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eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2020; 60(2): 289-295

Published online June 30, 2020

### Minkowski's Inequality for Variational Fractional Integrals

Azizollah Babakhani

Department of Mathematics, Faculty of Basic Science, Babol Noshirvani University of Technology, Shariati Ave., 47148-71167 Babol, Iran e-mail : babakhani@nit.ac.ir

Received: February 4, 2016; Revised: February 1, 2019; Accepted: March 4, 2019

### Abstract

Minkowski’s inequality is one of the most famous inequalities in mathematics, and has many applications. In this paper, we give Minkowski’s inequality for generalized variational integrals that are based on a supermultiplicative function. Our results include previous results about fractional integral inequalities of Minkowski’s type.

Keywords: inequality, Riemann-Liouville fractional integral, variational integral

### 1. Introduction

Minkowski’s inequality is unequivocally one of the most famous inequalities in mathematics. The well-known Minkowski integral inequality is given as follows:

### Theorem 1.1.([1])

Let s ≥ 1 and $∫abfs(x) dx$and $∫abgs(x) dx$be finite. Then

$(∫ab(fs(x)+gs(x)) dx)1s≤(∫abfs(x) dx)1s+(∫abgs(x)dx)1s.$

The following reverse Minkowski integral inequality was obtained by Bougoffa [3] in 2006.

### Theorem 1.2.([3])

Let f and g be positive functions satisfying

$0

then

$(∫abfs(x) dx)1s+(∫abgs(x)dx)1s≤c(∫ab(fs(x)+gs(x)) dx)1s,$

where $c=M(m+1)+(M+1)(m+1)(M+1)$.

Variational integrals, which generalize fractional integrals, play an important role in all fields of mathematics [2, 5]. They are versatile, and have wide application in applied mathematics.

In 2010, Agrawal [2] introduced a generalized variational integral which generalizes the Riemann-Liouville fractional integral.

### Definition 1.3.([2])

A generalized variational integral of order α of a real function f is defined as

$S⟨a,t,b,p,q⟩αf(t)=p∫atkα(t,s)f(s)ds+q∫tbkα(s,t)f(s)ds=SPαf(t),$

where t ∈ (a, b), p, q ∈ ℝ, P = ⟨a, t, b, p, q⟩ and kα(t, s) is a kernel which is nonnegative and depends upon a parameter α > 0.

### Remark 1.4

Let $kα(t,s):=1Γ(α)(t-s)α-1$ and P = ⟨a, t, b, p, q⟩.

• (i) If P = P1 = ⟨a, t, b, 1, 0⟩, then the left Riemann-Liouville fractional integral yields i.e,

$SP1αf(t)=∫at1Γ(α)(t-s)α-1f(s)ds=Ja+αf(t).$

• (ii) If P = P2 = ⟨a, t, b, 0, 1⟩, then the right Riemann-Liouville fractional integral is concluded as

$SP2αf(t)=∫tb1Γ(α)(s-t)α-1f(s)ds=Jb-αf(t).$

• (iii) If $P=P3=⟨a,t,b,12,12⟩$, then we have the Riesz fractional integral as follows

$SP3αf(t)=12SP1αf(t)+12SP2αf(t)=12Ja+αf(t)+12Jb-αf(t).$

In 2010, Dahmani [4] proved the following inequalities related to Minkowski’s inequality for Riemann-Liouville fractional integrals; these generalize the results in [3].

### Theorem 1.5.([4])

Let α > 0, s ≥ 1 and f, g be positive on [0, ∞) such that t > 0, $J0+αfs(t)<∞$and $J0+αgs(t)<∞$. If $0, τ ∈ [0, t], then

$[J0+αfs(t)]1s+[J0+αgs(t)]1s≤1+M(m+2)(m+1)(M+1)[J0+α(f+g)s(t)]1s.$

### Theorem 1.6.([4])

Suppose that α > 0, s ≥ 1 and f, g are two positive functions on (0, ∞) such that t > 0, and. If $0, τ ∈ [0, t], then

$[Jαfs(t)]2s+Jαgs(t)]2s≥((M+1)(m+1)M-2)[Jαfs(t)]1s[Jαgs(t)]1s.$

In this article, we are going to extend these theorems for the generalized variational integral.

