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Article

On Further Inequalities for Convex Functions via Generalized Weighted-Type Fractional Operators

by
Çetin Yıldız
1,*,
Gauhar Rahman
2 and
Luminiţa-Ioana Cotîrlă
3,*
1
Department of Mathematics, K.K. Education Faculty, Atatürk University, Erzurum 25320, Turkey
2
Department of Mathematics and Statistic, Hazara University, Mansehra 21300, Pakistan
3
Department of Mathematics, Technical University of Cluj-Napoca, 400020 Cluj-Napoca, Romania
*
Authors to whom correspondence should be addressed.
Fractal Fract. 2023, 7(7), 513; https://doi.org/10.3390/fractalfract7070513
Submission received: 9 May 2023 / Revised: 20 June 2023 / Accepted: 22 June 2023 / Published: 28 June 2023
(This article belongs to the Special Issue Mathematical Inequalities in Fractional Calculus and Applications)

Abstract

:
Several inequalities for convex functions are derived in this paper using the monotonicity properties of functions and a generalized weighted-type fractional integral operator, which allows the integration of a function κ with respect to another function in fractional order. Additionally, it is clear that the results were generalizations of the previously presented findings. In addition, different types of inequalities are obtained using the basic features of mathematical analysis. Finally, we believe that the methodology used in this work will inspire additional research in this field.

1. Introduction

With the extensive content and daily addition of new fractional operators, particularly in recent years, fractional calculus plays a significant role in the subject of inequality theory. Some algebraic features, such as the semigroup property, are present in some of these fractional operators but not in others. In addition, while some of them do not, some of them occasionally have a singularity difficulty. Therefore, the application areas of the operators can differ. The generalizations of fractional integrals have been provided by scholars continuously using various methods. For us, providing a generalization of inequalities that includes all consequences that have so far been demonstrated for various fractional integrals is always interesting and inspiring.
Convex analysis has become one of the important application areas of fractional analysis (see [1,2,3,4,5,6,7,8,9]). In addition, several mathematicians have studied certain inequalities for convex functions using different types of integral operators (for example, the R-L fractional integral operator, the conformable fractional integral operator, tempered fractional integral operators, generalized proportional integral operators, and generalized proportional Hadamard integral operators). These studies have helped to develop different aspects of operator analysis [10,11,12,13,14,15,16]. Different from other mapping classes, convex functions have several applications in the areas of optimization theory, probability theory, statistics, mathematics, and applied sciences. Furthermore, its geometric formulation is very important. It is also one of the cornerstones of the theory of inequality and has developed into the main motivating reason behind one variety of inequalities. Convex functions may be used in many fields of mathematical analysis and statistics, but the one in which they have been most successfully used is inequality theory [17,18,19,20,21,22].
First, we recall the elementary notation in convex analysis:
Definition 1.
A set ϝ R is said to be convex if
φ θ + ( 1 φ ) γ ϝ
for each θ , γ ϝ and φ [ 0 , 1 ] .
Definition 2.
The mapping κ 1 : ϝ R , is said to be convex if the following inequality holds:
κ 1 ( φ θ + ( 1 φ ) γ ) φ κ 1 ( θ ) + ( 1 φ ) κ 1 ( γ )
for all θ , γ ϝ , and φ [ 0 , 1 ] . We say that κ 1 is concave if ( κ 1 ) is convex.
The properties and definitions of the convex functions have recently ascribed a significant role to its theory and practice in the field of fractional integral operators.
The following inequality was established in [23] by Ngo et al.:
0 1 g 1 ζ + 1 ( ρ ) d ρ 0 1 ρ ζ g 1 ζ ( ρ ) d ρ
and
0 1 g 1 ζ + 1 ( ρ ) d ρ 0 1 ρ g 1 ζ ( ρ ) d ρ ,
where ζ > 0 and g 1 > 0 and the continuous function on [ 0 , 1 ] is such that
ϰ 1 g 1 ( ρ ) d ρ ϰ 1 ρ d ρ , ϰ [ 0 , 1 ] .
Then, Liu et al. established the following inequalities in [24]:
θ γ g 1 ζ + ϑ ( ρ ) d ρ θ γ ( ρ θ ) ζ g 1 ϑ ( ρ ) d ρ ,
where ζ > 0 ,   ϑ > 0 , and g 1 > 0 and the continuous function on [ θ , γ ] is such that
θ γ g 1 ξ ( ρ ) d ρ 0 1 ( ρ θ ) ξ d ρ , ξ = min ( 1 , ϑ ) , ρ [ 0 , 1 ] .
The following two theorems were obtained by Liu in [25]:
Theorem 1.
Let κ 1 and κ 2 be continuous and positive functions with κ 1 κ 2 on [ θ , γ ] such that κ 1 is increasing and κ 1 κ 2 ( κ 2 0 ) is decreasing. If ξ is a convex function, then the inequality
θ γ κ 1 ( ϖ ) d ϖ θ γ κ 2 ( ϖ ) d ϖ θ γ ξ κ 1 ( ϖ ) d ϖ θ γ ξ κ 2 ( ϖ ) d ϖ .
holds where ξ ( 0 ) = 0 .
Theorem 2.
Let κ 1 ,   κ 2 , and κ 3 be continuous and positive functions with κ 1 κ 2 on [ θ , γ ] such that κ 1 and κ 3 are increasing and κ 1 κ 2 ( κ 2 0 ) is decreasing. If ξ is a convex function, then the inequality
θ γ κ 1 ( ϖ ) d ϖ θ γ κ 2 ( ϖ ) d ϖ θ γ ξ κ 1 ( ϖ ) κ 3 ( ϖ ) d ϖ θ γ ξ κ 2 ( ϖ ) κ 3 ( ϖ ) d ϖ
holds where ξ ( 0 ) = 0 .
For functions in L p [ θ , γ ] , defined as follows, several novel conclusions were obtained:
Definition 3.
For p 1 , , if the function κ holds
θ γ κ ( τ ) p d ϖ 1 p < ,
then it is said to be in L p [ θ , γ ] .
The well-known Minkowski inequality has been presented as follows in the mathematical literature (see [26]):
Theorem 3.
θ γ κ p ( ϖ ) d ϖ and θ γ p ( ϖ ) d ϖ are positive finite reals for p 1 . Then, the inequality
θ γ ( κ ( ϖ ) + ( ϖ ) ) p d ϖ 1 p θ γ κ p ( ϖ ) d ϖ 1 p + θ γ p ( ϖ ) d ϖ 1 p
holds.
L. Bougoffa in [27] derives the reverse Minkowski inequality for Riemann–Liouville fractional integrals, which is explained as follows:
Theorem 4.
Let κ , L p [ θ , γ ] and κ , > 0 , with 1 p < ,   0 < θ γ κ p ( ϖ ) d ϖ < and 0 < θ γ p ( ϖ ) d ϖ < . If 0 η κ ( ϖ ) ( ϖ ) for η , R + and every ϖ [ θ , γ ] , then the inequality
θ γ κ p ( ϖ ) d ϖ 1 p + θ γ p ( ϖ ) d ϖ 1 p c θ γ ( κ ( ϖ ) + ( ϖ ) ) p d ϖ 1 p
holds where c = ( η + 1 ) + ( + 1 ) ( η + 1 ) ( + 1 ) .
Z. Dahmani provided the reverse Minkowski and Hadamard inequalities utilizing the Riemann–Liouville fractional integral in [28]. Set et al. presented some inequalities of reverse Minkowski and Hermite–Hadamard involving two functions using the classical Riemann integral in [29]. Chinchane and Pachpatte established the reverse Minkowski inequality via the Saigo fractional integral operator in [30]. Vanterler et al. studied the reverse Minkowski inequalities and some other related inequalities by means of the Katugampola fractional integral operator in [31]. The reverse Minkowski inequality and other fractional inequalities in [32] were established by Rahman et al. using the generalized proportional fractional integral operators. By taking general kernels into account in [33], Iqbal et al. were able to achieve novel conclusions for Minkowski and associated inequalities. There are many studies on the reverse Minkowski inequality in the literature. The "Young’s inequality" theorem is as follows (see [34]):
Theorem 5.
Let 0 , k be an interval with k > 0 , and h be an increasing, continuous function on 0 , k . If γ 0 , ( k ) , θ 0 , k , ( 0 ) = 0 and 1 stands for the inverse function of h , then
0 θ ( ϖ ) d ϖ + 0 γ 1 ( ϖ ) d ϖ θ γ .
Different forms of Young’s inequality are defined as follows:
1 r θ r + 1 s γ s θ γ , θ , γ 0 , r 1 and 1 r + 1 s = 1 .
In other words, this inequality illustrates the relationship between the geometric and arithmetic means.

