# Hermite-Hadamard-Fejér Inequalities for Conformable Fractional Integrals via Preinvex Functions

Authors: Yousaf Khurshid; Muhammad Adil Khan; Yu-Ming Chu; Zareen Abdulhameed Khan
Journal: Journal of Function Spaces
 (2019)

---

## Abstract

In this paper, we present a Hermite-Hadamard-Fejér inequality for conformable fractional integrals by using symmetric preinvex functions. We also establish an identity associated with the right hand side of Hermite-Hadamard inequality for preinvex functions; then by using this identity and preinvexity of functions and some well-known inequalities, we find several new Hermite-Hadamard type inequalities for conformal fractional integrals.


---

## Body

## 1. Introduction

LetI
∈
R be an interval and h
:
I
→
R be a convex function defined on I such that κ
1, κ
2
∈
I with κ
1
<
κ
2
. Then the well-known H
H (Hermite-Hadamard) inequality [1] states that(1)
h
κ
1
+
κ
2
2
≤
1
κ
2
-
κ
1
∫
κ
1
κ
2
h
x
d
x
≤
h
κ
1
+
h
κ
2
2holds. If the function h is concave on I, then both inequalities in (1) hold in the reverse direction.In the last few years, many researchers have shown their extensive attention on the generalizations, extensions, variations, refinements, and applications of theH
H inequality (see [2–15]). The most well-known generalization of the H
H inequality is the Hermite-Hadamard-Fejér inequality [16]. In 1906, Fejér [16] established the following weighted generalization of the Hermite-Hadamard inequality for symmetric functions:(2)
h
κ
1
+
κ
2
2
∫
κ
1
κ
2
g
x
d
x
≤
1
κ
2
-
κ
1
∫
κ
1
κ
2
g
x
h
x
d
x
≤
h
κ
1
+
h
κ
2
2
∫
κ
1
κ
2
g
x
d
xfor all convex functions h
:
I
→
R, κ
1
,
κ
2
∈
I with κ
1
<
κ
2 and g
:
[
κ
1
,
κ
2
]
→
R
+ is symmetric with respect to κ
1
+
κ
2
/
2.It is well known that the convex sets and convex functions play important roles in the nonlinear programming and optimization theory. Many generalizations and extensions have been considered for the classical convexity in the last few decades. A significant generalization of convex functions is that of invex functions introduced by Hanson in [17]. The basic properties of the preinvex functions and their roles in optimization theory can be found in [18]. The H
H inequalities for preinvex and log-preinvex functions were established by Noor [19, 20].Now, we recall some notions and definitions in invexity analysis, which will be used throughout the paper (see [21, 22] and references therein).LetA
∈
R be a nonempty set and the functions h
:
A
→
R and Ψ
:
A
×
A
→
R be continuous.Definition 1.
The setA
⊆
R
n is said to be invex with respect to Ψ(.,.) if(3)
μ
1
+
s
Ψ
μ
2
,
μ
1
∈
Afor all μ
1, μ
2
∈
A and s
∈
[
0,1
].The invex setA is also called a Ψ-connected set. If Ψ
(
μ
2
,
μ
1
)
=
μ
2
-
μ
1, then the invex set is also a convex set, but some of the invex sets are not convex [21].Definition 2.
The functionh is said to be preinvex with respect to Ψ on the invex set A if(4)
h
μ
1
+
s
Ψ
μ
2
,
μ
1
≤
1
-
s
h
μ
1
+
s
h
μ
2for all μ
1, μ
2
∈
A and s
∈
[
0,1
]. The function h is called preconcave if -
h is preinvex.The following Condition C was introduced by Mohan and Neogy [23].Condition C.
SupposeA is an open invex subset of R
n with respect to Ψ
:
A
×
A
→
R and Ψ satisfies(5)
Ψ
μ
2
,
μ
2
+
s
Ψ
μ
1
,
μ
2
=
-
s
Ψ
μ
1
,
μ
2
,
Ψ
μ
1
,
μ
2
+
s
Ψ
μ
1
,
μ
2
=
1
-
s
Ψ
μ
1
,
μ
2for any μ
1, μ
2
∈
A and s
∈
[
0,1
].From Condition C, we clearly see that(6)
Ψ
μ
2
+
s
2
Ψ
μ
1
,
μ
2
,
μ
2
+
s
1
Ψ
μ
1
,
μ
2
=
s
2
-
s
1
Ψ
μ
1
,
μ
2for any μ
1, μ
2
∈
A and s
∈
[
0,1
].The followingH
H inequality for the preinvex functions was proved by Noor [20].Theorem 3.
