# Some New Riemann-Liouville Fractional Integral Inequalities

Authors: Jessada Tariboon; Sotiris K. Ntouyas; Weerawat Sudsutad
Journal: International Journal of Mathematics and Mathematical Sciences
 (2014)

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## Abstract

In this paper, some new fractional integral inequalities are established.


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## Body

## 1. Introduction

In [1] (see also [2]), the Grüss inequality is defined as the integral inequality that establishes a connection between the integral of the product of two functions and the product of the integrals. The inequality is as follows.Iff and g are two continuous functions on [a,b] satisfying m≤f(t)≤M and p≤g(t)≤P for all t∈[a,b], m,M,p,P∈ℝ, then
(1)|1b-a∫ab‍f(t)g(t)dt-1(b-a)2∫ab‍f(t)dt∫ab‍g(t)dt|≤14(M-m)(P-p).The literature on Grüss type inequalities is now vast, and many extensions of the classical inequality were intensively studied by many authors. In the past several years, by using the Riemann-Liouville fractional integrals, the fractional integral inequalities and applications have been addressed extensively by several researchers. For example, we refer the reader to [3–9] and the references cited therein. Dahmani et al. [10] gave the following fractional integral inequalities by using the Riemann-Liouville fractional integrals. Let f and g be two integrable functions on [0,∞) satisfying the following conditions:
(2)m≤f(t)≤M,p≤g(t)≤Phhm,M,p,P∈ℝ,t∈[0,∞).
For all t>0, α>0, β>0, then
(3)|tαΓ(α+1)Jα(fg)(t)-Jαf(t)Jαg(t)|≤(tαΓ(α+1))2(M-m)(P-p),(tαΓ(α+1)Jβ(fg)(t)+tβΓ(β+1)Jα(fg)(t)nn-Jαf(t)Jβg(t)-Jβf(t)Jαg(t)tβΓ(β+1))2≤{(MtαΓ(α+1)-Jαf(t))(Jβf(t)-mtβΓ(β+1))nnnnn+(Jαf(t)-mtαΓ(α+1))nnnnn×(MtβΓ(β+1)-Jβf(t))}×{(PtαΓ(α+1)-Jαg(t))(Jβg(t)-ptβΓ(β+1))nnnnnn+(Jαg(t)-ptαΓ(α+1))nnnnnn×(PtβΓ(β+1)-Jβg(t))}.In this paper, we use the Riemann-Liouville fractional integrals to establish some new fractional integral inequalities of Grüss type. We replace the constants appeared as bounds of the functionsf and g, by four integrable functions. From our results, the above inequalities of [10] and the classical Grüss inequalities can be deduced as some special cases.In Section2 we briefly review the necessary definitions. Our results are given in Section 3. The proof technique is close to that presented in [10]. But the obtained results are new and also can be applied to unbounded functions as shown in examples.


## 2. Preliminaries

Definition 1.
The Riemann-Liouville fractional integral of orderα≥0 of a function g∈L1((0,∞),ℝ) is defined by
(4)Jαf(t)=∫0t‍(t-s)α-1Γ(α)f(s)ds,J0f(t)=f(t),
where Γ is the gamma function.For the convenience of establishing our results, we give the semigroup property:(5)JαJβf(t)=Jα+βf(t),α≥0,β≥0,
which implies the commutative property
(6)JαJβf(t)=JβJαf(t).
From Definition 1, if f(t)=tγ, then we have
(7)Jαtγ=Γ(γ+1)Γ(α+γ+1)tα+γ,α>0,γ>-1,t>0.


