum _{n=0}^{ \infty }x^{n}((1-x^{n})a)^{3}]. \end{aligned}$$

If $h=0$, then we get

$$\begin{aligned} \begin{aligned} \int ^{x}_{a}x^{3}\,_{0}d_{x}x&=(1-x)(x-a)\sum _{n=0}^{\infty }x^{n}( x^{n}.x+(1-x^{n})a)^{3}\\ {}&=(1-x)(x-a)[x^{3}\sum _{n=0}^{\infty }x^{4n}+3x^{2}a\sum _{n=0}^{ \infty }(x^{3n}-x^{4n})\\ {}&\quad +3xa^{2}\sum _{n=0}^{ \infty }x^{2n}(1-x^{n})^{2}+a^{3}\sum _{n=0}^{ \infty }x^{n}(1-x^{n})^{3}]. \end{aligned} \end{aligned}$$

As a special case, if we take $a=0$ and $\phi (x)=x=q$, then we have

$$\begin{aligned} \begin{aligned}&\int ^{x}_{0}x^{3}\,_{0}d_{q}x =x^{4}\frac{1-q}{1-q^{4}}=\frac{x^{4}}{[4]_{q}}. \end{aligned} \end{aligned}$$

(2.9)

In the next section, we give new extensions of the q-Hermite–Hadamard inequality. Here, we will employ $(\phi -h)$-integrals to demonstrate inequalities for convex, m-convex, and $\hbar$-convex functions. Additionally, we will discuss theorems that have been proved using q-integrals.

3 Extended forms of the q-Hermite–Hadamard inequalities

Theorem 7

Let $f:[a,b]\rightarrow \mathbb {R}$ be a convex function, $\Omega$ an interval containing (0, 1) and $\phi :\Omega \rightarrow {\Omega }$ a function such that $\phi (x)(x+h)\in [a, b]$, $\forall x\in {\Omega }$, $\text {and}\,h\in \mathbb {R}$. Then the following inequality holds for $(\phi -h)$-integrals:

$$\begin{aligned}&f\left( \frac{a+\phi (x)x}{1+\phi (x)}+\frac{h\phi ^{2}(x)}{(1- \phi ^{2}(x))^{2}}(1-\phi (x))\right) \nonumber \\&\qquad +f\left( \frac{x+\phi (x)b}{1+\phi (x)}+ \frac{h\phi ^{2}(x)}{(1-\phi ^{2}(x))^{2}}(1-\phi (x))\right) \nonumber \\&\quad \le \frac{(1-\phi (x))}{(1-\phi (x))(x-a)+\phi (x)h}\int _{a}^{x}f(x) _{h}d_{\phi }x\nonumber \\&\qquad + \frac{(1-\phi (x))}{(1-\phi (x))(b-x)+\phi (x)h}\int _{x}^{b}f(x) _{h}d_{\phi }x\nonumber \\&\quad \le \frac{(f(a)+\phi (x)f(b))(b-a)+(1+\phi (x))[(x-a)f(b)+(b-x)f(a)]}{(b-a)(1+ \phi (x))}\nonumber \\&\qquad +\left( \frac{f(b)-f(a)}{b-a}\right) \frac{2h\phi ^{2}(x)}{(1-\phi ^{2}(x))^{2}}(1-\phi (x)),\, \text {where}\,\phi (x)\,\in (0, 1). \end{aligned}$$

(3.1)

Proof

Using the Jensen’s inequality, we have

$$\begin{aligned} \begin{aligned}&f\left( \sum _{n=0}^{\infty }(1-\phi (x))\phi ^{n}(x)\left( \phi ^{n}(x)a+(1- \phi ^{n}(x))x+nh\phi ^{n}(x)\right) \right) \\&\quad \le \sum _{n=0}^{\infty }(1-\phi (x))\phi ^{n}(x)f\left( \phi ^{n}(x)a+(1- \phi ^{n}(x))x+nh\phi ^{n}(x)\right) , \end{aligned} \end{aligned}$$

