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Article

Some Double Generalized Weighted Fractional Integral Inequalities Associated with Monotone Chebyshev Functionals

by
Gauhar Rahman
1,
Saud Fahad Aldosary
2,
Muhammad Samraiz
3 and
Kottakkaran Sooppy Nisar
2,*
1
Department of Mathematics and Statistics, Hazara University, Mansehra 21300, Pakistan
2
Department of Mathematics, College of Arts and Sciences, Prince Sattam Bin Abdulaziz University, Wadi Aldawaser 11991, Saudi Arabia
3
Department of Mathematics, University of Sargodha, Sargodha 40100, Pakistan
*
Author to whom correspondence should be addressed.
Fractal Fract. 2021, 5(4), 275; https://doi.org/10.3390/fractalfract5040275
Submission received: 12 November 2021 / Revised: 5 December 2021 / Accepted: 8 December 2021 / Published: 15 December 2021
(This article belongs to the Special Issue Fractional Dynamical Systems: Applications and Theoretical Results)

Abstract

:
In this manuscript, we study the unified integrals recently defined by Rahman et al. and present some new double generalized weighted type fractional integral inequalities associated with increasing, positive, monotone and measurable function F . Also, we establish some new double-weighted inequalities, which are particular cases of the main result and are represented by corollaries. These inequalities are further refinement of all other inequalities associated with increasing, positive, monotone and measurable function existing in literature. The existing inequalities associated with increasing, positive, monotone and measurable function are also restored by applying specific conditions as given in Remarks. Many other types of fractional integral inequalities can be obtained by applying certain conditions on F and Ψ given in the literature.

1. Introduction

In the context of fractional differential equations, integral inequalities are very significant. This field has gained popularity during the last few decades. Various researchers, such as [1,2,3], have investigated the significant developments in this domain. By employing Riemann-Liouville (R-L) fractional integrals, the authors presented Grüss type and several other new inequalities in [4,5]. Certain inequalities for the generalised ( k , ρ ) -fractional integral operator are proposed in [6]. In [7], the modified Hermite-Hadamard type inequalities can be found. Dahmani [8] discovered various fractional integral inequalities employing a family of n positive functions. In [9], Srivastava et al. presented the Chebyshev inequality by employing general family of fractional integral operators. Some remarkable inequalities and their applications can be found in nthe work of [10,11,12,13,14,15].
In [16,17], the Chebyshev functional for the integrable functions Z 1 and Z 1 on [ v 1 , v 2 ] , is given by
H ( Z 1 , Z 2 , ħ 1 ) = v 1 v 2 ħ 1 ( θ ) d θ v 1 v 2 ħ 1 ( θ ) Z 1 ( θ ) Z 2 ( θ ) d θ v 1 v 2 ħ 1 ( θ ) Z 1 ( θ ) d θ v 1 v 2 ħ 1 ( θ ) Z 2 ( θ ) d θ ,
where the function ħ 1 is a positive and integrable on [ v 1 , v 2 ] .
The following extended Chebyshev functional for the integrable functions Z 1 and Z 1 on [ v 1 , v 2 ] can be found in [5,18] by
H ( Z 1 , Z 2 , ħ 1 , ħ 2 ) = v 1 v 2 ħ 2 ( θ ) d θ v 1 v 2 ħ 1 ( θ ) Z 1 ( θ ) Z 2 ( θ ) d θ + v 1 v 2 ħ 1 ( θ ) d θ v 1 v 2 ħ 2 ( θ ) Z 1 ( θ ) Z 2 ( θ ) d θ v 1 v 2 ħ 1 ( θ ) Z 1 ( θ ) d θ v 1 v 2 ħ 2 ( θ ) Z 2 ( θ ) d θ v 1 v 2 ħ 2 ( θ ) Z 1 ( θ ) d θ v 1 v 2 ħ 1 ( θ ) Z 2 ( θ ) d θ .
where the two functions ħ 1 and ħ 2 are positive and integrable on [ v 1 , v 2 ] .
Kuang [19] and Mitrinovic [18] proved that H Z 1 , Z 2 , ħ 1 0 and H Z 1 , Z 2 , ħ 1 , ħ 2 0 if the functions Z 1 and Z 2 are synchronous on [ v 1 , v 2 ] .
Remark 1.
If we take ħ 1 ( θ ) = ħ 2 ( θ ) = 1 , θ [ v 1 , v 2 ] , then
H Z 1 , Z 2 , ħ 1 = 1 2 H Z 1 , Z 2 , ħ 1 , ħ 2 .
Certain remarkable integral inequalities associated with the Chebyshev’s functionals (1) and (2) can be found in the work of [20,21,22,23,24,25].
Awan et al. [26] proposed the following inequality by:
Theorem 1.
Let the function g be an absolutely continuous on [ v 1 , v 2 ] , and ħ 1 be integrable and positive function on [ v 1 , v 2 ] and g 2 L 1 [ v 1 , v 2 ] , then the following inequality holds;
H Φ , Φ , ħ 1 1 G 2 ( v 2 ) v 1 v 2 v 1 θ ħ 1 ( ϱ ) d ϱ v 1 v 2 ϱ ħ 1 ( ϱ ) d ϱ v 1 v 2 ħ 1 ( ϱ ) d ϱ v 1 θ ϱ ħ 1 ( ϱ ) d ϱ Φ ( θ ) 2 d θ ,
where G ( v 2 ) = v 1 v 2 ħ 1 ( ϱ ) d ϱ .
In [27], Bezziou et al. proposed the below result for Riemann-Liouville fractional integral as follows:
Theorem 2.
Assume that the function g : [ v 1 , v 2 ] R be an absolutely continuous function, and the function ħ 1 : [ v 1 , v 2 ] R + be an integrable, and g 2 L 1 [ v 1 , v 2 ] . Then the following inequality for κ > 0 holds:
I v 1 + κ ħ 1 ( v 2 ) I v 1 + κ ħ 1 g 2 ( v 2 ) I v 1 + κ ħ 1 g ( v 2 ) 2 v 1 v 2 Υ ( θ ) g ( θ ) 2 d θ ,
where
Υ ( θ ) = 1 2 I v 1 + κ v 2 ħ 1 ( v 2 ) v 1 θ ( v 2 ϱ ) κ 1 ħ 1 ( ϱ ) d ϱ I v 1 + κ ħ 1 ( v 2 ) v 1 θ ϱ ( v 2 ϱ ) κ 1 ħ 1 ( ϱ ) d ϱ .
Dahmani and Bounoua [28] proposed the following inequality for Riemann-Liouville fractional integral by:
Theorem 3.
