International Journal of Analysis and Applications
ISSN 2291-8639
Volume 5, Number 1 (2014), 45-55
http://www.etamaths.com

INEQUALITIES FOR CO-ORDINATED m−CONVEX FUNCTIONS
VIA RIEMANN-LIOUVILLE FRACTIONAL INTEGRALS

ÇETİN YILDIZ1,∗, MEVLÜT TUNÇ2, AND HAVVA KAVURMACI1

Abstract. In this paper, we prove some new inequalities of Hadamard-type

for m−convex functions on the co-ordinates via Riemann-Liouville fractional
integrals.

1. INTRODUCTION

Let f : I ⊆ R → R be a convex function defined on the interval I of real numbers
and a < b. The following double inequality;

f

(
a + b

2

)
≤

1

b−a

b∫
a

f(x)dx ≤
f(a) + f(b)

2

is well known in the literature as Hadamard’s inequality. Both inequalities hold
in the reversed direction if f is concave.

In [7], Dragomir defined convex functions on the co-ordinates as following:

Definition 1. Let us consider the bidimensional interval ∆ = [a,b] × [c,d] in
R2 with a < b, c < d. A function f : ∆ → R will be called convex on the co-
ordinates if the partial mappings fy : [a,b] → R, fy(u) = f(u,y) and fx : [c,d] → R,
fx(v) = f(x,v) are convex where defined for all y ∈ [c,d] and x ∈ [a,b]. Recall that
the mapping f : ∆ → R is convex on ∆ if the following inequality holds,

f(λx + (1 −λ)z,λy + (1 −λ)w) ≤ λf(x,y) + (1 −λ)f(z,w)

for all (x,y), (z,w) ∈ ∆ and λ ∈ [0, 1].

In [7], Dragomir established the following inequalities of Hadamard’s type for
co-ordinated convex functions on a rectangle from the plane R2.

2000 Mathematics Subject Classification. 26A15, 26A51, 26D10.
Key words and phrases. Riemann-Liouville fractional integrals, co-ordinates, m−convex

functions.

c©2014 Authors retain the copyrights of their papers, and all
open access articles are distributed under the terms of the Creative Commons Attribution License.

45



46 ÇETİN YILDIZ1,∗, MEVLÜT TUNÇ2, AND HAVVA KAVURMACI1

Theorem 1. Suppose that f : ∆ = [a,b] × [c,d] → R is convex on the co-ordinates
on ∆. Then one has the inequalities;

f

(
a + b

2
,
c + d

2

)
(1.1)

≤
1

2

[
1

b−a

∫ b
a

f

(
x,
c + d

2

)
dx +

1

d− c

∫ d
c

f

(
a + b

2
,y

)
dy

]

≤
1

(b−a)(d− c)

∫ b
a

∫ d
c

f(x,y)dxdy

≤
1

4

[
1

(b−a)

∫ b
a

f(x,c)dx +
1

(b−a)

∫ b
a

f(x,d)dx

+
1

(d− c)

∫ d
c

f(a,y)dy +
1

(d− c)

∫ d
c

f(b,y)dy

]

≤
f(a,c) + f(a,d) + f(b,c) + f(b,d)

4
.

The above inequalities are sharp.

Similar results can be found in [7]-[12].
In [17], Toader defined m−convex functions as following:

Definition 2. The function f : [0,b] → R, b > 0 is said to be m−convex, where
m ∈ [0, 1], if we have

f(tx + m(1 − t)y) ≤ tf(x) + m(1 − t)f(y)
for all x,y ∈ [0,b] and t ∈ [0, 1].

Denote by Km(b) the class of all m−convex functions on [0,b] for which f(0) ≤ 0.
Obviously, if we choose m = 1, we have ordinary convex functions on [0,b].

In [10], Özdemir et al. defined co-ordinated m−convex functions as following:

Definition 3. Let us consider the bidimensional interval ∆ = [0,b] × [0,d] in
[0,∞)2. The mapping f : ∆ → R is m−convex on ∆ if

f(tx + (1 − t)z,ty + m(1 − t)w) ≤ tf(x,y) + m(1 − t)f(z,w)
holds for all (x,y), (z,w) ∈ ∆ and t ∈ [0, 1], b,d > 0 and for some fixed m ∈ [0, 1].