2. Preliminaries

The application of fractional integral operators is another method for obtaining extensions of the traditional integral inequalities that are known from the literature. By using different versions of fractional integrals, many extensions, generalizations, and variations of inequalities, such as those of Hermite–Hadamard, Simpson, Ostrowski, Minkowski, Chebyshev, and Grüss, have been obtained.
Now, some fractional integral operators used to obtain integral inequalities will be given. First of them is the Riemann–Liouville fractional integral operator (see [35]), which is widely used in fractional calculus.
Definition 4.
Let κ L 1 [ θ , γ ] .   J θ + κ and J γ κ (Riemann–Liouville fractional integral operators) of order > 0 with θ 0 are defined by
J θ + κ ( ϰ ) = 1 Γ ( ) θ ϰ ϰ ϖ 1 κ ( ϖ ) d ϖ , ϰ > θ
and
J γ κ ( ϰ ) = 1 Γ ( ) ϰ γ ϖ ϰ 1 κ ( ϖ ) d ϖ , ϰ < γ ,
respectively, where Γ ( ) = 0 e u u 1 d u . Here, J θ + 0 κ ( ϰ ) = J γ 0 κ ( ϰ ) = κ ( ϰ ) . The fractional integral becomes the classical integral when = 1 .
Definition 5.
([36]). Suppose that the function Ψ : [ 0 , ) [ 0 , ) satisfies the conditions given below:
0 1 Ψ ϖ ϖ d ϖ < ,
1 ρ Ψ 1 Ψ ( 2 ) ρ , 1 2 1 2 2 ,
Ψ ( 2 ) 2 2 φ Ψ ( 1 ) 1 2 , 1 2 ,
Ψ ( 2 ) 2 2 Ψ ( 1 ) 1 2 ζ 2 1 Ψ ( 2 ) 2 2 , 1 2 1 2 2 ,
where ρ , φ , ζ > 0 are independent of 1 , 2 > 0 . If Ψ ( 2 ) 2 α is increasing for some α > 0 and Ψ ( 2 ) 2 μ is decreasing for some μ > 0 , then Ψ satisfies (2)–(5).
It became important to obtain more general versions of the new results when fractional integral operators were utilized more frequently. As a result, weighted integral operators started to be introduced. While generalized versions of the findings in the literature may also be obtained, novel results are also produced with these operators. The following list includes one of the most useful weighted integral operators lately presented:
Definition 6.
([37]). The generalized weighted-type fractional integral operators, on both the right and left side, are respectively defined by:
ω θ + Ψ κ ϰ = ω 1 ( ϰ ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) κ ( ϖ ) d ϖ , θ < ϰ
ω γ Ψ κ ϰ = ω 1 ( ϰ ) ϰ γ Ψ ( ϖ ) ( x ) ( ϖ ) ( x ) ω ( ϖ ) ( ϖ ) κ ( ϖ ) d ϖ , γ > ϰ
where ω 1 ( ϰ ) = 1 ω ( ϰ ) ,   ω ( ϰ ) 0 .
These important special cases of the generalized weighted-type fractional integral operators are mentioned below:
Remark 1.
In Definition 6:
  • If we consider Ψ ( ϰ ) = ( ϰ ) , we obtain
    ω θ + κ ϰ = ω 1 ( ϰ ) θ ϰ ω ( ϖ ) ( ϖ ) κ ( ϖ ) d ϖ , θ < ϰ ω γ κ ϰ = ω 1 ( ϰ ) ϰ γ ω ( ϖ ) ( ϖ ) κ ( ϖ ) d ϖ , γ > ϰ .
  • If we consider ϰ = ϰ , then
    ω θ + κ ϰ = ω 1 ( ϰ ) θ ϰ Ψ ϰ ϖ ϰ ϖ ω ( ϖ ) κ ( ϖ ) d ϖ , θ < ϰ ω γ κ ϰ = ω 1 ( ϰ ) ϰ γ Ψ ϖ ϰ ϖ ϰ ω ( ϖ ) κ ( ϖ ) d ϖ , γ > ϰ .
    In addition, if we take ω ( ϰ ) = 1 , we obtain the generalized fractional operators in [38].
  • If we consider Ψ ( ϰ ) = ( ϰ ) Γ ( k ) , then we obtain
    ω θ + κ ϰ = ω 1 ( ϰ ) Γ ( ) θ ϰ ( ϰ ) ( ϖ ) 1 ω ( ϖ ) ( ϖ ) κ ( ϖ ) d ϖ , θ < ϰ ω γ κ ϰ = ω 1 ( ϰ ) Γ ( ) ϰ γ ( ϖ ) ( ϰ ) 1 ω ( ϖ ) ( ϖ ) κ ( ϖ ) d ϖ , γ > ϰ
    where C with ( ) > 0 [39].
  • If we consider ϰ = ϰ and Ψ ( ϰ ) = ϰ Γ ( ) , then
    ω θ + κ ϰ = ω 1 ( ϰ ) Γ ( ) θ ϰ ϰ ϖ 1 ω ( ϖ ) κ ( ϖ ) d ϖ , θ < ϰ ω γ κ ϰ = ω 1 ( ϰ ) Γ ( ) ϰ γ ϖ ϰ 1 ω ( ϖ ) κ ( ϖ ) d ϖ , γ > ϰ .
    In addition to the above identities, if we choose ω ( ϰ ) = 1 , we obtain classical Riemann–Liouville fractional integral operators.
  • If we consider ϰ = ln ϰ and Ψ ( ϰ ) = ( ln ϰ ) Γ ( ) , then the operators reduce to the weighted Hadamard fractional integrals as follows:
    ω θ + κ ϰ = ω 1 ( ϰ ) Γ ( ) θ ϰ ln ϰ ln ϖ 1 ω ( ϖ ) κ ( ϖ ) d ϖ ϖ , θ < x ω γ κ ϰ = ω 1 ( ϰ ) Γ ( ) ϰ γ ln ϖ ln ϰ 1 ω ( ϖ ) κ ( ϖ ) d ϖ ϖ , γ > x .
  • If we consider ϰ = ϰ τ and Ψ ( ϰ ) = ϰ τ τ , then the operators reduce to the weighted Katugampola fractional integrals as follows:
    ω θ + κ ϰ = ω 1 ( ϰ ) Γ ( ) θ ϰ ϰ τ ϖ τ τ 1 ω ( ϖ ) κ ( ϖ ) d ϖ ϖ 1 τ , θ < ϰ ω γ κ ϰ = ω 1 ( ϰ ) Γ ( ) ϰ γ ϖ τ ϰ τ τ 1 ω ( ϖ ) κ ( ϖ ) d ϖ ϖ 1 τ , γ > ϰ .
  • In addition, to obtain the different version of the fractional integral operator that is defined in [9,37], one can choose ω ( ϰ ) = 1 in (6).
This article discussed several important inequalities (such as Hadamard, Grüss, Chebyshev, Fejer, and Minkowski types) via the generalized weighted-type fractional integral operators (6) with increasing, decreasing, positive, continuous, and convex functions. The existing inequalities associated with increasing, decreasing, positive, continuous, and convex functions are also restored by applying specific conditions as given in the remarks. By applying certain conditions on and Ψ described in the literature, several other varieties of fractional integral inequalities can be established.
The main purpose of this article is to generalize some classical integral inequalities using the generalized weighted-type fractional integral operators. In addition, we used the basic features of mathematical analysis to achieve our main results.
The article is arranged as follows: We recall some of the notations, definitions, results, and introductory facts that were used in Section 1 and Section 2 and are utilized throughout the remaining chapters of this work. In Section 3, we introduce some inequalities for convex functions utilizing the generalized weighted-type fractional integral operators. In Section 4, we give results of the reverse Minkowski inequality, which is the first of our main results. Finally, in Section 5, we present some other related results involving constant generalized weighted-type fractional integral operators.

