On q-Hermite-Hadamard Inequalities for Differentiable Convex Functions

Abstract

In this paper, we establish some new results on the left-hand side of the q-Hermite–Hadamard inequality for differentiable convex functions with a critical point. Our work extends the results of Alp et. al (q-Hermite Hadamard inequalities and quantum estimates for midpoint type inequalities via convex and quasi-convex functions, J. King Saud Univ. Sci., 2018, 30, 193-203), by considering the critical point-type inequalities.

1. Introduction

Quantum calculus (also known as q-calculus) is the study of calculus without limits, where classical mathematical formulas are obtained as $q\to 1$. Firstly introduced by Euler (1707–1783) in the tracks of Newton’s infinite series, the study of q-calculus was established in the early Twentieth Century after the work of Jackson (1910) on defining an integral later known as the q-Jackson integral; see [1,2,3,4]. In q-calculus, the classical derivative is replaced by the q-difference operator in order to deal with non-differentiable functions; see [5,6] for more details. Applications of q-calculus can be found in various fields of mathematics and physics, and the interested readers are referred to [7,8,9,10].

The theory of convex functions has been widely studied and applied to various fields of science. Due to its close relation to the theory of inequalities, a rich literature on inequalities can be found in the study of convex functions; see [11,12,13,14,15,16,17,18]. This includes the Hermite–Hadamard inequality, introduced by Hermite and Hadamard independently, which has been studied extensively in recent years.

Let $J\subseteq \mathbb{R}$ be an interval and $f:J\to \mathbb{R}$ be a function from J to $\mathbb{R}$. Recall that f is said to be a convex function if it satisfies the inequality:

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

(1)

for all $x,y\in J$ and $\lambda \in [0,1]$. In addition, if an equality holds for all $x,y\in J$ and $\lambda \in [0,1]$, then f is said to be affine.

It is also well known that f is convex if and only if it satisfies the Hermite–Hadamard inequality, which is defined by:

$$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx\le \frac{f\left(a\right)+f\left(b\right)}{2},$$

(2)

for all $a,b\in J$ and $a<b$. One can estimate by the right-hand side of (2) by using Iyengar’s inequality, which is defined by:

$$\left|\frac{f\left(a\right)+f\left(b\right)}{2}-\frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx\right|\le \frac{M(b-a)}{4}-\frac{1}{4M(b-a)}{(f\left(b\right)-f\left(a\right))}^2,$$

where M denotes the Lipschitz constant, that is $M=sup\left\{\left|\frac{f\left(x\right)-f\left(y\right)}{x-y}\right|;x\ne y\right\}$.

This fundamental result of Hermite and Hadamard has attracted many mathematicians, and consequently, this inequality has been generalized and extended in many directions; see [19,20,21,22,23,24,25,26,27,28,29,30,31] and the references cited therein.

In 2018, Alp et al. [32] studied the q-analogue of Hermite–Hadamard’s inequality for increasing functions, that is,

$$f\left(\frac{qa+b}{1+q}\right)\le \frac{1}{b-a}{\int }_a^{b}f{\left(x\right)}_a{d}_{q}x\le \frac{qf\left(a\right)+f\left(b\right)}{1+q},$$

(3)

where q is a constant with $0<q<1.$ Moreover, they studied the generalized q-Hermite–Hadamard inequality for differentiable convex functions, that is,

$$max\{I_1,I_2,I_3\}\le \frac{1}{b-a}{\int }_a^{b}f\left(x\right){}_a{d}_{q}x\le \frac{qf\left(a\right)+f\left(b\right)}{1+q},$$

(4)

where:

$$\begin{array}{cc}\hfill I_1& =f\left(\frac{qa+b}{1+q}\right),\hfill \\ \hfill I_2& =f\left(\frac{a+qb}{1+q}\right)+\frac{(1-q)(b-a)}{1+q}{f}^{\prime }\left(\frac{a+qb}{1+q}\right),\hfill \\ \hfill I_3& =f\left(\frac{a+b}{2}\right)+\frac{(1-q)(b-a)}{2(1+q)}{f}^{\prime }\left(\frac{a+b}{2}\right).\hfill \end{array}$$

This paper aims to establish the generalized q-Hermite–Hadamard inequality for differentiable convex functions with a critical point.

