DOI 10.1515/apam-2013-0029 |
Adv. Pure Appl. Math. 2014; aop
Research Article İmdat İşcan
Some new general integral inequalities for ℎ-convex and ℎ-concave functions Abstract: In this paper, we derive new integral inequalities for functions with ℎ-convex and ℎ-concave first derivatives. As a consequence, we give new estimates for the remainder term of the midpoint, trapezoid, and Simpson formulae for functions whose derivatives in absolute value at certain power are ℎ-convex and ℎ-concave and we point out the results for some special classes of functions. Keywords: Hermite–Hadamard-type inequalities, Simpson-type inequality, ℎ-convex function, 𝑠-convex function, 𝑃-function MSC 2010: 26A51, 26D10 || İmdat İşcan: Department of Mathematics, Faculty of Arts and Sciences, Giresun University, Giresun, Turkey, e-mail:
[email protected]
1 Introduction Let 𝑓 : 𝐼 ⊆ ℝ → ℝ be a convex function defined on the interval 𝐼 of real numbers and 𝑎, 𝑏 ∈ 𝐼 with 𝑎 < 𝑏. The inequality 𝑏
𝑓(𝑎) + 𝑓(𝑏) 𝑎+𝑏 1 )≤ ∫ 𝑓(𝑥)𝑑𝑥 ≤ 𝑓( 2 𝑏−𝑎 2
(1.1)
𝑎
holds. This double inequality is known in the literature as the Hermite–Hadamard integral inequality for convex functions. Note that some of the classical inequalities for means can be derived from (1.1) for appropriate particular selections of the mapping 𝑓. Both inequalities hold in the reversed direction if 𝑓 is concave. For some results which generalize, improve and extend the inequalities in (1.1) we refer the reader to the papers (see [1, 2, 4, 5, 7–14]). In [3], Breckner introduced 𝑠-convex functions as a generalization of convex functions as follows: Definition 1.1. Let 𝑠 ∈ (0, 1] be a fixed real number. A function 𝑓 : [0, ∞) → [0, ∞) is said to be 𝑠-convex (in the second sense) or that 𝑓 belongs to the class 𝐾𝑠2 if 𝑓(𝑡𝑥 + (1 − 𝑡)𝑦) ≤ 𝑡𝑠 𝑓(𝑥) + (1 − 𝑡)𝑠 𝑓(𝑦)
(1.2)
for all 𝑥, 𝑦 ∈ [0, ∞) and 𝑡 ∈ [0, 1]. If inequality (1.2) is reversed, then 𝑓 is said to be 𝑠-concave (in the second sense). Of course, 𝑠-convexity means just convexity when 𝑠 = 1. In [5], Dragomir, Pečarić and Persson defined the concept of 𝑃-function as the following: Definition 1.2. We say that 𝑓 : 𝐼 → ℝ is a 𝑃-function or that 𝑓 belongs to the class 𝑃(𝐼) if 𝑓 is a non-negative function and for all 𝑥, 𝑦 ∈ 𝐼, 𝑡 ∈ [0, 1], we have 𝑓(𝑡𝑥 + (1 − 𝑡)𝑦) ≤ 𝑓(𝑥) + 𝑓(𝑦). Let 𝐼 and 𝐽 be intervals in ℝ, (0, 1) ⊆ 𝐽 and ℎ and 𝑓 be real non-negative functions defined on 𝐽 and 𝐼, respectively. In [15], Varošanec defined the concept of ℎ-convexity as follows: Definition 1.3. Let ℎ : 𝐽 → ℝ be a non-negative function, ℎ ≠ 0. We say that 𝑓 : 𝐼 → ℝ is an ℎ-convex function or that 𝑓 belongs to the class SX(ℎ, 𝐼) if 𝑓 is non-negative function and for all 𝑥, 𝑦 ∈ 𝐼 and 𝑡 ∈ (0, 1) we have 𝑓(𝑡𝑥 + (1 − 𝑡)𝑦) ≤ ℎ(𝑡)𝑓(𝑥) + ℎ(1 − 𝑡)𝑓(𝑦).
Authenticated |
[email protected] author's copy Download Date | 2/5/14 12:43 AM
(1.3)
2 | İ. İşcan, Some new general integral inequalities If inequality (1.3) is reversed, then 𝑓 is said to be ℎ-concave, i.e.𝑓 ∈ SV(ℎ, 𝐼). The notion of ℎ-convexity unifies and generalizes the known classes of functions, 𝑠-convex functions, Gudunova–Levin functions [6] and 𝑃-functions, which are obtained by putting in (1.3), ℎ(𝑡) = 𝑡, ℎ(𝑡) = 𝑡𝑠 , ℎ(𝑡) = 1𝑡 , and ℎ(𝑡) = 1, respectively. In [13], Sarikaya, Saglam and Yıldırım established a new Hadamard-type inequality for ℎ-convex functions. Theorem 1.4. Let 𝑓 ∈ SX(ℎ, 𝐼), 𝑎, 𝑏 ∈ 𝐼 with 𝑎 < 𝑏 and 𝑓 ∈ 𝐿([𝑎, 𝑏]). Then 𝑏
