Hindawi Publishing Corporation International Journal of Analysis Volume 2016, Article ID 4765691, 8 pages http://dx.doi.org/10.1155/2016/4765691
Research Article Hermite-Hadamard Type Inequalities Obtained via Fractional Integral for Differentiable 𝑚-Convex and (𝛼, 𝑚)-Convex Functions Erhan Set,1 Süleyman Sami KarataG,1 and Muhammad Adil Khan2 1
Department of Mathematics, Faculty of Sciences, Ordu University, 52200 Ordu, Turkey Department of Mathematics, University of Peshawar, Peshawar, Pakistan
2
Correspondence should be addressed to Erhan Set;
[email protected] Received 1 July 2016; Accepted 20 September 2016 Academic Editor: Ahmed Zayed Copyright © 2016 Erhan Set et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Some Hermite-Hadamard type inequalities involving fractional integrals for 𝑚-convex and (𝛼, 𝑚)-convex functions are obtained.
for all 𝑥, 𝑦 ∈ [0, 𝑏] and 𝑡 ∈ [0, 1]. One says that 𝑓 is 𝑚-concave if −𝑓 is 𝑚-convex. Denote by 𝐾𝑚 (𝑏) the class of all 𝑚-convex functions on [0, 𝑏] for which 𝑓(0) ≤ 0.
1. Introduction and Preliminaries The following definition is well known in the literature. Definition 1. A function 𝑓 : 𝐼 → R, 0 ≠ 𝐼 ⊆ R, is said to be convex on the interval 𝐼 if the inequality 𝑓 (𝑡𝑥 + (1 − 𝑡) 𝑦) ≤ 𝑡𝑓 (𝑥) + (1 − 𝑡) 𝑓 (𝑦)
(1)
holds for all 𝑥, 𝑦 ∈ 𝐼 and 𝑡 ∈ [0, 1]. Geometrically, this means that if 𝑃, 𝑄, and 𝑅 are three points on the graph of 𝑓 with 𝑄 between 𝑃 and 𝑅, then 𝑄 is on or below the chord 𝑃𝑅. Theorem 2 (Hermite-Hadamard inequality). Let 𝑓 : 𝐼 ⊂ R → R be a convex function and 𝑎, 𝑏 ∈ 𝐼 with 𝑎 < 𝑏. Then
Obviously, for 𝑚 = 1, Definition 3 recaptures the concept of standard convex functions on [𝑎, 𝑏] and for 𝑚 = 0 the concept of starshaped functions. The notion of 𝑚-convexity has been further generalized in [2] as it is stated in the following definition. Definition 4 (see [2]). The function 𝑓 : [0, 𝑏] → R, 𝑏 > 0, is said to be (𝛼, 𝑚)-convex, where (𝛼, 𝑚) ∈ [0, 1]2 , if one has 𝑓 (𝑡𝑥 + 𝑚 (1 − 𝑡) 𝑦) ≤ 𝑡𝛼 𝑓 (𝑥) + 𝑚 (1 − 𝑡𝛼 ) 𝑓 (𝑦)
(4)
(2)
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.
Definition 3 (see [1]). The function 𝑓 : [0, 𝑏] → R, 𝑏 > 0, is said to be 𝑚-convex, where 𝑚 ∈ [0, 1], if one has
It can be easily seen that when (𝛼, 𝑚) ∈ {(1, 1), (1, 𝑚)} one obtains the following classes of functions: convex and 𝑚convex, respectively. Note that 𝐾11 (b) is a proper subclass of 𝑚-convex and (𝛼, 𝑚)-functions [0, 𝑏]. The interested reader can find more about partial ordering of convexity in [3].
𝑏 𝑓 (𝑎) + 𝑓 (𝑏) 𝑎+𝑏 1 𝑓( . )≤ ∫ 𝑓 (𝑥) 𝑑𝑥 ≤ 2 𝑏−𝑎 𝑎 2
𝑚-convexity was defined by Toader as follows.
𝑓 (𝑡𝑥 + 𝑚 (1 − 𝑡) 𝑦) ≤ 𝑡𝑓 (𝑥) + 𝑚 (1 − 𝑡) 𝑓 (𝑦)
(3)
2
International Journal of Analysis
Definition 5 (see [4]). Let 𝑓 ∈ 𝐿 1 [𝑎, 𝑏]. Then Riemann-Liouville integrals 𝐽𝑎𝛼+ 𝑓 and 𝐽𝑏𝛼− 𝑓 of order 𝛼 > 0 with 𝑎 ≥ 0 are defined by 𝐽𝑎𝛼+ 𝑓 (𝑥) = 𝐽𝑏𝛼− 𝑓 (𝑥) =
𝑥 1 ∫ (𝑥 − 𝑡)𝛼−1 𝑓 (𝑡) 𝑑𝑡, Γ (𝛼) 𝑎
𝑥 > 𝑎,
𝑏 1 ∫ (𝑡 − 𝑥)𝛼−1 𝑓 (𝑡) 𝑑𝑡, Γ (𝛼) 𝑥
𝑥 < 𝑏,
(5)
where
where 𝐺 (𝑘; 𝑛; 𝑎, 𝑥, 𝑏) (𝑓) =
𝑛 + 1 (𝑥 − 𝑎)𝑘 + (𝑏 − 𝑥)𝑘 [ 𝑓 (𝑥) 2 𝑏−𝑎
+
(𝑥 − 𝑎)𝑘 𝑓 (𝑎) + (𝑏 − 𝑥)𝑘 𝑓 (𝑏) ] 𝑏−𝑎
−
𝑛 1 (𝑛 + 1)𝑘+1 Γ (𝑘 + 1) 𝑘 {𝐽𝑥− 𝑓 ( 𝑥+ 𝑎) 2 (𝑏 − 𝑎) 𝑛+1 𝑛+1
+ 𝐽𝑎𝑘+ 𝑓 (
1 𝑛 𝑥+ 𝑎) 𝑛+1 𝑛+1
is the Gamma function.
+ 𝐽𝑥𝑘+ 𝑓 (
𝑛 1 𝑥+ 𝑏) 𝑛+1 𝑛+1
We now give the definition of the hypergeometric series which will be used in obtaining some integrals.
