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EXTRACTA

MATHEMATICAE

Volumen 33, Número 2, 2018

instituto de investigación de matemáticas de la

universidad de extremadura

EXTRACTA MATHEMATICAE

Vol. 34, Num. 2 (2019), 285 – 301

doi:10.17398/2605-5686.34.2.285

Available online July 2, 2019

Fractional Ostrowski type inequalities for
functions whose derivatives are s-preinvex

B. Meftah 1,2, M. Merad 2, A. Souahi 3

1 Laboratoire des télécommunications, Faculté des Sciences et de la Technologie
University of 8 May 1945 Guelma, P.O. Box 401, 24000 Guelma, Algeria

2 Département des Mathématiques, Faculté des mathématiques, de l’informatique et des
sciences de la matière, Université 8 mai 1945 Guelma, Algeria

3 Laboratory of Advanced Materials, University Badji Mokhtar Annaba, Algeria

badrimeftah@yahoo.fr , mrad.meriem@gmail.com , arsouahi@yahoo.fr

Received January 16, 2019 Presented by Manuel Maestre
Accepted May 25, 2019

Abstract: In this paper, we establish a new integral identity, and then we derive some new fractional

Ostrowski type inequalities for functions whose derivatives are s-preinvex.

Key words: integral inequality, -preinvex functions, Hölder inequality, power mean inequality.

AMS Subject Class. (2010): 26D15, 26D20, 26A51.

1. Introduction

In 1938, A.M. Ostrowski proved an interesting integral inequality, esti-
mating the absolute value of the derivative of a differentiable function by its
integral mean as follows

Theorem 1. ([21]) Let I ⊆ R be an interval. Let f : I → R, be a
differentiable mapping in the interior I◦of I, and a,b ∈ I◦ with a < b. If
|f ′(x)| ≤ M for all x ∈ [a,b], then∣∣∣∣f(x) − 1b−a

∫ b
a
f(t)dt

∣∣∣∣ ≤ M (b−a)
[

1
4

+
(x−a+b2 )

2

(b−a)2

]
. (1.1)

The above inequality has attracted many researchers, various generaliza-
tions, refinements, extensions and variants have appeared in the literature see
for instance [6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 19, 20, 24, 25], and references
therein.

ISSN: 0213-8743 (print), 2605-5686 (online)

https://doi.org/10.17398/2605-5686.34.2.285
mailto:badrimeftah@yahoo.fr
mailto:mrad.meriem@gmail.com
mailto:arsouahi@yahoo.fr
https://www.eweb.unex.es/eweb/extracta/
https://creativecommons.org/licenses/by-nc/3.0/


286 b. meftah, m. merad, a. souahi

Set [25], gave the following fractional Ostrwski inequality for differentiable
s-convex functions∣∣∣((x−a)α+(b−x)αb−a )f (x) − Γ(α+1)b−a ((Iαx−f) (a) + (Iαx+f) (b))∣∣∣

≤ M
b−a

(
1 +

Γ(α+1)Γ(s+1)
Γ(α+s+1)

)(
(x−a)α+1+(b−x)α+1

α+s+1

)
and ∣∣∣((x−a)α+(b−x)αb−a )f (x) − Γ(α+1)b−a ((Iαx−f) (a) + (Iαx+f) (b))∣∣∣

≤ M
(1+pα)

1
p

(
2
s+1

)1
q
(

(x−a)α+(b−x)α
b−a

)
.

Kirmaci et al. [4], presented some results connected with inequality (1.1)∣∣∣∣ 1b−a
∫ b
a
f(x)dx−f

(
a+b

2

)∣∣∣∣ ≤ b−a8 (∣∣f ′ (a)∣∣ + ∣∣f ′ (b)∣∣).
Zhu et al. [27], established the following fractional midpoint inequality∣∣∣ Γ(α+1)2(b−a)α[ (Jαa+f) (b) + (Jαb−f) (a) ]−f (a+b2 )∣∣∣

≤ b−a
4(1+α)

(∣∣f ′ (a)∣∣ + ∣∣f ′ (b)∣∣)(α + 3 − 1
2α−1

)
.

Motivated by the above results, in this paper, we establish a new integral
identity, and then we derive some new fractional Ostrowski type inequalities
for functions whose derivatives are s-preinvex.

