# On Hermite-Hadamard Type Inequalities for Functions Whose Second Derivatives Absolute Values Ares-Convex

Authors: Feixiang Chen; Xuefei Liu
Journal: ISRN Applied Mathematics
 (2014)

---

## Abstract

Some new Hermite-Hadamard type inequalities are obtained for functions
whose second derivatives absolute values ares-convex.


---

## Body

## 1. Introduction

Letf
:
I
⊂
R
→
R be a convex function on the interval I; then for any a
,
b
∈
I with a
≠
b we have the following double inequality:
(1)
f
(
a
+
b
2
)
≤
1
b
-
a
∫
a
b
‍
f
(
t
)
d
t
≤
f
(
a
)
+
f
(
b
)
2
.
This remarkable result is well known in the literature as the Hermite-Hadamard inequality. Note that some of the classical inequalities for means can be derived from (1) for appropriate particular selections of the mapping f. Both inequalities hold in the reversed direction if f is concave. Some refinements of the Hermite-Hadamard inequality on convex functions have been extensively investigated by a number of authors (e.g., [1–9]).Definition 1 (see [10]).
A functionf
:
I
⊂
[
0
;
1
)
→
[
0
;
1
) is said to be s-convex on I, if the inequality
(2)
f
(
α
x
+
(
1
-
α
)
y
)
≤
α
s
f
(
x
)
+
(
1
-
α
)
s
f
(
y
)
holds for all x
,
y
∈
I with α
∈
[
0,1
] and for some fixed s
∈
(
0,1
].It can be easily seen that fors
=
1, s-convexity reduces to ordinary convexity of functions defined on [
0
,
∞
).In [11], Dragomir and Fitzpatrick proved a variant of Hermite-Hadamard inequality which holds for the s-convex functions.Theorem 2 (see [11]).
Suppose thatf
:
I
⊂
[
0
,
∞
)
→
[
0
,
∞
) is an s-convex function in the second sense, where s
∈
(
0,1
] and let a
,
b
∈
[
0
,
∞
), a
<
b. If f
∈
L
1
[
0,1
], then the following inequality holds:
(3)
2
s
-
1
f
(
a
+
b
2
)
≤
1
b
-
a
∫
a
b
‍
f
(
t
)
d
t
≤
f
(
a
)
+
f
(
b
)
s
+
1
.The constantk
=
1
/
(
s
+
1
) is the best possible in the second inequality in (3).Definition 3 (see [4]).
We say thatf
:
I
⊂
R
→
R is P-convex function or that f belongs to the class P
(
I
), if f is nonnegative and for all x
,
y
∈
I and α
∈
(
0,1
) one has
(4)
f
(
α
x
+
(
1
-
α
)
y
)
≤
f
(
x
)
+
f
(
y
)
.Remark 4.
Applying Definition1 for s
→
0
+, we get Definition 3.
Along this paper we consider a real intervalI
⊂
R, and we denote that I
0 is the interior of I.
In [12], Barani et al. introduced the following theorems for twice differentiable P-convex functions.Theorem 5 (see [12]).
Letf
:
I
→
R be a twice differentiable function on I
0 such that |
f
′′
| is a P-convex function on I. Suppose that a
,
b
∈
I
0 with a
<
b and f
′′
∈
L
[
a
,
b
]. Then, the following inequality holds:
(5)
|
f
(
a
)
+
f
(
b
)
2
-
1
b
-
a
∫
a
b
‍
f
(
t
)
d
t
|
≤
(
b
-
a
)
2
24
(
|
f
′′
(
a
)
|
+
2
|
f
′′
(
a
+
b
2
)
|
+
|
f
′′
(
b
)
|
)
.Theorem 6 (see [12]).
Letf
:
I
→
R be a twice differentiable function on I
0. Assume that p
∈
R, p
>
1 such that |
f
′′
|
p
/
(
p
-
1
) is a P-convex function on I. Suppose that a
,
b
∈
I
0 with a
<
b and f
′′
∈
L
[
a
,
b
]. Then, the following inequality holds:
(6)
|
f
(
a
)
+
f
(
b
)
2
-
1
b
-
a
∫
a
b
‍
f
(
t
)
d
t
|
≤
(
b
-
a
)
2
16
(
π
Γ
(
p
+
1
)
2
Γ
(
p
+
3
/
2
)
)
1
/
p
×
[
(
|
f
′′
(
a
)
|
q
+
|
f
′′
(
a
+
b
2
)
|
q
)
1
/
q
f
f
f
i
f
f
+
(
|
f
′′
(
b
)
|
q
+
|
f
′′
(
a
+
b
2
)
|
q
)
1
/
q
]
.Theorem 7 (see [12]).
Letf
:
I
→
R be a twice differentiable function on I
0. Assume that q
≥
1 such that |
f
′′
|
q is a P-convex function on I. Suppose that a
,
b
∈
I
0 with a
<
b and f
′′
∈
L
[
a
,
b
]. Then, the following inequality holds:
(7)
|
f
(
a
)
+
f
(
b
)
2
-
1
b
-
a
∫
a
b
‍
f
(
t
)
d
t
|
≤
(
b
-
a
)
2
24
[
(
|
f
′′
(
b
)
|
q
+
|
f
′′
(
a
+
b
2
)
|
q
)
1
/
q
(
|
f
′′
(
a
)
|
q
+
|
f
′′
(
a
+
b
2
)
|
q
)
1
/
q
f
f
f
f
f
f
d
d
f
+
(
|
f
′′
(
b
)
|
q
+
|
f
′′
(
a
+
b
2
)
|
q
)
1
/
q
]
.For recent results and generalizations concerning Hermite-Hadamard's inequality for twice differentiable functions see [10, 12–14] and the references given therein.In this paper, we establish some new inequalities of Hadamard's type for the class ofs-convex functions in the second sense.


