2 Description of the Fractional Calculus
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1 Int. Journal of Math. Analysis, Vol. 4, 21, no. 4, On Some New Fractional Integral Inequalities Somia Belarbi and Zoubir Dahmani 1 Department of Mathematics, Faculty of Sicence University of Mostaganem, Algeria Abstract In this paper, using the Riemann-Liouville fractional integral, we establish some new integral inequalities for the Chebyshev functional in the case of two synchronous functions. Mathematics Subject Classification: 26D1, 26A33 Keywords: Fractional integral inequalities, Riemann-Liouville fractional integral 1 Introduction Let us consider the functional [1]: T (f,g := 1 b a b a f (x g (x dx 1 ( b ( 1 f (x dx b a a b a b a g (x dx, (1 where ( f and g are two integrable functions which are synchronous on [a, b] i.e. (f(x f(y(g(x g(y, for any x, y [a, b]. Many researchers have given considerable attention to (1 and a number of inequalities have appeared in the literature, see [2, 4, 5, 6]. The main purpose of this paper is to establish some inequalities for the functional (1 using fractional integrals. 2 Description of the Fractional Calculus In the following, we will give the necessary notation and basic definitions. More details, one can consult [3, 7]. Definition1 A real valued function f(x,x is said to be in the space 1 zzdahmani@yahoo.fr
2 186 S. Belarbi and Z. Dahmani C μ,μ R if there exists a real number p>μsuch that f(x =x p f 1 (x, where f 1 (x C([, [. Definition2 A function f(x,x is said to be in the space Cμ n,n R, if f (n C μ. Definition3 The Riemann-Liouville fractional integral operator of order α, for a function f C μ, (μ 1 is defined as J α f(x = 1 x (x tα 1 f(tdt; α>,x>, J f(x =f(x, (2 where := e u u α 1 du. For the convenience of establishing the results, we give the semigroup property: which implies the commutative property J α J β f(x =J α+β f(x,α,β, (3 J α J β f(x =J β J α f(x. (4 For the expression (2, when f(x =x μ we get another expression that will be used later: J α x μ = 3 Main Results Γ(μ +1 Γ(α + μ +1 xα+μ,α>; μ> 1,x>. (5 Theorem 3.1 Let f and g be two synchronous functions on [, [. Then for all t>,α>, we have: J α Γ(α +1 (fg(t J α f(tj α g(t. (6 t α Proof. Since the functions f and g are synchronous on [, [, then for all τ,ρ, we have ( ( f(τ f(ρ g(τ g(ρ. (7 Therefore f(τg(τ+f(ρg(ρ f(τg(ρ+f(ρg(τ. (8 Multiplying both sides of (8 by (t τα 1,τ (,t, we get
3 On some new fractional integral inequalities 187 (t τ α 1 f(τg(τ+ (t τα 1 f(ρg(ρ Integrating (9 over (,t, we obtain: (t τα 1 f(τg(ρ+ (t τα 1 f(ρg(τ. (9 1 Consequently (t τ α 1 f(τg(τdτ + 1 (t τ α 1 f(ρg(ρdτ 1 t (t τα 1 f(τg(ρdτ + 1 t (t τα 1 f(ρg(τdτ. (1 So we have J α (fg(t+f (ρ g (ρ 1 (t τ α 1 dτ g(ρ (t τα 1 f (τ dτ + f(ρ (t τα 1 g (τ dτ. (11 J α (fg(t+f (ρ g (ρ J α (1 g (ρ J α (f(t+f (ρ J α (g(t. (12 Now multiplying both sides of (12 by (t ρα 1,ρ (,t, we obtain: (t ρ α 1 J α (fg(t+ (t ρ α 1 Integrating (13 over (, t, we get: (t ρα 1 f (ρ g (ρ J α (1 g (ρ J α f(t+ (t ρα 1 f (ρ J α g(t. (13 Therefore J α (fg(t J α f(t (t ρ α 1 dρ + J α (1 f(ρg(ρ(t ρ α 1 dρ (t ρα 1 g(ρdρ + J α g(t (t ρα 1 f(ρdρ. (14 J α (fg(t 1 J α f(tj α g(t, (15 J α (1 and this ends the proof. Our second result is:
4 188 S. Belarbi and Z. Dahmani Theorem 3.2 Let f and g be two synchronous functions on [, [. Then for all t>, α>, β>, we have: t α Γ(α+1 J β (fg(t+ tβ Γ(β+1 J α (fg(t J α f(tj β g(t+j β f(tj α g(t. (16 Proof. Using similar arguments as in the proof of Theorem3.1, we get (t ρ β 1 J α (fg(t+j α (1 (t ρβ 1 f (ρ g (ρ (t ρ β 1 g (ρ J α f (t+ (t ρβ 1 f (ρ J α g (t. Integrating (17 over (,t, we obtain (17 J α (fg(t J α f(t (t ρ β 1 dρ + J α (1 f (ρ g (ρ(t ρ β 1 dρ (t ρβ 1 g (ρ dρ + J α g(t (t ρβ 1 f (ρ dρ. (18 The theorem in proved. Remark 3.3 The inequalities (6 and (16 are reversed if the functions are monotone in the opposite sense. Remark 3.4 Applying Theorem3.2 for α = β, we obtain Theorem3.1. We continue with: Theorem 3.5 Let (f i i=1,...n be n positive increasing functions on [, [. Then for any t>,α>, we have J α (Π n i=1 f i(t (J α (1 1 n Π n i=1 J α f i (t. (19 Proof. Our proof is by induction. Clearly, for n =1, we have J α (f 1 (t J α (f 1 (t, for all t>,α>. For n =2, applying (6, we obtain: J α (f 1 f 2 (t (J α (1 1 J α (f 1 (t J α (f 2 (t, for all t>,α>. Now, suppose that ( induction hypothesis J α ( Π n 1 i=1 f i (t (J α (1 2 n Π n 1 i=1 J α f i (t, t >,α>. (2
