Some Integral Inequalities for Convex Functions via Riemann-Liouville ...

Applied Mathematical Sciences, Vol. 9, 2015, no. 27, 1341 - 1353 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ams.2015.5159

Some Integral Inequalities for Convex Functions via Riemann-Liouville Integrals Jaekeun Park Department of Mathematics Hanseo University Seosan, Choongnam, 356-706, Korea c 2015 Jaekeun Park. This is an open access article distributed under the Copyright Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract In this paper, by setting up a generalized integral identity for differentiable functions via Riemann-Liouville fractional integrals, the author derive new estimates on generalization of Hermite-Hadamard and Ostrowski types inequaalities for functions whose derivatives in the absolute value at certain powers are convex.

Mathematics Subject Classification: 26A51, 26A33, 26D10 Keywords: Hermite-Hadamard type inequality, Simpson inequality, Ostrowski inequality, Riemann-Liouville fractional integrals, Hölder’s inequality, mean-power integral inequality

1

Introduction

Let f : I ⊆ R → R be a function defined on the interval I of real numbers. Then f is called to be convex on I if the following inequality f (ta + (1 − t)b) ≤ tf (a) + (1 − t)f (b) for all a, b ∈ I and t ∈ [0, 1]. There are many results associated with convex functions in the area of inequalities, but some of those is the classical HermiteHadamard inequality and Ostrowski inequality, respectively [10, 13]:

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Theorem 1.1. Let f : I ⊆ R → R be a convex function defined on the interval I of real numbers and a, b ∈ I with a < b. Then following double inequality holds: Z b a + b 1 f (a) + f (b) ≤ f (x)dx ≤ . (1) f 2 b−a a 2 Theorem 1.2. [11] Let f : I ⊆ R → R be a differentiable function on the interior I 0 of an interval I and a, b ∈ I 0 with a < b. If | f 0 (x) |≤ M, x ∈ [a, b], then the following inequality holds: Z b 1 M h (x − a)2 + (b − x)2 i . (2) f (t)dt ≤ f (x) − b−a a b−a 2 Definition 1. The beta function, also called the Euler integral of the first kind, is a special function defined by Z 1 β(x, y) = tx−1 (1 − t)y−1 dt, x, y > 0, 0

and Z

a

β(a, x, y) =

tx−1 (1 − t)y−1 dt, 0 < a < 1, x, y > 0,

0

is incomplete Beta function. Definition 2. A function f : [a, b] → (0, ∞) is said to be s-convex in the second sense if the following inequality f (ta + (1 − t)b) ≤ ts f (a) + (1 − t)s f (b) holds for all a, b ∈ I, t ∈ [0, 1], and s ∈ (0, 1]. We give some necessary definitions and mathematical preliminaries of fractional calculus theory which are used throughout this paper. Definition 3. Let f ∈ L([a, b]). The symbols Jaα+ f and Jbα− f denote the leftside and right-side Riemann-Liouville integrals of the order α and are defined by Z x 1 α (x − t)α−1 f (t)dt (0 ≤ a < x), Ja+ f (x) = Γ(α) a and Jbα− f (x)

1 = Γ(α)

Z

x

(t − x)α−1 f (t)dt,

(0 < x < b),

a

respectively, where Γ(α) is the Gamma function defined by Γ(α) = and Ja0+ f (x) = Jb0− f (x) = f (x).