1
𝑎+𝑏 1 1 )≤ ∫ 𝑓(𝑥)𝑑𝑥 ≤ [𝑓(𝑎) + 𝑓(𝑏)] ∫ ℎ(𝑡)𝑑𝑡. 𝑓( 1 2 𝑏−𝑎 2ℎ( 2 ) 𝑎
(1.4)
0
The following inequality is well known in the literature as Simpson’s inequality: Let 𝑓 : [𝑎, 𝑏]→ ℝ be a four times continuously differentiable mapping on (𝑎, 𝑏) and ‖𝑓(4) ‖∞ = sup𝑥∈(𝑎,𝑏) |𝑓(4) (𝑥)| < ∞. Then the following inequality holds: 𝑏 1 𝑎+𝑏 1 1 𝑓(𝑎) + 𝑓(𝑏) 4 (4) [ ≤ )] ∫ + 2𝑓( − 𝑓(𝑥)𝑑𝑥 3 2880 ‖𝑓 ‖∞ (𝑏 − 𝑎) . 2 2 𝑏 − 𝑎 𝑎
In recent years many authors have studied error estimations for Simpson-type inequalities; for refinements, counterparts, generalizations and new Simpson-type inequalities, see [7–9, 12, 14]. In [9], Iscan obtained a new generalization of some integral inequalities for differentiable convex mapping and he used the following lemma to prove this. Lemma 1.5. Let 𝑓 : 𝐼 ⊆ ℝ → ℝ be a differentiable mapping on 𝐼∘ such that 𝑓 ∈ 𝐿[𝑎, 𝑏], where 𝑎, 𝑏 ∈ 𝐼 with 𝑎 < 𝑏 and 𝜃, 𝜆 ∈ [0, 1]. Then the following equality holds: 𝑏
1 ∫ 𝑓(𝑥)𝑑𝑥 (1 − 𝜃)(𝜆𝑓(𝑎) + (1 − 𝜆)𝑓(𝑏)) + 𝜃𝑓((1 − 𝜆)𝑎 + 𝜆𝑏) − 𝑏−𝑎 𝑎
1
= (𝑏 − 𝑎)[−𝜆2 ∫(𝑡 − 𝜃)𝑓 (𝑡𝑎 + (1 − 𝑡)[(1 − 𝜆)𝑎 + 𝜆𝑏])𝑑𝑡 0 1
+ (1 − 𝜆)2 ∫(𝑡 − 𝜃)𝑓 (𝑡𝑏 + (1 − 𝑡)[(1 − 𝜆)𝑎 + 𝜆𝑏])𝑑𝑡]. 0
The main inequalities that have been pointed out in [9] are as follows. Theorem 1.6. Let 𝑓 : 𝐼 ⊆ [0, ∞)→ ℝ be a differentiable mapping on 𝐼∘ such that 𝑓 ∈ 𝐿[𝑎, 𝑏], where 𝑎, 𝑏 ∈ 𝐼∘ with 𝑎 < 𝑏 and 𝜃, 𝜆 ∈ [0, 1]. If |𝑓 |𝑞 is 𝑠-convex on [𝑎, 𝑏], 𝑞 ≥ 1, then the following inequality holds: 𝑏 1 (1 − 𝜃)(𝜆𝑓(𝑎) + (1 − 𝜆)𝑓(𝑏)) + 𝜃𝑓((1 − 𝜆)𝑎 + 𝜆𝑏) − ∫ 𝑓(𝑥)𝑑𝑥 𝑏−𝑎 𝑎 1− 1
1
≤ (𝑏 − 𝑎)𝐴 1 𝑞 (𝜃){𝜆2 [|𝑓 (𝑎)|𝑞 𝐴 2 (𝜃, 𝑠) + |𝑓 (𝐶)|𝑞 𝐴 3 (𝜃, 𝑠)] 𝑞 1
+ (1 − 𝜆)2 [|𝑓 (𝑏)|𝑞 𝐴 2 (𝜃, 𝑠) + |𝑓 (𝐶)|𝑞 𝐴 3 (𝜃, 𝑠)] 𝑞 }, where
1 𝐴 1 (𝜃) = 𝜃2 − 𝜃 + , 2 2𝜃𝑠+2 𝜃 1 𝐴 2 (𝜃, 𝑠) = − + , (𝑠 + 1)(𝑠 + 2) 𝑠 + 1 𝑠 + 2 2(1 − 𝜃)𝑠+2 1−𝜃 1 𝐴 3 (𝜃, 𝑠) = − + (𝑠 + 1)(𝑠 + 2) 𝑠 + 1 𝑠 + 2
and 𝐶 = (1 − 𝜆)𝑎 + 𝜆𝑏.
Authenticated |
[email protected] author's copy Download Date | 2/5/14 12:43 AM
(1.5)
İ. İşcan, Some new general integral inequalities
|
3
Theorem 1.7. Let 𝑓 : 𝐼 ⊆ [0, ∞)→ ℝ be a differentiable mapping on 𝐼∘ such that 𝑓 ∈ 𝐿[𝑎, 𝑏], where 𝑎, 𝑏 ∈ 𝐼∘ with 𝑎 < 𝑏 and 𝜃, 𝜆 ∈ [0, 1]. If |𝑓 |𝑞 is 𝑠-convex on [𝑎, 𝑏], 𝑞 > 1, then the following inequality holds: 𝑏 1 (1 − 𝜃)(𝜆𝑓(𝑎) + (1 − 𝜆)𝑓(𝑏)) + 𝜃𝑓((1 − 𝜆)𝑎 + 𝜆𝑏) − ∫ 𝑓(𝑥)𝑑𝑥 𝑏 − 𝑎 𝑎 1
≤ (𝑏 − 𝑎)(
1
1
|𝑓 (𝑏)|𝑞 + |𝑓 (𝐶)|𝑞 𝑞 𝜃𝑝+1 + (1 − 𝜃)𝑝+1 𝑝 2 |𝑓 (𝑎)|𝑞 + |𝑓 (𝐶)|𝑞 𝑞 ) [𝜆 ( ) + (1 − 𝜆)2 ( ) ], 𝑝+1 𝑠+1 𝑠+1
where 𝐶 = (1 − 𝜆)𝑎 + 𝜆𝑏 and
1 𝑝
+
1 𝑞
(1.6)
= 1.
In [1] Alomari, Darus and Kirmaci obtained the following inequalities of the left-hand side of Hermite– Hadamard’s inequality for 𝑠-convex mappings. Theorem 1.8. Let 𝑓 : 𝐼 ⊆ [0, ∞) → ℝ be a differentiable mapping on 𝐼∘ such that 𝑓 ∈ 𝐿[𝑎, 𝑏], where 𝑎, 𝑏 ∈ 𝐼 with 𝑎 < 𝑏. If |𝑓 |𝑞 , 𝑞 ≥ 1, is 𝑠-convex on [𝑎, 𝑏], for some fixed 𝑠 ∈ (0, 1], then the following inequality holds: 𝑏 1 1 𝑞 1 2 𝑎 + 𝑏 𝑏 − 𝑎 𝑓( ≤ [{(21−𝑠 + 1)|𝑓 (𝑏)|𝑞 + 21−𝑠 |𝑓 (𝑎)|𝑞 } 𝑞 ) ( ) ∫ 𝑓(𝑥)𝑑𝑥 − 2 𝑏−𝑎 8 (𝑠 + 1)(𝑠 + 2) 𝑎 1
+ {(21−𝑠 + 1)|𝑓 (𝑎)|𝑞 + 21−𝑠 |𝑓 (𝑏)|𝑞 } 𝑞 ].
