# Ostrowski Inequalities for Functions Whose First Derivatives Are Logarithmically Preinvex

Authors: Badreddine Meftah
Journal: Chinese Journal of Mathematics
 (2016)

---

## Abstract

Some Ostrowski type inequalities for functions whose first derivatives are logarithmically preinvex are established.


---

## Body

## 1. Introduction

In 1938, A. M. Ostrowski proved the following important inequality.Theorem 1 (see [1]).
Letf
:
I
⊆
R
→
R be a differentiable mapping on I
∘ (interior of I), and let a
,
b
∈
I
∘with a
<
b. If f
′
≤
M for all x
∈
a
,
b, then(1)
f
x
-
1
b
-
a
∫
a
b
f
t
d
t
≤
M
b
-
a
1
4
+
x
-
a
+
b
/
2
2
b
-
a
2
,
∀
x
∈
a
,
b
.This is well known in the literature as Ostrowski’s inequality. Due to its wide range of applications in numerical analysis and in probability, many researchers have established generalizations, extensions, and variants of inequality (1); we refer readers to [2–10] and the references cited therein.In recent years, a lot of efforts have been made by many mathematicians to generalize classical convexity. Hanson [11] introduced a new class of generalized convexity, called invexity. In [12], the authors gave the concept of preinvex function which is a special case of invexity. Pini [13], Noor [14, 15], Yang and Li [16], and Weir and Mond [17] have studied the basic properties of the preinvex functions and their role in optimization, variational inequalities, and equilibrium problems.In [5], Işcan established some Ostrowski type inequalities for functions whose derivatives in absolute value are preinvex, by using the following identity.Lemma 2 (see [5]).
LetA
⊂
R be an open invex subset with respect to η
:
A
×
A
→
R and a
,
b
∈
A with a
<
a
+
η
(
b
,
a
). Suppose that f
:
A
→
R is a differentiable function. If f
′ is integrable on [
a
,
a
+
η
(
b
,
a
)
], then the following equality holds: (2)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
=
η
b
,
a
∫
0
x
-
a
/
η
b
,
a
t
f
′
a
+
t
η
b
,
a
d
t
+
∫
x
-
a
/
η
b
,
a
1
t
-
1
f
′
a
+
t
η
b
,
a
d
t
, for all x
∈
[
a
,
a
+
η
(
b
,
a
)
].Motivated by the results given in [5], in the present paper, we establish some new Ostrowski type inequalities for functions whose first derivatives in absolute value are logarithmically preinvex.


## 2. Preliminaries

In this section, we recall some concepts of convexity that are well known in the literature. Throughout this section,I is an interval of R.Definition 3 (see [18]).
A positive functionf
:
I
→
R is said to be logarithmically convex, if, for all x
,
y
∈
I and all t
∈
[
0,1
], we have (3)
f
t
x
+
1
-
t
y
≤
f
x
t
f
y
1
-
t
.Definition 4 (see [17]).
A setK is said to be invex at x with respect to η, if, for all x
,
y
∈
K and t
∈
0,1, we have (4)
x
+
t
η
y
,
x
∈
K
.
K is said to be an invex set with respect to η if K is invex at each x
∈
K.Definition 5 (see [14]).
A positive functionf on the invex set K is said to be l
o
g-preinvex with respect to η, if, for all x
,
y
∈
K and t
∈
0,1, we have (5)
f
x
+
t
η
y
,
x
≤
f
x
1
-
t
f
y
t
.Lemma 6 (see [19]).
Forα
>
0
,
k
>
0, and z
>
0, we have (6)
J
α
,
k
=
∫
0
1
1
-
t
α
-
1
k
t
d
t
=
∑
i
=
1
∞
ln
⁡
k
i
-
1
α
i
<
∞
H
α
,
k
,
z
=
∫
0
1
t
α
-
1
k
t
d
t
=
z
α
k
z
∑
i
=
1
∞
-
z
ln
⁡
k
i
-
1
α
i
<
∞
, where α
i
=
∏
j
=
0
i
-
1
α
+
j.