## 3. Main Results

Theorem 2.
Letf be an integrable function on [0,∞). Assume that (H1)
there exist two integrable functionsφ1, φ2 on [0,∞) such that
(8)φ1(t)≤f(t)≤φ2(t)∀t∈[0,∞).Then, for t>0, α,β>0, one has
(9)Jβφ1(t)Jαf(t)+Jαφ2(t)Jβf(t)≥Jαφ2(t)Jβφ1(t)+Jαf(t)Jβf(t).Proof.
From(H1), for all τ≥0, ρ≥0, we have
(10)(φ2(τ)-f(τ))(f(ρ)-φ1(ρ))≥0.
Therefore
(11)φ2(τ)f(ρ)+φ1(ρ)f(τ)≥φ1(ρ)φ2(τ)+f(τ)f(ρ).
Multiplying both sides of (11) by (t-τ)α-1/Γ(α), τ∈(0,t), we get
(12)f(ρ)(t-τ)α-1Γ(α)φ2(τ)+φ1(ρ)(t-τ)α-1Γ(α)f(τ)≥φ1(ρ)(t-τ)α-1Γ(α)φ2(τ)+f(ρ)(t-τ)α-1Γ(α)f(τ).
Integrating both sides of (12) with respect to τ on (0,t), we obtain
(13)f(ρ)∫0t‍(t-τ)α-1Γ(α)φ2(τ)dτ+φ1(ρ)∫0t‍(t-τ)α-1Γ(α)f(τ)dτ≥φ1(ρ)∫0t‍(t-τ)α-1Γ(α)φ2(τ)dτ+f(ρ)∫0t‍(t-τ)α-1Γ(α)f(τ)dτ,
which yields
(14)f(ρ)Jαφ2(t)+φ1(ρ)Jαf(t)≥φ1(ρ)Jαφ2(t)+f(ρ)Jαf(t).
Multiplying both sides of (14) by (t-ρ)β-1/Γ(β), ρ∈(0,t), we have
(15)Jαφ2(t)(t-ρ)β-1Γ(β)f(ρ)+Jαf(t)(t-ρ)β-1Γ(β)φ1(ρ)≥Jαφ2(t)(t-ρ)β-1Γ(β)φ1(ρ)+Jαf(t)(t-ρ)β-1Γ(β)f(ρ).
Integrating both sides of (15) with respect to ρ on (0,t), we get
(16)Jαφ2(t)∫0t‍(t-ρ)β-1Γ(β)f(ρ)dρ+Jαf(t)∫0t‍(t-ρ)β-1Γ(β)φ1(ρ)dρ≥Jαφ2(t)∫0t‍(t-ρ)β-1Γ(β)φ1(ρ)dρ+Jαf(t)∫0t‍(t-ρ)β-1Γ(β)f(ρ)dρ.
Hence, we deduce inequality (9) as requested. This completes the proof.As a special case of Theorem2, we obtain the following result.Corollary 3.
Letf be an integrable function on [0,∞) satisfying m≤f(t)≤M, for all t∈[0,∞) and m,M∈ℝ. Then, for t>0 and α,β>0, one has
(17)mtβΓ(β+1)Jαf(t)+MtαΓ(α+1)Jβf(t)≥mMtα+βΓ(α+1)Γ(β+1)+Jαf(t)Jβf(t).Example 4.
Letf be a function satisfying t≤f(t)≤t+1 for t∈[0,∞). Then, for t>0 and α>0, we have
(18)(2tα+1Γ(α+2)+tαΓ(α+1))Jαf(t)≥(tα+1Γ(α+2)+tαΓ(α+1))(tα+1Γ(α+2))+(Jαf(t))2.Theorem 5.
Letf and g be two integrable functions on [0,∞). Suppose that (H1) holds and moreover one assumes that(H2) there exist ψ1 and ψ2 integrable functions on [0,∞) such that
(19)ψ1(t)≤g(t)≤ψ2(t),∀t∈[0,∞).Then, for t>0, α,β>0, the following inequalities hold:
(20)(a)Jβψ1(t)Jαf(t)+Jαφ2(t)Jβg(t)≥Jβψ1(t)Jαφ2(t)+Jαf(t)Jβg(t),(b)Jβφ1(t)Jαg(t)+Jαψ2(t)Jβf(t)≥Jβφ1(t)Jαψ2(t)+Jβf(t)Jαg(t),(c)Jαφ2(t)Jβψ2(t)+Jαf(t)Jβg(t)≥Jαφ2(t)Jβg(t)+Jβψ2(t)Jαf(t),(d)Jαφ1(t)Jβψ1(t)+Jαf(t)Jβg(t)≥Jαφ1(t)Jβg(t)+Jβψ1(t)Jαf(t).Proof.
To prove(a), from (H1) and (H2), we have for t∈[0,∞) that
(21)(φ2(τ)-f(τ))(g(ρ)-ψ1(ρ))≥0.
Therefore
(22)φ2(τ)g(ρ)+ψ1(ρ)f(τ)≥ψ1(ρ)φ2(τ)+f(τ)g(ρ).
Multiplying both sides of (22) by (t-τ)α-1/Γ(α), τ∈(0,t), we get
(23)g(ρ)(t-τ)α-1Γ(α)φ2(τ)+ψ1(ρ)(t-τ)α-1Γ(α)f(τ)≥ψ1(ρ)(t-τ)α-1Γ(α)φ2(τ)+g(ρ)(t-τ)α-1Γ(α)f(τ).
Integrating both sides of (23) with respect to τ on (0,t), we obtain
(24)g(ρ)∫0t‍(t-τ)α-1Γ(α)φ2(τ)dτ+ψ1(ρ)∫0t‍(t-τ)α-1Γ(α)f(τ)dτ≥ψ1(ρ)∫0t‍(t-τ)α-1Γ(α)φ2(τ)dτ+g(ρ)∫0t‍(t-τ)α-1Γ(α)f(τ)dτ.
Then we have
(25)g(ρ)Jαφ2(t)+ψ1(ρ)Jαf(t)≥ψ1(ρ)Jαφ2(t)+g(ρ)Jαf(t).
Multiplying both sides of (25) by (t-ρ)β-1/Γ(β), ρ∈(0,t), we have
(26)Jαφ2(t)(t-ρ)β-1Γ(β)g(ρ)+Jαf(t)(t-ρ)β-1Γ(β)ψ1(ρ)≥Jαφ2(t)(t-ρ)β-1Γ(β)ψ1(ρ)+Jαf(t)(t-ρ)β-1Γ(β)g(ρ).
Integrating both sides of (26) with respect to ρ on (0,t), we get the desired inequality (a).
To prove(b)–(d), we use the following inequalities:
(27)(b)(ψ2(τ)-g(τ))(f(ρ)-φ1(ρ))≥0,(c)(φ2(τ)-f(τ))(g(ρ)-ψ2(ρ))≤0,(d)(φ1(τ)-f(τ))(g(ρ)-ψ1(ρ))≤0.As a special case of Theorem5, we have the following corollary.Corollary 6.
Letf and g be two integrable functions on [0,∞). Assume that(H3) there exist real constants m,M,n,N such that
(28)m≤f(t)≤M,n≤g(t)≤N∀t∈[0,∞).Then, for t>0, α,β>0, we have
(29)(a1)ntβΓ(β+1)Jαf(t)+MtαΓ(α+1)Jβg(t)≥nMtα+βΓ(α+1)Γ(β+1)+Jαf(t)Jβg(t),(b1)mtβΓ(β+1)Jαg(t)+NtαΓ(α+1)Jβf(t)≥mNtα+βΓ(α+1)Γ(β+1)+Jβf(t)Jαg(t),(c1)MNtα+βΓ(α+1)Γ(β+1)+Jαf(t)Jβg(t)≥MtαΓ(α+1)Jβg(t)+NtβΓ(β+1)Jαf(t),(d1)mntα+βΓ(α+1)Γ(β+1)+Jαf(t)Jβg(t)≥mtαΓ(α+1)Jβg(t)+ntβΓ(β+1)Jαf(t).Lemma 7.
Letf be an integrable function on [0,∞) and let φ1, φ2 be two integrable functions on [0,∞). Assume that the condition (H1) holds. Then, for t>0, α>0, we have
