Open Access

Integral inequalities via fractional quantum calculus

Journal of Inequalities and Applications20162016:81

https://doi.org/10.1186/s13660-016-1024-1

Received: 8 November 2015

Accepted: 18 February 2016

Published: 1 March 2016

Abstract

In this paper we prove several fractional quantum integral inequalities for the new q-shifting operator \({_{a}}\Phi_{q}(m) = qm + (1-q)a\) introduced in Tariboon et al. (Adv. Differ. Equ. 2015:18, 2015), such as: the q-Hölder inequality, the q-Hermite-Hadamard inequality, the q-Cauchy-Bunyakovsky-Schawrz integral inequality, the q-Grüss integral inequality, the q-Grüss-Čebyšev integral inequality, and the q-Pólya-Szegö integral inequality.

Keywords

fractional integral fractional integral inequalities quantum calculus

MSC

05A30 26D10 26A33

1 Introduction

The quantum calculus is known as the calculus without limits. It substitutes the classical derivative by a difference operator, which allows one to deal with sets of nondifferentiable functions. Quantum difference operators have an interesting role due to their applications in several mathematical areas, such as orthogonal polynomials, basic hyper-geometric functions, combinatorics, the calculus of variations, mechanics, and the theory of relativity. The book by Kac and Cheung [2] covers many of the fundamental aspects of quantum calculus.

In recent years, the topic of q-calculus has attracted the attention of several researchers and a variety of new results can be found in the papers [315] and the references cited therein.

In [16] the notions of \(q_{k}\)-derivative and \(q_{k}\)-integral of a continuous function \(f:[t_{k},t_{k+1}]\to{\mathbb {R}}\), have been introduced and their basic properties were proved. As applications existence and uniqueness results for initial value problems of first and second order impulsive \(q_{k}\)-difference equations were investigated. The q-calculus analogs of some classical integral inequalities, such as Hölder, Hermite-Hadamard, Trapezoid, Ostrowski, Cauchy-Bunyakovsky-Schwarz, Grüss and Grüss-Čebyšev were established in [17]. For recent results on quantum inequalities, see [1820].

In [1] new concepts of fractional quantum calculus were defined, by defining a new q-shifting operator \({_{a}}\Phi_{q}(m) = qm + (1-q)a\). After giving the basic properties the q-derivative and q-integral were defined. New definitions of the Riemann-Liouville fractional q-integral and the q-difference on an interval \([a,b]\) were given and their basic properties were discussed. As applications of the new concepts, one proved existence and uniqueness results for first and second order initial value problems for impulsive fractional q-difference equations.

In this paper we prove several integral inequalities for the new q-shifting operator \({_{a}}\Phi_{q}(m) = qm + (1-q)a\), such as: the q-Hölder inequality, the q-Hermite-Hadamard inequality, the q-Korkine integral equality, the q-Cauchy-Bunyakovsky-Schwarz integral inequality, the q-Grüss integral inequality, the q-Grüss-Čebyšev integral inequality, and the q-Polya-Szegö integral inequality.

2 Preliminaries

To make this paper self-contained, below we recall some well-known facts on fractional q-calculus. The presentation here can be found, for example, in [7, 8].

Let us define a q-shifting operator as
$$ {_{a}}\Phi_{q}(m) = qm+(1-q)a, $$
(2.1)
where \(0< q<1\), \(m,a\in\mathbb{R}\). For any positive integer k, we have
$$ {_{a}}\Phi_{q}^{k}(m) = {_{a}}\Phi_{q}^{k-1} \bigl({_{a}} \Phi_{q}(m) \bigr)\quad \mbox{and}\quad {_{a}} \Phi_{q}^{0}(m) = m. $$
(2.2)
The following results can be found in [1].

Property 2.1

For any \(m,n\in\mathbb{R}\) and for all positive integer k, j, the following properties hold:
  1. (i)

    \({_{a}}\Phi_{q}^{k}(m) = {_{a}}\Phi_{q^{k}}(m)\);

     
  2. (ii)

    \({_{a}}\Phi_{q}^{j}({_{a}}\Phi_{q}^{k}(m)) = {_{a}}\Phi _{q}^{k}({_{a}}\Phi_{q}^{j}(m)) = {_{a}}\Phi_{q}^{j+k}(m)\);

     
  3. (iii)

    \({_{a}}\Phi_{q}(a) = a\);

     
  4. (iv)

    \({_{a}}\Phi_{q}^{k}(m) - a = q^{k}(m-a)\);

     
  5. (v)

    \(m - {_{a}}\Phi_{q}^{k}(m) = (1-q^{k})(m-a)\);

     
  6. (vi)

    \({_{a}}\Phi_{q}^{k}(m) = m_{\frac{a}{m}}\Phi_{q}^{k}(1)\), for \(m\neq0\);

     
  7. (vii)

    \({_{a}}\Phi_{q}(m) - {_{a}}\Phi_{q}^{k}(n) = q(m-{_{a}}\Phi _{q}^{k-1}(n))\).

     
The q-analog of the Pochhammer symbol is defined by
$$ (m;q)_{0} = 1,\qquad (m;q)_{k} = \prod _{i=0}^{k-1} \bigl(1-q^{i}m \bigr), \quad k\in \mathbb{N}\cup\{\infty\}. $$
(2.3)
We also define the power of the q-shifting operator as
$$ {_{a}}(n-m)_{q}^{(0)} = 1,\qquad {_{a}}(n-m)_{q}^{(k)} = \prod _{i=0}^{k-1} \bigl(n-{_{a}} \Phi_{q}^{i}(m) \bigr),\quad k\in\mathbb {N}\cup\{\infty\}. $$
(2.4)
More generally, if \(\gamma\in\mathbb{R}\), then
$$ {_{a}}(n-m)_{q}^{(\gamma)} = n^{(\gamma)}\prod_{i=0}^{\infty} \frac {1-{_{\frac{a}{n}}}\Phi_{q}^{i}(m/n)}{1-{_{\frac{a}{n}}}\Phi _{q}^{\gamma +i}(m/n)},\quad n\neq0. $$
(2.5)
From the above definitions, the following results were proved in [1].

Property 2.2

For any \(\gamma,m,n\in\mathbb{R}\) with \(n\neq a\) and \(k\in\mathbb {N}\cup \{\infty\}\), the following properties hold:
  1. (i)

    \({_{a}}(n-m)_{q}^{(k)} = (n-a)^{k} (\frac {m-a}{n-a};q )_{k}\);

     
  2. (ii)

    \({_{a}}(n-m)_{q}^{(\gamma)} = (n-a)^{\gamma} \prod_{i=0}^{\infty}\frac{1-\frac{m-a}{n-a}q^{i}}{ 1-\frac{m-a}{n-a}q^{\gamma+i}} = (n-a)^{\gamma}\frac{ (\frac {m-a}{n-a};q )_{\infty}}{ (\frac{m-a}{n-a}q^{\gamma};q )_{\infty}}\);

     
  3. (iii)

    \({_{a}}(n-{_{a}}\Phi_{q}^{k}(n))_{q}^{(\gamma)} = (n-a)^{\gamma}\frac{(q^{k};q)_{\infty}}{(q^{\gamma+k};q)_{\infty}}\).

