International Journal of Analysis and Applications Volume 16, Number 1 (2018), 38-49 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-16-2018-38 INEQUALITIES OF FEJÉR TYPE RELATED TO GENERALIZED CONVEX FUNCTIONS WITH APPLICATIONS S. MOHAMMADI ASLANI1, M. ROSTAMIAN DELAVAR2,∗ AND S. M. VAEZPOUR3 1 Department of Mathematics, Karaj Branch, Islamic Azad University, Karaj, Iran 2Department of Mathematics, Faculty of Basic Sciences, University of Bojnord, Bojnord, Iran 3Department of Mathematics, Amirkabir University of Technology, Tehran, Iran ∗Corresponding author: m.rostamian@ub.ac.ir Abstract. This paper deals with some Fejér type inequalities related to (η1,η2)-convex functions. In fact the difference between the right and middle part of Fejér inequality is estimated without using Hölder’s inequality when the absolute value of the derivative of considered function is (η1,η2)-convex. Furthermore we give two estimation results when the derivative of considered function is bounded and satisfies a Lipschitz condition. 1. Introduction and Preliminaries The Fejér integral inequality for convex functions has been proved in [5]: Theorem 1.1. Let f : [a,b] → R be a convex function. Then f (a + b 2 )∫ b a g(x)dx ≤ ∫ b a f(x)g(x)dx ≤ f(a) + f(b) 2 ∫ b a g(x)dx, (1.1) where g : [a,b] → [0, +∞) is integrable and symmetric to x = a+b 2 ( g(x) = g(a + b−x),∀x ∈ [a,b] ) . Received 10th September, 2017; accepted 28th November, 2017; published 3rd January, 2018. 2010 Mathematics Subject Classification. 26A51, 26D15, 52A01. Key words and phrases. (η1,η2)-convex function; Fejér inequality. c©2018 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 38 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-16-2018-38 Int. J. Anal. Appl. 16 (1) (2018) 39 The estimation for difference of the right and middle part in (1.1) is an interesting problem. The following theorem has been proved in [12], that estimates the difference between the right and middle part in (1.1) using Hölder’s inequality when the absolute value of the derivative of considered function is convex. Theorem 1.2. Let f : I◦ ⊂ R → R be a differentiable mapping on I◦, a,b ∈ I◦ with a < b, and w : [a,b] → [0, +∞) be a differentiable mapping and symmetric to a+b 2 . If |f′| is convex on [a,b] , then the following inequality holds: ∣∣∣∣ 1b−a f(a) + f(b)2 ∫ b a w(x)dx− 1 b−a ∫ b a f(x)w(x)dx ∣∣∣∣ ≤ (1.2) 1 2 [∫ 1 0 ( g(x) )p dt ]1 p ( |f′(a)|q + |f′(b)|q 2 )1 q , where g(x) = ∣∣∣∣ ∫ b−(b−a)t a+(b−a)t w(x)dx ∣∣∣∣, for t ∈ [0, 1]. The preinvex functions as a generalization of convex functions was considered by Ben-Israel and Mond in [1] and Hanson and Mond in [6], but so named by Jeyakumar [7]. Definition 1.1. [1, 6] A set I ⊆ R is invex with respect to a real bifunction η : I × I → R, if x,y ∈ I,λ ∈ [0, 1] ⇒ y + λη(x,y) ∈ I. Also if I is an invex set with respect to η, then a function f : I → R is said to be preinvex if x,y ∈ I and λ ∈ [0, 1] implies f ( y + λη(x,y) ) ≤ λf(x) + (1 −λ)f(y). The following theorem has been proved in [8] which is preinvex version of Theorem 1.2. Theorem 1.3. Let K ⊂ R be an open invex subset with respect to η : K × K → R and a,b ∈ K with a < a + η(b,a). Suppose f : K → R is a differentiable mapping on K such that f′ ∈ L([a,a + a + η(b,a)]) and w : [a,a + η(b,a)] → [0, +∞) is an integrable mapping and symmetric to a + 1 2 η(b,a). If |f′|q, q > 1, is preinvex on K, then for every a,b ∈ K with η(b,a) 6= 0 we have the following inequality:∣∣∣∣f(a) + f ( a + η(b,a) ) 2 ∫ a+η(b,a) a w(x)dx− 1 η(b,a) ∫ a+η(b,a) a f(x)w(x)dx ∣∣∣∣ ≤ (1.3) 1 2 (∫ 1 0 gp(t) )1 p ( |f′(a)|q + |f′(b)|q 2 )1 q , where g(t) = ∣∣∣∣ ∫ a+(1−t)η(b,a) a+tη(b,a) w(x)dx ∣∣∣∣, t ∈ [0, 1] and 1p + 1q = 1. Int. J. Anal. Appl. 16 (1) (2018) 40 Furthermore the concept of η-convex functions (at the beginning was named by ϕ-convex functions), considered in [4], has been introduced as the following. Definition 1.2. Consider a convex set I ⊂ R and a bifunction η : f(I) ×f(I) → R. A function f : I → R is called convex with respect to η (briefly η-convex), if f ( λx + (1 −λ)y ) ≤ f(y) + λη ( f(x),f(y) ) , for all x,y ∈ I and λ ∈ [0, 1]. Geometrically it says that if a function is η-convex on I, then for any x,y ∈ I, its graph is on or under the path starting from ( y,f(y) ) and ending at ( x,f(y) + η(f(x),f(y) )) . If f(x) should be the end point of the path for every x,y ∈ I, then we have η(x,y) = x−y and the function reduces to a convex one. The following theorem has been proved in [3], where the absolute value of the derivative of considered function is η-convex. Theorem 1.4. Suppose that f : [a,b] → R is a differentiable function, g : [a,b] → [0, +∞) is a continuous function and symmetric to a+b 2 and |f′| is an η-convex function where η is bounded from above on [a,b]. Then ∣∣∣∣f(a) + f(b)2 ∫ b a g(x)dx− ∫ b a f(x)g(x)dx ∣∣∣∣ ≤ (b−a) 4 [ 2 ∣∣f′(b)∣∣ + η(|f′(a)|, |f′(b)|)]∫ 1 0 ∫ 1−t 2 a+ 1+t 2 b 1+t 2 a+ 1−t 2 b g(u)dudt. (1.4) For more results about η-convex functions see [3, 4, 10, 11]. Motivated by above works and references therein, we introduce the concept of (η1,η2)-convex functions as a generalization of preinvex and η-convex functions. Also we give some Fejér type trapezoid inequalities when the absolute value of