If the function g : [ v 1 , v 2 ] R be an absolutely continuous and let ħ 1 : [ v 1 , v 2 ] R + be an integrable function. If Φ 2 L 1 [ v 1 , v 2 ] , then for all κ > 0 , and θ [ v 1 , v 2 ] , the following inequality holds;
1 I v 1 κ ħ 1 ( θ ) I v 1 + θ ħ 1 g 2 ( θ ) 1 I v 1 + κ ħ 1 ( θ ) I v 1 + κ ħ 1 g ( θ ) 2 1 I v 1 + κ ħ 1 ( θ ) 2 v 1 θ P x ( θ ) g ( θ ) 2 d θ ,
with
P x ( θ ) = 1 Γ ( κ ) I v 1 + κ ( x ħ 1 ( x ) ) v 1 θ ħ 1 ( ϱ ) ( x ϱ ) κ 1 d ϱ J v 1 + κ ħ 1 ( x ) v 1 θ ϱ ħ 1 ( ϱ ) ( x ϱ ) κ 1 d ϱ .
Definition 1
([29]). Suppose that the function Ψ : [ 0 , ) [ 0 , ) be satisfying the conditions given below:
0 1 Ψ ( ϱ ) ϱ d ϱ < ,
1 P Ψ ( ħ 1 ) Ψ ( ħ 2 ) P , 1 2 ħ 1 ħ 2 2 ,
Ψ ( ħ 2 ) ħ 2 2 Q Ψ ( ħ 1 ) ħ 1 2 , ħ 1 ħ 2 ,
| Ψ ( ħ 2 ) ħ 2 2 Ψ ( ħ 1 ) ħ 1 2 | S | ħ 2 ħ 1 | Ψ ( ħ 2 ) ħ 2 2 , 1 2 ħ 1 ħ 2 2 ,
where P, Q, S > 0 and are independent of ħ 1 , ħ 2 > 0 . If Ψ ( ħ 2 ) ħ 2 α is increasing for some α > 0 and Ψ ( ħ 2 ) ħ 2 β is decreasing for some β > 0 , then Ψ satisfies (3)–(6).
Next, we recall the following generalized weighted type fractional integral operators recently proposed by Rahman et al. [30].
Definition 2.
The generalized weighted type fractional integral operators both left and right sided are respectively defined by:
ϖ F I v 1 + Ψ Z 1 ( θ ) = ϖ 1 ( θ ) v 1 θ Ψ F ( θ ) F ( ϱ ) F ( θ ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) Z 1 ( ϱ ) d ϱ , v 1 < θ
and
ϖ F I v 2 Ψ Z 1 ( θ ) = ϖ 1 ( θ ) θ v 2 Ψ F ( ϱ ) F ( θ ) F ( ϱ ) F ( θ ) ϖ ( ϱ ) F ( ϱ ) Z 1 ( ϱ ) d ϱ , v 2 > θ .
Remark 2.
1. If we consider Ψ F ( θ ) = F ( θ ) , the fractional integrals (7) and (8) reduce to the following:
ϖ F I v 1 + Z 1 ( θ ) = ϖ 1 ( θ ) v 1 θ ϖ ( ϱ ) F ( ϱ ) Z 1 ( ϱ ) d ϱ , v 1 < θ
and
ϖ F I v 2 Z 1 ( θ ) = ϖ 1 ( θ ) θ v 2 ϖ ( ϱ ) F ( ϱ ) Z 1 ( ϱ ) d ϱ , v 2 > θ ,
respectively.
2. If we consider F ( θ ) = θ , the fractional integrals (7) and (8) reduce to the following respectively
ϖ F I v 1 + Z 1 ( θ ) = ϖ 1 ( θ ) v 1 θ Ψ θ ϱ θ ϱ ϖ ( ϱ ) Z 1 ( ϱ ) d ϱ , v 1 < θ
and
ϖ F I v 2 Z 1 ( θ ) = ϖ 1 ( θ ) θ v 2 Ψ ϱ θ ϱ θ ϖ ( ϱ ) Z 1 ( ϱ ) d ϱ , v 2 > θ .
3. If we consider Ψ ( F ( θ ) ) = F ( θ ) κ Γ ( κ ) , the fractional integrals (7) and (8) reduce to the following respectively (see [31]):
ϖ F I v 1 κ Z 1 ( θ ) = ϖ 1 ( θ ) Γ ( κ ) v 1 θ F ( θ ) F ( ϱ ) κ 1 ϖ ( ϱ ) F ( ϱ ) Z 1 ( ϱ ) d ϱ , v 1 < θ
and
ϖ F I v 2 κ Z 1 ( θ ) = ϖ 1 ( θ ) Γ ( κ ) θ v 2 F ( ϱ ) F ( θ ) κ 1 ϖ ( ϱ ) F ( ϱ ) Z 1 ( ϱ ) d ϱ , v 2 > θ ,
where κ , C with ( κ ) > 0 .
4. If we consider F ( θ ) = θ and Ψ ( F ( θ ) ) = θ κ Γ ( κ ) , the fractional integrals (7) and (8) reduce to the following:
ϖ I v 1 + κ Z 1 ( θ ) = ϖ 1 ( θ ) Γ ( κ ) v 1 θ ( θ ϱ ) κ 1 ϖ ( ϱ ) Z 1 ( ϱ ) d ϱ , v 1 < θ
and
ϖ I v 2 κ Z 1 ( θ ) = ϖ 1 ( θ ) Γ ( κ ) θ v 2 ( ϱ θ ) κ 1 ϖ ( ϱ ) Z 1 ( ϱ ) d ϱ , v 2 > θ ,
respectively.
5. If we consider F ( θ ) = ln θ and Ψ ( F ( θ ) ) = ( ln θ ) κ Γ ( κ ) , the fractional integrals (7) and (8) reduce to the following weighted Hadamard fractional integrals:
ϖ I v 1 + κ Z 1 ( θ ) = ϖ 1 ( θ ) Γ ( κ ) v 1 θ ( ln θ ln ϱ ) κ 1 ϖ ( ϱ ) Z 1 ( ϱ ) d ϱ ϱ , v 1 < θ
and
ϖ I v 2 κ Z 1 ( θ ) = ϖ 1 ( θ ) Γ ( κ ) θ v 2 ( ln ϱ ln θ ) κ 1 ϖ ( ϱ ) Z 1 ( ϱ ) d ϱ ϱ , v 2 > θ .
6. If we consider F ( θ ) = θ η and Ψ ( F ( θ ) ) = θ η η , η > 0 , the fractional integrals (7) and (8) reduce to the following weighted Katugampola fractional integrals,
ϖ I v 1 κ Z 1 ( θ ) = ϖ 1 ( θ ) Γ ( κ ) v 1 θ θ η ϱ η η κ 1 ϖ ( ϱ ) Z 1 ( ϱ ) d ϱ ϱ 1 η , v 1 < θ
and
ϖ I v 2 κ Z 1 ( θ ) = ϖ 1 ( θ ) Γ ( κ ) θ v 2 ϱ η θ η η κ 1 ϖ ( ϱ ) Z 1 ( ϱ ) d ϱ ϱ 1 η , v 2 > θ .
