# Some Inequalities of Simpson Type forh-Convex Functions via Fractional Integrals

Authors: Marian Matłoka
Journal: Abstract and Applied Analysis
 (2015)

---

## Abstract

We establish some inequalities of Simpson type involving Riemann-Liouville fractional integrals for mappings whose first derivatives areh-convex.


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## Body

## 1. Introduction

The following inequality is well known in the literature as Simpson’s inequality.Theorem 1.
Letf
:
[
a
,
b
]
→
R be four times continuously differentiable mapping on (
a
,
b
) and f
4
∞
=
s
u
p
x
∈
(
a
,
b
)
f
4
(
x
)
<
∞.
Then, the following inequality holds:(1)
1
3
f
a
+
f
b
2
+
2
f
a
+
b
2
-
1
b
-
a
∫
a
b
f
x
d
x
≤
1
2880
f
4
∞
b
-
a
4
.In [1], Dragomir et al. proved the following inequality.Theorem 2.
Supposef
:
[
a
,
b
]
→
R is a differentiable mapping whose derivative is continuous on (
a
,
b
) and f
′
∈
L
1
[
a
,
b
]. Then the following inequality holds:(2)
1
3
f
a
+
f
b
2
+
2
f
a
+
b
2
-
1
b
-
a
∫
a
b
f
x
d
x
≤
b
-
a
3
f
′
1
,where f
′
1
=
∫
a
b
f
′
(
x
)
d
x.In [2], Sarikaya et al. obtained inequalities for differentiable convex mappings. The main inequality is as follows.Theorem 3.
Letf
:
I
⊂
R
→
R be a differentiable mapping on I
∘ such that f
′
∈
L
1
[
a
,
b
], where a
,
b
∈
I
∘ with a
<
b. If f
′
q is convex on [
a
,
b
], q
>
1, then the following inequality holds:(3)
1
6
f
a
+
4
f
a
+
b
2
+
f
b
-
1
b
-
a
∫
a
b
f
x
d
x
≤
b
-
a
12
1
+
2
p
+
1
3
p
+
1
1
/
p
·
3
f
′
b
q
+
f
′
a
q
4
1
/
q
+
f
′
b
q
+
3
f
′
a
q
4
1
/
q
,where 1
/
p
+
1
/
q
=
1.In [3], Sarikaya et al. obtained the following inequality for s-convex function.Theorem 4.
Letf
:
I
⊂
[
0
,
∞
]
→
R be a differentiable mapping on I
∘ such that f
′
∈
L
1
[
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:(4)
1
6
f
a
+
4
f
a
+
b
2
+
f
b
-
1
b
-
a
∫
a
b
f
x
d
x
≤
b
-
a
12
1
+
2
p
+
1
3
p
+
1
1
/
p
·
f
′
b
q
+
f
′
a
+
b
/
2
q
s
+
1
1
/
q
+
f
′
a
q
+
f
′
a
+
b
/
2
q
s
+
1
1
/
q
,where 1
/
p
+
1
/
q
=
1.For recent refinements, counterparts, generalizations, and inequalities of Simpson type, see [1–7].In 2007, Varošanec in [8] introduced a large class of functions, the so-called h-convex functions. This class contains several well-known classes of functions such as nonnegative convex functions and s-convex functions. This class is defined in the following way: a function f
:
I
→
R, ∅
≠
I
⊂
R being an interval, is called h-convex if(5)
f
t
x
+
1
-
t
y
≤
h
t
f
x
+
h
1
-
t
f
yholds for all x
,
y
∈
I, t
∈
(
0
,
1
), where h
:
J
→
R, h
≢
0, and J is an interval, (
0
,
1
)
⊆
J.The aim of this paper is to establish inequalities of Simpson type forh-convex mappings via fractional integrals which are defined in the following way: left-sided and right-sided Riemann-Liouville fractional integrals of the order α
>
0 are defined by(6)
I
a
+
α
f
x
=
1
Γ
α
∫
a
x
x
-
t
α
-
1
f
t
d
t
0
≤
a
<
x
≤
b
,
I
b
-
α
f
x
=
1
Γ
α
∫
x
b
t
-
x
α
-
1
f
t
d
t
0
≤
a
≤
x
<
b
,where Γ
α
=
∫
0
∞
e
-
u
u
α
-
1
d
u is the gamma function. Here is I
a
+
∘
f
x
=
I
b
-
∘
f
x
=
f
x.


