International Journal of Analysis and Applications ISSN 2291-8639 Volume 13, Number 1 (2017), 98-107 http://www.etamaths.com SOME GENERALIZED STEFFENSEN’S INEQUALITIES VIA A NEW IDENTITY FOR LOCAL FRACTIONAL INTEGRALS TUBA TUNÇ1,∗, MEHMET ZEKI SARIKAYA1 AND H. M. SRIVASTAVA2,3 Abstract. In this study, we first give an identity for local fractional integrals. We then make use of this identity in order to derive several generalizations of the celebrated Steffensen’s inequality associated with local fractional integrals. Relevant connections of the results presented in this paper with those that were proven in earlier works are also pointed out. 1. Introduction As long ago as 1919, Steffensen [25] established the following result which is known in the literature as Steffensen’s inequality. Theorem 1.1. Let a and b be real numbers such that a < b. Also let f,g : [a,b] → R be integrable functions such that f is nonincreasing and, for every x ∈ [a,b] , 0 5 g(x) 5 1. Then∫ b b−λ f(x)dx 5 ∫ b a f(x)g(x)dx 5 ∫ a+λ a f(x)dx, (1.1) where λ = ∫ b a g(x)dx. Steffensen’s inequality (1.1) happens to be the most basic inequality which deals with the comparison between integrals over a whole interval [a,b] and integrals over a subset of [a,b]. In fact, the inequality (1.1) has attracted considerable attention and interest from mathematicians and researchers. In this connection, the interested reader is referred to a number of works (see, for example, [3], [4], [7]- [10], [13], [14], [17] and [26]) for various related integral inequalities. Recently, Wu and Srivastava [27] proved the following inequality which is a weighted version of the inequality (1.1). Theorem 1.2. Let f, g and h be integrable functions defined on [a,b] with f nonincreasing. Also let 0 5 g(x) 5 h(x) for all x ∈ [a,b] . Then the following inequalities hold true:∫ b b−λ f(x)h(x)dx 5 ∫ b b−λ (f(x)h(x) − [f(x) −f(b−λ)] [h(x) −g(x)]) dx 5 ∫ b a f(x)g(x)dx 5 ∫ a+λ a ( f(x)h(x) − [f(x) −f(a + λ)] [h(x) −g(x)] ) dx 5 ∫ a+λ a f(x)h(x)dx, where λ is given by ∫ a+λ a h(x)dx = ∫ b a g(x)dx = ∫ b b−λ h(x)dx. Received 12th August, 2016; accepted 7th October, 2016; published 3rd January, 2017. 2010 Mathematics Subject Classification. 26D15, 26D99, 26A33. Key words and phrases. Steffensen’s inequality; local fractional integrals; fractal space; Fink’s identity. 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. 98 SOME GENERALIZED STEFFENSEN’S INEQUALITIES 99 We recall the following identity given by Fink [1]: 1 n ( f(x) + n−1∑ k=1 Fk(x) ) − 1 b−a ∫ b a f(t)dt = 1 n!(b−a) ∫ b a (x− t)n−1 κ(t.x)f(n)(t)dt, (1.2) where Fk(x) = ( n−k k! )( f(k−1)(a)(x−a)k −f(k−1)(b)(x− b)k b−a ) (1.3) and κ(t,x) =   t−a (a 5 t 5 x 5 b) t− b (a 5 x < t 5 b) Pečarić et al. [15] gave generalizations of Steffensen’s inequality (1.1) via Fink’s identity (1.2). Subsequently, Pečarić et al. [16] derived several new identities related to various generalizations of Steffensen’s inequality (1.1). 