# On Hermite-Hadamard Type Inequalities fors-Convex Functions on the Coordinates via Riemann-Liouville Fractional Integrals

Authors: Feixiang Chen
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

We obtain some Hermite-Hadamard type inequalities fors-convex functions on the coordinates via Riemann-Liouville integrals. Some integral inequalities with the right-hand side of the fractional Hermite-Hadamard type inequality are also established.


---

## Body

## 1. Introduction

Iff
:
I
→
R is 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.Since then, some refinements of the Hermite-Hadamard inequality on convex functions have been extensively investigated by a number of authors (e.g., [1–7]).Definition 1 (see [8]).
Lets
∈
(
0,1
] be a fixed real number. A function f
:
I
⊂
[
0
,
∞
)
→
[
0
,
∞
) is said to be s-convex in the second sense, or that f belongs to the class K
s
2, 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 and α
∈
[
0,1
].It can be easily seen that, fors
=
1, s-convexity reduces to ordinary convexity of functions defined on [
0
,
∞
).In [4], Dragomir defined convex functions on the coordinates as follows.Let us consider the bidimensional intervalΔ
:
=
[
a
,
b
]
×
[
c
,
d
] in R
2 with a
<
b and c
<
d; a mapping f
:
Δ
→
R is said to be convex on Δ if the inequality,
(3)
f
(
λ
x
+
(
1
-
λ
)
z
,
λ
y
+
(
1
-
λ
)
w
)
≤
λ
f
(
x
,
y
)
+
(
1
-
λ
)
f
(
z
,
w
)
,
holds for all (
x
,
y
), (
z
,
w
)
∈
Δ and λ
∈
[
0,1
].A functionf
:
Δ
→
R is said to be coordinated convex on Δ if the partial mappings f
y
:
[
a
,
b
]
→
R, f
y
(
u
)
=
f
(
u
,
y
) and f
x
:
[
c
,
d
]
→
R, f
x
(
v
)
=
f
(
x
,
v
) are convex for all y
∈
[
c
,
d
] and x
∈
[
a
,
b
].A formal definition for convex functions on the coordinates may be stated as follows.Definition 2.
A functionf
:
Δ
→
R is said to be convex on coordinates on Δ if the inequality,
(4)
f
(
λ
x
+
(
1
-
λ
)
z
,
t
y
+
(
1
-
t
)
w
)
≤
λ
t
f
(
x
,
y
)
+
λ
(
1
-
t
)
f
(
x
,
w
)
+
(
1
-
λ
)
t
f
(
z
,
y
)
+
(
1
-
t
)
(
1
-
λ
)
f
(
z
,
w
)
,
holds for all (
x
,
y
), (
z
,
y
), (
x
,
w
), (
z
,
w
)
∈
Δ, and t
,
λ
∈
[
0,1
].In [4], Dragomir established the following Hadamard-type inequalities for convex functions on the coordinates in a rectangle from the plane R
2.Theorem 3 (see [4]).
Suppose thatf
:
Δ
=
[
a
,
b
]
×
[
c
,
d
]
→
R is convex on the coordinates on Δ. Then one has the inequalities:
(5)
f
(
a
+
b
2
,
c
+
d
2
)
≤
1
(
b
-
a
)
(
d
-
c
)
∫
a
b
∫
c
d
f
(
x
,
y
)
d
y
d
x
‍
≤
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
.The concept ofs-convex functions on the coordinates in the second sense was introduced by Alomari and Darus in [9].Let us consider the bidimensional intervalΔ
:
=
[
a
,
b
]
×
[
c
,
d
] in R
2 with a
<
b and c
<
d; a mapping f
:
Δ
→
R is s-convex on Δ if the inequality,
(6)
f
(
λ
x
+
(
1
-
λ
)
z
,
λ
y
+
(
1
-
λ
)
w
)
≤
λ
s
f
(
x
,
y
)
+
(
1
-
λ
)
s
f
(
z
,
w
)
,
holds for all (
x
,
y
), (
z
,
w
)
∈
Δ with λ
∈
[
0,1
] and for some fixed s
∈
(
0,1
].A functionf
:
Δ
→
R is said to be s-convex on the coordinates on Δ in the second sense if the partial mappings f
y
:
[
a
,
b
]
→
R, f
y
(
u
)
=
f
(
u
,
y
), and f
x
:
[
c
,
d
]
→
R, f
x
(
v
)
=
f
(
x
,
v
) are s-convex in the second sense for all y
∈
[
c
,
d
] and x
∈
[
a
,
b
] with some fixed s
∈
(
0,1
].A formal definition for convex functions on the coordinates in the second sense may be stated as follows.Definition 4.
A functionf
:
Δ
→
R is said to be s-convex on coordinates in the second sense on Δ if the inequality,
(7)
f
(
λ
x
+
(
1
-
λ
)
z
,
t
y
+
(
1
-
t
)
w
)
≤
λ
s
t
s
f
(
x
,
y
)
+
λ
s
(
1
-
t
)
s
f
(
x
,
w
)