2. Preliminaries

In this sections we recall some definitions and lemmas

Definition 1. ([23]) A function f : I → R is said to be convex, if

f (tx + (1 − t) y) ≤ tf (x) + (1 − t) f(y)

holds for all x,y ∈ I and all t ∈ [0, 1].

Definition 2. ([1]) A nonnegative function f : I ⊂ [0,∞) → R is said
to be s-convex in the second sense for some fixed s ∈ (0, 1], if

f(tx + (1 − t)y) ≤ tsf(x) + (1 − t)sf(y)

holds for all x,y ∈ I and t ∈ [0, 1].



fractional ostrowski type inequalities 287

Definition 3. ([18]) A set K ⊆ Rn is said to be an invex with respect
to the bifunction η : K ×K → Rn, if for all x,y ∈ K, we have

x + tη (y,x) ∈ K.

In what follows we assume that K ⊆ R be an invex set with respect to the
bifunction η : K ×K → R.

Definition 4. ([26]) A function f : K → R is said to be preinvex with
respect to η, if

f (x + tη (y,x)) ≤ (1 − t) f (x) + tf(y)
holds for all x,y ∈ K and all t ∈ [0, 1].

Definition 5. ([5]) A nonnegative function f : K ⊂ [0,∞) → R is said
to be s-preinvex in the second sense with respect to η for some fixed s ∈ (0, 1],
if

f (x + tη (y,x)) ≤ (1 − t)sf(x) + tsf(y)
holds for all x,y ∈ K and t ∈ [0, 1].

Definition 6. ([2, 3, 17]) Let f ∈ L1[a,b]. The Riemann-Liouville frac-
tional integrals Iα

a+
f and Iα

b−
f of order α > 0 with a ≥ 0 are defined by

Iαa+f(x) =
1

Γ (α)

∫ x
a

(x− t)α−1 f(t)dt, x > a,

Iαb−f(x) =
1

Γ (α)

∫ b
x

(t−x)α−1 f(t)dt, b > x,

respectively, where Γ(α) =
∫∞

0
e−ttα−1dt, is the Gamma function and

I0
a+
f(x) = I0

b−
f(x) = f(x).

We also recall that the beta function for any complex numbers and non-
positive integers ρ,τ such that Re(ρ) > 0 and Re(τ) > 0 is defined by

B (ρ,τ) =

∫ 1
0
θρ−1 (1 −θ)τ−1 dθ =

Γ (ρ) Γ (τ)

Γ (ρ + τ)
.

The incomplete beta function is given by

Bt (ρ,τ) =

∫ t
0
θρ−1 (1 −θ)τ−1 dθ , 0 < t < 1 .

Lemma 1. ([22]) For any 0 ≤ a < b in R and fixed p ≥ 1, we have

(b−a)p ≤ bp −ap.



288 b. meftah, m. merad, a. souahi

3. Main results

In what follows, in order to simplify and lighten the writing, we note in all
the proofs η(b,a) by c.

Lemma 2. Let f : [a,a + η(b,a)] → R be a differentiable mapping on
(a,a + η(b,a)) with η(b,a) > 0, and assume that f ′ ∈ L

(
[a,a + η(b,a)]

)
. Then

the following equality holds

f (x) − Γ(α+1)
2ηα(b,a)

((
Iαa+f

)
(a + η (b,a)) +

(
Iα

(a+η(b,a))−
f
)

(a)

)
=

η(b,a)
2

(∫ 1
0
kf ′ (a + tη (b,a)) dt

+

∫ 1
0

(tα − (1 − t)α) f ′ (a + tη (b,a)) dt
)
,

(3.1)

where

k =


 1 if 0 ≤ t <

x−a
η(b,a)

,

−1 if x−a
η(b,a)

≤ t < 1 .
(3.2)

Proof. Let c = η (b,a). And let

I =

∫ 1
0
kf ′(a + tc)dt +

∫ 1
0

(tα − (1 − t)α)f ′(a + tc)dt

= I1 + I2 ,

(3.3)

where

I1 =

∫ 1
0
kf ′(a + tc)dt, (3.4)

I2 =

∫ 1
0

(tα − (1 − t)α)f ′(a + tc)dt, (3.5)

and k is defined by (3.2).
Clearly,

I1 =

∫ 1
0
kf ′(a + tc)dt =

∫ x−a
c

0
f ′(a + tc)dt−

∫ 1
x−a
c

f ′(a + tc)dt

= 2
c
f(x) − 1

c
[f(a) + f(a + c)].