Leth
:
K
=
[
κ
1
,
κ
1
+
Ψ
(
κ
2
,
κ
1
)
]
→
(
0
,
∞
) be a preinvex function on the interval K
∘ (the interior of K) and κ
1, κ
2
∈
K
∘ with κ
1
<
κ
1
+
Ψ
(
κ
2
,
κ
1
). Then the following inequality holds:(7)
h
2
κ
1
+
Ψ
κ
2
,
κ
1
2
≤
1
Ψ
κ
2
,
κ
1
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
h
x
d
x
≤
h
κ
1
+
h
κ
2
2
.Several important variants ofH
H inequality for preinvex functions have been provided in the literature [24]. Recently, the authors in [25] defined a new well-behaved simple fractional derivative called the “conformable fractional derivative”. Namely, the conformable fractional derivative of order 0
<
α
≤
1 at s
>
0 for the function h
:
[
0
,
∞
)
→
R is defined by (8)
D
α
h
s
=
l
i
m
ϵ
→
0
h
s
+
ϵ
s
1
-
α
-
h
s
ϵ
.If the conformable fractional derivative of h of order α exists, then we say that h is α-fractional differentiable. The fractional derivative at 0 is defined as h
α
(
0
)
=
lim
s
→
0
+
⁡
h
α
(
s
).Next, we present some basic results related to conformable fractional derivative in the following theorem.Theorem 4 (see [25]).
Letα
∈
(
0,1
] and h
1
,
h
2 be α-differentiable at a point s
>
0. Then (i)
d
α
/
d
α
s
s
n
=
n
s
n
-
α for all n
∈
R.
(ii)
d
α
/
d
α
s
(
c
)
=
0 for any constant c
∈
R
.
(iii)
d
α
/
d
α
s
κ
1
h
1
(
s
)
+
κ
2
h
2
(
s
)
=
κ
1
d
α
/
d
α
s
(
h
1
(
s
)
)
+
κ
2
d
α
/
d
α
s
(
h
2
(
s
)
) for all κ
1
,
κ
2
∈
R
.
(iv)
d
α
/
d
α
s
h
1
(
s
)
h
2
(
s
)
=
h
1
(
s
)
d
α
/
d
α
s
(
h
2
(
s
)
)
+
h
2
(
s
)
d
α
/
d
α
s
(
h
1
(
s
)
)
.
(v)
d
α
/
d
α
s
h
1
(
s
)
/
h
2
(
s
)
=
h
2
(
s
)
d
α
/
d
α
s
(
h
1
(
s
)
)
-
h
1
(
s
)
d
α
/
d
α
s
(
h
2
(
s
)
)
/
h
2
s
2
.
(vi)
d
α
/
d
α
s
h
1
∘
h
2
s
=
h
1
′
h
2
s
d
α
/
d
α
s
h
2
s if h
1 differentiable at h
2
(
s
)
.
If in additionh
1 is differentiable, then(9)
d
α
d
α
s
h
1
s
=
s
1
-
α
d
d
s
h
1
s
.Definition 5 (see [25] conformable fractional integral).
Letα
∈
(
0,1
] and 0
≤
κ
1
<
κ
2. A function h
1
:
[
κ
1
,
κ
2
]
→
R is α-fractional integrable on [
κ
1
,
κ
2
] if the integral(10)
∫
κ
1
κ
2
h
1
x
d
α
x
≔
∫
κ
1
κ
2
h
1
x
x
α
-
1
d
xexists and is finite. All α-fractional integrable functions on [
κ
1
,
κ
2
] are indicated by L
α
(
[
κ
1
,
κ
2
]
)
.Remark 6.
(11)
I
α
κ
1
h
1
s
=
I
1
κ
1
s
α
-
1
h
1
=
∫
κ
1
s
h
1
x
x
1
-
α
d
x
,where the integral is the usual Riemann improper integral and α
∈
(
0,1
].Recently, the conformable integrals and derivatives have been the subject of intensive research, and many remarkable properties and inequalities involving the conformable integrals and derivatives can be found in the literature [26–38].In [39], Anderson provided the conformable integral version of H
H inequality as follows.Theorem 7 (see [39]).
