# Hermite-Hadamard Type Inequalities for Superquadratic Functions via Fractional Integrals

Authors: Guangzhou Li; Feixiang Chen
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We use basic properties of superquadratic functions to obtain some new Hermite-Hadamard type inequalities via Riemann-Liouville fractional integrals. For superquadratic functions which are also convex, we get refinements of existing results.


---

## Body

## 1. Introduction

Letf
:
I
⊆
R
→
R be a convex function and a
,
b
∈
I with a
<
b; then
(1)
f
(
a
+
b
2
)
≤
1
b
-
a
∫
a
b
‍
f
(
t
)
d
t
≤
f
(
a
)
+
f
(
b
)
2
is known as the Hermite-Hadamard inequality.This remarkable result is well known in the literature as the Hermite-Hadamard inequality. Recently, the generalizations, refinements, and improvements of the classical Hermite-Hadamard inequality have been the subject of intensive research.Definition 1 (see [1]).
A functionf
:
[
0
,
∞
)
→
R is superquadratic provided that for all x
≥
0 there exists a constant C
x
∈
R such that
(2)
f
(
y
)
≥
f
(
x
)
+
C
x
(
y
-
x
)
+
f
(
|
y
-
x
|
)
,
for all y
≥
0.Theorem 2 (see [1]).
The inequality(3)
f
(
∫
‍
g
d
μ
)
≤
∫
‍
[
f
(
g
(
s
)
)
-
f
(
|
g
(
s
)
-
∫
‍
g
d
μ
|
)
]
d
μ
(
s
)
holds for all probability measures μ and all nonnegative, μ-integrable functions g, if and only if f is superquadratic.The discrete version of the above theorem is also used in the sequel.Lemma 3 (see [2]).
Suppose thatf is superquadratic. Let x
i
≥
0, 1
≤
i
≤
n, and let x
¯
=
∑
i
=
1
n
‍
λ
i
x
i, where λ
i
≥
0 and ∑
i
=
1
n
‍
λ
i
=
1. Then
(4)
∑
i
=
1
n
‍
λ
i
f
(
x
i
)
≥
f
(
x
¯
)
+
∑
i
=
1
n
‍
λ
i
f
(
|
x
i
-
x
¯
|
)
.Nonnegative superquadratic functions are much better behaved as we see next.Lemma 4 (see [1]).
Letf be a superquadratic function with C
x
∈
R as in Definition 1. Then one gets the following:(1)
f
(
0
)
≤
0;
(2)
iff
(
0
)
=
f
′
(
0
)
=
0, then C
x
=
f
′
(
x
) whenever f is differentiable at x
>
0;
(3)
iff
≥
0, then f is convex and f
(
0
)
=
f
′
(
0
)
=
0.The Hermite-Hadamard inequalities for superquadratic functions are established by Banic et al. in [3].Theorem 5 (see [3]).
Letf
:
[
0
,
∞
)
→
R be a superquadratic function and 0
≤
a
<
b; then
(5)
f
(
a
+
b
2
)
+
1
b
-
a
∫
a
b
‍
f
(
|
x
-
a
+
b
2
|
)
d
x
≤
1
b
-
a
∫
a
b
‍
f
(
x
)
d
x
≤
f
(
a
)
+
f
(
b
)
2
-
1
(
b
-
a
)
2
∫
a
b
‍
[
(
b
-
x
)
f
(
x
-
a
)
+
(
x
-
a
)
f
(
b
-
x
)
]
d
x
.It is remarkable that Sarikaya et al. [4] proved the following interesting inequalities of Hermite-Hadamard type involving Riemann-Liouville fractional integrals.Theorem 6 (see [4]).
Letf
:
[
a
,
b
]
→
R be a positive function with a
<
b and f
∈
L
1
[
a
,
b
]. If f is a convex function on [
a
,
b
], then the following inequalities for fractional integrals hold:
(6)
f
(
a
+
b
2
)
≤
Γ
(
α
+
1
)
2
(
b
-
a
)
α
[
J
a
+
α
f
(
b
)
+
J
b
-
α
f
(
a
)
]
≤
f
(
a
)
+
f
(
b
)
2
,
with α
>
0.We remark that the symbolsJ
a
+
α and J
b
-
α
f denote the left-sided and right-sided Riemann-Liouville fractional integrals of the order α
≥
0 with a
≥
0 which are defined by
(7)
J
a
+
α
f
(
x
)
=
1
Γ
(
α
)
∫
a
x
‍
(
x
-
t
)
α
-
1
f
(
t
)
d
t
,
x
>
a
,
J
b
-
α
f
(
x
)
=
1
Γ
(
α
)
∫
x
b
‍
(
t
-
x
)
α
-
1
f
(
t
)
d
t
,
x
<
b
,
respectively. Here, Γ
(
α
) is the gamma function defined by Γ
(
α
)
=
∫
0
∞
‍
e
-
t
t
α
-
1
d
t.Fractional integral operators are widely used to solve differential equations and integral equations. So a lot of work has been obtained on the theory and applications of fractional integral operators.For more results concerning the fractional integral operators, we refer the reader to [5–10] and references cited therein.In this paper, we establish some new Hermite-Hadamard type inequalities for superquadratic functions via Riemann-Liouville fractional integrals which refine the inequalities of (6) for superquadratic functions which are also convex.


