# Fractional Integral Inequalities via Hadamard’s Fractional Integral

Authors: Weerawat Sudsutad; Sotiris K. Ntouyas; Jessada Tariboon
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We establish new fractional integral inequalities, via Hadamard’s fractional integral. Several new integral inequalities are obtained, including a Grüss type Hadamard fractional integral inequality, by using Young and weighted AM-GM inequalities. Many special cases are also discussed.


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## Body

## 1. Introduction

Inequalities have proved to be one of the most powerful and far-reaching tools for the development of many branches of mathematics. The study of mathematical inequalities plays very important role in classical differential and integral equations which has applications in many fields. Fractional inequalities are important in studying the existence, uniqueness, and other properties of fractional differential equations. Recently many authors have studied integral inequalities on fractional calculus using Riemann-Liouville and Caputo derivative; see [1–7] and the references therein.Another kind of fractional derivative that appears in the literature is the fractional derivative due to Hadamard introduced in 1892 [8], which differs from the Riemann-Liouville and Caputo derivatives in the sense that the kernel of the integral contains logarithmic function of arbitrary exponent. Details and properties of Hadamard fractional derivative and integral can be found in [9–14]. Recently in the literature, there appeared some results on fractional integral inequalities using Hadamard fractional integral; see [15–17].In this paper we present some new fractional integral inequalities using the Hadamard fractional integral. Several new integral inequalities are obtained by using Young and weighted AM-GM inequalities. Many special cases are also discussed. Moreover, a Grüss type Hadamard fractional integral inequality is obtained.


## 2. Preliminaries

In this section we give some preliminaries and basic proposition used in our subsequent discussion. The necessary background details are given in the book by Kilbas et al. [9].Definition 1.
The Hadamard fractional integral of orderα∈ℝ+ of a function f(t), for all t>1, is defined as
(1)HJαf(t)=1Γ(α)∫1t‍(log⁡ts)α-1f(s)dss,
where Γ is the standard gamma function defined by Γ(α)=∫0∞‍e-ssα-1ds, provided the integral exists, where log⁡(·)=log⁡e(·).Definition 2.
The Hadamard fractional derivative of orderα∈[n-1,n), n∈ℤ+, of a function f(t) is given by
(2)HDαf(t)=1Γ(n-α)(tddt)n∫1t‍(log⁡ts)n-α-1f(s)dss.Next, we recall a proposition concerning a Hadamard integral and derivative.Proposition 3 (see [9]).
Ifα>0, the following relations hold:
(3)HJα(log⁡t)β-1=Γ(β)Γ(β+α)(log⁡t)β+α-1,(4)HDα(log⁡t)β-1=Γ(β)Γ(β-α)(log⁡t)β-α-1,
respectively.For the convenience of establishing our results, we give the semigroup property(5)HJαHJβf(t)=HJα+βf(t),α≥0,β≥0,
which implies the commutative property
(6)HJαHJβf(t)=HJβHJαf(t).


## 3. Main Results

Now, we are in a position to give our main results.Theorem 4.
Letf be an integrable function on [1,∞). Assume the following.
(H1) There exist two integrable functions φ1,φ2 on [1,∞) such that
(7)φ1(t)≤f(t)≤φ2(t),∀t∈[1,∞).
Then, for t>1, α,β>0, one has
(8)HJβφ1(t)HJαf(t)+HJαφ2(t)HJβf(t)≥HJαφ2(t)HJβφ1(t)+HJαf(t)HJβf(t).Proof.
From(H1), for all τ≥1, ρ≥1, we have
(9)(φ2(τ)-f(τ))(f(ρ)-φ1(ρ))≥0.
Therefore,
(10)φ2(τ)f(ρ)+φ1(ρ)f(τ)≥φ1(ρ)φ2(τ)+f(τ)f(ρ).
Multiplying both sides of (10) by (log⁡(t/τ))α-1/τΓ(α), τ∈(1,t), we get
(11)f(ρ)(log⁡(t/τ))α-1τΓ(α)φ2(τ)+φ1(ρ)(log⁡(t/τ))α-1τΓ(α)f(τ)≥φ1(ρ)(log⁡(t/τ))α-1τΓ(α)φ2(τ)+f(ρ)(log⁡(t/τ))α-1τΓ(α)f(τ).
Integrating both sides of (11) with respect to τ on (1,t), we obtain
(12)f(ρ)1Γ(α)∫1t‍(log⁡tτ)α-1φ2(τ)dττ+φ1(ρ)1Γ(α)∫1t‍(log⁡tτ)α-1f(τ)dττ≥φ1(ρ)1Γ(α)∫1t‍(log⁡tτ)α-1φ2(τ)dττ+f(ρ)1Γ(α)∫1t‍(log⁡tτ)α-1f(τ)dττ,
which yields
(13)f(ρ)HJαφ2(t)+φ1(ρ)HJαf(t)≥φ1(ρ)HJαφ2(t)+f(ρ)HJαf(t).
