Quantum Hermite–Hadamard-type inequalities for functions with convex absolute values of second $q^{b}$-derivatives

Abstract

In this paper, we obtain Hermite–Hadamard-type inequalities of convex functions by applying the notion of $q^{b}$-integral. We prove some new inequalities related with right-hand sides of $q^{b}$-Hermite–Hadamard inequalities for differentiable functions with convex absolute values of second derivatives. The results presented in this paper are a unification and generalization of the comparable results in the literature on Hermite–Hadamard inequalities.

1 Introduction

The Hermite–Hadamard inequality introduced by Hermite and Hadamard (see also [1] and [2, p. 137]) is one of the most well-known inequalities in the theory of convex functional analysis. It has an interesting geometrical interpretation with several applications.

These inequalities state that if $f:I\rightarrow \mathbb{R}$ is a convex function on an interval I of real numbers and $a,b\in I$ with $a< b$, then

$$ f \biggl( \frac{a+b}{2} \biggr) \leq \frac{1}{b-a} \int _{a}^{b}f(x)\,dx \leq \frac{f ( a ) +f ( b ) }{2}. $$

(1.1)

Both inequalities hold in the reversed manner if f is a concave function. Note that the Hermite–Hadamard inequalities may be viewed as a refinement of the concept of convexity and follows from Jensen’s inequality. Hermite–Hadamard inequalities for convex functions have received much attention in the recent years, and, consequently, a remarkable variety of refinements and generalizations have been obtained.

Many well-known integral inequalities such as the Hölder, Hermite–Hadamard, Ostrowski, Cauchy–Bunyakovsky–Schwarz, Gruss, Gruss-Chebyshev, and other integral inequalities have been studied in the setup of q-calculus using the concept of classical convexity. For more results in this direction, we refer to [3–18].

The purpose of this paper is to study Hermite–Hadamard-like inequalities for convex functions by applying the new concept of $q^{b}$-integral. We also discuss the relation of our results with comparable results existing in the literature.

The organization of this paper is as follows. In Sect. 2, we give a brief description of the concepts of q-calculus and some related works in this direction. In Sect. 3, we present the Hermite-Hadamard-type inequalities for the $q^{b}$-integrals. We also study the relation between the results presented herein and comparable results in the literature. Section 4 contains some conclusions and further directions for the future research. We believe that the study initiated in this paper may inspire new research in this area.

2 Preliminaries of q-calculus and some inequalities

In this section, we first present some known definitions and related inequalities in q-calculus. Set the following notation (see [19]):

$$ [ n ] _{q}=\frac{1-q^{n}}{1-q}=1+q+q^{2}+ \cdots+q^{n-1}, \quad q\in ( 0,1 ) . $$

Jackson [20] defined the q-Jackson integral of a given function f from 0 to b as follows:

$$ \int _{0}^{b}f ( x ) \,d_{q}x= ( 1-q ) b\sum_{n=0}^{\infty }q^{n}f \bigl( bq^{n} \bigr),\quad \text{where }0< q< 1, $$

(2.1)

provided that the sum converges absolutely.

Jackson [20] defined the q-Jackson integral of a given function over the interval $[a,b]$ as follows:

$$ \int _{a}^{b}f ( x ) \,d_{q}x= \int _{0}^{b}f ( x ) \,d_{q}x- \int _{0}^{a}f ( x )\, d_{q}x.$$

Definition 1

([21])

Let $f: [ a,b ] \rightarrow \mathbb{R}$ be a continuous function. Then the $q_{a}$-derivative of f at $x\in [ a,b ]$ is identified as

$$ {}_{a}D_{q}f ( x ) = \frac{f ( x ) -f ( qx+ ( 1-q ) a ) }{ ( 1-q ) ( x-a ) },\quad x\neq a. $$

(2.2)

Since $f: [ a,b ] \rightarrow \mathbb{R}$ is a continuous function, we can define

$$ {}_{a}D_{q}f ( a ) =\lim _{x\rightarrow a} {}_{a}D_{q}f ( x ) . $$

The function f is said to be $q_{a}$-differentiable on $[ a,b ]$ if ${}_{a}D_{q}f ( x )$ exists for all $x\in [ a,b ]$. If we take $a=0$ in (2.2), then we have ${}_{0}D_{q}f ( x ) =D_{q}f ( x )$, where $D_{q}f ( x )$ is the q-derivative of f at $x\in [ 0,b ]$ (see [19]) given by

$$ D_{q}f ( x ) = \frac{f ( x ) -f ( qx ) }{ ( 1-q ) x}, \quad x \neq 0. $$

Definition 2

([22])

Let $f: [ a,b ] \rightarrow \mathbb{R}$ be a continuous function. Then the $q^{b}$-derivative of f at $x\in [ a,b ]$ is given by

$$ {}^{b}D_{q}f ( x )= \frac{f ( qx+ ( 1-q ) b ) -f ( x ) }{ ( 1-q ) ( b-x ) },\quad x\neq b. $$

