# On Some Grüss-Type Fractional Inequalities Using Saigo Fractional Integral Operator

Authors: Vaijanath L. Chinchane; Deepak B. Pachpatte
Journal: Journal of Mathematics
 (2014)

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## Abstract

The main aim of this paper is to establish some new fractional integral inequalities of Grüss-type via the Saigo fractional integral operator.


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## Body

## 1. Introduction

Grüss inequality is an inequality which establishes a connection between the integral of the product of two functions and the product of the integrals of the two functions. In 1935, Grüss proved the following well-known classical integral inequality; see [1, 2].Theorem 1 (see [2]).
Letf,g:[a,b]→R be two integrable functions such that ϕ≤f(x)≤Φ and γ≤g(x)≤Γ for all x∈[a,b]; ϕ,Φ,γ, and Γ are constant; then
(1)1b-a∫abfxgxdxi-1b-a∫abfxdx·1b-a∫abgxdx≤14Γ-γΦ-ϕ,
where the constant 1/4 is sharp.In the past decade, many researchers had numerous generalizations, variants, and extensions of Grüss inequality done in literature; to mention a few, see [1–12] and the references cited therein.Recently, many authors have studied the fractional integral inequalities via Caputo, Riemann-Liouville, andq-fractional integral; see [6, 13–21]. Some authors have studied the Saigo fractional integral operator; for example, we refer the reader to [22–29] and references cited therein. In [19], Dahmani et al. gave the following fractional integral inequality using Riemann-Liouville fractional integral.Theorem 2 (see [19]).
Letf and g be two integrable functions on [0,∞) satisfying the condition
(2)m≤fx≤M,np≤gx≤P;m,M,p,P∈R,
on [0,∞). Then, for all t>0 and α>0, one has
(3)tαΓα+1Jαfgt-JαftJαgt≤tα2Γα+12M-mP-p.In literature few results have been obtained on some fractional integral inequalities using Hadamard fractional integral and Saigo fractional integral operator in [25, 26, 30–32]. Our purpose in this paper is to establish some new results using Saigo fractional integral.


## 2. Preliminaries

We give some definitions and properties which will be used later. For more details, see [28, 33].Definition 3.
A real-valued functionf(x) (x>1) is said to be in space Cμ(μ∈R), if there exists a real number p>μ such that f(x)=xpϕ(x), where ϕ(x)∈C(0,∞).Definition 4 (see [27, 28]).
Letα>0, β,η∈R; then the Saigo fractional integral I0,xα,β,η[f(x)] of order α for a real-valued continuous function f(x) is defined by
(4)I0,xα,β,ηfxW=x-α-βΓαWW·∫0xx-tα-1F21α+β,-η;α;1-txftdt,
where the function 2F1(-) in the right-hand side of (4) is the Gaussian hypergeometric function defined by
(5)F21a,b;c;x=∑n=0∞anbncnxnn!.
And (a)n is the Pochhammer symbol
(6)an=aa+1⋯a+n-1=Γa+nΓa,WWWWWWWa0=1.
For f(x)=xμ in (4), we have the known result [28] as
(7)I0,xα,β,ηxμ=Γμ+1Γμ+1-β+ηΓμ+1-βΓμ+1+α+ηxμ-β(α>0,min⁡(μ,μ-β+η)>-1,x>0).


