The Minkowski's inequality by means of a generalized fractional integral

TL;DR

This paper generalizes Minkowski's inequality using a fractional integral introduced by Katugampola, establishing new theorems and related inequalities in the context of fractional calculus.

Contribution

It introduces a novel generalization of reverse Minkowski's inequality via a recently proposed fractional integral, expanding the theoretical framework.

Findings

New theorems on fractional Minkowski's inequality

Generalization of reverse Minkowski's inequality

Additional inequalities related to fractional operators

Abstract

We use the definition of a fractional integral, recently proposed by Katugampola, to establish a generalization of the reverse Minkowski's inequality. We show two new theorems associated with this inequality, as well as state and show other inequalities related to this fractional operator.

Equations

$${\Vert f\Vert }_{X_c^p}={\left({\int }_a^b|x^{c}f(x){|}^p\frac{dx}{x}\right)}^{1/p}(1\le p<\infty )$$

$${\Vert f\Vert }_{X_c^{\infty }}=\sup {\text{ess}}_{x\in (a,b)}[x^c|f(x)|].$$

$${\left({\int }_a^b{f}^p\left(t\right)dt\right)}^{\frac{1}{p}}+{\left({\int }_a^b{g}^p\left(t\right)dt\right)}^{\frac{1}{p}}\le c_1{\left({\int }_a^b\left(f^p+g^p\right)\left(t\right)dt\right)}^{\frac{1}{p}},$$

$${\left({\int }_a^b{f}^p\left(t\right)dt\right)}^{\frac{2}{p}}+{\left({\int }_a^b{g}^p\left(t\right)dx\right)}^{\frac{2}{p}}\ge c_2{\left({\int }_a^b{f}^p\left(t\right)dx\right)}^{\frac{1}{p}}{\left({\int }_a^b{g}^p\left(t\right)dx\right)}^{\frac{1}{p}},$$

$$J_{a^{+}}^{\alpha }f\left(x\right):=\frac{1}{\Gamma \left(\alpha \right)}{\int }_a^x\frac{f\left(t\right)}{{\left(x-t\right)}^{1-\alpha }}dt,x>a$$

$$J_{b^{-}}^{\alpha }f\left(x\right):=\frac{1}{\Gamma \left(\alpha \right)}{\int }_x^b\frac{f\left(t\right)}{{\left(t-x\right)}^{1-\alpha }}dt,x<b,$$

$$H_{a^{+}}^{\alpha }f\left(x\right):=\frac{1}{\Gamma \left(\alpha \right)}{\int }_a^x{\left(\log \frac{x}{t}\right)}^{\alpha -1}\frac{f\left(t\right)}{t}dt,a<x<b$$

$$H_{b^{-}}^{\alpha }f\left(x\right):=\frac{1}{\Gamma \left(\alpha \right)}{\int }_x^b{\left(\log \frac{t}{x}\right)}^{\alpha -1}\frac{f\left(t\right)}{t}dt\alpha ,a<x<b,$$

$$I_{a^{+},\sigma ,\eta }^{\alpha }f\left(x\right):=\frac{\sigma x^{-\sigma \left(\alpha +\eta \right)}}{\Gamma \left(\alpha \right)}{\int }_a^x\frac{t^{\sigma \left(\eta +1\right)-1}}{{\left(x^{\sigma }-t^{\sigma }\right)}^{1-\alpha }}f\left(t\right)dt,0\le a<x<b\le \infty$$

$$I_{b^{-},\sigma ,\eta }^{\alpha }f\left(x\right):=\frac{\sigma x^{\sigma \eta }}{\Gamma \left(\alpha \right)}{\int }_x^b\frac{t^{\sigma \left(1-\eta -\alpha \right)-1}}{{\left(t^{\sigma }-x^{\sigma }\right)}^{1-\alpha }}f\left(t\right)dt,0\le a<x<b\le \infty ,$$

$${}^{\rho }{I}_{a^{+}}^{\alpha }f\left(x\right):=\frac{{\rho }^{1-\alpha }}{\Gamma \left(\alpha \right)}{\int }_a^x\frac{t^{\rho -1}}{{\left(x^{\rho }-t^{\rho }\right)}^{1-\alpha }}f\left(t\right)dt,x>a$$

$${}^{\rho }{I}_{b^{-}}^{\alpha }f\left(x\right):=\frac{{\rho }^{1-\alpha }}{\Gamma \left(\alpha \right)}{\int }_x^b\frac{t^{\rho -1}}{{\left(t^{\rho }-x^{\rho }\right)}^{1-\alpha }}f\left(t\right)dt,x<b,$$

$${}_x{W}_{\infty }^{\alpha }{=}_x{I}_{\infty }^{\alpha }f\left(x\right):=\frac{1}{\Gamma \left(\alpha \right)}{\int }_{-\infty }^x\frac{f\left(t\right)}{{\left(x-t\right)}^{1-\alpha }}dt$$

$${}_{-\infty }{W}_x^{\alpha }{=}_{-\infty }{I}_x^{\alpha }f\left(x\right):=\frac{1}{\Gamma \left(\alpha \right)}{\int }_x^{\infty }\frac{f\left(t\right)}{{\left(t-x\right)}^{1-\alpha }}dt,$$

$$I_{+}^{\alpha }f\left(x\right):=\frac{1}{\Gamma \left(\alpha \right)}{\int }_{-\infty }^x\frac{f\left(t\right)}{{\left(x-t\right)}^{1-\alpha }}dt$$