+ 𝐽𝑏𝑘− 𝑓 (
𝑛 1 𝑥+ 𝑎)} . 𝑛+1 𝑛+1
∞
Γ (𝛼) = ∫ 𝑒−𝑥 𝑥𝛼−1 𝑑𝑥
(6)
0
Definition 6 (see [4]). The integral representation of the hypergeometric functions is as follows: 2 𝐹1
[𝑎, 𝑏, 𝑐; 𝑧]
1 1 = ∫ 𝑡𝑏−1 (1 − 𝑡)𝑐−𝑏−1 (1 − 𝑧𝑡)−𝑎 𝑑𝑡, 𝐵 (𝑏, 𝑐 − 𝑏) 0
(7)
where |𝑧| < 1, 𝑐 > 𝑏 > 0, and 1
𝑥−1
𝐵 (𝑥, 𝑦) = ∫ 𝑡 0
(1 − 𝑡)
𝑦−1
𝑑𝑡
(8)
𝐵 (𝑥, 𝑦) =
Γ (𝑥) Γ (𝑦) . Γ (𝑥 + 𝑦)
Lemma 7 (see [5]). Let 𝑓 : [𝑎, 𝑏] ⊂ R → R be a differentiable function such that 𝑓 ∈ 𝐿[𝑎, 𝑏]. Then, for 𝑛 ∈ N, 𝑘 > 0, and 𝑥 ∈ [𝑎, 𝑏], one has
(𝑥 − 𝑎) 𝑏−𝑎
−∫
+
0
𝑡 𝑛+𝑡 1−𝑡 𝑓 ( 𝑥+ 𝑎) 𝑑𝑡 2 𝑛+1 𝑛+1
𝐺 (𝑘; 𝑛; 𝑎, 𝑥, 𝑏) (𝑓) ≤
+
(𝑏 − 𝑥) 𝑏−𝑎
{∫
(10)
1 𝑘 𝑡 (𝑥 − 𝑎)𝑘+1 [∫ 𝑏−𝑎 0 2
𝑡 𝑛+𝑡 1−𝑡 𝑓 ( 𝑥+ 𝑏) 𝑑𝑡 2 𝑛+1 𝑛+1
=
𝑡 1−𝑡 𝑛+𝑡 𝑓 ( 𝑥+ 𝑏) 𝑑𝑡} , 2 𝑛+1 𝑛+1
𝑡 2
1 𝑘
0
𝑡 2
+∫
1 𝑘
0
𝑡 2
𝑛 + 𝑡 1 − 𝑡 𝑓 ( 𝑥+ 𝑎) 𝑑𝑡 𝑛+1 𝑛 + 1
1 − 𝑡 𝑛 + 𝑡 𝑓 ( 𝑥+ 𝑎) 𝑑𝑡] 𝑛+1 𝑛 + 1 𝑛 + 𝑡 1 − 𝑡 𝑓 ( 𝑥+ 𝑏) 𝑑𝑡 𝑛+1 𝑛 + 1
1 − 𝑡 𝑛 + 𝑡 𝑓 ( 𝑥+ 𝑏) 𝑑𝑡] 𝑛+1 𝑛 + 1
1 𝑘 𝑡 (𝑥 − 𝑎)𝑘+1 [∫ 𝑏−𝑎 0 2
1 𝑘
0
1 𝑘
1 𝑘 𝑡 (𝑏 − 𝑥)𝑘+1 [∫ 𝑏−𝑎 0 2
+∫
1 𝑘
0
𝑏 (𝑏 − 𝑥)𝑘+1 [𝑓 (𝑥) + 𝑚 𝑓 ( )] , 2 (𝑏 − 𝑎) (𝑘 + 1) 𝑚
Proof. Using Lemma 7, taking modulus and the fact that |𝑓 | is an 𝑚-convex function, we have
0
{∫
(12)
where 𝑥 ∈ [𝑎, 𝑏].
1 𝑘
𝑡 1−𝑡 𝑛+𝑡 𝑓 ( 𝑥+ 𝑎) 𝑑𝑡} 2 𝑛+1 𝑛+1 𝑘+1
−∫
𝑎 (𝑥 − 𝑎)𝑘+1 [𝑓 (𝑥) + 𝑚 𝑓 ( )] 2 (𝑏 − 𝑎) (𝑘 + 1) 𝑚
1 𝑘
0
−
𝐺 (𝑘; 𝑛; 𝑎, 𝑥, 𝑏) (𝑓)
+∫
𝐺 (𝑘; 𝑛; 𝑎, 𝑥, 𝑏) (𝑓) =
Theorem 8. Let 𝐼 be on open real interval such that [0, ∞) ⊂ 𝐼. Let 𝑓 : 𝐼 → R be a differentiable function on 𝐼 such that 𝑛 ∈ N, 𝑘 > 0 and 𝑓 ∈ 𝐿[𝑎, 𝑏], where 0 ≤ 𝑎 < 𝑏 < ∞. If |𝑓 | is an 𝑚-convex function on [𝑎, 𝑏] for some fixed 𝑚 ∈ (0, 1], then
(9)
In the present paper, we establish some new HermiteHadamard’s type inequalities for the classes of 𝑚-convex and (𝛼, 𝑚)-convex functions via Riemann-Liouville fractional integrals. To prove our main results, we consider the following lemma.