     
The q-number is defined by
$$ [m]_{q} = \frac{1-q^{m}}{1-q},\quad m\in\mathbb{R}. $$
(2.6)
If \(a=0\) and \(m=n=1\), then (2.5) is reduced to
$$ {_{0}} \bigl(1-{_{0}}\Phi_{q}(1) \bigr)_{q}^{(\gamma)} =\prod_{i=0}^{\infty} \frac {1-q^{i+1}}{1-q^{\gamma+i+1}}. $$
(2.7)
The q-gamma function is defined by
$$ \Gamma_{q}(t) = \frac{{_{0}}(1-{_{0}}\Phi_{q}(1))_{q}^{(t-1)}}{(1-q)^{t-1}}, \quad t\in \mathbb{R} \backslash\{0, -1, -2, \ldots\}. $$
(2.8)
Obviously, \(\Gamma_{q}(t+1) = [t]_{q}\Gamma_{q}(t)\). For any \(s,t>0\), the q-beta function is defined by
$$ B_{q}(s,t) = \int_{0}^{1} u^{(s-1)} \bigl(1-{_{0}} \Phi_{q}(u) \bigr)^{(t-1)}\,d_{q}u. $$
(2.9)
The q-beta function in terms of the q-gamma function can be written as
$$ B_{q}(s,t) = \frac{\Gamma_{q}(s)\Gamma_{q}(t)}{\Gamma_{q}(s+t)}. $$
(2.10)

Let us give the definitions of Riemann-Liouville fractional q-integral and the q-derivative on the dense interval \([a,b]\).

Definition 2.3

Let \(\alpha\geq0\) and f be a continuous function defined on \([a,b]\). The fractional q-integral of Riemann-Liouville type is given by \(({_{a}}I_{q}^{0}f)(t) = f(t)\) and
$$\begin{aligned} \bigl({_{a}}I_{q}^{\alpha}f \bigr) (t) =& \frac{1}{\Gamma_{q}(\alpha)} \int _{a}^{t}{_{a}} \bigl(t-{_{a}} \Phi_{q}(s) \bigr)_{q}^{(\alpha-1)}f(s) {_{a}}\,d_{q}s \\ =& \frac{(1-q)(t-a)}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}{_{a}} \bigl(t-{_{a}}\Phi_{q}^{i+1}(t) \bigr)_{q}^{(\alpha-1)}f \bigl({_{a}}\Phi_{q}^{i}(t) \bigr). \end{aligned}$$

Definition 2.4

The fractional q-derivative of Riemann-Liouville type of order \(\alpha\geq0\) of a continuous function f on the interval \([a,b]\) is defined by \(({_{a}}D_{q}^{0}f)(t) = f(t)\) and
$$\begin{aligned} \bigl({_{a}}D_{q}^{\alpha}f \bigr) (t) = \bigl({_{a}}D_{q}^{\upsilon}{_{a}}I_{q}^{\upsilon -\alpha}f \bigr) (t),\quad \alpha>0, \end{aligned}$$
where υ is the smallest integer greater than or equal to α.

Lemma 2.5

[1]

Let \(\alpha, \beta\geq0\), and f be a continuous function on \([a,b]\). The Riemann-Liouville fractional q-integral has the following semi-group properties:
$$ {_{a}}I^{\beta}_{q} \bigl({_{a}}I^{\alpha}_{q}f \bigr) (t)={_{a}}I^{\alpha }_{q} \bigl({_{a}}I^{\beta}_{q}f \bigr) (t)= \bigl({_{a}}I^{\alpha+\beta}_{q}f \bigr) (t). $$
(2.11)
Throughout this paper, in some places, the variable s will be shown inside the fractional integral notation as \(({_{a}}I^{\alpha }_{q}f(s) )(t)\), which means
$$ \bigl({_{a}}I^{\alpha}_{q}f(s) \bigr) (t)= \frac{1}{\Gamma_{q}(\alpha )} \int _{a}^{t}{_{a}} \bigl(t-{_{a}} \Phi_{q}(s) \bigr)_{q}^{(\alpha-1)}f(s) {_{a}}\,d_{q}s. $$

Lemma 2.6

If \(\alpha,\beta\geq0\), then, for \(t \in[a,b]\), the following relation holds:
$$ \bigl({_{a}}I_{q}^{\alpha}(s-a)^{\beta} \bigr) (t) = \frac{\Gamma_{q}(\beta +1)}{\Gamma _{q}(\beta+\alpha+1)}(t-a)^{\beta+\alpha}. $$
(2.12)

Proof

From Definition 2.3 and applying Property 2.1(iv), Property 2.2(iii), it follows that
$$\begin{aligned} \bigl({_{a}}I_{q}^{\alpha}(s-a)^{\beta} \bigr) (t) =& \frac{1}{\Gamma_{q}(\alpha)} \int_{a}^{t}{_{a}} \bigl(t-{_{a}} \Phi _{q}(s) \bigr)_{q}^{(\alpha-1)}(s-a)^{\beta}{_{a}}\,d_{q}s \\ =& \frac{(1-q)(t-a)}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}{_{a}} \bigl(t-{_{a}}\Phi_{q}^{i+1}(t) \bigr)_{q}^{(\alpha-1)} \bigl({_{a}}\Phi_{q}^{i}(t)-a \bigr)^{\beta} \\ =& \frac{(1-q)(t-a)}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i} (t-a )^{\alpha-1}\frac{(q^{i+1};q)_{\infty}}{(q^{\alpha +i};q)_{\infty}} \bigl(q^{i}(t-a) \bigr)^{\beta} \\ =& \frac{(1-q)(t-a)^{\beta+\alpha}}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty}q^{i} \frac{(q^{i+1};q)_{\infty}}{(q^{\alpha+i};q)_{\infty}} q^{\beta i} \\ =& \frac{(1-q)(t-a)^{\beta+\alpha}}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty}q^{i} \prod_{i=0}^{\infty}\frac{(1-q^{i+1} q^{i})}{(1-q^{i+1} q^{\alpha -1+i})} q^{\beta i} \\ =& \frac{(1-q)(t-a)^{\beta+\alpha}}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty}q^{i}{_{0}} \bigl(1-{_{0}}\Phi_{q}^{i+1}(1) \bigr)_{q}^{(\alpha -1)}q^{\beta i} \\ =& \frac{(t-a)^{\beta+\alpha}}{\Gamma_{q}(\alpha)}(1-q)\sum_{i=0}^{\infty}q^{i}{_{0}} \bigl(1-{_{0}}\Phi_{q}^{i+1}(1) \bigr)_{q}^{(\alpha -1)} \bigl(q^{i} \bigr)^{(\beta)} \\ =& \frac{(t-a)^{\beta+\alpha}}{\Gamma_{q}(\alpha)} \int _{0}^{1}s^{(\beta )} \bigl(1-{_{0}} \Phi_{q}(s) \bigr)^{(\alpha-1)}{_{0}}\,d_{q}s \\ =& \frac{(t-a)^{\beta+\alpha}}{\Gamma_{q}(\alpha)} B_{q}(\beta +1,\alpha ) \\ =& \frac{\Gamma_{q}(\beta+1)}{\Gamma_{q}(\beta+\alpha +1)}(t-a)^{\beta +\alpha}, \end{aligned}$$
which leads to (2.12) as required. □

Corollary 2.7

Let \(f(t)=t\) and \(g(t)=t^{2}\) for \(t \in[a,b]\), and \(\alpha>0\). Then we have
  1. (i)

    \(({_{a}}I_{q}^{\alpha}f(s) )(t) =\frac{(t-a)^{\alpha}}{\Gamma_{q}(\alpha+2)} (t+ ([\alpha +1]_{q}-1 )a )\);

     
  2. (ii)

    \(({_{a}}I_{q}^{\alpha}g(s) )(t) =\frac{(t-a)^{\alpha}}{\Gamma_{q}(\alpha+3)} ((1+q)(t-a)^{2}+2a(t-a)[\alpha+2]_{q} +a^{2}[\alpha+1]_{q}[\alpha+2]_{q} )\).

     

3 Main results

Let us start with the fractional q-Hölder inequality on the interval \([a,b]\).