In [16], Sarıkaya et al. proved some Hadamard’s type inequalities for co-ordinated
convex functions as followings:

Theorem 2. Let f : ∆ ⊂ R2 → R be a partial differentiable mapping on ∆ :=
[a,b] × [c,d] in R2 with a < b and c < d. If

∣∣∣ ∂2f∂t∂s∣∣∣ is a convex function on the
co-ordinates on ∆, then one has the inequalities:

(1.2) ∣∣∣∣∣f(a,c) + f(a,d) + f(b,c) + f(b,d)4 + 1(b−a)(d− c)
∫ b
a

∫ d
c

f(x,y)dxdy −A

∣∣∣∣∣
≤

(b−a)(d− c)
16



∣∣∣ ∂2f∂t∂s∣∣∣ (a,c) + ∣∣∣ ∂2f∂t∂s∣∣∣ (a,d) + ∣∣∣ ∂2f∂t∂s∣∣∣ (b,c) + ∣∣∣ ∂2f∂t∂s∣∣∣ (b,d)

4






SOME FRACTIONAL INTEGRAL INEQUALITIES 47

where

A =
1

2

[
1

(b−a)

∫ b
a

[f(x,c) + f(x,d)] dx +
1

(d− c)

∫ d
c

[f(a,y)dy + f(b,y)] dy

]
.

Theorem 3. Let f : ∆ ⊂ R2 → R be a partial differentiable mapping on ∆ :=
[a,b] × [c,d] in R2 with a < b and c < d. If

∣∣∣ ∂2f∂t∂s∣∣∣q , q > 1, is a convex function on
the co-ordinates on ∆, then one has the inequalities:

(1.3)∣∣∣∣∣f(a,c) + f(a,d) + f(b,c) + f(b,d)4 + 1(b−a)(d− c)
∫ b
a

∫ d
c

f(x,y)dxdy −A

∣∣∣∣∣
≤

(b−a)(d− c)
4 (p + 1)

2
p



∣∣∣ ∂2f∂t∂s∣∣∣q (a,c) + ∣∣∣ ∂2f∂t∂s∣∣∣q (a,d) + ∣∣∣ ∂2f∂t∂s∣∣∣q (b,c) + ∣∣∣ ∂2f∂t∂s∣∣∣q (b,d)

4




1
q

where

A =
1

2

[
1

(b−a)

∫ b
a

[f(x,c) + f(x,d)] dx +
1

(d− c)

∫ d
c

[f(a,y)dy + f(b,y)] dy

]
and 1

p
+ 1

q
= 1.

Theorem 4. Let f : ∆ ⊂ R2 → R be a partial differentiable mapping on ∆ :=
[a,b] × [c,d] in R2 with a < b and c < d. If

∣∣∣ ∂2f∂t∂s∣∣∣q , q ≥ 1, is a convex function on
the co-ordinates on ∆, then one has the inequalities:

(1.4)∣∣∣∣∣f(a,c) + f(a,d) + f(b,c) + f(b,d)4 + 1(b−a)(d− c)
∫ b
a

∫ d
c

f(x,y)dxdy −A

∣∣∣∣∣
≤

(b−a)(d− c)
16



∣∣∣ ∂2f∂t∂s∣∣∣q (a,c) + ∣∣∣ ∂2f∂t∂s∣∣∣q (a,d) + ∣∣∣ ∂2f∂t∂s∣∣∣q (b,c) + ∣∣∣ ∂2f∂t∂s∣∣∣q (b,d)

4




1
q

where

A =
1

2

[
1

(b−a)

∫ b
a

[f(x,c) + f(x,d)] dx +
1

(d− c)

∫ d
c

[f(a,y)dy + f(b,y)] dy

]
.

We give some necessary definitions and mathematical preliminaries of fractional
calculus theory which are used throughout this paper.