𝑘+1
2. Generalized Inequalities for 𝑚-Convex Functions
≤
is Beta function with
(11)
𝑛 + 𝑡 1 − 𝑡 𝑎 𝑓 ( 𝑥+𝑚 ) 𝑑𝑡 𝑛+1 𝑛 + 1 𝑚
1 − 𝑡 𝑛 + 𝑡 𝑎 𝑓 ( 𝑥+𝑚 ) 𝑑𝑡] 𝑛+1 𝑛 + 1 𝑚
International Journal of Analysis +
1 𝑘 𝑡 (𝑏 − 𝑥)𝑘+1 [∫ 𝑏−𝑎 0 2
+∫
1 𝑘
0
𝑡 2
3
1 − 𝑡 𝑏 𝑛 + 𝑡 𝑥+𝑚 ) 𝑑𝑡 𝑓 ( 𝑛+1 𝑛 + 1 𝑚
𝑛 + 𝑡 𝑏 1 − 𝑡 𝑥+𝑚 ) 𝑑𝑡] 𝑓 ( 𝑛+1 𝑛 + 1 𝑚
1 𝑘
+∫
0
+
≤
𝑎 (𝑥 − 𝑎)𝑘+1 [𝑓 (𝑥) + 𝑚 𝑓 ( )] 2 (𝑏 − 𝑎) (𝑘 + 1) 𝑚
+
𝑏 (𝑏 − 𝑥) [𝑓 (𝑥) + 𝑚 𝑓 ( )] . 2 (𝑏 − 𝑎) (𝑘 + 1) 𝑚
1 𝑘
+∫
0
𝑘+1
Theorem 10. Let 𝐼 be on open real interval such that [0, ∞) ⊂ 𝐼. Let 𝑓 : 𝐼 → R be a differentiable function on 𝐼 such that 𝑛 ∈ N, 𝑘 > 0 and 𝑓 ∈ 𝐿[𝑎, 𝑏], where 0 ≤ 𝑎 < 𝑏 < ∞. If |𝑓 |𝑞 is an 𝑚-convex function on [𝑎, 𝑏] for some fixed 𝑚 ∈ (0, 1] and 1/𝑝 + 1/𝑞 = 1, 𝑞 > 1, then 1/𝑝 1 (𝑥 − 𝑎)𝑘+1 ( ) 𝐺 (𝑘; 𝑛; 𝑎, 𝑥, 𝑏) (𝑓) ≤ [[ 2 (𝑏 − 𝑎) 𝑘𝑝 + 1 [[
)
(14)
𝑞 𝑞 1/𝑞 (2𝑛 + 1) 𝑓 (𝑥) + 𝑚 𝑓 (𝑏/𝑚) [ × ( ) 2 [ 𝑞 𝑞 1/𝑞 𝑓 (𝑥) + 𝑚 (2𝑛 + 1) 𝑓 (𝑏/𝑚) ) ]]] , +( 2 ]]] where 𝑥 ∈ [𝑎, 𝑏]. Proof. Using Lemma 7, H¨older’s inequality, and the fact that |𝑓 |𝑞 is an 𝑚-convex function, 𝐺 (𝑘; 𝑛; 𝑎, 𝑥, 𝑏) (𝑓) ≤
(𝑥 − 𝑎) 𝑏−𝑎
[∫
1 𝑘
0
𝑡 2
𝑛 + 𝑡 1 − 𝑡 𝑓 ( 𝑥+ 𝑎) 𝑑𝑡 𝑛+1 𝑛 + 1
1/𝑞 1 1−𝑡 𝑛 + 1 𝑞 + (∫ 𝑓 ( 𝑥+ 𝑎) 𝑑𝑡) ] 𝑛+1 𝑛+1 0 𝑝
+
1 𝑡𝑘 (𝑏 − 𝑥)𝑘+1 (∫ ( ) 𝑑𝑡) 𝑏−𝑎 2 0
1/𝑝
1/𝑞 1 𝑛+𝑡 1 − 𝑡 𝑞 ⋅ [(∫ 𝑓 ( 𝑥+ 𝑏) 𝑑𝑡) 𝑛+1 𝑛+1 0
1/𝑝 1 1 1/𝑞 1 (𝑥 − 𝑎)𝑘+1 ( ) ( ) [(∫ (𝑛 + 𝑡) 2 (𝑏 − 𝑎) 𝑘𝑝 + 1 𝑛+1 0
1/𝑞 𝑎 𝑞 𝑞 ⋅ 𝑓 (𝑥) + 𝑚 (1 − 𝑡) 𝑓 ( ) 𝑑𝑡) 𝑚
1/𝑝 1 1 1/𝑞 (𝑏 − 𝑥)𝑘+1 +[ ( ) ( ) 2 (𝑏 − 𝑎) 𝑘𝑝 + 1 𝑛+1 [
𝑘+1
1/𝑞 1 𝑛+𝑡 1 − 𝑡 𝑞 ⋅ [(∫ 𝑓 ( 𝑥+ 𝑎) 𝑑𝑡) 𝑛+1 𝑛+1 0
≤
1/𝑞
𝑞 𝑞 1/𝑞 𝑓 (𝑥) + 𝑚 (2𝑛 + 1) 𝑓 (𝑎/𝑚) ) ]] +( 2 ]]
1/𝑝
1/𝑞 1 1−𝑡 𝑛 + 1 𝑞 + (∫ 𝑓 ( 𝑥+ 𝑏) 𝑑𝑡) ] 𝑛+1 𝑛+1 0
1 1/𝑞 ) 𝑛+1 2
1 − 𝑡 𝑛 + 𝑡 𝑓 ( 𝑥+ 𝑏) 𝑑𝑡] 𝑛+1 𝑛 + 1 𝑝
Remark 9. Observe that if in Theorem 8 we have 𝑚 = 𝑛 = 1, the statement of Theorem 8 becomes the statement of Theorem 1 in [6].
× [( [
𝑡 2
𝑛 + 𝑡 1 − 𝑡 𝑓 ( 𝑥+ 𝑏) 𝑑𝑡 𝑛+1 𝑛 + 1
1 𝑡𝑘 (𝑥 − 𝑎)𝑘+1 (∫ ( ) 𝑑𝑡) 𝑏−𝑎 2 0
≤
This completes the proof.
𝑞 𝑞 (2𝑛 + 1) 𝑓 (𝑥) + 𝑚 𝑓 (𝑎/𝑚)
1 − 𝑡 𝑛 + 𝑡 𝑓 ( 𝑥+ 𝑎) 𝑑𝑡] 𝑛+1 𝑛 + 1
1 𝑘 𝑡 (𝑏 − 𝑥)𝑘+1 [∫ 𝑏−𝑎 0 2
(13)
⋅(
𝑡 2
1 𝑞 + (∫ (1 − 𝑡) 𝑓 (𝑥) + 𝑚 (𝑛 + 𝑡) 0
