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 [3‐15] and the references cited therein.
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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 [18‐20].
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].
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Let us define a q-shifting operator as where \(0< q<1\), \(m,a\in\mathbb{R}\). For any positive integer k, we have The following results can be found in [1].
$$ {_{a}}\Phi_{q}(m) = qm+(1-q)a, $$
(2.1)
$$ {_{a}}\Phi_{q}^{k}(m) = {_{a}}\Phi_{q}^{k-1} \bigl({_{a}} \Phi_{q}(m) \bigr)\quad \mbox{and}\quad {_{a}} \Phi_{q}^{0}(m) = m. $$
(2.2)
Property 2.1
For any \(m,n\in\mathbb{R}\) and for all positive integer k, j, the following properties hold:
(i)
\({_{a}}\Phi_{q}^{k}(m) = {_{a}}\Phi_{q^{k}}(m)\);
(ii)
\({_{a}}\Phi_{q}^{j}({_{a}}\Phi_{q}^{k}(m)) = {_{a}}\Phi _{q}^{k}({_{a}}\Phi_{q}^{j}(m)) = {_{a}}\Phi_{q}^{j+k}(m)\);
(iii)
\({_{a}}\Phi_{q}(a) = a\);
(iv)
\({_{a}}\Phi_{q}^{k}(m) - a = q^{k}(m-a)\);
(v)
\(m - {_{a}}\Phi_{q}^{k}(m) = (1-q^{k})(m-a)\);
(vi)
\({_{a}}\Phi_{q}^{k}(m) = m_{\frac{a}{m}}\Phi_{q}^{k}(1)\), for \(m\neq0\);
(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 We also define the power of the q-shifting operator as More generally, if \(\gamma\in\mathbb{R}\), then From the above definitions, the following results were proved in [1].
$$ (m;q)_{0} = 1,\qquad (m;q)_{k} = \prod _{i=0}^{k-1} \bigl(1-q^{i}m \bigr), \quad k\in \mathbb{N}\cup\{\infty\}. $$
(2.3)
$$ {_{a}}(n-m)_{q}^{(0)} = 1,\qquad {_{a}}(n-m)_{q}^{(k)} = \prod _{i=0}^{k-1} \bigl(n-{_{a}} \Phi_{q}^{i}(m) \bigr),\quad k\in\mathbb {N}\cup\{\infty\}. $$
(2.4)
$$ {_{a}}(n-m)_{q}^{(\gamma)} = n^{(\gamma)}\prod_{i=0}^{\infty} \frac {1-{_{\frac{a}{n}}}\Phi_{q}^{i}(m/n)}{1-{_{\frac{a}{n}}}\Phi _{q}^{\gamma +i}(m/n)},\quad n\neq0. $$
(2.5)
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:
(i)
\({_{a}}(n-m)_{q}^{(k)} = (n-a)^{k} (\frac {m-a}{n-a};q )_{k}\);
(ii)
\({_{a}}(n-m)_{q}^{(\gamma)} = (n-a)^{\gamma} \prod_{i=0}^{\infty}\frac{1-\frac{m-a}{n-a}q^{i}}{ 1-\frac{m-a}{n-a}q^{\gamma+i}} = (n-a)^{\gamma}\frac{ (\frac {m-a}{n-a};q )_{\infty}}{ (\frac{m-a}{n-a}q^{\gamma};q )_{\infty}}\);
(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 If \(a=0\) and \(m=n=1\), then (2.5) is reduced to The q-gamma function is defined by Obviously, \(\Gamma_{q}(t+1) = [t]_{q}\Gamma_{q}(t)\). For any \(s,t>0\), the q-beta function is defined by The q-beta function in terms of the q-gamma function can be written as
$$ [m]_{q} = \frac{1-q^{m}}{1-q},\quad m\in\mathbb{R}. $$
(2.6)
$$ {_{0}} \bigl(1-{_{0}}\Phi_{q}(1) \bigr)_{q}^{(\gamma)} =\prod_{i=0}^{\infty} \frac {1-q^{i+1}}{1-q^{\gamma+i+1}}. $$
(2.7)
$$ \Gamma_{q}(t) = \frac{{_{0}}(1-{_{0}}\Phi_{q}(1))_{q}^{(t-1)}}{(1-q)^{t-1}}, \quad t\in \mathbb{R} \backslash\{0, -1, -2, \ldots\}. $$
(2.8)
$$ B_{q}(s,t) = \int_{0}^{1} u^{(s-1)} \bigl(1-{_{0}} \Phi_{q}(u) \bigr)^{(t-1)}\,d_{q}u. $$
(2.9)
$$ B_{q}(s,t) = \frac{\Gamma_{q}(s)\Gamma_{q}(t)}{\Gamma_{q}(s+t)}. $$
(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 where υ is the smallest integer greater than or equal to α.
