Vol.8, No.2, pp.137-144 (2018)
http://doi.org/10.11121/ijocta.01.2018.00541
RESEARCH ARTICLE
New extensions of Chebyshev type inequalities using generalized
Katugampola integrals via P´
olya-Szeg¨
o inequality
Erhan Seta*, Zoubir Dahmanib and ˙Ilker Mumcua a
Department of Mathematics, Faculty of Science and Arts, Ordu University, Ordu, Turkey
b
Laboratory of Pure and Applied Mathematics (LPAM), Faculty of Exact Sciences, University of Mostaganem Abdelhamid Ben Badis (UMAB), Mostaganem, Algeria
erhanset@yahoo.com, zzdahmani@yahoo.fr, mumcuilker@msn.com
ARTICLE INFO ABSTRACT
Article History:
Received 15 September 2017 Accepted 18 February 2018 Available 09 March 2018
A number of Chebyshev type inequalities involving various fractional integral operators have, recently, been presented. In this work, motivated essentially by the earlier works and their applications in diverse research subjects, we es-tablish some new P´olya-Szeg¨o inequalities involving generalized Katugampola fractional integral operator and use them to prove some new fractional Cheby-shev type inequalities which are extensions of the results in the paper: [On P´olya-Szeg¨o and Chebyshev type inequalities involving the Riemann-Liouville fractional integral operators, J. Math. Inequal, 10(2) (2016)].
Keywords: P´olya-Szeg¨o inequality Chebyshev inequality Beta function Katugampola fractional integral operators AMS Classification 2010: 26A33, 26D10, 33B20
1. Introduction and preliminaries
This article is based on the well known Chebyshev functional [1]: 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 ,
where f and g are two integrable functions which are synchronous on [a, b], i.e.
(f (x) − f(y))(g(x) − g(y)) ≥ 0
for any x, y ∈ [a, b], then the Chebyshev inequal-ity states that T (f, g) ≥ 0.
For some recent counterparts, generalizations of
Chebyshev inequality, the reader is refer to [2–6]. We also need to introduce the P´olya and Szeg¨o in-equality [7]: Rb af2(x)dx Rb ag2(x)dx (Rb af (x)g(x)dx)2 ≤ 14 r M N mn + r mn M N !2 .
Using the above P´olya-Szeg¨o inequality, Dragomir and Diamond [8] established the following Gr¨uss type inequality:
Theorem 1. Let f, g : [a, b] → R+ be two
inte-grable functions so that
0 < m ≤ f(x) ≤ M < ∞ and
0 < n ≤ g(x) ≤ N < ∞
*Corresponding Author
for a.e. x ∈ [a, b]. Then, we have |T (f, g; a, b)| ≤ 14(M − m)(N − n)√ mnM N (1) × 1 b − a Z b a f (x)dx 1 b − a Z b a g(x)dx.
The constant 14 is best possible in (1) in the sense it can not be replaced by a smaller constant. For our purpose, we recall some other preliminar-ies: We note that the beta function B(α, β) is defined by (see, e.g. [9, Section 1.1])
B(α, β) (2) = Z 1 0 tα−1(1 − t)β−1dt (ℜ(α) > 0; ℜ(β) > 0) Γ(α) Γ(β) Γ(α + β) α, β ∈ C \ Z − 0 ,
where Γ is the familiar Gamma function. Here and in the following, let C, R, R+ and Z−
0 be the
sets of complex numbers, real numbers, positive real numbers and non-positive integers, respec-tively, and let R+0 := R+∪ {0}.
Definition 1. (see, e.g., [10], [11]) Let [a, b] (−∞ < a < b < ∞) be a finite interval on the real axis R. The Riemann-Liouville fractional integrals(left-sided) of order α ∈ C, Re(α) > 0 of a real function f ∈ L(a, b), is defined:
Ja+α f (x) (3) := 1 Γ(α) Z x a f (t) (x − t)1−αdt (x > a).
