k-fractional integral inequalities of hadamard type for

$: k-fractional integral inequalities of Hadamard type for$
Computational Methods for Differential Equationshttp://cmde.tabrizu.ac.ir

Vol. 8, No. 1, 2020, pp. 119-140

DOI:10.22034/cmde.2019.9462

k-fractional integral inequalities of Hadamard type for (h−m)−convexfunctions

Ghulam Farid∗COMSATS University Islamabad, Attock Campus,Kamra Road, Attock 43600, Pakistan.E-mail: [email protected], [email protected]

Atiq Ur RehmanCOMSATS University Islamabad, Attock Campus,Kamra Road, Attock 43600, Pakistan.E-mail: [email protected]

Qurat Ul AinCOMSATS University Islamabad, Attock Campus,Kamra Road, Attock 43600, Pakistan.E-mail: [email protected]

Abstract In this paper, we establish Hadamard type fractional integral inequalities for a more

general class of functions that is the class of (h − m)−convex functions. Theseresults are due to Riemann-Liouville (RL) k-fractional integrals: a generalization of

RL fractional integrals. Several known results are special cases of proved results.

Keywords. Riemann-Liouville fractional integrals, Riemann-Liouville k−fractional integrals, (h−m)−convex

functions.

2010 Mathematics Subject Classification. 26B15, 26A51, 34L15.

1. Introduction and preliminaries

Mathematics is an art of giving things misleading names. The beautiful and atfirst glance mysterious name, the Fractional Calculus (FC) is just one of those mis-nomers. A misnomer for the theory of operators of integration and differentiationof arbitrary fractional order and their application. In 1871, Heaviside states that,there is universe of mathematics lying between the complete differentiation and inte-gration, and fractional operators push themselves forward sometimes and are just asreal as others. In recent decades, they have been found useful in various fields: rhe-ology, quantitative biology, electro-chemistry, scattering theory, diffusion, transporttheory, probability, statistics, potential theory and elasticity. Nowadays, there exitsa great number of articles, surveys and several books, entirely devoted to FC (see,[4, 5, 6, 9, 17, 21, 22, 23]). The definition of RL fractional differentiation played animportant role in the development of FC. Here we give definition of RL fractionalintegrals.

Received: 11 September 2018 ; Accepted: 23 February 2019.∗ Corresponding author.

119

120 G. FARID, A. U. REHMAN, AND Q. U. AIN

Definition 1.1. Let f ∈ L1[a, b]. We define left and right RL fractional integralIαa+f(x) and Iαb−f(x) of order α ∈ R (α > 0) by

Iαa+f(x) :=1

Γ(α)

∫ x

a

f(t)dt

(x− t)1−α, x > a,

and

Iαb−f(x) :=1

Γ(α)

∫ b

x

f(t)dt

(t− x)1−α, x < b

respectively. Here Γ(α) is the Euler’s Gamma function and I0a+f(x) = I0b−f(x) =f(x).

Many generalizations of RL fractional integral operators have been introduced. In[20], Mubeen and Habibullah gave a slight generalization as RL k-fractional integrals:

Definition 1.2. Let f ∈ L1[a, b]. Then we define left and right k-fractional RL

integrals Iα,ka+ f(x) and Iα,kb− f(x) of order α and k > 0 by

Iα,ka+ f(x) :=1

kΓk(α)

∫ x

a

f(t)dt

(x− t)αk−1, x > a,

Iα,kb− f(x) :=1

kΓk(α)

∫ b

x

f(t)dt

(t− x)αk−1

, x < b,

where Γk(α) is the k−Gamma function and I0,1a+ f(x) = I0,1b− f(x) = f(x).

In recent years the theory of inequalities in mathematical analysis via fractionalintegral operators of different kinds has been introduced in fractional calculus (FC)(see, [1, 4, 7, 8, 9, 12, 13, 11, 14, 15, 16, 26, 27] and references in there). Inequalitieshave a significant role in the field of convex analysis, while the classical Hadamardinequality is equivalent to the definition of convex functions.

Definition 1.3. A function f : I → R, where I is an interval in R, is said to beconvex function if

f(αx+ (1− α)y) ≤ αf(x) + (1− α)f(y), x, y ∈ I, α ∈ [0, 1].

Convex functions are generalized in many ways leading to very attractive results. In[24], Ozdemir et el. introduced a generalized class of functions namely (h−m)−convexfunctions.

Definition 1.4. Let J ⊆ R be an interval containing (0, 1) and let h : J → R be anon-negative function. We say that f : [0, b] → R is a (h −m)−convex function if fis non-negative satisfying

f(αx+m(1− α)y) ≤ h(α)f(x) +mh(1− α)f(y).

and for suitable choice of h(α) and m = 1, the class of (h − m)−convex functionsreduces to the different known classes of convex functions.

CMDE Vol. 8, No. 1, 2020, pp. 119-140 121

In Section 2, we prove k−fractional integral inequality of Hadamard type for (h−m)−convex functions and deduce some related results. In Section 3, we prove a versionof k−fractional integral inequality of Hadamard type for functions like f so that |f ′| is(h−m)−convex whose derivative in absolute values are (h−m)−convex. In Section4, we prove k−fractional integral inequality of Hadamard type for product of two(h −m)−convex functions and also for the product of (α, h −m)−convex functions.Also we find connection with some well known results.

2. Hadamard type inequalities for (h−m)−convex functions via RLk−fractional integrals

In the following we give k−fractional integral inequalities of Hadamard type for(h−m)−convex functions.

