Fractional Hermite-Hadamard Inequalities 9783110523621, 9783110522204

Citation preview

JinRong Wang, Michal Fečkan Fractional Hermite-Hadamard Inequalities

Fractional Calculus in Applied Sciences and Engineering

|

Editor-in Chief Changpin Li Editorial Board Virginia Kiryakova Francesco Mainardi Dragan Spasic Bruce Ian Henry YangQuan Chen

Volume 5

JinRong Wang, Michal Fečkan

Fractional Hermite-Hadamard Inequalities |

Mathematics Subject Classification 2010 26A33, 26A51, 26D15 Authors Prof. Dr JinRong Wang Department of Mathematics Guizhou University Guizhou 550025 P.R. China [email protected] Prof. Dr Michal Fečkan Department of Mathematical Analysis and Numerical Mathematics Comenius University in Bratislava 842 48 Bratislava Slovak Republic [email protected]

ISBN 978-3-11-052220-4 e-ISBN (PDF) 978-3-11-052362-1 e-ISBN (EPUB) 978-3-11-052244-0 ISSN 2509-7210 Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.dnb.de. © 2018 Walter de Gruyter GmbH, Berlin/Boston Typesetting: VTeX UAB, Lithuania Printing and binding: CPI books GmbH, Leck Cover image: naddi/iStock/thinkstock www.degruyter.com

|

To our beloved families

Acknowledgment JinRong Wang acknowledges the support by the National Natural Science Foundation of China (No. 11661016) and Training Object of High Level and Innovative Talents of Guizhou Province [(2016)4006]. Michal Fečkan acknowledges the support by the Slovak Research and Development Agency under the contract No. APVV-14-0378 and by the Slovak Grant Agency VEGA No. 1/0078/17.

https://doi.org/10.1515/9783110523621-201

Preface It was a pleasure for us when Professor Changpin Li invited us to write a monograph in his series Fractional Calculus in Applied Sciences and Engineering published by De Gruyter. Fractional Hermite-Hadamard type inequality, an extension of classical HermiteHadamard type inequality, which provides lower and upper estimations for the fractional integral average of any convex function, generalizations defined on a compact interval, involving fractional order and midpoint and endpoints of the domain. This book is devoted to Hermite-Hadamard type inequalities involving Riemann-Liouville and Hadamard fractional integrals not limited to integer integrals, which have not yet been presented in any other book. The broad variety of achieved results with rigorous proofs and many applications to special means makes this book unique. JinRong Wang and Michal Fečkan Guiyang, China and Bratislava, Slovakia November 2017

https://doi.org/10.1515/9783110523621-202

Contents Acknowledgment | VII Preface | IX 1 1.1 1.2 1.3

Introduction | 1 Fractional calculus via application and computation | 1 Motivation of fractional Hermite-Hadamard’s inequality | 6 Main contents | 8

2 2.1 2.2 2.3 2.4 2.5

Preliminaries | 11 Definitions of special functions and fractional integrals | 11 Definitions of convex functions | 13 Singular integrals via series | 16 Elementary inequalities | 18 Special mean | 19

3 3.1 3.2

Fractional integral identities | 21 Identities involving Riemann-Liouville fractional integrals | 21 Identities involving Hadamard fractional integrals | 45

4

Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals | 65 Inequalities via convex functions | 65 Main results | 65 Applications to some special means | 79 Inequalities via r-convex functions | 80 Main results | 80 Applications | 114 Inequalities via s-convex functions | 119 Main results | 119 Applications to some special means | 135 Inequalities via m-convex functions | 138 Inequalities via (s, m)-convex functions | 154 Main results | 154 Applications to some special means (please disconnect) | 171 Inequalities via preinvex convex functions | 171 Right-hand side inequalities via preinvex convex functions | 171 Left-hand side inequalities via Preinvex convex functions | 177 Inequalities via (β, m)-geometrically convex functions | 181

4.1 4.1.1 4.1.2 4.2 4.2.1 4.2.2 4.3 4.3.1 4.3.2 4.4 4.5 4.5.1 4.5.2 4.6 4.6.1 4.6.2 4.7

XII | Contents 4.8 4.8.1 4.8.2 4.9 4.10 4.10.1 4.10.2 4.11 4.11.1 4.11.2 4.12 5 5.1 5.1.1 5.1.2 5.1.3 5.1.4 5.2 5.3 5.4 5.4.1 5.4.2 5.5 5.5.1 5.5.2

Inequalities via geometrical-arithmetically s-convex functions | 185 Main results | 185 Applications to some special means | 213 Inequalities via (α, m)-logarithmically convex functions | 215 Inequalities via s-Godunova-Levin functions | 240 Main results | 240 Applications to some special means | 246 Inequalities via (s, m)-Godunova-Levin functions | 247 Main results | 247 Applications to some special means | 249 Inequalities via AG(log)-convex functions | 249 Hermite-Hadamard inequalities involving Hadamard fractional integrals | 253 Inequalities via convex functions | 253 Hermite-Hadamard’s inequalities | 253 Applications to some special means | 272 Hadamard fractional Fejér inequalities | 280 Further results | 290 Inequalities via s-e-condition functions | 297 Inequalities via geometric-geometric coordinated convex function | 308 Inequalities via geometric-geometric-convex functions | 320 Main results | 320 Applications to special means | 349 Inequalities via geometric-arithmetic-convex functions | 355 Main results | 355 Applications to special means | 359

Bibliography | 361 About the authors | 373 Index | 375

1 Introduction 1.1 Fractional calculus via application and computation Fractional calculus was introduced at the end of the nineteenth century by Liouville and Riemann, but the concept of non-integer calculus, as a generalization of the traditional integer order calculus, was mentioned already in 1695 by Leibnitz and L’Hospital. Thereafter, many famous mathematicians developed this area, for example, Euler, Lagrange, Laplace, Fourier, Abel, Weyl and Riesz. In this subsection, we would like to use two examples to explain why we need to employ fractional calculus. The first example will show that how fractional calculus helps to solve Abel’s mechanical problem. The second one will show how fractional calculus helps to make uniform Hooke’s and Newton’s laws. Example 1. Fractional calculus helps to solve Abel’s mechanical problem [113, 144]. In 1823, Abel first solved an integral equation by using fractional integral operations, which directly explain that fractional calculus arises in the formulation of physical problems, as put forward by Liouville, and made the first major study of fractional calculus that applied his definitions to the potential-theory problem. In rectangular coordinates, we place a thin wire C in the first quadrant of the vertical plane and a frictionless bead slides along the wire under the action of gravity. Assume that the initial velocity of the bead is zero. Now we are ready to formulate a generalized version of the tautochrone problem, namely, Abel’s mechanical problem: Finding the shape of the tautochrone curve C for which the time of descent T(y) from point P(x, y) to the origin point O(0, 0) is independent of the starting point. Abel’s solution begins with the principle of conservation of energy, since the particle is frictionless, and thus loses no energy to heat, so its kinetic energy at any point is exactly equal to the difference in potential energy from its starting point. Note that the following fact: d2 s = −g cos α, dt 2

(1.1)

dη = cos α, ds

(1.2)

and

where we denote s as the arc length measured along C from point O to arbitrary point Q on tautochrone curve C, α as the angle of inclination, and g as the gravitational constant. Further, we substitute (1.2) into (1.1) to derive that dη dη dt d2 s = −g = −g . ds dt ds dt 2 https://doi.org/10.1515/9783110523621-001

(1.3)

2 | 1 Introduction Integrating both sides of (1.3) for

ds dt

via initial conditions, we have 2

(

ds ) = −gη + 2gy. dt

(1.4)

(1.4) can also be derived by means of the principle of conservation of energy due to the gravitational potential energy gained in falling from an initial height y to a height η, as mg(y − η). Although the following procedure seems a very standard exercise, the details will explain why fractional calculus is a powerful tool for solving the tautochrone problem and how to apply fractional calculus to find the solution and curve. In fact, Abel’s work possessed very great mathematical significance and advanced the development of fractional calculus. Then, we rewrite (1.4) as ds = −√2g(y − η), dt which implies that dt = −

1 ds. √2g(y − η)

(1.5)

Integrating both sides of (1.5) from P to O, we derive 0

1 1 T(y) = − ds, ∫ −η √y √2g

(1.6)

y

where s := h(η) depends on the shape of the curve C and h󸀠 (η) = rewritten as

ds . Thus, (1.6) can be dη

0

T(y) = −

1 1 dh󸀠 (η) , ∫ √2g √y − η dη

(1.7)

y

which also can be written as the following formula involving Liouville fractional integral √2gT(y) Γ( 21 )

1 2

󸀠

= I0,y h (y) :=

1

Γ( 21 )

y

∫ 0

h󸀠 (y)

1

(y − η)1− 2

dη.

1 -order 2

Riemann-

(1.8)

Recalling Abel’s method, one can apply 21 -order Riemann-Liouville fractional derivative to both sides of (1.8), and then one can obtain 1

2 D0,y

√2gT(y) Γ( 21 )

= h󸀠 (y).

(1.9)

1.1 Fractional calculus via application and computation

| 3

If h󸀠 and T are of the class described in [113, Theorem 3, p. 105], then one can express (1.9) as 1 √2gT(y) ds , = h󸀠 (y) = dy √y Γ( 21 )

(1.10)

which implies that we do find the solution of (1.8). If T(y) = T, then one can proceed with Abel’s method to obtain the parametric equation of C. Let’s give more details. Note that 2

ds dx = sec θ = √1 + tan2 θ = √1 + ( ) dy dy

(1.11)

Set Ψ(y) = h󸀠 (y). Equating (1.10) and (1.11), we have 2

x 󸀠 = √(Ψ(y)) − 1 which yields t

x = ∫√ 0 2gT 2 π2

Substituting η =

2gT 2 − 1dη. π2η

(1.12)

sin2 ζ to (1.12), then one can obtain

z

x=

4gT 2 2gT 2 1 gT 2 1 (z + sin 2z) = 2 (2z + sin 2z), ∫ cos2 ζdζ = 2 2 2 2 π π π 0

and z = arcsin √

yπ 2 2gT 2

⇐⇒

y=

2gT 2 gT 2 2 sin z = (1 − cos 2z) π2 π2

Summarizing, the target curve C satisfies the parametric equations x=

gT 2 (v + sin v), π2

y=

gT 2 (1 − sin v), π2

v = 2z,

(1.13)

which implies that C is a cycloid. More precisely, C is the locus of the point O on a circle 2 2 of radius gT rotating without slippage on the line y = 2gT . π2 π2 Example 2. Fractional calculus help to unify Hooke’s and Newton’s laws [35]. In mechanics, the classical Hooke’s and Newton’s laws describe the behavior of certain materials under the influence of external forces. However, Nutting [127, 128]

4 | 1 Introduction and Blair and Reiner [22] find that it is not uncommon to encounter so-called viscoelastic materials that exhibit a behavior somewhere between the pure viscous liquid and the pure elastic solid. Let the function F(t) denotes the stress, the function E(t) denotes the strain, the constant ρ denotes the viscosity of the material and the constant σ denotes the modulus of elasticity of the material. Hooke’s law for the stress-strain relationship for elastic solids is given by F(t) = σE(t) := σD00,t E(t).

(1.14)

Newton’s law for viscous liquids is given by F(t) = ρE 󸀠 (t) := ρD10,t E(t).

(1.15)

Both Hooke’s law (1.14) and Newton’s law (1.15) of viscosity are constitutive equations that describe how a fluid resists attempts to move through it. The key is how to measure the constants ρ and σ, which is very difficult and even impossible sometimes. Thus, (1.14) and (1.15) hold true for some liquids and elastic solids and fails for others. Therefore, they are not considered as uniformly fundamental laws for viscoelastic materials that exist between pure viscous liquids and pure elastic solids. Next, we make use of an example to illustrate that fractional derivatives can be used to consolidate (1.14) and (1.15) to deal with a new class of modelling involving viscoelastic materials. For simplicity, we suppose that E(t) = t for t ∈ [0, 1]. Then (1.14) becomes F(t) = σE(t) := σt

(1.16)

F(t) = ρ.

(1.17)

and (1.15) becomes to

Motivated by the same idea as in [113, Theorem 3, p. 105], we can consolidate (1.16) and (1.17) to become a new equation, such as F(t) = cj t −j E(t).

(1.18)

If we set j = 0, 1 and c0 = σ and c1 = ρ, then (1.18) can be reduced to Newton’s law and Hooke’s law respectively. Next, we observe (1.18) involving (1.16) and (1.17), moreover, the constant j may be a noninteger. In this case, we consider j as constant for a viscoelastic material. For example, set j = 21 and cj = j = 21 , then (1.18) becomes F(t) = 1

1 1 21 := D0,t 1, 2 2√t

(1.19)

2 where D0,t is understood as being the 21 -order Riemann-Liouville fractional derivative.

1.1 Fractional calculus via application and computation

as

| 5

In general, by using suitable fractional derivative operators, we can rewrite (1.18) j

F(t) = c̃j D0,t E(t),

j may be a non-integer

(1.20)

where c̃j is another constant that depends on t −j E(t). Note that (1.20) was originally proposed by Nutting [127, 128], consolidates (1.14) and (1.15), but also extends them to describe some models of viscoelastic material that exist between pure viscous liquids and pure elastic solids. The above two examples may seem to be a trivial exercise in elementary physics, mechanics and differential equations, and they may be redundant to specialists in this field. However, we have emphasize that, whether called Abel’s mechanical problem or Nutting’s law, they demonstrate that fractional calculus plays an important role in applications in engineering and physics. In recent years, fractional calculus has been proven a powerful tool for the study of dynamical properties of many interesting systems in physics, chemistry and engineering. It serves as a powerful application in nonlinear oscillations of earthquakes, many physical phenomena such as seepage flow in porous media and in fluid-dynamical traffic modelling. For more recent developments on fractional calculus and applications to fractional differential equations, refer to the monographs of Baleanu et al. [19], Diethelm [35], Kilbas et al. [71], Lakshmikantham et al. [76], Miller and Ross [113], Michalski [111], Podlubny [144], Tarasov [173], Anastassiou [11], Baleanu et al. [19] and Zhou [248, 249]. We also note that fractional calculus is widely applied to numerical computation, see Li and Zeng [91]. For more works on Hadamard fractional calculus and their application, see [54, 108, 109, 84, 93, 94, 87, 92, 241, 85, 88, 86]. It is also remarkable that interest in the study of differential equations of fractional order centers on the fact that fractional derivatives provide an excellent tool for the description of memory and hereditary properties of various materials and processes. With this advantage, the fractional order models become more realistic and practical than the classical integer order models, in which such effects are not taken into account. As a matter of fact, fractional differential equations arise in many engineering and scientific disciplines, such as physics, chemistry, biology, economics, control theory, signal and image processing, biophysics, blood flow phenomena, aerodynamics and fitting experimental data. For more recent development on this hot topic, refer to the monographs of Baleanu et al. [19], Diethelm [35], Kilbas et al. [71], Lakshmikantham et al. [76], Miller and Ross [113], Michalski [111], Podlubny [144], Tarasov [173], Zhou et al. [250] and Fečkan et al. [49], Liu et al. [103], Sun and Gao [172], and the contributions [212, 225, 199, 200, 219, 47, 188, 189, 190, 203, 209, 201, 210, 218, 226, 213, 46, 194, 207, 222, 208, 180, 202, 181, 186, 195, 205, 183, 197, 191, 198, 107, 224, 206, 235, 214, 196, 204, 193, 192, 215, 185, 230, 182, 187, 227, 228, 223, 36, 48, 31, 244, 90, 234, 106, 50, 102, 229, 142, 245, 89] and the references therein.

6 | 1 Introduction

1.2 Motivation of fractional Hermite-Hadamard’s inequality Let f : [a, b] ⊂ ℝ → ℝ be a convex function, i. e., f (λx+(1−λ)y) ≤ λf (x)+(1−λ)f (y) for all x, y ∈ [a, b] and λ ∈ [0, 1]. The classical Hermite-Hadamard type inequality provides lower and upper estimates for the integral average of any convex function defined on a compact interval, involving the midpoint and the endpoints of the domain. More precisely, if f : [a, b] → ℝ is a convex function, then it is integrable in the Riemannian sense and b

f(

1 f (a) + f (b) a+b )≤ . ∫ f (t)dt ≤ 2 b−a 2 a

This interesting inequality was first discovered by Hermite in 1881 in the journal Mathesis (see Mitrinović and Lacković [115]). However, this beautiful result was nowhere mentioned in the mathematical literature and was not widely known as Hermite’s result (see Pec̆ arić et al. [141]). For more recent results that generalize, improve and extend this classical Hermite-Hadamard inequality, refer to Abramovich et al. [2], Cal et al. [25], Ödemir et al. [129, 130], Dragomir [38, 39], Sarikaya et al. [150], Xiao et al. [240], Bessenyei [21], Tseng et al. [176], Niculescu [118] and references therein. Due to the wide application of Hermite-Hadamard type inequality and fractional calculus, it is natural to seek to study Hermite-Hadamard type inequalities involving fractional integrals. It is remarkable that Sarikaya et al. [159] initially studied inequalities of HermiteHadamard type involving Riemann-Liouville fractional integrals and produced a new fractional Hermite-Hadamard type inequality as follows: f(

a+b Γ(α + 1) f (a) + f (b) )≤ [ J α+ f (b) + RL Jbα− f (a)] ≤ , 2 2(b − a)α RL a 2

where the symbol Jaα+ f and Jbα− f denote the left-sided and right-sided Riemann-Liouville fractional integrals of the order α ∈ ℝ+ := [0, ∞) are defined by (RL Jaα+ f )(x)

x

1 = ∫(x − t)α−1 f (t)dt, Γ(α)

(0 ≤ a < x ≤ b)

a

and (RL Jbα− f )(x) =

b

1 ∫(t − x)α−1 f (t)dt, Γ(α)

(0 ≤ a < x ≤ b),

x

respectively. Here Γ(⋅) is the Gamma function. Moreover, Sarikaya et al. [159] initially established the following fundamental fractional integral identities: f (a) + f (b) Γ(α + 1) − [ J α+ f (b) + RL Jbα− f (a)] 2 2(b − a)α RL a

1.2 Motivation of fractional Hermite-Hadamard’s inequality | 7 1

b−a = ∫[(1 − t)α − t α ]f 󸀠 (ta + (1 − t)b)dt, 2 0

and give the following basic estimation: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 b−a Γ(α + 1) 1 󵄨󵄨 󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 ≤ (1 − α )|f 󸀠 (a) + f 󸀠 (b)| 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2(α + 1) 2 2(b − a) 2 when f : [a, b] → ℝ is a differentiable mapping on (a, b) and |f 󸀠 | is convex on [a, b]. Next, Wang et al. [216] established another two fundamental integral identities including the second-order derivatives of a given function and proved some HermiteHadamard type inequalities involving left-sided and right-sided Riemann-Liouville fractional integrals for m-convex and (s, m)-convex functions, respectively. Although Hermite-Hadamard inequality involving Riemann-Liouville fractional integrals have been paid increasingly more attention, there are few works on the Hadamard fractional integrals, even if though it had been reported many years ago. Thus, it is natural to seek to study Hermite-Hadamard type inequalities involving Hadamard fractional integrals. In fact, Wang et al. [217] initially established a new Hermite-Hadamard’s inequality involving Hadamard fractional integrals as follows: f (√ab) ≤

Γ(α + 1) [ J α+ f (b) + H Jbα− f (a)] ≤ f (b), 2(ln b − ln a)α H a

where f : [a, b] → ℝ is an integral, nondecreasing, convex and positive function, α α the symbol Ja+ f and Jb− f denote the left-sided and right-sided Hadamard fractional integrals of the order α ∈ ℝ+ are defined by (H Jaα+ f )(x) =

x

α−1

1 x ∫(ln ) Γ(α) t

f (t)

dt , t

(0 < a < x ≤ b)

f (t)

dt , t

(0 < a ≤ x < b).

a

and (H Jbα− f )(x)

b

α−1

1 t = ∫(ln ) Γ(α) x x

Further, Wang et al. [217] established the following fundamental fractional integral identities: f (a) + f (b) Γ(α + 1) − [ J α+ f (b) + H Jbα− f (a)] 2 2(ln b − ln a)α H a 1

ln b − ln a = ∫[(1 − t)α − t α ]et ln a+(1−t) ln b f 󸀠 (et ln a+(1−t) ln b )dt, 2 0

8 | 1 Introduction and give the following basic estimations: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [H Jaα+ f (b) + H Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(ln b − ln a) 󵄨󵄨 −1

≤

b(ln b − ln a) α + 2 ln b − ln a [ ( ) 2 α+1 2

+

√ ba

2(α + 1)

]|f 󸀠 (b)|.

The above contributions have inspired yet more researchers to do further work on this topic. Because of the wide application of Hermite-Hadamard type inequalities and fractional integrals, many researchers have extended their studies to Hermite-Hadamard type inequalities involving fractional integrals not limited to integer integrals. In fact, Hermite-Hadamard inequality involving Riemann-Liouville/Hadamard fractional integrals have been paid more and more attention. For further reading, the broad literature on the Riemann-Liouville case includes, for instance: [157, 104, 163, 170, 165, 169, 63, 62, 67, 68, 184, 26, 28, 59, 14, 133, 134, 135, 136, 17, 30, 32, 124, 123, 120, 161, 239, 116, 231, 146, 45, 154, 131, 24, 159, 3, 151, 1, 33, 23, 75, 4, 13, 69, 73, 5, 16, 119, 65, 164, 160, 112]. On the Hadamard case, one can refer, for instance to: [121, 152, 53, 67, 155, 40, 28, 59, 44, 163, 137, 164, 175, 41, 82, 12, 10, 247, 233, 125, 74, 114, 166, 168, 126, 110, 42, 43, 20, 70, 6, 29, 83, 27, 77, 104, 57, 238, 80, 179, 153, 64, 122, 52, 124, 167, 242].

1.3 Main contents Now, we are ready to outline the contents of this book. Chapter 2 is devoted to preliminaries. In Section 2.1, we summarize the definitions of special functions and Riemann-Liouville fractional integrals and Hadamard fractional integrals. In Section 2.2, we collect the definitions of all kinds of convex functions. In Section 2.3, we give some basic results for singular integrals by using series. Some elementary inequalities are also collected in Section 2.4. Chapter 3 investigates fractional integral identities for a differentiable mapping involving Riemann-Liouville fractional integrals and Hadamard fractional integrals. Section 3.1 gives identities involving Riemann-Liouville fractional integrals. Identities involving Hadamard fractional integrals are given in Section 3.2. Chapter 4 studies Hermite-Hadamard’s inequalities involving Riemann-Liouville fractional integrals by using several different concepts of convex functions. In Section 4.1, we give some inequalities via standard convex functions. In Section 4.2, we give some inequalities via r-convex functions. In Section 4.3, we give some inequalities via s-convex functions. In Section 4.4, we give some inequalities via m-convex functions. In Section 4.5, we give some inequalities via (s, m)-convex functions. In Section 4.6, we give some inequalities via preinvex convex functions. In Section 4.7, we give some inequalities via (β, m)-convex functions. In Section 4.8, we give some inequalities via geometrical-arithmetically s-convex functions. In Section 4.9, we give

1.3 Main contents | 9

some inequalities via (α, m)-logarithmically convex functions. In Section 4.10, we give some inequalities via s-Godunova-Levin convex functions. In Section 4.11, we give some inequalities via (s, m)-Godunova-Levin convex functions. In Section 4.12, we give some inequalities via AG(log)-convex functions. In Chapter 5, we examine Hermite-Hadamard’s inequalities involving Hadamard fractional integrals by using several different concepts of convex functions. In Section 5.1, we give some inequalities via standard convex functions. In Section 5.2, we give some inequalities via s-e-condition functions. In Section 5.3, we give some inequalities via geometric-geometric co-ordinated functions. In Section 5.4, we give some inequalities via geometric-geometric-convex functions. In Section 5.5, we give some inequalities via geometric-arithmetic-convex functions. The above obtained results extend the classical Hermite-Hadamard type inequalities to fractional cases, which will be used to find the lower and upper bounded for fractional integral for given functions. This monograph is intended for anyone who is interested in the theory of fractional Hermite-Hadamard’s inequalities and also their applications to numerical calculations.

2 Preliminaries 2.1 Definitions of special functions and fractional integrals Definition 1 (see [71]). The Gamma function is defined as follows: ∞

Γ(z) = ∫ t z−1 e−t dt, 0

where z > 0. Definition 2 (see [71]). The Beta function is defined as follows: 1

B(z, w) = ∫ t z−1 (1 − t)w−1 dt, 0

where z, w > 0. Definition 3 (see [140]). The incomplete Beta function is defined as follows: x

Bx (a, b) = ∫ t a−1 (1 − t)b−1 dt, 0

where x ∈ [0, 1] and a, b > 0. Definition 4 (see [71]). Let f ∈ L[a, b]. The symbol RL Jaα+ f and RL Jbα− f denote the leftsided and right-sided Riemann-Liouville fractional integrals of the order α ∈ ℝ+ are defined by x

(RL Jaα+ f )(x) =

1 ∫(x − t)α−1 f (t)dt, Γ(α)

(RL Jbα− f )(x) =

1 ∫(t − x)α−1 f (t)dt, Γ(α)

(0 ≤ a < x ≤ b),

a

and b

(0 ≤ a ≤ x < b),

x

respectively, where Γ(⋅) is the Gamma function. Definition 5 (see [149]). Let f ∈ L[a, b], y ∈ (a, b). The symbols RL Jaα+ f (x, y) and α RL Jb− f (x, y) denote the left-sided and right-sided Riemann-Liouville fractional integrals of the order α ∈ R+ are defined by (RL Jaα+ f )(x, y)

x

1 = ∫(x − t)α−1 f (t, y)dt, Γ(α)

https://doi.org/10.1515/9783110523621-002

a

(0 ≤ a < x ≤ b),

12 | 2 Preliminaries and (RL Jbα− f )(x, y) =

b

1 ∫(t − x)α−1 f (t, y)dt, Γ(α)

(0 ≤ a ≤ x < b),

x

respectively, where Γ(⋅) is the Gamma function. α α Definition 6 (see [71]). For f ∈ L[a, b], the symbol Ja+ f and Jb− f denote the left-sided and right-sided Hadamard fractional integrals of the order α ∈ ℝ+ are defined by x

α−1

(H Jaα+ f )(x)

x 1 = ∫(ln ) Γ(α) t

(H Jbα− f )(x)

1 t = ∫(ln ) Γ(α) x

f (t)

dt , t

(0 < a < x ≤ b)

f (t)

dt , t

(0 < a ≤ x < b).

a

and b

α−1

x

Definition 7 (see [71]). Δ =: [a, b] × [c, d], f ∈ L1 (Δ). The double Hadamard fractional integrals of order α, β ∈ R+ of function f with a, c > 0 are defined by α,β (RL Ja+ ,c+ f )(x, y)

x y

1 1 = ∫ ∫(x − t)α−1 (y − s)β−1 f (t, s)dtds, Γ(α) Γ(β) a c

x d

α,β

(RL Ja+ ,d− f )(x, y) =

1 1 ∫ ∫(x − t)α−1 (s − y)β−1 f (t, s)dtds, Γ(α) Γ(β) a y

b y

α,β (RL Jb− ,c+ f )(x, y)

1 1 = ∫ ∫(t − x)α−1 (y − s)β−1 f (t, s)dtds Γ(α) Γ(β)

α,β (RL Jb− ,d− f )(x, y)

1 1 = ∫ ∫(t − x)α−1 (s − y)β−1 f (t, s)dtds, Γ(α) Γ(β)

x c

and b d

x y

where (x, y) ∈ Δ, Γ(⋅) is the Gamma function. Definition 8 (see [71]). Δ =: [a, b] × [c, d], f ∈ L1 (Δ). The double Hadamard fractional integrals of order α, β ∈ ℝ+ of function f with a, c > 0 are defined by α,β (H Ja+ ,c+ f )(x, y)

x y

α−1

1 1 x = ∫ ∫(ln ) Γ(α) Γ(β) t a c

β−1

y (ln ) s

f (t, s)

dtds , ts

2.2 Definitions of convex functions | 13 x d

α−1

α,β (H Ja+ ,d− f )(x, y)

1 1 x = ∫ ∫(ln ) Γ(α) Γ(β) t

α,β (H Jb− ,c+ f )(x, y)

1 1 t = ∫ ∫(ln ) Γ(α) Γ(β) x

a y

b y

α−1

x c

β−1

s (ln ) y

β−1

y (ln ) s

f (t, s)

dtds , ts

f (t, s)

dtds ts

and α,β

b d

(H Jb− ,d− f )(x, y) =

α−1

t 1 1 ∫ ∫(ln ) Γ(α) Γ(β) x x y

β−1

s (ln ) y

f (t, s)

dtds , ts

where (x, y) ∈ Δ, Γ(⋅) is the Gamma function.

2.2 Definitions of convex functions Definition 9 (see [66]). The function f : I ⊂ ℝ → ℝ is said to be convex, if for every x, y ∈ I and λ ∈ [0, 1], we have f (λx + (1 − λ)y) ≤ λf (x) + (1 − λ)f (y). Definition 10 (see [9]). The function f : I ⊂ ℝ → ℝ is said to be quasi-convex, if for every x, y ∈ I and λ ∈ [0, 1], we have f (λx + (1 − λ)y) ≤ max{f (x), f (y)}. Definition 11 (see [60]). The function f : I ⊂ ℝ → ℝ is said to be quasi-geometrically convex, if for every x, y ∈ I and λ ∈ [0, 1], we have f (x λ y(1−λ) ) ≤ sup{f (x), f (y)}. Definition 12 (see [58]). The function f : I ⊆ ℝ+ → ℝ is said to be s-convex in the first sense, where s ∈ (0, 1], if for every x, y ∈ I and λ ∈ [0, 1], we have 1

f (λx + (1 − λs ) s y) ≤ λs f (x) + (1 − λs )f (y). Definition 13 (see [58]). The function f : I ⊆ ℝ+ → ℝ is said to be s-convex in the second sense, where s ∈ (0, 1], if for every x, y ∈ I and λ ∈ [0, 1], we have f (λx + (1 − λ)y) ≤ λs f (x) + (1 − λ)s f (y). Definition 14 (see [18]). The function f : I ⊆ ℝ+ → ℝ is said to be m-convex, where m ∈ [0, 1], if for every x, y ∈ I and λ ∈ [0, 1], we have f (λx + m(1 − λ)y) ≤ λf (x) + m(1 − λ)f (y).

14 | 2 Preliminaries Definition 15 (see [18]). The function f : I ⊆ ℝ+ → ℝ is said to be (s, m)-convex, where (s, m) ∈ [0, 1] × [0, 1], if for every x, y ∈ I and λ ∈ [0, 1], we have f (λx + m(1 − λ)y) ≤ λs f (x) + m(1 − λs )f (y). Definition 16 (see [177]). The function f : I ⊆ ℝ+ → ℝ+ is said to be h-convex, where h : J ⊆ ℝ+ → ℝ+ , if for every x, y ∈ I and λ ∈ [0, 1], we have f (λx + m(1 − λ)y) ≤ h(λ)f (x) + h(1 − λ)f (y). Definition 17 (see [139]). The function f : I ⊆ ℝ+ → ℝ+ is said to be r-convex, where r ≥ 0, if for every x, y ∈ I and λ ∈ [0, 1], we have r

r

1

f (λx + (1 − λ)y) ≤ [λ(f (x)) + (1 − λ)(f (y)) ] r , λ

f (λx + (1 − λ)y) ≤ (f (x)) (f (y))

1−λ

,

r ≠ 0, r = 0.

Definition 18 (see [117]). A function f : I ⊆ ℝ+ → ℝ+ is said to be arithmetic-geometric convex (or log-convex), if for every x, y ∈ I and λ ∈ [0, 1], we have f ((1 − λ)x + λy) ≤ f (x)1−λ f (y)λ . Definition 19 (see [117]). A function f : I ⊆ ℝ+ → ℝ+ is said to be geometric-geometric convex, if for every x, y ∈ I and λ ∈ [0, 1], we have f (x1−λ yλ ) ≤ f (x)1−λ f (y)λ . Definition 20 (see [148]). A function f : I ⊆ ℝ+ → ℝ is said to be geometric-arithmetically convex, if for every x, y ∈ I and λ ∈ [0, 1], we have f (x 1−λ yλ ) ≤ (1 − λ)f (x) + λf (y). Definition 21 (see [171]). The function f : I ⊆ ℝ+ → ℝ is said to be geometric-arithmetically s-convex, where s ∈ (0, 1], if for every x, y ∈ I and λ ∈ [0, 1], we have f (xλ y1−λ ) ≤ λs (f (x)) + (1 − λ)s f (y). Definition 22 (see [237]). The function f : I ⊆ ℝ+ → ℝ+ is said to be m-geometrically convex, where m ∈ (0, 1], if for every x, y ∈ [0, b] and λ ∈ [0, 1], we have λ

f (x λ ym(1−λ) ) ≤ [f (x)] [f (y)]

m(1−λ)

.

Definition 23 (see [237]). The function f : I ⊆ ℝ+ → ℝ+ is said to be (β, m)-geometrically convex, where (β, m) ∈ (0, 1] × (0, 1], if for every x, y ∈ [0, b] and λ ∈ [0, 1], we have λβ

f (x λ ym(1−λ) ) ≤ [f (x)] [f (y)]

m(1−λβ )

.

2.2 Definitions of convex functions | 15

Definition 24 (see [15]). The function I ⊆ ℝ+ → ℝ+ is said to be (α, m)-logarithmically convex, where (α, m) ∈ (0, 1] × (0, 1], if for every x, y ∈ [0, b] and λ ∈ [0, 1], we have λα

m(1−λα )

f (λx + m(1 − λ)y) ≤ (f (x)) (f (y))

.

Definition 25 (see [221]). The function f : I ⊆ ℝ+ → ℝ is said to satisfy s-e-condition, where s ∈ (0, 1], if for every x, y ∈ I and λ ∈ [0, 1], we have f (eλx+(1−λ)y ) ≤ λs f (ex ) + (1 − λ)s f (ey ). Remark 26 (see Remark 1.4, [220]). If f : I ⊂ (0, ∞) → ℝ is a nondecreasing and convex function, then f satisfies f (eλx+(1−λ)y ) ≤ f (λex + (1 − λ)ey ) ≤ λf (ex ) + (1 − λ)f (ey ) for all x, y ∈ I, λ ∈ [0, 1], which implies the above s-e-condition. Remark 27 (see Remark 1.5, [221]). Let f : I ⊂ (0, ∞) → R be a nondecreasing and convex function. Then f satisfies the above s-e-condition. Definition 28 (see [55]). The set S ⊆ ℝn is said to be invex with respect to the map η : S × S → ℝn , if for every x, y ∈ S and λ ∈ [0, 1], we have x + λη(y, x) ∈ S. Definition 29 (see [236]). Let S ⊆ ℝn is invex with respect to the map η : S × S → ℝn , the function f : S → ℝ is said to be preinvex with respect to η, if for every x, y ∈ S and λ ∈ [0, 1], f (x + λη(y, x)) ≤ (1 − λ)f (x) + tf (y). Let us now consider a bidimensional interval Δ := [a, b] × [c, d] in ℝ2 with a < b and c < d. A mapping f : Δ → ℝ is said to be convex on the co-ordinates Δ 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) ∈ Δ and t ∈ [0, 1]. A function f : Δ → ℝ is said to be convex on the co-ordinates on Δ if the partial mapping 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 [37]). A formal definition of a coordinated convex function may be stated as follows: Definition 30 (see [149]). The function f : Δ → ℝ is said to be coordinated convex, if, for every t, s ∈ [0, 1] × [0, 1] and (x, y), (u, w) ∈ Δ, we have f (tx + (1 − t)y, su + (1 − s)w) ≤ ts(f (x, u)) + s(1 − t)(f (y, u)) + t(1 − s)f (x, w) + (1 − t)(1 − s)f (y, w).

16 | 2 Preliminaries Clearly, every convex function is coordinated convex. Furthermore, there exists a coordinated convex function which is not convex, (see [37]). For several results concerning Hermite-Hadamard’s inequality for some convex function on the co-ordinates on a rectangle from the plane, we refer the reader to [37, 7, 8, 78, 79, 81, 132, 158, 156]. Definition 31. The function f : Δ → ℝ is said to be geometric-geometric coordinated convex, if for every t, s ∈ [0, 1] × [0, 1] and (x, y), (u, w) ∈ Δ, we have ts

f (x t y1−t , us w1−s ) ≤ (f (x, u)) (f (x, w))

t(1−s)

(1−t)s

(f (y, u))

(1−t)(1−s)

(f (y, w))

2.3 Singular integrals via series The results in this section are derived from [211]. Lemma 32. For α > 0 and k > 0, we have 1

∞

0

i=1

I(α, k) := ∫ t α−1 k t dt = k ∑(−1)i−1

(ln k)i−1 < +∞, (α)i

where (α)i = α(α + 1)(α + 2) ⋅ ⋅ ⋅ (α + i − 1). Moreover, it holds 󵄨󵄨 m−1 m i−1 󵄨󵄨󵄨 󵄨󵄨 |ln k|e |ln k| 󵄨󵄨I(α, k) − k ∑ (− ln k) 󵄨󵄨󵄨 ≤ ( ) . 󵄨 󵄨󵄨 (α)i 󵄨󵄨󵄨 α√2π(m − 1) m − 1 󵄨󵄨 i=1 Proof. Using integration by parts, one can obtain I(α, k) =

k ln k − I(α + 1, k). α α

Further, we find I(α + 1, k) =

k ln k − I(α + 2, k). α+1 α+1

As a result, I(α, k) =

k k(ln k) ln2 k − + I(α + 2, k). α α(α + 1) α(α + 1)

Repeating the same steps as above, we can obtain m

I(α, k) = k ∑ i=1

(− ln k)i−1 (− ln k)m + I(α + m, k). (α)i (α)m

.

2.3 Singular integrals via series | 17

Let m → ∞, then the proof is completed due to m

∑ i=1

(− ln k)i−1 (α)i

is the standard Leibnitz series, which is convergent. As a matter of fact, this series absolutely converges. Moreover, since I(α + m, k) ≤ max{1, k} and by Stirling’s formula m−1 󵄨󵄨 (− ln k)m 󵄨󵄨 |ln k| | ln k|e |ln k|m 󵄨󵄨 󵄨󵄨 = ( ) , 󵄨󵄨 󵄨󵄨 ≤ 󵄨󵄨 (α)m 󵄨󵄨 α(m − 1)! α√2π(m − 1) m − 1

we get 󵄨󵄨 m−1 m m i−1 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨I(α, k) − k ∑ (− ln k) 󵄨󵄨󵄨 ≤ |ln k| I(α + m, k) ≤ |ln k| max{1, k} ( |ln k|e ) . 󵄨󵄨 (α)i 󵄨󵄨󵄨󵄨 (α)m α√2π(m − 1) m − 1 󵄨󵄨 i=1 Clearly, the convergence is rapid as m → ∞. The proof is completed. Lemma 33. For α > 0 and k > 0, z > 0, we have 1

∞

J(α, k) := ∫(1 − t)α−1 k t dt = ∑ i=1

0

(ln k)i−1 < +∞, (α)i

z

∞

0

i=1

H(α, k, z) := ∫ t α−1 k t dt = z α k z ∑

(−z ln k)i−1 < +∞. (α)i

Proof. By using Lemma 32, we obtain 1

1

∞

0

i=1

J(α, k) := ∫(1 − t)α−1 k t dt = ∫ t α−1 k 1−t dt = kI(α, k −1 ) = ∑ 0

z

1

0

0

(ln k)i−1 < +∞, (α)i ∞

T

H(α, k, z) := ∫ t α−1 k t dt = z α ∫ T α−1 (k z ) dT = z α I(α, k z ) = z α k z ∑ i=1

(−z ln k)i−1 < +∞. (α)i

The proof is completed. Lemma 34. For α > 0 and k > 0, 1 ≥ z > 0, we have z

∞

0

i=1

R(α, k, z) := ∫(1 − t)α−1 k t dt = ∑

(ln k)i−1 (1 − k z (1 − z)α+i−1 ). (α)i

Proof. By using Lemma 32 and Lemma 33, we obtain 1

R(α, k, z) = ∫(1 − t)

α−1 t

0

1

k dt − ∫(1 − t)α−1 k t dt z

18 | 2 Preliminaries 1−z

= J(α, k) − ∫ t α−1 k 1−t dt = J(α, k) − kH(α, k −1 , 1 − z) ∞

=∑ i=1

0 i−1

(ln k) (α)i

∞

− k(1 − z)α k z−1 ∑ i=1

(1 − z)i−1 (ln k)i−1 . (α)i

The proof is completed.