Definition 2. (see, e.g., [10], [11]) Let (a, b) (0 ≤ a < b ≤ ∞) be a finite or infinite inter-val on the half-axis R+. The Hadamard fractional
integrals(left-sided) of order α ∈ C, Re(α) > 0 of a real function f ∈ L(a, b) are defined by
Ha+α f (x) (4) := 1 Γ(α) Z x a logx t α−1 f (t) t dt (a < x < b) Definition 3. (see, e.g., [10], [11]) Let (a, b) (−∞ ≤ a < b ≤ ∞) be a finite or infinite interval on the half-axis R+. Also let Re(α) > 0, σ > 0
and η ∈ C. The Erdelyi-Kober fractional inte-grals (left-sided) of order α ∈ C of a real function f ∈ L(a, b) are defined by
Ia+,σ,ηα f (x) (5) = σx −σ(α+η) Γ(α) Z x a tσ(η+1)−1 (xα− tα)1−αf (t) dt (0 ≤ a < x < b ≤ ∞).
Definition 4. [12] Let [a, b] ⊂ R be a finite in-terval. The Katugampola fractional integrals (left-sided) of order α ∈ C, ρ > 0 Re(α > 0) of a real function f ∈ Xcp(a, b) are defined by
ρIα a+f (x) (6) := ρ 1−α Γ(α) Z x a tρ−1 (xρ− tρ)1−αf (t) dt. (x > a)
Definition 5. (see, e.g., [10], [11]) Let a con-tinuous function by parts in R = (−∞, ∞). The Liouville fractional integrals (left-sided) of order α ∈ C, ℜ(α) > 0 of a real function f, are defined by I+αf (x) (7) := 1 Γ(α) Z x −∞ f (t) (x − t)1−αdt (x ∈ R) .
Here, the space Xcp(a, b) (c ∈ R, 1 ≤ p ≤ ∞)
con-sists of those complex-valued Lebesgue measur-able functions ϕ on (a, b) for which kϕkXp
c < ∞, with kϕkXp c = Z b a |x c ϕ(x)|pdxx 1/p (1 ≤ p < ∞) and kϕkXp c = esssupx∈(a,b)[x c|ϕ(x)|].
In particular, when c = 1/p (1 ≤ p < ∞), the space Xcp(a, b) coincides with the classical Lp(a, b)
space.
Let 0 ≤ a < x < b ≤ ∞. Also, let ϕ ∈ Xcp(a, b),
α, ρ ∈ R+, and β, η, κ ∈ R. Then, the fractional integrals (left-sided and right-sided) of a function ϕ are defined, respectively, by (see [13])
ρIα,β a+,η,κϕ (x) (8) := ρ 1−βxκ Γ(α) Z x a τρ(η+1)−1 (xρ− τρ)1−αϕ(τ ) dτ and
ρIα,β b−,η,κϕ (x) (9) := ρ 1−βxρη Γ(α) Z b x τκ+ρ−1 (τρ− xρ)1−αϕ(τ ) dτ.
Remark 1. The fractional integral (8) contains five well-known fractional integrals as its particu-lar cases (see also [13–15]):
(i) Setting κ = 0, η = 0 and ρ = 1 in (8), the integral operator (8) reduces to the Riemann-Liouville fractional integral (3) (see also [10, p. 69]).
(ii) Setting κ = 0, η = 0, a = −∞ and ρ = 1 in (8), the integral operator (8) reduces to the Liouville fractional integral (7) (see also [10, p.79]).
(iii) Setting β = α, κ = 0, η = 0, and taking the limit ρ → 0+with L’Hˆospital’s rule in
(8), the integral operator (8) reduces to the Hadamard fractional integral (4) (see also [10, p. 110]).
(iv) Setting β = 0 and κ = −ρ(α + η) in (8), the integral operator (8) reduces to the Erd´elyi-Kober fractional integral (5) (see also [10, p. 105]).