Theorem 2.1. Let f : [0,∞) → R be a (h − m)−convex function with m ∈ (0, 1],f ∈ L1[a, b], a, b ∈ [0,∞) where a

m ,mb ∈ [a, b]. Then we will have

f

(bm+ a

2

)≤ h

(1

2

)Γk(α+ k)

(mb− a)αk

[Iα,ka+ f(mb) + Iα,kb− f

( am

)]≤ α

kh

(1

2

){[m2f

( a

m2

)+mf(b)

] ∫ 1

0

tαk−1h(1− t)dt

+ [mf(b) + f(a)]

∫ 1

0

tαk−1h(t)dt

}.

(2.1)

Proof. Since f is (h−m)−convex on [a, b], then

f

(um+ v

2

)≤ h

(1

2

)(mf(u) + f(v)), u, v ∈ [a, b].

Since am ,mb ∈ [a, b], then (1 − t) am + tb ≤ b and (1 − t)mb + ta ≥ a. By setting

u = (1 − t) am + tb and v = m(1 − t)b + ta in the above inequality for t ∈ [0, 1], then

by integrating over [0, 1] after multiplying with tαk−1 we have

f

(bm+ a

2

)∫ 1

0

tαk−1dt

≤ h(

1

2

)[∫ 1

0

tαk−1mf

((1− t) a

m+ tb

)dt

+

∫ 1

0

tαk−1f(m(1− t)b+ ta)dt

].

Now if we let w = (1−t) am +tb and z = m(1−t)b+ta in right side of above inequality,we get

f

(bm+ a

2

)≤

h

(1

2

)Γk(α+ k)

(mb− a)αk

[mα+1Iα,kb− f

( am

)+ Iα,ka+ f(mb)

].

(2.2)


Which completes the proof of first inequality in (2.1). On the other hand by usingthe (h−m)−convexity of f , we have

mf(

(1− t) am

+ tb)

+ f(m(1− t)b+ ta)

≤ m2h(1− t)f( a

m2

)+mh(t)f(b) +mh(1− t)f(b) + h(t)f(a).

By multiplying both sides of above inequality with αh(12

)tαk−1 and integrating over

[0, 1], after some calculations we get

h

(1

2

)Γk(α+ k)

(mb− a)αk

[mα+1Iα,kb− f

( am

)+ Iα,ka+ f(mb)

]≤ α

kh

(1

2

){[m2f

( a

m2

)+mf(b)

] ∫ 1

0

tαk−1h(1− t)dt

+ [mf(b) + f(a)]

∫ 1

0

tαk−1h(t)dt

}Which is the other side of the inequality (2.1). �

Corollary 2.2. In Theorem 2.1 with k = 1, leads to the following inequality for(h−m)−convex function via RL fractional integrals

f

(bm+ a

2

)≤ h

(1

2

)Γ(α+ 1)

(mb− a)α

[Iαa+f(mb) + Iαb−f

( am

)]≤ α

kh

(1

2

){[m2f

( a

m2

)+mf(b)

] ∫ 1

0

tα−1h(1− t)dt

+ [mf(b) + f(a)]

∫ 1

0

tα−1h(t)dt

}Which leads to in above theorem we get [26, Theorem 2] stated in following corol-

lary if we take k = 1, h(t) = t and m = 1.

Corollary 2.3. 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 we will have

f

(a+ b

2

)≤ Γ(α+ 1)

2(b− a)α[Iαa+f(b) + Iαb−f (a)] ≤ f(a) + f(b)

2.

Another k−fractional integral inequality of Hadamard type for (h − m)−convexfunction is obtained as follows.

Theorem 2.4. Let f : [0,∞) → R be a (h − m)−convex function with m ∈ (0, 1],also let f ∈ L1[a, b], a, b ∈ [0,∞). Then we will have

kΓk(α)

(b− a)αk

[Iα,ka+

f(b) + Iα,kb−

f(a)] (2.3)

≤ [f(a) + f(b)]

∫ 1

0tαk−1h(t)dt+m

[f

(b

m

)+ f

( am

)]∫ 1

0tαk−1h(1− t)dt

≤1

(αkp− p+ 1)

1p

(∫ 1

0(h(t))qdt

) 1q[f(a) + f(b) +m

(f

(b

m

)+ f

( am

))],

CMDE Vol. 8, No. 1, 2020, pp. 119-140 123

where p−1 + q−1 = 1 and p > 1.

Proof. Since f is (h−m)−convex on [a, b], then for m ∈ (0, 1] and t ∈ [0, 1], we have

f(ta+ (1− t)b) + f((1− t)a+ tb)

≤ h(t)[f(a) + f(b)] +mh(1− t)[f

(b

m

)+ f

( am

)].

from which multiplying both sides with tαk−1 and integrating over [0, 1], we will have∫ 1

0tαk−1[f(ta+ (1− t)b) + f((1− t)a+ tb)]dt

≤ [f(a) + f(b)]

∫ 1

0tαk−1h(t)dt+m

[f

(b

m

)+ f

( am

)]∫ 1

0tαk−1h(1− t)dt.

By changing variables we will have

kΓk(α)

(b− a)αk

[Iα,ka+

f(b) + Iα,kb−

f(a)] (2.4)

≤ [f(a) + f(b)]

∫ 1

0tαk−1h(t)dt+m

[f

(b

m

)+ f

( am

)]∫ 1

0tαk−1h(1− t)dt

Which completes the proof of first inequality in (2.3). The second inequality in (2.3)follows by using the Holder’s inequality∫ 1

0

tαk−1(h(t))dt ≤ 1

(αk p− p+ 1)1p

(∫ 1

0

(h(t))qdt

) 1q

.