Theorem 3.1

Let \(0< q<1\), \(\alpha>0\), \(p_{1}, p_{2}>1\), such that \(\frac{1}{p_{1}}+\frac {1}{p_{2}}=1\). Then for \(t\in[a,b]\) we have
$$ \bigl({_{a}}I_{q}^{\alpha} \bigl|f(s) \bigr| \bigl|g(s) \bigr| \bigr) (t) \leq \bigl( \bigl({_{a}}I_{q}^{\alpha} \bigl|f(s) \bigr|^{p_{1}} \bigr) (t) \bigr)^{\frac{1}{p_{1}}} \bigl( \bigl({_{a}}I_{q}^{\alpha} \bigl|g(s) \bigr|^{p_{2}} \bigr) (t) \bigr)^{\frac{1}{p_{2}}}. $$
(3.1)

Proof

From Definition 2.3 and the discrete Hölder inequality, we have
$$\begin{aligned} & \bigl({_{a}}I_{q}^{\alpha} \bigl|f(s) \bigr| \bigl|g(s) \bigr| \bigr) (t) \\ &\quad= \frac{1}{\Gamma_{q}(\alpha)} \int_{a}^{t}{_{a}} \bigl(t-{_{a}} \Phi _{q}(s) \bigr)_{q}^{(\alpha-1)} \bigl|f(s) \bigr| \bigl|g(s) \bigr|{_{a}}\,d_{q}s \\ &\quad= \frac{(1-q)(t-a)}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}{_{a}} \bigl(t-{_{a}}\Phi_{q}^{i+1}(t) \bigr)_{q}^{(\alpha-1)} \bigl\vert f \bigl({_{a}} \Phi_{q}^{i}(t) \bigr) \bigr\vert \bigl\vert g \bigl({_{a}}\Phi _{q}^{i}(t) \bigr) \bigr\vert \\ &\quad= \frac{(1-q)(t-a)}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }{_{a}} \bigl(t-{_{a}}\Phi_{q}^{i+1}(t) \bigr)_{q}^{(\alpha-1)} \bigl(q^{i} \bigr)^{\frac{1}{p_{1}}} \bigl\vert f \bigl({_{a}}\Phi_{q}^{i}(t) \bigr) \bigr\vert \bigl(q^{i} \bigr)^{\frac{1}{p_{2}}} \bigl\vert g \bigl({_{a}}\Phi_{q}^{i}(t) \bigr) \bigr\vert \\ &\quad\leq \Biggl(\frac{(1-q)(t-a)}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}{_{a}} \bigl(t-{_{a}}\Phi_{q}^{i+1}(t) \bigr)_{q}^{(\alpha-1)} \bigl\vert f \bigl({_{a}} \Phi_{q}^{i}(t) \bigr) \bigr\vert ^{p_{1}} \Biggr)^{\frac {1}{p_{1}}} \\ &\qquad{} \times \Biggl(\frac{(1-q)(t-a)}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}{_{a}} \bigl(t-{_{a}}\Phi_{q}^{i+1}(t) \bigr)_{q}^{(\alpha-1)} \bigl\vert g \bigl({_{a}} \Phi_{q}^{i}(t) \bigr) \bigr\vert ^{p_{2}} \Biggr)^{\frac {1}{p_{2}}} \\ &\quad= \biggl(\frac{1}{\Gamma_{q}(\alpha)} \int_{a}^{t}{_{a}} \bigl(t-{_{a}} \Phi _{q}(s) \bigr)_{q}^{(\alpha-1)} \bigl|f(s) \bigr|^{p_{1}}{_{a}}\,d_{q}s \biggr)^{\frac {1}{p_{1}}} \\ &\qquad{}\times \biggl(\frac{1}{\Gamma_{q}(\alpha)} \int _{a}^{t}{_{a}} \bigl(t-{_{a}} \Phi _{q}(s) \bigr)_{q}^{(\alpha-1)} \bigl|g(s) \bigr|^{p_{2}}{_{a}}\,d_{q}s \biggr)^{\frac {1}{p_{2}}} \\ &\quad= \bigl( \bigl({_{a}}I_{q}^{\alpha} \bigl|f(s) \bigr|^{p_{1}} \bigr) (t) \bigr)^{\frac {1}{p_{1}}} \bigl( \bigl({_{a}}I_{q}^{\alpha} \bigl|g(s) \bigr|^{p_{2}} \bigr) (t) \bigr)^{\frac{1}{p_{2}}}. \end{aligned}$$
Therefore, inequality (3.1) holds. □

Remark 3.2

If \(\alpha= 1\) and \(a=0\), then (3.1) is reduced to the q-Hölder inequality in [21].

The fractional q-Hermite-Hadamard integral inequality on the interval \([a,b]\) will be proved as follows.

Theorem 3.3

Let \(f:[a,b]\to\mathbb{R}\) be a convex continuous function, \(0< q<1\) and \(\alpha>0\). Then we have
$$\begin{aligned} &\frac{2}{\Gamma_{q}(\alpha+1)}f \biggl(\frac {a+b}{2} \biggr)- \frac{1}{(b-a)^{\alpha}}\bigl({_{a}}I_{q}^{\alpha}f(a+b-s) \bigr) (b) \\ &\quad\leq\frac{1}{(b-a)^{\alpha}}\bigl({_{a}}I_{q}^{\alpha}f(s) \bigr) (b) \\ &\quad\leq\frac{1}{\Gamma_{q}(\alpha+2)} \bigl( \bigl( [\alpha+1 ]_{q}-1 \bigr)f(a)+f(b) \bigr). \end{aligned}$$
(3.2)