(1.7)
Theorem 1.9. Let𝑝 𝑓 : 𝐼 ⊆ [0, ∞) → ℝ be a differentiable mapping on 𝐼∘ such that 𝑓 ∈ 𝐿[𝑎, 𝑏], where 𝑎, 𝑏 ∈ 𝐼 with 𝑎 < 𝑏. If |𝑓 | 𝑝−1 , 𝑝 > 1, is 𝑠-convex on [𝑎, 𝑏], for some fixed 𝑠 ∈ (0, 1], then the following inequality holds: 𝑏 1 2 1 𝑝 𝑞 1 1 1 𝑏−𝑎 𝑎 + 𝑏 𝑓( )− )( ) ( ) [((21−𝑠 + 𝑠 + 1)|𝑓 (𝑎)|𝑞 + 21−𝑠 |𝑓 (𝑏)|𝑞 ) 𝑞 ∫ 𝑓(𝑥)𝑑𝑥 ≤ ( 2 𝑏 − 𝑎 4 𝑝 + 1 𝑠 + 1 𝑎 1
× ((21−𝑠 + 𝑠 + 1)|𝑓 (𝑏)|𝑞 + 21−𝑠 |𝑓 (𝑎)|𝑞 ) 𝑞 ], (1.8) 𝑝 . 𝑝−1
where 𝑝 is the conjugate of 𝑞, 𝑞 =
In [14], Sarikaya, Set and Özdemir obtained a new upper bound for the right-hand side of Simpson’s inequality for 𝑠-convex mapping as follows: Theorem 1.10. Let 𝑓 : 𝐼 ⊆ [0, ∞) → ℝ be a differentiable mapping on 𝐼∘ such that 𝑓 ∈ 𝐿[𝑎, 𝑏], where 𝑎, 𝑏 ∈ 𝐼∘ with 𝑎 < 𝑏. If |𝑓 |𝑞 is 𝑠-convex on [𝑎, 𝑏], for some fixed 𝑠 ∈ (0, 1] and 𝑞 > 1, then the following inequality holds: 𝑏 𝑎+𝑏 1 1 [𝑓(𝑎) + 4𝑓( ) ∫ + 𝑓(𝑏)] − 𝑓(𝑥)𝑑𝑥 6 2 𝑏 − 𝑎 𝑎 1
1
|𝑓 ( 𝑎+𝑏 )|𝑞 + |𝑓 (𝑎)|𝑞 𝑞 |𝑓 ( 𝑎+𝑏 )|𝑞 + |𝑓 (𝑏)|𝑞 𝑞 𝑏 − 𝑎 1 + 2𝑝+1 𝑝 2 2 ( ) {( ) +( ) }, ≤ 12 3(𝑝 + 1) 𝑠+1 𝑠+1 1
where
1 𝑝
+
1 𝑞
(1.9)
= 1.
In [10], Kirmaci, Bakula, Özdemir and Pečarić proved the following trapezoid inequality. Theorem 1.11. Let 𝑓 : 𝐼 ⊆ [0, ∞) → ℝ be a differentiable mapping on 𝐼∘ such that 𝑓 ∈ 𝐿[𝑎, 𝑏], where 𝑎, 𝑏 ∈ 𝐼∘ with 𝑎 < 𝑏. If |𝑓 |𝑞 is 𝑠-convex on [𝑎, 𝑏], for some fixed 𝑠 ∈ (0, 1) and 𝑞 > 1, then 𝑏 1 𝑓(𝑎) + 𝑓(𝑏) ∫ − 𝑓(𝑥)𝑑𝑥 2 𝑏−𝑎 𝑎
≤
𝑞−1 𝑏−𝑎 ( ) 2 2(2𝑞 − 1)
𝑞−1 𝑞
(
1 1 1 𝑞 𝑞 𝑞 1 𝑎 + 𝑏 𝑞 𝑎 + 𝑏 𝑞 ) {(𝑓 ( ) + |𝑓 (𝑎)|𝑞 ) + (𝑓 ( ) + |𝑓 (𝑏)|𝑞 ) }. 𝑠+1 2 2
Authenticated |
[email protected] author's copy Download Date | 2/5/14 12:43 AM
(1.10)
4 | İ. İşcan, Some new general integral inequalities
2 Main results Let 𝑓 : 𝐼 ⊆ ℝ → ℝ be a differentiable function on 𝐼∘ , the interior of 𝐼. Throughout this section we will take 𝑏
𝑆𝑓 (𝜆, 𝜃, 𝑎, 𝑏) = (1 − 𝜃)(𝜆𝑓(𝑎) + (1 − 𝜆)𝑓(𝑏)) + 𝜃𝑓((1 − 𝜆)𝑎 + 𝜆𝑏) −
1 ∫ 𝑓(𝑥)𝑑𝑥, 𝑏−𝑎 𝑎
where 𝑎, 𝑏 ∈ 𝐼∘ with 𝑎 < 𝑏 and 𝜃, 𝜆 ∈ [0, 1]. The following theorems give new integral inequalities for ℎ-convex functions. In the sequel of the paper, 𝐽 is an interval in ℝ, (0, 1) ⊂ 𝐽, ℎ is a real non-negative function defined on 𝐽 and ℎ ∈ 𝐿[0, 1], ℎ ≠ 0. Theorem 2.1. Let 𝑓 : 𝐼 ⊆ ℝ → ℝ be a differentiable mapping on 𝐼∘ such that 𝑓 ∈ 𝐿[𝑎, 𝑏], where 𝑎, 𝑏 ∈ 𝐼∘ with 𝑎 < 𝑏 and 𝜃, 𝜆 ∈ [0, 1]. If |𝑓 |𝑞 is ℎ-convex on [𝑎, 𝑏], 𝑞 ≥ 1, then the following inequality holds: 1− 1
1
|𝑆𝑓 (𝜆, 𝜃, 𝑎, 𝑏)| ≤ (𝑏 − 𝑎)𝐴 1 𝑞 (𝜃){𝜆2 [|𝑓 (𝑎)|𝑞 𝐴 2 (𝜃, ℎ) + |𝑓 (𝐶)|𝑞 𝐴 3 (𝜃, ℎ)] 𝑞 1
+ (1 − 𝜆)2 [|𝑓 (𝑏)|𝑞 𝐴 2 (𝜃, ℎ) + |𝑓 (𝐶)|𝑞 𝐴 3 (𝜃, ℎ)] 𝑞 }, where
(2.1)
1 𝐴 1 (𝜃) = 𝜃2 − 𝜃 + , 2 1
𝐴 2 (𝜃, ℎ) = ∫ |𝑡 − 𝜃|ℎ(𝑡)𝑑𝑡, 0 1
𝐴 3 (𝜃, ℎ) = ∫ |𝑡 − 𝜃|ℎ(1 − 𝑡)𝑑𝑡 = 𝐴 2 (1 − 𝜃, ℎ) 0
and 𝐶 = (1 − 𝜆)𝑎 + 𝜆𝑏. Proof. Suppose that 𝑞 ≥ 1 and 𝐶 = (1 − 𝜆)𝑎 + 𝜆𝑏. From Lemma 1.5 and using the well-known power mean inequality, we have 1
1
2
2
|𝑆𝑓 (𝜆, 𝜃, 𝑎, 𝑏)| ≤ (𝑏 − 𝑎)[𝜆 ∫ |𝑡 − 𝜃||𝑓 (𝑡𝑎 + (1 − 𝑡)𝐶)|𝑑𝑡 + (1 − 𝜆) ∫ |𝑡 − 𝜃||𝑓 (𝑡𝑏 + (1 − 𝑡)𝐶)|𝑑𝑡] 0
0 1
1− 𝑞1
≤ (𝑏 − 𝑎){𝜆2 (∫ |𝑡 − 𝜃|𝑑𝑡)
1
(∫ |𝑡 − 𝜃||𝑓 (𝑡𝑎 + (1 − 𝑡)𝐶)|𝑞 𝑑𝑡)
0
1 𝑞
0 1
1− 𝑞1
2
+ (1 − 𝜆) (∫ |𝑡 − 𝜃|𝑑𝑡) 0
1
𝑞
1 𝑞
(∫ |𝑡 − 𝜃||𝑓 (𝑡𝑏 + (1 − 𝑡)𝐶)| 𝑑𝑡) }.