2.4 Elementary inequalities Lemma 35 (see [56]). If p, q ≥ 1, then

1 p

1 q

+

= 1, a, b ∈ ℝ, a < b and f ∈ Lp [a, b], g ∈ Lq [a, b],

b

b

a

a

1 p

1 q

b

∫ |f (t)g(t)|dt ≤ (∫ |f (t)|p dt) (∫ |g(t)|q dt) . a

Lemma 36 (see Lemma 3.3, [237]). For x, y ∈ [0, ∞) and m, t ∈ (0, 1], if x < y and y ≥ 1, then xt ym(1−t) ≤ tx + (1 − t)y. Lemma 37 (see [211, 216]). For A ≥ 0, B ≥ 0, it holds that Aθ + Bθ ≤ (A + B)θ ≤ 2θ−1 (Aθ + Bθ ) 2

θ−1

θ

θ

θ

θ

θ

(A + B ) ≤ (A + B) ≤ A + B

when θ ≥ 1, when 0 < θ ≤ 1.

Lemma 38 (see [216]). For A > B > 0, it holds that (A − B)θ ≤ Aθ − Bθ

when θ ≥ 1,

(A − B) ≥ A − B

when 0 < θ ≤ 1.

θ

θ

θ

Lemma 39 (see Lemma 2.5, [34]). For t ∈ [0, 1], we have (1 − t)n ≤ 21−n − t n

for n ∈ [0, 1],

(1 − t) ≤ 2

for n ∈ [1, ∞).

n

1−n

−t

n

Proof. Let f (t) = t n + (1 − t)n for n, t ∈ [0, 1]. Clearly, f (⋅) is increasing on the interval [0, 21 ] and decreasing on the interval [ 21 , 1]. So f (t) ≤ f ( 21 ) = 21−n for all t ∈ [0, 1]. Then we have the first statement. Similarly one can obtain the second one. The proof is completed. Lemma 40 (see [145]). For 0 < σ ≤ 1 and 0 < A < B, we have |Aσ − Bσ | ≤ (B − A)σ .

2.5 Special mean

| 19

Lemma 41 (see [72]). For all λ, v, w > 0, then for any t > 0, we have t

λ t 1−v ∫(t − s)v−1 sλ−1 e−ws ds ≤ max{1, 21−v }Γ(λ)(1 + )w−λ . v 0

2.5 Special mean Consider the following special means (see Pearce and Pec̆ arić [138]) for arbitrary real numbers x, y, x ≠ y as follows: (1) The arithmetic mean: x+y , x, y ∈ ℝ. 2

A(x, y) = (2) The harmonic mean: H(x, y) =

1 x

2 +

1 y

, x, y ∈ ℝ \ {0}.

(3) The geometric mean: G(x, y) = √xy. (4) The logarithmic mean: L = L(a, b) := { a

(5) The identric mean: I = I(a, b) := { 1

b

a

b−a ln b−ln a 1

( b ) b−a e aa

if a = b if a ≠ b

if a = b if a ≠ b a

(6) The p-logarithmic mean: Lp = Lp (a, b) := { a, b > 0.

a, b > 0.

a, b > 0. 1

bp+1 −ap+1 p [ (p+1)(b−a) ]

if a = b if a ≠ b

, p ∈ ℝ\{−1, 0};

3 Fractional integral identities 3.1 Identities involving Riemann-Liouville fractional integrals The results in this section are taken from [149, 216, 159, 34, 251, 162, 243, 147, 99, 61, 98, 105, 95, 51]. Lemma 42. Let f : [a, b] → ℝ be a differentiable mapping on (a, b). If f 󸀠 ∈ L1 [a, b], then the following equality for fractional integrals holds: f (a) + f (b) Γ(α + 1) − [ J α+ f (b) + RL Jbα− f (a)] 2 2(b − a)α RL a 1

=

b−a ∫[(1 − t)α − t α ]f 󸀠 (ta + (1 − t)b)dt. 2

(3.1)

0

Proof. It suffices to note that 1

I = ∫[(1 − t)α − t α ]f 󸀠 (ta + (1 − t)b)dt 0

1

1

α 󸀠

= [∫(1 − t) f (ta + (1 − t)b)dt] + [− ∫ t α f 󸀠 (ta + (1 − t)b)dt] 0

0

= I1 + I2 .

(3.2) (3.3)

Integrating by parts, 1

I1 = ∫(1 − t)α f 󸀠 (ta + (1 − t)b)dt 0 α f (ta

= (1 − t)

1 1 + (1 − t)b) 󵄨󵄨󵄨󵄨 α f (ta + (1 − t)b) dt 󵄨󵄨 + ∫ α(1 − t) 󵄨󵄨0 a−b a−b a

=

0

α−1

f (b) α a−x − ) ∫( b−a b−a a−b b

f (x) dx a−b

f (b) Γ(α + 1) = − J α− f (a), b − a (b − a)α+1 RL b and similarly we get 1

I2 = − ∫ t α f 󸀠 (ta + (1 − t)b)dt 0 https://doi.org/10.1515/9783110523621-003

(3.4)

22 | 3 Fractional integral identities 1 1 t α f (ta + (1 − t)b) 󵄨󵄨󵄨󵄨 α−1 f (ta + (1 − t)b) dt =− 󵄨󵄨 + α ∫ t 󵄨󵄨0 a−b a−b 0

a

=

α−1

α b−x f (a) − ) ∫( b−a b−a b−a b

f (x) dx a−b

f (a) Γ(α + 1) = − J α+ f (b). b − a (b − a)α+1 RL a

(3.5)

Using (3.4) and (3.5) in (3.2), it follows that: I=

Γ(α + 1) f (a) + f (b) − [ J α+ f (b) + RL Jbα− f (a)]. b−a (b − a)α+1 RL a

Thus, by multiplying both sides by proof is completed.

b−a , 2

we have reached the conclusion (3.1). The

Lemma 43. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with a < mb ≤ b. If f 󸀠 ∈ L1 [a, b], then the following equality for fractional integrals holds: Γ(α + 1) f (a) + f (mb) α − [ J α+ f (mb) + RL Jmb − f (a)] 2 2(mb − a)α RL a 1

=

mb − a ∫[(1 − t)α − t α ]f 󸀠 (ta + m(1 − t)b)dt. 2

(3.6)

0

Proof. This is just Lemma 42 on the interval [a, mb] ⊂ [a, b]. Lemma 44. Suppose that f : [a, b] → ℝ be a differentiable mapping. If f 󸀠 ∈ L[a, a + η(b, a)], then the following equality for fractional integrals holds: f (a) + f (a + η(b, a)) Γ(α + 1) α − α [ J α+ f (a + η(b, a)) + RL J(a+η(b,a)) − f (a)] 2 2η (b, a) RL a 1

η(b, a) = ∫[t α − (1 − t)α ]f 󸀠 (a + tη(b, a))dt. 2 0

Proof. This is just Lemma 42 on the interval [a, a + η(b, a)] ⊂ [a, b]. Lemma 45. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with a < b. If f 󸀠 ∈ L1 [a, b], then the following equality for fractional integrals holds: f (a) + f (b) Γ(α + 1) − [ J α+ f (b) + RL Jbα− f (a)] 2 2(b − a)α RL a 1

=

α

α

b−a t t a+b {∫[( ) − (1 − ) ]f 󸀠 (t + (1 − t)a)dt 4 2 2 2 1

0

α

α

1 t 1 t a+b + ∫[( + ) − ( − ) ]f 󸀠 (tb + (1 − t) )dt}. 2 2 2 2 2 0

(3.7)

3.1 Identities involving Riemann-Liouville fractional integrals | 23

Proof. It suffices to note that 1

α

α

t a+b t + (1 − t)a)dt I := ∫[( ) − (1 − ) ]f 󸀠 (t 2 2 2 0

1

α

α

1 t 1 t a+b + ∫[( + ) − ( − ) ]f 󸀠 (tb + (1 − t) )dt 2 2 2 2 2 0

1

1

α

α

t a+b t a+b = ∫( ) f 󸀠 (t + (1 − t)a)dt − ∫(1 − ) f 󸀠 (t + (1 − t)a)dt 2 2 2 2 0

0

1

α

1 t a+b + ∫( + ) f 󸀠 (tb + (1 − t) )dt 2 2 2 0

1

α

1 t a+b − ∫( − ) f 󸀠 (tb + (1 − t) )dt 2 2 2 0

= I1 − I2 + I3 − I4 .

(3.8)

+ (1 − t)a, we have Calculating I1 , I2 , I3 and I4 and put x = t a+b 2 1

α

t a+b I1 = ∫( ) f 󸀠 (t + (1 − t)a)dt 2 2 0

=

1

α 󵄨󵄨1 2 t a+b a+b α 󵄨 [( ) f (t + (1 − t)a)󵄨󵄨󵄨 − α ∫ t α−1 f (t + (1 − t)a)dt] 󵄨󵄨0 2 b−a 2 2 2 0

α

=

a+b 2

2 1 a+b α 2 2(x − a) {( ) f ( )− α ] ∫[ b−a 2 2 2 b−a b−a a

α

=

f (x)dx}

a+b 2

2 1 a+b α 2 2 [( ) f ( )− α ∫ (x − a)α−1 f (x)dx] b−a 2 2 2 b − a (b − a)α−1 a

α

=

α−1

α−1

a+b 2

2 1 a+b α [( ) f ( )− ∫ (x − a)α−1 f (x)dx], b−a 2 2 (b − a)α a

and 1

α

t a+b I2 = ∫(1 − ) f 󸀠 (t + (1 − t)a)dt 2 2 0

=

α 󵄨󵄨1 2 t a+b 󵄨 [(1 − ) f (t + (1 − t)a)󵄨󵄨󵄨 󵄨󵄨0 b−a 2 2

(3.9)

24 | 3 Fractional integral identities 1

α−1

t α + ∫(1 − ) 2 2

f (t

0

a+b + (1 − t)a)dt] 2 a+b 2

α

α−1

2 x−a 1 a+b α 2 [( ) f ( ) − f (a) + ) ∫ (1 − b−a 2 2 2 b−a b−a

=

a

a+b 2

α

α−1

b−x 2 1 a+b α 2 = [( ) f ( ) − f (a) + ) ∫( b−a 2 2 2 b−a b−a a

f (x)dx]

f (x)dx]

a+b 2

α

1 a+b α 2 [( ) f ( ) − f (a) + = ∫ (b − x)α−1 f (x)dx]. b−a 2 2 (b − a)α

(3.10)

a

Put x = tb + (1 − t) a+b , we have 2 1

α

1 t a+b )dt I3 = ∫( + ) f 󸀠 (tb + (1 − t) 2 2 2 0

=

α 1 2 1 t a + b 󵄨󵄨󵄨󵄨 [( + ) f (tb + (1 − t) )󵄨󵄨 󵄨󵄨0 b−a 2 2 2 1

α−1

1 t α − ∫( + ) 2 2 2

f (tb + (1 − t)

0

a+b )dt] 2 b

α

α−1

1 2x − (a + b) 2 1 a+b α 2 = [f (b) − ( ) f ( )− ) ∫( + b−a 2 2 2 b−a 2 2(b − a)

f (x)dx]

a+b 2

b

α

α−1

1 a+b α 2 x−a 2 [f (b) − ( ) f ( )− ) = ∫( b−a 2 2 2 b−a b−a

f (x)dx]

a+b 2

b

α

2 a+b 1 α = [f (b) − ( ) f ( )− ∫ (x − a)α−1 f (x)dx], b−a 2 2 (b − a)α a+b 2

and 1

α

1 t a+b I4 = ∫( − ) f 󸀠 (tb + (1 − t) )dt 2 2 2 0

=

α 1 2 1 t a + b 󵄨󵄨󵄨󵄨 [( − ) f (tb + (1 − t) )󵄨󵄨 󵄨󵄨0 b−a 2 2 2 1

α−1

α 1 t + ∫( − ) 2 2 2 0

f (tb + (1 − t)

a+b )dt] 2

(3.11)

3.1 Identities involving Riemann-Liouville fractional integrals | 25 b

α

α−1

2 1 a+b α 2 1 2x − (a + b) = [−( ) f ( )+ ) ∫( − b−a 2 2 2 b−a 2 2(b − a)

f (x)dx]

a+b 2

b

α

α−1

2 1 a+b α 2 b−x = [−( ) f ( )+ ) ∫( b−a 2 2 2 b−a b−a

f (x)dx]

a+b 2

b

α

2 1 a+b α [−( ) f ( )+ = ∫ (b − x)α−1 f (x)dx]. b−a 2 2 (b − a)α a+b 2

Submitting (3.9), (3.10), (3.11) and (3.12) to (3.8), we obtain I=

a+b 2

α

1 a+b α 2 [( ) f ( )− ∫ (x − a)α−1 f (x)dx b−a 2 2 (b − a)α a

a+b 2

α

1 a+b α − ( ) f( ) + f (a) − ∫ (b − x)α−1 f (x)dx 2 2 (b − a)α a

b

α

1 a+b α + f (b) − ( ) f ( )− ∫ (x − a)α−1 f (x)dx 2 2 (b − a)α a+b 2

b

α

1 a+b α + ( ) f( )− ∫ (b − x)α−1 f (x)dx] 2 2 (b − a)α a+b 2

b

α 2 [f (b) + f (a) − = ∫(x − a)α−1 f (x)dx b−a (b − a)α a

b

−

α ∫(b − x)α−1 f (x)dx] (b − a)α a

b

b

a

a

2 α = {f (b) + f (a) − [∫(x − a)α−1 f (x)dx + ∫(b − x)α−1 f (x)dx]} b−a (b − a)α 2 αΓ(α) = {f (b) + f (a) − [ J α+ f (b) + RL Jbα− f (a)]} b−a (b − a)α RL a Γ(α + 1) 2 = {[f (b) + f (a)] − [ J α+ f (b) + RL Jbα− f (a)]}. b−a (b − a)α RL a In summary, we get 2 2Γ(α + 1) [f (b) + f (a)] − [ J α+ f (b) + RL Jbα− f (a)] b−a (b − a)α+1 RL a

(3.12)

26 | 3 Fractional integral identities 1

α

α

t a+b t + (1 − t)a)dt = ∫[( ) − (1 − ) ]f 󸀠 (t 2 2 2 0

1

α

α

1 t 1 t a+b + ∫[( + ) − ( − ) ]f 󸀠 (tb + (1 − t) )dt. 2 2 2 2 2

(3.13)

0

Multiplying both sides of (3.13) by pleted.

b−a , 4

we obtain the equality (3.7). The proof is com-

Lemma 46. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with a < b. If f 󸀠 ∈ L1 [a, b], then the following equality for fractional integrals holds: f (a) + f (b) Γ(α + 1) α [ J α f (b) + RL Jb− − f (a)] 2 2(b − a)α RL a+ 1

b−a 1+t 1−t = α+2 [∫[(1 − t)α − (1 + t)α ]f 󸀠 ( a+ b)dt 2 2 2 0

1

+ ∫[(1 + t)α − (1 − t)α ]f 󸀠 ( 0

1+t 1−t b+ a)dt]. 2 2

Proof. Note that 1

I := ∫[(1 − t)α − (1 + t)α ]f 󸀠 ( 0

1

1+t 1−t a+ b)dt 2 2

+ ∫[(1 + t)α − (1 − t)α ]f 󸀠 ( 1

0

= ∫(1 − t)α f 󸀠 ( 0

1

1+t 1−t b+ a)dt 2 2 1

1+t 1+t 1−t 1−t b+ a)dt + ∫(1 + t)α f 󸀠 ( b+ a)dt 2 2 2 2 0

1

1+t 1−t 1−t 1+t − ∫(1 − t) f ( a+ b)dt − ∫(1 + t)α f 󸀠 ( a+ b)dt 2 2 2 2 α 󸀠

0

0

:= I1 + I2 − I3 − I4 . Integrating by parts, we have 1

I1 = ∫(1 − t)α f 󸀠 ( 0 −1

1−t 1+t b+ a)dt 2 2

= − ∫ (1 + t)α f 󸀠 ( 0

1+t 1−t b+ a)dt 2 2

(3.14)

3.1 Identities involving Riemann-Liouville fractional integrals | 27 0

= ∫ (1 + t)α f 󸀠 ( −1

1+t 1−t b+ a)dt, 2 2

and 1

I1 + I2 = ∫ (1 + t)α f 󸀠 ( −1

Put s =

1+t b 2

+

1−t a, 2

1−t 1+t b+ a)dt. 2 2

we have α+1 b

2 I1 + I2 = ( ) b−a

∫(s − a)α f 󸀠 (s)ds. a

Note that 1

I3 = ∫(1 − t)α f 󸀠 ( 0

1+t 1−t a+ b)dt 2 2

−1

= − ∫ (1 + t)α f 󸀠 ( 0

0

= ∫ (1 + t)α f 󸀠 ( −1

1+t 1−t a+ b)dt 2 2

1−t 1+t a+ b)dt, 2 2

and 1

I3 + I4 = ∫ (1 + t)α f 󸀠 ( −1

Put s =

1+t a 2

+

1−t b, 2

1+t 1−t a+ b)dt. 2 2

we have I3 + I4 = (

α+1 b

2 ) b−a

∫(b − s)α f 󸀠 (s)ds. a

Thus, α+1

2 I=( ) b−a

b

α 󸀠

[∫(s − a) f (s)ds − ∫(b − s)α f 󸀠 (s)ds] a

2 =( ) b−a

α+1

b

a

b

Γ(α + 1) [(b − a) (f (a) + f (b)) − ∫(s − a)α−1 f (s)ds Γ(α) α

a

28 | 3 Fractional integral identities b

Γ(α + 1) − ∫(b − s)α−1 f (s)ds] Γ(α) a α+1

2 ) [(b − a)α (f (a) + f (b)) b−a α α − Γ(α + 1)(RL Jb− f (a) + RL Ja+ f (b))].

=(

(3.15)

Substituting (3.15) into the right side of (3.14), the desired equation is reached. The proof is completed. Lemma 47. Let f : [a, b] → ℝ be a twice-differentiable mapping on (a, b) with a < b. If f 󸀠󸀠 ∈ L1 [a, b], then the following equality for fractional integrals holds: Γ(α + 1) f (a) + f (b) − [ J α+ f (b) + RL Jbα− f (a)] 2 2(b − a)α RL a 1

=

(b − a)2 1 − (1 − t)α+1 − t α+1 󸀠󸀠 f (ta + (1 − t)b)dt. ∫ 2 α+1

(3.16)

0

Proof. By using (3.1), it suffices to verify that 1

∫[(1 − t)α − t α ]f 󸀠 (ta + (1 − t)b)dt 0

1

= (b − a) ∫ 0

1 − (1 − t)α+1 − t α+1 󸀠󸀠 f (ta + (1 − t)b)dt. α+1

(3.17)

Note that 1

∫[(1 − t)α − t α ]f 󸀠 (ta + (1 − t)b)dt 0

1

= − ∫ f 󸀠 (ta + (1 − t)b)d 0

(1 − t)α+1 + t α+1 α+1

󵄨󵄨1 (1 − t)α+1 + t α+1 󸀠 󵄨 =− f (ta + (1 − t)b)󵄨󵄨󵄨 󵄨󵄨0 α+1 1

− (b − a) ∫ 0

(1 − t)α+1 + t α+1 󸀠󸀠 f (ta + (1 − t)b)dt α+1 1

f 󸀠 (b) − f 󸀠 (a) (1 − t)α+1 + t α+1 󸀠󸀠 = − (b − a) ∫ f (ta + (1 − t)b)dt, α+1 α+1 0

(3.18)

3.1 Identities involving Riemann-Liouville fractional integrals | 29

and b

1

f 󸀠 (b) − f 󸀠 (a) = ∫ f 󸀠󸀠 (x)dx = (b − a) ∫ f 󸀠󸀠 (ta + (1 − t)b)dt. a

(3.19)

0

Substituting (3.19) into (3.18), we obtain (3.17). The proof is completed. Lemma 48. Let f : [a, b] → ℝ be a twice-differentiable mapping on (a, b) with a < mb ≤ b. If f 󸀠󸀠 ∈ L1 [a, b], then the following equality for fractional integrals holds: f (a) + f (mb) Γ(α + 1) α [ J α+ f (mb) + RL Jmb − − f (a)] 2 2(mb − a)α RL a 1

(mb − a)2 1 − (1 − t)α+1 − t α+1 󸀠󸀠 = f (ta + m(1 − t)b)dt. ∫ 2 α+1 0

Proof. This is just Lemma 47 on the interval [a, mb] ⊂ [a, b]. Lemma 49. Suppose that f : [a, b] → ℝ be a twice-differentiable mapping. If f 󸀠󸀠 ∈ L[a, a + η(b, a)], then the following equality for fractional integrals holds: f (a) + f (a + η(b, a)) Γ(α + 1) α − α [ J α+ f (a + η(b, a)) + RL J(a+η(b,a)) − f (a)] 2 2η (b, a) RL a 1

=

η2 (b, a) 1 − (1 − t)α+1 − t α+1 󸀠󸀠 f (a + tη(b, a))dt. ∫ 2 α+1

(3.20)

0

Proof. This is just Lemma 47 on the interval [a, a + η(b, a)] ⊂ [a, b]. Lemma 50. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with a < b. If f 󸀠 ∈ L1 [a, b], then the following equality for fractional integrals holds: a+b Γ(α + 1) α α [RL Ja+ f (b) + RL Jb− f (a)] − f ( ) α 2(b − a) 2 1

1

0

0

b−a = [∫ kf 󸀠 (ta + (1 − t)b)dt − ∫[(1 − t)α − t α ]f 󸀠 (ta + (1 − t)b)dt] 2 where 1, k={ −1,

0 ≤ t < 21 , 1 2

≤ t < 1.

Proof. It suffices to note that 1 2

1

I = ∫ f (ta + (1 − t)b)dt − ∫ f 󸀠 (ta + (1 − t)b)dt 0

󸀠

1 2

(3.21)

30 | 3 Fractional integral identities 1

− ∫[(1 − t)α − t α ]f 󸀠 (ta + (1 − t)b)dt 0 1 2

1

= [∫ f 󸀠 (ta + (1 − t)b)dt] + [− ∫ f 󸀠 (ta + (1 − t)b)dt] 0

1 2

1

1

α 󸀠

+ [− ∫(1 − t) f (ta + (1 − t)b)dt] + [∫ t α f 󸀠 (ta + (1 − t)b)dt] 0

0

:= I1 + I2 + I3 + I4 .

(3.22)

Integrating by parts, we have 1 2

1

󵄨󵄨 2 1 󵄨 I1 = ∫ f (ta + (1 − t)b)dt = f (ta + (1 − t)b)󵄨󵄨󵄨 󵄨󵄨0 a−b 󸀠

0

=

1 a+b [−f ( ) + f (b)], b−a 2 1

I2 = − ∫ f 󸀠 (ta + (1 − t)b)dt = 1 2

=

(3.23)

󵄨󵄨1 −1 󵄨 f (ta + (1 − t)b)󵄨󵄨󵄨 󵄨󵄨 1 a−b 2

1 a+b [f (a) − f ( )]. b−a 2

(3.24)

Putting x = ta + (1 − t)b, we have 1

I3 = − ∫(1 − t)α f 󸀠 (ta + (1 − t)b)dt 0

=−

1

󵄨󵄨1 (1 − t)α 1 󵄨 f (ta + (1 − t)b)󵄨󵄨󵄨 − ∫ α(1 − t)α−1 f (ta + (1 − t)b)dt 󵄨 a−b 󵄨0 a − b a

=−

f (b) α a−x + ) ∫( b−a b−a a−b b

and

0

α−1

f (x) dx a−b

f (b) Γ(α + 1) =− + J α f (a), b − a (b − a)α+1 RL b−

(3.25)

1

I4 = ∫ t α f 󸀠 (ta + (1 − t)b)dt 0

1 󵄨󵄨1 tα 1 󵄨󵄨 = f (ta + (1 − t)b)󵄨󵄨 − ∫ αt α−1 f (ta + (1 − t)b)dt 󵄨󵄨0 a − b a−b 0

3.1 Identities involving Riemann-Liouville fractional integrals | 31 a

=−

α−1

α b−x f (a) + ) ∫( b−a b−a b−a b

f (x) dx a−b

f (a) Γ(α + 1) =− + J α f (b). b − a (b − a)α+1 RL a+

(3.26)

Substituting (3.23), (3.24), (3.25), (3.26) into (3.22), it follows that: I=−

a+b Γ(α + 1) 2 α [ J α f (b) + RL Jb− f (a)]. f( )+ b−a 2 (b − a)α+1 RL a+

Thus, by multiplying both sides by proof is completed.

b−a , 2

we have reached the conclusion (3.21). The

Remark 51. In Lemma 50, if we put α = 1, then the equality (3.21) becomes b

a+b 1 ) ∫ f (x)dx − f ( b−a 2 a

1 2

=

1

b−a [∫ f 󸀠 (ta + (1 − t)b)dt − ∫ f 󸀠 (ta + (1 − t)b)dt 2 0

1 2

1

− ∫(1 − 2t)f 󸀠 (ta + (1 − t)b)dt].

(3.27)

0

Lemma 52. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with a < mb ≤ b. If f 󸀠 ∈ L1 [a, b], then the following equality for fractional integrals holds: Γ(α + 1) a + mb α [ J α f (mb) + RL Jmb− f (a)] − f ( ) 2(mb − a)α RL a+ 2 1

=

mb − a ∫[h(t) − (1 − t)α + t α ]f 󸀠 (ta + m(1 − t)b)dt, 2 0

where 1, h(t) = { −1,

0 ≤ t < 21 , 1 2

≤ t < 1.

Proof. This is just Lemma 50 on the interval [a, mb] ⊂ [a, b]. Lemma 53. Let f : [a, b] → ℝ be twice-differentiable mapping on (a, b) with a < b. If f 󸀠󸀠 ∈ L1 [a, b], then Γ(α + 1) a+b [ J α+ f (b) + RL Jbα− f (a)] − f ( ) 2(b − a)α RL a 2

32 | 3 Fractional integral identities 1

(b − a)2 = ∫ m(t)f 󸀠󸀠 (ta + (1 − t)b)dt, 2

(3.28)

0

where 1−(1−t)α+1 −t α+1 , α+1

{t − m(t) = { {1 − t −

Proof. Denote 1 2

t ∈ [0, 21 ),

1−(1−t)α+1 −t α+1 , α+1

t ∈ [ 21 , 1).

1

J = ∫ tf 󸀠󸀠 (ta + (1 − t)b)dt + ∫(1 − t)f 󸀠󸀠 (ta + (1 − t)b)dt := J1 + J2 . 0

(3.29)

1 2

Integrating by parts, we have 1 2

J1 = ∫ tf 󸀠󸀠 (ta + (1 − t)b)dt 0 1

1

2 󵄨󵄨 2 1 1 󵄨 = tf 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 − ∫ f 󸀠 (ta + (1 − t)b)dt 󵄨󵄨0 a − b a−b

0

a+b 1 a+b 1 [f ( = f 󸀠( )− ) − f (b)], 2(a − b) 2 2 (a − b)2

(3.30)

and 1

J2 = ∫(1 − t)f 󸀠󸀠 (ta + (1 − t)b)dt 1 2

1 󵄨󵄨1 1 1 󵄨󵄨 󸀠 = (1 − t)f (ta + (1 − t)b)󵄨󵄨 + ∫ f 󸀠 (ta + (1 − t)b)dt 󵄨󵄨 1 a − b a−b 2 1 2

=−

a+b 1 1 a+b f 󸀠( )+ [f (a) − f ( )]. 2(a − b) 2 2 (a − b)2

(3.31)

Submitting (3.30) and (3.31) to (3.29), it follows that: a+b f (a) + f (b) 2f ( 2 ) J= − . (b − a)2 (b − a)2

Thus, by multiplying both sides of (3.32) by 1

(b−a)2 , 2

we have

(b − a)2 ∫ m(t)f 󸀠󸀠 (ta + (1 − t)b)dt 2 0

(3.32)

3.1 Identities involving Riemann-Liouville fractional integrals | 33

=

a+b f (a) + f (b) − f( ) 2 2 1

−

(b − a)2 1 − (1 − t)α+1 − t α+1 󸀠󸀠 f (ta + (1 − t)b)dt. ∫ 2 α+1

(3.33)

0

On the other hand, from Γ(α + 1) f (a) + f (b) − [ J α+ f (b) + RL Jbα− f (a)] 2 2(b − a)α RL a 1

=

(b − a)2 1 − (1 − t)α+1 − t α+1 󸀠󸀠 f (ta + (1 − t)b)dt, ∫ 2 α+1 0

we obtain 1

(b − a)2 1 − (1 − t)α+1 − t α+1 󸀠󸀠 f (ta + (1 − t)b)dt ∫ 2 α+1 0

f (a) + f (b) Γ(α + 1) [ J α+ f (b) + RL Jbα− f (a)]. = − 2 2(b − a)α RL a

(3.34)

Combining (3.33) and (3.34), we obtain the conclusion (3.28). The proof is completed. Lemma 54. Let f : [a, b] → ℝ be twice-differentiable mapping on (a, b) with a < mb ≤ b. If f 󸀠󸀠 ∈ L1 [a, b], then Γ(α + 1) a + mb α [ J α+ f (mb) + RL Jmb ) − f (a)] − f ( 2(mb − a)α RL a 2 1

=

(mb − a)2 ∫ g(t)f 󸀠󸀠 (ta + m(1 − t)b)dt, 2 0

where 1−(1−t)α+1 −t α+1 , α+1

{t − g(t) = { {1 − t −

1−(1−t)α+1 −t α+1 , α+1

t ∈ [0, 21 ), t ∈ [ 21 , 1).

Proof. This is just Lemma 53 on the interval [a, mb] ⊂ [a, b]. Lemma 55. Suppose that f : [a, b] → ℝ is a twice-differentiable mapping. If f 󸀠󸀠 ∈ L[a, a+ η(b, a)], then the following equality for fractional integrals holds: 2a + η(b, a) Γ(α + 1) α [ J α+ f (a + η(b, a)) + RL J(a+η(b,a)) ) − f (a)] − f ( 2ηα (b, a) RL a 2 1

η2 (b, a) = ∫ m(t)f 󸀠󸀠 (a + tη(b, a))dt, 2 0

34 | 3 Fractional integral identities where 1−(1−t)α+1 −t α+1 , α+1

{t − m(t) = { {1 − t −

1−(1−t)α+1 −t α+1 , α+1

t ∈ [0, 21 ),

t ∈ [ 21 , 1).

Proof. This is just Lemma 53 on the interval [a, a + η(b, a)] ⊂ [a, b]. Lemma 56. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with a < b. If f 󸀠 ∈ L1 [a, b], then the following equality for fractional integrals holds: (x − a)α + (b − x)α Γ(α + 1) f (x) − [ J α− f (a) + RL Jxα+ f (b)] b−a b − a RL x 1

1

0

0

(b − x)α+1 (x − a)α+1 = ∫ t α f 󸀠 (tx + (1 − t)a)dt − ∫ t α f 󸀠 (tx + (1 − t)b)dt. b−a b−a Proof. Integrating by part, we can state 1

α 󸀠

∫ t f (tx + (1 − t)a)dt = t

α f (tx

0

1 1 + (1 − t)a) 󵄨󵄨󵄨󵄨 α−1 f (tx + (1 − t)a) dt 󵄨󵄨 − ∫ αt 󵄨󵄨0 x−a x−a x

=

0

f (x) α (a − u)α−1 f (u) − du ∫ x − a x − a (a − x)α−1 x − a a

x

αΓ(α) f (x) 1 − = ∫(u − a)α−1 f (u)du, α+1 x − a (x − a) Γ(α)

(3.35)

a

and 1

α 󸀠

∫ t f (tx + (1 − t)b)dt = t 0

α f (tx

1 1 + (1 − t)b) 󵄨󵄨󵄨󵄨 α−1 f (tx + (1 − t)b) dt 󵄨󵄨 − ∫ αt 󵄨󵄨0 x−b x−b x

=

0

f (x) α (b − u)α−1 f (u) − du ∫ x − b x − b (b − x)α−1 x − b b

b

f (x) αΓ(α) 1 = + ∫(b − u)α−1 f (u)du. α+1 x − b (b − x) Γ(α)

(3.36)

x

Multiplying both sides of (3.35) and (3.36) by

(x−a)α+1 b−a

and

(b−x)α+1 , respectively, we have b−a

1

(x − a)α+1 (x − a)α f (x) Γ(α + 1) α − J − f (a), ∫ t α f 󸀠 (tx + (1 − t)a)dt = b−a b−a b−a x 0

(3.37)

3.1 Identities involving Riemann-Liouville fractional integrals | 35

and 1

(b − x)α+1 (b − x)α f (x) Γ(α + 1) α − J + f (b). ∫ t α f 󸀠 (tx + (1 − t)b)dt = b−a b−a b−a x

(3.38)

0

From (3.37) to (3.38), we obtain the desired result. The proof is completed. Lemma 57. Let f : [a, b] → ℝ be twice-differentiable mapping on (a, b) with a < b. If f 󸀠󸀠 ∈ L1 [a, b], r > 0, then f (a) + f (b) 2 a+b Γ(α + 1) + f( )− [ J α+ f (b) + RL Jbα− f (a)] r(r + 1) r+1 2 r(b − a)α RL a 1

= (b − a)2 ∫ k(t)f 󸀠󸀠 (ta + (1 − t)b)dt,

(3.39)

0

where α+1

α+1

1−(1−t) −t − { r(α+1) k(t) = { α+1 α+1 1−(1−t) −t − { r(α+1)

Proof. By multiplying both sides of (3.34) by

t , r+1

t ∈ [0, 21 ),

1−t , r+1

t ∈ [ 21 , 1).

2 , r(r+1)

(3.40)

we have

f (a) + f (b) Γ(α + 1) − [ J α+ f (b) + RL Jbα− f (a)] r(r + 1) r(r + 1)(b − a)α RL a 1

(b − a)2 1 − (1 − t)α+1 − t α+1 󸀠󸀠 = f (ta + (1 − t)b)dt. ∫ r(r + 1) α+1

(3.41)

0

2 , we have By multiplying both sides of (3.28) by − r+1

2 a+b Γ(α + 1) f( )− [ J α+ f (b) + RL Jbα− f (a)] r+1 2 (r + 1)(b − a)α RL a 1

(b − a)2 =− ∫ m(t)f 󸀠󸀠 (ta + m(1 − t)b)dt. r+1 0

Hence, it follows (3.41) and (3.42) we obtain f (a) + f (b) 2 a+b Γ(α + 1) + f( )− [ J α+ f (b) + RL Jbα− f (a)] r(r + 1) r+1 2 r(b − a)α RL a 1

(b − a)2 1 − (1 − t)α+1 − t α+1 󸀠󸀠 = f (ta + (1 − t)b)dt ∫ r(r + 1) α+1 0

(3.42)

36 | 3 Fractional integral identities 1

(b − a)2 − ∫ m(t)f 󸀠󸀠 (ta + m(1 − t)b)dt r+1 1

0

= (b − a)2 ∫ k(t)f 󸀠󸀠 (ta + (1 − t)b)dt, 0

where k(t) is defined in (3.40). The proof is completed. Lemma 58. Let f : [a, b] → ℝ be a twice-differentiable mapping on (a, b) with a < mb ≤ b. If f 󸀠󸀠 ∈ L1 [a, b], r > 0, then f (a) + f (mb) 2 a + mb Γ(α + 1) α [ J α+ f (mb) + RL Jmb + f( )− − f (a)] r(r + 1) r+1 2 r(mb − a)α RL a 1

= (mb − a)2 ∫ k(t)f 󸀠󸀠 (ta + m(1 − t)b)dt, 0

where k(t) is defined in (3.40). Proof. This is just Lemma 57 on the interval [a, mb] ⊂ [a, b]. Lemma 59. Let f : Δ ⊂ ℝ2 → ℝ be a partially differentiable mapping on Δ := [a, b] × 𝜕2 f [c, d] in ℝ2 with 0 ≤ a < b, 0 ≤ c < d. If 𝜕t𝜕s ∈ L1 (Δ), then the following equality holds Γ(α + 1)Γ(β + 1) f (a, c) + f (a, d) + f (b, c) + f (b, d) + 4 4(b − a)α (d − c)β α,β

α,β

α,β

α,β

× [(RL Ja+ ,c+ f )(b, d) + (RL Ja+ ,d− f )(b, c) + (RL Jb− ,c+ f )(a, d) + (RL Jb− ,d− f )(a, c)] − A 1 1

𝜕2 f (b − a)(d − c) {∫ ∫ t α sβ (ta + (1 − t)b, sc + (1 − s)d)dsdt = 4 𝜕t𝜕s 0 0

1 1

− ∫ ∫(1 − t)α sβ 0 0

1 1

− ∫ ∫ t α (1 − s)β 0 0

1 1

𝜕2 f (ta + (1 − t)b, sc + (1 − s)d)dsdt 𝜕t𝜕s 𝜕2 f (ta + (1 − t)b, sc + (1 − s)d)dsdt 𝜕t𝜕s

+ ∫ ∫(1 − t)α (1 − s)β 0 0

𝜕2 f (ta + (1 − t)b, sc + (1 − s)d)dsdt}, 𝜕t𝜕s

where A=

Γ(β + 1) β β [(RL Jc+ f )(a, d) + (RL Jc+ f )(b, d) 4(d − c)β Γ(α + 1) β β + (RL Jd− f )(a, c) + (RL Jd− f )(b, c)] + 4(b − a)α

(3.43)

3.1 Identities involving Riemann-Liouville fractional integrals | 37

× [(RL Jaα+ f )(b, c) + (RL Jaα+ f )(b, d) + (RL Jbα− f )(a, c) + (RL Jbα− f )(a, d)].