## 2. Main Results

To prove our main results, we consider the following lemma.Lemma 5.
Letf
:
I
⊂
R
→
R be an absolutely continuous mapping on I
∘ such that f
′
∈
L
1
[
a
,
b
], where a
,
b
∈
I
∘ with a
<
b. Then the following inequality holds:(7)
1
6
f
a
+
4
f
a
+
b
2
+
f
b
-
2
α
-
1
Γ
α
+
1
b
-
a
α
I
a
+
α
f
a
+
b
2
+
I
b
-
α
f
a
+
b
2
=
b
-
a
2
∫
0
1
t
α
2
-
1
3
f
′
1
+
t
2
b
+
1
-
t
2
a
d
t
+
∫
0
1
1
3
-
t
α
2
f
′
1
+
t
2
a
+
1
-
t
2
b
d
t
.Proof.
By integration by parts and by the change of the variables, we have(8)
∫
0
1
t
α
2
-
1
3
f
′
1
+
t
2
b
+
1
-
t
2
a
d
t
=
2
b
-
a
t
α
2
-
1
3
f
1
+
t
2
b
+
1
-
t
2
a
0
1
-
α
b
-
a
∫
0
1
t
α
-
1
f
1
+
t
2
b
+
1
-
t
2
a
d
t
=
1
3
b
-
a
f
b
+
2
3
b
-
a
f
a
+
b
2
-
2
α
·
α
b
-
a
α
+
1
∫
a
+
b
/
2
b
u
-
a
+
b
2
α
-
1
f
u
d
u
=
1
3
b
-
a
f
b
+
2
3
b
-
a
f
a
+
b
2
-
2
α
Γ
α
+
1
b
-
a
α
+
1
I
b
-
α
f
a
+
b
2
.Similarly, we have (9)
∫
0
1
1
3
-
t
α
2
f
′
1
+
t
2
a
+
1
-
t
2
b
d
t
=
2
a
-
b
1
3
-
t
α
2
f
1
+
t
2
a
+
1
-
t
2
b
0
1
+
α
a
-
b
∫
0
1
t
α
-
1
f
1
+
t
2
a
+
1
-
t
2
b
d
t
=
1
3
b
-
a
f
a
+
2
3
b
-
a
f
a
+
b
2
+
2
α
·
α
b
-
a
α
+
1
∫
a
+
b
/
2
a
a
+
b
2
-
u
α
-
1
f
u
d
u
=
1
3
b
-
a
f
a
+
2
3
b
-
a
f
a
+
b
2
-
2
α
Γ
α
+
1
b
-
a
α
+
1
I
a
+
α
f
a
+
b
2
.From (8) and (9), we get (7). This completes the proof.The following theorems give a new result of Simpson’s inequality forh-convex functions via fractional integrals.Theorem 6.
Letf
:
I
⊂
R
→
R be a differentiable mapping on I
∘ such that f
′
∈
L
1
[
a
,
b
], where a
,
b
∈
I
∘ with a
<
b. If f
′ is h-convex on [
a
,
b
], then the following inequality holds:(10)
1
6
f
a
+
4
f
a
+
b
2
+
f
b
-
2
α
-
1
Γ
α
+
1
b
-
a
α
I
a
+
α
f
a
+
b
2
+
I
b
-
α
f
a
+
b
2
≤
b
-
a
3
f
′
a
+
f
′
b
∫
0
1
h
t
d
t
.Proof.
From Lemma5 and since f
′ is h-convex on [
a
,
b
], we get(11)
1
6
f
a
+
4
f
a
+
b
2
+
f
b
-
2
α
-
1
Γ
α
+
1
b
-
a
α
I
a
+
α
f
a
+
b
2
+
I
b
-
α
f
a
+
b
2
≤
b
-
a
2
∫
0
1
t
α
2
-
1
3
f
′
1
+
t
2
b
+
1
-
t
2
a
+
1
3
-
t
α
2
f
′
1
+
t
2
a
+
1
-
t
2
b
d
t
≤
b
-
a
2
·
∫
0
1
t
α
2
-
1
3
h
1
+
t
2
f
′
b
+
h
1
-
t
2
f
′
a
+
h
1
+
t
2
f
′
a
+
h
1
-
t