(3.6)



fractional ostrowski type inequalities 289

Now, by integration by parts, I2 gives

I2 =
1
c
f (a + c) + 1

c
f (a)

− α
c

(∫ 1
0
tα−1f (a + tc) dt +

∫ 1
0

(1 − t)α−1 f (a + tc) dt
)

= 1
c
f (a + c) + 1

c
f (a) (3.7)

− α
cα+1

(∫ a+c
a

(u−a)α−1 f (u) du +
∫ a+c
a

(c + a−u)α−1 f (u) du
)

= 1
c
f (a + c) + 1

c
f (a) − Γ(α+1)

cα+1

(
(Iαa+f) (a + c) +

(
Iα

(a+c)−
f
)

(a)
)
.

Substituting (3.6) and (3.7) in (3.3). Multiplying the resulting equality by
c
2
, and then replacing c by η(a,b), we get the desired result.

Theorem 2. Let f : [a,a + η(b,a)] → R be a positive function on [a,b]
with η(b,a) > 0 and f ∈ L[a,b]. If |f ′| is s-preinvex function with s ∈ (0, 1],
then the following fractional inequality holds∣∣∣∣f (x) − Γ(α+1)2ηα(b,a) ((Iαa+f) (a + η (b,a)) + (Iα(a+η(b,a))−f) (a))

∣∣∣∣
≤ η(b,a)

2

(∣∣f ′(a)∣∣ + ∣∣f ′ (b)∣∣) (3.8)
×
(

1
s+1

+
1−( 12 )

α+s

α+s+1
+ B1

2
(s + 1,α + 1) −B1

2
(α + 1,s + 1)

)
.

where B1
2
(·, ·) is the incomplete beta function.

Proof. Let c = η(a,b). From Lemma 2, and properties of modulus, we
have∣∣∣∣f (x) − Γ(α+1)2cα ((Iαa+f) (a + c) + (Iα(a+c)−f) (a))

∣∣∣∣
≤ c

2

[∫ x−a
c

0

∣∣f ′ (a + tc)∣∣dt + ∫ 1
x−a
c

∣∣f ′ (a + tc)∣∣dt
+

∫ 1
2

0
((1 − t)α − tα)

∣∣f ′ (a + tc)∣∣dt +∫ 1
1
2

(tα − (1 − t)α)
∣∣f ′ (a + tc)∣∣dt

]
.



290 b. meftah, m. merad, a. souahi

Using the s-preinvexity of |f ′|, the above inequality gives∣∣∣∣f(x) − Γ(α+1)2cα ((Iαa+f) (a + c) + (Iα(a+c)−f) (a))
∣∣∣∣

≤ c
2

(∫ x−a
c

0

(
(1 − t)s|f ′(a)| + ts|f ′(b)|

)
dt

+

∫ 1
x−a
c

(
(1 − t)s|f ′(a)| + ts|f ′(b)|

)
dt

+

∫ 1
2

0

(
(1 − t)α − tα

)(
(1 − t)s|f ′(a)| + ts|f ′(b)|

)
dt

+

∫ 1
1
2

(
(1 − t)α − tα

)(
(1 − t)s|f ′(a)| + ts|f ′(b)|

)
dt

)

= c
2

[((
1−(1−x−ac )

s+1
)
|f′(a)|+( x−ac )

s+1
|f′(b)|

s+1

)

+

(
(1−x−ac )

s+1
|f′(a)|+

(
1−( x−ac )

s+1
)
|f′(b)|

s+1

)

+
∣∣f ′(a)∣∣

(∫ 1
2

0
((1 − t)α+s − tα(1 − t)s)dt

+

∫ 1
1
2

(
tα(1 − t)s − (1 − t)α+s

)
dt

)

+
∣∣f ′ (b)∣∣

(∫ 1
2

0
(ts (1 − t)α − tα+s)dt +

∫ 1
1
2

(tα+s − ts(1 − t)α)dt

)]
= c

2

(∣∣f ′(a)∣∣ + ∣∣f ′ (b)∣∣)
×

(
1
s+1

+

∫ 1
2

0
((1 − t)α+s − tα(1 − t)s) dt +

∫ 1
2

0
(ts (1 − t)α − tα+s)dt

)
= c

2

(∣∣f ′(a)∣∣ + ∣∣f ′ (b)∣∣)
×
(

1
s+1

+
1−( 12 )

α+s

α+s+1
+ B1

2
(s + 1,α + 1) −B1

2
(α + 1,s + 1)

)
.