3. New Inequalities Involving Generalized Weighted-Type Fractional Operators

This section introduces some inequalities for convex functions using the generalized weighted-type fractional integral operators.
Theorem 6.
Let κ 1 and κ 2 be two positive continuous functions on [ θ , γ ] and κ 1 κ 2 on [ θ , γ ] . If κ 1 κ 2 is decreasing and κ 1 is increasing on [ θ , γ ] , then for a convex function ξ with ξ ( 0 ) = 0 , the generalized weighted-type fractional integral operator given by (6) satisfies the following inequality:
ω θ + Ψ κ 1 ϰ ω θ + Ψ κ 2 ϰ ω θ + Ψ ξ κ 1 ϰ ω θ + Ψ ξ κ 2 ϰ ,
where ϰ > θ > 0 .
Proof. 
ξ ( ϰ ) ϰ is increasing since ξ is defined as convex function satisfying ξ ( 0 ) = 0 . Moreover, the function ξ ( κ 1 ( ϰ ) ) κ 1 ( ϰ ) is also increasing as κ 1 is increasing. Clearly, the function κ 1 ( ϰ ) κ 2 ( ϰ ) is decreasing. Thus, for all [ θ , ϰ ] ,   θ < ϰ γ , it can be written
ξ ( κ 1 ( ϖ ) ) κ 1 ( ϖ ) ξ ( κ 1 ( φ ) ) κ 1 ( φ ) κ 1 ( φ ) κ 2 ( φ ) κ 1 ( ϖ ) κ 2 ( ϖ ) 0 .
It follows that
ξ ( κ 1 ( ϖ ) ) κ 1 ( ϖ ) κ 1 ( φ ) κ 2 ( φ ) + ξ ( κ 1 ( φ ) ) κ 1 ( φ ) κ 1 ( ϖ ) κ 2 ( ϖ )
ξ ( κ 1 ( φ ) ) κ 1 ( φ ) κ 1 ( φ ) κ 2 ( φ ) ξ ( κ 1 ( ϖ ) ) κ 1 ( ϖ ) κ 1 ( ϖ ) κ 2 ( ϖ ) 0 .
Multiplying (8) by κ 2 ( ϖ ) κ 2 ( φ ) , we have
ξ ( κ 1 ( ϖ ) ) κ 1 ( ϖ ) κ 1 ( φ ) κ 2 ( ϖ ) + ξ ( κ 1 ( φ ) ) κ 1 ( φ ) κ 1 ( ϖ ) κ 2 ( φ )
ξ ( κ 1 ( φ ) ) κ 1 ( φ ) κ 1 ( φ ) κ 2 ( ϖ ) ξ ( κ 1 ( ϖ ) ) κ 1 ( ϖ ) κ 1 ( ϖ ) κ 2 ( φ ) 0 .
Now, multiplying both sides of (9) by ω 1 ( ϰ ) Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) and then integrating with regard to the variable ϖ from θ to ϰ , we obtain
ω 1 ( ϰ ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ξ ( κ 1 ( ϖ ) ) κ 1 ( ϖ ) ω ( ϖ ) ( ϖ ) κ 1 ( φ ) κ 2 ( ϖ ) d ϖ + ω 1 ( ϰ ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ξ ( κ 1 ( φ ) ) κ 1 ( φ ) ω ( ϖ ) ( ϖ ) κ 1 ( ϖ ) κ 2 ( φ ) d ϖ ω 1 ( ϰ ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ξ ( κ 1 ( φ ) ) κ 1 ( φ ) ω ( ϖ ) ( ϖ ) κ 1 ( φ ) κ 2 ( ϖ ) d ϖ
ω 1 ( ϰ ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ξ ( κ 1 ( ϖ ) ) κ 1 ( ϖ ) ω ( ϖ ) ( ϖ ) κ 1 ( ϖ ) κ 2 ( φ ) d ϖ 0 .
Then, it follows that
κ 1 ( φ ) ω θ + Ψ ξ κ 1 κ 1 κ 2 ϰ + ξ ( κ 1 ( φ ) ) κ 1 ( φ ) κ 2 ( φ ) ω θ + Ψ κ 1 ϰ
ξ ( κ 1 ( φ ) ) κ 1 ( φ ) κ 1 ( φ ) ω θ + Ψ κ 2 ϰ κ 2 ( φ ) ω θ + Ψ ξ κ 1 κ 1 κ 1 ϰ 0 .
Again, by multiplying both sides of (10) by ω 1 ( ϰ ) Ψ ( ϰ ) ( φ ) ( ϰ ) ( φ ) ω ( φ ) ( φ ) and then integrating from θ to ϰ with regard to φ , we have
ω θ + Ψ κ 1 ϰ ω θ + Ψ ξ κ 1 κ 1 κ 2 ϰ + ω θ + Ψ ξ κ 1 κ 1 κ 2 ϰ ω θ + Ψ κ 1 ϰ
ω θ + Ψ ξ κ 1 ϰ ω θ + Ψ κ 2 ϰ + ω θ + Ψ κ 2 ϰ ω θ + Ψ ξ κ 1 ϰ .
It follows that
ω θ + Ψ κ 1 ϰ ω θ + Ψ κ 2 ϰ ω θ + Ψ ξ κ 1 ϰ ω θ + Ψ ξ κ 1 κ 1 κ 2 ϰ .
Now, since ξ ( ϰ ) ϰ is an increasing function and κ 1 κ 2 on [ θ , γ ] , we obtain
ξ ( κ 1 ( ϖ ) ) κ 1 ( ϖ ) ξ ( κ 2 ( ϖ ) ) κ 2 ( ϖ )
for [ θ , ϰ ] .
Multiplying both sides of (13) by ω 1 ( ϰ ) Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) κ 2 ( ϖ ) and then integrating with regard to the variable ϖ from θ to ϰ , we have
ω 1 ( ϰ ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ξ ( κ 1 ( ϖ ) ) κ 1 ( ϖ ) ω ( ϖ ) ( ϖ ) κ 2 ( ϖ ) d ϖ ω 1 ( ϰ ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ξ ( κ 2 ( ϖ ) ) κ 2 ( ϖ ) ω ( ϖ ) ( ϖ ) κ 2 ( ϖ ) d ϖ ,
which yields
ω θ + Ψ ξ κ 1 κ 1 κ 2 ϰ ω θ + Ψ ξ κ 2 ϰ .