## 3. Grüss-Type Fractional Integral Inequality

In this section, we establish Grüss-type fractional integral inequality involving Saigo fractional integer operator (4), for which we assume the following.(
A
1
) There exist two integrable functions Φ1(x) and Φ2(x) on [0,∞[, such that
(8)Φ1x≤ux≤Φ2x,∀x∈0,∞.(
A
2
) There exist two integrable functions Ψ1(x) and Ψ2(x) on [0,∞[, such that
(9)Ψ1x≤vx≤Ψ2x,∀x∈0,∞.Lemma 5.
Suppose thatu is an integrable function on [0,∞) and Φ1(x) and Φ2(x) are two integrable functions on [0,∞). Assume that the condition (A1) holds.
Then, for allx>0, α>max⁡{0,-β}, β<1, β-1<η<0, one has
(10)Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηu2x-I0,xα,β,ηux2W=I0,xα,β,ηΦ2x-I0,xα,β,ηuxWW·I0,xα,β,ηux-I0,xα,β,ηΦ1xWW-Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηWW·Φ2x-uxux-Φ1xWW+Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηΦ1uxWW-I0,xα,β,ηΦ1xI0,xα,β,ηuxWW+Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηΦ2uxWW-I0,xα,β,ηΦ2xI0,xα,β,ηuxWW+I0,xα,β,ηΦ1xI0,xα,β,ηΦ2xWW-Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηΦ1Φ2x.Proof.
Letu be an integrable function on [0,∞). For all ρ,σ>0, we have
(11)Φ2σ-uσuρ-Φ1ρWW+Φ2ρ-uρuσ-Φ1σWW-Φ2ρ-uρuρ-Φ1ρWW-Φ2σ-uσuσ-Φ1σ=u2ρ+u2σ-2uσuρWW+Φ2σuρ+Φ1ρuσ-Φ1ρΦ2σWW+Φ1ρuσ+Φ1σuρ-Φ1σΦ2ρWW-Φ2ρuρ+Φ1ρΦ2ρ-Φ1ρuρWW-Φ2σuσ+Φ1σΦσ-Φ1σuσ.
Consider
(12)Gx,ρ=x-α-βx-ρα-1Γα·F21α+β,-η;α;1-ρx=1Γαx-ρα-1xα+β+α+β-ηΓα+1x-ραxα+β+1+α+βα+β+1-η-η+1Γα+2x-ρα+1xα+β+2+⋯,ρ∈0,x;x>0.
Clearly, function G(x,ρ) positive since for all ρ∈(0,x)(x>0). Multiplying both sides of (11) by G(x,ρ), then integrating resulting identity with respect to ρ from 0 to x, we get
(13)Φ2σ-uσI0,xα,β,ηux-I0,xα,β,ηΦ1xWW+I0,xα,β,ηΦ2x-I0,xα,β,ηuxuσ-Φ1σWW-I0,xα,β,ηΦ2x-uxux-Φ1xWW-Γ1-β+ηΓ1-βΓ1+α+ηxβWW·Φ2σ-uσuσ-Φ1σW=I0,xα,β,ηu2x+u2σΓ1-β+ηΓ1-βΓ1+α+ηxβWW-2uσI0,xα,β,ηux+Φ2σI0,xα,β,ηuxWW+uσI0,xα,β,ηΦ1x-Φ2σI0,xα,β,ηΦ1xWW+uσI0,xα,β,ηΦ2x+Φ1σI0,xα,β,ηuxWW-Φ1σI0,xα,β,ηΦ2x-I0,xα,β,ηΦ2uxWW+I0,xα,β,ηΦ1Φ2x-I0,xα,β,ηΦ1uxWW-Φ2σuσΓ1-β+ηΓ1-βΓ1+α+ηxβWW+Φ1σΦ2σΓ1-β+ηΓ1-βΓ1+α+ηxβWW-Φ1σuσΓ1-β+ηΓ1-βΓ1+α+ηxβ.
Multiplying both sides of (13) by G(x,σ) and is positive since for all σ∈(0,x)  (x>0), and defined as in (12). Then integrating resulting identity with respect to σ from 0 to x, we have