1/𝑞 1/𝑝 𝑎 𝑞 1 (𝑏 − 𝑥)𝑘+1 ⋅ 𝑓 ( ) 𝑑𝑡) ] + ( ) 𝑚 2 (𝑏 − 𝑎) 𝑘𝑝 + 1
⋅(
1 1 1/𝑞 𝑞 ) [(∫ (𝑛 + 𝑡) 𝑓 (𝑥) + 𝑚 (1 − 𝑡) 𝑛+1 0
1/𝑞 1 𝑏 𝑞 𝑞 ⋅ 𝑓 ( ) 𝑑𝑡) + (∫ (1 − 𝑡) 𝑓 (𝑥) 𝑚 0 1/𝑞 𝑏 𝑞 + 𝑚 (𝑛 + 𝑡) 𝑓 ( ) 𝑑𝑡) ] 𝑚 1/𝑝 1 1 1/𝑞 (𝑥 − 𝑎)𝑘+1 ( =[ ) ( ) 2 (𝑏 − 𝑎) 𝑘𝑝 + 1 𝑛+1 [
× [( [
𝑞 𝑞 (2𝑛 + 1) 𝑓 (𝑥) + 𝑚 𝑓 (𝑎/𝑚) 2
1/𝑞
)
4
International Journal of Analysis 𝑞 𝑞 1/𝑞 𝑓 (𝑥) + 𝑚 (2𝑛 + 1) 𝑓 (𝑎/𝑚) ) ]] + ( 2 ]]
Proof. Using Lemma 7, Power’s mean inequality, and the fact that |𝑓 |𝑞 is an 𝑚-convex function, 𝐺 (𝑘; 𝑛; 𝑎, 𝑥, 𝑏) (𝑓)
1/𝑝 1 1 1/𝑞 (𝑏 − 𝑥)𝑘+1 +[ ( ) ( ) 2 (𝑏 − 𝑎) 𝑘𝑝 + 1 𝑛+1 [
× [( [
𝑞 𝑞 (2𝑛 + 1) 𝑓 (𝑥) + 𝑚 𝑓 (𝑏/𝑚) 2
≤ 1/𝑞
+∫
+
𝑞 𝑞 1/𝑞 𝑓 (𝑥) + 𝑚 (2𝑛 + 1) 𝑓 (𝑏/𝑚) ) ]] . +( 2 ]]
≤
Remark 11. Observe that if in Theorem 10 we have 𝑚 = 𝑛 = 1, the statement of Theorem 10 becomes the statement of Theorem 2 in [6]. Theorem 12. Let 𝐼 be on open real interval such that [0, ∞) ⊂ 𝐼. Let 𝑓 : 𝐼 → R be a differentiable function on 𝐼 such that 𝑛 ∈ N, 𝑘 > 0 and 𝑓 ∈ 𝐿[𝑎, 𝑏], where 0 ≤ 𝑎 < 𝑏 < ∞. If |𝑓 |𝑞 is an 𝑚-convex function on [𝑎, 𝑏] for some fixed 𝑚 ∈ (0, 1] and 𝑞 ≥ 1, then 𝐺 (𝑘; 𝑛; 𝑎, 𝑥, 𝑏) (𝑓)
1 − 𝑡 𝑛 + 𝑡 𝑓 ( 𝑥+ 𝑎) 𝑑𝑡] 𝑛+1 𝑛 + 1
𝑡 2
1 𝑘
𝑡 2
+ (∫
0
1 𝑘
𝑡 2
1 𝑘
0
𝑡 2
1−1/𝑞
1/𝑞 𝑛 + 𝑡 1 − 𝑡 𝑞 𝑓 ( 𝑑𝑡) 𝑥 + 𝑎) 𝑛+1 𝑛 + 1 1/𝑞 1 − 𝑡 𝑛 + 1 𝑞 𝑓 ( 𝑥+ 𝑎) 𝑑𝑡) ] 𝑛+1 𝑛+1
1 𝑘 𝑡 (𝑏 − 𝑥)𝑘+1 (∫ 𝑑𝑡) 𝑏−𝑎 0 2
+ (∫ ≤
𝑛 + 𝑡 1 − 𝑡 𝑓 ( 𝑥+ 𝑏) 𝑑𝑡 𝑛+1 𝑛 + 1
1 − 𝑡 𝑛 + 𝑡 𝑓 ( 𝑥+ 𝑏) 𝑑𝑡] 𝑛+1 𝑛 + 1
1 𝑘
0
1/𝑞
1−1/𝑞
1/𝑞 𝑛 + 𝑡 1 − 𝑡 𝑞 𝑓 ( 𝑥+ 𝑏) 𝑑𝑡) 𝑛+1 𝑛+1 1/𝑞 1 − 𝑡 𝑛 + 1 𝑞 𝑓 ( 𝑑𝑡) ] 𝑥 + 𝑏) 𝑛+1 𝑛 + 1
1 1−1/𝑞 1 1/𝑞 (𝑥 − 𝑎)𝑘+1 ( ( ) ) 2 (𝑏 − 𝑎) 𝑘 + 1 𝑛+1
1/𝑞 1 𝑎 𝑞 𝑞 ⋅ {(∫ 𝑡𝑘 [(𝑛 + 𝑡) 𝑓 (𝑥) + 𝑚 (1 − 𝑡) 𝑓 ( ) ] 𝑑𝑡) 𝑚 0
𝑎 𝑞 1/𝑞 𝑞 + 𝑚 𝑓 ( ) ) + (𝑓 (𝑥) 𝑚 𝑎 𝑞 1/𝑞 + 𝑚 (𝑛 (𝑘 + 2) + 𝑘 + 1) 𝑓 ( ) ) ] 𝑚 1/𝑞
𝑞 ⋅ [((𝑛 (𝑘 + 2) + 𝑘 + 1) 𝑓 (𝑥) 𝑞 1/𝑞 𝑞 𝑏 + 𝑚 𝑓 ( ) ) + (𝑓 (𝑥) 𝑚 1/𝑞 𝑏 𝑞 + 𝑚 (𝑛 (𝑘 + 2) + 𝑘 + 1) 𝑓 ( ) ) ] , 𝑚
where 𝑥 ∈ [𝑎, 𝑏].
⋅ [(∫
+
𝑛 + 𝑡 1 − 𝑡 𝑓 ( 𝑥+ 𝑎) 𝑑𝑡 𝑛+1 𝑛 + 1
1 𝑘 𝑡 (𝑥 − 𝑎)𝑘+1 (∫ 𝑑𝑡) 𝑏−𝑎 0 2
0
1 (𝑥 − 𝑎) ) ( 2 (𝑏 − 𝑎) (𝑘 + 1) (𝑘 + 2) (𝑛 + 1)
1 (𝑏 − 𝑥) ( ) 2 (𝑏 − 𝑎) (𝑘 + 1) (𝑘 + 2) (𝑛 + 1)
𝑡 2
⋅ [(∫
𝑞 ⋅ [((𝑛 (𝑘 + 2) + 𝑘 + 1) 𝑓 (𝑥)
+
1 𝑘
0
This completes the proof.