(3.44)

Proof. Integrating by parts, we get 1 1

I1 = ∫ ∫ t α sβ 0 0 1

= ∫ sβ {t α 0

𝜕2 f (ta + (1 − t)b, sc + (1 − s)d)dsdt 𝜕t𝜕s 󵄨󵄨1 1 𝜕f 󵄨 (ta + (1 − t)b, sc + (1 − s)d)󵄨󵄨󵄨 󵄨󵄨0 a − b 𝜕s 1

−

𝜕f α ∫ t α−1 (ta + (1 − t)b, sc + (1 − s)d)dt}ds a−b 𝜕s 0

1

= ∫ sβ {− 0

󵄨󵄨1 1 𝜕f 󵄨 (a, sc + (1 − s)d)󵄨󵄨󵄨 󵄨󵄨0 b − a 𝜕s 1

𝜕f α + ∫ t α−1 (ta + (1 − t)b, sc + (1 − s)d)dt}ds b−a 𝜕s 0

1

=−

1 𝜕f ∫ sβ (a, sc + (1 − s)d)ds b−a 𝜕s 0

1

1

0

0

α 𝜕f + ∫ t α−1 [∫ sβ (ta + (1 − t)b, sc + (1 − s)d)dt]ds b−a 𝜕s =

1

β 1 f (a, c) − ∫ sβ−1 f (a, sc + (1 − s)d)ds (b − a)(d − c) (b − a)(d − c) 1

−

0

α ∫ t α−1 f (ta + (1 − t)b, c)dt (b − a)(d − c) 0

1 1

αβ + ∫ ∫ t α−1 sβ−1 f (ta + (1 − t)b, sc + (1 − s)d)dsdt. (b − a)(d − c) 0 0

Thus, similarly, integrating by parts, it follows that: 1 1

I2 = ∫ ∫(1 − t)α sβ 0 0

𝜕2 f (ta + (1 − t)b, sc + (1 − s)d)dsdt 𝜕t𝜕s 1

β 1 =− f (b, c) + ∫ sβ−1 f (b, sc + (1 − s)d)ds (b − a)(d − c) (b − a)(d − c) 0

(3.45)

38 | 3 Fractional integral identities 1

αβ α + ∫(1 − t)α−1 f (ta + (1 − t)b, c)dt + (b − a)(d − c) (b − a)(d − c) 0

1 1

× ∫ ∫(1 − t)α−1 sβ−1 f (ta + (1 − t)b, sc + (1 − s)d)dsdt,

(3.46)

0 0

and 1 1

I3 = ∫ ∫ t α (1 − s)β 0 0

=−

𝜕2 f (ta + (1 − t)b, sc + (1 − s)d)dsdt 𝜕t𝜕s 1

β 1 f (a, d) + ∫(1 − s)β−1 f (a, sc + (1 − s)d)ds (b − a)(d − c) (b − a)(d − c) 0

1

αβ α ∫ t α−1 f (ta + (1 − t)b, d)dt − (b − a)(d − c) (b − a)(d − c)

+

0

1 1

× ∫ ∫ t α−1 (1 − s)β−1 f (ta + (1 − t)b, sc + (1 − s)d)dsdt,

(3.47)

0 0

and 1 1

I4 = ∫ ∫(1 − t)α (1 − s)β 0 0

𝜕2 f (ta + (1 − t)b, sc + (1 − s)d)dsdt 𝜕t𝜕s 1

β 1 = f (b, d) − ∫(1 − s)β−1 f (b, sc + (1 − s)d)ds (b − a)(d − c) (b − a)(d − c) 0

1

−

αβ α ∫(1 − t)α−1 f (ta + (1 − t)b, d)dt + (b − a)(d − c) (b − a)(d − c) 0

1 1

× ∫ ∫(1 − t)α−1 (1 − s)β−1 f (ta + (1 − t)b, sc + (1 − s)d)dsdt.

(3.48)

0 0

From (3.45), (3.46), (3.47), (3.48), using changes of the variables x = ta + (1 − t)b and y = sc + (1 − s)d for t, s ∈ [0, 1]2 , we can write I1 − I2 − I3 + I4 =

Γ(β + 1) f (a, c) + f (a, d) + f (b, c) + f (b, d) − 4 (b − a)(d − c)β−1 β

β

β

β

× [(RL Jc+ f )(a, d) + (RL Jc+ f )(b, d) + (RL Jd− f )(a, c) + (RL Jd− f )(b, c)] −

Γ(β + 1) [( J α+ f )(b, c) + (RL Jaα+ f )(b, d) (b − a)α+1 (d − c) RL a

3.1 Identities involving Riemann-Liouville fractional integrals | 39

+ (RL Jbα− f )(a, c) + (RL Jbα− f )(a, d)] Γ(α + 1)Γ(β + 1) α,β α,β [(RL Ja+ ,c+ f )(b, d) + (RL Ja+ ,d− f )(b, c) + (b − a)α (d − c)β α,β

α,β

+ (RL Jb− ,c+ f )(a, d) + (RL Jb− ,d− f )(a, c)].

Multiplying the both sides of (3.49) by pleted.

(b−a)(d−c) , 4

(3.49)

we obtain (3.43). The proof is com-

Lemma 60. Let f : [a, b] → ℝ be a differentiable function and f 󸀠 ∈ L[a, b]. For any 0 < α ≤ 1 and 0 < λ ≤ 1, the following equality for fractional integrals holds: (1 + λ(1 − α))f (b) + (1 − λα)f (a) + λ(2α − 1)f ((1 − α)b + αa) Γ(α + 1) − ( J α+ f (b) + RL Jbα− f (a)) (b − a)α RL a 1−α

= (b − a){ ∫ (t α − (1 − t)α + αλ)f 󸀠 (tb + (1 − t)a)dt 0

1

+ ∫ (t α − (1 − t)α + λ(1 − α))f 󸀠 (tb + (1 − t)a)dt}. 1−α

Proof. It suffices to note that 1−α

I := { ∫ (t α − (1 − t)α + αλ)f 󸀠 (tb + (1 − t)a)dt 0

1

+ ∫ (t α − (1 − t)α + λ(1 − α))f 󸀠 (tb + (1 − t)a)dt} 1−α

1−α

1

= ∫ αλf (tb + (1 − t)a)dt + ∫ λ(1 − α)f 󸀠 (tb + (1 − t)a)dt 0

󸀠

1−α

1

+ ∫(t α − (1 − t)α )f 󸀠 (tb + (1 − t)a)dt 0

:= I1 + I2 + I3 .

(3.50)

Integrating by parts, 1−α

1−α

αλ I1 := ∫ αλf (tb + (1 − t)a)dt = ∫ df (tb + (1 − t)a) b−a 0

󸀠

αλ = [f ((1 − α)b + αa) − f (a)], b−a

0

(3.51)

40 | 3 Fractional integral identities and similarly we get 1

1

λ(1 − α) I2 := ∫ λ(1 − α)f (tb + (1 − t)a)dt = ∫ df (tb + (1 − t)a) b−a 󸀠

1−α

1−α

λ(1 − α) = [f (b) − f ((1 − α)b + αa)], b−a

(3.52)

1

I3 := ∫(t α − (1 − t)α )f 󸀠 (tb + (1 − t)a)dt 0

1

1

1 1 = ∫ t α df (tb + (1 − t)a) − ∫(1 − t)α df (tb + (1 − t)a) b−a b−a 0

0

b

=

α−1

f (b) α x−a − ) ∫( b − a (b − a)2 b−a a

b

+ =

f (x)dx

α−1

f (a) α b−u − ) ∫( b − a (b − a)2 b−a a

f (u)du

f (b) + f (a) Γ(α + 1) Γ(α + 1) α J α− f (a). − RL Ja+ f (b) − α+1 b−a (b − a) (b − a)α+1 RL b

(3.53)

Substituting (3.51), (3.52) and (3.53) into (3.50), we have the result: I :=

(1 + λ(1 − α))f (b) (1 − λα)f (a) λ(2α − 1)f ((1 − α)b + αa) + + b−a b−a b−a Γ(α + 1) α α − ( J + f (b) + RL Jb− f (a)). (b − a)α+1 RL a

(3.54)

Next, by multiplying both sides by (b − a) for (3.54), we have reached the conclusion. The proof is completed. Lemma 61. Let A > 0, B > 0, α ∈ (0, 1). The following equality holds: 1−α

∫ Bt A1−t dt = 0

1−α

A B [( ) ln B − ln A A

− 1].

Proof. 1−α

t 1−t

∫ BA 0

1−α

1−α

t

B dt = A ∫ B A dt = A ∫ ( ) dt A 0

=

t −t

1−α

0

t

1−α

A B A B [( ) ∫ d( ) = ln(B) − ln(A) A ln(B) − ln(A) A 0

The proof is completed.

− 1].

3.1 Identities involving Riemann-Liouville fractional integrals |

41

Lemma 62. Let f : [a, b] → ℝ be a differentiable function and f 󸀠 ∈ L[a, b]. For any 0 < α ≤ 1 and 0 < λ ≤ 1, the following equality for fractional integrals holds: f (b)(1 + α(1 − λ)) + f (a)(1 − α(1 − λ)) Γ(α + 1) − ( J α+ f (b) + RL Jbα− f (a)) 2 2(b − a)α RL a 1

=

b−a ∫(t α + α(1 − λ) − (1 − t)α )f 󸀠 (tb + (1 − t)a)dt. 2 0

Proof. It suffices to note that 1

I = ∫(t α + α(1 − λ) − (1 − t)α )f 󸀠 (tb + (1 − t)a)dt 0

1

1

0

0

= ∫(t α + α(1 − λ))f 󸀠 (tb + (1 − t)a)dt − ∫(1 − t)α f 󸀠 (tb + (1 − t)a)dt := I1 + I2 .

(3.55)

Integrating by parts, 1

I1 := ∫(t α + α(1 − λ))f 󸀠 (tb + (1 − t)a)dt 0

1

f (b)(1 + α(1 − λ)) − f (a)α(1 − λ) α = − ∫ t α−1 f (tb + (1 − t)a)dt b−a b−a 0

b

α−1

=

f (b)(1 + α(1 − λ)) − f (a)α(1 − λ) u−a α − ) ∫( b−a b−a (b − a)2

=

f (b)(1 + α(1 − λ)) − f (a)α(1 − λ) Γ(α + 1) − J α− f (a), b−a (b − a)α+1 RL b

a

f (u)du (3.56)

and similarly we get 1

I2 := − ∫(1 − t)α f 󸀠 (tb + (1 − t)a)dt 0 b

α−1

f (a) α b−u − ) = ∫( b − a (b − a)2 b−a a

=

f (u)du

f (a) Γ(α + 1) − J α+ f (b). b − a (b − a)α+1 RL a

Substituting (3.56) and (3.57) into (3.55), we have I :=

f (b)(1 + α(1 − λ)) + f (a)(1 − α(1 − λ)) b−a

(3.57)

42 | 3 Fractional integral identities

−

Γ(α + 1) ( J α+ f (b) + RL Jbα− f (a)). (b − a)α+1 RL a

Next, multiplying both sides by proof is completed.

b−a 2

(3.58)

for (3.58), we have reached the conclusion. The

Lemma 63. Let f : [a, b] → ℝ be twice-differentiable function and f 󸀠󸀠 ∈ L[a, b]. For any 0 < α ≤ 1 and 0 < m ≤ 1, the following equality for fractional integrals holds: f (mb) + f (a) Γ(α + 1) α ( J α+ f (mb) + RL Jmb − − f (a)) 2 2(mb − a)α RL a 1

(mb − a)2 (1 − (1 − t)α+1 − t α+1 ) 󸀠󸀠 = f (ta + m(1 − t)b)dt. ∫ 2 α+1 0

Proof. It suffices to note that 1

I = ∫((1 − t)α+1 + t α+1 )f 󸀠󸀠 (ta + m(1 − t)b)dt 0

=

1

f 󸀠 (mb) − f 󸀠 (a) α+1 + ∫(t α − (1 − t)α )f 󸀠 (ta + m(1 − t)b)dt. mb − a mb − a

(3.59)

0

By using Lemma 62 with λ = 1, it suffices to verify that 1

∫(t α − (1 − t)α )f 󸀠 (ta + m(1 − t)b)dt 0

=−

f (mb) + f (a) Γ(α + 1) α ( J α+ f (mb) + RL Jmb + − f (a)). mb − a (mb − a)α+1 RL a

(3.60)

Substituting (3.60) to (3.59), we have I=

f 󸀠 (mb) − f 󸀠 (a) (α + 1)(f (mb) + f (a)) − mb − a (mb − a)2 Γ(α + 1) α + ( J α+ f (mb) + RL Jmb − f (a)). (mb − a)α+2 RL a

(3.61)

Note that mb

1

f 󸀠 (mb) − f 󸀠 (a) = ∫ f 󸀠󸀠 (x)dx = (mb − a) ∫ f 󸀠󸀠 (ta + m(1 − t)b)dt. a

(3.62)

0

Substituting (3.62) to (3.61) and multiplying both sides by the result. The proof is completed.

(mb−a)2 2

for (3.61), we obtain

3.1 Identities involving Riemann-Liouville fractional integrals |

43

Lemma 64. Let f : [a, b] → ℝ be a differentiable function and f 󸀠 ∈ L[a, b]. For any 0 < α ≤ 1 and a < x < b, the following equality for fractional integrals holds: (x − a)[(a + x)f (x) − xf (a)] + (b − x)[(x + b)f (x) − xf (b)] (b − a)2 Γ(α + 1) a b − ( J α− f (a) + J α+ f (b)) (b − x)α−1 RL x (b − a)2 (x − a)α−1 RL x 1

=

(x − a)2 ∫(at α + x)f 󸀠 (tx + (1 − t)a)dt (b − a)2 0

1

(b − x)2 − ∫(b(1 − t)α + x)f 󸀠 (tb + (1 − t)x)dt, (b − a)2 0

1

(x − a)2 I = ∫(at α + x)f 󸀠 (tx + (1 − t)a)dt (b − a)2 0

1

(b − x)2 − ∫(b(1 − t)α + x)f 󸀠 (tb + (1 − t)x)dt (b − a)2 0

:= I1 − I2 .

(3.63)

Proof. Integrating by parts, 1

I1 :=

(x − a)2 ∫(at α + x)f 󸀠 (tx + (1 − t)a)dt (b − a)2 0

1

(x − a)[(a + x)f (x) − xf (a)] aα(x − a) α−1 = − ∫ t f (tx + (1 − t)a)dt (b − a)2 (b − a)2 x

0

α−1

=

(x − a)[(a + x)f (x) − xf (a)] aα τ−a − ) ∫( x−a (b − a)2 (b − a)2

=

(x − a)[(a + x)f (x) − xf (a)] aΓ(α + 1) − J α− f (a), 2 (b − a) (b − a)2 (x − a)α−1 RL x

a

f (τ)dτ (3.64)

and similarly we get 1

(b − x)2 I2 := ∫(b(1 − t)α + x)f 󸀠 (tb + (1 − t)x)dt (b − a)2 0

b

α−1

(b − x)[xf (b) − (x + b)f (x)] αb b−τ = + ) ∫( b−x (b − a)2 (b − a)2 x

=

f (τ)dτ

(b − x)[xf (b) − (x + b)f (x)] bΓ(α + 1) + J α+ f (b). (b − a)2 (b − a)2 (b − x)α−1 RL x

(3.65)

44 | 3 Fractional integral identities Substituting (3.64) and (3.65) into (3.63), we have I :=

(x − a)[(a + x)f (x) − xf (a)] + (b − x)[(x + b)f (x) − xf (b)] (b − a)2 Γ(α + 1) a b α − ( J α+ f (b)). RL Jx − f (a) + α−1 2 (b − x)α−1 RL x (b − a) (x − a)

The proof is completed. Lemma 65. Let f : [a, b] → ℝ be twice-differentiable function and f 󸀠󸀠 ∈ L[a, b]. For any 0 < α ≤ 1 and a < x < b, the following equality for fractional integrals holds: (x(α + 1) + b)f (x) − bf (b) (b(α + 1) + x)f (x) − xf (a) 1 [ + ] b−x x−a (b − a)2 x b (α + 1)Γ(α + 1) ( J α+ f (b) + J α− f (a)) − (b − x)α+1 RL x (x − a)α+1 RL x (b − a)2 1

=

(x − b) ∫(t α+1 x + tb)f 󸀠󸀠 (tx + (1 − t)b)dt (b − a)2 0

1

(a − x) + ∫((1 − t)α+1 b + (1 − t)x)f 󸀠󸀠 (ta + (1 − t)x)dt. (b − a)2 0

Proof. It suffices to note that 1

(x − b) I = ∫(t α+1 x + tb)f 󸀠󸀠 (tx + (1 − t)b)dt (b − a)2 0

1

(a − x) + ∫((1 − t)α+1 b + (1 − t)x)f 󸀠󸀠 (ta + (1 − t)x)dt (b − a)2 0

:= I3 + I4 .

(3.66)

Integrating by parts, 1

(x − b) I3 := ∫(t α+1 x + tb)f 󸀠󸀠 (tx + (1 − t)b)dt (b − a)2 0

=

1

(x + b)f 󸀠 (x) 1 − ∫((α + 1)xt α + b)f 󸀠 (tx + (1 − t)b)dt (b − a)2 (b − a)2 0

(x + b)f 󸀠 (x) (x(α + 1) + b)f (x) − bf (b) = − (b − a)2 (x − b)(b − a)2 x

α−1

α(α + 1)x τ−b + ) ∫( x−b (x − b)(b − a)2 b

f (τ)

1 dτ x−b

3.2 Identities involving Hadamard fractional integrals | 45

=

(x + b)f 󸀠 (x) (x(α + 1) + b)f (x) − bf (b) + (b − a)2 (b − x)(b − a)2 x(α + 1)Γ(α + 1) − J α+ f (b), (b − x)α+1 (b − a)2 RL x

(3.67)

and similarly we get

1

(a − x) I4 := ∫((1 − t)α+1 b + (1 − t)x)f 󸀠󸀠 (ta + (1 − t)x)dt (b − a)2 0

=

1

1 −(x + b)f 󸀠 (x) + ∫((α + 1)b(1 − t)α + x)f 󸀠 (ta + (1 − t)x)dt (b − a)2 (b − a)2 0

−(x + b)f 󸀠 (x) xf (a) − (b(α + 1) + x)f (x) + = (b − a)2 (a − x)(b − a)2 a

α−1

a−τ bα(α + 1) + ) ∫( a−x (a − x)(b − a)2 x

f (τ)

1 dτ a−x

−(x + b)f 󸀠 (x) xf (a) − (b(α + 1) + x)f (x) = + (b − a)2 (a − x)(b − a)2 b(α + 1)Γ(α + 1) − J α− f (a). (x − a)α+1 (b − a)2 RL x

(3.68)

Substituting (3.67) and (3.68) into (3.66), we have I :=

1 (x(α + 1) + b)f (x) − bf (b) (b(α + 1) + x)f (x) − xf (a) [ + ] 2 b−x x−a (b − a) (α + 1)Γ(α + 1) x b − ( J α+ f (b) + J α− f (a)). (b − x)α+1 RL x (x − a)α+1 RL x (b − a)2

The proof is completed.

3.2 Identities involving Hadamard fractional integrals The results in this section are taken from [221, 217, 220, 246, 174]. Lemma 66. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with 0 < a < b. If f 󸀠 ∈ L1 [a, b], then the following equality for fractional integrals holds: f (a) + f (b) Γ(α + 1) − [ J α+ f (b) + H Jbα− f (a)] 2 2(ln b − ln a)α H a 1

=

ln b − ln a ∫[(1 − t)α − t α ]et ln a+(1−t) ln b f 󸀠 (et ln a+(1−t) ln b )dt 2 0

1

ln b − ln a = ∫[(1 − t)α − t α ]at b1−t f 󸀠 (at b1−t )dt. 2 0

(3.69)

46 | 3 Fractional integral identities Proof. Denote 1

I = ∫[(1 − t)α − t α ]eln b−t(ln b−ln a) f 󸀠 (eln b−t(ln b−ln a) )dt 0

= I1 + I2 ,

(3.70)

where 1

I1 = ∫(1 − t)α eln b−t(ln b−ln a) f 󸀠 (eln b−t(ln b−ln a) )dt, 0

1

I2 = − ∫ t α eln b−t(ln b−ln a) f 󸀠 (eln b−t(ln b−ln a) )dt. 0

Integrating the term I1 with t over [0, 1], we have 1

I1 = ∫(1 − t)α eln b−t(ln b−ln a) f 󸀠 (eln b−t(ln b−ln a) )dt 0 α f (e

= (1 − t)

ln b−t(ln b−ln a)

1

1

α ) 󵄨󵄨󵄨󵄨 ∫(1 − t)α−1 f (eln b−t(ln b−ln a) )dt 󵄨 − ln a − ln b 󵄨󵄨󵄨0 ln b − ln a 0

a

=

α−1

f (b) ln u − ln a α + ) ∫( 2 ln b − ln a (ln a − ln b) ln b − ln a b

f (u)

du u

b

α du f (b) − = ∫(ln u − ln a)α−1 f (u) ln b − ln a (ln b − ln a)α+1 u a

f (b) Γ(α + 1) = − J α− f (a). ln b − ln a (ln b − ln a)α+1 H b

(3.71)

Similarly, we get 1

I2 = − ∫ t α eln b−t(ln b−ln a) f 󸀠 (eln b−t(ln b−ln a) )dt 0

=t

α f (e

ln b−t(ln b−ln a)

1

1

) 󵄨󵄨󵄨󵄨 α ∫ t α−1 f (eln b−t(ln b−ln a) )dt 󵄨󵄨 − 󵄨 ln b − ln a 󵄨0 ln b − ln a 0

b

=

f (a) α du − ∫(ln b − ln u)α−1 f (u) ln b − ln a (ln b − ln a)α+1 u a

f (a) Γ(α + 1) = − J α+ f (b). ln b − ln a (ln b − ln a)α+1 H a

(3.72)

3.2 Identities involving Hadamard fractional integrals | 47

Substituting (3.71) and (3.72) into (3.70), it follows that: I=

f (a) + f (b) Γ(α + 1) [ J α+ f (b) + H Jbα− f (a)]. − ln b − ln a (ln b − ln a)α+1 H a

Thus, by multiplying both sides by diately. The proof is completed.

ln b−ln a 2

(3.73)

in (3.73), we have conclusion (3.69) imme-

Example 67. Let a = 1, b = e, α = 2, f (x) = x2 . Then all the assumptions in Lemma 66 are satisfied. Clearly, the left-sided term of (3.69)

⇐⇒

1 + e2 1 2 1 − e + = 1, 2 2 2

the right-sided term of (3.69)

⇐⇒

∫(1 − 2t)e2(1−t) dt

1

0

1

1

0

0

= ∫ e2(1−t) dt − 2 ∫ te2(1−t) dt 1 1 3 = (e2 − 1) − 2( e2 − ) 2 4 4 = 1. Lemma 68. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with 0 < a < b. If f 󸀠 ∈ L1 [a, b], then the following equality for fractional integrals holds: Γ(α + 1) (ln x − ln a)α + (ln b − ln x)α f (x) − [ J α− f (a) + H Jxα+ f (b)] ln b − ln a ln b − ln a H x 1

(ln x − ln a)α+1 = ∫ t α et ln x+(1−t) ln a f 󸀠 (et ln x+(1−t) ln a )dt ln b − ln a 0

1

(ln b − ln x)α+1 − ∫ t α et ln x+(1−t) ln b f 󸀠 (et ln x+(1−t) ln b )dt ln b − ln a 1

=

0

(ln x − ln a)α+1 ∫ t α xt a1−t f 󸀠 (xt a1−t )dt ln b − ln a 0

α+1

(ln b − ln x) − ln b − ln a

1

∫ t α xt b1−t f 󸀠 (xt b1−t )dt 0

for any x ∈ (a, b). Proof. Integrating by parts, we can state 1

∫ t α et ln x+(1−t) ln a f 󸀠 (et ln x+(1−t) ln a )dt 0

48 | 3 Fractional integral identities

=t

t ln x+(1−t) ln a

1

1

α ) 󵄨󵄨󵄨󵄨 ∫ t α−1 f (et ln x+(1−t) ln a )dt 󵄨 − ln x − ln a 󵄨󵄨󵄨0 ln x − ln a

α f (e

x

=

0

du f (x) α − ∫(ln u − ln a)α−1 f (u) α+1 ln x − ln a (ln x − ln a) u a

Γ(α + 1) f (x) − = J α− f (a), ln x − ln a (ln x − ln a)α+1 H x

(3.74)

and 1

∫ t α et ln x+(1−t) ln b f 󸀠 (et ln x+(1−t) ln b )dt 0

= tα

1

1 f (et ln x+(1−t) ln b ) 󵄨󵄨󵄨󵄨 α ∫ t α−1 f (et ln x+(1−t) ln b )dt 󵄨󵄨 − ln x − ln b 󵄨󵄨0 ln x − ln b x

=

0

du f (x) α − ∫(ln b − ln u)α−1 f (u) ln x − ln b (ln b − ln x)α+1 u b

f (x) Γ(α + 1) = + J α+ f (b). ln x − ln b (ln b − ln x)α+1 H x Multiplying both sides of (3.74) and (3.75) by we have

(ln x−ln a)α+1 ln b−ln a

(3.75) and

(ln b−ln x)α+1 , ln b−ln a

respectively,

1

(ln x − ln a)α+1 ∫ t α et ln x+(1−t) ln a f 󸀠 (et ln x+(1−t) ln a )dt ln b − ln a 0

Γ(α + 1) (ln x − ln a)α f (x) − J α− f (a), = ln b − ln a ln b − ln a H x

(3.76)

and 1

(ln b − ln x)α+1 ∫ t α et ln x+(1−t) ln b f 󸀠 (et ln x+(1−t) ln b )dt ln b − ln a 0

Γ(α + 1) (ln b − ln x)α f (x) + J α+ f (b). =− ln b − ln a ln b − ln a H x

(3.77)

From (3.76) and (3.77), we obtain the desired result. The proof is completed. Lemma 69. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with 0 < a < b. If f 󸀠 ∈ L1 [a, b], then the following equality for fractional integrals holds: Γ(α + 1) a+b [ J α+ f (b) + H Jbα− f (a)] − f ( ) 2(ln b − ln a)α H a 2

3.2 Identities involving Hadamard fractional integrals | 49 1

b−a = ∫ kf 󸀠 (ta + (1 − t)b)dt 2 0

−

1

ln b − ln a ∫[(1 − t)α − t α ]et ln a+(1−t) ln b f 󸀠 (et ln a+(1−t) ln b )dt 2 0

1

=

b−a ∫ kf 󸀠 (ta + (1 − t)b)dt 2 0

−

1

ln b − ln a ∫[(1 − t)α − t α ]at b1−t f 󸀠 (at b1−t )dt, 2

(3.78)

0

where 1, k={ −1,

0 ≤ t < 21 , 1 2

≤ t < 1.

Proof. Denote 1 2

1

b−a b−a I = ∫ f 󸀠 (ta + (1 − t)b)dt − ∫ f 󸀠 (ta + (1 − t)b)dt 2 2 0

1 2

1

−

ln b − ln a ∫[(1 − t)α − t α ]et ln a+(1−t) ln b f 󸀠 (et ln a+(1−t) ln b )dt 2 0

:= I1 + I2 + I3 + I4 ,

(3.79)

where 1 2

b−a I1 = ∫ f 󸀠 (ta + (1 − t)b)dt, 2 0

I2 = −

1

b−a ∫ f 󸀠 (ta + (1 − t)b)dt, 2 1 2

1

I3 = −

ln b − ln a ∫(1 − t)α eln b−t(ln b−ln a) f 󸀠 (eln b−t(ln b−ln a) )dt, 2 1

I4 =

0

ln b − ln a α ln b−t(ln b−ln a) 󸀠 ln b−t(ln b−ln a) f (e )dt. ∫t e 2 0

50 | 3 Fractional integral identities Integrating by parts, we have 1 2

b−a I1 = ∫ f 󸀠 (ta + (1 − t)b)dt 2 0

1

f (ta + (1 − t)b) 󵄨󵄨󵄨󵄨 2 1 a+b =− )], 󵄨󵄨 = [f (b) − f ( 󵄨󵄨0 2 2 2

(3.80)

and 1

I2 = −

b−a ∫ f 󸀠 (ta + (1 − t)b)dt 2 1 2

=

1 a+b f (ta + (1 − t)b) 󵄨󵄨󵄨󵄨 1 )], 󵄨󵄨 = [f (a) − f ( 1 󵄨 2 2 󵄨2 2

(3.81)

and 1

ln b − ln a I3 = − ∫(1 − t)α eln b−t(ln b−ln a) f 󸀠 (eln b−t(ln b−ln a) )dt 2 0

α f (e

= (1 − t)

ln b−t(ln b−ln a)

2

1

1

) 󵄨󵄨󵄨󵄨 α α−1 ln b−t(ln b−ln a) )dt 󵄨󵄨 + ∫(1 − t) f (e 󵄨󵄨0 2 0

a

=−

α−1

ln u − ln a α f (b) + ) ∫( 2 2(ln a − ln b) ln b − ln a

f (u)

b

du u

b

f (b) α du =− + ∫(ln u − ln a)α−1 f (u) 2 2(ln b − ln a)α u a

Γ(α + 1) f (b) + =− J α− f (a), 2 2(ln b − ln a)α H b

(3.82)

and 1

ln b − ln a α ln b−t(ln b−ln a) 󸀠 ln b−t(ln b−ln a) I4 = f (e )dt ∫t e 2 0

= −t α

1

1 f (eln b−t(ln b−ln a) ) 󵄨󵄨󵄨󵄨 α α−1 ln b−t(ln b−ln a) )dt 󵄨󵄨 + ∫ t f (e 2 󵄨󵄨0 2 a

=−

0

α−1

f (a) α ln b − ln u + ) ∫( 2 2(ln a − ln b) ln b − ln a b

f (u)

du u

3.2 Identities involving Hadamard fractional integrals | 51 b

f (a) α du =− + ∫(ln b − ln u)α−1 f (u) 2 2(ln b − ln a)α u a

f (a) Γ(α + 1) =− + J α+ f (b). 2 2(ln b − ln a)α H a

(3.83)

Submitting (3.80), (3.81), (3.82) and (3.83) into (3.79), it follows that I=

a+b Γ(α + 1) [H Jaα+ f (b) + H Jbα− f (a)] − f ( ). α 2(ln b − ln a) 2

The proof is completed. Lemma 70. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with 0 < a < b. If f 󸀠 ∈ L1 [a, b], then the following equality for fractional integrals holds: Γ(α + 1) [ J α+ f (b) + H Jbα− f (a)] − f (√ab) 2(ln b − ln a)α H a 1

=

ln b − ln a [∫ ket ln a+(1−t) ln b f 󸀠 (et ln a+(1−t) ln b )dt 2 1

0

− ∫[(1 − t)α − t α ]et ln a+(1−t) ln b f 󸀠 (et ln a+(1−t) ln b )dt] 0

=

1

ln b − ln a [∫ kat b1−t f 󸀠 (at b1−t )dt 2 1

0

− ∫[(1 − t)α − t α ]at b1−t f 󸀠 (at b1−t )dt]

(3.84)

0

where 1, k={ −1,

0 ≤ t < 21 , 1 2

≤ t < 1.

Proof. It is not difficult to verify that 1 2

1

I = ∫ eln b−t(ln b−ln a) f 󸀠 (eln b−t(ln b−ln a) )dt − ∫ eln b−t(ln b−ln a) f 󸀠 (eln b−t(ln b−ln a) )dt 0

1 2

1

− ∫[(1 − t)α − t α ]et ln a+(1−t) ln b f 󸀠 (et ln a+(1−t) ln b )dt 0

=

f (b) − f (√ab) f (a) − f (√ab) f (b) Γ(α + 1) + − + J α− f (a) ln b − ln a ln b − ln a ln b − ln a (ln b − ln a)α+1 H b

52 | 3 Fractional integral identities

− =

f (a) Γ(α + 1) + J α+ f (b) ln b − ln a (ln b − ln a)α+1 H a

Γ(α + 1) 2f (√ab) [H Jaα+ f (b) + H Jbα− f (a)] − . α+1 ln b − ln a (ln b − ln a)

(3.85)

a in (3.85), we have the conclusion (3.84) Thus, by multiplying both sides by ln b−ln 2 immediately. The proof is completed.

Lemma 71. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with 0 < a < b. If f 󸀠 ∈ L1 [a, b], then the following equality for fractional integrals holds Γ(α + 1)[H Jxα− f (a) + H Jxα+ f (b)] − [f (a)(ln x − ln a)α + f (b)(ln b − ln x)α ] α+1

= (ln b − ln x)

1

∫(t α − 1)et ln x+(1−t) ln b f 󸀠 (et ln x+(1−t) ln b )dt 0

1

− (ln x − ln a)α+1 ∫(t α − 1)et ln x+(1−t) ln a f 󸀠 (et ln x+(1−t) ln a )dt 1

0

= (ln b − ln x)α+1 ∫(t α − 1)xt b1−t f 󸀠 (xt b1−t )dt 0 α+1

− (ln x − ln a)

1

∫(t α − 1)xt a1−t f 󸀠 (xt a1−t )dt.

(3.86)

0

Proof. Integrating by parts, we have 1

∫(t α − 1)et ln x+(1−t) ln b f 󸀠 (et ln x+(1−t) ln b )dt 0

= (t α − 1)

1 1 f (et ln x+(1−t) ln b ) 󵄨󵄨󵄨󵄨 α t α−1 f (et ln x+(1−t) ln b )dt − ∫ 󵄨 ln x − ln b 󵄨󵄨󵄨0 ln x − ln b x

=

0

f (b) α du − ∫(ln b − ln u)α−1 f (u) ln x − ln b (ln b − ln x)α+1 u b

f (b) Γ(α + 1) =− + J α+ f (b), ln b − ln x (ln b − ln x)α+1 H x

(3.87)

and 1

∫(t α − 1)et ln x+(1−t) ln a f 󸀠 (et ln x+(1−t) ln a )dt 0

1 1 f (et ln x+(1−t) ln a ) 󵄨󵄨󵄨󵄨 α = (t − 1) ∫ t α−1 f (et ln x+(1−t) ln a )dt 󵄨 − ln x − ln a 󵄨󵄨󵄨0 ln x − ln a α

0

3.2 Identities involving Hadamard fractional integrals | 53 x

=

du f (a) α − ∫(ln u − ln a)α−1 f (u) ln x − ln a (ln x − ln a)α+1 u a

Γ(α + 1) f (a) − = J α− f (a). ln x − ln a (ln x − ln a)α+1 H x

(3.88)

Multiplying both sides of (3.87) and (3.88) by (ln b − ln x)α+1 and −(ln x − ln a)α+1 respectively, we have α+1

(ln b − ln x)

1

∫(t α − 1)et ln x+(1−t) ln b f 󸀠 (et ln x+(1−t) ln b )dt 0

= −f (b)(ln b − ln x)α + Γ(α + 1)H Jxα+ f (b),

(3.89)

and α+1

−(ln x − ln a)

1

∫(t α − 1)et ln x+(1−t) ln a f 󸀠 (et ln x+(1−t) ln a )dt 0 α

= −f (a)(ln x − ln a) + Γ(α + 1)H Jxα− f (a).