Hence, from (12) and (14), we have (7). □
Remark 2.
If we take ω ( ϰ ) = ϰ ,   ( ϰ ) = ϰ ,   ϰ = γ , and Ψ ( ϰ ) = ( ϰ ) in Theorem 6, we obtain Theorem 1.
Theorem 7.
Let κ 1 , κ 2 , and κ 3 be positive continuous functions and κ 1 κ 2 on [ θ , γ ] . If κ 1 κ 2 is decreasing and κ 1 and κ 3 are increasing on [ θ , γ ] , then, for a convex function ξ with ξ ( 0 ) = 0 , the weighted fractional operator shows the following inequality (6):
ω θ + Ψ κ 1 ϰ ω θ + Ψ κ 2 ϰ ω θ + Ψ ( ξ κ 1 ) κ 3 ϰ ω θ + Ψ ( ξ κ 2 ) κ 3 ϰ ,
where ϰ > θ > 0 .
Proof. 
Since κ 1 κ 2 on [ θ , γ ] and ξ ( ϰ ) ϰ is increasing for ϖ , φ [ θ , ϰ ] ,   θ < ϰ γ , we obtain
ξ ( κ 1 ( ϖ ) ) κ 1 ( ϖ ) ξ ( κ 2 ( ϖ ) ) κ 2 ( ϖ ) .
Multiplying both sides of (15) by ω 1 ( ϰ ) Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) κ 2 ( ϖ ) κ 3 ( ϖ ) and then integrating with regard to the variable ϖ from θ to ϰ , we have
ω 1 ( ϰ ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ξ ( κ 1 ( ϖ ) ) κ 1 ( ϖ ) ω ( ϖ ) ( ϖ ) κ 2 ( ϖ ) κ 3 ( ϖ ) d ϖ ω 1 ( ϰ ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ξ ( κ 2 ( ϖ ) ) κ 2 ( ϖ ) ω ( ϖ ) ( ϖ ) κ 2 ( ϖ ) κ 3 ( ϖ ) d ϖ
which, by virtue of (6), can be written as
ω θ + Ψ ξ κ 1 κ 1 κ 2 κ 3 ϰ ω θ + Ψ ( ξ κ 2 ) κ 3 ϰ .
Moreover, ξ ( ϖ ) ϖ is increasing, since ξ ( 0 ) = 0 and the function ξ is convex. Since κ 1 is increasing, so is ξ ( κ 1 ( ϖ ) ) κ 1 ( ϖ ) . Obviously, the function κ 1 ( ϖ ) κ 2 ( ϖ ) is decreasing for ϖ , φ [ θ , ϰ ] ,   θ < ϰ γ . Thus,
ξ ( κ 1 ( ϖ ) ) κ 1 ( ϖ ) κ 3 ( ϖ ) ξ ( κ 1 ( φ ) ) κ 1 ( φ ) κ 3 ( φ ) κ 1 ( φ ) κ 2 ( ϖ ) κ 1 ( ϖ ) κ 2 ( φ ) 0 .
It becomes
ξ ( κ 1 ( ϖ ) ) κ 3 ( ϖ ) κ 1 ( ϖ ) κ 1 ( φ ) κ 2 ( ϖ ) + ξ ( κ 1 ( φ ) ) κ 3 ( φ ) κ 1 ( φ ) κ 1 ( ϖ ) κ 2 ( φ )
ξ ( κ 1 ( φ ) ) κ 3 ( φ ) κ 1 ( φ ) κ 1 ( φ ) κ 2 ( ϖ ) ξ ( κ 1 ( ϖ ) ) κ 3 ( ϖ ) κ 1 ( ϖ ) κ 1 ( ϖ ) κ 2 ( φ ) 0 .
Multiplying both sides of (17) by ω 1 ( ϰ ) Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) and then integrating with regard to the variable ϖ from θ to ϰ , we obtain
ω 1 ( ϰ ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ξ ( κ 1 ( ϖ ) ) κ 3 ( ϖ ) κ 1 ( ϖ ) ω ( ϖ ) ( ϖ ) κ 1 ( φ ) κ 2 ( ϖ ) d ϖ
+ ω 1 ( ϰ ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ξ ( κ 1 ( φ ) ) κ 3 ( φ ) κ 1 ( φ ) ω ( ϖ ) ( ϖ ) κ 1 ( ϖ ) κ 2 ( φ ) d ϖ
ω 1 ( ϰ ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ξ ( κ 1 ( φ ) ) κ 3 ( φ ) κ 1 ( φ ) ω ( ϖ ) ( ϖ ) κ 1 ( φ ) κ 2 ( ϖ ) d ϖ
ω 1 ( ϰ ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ξ ( κ 1 ( ϖ ) ) κ 3 ( ϖ ) κ 1 ( ϖ ) ω ( ϖ ) ( ϖ ) κ 1 ( ϖ ) κ 2 ( φ ) d ϖ 0 .
This follows that
κ 1 ( φ ) ω θ + Ψ ξ κ 1 κ 1 κ 2 κ 3 ϰ + ξ ( κ 1 ( φ ) ) κ 3 ( φ ) κ 1 ( φ ) κ 2 ( φ ) ω θ + Ψ κ 1 ϰ
ξ ( κ 1 ( φ ) ) κ 3 ( φ ) κ 1 ( φ ) κ 1 ( φ ) ω θ + Ψ κ 2 ϰ κ 2 ( φ ) ω θ + Ψ ξ κ 1 κ 3 ϰ 0 .
Again, multiplying both sides of (18) by ω 1 ( ϰ ) Ψ ( ϰ ) ( φ ) ( ϰ ) ( φ ) ω ( φ ) ( φ ) and then integrating with regard to the variable φ from θ to ϰ , we have
ω θ + Ψ κ 1 ϰ ω θ + Ψ ξ κ 1 κ 1 κ 2 κ 3 ϰ + ω θ + Ψ ξ κ 1 κ 1 κ 2 κ 3 ϰ ω θ + Ψ κ 1 ϰ ω θ + Ψ κ 2 ϰ ω θ + Ψ ξ κ 1 κ 3 ϰ + ω θ + Ψ κ 2 ϰ ω θ + Ψ ξ κ 1 κ 3 ϰ .
Therefore, we can write
ω θ + Ψ κ 1 ϰ ω θ + Ψ κ 2 ϰ ω θ + Ψ ( ξ κ 1 ) κ 3 ϰ ω θ + Ψ ξ κ 1 κ 1 κ 2 κ 3 ϰ .
Hence, from (16) and (19), we obtain the required result. □
Remark 3.
If we take ω ( ϰ ) = ϰ ,   ( ϰ ) = ϰ ,   ϰ = γ , and Ψ ( ϰ ) = ( ϰ ) in Theorem 7, we obtain Theorem 2.

