International Journal of Analysis and Applications
ISSN 2291-8639
Volume 14, Number 1 (2017), 88-98
http://www.etamaths.com

GENERALIZED STEFFENSEN INEQUALITIES FOR LOCAL FRACTIONAL

INTEGRALS

MEHMET ZEKI SARIKAYA1, TUBA TUNÇ1,∗ AND SAMET ERDEN2

Abstract. Firstly we give a important integral inequality which is generalized Steffensen’s inequality.

Then, we establish weighted version of generalized Steffensen’s inequality for local fractional integrals.

Finally, we obtain several inequalities related these inequalities using the local fractional integral.

1. Introduction

In [17], J. S. Steffensen established the following result which is known as Steffensen’s inequality in
the literature.

Theorem 1.1. Let a and b be real numbers such that a < b, f,g : [a,b] → R be integrable functions
such that f is nonincreasing and for every x ∈ [a,b] , 0 ≤ g(x) ≤ 1. Then

b∫
b−λ

f(x)dx ≤
b∫
a

f(x)g(x)dx ≤
a+λ∫
a

f(x)dx (1.1)

where

λ =

b∫
a

g(x)dx.

The most basic inequality which deals with the comparison between integrals over a whole interval
[a,b] and integrals over a subset of [a,b] is the following inequality. The inequality (1.1) has attracted
considerable attention and interest from mathematicans and researchers. Due to this, over the years,
the interested reader is also refered to ( [1], [2], [4]- [7], [10], [11] and [18]) for integral inequalities.

In [19], Wu and Srivastava proved the following inequality which is weighted version of the inequality
(1.1).

Theorem 1.2. Let f, g and h be integrable functions defined on [a,b] with f nonincreasin. Also let
0 ≤ g(x) ≤ h(x) for all x ∈ [a,b] . Then, the following inequalities hold:

b∫
b−λ

f(x)h(x)dx

≤
b∫

b−λ

(f(x)h(x) − [f(x) −f(b−λ)] [h(x) −g(x)]) dx ≤
b∫
a

f(x)g(x)dx

≤
a+λ∫
a

(f(x)h(x) − [f(x) −f(a + λ)] [h(x) −g(x)]) dx ≤
a+λ∫
a

f(x)h(x)dx

2010 Mathematics Subject Classification. 26D15, 26A33.
Key words and phrases. Steffensen’s inequality; local fractional integral; fractal space; generalized convex function.

c©2017 Authors retain the copyrights of
their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

88



GENERALIZED STEFFENSEN INEQUALITIES 89

where λ is given by
a+λ∫
a

h(x)dx =

b∫
a

g(x)dx =

b∫
b−λ

h(x)dx.

2. Preliminaries

Recall the set Rα of real line numbers and use the Gao-Yang-Kang’s idea to describe the definition
of the local fractional derivative and local fractional integral, see [20, 21] and so on.

Recently, the theory of Yang’s fractional sets [20] was introduced as follows.
For 0 < α ≤ 1, we have the following α-type set of element sets:
Zα : The α-type set of integer is defined as the set {0α,±1α,±2α, ...,±nα, ...} .
Qα : The α-type set of the rational numbers is defined as the set {mα =

(
p
q

)α
: p,q ∈ Z, q 6= 0}.

Jα : The α-type set of the irrational numbers is defined as the set {mα 6=
(
p
q

)α
: p,q ∈ Z, q 6= 0}.

Rα : The α-type set of the real line numbers is defined as the set Rα = Qα ∪Jα.
If aα,bα and cα belongs the set Rα of real line numbers, then
(1) aα + bα and aαbα belongs the set Rα;
(2) aα + bα = bα + aα = (a + b)

α
= (b + a)

α
;

(3) aα + (bα + cα) = (a + b)
α

+ cα;
(4) aαbα = bαaα = (ab)

α
= (ba)

α
;

(5) aα (bαcα) = (aαbα) cα;
(6) aα (bα + cα) = aαbα + aαcα;
(7) aα + 0α = 0α + aα = aα and aα1α = 1αaα = aα.
The definition of the local fractional derivative and local fractional integral can be given as follows.