(3.2)

where $\sum _{n=0}^{\infty }(1-\phi (x))\phi ^{n}(x)=1$.On simplifying the left hand side and applying the definition of left $(\phi -h)$-integral on right hand side of the inequality, we obtain that

$$\begin{aligned}&f\left( \frac{a+\phi (x)x}{1+\phi (x)}+\frac{h\phi ^{2}(x)}{(1- \phi ^{2}(x))^{2}}(1-\phi (x))\right) \nonumber \\&\quad \le \frac{(1-\phi (x))}{(1-\phi (x))(x-a)+h \phi (x)} \sum _{n=0}^{\infty }(\left( 1-\phi (x))(x-a)+h\phi (x)\right) \phi ^{n}(x)\nonumber \\&\qquad \times f\left( \phi ^{n}(x)a+(1-\phi ^{n}(x))x+nh\phi ^{n}(x)\right) \nonumber \\&\quad =\frac{(1-\phi (x))}{(1-\phi (x))(x-a)+h\phi (x)}\int _{a}^{x}f(x) _{h}d_{\phi }x. \end{aligned}$$

(3.3)

Again, by using the Jensen’s inequality, we get

$$\begin{aligned} \begin{aligned}&f\left( \sum _{n=0}^{\infty }(1-\phi (x))\phi ^{n}(x)\left( \phi ^{n}(x).x+(1- \phi ^{n}(x))b+nh\phi ^{n}(x)\right) \right) \\&\quad \le \sum _{n=0}^{\infty }f\left( (1-\phi (x))\phi ^{n}(x)\left( \phi ^{n}(x).x+(1- \phi ^{n}(x))b+nh\phi ^{n}(x)\right) \right) . \end{aligned} \end{aligned}$$

(3.4)

On simplification the left side of the above inequality and using the definition of right $(\phi -h)$-integral on the right side, we get

$$\begin{aligned}&f\left( \frac{x+\phi (x)b}{1+\phi (x)}+\frac{h\phi ^{2}(x)}{(1- \phi ^{2}(x))^{2}}(1-\phi (x))\right) \nonumber \\&\quad \le \frac{(1-\phi (x))}{(1-\phi (x))(b-x)+h \phi (x)} \sum _{n=0}^{\infty }(\left( 1-\phi (x))(b-x)+h\phi (x)\right) \phi ^{n}(x)\nonumber \\&\qquad \times f\left( \phi ^{n}(x)x+(1-\phi ^{n}(x))b+nh\phi ^{n}(x)\right) \nonumber \\&\quad =\frac{(1-\phi (x))}{(1-\phi (x))(b-x)+h\phi (x)}\int _{x}^{b}f(x) _{h}d_{\phi }x. \end{aligned}$$

(3.5)

Note that, the left inequality of (3.1) is obtained by adding (3.3 ) and (3.5). Let Z(x) be the chord joining two points (a, f(a)) and (b, f(b)). As f is a convex function, $f(x)\le Z(x)$ which can be stated as:

$$\begin{aligned} Z(x)=\frac{f(b)-f(a)}{b-a}x+\frac{bf(a)-af(b)}{b-a}. \end{aligned}$$

(3.6)

That is,

$$\begin{aligned} f(x)\le \frac{bf(a)-af(b)}{b-a}+\frac{f(b)-f(a)}{b-a}x. \end{aligned}$$

(3.7)

The left $(\phi -h)$ integration of (3.7) over the interval [a, x) gives

$$\begin{aligned} \int _{a}^{x}f(x) _{h}d_{\phi }x&\le \int _{a}^{x}\left( \frac{bf(a)-af(b)}{b-a}+\frac{f(b)-f(a)}{b-a}x\right) \, _{h}d_{\phi }x\nonumber \\&=\left( (1-\phi (x))(x-a)+h\phi (x)\right) \nonumber \\&\quad \times \left( \frac{(b-a)f(a)+\phi (x)f(a)(b-x)+\phi (x)f(b)(x-a)}{(b-a)(1- \phi ^{2}(x))}\right. \nonumber \\&\quad \left. +\left( \frac{f(b)-f(a)}{b-a}\right) \frac{h\phi ^{2}(x)}{(1-\phi ^{2}(x))^{2}} \right) , \end{aligned}$$