## 3. Main Results

Theorem 7.
LetK
⊆
0
,
∞ be an invex subset with respect to η
:
K
×
K
→
R and a
,
b
∈
K
∘ (K
∘ interior of K) with η
(
b
,
a
)
>
0 and a
,
a
+
η
b
,
a
⊂
K. Let f
:
a
,
a
+
η
b
,
a
→
0
,
∞ be a differentiable function such that f
′
∈
L
a
,
a
+
η
b
,
a and f
′
(
a
)
≠
0. If f
′ is logarithmically preinvex function, then the following inequality holds: (7)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
2
x
-
a
η
b
,
a
2
+
1
-
x
-
a
η
b
,
a
2
i
f
A
=
1
,
2
2
x
-
a
η
b
,
a
-
1
A
x
-
a
/
η
b
,
a
ln
⁡
A
+
1
-
2
A
x
-
a
/
η
b
,
a
+
A
ln
2
⁡
A
i
f
A
≠
1
,for all x
∈
a
,
a
+
η
(
b
,
a
), where A
=
f
′
(
b
)
/
f
′
(
a
).Proof.
From Lemma2 and properties of modulus, we have(8)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
∫
0
x
-
a
/
η
b
,
a
t
f
′
a
+
t
η
b
,
a
d
t
+
∫
x
-
a
/
η
b
,
a
1
1
-
t
f
′
a
+
t
η
b
,
a
d
t
.Since f
′ is a logarithmically preinvex function, we deduce(9)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
∫
0
x
-
a
/
η
b
,
a
t
f
′
a
1
-
t
f
′
b
t
d
t
+
∫
x
-
a
/
η
b
,
a
1
1
-
t
f
′
a
1
-
t
f
′
b
t
d
t
=
η
b
,
a
f
′
a
∫
0
x
-
a
/
η
b
,
a
t
A
t
d
t
+
∫
x
-
a
/
η
b
,
a
1
1
-
t
A
t
d
t
.If A
=
1, then (9) gives(10)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
2
x
-
a
η
b
,
a
2
+
1
-
x
-
a
η
b
,
a
2
.In the case where A
≠
1, (9) gives(11)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
2
x
-
a
η
b
,
a
-
1
A
x
-
a
/
η
b
,
a
ln
⁡
A
+
1
-
2
A
x
-
a
/
η
b
,
a
+
A
ln
⁡
A
2
,where we use the fact that (12)
∫
0
x
-
a
/
η
b
,
a
t
A
t
=
x
-
a
η
b
,
a
A
x
-
a
/
η
b
,
a
ln
⁡
A
-
A
x
-
a
/
η
b
,
a
-
1
ln
⁡
A
2
,
∫
x
-
a
/
η
b
,
a
1
1
-
t
A
t
d
t
=
-
1
-
x
-
a
η
b
,
a
A
x
-
a
/
η
b
,
a
ln
⁡
A
+
A
-
A
x
-
a
/
η
b
,
a
ln
⁡
A
2
.The desired result follows from (10) and (11).Corollary 8.
In Theorem7, if we choose x
=
2
a
+
η
b
,
a
/
2, we obtain the following midpoint inequality: (13)
f
2
a
+
η
b
,
a
2
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
4
i
f
A
=
1
,
η
b
,
a
f
′
a
A
-
1
ln
⁡
A
2
i
f
A
≠
1
.Corollary 9.
Letf
:
a
,
b
→
0
,
∞ be a differentiable function such that f
′
∈
L
a
,
b and f
′
(
a
)
≠
0. If f
′ is a logarithmically convex function, then the following inequality holds: (14)
f
x
-
1
b
-
a
∫
a
b
f
u
d
u
≤
b
-
a
f
′
a
2