Definition 2.1. [20] A non-differentiable function f : R → Rα, x → f(x) is called to be local
fractional continuous at x0, if for any ε > 0, there exists δ > 0, such that

|f(x) −f(x0)| < εα

holds for |x−x0| < δ, where ε,δ ∈ R. If f(x) is local continuous on the interval (a,b) , we denote
f(x) ∈ Cα(a,b).

Definition 2.2. [20] The local fractional derivative of f(x) of order α at x = x0 is defined by

f(α)(x0) =
dαf(x)

dxα

∣∣∣∣
x=x0

= lim
x→x0

∆α (f(x) −f(x0))
(x−x0)

α ,

where ∆α (f(x) −f(x0)) =̃Γ(α + 1) (f(x) −f(x0)) .

If there exists f(k+1)α(x) =

k+1 times︷ ︸︸ ︷
Dαx ...D

α
xf(x) for any x ∈ I ⊆ R, then we denoted f ∈ D(k+1)α(I),

where k = 0, 1, 2, ...

Definition 2.3. [20] Let f(x) ∈ Cα [a,b] . Then the local fractional integral is defined by,

aI
α
b f(x) =

1

Γ(α + 1)

b∫
a

f(t)(dt)α =
1

Γ(α + 1)
lim

∆t→0

N−1∑
j=0

f(tj)(∆tj)
α,

with ∆tj = tj+1 − tj and ∆t = max{∆t1, ∆t2, ..., ∆tN−1} , where [tj, tj+1] , j = 0, ...,N − 1 and
a = t0 < t1 < ... < tN−1 < tN = b is partition of interval [a,b] .

Here, it follows that aI
α
b f(x) = 0 if a = b and aI

α
b f(x) = −bI

α
a f(x) if a < b. If for any x ∈ [a,b] ,

there exists aI
α
x f(x), then we denoted by f(x) ∈ Iαx [a,b] .

Lemma 2.1. [20] We have

i)
dαxkα

dxα
=

Γ(1 + kα)

Γ(1 + (k − 1) α)
x(k−1)α;

ii)
1

Γ(α + 1)

b∫
a

xkα(dx)α =
Γ(1 + kα)

Γ(1 + (k + 1) α)

(
b(k+1)α −a(k+1)α

)
, k ∈ R.



90 SARIKAYA

The interested reader is able to look over the references [3], [8], [9], [12]- [16], [20]- [24] for local
freactional theory.

In this study, generalized Steffensen’s inequality is established. Then, some inequalities related
generalized this inequality are given by using local fractional integrals.

3. Main Results

We start the following important inequality for local fractional integrals:

Theorem 3.1 (Generalized Steffensen’s Inequality). Let f(x),g(x) ∈ Iαx [a,b] such that f never in-
creases and 0 ≤ g(x) ≤ 1 on [a,b] with a < b. Then

b−λI
α
b f(x) ≤ aI

α
b f(x)g(x) ≤ a I

α
a+λf(x) (3.1)

where

λα = Γ(α + 1) aI
α
b g(x). (3.2)

Proof. For the proof of theorem, we give two different methods:
First method: By direct computation, we get

1

Γ(α + 1)

a+λ∫
a

f(x)(dx)α − aIαb f(x)g(x) (3.3)

=
1

Γ(α + 1)

a+λ∫
a

[f(x) −f (a + λ)] [1 −g(x)] (dx)α

+
1

Γ(α + 1)

b∫
a+λ

[f (a + λ) −f(x)] g(x)(dx)α.

Using the equality (3.3), because f is nonincreasing, we obtain the second inequality of (3.1).
Similarly, we have

aI
α
b f(x)g(x) −

1

Γ(α + 1)

b∫
b−λ

f(x)(dx)α (3.4)

=
1

Γ(α + 1)

b−λ∫
a

[f(x) −f (b−λ)] g(x)(dx)α

+
1

Γ(α + 1)

b∫
b−λ

[f (b−λ) −f(x)] [1 −g(x)] (dx)α.