2. Definitions, Notations and Preliminaries The concepts of fractional calculus [8] and local fractional calculus (also called fractal calculus) (see, for details, [28] and [32]) are becoming increasingly useful in a wide variety of problems in mathematical, physical and engineering sciences (see, for example, the recent works [29] to [39]). With a view to introducing the definition of the local fractional derivative and the local fractional integral, we need the following notations and preliminaries (see [28] and [32]). For 0 < α 5 1, we have the following α-type sets of elements: Zα : The α-type set of integers defined by Zα := {0α,±1α,±2α, · · · ,±nα, · · · } . Qα : The α-type set of the rational numbers defined by Qα := { mα : mα = ( p q )α (p,q ∈ Z; q 6= 0) } . Jα : The α-type set of the irrational numbers defined by Jα := { mα : mα 6= ( p q )α (p,q ∈ Z; q 6= 0) } . Rα : The α-type set of the real line numbers defined by Rα := Qα ∪Jα. Proposition. Let aα,bα and cα belong to the set Rα of real line numbers. Then (1) aα + bα and aαbα belong to 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 definitions of the local fractional derivative and the local fractional integral can now be given as follows. Definition 2.1. (see [28] and [32]) A non-differentiable function f : R → Rα ( x → f(x) ) is said to be local fractional continuous at x = x0 if, for any ε > 0, there exists δ > 0 such that |f(x) −f(x0)| < εα holds true for |x−x0| < δ (ε,δ ∈ R). 100 TUNÇ,SARIKAYA AND SRIVASTAVA If the function f(x) is local continuous on the interval (a,b) , we denote this property as follows: f(x) ∈ Cα(a,b). Definition 2.2. (see [28] and [32]) The local fractional derivative of f(x) of order α (0 < α 5 1) 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 , (2.1) 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) (x ∈ I ⊆ R), then we write f ∈ D(k+1)α(I) (k ∈ N0 := {0, 1, 2, · · ·} = N∪{0}). Definition 2.3. (see [28] and [32]) Let f(x) ∈ Cα [a,b] . Then the local fractional integral of f(x) of order α (0 < α 5 1) is defined by aI α b f(x) = 1 Γ(1 + α) ∫ b a f(t)(dt)α = 1 Γ(1 + α) lim ∆t→0 N−1∑ j=0 f(tj)(∆tj) α (2.2) with ∆tj = tj+1 − tj and ∆t = max{∆t1, ∆t2, · · · , ∆tN−1} , where [tj, tj+1] (j = 0, 1, · · · ,N − 1) and a = t0 < t1 < · · · < tN−1 < tN = b is a partition of the interval [a,b] . Clearly, we find from Definition 2.3 that aI α b f(x) =   0 (a = b) − bIαa f(x) (a < b). (2.3) If, for any x ∈ [a,b] , there exists aIαx f(x), then we denote it simply as follows: f(x) ∈ Iαx [a,b] . Lemma 2.1. (see [28] and [32]) It is asserted that (i) dαxkα dxα = Γ(1 + kα) Γ(1 + (k − 1) α) x(k−1)α; (ii) 1 Γ(1 + α) ∫ b a xkα(dx)α = Γ(1 + kα) Γ(1 + (k + 1) α) ( b(k+1)α −a(k+1)α ) (k ∈ R). Lemma 2.2. (see [28] and [32]) Each of the following assertions holds true. (1) Local fractional integration is anti-differentiation: Suppose that f(x) = g(α)(x) ∈ Cα [a,b] . Then 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 aI α b f(x)g (α)(x) = f(x)g(x)|ba −a I α b f (α)(x)g(x). We next recall that Sarikaya et al. [23] proved the following generalized Steffensen inequality for local fractional integrals. SOME GENERALIZED STEFFENSEN’S INEQUALITIES 101 Theorem 2.1. Let f(x),g(x) ∈ Iαx [a,b] such that the function f(x) is non-increasing and 0 5 g(x) 5 1 on [a,b] (a < b). Then b−λI α b f(x) 5 aI α b f(x)g(x) 5 aI α a+λf(x), (2.4) where λα = Γ(1 + α) aI α b g(x). Sarikaya et al. [23] also stated the following identities which we shall use in order to prove our main results in this paper. 