International Journal of Analysis and Applications ISSN 2291-8639 Volume 14, Number 1 (2017), 9-19 http://www.etamaths.com SOME INTEGRAL INEQUALITIES FOR LOCAL FRACTIONAL INTEGRALS M. ZEKI SARIKAYA1,∗, SAMET ERDEN2 AND HÜSEYIN BUDAK1 Abstract. In this paper, firstly we extend some generalization of the Hermite-Hadamard inequal- ity and Bullen inequality to generalized convex functions. Then, we give some important integral inequalities related to these inequalities. 1. Introduction Definition 1.1 (Convex function). The function f : [a,b] ⊂ R → R, is said to be convex if the following inequality holds f(tx + (1 − t)y) ≤ tf(x) + (1 − t)f(y) for all x,y ∈ [a,b] and t ∈ [0, 1] . We say that f is concave if (−f) is convex. The classical Hermite-Hadamard inequality which was first published in [8] gives us an estimate of the mean value of a convex function f : I → R, f ( a + b 2 ) ≤ 1 b−a ∫ b a f(x)dx ≤ f(a) + f(b) 2 (1.1) In [1], Bullen proved the following inequality which is known as Bullen’s inequality for convex function. Let f : I ⊂ R → R be a convex function on the interval I of real numbers and a,b ∈ I with a < b. The inequality 1 b−a ∫ b a f(x)dx ≤ 1 2 [ f ( a + b 2 ) + f(a) + f(b) 2 ] . An account the history of this inequality can be found in [3]. Surveys on various generalizations and developments can be found in [12] and [2]. Recently in [5], the author established this inequality for twice differentiable functions. In the case where f is convex then there exists an estimation better than (1.1). In [6], Farissi gave the refinement of the inequality (1.1) as follows: Theorem 1.1. Assume that f : I → R is a convex function on I. Then for all λ ∈ [0, 1], we have f ( a + b 2 ) ≤ l (λ) ≤ 1 b−a b∫ a f (x) dx ≤ L (λ) ≤ f (a) + f (b) 2 , where l (λ) := λf ( λb + (2 −λ) a 2 ) + (1 −λ) f ( (1 + λ) b + (1 −λ) a 2 ) and L (λ) := 1 2 (f (λb + (1 −λ) a) + λf (a) + (1 −λ) f (b)) . For more information recent developments to above inequalities, please refer to [2]- [7], [9]- [11], [14] and so on. Received 1st December, 2016; accepted 9th February, 2017; published 2nd May, 2017. 2010 Mathematics Subject Classification. 26D15, 26A51, 52A40, 52A41. Key words and phrases. Hermite-Hadamard inequality; local fractional integral; fractal space; generalized convex function. c©2017 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 9 10 SARIKAYA, ERDEN AND BUDAK 2. Preliminaries Recall the set Rα of real line numbers and use the Gao-Yang-Kang’s idea to describe the definition of the local fractional derivative and local fractional integral, see [15, 16] and so on. Recently, the theory of Yang’s fractional sets [15] was introduced as follows. For 0 < α ≤ 1, we have the following α-type set of element sets: Zα : The α-type set of integer is defined as the set {0α,±1α,±2α, ...,±nα, ...