Replacing c by η (b,a) in the above inequality, we get the desired result. The
proof is completed.



fractional ostrowski type inequalities 291

Corollary 1. In Theorem 2 if we choose x =
2a+η(b,a)

2
, we obtain the

following fractional midpoint inequality∣∣∣∣f (2a+η(b,a)2 )− Γ(α+1)2ηα(b,a) ((Iαa+f) (a + η (b,a)) + (Iα(a+η(b,a))−f) (a))
∣∣∣∣

≤ η(b,a)
2

(
∣∣f ′(a)∣∣ + ∣∣f ′ (b)∣∣)

×
(

1
s+1

+
1−( 12 )

α+s

α+s+1
+ B1

2
(s + 1,α + 1) −B1

2
(α + 1,s + 1)

)
.

Moreover if we take η(b,a) = b−a, we obtain∣∣∣∣f (a+b2 )− Γ(α+1)2(b−a)α ((Iαa+f) (b) + (Iαb−f) (a))
∣∣∣∣

≤ b−a
2

(
∣∣f ′(a)∣∣ + ∣∣f ′ (b)∣∣)

×
(

1
s+1

+
1−( 12 )

α+s

α+s+1
+ B1

2
(s + 1,α + 1) −B1

2
(α + 1,s + 1)

)
.

Corollary 2. In Theorem 2 if we put s = 1, we obtain∣∣∣∣f (x) − Γ(α+1)2ηα(b,a) ((Iαa+f) (a + η (b,a)) + (Iα(a+η(b,a))−f) (a))
∣∣∣∣

≤ η(b,a)
4(α+1)

(
∣∣f ′(a)∣∣ + ∣∣f ′ (b)∣∣) (α + 3 −(1

2

)α−1)
.

Moreover if we choose x =
2a+η(b,a)

2
, we obtain the following fractional

midpoint inequality∣∣∣∣f (2a+η(b,a)2 )− Γ(α+1)2ηα(b,a) ((Iαa+f) (a + η (b,a)) + (Iα(a+η(b,a))−f) (a))
∣∣∣∣

≤ η(b,a)
4(α+1)

(
∣∣f ′(a)∣∣ + ∣∣f ′ (b)∣∣) 1

2(α+1)

(
α + 3 −

(
1
2

)α−1)
.

Remark 1. Theorem 2 will be reduced to Theorem 2.3 from [27], if we
choose η(b,a) = b−a, x = a+b

2
and s = 1.

Theorem 3. Let f : [a,b] ⊂ [0,∞) → R be a positive function on [a,b]
with a < b and f ∈ L[a,b]. If |f ′|q is s-preinvex function, where s ∈ (0, 1] and



292 b. meftah, m. merad, a. souahi

q ≥ 1, then the following fractional inequality holds∣∣∣∣f(x) − Γ(α+1)2ηα(b,a) ((Iαa+f) (a + η (b,a)) + (Iα(a+η(b,a))−f) (a))
∣∣∣∣

≤ η(b,a)
2


(

(
1−

(
1− x−a

η(b,a)

)s+1)
|f′(a)|q+

(
x−a
η(b,a)

)s+1
|f′(b)|qd

s+1

)1
q

+

((
1− x−a

η(b,a)

)s+1
|f′(a)|q+

(
1−

(
x−a
η(b,a)

)s+1)
|f′(b)|qd

s+1

)1
q

+
(

2α−1
(α+1)2α

)1−1
q

((
µ1α,s

∣∣f ′(a)∣∣q + µ2α,s ∣∣f ′(b)∣∣q)1q
+
(
µ2α,s

∣∣f ′(a)∣∣q + µ1α,s ∣∣f ′(b)∣∣q)1q )

 ,

(3.9)

where

µ1α,s =
1−( 12 )