(14)I0,xα,β,ηΦ2x-I0,xα,β,ηuxW·I0,xα,β,ηux-I0,xα,β,ηΦ1xW+I0,xα,β,ηΦ2x-I0,xα,β,ηuxW·I0,xα,β,ηux-I0,xα,β,ηΦ1xW-Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηW·Φ2x-uxux-Φ1xW-Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηW·Φ2x-uxux-Φ1x=I0,xα,β,ηu2xΓ1-β+ηΓ1-βΓ1+α+ηxβW+I0,xα,β,ηu2xΓ1-β+ηΓ1-βΓ1+α+ηxβW-2I0,xα,β,ηuxI0,xα,β,ηux+I0,xα,β,ηΦ2xI0,xα,β,ηuxW+I0,xα,β,ηuxI0,xα,β,ηΦ1xW-I0,xα,β,ηΦ2xI0,xα,β,ηΦ1xW+I0,xα,β,ηuxI0,xα,β,ηΦ2x+I0,xα,β,ηΦ1xI0,xα,β,ηuxW-I0,xα,β,ηΦ1xI0,xα,β,ηΦ2xW-I0,xα,β,ηΦ2uxΓ1-β+ηΓ1-βΓ1+α+ηxβW+I0,xα,β,ηΦ1Φ2xΓ1-β+ηΓ1-βΓ1+α+ηxβW-I0,xα,β,ηΦ1uxΓ1-β+ηΓ1-βΓ1+α+ηxβW-I0,xα,β,ηΦ2uxΓ1-β+ηΓ1-βΓ1+α+ηxβW+I0,xα,β,ηΦ1Φ2xΓ1-β+ηΓ1-βΓ1+α+ηxβW-I0,xα,β,ηΦ1uxΓ1-β+ηΓ1-βΓ1+α+ηxβ.
It follows that
(15)2I0,xα,β,ηΦ2x-I0,xα,β,ηuxWW·I0,xα,β,ηux-I0,xα,β,ηΦ1xWW-2Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηWW·Φ2x-uxux-Φ1xW=2I0,xα,β,ηu2xΓ1-β+ηΓ1-βΓ1+α+ηxβWW-2I0,xα,β,ηux2+2I0,xα,β,ηΦ2xI0,xα,β,ηuxWW+2I0,xα,β,ηuxI0,xα,β,ηΦ1xWW-2I0,xα,β,ηΦ2xI0,xα,β,ηΦ1xWW-2I0,xα,β,ηΦ2uxΓ1-β+ηΓ1-βΓ1+α+ηxβWW+2I0,xα,β,ηΦ1Φ2xΓ1-β+ηΓ1-βΓ1+α+ηxβWW-2I0,xα,β,ηΦ1uxΓ1-β+ηΓ1-βΓ1+α+ηxβ,
which gives (10).IfΦ1(x)=p and Φ2x=P,p,P∈R for all x∈[0,∞[, then inequality (10) reduces to following lemma.Lemma 6.
Letp,P∈R, and u(x) be an integrable function on [0,∞) satisfying the condition p≤u(x)≤P. Then for all x>0, α>max⁡{0,-β}, β<1, β-1<η<0, one has
(16)Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηu2x-I0,xα,β,ηux2=PΓ1-β+ηΓ1-βΓ1+α+ηxβ-I0,xα,β,ηux·I0,xα,β,ηux-pΓ1-β+ηΓ1-βΓ1+α+ηxβ-Γ1-β+ηΓ1-βΓ1+α+ηxβ×I0,xα,β,ηP-uxux-p.Theorem 7.
Letu and v be two integrable functions on [0,∞), and Φ1(x), Φ2(x), Ψ1(x), and Ψ2(x) are four integrable functions on [0,∞) satisfying the conditions A1 and A2 on [0,∞). Then for all x>0, α>max⁡{0,-β}, β<1, β-1<η<0, one has
(17)Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηuvxnΓ1-β+ηΓ1-βΓ1+α+ηxβ-I0,xα,β,ηuxI0,xα,β,ηvx≤Tu,Φ1x,Φ2xTv,Ψ1x,Ψ2x,
where T(a,b,c) is defined by
(18)Ta,b,c=I0,xα,β,ηcx-I0,xα,β,ηaxI0,xα,β,ηax-I0,xα,β,ηbx+Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηbax-I0,xα,β,ηbxI0,xα,β,ηax+Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηcax-I0,xα,β,ηcxI0,xα,β,ηax+I0,xα,β,ηbxI0,xα,β,ηcx+Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηbcx.Proof.
Letu and v be two functions defined on [0,∞[ satisfying the conditions A1 and A2. Define
(19)Hρ,σ≔uρ-uσvρ-vσ;MMMMMMM1Mρ,σ∈0,x,x>0.
It follows that