(2.2)
0
Since |𝑓 |𝑞 is ℎ-convex on [𝑎, 𝑏], we know that for 𝑡 ∈ (0, 1) |𝑓 (𝑡𝑎 + 𝐶(1 − 𝑡))|𝑞 ≤ ℎ(𝑡)|𝑓 (𝑎)|𝑞 + ℎ(1 − 𝑡)|𝑓 (𝐶)|𝑞 and |𝑓 (𝑡𝑏 + 𝐶(1 − 𝑡))|𝑞 ≤ ℎ(𝑡)|𝑓 (𝑏)|𝑞 + ℎ(1 − 𝑡)|𝑓 (𝐶)|𝑞 . Hence, by simple computation 1
1
𝑞
𝑞
𝑞
1
𝑞
∫ |𝑡 − 𝜃|[ℎ(𝑡)|𝑓 (𝑎)| + ℎ(1 − 𝑡)|𝑓 (𝐶)| ]𝑑𝑡 = |𝑓 (𝑎)| ∫ |𝑡 − 𝜃|ℎ(𝑡)𝑑𝑡 + |𝑓 (𝐶)| ∫ |𝑡 − 𝜃|ℎ(1 − 𝑡)𝑑𝑡, 0 1
0
0
1
1
∫ |𝑡 − 𝜃|[ℎ(𝑡)|𝑓 (𝑏)|𝑞 + ℎ(1 − 𝑡)|𝑓 (𝐶)|𝑞 ]𝑑𝑡 = |𝑓 (𝑏)|𝑞 ∫ |𝑡 − 𝜃|ℎ(𝑡)𝑑𝑡 + |𝑓 (𝐶)|𝑞 ∫ |𝑡 − 𝜃|ℎ(1 − 𝑡)𝑑𝑡 0
0
Authenticated |
[email protected] author's copy Download Date | 2/5/14 12:43 AM
0
(2.3)
(2.4)
İ. İşcan, Some new general integral inequalities
and
|
5
1
1 ∫ |𝑡 − 𝜃|𝑑𝑡 = 𝜃2 − 𝜃 + . 2
(2.5)
0
Thus, using (2.3)–(2.5) in (2.2), we obtain inequality (2.1). This completes the proof. Corollary 2.2. Under the assumptions of Theorem 2.1 with 𝑞 = 1, we have |𝑆𝑓 (𝜆, 𝜃, 𝑎, 𝑏)| ≤ (𝑏 − 𝑎){𝐴 2 (𝜃, ℎ)(𝜆2 |𝑓 (𝑎)| + (1 − 𝜆)2 |𝑓 (𝑏)|) + 𝐴 3 (𝜃, ℎ)(2𝜆2 − 2𝜆 + 1)|𝑓 (𝐶)|}. Corollary 2.3. Under the assumptions of Theorem 2.1 with 𝐼 ⊆ [0, ∞), ℎ(𝑡) = 𝑡𝑠 , 𝑠 ∈ (0, 1], we get the following inequality which is identical to inequality (1.5): 1− 1
1
|𝑆𝑓 (𝜆, 𝜃, 𝑎, 𝑏)| ≤ (𝑏 − 𝑎)𝐴 1 𝑞 (𝜃){𝜆2 [|𝑓 (𝑎)|𝑞 𝐴 2 (𝜃, ℎ) + |𝑓 (𝐶)|𝑞 𝐴 3 (𝜃, ℎ)𝑟] 𝑞 1
+ (1 − 𝜆)2 [|𝑓 (𝑏)|𝑞 𝐴 2 (𝜃, ℎ) + |𝑓 (𝐶)|𝑞 𝐴 3 (𝜃, ℎ)] 𝑞 }, where
(2.6)
1 𝐴 1 (𝜃) = 𝜃2 − 𝜃 + , 2 2𝜃𝑠+2 𝜃 1 𝐴 2 (𝜃, ℎ) = − + , (𝑠 + 1)(𝑠 + 2) 𝑠 + 1 𝑠 + 2 2(1 − 𝜃)𝑠+2 1−𝜃 1 𝐴 3 (𝜃, ℎ) = − + . (𝑠 + 1)(𝑠 + 2) 𝑠 + 1 𝑠 + 2
Corollary 2.4. Under the assumptions of Theorem 2.1 with ℎ(𝑡) = 1, we have 1
1
|𝑆𝑓 (𝜆, 𝜃, 𝑎, 𝑏)| ≤ (𝑏 − 𝑎)𝐴 1 (𝜃){𝜆2 [|𝑓 (𝑎)|𝑞 + |𝑓 (𝐶)|𝑞 ] 𝑞 + (1 − 𝜆)2 [|𝑓 (𝑏)|𝑞 + |𝑓 (𝐶)|𝑞 ] 𝑞 }. Remark 2.5. In Corollary 2.3 , if we take 𝜆 =
1 2
and 𝜃 = 23 , then we have the following Simpson-type inequality:
𝑏 𝑎+𝑏 1 1 [𝑓(𝑎) + 4𝑓( ) ∫ + 𝑓(𝑏)] − 𝑓(𝑥)𝑑𝑥 6 2 𝑏 − 𝑎 𝑎
≤