x
-
a
b
-
a
2
+
b
-
x
b
-
a
2
i
f
A
=
1
2
2
x
-
a
b
-
a
-
1
A
x
-
a
/
b
-
a
ln
⁡
A
+
1
-
2
A
x
-
a
/
b
-
a
+
A
l
n
2
A
i
f
A
≠
1
,for all x
∈
a
,
b, where A
=
f
′
(
b
)
/
f
′
(
a
).Example 10.
In Theorem7, if we choose η
(
b
,
a
)
=
a
b, the geometric mean, we obtain the following inequality: (15)
f
x
-
1
η
b
,
a
∫
a
a
+
a
b
f
u
d
u
≤
f
′
a
a
b
2
x
-
a
a
b
2
+
1
-
x
-
a
a
b
2
if
A
=
1
,
2
2
x
-
a
a
b
-
1
A
x
-
a
/
a
b
ln
⁡
A
+
1
-
2
A
x
-
a
/
a
b
+
A
l
n
2
A
if
A
≠
1
.Theorem 11.
LetK
⊆
0
,
∞ be an invex subset with respect to η
:
K
×
K
→
R and a
,
b
∈
K
∘ (K
∘ interior of K) with η
(
b
,
a
)
>
0 and a
,
a
+
η
b
,
a
⊂
K. Let f
:
a
,
a
+
η
b
,
a
→
0
,
∞ be a differentiable function such that f
′
∈
L
a
,
a
+
η
b
,
a and f
′
(
a
)
≠
0; let q
>
1 with 1
/
p
+
1
/
q
=
1. If f
′
q is a logarithmically preinvex function, then the following inequality holds: (16)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
p
+
1
1
/
p
x
-
a
η
b
,
a
2
+
1
-
x
-
a
η
b
,
a
2
i
f
A
=
1
,
x
-
a
η
b
,
a
1
+
1
/
p
A
q
x
-
a
/
η
b
,
a
-
1
q
ln
⁡
A
1
/
q
+
1
-
x
-
a
η
b
,
a
1
+
1
/
p
A
q
-
A
q
x
-
a
/
η
b
,
a
q
ln
⁡
A
1
/
q
i
f
A
≠
1
,for all x
∈
a
,
a
+
η
(
b
,
a
), where A
=
f
′
(
b
)
/
f
′
(
a
).Proof.
From Lemma2, properties of modulus, and Hölder’s inequality, we have(17)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
∫
0
x
-
a
/
η
b
,
a
t
p
d
t
1
/
p
∫
0
x
-
a
/
η
b
,
a
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
+
∫
x
-
a
/
η
b
,
a
1
1
-
t
p
d
t
1
/
p
∫
x
-
a
/
η
b
,
a
1
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
=
η
b
,
a
p
+
1
1
/
p
x
-
a
η
b
,
a
1
+
1
/
p
∫
0
x
-
a
/
η
b
,
a
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
+
1
-
x
-
a
η
b
,
a
1
+
1
/
p
∫
x
-
a
/
η
b
,
a
1
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
.Since f
′
q is a logarithmically preinvex function, we deduce(18)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
p
+
1
1
/
p
x
-
a
η
b
,
a
1
+
1
/
p
∫
0
x
-
a
/
η
b
,
a
f
′
a
q
1
-
t
f
′
b
q
t
d
t
1
/
q
+
1
-
x
-
a
η
b
,
a
1
+
1
/
p
∫
x
-
a
/
η
b
,
a
1
f
′
a
q
1
-
t
f
′
b
q
t
d
t
1
/
q
=
η
b
,
a
f
′
a
p
+
1
1
/
p
x
-
a
η
b
,
a
1
+
1
/
p
∫
0
x
-
a
/
η
b
,
a
A
q
t
d
t
1
/
q
+
1
-
x
-
a
η
b
,
a
1
+
1
/
p
∫
x
-
a
/
η
b
,
a
1
A
q
t
d
t