7. If we consider F ( θ ) = θ and Ψ ( F ( θ ) ) = θ η exp 1 η η θ , η ( 0 , 1 ) , the fractional integrals (7) and (8) reduce to the following weighted fractional integrals,
ϖ I v 1 + η Z 1 ( θ ) = ϖ 1 ( θ ) η v 1 θ exp 1 η η ( θ ϱ ) ϖ ( ϱ ) Z 1 ( ϱ ) , v 1 < θ
and
ϖ I v 2 η Z 1 ( θ ) = ϖ 1 ( θ ) η θ v 2 exp 1 η η ( ϱ θ ) ϖ ( ϱ ) Z 1 ( ϱ ) d ϱ , v 2 > θ .
Also, one can derive the weighted form of conformable fractional integrals introduced by [32,33,34,35].
The following special cases can be easily obtained by applying the conditions on ϖ ( θ ) and Ψ F ( θ ) .
Remark 3.
1. If we consider ϖ ( θ ) = 1 and Ψ F ( θ ) = F ( θ ) , the fractional integrals (7) and (8) reduce to the following:
F I v 1 + Z 1 ( θ ) = v 1 θ F ( ϱ ) Z 1 ( ϱ ) d ϱ , v 1 < θ
and
F I v 2 Z 1 ( θ ) = θ v 2 F ( ϱ ) Z 1 ( ϱ ) d ϱ , v 2 > θ ,
respectively.
2. If we consider ϖ ( θ ) = 1 and F ( θ ) = θ , the fractional integrals (7) and (8) reduce to the following respectively (see [36]) as follows:
F I v 1 + Z 1 ( θ ) = v 1 θ Ψ θ ϱ θ ϱ Z 1 ( ϱ ) d ϱ , v 1 < θ
and
F I v 2 Z 1 ( θ ) = θ v 2 Ψ ϱ θ ϱ θ Z 1 ( ϱ ) d ϱ , v 2 > θ .
3. If we consider ϖ ( θ ) = 1 and Ψ ( F ( θ ) ) = F ( θ ) κ Γ ( κ ) , the fractional integrals (7) and (8) reduce to the following respectively (see [37,38]):
F I v 1 κ Z 1 ( θ ) = 1 Γ ( κ ) v 1 θ F ( θ ) F ( ϱ ) κ 1 F ( ϱ ) Z 1 ( ϱ ) d ϱ , v 1 < θ
and
F I v 2 κ Z 1 ( θ ) = 1 Γ ( κ ) θ v 2 F ( ϱ ) F ( θ ) κ 1 F ( ϱ ) Z 1 ( ϱ ) d ϱ , v 2 > θ ,
where κ , C with ( κ ) > 0 .
4. If we consider ϖ ( θ ) = 1 , F ( θ ) = θ and Ψ ( F ( θ ) ) = θ κ Γ ( κ ) , the fractional integrals (7) and (8) reduce to the following (see [37,38]):
I v 1 + κ Z 1 ( θ ) = 1 Γ ( κ ) v 1 θ ( θ ϱ ) κ 1 Z 1 ( ϱ ) d ϱ , v 1 < θ
and
I v 2 κ Z 1 ( θ ) = 1 Γ ( κ ) θ v 2 ( ϱ θ ) κ 1 Z 1 ( ϱ ) d ϱ , v 2 > θ ,
respectively.
5. If we consider ϖ ( θ ) = 1 , F ( θ ) = ln θ and Ψ ( F ( θ ) ) = ( ln θ ) κ Γ ( κ ) , the fractional integrals (7) and (8) reduce to the following weighted Hadamard fractional integrals (see [37,38]):
I v 1 + κ Z 1 ( θ ) = 1 Γ ( κ ) v 1 θ ( ln θ ln ϱ ) κ 1 Z 1 ( ϱ ) d ϱ ϱ , v 1 < θ
and
I v 2 κ Z 1 ( θ ) = 1 Γ ( κ ) θ v 2 ( ln ϱ ln θ ) κ 1 Z 1 ( ϱ ) d ϱ ϱ , v 2 > θ .
6. If we consider ϖ ( θ ) = 1 , F ( θ ) = θ η and Ψ ( F ( θ ) ) = θ η η , η > 0 , the fractional integrals (7) and (8) reduce to the following Katugampola [39] fractional integrals respectively,
I v 1 κ Z 1 ( θ ) = 1 Γ ( κ ) v 1 θ θ η ϱ η η κ 1 Z 1 ( ϱ ) d ϱ ϱ 1 η , v 1 < θ
and
I v 2 κ Z 1 ( θ ) = 1 Γ ( κ ) θ v 2 ϱ η θ η η κ 1 Z 1 ( ϱ ) d ϱ ϱ 1 η , v 2 > θ .
7. If we consider ϖ ( θ ) = 1 F ( θ ) = θ and Ψ ( F ( θ ) ) = θ η exp 1 η η θ , η ( 0 , 1 ) , the fractional integrals (7) and (8) reduce to the following weighted fractional integrals,
I v 1 + η Z 1 ( θ ) = 1 η v 1 θ exp 1 η η ( θ ϱ ) Z 1 ( ϱ ) , v 1 < θ
and
I v 2 η Z 1 ( θ ) = 1 η θ v 2 exp 1 η η ( ϱ θ ) Z 1 ( ϱ ) d ϱ , v 2 > θ .
Similarly, (7) and (8) will lead to the fractional integrals defined by [32,33,34,35].

2. Some Double-Weighted Generalized Fractional Integral Inequalities

In this section, some double-weighted generalized fractional integral inequalities are presented. To this end, we begin by proving the following Lemma.
Lemma 1.
Let the function F be measurable, increasing, positive and monotone function on ( v 1 , v 2 ) , and has a continuous derivative F on [ v 1 , v 2 ] . If Φ : [ v 1 , v 2 ] R is continuous on [ v 1 , v 2 ] , and ħ 1 , ħ 2 : [ u , v ] R + are positive integrable. Then, we have
ϖ F I v 1 + Ψ ħ 1 ( v 2 ) ϖ F I v 1 + Ψ v 2 ( ħ 2 Φ ) ( v 2 ) + ϖ F I v 1 + Ψ Φ ( v 2 ) ϖ F I v 1 + Ψ v 2 ( ħ 1 Φ ) ( v 2 ) ϖ F I v 1 + Ψ v 2 ħ 1 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 Φ ) ( v 2 ) ϖ F I v 1 + Ψ v 2 ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 Φ ) ( v 2 ) = 1 ϖ ( θ ) v 1 v 2 [ ϖ F I v 1 + v 2 ħ 1 ( v 2 ) v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ϖ F I v 1 + Ψ ħ 1 ( v 2 ) v 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ] Φ ( θ ) d θ .
Proof. 
