International Journal of Analysis and Applications

Volume 17, Number 5 (2019), 892-903

URL: https://doi.org/10.28924/2291-8639

DOI: 10.28924/2291-8639-17-2019-892

HADAMARD AND FEJÉR-HADAMARD TYPE INEQUALITIES FOR CONVEX AND

RELATIVE CONVEX FUNCTIONS VIA AN EXTENDED GENERALIZED

MITTAG-LEFFLER FUNCTION

GHULAM FARID1, VISHNU NARAYAN MISHRA2,3,∗, SAJID MEHMOOD4

1Department of Mathematics, COMSATS University Islamabad, Attock Campus, Pakistan

2Department of Mathematics, Indira Gandhi National Tribal University, Lalpur, Amarkantak-484 887,

Madhya Pradesh, India

3L. 1627 Awadh Puri Colony, Beniganj, Phase-III, Opposite Industrial Training Institute (I.T.I.),

Ayodhya-224 001, Uttar Pradesh, India

4Govt Boys Primary school Sherani, Hazro, Attock, Pakistan

∗Corresponding author: vishnunarayanmishra@gmail.com

Abstract. In this paper, we will prove the Hadamard and the Fejér-Hadamard type integral inequalities for

convex and relative convex functions due to an extended generalized Mittag-Leffler function. These results

contain several fractional integral inequalities for the well known fractional integral operators.

1. Introduction

Convex functions are very useful in the field of mathematical inequalities.

Definition 1.1. Let I be an interval of real numbers. Then a function f : I → R is said to be convex

function, if for all x,y ∈ I and 0 ≤ λ ≤ 1, the following inequality holds:

f(xλ + (1 −λ)y) ≤ λf(x) + (1 −λ)f(y).

Received 2019-01-22; accepted 2019-04-09; published 2019-09-02.

2010 Mathematics Subject Classification. 26B25, 26A51, 26A33, 33E12.

Key words and phrases. Convex functions; Hadamad inequality; Fejér-Hadamard inequality; Mittag-Leffler function; Frac-

tional integral operators.

c©2019 Authors retain the copyrights

of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

892

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-17-2019-892


Int. J. Anal. Appl. 17 (5) (2019) 893

Convex functions are equivalently defined by the following inequality which is well known as the Hadamard

inequality:

f

(
a + b

2

)
≤

1

b−a

∫ b
a

f(x)dx ≤
f(a) + f(b)

2

where f : I → R is a convex function on I and a,b ∈ I,a < b.

Following definitions are given in [8].

Definition 1.2. Let Tg be a set of real numbers. This set Tg is said to be relative convex with respect to an

arbitrary function g : R → R, if

(1 − t)x + tg(y) ∈ Tg

where x,y ∈ R such that x,g(y) ∈ Tg, 0 ≤ t ≤ 1.

Definition 1.3. A function f : Tg → R is said to be relative convex, if there exists an arbitrary function

g : R → R, such that

f((1 − t)x + tg(y)) ≤ (1 − t)f(x) + tf(g(y)),

holds, where x,y ∈ R such that x,g(y) ∈ Tg, 0 ≤ t ≤ 1.

Noor et al. proved the following Hadamard type integral inequality in [8] for relative convex functions via

Riemann-Liouville fractional integral operators.

Theorem 1.1. Let f be a positive relative convex function and integralable on [a,g(b)]. Then for α > 0, the

following inequalities hold:

f

(
a + g(b)

2

)
≤

Γ(α + 1)

2(g(b) −a)α
[Iαa+f(g(b)) + I

α
b−
f(a)]

≤
f(a) + f(g(b))

2
.