Ifα
∈
(
0,1
] and h
1
:
[
κ
1
,
κ
2
]
→
R is an α-fractional differentiable function such that D
α
h is increasing, then we have the following inequality:(12)
α
κ
2
α
-
κ
1
α
∫
κ
1
κ
2
h
x
d
α
x
≤
h
κ
1
+
h
κ
2
2
.Moreover if the function h is decreasing on [
κ
1
,
κ
2
], then we have(13)
h
κ
1
+
κ
2
2
≤
α
κ
2
α
-
κ
1
α
∫
κ
1
κ
2
h
x
d
α
x
.If α
=
1, then this reduces to the classical H
H inequality.In this paper, we first establish the Hermite-Hadamard-Fejér inequality for conformable fractional integrals by using symmetric preinvex functions; then we presentH
H inequalities as their special cases (see Corollary 9). Secondly, we give an identity associated with the right side of H
H inequality for preinvex functions using the conformable fractional integrals; then we establish H
H inequalities for preinvex functions by use of Hölder inequality, power mean inequality, and preinvexity of functions.


## 2. Hermite-Hadamard-Fejér Inequalities for Conformable Fractional Integrals

The preinvex version of Fejer-Hermite-Hadamard inequality can be represented in conformable fractional integrals forms as follows.Theorem 8.
Suppose thatκ
1, κ
2
∈
K such that Ψ
(
κ
2
,
κ
1
)
>
0, h
:
K
=
[
κ
1
,
κ
1
+
Ψ
(
κ
2
,
κ
1
)
]
→
(
0
,
∞
) is a preinvex function and symmetric with respect to 2
κ
1
+
Ψ
(
κ
2
,
κ
1
)
/
2, and g
:
[
κ
1
,
κ
2
]
→
R is a nonnegative integrable function. Also assume that Ψ satisfies Condition C; then the inequality(14)
h
2
κ
1
+
Ψ
κ
2
,
κ
1
2
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
g
x
d
α
x
≤
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
h
x
g
x
d
α
x
≤
h
κ
1
+
h
κ
1
+
Ψ
κ
2
,
κ
1
2
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
g
x
d
α
xholds for any α
∈
(
0,1
].Proof.
Sinceh
:
K
→
R is preinvex function and is symmetric with respect to 2
κ
1
+
Ψ
(
κ
2
,
κ
1
)
/
2, then for any x, y
∈
K and t
=
1
/
2, we have(15)
h
x
+
Ψ
y
,
x
2
≤
h
x
+
h
y
2i.e., with x
=
κ
1
+
(
1
-
s
)
Ψ
(
κ
2
,
κ
1
) and y
=
κ
1
+
s
Ψ
(
κ
2
,
κ
1
), inequality (15) becomes (16)
h
κ
1
+
1
-
s
Ψ
κ
2
,
κ
1
+
Ψ
κ
1
+
s
Ψ
κ
2
,
κ
1
,
κ
1
+
1
-
s
Ψ
κ
2
,
κ
1
2
=
h
κ
1
+
1
-
s
Ψ
κ
2
,
κ
1
+
s
-
1
+
s
Ψ
κ
2
,
κ
1
2
u
s
i
n
g
C
o
n
d
i
t
i
o
n
C
=
h
κ
1
+
1
-
s
Ψ
κ
2
,
κ
1
+
2
s
-
1
Ψ
κ
2
,
κ
1
2
=
h
2
κ
1
+
Ψ
κ
2
,
κ
1
2
≤
h
κ
1
+
1
-
s
Ψ
κ
2
,
κ
1
+
h
κ
1
+
s
Ψ
κ
2
,
κ
1
2
=
h
κ
1
+
s
Ψ
κ
2
,
κ
1
.
h
i
s
s
y
m
m
e
t
r
i
cBy change of variables, we have (17)
h
2
κ
1
+
Ψ
κ
2
,
κ
1
2
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
g
x
d
α
x
=
Ψ
κ
2
,
κ
1
h
2
κ
1
+
Ψ
κ
2
,
κ
1
2
∫
0
1
g
κ
1
+
s
Ψ
κ
2
,
κ
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
1
d
s
≤
Ψ
κ
2
,
κ
1
∫
0
1
h
κ
1
+
s
Ψ
κ
2
,
κ
1
g
κ
1
+
s
Ψ
κ
2
,
κ
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
1
d
s
=
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
h
x
g
x
d
α
x
.So we can write(18)
h
2
κ
1
+
Ψ
κ
2
,
κ
1
2
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
g
x
d
α
x
≤
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
h
x
g
x
d
α
x
.