Assume that Z 1 : [ v 1 , v 2 ] R is a continuous function on [ v 1 , v 2 ] . Then, one may gets
ϖ F I v 1 + Ψ ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 Z 1 Φ ) ( v 2 ) + ϖ F I v 1 + Ψ Φ ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 Z 1 Φ ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 Z 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 Φ ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 Z 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 Φ ) ( v 2 ) = 1 ϖ 2 ( θ ) v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) ϖ ( ξ ) F ( ξ ) × Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 1 ( ξ ) ħ 2 ( ϱ ) Z 1 ( ξ ) Z 1 ( ϱ ) Φ ( ξ ) Φ ( ϱ ) d ξ d ϱ .
Consequently, it follows
ϖ F I v 1 + Ψ ħ 1 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 Z 1 Φ ) ( v 2 ) + ϖ F I v 1 + Ψ Φ ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 Z 1 Φ ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 Z 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 Φ ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 Z 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 Φ ) ( v 2 ) = 1 ϖ 2 ( θ ) v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) ϖ ( ξ ) F ( ξ ) × Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 1 ( ξ ) ħ 2 ( ϱ ) Z 1 ( ξ ) Z 1 ( ϱ ) ϱ ξ Φ ( ϑ ) d ϑ d ξ d ϱ .
By utilizing the given condition v 1 ϱ θ ξ v 2 , we get
ϖ F I v 1 + Ψ ħ 1 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 Z 1 Φ ) ( v 2 ) + ϖ F I v 1 + Ψ Φ ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 Z 1 Φ ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 Z 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 Φ ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 Z 1 ) ( x 2 ) ϖ F I v 1 + Ψ ( ħ 1 Φ ) ( x 2 ) = 1 ϖ 2 ( θ ) v 1 v 2 [ v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) × v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) Z 1 ( ξ ) Z 1 ( ϱ ) ϖ ( ξ ) F ( ξ ) ħ 1 ( ξ ) d ξ d ϱ ] Φ ( θ ) d θ .
Applying (11) for the particular case when Z 1 ( v ) = v , then we can write
ϖ F I v 1 + Ψ ħ 1 ( v 2 ) ϖ F I v 1 + Ψ v 2 ( ħ 2 Φ ) ( v 2 ) + ϖ F I v 1 + Ψ Φ ( v 2 ) ϖ F I v 1 + Ψ v 2 ( ħ 1 Φ ) ( v 2 ) ϖ F I v 1 + Ψ v 2 ħ 1 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 Φ ) ( v 2 ) ϖ F I v 1 + Ψ v 2 ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 Φ ) ( v 2 ) = 1 ϖ 2 ( θ ) v 1 v 2 [ v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) × v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) ξ ϱ ϖ ( ξ ) F ( ξ ) ħ 1 ( ξ ) d ξ d ϱ ] Φ ( ϑ ) d ϑ = 1 ϖ ( θ ) v 1 v 2 [ 1 ϖ ( θ ) v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) ϖ ( ξ ) F ( ξ ) ξ ħ 1 ( ξ ) d ξ × v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ 1 ϖ ( θ ) × v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) ϖ ( ξ ) F ( ξ ) ħ 1 ( ξ ) d ξ × v 1 θ × Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ϱ ħ 2 ( ϱ ) d ϱ ] Φ ( θ ) d θ .
Thus with the aid of (7), the above equation gives,
ϖ F I v 1 + Ψ ħ 1 ( v 2 ) ϖ F I v 1 + Ψ v 2 ( ħ 2 Φ ) ( v 2 ) + ϖ F I v 1 + Ψ Φ ( v 2 ) ϖ F I v 1 + Ψ v 2 ( ħ 1 Φ ) ( v 2 ) ϖ F I v 1 + Ψ v 2 ħ 1 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 Φ ) ( v 2 ) ϖ F I v 1 + Ψ v 2 ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 Φ ) ( v 2 ) = 1 ϖ ( θ ) v 1 v 2 [ ϖ F I v 1 + Ψ v 2 ħ 1 ( v 2 ) v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ϱ ħ 2 ( ϱ ) d ϱ ϖ F I v 1 + Ψ ħ 1 ( v 2 ) x 1 ϑ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ϱ ħ 2 ( ϱ ) d ϱ ] Φ ( θ ) d θ ,
which completes the proof. □
Based on Lemma 1, we prove the following theorem.
Theorem 4.
Suppose that the function F be measurable, increasing, positive and monotone function on ( v 1 , v 2 ) , and has a continuous derivative F on ( v 1 , v 2 ) . Assume that Φ : [ v 1 , v 2 ] R is absolutely continuous on [ v 1 , v 2 ] , and ħ 1 , ħ 2 : [ v 1 , v 2 ] R + are positive integrable functions. If ( Φ ) 2 L 1 [ v 1 , v 2 ] . Then, we have
ϖ F I v 1 + Ψ ħ 1 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 Φ 2 ) ( v 2 ) + ϖ F I v 1 + Ψ ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 Φ 2 ) ( v 2 ) 2 ϖ F I v 1 + Ψ ( ħ 1 Φ ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 Φ ) ( v 2 ) 1 ϖ ( θ ) v 1 v 2 [ ϖ F I v 1 + Ψ v 2 ħ 1 ( v 2 ) v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ϖ F I v 1 + Ψ ħ 1 ( v 2 ) v 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ] Φ ( θ ) 2 d θ .
Proof. 
By employing the definition (7) and Lemma 1, we obtain
ϖ F I v 1 + Ψ ħ 1 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 Φ 2 ) ( v 2 ) + ϖ F I v 1 + Ψ ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 Φ 2 ) ( v 2 ) 2 x 1 Ψ I κ θ ( ħ 1 Φ ) ( x 2 ) x 1 Ψ I κ θ ( ħ 2 Φ ) ( x 2 ) = 1 ϖ 2 ( θ ) v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) × Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ξ ) F ( ξ ) ħ 1 ( ξ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) Φ ( ξ ) Φ ( ϱ ) 2 d ξ d ϱ = 1 ϖ 2 ( θ ) v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) × Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ξ ) F ( ξ ) ħ 1 ( ξ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) ξ ϱ 2 Φ ( ξ ) Φ ( ϱ ) ξ ϱ 2 d ξ d ϱ .
Consequently, it follows that
ϖ F I v 1 + Ψ ħ 1 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 Φ 2 ) ( v 2 ) + ϖ F I v 1 + Ψ ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 Φ 2 ) ( v 2 ) 2 x 1 Ψ I κ θ ( ħ 1 Φ ) ( x 2 ) x 1 Ψ I κ θ ( ħ 2 Φ ) ( x 2 ) = 1 ϖ 2 ( θ ) v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) × Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ξ ) F ( ξ ) ħ 1 ( ξ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) ξ ϱ 2 ϱ ξ Φ ( θ ) d θ ξ ϱ 2 d ξ d ϱ .
