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Hadamard Type Inequalities for m-Convex and (α,m)-Convex Functions via Fractional Integrals PDF

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Preview Hadamard Type Inequalities for m-Convex and (α,m)-Convex Functions via Fractional Integrals

HADAMARD TYPE INEQUALITIES FOR m−CONVEX AND (α,m)−CONVEX FUNCTIONS VIA FRACTIONAL INTEGRALS 2 1 M.EMINO¨ZDEMIR(cid:7),MERVEAVCI♣,⋆,AHMETOCAKAKDEMIR♠, 0 AND♠ALPEREKINCI 2 n Abstract. In this paper, we established some new Hadamard-type integral a inequalities for functions whose derivatives of absolute values are m−convex J and(α,m)−convex functionsviaRiemann-Liouvillefractionalintegrals. 7 2 ] A 1. Introduction F . Letf :I ⊂R→RbeaconvexfunctiondefinedontheintervalI ofrealnumbers h t and a,b∈I with a<b. The following inequality is well known in the literature as a the Hermite–Hadamard inequality: m [ a+b 1 b f(a)+f(b) f ≤ f(x)dx≤ . 1 (cid:18) 2 (cid:19) b−aZa 2 v 3 In [1], G.Toader defined the concept of m-convexity as the following: 8 7 Definition1. Thefunction f :[0,b]→Ris saidtobem−convex,wherem∈[0,1], 5 if for every x,y ∈[0,b] and t∈[0,1] we have: . 1 f(tx+m(1−t)y)≤tf(x)+m(1−t)f(y). 0 2 1 Denote by Km(b) the set of the m−convex functions on [0,b] for which f(0)≤0. : v Severalpapershavebeenwrittenonm−convexfunctionsandwereferthepapers i X [2]-[11]. In [11], the following inequality of Hermite-Hadamard type for m−convex func- r a tions holds: Theorem 1. Let f : [0,∞) → R be a m−convex function with m ∈ (0,1]. If 0≤a<b<∞ and f ∈L [a,b], then one has the inequality: 1 1 b f(a)+mf(b) f(b)+mf(a) (1.1) f(x)dx≤min m , m . b−a 2 2 Za ( ) In [19], S.S. Dragomir proved the following theorem. Key words and phrases. m−convex functions, (α,m)−convex functions, Riemann-Liouville fractionalintegral. ⋆CorrespondingAuthor. 1 2M.EMINO¨ZDEMIR(cid:7),MERVEAVCI♣,⋆,AHMETOCAKAKDEMIR♠,AND♠ALPEREKINCI Theorem 2. Let f : [0,∞) → R be a m−convex function with m ∈ (0,1]. If f ∈L [am,b] where 0≤a<b, then one has the inequality: 1 1 1 mb 1 b (1.2) f(x)dx+ f(x)dx m+1 mb−a b−ma " Za Zma # f(a)+f(b) ≤ . 2 In [18], Mihe¸san gave definition of (α,m)−convexity as following; Definition 2. The function f : [0,b] → R, b > 0 is said to be (α,m)−convex, where (α,m)∈[0,1]2, if we have f(tx+m(1−t)y)≤tαf(x)+m(1−tα)f(y) for all x,y ∈[0,b] and t∈[0,1]. Denote by Kα(b) the class of all (α,m)−convex functions on [0,b] for which m f(0)≤0. If we choose (α,m)=(1,m), it can be easily seen that (α,m)−convexity reducestom−convexityandfor(α,m)=(1,1), wehaveordinaryconvexfunctions on[0,b]. For the recentresults based onthe above definition see the papers [2], [9], [6], [7], [4], and [5]. We give some necessarydefinitions andmathematicalpreliminaries of fractional calculus theory which are used throughout this paper. Definition 3. Let f ∈L [a,b]. The Riemann-Liouville integrals Jα f and Jα f of 1 a+ b− order α>0 with a≥0 are defined by 1 x Jα f(x)= (x−t)α−1f(t)dt, x>a a+ Γ(α) Za and 1 b Jα f(x)= (t−x)α−1f(t)dt, x<b b− Γ(α) Zx where Γ(α)= ∞e−tuα−1du, here is J0 f(x)=J0 f(x)=f(x). 0 a+ b− In the caseRof α = 1, the fractional integral reduces to the classical integral. Properties of this operator can be found in the references [12]-[14]. In[15],Sarikayaetal. provedavariantoftheidentityisestablishedbyDragomir and Agarwal in [16, Lemma 2.1] for fractional integrals as the following. Lemma 1. Let f : [a,b] → R be a differentiable mapping on (a,b) with a < b. If f′ ∈L[a,b], then the following equality for fractional integrals holds: f(a)+f(b) Γ(α+1) − [Jα f(b)+Jα f(a)] 2 2(b−a)α a+ b− b−a 1 = [(1−t)α−tα]f′(ta+(1−t)b)dt. 2 Z0 The aim of this paper is to establish Hadamard type inequalities for m−convex and (α,m)−convex functions via Riemann-Liouville fractional integrals. INEQUALITIES VIA FRACTIONAL INTEGRALS 3 2. Inequalities for m−convex functions Theorem 3. Let f : [a,b] → R be a positive function with 0 ≤ a < b and f ∈ L [a,b]. If f is a m−convex function on [a,b], then the following inequalities for 1 fractional integrals with α>0 and m∈(0,1] hold: (2.1) Γ(α) f(a) b Γ(α)Γ(2) Jα f(b) ≤ +mf , (b−a)α a+ α+1 m Γ(α+2) (cid:18) (cid:19) Γ(α) f(b) a Γ(α)Γ(2) Jα f(a) ≤ +mf . (b−a)α b− α+1 m Γ(α+2) (cid:16) (cid:17) where Γ is Euler Gamma function. Proof. Since f is a m−convex function on [a,b], we know that for any t∈[0,1] b (2.2) f(ta+(1−t)b)≤tf(a)+m(1−t)f m (cid:18) (cid:19) and a (2.3) f(tb+(1−t)a)≤tf(b)+m(1−t)f . m By multiplying both sides of (2.2) by tα−1, then by int(cid:16)egra(cid:17)ting the resulting in- equality with respect to t over [0,1], we obtain 1 1 b tα−1f(ta+(1−t)b)dt ≤ tα−1 tf(a)+m(1−t)f m Z0 Z0 (cid:20) (cid:18) (cid:19)(cid:21) f(a) b = +mf β(α,2). α+1 m (cid:18) (cid:19) It is easy to see that 1tα−1f(ta+(1−t)b)dt = Γ(α) Jα f(b) and we note that, 0 (b−a)α a+ the Beta and the Gamma function (see [17, pp 908-910]) R 1 ∞ β(x,y)= tx−1(1−t)y−1dt, x,y >0, Γ(x)= e−ttx−1dt, x>0 Z0 Z0 are used to evaluate the integral 1 tα−1(1−t)dt, Z0 where Γ(x)Γ(y) β(x,y)= , Γ(x+y) thus we can obtain that Γ(α)Γ(2) β(α,2)= Γ(α+2) which completes the proof. For the proofofthe secondinequality in (2.1) wemultiply both sides of(2.3) by tα−1, then integrate the resulting inequality with respect to t over [0,1]. (cid:3) Remark 1. If we choose α = 1 in Theorem 3, then the inequalities (2.1) become the inequality in (1.1). 