Some new fractional q-integral Grüss-type inequalities and other inequalities

Abstract

In this paper, we employ a fractional q-integral on the specific time scale, ${\mathbb{T}}_{t_0}=\{t:t=t_0{q}^n,n\text{ a nonnegative integer}\}\cup \{0\}$, where $t_0\in \mathbb{R}$ and $0<q<1$, to establish some new fractional q-integral Grüss-type inequalities by using one or two fractional parameters. Furthermore, other fractional q-integral inequalities are also obtained.

MSC:26D10, 26A33.

1 Introduction

In the past several years, by using the Riemann-Liouville fractional integrals, the fractional integral inequalities and applications have been addressed extensively by several researchers. For example, we refer the reader to [1–6] and the references cited therein. Dahmani et al. [7] gave the following fractional integral inequalities by using the Riemann-Liouville fractional integrals. Let f and g be two integrable functions on $[0,\mathrm{\infty })$ satisfying the following conditions:

$${\phi }_1\le f(x)\le {\phi }_2,{\psi }_1\le g(x)\le {\psi }_2,{\phi }_1,{\phi }_2,{\psi }_1,{\psi }_2\in \mathbb{R},x\in [0,\mathrm{\infty }).$$

For all $t>0$, $\alpha >0$ and $\beta >0$, then

$$|\frac{t^{\alpha }}{\mathrm{\Gamma }(\alpha +1)}{J}^{\alpha }(fg)(t)-J^{\alpha }f(t)J^{\alpha }g(t)|\le {\left(\frac{t^{\alpha }}{\mathrm{\Gamma }(\alpha +1)}\right)}^2({\phi }_2-{\phi }_1)({\psi }_2-{\psi }_1)$$

and

To the best of authors’ knowledge, only some fractional q-integral inequalities have been established in recent years. That is, only Öğünmez and Özkan [8], Bohner and Ferreira [9] and Yang [10] obtained some fractional q-integral inequalities. With motivation from the papers [7, 11, 12], the main purpose of this article is to establish some new fractional q-integral inequalities. First of all, by using one or two fractional parameters, we establish some new fractional q-integral Grüss-type inequalities on the specific time scale ${\mathbb{T}}_{t_0}=\{t:t=t_0{q}^n,n\text{ a nonnegative integer}\}\cup \{0\}$, where $t_0\in \mathbb{R}$ and $0<q<1$. In general, a time scale is an arbitrary nonempty closed subset of real numbers [13]. Furthermore, other fractional q-integral inequalities are also obtained.

2 Description of fractional q-calculus

In this section, we introduce the basic definitions on fractional q-calculus. More results concerning fractional q-calculus can be found in [14–17].

Let $t_0\in \mathbb{R}$ and define ${\mathbb{T}}_{t_0}=\{t:t=t_0{q}^n,n\text{ a nonnegative integer}\}\cup \{0\}$, $0<q<1$. For a function $f:{\mathbb{T}}_{t_0}\to \mathbb{R}$, the nabla q-derivative of f

$${\mathrm{\nabla }}_{q}f(t)=\frac{f(qt)-f(t)}{(q-1)t}$$

for all $t\in {\mathbb{T}}_{t_0}\mathrm{\setminus }\{0\}$. The q-integral of f is

$${\int }_0^{t}f(s)\mathrm{\nabla }s=(1-q)t\sum _{i=0}^{\mathrm{\infty }}{q}^{i}f\left(t{q}^i\right).$$

The q-factorial function is defined in the following way: if n is a positive integer, then

$${(t-s)}^{\underline{(n)}}=(t-s)(t-qs)(t-q^{2}s)\cdots (t-q^{n-1}s).$$

If n is not a positive integer, then

$${(t-s)}^{\underline{(n)}}=t^n\prod _{k=0}^{\mathrm{\infty }}\frac{1-(s/t)q^k}{1-(s/t)q^{n+k}}.$$

The q-derivative of the q-factorial function with respect to t is

$${\mathrm{\nabla }}_q{(t-s)}^{\underline{(n)}}=\frac{1-q^n}{1-q}{(t-s)}^{\underline{(n-1)}},$$

and the q-derivative of the q-factorial function with respect to s is

$${\mathrm{\nabla }}_q{(t-s)}^{\underline{(n)}}=-\frac{1-q^n}{1-q}{(t-qs)}^{\underline{(n-1)}}.$$

The q-exponential function is defined as

$$e_q(t)=\prod _{k=0}^{\mathrm{\infty }}(1-q^{k}t),e_q(0)=1.$$

Define the q-gamma function by

$${\mathrm{\Gamma }}_q(\nu )=\frac{1}{1-q}{\int }_0^1{\left(\frac{t}{1-q}\right)}^{\nu -1}{e}_q(qt)\mathrm{\nabla }t,\nu \in {\mathbb{R}}^{+}.$$

Note that

$${\mathrm{\Gamma }}_q(\nu +1)={[\nu ]}_q{\mathrm{\Gamma }}_q(\nu ),\nu \in {\mathbb{R}}^{+},$$

where ${[\nu ]}_q:=(1-q^{\nu })/(1-q)$. The fractional q-integral is defined as

$${\mathrm{\nabla }}_q^{-\nu }f(t)=\frac{1}{{\mathrm{\Gamma }}_q(\nu )}{\int }_0^t{(t-qs)}^{\underline{(\nu -1)}}f(s)\mathrm{\nabla }s.$$

Note that

$${\mathrm{\nabla }}_q^{-\nu }(1)=\frac{1}{{\mathrm{\Gamma }}_q(\nu )}\frac{q-1}{q^{\nu }-1}{t}^{\underline{(\nu )}}=\frac{1}{{\mathrm{\Gamma }}_q(\nu +1)}{t}^{\underline{(\nu )}}.$$