𝑘+1
𝑡 2
1 𝑘 𝑡 (𝑏 − 𝑥)𝑘+1 [∫ 𝑏−𝑎 0 2
+∫
(15)
≤
1 𝑘
0
)
𝑘+1
1 𝑘 𝑡 (𝑥 − 𝑎)𝑘+1 [∫ 𝑏−𝑎 0 2
1/𝑞 1 𝑎 𝑞 𝑞 + (∫ 𝑡𝑘 [(1 − 𝑡) 𝑓 (𝑥) + 𝑚 (𝑛 + 𝑡) 𝑓 ( ) ] 𝑑𝑡) } 𝑚 0
(16)
+
1 1/𝑞 1 1−1/𝑞 (𝑏 − 𝑥)𝑘+1 ( ( ) ) 2 (𝑏 − 𝑎) 𝑘 + 1 𝑛+1
1/𝑞 1 𝑏 𝑞 𝑞 ⋅ {(∫ 𝑡𝑘 [(𝑛 + 𝑡) 𝑓 (𝑥) + 𝑚 (1 − 𝑡) 𝑓 ( ) ] 𝑑𝑡) 𝑚 0 1/𝑞 1 𝑏 𝑞 𝑞 + (∫ 𝑡𝑘 [(1 − 𝑡) 𝑓 (𝑥) + 𝑚 (𝑛 + 𝑡) 𝑓 ( ) ] 𝑑𝑡) } 𝑚 0
=
1/𝑞 1 (𝑥 − 𝑎)𝑘+1 ( ) 2 (𝑏 − 𝑎) (𝑘 + 1) (𝑘 + 2) (𝑛 + 1)
𝑎 𝑞 1/𝑞 𝑞 ⋅ [((𝑛 (𝑘 + 2) + 𝑘 + 1) 𝑓 (𝑥) + 𝑚 𝑓 ( ) ) 𝑚 𝑎 𝑞 1/𝑞 𝑞 + (𝑓 (𝑥) + 𝑚 (𝑛 (𝑘 + 2) + 𝑘 + 1) 𝑓 ( ) ) ] 𝑚
International Journal of Analysis +
5
1/𝑞 1 (𝑏 − 𝑥)𝑘+1 ) ( 2 (𝑏 − 𝑎) (𝑘 + 1) (𝑘 + 2) (𝑛 + 1)
+
1 𝑘 𝑡 (𝑏 − 𝑥)𝑘+1 [∫ 𝑏−𝑎 0 2
1/𝑞 𝑏 𝑞 𝑞 ⋅ [((𝑛 (𝑘 + 2) + 𝑘 + 1) 𝑓 (𝑥) + 𝑚 𝑓 ( ) ) 𝑚
+∫
1/𝑞 𝑏 𝑞 𝑞 + (𝑓 (𝑥) + 𝑚 (𝑛 (𝑘 + 2) + 𝑘 + 1) 𝑓 ( ) ) ] . 𝑚
≤
0
(17)
𝑡 2
1 − 𝑡 𝑛 + 𝑡 𝑓 ( 𝑥+ 𝑏) 𝑑𝑡] 𝑛+1 𝑛 + 1
1 𝑘 𝑡 𝑛 + 𝑡 𝛼 (𝑥 − 𝑎)𝑘+1 [∫ (( ) 𝑓 (𝑥) 𝑏−𝑎 𝑛+1 0 2
+ 𝑚 (1 − (
This completes the proof. Remark 13. Observe that if in Theorem 12 we have 𝑚 = 𝑛 = 1, the statement of Theorem 12 becomes the statement of Theorem 3 in [6].
3. Generalized Inequalities for (𝛼, 𝑚)-Convex Functions Theorem 14. Let 𝐼 be on open real interval such that [0, ∞) ⊂ 𝐼. Let 𝑓 : 𝐼 → R be a differentiable function on 𝐼 such that 𝑛 ∈ N, 𝑘 > 0, and 𝑓 ∈ 𝐿[𝑎, 𝑏], where 0 ≤ 𝑎 < 𝑏 < ∞. If |𝑓 | is (𝛼, 𝑚)-convex function on [𝑎, 𝑏] for some fixed (𝛼, 𝑚) ∈ (0, 1]2 , then (𝑥 − 𝑎)𝑘+1 𝐺 (𝑘; 𝑛; 𝑎, 𝑥, 𝑏) (𝑓) ≤ 𝛼 [𝐴 𝑓 (𝑥) 2 (𝑏 − 𝑎) (𝑛 + 1) 𝑎 + 𝑚 (𝐵 − 𝐴) 𝑓 ( )] 𝑚 (𝑏 − 𝑥)𝑘+1 [𝐴 𝑓 (𝑥) 2 (𝑏 − 𝑎) (𝑛 + 1)𝛼 𝑏 + 𝑚 (𝐵 − 𝐴) 𝑓 ( )] , 𝑚
+∫
𝑛 + 𝑡 𝛼 𝑎 ) ) 𝑓 ( )) 𝑑𝑡 𝑛+1 𝑚
1 𝑘
0
1−𝑡 𝛼 𝑡 1 − 𝑡 𝛼 (( ) 𝑓 (𝑥) + 𝑚 (1 − ( ) ) 2 𝑛+1 𝑛+1
𝑎 ⋅ 𝑓 ( )) 𝑑𝑡] 𝑚 +
1 𝑘 𝑡 𝑛 + 𝑡 𝛼 (𝑏 − 𝑥)𝑘+1 [∫ (( ) 𝑓 (𝑥) 𝑏−𝑎 𝑛+1 0 2
+ 𝑚 (1 − ( +∫
𝑛 + 𝑡 𝛼 𝑏 ) ) 𝑓 ( )) 𝑑𝑡 𝑛+1 𝑚
1 𝑘
0
1−𝑡 𝛼 𝑡 1 − 𝑡 𝛼 (( ) 𝑓 (𝑥) + 𝑚 (1 − ( ) ) 2 𝑛+1 𝑛+1
𝑏 ⋅ 𝑓 ( )) 𝑑𝑡] 𝑚 (18)
+
=
(𝑥 − 𝑎)𝑘+1 𝛼 [𝐴 𝑓 (𝑥) + 𝑚 (𝐵 − 𝐴) 2 (𝑏 − 𝑎) (𝑛 + 1)
𝑎 (𝑏 − 𝑥)𝑘+1 ⋅ 𝑓 ( )] + [𝐴 𝑓 (𝑥) 𝑚 2 (𝑏 − 𝑎) (𝑛 + 1)𝛼 𝑏 + 𝑚 (𝐵 − 𝐴) 𝑓 ( )] . 𝑚
where 𝑥 ∈ [𝑎, 𝑏] and 𝐴=
1 𝑘
𝑛 + 𝑡 1 − 𝑡 𝑓 ( 𝑥+ 𝑏) 𝑑𝑡 𝑛+1 𝑛 + 1
(20)
𝑛𝛼 1 𝐹 (−𝛼, 𝑘 + 1, 𝑘 + 2; − ) 𝑘+12 1 𝑛
This completes the proof.
(19)
Remark 15. Observe that if in Theorem 14 we have 𝛼 = 1, the statement of Theorem 14 becomes the statement of Theorem 8.