Thus from (2.4) we will get (2.3). �

from this theorem, if we will put k = 1 and m = 1, then we get [27, Theorem 2.1],which is stated in the following corollary.

Corollary 2.5. Let f : [0,∞)→ R be h−convex function and f ∈ L1[a, b]. Then forRL fractional integrals, we have

Γ(α)

(b− a)α[Iαa+f(b) + Iα

b−f(a)]

≤ [f(a) + f(b)]

∫ 1

0tα−1[h(t) + h(1− t)]dt ≤

2 [f(a) + f(b)]

(αp− p+ 1)1p

(∫ 1

0(h(t))qdt

) 1q

.

Theorem 2.6. Let f : [0,∞) → R be a (h−m)−convex function with m ∈ (0, 1], hbe superaddtive and f ∈ L1[a, b], a, b ∈ [0,∞). Then for RL k−fractional integrals wehave

Γk(α+ k)

(b− a)αk

[Iα,ka+ f(b) + Iα,kb− f(a)]

≤ h(1)

[f (a) + f (b)

2+m

(f(am

)+ f

(bm

)2

)].

(2.5)


Proof. Since f is (h−m)−convex on [a, b], then for t ∈ [0, 1], we get

f(ta+ (1− t)b) + f((1− t)a+ tb)

≤ [h(t) + h(1− t)]

[f (a) + f (b)

2+m

(f(am

)+ f

(bm

)2

)].

Since h is superadditive, therefore

f(ta+ (1− t)b) + f((1− t)a+ tb)

≤ h(1)

[f (a) + f (b)

2+m

(f(am

)+ f

(bm

)2

)].

By multiplying both sides of above inequality with tαk−1 and integrating over [0, 1],

we will have∫ 1

0

tαk−1[f(ta+ (1− t)b) + f((1− t)a+ tb)]dt

≤ h(1)

[f (a) + f (b)

2+m

(f(am

)+ f

(bm

)2

)]∫ 1

0

tαk−1dt.

By substituting w = ta+ (1− t)b in left side of above inequality leads to (2.5). �

Corollary 2.7. In Theorem 2.6, if we take k = 1, then for (h−m)−convex functionvia RL fractional integrals we get

Γ(α+ 1)

(b− a)α[Iαa+f(b) + Iαb−f(a)]

≤ h(1)

[f (a) + f (b)

2+m

(f(am

)+ f

(bm

)2

)].

3. k−fractional integral inequality of Hadamard type for functionswhose derivatives absolute values are (h−m)−convex

In the following k−fractional integral inequalities of Hadamard type for (h −m)−convex function in terms of the first derivatives have been obtained. For nextresult we use the following lemma.

Lemma 3.1. [9] Let function f : [a, b] → R be differentiable on interval (a, b). Iff ′ ∈ L[a, b], then we will have

f(a) + f(b)

2− Γk(α+ k)

2(b− a)αk

(Iα,ka+ f(b) + Iα,kb− f(a))

=b− a

2

∫ 1

0

[(1− t)αk − tαk ]f ′(ta+ (1− t)b)dt.

Theorem 3.2. Let f : [0,∞) → R be a function such that [a, b] ⊂ [0,∞) and f ∈L1[a, b]. If |f ′| is an (h−m)−convex with m ∈ (0, 1] and hq ∈ [0, 1], q > 1. Then forRL k−fractional integrals we will have

CMDE Vol. 8, No. 1, 2020, pp. 119-140 125

∣∣∣∣∣f(a) + f(b)

2−

Γk(α+ k)

2(b− a)αk

[Iα,ka+

f(b) + Iα,kb−

f(a)]

∣∣∣∣∣ (3.1)

≤(b− a)

[|f ′(a)|+m

∣∣∣f ′ ( bm

)∣∣∣]2

[ 2αkp+1 − 1

2αkp+1(α

kp+ 1)

] 1p

−[

1

2αkp+1(α

kp+ 1)

] 1p

×

[∫ 12

0(h(t))qdt

] 1q

+

[∫ 1

12

(h(t))qdt

] 1q

,where 1

p + 1q = 1.

Proof. By using (h−m)−convexity of |f ′| and Lemma 3.1, we have∣∣∣∣f(a) + f(b)

2− Γk(α+ k)

2(b− a)αk

[Iα,ka+ f(b) + Iα,kb− f(a)]

∣∣∣∣≤ b− a

2

∫ 1

0

∣∣(1− t)αk − tαk ∣∣ |f ′(ta+ (1− t)b)| dt.

Furthermore, we have∣∣∣∣∣f(a) + f(b)

2−

Γk(α+ k)

2(b− a)αk

[Iα,ka+

f(b) + Iα,kb−

f(a)]

∣∣∣∣∣ (3.2)

≤b− a

2

∫ 12

0

[(1− t)

αk − t

αk

] [h(t)

∣∣f ′(a)∣∣+mh(1− t)

∣∣∣∣f ′ ( b

m

)∣∣∣∣] dt+

∫ 1

12

[(1− t)

αk − t

αk

] [h(t)

∣∣f ′(a)∣∣+mh(1− t)

∣∣∣∣f ′ ( b

m

)∣∣∣∣] dt=b− a

2

{∣∣f ′(a)∣∣ ∫ 1

2

0(1− t)

αk h(t)dt−

∣∣f ′(a)∣∣ ∫ 1

2

0tαk h(t)dt

+m

∣∣∣∣f ′ ( b

m

)∣∣∣∣ ∫ 12

0(1− t)

αk h(1− t)dt−m

∣∣∣∣f ′ ( b

m

)∣∣∣∣ ∫ 12

0tαk h(1− t)dt

+∣∣f ′(a)

∣∣ ∫ 1

12

tαk h(t)dt−

∣∣f ′(a)∣∣ ∫ 1

12

(1− t)αk h(t)dt

+m

∣∣∣∣f ′ ( b

m

)∣∣∣∣ ∫ 1

12

tαk h(1− t)dt−m

∣∣∣∣f ′ ( b

m

)∣∣∣∣ ∫ 1

12

(1− t)αk h(1− t)dt

}.

