International Journal of Analysis and Applications ISSN 2291-8639 Volume 11, Number 1 (2016), 61-69 http://www.etamaths.com INTEGRAL INEQUALITIES OF HERMITE-HADAMARD TYPE FOR HARMONIC (h, s)-CONVEX FUNCTIONS MUHAMMAD ASLAM NOOR∗, KHALIDA INAYAT NOOR AND SABAH IFTIKHAR Abstract. In this paper, we introduce a new concept of harmonic (h, s)-convex functions in the second sense which generalizes the harmonic convex functions. Some Hermite-Hadamard- Fejer type integral inequalities are derived. Some special cases also discussed. Results derived in this paper represent significant refinement and improvement of the known results. 1. Introduction Convexity theory has appeared as a powerful technique to study a wide class of unrelated problems in pure and applied sciences. For recent applications, generalizations and other aspects of convex functions and their variant forms, see [1, 14–16] and the references therein. Varosanec [17] introduced a class of convex functions with respective to an arbitrary non-negative function h, which is known as h-convex function.. This class of functions unifies various classes of convex functions and is being used to discuss several concepts in a unified manners. An other important class of convex functions is known as harmonic convex functions, was investigated by Anderson et al. [1] and Iscan [8]. For the recent developments, see [8,9, 11,12,13,15 and the references therein. Nooe et al. . [13] introduced and investigated a new class of convex functions. It has been shown a number new and known classes of convex functions can be obtained as special cases. Motivated and inspired by ongoing research in this filed, we introduce and study a new class of convex functions, which is called harmonic (h, s)-convex functions. One can easily show that harmonic (h, s)-convex functions include Godunova-Levin harmonic convex functions and harmonic s-convex functions as special cases. This is the main motivation of this paper. We also obtain several new Hermite-Hadamard-Fejer type inequalities for harmonic (h, s)-convex functions. Our results include several previously known and new results as special cases. The ideas and technique of this paper may be a starting point for further research in this dynamic field. 2. Preliminaries First of all, we recall the following basic concepts. Definition 2.1. [15]. A set I = [a, b] ⊆ R\{0} is said to be a harmonic convex set, if xy tx + (1 − t)y ∈ I, ∀x, y ∈ I, t ∈ [0, 1]. 2010 Mathematics Subject Classification. 