-9/--4)15)615./47551-37)16;751/41-)1781-.4)+61)16-/4)5

Bulletin of Mathematical Analysis and Applications ISSN: 1821-1291, URL: http://www.bmathaa.org Volume 2 Issue 3(2010), Pages 93-99.

NEW GENERALISATIONS OF GRUSS INEQUALITY USING RIEMANN-LIOUVILLE FRACTIONAL INTEGRALS

ZOUBIR DAHMANI, LOUIZA TABHARIT, SABRINA TAF

Abstract. In this paper, we use the Riemann-Liouville fractional integrals to establish some new integral inequalities of Gruss type. We give two main re- sults; the first one deals with some inequalities using one fractional parameter.

The second result concerns others inequalities using two fractional parameters.

1. Introduction In 1935, G. Gruss [3] proved the well known inequality:

1 𝑏−𝑎

∫ 𝑏

𝑎

𝑓(𝑥)𝑔(𝑥)𝑑𝑥− 1 𝑏−𝑎

(∫ 𝑏

𝑎

𝑓(𝑥)𝑑𝑥 ) ( 1

𝑏−𝑎

∫ 𝑏

𝑎

𝑔(𝑥)𝑑𝑥 )

≤ ^{(𝑀−𝑚)(𝑃−𝑝)}₄

(1.1)

provided that 𝑓 and 𝑔 are two integrable functions on [𝑎, 𝑏] and satisfying the conditions

𝑚≤𝑓(𝑥)≤𝑀, 𝑝≤𝑔(𝑥)≤𝑃; 𝑚, 𝑀, 𝑝, 𝑃 ∈ℝ, 𝑥∈[𝑎, 𝑏]. (1.2) The inequality (1.1) has evoked the interest of many researchers and numerous generalizations, variants and extensions have appeared in the literature, to mention a few, see [1, 4, 5, 6, 7] and the references cited therein.

The main aim of this paper is to establish some new generalizations for (1.1) by using the Riemann-Liouville fractional integrals. We give two main results; the first one deals with some inequalities using one fractional parameter. The second result concerns another class of inequalities using two fractional parameters.

2. Basic Definitions of the Fractional Calculus

Definition 1. A real valued function 𝑓(𝑡), 𝑡 ≥ 0 is said to be in the space 𝐶𝜇, 𝜇 ∈ ℝ if there exists a real number 𝑝 > 𝜇 such that 𝑓(𝑡) = 𝑡^𝑝𝑓1(𝑡), where 𝑓1(𝑡)∈𝐶([0,∞[).

Definition 2. A function 𝑓(𝑡), 𝑡 ≥ 0 is said to be in the space 𝐶_𝜇^𝑛, 𝜇 ∈ ℝ, if

2000Mathematics Subject Classification. 26D10, 26A33.

Key words and phrases. Integral inequalities; Riemann-Liouville fractional integral; Gruss inequality.

c

⃝2010 Universiteti i Prishtin¨es, Prishtin¨e, Kosov¨e.

Submitted January 30, 2010. Published August 09, 2010.

93

𝑓^(𝑛)∈𝐶𝜇.

Definition 3. The Riemann-Liouville fractional integral operator of order 𝛼≥0, for a function𝑓 ∈𝐶_𝜇,(𝜇≥ −1) is defined as

𝐽^𝛼𝑓(𝑡) =_Γ(𝛼)¹ ∫𝑡

0(𝑡−𝜏)^𝛼−1𝑓(𝜏)𝑑𝜏; 𝛼 >0, 𝑡 >0,

𝐽⁰𝑓(𝑡) =𝑓(𝑡), (2.1)

where Γ(𝛼) :=∫∞

0 𝑒^−𝑢𝑢^𝛼−1𝑑𝑢.

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

𝐽^𝛼𝐽^𝛽𝑓(𝑡) =𝐽^𝛼+𝛽𝑓(𝑡), 𝛼≥0, 𝛽≥0, (2.2) which implies the commutative property

𝐽^𝛼𝐽^𝛽𝑓(𝑡) =𝐽^𝛽𝐽^𝛼𝑓(𝑡). (2.3) More details, one can consult [2].

