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Some new Riemann-Liouville fractional integralinequalities

Jessada Tariboona,1

,Sotiris K. Ntouyasb,2

and Weerawat Sudsutada

a Department of Mathematics, Faculty of Applied Science,King Mongkuts University of Technology,

North Bangkok, Bangkok, ThailandE-mail: [email protected], [email protected]

b Department of Mathematics, University of Ioannina,451 10 Ioannina, Greece

E-mail: [email protected]

Abstract

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

Keywords: fractional integral; fractional integral inequalities; Gruss inequality

(2010) Mathematics Subject Classifications: 26D10; 26A33

1 Introduction

In [2] (see also [3]), the Gruss 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 andg are two continuous functions on [a, b] satisfyingm f(t) M

and p g(t) P for all t [a, b], m, M,p,P R, then

1

b a b

a

f(t)g(t)dt1

(b a)2 b

a

f(t)dt b

a

g(t)dt 1

4(Mm)(Pp).

The literature on Gruss type inequalities is now vast, and many extensions of

the classical inequality were intensively studied by many authors. In the past sev-

eral 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 [4], [5], [6], [7], [8], [9], [10] and the references

1Corresponding author2Member of Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group at King

Abdulaziz University, Jeddah, Saudi Arabia

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2 J. Tariboon, S. K. Ntouyas and W. Sudsutad

cited therein. Dahmani et al. [1] 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:

m f(t) M, p g(t) P, m, M , p, P R, t [0,).

For all t >0, >0, >0,then t(+ 1) J(f g)(t) Jf(t)Jg(t)

t

(+ 1)

2(Mm)(Pp)

and

t

(+ 1)

J(f g)(t) + t

(+ 1)

J(f g)(t) Jf(t)Jg(t) Jf(t)Jg(t)2

M

t

(+ 1) Jf(t)

Jf(t)m

t

(+ 1)

+

Jf(t)m t

(+ 1)

M

t

(+ 1) Jf(t)

P

t

(+ 1) Jg(t)

Jg(t)p

t

(+ 1)

+

J

g(t)p

t

(+ 1)

P

t

(+ 1) J

g(t)

.

In this paper, we use the Riemann-Liouville fractional integrals to establish some

new fractional integral inequalities of Gruss 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 [1] and the classical Gruss 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 [1]. But the obtained

results are new and also can be applied to unbounded functions as shown in examples.

2 Preliminaries

Definition 2.1. The Riemann-Liouville fractional integral of order 0 of a func-

tiong L1((0,),R) is defined by

Jf(t) =

t

0

(t s)1

() f(s)ds,

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Some new Riemann-Liouville fractional integral inequalities 3

and

J0f(t) =f(t),

where is the Gamma function.

For the convenience of establishing our results, we give the semigroup property:

JJf(t) =J+f(t), 0, 0,

which implies the commutative property:

JJf(t) =JJf(t).

From Definition 2.1, iff(t) =t, then we have

Jt = (+ 1)

(++ 1)t+, >0, > 1, t >0.

3 Main Results

Theorem 3.1. Letf be an integrable function on [0,). Assume that:

(H1) There exist two integrable functions1, 2 on [0,) such that

1(t) f(t) 2(t) for all t [0,).

Then, fort >0, , >0, we have

J1(t)Jf(t) +J2(t)J

f(t) J2(t)J1(t) +J

f(t)Jf(t). (3.1)

Proof. From (H1), for all 0, 0, we have

(2() f()) (f() 1()) 0.

Therefore

2()f() +1()f() 1()2() +f()f(). (3.2)

Multiplying both sides of (3.2) by (t )1/(), (0, t), we get

f()(t )1

() 2() +1()

(t )1

() f()

1()(t )1

() 2() +f()

(t )1

() f(). (3.3)

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Integrating both sides of (3.3) with respect to on (0, t), we obtain

f()

t0

(t )1

() 2()d+1()

t0

(t )1

() f()d

1() t0

(t )

1

() 2()d+f()

t0

(t )

1

() f()d,

which yields

f()J2(t) +1()Jf(t) 1()J

2(t) +f()Jf(t). (3.4)

Multiplying both sides of (3.4) by (t )1/(), (0, t), we have

J2(t)(t )1

() f() +Jf(t)

(t )1

() 1()

J2(t) (t )

1

() 1() +J

f(t) (t )

1

() f(). (3.5)

Integrating both sides of (3.5) with respect to on (0, t), we get

J2(t)

t0

(t )1

() f()d+Jf(t)

t0

(t )1

() 1()d

J2(t)

t0

(t )1

() 1()d+J

f(t)

t0

(t )1

() f()d.