Using the equality (3.4), because f is nonincreasing, we obtain the first inequality of (3.1). Thus, the
proof is completed.

Second method: Now, we prove the same of above Theorem in a deffirent way.



GENERALIZED STEFFENSEN INEQUALITIES 91

Because f is nonincreasing, the second inequality of (3.1) may be derived as follows:

1

Γ(α + 1)

a+λ∫
a

f(x)(dx)α − aIαb f(x)g(x)

=
1

Γ(α + 1)

a+λ∫
a

f(x) [1 −g(x)] (dx)α −
1

Γ(α + 1)

b∫
a+λ

f(x)g(x)(dx)α

≥
f(a + λ)

Γ(α + 1)

a+λ∫
a

[1 −g(x)] (dx)α −
1

Γ(α + 1)

b∫
a+λ

f(x)g(x)(dx)α

=
f(a + λ)

Γ(α + 1)


λα − a+λ∫

a

g(x)(dx)α


− 1

Γ(α + 1)

b∫
a+λ

f(x)g(x)(dx)α

=
1

Γ(α + 1)

b∫
a+λ

[f(a + λ) −f(x)] g(x)(dx)α ≥ 0.

The first inequality of (3.1) can be proved in a similar way. However, the second inequality implies
the first.

Indeed, let

G(x) = 1 −g(x)

and

Λα = Γ(α + 1) aI
α
b G(x).

Note that 0 ≤ G(x) ≤ 1 if 0 ≤ g(x) ≤ 1 in (a,b) .
Suppose the second inequality of (3.1) holds. Then, we obtain

1

Γ(α + 1)

b∫
a

f(x)G(x)(dx)α ≤
1

Γ(α + 1)

a+Λ∫
a

f(x)(dx)α

1

Γ(α + 1)

b∫
a

f(x)(dx)α −
1

Γ(α + 1)

a+Λ∫
a

f(x)(dx)α ≤
1

Γ(α + 1)

b∫
a

f(x)g(x)(dx)α

1

Γ(α + 1)

b∫
a+Λ

f(x)(dx)α ≤
1

Γ(α + 1)

b∫
a

f(x)g(x)(dx)α.

Because of

Λα = Γ(α + 1) aI
α
b G(x) = b

α −aα −λα

we have the identity

Λ + a = b−λ. (3.5)

From (3.5), we get the inequality

1

Γ(α + 1)

b∫
b−λ

f(x)(dx)α ≤ aIαb f(x)g(x)

which is the first inequality of (3.1). The proof is thus completed. �

In order to prove weighted version of generalized Steffensen’s inequality we need the following lemma:



92 SARIKAYA

Lemma 3.1. Let f, g and h belong to Iαx [a,b] . Suppose also that λ is a real number such that

aI
α
a+λh(x) = aI

α
b g(x) = b−λI

α
b h(x).

Then, we have

aI
α
b f(x)g(x) (3.6)

=
1

Γ(α + 1)

a+λ∫
a

(f(x)h(x) − [f(x) −f(a + λ)] [h(x) −g(x)]) (dx)α

+
1

Γ(α + 1)

b∫
a+λ

[f(x) −f(a + λ)] g(x)(dx)α

and

aI
α
b f(x)g(x) (3.7)

=
1

Γ(α + 1)

b∫
b−λ

(f(x)h(x) − [f(x) −f(b−λ)] [h(x) −g(x)]) (dx)α

+
1

Γ(α + 1)

b−λ∫
a

[f(x) −f(b−λ)] g(x)(dx)α.