(v) Setting β = α, κ = 0 and η = 0 in (8), the integral operator (8) reduces to the Katugampola fractional integral (6) (see also [12]).
The principle aim of the present paper is to es-tablish new P´olya-Szeg¨o inequalities and other of Chebyshev type by using generalized Katugam-pola fractional integration theory.
2. Main Results
In this section, we establish some new Chebyshev type inequalities involving the Katugampola frac-tional integration approach. Thanks to (2), we obtain (see [15, Eq. (3.1)])
ρ1−βxκ Γ(α) Z x 0 τρ(η+1)−1 (xρ− τρ)1−αdτ = x κ+ρ(η+α)Γ(η + 1) ρβΓ(α + η + 1) (10) = Λρ,βx,κ(α, η) α, x ∈ R+; β, ρ, η, κ ∈ R . We also let ρIα,β 0+,η,κϕ (x) :=ρIη,κα,βϕ(x).
Lemma 1. Let β, κ ∈ R, x, α, ρ ∈ R+, and η ∈ R+0. Let f and g be two positive integrable
functions on [0, ∞). Assume that there exist four positive integrable functions υ1, υ2, ω1 and ω2,
such that:
0 < υ1(τ ) ≤ f(τ) ≤ υ2(τ )
0 < ω1(τ ) ≤ g(τ) ≤ ω2(τ ) (11)
(τ ∈ [0, x], x > 0)
Then the following inequality holds:
ρIα,β η,κ ω1ω2f2 (x)ρIη,κα,βυ1υ2g2 (x) ρIα,β η,κ {(υ1ω1+ υ2ω2)f g} (x) 2 ≤ 1 4. (12)
Proof. From (11), for τ ∈ [0, x], x > 0, we can write υ2(τ ) ω1(τ )− f (τ ) g(τ ) ≥ 0 (13) and f (τ ) g(τ ) − υ1(τ ) ω2(τ ) ≥ 0. (14)
Multiplying (13) and (14), we get υ2(τ ) ω1(τ )− f (τ ) g(τ ) f (τ ) g(τ ) − υ1(τ ) ω2(τ ) ≥ 0. From the above inequality, we can write
(υ1(τ )ω1(τ ) + υ2(τ )ω2(τ )) f (τ )g(τ ) (15)
≥ ω1(τ )ω2(τ )f2(τ ) + υ1(τ )υ2(τ )g2(τ ).
Multiplying both sides of (15) by ρ1−βxκ
Γ(α)
τρ(η+1)−1 (xρ− τρ)1−α
and integrating the resulting inequality with re-spect to τ over (0, x), we get
ρIα,β
η,κ {(υ1ω1+ υ2ω2)f g} (x)
Applying the AM-GM inequality, i.e. (a + b ≥ 2√ab, a, b ∈ R+), we have ρIα,β η,κ {(υ1ω1+ υ2ω2)f g} (x) ≥ 2 q ρIα,β η,κ {ω1ω2f2} (x)ρIη,κα,β{υ1υ2g2} (x)
which implies that
ρIα,β
η,κ ω1ω2f2 (x)ρIη,κα,βυ1υ2g2 (x)
≤ 14ρIη,κα,β{(υ1ω1+ υ2ω2)f g} (x)
2 .
So, we get the desired result.
Corollary 1. If υ1 = m, υ2 = M , ω1 = n and ω2 = N , then we have ρIα,β η,κf2 (x)ρIα,β η,κg2 (x) ρIα,β η,κf g (x)2 ≤ 14 r mnM N +r MN mn !2
Remark 2. Setting κ = 0, η = 0 and ρ = 1 in Lemma 1, yields the inequality in [16, Lemma 3.1].