Proof

The convexity of f on \([a,b]\) means that
$$ f \bigl((1-s)a+sb \bigr) \leq(1-s)f(a) + s f(b),\quad s \in[0,1]. $$
(3.3)
Multiplying both sides of (3.3) by \({_{0}}(1-{_{0}}\Phi _{q}(s))_{q}^{(\alpha-1)}/\Gamma_{q}(\alpha)\), \(s \in(0,1)\), we get
$$\begin{aligned} &\frac{1}{\Gamma_{q}(\alpha)}{_{0}}\bigl(1-{_{0}}\Phi _{q}(s)\bigr)_{q}^{(\alpha-1)}f \bigl((1-s)a+sb \bigr) \\ &\quad\leq\frac{f(a)}{\Gamma_{q}(\alpha)}{_{0}}\bigl(1-{_{0}}\Phi _{q}(s)\bigr)_{q}^{(\alpha-1)}(1-s) + \frac{f(b)}{\Gamma_{q}(\alpha )}{_{0}}\bigl(1-{_{0}}\Phi_{q}(s) \bigr)_{q}^{(\alpha-1)}s. \end{aligned}$$
(3.4)
Taking q-integration of order \(\alpha>0\) for (3.4) with respect to s on \([0,1]\), we have
$$\begin{aligned} &\frac{1}{\Gamma_{q}(\alpha)} \int _{0}^{1}{_{0}}\bigl(1-{_{0}} \Phi _{q}(s)\bigr)_{q}^{(\alpha-1)}f \bigl((1-s)a+sb \bigr){_{0}}\,d_{q}s \\ &\quad\leq\frac{f(a)}{\Gamma_{q}(\alpha)} \int _{0}^{1}{_{0}}\bigl(1-{_{0}} \Phi_{q}(s)\bigr)_{q}^{(\alpha-1)}(1-s){_{0}}\,d_{q}s + \frac {f(b)}{\Gamma_{q}(\alpha)} \int_{0}^{1}{_{0}}\bigl(1-{_{0}} \Phi _{q}(s)\bigr)_{q}^{(\alpha-1)}s{_{0}}\,d_{q}s, \end{aligned}$$
(3.5)
which means that
$$ \bigl({_{0}}I_{q}^{\alpha}f \bigl((1-s)a+sb \bigr)\bigr) (1) \leq f(a) \bigl({_{0}}I_{q}^{\alpha}(1-s) \bigr) (1)+f(b) \bigl({_{0}}I_{q}^{\alpha}s\bigr) (1). $$
(3.6)
From Corollary 2.7(i), we have
$$ \bigl({_{0}}I_{q}^{\alpha}s \bigr) (1) = \frac{1}{\Gamma_{q}(\alpha+2)} \quad\mbox{and}\quad \bigl({_{0}}I_{q}^{\alpha}(1-s) \bigr) (1) = \frac{1}{\Gamma _{q}(\alpha+1)}-\frac{1}{\Gamma_{q}(\alpha+2)}. $$
Using the definition of fractional q-integration on \([a,b]\), we have
$$\begin{aligned} & \bigl({_{0}}I_{q}^{\alpha}f \bigl((1-s)a+sb \bigr) \bigr) (1) \\ &\quad= \frac{1}{\Gamma_{q}(\alpha)} \int_{0}^{1}{_{0}} \bigl(1-{_{0}} \Phi _{q}(s) \bigr)_{q}^{(\alpha-1)}f \bigl((1-s)a+sb \bigr){_{0}}\,d_{q}s \\ &\quad= \frac{1-q}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}{_{0}} \bigl(1-{_{0}}\Phi_{q}^{i+1}(1) \bigr)_{q}^{(\alpha-1)}f \bigl( \bigl(1-q^{i} \bigr)a+q^{i}b \bigr) \\ &\quad= \frac{1-q}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty}q^{i} \prod_{i=0}^{\infty}\frac{1-q^{i} q^{i+1}}{1-q^{i} q^{\alpha +i}}f \bigl({_{a}}\Phi_{q}^{i}(b) \bigr) \\ &\quad= \frac{1-q}{(b-a)^{\alpha-1}\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}(b-a)^{\alpha-1} \frac{(q^{i+1};q)_{\infty}}{(q^{i+\alpha};q)_{\infty}}f \bigl({_{a}}\Phi _{q}^{i}(b) \bigr) \\ &\quad= \frac{1}{(b-a)^{\alpha}} \biggl(\frac{1}{\Gamma_{q}(\alpha )} \int _{a}^{b}{_{a}} \bigl(b-{_{a}} \Phi_{q}(s) \bigr)_{q}^{(\alpha-1)}f (s ){_{a}}\,d_{q}s \biggr) \\ &\quad= \frac{1}{(b-a)^{\alpha}} \bigl({_{a}}I_{q}^{\alpha}f \bigr) (b), \end{aligned}$$
which gives the second part of (3.2) by using (3.6).
To prove the first part of (3.2), we use the convex property of f as follows:
$$\begin{aligned} \frac{1}{2} \bigl[f \bigl((1-s)a+sb \bigr)+f\bigl(sa+(1-s)b \bigr) \bigr] \geq& f \biggl(\frac{(1-s)a+sb+sa+(1-s)b}{2} \biggr) \\ =& f \biggl(\frac{a+b}{2} \biggr). \end{aligned}$$
(3.7)
Multiplying both sides of (3.7) by \({_{0}}(1-{_{0}}\Phi _{q}(s))_{q}^{(\alpha-1)}/\Gamma_{q}(\alpha)\), \(s \in(0,1)\), we get
$$\begin{aligned} &f \biggl(\frac{a+b}{2} \biggr)\frac{1}{\Gamma_{q}(\alpha )}{_{0}} \bigl(1-{_{0}}\Phi _{q}(s) \bigr)_{q}^{(\alpha-1)} \\ &\quad\leq \frac{1}{2\Gamma_{q}(\alpha)}{_{0}} \bigl(1-{_{0}}\Phi _{q}(s) \bigr)_{q}^{(\alpha -1)}f \bigl((1-s)a+sb \bigr) \\ &\qquad{} + \frac{1}{2\Gamma_{q}(\alpha)}{_{0}} \bigl(1-{_{0}}\Phi _{q}(s) \bigr)_{q}^{(\alpha -1)}f \bigl(sa+(1-s)b \bigr). \end{aligned}$$
Again on fractional q-integration of order \(\alpha>0\) to the above inequality with respect to t on \([0,1]\) and changing variables, we get
$$\begin{aligned} f \biggl(\frac{a+b}{2} \biggr) \leq\frac{\Gamma_{q}(\alpha +1)}{2(b-a)^{\alpha}} \bigl({_{a}}I_{q}^{\alpha}f(s) \bigr) (b)+ \frac{\Gamma _{q}(\alpha +1)}{2(b-a)^{\alpha}} \bigl({_{a}}I_{q}^{\alpha}f(a+b-s) \bigr) (b). \end{aligned}$$
(3.8)
By a direct computation, we have
$$\begin{aligned} & \bigl({_{0}}I_{q}^{\alpha}f \bigl((1-s)b+sa \bigr) \bigr) (1) \\ &\quad= \frac{1-q}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}{_{0}} \bigl(1-q^{i+1} \bigr)_{q}^{(\alpha-1)}f \bigl( \bigl(1-q^{i} \bigr)b+q^{i}a \bigr) \\ &\quad= \frac{1-q}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}{_{0}} \bigl(1-q^{i+1} \bigr)_{q}^{(\alpha-1)}f \bigl(a+b-{_{a}}\Phi _{q}^{i}(b) \bigr) \\ &\quad= \frac{1-q}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty}q^{i} \frac{(q^{i+1};q)_{\infty}}{(q^{\alpha+i};q)_{\infty}}f \bigl(a+b-{_{a}}\Phi_{q}^{i}(b) \bigr) \\ &\quad= \frac{1}{(b-a)^{\alpha}} \biggl(\frac{1}{\Gamma_{q}(\alpha )} \int _{a}^{b}{_{a}} \bigl(b-{_{a}} \Phi_{q}(s) \bigr)_{q}^{(\alpha-1)}f (a+b-s ){_{a}}\,d_{q}s \biggr) \\ &\quad= \frac{1}{(b-a)^{\alpha}} \bigl({_{a}}I_{q}^{\alpha}f(a+b-s) \bigr) (b), \end{aligned}$$
together with (3.8), we derive the first part of inequality (3.2) as requested. The proof is completed. □

Remark 3.4

If \(\alpha= 1\) and \(q \to1\), then inequality (3.2) is reduced to the classical Hermite-Hadamard integral inequality as
$$\begin{aligned} f \biggl(\frac{a+b}{2} \biggr) \leq\frac{1}{b-a} \int_{a}^{b} f(s)\,ds \leq \frac {f(a)+f(b)}{2}. \end{aligned}$$
See also [22, 23].

Let us prove the fractional q-Korkine equality on the interval \([a,b]\).

Lemma 3.5

Let \(f, g:[a,b]\to\mathbb{R}\) be continuous functions, \(0< q<1\), and \(\alpha>0\). Then we have
$$\begin{aligned} &\frac{1}{2} \bigl({_{a}}I_{q}^{2\alpha } \bigl(f(s)-f(r)\bigr) \bigl(g(s)-g(r)\bigr)\bigr) (b) \\ &\quad = \frac{(b-a)^{\alpha}}{\Gamma_{q}(\alpha +1)} \bigl({_{a}}I_{q}^{\alpha}f(s)g(s) \bigr) (b) - \bigl({_{a}}I_{q}^{\alpha}f(s) \bigr) (b) \bigl({_{a}}I_{q}^{\alpha }g(s) \bigr) (b). \end{aligned}$$
(3.9)