the derivative of considered function is (η1,η2)-convex but with new face without using of Hölder’s inequality. Furthermore we obtain two estimation results when the derivative of considered function is bounded and satisfies a Lipschitz condition. Definition 1.3. Let I ⊂ R be an invex set with respect to η1 : I × I → R. Consider f : I → R and η2 : f(I) ×f(I) → R. The function f is said to be (η1,η2)-convex if f ( x + λη1(y,x) ) ≤ f(x) + λη2 ( f(y),f(x) ) , for all x,y ∈ I and λ ∈ [0, 1]. Note. An (η1,η2)-convex function reduces to (i) an η-convex function if we consider η1(x,y) = x−y for all x,y ∈ I. Int. J. Anal. Appl. 16 (1) (2018) 41 (ii) a preinvex function if we consider η2(x,y) = x−y for all x,y ∈ f(I). (iii) a convex function if satisfies (i) and (ii). We can find an (η1,η2)-convex function which is not convex. Example 1.1. Consider the function f : [0, +∞) → [0, +∞) by f(x) =   x, 0 ≤ x ≤ 1;1, x > 1. Define two bifunction η1 : [0, +∞) × [0, +∞) −→ R and η2 : [0, +∞) × [0, +∞) −→ [0, +∞) by η1(x,y) =   −y, 0 ≤ y ≤ 1;x + y, y > 1, and η2(x,y) =   x + y, x ≤ y;2(x + y), x > y. Then f is an (η1,η2)-convex function. But f is not preinvex with respect to η1 and also it is not convex. (consider x = 0, y = 2 and λ > 0). 2. Main results In this section without using Hölder’s inequality we obtain a trapezoid type inequality related to (1.1). The obtained results are different from (1.2), (1.3) and (1.4) in the face and proof. The following lemma is of importance: Lemma 2.1. Suppose that I ⊂ R is an invex set with respect to η1. If for a,b ∈ I with η1(b,a) > 0 the function g : [a,a + η1(b,a)] → R is integrable and symmetric to a + 12η1(b,a), then for any 0 ≤ t ≤ 1 2 we have∫ 1 t g ( a + sη1(b,a) ) ds− ∫ t 0 g ( a + sη1(b,a) ) ds = 2 ∫ 1 2 t g ( a + sη1(b,a) ) ds. (2.1) Proof. Using the change of variable x = a + sη1(b,a), for 0 ≤ t ≤ 12 we get∫ 1 t g ( a + sη1(b,a) ) ds− ∫ t 0 g ( a + sη1(b,a) ) ds = (2.2) 1 η1(b,a) [∫ a+η1(b,a) u g(x)dx− ∫ u a g(x)dx ] , where a ≤ u ≤ a + 1 2 η1(b,a). Since g is symmetric to a + 1 2 η1(b,a) we have∫ a+η1(b,a) a+ 1 2 η1(b,a) g(x)dx = ∫ a+ 1 2 η1(b,a) a g(x)dx. Int. J. Anal. Appl. 16 (1) (2018) 42 Then ∫ a+η1(b,a) u g(x)dx = ∫ a+ 1 2 η1(b,a) u g(x)dx + ∫ a+η1(b,a) a+ 1 2 η1(b,a) g(x)dx = ∫ a+ 1 2 η1(b,a) u g(x)dx + ∫ a+ 1 2 η1(b,a) a g(x)dx. Also ∫ a+ 1 2 η1(b,a) a g(x)dx = ∫ u a g(x)dx + ∫ a+ 1 2 η1(b,a) u g(x)dx. So 1 η1(b,a) [∫ a+η1(b,a) u g(x)dx− ∫ u a g(x)dx ] = 2 η1(b,a) ∫ a+ 1 2 η1(b,a) u g(x)dx = 2 ∫ 1 2 t g ( a + sη1(b,a) ) ds. (2.3) Using (2.3) in (2.2) we get (2.1). � With the same argument used in the proof of Lemma 2.1 we can drive the