Multiplying both sides of (13) by (log⁡(t/ρ))β-1/ρΓ(β), ρ∈(1,t), we have
(14)HJαφ2(t)(log⁡(t/ρ))β-1ρΓ(β)f(ρ)+HJαf(t)(log⁡(t/ρ))β-1ρΓ(β)φ1(ρ)≥HJαφ2(t)(log⁡(t/ρ))β-1ρΓ(β)φ1(ρ)+HJαf(t)(log⁡(t/ρ))β-1ρΓ(β)f(ρ).
Integrating both sides of (14) with respect to ρ on (1,t), we get
(15)HJαφ2(t)1Γ(β)∫1t‍(log⁡tρ)β-1f(ρ)dρρ+HJαf(t)1Γ(β)∫1t‍(log⁡tρ)β-1φ1(ρ)dρρ≥HJαφ2(t)1Γ(β)∫1t‍(log⁡tρ)β-1φ1(ρ)dρρ+HJαf(t)1Γ(β)∫1t‍(log⁡tρ)β-1f(ρ)dρρ.
Hence, we deduce inequality (8) as requested. This completes the proof.As special cases of Theorems4, we obtain the following results.Corollary 5.
Letf be an integrable function on [1,∞) satisfying m≤f(t)≤M, for all t∈[1,∞) and m,M∈ℝ. Then for t>1 and α,β>0, one has
(16)m(log⁡t)βΓ(β+1)HJαf(t)+M(log⁡t)αΓ(α+1)HJβf(t)≥mM(log⁡t)α+βΓ(α+1)Γ(β+1)+HJαf(t)HJβf(t).Corollary 6.
Letf be an integrable function on [1,∞). Assume that there exists an integrable function φ(t) on [1,∞) and a constant M>0 such that
(17)φ(t)-M≤f(t)≤φ(t)+M,
for all t∈[1,∞). Then for t>1 and α,β>0, one has
(18)HJβφ(t)HJαf(t)+HJαφ(t)HJβf(t)+M(log⁡t)βΓ(β+1)HJαφ(t)+M(log⁡t)αΓ(α+1)HJβf(t)+M2(log⁡t)α+βΓ(α+1)Γ(β+1)≥HJαφ(t)HJβφ(t)+HJαf(t)HJβf(t)+M(log⁡t)βΓ(β+1)HJαf(t)+M(log⁡t)αΓ(α+1)HJβφ(t).Example 7.
Letf be a function satisfying log⁡t≤f(t)≤1+log⁡t for t∈[1,∞). Then for t>1 and α>0, we have
(19)(2(log⁡t)α+1Γ(α+2)+(log⁡t)αΓ(α+1))HJαf(t)≥((log⁡t)α+1Γ(α+2)+(log⁡t)αΓ(α+1))((log⁡t)α+1Γ(α+2))+(HJαf(t))2.Theorem 8.
Letf be an integrable function on [1,∞) and θ1,θ2>0 satisfying 1/θ1+1/θ2=1. Suppose that (H1) holds. Then, for t>1, α,β>0, one has
(20)1θ1(log⁡t)βΓ(β+1)HJα((φ2-f)θ1)(t)+1θ2(log⁡t)αΓ(α+1)HJβ((f-φ1)θ2)(t)+HJαφ2(t)HJβφ1(t)+HJαf(t)HJβf(t)≥HJαφ2(t)HJβf(t)+HJαf(t)HJβφ1(t).Proof.
According to the well-known Young’s inequality(21)1θ1xθ1+1θ2yθ2≥xy,∀x,y≥0,θ1,θ2>0,1θ1+1θ2=1,
setting x=φ2(τ)-f(τ) and y=f(ρ)-φ1(ρ), τ,ρ>1, we have
(22)1θ1(φ2(τ)-f(τ))θ1+1θ2(f(ρ)-φ1(ρ))θ2≥(φ2(τ)-f(τ))(f(ρ)-φ1(ρ)).
Multiplying both sides of (22) by (log⁡(t/τ))α-1(log⁡(t/ρ))β-1/τρΓ(α)Γ(β), τ,ρ∈(1,t), we get
(23)1θ1(log⁡(t/τ))α-1(log⁡(t/ρ))β-1τρΓ(α)Γ(β)(φ2(τ)-f(τ))θ1+1θ2(log⁡(t/τ))α-1(log⁡(t/ρ))β-1τρΓ(α)Γ(β)(f(ρ)-φ2(ρ))θ2≥(log⁡(t/τ))α-1τΓ(α)(φ2(τ)-f(τ))(log⁡(t/ρ))β-1ρΓ(β)×(f(ρ)-φ1(ρ)).
Integrating the above inequality with respect to τ and ρ from 1 to t, we have
(24)1θ1HJβ(1)(t)HJα(φ2-f)θ1(t)+1θ2HJα(1)(t)HJβ(f-φ1)θ2(t)≥HJα(φ2-f)(t)HJβ(f-φ1)(t),
which implies (20).Corollary 9.