Proof

From Definition 2.3, we have
$$\begin{aligned} & \bigl({_{a}}I_{q}^{2\alpha} \bigl(f(s)-f(r) \bigr) \bigl(g(s)-g(r) \bigr) \bigr) (b) \\ &\quad=\frac{1}{\Gamma_{q}^{2}(\alpha)} \int_{a}^{b} \int _{a}^{b}{_{a}} \bigl(b-{_{a}} \Phi _{q}(s) \bigr)_{q}^{(\alpha-1)}{_{a}} \bigl(b-{_{a}}\Phi_{q}(r) \bigr)_{q}^{(\alpha-1)} \\ &\qquad{}\times \bigl(f(s)-f(r) \bigr) \bigl(g(s)-g(r) \bigr){_{a}}\,d_{q}s{_{a}}\,d_{q}r \\ &\quad=\frac{1}{\Gamma_{q}^{2}(\alpha)} \int_{a}^{b} \int_{a}^{b} \bigl(b-{_{a}} \Phi_{q}(s) \bigr)_{a}^{(\alpha-1)} \bigl(b-{_{a}} \Phi_{q}(r) \bigr)_{a}^{(\alpha-1)} \\ &\qquad{}\times \bigl(f(s)g(s)-f(s)g(r)-f(r)g(s)+f(r)g(r) \bigr){_{a}}\,d_{q}s{_{a}}\,d_{q}r \\ &\quad=\frac{(b-a)^{\alpha}}{\Gamma_{q}(\alpha+1)} \Biggl(\frac{(1-q)(b-a)}{\Gamma_{q}(\alpha)}\sum _{i=0}^{\infty }q^{i}{_{a}} \bigl(b-{_{a}}\Phi_{q}^{i+1}(b) \bigr)_{q}^{(\alpha-1)} f \bigl({_{a}} \Phi_{q}^{i}(b) \bigr)g \bigl({_{a}} \Phi_{q}^{i}(b) \bigr) \Biggr) \\ &\qquad{}- \Biggl(\frac{(1-q)(b-a)}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}{_{a}} \bigl(b-{_{a}}\Phi_{q}^{i+1}(b) \bigr)_{q}^{(\alpha-1)} f \bigl({_{a}} \Phi_{q}^{i}(b) \bigr) \Biggr) \\ &\qquad{}\times \Biggl(\frac{(1-q)(b-a)}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}{_{a}} \bigl(b-{_{a}}\Phi_{q}^{i+1}(b) \bigr)_{q}^{(\alpha-1)} g \bigl({_{a}} \Phi_{q}^{i}(b) \bigr) \Biggr) \\ &\qquad{}- \Biggl(\frac{(1-q)(b-a)}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}{_{a}} \bigl(b-{_{a}}\Phi_{q}^{i+1}(b) \bigr)_{q}^{(\alpha-1)} g \bigl({_{a}} \Phi_{q}^{i}(b) \bigr) \Biggr) \\ &\qquad{}\times \Biggl(\frac{(1-q)(b-a)}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}{_{a}} \bigl(b-{_{a}}\Phi_{q}^{i+1}(b) \bigr)_{q}^{(\alpha-1)} f \bigl({_{a}} \Phi_{q}^{i}(b) \bigr) \Biggr) \\ &\qquad{}+\frac{(b-a)^{\alpha}}{\Gamma_{q}(\alpha+1)} \Biggl(\frac{(1-q)(b-a)}{\Gamma_{q}(\alpha)}\sum _{i=0}^{\infty }q^{i}{_{a}} \bigl(b-{_{a}}\Phi_{q}^{i+1}(b) \bigr)_{q}^{(\alpha-1)} f \bigl({_{a}} \Phi_{q}^{i}(b) \bigr)g \bigl({_{a}} \Phi_{q}^{i}(b) \bigr) \Biggr) \\ &\quad=\frac{2(b-a)^{\alpha}}{\Gamma_{q}(\alpha+1)} \Biggl(\frac{(1-q)(b-a)}{\Gamma_{q}(\alpha)}\sum _{i=0}^{\infty }q^{i}{_{a}} \bigl(b-{_{a}}\Phi_{q}^{i+1}(b) \bigr)_{q}^{(\alpha-1)} f \bigl({_{a}} \Phi_{q}^{i}(b) \bigr)g \bigl({_{a}} \Phi_{q}^{i}(b) \bigr) \Biggr) \\ &\qquad{}-2 \Biggl(\frac{(1-q)(b-a)}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}{_{a}} \bigl(b-{_{a}}\Phi_{q}^{i+1}(b) \bigr)_{q}^{(\alpha-1)} f \bigl({_{a}} \Phi_{q}^{i}(b) \bigr) \Biggr) \\ &\qquad{}\times \Biggl(\frac{(1-q)(b-a)}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}{_{a}} \bigl(b-{_{a}}\Phi_{q}^{i+1}(b) \bigr)_{q}^{(\alpha-1)} g \bigl({_{a}} \Phi_{q}^{i}(b) \bigr) \Biggr) \\ &\quad=\frac{2(b-a)^{\alpha}}{\Gamma_{q}(\alpha+1)} \biggl(\frac{1}{\Gamma_{q}(\alpha)} \int_{a}^{b}{_{a}} \bigl(b-{_{a}} \Phi _{q}(s) \bigr)_{q}^{(\alpha-1)}f(s)g(s){_{a}}\,d_{q}s \biggr) \\ &\qquad{}-2 \biggl(\frac{1}{\Gamma_{q}(\alpha)} \int_{a}^{b}{_{a}} \bigl(b-{_{a}} \Phi _{q}(s) \bigr)_{q}^{(\alpha-1)}f(s){_{a}}\,d_{q}s \biggr)\\ &\qquad{}\times \biggl(\frac{1}{\Gamma_{q}(\alpha)} \int_{a}^{b}{_{a}} \bigl(b-{_{a}} \Phi _{q}(s) \bigr)_{q}^{(\alpha-1)}g(s){_{a}}\,d_{q}s \biggr) \\ &\quad=\frac{2(b-a)^{\alpha}}{\Gamma_{q}(\alpha+1)} \bigl({_{a}}I_{q}^{\alpha}fg \bigr) (b)-2 \bigl({_{a}}I_{q}^{\alpha }f \bigr) (b) \bigl({_{a}}I_{q}^{\alpha}g \bigr) (b), \end{aligned}$$
from which one deduces (3.9). □

Remark 3.6

If \(\alpha= 1\), then Lemma 3.5 is reduced to Lemma 3.1 in [17].

Next, we will prove the fractional q-Cauchy-Bunyakovsky-Schwarz integral inequality on the interval \([a,b]\).

Theorem 3.7

Let \(f, g:[a,b]\to\mathbb{R}\) be continuous functions, \(0< q<1\), and \(\alpha,\beta>0\). Then we have
$$ \bigl\vert \bigl({_{a}}I_{q}^{\beta+\alpha}f(s,r)g(s,r) \bigr) (b)\bigr\vert \leq\sqrt{\bigl({_{a}}I_{q}^{\beta+\alpha}f^{2}(s,r) \bigr) (b)}\sqrt {\bigl({_{a}}I_{q}^{\beta+\alpha}g^{2}(s,r) \bigr) (b)}. $$
(3.10)