The paper is organized as follows. Some basic concepts are recalled in Section 2. Section 3 contain the main results, while conclusions are given in Section 4.

2. Preliminaries

In this section, some basic results are mentioned. Throughout this section, we let $J=[a,b]\subseteq \mathbb{R}$ be an interval and q be a constant with $0<q<1$.

Definition 1.

[33] The q-derivative of a continuous function$f:J\to \mathbb{R}$at x is defined as:

$${}_a{D}_{q}f\left(x\right)=\frac{f\left(x\right)-f(qx+(1-q\left)a\right)}{(1-q)(x-a)},\mathrm{for}x\ne a.$$

(5)

For$x=a$, we define${\displaystyle {}_a{D}_{q}f\left(a\right)=\underset{x\to a}{lim}}$${}_a{D}_{q}f\left(x\right)$.

If ${}_a{D}_{q}f\left(x\right)$ exists for all $x\in J$, then f is q-differentiable on $J.$ Moreover, if $a=0$, then (5) reduces to ${}_0{D}_{q}f=D_{q}f$, where $D_q$ is the q-derivative of f, which is defined by:

$$D_{q}f\left(x\right)=\frac{f\left(x\right)-f\left(qx\right)}{(1-q)x}.$$

For more details, see [4].

The higher-order q-derivatives of functions on J are also defined.

Definition 2.

[33] For a continuous function$f:J\to \mathbb{R}$, the second-order q-derivative of f on J, if${}_a{D}_{q}f$is q-differentiable on J, denoted by${}_a{D}_q^{2}f$and defined by:

$${}_a{D}_q^{2}f={}_a{D}_q{(}_a{D}_q).$$

Similarly, provided that${}_a{D}_q^{n-1}f$is the q-derivative on J for some integer$n>2$, the$n^{\mathit{th}}$-order q-derivative of f on J is the function from$J\to \mathbb{R}$defined by:

$${}_a{D}_q^{n}f={}_a{D}_q{(}_a{D}_q^{n-1}f).$$

Example 1.

Let$f:J\to \mathbb{R}$with$f\left(x\right)=x^2+1.$Let q be a constant with$0<q<1$. Then, for$x\ne a$, we have:

$$\begin{array}{cc}{}_a{D}_q(x^2+1)\hfill & =\frac{(x^2+1)-\left[{(qx+(1-q)a)}^2+1\right]}{(1-q)(x-a)}\hfill \\ & =\frac{(1+q)x^2-2qax-(1-q)a^2}{(x-a)}\hfill \\ & =(1+q)x+(1-q)a.\hfill \end{array}$$

(6)

For$x=a$, ${\displaystyle {}_a{D}_{q}f\left(a\right)=\underset{x\to a}{lim}}$${}_a{D}_{q}f\left(x\right)=2a$.

Definition 3.

[33] The q-integral of a continuous function$f:J\to \mathbb{R}$is defined as:

$${\int }_a^{x}f\left(t\right){}_a{d}_{q}t=(1-q)(x-a)\sum _{n=0}^{\infty }{q}^{n}f(q^{n}x+(1-q^n)a),$$

(7)

for$x\in J$.

Note that if$a=0$, then (7) becomes the classical q-integral of f, that is,

$${\int }_0^{x}f\left(t\right){}_0{d}_{q}t=(1-q)x\sum _{n=0}^{\infty }{q}^{n}f\left(q^{n}x\right)$$

for$x\in [0,\infty )$; see [4] for more details.

Example 2.

Let$f:[a,b]\to \mathbb{R}$with$f\left(x\right)=2x.$Let q be a constant with$0<q<1$. Then, we have:

$$\begin{array}{cc}\hfill {\int }_a^{b}f\left(x\right){}_a{d}_{q}x& ={\int }_a^{b}2x{}_a{d}_{q}x\hfill \\ & =2(1-q)(b-a)\sum _{n=0}^{\infty }{q}^n(q^{n}b+(1-q^n)a)\hfill \\ & =\frac{2(b-a)(b+qa)}{1+q}.\hfill \end{array}$$

Note that if$q\to 1$, we obtain the classical integration:

$${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}2xdx=b^2-a^2.$$

Theorem 1.