## 2. The New Hermite-Hadamard Type Inequalities

Lemma 8 (see [13]).
Letf
:
I
⊂
R
→
R be a twice differentiable mapping on I
0, where a
,
b
∈
I with a
<
b. If f
′′
∈
L
[
a
,
b
], then the following equality holds:
(8)
f
(
a
)
+
f
(
b
)
2
-
1
b
-
a
∫
a
b
‍
f
(
t
)
d
t
=
(
b
-
a
)
2
16
∫
0
1
‍
(
1
-
t
2
)
(
f
′′
(
1
+
t
2
a
+
1
-
t
2
b
)
ddddddddddddddddddd
+
f
′′
(
1
-
t
2
a
+
1
+
t
2
b
)
)
d
t
.Theorem 9.
Letf
:
I
⊂
R
→
R be a differentiable mapping on I
0, such that f
′′
∈
L
[
a
,
b
], where a
,
b
∈
I with a
<
b. If |
f
′′
| is s-convex on [
a
,
b
] for some fixed s
∈
(
0,1
]; then the following inequality holds:
(9)
|
f
(
a
)
+
f
(
b
)
2
-
1
b
-
a
∫
a
b
‍
f
(
t
)
d
t
|

≤

(
b
-
a
)
2
8

(
|
f
′′
(
a
)
|
+
|
f
′′
(
b
)
|
(
s
+
1
)
(
s
+
3
)

+

(
s
+
4
)
|
f
′′
(
(
a
+
b
)
/
2
)
|
(
s
+
2
)
(
s
+
3
)
)
.Proof.
From Lemma8, we have
(10)
|
f
(
a
)
+
f
(
b
)
2
-
1
b
-
a
∫
a
b
‍
f
(
t
)
d
t
|
≤
(
b
-
a
)
2
16
∫
0
1
‍
(
1
-
t
2
)
(
|
f
′′
(
1
+
t
2
a
+
1
-
t
2
b
)
|
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
f
+
|
f
′′
(
1
-
t
2
a
+
1
+
t
2
b
)
|
)
d
t
.
Because |
f
′′
| is s-convex, we have
(11)
|
f
(
a
)
+
f
(
b
)
2
-
1
b
-
a
∫
a
b
‍
f
(
t
)
d
t
|
≤
(
b
-
a
)
2
16
×
∫
0
1
‍
(
1
-
t
2
)
f
f
f
f
f

×

(
t
s
|
f
′′
(
a
)
|
+
2
(
1
-
t
)
s
|
f
′′
(
a
+
b
2
)
|
+
t
s
|
f
′′
(
b
)
|
)
d
t

=

(
b
-
a
)
2
8
(
|
f
′′
(
a
)
|
+
|
f
′′
(
b
)
|
(
s
+
1
)
(
s
+
3
)