Proof

From Definition 2.3, we have
$$\begin{aligned} & \bigl({_{a}}I_{q}^{\beta+\alpha}f(s,r) \bigr) (b) \\ &\quad=\frac{1}{\Gamma_{q}(\alpha)\Gamma_{q}(\beta)} \int_{a}^{b} \int _{a}^{b}{_{a}} \bigl(b-{_{a}} \Phi_{q}(s) \bigr)_{q}^{(\alpha-1)} {_{a}} \bigl(b-{_{a}}\Phi_{q}(r) \bigr)_{q}^{(\beta-1)}f(s,r){_{a}}\,d_{q}s{_{a}}\,d_{q}r \\ &\quad=\frac{(1-q)^{2}(b-a)^{2}}{\Gamma_{q}(\alpha)\Gamma_{q}(\beta)}\sum_{i=0}^{\infty} \sum_{n=0}^{\infty}q^{i+n} {_{a}} \bigl(b-{_{a}}\Phi_{q}^{i+1}(b) \bigr)_{q}^{(\alpha-1)}{_{a}} \bigl(b-{_{a}} \Phi _{q}^{i+1}(b) \bigr)_{q}^{(\beta-1)} \\ &\qquad{}\times f \bigl({_{a}}\Phi_{q}^{i}(b),{_{a}} \Phi_{q}^{n}(b) \bigr). \end{aligned}$$
Using the classical discrete Cauchy-Schwarz inequality, we have
$$\begin{aligned} & \bigl( \bigl({_{a}}I_{q}^{\beta+\alpha}f(s,r)g(s,r) \bigr) (b) \bigr)^{2} \\ &\quad= \Biggl(\frac{(1-q)^{2}(b-a)^{2}}{\Gamma_{q}(\alpha)\Gamma_{q}(\beta )}\sum_{i=0}^{\infty} \sum_{n=0}^{\infty}q^{i+n} {_{a}} \bigl(b-{_{a}}\Phi_{q}^{i+1}(b) \bigr)_{q}^{(\alpha-1)}{_{a}} \bigl(b-{_{a}} \Phi _{q}^{i+1}(b) \bigr)_{q}^{(\beta-1)} \\ &\qquad{}\times f \bigl({_{a}}\Phi_{q}^{i}(b),{_{a}} \Phi_{q}^{n}(b) \bigr)g \bigl({_{a}} \Phi_{q}^{i}(b),{_{a}}\Phi_{q}^{n}(b) \bigr) \Biggr)^{2} \\ &\quad\leq \Biggl(\frac{(1-q)^{2}(b-a)^{2}}{\Gamma_{q}(\alpha)\Gamma _{q}(\beta )}\sum_{i=0}^{\infty} \sum_{n=0}^{\infty}q^{i+n} {_{a}} \bigl(b-{_{a}}\Phi_{q}^{i+1}(b) \bigr)_{q}^{(\alpha-1)}{_{a}} \bigl(b-{_{a}} \Phi _{q}^{i+1}(b) \bigr)_{q}^{(\beta-1)} \\ &\qquad{}\times f^{2} \bigl({_{a}}\Phi_{q}^{i}(b),{_{a}} \Phi_{q}^{n}(b) \bigr) \Biggr) \Biggl(\frac{(1-q)^{2}(b-a)^{2}}{\Gamma_{q}(\alpha)\Gamma_{q}(\beta )} \sum_{i=0}^{\infty} \sum _{n=0}^{\infty}q^{i+n}{_{a}} \bigl(b-{_{a}}\Phi_{q}^{i+1}(b) \bigr)_{q}^{(\alpha -1)} \\ &\qquad{}\times{_{a}} \bigl(b-{_{a}}\Phi_{q}^{i+1}(b) \bigr)_{q}^{(\beta-1)}g^{2} \bigl({_{a}}\Phi _{q}^{i}(b),{_{a}}\Phi_{q}^{n}(b) \bigr) \Biggr) \\ &\quad= \bigl( \bigl({_{a}}I_{q}^{\beta+\alpha}f^{2}(s,r) \bigr) (b) \bigr) \bigl( \bigl({_{a}}I_{q}^{\beta+\alpha}g^{2}(s,r) \bigr) (b) \bigr). \end{aligned}$$
Therefore, inequality (3.10) holds. □

Remark 3.8

If \(\alpha= 1\), then inequality (3.10) is reduced to the q-Cauchy-Bunyakovsky-Schwarz integral inequality in [17].

Now, we will prove the fractional q-Grüss integral inequality on the interval \([a,b]\).

Theorem 3.9

Let \(f, g:[a,b]\to\mathbb{R}\) be continuous functions satisfying
$$ \phi\leq f(s) \leq\Phi, \qquad\psi\leq g(s) \leq\Psi, \quad \textit{for all } s \in[a,b], \phi, \Phi, \psi, \Psi\in\mathbb{R}. $$
(3.11)
For \(0< q<1\) and \(\alpha>0\), we have the inequality
$$\begin{aligned} &\biggl|\frac{\Gamma_{q}(\alpha+1)}{(b-a)^{\alpha}} \bigl({_{a}}I_{q}^{\alpha}f(s)g(s) \bigr) (b) - \biggl(\frac{\Gamma_{q}(\alpha+1)}{(b-a)^{\alpha}} \bigl({_{a}}I_{q}^{\alpha}f(s) \bigr) (b) \biggr) \\ &\quad{}\times \biggl(\frac{\Gamma _{q}(\alpha +1)}{(b-a)^{\alpha}} \bigl({_{a}}I_{q}^{\alpha}g(s) \bigr) (b) \biggr) \biggr| \leq\frac{1}{4} (\Phi-\phi ) (\Psi-\psi ). \end{aligned}$$
(3.12)

Proof

Applying Theorem 3.7, we have
$$\begin{aligned} &\bigl\vert \bigl({_{a}}I_{q}^{2\alpha } \bigl(f(s)-f(r)\bigr) \bigl(g(s)-g(r)\bigr) \bigr) (b)\bigr\vert \\ &\quad\leq \bigl( \bigl({_{a}}I_{q}^{2\alpha} \bigl(f(s)-f(r)\bigr)^{2} \bigr) (b) \bigr)^{\frac{1}{2}} \bigl( \bigl({_{a}}I_{q}^{2\alpha}\bigl(g(s)-g(r) \bigr)^{2} \bigr) (b) \bigr)^{\frac{1}{2}}. \end{aligned}$$
(3.13)
From Lemma 3.5, it follows that
$$ \frac{1}{2} \bigl({_{a}}I_{q}^{2\alpha} \bigl(f(s)-f(r)\bigr)^{2}\bigr) (b) = \frac{(b-a)^{\alpha}}{\Gamma_{q}(\alpha+1)} \bigl({_{a}}I_{q}^{\alpha }f^{2}(s) \bigr) (b) - \bigl( \bigl({_{a}}I_{q}^{\alpha}f(s) \bigr) (b) \bigr)^{2}. $$
(3.14)
By a simple computation, we have
$$\begin{aligned} &\frac{\Gamma_{q}(\alpha+1)}{(b-a)^{\alpha }}\bigl({_{a}}I_{q}^{\alpha}f^{2}(s) \bigr) (b) - \biggl(\frac{\Gamma_{q}(\alpha+1)}{(b-a)^{\alpha }}\bigl({_{a}}I_{q}^{\alpha }f(s) \bigr) (b) \biggr)^{2} \\ &\quad= \biggl(\Phi-\frac{\Gamma_{q}(\alpha +1)}{(b-a)^{\alpha }}\bigl({_{a}}I_{q}^{\alpha}f(s) \bigr) (b) \biggr) \biggl(\frac{\Gamma_{q}(\alpha+1)}{(b-a)^{\alpha }}\bigl({_{a}}I_{q}^{\alpha }f(s) \bigr) (b)-\phi \biggr) \\ &\qquad{} -\frac{\Gamma_{q}(\alpha+1)}{(b-a)^{\alpha}} \bigl({_{a}}I_{q}^{\alpha} \bigl(f(s)-\phi \bigr) \bigl(\Phi-f(s) \bigr) \bigr) (b), \end{aligned}$$
(3.15)
and an analogous identity for g.
By assumption (3.11) we have \((f(s)-\phi)(\Phi-f(s)) \geq 0\) for all \(s\in[a,b]\), which implies
$$ \bigl({_{a}}I_{q}^{\alpha} \bigl(f(s)-\phi \bigr) \bigl(\Phi -f(s) \bigr) \bigr) (b) \geq0. $$
From (3.15) and using the fact that \((\frac{A+B}{2} )^{2} \geq AB\), \(A, B\in\mathbb{R}\), we have
$$\begin{aligned} &\frac{\Gamma_{q}(\alpha+1)}{(b-a)^{\alpha }}\bigl({_{a}}I_{q}^{\alpha}f^{2}(s) \bigr) (b) - \biggl(\frac{\Gamma_{q}(\alpha+1)}{(b-a)^{\alpha }}\bigl({_{a}}I_{q}^{\alpha }f(s) \bigr) (b) \biggr)^{2} \\ &\quad\leq \biggl(\Phi-\frac{\Gamma_{q}(\alpha +1)}{(b-a)^{\alpha }}\bigl({_{a}}I_{q}^{\alpha}f(s) \bigr) (b) \biggr) \biggl(\frac{\Gamma_{q}(\alpha+1)}{(b-a)^{\alpha }}\bigl({_{a}}I_{q}^{\alpha }f(s) \bigr) (b)-\phi \biggr) \\ &\quad\leq\frac{1}{4} \biggl[ \biggl(\Phi-\frac{\Gamma _{q}(\alpha +1)}{(b-a)^{\alpha}} \bigl({_{a}}I_{q}^{\alpha}f(s)\bigr) (b) \biggr)+ \biggl(\frac{\Gamma_{q}(\alpha+1)}{(b-a)^{\alpha }}\bigl({_{a}}I_{q}^{\alpha }f(s) \bigr) (b)-\phi \biggr) \biggr]^{2} \\ &\quad\leq\frac{1}{4} (\Phi-\phi )^{2}. \end{aligned}$$
(3.16)
A similar argument gives
$$ \bigl({_{a}}I_{q}^{2\alpha} \bigl(g(s)-g(r)\bigr)^{2}\bigr) (b) \leq\frac{1}{4} (\Psi-\psi )^{2}. $$
(3.17)
Using inequality (3.13) via (3.14) and the estimations (3.16) and (3.17), we get
$$ \biggl\vert \frac{1}{2} \bigl({_{a}}I_{q}^{2\alpha } \bigl(f(s)-f(r)\bigr) \bigl(g(s)-g(r)\bigr)\bigr) (b)\biggr\vert \leq \frac{(b-a)^{\alpha}}{4} (\Phi-\phi ) (\Psi -\psi ). $$
Therefore, inequality (3.12) holds, as desired. □