+
(
1
-
λ
)
s
t
s
f
(
z
,
y
)
+
(
1
-
t
)
s
(
1
-
λ
)
s
f
(
z
,
w
)
,
holds for all (
x
,
y
), (
z
,
y
), (
x
,
w
), (
z
,
w
)
∈
Δ with t
,
λ
∈
[
0,1
] and some fixed s
∈
(
0,1
].In [10], Alomari and Darus proved the following inequalities based on the above definition.Theorem 5 (see [10]).
Suppose thatf
:
Δ
=
[
a
,
b
]
×
[
c
,
d
]
⊆
[
0
,
∞
)
2
→
[
0
,
∞
) is s-convex on the coordinates in the second sense on Δ. Then one has the inequalities:
(8)
4
s
-
1
f
(
a
+
b
2
,
c
+
d
2
)
≤
1
(
b
-
a
)
(
d
-
c
)
∫
a
b
∫
c
d
f
(
x
,
y
)
d
y
d
x
‍
‍
≤
(
+
f
(
b
,
d
)
)
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
)
(
(
s
+
1
)
2
)
-
1
.It is remarkable that Sarikaya et al. [11] proved the following interesting inequalities of Hermite-Hadamard type involving Riemann-Liouville fractional integrals.Theorem 6 (see [11]).
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
(9)
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 symbolJ
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
(10)
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.Definition 7 (see [12]).
Letf
∈
L
1
[
a
,
b
]
×
[
c
,
d
]. The Riemann-Liouville fractional integrals J
a
+
,
c
+
α
,
β, J
a
+
,
d
-
α
,
β, J
b
-
,
c
+
α
,
β and J
b
-
,
d
-
α
,
β of order α
,
β
>
0 with a
,
c
≥
0 are defined by
(11)
J
a
+
,
c
+
α
,
β
f
(
x
,
y
)
=
1
Γ
(
α
)
Γ
(
β
)
∫
a
x
∫
c
y
(
x
-
t
)
α
-
1
(
y
-
s
)
β
-
1
f
(
t
,
s
)
d
s
d
t
,
‍
‍
x
>
a
,
y
>
c
,
J
a
+
,
d
-
α
,
β
f
(
x
,
y
)
=
1
Γ
(
α
)
Γ
(
β
)
∫
a
x
∫
y
d
‍
(
x
-
t
)
α
-
1
(
s
-
y
)
β
-
1
f
(
t
,
s
)
d
s
d
t
,
‍
x
>
a
,
y
<
d
,
J
b
-
,
c
+
α
,
β
f
(
x
,
y
)
=
1
Γ
(
α
)
Γ
(
β
)
∫
x
b
∫
c
y
(
t
-
x
)
α
-
1
(
y
-
s
)
β
-
1
f
(
t
,
s
)
d
s
d
t
,
‍
‍
x
<
b
,
y
>
c
,
J
b
-
,
d
-
α
,
β
f
(
x
,
y
)
=
1
Γ
(
α
)
Γ
(
β
)
∫
x
b
∫
y
d
(
t
-
x
)
α
-
1
(
s
-
y
)
β
-
1
f
(
t
,
s
)
d
s
d
t
,
‍
‍
x
<
b
,
y
<
d
.In [12], Sarıkaya proposed the following Hermite-Hadamard-type inequalities for Riemann-Liouville fractional integrals by using convex functions of two variables on the coordinates.Theorem 8 (see [12]).
Letf
:
Δ
⊆
R
2
→
R be convex functions on the coordinates on Δ
=
[
a
,
b
]
×
[
c
,
d
] in R
2 with 0
≤
a
<
b, 0
≤
c
<
d, and f
∈
L
1
(
Δ
). Then one has the inequalities:
(12)
f
(
a
+
b
2
,
c
+
d
2
)
≤
Γ
(
α
+
1
)
Γ
(
β
+
1
)
4
(
b
-
a
)
α
(
d
-
c
)
β
×
[
J
a
+
,
c
+
α
,
β
f
(
b
,
d
)
+
J
a
+
,
d
-
α
,
β
f
(
b
,
c
)
+
J
b
-
,
c
+
α
,
β
f
(
a
,
d
)
+
J
b
-
,
d
-
α
,
β
f
(
a
,
c
)
]
≤
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
.In [12], Sarıkaya established some Hermite-Hadamard inequalities for convex functions on the coordinates in the second sense via fractional integrals based on the following lemma.Lemma 9 (see [12]).
Letf
:
Δ
⊆
R
2
→
R be a partial differentiable mapping on Δ
:
=
[
a
,
b
]
×
[
c
,
d
] in R
2 with 0
≤
a
<
b, 0
≤
c
<
d. If ∂
2
f
/
∂
t
∂
s
∈
L
(
Δ
), then the following equality holds:
(13)
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
+
Γ
(
α
+
1
)
Γ
(
β
+
1
)
4
(
b
-
a
)
α
(
d
-
c
)
β
×
[
J
a
+
,
c
+
α
,
β
f
(
b
,
d
)
+
J
a
+
,
d
-
α
,
β
f
(
b
,
c
)
+
J
b
-
,
c
+
α
,
β
f
(
a
,
d
)
+
J
b
-
,
d
-
α
,
β
f
(
a
,
c
)
]
-
A
=
(
b
-
a
)
(
d
-
c
)
4
×
{
∬
0
1
t
α
s
β
∂
2
f
∂
t
∂
s
‍
(
t
a
+
(
1
-
t
)
b
,
s
c
+
(
1
-
s
)
d
)
d
s
d
t
-
∬
0
1
(
1
-
t
)
α
s
β
∂
2
f
∂
t
∂
s
‍
×
(
t
a
+
(
1
-
t
)
b
,
s
c
+
(
1
-
s
)
d
)
d
s
d
t
-
∬
0
1
t
α
(
1
-
s
)
β
‍
∂
2
f
∂
t
∂
s
×
(
t
a
+
(
1
-
t
)
b
,
s
c
+
(
1
-
s
)
d
)
d
s
d
t
+
∬
0
1
(
1
-
t
)
α
(
1
-
s
)
β
‍
∂
2
f
∂
t
∂
s
×
(
t
a
+
(
1
-
t
)
b
,
s
c
+
(
1
-
s
)
d
)
d
s
d
t
∫
0
1
∫
0
1
t
α
s
β
∂
2
f
∂
t
∂
s
‍
}
,
where
(14)
A
=
Γ
(
β
+
1
)
4
(
d
-
c
)
β
[
J
c
+
β
f
(
a
,
d
)