Letf be an integrable function on [1,∞) satisfying m≤f(t)≤M, for all t∈[1,∞) and m,M∈ℝ. Then for t>1 and α,β>0, one has
(25)(m+M)2(log⁡t)α+βΓ(α+1)Γ(β+1)+(log⁡t)βΓ(β+1)HJαf2(t)+(log⁡t)αΓ(α+1)HJβf2(t)+2HJαf(t)HJβf(t)≥2(m+M)×((log⁡t)βΓ(β+1)HJαf(t)+(log⁡t)αΓ(α+1)HJβf(t)).Example 10.
Letf be a function satisfying log⁡t≤f(t)≤1+log⁡t for t∈[1,∞). Then for t>1 and α>0, we have
(26)12(log⁡t)αΓ(α+1)×((log⁡t)αΓ(α+1)+2(log⁡t)α+1Γ(α+2)+4(log⁡t)α+2Γ(α+3)+2HJαf2(t))+((log⁡t)αΓ(α+1)+(log⁡t)α+1Γ(α+2))(log⁡t)α+1Γ(α+2)+(HJαf(t))2≥2((log⁡t)αΓ(α+1)+(log⁡t)α+1Γ(α+2))HJαf(t)+2(log⁡t)αΓ(α+1)HJα(flog⁡t)(t).Theorem 11.
Letf be an integrable function on [1,∞) and θ1,θ2>0 satisfying θ1+θ2=1. In addition, suppose that (H1) holds. Then, for t>1, α,β>0, one has
(27)θ1(log⁡t)βΓ(β+1)HJαφ2(t)+θ2(log⁡t)αΓ(α+1)HJβf(t)≥HJα(φ2-f)θ1(t)HJβ(f-φ1)θ2(t)+θ1(log⁡t)βΓ(β+1)HJαf(t)+θ2(log⁡t)αΓ(α+1)HJβφ1(t).Proof.
From the well-known weighted AM-GM inequality(28)θ1x+θ2y≥xθ1yθ2,∀x,y≥0,θ1,θ2>0,θ1+θ2=1,
by setting x=φ2(τ)-f(τ) and y=f(ρ)-φ1(ρ), τ,ρ>1, we have
(29)θ1(φ2(τ)-f(τ))+θ2(f(ρ)-φ1(ρ))≥(φ2(τ)-f(τ))θ1(f(ρ)-φ1(ρ))θ2.
Multiplying both sides of (29) by (log⁡(t/τ))α-1(log⁡(t/ρ))β-1/τρΓ(α)Γ(β), τ,ρ∈(1,t), we get
(30)θ1(log⁡(t/τ))α-1(log⁡(t/ρ))β-1τρΓ(α)Γ(β)(φ2(τ)-f(τ))+θ2(log⁡(t/τ))α-1(log⁡(t/ρ))β-1τρΓ(α)Γ(β)(f(ρ)-φ2(ρ))≥(log⁡(t/τ))α-1τΓ(α)(φ2(τ)-f(τ))θ1×(log⁡(t/ρ))β-1ρΓ(β)(f(ρ)-φ1(ρ))θ2.
Integrating the above inequality with respect to τ and ρ from 1 to t, we have
(31)θ1HJβ(1)(t)HJα(φ2-f)(t)+θ2HJα(1)(t)HJβ(f-φ1)(t)≥HJα(φ2-f)θ1(t)HJβ(f-φ1)θ2(t).
Therefore, we deduce inequality (27).Corollary 12.
Letf be an integrable function on [1,∞) satisfying m≤f(t)≤M, for all t∈[1,∞) and m,M∈ℝ. Then for t>1 and α,β>0, one has
(32)M(log⁡t)α+βΓ(α+1)Γ(β+1)+(log⁡t)αΓ(α+1)HJβf(t)≥m(log⁡t)α+βΓ(α+1)Γ(β+1)+(log⁡t)βΓ(β+1)HJαf(t)+2HJα(M-f)1/2(t)HJβ(f-m)1/2(t).Example 13.
Letf be a function satisfying log⁡t≤f(t)≤1+log⁡t for t∈[1,∞). Then for t>1 and α>0, we have
(33)(log⁡t)2αΓ2(α+1)≥2HJα(1+log⁡t-f)(t)HJα(f-log⁡t)(t).Lemma 14 (see [18]).
Assume thata≥0, p≥q≥0, and p≠0. Then
(34)aq/p≤(qpkq-p/pa+p-qpkq/p),foranyk>0.Theorem 15.
Letf be an integrable function on [1,∞) and constants p≥q≥0, p≠0. In addition, assume that (H1) holds. Then for any k>0, t>1, α, β>0, the following two inequalities hold:
(35)(A1)HJα(φ2-f)q/p(t)+qpk(q-p)/pHJαf(t)≤qpk(q-p)/pHJαφ2(t)+p-qpkq/p(log⁡t)αΓ(α+1),(B1)HJα(f-φ1)q/p(t)+qpkq-p/pHJαφ1(t)11.111111≤qpk(q-p)/pHJαf(t)+p-qpkq/p(log⁡t)αΓ(α+1).Proof.