Remark 3.10

If \(\alpha= 1\) and \(q \to1\), then inequality (3.12) is reduced to the classical Grüss integral inequality as
$$\begin{aligned} & \biggl\vert \frac{1}{b-a} \int_{a}^{b} f(s)g(s)\,ds- \biggl(\frac{1}{b-a} \int_{a}^{b} f(s)\,ds \biggr) \biggl( \frac{1}{b-a} \int_{a}^{b} g(s)\,ds \biggr) \biggr\vert \\ &\quad \leq\frac{1}{4} (\Phi-\phi ) (\Psi-\psi ). \end{aligned}$$
See also [22, 23].

Next, we are going to prove the fractional q-Grüss-Čebyšev integral inequality on the interval \([a,b]\).

Theorem 3.11

Let \(f, g:[a,b]\to\mathbb{R}\) be \(L_{1}\)-, \(L_{2}\)-Lipschitzian continuous functions, so that
$$ \bigl|f(s)-f(r)\bigr| \leq L_{1}|s-r|,\qquad \bigl|g(s)-g(r)\bigr| \leq L_{2}|s-r|, $$
(3.18)
for all \(s,r\in[a,b]\), \(0< q<1\), \(L_{1}, L_{2}>0\), and \(\alpha>0\). Then we have the inequality
$$\begin{aligned} &\biggl\vert \frac{(b-a)^{\alpha}}{\Gamma_{q}(\alpha +1)} \bigl({_{a}}I_{q}^{\alpha}f(s)g(s) \bigr) (b) - \bigl({_{a}}I_{q}^{\alpha}f(s) \bigr) (b) \bigl({_{a}}I_{q}^{\alpha }g(s) \bigr) (b)\biggr\vert \\ &\quad\leq\frac{L_{1}L_{2}(b-a)^{2\alpha+2}}{\Gamma_{q}(\alpha +2)\Gamma_{q}(\alpha+3)} \bigl((1+q)[\alpha+1]_{q}-[ \alpha+2]_{q} \bigr). \end{aligned}$$
(3.19)

Proof

Recall the fractional q-Korkine equality as
$$\begin{aligned} &\frac{(b-a)^{\alpha}}{\Gamma_{q}(\alpha+1)} \bigl({_{a}}I_{q}^{\alpha}f(s)g(s) \bigr) (b) - \bigl({_{a}}I_{q}^{\alpha}f(s) \bigr) (b) \bigl({_{a}}I_{q}^{\alpha }g(s) \bigr) (b) \\ &\quad = \frac{1}{2} \bigl({_{a}}I_{q}^{2\alpha } \bigl(f(s)-f(r)\bigr) \bigl(g(s)-g(r)\bigr)\bigr) (b). \end{aligned}$$
(3.20)
It follows from (3.18) that
$$ \bigl\vert \bigl(f(s)-f(r) \bigr) \bigl(g(s)-g(r) \bigr) \bigr\vert \leq L_{1}L_{2}(s-r)^{2}, $$
(3.21)
for all \(s,r \in[a,b]\). Taking the double fractional q-integration of order α with respect to \(s, r\in[a,b]\), we get
$$\begin{aligned} & \bigl({_{a}}I_{q}^{2\alpha} \bigl\vert \bigl(f(s)-f(r) \bigr) \bigl(g(s)-g(r) \bigr) \bigr\vert \bigr) (b) \\ &\quad= \frac{1}{\Gamma_{q}^{2}(\alpha)} \int_{a}^{b} \int _{a}^{b}{_{a}} \bigl(b-{_{a}} \Phi_{q}(s) \bigr)_{q}^{(\alpha-1)} {_{a}} \bigl(b-{_{a}}\Phi_{q}(r) \bigr)_{q}^{(\alpha-1)} \\ &\qquad{}\times \bigl\vert \bigl(f(s)-f(r) \bigr) \bigl(g(s)-g(r) \bigr) \bigr\vert {_{a}}\,d_{q}s{_{a}}\,d_{q}r \\ &\quad\leq\frac{L_{1}L_{2}}{\Gamma_{q}^{2}(\alpha)} \int_{a}^{b} \int _{a}^{b}{_{a}} \bigl(b-{_{a}} \Phi_{q}(s) \bigr)_{q}^{(\alpha-1)} {_{a}} \bigl(b-{_{a}}\Phi_{q}(r) \bigr)_{q}^{(\alpha-1)} (s-r )^{2}{_{a}}\,d_{q}s{_{a}}\,d_{q}r \\ &\quad=\frac{L_{1}L_{2}}{\Gamma_{q}^{2}(\alpha)} \int_{a}^{b} \int _{a}^{b}{_{a}} \bigl(b-{_{a}} \Phi_{q}(s) \bigr)_{q}^{(\alpha-1)} {_{a}} \bigl(b-{_{a}}\Phi_{q}(r) \bigr)_{q}^{(\alpha -1)}s^{2}{_{a}}\,d_{q}s{_{a}}\,d_{q}r \\ &\qquad{}-\frac{2L_{1}L_{2}}{\Gamma_{q}^{2}(\alpha)} \int_{a}^{b} \int _{a}^{b}{_{a}} \bigl(b-{_{a}} \Phi_{q}(s) \bigr)_{q}^{(\alpha-1)} {_{a}} \bigl(b-{_{a}}\Phi_{q}(r) \bigr)_{q}^{(\alpha -1)}sr{_{a}}\,d_{q}s{_{a}}\,d_{q}r \\ &\qquad{}+\frac{L_{1}L_{2}}{\Gamma_{q}^{2}(\alpha)} \int_{a}^{b} \int _{a}^{b}{_{a}} \bigl(b-{_{a}} \Phi_{q}(s) \bigr)_{q}^{(\alpha-1)} {_{a}} \bigl(b-{_{a}}\Phi_{q}(r) \bigr)_{q}^{(\alpha -1)}r^{2}{_{a}}\,d_{q}s{_{a}}\,d_{q}r \\ &\quad=2L_{1}L_{2} \biggl(\frac{(b-a)^{\alpha}}{\Gamma_{q}(\alpha+1)} \bigl({_{a}}I_{q}^{\alpha}s^{2} \bigr) (b) - \bigl( \bigl({_{a}}I_{q}^{\alpha}s \bigr) (b) \bigr)^{2} \biggr). \end{aligned}$$
(3.22)
From Corollary 2.7(ii), with \(t=b\), we have
$$ \bigl({_{a}}I_{q}^{\alpha}s^{2} \bigr) (b) = \frac{(b-a)^{\alpha }}{\Gamma _{q}(\alpha+3)} \bigl((1+q) (b-a)^{2}+2a(b-a)[ \alpha+2]_{q} +a^{2}[\alpha+1]_{q}[ \alpha+2]_{q} \bigr). $$
By direct computation, we have
$$\begin{aligned} & \frac{(b-a)^{\alpha}}{\Gamma_{q}(\alpha+1)} \bigl({_{a}}I_{q}^{\alpha }s^{2} \bigr) (b) - \bigl( \bigl({_{a}}I_{q}^{\alpha}s \bigr) (b) \bigr)^{2} \\ &\quad=\frac{(b-a)^{2\alpha}}{\Gamma_{q}(\alpha+1)\Gamma_{q}(\alpha +3)} \bigl((1+q) (b-a)^{2}+2a(b-a)[ \alpha+2]_{q}+a^{2}[\alpha+1]_{q}[\alpha +2]_{q} \bigr) \\ &\qquad{} -\frac{(b-a)^{2\alpha}}{\Gamma_{q}^{2}(\alpha+2)} \bigl(b+ \bigl([\alpha +1]_{q}-1 \bigr)a \bigr)^{2} \\ &\quad=\frac{(b-a)^{2\alpha+2}}{\Gamma_{q}(\alpha+2)\Gamma_{q}(\alpha+3)} \bigl((1+q)[\alpha+1]_{q}-[ \alpha+2]_{q} \bigr). \end{aligned}$$
(3.23)
Thus, from (3.22) and (3.23), we have
$$\begin{aligned} &\bigl({_{a}}I_{q}^{2\alpha} \bigl\vert \bigl(f(s)-f(r) \bigr) \bigl(g(s)-g(r) \bigr)\bigr\vert \bigr) (b) \\ &\quad \leq\frac{2L_{1}L_{2}(b-a)^{2\alpha+2}}{\Gamma _{q}(\alpha +2)\Gamma_{q}(\alpha+3)} \bigl((1+q)[\alpha+1]_{q}-[ \alpha+2]_{q} \bigr). \end{aligned}$$
(3.24)