1
/
q
.If A
=
1, then (18) gives(19)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
p
+
1
1
/
p
x
-
a
η
b
,
a
2
+
1
-
x
-
a
η
b
,
a
2
.In the case where A
≠
1, (18) becomes(20)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
p
+
1
1
/
p
x
-
a
η
b
,
a
1
+
1
/
p
A
q
x
-
a
/
η
b
,
a
-
1
q
ln
⁡
A
1
/
q
+
1
-
x
-
a
η
b
,
a
1
+
1
/
p
A
q
-
A
q
x
-
a
/
η
b
,
a
q
ln
⁡
A
1
/
q
,where we use the fact that(21)
∫
0
x
-
a
/
η
b
,
a
A
q
t
d
t
=
A
q
x
-
a
/
η
b
,
a
-
1
q
ln
⁡
A
,
∫
x
-
a
/
η
b
,
a
1
A
q
t
d
t
=
A
q
-
A
q
x
-
a
/
η
b
,
a
q
ln
⁡
A
.From (19) and (20), we get the desired result.Corollary 12.
In Theorem11, if we choose x
=
2
a
+
η
b
,
a
/
2, we obtain the following midpoint inequality: (22)
f
2
a
+
η
b
,
a
2
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
2
1
+
1
/
p
p
+
1
1
/
p
2
1
/
p
i
f
A
=
1
,
1
+
A
A
q
-
1
q
ln
⁡
A
1
/
q
i
f
A
≠
1
.Corollary 13.
Letf
:
a
,
b
→
0
,
∞ be a differentiable function such that f
′
∈
L
a
,
b and f
′
(
a
)
≠
0; let q
>
1 with 1
/
p
+
1
/
q
=
1. If f
′
q is a logarithmically convex function, then the following inequality holds: (23)
f
x
-
1
b
-
a
∫
a
b
f
u
d
u
≤
b
-
a
f
′
a
p
+
1
1
/
p
x
-
a
b
-
a
2
+
b
-
x
b
-
a
2
i
f
A
=
1
,
x
-
a
b
-
a
1
+
1
/
p
A
q
x
-
a
/
b
-
a
-
1
q
ln
⁡
A
1
/
q
+
b
-
x
b
-
a
1
+
1
/
p
A
q
-
A
q
x
-
a
/
b
-
a
q
ln
⁡
A
1
/
q
i
f
A
≠
1
,for all x
∈
a
,
b, where A
=
f
′
(
b
)
/
f
′
(
a
).Example 14.
In Theorem11, if we choose η
(
b
,
a
)
=
a
+
b
/
2, the arithmetic mean, we obtain the following inequality: (24)
f
x
-
2
a
+
b
∫
a
a
+
a
+
b
/
2
f
u
d
u
≤
a
+
b
f
′
a
2
p
+
1
1
/
p
4
a
+
b
2
x
-
a
2
+
b
+
3
a
-
2
x
2
2
if
A
=
1
,
2
x
-
a
a
+
b
1
+
1
/
p
A
2
q
x
-
a
/
a
+
b
-
1
q
ln
⁡
A
1
/
q
+
b
+
3
a
-
2
x
2
1
+
1
/
p
A
q
-
A
2
q
x
-
a
/
a
+
b
q
ln
⁡
A
1
/
q
if
A
≠
1
.Theorem 15.
LetK
⊆
0
,
∞ be an invex subset with respect to η
:
K
×
K
→
R and  a
,
b
∈
K
∘ (K
∘ interior of K) with η
(
b
,
a
)
>
0 and a
,
a
+
η
b
,
a
⊂
K. Let f
:
a
,
a
+
η
b
,
a
→
0
,
∞ be a differentiable function such that f
′
∈
L
a
,
a
+
η
b
,
a and f
′
(
a
)
≠
0; let q
>
1
. If f
′
q is a logarithmically preinvex function, then the following inequality holds:(25)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
2
1
-
1
/
q
f
′
a
1
2
1
/
q