26D15, 26D10, 90C23. Key words and phrases. harmonic convex functions; harmonic h-convex functions; harmonic s-convex func- tions; Hermite-Hadamard type inequality. c©2016 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 61 62 NOOR, NOOR AND IFTIKHAR Definition 2.2. [8]. A function f : I = [a, b] ⊆ R \ {0} → R is said to be harmonic convex function, if f ( xy tx + (1 − t)y ) ≤ (1 − t)f(x) + tf(y), ∀x, y ∈ I, t ∈ [0, 1]. We now introduce a new class of harmonic convex function in second sense, which is called the (h, s)-harmonic convex function. Definition 2.3. Let h : J = [0, 1] → R a nonnegative function. A function f : I = [a, b] ⊆ R\{0}→ R is said to be harmonic (h, s)-convex function in second sense, where s ∈ [−1, 1], if f ( xy tx + (1 − t)y ) ≤ h((1 − t)s)f(x) + h(ts)f(y), ∀x, y ∈ I, t ∈ (0, 1). For t = 1 2 , we have f ( 2xy x + y ) ≤ h ( 1 2s ) [f(x) + f(y)], which is called Jensen type harmonic (h, s)-convex function. We now discuss some special cases of harmonic (h, s) convex function. I. If we take h(ts) = ts and s = −1 in Definition 2.3, then it reduces to Godunova-Levin harmonic convex functions. Definition 2.4. [9]. A function f : I = [a, b] ⊆ R \ {0} → R is said to be Godunova-Levin harmonic convex, if f ( xy tx + (1 − t)y ) ≤ 1 1 − t f(x) + 1 t f(y), ∀x, y ∈ I, t ∈ (0, 1). II. If we take h(ts) = ts in Definition 2.3, then it reduces to extended harmonic s-convex functions. Definition 2.5. A function f : I = [a, b] ⊆ R \ {0} → R is said to be extended harmonic s-convex function in second sense, where s ∈ [−1, 1], if f ( xy tx + (1 − t)y ) ≤ (1 − t)sf(x) + tsf(y), ∀x, y ∈ I, t ∈ [0, 1]. III. If s = 1 in Definition 2.3, then it reduces to the harmonic h-convex functions. Definition 2.6. [11]. A function f : I = [a, b] ⊆ R\{0}→ R is said to be harmonic h-convex function, if f ( xy tx + (1 − t)y ) ≤ h(1 − t)f(x) + h(t)f(y), ∀x, y ∈ I, t ∈ [0, 1]. Definition 2.7. [14]. Two functions f, g are said to be similarly ordered (f is g-monotone), if and only if, 〈f(x) −f(y), g(x) −g(y)〉≥ 0, ∀x, y ∈ Rn. We now show that the product of two harmonic (h, s)-convex functions is again harmonic (h, s)-convex function. Lemma 2.1. If h(ts) + h((1 − t)s) ≤ 1, then the product of two similarly ordered harmonic (h, s)-convex functions is harmonic ((h, s)-convex function. INTEGRAL INEQUALITIES 63 Proof. Let f, g be two (h, s)-harmonic convex functions. Then f ( xy tx + (1 − t)y ) g ( xy tx + (1 − t)y ) ≤ [h((1 − t)s)f(x) + h(ts)f(y)][h((1 − t)s)g(x) + h(ts)g(y))] = [h((1 − t)s)]2f(x)g(x) + h(ts)h((1 − t)s)[f(x)g(y) + f(y)g(x)] +[h(ts)]2f(y)g(y) ≤ [h((1 − t))s]2f(x)g(x) + h(ts)h((1 − t)s)[f(x)g(x) + f(y)g(y)] +[h(ts)]2f(y)g(y) = [h((1 − t)s)f(x)g(x) + h(ts)f(y)g(y)][h(ts) + h((1 − t)s)] ≤ h((1 − t)s)f(x)g(x) + h(ts)f(y)g(y).