Hence, we deduce inequality (3.1) as requested. This completes the proof.

As a special case of Theorems 3.1, we obtain the following result:

Corollary 3.1. Letfbe an integrable function on [0,) satisfyingm f(t) M,

for allt [0,) andm, M R. Then fort >0 and, >0, we have

m t

(+ 1)Jf(t) +M

t

(+ 1)Jf(t) mM

t+

(+ 1)(+ 1)+Jf(t)Jf(t).

Example 3.1. Letfbe a function satisfying t f(t) t+ 1 for t [0,). Then

fort >0 and >0, we have 2t+1

(+ 2)+

t

(+ 1)

Jf(t)

t+1

(+ 2)+

t

(+ 1)

t+1

(+ 2)

+ (Jf(t))2 .

Theorem 3.2. Letf andg be two integrable functions on[0,). Suppose that(H1)

holds and moreover we assume that:

(H2) There exist1 and2 integrable functions on [0,) such that

1(t) g(t) 2(t) for all t [0,).

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Then, fort >0, , >0, the following inequalities hold:

(a) J1(t)Jf(t) +J2(t)J

g(t) J1(t)J2(t) +J

f(t)Jg(t).(b) J1(t)J

g(t) +J2(t)Jf(t) J1(t)J

2(t) +Jf(t)Jg(t).

(c) J

2(t)J

2(t) +J

f(t)J

g(t) J

2(t)J

g(t) +J

2(t)J

f(t).(d) J1(t)J1(t) +J

f(t)Jg(t) J1(t)Jg(t) +J1(t)J

f(t).

Proof. To prove (a), from (H1) and (H2), we have for t [0,) that

(2() f()) (g() 1()) 0.

Therefore

2()g() +1()f() 1()2() +f()g(). (3.6)

Multiplying both sides of (3.6) by (t )1/(), (0, t), we get

g()(t )1

() 2() +1()

(t )1

() f()

1()(t )1

() 2() +g()

(t )1

() f(). (3.7)

Integrating both sides of (3.7) with respect to on (0, t), we obtain

g()

t0

(t )1

() 2()d+1()

t0

(t )1

() f()d

1() t0

(t )1

() 2()d+g() t0

(t )1

() f()d.

Then we have

g()J2(t) +1()Jf(t) 1()J

2(t) +g()Jf(t). (3.8)

Multiplying both sides of (3.8) by (t )1/(), (0, t), we have

J2(t)(t )1

() g() +Jf(t)

(t )1

() 1()

J2(t)(t )1

() 1() +J

f(t)(t )1

() g(). (3.9)

Integrating both sides of (3.9) with respect to on (0, t), we get the desired inequality

(a).

To prove (b)-(d), we use the following inequalities

(b) (2() g()) (f() 1()) 0.

(c) (2() f()) (g() 2()) 0.

(d) (1() f()) (g() 1()) 0.

As a special case of Theorem 3.2, we have the following Corollary.

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Corollary 3.2. Letf andg be two integrable functions on[0,). Assume that:

(H3) There exist real constantsm, M,n,N such that

m f(t) M and n g(t) N for all t [0,).

Then, fort >0, , >0, we have

(a1) nt

(+ 1)Jf(t) +

Mt

(+ 1)Jg(t)

nMt+

(+ 1)(+ 1)+Jf(t)Jg(t).

(b1) mt

(+ 1)Jg(t) +

N t

(+ 1)Jf(t)

mNt+

(+ 1)(+ 1)+Jf(t)Jg(t).

