Chebyshev Type Inequalities for Conformable Fractional Integrals

Abstract

In this article, firstly some necessary definitions and results involving fractional integrals are given. Secondly, a new identity involving conformable fractional integrals is given. Then, by using this identity, we establish new Chebyshev inequalities for the Chebyshev functional via conformable fractional integral.

Full text

Chebyshev Type Inequalities for Conformable Fractional
Integrals
Erhan Seta∗Ahmet Ocak Akdemirb˙
Ilker Mumcuc
a,cDepartment of Mathematics, Faculty of Arts and Sciences, Ordu University, 52200 Ordu, Turkey
bDepartment of Mathematics, Faculty of Science and Arts, Aˇgrı ˙
Ibrahim C¸e¸cen University, 04100 Aˇgrı,Turkey
aerhanset@yahoo.com ,bahmetakdemir@agri.edu.tr, cmumcuilker@msn.com
Abstract
In this article, firstly some necessary definitions and results involving fractional integrals are given. Secondly,
a new identity involving conformable fractional integrals is given. Then, by using this identity, we establish
new Chebyshev inequalities for the Chebyshev functional via conformable fractional integral.
Keywords: Chebyshev Inequality, Riemann-Liouville Integral, Conformable fractional integrals.
1 Introduction
There are many different type integral inequalities in the literature such as classical and analytic inequalities.
We will start with the most prominent one of classical inequalities that is called Chebyshev inequality.
This inequality was established by ˇ
Cebyˇsev in [?] as following;
|T(f, g)| ≤ 1
12 (b−a)2kf0k∞kg0k∞,(1)
where f, g : [a, b]→Rare absolutely continuous functions whose derivatives f0, g0∈L∞[a, b] and
T(f, g) = 1
b−a
b
Z
a
f(x)g(x)dx −

1
b−a
b
Z
a
f(x)dx


1
b−a
b
Z
a
g(x)dx
,(2)
which is called the ˇ
Cebyˇsev functional, provided the integrals in (??) exist.
We will remind some well-known concepts as followings:
The beta function defined as follows [?, p18]:
B(a, b) = Γ(a)Γ(b)
Γ(a+b)=Z1
0
ta−1(1 −t)b−1dt, a, b > 0,
where Γ (α) = R∞
0e−tuα−1du is Gamma function.
Definition 1.1 Let f∈L1[a, b]. The Riemann-Liouville integrals Jα
a+fand Jα
b−fof order α > 0are defined
by
Jα
a+f(x) = 1
Γ(α)Zx
a
(x−t)α−1f(t)dt, x > a
and
Jα
b−f(x) = 1
Γ(α)Zb
x
(t−x)α−1f(t)dt, x < b
respectively. Here J0
a+f(x) = J0
b−f(x) = f(x).
∗corresponding author
1