$$\begin{aligned} \bigl({_{a}}D_{q}^{\alpha}f \bigr) (t) = \bigl({_{a}}D_{q}^{\upsilon}{_{a}}I_{q}^{\upsilon -\alpha}f \bigr) (t),\quad \alpha>0, \end{aligned}$$
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 which leads to (2.12) as required. □
$$\begin{aligned} \bigl({_{a}}I_{q}^{\alpha}(s-a)^{\beta} \bigr) (t) =& \frac{1}{\Gamma_{q}(\alpha)} \int_{a}^{t}{_{a}} \bigl(t-{_{a}} \Phi _{q}(s) \bigr)_{q}^{(\alpha-1)}(s-a)^{\beta}{_{a}}\,d_{q}s \\ =& \frac{(1-q)(t-a)}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}{_{a}} \bigl(t-{_{a}}\Phi_{q}^{i+1}(t) \bigr)_{q}^{(\alpha-1)} \bigl({_{a}}\Phi_{q}^{i}(t)-a \bigr)^{\beta} \\ =& \frac{(1-q)(t-a)}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i} (t-a )^{\alpha-1}\frac{(q^{i+1};q)_{\infty}}{(q^{\alpha +i};q)_{\infty}} \bigl(q^{i}(t-a) \bigr)^{\beta} \\ =& \frac{(1-q)(t-a)^{\beta+\alpha}}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty}q^{i} \frac{(q^{i+1};q)_{\infty}}{(q^{\alpha+i};q)_{\infty}} q^{\beta i} \\ =& \frac{(1-q)(t-a)^{\beta+\alpha}}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty}q^{i} \prod_{i=0}^{\infty}\frac{(1-q^{i+1} q^{i})}{(1-q^{i+1} q^{\alpha -1+i})} q^{\beta i} \\ =& \frac{(1-q)(t-a)^{\beta+\alpha}}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty}q^{i}{_{0}} \bigl(1-{_{0}}\Phi_{q}^{i+1}(1) \bigr)_{q}^{(\alpha -1)}q^{\beta i} \\ =& \frac{(t-a)^{\beta+\alpha}}{\Gamma_{q}(\alpha)}(1-q)\sum_{i=0}^{\infty}q^{i}{_{0}} \bigl(1-{_{0}}\Phi_{q}^{i+1}(1) \bigr)_{q}^{(\alpha -1)} \bigl(q^{i} \bigr)^{(\beta)} \\ =& \frac{(t-a)^{\beta+\alpha}}{\Gamma_{q}(\alpha)} \int _{0}^{1}s^{(\beta )} \bigl(1-{_{0}} \Phi_{q}(s) \bigr)^{(\alpha-1)}{_{0}}\,d_{q}s \\ =& \frac{(t-a)^{\beta+\alpha}}{\Gamma_{q}(\alpha)} B_{q}(\beta +1,\alpha ) \\ =& \frac{\Gamma_{q}(\beta+1)}{\Gamma_{q}(\beta+\alpha +1)}(t-a)^{\beta +\alpha}, \end{aligned}$$
Corollary 2.7
Let
\(f(t)=t\)
and
\(g(t)=t^{2}\)
for
\(t \in[a,b]\), and
\(\alpha>0\). Then we have
(i)
\(({_{a}}I_{q}^{\alpha}f(s) )(t) =\frac{(t-a)^{\alpha}}{\Gamma_{q}(\alpha+2)} (t+ ([\alpha +1]_{q}-1 )a )\);
(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 Therefore, inequality (3.1) holds. □
$$\begin{aligned} & \bigl({_{a}}I_{q}^{\alpha} \bigl|f(s) \bigr| \bigl|g(s) \bigr| \bigr) (t) \\ &\quad= \frac{1}{\Gamma_{q}(\alpha)} \int_{a}^{t}{_{a}} \bigl(t-{_{a}} \Phi _{q}(s) \bigr)_{q}^{(\alpha-1)} \bigl|f(s) \bigr| \bigl|g(s) \bigr|{_{a}}\,d_{q}s \\ &\quad= \frac{(1-q)(t-a)}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}{_{a}} \bigl(t-{_{a}}\Phi_{q}^{i+1}(t) \bigr)_{q}^{(\alpha-1)} \bigl\vert f \bigl({_{a}} \Phi_{q}^{i}(t) \bigr) \bigr\vert \bigl\vert g \bigl({_{a}}\Phi _{q}^{i}(t) \bigr) \bigr\vert \\ &\quad= \frac{(1-q)(t-a)}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }{_{a}} \bigl(t-{_{a}}\Phi_{q}^{i+1}(t) \bigr)_{q}^{(\alpha-1)} \bigl(q^{i} \bigr)^{\frac{1}{p_{1}}} \bigl\vert f \bigl({_{a}}\Phi_{q}^{i}(t) \bigr) \bigr\vert \bigl(q^{i} \bigr)^{\frac{1}{p_{2}}} \bigl\vert