## 2. Main Results

From Lemma3 for n
=
2, we get for 0
≤
m
≤
z
≤
M, m
≤
M, for the superquadratic function f that
(8)
f
(
z
)
≤
M
-
z
M
-
m
f
(
m
)
+
z
-
m
M
-
m
f
(
M
)
-
M
-
z
M
-
m
f
(
z
-
m
)
-
z
-
m
M
-
m
f
(
M
-
z
)
,
f
(
M
+
m
-
z
)
≤
z
-
m
M
-
m
f
(
m
)
+
M
-
z
M
-
m
f
(
M
)
-
M
-
z
M
-
m
f
(
z
-
m
)
-
z
-
m
M
-
m
f
(
M
-
z
)
hold, and therefore
(9)
f
(
z
)
+
f
(
M
+
m
-
z
)
≤
f
(
m
)
+
f
(
M
)
-
2
M
-
z
M
-
m
f
(
z
-
m
)
-
2
z
-
m
M
-
m
f
(
M
-
z
)
.
Let 0
≤
a
≤
x
≤
(
a
+
b
)
/
2; we get that
(10)
a
≤
x
≤
a
+
b
2
≤
a
+
b
-
x
≤
b
.
Therefore, by replacing in (9)
(11)
z
=
a
+
b
2
,
M
=
a
+
b
-
x
,
m
=
x
,
we get that
(12)
2
f
(
a
+
b
2
)
≤
f
(
x
)
+
f
(
a
+
b
-
x
)
-
2
f
(
a
+
b
2
-
x
)
.Theorem 7.
Letf be a superquadratic integrable function on [
0
,
b
] with 0
≤
a
<
b. Then
(13)
f
(
a
+
b
2
)
≤
Γ
(
α
+
1
)
2
(
b
-
a
)
α
[
J
a
+
α
f
(
b
)
+
J
b
-
α
f
(
a
)
]
-
α
2
(
b
-
a
)
α
∫
a
b
‍
f
|
a
+
b
2
-
x
|
×
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
with α
>
0.Proof.
By inequality (12), we get that
(14)
f
(
a
+
b
2
)
=
f
(
a
+
b
2
)
α
2
(
b
-
a
)
α
×
∫
a
b
‍
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
=
f
(
a
+
b
2
)
α
(
b
-
a
)
α
×
∫
a
(
a
+
b
)
/
2
‍
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
≤
α
2
(
b
-
a
)
α
∫
a
(
a
+
b
)
/
2
‍
[
f
(
x
)
+
f
(
a
+
b
-
x
)
]
×
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
-
α
(
b
-
a
)
α
∫
a
(
a
+
b
)
/
2
‍
f
(
a
+
b
2
-
x
)
×
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
.
By the change of variable x
→
a
+
b
-
x, we get
(15)
α
2
(
b
-
a
)
α
∫
a
(
a
+
b
)
/
2
‍
f
(
a
+
b
-
x
)
×
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
=
α
(
b
-
a
)
α
∫
(
a
+
b
)
/
2
b
‍
f
(
x
)
×
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
,
α
(
b
-
a
)
α
∫
a
(
a
+
b
)
/
2
‍
f
(
a
+
b
2
-
x
)
×
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
=
α
(
b
-
a
)
α
∫
(
a
+
b
)
/
2
b
‍
f
(
x
-
a
+
b
2
)
×
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
.
Therefore
(16)
f
(
a
+
b
2
)
=
f
(
a
+
b
2
)
α
2
(
b
-
a
)
α
×
∫
a
b
‍
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
=
f
(
a
+
b
2
)
α
(
b
-
a
)
α
×
∫
a
(
a
+
b
)
/
2
‍
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
≤
α
2
(
b
-
a
)
α
∫
a
b
‍
f
(
x
)
×
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
-
α
2
(
b
-
a
)
α
∫
a
b
‍
f
|
a
+
b
2
-
x
|
×
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
=
Γ
(
α
+
1
)
2
(
b
-
a
)
α
[
J
a
+
α
f
(
b
)
+
J
b
-
α
f
(
a
)
]
-
α
2
(
b
-
a
)
α
∫
a
b
‍
f
|
a
+
b
2
-
x
|
×
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
.
We have completed the proof.Corollary 8.
Puttingα
=
1 in Theorem 7 gives
(17)
f
(
a
+
b
2
)
≤
1
b
-
a
∫
a
b
‍
f
(
x
)
d
x
-
1
b
-
a
∫
a
b
‍
f
|
a
+
b
2
-
x
|
d
x
.Let0
≤
a
≤
x
≤
(
a
+
b
)
/
2; we get that
(18)
a
≤
x
≤
a
+
b
-
x
≤
b
.
Therefore, by replacing in (9)
(19)
z
=
x
,
M
=
b
,
m
=
a
,
we get that
(20)
f
(
x
)
+
f
(
a
+
b
-
x
)
≤
f
(
a
)
+
f
(
b