(3.8)

which takes the following form:

$$\begin{aligned} \int _{a}^{x}f(x) _{h}d_{\phi }x&\le \left( (1-\phi (x))(x-a)+h\phi (x)\right) \nonumber \\&\quad \times \left( \frac{(b-a)f(a)+\phi (x)f(a)(b-x)+\phi (x)f(b)(x-a)}{(b-a)(1- \phi ^{2}(x))} \right. \nonumber \\&\quad \left. +\left( \frac{f(b)-f(a)}{b-a}\right) \frac{h\phi ^{2}(x)}{(1-\phi ^{2}(x))^{2}} \right) . \end{aligned}$$

(3.9)

Moreover, the right $(\phi -h)$ integration of (3.7) over the interval (x, b] gives

$$\begin{aligned} \int _{x}^{b}f(x) _{h}d_{\phi }x&\le \int _{x}^{b}\left( \frac{bf(a)-af(b)}{b-a}+\frac{f(b)-f(a)}{b-a}x\right) \, _{h}d_{\phi }x\nonumber \\&=\left( (1-\phi (x))(b-x)+h\phi (x)\right) \nonumber \\&\quad \times \left( \frac{(b-a)\phi (x)f(b)+f(a)(b-x)+f(b)(x-a)}{(b-a)(1-\phi ^{2}(x))} \right. \nonumber \\&\quad \left. +\left( \frac{f(b)-f(a)}{b-a}\right) \frac{h\phi ^{2}(x)}{(1-\phi ^{2}(x))^{2}} \right) , \end{aligned}$$

(3.10)

which takes the following form:

$$\begin{aligned}&\int _{x}^{b}f(x) _{h}d_{\phi }x\le \left( (1-\phi (x))(b-x)+h\phi (x)\right) \nonumber \\&\quad \times \left( \frac{(b-a)\phi (x)f(b)+f(a)(b-x)+f(b)(x-a)}{(b-a)(1-\phi ^{2}(x))} \right. \nonumber \\&\quad \left. +\left( \frac{f(b)-f(a)}{b-a}\right) \frac{h\phi ^{2}(x)}{(1-\phi ^{2}(x))^{2}} \right) . \end{aligned}$$

(3.11)

By combining (3.9) and (3.11), we obtain the second inequality of ( 3.1). $\square$

Corollary 1

Take $h=0$ in (3.1) and obtain the inequality for $\phi$- integrals:

$$\begin{aligned} \begin{aligned}&f\left( \frac{a+\phi (x)x}{1+\phi (x)}\right) +f\left( \frac{x+\phi (x)b}{1+ \phi (x)}\right) \\&\quad \le \frac{1}{x-a}\int _{a}^{x}f(x) d_{\phi }x+ \frac{1}{b-x}\int _{x}^{b}f(x) d_{\phi }x\\&\quad \le \frac{(f(a)+\phi (x)f(b))(b-a)+(1+\phi (x))[(x-a)f(b)+(b-x)f(a)]}{(b-a)(1+ \phi (x))}. \end{aligned}\nonumber \\ \end{aligned}$$

(3.12)

Remark 1

Setting $x=b$ in (3.3), $x=a$ in (3.5), and $x=b$ in ( 3.9), $x=a$, in (3.11), we get the following inequalities:

$$\begin{aligned}&f\left( \frac{a+\phi (x)b}{1+\phi (x)}+\frac{h\phi ^{2}(x)}{(1- \phi ^{2}(x))^{2}}(1-\phi (x))\right) \nonumber \\&\quad \le \frac{1-\phi (x)}{(1-\phi (x))(b-a)+h\phi (x)}\int _{a}^{b}f(x) _{h}d_{\phi }x\nonumber \\&\quad \le \frac{f(a)+\phi (x)f(b)}{1+\phi (x)} +\left( \frac{f(b)-f(a)}{b-a}\right) \frac{h\phi ^{2}(x)}{(1- \phi ^{2}(x))^{2}}(1-\phi (x)). \end{aligned}$$

(3.13)

Corollary 2

Taking $h=0$ in the Remark 1, the Hermite–Hadamard inequality for $\phi$-integrals is given below:

$$\begin{aligned} \begin{aligned}&f\left( \frac{a+\phi (x)b}{1+\phi (x)}\right) \le \frac{1}{b-a}\int _{a}^{b}f(x) d_{\phi }x \le \frac{f(a)+\phi (x)f(b)}{1+\phi (x)}. \end{aligned} \end{aligned}$$

(3.14)

If we set, $\phi (x)=q$, we get q-Hermite–Hadamard inequality which reduces classical Hermite–Hadamard inequality on taking $q=1$.

