Hacettepe Journal of Mathematics and Statistics Volume 40 (2) (2011), 219 – 229 SOME NEW HADAMARD TYPE INEQUALITIES FOR CO-ORDINATED m-CONVEX AND (α, m)-CONVEX FUNCTIONS M. Emin Özdemir∗†, Erhan Set‡ and Mehmet Zeki Sarıkaya∗ Received 15 : 06 : 2010 : Accepted 21 : 11 : 2010 Abstract In this paper, we establish some new Hermite-Hadamard type inequalities for m-convex and (α, m)-convex functions of 2-variables on the co-ordinates. Keywords: m-convex function, (α, m)-convex function, co-ordinated convex mapping, Hermite-Hadamard inequality. 2010 AMS Classification: 26 A 51, 26 D 15. Communicated by Alex Goncharov 1. Introduction Let f : I ⊆ R → R be a convex mapping defined on the interval I of real numbers, and a, b ∈ I with a < b. The following double inequality is well known in the literature as the Hermite-Hadamard inequality [5]: Z b a+b 1 f (a) + f (b) f ≤ f (x) dx ≤ . 2 b−a a 2 In [8], the notion of m-convexity was introduced by G.Toader as the following: 1.1. Definition. The function f : [0, b] → R, b > 0 is said to be m-convex, where m ∈ [0, 1], if we have f (tx + m (1 − t) y) ≤ tf (x) + m (1 − t) f (y) for all x, y ∈ [0, b] and t ∈ [0, 1]. We say that f is m-concave if −f is m-convex. ∗Atatürk University, K. K. Education Faculty, Department of Mathematics, 25240 Campus, Erzurum, Turkey. E-mail: [email protected] † Corresponding Author. ‡ Department of Mathematics, Faculty of Science and Arts, Düzce University, Düzce, Turkey. E-mail: (E. Set) [email protected] (M. Z. Sarıkaya) [email protected] 220 M. E. Özdemir, E. Set, M. Z. Sarıkaya Denote by Km (b) the class of all m-convex functions on [0, b] for which f (0) ≤ 0. Obviously, if we choose m = 1, Definition 1.1 recaptures the concept of standard convex functions on [0, b]. In [6], S. S. Dragomir and G. Toader proved the following Hadamard type inequalities for m-convex functions. 1.2. Theorem. Let f : [0, ∞) → R be an m-convex function with m ∈ (0, 1]. If 0 ≤ a < b < ∞ and f ∈ L1 [a, b], then the following inequality holds: Z b b a f (a) + mf m f (b) + mf m 1 f (x) dx ≤ min , . (1.1) b−a a 2 2 Some generalizations of this result can be found in [2, 3]. 1.3. Theorem. Let f : [0, ∞) → R be an m-convex differentiable function with m ∈ (0, 1]. Then for all 0 ≤ a < b the following inequality holds: Z b f (mb) b−a ′ 1 − f (mb) ≤ f (x) dx m 2 b−a a (1.2) (b − ma) f (b) − (a − mb) f (a) ≤ . 2 (b − a) Also, in [5], Dragomir and Pearce proved the following Hadamard type inequality for m-convex functions. 1.4. Theorem. Let f : [0, ∞) → R be an m-convex function with m ∈ (0, 1] and 0 ≤ a < b. If f ∈ L1 [a, b], then one has the inequality: Z b x f (x) + mf m a+b 1 (1.3) f ≤ dx. 2 b−a a 2 In [7], the definition of (α, m)-convexity was introduced by V. G. Miheşan as the following: 1.5. Definition. 