Proof. Using Lemma 7 and taking modulus and the fact that |𝑓 | is (𝛼, 𝑚)-convex function, we have
Theorem 16. Let 𝐼 be on open real interval such that [0, ∞) ⊂ 𝐼. Let 𝑓 : 𝐼 → R be a differentiable function on 𝐼 such that 𝑛 ∈ N, 𝑘 > 0, and 𝑓 ∈ 𝐿[𝑎, 𝑏], where 0 ≤ 𝑎 < 𝑏 < ∞. If |𝑓 |𝑞 is (𝛼, 𝑚)-convex function on [𝑎, 𝑏] for some fixed (𝛼, 𝑚) ∈ (0, 1]2 and 1/𝑝 + 1/𝑞 = 1, 𝑞 > 1, then
Γ (𝛼 + 1) Γ (𝑘 + 1) + Γ (𝛼 + 𝑘 + 2) 𝐵=
2 (𝑛 + 1)𝛼 . 𝑘+1
1/𝑝 𝑘+1 1 (𝑥 − 𝑎) ( ) 𝐺 (𝑘; 𝑛; 𝑎, 𝑥, 𝑏) (𝑓) ≤ 𝑏−𝑎 𝑝𝑘 + 1
𝐺 (𝑘; 𝑛; 𝑎, 𝑥, 𝑏) (𝑓) ≤
1 𝑘 𝑡 (𝑥 − 𝑎)𝑘+1 [∫ 𝑏−𝑎 0 2
+∫
1 𝑘
0
𝑡 2
𝑛 + 𝑡 1 − 𝑡 𝑓 ( 𝑥+ 𝑎) 𝑑𝑡 𝑛+1 𝑛 + 1
1 − 𝑡 𝑛 + 𝑡 𝑓 ( 𝑥+ 𝑎) 𝑑𝑡] 𝑛+1 𝑛 + 1
⋅(
1 𝛼/𝑞 ) 𝑛+1
𝑎 𝑞 1/𝑞 𝑞 ⋅ [(𝐶 𝑓 (𝑥) + 𝑚 (𝐷 − 𝐶) 𝑓 ( ) ) 𝑚
6
International Journal of Analysis 𝑎 𝑞 1/𝑞 𝑞 + ((𝐷 − 𝐶) 𝑓 (𝑥) + 𝑚𝐶 𝑓 ( ) ) ] 𝑚
where 𝑥 ∈ [𝑎, 𝑏] and
1/𝑝 1 𝛼/𝑞 1 (𝑏 − 𝑥)𝑘+1 ( ) ( ) 𝑏−𝑎 𝑝𝑘 + 1 𝑛+1
+
𝐶=
1/𝑞 𝑏 𝑞 𝑞 ⋅ [(𝐶 𝑓 (𝑥) + 𝑚 (𝐷 − 𝐶) 𝑓 ( ) ) 𝑚
𝐷 = (𝑛 + 1) .
(21)
Proof. Using Lemma 7, H¨older’s inequality, and the fact that |𝑓 |𝑞 is (𝛼, 𝑚)-convex function,
1 𝑘 𝑛 + 𝑡 𝑡 1 − 𝑡 1 − 𝑡 𝑛 + 𝑡 𝑓 ( 𝑓 ( 𝑥+ 𝑎) 𝑑𝑡 + ∫ 𝑥+ 𝑎) 𝑑𝑡] 𝑛+1 𝑛+1 𝑛+1 𝑛 + 1 0 2
1 𝑘 𝑛 + 𝑡 𝑡 1 − 𝑡 1 − 𝑡 𝑛 + 𝑡 𝑓 ( 𝑓 ( 𝑥+ 𝑏) 𝑑𝑡 + ∫ 𝑥 + 𝑏) 𝑑𝑡] 𝑛+1 𝑛+1 𝑛+1 𝑛 + 1 0 2
+
1 𝑘 𝑡 (𝑏 − 𝑥)𝑘+1 [∫ 𝑏−𝑎 0 2
≤
1 𝑡𝑘 (𝑥 − 𝑎)𝑘+1 (∫ ( ) 𝑑𝑡) 𝑏−𝑎 2 0
+
1 𝑡𝑘 (𝑏 − 𝑥)𝑘+1 (∫ ( ) 𝑑𝑡) 𝑏−𝑎 2 0
≤
1/𝑞 1/𝑝 1 𝑛 + 𝑡 𝛼 𝑞 𝑛 + 𝑡 𝛼 𝑎 𝑞 1 (𝑥 − 𝑎)𝑘+1 ( ) {[∫ (( ) 𝑓 (𝑥) + 𝑚 (1 − ( ) ) 𝑓 ( ) ) 𝑑𝑡] 𝑏−𝑎 𝑝𝑘 + 1 𝑛+1 𝑛+1 𝑚 0
𝑝
1/𝑝
1/𝑞 1/𝑞 1 1 𝑛+𝑡 1−𝑡 1 − 𝑡 𝑞 𝑛 + 1 𝑞 [∫ (𝑓 ( 𝑥+ 𝑎) 𝑑𝑡) + ∫ (𝑓 ( 𝑥+ 𝑎) 𝑑𝑡) ] 𝑛+1 𝑛+1 𝑛+1 𝑛+1 0 0
𝑝
1/𝑝
1/𝑞 1/𝑞 1 1 𝑛+𝑡 1−𝑡 1 − 𝑡 𝑞 𝑛 + 1 𝑞 [∫ (𝑓 ( 𝑥+ 𝑏) 𝑑𝑡) + ∫ (𝑓 ( 𝑥+ 𝑏) 𝑑𝑡) ] 𝑛+1 𝑛+1 𝑛+1 𝑛+1 0 0
1/𝑞 1/𝑝 1 − 𝑡 𝑞 1−𝑡 𝑎 𝑞 1 (𝑏 − 𝑥)𝑘+1 + [∫ (( ) 𝑓 (𝑥) + 𝑚 (1 − ( ) ) 𝑓 ( ) ) 𝑑𝑡] } + ( ) 𝑛+1 𝑛+1 𝑚 𝑏−𝑎 𝑝𝑘 + 1 0 𝛼
1
1
⋅ {[∫ (( 0
1
+ [∫ (( 0
(22)
𝛼
1/𝑞 𝑏 𝑞 𝑞 + ((𝐷 − 𝐶) 𝑓 (𝑥) + 𝑚𝐶 𝑓 ( ) ) ] , 𝑚
𝑘+1 1 𝑘 𝑡 (𝑥 − 𝑎) [∫ 𝐺 (𝑘; 𝑛; 𝑎, 𝑥, 𝑏) (𝑓) ≤ 𝑏−𝑎 0 2
(𝑛 + 1)𝛼 − 𝑛𝛼 𝛼+1
(23)
𝛼
1/𝑞 𝑛 + 𝑡 𝛼 𝑞 𝑛 + 𝑡 𝛼 𝑏 𝑞 ) 𝑓 (𝑥) + 𝑚 (1 − ( ) ) 𝑓 ( ) ) 𝑑𝑡] 𝑛+1 𝑛+1 𝑚
1/𝑞 1/𝑝 1 − 𝑡 𝛼 𝑞 1 − 𝑡 𝛼 𝑏 𝑞 1 𝛼/𝑞 1 (𝑥 − 𝑎)𝑘+1 ) 𝑓 (𝑥) + 𝑚 (1 − ( ) ) 𝑓 ( ) ) 𝑑𝑡] } = ( ) ( ) 𝑛+1 𝑛+1 𝑚 𝑏−𝑎 𝑝𝑘 + 1 𝑛+1
1/𝑝 𝑎 𝑞 1/𝑞 𝑎 𝑞 1/𝑞 1 𝑞 𝑞 (𝑏 − 𝑥)𝑘+1 ⋅ [(𝐶 𝑓 (𝑥) + 𝑚 (𝐷 − 𝐶) 𝑓 ( ) ) + ((𝐷 − 𝐶) 𝑓 (𝑥) + 𝑚𝐶 𝑓 ( ) ) ] + ( ) 𝑚 𝑚 𝑏−𝑎 𝑝𝑘 + 1
⋅(
1/𝑞 1/𝑞 1 𝛼/𝑞 𝑏 𝑞 𝑏 𝑞 𝑞 𝑞 ) [(𝐶 𝑓 (𝑥) + 𝑚 (𝐷 − 𝐶) 𝑓 ( ) ) + ((𝐷 − 𝐶) 𝑓 (𝑥) + 𝑚𝐶 𝑓 ( ) ) ] . 𝑛+1 𝑚 𝑚
This completes the proof. Remark 17. Observe that if in Theorem 16 we have 𝛼 = 1, the statement of Theorem 16 becomes the statement of Theorem 10. Theorem 18. Let 𝐼 be on open real interval such that [0, ∞) ⊂ 𝐼. Let 𝑓 : 𝐼 → R be a differentiable function on 𝐼 such that 𝑛 ∈ N, 𝑘 > 0, and 𝑓 ∈ 𝐿[𝑎, 𝑏], where 0 ≤ 𝑎 < 𝑏 < ∞. If |𝑓 |𝑞 is