+

J
c
+
β
f
(
b
,
d
)
+
J
d
-
β
f
(
a
,
c
)
+
J
d
-
β
f
(
b
,
c
)
]
+
Γ
(
α
+
1
)
4
(
b
-
a
)
α
[
J
a
+
α
f
(
b
,
c
)
+
J
a
+
α
f
(
b
,
d
)
+
J
b
-
α
f
(
a
,
c
)
+
J
b
-
α
f
(
a
,
d
)
[
J
a
+
α
f
(
b
,
c
)
+
J
a
+
α
f
(
b
,
d
)
+
J
b
-
α
f
(
a
,
c
)
]
.In this paper, we establish some Hermite-Hadamard type inequalities fors-convex functions on the coordinates functions via Riemann-Liouville integrals. Some integral inequalities with the right-hand side of the fractional Hermite-Hadamard type inequality are also given.


## 2. Fractional Inequalities fors-Convex Functions on the Coordinates

Theorem 10.
Suppose thatf
:
Δ
=
[
a
,
b
]
×
[
c
,
d
]
⊆
[
0
,
∞
)
2
→
[
0
,
∞
) is s-convex function on the coordinates in the second sense on Δ. Then one has the inequalities:
(15)
4
s
-
1
f
(
a
+
b
2
,
c
+
d
2
)
≤
Γ
(
α
+
1
)
Γ
(
β
+
1
)
4
(
b
-
a
)
α
(
d
-
c
)
β
×
[
J
a
+
,
c
+
α
,
β
f
(
b
,
d
)
+
J
a
+
,
d
-
α
,
β
f
(
b
,
c
)
+
J
b
-
,
c
+
α
,
β
f
(
a
,
d
)
+
J
b
-
,
d
-
α
,
β
f
(
a
,
c
)
]
≤
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
×
(
1
(
α
+
s
)
(
β
+
s
)
+
1
α
+
1
B
(
β
,
s
+
1
)
+
1
β
+
1
×
B
(
α
,
s
+
1
)
+
B
(
β
,
s
+
1
)
B
(
α
,
s
+
1
)
1
(
α
+
s
)
(
β
+
s
)
)
.Proof.
From (7), with x
=
t
1
a
+
(
1
-
t
1
)
b, y
=
(
1
-
t
1
)
a
+
t
1
b, z
=
t
2
c
+
(
1
-
t
2
)
d, w
=
(
1
-
t
2
)
c
+
t
2
d, and t
=
λ
=
1
/
2, we get
(16)
f
(
a
+
b
2
,
c
+
d
2
)
≤
1
4
s
[
+
f
(
(
1
-
t
1
)
a
+
t
1
b
,
(
1
-
t
2
)
c
+
t
2
d
)
]
f
(
t
1
a
+
(
1
-
t
1
)
b
,
t
2
c
+
(
1
-
t
2
)
d
)
+
f
(
t
1
a
+
(
1
-
t
1
)
b
,
(
1
-
t
2
)
c
+
t
2
d
)
+
f
(
(
1
-
t
1
)
a
+
t
1
b
,
t
2
c
+
(
1
-
t
2
)
d
)
+
f
(
(
1
-
t
1
)
a