α+s+1

α+s+1
−B1

2
(α + 1,s + 1) (3.10)

and

µ2α,s = B1
2

(s + 1,α + 1) − (
1
2 )
α+s+1

α+s+1
. (3.11)

Proof. Let c = η (b,a). From Lemma 2, properties of modulus, and power
mean inequality, we have∣∣∣∣f(x) − Γ(α+1)2cα ((Iαa+f) (a + c) + (Iα(a+c)−f) (a))

∣∣∣∣
≤ c

2


(∫ x−ac

0

∣∣f ′ (a + tc)∣∣q dt
)1

q

+

(∫ 1
x−a
c

∣∣f ′ (a + tc)∣∣q dt
)1

q

+

(∫ 1
2

0
((1 − t)α − tα) dt

)1−1
q
(∫ 1

2

0
((1 − t)α − tα)

∣∣f ′(a + tc)∣∣q dt
)1

q

+

(∫ 1
1
2

(tα − (1 − t)α) dt

)1−1
q
(∫ 1

1
2

(tα − (1 − t)α)
∣∣f ′ (a + tc)∣∣q dt

)1
q


 .



fractional ostrowski type inequalities 293

Since |f ′|q is s-preinvex function, we deduce∣∣∣∣f(x) − Γ(α+1)2cα ((Iαa+f) (a + c) + (Iα(a+c)−f) (a))
∣∣∣∣

≤ c
2


(∫ x−ac

0
(1 − t)s

∣∣f ′(a)∣∣q + ts ∣∣f ′ (b)∣∣q dt
)1

q

+

(∫ 1
x−a
c

(1 − t)s
∣∣f ′(a)∣∣q + ts ∣∣f ′ (b)∣∣q dt

)1
q

+

(
1−( 12 )

α

α+1

)1−1
q



(∫ 1

2

0
((1 − t)α − tα) ((1 − t)s

∣∣f ′(a)∣∣q + ts ∣∣f ′ (b)∣∣q)dt
)1
q

+

(∫ 1
2

0
((1 − t)α − tα) (ts

∣∣f ′(a)∣∣q + (1 − t)s ∣∣f ′(b)∣∣q)dt
)1

q






= c
2


((1−(1−x−ac )s+1)|f′(a)|q+( x−ac )s+1|f′(b)|q

s+1

)1
q

+

(
(1−x−ac )

s+1
|f′(a)|q+

(
1−( x−ac )

s+1
)
|f′(b)|q

s+1

)1
q

+

(
1−( 12 )

α

α+1

)1−1
q

({(
1−( 12 )

α+s+1

α+s+1
−B1

2
(α + 1,s + 1)

)∣∣f ′(a)∣∣q
+

(
B1

2
(s + 1,α + 1) − (

1
2 )
α+s+1

α+s+1

)∣∣f ′ (b)∣∣q}1q
+

{(
B1

2
(s + 1,α + 1) − (

1
2 )
α+s+1

α+s+1

)∣∣f ′(a)∣∣q
+

(
1−( 12 )

α+s+1

α+s+1
−B1

2
(α + 1,s + 1)

)∣∣f ′ (b)∣∣q}1q
)



294 b. meftah, m. merad, a. souahi

= c
2

[((
1−(1−x−ac )

s+1
)
|f′(a)|q+( x−ac )

s+1
|f′(b)|q

s+1

)1
q

+

(
−(1−x−ac )

s+1
|f′(a)|q+

(
1−( x−ac )

s+1
)
|f′(b)|q

s+1

)1
q

+
(

2α−1
(α+1)2α

)1−1
q

{(
µ1α,s

∣∣f ′(a)∣∣q + µ2α,s ∣∣f ′ (b)∣∣q )1q
+
(
µ2α,s

∣∣f ′(a)∣∣q + µ1α,s ∣∣f ′ (b)∣∣q )1q}
]
,

where µ1α,s and µ
2
α,s are defined as in (3.10) and (3.11) respectively. By re-

placing c by η (b,a) in the above inequality, we get the desired result. The
prove is completed.