To prove the second inequality in (14), we know that h is preinvex and Ψ satisfies Condition C, so we have(19)
h
κ
1
+
s
Ψ
κ
2
,
κ
1
=
h
κ
1
+
Ψ
κ
2
,
κ
1
+
1
-
s
Ψ
κ
1
,
κ
1
+
Ψ
κ
2
,
κ
1
≤
s
h
κ
1
+
Ψ
κ
2
,
κ
1
+
1
-
s
h
κ
1
,and similarly(20)
h
κ
1
+
1
-
s
Ψ
κ
2
,
κ
1
=
h
κ
1
+
Ψ
κ
2
,
κ
1
+
s
Ψ
κ
1
,
κ
1
+
Ψ
κ
2
,
κ
1
≤
s
h
κ
1
+
1
-
s
h
κ
1
+
Ψ
κ
2
,
κ
1
.Now with x
=
κ
1
+
s
Ψ
(
κ
2
,
κ
1
), we have(21)
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
h
x
g
x
x
α
-
1
d
x
=
∫
0
1
h
κ
1
+
s
Ψ
κ
2
,
κ
1
g
κ
1
+
s
Ψ
κ
2
,
κ
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
1
d
s
≤
Ψ
κ
2
,
κ
1
∫
0
1
s
h
κ
1
+
Ψ
κ
2
,
κ
1
+
1
-
s
h
κ
1
×
g
κ
1
+
s
Ψ
κ
2
,
κ
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
1
d
s
u
s
i
n
g
19
.
Also(22)
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
h
x
g
x
x
α
-
1
d
x
=
∫
0
1
h
κ
1
+
s
Ψ
κ
2
,
κ
1
g
κ
1
+
s
Ψ
κ
2
,
κ
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
1
d
s
=
∫
0
1
h
κ
1
+
1
-
s
Ψ
κ
2
,
κ
1
g
κ
1
+
s
Ψ
κ
2
,
κ
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
1
d
s
h
i
s
s
y
m
m
e
t
r
i
c
≤
Ψ
κ
2
,
κ
1
∫
0
1
s
h
κ
1
+
1
-
s
h
κ
1
+
Ψ
κ
2
,
κ
1
×
g
κ
1
+
s
Ψ
κ
2
,
κ
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
1
d
s
u
s
i
n
g
20
.If we add (21) and (22), we obtain(23)
2
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
h
x
g
x
d
α
x
≤
Ψ
κ
2
,
κ
1
h
κ
1
+
h
κ
1
+
Ψ
κ
2
,
κ
1
∫
0
1
g
κ
1
+
s
Ψ
κ
2
,
κ
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
1
d
s
=
h
κ
1
+
h
κ
1
+
Ψ
κ
2
,
κ
1
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
g
x
x
α
-
1
d
x
.So we can write(24)
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
h
x
g
x
d
α
x
≤
h
κ
1
+
h
κ
1
+
Ψ
κ
2
,
κ
1
2
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
g
x
d
α
x
.From inequalities (18) and (24), we obtain over required result.Corollary 9.
If we putg
(
x
)
=
1 in (14), then we get(25)
h
2
κ
1
+
Ψ
κ
2
,
κ
1
2
≤
α
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
α
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
h
x
d
α
x
≤
h
κ
1
+
h
κ
1
+
Ψ
κ
2
,
κ
1
2
.


## 3.H
H Type Inequalities for Conformable Fractional Integrals

Lemma 10.
Letκ
1, κ
2
∈
K with Ψ
(
κ
2
,
κ
1
)
>
0 and h
:
K
=
[
κ
1
,
κ
1
+
Ψ
(
κ
2
,
κ
1
)
]
→
(
0
,
∞
) be an α-fractional differentiable function on (
κ
1
,
κ
1
+
Ψ
(
κ
2
,
κ
1
)
) for α
∈
(
0,1
]. If D
α
(
h
)
∈
L
α
(
[
κ
1
,
κ
1
+
Ψ
(
κ
2
,
κ
1
)
]
), then the following identity holds:(26)
h
κ
1
+
h
κ
1
+
Ψ
κ
2
,
κ
1
2
-
α
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
α
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
h
x
d
α
x
=
Ψ
κ
2
,
κ
1
2
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
α
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
2
α
-
1
-
κ
1
α
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
1
×
D
α
h
κ
1
+
s
Ψ
κ
2
,
κ
1
s
1
-
α
d
α
s
+
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
2
α
-
1
-
κ
1
+
Ψ
κ
2
,
κ
1
α
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
1
×
D
α
h
κ
1
+
s
Ψ
κ
2
,
κ
1
s
1
-
α
d
α
s
.Proof.