Assume that the function$f:J\to \mathbb{R}$is continuous. Then, we have the following:

(i) 

${}_a{D}_q{\int }_a^{x}f\left(t\right){}_a{d}_{q}t=f\left(x\right)-f\left(a\right)$;

(ii) 

${\int }_c^x{}_a{D}_{q}f\left(t\right){}_a{d}_{q}t=f\left(x\right)-f\left(c\right)$for$c\in (a,x)$.

Theorem 2.

Assume that the functions$f,g:J\to \mathbb{R}$are continuous and$\alpha \in \mathbb{R}$. Then, we have the following:

(i) 

${\int }_a^x\left[f\right(t)+g(t\left)\right]{}_a{d}_{q}t={\int }_a^{x}f\left(t\right){}_a{d}_{q}t+{\int }_a^{x}g\left(t\right){}_a{d}_{q}t$;

(ii) 

${\int }_a^x\left(\alpha f\right)\left(t\right){}_a{d}_{q}t=\alpha {\int }_a^{x}f\left(t\right){}_a{d}_{q}t$;

(iii) 

${\int }_c^{x}f\left(t\right){}_a{D}_{q}g\left(t\right){}_a{d}_q{t=\left(fg\right)|}_c^x-{\int }_c^{x}g{(qt+(1-q)a)}_a{D}_{q}f\left(t\right){}_a{d}_{q}t$for$c\in (a,x)$.

For the proofs of the properties in Theorems 1 and 2, see [34].

3. Main Results

In this section, we present our main results on the left-hand side of the q-Hermite–Hadamard inequality for differentiable convex functions with a critical point.

Theorem 3.

Suppose that$f:[a,b]\to \mathbb{R}$is a differentiable convex function on$(a,b)$such that$f^{\prime }\left(c\right)=0$for$c\in (a,b)$, and let q be a constant with$0<q<1$. Then, we have:

$$\begin{array}{cc}f\left(\frac{q(a+c)+(1-q)b}{1+q}\right)+\hfill & f^{\prime }\left(\frac{q(a+c)+(1-q)b}{1+q}\right)\left(\frac{q(b-c)}{1+q}\right)\hfill \\ & \le \frac{1}{b-a}{\int }_a^{b}f{\left(x\right)}_a{d}_{q}x\hfill \\ & \le \frac{qf\left(a\right)+f\left(b\right)}{1+q}.\hfill \end{array}$$

(8)

Proof. 

Since the function f is differentiable on $(a,b)$, there exists a tangent line at the point $\frac{q(a+c)+(1-q)b}{1+q}\in (a,b)$, given by:

$$h\left(x\right)=f\left(\frac{q(a+c)+(1-q)b}{1+q}\right)+f^{\prime }\left(\frac{q(a+c)+(1-q)b}{1+q}\right)\left(x-\frac{q(a+c)+(1-q)b}{1+q}\right).$$

(9)

Since f is a convex function on $[a,b]$, it follows that $h\left(x\right)\le f\left(x\right)$ for all $x\in [a,b]$. After q-integrating of (9) on $[a,b]$, we have:

$$\begin{array}{c}{\int }_a^{b}h\left(x\right){}_a{d}_{q}x\hfill \\ ={\int }_a^b\left[f\left(\frac{q(a+c)+(1-q)b}{1+q}\right)+f^{\prime }\left(\frac{q(a+c)+(1-q)b}{1+q}\right)\right.\hfill \\ \times \left.\left(x-\frac{q(a+c)+(1-q)b}{1+q}\right)\right]{}_a{d}_{q}x\hfill \\ =(b-a)f\left(\frac{q(a+c)+(1-q)b}{1+q}\right)\hfill \\ +f^{\prime }\left(\frac{q(a+c)+(1-q)b}{1+q}\right)\left({\int }_a^{b}x{}_a{d}_{q}x-(b-a)\frac{q(a+c)+(1-q)b}{1+q}\right)\hfill \\ =(b-a)f\left(\frac{q(a+c)+(1-q)b}{1+q}\right)\hfill \\ +f^{\prime }\left(\frac{q(a+c)+(1-q)b}{1+q}\right)\left((b-a)\frac{(qa+b)}{1+q}-(b-a)\frac{q(a+c)+(1-q)b}{1+q}\right)\hfill \\ =(b-a)\left[f\left(\frac{q(a+c)+(1-q)b}{1+q}\right)+f^{\prime }\left(\frac{q(a+c)+(1-q)b}{1+q}\right)\left(\frac{q(b-c)}{1+q}\right)\right]\hfill \\ \le {\int }_a^{b}f\left(x\right){}_a{d}_{q}x.\hfill \end{array}$$

On the other hand, since f is a convex function, we obtain:

$$\begin{array}{cc}\frac{1}{(b-a)}{\int }_a^{b}f{\left(x\right)}_a{d}_{q}x\hfill & =\frac{1}{(b-a)}\left[(1-q)(b-a)\sum _{n=0}^{\infty }{q}^{n}f(q^{n}b+(1-q^n)a)\right]\hfill \\ & =(1-q)\sum _{n=0}^{\infty }{q}^{n}f(q^{n}b+(1-q^n)a)\hfill \\ & \le (1-q)\sum _{n=0}^{\infty }{q}^n[q^{n}f\left(b\right)+(1-q^n)f\left(a\right)]\hfill \\ & =(1-q)\left[\frac{f\left(b\right)}{1-q^2}+\frac{f\left(a\right)}{1-q}-\frac{f\left(a\right)}{1-q^2}\right]\hfill \\ & =\frac{qf\left(a\right)+f\left(b\right)}{1+q}.\hfill \end{array}$$

The proof is complete. □

Remark 1.

In Theorem 3, if$q\in (0,\frac{c-b}{a-b}]$, then$\frac{q(a+c)+(1-q)b}{1+q}\in [c,b)$. We can reduce the left-hand side of Theorem 3 as:

$$\begin{array}{cc}\hfill f\left(\frac{q(a+c)+(1-q)b}{1+q}\right)& \le \frac{1}{(b-a)}{\int }_a^{b}f\left(x\right){}_a{d}_{q}x\le \frac{qf\left(a\right)+f\left(b\right)}{1+q},\hfill \end{array}$$

since$f^{\prime }\left(\frac{q(a+c)+(1-q)b}{1+q}\right)\left(\frac{q(b-c)}{1+q}\right)\ge 0$.

Remark 2.

In Remark 1, if$c\to a^{+}$, then$\frac{c-b}{a-b}\to 1^{-}$. Since$q\in (0,1)$, we have$\frac{q(a+c)+(1-q)b}{1+q}\in (a,b)$. We can reduce the left-hand side of Theorem 3 as:

$$\begin{array}{cc}\hfill f\left(qa+(1-q)b\right)& \le \frac{1}{(b-a)}{\int }_a^{b}f\left(x\right){}_a{d}_{q}x\le \frac{qf\left(a\right)+f\left(b\right)}{1+q},\hfill \end{array}$$

since$f^{\prime }\left(\frac{q(a+c)+(1-q)b}{1+q}\right)\left(\frac{q(b-c)}{1+q}\right)\ge 0$.

Corollary 1.

Assume that$f:[a,b]\to \mathbb{R}$is a differentiable convex function on$(a,b)$such that$f^{\prime }\left(\frac{a+b}{2}\right)=0$, for$0<q<1$. Then, we have:

$$\begin{array}{cc}f\left(\frac{q(a+\frac{a+b}{2})+(1-q)b}{1+q}\right)\hfill & +f^{\prime }\left(\frac{q(a+\frac{a+b}{2})+(1-q)b}{1+q}\right)\frac{q(b-a)}{2(1+q)}\hfill \\ & \le \frac{1}{b-a}{\int }_a^{b}f{\left(x\right)}_a{d}_{q}x\hfill \\ & \le \frac{qf\left(a\right)+f\left(b\right)}{1+q}.\hfill \end{array}$$

(10)

Corollary 2.

