EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 16, No. 1, 2023, 503-522
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Hermite Hadamard Type Inequalities Involving (k-p)
Fractional Operator with (α, h−m)− p convexity

Vuk Stojiljković

Faculty of Science, University of Novi Sad, Novi Sad, Serbia, Serbia

Abstract. We establish various fractional convex inequalities of the Hermite-Hadamard type
which generalize the previously obtained results in the literature. Various types of such inequalities
are obtained and given as corollaries. The main motivation of the paper is to generalize the recently
published results in terms of the (α, h − m) − p convexity with k-p Riemann Liouville fractional
operator. The application of Hölders inequality is given in tandem with the k-p fractional operator
of the convex type.

2020 Mathematics Subject Classifications: 26D10, 26A33

Key Words and Phrases: Hermite-Hadamard inequality, (α, h−m)−p-convex function, Hölder
inequality, Fractional Inequality

1. Introduction

Convex inequalities in mathematics have been an ongoing topic of research since the
introduction of the first convex inequality by Jensen. Many inequalities followed as a
consequence of the said inequality , see books [24, 31]. Inequalities have applications in
many fields , such as analysis, optimization and the probability theory. For further infor-
mation, we refer the reader to the papers [8, 9, 12, 17, 25, 26, 35]. The inequality that has
attracted the most attention in the math community is the Hermite-Hadamard inequality
[16]. The said inequality has been generalized in various forms by many mathematicians
throughout the years. The inequality was proved independently by Charles Hermite and
Jacques Hadamard. This inequality is stated as follows:
Let F : I → R be a convex function on I in R, where I is a bounded subset of R and
ρ1, ρ2 ∈ I with ρ1 < ρ2, then

F
(
ρ1 + ρ2

2

)
⩽

1

ρ2 − ρ1

∫ ρ2

ρ1

F(t)dt ⩽
F(ρ1) + F(ρ2)

2
.

Lately, various types of Hermite-Hadamard type inequalities have been studied and gen-
eralized for different types of convex functions under different conditions and parameters.

DOI: https://doi.org/10.29020/nybg.ejpam.v16i1.4689

Email address: vuk.stojiljkovic999@gmail.com (V. Stojiljković)

https://www.ejpam.com 503 © 2023 EJPAM All rights reserved.



V. Stojiljković / Eur. J. Pure Appl. Math, 16 (1) (2023), 503-522 504

See the following papers for more information and references therein [2–7, 10, 11, 14, 15,
22, 23, 32–34, 37–39, 41–43].In 1695, l’Hospital sent a letter to Leibniz. In his message an
important question about the order of the derivative emerged, what might be a derivative
of order 1

2? That letter sparked the interest of many upcoming mathematicians to inves-
tigate further into the matter of fractional derivatives. Then came Fourier in 1822 who
suggested an integral representation to define the derivative, and his version can be consid-
ered the first definition of the derivative of the arbitrary positive order. Abel in 1826 solved
an integral equation associated with tautochrone problem, which was the first application
of FC(fractional calculus). After Abel, many mathematicians proceeded to work in the
field, some of the names: Riemann, Grünwald and Letnikov, Hadamard, Weyl, and many
more. In the late upper half of the 20th century, Caputo formulated a definition, more
restrictive than the Riemann-Liouville but more appropriate to discuss problems involving
fractional differential equations with initial conditions. Fractional calculus was found to
be useful in physics as well, for example Whatcraft and Meerschaert (2008) described a
fractional conservation of mass, Fractional Schrödinger equation in quantum theory, and
many others. Different types of fractional integrals and derivatives were defined through-
out the years, we refer the interested reader to the following books [18, 28, 44] for more
information on the matter.
The motivation for this paper comes from the recently published paper by Stojiljković et
al.[40] where the authors established some Theorems regarding k − p fractional inequali-
ties. In this paper, we generalize the obtained inequalities.
The goal of this paper is to provide various convex inequalities with the usage of the
(α, h−m)− p convexity in addition to the usage of the fractional calculus.
We start by defining various types of convex-inequalities. From Jensen’s inequality which
was the first inequality of its type to the (α, h −m) − p convexity which will be used in
the paper.

Definition 1. For an interval I in R, a function F : I → R is said to be convex on I if,

F(ζρ1 + (1− ζ)ρ2) ⩽ ζF(ρ1) + (1− ζ)F(ρ2)

for all ρ1, ρ2 ∈ I and ζ ∈ [0, 1] holds and is said to be a concave function if the inequality
is reversed.

Among the first generalizations of the convex function was given by Hudzik and Ma-
ligranda , in their paper [19].

Definition 2. A function F : [0,+∞) → R is said to be s-convex in the second sense if

F(tx+ (1− t)y) ⩽ tsF(x) + (1− t)sF(y)

holds for all x, y ∈ [0,+∞), t ∈ [0, 1] and for some fixed s ∈ (0, 1].

The (s,m) convexity generalized the s convexity, J. Park asserted a new definition given
in the following and gave some properties about this class of functions in [30].



V. Stojiljković / Eur. J. Pure Appl. Math, 16 (1) (2023), 503-522 505

Definition 3. For some fixed s ∈ (0, 1] and m ∈ [0, 1] a mapping F : [0,+∞) → R is said
to be (s,m)-convex in the second sense on I if

F(tρ1 +m(1− t)ρ2) ⩽ tsF(ρ1) +m(1− t)sF(ρ2)

holds for all ρ1, ρ2 ∈ I and t ∈ [0, 1].

The following definition was introduced by Zhong Fang which generalizes the p-convexity.
More about the property of the class of (p, h) convex functions can be found here [13].