By condition(H1) and Lemma 14, for p≥q≥0, p≠0, it follows that
(36)(φ2(τ)-f(τ))q/p≤qpk(q-p)/p(φ2(τ)-f(τ))+p-qpkq/p,
for any k>0. Multiplying both sides of (36) by (log⁡(t/τ))α-1/τΓ(α), τ∈(1,t), and integrating the resulting identity with respect to τ from 1 to t, one has
(37)HJα(f-φ1)q/p≤qpk(q-p)/p(HJαφ2(t)-HJαf(t))+p-qpkq/p(log⁡t)αΓ(α+1),
which leads to inequality (A1). Inequality (B1) is proved by similar arguments.Corollary 16.
Letf be an integrable function on [1,∞) satisfying m≤f(t)≤M, for all t∈[1,∞) and m,M∈ℝ. Then for t>1 and α>0, one has
(38)(A2)2HJα(M-f)1/2(t)+HJαf(t)≤(M+1)(log⁡t)αΓ(α+1),(B2)2HJα(f-m)1/2(t)+m(log⁡t)αΓ(α+1)1111111111.1111111≤HJαf(t)+(log⁡t)αΓ(α+1).Example 17.
Letf be a function satisfying log⁡t≤f(t)≤1+log⁡t for t∈[1,∞). Then for t>1 and α>0, we have
(39)2HJα(1+log⁡t-f)(t)+HJαf(t)≤2(log⁡t)αΓ(α+1)+(log⁡t)α+1Γ(α+2),2HJα(f-log⁡t)(t)+(log⁡t)α+1Γ(α+2)≤HJαf(t)+(log⁡t)αΓ(α+1).Theorem 18.
Letf and g be two integrable functions on [1,∞). Suppose that (H1) holds and moreover one assumes the following.
(
H
2
) There exist ψ1 and ψ2 integrable functions on [1,∞) such that
(40)ψ1(t)≤g(t)≤ψ2(t)∀t∈[1,∞).
Then, for t>0, α, β>0, the following inequalities hold:
(41)(A3)HJβψ1(t)HJαf(t)+HJαφ2(t)HJβg(t)≥HJβψ1(t)HJαφ2(t)+HJαf(t)HJβg(t),(B3)HJβφ1(t)HJαg(t)+HJαψ2(t)HJβf(t)≥HJβφ1(t)HJαψ2(t)+HJβf(t)HJαg(t),(C3)HJβψ2(t)HJαφ2(t)+HJαf(t)HJβg(t)≥HJαφ2(t)HJβg(t)+HJβψ2(t)HJαf(t),(D3)HJαφ1(t)HJβψ1(t)+HJαf(t)HJβg(t)≥HJαφ1(t)HJβg(t)+HJβψ1(t)HJαf(t).Proof.
To prove(A3), from (H1) and (H2), we have for t∈[1,∞) that
(42)(φ2(τ)-f(τ))(g(ρ)-ψ1(ρ))≥0.
Therefore,
(43)φ2(τ)g(ρ)+ψ1(ρ)f(τ)≥ψ1(ρ)φ2(τ)+f(τ)g(ρ).
Multiplying both sides of (43) by (log⁡(t/τ))α-1/τΓ(α), τ∈(1,t), we get
(44)g(ρ)(log⁡(t/τ))α-1τΓ(α)φ2(τ)+ψ1(ρ)(log⁡(t/τ))α-1τΓ(α)f(τ)≥ψ1(ρ)(log⁡(t/τ))α-1τΓ(α)φ2(τ)+g(ρ)(log⁡(t/τ))α-1τΓ(α)f(τ).
Integrating both sides of (44) with respect to τ on (1,t), we obtain
(45)g(ρ)1Γ(α)∫1t‍(log⁡tτ)α-1φ2(τ)dττ+ψ1(ρ)1Γ(α)∫1t‍(log⁡tτ)α-1f(τ)dττ≥ψ1(ρ)1Γ(α)∫1t‍(log⁡tτ)α-1φ2(τ)dττ+g(ρ)1Γ(α)∫1t‍(log⁡tτ)α-1f(τ)dττ.
Then we have
(46)g(ρ)HJαφ2(t)+ψ1(ρ)HJαf(t)≥ψ1(ρ)HJαφ2(t)+g(ρ)HJαf(t).
Multiplying both sides of (46) by (log⁡(t/ρ))β-1/ρΓ(β), ρ∈(1,t), we have
(47)HJαφ2(t)(log⁡(t/ρ))β-1ρΓ(β)g(ρ)+HJαf(t)(log⁡(t/ρ))β-1ρΓ(β)ψ1(ρ)≥HJαφ2(t)(log⁡(t/ρ))β-1ρΓ(β)ψ1(ρ)+HJαf(t)(log⁡(t/ρ))β-1ρΓ(β)g(ρ).
Integrating both sides of (47) with respect to ρ on (1,t), we get the desired inequality (A3).
To prove(B3)–(D3), we use the following inequalities:
(48)(B3)(ψ2(τ)-g(τ))(f(ρ)-φ1(ρ))≥0,(C3)(φ2(τ)-f(τ))(g(ρ)-ψ2(ρ))≤0,(D3)(φ1(τ)-f(τ))(g(ρ)-ψ1(ρ))≤0.As a special case of Theorem18, we have the following corollary.Corollary 19.