} . Qα : The α-type set of the rational numbers is defined as the set {mα = ( p q )α : p,q ∈ Z, q 6= 0}. Jα : The α-type set of the irrational numbers is defined as the set {mα 6= ( p q )α : p,q ∈ Z, q 6= 0}. Rα : The α-type set of the real line numbers is defined as the set Rα = Qα ∪Jα. If aα,bα and cα belongs the set Rα of real line numbers, then (1) aα + bα and aαbα belongs the set Rα; (2) aα + bα = bα + aα = (a + b) α = (b + a) α ; (3) aα + (bα + cα) = (a + b) α + cα; (4) aαbα = bαaα = (ab) α = (ba) α ; (5) aα (bαcα) = (aαbα) cα; (6) aα (bα + cα) = aαbα + aαcα; (7) aα + 0α = 0α + aα = aα and aα1α = 1αaα = aα. The definition of the local fractional derivative and local fractional integral can be given as follows. Definition 2.1. [15] A non-differentiable function f : R → Rα, x → f(x) is called to be local fractional continuous at x0, if for any ε > 0, there exists δ > 0, such that |f(x) −f(x0)| < εα holds for |x−x0| < δ, where ε,δ ∈ R. If f(x) is local continuous on the interval (a,b) , we denote f(x) ∈ Cα(a,b). Definition 2.2. [15] The local fractional derivative of f(x) of order α at x = x0 is defined by f(α)(x0) = dαf(x) dxα ∣∣∣∣ x=x0 = lim x→x0 ∆α (f(x) −f(x0)) (x−x0) α , where ∆α (f(x) −f(x0)) =̃Γ(α + 1) (f(x) −f(x0)) . If there exists f(k+1)α(x) = k+1 times︷ ︸︸ ︷ Dαx ...D α xf(x) for any x ∈ I ⊆ R, then we denoted f ∈ D(k+1)α(I), where k = 0, 1, 2, ... Definition 2.3. [15] Let f(x) ∈ Cα [a,b] . Then the local fractional integral is defined by, aI α b f(x) = 1 Γ(α + 1) b∫ a f(t)(dt)α = 1 Γ(α + 1) lim ∆t→0 N−1∑ j=0 f(tj)(∆tj) α, with ∆tj = tj+1 − tj and ∆t = max{∆t1, ∆t2, ..., ∆tN−1} , where [tj, tj+1] , j = 0, ...,N − 1 and a = t0 < t1 < ... < tN−1 < tN = b is partition of interval [a,b] . Here, it follows that aI α b f(x) = 0 if a = b and aI α b f(x) = −bI α a f(x) if a < b. If for any x ∈ [a,b] , there exists aI α x f(x), then we denoted by f(x) ∈ Iαx [a,b] . Definition 2.4 (Generalized convex function). [15] Let f : I ⊆ R → Rα. For any x1,x2 ∈ I and λ ∈ [0, 1] , if the following inequality f(λx1 + (1 −λ)x2) ≤ λαf(x1) + (1 −λ)αf(x2) holds, then f is called a generalized convex function on I. Here are two basic examples of generalized convex functions: (1) f(x) = xαp, x ≥ 0, p > 1; (2) f(x) = Eα(x α), x ∈ R where Eα(xα) = ∞∑ k=0 xαk Γ(1+kα) is the Mittag-Lrffer function. SOME INTEGRAL INEQUALITIES FOR LOCAL FRACTIONAL INTEGRALS 11 Theorem 2.1. [13] Let f ∈ Dα(I), then the following