2
f
′
b
d
t
=
b
-
a
2
f
′
a
+
f
′
b
∫
0
1
t
α
2
-
1
3
h
1
+
t
2
+
h
1
-
t
2
d
t
≤
b
-
a
3
f
′
a
+
f
′
b
∫
0
1
h
t
d
t
,where we used the fact that t
α
/
2
-
1
/
3
≤
1
/
3 for all x
∈
[
0
,
1
].
The proof is completed.Corollary 7.
If in Theorem6 one takes h
t
=
t then inequality (10) reduces to the following inequality for the convex function:(12)
1
6
f
a
+
4
f
a
+
b
2
+
f
b
-
2
α
-
1
Γ
α
+
1
b
-
a
α
I
a
+
α
f
a
+
b
2
+
I
b
-
α
f
a
+
b
2
≤
b
-
a
6
f
′
a
+
f
′
b
.Corollary 8.
If in Theorem6 one takes h
t
=
t
s then inequality (10) reduces to the following inequality for the s-convex function:(13)
1
6
f
a
+
4
f
a
+
b
2
+
f
b
-
2
α
-
1
Γ
α
+
1
b
-
a
α
I
a
+
α
f
a
+
b
2
+
I
b
-
α
f
a
+
b
2
≤
b
-
a
3
s
+
1
f
′
a
+
f
′
b
.Corollary 9.
If in Theorem6 one takes h
t
=
t and α
=
1 then from the proof of Theorem 6 it follows that the following inequality holds:(14)
1
6
f
a
+
4
f
a
+
b
2
+
f
b
-
1
b
-
a
∫
a
b
f
x
d
x
≤
5
b
-
a
72
f
′
a
+
f
′
b
.Theorem 10.
Letf
:
I
⊂
R
→
R be a differentiable mapping on I
∘ such that f
′
∈
L
1
[
a
,
b
], where a
,
b
∈
I
∘ with a
<
b. If f
′
q is h-convex on a
,
b and q
>
1, then the following inequality holds:(15)
1
6
f
a
+
4
f
a
+
b
2
+
f
b
-
2
α
-
1
Γ
α
+
1
b
-
a
α
I
a
+
α
f
a
+
b
2
+
I
b
-
α
f
a
+
b
2
≤
b
-
a
6
f
′
b
q
∫
0
1
h
1
+
t
2
d
t
+
f
′
a
q
∫
0
1
h
1
-
t
2
d
t
1
/
q
+
f
′
a
q
∫
0
1
h
1
+
t
2
d
t
+
f
′
b
q
∫
0
1
h
1
-
t
2
d
t
1
/
q
.Proof.
From Lemma5 and the Hőlder inequality, we have(16)
1
6
f
a
+
4
f
a
+
b
2
+
f
b
-
2
α
-
1
Γ
α
+
1
b
-
a
α
I
a
+
α
f
a
+
b
2
+
I
b
-
α
f
a
+
b
2
≤
b
-
a
2
∫
0
1
t
α
2
-
1
3
p
d
t
1
/
p
·
∫
0
1
f
′
1
+
t
2
b
+
1
-
t
2
a
q
d
t
1
/
q
+
∫
0
1
1
3
-
t
α
2
p
d
t
1
/
p
·
∫
0
1
f
′
1
+
t
2
a
+
1
-
t
2
b
q
d
t
1
/
q
,where 1
/
p
+
1
/
q
=
1.
Becausef
′
q is h-convex, we have(17)
∫
0
1
f
′
1
+
t
2
b
+
1
-
t
2
a
q
d
t
≤
f
′
b
q
∫
0
1
h
1
+
t
2
d
t
+
f
′
a
q
∫
0
1
h
1
-
t
2
d
t
,
∫
0
1
f
′
1
+
t
2
a
+
1
-
t
2
b
q
d
t
≤
f
′
a
q
∫
0
1
h
1
+
t
2
d
t
+
f
′
b
q
∫
0
1
h
1
-
t
2
d
t
.Using the fact that t
α
/
2
-
1
/
3
=
1
/
3
-
t
α
/
2
≤
1
/
3 for all t
∈
[
0
,
1