+

t
1
b
,
(
1
-
t
2
)
c

+

t
2
d
)
]
.
Multiplying both sides of above inequality by t
1
α
-
1
t
2
β
-
1 then integrating the resulting inequality with respect to t
1
,
t
2 over [
0,1
]
×
[
0,1
], we obtain
(17)
1
α
β
f
(
a
+
b
2
,
c
+
d
2
)
≤
1
4
s
[
∬
0
1
t
1
α
-
1
t
2
β
-
1
f
(
(
(
1
-
t
1
)
a
+
t
1
b
,
(
1
-
t
2
)
c
t
1
a
+
(
1
-
t
1
)
b
,
t
2
c
‍
+
(
1
-
t
2
)
d
(
(
1
-
t
1
)
a
+
t
1
b
,
(
1
-
t
2
)
c
)
d
t
2
d
t
1
+
∬
0
1
t
1
α
-
1
t
2
β
-
1
f
(
(
(
1
-
t
1
)
a
+
t
1
b
,
(
1
-
t
2
)
c
t
1
a
+
(
1
-
t
1
)
b
,
(
1
-
t
2
)
c
‍
+
t
2
d
(
(
1
-
t
1
)
a
+
t
1
b
,
(
1
-
t
2
)
c
)
d
t
2
d
t
1
+
∬
0
1
t
1
α
-
1
t
2
β
-
1
f
(
(
1
-
t
1
)
a
+
t
1
b
,
t
2
c
‍
‍
+
(
1
-
t
2
)
d
(
(
1
-
t
1
)
a
+
t
1
b
,
(
1
-
t
2
)
c
)
d
t
2
d
t
1
+
∬
0
1
t
1
α
-
1
t
2
β
-
1
f
(
(
1
-
t
1
)
a
+
t
1
b
,
(
1
-
t
2
)
c
‍
‍
+
t
2
d
(
(
1
-
t
1
)
a
+
t
1
b
,
(
1
-
t
2
)
c
)
d
t
2
d
t
1
[
∬
0
1
t
1
α
-
1
t
2
β
-
1
f
(
(
(
1
-
t
1
)
a
+
t
1
b
,
(
1
-
t
2
)
c
t
1
a
+
(
1
-
t
1
)
b
,
t
2
c
]
.
Using the change of the variable, we get
(18)
1
α
β
f
(
a
+
b
2
,
c
+
d
2
)
≤
1
4
s
[
∫
a
b
∫
c
d
(
b
-
x
)
α
-
1
(
d
-
y
)
β
-
1
f
(
x
,
y
)
d
y
d
x
‍
‍
+
∫
a
b
∫
c
d
‍
(
b
-
x
)
α
-
1
(
y
-
c
)
β
-
1
f
(
x
,
y
)
d
y
d
x
‍
+
∫
a
b
∫
c
d
‍
(
x
-
a
)
α
-
1
(
d
-
y
)
β
-
1
f
(
x
,
y
)
d
y
d
x
‍
+
∫
a
b
∫
c
d
(
x
-
a
)
α
-
1
(
y
-
c
)
β
-
1
f
(
x
,
y
)
d
y
d
x
‍
]
,
by which the first inequality is proved. For the proof of the second inequality, we note that f is s-convex on coordinates, then
(19)
f
(
t
1
a
+
(
1
-
t
1
)
b
,
t
2
c
+
(
1
-
t
2
)
d
)
≤
t
1
s
t
2
s
f
(
a
,
c
)
+
t
1
s
(
1
-
t
2
)
s
f
(
a
,
d
)
+
(
1
-
t
1
)
s
t
2
s
f
(
b
,
c
)
+
(
1
-
t
1
)
s
(
1
-
t
2
)
s
f
(
b
,
d
)
,
f
(
t
1
a
+
(
1
-
t
1
)
b
,
(
1
-
t
2
)
c
+
t
2
d
)
≤
t
1
s
(
1
-
t
2
)
s
f
(
a
,
c
)
+
t
1
s
t
2
s
f
(
a
,
d
)
+
(
1
-
t
1
)
s
(
1
-
t
2
)
s
f
(
b
,
c
)
+
(
1
-
t
1
)
s
t
2
s
f
(
b
,
d
)
,
f
(
(
1
-
t
1
)
a
+
t
1
b
,
t
2
c
+
(
1
-
t
2
)
d
)
≤
(
1
-
t
1
)
s
t
2
s
f
(
a
,
c
)
+
(
1
-
t
1
)
s
(
1
-
t
2
)
s
f
(
a
,
d
)
+
t
1
s
t
2
s
f
(
b
,
c
)
+
t
1
s
(
1
-
t
2
)
s
f
(
b
,
d
)
,
f
(
(
1
-
t
1
)
a
+
t
1
b
,
(
1
-
t
2
)
c
+
t
2
d
)
≤
(
1
-
t
1
)
s
(
1
-
t
2
)
s
f
(
a
,
c
)
+
(
1
-
t
1
)
s
t
2
s
f
(
a
,
d
)
+
t
1
s
(
1
-
t
2
)
s
f
(
b
,
c
)
+
t
1
s
t
2
s
f
(
b
,
d
)
.
Multiplying both sides of above inequalities by t
1
α
-
1
t
2
β
-
1, then integrating the resulting inequality with respect to t
1
,
t
2 over [
0,1
]
×
[
0,1
], we obtain
(20)
∬
0
1
t
1
α
-
1
t
2
β
-
1
[
+
t
1
b
,
(
1
-
t
2
)
c
+
t
2
d
(
(
1
-
t
1
)
a
)
]
f
(
t
1
a
+
(
1
-
t
1
)
b
,
t
2
c
+
(
1
-
t
2
)
d
)
‍
+
f
(
t
1
a
+
(
1
-
t
1
)
b
,
(
1
-
t
2
)
c
+
t
2
d
)
+
f
(
(
1
-
t
1
)
a
+
t
1
b
,
t
2
c
+
(
1
-
t
2
)
d
)
+
f
(
(
1
-
t
1
)
a
+
t
1
b
,
(
1
-
t
2
)
c
+
t
2
d
(
(
1
-
t
1
)
a
)
]
d
t
2
d
t
1
≤
[
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
]
×
∬
0
1
t
1
α
-
1
t
2
β
-
1
[
t
1
s
t
2
s
+
t
1
s
(
1
-
t
2
)
s
+
(
1
-
t
1
)
s
t
2
s
‍
‍
+
(
1
-
t
1
)
s
(
1
-
t
2
)
s
]
d
t
2
d
t
1
=
[
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
]
×
(
1
(
α
+
s
)
(
β
+
s
)
+
1
α
+
1
B
(
β
,
s
+
1
)
+
1
β
+
1
B
(
α
,
s
+
1
)
+
B
(
β
,
s
+
1
)
B
(
α
,
s
+
1
)
1
(
α
+
s
)
(
β
+
s
)
)
.
Here, using the change of the variable, we have
(21)
Γ
(
α
+
1
)
Γ
(
β
+
1
)
(
b
-
a
)
α
(
d
-
c
)
β
[
J
a
+
,
c
+
α
,
β
f
(
b
,
d
)
+
J
a
+
,
d
-
α
,
β
f
(
b
,
c
)
+
J
b
-
,
c
+
α
,
β
f
(
a
,
d
)
+
J
b
-
,
d
-
α
,
β
f
(
a
,
c
)
]
≤
[
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
]
×
(
1
(
α
+
s
)
(
β
+
s
)
+
1
α
+
1
B
(
β
,
s
+
1
)
+
1
β
+
1
B
(
α
,
s
+
1
)
+
B
(
β
,
s
+
1
)
B
(
α
,
s
+
1
)
1
(
α
+
s
)
(
β
+
s
)
)
.
The proof is completed.Remark 11.
Applying Theorem10 for s
=
1, we get Theorem 8.
We note that the Beta functions is defined by(22)
B
(
x
,
y
)
=
∫
0
1
(
1
-
t
)
y
-
1
t
x
-
1
d
t
,
x
>
0
,
y
>
0
.