### 2. Main Results

In this section, we give a generalized Minkowski type inequality for the generalized variational integral. Before we begin our results we need the following definition and lemma.

### Definition 2.1.([6, 7, 8])

A function φ: (0, ∞) → (0, ∞) is called C-submultiplicative with C > 0 if

$φ(xy)≤Cφ(x)φ(y),$

for all x, y ∈ (0, ∞). If inequality (2.1) is reversed, then φ will be called C-supermultiplicative.

In the following theorem, we give a more general version of Theorem 1.5 based on a supermultiplicative function.

### Theorem 2.2

Let f and g be positive functions. Let two functions φi: (0, ∞) → (0, ∞), i = 1, 2 be increasing such that φ1is M1-supermultiplicative and φ2is M2-submultiplicative and $SPαφi(f)(t)$and $SPαφi(g)(t)$are finite. If there exist C1, C2 ∈ (0, ∞) such that

$(f+g) (t)≥max{C1f(t),C2g(t)},$

then

$φ2[SPαφ1(f)(t)]+φ2[SPαφ1(g)(t)]⩽M2(φ2(1M1φ1(C1))+φ2(1M1φ1(C2)))φ2[SPαφ(f+g)(t)]$
Proof

Since (f + g) (t) ≥ max{C1f(t), C2g(t)} and φ1 is increasing, then

$φ1(C1f (t))≤φ1(f+g) (t),$$φ1(C2g (t))≤φ1(f+g) (t),$

If φ1 is M1-supermultiplicative, then by Definition 2.1, (2.2) and (2.3), we obtain

$M1φ1(C1)φ1(f(t))≤φ1(f+g)(t),$$M1φ1(C2)φ1(g(t))≤φ1(f+g)(t).$

Multiplying both sides of (2.4) and (2.5) by pkα(t, τ) and integrating respect to τ on [a, t], we have

$pM1φ1(C1)∫atkα(t,τ)φ1(f) (τ) dτ≤p∫atkα(t,τ)φ1(f+g)(τ)dτ,$$pM1φ1(C2)∫atkα(t,τ)φ1(g) (τ) dτ≤p∫atkα(t,τ)φ1(f+g)(τ)dτ.$

Similarly,

$qM1φ1(C1)∫tbkα(τ,t)φ1(f)dτ≤q∫tbkα(τ,t)φ1(f+g)(τ)dτ.$$qM1φ1(C2)∫tbkα(τ,t)φ1(g)dτ≤q∫tbkα(τ,t)φ1(f+g)(τ)dτ.$

Now by adding (2.6) and (2.8), we have

$pM1φ1(C1)∫atkα(t,τ)φ1(f) (τ) dτ+qM1φ1(C1)∫tbkα(τ,t)φ1(f)dτ≤p∫atkα(t,τ)φ1(f+g)(τ)dτ+q∫tbkα(τ,t)φ1(f+g)(τ)dτ,$

which is equivalent to

$M1φ1(C1)[SPαφ1(f) (t)]≤SPαφ1(f+g) (t)SPαφ1(f)(t)≤1M1φ1(C1)[SPαφ1(f+g)(t)]$

Since φ2 is M2-submultiplicative, then

$φ2[SPαφ1(f)(t)]≤M2φ2(1M1φ1(C1))φ2[SPαφ1(f+g)(t)].$

Now by adding (2.7) and (2.9), we have

$pM1φ1(C2)∫atkα(t,τ)φ1(g) (τ) dτ+qM1φ1(C2)∫tbkα(τ,t)φ1(g)dτ≤p∫atkα(t,τ)φ1(f+g)(τ)dτ+q∫tbkα(τ,t)φ1(f+g)(τ)dτ,$