3 Fractional q-integral Grüss-type inequalities

To state the main results in this paper, we employ the following lemmas. For the sake of convenience, we use the following assumption (A) in this section:

$${\phi }_1\le f(x)\le {\phi }_2,{\psi }_1\le g(x)\le {\psi }_2,{\phi }_1,{\phi }_2,{\psi }_1,{\psi }_2\in \mathbb{R},x\in {\mathbb{T}}_{t_0}.$$

Lemma 1 Let ${\phi }_1,{\phi }_2\in \mathbb{R}$ and f be a function defined on ${\mathbb{T}}_{t_0}$. Then, for all $t>0$ and $\nu >0$, we have

$$\begin{array}{rcl}\frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)}{\mathrm{\nabla }}_q^{-\nu }{f}^2(t)-{({\mathrm{\nabla }}_q^{-\nu }f(t))}^2& =& ({\phi }_2\frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)}-{\mathrm{\nabla }}_q^{-\nu }f(t))({\mathrm{\nabla }}_q^{-\nu }f(t)-{\phi }_1\frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)})\\ -\frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)}{\mathrm{\nabla }}_q^{-\nu }({\phi }_2-f(t))(f(t)-{\phi }_1).\end{array}$$

(1)

Proof Let ${\phi }_1,{\phi }_2\in \mathbb{R}$ and f be a function defined on ${\mathbb{T}}_{t_0}$. For any $\tau >0$ and $\rho >0$, we have

(2)

Multiplying both sides of (2) by ${(t-q\tau )}^{\underline{(\nu -1)}}/{\mathrm{\Gamma }}_q(\nu )$ and integrating the resulting identity with respect to τ from 0 to t, we get

(3)

Multiplying both sides of (3) by ${(t-q\rho )}^{\underline{(\nu -1)}}/{\mathrm{\Gamma }}_q(\nu )$ and integrating the resulting identity with respect to ρ from 0 to t, we obtain

which implies (1). □

Lemma 2 Let f and g be two functions defined on ${\mathbb{T}}_{t_0}$. Then, for all $t>0$, $\mu >0$ and $\nu >0$, we have

(4)

Proof In order to prove Lemma 2, we firstly prove that the following inequality (i.e., Cauchy-Schwarz inequality for double q-integrals) holds. Let $f(x,y)$, $g(x,y)$ and $h(x,y)$ be three functions defined on ${\mathbb{T}}_{t_0}^2$ with $h(x,y)\ge 0$. Then we have

According to the definition of q-integral, it is easy to obtain that double q-integral is

Due to discrete Cauchy-Schwarz inequality with weight coefficient, we have

Next, we prove that Lemma 2 holds. Let $H(\tau ,\rho )$ be defined by

$$H(\tau ,\rho )=(f(\tau )-f(\rho ))(g(\tau )-g(\rho )),t>0,\tau >0,\rho >0.$$

(5)

Multiplying both sides of (5) by ${(t-q\tau )}^{\underline{(\nu -1)}}{(t-q\rho )}^{\underline{(\mu -1)}}/({\mathrm{\Gamma }}_q(\nu ){\mathrm{\Gamma }}_q(\mu ))$ and integrating the resulting identity with respect to τ and ρ from 0 to t, then applying the Cauchy-Schwarz inequality for double q-integrals, we obtain (4). □

Lemma 3 Let ${\phi }_1,{\phi }_2\in \mathbb{R}$ and f be a function defined on ${\mathbb{T}}_{t_0}$. Then, for all $t>0$ and $\nu >0$, we have

(6)

Proof Multiplying both sides of (3) by ${(t-q\rho )}^{\underline{(\mu -1)}}/{\mathrm{\Gamma }}_q(\mu )$ and integrating the resulting identity with respect to ρ from 0 to t, we obtain

which implies (6). □

Theorem 1 Let f and g be two functions defined on ${\mathbb{T}}_{t_0}$ satisfying (A). Then, for all $t>0$ and $\nu >0$, we have

$$|\frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)}{\mathrm{\nabla }}_q^{-\nu }(fg)(t)-{\mathrm{\nabla }}_q^{-\nu }f(t){\mathrm{\nabla }}_q^{-\nu }g(t)|\le {\left(\frac{t^{\underline{(\nu )}}}{2{\mathrm{\Gamma }}_q(\nu +1)}\right)}^2({\phi }_2-{\phi }_1)({\psi }_2-{\psi }_1).$$

(7)

Proof Let f and g be two functions defined on ${\mathbb{T}}_{t_0}$ satisfying (A). Multiplying both sides of (6) by ${(t-q\tau )}^{\underline{(\nu -1)}}{(t-q\rho )}^{\underline{(\nu -1)}}/{\mathrm{\Gamma }}_q^2(\nu )$ and integrating the resulting identity with respect to τ and ρ from 0 to t, we can state that

(8)

Applying the Cauchy-Schwarz inequality for double q-integrals, we have

(9)

Since $({\phi }_2-f(x))(f(x)-{\phi }_1)\ge 0$ and $({\psi }_2-g(x))(g(x)-{\psi }_1)\ge 0$, we have

Thus,

$$\begin{array}{r}\frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)}{\mathrm{\nabla }}_q^{-\nu }{f}^2(t)-{({\mathrm{\nabla }}_q^{-\nu }f(t))}^2\\ \le ({\phi }_2\frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)}-{\mathrm{\nabla }}_q^{-\nu }f(t))({\mathrm{\nabla }}_q^{-\nu }f(t)-{\phi }_1\frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)}),\\ \frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)}{\mathrm{\nabla }}_q^{-\nu }{g}^2(t)-{({\mathrm{\nabla }}_q^{-\nu }g(t))}^2\\ \le ({\psi }_2\frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)}-{\mathrm{\nabla }}_q^{-\nu }g(t))({\mathrm{\nabla }}_q^{-\nu }g(t)-{\psi }_1\frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)}).\end{array}$$

(10)

Combining (9) and (10), from Lemma 1, we deduce that

(11)