R∞ 0

e−t tα−1 dt

Some integral inequalities

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In the caes of α = 1, the fractional integrals reduces to the classical integral. Recently, many authors have studied a number of inequalities by used the Riemann-Liouville fractional integrals, see [1-9,12,14-17] and the references cited therein. Especially, in [2, 12], Imdat I¸scan, Noor, and Awan proved a variant of Hermite-Hadamard and Ostrowski type inequalities which hold for the convex functions via Riemann-Liouville fractional integrals. Theorem 1.3. Let f : [a, b] → R be twice differentiable function on (a, b) with a < b. If f 00 ∈ L([a, b]) and | f 00 | is convex on [a, b], then we have the following inequality for fractional integrals: 2α−1 Γ(α + 1) n o a + b α α J ) f (a) + J f (b) − f ( ( a+b )− )+ ( a+b 2 2 (b − a)α 2 (b − a)2 n α+3 1 Γ(α + 3) o 00 ≤ 4 2 β( , α + 1, 2) + [| f (a) | + | f 00 (b) |]. 2 (α + 1) 2 Γ(α + 4) Theorem 1.4. Let f : [a, b] → R be twice differentiable function on (a, b) with a < b. If f 00 ∈ L([a, b]) and | f 00 | is convex on [a, b], then we have the following inequality for fractional integrals: 2α−1 Γ(α + 1) n o a + b α α J f (a) + J f (b) − f ( ) ( a+b )− ( a+b )+ 2 2 (b − a)α 2 p1 h 1 (b − a)2 1 ≤ 3+ 2 3 | f 00 (a) |q + | f 00 (b) |q q 2 q (α + 1) p(α + 1) + 1 1 i + | f 00 (a) |q +3 | f 00 (b) |q q .

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In this paper, we give some generalized inequalities connected with HermiteHadamard-like type for differentiable functions whose derivatives in the absolute value are convex via fractional integrals.

2

Lemmas

Now we turn our attention to establish integral inequalities of HermiteHadamard and Ostrowski inequality type for convex functions via RiemannLiouville fractional integrals, we need the lemmas below: Lemma 1. Let f : I ⊆ R → R be a differentiable function on the interior I 0 of an interval I such that f 0 ∈ L([a, b]), where a, b ∈ I with a < b. Then for

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any λ, µ ∈ R and x ∈ [a, b] the following identity holds: I(f ; α; λ, µ; x) λ 1 − λ µ 1 − µ ≡ + f (x) + f (a) + f (b) x−a b−x x−a b−x o n 1 1 α α f (a) + f (b) − Γ(α + 1) J J − + (x − a)α+1 x (b − x)α+1 x Z 1 Z 1 α 0 (µ − tα )f 0 tx + (1 − t)b dt. (4) (t − 1 + λ)f tx + (1 − t)a dt + = 0

0

Proof. Integrating by parts and changing variable of definite integral, we have: Z 1 (tα − 1 + λ)f 0 tx + (1 − t)a dt 0

=λ

f (x) f (a) Γ(α + 1) α + (1 − λ) − J − f (a). x−a x − a (x − a)α+1 x

Similarly, we have Z

1

(µ − tα )f 0 tx + (1 − t)b dt

0

= (1 − µ)λ

f (x) f (b) Γ(α + 1) α +µ − J + f (b). b−x b − x (b − x)α+1 x

Adding these two equalities leads to Lemma 1. Lemma 2. For 0 ≤ ξ ≤ 1, one has Z (a)

1

ξ − tα q dt ≡ δ1 (α, ξ, q)

0 1 o ξ q+ α n 1 1 1 = β( , 1 + q) + β(−q − , 1 + q) − β(ξ, −q − , 1 + q) , α α α Z 1 α ξ − tα q tdt ≡ δ2 (α, ξ, q) (b)

0 2 o ξ q+ α n 2 2 2 = β( , 1 + q) + β(−q − , 1 + q) − β(ξ, −q − , 1 + q) . (5) α α α α

Proof. These equalities follows from a straightfoward computation of definite integrals.