x
-
a
η
b
,
a
2
+
1
-
x
-
a
η
b
,
a
2
i
f
A
=
1
,
x
-
a
η
b
,
a
2
-
2
/
q
x
-
a
η
b
,
a
A
q
x
-
a
/
η
b
,
a
ln
⁡
A
+
1
-
A
q
x
-
a
/
η
b
,
a
l
n
2
A
1
/
q
+
1
-
x
-
a
η
b
,
a
2
-
2
/
q
A
q
-
A
q
x
-
a
/
η
b
,
a
l
n
2
A
-
1
-
x
-
a
η
b
,
a
A
q
x
-
a
/
η
b
,
a
ln
⁡
A
1
/
q
i
f
A
≠
1
,for all x
∈
a
,
a
+
η
(
b
,
a
), where A
=
f
′
(
b
)
/
f
′
(
a
).Proof.
From Lemma2, properties of modulus, and power mean inequality, we have(26)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
∫
0
x
-
a
/
η
b
,
a
t
d
t
1
-
1
/
q
∫
0
x
-
a
/
η
b
,
a
t
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
+
∫
x
-
a
/
η
b
,
a
1
1
-
t
d
t
1
-
1
/
q
∫
x
-
a
/
η
b
,
a
1
1
-
t
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
=
η
b
,
a
2
1
-
1
/
q
x
-
a
η
b
,
a
2
1
-
1
/
q
∫
0
x
-
a
/
η
b
,
a
t
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
+
1
-
x
-
a
η
b
,
a
2
1
-
1
/
q
∫
x
-
a
/
η
b
,
a
1
1
-
t
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
.Since f
′
q is a logarithmically preinvex function, we deduce(27)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
2
1
-
1
/
q
f
′
a
x
-
a
η
b
,
a
2
-
2
/
q
∫
0
x
-
a
/
η
b
,
a
t
A
q
t
d
t
1
/
q
+
1
-
x
-
a
η
b
,
a
2
-
2
/
q
∫
x
-
a
/
η
b
,
a
1
1
-
t
A
q
t
d
t
1
/
q
.In the case where A
=
1, (27) gives(28)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
2
f
′
a
x
-
a
η
b
,
a
2
+
1
-
x
-
a
η
b
,
a
2
.For A
≠
1, (27) gives(29)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
2
1
-
1
/
q
f
′
a
x
-
a
η
b
,
a
2
-
2
/
q
x
-
a
η
b
,
a
A
q
x
-
a
/
η
b
,
a
ln
⁡
A
+
1
-
A
q
x
-
a
/
η
b
,
a
l
n
2
A
1
/
q
+
1
-
x
-
a
η
b
,
a
2
-
2
/
q
A
q
-
A
q
x
-
a
/
η
b
,
a
l
n
2
A
-
1
-
x
-
a
η
b
,
a
A
q
x
-
a
/
η
b
,
a
ln
⁡
A
1
/
q
,where we use the fact that (30)
∫
0
x
-
a
/
η
b
,
a
t
A
q
t
d
t
=
x
-
a
η
b
,
a
A
q
x
-
a
/
η
b
,
a
ln
⁡
A
+
1
-
A
q
x
-
a
/
η
b
,
a
l
n
2
A
,
∫
x
-
a
/
η
b
,
a
1
1
-
t
A
q
t
d
t
=
A
q
-
A
q
x
-
a
/
η
b
,
a
l
n
2
A
-
1
-
x
-
a
η
b
,
a
A
q
x
-
a
/
η
b
,
a
ln
⁡
A
.From (28) and (29), we obtain the desired result.Corollary 16.
In Theorem15, if we choose x
=
2
a
+
η
b
,
a
/
2, we obtain the following midpoint inequality: (31)
f
2
a
+
η
b
,
a
2
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
4
f
′
a
1
i
f
A
=
1
,
1
2
1
-