g \bigl({_{a}}\Phi_{q}^{i}(t) \bigr) \bigr\vert \\ &\quad\leq \Biggl(\frac{(1-q)(t-a)}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}{_{a}} \bigl(t-{_{a}}\Phi_{q}^{i+1}(t) \bigr)_{q}^{(\alpha-1)} \bigl\vert f \bigl({_{a}} \Phi_{q}^{i}(t) \bigr) \bigr\vert ^{p_{1}} \Biggr)^{\frac {1}{p_{1}}} \\ &\qquad{} \times \Biggl(\frac{(1-q)(t-a)}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}{_{a}} \bigl(t-{_{a}}\Phi_{q}^{i+1}(t) \bigr)_{q}^{(\alpha-1)} \bigl\vert g \bigl({_{a}} \Phi_{q}^{i}(t) \bigr) \bigr\vert ^{p_{2}} \Biggr)^{\frac {1}{p_{2}}} \\ &\quad= \biggl(\frac{1}{\Gamma_{q}(\alpha)} \int_{a}^{t}{_{a}} \bigl(t-{_{a}} \Phi _{q}(s) \bigr)_{q}^{(\alpha-1)} \bigl|f(s) \bigr|^{p_{1}}{_{a}}\,d_{q}s \biggr)^{\frac {1}{p_{1}}} \\ &\qquad{}\times \biggl(\frac{1}{\Gamma_{q}(\alpha)} \int _{a}^{t}{_{a}} \bigl(t-{_{a}} \Phi _{q}(s) \bigr)_{q}^{(\alpha-1)} \bigl|g(s) \bigr|^{p_{2}}{_{a}}\,d_{q}s \biggr)^{\frac {1}{p_{2}}} \\ &\quad= \bigl( \bigl({_{a}}I_{q}^{\alpha} \bigl|f(s) \bigr|^{p_{1}} \bigr) (t) \bigr)^{\frac {1}{p_{1}}} \bigl( \bigl({_{a}}I_{q}^{\alpha} \bigl|g(s) \bigr|^{p_{2}} \bigr) (t) \bigr)^{\frac{1}{p_{2}}}. \end{aligned}$$
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 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 Taking q-integration of order \(\alpha>0\) for (3.4) with respect to s on \([0,1]\), we have which means that From Corollary 2.7(i), we have Using the definition of fractional q-integration on \([a,b]\), we have which gives the second part of (3.2) by using (3.6).
$$ f \bigl((1-s)a+sb \bigr) \leq(1-s)f(a) + s f(b),\quad s \in[0,1]. $$
(3.3)
$$\begin{aligned} &\frac{1}{\Gamma_{q}(\alpha)}{_{0}}\bigl(1-{_{0}}\Phi _{q}(s)\bigr)_{q}^{(\alpha-1)}f \bigl((1-s)a+sb \bigr) \\ &\quad\leq\frac{f(a)}{\Gamma_{q}(\alpha)}{_{0}}\bigl(1-{_{0}}\Phi _{q}(s)\bigr)_{q}^{(\alpha-1)}(1-s) + \frac{f(b)}{\Gamma_{q}(\alpha )}{_{0}}\bigl(1-{_{0}}\Phi_{q}(s) \bigr)_{q}^{(\alpha-1)}s. \end{aligned}$$
(3.4)
$$\begin{aligned} &\frac{1}{\Gamma_{q}(\alpha)} \int _{0}^{1}{_{0}}\bigl(1-{_{0}} \Phi _{q}(s)\bigr)_{q}^{(\alpha-1)}f \bigl((1-s)a+sb \bigr){_{0}}\,d_{q}s \\ &\quad\leq\frac{f(a)}{\Gamma_{q}(\alpha)} \int _{0}^{1}{_{0}}\bigl(1-{_{0}} \Phi_{q}(s)\bigr)_{q}^{(\alpha-1)}(1-s){_{0}}\,d_{q}s + \frac {f(b)}{\Gamma_{q}(\alpha)} \int_{0}^{1}{_{0}}\bigl(1-{_{0}} \Phi _{q}(s)\bigr)_{q}^{(\alpha-1)}s{_{0}}\,d_{q}s, \end{aligned}$$
(3.5)
$$ \bigl({_{0}}I_{q}^{\alpha}f \bigl((1-s)a+sb \bigr)\bigr) (1) \leq f(a) \bigl({_{0}}I_{q}^{\alpha}(1-s) \bigr) (1)+f(b) \bigl({_{0}}I_{q}^{\alpha}s\bigr) (1). $$
(3.6)
$$ \bigl({_{0}}I_{q}^{\alpha}s \bigr) (1) = \frac{1}{\Gamma_{q}(\alpha+2)} \quad\mbox{and}\quad \bigl({_{0}}I_{q}^{\alpha}(1-s) \bigr) (1) = \frac{1}{\Gamma _{q}(\alpha+1)}-\frac{1}{\Gamma_{q}(\alpha+2)}. $$