Definition 4. Let f ∈ L1[a,b]. The Riemann-Liouville integrals Jαa+f and J
α
b−
f of

order α > 0 with a ≥ 0 are defined by

Jαa+f(x) =
1

Γ(α)

∫ x
a

(x− t)α−1f(t)dt, x > a

and

Jαb−f(x) =
1

Γ(α)

∫ b
x

(t−x)α−1f(t)dt, x < b

where Γ(α) =
∫∞

0
e−tuα−1du, here is J0

a+
f(x) = J0

b−
f(x) = f(x).



48 ÇETİN YILDIZ1,∗, MEVLÜT TUNÇ2, AND HAVVA KAVURMACI1

In the case of α = 1, the fractional integral reduces to the classical integral.
Properties of this operator can be found in the references [3]-[?].

Throughout of this paper, we will use the following notation:

B =
Γ (α + 1) Γ (β + 1)

4 (b−a)α (d− c)β
[
J
α,β
b−,d−

f (a,c) + J
α,β
a+,d−

f (b,c) + J
α,β
b−,c+

f (a,d) + J
α,β
a+,c+

f (b,d)
]

−
Γ (β + 1)

4 (d− c)β
[
J
β
d−
f (a,c) + J

β
d−
f (b,c) + J

β
c+
f (b,d) + J

β
c+
f (a,d)

]
−

Γ (α + 1)

4 (b−a)α
[Jαb−f (a,d) + J

α
b−f (a,c) + J

α
a+f (b,d) + J

α
a+f (b,c)]

where

J
α,β
b−,d−

f (a,c) =
1

Γ (α) Γ (β)

b∫
a

d∫
c

(x−a)α−1 (y − c)β−1 f (x,y) dydx

J
α,β
a+,d−

f (b,c) =
1

Γ (α) Γ (β)

b∫
a

d∫
c

(x−a)α−1 (d−y)β−1 f (x,y) dydx

J
α,β
b−,c+

f (a,d) =
1

Γ (α) Γ (β)

b∫
a

d∫
c

(b−x)α−1 (y − c)β−1 f (x,y) dydx

J
α,β
a+,c+

f (b,d) =
1

Γ (α) Γ (β)

b∫
a

d∫
c

(b−x)α−1 (d−y)β−1 f (x,y) dydx.

The main purpose of this paper is to establish inequalities of Hadamard-type
inequalities for m−convex functions on the co-ordinates via Riemann-Liouville frac-
tional integrals by using a new Lemma and fairly elemantery analysis.

2. MAIN RESULTS

To prove our main result, we need the following Lemma:

Lemma 1. Let f : ∆ = [a,b] × [c,d] → R be a twice partial differentiable mapping
on ∆ = [a,b] × [c,d] . If ∂

2f
∂t∂s

∈ L (∆) and α,β > 0, a,c ≥ 0, then the following
equality holds:

(2.1)

f(a,c) + f(a,d) + f(b,c) + f(b,d)

4
+ B

=
(b−a) (d− c)

4

1∫
0

1∫
0

[(1 − t)α − tα]
[
(1 −s)β −sβ

] ∂2f
∂t∂s

(ta + (1 − t) b,sc + (1 −s) d) dsdt.



SOME FRACTIONAL INTEGRAL INEQUALITIES 49

Proof. Integration by parts, we can write

K =

1∫
0

1∫
0

[(1 − t)α − tα]
[
(1 −s)β −sβ

] ∂2f
∂t∂s

(ta + (1 − t) b,sc + (1 −s) d) dsdt

=

1∫
0

[
(1 −s)β −sβ

] 1∫
0

(1 − t)α
∂2f

∂t∂s
(ta + (1 − t) b,sc + (1 −s) d) dt

−
1∫

0

tα
∂2f

∂t∂s
(ta + (1 − t) b,sc + (1 −s) d) dt


ds

=
1

b−a




1∫
0

[
(1 −s)β −sβ

][∂f
∂s

(b,sc + (1 −s) d) +
∂f

∂s
(a,sc + (1 −s) d)

−α
1∫

0

(1 − t)α−1
∂f

∂s
(ta + (1 − t) b,sc + (1 −s) d) dt

−α
1∫

0

tα−1
∂f

∂s
(ta + (1 − t) b,sc + (1 −s) d) dt


ds


 .