which is equivalent to

$M1φ1(C2)[SPαφ1(g) (t)]≤SPαφ1(f+g) (t).$

Then

$SPαφ1(g)(t)≤1M1φ1(C2)[SPαφ1(f+g)(t)]$

Hence,

$φ2[SPαφ(g)(t)]≤M2φ2(1M1φ1(C2))φ2[SPαφ1(f+g) (t)].$

$φ2[SPαφ1(f)(t)]+φ2[SPαφ1(g)(t)]⩽M2(φ2(1M1φ1(C1))+φ2(1M1φ1(C2)))φ2[SPαφ1(f+g)(t)]$

we obtain the desired result.

If φ1(x) = xp, $φ2(x)=x1p$, p ≥ 1, M1 = M2 = 1, $C1=M+1M$ and C2 = m + 1 in Theorem 2.2, then the following corollary is achieved.

### Corollary 2.3

Let α > 0, p ≥ 1 and f, g be two positive functions on [0, ∞) such that $SPαfp(t)$and $SPαgp(t)$are finite. If $0, τ ∈ [0, t], then

$[SPαfp(t)]1p+[SPαgp(t)]1p≤1+M(m+2)(m+1)(M+1)[SPα(f+g)p(t)]1p.$

If P = P1 = ⟨a, t, b, 1, 0⟩ in Corollary 2.3, then the following result holds.

### Corollary 2.4.([4])

Let α > 0, s ≥ 1 and f, g be positive on [0, ∞) such that t > 0, $J0+αfs(t)<∞$and $J0+αgs(t)<∞$. If $0, τ ∈ [0, t], then

$[J0+αfs(t)]1s+[J0+αgs(t)]1s≤1+M(m+2)(m+1)(M+1)[J0+α(f+g)s(t)]1s.$

### Theorem 2.5

Let α > 0, f and g be two nonnegative functions on [0, ∞). Let two functions φi: (0, ∞) → (0, ∞), i = 1, 2 be increasing such that φ1is M1-supermultiplicative and φ2 is M2-submultiplicative and $SPαφi(f)(t)$ and $SPαφi(g)(t)$ are finite. If there exist C1, C2 ∈ (0, ∞) such that

$(f+g) (t)≥max{C1f(t),C2g(t)},$

then

$φ2[SPαφ1(f)(t)]φ2[SPαφ1(g)(t)]⩽M2φ2(1M1φ1(C1))φ2(1M1φ1(C2)) (φ2[SPαφ1(f+g)(t)])2.$
Proof

Multiplying (2.10) and (2.11), we have

$φ2[SPαφ1(f)(t)]φ2[SPαφ1(g)(t)]⩽M2φ2(1M1φ1(C1))φ2(1M1φ1(C2)) (φ2[SPαφ1(f+g)(t)])2.$

Then

$φ2[SPαφ1(f)(t)]φ2[SPαφ1(g)(t)]M2φ2(1M1φ1(C1))φ2(1M1φ1(C2))≤(φ2[SPαφ1(f+g)(t)])2.$

If P = P1 = ⟨a, t, b, 1, 0⟩, φ1(x) = xp, $φ2(x)=x1p$, p ≥ 1, M1 = M2 = 1, $C1=M+1M$ and C2 =m + 1 in Theorem 2.5, then following result holds by using Minkowski inequality.

### Corollary 2.6.([4])

Suppose that α > 0, s ≥ 1 and f, g are two positive functions on (0, ∞) such that t > 0, and. If $0, τ ∈ [0, t], then

$[Jαfs(t)]2s+Jαgs(t)]2s≥((M+1)(m+1)M-2)[Jαfs(t)]1s[Jαgs(t)]1s.$

### 3. Conclusions

In this paper, we have proven a Minkowski type inequality for the generalized variational integral. We have also observed that the results obtained in this paper are generalizations of some earlier results. An interesting problem would be to study the methods in this paper to establish the Hermite-Hadamard inequalities for convex functions via the generalized variational integral.

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