+

(
s
+
4
)
|
f
′′
(
(
a
+
b
)
/
2
)
|
(
s
+
2
)
(
s
+
3
)
)
,
which completes the proof.The corresponding version for powers of the absolute value of the second derivative is incorporated in the following theorems.Theorem 10.
Letf
:
I
⊂
R
→
R be a differentiable mapping on I
0, such that f
′′
∈
L
[
a
,
b
], where a
,
b
∈
I with a
<
b. If |
f
′′
|
p
/
(
p
-
1
) is s-convex on [
a
,
b
] for some fixed s
∈
(
0,1
] and p
>
1, then the following inequality holds:
(12)
|
f
(
a
)
+
f
(
b
)
2
-
1
b
-
a
∫
a
b
‍
f
(
t
)
d
t
|
≤
(
b
-
a
)
2
16
(
π
Γ
(
p
+
1
)
2
Γ
(
p
+
3
/
2
)
)
1
/
p
×
[
(
|
f
′′
(
a
)
|
q
+
|
f
′′
(
(
a
+
b
)
/
2
)
|
q
s
+
1
)
1
/
q
d
f
f
f
f
f
+
(
|
f
′′
(
b
)
|
q
+
|
f
′′
(
(
a
+
b
)
/
2
)
|
q
s
+
1
)
1
/
q
]
,
where (
1
/
p
)
+
(
1
/
q
)
=
1.Proof.
From Lemma8 and using the Hölder inequality, we have
(13)
|
f
(
a
)
+
f
(
b
)
2
-
1
b
-
a
∫
a
b
‍
f
(
t
)
d
t
|
≤
(
b
-
a
)
2
16
∫
0
1
‍
(
1
-
t
2
)
(
|
f
′′
(
1
+
t
2
a
+
1
-
t
2
b
)
|
f
f
f
f
f
f
f
f
f
f
f
f
f
f
+
|
f
′′
(
1
-
t
2
a
+
1
+
t
2
b
)
|
)
d
t
≤
(
b
-
a
)
2
16
(
∫
0
1
‍
(
1
-
t
2
)
p
d
t
)
1
/
p
×
[
(
∫
0
1
‍
|
f
′′
(
1
+
t
2
a
+
1
-
t
2
b
)
|
q
d
t
)
1
/
q
dddfd
+
(
∫
0
1
‍
|
f
′′
(
1
-
t
2
a
+
1
+
t
2
b
)
|
q
d
t
)
1
/
q
]
≤
(
b
-
a
)
2
16
(
∫
0
1
‍
(
1
-
t
2
)
p
d
t
)
1
/
p
×
[
(
∫
0
1
‍
(
t
s
|
f
′′
(
a
)
|
q
+
(
1
-
t
)
s
|
f
′′
(
a
+
b
2
)
|
q
)
d
t
)
1
/
q
ddfffd
+
(
∫
0
1
‍
(
t
s
|
f
′′
(
b
)
|
q
+
(
1
-
t
)
s
|
f
′′
(
a
+
b
2
)
|
q
)
d
t
)
1
/
q
]
=
(
b
-
a
)
2
16
(
π
Γ
(
p
+
1
)
2
Γ
(
p
+
3
/
2
)
)
1
/
p
×
[
(
|
f
′′
(
a
)
|
q
+
|
f
′′
(
(
a
+
b
)
/
2
)
|
q
s
+
1
)
1
/
q
ddfffd
+
(
|
f
′′
(
b
)
|
q
+
|
f
′′
(
(
a
+
b
)
/
2
)
|
q
s
+
1
)
1
/
q
]
.
We note that the Beta and Gamma functions are defined, respectively, as follows:
(14)
Γ
(
x
)
=
∫
0
1
‍
e
-
x
t
x
-
1
d
t
,
x
>
0
,
β
(
x
,
y
)
=
∫
0
1
‍
(
1
-
t
)
y
-
1
t
x
-
1
d
t
,
x
>
0
,
y
>
0
,
β
(
x
,
y
)
=
Γ
(
x
)
Γ
(
y
)
Γ
(
x
+
y
)
.
Then
(15)
∫
0
1
‍
(
1
-
t
2
)
p
d
t
=
π
Γ
(
p
+
1
)
2
Γ
(
p
+
3
/
2
)
,
and we get the desired result.Theorem 11.