(20)Hρ,σ≔uρvρ-uρvσ-uσvρW+uσvσ.
Then, multiplying both sides of (20) by G(x,ρ) which is defined by  (12) and is positive because for all ρ∈(0,x)(x>0). Then integrating resulting identity with respect to ρ from 0 to x, we have
(21)x-α-βΓα∫0xx-ρα-1F21α+β,-η;α;1-ρxHρ,σdρW=I0,xα,β,ηuvx-uσI0,xα,β,ηuxWW-uσI0,xα,β,ηvx+uσvσI0,xα,β,η.
Again, multiplying both sides of (21) by G(x,σ) which is defined by  (12) and is positive because for all σ∈(0,x)(x>0) since each term of (12) is positive. Then integrating resulting identity with respect to σ from 0 to x, we have
(22)x-α-βx-α-βΓ2αWW·∬0xx-ρα-1x-σα-1F21α+β,-η;α;1-ρxWWWWW·F21α+β,-η;α;1-σxHρ,σdρdσW=2Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηuvxWWWWiΓ1-β+ηΓ1-βΓ1+α+ηxβ-I0,xα,β,ηuxI0,xα,β,ηvx.
Applying the Cauchy-Schwarz inequality to (22), we have
(23)Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηI0,xα,β,ηuvxΓ1-β+ηΓ1-βΓ1+α+ηxβ-I0,xα,β,ηuxI0,xα,β,ηvx2W≤Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηI0,xα,β,ηu2xWWWΓ1-β+ηΓ1-βΓ1+α+ηxβ-I0,xα,β,ηux2WW×Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηI0,xα,β,ηv2xWWWWΓ1-β+ηΓ1-βΓ1+α+ηxβ-I0,xα,β,ηvx2.
Since (Φ2(x)-u(t))(u(t)-Φ1(x))≥0 and (Ψ2(x)-v(t))(v(t)-Ψ1(x))≥0, we have
(24)Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηWW·Φ2x-utut-Φ1x≥0,Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηWW·Ψ2x-vtvt-Ψ1x≥0.
Thus we have
(25)Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηu2x-I0,xα,β,ηux2≤I0,xα,β,ηΦ2x-I0,xα,β,ηux·I0,xα,β,ηux-I0,xα,β,ηΦ1x+Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηΦ1ux-I0,xα,β,ηΦ1xI0,xα,β,ηux+Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηΦ2ux-I0,xα,β,ηΦ2xI0,xα,β,ηux+I0,xα,β,ηΦ1xI0,xα,β,ηΦ2x-Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηΦ1Φ2x=Tu,Φ1,Φ2,Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηu2x-I0,xα,β,ηux2≤I0,xα,β,ηΨ2x-I0,xα,β,ηux·I0,xα,β,ηux-I0,xα,β,ηΨ1x+Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηΨ1ux-I0,xα,β,ηΨ1xI0,xα,β,ηux+Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηΨ2ux-I0,xα,β,ηΨ2xI0,xα,β,ηux+I0,xα,β,ηΨ1xI0,xα,β,ηΨ2x-Γ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηΨ1Ψ2x=Tu,Ψ1,Ψ2.
Combining (23) and (25) we get the required inequality (17).Theorem 8.