4M.EMINO¨ZDEMIR(cid:7),MERVEAVCI♣,⋆,AHMETOCAKAKDEMIR♠,AND♠ALPEREKINCI Theorem 4. Let f : I◦ ⊂ [0,∞) → R, be a differentiable function on I◦ such that f′ ∈ L[a,b] where a,b ∈ I,a < b. If |f′|q is m−convex on [a,b] for some fixed m∈(0,1] and q ≥1, then the following inequality for fractional integrals holds: f(a)+f(b) Γ(α+1) − [Jα f(b)+Jα f(a)] 2 2(b−a)α a+ b− (cid:12) (cid:12) ≤ (cid:12)(cid:12)(cid:12)b−a21−q1 2α−1 |f′(a)|q +m f′ b (cid:12)(cid:12)(cid:12) q 1q . 2 2α(α+1) m (cid:20) (cid:21)(cid:20) (cid:12) (cid:18) (cid:19)(cid:12) (cid:21) (cid:12) (cid:12) Proof. Supposethatq =1.FromLemma1andbyusing(cid:12)theprope(cid:12)rtiesofmodulus, (cid:12) (cid:12) we have f(a)+f(b) Γ(α+1) − [Jα f(b)+Jα f(a)] 2 2(b−a)α a+ b− (cid:12) (cid:12) ≤ (cid:12)(cid:12)(cid:12)b−a 1|(1−t)α−tα||f′(ta+(1−t)b)|dt. (cid:12)(cid:12)(cid:12) 2 Z0 Since |f′| is m−convex on [a,b], we have f(a)+f(b) Γ(α+1) − [Jα f(b)+Jα f(a)] 2 2(b−a)α a+ b− (cid:12) (cid:12) ≤ (cid:12)(cid:12)(cid:12)b−a 1|(1−t)α−tα| t|f′(a)|+m(1−t) f(cid:12)(cid:12)(cid:12)′ b dt 2 m Z0 (cid:20) (cid:12) (cid:18) (cid:19)(cid:12)(cid:21) 1 (cid:12) (cid:12) = b−a 2 [(1−t)α−tα] t|f′(a)|+m(1−(cid:12)(cid:12)t) f′ b(cid:12)(cid:12) dt 2 m (Z0 (cid:20) (cid:12) (cid:18) (cid:19)(cid:12)(cid:21) (cid:12) (cid:12) 1 b(cid:12) (cid:12) + [tα−(1−t)α] t|f′(a)|+m(1−t) f′ (cid:12) dt (cid:12) m Z21 (cid:20) (cid:12) (cid:18) (cid:19)(cid:12)(cid:21) ) (cid:12) (cid:12) b−a 2α−1 ′ ′ b (cid:12) (cid:12) = |f (a)|+m f (cid:12) (cid:12) 2 2α(α+1) m (cid:20) (cid:21)(cid:20) (cid:12) (cid:18) (cid:19)(cid:12)(cid:21) (cid:12) (cid:12) (cid:12) (cid:12) where we use the facts that (cid:12) (cid:12) 12 (1−t)αtdt= 1tα(1−t)dt= 1 − α+3 , Z0 Z21 (α+1)(α+2) 2α+2(α+1)(α+2) 21 tα+1dt= 1(1−t)α+1dt= 1 , Z0 Z21 2α+2(α+2) 21 (1−t)α+1dt= 1tα+1dt= 1 − 1 Z0 Z12 (α+2) 2α+2(α+2) and 12 tα(1−t)dt= 1(1−t)αtdt= α+3 Z0 Z21 2α+2(α+1)(α+2) INEQUALITIES VIA FRACTIONAL INTEGRALS 5 which completes the proof for this case. Suppose now that q >1. From Lemma 1, m−convexityof|f′|q andusing the well-knownHo¨lder’s inequality we havesucces- sively f(a)+f(b) Γ(α+1) − [Jα f(b)+Jα f(a)] 2 2(b−a)α a+ b− (cid:12) (cid:12) ≤ (cid:12)(cid:12)(cid:12)b−a 1|(1−t)α−tα||f′(ta+(1−t)b)|dt (cid:12)(cid:12)(cid:12) 2 Z0 = b−a 1|(1−t)α−tα|1−q1 |(1−t)α−tα|q1 |f′(ta+(1−t)b)|dt 2 Z0 q−1 1 b−a 1 q 1 q ≤ |(1−t)α−tα|dt |(1−t)α−tα||f′(ta+(1−t)b)|qdt 2 (cid:18)Z0 (cid:19) (cid:18)Z0 (cid:19) ≤ b−a21−q1 2α−1 |f′(a)|q+m f′ b q q1 2 2α(α+1) m (cid:20) (cid:21)(cid:20) (cid:12) (cid:18) (cid:19)(cid:12) (cid:21) (cid:12) (cid:12) where we use the fact that (cid:12) (cid:12) 1 1 (cid:12) (cid:12) 1 |(1−t)α−tα|dt = 2 [(1−t)α−tα]dt+ [tα−(1−t)α]dt Z0 Z0 Z12 2 1 = 1− . α+1 2α (cid:18) (cid:19) (cid:3) Corollary 1. Under the assumptions of Theorem 4 with α ∈ (0,1], the following inequality holds: f(a)+f(b) Γ(α+1) − [Jα f(b)+Jα f(a)] 2 2(b−a)α a+ b− (cid:12) (cid:12) ≤ (cid:12)(cid:12)(cid:12)b−a 1 p1 |f′(a)|q+m f′ mb q q1(cid:12)(cid:12)(cid:12). 