Now, by using the Holder’s inequality, we have

∫ 12

0

(1− t)αk h(t)dt =

∫ 1

12

tαk h(1− t)dt

≤[

2αk p+1 − 1

2αk p+1(αk p+ 1)

] 1p

[∫ 12

0

[h(t)]qdt

] 1q

,


∫ 12

0

(1− t)αk h(1− t)dt =

∫ 1

12

tαk h(t)dt

≤[

2αk p+1 − 1

2αk p+1(αp+ 1)

] 1p

[∫ 1

12

[h(t)]qdt

] 1q

,

∫ 12

0

tαk h(1− t)dt =

∫ 1

12

(1− t)αk h(t)dt

≤[

1

2αk p+1(αk p+ 1)

] 1p

[∫ 1

12

[h(t)]qdt

] 1q

,

and

∫ 12

0

tαk h(t)dt =

∫ 1

12

(1− t)αk h(1− t)dt

≤[

1

2αk p+1(αk p+ 1)

] 1p

[∫ 12

0

[h(t)]qdt

] 1q

.

By using the above inequalities in the right hand side of (3.2), we get

∣∣∣∣∣f(a) + f(b)

2−

Γk(α+ k)

2(b− a)αk

[Iα,ka+

f(b) + Iα,kb−

f(a)]

∣∣∣∣∣≤b− a

2

∣∣f ′(a)∣∣ [ 2

αkp+1 − 1

2αkp+1(α

kp+ 1)

] 1p

−[

1

2αkp+1(α

kp+ 1)

] 1p

(∫ 12

0[h(t)]qdt

) 1q

+

[ 2αkp+1 − 1

2αkp+1(α

kp+ 1)

] 1p

−[

1

2αkp+1(α

kp+ 1)

] 1p

(∫ 1

12

[h(t)]qdt

) 1q

+m

∣∣∣∣f ′ ( b

m

)∣∣∣∣[ 2

αkp+1 − 1

2αkp+1(α

kp+ 1)

] 1p

−[

1

2αkp+1(α

kp+ 1)

] 1p

(∫ 1

12

[h(t)]qdt

) 1q

+

[ 2αkp+1 − 1

2αkp+1(α

kp+ 1)

] 1p

−[

1

2αkp+1(α

kp+ 1)

] 1p

(∫ 12

0[h(t)]qdt

) 1q

.

After a little computation one can get inequality (3.1). �

CMDE Vol. 8, No. 1, 2020, pp. 119-140 127

Corollary 3.3. Theorem 3.2 with k = 1 gives the result for (h-m)-convex functionsvia RL fractional integrals, we will have∣∣∣∣f(a) + f(b)

2−

Γ(α+ 1)

2(b− a)α[Iαa+f(b) + Iα

b−f(a)]

∣∣∣∣≤

(b− a)[|f ′(a)|+m

∣∣∣f ′ ( bm

)∣∣∣]2

[ 2αp+1 − 1

2αp+1(αp+ 1)

] 1p

−[

1

2αp+1(αp+ 1)

] 1p

×[∫ 1

2

0(h(t))qdt

] 1q

+

[∫ 1

12

(h(t))qdt

] 1q

.Corollary 3.4. If we put m = 1 and k = 1 in Theorem 3.2, then we have thefollowing inequality∣∣∣∣f(a) + f(b)

2−

Γ(α)

2(b− a)α[Iαa+f(b) + Iα

b−f(a)]

∣∣∣∣≤

(b− a) [|f ′(a)|+ |f ′(b)|]2

[ 2αp+1 − 1

2αp+1(αp+ 1)

] 1p

−[

1

2αp+1(αp+ 1)

] 1p

×

(∫ 12

0[h(t)]qdt

) 1q

+

(∫ 1

12

[h(t)]qdt

) 1q

.which is given in [27, Theorem 2.6]. Before, we prove next theorem, we need the

following lemma.

Lemma 3.5. Let f : [a,mb]→ R be a differentiable mapping on interval (a,mb) witha < mb. If f ′ ∈ L1[a,mb], then for k−fractional integrals we will have

f(mb) + f(a)

2−kΓk(αk + k)

2(mb− a)αk

(Iα,ka+ f(mb) + Iα,kmb−f(a))

=mb− a

2

∫ 1

0

[(1− t)αk − tαk ]f ′(m(1− t)b+ ta)dt.

Proof. Consider the right hand side

mb− a2

∫ 1

0

[(1− t)αk − tαk ]f ′(m(1− t)b+ ta)

=mb− a

2

[∫ 1

0

(1− t)αk f ′(m(1− t)b+ ta)dt

−∫ 1

0

tαk f ′(m(1− t)b+ ta)dt

].

One has

mb− a2

∫ 1

0

(1− t)αk f ′(m(1− t)b+ ta)dt

=f(mb)

2−kΓk(αk + k)

2(mb− a)αkIα,kmb−f(a),


and

mb− a2

∫ 1

0

(t)αk f ′(m(1− t)b+ ta)dt

=f(b)

2−kΓk(αk + k)

2(mb− a)αkIα,ka+ f(mb).

