IJMMS International Journal of Mathematics and Mathematical Sciences 1687-0425 0161-1712 Hindawi Publishing Corporation 869434 10.1155/2014/869434 869434 Research Article Some New Riemann-Liouville Fractional Integral Inequalities http://orcid.org/0000-0001-8185-3539 Tariboon Jessada 1 http://orcid.org/0000-0002-7695-2118 Ntouyas Sotiris K. 2 http://orcid.org/0000-0002-3776-9716 Sudsutad Weerawat 1 Mohapatra Ram N. 1 Department of Mathematics Faculty of Applied Science King Mongkut's University of Technology North Bangkok Bangkok Thailand kmutt.ac.th 2 Department of Mathematics University of Ioannina 451 10 Ioannina Greece uoi.gr 2014 1032014 2014 12 12 2013 23 01 2014 10 3 2014 2014 Copyright © 2014 Jessada Tariboon et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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

1. Introduction

In  (see also ), 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.

If f and g are two continuous functions on [a,b] satisfying mf(t)M and pg(t)P for all t[a,b], m,M,p,P, then (1)|1b-aabf(t)g(t)dt-1(b-a)2abf(t)dtabg(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  and the references cited therein. Dahmani et al.  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)mf(t)M,pg(t)Phhm,M,p,P,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 functions f and g, by four integrable functions. From our results, the above inequalities of  and the classical Grüss inequalities can be deduced as some special cases.

In Section 2 we briefly review the necessary definitions. Our results are given in Section 3. The proof technique is close to that presented in . 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 order α0 of a function gL1((0,),) 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),α0,β0, 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.

Let f be an integrable function on [0,). Assume that

there exist two integrable functions φ1, 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 τ0, ρ0, 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)dτφ1(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)dρJαφ2(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 Theorem 2, we obtain the following result.

Corollary 3.

Let f be an integrable function on [0,) satisfying mf(t)M, for all t[0,) and m,M. 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.

Let f be a function satisfying tf(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.

Let f 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)dτψ1(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 Theorem 5, we have the following corollary.

Corollary 6.

Let f and g be two integrable functions on [0,). Assume that

(H3) there exist real constants m,M,n,N such that (28)mf(t)M,ng(t)Nt[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.

Let f 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 any τ>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).

If φ1(t)m and φ2(t)M, m,M, for all t[0,), then inequality (30) reduces to the following corollary [10, Lemma 3.2].

Corollary 8.

Let f be an integrable function on [0,) satisfying mf(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.

Let f 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.

Let f 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.

If T(f,φ1,φ2)=T(f,m,M) and T(g,ψ1,ψ2)=T(g,p,P), m,M,p,P, 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.

Let f and g be two functions satisfying tf(t)t+1 and t-1g(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)).

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This research was funded by King Mongkut’s University of Technology North Bangkok, Thailand. Project code: KMUTNB-GRAD-56-02. Sotiris K. Ntouyas is a member of Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group at King Abdulaziz University, Jeddah, Saudi Arabia.

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