International Journal of Analysis and Applications

Volume 16, Number 4 (2018), 605-613

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

DOI: 10.28924/2291-8639-16-2018-605

HERMITE-HADAMARD TYPE INEQUALITIES FOR QUASI-CONVEX FUNCTIONS

VIA KATUGAMPOLA FRACTIONAL INTEGRALS

ERHAN SET∗, İLKER MUMCU

Department of Mathematics, Faculty of Science and Arts, Ordu University, Ordu, Turkey

∗Corresponding author: erhanset@yahoo.com

Abstract. The paper deals with quasi-convex functions, Katugampola fractional integrals and Hermite-

Hadamard type integral inequalities. The main idea of this paper is to present new Hermite-Hadamard type

inequalities for quasi-convex functions using Katugampola fractional integrals, Hölder inequality and the

identities in the literature.

1. Introduction

A function f : I ⊆ R → R is said to be convex if the inequality

f (λu + (1 −λ) v) ≤ λf (u) + (1 −λ) f (v)

holds for all u,v ∈ I and λ ∈ [0, 1]. This definition has been used in the following inequality that is called

Hermite-Hadamard inequality:

Let f : I ⊆ R → R be a convex function and a,b ∈ I with a < b, then

f

(
a + b

2

)
≤

1

b−a

∫ b
a

f (x) dx ≤
f (a) + f (b)

2
. (1.1)

This inequality has attracted many mathematicians. Especially, in the last three decades, numerous gener-

alizations, variants, and extensions of this inequality have been presented (see, e.g., [1, 3, 13, 14, 20] and the

references cited therein).

Received 2017-10-26; accepted 2018-01-03; published 2018-07-02.

2010 Mathematics Subject Classification. 26A33, 26D10, 33B20.

Key words and phrases. Hermite-Hadamard inequality; Riemann-Liouville fractional integrals; Katugampola fractional

integrals.

c©2018 Authors retain the copyrights

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

605

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-16-2018-605


Int. J. Anal. Appl. 16 (4) (2018) 606

The notion of quasi-convex functions generalizes the notion of convex functions. More precisely, a function

f : [a,b] → R is said quasi-convex on [a,b] if

f (λu + (1 −λ) v) ≤ max{f(x),f(y)},

for any x,y ∈ [a,b] and λ ∈ [0, 1]. Clearly, any convex function is a quasi-convex function. Furthermore,

there exist quasi-convex functions which are not convex (see [4]).

Let f ∈ L1[a,b] := L(a,b). The Riemann-Liouville integrals Jαa+f and Jαb−f of order α ∈ R
+ are defined,

respectively, by

Jαa+f(x) =
1

Γ(α)

∫ x
a

(x− t)α−1 f(t) dt (x > a)

and

Jαb−f(x) =
1

Γ(α)

∫ b
x

(t−x)α−1 f(t) dt (x < b)

where Γ is the familiar Gamma function (see, e.g., [21, Section 1.1]). It is noted that J1a+f(x) and J
1
b−f(x)

become the usual Riemann integrals.

In the case of α = 1, the fractional integral reduces to classical integral.

For further results related to Hermite-Hadamard type inequalities involving fractional integrals on can see

[7, 11–19].

The beta function B(α,β) is defined by (see, e.g., [21, Section 1.1] [10, p18])

B(α, β) =



∫ 1
0

tα−1(1 − t)β−1 dt (<(α) > 0; <(β) > 0)

Γ(α) Γ(β)

Γ(α + β)

(
α, β ∈ C\Z−0

)
.

(1.2)

The hypergeometric function [6]:

2F1(a,b; c; z) =
1

β(b,c− b)

∫ 1
0

tb−1(1 − t)c−b−1(1 −zt)−adt, c > b > 0, |z| < 1.