Definition 3

Let $f: [ a,b ] \rightarrow \mathbb{R}$ be a continuous function. Then the second $q^{b}$-derivative of f at $x\in [ a,b ]$ is given by

$$\begin{aligned} {}^{b}D_{q}^{2}f ( x ) =&{}^{b}D_{q} \bigl( ~^{b}D_{q}f ( x ) \bigr) \\ =&\frac{f ( q^{2}ta+ ( 1-tq^{2} ) b ) - ( 1+q ) f ( qta+ ( 1-qt ) b ) +qf ( ta+ ( 1-t ) b ) }{ ( 1-q ) ^{2}q ( b-a ) ^{2}t^{2}}. \end{aligned}$$

Definition 4

([21])

Let $f: [ a,b ] \rightarrow \mathbb{R}$ be a continuous function. Then the $q_{a}$-definite integral on $[ a,b ]$ is defined by

$$\begin{aligned} \int _{a}^{b}f ( x ) \,_{a}d_{q}x =& ( 1-q ) ( b-a ) \sum_{n=0}^{\infty }q^{n}f \bigl( q^{n}b+ \bigl( 1-q^{n} \bigr) a \bigr) \\ =& ( b-a ) \int _{0}^{1}f \bigl( ( 1-t ) a+tb \bigr) \,d_{q}t. \end{aligned}$$

Alp et al. [3] proved the following $q_{a}$-Hermite–Hadamard inequalities for convex functions in the setting of quantum calculus.

Theorem 1

If $f: [ a,b ] \rightarrow \mathbb{R}$ is a convex differentiable function on $[ a,b ]$ and $0< q<1$, then we have

$$ f \biggl( \frac{qa+b}{1+q} \biggr) \leq \frac{1}{b-a} \int _{a}^{b}f ( x ) \,_{a}d_{q}x\leq \frac{qf ( a ) +f ( b ) }{1+q}. $$

(2.3)

In [3] and [23] authors established some bounds for the left- and right-hand sides of inequality (2.3).

On the other hand, Bermudo et al. [22] gave the following definition and obtained the related Hermite–Hadamard-type inequalities.

Definition 5

([22])

Let $f: [ a,b ] \rightarrow \mathbb{R}$ be a continuous function. Then the $q^{b}$-definite integral on $[ a,b ]$ is given by

$$\begin{aligned} \int _{a}^{b}f ( x ) \,^{b}d_{q}x =& ( 1-q ) ( b-a ) \sum_{n=0}^{\infty }q^{n}f \bigl( q^{n}a+ \bigl( 1-q^{n} \bigr) b \bigr) \\ =& ( b-a ) \int _{0}^{1}f \bigl( ta+ ( 1-t ) b \bigr) \,d_{q}t. \end{aligned}$$

Theorem 2

([22])

If $f: [ a,b ] \rightarrow \mathbb{R}$ is a convex differentiable function on $[ a,b ]$ and $0< q<1$, then we have the following q-Hermite-Hadamard inequalities:

$$ f \biggl( \frac{a+qb}{1+q} \biggr) \leq \frac{1}{b-a} \int _{a}^{b}f ( x ) {}^{b}d_{q}x\leq \frac{f ( a ) +qf ( b ) }{1+q}. $$

(2.4)

From Theorems 1 and 2 we obtain the following inequalities.

Corollary 1

[22] For any convex function $f: [ a,b ] \rightarrow \mathbb{R}$ and $0< q<1$, we have

$$ f \biggl( \frac{qa+b}{1+q} \biggr) +f \biggl( \frac{a+qb}{1+q} \biggr) \leq \frac{1}{b-a} \biggl\{ \int _{a}^{b}f ( x ) \,_{a}d_{q}x+ \int _{a}^{b}f ( x ) \,^{b}d_{q}x \biggr\} \leq f ( a ) +f ( b ) $$

(2.5)

and

$$ f \biggl( \frac{a+b}{2} \biggr) \leq \frac{1}{2 ( b-a ) } \biggl\{ \int _{a}^{b}f ( x ) \,_{a}d_{q}x+ \int _{a}^{b}f ( x ) \,^{b}d_{q}x \biggr\} \leq \frac{f ( a ) +f ( b ) }{2}. $$

(2.6)

Theorem 3

(Hölder’s inequality, [24, p. 604])

Suppose that $x>0$, $0< q<1$, $p_{1}>1$. If $\frac{1}{p_{1}}+\frac{1}{r_{1}}=1$, then

$$ \int _{0}^{x} \bigl\vert f ( x ) g ( x ) \bigr\vert d_{q}x\leq \biggl( \int _{0}^{x} \bigl\vert f ( x ) \bigr\vert ^{p_{1}}d_{q}x \biggr) ^{\frac{1}{p_{1}}} \biggl( \int _{0}^{x} \bigl\vert g ( x ) \bigr\vert ^{r_{1}}d_{q}x \biggr) ^{\frac{1}{r_{1}}}. $$

In this paper, we will also find some bounds for right-hand side of inequality (2.4).

3 New Hermite–Hadamard-type inequalities for quantum integrals

We now give some new Hermite–Hadamard-type inequalities for functions whose second $q^{b}$-derivatives in absolute value are convex.

We start with the following useful lemma.

Lemma 1

If $f: [ a,b ] \subset \mathbb{R} \rightarrow \mathbb{R}$ is a twice $q^{b}$-differentiable function on $( a,b )$ such that $^{b}D_{q}^{2}f$ is continuous and integrable on $[ a,b ]$, then we have:

$$\begin{aligned}& \frac{f ( a ) +qf ( b ) }{1+q}-\frac{1}{b-a} \int _{a}^{b}f ( x )\,^{b}d_{q}x \\& \quad = \frac{q^{2} ( b-a ) ^{2}}{1+q} \int _{0}^{1}t ( 1-qt ) ~^{b}D_{q}^{2}f \bigl( ta+ ( 1-t ) b \bigr)\,d_{q}t, \end{aligned}$$

(3.1)

where $0< q<1$.