3. Main Results

Theorem 3.1. Let 𝑓 and 𝑔 be two integrable functions on [0,∞[ satisfying the condition (1.2) on[0,∞.[ Then for all𝑡 >0, 𝛼 >0,we have:

𝑡^𝛼

Γ(𝛼+ 1)𝐽^𝛼𝑓 𝑔(𝑡)−𝐽^𝛼𝑓(𝑡)𝐽^𝛼𝑔(𝑡)

≤( 𝑡^𝛼 2Γ(𝛼+ 1)

)2

(𝑀−𝑚)(𝑃−𝑝). (3.1) We need the following lemma

Lemma 3.2. Let 𝑢 be an integrable function on [0,∞[ satisfying the condition (1.2) on [0,∞[. Then for all𝑡 >0, 𝛼 >0, we have:

𝑡^𝛼

Γ(𝛼+ 1)𝐽^𝛼𝑢²(𝑡)−(

𝐽^𝛼𝑢(𝑡))2

=(

𝑀_Γ(𝛼+1)^𝑡𝛼 −𝐽^𝛼𝑢(𝑡))(

𝐽^𝛼𝑢(𝑡)−𝑚_Γ(𝛼+1)^𝑡𝛼 )

−_Γ(𝛼+1)^𝑡𝛼 𝐽^𝛼(𝑀 −𝑢(𝑡))(𝑢(𝑡)−𝑚).

(3.2)

Proof. Let𝑢 be an integrable function on [0,∞[ satisfying the condition (1.2) on [0,∞[. For any𝜏, 𝜌∈[0,∞[,we have

(

𝑀−𝑢(𝜌))(

𝑢(𝜏)−𝑚) +(

𝑀 −𝑢(𝜏))(

𝑢(𝜌)−𝑚)

−(

𝑀−𝑢(𝜏))(

𝑢(𝜏)−𝑚)

−(

𝑀 −𝑢(𝜌))(

𝑢(𝜌)−𝑚)

=𝑢²(𝜏) +𝑢²(𝜌)−2𝑢(𝜏)𝑢(𝜌).

(3.3)

Multiplying (3.3) by ^{(𝑡−𝜏)}_Γ(𝛼)^𝛼−1;𝜏 ∈(0, 𝑡), 𝑡 >0 and integrating the resulting identity with respect to𝜏 from 0 to𝑡,we get

(

𝑀−𝑢(𝜌))(

𝐽^𝛼𝑢(𝑡)−𝑚 𝑡^𝛼 Γ(𝛼+ 1)

) +(

𝑀 𝑡^𝛼

Γ(𝛼+ 1)−𝐽^𝛼𝑢(𝑡))(

𝑢(𝜌)−𝑚)

−𝐽^𝛼(

(𝑀 −𝑢(𝑡))(𝑢(𝑡)−𝑚))

−(

𝑀−𝑢(𝜌))(

𝑢(𝜌)−𝑚)

𝑡^𝛼 Γ(𝛼+1)

=𝐽^𝛼𝑢²(𝑡) +𝑢²(𝜌)_Γ(𝛼+1)^𝑡𝛼 −2𝑢(𝜌)𝐽^𝛼𝑢(𝑡).

(3.4) Now, multiplying (3.4) by ^{(𝑡−𝜌)}_Γ(𝛼)^𝛼−1; 𝜌∈(0, 𝑡) and integrating the resulting identity with respect to𝜌from 0 to𝑡,we have

(

𝐽^𝛼𝑢(𝑡)−𝑚 𝑡^𝛼 Γ(𝛼+ 1)

) 1 Γ(𝛼)

∫ 𝑡

0

(𝑡−𝜌)^𝛼−1(𝑀 −𝑢(𝜌))𝑑𝜌 +(

𝑀_Γ(𝛼+1)^𝑡𝛼 −𝐽^𝛼𝑢(𝑡))

1 Γ(𝛼)

∫𝑡

0(𝑡−𝜌)^𝛼−1(𝑢(𝜌)−𝑚)𝑑𝜌

−𝐽^𝛼(

(𝑀−𝑢(𝑡))(𝑢(𝑡)−𝑚))

1 Γ(𝛼)

∫𝑡

0(𝑡−𝜌)^𝛼−1𝑑𝜌

−_Γ(𝛼+1)^𝑡𝛼 _Γ(𝛼)¹ ∫𝑡

0(𝑡−𝜌)^𝛼−1(𝑀 −𝑢(𝜌))(𝑢(𝜌)−𝑚)𝑑𝜌

=_Γ(𝛼+1)^𝑡𝛼 𝐽^𝛼𝑢²(𝑡) +𝐽^𝛼𝑢²(𝑡)_Γ(𝛼+1)^𝑡𝛼 −2𝐽^𝛼𝑢(𝑡)𝐽^𝛼𝑢(𝑡)

(3.5)

which gives (3.2) and proves the lemma.

□ Proof of Theorem 3.1. Let 𝑓 and 𝑔 be two functions satisfying the conditions of Theorem 3.1.