Proof. The essumptions of the Lemma imply that

a ≤ a + λ ≤ b and a ≤ b−λ ≤ b.
Firstly, we prove the validity of the equality (3.6). Indeed, by direct computation, we find that

aI
α
b f(x)g(x) (3.8)

=
1

Γ(α + 1)

a+λ∫
a

f(x)g(x)(dx)α +
1

Γ(α + 1)

b∫
a+λ

f(x)g(x)(dx)α

+
f(a + λ)

Γ(α + 1)


 b∫
a

g(x)(dx)α −
a+λ∫
a

g(x)(dx)α −
b∫

a+λ

g(x)(dx)α




Now, if we apply the following assumption of the Lemma:

aI
α
a+λh(x) = aI

α
b g(x)

to (3.8), we obtain

aI
α
b f(x)g(x) (3.9)

=
1

Γ(α + 1)

a+λ∫
a

(f(x)g(x) + f(a + λ) [h(x) −g(x)]) (dx)α

+
1

Γ(α + 1)

b∫
a+λ

[f(x) −f(a + λ)] g(x)(dx)α.

If we add

1

Γ(α + 1)

a+λ∫
a

f(x)h(x)(dx)α −
1

Γ(α + 1)

a+λ∫
a

f(x)h(x)(dx)α

to right side of (3.9) and also we use elementary analysis, then we easily get the equality (3.6).
Secondly, if we apply above the operations for the following assumption of the Lemma

aI
α
b g(x) = b−λI

α
b h(x)



GENERALIZED STEFFENSEN INEQUALITIES 93

and also we consider the case a ≤ b − λ ≤ b, then we obtain the equality (3.7). Thus, the proof is
completed. �

Now, we prove weighted version generalized Steffensen’s inequality using local fractional integrals.

Theorem 3.2. Let f, g and h belong to Iαx [a,b] with f nonincreasing. Suppose also that 0 ≤ g(x) ≤
h(x) for all x ∈ [a,b] . Then, we have the following inequalities

b−λI
α
b f(x)h(x) (3.10)

≤
1

Γ(α + 1)

b∫
b−λ

(f(x)h(x) − [f(x) −f(b−λ)] [h(x) −g(x)]) (dx)α

≤ aIαb f(x)g(x)

≤
1

Γ(α + 1)

a+λ∫
a

(f(x)h(x) − [f(x) −f(a + λ)] [h(x) −g(x)]) (dx)α

≤ aIαa+λf(x)h(x)
where λ is given by

aI
α
a+λh(x) = aI

α
b g(x) = b−λI

α
b h(x).

Proof. In view of the assumptions that the function f is nonincreasing on [a,b] and that 0 ≤ g(x) ≤ h(x)
for all x ∈ [a,b] , we find that

1

Γ(α + 1)

b−λ∫
a

[f(x) −f(b−λ)] g(x)(dx)α ≥ 0, (3.11)

1

Γ(α + 1)

b∫
b−λ

[f(b−λ) −f(x)] [h(x) −g(x)] (dx)α ≥ 0, (3.12)

1

Γ(α + 1)

b∫
a+λ

[f(x) −f(a + λ)] g(x)(dx)α ≤ 0, (3.13)

and

1

Γ(α + 1)

a+λ∫
a

[f(a + λ) −f(x)] [h(x) −g(x)] (dx)α ≤ 0. (3.14)

Using the equality (3.7) together with the inequalities (3.11) and (3.12), we obtain that

b−λI
α
b f(x)h(x) (3.15)

≤
1

Γ(α + 1)

b∫
b−λ

(f(x)h(x) − [f(x) −f(b−λ)] [h(x) −g(x)]) (dx)α

≤ aIαb f(x)g(x).
Using the equality (3.6) together with the inequalities (3.13) and (3.14) either, we get that

aI
α
b f(x)g(x) (3.16)

≤
1

Γ(α + 1)

a+λ∫
a

(f(x)h(x) − [f(x) −f(a + λ)] [h(x) −g(x)]) (dx)α

≤ aIαa+λf(x)h(x).