Assume that$f:[a,b]\to \mathbb{R}$is a differentiable convex function on$(a,b)$such that$f^{\prime }\left(0\right)=0$, for$0\in (a,b)$and$0<q<1$. Then, we have:

$$\begin{array}{cc}f\left(\frac{qa+(1-q)b}{1+q}\right)\hfill & +f^{\prime }\left(\frac{qa+(1-q)b}{1+q}\right)\frac{qb}{(1+q)}\hfill \\ & \le \frac{1}{b-a}{\int }_a^{b}f{\left(x\right)}_a{d}_{q}x\le \frac{qf\left(a\right)+f\left(b\right)}{1+q}.\hfill \end{array}$$

(11)

Theorem 4.

Let$f:[a,b]\to \mathbb{R}$be a differentiable convex function on$(a,b)$such that$f^{\prime }\left(c\right)=0$for$c\in (a,b)$and$0<q<1$. Then, we have:

$$\begin{array}{cc}f\left(\frac{(1-q)a+q(c+b)}{1+q}\right)\hfill & +f^{\prime }\left(\frac{(1-q)a+q(c+b)}{1+q}\right)\left(\frac{q(2a-b-c)+b-a}{1+q}\right)\hfill \\ & \le \frac{1}{b-a}{\int }_a^{b}f{\left(x\right)}_a{d}_{q}x\hfill \\ & \le \frac{qf\left(a\right)+f\left(b\right)}{1+q}.\hfill \end{array}$$

(12)

Proof. 

Since the function f is differentiable on $(a,b)$, there exists a tangent line at the point $\frac{(1-q)a+q(c+b)}{1+q}\in (a,b)$, which is given by:

$$k\left(x\right)=f\left(\frac{(1-q)a+q(c+b)}{1+q}\right)+f^{\prime }\left(\frac{(1-q)a+q(c+b)}{1+q}\right)\left(x-\frac{(1-q)a+q(c+b)}{1+q}\right).$$

(13)

Since f is convex on $[a,b]$, it follows that $k\left(x\right)\le f\left(x\right)$ for all $x\in [a,b]$. After q-integrating (13), we obtain:

$$\begin{array}{c}{\int }_a^{b}k\left(x\right){}_a{d}_{q}x\hfill \\ ={\int }_a^b\left[f\left(\frac{(1-q)a+q(c+b)}{1+q}\right)\right.\hfill \\ \left.+f^{\prime }\left(\frac{(1-q)a+q(c+b)}{1+q}\right)\left(x-\frac{(1-q)a+q(c+b)}{1+q}\right)\right]{}_a{d}_{q}x\hfill \\ =(b-a)f\left(\frac{(1-q)a+q(c+b)}{1+q}\right)\hfill \\ +f^{\prime }\left(\frac{1-q)a+q(c+b)}{1+q}\right)\left({\int }_a^{b}x{}_a{d}_{q}x-(b-a)\frac{(1-q)a+q(c+b)}{1+q}\right)\hfill \\ =(b-a)f\left(\frac{(1-q)a+q(c+b)}{1+q}\right)\hfill \\ +f^{\prime }\left(\frac{(1-q)a+q(c+b)}{1+q}\right)\left[(b-a)\left(\left(\frac{aq+b}{1+q}\right)-\frac{(1-q)a+q(c+b)}{1+q}\right)\right]\hfill \\ =(b-a)\left[f\left(\frac{(1-q)a+q(c+b)}{1+q}\right)\right.\hfill \\ \left.+f^{\prime }\left(\frac{(1-q)a+q(c+b)}{1+q}\right)\left(\frac{q(2a-b-c)+b-a}{1+q}\right)\right]\hfill \\ \le {\int }_a^b{f\left(x\right)}_a{d}_{q}x.\hfill \end{array}$$

(14)

The proof is complete. □

Remark 3.