3
/
q
1
-
A
q
/
2
l
n
2
A
+
A
q
/
2
2
ln
⁡
A
1
/
q
+
A
1
/
2
A
q
/
2
-
1
l
n
2
A
-
1
2
ln
⁡
A
1
/
q
i
f
A
≠
1
.Corollary 17.
Letf
:
a
,
a
+
η
b
,
a
→
0
,
∞ be a differentiable function such that f
′
∈
L
a
,
a
+
η
b
,
a and f
′
(
a
)
≠
0; let q
>
1
. If f
′
q is a logarithmically convex function, then the following inequality holds: (32)
f
x
-
1
b
-
a
∫
a
b
f
u
d
u
≤
b
-
a
2
1
-
1
/
q
f
′
a
1
2
1
/
q
x
-
a
b
-
a
2
+
b
-
x
b
-
a
2
i
f
A
=
1
,
x
-
a
b
-
a
2
-
2
/
q
x
-
a
b
-
a
A
q
x
-
a
/
b
-
a
ln
⁡
A
+
1
-
A
q
x
-
a
/
b
-
a
l
n
2
A
1
/
q
+
b
-
x
b
-
a
2
-
2
/
q
A
q
-
A
q
x
-
a
/
b
-
a
l
n
2
A
-
b
-
x
b
-
a
A
q
x
-
a
/
b
-
a
ln
⁡
A
1
/
q
i
f
A
≠
1
,for all x
∈
a
,
b, where A
=
f
′
(
b
)
/
f
′
(
a
).Example 18.
In Theorem15, if we choose η
(
b
,
a
)
=
b
-
a
/
ln
⁡
b
-
ln
⁡
a with a
≠
b, the logarithmic mean, we obtain the following inequality: (33)
f
x
-
ln
⁡
b
-
ln
⁡
a
b
-
a
∫
a
a
+
b
-
a
/
ln
⁡
b
-
ln
⁡
a
f
u
d
u
≤
b
-
a
2
1
-
1
/
q
ln
⁡
b
-
ln
⁡
a
f
′
a
1
2
1
/
q
θ
2
+
1
-
θ
2
i
f
A
=
1
,
θ
2
-
2
/
q
θ
A
q
θ
ln
⁡
A
+
1
-
A
q
θ
l
n
2
A
1
/
q
+
1
-
θ
2
-
2
/
q
A
q
-
A
q
θ
l
n
2
A
-
1
-
θ
A
q
θ
ln
⁡
A
1
/
q
i
f
A
≠
1
,where θ
=
x
-
a
ln
⁡
b
-
ln
⁡
a
/
b
-
a.Theorem 19.
Suppose that all the assumptions of Theorem15 are satisfied, then the following inequality holds:(34)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
1
q
+
1
1
/
q
x
-
a
η
b
,
a
2
+
1
-
x
-
a
η
b
,
a
2
i
f
A
=
1
,
A
x
-
a
/
η
b
,
a
1
-
x
-
a
η
b
,
a
x
-
a
η
b
,
a
∑
i
=
1
∞
ln
⁡
A
q
1
-
x
-
a
/
η
b
,
a
i
-
1
α
i
1
/
q
+
x
-
a
η
b
,
a
2
∑
i
=
1
∞
-
ln
⁡
A
q
x
-
a
/
η
b
,
a
i
-
1
α
i
1
/
q
i
f
A
≠
1
,for all x
∈
a
,
a
+
η
(
b
,
a
), where A
=
f
′
(
b
)
/
f
′
(
a
) and q
+
1
i
=
∏
j
=
0
i
-
1
q
+
1
+
j.Proof.
From Lemma2, properties of modulus, and power mean inequality, we have(35)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
∫
0
x
-
a
/
η
b
,
a
d
t
1
-
1
/
q
∫
0
x
-
a
/
η
b
,
a
t
q
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
+
∫
x
-
a
/
η
b
,
a
1
d
t
1
-
1
/
q
∫
x
-
a
/
η
b
,
a
1
1
-
t
q
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
=
η
b
,
a
x
-
a
η
b
,
a
1
-
1
/
q
∫
0
x
-
a
/
η
b
,
a
t
q
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
+
1
-
x
-
a
η
b
,
a
1
-
1
/
q
∫
x
-
a
/
η
b
