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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.

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 [318].

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.

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).

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

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

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.

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Acknowledgements

The authors would like to express their sincere thanks to the editor and the anonymous reviewers for their helpful comments and suggestions.

Funding

The work was supported by the Natural Science Foundation of China (Grant Nos. 61673169, 11301127, 11701176, 11626101, 11601485, 11971241).

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Correspondence to Yu-Ming Chu.

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Ali, M.A., Budak, H., Abbas, M. et al. Quantum Hermite–Hadamard-type inequalities for functions with convex absolute values of second \(q^{b}\)-derivatives. Adv Differ Equ 2021, 7 (2021). https://doi.org/10.1186/s13662-020-03163-1

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Keywords

  • Hermite–Hadamard inequality
  • q-integral
  • Quantum calculus
  • Convex function