Corollary 3. In Theorem 3 if we choose x =
2a+η(b,a)

2
, we obtain the

following fractional midpoint inequality

∣∣∣∣f (2a+η(b,a)2 )− Γ(α+1)2ηα(b,a) ((Iαa+f) (a + η(b,a)) + (Iα(a+η(b,a))−f) (a))
∣∣∣∣

≤ η(b,a)
2


((1−( 12 )s+1)|f′(a)|q+( 12 )s+1|f′(b)|qd

s+1

)1
q

+

(
(( 12 ))

s+1
|f′(a)|q+

(
1−( 12 )

s+1
)
|f′(b)|qd

s+1

)1
q

+
(

2α−1
(α+1)2α

)1−1
q

{(
µ1α,s

∣∣f ′(a)∣∣q + µ2α,s ∣∣f ′ (b)∣∣q )1q
+
(
µ2α,s

∣∣f ′(a)∣∣q + µ1α,s ∣∣f ′ (b)∣∣q )1q}

 .

Moreover if we take η (b,a) = b−a, we obtain



fractional ostrowski type inequalities 295

∣∣∣∣f (a+b2 )− Γ(α+1)2(b−a)α ((Iαa+f) (b) + (Iαb−f) (a))
∣∣∣∣

≤ b−a
2


((1−( 12 )s+1)|f′(a)|q+( 12 )s+1|f′(b)|qd

s+1

)1
q

+

(
(( 12 ))

s+1
|f′(a)|q+

(
1−( 12 )

s+1
)
|f′(b)|qd

s+1

)1
q

+
(

2α−1
(α+1)2α

)1−1
q

{(
µ1α,s

∣∣f ′(a)∣∣q + µ2α,s ∣∣f ′ (b)∣∣q )1q
+
(
µ2α,s

∣∣f ′(a)∣∣q + µ1α,s ∣∣f ′ (b)∣∣q )1q}

 .

Corollary 4. In Theorem 3 if we put s = 1, we obtain∣∣∣∣f (x) − Γ(α+1)2ηα(b,a) ((Iαa+f) (a + η (b,a)) + (Iα(a+η(b,a))−f) (a))
∣∣∣∣

≤ η(b,a)
2


(

(
1−

(
1− x−a

η(b,a)

)2)
|f′(a)|q+

(
x−a
η(b,a)

)2
|f′(b)|qd

2

)1
q

+

((
1− x−a

η(b,a)

)2
|f′(a)|q+

(
1−

(
x−a
η(b,a)

)2)
|f′(b)|qd

2

)1
q

+
(

2α−1
(α+1)2α

)1−1
q

{(
µ1α,s

∣∣f ′(a)∣∣q + µ2α,s ∣∣f ′ (b)∣∣q )1q
+
(
µ2α,s

∣∣f ′(a)∣∣q + µ1α,s ∣∣f ′ (b)∣∣q )1q}

 ,

where
µ1α,1 =

1
α+2
− 1

α+1

(
1
2

)α+1
(3.12)

and
µ2α,1 =

1
(α+1)(α+2)

− 1
α+1

(
1
2

)α+1
. (3.13)



296 b. meftah, m. merad, a. souahi

Moreover if we choose x =
2a+η(b,a)

2
, we obtain the following fractional

midpoint inequality∣∣∣∣f (2a+η(b,a)2 )− Γ(α+1)2ηα(b,a) ((Iαa+f) (a + η (b,a)) + (Iα(a+η(b,a))−f) (a))
∣∣∣∣

≤ η(b,a)
2

[(
3|f′(a)|q+|f′(b)|qd

8

)1
q

+
(
|f′(a)|q+3|f′(b)|qd

8

)1
q

+
(

2α−1
(α+1)2α

)1−1
q

{(
µ1α,1

∣∣f ′(a)∣∣q + µ2α,1 ∣∣f ′ (b)∣∣q )1q
+
(
µ2α,1

∣∣f ′(a)∣∣q + µ1α,1 ∣∣f ′ (b)∣∣q )1q}
]
.

Additionally if we take η (b,a) = b−a, we obtain∣∣∣∣f (a+b2 )− Γ(α+1)2(b−a)α ((Iαa+f) (b) + (Iαb−f) (a))
∣∣∣∣

≤ b−a
2

[(
3|f′(a)|q+|f′(b)|qd

8

)1
q

+
(
|f′(a)|q+3|f′(b)|qd

8

)1
q

+
(

2α−1
(α+1)2α

)1−1
q

{(
µ1α,1

∣∣f ′(a)∣∣q + µ2α,1 ∣∣f ′ (b)∣∣q )1q
+
(
µ2α,1

∣∣f ′(a)∣∣q + µ1α,1 ∣∣f ′ (b)∣∣q )1q}
]
,

where µ1α,1 and µ
2
α,1 are defined as in (3.12) and (3.13) respectively.