conditions are equivalent a) f is a generalized convex function on I b) f(α) is an increasing function on I c) for any x1,x2 ∈ I, f(x2) −f(x1) ≥ f(α)(x1) Γ (1 + α) (x2 −x1) α . Corollary 2.1. [13] Let f ∈ D2α(a,b). Then f is a generalized convex function ( or a generalized concave function) if and only if f(2α)(x) ≥ 0 ( or f(2α)(x) ≤ 0 ) for all x ∈ (a,b) . Lemma 2.1. [15] (1) (Local fractional integration is anti-differentiation) Suppose that f(x) = g(α)(x) ∈ Cα [a,b] , then we have aI α b f(x) = g(b) −g(a). (2) (Local fractional integration by parts) Suppose that f(x),g(x) ∈ Dα [a,b] and f(α)(x), g(α)(x) ∈ Cα [a,b] , then we have aI α b f(x)g (α)(x) = f(x)g(x)|ba −a I α b f (α)(x)g(x). Lemma 2.2. [15] We have i) dαxkα dxα = Γ(1 + kα) Γ(1 + (k − 1) α) x(k−1)α; ii) 1 Γ(α + 1) b∫ a xkα(dx)α = Γ(1 + kα) Γ(1 + (k + 1) α) ( b(k+1)α −a(k+1)α ) , k ∈ R. Lemma 2.3 (Generalized Hölder’s inequality). [15] Let f,g ∈ Cα [a,b] , p,q > 1 with 1p + 1 q = 1, then 1 Γ(α + 1) b∫ a |f(x)g(x)|(dx)α ≤   1 Γ(α + 1) b∫ a |f(x)|p (dx)α   1 p   1 Γ(α + 1) b∫ a |g(x)|q (dx)α   1 q . In [13], Mo et al. proved the following generalized Hermite-Hadamard inequality for generalized convex function: Theorem 2.2 (Generalized Hermite-Hadamard inequality). Let f(x) ∈ I(α)x [a,b] be a generalized convex function on [a,b] with a < b. Then f ( a + b 2 ) ≤ Γ (1 + α) (b−a)α a Iαb f(x) ≤ f (a) + f (b) 2α . (2.1) The aim of this paper is to extend the generalized Hermite-Hadamard inequalities and generalized Bullen inequalities to generalized convex functions. 3. Main Results Theorem 3.1 (Generalized Hermite–Hadamard-type inequality). Let f(x) ∈ I(α)x [a,b] be a generalized convex function on [a,b] with a < b. Then f ( a + b 2 ) ≤ h (λ) ≤ Γ (1 + α) (b−a)α a Iαb f(x) ≤ H (λ) ≤ f (a) + f (b) 2α , (3.1) where h (λ) := λαf ( λb + (2 −λ) a 2 ) + (1 −λ)α f ( (1 + λ) b + (1 −λ) a 2 ) and H (λ) := 1 2α [f (λb + (1 −λ) a) + λαf (a) + (1 −λ)α f (b)] . 12 SARIKAYA, ERDEN AND BUDAK Proof. Let f be a generalized convex. Then, applying (2.1) on the subinterval [a,λb + (1 −λ) a], with λ 6= 0, we have f ( λb + (2 −λ) a 2 ) (3.2) ≤ 1 λα (b−a)α λb+(1−λ)a∫ a f (t) (dt) α ≤ f (a) + f (λb + (1 −λ) a) 2α . Applying (2.1) again on [λb + (1 −λ) a,b], with λ 6= 1, we get f ( (1 + λ) b + (1 −λ) a 2 ) (3.3) ≤ 1 (1 −λ)α (b−a)α b∫ λb+(1−λ)a f (t) (dt) α ≤ f (λb + (1 −λ) a) + f (b) 2α . Multiplying (3.2) by λα, (3.3) by (1 −λ)α, and adding the resulting inequalities, we get: h (λ) ≤ Γ (1 + α) (b−a)α a Iαb f(x) ≤ H (λ) (3.4) where h (λ) and H (λ) are defined as in Theorem 3.1. Using the fact that f is a generalized convex function, we obtain f ( a + b 2 ) (3.5) = f ( λ λb + (2 −λ) a 2 + (1 −λ) (1 + λ) b + (1 −λ) a 2 ) ≤ λαf ( λv + (2 −λ) a 2 ) + (1 −λ)α f ( (1 + λ) b + (1 −λ) a 2 ) ≤ λα 2α [f (λb + (1 −λ) a) + f (a)] + (1 −λ)α 2α [f (b) + f (λb + (1 −λ) a)] = 1 2α [f (λb + (1 −λ) a) + λαf (a) + (1 −λ)α f (b)] ≤ f (a) + f (b) 2α . Then by (3.4) and (3.5), we get (3.1). � Theorem 3.2. Let g(x) ∈ D2α [a,b] such that there exist constants m,M ∈ Rα so that m ≤ g(2α) (x) ≤ M for x ∈ [a,b]. Then m (bα + aαbα + aα) Γ (1 + 3α) − m Γ (1 + 2α) ( a2α + b2α 2α ) (3.6) ≤ Γ (1 + α) (b−a)α a Iαb g(x) −g ( a + b 2 ) ≤ M Γ (1 + 2α) ( a2α + b2α 2α ) − M (bα + aαbα + aα) Γ (1 + 3α) . SOME INTEGRAL INEQUALITIES FOR LOCAL FRACTIONAL INTEGRALS 13 and m Γ (1 + 2α) ( a2α + b2α 2α ) − m (bα + aαbα + aα) Γ (1 + 3α) (3.7) ≤ g(a) + g(b) 2α − Γ (1 + α) (b−a)α a Iαb g(x) ≤ M (bα + aαbα + aα) Γ (1 + 3α) − M Γ (1 + 2α) ( a2α + b2α 2α ) . Proof. Let f(x) = g(x) − m Γ(1+2α) x2α, then f(2α) (x) = g(2α) (x) − m ≥ 0, which shows that f is generalized convex on (a,b). Appliying ineqaulity (2.1) for f , then we have g ( a + b 2 ) − m Γ (1 + 2α) ( a + b 2 )2α = f ( a + b 2 ) ≤ Γ (1 + α) (b−a)α a Iαb f(x) = 1 (b−a)α ∫ b a [ g(x) − m Γ (1 + 2α) x2α ] (dx) α = Γ (1 + α) (b−a)α a Iαb g(x) − 1 (b−a)α m Γ (1 + 2α) Γ (1 + 2α) Γ (1 + 3α) ( b3α −a3α ) . This implies that m (bα + aαbα + aα) Γ (1 + 3α) − m Γ (1 + 2α) ( a + b 2 )2α ≤ Γ (1 + α) (b−a)α a Iαb g(x) −g ( a + b 2 ) which proves the first inequality in (3.6). To prove the second part of (3.6), we apply the same argument for the generalized convex mapping f(x) = M Γ(1+2α) x2α −g(x); x ∈ [a,b]. By applying the second part of the generalized Hermite-Hadamard inequality (2.1) for the mapping f(x) = g(x) − m Γ(1+2α) x2α as follows g(a) + g(b) 2α − m Γ (1 + 2α) ( a2α + b2α 2α ) = f(a) + f(b) 2α ≥ Γ (1 + α) (b−a)α a Iαb f(x) = 1 (b−a)α ∫ b a [ g(x) − m Γ (1 + 2α) x2α ] (dx) α = Γ (1 + α) (b−a)α a Iαb g(x) − 1 (b−a)α m Γ (1 + 2α) Γ (1 + 2α) Γ (1 + 3α) ( b3α −a3α ) . 14 SARIKAYA, ERDEN AND BUDAK This is equivalent to m Γ (1 + 2α) ( a2α + b2α 2α ) − m (bα + aαbα + aα) Γ (1 + 3α) ≤ g(a) + g(b) 2α − Γ (1 + α) (b−a)α a Iαb g(x) which proves the rest part of (3.7). The second part is established in a similar manner; and we omit the details which completes the proof. � We prove the following inequality which is generalized Bullen inequality for generalized convex function. Theorem 3.3 (Generalized Bullen inequality). Let f(x) ∈ I(α)x [a,b] be a generalized convex function on [a,b] with a < b. Then we have the inequality Γ (1 + α) (b−a)α a Iαb f(x) ≤ 1 2α [ f ( a + b 2 ) + f (a) + f (b) 2α ] . (3.8) Proof. Using the Theorem 2.2, we