A hypergeometric function can be written using Euler’s hypergeometric transformations

(t → 1 − t) in equivalent form:

2F1(a,b; c; z) = (1 −z)−a2F1
(
a,c− b; c;

z

z − 1

)
(1.3)

Lemma 1.1. [9] For 0 < α ≤ 1 and 0 ≤ a < b, we have

|aα − bα| ≤ (b−a)α.

In [11], Sarıkaya et. al. proved a new version of Hermite-Hadamard’s inequalities in Riemann-Liouville

fractional integral form as follows:



Int. J. Anal. Appl. 16 (4) (2018) 607

Theorem 1.1. Let f : [a,b] → R be a positive function with 0 ≤ a < b and f ∈ L1[a,b]. If f is a convex

function on [a,b], then the following inequalities for fractional integrals holds:

f

(
a + b

2

)
≤

Γ(α + 1)

2(b−a)α
[Jαa+f(b) + J

α
b−f(a)] ≤

f(a) + f(b)

2
(1.4)

with α > 0.

In [8], Özdemir et. al. gave following results for quasi-convex functions via Riemann-Liouville fractional

integrals.

Theorem 1.2. Let f : [a,b] → R, be a positive function with 0 ≤ a < b and f ∈ L1[a,b]. If f is a

quasi-convex function [a,b], then the following inequality for fractional integrals holds:

Γ(α + 1)

2(b−a)α
[Jαa+f(b) + J

α
b−f(a)] ≤ max{f(a),f(b)} (1.5)

with α > 0.

Theorem 1.3. Let f : [a,b] → R, be a differentiable mapping on (a,b) with a < b. If |f′| is quasi-convex on

[a,b] with α > 0, then the following inequality holds:∣∣∣∣f(a) + f(b)2 − Γ(α + 1)2(b−a)α [Jαa+f(b) + Jαb−f(a)]
∣∣∣∣

≤
b−a
α + 1

(
1 −

1

2α

)
max{f′(a),f′(b)}. (1.6)

Theorem 1.4. Let f : [a,b] → R, be a differentiable mapping on (a,b) such that f′ ∈ L1[a,b]. If |f′|q is

quasi-convex on [a,b], and p > 1, then the following inequality for fractional integrals holds:∣∣∣∣f(a) + f(b)2 − Γ(α + 1)2(b−a)α [Jαa+f(b) + Jαb−f(a)]
∣∣∣∣

≤
b−a

2(αp + 1)1/p

(
1 −

1

2α

)
(max{|f′(a)|q, |f′(b)|q})1/q . (1.7)

where 1
p

+ 1
q

= 1 and α ∈ [0, 1].

Theorem 1.5. Let f : [a,b] → R, be a differentiable mapping on (a,b) such that f′ ∈ L1[a,b]. If |f′|q is

quasi-convex on [a,b], and q ≥ 1, then the following inequality for fractional integrals holds:∣∣∣∣f(a) + f(b)2 − Γ(α + 1)2(b−a)α [Jαa+f(b) + Jαb−f(a)]
∣∣∣∣

≤
b−a
α + 1

(
1 −

1

2α

)
(max{|f′(a)|q, |f′(b)|q})1/q . (1.8)

with α ∈ [0, 1].

Katugampola gave a new fractional integral that generalizes the Riemann-Liouville and the Hadamard

fractional integrals into a single form.



Int. J. Anal. Appl. 16 (4) (2018) 608

Definition 1.1. [5] Let [a,b] ⊂ R be a finite interval. Then, the left- and right-side Katugampola fractional

integrals of order (α > 0) of f ∈ Xpc (a,b) are defined:

ρIαa+f(x) =
ρ1−α

Γ(α)

∫ x
a

tρ−1

(xρ − tρ)1−α
f(t)dt

and

ρIαb−f(x) =
ρ1−α

Γ(α)

∫ b
x

tρ−1

(tρ −xρ)1−α
f(t)dt

with a < x < b and ρ > 0, if the integral exist.