By integrating again, we get

K =
1

(b−a)(d− c)
{f(a,c) + f(a,d) + f(b,c) + f(b,d)

−β
1∫

0

(1 −s)β−1 f (b,sc + (1 −s) d) ds−β
1∫

0

sβ−1f (a,sc + (1 −s) d) ds

−β
1∫

0

(1 −s)β−1 f (a,sc + (1 −s) d) ds−β
1∫

0

sβ−1f (b,sc + (1 −s) d) ds

−α
1∫

0

(1 − t)α−1 f (ta + (1 − t) b,d) dt−α
1∫

0

tα−1f (ta + (1 − t) b,d) dt

−α
1∫

0

(1 − t)α−1 f (ta + (1 − t) b,c) dt−α
1∫

0

tα−1f (ta + (1 − t) b,c) dt

+αβ

1∫
0

1∫
0

(1 − t)α−1 (1 −s)β−1 f (ta + (1 − t) b,sc + (1 −s) d) dsdt

+αβ

1∫
0

1∫
0

tα−1 (1 −s)β−1 f (ta + (1 − t) b,sc + (1 −s) d) dsdt



50 ÇETİN YILDIZ1,∗, MEVLÜT TUNÇ2, AND HAVVA KAVURMACI1

+αβ

1∫
0

1∫
0

(1 − t)α−1 sβ−1f (ta + (1 − t) b,sc + (1 −s) d) dsdt

+αβ

1∫
0

1∫
0

tα−1sβ−1f (ta + (1 − t) b,sc + (1 −s) d) dsdt


 .

By using the change of the variables, we can get

x = ta + (1 − t) b and y = sc + (1 −s) d,

that is

t =
x− b
a− b

and s =
y −d
c−d

.

Taking into account these equalities, we obtain

(2.2)

K =
1

(b−a)(d− c)
{f(a,c) + f(a,d) + f(b,c) + f(b,d)

−
β

(d− c)β−1


 d∫
c

(y − c)β−1 f (a,y) dy +
d∫
c

(d−y)β−1 f (a,y) dy

+

d∫
c

(y − c)β−1 f (b,y) dy +
d∫
c

(d−y)β−1 f (b,y) dy




−
α

(b−a)α−1


 b∫
a

(x−a)α−1 f (x,d) dx +
b∫
a

(x−a)α−1 f (x,c) dx

+

b∫
a

(b−x)α−1 f (x,d) dx +
b∫
a

(b−x)α−1 f (x,c) dx


 + αβ

(b−a)α−1 (d− c)β−1

×


 b∫
a

d∫
c

(x−a)α−1 (y − c)β−1 f (x,y) dydx +
b∫
a

d∫
c

(x−a)α−1 (d−y)β−1 f (x,y) dydx

+

b∫
a

d∫
c

(b−x)α−1 (y − c)β−1 f (x,y) dydx +
b∫
a

d∫
c

(b−x)α−1 (d−y)β−1 f (x,y) dydx




 .

Multiplying both sides of (2.2) by
(b−a)(d−c)

4
and using the Riemann-Liouville

integrals, we obtain equality (2.1). This completes the proof.

Theorem 5. Let f : ∆ = [0,b] × [0,d] → R be a partial differentiable mapping on
∆ and ∂

2f
∂t∂s

∈ L (∆), α,β > 0. If
∣∣∣ ∂2f∂t∂s∣∣∣ is m−convex function on the co-ordinates



SOME FRACTIONAL INTEGRAL INEQUALITIES 51

on ∆ where 0 ≤ a < b < ∞ and 0 ≤ c < d < ∞, then the following inequality holds;∣∣∣∣f(a,c) + f(a,d) + f(b,c) + f(b,d)4 + B
∣∣∣∣

≤
(b−a) (d− c)

4
MαMβ

×
(∣∣∣∣ ∂2f∂t∂s (a,c)

∣∣∣∣ +
∣∣∣∣ ∂2f∂t∂s (b,c)

∣∣∣∣ + m
∣∣∣∣ ∂2f∂t∂s

(
a,
d

m

)∣∣∣∣ + m
∣∣∣∣ ∂2f∂t∂s

(
b,
d

m

)∣∣∣∣
)

where

Mα =

[
1

α + 1
−
(

1
2

)α
α + 1

]

Mβ =

[
1

β + 1
−
(

1
2

)β
β + 1

]
.