find that 2αΓ (1 + α) (b−a)α 1 Γ (1 + α) b∫ a f (x) (dx) α = 2αΓ (1 + α) (b−a)α   1 Γ (1 + α) a+b 2∫ a f (x) (dx) α + 1 Γ (1 + α) b∫ a+b 2 f (x) (dx) α   ≤ f ( a+b 2 ) + f (a) 2α + f (b) + f ( a+b 2 ) 2α = f ( a + b 2 ) + f (a) + f (b) 2α . Hence, the proof is completed. � Theorem 3.4. Let I ⊆ R be an interval, f : I0 ⊆ R → Rα (I0 is the interior of I) such that f ∈ D2α(I0) and f(α) ∈ Cα [a,b] for a,b ∈ I0 with a < b. Then, for all x ∈ [a,b] , we have the following identity 1 2α (b−a)α (Γ (1 + α))2 b∫ a ( x− a + b 2 )α p(x)f(2α) (x) (dx) α (3.9) = 1 2α [ f ( a + b 2 ) + f (a) + f (b) 2α ] − Γ (1 + α) (b−a)α a Iαb f(x) where p(x) =   (a−x)α , [ a, a+b 2 ) (b−x)α , [ a+b 2 ,b ] . SOME INTEGRAL INEQUALITIES FOR LOCAL FRACTIONAL INTEGRALS 15 Proof. Using the local fractional integration by parts, we have 1 Γ (1 + α) b∫ a ( x− a + b 2 )α p(x)f(2α) (x) (dx) α = 1 Γ (1 + α) a+b 2∫ a ( x− a + b 2 )α (a−x)α f(2α) (x) (dx)α + 1 Γ (1 + α) b∫ a+b 2 ( x− a + b 2 )α (b−x)α f(2α) (x) (dx)α = ( x− a + b 2 )α (a−x)α f(α) (x) ∣∣∣∣ a+b 2 a − Γ (1 + α) Γ (1 + α) a+b 2∫ a ( 3a + b 2 − 2x )α f(α) (x) (dx) α + ( x− a + b 2 )α (b−x)α f(α) (x) ∣∣∣∣b a+b 2 − Γ (1 + α) Γ (1 + α) b∫ a+b 2 ( a + 3b 2 − 2x )α f(α) (x) (dx) α . Using the local fractional integration by parts again, we find that 1 Γ (1 + α) b∫ a ( x− a + b 2 )α p(x)f(2α) (x) (dx) α = Γ (1 + α) (b−a)α f ( a + b 2 ) + Γ (1 + α) (b−a)α f (a) + f (b) 2α − 2α (Γ (1 + α)) 2 Γ (1 + α) b∫ a f (x) (dx) α . If we devide the resulting equality with 2αΓ (1 + α) (b−a)α, then we complete the proof. � Theorem 3.5. Suppose that the assumptions of Theorem 3.4 are satisfied, then we have the following inequality ∣∣∣∣ 12α [ f ( a + b 2 ) + f (a) + f (b) 2α ] − Γ (1 + α) (b−a)α a Iαb f(x) ∣∣∣∣ ≤ (b−a)(1+ 1 p )α 8αΓ (1 + α) (B(p + 1,p + 1)) 1 p ∥∥∥f(2α) (x)∥∥∥ q where, p,q > 1, 1 p + 1 q = 1, ∥∥f(2α)∥∥ q is defined by ∥∥∥f(2α)∥∥∥ q =   1 Γ (1 + α) b∫ a ∣∣∣f(2α)(x)∣∣∣q (dx)α   1 q 16 SARIKAYA, ERDEN AND BUDAK and B (x,y) is defined by B (x,y) = 1 Γ (1 + α) 1∫ 0 t(x−1)α (1 − t)(y−1)α (dt)α . Proof. Taking madulus in (3.9) and using generalized Hölder inequality, we have∣∣∣∣ 12α [ f ( a + b 2 ) + f (a) + f (b) 2α ] − Γ (1 + α) (b−a)α a Iαb f(x) ∣∣∣∣ (3.10) ≤ 1 2α (b−a)α (Γ (1 + α))2 b∫ a ∣∣∣∣x− a + b2 ∣∣∣∣α |p(x)| ∣∣∣f(2α) (x)∣∣∣ (dx)α ≤ 1 2α (b−a)α Γ (1 + α)   1 Γ (1 + α) b∫ a ∣∣∣f(2α) (x)∣∣∣q (dx)α   1 q ×   1 Γ (1 + α) b∫ a ∣∣∣∣x− a + b2 ∣∣∣∣pα |p(x)|p (dx)α   1 p = ∥∥f(2α)∥∥ q 2α (b−a)α Γ (1 + α)   1 Γ (1 + α) a+b 2∫ a ( a + b 2 −x )pα (x−a)pα (dx)α + 1 Γ (1 + α) b∫ a+b 2 ( x− a + b 2 )pα (b−x)pα (dx)α   1 p = ∥∥f(2α)∥∥ q 2α (b−a)α Γ (1 + α) (K1 + K2) 1 p . For