)
-
2
b
-
x
b
-
a
f
(
x
-
a
)
-
2
x
-
a
b
-
a
f
(
b
-
x
)
.Theorem 9.
Letf be a superquadratic integrable function on [
0
,
b
] with 0
≤
a
<
b. Then
(21)
Γ
(
α
+
1
)
2
(
b
-
a
)
α
[
J
a
+
α
f
(
b
)
+
J
b
-
α
f
(
a
)
]
≤
f
(
a
)
+
f
(
b
)
2
-
α
2
(
b
-
a
)
α
∫
a
b
‍
[
b
-
x
b
-
a
f
(
x
-
a
)
+
x
-
a
b
-
a
f
(
b
-
x
)
]
×
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
with α
>
0.Proof.
By inequality (20), we get that
(22)
f
(
a
)
+
f
(
b
)
2
=
f
(
a
)
+
f
(
b
)
2
α
2
(
b
-
a
)
α
×
∫
a
b
‍
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
=
[
f
(
a
)
+
f
(
b
)
]
α
2
(
b
-
a
)
α
×
∫
a
(
a
+
b
)
/
2
‍
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
,
Γ
(
α
+
1
)
2
(
b
-
a
)
α
[
J
a
+
α
f
(
b
)
+
J
b
-
α
f
(
a
)
]
=
α
2
(
b
-
a
)
α
∫
a
b
‍
f
(
x
)
×
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
=
α
2
(
b
-
a
)
α
∫
a
(
a
+
b
)
/
2
‍
[
f
(
x
)
+
f
(
a
+
b
-
x
)
]
×
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
≤
α
2
(
b
-
a
)
α
∫
a
(
a
+
b
)
/
2
‍
[
f
(
a
)
+
f
(
b
)
]
×
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
-
α
(
b
-
a
)
α
∫
a
(
a
+
b
)
/
2
‍
b
-
x
b
-
a
f
(
x
-
a
)
×
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
-
α
(
b
-
a
)
α
∫
a
(
a
+
b
)
/
2
‍
x
-
a
b
-
a
f
(
b
-
x
)
×
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
=
f
(
a
)
+
f
(
b
)
2
-
α
2
(
b
-
a
)
α
∫
a
b
[
b
-
x
b
-
a
f
(
x
-
a
)
‍
+
x
-
a
b
-
a
f
(
b
-
x
)
]
×
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
.
The proof is completed.Corollary 10.
Choosingα
=
1 in Theorem 9, one has
(23)
1
b
-
a
∫
a
b
‍
f
(
x
)
d
x
≤
f
(
a
)
+
f
(
b
)
2
-
1
(
b
-
a
)
2
∫
a
b
‍
[
(
b
-
x
)
f
(
x
-
a
)
+
(
x
-
a
)
f
(
b
-
x
)
]
d
x
.Corollary 11.
Letf be defined as in Theorem 7; one gets
(24)
f
(
a
+
b
2
)
+
α
2
(
b
-
a
)
α
×
∫
a
b
‍
f
|
a
+
b
2
-
x
|
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
≤
Γ
(
α
+
1
)
2
(
b
-
a
)
α
[
J
a
+
α
f
(
b
)
+
J
b
-
α
f
(
a
)
]
≤
f
(
a
)
+
f
(
b
)
2
-
α
2
(
b
-
a
)
α
×
∫
a
b
‍
[
b
-
x
b
-
a
f
(
x
-
a
)
+
x
-
a
b
-
a
f
(
b
-
x
)
]
×
(
(
b
-
x
)
α
-
1
+
(
x
-
a
)
α
-
1
)
d
x
.Corollary 12.
Takingα
=
1 in Corollary 11, one obtains
(25)
f
(
a
+
b
2
)
+
1
b
-
a
∫
a
b
‍
f
|
a
+
b
2
-
x
|
d
x
≤
∫
a
b
‍
f
(
x
)
d
x
≤
f
(
a
)
+
f
(
b
)
2
-
1
(
b
-
a
)
2
×
∫
a
b
‍
[
(
b
-
x
)
f
(
x
-
a
)
+
(
x
-
a
)
f
(
b
-
x
)
]
d
x
.


## 3. Conclusion

In this note, we obtain some new Hermite-Hadamard type inequalities for superquadratic functions via Riemann-Liouville fractional integrals. For superquadratic functions which are also convex, we get refinements of known results. The concept of superquadratic functions in several variables is introduced in [11]. An interesting topic is whether we can use the methods in this paper to establish the Hermite-Hadamard inequalities for superquadratic functions in several variables via Riemann-Liouville integrals.



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*Source: 851271-2014-08-14.xml*