Proof. From Lemma 1 and using the property of modulus, we have∣∣∣∣f(a,c) + f(a,d) + f(b,c) + f(b,d)4 + B
∣∣∣∣

≤
(b−a) (d− c)

4

1∫
0

1∫
0

|(1 − t)α − tα|
∣∣∣(1 −s)β −sβ∣∣∣ ∣∣∣∣ ∂2f∂t∂s (ta + (1 − t) b,sc + (1 −s) d)

∣∣∣∣dsdt.
Since

∣∣∣ ∂2f∂t∂s∣∣∣ is co-ordinated m−convex, we can write∣∣∣∣f(a,c) + f(a,d) + f(b,c) + f(b,d)4 + B
∣∣∣∣

≤
(b−a) (d− c)

4

1∫
0

1∫
0

∣∣∣(1 −s)β −sβ∣∣∣ |(1 − t)α − tα|{ts∣∣∣∣ ∂2f∂t∂s (a,c)
∣∣∣∣ + mt(1 −s)

∣∣∣∣ ∂2f∂t∂s
(
a,
d

m

)∣∣∣∣
+(1 − t)s

∣∣∣∣ ∂2f∂t∂s (b,c)
∣∣∣∣ + m(1 − t)(1 −s)

∣∣∣∣ ∂2f∂t∂s
(
b,
d

m

)∣∣∣∣
}
dtds

By computing these integrals, we obtain∣∣∣∣f(a,c) + f(a,d) + f(b,c) + f(b,d)4 + B
∣∣∣∣

≤
(b−a) (d− c)

4

[
1

α + 1
−
(

1
2

)α
α + 1

]

×
1∫

0

∣∣∣(1 −s)β −sβ∣∣∣(s∣∣∣∣ ∂2f∂t∂s (a,c)
∣∣∣∣ + m(1 −s)

∣∣∣∣ ∂2f∂t∂s
(
a,
d

m

)∣∣∣∣
+s

∣∣∣∣ ∂2f∂t∂s (b,c)
∣∣∣∣ + m(1 −s)

∣∣∣∣ ∂2f∂t∂s
(
b,
d

m

)∣∣∣∣
)
ds.



52 ÇETİN YILDIZ1,∗, MEVLÜT TUNÇ2, AND HAVVA KAVURMACI1

Using co-ordinated m−convexity of
∣∣∣ ∂2f∂t∂s∣∣∣ again, we get∣∣∣∣f(a,c) + f(a,d) + f(b,c) + f(b,d)4 + B

∣∣∣∣
≤

(b−a) (d− c)
4

[
1

α + 1
−
(

1
2

)α
α + 1

][
1

β + 1
−
(

1
2

)β
β + 1

]

×
(∣∣∣∣ ∂2f∂t∂s (a,c)

∣∣∣∣ +
∣∣∣∣ ∂2f∂t∂s (b,c)

∣∣∣∣ + m
∣∣∣∣ ∂2f∂t∂s

(
a,
d

m

)∣∣∣∣ + m
∣∣∣∣ ∂2f∂t∂s

(
b,
d

m

)∣∣∣∣
)

Thus, the proof is completed.

Remark 1. Suppose that all the assumptions of Theorem 5 are satisfied. If we
choose α = β = m = 1, we obtain the inequality (1.2) .