1 Γ(α + 1) ∫ a+λ a f(x)(dx)α − aIαb f(x)g(x) = 1 Γ(α + 1) ∫ a+λ a [f(x) −f (a + λ)] [1 −g(x)] (dx)α + 1 Γ(α + 1) ∫ b a+λ [f (a + λ) −f(x)] g(x)(dx)α (2.5) and aI α b f(x)g(x) − 1 Γ(α + 1) ∫ b b−λ f(x)(dx)α = 1 Γ(α + 1) ∫ b−λ a [f(x) −f (b−λ)] g(x)(dx)α + 1 Γ(α + 1) ∫ b b−λ [f (b−λ) −f(x)] [1 −g(x)] (dx)α. (2.6) The interested reader is referred to several other related works including (for example) [2], [5], [6], [11], [12], [18] to [24] and [28] to [39] for the theory and applications of local fractional calculus. In Section 3 of this paper, we give several inequalities which provide generalizations Steffensen’s inequality (1.1) for local fractional integrals. 3. Main Results We start with the following important identity for our work. Throughout this paper, Tk(x) is defined by Tk(x) = (n− 1 −k)α Γ(1 + kα) ( f(kα)(a)(x−a)kα −f(kα)(b)(x− b)α (b−a)α ) . (3.1) Lemma 3.1. Let f(n−1)α(t) be absolutely continuous on [a,b] with f(n)α ∈ Iαx [a,b]. Then 1 nα ( f(x) Γ(1 + α) + n−1∑ k=1 Fk ) − 1 (b−a)αΓ(1 + α) ∫ b a f(y)(dy)α = 1 nα (b−a)αΓ ( 1 + (n− 1)α ) [Γ(1 + α)]2 × ∫ b a (x− t)(n−1)α κα(t,x)f(nα)(t)dt, where Fk(x) = (n−k)α Γ(1 + kα) ( f(k−1)α(a)(x−a)kα −f(k−1)α(b)(x− b)kα (b−a)α ) and κα(t,x) =   (t−a)α (a 5 t 5 x 5 b) (t− b)α (a 5 x < t 5 b). 102 TUNÇ,SARIKAYA AND SRIVASTAVA Proof. We begin by recalling the following local fractional Taylor’s formula: f(x) = f(y) + n−1∑ k=1 f(kα)(y)(x−y)kα Γ(1 + kα) + 1 Γ(1 + α) ∫ b a f(nα)(t)(x− t)(n−1)α Γ ( 1 + (n− 1)α ) (dt)α, which, upon integration with respect to y from y = a to y = b, yields∫ b a (dy)α ∫ x y (dt)α = ∫ x a (dy)α ∫ x y (dt)α + ∫ b x (dy)α ∫ x y (dt)α = ∫ x a (dt)α ∫ t a (dy)α − ∫ b x (dy)α ∫ y x (dt)α = ∫ x a (dt)α ∫ t a (dy)α − ∫ b x (dt)α ∫ b t (dy)α. Thus, by evaluating the last integral, we have f(x)(b−a)α Γ(1 + α) = 1 Γ(1 + α) ∫ b a f(y)(dy)α + n−1∑ k=1 Ik + 1 Γ ( 1 + (n− 1)α ) [Γ(1 + α)]2 × ∫ b a f(nα)(t)(x− t)(n−1)ακα(t,x)(dt)α, (3.2) where Ik(x) = 1 Γ(1 + α) ∫ b a f(kα)(y)(x−y)kα Γ(1 + kα) (dy)α. By local fractional integration by parts in the above expression for Ik(x), we get Ik(x) = Ik−1(x) − (b−a)αFk(x)(n−k)−α (1 5 k 5 n− 1). Hence (n−k)α [Ik(x) − Ik−1(x)] = −(b−a)αFk(x). (3.3) The sum from k = 1 to k = n− 1 in (3.3) is given by n−1∑ k=1 Ik = −(b−a)α n−1∑ k=1 Fk(x) + (n− 1)αI0. (3.4) Substituting from (3.4) into (3.2) and rearranging the resulting equation, we have desired inequality asserted by Lemma 3.1. � Theorem 3.1. Let f : [a,b] 7→ Rα be such that f(n−1)α is absolutely continuous for some n ∈ N\{1}. Suppose that the functions g,h ∈ Iαx [a,b] are such that h is positive and 0α 5 g 5 1α on [a,b]. Also let aIa+λh(t) = aIbg(t)h(t) and the function S1 be defined by S1(x) =   1 Γ(1 + α) ∫x a [1 −g(t)] h(t)(dt)α (x ∈ [a,a + λ]) 1 Γ(1 + α) ∫ b x