1 1 2 2 𝑎 + 𝑏 𝑞 𝑞 𝑏 − 𝑎 5 1− 𝑞 ( ) {(𝐴 2 ( , ℎ)|𝑓 (𝑎)|𝑞 + 𝐴 3 ( , ℎ)𝑓 ( ) ) 4 18 3 3 2
𝑎 + 𝑏 𝑞 𝑞1 2 2 𝑞 ) ) }, + (𝐴 2 ( , ℎ)|𝑓 (𝑏)| + 𝐴 3 ( , ℎ)𝑓 ( 3 3 2 where 2 𝐴 2 ( , ℎ) = 3 2 𝐴 3 ( , ℎ) = 3 Remark 2.6. In Corollary 2.3 , if we take 𝜆 =
1 2
2𝑠+3 + 3𝑠+1 (𝑠 − 1) , 3𝑠+2 (𝑠 + 1)(𝑠 + 2) 2 + 3𝑠+1 (2𝑠 + 1) . 3𝑠+2 (𝑠 + 1)(𝑠 + 2)
and 𝜃 = 1, then we have following midpoint inequality:
𝑏 1 𝑎 + 𝑏 𝑞 𝑞1 𝑞 1 2 𝑎 + 𝑏 𝑏 − 𝑎 𝑞 𝑓 ( 𝑓( ≤ ) ( ) {(|𝑓 ) ) ∫ − 𝑓(𝑥)𝑑𝑥 (𝑎)| + (𝑠 + 1) 2 𝑏−𝑎 8 (𝑠 + 1)(𝑠 + 2) 2 𝑎 1 𝑎 + 𝑏 𝑞 𝑞 ) ) }. + (|𝑓 (𝑏)|𝑞 + (𝑠 + 1)𝑓 ( 2
Remark 2.7. In Corollary 2.3 , if we take 𝜆 =
1 2
and 𝜃 = 0, then we get the following trapezoid inequality:
𝑏 1 𝑎 + 𝑏 𝑞 𝑞1 𝑞 1 2 𝑓(𝑎) + 𝑓(𝑏) 𝑏 − 𝑎 𝑞 𝑓 ( ≤ ( ) {((𝑠 ) ) ∫ − 𝑓(𝑥)𝑑𝑥 + 1)|𝑓 (𝑎)| + 2 𝑏 − 𝑎 8 (𝑠 + 1)(𝑠 + 2) 2 𝑎 1 𝑎 + 𝑏 𝑞 𝑞 ) ) }. + ((𝑠 + 1)|𝑓 (𝑏)|𝑞 + 𝑓 ( 2
Authenticated |
[email protected] author's copy Download Date | 2/5/14 12:43 AM
6 | İ. İşcan, Some new general integral inequalities Using Lemma 1.5, we shall give another result for convex functions as follows. Theorem 2.8. Let 𝑓 : 𝐼 ⊆ ℝ → ℝ be a differentiable mapping on 𝐼∘ such that 𝑓 ∈ 𝐿[𝑎, 𝑏], where 𝑎, 𝑏 ∈ 𝐼∘ with 𝑎 < 𝑏 and 𝜃, 𝜆 ∈ [0, 1]. If |𝑓 |𝑞 is ℎ-convex on [𝑎, 𝑏], 𝑞 > 1, then the following inequality holds: 1
1
1
𝑞 𝜃𝑝+1 + (1 − 𝜃)𝑝+1 𝑝 ) (∫ ℎ(𝑡)𝑑𝑡) |𝑆𝑓 (𝜆, 𝜃, 𝑎, 𝑏)| ≤ (𝑏 − 𝑎)( 𝑝+1
0
1
1
× [𝜆 (|𝑓 (𝑎)| + |𝑓 (𝐶)|𝑞 ) 𝑞 + (1 − 𝜆)2 (|𝑓 (𝑏)|𝑞 + |𝑓 (𝐶)|𝑞 ) 𝑞 ], 2
where 𝐶 = (1 − 𝜆)𝑎 + 𝜆𝑏 and
1 𝑝
+
1 𝑞
𝑞
(2.7)
= 1.
Proof. Suppose that 𝐶 = (1 − 𝜆)𝑎 + 𝜆𝑏. From Lemma 1.5 and by Hölder’s integral inequality, we have 1
1
|𝑆𝑓 (𝜆, 𝜃, 𝑎, 𝑏)| ≤ (𝑏 − 𝑎)[𝜆2 ∫ |𝑡 − 𝜃||𝑓 (𝑡𝑎 + (1 − 𝑡)𝐶)|𝑑𝑡 + (1 − 𝜆)2 ∫ |𝑡 − 𝜃||𝑓 (𝑡𝑏 + (1 − 𝑡)𝐶)|𝑑𝑡] 0
0 1
2
1
1 𝑝
𝑝
𝑞
1 𝑞
≤ (𝑏 − 𝑎){𝜆 (∫ |𝑡 − 𝜃| 𝑑𝑡) (∫ |𝑓 (𝑡𝑎 + (1 − 𝑡)𝐶)| 𝑑𝑡) 0
0 1
1
1 𝑝
1 𝑞
+ (1 − 𝜆)2 (∫ |𝑡 − 𝜃|𝑝 𝑑𝑡) (∫ |𝑓 (𝑡𝑏 + (1 − 𝑡)𝐶)|𝑞 𝑑𝑡) }.