(30)tαΓ(α+1)Jαf2(t)-(Jαf(t))2=(Jαφ2(t)-Jαf(t))(Jαf(t)-Jαφ1(t))-tαΓ(α+1)Jα((φ2(t)-f(t))(f(t)-φ1(t)))+tαΓ(α+1)Jαφ1f(t)-Jαφ1(t)Jαf(t)+tαΓ(α+1)Jαφ2f(t)-Jαφ2(t)Jαf(t)+Jαφ1(t)Jαφ2(t)-tαΓ(α+1)Jαφ1φ2(t).Proof.
For anyτ>0 and ρ>0, we have
(31)(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 (31) by (t-τ)α-1/Γ(α), τ∈(0,t), t>0 and integrating the resulting identity with respect to τ, from 0 to t, we get
(32)(φ2(ρ)-f(ρ))(Jαf(t)-Jαφ1(t))+(Jαφ2(t)-Jαf(t))(f(ρ)-φ1(ρ))-Jα((φ2(t)-f(t))(f(t)-φ1(t)))-(φ2(ρ)-f(ρ))(f(ρ)-φ1(ρ))tαΓ(α+1)=Jαf2(t)+f2(ρ)tαΓ(α+1)-2f(ρ)Jαf(t)+φ2(ρ)Jαf(t)+f(ρ)Jαφ1(t)-φ2(ρ)Jαφ1(t)+f(ρ)Jαφ2(t)+φ1(ρ)Jαf(t)-φ1(ρ)Jαφ2(t)-Jαφ2f(t)+Jαφ1φ2(t)-Jαφ1f(t)-φ2(ρ)f(ρ)tαΓ(α+1)+φ1(ρ)φ2(ρ)tαΓ(α+1)-φ1(ρ)f(ρ)tαΓ(α+1).
Multiplying (32) by (t-ρ)α-1/Γ(α), ρ∈(0,t), t>0 and integrating the resulting identity with respect to ρ, from 0 to t, we have
(33)(Jαφ2(t)-Jαf(t))(Jαf(t)-Jαφ1(t))+(Jαφ2(t)-Jαf(t))(Jαf(t)-Jαφ1(t))-Jα((φ2(t)-f(t))(f(t)-φ1(t)))tαΓ(α+1)-Jα((φ2(t)-f(t))(f(t)-φ1(t)))tαΓ(α+1)=tαΓ(α+1)Jαf2(t)+tαΓ(α+1)Jαf2(t)-2Jαf(t)Jαf(t)+Jαφ2(t)Jαf(t)+Jαφ1(t)Jαf(t)-Jαφ1(t)Jαφ2(t)+Jαφ2(t)Jαf(t)+Jαφ1(t)Jαf(t)-Jαφ1(t)Jαφ2(t)-tαΓ(α+1)Jαφ2f(t)+tαΓ(α+1)Jαφ1φ2(t)-tαΓ(α+1)Jαφ1f(t)-tαΓ(α+1)Jαφ2f(t)+tαΓ(α+1)Jαφ1φ2(t)-tαΓ(α+1)Jαφ1f(t),
which implies (30).Ifφ1(t)≡m and φ2(t)≡M, m,M∈ℝ, for all t∈[0,∞), then inequality (30) reduces to the following corollary [10, Lemma 3.2].Corollary 8.
Letf be an integrable function on [0,∞) satisfying m≤f(t)≤M, for all t∈[0,∞). Then, for all t>0, α>0, one has
(34)tαΓ(α+1)Jαf2(t)-(Jαf(t))2=(MtαΓ(α+1)-Jαf(t))(Jαf(t)-mtαΓ(α+1))-tαΓ(α+1)Jα((M-f(t))(f(t)-m)).Theorem 9.
Letf and g be two integrable functions on [0,∞) and let φ1, φ2, ψ1, and ψ2 be four integrable functions on [0,∞) satisfying the conditions (H1) and (H2) on [0,∞). Then, for all t>0, α>0, one has
(35)|tαΓ(α+1)Jαfg(t)-Jαf(t)Jαg(t)|≤T(f,φ1,φ2)T(g,ψ1,ψ2),
where T(u,v,w) is defined by
(36)T(u,v,w)=(Jαw(t)-Jαu(t))(Jαu(t)-Jαv(t))+tαΓ(α+1)Jαvu(t)-Jαv(t)Jαu(t)+tαΓ(α+1)Jαwu(t)-Jαw(t)Jαu(t)+Jαv(t)Jαw(t)-tαΓ(α+1)Jαvw(t).Proof.