Proof

From Definition 2 it follows that

$$\begin{aligned}& {}^{b}D_{q}^{2}f \bigl( ta+ ( 1-t ) b \bigr) \\& \quad =~^{b}D_{q} \bigl( ~^{b}D_{q} \bigl( f \bigl( ta+ ( 1-t ) b \bigr) \bigr) \bigr) \\& \quad =~^{b}D_{q} \biggl( \frac{f ( qta+ ( 1-qt ) b ) -f ( ta+ ( 1-t ) b ) }{ ( 1-q ) ( b-a ) t} \biggr) \\& \quad =\frac{1}{ ( 1-q ) ( b-a ) t} \biggl[ \frac{f ( q^{2}ta+ ( 1-tq^{2} ) b ) -f ( qta+ ( 1-qt ) b ) }{ ( 1-q ) q ( b-a ) t} \\& \qquad {} - \frac{f ( qta+ ( 1-qt ) b ) -f ( ta+ ( 1-t ) b ) }{ ( 1-q ) ( b-a ) t} \biggr] \\& \quad =\frac{f ( q^{2}ta+ ( 1-tq^{2} ) b ) -f ( qta+ ( 1-qt ) b ) }{ ( 1-q ) ^{2}q ( b-a ) ^{2}t^{2}} \\& \qquad {}- \frac{f ( qta+ ( 1-qt ) b ) -f ( ta+ ( 1-t ) b ) }{ ( 1-q ) ^{2} ( b-a ) ^{2}t^{2}} \\& \quad =\frac{f ( q^{2}ta+ ( 1-tq^{2} ) b ) - ( 1+q ) f ( qta+ ( 1-qt ) b ) +qf ( ta+ ( 1-t ) b ) }{ ( 1-q ) ^{2}q ( b-a ) ^{2}t^{2}}. \end{aligned}$$

(3.2)

Also,

$$\begin{aligned}& \int _{0}^{1}t ( 1-qt ) ~^{b}D_{q}^{2}f \bigl( ta+ ( 1-t ) b \bigr)\,d_{q}t \\& \quad = \int _{0}^{1} \frac{f ( q^{2}ta+ ( 1-tq^{2} ) b ) - ( 1+q ) f ( qta+ ( 1-qt ) b ) +qf ( ta+ ( 1-t ) b ) }{ ( 1-q ) ^{2}q ( b-a ) ^{2}t} \,d_{q}t \\& \qquad {}- \int _{0}^{1}q \biggl[ \frac{f ( q^{2}ta+ ( 1-tq^{2} ) b ) - ( 1+q ) f ( qta+ ( 1-qt ) b ) +qf ( ta+ ( 1-t ) b ) }{ ( 1-q ) ^{2}q ( b-a ) ^{2}} \biggr]\,d_{q}t. \end{aligned}$$

(3.3)

By equality (2.1) we obtain that

$$\begin{aligned}& \int _{0}^{1} \frac{f ( q^{2}ta+ ( 1-tq^{2} ) b ) - ( 1+q ) f ( qta+ ( 1-qt ) b ) +qf ( ta+ ( 1-t ) b ) }{ ( 1-q ) ^{2}q ( b-a ) ^{2}t} \,d_{q}t \\& \quad = ( 1-q ) \sum_{n=0}^{\infty } \frac{f ( q^{n+2}a+ ( 1-q^{n+2} ) b ) }{ ( 1-q ) ^{2}q ( b-a ) ^{2}}- ( 1-q ) ( 1+q ) \sum_{n=0}^{\infty } \frac{f ( q^{n+1}a+ ( 1-q^{n+1} ) b ) }{ ( 1-q ) ^{2}q ( b-a ) ^{2}} \\& \qquad {}+q ( 1-q ) \sum_{n=0}^{\infty } \frac{f ( q^{n}a+ ( 1-q^{n} ) b ) }{ ( 1-q ) ^{2}q ( b-a ) ^{2}} \\& \quad =\sum_{n=0}^{\infty } \frac{f ( q^{n+2}a+ ( 1-q^{n+2} ) b ) }{ ( 1-q ) q ( b-a ) ^{2}}- \sum_{n=0}^{\infty } \frac{f ( q^{n+1}a+ ( 1-q^{n+1} ) b ) }{ ( 1-q ) q ( b-a ) ^{2}} \\& \qquad {}-q \Biggl[ \sum_{n=0}^{\infty } \frac{f ( q^{n+1}a+ ( 1-q^{n+1} ) b ) }{ ( 1-q ) q ( b-a ) ^{2}}- \sum_{n=0}^{\infty } \frac{f ( q^{n}a+ ( 1-q^{n} ) b ) }{ ( 1-q ) q ( b-a ) ^{2}} \Biggr] \\& \quad =\frac{f ( b ) -f ( qa+ ( 1-q ) b ) }{ ( 1-q ) q ( b-a ) ^{2}}-q \biggl[ \frac{f ( b ) -f ( a ) }{ ( 1-q ) q ( b-a ) ^{2}} \biggr] . \end{aligned}$$