$$\begin{aligned} & \bigl({_{0}}I_{q}^{\alpha}f \bigl((1-s)a+sb \bigr) \bigr) (1) \\ &\quad= \frac{1}{\Gamma_{q}(\alpha)} \int_{0}^{1}{_{0}} \bigl(1-{_{0}} \Phi _{q}(s) \bigr)_{q}^{(\alpha-1)}f \bigl((1-s)a+sb \bigr){_{0}}\,d_{q}s \\ &\quad= \frac{1-q}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}{_{0}} \bigl(1-{_{0}}\Phi_{q}^{i+1}(1) \bigr)_{q}^{(\alpha-1)}f \bigl( \bigl(1-q^{i} \bigr)a+q^{i}b \bigr) \\ &\quad= \frac{1-q}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty}q^{i} \prod_{i=0}^{\infty}\frac{1-q^{i} q^{i+1}}{1-q^{i} q^{\alpha +i}}f \bigl({_{a}}\Phi_{q}^{i}(b) \bigr) \\ &\quad= \frac{1-q}{(b-a)^{\alpha-1}\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}(b-a)^{\alpha-1} \frac{(q^{i+1};q)_{\infty}}{(q^{i+\alpha};q)_{\infty}}f \bigl({_{a}}\Phi _{q}^{i}(b) \bigr) \\ &\quad= \frac{1}{(b-a)^{\alpha}} \biggl(\frac{1}{\Gamma_{q}(\alpha )} \int _{a}^{b}{_{a}} \bigl(b-{_{a}} \Phi_{q}(s) \bigr)_{q}^{(\alpha-1)}f (s ){_{a}}\,d_{q}s \biggr) \\ &\quad= \frac{1}{(b-a)^{\alpha}} \bigl({_{a}}I_{q}^{\alpha}f \bigr) (b), \end{aligned}$$
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 By a direct computation, we have together with (3.8), we derive the first part of inequality (3.2) as requested. The proof is completed. □
$$\begin{aligned} f \biggl(\frac{a+b}{2} \biggr) \leq\frac{\Gamma_{q}(\alpha +1)}{2(b-a)^{\alpha}} \bigl({_{a}}I_{q}^{\alpha}f(s) \bigr) (b)+ \frac{\Gamma _{q}(\alpha +1)}{2(b-a)^{\alpha}} \bigl({_{a}}I_{q}^{\alpha}f(a+b-s) \bigr) (b). \end{aligned}$$
(3.8)
$$\begin{aligned} & \bigl({_{0}}I_{q}^{\alpha}f \bigl((1-s)b+sa \bigr) \bigr) (1) \\ &\quad= \frac{1-q}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}{_{0}} \bigl(1-q^{i+1} \bigr)_{q}^{(\alpha-1)}f \bigl( \bigl(1-q^{i} \bigr)b+q^{i}a \bigr) \\ &\quad= \frac{1-q}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty }q^{i}{_{0}} \bigl(1-q^{i+1} \bigr)_{q}^{(\alpha-1)}f \bigl(a+b-{_{a}}\Phi _{q}^{i}(b) \bigr) \\ &\quad= \frac{1-q}{\Gamma_{q}(\alpha)}\sum_{i=0}^{\infty}q^{i} \frac{(q^{i+1};q)_{\infty}}{(q^{\alpha+i};q)_{\infty}}f \bigl(a+b-{_{a}}\Phi_{q}^{i}(b) \bigr) \\ &\quad= \frac{1}{(b-a)^{\alpha}} \biggl(\frac{1}{\Gamma_{q}(\alpha )} \int _{a}^{b}{_{a}} \bigl(b-{_{a}} \Phi_{q}(s) \bigr)_{q}^{(\alpha-1)}f (a+b-s ){_{a}}\,d_{q}s \biggr) \\ &\quad= \frac{1}{(b-a)^{\alpha}} \bigl({_{a}}I_{q}^{\alpha}f(a+b-s) \bigr) (b), \end{aligned}$$
Remark 3.4
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 from which one deduces (3.9). □
$$\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}$$
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 Using the classical discrete Cauchy-Schwarz inequality, we have Therefore, inequality (3.10) holds. □
$$\begin{aligned} & \bigl({_{a}}I_{q}^{\beta+\alpha}f(s,r) \bigr) (b) \\ &\quad=\frac{1}{\Gamma_{q}(\alpha)\Gamma_{q}(\beta)} \int_{a}^{b} \int _{a}^{b}{_{a}} \bigl(b-{_{a}} \Phi_{q}(s) \bigr)_{q}^{(\alpha-1)} {_{a}} \bigl(b-{_{a}}\Phi_{q}(r) \bigr)_{q}^{(\beta-1)}f(s,r){_{a}}\,d_{q}s{_{a}}\,d_{q}r \\ &\quad=\frac{(1-q)^{2}(b-a)^{2}}{\Gamma_{q}(\alpha)\Gamma_{q}(\beta)}\sum_{i=0}^{\infty} \sum_{n=0}^{\infty}q^{i+n} {_{a}} \bigl(b-{_{a}}\Phi_{q}^{i+1}(b) \bigr)_{q}^{(\alpha-1)}{_{a}} \bigl(b-{_{a}} \Phi _{q}^{i+1}(b) \bigr)_{q}^{(\beta-1)} \\ &\qquad{}\times f \bigl({_{a}}\Phi_{q}^{i}(b),{_{a}} \Phi_{q}^{n}(b) \bigr). \end{aligned}$$