$$I_{-}^{\alpha }f\left(x\right):=\frac{1}{\Gamma \left(\alpha \right)}{\int }_x^{\infty }\frac{f\left(t\right)}{{\left(t-x\right)}^{1-\alpha }}dt,$$

$${\left(J^{\alpha }{f}^p\left(x\right)\right)}^{\frac{1}{p}}+{\left(J^{\alpha }{g}^p\left(x\right)\right)}^{\frac{1}{p}}\le c_1{\left(J^{\alpha }{\left(f+g\right)}^p\left(x\right)\right)}^{\frac{1}{p}},$$

$${\left(J^{\alpha }{f}^p\left(x\right)\right)}^{\frac{2}{p}}+{\left(J^{\alpha }{g}^p\left(x\right)\right)}^{\frac{2}{p}}\ge c_2{\left(J^{\alpha }{f}^p\left(x\right)\right)}^{\frac{1}{p}}{\left(J^{\alpha }{g}^p\left(x\right)\right)}^{\frac{1}{p}},$$

$${\left(H_1^{\alpha }{f}^p\left(x\right)\right)}^{\frac{1}{p}}+{\left(H_1^{\alpha }{g}^p\left(x\right)\right)}^{\frac{1}{p}}\le c_1{\left(H_1^{\alpha }{\left(f+g\right)}^p\left(x\right)\right)}^{\frac{1}{p}},$$

$${\left(H_1^{\alpha }{f}^p\left(x\right)\right)}^{\frac{2}{p}}+{\left(H_1^{\alpha }{g}^p\left(x\right)\right)}^{\frac{2}{p}}\ge c_2{\left(H_1^{\alpha }{f}^p\left(x\right)\right)}^{\frac{1}{p}}{\left(H_1^{\alpha }{g}^p\left(x\right)\right)}^{\frac{1}{p}}$$

$${}^{\rho }{\mathcal{I}}_{a+,\eta ,\kappa }^{\alpha ,\beta }\phi \left(x\right):=\frac{{\rho }^{1-\beta }{x}^{\kappa }}{\Gamma \left(\alpha \right)}{\int }_a^x\frac{{\tau }^{\rho \left(\eta +1\right)-1}}{{\left(x^{\rho }-{\tau }^{\rho }\right)}^{1-\alpha }}\phi \left(\tau \right)d\tau ,0\le a<x<b\le \infty$$

$${}^{\rho }{\mathcal{I}}_{b-,\eta ,\kappa }^{\alpha ,\beta }\phi \left(x\right):=\frac{{\rho }^{1-\beta }{x}^{\rho \eta }}{\Gamma \left(\alpha \right)}{\int }_x^b\frac{{\tau }^{\kappa +\rho -1}}{{\left({\tau }^{\rho }-x^{\rho }\right)}^{1-\alpha }}\phi \left(\tau \right)d\tau ,0\le a<x<b\le \infty$$

$${\left({}^{\rho }{\mathcal{I}}_{a+,\eta ,\kappa }^{\alpha ,\beta }{f}^p\left(x\right)\right)}^{\frac{1}{p}}+{\left({}^{\rho }{\mathcal{I}}_{a+,\eta ,\kappa }^{\alpha ,\beta }{g}^p\left(x\right)\right)}^{\frac{1}{p}}\le c_1{\left({}^{\rho }{\mathcal{I}}_{a+,\eta ,\kappa }^{\alpha ,\beta }{\left(f+g\right)}^p\left(x\right)\right)}^{\frac{1}{p}},$$

$$f\left(t\right)\le M\left(f\left(t\right)+g\left(t\right)\right)-Mf\left(t\right),$$

$${\left(M+1\right)}^p{f}^p\left(t\right)\le M^p{\left(f\left(t\right)+g\left(t\right)\right)}^p.$$

$$\frac{{\left(M+1\right)}^p{\rho }^{1-\beta }{x}^{\kappa }}{\Gamma \left(\alpha \right)}{\int }_a^x\frac{t^{\rho \left(\eta +1\right)-1}}{{\left(x^{\rho }-t^{\rho }\right)}^{1-\alpha }}{f}^p\left(t\right)dt\le \frac{M^p{\rho }^{1-\beta }{x}^{\kappa }}{\Gamma \left(\alpha \right)}{\int }_a^x\frac{t^{\rho \left(\eta +1\right)-1}}{{\left(x^{\rho }-t^{\rho }\right)}^{1-\alpha }}{\left(f+g\right)}^p\left(t\right)dt.$$

$${\left({}^{\rho }{\mathcal{I}}_{a+,\eta ,\kappa }^{\alpha ,\beta }{f}^p\left(x\right)\right)}^{\frac{1}{p}}\le \frac{M}{M+1}{\left({}^{\rho }{\mathcal{I}}_{a+,\eta ,\kappa }^{\alpha ,\beta }{\left(f+g\right)}^p\left(x\right)\right)}^{\frac{1}{p}}.$$

$${\left(1+\frac{1}{m}\right)}^p{g}^p\left(t\right)\le {\left(\frac{1}{m}\right)}^p{\left(f\left(t\right)+g\left(t\right)\right)}^p.$$

$${\left({}^{\rho }{\mathcal{I}}_{a+,\eta ,\kappa }^{\alpha ,\beta }{g}^p\left(t\right)\right)}^{\frac{1}{p}}\le \frac{1}{m+1}{\left({}^{\rho }{\mathcal{I}}_{a+,\eta ,\kappa }^{\alpha ,\beta }{\left(f+g\right)}^p\left(t\right)\right)}^{\frac{1}{p}}.$$

$${\left({}^{\rho }{\mathcal{I}}_{a+,\eta ,\kappa }^{\alpha ,\beta }{f}^p\left(x\right)\right)}^{\frac{2}{p}}+{\left({}^{\rho }{\mathcal{I}}_{a+,\eta ,\kappa }^{\alpha ,\beta }{g}^p\left(x\right)\right)}^{\frac{2}{p}}\ge c_2{\left({}^{\rho }{\mathcal{I}}_{a+,\eta ,\kappa }^{\alpha ,\beta }{f}^p\left(x\right)\right)}^{\frac{1}{p}}{\left({}^{\rho }{\mathcal{I}}_{a+,\eta ,\kappa }^{\alpha ,\beta }{g}^p\left(x\right)\right)}^{\frac{1}{p}}$$

Full text

Keywords: Minkowski’s inequality, Generalized fractional integral.