(3.4)

From (2.1) and Definition 5 it follows that

$$\begin{aligned}& \int _{0}^{1}q \biggl[ \frac{f ( q^{2}ta+ ( 1-tq^{2} ) b ) - ( 1+q ) f ( qta+ ( 1-qt ) b ) +qf ( ta+ ( 1-t ) b ) }{ ( 1-q ) ^{2}q ( b-a ) ^{2}} \biggr]\, d_{q}t \\& \quad =q \Biggl[ ( 1-q ) ( b-a ) \sum_{n=0}^{ \infty }\frac{q^{n+2}f ( q^{n+2}a+ ( 1-q^{n+2} ) b ) }{ ( 1-q ) ^{2}q^{3} ( b-a ) ^{3}} \\& \qquad {}- ( 1-q ) ( 1+q ) ( b-a ) \sum_{n=0}^{\infty } \frac{q^{n+1}f ( q^{n+1}a+ ( 1-q^{n+1} ) b ) }{ ( 1-q ) ^{2}q^{2} ( b-a ) ^{3}} \\& \qquad {} +q ( 1-q ) ( b-a ) \sum_{n=0}^{ \infty }\frac{q^{n}f ( q^{n}a+ ( 1-q^{n} ) b ) }{ ( 1-q ) ^{2}q ( b-a ) ^{3}} \Biggr] \\& \quad =q \biggl[ \frac{1}{ ( 1-q ) ^{2}q^{3} ( b-a ) ^{3}} \\& \qquad {}\times \biggl( \int _{a}^{b}f ( x ) ~^{b}d_{q}x- ( 1-q ) ( b-a ) f ( a ) - ( 1-q ) ( b-a ) qf \bigl( qa+ ( 1-q ) b \bigr) \biggr) \\& \qquad {}-\frac{1+q}{ ( 1-q ) ^{2}q^{2} ( b-a ) ^{3}} \biggl( \int _{a}^{b}f ( x )\,^{b}d_{q}x- ( 1-q ) ( 1+q ) ( b-a ) f ( a ) \biggr) \\& \qquad {} + \frac{1}{ ( 1-q ) ^{2} ( b-a ) ^{3}}\int _{a}^{b}f ( x )\,^{b}d_{q}x \biggr] \\& \quad =\frac{1+q}{ ( b-a ) ^{2}q^{2}} \int _{a}^{b}f ( x )\,^{b}d_{q}x+ \frac{q^{2}+q-1}{ ( 1-q ) q^{2} ( b-a ) ^{2}}f ( a ) - \frac{f ( qa+ ( 1-q ) b ) }{ ( 1-q ) q ( b-a ) ^{2}} \end{aligned}$$

(3.5)

Using (3.4) and (3.5) in (3.3), we have

$$\begin{aligned}& \int _{0}^{1}t ( 1-qt ) ~^{b}D_{q}^{2}f \bigl( ta+ ( 1-t ) b \bigr)\,d_{q}t \\& \quad =\frac{f ( b ) -f ( qa+ ( 1-q ) b ) }{ ( 1-q ) q ( b-a ) ^{2}}-q \biggl[ \frac{f ( b ) -f ( a ) }{ ( 1-q ) q ( b-a ) ^{2}} \biggr] \\& \qquad {}-\frac{1+q}{ ( b-a ) ^{2}q^{2}} \int _{a}^{b}f ( x )\,^{b}d_{q}x- \frac{q^{2}+q-1}{ ( 1-q ) q^{2} ( b-a ) ^{2}}f ( a ) + \frac{f ( qa+ ( 1-q ) b ) }{ ( 1-q ) q ( b-a ) ^{2}} \\& \quad =\frac{f ( a ) +qf ( b ) }{ ( b-a ) ^{2}q^{2}}-\frac{1+q}{ ( b-a ) ^{2}q^{2}} \int _{a}^{b}f ( x )\,^{b}d_{q}x. \end{aligned}$$

(3.6)

Multiplying both sides of (3.6) by $\frac{ ( b-a ) ^{2}q^{2}}{1+q}$, we obtain the required identity (3.1) and hence we complete the proof of Lemma 1. □

Remark 1

If we take the limit as $q\rightarrow 1^{-}$ in Lemma 1, then we have

$$ \frac{f ( a ) +f ( b ) }{2}-\frac{1}{b-a} \int _{a}^{b}f ( x )\,dx= \frac{ ( b-a ) ^{2}}{2} \int _{0}^{1}t ( 1-t ) f^{\prime \prime } \bigl( ta+ ( 1-t ) b \bigr) \,dt, $$

as given in [25].

Theorem 4

If $f: [ a,b ] \subset \mathbb{R} \rightarrow \mathbb{R}$ is a twice $q^{b}$-differentiable function on $( a,b )$ such that $^{b}D_{q}^{2}f$ is continuous and integrable on $[ a,b ]$, then we have the following inequality, provided that $\vert ^{b}D_{q}^{2}f \vert$ is convex on $[ a,b ]$:

$$\begin{aligned}& \biggl\vert \frac{f ( a ) +qf ( b ) }{1+q}- \frac{1}{b-a}\int _{a}^{b}f ( x )\,^{b}d_{q}x \biggr\vert \\& \quad \leq \frac{q^{2} ( b-a ) ^{2}}{ ( 1+q ) ( q^{2}+q+1 ) ( q^{3}+q^{2}+q+1 ) } \bigl[ \bigl\vert {}^{b}D_{q}^{2}f ( a ) \bigr\vert +q^{2} \bigl\vert {}^{b}D_{q}^{2}f ( b ) \bigr\vert \bigr] , \end{aligned}$$

where $0< q<1$.