Definition 4. Let h : J → R be a non-negative and non-zero function and it is also
assumed that (0, 1) ⊂ J . We say that F : I → R is a (p, h)-convex function or that F
belongs to the class ghx(h, p, I), if F is non-negative and

F([αρp1 + (1− α)ρp2]
1
p ) ⩽ h(α)F(ρ1) + h(1− α)F(ρ2)

for all ρ1, ρ2 ∈ I and α ∈ (0, 1). Similarly, if the inequality is reversed, then F is said to
be a (p, h)-concave function or belong to the class ghv(h, p, I).

The following definition is due to M. Emin Ozdemir et al. [29], it generalizes the
definition of h- convex functions.

Definition 5. Let J ⊂ R be an interval containing (0, 1) and let h : J → R be a non-
negative function. If F : [0, b] → R is a (h-m)-convex function, if F is non-negative and,
for all ρ1, ρ2 ∈ [0, b],m ∈ [0, 1] and α ∈ (0, 1), one has

F(αρ1 +m(1− α)ρ2) ⩽ h(α)F(ρ1) +mh(1− α)F(ρ2).

For suitable choices of h and m, the class of (h–m)-convex functions is reduced to differ-
ent known classes of convex and related functions defined on [0, b] given in the following
remark.

In the following cases, we fix various parameters in the (h-m)-convexity to obtain various
other types of convexity:
1. If m = 1, then we get an h-convex function.
2. If h(α) = α, then we get an m-convex function.
3. If h(α) = α and m = 1, then we get a convex function.
4. If h(α) = 1 and m = 1, then we get a p-function.
5. If h(α) = αs and m = 1, then we get an s-convex function in the second sense.
6. If h(α) = 1

α and m = 1, then we get a Godunova–Levin function.
7. If h(α) = 1

αs and m = 1, then we get an s–Godunova–Levin function of the second
kind.

Motivation behind defining the following class of convex functions comes from the last two
defined convex classes, as this one unifies them all.
The following definition given by Jia et al. [20] generalizes all the previously defined types
of convex functions.



V. Stojiljković / Eur. J. Pure Appl. Math, 16 (1) (2023), 503-522 506

Definition 6. Let J ⊂ R be an interval containing (0, 1) and let h : J → R be a non-
negative function. Let I ⊂ (0,+∞) be an interval and p ∈ R \ {0}. A function F : I → R
is said to be (α, h−m)-p convex, if

F
(
(tap +m(1− t)bp)

1
p

)
⩽ h(tα)F(a) +mh(1− tα)F(b)

holds provided (tap +m(1− t)bp)
1
p ∈ I for t ∈ [0, 1] and (α,m) ∈ [0, 1]2.

Before we introduce the fractional type integrals, we need the following definitions.

The Pochammer k-symbol (y)m,k is defined as (see [1])

(y)m,k = y(y + k)(y + 2k)...(y + (m− 1)k),

where m ∈ N ∪ 0, k > 0.
The k−gamma function Γk is given by (see [1]).

Γk(y) = lim
m→+∞

m!km(mk)
y
k
−1

(y)m,k

where k > 0, y ∈ C \ kZ− ∪ 0.
When k = 1 the above definitions reduce to the Pochammer symbol (y)m

(y)m =

{∏m
r=1(y + r − 1),m ∈ N

1,m = 0

and Γ function defined as

Γ(t) =

∫ +∞

0
e−zzt−1dz.

In the following we will introduce the fractional type integrals which will be used
throughout the paper.

Definition 7. The Riemann–Liouville fractional integral is defined by [18, 28, 44] where
ℜ(α) > 0 and F is locally integrable.

aI
α
t F(t) =

1

Γ(α)

∫ t

a
(t− x)α−1F(x)dx.

The following definition represents the Riemann-Liouville k fractional integral which was
defined by Mubeen and Habibullah [27].

Definition 8. Let G ∈ L1[a, b]. Then the k-fractional integrals of order α, k > 0 with
a ⩾ 0 are defined as:

Iα,k
a+ G(x) = 1

kΓk(α)

∫ x

a
(x− t)

α
k
−1G(t)dt, x > a



V. Stojiljković / Eur. J. Pure Appl. Math, 16 (1) (2023), 503-522 507

and

Iα,k
b− G(x) = 1

kΓk(α)

∫ b

x
(t− x)

α
k
−1G(t)dt, x < b

where Γk(.) is the k-Gamma function.

The following definition is due to Udita Katugampola [21] of Katugampola Fractional
integrals, which generalizes the Riemann-Liouville fractional integrals.

Definition 9. Let [a, b] ⊂ R be a finite interval. Then, the left-and right-sided Katugam-
pola fractional integrals of order α > 0 of F ∈ [a, b] are defined by

pIαa+F(x) :=
p1−α

Γ(α)

∫ x

a

tp−1

(xp − tp)1−α
F(t)dt

and

pIαb−F(x) :=
p1−α

Γ(α)

∫ b

x

tp−1

(tp − xp)1−α
F(t)dt

with a < x < b and p > 0, if the integrals exist.

The following definition [36] generalizes all the previously defined fractional integrals.

Definition 10. The (k − p) Riemann-Liouville fractional integral operator p
kJ

α
c of order

α > 0 for a real valued function G(t) is defined as

p
kJ

α
c G(x) =

(p+ 1)1−
α
k

kΓk(θ)

∫ x

c
[xp+1 − tp+1]

α
k
−1tpG(t)dt

where k > 0, p ∈ R, p ̸= −1.
The left and right sided (k − p) Riemann-Liouville fractional integral operators are

given by

p
kJ

α
c+G(x) =

(p+ 1)1−
α
k

kΓk(α)

∫ x

c
[xp+1 − tp+1]

α
k
−1tpG(t)dt

p
kJ

α
d−G(x) =

(p+ 1)1−
α
k

kΓk(α)

∫ d

x
[tp+1 − xp+1]

α
k
−1tpG(t)dt

Special cases

1. When p = 0 the (k − p) Riemann-Liouville fractional integral reduces to
k-Riemann-Liouville fractional integral.
2. When k=1 the (k − p) Riemann-Liouville fractional integral reduces to Katugampola
fractional integral.
3. When k = 1, p = 0 the (k − p) Riemann-Liouville fractional integral reduces to
Riemann-Liouville fractional integral.