(2.1) This shows that product of two similarly ordered harmonic (h, s)-convex functions is again a harmonic (h, s)-convex function. � We need the following well-known fact, which establishes a relationship between convex functions and harmonic convex functions. This fact plays a crucial part in deriving our results. Remark 2.1. Let I = [a, b] ⊆ R \ {0} and consider the function g : [ 1 b , 1 a ] → R defined by g(x) = f ( 1 x ) , then f is harmonic (h, s)-convex on [a, b], if and only if, g is (h, s)-convex in the usual sense on [ 1 b , 1 a ] . 3. Main results In this section, we obtain Hermite-Hadamard inequalities for harmonic (h, s)-convex function. Theorem 3.1. Let f : I = [a, b] ⊆ R \ {0} −→ R be harmonic (h, s)-convex function, where s ∈ (−1, 1]. If f ∈ L[a, b], then 1 2h ( 1 2s )f( 2ab a + b ) ≤ ab b−a ∫ b a f(x) x2 dx ≤ [f(a) + f(b)] ∫ 1 0 h(ts)dt. Proof. Let f be harmonic (h, s)-convex function with t = 1 2 . Then f ( 2xy x + y ) ≤ h ( 1 2s ) [f(x) + f(y)]. Taking x = ab ta+(1−t)b and y = ab (1−t)a+tb , we have f ( 2ab a + b ) ≤ h ( 1 2s )[ f ( ab ta + (1 − t)b ) + f ( ab (1 − t)a + tb )] = h ( 1 2s )[∫ 1 0 f ( ab ta + (1 − t)b ) dt + ∫ 1 0 f ( ab (1 − t)a + tb ) dt ] ≤ h ( 1 2s )∫ 1 0 [ h((1 − t)s)f(a) + h(ts)f(b) + h((1 − t)s)f(b) +h(ts)f(a) ] dt = 2h ( 1 2s ) [f(a) + f(b)] ∫ 1 0 h(ts)dt. Using the fact that ∫ 1 0 f ( ab ta + (1 − t)b ) dt = ab b−a ∫ b a f(x) x2 dx, 64 NOOR, NOOR AND IFTIKHAR we have 1 2h ( 1 2s )f( 2ab a + b ) ≤ ab b−a ∫ b a f(x) x2 dx ≤ [f(a) + f(b)] ∫ 1 0 h(ts)dt. � Theorem 3.2. Let f : I = [a, b] ⊆ R \ {0} −→ R be harmonic (h, s)-convex function, where s ∈ (−1, 1]. If f ∈ L[a, b], then ab b−a ∫ b a f(x) x2 dx ≤ 1 2 [f(a) + f(b)] ∫ 1 0 [h((1 − t)s) + h(ts)]dt. Proof. Let f be harmonic (h, s)-convex function. Then f ( ab ta + (1 − t)b ) ≤ h((1 − t)s)f(a) + h(ts)f(b) f ( ab (1 − t)a + tb ) ≤ h(ts)f(a) + h((1 − t)s)f(b) f ( ab ta + (1 − t)b ) ≤ h((1 − t)s)f(a) + h(ts)f(b) and f ( ab (1 − t)a + tb ) ≤ h(ts)f(a) + h((1 − t)s)f(b). Adding the above inequalities, we have f ( ab ta + (1 − t)b ) + f ( ab (1 − t)a + tb ) + f ( ab ta + (1 − t)b ) +f ( ab (1 − t)a + tb ) ≤ 2[f(a) + f(b)][h((1 − t)s) + h(ts)] Integrating the above inequality over t ∈ [0, 1], we obtain∫ 1 0 f ( ab ta + (1 − t)b ) dt + ∫ 1 0 f ( ab (1 − t)a + tb ) dt + ∫ 1 0 f ( ab ta + (1 − t)b ) dt + ∫ 1 0 f ( ab (1 − t)a + tb ) dt ≤ 2[f(a) + f(b)] ∫ 1 0 [h((1 − t)s) + h(ts)]dt, which implies ab b−a ∫ b a f(x) x2 dx ≤ 1 2 [f(a) + f(b)] ∫ 1 0 [h((1 − t)s) + h(ts)]dt, which is the required result. � Theorem 3.3. Let f, g : I = [a, b] ⊆ R \ {0} −→ R be two harmonic (h, s)-convex functions, where s ∈ (−1, 1]. If f ∈ L[a, b], then ab b−a ∫ b a f(x)g(x) x2 dx ≤ M(a, b) ∫ 1 0 [h(ts)]2dt + N(a, b) ∫ 1 0 h(ts)h((1 − t)s)dt, where M(a, b) = f(a)g(a) + f(b)g(b),(3.1) N(a, b) = f(a)g(b) + f(b)g(a).