Now we define the extended generalized Mittag-Leffler function E
γ,δ,k,c
µ,α,l (.; p) as follows:

Definition 1.4. [2] Let µ,α,l,γ,c ∈ C, <(µ),<(α),<(l) > 0, <(c) > <(γ) > 0 with p ≥ 0, δ > 0 and

0 < k ≤ δ + <(µ). Then the extended generalized Mittag-Leffler function Eγ,δ,k,cµ,α,l (t; p) is defined by:

E
γ,δ,k,c
µ,α,l (t; p) =

∞∑
n=0

βp(γ + nk,c−γ)
β(γ,c−γ)

(c)nk
Γ(µn + α)

tn

(l)nδ
, (1.1)

where βp is the generalized beta function defined by:

βp(x,y) =

∫ 1
0

tx−1(1 − t)y−1e−
p

t(1−t) dt

and (c)nk is the Pochhammer symbol defined as (c)nk =
Γ(c+nk)

Γ(c)
.

In [2], properties of the generalized Mittag-Leffler function are discussed and it is given that E
γ,δ,k,c
µ,α,l (t; p) is

absolutely convergent for k < δ + <(µ). Let S be the sum of series of absolute terms of the Mittag-Leffler



Int. J. Anal. Appl. 17 (5) (2019) 894

function E
γ,δ,k,c
µ,α,l (t; p), then we have

∣∣∣Eγ,δ,k,cµ,α,l (t; p)∣∣∣ ≤ S. We use this property of Mittag-Leffler function to
prove the results of this paper.

Remark 1.2. Mittag-Leffler function (1.1) is the generalization of following functions:

(i) By setting p = 0, it reduces to the Salim-Faraj function E
γ,δ,k,c
µ,α,l (t) defined in [12].

(ii) By setting l = δ = 1, it reduces to the function Eγ,k,cµ,α (t; p) defined by Rahman et al. in [11].

(iii) By setting p = 0 and l = δ = 1, it reduces to the Shukla-Prajapati function Eγ,kµ,α(t) defined in [13] (see

also [14]).

(iv) By setting p = 0 and l = δ = k = 1, it reduces to the Prabhakar function Eγµ,α(t) defined in [10].

The corresponding left and right sided generalized fractional integral operators �
γ,δ,k,c
µ,α,l,ω,a+

and �
γ,δ,k,c
µ,α,l,ω,b−

are defined as follows:

Definition 1.5. [2] Let ω,µ,α,l,γ,c ∈ C, <(µ),<(α),<(l) > 0, <(c) > <(γ) > 0 with p ≥ 0, δ > 0

and 0 < k ≤ δ + <(µ). Let f ∈ L1[a,b] and x ∈ [a,b]. Then the generalized fractional integral operators

�
γ,δ,k,c
µ,α,l,ω,a+

f and �
γ,δ,k,c
µ,α,l,ω,b−

f are defined by:(
�
γ,δ,k,c
µ,α,l,ω,a+

f
)

(x; p) =

∫ x
a

(x− t)α−1Eγ,δ,k,cµ,α,l (ω(x− t)
µ; p)f(t)dt, (1.2)

and (
�
γ,δ,k,c
µ,α,l,ω,b−

f
)

(x; p) =

∫ b
x

(t−x)α−1Eγ,δ,k,cµ,α,l (ω(t−x)
µ; p)f(t)dt. (1.3)

Remark 1.3. Operators in (1.2) and (1.3) are the generalization of the following fractional integral opera-

tors:

(i) By setting p = 0, they reduce to the fractional integral operators defined by Salim-Faraj in [12].

(ii) By setting l = δ = 1, they reduce to the fractional integral operators defined by Rahman et al. in [11].

(iii) By setting p = 0 and l = δ = 1, they reduce to the fractional integral operators defined by Srivastava-

Tomovski in [14].

(iv) By setting p = 0 and l = δ = k = 1, they reduce to the fractional integral operators defined by Prabhakar

in [10].

(v) By setting p = ω = 0, they reduces to the Riemann-Liouville fractional integrals.

In [5] the Hadamard and the Fejér-Hadamard inequalities for convex functions via generalized fractional

integral operator containing the Mittag-Leffler function defined in [12] have been proved.