following lemma. Lemma 2.2. Suppose that I ⊂ R is an invex set with respect to η1. If for a,b ∈ I with η1(b,a) > 0 the function g : [a,a + η1(b,a)] → R is integrable and symmetric to a + 12η1(b,a), then for any 1 2 ≤ t ≤ 1 we have∫ t 0 g ( a + sη1(b,a) ) ds− ∫ 1 t g ( a + sη1(b,a) ) ds = 2 ∫ t 1 2 g ( a + sη1(b,a) ) ds. (2.4) From Lemma 2.1 and Lemma 2.2, if g is symmetric nonnegative function, we can obtain two integral inequalities that are useful for our next results. Corollary 2.1. Suppose that I ⊂ R is an invex set with respect to η1. If for a,b ∈ I with η1(b,a) > 0 the function g : [a,a + η1(b,a)] → [0, +∞) is integrable and symmetric to a + 12η1(b,a), then∫ 1 t g ( a + sη1(b,a) ) ds− ∫ t 0 g ( a + sη1(b,a) ) ds ≥ 0, 0 ≤ t ≤ 1 2 (2.5) and ∫ t 0 g ( a + sη1(b,a) ) ds− ∫ 1 t g ( a + sη1(b,a) ) ds ≥ 0, 1 2 ≤ t ≤ 1. (2.6) Also the following lemma has been proved in [8] which is needed. Lemma 2.3. Suppose that I◦ ⊂ R is an invex set with respect to η1 and f : I◦ → R is a differentiable mapping on I◦. For any a,b ∈ I◦ with η1(b,a) > 0, if g : [a,a + η1(b,a)] → [0, +∞) is differentiable mapping on I◦ and f′ ∈ L1[a,a + η1(b,a)], then 1 η1(b,a) (∫ a+η1(b,a) a f(x)g(x)dx− f(a) + f ( a + η1(b,a) ) 2 ∫ a+η1(b,a) a g(x)dx ) = η1(b,a) 2 ∫ 1 0 P(t)f′ ( a + tη1(b,a) ) dt, Int. J. Anal. Appl. 16 (1) (2018) 43 where P(t) = ∫ 1 t g ( a + sη1(b,a) ) ds− ∫ t 0 g ( a + sη1(b,a) ) ds. Now we are ready to give our main result of the paper which is a trapezoid type inequality related to (1.1) with a new face. Theorem 2.1. Suppose that I◦ ⊂ R is an invex set with respect to η1 and f : I◦ → R is a differentiable mapping on I◦. For any a,b ∈ I◦ with η1(b,a) > 0, let g : [a,a + η1(b,a)] → [0, +∞) be a differentiable mapping on I◦ and f′ ∈ L1[a,a + η1(b,a)]. If |f′| is a (η1,η2)-convex mapping on [a,a + η1(b,a)], then ∣∣∣∣f(a) + f ( a + η1(b,a) ) 2 ∫ a+η1(b,a) a g(x)dx− ∫ a+η1(b,a) a f(x)g(x)dx ∣∣∣∣ ≤ (2.7) [ 2|f′(a)| + η2 ( |f′(b)|, |f′(a)| )]∫ a+η1(b,a) a+ 1 2 η1(b,a) g(x) ( a + η1(b,a) −x ) dx. Proof. From Lemma 2.3, Corollary 2.1 and (η1,η2)-convexity of |f′| we have ∣∣∣∣f(a) + f ( a + η1(b,a) ) 2 ∫ a+η1(b,a) a g(x)dx− ∫ a+η1(b,a) a f(x)g(x)dx ∣∣∣∣ = η21(b,a) 2 ∣∣∣∣ ∫ 1 0 [∫ 1 t g ( a + sη1(b,a) ) ds− ∫ t 0 g ( a + sη1(b,a) ) ds ] f′ ( a + tη1(b,a) ) dt ∣∣∣∣ ≤ η21(b,a) 2 {∫ 1 2 0 ∣∣∣∫ 1 t g ( a + sη1(b,a) ) ds− ∫ t 0 g ( a + sη1(b,a) ) ds ∣∣∣∣∣f′(a + tη1(b,a))∣∣dt+∫ 1 1 2 ∣∣∣∫ 1 t g ( a + sη1(b,a) ) ds− ∫ t 0 g ( a + sη1(b,a) ) ds ∣∣∣∣∣f′(a + tη1(b,a))∣∣dt} ≤ η21(b,a) 2 {∫ 1 2 0 (∫ 1 t g ( a + sη1(b,a) ) ds− ∫ t 0 g ( a + sη1(b,a) ) ds ) × [ |f′(a)| + tη2 ( |f′(b)|, |f′(a)| )] dt+∫ 1 1 2 (∫ t 0 