(𝛼, 𝑚)-convex function on [𝑎, 𝑏] for some fixed (𝛼, 𝑚) ∈ (0, 1]2 and 𝑞 ≥ 1, then 𝑘+1 1 1−1/𝑞 (𝑥 − 𝑎) ( ) 𝐺 (𝑘; 𝑛; 𝑎, 𝑥, 𝑏) (𝑓) = 2 (𝑏 − 𝑎) 𝑘 + 1
⋅(
1 𝛼/𝑞 Γ (𝛼 + 1) Γ (𝑘 + 1) 𝑞 ) 𝑓 (𝑥) ) {[(𝐴 − 𝑛+1 Γ (𝛼 + 𝑘 + 2)
International Journal of Analysis + 𝑚( +[
𝐵 Γ (𝛼 + 1) Γ (𝑘 + 1) 𝑎 𝑞 1/𝑞 ) 𝑓 ( ) ] −𝐴+ 2 Γ (𝛼 + 𝑘 + 2) 𝑚
Γ (𝛼 + 1) Γ (𝑘 + 1) 𝑞 𝑓 (𝑥) Γ (𝛼 + 𝑘 + 2)
+ 𝑚(
+
7
+[
𝐵 Γ (𝛼 + 1) Γ (𝑘 + 1) 𝑎 𝑞 1/𝑞 ) 𝑓 ( ) ] } − 2 Γ (𝛼 + 𝑘 + 2) 𝑚
1 𝑘 𝑡 (𝑏 − 𝑥)𝑘+1 + [∫ 𝑏−𝑎 0 2
⋅ [(∫
1 𝑘
0
+
𝑡 2
1/𝑞 𝐵 Γ (𝛼 + 1) Γ (𝑘 + 1) 𝑏 𝑞 ) 𝑓 ( ) ] } , − 2 Γ (𝛼 + 𝑘 + 2) 𝑚
(24) where 𝐴 and 𝐵 are given by (19) and 𝑥 ∈ [𝑎, 𝑏].
Γ (𝛼 + 1) Γ (𝑘 + 1) 𝑞 ) 𝑓 (𝑥) Γ (𝛼 + 𝑘 + 2)
𝑘+1 1 𝑘 𝑡 (𝑥 − 𝑎) [∫ 𝐺 (𝑘; 𝑛; 𝑎, 𝑥, 𝑏) (𝑓) ≤ 𝑏−𝑎 0 2
1/𝑞 𝐵 Γ (𝛼 + 1) Γ (𝑘 + 1) 𝑏 𝑞 −𝐴+ ) 𝑓 ( ) ] 2 Γ (𝛼 + 𝑘 + 2) 𝑚
Γ (𝛼 + 1) Γ (𝑘 + 1) 𝑞 𝑓 (𝑥) Γ (𝛼 + 𝑘 + 2)
+ 𝑚(
1 𝛼/𝑞 1 1−1/𝑞 (𝑏 − 𝑥)𝑘+1 ( ( ) ) 2 (𝑏 − 𝑎) 𝑘 + 1 𝑛+1
⋅ {[(𝐴 −
+ 𝑚(
Proof. Using Lemma 7, Power’s mean inequality, and the fact that |𝑓 |𝑞 is (𝛼, 𝑚)-convex function,
1 𝑘 𝑛 + 𝑡 𝑡 1 − 𝑡 1 − 𝑡 𝑛 + 𝑡 𝑓 ( 𝑓 ( 𝑥+ 𝑎) 𝑑𝑡 + ∫ 𝑥 + 𝑎) 𝑑𝑡] 𝑛+1 𝑛+1 𝑛+1 𝑛 + 1 0 2
1−1/𝑞 1 𝑘 1 𝑘 𝑛 + 𝑡 𝑡 1 − 𝑡 𝑡 1 − 𝑡 𝑛 + 𝑡 (𝑥 − 𝑎)𝑘+1 𝑓 ( 𝑓 ( 𝑥+ 𝑏) 𝑑𝑡 + ∫ 𝑥+ 𝑏) 𝑑𝑡] ≤ (∫ 𝑑𝑡) 𝑛+1 𝑛 + 1 𝑛+1 𝑛 + 1 𝑏−𝑎 0 2 0 2
1/𝑞 1/𝑞 1 𝑘 𝑛 + 𝑡 𝑡 1 − 𝑡 1 − 𝑡 𝑞 𝑛 + 1 𝑞 𝑓 ( 𝑓 ( 𝑑𝑡) ] 𝑥+ 𝑎) 𝑑𝑡) + (∫ 𝑥 + 𝑎) 𝑛+1 𝑛+1 𝑛+1 𝑛 + 1 0 2
1 𝑘 𝑡 (𝑏 − 𝑥)𝑘+1 (∫ 𝑑𝑡) 𝑏−𝑎 0 2
1−1/𝑞
1 𝑘
[(∫
0
𝑡 2
1/𝑞 1/𝑞 1 𝑘 𝑛 + 𝑡 𝑡 1 − 𝑡 1 − 𝑡 𝑞 𝑛 + 1 𝑞 𝑓 ( 𝑓 𝑑𝑡) + (∫ ( 𝑑𝑡) ] 𝑥 + 𝑏) 𝑥 + 𝑏) 𝑛+1 𝑛 + 1 𝑛+1 𝑛 + 1 0 2
1/𝑞 1−1/𝑞 1 𝑘 𝑡 𝑛 + 𝑡 𝛼 𝑎 𝑞 1 𝑛 + 𝑡 𝛼 𝑞 (𝑥 − 𝑎)𝑘+1 ≤ {[∫ ( ) (( ) 𝑓 (𝑥) + 𝑚 (1 − ( ) ) 𝑓 ( ) ) 𝑑𝑡] 𝑏−𝑎 2 (𝑘 + 1) 𝑛+1 𝑛+1 𝑚 0 2
+ [∫
1/𝑞 1−1/𝑞 𝑡 1 − 𝑡 𝛼 𝑎 𝑞 1 − 𝑡 𝛼 𝑞 1 (𝑏 − 𝑥)𝑘+1 ) (( ) 𝑓 (𝑥) + 𝑚 (1 − ( ) ) 𝑓 ( ) ) 𝑑𝑡] } + ( 2 𝑛+1 𝑛+1 𝑚 𝑏−𝑎 2 (𝑘 + 1)
1 𝑘
0
(25) 1/𝑞 𝑞 𝑡 𝑛+𝑡 𝑛 + 𝑡 𝑞 𝑏 ⋅ {[∫ (( ) 𝑓 (𝑥) + 𝑚 (1 − ( ) ) 𝑓 ( ) ) 𝑑𝑡] 𝑛+1 𝑛+1 𝑚 0 2 1 𝑘
+ [∫
1/𝑞 1 1−1/𝑞 𝑡 1 − 𝑡 𝛼 𝑏 𝑞 1 𝛼/𝑞 1 − 𝑡 𝛼 𝑞 (𝑥 − 𝑎)𝑘+1 ( ( (( ) 𝑓 (𝑥) + 𝑚 (1 − ( ) ) 𝑓 ( ) ) 𝑑𝑡] } = ) ) 2 𝑛+1 𝑛+1 𝑚 2 (𝑏 − 𝑎) 𝑘 + 1 𝑛+1
⋅ {[(𝐴 −
⋅(
𝛼
1 𝑘
0
+[
𝛼
Γ (𝛼 + 1) Γ (𝑘 + 1) 𝑞 𝐵 Γ (𝛼 + 1) Γ (𝑘 + 1) 𝑎 𝑞 1/𝑞 ) 𝑓 (𝑥) + 𝑚 ( − 𝐴 + ) 𝑓 ( ) ] Γ (𝛼 + 𝑘 + 2) 2 Γ (𝛼 + 𝑘 + 2) 𝑚
Γ (𝛼 + 1) Γ (𝑘 + 1) 𝑞 𝐵 Γ (𝛼 + 1) Γ (𝑘 + 1) 𝑎 𝑞 1/𝑞 1 1−1/𝑞 (𝑏 − 𝑥)𝑘+1 𝑓 ( ) ] } + 𝑓 (𝑥) + 𝑚 ( − ) ( ) Γ (𝛼 + 𝑘 + 2) 2 Γ (𝛼 + 𝑘 + 2) 𝑚 2 (𝑏 − 𝑎) 𝑘 + 1
1/𝑞 1 𝛼/𝑞 Γ (𝛼 + 1) Γ (𝑘 + 1) 𝑞 𝐵 Γ (𝛼 + 1) Γ (𝑘 + 1) 𝑏 𝑞 ) 𝑓 (𝑥) + 𝑚 ( − 𝐴 + ) 𝑓 ( ) ] ) {[(𝐴 − 𝑛+1 Γ (𝛼 + 𝑘 + 2) 2 Γ (𝛼 + 𝑘 + 2) 𝑚
+[
1/𝑞 Γ (𝛼 + 1) Γ (𝑘 + 1) 𝑞 𝐵 Γ (𝛼 + 1) Γ (𝑘 + 1) 𝑏 𝑞 𝑓 (𝑥) + 𝑚 ( − 𝑓 ( ] }. ) ) Γ (𝛼 + 𝑘 + 2) 2 Γ (𝛼 + 𝑘 + 2) 𝑚
8
International Journal of Analysis This completes the proof.
Remark 19. Observe that if in Theorem 18 we have 𝛼 = 1, the statement of Theorem 18 becomes the statement of Theorem 12.