In [1, 6, 8], the Hadamard and the Fejér-Hadamard type inequalities for convex and relative convex functions

via Riemann-Liouville fractional integral operators and extended generalized fractional integral operators

have been proved. In this paper, we give fractional integral inequalities of the Hadamard and the Fejér-

Hadamard type for convex and relative convex functions by using the extended generalized Mittag-Leffler

function. We also produce the results which are given in [1, 6, 8] by setting particular values of parameters.



Int. J. Anal. Appl. 17 (5) (2019) 895

2. Main Results

Following lemmas are useful to establish the main results.

Lemma 2.1. Let f : [a,b] → R be a function such that f ∈ L1[a,b] and symmetric about a+b2 with a < b.

Then for extended generalized fractional integral operators (1.2) and (1.3), the following equality holds:

(
�
γ,δ,k,c
µ,α,l,ω,a+

f
)

(b; p) =
(
�
γ,δ,k,c
µ,α,l,ω,b−

f
)

(a; p) (2.1)

=

(
�
γ,δ,k,c
µ,α,l,ω,a+

f
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

f
)

(a; p)

2
.

Proof. Using symmetricity of f we have f(a + b− t) = f(t), therefore by (1.2) of Definition 1.5, we have

(
�
γ,δ,k,c
µ,α,l,ω,a+

f
)

(b; p) =

∫ b
a

(b− t)α−1Eγ,δ,k,cµ,α,l (ω(b− t)
µ; p)f(t)dt, (2.2)

putting t = a + b− t in above, we get

(
�
γ,δ,k,c
µ,α,l,ω,a+

f
)

(b; p) =

∫ b
a

(t−a)α−1Eγ,δ,k,cµ,α,l (ω(t−a)
µ; p)f(t)dt.

By using Definition 1.5, we get

(
�
γ,δ,k,c
µ,α,l,ω,a+

f
)

(b; p) =
(
�
γ,δ,k,c
µ,α,l,ω,b−

f
)

(a; p). (2.3)

Therefore we get (2.1). �

Lemma 2.2. Let f : [a,b] → R be a function such that f′ ∈ L1[a,b] with a < b. If g : [a,b] → R is integrable

and symmetric about a+b
2

, then for extended generalized fractional integral operators (1.2) and (1.3), the

following equality holds:

(
f(a) + f(b)

2

)[(
�
γ,δ,k,c
µ,α,l,ω,a+

g
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

g
)

(a; p)
]

(2.4)

−
[(
�
γ,δ,k,c
µ,α,l,ω,a+

fg
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

fg
)

(a; p)
]

=

∫ b
a

[∫ t
a

(b−s)α−1Eγ,δ,k,cµ,α,l (ω(b−s)
µ; p)g(s)ds

−
∫ b
t

(s−a)α−1Eγ,δ,k,cµ,α,l (ω(s−a)
µ; p)g(s)ds

]
f′(t)dt.



Int. J. Anal. Appl. 17 (5) (2019) 896

Proof. One can note that

∫ b
a

[∫ t
a

(b−s)α−1Eγ,δ,k,cµ,α,l (ω(b−s)
µ; p)g(s)ds

−
∫ b
t

(s−a)α−1Eγ,δ,k,cµ,α,l (ω(s−a)
µ; p)g(s)ds

]
f′(t)dt

=

∫ b
a

[∫ t
a

(b−s)α−1Eγ,δ,k,cµ,α,l (ω(b−s)
µ; p)g(s)ds

]
f′(t)dt

+

∫ b
a

[
−
∫ b
t

(s−a)α−1Eγ,δ,k,cµ,α,l (ω(s−a)
µ; p)g(s)ds

]
f′(t)dt.