Letf and g be two integrable functions on [1,∞). Assume the following.
(
H
3
) There exist real constants m,M,n,N such that
(49)m≤f(t)≤M,n≤g(t)≤N∀t∈[1,∞).
Then, for t>1, α, β>0, one has
(50)(A4)n(log⁡t)βΓ(β+1)HJαf(t)+M(log⁡t)αΓ(α+1)HJβg(t)≥nM(log⁡t)α+βΓ(α+1)Γ(β+1)+HJαf(t)HJβg(t),(B4)m(log⁡t)βΓ(β+1)HJαg(t)+N(log⁡t)αΓ(α+1)HJβf(t)≥mN(log⁡t)α+βΓ(α+1)Γ(β+1)+HJβf(t)HJαg(t),(C4)MN(log⁡t)α+βΓ(α+1)Γ(β+1)+HJαf(t)HJβg(t)≥M(log⁡t)αΓ(α+1)HJβg(t)+N(log⁡t)βΓ(β+1)HJαf(t),(D4)mn(log⁡t)α+βΓ(α+1)Γ(β+1)+HJαf(t)HJβg(t)≥m(log⁡t)αΓ(α+1)HJβg(t)+n(log⁡t)βΓ(β+1)HJαf(t).Theorem 20.
Letf and g be two integrable functions on [1,∞) and θ1,θ2>0 satisfying 1/θ1+1/θ2=1. Suppose that (H1) and (H2) hold. Then, for t>1, α,β>0, the following inequalities hold:
(51)(A5)1θ1(log⁡t)βΓ(β+1)HJα(φ2-f)θ1(t)+1θ2(log⁡t)αΓ(α+1)HJβ(ψ2-g)θ2(t)≥HJα(φ2-f)(t)HJβ(ψ2-g)(t),(B5)1θ1HJα(φ2-f)θ1(t)HJβ(ψ2-g)θ1(t)+1θ2HJα(ψ2-g)θ2(t)HJβ(φ2-f)θ2(t)≥HJα(φ2-f)(ψ2-g)(t)HJβ(φ2-f)(ψ2-g)(t),(C5)1θ1(log⁡t)βΓ(β+1)HJα(f-φ1)θ1(t)+1θ2(log⁡t)αΓ(α+1)HJβ(g-ψ1)θ2(t)≥HJα(f-φ1)(t)HJβ(g-ψ1)(t),(D5)1θ1HJα(f-φ1)θ1(t)HJβ(g-ψ1)θ1(t)+1θ2HJα(g-ψ1)θ2(t)HJβ(f-φ1)θ2(t)≥HJα(f-φ1)(g-ψ1)(t)HJβ(f-φ1)(g-ψ1)(t).Proof.
The inequalities(A5)–(D5) can be proved by choosing of the parameters in the Young inequality
(52)(A5)x=φ2(τ)-f(τ),y=ψ2(ρ)-g(ρ),(B5)x=(φ2(τ)-f(τ))(ψ2(ρ)-g(ρ)),1111.y=(ψ2(τ)-g(τ))(φ2(ρ)-f(ρ)),(C5)x=f(τ)-φ1(τ),y=g(ρ)-ψ1(ρ),(D5)x=(f(τ)-φ1(τ))(g(ρ)-ψ1(ρ)),11.11y=(g(τ)-ψ1(τ))(f(ρ)-φ1(ρ)).Theorem 21.
Letf and g be two integrable functions on [1,∞) and θ1,θ2>0 satisfying θ1+θ2=1. Suppose that (H1) and (H2) hold. Then, for t>1, α,β>0, the following inequalities hold:
(53)(A6)θ1(log⁡t)βΓ(β+1)HJαφ2(t)+θ2(log⁡t)αΓ(α+1)HJβψ2(t)≥HJα(φ2-f)θ1(t)HJβ(ψ2-g)θ2(t)+θ1(log⁡t)βΓ(β+1)HJαf(t)+θ2(log⁡t)αΓ(α+1)HJβg(t),(B6)θ1HJαφ2(t)HJβψ2(t)+θ1HJαf(t)HJβg(t)+θ2HJαψ2(t)HJβφ2(t)+θ2HJαg(t)HJβf(t)≥HJα(φ2-f)θ1(ψ2-g)θ2(t)HJβ(ψ2-g)θ1×(φ2-f)θ2(t)+θ1HJαφ2(t)HJβg(t)+θ1HJαf(t)HJβψ2(t)+θ2HJαψ2(t)HJβf(t)+θ2HJαg(t)HJβφ2(t),(C6)θ1(log⁡t)βΓ(β+1)HJαf(t)+θ2(log⁡t)αΓ(α+1)HJβg(t)≥HJα(f-φ1)θ1(t)HJβ(g-ψ1)θ2(t)+θ1(log⁡t)βΓ(β+1)HJαφ1(t)+θ2(log⁡t)αΓ(α+1)HJβψ1(t),(D6)θ1HJαf(t)HJβg(t)+θ1HJαφ1(t)HJβψ1(t)+θ2HJαg(t)HJβf(t)+θ2HJαψ1(t)HJβφ1(t)≥HJα(f-φ1)θ1(g-ψ1)θ2(t)HJβ(g-ψ1)θ1×(f-φ1)θ2(t)+θ1HJαf(t)HJβψ1(t)+θ1HJαφ1(t)HJβg(t)+θ2HJαg(t)HJβφ1(t)+θ2HJαψ1(t)HJβf(t).Proof.