V. Stojiljković / Eur. J. Pure Appl. Math, 16 (1) (2023), 503-522 508

Recently published paper by Stojiljković. et al [40] proved some inequalities regarding
k−p fractional operators. We state two of them for the completeness because the Theorems
in this paper generalize the results in the recently published paper.
Let J ⊂ R be an interval containing (0, 1) and let h : J → R be a non-negative function.
If F : [a, b] → R is an (h-m)-convex function, such that the Riemann–Liouville k-fractional
integral is defined, ζ ∈ (0, 1), ℜ(αk ) > 0, α ̸= 0, and in one of the cases, the following
inequality holds:

(i) a > 0, b > a, 0 < m < a
b

(ii) a > 0, b < a, 0 < m ⩽ 1

F
(
a+bm

2

)
h(12)

⩽ αΓk(α)

(Iα,k
(a)−F(mb)

(a− bm)
α
k

+
mIα,k

(b)+
F( a

m)

( a
m − b)

α
k

)

⩽
α(F(a) +mF(b))

k

∫ 1

0
t
α
k
−1h(t)dt+

α(F(a) +mF(b))

k

∫ 1

0
t
α
k
−1h(1− t)dt.

Let h : J → R be a non-negative and non-zero function. Let F : [ap, bp] → R be a (p, h)-
convex function, p > 0, ζ ∈ (0, 1). Then, the following inequality holds

F
(
[a

p+bp

2 ]
1
p

)
h(12)

⩽
2

α
k p

α
k αΓk(α)

(bp − ap)
α
k

(
p−1
k Jα

((a
p+bp

2
)
1
p )+

F(b) + p−1
k Jα

((a
p+bp

2
)
1
p )−

F(a)

)

⩽
αp

k

(
F(a) + F(b)

)(∫ 1

0
t
αp
k
−1

(
h

(
tp

2

)
+ h

(
1− tp

2

))
dt

)
.

In our analysis, we will need the integral version of the Hölder’s inequality. If
f, g ∈ C([r, s],R) and λ, α ∈ R with λ > 1 and 1

λ + 1
α = 1, then∫ b

a
|f(t)g(t)dt| ⩽

(∫ b

a
|f(t)|λdt

) 1
λ
(∫ b

a
|g(t)|αdt

) 1
α

.

2. Main results

The following Theorem generalizes the Theorem 1 from the recently published paper [40]
about k − p fractional inequalities.
Theorem 1. Let F : [ap, bp] → R. If F is (α, h − m) − p convex on [ap, bp], then the
inequality holds in one of the following cases
1.a > 0, b > a, 0 < m ⩽ a

b
2.a > 0, b < a, 0 < m ⩽ 1

F
(
[
ap +mpbp

2
]
1
p

)
⩽

h
(

1
2α

)
θΓk(θ)p

θ
k

(ap −mpbp)
θ
k

p−1
k Jθ

a−F(mb)



V. Stojiljković / Eur. J. Pure Appl. Math, 16 (1) (2023), 503-522 509

+mp θΓk(θ)h
(
2α−1
2α

)
p

θ
k

( ap

mp − bp)
θ
k

p−1
k Jθ

(b)+F
( a

m

)
⩽

((
h

(
1

2α

)
F(a) +mph

(
2α − 1

2α

)
F(b)

)∫ 1

0
h(tlp)t

θp
k
−1dt

)
· θp
k

+

((
mph

(
1

2α

)
F(b) + h

(
2α − 1

2α

)
F(a)

)∫ 1

0
h(1− tlp)t

θp
k
−1dt

)
· θp
k
.

Proof. Using the definition of a (α, h,m)-p convex function we have

F
(
(txp +m(1− t)yp)

1
p

)
⩽ h(tα)F(x) +mh(1− tα)F(y).

Setting t = 1
2 and xp = mp(1 − tp)bp + tpap, yp = (1 − tp) ap

mp + bptp in the inequality, we
get the following

F
(
[
ap + bpmp

2
]
1
p

)
⩽ h

(
1

2α

)
F
(
[mp(1− tp)bp + tpap]

1
p

)
+mph

(
2α − 1

2α

)
F
(
[(1− tp)

ap

mp
+ bptp]

1
p

)
.

Multiplying both sides by t
θp
k
−1 and integrating with respect to t from 0 to 1 we get∫ 1

0
F
(
[
ap + bpmp

2
]
1
p

)
t
θp
k
−1dt ⩽

∫ 1

0
t
θp
k
−1h

(
1

2α

)
F
(
[mp(1− tp)bp + tpap]

1
p

)
dt

+

∫ 1

0
t
θp
k
−1mph

(
2α − 1

2α

)
F
(
[(1− tp)

ap

mp
+ bptp]

1
p

)
dt

Integrating the left hand side is easy, therefore we focus on the right hand side. In the first
integral we introduce a substitution mp(1− tp)bp + (ta)p = yp. From which we get that∫ 1

0
t
θp
k
−1h

(
1

2α

)
F
(
[mp(1− tp)bp + tpap]

1
p

)
dt

=
1

(ap −mpbp)
θ
k

∫ a

mb
F(y)(yp −mpbp)

θ
k
−1yp−1dy.