4. Reverse Minkowski Inequalities for Generalized Weighted-Type Fractional Integral Operators

In this section, the results of the reverse Minkowski inequality using the generalized weighted-type fractional integral operators are given.
Theorem 8.
Let κ , L [ θ , ϰ ] be two positive functions on [ 0 , ) , such that ω θ + Ψ κ p ϰ and ω θ + Ψ p ϰ are finite reals for ϰ > θ > 0 ,   p 1 . If 0 η κ ( ϖ ) ( ϖ ) holds for η , R + and ϖ [ θ , ϰ ] , then
ω θ + Ψ κ p 1 p ϰ + ω θ + Ψ p 1 p ϰ c 1 ω θ + Ψ ( κ + ) p 1 p ϰ ,
with c 1 = ( η + 1 ) + ( N + 1 ) ( η + 1 ) ( + 1 ) .
Proof. 
Assuming the given condition κ ( ϖ ) ( ϖ ) ,   ϖ [ θ , ϰ ] , it can be written as
κ ( ϖ ) ( κ ( ϖ ) + ( ϖ ) ) κ ( ϖ )
which implies that
( + 1 ) p κ p ( ϖ ) p ( κ ( ϖ ) + ( ϖ ) ) p .
Multiplying both sides of (21) by ω 1 ( ϰ ) Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) and then integrating with respect to ϖ from θ to ϰ , we have
( + 1 ) p ω 1 ( ϰ ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) κ p ( ϖ ) d ϖ p ω 1 ( ϰ ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) ( κ + ) p ( ϖ ) d ϖ .
Consequently, we can write
ω θ + Ψ κ p 1 p ϰ + 1 ω θ + Ψ ( κ + ) p 1 p ϰ .
Nevertheless, as η ( ϖ ) κ ( ϖ ) , it follows
1 + 1 η p p ( ϖ ) 1 η p ( κ ( ϖ ) + ( ϖ ) ) p .
Now, multiplying both sides of (23) by ω 1 ( ϰ ) Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) and then integrating with respect to ϖ from θ to ϰ , we obtain
ω θ + Ψ p 1 p ϰ 1 η + 1 ω θ + Ψ ( κ + ) p 1 p ϰ .
From (22) and (24), the required result follows. □
Remark 4.
If we choose ω ( ϰ ) = ϰ ,   ( ϰ ) = ϰ ,   ϰ = γ , and Ψ ( ϰ ) = ( ϰ ) in Theorem 8, we obtain Theorem 4.
With the help of generalized weighted-type fractional operators, inequality (20) is a variant of the reverse Minkowski inequality.
Theorem 9.
Let κ , L [ θ , ϰ ] be two positive functions on [ 0 , ) , such that ω θ + Ψ κ p ϰ and ω θ + Ψ p ϰ are finite reals for ϰ > θ > 0 ,   p 1 . If 0 η κ ( ϖ ) ( ϖ ) holds for η , R + and ϖ [ θ , ϰ ] , then
ω θ + Ψ κ p 2 p ϰ + ω θ + Ψ p 2 p ϰ c 2 ω θ + Ψ κ p 1 p ϰ ω θ + Ψ p 1 p ϰ ,
with c 2 = ( + 1 ) ( η + 1 ) 2 .
Proof. 
Multiplying inequality (22) by inequality (24), we have
( + 1 ) ( η + 1 ) ω θ + Ψ κ p 1 p ϰ ω θ + Ψ p 1 p ϰ ω θ + Ψ ( κ + ) p 2 p ϰ .
On the right side of (25), using the Minkowski inequality, we obtain
( + 1 ) ( η + 1 ) ω θ + Ψ κ p 1 p ϰ ω θ + Ψ p 1 p ϰ ω θ + Ψ κ p 1 p ϰ + ω θ + Ψ p 1 p ϰ 2 .
Then, we have
ω θ + Ψ κ p 2 p ϰ + ω θ + Ψ p 2 p ϰ ( + 1 ) ( η + 1 ) 2 ω θ + Ψ κ p 1 p ϰ ω θ + Ψ p 1 p x
which is the desired result. □