The inequalities(A6)–(D6) can be proved by choosing of the parameters in the weighted AM-GM:
(54)(A6)x=φ2(τ)-f(τ),y=ψ2(ρ)-g(ρ),(B6)x=(φ2(τ)-f(τ))(ψ2(ρ)-g(ρ)),22y=(ψ2(τ)-g(τ))(φ2(ρ)-f(ρ)),(C6)x=f(τ)-φ1(τ),y=g(ρ)-ψ1(ρ),(D6)x=(f(τ)-φ1(τ))(g(ρ)-ψ1(ρ)),22y=(g(τ)-ψ1(τ))(f(ρ)-φ1(ρ)).Theorem 22.
Letf and g be two integrable functions on [1,∞) and constants p≥q≥0, p≠0. Assume that (H1) and (H2) hold. Then, for any k>0, t>1, α,β>0, the following inequalities hold:
(55)(A7)HJα(φ2-f)q/p(ψ2-g)q/p(t)+qpk(q-p)/pHJαφ2g(t)+qpk(q-p)/pHJαfψ2(t)≤qpk(q-p)/pHJαφ2ψ2(t)+qpk(q-p)/pHJαfg(t)+p-qpkq/p(log⁡t)αΓ(α+1),(B7)HJα(φ2-f)q/p(t)HJβ(ψ2-g)q/p(t)+qpk(q-p)/pHJαφ2(t)HJβg(t)+qpk(q-p)/pHJαf(t)HJβψ2(t)≤qpk(q-p)/pHJαφ2(t)HJβψ2(t)+qpk(q-p)/pHJαf(t)HJβg(t)+p-qpkq/p(log⁡t)α+βΓ(α+1)Γ(β+1),(C7)HJα(f-φ1)q/p(g-ψ1)q/p(t)+qpk(q-p)/pHJαfψ1(t)+qpk(q-p)/pHJαφ1g(t)≤qpk(q-p)/pHJαfg(t)+qpk(q-p)/pHJαφ1ψ1(t)+p-qpkq/p(log⁡t)αΓ(α+1),(D7)HJα(f-φ1)q/p(t)HJβ(g-ψ1)q/p(t)+qpk(q-p)/pHJαf(t)HJβψ1(t)+qpk(q-p)/pHJαφ1(t)HJβg(t)≤qpk(q-p)/pHJαf(t)HJβg(t)+qpk(q-p)/pHJαφ1(t)HJβψ1(t)+p-qpkq/p(log⁡t)α+βΓ(α+1)Γ(β+1).Proof.
The inequalities(A7)–(D7) can be proved by choosing of the parameters in Lemma 14:
(56)(A7)a=(φ2(τ)-f(τ))(ψ2(τ)-g(τ)),(B7)a=(φ2(τ)-f(τ))(ψ2(ρ)-g(ρ)),(C7)a=(f(τ)-φ1(τ))(g(τ)-ψ1(τ)),(D7)a=(f(τ)-φ1(τ))(g(ρ)-ψ1(ρ)).Lemma 23.
Letf be an integrable function on [1,∞) and φ1, φ2 are two integrable functions on [1,∞). Assume that the condition (H1) holds. Then, for t>1, α>0, one has
(57)(log⁡t)αΓ(α+1)HJαf2(t)-(HJαf(t))2=(HJαφ2(t)-HJαf(t))(HJαf(t)-HJαφ1(t))-(log⁡t)αΓ(α+1)HJα((φ2-f)(f-φ1)(t))+(log⁡t)αΓ(α+1)HJαφ1f(t)-HJαφ1(t)HJαf(t)+(log⁡t)αΓ(α+1)HJαφ2f(t)-HJαφ2(t)HJαf(t)+HJαφ1(t)HJαφ2(t)-(log⁡t)αΓ(α+1)HJαφ1φ2(t).Proof.
For anyτ>1 and ρ>1, we have
(58)(φ2(ρ)-f(ρ))(f(τ)-φ1(τ))+(φ2(τ)-f(τ))(f(ρ)-φ1(ρ))-(φ2(τ)-f(τ))(f(τ)-φ1(τ))-(φ2(ρ)-f(ρ))(f(ρ)-φ1(ρ))=f2(τ)+f2(ρ)-2f(τ)f(ρ)+φ2(ρ)f(τ)+φ1(τ)f(ρ)-φ1(τ)φ2(ρ)+φ2(τ)f(ρ)+φ1(ρ)f(τ)-φ1(ρ)φ2(τ)-φ2(τ)f(τ)+φ1(τ)φ2(τ)-φ1(τ)f(τ)-φ2(ρ)f(ρ)+φ1(ρ)φ2(ρ)-φ1(ρ)f(ρ).