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) ≤ tα f (x) + m(1 − tα )f (y) for all x, y ∈ [0, b] and t ∈ [0, 1]. α Denote by Km (b) the class of all (α, m)-convex functions on [0, b] for which f (0) ≤ 0. If we take (α, m) = (1, m), it can be easily seen that (α, m)-convexity reduces to mconvexity, and for (α, m) = (1, 1), (α, m)-convexity reduces to the usual concept of convexity defined on [0, b], b > 0. In [9], E. Set, M. Sardari, M. E. Ozdemir and J. Rooin proved the following Hadamard type inequalities for (α, m)-convex functions. 1.6. Theorem. Let f : [0, ∞) → R be an (α,m)-convex function with (α, m) ∈ (0, 1]2 . b a If 0 ≤ a < b < ∞ and f ∈ L1 [a, b] ∩ L1 m ,m , then the following inequality holds: b α Z f (x) + m(2 − 1)f x a+b 1 m dx. (1.4) f ≤ 2 b−a 2α a 1.7. Theorem. Let f : [0, ∞) → R be an (α, m)-convex function with (α, m) ∈ (0, 1]2 . If 0 ≤ a < b < ∞ and f ∈ L1 [a, b], then the following inequality holds: Z b b a f (a) + mαf m f (b) + mαf m 1 (1.5) f (x)dx ≤ min , . b−a a α+1 α+1 Co-ordinated m-Convex and (α, m)-Convex Functions 221 1.8. Theorem. Let f : [0, ∞) → R be an (α, m)-convex function with (α, m) ∈ (0, 1]2 . If 0 ≤ a < b < ∞ and f ∈ L1 [a, b], then the following inequality holds: " # Z b a b 1 1 f (a) + f (b) + mαf m + mαf m (1.6) f (x)dx ≤ . b−a a 2 α+1 Let us now consider a bidimensional interval ∆ =: [a, b] × [c, d] in R2 , with a < b and c < d. A function f : ∆ → R is said to be convex on ∆ if the following inequality: f (tx + (1 − t) z, ty + (1 − t) w) ≤ tf (x, y) + (1 − t) f (z, w) holds, for all (x, y) , (z, w) ∈ ∆ and t ∈ [0, 1]. A function f : ∆ → R is said to be convex on the co-ordinates on ∆ if the partial mappings fy : [a, b] → R, fy (u) = f (u, y) and fx : [c, d] → R, fx (v) = f (x, v) are convex where defined for all x ∈ [a, b] and y ∈ [c, d] (see [5, p. 317]). Also, in [4], Dragomir proved the following similar inequalities of Hadamard’s type for a co-ordinated convex mapping on a rectangle in the plane R2 . 1.9. Theorem. Suppose that f : ∆ → R is co-ordinated convex on ∆. Then one has the inequalities: a+b c+d f , 2 2 Z b Z d 1 c+d 1 a+b 1 f x, dx + f , y dy ≤ 2 b−a a 2 d−c c 2 Z bZ d 1 ≤ f (x, y) dx dy (b − a) (d − c) a c (1.7) Z b Z b 1 1 1 ≤ f (x, c) dx + f (x, d) dx 4 b−a a b−a a Z d Z d 1 1 f (a, y) dy + f (b, y) dy + d−c c d−c c f (a, c) + f (a, d) + f (b, c) + f (b, d) ≤ 4 The above inequalities are sharp. For co-ordinated s-convex functions, another version of this result can be found in [1]. The main purpose of this paper is to establish new Hadamard-type inequalities for functions of 2-variables which are m-convex or (α, m)-convex on the co-ordinates. 2. Inequalities for co-ordinated m-convex functions Firstly, we can define co-ordinated m-convex functions as follows: 2.1. Definition. Consider the bidimensional interval ∆ := [0, b] × [0, d] in [0, ∞)2 . The mapping f : ∆ → R is m-convex on ∆ if f (tx + (1 − t) z, ty + m (1 − t) w) ≤ tf (x, y) + m (1 − t) f (z, w) holds for all (x, y) , (z, w) ∈ ∆ with t ∈ [0, 1], b, d > 0, and for some fixed m ∈ [0, 1]. A function f : ∆ → R which is m-convex on ∆ is called co-ordinated m-convex on ∆ if the partial mappings fy : [0, b] → R, fy (u) = f (u, y) 222 M. E. Özdemir, E. Set, M. Z. Sarıkaya and fx : [0, d] → R, fx (v) = f (x, v) are m-convex for all y ∈ [0, d] and x ∈ [0, b] with b, d > 0, and for some fixed m ∈ [0, 1]. We also need the following Lemma for our main results. 