5 On some new fractional integral inequalities 189 Since (f i i=1,...n are positive increasing functions, then ( Π n 1 i=1 f i (t is an increasing function. Hence we can apply Theorem3.1 to the functions Π n 1 i=1 f i = g, f n = f. We obtain: J α (Π n i=1 f i(t =J α (fg(t (J α (1 1 J α ( Π n 1 i=1 f i (t J α (f n (t. (21 Taking into account the hypothesis (2, we obtain: J α (Π n i=1 f i(t (J α (1 1 ((J α (1 2 n ( Π n 1 i=1 J α f i (tj α (f n (t, (22 and this ends the proof. We further have: Theorem 3.6 Let f and g be two functions defined on [, + [, such that f is increasing and g is differentiable and there exist a real number m := inf t g (t. Then the inequality J α (fg(t (J α (1 1 J α f(tj α g(t is valid for all t>,α>. mt α +1 J α f(t+mj α (tf(t (23 Proof. Let us consider the function h (t :=g (t mt. It is evident that h is differentiable and it is increasing on [, + [. Then using the Theorem3.1, we find that: ( ( J α (g mt f (t (J α (1 1 J α f(t J α g(t mj α (t (J α (1 1 J α f(tj α g(t m(j α (1 1 t α+1 J α f(t Γ(α+2 (J α (1 1 J α f(tj α g(t mγ(α+1t J α f(t Γ(α+2 (24 (J α (1 1 J α f(tj α g(t mt α+1 J α f(t. Hence J α (fg(t (J α (1 1 J α f(tj α g(t mt α +1 J α f(t+mj α (tf(t,t>,α>. (25 The Theorem 3.6 is proved.
6 19 S. Belarbi and Z. Dahmani Corollary 3.7 Let f and g be two functions defined on [, + [. (1 Suppose that f is decreasing, g is differentiable and there exist a real number M := sup t g (t, then for all t>, α>, we have: J α (fg(t (J α (1 1 J α f(tj α g(t Mt α +1 J α f(t+mj α (tf(t. (26 (2 Suppose that f and g are differentiable and there exist m 1 := inf t f (x, m 2 := inf t g (t, then we have J α (fg(t m 1 J α tg(t m 2 J α tf(t+m 1 m 2 J α t 2 ( (J α (1 1 J α f(tj α g(t m 1 J α tj α g(t m 2 J α tj α f(t+m 1 m 2 (J α t. 2 (27 (3 Suppose that f and g are differentiable and there exist M 1 := sup t f (t, M 2 := sup t g (t, then the inequality J α (fg(t M 1 J α tg(t M 2 J α tf(t+m 1 M 2 J α t 2 ( (J α (1 1 J α f(tj α g(t M 1 J α tj α g(t M 2 J α tj α f(t+m 1 M 2 (J α t. 2 (28 is valid. Proof. (1 : Applying the Theorem 3.1 to the decreasing functions f and G such that: G(t :=g(t m 2 t. Using the Theorem 3.1 to the increasing functions F and G, where: F (t :=f(t m 1 t, G(t :=g(t m 2 t, we can prove (2. To prove (3, it suffice to consider the two decreasing functions F (t :=f(t M 1 t, G(t :=g(t M 2 t. Acknowledgements: The authors would like to thank Professor A. El Farissi for his helpful. References [1] P. L. Chebyshev, Sur les expressions approximatives des integrales definies par les autres prises entre les memes limites, Proc. Math. Soc. Charkov 2 (1882,
7 On some new fractional integral inequalities 191 [2] S. S. Dragomir, C. E. M. Pearse, Selected topic in Hermite-Hadamard inequalities, Monographs: hadamard.html, Victoria University, 2. [3] R. Gorenflo, F. Mainardi, Fractional calculus: integral and differential equations of fractional order, Springer Verlag, Wien (1997, [4] S. M. Malamud, Some complements to the Jenson and Chebyshev inequalities and a problem of W. Walter, Procc. Amer. Math. Soc. 129, 9 (21, [5] S. Marinkovic, P. Rajkovic, M. Stankovic, The inequalities for some types q-integrals, Comput. Math. Appl. 56 (28, [6] B. G. Pachpatte, A note on Chebyshev-Gruss type inequalities for differential functions, Tamsui Oxford Journal of Mathematical Sciecnes 22 (1 (26, [7] I. Podlubni, Fractional Differential Equations, Academic Press, San Diego, Received: February, 29
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