(c1) M Nt+

(+ 1)(+ 1)+Jf(t)Jg(t)

Mt

(+ 1)Jg(t) +

N t

(+ 1)Jf(t).

(d1) mnt+

(+ 1)(+ 1)+ Jf(t)Jg(t) mt

(+ 1)Jg(t) + nt

(+ 1)Jf(t).

Lemma 3.1. Letfbe an integrable function on[0,) and1, 2 are two integrable

functions on [0,). Assume that the condition(H1) holds. Then, for t >0, >0,

we have

t

(+1)Jf2(t) (Jf(t))2 = (J2(t) J

f(t)) (Jf(t) J1(t))

t

(+ 1)J ((2(t) f(t))(f(t) 1(t)))

+ t

(+ 1)J1f(t) J

1(t)Jf(t)

+ t

(+ 1)J2f(t) J

2(t)Jf(t)

+J1(t)J2(t)

t

(+ 1)J12(t).

(3.10)

Proof. For any >0 and >0, we have

(2() f()) (f() 1()) + (2() f()) (f() 1())

(2() f()) (f() 1()) (2() f()) (f() 1())=f2() +f2() 2f()f() +2()f() +1()f() 1()2()

+2()f() +1()f() 1()2() 2()f() +1()2()

1()f() 2()f() +1()2() 1()f(). (3.11)

Multiplying (3.11) by (t )1/(), (0, t), t >0 and integrating the resulting

identity with respect to from 0 to t, we get

(2() f()) (Jf(t) J1(t)) + (J

2(t) Jf(t)) (f() 1())

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J ((2(t) f(t)) (f(t) 1(t))) (2() f()) (f() 1()) t

(+ 1)

=Jf2(t) +f2() t

(+ 1) 2f()Jf(t) +2()J

f(t) +f()J1(t)

2()J1(t) +f()J2(t) +1()Jf(t) 1()J2(t) J2f(t)

+J12(t) J1f(t) 2()f()

t

(+ 1)+ 1()2()

t

(+ 1)

1()f() t

(+ 1). (3.12)

Multiplying (3.12) by (t )1/(), (0, t), t >0 and integrating the resulting

identity with respect to from 0 to t, we have

(J2(t) Jf(t)) (Jf(t) J1(t)) + (J

2(t) Jf(t)) (Jf(t) J1(t))

J ((2(t) f(t)) (f(t) 1(t))) t

(+ 1)

J ((2(t) f(t)) (f(t) 1(t))) t

(+ 1)

= t

(+ 1)Jf2(t) +

t

(+ 1)Jf2(t) 2Jf(t)Jf(t) +J2(t)J

f(t)

+J1(t)Jf(t) J1(t)J

2(t) +J2(t)J

f(t) +J1(t)Jf(t)

J1(t)J2(t)

t

(+ 1)J2f(t) +

t

(+ 1)J12(t)

t

(+ 1) J1f(t) t

(+ 1) J2f(t) + t

(+ 1) J12(t)

t

(+ 1)J1f(t), (3.13)

which implies (3.10).

If1(t) m and 2(t) M, m, M R, for all t [0,), then inequality (3.10)

reduces to the following corollary [1, Lemma 3.2].

Corollary 3.3. Letfbe an integrable function on [0,) satisfyingm f(t) M,

for allt [0,). The for allt >0, >0 we have

t

(+ 1)Jf2(t) (Jf(t))2

=

M

t

(+ 1) Jf(t)

Jf(t)m

t

(+ 1)

t

(+ 1)J ((M f(t))(f(t)m)) .