Multiplying the integral with the needed constants for the k−p Riemann Liouville fractional
integral, we get the following∫ 1

0
t
θp
k
−1h

(
1

2α

)
F
(
[mp(1− tp)bp + tpap]

1
p

)
dt

= h

(
1

2α

)
kΓk(θ)

p1−
θ
k (ap −mpbp)

θ
k

p−1
k Jθ

a−F(mb).



V. Stojiljković / Eur. J. Pure Appl. Math, 16 (1) (2023), 503-522 510

Similar procedure can be applied to the second integral introducing a substitution ap

mp −
tpap

mp + bptp = yp while noting that ap

mp > bp. From which we get that∫ 1

0
t
θp
k
−1mph

(
2α − 1

2α

)
F
(
[(1− tp)

ap

mp
+ bptp]

1
p

)
dt

= h

(
2α − 1

2α

)
m

pθ
k Γk(θ)

p1−
θ
k (ap −mpbp)

θ
k

p−1
k Jθ

b+F
( a

m

)
.

Now we focus on the right hand side inequality

h

(
1

2α

)
F
(
[mp(1− tp)bp + tpap]

1
p

)
+mph

(
2α − 1

2α

)
F
(
[(1− tp)

ap

mp
+ bptp]

1
p

)
.

Using the definition of (α, h,m)-p convexity and multiplying both sides by t
θp
k
−1 and inte-

grating with respect to t from 0 to 1 we get that

h

(
1

2α

)
kΓk(θ)

p1−
θ
k (ap −mpbp)

θ
k

p−1
k Jθ

a−F(mb)

+mph

(
2α − 1

2α

)
m

pθ
k Γk(θ)

p1−
θ
k (ap −mpbp)

θ
k

p−1
k Jθ

b+F
( a

m

)
⩽

((
h

(
1

2α

)
F(a) +mph

(
2α − 1

2α

)
F(b)

)∫ 1

0
h(tlp)t

θp
k
−1dt

)
θp

k

+

(
mph

(
1

2α

)
F(b) + h

(
2α − 1

2α

)
F(a)

)∫ 1

0
h(1− tlp)t

θp
k
−1dt

)
θp

k
.

Connecting the left hand side inequality with the right hand side inequality, we get the
desired inequality.

Corollary 1. Setting p = 1 and l, α = 1 in the previously derived Theorem, we obtain
Theorem 1 from the paper [40]

F
(
a+bm

2

)
h(12)

⩽ αΓk(α)

(Iα,k
(a)−F(mb)

(a− bm)
α
k

+
mIα,k

(b)+
F( a

m)

( a
m − b)

α
k

)

⩽
α(F(a) +mF(b))

k

∫ 1

0
t
α
k
−1h(t)dt+

α(F(a) +mF(b))

k

∫ 1

0
t
α
k
−1h(1− t)dt.

Corollary 2. Setting p = 3 in the previously derived Theorem, we obtain a new inequality
of the k − p Riemann Liouville fractional type

F
(
[
a3 +m3b3

2
]
1
3

)
⩽

h
(

1
2α

)
θΓk(θ)3

θ
k

(a3 −m3b3)
θ
k

2
kJ

θ
a−F(mb)



V. Stojiljković / Eur. J. Pure Appl. Math, 16 (1) (2023), 503-522 511

+m3 θΓk(θ)h
(
2α−1
2α

)
3

θ
k

( a3

m3 − b3)
θ
k

2
kJ

θ
(b)+F

( a

m

)

⩽

((
h

(
1

2α

)
F(a) +m3h

(
2α − 1

2α

)
F(b)

)∫ 1

0
h(t3l)t

3θ
k
−1dt

)
· 3θ
k

+

((
m3h

(
1

2α

)
F(b) +m3h

(
2α − 1

2α

)
F(a)

m3

)∫ 1

0
h(1− t3l)t

3θ
k
−1dt

)
· 3θ
k
.

Theorem 2. Let F : [ap, bp] → R. If F is (α, h − m) − p convex on [ap, bp], then the
inequality holds in the following case a ⩾ 0, b > a, ab < m ⩽ 1

F
(
[
ap +mpbp

2
]
1
p

)
⩽ h

(
1

2α

)
m

pθ
k θΓk(θ)2

θ
k

p−
θ
k (mpbp − ap)

θ
k

p−1
k Jθ(

( ap

2mp+
bp

2 )
1
p

)−F
( a

m

)

+h

(
2α − 1

2α

)
2

θ
k θΓk(θ)

p−
θ
k (mpbp − ap)

θ
k

p−1
k Jθ(

(ap

2
+ bpmp

2 )
1
p

)+F(mb)

⩽
θp

k

(
h

(
1

2α

)
F(a) +mph

(
2α − 1

2α

)
F(b)

)∫ 1

0
h

((
tp

2

)l
)
t
θp
k
−1dt

+
θp

k

(
h

(
1

2α

)
mpF(b) + h

(
2α − 1

2α

)
F(a)

)∫ 1

0
h

(
1−

(
tp

2

)l
)
t
θp
k
−1dt.

Proof. Using the definition of a (α, h,m)-p convex function we have

F
(
(txp +m(1− t)yp)

1
p

)
⩽ h(tα)F(x) +mh(1− tα)F(y).

Setting t = 1
2 and xp = (at)p

2 + mp(2−tp)
2 bp, yp = (bt)p

2 + (2−tp)
2 ( a

m)p in the inequality, we get
the following

F
(
[
ap + bpmp

2
]
1
p

)
⩽ h

(
1

2α

)
F
(
(at)p

2
+

mp(2− tp)

2
bp]

1
p

)

+mph

(
2α − 1

2α

)
F
(
[
(bt)p

2
+

(2− tp)

2
(
a

m
)p]

1
p

)
.