5. Other Related Inequalities via Generalized Weighted-Type Fractional Operators

Finally, the generalized weighted-type fractional operators are used in this section to derive a number of related inequalities.
Theorem 10.
Let κ , L [ θ , ϰ ] and κ , > 0 on [ 0 , ) , such that ω θ + Ψ κ p ϰ and ω θ + Ψ p ϰ are finite reals for ϰ > θ > 0 ,   p , q 1 and 1 p + 1 q = 1 . If 0 η κ ( ϖ ) ( ϖ ) holds for η , R + and ϖ [ θ , ϰ ] , then the inequality shown below holds for generalized weighted-type fractional operators:
ω θ + Ψ κ 1 p ϰ ω θ + Ψ 1 q τ η 1 q p ω θ + Ψ κ 1 p . 1 q ϰ .
Proof. 
Applying the specified condition κ ( ϖ ) ( ϖ ) ,   ϖ [ θ , ϰ ] , it can be written
κ ( ϖ ) ( ϖ ) 1 q κ 1 q ( ϖ ) 1 q ( ϖ ) .
Multiplying both sides of (26) by κ 1 p ( ϖ ) , we can rewrite it as
1 q κ ( ϖ ) κ 1 p ( ϖ ) 1 q ( ϖ )
where 1 p + 1 q = 1 .
Multiplying both sides of (27) by ω 1 ( ϰ ) Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) and then integrating, we obtain
1 q ω 1 ( ϰ ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) κ ( ϖ ) d ϖ ω 1 ( ϰ ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) κ 1 p ( ϖ ) 1 q ( ϖ ) d ϖ .
From generalized weighted-type fractional operators, we then have
1 p q ω θ + Ψ κ 1 p ϰ ω θ + Ψ κ 1 p . 1 q 1 p ϰ .
On the other hand, as η κ ( ϖ ) ( ϖ ) , it follows that
η 1 p 1 p ( ϖ ) κ 1 p ( ϖ ) .
Multiplying both sides of (29) by 1 q ( ϖ ) and using the relation 1 p + 1 q = 1 , we obtain
η 1 p ( ϖ ) κ 1 p ( ϖ ) 1 q ( ϖ ) .
Multiplying both sides of (30) by ω 1 ( ϰ ) Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) and then integrating, we have
η 1 p q ω θ + Ψ 1 q ϰ ω θ + Ψ κ 1 p . 1 q 1 q ϰ .
Conducting the product between (28) and (31), we have
ω θ + Ψ κ 1 p ϰ ω θ + Ψ 1 q ϰ η 1 q p ω θ + Ψ κ 1 p 1 q ϰ .
where 1 p + 1 q = 1 . Thus, the proof is completed. □
Theorem 11.
For p , q 1 and 1 p + 1 q = 1 . Let κ , L [ θ , ϰ ] and κ , > 0 on [ 0 , ) , such that ω θ + Ψ κ p ϰ and ω θ + Ψ p ϰ are finite reals for ϰ > θ > 0 . If 0 η κ ( ϖ ) ( ϖ ) for η , R + and for all ϖ [ θ , ϰ ] , then
ω θ + Ψ κ ϰ c 3 ω θ + Ψ ( κ p + p ) ( ϰ ) + c 4 ω θ + Ψ ( κ q + q ) ( ϰ )
with c 3 = 2 p 1 p p ( + 1 ) p and c 4 = 2 q 1 q ( η + 1 ) q .
Proof. 
The following inequality is obtained using the following hypothesis:
( + 1 ) p κ p ( ϖ ) p ( κ + ) p ( ϖ ) .
Multiplying both sides of (32) by ω 1 ( ϰ ) Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) and then integrating, we have
ω θ + Ψ κ p ϰ p ( + 1 ) p ω θ + Ψ ( κ + ) p ( ϰ ) .
For ϖ [ θ , ϰ ] , since 0 η κ ( ϖ ) ( ϖ ) holds, we obtain
( η + 1 ) q q ( ϖ ) ( κ + ) q ( ϖ ) .
Similarly, multiplying both sides of (34) by ω 1 ( ϰ ) Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) and then integrating, we can write
ω θ + Ψ q ϰ 1 ( η + 1 ) q ω θ + Ψ ( κ + ) q ( ϰ ) .
Using Young’s inequality, we have
κ ( ϖ ) ( ϖ ) 1 p κ p ( ϖ ) + 1 q q ( ϖ ) ,
again multiplying both sides of (36) by ω 1 ( ϰ ) Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) and then integrating, we obtain
ω θ + Ψ κ ϰ 1 p ω θ + Ψ κ p ϰ + 1 q ω θ + Ψ q ϰ .
Using (33) and (35) in (37), we obtain
ω θ + Ψ κ ϰ p p ( + 1 ) p ω θ + Ψ ( κ + ) p ( ϰ ) + 1 q ( η + 1 ) q ω θ + Ψ ( κ + ) q ( ϰ ) .
Utilizing the identity ( ς + ρ ) r 2 r 1 ( ς r + ρ r ) ,   r > 1 ,   ς , ρ > 0 in (38), we obtain
ω θ + Ψ κ ϰ 2 p 1 p p ( + 1 ) p ω θ + Ψ ( κ p + p ) ( ϰ ) + 2 q 1 q ( η + 1 ) q ω θ + Ψ ( κ q + q ) ( ϰ ) .
This is the required result. □
Theorem 12.
Let κ , L [ θ , ϰ ] and κ , > 0 on [ 0 , ) , such that ω θ + Ψ κ p ϰ and ω θ + Ψ p ϰ are finite reals for ϰ > θ > 0 . If 0 < c < η κ ( ϖ ) ( ϖ ) for η , R + and for all ϖ [ θ , ϰ ] , then the following inequalities hold:
+ 1 c ω θ + Ψ κ c p 1 p ϰ ( ω θ + Ψ κ p ) 1 p ( ϰ ) + ( ω θ + Ψ p ) 1 p ( ϰ ) η + 1 η c ω θ + Ψ κ c p 1 p ϰ
for p 1 .
Proof. 
By using the hypothesis 0 < c < η , we obtain
η c c η c + η η c + c + ( + 1 ) ( η c ) ( η + 1 ) ( c ) .
The conclusion is that
+ 1 c η + 1 η c .
In addition,
η κ ( ϖ ) ( ϖ ) η c κ ( ϖ ) c ( ϖ ) ( ϖ ) c
κ ( ϖ ) c ( ϖ ) p ( c ) p p ( ϖ ) κ ( ϖ ) c ( ϖ ) p ( η c ) p .
Multiplying both sides of (39) by ω 1 ( ϰ ) Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) and then integrating, we obtain