$$\begin{aligned} & \bigl( \bigl({_{a}}I_{q}^{\beta+\alpha}f(s,r)g(s,r) \bigr) (b) \bigr)^{2} \\ &\quad= \Biggl(\frac{(1-q)^{2}(b-a)^{2}}{\Gamma_{q}(\alpha)\Gamma_{q}(\beta )}\sum_{i=0}^{\infty} \sum_{n=0}^{\infty}q^{i+n} {_{a}} \bigl(b-{_{a}}\Phi_{q}^{i+1}(b) \bigr)_{q}^{(\alpha-1)}{_{a}} \bigl(b-{_{a}} \Phi _{q}^{i+1}(b) \bigr)_{q}^{(\beta-1)} \\ &\qquad{}\times f \bigl({_{a}}\Phi_{q}^{i}(b),{_{a}} \Phi_{q}^{n}(b) \bigr)g \bigl({_{a}} \Phi_{q}^{i}(b),{_{a}}\Phi_{q}^{n}(b) \bigr) \Biggr)^{2} \\ &\quad\leq \Biggl(\frac{(1-q)^{2}(b-a)^{2}}{\Gamma_{q}(\alpha)\Gamma _{q}(\beta )}\sum_{i=0}^{\infty} \sum_{n=0}^{\infty}q^{i+n} {_{a}} \bigl(b-{_{a}}\Phi_{q}^{i+1}(b) \bigr)_{q}^{(\alpha-1)}{_{a}} \bigl(b-{_{a}} \Phi _{q}^{i+1}(b) \bigr)_{q}^{(\beta-1)} \\ &\qquad{}\times f^{2} \bigl({_{a}}\Phi_{q}^{i}(b),{_{a}} \Phi_{q}^{n}(b) \bigr) \Biggr) \Biggl(\frac{(1-q)^{2}(b-a)^{2}}{\Gamma_{q}(\alpha)\Gamma_{q}(\beta )} \sum_{i=0}^{\infty} \sum _{n=0}^{\infty}q^{i+n}{_{a}} \bigl(b-{_{a}}\Phi_{q}^{i+1}(b) \bigr)_{q}^{(\alpha -1)} \\ &\qquad{}\times{_{a}} \bigl(b-{_{a}}\Phi_{q}^{i+1}(b) \bigr)_{q}^{(\beta-1)}g^{2} \bigl({_{a}}\Phi _{q}^{i}(b),{_{a}}\Phi_{q}^{n}(b) \bigr) \Biggr) \\ &\quad= \bigl( \bigl({_{a}}I_{q}^{\beta+\alpha}f^{2}(s,r) \bigr) (b) \bigr) \bigl( \bigl({_{a}}I_{q}^{\beta+\alpha}g^{2}(s,r) \bigr) (b) \bigr). \end{aligned}$$
Remark 3.8
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
For
\(0< q<1\)
and
\(\alpha>0\), we have the inequality
$$ \phi\leq f(s) \leq\Phi, \qquad\psi\leq g(s) \leq\Psi, \quad \textit{for all } s \in[a,b], \phi, \Phi, \psi, \Psi\in\mathbb{R}. $$
(3.11)
$$\begin{aligned} &\biggl|\frac{\Gamma_{q}(\alpha+1)}{(b-a)^{\alpha}} \bigl({_{a}}I_{q}^{\alpha}f(s)g(s) \bigr) (b) - \biggl(\frac{\Gamma_{q}(\alpha+1)}{(b-a)^{\alpha}} \bigl({_{a}}I_{q}^{\alpha}f(s) \bigr) (b) \biggr) \\ &\quad{}\times \biggl(\frac{\Gamma _{q}(\alpha +1)}{(b-a)^{\alpha}} \bigl({_{a}}I_{q}^{\alpha}g(s) \bigr) (b) \biggr) \biggr| \leq\frac{1}{4} (\Phi-\phi ) (\Psi-\psi ). \end{aligned}$$
(3.12)
Proof
Applying Theorem 3.7, we have From Lemma 3.5, it follows that
$$\begin{aligned} &\bigl\vert \bigl({_{a}}I_{q}^{2\alpha } \bigl(f(s)-f(r)\bigr) \bigl(g(s)-g(r)\bigr) \bigr) (b)\bigr\vert \\ &\quad\leq \bigl( \bigl({_{a}}I_{q}^{2\alpha} \bigl(f(s)-f(r)\bigr)^{2} \bigr) (b) \bigr)^{\frac{1}{2}} \bigl( \bigl({_{a}}I_{q}^{2\alpha}\bigl(g(s)-g(r) \bigr)^{2} \bigr) (b) \bigr)^{\frac{1}{2}}. \end{aligned}$$
(3.13)
$$ \frac{1}{2} \bigl({_{a}}I_{q}^{2\alpha} \bigl(f(s)-f(r)\bigr)^{2}\bigr) (b) = \frac{(b-a)^{\alpha}}{\Gamma_{q}(\alpha+1)} \bigl({_{a}}I_{q}^{\alpha }f^{2}(s) \bigr) (b) - \bigl( \bigl({_{a}}I_{q}^{\alpha}f(s) \bigr) (b) \bigr)^{2}. $$
(3.14)
By a simple computation, we have and an analogous identity for g.