MSC 2010 subject classifications. 26A33; 26A39; 26A42.

1. Introduction

Studies involving integral inequalities are important in several areas of science: mathematics, physics, engineering, among others, in particular we mention: initial value problem, linear transformation stability, integral-differential equations, and impulse equations [1, 2].

The space of $p$-integrable functions $L^{p}(a,b)$ play a relevant role in the study of inequalities involving integrals and sums. Further, it is possible to extend this space of $p$-integrable functions, to the space of the measurable Lebesgue functions, denoted by $X_{c}^{p}(a,b)$, in which the space $L^{p}(a,b)$ is contained [3]. Thus, new results involving integral inequalities have been possible and consequently, some applications have been made [1, 2]. We mention few of them, the inequalities of: Minkowski, Hölder, Hardy, Hermite-Hadamard, Jensen, among others [4, 5, 6, 7, 8, 9, 10].

On the other hand, non-integer order calculus, usually referred to as fractional calculus, is used to generalizes of integrals and derivatives, in particular integrals involving inequalities. There are many definitions of fractional integrals, for example: Riemann-Liouville, Hadamard, Liouville, Weyl, Erdéryi-Kober and Katugampola [3, 11, 12, 13]. Recently, Khalil et al. [14] and Adeljawad [15], introduced the local conformable fractional integrals and derivatives. From such fractional integrals, one obtains generalizations of the inequalities: Hadamard, Hermite-Hadamard, Opial, Gruss, Ostrowski, among others [16, 17, 18, 19, 20, 21, 22].

Recently, Katugampola [23] proposed a fractional integral unifying other well known ones: Riemann-Liouville, Hadamard, Weyl, Liouville and Erdélyi-Kober. Motivated by this formulation, we present a generalization of the reverse Minkowski’s inequality [24, 25, 26], using the fractional integral introduced by Katugampola. We point out that studies in this direction, involving fractional integrals, are growing in several branches of mathematics [18, 27, 28].

The work is organized as follows: In section 2, we present the definition of the fractional integral, as well as its particular cases. We present the main theorems involving the reverse Minkowski’s inequality, as well as the suitable spaces for such definitions. In section 3, our main result, we propose the reverse Minkowski’s inequality using the fractional integral. In section 4, we discuss other inequalities involving this fractional integral. Concluding remarks close the article.

2. Prelimiaries

In this section, we present the reverse Minkowski’s inequality theorem associated with the classical Riemann integral and its respective generalization via Riemann-Liouville and Hadamard fractional integrals. In addition, we present the fractional integral introduced by Katugampola, and we conclude with a theorem in order to recover particular cases.

Erhan et al. [5] address the inequalities of Hermite-Hadamard and reverse Minkowski for two functions $f$ and $g$ by means of the classical Riemann integral. On the other hand, Lazhar [7] also proposed a work related to the inequality involving integrals, that is, Hardy’s inequality and the reverse Minkowski’s inequality. Two theorems below were motivation for the works performed so far, via the Riemann-Liouville and Hadamard integrals, involving the reverse Minkowski’s inequality.

Definition 1**.**

The space ${X}_{c}^{p}(a,b)$ $(c\in\mathbb{R},1\leq{p}\leq{\infty})$ consists of those complex-valued Lebesgue measurable functions $f$ on $(a,b)$, for which ${\lVert f\rVert}_{{X_{c}^{p}}}<\infty$ with

[TABLE]

and

[TABLE]

In particular, when $c=1/p$ the space ${X}_{c}^{p}(a,b)$ coincides with the space $L^{p}(a,b)$ [3].

Theorem 1**.**

Let $f,g\in L^{p}(a,b)$ be two positive functions, with $1\leq{p}\leq{\infty}$, $0<\displaystyle\int_{a}^{b}f^{p}\left(t\right)dt<\infty$ and $0<\displaystyle\int_{a}^{b}g^{p}\left(t\right)dt<\infty$. If $0<m\leq\displaystyle\frac{f\left(t\right)}{g\left(t\right)}\leq M$, for $m,M\in\mathbb{R}^{*}_{+}$ and $\forall t\in[a,b]$, then

[TABLE]

with $c_{1}=\displaystyle\frac{M\left(m+1\right)+\left(M+1\right)}{\left(m+1\right)\left(M+1\right)}$ [5].

Theorem 2**.**

Let $f,g\in L^{p}(a,b)$ be two positive functions, with $1\leq{p}\leq{\infty}$, $0<\displaystyle\int_{a}^{b}f^{p}\left(t\right)dt<\infty$ and $0<\displaystyle\int_{a}^{b}g^{p}\left(t\right)dt<\infty$. If $0<m\leq\displaystyle\frac{f\left(t\right)}{g\left(t\right)}\leq M$, for $m,M\in\mathbb{R}^{*}_{+}$ and $\forall t\in[a,b]$, then

[TABLE]

with $c_{2}=\displaystyle\frac{\left(M+1\right)\left(m+1\right)}{M}-2$ [5].