By simple calculation, we get

∫ b
a

[∫ t
a

(b−s)α−1Eγ,δ,k,cµ,α,l (ω(b−s)
µ; p)g(s)ds

]
f′(t)dt

= f(b)

(∫ b
a

(b−s)α−1Eγ,δ,k,cµ,α,l (ω(b−s)
µ; p)g(s)ds

)

−
∫ b
a

(
(b− t)α−1Eγ,δ,k,cµ,α,l (ω(b− t)

µ; p
)
fg(t)dt.

By using Definition 1.5, we get

f(b)
(
�
γ,δ,k,c
µ,α,l,ω,a+

g
)

(b; p) −
(
�
γ,δ,k,c
µ,α,l,ω,a+

fg
)

(b; p).

Now by using Lemma 2.1, we have

∫ b
a

[∫ t
a

(b−s)α−1Eγ,δ,k,cµ,α,l (ω(b−s)
µ; p)g(s)ds

]
f′(t)dt (2.5)

=
f(b)

2

[(
�
γ,δ,k,c
µ,α,l,ω,a+

g
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

g
)

(a; p)
]
−
(
�
γ,δ,k,c
µ,α,l,ω,a+

fg
)

(b; p)

and

∫ b
a

[
−
∫ b
t

(s−a)α−1Eγ,δ,k,cµ,α,l (ω(s−a)
µ; p)g(s)ds

]
f′(t)dt (2.6)

=
f(a)

2

[(
�
γ,δ,k,c
µ,α,l,ω,a+

g
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

g
)

(a; p)
]
−
(
�
γ,δ,k,c
µ,α,l,ω,b−

fg
)

(a; p).

By adding (2.6) and (2.5), we get (2.4). �

In the following we give integral inequality of the Hadamard type.

Theorem 2.3. Let f : [a,b] → R be a differentiable function such that f′ ∈ L1[a,b] with a < b. If |f′| is

convex on [a,b] and g : I → R is continuous and symmetric function about a+b
2

, then for extended generalized



Int. J. Anal. Appl. 17 (5) (2019) 897

fractional integral operators (1.2) and (1.3), the following inequality holds:∣∣∣∣
(
f(a) + f(b)

2

)[(
�
γ,δ,k,c
µ,α,l,ω,a+

g
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

g
)

(a; p)
]

−
[(
�
γ,δ,k,c
µ,α,l,ω,a+

fg
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

fg
)

(a; p)
]∣∣∣

≤
‖ g ‖∞ S(b−a)α+1

α(α + 1)

(
1 −

1

2α

)
[|f′(a) + f′(b)|],

for k < δ + <(µ) and ‖ g ‖∞= sup
t∈[a,b]

|g(t)|.

Proof. By using Lemma 2.2, we have∣∣∣∣
(
f(a) + f(b)

2

)[(
�
γ,δ,k,c
µ,α,l,ω,a+

g
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

g
)

(a; p)
]

(2.7)

−
[(
�
γ,δ,k,c
µ,α,l,ω,a+

fg
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

fg
)

(a; p)
]∣∣∣

≤
∫ b
a

∣∣∣∣
[∫ t

a

(b−s)α−1Eγ,δ,k,cµ,α,l (ω(b−s)
µ; p)g(s)ds

−
∫ b
t

(s−a)α−1Eγ,δ,k,cµ,α,l (ω(s−a)
µ; p)g(s)ds

]∣∣∣∣∣ |f′(t)|dt.
Since |f′| is convex, so we have

|f′(t)| ≤
b− t
b−a

|f′(a)| +
t−a
b−a

|f′(b)| (2.8)

where t ∈ [a,b].

From symmetricity of g, we have∫ b
t

(s−a)α−1Eγ,δ,k,cµ,α,l (ω(s−a)
µ; p)g(s)ds

=

∫ a+b−t
a

(b−s)α−1Eγ,δ,k,cµ,α,l (ω(b−s)
µ; p)g(a + b−s)ds

=

∫ a+b−t
a

(b−s)α−1Eγ,δ,k,cµ,α,l (ω(b−s)
µ; p)g(s)ds.