$$\begin{aligned} &\frac{\Gamma_{q}(\alpha+1)}{(b-a)^{\alpha }}\bigl({_{a}}I_{q}^{\alpha}f^{2}(s) \bigr) (b) - \biggl(\frac{\Gamma_{q}(\alpha+1)}{(b-a)^{\alpha }}\bigl({_{a}}I_{q}^{\alpha }f(s) \bigr) (b) \biggr)^{2} \\ &\quad= \biggl(\Phi-\frac{\Gamma_{q}(\alpha +1)}{(b-a)^{\alpha }}\bigl({_{a}}I_{q}^{\alpha}f(s) \bigr) (b) \biggr) \biggl(\frac{\Gamma_{q}(\alpha+1)}{(b-a)^{\alpha }}\bigl({_{a}}I_{q}^{\alpha }f(s) \bigr) (b)-\phi \biggr) \\ &\qquad{} -\frac{\Gamma_{q}(\alpha+1)}{(b-a)^{\alpha}} \bigl({_{a}}I_{q}^{\alpha} \bigl(f(s)-\phi \bigr) \bigl(\Phi-f(s) \bigr) \bigr) (b), \end{aligned}$$
(3.15)
By assumption (3.11) we have \((f(s)-\phi)(\Phi-f(s)) \geq 0\) for all \(s\in[a,b]\), which implies From (3.15) and using the fact that \((\frac{A+B}{2} )^{2} \geq AB\), \(A, B\in\mathbb{R}\), we have
$$ \bigl({_{a}}I_{q}^{\alpha} \bigl(f(s)-\phi \bigr) \bigl(\Phi -f(s) \bigr) \bigr) (b) \geq0. $$
$$\begin{aligned} &\frac{\Gamma_{q}(\alpha+1)}{(b-a)^{\alpha }}\bigl({_{a}}I_{q}^{\alpha}f^{2}(s) \bigr) (b) - \biggl(\frac{\Gamma_{q}(\alpha+1)}{(b-a)^{\alpha }}\bigl({_{a}}I_{q}^{\alpha }f(s) \bigr) (b) \biggr)^{2} \\ &\quad\leq \biggl(\Phi-\frac{\Gamma_{q}(\alpha +1)}{(b-a)^{\alpha }}\bigl({_{a}}I_{q}^{\alpha}f(s) \bigr) (b) \biggr) \biggl(\frac{\Gamma_{q}(\alpha+1)}{(b-a)^{\alpha }}\bigl({_{a}}I_{q}^{\alpha }f(s) \bigr) (b)-\phi \biggr) \\ &\quad\leq\frac{1}{4} \biggl[ \biggl(\Phi-\frac{\Gamma _{q}(\alpha +1)}{(b-a)^{\alpha}} \bigl({_{a}}I_{q}^{\alpha}f(s)\bigr) (b) \biggr)+ \biggl(\frac{\Gamma_{q}(\alpha+1)}{(b-a)^{\alpha }}\bigl({_{a}}I_{q}^{\alpha }f(s) \bigr) (b)-\phi \biggr) \biggr]^{2} \\ &\quad\leq\frac{1}{4} (\Phi-\phi )^{2}. \end{aligned}$$
(3.16)
A similar argument gives Using inequality (3.13) via (3.14) and the estimations (3.16) and (3.17), we get Therefore, inequality (3.12) holds, as desired. □
$$ \bigl({_{a}}I_{q}^{2\alpha} \bigl(g(s)-g(r)\bigr)^{2}\bigr) (b) \leq\frac{1}{4} (\Psi-\psi )^{2}. $$
(3.17)
$$ \biggl\vert \frac{1}{2} \bigl({_{a}}I_{q}^{2\alpha } \bigl(f(s)-f(r)\bigr) \bigl(g(s)-g(r)\bigr)\bigr) (b)\biggr\vert \leq \frac{(b-a)^{\alpha}}{4} (\Phi-\phi ) (\Psi -\psi ). $$
Remark 3.10
If \(\alpha= 1\) and \(q \to1\), then inequality (3.12) is reduced to the classical Grüss integral inequality as See also [22, 23].
$$\begin{aligned} & \biggl\vert \frac{1}{b-a} \int_{a}^{b} f(s)g(s)\,ds- \biggl(\frac{1}{b-a} \int_{a}^{b} f(s)\,ds \biggr) \biggl( \frac{1}{b-a} \int_{a}^{b} g(s)\,ds \biggr) \biggr\vert \\ &\quad \leq\frac{1}{4} (\Phi-\phi ) (\Psi-\psi ). \end{aligned}$$
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
for all
\(s,r\in[a,b]\), \(0< q<1\), \(L_{1}, L_{2}>0\), and
\(\alpha>0\). Then we have the inequality
$$ \bigl|f(s)-f(r)\bigr| \leq L_{1}|s-r|,\qquad \bigl|g(s)-g(r)\bigr| \leq L_{2}|s-r|, $$
(3.18)
$$\begin{aligned} &\biggl\vert \frac{(b-a)^{\alpha}}{\Gamma_{q}(\alpha +1)} \bigl({_{a}}I_{q}^{\alpha}f(s)g(s) \bigr) (b) - \bigl({_{a}}I_{q}^{\alpha}f(s) \bigr) (b) \bigl({_{a}}I_{q}^{\alpha }g(s) \bigr) (b)\biggr\vert \\ &\quad\leq\frac{L_{1}L_{2}(b-a)^{2\alpha+2}}{\Gamma_{q}(\alpha +2)\Gamma_{q}(\alpha+3)} \bigl((1+q)[\alpha+1]_{q}-[ \alpha+2]_{q} \bigr). \end{aligned}$$