In Theorem 4, if$q\in (\frac{1}{2},\frac{c-a}{b-a}],$then$f^{\prime }\left(\frac{(1-q)a+q(c+b)}{1+q}\right)\le 0$and$\frac{q(2a-b-c)+b-a}{1+q}<0$. We can reduce the left-hand side of Theorem 4 as:

$$\begin{array}{cc}\hfill f\left(\frac{(1-q)a+q(c+b)}{1+q}\right)& \le \frac{1}{(b-a)}{\int }_a^{b}f\left(x\right){}_a{d}_{q}x\le \frac{qf\left(a\right)+f\left(b\right)}{1+q},\hfill \end{array}$$

since$f^{\prime }\left(\frac{(1-q)a+q(c+b)}{1+q}\right)\left(\frac{q(2a-b-c)+b-a}{1+q}\right)\ge 0$.

Remark 4.

In Remark 3, if $c\to b^{-},$ then $q\to 1^{-}$. Since $q\in (\frac{1}{2},1)$, we have $\frac{(1-q)a+q(b+c)}{1+q}\in (\frac{a+2b}{3},b)$. We can reduce the left-hand side of Theorem 4 as:

$$\begin{array}{cc}\hfill f\left((1-q)\left(\frac{2a+b}{3}\right)+qb)\right)& \le \frac{1}{(b-a)}{\int }_a^{b}f\left(x\right){}_a{d}_{q}x\le \frac{qf\left(a\right)+f\left(b\right)}{1+q},\hfill \end{array}$$

since $f^{\prime }\left(\frac{(1-q)a+q(c+b)}{1+q}\right)\left(\frac{q(2a-b-c)+b-a}{1+q}\right)\ge 0$.

Theorem 5.

[Generalized q-Hermite–Hadamard inequality for convex differentiable functions]. Let $f:[a,b]\to \mathbb{R}$ be a differentiable convex function on $(a,b)$ such that $f^{\prime }\left(c\right)=0$ for $c\in (a,b)$ and $0<q<1$. Then, we have:

$$max\{I_1,I_2,\}\le \frac{1}{b-a}{\int }_a^{b}f\left(x\right){}_a{d}_{q}x\le \frac{qf\left(a\right)+f\left(b\right)}{1+q},$$

(15)

where

$$\begin{array}{cc}{I}_1\hfill & =f\left(\frac{q(a+c)+(1-q)b}{1+q}\right)+f^{\prime }\left(\frac{q(a+c)+(1-q)b}{1+q}\right)\left(\frac{q(b-c)}{1+q}\right),\hfill \\ I_2\hfill & =f\left(\frac{(1-q)a+q(c+b)}{1+q}\right)+f^{\prime }\left(\frac{(1-q)a+q(c+b)}{1+q}\right)\left(\frac{q(2a-b-c)+b-a}{1+q}\right).\hfill \end{array}$$

Proof. 

A combination of (8) and (12) yields (15). Thus, the proof is complete. □

Example 3.

Define the function $f\left(x\right)=x^2$ on $[-1,3]$, and let $q\in (0,1)$. Applying Theorem 3 with $a=-1$, $b=3$, and $c=0$, the left-hand side becomes:

$$\begin{array}{cc}\hfill f& \left(\frac{q(a+c)+(1-q)b}{1+q}\right)+f^{\prime }\left(\frac{q(a+c)+(1-q)b}{1+q}\right)\frac{q(b-c)}{1+q}-\frac{1}{b-a}{\int }_a^{b}f{\left(x\right)}_a{d}_{q}x\hfill \\ & =f\left(\frac{3-4q}{1+q}\right)+f^{\prime }\left(\frac{3-4q}{1+q}\right)\left(\frac{3q}{1+q}\right)-\frac{1}{4}\left[4(1-q)\sum _{n=0}^{\infty }{q}^{n}f(3q^n-(1-q^n))\right]\hfill \\ & =\frac{-9q^4-9q^3-9q^2-16q}{{(1+q)}^2(1+q+q^2)}\le 0.\hfill \end{array}$$

For the right-hand side, we have:

$$\begin{array}{cc}\hfill \frac{1}{3-(-1)}{\int }_{-1}^3{x^2}_a{d}_{q}x-\frac{qf(-1)+f\left(3\right)}{1+q}& =\frac{16}{1+q+q^2}-\frac{8}{1+q}+1-\frac{9+q}{1+q}\le 0.\hfill \end{array}$$

Example 4.