We present the definitions of the fractional integrals that will be useful in the development of the article: Riemann-Liouville fractional integral, Hadamard integral, Erdélyi-Kober integral, Katugampola integral, Weyl integral and Liouville integral.

Definition 2**.**

Let $[a,b]$ $(-\infty<a<b<\infty)$ be a finite interval on the real-axis $\mathbb{R}$. The Riemann-Liouville fractional integrals $(\mbox{left-sided and right-sided})$ of order $\alpha\in\mathbb{C}$, $\rm Re(\alpha)>0$, are defined by

[TABLE]

and

[TABLE]

respectively [3, 12].

Definition 3**.**

Let $(a,b)$ $(0\leq a<b<\infty)$ be a finite or infinite interval on the half-axis $\mathbb{R}^{+}$. The Hadamard fractional integrals $(\mbox{left-sided and right-sided})$ of order $\alpha\in\mathbb{C}$, $\rm Re(\alpha)>0$ of a real function $f\in L^{p}(a,b)$ are defined by

[TABLE]

and

[TABLE]

respectively [3, 12].

Definition 4**.**

Let $(a,b)$ $(-\infty\leq a<b\leq\infty)$ be a finite or infinite interval or half-axis $\mathbb{R}^{+}$. Also let $\rm Re(\alpha)>0$, $\sigma>0$ and $\eta\in\mathbb{C}$. The Erdélyi-Kober fractional integrals $(\mbox{left-sided and right-sided})$ of order $\alpha\in\mathbb{C}$ of a real function $f\in L^{p}(a,b)$ are defined by

[TABLE]

and

[TABLE]

respectively [3, 12].

Definition 5**.**

Let $[a,b]\subset\mathbb{R}$ be a finite interval. Then the Katugampola fractional integrals $(\mbox{left-sided and right-sided})$ of order $\alpha\in\mathbb{C}$, $\rho>0$, $\rm Re(\alpha)>0$ of a real function $f\in X^{p}_{c}(a,b)$ are defined by

[TABLE]

and

[TABLE]

respectively [13].

Definition 6**.**

The Weyl fractional integrals of order $\alpha\in\mathbb{C}$, $\rm Re(\alpha)>0$ of a real function $f$ locally integrated into $(-\infty,\infty)$ being $-\infty\leq x\leq\infty$ are defined by

[TABLE]

and

[TABLE]

respectively [29].

Definition 7**.**

Let a continuous function by parts in $\mathbb{R}=(-\infty,\infty)$. The Liouville fractional integrals $(\mbox{left-sided and right-sided})$ of order $\alpha\in\mathbb{C}$, $\rm Re(\alpha)>0$ of a real function $f$, are defined by

[TABLE]

and

[TABLE]

respectively [3, 12].

Zoubir [25] established the reverse Minkowski’s inequality and another result that refers to the inequality via Riemann-Liouville fractional integral according to the following two theorems.

Theorem 3**.**

Let $\alpha>0$, $p\geq 1$ and $f,g$ two positive functions in $[0,\infty)$, such that $\forall x>0$, $J^{\alpha}f^{p}\left(x\right)<\infty$ and $J^{\alpha}g^{p}\left(x\right)<\infty$. If $0<m\leq\displaystyle\frac{f\left(t\right)}{g\left(t\right)}\leq M$, for $m,M\in\mathbb{R}^{*}_{+}$ and $\forall t\in[0,x]$, then

[TABLE]

where $c_{1}=\displaystyle\frac{M\left(m+1\right)+\left(M+1\right)}{\left(m+1\right)\left(M+1\right)}$ [25].

Theorem 4**.**

Let $\alpha>0$, $p\geq 1$ and $f,g$ two positive functions in $[0,\infty)$, such that $\forall x>0$, $J^{\alpha}f^{p}\left(x\right)<\infty$ and $J^{\alpha}g^{p}\left(x\right)<\infty$. If $0<m\leq\displaystyle\frac{f\left(t\right)}{g\left(t\right)}\leq M$, for $m,M\in\mathbb{R}^{*}_{+}$ e $\forall t\in[0,x]$, then

[TABLE]

where $c_{2}=\displaystyle\frac{\left(M+1\right)\left(m+1\right)}{M}-2$ [25].

In 2014, Chinchane et al. [26] and Sabrina et al. [30] also established the reverse Minkowski’s inequality via Hadamard fractional integral as in two theorems below.

Theorem 5**.**

Let $\alpha>0$, $p\geq 1$ and $f,g$ two positive functions in $[0,\infty)$, such that $\forall x>0$, $H_{1}^{\alpha}f^{p}\left(x\right)<\infty$ and $H_{1}^{\alpha}g^{p}\left(x\right)<\infty$. If $0<m\leq\displaystyle\frac{f\left(t\right)}{g\left(t\right)}\leq M$, for $m,M\in\mathbb{R}^{*}_{+}$ e $\forall t\in[0,x]$, then

[TABLE]

where $c_{1}=\displaystyle\frac{M\left(m+1\right)+\left(M+1\right)}{\left(m+1\right)\left(M+1\right)}$ [26, 30].