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3

Some inequalities of Hermite-Hadamard and Ostrowski type

Now we turn our attention to establish new integral inequalities of HermiteHadamard and Ostrowski type for convex functions via fractional integrals. Theorem 3.1. Let f : I ⊆ R → R be a differentiable function on the interior I 0 of an interval I and f 0 ∈ L([a, b]), where a, b ∈ I with a < b and λ, µ ∈ [0, 1]. If |f 0 |q is convex on [a, b] for q ≥ 1 with p1 + 1q = 1, then the following inequality holds: If (f ; α; λ, µ; x) 1

≤ δ1p (α, 1 − λ, 1) n 0 q 0 q o 1q × δ2 (α, 1 − λ, 1) f (x) + δ1 (α, 1 − λ, 1) − δ2 (α, 1 − λ, 1) f (a) n 1 q o 1q q + δ1p (α, µ, 1) δ2 (α, µ, 1) f 0 (x) + δ1 (α, µ, 1) − δ2 (α, µ, 1) f 0 (b) . Proof. Suppose that q > 1. From Lemma 1, the convexity of | f 0 |q on [a, b], and the noted power-mean integral inequality, we have If (f ; α; λ, µ; x) Z 1 α t − 1 + λ f 0 tx + (1 − t)a dt ≤ 0 Z 1 µ − tα f 0 tx + (1 − t)b dt + 0 Z Z 1 p1 1 α 1 α t − 1 + λ f 0 tx + (1 − t)a q dt q t − 1 + λ dt ≤ 0 0 p1 Z 1 Z 1 1 µ − tα dt µ − tα dt f 0 tx + (1 − t)b q dt q . + (6) 0

0

In virtue of Lemma 2, a direct calculation yields Z 1 α t − 1 + λ f 0 tx + (1 − t)a q dt 0 Z 1 α t − 1 + λ t f 0 x q + (1 − t) f 0 (a) q dt ≤ 0 q q = δ2 (α, 1 − λ, 1) f 0 (x) + δ1 (α, 1 − λ, 1) − δ2 (α, 1 − λ, 1) f 0 (a) , (7) and Z

1

µ − tα f 0 tx + (1 − t)b q dt 0 q q ≤ δ2 (α, µ, 1) f 0 (x) + δ1 (α, µ, 1) − δ2 (α, µ, 1) f 0 (b) .

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By substituting the above inequalities (7) and (8) into (6), we get the desired result for q > 1. Suppose that q = 1. From Lemma 1 and 2 it follows that If (f ; α; λ, µ; x) Z 1 α t − 1 + λ f 0 tx + (1 − t)a dt ≤ 0 1

Z + Z

µ − tα f 0 tx + (1 − t)b dt

0 1

α t − 1 + λ t f 0 x + (1 − t) f 0 (a) dt

≤ 0

Z

1

µ − tα t f 0 x + (1 − t) f 0 (b) dt 0 = δ2 (α, 1 − λ, 1) f 0 (x) + δ1 (α, 1 − λ, 1) − δ2 (α, 1 − λ, 1) f 0 (a) + δ2 (α, µ, 1) f 0 (x) + δ1 (α, µ, 1) − δ2 (α, µ, 1) f 0 (b) , +

which completes the proof. If taking x =

a+b 2

in Theorem 3.1, we derive the following corollary:

Corollary 3.1. Let f : I ⊆ R → R be a differentiable function on the interior I 0 of an interval I and f 0 ∈ L([a, b]), where a, b ∈ I with a < b and λ, µ ∈ [0, 1]. If |f 0 |q is convex on [a, b] for q ≥ 1 with p1 + 1q = 1, then the following inequality holds: a + b ) If (f ; α; λ, µ; 2 a+b b − a ) = (1 − λ)f (a) + µf (b) + (1 − µ + λ)f ( 2 2 o n 2 α − Γ(α + 1) J(αa+b )− f (a) + J(αa+b )+ f (b) 2 2 b−a n 1 a + b q p ≤ δ1 (α, 1 − λ, 1) δ2 (α, 1 − λ, 1) f 0 ( ) 2 q o 1q + δ1 (α, 1 − λ, 1) − δ2 (α, 1 − λ, 1) f 0 (a) n 1 a + b q + δ1p (α, µ, 1) δ2 (α, µ, 1) f 0 ( ) 2 0 q o 1q + δ1 (α, µ, 1) − δ2 (α, µ, 1) f (b) .