Multiplying the inequality with t
θp
k
−1 and integrating with respect to t from 0 to 1 we get∫ 1

0
t
θp
k
−1F

(
[
ap + bpmp

2
]
1
p

)
dt ⩽

∫ 1

0
t
θp
k
−1h

(
1

2α

)
F
(
(at)p

2
+

mp(2− tp)

2
bp]

1
p

)
dt

+

∫ 1

0
t
θp
k
−1mph

(
2α − 1

2α

)
F
(
[
(bt)p

2
+

(2− tp)

2

( a

m

)p
]
1
p

)
dt.



V. Stojiljković / Eur. J. Pure Appl. Math, 16 (1) (2023), 503-522 512

Integrating the left hand side is easy. Let us focus on the right hand side. Introducing a
substitution (at)p

2 + mp(2−tp)
2 bp = zp we get the following equality∫ 1

0
t
θp
k
−1h

(
1

2α

)
F
(
(at)p

2
+

mp(2− tp)

2
bp]

1
p

)
dt

=

∫ mb

(ap

2
+mpbp

2 )
1
p
F(z)(mpbp − zp)

θ
k
−1zp−1dz · 2

θ
k

(mpbp − ap)
θ
k

.

Where we used the condition a ⩾ 0, b > a, ab < m ⩽ 1 to swap the upper and lower
boundary. Which clearly can be seen to be of the k − p Riemann Liouville fractional
integral form, therefore we obtain∫ mb

(ap

2
+mpbp

2 )
1
p
F(z)(mpbp − zp)

θ
k
−1zp−1dz

· 2
θ
k

(mpbp − ap)
θ
k

=
2

θ
k kΓk(θ)

p1−
θ
k (mpbp − ap)

θ
k

p−1
k Jθ(

(ap

2
+mpbp

2 )
1
p

)+F(mb).

Applying the similar technique while using the substitution in the second integral (bt)p

2 +
(2−tp)

2 ( a
m)p = yp we obtain the following equality∫ 1

0
t
θp
k
−1mph

(
2α − 1

2α

)
F
(
[
(bt)p

2
+

(2− tp)

2

( a

m

)p
]
1
pdt

)

=

∫ (
bp

2
+ ap

2mp

) 1
p

a
m

F(y)

(
yp − ap

mp

) θ
k
−1

yp−1dy · 1

( b
p

2 − ap

2mp )
θ
k

.

Which can be seen to be of the form of the k − p Riemann Liouville fractional integral,
therefore we obtain

∫ (
bp

2
+ ap

2mp

) 1
p

a
m

F(y)

(
yp − ap

mp

) θ
k
−1

yp−1dy · 1

( b
p

2 − ap

2mp )
θ
k

=
m

pθ
k kΓk(θ)2

θ
k

p1−
θ
k (mpbp − ap)

θ
k

p−1
k Jθ(

( ap

2mp+
bp

2 )
1
p

)−F
( a

m

)
.

Now we focus on obtaining the right hand side inequality. Using the definition of the
(α, h−m)− p convex function on the following expression, we obtain

h

(
1

2α

)
F
(
(at)p

2
+

mp(2− tp)

2
bp]

1
p

)



V. Stojiljković / Eur. J. Pure Appl. Math, 16 (1) (2023), 503-522 513

+mph

(
2α − 1

2α

)
F
(
[
(bt)p

2
+

(2− tp)

2
(
a

m
)p]

1
p

)
⩽

(
h

(
1

2α

)
F(a) +mph

(
1− 1

2α

)
F(b)

)
h

((
tp

2

)l
)

+

(
h

(
1

2α

)
mpF(b) + h

(
1− 1

2α

)
F(a)

)
h

(
1−

(
tp

2

)l
)
.

Multiplying the inequality with t
θp
k
−1 and integrating with respect to t from 0 to 1 we obtain

kΓk(θ)2
θ
k

p1−
θ
k (mpbp − ap)

θ
k

m
pθ
k
p−1
k Jθ(

( ap

2mp+
bp

2 )
1
p

)−F
( a

m

)
+ p−1

k Jθ(
(ap

2
+mpbp

2 )
1
p

)+F(mb)


⩽(

h

(
1

2α

)
F(a) +mph

(
1− 1

2α

)
F(b)

)∫ 1

0
t
θp
k
−1h

((
tp

2

)l
)
dt

+

(
h

(
1

2α

)
mpF(b) + h

(
1− 1

2α

)
F(a)

)∫ 1

0
t
θp
k
−1h

(
1−

(
tp

2

)l
)
dt.

Connecting the left and right hand side inequality and multiplying everything with the
constant from the left hand side, we obtain the desired inequality.

Corollary 3. Setting α, l,m = 1 in the previously derived inequality, we obtain Theorem
4 from the paper [40] , namely we obtain

F
(
[a

p+bp

2 ]
1
p

)
h(12)

⩽
2

α
k p

α
k αΓk(α)

(bp − ap)
α
k

(
p−1
k Jα

((a
p+bp

2
)
1
p )+

F(b) + p−1
k Jα

((a
p+bp

2
)
1
p )−

F(a)

)

⩽
αp

k

(
F(a) + F(b)

)(∫ 1

0
t
αp
k
−1

(
h

(
tp

2

)
+ h

(
1− tp

2

))
dt

)
.