Theorem 6**.**

Let $\alpha>0$, $p\geq 1$ and $f,g$ two positive functions in $[0,\infty)$, such that $\forall x>0$, $H_{1}^{\alpha}f^{p}\left(x\right)<\infty$ and $H_{1}^{\alpha}g^{p}\left(x\right)<\infty$. If $0<m\leq\displaystyle\frac{f\left(t\right)}{g\left(t\right)}\leq M$, for $m,M\in\mathbb{R}^{*}_{+}$ e $\forall t\in[0,x]$, then

[TABLE]

where $c_{2}=\displaystyle\frac{\left(M+1\right)\left(m+1\right)}{M}-2$ [26, 30].

In 2014 Chinchane et al. [31] and recently Chinchane [32], established the reverse Minkowski’s inequality via fractional integral of Saigo and the $k$-fractional integral, respectively.

In 2017, Katugampola [23] introduced a fractional integral that unifies the six fractional integrals above mentioned. Finally, we introduce this integral and with a theorem we study their respective particular cases.

Definition 8**.**

Let $\varphi\in X^{p}_{c}(a,b)$, $\alpha>0$ and $\beta,\rho,\eta,\kappa\in\mathbb{R}$. Then, the fractional integrals of a function $f$, left and right, are given by

[TABLE]

and

[TABLE]

respectively, if integrals exist [23].

From now on, let’s work only with the integral on the left, Eq.(2.19), because with the right integral we have a similar treatment.

Theorem 7**.**

Let $\alpha>0$ and $\beta,\rho,\eta,\kappa\in\mathbb{R}$. Then for $f\in X^{p}_{c}(a,b)$, with $a<x<b$, we have [23]:

1. (1)

For $\kappa=0$, $\eta=0$ and the limit $\rho\rightarrow 1$, at Eq.(2.19), we get the Riemann-Liouville fractional integral, i.e; Eq.(2.3). 2. (2)

With $\beta=\alpha$, $\kappa=0$, $\eta=0$, we take the limit $\rho\rightarrow 0^{+}$ and using the $\ell$’Hospital role, at Eq.(2.19), we get the Hadamard fractional integral, i.e; Eq.(2.5). 3. (3)

In the case $\beta=0$ and $\kappa=-\rho(\alpha+\eta)$, at Eq.(2.19), we get the Erdélyi-Kober fractional integral, i.e; Eq.(2.7). 4. (4)

For $\beta=\alpha$, $\kappa=0$ and $\eta=0$, at Eq.(2.19), we get Katugampola fractional integral, i.e; Eq.(2.9). 5. (5)

With $\kappa=0$, $\eta=0$, $a=-\infty$ and take the limit $\rho\rightarrow 1$, at Eq.(2.19), we get Weyl fractional integral, i.e; Eq.(2.11). 6. (6)

With $\kappa=0$, $\eta=0$, $a=0$ and take the limit $\rho\rightarrow 1$, at Eq.(2.19), we get Liouville fractional integral, i.e; Eq.(2.13).

3. Reverse Minkowski fractional integral inequality

In this section, our main contribution, we establish and prove the reverse Minkowski’s inequality via generalized fractional integral Eq.(2.19) and a theorem that refers to the reverse Minkowski’s inequality.

Theorem 1**.**

Let $\alpha>0$, $\rho,\eta,\kappa,\beta\in\mathbb{R}$ and $p\geq 1$. Let $f,g\in X^{p}_{c}(a,x)$ two positive functions in $[0,\infty)$, such that $\forall x>a$, ${}^{\rho}\mathcal{I}_{a+,\eta,\kappa}^{\alpha,\beta}f^{p}\left(x\right)<\infty$ and ${}^{\rho}\mathcal{I}_{a+,\eta,\kappa}^{\alpha,\beta}g^{p}\left(x\right)<\infty$. If $0<m\leq\displaystyle\frac{f\left(t\right)}{g\left(t\right)}\leq M$, for $m,M\in\mathbb{R}^{*}_{+}$ and $\forall t\in\left[a,x\right]$, then

[TABLE]

with $c_{1}=\displaystyle\frac{M\left(m+1\right)+\left(M+1\right)}{\left(m+1\right)\left(M+1\right)}$.

Proof.

Using the condition $\displaystyle\frac{f\left(t\right)}{g\left(t\right)}\leq M$, $t\in[a,x]$, we can write

[TABLE]

which implies,

[TABLE]

Multiplying by $\displaystyle\frac{\rho^{1-\beta}x^{\kappa}t^{\rho\left(\eta+1\right)-1}}{\Gamma\left(\alpha\right)\left(x^{\rho}-t^{\rho}\right)^{1-\alpha}}$ both sides of Eq.(3.2) and integrating with respect to the variable $t$, we have

[TABLE]

Consequently, we can write

[TABLE]

On the other hand, as $mg\left(t\right)\leq f\left(t\right)$, follow

[TABLE]

Further, multiplying by $\displaystyle\frac{\rho^{1-\beta}x^{\kappa}t^{\rho\left(\eta+1\right)-1}}{\Gamma\left(\alpha\right)\left(x^{\rho}-t^{\rho}\right)^{1-\alpha}}$ both sides of Eq.(3.5) and integrating with respect to the variable $t$, we have

[TABLE]

From Eq.(3.4) and Eq.(3.6), the result follows.

Eq.(3.1) is the so-called reverse Minkowski’s inequality associated with the Katugampola fractional integral.