Lemma 2. Let β, κ ∈ R, x, α, θ, ρ ∈ R+, and η ∈ R+0. Let f and g be two positive integrable
functions on [0, ∞). Assume that there exist four positive integrable functions υ1, υ2, ω1 and ω2
satisfying condition (11). Then the following in-equality holds: ρIα,β η,κ {υ1υ2} (x)ρIη,κθ,β{ω1ω2} (x) ×ρIη,κα,βf2 (x)ρIθ,β η,κg2 (x) ≤ 1 4 ρIα,β η,κ {υ1f } (x)ρIη,κθ,β{ω1g} (x) (16) +ρIη,κα,β{υ2f } (x)ρIη,κθ,β{υ2g} (x) 2 .
Proof. From (11), we get υ2(τ ) ω1(ξ)− f (τ ) g(ξ) ≥ 0 and f (τ ) g(ξ) − υ1(τ ) ω2(ξ) ≥ 0 which lead to υ1(τ ) ω2(ξ) + υ2(τ ) ω1(ξ) f (τ ) g(ξ) ≥ f 2(τ ) g2(ξ) + υ1(τ )υ2(τ ) ω1(ξ)ω2(ξ) . (17)
Multiplying both sides of (17) by ω1(ξ)ω2(ξ)g2(ξ),
we get
υ1(τ )f (τ )ω1(ξ)g(ξ) + υ2(τ )f (τ )ω2(ξ)g(ξ)
≥ ω1(ξ)ω2(ξ)f2(τ ) + υ1(τ )υ2(τ )g2(ξ). (18)
Multiplying both sides of (18) by ρ2(1−β)x2κ Γ(α)Γ(θ) τρ(η+1)−1 (xρ− τρ)1−α ξρ(η+1)−1 (xρ− ξρ)1−θ
and integrating the resulting inequality with re-spect to τ and ξ over (0, x)2, we get
ρIα,β η,κ {υ1f } (x)ρIη,κθ,β{ω1g} (x) +ρIη,κα,β{υ2f } (x)ρIη,κθ,β{υ2g} (x) ≥ρIη,κα,βf2 (x)ρIθ,β η,κ{ω1ω2} (x) +ρIη,κα,β{υ1υ2} (x)ρIη,κθ,βg2 (x).
Applying the AM-GM inequality, we have
ρIα,β η,κ {υ1f } (x)ρIη,κθ,β{ω1g} (x) +ρIη,κα,β{υ2f } (x)ρIη,κθ,β{υ2g} (x) ≥ 2 q ρIη,κα,β{f2} (x)ρIη,κθ,β{ω 1ω2} (x) × q ρIα,β η,κ {υ1υ2} (x)ρIη,κθ,β{g2} (x).
So, we get the desired inequality of (16). Corollary 2. If υ1 = m, υ2 = M , ω1 = n and ω2 = N , then we have Λρ,βx,κ(α, η)Λρ,βx,κ(θ, η) × ρIα,β η,κf2 (x)ρIη,κα,βg2 (x) ρIα,β η,κf (x)ρIθ,β η,κg (x)2 ≤ 14 r mnM N +r MN mn !2
Remark 3. Setting κ = 0, η = 0 and ρ = 1 in Lemma 2 yields the inequality in [16, Lemma 3.3].
Lemma 3. Suppose that all assumptions of Lemma 2 are satisfied. Then, we have:
ρIα,β η,κ f2 (x)ρIη,κθ,βg2 (x) (19) ≤ρIη,κα,β υ2f g ω1 (x)ρIη,κθ,β ω2f g υ1 (x).