g ( a + sη1(b,a) ) ds− ∫ 1 t g ( a + sη1(b,a) ) ds ) × [ |f′(a)| + tη2 ( |f′(b)|, |f′(a)| )] dt } = I. If we change the order of integration in I, then I = η21(b,a) 2 {∫ 1 2 0 ∫ s 0 g ( a + sη1(b,a) )[ |f′(a)| + tη2 ( |f′(b)|, |f′(a)| )] dtds+ ∫ 1 1 2 ∫ 1 2 0 g ( a + sη1(b,a) )[ |f′(a)| + tη2 ( |f′(b)|, |f′(a)| )] dtds− ∫ 1 2 0 ∫ 1 2 s g ( a + sη1(b,a) )[ |f′(a)| + tη2 ( |f′(b)|, |f′(a)| )] dtds+ Int. J. Anal. Appl. 16 (1) (2018) 44 ∫ 1 1 2 ∫ 1 s g ( a + sη1(b,a) )[ |f′(a)| + tη2 ( |f′(b)|, |f′(a)| )] dtds+ ∫ 1 2 0 ∫ 1 1 2 g ( a + sη1(b,a) )[ |f′(a)| + tη2 ( |f′(b)|, |f′(a)| )] dtds− ∫ 1 1 2 ∫ s 1 2 g ( a + sη1(b,a) )[ |f′(a)| + tη2 ( |f′(b)|, |f′(a)| )] dtds } . Calculating all inner integrals in I we get I = η21(b,a) 2 {∫ 1 2 0 g ( a + sη1(b,a) )( s|f′(a)| + 1 2 s2η2 ( |f′(b)|, |f′(a)| )) ds+ ∫ 1 1 2 g ( a + sη1(b,a) )(1 2 |f′(a)| + 1 8 η2 ( |f′(b)|, |f′(a)| )) ds− ∫ 1 2 0 g ( a + sη1(b,a) )( ( 1 2 −s)|f′(a)| + ( 1 8 − 1 2 s2)η2 ( |f′(b)|, |f′(a)| )) ds+ ∫ 1 1 2 g ( a + sη1(b,a) )( (1 −s)|f′(a)| + ( 1 2 − 1 2 s2)η2 ( |f′(b)|, |f′(a)| )) ds+ ∫ 1 2 0 g ( a + sη1(b,a) )(1 2 |f′(a)| + 3 8 η2 ( |f′(b)|, |f′(a)| )) ds− ∫ 1 1 2 g ( a + sη1(b,a) )( (s− 1 2 )|f′(a)| + ( 1 2 s2 − 1 8 )η2 ( |f′(b)|, |f′(a)| )) ds } . Simple form of I can be obtained as the following. I = η21(b,a) 2 {∫ 1 2 0 g ( a + sη1(b,a) )( 2s|f′(a)| + (s2 + 1 4 )η2 ( |f′(b)|, |f′(a)| )) ds+ ∫ 1 1 2 g ( a + sη1(b,a) )( (−2s + 2)|f′(a)| + ( 3 4 −s2)η2 ( |f′(b)|, |f′(a)| )) ds } . If we apply the change of variable x = a + sη1(b,a) in I, we get I = η1(b,a) 2 {∫ a+ 1 2 η1(b,a) a g(x) ([ 2 ( x−a η1(b,a) )] |f′(a)|+ [( x−a η1(b,a) )2 + 1 4 ] η2 ( |f′(b)|, |f′(a)| )) dx+ ∫ a+η1(b,a) a+ 1 2 η1(b,a) g(x) ([ − 2 ( x−a η1(b,a) ) + 2 ] |f′(a)|+ [3 4 − ( x−a η1(b,a) )2] η2 ( |f′(b)|, |f′(a)| )) dx } . Int. J. Anal. Appl. 16 (1) (2018) 45 On the other hand since g is symmetric to a + 1 2 η1(b,a) then we have ∫ a+ 1 2 η1(b,a) a g(x) ([ 2 ( x−a η1(b,a) )] |f′(a)|+ [( x−a η1(b,a) )2 + 1 4 ] η2 ( |f′(b)|, |f′(a)| )) dx = ∫ a+η1(b,a) a+ 1 2 η1(b,a) g(x) ([ 2 (a + η1(b,a) −x η1(b,a) )] |f′(a)|+ [(a + η1(b,a) −x η1(b,a) )2 + 1 4 ] η2 ( |f′(b)|, |f′(a)| )) dx. So I = η1(b,a) 2 {∫ a+η1(b,a) a+ 1 2 η1(b,a) g(x) ([ 2 ( x−a η1(b,a) ) + 2 (a + η1(b,a) −x η1(b,a) )] |f′(a)| + [( x−a η1(b,a) )2 + 1 4 + (a + η1(b,a) −x η1(b,a) )2 + 1 4 ] η2 ( |f′(b)|, |f′(a)| )) dx } = η1(b,a) 2 {∫ a+η1(b,a) a+ 1 2 η1(b,a) g(x) ( 4 [a + η1(b,a) −x η1(b,a) ] |f′(a)| + 2 [a + η1(b,a) −x η1(b,a) ] η2 ( |f′(b)|, |f′(a)| )) dx } = [ 2|f′(a)| + η2 ( |f′(b)|, |f′(a)| )]∫ a+η1(b,a) a+ 1 2 η1(b,a) g(x) ( a + η1(b,a) −x ) dx. � Corollary 2.2. If in Theorem 2.1 we consider