Theorem 2**.**

Let $\alpha>0$, $\rho,\eta,\kappa,\beta\in\mathbb{R}$ and $p\geq 1$. Let $f,g\in X^{p}_{c}(a,x)$ be two positive functions in $[0,\infty)$, such that $\forall x>a$, ${}^{\rho}\mathcal{I}_{a+,\eta,\kappa}^{\alpha,\beta}f^{p}\left(x\right)<\infty$ and ${}^{\rho}\mathcal{I}_{a+,\eta,\kappa}^{\alpha,\beta}g^{p}\left(x\right)<\infty$. If $0<m\leq\displaystyle\frac{f\left(t\right)}{g\left(t\right)}\leq M$, for $m,M\in\mathbb{R}^{*}_{+}$ and $\forall t\in\left[a,x\right]$, then

[TABLE]

with $c_{2}=\displaystyle\frac{\left(M+1\right)\left(m+1\right)}{M}-2$.

Proof.

Carrying out the product between Eq.(3.4) and Eq.(3.6), we have

[TABLE]

Using the Minkowski’s inequality, on the right side of Eq.(3.8), we have

[TABLE]

So, from Eq.(3), we conclude that

[TABLE]

Note that, if $\beta=\alpha$, $\kappa=0$, $\eta=0$ and the limit $\rho\rightarrow 1$, in Eq.(2.19), we recover Riemann-Liouville fractional integral, Eq.(2.3). In this sense, choosing $a+=0$, and substituting in Theorem 8 and Theorem 9, we obtain, as particular cases, the respective Theorem 3 and Theorem 4, which correspond to the inequality via Riemann-Liouville fractional integral. On the other hand, if $\beta=\alpha$, $\kappa=0$, $\eta=0$, and the limit $\rho\rightarrow 0+$ and using the $\ell$’Hospital rule, in Eq.(2.19), we obtain the Hadamard fractional integral, Eq.(2.5). Similarly, choosing $a=1$ and substituting in Theorem 8 and Theorem 9, we obtain, as particular cases, the Theorem 5 and Theorem 6, respectively.

4. Other fractional integral inequalities

In this section we generalize the results discussed by Chinchane [32], Sulaiman [33] and Sroysang [34] on the reverse Minkowski’s inequality via Riemann integral, using the fractional integral proposed by Katugampola [23].

Theorem 1**.**

Let $\alpha>0$, $\rho,\eta,\kappa,\beta\in\mathbb{R}$, $p\geq 1$ and $\frac{1}{p}+\frac{1}{q}=1$. Let $f,g\in X^{p}_{c}(a,x)$ be two positive functions in $[0,\infty)$, such that $\forall x>a$, ${}^{\rho}\mathcal{I}_{a+,\eta,\kappa}^{\alpha,\beta}f\left(x\right)<\infty$ and ${}^{\rho}\mathcal{I}_{a+,\eta,\kappa}^{\alpha,\beta}g\left(x\right)<\infty$. If $0<m\leq\displaystyle\frac{f\left(t\right)}{g\left(t\right)}\leq M$, for $m,M\in\mathbb{R}^{*}_{+}$ and $\forall t\in\left[a,x\right]$, then

[TABLE]

Proof.

Using the condition $\displaystyle\frac{f\left(t\right)}{g\left(t\right)}\leq M$, $t\in[a,x]$ with $x>a$, we have

[TABLE]

Multiplying by $f^{\frac{1}{p}}\left(t\right)$ both sides of Eq.(4.2), we can rewrite it as follows

[TABLE]

Now, multiplying by $\displaystyle\frac{\rho^{1-\beta}x^{\kappa}t^{\rho\left(\eta+1\right)-1}}{\Gamma\left(\alpha\right)\left(x^{\rho}-t^{\rho}\right)^{1-\alpha}}$ both sides of Eq.(4.3) and integrating with respect to the variable $t$, we have

[TABLE]

So, the inequality follows

[TABLE]

On the order hand, we have

[TABLE]

Multiplying by $g^{\frac{1}{q}}(t)$ both sides of Eq.(4.6) and using the relation $\frac{1}{p}+\frac{1}{q}=1$, we have

[TABLE]

Multiplying by $\displaystyle\frac{\rho^{1-\beta}x^{\kappa}t^{\rho\left(\eta+1\right)-1}}{\Gamma\left(\alpha\right)\left(x^{\rho}-t^{\rho}\right)^{1-\alpha}}$ both sides of Eq.(4.7) and integrating with respect to the variable $t$, we have

[TABLE]

Evaluating the product between Eq.(4.5) and Eq.(4.8) and using the relation $\frac{1}{p}+\frac{1}{q}=1$, we conclude that

[TABLE]

Theorem 2**.**

Let $\alpha>0$, $\rho,\eta,\kappa,\beta\in\mathbb{R}$, $p\geq 1$ and $\frac{1}{p}+\frac{1}{q}=1$. Let $f,g\in X^{p}_{c}(a,x)$ be two positive functions in $[0,\infty)$, such that $\forall x>a$, ${}^{\rho}\mathcal{I}_{a+,\eta,\kappa}^{\alpha,\beta}f^{p}\left(x\right)<\infty$, ${}^{\rho}\mathcal{I}_{a+,\eta,\kappa}^{\alpha,\beta}f^{q}\left(x\right)<\infty$, ${}^{\rho}\mathcal{I}_{a+,\eta,\kappa}^{\alpha,\beta}g^{p}\left(x\right)<\infty$ and ${}^{\rho}\mathcal{I}_{a+,\eta,\kappa}^{\alpha,\beta}g^{q}\left(x\right)<\infty$. If $0<m\leq\displaystyle\frac{f\left(t\right)}{g\left(t\right)}\leq M$, for $m,M\in\mathbb{R}^{*}_{+}$ and $\forall t\in\left[a,x\right]$, then

[TABLE]

with $c_{3}=\displaystyle\frac{2^{p-1}M^{p}}{p\left(M+1\right)^{p}}$ and $c_{4}=\displaystyle\frac{2^{p-1}}{q\left(m+1\right)^{q}}$.