(3.90)

Following from the equalities (3.89) and (3.90), we obtain the inequality (3.86). The proof is completed. Lemma 72. Let a > 0 and f : [a, b] → ℝ be a once-differentiable mapping such that f 󸀠 is integrable. For 0 ≤ λ ≤ 1, the following equality for fractional integrals holds: 1

(ln b − ln x)α If (x, λ, α, a, b) = ∫(sα + λ)f 󸀠 (e(1−s) ln b+s ln x )e(1−s) ln b+s ln x ds ln b − ln a 0

1

(ln x − ln a)α − ∫(sα + λ)f 󸀠 (e(1−s) ln a+s ln x )e(1−s) ln a+s ln x ds ln b − ln a 1

=

0

(ln b − ln x)α ∫(sα + λ)f 󸀠 (b1−s x s )b1−s x s ds ln b − ln a 0

1

(ln x − ln a)α − ∫(sα + λ)(a1−s x s )a1−s x s ds ln b − ln a 0

where If (x, λ, α, a, b) = λ[

(ln b − ln x)α−1 f (b) + (ln x − ln a)α−1 f (a) ] ln b − ln a

(ln b − ln x)α−1 + (ln x − ln a)α−1 ]f (x) ln b − ln a Γ(α + 1) + [ J α+ f (b) + H Jxα− f (a)]. ln b − ln a H x − (1 + λ)[

54 | 3 Fractional integral identities Proof. Denote 1

(ln b − ln x)α Uf (x, λ, α, a, b) = ∫(sα + λ)f 󸀠 (e(1−s) ln b+s ln x )e(1−s) ln b+s ln x ds, ln b − ln a 0

Vf (x, λ, α, a, b) = −

1

(ln x − ln a)α ∫(sα + λ)f 󸀠 (e(1−s) ln a+s ln x )e(1−s) ln a+s ln x ds. ln b − ln a 0

Integrating by parts twice and changing the variable, we have Uf (x, λ, α, a, b) =

(ln b − ln x)α−1 (ln b − ln x)α−1 λf (b) − (1 + λ)f (x) ln b − ln a ln b − ln a 1

+

(ln b − ln x)α−1 ∫ αsα−1 f (e(1−s) ln b+s ln x )ds ln b − ln a 0

(ln b − ln x)α−1 (ln b − ln x)α−1 =λ [f (b) − f (x)] − f (x) ln b − ln a ln b − ln a 1

α−1

αΓ(α) b + ∫(ln ) Γ(α)(ln b − ln a) x

sα−1 f (e(1−s) ln b+s ln x )ds

0

(ln b − ln x)α−1 (ln b − ln x)α−1 =λ [f (b) − f (x)] − f (x) ln b − ln a ln b − ln a b

α−1

b Γ(α + 1) + ∫(ln ) Γ(α)(ln b − ln a) t x

f (t) dt t

(ln b − ln x)α−1 (ln b − ln x)α−1 Γ(α + 1) =λ [f (b) − f (x)] − f (x) + J α+ (b). ln b − ln a ln b − ln a ln b − ln a H x

Similarly, we get

Vf (x, λ, α, a, b) =−

(ln x − ln a)α−1 (ln x − ln a)α−1 (1 + λ)f (x) + λ f (a) ln b − ln a ln b − ln a 1

+

(ln x − ln a)α−1 ∫ αsα−1 f (e(1−s) ln a+s ln x )ds ln b − ln a 0

(ln x − ln a)α−1 (ln x − ln a)α−1 =− (1 + λ)f (x) + λ f (a) ln b − ln a ln b − ln a 1

+

α−1

αΓ(α) x ∫(ln ) Γ(α)(ln b − ln a) a 0

sα−1 f (e(1−s) ln a+s ln x )ds

(ln x − ln a)α−1 (ln x − ln a)α−1 =− (1 + λ)f (x) + λ f (a) ln b − ln a ln b − ln a

3.2 Identities involving Hadamard fractional integrals | 55 x

+

α−1

Γ(α + 1) t ∫(ln ) Γ(α)(ln b − ln a) a a

f (t) dt t

(ln x − ln a)α−1 Γ(α + 1) (ln x − ln a) (1 + λ)f (x) + λ f (a) + J α− f (a) ln b − ln a ln b − ln a ln b − ln a H x (ln x − ln a)α−1 (ln x − ln a)α−1 Γ(α + 1) = −λ [f (x) − f (a)] − f (x) + J α− f (a). ln b − ln a ln b − ln a ln b − ln a H x α−1

=−

From these two equalities, we have If (x, λ, α, a, b) = Uf (x, λ, α, a, b) + Vf (x, λ, α, a, b). The proof is completed. Lemma 73. Let f : Δ ⊂ ℝ2 → ℝ be a partially differentiable mapping on Δ := [a, b] × 𝜕2 f [c, d] in ℝ2 . If 𝜕t𝜕s ∈ L1 (Δ), then the following equality holds: Γ(α + 1)Γ(β + 1) α,β α,β [(H Jb− ,d− f )(a, c) + (H Ja+ ,d− f )(b, c) bd(ln b − ln a)α+1 (ln d − ln c)β+1 α,β

α,β

+ (H Ja+ ,c+ f )(b, d) + (H Jb− ,c+ f )(a, d)] − A − B 1 1

t

s

a 1 c 1 𝜕2 f (at1 b1−t1 , cs1 d1−s1 )dt1 ds1 = ∫ ∫(1 − t1 )α (1 − s1 )β ( ) ( ) b d 𝜕t1 𝜕s1 0 0

1 1

t

s

a 1 c 1 𝜕2 f − ∫ ∫ t1α (1 − s1 )β ( ) ( ) (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 b d 𝜕s1 𝜕t1 0 0

1 1

t

s

1 c 1 𝜕2 f β a + ∫ ∫ t1α s1 ( ) ( ) (at1 b1−t1 , cs1 d1−s1 )dt1 ds1 b d 𝜕t1 𝜕s1

0 0

1 1

t

s

1 c 1 𝜕2 f β a − ∫ ∫(1 − t1 )α s1 ( ) ( ) (at1 b1−t1 , cs1 d1−s1 )dt1 ds1 , b d 𝜕t1 𝜕s1

0 0

where 1

A=

α [∫(1 − t1 )α−1 f (at1 b1−t1 , d)dt1 bd(ln a − ln b)(ln c − ln d) 0

1

+ ∫(1 − t1 )α−1 f (at1 b1−t1 , c)dt1 0

1

+

∫ t1α−1 f (at1 b1−t1 , c)dt1 0

1

+ ∫ t1α−1 f (at1 b1−t1 , d)dt1 ], 0

56 | 3 Fractional integral identities and 1

s

c 1 𝜕f 1 [∫(1 − s1 )β ( ) (a, cs1 d1−s1 )ds1 B= b(ln a − ln b) d 𝜕s1 0

1

s1

𝜕f β c (a, cs1 d1−s1 )ds1 − ∫ s1 ( ) d 𝜕s1 0

1

s

1 𝜕f β c (b, cs1 d1−s1 )ds1 − ∫ s1 ( ) d 𝜕s1

0

1

s

c 1 𝜕f + ∫(1 − s1 )β ( ) (b, cs1 d1−s1 )ds1 ]. d 𝜕s1 0

Proof. Clearly, 1 1

β−1

∫ ∫ t1α−1 s1 f (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 0 0

Γ(α)Γ(β) α,β (H Ja+ ,c+ f )(b, d), (ln b − ln a)α (ln d − ln c)β

=

1 1

β−1

∫ ∫(1 − t1 )α−1 s1 f (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 0 0

Γ(α)Γ(β) α,β (H Jb− ,c+ f )(a, d), (ln b − ln a)α (ln d − ln c)β

=

1 1

∫ ∫ t1α−1 (1 − s1 )β−1 f (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 0 0

= and

Γ(α)Γ(β) α,β (H Ja+ ,d− f )(b, c) (ln b − ln a)α (ln d − ln c)β

1 1

∫ ∫(1 − t1 )α−1 (1 − s1 )β−1 f (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 0 0

= Thus, we have 1 1

Γ(α)Γ(β) α,β (H Jb− ,d− f )(a, c). (ln b − ln a)α (ln d − ln c)β

∫ ∫(1 − t1 )α−1 (1 − s1 )β−1 f (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 0 0

3.2 Identities involving Hadamard fractional integrals | 57 1 1

+ ∫ ∫ t1α−1 (1 − s1 )β−1 f (at1 b1−t1 , cs1 d1−s1 )dt1 ds1 0 0

1 1

β−1

+ ∫ ∫ t1α−1 s1 f (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 0 0

1 1

β−1

+ ∫ ∫(1 − t1 )α−1 s1 f (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 0 0

=

Γ(α)Γ(β) α,β α,β [(H Jb− ,d− f )(a, c) + (H Ja+ ,d− f )(b, c) (ln b − ln a)α (ln d − ln c)β α,β

α,β

+ (H Ja+ ,c+ f )(b, d) + (H Jb− ,c+ f )(a, d)]. Note that 1 1

∫ ∫(1 − t1 )α−1 (1 − s1 )β−1 f (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 0 0 1

α−1

= ∫(1 − t1 )

1

dt1 [∫(1 − s1 )β−1 f (at1 b1−t1 , cs1 d1−s1 )ds1 ] 0

0 1

1

0

0

1 = − ∫(1 − t1 )α−1 dt1 [∫ f (at1 b1−t1 , cs1 d1−s1 )d(1 − s1 )β ] β 1

󵄨󵄨s1 =1 1 󵄨 = − ∫(1 − t1 )α−1 dt1 [f (at1 b1−t1 , cs1 d1−s1 )(1 − s1 )β 󵄨󵄨󵄨 ] 󵄨󵄨s1 =0 β 0

+

1

1

0

0

1 ∫(1 − t1 )α−1 dt1 [∫(1 − s1 )β df (at1 b1−t1 , cs1 d1−s1 )] β 1

=

1 ∫(1 − t1 )α−1 dt1 f (at1 b1−t1 , d) β 0

+

1

1

0

0

𝜕f 1 𝜕(cs1 d1−s1 ) (at1 b1−t1 , cs1 d1−s1 ) ds1 ] ∫(1 − t1 )α−1 dt1 [∫(1 − s1 )β s 1−s β 𝜕s1 𝜕(c 1 d 1 ) 1

1

1 d c = ∫(1 − t1 )α−1 f (at1 b1−t1 , d)dt1 + ln( ) ∫(1 − t1 )α−1 dt1 β β d 0

0

1

× [∫(1 − s1 )β 0

s

𝜕f c 1 t1 1−t1 s1 1−s1 ) ds1 ] (a b , c d )( d 𝜕(cs1 d1−s1 )

58 | 3 Fractional integral identities 1

1 = ∫(1 − t1 )α−1 f (at1 b1−t1 , d)dt1 β 0

1

1

0

0

s

d c 𝜕f t1 1−t1 s1 1−s1 c 1 + ln( ) ∫(1 − t1 )α−1 dt1 [∫(1 − s1 )β (a b , c d )( ) ds1 ] β d 𝜕s1 d 1

=

1 ∫(1 − t1 )α−1 f (at1 b1−t1 , d)dt1 β 0

1

1

s

c 1 c 𝜕f t1 1−t1 s1 1−s1 d (a b , c d )dt1 ] + ln( ) ∫(1 − s1 )β ( ) ds1 [∫(1 − t1 )α−1 β d d 𝜕s1 0

0

1

=

1 ∫(1 − t1 )α−1 f (at1 b1−t1 , d)dt1 β 0

1

−

1

s

d c c 1 𝜕f t1 1−t1 s1 1−s1 ln( ) ∫(1 − s1 )β ( ) ds1 [∫ (a b , c d )d(1 − t1 )α ] αβ d d 𝜕s1 0

0

1

=

1 ∫(1 − t1 )α−1 f (at1 b1−t1 , d)dt1 β 0

−

1 s 󵄨󵄨t1 =1 d c c 1 𝜕f t1 1−t1 s1 1−s1 󵄨 (a b , c d )(1 − t1 )α 󵄨󵄨󵄨 ] ln( ) ∫(1 − s1 )β ( ) ds1 [ 󵄨󵄨t1 =0 αβ d d 𝜕s1 0

1

1

s

d c c 1 𝜕f t1 1−t1 s1 1−s1 (a b , c d ))]. + ln( ) ∫(1 − s1 )β ( ) ds1 [∫(1 − t1 )α d( αβ d d 𝜕s1 0

0

So we have 1 1

∫ ∫(1 − t1 )α−1 (1 − s1 )β−1 f (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 0 0 1

=

1 ∫(1 − t1 )α−1 f (at1 b1−t1 , d)dt1 β 0

1 s 󵄨󵄨t1 =1 d c c 1 𝜕f t1 1−t1 s1 1−s1 󵄨 − ln( ) ∫(1 − s1 )β ( ) ds1 [ (a b , c d )(1 − t1 )α 󵄨󵄨󵄨 ] 󵄨󵄨t1 =0 αβ d d 𝜕s1 0

1

s

1

d c c 1 𝜕f t1 1−t1 s1 1−s1 ln( ) ∫(1 − s1 )β ( ) ds1 [∫(1 − t1 )α d( (a b , c d ))] + αβ d d 𝜕s1 0

0

3.2 Identities involving Hadamard fractional integrals | 59 1

1 = ∫(1 − t1 )α−1 f (at1 b1−t1 , d)dt1 β 0

+

1

s

d c c 1 𝜕f ln( ) ∫(1 − s1 )β ( ) (b, cs1 d1−s1 )ds1 αβ d d 𝜕s1 0

1

s

c c 1 d ln( ) ∫(1 − s1 )β ( ) ds1 + αβ d d 1

× [∫(1 − t1 )α 1

=

0

0

𝜕2 f

𝜕(at1 b1−t1 )𝜕s1

(at1 b1−t1 , cs1 d1−s1 )

𝜕(at1 b1−t1 ) dt1 ] 𝜕t1

1 ∫(1 − t1 )α−1 f (at1 b1−t1 , d)dt1 β 0

1

s

d c c 1 𝜕f + (b, cs1 d1−s1 )ds1 ln( ) ∫(1 − s1 )β ( ) αβ d d 𝜕s1 0

+

1

× [∫(1 − t1 )α 1

=

1

s

bd c 1 a c ln( ) ln( ) ∫(1 − s1 )β ( ) ds1 αβ b d d

0

0

t

2

a 1 𝜕f (at1 b1−t1 , cs1 d1−s1 )( ) dt1 ] 𝜕t1 𝜕s1 b

1 ∫(1 − t1 )α−1 f (at1 b1−t1 , d)dt1 β 0

1

s

c c 1 𝜕f bd a c d ln( ) ∫(1 − s1 )β ( ) (b, cs1 d1−s1 )ds1 + ln( ) ln( ) + αβ d d 𝜕s1 αβ b d 1 1

0

t

s

a 1 c 1 𝜕2 f × ∫ ∫(1 − t1 )α (1 − s1 )β ( ) ( ) (at1 b1−t1 , cs1 d1−s1 )dt1 ds1 , b d 𝜕t1 𝜕s1 0 0

that is, 1 1

∫ ∫(1 − t1 )α−1 (1 − s1 )β−1 f (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 0 0 1

=

1 ∫(1 − t1 )α−1 f (at1 b1−t1 , d)dt1 β 0

1

s

d c c 1 𝜕f bd a c ln( ) ∫(1 − s1 )β ( ) (b, cs1 d1−s1 )ds1 + ln( ) ln( ) + αβ d d 𝜕s1 αβ b d 0

60 | 3 Fractional integral identities 1 1

t

s

a 1 c 1 𝜕2 f (at1 b1−t1 , cs1 d1−s1 )dt1 ds1 . × ∫ ∫(1 − t1 )α (1 − s1 )β ( ) ( ) b d 𝜕t1 𝜕s1 0 0

Similarly, we have 1 1

β−1

∫ ∫(1 − t1 )α−1 s1 f (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 0 0

1

1 = ∫(1 − t1 )α−1 f (at1 b1−t1 , c)dt1 β 0

1

s

1 bd d c 𝜕f a c β c − (b, cs1 d1−s1 )ds1 − ln( ) ∫ s1 ( ) ln( ) ln( ) αβ d d 𝜕s1 αβ b d

0

1 1

t

s

1 c 1 𝜕2 f β a × ∫ ∫(1 − t1 )α s1 ( ) ( ) (at1 b1−t1 , cs1 d1−s1 )dt1 ds1 , b d 𝜕t1 𝜕s1

0 0

and 1 1

β−1

∫ ∫ t1α−1 s1 f (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 0 0

1

1

s

1 d 1 c 𝜕f β c = ∫ t1α−1 f (at1 b1−t1 , c)dt1 − (a, cs1 d1−s1 )ds1 ln( ) ∫ s1 ( ) β αβ d d 𝜕s1

0

+

0

1 1

t

s

1 bd a c c 1 𝜕2 f β a ln( ) ln( ) ∫ ∫ t1α s1 ( ) ( ) (at1 b1−t1 , cs1 d1−s1 )dt1 ds1 , αβ b d b d 𝜕t1 𝜕s1

0 0

and 1 1

∫ ∫ t1α−1 (1 − s1 )β−1 f (at1 b1−t1 , cs1 d1−s1 )dt1 ds1 0 0

1

=

1 ∫ t α−1 f (at1 b1−t1 , d)dt1 β 1 0

+

1

s

d c c 1 𝜕f bd a c ln( ) ∫(1 − s1 )β ( ) (a, cs1 d1−s1 )ds1 − ln( ) ln( ) αβ d d 𝜕s1 αβ b d 1 1

0

t

s

a 1 c 1 𝜕2 f × ∫ ∫ t1α (1 − s1 )β ( ) ( ) (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 . b d 𝜕s1 𝜕t1 0 0

3.2 Identities involving Hadamard fractional integrals | 61

From these equations, we can derive that Γ(α)Γ(β) α,β α,β [(H Jb− ,d− f )(a, c) + (H Ja+ ,d− f )(b, c) α β (ln b − ln a) (ln d − ln c) α,β

α,β

+ (H Ja+ ,c+ f )(b, d) + (H Jb− ,c+ f )(a, d)]

=

bd a c ln( ) ln( ) αβ b d 1 1

t

s

a 1 c 1 𝜕2 f × ∫ ∫(1 − t1 )α (1 − s1 )β ( ) ( ) (at1 b1−t1 , cs1 d1−s1 )dt1 ds1 b d 𝜕t1 𝜕s1 0 0

−

bd a c ln( ) ln( ) αβ b d 1 1

t

s

a 1 c 1 𝜕2 f × ∫ ∫ t1α (1 − s1 )β ( ) ( ) (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 b d 𝜕s1 𝜕t1 0 0

1 1

t

s

1 bd a c c 1 𝜕2 f β a + ln( ) ln( ) ∫ ∫ t1α s1 ( ) ( ) (at1 b1−t1 , cs1 d1−s1 )dt1 ds1 αβ b d b d 𝜕t1 𝜕s1

bd a c − ln( ) ln( ) αβ b d 1 1

0 0

t

s

1 c 1 𝜕2 f β a (at1 b1−t1 , cs1 d1−s1 )dt1 ds1 × ∫ ∫(1 − t1 )α s1 ( ) ( ) b d 𝜕t1 𝜕s1

0 0

1

1

1 1 + ∫(1 − t1 )α−1 f (at1 b1−t1 , d)dt1 + ∫(1 − t1 )α−1 f (at1 b1−t1 , c)dt1 β β 0

1

+

0

1 1 ∫ t α−1 f (at1 b1−t1 , c)dt1 + ∫ t1α−1 f (at1 b1−t1 , d)dt1 β 1 β 0

+

1

0

1

s1

d c 𝜕f c ln( ) ∫(1 − s1 )β ( ) (a, cs1 d1−s1 )ds1 αβ d d 𝜕s1 0

1

s

1 c 𝜕f d β c − ln( ) ∫ s1 ( ) (a, cs1 d1−s1 )ds1 αβ d d 𝜕s1

0

1

−

s

1 d c 𝜕f β c ln( ) ∫ s1 ( ) (b, cs1 d1−s1 )ds1 αβ d d 𝜕s1

0

1

s

d c c 1 𝜕f + ln( ) ∫(1 − s1 )β ( ) (b, cs1 d1−s1 )ds1 . αβ d d 𝜕s1 0

The proof is completed.

62 | 3 Fractional integral identities Lemma 74. If α ∈ R+ , g : [a, b] → ℝ is integrable and symmetric to √ab, 0 < a < b, then 1 (H Jaα+ g)(b) = (H Jbα− g)(a) = [(H Jaα+ g)(b) + (H Jbα− g)(a)]. 2 Proof. Letting t = (H Jaα+ g)(b)

ab , x

we have b

α−1

1 b = ∫(ln ) Γ(α) t a

b

=

α−1

x 1 ∫(ln ) Γ(α) a

a

α−1

dt 1 x g(t) = ∫(ln ) t Γ(α) a

g(

b

g(x)

a

ab 1 )xd( ) x x

dx = (H Jbα− g)(a). x

The proof is completed. Lemma 75. Let f : [a, b] → ℝ be a differentiable mapping on [a, b] with 0 ≤ a < b and f 󸀠 ∈ L[a, b]. If f is a convex function on [a, b], then for all x ∈ [a, b], λ ∈ [0, 1] and α > 0, we have If (x, λ, α; a, b) = (x − a)

α+1

1

∫(t α − λ)f 󸀠 (tx + (1 − t)a)dt 0

α+1

− (b − x)

1

∫(t α − λ)f 󸀠 (tx + (1 − t)b)dt.

(3.91)

0

Proof. For x ≠ a, we have 1

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

1

󵄨1 = (t − λ)f (tx + (1 − t)a)󵄨󵄨󵄨0 − α ∫ t α−1 f (tx + (1 − t)a)dt α

0

ex

α du = (1 − λ)f (x) + λf (a) − ∫(ln u − a)α−1 f (ln u) (x − a)α u ea

Γ(α + 1) α = (1 − λ)f (x) + λf (a) − J x − (f ∘ ln)(ea ). (x − a)α H (e ) For x ≠ b, we get 1

−(b − x) ∫(t α − λ)f 󸀠 (tx + (1 − t)b)dt 0

(3.92)

3.2 Identities involving Hadamard fractional integrals | 63 1

󵄨1 = (t − λ)f (tx + (1 − t)b)󵄨󵄨󵄨0 − α ∫ t α−1 f (tx + (1 − t)b)dt α

0

eb

α du = (1 − λ)f (x) + λf (b) − ∫(b − ln u)α−1 f (ln u) (b − x)α x u e

Γ(α + 1) α = (1 − λ)f (x) + λf (b) − J x + (f ∘ ln)(eb ). (b − x)α H (e )

(3.93)

Multiplying both sides of (3.92) and (3.93) by (x − a)α and (b − x)α , respectively, one can obtain the desired result immediately. The proof is completed. In this section, we establish a new fractional integral identity which will be used in the next. Concerning (3.91), if we set x = b and x = a, one has If (b, λ, α; a, b) = (b − a)

α+1

1

∫(t α − λ)f 󸀠 (tb + (1 − t)a)dt, 0

and 1

If (a, λ, α; a, b) = −(b − a)α+1 ∫(t α − λ)f 󸀠 (ta + (1 − t)b)dt. 0

In what follows, we give the corresponding results for twice-differential functions. Lemma 76. Let f : [a, b] → ℝ be a twice-differentiable mapping on (a, b) with a < b. If f 󸀠󸀠 ∈ L[a, b], then the following equality for fractional integrals holds: If (x, λ, α; a, b) = (

1 − λ)f 󸀠 (x)[(x − a)α+1 − (b − x)α+1 ] α+1 1

− [(x − a)α+2 ∫ 1

0

+ (b − x)α+2 ∫ 0

t α+1 − λ(α + 1)t 󸀠󸀠 f (tx + (1 − t)a)dt α+1

t α+1 − λ(α + 1)t 󸀠󸀠 f (tx + (1 − t)b)dt]. α+1

Proof. Note that 1

∫ 0

t α+1 − λ(α + 1)t 󸀠 df (tx + (1 − t)a) α+1 1

= (x − a) ∫ 0

t α+1 − λ(α + 1)t 󸀠󸀠 f (tx + (1 − t)a)dt α+1

(3.94)

64 | 3 Fractional integral identities 󵄨󵄨1 1 t α+1 − λ(α + 1)t 󸀠 󵄨 f (tx + (1 − t)a)󵄨󵄨󵄨 − ∫ f 󸀠 (tx + (1 − t)a)(t α − λ)dt] =[ 󵄨󵄨0 α+1 0

1

=(

1 − λ)f 󸀠 (x) − ∫ f 󸀠 (tx + (1 − t)a)(t α − λ)dt α+1

(3.95)

0

and 1

∫ 0

t α+1 − λ(α + 1)t 󸀠 df (tx + (1 − t)b) α+1 1

= −(b − x) ∫ 0

=

t α+1 − λ(α + 1)t 󸀠󸀠 f (tx + (1 − t)b)dt α+1 1

󵄨󵄨1 t α+1 − λ(α + 1)t 󸀠 󵄨 f (tx + (1 − t)b)󵄨󵄨󵄨 − ∫ f 󸀠 (tx + (1 − t)b)(t α − λ)dt 󵄨󵄨0 α+1 0

1

=(

1 − λ)f 󸀠 (x) − ∫ f 󸀠 (tx + (1 − t)b)(t α − λ)dt. α+1

(3.96)

0

Multiplying both sides of (3.95) and (3.96) by (x − a)α+1 and −(b − x)α+1 , respectively, we have 1

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

= (x − a)

(

1 − λ)f 󸀠 (x) α+1 1

α+2

− (x − a)

∫ 0

t α+1 − λ(α + 1)t 󸀠󸀠 f (tx + (1 − t)a)dt, α+1

(3.97)

and 1

−(b − x)α+1 ∫ f 󸀠 (tx + (1 − t)b)(t α − λ)dt 0

= −(b − x)α+1 ( α+2

− (b − x)

1 − λ)f 󸀠 (x) α+1 1

∫ 0

t α+1 − λ(α + 1)t 󸀠󸀠 f (tx + (1 − t)b)dt. α+1

(3.98)

By adding the results of (3.97) and (3.98), we obtain the result. The proof is completed.

4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 4.1 Inequalities via convex functions The results in this section are taken from [159, 251, 105].

4.1.1 Main results Theorem 77. Let f : [a, b] → ℝ be a positive function with 0 ≤ a < b and f ∈ L1 [a, b]. If f is a convex function on [a, b], then the following inequality for fractional integrals hold f(

a+b f (a) + f (b) Γ(α + 1) [ J α+ f (b) + RL Jbα− f (a)] ≤ )≤ . 2 2(b − a)α RL a 2

Proof. Since f is a convex function on [a, b], we have for x, y ∈ [a, b] with λ = f(

(4.1) 1 2

x+y f (x) + f (y) )≤ , 2 2

i. e., with x = ta + (1 − t)b, y = (1 − t)a + tb, 2f (

a+b ) ≤ f (ta + (1 − t)b) + f ((1 − t)a + tb). 2

(4.2)

Multiplying both sides of (4.2) by t α−1 , then integrating the resulting inequality with respect to t over [0, 1], we obtain 1

1

0

0

2 a+b f( ) ≤ ∫ t α−1 f (ta + (1 − t)b)dt + ∫ t α−1 f ((1 − t)a + tb)dt α 2 Γ(α) = [ J α+ f (b) + RL Jbα− f (a)], (b − a)α RL a i. e., f(

a+b Γ(α + 1) )≤ [ J α+ f (b) + RL Jbα− f (a)], 2 2(b − a)α RL a

and the first inequality in (4.1) is proved. For the proof of the second inequality in (4.1), we first note that if f is a convex function, then, for t ∈ [0, 1], it yields f (ta + (1 − t)b) ≤ tf (a) + (1 − t)f (b) https://doi.org/10.1515/9783110523621-004

66 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals and f ((1 − t)a + tb) ≤ (1 − t)f (a) + tf (b). By adding these inequalities, we have f (ta + (1 − t)b) + f ((1 − t)a + tb)

≤ tf (a) + (1 − t)f (b) + (1 − t)f (a) + tf (b).

(4.3)

Then multiplying both sides of (4.3) by t α−1 and integrating the resulting inequality with respect to t over [0, 1], we obtain 1

∫t

α−1

1

f (ta + (1 − t)b)dt + ∫ t

α−1

f ((1 − t)a + tb)dt ≤ [f (a) + f (b)] ∫ t α−1 dt,

0

0

1

0

that is, Γ(α) f (a) + f (b) [RL Jaα+ f (b) + RL Jbα− f (a)] ≤ . α (b − a) α The proof is completed. Theorem 78. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with a < b, such that f ∈ L1 [a, b]. If f 󸀠 is convex on [a, b], then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) b−a 1 ≤ (1 − α )(|f 󸀠 (a)| + |f 󸀠 (b)|). 2(α + 1) 2 Proof. Using Lemma 42 and the convexity of f , we find 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) 󵄨 󵄨󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) 1

≤

b−a 󵄨 󵄨 ∫ |(1 − t)α − t α |󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 0

1

≤

b−a ∫ |(1 − t)α − t α |[t|f 󸀠 (a)| + (1 − t)|f 󸀠 (b)|]dt 2 0

1 2

b−a = {∫[(1 − t)α − t α ][t|f 󸀠 (a)| + (1 − t)|f 󸀠 (b)|]dt 2 0

(4.4)

4.1 Inequalities via convex functions | 67 1

+ ∫[t α − (1 − t)α ][t|f 󸀠 (a)| + (1 − t)|f 󸀠 (b)|]dt} 1 2

:=

b−a (K1 + K2 ). 2

(4.5)

Calculating K1 and K2 , we have 1 2

1 2

K1 = |f 󸀠 (a)|(∫(1 − t)α tds − ∫ t α+1 ds) 0

0 1 2

1 2

+ |f 󸀠 (b)|(∫(1 − t)α+1 ds − ∫ t α (1 − t)ds) 0

= |f 󸀠 (a)|(

0

( 21 )α+1

( 1 )α+1 1 1 − ) + |f 󸀠 (b)|( − 2 ) (α + 1)(α + 2) α+1 α+2 α+1

(4.6)

and 1

K2 = |f (a)|[∫ t 󸀠

α+1

1

ds − ∫(1 − t)α tds]

1 2

1 2

1

1 2

= |f 󸀠 (a)|(

α

1

+ |f (b)|[∫ t (1 − t)ds − ∫(1 − t)α+1 ds] 󸀠

1 2

( 1 )α+1 ( 1 )α+1 1 1 − 2 ) + |f 󸀠 (b)|( − 2 ). α+2 α+1 (α + 1)(α + 2) α+1

(4.7)

Thus, if we use (4.6) and (4.7) in (4.5), we obtain the inequality of (4.4). The proof is completed. In a similar vein, some fractional Hermite-Hadamard’s inequalities via convex functions are presented in Zhu, Fečkan and Wang [251]. Theorem 79. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with a < b. If |f 󸀠 | is convex on [a, b], then the following inequality for fractional integrals holds: 󵄨󵄨 Γ(α + 1) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 α α )󵄨󵄨 [RL Ja+ f (b) + RL Jb− f (a)] − f ( 󵄨󵄨 α 󵄨󵄨 2 󵄨󵄨 2(b − a) 1 b−a (α + 3 − α−1 )[|f 󸀠 (a)| + |f 󸀠 (b)|]. ≤ 4(α + 1) 2 Proof. Using Lemma 50 and the convexity of |f 󸀠 |, we have 󵄨󵄨 Γ(α + 1) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 α α [RL Ja+ f (b) + RL Jb− f (a)] − f ( )󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 󵄨󵄨 2(b − a)

(4.8)

68 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1 2

1

b−a 󵄨 󵄨 󵄨 󵄨 ≤ [∫ 󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt + ∫ 󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 0

1 2

1

󵄨 󵄨 + ∫ |(1 − t)α − t α |󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt] 0

1 2

1

b−a ≤ [∫[t|f 󸀠 (a)| + (1 − t)|f 󸀠 (b)|]dt + ∫[t|f 󸀠 (a)| + (1 − t)|f 󸀠 (b)|]dt 2 0

1 2

1

󵄨 󵄨 + ∫ |(1 − t)α − t α |󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt] 0

1 2

≤

1

b−a [∫[t|f 󸀠 (a)| + (1 − t)|f 󸀠 (b)|]dt + ∫[t|f 󸀠 (a)| + (1 − t)|f 󸀠 (b)|]dt 2 0

1 2

1 2

+ ∫[(1 − t)α − t α ][t|f 󸀠 (a)| + (1 − t)|f 󸀠 (b)|]dt 0 1

+ ∫[t α − (1 − t)α ][t|f 󸀠 (a)| + (1 − t)|f 󸀠 (b)|]dt] 1 2

b−a (K1 + K2 + K3 + K4 ). 2 After some calculation, we obtain

(4.9)

:=

1 2

1

1

|f 󸀠 (a)| 2 󵄨󵄨󵄨󵄨 2 t 2 󵄨󵄨󵄨 2 K1 = ∫[t|f (a)| + (1 − t)|f (b)|]dt = t 󵄨󵄨 + |f 󸀠 (b)|(t − )󵄨󵄨󵄨 󵄨󵄨0 2 2 󵄨󵄨0 󸀠

󸀠

0

=

1 󸀠 3 |f (a)| + |f 󸀠 (b)|, 8 8

(4.10)

1

K2 = ∫[t|f 󸀠 (a)| + (1 − t)|f 󸀠 (b)|]dt = 1 2

=

1 1 |f 󸀠 (a)| 2 󵄨󵄨󵄨󵄨 t 2 󵄨󵄨󵄨 t 󵄨󵄨 + |f 󸀠 (b)|(t − )󵄨󵄨󵄨 󵄨󵄨 1 2 2 󵄨󵄨 21 2

3 󸀠 1 |f (a)| + |f 󸀠 (b)|, 8 8

(4.11)

1 2

K3 = ∫[(1 − t)α − t α ][t|f 󸀠 (a)| + (1 − t)|f 󸀠 (b)|]dt 0

1 2

1 2

0

0

= |f 󸀠 (a)|[∫ t(1 − t)α dt − ∫ t α+1 dt]

4.1 Inequalities via convex functions | 69 1 2

α+1

+ |f (b)|[∫(1 − t) 󸀠

0

= |f 󸀠 (a)|[

1 2

dt − ∫(1 − t)t α dt] 0

( 21 )α+1

( 1 )α+1 1 1 − ] + |f 󸀠 (b)|[ − 2 ] (α + 1)(α + 2) α+1 (α + 2) α+1

(4.12)

and 1

K4 = ∫[t α − (1 − t)α ][t|f 󸀠 (a)| + (1 − t)|f 󸀠 (b)|]dt 1 2

1

1

= |f 󸀠 (a)|[∫ t α+1 dt − ∫ t(1 − t)α dt] 1 2

1 2

1

1 2

= |f 󸀠 (a)|[

α

1

+ |f (b)|[∫(1 − t)t dt − ∫(1 − t)α+1 dt] 󸀠

1 2

( 1 )α+1 ( 1 )α+1 1 1 − 2 ] + |f 󸀠 (b)|[ − 2 ]. (α + 2) α+1 (α + 1)(α + 2) α+1

(4.13)

Substituting (4.10), (4.11), (4.12), (4.13) into (4.9), we obtain the inequality (4.8). The proof is completed. Remark 80. If we take α = 1 in Theorem 79, then the inequality (4.8) becomes the following inequality: b 󵄨󵄨 󵄨 󵄨󵄨 1 a + b 󵄨󵄨󵄨󵄨 3(b − a) 󸀠 󵄨󵄨 f (x)dx − f ( ) (|f (a)| + |f 󸀠 (b)|). ∫ 󵄨󵄨 ≤ 󵄨󵄨 b − a 󵄨󵄨 2 8 󵄨󵄨 󵄨 a

(4.14)

First, we treat the inequalities for convex functions. Theorem 81. Let f : [a, b] → ℝ be a differentiable mapping. If |f 󸀠 | ∈ L[a, b] and |f 󸀠 | is a convex function, then for any 0 < α ≤ 1, 0 < λ ≤ 1, the following inequalities for fractional integrals hold: Case 1: 21 ≤ α ≤ 1. 󵄨󵄨 󵄨󵄨 󵄨󵄨(1 + λ(1 − α))f (b) + (1 − λα)f (a) + λ(2α − 1)f ((1 − α)b + αa) 󵄨󵄨 󵄨󵄨 Γ(α + 1) 󵄨 (RL Jaα+ f (b) + RL Jbα− f (a))󵄨󵄨󵄨 ≤ (b − a){|f 󸀠 (b)|φ(α, λ) + |f 󸀠 (a)|ϕ(α, λ)}, − α 󵄨󵄨 (b − a) where φ(α, λ) =

α+1

1 1 1 αλ (1 − α)(3 − 2α) + − ( ) 2 α+1 α+2 2

,

70 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals

ϕ(α, λ) =

α

αλ 1 1 (1 − α)(2α + 1) + (1 − ( ) ). 2 α+1 2

Case 2: 0 < α < 21 . 󵄨󵄨 󵄨󵄨 󵄨󵄨(1 + λ(1 − α))f (b) + (1 − λα)f (a) + λ(2α − 1)f ((1 − α)b + αa) 󵄨󵄨 󵄨󵄨 Γ(α + 1) 󵄨󵄨 α 󸀠 󸀠 󸀠 󸀠 α − ( J f (b) + J f (a)) + − 󵄨󵄨 ≤ (b − a){|f (b)|φ (α, λ) + |f (a)|ϕ (α, λ)}, RL b 󵄨󵄨 (b − a)α RL a where α

1 1 1 αα+1 (3α + 2) λα ( −( ) )+ + (1 − α)(3 − 2α), α+1 α+2 2 (α + 1)(α + 2) 2

φ󸀠 (α, λ) =

α

1 1 1 − (1 − α)α+2 λα (1 − ( ) ) + + (1 + 2α)(1 − α). α+1 2 (α + 1)(α + 2) 2

ϕ󸀠 (α, λ) =

Proof. Step 1. We check the results for Case 1. Using Lemma 60, |f 󸀠 | ∈ L[a, b] and |f 󸀠 | is a convex function, we have 󵄨󵄨 󵄨󵄨 󵄨󵄨(1 + λ(1 − α))f (b) + (1 − λα)f (a) + λ(2α − 1)f ((1 − α)b + αa) 󵄨󵄨 󵄨󵄨 Γ(α + 1) 󵄨 (RL Jaα+ f (b) + RL Jbα− f (a))󵄨󵄨󵄨 − α 󵄨󵄨 (b − a) 󵄨󵄨 1−α 󵄨󵄨 󵄨󵄨 󵄨󵄨 ≤ (b − a){󵄨󵄨󵄨 ∫ (t α − (1 − t)α + αλ)f 󸀠 (tb + (1 − t)a)dt 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨 0 1

󵄨󵄨 󵄨󵄨 2 󵄨󵄨 󵄨󵄨 + 󵄨󵄨󵄨 ∫ (t α − (1 − t)α + λ(1 − α))f 󸀠 (tb + (1 − t)a)dt 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 1−α

󵄨󵄨 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 α α 󸀠 󵄨 + 󵄨󵄨 ∫(t − (1 − t) + λ(1 − α))f (tb + (1 − t)a)dt 󵄨󵄨󵄨} 󵄨󵄨 󵄨󵄨 󵄨 󵄨1 2

1−α

≤ (b − a){ ∫ ((1 − t)α − t α + αλ)(t|f 󸀠 (b)| + (1 − t)|f 󸀠 (a)|)dt 0 1 2

+ ∫ ((1 − t)α − t α + λ(1 − α))(t|f 󸀠 (b)| + (1 − t)|f 󸀠 (a)|)dt 1−α 1

+ ∫(t α − (1 − t)α + λ(1 − α))(t|f 󸀠 (b)| + (1 − t)|f 󸀠 (a)|)dt} 1 2

= (b − a)[K1 + K2 + K3 ].

(4.15)

4.1 Inequalities via convex functions | 71

Integrating by parts, 1−α

K1 := ∫ ((1 − t)α − t α + αλ)(t|f 󸀠 (b)| + (1 − t)|f 󸀠 (a)|)dt 0

1−α

= |f (b)|{ ∫ ((1 − t)α t − t α+1 + αλt)dt} 󸀠

0

1−α

+ |f 󸀠 (a)|{ ∫ ((1 − t)α+1 − t α (1 − t) + αλ(1 − t))dt} 0

1 − αα+2 (1 − α)α+2 (1 − α)αα+1 + − = |f 󸀠 (b)|{− α+1 (α + 1)(α + 2) α+2 + + and similarly we get

λα(1 − α)2 1 − αα+2 } + |f 󸀠 (a)|{ 2 α+2

α(1 − α)α+1 (1 − α)α+2 λα(1 − α2 ) − + }, α+1 (α + 1)(α + 2) 2

(4.16)

1 2

K2 := ∫ ((1 − t)α − t α + λ(1 − α))(t|f 󸀠 (b)| + (1 − t)|f 󸀠 (a)|)dt 1−α

1 2

= |f 󸀠 (b)|{ ∫ ((1 − t)α t − t α+1 + λ(1 − α)t)dt} 1−α

1 2

+ |f (a)|{ ∫ ((1 − t)α+1 − t α (1 − t) + λ(1 − α)(1 − t))dt} 󸀠

1−α

= |f 󸀠 (b)|{− + + and

α+1

1 1 ( ) α+1 2

+

αα+1 (2 − α2 ) (1 − α)α+2 + (α + 1)(α + 2) α+2

α+1

λ(1 − α)(1 − 4(1 − α)2 ) 1 1 } + |f 󸀠 (a)|{− ( ) 8 α+1 2

α2 + α + 1 αα+2 λ(1 − α)(1 − 4α2 ) (1 − α)α+1 + + } (α + 1)(α + 2) α+2 8

1

K3 := ∫(t α − (1 − t)α + λ(1 − α))(t|f 󸀠 (b)| + (1 − t)|f 󸀠 (a)|)dt 1 2

1

= |f (b)|{∫(t α+1 − (1 − t)α t + λ(1 − α)t)dt} 󸀠

1 2

(4.17)

72 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

+ |f (a)|{∫(t α (1 − t) − (1 − t)α+1 + λ(1 − α)(1 − t))dt} 󸀠

1 2

= |f 󸀠 (b)|{

α+1

1 1 1 + ( ) α + 2 (α + 1)(α + 2) 2

+ |f 󸀠 (a)|{

+ α+1

1 1 1 − ( ) (α + 1)(α + 2) α + 1 2

3λ(1 − α) } 8 +

λ(1 − α) }. 8

(4.18)

Substituting (4.16), (4.17) and (4.18) into (4.15), we have the results for Case 1. Step 2. We check the results for Case 2. In fact, we have 󵄨󵄨 󵄨󵄨 󵄨󵄨(1 + λ(1 − α))f (b) + (1 − λα)f (a) + λ(2α − 1)f ((1 − α)b + αa) 󵄨󵄨 󵄨󵄨 Γ(α + 1) 󵄨 (RL Jaα+ f (b) + RL Jbα− f (a))󵄨󵄨󵄨 − α (b − a) 󵄨󵄨 1 2

󵄨 󵄨 ≤ (b − a)[∫((1 − t)α − t α + αλ)󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt 1−α

0

󵄨 󵄨 + ∫ (t α − (1 − t)α + λ(1 − α))󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt 1 2

1

󵄨 󵄨 + ∫ (t α − (1 − t)α + λ(1 − α))󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt] 1−α

1 2

≤ (b − a)[∫((1 − t)α − t α + αλ)(t|f 󸀠 (b)| + (1 − t)|f 󸀠 (a)|)dt 1−α

0

+ ∫ (t α − (1 − t)α + λα)(t|f 󸀠 (b)| + (1 − t)|f 󸀠 (a)|)dt 1 2

1

+ ∫ (t α − (1 − t)α + λ(1 − α))(t|f 󸀠 (b)| + (1 − t)|f 󸀠 (a)|)dt] 1−α

= (b − a)[K4 + K5 + K6 ]. Integrating by parts, 1 2

K4 := ∫((1 − t)α − t α + αλ)(t|f 󸀠 (b)| + (1 − t)|f 󸀠 (a)|)dt 0 1 2

= |f 󸀠 (b)| ∫((1 − t)α t − t α+1 + αλt)dt 0

(4.19)

4.1 Inequalities via convex functions | 73 1 2

+ |f (a)| ∫((1 − t)α+1 − t α (1 − t) + αλ(1 − t))dt 󸀠

0

= |f 󸀠 (b)|{−

α+1

1 1 ( ) α+1 2

+ |f 󸀠 (a)|{

+

1 αλ + } (α + 1)(α + 2) 8 α+1

1 1 1 − ( ) α+2 α+1 2

+

3αλ }, 8

(4.20)

and similarly we get 1−α

K5 := ∫ (t α − (1 − t)α + αλ)(t|f 󸀠 (b)| + (1 − t)|f 󸀠 (a)|)dt 1 2

1−α

= |f (b)| ∫ (t α+1 − (1 − t)α t + αλt)dt 󸀠

1 2

1−α

+ |f (a)| ∫ (t α (1 − t) − (1 − t)α+1 + αλ(1 − t))dt 󸀠

1 2

= |f 󸀠 (b)|{− + +

α+1

1 1 ( ) α+1 2

+

αα+1 (2 − α2 ) (1 − α)α+2 + (α + 1)(α + 2) α+2 α+1

αλ 1 1 1 ((1 − α)2 − )} + |f 󸀠 (a)|{− ( ) 2 4 α+1 2 1 λα 1 αα+2 + + ( − α2 )} α + 2 (α + 1)(α + 2) 2 4

+

α(1 − α)α+1 α+1 (4.21)

and 1

K6 := ∫ (t α − (1 − t)α + λ(1 − α))(t|f 󸀠 (b)| + (1 − t)|f 󸀠 (a)|)dt 1−α

1

= |f (b)| ∫ (t α+1 − (1 − t)α t + λ(1 − α)t)dt 󸀠

1−α

1

+ |f 󸀠 (a)| ∫ (t α (1 − t) − (1 − t)α+1 + λ(1 − α)(1 − t))dt 1−α

1 − (1 − α)α+2 αα+1 (α2 + 2α − 2) = |f 󸀠 (b)|{ + α+2 (α + 1)(α + 2) + −

λ(1 − α) 1 − (1 + α)αα+2 (1 − (1 − α)2 )} + |f 󸀠 (a)|{ 2 (α + 1)(α + 2) (1 − α)α+1 λ(1 − α) (α2 + α + 1) + α2 }. (α + 1)(α + 2) 2

(4.22)

74 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals Substituting (4.20), (4.21) and (4.22) into (4.19), we derive the results for Case 2. The proof is completed. Second, we develop the inequalities for s-convex functions. Theorem 82. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with a < b. If |f 󸀠 | ∈ L[a, b] and |f 󸀠 | is an s-convex function, then for any 0 < α ≤ 1 and 0 < λ ≤ 1, the following inequality for fractional integrals holds: 󵄨󵄨󵄨 󵄨󵄨(1 + λ(1 − α))f (b) + (1 − λα)f (a) + λ(2α − 1)f ((1 − α)b + αa) 󵄨󵄨 󵄨 󵄨󵄨 Γ(α + 1) 󵄨󵄨 α α ̃ λ)}, ̃(α, λ) + ϕ(α, − ( J f (b) + J f (a)) + − 󵄨󵄨 ≤ (b − a){φ RL RL a b 󵄨󵄨 (b − a)α

where

̃(α, λ) = φ

αλ 󵄨󵄨 󵄨 λ(1 − α) 󵄨󵄨 󵄨 󵄨f ((1 − α)b + αa) − f (a)󵄨󵄨󵄨 + 󵄨f (b) − f ((1 − α)b + αa)󵄨󵄨󵄨, b − a󵄨 b−a 󵄨

̃ λ) = ( ϕ(α,

1 − ( 21 )αp αp + 1

1 p

) {[

1/q

|f 󸀠 (b)|q 3|f 󸀠 (a)|q + ] 8 8

+[

3|f 󸀠 (b)|q |f 󸀠 (a)|q + ] 8 8

1/q

},

and 1/p + 1/q = 1.