Theorem 1.6. [5] Let α > 0 and ρ > 0. Then for x > a,

1. limρ→1
ρIαa+f(x) = Jαa+f(x),

2. limρ→0+
ρIαa+f(x) = Hαa+f(x).

Similar results also hold for right-sided operators.

In [2], Chen and Katugampola proved the following lemma:

Lemma 1.2. Let f : [aρ,bρ] → R be a differentiable mapping on (aρ,bρ) with 0 ≤ a < b. Then the following

equality holds if the fractional integrals exist:

f(aρ) + f(bρ)

2
−
αραΓ(α + 1)

2(bρ −aρ)α
[
ρIαa+(f ◦g)(b) +

ρIαb−(f ◦g)(a)
]

=
bρ −aρ

2

∫ 1
0

[(1 − tρ)α − tρα]tρ−1f′(tρaρ + (1 − tρ)bρ)dt (1.9)

where g(x) = xρ.

The main purpose of this paper is to establish Hermite-Hadamard’s inequalities for quasi-convex functions

via Katugampola fractional integral. We also obtain Hermite-Hadamard type inequalities of these classes

functions.

2. Main Results

Theorem 2.1. Let α > 0 and ρ > 0. Let f : [aρ,bρ] → R be a positive function with 0 ≤ a < b and

f ∈ Xpc (aρ,bρ). If f is a quasi-convex function on [aρ,bρ], then the following inequalities holds:

ραΓ(α + 1)

2(bρ −aρ)α
[
ρIαa+(f ◦g)(b) +

ρIαb−(f ◦g)(a)
]
≤ max{f(aρ),f(bρ)} (2.1)

where g(x) = xρ.

Proof. Since f is quasi-convex function on [aρ,bρ], we get

f(tρaρ + (1 − tρ)bρ) ≤ max{f(aρ),f(bρ)}



Int. J. Anal. Appl. 16 (4) (2018) 609

and

f((1 − tρ)aρ + tρbρ) ≤ max{f(aρ),f(bρ)}.

By adding these inequalities we have

1

2
[f(tρaρ + (1 − tρ)bρ) + f((1 − tρ)aρ + tρbρ)] ≤ max{f(aρ),f(bρ)}. (2.2)

Multiplying both sides of (2.2) by tαρ−1 and integrating the resulting inequality with respect to t over [aρ,bρ],

we obtain

∫ 1
0

tαρ−1f(tρaρ + (1 − tρ)bρ)dt +
∫ 1
0

tαρ−1f((1 − tρ)aρ + tρbρ)dt

=

∫ b
a

(
bρ −xρ

bρ −aρ

)α−1
f(xρ)

xρ−1

bρ −aρ
dx +

∫ b
a

(
xρ −aρ

bρ −aρ

)α−1
f(xρ)

xρ−1

bρ −aρ
dx

=
1

(bρ −aρ)α

∫ b
a

xρ−1

(bρ −xρ)1−α
f(xρ)dx +

1

(bρ −aρ)α

∫ b
a

xρ−1

(xρ −aρ)1−α
f(xρ)dx

=
Γ(α)

ρ1−α(bρ −aρ)α
[
ρIαa+(f ◦g)(b) +

ρIαb−(f ◦g)(a)
]

≤
2

ρα
max{f(aρ),f(bρ)}

So we get desired result. The proof is completed. �

Remark 2.1. In Theorem 2.1, taking limit ρ → 1 we obtain inequality of (1.5).

Theorem 2.2. Let α > 0 and ρ > 0. Let f : [aρ,bρ] → R be a differentiable mapping on [aρ,bρ] with

0 ≤ a < b. If |f′| is a quasi-convex function on [aρ,bρ], then the following inequalities holds:

∣∣∣∣f(aρ) + f(bρ)2 − αρ
αΓ(α + 1)

2(bρ −aρ)α
[
ρIαa+(f ◦g)(b) +

ρIαb−(f ◦g)(a)
]∣∣∣∣

=
bρ −aρ

ρ(α + 1)

(
1 −

1

2ρ(α+1)

)
max{|f′(aρ)| |f′(bρ)|} (2.3)

where g(x) = xρ.