Proof.

Using the hypothesis, we have the following identity

[TABLE]

Multiplying by $\displaystyle\frac{\rho^{1-\beta}x^{\kappa}t^{\rho\left(\eta+1\right)-1}}{\Gamma\left(\alpha\right)\left(x^{\rho}-t^{\rho}\right)^{1-\alpha}}$ both sides of Eq.(4.10) and integrating with respect to the variable $t$, we get

[TABLE]

In this way, we have

[TABLE]

On the other hand, as $0<m<\frac{f\left(t\right)}{g\left(t\right)}$, $t\in\left(a,x\right)$, we have

[TABLE]

Again, multiplying by $\displaystyle\frac{\rho^{1-\beta}x^{\kappa}t^{\rho\left(\eta+1\right)-1}}{\Gamma\left(\alpha\right)\left(x^{\rho}-t^{\rho}\right)^{1-\alpha}}$ both sides of Eq.(4.12) and integrating with respect to the variable $t$, we get

[TABLE]

Considering Young’s inequality, [35]

[TABLE]

multiplying by $\displaystyle\frac{\rho^{1-\beta}x^{\kappa}t^{\rho\left(\eta+1\right)-1}}{\Gamma\left(\alpha\right)\left(x^{\rho}-t^{\rho}\right)^{1-\alpha}}$ both sides of Eq.(4.14) and integrating with respect to the variable $t$, we have

[TABLE]

Thus, using Eq.(4.11), Eq.(4.13) and Eq.(4.15), we get

[TABLE]

Using the following inequality, $\left(a+b\right)^{r}\leq 2^{p-1}\left(a^{r}+b^{r}\right)$, $r>1$, $a,b\geq 0$, we get

[TABLE]

and

[TABLE]

Thus, replacing Eq.(4.17) and Eq.(4.18) at Eq.(4.16), we conclude that

[TABLE]

Theorem 3**.**

Let $\alpha>0$, $\rho,\eta,\kappa,\beta\in\mathbb{R}$ and $p\geq 1$. Let $f,g\in X^{p}_{c}(a,x)$ be two positive functions in $[0,\infty)$, such that $\forall x>a$, ${}^{\rho}\mathcal{I}_{a+,\eta,\kappa}^{\alpha,\beta}f^{p}\left(x\right)<\infty$ and ${}^{\rho}\mathcal{I}_{a+,\eta,\kappa}^{\alpha,\beta}g^{p}\left(x\right)<\infty$. If $0<m\leq\displaystyle\frac{f\left(t\right)}{g\left(t\right)}\leq M$, for $m,M\in\mathbb{R}^{*}_{+}$ and $\forall t\in\left[a,x\right]$, then

[TABLE]

Proof.

By hypothesis $0<c<m\leq M$, so

[TABLE]

Thus, we conclude that

[TABLE]

Also, we have

[TABLE]

which implies,

[TABLE]

Again, we have

[TABLE]

which implies,

[TABLE]

Multiplying by $\displaystyle\frac{\rho^{1-\beta}x^{\kappa}t^{\rho\left(\eta+1\right)-1}}{\Gamma\left(\alpha\right)\left(x^{\rho}-t^{\rho}\right)^{1-\alpha}}$ both sides of Eq.(4.20) and integrating with respect to the variable $t$, we have

[TABLE]

In this way, we obtain

[TABLE]

Realizing the same procedure as in Eq.(4.21), we have

[TABLE]

Adding Eq.(4) and Eq.(4), we conclude that

[TABLE]

Theorem 4**.**

Let $\alpha>0$, $\rho,\eta,\kappa,\beta\in\mathbb{R}$ and $p\geq 1$. Let $f,g\in X^{p}_{c}(a,x)$ be two positive functions in $[0,\infty)$, such that $\forall x>a$, ${}^{\rho}\mathcal{I}_{a+,\eta,\kappa}^{\alpha,\beta}f^{p}\left(x\right)<\infty$ and ${}^{\rho}\mathcal{I}_{a+,\eta,\kappa}^{\alpha,\beta}g^{p}\left(x\right)<\infty$. If $0\leq a\leq f(t)\leq A$ and $0\leq b\leq g(t)\leq B$, $\forall t\in\left[a,x\right]$, then

[TABLE]

with $c_{5}=\displaystyle\frac{A\left(a+B\right)+B\left(A+b\right)}{\left(A+b\right)\left(a+B\right)}$.

Proof.

By hypothesis, it follows that

[TABLE]

Realizing the product between Eq.(4.25) and $0<a\leq f\left(t\right)\leq A$, we have

[TABLE]

From Eq.(4.26), we get

[TABLE]

and

[TABLE]

Multiplying by $\displaystyle\frac{\rho^{1-\beta}x^{\kappa}t^{\rho\left(\eta+1\right)-1}}{\Gamma\left(\alpha\right)\left(x^{\rho}-t^{\rho}\right)^{1-\alpha}}$ both sides of Eq.(4.27) and integrating with respect to the variable $t$, we have

[TABLE]

Thus, it follows that

[TABLE]

Similarly, we performe the calculations for Eq.(4.28), we get

[TABLE]

Adding Eq.(4.29) and Eq.(4.30), we conclude that

[TABLE]

Theorem 5**.**

Let $\alpha>0$ and $\rho,\eta,\kappa,\beta\in\mathbb{R}$. Let $f,g\in X^{p}_{c}(a,x)$ be two positive functions in $[0,\infty)$, such that $\forall x>a$, ${}^{\rho}\mathcal{I}_{a+,\eta,\kappa}^{\alpha,\beta}f\left(x\right)<\infty$ and ${}^{\rho}\mathcal{I}_{a+,\eta,\kappa}^{\alpha,\beta}g\left(x\right)<\infty$. If $0<m\leq\displaystyle\frac{f\left(t\right)}{g\left(t\right)}\leq M$, for $m,M\in\mathbb{R}^{*}_{+}$ and $\forall t\in\left[a,x\right]$, then

[TABLE]

Proof.