In [?], Belarbi and Dahmani established following theorems for the Chebyshev inequalities.
Theorem 1.1 Let fand gbe two synchronous functions on [0,∞). Then for all t > 0, α > 0, we have:
Jα(fg)≥Γ(α+ 1)
tαJαf(t)Jαg(t).(3)
Theorem 1.2 Let fand gbe two synchronous functions on [0,∞). Then for all t > 0, α > 0, β > 0, we have:
tα
Γ(α+ 1)Jβ(f g)(t) + tβ
Γ(β+ 1)Jα(f g)(t)≥Jαf(t)Jβg(t) + Jβf(t)Jαg(t) (4)
Theorem 1.3 Let (fi)i=1,...,n be npositive inreasing functions on [0,∞). Then for any t > 0, α > 0, we have
Jα n
Y
i=1
fi!(t)≥(Jα(1))1−n
n
Y
i=1
Jαfi(t.) (5)
Theorem 1.4 Let fand gbe two functions defined on [0,∞), such that fis increrasing, gis differentiable and
there exist a real number m:= inft≥0g0(t). Then the inequality
Jα(fg)(t)≥(Jα(1))−1Jαf(t)Jαg(t)−mt
α+ 1Jαf(t) + mJ α(tf (t)) (6)
is valid for all t > 0, α > 0.
In [?], Khalil et al. define a new well-behaved simple fractional derivative called conformable fractional
derivative depending just on the basic limit definition of the derivative. They also defined the fractional integral
of order 0 < α ≤1 only.
In [?], Abdeljawad gave the definition of left and right conformable fractional integrals of any order α > 0.
Definition 1.2 Let α∈(n, n + 1] and set β=α−nthen the left conformable fractional integral starting at a
if order αis defined by
(Ia
αf)(t) = 1
n!Zt
a
(t−x)n(x−a)β−1f(x)dx
Analogously, the right conformable fractional integral is defined by
(bIαf)(t) = 1
n!Zb
t
(x−t)n(b−x)β−1f(x)dx.
In [?], Set and Mumcu gave following Lemma
Lemma 1.1 If g: [a, b]→Ris integrable and symmetric to (a+b)/2with a<b, then
Ia
αg(b) = bIαg(a) = 1
2[bIαg(a) + Ia
αg(b)]
with α∈(n, n + 1].
Notice that if α=n+ 1 then β=α−n=n+ 1 −n= 1 and hence (Ia
αf)(t)=(Ja
n+1f)(t).
For some recent results related to conformable fractional integrals, see, e.g., [?,?].
The main purpose of this paper is to establish Chebyshev type inequalities for conformable fractional integral.
2 Chebyshev Type Inequalities
Lemma 2.1 Let f: [a, b]→R+be convex, where 0≤a < b. If f∈L[a, b]then we get
Z1
0
tn(1 −t)α−n−1f(ta + (1 −t)b)dt +Z1
0
tn(1 −t)α−n−1f((1 −t)a+tb)dt
=n!
(b−a)α[bIαf(a) + Ia
αf(b)].
for α∈(n, n + 1], n = 0,1,2, ...
2

Proof. By integration by parts, we can state
Z1
0
tn(1 −t)α−n−1f(ta + (1 −t)b)dt =1
b−aZb
ab−x
b−anx−a
b−aα−n−1
f(x)dx
=1
(b−a)αZb
a
(b−x)n(x−a)α−n−1f(x)dx
=n!
(b−a)α(Ia
αf)(b) (7)
and
Z1
0
tn(1 −t)α−n−1f((1 −t)a+tb)dt =1
b−aZb
ax−a
b−anb−x
b−aα−n−1
f(x)dx
=1
(b−a)αZb
a
(x−a)n(b−x)α−n−1f(x)dx
=n!
(b−a)α(bIαf)(a).(8)
Adding (??) and (??), we get desired result.
Theorem 2.1 Let f, g : [a, b]→Rare two monotonic functions of the same monotonicity, f, g ∈L[a, b]and
α∈(n, n + 1] (n= 0,1,2, ...), then the following inequality holds:
Γ(α+ 1)Γ(α−n)
2(b−a)α)[ bIα(fg)(a) + Ia
α(fg)(b)]
≥Γ(α+ 1)
2(b−a)α[bIαf(a) + Ia
αf(b)] Γ(α+ 1)
2(b−a)α[bIαg(a) + Ia
αg(b)].(9)
Proof. Our proof is based on the property of positivity of the integral. The basic remark is the inequality
(f(x)−f(y)) (g(x)−g(y)) ≥0
for all x, y ∈[a, b]. For x=ta + (1 −t)b,t∈[0,1] we obtain
(f(ta + (1 −t)b)−f(y)) (g(ta + (1 −t)b)−g(y)) ≥0.(10)
Similarly, for x= (1 −t)a+tb,t∈[0,1] we have
(f((1 −t)a+tb)−f(y)) (g((1 −t)a+tb)−g(y)) ≥0.(11)
Adding the inequalities (??) and (??) , then multiplying both sides with tn(1 −t)α−n−1and integrating with
respect to tover [0,1] we have:
Z1
0
tn(1 −t)α−n−1(fg)(ta + (1 −t)b)dt
+Z1
0
tn(1 −t)α−n−1(fg)((1 −t)a+tb)dt + 2B(n+ 1, α −n)f(y)g(y)
≥g(y)Z1
0
tn(1 −t)α−n−1f(ta + (1 −t)b)dt +Z1
0
tn(1 −t)α−n−1f((1 −t)a+tb)dt.
+f(y)Z1
0
tn(1 −t)α−n−1g(ta + (1 −t)b)dt +Z1
0
tn(1 −t)α−n−1g((1 −t)a+tb)dt(12)
We apply Lemma ?? and multiplying both sides by 1
n!
1
(b−a)α[bIα(fg)(a) + Ia
α(fg)(b)] + 2
n!B(n+ 1, α −n−1)f(y)g(y)
≥g(y)
(b−a)α[bIαf(a) + Ia
αf(b)] + f(y)
(b−a)α[bIαg(a) + Ia
αg(b)].(13)
3