2.2. Lemma. Every m-convex mapping f : ∆ ⊂ [0, ∞)2 → R is m-convex on the coordinates, where ∆ = [0, b] × [0, d] and m ∈ [0, 1]. Proof. Suppose that f : ∆ = [0, b] × [0, d] → R is m-convex on ∆. Consider the function fx : [0, d] → R, fx (v) = f (x, v) , (x ∈ [0, b]) . Then for t, m ∈ [0, 1] and v1 , v2 ∈ [0, d], we have fx (tv1 + m (1 − t) v2 ) = f (x, tv1 + m (1 − t) v2 ) = f (tx + (1 − t) x, tv1 + m (1 − t) v2 ) ≤ tf (x, v1 ) + m (1 − t) f (x, v2 ) = tfx (v1 ) + m (1 − t) fx (v2 ) . Therefore, fx (v) = f (x, v) is m-convex on [0, d]. The fact that fy : [0, b] → R, fy (u) = f (u, y) is also m-convex on [0, b] for all y ∈ [0, d] goes likewise, and we shall omit the details. 2.3. Theorem. Suppose that f : ∆ = [0, b] × [0, d] → R is an m-convex function on the co-ordinates on ∆. If 0 ≤ a < b < ∞ and 0 ≤ c < d < ∞ with m ∈ (0, 1], then one has the inequality: Z dZ b 1 f (x, y) dx dy (d − c) (b − a) c a (2.1) 1 1 ≤ min {v1 , v2 } + min {v3 , v4 } , 4 (b − a) 4 (d − c) where v1 = Z b f (x, c) dx + m a Z b Z b f a Z b d x, m dx c dx f x, m a a Z d Z d b v3 = f (a, y) dy + m f , y dy m c c Z d Z d a v4 = , y dy. f (b, y) dy + m f m c c v2 = f (x, d) dx + m Proof. Since f : ∆ → R is co-ordinated m-convex on ∆ it follows that the mapping gx : [0, d] → R, gx (y) = f (x, y) is m-convex on [0, d] for all x ∈ [0, b]. Then by the inequality (1.1) one has: ( ) Z d d c gx (c) + mgx m gx (d) + mgx m 1 gx (y) dy ≤ min , , d−c c 2 2 or 1 d−c Z d f (x, y) dy ≤ min c ( where 0 ≤ c < d < ∞ and m ∈ (0, 1]. f (x, c) + mf x, 2 d m c f (x, d) + mf x, m , 2 ) , Co-ordinated m-Convex and (α, m)-Convex Functions 223 Dividing both sides by (b − a) and integrating this inequality over [a, b] with respect to x, we have (2.2) Z bZ d 1 f (x, y) dy dx (b − a) (d − c) a c Z b Z b 1 m d ≤ min f (x, c) dx + f x, dx, 2 (b − a) a 2 (b − a) a m Z b Z b 1 m c f (x, d) dx + f x, dx 2 (b − a) a 2 (b − a) a m Z b Z b 1 d = min f (x, c) dx + m f x, dx, 2 (b − a) m a a Z b Z b c dx f (x, d) dx + m f x, m a a where 0 ≤ a < b < ∞. By a similar argument applied to the mapping gy : [0, b] → R, gy (x) = f (x, y) with 0 ≤ a < b < ∞, we get 1 (d − c) (b − a) (2.3) ≤ Z d b f (x, y) dx dy Z d Z min f (a, y) dy + m c 1 2 (d − c) Z a c Z c d d f b ,y m dy, f (b, y) dy + m c Z d f c a , y dy . m Summing the inequalities (2.2) and (2.3), we get the inequality (2.1). 2.4. Corollary. With the above assumptions, and provided that the partial mappings fy : [0, b] → R, fy (u) = f (u, y) and fx : [0, d] → R, fx (v) = f (x, v) are differentiable on (0, b) and (0, d), respectively, we have the inequalities 1 (b − a) (d − c) (2.4) and Z b Z d f (x, y) dy dx 1 d ≤ min (b − ma) f (b, c) + mf b, 4 (b − a) m d − (a − mb) f (a, c) + mf a, , m h c i (b − ma) f (b, d) + mf b, m h c i − (a − mb) f (a, d) + mf a, , m a c 224 M. E. Özdemir, E. Set, M. Z. Sarıkaya 1 (d − c) (b − a) (2.5) Z d Z b