Hence the required equality can be established. �

Theorem 3.6. Let f : [a,mb] → R be a function such that [a,mb] ⊆ [0,∞) andf ∈ L1[a,mb]. If |f ′| is an (h − m)−convex with m ∈ (0, 1] and hq ∈ [0, 1], q > 1.Then for RL k−fractional integrals we will have

∣∣∣∣∣f(mb) + f(a)

2−kΓk(α

k+ k)

2(mb− a)αk

(Iα,ka+

f(mb) + Iα,kmb−

f(a))

∣∣∣∣∣ (3.3)

≤(mb− a) [|f ′(a)|+m |f ′ (b)|]

2

[ 2αkp+1 − 1

2αkp+1(α

kp+ 1)

] 1p

−[

1

2αkp+1(α

kp+ 1)

] 1p

×

(∫ 12

0(h(t))qdt

) 1q

+

(∫ 1

12

(h(t))qdt

) 1q

,

where 1p + 1

q = 1.

Proof. By using the property of modulus from Lemma 3.5, we will get

∣∣∣∣f(mb) + f(a)

2−kΓk(αk + k)

2(mb− a)αk

(Iα,kb+ f(mb) + Iα,kmb−f(a))

∣∣∣∣≤ mb− a

2

∫ 1

0

∣∣(1− t)αk − tαk ∣∣ |f ′(m(1− t)b+ ta)| dt.

CMDE Vol. 8, No. 1, 2020, pp. 119-140 129

By (h−m)−convexity of |f ′|, we will have

∣∣∣∣f(mb) + f(a)

2−kΓk(αk + k)

2(mb− a)αk

(Iα,ka+ f(mb) + Iα,kmb−f(a))

∣∣∣∣ (3.4)

≤ mb− a2

∫ 12

0

[(1− t)αk − tαk

][mh(1− t) |f ′(b)|+ h(t) |f ′ (a)|] dt

+

∫ 1

12

[(1− t)αk − tαk

][mh(1− t) |f ′(b)|+ h(t) |f ′ (a)|] dt

=mb− a

2

{|f ′(a)|

[∫ 12

0

(1− t)αk h(1− t)dt−∫ 1

2

0

tαk h(1− t)dt

]

+m |f ′ (b)|

[∫ 12

0

(1− t)αk h(t)dt−∫ 1

2

0

tαk h(t)dt

]

+ |f ′(a)|

[∫ 1

12

tαk h(1− t)dt−

∫ 1

12

(1− t)αk h(1− t)dt

]

+m |f ′ (b)|

[∫ 1

12

tαk h(1− t)dt−

∫ 1

12

(1− t)αk h(1− t)dt

]}.

Now, by using the Holder’s inequality in the right hand side of (3.4), we will get

∣∣∣∣∣f(mb) + f(a)

2−kΓk(α

k+ k)

2(mb− a)αk

[Iα,ka+

f(mb) + Iα,kmb−

f(a)]

∣∣∣∣∣≤mb− a

2

∣∣f ′(a)∣∣ [ 2

αkp+1 − 1

2αkp+1(α

kp+ 1)

] 1p

−[

1

2αkp+1(α

kp+ 1)

] 1p

(∫ 12

0[h(t)]qdt

) 1q

+

[ 2αkp+1 − 1

2αkp+1(α

kp+ 1)

] 1p

−[

1

2αkp+1(α

kp+ 1)

] 1p

(∫ 1

12

[h(t)]qdt

) 1q

+m

∣∣f ′ (b)∣∣[ 2

αkp+1 − 1

2αkp+1(α

kp+ 1)

] 1p

−[

1

2αkp+1(α

kp+ 1)

] 1p

(∫ 1

12

[h(t)]qdt

) 1q

+

[ 2αkp+1 − 1

2αkp+1(α

kp+ 1)

] 1p

−[

1

2αkp+1(α

kp+ 1)

] 1p

(∫ 12

0[h(t)]qdt

) 1q

.

By some manipulation one can get inequality (3.3). �


Corollary 3.7. In Theorem 3.6 if we take k = 1, we get the following inequality for(h−m)−convex function via RL fractional integrals∣∣∣∣f(mb) + f(a)

2−

Γ(α+ 1)

2(mb− a)α(Iαa+f(mb) + Iα

mb−f(a))

∣∣∣∣≤

(mb− a) [|f ′(a)|+m |f ′ (b)|]2

[ 2αp+1 − 1

2αp+1(αp+ 1)

] 1p

−[

1

2αp+1(αp+ 1)

] 1p

×

(∫ 12

0[h(t)]qdt

) 1q

+

(∫ 1

12

[h(t)]qdt

) 1q

,

4. RL k-fractional integral inequalities of Hadamard type forproduct of two (h−m)−convex functions

Now, we obtain some Hadamard type inequalities for products of two (h−m)−convexfunctions via RL k−fractional integrals.