Proof. Using the condition (11), we get ρ1−βxκ Γ(α) Z x 0 τρ(η+1)−1 (xρ− τρ)1−αf 2(τ )dτ ≤ ρ 1−βxκ Γ(α) Z x 0 τρ(η+1)−1 (xρ− τρ)1−α υ2(τ ) ω1(τ ) f (τ )g(τ )dτ which leads to ρIα,β η,κ f2 (x) ≤ρIη,κα,β υ2f g ω1 (x). (20) Similarly, we have ρ1−βxκ Γ(θ) Z x 0 ξρ(η+1)−1 (xρ− ξρ)1−θg 2(ξ)dξ ≤ ρ 1−βxκ Γ(θ) Z x 0 ξρ(η+1)−1 (xρ− ξρ)1−θ ω2(ξ) υ1(ξ) f (ξ)g(ξ)dξ, which implies ρIθ,β η,κg2 (x) ≤ρIη,κθ,β ω2f g υ1 (x). (21)
Multiplying (20) and (21), we get the inequality
of (19). Corollary 3. If υ1 = m, υ2 = M , ω1 = n and ω2 = N , then we have ρIα,β η,κf2 (x)ρIα,β η,κg2 (x) ρIα,β η,κf g (x)ρIθ,β η,κf g (x) ≤ M N mn .
Remark 4. Setting κ = 0, η = 0 and ρ = 1 in Lemma 3 yields the inequality in [16, Lemma 3.4]. Theorem 2. Let β, κ ∈ R, x, α, θ, ρ ∈ R+, and η ∈ R+0. Let f and g be two positive integrable
functions on [0, ∞). Assume also that there exist four positive integrable functions υ1, υ2, ω1 and
ω2 satisfying the condition (11). Then the
follow-ing inequality holds:
Λρ,βx,κ(α, η)ρIη,κθ,βf g(x) +Λρ,βx,κ(θ, η)ρIη,κα,βf g(x) −ρIη,κα,βf(x)ρIη,κθ,βg(x) (22) −ρIη,κθ,βf(x)ρIη,κα,βg(x) ≤ |G1(f, υ1, υ2)(x) + G2(f, υ1, υ2)(x)|1/2 ×|G2(g, ω1, ω2)(x) + G2(g, ω1, ω2)(x)|1/2 where G1(f, υ1, υ2)(x) = Λ ρ,β x,κ(θ, η) 4 × ρIα,β η,κ {(υ1+ υ2)f } (x) 2 ρIη,κα,β{υ 1υ2} (x) −ρIη,κα,βf(x)ρIη,κθ,βf(x) and G2(f, ω1, ω2)(x) = Λ ρ,β x,κ(α, η) 4 × ρIθ,β η,κ{(ω1+ ω2)f } (x) 2 ρIθ,β η,κ{ω1ω2} (x) −ρIη,κα,βf(x)ρIη,κθ,βf(x).
Proof. Let f and g be two positive integrable functions on [0, ∞). For τ, ξ ∈ (0, x) with x > 0, we define H(τ, ξ) as
H(τ, ξ) = (f (τ ) − f(ξ)) (g(τ) − g(ξ)) , equivalently,
H(τ, ξ) (23)
= f (τ )g(τ ) + f (ξ)g(ξ) − f(τ)g(ξ) − f(ξ)g(τ). Multiplying both sides of (23) by
ρ2(1−β)x2κ Γ(α)Γ(θ) τρ(η+1)−1 (xρ− τρ)1−α ξρ(η+1)−1 (xρ− ξρ)1−θ
and double integrating the resulting inequality with respect to τ and ξ over (0, x)2, we get
