Ann. Funct. Anal. 1 (2010), no. 1, A nnals of F unctional A nalysis ISSN: (electronic) URL:

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1 Ann. Funct. Anal. 1 (1), no. 1, A nnals of F unctional A nalysis ISSN: (electronic) URL: ON MINKOWSKI AND HERMITE HADAMARD INTEGRAL INEQUALITIES VIA FRACTIONAL INTEGRATION ZOUBIR DAHMANI 1 Communicated by C. P. Niculescu Abstract. In this paper, we use the the Riemann Liouville fractional integral to develop some new results related to the Hermite Hadamard inequality. Other integral inequalities related to the Minkowsky inequality are also established. Our results have some relationships with [E. Set, M. E. Ozdemir S.S. Dragomir, J. Inequal. Appl. 1, Art. ID 1481, 9 pp.] [L. Bougoffa, J. Inequal. Pure Appl. Math. 7 (6), no., Article 6, 3 pp.]. Some interested inequalities of these references can be deduced as some special cases. 1. Introduction preliminaries In recent years, inequalities are playing a very significant role in all fields of mathematics, present a very active attractive field of research. As example, let us cite the field of integration which is dominated by inequalities involving functions their integrals [, 9, 1]. One of the famous integral inequalities is f(a + b) b a b a f(x)dx f(a) + f(b) where f is a convex function [7]. The history of this inequality begins with the paper of Ch. Hermite [8] J. Hadamard [7] in the years , see C.P. Niculescu L.E. Persson [11] the references therein for some historical notes of Hermite Hadamard inequality. Many researchers have given considerable attention to (1) a number of extensions generalizations have appeared in the literature, see [1, 4, 5]. Date: Received: 3 September 1; Accepted: 8 October 1. 1 Mathematics Subject Classification. Primary. 6D1; Secondary. 6A33. Key words phrases. Concave function, Hermite Hadamard inequality, Minkowski inequality, Riemann Liouville fractional integral. 51,

2 5 Z. DAHMANI The aim of this paper is to establish several new integral inequalities for nonnegative integrable functions that are related to the Hermite Hadamard result using the Riemann Liouville fractional integral. Other integral inequalities related to the Minkowski inequality are also established. Our results have some relationships with [3, 1]. Some interested inequalities of these references can be deduced as some special cases. We shall introduce the following definitions properties which are used throughout this paper. Definition 1.1. A real valued function f(t), t is said to be in the space C µ, µ R if there exists a real number p > µ such that f(t) = t p f 1 (t), where f 1 (t) C([, [). Definition 1.. A function f(t), t is said to be in the space C n µ, µ R, if f (n) C µ. Definition 1.3. The Riemann Liouville fractional integral operator of order α, for a function f C µ, (µ 1) is defined as J α f(t) = 1 (t τ) α 1 f(τ)dτ; α >, t >, J f(t) = f(t), where := e u u α 1 du. For the convenience of establishing the results, we give the semigroup property: More details, one can consult [6]. J α J β f(t) = J α+β f(t), α, β.. Main Results Our first result is the following reverse Minkowski fractional integral inequality Theorem.1. Let α >, p 1 let f, g be two positive functions on [, [, such that for all t >, J α f p (t) <, J α g p (t) <. If < m f(τ) M, τ g(τ) [, t], then we have [ [ (t) + J α g p p 1 + M(m + ) [ (t) J α (f + g) p p (t). (.1) (m + 1)(M + 1) Proof. Using the condition f(τ) g(τ) < M, τ [, t], t >, we can write (M + 1) p f p (τ) M p (f + g) p (τ). (.) Multiplying both sides of (.) by (t τ)α 1 ; τ (, t), then integrating the resulting inequalities with respect to τ over (, t), we obtain (M + 1) p (t τ) α 1 f p (τ)dτ M p t (t τ)α 1 (f + g) p (τ)dτ,

3 which is equivalent to Hence, we can write FRACTIONAL INTEGRAL INEQUALITIES 53 J α f p (t) M p (M + 1) p J α (f + g) p (t). [ (t) M [ J α (f + g) p p (t). (.3) M + 1 On the other h, using the condition mg(τ) f(τ), we can write (1 + 1 m )g(τ) 1 (f(τ) + g(τ)). m Therefore, ( 1 + m) 1 pg ( 1 p ( p. p (τ) f(τ) + g(τ)) (.4) m) Now, multiplying both sides of (.4) by (t τ)α 1 ; τ (, t), then integrating the resulting inequalities with respect to τ over (, t), we obtain [ J α g p p (t) 1 [ J α (f + g) p p (t). m + 1 (.5) Adding the inequalities (.3) (.5), we obtain the inequality (.1). Remark.. Applying Theorem.1 for α = 1, we obtain [3, Theorem 1.] on [, t]. Our second result is the following Theorem.3. Let α >, p 1 let f, g be two positive functions on [, [, such that for all t >, J α f p (t) <, J α g p (t) <. If < m f(τ) M, τ g(τ) [, t], then we have [ ] [ ] (t) + J α g p p (M + 1)(m + 1) [ [ (t) ( ) (t) J α g p p (t). M (.6) Proof. Multiplying the inequalities (.3) (.5), we obtain (M + 1)(m + 1) [ [ (t) J α g p p (t) ([J ) α (f(t) + g(t)) p p. (.7) M Applying Minkowski inequality to the right h side of (.7), we get ([J ) α (f(t) + g(t)) p ([ ) 1 ( ) p (t) + J α g p p (t). It follows then that, [J α (f(t) + g(t)) p ] p Using (.7) (.8), we obtain (.6). Theorem.3 is thus proved. [ [ ] [ [ (t) + J α g p p (t) + (t) J α g p (t) p. (.8)