(3.19)
Proof
Recall the fractional q-Korkine equality as It follows from (3.18) that 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 From Corollary 2.7(ii), with \(t=b\), we have
$$\begin{aligned} &\frac{(b-a)^{\alpha}}{\Gamma_{q}(\alpha+1)} \bigl({_{a}}I_{q}^{\alpha}f(s)g(s) \bigr) (b) - \bigl({_{a}}I_{q}^{\alpha}f(s) \bigr) (b) \bigl({_{a}}I_{q}^{\alpha }g(s) \bigr) (b) \\ &\quad = \frac{1}{2} \bigl({_{a}}I_{q}^{2\alpha } \bigl(f(s)-f(r)\bigr) \bigl(g(s)-g(r)\bigr)\bigr) (b). \end{aligned}$$
(3.20)
$$ \bigl\vert \bigl(f(s)-f(r) \bigr) \bigl(g(s)-g(r) \bigr) \bigr\vert \leq L_{1}L_{2}(s-r)^{2}, $$
(3.21)
$$\begin{aligned} & \bigl({_{a}}I_{q}^{2\alpha} \bigl\vert \bigl(f(s)-f(r) \bigr) \bigl(g(s)-g(r) \bigr) \bigr\vert \bigr) (b) \\ &\quad= \frac{1}{\Gamma_{q}^{2}(\alpha)} \int_{a}^{b} \int _{a}^{b}{_{a}} \bigl(b-{_{a}} \Phi_{q}(s) \bigr)_{q}^{(\alpha-1)} {_{a}} \bigl(b-{_{a}}\Phi_{q}(r) \bigr)_{q}^{(\alpha-1)} \\ &\qquad{}\times \bigl\vert \bigl(f(s)-f(r) \bigr) \bigl(g(s)-g(r) \bigr) \bigr\vert {_{a}}\,d_{q}s{_{a}}\,d_{q}r \\ &\quad\leq\frac{L_{1}L_{2}}{\Gamma_{q}^{2}(\alpha)} \int_{a}^{b} \int _{a}^{b}{_{a}} \bigl(b-{_{a}} \Phi_{q}(s) \bigr)_{q}^{(\alpha-1)} {_{a}} \bigl(b-{_{a}}\Phi_{q}(r) \bigr)_{q}^{(\alpha-1)} (s-r )^{2}{_{a}}\,d_{q}s{_{a}}\,d_{q}r \\ &\quad=\frac{L_{1}L_{2}}{\Gamma_{q}^{2}(\alpha)} \int_{a}^{b} \int _{a}^{b}{_{a}} \bigl(b-{_{a}} \Phi_{q}(s) \bigr)_{q}^{(\alpha-1)} {_{a}} \bigl(b-{_{a}}\Phi_{q}(r) \bigr)_{q}^{(\alpha -1)}s^{2}{_{a}}\,d_{q}s{_{a}}\,d_{q}r \\ &\qquad{}-\frac{2L_{1}L_{2}}{\Gamma_{q}^{2}(\alpha)} \int_{a}^{b} \int _{a}^{b}{_{a}} \bigl(b-{_{a}} \Phi_{q}(s) \bigr)_{q}^{(\alpha-1)} {_{a}} \bigl(b-{_{a}}\Phi_{q}(r) \bigr)_{q}^{(\alpha -1)}sr{_{a}}\,d_{q}s{_{a}}\,d_{q}r \\ &\qquad{}+\frac{L_{1}L_{2}}{\Gamma_{q}^{2}(\alpha)} \int_{a}^{b} \int _{a}^{b}{_{a}} \bigl(b-{_{a}} \Phi_{q}(s) \bigr)_{q}^{(\alpha-1)} {_{a}} \bigl(b-{_{a}}\Phi_{q}(r) \bigr)_{q}^{(\alpha -1)}r^{2}{_{a}}\,d_{q}s{_{a}}\,d_{q}r \\ &\quad=2L_{1}L_{2} \biggl(\frac{(b-a)^{\alpha}}{\Gamma_{q}(\alpha+1)} \bigl({_{a}}I_{q}^{\alpha}s^{2} \bigr) (b) - \bigl( \bigl({_{a}}I_{q}^{\alpha}s \bigr) (b) \bigr)^{2} \biggr). \end{aligned}$$
(3.22)
$$ \bigl({_{a}}I_{q}^{\alpha}s^{2} \bigr) (b) = \frac{(b-a)^{\alpha }}{\Gamma _{q}(\alpha+3)} \bigl((1+q) (b-a)^{2}+2a(b-a)[ \alpha+2]_{q} +a^{2}[\alpha+1]_{q}[ \alpha+2]_{q} \bigr). $$
By direct computation, we have Thus, from (3.22) and (3.23), 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)
$$\begin{aligned} &\bigl({_{a}}I_{q}^{2\alpha} \bigl\vert \bigl(f(s)-f(r) \bigr) \bigl(g(s)-g(r) \bigr)\bigr\vert \bigr) (b) \\ &\quad \leq\frac{2L_{1}L_{2}(b-a)^{2\alpha+2}}{\Gamma _{q}(\alpha +2)\Gamma_{q}(\alpha+3)} \bigl((1+q)[\alpha+1]_{q}-[ \alpha+2]_{q} \bigr). \end{aligned}$$
(3.24)
Remark 3.12
If \(\alpha= 1\) and \(q \to1\), then inequality (3.19) is reduced to the classical Grüss-Čebyšev integral inequality as See also [22, 23].