Integrating by parts, we have(27)
I
=
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
2
α
-
1
-
κ
1
α
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
1
D
α
h
κ
1
+
s
Ψ
κ
2
,
κ
1
d
s
+
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
2
α
-
1
-
κ
1
+
Ψ
κ
2
,
κ
1
α
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
1
×
D
α
h
κ
1
+
s
Ψ
κ
2
,
κ
1
d
s
=
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
κ
1
α
h
′
κ
1
+
s
Ψ
κ
2
,
κ
1
d
s
+
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
κ
1
+
Ψ
κ
2
,
κ
1
α
h
′
κ
1
+
s
Ψ
κ
2
,
κ
1
d
s
=
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
κ
1
α
h
κ
1
+
s
Ψ
κ
2
,
κ
1
Ψ
κ
2
,
κ
1
0
1
-
∫
0
1
α
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
1
Ψ
κ
2
,
κ
1
h
κ
1
+
s
Ψ
κ
2
,
κ
1
Ψ
κ
2
,
κ
1
d
s
+
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
κ
1
+
Ψ
κ
2
,
κ
1
α
h
κ
1
+
s
Ψ
κ
2
,
κ
1
Ψ
κ
2
,
κ
1
0
1
-
∫
0
1
α
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
1
Ψ
κ
2
,
κ
1
h
κ
1
+
s
Ψ
κ
2
,
κ
1
Ψ
κ
2
,
κ
1
d
s
=
1
Ψ
κ
2
,
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
α
h
κ
1
+
Ψ
κ
2
,
κ
1
-
α
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
h
x
d
α
x
+
1
Ψ
κ
2
,
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
α
h
κ
1
-
α
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
h
x
d
α
x
=
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
α
Ψ
κ
2
,
κ
1
h
κ
1
+
h
κ
1
+
Ψ
κ
2
,
κ
1
-
2
α
Ψ
κ
2
,
κ
1
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
h
x
d
α
x
,where we have used the change of variable x
=
κ
1
+
Ψ
(
κ
2
,
κ
1
) and then multiplied both sides by Ψ
(
κ
2
,
κ
1
)
/
2
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
α to get the desired result in (26).Remark 11.
If we setα
=
1 in (26), then we obtain the result which is proved by Barani et al. in [40] (28)
-
h
κ
1
+
h
κ
1
+
Ψ
κ
2
,
κ
1
2
+
1
Ψ
κ
2
,
κ
1
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
h
x
d
x
=
Ψ
κ
2
,
κ
1
2
∫
0
1
1
-
2
s
h
′
κ
1
+
s
Ψ
κ
2
,
κ
1
d
s
.Theorem 12.
Letκ
1, κ
2
∈
K such that Ψ
(
κ
2
,
κ
1
)
>
0 and h
:
K
=
[
κ
1
,
κ
1
+
Ψ
(
κ
2
,
κ
1
)
]
→
(
0
,
∞
) be an α-differentiable function on κ
1
,
κ
1
+
Ψ
κ
2
,
κ
1 for α
∈
(
0,1
] such that D
α
(
h
)
∈
L
α
κ
1
,
κ
1
+
Ψ
κ
2
,
κ
1. If |
h
′
| is preinvex, then we have the following inequality:(29)
h
κ
1
+
h
κ
1
+
Ψ
κ
2
,
κ
1
2
-
α
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
α
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
h
x
d
α
x
≤
Ψ
κ
2
,
κ
1
4
h
′
κ
1
+
h
′
κ
2
.Proof.