Using the Hermite–Hadamard inequality for $\hbar$-convex functions, we can now demonstrate the following generalization.

Theorem 8

Let $f:[a,b]\rightarrow \mathbb {R}$ be $\hbar$-convex function such that $\hbar (\frac{1}{2})\ne 0$, $\Omega$ an interval containing (0, 1) and $\phi :\Omega \rightarrow {\Omega }$ be a function such that $\phi (x)(x+h)\in [a, b]$, $\forall x\in {\Omega }$, $\text { and}\,$ $h\in \mathbb {R}$. Then the following inequality holds for $(\phi -h)$-integral. (i) The following inequality is applicable to the left $(\phi -h)$ -integral if f is symmetric about $\frac{a+x}{2},\forall x\in (a,b)$.

$$\begin{aligned} \begin{aligned} \frac{1}{2\hbar \left( \frac{1}{2}\right) } f\left( \frac{a+x}{2}\right)&\le \frac{1-\phi (x)}{(1-\phi (x))(x-a)+\phi (x) \,h_{1}}\int _{a}^{x}f(\lambda )_{h_{1}}d_{\phi }\lambda \\ {}&\le f(x)\int _{0}^{1}\hbar (\lambda )\, _{h}d_{\phi }\lambda +f(a)\int _{0}^{1}\hbar (1-\lambda )\, _{h}d_{\phi }\lambda ,\\ {}&\text {where}\, h_{1}=(x-a)h\,\text {and}\,\phi (x)\in (0,1). \end{aligned} \end{aligned}$$

(3.15)

(ii) The following inequality is applicable to the right $(\phi -h)$ -integral if f is symmetric about $\frac{x+b}{2},\forall x\in (a,b)$.

$$\begin{aligned} \begin{aligned} \frac{1}{2\hbar \left( \frac{1}{2}\right) } f\left( \frac{x+b}{2}\right)&\le \frac{1-\phi (x)}{(1-\phi (x))(b-x)+\phi (x) \,h_{2}}\int _{x}^{b}f(\lambda )_{h_{2}}d_{\phi }\lambda \\ {}&\le f(b)\int _{0}^{1}\hbar (\lambda )\, _{h}d_{\phi }\lambda +f(x)\int _{0}^{1}\hbar (1-\lambda )\, _{h}d_{\phi }\lambda ,\\ {}&\text {where}\, h_{2}=(b-x)h\, \text {and}\,\phi (x)\in (0,1). \end{aligned} \end{aligned}$$

(3.16)

Proof

We demonstrate (i) and (ii) in the following way:

(i) Since f is $\hbar$-convex function, we get:

$$\begin{aligned} \frac{1}{\hbar \left( \frac{1}{2}\right) }f\left( \frac{a+x}{2}\right) \le f(\lambda a+(1-\lambda )x)+f(\lambda x+(1-\lambda )a),\forall \lambda \in [0,1]. \nonumber \\ \end{aligned}$$

(3.17)

Applying the $(\phi -h)$-integral on both sides over the interval [0, 1]:

$$\begin{aligned} \begin{aligned} \frac{1}{\hbar \left( \frac{1}{2}\right) } f\left( \frac{a+x}{2}\right)&\le \frac{1-\phi (x)}{(1-\phi (x))+\phi (x)h} \left( \int _{0}^{1}f(x-\lambda (x-a)) _{h}d_{\phi }\lambda \right. \\&\quad \left. +\int _{0}^{1}f(a+\lambda (x-a)) _{h}d_{\phi }\lambda \right) . \end{aligned} \end{aligned}$$

(3.18)

Since $f(a+x-u)=f(u)$, $\forall u\in (a,x)$, we have $f(x-\lambda (x-a))=f(a+\lambda (x-a))$. Then, the inequality (3.18) takes the following form:

$$\begin{aligned} \begin{aligned}&\frac{1}{2\hbar \left( \frac{1}{2}\right) } f\left( \frac{a+x}{2}\right) \le \frac{1-\phi (x)}{(1-\phi (x))+\phi (x)h} \int _{0}^{1}f(a+\lambda (x-a)) _{h}d_{\phi }\lambda .&\end{aligned} \end{aligned}$$

(3.19)