Multiplying (58) by (log⁡(t/τ))α-1/τΓ(α), τ∈(1,t), t>1, and integrating the resulting identity with respect to τ from 1 to t, we get
(59)(φ2(ρ)-f(ρ))(HJαf(t)-HJαφ1(t))+(HJαφ2(t)-HJαf(t))(f(ρ)-φ1(ρ))-HJα((φ2-f)(f-φ1)(t))-(φ2(ρ)-f(ρ))×(f(ρ)-φ1(ρ))(log⁡t)αΓ(α+1)=HJαf2(t)+f2(ρ)(log⁡t)αΓ(α+1)-2f(ρ)HJαf(t)+φ2(ρ)HJαf(t)+f(ρ)HJαφ1(t)-φ2(ρ)HJαφ1(t)+f(ρ)HJαφ2(t)+φ1(ρ)HJαf(t)-φ1(ρ)HJαφ2(t)-HJαφ2f(t)+HJαφ1φ2(t)-HJαφ1f(t)-φ2(ρ)f(ρ)(log⁡t)αΓ(α+1)+φ1(ρ)φ2(ρ)(log⁡t)αΓ(α+1)-φ1(ρ)f(ρ)(log⁡t)αΓ(α+1).
Multiplying (59) by (log⁡(t/ρ))α-1/ρΓ(α), ρ∈(1,t), t>1, and integrating the resulting identity with respect to ρ from 1 to t, we have
(60)(HJαφ2(t)-HJαf(t))(HJαf(t)-HJαφ1(t))+(HJαφ2(t)-HJαf(t))(HJαf(t)-HJαφ1(t))-HJα((φ2-f)(f-φ1)(t))(log⁡t)αΓ(α+1)-HJα((φ2-f)(f-φ1)(t))(log⁡t)αΓ(α+1)=(log⁡t)αΓ(α+1)HJαf2(t)+(log⁡t)αΓ(α+1)HJαf2(t)-2HJαf(t)HJαf(t)+HJαφ2(t)HJαf(t)+HJαφ1(t)HJαf(t)-HJαφ1(t)HJαφ2(t)+HJαφ2(t)HJαf(t)+HJαφ1(t)HJαf(t)-HJαφ1(t)HJαφ2(t)-(log⁡t)αΓ(α+1)HJαφ2f(t)+(log⁡t)αΓ(α+1)HJαφ1φ2(t)-(log⁡t)αΓ(α+1)HJαφ1f(t)-(log⁡t)αΓ(α+1)HJαφ2f(t)+(log⁡t)αΓ(α+1)HJαφ1φ2(t)-(log⁡t)αΓ(α+1)HJαφ1f(t),
which implies (57).Corollary 24.
Letf be an integrable function on [1,∞) satisfying m≤f(t)≤M, for all t∈[1,∞). Then for all t>1, α>0 one has
(61)(log⁡t)αΓ(α+1)HJαf2(t)-(HJαf(t))2=(M(log⁡t)αΓ(α+1)-HJαf(t))(HJαf(t)-m(log⁡t)αΓ(α+1))-(log⁡t)αΓ(α+1)HJα((M-f(t))(f(t)-m)).Theorem 25.
Letf and g be two integrable functions on [1,∞) and φ1, φ2, ψ1, and ψ2 are four integrable functions on [1,∞) satisfying the conditions (H1) and (H2) on [1,∞). Then for all t>1, α>0, one has
(62)|(log⁡t)αΓ(α+1)HJαfg(t)-HJαf(t)HJαg(t)|≤|T(f,φ1,φ2)|1/2|T(g,ψ1,ψ2)|1/2,
where T(u,v,w) is defined by
(63)T(u,v,w)=(HJαw(t)-HJαu(t))(HJαu(t)-HJαv(t))+(log⁡t)αΓ(α+1)HJαvu(t)-HJαv(t)HJαu(t)+(log⁡t)αΓ(α+1)HJαwu(t)-HJαw(t)HJαu(t)+HJαv(t)HJαw(t)-(log⁡t)αΓ(α+1)HJαvw(t).Proof.
Letf and g be two integrable functions defined on [1,∞) satisfying (H1) and (H2). Define
(64)H(τ,ρ):=(f(τ)-f(ρ))(g(τ)-g(ρ)),22222222222222222τ,ρ∈(1,t),t>1.
Multiplying both sides of (64) by (log⁡(t/τ))α-1(log⁡(t/ρ))α-1/τρΓ2(α), τ,ρ∈(1,t), and integrating the resulting identity with respect to τ and ρ from 1 to t, we can state that
(65)12Γ2(α)∬1t(log⁡tτ)α-1(log⁡tρ)α-1H(τ,ρ)dττdρρ=(log⁡t)αΓ(α+1)HJαfg(t)-HJαf(t)HJαg(t).