2 αp+1 2 (cid:18) (cid:19) (cid:12) (cid:0) (cid:1)(cid:12) ! (cid:12) (cid:12) Proof. It is similar the proof of Theorem 4. In addition, we used the following inequality |tα−tα|≤|t −t |α, 1 2 1 2 where α∈(0,1] and t ,t ∈[0,1]. (cid:3) 1 2 Theorem 5. Let f : [0,∞) → R be a m−convex function with m ∈ (0,1]. If f ∈L [am,b], where 0≤a<b, then the following inequality for fractional integrals 1 with α>0 holds: (2.4) Γ(α) 1 1 [Jα f(mb)+Jα f(mb)]+ [Jα f(mb)+Jα f(mb)] m+1 (mb−a)α a+ mb− (b−ma)α ma+ b− (cid:26) (cid:27) f(a)+f(b) ≤ . α Proof. By the m−convexity of f we can write f(ta+m(1−t)b)≤tf(a)+m(1−t)f(b), f((1−t)a+mtb)≤(1−t)f(a)+mtf(b), 6M.EMINO¨ZDEMIR(cid:7),MERVEAVCI♣,⋆,AHMETOCAKAKDEMIR♠,AND♠ALPEREKINCI f(tb+m(1−t)a)≤tf(b)+m(1−t)f(a) and f((1−t)b+mta)≤(1−t)f(b)+mtf(a) for all t∈[0,1] and 0≤a<b. If we add the above inequalities we get (2.5) f(ta+m(1−t)b)+f((1−t)a+mtb)+f(tb+m(1−t)a)+f((1−t)b+mta) ≤ (m+1)(f(a)+f(b)). Bymultiplyingbothsidesof(2.5)withtα−1,thenintegratingtheresultinginequal- ity with respect to t over [0,1], we obtain 1 tα−1[f(ta+m(1−t)b)+f((1−t)a+mtb)+f(tb+m(1−t)a)+f((1−t)b+mta)]dt Z0 1 ≤ (m+1)(f(a)+f(b)) tα−1dt. Z0 It is easy to see that 1 Γ(α) tα−1f(ta+m(1−t)b)dt= Jα f(mb), (mb−a)α a+ Z0 1 Γ(α) tα−1f((1−t)a+mtb)dt= Jα f(mb), (mb−a)α mb− Z0 1 Γ(α) tα−1f(tb+m(1−t)a)dt= Jα f(mb) (b−ma)α b− Z0 and 1 Γ(α) tα−1f((1−t)b+mta)dt= Jα f(mb). (b−ma)α ma+ Z0 So the proof is completed. (cid:3) Remark 2. If we choose α = 1 in Theorem 5, then the inequality (2.4) becomes the inequality in (1.2). 3. Inequalities for (α,m)−convex functions Theorem 6. Let f : [a,b] → R be a positive function with 0 ≤ a < b and f ∈ L [a,b]. If f is an (α ,m)−convex function on [a,b], then the following inequalities 1 1 hold for fractional integrals with α>0 and (α ,m)∈(0,1]2: 1 (3.1) Γ(α) 1 mα b Jα f(b) ≤ f(a)+ 1 f , (b−a)α a+ α+α α(α+α ) m 1 1 (cid:18) (cid:19) Γ(α) 1 mα a Jα f(a) ≤ f(b)+ 1 f . (b−a)α b− α+α α(α+α ) m 1 1 (cid:16) (cid:17) Proof. Since f is (α ,m)−convexfunction on [a,b], we know that for any t∈[0,1] 1 b (3.2) f(ta+(1−t)b)≤tα1f(a)+m(1−tα1)f m (cid:18) (cid:19) INEQUALITIES VIA FRACTIONAL INTEGRALS 7 and a (3.3) f(tb+(1−t)a)≤tα1f(b)+m(1−tα1)f . m (cid:16) (cid:17) By multiplying both sides of (3.2) by tα−1, then by integrating the resulting