Applying the left $(\phi -h)$-integral definition, we get:

$$\begin{aligned}&\frac{(1-\phi (x))+\phi (x)h}{(1-\phi (x))(x-a)+\phi (x)\,h_{1}}\int _{a}^{x}f( \lambda )_{h_{1}}d_{\phi }\lambda \nonumber \\&\quad =((1-\phi (x))+\phi (x)h) \sum _{n=0}^{\infty }\phi ^{n}(x)f\left( \phi ^{n}(x)a+(1- \phi ^{n}(x)).x+nh_{1}\phi ^{n}(x)\right) \nonumber \\&\quad =\int _{0}^{1}f\left( a+(x-a)\lambda \right) \, _{h}d_{\phi }\lambda . \end{aligned}$$

(3.20)

Now, using the $\hbar$-convexity of f on the right hand side of the equality (3.20), we have

$$\begin{aligned} \begin{aligned}&\int _{0}^{1}f\left( \lambda x+(1-\lambda )a\right) \, _{h}d_{\phi }\lambda \le f(x)\int _{0}^{1}\hbar (\lambda )\, _{h}d_{\phi }\lambda +f(a)\int _{0}^{1}\hbar (1-\lambda )\, _{h}d_{\phi }\lambda .&\end{aligned} \end{aligned}$$

Therefore, (3.20) becomes

$$\begin{aligned} \begin{aligned}&\frac{(1-\phi (x))+\phi (x)h}{(1-\phi (x))(x-a)+\phi (x)\,h_{1}}\int _{a}^{x}f( \lambda )\,_{h_{1}}d_{\phi }\lambda \\&\quad \le f(x)\int _{0}^{1}\hbar (\lambda )\, _{h}d_{\phi }\lambda +f(a)\int _{0}^{1}\hbar (1-\lambda )\, _{h}d_{\phi }\lambda .&\end{aligned} \end{aligned}$$

(3.21)

Finally, we obtain the inequality (3.15) from (3.19), (3.20) and (3.21):(ii) By using the $\hbar$-convexity, we get

$$\begin{aligned} \frac{1}{\hbar \left( \frac{1}{2}\right) }f\left( \frac{x+b}{2}\right) \le f(\lambda x+(1-\lambda )b)+f(\lambda b+(1-\lambda )x),\forall \lambda \in [0,1]. \nonumber \\ \end{aligned}$$

(3.22)

Applying the $(\phi -h)$-integral on the both sides over the interval [0, 1], we have

$$\begin{aligned} \begin{aligned} \frac{1}{\hbar \left( \frac{1}{2}\right) } f\left( \frac{x+b}{2}\right)&\le \frac{1-\phi (x)}{(1-\phi (x))+\phi (x)h}\times \left( \int _{0}^{1}f(b-\lambda (b-x)) _{h}d_{\phi }\lambda \right. \\&\quad \left. +\int _{0}^{1}f(x+\lambda (b-x)) _{h}d_{\phi }\lambda \right) .&\end{aligned} \end{aligned}$$

(3.23)

Since $f(x+b-u)=f(u)$, $\forall u\in (x,b),$ we have $f(b-\lambda (b-x))=f(x+\lambda (b-x))$. The inequality (3.23) becomes:

$$\begin{aligned} \begin{aligned}&\frac{1}{2\hbar \left( \frac{1}{2}\right) } f\left( \frac{x+b}{2}\right) \le \frac{1-\phi (x)}{(1-\phi (x))+\phi (x)h} \int _{0}^{1}f(x+\lambda (b-x)) _{h}d_{\phi }\lambda .&\end{aligned} \end{aligned}$$

(3.24)

We now apply the right $(\phi -h)$-integral to obtain the following

$$\begin{aligned}&\frac{(1-\phi (x))+\phi (x)h}{(1-\phi (x))(b-x)+\phi (x)\,h_{2}}\int _{x}^{b}f( \lambda )_{h_{2}}d_{\phi }\lambda \nonumber \\&\quad =((1-\phi (x))+\phi (x)h)\sum _{n=0}^{ \infty }\phi ^{n}(x)f\left( \phi ^{n}(x).x+(1-\phi ^{n}(x))b+nh_{2}\phi ^{n}(x) \right) \nonumber \\&\quad =\int _{0}^{1}f\left( x+\lambda (b-x)\right) \, _{h}d_{\phi }\lambda . \end{aligned}$$