(3.14)

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Theorem 3.3. Letf andg be two integrable functions on[0,) and1, 2, 1 and

2 are four integrable functions on[0,) satisfying the conditions(H1) and(H2) on

[0,). Then for allt >0, >0, we have

t(+ 1) Jf g(t) Jf(t)Jg(t) T(f, 1, 2)T(g, 1, 2), (3.15)

whereT(u,v,w) is defined by

T(u,v,w) = (Jw(t) Ju(t)) (Ju(t) Jv(t))

+ t

(+ 1)Jvu(t) Jv(t)Ju(t)

+ t

(+ 1)Jwu(t) Jw(t)Ju(t)

+Jv(t)Jw(t)t

(+ 1) Jvw(t). (3.16)

Proof. Letfandg be two integrable functions defined on [0,) satisfying (H1) and

(H2). Define

H(, ) := (f() f()) (g() g()) , , (0, t), t >0. (3.17)

Multiplying both sides of (3.17) by (t )1(t )1/2(), , (0, t) and

integrating the resulting identity with respect to and from 0 to t, we can state

that

1

22()

t0

t0

(t )1(t )1H(, )dd

= t

(+ 1)Jf g(t) Jf(t)Jg(t). (3.18)

Applying the Cauchy-Schwarz inequality to (3.18), we have t

(+ 1)Jf g(t) Jf(t)Jg(t)

2

t

(+ 1) J

f2

(t) (J

f(t))

2 t(+ 1) J

g2

(t) (J

g(t))

2. (3.19)

Since (2(t)f(t))(f(t)1(t)) 0 and (2(t)g(t))(g(t)1(t)) 0 fort [0,),

we have

t

(+ 1)J ((2(t) f(t))(f(t) 1(t))) 0

and

t

(+ 1)J ((2(t) g(t))(g(t) 1(t))) 0.

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Thus, from Lemma 3.1, we get

t

(+ 1)Jf2(t) (Jf(t))2 (J2(t) J

f(t)) (Jf(t) J1(t))

+ t

(+ 1) J1f(t) J1(t)Jf(t)

+ t

(+ 1)J2f(t) J

2(t)Jf(t)

+J1(t)J2(t)

t

(+ 1)J12(t)

= T(f, 1, 2), (3.20)

and

t

(+ 1) J

g2

(t) (J

g(t))

2

(J

2(t) J

g(t)) (J

g(t) J

1(t))

+ t

(+ 1)J1g(t) J

1(t)Jg(t)

+ t

(+ 1)J2g(t) J

2(t)Jg(t)

+J1(t)J2(t)

t

(+ 1)J12(t)

= T(g, 1, 2). (3.21)

From (3.19), (3.20) and (3.21), we obtain (3.15).

Remark 3.1. IfT(f, 1, 2) =T(f , m , M)andT(g, 1, 2) =T(g,p,P),m, M,p,P

R then inequality(3.15) reduces to t(+ 1) Jf g(t) Jf(t)Jg(t)

t

2(+ 1)

2(Mm)(Pp), (3.22)

see [1, Theorem 3.1].

Example 3.2. Letf andg be two functions satisfyingt f(t) t+ 1 and t 1

g(t) t fort [0,). Then fort >0 and >0, we have t(+ 1) Jf g(t) Jf(t)Jg(t) T(f,t,t+ 1)T(g, t 1, t),

where

T(f , t , t+ 1) =

t+1

(+ 2)+

t

(+ 1) Jf(t)

Jf(t)

t+1

(+ 2)

+ t

(+ 1)J(tf)(t)

t+1

(+ 2)Jf(t)

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+ t

(+ 1)J((t+ 1)f)(t)

t+1

(+ 2)+

t

(+ 1)

Jf(t)

+ t+1

(+ 2) t+1

(+ 2)+

t

(+ 1)

t

(+ 1)

2t+2

(+ 3)+

t+1

(+ 2)

,

and

T(g, t 1, t) =

t+1

(+ 2) Jg(t)

Jg(t)

t+1

(+ 2)+

t

(+ 1)

+ t

(+ 1)J((t 1)g)(t)

t+1

(+ 2)

t

(+ 1)

Jg(t)

+ t

(+ 1)J(tg)(t) t

+1

(+ 2) Jg(t)

+

t+1

(+ 2)

t

(+ 1)

t+1

(+ 2)

t

(+ 1)

2t+2

(+ 3)

t+1

(+ 2)

.

Conflict of Interests

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

this article.

Acknowledgement:

This research was funded by King Mongkuts University of Technology North Bangkok,

Thailand. Project Code: KMUTNB-GRAD-56-02.

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