Int. J. Anal. Appl. 16 (4) (2018) 610

Proof. Using Lemma 1.2 and quasi-convex of |f′| with modulus, we get
∣∣∣∣f(aρ) + f(bρ)2 − αρ

αΓ(α + 1)

2(bρ −aρ)α
[
ρIαa+(f ◦g)(b) +

ρIαb−(f ◦g)(a)
]∣∣∣∣

≤
bρ −aρ

2

∫ 1
0

|(1 − tρ)α − tρα|tρ−1|f′(tρaρ + (1 − tρ)bρ)|dt

≤
bρ −aρ

2

∫ 1
0

|(1 − tρ)α − tρα|tρ−1max{|f′(aρ)| |f′(bρ)|}dt

=
bρ −aρ

2
max{|f′(aρ)| |f′(bρ)|}

×
{∫ 1/21/ρ

0

[(1 − tρ)α − tρα]tρ−1dt +
∫ 1
1/21/ρ

[tρα + (1 − tρ)α]tρ−1dt
}

=
bρ −aρ

ρ(α + 1)

(
1 −

1

2α

)
max{|f′(aρ)| |f′(bρ)|}

where

∫ 1/21/ρ
0

[(1 − tρ)α − tρα]tρ−1dt +
∫ 1
1/21/ρ

[tρα + (1 − tρ)α]tρ−1dt

=
1

ρ

{∫ 1/2
0

[(1 −u)α −uα] du +
∫ 1
1/2

[uα − (1 −u)α] du
}

=
2

ρ(α + 1)

(
1 −

1

2α

)
. (2.4)

The proof is completed. �

Remark 2.2. In Theorem 2.2, taking limit ρ → 1 we obtain inequality of (1.6).

Theorem 2.3. Let α > 0 and ρ > 0. Let f : [aρ,bρ] → R be a differentiable mapping on [aρ,bρ] with

0 ≤ a < b. If |f′|q is a quasi-convex function on [aρ,bρ] and s > 1, then the following inequalities holds:
∣∣∣∣f(aρ) + f(bρ)2 − αρ

αΓ(α + 1)

2(bρ −aρ)α
[
ρIαa+(f ◦g)(b) +

ρIαb−(f ◦g)(a)
]∣∣∣∣

=
bρ −aρ

2
(max{|f′(aρ)|q, |f′(bρ)|q})1/q (K1 + K2)1/s

where

K1 =
1

ρ2s+
1−s
ρ

B(s +
1 −s
ρ

,αs + 1),

K2 =
αs + 1

2ρ
2F1

(
1 −s +

s− 1
ρ

, 1; αs + 2;
1

2

)
,

1
s

+ 1
q

= 1 and g(x) = xρ.



Int. J. Anal. Appl. 16 (4) (2018) 611

Proof. From Lemma 1.1, Lemma 1.2, Hölder inequality and quasi-convex of |f′| with proporties of modulus,

we have

∣∣∣∣f(aρ) + f(bρ)2 − αρ
αΓ(α + 1)

2(bρ −aρ)α
[
ρIαa+(f ◦g)(b) +

ρIαb−(f ◦g)(a)
]∣∣∣∣

≤
bρ −aρ

2

∫ 1
0

|(1 − tρ)α − tρα|tρ−1|f′(tρaρ + (1 − tρ)bρ)|dt

≤
bρ −aρ

2

(∫ 1
0

|(1 − tρ)α − tρα|sts(ρ−1)dt
)1/s (∫ 1

0

|f′(tρaρ + (1 − tρ)bρ)|qdt
)1/q

≤
bρ −aρ

2

(∫ 1
0

|1 − 2tρ|αsts(ρ−1)dt
)1/s

(max{|f′(aρ)|q, |f′(bρ)|q})1/q

=
bρ −aρ

2
(max{|f′(aρ)|q, |f′(bρ)|q})1/q

×
{∫ 1/21/ρ

0

(1 − 2tρ)αsts(ρ−1)dt +
∫ 1
1/21/ρ

(2tρ − 1)αsts(ρ−1)dt
}1/s

=
bρ −aρ

2
(max{|f′(aρ)|q, |f′(bρ)|q})1/q (K1 + K2)1/s (2.5)