Applying the Cauchy-Schwarz inequality to (65), we have
(66)((log⁡t)αΓ(α+1)HJαfg(t)-HJαf(t)HJαg(t))2=(12Γ2(α)∬1t(log⁡tτ)α-1(log⁡tρ)α-1×H(τ,ρ)dττdρρ(log⁡tρ)α-1)2≤(12Γ2(α)∬1t(log⁡tτ)α-1(log⁡tρ)α-1×(f(τ)-f(ρ))2dττdρρ(log⁡tρ)α-112Γ2(α)∬1t(log⁡tτ)α-1(log⁡tρ)α-1)×(12Γ2(α)∬1t(log⁡tτ)α-1(log⁡tρ)α-1×(g(τ)-g(ρ))2dττdρρ12Γ2(α)∬1t(log⁡tτ)α-1(log⁡tρ)α-1)=((log⁡t)αΓ(α+1)HJαf2(t)-(HJαf(t))2)×((log⁡t)αΓ(α+1)HJαg2(t)-(HJαg(t))2).
Since (φ2(t)-f(t))(f(t)-φ1(t))≥0 and (ψ2(t)-g(t))(g(t)-ψ1(t))≥0 for t∈[1,∞), we have
(67)(log⁡t)αΓ(α+1)HJα(φ2-f)(f-φ1)(t)≥0,(log⁡t)αΓ(α+1)HJα(ψ2-g)(g-ψ1)(t)≥0.
Thus, from Lemma 23, we get
(68)(log⁡t)αΓ(α+1)HJαf2(t)-(HJαf(t))2≤(HJαφ2(t)-HJαf(t))(HJαf(t)-HJαφ1(t))+(log⁡t)αΓ(α+1)HJαφ1f(t)-HJαφ1(t)HJαf(t)+(log⁡t)αΓ(α+1)HJαφ2f(t)-HJαφ2(t)HJαf(t)+HJαφ1(t)HJαφ2(t)-(log⁡t)αΓ(α+1)HJαφ1φ2(t)=T(f,φ1,φ2),(log⁡t)αΓ(α+1)HJαg2(t)-(HJαg(t))2≤(HJαψ2(t)-HJαg(t))×(HJαg(t)-HJαψ1(t))+(log⁡t)αΓ(α+1)HJαψ1g(t)-HJαψ1(t)HJαg(t)+(log⁡t)αΓ(α+1)HJαψ2g(t)-HJαψ2(t)HJαg(t)+HJαψ1(t)HJαψ2(t)-(log⁡t)αΓ(α+1)HJαψ1ψ2(t)=T(g,ψ1,ψ2).
From (66) and (68), we obtain (62).Remark 26.
IfT(f,φ1,φ2)=T(f,m,M) and T(g,ψ1,ψ2)=T(g,p,P), m,M,p,P∈ℝ, then inequality (62) reduces to the following Grüss type Hadamard fractional integral inequality:
(69)|(log⁡t)αΓ(α+1)HJαfg(t)-HJαf(t)HJαg(t)|≤(12(log⁡t)αΓ(α+1))2(M-m)(P-p).Example 27.
Letf and g be two functions satisfying log⁡t≤f(t)≤1+log⁡t and -1+log⁡t≤g(t)≤log⁡t for t∈[1,∞). Then for t>1 and α>0, we have
(70)|(log⁡t)αΓ(α+1)HJαfg(t)-HJαf(t)HJαg(t)|≤|T(f,log⁡t,1+log⁡t)|1/2|T(g,-1+log⁡t,log⁡t)|1/2,
where
(71)T(f,log⁡t,1+log⁡t)=((log⁡t)αΓ(α+1)+(log⁡t)α+1Γ(α+2)-HJαf(t))×(HJαf(t)-(log⁡t)α+1Γ(α+2))+(log⁡t)αΓ(α+1)HJα(flog⁡t)(t)-(log⁡t)α+1Γ(α+2)HJαf(t)+(log⁡t)αΓ(α+1)HJα((1+log⁡t)f)(t)-((log⁡t)αΓ(α+1)+(log⁡t)α+1Γ(α+2))HJαf(t)+(log⁡t)α+1Γ(α+2)((log⁡t)αΓ(α+1)+(log⁡t)α+1Γ(α+2))-(log⁡t)αΓ(α+1)((log⁡t)α+1Γ(α+2)+2(log⁡t)α+2Γ(α+3)),T(g,-1+log⁡t,log⁡t)=((log⁡t)α+1Γ(α+2)-HJαg(t))×(HJαg(t)+(log⁡t)αΓ(α+1)-(log⁡t)α+1Γ(α+2))+(log⁡t)αΓ(α+1)HJα((-1+log⁡t)g)(t)+((log⁡t)αΓ(α+1)-(log⁡t)α+1Γ(α+2))HJαg(t)+(log⁡t)αΓ(α+1)HJα(glog⁡t)(t)-((log⁡t)α+1Γ(α+2))HJαg(t)-((log⁡t)αΓ(α+1)-(log⁡t)α+1Γ(α+2))(log⁡t)α+1Γ(α+2)+(log⁡t)αΓ(α+1)((log⁡t)α+1Γ(α+2)-2(log⁡t)α+2Γ(α+3)).



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*Source: 563096-2014-04-22.xml*