By applying Cauchy-Schwartz inequality [40], we get
ϖ F I v 1 + Ψ ħ 1 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 Φ 2 ) ( v 2 ) + ϖ F I v 1 + Ψ ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 Φ 2 ) ( v 2 ) 2 x 1 Ψ I κ θ ( ħ 1 Φ ) ( x 2 ) x 1 Ψ I κ θ ( ħ 2 Φ ) ( x 2 ) = 1 ϖ 2 ( θ ) v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) × Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ξ ) F ( ξ ) ħ 1 ( ξ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) ξ ϱ 2 ϱ ξ Φ ( θ ) d θ ξ ϱ 2 d ξ d ϱ 1 ϖ 2 ( θ ) v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) × Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ξ ) F ( ξ ) ħ 1 ( ξ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) ξ ϱ 2 ϱ ξ d θ 1 2 ϱ ξ Φ ( θ ) 2 d θ 1 2 ξ ϱ 2 d ξ d ϱ 1 ϖ 2 ( θ ) v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) × Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ξ ) F ( ξ ) ħ 1 ( ξ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) ξ ϱ 2 ϱ ξ Φ ( θ ) 2 d θ ξ ϱ d ξ d ϱ 1 ϖ 2 ( θ ) v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) × Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ξ ) F ( ξ ) ħ 1 ( ξ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) ξ ϱ ϱ ξ Φ ( θ ) 2 d θ d ξ d ϱ
Hence, using (11) and (13) concludes the proof. □
The following new particular results of Theorem 4 can be easily obtained.
Corollary 1.
Suppose that the function F be measurable, increasing, positive and monotone function on ( v 1 , v 2 ) , and has a continuous derivative F on ( v 1 , v 2 ) . Assume that Φ : [ v 1 , v 2 ] R is absolutely continuous on [ v 1 , v 2 ] , and ħ 1 , ħ 2 : [ v 1 , v 2 ] R + are positive integrable functions. If ( Φ ) 2 L 1 [ v 1 , v 2 ] . Then, we have
ϖ F I v 1 + Ψ 1 ϖ F I v 1 + Ψ ( ħ 2 Φ 2 ) ( v 2 ) + ϖ F I v 1 + Ψ ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( Φ 2 ) ( v 2 ) 2 ϖ F I v 1 + Ψ ( Φ ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 Φ ) ( v 2 ) 1 ϖ ( θ ) v 1 v 2 [ ϖ F I v 1 + Ψ v 2 v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ϖ F I v 1 + Ψ 1 v 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ] Φ ( θ ) 2 d θ .
Proof. 
By considering ħ 1 ( θ ) = 1 , θ [ v 1 , v 2 ] in Theorem 4, the desired result is obtained. □
Corollary 2.
Suppose that the function F be measurable, increasing, positive and monotone function on ( v 1 , v 2 ) , and has a continuous derivative F on ( v 1 , v 2 ) . Assume that Φ : [ v 1 , v 2 ] R is absolutely continuous on [ v 1 , v 2 ] , and ħ 1 , ħ 2 : [ v 1 , v 2 ] R + are positive integrable functions. If ( Φ ) 2 L 1 [ v 1 , v 2 ] . Then, we have
ϖ F I v 1 + Ψ ħ 1 ( v 2 ) ϖ F I v 1 + Ψ ( Φ 2 ) ( v 2 ) + ϖ F I v 1 + Ψ 1 ϖ F I v 1 + Ψ ( ħ 1 Φ 2 ) ( v 2 ) 2 ϖ F I v 1 + Ψ ( ħ 1 Φ ) ( v 2 ) ϖ F I v 1 + Ψ ( Φ ) ( v 2 ) 1 ϖ ( θ ) v 1 v 2 [ ϖ F I v 1 + Ψ v 2 ħ 1 ( v 2 ) v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) d ϱ ϖ F I v 1 + Ψ ħ 1 ( v 2 ) v 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) d ϱ ] Φ ( θ ) 2 d θ .
Proof. 
By considering ħ 2 ( θ ) = 1 , θ [ v 1 , v 2 ] in Theorem 4, desired corollary is proven. □
Corollary 3.
Suppose that the function F be measurable, increasing, positive and monotone function on ( v 1 , v 2 ) , and has a continuous derivative F on ( v 1 , v 2 ) . Assume that Φ : [ v 1 , v 2 ] R is absolutely continuous on [ v 1 , v 2 ] , and ħ 1 , ħ 2 : [ v 1 , v 2 ] R + are positive integrable functions. If ( Φ ) 2 L 1 [ v 1 , v 2 ] . Then, we have
ϖ F I v 1 + Ψ 1 ϖ F I v 1 + Ψ ( Φ 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( Φ ) ( v 2 ) 2 1 ϖ ( θ ) v 1 v 2 [ ϖ F I v 1 + Ψ v 2 v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) d ϱ ϖ F I v 1 + Ψ 1 v 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) d ϱ ] Φ ( θ ) 2 d θ .
Proof. 
Taking ħ 1 ( θ ) = ħ 2 ( θ ) = 1 , ϑ [ v 1 , v 2 ] in Theorem 4, the desired result is obtained. □
Remark 4.
If we consider ϖ ( θ ) = 1 and Ψ ( F ( θ ) ) = F ( θ ) κ Γ ( κ ) , then Theorem 4 and Corollaries 1–3 will reduce to the work of Bezziou et al. [27].
Remark 5.
If we consider ϖ ( θ ) = 1 and Ψ ( F ( θ ) ) = F ( θ ) κ η Γ η ( κ ) , then Theorem 4 and Corollaries 1–3 will reduce to the work of Rahman et al. [41].
Theorem 5.
Let F be measurable, increasing, positive and monotone function on ( v 1 , v 2 ) , and having a continuous derivative F on ( v 1 , v 2 ) . Assume that f 1 , f 2 : [ v 1 , v 2 ] R are absolutely continuous on [ v 1 , v 2 ] , and ħ 1 , ħ 2 : [ v 1 , v 2 ] R + are positive integrable. If ( f 1 ) 2 L 1 [ v 1 , v 2 ] and ( f 2 ) 2 L 1 [ v 1 , v 2 ] . Then, we have
| ϖ F I v 1 + Ψ ħ 1 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 1 f 2 ) ( v 2 ) + ϖ F I v 1 + Ψ ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 1 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 2 ) ( v 2 ) | 1 ϖ ( θ ) ( v 1 v 2 [ ϖ F I v 1 + Ψ v 2 ħ 1 ( v 2 ) v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ϖ F I v 1 + Ψ ħ 1 ( v 2 ) v 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ] f 1 ( θ ) 2 d θ ) 1 2 × ( v 1 v 2 [ ϖ F I v 1 + Ψ v 2 ħ 1 ( v 2 ) v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ϖ F I v 1 + Ψ ħ 1 ( v 2 ) v 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ] f 2 ( θ ) 2 d θ ) 1 2
Proof. 