Proof. Using Lemma 60 via f 󸀠 ∈ L[a, b] and if |f 󸀠 | is a convex function, we have 󵄨󵄨 󵄨󵄨 󵄨󵄨(1 + λ(1 − α))f (b) + (1 − λα)f (a) + λ(2α − 1)f ((1 − α)b + αa) 󵄨󵄨 󵄨󵄨 Γ(α + 1) 󵄨 − (RL Jaα+ f (b) + RL Jbα− f (a))󵄨󵄨󵄨 α 󵄨󵄨 (b − a) 1 󵄨󵄨 1−α 󵄨󵄨 󵄨󵄨 󵄨󵄨 ≤ (b − a){λα󵄨󵄨󵄨 ∫ f 󸀠 (tb + (1 − t)a)dt| + λ(1 − α)| ∫ f 󸀠 (tb + (1 − t)a)dt 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨 1−α 0 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 󵄨󵄨 + 󵄨󵄨󵄨 ∫(t α − (1 − t)α )f 󸀠 (tb + (1 − t)a)dt 󵄨󵄨󵄨} 󵄨󵄨 󵄨󵄨 󵄨0 󵄨

≤ (b − a)[S1 + S2 + S3 ].

(4.23)

Integrating by parts, 󵄨 (1−α)b+αa 󵄨󵄨 󵄨󵄨 1−α 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 λα 󵄨󵄨󵄨󵄨 󸀠 󸀠 󵄨 󵄨 S1 := λα󵄨󵄨 ∫ f (tb + (1 − t)a)dt 󵄨󵄨 = 󵄨 ∫ f (x)dx 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 b − a 󵄨󵄨󵄨 󵄨 󵄨 󵄨 0 󵄨 a = and similarly we get

αλ 󵄨󵄨 󵄨 󵄨f ((1 − α)b + αa) − f (a)󵄨󵄨󵄨, b − a󵄨

b 󵄨 󵄨󵄨 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 󵄨󵄨 λ(1 − α) 󵄨󵄨󵄨 󵄨 󵄨󵄨 ∫ f 󸀠 (x)dx󵄨󵄨󵄨 S2 := λ(1 − α)󵄨󵄨󵄨 ∫ f 󸀠 (tb + (1 − t)a)dt 󵄨󵄨󵄨 = 󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 b − a 󵄨󵄨󵄨 󵄨󵄨 󵄨 1−α 󵄨 (1−α)b+αa

(4.24)

4.1 Inequalities via convex functions | 75

=

λ(1 − α) 󵄨󵄨 󵄨 󵄨f (b) − f ((1 − α)b + αa)󵄨󵄨󵄨, b−a 󵄨

(4.25)

and 󵄨󵄨 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 S3 := 󵄨󵄨󵄨 ∫(t α − (1 − t)α )f 󸀠 (tb + (1 − t)a)dt 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨0 1

󵄨󵄨 2 󵄨󵄨 󵄨󵄨 󵄨󵄨 ≤ 󵄨󵄨󵄨 ∫(t α − (1 − t)α )f 󸀠 (tb + (1 − t)a)dt 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨0 󵄨

󵄨󵄨 1 󵄨󵄨 󵄨󵄨 󵄨󵄨 + 󵄨󵄨󵄨 ∫(t α − (1 − t)α )f 󸀠 (tb + (1 − t)a)dt 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨1 󵄨 2

1 2

1 2

1 p

p 󵄨 󵄨q ≤ (∫((1 − t)α − t α ) dt) (∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨 dt) 0

1 q

0

1

1 p

α p

1

󵄨 󵄨q + (∫(t − (1 − t) ) dt) (∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨 dt) α

1 2

1 q

1 2

1 2

1 2

1 p

≤ (∫((1 − t)αp − t αp )dt) (∫[t|f 󸀠 (b)|q + (1 − t)|f 󸀠 (a)|q ]dt) 0

1 q

0

1 p

1

1

+ (∫(t αp − (1 − t)αp )dt) (∫[t|f 󸀠 (b)|q + (1 − t)|f 󸀠 (a)|q ]dt) 1 2

≤(

1 − ( 21 )αp αp + 1

+[

1 q

1 2 1 p

) {[

|f 󸀠 (b)|q 3|f 󸀠 (a)|q + ] 8 8

1/q

1/q

3|f 󸀠 (b)|q |f 󸀠 (a)|q + ] 8 8

}.

(4.26)

Substituting (4.24), (4.25) and (4.26) into (4.23), we have the desired result. The proof is completed. Theorem 83. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with a ≥ 0, f 󸀠 ∈ L[a, b]. If |f 󸀠 | is an s-convex function, then for any 0 < α ≤ 1, we have 󵄨󵄨 󵄨󵄨 󵄨󵄨(1 + λ(1 − α))f (b) + (1 − λα)f (a) + λ(2α − 1)f ((1 − α)b + αa) 󵄨󵄨 󵄨󵄨 Γ(α + 1) 󵄨 − (RL Jaα+ f (b) + RL Jbα− f (a))󵄨󵄨󵄨 α 󵄨󵄨 (b − a) 󸀠 |f (b)| |f 󸀠 (a)| ≤ (b − a){ (λ(1 − α)s+1 (2α − 1) + λ(1 − α)) + (λαs+1 − λα) s+1 s+1

76 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1 1 − 2s+1 (s + 1) 2α+1 (α + 1) 1 1 + )}. − α+s 2 (α + s + 1) α + s + 1

(|f 󸀠 (b)| + |f 󸀠 (a)|)(

Proof. Using Lemma 60 via |f 󸀠 | is an s-convex function, we have 󵄨󵄨 󵄨󵄨 󵄨󵄨(1 + λ(1 − α))f (b) + (1 − λα)f (a) + λ(2α − 1)f ((1 − α)b + αa) 󵄨󵄨 󵄨󵄨 Γ(α + 1) 󵄨 − (RL Jaα+ f (b) + RL Jbα− f (a))󵄨󵄨󵄨 α 󵄨󵄨 (b − a) 󵄨󵄨 1−α 󵄨󵄨 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󸀠 󵄨 󵄨 ≤ (b − a){λα󵄨󵄨 ∫ f (tb + (1 − t)a)dt 󵄨󵄨 + λ(1 − α)󵄨󵄨󵄨 ∫ f 󸀠 (tb + (1 − t)a)dt 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 0 󵄨 󵄨 1−α 󵄨 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 󵄨󵄨 + 󵄨󵄨󵄨 ∫(t α − (1 − t)α )f 󸀠 (tb + (1 − t)a)dt 󵄨󵄨󵄨} 󵄨󵄨 󵄨󵄨 󵄨0 󵄨 1−α

1

󵄨 󵄨 󵄨 󵄨 ≤ (b − a){λα ∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt + λ(1 − α) ∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt 1−α

0

1

󵄨 󵄨 + ∫ 󵄨󵄨󵄨(t α − (1 − t)α )f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt} 0

= (b − a){H1 + H2 + H3 }.

(4.27)

Integrating by parts, 1−α

1−α

0

0

󵄨 󵄨 H1 := λα ∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt ≤ λα ∫ (t s |f 󸀠 (b)| + (1 − t)s |f 󸀠 (a)|)dt 1 − αs+1 󸀠 (1 − α)s+1 󸀠 |f (b)| − |f (a)|}, ≤ λα{ s+1 s+1

(4.28)

and similarly we get

1

󵄨 󵄨 H2 := λ(1 − α) ∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt 1−α 1

≤ λ(1 − α) ∫ (t s |f 󸀠 (b)| + (1 − t)s |f 󸀠 (a)|)dt 1−α

≤ λ(1 − α){ and 1

αs+1 󸀠 1 − (1 − α)s+1 󸀠 |f (b)| + |f (a)|}, s+1 s+1

󵄨 󵄨 H3 := ∫ 󵄨󵄨󵄨(t α − (1 − t)α )f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt 0

(4.29)

4.1 Inequalities via convex functions | 77 1 2

≤ ∫((1 − t)α − t α )(t s |f 󸀠 (b)| + (1 − t)s |f 󸀠 (a)|)dt 0

1

+ ∫(t α − (1 − t)α )(t s |f 󸀠 (b)| + (1 − t)s |f 󸀠 (a)|)dt 1 2

1 2

s

≤ ∫(t − t 0

α+s

1 2

)|f (b)|dt + ∫((1 − t)α+s − t α )|f 󸀠 (a)|dt 󸀠

0

1

1

+ ∫(t α+s − (1 − t)α )|f 󸀠 (b)|dt + ∫((1 − t)s − (1 − t)α+s )|f 󸀠 (a)|dt 1 2

≤ (|f 󸀠 (b)| + |f 󸀠 (a)|)(

1

1 2

2s+1 (s

+ 1) 1 1 1 − α+1 − α+s + ). 2 (α + 1) 2 (α + s + 1) α + s + 1

(4.30)

Substituting (4.28), (4.29) and (4.30) into (4.27), we have the result. The proof is completed. Now, we give the inequalities for AG(log)-convex functions. Theorem 84. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with a < b. If f 󸀠 ∈ L[a, b] and |f 󸀠 | are AG(log)-convex functions, |f 󸀠 (a)| > 0, |f 󸀠 (b)| > 0, |f 󸀠 (a)| ≠ |f 󸀠 (b)|, then for any 0 < α ≤ 1 and 0 < λ ≤ 1, p > 1, q > 1, p1 + q1 = 1, the following inequality for fractional integrals holds: 󵄨󵄨 󵄨󵄨 󵄨󵄨(1 + λ(1 − α))f (b) + (1 − λα)f (a) + λ(2α − 1)f ((1 − α)b + αa) 󵄨󵄨 󵄨󵄨 Γ(α + 1) 󵄨 (RL Jaα+ f (b) + RL Jbα− f (a))󵄨󵄨󵄨 ≤ (b − a){Q󸀠1 + Q󸀠2 + Q󸀠3 }, − α (b − a) 󵄨󵄨 where

Q󸀠1 = Q󸀠2 = Q󸀠3

1−α

|f 󸀠 (b)| λα‖f 󸀠 (a)| {( 󸀠 ) 󸀠 󸀠 ln |f (b)| − ln |f (a)| |f (a)|

− 1},

1−α

λ(1 − α)|f 󸀠 (a)| |f 󸀠 (b)| ( 󸀠 ) 󸀠 󸀠 ln |f (b)| − ln |f (a)| |f (a)| 1 − ( 21 )αp

1 p

1

|f 󸀠 (b)| 2 =( ) {1 + ( 󸀠 ) } αp + 1 |f (a)|

α

{(

|f 󸀠 (b)| ) − 1}, |f 󸀠 (a)|

q

1

q |f 󸀠 (a)|q |f 󸀠 (b)| 2 [ (( 󸀠 ) − 1)] . 󸀠 󸀠 q(ln |f (b)| − ln |f (a)|) |f (a)|

Proof. Using Lemma 60 via f 󸀠 ∈ L[a, b] and |f 󸀠 | are AG(log)-convex functions, we have 󵄨󵄨 󵄨󵄨 󵄨󵄨(1 + λ(1 − α))f (b) + (1 − λα)f (a) + λ(2α − 1)f ((1 − α)b + αa) 󵄨󵄨

78 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals

−

󵄨󵄨 Γ(α + 1) 󵄨 (RL Jaα+ f (b) + RL Jbα− f (a))󵄨󵄨󵄨 α 󵄨󵄨 (b − a)

1

1−α

󵄨 󵄨 󵄨 󵄨 ≤ (b − a){λα ∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt + λ(1 − α) ∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt 1−α

0

1

󵄨 󵄨 + ∫ 󵄨󵄨󵄨(t α − (1 − t)α )f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt} 0

≤ (b − a){Q󸀠1 + Q󸀠2 + Q󸀠3 }.

(4.31)

Integrating by parts and using Lemma 61, 1−α

󵄨 󵄨 Q󸀠1 := λα ∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt 0

1−α

≤ λα ∫ |f 󸀠 (b)|t |f 󸀠 (a)|1−t dt 0

≤

1−α

λα|f 󸀠 (a)| |f 󸀠 (b)| {( ) ln |f 󸀠 (b)| − ln |f 󸀠 (a)| |f 󸀠 (a)|

− 1},

(4.32)

and similarly we get 1−α

λ(1 − α)|f 󸀠 (a)| |f 󸀠 (b)| ( 󸀠 ) 󸀠 󸀠 ln |f (b)| − ln |f (a)| |f (a)|

Q󸀠2 ≤

α

{(

|f 󸀠 (b)| ) − 1}. |f 󸀠 (a)|

(4.33)

Next, using Hölder inequality, we have 1

Q󸀠3

󵄨 󵄨 := ∫ 󵄨󵄨󵄨(t α − (1 − t)α )f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt 0

1 2

󵄨 󵄨 ≤ ∫((1 − t)α − t α )󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt 0

1

󵄨 󵄨 + ∫(t α − (1 − t)α )󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt 1 2

1 2

α p

1 2

1 p

󵄨 󵄨q ≤ (∫((1 − t) − t ) dt) (∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨 dt) α

0

1 q

0

1

1

󵄨 󵄨q + (∫((1 − t) − t ) dt) (∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨 dt) 1 2

α

α p

1 p

1 2

1 q

4.1 Inequalities via convex functions | 79

≤(

1 − ( 21 )αp αp + 1 1 󸀠

1 2

1 p

qt

󸀠

󸀠

q(1−t)

) {(∫ |f (b)| |f (a)| 0 qt

󸀠

q(1−t)

+ (∫ |f (b)| |f (a)|

dt)

1 q

1 q

dt) }

1 2

1 − ( 21 )αp

1 p

1

|f 󸀠 (b)| 2 ≤( ) {1 + ( 󸀠 ) } αp + 1 |f (a)|

q

1

q |f 󸀠 (a)|q |f 󸀠 (b)| 2 [ (( ) − 1)] . 󸀠 󸀠 󸀠 q(ln |f (b)| − ln |f (a)|) |f (a)|

(4.34)

Substituting (4.32), (4.33) and (4.34) into (4.31), we have the result. The proof is completed. After we present these inequalities, applications of them are treated in the following section.

4.1.2 Applications to some special means Now, using the obtained results in above section, we examine some applications to special means of real numbers (see Proposition 3.1–3.3, [251]). Proposition 85. Let a, b ∈ ℝ, a < b, 0 does not belong to [a, b] and n ∈ ℤ, |n| ≥ 2. Then, 󵄨󵄨 n 󵄨 3 n−1 n−1 n 󵄨󵄨Ln (a, b) − A (a, b)󵄨󵄨󵄨 ≤ |n|(b − a)A(|a| , |b| ). 4

(4.35)

Proof. Applying Remark 80 for f (x) = xn , one can obtain the result immediately. Proposition 86. Let a, b ∈ ℝ, a < b, 0 does not belong to [a, b]. Then, |L−1 (a, b) − A−1 (a, b)| ≤

3 (b − a)A(|a|−2 , |b|−2 ). 4

(4.36)

Proof. The assertion follows from Remark 80 applied to f (x) = x1 . Proposition 87. Let a, b ∈ ℝ \ {0}, a < b, a−1 > b−1 , 0 does not belong to [a, b] and n ∈ ℤ, |n| ≥ 2. Then we have 󵄨󵄨 n −1 −1 󵄨 3 −n −1 −1 −1 n−1 n−1 󵄨󵄨Ln (b , a ) − H (b, a)󵄨󵄨󵄨 ≤ |n|(a − b )H (|a| , |b| ), 4

(4.37)

󵄨󵄨 −1 −1 −1 󵄨 3 −1 −1 −1 −2 −2 󵄨󵄨L (b , a ) − H(b, a)󵄨󵄨󵄨 ≤ (a − b )H (|a| , |b| ). 4

(4.38)

and

80 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals Proof. Making the substitutions a → b−1 , b → a−1 in the inequalities (4.35) and (4.36), one can obtain the desired inequalities (4.37) and (4.38), respectively, where A−1 (a−1 , b−1 ) = H(a, b) = 1 2 1 , b−1 < a−1 . a

+b

4.2 Inequalities via r-convex functions The results in this section are adopted from [211, 100, 147].

4.2.1 Main results In [211], some new Hermite-Hadamard type inequalities for differentiable r-convex functions and twice-differentiable r-convex functions involving Riemann-Liouville fractional integrals were introduced, respectively. Now, inequalities via differentiable r-convex functions are initially presented. Theorem 88. Let f : [0, b∗ ] → ℝ be a differentiable mapping with b∗ > 0. If |f 󸀠 | is measurable and r-convex on [a, b] for some fixed 0 ≤ r < ∞, 0 ≤ a < b. Then the following inequality for fractional integrals holds: 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J + f (b) + RL Jb− f (a)]󵄨 ≤ Kr , 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 2 2(b − a)α 󵄨 where 1 − 2−α (b − a)(|f 󸀠 (a)| + |f 󸀠 (b)|) 1+α −α 1 1 − 2 (b − a)(|f 󸀠 (a)| + |f 󸀠 (b)|) Kr : = 2− r 1+α (b − a)|f 󸀠 (b)| ∞ (ln k)2i−1 (1 − k) + K0 : = ∑[ 2 (α + 1)2i i=1 1

Kr : = 2 r −2

and k =

for 0 < r ≤ 1, for r > 1, √k (ln k)2i−2 (k + 1 − α+2i−3 )], (α + 1)2i−1 2

|f 󸀠 (a)| . |f 󸀠 (b)|

Proof. (i) Case 1: 0 < r ≤ 1. By Definition 17, Lemma 37 and Lemma 42, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) 1

≤

b−a 󵄨 󵄨 ∫ |(1 − t)α − t α |󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 0

1

1 b−a ≤ ∫ |(1 − t)α − t α |[t|f 󸀠 (a)|r + (1 − t)|f 󸀠 (b)|r ] r dt 2

0

4.2 Inequalities via r-convex functions | 81 1

1 1 1 b−a ≤ ∫ |(1 − t)α − t α |2 r −1 [t r |f 󸀠 (a)| + (1 − t) r |f 󸀠 (b)|]dt 2

0

1

1

1

= 2 r −2 (b − a)(|f 󸀠 (a)| ∫ |(1 − t)α − t α |t r dt 0

1

1

+ |f 󸀠 (b)| ∫ |(1 − t)α − t α |(1 − t) r dt) 0

1

1

1

= 2 r −2 (b − a)(|f 󸀠 (a)| + |f 󸀠 (b)|) ∫ |(1 − t)α − t α |t r dt 0

=2

1 −2 r

1 2

1

1

(b − a)(|f 󸀠 (a)| + |f 󸀠 (b)|) ∫((1 − t)α − t α )(t r + (1 − t) r )dt 0

≤2

1 −2 r

1 2

(b − a)(|f (a)| + |f (b)|) ∫((1 − t)α − t α )dt 󸀠

󸀠

0

=2

1 −2 r

1 − 2−α (b − a)(|f 󸀠 (a)| + |f 󸀠 (b)|). 1+α

(ii) Case 2: 1 < r. As in Case 1, we have 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J + f (b) + RL Jb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 2 2(b − a)α 󵄨 1

1 b−a ≤ ∫ |(1 − t)α − t α |[t|f 󸀠 (a)|r + (1 − t)|f 󸀠 (b)|r ] r dt 2

0

1

≤

1 1 b−a ∫ |(1 − t)α − t α |[t r |f 󸀠 (a)| + (1 − t) r |f 󸀠 (b)|]dt 2

0

1

1 b−a = (|f 󸀠 (a)| ∫ |(1 − t)α − t α |t r dt 2

1

0

1

+ |f 󸀠 (b)| ∫ |(1 − t)α − t α |(1 − t) r dt) 0

=

1

1 b−a 󸀠 (|f (a)| + |f 󸀠 (b)|) ∫ |(1 − t)α − t α |t r dt 2

0

1 2

1 1 b−a 󸀠 = (|f (a)| + |f 󸀠 (b)|) ∫((1 − t)α − t α )(t r + (1 − t) r )dt 2

0

82 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1 2

− 1r

≤ 2 (b − a)(|f (a)| + |f (b)|) ∫((1 − t)α − t α )dt 󸀠

󸀠

0

− 1r

=2

1 − 2−α (b − a)(|f 󸀠 (a)| + |f 󸀠 (b)|). 1+α

(iii) Case 3: r = 0. By using Definition 17 and Lemmas 32, 34, 42, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J + f (b) + RL Jb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 2 2(b − a)α 󵄨 1

b−a 󵄨 󵄨 ≤ ∫ |(1 − t)α − t α |󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 0

1

≤

b−a ∫ |(1 − t)α − t α ||f 󸀠 (a)|t |f 󸀠 (b)|1−t dt 2 0

1

(b − a)|f 󸀠 (b)| = ∫ |(1 − t)α − t α |k t dt 2 0

1 2

1 2

1

1

(b − a)|f 󸀠 (b)| = (∫(1 − t)α k t dt − ∫ t α k t dt + ∫ t α k t dt − ∫(1 − t)α k t dt) 2 0

=

0

1 2

1 2

1 2

1 2

1

1

0

0

0

0

(b − a)|f 󸀠 (b)| (2 ∫(1 − t)α k t dt − 2 ∫ t α k t dt + ∫ t α k t dt − ∫(1 − t)α k t dt) 2

√k (ln k)2i−2 (b − a)|f 󸀠 (b)| ∞ (ln k)2i−1 (1 − k) + (k + 1 − α+2i−3 )]. = ∑[ 2 (α + 1) (α + 1) 2 2i 2i−1 i=1 The proof is completed. Theorem 89. Let f : [0, b∗ ] → ℝ be a differentiable mapping with b∗ > 0. If |f 󸀠 |q (q > 1) is measurable and r-convex on [a, b] for some fixed, 0 ≤ r < ∞, 0 ≤ a < b, then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 ≤ Kr , 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 where Kr := (b − a)2

1−2r qr

1

1

1 − 2−pα p r(|f 󸀠 (a)|q + |f 󸀠 (b)|q ) q ( ) [ ] pα + 1 r+1 1

for 0 < r ≤ 1,

1

b − a 1 − 2−pα p r(|f 󸀠 (a)|q + |f 󸀠 (b)|q ) q Kr := ( ) [ ] 1− 1 pα + 1 r+1 2 p

for r > 1,

4.2 Inequalities via r-convex functions | 83 1

1

q (b − a) 1 − 2−pα p |f 󸀠 (a)|q − |f 󸀠 (b)|q K0 := ( ) ( ) 1 󸀠 󸀠 1− pα + 1 q ln |f (a)| − q ln |f (b)| 2 p

when |f 󸀠 (a)| ≠ |f 󸀠 (b)|,

1

(b − a) 1 − 2−pα p 󸀠 ( ) |f (a)| when |f 󸀠 (a)| = |f 󸀠 (b)|, K0 := 1− p1 pα + 1 2 and

1 p

+

1 q

= 1.

Proof. (i) Case 1: 0 < r ≤ 1. By Definition 17, Lemmas 37, 42 and using Hölder inequality, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) 1

b−a 󵄨 󵄨 ≤ ∫ |(1 − t)α − t α |󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 0

1 p

1

1

b−a 󵄨 󵄨q ≤ [∫ |(1 − t)α − t α |p dt] [∫ 󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt] 2

1 q

0

0

1 2

1 p

1

b−a p 󵄨 󵄨q = [2 ∫((1 − t)α − t α ) dt] [∫ 󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt] 2 0

≤

b−a 1− p1

2

1 q

0

1 2

pα p

1 p

1

󵄨q

󵄨 [∫((1 − t)pα − t ) dt] [∫ 󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt] 0

0

1 p

1

b − a 1 − 2−pα 󵄨 󵄨q ( ) [∫ 󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt] = 1− p1 pα + 1 2

1 q

0

1 p

1 q

1

1 b − a 1 − 2−pα ≤ ( ) [∫[t|f 󸀠 (a)|qr + (1 − t)|f 󸀠 (b)|qr ] r dt] 1− p1 pα + 1 2

1 q

0

1

1 p

1 1 1 b − a 1 − 2−pα ≤ ( ) [2 r −1 ∫[t r |f 󸀠 (a)|q + (1 − t) r |f 󸀠 (b)|q ]dt] 1− p1 pα + 1 2

= (b − a)2

1−2r qr

1 p

0

1 q

1

1 − 2−pα r(|f 󸀠 (a)|q + |f 󸀠 (b)|q ) q ( ) [ ] . pα + 1 r+1

(ii) Case 2: 1 < r. By Definition 17, Lemmas 37, 42 and using Hölder inequality, we have, as we did previously 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨

84 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

1

1 b − a 1 − 2−pα p ≤ ( ) [∫[t|f 󸀠 (a)|qr + (1 − t)|f 󸀠 (b)|qr ] r dt] 1− p1 pα + 1 2

0

1

1 p

1 1 b − a 1 − 2−pα ≤ ( ) [∫[t r |f 󸀠 (a)|q + (1 − t) r |f 󸀠 (b)|q ]dt] 1 1− pα + 1 2 p

0

1 q

1

b−a 1−2 r(|f 󸀠 (a)|q + |f 󸀠 (b)|q ) q ( ) [ ] . 1 1− pα + 1 r+1 2 p −pα

=

1 p

1 q

(iii) Case 3: r = 0. By Definition 17, Lemma 42 and using Hölder inequality, we have, as we did previously 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 1

1

(b − a) 1 − 2−pα p 󵄨 󵄨q ( ) [∫ 󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt] ≤ 1− p1 pα + 1 2 0

1 q

1

1 p

1 q

(b − a) 1 − 2−pα ≤ ( ) [∫ |f 󸀠 (a)|qt |f 󸀠 (b)|q(1−t) dt] . 1− p1 pα + 1 2 0

Since 1

q

󸀠

󸀠

q

1

−|f (b)| q q ( q ln|f|f(a)| |f 󸀠 (a)|q − |f 󸀠 (b)|q 󸀠 (a)|−q ln |f 󸀠 (b)| ) ( ) = 󸀠 󸀠 q ln |f (a)| − q ln |f (b)| |f 󸀠 (a)|q

when |f 󸀠 (a)| ≠ |f 󸀠 (b)|, when |f 󸀠 (a)| = |f 󸀠 (b)|.

The proof is completed. Inequalities via twice-differentiable r-convex functions are then developed. Theorem 90. Let f : [0, b∗ ] → ℝ be a twice-differentiable mapping with b∗ > 0. If |f 󸀠󸀠 |q (q > 1) is measurable and r-convex on [a, b] for some fixed 0 ≤ r < ∞, 0 ≤ a < b, then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J + f (b) + RL Jb− f (a)]󵄨 ≤ Ir 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 2 2(b − a)α 󵄨 where 1

1−r−qr qr

1

1 p q (b − a)2 2 r Ir = (1 − ) (|f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q ) q ( ) , (α + 1) pα + p + 1 r+1 for 0 < r ≤ 1,

2

1

1

1 p q (b − a)2 r 2 Ir = (1 − ) (|f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q ) q ( ) 2(α + 1) pα + p + 1 r+1

for 1 < r,

4.2 Inequalities via r-convex functions | 85 1

1

p q (b − a)2 2 |f 󸀠󸀠 (a)|q − |f 󸀠󸀠 (b)|q I0 = (1 − ) ( ) , 󸀠󸀠 󸀠󸀠 2(α + 1) pα + p + 1 q ln |f (a)| − q ln |f (b)|

when |f 󸀠󸀠 (a)| ≠ |f 󸀠󸀠 (b)|,

1

p (b − a)2 |f 󸀠󸀠 (a)| 2 I0 = (1 − ) 2(α + 1) pα + p + 1

and

1 p

+

1 q

when |f 󸀠󸀠 (a)| = |f 󸀠󸀠 (b)|,

= 1.

Proof. (i) Case 1: 0 < r ≤ 1. By Definition 17, Lemmas 37, 47 and using Hölder inequality, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 − 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 1

≤

(b − a)2 1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 ∫ 󵄨󵄨f (ta + (1 − t)b)󵄨󵄨󵄨dt 2 α+1 0

1 p

1

1

(b − a)2 p 󵄨 󵄨q ≤ (∫(1 − (1 − t)α+1 − t α+1 ) dt) (∫ 󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) 2(α + 1) 0

1 q

0

1

(b − a)2 (∫(1 − (1 − t)p(α+1) − t p(α+1) )dt) ≤ 2(α + 1) 0

1

qr

1 r

qr

× (∫[t|f (a)| + (1 − t)|f (b)| ] dt) 󸀠󸀠

󸀠󸀠

1 p

1 q

0 1

1

p 1 1 1 2 (b − a)2 (1 − ) (2 r −1 ∫[t r |f 󸀠󸀠 (a)|q + (1 − t) r |f 󸀠󸀠 (b)|q ]dt) ≤ 2(α + 1) pα + p + 1

=

2

1−r−qr qr

1 p

2

0

1 q

1

1 q 2 r (b − a) (1 − ) (|f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q ) q ( ) . (α + 1) pα + p + 1 r+1

(ii) Case 2: 1 < r. By Definition 17, Lemmas 37, 47 and using Hölder inequality, as previously we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 1

1

1 p (b − a)2 2 ≤ (1 − ) (∫[t|f 󸀠󸀠 (a)|qr + (1 − t)|f 󸀠󸀠 (b)|qr ] r dt) 2(α + 1) pα + p + 1

0

1 p

1

1 1 (b − a)2 2 ≤ (1 − ) (∫[t r |f 󸀠󸀠 (a)|q + (1 − t) r |f 󸀠󸀠 (b)|q ]dt) 2(α + 1) pα + p + 1

0

1 q

1 q

86 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

1

1 p q (b − a)2 2 r = (1 − ) (|f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q ) q ( ) . 2(α + 1) pα + p + 1 r+1

(iii) Case 3: r = 0. By Definition 17, Lemma 47 and using Hölder inequality, as previously we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) 1

1

p (b − a)2 2 󵄨 󵄨q ≤ (1 − ) (∫ 󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) 2(α + 1) pα + p + 1

0

1 q

1

1 p

1 q

(b − a)2 2 ≤ (1 − ) (∫ |f 󸀠󸀠 (a)|qt |f 󸀠󸀠 (b)|q(1−t) dt) . 2(α + 1) pα + p + 1 0

Since 1

qt

󸀠󸀠

q(1−t)

󸀠󸀠

∫ |f (a)| |f (b)| 0

dt =

|f 󸀠󸀠 (a)|q −|f 󸀠󸀠 (b)|q q ln |f 󸀠󸀠 (a)|−q ln |f 󸀠󸀠 (b)|

|f 󸀠󸀠 (a)|q

when |f 󸀠󸀠 (a)| ≠ |f 󸀠󸀠 (b)|, when |f 󸀠󸀠 (a)| = |f 󸀠󸀠 (b)|.

The proof is completed. Corollary 91. Suppose that the assumptions of Theorem 90 hold. Moreover, |f 󸀠󸀠 (x)| ≤ M on [a, b]. Then, 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 ≤ Ir󸀠 , 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 where Ir󸀠

=

I2󸀠 =

2 2

1−qr qr

1−q q

1

1

p q M(b − a)2 2 r (1 − ) ( ) (α + 1) pα + p + 1 r+1 1

for 0 < r ≤ 1,

1

p q M(b − a)2 2 r (1 − ) ( ) (α + 1) pα + p + 1 r+1

for 1 < r,

1

I0󸀠 and

1 p

+

1 q

p M(b − a)2 2 = (1 − ) , 2(α + 1) pα + p + 1

= 1.

Theorem 92. Let f : [0, b∗ ] → ℝ be a twicedifferentiable mapping with b∗ > 0. If |f 󸀠󸀠 |q (q > 1) is measurable and r-convex on [a, b] for some fixed 0 ≤ r < ∞, 0 ≤ a < b, then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 ≤ Ir , 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨

4.2 Inequalities via r-convex functions | 87

where 2

Ir =

1−r−qr qr

1 (b − a)2 󸀠󸀠 (|f (a)|q + |f 󸀠󸀠 (b)|q ) q (α + 1)

1

q r r r ×( − β( , αq + q + 1) − ) r+1 r+1 αqr + qr + r + 1

1 (b − a)2 󸀠󸀠 (|f (a)|q + |f 󸀠󸀠 (b)|q ) q 2(α + 1)

Ir =

×(

for 0 < r ≤ 1,

1

q r r r − β( , αq + q + 1) − ) r+1 r+1 αqr + qr + r + 1

for 1 < r,

|f 󸀠󸀠 (a)|q − |f 󸀠󸀠 (b)|q (b − a)2 ( 2(α + 1) q(ln |f 󸀠󸀠 (a)| − ln |f 󸀠󸀠 (b)|)

I0 =

(ln |f 󸀠󸀠 (a)| − ln |f 󸀠󸀠 (b)|)i−1 i−1 󸀠󸀠 −∑ [q |f (b)|q + (−q)i−1 |f 󸀠󸀠 (a)|q ]) (αq + q + 1) i i=1 ∞

when |f 󸀠󸀠 (a)| ≠ |f 󸀠󸀠 (b)|,

and

1 p

1

q 2 (b − a)2 |f 󸀠󸀠 (a)| (1 − ) I0 = 2(α + 1) αq + q + 1

+

1 q

1 q

when |f 󸀠󸀠 (a)| = |f 󸀠󸀠 (b)|,

= 1.

Proof. (i) Case 1: 0 < r ≤ 1. By Definition 17, Lemmas 37, 47 and using Hölder inequality, we have 󵄨󵄨 󵄨󵄨󵄨 f (a) + f (b) Γ(α + 1) 󵄨 󵄨󵄨 [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 − 󵄨󵄨 α 2 2(b − a) 󵄨󵄨 󵄨 1

≤

(b − a)2 1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 ∫ 󵄨󵄨f (ta + (1 − t)b)󵄨󵄨󵄨dt 2 α+1 0

1 p

1

1

(b − a)2 q󵄨 󵄨q (∫ 1dt) (∫(1 − (1 − t)α+1 − t α+1 ) 󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) ≤ 2(α + 1) 0

1 q

0

1

1 (b − a)2 ≤ (∫(1 − (1 − t)(α+1)q − t (α+1)q )[t|f 󸀠󸀠 (a)|qr + (1 − t)|f 󸀠󸀠 (b)|qr ] r dt) 2(α + 1)

1 q

0

≤

=

2 2

1

1−r−qr qr

1 1 (b − a)2 (∫(1 − (1 − t)(α+1)q − t (α+1)q )[t r |f 󸀠󸀠 (a)|q + (1 − t) r |f 󸀠󸀠 (b)|q ]dt) (α + 1)

1−r−qr qr

0

2

1 (b − a) (|f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q ) q (α + 1)

1

q r r r ×( − β( , αq + q + 1) − ) . r+1 r+1 αqr + qr + r + 1

1 q

88 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals (ii) Case 2: 1 < r. By Definition 17, Lemmas 37, 47 and using Hölder inequality, as previously we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J + J f (a)] f (b) + − 󵄨󵄨 󵄨󵄨 RL RL b a 󵄨󵄨 󵄨󵄨 2 2(b − a)α

1 q

1

1 (b − a)2 ≤ (∫(1 − (1 − t)(α+1)q − t (α+1)q )[t|f 󸀠󸀠 (a)|qr + (1 − t)|f 󸀠󸀠 (b)|qr ] r dt) 2(α + 1)

0

1

1 1 (b − a)2 ≤ (∫(1 − (1 − t)(α+1)q − t (α+1)q )[t r |f 󸀠󸀠 (a)|q + (1 − t) r |f 󸀠󸀠 (b)|q ]dt) 2(α + 1)

1 q

0

1 (b − a)2 󸀠󸀠 = (|f (a)|q + |f 󸀠󸀠 (b)|q ) q 2(α + 1)

1

q r r r ( − β( , αq + q + 1) − ) . r+1 r+1 αqr + qr + r + 1

(iii) Case 3: r = 0. According to Lemmas 32, 33, 47 via Hölder inequality, as before, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J + f (b) + RL Jb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 2 2(b − a)α 󵄨 1

2

≤

(b − a) q󵄨 󵄨q (∫(1 − (1 − t)α+1 − t α+1 ) 󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) 2(α + 1) 0

1 q

1 q

1

(b − a)2 ≤ (∫(1 − (1 − t)(α+1)q − t (α+1)q )|f 󸀠󸀠 (a)|qt |f 󸀠󸀠 (b)|q(1−t) dt) . 2(α + 1) 0

Since 1

∫(1 − (1 − t)(α+1)q − t (α+1)q )|f 󸀠󸀠 (a)|qt |f 󸀠󸀠 (b)|q(1−t) dt 0 󸀠󸀠

=

q

󸀠󸀠

q

∞ −|f (b)| ( q(ln|f |f(a)| 󸀠󸀠 (a)|−ln |f 󸀠󸀠 (b)|) − ∑i=1

(ln |f 󸀠󸀠 (a)|−ln |f 󸀠󸀠 (b)|)i−1 i−1 󸀠󸀠 [q |f (b)|q (αq+q+1)i 󸀠󸀠

q

|f (a)|

+ (−q)i−1 |f 󸀠󸀠 (a)|q ])

when |f 󸀠󸀠 (a)| ≠ |f 󸀠󸀠 (b)|, when |f 󸀠󸀠 (a)| = |f 󸀠󸀠 (b)|.