## 3. Inequalities for Differentiable Functions

Theorem 12.
Letf
:
Δ
=
[
a
,
b
]
×
[
c
,
d
]
⊆
[
0
,
∞
)
2
→
[
0
,
∞
) be a partial differentiable mapping with 0
≤
a
<
b, 0
≤
c
<
d. If |
∂
2
f
/
∂
t
1
∂
t
2
| is s-convex on the coordinates in the second sense on Δ for some fixed s
∈
(
0,1
]. Then one has the inequalities:
(23)
|
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
+
Γ
(
α
+
1
)
Γ
(
β
+
1
)
4
(
b
-
a
)
α
(
d
-
c
)
β
×
[
J
a
+
,
c
+
α
,
β
f
(
b
,
d
)
+
J
a
+
,
d
-
α
,
β
f
(
b
,
c
)
+
J
b
-
,
c
+
α
,
β
f
(
a
,
d
)
+
J
b
-
,
d
-
α
,
β
f
(
a
,
c
)
]
-
A
|
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
+
Γ
(
α
+
1
)
Γ
(
β
+
1
)
4
(
b
-
a
)
α
(
d
-
c
)
β
|
=
(
b
-
a
)
(
d
-
c
)
4
[
1
α
+
s
+
1
+
B
(
s
+
1
,
α
+
1
)
]
×
[
1
β
+
s
+
1
+
B
(
s
+
1
,
β
+
1
)
]
×
[
|
∂
2
f
∂
t
1
∂
t
2
(
a
,
c
)
|
+
|
∂
2
f
∂
t
1
∂
t
2
(
b
,
c
)
|
+
|
∂
2
f
∂
t
1
∂
t
2
(
a
,
d
)
|
+
|
∂
2
f
∂
t
1
∂
t
2
(
b
,
d
)
|
]
,
where A is as given in Lemma 9.Proof.
From Lemma9, we obtain
(24)
|
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
+
Γ
(
α
+
1
)
Γ
(
β
+
1
)
4
(
b
-
a
)
α
(
d
-
c
)
β
[
J
a
+
,
c
+
α
,
β
f
(
b
,
d
)
+
J
a
+
,
d
-
α
,
β
f
(
b
,
c
)
+
J
b
-
,
c
+
α
,
β
f
(
a
,
d
)
+
J
b
-
,
d
-
α
,
β
f
(
a
,
c
)
]
-
A
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
|
≤
(
b
-
a
)
(
d
-
c
)
4
×
{
∬
0
1
t
1
α
t
2
β
|
∂
2
f
∂
t
1
∂
t
2
(
t
1
a
+
(
1
-
t
1
)
b
,
t
2
c
‍
‍
+
(
1
-
t
2
)
d
)
∂
2
f
∂
t
1
∂
t
2
|
d
t
2
d
t
1
+
∬
0
1
(
1
-
t
1
)
α
t
2
β
‍
×
|
∂
2
f
∂
t
1
∂
t
2
×
(
t
1
a
+
(
1
-
t
1
)
b
,
t
2
c
+
(
1
-
t
2
)
d
)
∂
2
f
∂
t
1
∂
t
2
|
d
t
2
d
t
1
+
∬
0
1
t
1
α
(
1
-
t
2
)
β
‍
×
|
∂
2
f
∂
t
1
∂
t
2
×
(
t
1
a
+
(
1
-
t
1
)
b
,
t
2
c
+
(
1
-
t
2
)
d
)
∂
2
f
∂
t
1
∂
t
2
|
d
t
2
d
t
1
+
∬
0
1
(
1
-
t
1
)
α
(
1
-
t
2
)
β
‍
×
|
∂
2
f
∂
t
1
∂
t
2
(
t
1
a
+
(
1
-
t
1
)
b
,
t
2
c
‍
+
(
1
-
t
2
)
d
)
∂
2
f
∂
t
1
∂
t
2
|
d
t
2
d
t
1
}
.
Because |
∂
2
f
/
∂
u
∂
v
| is s-convex function on the coordinates on Δ, we obtain
(25)
|
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
+
Γ
(
α
+
1
)
Γ
(
β
+
1
)
4
(
b
-
a
)
α
(
d
-
c
)
β
[
J
a
+
,
c
+
α
,
β
f
(
b
,
d
)
+
J
a
+
,
d
-
α
,
β
f
(
b
,
c
)
+
J
b
-
,
c
+
α
,
β
f
(
a
,
d
)
+
J
b
-
,
d
-
α
,
β
f
(
a
,
c
)
]
-
A
|
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
|
≤
(
b
-
a
)
(
d
-
c
)
4
×
∬
0
1
[
t
1
α
t
2
β
+
(
1
-
t
1
)
α
t
2
β
+
t
1
α
(