Corollary 4. Setting p = 3 in the previously derived inequality, we obtain the new in-
equality of the fractional k − p Riemann Liouville type

F
(
[
a3 +m3b3

2
]
1
3

)
⩽ h

(
1

2α

)
m

3θ
k θΓk(θ)2

θ
k

3−
θ
k (m3b3 − a3)

θ
k

2
kJ

θ((
a3

2m3+
b3

2

) 1
3

)−F
( a

m

)

+h

(
2α − 1

2α

)
2

θ
k θΓk(θ)

3−
θ
k (m3b3 − a3)

θ
k

2
kJ

θ((
a3

2
+ b3m3

2

) 1
3

)+F(mb)



V. Stojiljković / Eur. J. Pure Appl. Math, 16 (1) (2023), 503-522 514

⩽
3θ

k

(
h

(
1

2α

)
F(a) +m3h

(
2α − 1

2α

)
F(b)

)∫ 1

0
h

((
t3

2

)l
)
t
3θ
k
−1dt

+
3θ

k

(
h

(
1

2α

)
m3F(b) + h

(
2α − 1

2α

)
F(a)

)∫ 1

0
h

(
1−

(
t3

2

)l
)
t
3θ
k
−1dt.

In the following we present a new Theorem of the generalized q type.

Theorem 3. Let F : [ap, bp] → R. If f is (α, h−m)− p convex on [ap, bp] and a ⩾ 0, b >
a, ab < m ⩽ 1, then the following inequality holds

F
(
[
ap +mpbp

2
]
1
p

)
⩽

h
(

1
2α

)
2

θ
k p

θ
k θΓk(θ)

(bpmp − ap)
θ
k

p−1
k Jθ(

( q
2
(ap−mpbp)+ap

2
+ bpmp

2 )
1
p

)+F
((q

2
(ap − bpmp) +mpbp

) 1
p

)

+
h
(
1− 1

2α

)
mp+ θp

k 2
θ
k p

θ
k θΓk(θ)

(bpmp − ap)
θ
k

p−1
k J(

( q
2
(bp− ap

mp )+
ap

2mp+
bp

2 )
1
p

)−F

((
q

2
(bp − ap

mp
) +

ap

mp

) 1
p

)

⩽

((
h

(
1

2α

)
F(a) + h

(
1− 1

2α

)
mpF(b)

)∫ 1

0
h

((
q + tp

2

)l
)
t
θp
k
−1dt

)
θp

k

+

((
h

(
1

2α

)
F(b)mp + h

(
1− 1

2α

)
F(a)

)∫ 1

0
h

(
1−

(
q + tp

2

)l
)
t
θp
k
−1dt

)
θp

k
.

Proof. Since F is (α, h−m)− p convex, we have the following inequality

F
(
(tap +m(1− t)bp)

1
p

)
⩽ h(tα)F(a) +mh(1− tα)F(b).

Setting t = 1
2 and xp = q+tp

2 ap +
(
1− q+tp

2

)
bpmp, yp = q+tp

2 bp +
(
1− q+tp

2

)
ap

mp in the

inequality, we obtain

F
(
[
ap +mpbp

2
]
1
p

)
⩽ h

(
1

2α

)
F
(
(
q + tp

2
ap +

(
1− q + tp

2

)
bpmp)

1
p

)

+h

(
1− 1

2α

)
mpF

(
(
q + tp

2
bp +

(
1− q + tp

2

)
ap

mp
)
1
p

)
.

Multiplying the inequality with t
θp
k
−ptp−1 and integrating with respect to t from 0 to 1 we

get ∫ 1

0
t
θp
k
−1F

(
[
ap +mpbp

2
]
1
p

)
dt ⩽



V. Stojiljković / Eur. J. Pure Appl. Math, 16 (1) (2023), 503-522 515∫ 1

0
t
θp
k
−1h

(
1

2α

)
F
(
(
q + tp

2
ap +

(
1− q + tp

2

)
bpmp)

1
p

)
dt

+

∫ 1

0
t
θp
k
−1h

(
1− 1

2α

)
mpF

(
(
q + tp

2
bp +

(
1− q + tp

2

)
ap

mp
)
1
p

)
dt.

The left hand side is easy to integrate, therefore we focus our attention to the other two

integrals. Introducing a substitution xp = q+tp

2 ap +
(
1− q+tp

2

)
bpmp while noting that

a ⩾ 0, b > a, ab < m ⩽ 1, we obtain∫ 1

0
t
θp
k
−1h

(
1

2α

)
F
(
(
q + tp

2
ap +

(
1− q + tp

2

)
bpmp)

1
p

)
dt

=

∫ ( q
2
(ap−mpbp)+bpmp)

1
p

( q
2
(ap−bpmp)+ap

2
+mpbp

2 )
1
p
F(z)(bpmp +

q

2
(ap − bpmp)− zp)

θ
k
−1zp−1dz · 2

θ
k

(bpmp − ap)
θ
k

.

Which can be seen to be of the k− p Riemann Liouville integral form, therefore we obtain
the following equality

∫ ( q
2
(ap−mpbp)+bpmp)

1
p

( q
2
(ap−bpmp)+ap

2
+mpbp

2 )
1
p
F(z)(bpmp +

q

2
(ap − bpmp)− zp)

θ
k
−1zp−1dz · 2

θ
k

(bpmp − ap)
θ
k

=
h
(

1
2α

)
2

θ
k kΓk(θ)

p1−
θ
k (bpmp − ap)

θ
k

p−1
k Jθ(

( q
2
(ap−mpbp)+ap

2
+ bpmp

2 )
1
p

)+F
((q

2
(ap − bpmp) +mpbp

) 1
p

)
.