in- equality with respect to t over [0,1], we get 1 1 b tα−1f(ta+(1−t)b)dt ≤ tα−1 tα1f(a)+m(1−tα1)f m Z0 Z0 (cid:20) (cid:18) (cid:19)(cid:21) 1 mα b = f(a)+ 1 f . α+α α(α+α ) m 1 1 (cid:18) (cid:19) It is easy to see that 1tα−1f(ta+(1−t)b)dt= Γ(α) Jα f(b), by using this fact 0 (b−a)α a+ the proof of the first inequality is completed. R Similarly, by multiplying both sides of (3.3) by tα−1, then by integration, the proof of the second inequality is completed. (cid:3) Corollary 2. If we choose α = α with α,α ∈ (0,1] in Theorem 6, we have the 1 1 following inequalities; Γ(α) 1 b Jα f(b) ≤ f(a)+mf , (b−a)α a+ 2α m (cid:20) (cid:18) (cid:19)(cid:21) Γ(α) 1 a Jα f(a) ≤ f(b)+mf . (b−a)α b− 2α m h (cid:16) (cid:17)i Remark 3. If we choose α = α = 1 in Theorem 6, then the inequalities reduces 1 to the inequality (1.1). Theorem 7. Let f :I◦ ⊂[0,∞)→R, be a differentiable function on I◦ such that f′ ∈L[a,b] where a,b∈I, a<b. If |f′|q is an (α ,m)−convex function and |f′| is 1 decreasing on [a,b], then the following inequalities hold for fractional integrals with α>0, (α ,m)∈(0,1]2 ; 1 f(a)+f(b) Γ(α+1) − [Jα f(b)+Jα f(a)] 2 2(b−a)α a+ b− (cid:12) (cid:12) ≤ (cid:12)(cid:12)(cid:12)b−a 2α−1 q−q1 2α+α1 −1 (cid:12)(cid:12)(cid:12) |f′(a)|q−m f′ b q 2 (cid:20)2α−1(α+1)(cid:21) (cid:20)(cid:18)2α+α1(α+α1+1)(cid:19)(cid:18) (cid:12) (cid:18)m(cid:19)(cid:12) (cid:19) m b q 1 q1 (cid:12)(cid:12) (cid:12)(cid:12) + f′ 1− . (cid:12) (cid:12) α+1 m 2α (cid:12) (cid:18) (cid:19)(cid:12) (cid:18) (cid:19)(cid:21) (cid:12) (cid:12) Proof. Suppose(cid:12)thatq =(cid:12)1.FromLemma1andbyusingthepropertiesofmodulus, (cid:12) (cid:12) we have f(a)+f(b) Γ(α+1) − [Jα f(b)+Jα f(a)] 2 2(b−a)α a+ b− (cid:12) (cid:12) ≤ (cid:12)(cid:12)(cid:12)b−a 1|(1−t)α−tα||f′(ta+(1−t)b)|dt. (cid:12)(cid:12)(cid:12) 2 Z0 8M.EMINO¨ZDEMIR(cid:7),MERVEAVCI♣,⋆,AHMETOCAKAKDEMIR♠,AND♠ALPEREKINCI Since |f′| is (α ,m)−convex on [a,b], we have 1 f(a)+f(b) Γ(α+1) − [Jα f(b)+Jα f(a)] 2 2(b−a)α a+ b− (cid:12) (cid:12) ≤ (cid:12)(cid:12)(cid:12)b−a 1|(1−t)α−tα| tα|f′(a)|+m(1−tα(cid:12)(cid:12)(cid:12)) f′ b dt 2 m Z0 (cid:20) (cid:12) (cid:18) (cid:19)(cid:12)(cid:21) 1 (cid:12) (cid:12) = b−a 2 [(1−t)α−tα] tα|f′(a)|+m(1−tα(cid:12)(cid:12)) f′ b(cid:12)(cid:12) dt 2 m (Z0 (cid:20) (cid:12) (cid:18) (cid:19)(cid:12)(cid:21) (cid:12) (cid:12) 1 b(cid:12) (cid:12) + [tα−(1−t)α] tα|f′(a)|+m(1−tα) f′ (cid:12) dt (cid:12). m Z21 (cid:20) (cid:12) (cid:18) (cid:19)(cid:12)(cid:21) ) (cid:12) (cid:12) (cid:12) (cid:12) By computing the above integrals, we obtain (cid:12) (cid:12) f(a)+f(b) Γ(α+1) − [Jα f(b)+Jα f(a)] 2 2(b−a)α a+ b− (cid:12) (cid:12) (cid:12)(cid:12)b−a 2α+α1 −1 ′ (cid:12)(cid:12)′ b ≤ (cid:12) |f (a)|−m f(cid:12) 2 (cid:20)(cid:18)2α+α1(α+α1+1)(cid:19)(cid:18) (cid:12) (cid:18)m(cid:19)(cid:12)(cid:19) m b 1 (cid:12) (cid:12) + f′ 1− (cid:12)(cid:12) (cid:12)(cid:12) α+1 m 2α (cid:12) (cid:18) (cid:19)(cid:12)(cid:18) (cid:19)(cid:21) (cid:12) (cid:12) (cid:12) (cid:12) where we used the fact th(cid:12)at (cid:12) 21 1 1 tα1(1−t)αdt= tα1(1−t)αdt=β( ;α +1,α+1). 1 Z0 Z12 2 Thiscompletestheproofofthiscase. Supposenowthatq >1.Again,fromLemma 1 and by applying well-known Ho¨lder’s inequality, we have f(a)+f(b) Γ(α+1) − [Jα f(b)+Jα f(a)] 2 2(b−a)α a+ b− (cid:12) (cid:12) (cid:12) q−1 (cid:12) 1 (cid:12)b−a 1 q 1 (cid:12) q ≤ (cid:12) |(1−t)α−tα|dt |(1−t)(cid:12)α−tα||f′(ta+(1−t)b)|qdt 2 (cid:18)Z0 (cid:19) (cid:18)Z0 (cid:19) By computing the above integrals and by using (α ,m)−convexity of |f′|q, we 1 deduce f(a)+f(b) Γ(α+1) − [Jα f(b)+Jα f(a)] 2 2(b−a)α a+ b− (cid:12) (cid:12) ≤ (cid:12)(cid:12)(cid:12)b−a 2α−1 q−q1 2α+α1 −1 (cid:12)(cid:12)(cid:12) |f′(a)|q−m f′ b q 2 (cid:20)2α−1(α+1)(cid:21) (cid:20)(cid:18)2α+α1(α+α1+1)(cid:19)(cid:18) (cid:12) (cid:18)m(cid:19)(cid:12) (cid:19) m b q 1 q1 (cid:12)(cid:12) (cid:12)(cid:12) + f′ 1− . (cid:12) (cid:12) α+1 m 2α (cid:12) (cid:18) (cid:19)(cid:12) (cid:18) (cid:19)(cid:21) (cid:12) (cid:12) Which complet(cid:12)(cid:12)es the pr(cid:12)(cid:12)oof. (cid:3) INEQUALITIES VIA FRACTIONAL INTEGRALS 9 Corollary 3. If we choose α = α with α,α ∈ (0,1] in Theorem 7, we have the 1 1 inequality; f(a)+f(b) Γ(α+1) − [Jα f(b)+Jα f(a)] 2 2(b−a)α a+ b− (cid:12) (cid:12) ≤ (cid:12)(cid:12)(cid:12)b−a 2α−1 q−q1 22α−1 |f(cid:12)(cid:12)(cid:12)′(a)|q−m f′ b q 2 2α−1(α+1) 22α(2α+1) m (cid:20) (cid:21) (cid:20)(cid:18) (cid:19)(cid:18) (cid:12) (cid:18) (cid:19)(cid:12) (cid:19) m b q 1 q1 (cid:12)(cid:12) (cid:12)(cid:12) + f′ 1− . (cid:12) (cid:12) α+1 m 2α (cid:12) (cid:18) (cid:19)(cid:12) (cid:18) (cid:19)(cid:21) (cid:12) (cid:12) Remark 4. Ident(cid:12)ical resul(cid:12)t of Theorem 5, can be stated for (α,m)−convex func- (cid:12) (cid:12) tions, but we omit the details. References [1] G.H.Toader,Somegeneralisationsoftheconvexity,Proc. Colloq.Approx. Optim (1984)329 338. [2] M.K. Bakula, M. E O¨zdemir, J. Peˇcari´c, Hadamard type inequalities for m−convex and (α,m)−convex functions,J. Inequal. Pure Appl. Math.9(2008), Article96. [3] M.E. O¨zdemir, M. Avci, E. Set, On some inequalities of Hermite–Hadamard type via m−convexity, Appl. Math. Lett.23(9)(2010)1065–1070. [4] M.E.O¨zdemir,H.Kavurmaci,E.Set, Ostrowski’stypeinequalitiesfor(α,m)−convex func- tions,Kyungpook Math. J. 50(2010)371–378. [5] M.E. O¨zdemir, M. Avcı and