$$\begin{aligned} \biggl\vert \frac{1}{b-a} \int_{a}^{b} f(s)g(s)\,ds- \biggl(\frac{1}{b-a} \int_{a}^{b} f(s)\,ds \biggr) \biggl( \frac{1}{b-a} \int_{a}^{b} g(s)\,ds \biggr) \biggr\vert \leq \frac {L_{1}L_{2}}{12} (b-a )^{2}. \end{aligned}$$
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
Then for
\(0< q<1\)
and
\(\alpha>0\), we have the inequality
$$ 0< \phi\leq f(s) \leq\Phi,\qquad 0< \psi\leq g(s) \leq\Psi, \quad \textit{for all } s\in[a,b], \phi, \Phi, \psi, \Psi\in\mathbb{R}^{+}. $$
(3.25)
$$ \frac{ ({_{a}}I_{q}^{\alpha}(f^{2}(s)) )(b) ({_{a}}I_{q}^{\alpha}(g^{2}(s)) )(b)}{ ( ({_{a}}I_{q}^{\alpha}(f(s)g(s) )(b) )^{2}} \leq\frac{1}{4} \biggl( \sqrt{\frac{\phi\psi}{\Phi\Psi}}+\sqrt {\frac{\Phi \Psi}{\phi\psi}} \biggr)^{2}. $$
(3.26)
Proof
From (3.25), for \(s\in[a,b]\), we have which yields and Multiplying (3.27) and (3.28), we obtain or Inequality (3.29) can be written as 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 Applying the AM-GM inequality, \(A+B\geq2\sqrt{AB}\), \(A,B\in\mathbb {R}^{+}\), we have which leads to Therefore, inequality (3.26) is proved. □
$$ \frac{\phi}{\Psi} \leq\frac{f(s)}{g(s)} \leq\frac{\Phi}{\psi}, $$
$$ \biggl(\frac{\Phi}{\psi}-\frac{f(s)}{g(s)} \biggr) \geq0 $$
(3.27)
$$ \biggl(\frac{f(s)}{g(s)}-\frac{\phi}{\Psi} \biggr) \geq0. $$
(3.28)
$$ \biggl(\frac{\Phi}{\psi}-\frac{f(s)}{g(s)} \biggr) \biggl(\frac {f(s)}{g(s)}- \frac{\phi}{\Psi} \biggr) \geq0, $$
$$ \biggl(\frac{\Phi}{\psi}+\frac{\phi}{\Psi} \biggr) \frac {f(s)}{g(s)} \geq \frac{f^{2}(s)}{g^{2}(s)}+\frac{\phi\Psi}{\psi\Phi}. $$
(3.29)
$$ (\phi\psi+\Phi\Psi )f(s)g(s) \geq\psi\Psi f^{2}(s)+\phi \Phi g^{2}(s). $$
(3.30)
$$ (\phi\psi+\Phi\Psi ) \bigl({_{a}}I_{q}^{\alpha } \bigl(f(s)g(s) \bigr)\bigr) (b) \geq\psi\Psi \bigl({_{a}}I_{q}^{\alpha} \bigl(f^{2}(s) \bigr)\bigr) (b)+\phi\Phi \bigl({_{a}}I_{q}^{\alpha} \bigl(g^{2}(s) \bigr)\bigr) (b). $$
$$ (\phi\psi+\Phi\Psi ) \bigl({_{a}}I_{q}^{\alpha } \bigl(f(s)g(s) \bigr)\bigr) (b) \geq2\sqrt{\phi\psi\Phi\Psi \bigl({_{a}}I_{q}^{\alpha } \bigl(f^{2}(s) \bigr)\bigr) (b) \bigl({_{a}}I_{q}^{\alpha} \bigl(g^{2}(s) \bigr)\bigr) (b)}, $$
$$ \phi\psi\Phi\Psi \bigl({_{a}}I_{q}^{\alpha} \bigl(f^{2}(s) \bigr) \bigr) (b) \bigl({_{a}}I_{q}^{\alpha} \bigl(g^{2}(s) \bigr) \bigr) (b) \leq\frac{1}{4} ( (\phi\psi+ \Phi \Psi ) \bigl({_{a}}I_{q}^{\alpha} \bigl(f(s)g(s) \bigr) (b) \bigr)^{2}. $$
Remark 3.14
If \(\alpha= 1\) and \(q \to1\), then inequality (3.26) is reduced to the classical Pólya-Szegö integral inequality as See also [24].
$$\begin{aligned} \frac{\int_{a}^{b} f^{2}(s)\,ds\int_{a}^{b} g^{2}(s)\,ds}{ (\int_{a}^{b} f(s)g(s)\,ds )^{2}} \leq\frac{1}{4} \biggl(\sqrt{\frac{\phi\psi }{\Phi \Psi}}+ \sqrt{\frac{\Phi\Psi}{\phi\psi}} \biggr)^{2}. \end{aligned}$$
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.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to this article. They read and approved the final manuscript.