Now by using the elementary inequality $4xy\le {(x+y)}^2$, $x,y\in \mathbb{R}$, we can state that

$$\begin{array}{r}4({\phi }_2\frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)}-{\mathrm{\nabla }}_q^{-\nu }f(t))({\mathrm{\nabla }}_q^{-\nu }f(t)-{\phi }_1\frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)})\\ \le {(\frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)}({\phi }_2-{\phi }_1))}^2,\\ 4({\psi }_2\frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)}-{\mathrm{\nabla }}_q^{-\nu }g(t))({\mathrm{\nabla }}_q^{-\nu }g(t)-{\psi }_1\frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)})\\ \le {(\frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)}({\psi }_2-{\psi }_1))}^2.\end{array}$$

(12)

From (11) and (12), we obtain (7). □

Theorem 2 Let f and g be two functions defined on ${\mathbb{T}}_{t_0}$ satisfying (A). Then, for all $t>0$, $\mu >0$ and $\nu >0$, we have

Proof Since $({\phi }_2-f(x))(f(x)-{\phi }_1)\ge 0$ and $({\psi }_2-g(x))(g(x)-{\psi }_1)\ge 0$, then we can write

$$\begin{array}{r}-\frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)}{\mathrm{\nabla }}_q^{-\mu }({\phi }_2-f(x))(f(x)-{\phi }_1)-\frac{t^{\underline{(\mu )}}}{{\mathrm{\Gamma }}_q(\mu +1)}{\mathrm{\nabla }}_q^{-\nu }({\phi }_2-f(x))(f(x)-{\phi }_1)\\ \le 0,\\ -\frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)}{\mathrm{\nabla }}_q^{-\mu }({\psi }_2-g(x))(g(x)-{\psi }_1)-\frac{t^{\underline{(\mu )}}}{{\mathrm{\Gamma }}_q(\mu +1)}{\mathrm{\nabla }}_q^{-\nu }({\psi }_2-g(x))(g(x)-{\psi }_1)\\ \le 0.\end{array}$$

(13)

Applying Lemma 3 to f and g, then by using Lemma 2 and the formula (13), we obtain Theorem 2. □

4 The other fractional q-integral inequalities

For the sake of simplicity, we always assume that ${\mathrm{\nabla }}_q^{\nu }\varphi$ denotes ${\mathrm{\nabla }}_q^{\nu }\varphi (t)$ and all of fractional q-integrals are finite in this section.

Theorem 3 Let f and g be two functions defined on ${\mathbb{T}}_{t_0}$ and $\alpha ,\beta >1$ satisfying $1/\alpha +1/\beta =1$. Then the following inequalities hold:

• (a) $\frac{1}{\alpha }{\mathrm{\nabla }}_q^{-\nu }({|f|}^{\alpha })+\frac{1}{\beta }{\mathrm{\nabla }}_q^{-\nu }({|g|}^{\beta })\ge \frac{{\mathrm{\Gamma }}_q(\nu +1)}{t^{\underline{(\nu )}}}{\mathrm{\nabla }}_q^{-\nu }(|f|){\mathrm{\nabla }}_q^{-\nu }(|g|)$.

• (b) $\frac{1}{\alpha }{\mathrm{\nabla }}_q^{-\nu }({|f|}^{\alpha }){\mathrm{\nabla }}_q^{-\nu }({|g|}^{\alpha })+\frac{1}{\beta }{\mathrm{\nabla }}_q^{-\nu }({|f|}^{\beta }){\mathrm{\nabla }}_q^{-\nu }({|g|}^{\beta })\ge {({\mathrm{\nabla }}_q^{-\nu }(|fg|))}^2$.

• (c) $\frac{1}{\alpha }{\mathrm{\nabla }}_q^{-\nu }({|f|}^{\alpha }){\mathrm{\nabla }}_q^{-\nu }({|g|}^{\beta })+\frac{1}{\beta }{\mathrm{\nabla }}_q^{-\nu }({|f|}^{\beta }){\mathrm{\nabla }}_q^{-\nu }({|g|}^{\alpha })\ge {\mathrm{\nabla }}_q^{-\nu }(|f|{|g|}^{\alpha -1}){\mathrm{\nabla }}_q^{-\nu }(|f|{|g|}^{\beta -1})$.

• (d) ${\mathrm{\nabla }}_q^{-\nu }({|f|}^{\alpha }){\mathrm{\nabla }}_q^{-\nu }({|g|}^{\beta })\ge {\mathrm{\nabla }}_q^{-\nu }(|fg|){\mathrm{\nabla }}_q^{-\nu }({|f|}^{\alpha -1}{|g|}^{\beta -1})$.

Proof According to the well-known Young inequality,

$$\frac{1}{\alpha }{x}^{\alpha }+\frac{1}{\beta }{y}^{\beta }\ge xy,\mathrm{\forall }x,y\ge 0,\alpha ,\beta >1,\frac{1}{\alpha }+\frac{1}{\beta }=1.$$

Putting $x=f(\tau )$ and $y=g(\rho )$, $\tau ,\rho >0$, we have

$$\frac{1}{\alpha }{|f(\tau )|}^{\alpha }+\frac{1}{\beta }{|g(\rho )|}^{\beta }\ge |f(\tau )||g(\rho )|,\mathrm{\forall }\tau ,\rho >0.$$

(14)

Multiplying both sides of (6) by ${(t-q\tau )}^{\underline{(\nu -1)}}{(t-q\rho )}^{\underline{(\nu -1)}}/{\mathrm{\Gamma }}_q^2(\nu )$, we obtain

Integrating the preceding identity with respect to τ and ρ from 0 to t, we can state that

$$\frac{1}{\alpha }\frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)}{\mathrm{\nabla }}_q^{-\nu }\left({|f(t)|}^{\alpha }\right)+\frac{1}{\beta }\frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)}{\mathrm{\nabla }}_q^{-\nu }\left({|g(t)|}^{\beta }\right)\ge {\mathrm{\nabla }}_q^{-\nu }\left(|f(t)|\right){\mathrm{\nabla }}_q^{-\nu }\left(|g(t)|\right),$$

which implies (a). The rest of inequalities can be proved in the same manner by the next choice of the parameters in the Young inequality:

• (b) $x=|f(\tau )||g(\rho )|$, $y=|f(\rho )||g(\tau )|$.