We put y=ta + (1 −t)b,t∈[0,1] in (??), whence it follows
1
(b−a)α[bIα(fg)(a) + Ia
α(fg)(b)] + 2
n!B(n+ 1, α −n−1)(f g)(ta + (1 −t)b)
≥g(ta + (1 −t)b)
(b−a)α[bIαf(a) + Ia
αf(b)] + f(ta + (1 −t)b)
(b−a)α[bIαg(a) + Ia
αg(b)].(14)
On the other hand, for y= (1 −t)a+tb,t∈[0,1] in (??) we have
1
(b−a)α[bIα(fg)(a) + Ia
α(fg)(b)] + 2
n!B(n+ 1, α −n−1)(f g)((1 −t)a+tb)
≥g((1 −t)a+tb)
(b−a)α[bIαf(a) + Ia
αf(b)] + f((1 −t)a+tb
(b−a)α[bIαg(a) + Ia
αg(b)].(15)
Adding the inequalities (??) and (??), then the multiplying both sides tn(1 −t)α−n−1and integrating with
respect to tover [0,1], one has
2
(b−a)αB(n+ 1, α −n)[ bIα(f g)(a) + Ia
α(fg)(b)] (16)
+2
n!B(n+ 1, α −n)Z1
0
tn(1 −t)α−n−1(fg)(ta + (1 −t)b)
+Z1
0
tn(1 −t)α−n−1(fg)((1 −t)a+tb)
≥1
(b−a)α[bIαf(a) + Ia
αf(b)]
×Z1
0
tn(1 −t)α−n−1g(ta + (1 −t)b)Z1
0
tn(1 −t)α−n−1g((1 −t)a+tb)
+1
(b−a)α[bIαg(a) + Ia
αg(b)]
×Z1
0
tn(1 −t)α−n−1f(ta + (1 −t)b)Z1
0
tn(1 −t)α−n−1f((1 −t)a+tb).
We apply Lemma ??
4
(b−a)αB(n+ 1, α −n)[ bIα(f g)(a) + Ia
α(fg)(b)]
≥2n!
(b−a)2α[bIαf(a) + Ia
αf(b)][ bIαg(a) + Ia
αg(b)].
Multiplying both sides of the last inequality with (Γ(α+1))2
8and using the properties of the beta functions, we
obtain desired inequality.
Remark 2.1 If we take α=n+ 1 in above theorem, then the inequality (??) becomes:
Γ(α+ 1)
2(b−a)α)[Jα
a+(fg)(b) + Jα
b−(fg)(a)]
≥Γ(α+ 1)
2(b−a)α[Jα
a+f(b) + Jα
b−f(a)] Γ(α+ 1)
2(b−a)α[Jα
a+g(b) + Jα
b−g(a)].(17)
Remark 2.2 In above theorem, if we take α=n+ 1 and f, g is symmetric to (a+b)/2, and use Lemma ??
then inequality (??) becomes inequality (??).
4