4 54 Z. DAHMANI Remark.4. Applying Theorem.3 for α = 1, we obtain [1, Theorem.] on [, t]. We further have Theorem.5. Let α >, p > 1, q > 1 let f, g be two positive functions on [, [. If f p, g q are two concave functions on [, [, then we have p q (f() + f(t)) p (g() + g(t)) q ( J α ( t α 1)) J α (t α 1 f p (t)) J α (t α 1 g q (t)). (.9) To prove this theorem, we need the following lemma. Lemma.6. Let h be a concave function on [a, b]. Then we have h(a) + h(b) h(b + a x) + h(x) h( a + b ). (.1) Proof. Let h be a concave function on [a, b]. Then we can write h( a + b + x ) = h( a + b If we choose x = λa + (1 λ)b, then we have h(b + a x) + h(x) ). (.11) 1 ( ) h(a + b λa (1 λ)b) + h(λa + (1 λ)b) ( ) h(λb (1 λ)a) + h(λa + (1 λ)b). = 1 Using the concavity of h, we obtain 1 ( ) h(λb (1 λ)a) + h(λa + (1 λ)b) 1 ( ) h(a) + h(b) By (.11) (.1), we get (.1). (.1) Proof of Theorem.5. Since the f p g q are concave functions on [, [, then by Lemma.6, for any t >, we have f p () + f p (t) f p (t τ) + f p (τ) f p ( t ) (.13) g q () + g q (t) f q (t τ) + g q (τ) g q ( t ). (.14)

5 FRACTIONAL INTEGRAL INEQUALITIES 55 Multiplying both sides of (.13) (.14) by (t τ)α 1 τ α 1 ; τ (, t), then integrating the resulting inequalities with respect to τ over (, t), we obtain f p () + f p (t) (t τ) α 1 τ α 1 dτ 1 t (t τ)α 1 τ α 1 f p (t τ)dτ + 1 t (t τ)α 1 τ α 1 f p (τ)dτ (.15) f p ( t ) g q () + g q (t) (t τ)α 1 τ α 1 dτ (t τ) α 1 τ α 1 dτ 1 t (t τ)α 1 τ α 1 g q (t τ)dτ + 1 t (t τ)α 1 τ α 1 g q (τ)dτ (.16) gq ( t ) (t τ)α 1 τ α 1 dτ. Using the change of variables t τ = y, we can write 1 1 Now, using (.15) (.17), we get (t τ) α 1 τ α 1 f p (t τ)dτ = J α ( t α 1 f p (t) ) (.17) (t τ) α 1 τ α 1 g q (t τ)dτ = J α ( t α 1 g q (t) ). (.18) (f p () + f p (t)) ( J α ( t α 1)) J α ( t α 1 f p (t) ) f p ( t ) ( J α ( t α 1)). (.19) For g, we use (.16) (.18). We obtain (g q () + g q (t)) ( J α ( t α 1)) J α ( t α 1 g q (t) ) g q ( t ) ( J α ( t α 1)). (.) The inequalities (.19) (.) imply that (f p () + f p (t)) (g q () + g q (t)) (J ( α t α 1) ) 4J ( α t α 1 f p (t) ) J ( α t α 1 g q (t) ). (.1) On the other h, since f g are positive functions, then for any t >, p 1, q 1, we have ( (f p () + f p (t)) ( (g q () + g q (t)) ) 1 p ) 1 q 1 (f() + f(t)) 1 (g() + g(t)).

6 56 Z. DAHMANI Hence, we obtain (f p () + f p (t)) J ( α t α 1) p (f() + f(t)) p J ( α t α 1) (.) (g q () + g q (t)) J ( α t α 1) q (g() + g(t)) q J ( α t α 1). (.3) The inequalities (.) (.3) imply that (f p () + f p (t))(g q () + g q (t)) [ ] J α (t α 1 ) 4 (.4) ]. p q (f() + f(t)) p (g() + g(t)) [J q α (t α 1 ) Combining (.1) (.4), we obtain the desired inequality (.9). Remark.7. Applying Theorem.5 for α = 1, we obtain [1, Theorem.3] on [, t]. Theorem.8. Let α >, β >, p > 1, q > 1 let f, g be two positive functions on [, [. If f p, g q are two concave functions on [, [, then we have p q (f() + f(t)) p (g() + g(t)) q [J α (t β 1 ) ] [ Γ(β) J β (t α 1 f p (t)) + J α ( t β 1 f p (t) ) ][ Γ(β) J β (t α 1 g q (t)) + J α ( t β 1 g q (t) ) ]. (.5) Proof. Multiplying both sides of (.13) (.14) by (t τ)α 1 τ β 1 ; τ (, t), then integrating the resulting inequalities with respect to τ over (, t), we obtain f p () + f p (t) (t τ) α 1 τ β 1 dτ 1 t (t τ)α 1 τ β 1 f p (t τ)dτ + 1 t (t τ)α 1 τ β 1 f p (τ)dτ (.6) f p ( t ) g q () + g q (t) (t τ)α 1 τ β 1 dτ (t τ) α 1 τ β 1 dτ 1 t (t τ)α 1 τ β 1 g q (t τ)dτ + 1 t (t τ)α 1 τ β 1 g q (τ)dτ (.7) gq ( t ) (t τ)α 1 τ β 1 dτ.