• (c) $x=|f(\tau )|/|g(\tau )|$, $y=|f(\rho )|/|g(\rho )|$, ($g(\tau )g(\rho )\ne 0$).

• (d) $x=|f(\rho )|/|f(\tau )|$, $y=|g(\rho )|/|g(\tau )|$, ($f(\tau )g(\rho )\ne 0$).

Repeating the foregoing arguments, we obtain (b)-(d). □

Theorem 4 Let f and g be two functions defined on ${\mathbb{T}}_{t_0}$ and $\alpha ,\beta >1$ satisfying $1/\alpha +1/\beta =1$. Then the following inequalities hold:

• (a) $\frac{1}{\alpha }{\mathrm{\nabla }}_q^{-\nu }({|f|}^{\alpha }){\mathrm{\nabla }}_q^{-\nu }({|g|}^2)+\frac{1}{\beta }{\mathrm{\nabla }}_q^{-\nu }({|f|}^2){\mathrm{\nabla }}_q^{-\nu }({|g|}^{\beta })\ge {\mathrm{\nabla }}_q^{-\nu }(|fg|){\mathrm{\nabla }}_q^{-\nu }({|f|}^{2/\beta }{|g|}^{2/\alpha })$.

• (b) $\frac{1}{\alpha }{\mathrm{\nabla }}_q^{-\nu }({|f|}^2){\mathrm{\nabla }}_q^{-\nu }({|g|}^{\beta })+\frac{1}{\beta }{\mathrm{\nabla }}_q^{-\nu }({|f|}^{\beta }){\mathrm{\nabla }}_q^{-\nu }({|g|}^2)\ge {\mathrm{\nabla }}_q^{-\nu }({|f|}^{2/\alpha }{|g|}^{2/\beta }){\mathrm{\nabla }}_q^{-\nu }({|f|}^{\alpha -1}{|g|}^{\beta -1})$.

• (c) ${\mathrm{\nabla }}_q^{-\nu }({|f|}^2){\mathrm{\nabla }}_q^{-\nu }(\frac{1}{\alpha }{|g|}^{\alpha }+\frac{1}{\beta }{|g|}^{\beta })\ge {\mathrm{\nabla }}_q^{-\nu }({|f|}^{2/\alpha }|g|){\mathrm{\nabla }}_q^{-\nu }({|f|}^{2/\beta }|g|)$.

Proof As a previous one, the proof is based on the Young inequality with the following appropriate choice of parameters:

• (a) $x=|f(\tau )|{|g(\rho )|}^{2/\alpha }$, $y={|f(\rho )|}^{2/\beta }|g(\tau )|$.

• (b) $x={|f(\tau )|}^{2/\alpha }/|f(\rho )|$, $y={|g(\tau )|}^{2/\beta }/|g(\rho )|$, $(f(\rho )g(\rho )\ne 0)$.

• (c) $x={|f(\tau )|}^{2/\alpha }/|g(\rho )|$, $y={|f(\rho )|}^{2/\beta }/|g(\tau )|$, $(g(\tau )g(\rho )\ne 0)$.

 □

Theorem 5 Let f and g be two positive functions defined on ${\mathbb{T}}_{t_0}$ such that for all $t>0$,

$$m=\underset{0\le \tau \le t}{min}\frac{f(\tau )}{g(\tau )},M=\underset{0\le \tau \le t}{max}\frac{f(\tau )}{g(\tau )}.$$

(15)

Then the following inequalities hold:

• (a) $0\le {\mathrm{\nabla }}_q^{-\nu }(f^2){\mathrm{\nabla }}_q^{-\nu }(g^2)\le \frac{{(m+M)}^2}{4mM}{({\mathrm{\nabla }}_q^{-\nu }(fg))}^2$.

• (b) $0\le \sqrt{{\mathrm{\nabla }}_q^{-\nu }(f^2){\mathrm{\nabla }}_q^{-\nu }(g^2)}-{\mathrm{\nabla }}_q^{-\nu }(fg)\le \frac{{(\sqrt{M}-\sqrt{m})}^2}{2\sqrt{mM}}{\mathrm{\nabla }}_q^{-\nu }(fg)$.

• (c) $0\le {\mathrm{\nabla }}_q^{-\nu }(f^2){\mathrm{\nabla }}_q^{-\nu }(g^2)-{({\mathrm{\nabla }}_q^{-\nu }(fg))}^2\le \frac{{(M-m)}^2}{4mM}{({\mathrm{\nabla }}_q^{-\nu }(fg))}^2$.

Proof It follows from (15) and

$$(\frac{f(\tau )}{g(\tau )}-m)(M-\frac{f(\tau )}{g(\tau )})g^2(\tau )\ge 0,0\le \tau \le t.$$

(16)

Multiplying both sides of (15) by ${(t-q\tau )}^{\underline{(\nu -1)}}/{\mathrm{\Gamma }}_q(\nu )$ and integrating the resulting identity with respect to τ from 0 to t, we can get

$${\mathrm{\nabla }}_q^{-\nu }\left(f^2\right)+mM{\mathrm{\nabla }}_q^{-\nu }\left(g^2\right)\le (m+M){\mathrm{\nabla }}_q^{-\nu }(fg).$$

(17)

On the other hand, it follows from $mM>0$ and ${(\sqrt{{\mathrm{\nabla }}_q^{-\nu }(f^2)}-\sqrt{mM{\mathrm{\nabla }}_q^{-\nu }(g^2)})}^2\ge 0$ that

$$2\sqrt{{\mathrm{\nabla }}_q^{-\nu }\left(f^2\right)}\sqrt{mM{\mathrm{\nabla }}_q^{-\nu }\left(g^2\right)}\le {\mathrm{\nabla }}_q^{-\nu }\left(f^2\right)+mM{\mathrm{\nabla }}_q^{-\nu }\left(g^2\right).$$