Theorem 2.2 Let f, g : [a, b]→Rare two monotonic functions of the same monotonicity, f, g ∈L[a, b],
α∈[n, n + 1] (n= 0,1,2, ...)and β∈(k, k + 1] (k= 0,1,2, ...), then
2
(b−a)βB(k+ 1, β −k)[ bIα(fg)(a) + Ia
α(fg)(b)]
+2
(b−a)αB(n+ 1, α −n)[ bIβ(f g)(a) + Ia
β(fg)(b)]
≥1
(b−a)α+β[bIαf(a) + Ia
αf(b)][ bIβg(a) + Ia
βg(b)]
+1
(b−a)α+β[bIαg(a) + Ia
αg(b)][ bIβf(a) + Ia
βf(b)].(18)
Proof. Adding the inequalities (??) and (??),and then multiplying both sides of obtained inequality with
tk(1 −t)β−k−1and integrating with respect to tover [0,1], we get
2
(b−a)βB(n+ 1, β −n−1)[ bIα(fg)(a) + Ia
α(fg)(b)] (19)
+2
n!B(n+ 1, α −n−1)Z1
0
tk(1 −t)β−k−1(fg)(ta + (1 −t)b)
+Z1
0
tk(1 −t)β−k−1(fg)((1 −t)a+tb)
≥1
(b−a)α[bIαf(a) + Ia
αf(b)]
×Z1
0
tk(1 −t)β−k−1g(ta + (1 −t)b)Z1
0
tk(1 −t)β−k−1g((1 −t)a+tb)
+1
(b−a)α[bIαg(a) + Ia
αg(b)]
×Z1
0
tk(1 −t)β−k−1f(ta + (1 −t)b)Z1
0
tk(1 −t)β−k−1f((1 −t)a+tb).
So, if we use Lemma ??, we have desired result.
Remark 2.3 Applying Theorem ?? for α=β, we obtain Theorem ??.
Remark 2.4 If we take α=n+ 1 and β=k+ 1 in theorem ??, then inequality (??) becomes:
2
(b−a)k+1(k+ 1) [Jα
a+fg(b) + Jα
b−fg(a)]
+2
(b−a)n+1(n+ 1) [Jβ
a+fg(b) + Jβ
b−fg(a)]
≥1
(b−a)n+k+2 [Jα
a+f(b) + Jα
b−f(a)][Jβ
a+g(b) + Jβ
b−g(a)]
+1
(b−a)n+k+2 [Jα
a+g(b) + Jα
b−g(a)][Jβ
a+f(b) + Jβ
b−f(a)].
Remark 2.5 In above theorem, if we take α=n+ 1, and f, g is symmetric to (a+b)/2, and use Lemma ??,
then inequality (??) becomes inequality (??).
Theorem 2.3 Let (fi)i=1,...,n be a positive increasing functions on [a,b]. For α∈[n, n + 1] n= 0,1,2, ..., we
have
bIα n
Y
i=1
fi!(a) + Ia
α n
Y
i=1
fi!(b)
≥Γ(α+ 1)
2(b−a)αΓ(α−n)n−1n
Y
i=1
[bIα(fi)(a) + Ia
α(fi)(b)].(20)
5

Proof. We prove this theorem by induction. For n= 1, we have
[bIαf1(a) + Ia
αf1(b)] ≥[bIαf1(a) + Ia
αf1(b)].
For n= 2, applying Theorem ??, we get
[bIα(f1f2)(a) + Ia
α(f1f2)(b)]
≥Γ(α+ 1)
2(b−a)αΓ(α−n)[bIα(f1)(a) + Ia
α(f1)(b)][ bIαf2)(a) + Ia
α(f2)(b)].
Now suppose that
bIα n−1
Y
i=1
fi!(a) + Ia
α n−1
Y
i=1
fi!(b)
≥Γ(α+ 1)
2(b−a)αΓ(α−n)n−2n−1
Y
i=1
[bIα(fi)(a) + Ia
α(fi)(b)].(21)
Since (fi)i=1,...,n are positive increasing functions, then Qn−1
i=1 fi(t) is an increasing function. So we can apply
Theorem ?? to the functions Qn−1
i=1 fi=g, fn=f, we obtain
bIα n
Y
i=1
fi!(a) + Ia
α n
Y
i=1
fi!(b)
≥Γ(α+ 1)
2(b−a)αΓ(α−n)
×bIα n−1
Y
i=1
fi!(a) + Ia
α n−1
Y
i=1
fi!(b)[bIα(fn)(a) + Ia
α(fn)(b)].
Using (??), we get desired result.
Remark 2.6 If we take α=n+ 1 and β=k+ 1 in Theorem ??, then inequality (??) becomes:
Jα
b− n
Y
i=1
fi!(a) + Jα
a+ n
Y
i=1
fi!(b)
≥(n+ 1)!
2(b−a)αn−1n
Y
i=1
[Jα
a+fi(b) + Jα
b−fi(a)].(22)
Remark 2.7 If we take α=n+ 1,fi;i= 1, ..., n is symmetric to (a+b)/2, then, using Lemma ??, inequality
(??) becomes inequality (??).
Theorem 2.4 Let fand gbe two functions defined on [0,∞), such that fis increrasing, gis differentiable and
there exist a real number m:= inft≥0g0(t). Then the inequality
Γ(α−n)[ bIα(fg)(a) + Ia
α(fg)(b)]
≥Γ(α+ 1)
2(b−a)α[bIαf(a) + Ia
αf(b)][ bIαg(a) + Ia
αg(b)]
−Γ(n+ 2)Γ(β−n)
2(b−a)α(α+ 1) [bIαf(a) + Ia
αf(b)] + m[bIα(tf)(a) + Ia
α(tf)(b)].(23)
is valid for all t > 0, α ∈(n, n + 1], n = 0,1,2, ....
6