7 FRACTIONAL INTEGRAL INEQUALITIES 57 Using the change of variables t τ = y, we obtain Γ(β) Γ(β) Γ(β) Γ(β) (t τ) α 1 τ β 1 f p (t τ)dτ = Γ(β) J β ( t α 1 f p (t) ) (.8) (t τ) α 1 τ β 1 g q (t τ)dτ = Γ(β) J β ( t α 1 g q (t) ) (.9) By the relations (.6) (.8), we can state that (f p () + f p (t)) ( J α ( t β 1)) Γ(β) J β ( t α 1 f p (t) ) + J α ( t β 1 f p (t) ) with (.7) (.9), we can write f p ( t ) ( J α ( t β 1)) (.3) (g q () + g q (t)) ( J α ( t β 1)) Γ(β) J β ( t α 1 g q (t) ) + J α ( t β 1 g q (t) ) The inequalities (.3) (.31) imply that g q ( t ) ( J α ( t β 1)). (.31) (f p () + f p (t)) (g q () + g q (t)) (J α ( t β 1) ) [ Γ(β) J β (t α 1 f p (t)) + J ( α t β 1 f p (t) ) ][ Γ(β) J β (t α 1 g q (t)) + J ( α t β 1 g q (t) ) ]. (.3) As before, since f g are positive functions, then for any t >, p 1, q 1, we have (f p () + f p (t)) J ( α t β 1) p (f() + f(t)) p J ( α t β 1) (.33) (g q () + g q (t)) J ( α t β 1) q (g() + g(t)) q J ( α t β 1). (.34) The inequalities (.33) (.34) imply that (f p () + f p (t))(g q () + g q (t)) [ ] J α (t β 1 ) 4 (.35) ]. p q (f() + f(t)) p (g() + g(t)) [J q α (t β 1 ) Combining (.3) (.35), we obtain the desired inequality (.5). Remark.9. Applying Theorem.8 for α = β, we obtain Theorem.5.

8 58 Z. DAHMANI References 1. A. El Farissi, Z. Latreuch B. Belaidi, Hadamard-type inequalities for twice differentiable functions, RGMIA 1 (9), no. 1, 7 pp.. S. Belarbi Z. Dahmani, On some new fractional integral inequalities, J. Inequal. Pure Appl. Math. 1 (9), no. 3, Article 86, 5 pp. 3. L. Bougoffa, On Minkowski Hardy integral inequality, J. Inequal. Pure Appl. Math. 7 (6), no., Article 6, 3 pp. 4. S.S. Dragomir, C.E.M. Pearse, Selected Topic in Hermite Hadamard Inequalities, Monographs: hadamard.html, Victoria University,. 5. A. Florea C.P. Niculescu, A Hermite Hadamard inequality for convex-concave symmetric functions, Bull. Math. Soc. Sci. Math. Roumanie (N.S.) 5(98) (7), no., R. Gorenflo F. Mainardi, Fractional calculus: integral differential equations of fractional order, Fractals fractional calculus in continuum mechanics (Udine, 1996), 3 76, CISM Courses Lectures, 378, Springer, Vienna, J. Hadamard, Etude sur les proprietes des fonctions entieres et en particulier d une fonction considree par Riemann, J. Math. Pures et Appl. 58 (1893), Ch. Hermite, Sur deux limites d une integrale definie, Mathesis 3 (1883), A.W. Marshall I. Olkin, Inequalities: Theory of Majoration Applications, Academic Press, S. Marinkovic, P. Rajkovic M. Stankovic, The inequalities for some types q-integrals, Comput. Math. Appl. 56 (8), C.P. Niculescu L.E. Persson, Convex functions their applications, A comtemporary approach, CMS Books in Mathematics, vol. 3, Springer Verlag, New York, E. Set, M. E. Ozdemir S.S. Dragomir, On the Hermite Hadamard Inequality Other Integral Inequalities Involving Two Functions, J. Inequal. Appl. 1, Art. ID 1481, 9 pp. 1 Laboratory of Pure Applied Mathematics, Faculty of SESNV, University of Mostaganem Abdelhamid Ben Badis, UMAB, Mostaganem, Algeria. address: zzdahmani@yahoo.fr