f (x, y) dx dy 1 b ≤ min (d − mc) f (a, d) + mf ,d 4 (d − c) m b ,c , − (c − md) f (a, c) + mf m h a i (d − mc) f (b, d) + mf ,d m h a i − (c − md) f (b, c) + mf ,c . m c a Proof. Since the partial mappings fx : [0, d] → R, fx (v) = f (x, v) are differentiable on [0, d], by the inequality (1.2) we have Z b 1 (b − ma) f (b, c) − (a − mb) f (a, c) f (x, c) dx ≤ , (b − a) a 2 (b − a) Z b d d (b − ma) f (b, m ) − (a − mb) f (a, m ) 1 d f x, dx ≤ , (b − a) a m 2 (b − a) Z b 1 (b − ma) f (b, d) − (a − mb) f (a, d) f (x, d) dx ≤ , and (b − a) a 2 (b − a) Z b c c (b − ma) f (b, m ) − (a − mb) f (a, m ) 1 c f x, dx ≤ . (b − a) a m 2 (b − a) Hence, using (2.2), we get the inequality (2.4). Analogously, Since the partial mappings fy : [0, b] → R, fy (u) = f (u, y) are differentiable on [0, b], using the inequality (2.3), we get the inequality (2.5). The proof is completed. 2.5. Remark. Choosing m = 1 in (2.4) or (2.5), we get the relationship between the third and fourth inequalities in (1.7). 2.6. Theorem. Suppose that f : ∆ = [0, b] × [0, d] → R is an m-convex function on the co-ordinates on ∆. If 0 ≤ a < b < ∞ and 0 ≤ c < d < ∞, m ∈ (0, 1] with fx ∈ L1 [0, d] and fy ∈ L1 [0, b], then one has the inequality: Z b Z d 1 c+d 1 a+b f x, dx + f , y dy b−a a 2 d−c 2 "Z Z c y b d f (x, y) + mf x, m 1 ≤ dy dx (2.6) (b − a) (d − c) a c 2 # Z dZ b x f (x, y) + mf m ,y dx dy . + 2 c a Proof. Since f : ∆ → R is co-ordinated m-convex on ∆ it follows that the mapping gx : [0, d] → R, gx (y) = f (x, y) is m-convex on [0, d] for all x ∈ [0, b]. Then by the Co-ordinated m-Convex and (α, m)-Convex Functions inequality (1.3) one has: Z d gx (y) + mgx c+d 1 gx ≤ 2 d−c c 2 y m 225 dy, or f x, c+d 2 ≤ 1 d−c Z d c y f (x, y) + mf x, m dy 2 for all x ∈ [0, b]. Integrating this inequality on [a, b], we have Z b 1 c+d f x, dx b−a a 2 (2.7) Z bZ d y f (x, y) + mf x, m 1 ≤ dy dx. (b − a) (d − c) a c 2 By a similar argument applied to the mapping gy : [0, b] → R, gy (x) = f (x, y), we get Z d 1 a+b f , y dy d−c c 2 (2.8) Z dZ b x f (x, y) + mf m ,y 1 ≤ dx dy. (d − c) (b − a) c a 2 Summing the inequalities (2.7) and (2.8), we get the inequality (2.6). 2.7. Remark. Choosing m = 1 in (2.6), we get the second inequality of (1.7). 3. Inequalities for co-ordinated (α, m)-convex functions 3.1. Definition. Consider the bidimensional interval ∆ := [0, b] × [0, d] in [0, ∞)2 . The mapping f : ∆ → R is (α, m)-convex on ∆ if (3.1) f (tx + (1 − t) z, ty + m (1 − t) w) ≤ tα f (x, y) + m (1 − tα ) f (z, w) holds for all (x, y) , (z, w) ∈ ∆ and (α, m) ∈ [0, 1]2 , with t ∈ [0, 1]. A function f : ∆ → R which is (α, m)-convex on ∆ is called co-ordinated (α, m)-convex on ∆ if the partial mappings fy : [0, b] → R, fy (u) = f (u, y) and fx : [0, d] → R, fx (v) = f (x, v) are (α, m)-convex for all y ∈ [0, d] and x ∈ [0, b] with some fixed (α, m) ∈ [0, 1]2 . Note that for (α, m) = (1, 1) and (α, m) = (1, m), one obtains the class of co-ordinated convex and of co-ordinated m-convex functions on ∆, respectively. 