If taking λ = µ in Theorem 3.1, we derive the following corollary:

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Corollary 3.2. Let f : I ⊆ R → R be a differentiable function on the interior I 0 of an interval I and f 0 ∈ L([a, b]), where a, b ∈ I with a < b and λ ∈ [0, 1]. If |f 0 |q is convex on [a, b] for q ≥ 1 with p1 + 1q = 1, then the following inequality holds: If (f ; α; λ, λ; x) 1

≤ δ1p (α, 1 − λ, 1) n 0 q 0 q o 1q × δ2 (α, 1 − λ, 1) f (x) + δ1 (α, 1 − λ, 1) − δ2 (α, 1 − λ, 1) f (a) n 1 0 q 0 q o 1q p + δ1 (α, λ, 1) δ2 (α, λ, 1) f (x) + δ1 (α, λ, 1) − δ2 (α, λ, 1) f (b) . Corollary 3.3. Let f : I ⊆ R → R be a differentiable function on the interior I 0 of an interval I and f 0 ∈ L([a, b]), where a, b ∈ I with a < b. If |f 0 |q is convex on [a, b] for q ≥ 1 with p1 + 1q = 1, then the following inequality holds: 1 1 If (f ; α; , ; x) 2 2 1 q o 1q q 1 n 1 1 1 ≤ δ1p (α, , 1) δ2 (α, , 1) f 0 (x) + δ1 (α, , 1) − δ2 (α, , 1) f 0 (a) 2 2 2 2 n 1 1 1 1 0 q o 1q 1 q p 0 + δ1 (α, , 1) δ2 (α, , 1) f (x) + δ1 (α, , 1) − δ2 (α, , 1) f (b) , 2 2 2 2 where 1 n 1 α δ1 (α, , 1) = 2 1+α 1 1 n δ2 (α, , 1) = α 2 2+α

1 α1 1 − α o , + 2 2 1 1+ α2 2 − α o + . 2 4

Corollary 3.4. Let f : I ⊆ R → R be a differentiable function on the interior I 0 of an interval I and f 0 ∈ L([a, b]), where a, b ∈ I with a < b. If |f 0 |q is convex on [a, b] for q ≥ 1 with p1 + 1q = 1, then the following inequality holds: 2 2 If (f ; α; , ; x) 3 3 1 1 n 1 0 q 1 1 0 q o 1q p ≤ δ1 (α, , 1) δ2 (α, , 1) f (x) + δ1 (α, , 1) − δ2 (α, , 1) f (a) 3 3 3 3 n 1 q o 1q 2 2 2 2 q + δ1p (α, , 1) δ2 (α, , 1) f 0 (x) + δ1 (α, , 1) − δ2 (α, , 1) f 0 (b) , 3 3 3 3

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where 1 δ1 (α, , 1) = 3 2 δ1 (α, , 1) = 3 1 δ2 (α, , 1) = 3 2 δ2 (α, , 1) = 3

α n 1 1+ α1 2 − α o 2α , + 1+α 3 3 α n 2 1+ α1 1 − 2α o 2α + , 1+α 3 3 1 n 1 1+ α2 4 − α o α + , 2+α 3 6 1 n 2 1+ α2 1 − α o α + . 2+α 3 3

Corollary 3.5. Let f : I ⊆ R → R be a differentiable function on the interior I 0 of an interval I and f 0 ∈ L([a, b]), where a, b ∈ I with a < b. If |f 0 |q is convex on [a, b] for q ≥ 1 with p1 + 1q = 1, then the following inequality holds: 1 1 I (f ; α; , ; x) f 3 3 1 2 n 2 2 0 q 2 0 q o 1q p ≤ δ1 (α, , 1) δ2 (α, , 1) f (x) + δ1 (α, , 1) − δ2 (α, , 1) f (a) 3 3 3 3 n 1 q o 1q 1 1 1 1 q + δ1p (α, , 1) δ2 (α, , 1) f 0 (x) + δ1 (α, , 1) − δ2 (α, , 1) f 0 (b) , 3 3 3 3 where δ1 and δ2 are defined as in Corollary 3.4.