ω 1 ( ϰ ) ( c ) p θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) κ ( ϖ ) c ( ϖ ) p d ϖ ω 1 ( ϰ ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) p ( ϖ ) d ϖ ω 1 ( ϰ ) ( η c ) p θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) κ ( ϖ ) c ( ϖ ) p d ϖ
Then, we can write
1 c ω θ + Ψ κ c p 1 p ϰ ( ω θ + Ψ p ) 1 p ( ϰ ) 1 η c ω θ + Ψ κ c p 1 p ϰ .
Again, we have
1 ( ϖ ) κ ( ϖ ) 1 η η c η c κ ( ϖ ) c ( ϖ ) c κ ( ϖ ) c c
which implies
c p κ ( ϖ ) c ( ϖ ) p κ p ( ϖ ) η η c p κ ( ϖ ) c ( ϖ ) p .
By using the same procedures with (41), we obtain
c ω θ + Ψ κ c p 1 p ϰ ( ω θ + Ψ κ p ) 1 p ( ϰ ) η η c ω θ + Ψ κ c p 1 p ϰ .
Adding (40) and (42), the required result is obtained. □
Theorem 13.
Let κ , L [ θ , ϰ ] and κ , > 0 on [ 0 , ) , such that ω θ + Ψ κ p ϰ and ω θ + Ψ p ϰ are finite reals for ϰ > θ > 0 . If 0 a κ ( ϖ ) A and 0 b ( ϖ ) B , ϖ [ θ , ϰ ] , then
( ω θ + Ψ κ p ) 1 p ( ϰ ) + ( ω θ + Ψ p ) 1 p ( ϰ ) c 5 ω θ + Ψ κ + p 1 p ϰ
with c 5 = A ( a + B ) + B ( b + A ) ( a + B ) ( b + A ) and p 1 .
Proof. 
As a result of the conditions, it follows that
1 B 1 ( ϖ ) 1 b .
Considering the product of (44) and 0 a κ ( ϖ ) A , we obtain
a B κ ( ϖ ) ( ϖ ) A b .
From (45), we obtain
p ( ϖ ) B a + B p ( κ ( ϖ ) + ( ϖ ) ) p
and
κ p ( ϖ ) A b + A p ( κ ( ϖ ) + ( ϖ ) ) p .
Multiplying both sides of (46) and (47) by ω 1 ( ϰ ) Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) and then integrating, we obtain
( ω θ + Ψ p ) 1 p ( ϰ ) B a + B ω θ + Ψ κ + p 1 p ϰ
and
( ω θ + Ψ κ p ) 1 p ( ϰ ) A b + A ω θ + Ψ κ + p 1 p ϰ ,
respectively. Adding (48) and (49) completes the proof of (43). □
Theorem 14.
Let κ , L [ θ , ϰ ] and κ , > 0 on [ 0 , ) , such that ω θ + Ψ κ p ϰ and ω θ + Ψ p ϰ are finite reals for ϰ > θ > 0 . If 0 η κ ( ϖ ) ( ϖ ) for η , R + and for all ϖ [ θ , ϰ ] , then we have
1 ( ω θ + Ψ κ ) ( ϰ ) 1 ( η + 1 ) ( + 1 ) ω θ + Ψ κ + 2 ϰ 1 η ( ω θ + Ψ κ ) ( ϰ ) .
Proof. 
Using 0 η κ ( ϖ ) ( ϖ ) , we have
( ϖ ) ( η + 1 ) ( ϖ ) + κ ( ϖ ) ( ϖ ) ( + 1 ) .
In addition, it follows that 1 ( ϖ ) κ ( ϖ ) 1 η , which yields
κ ( ϖ ) + 1 ( ϖ ) + κ ( ϖ ) κ ( ϖ ) η + 1 η .
Evaluating the product between (50) and (51), we obtain
κ ( ϖ ) ( ϖ ) ( ϖ ) + κ ( ϖ ) 2 ( η + 1 ) ( + 1 ) κ ( ϖ ) ( ϖ ) η .
Multiplying both sides of (52) by ω 1 ( ϰ ) Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) and then integrating, we have
ω 1 ( ϰ ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) κ ( ϖ ) ( ϖ ) d ϖ ω 1 ( ϰ ) ( η + 1 ) ( + 1 ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) ( ϖ ) + κ ( ϖ ) 2 d ϖ ω 1 ( ϰ ) η θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) κ ( ϖ ) ( ϖ ) d ϖ .
Hence,
1 ( ω θ + Ψ κ ) ( ϰ ) 1 ( η + 1 ) ( + 1 ) ω θ + Ψ κ + 2 ϰ 1 η ( ω θ + Ψ κ ) ( ϰ ) .
This completes the proof. □
Theorem 15.
Let κ , L [ θ , ϰ ] and κ , > 0 on [ 0 , ) , such that ω θ + Ψ κ p ϰ and ω θ + Ψ p ϰ are finite reals for ϰ > θ > 0 . If 0 < η κ ( ϖ ) ( ϖ ) holds for η , R + and for all ϖ [ θ , ϰ ] , then
( ω θ + Ψ κ p ) 1 p ( ϰ ) + ( ω θ + Ψ p ) 1 p ( ϰ ) 2 ω θ + Ψ F p ( κ , ) 1 p ϰ
holds. F ( κ ( ϖ ) , ( ϖ ) ) = max η + 1 κ ( ϖ ) ( ϖ ) , ( + η ) ( ϖ ) κ ( ϖ ) η .
Proof. 
From the hypothesis 0 < η κ ( ϖ ) ( ϖ ) , we obtain
0 < η + η κ ( ϖ ) ( ϖ )
and
+ η κ ( ϖ ) ( ϖ ) .
Hence, using (54) and (55), we obtain
( ϖ ) < ( + η ) ( ϖ ) κ ( ϖ ) η F ( κ ( ϖ ) , ( ϖ ) ) ,
where F ( κ ( ϖ ) , ( ϖ ) ) = max η + 1 κ ( ϖ ) ( ϖ ) , ( + η ) ( ϖ ) κ ( ϖ ) η .
Using the hypothesis, it follows that 0 < 1 ( ϖ ) κ ( ϖ ) 1 η . In this way, we have
1 1 + 1 η ( ϖ ) κ ( ϖ )
and
1 + 1 η ( ϖ ) κ ( ϖ ) 1 η .
From (57) and (58), we obtain
1 1 + 1 η κ ( ϖ ) ( ϖ ) κ ( ϖ ) 1 η ,
which can be rewritten as
κ ( ϖ ) 1 + 1 η κ ( ϖ ) ( ϖ ) = η + 1 κ ( ϖ ) ( ϖ ) η + 1 κ ( ϖ ) ( ϖ ) F ( κ ( ϖ ) , ( ϖ ) ) .
We can write from (56) and (59)
κ p ( ϖ ) F p ( κ ( ϖ ) , ( ϖ ) )
p ( ϖ ) F p ( κ ( ϖ ) , ( ϖ ) ) .
Multiplying both sides of (60) by ω 1 ( ϰ ) Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) and then integrating, we have
ω 1 ( ϰ ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) κ p ( ϖ ) d ϖ ω 1 ( ϰ ) θ ϰ Ψ ( ϰ ) ( ϖ ) ( ϰ ) ( ϖ ) ω ( ϖ ) ( ϖ ) F p ( κ ( ϖ ) , ( ϖ ) ) d ϖ .
Accordingly, it can be written as
( ω θ + Ψ κ p ) 1 p ( ϰ ) ω θ + Ψ F p ( κ , ) 1 p ϰ .
Using the same procedure as above, for (61), we have
( ω θ + Ψ p ) 1 p ( ϰ ) ω θ + Ψ F p ( κ , ) 1 p ϰ .
The required result (53) follows from (62) and (63). □