By applying (3.24) to (3.20), we get the desired inequality in (3.19). □

Remark 3.12

If \(\alpha= 1\) and \(q \to1\), then inequality (3.19) is reduced to the classical Grüss-Čebyšev integral inequality as
$$\begin{aligned} \biggl\vert \frac{1}{b-a} \int_{a}^{b} f(s)g(s)\,ds- \biggl(\frac{1}{b-a} \int_{a}^{b} f(s)\,ds \biggr) \biggl( \frac{1}{b-a} \int_{a}^{b} g(s)\,ds \biggr) \biggr\vert \leq \frac {L_{1}L_{2}}{12} (b-a )^{2}. \end{aligned}$$
See also [22, 23].

For the final result, we establish the fractional q-Pólya-Szegö integral inequality on the interval \([a,b]\).

Theorem 3.13

Let \(f,g:[a,b]\to\mathbb{R}\) be two positive integrable functions satisfying
$$ 0< \phi\leq f(s) \leq\Phi,\qquad 0< \psi\leq g(s) \leq\Psi, \quad \textit{for all } s\in[a,b], \phi, \Phi, \psi, \Psi\in\mathbb{R}^{+}. $$
(3.25)
Then for \(0< q<1\) and \(\alpha>0\), we have the inequality
$$ \frac{ ({_{a}}I_{q}^{\alpha}(f^{2}(s)) )(b) ({_{a}}I_{q}^{\alpha}(g^{2}(s)) )(b)}{ ( ({_{a}}I_{q}^{\alpha}(f(s)g(s) )(b) )^{2}} \leq\frac{1}{4} \biggl( \sqrt{\frac{\phi\psi}{\Phi\Psi}}+\sqrt {\frac{\Phi \Psi}{\phi\psi}} \biggr)^{2}. $$
(3.26)

Proof

From (3.25), for \(s\in[a,b]\), we have
$$ \frac{\phi}{\Psi} \leq\frac{f(s)}{g(s)} \leq\frac{\Phi}{\psi}, $$
which yields
$$ \biggl(\frac{\Phi}{\psi}-\frac{f(s)}{g(s)} \biggr) \geq0 $$
(3.27)
and
$$ \biggl(\frac{f(s)}{g(s)}-\frac{\phi}{\Psi} \biggr) \geq0. $$
(3.28)
Multiplying (3.27) and (3.28), we obtain
$$ \biggl(\frac{\Phi}{\psi}-\frac{f(s)}{g(s)} \biggr) \biggl(\frac {f(s)}{g(s)}- \frac{\phi}{\Psi} \biggr) \geq0, $$
or
$$ \biggl(\frac{\Phi}{\psi}+\frac{\phi}{\Psi} \biggr) \frac {f(s)}{g(s)} \geq \frac{f^{2}(s)}{g^{2}(s)}+\frac{\phi\Psi}{\psi\Phi}. $$
(3.29)
Inequality (3.29) can be written as
$$ (\phi\psi+\Phi\Psi )f(s)g(s) \geq\psi\Psi f^{2}(s)+\phi \Phi g^{2}(s). $$
(3.30)
Multiplying both sides of (3.30) by \({_{0}}(b-{_{0}}\Phi _{q}(s))_{q}^{(\alpha-1)}/\Gamma_{q}(\alpha)\) and integrating with respect to s from a to b, we get
$$ (\phi\psi+\Phi\Psi ) \bigl({_{a}}I_{q}^{\alpha } \bigl(f(s)g(s) \bigr)\bigr) (b) \geq\psi\Psi \bigl({_{a}}I_{q}^{\alpha} \bigl(f^{2}(s) \bigr)\bigr) (b)+\phi\Phi \bigl({_{a}}I_{q}^{\alpha} \bigl(g^{2}(s) \bigr)\bigr) (b). $$
Applying the AM-GM inequality, \(A+B\geq2\sqrt{AB}\), \(A,B\in\mathbb {R}^{+}\), we have
$$ (\phi\psi+\Phi\Psi ) \bigl({_{a}}I_{q}^{\alpha } \bigl(f(s)g(s) \bigr)\bigr) (b) \geq2\sqrt{\phi\psi\Phi\Psi \bigl({_{a}}I_{q}^{\alpha } \bigl(f^{2}(s) \bigr)\bigr) (b) \bigl({_{a}}I_{q}^{\alpha} \bigl(g^{2}(s) \bigr)\bigr) (b)}, $$
which leads to
$$ \phi\psi\Phi\Psi \bigl({_{a}}I_{q}^{\alpha} \bigl(f^{2}(s) \bigr) \bigr) (b) \bigl({_{a}}I_{q}^{\alpha} \bigl(g^{2}(s) \bigr) \bigr) (b) \leq\frac{1}{4} ( (\phi\psi+ \Phi \Psi ) \bigl({_{a}}I_{q}^{\alpha} \bigl(f(s)g(s) \bigr) (b) \bigr)^{2}. $$
Therefore, inequality (3.26) is proved. □

Remark 3.14

If \(\alpha= 1\) and \(q \to1\), then inequality (3.26) is reduced to the classical Pólya-Szegö integral inequality as
$$\begin{aligned} \frac{\int_{a}^{b} f^{2}(s)\,ds\int_{a}^{b} g^{2}(s)\,ds}{ (\int_{a}^{b} f(s)g(s)\,ds )^{2}} \leq\frac{1}{4} \biggl(\sqrt{\frac{\phi\psi }{\Phi \Psi}}+ \sqrt{\frac{\Phi\Psi}{\phi\psi}} \biggr)^{2}. \end{aligned}$$
See also [24].

4 Conclusion

In this work, some important integral inequalities involving the new q-shifting operator \({_{a}}\Phi_{q}(m) = qm + (1-q)a\), introduced in [1], are established in the context of fractional quantum calculus. The derived results constitute contributions to the theory of integral inequalities and fractional calculus and can be specialized to yield numerous interesting fractional integral inequalities including some known results. Furthermore, they are expected to lead to some applications in fractional boundary value problems.

Declarations

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Authors’ Affiliations

(1)
Nonlinear Dynamic Analysis Research Center, Department of Mathematics, Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok
(2)
Department of Mathematics, University of Ioannina
(3)
Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University

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© Sudsutad et al. 2016