g(t)h(t)(dt)α (x ∈ [a + λ,b]). (3.5) SOME GENERALIZED STEFFENSEN’S INEQUALITIES 103 Then 1 Γ(1 + α) ∫ a+λ a f(t)h(t)(dt)α − 1 Γ(1 + α) ∫ b a f(t)g(t)h(t)(dt)α − Γ(1 + α) n−2∑ k=0 1 Γ(1 + α) ∫ b a S1(x)Tk(x)(dx) α = − 1 (b−a)αΓ ( 1 + (n− 2)α ) [Γ(1 + α)]2 × ∫ b a (∫ b a S1(x)(x− t)(n−2)α κα(t,x)(dx)α ) f(nα)(t)(dt)α. (3.6) Proof. It is easily seen that 1 Γ(1 + α) ∫ a+λ a f(t)h(t)(dt)α − 1 Γ(1 + α) ∫ b a f(t)g(t)h(t)(dt)α = 1 Γ(1 + α) ∫ a+λ a [f(t) −f (a + λ)] [1 −g(t)] h(t)(dt)α + 1 Γ(1 + α) ∫ b a+λ [f (a + λ) −f(t)] g(t)h(t)(dt)α = ( 1 Γ(1 + α) ∫ t a [1 −g(x)] h(x)(dx)α ) [f(t) −f (a + λ)] ∣∣∣∣a+λ a − 1 Γ(1 + α) ∫ a+λ a ( 1 Γ(1 + α) ∫ t a [1 −g(x)] h(x)(dx)α ) (df(t))α + ( 1 Γ(1 + α) ∫ t b g(x)h(x)(dx)α ) [f (a + λ) −f(t)] ∣∣∣∣b a+λ − 1 Γ(1 + α) ∫ b a+λ ( 1 Γ(1 + α) ∫ b t g(x)h(x)(dx)α ) (df(t))α = − 1 Γ(1 + α) ∫ a+λ a ( 1 Γ(1 + α) ∫ t a [1 −g(x)] h(x)(dx)α ) (df(t))α − 1 Γ(1 + α) ∫ b a+λ ( 1 Γ(1 + α) ∫ b t g(x)h(x)(dx)α ) (df(t))α = − 1 Γ(1 + α) ∫ b a S1(t)d(f(t)) α = − 1 Γ(1 + α) ∫ b a S1(x)f (α)(x)(dx)α. By applying Lemma 3.1 with f(α) and replacing n by n− 1 (n ∈ N\{1}), we obtain f(α)(x) = −Γ(1 + α) n−2∑ k=0 Tk(x) + 1 (b−a)αΓ ( 1 + (n− 2)α ) Γ(1 + α) × ∫ b a (x− t)(n−2)α κα(t,x)f(nα)(t)(dt)α. (3.7) Moreover, by using the equation (3.7), we get 104 TUNÇ,SARIKAYA AND SRIVASTAVA 1 Γ(1 + α) ∫ b a S1(x)f (α)(x)(dx)α = −Γ(1 + α) n−2∑ k=0 1 Γ(1 + α) ∫ b a S1(x)Tk(x)(dx) α + 1 (b−a)αΓ ( 1 + (n− 2)α ) [Γ(1 + α)]2 × ∫ b a S1(x) (∫ b a (x− t)(n−2)α κα(t,x)f(nα)(t)(dt)α ) (dx)α. (3.8) Finally, by applying Fubini’s theorem for local fractional double integrals in the last term in (3.8), we arrive at the assertion (3.6) of Theorem 3.1. � Theorem 3.2. Let f : [a,b] 7→ Rα be such that f(n−1)α is absolutely continuous for some n ∈ N\{1}. Suppose that the functions g,h ∈ Iαx [a,b] are such that h is positive and 0α 5 g 5 1α on [a,b]. Also let b−λIbh(t) = aIbg(t)h(t) and the function S2 be defined by S2(x) =   1 Γ(1 + α) ∫x a g(t)h(t) (dt)α (x ∈ [a,b−λ]) 1 Γ(1 + α) ∫ b x [1 −g(t)] h(t)(dt)α (x ∈ [b−λ,b]). (3.9) Then 1 Γ(1 + α) ∫ b a f(t)g(t)h(t)(dt)α − 1 Γ(1 + α) ∫ b b−λ f(t)h(t)(dt)α − Γ(1 + α) n−2∑ k=0 1 Γ(1 + α) ∫ b a S2(x)Tk(x)(dx) α = − 1 (b−a)αΓ ( 1 + (n− 2)α ) [Γ(1 + α)]2 × ∫ b a (∫ b a S2(x)(x− t)(n−2)α κα(t,x)(dx)α ) f(nα)(t)(dt)α. (3.10) SOME GENERALIZED STEFFENSEN’S INEQUALITIES 105 Proof. We observe that 1 Γ(1 + α) ∫ b a f(t)g(t)h(t)(dt)α − 1 Γ(1 + α) ∫ b b−λ f(t)h(t)(dt)α = 1 Γ(1 + α) ∫ b−λ a [f(t) −f (b−λ)] g(t)h(t)(dt)α + 1 Γ(1 + α) ∫ b b−λ [f (b−λ) −f(t)] [1 −g(t)] h(t)(dt)α = ( 1 Γ(1 + α) ∫ t a g(x)h(x)(dx)α ) [f(t) −f (b−λ)] ∣∣∣∣b−λ a − 1 Γ(1 + α) ∫ b−λ a ( 1 Γ(1 + α) ∫ t a g(x)h(x)(dx)α ) (df(t))α + ( 1 Γ(1 + α) ∫ t b [1 −g(x)] h(x)(dx)α ) [f (b−λ) −f(t)] ∣∣∣∣b b−λ − 1 Γ(1 + α) ∫ b b−λ ( 1 Γ(1 + α) ∫ b t [1 −g(x)] h(x)(dx)α )( df(t) )α = − 1 Γ(1 + α) ∫ b−λ a ( 1 Γ(1 + α) ∫ t a g(x)h(x)(dx)α )( df(t) )α − 1 Γ(1 + α) ∫ b b−λ ( 1 Γ(1 + α) ∫ b t [1 −g(x)] h(x)(dx)α )( df(t) )α = − 1 Γ(1 + α) ∫ b a S2(t)d ( f(t) )α = − 1 Γ(1 + α) ∫ b a S2(x)f (α)(x)(dx)α. By making use of Lemma 3.1 with f(α) and replacing n by n− 1 (n ∈ N\{1}), we