1
-
t
2
)
β
+
(
1
-
t
1
)
α
(
1
-
t
2
)
β
[
t
1
α
t
2
β
+
(
1
-
t
1
)
α
t
2
β
+
t
1
α
(
1
-
t
2
)
β
]
×
|
∂
2
f
∂
t
1
∂
t
2
(
t
1
a
+
(
1
-
t
1
)
b
,
t
2
c
+
(
1
-
t
2
)
d
)
|
d
t
2
d
t
1
≤
(
b
-
a
)
(
d
-
c
)
4
×
∬
0
1
[
t
1
α
t
2
β
+
(
1
-
t
1
)
α
t
2
β
+
t
1
α
(
1
-
t
2
)
β
+
(
1
-
t
1
)
α
(
1
-
t
2
)
β
[
t
1
α
t
2
β
+
(
1
-
t
1
)
α
t
2
β
+
t
1
α
(
1
-
t
2
)
β
]
×
{
t
1
s
t
2
s
|
∂
2
f
∂
t
1
∂
t
2
(
a
,
c
)
|
+
(
1
-
t
1
)
s
t
2
s
|
∂
2
f
∂
t
1
∂
t
2
(
b
,
c
)
|
+
t
1
s
(
1
-
t
2
)
s
|
∂
2
f
∂
t
1
∂
t
2
(
a
,
d
)
|
+
(
1
-
t
1
)
s
(
1
-
t
2
)
s
|
∂
2
f
∂
t
1
∂
t
2
(
b
,
d
)
|
}
d
t
2
d
t
1
=
(
b
-
a
)
(
d
-
c
)
4
[
1
α
+
s
+
1
+
B
(
s
+
1
,
α
+
1
)
]
×
[
1
β
+
s
+
1
+
B
(
s
+
1
,
β
+
1
)
]
×
[
|
∂
2
f
∂
t
1
∂
t
2
(
a
,
c
)
|
+
|
∂
2
f
∂
t
1
∂
t
2
(
b
,
c
)
|
+
|
∂
2
f
∂
t
1
∂
t
2
(
a
,
d
)
|
+
|
∂
2
f
∂
t
1
∂
t
2
(
b
,
d
)
|
]
,which completes the proof.Theorem 13.
Letf
:
Δ
=
[
a
,
b
]
×
[
c
,
d
]
⊆
[
0
,
∞
)
2
→
[
0
,
∞
) be a partial differentiable mapping with 0
≤
a
<
b, 0
≤
c
<
d. If |
∂
2
f
/
∂
u
∂
v
|
q is s-convex on the coordinates in the second sense on Δ for some fixed s
∈
(
0,1
] and q
>
1. Then, one has the inequalities:
(26)
|
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
+
Γ
(
α
+
1
)
Γ
(
β
+
1
)
4
(
b
-
a
)
α
(
d
-
c
)
β
×
[
J
a
+
,
c
+
α
,
β
f
(
b
,
d
)
+
J
a
+
,
d
-
α
,
β
f
(
b
,
c
)
+
J
b
-
,
c
+
α
,
β
f
(
a
,
d
)
∑
.
.
∑
.
.
+
J
b
-
,
d
-
α
,
β
f
(
a
,
c
)
]
-
A
Γ
(
α
+
1
)
Γ
(
β
+
1
)
4
(
b
-
a
)
α
(
d
-
c
)
β
|
≤
(
b
-
a
)
(
d
-
c
)
[
(
α
p
+
1
)
(
β
p
+
1
)
]
1
/
p
×
(
|
(
∂
2
f
/
∂
t
1
∂
t
2
)
(
a
,
c
)
|
q
+
|
(
∂
2
f
/
∂
t
1
∂
t
2
)
(
b
,
c
)
|
q
+
|
(
∂
2
f
/
∂
t
1
∂
t
2
)
(
a
,
d
)
|
q
+
|
(
∂
2
f
/
∂
t
1
∂
t
2
)
(
b
,
d
)
|
q
(
s
+
1
)
2
)
1
/
q
,
where (
1
/
p
)
+
(
1
/
q
)
=
1 and A are as given in Lemma 9.Proof.
From Lemma9, we obtain
(27)
|
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
+
Γ
(
α
+
1
)
Γ
(
β
+
1
)
4
(
b
-
a
)
α
(
d
-
c
)
β
×
[
J
a
+
,
c
+
α
,
β
f
(
b
,
d
)
+
J
a
+
,
d
-
α
,
β
f
(
b
,
c
)
+
J
b
-
,
c
+
α
,
β
f
(
a
,
d
)
∑
.
.
∑
.
.
+
J
b
-
,
d
-
α
,
β
f
(
a
,
c
)
]
-
A
|
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
+
Γ
(
α
+
1
)
Γ
(
β
+
1
)
4
(
b
-
a
)
α
(
d
-
c
)
β
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
|
≤
(
b
-
a
)
(
d
-
c
)
4
×
{
∬
0
1
t
1
α
t
2
β
×
|
∂
2
f
∂
t
1
∂
t
2
(
t
1
a
+
(
1
-
t
1
)
b
,
t
2
c