Define function $f\left(x\right)=x^2$ on $[-1,1]$, and let $q\in (0,1)$. Applying Corollary 2 with $a,b=-1$ and $c=0$, the left hand-side becomes:

$$\begin{array}{c}f\left(\frac{qa+(1-q)b}{1+q}\right)+f^{\prime }\left(\frac{qa+(1-q)b}{1+q}\right)\frac{\left(qb\right)}{1+q}-\frac{1}{b-a}{\int }_a^{b}f{\left(x\right)}_a{d}_{q}x\\ =f\left(\frac{1-2q}{1+q}\right)+f^{\prime }\left(\frac{1-2q}{1+q}\right)\left(\frac{q}{1+q}\right)-(1-q)\sum _{n=0}^{\infty }{q}^{n}f(2q^n-1)\end{array}$$

$$\begin{array}{c}=\frac{4q^2-4q+1}{{(1+q)}^2}+\frac{2q(1-2q)}{{(1+q)}^2}-\frac{1+2q-2q^2+q^3}{(1+q+q^2)(1+q)}\le 0.\hfill \end{array}$$

For the right-hand side, we have:

$$\begin{array}{cc}\hfill \frac{1}{1-(-1)}{\int }_{-1}^1{x^2}_a{d}_{q}x& -\frac{qf(-1)+f\left(1\right)}{1+q}\hfill \\ & =\frac{1}{2}\left[(1-q)\left(2\right)\sum _{n=0}^{\infty }{q}^{n}f(q^n-(1-q^n))\right]-\frac{1+q}{1+q}\hfill \\ & =\frac{4}{1+q+q^2}-\frac{4}{1+q}+1-1\le 0.\hfill \end{array}$$

Example 5.

Define functions $f\left(x\right)=x^2$ on $[-3,1]$, and let $q\in (0,1)$. Applying Theorem 4 with $a=-3$, $b=1$, and $c=0$, the left-hand side becomes:

$$\begin{array}{c}f\left(\frac{(1-q)a+q(c+b)}{1+q}\right)+f^{\prime }\left(\frac{(1-q)a+q(c+b)}{1+q}\right)\left(\frac{q(2a-b-c)+b-a}{1+q}\right)\hfill \\ -\frac{1}{b-a}{\int }_a^{b}f{\left(x\right)}_a{d}_{q}x\hfill \\ =f\left(\frac{4q-3}{1+q}\right)+f^{\prime }\left(\frac{4q-3}{1+q}\right)\left(\frac{4-7q}{1+q}\right)-\frac{1}{4}\left[4(1-q)\sum _{n=0}^{\infty }{q}^{n}f(4q^n-3)\right]\hfill \\ =\frac{16q^2-24q+9}{{(1+q)}^2}+\frac{-56q^2+74q-24}{{(1+q)}^2}-\frac{16}{1+q+q^2}+\frac{24}{1+q}-9\le 0.\hfill \end{array}$$

For the right-hand side, we have:

$$\begin{array}{cc}\hfill \frac{1}{3-(-1)}{\int }_{-1}^3{x^2}_a{d}_{q}x-\frac{qf(-3)+f\left(1\right)}{1+q}& =\frac{16}{1+q+q^2}-\frac{24}{1+q}+9-\frac{9q+1}{1+q}\le 0.\hfill \end{array}$$

4. Conclusions

In this paper, we considered and investigated the class of differentiable convex functions, which has a critical point in the setting of q-calculus. We used the approach of q-calculus to derive some new results on the left-hand side of q-Hermite–Hadamard inequalities. It is expected that the ideas and techniques presented in this paper will stimulate further research in this field.