ρ2(1 − β)x2κ Γ(α)Γ(θ) Z x 0 Z x 0 τρ(η+1)−1 (xρ− τρ)1−α × ξ ρ(η+1)−1 (xρ− ξρ)1−θH(τ, ξ)dτ dξ = Λρ,βx,κ(α, η)ρIη,κθ,βf g(x) +Λρ,βx,κ(θ, η)ρIη,κα,βf g(x) −ρIη,κα,βf(x)ρIη,κθ,βg(x) −ρIη,κθ,βf(x)ρIη,κα,βg(x). (24)
Applying the Cauchy-Schwarz inequality for dou-ble integrals, we can write
ρ2(1 − β)x2κ Γ(α)Γ(θ) Z x 0 Z x 0 τρ(η+1)−1 (xρ− τρ)1−α × ξ ρ(η+1)−1 (xρ− ξρ)1−θH(τ, ξ)dτ dξ ≤ ρ 2(1 − β)x2κ Γ(α)Γ(θ) Z x 0 Z x 0 τρ(η+1)−1 (xρ− τρ)1−α × ξ ρ(η+1)−1 (xρ− ξρ)1−θf 2(τ )dτ dξ +ρ2(1 − β)x2κ Γ(α)Γ(θ) × Z x 0 Z x 0 τρ(η+1)−1 (xρ− τρ)1−α ξρ(η+1)−1 (xρ− ξρ)1−θf 2(ξ)dτ dξ −2ρ 2(1 − β)x2κ Γ(α)Γ(θ) Z x 0 Z x 0 τρ(η+1)−1 (xρ− τρ)1−α × ξ ρ(η+1)−1 (xρ− ξρ)1−θf (τ )f (ξ)dτ dξ 1/2 × ρ 2(1 − β)x2κ Γ(α)Γ(θ) Z x 0 Z x 0 τρ(η+1)−1 (xρ− τρ)1−α × ξ ρ(η+1)−1 (xρ− ξρ)1−θg 2(τ )dτ dξ + ρ2(1 − β)x2κ Γ(α)Γ(θ) Z x 0 Z x 0 τρ(η+1)−1 (xρ− τρ)1−α ξρ(η+1)−1 (xρ− ξρ)θg 2(ξ)dτ dξ −2ρ 2(1 − β)x2κ Γ(α)Γ(θ) Z x 0 Z x 0 τρ(η+1)−1 (xρ− τρ)1−α × ξ ρ(η+1)−1 (xρ− ξρ)1−θg(τ )g(ξ)dτ dξ 1/2 Therefore, ρ2(1 − β)x2κ Γ(α)Γ(θ) Z x 0 Z x 0 τρ(η+1)−1 (xρ− τρ)1−α × ξ ρ(η+1)−1 (xρ− ξρ)1−θH(τ, ξ)dτ dξ ≤ Λρ,βx,κ(α, η)ρIη,κθ,βf2(x) +Λρ,βx,κ(θ, η)ρIη,κα,βf2(x) −2ρIη,κα,βf(x)ρIη,κθ,βf(x) 1/2 × Λρ,βx,κ(α, η)ρIη,κθ,βg2(x) +Λρ,βx,κ(θ, η)ρIη,κα,βg2(x) −2ρIη,κα,βg(x)ρIη,κθ,βg(x) 1/2 . (25)
Applying Lemma 1 with ω1(τ ) = ω2(τ ) = g(τ ) =
1, we get Λρ,βx,κ(θ, η)ρIη,κα,βf2(x) ≤ Λ ρ,β x,κ(θ, η) 4 ρIα,β η,κ {(υ1+ υ2)f } (x) 2 ρIη,κα,β{υ 1υ2} (x) . This implies that
Λρ,βx,κ(θ, η)ρIη,κα,βf2(x) −ρIη,κα,βf(x)ρIη,κθ,βf(x) ≤ Λ ρ,β x,κ(θ, η) 4 ρIα,β η,κ {(υ1+ υ2)f } (x) 2 ρIα,β η,κ {υ1υ2} (x) −ρIη,κα,βf(x)ρIη,κθ,βf(x) = G1(f, υ1, υ2)(x). (26) and Λρ,βx,κ(α, η)ρIη,κθ,βf2(x) −ρIη,κα,βf(x)ρIη,κθ,βf(x) ≤ Λ ρ,β x,κ(α, η) 4 ρIθ,β η,κ{(ω1+ ω2)f } (x) 2 ρIθ,β η,κ{ω1ω2} (x) −ρIη,κα,βf(x)ρIη,κθ,βf(x) = G2(f, ω1, ω2)(x). (27)
Similarly, applying Lemma 1 with υ1(τ ) =
Λρ,βx,κ(θ, η)ρIη,κα,βg2(x) −ρIη,κα,βg(x)ρIη,κθ,βg(x) ≤ G1(g, ω1, ω2)(x) (28) and Λρ,βx,κ(α, η)ρIη,κθ,βg2(x) −ρIη,κα,βg(x)ρIη,κθ,βg(x) ≤ G2(g, ω1, ω2)(x). (29)
Using (26)-(29), we conclude the desired
re-sult.