(3.25)

As, f is $\hbar$-convex, the right hand side of equality (3.25) becomes

$$\begin{aligned} \begin{aligned}&\int _{0}^{1}f\left( x+\lambda (b-x)\right) \, _{h}d_{\phi }\lambda \le f(b)\int _{0}^{1}\hbar (\lambda )\, _{h}d_{\phi }\lambda +f(x)\int _{0}^{1}\hbar (1-\lambda )\, _{h}d_{\phi }\lambda .&\end{aligned} \end{aligned}$$

So, (3.25) reduces to

$$\begin{aligned} \begin{aligned}&\frac{(1-\phi (x))+\phi (x)h}{(1-\phi (x))(b-x)+\phi (x)\,h_{2}}\int _{x}^{b}f( \lambda )\,_{h_{2}}d_{\phi }\lambda \\&\quad \le f(b)\int _{0}^{1}\hbar (\lambda )\, _{h}d_{\phi }\lambda +f(x)\int _{0}^{1}\hbar (1-\lambda )\, _{h}d_{\phi }\lambda . \end{aligned} \end{aligned}$$

(3.26)

We obtain the inequality (3.16) from (3.24), (3.25) and (3.26). $\square$

Corollary 3

Taking $h=0$ in (3.15) and (3.16), we obtain the inequalities for the left and right $\phi$-integrals as follows:

$$\begin{aligned} \begin{aligned} \frac{1}{2\hbar \left( \frac{1}{2}\right) } f\left( \frac{a+x}{2}\right)&\le \frac{1}{x-a}\int _{a}^{x}f(\lambda )d_{\phi } \lambda \\ {}&\le f(x)\int _{0}^{1}\hbar (\lambda ) d_{\phi }\lambda +f(a)\int _{0}^{1}\hbar (1-\lambda ) d_{\phi }\lambda \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \frac{1}{2\hbar \left( \frac{1}{2}\right) } f\left( \frac{x+b}{2}\right)&\le \frac{1}{b-x}\int _{x}^{b}f(\lambda )d_{\phi } \lambda \\ {}&\le f(b)\int _{0}^{1}\hbar (\lambda ) d_{\phi }\lambda +f(x)\int _{0}^{1}\hbar (1-\lambda )d_{\phi }\lambda . \end{aligned} \end{aligned}$$

Remark 2

By setting $x=b$ in (3.15) and $x=a$ in (3.16), we get the following inequality:

$$\begin{aligned} \begin{aligned} \frac{1}{2\hbar \left( \frac{1}{2}\right) } f\left( \frac{a+b}{2}\right)&\le \frac{1-\phi (x)}{(1-\phi (x))(b-a)+\phi (x) \,h_{3}}\int _{a}^{b}f(\lambda )_{h_{3}}d_{\phi }\lambda \\ {}&\le f(b)\int _{0}^{1}\hbar (\lambda )\, _{h}d_{\phi }\lambda +f(a)\int _{0}^{1}\hbar (1-\lambda )\, _{h}d_{\phi }\lambda , \end{aligned} \end{aligned}$$

(3.27)

where $h_{3}=(b-a)h.$

Corollary 4

Using $h=0$ in the Remark 2, the inequality for ${\phi }$ -integrals is given as:

$$\begin{aligned} \begin{aligned} \frac{1}{2\hbar \left( \frac{1}{2}\right) } f\left( \frac{a+b}{2}\right)&\le \frac{1}{b-a}\int _{a}^{b}f(\lambda )d_{\phi } \lambda \\ {}&\le f(b)\int _{0}^{1}\hbar (\lambda )\, d_{\phi }\lambda +f(a)\int _{0}^{1}\hbar (1-\lambda )d_{\phi }\lambda . \end{aligned} \end{aligned}$$

(3.28)

Theorem 9

Let $f:[0,y]\rightarrow \mathbb {R}$ be m-convex function, $\Omega$ an interval containing (0, 1) and $\phi :\Omega \rightarrow { \Omega }$ be a function such that $\phi (x)(x+h)\in [0, y]$, $\forall x\in { \Omega }$, $h\in \mathbb {R}$ and $a,b\in [0,y]$ with $a<b$. Then,