η2(x,y) = x−y for all x,y ∈ f(I◦), then ∣∣∣∣f(a) + f ( a + η1(b,a) ) 2 ∫ a+η1(b,a) a g(x)dx− ∫ a+η1(b,a) a f(x)g(x)dx ∣∣∣∣ ≤ (2.8) [ |f′(a)| + |f′(b)| ]∫ a+η1(b,a) a+ 1 2 η1(b,a) g(x) ( a + η1(b,a) −x ) dx. Also if we put η1(x,y) = x−y for all x,y ∈ f(I◦) in (2.8) we get ∣∣∣∣f(a) + f(b)2 ∫ b a g(x)dx− ∫ b a f(x)g(x)dx ∣∣∣∣ ≤ [|f′(a)| + |f′(b)|] ∫ b a+b 2 g(x)(b−x)dx. (2.9) Furthermore if in (2.9) we set g ≡ 1, then we recapture Theorem 2.2 in [2]. ∣∣∣∣f(a) + f(b)2 (b−a) − ∫ b a f(x)dx ∣∣∣∣ ≤ (b−a)28 (|f′(a)| + |f′(b)|). Remark 2.1. Inequalities (2.8) and (2.9), obtained in Corollary 2.2 are new inequalities in the literature. Int. J. Anal. Appl. 16 (1) (2018) 46 3. Estimation type results In this section we give two estimation results when the derivative of considered function is bounded and satisfies a Lipschitz condition respectively. If the derivative of the considered function is bounded from below and above, then we can drive an estimation type result related to Fejér inequality. Theorem 3.1. Suppose that I◦ ⊂ R is an invex set with respect to η1 and f : I◦ → R is a differentiable mapping on I◦. For any a,b ∈ I◦ with η1(b,a) > 0, let g : [a,a + η1(b,a)] → [0, +∞) be a differentiable mapping on I◦. Assume that f′ ∈ L1[a,a + η1(b,a)] and there exist constants m < M such that −∞ < m ≤ f′(x) ≤ M < +∞ for all x ∈ [a,a + η1(b,a)]. Then 1 η1(b,a) ∣∣∣∣∣f(a) + f ( a + η1(b,a) ) 2 ∫ a+η1(b,a) a g(x)dx− ∫ a+η1(b,a) a f(x)g(x)dx (3.1) − η1(b,a)(m + M) 4 ∫ 1 0 P(t)dt ∣∣∣∣∣ ≤ η1(b,a)(M −m)4 ∫ 1 0 |P(t)|dt, where P(t) is defined in Lemma 2.3. Proof. From Lemma 2.3 we have 1 η1(b,a) (∫ a+η1(b,a) a f(x)g(x)dx− f(a) + f ( a + η1(b,a) ) 2 ∫ a+η1(b,a) a g(x)dx ) = η1(b,a) 2 ∫ 1 0 P(t) [ f′ ( a + tη1(b,a) ) − m + M 2 + m + M 2 ] dt = (m + M)η1(b,a) 4 ∫ 1 0 P(t)dt + η1(b,a) 2 ∫ 1 0 P(t) [ f′ ( a + tη1(b,a) ) − m + M 2 ] dt. So I = 1 η1(b,a) (∫ a+η1(b,a) a f(x)g(x)dx− f(a) + f ( a + η1(b,a) ) 2 ∫ a+η1(b,a) a g(x)dx ) − (m + M)η1(b,a) 4 ∫ 1 0 P(t)dt = η1(b,a) 2 ∫ 1 0 P(t) [ f′ ( a + tη1(b,a) ) − m + M 2 ] dt. Taking the modulus on I we obtain |I| ≤ η1(b,a) 2 ∫ 1 0 ∣∣∣P(t)∣∣∣∣∣∣f′(a + tη1(b,a))− m + M 2 ∣∣∣dt ≤ (M −m)η1(b,a) 4 ∫ 1 0 |P(t)|dt, since from m ≤ f′ ( a + tη1(b,a) ) ≤ M we have m− m + M 2 ≤ f′ ( a + tη1(b,a) ) − m + M 2 ≤ M − m + M 2 , which implies that ∣∣∣∣f′(a + tη1(b,a))− m + M2 ∣∣∣∣ ≤ M −m2 . Int. J. Anal. Appl. 16 (1) (2018) 47 � Corollary 3.1. In Theorem 3.1 if we set η1(x,y) = x−y for all x,y ∈ I◦ and g ≡ 1, then∣∣∣∣f(a) + f(b)2 − 1b−a ∫ b a f(x)dx ∣∣∣∣ ≤ m(1 + a− b) + M(1 + b−a)8 . Proof. If we consider g ≡ 1, then the relations ||g||∞ = 1 and ∫ 1 0 |p(t)|dt ≤ 1 2 imply that 