Remark 5. Setting κ = 0, η = 0 and ρ = 1 in Theorem 2, yields the inequality in [16, Theorem 3.6].
Theorem 3. Assume that all conditions of The-orem (2) are fulfilled. Then, we have:
Λρ,βx,κ(α, η)ρIη,κα,βf g(x) −ρIη,κα,βf(x)ρIη,κα,βg(x) ≤ |G(f, υ1, υ2)(x)G(g, ω1, ω2)(x)|1/2 (30) where G(u, v, w)(x) = Λ ρ,β x,κ(α, η) 4 × ρIα,β η,κ {(v + w)u} (x) 2 ρIη,κα,β{vw} (x) −ρIη,κα,βu(x)2.
Proof. Setting α = θ in (22), we obtain (30). Corollary 4. If υ1 = m, υ2 = M , ω1 = n and ω2 = N , then we have G(f, m, M )(x) = (M − m) 2 4M m ρIα,β η,κf (x)2, G(g, n, N )(x) = (N − n) 2 4N n ρIα,β η,κg (x)2.
Remark 6. We consider some particular cases of the result in Theorem 3.
(i) Setting κ = 0, η = 0 and ρ = 1 in the result in Theorem 3 yields the inequality in [16, Theorem 3.7],
(ii) Setting β = 0 and κ = −ρ(α + η) in the result in inequality 30 yields to
Γ(η + 1) Γ(α + η + 1) I α 0+,ρ,ηf g (x) − I0+,ρ,ηα f (x) I0+,ρ,ηα g (x) ≤ |G(f, υ1, υ2)(x)G(g, ω1, ω2)(x)|1/2 where G(u, v, w)(x) = Γ(η + 1) 4Γ(α + η + 1) × I α 0+,ρ,η{(v + w)u} (x) 2 I0+,ρ,ηα {vw} (x) − I0+,ρ,ηα u (x) 2
(iii) Setting β = α, κ = 0 and η = 0 in the re-sult in Theorem 3, under the correspond-ing reduced assumption, we obtain
xρα Γ(α + 1) ρIα 0+f g (x) − ρI0+α f (x) ρIα 0+g (x) ≤ |G(f, υ1, υ2)(x)G(g, ω1, ω2)(x)|1/2 where G(u, v, w)(x) = x ρα 4Γ(α + 1) × ρIα 0+{(v + w)u} (x) 2 ρIα 0+{vw} (x) − ρI0+α u (x)2. References
[1] ˇCebyˇsev, P.L. (1982). Sur les expressions approxima-tives des integrales definies par les autres prises entre les memes limites. Proc. Math. Soc. Charkov, 2, 93-98. [2] Belarbi, S. and Dahmani, Z. (2009). On some new fractional integral inequalities. J. Ineq. Pure Appl. Math., 10(3) Article 86, 5 pages.
[3] Dahmani, Z., Mechouar, O. and Brahami, S. (2011). Certain inequalities related to the Chebyshevs func-tional involving a type Riemann-Liouville operator. Bull. of Math. Anal. and Appl., 3(4), 38-44.
[4] Dahmani, Z. (2010). New inequalities in fractional in-tegrals. Int. J. of Nonlinear Sci., 9(4), 493-497. [5] Set, E., Akdemir, O.A. and Mumcu, ˙I. Chebyshev
type inequalities for conformable fractional integrals. Submitted.
[6] ¨Ozdemir, M.E., Set, E., Akdemir, O.A. and Sarıkaya, M.Z. (2015). Some new Chebyshev type inequalities for functions whose derivatives belongs to Lpspaces.