Letf
:
I
⊂
R
→
R be a differentiable mapping on I
0, such that f
′′
∈
L
[
a
,
b
], where a
,
b
∈
I with a
<
b. If |
f
′′
|
q is s-convex on [
a
,
b
] for some fixed s
∈
(
0,1
] and q
≥
1, then the following inequality holds:
(16)
|
f
(
a
)
+
f
(
b
)
2
-
1
b
-
a
∫
a
b
‍
f
(
t
)
d
t
|
≤
(
b
-
a
)
2
16
(
2
3
)
1
-
1
/
q
×
[
(
2
|
f
′′
(
a
)
|
q
(
s
+
1
)
(
s
+
3
)
+
(
s
+
4
)
|
f
′′
(
(
a
+
b
)
/
2
)
|
q
(
s
+
2
)
(
s
+
3
)
)
1
/
q
f
f
f
f
f
+
(
2
|
f
′′
(
b
)
|
q
(
s
+
1
)
(
s
+
3
)
+
(
s
+
4
)
|
f
′′
(
(
a
+
b
)
/
2
)
|
q
(
s
+
2
)
(
s
+
3
)
)
1
/
q
]
.Proof.
From Lemma8 and using the well-known power-mean inequality, we have
(17)
|
f
(
a
)
+
f
(
b
)
2
-
1
b
-
a
∫
a
b
‍
f
(
t
)
d
t
|
≤
(
b
-
a
)
2
16
∫
0
1
‍
(
1
-
t
2
)
×
(
|
f
′′
(
1
+
t
2
a
+
1
-
t
2
b
)
|
+
|
f
′′
(
1
-
t
2
a
+
1
+
t
2
b
)
|
)
d
t
≤
(
b
-
a
)
2
16
(
∫
0
1
‍
(
1
-
t
2
)
d
t
)
1
-
1
/
q
×
[
(
∫
0
1
‍
(
1
-
t
2
)
|
f
′′
(
1
+
t
2
a
+
1
-
t
2
b
)
|
q
d
t
)
1
/
q
d
d
d
d
d
+
(
∫
0
1
‍
(
1
-
t
2
)
|
f
′′
(
1
-
t
2
a
+
1
+
t
2
b
)
|
q
d
t
)
1
/
q
]
≤
(
b
-
a
)
2
16
(
2
3
)
1
-
1
/
q
×
[
+
(
1
-
t
)
s
|
f
′′
(
a
+
b
2
)
|
q
)
d
t
)
1
/
q
(
∫
0
1
‍
(
1
-
t
2
)
(
|
f
′′
(
a
+
b
2
)
|
q
t
s
|
f
′′
(
a
)
|
q
f
f
f
f
f
f
f
f
f
f
f
f
+
(
1
-
t
)
s
|
f
′′
(
a
+
b
2
)
|
q
)
d
t
)
1
/
q
f
f
f
f
f
+
(
∫
0
1
‍
(
1
-
t
2
)
(
∑
t
s
|
f
′′
(
b
)
|
q
f
f
f
f
f
f
f
f
f
f
f
f
f
f
+
(
1
-
t
)
s
|
f
′′
(
a
+
b
2
)
|
q
)
d
t
)
1
/
q
]
=
(
b
-
a
)
2
16
(
2
3
)
1
-
1
/
q
×
[
(
2
|
f
′′
(
a
)
|
q
(
s
+
1
)
(
s
+
3
)
+
(
s
+
4
)
|
f
′′
(
(
a
+
b
)
/
2
)
|
q
(
s
+
2
)
(
s
+
3
)
)
1
/
q
ddddd
+
(
2
|
f
′′
(
b
)
|
q
(
s
+
1
)
(
s
+
3
)
+
(
s
+
4
)
|
f
′′
(
(
a
+
b
)
/
2
)
|
q
(
s
+
2
)
(
s
+
3
)
)
1
/
q
]
.
This proves the theorem.Remark 12.
Applying Theorem9 for s
→
0
+, we get the result of Theorem 5.Remark 13.
Applying Theorem10 for s
→
0
+, we get the result of Theorem 6.Remark 14.
Applying Theorem11 for s
→
0
+, we get the result in Theorem 7.



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*Source: 829158-2014-02-05.xml*