Theorem 4.1. Let f, g : [0,∞) → R be functions such that fg ∈ L1[a, b], a, b ∈[0,∞), a < b. If function f is (h1 −m)−convex and function g is (h2 −m)−convexon [0,∞) with m ∈ (0, 1], then for RL k−fractional integrals we have

k

αh1(12

)h2(12

)f (a+ b

2

)g

(a+ b

2

)(4.1)

− Γk(α)

(mb− a)αk

[Iα,ka+ f(mb)g(mb) +mα+1Iα,kb− f

( am

)g( am

)]≤ m

[f(a)g

( a

m2

)+ f (b) g (b)

] ∫ 1

0

tαk−1h1(t)h2(1− t)dt

+m2[f (b) g

( a

m2

)+ f

( a

m2

)g (b)

] ∫ 1

0

tαk−1h1(1− t)h2(1− t)dt

+ [f (b) g(a) + f(a)g (b)]

∫ 1

0

tαk−1h1(t)h2(t)dt

+m[f (b) g (b) + f

( a

m2

)g(a)

] ∫ 1

0

tαk−1h2(t)h1(1− t)dt.

Proof. We can write

f

(a+mb

2

)g

(a+mb

2

)= f

(at+m(1− t)b

2+

(1− t)a+mtb

2

)× g

(at+m(1− t)b

2+

(1− t)a+mtb

2

).

CMDE Vol. 8, No. 1, 2020, pp. 119-140 131

From (h1 −m)−convexity of f and (h2 −m)−convexity of g, we will have

f

(a+mb

2

)g

(a+mb

2

)≤ h1

(1

2

)[f (at+m(1− t)b) +mf

((1− t) a

m+ tb

)]×h2

(1

2

)[g (at+m(1− t)b) +mg

((1− t) a

m+ tb

)]

=h1

(1

2

)h2

(1

2

){f (at+m(1− t)b) g (at+m(1− t)b)

+m2f(

(1− t) am

+ tb)g(

(1− t) am

+ tb)

+mf (at+m(1− t)b) g(

(1− t) am

+ tb)

+ mf(

(1− t) am

+ tb)g (at+m(1− t)b)

}.

Thus we obtain

f

(a+mb

2

)g

(a+mb

2

)≤ h1

(1

2

)h2

(1

2

){f (at+m(1− t)b) g (at+m(1− t)b)

+m2f(

(1− t) am

+ tb)g(

(1− t) am

+ tb)

+m2h1(t)h2(1− t)[f(a)g

( a

m2

)+ f

(b

m

)g

(b

m

)]+m2h1(1− t)h2(1− t)

[f

(b

m

)g( a

m2

)+ f

( a

m2

)g( am

)]+ h1(t)h2(t)

[f

(b

m

)g(a) + f(a)g

(b

m

)]+ mh2(t)h1(1− t)

[f

(b

m

)g

(b

m

)+ f

( a

m2

)g(a)

]}.



we have

1

h1(12

)h2(12

)f (a+ b

2

)g

(a+ b

2

)∫ 1

0

tαk−1dt

− Γk(α)

(mb− a)αk

[Iα,ka+ f(mb)g(mb) +mα+1Iα,kb− f

( am

)g( am

)]≤ m

[f(a)g

( a

m2

)+ f (b) g (b)

] ∫ 1

0

tαk−1h1(t)h2(1− t)dt

+m2[f (b) g

( a

m2

)+ f

( a

m2

)g (b)

] ∫ 1

0

tαk−1h1(1− t)h2(1− t)dt

+ [f (b) g(a) + f(a)g (b)]

∫ 1

0

tαk−1h1(t)h2(t)dt

+m[f (b) g (b) + f

( a

m2

)g(a)

] ∫ 1

0

tαk−1h2(t)h1(1− t)dt.

After a little computation one can have the required result. �

Corollary 4.2. For k = 1, theorem 4.1 gives the following inequality for (h −m)−convex function via RL fractional integrals

1

αh1(12

)h2(12

)f (a+ b

2

)g

(a+ b

2

)− Γ(α)

(mb− a)α

[Iαa+f(mb)g(mb) +mα+1Iαb−f

( am

)g( am

)]≤ m

[f(a)g

( a

m2

)+ f (b) g (b)

] ∫ 1

0

tα−1h1(t)h2(1− t)dt

+m2[f (b) g

( a

m2

)+ f

( a

m2

)g (b)

] ∫ 1

0

tα−1h1(1− t)h2(1− t)dt

+ [f (b) g(a) + f(a)g (b)]

∫ 1

0

tα−1h1(t)h2(t)dt

+m[f (b) g (b) + f

( a

m2

)g(a)

] ∫ 1

0

tα−1h2(t)h1(1− t)dt.

Theorem 4.3. Let f, g : [0,∞) → R be functions such that fg ∈ L1[a, b], a, b ∈[0,∞), a < b. If function f is (h −m1)−convex and function g is (h −m2)−convexon [0,∞) with m1,m2 ∈ (0, 1], then the following inequalities hold for RL k−fractionalintegrals

CMDE Vol. 8, No. 1, 2020, pp. 119-140 133

kΓk(α)

(b− a)αk

[Iα,ka+

f(b)g(b) + Iα,kb−

f(a)g(a)]

(4.2)

≤ [f(a)g(a) + f(b)g(b)]

∫ 1

0tαk−1h2(t)dt

+m1m2

[f

(b

m1

)g

(b

m2

)+ f

(a

m1

)g

(a

m2

)]∫ 1

0tαk−1h2(1− t)dt

+

{m2

[f(a)g

(b

m2

)+ f(b)g

(a

m2

)]+m1

[g(a)f

(b

m1

)g(b)f

(a

m1

)]}×∫ 1

0tαk−1h(t)h(1− t)dt

≤1

(αkp− p+ 1)

1p

(∫ 1

0h2q(t)dt

) 1q[f(a)g(a) +m1m2f

(b

m1

)g

(b

m2

)

+f(b)g(b) +m1m2f

(a

m1

)g

(a

m2

)]+

(∫ 1

0(h(t)h(1− t))qdt

) 1q

×[m2f(a)g

(b

m2

)+m1g(a)f

(b

m1

)+m2f(b)g

(a

m2

)+m1g(b)f

(a

m1

)]},

where p > 1 and 1p + 1

q = 1.