The proof is completed. Theorem 93. Let f : [0, b∗ ] → ℝ be a twice-differentiable mapping with b∗ > 0. If |f 󸀠󸀠 |q (q > 1) is measurable and r-convex on [a, b] for some fixed 0 ≤ r < ∞, 0 ≤ a < b, then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 ≤ Ir , 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨

4.2 Inequalities via r-convex functions | 89

where Ir = 2

n−r−nqr qnr

1

αp

1 + qm

(b − a)2

(α + 1)(α + 2)

1

1 + 1 p qm

(|f 󸀠󸀠 (a)|qn + |f 󸀠󸀠 (b)|qn ) qn 1

qn r n+r r ×( − β( , α + 2) − ) n+r r αr + 2r + n 1

Ir =

αp

1 + qm

(b − a)2

2(α + 1)(α + 2)

1

1 + 1 p qm

(|f 󸀠󸀠 (a)|qn + |f 󸀠󸀠 (b)|qn ) qn 1

qn r n+r r ×( − β( , α + 2) − ) n+r r αr + 2r + n 1

I0 =

αp

1 + qm

(b − a)2

2(α + 1)(α + 2)

for 0 < r ≤ n,

1 + 1 p qm

(

for n < r,

|f 󸀠󸀠 (a)|nq − |f 󸀠󸀠 (b)|nq nq[ln |f 󸀠󸀠 (a)| − ln |f 󸀠󸀠 (b)|]

(nq)i−1 (ln |f 󸀠󸀠 (a)| − ln |f 󸀠󸀠 (b)|)i−1 [|f 󸀠󸀠 (b)|nq + |f 󸀠󸀠 (a)|nq (−1)i−1 ] −∑ ) (α + 2)i i=1 ∞

1 qn

when |f 󸀠󸀠 (a)| ≠ |f 󸀠󸀠 (b)|,

I0 = and

1 p

+

1 q

α

1 + 1 + 1 p qm qn

(b − a)2 |f 󸀠󸀠 (a)| 1

2(α + 1)(α + 2) p 1 m

= 1,

1 n

+

1 1 + qm + qn

when |f 󸀠󸀠 (a)| = |f 󸀠󸀠 (b)|,

= 1, n > 1.

Proof. (i) Case 1: 0 < r ≤ n. By Lemmas 37, 47 and using Hölder inequality, we have 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J + f (b) + RL Jb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 2 2(b − a)α 󵄨 1

(b − a)2 1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 ≤ ∫ 󵄨󵄨f (ta + (1 − t)b)󵄨󵄨󵄨dt 2 α+1 0

1

(b − a)2 (∫(1 − (1 − t)α+1 − t α+1 )dt) ≤ 2(α + 1)

1 p

0

1

󵄨q

󵄨 × (∫(1 − (1 − t)α+1 − t α+1 )󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) 0

1

1

≤

α p (b − a)2 2(α + 1)(α + 2) qr

1 p

(∫(1 − (1 − t)(α+1) − t (α+1) ) 0 qr

1 r

[t|f (a)| + (1 − t)|f (b)| ] dt) 󸀠󸀠

󸀠󸀠

1 q

1 q

90 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

1

≤

α p (b − a)2 2(α + 1)(α + 2)

1 p

(∫(1 − (1 − t)(α+1) − t (α+1) )dt) 0

1

× (∫(1 − (1 − t)

−t

(α+1)

0 1

≤

αp

1 + qm

1

× (2 =2

×(

qr

qr

n r

n r

󸀠󸀠

)[t|f (a)| + (1 − t)|f (b)| ] dt) 󸀠󸀠

󸀠󸀠

1 qn

1 + qm

1

∫(1 − (1 − t)

(α+1)

0 n−r−nqr qnr

(α+1)

(b − a)2

2(α + 1)(α + 2) p n −1 r

1 qm

1

αp

1 + qm

(b − a)2

(α + 1)(α + 2)

−t

(α+1)

n r

󸀠󸀠

)[t |f (a)|

qn

qn

+ (1 − t) |f (b)| ]dt) 1

1 + 1 p qm

(|f 󸀠󸀠 (a)|qn + |f 󸀠󸀠 (b)|qn ) qn 1

qn n+r r r − β( , α + 2) − ) . n+r r αr + 2r + n

(ii) Case 2: For 1 < n < r, as in Case 1, we have 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J + f (b) + RL Jb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 2 2(b − a)α 󵄨 1

≤

αp

1 + qm

(b − a)2 1

2(α + 1)(α + 2) p

1 + qm

1

× (∫(1 − (1 − t)

(α+1)

0 1

≤

αp

1 + qm

−t

(α+1)

−t

(α+1)

1

× (∫(1 − (1 − t)

(α+1)

1

αp

1 + qm

−t

(α+1)

1

× (∫(1 − (1 − t)

(α+1)

1

≤

1 + qm

n r

󸀠󸀠

qn

n r

󸀠󸀠

qn

)[t |f (a)|

n r

󸀠󸀠

n r

󸀠󸀠

qn

+ (1 − t) |f (b)| ]dt)

1 qn

1 + qm

1

αp

)[t|f (a)| + (1 − t)|f (b)| ] dt)

1 qn

(b − a)2

2(α + 1)(α + 2) p

0

n r

1 + qm

1

≤

qr

󸀠󸀠

(b − a)2

2(α + 1)(α + 2) p

0

qr

󸀠󸀠

(b − a)2

2(α + 1)(α + 2)

1 + 1 p qm

)[t |f (a)|

qn

+ (1 − t) |f (b)| ]dt) 1

(|f 󸀠󸀠 (a)|qn + |f 󸀠󸀠 (b)|qn ) qn 1

qn r n+r r ×( − β( , α + 2) − ) . n+r r αr + 2r + n

1 qn

1 qn

4.2 Inequalities via r-convex functions | 91

(iii) Case 3: r = 0. Denote k=(

nq

|f 󸀠󸀠 (a)| ) . |f 󸀠󸀠 (a)|

If |f 󸀠󸀠 (a)| ≠ |f 󸀠󸀠 (b)|, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 1

≤

(b − a)2 1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 ∫ 󵄨󵄨f (ta + (1 − t)b)󵄨󵄨󵄨dt 2 α+1 0

1

(b − a)2 (∫(1 − (1 − t)α+1 − t α+1 )dt) ≤ 2(α + 1)

1 p

0

1

α+1

× (∫(1 − (1 − t)

−t

α+1

0

≤

1

2(α + 1)(α + 2) p

≤

(∫(1 − (1 − t)

(α+1)

−t

(α+1)

−t

(α+1)

1

2(α + 1)(α + 2) p

(∫(1 − (1 − t)

× (∫(1 − (1 − t)

(α+1)

−t

(α+1)

󸀠󸀠

1

=

αp

1

1 + qm

αp

1

1 + qm

1 + qm

(b − a)2

2(α + 1)(α + 2) p 1

=

(b − a)2

2(α + 1)(α + 2) p 1

=

1 + qm

qt

󸀠󸀠

)[|f (a)| |f (b)|

q(1−t) n

] dt)

1 + qm

(b − a)2 1

2(α + 1)(α + 2) p

1 + qm

1

|f (b)| × (∫(1 − (1 − t) 󸀠󸀠

]dt)

)dt) 1 qn

0

αp

1 q

1 qm

0

1

q(1−t)

󸀠󸀠

)[|f (a)| |f (b)|

0 (α+1)

qt

󸀠󸀠

1

1 p

α (b − a)2

1 q

1

1

α p (b − a)2

󵄨 󵄨q )󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt)

(α+1)

−t

(α+1)

t

)k dt)

1 qn

0 ∞ k − 1 ∞ (ln k)i−1 (ln k)i−1 |f (b)| × ( −∑ − k ∑(−1)i−1 ) ln k i=1 (α + 2)i (α + 2)i i=1 󸀠󸀠

[

|f 󸀠󸀠 (a)|nq − |f 󸀠󸀠 (b)|nq nq[ln |f 󸀠󸀠 (a)| − ln |f 󸀠󸀠 (b)|] 1 qn

(nq)i−1 (ln |f 󸀠󸀠 (a)| − ln |f 󸀠󸀠 (b)|)i−1 (|f 󸀠󸀠 (b)|nq + |f 󸀠󸀠 (a)|nq (−1)i−1 ) −∑ ] . (α + 2)i i=1 ∞

If |f 󸀠󸀠 (a)| = |f 󸀠󸀠 (b)|, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨

1 qn

92 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

≤

=

αp

1 + qm

(b − a)2

2(α + 1)(α + 2) α

1 + 1 + 1 p qm qn

1

1 + 1 p qm

× (∫(1 − (1 − t)(α+1) − t (α+1) )|f 󸀠󸀠 (a)|qn dt) 0

(b − a)2 |f 󸀠󸀠 (a)| 1

2(α + 1)(α + 2) p

1 qn

1 1 + qm + qn

.

The proof is completed. Theorem 94. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with 0 ≤ a < b. If |f 󸀠󸀠 | is integrable and r-convex on [a, b] for some fixed 0 ≤ r < ∞, then the following inequality for fractional integrals holds: 󵄨󵄨 Γ(α + 1) f (a) + f (b) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 (Jaα+ f (b) + Jbα− f (a)) − (1 + λ) − (1 − λ)f ( )󵄨󵄨 ≤ Kr , 󵄨󵄨 α 󵄨󵄨 2 2 󵄨󵄨 (b − a)

(4.39)

where 1

Kr := 2 r −1 (b − a)2 {

|f 󸀠󸀠 (a)| + |f 󸀠󸀠 (b)| r r 1 [ − − B( + 1, α + 2)] α+1 r + 1 αr + 2r + 1 r 1

+

r r 1 r 1 − λ 󸀠󸀠 |f (a)|( − )[1 − ( ) ] 2 r + 1 2r + 1 2 +1

1

r 1 r 1 − λ 󸀠󸀠 1 |f (b)|[B 1 (2, + 1) + ( ) ]} for 0 < r ≤ 1, + 2 2 r 2r + 1 2 󸀠󸀠 󸀠󸀠 |f (a)| + |f (b)| r r 1 Kr := (b − a)2 { [ − − B( + 1, α + 2)] α+1 r + 1 αr + 2r + 1 r 1

+2

1 − λ 󸀠󸀠 r r 1 r + |f (a)|( − )[1 − ( ) ] 2 r + 1 2r + 1 2 +1

1

+

1 − λ 󸀠󸀠 1 r 1 r |f (b)|[B 1 (2, + 1) + ( ) ]} 2 2 r 2r + 1 2

Kr := (b − a)2 |f 󸀠󸀠 (b)|{

for r > 1,

∞ |k| − 1 (ln |k|)i−1 ∞ (ln |k|)i−1 1 [ − |k| ∑(−1)i−1 −∑ ] α + 1 ln |k| (α + 2)i (α + 2)i i=1 i=1 1

+

+2

1 − λ 21 ∞ (− 2 ln |k|) |k| ∑ 4 (2)i i=1 for r = 0, and k =

i−1

f 󸀠󸀠 (a) . f 󸀠󸀠 (b)

1

+

∞ 1 − λ |k| − |k| 2 (ln |k|)i−1 [ − |k| ∑(−1)i−1 ]} 2 ln |k| (2)i i=1

Proof. (i) Case 1: 0 < r ≤ 1. 󵄨󵄨 Γ(α + 1) f (a) + f (b) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 α α (J − (1 − λ)f ( )󵄨󵄨 + f (b) + Jb− f (a)) − (1 + λ) 󵄨󵄨 a 󵄨󵄨 (b − a)α 󵄨󵄨 2 2 2

1

1

≤ (b − a) ∫ |k(t, λ)|(t|f 󸀠󸀠 (a)|r + (1 − t)|f 󸀠󸀠 (b)|r ) r dt 0

4.2 Inequalities via r-convex functions | 93

≤2

1 −1 r

2

1

1

1

(b − a) ∫ |k(t, λ)|(t r |f 󸀠󸀠 (a)| + (1 − t) r |f 󸀠󸀠 (b)|)dt 0

=2

1 −1 r

1 2

󵄨󵄨 t α+1 + (1 − t)α+1 − 1 1 − λ 󵄨󵄨 1 1 󵄨 󵄨 (b − a)2 [∫ 󵄨󵄨󵄨 + t 󵄨󵄨󵄨(|f 󸀠󸀠 (a)|t r + |f 󸀠󸀠 (b)|(1 − t) r )dt 󵄨󵄨 α+1 2 󵄨󵄨 0

1

󵄨󵄨 t α+1 + (1 − t)α+1 − 1 1 − λ 󵄨󵄨 1 1 󵄨 󵄨 + ∫ 󵄨󵄨󵄨 + (1 − t)󵄨󵄨󵄨(|f 󸀠󸀠 (a)|t r + |f 󸀠󸀠 (b)|(1 − t) r )dt] 󵄨󵄨 󵄨󵄨 α+1 2 1 2

≤2

1 −1 r

1 2

󵄨󵄨 t α+1 + (1 − t)α+1 − 1 󵄨󵄨 󵄨󵄨 1 − λ 󵄨󵄨 󵄨 󵄨󵄨 󵄨󵄨 󵄨 (b − a) [∫(󵄨󵄨󵄨 t 󵄨󵄨) 󵄨󵄨 + 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 󵄨󵄨󵄨 α+1 2

1 r

0

1

× (|f (a)|t + |f 󸀠󸀠 (b)|(1 − t) r )dt 󸀠󸀠

1

󵄨󵄨 t α+1 + (1 − t)α+1 − 1 󵄨󵄨 󵄨󵄨 1 − λ 󵄨󵄨 1 1 󵄨 󵄨󵄨 󵄨󵄨 󵄨 (1 − t)󵄨󵄨󵄨)(|f 󸀠󸀠 (a)|t r + |f 󸀠󸀠 (b)|(1 − t) r )dt] + ∫(󵄨󵄨󵄨 󵄨󵄨 + 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 󵄨󵄨 α+1 1 2

=2

1 −1 r

1 2

2

(b − a) [∫( 1 r

0

1 − t α+1 − (1 − t)α+1 1 − λ + t) α+1 2 1

× (|f (a)|t + |f 󸀠󸀠 (b)|(1 − t) r )dt 󸀠󸀠

1

+ ∫( 1 2

=2

1 −1 r

1 1 1 − t α+1 − (1 − t)α+1 1 − λ + (1 − t))(|f 󸀠󸀠 (a)|t r + |f 󸀠󸀠 (b)|(1 − t) r )dt] α+1 2 1 2

(b − a)2 [∫( 0 1

1 1 − t α+1 − (1 − t)α+1 󸀠󸀠 1 − t α+1 − (1 − t)α+1 |f (a)|t r + α+1 α+1

× |f (b)|(1 − t) r + 󸀠󸀠

1

+ ∫( 1 2

+

1 1 1 − t α+1 − (1 − t)α+1 󸀠󸀠 1 − t α+1 − (1 − t)α+1 󸀠󸀠 |f (a)|t r + |f (b)|(1 − t) r α+1 α+1

1 1 1 − λ 󸀠󸀠 1 − λ 󸀠󸀠 |f (a)|(1 − t)t r + |f (b)|(1 − t) r +1 )dt] 2 2

1

= 2 r −1 (b − a)2 { 󸀠󸀠

+

1 1 1 − λ 󸀠󸀠 1 − λ 󸀠󸀠 |f (a)|t r +1 + |f (b)|t(1 − t) r )dt 2 2

1

1

1 1 1 |f 󸀠󸀠 (a)| ∫[t r − t α+ r +1 − t r (1 − t)α+1 ]dt α+1

0

1 1 1 |f (b)| ∫[(1 − t) r − t α+1 (1 − t) r − (1 − t)α+ r +1 ]dt α+1

0

94 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1 2

1

1 1 1 − λ 󸀠󸀠 + |f (a)|[∫ t r +1 dt + ∫(1 − t)t r dt] 2

0

1 2

1 2

+

1

1 1 1 − λ 󸀠󸀠 |f (b)|[∫ t(1 − t) r dt + ∫(1 − t) r +1 dt]} 2

0

1 2

r r 1 |f 󸀠󸀠 (a)| [ − − B( + 1, α + 2)] α + 1 r + 1 αr + 2r + 1 r 󸀠󸀠 |f (b)| r 1 r + [ − B( + 1, α + 2) − ] α+1 r+1 r αr + 2r + 1 1

= 2 r −1 (b − a)2 {

1

r 1 r 1 − λ 󸀠󸀠 |f (a)|[ ( ) + 2 2r + 1 2 1

+2

1

1 r r (1 − ( ) ) + r+1 2 +1

1

1 r r 1 r r 1 − λ 󸀠󸀠 1 (1 − ( ) )] + |f (b)|[B 1 (2, + 1) + ( ) ]} − 2 2r + 1 2 2 r 2r + 1 2 󸀠󸀠 󸀠󸀠 1 |f (a)| + |f (b)| r r 1 = 2 r −1 (b − a)2 { [ − − B( + 1, α + 2)] α+1 r + 1 αr + 2r + 1 r +2

1

+

+2

1 − λ 󸀠󸀠 r r 1 r |f (a)|( − )[1 − ( ) ] 2 r + 1 2r + 1 2 +1

1

r 1 r 1 − λ 󸀠󸀠 1 |f (b)|[B 1 (2, + 1) + ( ) ]}. + 2 2 r 2r + 1 2 +2

(ii) Case 2: r > 1. As in Case 1, we have 󵄨󵄨 Γ(α + 1) f (a) + f (b) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 α α (J f (b) + J f (a)) − (1 + λ) − (1 − λ)f ( )󵄨󵄨 + − 󵄨󵄨 b 󵄨󵄨 (b − a)α a 󵄨󵄨 2 2 2

1

1

1

≤ (b − a) ∫ |k(t, λ)|(t r |f 󸀠󸀠 (a)| + (1 − t) r |f 󸀠󸀠 (b)|)dt 0 1 2

󵄨󵄨 t α+1 + (1 − t)α+1 − 1 1 − λ 󵄨󵄨 1 1 󵄨 󵄨 = (b − a) [∫ 󵄨󵄨󵄨 + t 󵄨󵄨(|f 󸀠󸀠 (a)|t r + |f 󸀠󸀠 (b)|(1 − t) r )dt 󵄨󵄨 α+1 2 󵄨󵄨󵄨 2

1

0

󵄨󵄨 t α+1 + (1 − t)α+1 − 1 1 − λ 󵄨󵄨 1 1 󵄨 󵄨 + ∫ 󵄨󵄨󵄨 + (1 − t)󵄨󵄨󵄨(|f 󸀠󸀠 (a)|t r + |f 󸀠󸀠 (b)|(1 − t) r )dt] 󵄨󵄨 󵄨󵄨 α+1 2 1 2

≤ (b − a)2 {

|f 󸀠󸀠 (a)| + |f 󸀠󸀠 (b)| r r 1 [ − − B( + 1, α + 2)] α+1 r + 1 αr + 2r + 1 r 1

1 − λ 󸀠󸀠 r r 1 r + |f (a)|( − )[1 − ( ) ] 2 r + 1 2r + 1 2 +1

4.2 Inequalities via r-convex functions | 95 1

1 1 − λ 󸀠󸀠 r 1 r + |f (b)|[B 1 (2, + 1) + ( ) ]}. 2 2 r 2r + 1 2 +2

(iii) Case 3: r = 0. We have 󵄨󵄨 Γ(α + 1) a + b 󵄨󵄨󵄨󵄨 f (a) + f (b) 󵄨󵄨 (J α+ f (b) + Jbα− f (a)) − (1 + λ) − (1 − λ)f ( )󵄨󵄨 󵄨󵄨 󵄨󵄨 (b − a)α a 󵄨󵄨 2 2 1

󵄨 󵄨 ≤ (b − a)2 ∫ |k(t, λ)|󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 0

1

≤ (b − a) ∫ |k(t, λ)|(|f 󸀠󸀠 (a)|t |f 󸀠󸀠 (b)|1−t )dt 2

0 1 2

󵄨󵄨 t α+1 + (1 − t)α+1 − 1 1 − λ 󵄨󵄨 󵄨 󵄨 = (b − a)2 [∫ 󵄨󵄨󵄨 + t 󵄨󵄨(|f 󸀠󸀠 (a)|t |f 󸀠󸀠 (b)|1−t )dt 󵄨󵄨 α+1 2 󵄨󵄨󵄨 0

1

󵄨󵄨 t α+1 + (1 − t)α+1 − 1 1 − λ 󵄨󵄨 󵄨 󵄨 + (1 − t)󵄨󵄨󵄨(|f 󸀠󸀠 (a)|t |f 󸀠󸀠 (b)|1−t )dt] + ∫ 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 α+1 2 1 2

1 2

≤ (b − a)2 [∫( 1

+ ∫( 1 2

0

1 − t α+1 − (1 − t)α+1 1 − λ + t)(|f 󸀠󸀠 (a)|t |f 󸀠󸀠 (b)|1−t )dt α+1 2

1 − t α+1 − (1 − t)α+1 1 − λ + (1 − t))(|f 󸀠󸀠 (a)|t |f 󸀠󸀠 (b)|1−t )dt] α+1 2 1 2

= (b − a)2 |f 󸀠󸀠 (b)|[∫( 1

+ ∫( 1 2

0

t 1 − t α+1 − (1 − t)α+1 1 − λ 󵄨󵄨󵄨󵄨 f 󸀠󸀠 (a) 󵄨󵄨󵄨󵄨 + t)󵄨󵄨 󸀠󸀠 󵄨󵄨 dt 󵄨󵄨 f (b) 󵄨󵄨 α+1 2

󵄨󵄨 f 󸀠󸀠 (a) 󵄨󵄨t 1 − t α+1 − (1 − t)α+1 1 − λ 󵄨 󵄨 + (1 − t))󵄨󵄨󵄨 󸀠󸀠 󵄨󵄨󵄨 dt] 󵄨󵄨 f (b) 󵄨󵄨 α+1 2

= (b − a)2 |f 󸀠󸀠 (b)|[

1 󵄨󵄨 f 󸀠󸀠 (a) 󵄨󵄨t 1 󵄨 󵄨 ∫[1 − t α+1 − (1 − t)α+1 ]󵄨󵄨󵄨 󸀠󸀠 󵄨󵄨󵄨 dt 󵄨󵄨 f (b) 󵄨󵄨 α+1 0

1 2

1 󵄨󵄨 f 󸀠󸀠 (a) 󵄨󵄨t 󵄨󵄨 f 󸀠󸀠 (a) 󵄨󵄨t 1−λ 1−λ 󵄨󵄨 󵄨󵄨 󵄨 󵄨 + ∫ t 󵄨󵄨 󸀠󸀠 󵄨󵄨 dt + ∫(1 − t)󵄨󵄨󵄨 󸀠󸀠 󵄨󵄨󵄨 dt] 󵄨󵄨 f (b) 󵄨󵄨 󵄨󵄨 f (b) 󵄨󵄨 2 2 1 0

2

(a) |−1 | ff 󸀠󸀠 (b)

(a) i−1 󵄨󵄨 f 󸀠󸀠 (a) 󵄨󵄨 ∞ |) (ln | ff 󸀠󸀠 (b) 1 󵄨󵄨 󵄨󵄨 i−1 = (b − a) |f (b)|{ [ − 󵄨󵄨 󸀠󸀠 󵄨󵄨 ∑(−1) 󸀠󸀠 α + 1 ln | f 󸀠󸀠 (a) | 󵄨󵄨 f (b) 󵄨󵄨 i=1 (α + 2)i f (b) 󸀠󸀠

2

󸀠󸀠

󸀠󸀠

96 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals

∞

−∑

(a) i−1 (ln | ff 󸀠󸀠 (b) |) 󸀠󸀠

(α + 2)i

i=1

f (a) i−1 1 1 − λ 󵄨󵄨󵄨󵄨 f 󸀠󸀠 (a) 󵄨󵄨󵄨󵄨 2 ∞ (− 2 ln | f 󸀠󸀠 (b) |) ]+ 󵄨 󵄨 ∑ 8 󵄨󵄨󵄨 f 󸀠󸀠 (b) 󵄨󵄨󵄨 i=1 (2)i 1

󸀠󸀠

1

(a) i−1 f (a) f (a) (ln | ff 󸀠󸀠 (b) |) 1 − λ | f 󸀠󸀠 (b) | − | f 󸀠󸀠 (b) | 2 󵄨󵄨󵄨󵄨 f 󸀠󸀠 (a) 󵄨󵄨󵄨󵄨 ∞ i−1 + [ − 󵄨󵄨 󸀠󸀠 󵄨󵄨 ∑(−1) 󸀠󸀠 󵄨󵄨 f (b) 󵄨󵄨 2 (2)i ln | f 󸀠󸀠 (a) | i=1 󸀠󸀠

󸀠󸀠

󸀠󸀠

f (b) 1

1 2

f 󸀠󸀠 (a) i−1

1 󵄨󵄨󵄨 f 󸀠󸀠 (a) 󵄨󵄨󵄨 ∞ (− 2 ln | f 󸀠󸀠 (b) |) + 󵄨󵄨󵄨 󸀠󸀠 󵄨󵄨󵄨 ∑ 4 󵄨󵄨 f (b) 󵄨󵄨 i=1 (2)i

]}

(a) (a) i−1 | ff 󸀠󸀠 (b) (ln | ff 󸀠󸀠 (b) | − 1 󵄨󵄨󵄨 f 󸀠󸀠 (a) 󵄨󵄨󵄨 ∞ |) 1 i−1 󵄨 󵄨 [ = (b − a) |f (b)|{ − (−1) ∑ 󵄨 󵄨 󸀠󸀠 (a) 󵄨 󵄨 󸀠󸀠 f α + 1 ln | 󸀠󸀠 | 󵄨󵄨 f (b) 󵄨󵄨 i=1 (α + 2)i 󸀠󸀠

2

󸀠󸀠

󸀠󸀠

f (b)

󸀠󸀠 ∞ (ln | f 󸀠󸀠 (a) |)i−1 f (b)

−∑ i=1

(α + 2)i

f (a) i−1 1 1 − λ 󵄨󵄨󵄨󵄨 f 󸀠󸀠 (a) 󵄨󵄨󵄨󵄨 2 ∞ (− 2 ln | f 󸀠󸀠 (b) |) ]+ 󵄨 󵄨 ∑ 4 󵄨󵄨󵄨 f 󸀠󸀠 (b) 󵄨󵄨󵄨 i=1 (2)i 1

󸀠󸀠

1

f (a) f (a) (a) i−1 |) (ln | ff 󸀠󸀠 (b) 1 − λ | f 󸀠󸀠 (b) | − | f 󸀠󸀠 (b) | 2 󵄨󵄨󵄨󵄨 f 󸀠󸀠 (a) 󵄨󵄨󵄨󵄨 ∞ i−1 + ]}. [ − 󵄨󵄨 󸀠󸀠 󵄨󵄨 ∑(−1) 󸀠󸀠 󵄨󵄨 f (b) 󵄨󵄨 2 (2)i ln | f 󸀠󸀠 (a) | i=1 󸀠󸀠

󸀠󸀠

󸀠󸀠

f (b)

The proof is completed. Remark 95. If we take α = 1 in Theorem 94, then the equality (4.39) becomes the following inequality: b 󵄨 󵄨󵄨 󵄨󵄨 2 f (a) + f (b) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 f (t)dt − (1 + λ) − (1 − λ)f ( ) ∫ 󵄨󵄨 ≤ Kr , 󵄨󵄨 b − a 󵄨󵄨 2 2 󵄨󵄨 󵄨 a

where 1

Kr := 2 r −1 (b − a)2 {

r r 1 |f 󸀠󸀠 (a)| + |f 󸀠󸀠 (b)| [ − − B( + 1, 3)] 2 r + 1 3r + 1 r 1

r r 1 r 1 − λ 󸀠󸀠 + |f (a)|( − )[1 − ( ) ] 2 r + 1 2r + 1 2 +1

1

1 r 1 r 1 − λ 󸀠󸀠 |f (b)|[B 1 (2, + 1) + ( ) ]} for 0 < r ≤ 1, + 2 2 r 2r + 1 2 󸀠󸀠 󸀠󸀠 r r 1 2 |f (a)| + |f (b)| Kr := (b − a) { [ − − B( + 1, 3)] 2 r + 1 3r + 1 r 1

+

+2

1 − λ 󸀠󸀠 r r 1 r |f (a)|( − )[1 − ( ) ] 2 r + 1 2r + 1 2 +1

1

1 − λ 󸀠󸀠 1 r 1 r + |f (b)|[B 1 (2, + 1) + ( ) ]} 2 2 r 2r + 1 2 +2

for r > 1,

∞ 1 |k| − 1 (ln |k|)i−1 ∞ (ln |k|)i−1 Kr := (b − a)2 |f 󸀠󸀠 (b)|{ [ − |k| ∑(−1)i−1 −∑ ] 2 ln |k| (3)i (3)i i=1 i=1

4.2 Inequalities via r-convex functions | 97 1

1 − λ 21 ∞ (− 2 ln |k|) |k| ∑ 4 (2)i i=1

+

for r = 0, and k =

i−1

1

∞ 1 − λ |k| − |k| 2 (ln |k|)i−1 [ − |k| ∑(−1)i−1 ]} 2 ln |k| (2)i i=1

+

f 󸀠󸀠 (a) . f 󸀠󸀠 (b)

Theorem 96. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with 0 ≤ a < b. If |f 󸀠󸀠 |q , q > 1 is integrable and r-convex on [a, b] for some fixed 0 ≤ r < ∞, then the following inequality for fractional integrals holds:

where

󵄨󵄨 Γ(α + 1) f (a) + f (b) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 (Jaα+ f (b) + Jbα− f (a)) − (1 + λ) − (1 − λ)f ( )󵄨󵄨 ≤ Kr , 󵄨󵄨 α 󵄨󵄨 2 2 󵄨󵄨 (b − a) q

q

(4.40)

1− q1

q−1 2(q − 1) 1 1 − λ q−1 q − 1 Kr := 2 (b − a) [( ) (1 − )+( ) ] α+1 qα + 2q − 1 4 2q − 1 1 qr

2

1

× [(|f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q ) q

q r ] r+1

for 0 < r ≤ 1, q

1− q1

q−1 1 2(q − 1) 1 − λ q−1 q − 1 Kr := 2 (b − a) [( ) (1 − )+( ) ] α+1 qα + 2q − 1 4 2q − 1 1 q

2

1

q r × [(|f (a)| + |f (b)| ) ] r+1

q

󸀠󸀠

󸀠󸀠

q

for r > 1,

q

q

1− q1

q−1 1 2(q − 1) 1 − λ q−1 q − 1 Kr := 2 (b − a) |f (b)|[( ) (1 − )+( ) ] α+1 qα + 2q − 1 4 2q − 1 1 q

2

󸀠󸀠

1

×(

|k|q − 1 q ) q ln |k|

for r = 0, and k =

f 󸀠󸀠 (a) . f 󸀠󸀠 (b)

Proof. (i) Case 1: 0 < r ≤ 1. Like Theorem 94, we have 󵄨󵄨 Γ(α + 1) a + b 󵄨󵄨󵄨󵄨 f (a) + f (b) 󵄨󵄨 (Jaα+ f (b) + Jbα− f (a)) − (1 + λ) − (1 − λ)f ( )󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2 󵄨󵄨 (b − a) 1

≤ (b − a)2 (∫ |k(t, λ)| 0

q q−1

1− q1

dt)

1

󵄨q

󵄨 (∫ 󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) 0

1 2

q

󵄨󵄨 t α+1 + (1 − t)α+1 − 1 1 − λ 󵄨󵄨 q−1 󵄨 󵄨 + t 󵄨󵄨 dt ≤ (b − a)2 (∫ 󵄨󵄨󵄨 󵄨󵄨 α+1 2 󵄨󵄨󵄨 0

1

q

1− q1

󵄨󵄨 t α+1 + (1 − t)α+1 − 1 1 − λ 󵄨󵄨 q−1 󵄨 󵄨 + ∫ 󵄨󵄨󵄨 + (1 − t)󵄨󵄨󵄨 dt) 󵄨󵄨 󵄨󵄨 α+1 2 1 2

1

1 r

× (∫[t|f 󸀠󸀠 (a)|qr + (1 − t)|f 󸀠󸀠 (b)|qr ] dt) 0

1 q

1 q

98 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1 2

q

󵄨󵄨 t α+1 + (1 − t)α+1 − 1 󵄨󵄨 󵄨󵄨 1 − λ 󵄨󵄨 q−1 󵄨 󵄨󵄨 󵄨󵄨 󵄨 ≤ (b − a) [∫(󵄨󵄨󵄨 t 󵄨󵄨) dt 󵄨󵄨 + 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 󵄨󵄨󵄨 α+1 2

0

1

1− q1

q

󵄨󵄨 t α+1 + (1 − t)α+1 − 1 󵄨󵄨 󵄨󵄨 1 − λ 󵄨󵄨 q−1 󵄨 󵄨󵄨 󵄨󵄨 󵄨 (1 − t)󵄨󵄨󵄨) dt] + ∫(󵄨󵄨󵄨 󵄨󵄨 + 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 󵄨󵄨 α+1 1 2

× (2

1 −1 r

1

q

1 r

1 r

q

∫[|f (a)| t + |f (b)| (1 − t) ]dt) 󸀠󸀠

0

󸀠󸀠

1 2

1 q

q

1 − t α+1 − (1 − t)α+1 1 − λ q−1 = (b − a) [∫( + t) dt α+1 2 2

0

1

1− q1

q

q−1 1 − t α+1 − (1 − t)α+1 1 − λ + ∫( + (1 − t)) dt] α+1 2 1 2

× [2

1 −1 r

1

q r ] (|f (a)| + |f (b)| ) r+1

q

󸀠󸀠

2

≤ (b − a) {2

1 q−1

1 2

q

+2

q

1 − t α+1 − (1 − t)α+1 q−1 1 − λ q−1 ) +( t) ]dt α+1 2

∫[( 0

1 q−1

q

󸀠󸀠

1

q

1 2

1

1

× [2 r −1 (|f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q ) 1 q

= 2 (b − a)2 {( 1

1 ) α+1

q q−1

1− q1

q

+ ∫(1 − t) q−1 dt]}

1

∫[1 − t α+1 − (1 − t)α+1 ] 1

1−λ +( ) 2

dt + (

r ] r+1

1 2

1 ( ) 2

q q−1

1− q1

q−1 ] 2q − 1 q

1

× [2

1 −1 r

0

1 q

1

q r (|f (a)| + |f (b)| ) ] r+1

󸀠󸀠

q

0

󸀠󸀠

q

q

[∫ t q−1 dt

1

q−1 q(α+1) q(α+1) 1 ≤ 2 (b − a) [( ) ∫(1 − [t q−1 + (1 − t) q−1 ])dt α+1

2

1−λ ) 2

q q−1

q q−1 1 ) ∫(1 − [t α+1 + (1 − t)α+1 ] q−1 )dt α+1

0

q q−1

q q−1

[2 r −1 (|f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q )

q

1

≤ 2 q (b − a)2 [(

q r ] r+1

0

1 2

1 q

1− q1

q

q−1 1−λ 1 − t α+1 − (1 − t)α+1 q−1 ) +( (1 − t)) ]dt} ∫[( α+1 2

4.2 Inequalities via r-convex functions | 99

q

1− q1

q

1 − λ q−1 1 q−1 q − 1 +( ) ( ) ] 2 2 2q − 1

× [2

1 −1 r

1

q r (|f (a)| + |f (b)| ) ] r+1

q

󸀠󸀠

q

󸀠󸀠

q

q

q−1 1 2(q − 1) 1 − λ q−1 q − 1 ) (1 − )+( ) ] α+1 qα + 2q − 1 4 2q − 1

1

= 2 qr (b − a)2 [(

1− q1

1

q r ] . × [(|f (a)| + |f (b)| ) r+1

q

󸀠󸀠

q

󸀠󸀠

(ii) Case 2: r > 1. As in Case 1, we have 󵄨󵄨 Γ(α + 1) f (a) + f (b) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 α α (J − (1 − λ)f ( )󵄨󵄨 + f (b) + Jb− f (a)) − (1 + λ) 󵄨󵄨 a 󵄨󵄨 (b − a)α 󵄨󵄨 2 2 1

2

≤ (b − a) (∫ |k(t, λ)|

q q−1

1− q1

dt)

1

󵄨 󵄨q (∫ 󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt)

1 q

0

0 1 2

q

󵄨󵄨 t α+1 + (1 − t)α+1 − 1 1 − λ 󵄨󵄨 q−1 󵄨 󵄨 ≤ (b − a) (∫ 󵄨󵄨󵄨 + t 󵄨󵄨 dt 󵄨󵄨 α+1 2 󵄨󵄨󵄨 2

0

1

1− q1

q

󵄨󵄨 q−1 󵄨󵄨 t α+1 + (1 − t)α+1 − 1 1 − λ 󵄨 󵄨 + ∫ 󵄨󵄨󵄨 + (1 − t)󵄨󵄨󵄨 dt) 󵄨󵄨 󵄨󵄨 α+1 2 1 2

1

1 r

× (∫[t|f 󸀠󸀠 (a)|qr + (1 − t)|f 󸀠󸀠 (b)|qr ] dt)

1 q

0 1 2

q

󵄨󵄨 t α+1 + (1 − t)α+1 − 1 󵄨󵄨 󵄨󵄨 1 − λ 󵄨󵄨 q−1 󵄨 󵄨󵄨 󵄨󵄨 󵄨 ≤ (b − a) [∫(󵄨󵄨󵄨 t 󵄨󵄨) dt 󵄨󵄨 + 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 󵄨󵄨󵄨 α+1 2

0

1

q

1− q1

󵄨󵄨 q−1 󵄨󵄨 t α+1 + (1 − t)α+1 − 1 󵄨󵄨 󵄨󵄨 1 − λ 󵄨 󵄨󵄨 󵄨󵄨 󵄨 + ∫(󵄨󵄨󵄨 (1 − t)󵄨󵄨󵄨) dt] 󵄨󵄨 + 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 󵄨󵄨 α+1 1 2

1

1 r

1 r

× (∫[|f 󸀠󸀠 (a)|q t + |f 󸀠󸀠 (b)|q (1 − t) ]dt) 0

q

1 q

q

q−1 1 2(q − 1) 1 − λ q−1 q − 1 ≤ 2 (b − a) [( ) (1 − )+( ) ] α+1 qα + 2q − 1 4 2q − 1 1 q

2

1− q1

1

q r × [(|f (a)| + |f (b)| ) ] . r+1

󸀠󸀠

q

󸀠󸀠

q

(iii) Case 3: r = 0. We have 󵄨󵄨 Γ(α + 1) f (a) + f (b) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 (Jaα+ f (b) + Jbα− f (a)) − (1 + λ) − (1 − λ)f ( )󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2 󵄨󵄨 (b − a)

100 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

≤ (b − a)2 (∫ |k(m, λ)|

q q−1

1− q1

dt)

0

1

󵄨q 󵄨 (∫ 󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt)

1 q

0

1 2

q

󵄨󵄨 t α+1 + (1 − t)α+1 − 1 1 − λ 󵄨󵄨 q−1 󵄨 󵄨 ≤ (b − a) (∫ 󵄨󵄨󵄨 + t 󵄨󵄨 dt 󵄨󵄨 α+1 2 󵄨󵄨󵄨 2

0

1

1− q1

q

󵄨󵄨 q−1 󵄨󵄨 t α+1 + (1 − t)α+1 − 1 1 − λ 󵄨 󵄨 + ∫ 󵄨󵄨󵄨 + (1 − t)󵄨󵄨󵄨 dt) 󵄨󵄨 󵄨󵄨 α+1 2 1 2

1

(∫ |f 󸀠󸀠 (a)|qt |f 󸀠󸀠 (b)|q(1−t) dt)

1 q

0 1 2

q

󵄨󵄨 t α+1 + (1 − t)α+1 − 1 󵄨󵄨 󵄨󵄨 1 − λ 󵄨󵄨 q−1 󵄨 󵄨󵄨 󵄨󵄨 󵄨 ≤ (b − a) [∫(󵄨󵄨󵄨 t 󵄨󵄨) dt 󵄨󵄨 + 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 󵄨󵄨󵄨 α+1 2

0

1

1− q1

q

󵄨󵄨 t α+1 + (1 − t)α+1 − 1 󵄨󵄨 󵄨󵄨 1 − λ 󵄨󵄨 q−1 󵄨 󵄨󵄨 󵄨󵄨 󵄨 + ∫(󵄨󵄨󵄨 (1 − t)󵄨󵄨󵄨) dt] 󵄨󵄨 + 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 󵄨󵄨 α+1 1 2

× (|f

󸀠󸀠

󸀠󸀠 1 | f (a) |q − 1 q q f 󸀠󸀠 (b) ) (b)| 󸀠󸀠 (a) q ln | ff 󸀠󸀠 (b) |

1

≤ 2 q (b − a)2 |f 󸀠󸀠 (b)|[( q

q

q−1 2(q − 1) 1 ) (1 − ) α+1 qα + 2q − 1

1− q1

1 − λ q−1 q − 1 ) ] +( 4 2q − 1

(a) q | ff 󸀠󸀠 (b) | −1 󸀠󸀠

(

(a) q ln | ff 󸀠󸀠 (b) | 󸀠󸀠

1 q

) .