(18)

According to (17) and (18), we have

$$4mM{\mathrm{\nabla }}_q^{-\nu }\left(f^2\right){\mathrm{\nabla }}_q^{-\nu }\left(g^2\right)\le {(m+M)}^2{({\mathrm{\nabla }}_q^{-\nu }(fg))}^2,$$

which implies (a). By a few transformations of (a), similarly, we obtain (b) and (c). □

Corollary 1 Under the conditions of Theorem 5, if $\alpha ,\beta \in (0,1)$, $\alpha +\beta =1$, then it follows from the arithmetric-geometric mean inequality that

$${\left(\frac{1}{\alpha }{\mathrm{\nabla }}_q^{-\nu }\left(f^2\right)\right)}^{\alpha }{\left(\frac{mM}{\beta }{\mathrm{\nabla }}_q^{-\nu }\left(g^2\right)\right)}^{\beta }\le {\mathrm{\nabla }}_q^{-\nu }\left(f^2\right)+mM{\mathrm{\nabla }}_q^{-\nu }\left(g^2\right)\le (m+M){\mathrm{\nabla }}_q^{-\nu }(fg),$$

which implies that

$${\left({\mathrm{\nabla }}_q^{-\nu }\left(f^2\right)\right)}^{\alpha }{\left({\mathrm{\nabla }}_q^{-\nu }\left(g^2\right)\right)}^{\beta }\le {\alpha }^{\alpha }{\beta }^{\beta }\frac{m+M}{{(mM)}^{\beta }}{\mathrm{\nabla }}_q^{-\nu }(fg).$$

Theorem 6 Let f and g be two positive functions on ${\mathbb{T}}_{t_0}$ and

$$0<{\mathrm{\Phi }}_1\le f(\tau )\le {\mathrm{\Phi }}_2<\mathrm{\infty },0<{\mathrm{\Psi }}_1\le g(\tau )\le {\mathrm{\Psi }}_2<\mathrm{\infty }.$$

(19)

Then the following inequalities hold:

• (a) $0\le {\mathrm{\nabla }}_q^{-\nu }(f^2){\mathrm{\nabla }}_q^{-\nu }(g^2)\le \frac{{({\mathrm{\Phi }}_1{\mathrm{\Psi }}_1+{\mathrm{\Phi }}_2{\mathrm{\Psi }}_2)}^2}{4{\mathrm{\Phi }}_1{\mathrm{\Psi }}_1{\mathrm{\Phi }}_2{\mathrm{\Psi }}_2}{({\mathrm{\nabla }}_q^{-\nu }(fg))}^2$.

• (b) $0\le \sqrt{{\mathrm{\nabla }}_q^{-\nu }(f^2){\mathrm{\nabla }}_q^{-\nu }(g^2)}-{\mathrm{\nabla }}_q^{-\nu }(fg)\le \frac{{(\sqrt{{\mathrm{\Phi }}_2{\mathrm{\Psi }}_2}-\sqrt{{\mathrm{\Phi }}_1{\mathrm{\Psi }}_1})}^2}{2\sqrt{{\mathrm{\Phi }}_1{\mathrm{\Psi }}_1{\mathrm{\Phi }}_2{\mathrm{\Psi }}_2}}{\mathrm{\nabla }}_q^{-\nu }(fg)$.

• (c) $0\le {\mathrm{\nabla }}_q^{-\nu }(f^2){\mathrm{\nabla }}_q^{-\nu }(g^2)-{({\mathrm{\nabla }}_q^{-\nu }(fg))}^2\le \frac{{({\mathrm{\Phi }}_2{\mathrm{\Psi }}_2-{\mathrm{\Phi }}_1{\mathrm{\Psi }}_1)}^2}{4{\mathrm{\Phi }}_1{\mathrm{\Psi }}_1{\mathrm{\Phi }}_2{\mathrm{\Psi }}_2}{({\mathrm{\nabla }}_q^{-\nu }(fg))}^2$.

Proof Under the conditions satisfied by the functions f and g, we have

$$\frac{{\mathrm{\Phi }}_1}{{\mathrm{\Psi }}_2}\le \frac{f(\tau )}{g(\tau )}\le \frac{{\mathrm{\Phi }}_2}{{\mathrm{\Psi }}_1}.$$

Applying Theorem 6, we get the inequality (a) and using it, we have (b) and (c). □

Corollary 2 Let f be a positive function on ${\mathbb{T}}_{t_0}$ satisfying (19). Then the following inequality holds:

$${\mathrm{\nabla }}_q^{-\nu }\left(f^2\right)\le \frac{{\mathrm{\Gamma }}_q(\nu +1){({\mathrm{\Phi }}_1+{\mathrm{\Phi }}_2)}^2}{4t^{\underline{(\nu )}}{\mathrm{\Phi }}_1{\mathrm{\Phi }}_2}{({\mathrm{\nabla }}_q^{-\nu }(f))}^2.$$

Theorem 7 Let f and g be two positive functions on ${\mathbb{T}}_{t_0}$ and

$$0<m\le \frac{g(\tau )}{f(\tau )}\le M<\mathrm{\infty }$$

(20)

and $p\ne 0$ be a real number, then the following inequality holds:

$${\mathrm{\nabla }}_q^{-\nu }\left(f^{2-p}{g}^p\right)+\frac{mM(M^{p-1}-m^{p-1})}{M-m}{\mathrm{\nabla }}_q^{-\nu }\left(f^p\right)\le \frac{M^p-m^p}{M-m}{\mathrm{\nabla }}_q^{-\nu }(fg)$$

for $p\notin (0,1)$, or reverse for $p\in (0,1)$. Especially, for $p=2$, we have

$${\mathrm{\nabla }}_q^{-\nu }\left(g^2\right)+mM{\mathrm{\nabla }}_q^{-\nu }\left(f^2\right)\le (m+M){\mathrm{\nabla }}_q^{-\nu }(fg).$$

Proof The inequality is based on the Lah-Ribaric inequality [[18], p.9] and [[19], p.123]. □

Theorem 8 Let f and g be two positive functions on ${\mathbb{T}}_{t_0}$ and $p\ne 0$ be a real number. Then the following inequality holds:

$${({\mathrm{\nabla }}_q^{-\nu }(fg))}^p\le {\left({\mathrm{\nabla }}_q^{-\nu }\left(f^2\right)\right)}^{p-1}{\mathrm{\nabla }}_q^{-\nu }\left(f^{2-p}{g}^p\right)$$

for $p\notin (0,1)$, or reverse for $p\in (0,1)$.