3.2. Lemma. Every (α, m)-convex mapping f : ∆ → R is (α, m)-convex on the coordinates, where ∆ = [0, b] × [0, d] and α, m ∈ [0, 1]. Proof. Suppose that f : ∆ → R is (α, m)-convex on ∆. Consider the function fx : [0, d] → R, fx (v) = f (x, v) , (x ∈ [0, b]) . 226 M. E. Özdemir, E. Set, M. Z. Sarıkaya Then for t ∈ [0, 1], (α, m) ∈ [0, 1]2 and v1 , v2 ∈ [0, d], one has fx (tv1 + m (1 − t) v2 ) = f (x, tv1 + m (1 − t) v2 ) = f (tx + (1 − t) x, tv1 + m (1 − t) v2 ) ≤ tα f (x, v1 ) + m (1 − tα ) f (x, v2 ) = tα fx (v1 ) + m (1 − tα ) fx (v2 ) . Therefore, fx (v) = f (x, v) is (α, m)-convex on [0, d]. The fact that fy : [0, b] → R, fy (u) = f (u, y) is also (α, m)-convex on [0, b] for all y ∈ [0, d] goes likewise, and we shall omit the details. 3.3. Theorem. Suppose that f : ∆ = [0, b] × [0, d] → R is an (α, m)-convex function on the co-ordinates on ∆, where (α, m) ∈ (0, 1]2 . If 0 ≤ a < b < ∞, 0 ≤ c < d < ∞ and fx ∈ L1 [0, d], fy ∈ L1 [0, b], then the following inequalities hold: Z b Z d 1 c+d 1 a+b f x, dx + f , y dy b−a a 2 d−c c 2 1 ≤ (3.2) (d − c) (b − a) Z d Z b 2f (x, y) + m(2α − 1) f x, y + f x , y m m dx dy, × 2α c a and (3.3) 1 (d − c) (b − a) ≤ Z d c Z b f (x, y) dx dy a 1 1 min {w1 , w2 } + min {w3 , w4 } , 2 (α + 1) (b − a) 2 (α + 1) (d − c) where w1 = Z Z b f (x, c) dx + αm a b Z b f a Z b x, d m dx c dx f x, m a a Z d Z d b , y dy w3 = f (a, y) dy + αm f m c c Z d Z d a w4 = f (b, y) dy + αm f , y dy. m c c w2 = f (x, d) dx + αm Proof. Since f : ∆ → R is co-ordinated (α, m)-convex on ∆ it follows that the mapping gx : [0, d] → R, gx (y) = f (x, y) is (α, m)-convex on [0, d] for all x ∈ [0, b]. Then by the inequality (1.4) one has: Z d y gx (y) + m(2α − 1)gx m c+d 1 gx ≤ dy, 2 d−c c 2α that is f c+d x, 2 1 ≤ d−c Z d c f (x, y) + m(2α − 1)f x, 2α y m dy, Co-ordinated m-Convex and (α, m)-Convex Functions 227 where 0 ≤ c < d < ∞ and (α, m) ∈ (0, 1]2 . Integrating this inequality on [a, b], we have (3.4) 1 b−a Z b f a x, c+d 2 dx 1 ≤ (d − c) (b − a) Z b a Z d c f (x, y) + m(2α − 1)f x, 2α y m dy dx, where 0 ≤ a < b < ∞. By a similar argument applied for the mapping gy : [0, b] → [0, ∞), gy (x) = f (x, y) with 0 ≤ a < b < ∞, we get Z d 1 a+b f , y dy d−c c 2 (3.5) Z bZ d x f (x, y) + m(2α − 1)f m ,y 1 ≤ dy dx. (d − c) (b − a) a c 2α Summing the inequalities (3.4) and (3.5), we get the inequality (3.2). The inequality (3.3) can be obtained in a similar way to the proof of Theorem 2.3 by using (1.5). 3.4. Remark. If we take α = 1, (3.2) and (3.3) reduce to (2.6) and (2.1), respectively. 3.5. Theorem. Suppose that f : ∆ = [0, b] × [0, d] → R is (α, m)-convex function on the co-ordinates on ∆, where (α, m) ∈ (0, 1]2 . If 0 ≤ a < b < ∞, 0 ≤ c < d < ∞ and fx ∈ L1 [0, d], fy ∈ L1 [0, b], then the following inequality holds: 1 (b − a) (d − c) ≤ (3.6) Z b a Z d f (x, y) dy dx c Z b Z b 1 1 1 f (x, c) dx + f (x, d) dx 4 (α + 1) b − a a b−a a Z b Z b c mα d mα f x, dx + f x, dx + b−a a m b−a a m Z d Z d 1 1 + f (a, y) dy + f (b, y) dy d−c c d−c c Z d Z d mα a mα b + f , y dy + f , y dy . d−c c m d−c c m Proof. Since