From Lemma10, using the property of the modulus and preinvexity of h
′, we have (30)
h
κ
1
+
h
κ
1
+
Ψ
κ
2
,
κ
1
2
-
α
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
α
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
h
x
d
α
x
=
Ψ
κ
2
,
κ
1
2
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
α
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
2
α
-
1
-
κ
1
α
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
1
×
D
α
h
κ
1
+
s
Ψ
κ
2
,
κ
1
d
s
+
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
2
α
-
1
-
κ
1
+
Ψ
κ
2
,
κ
1
α
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
1
×
D
α
h
κ
1
+
s
Ψ
κ
2
,
κ
1
d
s
≤
Ψ
κ
2
,
κ
1
2
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
α
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
κ
1
α
h
′
κ
1
+
s
Ψ
κ
2
,
κ
1
d
s
+
∫
0
1
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
+
s
Ψ
κ
2
,
κ
1
α
h
′
κ
1
+
s
Ψ
κ
2
,
κ
1
d
s
≤
Ψ
κ
2
,
κ
1
2
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
α
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
κ
1
α
1
-
s
h
′
κ
1
+
s
h
′
κ
2
d
t
+
∫
0
1
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
+
s
Ψ
κ
2
,
κ
1
α
1
-
s
h
′
κ
1
+
s
h
′
κ
2
d
s
=
Ψ
κ
2
,
κ
1
4
h
′
κ
1
+
h
′
κ
2
.Theorem 13.
Letκ
1, κ
2
∈
K such that Ψ
(
κ
2
,
κ
1
)
>
0 and h
:
K
=
[
κ
1
,
κ
1
+
Ψ
(
κ
2
,
κ
1
)
]
→
(
0
,
∞
) be an α-differentiable function on (
κ
1
,
κ
1
+
Ψ
(
κ
2
,
κ
1
)
) for α
∈
(
0,1
] such that D
α
(
h
)
∈
L
α
(
[
κ
1
,
κ
1
+
Ψ
(
κ
2
,
κ
1
)
]
). If h
′
q is preinvex for q
>
1 and q
-
1
+
p
-
1
=
1, then we have the following inequality:(31)
h
κ
1
+
h
κ
1
+
Ψ
κ
2
,
κ
1
2
-
α
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
α
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
h
x
d
α
x
≤
Ψ
κ
2
,
κ
1
2
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
α
A
1
α
,
p
1
/
p
h
′
κ
1
q
+
h
′
κ
2
q
2
1
/
q
+
A
2
α
,
p
1
/
p
h
′
κ
1
q
+
h
′
κ
2
q
2
1
/
q
,where(32)
A
1
α
,
p
=
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
κ
1
α
p
d
s
,
A
2
α
,
p
=
∫
0
1
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
+
s
Ψ
κ
2
,
κ
1
α
p
d
s
.Proof.
From Lemma10, using the property of the modulus and preinvexity of h
′
q, we have (33)
h
κ
1
+
h
κ
1
+
Ψ
κ
2
,
κ
1
2
-
α
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
α
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
h
x
d
α
x
≤
Ψ
κ
2
,
κ
1
2
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
α
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
κ
1
α
h
′
κ
1
+
s
Ψ
κ
2
,
κ
1
d
s
+
∫
0
1
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
+
s
Ψ
κ
2
,
κ
1
α
h
′
κ
1
+
s
Ψ
κ
2
,
κ
1
d
s
.Now by Hölder’s inequality (34)
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
κ
1
α
h
′
κ
1
+
s
Ψ
κ
2
,
κ
1
d
s
≤
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
κ
1
α
p
d
s
1
/
p
∫
0
1
h
′
κ
1
+
s
Ψ
κ
2
,
κ
1
q
d
s
1
/
q
≤
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
κ
1
α
p
d
s
1
/
p
∫
0
1
1
-
s
h
′
κ
1
q
+
s
h
′
κ
2
q
d
s
1
/
q
=
A
1
α
,
p
1
/
p
h
′
κ
1
q
+
h
′
κ
2
q
2
1
/
q
.Similarly, we have(35)
∫
0
1
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
+
s
Ψ
κ
2
,
κ
1
α
h
′
κ
1
+
s
Ψ
κ
2
,
κ
1
d
s
≤
∫
0
1
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
+
s
Ψ
κ
2
,
κ
1
α
p
d
s
1
/
p
∫
0
1
h
′
κ
1
+
s
Ψ
κ
2
,
κ
1
q
d
s
1
/
q
≤
∫
0
1
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
+
s
Ψ
κ
2
,
κ
1
α
p
d
s
1
/
p
∫
0
1
1
-
s
h
′
κ
1
q
+
s
h
′
κ
2
q
d
s
1
/
q
=
A
2
α
,
p
1
/
p
h
′
κ
1
q
+
h
′
κ
2
q
2
1
/
q
.Hence, we have the result in (31).Remark 14.
If we setα
=
1 in (31), then we have the following inequality: (36)
h
κ
1
+
h
κ
1
+
Ψ
κ
2
,
κ
1
2
-
1
Ψ
κ
2
,
κ
1
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
h
x
d
x
≤
Ψ
κ
2
,
κ
1
1
+
p
1
/
p
h
′
κ
1
q
+
h
′
κ
2
q
2
1
/
q
.Theorem 15.