(i) The following inequality holds for the left $(\phi -h)$ -integral provided that $f(\frac{a+x-u}{m})=f(u),\forall {u}\in (a,x)$

$$\begin{aligned} f\left( \frac{a+x}{2}\right)&\le \frac{(1-\phi (x))(1+m)}{2((1-\phi (x))(x-a)+ \phi (x)\,h_{1})}\int _{a}^{x}f(\lambda )_{h_{1}}d_{\phi }\lambda \nonumber \\&\le \frac{(1-\phi (x))+\phi (x)h}{2}\left( f(x)\left( \frac{\phi (x)}{1- \phi ^{2}(x)}+\frac{\phi ^{2}(x)h}{(1-\phi ^{2}(x))^{2}}\right) \right. \nonumber \\&\quad \left. +mf\left( \frac{a}{m}\right) \left( \frac{1}{1-\phi ^{2}(x)}+\frac{\phi ^{2}(x)h}{(1- \phi ^{2}(x))^{2}}\right) \right) ,\nonumber \\&\quad \text {where}\, h_{1}=(x-a)h\, \text {and}\,\phi (x)\in (0,1). \end{aligned}$$

(3.29)

(ii) The following inequality holds for the right $(\phi -h)$ -integral if $f(\frac{x+b-u}{m})=f(u),\forall {u}\in (x,b)$

$$\begin{aligned} f\left( \frac{x+b}{2}\right)&\le \frac{(1-\phi (x))(1+m)}{2((1-\phi (x))(b-x)+ \phi (x)\,h_{2})}\int _{x}^{b}f(\lambda )_{h_{2}}d_{\phi }\lambda \nonumber \\&\le \frac{(1-\phi (x))+\phi (x)h}{2} \left( f(b)\left( \frac{\phi (x)}{1- \phi ^{2}(x)}+\frac{\phi ^{2}(x)h}{(1-\phi ^{2}(x))^{2}}\right) \right. \nonumber \\&\quad \left. +mf\left( \frac{x}{m}\right) \left( \frac{1}{1-\phi ^{2}(x)}-\frac{\phi ^{2}(x)h}{(1- \phi ^{2}(x))^{2}}\right) \right) ,\nonumber \\&\quad \text {where}\, h_{2}=(b-x)h\,\text {and}\,\phi (x)\in (0,1). \end{aligned}$$

(3.30)

Proof

We now prove (i) and (ii) as follows:

(i) It follows from the m-convexity of f that

$$\begin{aligned} f\left( \frac{a+x}{2}\right) \le \frac{1}{2}f(a+\lambda (x-a))+\frac{m}{2} f\left( \frac{x-\lambda (x-a)}{m}\right) ,\forall \lambda \in [0,1]. \nonumber \\ \end{aligned}$$

(3.31)

Using the fact: $f(\frac{x+a-u}{m})=f(u),\forall {u}\in (a,x)$, we have $f\left( \frac{x-\lambda (x-a)}{m}\right) =f(a+\lambda (x-a))$. The above inequality becomes

$$\begin{aligned} f\left( \frac{a+x}{2}\right) \le \frac{1+m}{2}f(a+\lambda (x-a)),\forall \lambda \in [0,1]. \end{aligned}$$

(3.32)

Applying the left $(\phi -h)$-integral on the both sides over the interval [0, 1], we have

$$\begin{aligned} \begin{aligned}&f\left( \frac{a+x}{2}\right) \le \frac{(1-\phi (x))(1+m)}{2((1-\phi (x))+ \phi (x)h)}\int _{0}^{1}f(a+\lambda (x-a)) _{h}d_{\phi }\lambda .&\end{aligned} \end{aligned}$$

(3.33)

By using the $\hbar$-convexity of f,  the right term of inequality (3.33) reduces to

$$\begin{aligned} \int _{0}^{1}f(a+\lambda (x-a))\,_{h}d_{\phi }\lambda \le f(x)\int _{0}^{1}\lambda \,_{h}d_{\phi }\lambda +mf\left( \frac{a}{m}\right) \int _{0}^{1}(1-\lambda )\,_{h}d_{\phi }\lambda \nonumber \\ \end{aligned}$$

(3.34)