Suppose thatu is an integrable function defined on [0,∞[ and condition (A1) holds. Then for all x>0, α>max⁡{0,-β}, ψ>max⁡{0,-ϕ}, β<1, β-1<η<0, ϕ<1, ϕ-1<ζ<0,
(26)I0,xψ,ϕ,ζΦ1xI0,xα,β,ηuxWW+I0,xα,β,ηΦ2xI0,xψ,ϕ,ζuxW≥I0,xα,β,ηΦ2xI0,xψ,ϕ,ζΦ1xWW+I0,xα,β,ηuxI0,xψ,ϕ,ζux.Proof.
From the condition(A1), for all ρ,σ≥0, we have
(27)Φ2ρ-uρuσ-Φ1σ≥0,Φ2ρuσ-Φ2ρΦ1σ-uρuσ+uρΦ1σ≥0,
which implies that
(28)Φ2ρuσ+uρΦ1σ≥Φ2ρΦ1σ+uρuσ.
Multiplying both sides of (28) by G(x,ρ) which is defined by (12) and remain positive because for all ρ∈(0,x)(x>0) since each term of (12) is positive. Then integrating resulting identity with respect to ρ from 0 to x, we get
(29)uσx-α-βΓαW·∫0xx-ρα-1F21α+β,-η;α;1-ρxΦ2ρdρW+Φ1σx-α-βΓαW·∫0xx-ρα-1F21α+β,-η;α;1-ρxuρdρ≥Φ1σx-α-βΓαW·∫0xx-ρα-1F21α+β,-η;α;1-ρxΦ2ρdρW·uσx-α-βΓαW·∫0xx-ρα-1F21α+β,-η;α;1-ρxuρdρ.
Therefore,
(30)uσI0,xα,β,ηΦ2x+Φ1σI0,xα,β,ηux≥Φ1σI0,xα,β,ηΦ2x+uσI0,xα,β,ηux.
Multiplying (30) by x-ψ-ϕ/Γ(ψ)(x-σ)ψ-1F21(ψ+ϕ,-ζ;ψ;1-σ/x)(σ∈(0,x), x>0), which remains positive. Then integrating the resulting identity with respect to σ from 0 to x, we have
(31)I0,xα,β,ηΦ2xx-ψ-ϕΓψWW·∫0xx-σα-1F21ψ+ϕ,-ζ;ψ;1-σxuσdσWW·I0,xα,β,ηuxx-ψ-ϕΓψWW·∫0xx-σα-1F21ψ+ϕ,-ζ;ψ;1-σxΦ1σdσW≥I0,xα,β,ηΦ2xx-ψ-ϕΓψWW·∫0xx-σα-1F21ψ+ϕ,-ζ;ψ;1-σxΦ1σdσWW·I0,xα,β,ηuxx-ψ-ϕΓψWW·∫0xx-σα-1F21ψ+ϕ,-ζ;ψ;1-σxuσdσ,
which gives (26) and proves the theorem.Remark 9.
Suppose thatu is an integrable function defined on [0,∞[, such that p≤u(x)≤P, for all x∈[0,∞[ and p,P∈R. Then for all x>0,  α>max⁡{0,-β},  ψ>max⁡{0,-ϕ},  β<1,  β-1<η<0,  ϕ<1,  ϕ-1<ζ<0, we have
(32)pΓ1-ϕ+ζΓ1-ϕΓ1+ψ+ζxϕI0,xα,β,ηuxWW+PΓ1-β+ηΓ1-βΓ1+α+ηxβI0,xψ,ϕ,ζuxW≥PpΓ1-β+ηΓ1-ϕ+ζΓ1-βΓ1+α+ηΓ1-ϕΓ1+ψ+ζxβxϕWW+I0,xα,β,ηuxI0,xψ,ϕ,ζux.Theorem 10.
Suppose thatu and v are two integrable functions defined on [0,∞[ and suppose that (A1) and (A2) hold. Then for all x>0, α>max⁡{0,-β},  ψ>max⁡{0,-ϕ}, β<1, β-1<η<0, ϕ<1, ϕ-1<ζ<0, the following inequalities are satisfied:
(33)aI0,xψ,ϕ,ζΨ1xI0,xα,β,ηux+I0,xα,β,ηΦ2xI0,xψ,ϕ,ζvxWwW≥I0,xψ,ϕ,ζΨ1xI0,xα,β,ηΦ2xWwWW+I0,xα,β,ηuxI0,xψ,ϕ,ζvx.bI0,xψ,ϕ,ζΦ1xI0,xα,β,ηvx+I0,xα,β,ηΨ2xI0,xψ,ϕ,ζuxWwW≥I0,xψ,ϕ,ζΦ1xI0,xα,β,ηΨ2xWwWW+I0,xψ,ϕ,ζuxI0,xα,β,ηvx.cI0,xα,β,ηΦ2xI0,xψ,ϕ,ζΨ2x+I0,xα,β,ηuxI0,xψ,ϕ,ζvxWwW≥I0,xα,β,ηΦ2xI0,xψ,ϕ,ζvxWwWW+I0,xψ,ϕ,ζΨ2xI0,xα,β,ηux.dI0,xα,β,ηΦ1xI0,xψ,ϕ,ζΨ1x+I0,xα,β,ηuxI0,xψ,ϕ,ζvxWwW≥I0,xα,β,ηΦ1xI0,xψ,ϕ,ζvxWwWW+I0,xψ,ϕ,ζΨ1xI0,xα,β,ηux.Proof.