This gives ∣∣∣∣
∫ t
a

(b−s)α−1Eγ,δ,k,cµ,α,l (ω(b−s)
µ; p)g(s)ds (2.9)

−
∫ b
t

(s−a)α−1Eγ,δ,k,cµ,α,l (ω(s−a)
µ; p)g(s)ds

∣∣∣∣∣
=

∣∣∣∣∣
∫ a+b−t
t

(b−s)α−1Eγ,δ,k,cµ,α,l (ω(b−s)
µ; p)g(s)ds

∣∣∣∣∣
≤



∫a+b−t
t

|(b−s)α−1Eγ,δ,k,cµ,α,l (ω(b−s)
µ; p)g(s)|ds,t ∈ [a, a+b

2
]∫ t

a+b−t |(b−s)
α−1E

γ,δ,k,c
µ,α,l (ω(b−s)

µ; p)g(s)|ds,t ∈ [a+b
2
,b].



Int. J. Anal. Appl. 17 (5) (2019) 898

From (2.7), (2.8), (2.9) and absolute convergence of Mittag-Leffler function, we get∣∣∣∣
(
f(a) + f(b)

2

)[(
�
γ,δ,k,c
µ,α,l,ω,a+

g
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

g
)

(a; p)
]

(2.10)

−
[(
�
γ,δ,k,c
µ,α,l,ω,a+

fg
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

fg
)

(a; p)
]∣∣∣

≤
∫ a+b

2

a

(∫ a+b−t
a

|(b−s)α−1Eγ,δ,k,cµ,α,l (ω(b−s)
µ; p)g(s)|ds

)(
b− t
b−a

|f′(a)|

+
t−a
b−a

|f′(b)|
)
dt +

∫ b
a+b
2

(∫ t
a+b−t

|(b−s)α−1Eγ,δ,k,cµ,α,l (ω(b−s)
µ; p)g(s)|ds

)
×
(
b− t
b−a

|f′(a)| +
t−a
b−a

|f′(b)|
)
dt

≤
‖ g ‖∞ S
α(b−a)

[∫ a+b
2

a

((b− t)α − (t−a)α(b− t)|f′(a)|)dt

+

∫ a+b
2

a

((b− t)α − (t−a)α(t−a)|f′(b)|)dt

+

∫ b
a+b
2

((t−a)α − (b− t)α(b− t)|f′(a)|)dt

+

∫ b
a+b
2

((t−a)α − (b− t)α(t−a)|f′(b)|)dt

]
.

As we have ∫ a+b
2

a

((b− t)α − (t−a)α) (b− t)dt =
(b−a)α+2

α + 1

(
α + 1

α + 2
−

1

2α+1

)
and ∫ a+b

2

a

((b− t)α − (t−a)α) (t−a)dt =
(b−a)α+2

α + 1

(
1

α + 2
−

1

2α+1

)
.

By using the values of above integrals in (2.10), we have∣∣∣∣
(
f(a) + f(b)

2

)[(
�
γ,δ,k,c
µ,α,l,ω,a+

g
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

g
)

(a; p)
]

−
[(
�
γ,δ,k,c
µ,α,l,ω,a+

fg
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

fg
)

(a; p)
]∣∣∣

≤
‖ g ‖∞ S
α(b−a)

(b−a)α+2

α + 1

[(
α + 1

α + 2
−

1

2α+1

)
+

(
1

α + 2
−

1

2α+1

)]
[|f′(a)| + |f′(b)|]

=
‖ g ‖∞ S
α(α + 1)

(b−a)α+1
(

1 −
1

2α

)
[|f′(a)| + |f′(b)|].

�

Remark 2.4. (i) If we put p = 0 in Theorem 2.3, then we get [1, Theorem 2.3].

(ii) If we put ω = p = 0 in Theorem 2.3, then we get [6, Theorem 2.6].