Letκ
1, κ
2
∈
K such that Ψ
(
κ
2
,
κ
1
)
>
0 and h
:
K
=
[
κ
1
,
κ
1
+
Ψ
(
κ
2
,
κ
1
)
]
→
(
0
,
∞
) be an α-differentiable function on (
κ
1
,
κ
1
+
Ψ
(
κ
2
,
κ
1
)
) for α
∈
(
0,1
] such that D
α
(
h
)
∈
L
α
(
[
κ
1
,
κ
1
+
Ψ
(
κ
2
,
κ
1
)
]
). If h
′
q is preinvex for q
>
1 and q
-
1
+
p
-
1
=
1, then we have the following inequality:(37)
h
κ
1
+
h
κ
1
+
Ψ
κ
2
,
κ
1
2
-
α
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
α
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
h
x
d
α
x
≤
Ψ
κ
2
,
κ
1
2
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
α
A
1
α
1
-
1
/
q
A
2
α
h
′
κ
1
q
+
A
3
α
h
′
κ
2
q
1
/
q
+
B
1
α
1
-
1
/
q
B
2
α
h
′
κ
1
q
+
B
3
α
h
′
κ
2
q
1
/
q
,where(38)
A
1
α
=
κ
1
+
Ψ
κ
2
,
κ
1
α
+
1
-
κ
1
α
+
1
α
+
1
Ψ
κ
2
,
κ
1
-
κ
1
α
,
B
2
α
=
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
+
Ψ
κ
2
,
κ
1
α
+
1
-
κ
1
α
+
1
α
+
1
Ψ
κ
2
,
κ
1
,
A
2
α
=
-
κ
1
α
+
1
α
+
1
Ψ
κ
2
,
κ
1
α
+
2
Ψ
κ
2
,
κ
1
+
κ
1
α
+
2
Ψ
κ
2
,
κ
1
+
κ
1
+
Ψ
κ
2
,
κ
1
α
+
2
α
+
1
Ψ
κ
2
,
κ
1
2
α
+
2
-
κ
1
α
2
,
A
3
α
=
κ
1
+
Ψ
κ
2
,
κ
1
α
2
-
κ
1
+
Ψ
κ
2
,
κ
1
α
+
1
α
+
1
Ψ
κ
2
,
κ
1
α
+
2
Ψ
κ
2
,
κ
1
-
κ
1
+
Ψ
κ
2
,
κ
1
α
+
2
Ψ
κ
2
,
κ
1
-
κ
1
α
+
2
α
+
1
Ψ
κ
2
,
κ
1
2
α
+
2
,
B
2
α
=
κ
1
+
Ψ
κ
2
,
κ
1
α
2
+
κ
1
α
+
1
α
+
1
Ψ
κ
2
,
κ
1
α
+
2
Ψ
κ
2
,
κ
1
+
κ
1
α
+
2
Ψ
κ
2
,
κ
1
,
-
κ
1
+
Ψ
κ
2
,
κ
1
α
+
2
α
+
1
Ψ
κ
2
,
κ
1
2
α
+
2
B
3
α
=
κ
1
+
Ψ
κ
2
,
κ
1
α
2
-
κ
1
+
Ψ
κ
2
,
κ
1
α
+
1
α
+
1
Ψ
κ
2
,
κ
1
α
+
2
Ψ
κ
2
,
κ
1
-
κ
1
+
Ψ
κ
2
,
κ
1
α
+
2
Ψ
κ
2
,
κ
1
-
κ
1
α
+
2
α
+
1
Ψ
κ
2
,
κ
1
2
α
+
2
.Proof.