Theorem 3.2. Let f : I ⊆ R → R be a differentiable function on the interior I 0 of an interval I and f 0 ∈ L([a, b]), where a, b ∈ I with a < b and λ, µ ∈ [0, 1]. If |f 0 |q is convex on [a, b] for q ≥ 1 with p1 + 1q = 1, then the following inequality holds: I (f ; α; λ, µ; x) f n 0 q 0 q o 1q ≤ δ2 (α, 1 − λ, q) f (x) + δ1 (α, 1 − λ, q) − δ2 (α, 1 − λ, q) f (a) n q q o 1q + δ2 (α, µ, q) f 0 (x) + δ1 (α, µ, q) − δ2 (α, µ, q) f 0 (b) .

Proof. Suppose that q > 1. By Lemma 1, the convexity of | f 0 |q on [a, b],


and Hölder’s integral inequality, it follows that If (f ; α; λ, µ; x) Z 1 α t − 1 + λ f 0 tx + (1 − t)a dt ≤ 0 Z 1 µ − tα f 0 tx + (1 − t)b dt + 0 Z 1 p1 Z 1 1 tα − 1 + λ q f 0 tx + (1 − t)a q dt q ≤ dt 0 0 Z 1 p1 Z 1 1 µ − tα q f 0 tx + (1 − t)b q dt q + dt 0 0 Z 1 q 1q α q 0 = | t − 1 + λ | | f tx + (1 − t)a | dt 0 Z 1 q 1q α q 0 | µ − t | | f tx + (1 − t)b | dt . +

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(9)

0

In virtue of Lemma 2, a direct calculation yields Z 1 α t − 1 + λ q f 0 tx + (1 − t)a q dt 0 Z 1 α t − 1 + λ q t f 0 x q + (1 − t) f 0 (a) q dt ≤ 0

q q = δ2 (α, 1 − λ, q) f 0 (x) + δ1 (α, 1 − λ, q) − δ2 (α, 1 − λ, q) f 0 (a) , (10) and Z

1

µ − tα q f 0 tx + (1 − t)b q dt 0 q q ≤ δ2 (α, µ, q) f 0 (x) + δ1 (α, µ, q) − δ2 (α, µ, q) f 0 (b) .

(11)

By substituting the above inequalities (10) and (11) into (9), we get the desired result for q > 1. For q = 1, from Lemma 1 and 2 it follows that If (f ; α; λ, µ) Z 1 α t − 1 + λ t f 0 x + (1 − t) f 0 (a) dt ≤ 0 Z 1 µ − tα t f 0 x + (1 − t) f 0 (b) dt + 0 = δ2 (α, 1 − λ, 1) f 0 (x) + δ1 (α, 1 − λ, 1) − δ2 (α, 1 − λ, 1) f 0 (a) + δ2 (α, µ, 1) f 0 (x) + δ1 (α, µ, 1) − δ2 (α, µ, 1) f 0 (b) , which completes the proof.