6. Conclusions

Mathematical inequalities are of significant importance in the study of mathematics and related fields. Future research on integral inequalities will now be encouraged by these recommendations to examine how integral inequalities can be extended using fractional calculus operators. In this study, first of all, we presented several definitions of fractional integral operators, and then we showed some inequalities utilizing the monotonicity properties of functions for the generalized weighted-type fractional operators. The acquired results reflect an expansion of certain previously published results. Different types of inequalities were also obtained using this operator. We would like to emphasize, in particular, that these operators can be used to obtain new types of integral inequalities and results.

Author Contributions

Conceptualization, Ç.Y., G.R. and L.-I.C.; methodology, Ç.Y., G.R. and L.-I.C.; validation, Ç.Y., G.R. and L.-I.C.; formal analysis, Ç.Y., G.R. and L.-I.C.; investigation, Ç.Y., G.R. and L.-I.C.; resources: Ç.Y.; writing—original draft preparation, Ç.Y.; writing—review and editing, Ç.Y.; visualization, Ç.Y., G.R. and L.-I.C.; supervision: Ç.Y.; project administration, Ç.Y.; funding acquisition: L.-I.C. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Data Availability Statement

Not applicable.

Conflicts of Interest

The authors declare no conflict of interest.

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MDPI and ACS Style

Yıldız, Ç.; Rahman, G.; Cotîrlă, L.-I. On Further Inequalities for Convex Functions via Generalized Weighted-Type Fractional Operators. Fractal Fract. 2023, 7, 513. https://doi.org/10.3390/fractalfract7070513

AMA Style

Yıldız Ç, Rahman G, Cotîrlă L-I. On Further Inequalities for Convex Functions via Generalized Weighted-Type Fractional Operators. Fractal and Fractional. 2023; 7(7):513. https://doi.org/10.3390/fractalfract7070513

Chicago/Turabian Style

Yıldız, Çetin, Gauhar Rahman, and Luminiţa-Ioana Cotîrlă. 2023. "On Further Inequalities for Convex Functions via Generalized Weighted-Type Fractional Operators" Fractal and Fractional 7, no. 7: 513. https://doi.org/10.3390/fractalfract7070513

APA Style

Yıldız, Ç., Rahman, G., & Cotîrlă, L. -I. (2023). On Further Inequalities for Convex Functions via Generalized Weighted-Type Fractional Operators. Fractal and Fractional, 7(7), 513. https://doi.org/10.3390/fractalfract7070513

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