Being $0<m\leq\displaystyle\frac{f\left(t\right)}{g\left(t\right)}\leq M$, $\forall t\in[a,x]$, we have

[TABLE]

Also, it follows that $\displaystyle\frac{1}{M}\leq\displaystyle\frac{g\left(t\right)}{f\left(t\right)}\leq\displaystyle\frac{1}{m}$, which implies,

[TABLE]

Evaluating the product between Eq.(4.32) and Eq.(4.33), we have

[TABLE]

Multiplying by $\displaystyle\frac{\rho^{1-\beta}x^{\kappa}t^{\rho\left(\eta+1\right)-1}}{\Gamma\left(\alpha\right)\left(x^{\rho}-t^{\rho}\right)^{1-\alpha}}$ both sides of Eq.(4.34) and integrating with respect to the variable $t$, we have

[TABLE]

with $c_{6}=\displaystyle\frac{1}{\left(m+1\right)\left(M+1\right)}$.

Thus, we conclude that

[TABLE]

Theorem 6**.**

Let $\alpha>0$, $\rho,\eta,\kappa,\beta\in\mathbb{R}$ and $p\geq 1$. Let $f,g\in X^{p}_{c}(a,x)$ be two positive functions in $[0,\infty)$, such that $\forall x>a$, ${}^{\rho}\mathcal{I}_{a+,\eta,\kappa}^{\alpha,\beta}f^{p}\left(x\right)<\infty$ and ${}^{\rho}\mathcal{I}_{a+,\eta,\kappa}^{\alpha,\beta}g^{p}\left(x\right)<\infty$. If $0<m\leq\displaystyle\frac{f\left(t\right)}{g\left(t\right)}\leq M$, for $m,M\in\mathbb{R}^{*}_{+}$ and $\forall t\in\left[a,x\right]$, then

[TABLE]

with $h\left(f\left(x\right),g\left(x\right)\right)=\max\left\{M\left[\left(\displaystyle\frac{M}{m}+1\right)f\left(x\right)-Mg\left(x\right)\right],\displaystyle\frac{\left(m+M\right)g\left(x\right)-f\left(x\right)}{m}\right\}$.

Proof.

From the hypothesis, $0<m\leq\displaystyle\frac{f\left(t\right)}{g\left(t\right)}\leq M$, $\forall t\in[a,x]$, we have

[TABLE]

and

[TABLE]

Thus, using Eq.(4.35) and Eq.(4.36), we get

[TABLE]

where $h\left(f\left(t\right),g\left(t\right)\right)=\max\left\{M\left[\left(\frac{M}{m}+1\right)f\left(t\right)-Mg\left(t\right)\right],\frac{\left(M+m\right)g\left(t\right)-f\left(t\right)}{m}\right\}.$

Using the hypothesis, it follows that $0<\displaystyle\frac{1}{M}\leq\displaystyle\frac{g\left(t\right)}{f\left(t\right)}\leq\displaystyle\frac{1}{m}$. In this way, we obtain

[TABLE]

and

[TABLE]

Then, from Eq.(4.38) and Eq.(4.39), we have

[TABLE]

which can be rewrite as

[TABLE]

Thus, using Eq.(4.37) and Eq.(4.40), we can write

[TABLE]

and

[TABLE]

Multiplying by $\displaystyle\frac{\rho^{1-\alpha}x^{\kappa}t^{\rho\left(\eta+1\right)-1}}{\Gamma\left(\alpha\right)\left(x^{\rho}-t^{\rho}\right)^{1-\alpha}}$ both sides of Eq.(4.41) and integrating with respect to the variable $t$, we have

[TABLE]

In this way, we obtain

[TABLE]

Using the same procedure as above, for Eq.(4.42), we have

[TABLE]

Thus, using Eq.(4.43) and Eq.(4.44), we conclude that

[TABLE]

Using Eq.(2.19) and Theorem 7 with the convenient conditions for each respective fractional integral, we have the previous theorems, that is, Theorem 10 to Theorem 15 introduced and demonstrated above, contain as particular cases, each result involving the following fractional integrals: Riemann-Liouville, Hadamard, Liouville, Weyl, Edérlyi-Kober, and Katugampola.

5. Concluding remarks

After a brief introduction to the fractional integral, proposed by Katugampola and fractional integrals in the sense of Riemann-Liouville and Hadamard, we generalize the reverse Minkowski’s inequality obtaining, as a particular case, the inequality involving the fractional integral in the Riemann-Liouville sense and Hadamard sense [23]. We also show other inequalities using the Katugampola fractional integral. The application of this fractional integral can be used to generalize several inequalities, among them, we mention the Gruss-type inequality, recently introduced and proved [36]. A continuation of this work, with this formulation of fractional integral, consists in generalize the inequalities of Hermite-Hadamard and Hermite-Hadamard-Féjer. Moreover, we will discuss inequalities via $M$-fractional integral according to [37].