From (3.33), (3.20) and (3.34), we obtain the following:

$$\begin{aligned} \begin{aligned} f\left( \frac{a+x}{2}\right)&\le \frac{(1-\phi (x))(1+m)}{2((1-\phi (x))(x-a)+ \phi (x)h_{1})}\int _{a}^{x}f(\lambda )\,_{h_{1}}d_{\phi }\lambda \\ {}&\le f(x)\int _{0}^{1}\lambda \, _{h}d_{\phi }\lambda +mf\left( \frac{a}{m}\right) \int _{0}^{1}(1-\lambda )\, _{h}d_{\phi }\lambda .&\end{aligned} \end{aligned}$$

(3.35)

On $(\phi -h)$-integrating (3.35) over the interval [0, 1], we get the inequality (3.29). Following arguments similar to those given above, we obtain (3.30) inequality from (3.33) by using the definition of right $(\phi -h)$ -integral.

$$\begin{aligned} \begin{aligned} f\left( \frac{x+b}{2}\right)&\le \frac{(1-\phi (x))(1+m)}{2((1-\phi (x))(b-x)+ \phi (x)h_{2})}\int _{x}^{b}f(\lambda )\,_{h_{2}}d_{\phi }\lambda \\ {}&\le f(b)\int _{0}^{1}\lambda \, _{h}d_{\phi }\lambda +mf\left( \frac{x}{m}\right) \int _{0}^{1}(1-\lambda )\, _{h}d_{\phi }\lambda .&\end{aligned} \end{aligned}$$

(3.36)

$(\phi -h)$-integrating (3.36) over the interval [0, 1] gives the inequality (3.30). $\square$

Corollary 5

Taking $h=0$ in (3.29) and (3.30), we obtain the inequalities for the left and right $\phi$-integrals as follows:

$$\begin{aligned} \begin{aligned}&f\left( \frac{a+x}{2}\right) \le \frac{(1+m)}{2(x-a)}\int _{a}^{x}f(\lambda )d_{ \phi }\lambda \le \frac{1}{2(1+\phi (x))}\left( \phi (x)f(x)+mf\left( \frac{a}{m}\right) \right) , \end{aligned}\nonumber \\ \end{aligned}$$

(3.37)

$$\begin{aligned} \begin{aligned}&f\left( \frac{b+x}{2}\right) \le \frac{(1+m)}{2(b-x)}\int _{x}^{b}f(\lambda )d_{ \phi }\lambda \le \frac{1}{2(1+\phi (x))}\left( \phi (x)f(b)+mf\left( \frac{x}{m}\right) \right) . \end{aligned}\nonumber \\ \end{aligned}$$

(3.38)

Remark 3

By setting $x=b$ in (3.29) and $x=a$ in (3.30), we get the following inequality:

$$\begin{aligned} f\left( \frac{a+b}{2}\right)&\le \frac{(1-\phi (x))(1+m)}{2((1-\phi (x))(b-a)+ \phi (x)\,h_{3})}\int _{a}^{b}f(\lambda )_{h_{3}}d_{\phi }\lambda \nonumber \\&\le \frac{(1-\phi (x))+\phi (x)h}{2} \left( f(b)\left( \frac{\phi (x)}{1- \phi ^{2}(x)}+\frac{\phi ^{2}(x)h}{(1-\phi ^{2}(x))^{2}}\right) \right. \nonumber \\&\quad \left. +mf\left( \frac{a}{m}\right) \left( \frac{1}{1-\phi ^{2}(x)}+\frac{\phi ^{2}(x)h}{(1- \phi ^{2}(x))^{2}}\right) \right) , \end{aligned}$$

(3.39)

where $h_{3}=(b-a)h.$

Corollary 6

If we take $h=0$ in (3.40), then we have the inequality for $\phi$ -integral:

$$\begin{aligned} \begin{aligned}&f\left( \frac{a+b}{2}\right) \le \frac{1+m}{2(b-a)}\int _{a}^{b}f(\lambda )d_{ \phi }\lambda \le \frac{1}{2(1+\phi (x))}\left( \phi (x)f(b)+mf\left( \frac{a}{m}\right) \right) , \end{aligned}\nonumber \\ \end{aligned}$$

(3.40)

where $h_{3}=(b-a)h.$

Remark 4

If we take $\phi (x)=q$; constant function, then the results in [7] are obtained.