The proof is completed. Remark 97. If we take α = 1 in Theorem 96, then the equality (4.40) becomes the following inequality: b 󵄨 󵄨󵄨 󵄨󵄨 2 f (a) + f (b) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 f (t)dt − (1 + λ) − (1 − λ)f ( ) ∫ 󵄨󵄨 ≤ Kr , 󵄨󵄨 b − a 󵄨󵄨 2 2 󵄨󵄨 󵄨 a

where q

q

1 q−1 2(q − 1) 1 − λ q−1 q − 1 )+( ) ] Kr := 2 (b − a) [( ) (1 − 2 3q − 1 4 2q − 1 1 qr

2

1

q r × [(|f (a)| + |f (b)| ) ] r+1

󸀠󸀠

q

󸀠󸀠

q

for 0 < r ≤ 1,

1− q1

4.2 Inequalities via r-convex functions | 101 q

q

1− q1

1 q−1 2(q − 1) 1 − λ q−1 q − 1 Kr := 2 (b − a) [( ) (1 − )+( ) ] 2 3q − 1 4 2q − 1 1 q

2

1

q r × [(|f (a)| + |f (b)| ) ] r+1

q

󸀠󸀠

󸀠󸀠

q

for r > 1,

q

q

1− q1

1 q−1 1 − λ q−1 q − 1 2(q − 1) )+( ) ] Kr := 2 (b − a) |f (b)|[( ) (1 − 2 3q − 1 4 2q − 1 1 q

2

󸀠󸀠

1

|k|q − 1 q ) ×( q ln |k|

for r = 0, and k =

f 󸀠󸀠 (a) . f 󸀠󸀠 (b)

Lemma 98. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with a < b. If |f 󸀠 | is convex on [a, b], then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 1 b−a 󵄨󵄨 󵄨 − [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 ≤ (1 − α )[|f 󸀠 (a)| + |f 󸀠 (b)|]. 󵄨󵄨 α 󵄨󵄨 2(α + 1) 2 2(b − a) 2 󵄨󵄨 Lemma 99. Let f : I ⊂ ℝ → ℝ be a differentiable mapping on I, a, b ∈ I with a < b. If |f 󸀠 | is convex on [a, b], then the following inequality holds: b 󵄨󵄨 󵄨󵄨 󵄨󵄨 f (a) + f (b) 󵄨󵄨 b − a 󸀠 1 󵄨󵄨 󵄨󵄨 ≤ − f (x)dx (|f (a)| + |f 󸀠 (b)|). ∫ 󵄨󵄨 󵄨󵄨 2 b−a 8 󵄨󵄨 󵄨 󵄨 a

Theorem 100. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with a < b. If |f 󸀠 |q is convex on [a, b] for some fixed q ≥ 1, then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨 − [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 ≤

b−a

2

α+1+ q1

1− q1

2α − 1 ( ) α+1

1

q 2α+1 − (α + 2) 󸀠 2α+1 (α + 1) − (α + 2) 󸀠 {[ |f (a)|q + |f (b)|q ] (α + 1)(α + 2) (α + 1)(α + 2) 1

q 2α+1 (α + 1) − (α + 2) 󸀠 2α+1 − (α + 2) 󸀠 +[ |f (a)|q + |f (b)|q ] }. (α + 1)(α + 2) (α + 1)(α + 2)

(4.41)

Proof. Firstly, we suppose that q = 1. Using Lemma 45 and the convexity of |f 󸀠 |, for any t ∈ [0, 1], we find 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨 − [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) α 1 󵄨󵄨 󵄨󵄨󵄨 b−a 1 a+b 󵄨 ≤ ( ) [∫ |t α − (2 − t)α |󵄨󵄨󵄨f 󸀠 (t + (1 − t)a)󵄨󵄨󵄨dt 󵄨 󵄨󵄨 4 2 2 󵄨 1

0

󵄨󵄨 a + b 󵄨󵄨󵄨󵄨 󵄨 + ∫ |(1 + t)α − (1 − t)α 󵄨󵄨󵄨f 󸀠 (tb + (1 − t) )󵄨󵄨dt] 󵄨󵄨 󵄨󵄨 2 0

102 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals α 1 󵄨󵄨 a + b 󵄨󵄨󵄨󵄨 b−a 1 󵄨 ( ) {∫[(2 − t)α − t α ](t 󵄨󵄨󵄨f 󸀠 ( )󵄨󵄨 + (1 − t)|f 󸀠 (a)|)dt ≤ 󵄨󵄨 󵄨󵄨 4 2 2 0

1

󵄨󵄨 a + b 󵄨󵄨󵄨󵄨 󵄨 + ∫[(1 + t)α − (1 − t)α ](t|f 󸀠 (b)| + (1 − t)󵄨󵄨󵄨f 󸀠 ( )󵄨󵄨)dt} 󵄨󵄨 󵄨󵄨 2 0

α 1 󵄨󵄨 a + b 󵄨󵄨󵄨󵄨 b−a 1 󵄨 = ( ) {∫[(2 − t)α t − t α+1 + (1 + t)α (1 − t) − (1 − t)α+1 ]󵄨󵄨󵄨f 󸀠 ( )󵄨󵄨dt 󵄨󵄨 󵄨󵄨 4 2 2 0

1

1

+ ∫[(2 − t)α (1 − t) − t α (1 − t)]|f 󸀠 (a)|dt + ∫[t(1 + t)α − t(1 − t)α ]|f 󸀠 (b)|dt} 0

0

α

b−a 1 = ( ) (K1 + K2 + K3 ). 4 2

(4.42)

Integrating by parts, we have 1

󵄨󵄨 a + b 󵄨󵄨󵄨󵄨 󵄨 )󵄨󵄨dt K1 = ∫[(2 − t)α t − t α+1 + (1 + t)α (1 − t) − (1 − t)α+1 ]󵄨󵄨󵄨f 󸀠 ( 󵄨󵄨 󵄨󵄨 2 0

= (−

1 2α+2 1 2α+1 1 − + − + α + 1 (α + 1)(α + 2) (α + 1)(α + 2) α + 2 (α + 1)

󵄨󵄨 1 2α+1 2α+2 1 1 a + b 󵄨󵄨󵄨󵄨 󵄨 − + − − )󵄨󵄨󵄨f 󸀠 ( )󵄨󵄨 󵄨󵄨 α + 1 (α + 1) (α + 1)(α + 2) (α + 1)(α + 2) α + 2 󵄨󵄨 2 −4α − 8 + 2α+3 󵄨󵄨󵄨󵄨 󸀠 a + b 󵄨󵄨󵄨󵄨 )󵄨󵄨, (4.43) = 󵄨f ( 󵄨󵄨 (α + 1)(α + 2) 󵄨󵄨󵄨 2 −

and

1

K2 = ∫[(2 − t)α (1 − t) − t α (1 − t)]|f 󸀠 (a)|dt 0

=( =

1 2α+2 1 2α+1 + − − )|f 󸀠 (a)| α + 1 (α + 1)(α + 2) (α + 1)(α + 2) (α + 1)(α + 2)

α2α+1 |f 󸀠 (a)|, (α + 1)(α + 2)

(4.44)

and 1

K3 = ∫[t(1 + t)α − t(1 − t)α ]|f 󸀠 (b)|dt 0

=( =

2α+1 2α+2 − 1 1 − − )|f 󸀠 (b)| α + 1 (α + 1)(α + 2) (α + 1)(α + 2)

α2α+1 |f 󸀠 (b)| (α + 1)(α + 2)

(4.45)

4.2 Inequalities via r-convex functions | 103

and 󸀠 󸀠 󵄨󵄨 󵄨 󵄨󵄨 󸀠 a + b 󵄨󵄨󵄨 |f (a)| + |f (b)| )󵄨󵄨 ≤ . 󵄨󵄨f ( 󵄨󵄨 󵄨󵄨 2 2

(4.46)

In view of (4.42), (4.43), (4.44), (4.45) and (4.46), we get 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) α 󵄨󵄨 󵄨 − [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) α α+3 󸀠 󸀠 b−a 1 −4α − 8 + 2 |f (a)| + |f (b)| α2α+1 ≤ ( ) [ + (|f 󸀠 (a)| + |f 󸀠 (b)|)] 4 2 (α + 1)(α + 2) 2 (α + 1)(α + 2) α

=

b−a 1 b − a 1 2α+1 − 2 ( ) = (1 − α )(|f 󸀠 (a)| + |f 󸀠 (b)|), 4 2 α+1 2(α + 1) 2

(4.47)

which is just the result of Lemma 98. Second, we suppose that q > 1. Because |f 󸀠 |q is convex on [a, b], for any t ∈ [0, 1], then we have q 󵄨󵄨q 󵄨󵄨 󵄨󵄨 a + b 󵄨󵄨󵄨󵄨 󵄨 󵄨󵄨 󸀠 a + b 󵄨 ) + (1 − t)a󵄨󵄨󵄨 ≤ t 󵄨󵄨󵄨f 󸀠 ( )󵄨󵄨 + (1 − t)|f 󸀠 (a)|q , 󵄨󵄨f (t 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 2 q 󵄨󵄨 󵄨 󵄨󵄨q 󵄨󵄨 󵄨 󵄨󵄨󵄨f 󸀠 (tb + (1 − t) a + b )󵄨󵄨󵄨 ≤ t|f 󸀠 (b)|q + (1 − t)󵄨󵄨󵄨f 󸀠 ( a + b )󵄨󵄨󵄨 , 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 2 󵄨 󵄨 󵄨 󵄨

(4.48) (4.49)

and 󵄨q |f 󸀠 (a)|q + |f 󸀠 (b)|q 󵄨󵄨 󵄨󵄨 󸀠 a + b 󵄨󵄨󵄨 )󵄨󵄨 ≤ . 󵄨󵄨f ( 󵄨󵄨 󵄨󵄨 2 2

(4.50)

Using Lemma 45 and the power mean inequality for q, we obtain 1

∫[(2 − t)α − t α ]f 󸀠 (t 0 1

α

a+b + (1 − t)a)dt 2 1− q1

α

≤ {∫[(2 − t) − t ]dt} 0 1

α

α

≤ {∫[(2 − t) − t ]dt}

1− q1

0

2

1− q1

α+1

−2 =( ) α+1

1− q1

2α+1 − 2 ≤( ) α+1

1− q1

2α+1 − 2 =( ) α+1

1

1

q 󵄨󵄨 󵄨󵄨q a+b 󵄨 󵄨 {∫[(2 − t) − t ]󵄨󵄨󵄨f 󸀠 (t + (1 − t)a)󵄨󵄨󵄨 dt} 󵄨󵄨 󵄨󵄨 2

α

α

0

1

1

q q 󵄨󵄨 a + b 󵄨󵄨󵄨󵄨 󵄨 {∫[(2 − t) − t ](t 󵄨󵄨󵄨f 󸀠 ( )󵄨󵄨 + (1 − t)|f 󸀠 (a)|q dt)} 󵄨󵄨 󵄨󵄨 2

0

α

α

1

q q −2α − 4 + 2α+2 󵄨󵄨󵄨󵄨 󸀠 a + b 󵄨󵄨󵄨󵄨 α2α+1 [ )󵄨󵄨 + |f 󸀠 (a)|q ] 󵄨󵄨f ( 󵄨 󵄨 (α + 1)(α + 2) 󵄨 2 (α + 1)(α + 2) 󵄨

1

q −2α − 4 + 2α+2 |f 󸀠 (a)|q + |f 󸀠 (b)|q α2α+1 [ + |f 󸀠 (a)|q ] (α + 1)(α + 2) 2 (α + 1)(α + 2) 1

q (α + 1)2α+1 − (α + 2) 󸀠 2α+1 − (α + 2) 󸀠 [ |f (a)|q + |f (b)|q ] (α + 1)(α + 2) (α + 1)(α + 2)

(4.51)

104 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals and 1

∫[(1 + t)α − (1 − t)α ]f 󸀠 (tb + (1 − t) 0 1

α

a+b )dt 2

1− q1

α

≤ {∫[(1 + t) − (1 − t) ]dt} 0

1

1

q q 󵄨󵄨 a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 󸀠 × {∫[(1 + t) − (1 − t) ]󵄨󵄨f (tb + (1 − t) )󵄨󵄨 dt} 󵄨󵄨 󵄨󵄨 2

α

α

0

1− q1

2α+1 − 2 =( ) α+1 2

1− q1

α+1

−2 =( ) α+1

1− q1

2α+1 − 2 ) ≤( α+1

1− q1

=(

2α+1 − 2 ) α+1

1

1

q 󵄨󵄨 a + b 󵄨󵄨q 󵄨 󵄨 {∫[(1 + t) − (1 − t) ](t|f (b)| + (1 − t)󵄨󵄨󵄨f ( )󵄨󵄨󵄨 )dt} 󵄨󵄨 󵄨󵄨 2

α

α

󸀠

q

0

1

q α2α+1 −2α − 4 + 2α+2 󵄨󵄨󵄨󵄨 a + b 󵄨󵄨󵄨󵄨 q [ |f 󸀠 (b)|q + )󵄨󵄨 ] 󵄨󵄨f ( 󵄨󵄨 (α + 1)(α + 2) (α + 1)(α + 2) 󵄨󵄨 2

1

−2α − 4 + 2α+2 |f 󸀠 (a)|q + |f 󸀠 (b)|q q α2α+1 |f 󸀠 (b)|q + ] [ (α + 1)(α + 2) (α + 1)(α + 2) 2 1

[

q (α + 1)2α+1 − (α + 2) 󸀠 2α+1 − (α + 2) 󸀠 |f (b)|q + |f (a)|q ] . (α + 1)(α + 2) (α + 1)(α + 2)

(4.52)

By combining (4.51) and (4.52), we get the inequality (4.41). The proof is completed. Theorem 100 may be extended as follows: Corollary 101. If we take α = 1, q > 1 in theorem 100, then the following inequality for fractional integrals holds: 1 1 b 󵄨󵄨 󵄨󵄨 󵄨󵄨 f (a) + f (b) 󵄨󵄨 b − a 1 q 1 5 q 󵄨󵄨 󵄨 − [( ) + ( ) ][|f 󸀠 (a)| + |f 󸀠 (b)|]. ∫ f (x)dx󵄨󵄨 ≤ 󵄨󵄨 󵄨󵄨 2 b−a 8 6 6 󵄨󵄨 󵄨 a

Proof. If we consider inequality (4.41) and take α = 1, then we have b 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 f (a) + f (b) 1 󵄨󵄨 − ∫ f (x)dx󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 2 b − a 󵄨󵄨 󵄨 a

1

1

q q b−a 1 󸀠 5 5 1 ≤ {[ |f (a)|q + |f 󸀠 (b)|q ] + [ |f 󸀠 (a)|q + |f 󸀠 (b)|q ] } 8 6 6 6 6

Let a1 = 61 |f 󸀠 (a)|q , b1 = 65 |f 󸀠 (b)|q , a2 = 65 |f 󸀠 (a)|q and b1 = 61 |f 󸀠 (b)|q . Here, 0 < q > 1. Using the fact that n

n

n

i=1

i=1

i=1

∑(ai + bi )r ≤ ∑(ai )r + ∑(bi )r

1 q

< 1, for

4.2 Inequalities via r-convex functions | 105

for 0 < r < 1, a1 , a2 , . . . , an ≥ 0, and b1 , b2 , . . . , bn ≥ 0, we obtain the inequalities cited. The proof is completed. Remark 102. If we take α = 1, q = 1 in (4.41), then the inequality (4.41) becomes the conclusion of Lemma 99. Next, we begin to consider the second theorem containing Hermite-Hadmard type inequality. Theorem 103. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with a < b. If |f 󸀠 |q is convex on [a, b] for some fixed q > 1, then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨󵄨 α − [J f (b) + J f (a)] + − 󵄨󵄨 󵄨󵄨 b 󵄨󵄨 󵄨󵄨 2 2(b − a)α a 1

1

(b − a) 2αp − 1 p 3|f 󸀠 (a)|q + |f 󸀠 (b)|q q ) {[ ] ≤ ( α+2− p1 αp + 1 4 2 1

|f 󸀠 (a)|q + 3|f 󸀠 (b)|q q +[ ] }, 4

where

1 p

+

1 q

(4.53)

= 1.

Proof. From Lemma 45 and using the well-known Hölder’s inequality, for any t ∈ [0, 1], we have 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) α 󵄨󵄨 󵄨󵄨 α [J − + f (b) + Jb− f (a)]󵄨 󵄨󵄨 a 󵄨󵄨 󵄨󵄨 2 2(b − a)α 󵄨 α 1 󵄨 󵄨󵄨 b−a 1 a+b 󵄨󵄨 󵄨 ≤ ( ) [∫ |t α − (2 − t)α |󵄨󵄨󵄨f 󸀠 (t + (1 − t)a)󵄨󵄨󵄨dt 󵄨󵄨 󵄨󵄨 4 2 2 1

0

󵄨󵄨 a + b 󵄨󵄨󵄨󵄨 󵄨 )󵄨󵄨dt] + ∫ |(1 + t)α − (1 − t)α 󵄨󵄨󵄨f 󸀠 (tb + (1 − t) 󵄨󵄨 󵄨󵄨 2 0

α 1 󵄨󵄨 󵄨󵄨 b−a 1 a+b 󵄨 󵄨 ( ) {∫[(2 − t)α − t α ]󵄨󵄨󵄨f 󸀠 (t + (1 − t)a)󵄨󵄨󵄨dt = 󵄨󵄨 󵄨󵄨 4 2 2 1

0

󵄨󵄨 a + b 󵄨󵄨󵄨󵄨 󵄨 )󵄨󵄨dt} + ∫[(1 + t)α − (1 − t)α ]󵄨󵄨󵄨f 󸀠 (tb + (1 − t) 󵄨󵄨 󵄨󵄨 2 0

1

1

1 p q α 1 󵄨󵄨 󵄨󵄨q b−a 1 a+b 󵄨 󵄨 α α p ≤ ( ) {∫[(2 − t) − t ] dt} {∫ 󵄨󵄨󵄨f 󸀠 (t + (1 − t)a)󵄨󵄨󵄨 dt} 󵄨󵄨 󵄨󵄨 4 2 2 0

α

0

1

b−a 1 p + ( ) {∫[(1 + t)α − (1 − t)α ] dt} 4 2 0

1 p

106 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

1

q q 󵄨󵄨 a + b 󵄨󵄨󵄨󵄨 󵄨 × {∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t) )󵄨󵄨 dt} . 󵄨󵄨 󵄨󵄨 2

(4.54)

0

Because |f 󸀠 |q is convex on [a, b], for any t ∈ [0, 1], then we have 1 1 q 󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨q a + b 󵄨󵄨󵄨󵄨 a+b 󵄨 ) + (1 − t)a󵄨󵄨󵄨 dt ≤ ∫[t 󵄨󵄨󵄨f 󸀠 ( )󵄨󵄨 + (1 − t)|f 󸀠 (a)|q ]dt ∫ 󵄨󵄨󵄨f 󸀠 (t 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 2 0

0

=

|f 󸀠 ( a+b )|q + |f 󸀠 (a)|q 2

(4.55)

2

and 1

1 q q 󵄨󵄨 󵄨󵄨 a + b 󵄨󵄨󵄨󵄨 a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 󸀠 󵄨 )󵄨󵄨 dt ≤ ∫[t|f 󸀠 (b)|q + (1 − t)󵄨󵄨󵄨f 󸀠 ( )󵄨󵄨 ]dt ∫ 󵄨󵄨f (tb + (1 − t) 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 2 0

0

=

)|q + |f 󸀠 (b)|q |f 󸀠 ( a+b 2 2

.

(4.56)

On the other hand, 1

1

p

∫[(2 − t)α − t α ] dt ≤ ∫[(2 − t)αp − t αp ]dt 0

0

=

2αp+1 − 2 , αp + 1

(4.57)

and 1

1

α p

α

∫[(1 + t) − (1 − t) ] dt ≤ ∫[(1 + t)αp − (1 − t)αp ]dt 0

0

=

2αp+1 − 2 . αp + 1

(4.58)

Here, we use the following facts: p

[(2 − t)α − t α ] ≤ (2 − t)αp − t αp , p

[(1 + t)α − (1 − t)α ] ≤ (1 + t)αp − (1 − t)αp , for any t ∈ [0, 1], which follows from (A − B)q ≤ Aq − Bq , for any A > B ≥ 0 and q ≥ 1. Thus if we substitute (4.50), (4.55), (4.56), (4.57), and (4.58) into (4.54), we obtain the inequality (4.53). The proof is completed.

4.2 Inequalities via r-convex functions | 107

By Lemma 46 and Definition 17 via Lemma 37 and Lemma 38, we have the following main results. Theorem 104. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with 0 ≤ a < b. If |f 󸀠 | is measurable and r-convex on [a, b] for some fixed 0 ≤ r < ∞, then the following inequality for fractional integrals holds: 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) α 󵄨󵄨 󵄨 α − [Ja+ f (b) + Jb− f (a)]󵄨󵄨󵄨 ≤ Kr , 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨

(4.59)

where Kr : = 2

1−r−r 2 r

−α

1−r

(

1−r ) α+1−r

1−r

α

(2 1−r − 1)

r

(

r2 ) 1 + r2

× (|f 󸀠 (a)| + |f 󸀠 (b)|)(b − a) for 0 < r < 1, Kr : = Kr : =

2α − 1 (|f 󸀠 (a)| + |f 󸀠 (b)|)(b − a) for r = 1, + 1)

2α+1 (α 1

2

2α+ r

Kr : = 2

1− 1r

αr

(r − 1)(2 r−1 − 1) [ ] αr + r − 1

1 −α−1 p

1

(|f 󸀠 (a)| + |f 󸀠 (b)|)(b − a) for r > 1, 1− p1

2pα − 1 p p − 1 ( ) ( ) pα + 1 p ln |k| 1

p

p 1− p1 − 2(p−1)

(|k| 2(p−1) − |k|

)

1

× |f 󸀠 (a)| 2 |f 󸀠 (b)| 2 (b − a) for r = 0, and p > 1, k =

f 󸀠 (a) . f 󸀠 (b)

Proof. (i) Case 1: 0 < r < 1. From Definition 17, Lemma 37 and Lemma 38, we have 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) α 󵄨 󵄨󵄨 α [Ja+ f (b) + Jb− − f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 󵄨󵄨 1 b − a 󵄨󵄨 1−t 1+t ≤ α+2 󵄨󵄨󵄨 ∫[(1 − t)α − (1 + t)α ]f 󸀠 ( a+ b)dt 󵄨󵄨 2 2 2 󵄨0 1

+ ∫[(1 + t)α − (1 − t)α ]f 󸀠 ( 0

≤

󵄨󵄨 󵄨󵄨 1+t 1−t b+ a)dt 󵄨󵄨󵄨 󵄨󵄨 2 2 󵄨

1 󵄨󵄨 1 − t 󵄨󵄨󵄨󵄨 b−a 1+t 󵄨 [ |(1 − t)α − (1 + t)α |󵄨󵄨󵄨f 󸀠 ( a+ b)󵄨󵄨dt ∫ α+2 󵄨󵄨 󵄨󵄨 2 2 2 1

0

󵄨󵄨 1 − t 󵄨󵄨󵄨󵄨 1+t 󵄨 b+ a)󵄨󵄨dt] + ∫ |(1 + t)α − (1 − t)α |󵄨󵄨󵄨f 󸀠 ( 󵄨󵄨 󵄨󵄨 2 2 0

1

1

r b−a 1+t 󸀠 1−t 󸀠 ≤ α+2 [∫[(1 + t)α − (1 − t)α ]( |f (a)|r + |f (b)|r ) dt 2 2 2

0

108 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

1

r 1−t 󸀠 1+t 󸀠 |f (b)|r + |f (a)|r ) dt] + ∫[(1 + t) − (1 − t) ]( 2 2

α

0

α

1

1

1

1 1+t r 󸀠 b−a 1−t r 󸀠 ≤ α+2 {∫[(1 + t)α − (1 − t)α ]2 r −1 [( ) |f (a)| + ( ) |f (b)|]dt 2 2 2

0

1

α

α

+ ∫[(1 + t) − (1 − t) ]2 0

=

1 −1 r

1

1

1+t r 󸀠 1−t r 󸀠 [( ) |f (b)| + ( ) |f (a)|]dt} 2 2

1

1 1 b−a {∫[(1 + t)α − (1 − t)α ][(1 + t) r |f 󸀠 (a)| + (1 − t) r |f 󸀠 (b)|]dt 2α+3

0

1

1

1

+ ∫[(1 + t)α − (1 − t)α ][(1 + t) r |f 󸀠 (b)| + (1 − t) r |f 󸀠 (a)|]dt} 0

1

1 1 b−a = α+3 (|f 󸀠 (a)| + |f 󸀠 (b)|) ∫[(1 + t)α − (1 − t)α ][(1 + t) r + (1 − t) r ]dt 2

0

≤

1−r

1

1 b−a 󸀠 (|f (a)| + |f 󸀠 (b)|)(∫[(1 + t)α − (1 − t)α ] 1−r dt) α+3 2

0

1

1 r

1 r

r

1 r

× (∫[(1 + t) + (1 − t) ] dt) 0

1−r

1

α α b−a ≤ α+3 (|f 󸀠 (a)| + |f 󸀠 (b)|)(∫[(1 + t) 1−r − (1 − t) 1−r ]dt) 2

× (2

1 −1 r

1

∫[(1 + t)

0

1 r2

r

1 r2

+ (1 − t) ]dt)

0

=

1−r+α b−a 󸀠 1−r 1−r (|f (a)| + |f 󸀠 (b)|)[ (2 1−r − 1) − ] α+3 1−r+α 1−r+α 2

× 21−r [ =2

1−r−r 2 r

−α

1−r

r

1 r2 r2 +1 r2 (2 − 1) + ] 1 + r2 1 + r2

1−r

(

1−r ) α+1−r

α

1−r

(2 1−r − 1)

r

(

r2 ) (|f 󸀠 (a)| + |f 󸀠 (b)|)(b − a). 1 + r2

(ii) Case 2: r > 1. As in Case 1, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨󵄨 α − [J f (b) + J f (a)] 󵄨󵄨 󵄨󵄨 a+ b− 󵄨󵄨 󵄨󵄨 2 2(b − a)α 1

1

r b−a 1+t 󸀠 1−t 󸀠 ≤ α+2 [∫[(1 + t)α − (1 − t)α ]( |f (a)|r + |f (b)|r ) dt 2 2 2

0

4.2 Inequalities via r-convex functions | 109 1

1

r 1−t 󸀠 1+t 󸀠 |f (b)|r + |f (a)|r ) dt] + ∫[(1 + t) − (1 − t) ]( 2 2

α

0

α

1

1

1

b−a 1+t r 󸀠 1−t r 󸀠 ≤ α+2 {∫[(1 + t)α − (1 − t)α ][( ) |f (a)| + ( ) |f (b)|]dt 2 2 2 0

1

1

1

1+t r 󸀠 1−t r 󸀠 + ∫[(1 + t) − (1 − t) ][( ) |f (b)| + ( ) |f (a)|]dt} 2 2 α

0

=

b−a

α+ 1r +2

2

1

α

1

1

1

{∫[(1 + t)α − (1 − t)α ][(1 + t) r |f 󸀠 (a)| + (1 − t) r |f 󸀠 (b)|]dt 0 1

1

+ ∫[(1 + t)α − (1 − t)α ][(1 + t) r |f 󸀠 (b)| + (1 − t) r |f 󸀠 (a)|]dt} 0

=

≤

b−a

α+ 1r +2

2

b−a 1

2α+ r +2

1

󸀠

0 1

(|f 󸀠 (a)| + |f 󸀠 (b)|)(∫[(1 + t)α − (1 − t)α ] 0

1

1 r

1 r

r

× (∫[(1 + t) + (1 − t) ] dt)

r r−1

≤

1

2α+ r +2

1 r

1

(|f 󸀠 (a)| + |f 󸀠 (b)|)(∫[(1 + t) 0 1

× (2r−1 ∫[(1 + t) + (1 − t)]dt) 0

=

1

α+ 2r

2

αr

(r − 1)(2 r−1 − 1) ] [ αr + r − 1

1− 1r

1− 1r

dt)

0

b−a

1

1

(|f (a)| + |f (b)|) ∫[(1 + t)α − (1 − t)α ][(1 + t) r + (1 − t) r ]dt 󸀠

αr r−1

− (1 − t)

αr r−1

1− 1r

]dt)

1 r

(|f 󸀠 (a)| + |f 󸀠 (b)|)(b − a).

(iii) Case 3: r = 1. We have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨 α − f (a)]󵄨󵄨󵄨 [Ja+ f (b) + Jb− 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) 1

≤

b−a 1+t 󸀠 1−t 󸀠 [∫[(1 + t)α − (1 − t)α ]( |f (a)| + |f (b)|)dt 2 2 2α+2 1

0

+ ∫[(1 + t)α − (1 − t)α ]( 0

1+t 󸀠 1−t 󸀠 |f (b)| + |f (a)|)dt] 2 2

110 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

b−a = α+3 {∫[(1 + t)α − (1 − t)α ][(1 + t)|f 󸀠 (a)| + (1 − t)|f 󸀠 (b)|]dt 2 0

1

+ ∫[(1 + t)α − (1 − t)α ][(1 + t)|f 󸀠 (b)| + (1 − t)|f 󸀠 (a)|]dt} 0

=

1

b−a 󸀠 (|f (a)| + |f 󸀠 (b)|) ∫[(1 + t)α − (1 − t)α ][(1 + t) + (1 − t)]dt 2α+3 0

2α − 1 = α+1 (|f 󸀠 (a)| + |f 󸀠 (b)|)(b − a). 2 (α + 1) (iv) Case 4: r = 0. We have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨󵄨 α − [J f (b) + J f (a)] 󵄨󵄨 󵄨󵄨 b− 󵄨󵄨 󵄨󵄨 2 2(b − a)α a+ 1 󵄨󵄨 1 − t 󵄨󵄨󵄨󵄨 b−a 1+t 󵄨 a+ b)󵄨󵄨dt ≤ α+2 [∫ |(1 − t)α − (1 + t)α |󵄨󵄨󵄨f 󸀠 ( 󵄨󵄨 󵄨󵄨 2 2 2 0

1

󵄨󵄨 1 − t 󵄨󵄨󵄨󵄨 1+t 󵄨 b+ a)󵄨󵄨dt] + ∫ |(1 + t)α − (1 − t)α |󵄨󵄨󵄨f 󸀠 ( 󵄨󵄨 󵄨󵄨 2 2 0

≤

1

1+t 1−t b−a [∫[(1 + t)α − (1 − t)α ](|f 󸀠 (a)| 2 |f 󸀠 (b)| 2 )dt α+2 2

0

1

+ ∫[(1 + t)α − (1 − t)α ](|f 󸀠 (b)| 0

=

1+t 2

|f 󸀠 (a)|

1−t 2

)dt]

1

1+t 1−t 1+t 1−t b−a ∫[(1 + t)α − (1 − t)α ][|f 󸀠 (a)| 2 |f 󸀠 (b)| 2 + |f 󸀠 (b)| 2 |f 󸀠 (a)| 2 ]dt α+2 2

0

=

1 󵄨 󸀠 󵄨 2 󵄨󵄨 f 󸀠 (b) 󵄨󵄨 2 1 1 b−a 󸀠 󵄨󵄨 󵄨󵄨 󸀠 α α 󵄨󵄨 f (a) 󵄨󵄨󵄨 2 |f (b)| 2 ∫[(1 + t) − (1 − t) ][󵄨󵄨 |f (a)| 󵄨󵄨 f 󸀠 (b) 󵄨󵄨󵄨 + 󵄨󵄨󵄨 f 󸀠 (a) 󵄨󵄨󵄨 ]dt α+2 2 󵄨 󵄨 󵄨 󵄨 t

1 1 b−a ≤ α+2 |f 󸀠 (a)| 2 |f 󸀠 (b)| 2 2

1

t

0

1 p

1

t

t

p

1− p1

󵄨󵄨 f 󸀠 (a) 󵄨󵄨 2 󵄨󵄨 f 󸀠 (b) 󵄨󵄨 2 p−1 󵄨 󵄨 󵄨 󵄨 × (∫[(1 + t) − (1 − t) ] dt) (∫[󵄨󵄨󵄨 󸀠 󵄨󵄨󵄨 + 󵄨󵄨󵄨 󸀠 󵄨󵄨󵄨 ] dt) 󵄨󵄨 f (b) 󵄨󵄨 󵄨󵄨 f (a) 󵄨󵄨 α p

α

0

0

1

1 1 b−a ≤ α+2 |f 󸀠 (a)| 2 |f 󸀠 (b)| 2 (∫[(1 + t)pα − (1 − t)pα ]dt) 2

0

× (2

1 p−1

1

󵄨󵄨 f 󸀠 (a) 󵄨󵄨 󵄨 󵄨 ∫[󵄨󵄨󵄨 󸀠 󵄨󵄨󵄨 󵄨󵄨 f (b) 󵄨󵄨 0

pt 2(p−1)

pt

󵄨󵄨 f 󸀠 (b) 󵄨󵄨 2(p−1) 󵄨 󵄨 + 󵄨󵄨󵄨 󸀠 󵄨󵄨󵄨 ]dt) 󵄨󵄨 f (a) 󵄨󵄨

1− p1

1 p

4.2 Inequalities via r-convex functions | 111

=2

1 −α−1 p

1

1− p1

2pα − 1 p p−1 ( ) ( ) 󸀠 pα + 1 p ln | f 󸀠 (a) | 1 2

f (b)

1 2

p

p

1

󵄨󵄨 f 󸀠 (a) 󵄨󵄨 2(p−1) 󵄨󵄨 f 󸀠 (b) 󵄨󵄨 2(p−1) 1− p 󵄨 󵄨 󵄨 󵄨 (󵄨󵄨󵄨 󸀠 󵄨󵄨󵄨 − 󵄨󵄨󵄨 󸀠 󵄨󵄨󵄨 ) 󵄨󵄨 f (a) 󵄨󵄨 󵄨󵄨 f (b) 󵄨󵄨

× |f (a)| |f (b)| (b − a). 󸀠

󸀠

The proof is completed. Remark 105. If we take α = 1 in Theorem 104, then the equality (4.59) becomes the following inequality: b 󵄨󵄨 󵄨󵄨 󵄨󵄨 f (a) + f (b) 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 ≤ Kr , − f (x)dx ∫ 󵄨󵄨 󵄨󵄨 2 b−a 󵄨󵄨 󵄨󵄨 a

where Kr : = 2

1−2r−r 2 r

1−r

(

1−r ) 2−r

1−r

1

(2 1−r − 1)

× (|f 󸀠 (a)| + |f 󸀠 (b)|)(b − a)

Kr : = Kr : =

r

(

r2 ) 1 + r2

for 0 < r < 1,

1 󸀠 (|f (a)| + |f 󸀠 (b)|)(b − a) for r = 1, 8 1+ 2r

2

1

1− 1r

r

1

(r − 1)(2 r−1 − 1) [ ] 2r − 1 1

Kr : = 2 p ( −2

1− p1

2p − 1 p p − 1 ) ( ) p+1 p ln |k| 1

(|f 󸀠 (a)| + |f 󸀠 (b)|)(b − a) for r > 1, p

p 1− p1 − 2(p−1)

(|k| 2(p−1) − |k|

)

1

× |f 󸀠 (a)| 2 |f 󸀠 (b)| 2 (b − a) for r = 0, and p > 1, k =

f 󸀠 (a) . f 󸀠 (b)

Theorem 106. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with 0 ≤ a < b. If |f 󸀠 |q , q > 1 is measurable and r-convex on [a, b] for some fixed 0 ≤ r < ∞, then the following inequality for fractional integrals holds: 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) α 󵄨 󵄨󵄨 α − f (a)]󵄨󵄨󵄨 ≤ Kr , [Ja+ f (b) + Jb− 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨

(4.60)

where b−a

qα

1− q1

(q − 1)(2 q−1 − 1) Kr : = [ ] 1 α+ +1 qα + q − 1 2 q

1

1

3|f 󸀠 (a)|qr + |f 󸀠 (b)|qr q 3|f 󸀠 (b)|qr + |f 󸀠 (a)|qr q × [( ) +( ) ], 4 4

for r > 0,

112 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1− q1

qα

b − a (q − 1)(2 q−1 − 1) Kr : = α+1 [ ] qα + q − 1 2 1

q

1

q

1 1 |k| 2 − 1 q 1 − |k|− 2 q 󸀠 × [( ) +( ) ]|f (a)| 2 |f 󸀠 (b)| 2 , q ln |k| q ln |k|

and k =

for r = 0,

f 󸀠 (a) . f 󸀠 (b)

Proof. (i) Case 1: r > 0. We have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨 α − [Ja+ f (b) + Jb− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) 󵄨󵄨 1 1−t 1+t b − a 󵄨󵄨 a+ b)dt ≤ α+2 󵄨󵄨󵄨 ∫[(1 − t)α − (1 + t)α ]f 󸀠 ( 󵄨󵄨 2 2 2 󵄨0 1

+ ∫[(1 + t)α − (1 − t)α ]f 󸀠 ( 0

󵄨󵄨 󵄨󵄨 1+t 1−t b+ a)dt 󵄨󵄨󵄨 󵄨󵄨 2 2 󵄨

1 󵄨󵄨 1 − t 󵄨󵄨󵄨󵄨 1+t b−a 󵄨 a+ b)󵄨󵄨dt ≤ α+2 [∫ |(1 − t)α − (1 + t)α |󵄨󵄨󵄨f 󸀠 ( 󵄨󵄨 󵄨󵄨 2 2 2 0

1

󵄨󵄨 1 − t 󵄨󵄨󵄨󵄨 1+t 󵄨 b+ a)󵄨󵄨dt] + ∫ |(1 + t)α − (1 − t)α |󵄨󵄨󵄨f 󸀠 ( 󵄨󵄨 󵄨󵄨 2 2 0

1

q b−a ≤ α+2 [(∫ |(1 − t)α − (1 + t)α | q−1 dt) 2

0

1

α

α

+ (∫ |(1 + t) − (1 − t) |

q q−1

1− q1

dt)

0

1− q1

1

1

q q 󵄨󵄨 1 − t 󵄨󵄨󵄨󵄨 1+t 󵄨 (∫ 󵄨󵄨󵄨f 󸀠 ( a+ b)󵄨󵄨 dt) 󵄨󵄨 󵄨󵄨 2 2

0

1

1

q q 󵄨󵄨 1 − t 󵄨󵄨󵄨󵄨 󵄨󵄨 󸀠 1 + t (∫ 󵄨󵄨f ( b+ a)󵄨󵄨 dt) ] 󵄨󵄨 󵄨󵄨 2 2

0

1− q1

1

qα qα b−a ≤ α+2 {(∫[(1 + t) q−1 − (1 − t) q−1 ]dt) 2

0

1

1+t 󸀠 1−t 󸀠 |f (a)|qr + |f (b)|qr )dt] × [∫( 2 2 0

1

+ (∫[(1 + t)

qα q−1

− (1 − t)

0

b−a

qα q−1

qα q−1

1− q1

]dt)

1− q1

(q − 1)(2 − 1) = [ ] α+ q1 +1 qα + q − 1 2

1

3|f 󸀠 (b)|qr + |f 󸀠 (a)|qr q +( ) ]. 4

1 q

1 q

1

1+t 󸀠 1−t 󸀠 [∫( |f (b)|qr + |f (a)|qr )dt] } 2 2 0

1

3|f (a)|qr + |f 󸀠 (b)|qr q [( ) 4 󸀠

4.2 Inequalities via r-convex functions | 113

(ii) Case 2: r = 0. As in Case 1, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨 α − [Ja+ f (b) + Jb− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) 1

q b−a ≤ α+2 [(∫ |(1 − t)α − (1 + t)α | q−1 dt) 2

0

1

α

q q−1

α

+ (∫ |(1 + t) − (1 − t) |

1− q1

1− q1

0

1

q q 󵄨󵄨 1+t 1 − t 󵄨󵄨󵄨󵄨 󵄨 (∫ 󵄨󵄨󵄨f 󸀠 ( b+ a)󵄨󵄨 dt) ] 󵄨󵄨 󵄨󵄨 2 2

dt)

0

1− q1

1

qα qα b−a ≤ α+2 [(∫[(1 + t) q−1 − (1 − t) q−1 ]dt) 2

0

1

+ (∫[(1 + t)

qα q−1

qα q−1

− (1 − t)

1

(∫ |f 󸀠 (a)|

1− q1

1

(∫ |f 󸀠 (b)|

]dt)

q(1+t) 2

|f 󸀠 (b)|

q(1+t) 2

|f 󸀠 (a)|

q(1−t) 2

0

qα b − a 2(q − 1) = α+2 [ (2 q−1 − 1)] qα + q − 1 2

1 2

1 2

+ 2 |f (a)| |f (b)| ( 󸀠

q ln | ff 󸀠 (b) | (a)

q

󸀠 (a) q ln | ff 󸀠 (b) |

1 2

1 2

󸀠

󸀠

1 q

q

(a) q ln | ff 󸀠 (b) | 󸀠

1 q

)

1 q

) ]

1− q1

qα

󸀠

󸀠

[2 |f (a)| |f (b)| (

q

b − a (q − 1)(2 q−1 − 1) = α+1 [ ] qα + q − 1 2 (a) 2 | ff 󸀠 (b) | −1

1 q

dt)

1 q

dt) ]

(a) 2 | ff 󸀠 (b) | −1 󸀠

| ff 󸀠 (b) |2 − 1 (a) 󸀠

󸀠

1− q1

q(1−t) 2

0

0

× [(

q q 󵄨󵄨 1 − t 󵄨󵄨󵄨󵄨 󵄨󵄨 󸀠 1 + t a+ b)󵄨󵄨 dt) (∫ 󵄨󵄨f ( 󵄨󵄨 󵄨󵄨 2 2

1

0

1 q

1

1

1 q

󸀠

) +(

q

| ff 󸀠 (b) |2 − 1 (a) 󸀠 q ln | ff 󸀠 (b) | (a)

1 q

1

1

) ]|f 󸀠 (a)| 2 |f 󸀠 (b)| 2 .