Theorem 6. Let f : ∆ → R be a partial differentiable mapping on ∆ and ∂
2f

∂t∂s
∈

L (∆), α,β ∈ (0, 1]. If
∣∣∣ ∂2f∂t∂s∣∣∣q , q > 1, is m−convex function on the co-ordinates on

∆ where 0 ≤ a < b < ∞ and 0 ≤ c < d < ∞, then the following inequality holds;∣∣∣∣f(a,c) + f(a,d) + f(b,c) + f(b,d)4 + B
∣∣∣∣

≤
(b−a) (d− c)

4 (αp + 1)
1
p (βp + 1)

1
p

×



∣∣∣ ∂2f∂t∂s (a,c)∣∣∣q + m ∣∣∣ ∂2f∂t∂s (a, dm)∣∣∣q + ∣∣∣ ∂2f∂t∂s (b,c)∣∣∣q + m ∣∣∣ ∂2f∂t∂s (b, dm)∣∣∣q

4




1
q

.

where p−1 + q−1 = 1.

Proof. From Lemma 1 and by applying the well-known Hölder inequality for double
integrals, then one has∣∣∣∣f(a,c) + f(a,d) + f(b,c) + f(b,d)4 + B

∣∣∣∣
≤

(b−a) (d− c)
4


 1∫

0

1∫
0

[
|(1 − t)α − tα|

∣∣∣(1 −s)β −sβ∣∣∣]p dsdt



1
p

×


 1∫

0

1∫
0

∣∣∣∣ ∂2f∂t∂s (ta + (1 − t) b,sc + (1 −s) d)
∣∣∣∣q dsdt




1
q

.

By using the fact that
|tα1 − t

α
2 | ≤ |t1 − t2|

α

for α ∈ (0, 1] and t1, t2 ∈ [0, 1] , we get
1∫

0

|(1 − t)α − tα|p dt ≤
1∫

0

|1 − 2t|αp dt

=
1

αp + 1



SOME FRACTIONAL INTEGRAL INEQUALITIES 53

and

1∫
0

∣∣∣∣∣∣(1 −s)β −sβ∣∣∣∣∣∣p dt ≤ 1∫
0

|1 − 2s|βp dt

=
1

βp + 1
.

Since
∣∣∣ ∂2f∂t∂s∣∣∣q is co-ordinated m−convex, we can write∣∣∣∣f(a,c) + f(a,d) + f(b,c) + f(b,d)4 + B

∣∣∣∣
≤

(b−a) (d− c)
4 (αp + 1)

1
p (βp + 1)

1
p

×


 1∫

0

1∫
0

[
ts

∣∣∣∣ ∂2f∂t∂s (a,c)
∣∣∣∣q + mt (1 −s)

∣∣∣∣ ∂2f∂t∂s
(
a,
d

m

)∣∣∣∣q
]

+ (1 − t) s
∣∣∣∣ ∂2f∂t∂s (b,c)

∣∣∣∣q + m (1 − t) (1 −s)
∣∣∣∣ ∂2f∂t∂s

(
b,
d

m

)∣∣∣∣q dsdt
)1

q

.

By computing these integrals, we obtain∣∣∣∣f(a,c) + f(a,d) + f(b,c) + f(b,d)4 + B
∣∣∣∣

≤
(b−a) (d− c)

4 (αp + 1)
1
p (βp + 1)

1
p

×



∣∣∣ ∂2f∂t∂s (a,c)∣∣∣q + m ∣∣∣ ∂2f∂t∂s (a, dm)∣∣∣q + ∣∣∣ ∂2f∂t∂s (b,c)∣∣∣q + m ∣∣∣ ∂2f∂t∂s (b, dm)∣∣∣q

4




1
q

.

Which completes the proof.

Remark 2. Suppose that all the assumptions of Theorem 6 are satisfied. If we
choose α = β = m = 1, we obtain the inequality (1.3) .