To show(a), we use conditions (A1) and (A2), for all x∈[0,∞[; we have
(34)Φ2ρ-uρvσ-Ψ1σ≥0,
which implies that
(35)Φ2ρvσ+uρΨ1σ≥Φ2ρΨ1σ+uρuσ.
Multiplying both sides of (35) by G(x,ρ), which is defined by (12) and remain positive because for all ρ∈(0,x)(x>0), since each term of (12) is positive. Then integrating the resulting identity with respect to ρ from 0 to x, we get
(36)vσx-α-βΓαWW·∫0xx-ρα-1F21α+β,-η;α;1-ρxΦ2ρdρWW+Ψ1σx-α-βΓαWW·∫0xx-ρα-1F21α+β,-η;α;1-ρxuρdρW≥Ψ1σx-α-βΓαWW·∫0xx-ρα-1F21α+β,-η;α;1-ρxΦ2ρdρWW·vσx-α-βΓαWW·∫0xx-ρα-1F21α+β,-η;α;1-ρxuρdρ.
Therefore,
(37)vσI0,xα,β,ηΦ2x+Ψ1σI0,xα,β,ηux≥Ψ1σI0,xα,β,ηΦ2x+vσI0,xα,β,ηux.
Multiplying (37) by x-ψ-ϕ/Γ(ψ)(x-σ)ψ-1F21(ψ+ϕ,-ζ;ψ;1-σ/x)(σ∈(0,x), x>0), which remains positive. Then by integrating the resulting identity with respect to σ from 0 to x, we have
(38)I0,xα,β,ηΦ2xx-ψ-ϕΓψWW·∫0xx-σα-1F21ψ+ϕ,-ζ;ψ;1-σxvσdσWW·I0,xα,β,ηuxx-ψ-ϕΓψWW·∫0xx-σα-1F21ψ+ϕ,-ζ;ψ;1-σxΨ1σdσW≥I0,xα,β,ηΦ2xx-ψ-ϕΓψWW·∫0xx-σα-1F21ψ+ϕ,-ζ;ψ;1-σxΨ1σdσWW·I0,xα,β,ηuxx-ψ-ϕΓψWW·∫0xx-σα-1F21ψ+ϕ,-ζ;ψ;1-σxvσdσ.
This gives the desired inequality(a). Using the following condition we prove (b)–(d):
(39)Ψ2ρ-vρuσ-Φ1σ≥0,Φ2ρ-uρvσ-Ψ2σ≤0,Φ2ρ-uρvσ-Ψ1σ≤0.Remark 11.
Suppose thatu and v are two integrable functions defined on [0,∞[. Assume that there exist real constants P,p,Q, and q such that
(40)p≤ux≤P,q≤vx≤Q,ii∀x∈0,∞.
Then for all x>0, α>max⁡{0,-β}, ψ>max⁡{0,-ϕ}, β<1, β-1<η<0, ϕ<1, ϕ-1<ζ<0, the following inequalities are satisfied:
(41)a0qΓ1-ϕ+ζΓ1-ϕΓ1+ψ+ζxϕI0,xα,β,ηuxWiWw+PI0,xα,β,ηI0,xψ,ϕ,ζvxWiw≥qPΓ1-β+ηΓ1-ϕ+ζΓ1-βΓ1+α+ηΓ1-ϕΓ1+ψ+ζxβxϕWiWw+I0,xα,β,ηuxI0,xψ,ϕ,ζvx.b0pΓ1-β+ηΓ1-βΓ1+α+ηxβI0,xα,β,ηvxWiWw+QΓ1-β+ηΓ1-βΓ1+α+ηxβI0,xψ,ϕ,ζuxWiw≥pQΓ1-β+ηΓ1-ϕ+ζΓ1-βΓ1+α+ηΓ1-ϕΓ1+ψ+ζxβxϕWiWw+I0,xψ,ϕ,ζuxI0,xα,β,ηvx.c0PQΓ1-β+ηΓ1-ϕ+ζΓ1-βΓ1+α+ηΓ1-ϕΓ1+ψ+ζxβxϕWiWw+I0,xα,β,ηuxI0,xψ,ϕ,ζvxWiw≥PΓ1-β+ηΓ1-βΓ1+α+ηxβI0,xψ,ϕ,ζvxWWiw+pΓ1-ϕ+ζΓ1-ϕΓ1+ψ+ζxϕI0,xα,β,ηux.d0pqΓ1-ϕ+ζΓ1-ϕΓ1+ψ+ζxϕWiWw+I0,xα,β,ηuxI0,xψ,ϕ,ζvxWiw≥pΓ1-ϕ+ζΓ1-ϕΓ1+ψ+ζxϕI0,xψ,ϕ,ζvxWiWw+qΓ1-ϕ+ζΓ1-ϕΓ1+ψ+ζxϕI0,xα,β,ηux.
It is noted that the results give some contributions to the theory of integral inequalities and fractional calculus. Moreover, they are expected to lead to some applications for establishing uniqueness of solutions in fractional differential equations.



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*Source: 527910-2014-08-24.xml*