Proof. Since function f is (h−m1)−convex and function g is (h−m2)−convex, thenfor t ∈ [0, 1], we have

f(ta+ (1− t)b)g(ta+ (1− t)b)

≤ h2(t)f(a)g(a) +m2f(a)g

(b

m2

)h(t)h(1− t)

+m1g(a)f

(b

m1

)h(t)h(1− t) +m1m2f

(b

m1

)g

(b

m2

)h2(1− t).


we have

∫ 1

0

tαk−1f(ta+ (1− t)b)g(ta+ (1− t)b)dt

≤ f(a)g(a)

∫ 1

0

tαk−1h2(t)dt+m2f(a)g

(b

m2

)∫ 1

0

tαk−1h(t)h(1− t)dt

+m1g(a)f

(b

m1

)∫ 1

0


+m1m2f

(b

m1

)g

(b

m2

)∫ 1

0

tαk−1h2(1− t)dt.


By substituting z = ta+ (1− t)b in left side of above inequality, we get

kΓk(α)

(b− a)αkIα,ka+ f(b)g(b)

= f(a)g(a)

∫ 1

0

tαk−1h2(t)dt+m2f(a)g

(b

m2

)∫ 1

0


+m1g(a)f

(b

m1

)∫ 1

0


+m1m2f

(b

m1

)g

(b

m2

)∫ 1

0

tαk−1h2(1− t)dt.

By using the Holder’s inequality, we will have

∫ 1

0

tαk−1h2(t)dt ≤ 1

(αk p− p+ 1)1p

(∫ 1

0

h2q(t)dt

) 1q

,

and similarly

∫ 1

0

tαk−1h(t)h(1− t)dt =

1

(αk p− p+ 1)1p

(∫ 1

0

(h(t)h(1− t))qdt) 1q

.

Thus we will get

kΓk(α)

(b− a)αk

Iα,ka+

f(b)g(b) (4.3)

≤ f(a)g(a)

∫ 1

0tαk−1h2(t)dt+m2f(a)g

(b

m2

)∫ 1


+m1g(a)f

(b

m1

)∫ 1


+m1m2f

(b

m1

)g

(b

m2

)∫ 1


≤1

(αkp− p+ 1)

1p

(∫ 1

0h2q(t)dt

) 1q[f(a)g(a) +m1m2f

(b

m1

)g

(b

m2

)]

+

(∫ 1

0(h(t)h(1− t))qdt

) 1q[m2f(a)g

(b

m2

)+m1g(a)f

(b

m1

)] .

CMDE Vol. 8, No. 1, 2020, pp. 119-140 135

Similarly, by changing the roles of a and b, after a little computation one can get

kΓk(α)

(b− a)αk

Iα,ka+

f(a)g(a) (4.4)

≤ f(b)g(b)

∫ 1

0tαk−1h2(t)dt+m2f(b)g

(a

m2

)∫ 1


+m1g(b)f

(a

m1

)∫ 1


+m1m2f

(a

m1

)g

(a

m2

)∫ 1


≤1

(αkp− p+ 1)

1p

(∫ 1

0h2q(t)dt

) 1q[f(b)g(b) +m1m2f

(a

m1

)g

(a

m2

)]

+

(∫ 1

0(h(t)h(1− t))qdt

) 1q[m2f(b)g

(a

m2

)+m1g(b)f

(a

m1

)] .

Adding (4.3) and(4.4), we get the required result. �

For k = 1, h(t) = t this theorem gives [25, Theorem 8], which is stated in thefollowing corollary.

Corollary 4.4. Let f, g : [0,∞) → [0,∞), [a, b] ⊆ [0,∞) be functions such thatfg ∈ L1[a, b]. If f is m1−convex and g is m2−convex on [a, b] with m1,m2 ∈ (0, 1],then one has

Γ(α)

(b− a)αIαa+f(b)g(b) ≤ f(a)g(a)

α+ 2+

m2

(α+ 1)(α+ 2)f(a)g

(b

m2

)(4.5)

+m1

(α+ 1)(α+ 2)g(a)f

(b

m1

)+

2m1m2

α(α+ 1)(α+ 2)f

(b

m1

)g

(b

m2

),

and

Γ(α)

(b− a)αIαb−f(a)g(a) ≤ f(b)g(b)

α+ 2+

m2

(α+ 1)(α+ 2)f(b)g

(a

m2

)(4.6)

+m1

(α+ 1)(α+ 2)g(b)f

(a

m1

)+

2m1m2

α(α+ 1)(α+ 2)f

(a

m1

)g

(a

m2

).

Proof. Taking (4.3) for k = 1 and h(t) = t, we get (4.5). Similarly using k = 1 andh(t) = t in (4.4), we get (4.6). �

In the following we give Hadamard type inequality for the product of two (α, h−m)−convex via RL k−fractional integral.The following definition is needed

Definition 4.5. Let J ⊆ R be an interval containing (0, 1) and let h : J → R be anon-negative function. We say that f : [0, b]→ R is a (α, h−m)−convex function, iff is non-negative and for all x, y ∈ [0, b], (α,m) ∈ [0, 1]2 and t ∈ (0, 1), one has

f(tx+m(1− t)y) ≤ h(tα)f(x) +mh(1− tα)f(y).