obtain f(α)(x) = −Γ(1 + α) n−2∑ k=0 Tk(x) + 1 (b−a)αΓ ( 1 + (n− 2)α ) Γ(1 + α) × ∫ b a (x− t)(n−2)α κα(t,x)f(nα)(t)(dt)α. (3.11) Thus, by using this last equation (3.11), we get 1 Γ(1 + α) ∫ b a S2(x)f (α)(x)(dx)α = −Γ(1 + α) n−2∑ k=0 1 Γ(1 + α) ∫ b a S2(x)Tk(x)(dx) α + 1 (b−a)αΓ ( 1 + (n− 2)α ) [Γ(1 + α)]2 × ∫ b a S2(x) (∫ b a (x− t)(n−2)α κα(t,x)f(nα)(t)(dt)α ) (dx)α. (3.12) Finally, by applying Fubini’s theorem of local fractional double integrals in the last term in (3.12), we deduce the result asserted by Theorem 3.2. � Theorem 3.3. Let f : [a,b] 7→ Rα be such that f(n−1)α is absolutely continuous for some n ∈ N\{1}. Suppose that the functions g,h ∈ Iαx [a,b] are such that h is positive and 0α 5 g 5 1α on [a,b]. Also let aIa+λh(t) = aIbg(t)h(t) 106 TUNÇ,SARIKAYA AND SRIVASTAVA and the function S1 be defined by (3.5). If the function f is n-convex for local fractional calculus and∫ b a S1(x)(x− t)(n−2)α κα(t,x)(dx)α 5 0 (t ∈ [a,b]), then the following inequality holds true: 1 Γ(1 + α) ∫ b a f(t)g(t)h(t)(dt)α 5 1 Γ(1 + α) ∫ a+λ a f(t)h(t)(dt)α − Γ(1 + α) n−2∑ k=0 1 Γ(1 + α) ∫ b a S1(x)Tk(x)(dx) α. Proof. Since the function f is n-convex, we can suppose that f is n times differentiable and f(nα) = 0. Using this property and the assumption (3.6) of Theorem 3.1, we get the required inequality asserted by Theorem 3.3. � Theorem 3.4. Let f : [a,b] 7→ Rα be such that f(n−1)α is absolutely continuous for some n ∈ N\{1}. Suppose that the functions g,h ∈ Iαx [a,b] are such that h is positive and 0α 5 g 5 1α on [a,b]. Also let b−λIbh(t) = aIbg(t)h(t) and the function S2 be defined by (3.9). If the function f is n-convex for local fractional calculus and∫ b a S2(x)(x− t)(n−2)α κα(t,x)(dx)α 5 0 (t ∈ [a,b]), then the following inequality holds true: 1 Γ(1 + α) ∫ b a f(t)g(t)h(t)(dt)α = 1 Γ(1 + α) ∫ b b−λ f(t)h(t)(dt)α + Γ(1 + α) n−2∑ k=0 1 Γ(1 + α) ∫ b a S2(x)Tk(x)(dx) α. Proof. Since the function f is n-convex, we can suppose that f is n times differentiable and f(nα) = 0. Using this property and the assumption (3.10) of Theorem 3.2, we obtain the required inequality asserted by Theorem 3.4. � 4. Concluding Remarks and Observations The present investigation is motivated essentially by widespread applications of fractional calculus and local fractional calculus (also called fractal calculus) in a large variety of problems in mathematical, physical and engineering sciences. Here, in this paper, we have first derived an identity for local fractional integrals. 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Engrg. 7 (10) (2015), 1–5. 1Department of Mathematics, Faculty of Science and Arts, Düzce University, Düzce-Turkey 2Department of Mathematics and Statistics, University of Victoria, Victoria, British Columbia V8W 3R4, Canada 3Center for General Education (Department of Science and Technology), China Medical University, Taichung 40402, Taiwan, Republic of China ∗Corresponding author: tubatunc@duzce.edu.tr 1. Introduction 2. Definitions, Notations and Preliminaries 3. Main Results 4. Concluding Remarks and Observations References