Proof

Taking the modulus in Lemma 1 and applying the convexity of $\vert ^{b}D_{q}^{2}f \vert$, we obtain

$$\begin{aligned}& \biggl\vert \frac{f ( a ) +qf ( b ) }{1+q}- \frac{1}{b-a}\int _{a}^{b}f ( x )\,^{b}d_{q}x \biggr\vert \\& \quad \leq \frac{q^{2} ( b-a ) ^{2}}{1+q} \int _{0}^{1} \bigl( t ( 1-qt ) \bigr) \bigl\vert {} ^{b}D_{q}^{2}f \bigl( ta+ ( 1-t ) b \bigr) \bigr\vert \,d_{q}t \\& \quad \leq \frac{q^{2} ( b-a ) ^{2}}{1+q} \int _{0}^{1} \bigl( t ( 1-qt ) \bigr) \bigl[ t \bigl\vert {}^{b}D_{q}^{2}f ( a ) \bigr\vert + ( 1-t ) \bigl\vert {} ^{b}D_{q}^{2}f ( b ) \bigr\vert \bigr]\,d_{q}t \\& \quad =\frac{q^{2} ( b-a ) ^{2}}{1+q} \biggl[ \bigl\vert {}^{b}D_{q}^{2}f ( a ) \bigr\vert \int _{0}^{1}t \bigl( t ( 1-qt ) \bigr) \,d_{q}t+ \bigl\vert {} ^{b}D_{q}^{2}f ( b ) \bigr\vert \int _{0}^{1} ( 1-t ) \bigl( t ( 1-qt ) \bigr) \,d_{q}t \biggr] \\& \quad =\frac{q^{2} ( b-a ) ^{2}}{1+q} \biggl[ \frac{ \vert {} ^{b}D_{q}^{2}f ( a ) \vert }{ ( q^{2}+q+1 ) ( q^{3}+q^{2}+q+1 ) }+ \frac{q^{2} \vert {} ^{b}D_{q}^{2}f ( b ) \vert }{ ( q^{2}+q+1 ) ( q^{3}+q^{2}+q+1 ) } \biggr], \end{aligned}$$

which completes the proof. □

Remark 2

Under the assumptions of Theorem 4 with the limit as $q\rightarrow 1^{-}$, we have the following trapezoidal inequality:

$$ \biggl\vert \frac{1}{b-a} \int _{a}^{b}f ( x ) \,dx- \frac{f ( a ) +f ( b ) }{2} \biggr\vert \leq \frac{ ( b-a ) ^{2}}{12} \biggl[ \frac{ \vert f^{\prime \prime } ( a ) \vert + \vert f^{\prime \prime } ( b ) \vert }{2} \biggr], $$

as given by Sarikaya and Aktan [26, Proposition 2].

Theorem 5

Suppose that $f: [ a,b ] \subset \mathbb{R} \rightarrow \mathbb{R}$ is a twice $q^{b}$-differentiable function on $( a,b )$ and $^{b}D_{q}^{2}f$ is continuous and integrable on $[ a,b ]$. If $\vert ^{b}D_{q}^{2}f \vert ^{p_{1}},p_{1}> 1$, is convex on $[ a,b ]$, then we have the following inequality:

$$\begin{aligned}& \biggl\vert \frac{f ( a ) +qf ( b ) }{1+q}- \frac{1}{b-a}\int _{a}^{b}f ( x )\,^{b}d_{q}x \biggr\vert \\& \quad \leq \frac{q^{2} ( b-a ) ^{2}}{ ( 1+q ) ^{2-\frac{1}{p_{1}}} ( 1+q+q^{2} ) } \biggl( \frac{1}{q^{3}+q^{2}+q+1} \biggr) ^{\frac{1}{p_{1}}} \bigl( \bigl\vert {} ^{b}D_{q}^{2}f ( a ) \bigr\vert ^{p_{1}}+q^{2} \bigl\vert {} ^{b}D_{q}^{2}f ( b ) \bigr\vert ^{p_{1}} \bigr) ^{\frac{1}{p_{1}}}, \end{aligned}$$

where $0< q<1$.