f : ∆ → R is co-ordinated (α, m)-convex on ∆ it follows that the mapping gx : [0, d] → R, gx (y) = f (x, y) is (α, m)-convex on [0, d] for all x ∈ [0, b]. Then by inequality (1.6) one has: " # Z d c d 1 1 gx (c) + gx (d) + mα gx m + gx m gx (y) dy ≤ , d−c c 2 α+1 that is 1 d−c Z d c 1 f (x, y) dy ≤ 2 " f (x, c) + f (x, d) + mα f x, α+1 c m d + f x, m # , 228 M. E. Özdemir, E. Set, M. Z. Sarıkaya where 0 ≤ c < d < ∞ and (α, m) ∈ (0, 1]2 . Integrating this inequality on [a, b], we have Z bZ d 1 f (x, y) dy dx (b − a) (d − c) a c Z b Z b 1 1 1 (3.7) ≤ f (x, c) dx + f (x, d) dx 2 (α + 1) b − a a b−a a Z b Z b mα c mα d + f x, dx + f x, dx b−a a m b−a a m where 0 ≤ a < b < ∞. By a similar argument applied to the mapping gy : [0, b] → [0, ∞), gy (x) = f (x, y) with 0 ≤ a < b < ∞, we get Z dZ b 1 f (x, y) dx dy (d − c) (b − a) c a Z d Z d 1 1 1 (3.8) ≤ f (a, y) dy + f (b, y) dy 2 (α + 1) d − c c d−c c Z d Z d mα a mα b + f , y dy + f , y dy . d−c c m d−c c m Summing the inequalities (3.7) and (3.8), we get the inequality (3.6). 3.6. Corollary. Choosing m = 1 in Theorem 3.5, we get the following inequality Z bZ d 1 f (x, y) dy dx (b − a) (d − c) a c Z b Z b 1 1 1 ≤ f (x, c) dx + f (x, d) dx 4 (α + 1) b − a a b−a a Z b Z b α α + f (x, c) dx + f (x, d) dx b−a a b−a a Z d Z d 1 1 + f (a, y) dy + f (b, y) dy d−c c d−c c Z d Z d α α f (a, y) dy + f (b, y) dy . + d−c c d−c c 3.7. Remark. Choosing (α, m) = (1, 1) in (3.6), we get the third inequality of (1.7). References [1] Alomari, M. and Darus, M. Co-ordinated s-convex function in the first sense with some Hadamard-type inequalities, Int. J. Contemp. Math. Sciences 3 (32), 1557–1567, 2008. [2] Bakula, M. K. Özdemir, M. E. and Pečarić, J. Hadamard type inequalities for m-convex and (α, m)-convex functions, J. Inequal. Pure and Appl. Math. 9 (4), Art. 96, 2008. [3] Bakula, M. K., Pečarić, J. and Ribičić, M. Companion inequalities to Jensen’s inequality for m-convex and (α, m)-convex functions, J. Inequal. Pure and Appl. Math. 7 (5), Art. 194, 2006. [4] Dragomir, S. S. On Hadamard’s inequality for convex functions on the co-ordinates in a rectangle from the plane, Taiwanese Journal of Mathematics 4, 775–788, 2001. [5] Dragomir, S. S. and Pearce, C. E. M. Selected Topics on Hermite-Hadamard Inequalities and Applications (RGMIA Monographs, Victoria University, 2000). [Online: http://www.staff.vu.edu.au/RGMIA/monographs/hermite-hadamard.html]. [6] Dragomir, S. S. and Toader, G. Some inequalities for m-convex functions, Studia Univ. BabesBolyai, Mathematica 38 (1), 21–28, 1993. Co-ordinated m-Convex and (α, m)-Convex Functions 229 [7] Miheşan, V. G. A generalization of the convexity (Seminar on Functional Equations, Approx. and Convex., Cluj-Napoca, Romania, 1993). [8] Toader, G. Some generalizations of the convexity, Proc. Colloq. Approx. Opt. Cluj-Napoca, 329–338, 1984. [9] Set, E., Sardari, M., Ozdemir, M. E. and Rooin, J. On generalizations of the Hadamard inequality for (α, m)-convex functions, RGMIA Res. Rep. Coll. 12 (4), Article 4, 2009.