Proof The above inequality is obtained via the Jensen inequality for the convex functions. □

Corollary 3 Let f be a positive function on ${\mathbb{T}}_{t_0}$ and $p\ne 0$ be a real number. Then the following inequality holds:

$${({\mathrm{\nabla }}_q^{-\nu }(f))}^p\le {\left(\frac{t^{\underline{(\nu )}}}{{\mathrm{\Gamma }}_q(\nu +1)}\right)}^{p-1}{\mathrm{\nabla }}_q^{-\nu }\left(f^p\right)$$

for $p\notin (0,1)$, or reverse for $p\in (0,1)$.

Theorem 9 Let p, f and g be three positive functions on ${\mathbb{T}}_{t_0}$ satisfying (19). If $0<\alpha \le \beta <1$, $\alpha +\beta =1$, then the following inequalities hold:

(21)

(22)

Proof Since $(\beta f(\tau )-\alpha {\mathrm{\Phi }}_1)(f(\tau )-{\mathrm{\Phi }}_2)\le 0$ on ${\mathbb{T}}_{t_0}$, we have

$$\beta f^2(\tau )-(\alpha {\mathrm{\Phi }}_1+\beta {\mathrm{\Phi }}_2)f(\tau )+\alpha {\mathrm{\Phi }}_1{\mathrm{\Phi }}_2\le 0.$$

(23)

Multiplying both sides of (23) by $p(\tau )/f(\tau )$, we get

$$\beta p(\tau )f(\tau )+\alpha {\mathrm{\Phi }}_1{\mathrm{\Phi }}_2\frac{p(\tau )}{f(\tau )}\le (\alpha {\mathrm{\Phi }}_1+\beta {\mathrm{\Phi }}_2)p(\tau ).$$

(24)

From (24) and arithmetric-geometric mean inequality, we obtain

(25)

which implies (21).

Replacing p and f by $pfg$ and $f/g$ in (25), respectively, and ${\mathrm{\Phi }}_1/{\mathrm{\Psi }}_2\le f(\tau )/g(\tau )\le {\mathrm{\Phi }}_2/{\mathrm{\Psi }}_1$, we get

which implies (22). □

Corollary 4 Let p, f and g be three positive functions on ${\mathbb{T}}_{t_0}$ satisfying (20). If $0<\alpha \le \beta <1$, $\alpha +\beta =1$, then the following inequality holds:

$$\alpha {\mathrm{\nabla }}_q^{-\nu }\left(p{g}^2\right)+\beta mM{\mathrm{\nabla }}_q^{-\nu }\left(p{f}^2\right)\le (\alpha m+\beta M){\mathrm{\nabla }}_q^{-\nu }(pfg).$$

(26)

Proof Replacing ${\mathrm{\Phi }}_1$, ${\mathrm{\Phi }}_2$ and $f(\tau )$ by m, M and $g(\tau )/f(\tau )$ in (24), and multiplying both sides by ${(t-q\tau )}^{\underline{(\nu -1)}}/{\mathrm{\Gamma }}_q(\nu )$ and integrating the resulting identity with respect to τ from 0 to t, we get (25). □

Theorem 10 Let p, f and g be three functions on ${\mathbb{T}}_{t_0}$ with $p(\tau )\ge 0$.

• (a) If there exist four constants ${\mathrm{\Phi }}_1,{\mathrm{\Phi }}_2,{\mathrm{\Psi }}_1,{\mathrm{\Psi }}_2\in \mathbb{R}$ such that $({\mathrm{\Phi }}_{2}g(\tau )-{\mathrm{\Psi }}_{1}f(\tau ))({\mathrm{\Psi }}_{2}f(\tau )-{\mathrm{\Phi }}_{1}g(\tau ))\ge 0$ for all $\tau >0$, then

  $$\begin{array}{rcl}{\mathrm{\Phi }}_1{\mathrm{\Phi }}_2{\mathrm{\nabla }}_q^{-\nu }\left(p{g}^2\right)+{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2{\mathrm{\nabla }}_q^{-\nu }\left(p{f}^2\right)& \le & ({\mathrm{\Phi }}_1{\mathrm{\Psi }}_1+{\mathrm{\Phi }}_2{\mathrm{\Psi }}_2){\mathrm{\nabla }}_q^{-\nu }(pfg)\\ \le & |{\mathrm{\Phi }}_1{\mathrm{\Psi }}_1+{\mathrm{\Phi }}_2{\mathrm{\Psi }}_2|({\mathrm{\nabla }}_q^{-\nu }\left(p{f}^2\right)+{\mathrm{\nabla }}_q^{-\nu }\left(p{g}^2\right)).\end{array}$$

  (27)

Moreover, if ${\mathrm{\Phi }}_1{\mathrm{\Phi }}_2{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2>0$, then

(28)

(29)

• (b) If there exist four constants ${\mathrm{\Phi }}_1,{\mathrm{\Phi }}_2,{\mathrm{\Psi }}_1,{\mathrm{\Psi }}_2\in \mathbb{R}$ such that $({\mathrm{\Phi }}_{2}g(\tau )-{\mathrm{\Psi }}_{1}f(\rho ))({\mathrm{\Psi }}_{2}f(\rho )-{\mathrm{\Phi }}_{1}g(\tau ))\ge 0$ for all $\tau ,\rho >0$, then