Proof

Taking the modulus in Lemma 1 and applying the well-known power mean inequality, we have

$$\begin{aligned}& \biggl\vert \frac{f ( a ) +qf ( b ) }{1+q}- \frac{1}{b-a}\int _{a}^{b}f ( x )\,^{b}d_{q}x \biggr\vert \\& \quad \leq \frac{q^{2} ( b-a ) ^{2}}{1+q} \int _{0}^{1} \bigl( t ( 1-qt ) \bigr) \bigl\vert {}^{b}D_{q}^{2}f \bigl( ta+ ( 1-t ) b \bigr) \bigr\vert \, d_{q}t \\& \quad \leq \frac{q^{2} ( b-a ) ^{2}}{1+q} \biggl( \int _{0}^{1} \bigl( t ( 1-qt ) \bigr) \,d_{q}t \biggr) ^{1- \frac{1}{p_{1}}} \\& \qquad {}\times\biggl( \int _{0}^{1} \bigl( t ( 1-qt ) \bigr) \bigl\vert {} ^{b}D_{q}^{2}f \bigl( ta+ ( 1-t ) b \bigr) \bigr\vert ^{p_{1}}\,d_{q}t \biggr) ^{ \frac{1}{p_{1}}}. \end{aligned}$$

By the convexity of $\vert ^{b}D_{q}^{2}f \vert ^{p_{1}}$ we have

$$\begin{aligned}& \biggl\vert \frac{f ( a ) +qf ( b ) }{1+q}- \frac{1}{b-a}\int _{a}^{b}f ( x )\,^{b}d_{q}x \biggr\vert \\& \quad \leq \frac{q^{2} ( b-a ) ^{2}}{1+q} \biggl( \int _{0}^{1} \bigl( t ( 1-qt ) \bigr) \,d_{q}t \biggr) ^{1- \frac{1}{p_{1}}} \\& \qquad {}\times \biggl( \int _{0}^{1} \bigl( t ( 1-qt ) \bigr) \bigl[ t \bigl\vert {} ^{b}D_{q}^{2}f ( a ) \bigr\vert ^{p_{1}}+ ( 1-t ) \bigl\vert {}^{b}D_{q}^{2}f ( b ) \bigr\vert ^{p_{1}} \bigr]\,d_{q}t \biggr) ^{\frac{1}{p_{1}}} \\& \quad =\frac{q^{2} ( b-a ) ^{2}}{1+q} \biggl( \int _{0}^{1} \bigl( t ( 1-qt ) \bigr) \,d_{q}t \biggr) ^{1- \frac{1}{p_{1}}} \\& \qquad {}\times \biggl( \bigl\vert {} ^{b}D_{q}^{2}f ( a ) \bigr\vert ^{p_{1}} \int _{0}^{1}t \bigl( t ( 1-qt ) \bigr) \,d_{q}t+ \bigl\vert {}^{b}D_{q}^{2}f ( b ) \bigr\vert ^{p_{1}} \int _{0}^{1} ( 1-t ) \bigl( t ( 1-qt ) \bigr) \,d_{q}t \biggr) ^{\frac{1}{p_{1}}} \\& \quad =\frac{q^{2} ( b-a ) ^{2}}{1+q} \biggl( \frac{1}{ ( 1+q ) ( 1+q+q^{2} ) } \biggr) ^{1-\frac{1}{p_{1}}} \\& \qquad {}\times \biggl( \frac{ \vert {} ^{b}D_{q}^{2}f ( a ) \vert ^{p_{1}}}{ ( q^{2}+q+1 ) ( q^{3}+q^{2}+q+1 ) }+ \frac{q^{2} \vert {} ^{b}D_{q}^{2}f ( b ) \vert ^{p_{1}}}{ ( q^{2}+q+1 ) ( q^{3}+q^{2}+q+1 ) } \biggr) ^{ \frac{1}{p_{1}}}, \end{aligned}$$

which completes the proof. □

Remark 3

If we take the limit as $q\rightarrow 1^{-}$ in Theorem 5, then we have

$$ \biggl\vert \frac{f ( a ) +f ( b ) }{2}- \frac{1}{b-a} \int _{a}^{b}f ( x ) \,dx \biggr\vert \leq \frac{ ( b-a ) ^{2}}{12.2^{\frac{1}{p_{1}}}} \bigl( \bigl\vert f^{\prime \prime } ( a ) \bigr\vert ^{p_{1}}+ \bigl\vert f^{\prime \prime } ( b ) \bigr\vert ^{p_{1}} \bigr) ^{\frac{1}{p_{1}}}. $$

Theorem 6

Suppose that $f: [ a,b ] \subset \mathbb{R} \rightarrow \mathbb{R}$ is a twice $q^{b}$-differentiable function on $( a,b )$ and $^{b}D_{q}^{2}f$ is continuous and integrable on $[ a,b ]$. If $\vert ^{b}D_{q}^{2}f \vert ^{p_{1}}$ is convex on $[ a,b ]$ for some $p_{1}>1$ and $\frac{1}{r_{1}}+\frac{1}{p_{1}}=1$, then we have

$$\begin{aligned}& \biggl\vert \frac{f ( a ) +qf ( b ) }{1+q}- \frac{1}{b-a}\int _{a}^{b}f ( x )\,^{b}d_{q}x \biggr\vert \\& \quad \leq \frac{q^{2} ( b-a ) ^{2}}{1+q} ( u_{1} ) ^{ \frac{1}{r_{1}}} \biggl( \frac{ \vert {}^{b}D_{q}^{2}f ( a ) \vert ^{p_{1}}+q \vert {}^{b}D_{q}^{2}f ( b ) \vert ^{p_{1}}}{q+1} \biggr) ^{\frac{1}{p_{1}}}, \end{aligned}$$

(3.7)

where $u_{1}= ( 1-q ) \sum_{n=0}^{\infty } ( q^{n} ) ^{r_{1}+1} ( 1-q^{n+1} ) ^{r_{1}}$ and $0< q<1$.