calculating integral K1, using changing variable with x = (1 − t)a + ta+b2 , we obtain K1 = 1 Γ (1 + α) a+b 2∫ a ( a + b 2 −x )pα (x−a)pα (dx)α (3.11) = ( b−a 2 )(2p+1)α 1 Γ (1 + α) 1∫ 0 (1 − t)pαtpα (dt)α = ( b−a 2 )(2p+1)α B(p + 1,p + 1). Similarliy, using changing variable with x = (1 − t)a+b 2 + tb, we have K2 = 1 Γ (1 + α) b∫ a+b 2 ( x− a + b 2 )pα (b−x)pα (dx)α (3.12) = ( b−a 2 )(2p+1)α B(p + 1,p + 1) SOME INTEGRAL INEQUALITIES FOR LOCAL FRACTIONAL INTEGRALS 17 Putting (3.11) and (3.12) in (3.10), we obtain∣∣∣∣ 12α [ f ( a + b 2 ) + f (a) + f (b) 2α ] − Γ (1 + α) (b−a)α a Iαb f(x) ∣∣∣∣ ≤ ∥∥f(2α)∥∥ q 2α (b−a)α Γ (1 + α) ( 2α (b−a)(2p+1)α 2(2p+1)α B(p + 1,p + 1) )1 p = (b−a)(1+ 1 p )α 8αΓ (1 + α) (B(p + 1,p + 1)) 1 p ∥∥∥f(2α)∥∥∥ q which completes the proof. � Theorem 3.6. The assumptions of Theorem 3.4 are satisfied. If the mapping ϕ(x) =   (a−x)α ( x− a+b 2 )α f(2α) (x) , [ a, a+b 2 ) (b−x)α ( x− a+b 2 )α f(2α) (x) , [ a+b 2 ,b ] . is a generalized convex, then we have the inequality (b−a)2α 64α (Γ (1 + α)) 2 [ f(2α) ( 3a + b 4 ) + f(2α) ( a + 3b 4 )] ≤ 1 2α [ f ( a + b 2 ) + f (a) + f (b) 2α ] − Γ (1 + α) (b−a)α a Iαb f(x) ≤ (b−a)2α 128α (Γ (1 + α)) 2 [ f(2α) ( 3a + b 4 ) + f(2α) ( a + 3b 4 )] . Proof. Applying the first inequality (2.1) for the mapping ϕ, we get Γ (1 + α) (b−a)α 2α Γ (1 + α) a+b 2∫ a ϕ (x) (dx) α (3.13) ≥ ϕ ( 3a + b 4 ) = (b−a)2α 16α f(2α) ( 3a + b 4 ) and Γ (1 + α) (b−a)α 2α Γ (1 + α) b∫ a+b 2 ϕ (x) (dx) α (3.14) ≥ ϕ ( a + 3b 4 ) = (b−a)2α 16α f(2α) ( a + 3b 4 ) . Applying the inequality (3.8) for the mapping ϕ, we have Γ (1 + α) (b−a)α 2α Γ (1 + α) a+b 2∫ a ϕ (x) (dx) α (3.15) ≤ 1 2α [ ϕ ( 3a + b 4 ) + ϕ (a) + ϕ ( a+b 2 ) 2α ] = (b−a)2α 32α f(2α) ( 3a + b 4 ) 18 SARIKAYA, ERDEN AND BUDAK and Γ (1 + α) (b−a)α 2α Γ (1 + α) b∫ a+b 2 ϕ (x) (dx) α (3.16) ≤ 1 2α [ ϕ ( a + 3b 4 ) + ϕ ( a+b 2 ) + ϕ (b) 2α ] = (b−a)2α 32α f(2α) ( a + 3b 4 ) . Adding the inequalities (3.13)-(3.16) and from Theorem 3.4, we write (b−a)2α 16α [ f(2α) ( 3a + b 4 ) + f(2α) ( a + 3b 4 )] ≤ Γ (1 + α) (b−a)α 2α Γ (1 + α) b∫ a+b 2 ϕ (x) (dx) α = 4α (Γ (1 + α)) 2 [ 1 2α ( f ( a + b 2 ) + f (a) + f (b) 2α ) − Γ (1 + α) (b−a)α a Iαb f(x) ] ≤ (b−a)2α 32α [ f(2α) ( 3a + b 4 ) + f(2α) ( a + 3b 4 )] . 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SOME INTEGRAL INEQUALITIES FOR LOCAL FRACTIONAL INTEGRALS 19 1Department of Mathematics, Faculty of Science and Arts, Düzce University, Konuralp Campus, Düzce- TURKEY 2Department of Mathematics, Faculty of Science, Bartın University, BARTIN-TURKEY ∗Corresponding author: sarikayamz@gmail.com 1. Introduction 2. Preliminaries 3. Main Results References