Theorem 4. Let f : [a,b] ⊂ [0,∞) → R be a positive function on [a,b]
with a < b and f ∈ L[a,b]. If |f ′|q is s-preinvex function, where s ∈ (0, 1],
and q > 1 with 1

p
+ 1

q
= 1, then the following fractional inequality holds∣∣∣∣f (x) − Γ(α+1)2ηα(b,a) ((Iαa+f) (a + η (b,a)) + (Iα(a+η(b,a))−f) (a))

∣∣∣∣
≤ η(b,a)

2(s+1)
1
q

[((
1 −

(
1 − x−a

η(b,a)

)s+1)∣∣f ′(a)∣∣q + ( x−a
η(b,a)

)s+1 ∣∣f ′(b)∣∣q)1q
+

((
1 − x−a

η(b,a)

)s+1 ∣∣f ′(a)∣∣q + (1 −( x−a
η(b,a)

)s+1)∣∣f ′(b)∣∣q)1q



fractional ostrowski type inequalities 297

+

(
1−( 12 )

αp

αp+1

)1
p

×

((
(2s+1−1)|f′(a)|q+|f′(b)|q

2s+1

)1
q

+

(
|f′(a)|q+(2s+1−1)|f′(b)|q

2s+1

)1
q

)]
.

Proof. Let c = η (b,a). From Lemma 2, properties of modulus, Hölder’s
inequality, and Lemma 1, we have∣∣∣∣f(x) − Γ(α+1)2cα ((Iαa+f) (a + c) + (Iα(a+c)−f) (a))

∣∣∣∣
≤ c

2


(∫ x−ac

0

∣∣f ′ (a + tc)∣∣q dt
)1

q

+

(∫ 1
x−a
c

∣∣f ′(a + tc)∣∣q dt
)1

q

+

(∫ 1
2

0
((1 − t)α − tα)p dt

)1
p
(∫ 1

2

0

∣∣f ′ (a + tc)∣∣q dt
)1

q

+

(∫ 1
1
2

(tα − (1 − t)α)p dt

)1
p
(∫ 1

1
2

|f ′(a + tc)|qdt

)1
q




≤ c
2


(∫ x−ac

0

∣∣f ′ (a + tc)∣∣q dt
)1

q

+

(∫ 1
x−a
c

|f ′(a + tc)|qdt

)1
q

+

(∫ 1
2

0
((1 − t)αp − tαp) dt

)1
p
(∫ 1

2

0
|f ′(a + tc)|qdt

)1
q

+

(∫ 1
1
2

(tαp − (1 − t)αp) dt

)1
p
(∫ 1

1
2

|f ′(a + tc)|qdt

)1
q




= c
2


(∫ x−ac

0
|f ′(a + tc)|qdt

)1
q

+

(∫ 1
x−a
c

|f ′(a + tc)|qdt

)1
q

+

(
1−( 12 )

αp

αp+1

)1
p

×


(∫ 12

0
|f ′(a + tc)|qdt

)1
q

+

(∫ 1
1
2

|f ′(a + tc)|qdt

)1
q




 .



298 b. meftah, m. merad, a. souahi

Since |f ′|q is s-preinvex function, we have

∣∣∣∣f(x) − Γ(α+1)2cα ((Iαa+f) (a + c) + (Iα(a+ct)−f) (a))
∣∣∣∣

≤ c
2


(∫ x−ac

0
((1 − t)s

∣∣f ′(a)∣∣q + ts|f ′(b)|q)dt
)1

q

+

(∫ 1
x−a
c

((1 − t)s|f ′(a)|q + ts|f ′(b)|q)dt

)1
q

+

(
1−( 12 )

αp

αp+1

)1
p



(∫ 1

2

0
((1 − t)s

∣∣f ′(a)∣∣q + ts ∣∣f ′(b)∣∣q)dt
)1

q

+

(∫ 1
1
2

((1 − t)s
∣∣f ′(a)∣∣q + ts ∣∣f ′(b)∣∣q)dt

)1
q






= c
2(s+1)