] and using the last two inequalities in (16) we obtain (15).
This completes the proof of the theorem.Corollary 11.
If in Theorem10 one takes h
t
=
t then inequality (15) reduces to the following inequality for the convex function:(18)
1
6
f
a
+
4
f
a
+
b
2
+
f
b
-
2
α
-
1
Γ
α
+
1
b
-
a
α
I
a
+
α
f
a
+
b
2
+
I
b
-
α
f
a
+
b
2
≤
b
-
a
2
3
4
f
′
b
q
+
1
4
f
′
a
q
1
/
q
+
1
4
f
′
b
q
+
3
4
f
′
a
q
1
/
q
.Corollary 12.
If in Theorem10 one takes h
t
=
t
s then inequality (15) reduces to the following inequality for the s-convex function:(19)
1
6
f
a
+
4
f
a
+
b
2
+
f
b
-
2
α
-
1
Γ
α
+
1
b
-
a
α
I
a
+
α
f
a
+
b
2
+
I
b
-
α
f
a
+
b
2
≤
b
-
a
2
2
s
+
1
1
/
q
·
f
′
b
q
1
-
1
2
s
+
1
+
f
′
a
q
1
2
s
+
1
1
/
q
+
f
′
b
q
1
2
s
+
1
+
f
′
a
q
1
-
1
2
s
+
1
1
/
q
.Theorem 13.
Letf
:
I
⊂
R
→
R be a differentiable mapping on I
∘ such that f
′
∈
L
1
[
a
,
b
], where a
,
b
∈
I
∘ with a
<
b. If f
′
q is h-convex on a
,
b and q
≥
1, then the following inequality holds:(20)
1
6
f
a
+
4
f
a
+
b
2
+
f
b
-
2
α
-
1
Γ
α
+
1
b
-
a
α
I
a
+
α
f
a
+
b
2
+
I
b
-
α
f
a
+
b
2
≤
b
-
a
2
f
′
b
q
∫
0
1
h
1
+
t
2
d
t
+
f
′
a
q
∫
0
1
h
1
-
t
2
d
t
1
/
q
+
f
′
a
q
∫
0
1
h
1
+
t
2
d
t
+
f
′
b
q
∫
0
1
h
1
-
t
2
d
t
1
/
q
.Proof.
From Lemma5 and the power mean inequality, we have that the following inequality holds:(21)
1
6
f
a
+
4
f
a
+
b
2
+
f
b
-
2
α
-
1
Γ
α
+
1
b
-
a
α
I
a
+
α
f
a
+
b
2
+
I
b
-
α
f
a
+
b
2
≤
b
-
a
2
∫
0
1
t
α
2
-
1
3
d
t
1
-
1
/
q
·
∫
0
1
t
α
2
-
1
3
f
′
1
+
t
2
b
+
1
-
t
2
a
q
d
t
1
/
q
+
∫
0
1
1
3
-
t
α
2
d
t
·
f
′
1
+
t
2
a
+
1
-
t
2
b
q
d
t
1
/
q
.By the h-convexity of f
′
q and using the fact that t
α
/
2
-
1
/
3
=
1
/
3
-
t
α
/
2
≤
1
/
3 for all t
∈
[
0
,
1
], we have(22)
∫
0
1
t
α
2
-
1
3
f
′
1
+
t
2
b
+
1
-
t
2
a
q
d
t
≤
1
3
f
′
b
q
∫
0
1
h
1
+
t
2
d
t
+
f
′
a
q
∫
0
1
h
1
-
t
2
d
t
,
∫
0
1
1
3
-
t
α
2
f
′
1
+
t
2
a
+
1
-
t
2
b
q
d
t
≤
1
3
f
′
a
q
∫
0
1
h
1
+
t
2
d
t
+
f
′
b
q
∫
0
1
h
1
-
t
2
d
t
.Using the last two inequalities in (21) we obtain (20). This completes the proof.



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*Source: 956850-2015-07-12.xml*