Afr. Mat., 26, 1609-1619.
[7] Polya, G. and Szeg¨o, G. (1925). Aufgaben und Lehrsatze aus der Analysis,Band 1, Die Grundlehren der mathmatischen Wissenschaften 19, J. Springer, Berlin.
[8] Dragomir, S.S. and Diamond, N.T. (2003). Integral inequalities of Gr¨uss type via P´olya-Szeg¨o and Shisha-Mond results. East Asian Math. J., 19, 27-39. [9] Srivastava, H.M. and Choi, J. (2012). Zeta and q-Zeta
Functions and Associated Series and Integrals, Else-vier Science Publishers, Amsterdam, London and New York.
[10] Kilbas, A.A., Srivastava, H.M. and Trujillo, J.J. (2006). Theory and applications of fractional differen-tial equations, North-Holland Mathematical Studies, Vol.204, Elsevier (North-Holland) Science Publishers, Amsterdam, London and New York.
[11] Podlubny, I. (1999). Fractional differential equations. An introduction to fractional derivatives, fractional differential equations, to methods of their Solution and some of their applications, Academic Press, San Diego, CA.
[12] Katugampola, U.N. (2014). A new approach to gener-alized fractional derivatives. Bull. Math. Anal. Appl., 6, 1-15.
[13] Katugampola, U.N. (2016). New fractional in-tegral unifying six existing fractional inin-tegrals. arXiv:1612.08596v1 [math.CA].
[14] Sousa, J. and Oliviera, E.C. The Minkowski’s inequal-ity by means of a generalized fractional integral. arXiv preprint arXiv:1705.07191.
[15] Sousa, J., Oliveira, D.S and Oliveira, E.C. Gr¨ uss-type inequality by mean of a fractional integral. arXiv preprint arXiv:1705.00965.
[16] Ntouyas, S.K., Agarwal, P. and Tariboon, J. (2016). On P´olya-Szeg¨o and Chebyshev type inequalities in-volving the Riemann-Liouville fractional integral op-erators. J. Math. Inequal, 10(2), 491-504.
Erhan Setis a Associate Professor at the Department of Mathematics in Ordu University, Ordu, Turkey. He received his Ph.D. degree in Analysis in 2010 from Atat¨urk University, Erzurum, Turkey. He is editor on the Turkish Journal of Inequalities. He has a lot of awards. He has many research papers about con-vex functions, the theory of inequalities and fractional calculus.
Zoubir Dahmaniis Professor, Labo. Pure and Appl. Maths., Faculty of Exact Sciences and Informatics, UMAB, University of Mostaganem, Algeria. He re-ceived his Ph.D. of Mathematics, USTHB (Algeria )and La ROCHELLE University (France ), 2009. He has many research papers about the Evolution tions,Inequality Theory, Fractional Differential Equa-tions,Fixed Point Theory, Numerical Analysis, Proba-bility Theory, Generalized Metric Spaces.
˙
Ilker Mumcu graduated from the department of mathematics teaching, Karadeniz Technical Univer-sity, Trabzon, Turkey in 2000. He received his mas-ter degree in Mathematics from Ordu University in 2014. Her research interests focus mainly in inequal-ity theory, especially Hermite- Hadamard, Gr¨uss type inequalities and fractional integral inequalities.
An International Journal of Optimization and Control: Theories & Applications (http://ijocta.balikesir.edu.tr)
This work is licensed under a Creative Commons Attribution 4.0 International License. The authors retain ownership of the copyright for their article, but they allow anyone to download, reuse, reprint, modify, distribute, and/or copy articles in IJOCTA, so long as the original authors and source are credited. To see the complete license contents, please visit http://creativecommons.org/licenses/by/4.0/.
![[PDF] Systèmes d’Information multimédias basés sur le Web](data:image/gif;base64,R0lGODlhAQABAIAAAP///wAAACH5BAEAAAAALAAAAAABAAEAAAICRAEAOw==)