1
q

[((
1 −

(
1 − x−a

c

)s+1)∣∣f ′(a)∣∣q + (x−a
c

)s+1 ∣∣f ′(b)∣∣q
)1

q

+

((
1 − x−a

c

)s+1 ∣∣f ′(a)∣∣q + (1 −(x−a
c

)s+1)∣∣f ′ (b)∣∣q
)1

q

+

(
1−( 12 )

αp

αp+1

)1
p

×

((
(2s+1−1)|f′(a)|q+|f′(b)|q

2s+1

)1
q

+

(
|f′(a)|q+(2s+1−1)|f′(b)|q

2s+1

)1
q

)]
.

Replacing c by η(b,a) in the above inequality, we get the desired result.

Corollary 5. In Theorem 4 if we choose x =
2a+η(b,a)

2
, we obtain the

following fractional midpoint inequality

∣∣∣∣f (2a+η(b,a)2 )− Γ(α+1)2ηα(b,a) ((Iαa+f) (a + η (b,a)) + (Iα(a+η(b,a))−f) (a))
∣∣∣∣



fractional ostrowski type inequalities 299

≤ η(b,a)
2(s+1)

1
q

(
1 +

(
1−( 12 )

αp

αp+1

)1
p

)

×

((
(2s+1−1)|f′(a)|q+|f′(b)|q

2s+1

)1
q

+

(
|f′(a)|q+(2s+1−1)|f′(b)|q

2s+1

)1
q

)
.

Moreover if we take η (b,a) = b−a, we obtain∣∣∣∣f (a+b2 )− Γ(α+1)2(b−a)α ((Iαa+f) (b) + (Iαb−f) (a))
∣∣∣∣

≤ b−a
2(s+1)

1
q

(
1 +

(
1−( 12 )

αp

αp+1

)1
p

)

×

((
(2s+1−1)|f′(a)1q+|f′(b)|q

2s+1

)1
q

+

(
|f′(a)|q+(2s+1−1)|f′(b)|q

2s+1

)1
q

)
.

Corollary 6. In Theorem 4 if we put s = 1, we obtain∣∣∣∣f (x) − Γ(α+1)2ηα(b,a) ((Iαa+f) (a + η (b,a)) + (Iα(a+η(b,a))−f) (a))
∣∣∣∣

≤ η(b,a)
2
1+ 1q

[((
1 −

(
1 − x−a

η(b,a)

)2)∣∣f ′(a)∣∣q + ( x−a
η(b,a)

)2 ∣∣f ′(b)∣∣q)1q

+

((
1 − x−a

η(b,a)

)2 ∣∣f ′(a)∣∣q + (1 −( x−a
η(b,a)

)2)∣∣f ′(b)∣∣q)1q
+

(
1−( 12 )

αp

αp+1

)1
p
((

3|f′(a)|q+|f′(b)|q
4

)1
q

+
(
|f′(a)|q+3|f′(b)|q

4

)1
q

)]
.

Moreover if we choose x =
2a+η(b,a)

2
, we obtain the following fractional

midpoint inequality∣∣∣∣f (2a+η(b,a)2 )− Γ(α+1)2ηα(b,a) ((Iαa+f) (a + η(b,a)) + (Iα(a+η(b,a))−f) (a))
∣∣∣∣

≤ η(b,a)
2
1+ 1q

(
1 +

(
1−( 12 )

αp

αp+1

)1
p

)((
3|f′(a)|q+|f′(b)|q

4

)1
q

+
(
|f′(a)|q+3|f′(b)|q

4

)1
q

)
.



300 b. meftah, m. merad, a. souahi

Additionally if we take η (b,a) = b−a, we obtain∣∣∣∣f (a+b2 )− Γ(α+1)2(b−a)α ((Iαa+f) (b) + (Iαb−f) (a))
∣∣∣∣

≤ b−a
2
1+ 1q

(
1 +

(
1−( 12 )

αp

αp+1

)1
p

)((
3|f′(a)|q+|f′(b)|q

4

)1
q

+
(
|f′(a)|q+3|f′(b)|q

4

)1
q

)
.

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	Introduction 
	Preliminaries
	Main results