Using the similar technique on the other integral, using the substitution

q + tp

2
bp +

(
1− q + tp

2

)
ap

mp
= zp

we get ∫ 1

0
t
θp
k
−1h

(
1− 1

2α

)
mpF

(
(
q + tp

2
bp +

(
1− q + tp

2

)
ap

mp
)
1
p

)
dt

=

∫ (
q
2

(
bp− ap

mp

)
+ ap

2mp+
bp

2

) 1
p

( q
2(bp−

ap

mp )+ ap

mp )
1
p

F(z)

(
zp − q

2
(bp − ap

mp
)− ap

mp

) θ
k

· 2
θ
km

θp
k

(mpbp − ap)
θ
k

.

Which can be seen to be of the k− p Riemann Liouville integral form, therefore we obtain

∫ (
q
2

(
bp− ap

mp

)
+ ap

2mp+
bp

2

) 1
p

( q
2(bp−

ap

mp )+ ap

mp )
1
p

F(z)

(
zp − q

2
(bp − ap

mp
)− ap

mp

) θ
k

· 2
θ
km

θp
k

(mpbp − ap)
θ
k

=
2

θ
km

pθ
k kΓk(θ)

p1−
θ
k (bpmp − ap)

θ
k

p−1
k Jθ(

( q
2
(bp− ap

mp )+
ap

2mp+
bp

2 )
1
p

)+F

((
q

2
(bp − ap

mp
) +

ap

mp

) 1
p

)
.



V. Stojiljković / Eur. J. Pure Appl. Math, 16 (1) (2023), 503-522 516

In order to obtain the right hand side inequality, we use the definition of the (α, h−m)−p
convex function and deduce the following

h

(
1

2α

)
F
(
(
q + tp

2
ap +

(
1− q + tp

2

)
bpmp)

1
p

)

+h

(
1− 1

2α

)
mpF

(
(
q + tp

2
bp +

(
1− q + tp

2

)
ap

mp
)
1
p

)
⩽

((
h

(
1

2α

)
F(a) + h

(
1− 1

2α

)
mpF(b)

)
h

((
q + tp

2

)l
)
t
θp
k
−1

)

+

((
h

(
1

2α

)
F(b)mp + h

(
1− 1

2α

)
F(a)

)
h

(
1−

(
q + tp

2

)l
)
t
θp
k
−1

)
.

Multiplying the inequality with t
θp
k
−1 and integrating with respect to t from 0 to 1, while

also multiplying everything with the constant from the left hand side, we obtain the original
inequality

F
(
[
ap +mpbp

2
]
1
p

)
⩽

h
(

1
2α

)
2

θ
k p

θ
k θΓk(θ)

(bpmp − ap)
θ
k

p−1
k Jθ(

( q
2
(ap−mpbp)+ap

2
+ bpmp

2 )
1
p

)+F
((q

2
(ap − bpmp) +mpbp

) 1
p

)

+
h
(
1− 1

2α

)
mp+ θp

k 2
θ
k p

θ
k θΓk(θ)

(bpmp − ap)
θ
k

p−1
k J(

( q
2
(bp− ap

mp )+
ap

2mp+
bp

2 )
1
p

)−F

((
q

2
(bp − ap

mp
) +

ap

mp

) 1
p

)

⩽

((
h

(
1

2α

)
F(a) + h

(
1− 1

2α

)
mpF(b)

)∫ 1

0
h

((
q + tp

2

)l
)
t
θp
k
−1dt

)
θp

k

+

((
h

(
1

2α

)
F(b)mp + h

(
1− 1

2α

)
F(a)

)∫ 1

0
h

(
1−

(
q + tp

2

)l
)
t
θp
k
−1dt

)
θp

k
.

Corollary 5. Setting p = 2 in the previously derived Theorem, we obtain the following
new inequality

F
(
[
a2 +m2b2

2
]
1
2

)
⩽

h
(

1
2α

)
2

θ
k 2

θ
k θΓk(θ)

(b2m2 − a2)
θ
k

1
kJ

θ((
q
2
(a2−m2b2)+a2

2
+ b2m2

2

) 1
2

)+F
((q

2
(a2 − b2m2) +m2b2

) 1
2

)

+
h
(
1− 1

2α

)
m2+ 2θ

k 2
θ
k 2

θ
k θΓk(θ)

(b2m2 − a2)
θ
k

1
kJ((

q
2
(b2− a2

m2 )+
a2

2m2+
b2

2

) 1
2

)−F

((
q

2
(b2 − a2

m2
) +

a2

m2

) 1
2

)



V. Stojiljković / Eur. J. Pure Appl. Math, 16 (1) (2023), 503-522 517

⩽

((
h

(
1

2α

)
F(a) + h

(
1− 1

2α

)
m2F(b)

)∫ 1

0
h

((
q + t2

2

)l
)
t
2θ
k
−1dt

)
2θ

k

+

((
h

(
1

2α

)
F(b)m2 + h

(
1− 1

2α

)
F(a)

)∫ 1

0
h

(
1−

(
q + t2

2

)l
)
t
2θ
k
−1dt

)
2θ

k
.

Corollary 6. Setting p = 7, l, α = 3 in the previously derived Theorem, we obtain a new
inequality as a consequence

F
(
[
a7 +m7b7

2
]
1
7

)
⩽

h
(

1
23

)
2

θ
k 7

θ
k θΓk(θ)

(b7m7 − a7)
θ
k

6
kJ

θ((
q
2
(a7−m7b7)+a7

2
+ b7m7

2

) 1
7

)+F
((q

2
(a7 − b7m7) +m7b7

) 1
7

)

+
h
(
1− 1

23

)
m7+ 7θ

k 2
θ
k 7

θ
k θΓk(θ)

(b7m7 − a7)
θ
k

6
kJ((

q
2
(b7− a7

m7 )+
a7

2m7+
b7

2

) 1
7

)−F

((
q

2
(b7 − a7

m7
) +

a7

m7

) 1
7

)

⩽

((
h

(
1

23

)
F(a) + h

(
1− 1

23

)
m7F(b)

)∫ 1

0
h

((
q + t7

2

)α)
t
7θ
k
−1dt

)
7θ

k

+

((
h

(
1

23

)
F(b)m7 + h

(
1− 1

23

)
F(a)

)∫ 1

0
h

(
1−

(
q + t7

2

)3
)
t
7θ
k
−1dt

)
7θ

k
.