Competing Interests The authors declare that they have no competing interests.
References [1] G. Toader, “On a generalization of the convexity,” Mathematica, vol. 30, no. 53, pp. 83–87, 1988. [2] V. G. Mihesan, A Generalization of the Convexity, Seminar of Functional Equations, Approx. and Convex, Cluj-Napoca, Romania, 1993. [3] J. E. Peˇcari´c, F. Proschan, and Y. L. Tong, Convex Functions, Partial Orderings, and Statistical Applications, Academic Press, 1992. [4] A. Kilbas, H. M. Srivastava, and J. J. Trujillo, Theory and Applications of Fractional Differential Equations, Elsevier, Amsterdam, The Netherlands, 2006. [5] M. A. Noor, K. I. Noor, M. V. Mihai, and M. U. Awan, “Fractional Hermite-Hadamard inequalities for differentiable sGodunova-Levin functions,” Filomat, In press. [6] M. V. Mihai and F.-C. Mitroi, “Hermite-Hadamard type inequalities obtained via Riemann-Liouville fractional calculus,” Acta Mathematica Universitatis Comenianae, vol. 83, no. 2, pp. 209–215, 2014.
Advances in
Operations Research Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Advances in
Decision Sciences Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Journal of
Applied Mathematics
Algebra
Hindawi Publishing Corporation http://www.hindawi.com
Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Journal of
Probability and Statistics Volume 2014
The Scientific World Journal Hindawi Publishing Corporation http://www.hindawi.com
Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
International Journal of
Differential Equations Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Volume 2014
Submit your manuscripts at http://www.hindawi.com International Journal of
Advances in
Combinatorics Hindawi Publishing Corporation http://www.hindawi.com
Mathematical Physics Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Journal of
Complex Analysis Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
International Journal of Mathematics and Mathematical Sciences
Mathematical Problems in Engineering
Journal of
Mathematics Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Volume 2014
Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Discrete Mathematics
Journal of
Volume 2014
Hindawi Publishing Corporation http://www.hindawi.com
Discrete Dynamics in Nature and Society
Journal of
Function Spaces Hindawi Publishing Corporation http://www.hindawi.com
Abstract and Applied Analysis
Volume 2014
Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
International Journal of
Journal of
Stochastic Analysis
Optimization
Hindawi Publishing Corporation http://www.hindawi.com
Hindawi Publishing Corporation http://www.hindawi.com
Volume 2014
Volume 2014