94 SARIKAYA

Combining the inequalities (3.15) and (3.16), we easily deduce required inequalities. �

In particular, if we chose h(t) = 1 in (3.10), we obtain the following refinement of generalized
Steffensen’s inequality.

Corollary 3.1. Let f(x),g(x) ∈ Iαx [a,b] such that f never increases and 0 ≤ g(x) ≤ 1 on [a,b] with
a < b. Then

b−λI
α
b f(x)

≤
1

Γ(α + 1)

b∫
b−λ

(f(x) − [f(x) −f(b−λ)] [1 −g(x)]) (dx)α

≤ aIαb f(x)g(x)

≤
1

Γ(α + 1)

a+λ∫
a

(f(x) − [f(x) −f(a + λ)] [1 −g(x)]) (dx)α

≤ aIαa+λf(x)

where

λα = Γ(α + 1) aI
α
b g(x).

Theorem 3.3. Let f, g and h belong to Iαx [a,b] with f nonincreasing. Also let

0 ≤ ψ(x) ≤ g(x) ≤ h(x) −ψ(x)

for all x ∈ [a,b] . Then we have the inequalities

b−λI
α
b f(x)h(x) +

1

Γ(α + 1)

b∫
a

|[f(x) −f(b−λ)] ψ(x)|(dx)α

≤ aIαb f(x)g(x)

≤ aIαa+λf(x)h(x) −
1

Γ(α + 1)

b∫
a

|[f(x) −f(a + λ)] ψ(x)|(dx)α

where λ is given by

aI
α
a+λh(x) = aI

α
b g(x) = b−λI

α
b h(x).

Proof. By the assumptions that the function f is nonincreasing on [a,b] and that

0 ≤ ψ(x) ≤ g(x) ≤ h(x) −ψ(x)



GENERALIZED STEFFENSEN INEQUALITIES 95

for all x ∈ [a,b] , it follows that

1

Γ(α + 1)

a+λ∫
a

[f(x) −f(a + λ)] [h(x) −g(x)] (dx)α (3.17)

+
1

Γ(α + 1)

b∫
a+λ

[f(a + λ) −f(x)] g(x)(dx)α

=
1

Γ(α + 1)

a+λ∫
a

|f(x) −f(a + λ)| [h(x) −g(x)] (dx)α

+
1

Γ(α + 1)

b∫
a+λ

|f(a + λ) −f(x)|g(x)(dx)α

≥
1

Γ(α + 1)

a+λ∫
a

|f(x) −f(a + λ)|ψ(x)(dx)α

+
1

Γ(α + 1)

b∫
a+λ

|f(a + λ) −f(x)|ψ(x)(dx)α

=
1

Γ(α + 1)

b∫
a

|[f(x) −f(a + λ)] ψ(x)|(dx)α.

Similarly, we find that

1

Γ(α + 1)

b∫
b−λ

[f(b−λ) −f(x)] [h(x) −g(x)] (dx)α (3.18)

+
1

Γ(α + 1)

b−λ∫
a

[f(x) −f(b−λ)] g(x)(dx)α

≥
1

Γ(α + 1)

b∫
a

|[f(x) −f(b−λ)] ψ(x)|(dx)α.

If we use the equalities (3.6) and (3.7) and the inequalities (3.17) and (3.18), we obtain required
inequalities. �

Corollary 3.2. Under the same assumptions of Theorem 3.3 with h(x) = 1 and ψ(x) = Mα, then the
following inequalities hold:

b−λI
α
b f(x) +

Mα

Γ(α + 1)

b∫
a

|[f(x) −f(b−λ)]|(dx)α

≤ aIαb f(x)g(x)

≤ a+λIαb f(x) −
Mα

Γ(α + 1)

b∫
a

|[f(x) −f(a + λ)]|(dx)α

where Mα ∈ Rα+ ∪{0α} and
λα = Γ(α + 1) aI

α
b g(x).



96 SARIKAYA

Finally, we give a general result on a considerably improved version of generalized Steffensen’s
inequality by introducing the additional paramaters λ1 and λ2.