1 b−a ∣∣∣∣f(a) + f(b)2 (b−a) − ∫ b a f(x)dx ∣∣∣∣ ≤∣∣∣∣m + M4 ∫ 1 0 p(t)dt ∣∣∣∣ + (M −m)(b−a)8 ≤ m + M 8 + (M −m)(b−a) 8 = m(1 + a− b) + M(1 + b−a) 8 . � Estimation for difference between the right and middle terms of (1.1) when the derivative of considered function satisfies a Lipschitz condition is our next aim. Definition 3.1. [9] A function f : [a,b] → R is said to satisfy the Lipschitz condition on [a,b] if there is a constant K so that for any two points x,y ∈ [a,b], |f(x) −f(y)| ≤ K|x−y|. Theorem 3.2. Suppose that I◦ ⊂ R is an invex set with respect to η1 and f : I◦ → R is a differentiable mapping on I◦. For any a,b ∈ I◦ with η1(b,a) > 0, let g : [a,a + η1(b,a)] → [0, +∞) be a differentiable mapping on I◦. Assume that f′ is integrable on [a,a + η1(b,a)] and satisfies a Lipschitz condition for some K > 0. Then ∣∣∣∣∣ 1η1(b,a) ( f(a) + f ( a + η1(b,a) ) 2 ∫ a+η1(b,a) a g(x)dx− ∫ a+η1(b,a) a f(x)g(x)dx ) − η1(b,a) 2 f′ (2a + η1(b,a) 2 )∫ 1 0 P(t)dt ∣∣∣∣∣ ≤ Kη1(b,a)2 ∫ 1 0 |t− 1 2 ||P(t)|dt, where P(t) is defined in Lemma 2.3. Proof. From Lemma 2.3 we have 1 η1(b,a) (∫ a+η1(b,a) a f(x)g(x)dx− f(a) + f ( a + η1(b,a) ) 2 ∫ a+η1(b,a) a g(x)dx ) = η1(b,a) 2 ∫ 1 0 P(t) [ f′ ( a + tη1(b,a) ) −f′ ( 2a + η1(b,a) 2 ) + f′ ( 2a + η1(b,a) 2 )] dt = η1(b,a) 2 ∫ 1 0 P(t) [ f′ ( a + tη1(b,a) ) −f′ ( 2a + η1(b,a) 2 )] dt+ η1(b,a) 2 f′ ( 2a + η1(b,a) 2 )∫ 1 0 P(t)dt. Int. J. Anal. Appl. 16 (1) (2018) 48 Then 1 η1(b,a) (∫ a+η1(b,a) a f(x)g(x)dx− f(a) + f ( a + η1(b,a) ) 2 ∫ a+η1(b,a) a g(x)dx ) − η1(b,a) 2 f′ ( 2a + η1(b,a) 2 )∫ 1 0 P(t)dt = η1(b,a) 2 ∫ 1 0 P(t) [ f′ ( a + tη1(b,a) ) −f′ ( 2a + η1(b,a) 2 )] dt. Since f′ satisfies a Lipschitz condition for K > 0, we have∣∣∣∣f′(a + tη1(b,a))−f′ ( 2a + η1(b,a) 2 )∣∣∣∣ ≤ K ∣∣∣∣a + tη1(b,a) − 2a + η1(b,a)2 ∣∣∣∣ = K ∣∣∣t− 1 2 ∣∣∣η1(b,a). So ∣∣∣∣ 1η1(b,a) (∫ a+η1(b,a) a f(x)g(x)dx− f(a) + f ( a + η1(b,a) ) 2 ∫ a+η1(b,a) a g(x)dx ) − η1(b,a) 2 f′ ( 2a + η1(b,a) 2 )∫ 1 0 P(t)dt ∣∣∣∣ ≤ Kη1(b,a)2 ∫ 1 0 ∣∣t− 1 2 ∣∣∣∣P(t)∣∣dt. � Corollary 3.2. If in Theorem 3.2 we consider g ≡ 1, then∣∣∣∣∣f(a) + f ( a + η1(b,a) ) 2 − 1 η1(b,a) ∫ a+η1(b,a) a f(x)dx ∣∣∣∣∣ ≤ η1(b,a) ( K 12 + 1 4 ∣∣∣f′(2a + η1(b,a) 2 )∣∣∣) . Furthermore if we consider η1(x,y) = x−y for all x,y ∈ I◦ we get∣∣∣∣∣f(a) + f(b)2 − 1b−a ∫ b a f(x)dx ∣∣∣∣∣ ≤ (b−a) ( K 12 + 1 4 ∣∣∣f′(a + b 2 )∣∣∣) . 4. Acknowledgements The authors are very grateful to professor S. S. Dragomir for his valuable suggestions and comments. References [1] A. Ben-Israel and B. Mond, What is invexity?, J. Aust. Math. Soc., Ser. B. 28 (1986) 1–9. [2] S. S. Dragomir and R. P. 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