  

  (30)

• (c) If ${\mathrm{\Phi }}_1{\mathrm{\Phi }}_2>0$ and ${\mathrm{\Psi }}_1{\mathrm{\Psi }}_2>0$, then

  $${\mathrm{\Phi }}_1{\mathrm{\Phi }}_2{({\mathrm{\nabla }}_q^{-\nu }(pg))}^2+{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2{({\mathrm{\nabla }}_q^{-\nu }(pf))}^2\le ({\mathrm{\Phi }}_1{\mathrm{\Psi }}_1+{\mathrm{\Phi }}_2{\mathrm{\Psi }}_2){\mathrm{\nabla }}_q^{-\nu }(p){\mathrm{\nabla }}_q^{-\nu }(pfg).$$

  (31)

• (d) If ${\mathrm{\Phi }}_1{\mathrm{\Phi }}_2>0$ and ${\mathrm{\Psi }}_1{\mathrm{\Psi }}_2>0$, then

  $${\mathrm{\Phi }}_1{\mathrm{\Phi }}_2{({\mathrm{\nabla }}_q^{-\nu }(pg))}^2+{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2{({\mathrm{\nabla }}_q^{-\nu }(pf))}^2\le ({\mathrm{\Phi }}_1{\mathrm{\Psi }}_1+{\mathrm{\Phi }}_2{\mathrm{\Psi }}_2){\mathrm{\nabla }}_q^{-\nu }(pf){\mathrm{\nabla }}_q^{-\nu }(pg).$$

  (32)

Proof Case (a). It follows from the assumption that

$$p(\tau )({\mathrm{\Phi }}_{2}g(\tau )-{\mathrm{\Psi }}_{1}f(\tau ))({\mathrm{\Psi }}_{2}f(\tau )-{\mathrm{\Phi }}_{1}g(\tau ))\ge 0$$

for all $\tau \ge 0$, which implies that

$${\mathrm{\Phi }}_1{\mathrm{\Phi }}_{2}p(\tau )g^2(\tau )+{\mathrm{\Psi }}_1{\mathrm{\Psi }}_{2}p(\tau )f^2(\tau )\le ({\mathrm{\Phi }}_1{\mathrm{\Psi }}_1+{\mathrm{\Phi }}_2{\mathrm{\Psi }}_2)p(\tau )f(\tau )g(\tau ).$$

(33)

Multiplying both sides of (33) by ${(t-q\tau )}^{\underline{(\nu -1)}}/{\mathrm{\Gamma }}_q(\nu )$ and integrating the resulting identity with respect to τ from 0 to t, we obtain the left-hand side of (27). Furthermore, by Cauchy’s inequality, we get the right-hand side of (27).

Multiplying both sides of the inequality

$${\mathrm{\Phi }}_1{\mathrm{\Phi }}_2{\mathrm{\nabla }}_q^{-\nu }\left(p{g}^2\right)+{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2{\mathrm{\nabla }}_q^{-\nu }\left(p{f}^2\right)\le ({\mathrm{\Phi }}_1{\mathrm{\Psi }}_1+{\mathrm{\Phi }}_2{\mathrm{\Psi }}_2){\mathrm{\nabla }}_q^{-\nu }(pfg)$$

by $1/\sqrt{{\mathrm{\Phi }}_1{\mathrm{\Phi }}_2{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2}$, we get (28).

On the other hand, it follows from ${\mathrm{\Phi }}_1{\mathrm{\Phi }}_2{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2>0$ and ${(\sqrt{{\mathrm{\Phi }}_1{\mathrm{\Phi }}_2{\mathrm{\nabla }}_q^{-\nu }(p{g}^2)}-\sqrt{{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2{\mathrm{\nabla }}_q^{-\nu }(p{f}^2)})}^2\ge 0$ that

$$2\sqrt{{\mathrm{\Phi }}_1{\mathrm{\Phi }}_2{\mathrm{\nabla }}_q^{-\nu }\left(p{g}^2\right)}\sqrt{{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2{\mathrm{\nabla }}_q^{-\nu }\left(p{f}^2\right)}\le {\mathrm{\Phi }}_1{\mathrm{\Phi }}_2{\mathrm{\nabla }}_q^{-\nu }\left(p{g}^2\right)+{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2{\mathrm{\nabla }}_q^{-\nu }\left(p{f}^2\right).$$

(34)

According to (27) and (34), we have

$$4{\mathrm{\Phi }}_1{\mathrm{\Phi }}_2{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2{\mathrm{\nabla }}_q^{-\nu }\left(p{g}^2\right){\mathrm{\nabla }}_q^{-\nu }\left(p{f}^2\right)\le {({\mathrm{\Phi }}_1{\mathrm{\Psi }}_1+{\mathrm{\Phi }}_2{\mathrm{\Psi }}_2)}^2{({\mathrm{\nabla }}_q^{-\nu }(pfg))}^2,$$

which implies (29).

Case (b). It follows from the assumption that

$$p(\tau )p(\rho )({\mathrm{\Phi }}_{2}g(\tau )-{\mathrm{\Psi }}_{1}f(\rho ))({\mathrm{\Psi }}_{2}f(\rho )-{\mathrm{\Phi }}_{1}g(\tau ))\ge 0$$

for all $\tau ,\rho >0$, which implies that

(35)

Multiplying both sides of (35) by ${(t-q\tau )}^{\underline{(\nu -1)}}{(t-q\rho )}^{\underline{(\nu -1)}}/{\mathrm{\Gamma }}_q^2(\nu )$ and integrating the resulting identity with respect to τ and ρ from 0 to t, respectively, we obtain (30).