Theorem 3.4. Let f(x),g(x) ∈ Iαx [a,b] such that f never increases on [a,b] . Also let

0α ≤ λα1 ≤ λ
α = Γ(α + 1) aI

α
b g(x) ≤ λ

α
2 ≤ (b−a)

α

and

0 ≤ Mα ≤ g(x) ≤ (1 −M)α

for all x ∈ [a,b] . Then, we have the inequalities

(3.19)

b−λ1I
α
b f(x) +

f(b)

Γ(α + 1)
(λ−λ1)

α
+

Mα

Γ(α + 1)

b∫
a

|[f(x) −f(b−λ)]|(dx)α

≤ aIαb f(x)g(x)

≤ aIαa+λ2f(x) +
f(b)

Γ(α + 1)
(λ2 −λ)

α −
Mα

Γ(α + 1)

b∫
a

|[f(x) −f(a + λ)]|(dx)α.

Proof. By direct computation, we obtain

aI
α
b f(x)g(x) −a I

α
a+λ2

f(x) +
f(b)

Γ(α + 1)


λα2 −

b∫
a

g(x)(dx)α


 (3.20)

=
1

Γ(α + 1)


 b∫
a

f(x)g(x)(dx)α −
a+λ2∫
a

f(x)(dx)α




+
1

Γ(α + 1)


 a+λ2∫

a

f(b)(dx)α −
b∫
a

f(b)g(x)(dx)α




≤
1

Γ(α + 1)

b∫
a

[f(x) −f(b)] g(x)(dx)α −
1

Γ(α + 1)

a+λ∫
a

[f(x) −f(b)] (dx)α.

Because of the following assumption of the Theorem

0α ≤ λα1 ≤ λ
α ≤ λα2 ≤ (b−a)

α
,

we find that

aα ≤ aα + λα ≤ aα + λα2 ≤ b
α

that is

a ≤ a + λ ≤ a + λ2 ≤ b.

Also, since f is nonincreasing, we have

f(x) −f(b) ≥ 0

for all x ∈ [a.b] .
Onthe other hand, since the hypothesis of the Theorem, we we conclude that the function f(x)−f(b)

belong to Iαx [a,b] and nonincreasing on [a,b] . Thus, substituting f(x)−f(b) instead of f(x) in Corollary



GENERALIZED STEFFENSEN INEQUALITIES 97

3.2, we find that

1

Γ(α + 1)

b∫
a

[f(x) −f(b)] g(x)(dx)α −
1

Γ(α + 1)

b∫
a+λ

[f(x) −f(b)] (dx)α (3.21)

≤ −
M

Γ(α + 1)

b∫
a

|[f(x) −f(a + λ)]|(dx)α.

Combining the inequalities (3.20) and (3.21), we obtain

aI
α
b f(x)g(x) −a I

α
a+λ2

f(x) +
f(b)

Γ(α + 1)
(λα2 −λ

α)

≤ −
Mα

Γ(α + 1)

b∫
a

|[f(x) −f(a + λ)]|(dx)α.

which is the second inequality of (3.19).
In a similar way, we can prove that

aI
α
b f(x)g(x) −b−λ1 I

α
b f(x) −

f(b)

Γ(α + 1)


 b∫
a

g(x)(dx)α −λα1




≥
1

Γ(α + 1)

b∫
a

[f(x) −f(b)] g(x)(dx)α +
1

Γ(α + 1)

b∫
b−λ

[f(b) −f(x)] (dx)α

≥
Mα

Γ(α + 1)

b∫
a

|[f(x) −f(b−λ)]|(dx)α

which is the first inequality of (3.19). The proof is thus completed. �

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1Department of Mathematics, Faculty of Science and Arts, Düzce University, Düzce-Turkey

2Department of Mathematics, Faculty of Science, Bartın University, Bartın-Turkey

∗Corresponding author: tubatunc03@gmail.com


	1. Introduction
	2. Preliminaries
	3. Main Results
	References