(3.2) INTEGRAL INEQUALITIES 65 Proof. Let f, g be two harmonic (h, s)-convex functions. Then f ( ab ta + (1 − t)b ) ≤ h((1 − t)s)f(a) + h(ts)f(b) g ( ab ta + (1 − t)b ) ≤ h((1 − t)s)g(a) + h(ts)g(b). Now f ( ab ta + (1 − t)b ) g ( ab (1 − t)a + tb ) ≤ [ h((1 − t)s)f(a) + h(ts)f(b) ][ h((1 − t)s)g(a) + h(ts)g(b) ] = [h((1 − t)s)]2[f(a)g(a)] + h(ts)h((1 − t)s)[f(a)g(b) + f(b)g(a)] +[h(ts)]2[f(b)g(b)] Integrating the above inequality over [0, 1], we have∫ 1 0 f ( ab ta + (1 − t)b ) g ( ab (1 − t)a + tb ) dt ≤ [f(a)g(a)] ∫ 1 0 [h((1 − t)s)]2dt + [f(a)g(b) + f(b)g(a)] ∫ 1 0 h(ts)h((1 − t)s)dt +[f(b)g(b)] ∫ 1 0 [h(ts)]2dt = [f(a)g(a) + f(b)g(b)] ∫ 1 0 [h(ts)]2dt +[f(a)g(b) + f(b)g(a)] ∫ 1 0 h(ts)h((1 − t)s)dt, Thus ab b−a ∫ b a f(x)g(x) x2 dx ≤ M(a, b) ∫ 1 0 [h(ts)]2dt + N(a, b) ∫ 1 0 h(ts)h((1 − t)s)dt, the required result. � Theorem 3.4. Let f, g : I = [a, b] ⊆ R \{0} −→ R be harmonic (h, s)-convex functions, where s ∈ (−1, 1]. If f, g ∈ L[a, b], then( ab b−a )s+1 ∫ 1 a 1 b h (( x− 1 b )s)[ f(a)g ( 1 x ) + g(a)f ( 1 x )] dx ( ab b−a )s+1 ∫ 1 a 1 b h (( 1 a −x )s)[ f(b)g ( 1 x ) + g(b)f ( 1 x )] dx ≤ M(a, b) ∫ 1 0 [h(ts)]2dt + N(a, b) ∫ 1 0 h(ts)h((1 − t)s)dt + ab b−a ∫ b a f(x)g(x) x2 dx, where M(a, b) and N(a, b) are given by (3.1) and (3.2) respectively. Proof. Let f, g be harmonic (h, s)-convex functions. Then f ( ab ta + (1 − t)b ) ≤ h((1 − t)s)f(a) + h(ts)f(b) g ( ab ta + (1 − t)b ) ≤ h((1 − t)s)g(a) + h(ts)g(b). 66 NOOR, NOOR AND IFTIKHAR Now, using 〈x1 −x2, x3 −x4〉≥ 0, (x1, x2, x3, x4 ∈ R) and x1 < x2, x3 < x4, we have f ( ab ta + (1 − t)b ) [h((1 − t)s)g(a) + h(ts)g(b)] +g ( ab ta + (1 − t)b ) [h((1 − t)s)f(a) + h(ts)f(b)] ≤ [h((1 − t)s)f(a) + h(ts)f(b)][h((1 − t)s)g(a) + h(ts)g(b)] +f ( ab ta + (1 − t)b ) g ( ab ta + (1 − t)b ) . Thus g(a)h((1 − t)s)f ( ab ta + (1 − t)b ) + g(b)h(ts)f ( ab ta + (1 − t)b ) +f(a)h((1 − t)s)g ( ab ta + (1 − t)b ) + f(b)h(ts)g ( ab ta + (1 − t)b ) ≤ [h((1 − t)s)]2[f(a)g(a)] + h(ts)h((1 − t)s)[f(b)g(a) +f(a)g(b)] + [h(ts)]2[f(b)g(b)] +f ( ab ta + (1 − t)b ) g ( ab ta + (1 − t)b ) Integrating the above inequality with respect to t over [0, 1], we have g(a) ∫ 1 0 h((1 − t)s)f ( ab ta + (1 − t)b ) dt + g(b) ∫ 1 0 h(ts)f ( ab ta + (1 − t)b ) dt +f(a) ∫ 1 0 h((1 − t)s)g ( ab ta + (1 − t)b ) dt + f(b) ∫ 1 0 h(ts)g ( ab ta + (1 − t)b ) dt ≤ [f(a)g(a)] ∫ 1 0 [h((1 − t)s)]2dt + [f(a)g(b) + f(b)g(a)] ∫ 1 0 h(ts)h((1 − t)s)dt +[f(b)g(b)] ∫ 1 0 [h(ts)]2dt + ∫ 1 0 f ( ab ta + (1 − t)b ) g ( ab ta + (1 − t)b ) dt, from which, it follows that( ab b−a )s+1 ∫ 1 a 1 b h (( x− 1 b )s)[ f(a)g ( 1 x ) + g(a)f ( 1 x )] dx ( ab b−a )s+1 ∫ 1 a 1 b h (( 1 a −x )s)[ f(b)g ( 1 x ) + g(b)f ( 1 x )] dx ≤ [f(a)g(a) + f(b)g(b)] ∫ 1 0 [h(ts)]2dt +[f(b)g(a) + f(a)g(b)] ∫ 1 0 h(ts)h((1 − t)s)dt + ab b−a ∫ b a f(x)g(x) x2 dx, which is the required result. � We need the following Lemma in order to obtain the Fejer type Hermite-Hadamard inequality for Harmonic (h, s)-convex functions. INTEGRAL INEQUALITIES 67 Lemma 3.1. Let f : I = [a, b] ⊆ R \ {0} −→ R be harmonic (h, s)-convex function, where s ∈ (−1, 1]. Then f ( abx (a + b)x−ab ) ≤ [h((1 − t)s) + h(ts)][f(a) + f(b)] −f(x). Proof. It is known that that x ∈ [a, b], can be represented as x = ab ta+(1−t)b , ∀t ∈ [0, 1]. Thus f ( abx (a + b)x−ab ) = f ( ab (1 − t)a + tb ) ≤ h(ts)f(a) + h((1 − t)s)f(b) = h(ts)[f(a) + f(b)] + h((1 − t)s)[f(a) + f(b)] −[h(ts)f(b) + h((1 − t)s)f(a)] ≤ h(ts)[f(a) + f(b)] + h((1 − t)s)[f(a) + f(b)] −f(x), the required result. � Theorem 3.5. Let f : I = [a, b] ⊆ R \ {0} −→ R be harmonic (h, s)-convex function, where s ∈ (−1, 1]. If f ∈ L[a, b], then 1 2h( 1 2s ) f ( 2ab a + b )∫ b a g(x) x2 dx ≤ ∫ b a f(x)g(x) x2 dx ≤ [f(a) + f(b)] 2 ∫ b a [ h ( b(x−a) x(b−a) )s + h ( a(b−x) x(b−a) )s] g(x) x2 dx, where g : [a, b] ⊆ R\{0} is a nonnegative, integrable function and satisfies g(x) = g ( abx [a + b]x−ab ) , ∀x ∈ [a, b]. Proof. Using the given fact and Lemma 3.1, we have 1 2h( 1 2s ) f ( 2ab a + b )∫ b a g(x) x2 dx = 1 2h( 1 2s ) ∫ b a f ( 2abx (a + b)x−ab + ab ) g(x) x2 dx ≤ 1 2h( 1 2s ) ∫ b a h ( 1 2s )[ f ( abx (a + b)x−ab ) + f(x) ] g(x) x2 dx = 1 2 ∫ b a f ( abx (a + b)x−ab ) g(x) x2 dx + 1 2 ∫ b a f(x)g(x) x2 dx = ∫ b a f(x)g(x) x2 dx 68 NOOR, NOOR AND IFTIKHAR To prove the other part of the inequality, we consider∫ b a f(x)g(x) x2 dx = 1 2 ∫ b a f ( abx (a + b)x−ab ) g(x) x2 dx + 1 2 ∫ b a f(x)g(x) x2 dx ≤ 1 2 ∫ b a [ [h((1 − t)s) + h(ts)][f(a) + f(b)] −f(x) ] g(x) x2 dx + 1 2 ∫ b a f(x)g(x) x2 dx ≤ [f(a) + f(b)] 2 ∫ b a [ h ( b(x−a) x(b−a) )s + h ( a(b−x) x(b−a) )s] g(x) x2 dx, This completes the proof. � Acknowledgements The authors would like to thank Dr. S. M. Junaid Zaidi, (S. I, H. I), Rector, COMSATS Institute of Information Technology, Pakistan, for providing excellent research and academic environments. References [1] G. D. Anderson, M. K. Vamanamurthy and M. 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Shi and Zhang, Some new judgement theorems of Schur geometric and Schur har- monic convexities for a class of symmetric functions, J. Inequal. Appl., 2013(2013), Article ID 527. INTEGRAL INEQUALITIES 69 [16] G. H. Toader, Some generalizations of the convexity, Proc. Colloq. Approx. Optim, Cluj- Napoca (Romania), 1984, 329-338. [17] S. Varoanec, On h-convexity, J. Math. Anal. Appl., 326(2007), 303-311. Department of Mathematics, COMSATS Institute of Information Technology, Park Road, Islam- abad, Pakistan ∗Corresponding author: noormaslam@gmail.com