Proof

Taking the modulus in Lemma 1 and applying well-known Hölder’s inequality, we obtain

$$\begin{aligned}& \biggl\vert \frac{f ( a ) +qf ( b ) }{1+q}- \frac{1}{b-a}\int _{a}^{b}f ( x )\,^{b}d_{q}x \biggr\vert \\& \quad \leq \frac{q^{2} ( b-a ) ^{2}}{1+q} \int _{0}^{1} \bigl( t ( 1-qt ) \bigr) \bigl\vert {} ^{b}D_{q}^{2}f \bigl( ta+ ( 1-t ) b \bigr) \bigr\vert \,d_{q}t \\& \quad \leq \frac{q^{2} ( b-a ) ^{2}}{1+q} \biggl( \int _{0}^{1} \bigl( t ( 1-qt ) \bigr) ^{r_{1}}\,d_{q}t \biggr) ^{\frac{1}{r_{1}}} \biggl( \int _{0}^{1} \bigl\vert ^{b}D_{q}^{2}f \bigl( ta+ ( 1-t ) b \bigr) \bigr\vert ^{p_{1}}\,d_{q}t \biggr) ^{\frac{1}{p_{1}}}. \end{aligned}$$

Since $\vert ^{b}D_{q}^{2}f \vert ^{p_{1}}$ is convex, we have

$$\begin{aligned}& \biggl\vert \frac{f ( a ) +qf ( b ) }{1+q}- \frac{1}{b-a}\int _{a}^{b}f ( x )\,^{b}d_{q}x \biggr\vert \\& \quad \leq \frac{q^{2} ( b-a ) ^{2}}{1+q} \biggl( \int _{0}^{1} \bigl( t ( 1-qt ) \bigr) ^{r_{1}}\,d_{q}t \biggr) ^{\frac{1}{r_{1}}} \\& \qquad {}\times\biggl( \bigl\vert {}^{b}D_{q}^{2}f ( a ) \bigr\vert ^{p_{1}} \int _{0}^{1}t\,d_{q}t+ \bigl\vert {} ^{b}D_{q}^{2}f ( b ) \bigr\vert ^{p_{1}} \int _{0}^{1} ( 1-t )\,d_{q}t \biggr) ^{\frac{1}{p_{1}}} \\& \quad =\frac{q^{2} ( b-a ) ^{2}}{1+q} ( u_{1} ) ^{ \frac{1}{r_{1}}} \biggl( \frac{ \vert {} ^{b}D_{q}^{2}f ( a ) \vert ^{p_{1}}+q \vert {} ^{b}D_{q}^{2}f ( b ) \vert ^{p_{1}}}{q+1} \biggr) ^{\frac{1}{p_{1}}}. \end{aligned}$$

Thus

$$ u_{1}= \int _{0}^{1} \bigl( t ( 1-qt ) \bigr) ^{r_{1}} \,d_{q}t= ( 1-q ) \sum _{n=0}^{\infty } \bigl( q^{n} \bigr) ^{r_{1}+1} \bigl( 1-q^{n+1} \bigr) ^{r_{1}}, $$

which completes the proof. □

Remark 4

If we take the limit as $q\rightarrow 1^{-}$ in Theorem 6, then we have

$$ u_{1}= \int _{0}^{1} \bigl( t ( 1-t ) \bigr) ^{r_{1}}dt=B(r_{1}+1,r_{1}+1), $$

where $B(x,y)$ is the Euler beta function. Moreover, inequality (3.7) reduces to

$$\begin{aligned}& \biggl\vert \frac{f ( a ) +f ( b ) }{2}- \frac{1}{b-a} \int _{a}^{b}f ( x )\,dx \biggr\vert \\& \quad \leq \frac{ ( b-a ) ^{2}}{2} \bigl( B(r_{1}+1,r_{1}+1) \bigr) ^{\frac{1}{r_{1}}} \biggl( \frac{ \vert f^{\prime \prime } ( a ) \vert ^{p_{1}}+ \vert f^{\prime \prime } ( b ) \vert ^{p_{1}}}{2} \biggr) ^{\frac{1}{p_{1}}}. \end{aligned}$$

We obtain another Hermite–Hadamard-type inequality for powers in terms of the second quantum derivatives.

Theorem 7

With assumptions of Theorem 6, we have the inequality

$$\begin{aligned}& \biggl\vert \frac{f ( a ) +qf ( b ) }{1+q}- \frac{1}{b-a}\int _{a}^{b}f ( x )\,^{b}d_{q}x \biggr\vert \\& \quad \leq \frac{q^{2} ( b-a ) ^{2}}{1+q} \biggl( \frac{1}{ [ r_{1}+1 ] _{q}} \biggr) ^{\frac{1}{r_{1}}} \bigl( u_{2} \bigl\vert {}^{b}D_{q}^{2}f ( a ) \bigr\vert ^{p_{1}}+u_{3} \bigl\vert {}^{b}D_{q}^{2}f ( b ) \bigr\vert ^{p_{1}} \bigr) ^{\frac{1}{p_{1}}}, \end{aligned}$$

(3.8)

where

$$ u_{2}= ( 1-q ) \sum_{n=0}^{\infty }q^{2n} \bigl( 1-q^{n+1} \bigr) ^{p_{1}}\quad \textit{and}\quad u_{3}= ( 1-q ) \sum_{n=0}^{ \infty }q^{n} \bigl( 1-q^{n} \bigr) \bigl( 1-q^{n+1} \bigr) ^{p_{1}}. $$