Letf and g be two integrable functions defined on [0,∞) satisfying (H1) and (H2). Define
(37)H(τ,ρ)≔(f(τ)-f(ρ))(g(τ)-g(ρ)),hhhhhhhhhhhhhhhhτ,ρ∈(0,t),t>0.
Multiplying both sides of (37) by (t-τ)α-1(t-ρ)α-1/Γ2(α), τ,ρ∈(0,t) and integrating the resulting identity with respect to τ and ρ, from 0 to t, we can state that
(38)12Γ2(α)∬0t(t-τ)α-1(t-ρ)α-1H(τ,ρ)dτdρ=tαΓ(α+1)Jαfg(t)-Jαf(t)Jαg(t).
Applying the Cauchy-Schwarz inequality to (38), we have
(39)(tαΓ(α+1)Jαfg(t)-Jαf(t)Jαg(t))2≤(tαΓ(α+1)Jαf2(t)-(Jαf(t))2)×(tαΓ(α+1)Jαg2(t)-(Jαg(t))2).
Since (φ2(t)-f(t))(f(t)-φ1(t))≥0 and (ψ2(t)-g(t))(g(t)-ψ1(t))≥0, for t∈[0,∞), we have
(40)tαΓ(α+1)Jα((φ2(t)-f(t))(f(t)-φ1(t)))≥0,tαΓ(α+1)Jα((ψ2(t)-g(t))(g(t)-ψ1(t)))≥0.
Thus, from Lemma 7, we get
(41)tαΓ(α+1)Jαf2(t)-(Jαf(t))2≤(Jαφ2(t)-Jαf(t))(Jαf(t)-Jαφ1(t))+tαΓ(α+1)Jαφ1f(t)-Jαφ1(t)Jαf(t)+tαΓ(α+1)Jαφ2f(t)-Jαφ2(t)Jαf(t)+Jαφ1(t)Jαφ2(t)-tαΓ(α+1)Jαφ1φ2(t)=T(f,φ1,φ2),(42)tαΓ(α+1)Jαg2(t)-(Jαg(t))2≤(Jαψ2(t)-Jαg(t))(Jαg(t)-Jαψ1(t))+tαΓ(α+1)Jαψ1g(t)-Jαψ1(t)Jαg(t)+tαΓ(α+1)Jαψ2g(t)-Jαψ2(t)Jαg(t)+Jαψ1(t)Jαψ2(t)-tαΓ(α+1)Jαψ1ψ2(t)=T(g,ψ1,ψ2).
From (39), (41), and (42), we obtain (35).Remark 10.
IfT(f,φ1,φ2)=T(f,m,M) and T(g,ψ1,ψ2)=T(g,p,P), m,M,p,P∈ℝ, then inequality (35) reduces to
(43)|tαΓ(α+1)Jαfg(t)-Jαf(t)Jαg(t)|≤(tα2Γ(α+1))2(M-m)(P-p).
See [10, Theorem 3.1].Example 11.
Letf and g be two functions satisfying t≤f(t)≤t+1 and t-1≤g(t)≤t for t∈[0,∞). Then, for t>0 and α>0, we have
(44)|tαΓ(α+1)Jαfg(t)-Jαf(t)Jαg(t)|≤T(f,t,t+1)T(g,t-1,t),
where
(45)T(f,t,t+1)=(tα+1Γ(α+2)+tαΓ(α+1)-Jαf(t))×(Jαf(t)-tα+1Γ(α+2))+tαΓ(α+1)Jα(tf)(t)-tα+1Γ(α+2)Jαf(t)+tαΓ(α+1)Jα((t+1)f)(t)-(tα+1Γ(α+2)+tαΓ(α+1))Jαf(t)+(tα+1Γ(α+2))(tα+1Γ(α+2)+tαΓ(α+1))-tαΓ(α+1)(2tα+2Γ(α+3)+tα+1Γ(α+2)),(46)T(g,t-1,t)=(tα+1Γ(α+2)-Jαg(t))×(Jαg(t)-tα+1Γ(α+2)+tαΓ(α+1))+tαΓ(α+1)Jα((t-1)g)(t)-(tα+1Γ(α+2)-tαΓ(α+1))Jαg(t)+tαΓ(α+1)Jα(tg)(t)-(tα+1Γ(α+2))Jαg(t)+(tα+1Γ(α+2)-tαΓ(α+1))(tα+1Γ(α+2))-tαΓ(α+1)(2tα+2Γ(α+3)-tα+1Γ(α+2)).



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