(2.8)
0
0
Since |𝑓 |𝑞 is ℎ-convex on [𝑎, 𝑏], we know that for 𝑡 ∈ (0, 1) |𝑓 (𝑡𝑎 + 𝐶(1 − 𝑡))|𝑞 ≤ ℎ(𝑡)|𝑓 (𝑎)|𝑞 + ℎ(1 − 𝑡)|𝑓 (𝐶)|𝑞 and |𝑓 (𝑡𝑏 + 𝐶(1 − 𝑡))|𝑞 ≤ ℎ(𝑡)|𝑓 (𝑏)|𝑞 + ℎ(1 − 𝑡)|𝑓 (𝐶)|𝑞 . Hence, by simple computation, 1
1
𝑞
𝑞
𝑞
𝑞
(2.9)
∫(ℎ(𝑡)|𝑓 (𝑎)| + ℎ(1 − 𝑡)|𝑓 (𝐶)| )𝑑𝑡 ≤ (|𝑓 (𝑎)| + |𝑓 (𝐶)| ) ∫ ℎ(𝑡)𝑑𝑡, 0
0 1
1
𝑞
𝑞
𝑞
𝑞
(2.10)
∫(ℎ(𝑡)|𝑓 (𝑏)| + ℎ(1 − 𝑡)|𝑓 (𝐶)| )𝑑𝑡 ≤ (|𝑓 (𝑏)| + |𝑓 (𝐶)| ) ∫ ℎ(𝑡)𝑑𝑡 0
and
0 1
∫ |𝑡 − 𝜃|𝑝 𝑑𝑡 = 0
𝜃𝑝+1 + (1 − 𝜃)𝑝+1 . 𝑝+1
(2.11)
Thus, using (2.9)–(2.11) in (2.8), we obtain inequality (2.7). This completes the proof. Corollary 2.9. Under the assumptions of Theorem 2.8 with 𝐼 ⊆ [0, ∞), ℎ(𝑡) = 𝑡𝑠 , 𝑠 ∈ (0, 1], we have the following inequality which is identical to inequality (1.6): 1
1
𝑞 𝜃𝑝+1 + (1 − 𝜃)𝑝+1 𝑝 1 ) ( ) |𝑆𝑓 (𝜆, 𝜃, 𝑎, 𝑏)| ≤ (𝑏 − 𝑎)( 𝑝+1 𝑠+1 1
1
× [𝜆2 (|𝑓 (𝑎)|𝑞 + |𝑓 (𝐶)|𝑞 ) 𝑞 + (1 − 𝜆)2 (|𝑓 (𝑏)|𝑞 + |𝑓 (𝐶)|𝑞 ) 𝑞 ].
(2.12)
Corollary 2.10. Under the assumptions of Theorem 2.8 with ℎ(𝑡) = 1, we have 1
|𝑆𝑓 (𝜆, 𝜃, 𝑎, 𝑏)| ≤ (𝑏 − 𝑎)(
1 1 𝜃𝑝+1 + (1 − 𝜃)𝑝+1 𝑝 2 ) [𝜆 (|𝑓 (𝑎)|𝑞 + |𝑓 (𝐶)|𝑞 ) 𝑞 + (1 − 𝜆)2 (|𝑓 (𝑏)|𝑞 + |𝑓 (𝐶)|𝑞 ) 𝑞 ]. (2.13) 𝑝+1
Authenticated |
[email protected] author's copy Download Date | 2/5/14 12:43 AM
İ. İşcan, Some new general integral inequalities
Remark 2.11. In Corollary 2.9, if we take 𝜆 =
1 2
|
7
and 𝜃 = 23 , then we have the following Simpson-type inequality: 𝑏
1 [𝑓(𝑎) + 4𝑓( 𝑎 + 𝑏 ) + 𝑓(𝑏)] − 1 ∫ 𝑓(𝑥)𝑑𝑥 2 𝑏−𝑎 6 𝑎
𝑝+1
1 𝑝
𝑏−𝑎 1+2 ( ) {( ≤ 12 3(𝑝 + 1)
)|𝑞 |𝑓 ( 𝑎+𝑏 2
+ |𝑓 (𝑎)|𝑞
𝑠+1
1 𝑞
) +(
)|𝑞 + |𝑓 (𝑏)|𝑞 |𝑓 ( 𝑎+𝑏 2 𝑠+1
1 𝑞
) },
which is identical to inequality (1.9). Remark 2.12. In Corollary 2.9, if we take 𝜆 = ity:
1 2
and 𝜃 = 1, then we have the following midpoint-type inequal-
1 1 𝑏 1 )|𝑞 + |𝑓 (𝑎)|𝑞 𝑞 )|𝑞 + |𝑓 (𝑏)|𝑞 𝑞 |𝑓 ( 𝑎+𝑏 |𝑓 ( 𝑎+𝑏 𝑝 1 1 𝑎 + 𝑏 𝑏 − 𝑎 2 2 𝑓( )− ( ) {( ) +( ) }. ∫ 𝑓(𝑥)𝑑𝑥 ≤ 2 𝑏−𝑎 4 𝑝+1 𝑠+1 𝑠+1 𝑎
We note that by the inequality
𝑎 + 𝑏 𝑞 |𝑓 (𝑎)|𝑞 + |𝑓 (𝑏)|𝑞 ) ≤ 2𝑠−1 𝑓 ( 2 𝑠+1
we have 𝑏 1 2 1 𝑝 𝑞 1 1 1 𝑏−𝑎 𝑎 + 𝑏 𝑓( ≤ ( ) )( ) ( ) [((21−𝑠 + 𝑠 + 1)|𝑓 (𝑎)|𝑞 + 21−𝑠 |𝑓 (𝑏)|𝑞 ) 𝑞 ∫ 𝑓(𝑥)𝑑𝑥 − 2 𝑏−𝑎 4 𝑝+1 𝑠+1 𝑎 1
× ((21−𝑠 + 𝑠 + 1)|𝑓 (𝑏)|𝑞 + 21−𝑠 |𝑓 (𝑎)|𝑞 ) 𝑞 ], which is identical to inequality (1.8). Remark 2.13. In Corollary 2.9, if we take 𝜆 = ity:
1 2
and 𝜃 = 0, then we have the following trapezoid-type inequal-
1 1 𝑏 1 |𝑓 ( 𝑎+𝑏 )|𝑞 + |𝑓 (𝑎)|𝑞 𝑞 |𝑓 ( 𝑎+𝑏 )|𝑞 + |𝑓 (𝑏)|𝑞 𝑞 𝑝 1 1 𝑓(𝑎) + 𝑓(𝑏) 𝑏 − 𝑎 2 2 ( ) {( ) +( ) }. (2.14) ∫ 𝑓(𝑥)𝑑𝑥 ≤ − 2 𝑏−𝑎 4 𝑝+1 𝑠+1 𝑠+1 𝑎
We note that the obtained midpoint inequality (2.14) is better than inequality (1.10). Theorem 2.14. Let 𝑓 : 𝐼 ⊆ ℝ → ℝ be a differentiable mapping on 𝐼∘ such that 𝑓 ∈ 𝐿[𝑎, 𝑏], where 𝑎, 𝑏 ∈ 𝐼∘ with 𝑎 < 𝑏 and 𝛼, 𝜆 ∈ [0, 1]. If |𝑓 |𝑞 is ℎ-concave on [𝑎, 𝑏], 𝑞 > 1, then the following inequality holds: 1
1
𝑞 𝜃𝑝+1 + (1 − 𝜃)𝑝+1 𝑝 1 ( ) ) |𝑆𝑓 (𝜆, 𝜃, 𝑎, 𝑏)| ≤ (𝑏 − 𝑎)( 𝑝+1 2ℎ( 12 ) (2 − 𝜆)𝑎 + 𝜆𝑏 (1 − 𝜆)𝑎 + (1 + 𝜆)𝑏 ) + (1 − 𝜆)2 𝑓 ( )}, × {𝜆2 𝑓 ( 2 2
where 𝐶 = (1 − 𝜆)𝑎 + 𝜆𝑏 and
1 𝑝
+
1 𝑞
(2.15)
= 1.