Define

𝐻(𝜏, 𝜌) := (𝑓(𝜏)−𝑓(𝜌))(𝑔(𝜏)−𝑔(𝜌));𝜏, 𝜌∈(0, 𝑡), 𝑡 >0. (3.6) Then, multiplying (3.6) by ^{(𝑡−𝜏)𝛼−1}_Γ2(𝛼)^{(𝑡−𝜌)𝛼−1}; 𝜏, 𝜌∈ (0, 𝑡) and integrating with re- spect to𝜏 and𝜌over (0, 𝑡)²,we can state that

1 Γ²(𝛼)

∫ 𝑡

0

∫ 𝑡

0

(𝑡−𝜏)^𝛼−1(𝑡−𝜌)^𝛼−1𝐻(𝜏, 𝜌)𝑑𝜏 𝑑𝜌

= 2_Γ(𝛼+1)^𝑡𝛼 𝐽^𝛼𝑓 𝑔(𝑡)−2𝐽^𝛼𝑓(𝑡)𝐽^𝛼𝑔(𝑡).

(3.7) Applying Cauchy Schwarz inequality, we have

( 𝑡^𝛼

Γ(𝛼+ 1)𝐽^𝛼𝑓 𝑔(𝑡)−𝐽^𝛼𝑓(𝑡)𝐽^𝛼𝑔(𝑡))²

≤(

𝑡^𝛼

Γ(𝛼+1)𝐽^𝛼𝑓²(𝑡)−(𝐽^𝛼𝑓(𝑡))²)(

𝑡^𝛼

Γ(𝛼+1)𝐽^𝛼𝑔²(𝑡)−(𝐽^𝛼𝑔(𝑡))²) .

(3.8)

Since (𝑀−𝑓(𝑥))(𝑓(𝑥)−𝑚)≥0 and (𝑃−𝑔(𝑥))(𝑔(𝑥)−𝑝)≥0,we have 𝑡^𝛼

Γ(𝛼+ 1)𝐽^𝛼(𝑀−𝑓(𝑡))(𝑓(𝑡)−𝑚)≥0 (3.9)

and

𝑡^𝛼

Γ(𝛼+ 1)𝐽^𝛼(𝑃−𝑔(𝑡))(𝑔(𝑡)−𝑝)≥0. (3.10) Therefore

𝑡^𝛼

Γ(𝛼+ 1)𝐽^𝛼𝑓²(𝑡)−(

𝐽^𝛼𝑓(𝑡))²

≤(

𝑀_Γ(𝛼+1)^𝑡𝛼 −𝐽^𝛼𝑓(𝑡))(

𝐽^𝛼𝑓(𝑡)−𝑚_Γ(𝛼+1)^𝑡𝛼 ) (3.11)

and

𝑡^𝛼

Γ(𝛼+ 1)𝐽^𝛼𝑔²(𝑡)−(

𝐽^𝛼𝑔(𝑡))²

≤(

𝑃_Γ(𝛼+1)^𝑡𝛼 −𝐽^𝛼𝑔(𝑡))(

𝐽^𝛼𝑔(𝑡)−𝑝_Γ(𝛼+1)^𝑡𝛼 ) .

(3.12)

By Lemma 3.2 and the inequalities (3.8), (3.11), (3.12), we deduce that

( 𝑡^𝛼

Γ(𝛼+ 1)𝐽^𝛼𝑓 𝑔(𝑡)−2𝐽^𝛼𝑓(𝑡)𝐽^𝛼𝑔(𝑡))²

≤(

𝑀_Γ(𝛼+1)^𝑡𝛼 −𝐽^𝛼𝑓(𝑡))(

𝐽^𝛼𝑓(𝑡)−𝑚_Γ(𝛼+1)^𝑡𝛼 )(

𝑃_Γ(𝛼+1)^𝑡𝛼 −𝐽^𝛼𝑔(𝑡))(

𝐽^𝛼𝑔(𝑡)−𝑝_Γ(𝛼+1)^𝑡𝛼 ) . (3.13) Now using the elementary inequality 4𝑟𝑠≤(𝑟+𝑠)², 𝑟, 𝑠∈ℝ,we can state that

4(

𝑀 𝑡^𝛼

Γ(𝛼+ 1)−𝐽^𝛼𝑓(𝑡))(

𝐽^𝛼𝑓(𝑡)−𝑚 𝑡^𝛼 Γ(𝛼+ 1)

)≤( 𝑡^𝛼

Γ(𝛼+ 1)(𝑀 −𝑚))2

(3.14) and

4( 𝑃 𝑡^𝛼

Γ(𝛼+ 1)−𝐽^𝛼𝑔(𝑡))(

𝐽^𝛼𝑔(𝑡)−𝑝 𝑡^𝛼 Γ(𝛼+ 1)

)≤( 𝑡^𝛼

Γ(𝛼+ 1)(𝑃−𝑝))2

. (3.15)

Using (3.13), (3.14) and (3.15) we get (3.1). □

Remark. Applying Theorem 3.1 for𝛼= 1,we obtain the inequality (1.1) on [0, 𝑡].