Int. J. Anal. Appl. 17 (5) (2019) 899

Theorem 2.5. Let f : [a,b] → R be a differentiable function such that f′ ∈ L1[a,b] with a < b. If |f′|q,

q ≥ 1 is convex on [a,b] and g : I → R is continuous and symmetric function about a+b
2

, then for extended

generalized fractional integral operators (1.2) and (1.3), the following inequality holds:

∣∣∣∣
(
f(a) + f(b)

2

)[(
�
γ,δ,k,c
µ,α,l,ω,a+

g
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

g
)

(a; p)
]

(2.11)

−
[(
�
γ,δ,k,c
µ,α,l,ω,a+

fg
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

fg
)

(a; p)
]∣∣∣

≤
2 ‖ g ‖∞ S(b−a)α+1

α(α + 1)

(
1 −

1

2α

)
(|f′(a)|q + |f′(b)|q)

1
q ,

for k < δ + <(µ) and ‖ g ‖∞= sup
t∈[a,b]

|g(t)|.

Proof. By using Lemma 2.2, power mean inequality, the inequality (2.9) takes the following form:

∣∣∣∣
(
f(a) + f(b)

2

)[(
�
γ,δ,k,c
µ,α,l,ω,a+

g
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

g
)

(a; p)
]

(2.12)

−
[(
�
γ,δ,k,c
µ,α,l,ω,a+

fg
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

fg
)

(a; p)
]∣∣∣

≤

[∫ b
a

∣∣∣∣∣
∫ a+b−t
t

(b−s)α−1Eγ,δ,k,cµ,α,l (ω(b−s)
µ; p)g(s)ds

∣∣∣∣∣dt
]1−1

q

[∫ b
a

∣∣∣∣∣
∫ a+b−t
t

(b−s)α−1Eγ,δ,k,cµ,α,l (ω(b−s)
µ; p)g(s)ds

∣∣∣∣∣ |f′(t)|qdt
]1
q

.

Using absolute convergence of Mittag-Leffler function and ‖ g ‖∞= sup
t∈[a,b]

|g(t)|, we have

∣∣∣∣
(
f(a) + f(b)

2

)[(
�
γ,δ,k,c
µ,α,l,ω,a+

g
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

g
)

(a; p)
]

−
[(
�
γ,δ,k,c
µ,α,l,ω,a+

fg
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

fg
)

(a; p)
]∣∣∣

≤‖ g ‖
1−1

q
∞ S

1−1
q

[∫ a+b
2

a

(∫ a+b−t
t

(b−s)α−1ds

)
dt

+

∫ b
a+b
2

(∫ t
a+b−t

(b−s)α−1ds
)
dt

]1−1
q

×‖ g ‖
1
q
∞ S

1
q

[∫ a+b
2

a

(∫ a+b−t
t

(b−s)α−1ds

)
|f′(t)|qdt

+

∫ b
a+b
2

(∫ t
a+b−t

(b−s)α−1ds
)
|f′(t)|qdt

]1
q

.



Int. J. Anal. Appl. 17 (5) (2019) 900

By some calculation, we have∣∣∣∣
(
f(a) + f(b)

2

)[(
�
γ,δ,k,c
µ,α,l,ω,a+

g
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

g
)

(a; p)
]

−
[(
�
γ,δ,k,c
µ,α,l,ω,a+

fg
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

fg
)

(a; p)
]∣∣∣

≤‖ g ‖∞ S
[

(b−a)α+1

α(α + 1)

(
1 −

1

2α

)
+

(b−a)α+1

α(α + 1)

(
1 −

1

2α

)]1−1
q

×

[∫ a+b
2

a

((b− t)α − (t−a)α) |f′(t)|qdt +
∫ b
a+b
2

((b− t)α − (t−a)α) |f′(t)|qdt

]1
q

.