Proof. Suppose that 𝐶 = (1 − 𝜆)𝑎 + 𝜆𝑏. We proceed similarly as in the proof Theorem 2.8. Since |𝑓 |𝑞 is ℎ-concave on [𝑎, 𝑏], 𝜆 ∈ (0, 1], from inequality (1.4) we get 𝐶
1
1 1 (2 − 𝜆)𝑎 + 𝜆𝑏 𝑞 𝑓 ( ) ∫ |𝑓 (𝑥)|𝑞 𝑑𝑥 = ∫ |𝑓 (𝑡𝑎 + (1 − 𝑡)𝐶)|𝑞 𝑑𝑡 ≤ 𝐶−𝑎 2 2ℎ( 21 ) 𝑎
(2.16)
0
and inequality (2.16) also holds for 𝜆 = 0. Similarly, for 𝜆 ∈ [0, 1), from inequality (1.4) we get 𝑏
1
1 1 (1 − 𝜆)𝑎 + (1 + 𝜆)𝑏 𝑞 𝑓 ( ) ∫ |𝑓 (𝑥)|𝑞 𝑑𝑥 = ∫ |𝑓 (𝑡𝑏 + (1 − 𝑡)𝐶)|𝑞 𝑑𝑡 ≤ 𝑏−𝐶 2 2ℎ( 21 ) 𝐶
(2.17)
0
and inequality (2.17) also holds for 𝜆 = 1. Thus using (2.11), (2.16) and (2.17) in (2.8), we obtain inequality (2.15). This completes the proof.
Authenticated |
[email protected] author's copy Download Date | 2/5/14 12:43 AM
8 | İ. İşcan, Some new general integral inequalities Corollary 2.15. Under the assumptions of Theorem 2.14 with 𝐼 ⊆ [0, ∞), ℎ(𝑡) = 𝑡𝑠 , 𝑠 ∈ (0, 1], we have 𝑏 1 (1 − 𝜃)(𝜆𝑓(𝑎) + (1 − 𝜆)𝑓(𝑏)) + 𝜃𝑓((1 − 𝜆)𝑎 + 𝜆𝑏) − ∫ 𝑓(𝑥)𝑑𝑥 𝑏 − 𝑎 𝑎 1
≤ (𝑏 − 𝑎)2
𝑠−1 𝑞
(1 − 𝜆)𝑎 + (1 + 𝜆)𝑏 𝜃𝑝+1 + (1 − 𝜃)𝑝+1 𝑝 2 (2 − 𝜆)𝑎 + 𝜆𝑏 ( ) {𝜆 𝑓 ( ) + (1 − 𝜆)2 𝑓 ( )}. 𝑝+1 2 2
Corollary 2.16. Under the assumptions of Theorem 2.14 with ℎ(𝑡) = 𝑡, we have 1
𝜃𝑝+1 + (1 − 𝜃)𝑝+1 𝑝 2 (2 − 𝜆)𝑎 + 𝜆𝑏 (1 − 𝜆)𝑎 + (1 + 𝜆)𝑏 ) {𝜆 𝑓 ( ) + (1 − 𝜆)2 𝑓 ( )}. |𝑆𝑓 (𝜆, 𝜃, 𝑎, 𝑏)| ≤ (𝑏 − 𝑎)( 𝑝+1 2 2 Corollary 2.17. Under the assumptions of Theorem 2.14 with ℎ(𝑡) = 1, we have 1
−1
|𝑆𝑓 (𝜆, 𝜃, 𝑎, 𝑏)| ≤ (𝑏 − 𝑎)2 𝑞 (
(1 − 𝜆)𝑎 + (1 + 𝜆)𝑏 𝜃𝑝+1 + (1 − 𝜃)𝑝+1 𝑝 2 (2 − 𝜆)𝑎 + 𝜆𝑏 ) {𝜆 𝑓 ( ) + (1 − 𝜆)2 𝑓 ( )}. 𝑝+1 2 2
Corollary 2.18. Under the assumptions of Theorem 2.14 with ℎ(𝑡) = 1𝑡 , 𝑡 ∈ (0, 1), we have 1
𝜃𝑝+1 + (1 − 𝜃)𝑝+1 𝑝 2 (2 − 𝜆)𝑎 + 𝜆𝑏 (1 − 𝜆)𝑎 + (1 + 𝜆)𝑏 ) {𝜆 𝑓 ( ) + (1 − 𝜆)2 𝑓 ( )}. |𝑆𝑓 (𝜆, 𝜃, 𝑎, 𝑏)| ≤ (𝑏 − 𝑎)4 ( 𝑝+1 2 2 − 𝑞1
Remark 2.19. In Corollary 2.15, if we take 𝜆 =
1 2
and 𝜃 = 0, then we have the following trapezoid inequality:
𝑏 1 1−𝑠 𝑝 1 1 1 𝑞 3𝑏 + 𝑎 3𝑎 + 𝑏 𝑏 − 𝑎 𝑓(𝑎) + 𝑓(𝑏) ≤ ( ) ( ) [𝑓 ( ) + 𝑓 ( )], ∫ 𝑓(𝑥)𝑑𝑥 − 2 𝑏−𝑎 4 𝑝+1 2 4 4 𝑎
which is identical to the inequality in [11, Theorem 8 (i)]. Remark 2.20. In Corollary 2.15, if we take 𝜆 =
1 2
and 𝜃 = 1, then we have the following midpoint inequality:
𝑏 1 1−𝑠 𝑝 1 1 1 𝑞 3𝑏 + 𝑎 3𝑎 + 𝑏 𝑏 − 𝑎 𝑎 + 𝑏 ≤ 𝑓( ) ( ) ) ) + 𝑓 ( )], ( [𝑓 ( ∫ − 𝑓(𝑥)𝑑𝑥 2 𝑏−𝑎 4 𝑝+1 2 4 4 𝑎
which is identical to the inequality in [11, Theorem 8 (ii)]. Remark 2.21. In Corollary 2.16, if we take 𝜆 =
1 2
and 𝜃 = 0, then we have the following trapezoid inequality:
𝑏 1 𝑝 1 1 3𝑏 + 𝑎 3𝑎 + 𝑏 𝑏 − 𝑎 𝑓(𝑎) + 𝑓(𝑏) ≤ ( ) [𝑓 ( ) + 𝑓 ( )], ∫ − 𝑓(𝑥)𝑑𝑥 2 𝑏−𝑎 4 𝑝+1 4 4
(2.18)
𝑎
which is identical to the inequality in [10, Theorem 2]. Remark 2.22. In Corollary 2.16, if we take 𝜆 =
1 2
and 𝜃 = 1, then we have the following trapezoid inequality:
𝑏 1 𝑝 1 1 3𝑏 + 𝑎 3𝑎 + 𝑏 𝑎 + 𝑏 𝑏 − 𝑎 𝑓( ≤ ) ( ) [𝑓 ( ) + 𝑓 ( )], ∫ − 𝑓(𝑥)𝑑𝑥 2 𝑏−𝑎 4 𝑝+1 4 4 𝑎
(2.19)
which is identical to the inequality in [1, Theorem 2.5]. Remark 2.23. In Corollary 2.16, since |𝑓 |𝑞 , 𝑞 > 1, is concave on [𝑎, 𝑏], using the power mean inequality, we have |𝑓 (𝜆𝑥 + (1 − 𝜆)𝑦)|𝑞 ≥ 𝜆|𝑓 (𝑥)|𝑞 + (1 − 𝜆)|𝑓 (𝑦)|𝑞 ≥ (𝜆|𝑓 (𝑥)| + (1 − 𝜆)|𝑓 (𝑦)|)𝑞 for all 𝑥, 𝑦 ∈ [𝑎, 𝑏] and 𝜆 ∈ [0, 1]. Hence |𝑓 (𝜆𝑥 + (1 − 𝜆)𝑦)| ≥ 𝜆|𝑓 (𝑥)| + (1 − 𝜆)|𝑓 (𝑦)|,