H. Kavurmacı, Hermite–Hadamard-type inequalities via (α,m)−convexity, Comput. Math. Appl.,61(2011), 2614–2620. [6] M.E. O¨zdemir, E. Set and M.Z. Sarıkaya, Some new Hadamard’s type inequalities for co- ordinatedm−convexand(α,m)−convexfunctions,Hacettepe J. of.Math. and Ist.,40,219- 229,(2011). [7] E.Set,M.Sardari,M.E.O¨zdemirandJ.Rooin,OngeneralizationsoftheHadamardinequal- ityfor(α,m)−convex functions,Kyungpook Math. J., Accepted. [8] M.Z.Sarıkaya,M.E.O¨zdemirandE.Set,InequalitiesofHermite–Hadamard’stypeforfunc- tions whose derivatives absolute values are m−convex, RGMIA Res. Rep. Coll. 13 (2010) Supplement,Article5. [9] M.Klariˇci´cBakula,J.Peˇcari´c,M.Ribiˇci´c,CompanioninequalitiestoJensen’sinequalityfor m−convexand(α,m)−convexfunctions,J.Inequal.PureAppl.Math.7(2006),Article194. [10] S.S.Dragomir,OnsomenewinequalitiesofHermite-Hadamardtypeform−convexfunctions, Tamkang J. Math.,3(1)2002. [11] S.S. Dragomir and G.H. Toader, Some inequalities for m−convex functions, Studia Univ. Babe¸s-Bolyai, Math.,38(1)(1993), 21-28. [12] R.Gorenflo,F.Mainardi,Fractionalcalculus: integralanddifferentialequationsoffractional order,SpringerVerlag, Wien (1997), 223-276. [13] S.MillerandB.Ross,AnintroductiontotheFractionalCalculusandFractionalDifferential Equations,John Wiley and Sons, USA,1993,p.2. [14] I.Podlubni,FractionalDifferentialEquations, Academic Press, San Diego,1999. [15] M.Z.Sarıkaya,E.Set,H.YaldızandN.Ba¸sak,Hermite-Hadamard’sinequalitiesforfractional integralsandrelatedfractionalinequalities,Submitted. [16] S.S. Dragomir and R.P. Agarwal, Two inequalities for differentiable mappings and applica- tionstospecialmeansofrealnumbersandtotrapezoidal formula,Appl. Math. Lett.,11(5) (1998), 91-95. [17] I.S. Gradshteyn, I.M. Ryzhik, Table of Integrals, Series, and Products, 7th ed., Academic Press, ElsevierInc.,2007. [18] V.G.Mihe¸san,Ageneralizationoftheconvexity, Seminar of Functional Equations, Approx. and Convex,Cluj-Napoca(Romania)(1993). [19] S.S.Dragomir,OnsomenewinequalitiesofHermite-Hadamardtypeform−convexfunctions, TamkangJournalofMathematics,33(1)(2002). 10M.EMINO¨ZDEMIR(cid:7),MERVEAVCI♣,⋆,AHMETOCAKAKDEMIR♠,AND♠ALPEREKINCI (cid:7)Ataturk University, K.K. Education Faculty, Department of Mathematics, 25240, Erzurum,Turkey E-mail address: [email protected] ♣Adıyaman University, Faculty of Science and Arts, Department of Mathematics, 02040,Adıyaman,Turkey E-mail address: [email protected] ♠Ag˘rıI˙brahimC¸ec¸enUniversity,FacultyofScienceandArts,DepartmentofMath- ematics,04100,Ag˘rı,Turkey E-mail address: [email protected] E-mail address: [email protected]

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