Consider the left-hand side of (14), we have
| ϖ F I v 1 + Ψ ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 1 f 2 ) ( v 2 ) + ϖ F I v 1 + Ψ ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 1 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 2 ) ( v 2 ) | 1 ϖ 2 ( θ ) ( v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) × ϖ ( ξ ) F ( ξ ) ϖ ( ϱ ) F ( ϱ ) ħ 1 ( ξ ) ħ 2 ( ϱ ) f 1 ( ξ ) f 1 ( ϱ ) 2 d ξ d ϱ ) 1 2 × ( v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) × Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ξ ) F ( ξ ) ϖ ( ϱ ) F ( ϱ ) ħ 1 ( ξ ) ħ 2 ( ϱ ) f 2 ( ξ ) f 2 ( ϱ ) 2 d ξ d ϱ ) 1 2 1 ϖ 2 ( θ ) ( v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ξ ) F ( ξ ) ϖ ( ϱ ) F ( ϱ ) ħ 1 ( ξ ) ħ 2 ( ϱ ) × ϱ ξ f 1 ( θ ) d θ 2 d ξ d ϱ ) 1 2 × ( v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) × Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ξ ) F ( ξ ) ϖ ( ϱ ) F ( ϱ ) ħ 1 ( ξ ) ħ 2 ( ϱ ) ϱ ξ f 2 ( θ ) d θ 2 d ξ d ϱ ) 1 2
Applying Cauchy-Schwartz inequality [40] to the above equation yields,
| ϖ F I v 1 + Ψ ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 1 f 2 ) ( v 2 ) + ϖ F I v 1 + Ψ ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 1 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 2 ) ( v 2 ) | 1 ϖ 2 ( θ ) { v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ξ ) F ( ξ ) ϖ ( ϱ ) F ( ϱ ) ħ 1 ( ξ ) ħ 2 ( ϱ ) × ϱ ξ d θ 1 2 ϱ ξ f 1 ( θ ) 2 d θ 1 2 2 d ξ d ϱ } 1 2 × { v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) × Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ξ ) F ( ξ ) ϖ ( ϱ ) F ( ϱ ) ħ 1 ( ξ ) ħ 2 ( ϱ ) ϱ ξ d θ 1 2 ϱ ξ f 2 ( θ ) 2 d θ 1 2 2 d ξ d ϱ } 1 2 1 ϖ 2 ( θ ) { v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ξ ) F ( ξ ) ϖ ( ϱ ) F ( ϱ ) ħ 1 ( ξ ) ħ 2 ( ϱ ) × ϖ ( ϱ ) F ( ϱ ) ħ 1 ( ξ ) ħ 2 ( ϱ ) ξ ϱ ϱ ξ f 1 ( θ ) 2 d θ d ξ d ϱ } 1 2 × { v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) × Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ξ ) F ( ξ ) ϖ ( ϱ ) F ( ϱ ) ħ 1 ( ξ ) ħ 2 ( ϱ ) ξ ϱ ϱ ξ f 2 ( θ ) 2 d θ d ξ d ϱ } 1 2 1 ϖ 2 ( θ ) { v 1 v 2 ( v 1 v 2 ξ Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) ϖ ( ξ ) F ( ξ ) ħ 1 ( ξ ) d ξ × v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) ϖ ( ξ ) F ( ξ ) ħ 1 ( ξ ) d ξ × x 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ) f 1 ( θ ) 2 } 1 2 { v 1 v 2 ( v 1 v 2 ξ Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) ϖ ( ξ ) F ( ξ ) ħ 1 ( ξ ) d ξ × v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) ϖ ( ξ ) F ( ξ ) ħ 1 ( ξ ) d ξ × x 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ) f 2 ( θ ) 2 ] 1 2 .
In view of (7), we get the desired proof of (14). □
Corollary 4.
Let F be measurable, increasing, positive and monotone function on ( v 1 , v 2 ) , and having a continuous derivative F on ( v 1 , v 2 ) . Assume that f 1 , f 2 : [ v 1 , v 2 ] R are absolutely continuous on [ v 1 , v 2 ] , and ħ 1 , ħ 2 : [ v 1 , v 2 ] R + are positive integrable. If ( f 1 ) 2 L 1 [ v 1 , v 2 ] and ( f 2 ) 2 L 1 [ v 1 , v 2 ] . Then, we have
| ϖ F I v 1 + Ψ ( 1 ) ϖ F I v 1 + Ψ ( ħ 2 f 1 f 2 ) ( v 2 ) + ϖ F I v 1 + Ψ ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( f 1 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 2 ) ( v 2 ) | 1 ϖ ( θ ) ( v 1 v 2 [ ϖ F I v 1 + Ψ v 2 v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ϖ F I v 1 + Ψ ( 1 ) v 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ] f 1 ( θ ) 2 d θ ) 1 2 × ( v 1 v 2 [ ϖ F I v 1 + Ψ v 2 v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ϖ F I v 1 + Ψ ( 1 ) v 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ] f 2 ( θ ) 2 d θ ) 1 2
Proof. 
Applying Theorem 5 for ħ 1 ( θ ) = 1 , θ [ v 1 , v 2 ] , the desired result is obtained. □
Corollary 5.
Let F be measurable, increasing, positive and monotone function on ( v 1 , v 2 ) , and having a continuous derivative F on ( v 1 , v 2 ) . Assume that f 1 , f 2 : [ v 1 , v 2 ] R are absolutely continuous on [ v 1 , v 2 ] , and ħ 1 , ħ 2 : [ v 1 , v 2 ] R + are positive integrable. If ( f 1 ) 2 L 1 [ v 1 , v 2 ] and ( f 2 ) 2 L 1 [ v 1 , v 2 ] . Then, we have
| ϖ F I v 1 + Ψ ħ 1 ( v 2 ) ϖ F I v 1 + Ψ ( f 1 f 2 ) ( v 2 ) + ϖ F I v 1 + Ψ ( 1 ) ϖ F I v 1 + Ψ ( ħ 1 f 1 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( f 2 ) ( v 2 ) | 1 ϖ ( θ ) ( v 1 v 2 [ ϖ F I v 1 + Ψ v 2 ħ 1 ( v 2 ) v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) d ϱ ϖ F I v 1 + Ψ ħ 1 ( v 2 ) v 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) d ϱ ] f 1 ( θ ) 2 d θ ) 1 2 × ( v 1 v 2 [ ϖ F I v 1 + Ψ v 2 ħ 1 ( v 2 ) v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) d ϱ ϖ F I v 1 + Ψ ħ 1 ( v 2 ) v 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) d ϱ ] f 2 ( θ ) 2 d θ ) 1 2
Proof. 
Applying Theorem 5 for ħ 2 ( θ ) = 1 , θ [ v 1 , v 2 ] , the desired result is obtained. □
Corollary 6.