Our next result is the following theorem, in which we use two real positive parameters.

Theorem 3.3. Let 𝑓 and 𝑔 be two integrable functions on [0,∞[ satisfying the condition (1.2) on[0,∞[. Then for all𝑡 >0, 𝛼 >0, 𝛽 >0,we have:

( 𝑡^𝛼

Γ(𝛼+ 1)𝐽^𝛽𝑓 𝑔(𝑡) + 𝑡^𝛽

Γ(𝛽+ 1)𝐽^𝛼𝑓 𝑔(𝑡)−𝐽^𝛼𝑓(𝑡)𝐽^𝛽𝑔(𝑡)−𝐽^𝛽𝑓(𝑡)𝐽^𝛼𝑔(𝑡))²

≤[(

𝑀_Γ(𝛼+1)^𝑡𝛼 −𝐽^𝛼𝑓(𝑡))(

𝐽^𝛽𝑓(𝑡)−𝑚_Γ(𝛽+1)^𝑡𝛽 ) +(

𝐽^𝛼𝑓(𝑡)−𝑚_Γ(𝛼+1)^𝑡𝛼 )(

𝑀_Γ(𝛽+1)^𝑡𝛽 −𝐽^𝛽𝑓(𝑡))]

×[(

𝑃_Γ(𝛼+1)^𝑡𝛼 −𝐽^𝛼𝑔(𝑡))(

𝐽^𝛽𝑔(𝑡)−𝑝_Γ(𝛽+1)^𝑡𝛽 ) +(

𝐽^𝛼𝑔(𝑡)−𝑝_Γ(𝛼+1)^𝑡𝛼 )(

𝑃_Γ(𝛽+1)^𝑡𝛽 −𝐽^𝛽𝑔(𝑡))]

. (3.16)

To prove Theorem 3.3 we need the following lemmas:

Lemma 3.4. Let 𝑓 and 𝑔 be two integrable functions on [0,∞.[ Then for all 𝑡 >0, 𝛼 >0, 𝛽 >0,we have:

( 𝑡^𝛼

Γ(𝛼+ 1)𝐽^𝛽𝑓 𝑔(𝑡) + 𝑡^𝛽

Γ(𝛽+ 1)𝐽^𝛼𝑓 𝑔(𝑡)−𝐽^𝛼𝑓(𝑡)𝐽^𝛽𝑔(𝑡)−𝐽^𝛽𝑓(𝑡)𝐽^𝛼𝑔(𝑡))2

≤(

𝑡^𝛼

Γ(𝛼+1)𝐽^𝛽𝑓²(𝑡) +_Γ(𝛽+1)^𝑡𝛽 𝐽^𝛼𝑓²(𝑡)−2𝐽^𝛼𝑓(𝑡)𝐽^𝛽𝑓(𝑡))

×(

𝑡^𝛼

Γ(𝛼+1)𝐽^𝛽𝑔²(𝑡) +_Γ(𝛽+1)^𝑡𝛽 𝐽^𝛼𝑔²(𝑡)−2𝐽^𝛼𝑔(𝑡)𝐽^𝛽𝑔(𝑡)) .

(3.17)

Proof. Multiplying (3.6) by ^{(𝑡−𝜏)}_{Γ(𝛼)Γ(𝛽)}^{𝛼−1(𝑡−𝜌)𝛽−1}; 𝜏, 𝜌∈(0, 𝑡),integrating with respect to 𝜏 and 𝜌 over (0, 𝑡)², then applying the Cauchy-Schwarz inequality for double

integrals, we obtain (3.17). □

Lemma 3.5. Let 𝑢 be an integrable function on [0,∞[ satisfying the condition (1.2) on [0,∞[. Then for all𝑡 >0, 𝛼 >0, 𝛽 >0,we have:

𝑡^𝛼

Γ(𝛼+ 1)𝐽^𝛽𝑢²(𝑡) + 𝑡^𝛽

Γ(𝛽+ 1)𝐽^𝛼𝑢²(𝑡)−2𝐽^𝛼𝑢(𝑡)𝐽^𝛽𝑢(𝑡)

=(

𝑀_Γ(𝛼+1)^𝑡𝛼 −𝐽^𝛼𝑢(𝑡))(

𝐽^𝛽𝑢(𝑡)−𝑚_Γ(𝛽+1)^𝑡𝛽 ) +(

𝑀_Γ(𝛼+1)^𝑡𝛽 −𝐽^𝛽𝑢(𝑡))(

𝐽^𝛼𝑢(𝑡)−𝑚_Γ(𝛼+1)^𝑡𝛼 )

−_Γ(𝛼+1)^𝑡𝛼 𝐽^𝛽(𝑀−𝑢(𝑡))(𝑢(𝑡)−𝑚)−_Γ(𝛽+1)^𝑡𝛽 𝐽^𝛼(𝑀−𝑢(𝑡))(𝑢(𝑡)−𝑚).