Authenticated |
[email protected] author's copy Download Date | 2/5/14 12:43 AM
İ. İşcan, Some new general integral inequalities
|
9
so |𝑓 | is also concave. Then by inequality (1.1), we have 3𝑏 + 𝑎 3𝑎 + 𝑏 𝑎 + 𝑏 𝑓 ( ) + 𝑓 ( ) ≤ 2𝑓 ( ). 4 4 2
(2.20)
Thus, using inequality (2.20) in (2.18) and (2.19), we get 𝑏 1 𝑝 1 1 𝑓(𝑎) + 𝑓(𝑏) 𝑏 − 𝑎 𝑎 + 𝑏 ≤ 𝑓 ( ( ) ), ∫ − 𝑓(𝑥)𝑑𝑥 2 𝑏−𝑎 2 𝑝 + 1 2 𝑎
𝑏 1 𝑝 1 1 𝑎 + 𝑏 𝑏 − 𝑎 𝑎 + 𝑏 𝑓( ≤ 𝑓 ( ) ( ) ), ∫ − 𝑓(𝑥)𝑑𝑥 2 𝑏−𝑎 2 𝑝 + 1 2 𝑎
where
1 𝑝
+
1 𝑞
= 1.
References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15]
M. W. Alomari, M. Darus and U. S. Kirmaci, Some inequalities of Hermite-Hadamard type for 𝑠-convex functions, Acta Math. Sci . Ser. B Engl. Ed. 31 (2011), no. 4, 1643–1652. M. Bombardelli and S. Varošanec, Properties of ℎ-convex functions related to the Hermite–Hadamard–Fejér inequalities, Comput. Math. Appl. 58 (2009), 1869–1877. W. W. Breckner, Stetigkeitsaussagen für eine Klasse verallgemeinerter konvexer Funktionen in topologischen linearen Räumen, Publ. Inst. Math. (Beograd) (N.S.) 23 (1978), 13–20. S. S. Dragomir and C. E. M. Pearce, Selected Topics on Hermite–Hadamard Inequalities and Applications, RGMIA Monographs, Victoria University, 2000. S. S. Dragomir, J. Pečarić and L. E. Persson, Some inequalities of Hadamard type, Soochow J. Math. 21 (1995), 335–341. E. K. Godunova and V. I. Levin, Neravenstva dlja funkcii širokogo klassa, soderžaščego vypuklye, monotonnye i neko-torye drugie vidy funkcii, in: Vyčislitel. Mat. i. Mat. Fiz. Mežvuzov. Sb. Nauč. Trudov, MGPI, Moskva (1985), 138–142. I. Iscan, A new generalization of some integral inequalities for (𝛼, 𝑚)-convex functions, Math. Sci. 7 (2013), Article 22. I. Iscan, New estimates on generalization of some integral inequalities for (𝛼, 𝑚)-convex functions, Contemp. Anal. Appl. Math. 1 (2013), no. 2, 253–264. I. Iscan, New estimates on generalization of some integral inequalities for 𝑠-convex functions and their applications, Int. J. Pure Appl. Math. 86 (2013), no. 4, 727–746. U. S. Kirmaci, M. K. Bakula, M. E. Özdemir and J. Pečarić, Hadamard-type inequalities for 𝑠-convex functions, Appl. Math. Comput. 193 (2007), 26–35. J. Park, Hermite–Hadamard-type inequalities for real 𝛼-star 𝑠-convex mappings, J. Appl. Math. Inform. 28 (2010), no. 5–6, 1507–1518. M. Z. Sarikaya and N. Aktan, On the generalization of some integral inequalities and their applications, Math. Comput. Modelling 54 (2011), 2175–2182. M. Z. Sarikaya, A. Saglam and H. Yıldırım, On some Hadamard–type inequalities for ℎ-convex functions, J. Math. Inequal. 2 (2008), no. 3, 335–341. M. Z. Sarikaya, E. Set and M. E. Özdemir, On new inequalities of Simpson’s type for 𝑠-convex functions, Comput. Math. Appl. 60 (2010), 2191–2199. S. Varošanec, On ℎ-convexity, J. Math. Anal. Appl. 326 (2007), 303–311.
Received September 3, 2013; revised January 10, 2014; accepted January 12, 2014.
Authenticated |
[email protected] author's copy Download Date | 2/5/14 12:43 AM