Theorem 4. Let F : [xp, yp] → R. If f is (α, h − m) − p convex on [xp, yp] and if the
following condition holds, then
0 < m < 1, x > 0, x < y < x

m ,

kΓk(θ)

p1−
θ
k

( p−1
k Jθ

x−F(my)

(xp −mpyp)
θ
k

+

p−1
k Jθ

y−F(mx)

(yp −mpxp)
θ
k

)

⩽ (f(x) + f(y))

∫ 1

0
h(tαp)t

θp
k
−1dt+m2p

(
f
( y

mp

)
+ f

( x

mp

))∫ 1

0
h(1− tαp)t

θp
k
−1dt.

⩽ (f(x) + f(y))

(∫ 1

0
(h(tαp))ldt

) 1
l
(∫ 1

0
(t

θp
k
−1)qdt

) 1
q

+m2p

(
f
( y

mp

)
+ f

( x

mp

))(∫ 1

0
(h(1− tαp))ldt

) 1
l
(∫ 1

0
(t

θp
k
−1)qdt

) 1
q

.

Proof. Since F is (α, h−m)− p convex, we have the following inequalities

F
(
[tpxp +m2p(1− tp)

yp

mp
]
1
p

)
⩽ h(tαp)f(x) +m2ph(1− tαp)f

( y

mp

)
,



V. Stojiljković / Eur. J. Pure Appl. Math, 16 (1) (2023), 503-522 518

F
(
[tpyp +m2p(1− tp)

xp

mp
]
1
p

)
⩽ h(tαp)f(y) +m2ph(1− tαp)f

( x

mp

)
.

Adding both inequalities and multiplying with t
θp
k
−1 and integrating with respect to t we

get∫ 1

0
t
θp
k
−1F

(
[tpxp +m2p(1− tp)

yp

mp
]
1
p

)
dt+

∫ 1

0
t
θp
k
−1F

(
[tpyp +m2p(1− tp)

xp

mp
]
1
p

)
dt

⩽ (f(x) + f(y))

∫ 1

0
h(tαp)t

θp
k
−1dt+m2p

(
f
( y

mp

)
+ f

( x

mp

))∫ 1

0
h(1− tαp)t

θp
k
−1dt.

Which when identified in terms of the k − p fractional operator, we get

kΓk(θ)

p1−
θ
k

( p−1
k Jθ

x−F(my)

(xp −mpyp)
θ
k

+

p−1
k Jθ

y−F(mx)

(yp −mpxp)
θ
k

)

⩽ (f(x) + f(y))

∫ 1

0
h(tαp)t

θp
k
−1dt+m2p

(
f
( y

mp

)
+ f

( x

mp

))∫ 1

0
h(1− tαp)t

θp
k
−1dt.

Now applying the Hölders inequality on the integrals, we get

(f(x) + f(y))

∫ 1

0
h(tαp)t

θp
k
−1dt+m2p

(
f
( y

mp

)
+ f

( x

mp

))∫ 1

0
h(1− tαp)t

θp
k
−1dt

⩽ (f(x) + f(y))

(∫ 1

0
(h(tαp))ldt

) 1
l
(∫ 1

0
(t

θp
k
−1)qdt

) 1
q

+m2p

(
f
( y

mp

)
+ f

( x

mp

))(∫ 1

0
(h(1− tαp))ldt

) 1
l
(∫ 1

0
(t

θp
k
−1)qdt

) 1
q

.

Connecting the left and right hand side, we obtain the inequality.

Corollary 7. Setting l, q = 1
2 we get a new (α, h−m)− p k − p fractional inequality

kΓk(θ)

p1−
θ
k

( p−1
k Jθ

x−F(my)

(xp −mpyp)
θ
k

+

p−1
k Jθ

y−F(mx)

(yp −mpxp)
θ
k

)

⩽ (f(x) + f(y))

∫ 1

0
h(tαp)t

θp
k
−1dt+m2p

(
f
( y

mp

)
+ f

( x

mp

))∫ 1

0
h(1− tαp)t

θp
k
−1dt.

⩽ (f(x) + f(y))

(∫ 1

0
(h(tαp))

1
2dt

)2(∫ 1

0
(t

θp
k
−1)

1
2dt

)2

+m2p

(
f
( y

mp

)
+ f

( x

mp

))(∫ 1

0
(h(1− tαp))

1
2dt

)2(∫ 1

0
(t

θp
k
−1)

1
2dt

)2

.



REFERENCES 519

3. Conclusions and Outlook

In this paper new fractional variations of the Hermite-Hadamard inequality have been
obtained, as well as an application of the Hölder’s inequality in the fractional setting. In
the generalizations, k−p fractional operator has been utilized in tandem with (α, h−m)−p
convexity to produce the results. Recently reported results in the literature have been given
as corollaries. Questions arise whether further generalizations of the obtained convex-
fractional inequalities are obtainable. A possible open problem for further investigation
is whether k − p Riemann Liouville fractional operator can be paired up with Raina’s
function to produce more fractional convex inequalities. Another interesting problem is
whether interval valued analysis can be used to generalize the obtained inequalities.

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