Proof. We consider the function h(t) := g(t)−mt. It is clear that h is differentiable and it is increasing on
[0,∞). Then by using Theorem ??, we get
Γ(α−n)[ bIα((g−mt)f)(a) + Ia
α((g−mt)f)(b)]
≥Γ(α+ 1)
2(b−a)α[bIαf(a) + Ia
αf(b)][ bIαg(a) + Ia
αg(b)−mbIαt(a)−mIa
αt(b)]
≥Γ(α+ 1)
2(b−a)α[bIαf(a) + Ia
αf(b)][ bIαg(a) + Ia
αg(b)]
−mB(n+ 2, α −n)Γ(α+ 1)
2(b−a)α[bIαf(a) + Ia
αf(b)]
≥Γ(α+ 1)
2(b−a)α[bIαf(a) + Ia
αf(b)][ bIαg(a) + Ia
αg(b)]
−Γ(n+ 2)Γ(β−n)
2(b−a)α(α+ 1) [bIαf(a) + Ia
αf(b)].
So
Γ(α−n)[ bIα(fg)(a) + Ia
α(fg)(b)]
≥Γ(α+ 1)
2(b−a)α[bIαf(a) + Ia
αf(b)][ bIαg(a) + Ia
αg(b)]
−Γ(n+ 2)Γ(β−n)
2(b−a)α(α+ 1) [bIαf(a) + Ia
αf(b)] + m[bIα(tf)(a) + Ia
α(tf)(b)].
which the proof is completed.
Remark 2.8 In Theorem ??, if we take α=n+ 1 and β=k+ 1, inequality (??) becomes:
[Jβ
a+(fg)(b) + Jβ
b−fg)(a)]
≥(n+ 2)!
2(b−a)α[Jβ
a+f(b) + Jβ
b−f(a)][Jβ
a+g(b) + Jβ
b−g(a)]
−Γ(n+ 2)Γ(β−n)
2(b−a)α(α+ 1) [Jβ
a+f(b) + Jβ
b−f(a)] + m[Jβ
a+f(b) + Jβ
b−g(a)] (24)
Remark 2.9 In Theorem ??, if we take α=n+ 1 and f, g is symmetric to (a+b)/2, then, using Lemma ??
inequality (??) becomes inequality (??).
References
[1] T. Abdeljawad, On conformable fractional calculus, Journal of Computational and Applied Mathematics,
279 (2015), 57-66.
[2] S. Belarbi, Z. Dahmani On some new fractional integral inequalities, J. Inequal. Pure Appl. Math. 10
(2009), no. 3, Article 86, 5 pp.
[3] P.L. ˇ
Cebyˇsev Sur les expressions approximatives des integrales definies par les autres prises entre les memes
limites Proc. Math. Soc. Charkov, 2 (1882), 93-98
[4] R. Khalil, M. Al Horani, A. Yousef, M. Sababheh A new definition of fractional derivative Journal of
Computational and Applied Mathematics, 264 (2014), 65-70.
[5] E.D. Rainville Special Functions The Mcmillan Company, New York
[6] E. Set, ˙
I.Mumcu Hermite-Hadamard-Fejer type inequalities for conformable fractional integrals Submitted.
[7] E. Set, A.O. Akdemir, ˙
I.Mumcu Hermite-Hadamard’s inequality and its extensions for conformable fraac-
tional integrals of any order α > 0 Submitted.
[8] E. Set, A.O. Akdemir, ˙
I .Mumcu Ostrowski type inequalities for functions whose derivatives are convex via
conformable fractional integrals Submitted.
7