For suitable choice of α, h(t) and m = 1, class of (α, h−m)−convex functions reducesto the different known classes of convex functions.


Theorem 4.6. Let f, g : [0,∞) → R be functions such that fg ∈ L1[a, b], a, b ∈[0,∞). If f is (α1, h − m1)−convex and function g is (α2, h − m2)−convex on[0,∞) with (α1,m1) and (α2,m2) ∈ (0, 1]2, then the following inequalities hold forRL k−fractional integrals

kΓk(α)

(b− a)αk

[Iα,ka+ f(b)g(b) + Iα,kb− f(a)g(a)

](4.7)

≤ [f(a)g(a) + f(b)g(b)]

∫ 1

0

tαk−1h(tα1)h(tα2)dt

+m1m2

[f

(b

m1

)g

(b

m2

)+ f

(a

m1

)g

(a

m2

)]×∫ 1

0

tαk−1h(1− tα1)h(1− tα2)dt+

{m2

[f(a)g

(b

m2

)+ f(b)g

(a

m2

)]+m1

[g(a)f

(b

m1

)+ g(b)f

(a

m1

)]}∫ 1

0

tαk−1h(tα2)h(1− tα1)dt.

Proof. Since function f is (α1, h−m1)−convex and function g is (α2, h−m2)−convex,therefore for t ∈ [0, 1], we have

f(ta+ (1− t)b)g(ta+ (1− t)b)

≤ h(tα1)h(tα2)f(a)g(a) +m2f(a)g

(b

m2

)h(tα1)h(1− tα2)

+m1g(a)f

(b

m1

)h(tα2)h(1− tα1)

+m1m2f

(b

m1

)g

(b

m2

)h(1− tα1)h(1− tα2).

By multiplying both sides of the above inequality with tαk−1 and integrating over

[0, 1], we will have

∫ 1

0

tαk−1f(ta+ (1− t)b)g(ta+ (1− t)b)dt

≤ f(a)g(a)

∫ 1

0

tαk−1h(tα1)h(tα2)dt+m2f(a)g

(b

m2

)∫ 1

0

tαk−1h(tα1)h(1− tα2)dt+m1g(a)f

(b

m1

)∫ 1

0

tαk−1h(tα2)h(1− tα1)dt

+m1m2f

(b

m1

)g

(b

m2

)∫ 1

0

tαk−1h(1− tα1)h(1− tα2)dt.

CMDE Vol. 8, No. 1, 2020, pp. 119-140 137

By substituting g = ta+ (1− t)b in left side of above inequality we will get

kΓk(α)

(b− a)αkIα,ka+ f(b)g(b) (4.8)

≤ f(a)g(a)

∫ 1

0


+m2f(a)g

(b

m2

)∫ 1

0


+m1g(a)f

(b

m1

)∫ 1

0


+m1m2f

(b

m1

)g

(b

m2

)∫ 1

0

tαk−1h(1− tα1)h(1− tα2)dt,

and similarly, changing the roles of a and b, after a little computation one can get

kΓk(α)

(b− a)αkIα,ka+ f(a)g(a) (4.9)

≤ f(b)g(b)

∫ 1

0


+m2f(b)g

(a

m2

)∫ 1

0


+m1g(b)f

(a

m1

)∫ 1

0


+m1m2f

(a

m1

)g

(a

m2

)∫ 1

0

tαk−1h(1− tα1)h(1− tα2)dt.

Adding (4.8) and(4.9), we get the required result. �

If we put k = 1, h(t) = t this theorem gives [25, Theorem 12], which is stated inthe following corollary.

Corollary 4.7. Let f, g : [0,∞)→ [0,∞), be functions such that fg ∈ L1[a, b], a, b ∈[0,∞), a < b. If f is (α1,m1)−convex and g is (α2,m2) − convex on [a, b] with


(α1,m1), (α2,m2) ∈ (0, 1]2 respectively, then we will have

Γ(α)

(b− a)αIαa+f(b)g(b) (4.10)

≤ 1

α1 + α2 + αf(a)g(a) +

α1

(α+ α1)(α1 + α2 + α)m2f(a)g

(b

m2

)+

α1

(α+ α1)(α1 + α2 + α)m1g(a)f

(b

m1

)+

(1

α− 1

α+ α1− 1

α+ α2+

1

α1 + α2 + α

)m1m2f

(b

m1

)g

(b

m2

),

and

Γ(α)

(b− a)αIαb−f(a)g(a) (4.11)

≤ 1

α1 + α2 + αf(b)g(b) +

α1

(α+ α1)(α1 + α2 + α)m2f(b)g

(b

m2

)+

α1

(α+ α1)(α1 + α2 + α)m1g(b)f

(b

m1

)+

(1

α− 1

α+ α1− 1

α+ α2+

1

α1 + α2 + α

)m1m2f

(a

m1

)g

(a

m2

).

Proof. From (4.8) for k = 1 and h(t) = t, we get (4.10). Similarly, using k = 1 andh(t) = t in (4.9), we get (4.11). �

Conclusion

In this study some of the general versions of Hadamard inequality are analyzedin fractional calculus. A generalization of convex functions; namely (h−m)−convexfunction is used to establish these results. Some identities have been established whichare further utilized in the formation of Hadamard type inequalities. Furthermore,Hadamard type inequalities for product of two (h−m)−convex functions have beenstudied and connection with already published results is investigated.

Acknowledgment

The research work of first and second authors is supported by Higher EducationCommission of Pakistan under NRPU Project 5421 and 7962 respectively.

CMDE Vol. 8, No. 1, 2020, pp. 119-140 139

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