+

(
1
-
t
2
)
d
)
|
d
t
2
d
t
1
+
∬
0
1
(
1
-
t
1
)
α
t
2
β
×
|
∂
2
f
∂
t
1
∂
t
2
c
(
t
1
a
+
(
1
-
t
1
)
b
,
t
2
c

+

(
1
-
t
2
)
d
)
|
d
t
2
d
t
1
+
∬
0
1
t
1
α
(
1
-
t
2
)
β
×
|
∂
2
f
∂
t
1
∂
t
2
(
t
1
a
+
(
1
-
t
1
)
b
,
t
2
c
(
1
-
t
2
)
d
)
|
d
t
2
d
t
1
+
∬
0
1
(
1
-
t
1
)
α
(
1
-
t
2
)
β
×
|
∂
2
f
∂
t
1
∂
t
2
×
(
t
1
a

+

(
1
-
t
1
)
b
,
t
2
c

+

(
1
-
t
2
)
d
)
∂
2
f
∂
t
1
∂
t
2
|
d
t
2
d
t
1
}
.
By using the well-known Hölder's inequality for double integrals, then one has(28)
|
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
+
Γ
(
α
+
1
)
Γ
(
β
+
1
)
4
(
b
-
a
)
α
(
d
-
c
)
β
×
[
J
a
+
,
c
+
α
,
β
f
(
b
,
d
)
+
J
a
+
,
d
-
α
,
β
f
(
b
,
c
)
∑
.
.
∑
.
.
+
J
b
-
,
c
+
α
,
β
f
(
a
,
d
)
+
J
b
-
,
d
-
α
,
β
f
(
a
,
c
)
]
-
A
|
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
|
≤
(
b
-
a
)
(
d
-
c
)
4
{
(
∬
0
1
t
1
p
α
t
2
p
β
d
t
2
d
t
1
)
1
/
p
+
(
∬
0
1
(
1
-
t
1
)
p
α
t
2
p
β
d
t
2
d
t
1
)
1
/
p
+
(
∬
0
1
t
1
p
α
(
1
-
t
2
)
p
β
d
t
2
d
t
1
)
1
/
p
+
(
∬
0
1
(
1
-
t
)
p
α
(
1
-
s
)
p
β
d
t
2
d
t
1
)
1
/
p
}
×
(
∬
0
1
|
∂
2
f
∂
t
1
∂
t
2
(
t
1
a
+
(
1
-
t
1
)
b
,
t
2
c
+
(
1
-
t
2
)
d
)
|
∂
2
f
∂
t
1
∂
t
2
|
q
d
t
2
d
t
1
∬
0
1
|
∂
2
f
∂
t
1
∂
t
2
|
q
)
1
/
q
.
Because |
∂
2
f
/
∂
t
1
∂
t
2
| is s-convex function on the coordinates on Δ, by (8), we have(29)
∬
0
1
|
∂
2
f
∂
t
1
∂
t
2
(
t
1
a
+
(
1
-
t
1
)
b
,
t
2
c
+
(
1
-
t
2
)
d
)
|
q
d
t
2
d
t
1
≤
|
(
∂
2
f
/
∂
t
1
∂
t
2
)
(
a
,
c
)
|
q
+
|
(
∂
2
f
/
∂
t
1
∂
t
2
)
(
b
,
c
)
|
q
+
|
(
∂
2
f
/
∂
t
1
∂
t
2
)
(
a
,
d
)
|
q
+
|
(
∂
2
f
/
∂
t
1
∂
t
2
)
(
b
,
d
)
|
q
(
s
+
1
)
2
.
So
(30)
|
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
+
Γ
(
α
+
1
)
Γ
(
β
+
1
)
4
(
b
-
a
)
α
(
d
-
c
)
β
×
[
J
a
+
,
c
+
α
,
β
f
(
b
,
d
)
+
J
a
+
,
d
-
α
,
β
f
(
b
,
c
)
+
J
b
-
,
c
+
α
,
β
f
(
a
,
d
)
+
J
b
-
,
d
-
α
,
β
f
(
a
,
c
)
]
-
A
|
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
+
Γ
(
α
+
1
)
Γ
(
β
+
1
)
4
(
b
-
a
)
α
(
d
-
c
)
β
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
|
≤
(
b
-
a
)
(
d
-
c
)
[
(
α
p
+
1
)
(
β
p
+
1
)
]
1
/
p
×
(
|
(
∂
2
f
/
∂
t
1
∂
t
2
)
(
a
,
c
)
|
q
+
|
(
∂
2
f
/
∂
t
1
∂
t
2
)
(
b
,
c
)
|
q
+
|
(
∂
2
f
/
∂
t
1
∂
t
2
)
(
a
,
d
)
|
q
+
|
(
∂
2
f
/
∂
t
1
∂
t
2
)
(
b
,
d
)
|
q
(
s
+
1
)
2
)
1
/
q
, which completes the proof.Theorem 14.
Letf
:
Δ
=
[
a
,
b
]
×
[
c
,
d
]
⊆
[
0
,
∞
)
2
→
[
0
,
∞
) be a partial differentiable mapping with 0
≤
a
<
b, 0
≤
c
<
d. If |
∂
2
f
/
∂
u
∂
v
|
q is s-concave on the coordinates in the second sense on Δ for some fixed s
∈