where

K1 =

∫ 1/21/ρ
0

(1 − 2tρ)αsts(ρ−1)dt =
1

ρ2s+
1−s
ρ

∫ 1
0

us−1+
1−s
ρ (1 −u)αsdu

=
1

ρ2s+
1−s
ρ

B

(
s +

1 −s
ρ

,αs + 1

)
(2.6)

K2 =

∫ 1
1/21/ρ

(2tρ − 1)αsts(ρ−1)dt =
1

2s+
1−s
ρ ρ

∫ 1
0

uαs(1 + u)s−1+
1−s
ρ du

=
αs + 1

2ρ
2F1

(
1 −s +

s− 1
ρ

, 1; αs + 2;
1

2

)
(2.7)

So, if we use (2.6), (2.7) in (2.5), we obtain desired result. �

Remark 2.3. In Theorem 2.3, taking limit ρ → 1 we obtain inequality of (1.7).

Theorem 2.4. Let α > 0 and ρ > 0. Let f : [aρ,bρ] → R be a differentiable mapping on [aρ,bρ] with

0 ≤ a < b. If |f′|q is a quasi-convex function on [aρ,bρ] and q ≥ 1, then the following inequalities holds:

∣∣∣∣f(aρ) + f(bρ)2 − αρ
αΓ(α + 1)

2(bρ −aρ)α
[
ρIαa+(f ◦g)(b) +

ρIαb−(f ◦g)(a)
]∣∣∣∣

≤
bρ −aρ

ρ(α + 1)

(
1 −

1

2α

)
(max{|f′(aρ)|q, |f′(bρ)|q})1/q

where g(x) = xρ.



Int. J. Anal. Appl. 16 (4) (2018) 612

Proof. From Lemma 1.2, quasi-convex of |f′| and using power-mean inequality with proporties of modulus,

we have ∣∣∣∣f(aρ) + f(bρ)2 − αρ
αΓ(α + 1)

2(bρ −aρ)α
[
ρIαa+(f ◦g)(b) +

ρIαb−(f ◦g)(a)
]∣∣∣∣

≤
bρ −aρ

2

∫ 1
0

|(1 − tρ)α − tρα|tρ−1|f′(tρaρ + (1 − tρ)bρ)|dt

≤
bρ −aρ

2

(∫ 1
0

|(1 − tρ)α − tρα|tρ−1dt
)1−1/q

×
(∫ 1

0

|(1 − tρ)α − tρα|tρ−1|f′(tρaρ + (1 − tρ)bρ)|qdt
)1/q

≤
bρ −aρ

2

(∫ 1
0

|(1 − tρ)α − tρα|tρ−1dt
)1−1/q

×(max{|f′(aρ)|q, |f′(bρ)|q})1/q
(∫ 1

0

|(1 − tρ)α − tρα|tρ−1dt
)1/q

=
bρ −aρ

2

(∫ 1
0

|(1 − tρ)α − tρα|tρ−1dt
)

(max{|f′(aρ)|q, |f′(bρ)|q})1/q

Using (2.4) we get desired result. �

Remark 2.4. In Theorem 2.4, taking limit ρ → 1 we obtain inequality of (1.8).

3. Acknowledgement

This research is supported by Ordu University Scientific Research Projects Coordination Unit (BAP).

Project Number: YKD-17224

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	1. Introduction
	2. Main Results
	3. Acknowledgement
	References