(3.18)

Proof. Multiplying (3.4) by ^{(𝑡−𝜌)}_Γ(𝛽)^𝛽−1;𝜌∈(0, 𝑡) and integrating the resulting iden- tity with respect to𝜌from 0 to𝑡,we have

(𝐽^𝛼𝑢(𝑡)−𝑚 𝑡^𝛼 Γ(𝛼+ 1)

) 1 Γ(𝛽)

∫ 𝑡

0

(𝑡−𝜌)^𝛽−1(𝑀−𝑢(𝜌))𝑑𝜌

+(

𝑀_Γ(𝛼+1)^𝑡𝛼 −𝐽^𝛼𝑢(𝑡))

1 Γ(𝛽)

∫𝑡

0(𝑡−𝜌)^𝛽−1(𝑢(𝜌)−𝑚)𝑑𝜌

−𝐽^𝛼(

(𝑀−𝑢(𝑡))(𝑢(𝑡)−𝑚))

1 Γ(𝛽)

∫𝑡

0(𝑡−𝜌)^𝛽−1𝑑𝜌

−_Γ(𝛼+1)^𝑡𝛼 _Γ(𝛽)¹ ∫𝑡

0(𝑡−𝜌)^𝛽−1(𝑀−𝑢(𝜌))(𝑢(𝜌)−𝑚)𝑑𝜌

= _Γ(𝛼+1)^𝑡𝛼 𝐽^𝛽𝑢²(𝑡) +𝐽^𝛽𝑢²(𝑡)_Γ(𝛼+1)^𝑡𝛼 −2𝐽^𝛼𝑢(𝑡)𝐽^𝛽𝑢(𝑡).

(3.19)

Lemma 3.5 is thus proved. □

Proof of Theorem 3.3. Since (𝑀−𝑓(𝑥))(𝑓(𝑥)−𝑚)≥0 and (𝑃−𝑔(𝑥))(𝑔(𝑥)−𝑝)≥0, then can write

− 𝑡^𝛼

Γ(𝛼+ 1)𝐽^𝛽(𝑀−𝑓(𝑡))(𝑓(𝑡)−𝑚)− 𝑡^𝛽

Γ(𝛽+ 1)𝐽^𝛼(𝑀−𝑓(𝑡))(𝑓(𝑡)−𝑚)≤0 (3.20) and

− 𝑡^𝛼

Γ(𝛼+ 1)𝐽^𝛽(𝑃−𝑔(𝑡))(𝑔(𝑡)−𝑝)− 𝑡^𝛽

Γ(𝛽+ 1)𝐽^𝛼(𝑃−𝑔(𝑡))(𝑔(𝑡)−𝑝)≤0.

(3.21) Applying Lemma 3.5 to 𝑓 and𝑔, then using Lemma 3.4 and the formulas (3.20),

(3.21), we obtain (3.16). □

Remark. (𝑖) Applying Theorem 3.3 for𝛼=𝛽 we obtain Theorem 3.1.

(𝑖𝑖) Applying Theorem 3.3 for𝛼=𝛽 = 1,we obtain the inequality (1.1) on [0, 𝑡].

Acknowledgments. The authors would like to thank the anonymous referee for his/her valuable comments.

References

[1] S.S. Dragomir, Some integral inequalities of Gruss type, Indian J. Pur. Appl. Math.31 (4)(2002), 397–415.

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𝑎𝑓(𝑥)𝑔(𝑥)𝑑𝑥 −

1 (𝑏−𝑎)²

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𝑎𝑔(𝑥)𝑑𝑥,Math.Z.39(1935), 215–226.

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Zoubir Dahmani, Louiza Tabharit, Sabrina Taf

Laboratory of Pure and Applied Mathematics (LPAM), Faculty of Exact Sciences, Health and Natural Sciences, University of Mostaganem Abdelhamid Ben Badis (UMAB), Mostaganem, Algeria

E-mail address:[email protected] E-mail address:[email protected] E-mail address:[email protected]