Theorem 7. Let f : ∆ → R be a partial differentiable mapping on ∆ and ∂
2f

∂t∂s
∈

L (∆), α,β ∈ (0, 1]. If
∣∣∣ ∂2f∂t∂s∣∣∣q , q ≥ 1, is m−convex function on the co-ordinates on

∆ where 0 ≤ a < b < ∞ and 0 ≤ c < d < ∞, then the following inequality holds;∣∣∣∣f(a,c) + f(a,d) + f(b,c) + f(b,d)4 + B
∣∣∣∣

≤
(b−a) (d− c)

4

([
1 −

(
1
2

)α
α + 1

][
1 −

(
1
2

)β
β + 1

])1−1
q

M
1
q
α M

1
q

β

×
(∣∣∣∣ ∂2f∂t∂s (a,c)

∣∣∣∣q + m
∣∣∣∣ ∂2f∂t∂s

(
a,
d

m

)∣∣∣∣q +
∣∣∣∣ ∂2f∂t∂s (b,c)

∣∣∣∣q + m
∣∣∣∣ ∂2f∂t∂s

(
b,
d

m

)∣∣∣∣q
)1

q

where Mα,Mβ are defined as in Theorem 5.



54 ÇETİN YILDIZ1,∗, MEVLÜT TUNÇ2, AND HAVVA KAVURMACI1

Proof. From Lemma 1 and by applying the well-known Power-mean inequality for
double integrals, then one has∣∣∣∣f(a,c) + f(a,d) + f(b,c) + f(b,d)4 + B

∣∣∣∣
≤

(b−a) (d− c)
4


 1∫

0

1∫
0

|(1 − t)α − tα|
∣∣∣(1 −s)β −sβ∣∣∣dsdt


1−

1
q

×


 1∫

0

1∫
0

|(1 − t)α − tα|
∣∣∣(1 −s)β −sβ∣∣∣ ∣∣∣∣ ∂2f∂t∂s (ta + (1 − t) b,sc + (1 −s) d)

∣∣∣∣q dsdt



1
q

.

Since
∣∣∣ ∂2f∂t∂s∣∣∣q is co-ordinated m−convex, we can write∣∣∣∣f(a,c) + f(a,d) + f(b,c) + f(b,d)4 + B

∣∣∣∣
≤

(b−a) (d− c)
4


 1∫

0

1∫
0

|(1 − t)α − tα|
∣∣∣(1 −s)β −sβ∣∣∣dsdt


1−

1
q

×


 1∫

0

1∫
0

|(1 − t)α − tα|
∣∣∣(1 −s)β −sβ∣∣∣[ts∣∣∣∣ ∂2f∂t∂s (a,c)

∣∣∣∣q + mt (1 −s)
∣∣∣∣ ∂2f∂t∂s

(
a,
d

m

)∣∣∣∣q
]

+ (1 − t) s
∣∣∣∣ ∂2f∂t∂s (b,c)

∣∣∣∣q + m (1 − t) (1 −s)
∣∣∣∣ ∂2f∂t∂s

(
b,
d

m

)∣∣∣∣q dsdt
)1

q

.

By computing these integrals, we obtain∣∣∣∣f(a,c) + f(a,d) + f(b,c) + f(b,d)4 + B
∣∣∣∣

≤
(b−a) (d− c)

4

([
1 −

(
1
2

)α
α + 1

][
1 −

(
1
2

)β
β + 1

])1−1
q

M
1
q
α M

1
q

β

×
(∣∣∣∣ ∂2f∂t∂s (a,c)

∣∣∣∣q +
∣∣∣∣m ∂2f∂t∂s

(
a,
d

m

)∣∣∣∣q +
∣∣∣∣ ∂2f∂t∂s (b,c)

∣∣∣∣q + m
∣∣∣∣ ∂2f∂t∂s

(
b,
d

m

)∣∣∣∣q
)1

q

which completes the proof.

Remark 3. Suppose that all the assumptions of Theorem 7 are satisfied. If we
choose α = β = m = 1, we obtain the inequality (1.4) .

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1ATATÜRK UNIVERSITY, K.K. EDUCATION FACULTY, DEPARTMENT OF MATH-

EMATICS, 25240, KAMPUS, ERZURUM, TURKEY

2DEPARTMENT OF MATHEMATICS, FACULTY OF ART AND SCIENCES,KİLİS 7

ARALIK UNIVERSITY, KİLİS, 79000, TURKEY

∗Corresponding author