Let F be measurable, increasing, positive and monotone function on ( v 1 , v 2 ) , and having a continuous derivative F on ( v 1 , v 2 ) . Assume that f 1 , f 2 : [ v 1 , v 2 ] R are absolutely continuous on [ v 1 , v 2 ] , and ħ 1 , ħ 2 : [ v 1 , v 2 ] R + are positive integrable. If ( f 1 ) 2 L 1 [ v 1 , v 2 ] and ( f 2 ) 2 L 1 [ v 1 , v 2 ] . Then, we have
| ϖ F I v 1 + Ψ ( 1 ) ϖ F I v 1 + Ψ ( f 1 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( f 2 ) ( v 2 ) | 1 ϖ ( θ ) ( v 1 v 2 [ ϖ F I v 1 + Ψ v 2 v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) d ϱ ϖ F I v 1 + Ψ ( 1 ) v 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) d ϱ ] f 1 ( θ ) 2 d θ ) 1 2 × ( v 1 v 2 [ ϖ F I v 1 + Ψ v 2 v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) d ϱ ϖ F I v 1 + Ψ ( 1 ) v 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) d ϱ ] f 2 ( θ ) 2 d θ ) 1 2
Proof. 
Applying Theorem 5 for ħ 1 ( θ ) = ħ 2 ( θ ) = 1 , θ [ v 1 , v 2 ] , the desired result is obtained. □
Theorem 6.
Let F be measurable, increasing, positive and monotone function on ( v 1 , v 2 ) , and having a continuous derivative F on ( v 1 , v 2 ) . Assume that f 1 : [ v 1 , v 2 ] R is absolutely continuous function on [ v 1 , v 2 ] , and f 2 : [ v 1 , v 2 ] R is non-decreasing on [ v 1 , v 2 ] . Moreover, suppose that both ħ 1 , ħ 2 : [ v 1 , v 2 ] R + are positive integrable. If ( f 1 ) L [ v 1 , v 2 ] , then we have
| ϖ F I v 1 + Ψ ħ 1 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 1 f 2 ) ( v 2 ) + ϖ F I v 1 + Ψ ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 1 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 2 ) ( v 2 ) | | | f 1 | | ϖ ( θ ) [ ϖ F I v 1 + Ψ v 2 ħ 1 ( v 2 ) v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ϖ F I v 1 + Ψ ħ 1 ( v 2 ) v 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ] v 1 v 2 f 2 ( θ ) d θ .
Proof. 
Consider the left-hand side of (15), we have
| ϖ F I v 1 + Ψ ħ 1 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 1 f 2 ) ( v 2 ) + ϖ F I v 1 + Ψ ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 1 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 2 ) ( v 2 ) | 1 ϖ 2 ( θ ) | v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ξ ) F ( ξ ) ϖ ( ϱ ) F ( ϱ ) ħ 1 ( ξ ) ħ 2 ( ϱ ) × f 1 ( ξ ) f 1 ( ϱ ) f 2 ( ξ ) f 2 ( ϱ ) d ξ d ϱ | 1 ϖ 2 ( θ ) | v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ξ ) F ( ξ ) ϖ ( ϱ ) F ( ϱ ) ħ 1 ( ξ ) ħ 2 ( ϱ ) × | f 1 ( ξ ) f 1 ( ϱ ) ξ ϱ | | ( ξ ϱ ) f 2 ( ξ ) f 2 ( ϱ ) | d ξ d ϱ | | f 1 | | ϖ 2 ( θ ) | v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ξ ) F ( ξ ) ϖ ( ϱ ) F ( ϱ ) ħ 1 ( ξ ) ħ 2 ( ϱ ) × ξ ϱ v 1 v 2 f 2 ( ϑ ) d ϑ d ξ d ϱ | | f 1 | | ϖ 2 ( θ ) [ v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) ξ F ( ξ ) ϖ ( ξ ) ħ 2 ( ξ ) d ξ v 1 θ Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ϱ ) F ( ϱ ) ϖ ( ϱ ) ħ 2 ( ϱ ) d ϱ v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) F ( ξ ) ϖ ( ϱ ) ħ 2 ( ξ ) d ξ v 1 θ ϱ Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ϱ ) F ( ϱ ) ϖ ( ϱ ) ħ 2 ( ϱ ) d ϱ ] v 1 v 2 f 2 ( θ ) d θ
Hence taking (7) into account, the proof of (15) is completed.
Corollary 7.
Let F be measurable, increasing, positive and monotone function on ( v 1 , v 2 ) , and having a continuous derivative F on ( v 1 , v 2 ) . Assume that f 1 : [ v 1 , v 2 ] R is absolutely continuous function on [ v 1 , v 2 ] , and f 2 : [ v 1 , v 2 ] R is non-decreasing on [ v 1 , v 2 ] . Moreover, suppose that both ħ 1 , ħ 2 : [ v 1 , v 2 ] R + are positive integrable. If ( f 1 ) L [ v 1 , v 2 ] , then we have
| ϖ F I v 1 + Ψ ( 1 ) ϖ F I v 1 + Ψ ( ħ 2 f 1 f 2 ) ( v 2 ) + ϖ F I v 1 + Ψ ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( f 1 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( f 2 ) ( v 2 ) | | | f 1 | | ϖ ( θ ) [ ϖ F I v 1 + Ψ v 2 v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ϖ F I v 1 + Ψ ( 1 ) v 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ] v 1 v 2 f 2 ( θ ) d θ .
Proof. 
Applying Theorem 6 for ħ 1 ( θ ) = 1 , θ [ v 1 , v 2 ] , the desired result is obtained. □
Corollary 8.
Let F be measurable, increasing, positive and monotone function on ( v 1 , v 2 ) , and having a continuous derivative F on ( v 1 , v 2 ) . Assume that f 1 : [ v 1 , v 2 ] R is absolutely continuous function on [ v 1 , v 2 ] , and f 2 : [ v 1 , v 2 ] R is non-decreasing on [ v 1 , v 2 ] . Moreover, suppose that both ħ 1 , ħ 2 : [ v 1 , v 2 ] R + are positive integrable. If ( f 1 ) L [ v 1 , v 2 ] , then we have
| ϖ F I v 1 + Ψ ħ 1 ( v 2 ) ϖ F I v 1 + Ψ ( f 1 f 2 ) ( v 2 ) + ϖ F I v 1 + Ψ ( 1 ) ϖ F I v 1 + Ψ ( f 1 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 2 ) ( v 2 ) | | | f 1 | | ϖ ( θ ) [ ϖ F I v 1 + Ψ v 2 ħ 1 ( v 2 ) v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) d ϱ ϖ F I v 1 + Ψ ħ 1 ( v 2 ) v 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) d ϱ ] v 1 v 2 f 2 ( θ ) d θ .
Proof. 
Applying Theorem 6 for ħ 2 ( θ ) = 1 , θ [ v 1 , v 2 ] , the desired result is obtained. □
Corollary 9.