Proof

Taking the modulus of the right-hand side of the equality in Lemma 1 and applying well-known Hölder’s inequality, we obtain

$$\begin{aligned}& \biggl\vert \frac{f ( a ) +qf ( b ) }{1+q}- \frac{1}{b-a}\int _{a}^{b}f ( x )\,^{b}d_{q}x \biggr\vert \\& \quad \leq \frac{q^{2} ( b-a ) ^{2}}{1+q} \int _{0}^{1} \bigl( t ( 1-qt ) \bigr) \bigl\vert {} ^{b}D_{q}^{2}f \bigl( ta+ ( 1-t ) b \bigr) \bigr\vert \,d_{q}t \\& \quad \leq \frac{q^{2} ( b-a ) ^{2}}{1+q} \biggl( \int _{0}^{1}t^{r_{1}} \,d_{q}t \biggr) ^{\frac{1}{r_{1}}} \biggl( \int _{0}^{1} ( 1-qt ) ^{p_{1}} \bigl\vert {} ^{b}D_{q}^{2}f \bigl( ta+ ( 1-t ) b \bigr) \bigr\vert ^{p_{1}}\,d_{q}t \biggr) ^{ \frac{1}{p_{1}}}. \end{aligned}$$

Since $\vert ^{b}D_{q}^{2}f \vert ^{p_{1}}$ is convex, we have

$$\begin{aligned}& \biggl\vert \frac{f ( a ) +qf ( b ) }{1+q}- \frac{1}{b-a}\int _{a}^{b}f ( x )\,^{b}d_{q}x \biggr\vert \\& \quad \leq \frac{q^{2} ( b-a ) ^{2}}{1+q} \biggl( \int _{0}^{1}t^{r_{1}} \,d_{q}t \biggr) ^{\frac{1}{r_{1}}} \\& \qquad {}\times \biggl( \bigl\vert {} ^{b}D_{q}^{2}f ( a ) \bigr\vert ^{p_{1}} \int _{0}^{1} ( 1-qt ) ^{p_{1}}t \,d_{q}t+ \bigl\vert ^{b}D_{q}^{2}f ( b ) \bigr\vert ^{p_{1}} \int _{0}^{1} ( 1-qt ) ^{p_{1}} ( 1-t ) \,d_{q}t \biggr) ^{\frac{1}{p_{1}}} \\& \quad =\frac{q^{2} ( b-a ) ^{2}}{1+q} \biggl( \frac{1}{ [ r_{1}+1 ] _{q}} \biggr) ^{\frac{1}{r_{1}}} \bigl( u_{2} \bigl\vert {}^{b}D_{q}^{2}f ( a ) \bigr\vert ^{p_{1}}+u_{3} \bigl\vert {} ^{b}D_{q}^{2}f ( b ) \bigr\vert ^{p_{1}} \bigr) ^{\frac{1}{p_{1}}}. \end{aligned}$$

We can easily see that

$$ u_{2}= \int _{0}^{1} ( 1-qt ) ^{p_{1}}t \,d_{q}t= ( 1-q ) \sum_{n=0}^{\infty }q^{2n} \bigl( 1-q^{n+1} \bigr) ^{p_{1}} $$

and

$$ u_{3}= \int _{0}^{1} ( 1-qt ) ^{p_{1}} ( 1-t ) \,d_{q}t= ( 1-q ) \sum_{n=0}^{\infty }q^{n} \bigl( 1-q^{n} \bigr) \bigl( 1-q^{n+1} \bigr) ^{p_{1}}. $$

This completes the proof. □

Remark 5

If we take the limit as $q\rightarrow 1^{-}$ in Theorem 7, then we have

$$ u_{2}= \int _{0}^{1} ( 1-t ) ^{p_{1}}t \,d_{q}t= \frac{1}{ ( p_{1}+1 ) ( p_{1}+2 ) } $$

and

$$ u_{3}= \int _{0}^{1} ( 1-t ) ^{p_{1}} ( 1-t ) \,dt= \frac{1}{p_{1}+2}. $$

Moreover, inequality (3.8) reduces to

$$\begin{aligned}& \biggl\vert \frac{f ( a ) +f ( b ) }{2}- \frac{1}{b-a}\int _{a}^{b}f ( x )\,dx \biggr\vert \\& \quad \leq \frac{ ( b-a ) ^{2}}{2} \biggl( \frac{1}{r_{1}+1} \biggr) ^{\frac{1}{r_{1}}} \biggl( \frac{1}{ ( p_{1}+1 ) ( p_{1}+2 ) } \biggr) ^{\frac{1}{p_{1}}} \bigl( ( p_{1}+2 ) \bigl\vert f^{\prime \prime } ( a ) \bigr\vert ^{p_{1}}+ \bigl\vert f^{\prime \prime } ( b ) \bigr\vert ^{p_{1}} \bigr) ^{\frac{1}{p_{1}}}. \end{aligned}$$

4 Conclusions

In this paper, we obtained Hermite–Hadamard-type inequalities for convex functions by applying the newly defined $q^{b}$-integral. The results proved in this paper are a potential generalization of the existing comparable results in the literature. As future directions, we can find similar inequalities through different types of convexities.