From Lemma10, using the property of the modulus and preinvexity of h
′
q, we have (39)
h
κ
1
+
h
κ
1
+
Ψ
κ
2
,
κ
1
2
-
α
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
α
∫
κ
1
κ
1
+
Ψ
κ
2
,
κ
1
h
x
d
α
x
≤
Ψ
κ
2
,
κ
1
2
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
α
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
κ
1
α
h
′
κ
1
+
s
Ψ
κ
2
,
κ
1
d
s
+
∫
0
1
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
+
s
Ψ
κ
2
,
κ
1
α
h
′
κ
1
+
s
Ψ
κ
2
,
κ
1
d
s
.Now by the power-mean inequality (40)
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
κ
1
α
h
′
κ
1
+
s
Ψ
κ
2
,
κ
1
d
s
≤
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
κ
1
α
d
s
1
-
1
/
q
×
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
κ
1
α
h
′
κ
1
+
s
Ψ
κ
2
,
κ
1
q
d
s
1
/
q
,and similarly, we have (41)
∫
0
1
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
+
s
Ψ
κ
2
,
κ
1
α
h
′
κ
1
+
s
Ψ
κ
2
,
κ
1
d
s
≤
∫
0
1
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
+
s
Ψ
κ
2
,
κ
1
α
d
s
1
-
1
/
q
∫
0
1
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
+
s
Ψ
κ
2
,
κ
1
α
h
′
κ
1
+
s
Ψ
κ
2
,
κ
1
q
d
s
1
/
q
.Now by the preinvexity of h
′
q from above, we have(42)
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
κ
1
α
1
-
s
h
′
κ
1
q
+
s
h
′
κ
2
q
d
s
=
h
′
κ
1
q
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
κ
1
α
1
-
s
d
s
+
h
′
κ
2
q
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
κ
1
α
s
d
s
=
h
′
κ
1
q
-
κ
1
α
+
1
α
+
1
Ψ
κ
2
,
κ
1
α
+
2
Ψ
κ
2
,
κ
1
+
κ
1
α
+
2
Ψ
κ
2
,
κ
1
+
κ
1
+
Ψ
κ
2
,
κ
1
α
+
2
α
+
1
Ψ
κ
2
,
κ
1
2
α
+
2
-
κ
1
α
2
+
h
′
κ
2
q
κ
1
+
Ψ
κ
2
,
κ
1
α
2
-
κ
1
+
Ψ
κ
2
,
κ
1
α
+
1
α
+
1
Ψ
κ
2
,
κ
1
α
+
2
Ψ
κ
2
,
κ
1
-
κ
1
+
Ψ
κ
2
,
κ
1
α
+
2
Ψ
κ
2
,
κ
1
-
κ
1
α
+
2
α
+
1
Ψ
κ
2
,
κ
1
2
α
+
2and(43)
∫
0
1
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
+
s
Ψ
κ
2
,
κ
1
α
1
-
s
h
′
κ
1
q
+
s
h
′
κ
2
q
d
s
=
h
′
κ
1
q
∫
0
1
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
+
s
Ψ
κ
2
,
κ
1
α
1
-
s
d
s
+
h
′
κ
2
q
∫
0
1
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
+
s
Ψ
κ
2
,
κ
1
α
s
d
s
=
h
′
κ
1
q
κ
1
+
Ψ
κ
2
,
κ
1
α
2
+
κ
1
α
+
1
α
+
1
Ψ
κ
2
,
κ
1
α
+
2
Ψ
κ
2
,
κ
1
+
κ
1
α
+
2
Ψ
κ
2
,
κ
1
-
κ
1
+
Ψ
κ
2
,
κ
1
α
+
2
α
+
1
Ψ
κ
2
,
κ
1
2
α
+
2
+
h
′
κ
2
q
κ
1
+
Ψ
κ
2
,
κ
1
α
2
-
κ
1
+
Ψ
κ
2
,
κ
1
α
+
1
α
+
1
Ψ
κ
2
,
κ
1
α
+
2
Ψ
κ
2
,
κ
1
-
κ
1
+
Ψ
κ
2
,
κ
1
α
+
2
Ψ
κ
2
,
κ
1
-
κ
1
α
+
2
α
+
1
Ψ
κ
2
,
κ
1
2
α
+
2
,where we have the following:(44)
A
1
α
=
∫
0
1
κ
1
+
s
Ψ
κ
2
,
κ
1
α
-
κ
1
α
d
s
=
κ
1
+
Ψ
κ
2
,
κ
1
α
+
1
-
κ
1
α
+
1
α
+
1
Ψ
κ
2
,
κ
1
-
κ
1
α
,
B
1
α
=
∫
0
1
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
+
s
Ψ
κ
2
,
κ
1
α
d
s
=
κ
1
+
Ψ
κ
2
,
κ
1
α
-
κ
1
+
Ψ
κ
2
,
κ
1
α
+
1
-
κ
1
α
+
1
α
+
1
Ψ
κ
2
,
κ
1
.Hence, we have the result in (37).



---

*Source: 3146210-2019-01-06.xml*