Let F be measurable, increasing, positive and monotone function on ( v 1 , v 2 ) , and having a continuous derivative F on ( v 1 , v 2 ) . Assume that f 1 : [ v 1 , v 2 ] R is absolutely continuous function on [ v 1 , v 2 ] , and f 2 : [ v 1 , v 2 ] R is non-decreasing on [ v 1 , v 2 ] . Moreover, suppose that both ħ 1 , ħ 2 : [ v 1 , v 2 ] R + are positive integrable. If ( f 1 ) L [ v 1 , v 2 ] , then we have
| ϖ F I v 1 + Ψ ( 1 ) ϖ F I v 1 + Ψ ( f 1 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( f 2 ) ( v 2 ) | | | f 1 | | 2 ϖ ( θ ) [ ϖ F I v 1 + Ψ v 2 v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) d ϱ ϖ F I v 1 + Ψ ( 1 ) v 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) d ϱ ] v 1 v 2 f 2 ( θ ) d θ .
Proof. 
Applying Theorem 6 for ħ 1 ( θ ) = ħ 2 ( θ ) = 1 , θ [ v 1 , v 2 ] , the desired result is obtained. □
Theorem 7.
Let F be measurable, increasing, positive and monotone function on ( v 1 , v 2 ) and having continuous derivative F on ( v 1 , v 2 ) . Assume that f 1 , f 2 : [ v 1 , v 2 ] R are absolutely continuous on [ v 1 , v 2 ] and f 2 : [ v 1 , v 2 ] R is nondecreasing on [ v 1 , v 2 ] . Suppose that ħ 1 , ħ 2 : [ v 1 , v 2 ] R + are positive integrable. If f 1 , f 2 L [ v 1 , v 2 ] , then we have
| ϖ F I v 1 + Ψ ħ 1 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 1 f 2 ) ( v 2 ) + ϖ F I v 1 + Ψ ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 1 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 2 ) ( x 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 2 ) ( v 2 ) | | | f 1 | | | | f 2 | | ϖ ( θ ) [ ϖ F I v 1 + Ψ v 2 ħ 1 ( v 2 ) v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ 2 ϖ F I v 1 + Ψ v 2 ħ 1 ( v 2 ) v 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ + ϖ F I v 1 + Ψ ħ 1 ( v 2 ) v 1 θ ϱ 2 Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ] .
Proof. 
Consider the left-hand side of (16), we have
| ϖ F I v 1 + Ψ ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 1 f 2 ) ( v 2 ) + ϖ F I v 1 + Ψ ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 1 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 2 ) ( x 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 2 ) ( v 2 ) | 1 ϖ 2 ( θ ) | v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) × ϖ ( ξ ) F ( ξ ) ħ 1 ( ξ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) f 1 ( ξ ) f 1 ( ϱ ) f 2 ( ξ ) f 2 ( ϱ ) d ξ d ϱ | 1 ϖ 2 ( θ ) v 1 v 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ξ ) F ( ξ ) ħ 1 ( ξ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) × | f 1 ( ξ ) f 1 ( ϱ ) ξ ϱ | | f 2 ( ξ ) f 2 ( ϱ ) ξ ϱ | ( ξ ϱ ) 2 d ξ d ϱ | | f 1 | | | | f 2 | | ϖ 2 ( θ ) v 1 v 2 v 1 θ Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ξ ) F ( ξ ) ħ 1 ( ξ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) × ξ 2 2 ξ ϱ + ϱ 2 d ξ d ϱ | | f 1 | | | | f 2 | | ϖ 2 ( θ ) [ v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) ξ 2 ϖ ( ξ ) F ( ξ ) ħ 1 ( ξ ) d ξ v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ 2 v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) ξ ϖ ( ξ ) F ( ξ ) ħ 1 ( ξ ) d ξ v 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ + v 1 v 2 Ψ F ( v 2 ) F ( ξ ) F ( v 2 ) F ( ξ ) ϖ ( ξ ) F ( ξ ) ħ 1 ( ξ ) d ξ v 1 θ ϱ 2 Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ] .
Hence, by using (7), the proof of the theorem is completed. □
Corollary 10.
Let F be measurable, increasing, positive and monotone function on ( v 1 , v 2 ) and having continuous derivative F on ( v 1 , v 2 ) . Assume that f 1 , f 2 : [ v 1 , v 2 ] R are absolutely continuous on [ v 1 , v 2 ] and f 2 : [ v 1 , v 2 ] R is nondecreasing on [ v 1 , v 2 ] . Suppose that ħ 1 , ħ 2 : [ v 1 , v 2 ] R + are positive integrable. If f 1 , f 2 L [ v 1 , v 2 ] , then we have
| ϖ F I v 1 + Ψ ( 1 ) ϖ F I v 1 + Ψ ( ħ 2 f 1 f 2 ) ( v 2 ) + ϖ F I v 1 + Ψ ħ 2 ( v 2 ) ϖ F I v 1 + Ψ ( f 1 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( f 2 ) ( x 2 ) ϖ F I v 1 + Ψ ( f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 2 f 2 ) ( v 2 ) | | | f 1 | | | | f 2 | | ϖ ( θ ) [ ϖ F I v 1 + Ψ v 2 v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ 2 ϖ F I v 1 + Ψ v 2 v 1 θ ϱ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ + ϖ F I v 1 + Ψ ( 1 ) v 1 θ ϱ 2 Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) ħ 2 ( ϱ ) d ϱ ] .
Proof. 
Setting ħ 1 ( θ ) = 1 , θ [ v 1 , v 2 ] in Theorem 7, then the desired result is obtained. □
Corollary 11.
Let F be measurable, increasing, positive and monotone function on ( v 1 , v 2 ) and having continuous derivative F on ( v 1 , v 2 ) . Assume that f 1 , f 2 : [ v 1 , v 2 ] R are absolutely continuous on [ v 1 , v 2 ] and f 2 : [ v 1 , v 2 ] R is nondecreasing on [ v 1 , v 2 ] . Suppose that ħ 1 , ħ 2 : [ v 1 , v 2 ] R + are positive integrable. If f 1 , f 2 L [ v 1 , v 2 ] , then we have
| ϖ F I v 1 + Ψ ħ 1 ( v 2 ) ϖ F I v 1 + Ψ ( f 1 f 2 ) ( v 2 ) + ϖ F I v 1 + Ψ ( 1 ) ϖ F I v 1 + Ψ ( ħ 1 f 1 f 2 ) ( v 2 ) ϖ F I v 1 + Ψ ( f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 2 ) ( x 2 ) ϖ F I v 1 + Ψ ( ħ 1 f 1 ) ( v 2 ) ϖ F I v 1 + Ψ ( f 2 ) ( v 2 ) | | | f 1 | | | | f 2 | | ϖ ( θ ) [ ϖ F I v 1 + Ψ v 2 ħ 1 ( v 2 ) v 1 θ Ψ F ( v 2 ) F ( ϱ ) F ( v 2 ) F ( ϱ ) ϖ ( ϱ ) F ( ϱ ) d ϱ 2 ϖ F I v 1 + Ψ v 2 ħ 1 ( v 2 )