(
0,1
] and q
>
1, then one has the inequalities:
(31)
|
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
+
Γ
(
α
+
1
)
Γ
(
β
+
1
)
4
(
b
-
a
)
α
(
d
-
c
)
β
×
[
J
a
+
,
c
+
α
,
β
f
(
b
,
d
)
+
J
a
+
,
d
-
α
,
β
f
(
b
,
c
)
+
J
b
-
,
c
+
α
,
β
f
(
a
,
d
)
+
J
b
-
,
d
-
α
,
β
f
(
a
,
c
)
]
-
A
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
|
≤
(
b
-
a
)
(
d
-
c
)
[
(
α
p
+
1
)
(
β
p
+
1
)
]
1
/
p
×
4
(
s
-
1
)
/
q
|
∂
2
f
∂
t
1
∂
t
2
(
a
+
b
2
,
c
+
d
2
)
|
,
where (
1
/
p
)
+
(
1
/
q
)
=
1 and A is as given in Lemma 9.Proof.
Similarly as in Theorem13, we obtain
(32)
|
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
+
Γ
(
α
+
1
)
Γ
(
β
+
1
)
4
(
b
-
a
)
α
(
d
-
c
)
β
×
[
J
a
+
,
c
+
α
,
β
f
(
b
,
d
)
+
J
a
+
,
d
-
α
,
β
f
(
b
,
c
)
+
J
b
-
,
c
+
α
,
β
f
(
a
,
d
)
+
J
b
-
,
d
-
α
,
β
f
(
a
,
c
)
]
-
A
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
|
≤
(
b
-
a
)
(
d
-
c
)
4
×
{
(
∬
0
1
t
1
p
α
t
2
p
β
d
t
2
d
t
1
)
1
/
p
+
(
∬
0
1
(
1
-
t
1
)
p
α
t
2
p
β
d
t
2
d
t
1
)
1
/
p
+
(
∬
0
1
t
1
p
α
(
1
-
t
2
)
p
β
d
t
2
d
t
1
)
1
/
p
+
(
∬
0
1
(
1
-
t
)
p
α
(
1
-
s
)
p
β
d
t
2
d
t
1
)
1
/
p
}
×
(
∬
0
1
|
∂
2
f
∂
t
1
∂
t
2
(
t
1
a
+
(
1
-
t
1
)
b
,
t
2
c
+
(
1
-
t
2
)
d
)
∂
2
f
∂
t
1
∂
t
2
|
q
d
t
2
d
t
1
∬
0
1
|
∂
2
f
∂
t
1
∂
t
2
(
t
1
a
+
(
1
-
t
1
)
b
,
t
2
c
)
1
/
q
.
Because |
∂
2
f
/
∂
t
1
∂
t
2
| is s-concave function on the coordinates on Δ, by the reversed direction of (8), then one has
(33)
∬
0
1
|
∂
2
f
∂
t
1
∂
t
2
(
t
1
a
+
(
1
-
t
1
)
b
,
t
2
c
+
(
1
-
t
2
)
d
)
|
q
d
t
2
d
t
1
≤
4
s
-
1
|
∂
2
f
∂
t
1
∂
t
2
(
a
+
b
2
,
c
+
d
2
)
|
q
.
Then
(34)
|
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4

+

Γ
(
α
+
1
)
Γ
(
β
+
1
)
4
(
b
-
a
)
α
(
d
-
c
)
β
×
[
J
a
+
,
c
+
α
,
β
f
(
b
,
d
)
+
J
a
+
,
d
-
α
,
β
f
(
b
,
c
)
+
J
b
-
,
c
+
α
,
β
f
(
a
,
d
)
+
J
b
-
,
d
-
α
,
β
f
(
a
,
c
)
]
-
A
|
f
(
a
,
c
)
+
f
(
a
,
d
)
+
f
(
b
,
c
)
+
f
(
b
,
d
)
4
+
Γ
(
α
+
1
)
Γ
(
β
+
1
)
4
(
b
-
a
)
α
(
d
-
c
)
β
|
≤
(
b
-
a
)
(
d
-
c
)
[
(
α
p
+
1
)
(
β
p
+
1
)
]
1
/
p
×
4
(
s
-
1
)
/
q
|
∂
2
f
∂
t
1
∂
t
2
(
a
+
b
2
,
c
+
d
2
)
|
.
We get the desired results.


## 4. Conclusion

In this paper, we obtain some Hermite-Hadamard type inequalities for coordinateds-convex functions via Riemann-Liouville integrals. An interesting topic is whether we can use the methods in this paper to establish the Hermite-Hadamard inequalities for other kinds of convex functions on the coordinates via Riemann-Liouville integrals.



---

*Source: 248710-2014-05-14.xml*