Since |f′|q is convex, so we have

|f′(t)|q ≤
b− t
b−a

|f′(a)|q +
t−a
b−a

|f′(b)|q. (2.13)

Hence ∣∣∣∣
(
f(a) + f(b)

2

)[(
�
γ,δ,k,c
µ,α,l,ω,a+

g
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

g
)

(a; p)
]

−
[(
�
γ,δ,k,c
µ,α,l,ω,a+

fg
)

(b; p) +
(
�
γ,δ,k,c
µ,α,l,ω,b−

fg
)

(a; p)
]∣∣∣

≤‖ g ‖∞ S
[
2

(b−a)α+1

α(α + 1)

(
1 −

1

2α

)]1−1
q

×

[∫ a+b
2

a

((b− t)α − (t−a)α)
(
b− t
b−a

|f′(a)|q +
t−a
b−a

|f′(b)|q
)
dt

+

∫ b
a+b
2

((b− t)α − (t−a)α)
(
b− t
b−a

|f′(a)|q +
t−a
b−a

|f′(b)|q
)
dt

]1
q

.

From it (2.11) can be obtained. �

Remark 2.6. (i) If we put p = 0 in Theorem 2.5, then we get [1, Theorem 2.5].

(ii) If we put ω = p = 0 in Theorem 2.5, then we get [6, Theorem 2.8].

In the following we give the Hadamard inequality for relative convex functions via generalized fractional

integral operators.

Theorem 2.7. Let f : [a,g(b)] → R be a function such that f ∈ L1[a,g(b)] with a < b. If f is relative

convex on [a,g(b)], then for extended generalized fractional integral operators (1.2) and (1.3), the following

inequalities hold:

f

(
a + g(b)

2

)(
�
γ,δ,k,c
µ,α,l,ω′,a+

1
)

(g(b); p) (2.14)

≤
1

2

[(
�
γ,δ,k,c
µ,α,l,ω′,a+

f
)

(g(b); p) +
(
�
γ,δ,k,c
µ,α,l,ω′,g(b)−

f
)

(a; p)
]

≤
f(a) + f(g(b))

2

(
�
γ,δ,k,c
µ,α,l,ω′,g(b)−

1
)

(a; p),



Int. J. Anal. Appl. 17 (5) (2019) 901

where ω′ = ω
(g(b)−a)µ .

Proof. Since f is relative convex, so we have

f

(
a + g(b)

2

)
= f

[(
1

2
(ta + (1 − t)g(b)

)
+

(
1 −

1

2

)
((1 − t)a + tg(b))

]
(2.15)

≤
1

2
f (ta + (1 − t)g(b)) +

1

2
f ((1 − t)a + tg(b)) .

Multiplying (2.15) by 2tα−1E
γ,δ,k,c
µ,α,l (ωt

µ; p) on both sides and then integrating over [0, 1], we have

2f

(
a + g(b)

2

)∫ 1
0

tα−1E
γ,δ,k,c
µ,α,l (ωt

µ; p)dt (2.16)

≤
∫ 1

0

tα−1E
γ,δ,k,c
µ,α,l (ωt

µ; p)f (ta + (1 − t)g(b)) dt

+

∫ 1
0

tα−1E
γ,δ,k,c
µ,α,l (ωt

µ; p)f ((1 − t)a + tg(b)) dt.

Putting x = ta + (1 − t)g(b) and y = (1 − t)a + tg(b) in above, we have

2f

(
a + g(b)

2

)∫ a
g(b)

(
g(b) −x
g(b) −a

)α−1
E
γ,δ,k,c
µ,α,l

(
ω

(
g(b) −x
g(b) −a

)µ
; p

)(
−dx

g(b) −a

)
(2.17)

≤
∫ a
g(b)

(
g(b) −x
g(b) −a

)α−1
E
γ,δ,k,c
µ,α,l

(
ω

(
g(b) −x
g(b) −a

)µ
; p

)
f(x)

(
−dx

g(b) −a

)

+

∫ g(b)
a

(
y −a
g(b) −a

)α−1
E
γ,δ,k,c
µ,α,l

(
ω

(
y −a
g(b) −a

)µ
; p

)
f(y)

(
dy

g(b) −a

)
.