,
a
1
1
-
t
q
f
′
a
+
t
η
b
,
a
q
d
t
1
/
q
.Since f
′
q is a logarithmically preinvex function, we deduce(36)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
x
-
a
η
b
,
a
1
-
1
/
q
∫
0
x
-
a
/
η
b
,
a
t
q
A
q
t
1
/
q
+
1
-
x
-
a
η
b
,
a
1
-
1
/
q
∫
x
-
a
/
η
b
,
a
1
1
-
t
q
A
q
t
d
t
1
/
q
.If A
=
1, (36) gives(37)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
q
+
1
1
/
q
x
-
a
η
b
,
a
2
+
1
-
x
-
a
η
b
,
a
2
.
In the case whereA
≠
1, we can restate (36) as(38)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
x
-
a
η
b
,
a
2
∫
0
1
t
q
A
q
x
-
a
/
η
b
,
a
t
d
t
1
/
q
+
1
-
x
-
a
η
b
,
a
x
-
a
η
b
,
a
A
x
-
a
/
η
b
,
a
∫
0
1
t
-
1
q
A
q
1
-
x
-
a
/
η
b
,
a
t
d
t
1
/
q
.Applying Lemma 6 with z
=
1, we get(39)
∫
0
1
t
q
A
q
x
-
a
/
η
b
,
a
t
d
t
=
A
q
x
-
a
/
η
b
,
a
∑
i
=
1
∞
-
ln
⁡
A
q
x
-
a
/
η
b
,
a
i
-
1
α
i
∫
0
1
t
-
1
q
A
q
1
-
x
-
a
/
η
b
,
a
t
d
t
=
∑
i
=
1
∞
ln
⁡
A
q
1
-
x
-
a
/
η
b
,
a
i
-
1
α
i
.Substituting (39) into (38), we obtain(40)
f
x
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
x
-
a
η
b
,
a
2
A
x
-
a
/
η
b
,
a
∑
i
=
1
∞
-
ln
⁡
A
q
x
-
a
/
η
b
,
a
i
-
1
α
i
1
/
q
+
1
-
x
-
a
η
b
,
a
x
-
a
η
b
,
a
A
x
-
a
/
η
b
,
a
∑
i
=
1
∞
ln
⁡
A
q
1
-
x
-
a
/
η
b
,
a
i
-
1
α
i
1
/
q
.The desired result follows from (37) and (40).Corollary 20.
In Theorem19, if we choose x
=
2
a
+
η
b
,
a
/
2, we obtain the following midpoint inequality: (41)
f
2
a
+
η
b
,
a
2
-
1
η
b
,
a
∫
a
a
+
η
b
,
a
f
u
d
u
≤
η
b
,
a
f
′
a
2
1
q
+
1
1
/
q
i
f
A
=
1
,
A
2
∑
i
=
1
∞
ln
⁡
A
q
/
2
i
-
1
α
i
1
/
q
+
∑
i
=
1
∞
-
ln
⁡
A
q
/
2
i
-
1
α
i
1
/
q
i
f
A
≠
1
.Corollary 21.
In Theorem19, if we choose η
b
,
a
=
b
-
a, we obtain the following inequality: (42)
f
x
-
1
b
-
a
∫
a
b
f
u
d
u
≤
b
-
a
f
′
a
1
q
+
1
1
/
q
x
-
a
b
-
a
2
+
b
-
x
b
-
a
2
i
f
A
=
1
,
A
x
-
a
/
b
-
a
b
-
x
x
-
a
b
-
a
2
∑
i
=
1
∞
ln
⁡
A
q
b
-
x
/
b
-
a
i
-
1
α
i
1
/
q
+
x
-
a
b
-
a
2
∑
i
=
1
∞
-
ln
⁡
A
q
x
-
a
/
b
-
a
i
-
1
α
i
1
/
q
i
f
A
≠
1
.Remark 22.
In all the above theorems, inequalities for nonconvex functions could be drawn by just replacingη
b
,
a by other means than those in the previously mentioned examples.



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*Source: 5292603-2016-09-06.xml*