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Corollary 3.6. Let f : I ⊆ R → R be a differentiable function on the interior I 0 of an interval I and f 0 ∈ L([a, b]), where a, b ∈ I with a < b and λ, µ ∈ [0, 1]. If |f 0 |q is convex on [a, b] for q ≥ 1 with p1 + 1q = 1, then the following inequality holds: a + b ) I (f ; α; λ, µ; f 2 b − a a+b = ) (1 − λ)f (a) + µf (b) + (1 − µ + λ)f ( 2 2 n o 2 α Γ(α + 1) J(αa+b )− f (a) + J(αa+b )+ f (b) − 2 2 b−a n a + b q ≤ δ2 (α, 1 − λ, q) f 0 ( ) 2 q o 1q + δ1 (α, 1 − λ, q) − δ2 (α, 1 − λ, q) f 0 (a) n a + b q + δ2 (α, µ, q) f 0 ( ) 2 0 q o 1q . + δ1 (α, µ, q) − δ2 (α, µ, q) f (b) If taking λ = µ in Theorem 3.1, we derive the following corollary: Corollary 3.7. Let f : I ⊆ R → R be a differentiable function on the interior I 0 of an interval I and f 0 ∈ L([a, b]), where a, b ∈ I with a < b and λ ∈ [0, 1]. If |f 0 |q is convex on [a, b] for q ≥ 1 with p1 + 1q = 1, then the following inequality holds: If (f ; α; λ, λ; x) n 0 q 0 q o 1q ≤ δ2 (α, 1 − λ, q) f (x) + δ1 (α, 1 − λ, q) − δ2 (α, 1 − λ, q) f (a) n q q o 1q + δ2 (α, λ, q) f 0 (x) + δ1 (α, λ, q) − δ2 (α, λ, q) f 0 (b) . Corollary 3.8. Let f : I ⊆ R → R be a differentiable function on the interior I 0 of an interval I and f 0 ∈ L([a, b]), where a, b ∈ I with a < b. If |f 0 |q is convex on [a, b] for q ≥ 1 with p1 + 1q = 1, then the following inequality holds: 1 1 If (f ; α; , ; x) 2 2 n 1 0 q 1 1 0 q o 1q ≤ δ2 (α, , q) f (x) + δ1 (α, , q) − δ2 (α, , q) f (a) 2 2 2 n q o 1q 1 0 q 1 1 + δ2 (α, , q) f (x) + δ1 (α, , q) − δ2 (α, , q) f 0 (b) . 2 2 2


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Corollary 3.9. Let f : I ⊆ R → R be a differentiable function on the interior I 0 of an interval I and f 0 ∈ L([a, b]), where a, b ∈ I with a < b. If |f 0 |q is convex on [a, b] for q ≥ 1 with p1 + 1q = 1, then the following inequality holds: 2 2 If (f ; α; , ; x) 3 3 n q q o 1q 1 1 1 ≤ δ2 (α, , q) f 0 (x) + δ1 (α, , q) − δ2 (α, , q) f 0 (a) 3 3 3 n 2 0 q 2 2 0 q o 1q + δ2 (α, , q) f (x) + δ1 (α, , q) − δ2 (α, , q) f (b) . 3 3 3 Theorem 3.3. Let f : I ⊆ R → R be a differentiable function on the interior I 0 of an interval I and f 0 ∈ L([a, b]), where a, b ∈ I with a < b and λ, µ ∈ [0, 1]. If |f 0 |q is convex on [a, b] for q > 1 with p1 + 1q = 1, then the following inequality holds: If (f ; α; λ, µ; x) n f 0 (x) q + f 0 (a) q o 1q 1 p ≤ δ1 (α, 1 − λ, p) 2 n f 0 (x) q + f 0 (b) q o 1q 1 + δ1p (α, µ, p) . 2 Proof. Suppose that q > 1. From Lemma 1, the convexity of | f 0 |q on [a, b], and the Hölder’s integral inequality, we have If (f ; α; λ, µ; x) Z Z 1 p p1 1 0 1 α f tx + (1 − t)a q dt q t − 1 + λ dt ≤ 0 0 Z 1 p1 Z 1 1 µ − tα p dt f 0 tx + (1 − t)b q dt q + 0 0 Z 1 1 q 0 q 1q p 0 t f (x) + (1 − t) f (a) dt ≤ δ1 (α, 1 − λ, p) 0 Z 1 1 q 0 q 1q p 0 t f (x) + (1 − t) f (b) dt + δ1 (α, µ, p) 0 f 0 (x) q + f 0 (a) q 1q 1 ≤ δ1p (α, 1 − λ, p) 2 f 0 (x) q + f 0 (b) q 1q 1 + δ1p (α, µ, p) , 2 which completes the proof.

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