By using Definition 1.5, we get

2f

(
a + g(b)

2

)(
�
γ,δ,k,c
µ,α,l,ω′,a+

1
)

(g(b); p) (2.18)

≤
[(
�
γ,δ,k,c
µ,α,l,ω′,a+

f
)

(g(b); p) +
(
�
γ,δ,k,c
µ,α,l,ω′,g(b)−

f
)

(a; p)
]
.

Again by using the relative convexity of f, we have

f(ta + (1 − t)g(b)) + f((1 − t)a + tg(b)) ≤ tf(a) + (1 − t)f(g(b)) + (1 − t)f(a) + tf(g(b)). (2.19)

Multiplying (2.19) by tα−1E
γ,δ,k,c
µ,α,l (ωt

µ; p) on both sides and then integrating over [0, 1], we have

∫ 1
0

tα−1E
γ,δ,k,c
µ,α,l (ωt

µ; p)f(ta + (1 − t)g(b))dt

+

∫ 1
0

tα−1E
γ,δ,k,c
µ,α,l (ωt

µ; p)f((1 − t)a + tg(b))dt

≤
∫ 1

0

tα−1E
γ,δ,k,c
µ,α,l (ωt

µ; p)tf(a) + (1 − t)f(g(b))dt

+

∫ 1
0

tα−1E
γ,δ,k,c
µ,α,l (ωt

µ; p)(1 − t)f(a) + tf(g(b))dt.



Int. J. Anal. Appl. 17 (5) (2019) 902

Putting x = ta + (1 − t)g(b) and y = (1 − t)a + tg(b) in above and then using Definition 1.5, we get[(
�
γ,δ,k,c
µ,α,l,ω′,a+

f
)

(g(b); p) +
(
�
γ,δ,k,c
µ,α,l,ω′,g(b)−

f
)

(a; p)
]

(2.20)

≤ [f(a) + f(g(b))]
(
�
γ,δ,k,c
µ,α,l,ω′,g(b)−

1
)

(a; p).

Combining it with (2.18), (2.14) is obtained. �

Remark 2.8. (i) If we put p = 0 in Theorem 2.7, then we get [1, Theorem 2.8].

(ii) If we put ω = p = 0 and k = 1 in Theorem 2.7, then we get Theorem 1.1.

In the upcoming theorem we give the generalization of previous result.

Theorem 2.9. Let f : [g(a),g(b)] → R be a function such that f ∈ L1[g(a),g(b)] with a < b. If f is

relative convex on [g(a),g(b)], then for extended generalized fractional integral operators (1.2) and (1.3), the

following inequalities hold:

f

(
g(a) + g(b)

2

)(
�
γ,δ,k,c
µ,α,l,ω′,g(a)+

1
)

(g(b); p)

≤
1

2

[(
�
γ,δ,k,c
µ,α,l,ω′,g(a)+

f
)

(g(b); p) +
(
�
γ,δ,k
µ,α,l,ω′,g(b)−

f
)

(a; p)
]

≤
f(g(a)) + f(g(b))

2

(
�
γ,δ,k,c
µ,α,l,ω′,g(b)−

1
)

(g(a); p),

where ω′ = ω
(g(b)−g(a))µ .

Proof. Proof of this theorem is on the same lines of the proof of Theorem 2.7. �

Remark 2.10. (i) If we put p = 0 in Theorem 2.9, then we get [1, Theorem 2.10].

(ii) If we put ω = p = 0 in Theorem 2.9, then we get [4, Corollary 1].

Acknowledgement

The research work of Ghulam Farid is supported by Higher Education Commission of Pakistan under

NRPU 2016, Project No. 5421.

References

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	1. Introduction
	2. Main Results
	Acknowledgement
	References