Case (c) and (d). It follows from Cauchy’s inequality that

$${({\mathrm{\nabla }}_q^{-\nu }(pf))}^2\le {\mathrm{\nabla }}_q^{-\nu }(p){\mathrm{\nabla }}_q^{-\nu }\left(p{f}^2\right),{({\mathrm{\nabla }}_q^{-\nu }(pg))}^2\le {\mathrm{\nabla }}_q^{-\nu }(p){\mathrm{\nabla }}_q^{-\nu }\left(p{g}^2\right).$$

Combining (a), (b) and the preceding two inequalities, we see that

$$\begin{array}{rcl}{\mathrm{\Phi }}_1{\mathrm{\Phi }}_2{({\mathrm{\nabla }}_q^{-\nu }(pg))}^2+{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2{({\mathrm{\nabla }}_q^{-\nu }(pf))}^2& \le & {\mathrm{\Phi }}_1{\mathrm{\Phi }}_2{\mathrm{\nabla }}_q^{-\nu }(p){\mathrm{\nabla }}_q^{-\nu }\left(p{f}^2\right)+{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2{\mathrm{\nabla }}_q^{-\nu }(p){\mathrm{\nabla }}_q^{-\nu }\left(p{g}^2\right)\\ \le & ({\mathrm{\Phi }}_1{\mathrm{\Psi }}_1+{\mathrm{\Phi }}_2{\mathrm{\Psi }}_2){\mathrm{\nabla }}_q^{-\nu }(p){\mathrm{\nabla }}_q^{-\nu }(pfg),\end{array}$$

which implies (31). Furthermore,

$$\begin{array}{rcl}{\mathrm{\Phi }}_1{\mathrm{\Phi }}_2{({\mathrm{\nabla }}_q^{-\nu }(pg))}^2+{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2{({\mathrm{\nabla }}_q^{-\nu }(pf))}^2& \le & {\mathrm{\Phi }}_1{\mathrm{\Phi }}_2{\mathrm{\nabla }}_q^{-\nu }(p){\mathrm{\nabla }}_q^{-\nu }\left(p{f}^2\right)+{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2{\mathrm{\nabla }}_q^{-\nu }(p){\mathrm{\nabla }}_q^{-\nu }\left(p{g}^2\right)\\ \le & ({\mathrm{\Phi }}_1{\mathrm{\Psi }}_1+{\mathrm{\Phi }}_2{\mathrm{\Psi }}_2){\mathrm{\nabla }}_q^{-\nu }(pf){\mathrm{\nabla }}_q^{-\nu }(pg),\end{array}$$

which implies (32). □

Theorem 11 Let p, f and g be three positive functions on ${\mathbb{T}}_{t_0}$ with $p(\tau )\ge 0$. Then we have

$$\begin{array}{rcl}{({\mathrm{\nabla }}_q^{-\nu }(p){\mathrm{\nabla }}_q^{-\nu }(pfg)+{\mathrm{\nabla }}_q^{-\nu }(pf){\mathrm{\nabla }}_q^{-\nu }(pg))}^2& \le & ({\mathrm{\nabla }}_q^{-\nu }(p){\mathrm{\nabla }}_q^{-\nu }\left(p{f}^2\right)+{({\mathrm{\nabla }}_q^{-\nu }(pf))}^2)\\ \times ({\mathrm{\nabla }}_q^{-\nu }(p){\mathrm{\nabla }}_q^{-\nu }\left(p{g}^2\right)+{({\mathrm{\nabla }}_q^{-\nu }(pg))}^2).\end{array}$$

(36)

Moreover, under the assumptions of (a) and (b) in Theorem 10, the following inequality holds:

(37)

Proof First of all, we give the proof of (36). By Cauchy’s inequality and the element inequality $2xy\sqrt{uv}\le x^{2}u+y^{2}v$, for all $x,y,u,v\ge 0$, we have

which implies (36).

Next, we prove that (37) holds. It follows from (a) and (b) in Theorem 10 that

$$\begin{array}{rcl}({\mathrm{\Phi }}_1{\mathrm{\Psi }}_1+{\mathrm{\Phi }}_2{\mathrm{\Psi }}_2){\mathrm{\nabla }}_q^{-\nu }(p){\mathrm{\nabla }}_q^{-\nu }(pfg)& \ge & {\mathrm{\Phi }}_1{\mathrm{\Phi }}_2{\mathrm{\nabla }}_q^{-\nu }(p){\mathrm{\nabla }}_q^{-\nu }\left(p{g}^2\right)+{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2{\mathrm{\nabla }}_q^{-\nu }(p){\mathrm{\nabla }}_q^{-\nu }\left(p{f}^2\right)\\ \ge & {\mathrm{\Phi }}_1{\mathrm{\Phi }}_2{\mathrm{\nabla }}_q^{-\nu }(p){\mathrm{\nabla }}_q^{-\nu }\left(p{g}^2\right)+{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2{({\mathrm{\nabla }}_q^{-\nu }(pf))}^2,\\ ({\mathrm{\Phi }}_1{\mathrm{\Psi }}_1+{\mathrm{\Phi }}_2{\mathrm{\Psi }}_2){\mathrm{\nabla }}_q^{-\nu }(pf){\mathrm{\nabla }}_q^{-\nu }(pg)& \ge & {\mathrm{\Phi }}_1{\mathrm{\Phi }}_2{\mathrm{\nabla }}_q^{-\nu }(p){\mathrm{\nabla }}_q^{-\nu }\left(p{g}^2\right)+{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2{\mathrm{\nabla }}_q^{-\nu }(p){\mathrm{\nabla }}_q^{-\nu }\left(p{f}^2\right)\\ \ge & {\mathrm{\Phi }}_1{\mathrm{\Phi }}_2{({\mathrm{\nabla }}_q^{-\nu }(pg))}^2+{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2{\mathrm{\nabla }}_q^{-\nu }(p){\mathrm{\nabla }}_q^{-\nu }\left(p{f}^2\right).\end{array}$$

Combining the preceding two inequalities and the element inequality ${(x+y)}^2\ge 4xy$, we see that

which implies (37). □