The proof is completed. Remark 107. If we take α = 1 in Theorem 106, then the equality (4.60) becomes the following inequality: b 󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨 f (a) + f (b) 1 󵄨󵄨 󵄨󵄨 − f (x)dx ∫ 󵄨󵄨 ≤ Kr , 󵄨󵄨 󵄨󵄨 2 b−a 󵄨󵄨 󵄨 a

where q

b − a (q − 1)(2 q−1 − 1) [ ] Kr : = 2+ 1 2q − 1 2 q

1− q1

1

1

3|f 󸀠 (a)|qr + |f 󸀠 (b)|qr q 3|f 󸀠 (b)|qr + |f 󸀠 (a)|qr q × [( ) +( ) ], 4 4

for r > 0,

114 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals q

b − a (q − 1)(2 q−1 − 1) Kr : = [ ] 4 2q − 1 1

1

× |f 󸀠 (a)| 2 |f 󸀠 (b)| 2 , and k =

1− q1

1

q

q

1

|k| 2 − 1 q 1 − |k|− 2 q [( ) +( ) ] q ln |k| q ln |k|

for r = 0,

f (a) . f 󸀠 (b) 󸀠

4.2.2 Applications 4.2.2.1 Applications to quadrature formulas In this section, we point out some specific inequalities (see Proposition 4.1–4.6, [100]). Proposition 108. Under the assumptions Theorem 94 with λ = 0 in Theorem 94, then we get the following inequality: 󵄨󵄨 Γ(α + 1) f (a) + f (b) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 (Jaα+ f (b) + Jbα− f (a)) − − f( )󵄨󵄨 ≤ Kr , 󵄨󵄨 α 2 2 󵄨󵄨 󵄨󵄨 (b − a)

where

1

Kr := 2 r −1 (b − a)2 {

|f 󸀠󸀠 (a)| + |f 󸀠󸀠 (b)| r r 1 [ − − B( + 1, α + 2)] α+1 r + 1 αr + 2r + 1 r 1

|f 󸀠󸀠 (a)| r r 1 r + ( − )[1 − ( ) ] 2 r + 1 2r + 1 2 +1

1

|f 󸀠󸀠 (b)| 1 r 1 r + [B 1 (2, + 1) + ( ) ]} for 0 < r ≤ 1, 2 2 r 2r + 1 2 󸀠󸀠 󸀠󸀠 |f (a)| + |f (b)| r r 1 Kr := (b − a)2 { [ − − B( + 1, α + 2)] α+1 r + 1 αr + 2r + 1 r 1

+2

|f 󸀠󸀠 (a)| r r 1 r + ( − )[1 − ( ) ] 2 r + 1 2r + 1 2 +1

1

1 r 1 r |f 󸀠󸀠 (b)| [B 1 (2, + 1) + ( ) ]} + 2 2 r 2r + 1 2 Kr := (b − a)2 |f 󸀠󸀠 (b)|{ 1

1

+2

for r > 1,

∞ 1 |k| − 1 (ln |k|)i−1 ∞ (ln |k|)i−1 [ − |k| ∑(−1)i−1 −∑ ] α + 1 ln |k| (α + 2)i (α + 2)i i=1 i=1

|k| 2 ∞ (− 2 ln |k|) + ∑ 4 i=1 (2)i for r = 0, and k =

i−1

1

∞ 1 |k| − |k| 2 (ln |k|)i−1 + [ − |k| ∑(−1)i−1 ]} 2 ln |k| (2)i i=1

f 󸀠󸀠 (a) . f 󸀠󸀠 (b)

Proposition 109 (Trapezoid inequality). Under the assumptions Theorem 94 with λ = 1 in Theorem 94, then we get the following inequality, 󵄨󵄨󵄨 Γ(α + 1) α f (a) + f (b) 󵄨󵄨󵄨󵄨 α 󵄨󵄨 󵄨󵄨 ≤ Kr , 󵄨󵄨 2(b − a)α (Ja+ f (b) + Jb− f (a)) − 2 󵄨 󵄨󵄨

4.2 Inequalities via r-convex functions | 115

where 1

2 r −2 (b − a)2 (|f 󸀠󸀠 (a)| + |f 󸀠󸀠 (b)|) r r 1 Kr := [ − − B( + 1, α + 2)] α+1 r + 1 αr + 2r + 1 r for 0 < r ≤ 1, Kr :=

Kr :=

r r 1 (b − a)2 (|f 󸀠󸀠 (a)| + |f 󸀠󸀠 (b)|) [ − − B( + 1, α + 2)] 2(α + 1) r + 1 αr + 2r + 1 r for r > 1, ∞ (b − a)2 |f 󸀠󸀠 (b)| |k| − 1 (ln |k|)i−1 ∞ (ln |k|)i−1 −∑ ] [ − |k| ∑(−1)i−1 2(α + 1) ln |k| (α + 2)i (α + 2)i i=1 i=1

for r = 0, and k =

f 󸀠󸀠 (a) . f 󸀠󸀠 (b)

Proposition 110 (Midpoint inequality). Under the assumptions Theorem 94 with λ = −1 in Theorem 94, then we get the following inequality: 󵄨󵄨 Γ(α + 1) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 (Jaα+ f (b) + Jbα− f (a)) − f ( )󵄨󵄨 ≤ Kr , 󵄨󵄨 α 󵄨󵄨 2 󵄨󵄨 2(b − a) where 1

Kr := 2 r −2 (b − a)2 {

|f 󸀠󸀠 (a)| + |f 󸀠󸀠 (b)| r r 1 [ − − B( + 1, α + 2)] α+1 r + 1 αr + 2r + 1 r 1

+ |f 󸀠󸀠 (a)|(

r 1 r r − )[1 − ( ) ] r + 1 2r + 1 2 +1

1

1 r 1 r + |f (b)|[B 1 (2, + 1) + ( ) ]} for 0 < r ≤ 1, 2 r 2r + 1 2 +2

󸀠󸀠

Kr :=

r r 1 (b − a)2 |f 󸀠󸀠 (a)| + |f 󸀠󸀠 (b)| { [ − − B( + 1, α + 2)] 2 α+1 r + 1 αr + 2r + 1 r 1

r r 1 r + |f (a)|( − )[1 − ( ) ] r + 1 2r + 1 2 +1

󸀠󸀠

1

+ |f 󸀠󸀠 (b)|[B 1 (2, 2

Kr :=

1 r 1 r + 1) + ( ) ]} for r > 1, r 2r + 1 2 +2

∞ (b − a)2 󸀠󸀠 1 |k| − 1 (ln |k|)i−1 ∞ (ln |k|)i−1 |f (b)|{ [ − |k| ∑(−1)i−1 −∑ ] 2 α + 1 ln |k| (α + 2)i (α + 2)i i=1 i=1 1

1 1 ∞ (− ln |k|) + |k| 2 ∑ 2 2 (2)i i=1 for r = 0, and k =

i−1

1

∞ 1 − λ |k| − |k| 2 (ln |k|)i−1 + [ − |k| ∑(−1)i−1 ]} 2 ln |k| (2)i i=1

f 󸀠󸀠 (a) . f 󸀠󸀠 (b)

116 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals Proposition 111. Under the assumptions Theorem 96 with λ = 0 in Theorem 96, then we get the following inequality:

where

󵄨󵄨 Γ(α + 1) a + b 󵄨󵄨󵄨󵄨 f (a) + f (b) 󵄨󵄨 (Jaα+ f (b) + Jbα− f (a)) − − f( )󵄨󵄨 ≤ Kr , 󵄨󵄨 α 󵄨󵄨 2 2 󵄨󵄨 (b − a) q

q

q−1 1 2(q − 1) 1 q−1 q − 1 Kr := 2 (b − a) [( ) (1 − )+( ) ] α+1 qα + 2q − 1 4 2q − 1 1 qr

2

1− q1

1

q r × [(|f (a)| + |f (b)| ) ] r+1

q

󸀠󸀠

1 Kr := 2 (b − a) [( ) α+1 1 q

q

󸀠󸀠

2

q q−1

for 0 < r ≤ 1, q

2(q − 1) 1 q−1 q − 1 (1 − )+( ) ] qα + 2q − 1 4 2q − 1 1

q r × [(|f (a)| + |f (b)| ) ] r+1

q

󸀠󸀠

󸀠󸀠

1− q1

q

for r > 1,

q

q

q−1 1 2(q − 1) 1 q−1 q − 1 Kr := 2 (b − a) |f (b)|[( ) (1 − )+( ) ] α+1 qα + 2q − 1 4 2q − 1 1 q

2

󸀠󸀠

1

|k|q − 1 q ×( ) q ln |k|

for r = 0, and k =

1− q1

f 󸀠󸀠 (a) . f 󸀠󸀠 (b)

Proposition 112. Under the assumptions Theorem 96 with λ = 1 in Theorem 96, then we get the following trapezoidal inequality: 󵄨󵄨 Γ(α + 1) f (a) + f (b) 󵄨󵄨󵄨󵄨 󵄨󵄨 α α (J f (b) + J f (a)) − + − 󵄨󵄨 󵄨󵄨 ≤ Kr , b 󵄨󵄨 2(b − a)α a 󵄨󵄨 2

where 1

Kr :=

1− q1

2(q − 1) 2 qr (b − a)2 [1 − ] α+1 qα + 2q − 1 for 0 < r ≤ 1, 1

−1

1− q1

2 q (b − a)2 2(q − 1) [1 − ] Kr := α+1 qα + 2q − 1 for r > 1, 1

−1

1

[(|f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q )

for r = 0, and k =

f 󸀠󸀠 (a) . f 󸀠󸀠 (b)

1

q r [(|f (a)| + |f (b)| ) ] r+1

2 q (b − a)2 |f 󸀠󸀠 (b)| 2(q − 1) Kr := [1 − ] α+1 qα + 2q − 1 −1

q r ] r+1

󸀠󸀠

1− q1

q

q

󸀠󸀠

1

|k|q − 1 q ( ) q ln |k|

Proposition 113. Under the assumptions Theorem 96 with λ = −1 in Theorem 96, then we get the following midpoint inequality: 󵄨󵄨 Γ(α + 1) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 (Jaα+ f (b) + Jbα− f (a)) − f ( )󵄨󵄨 ≤ Kr , 󵄨󵄨 α 󵄨󵄨 2 󵄨󵄨 2(b − a)

4.2 Inequalities via r-convex functions | 117

where Kr := 2

1 −1 qr

q

q

q−1 1 2(q − 1) 1 q−1 q − 1 (b − a) [( ) (1 − )+( ) ] α+1 qα + 2q − 1 2 2q − 1

2

1− q1

1

q r × [(|f (a)| + |f (b)| ) ] r+1

q

󸀠󸀠

Kr := 2

1 −1 q

󸀠󸀠

q

for 0 < r ≤ 1,

q

q

q−1 1 q−1 q − 1 1 2(q − 1) ) (1 − )+( ) ] (b − a) [( α+1 qα + 2q − 1 2 2q − 1

2

1− q1

1

q r × [(|f (a)| + |f (b)| ) ] r+1

q

󸀠󸀠

Kr := 2

1 −1 q

󸀠󸀠

q

for r > 1,

q

q

1− q1

q−1 1 2(q − 1) 1 q−1 q − 1 (b − a) |f (b)|[( ) (1 − )+( ) ] α+1 qα + 2q − 1 2 2q − 1

2

󸀠󸀠

1

|k|q − 1 q ×( ) q ln |k|

for r = 0, and k =

f 󸀠󸀠 (a) . f 󸀠󸀠 (b)

4.2.2.2 Applications to special means Clearly, Lp is monotonically nondecreasing over p ∈ ℝ with L−1 := L and L0 = I. In particular, H ≤ G ≤ L ≤ I ≤ A. Now, using the results of Section 4.2, some new inequalities are derived for the means already treated (see Proposition 5.1–5.3, [100]). Proposition 114. Let a, b ∈ ℝ, 0 < a < b and n ∈ ℕ, n > 2. Then, we have 󵄨󵄨 n 󵄨 n n n 󵄨󵄨2Ln (a, b) − A(a , b ) − A (a, b)󵄨󵄨󵄨 ≤ Kr , where 1

Kr := 2 r −2 n(n − 1)(b − a)2 [(an−2 + bn−2 )( 1

+a

n−2

r r 1 r − )(1 − ( ) ) ( r + 1 2r + 1 2 +1

1

n−2

+b Kr :=

1 r 1 r (B 1 (2, + 1) + ( ) )] for 0 < r ≤ 1, 2 r 2r + 1 2 +2

n(n − 1)(b − a)2 r r 1 [(an−2 + bn−2 )( − − B( + 1, 3)) 2 r + 1 3r + 1 r 1

+ an−2 (

r r 1 r − )(1 − ( ) ) r + 1 2r + 1 2 +1

1

n−2

+b Kr :=

r 1 r − − B( + 1, 3)) r + 1 3r + 1 r

1 r 1 r (B 1 (2, + 1) + ( ) )] for r > 1, 2 r 2r + 1 2 +2

∞ n(n − 1)bn−2 (b − a)2 k − 1 (ln k)i−1 ∞ (ln k)i−1 [ − k ∑(−1)i−1 −∑ 2 ln k (3)i (3)i i=1 i=1

118 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1 i−1 ∞ k − k2 (ln k)i−1 k 2 ∞ (− 2 ln k) + − k ∑(−1)i−1 ] ∑ 2 i=1 (2)i ln k (2)i i=1 1

1

+

n−2

a for r = 0, and k = ( ) b

.

Proof. Let α = 1 in Proposition 108 and apply it to r-convex mapping f (x) = x n , x ∈ ℝ. The proof is completed. Proposition 115. Let a, b ∈ ℝ, 0 < a < b. Then, for all q > 1, we have 󵄨󵄨 −1 󵄨 −1 󵄨󵄨L (a, b) − H (a, b)󵄨󵄨󵄨 ≤ Kr , where 1

Kr := 2 r −2 (b − a)2 ( Kr := Kr :=

(b − a)2 ( a13 + 2

1 1 r r 1 + )[ − − B( + 1, 3)] r a3 b3 r + 1 3r + 1 1 ) b3

[

for 0 < r ≤ 1,

r r 1 − − B( + 1, 3)] for r > 1, r + 1 3r + 1 r

∞ ∞ i−1 (b − a)2 k − 1 (ln k)i−1 i−1 (ln k) [ − k (−1) − ] ∑ ∑ ln k (3)i (3)i 2b3 i=1 i=1 3

b for r = 0, and k = ( ) . a Proof. Let α = 1 in Proposition 109 and apply it to r-convex mapping f (x) = [a, b]. The proof is completed. Proposition 116. Let a, b ∈ ℝ, 0 < a < b. Then, for all q > 1, we have 󵄨󵄨 −1 󵄨 −1 −1 󵄨󵄨2L (a, b) − H (a, b) − A (a, b)󵄨󵄨󵄨 ≤ Kr , where 1

q

Kr := 2 qr (b − a)2 [1 − −1

q

2(q − 1) 1 q−1 q − 1 +( ) ] 3q − 1 2 2q − 1 q

1

q 2 2 r × [(( 3 ) + ( 3 ) ) ] r+1 a b

Kr := 2

1 −1 q

q

2(q − 1) 1 q−1 q − 1 (b − a) [1 − +( ) ] 3q − 1 2 2q − 1 q

1

for r > 1, q

2 q (b − a)2 2(q − 1) 1 q−1 q − 1 [1 − +( ) ] 3 3q − 1 2 2q − 1 b 1

kq − 1 q ×( ) q ln k

1− q1

1

q 2 2 r × [(( 3 ) + ( 3 ) ) ] r+1 a b

Kr :=

for 0 < r ≤ 1,

2

q

1− q1

3

b for r = 0, and k = ( ) . a

1− q1

1 , x

x ∈

4.3 Inequalities via s-convex functions | 119

Proof. Let α = 1 in Proposition 111 and apply it to r-convex mapping f (x) = x1 , x ∈ [a, b]. The proof is completed. Now, we give some applications to special means of real numbers (see Proposition 3.1–3.3, [147]). Proposition 117. Let a, b ∈ R, a < b, 0 ∉ [a, b], and n ∈ Z, |n| ≥ 2. Then, 1 5 󵄨󵄨 󵄨 b−a n n n |n|(√ + √ )A(|a|n−1 , |b|n−1 ). 󵄨󵄨A(a , b ) − Ln (a, b)󵄨󵄨󵄨 ≤ 4 6 6 Proof. Applying Corollary 101 for f (x) = xn and q = 2, one can obtain the result immediately. The proof is completed. Proposition 118. Let a, b ∈ R, a < b, 0 ∉ [a, b]. Then, 󵄨󵄨 −1 󵄨 b − a √1 √5 −1 ( + )H −1 (|a|2 , |b|2 ). 󵄨󵄨H (a, b) − L (a, b)󵄨󵄨󵄨 ≤ 4 6 6 Proof. The assertion follows from Corollary 101 applied for f (x) = proof is completed.

1 x

and q = 2. The

Proposition 119. Let a, b ∈ R \ {0}, a < b, a−1 > b−1 , 0 ∉ [a, b], and n ∈ Z, |n| ≥ 2. Then, a−1 − b−1 1 5 󵄨󵄨 −1 n n n −1 −1 󵄨 |n|(√ + √ )H −1 (|a|n−1 , |b|n−1 ), 󵄨󵄨H (a , b ) − Ln (b , a )󵄨󵄨󵄨 ≤ 4 6 6 and 1 5 a−1 − b−1 󵄨󵄨 −1 −1 −1 󵄨 (√ + √ )A(|a|2 , |b|2 ). 󵄨󵄨A(b, a) − L (b , a )󵄨󵄨󵄨 ≤ 4 6 6 Proof. Making the substitutions a → b−1 , b → a−1 in the results of Proposition 117 and Proposition 118, one can obtain the Proposition 119 immediately. The proof is completed.

4.3 Inequalities via s-convex functions The results in this section are due to [232, 95, 51].

4.3.1 Main results An important Hermite-Hadamard inequality involving Riemann-Liouville fractional integrals (with α ≥ 1) can be represented as follows.

120 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals Theorem 120. Let α ≥ 1 and f : [a, b] → ℝ be a positive function with 0 ≤ a < b and f ∈ L1 [a, b]. If f is a s-convex function on [a, b], then the following inequality for fractional integrals holds: 2s−1 f (

a+b Γ(α + 1) α 3[f (a) + f (b)] )≤ [J + f (b) + Jbα− f (a)] ≤ . 2 2(b − a)α a 2(1 + s)

Proof. Since f is a s-convex function on [a, b], we have for x, y ∈ [a, b] with λ = f(

(4.61) 1 2

x+y 1 1 ) ≤ s f (x) + s f (y), 2 2 2

i. e., with x = ta + (1 − t)b, y = (1 − t)a + tb, 2s f (

a+b ) ≤ f (ta + (1 − t)b) + f ((1 − t)a + tb). 2

(4.62)

Multiplying both sides of (4.62) by t α−1 , then integrating the resulting inequality with respect to t over [0, 1], we obtain 1

1

0

0

2s a + b f( ) ≤ ∫ t α−1 f (ta + (1 − t)b)dt + ∫ t α−1 f ((1 − t)a + tb)dt α 2 Γ(α) [J α+ f (b) + Jbα− f (a)], = (b − a)α a i. e., 2s−1 f (

Γ(α + 1) α a+b [J + f (b) + Jbα− f (a)] )≤ 2 2(b − a)α a

and the first inequality in (4.61) is proved. For the proof the second inequality in (4.61), we first note that if f is a s-convex function, then, for t ∈ [0, 1], it yields f (ta + (1 − t)b) ≤ t s f (a) + (1 − t)s f (b) and f ((1 − t)a + tb) ≤ (1 − t)s f (a) + t s f (b). By adding these inequalities, we have f (ta + (1 − t)b) + f ((1 − t)a + tb) ≤ t s f (a) + (1 − t)s f (b) + (1 − t)s f (a) + t s f (b) = [t s + (1 − t)s ][f (a) + f (b)].

(4.63)

4.3 Inequalities via s-convex functions | 121

Then multiplying both sides of (4.63) by t α−1 and integrating the resulting inequality with respect to t over [0, 1], we obtain 1

1

0

0

∫ t α−1 f (ta + (1 − t)b)dt + ∫ t α−1 f ((1 − t)a + tb)dt Γ(α) [J α+ f (b) + Jbα− f (a)] = (b − a)α a 1

≤ [f (a) + f (b)] ∫ t α−1 [t s + (1 − t)s ]dt 0

1

≤ [f (a) + f (b)](∫ t

α+s−1

1

dt + ∫ t α−1 (1 − t)s dt)

0

0 1 2

≤ [f (a) + f (b)](

1

1 + ∫ t α−1 (1 − t)s dt + ∫ t α−1 (1 − t)s dt) α+s 0

1 2

1 2

≤ [f (a) + f (b)](

0

≤ [f (a) + f (b)][ ≤

1

1 + ∫(1 − t)α−1 (1 − t)s dt + ∫ t α−1 t s dt) α+s 1 2

1 2 1 + (1 − α+s )] α+s α+s 2

3[f (a) + f (b)] , α+s

that is, 3[f (a) + f (b)] Γ(α + 1) α [J + f (b) + Jbα− f (a)] ≤ . 2(b − a)α a 2(1 + s) The proof is completed. Theorem 121. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with a < b, such that f ∈ L1 [a, b]. If f 󸀠 is s-convex on [a, b], for some fixed s ∈ (0, 1], then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨 − [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 1 b−a ≤ (1 − α+s )(|f 󸀠 (a)| + |f 󸀠 (b)|). α+s+1 2 Proof. Using Lemma 42 and the s-convexity of f , we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨 − [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨

122 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

b−a 󵄨 󵄨 ≤ ∫ |(1 − t)α − t α |󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 0

1

≤

b−a ∫ |(1 − t)α − t α |[t s |f 󸀠 (a)| + (1 − t)s |f 󸀠 (b)|]dt 2 0

1 2

b−a = {∫[(1 − t)α − t α ][t s |f 󸀠 (a)| + (1 − t)s |f 󸀠 (b)|]dt 2 1

0

+ ∫[t α − (1 − t)α ][t s |f 󸀠 (a)| + (1 − t)s |f 󸀠 (b)|]dt} 1 2

:=

b−a (K1 + K2 ). 2

(4.64)

Calculating K1 and K2 , we have 1 2

1 2

K1 = |f 󸀠 (a)|(∫(1 − t)α t s ds − ∫ t α+s ds) 0

0 1 2

α+s

+ |f (b)|(∫(1 − t) 󸀠

1 2

ds − ∫ t α (1 − t)s ds)

0

0

1 2

1 2

≤ |f 󸀠 (a)|(∫(1 − t)α+s ds − ∫ t α+s ds) 0

0 1 2

α+s

+ |f (b)|(∫(1 − t) 󸀠

1 2

ds − ∫ t α+s ds)

0

0

|f 󸀠 (a)| + |f 󸀠 (b)| 1 = (1 − α+s ), α+s+1 2 and 1

1

K2 = |f 󸀠 (a)|[∫ t α+s ds − ∫(1 − t)α t s ds] 1 2

1 2

1

α

s

1

+ |f (b)|[∫ t (1 − t) ds − ∫(1 − t)α+s ds] 󸀠

1 2

1

≤ |f (a)|[∫ t 󸀠

1 2

α+s

1 2

1

ds − ∫(1 − t)α+s ds] 1 2

(4.65)

4.3 Inequalities via s-convex functions | 123 1

+ |f (b)|[∫ t 󸀠

α+s

1

ds − ∫(1 − t)α+s ds] 1 2

1 2

|f (a)| + |f (b)| 1 (1 − α+s ). α+s+1 2 󸀠

=

󸀠

(4.66)

Thus if we use (4.65) and (4.66) in (4.64), we obtain 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨 [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 − 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) b−a 1 ≤ (1 − α+s )(|f 󸀠 (a)| + |f 󸀠 (b)|). α+s+1 2 The proof is completed. The second theorem gives a new upper bound for the left-Hadamard inequality for s-convex mappings. Theorem 122. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with a < b, such that f ∈ L1 [a, b]. If |f 󸀠 |q (q > 1) is s-convex on [a, b], for some fixed s ∈ (0, 1], then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨 − [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 1

1

1

p q 1 1 p 1 b−a ) )( ) (1 − αp ) ( ≤( s+1 2 αp + 1 2 (s + 1)2

1

1

× [(|f 󸀠 (a)|q + (2s+1 − 1)|f 󸀠 (b)|q ) q + ((2s+1 − 1)|f 󸀠 (a)|q + |f 󸀠 (b)|q ) q ], where

1 p

(4.67)

= 1 − q1 .

Proof. Using Lemma 42, Hölder inequality and the s-convexity of |f 󸀠 |q (q > 1), we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨 − [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 1 2

b−a 󵄨 󵄨 {∫[(1 − t)α − t α ]󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt ≤ 2 1

0

󵄨 󵄨 + ∫[t α − (1 − t)α ]󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt} 1 2

1 2

1 p

1 2

b−a p 󵄨 󵄨q {(∫[(1 − t)α − t α ] dt) (∫ 󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) ≤ 2 0

0

1 q

124 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1 p

1

1 q

1

p 󵄨q 󵄨 + (∫[t α − (1 − t)α ] dt) (∫ 󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) } 1 2

1 2

1 2

1 2

1 p

1 q

b−a {(∫[(1 − t)αp − t αp ]dt) (∫(t s |f 󸀠 (a)|q + (1 − t)s |f 󸀠 (b)|q )dt) } ≤ 2 0

0

1 p

1

1

+ (∫[t αp − (1 − t)αp ]dt) (∫(t s |f 󸀠 (a)|q + (1 − t)s |f 󸀠 (b)|q )dt)

1 q

1 2

1 2

1 2

1

p b−a 1 1 = ( (1 − αp )) {(∫(t s |f 󸀠 (a)|q + (1 − t)s |f 󸀠 (b)|q )dt) 2 αp + 1 2

0

1

s

q

s

1 q

1 q

q

+ (∫(t |f (a)| + (1 − t) |f (b)| )dt) }. 󸀠

󸀠

1 2

We note that 1 2

∫ t s |f 󸀠 (a)|q + (1 − t)s |f 󸀠 (b)|q dt 0

=

1 1 1 |f 󸀠 (a)|q + (1 − s+1 )|f 󸀠 (b)|q , (s+1) s+1 2 (s + 1)2

(4.68)

and 1

∫(t s |f 󸀠 (a)|q + (1 − t)s |f 󸀠 (b)|q )dt 1 2

=

1 1 1 (1 − s+1 )|f 󸀠 (a)|q + |f 󸀠 (b)|q . s+1 2 (s + 1)2(s+1)

(4.69)

A combination of (4.68) and (4.69), we get 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨󵄨 α − [J + f (b) + Jb− f (a)]󵄨 󵄨󵄨 a 󵄨󵄨 󵄨󵄨 2 2(b − a)α 󵄨 1

≤(

1

1

p q 1 1 p 1 b−a )( ) (1 − αp ) ( ) s+1 2 αp + 1 2 (s + 1)2 1

1

× [(|f 󸀠 (a)|q + (2s+1 − 1)|f 󸀠 (b)|q ) q + ((2s+1 − 1)|f 󸀠 (a)|q + |f 󸀠 (b)|q ) q ]. The proof is completed.

(4.70)

4.3 Inequalities via s-convex functions | 125

It is not difficult to see that Theorem 122 can be extended to the following result: Corollary 123. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with a < b, such that f ∈ L1 [a, b]. If |f 󸀠 |q , (q > 1) is s-convex on [a, b], for some fixed s ∈ (0, 1], then the following inequality for fractional integrals holds: 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) α 󵄨 󵄨󵄨 − [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) 1

1

1

p q b−a 1 1 p 1 ≤( )( ) (1 − αp ) ( ) s+1 2 αp + 1 2 (s + 1)2 1

× (1 + (2s+1 − 1) q )(|f 󸀠 (a)| + |f 󸀠 (b)|), where

1 p

= 1 − q1 .

Proof. We consider the inequality (4.67), and we let a1 = |f 󸀠 (a)|q , b1 = (2s+1 − 1)|f 󸀠 (b)|q , a2 = (2s+1 − 1)|f 󸀠 (a)|q , b2 = |f 󸀠 (b)|q . Here, 0 < q1 < 1 for q > 1. Using the fact that n

n

n

i=1

i=1

i=1

∑(ai + bi )r ≤ ∑ ari + ∑ bri for 0 < r < 1, a1 , a2 , . . . , an > 0 and b1 , b2 , . . . , bn > 0, we obtain 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨 − [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 1

1

1

p q 1 b−a 1 p 1 )( ) (1 − αp ) ( ) ≤( s+1 2 αp + 1 2 (s + 1)2

1

1

× [(|f 󸀠 (a)|q + (2s+1 − 1)|f 󸀠 (b)|q ) q + ((2s+1 − 1)|f 󸀠 (a)|q + |f 󸀠 (b)|q ) q ] 1

1

1

p q b−a 1 1 p 1 ) ≤( )( ) (1 − αp ) ( 2 αp + 1 2 (s + 1)2s+1 1

× (1 + (2s+1 − 1) q )(|f 󸀠 (a)| + |f 󸀠 (b)|). The proof is completed. The following theorem is the third main result in this section. Theorem 124. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with a < b, such that f ∈ L1 [a, b]. If |f 󸀠 |q , (q > 1) is s-convex on [a, b], for some fixed s ∈ (0, 1], then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨 − [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 1− q1

1 ≤ (b − a)( ) α+1

1− q1

1 (1 − α ) 2 1

× (|f 󸀠 (a)|q + |f 󸀠 (b)|q ) q .

1

1

q 1 1 q ( ) (1 − α+s ) α+s+1 2

(4.71)

126 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals Proof. Using Lemma 42 and the power mean inequality, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨󵄨 α − [J + J f (a)] f (b) + − 󵄨󵄨 󵄨󵄨 b 󵄨󵄨 󵄨󵄨 2 2(b − a)α a 1

≤

b−a 󵄨 󵄨 ∫ |(1 − t)α − t α |󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 0

1 2

b−a 󵄨 󵄨 = {∫[(1 − t)α − t α ]󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 0

1

󵄨 󵄨 + ∫[t α − (1 − t)α ]󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt} 1 2

1 2

1− q1

b−a ≤ {(∫[(1 − t)α − t α ]dt) 2 0

1

α

α

1− q1

+ (∫[t − (1 − t) ]dt) 1 2

󵄨q

󵄨 (∫[(1 − t)α − t α ]󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt)

1 q

0

1

1 q

󵄨 󵄨q (∫[t − (1 − t) ]󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) } α

α

1 2

1− q1

b−a 1 ≤ ( ) 2 α+1 1

1 2

1− q1

1 (1 − α ) 2

1 2

󵄨 󵄨q {(∫[(1 − t)α − t α ]󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt)

1 q

0

1 q

󵄨 󵄨q + (∫[t − (1 − t) ]󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) }. α

α

1 2

Because the s-convex of |f 󸀠 |q , (q > 1), we have 1 2

󵄨 󵄨q ∫[(1 − t)α − t α ]󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt 0

1 2

α s

1 2

≤ ∫(1 − t) t dt|f (a)| + ∫(1 − t)α+s dt|f 󸀠 (b)|q 󸀠

q

0

0 1 2

1 2

0

0

− ∫ t α+s dt|f 󸀠 (a)|q − ∫ t α (1 − t)s dt|f 󸀠 (b)|q ≤

1 1 (1 − α+s )(|f 󸀠 (a)|q + |f 󸀠 (b)|q ), α+s+1 2

(4.72)

4.3 Inequalities via s-convex functions | 127

and 1

󵄨q 󵄨 ∫[t α − (1 − t)α ]󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt 1 2

1

1

≤ ∫ t α+s dt|f 󸀠 (a)|q + ∫ t α (1 − t)s dt|f 󸀠 (b)|q 1 2

1 2

1

α s

q

1

− ∫(1 − t) t dt|f (a)| − ∫(1 − t)α+s dt|f 󸀠 (b)|q 󸀠

1 2

1 2

≤

1 1 (1 − α+s )(|f 󸀠 (a)|q + |f 󸀠 (b)|q ). α+s+1 2

(4.73)

By combining (4.72) and (4.73), we get 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨 − [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) 1− q1

1 ≤ (b − a)( ) α+1

1− q1

1 (1 − α ) 2 1

(|f 󸀠 (a)|q + |f 󸀠 (b)|q ) q .

1

1

q 1 1 q ( ) (1 − α+s ) α+s+1 2

The proof is completed. Corollary 125. Let f be as in Theorem 124, then the following inequality holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨 − [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) 1− q1

1 ) ≤ (b − a)( α+1

(|f 󸀠 (a)| + |f 󸀠 (b)|).

1− q1

1 (1 − α ) 2

1

1

q 1 1 q ) (1 − α+s ) ( α+s+1 2

Theorem 126. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with a < b. If |f 󸀠 | ∈ L[a, b] and |f 󸀠 | is an s-convex function, then, for some 0 < α ≤ 1 and 0 < λ ≤ 1, the following inequality for fractional integrals holds: 󵄨󵄨 f (a)(1 − α(1 − λ)) + f (b)(1 + α(1 − λ)) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − (RL Jbα− f (a) + RL Jaα+ f (b))󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 b−a 󸀠 󸀠 ≤ (φ(α, λ, s)|f (b)| + ϕ(α, λ, s)|f (a)|), 2 where φ(α, λ, s) =

1 α(1 − λ) Γ(s + 1)Γ(α + 1) + − , α+s+1 s+1 Γ(α + s + 2)

128 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals and ϕ(α, λ, s) =

1 Γ(s + 1)Γ(α + 1) α(1 − λ) + − . Γ(α + s + 2) s+1 α+s+1

Proof. Using Lemma 62 via f 󸀠 ∈ L[a, b] and |f 󸀠 | is an s-convex function, we have 󵄨󵄨 f (a)(1 − α(1 − λ)) + f (b)(1 + α(1 − λ)) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − (RL Jbα− f (a) + RL Jaα+ f (b))󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) 1

b−a 󵄨 󵄨 ≤ ∫(t α + α(1 − λ) − (1 − t)α )󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt 2 0

1

≤

b−a ∫(t α + α(1 − λ) − (1 − t)α )(t s |f 󸀠 (b)| + (1 − t)s |f 󸀠 (a)|)dt 2 0

b−a = (φ(α, λ, s)|f 󸀠 (b)| + ϕ(α, λ, s)|f 󸀠 (a)|). 2 The proof is completed. Theorem 127. Let f : [a, b] → ℝ be a differentiable mapping. If |f 󸀠 |q ∈ L[a, b] and |f 󸀠 |q (q ≥ 1) is an s-convex function. Then, for some 0 < α ≤ 1, 0 < λ ≤ 1, the following inequality for fractional integrals holds: 󵄨󵄨 f (a)(1 − α(1 − λ)) + f (b)(1 + α(1 − λ)) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 (RL Jbα− f (a) + RL Jaα+ f (b))󵄨󵄨󵄨 − 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 1

|f 󸀠 (a)|q + |f 󸀠 (b)|q q b−a ψ(α, λ, p)( ) , ≤ 2 1 + sq where

1 p

+

1 q

= 1 and 1

p 1 ) . ψ(α, λ, p) = α(1 − λ) + 2( αp + 1

Proof. Using Lemma 62, Hölder’s inequality via |f 󸀠 |q ∈ L[a, b] and |f 󸀠 |q is an s-convex function, we have 󵄨󵄨 f (a)(1 − α(1 − λ)) + f (b)(1 + α(1 − λ)) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − ( J − f (a) + RL Ja+ f (b))󵄨 󵄨󵄨 RL b 󵄨󵄨 󵄨󵄨 2 2(b − a)α 󵄨 1

1

b−a 󵄨 󵄨 󵄨 󵄨 ≤ (∫(t α + α(1 − λ))󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt + ∫(1 − t)α 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt) 2 0

1

1 p

0

1

b−a p 󵄨 󵄨q [(∫(t α + α(1 − λ)) dt) (∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨 dt) ≤ 2 0

0

1 q

4.3 Inequalities via s-convex functions | 129 1 p

1

1 q

1

󵄨q 󵄨 + (∫(1 − t)αp dt) (∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨 dt) ] 0

1 p

1

0

1 p

1

1 p

1

b−a [(∫ t αp dt) + (∫ αp (1 − λ)p dt) + (∫(1 − t)αp dt) ] ≤ 2 0

1

0

qs

0

1 q

q

1

qs

1 q

q

× [(∫(t |f (b)| dt)) + (∫(1 − t) |f (a)| dt) ] 󸀠

0

0

=

1

|f (b)| + |f (a)|q q b−a ψ(α, λ, p)( ) . 2 1 + qs 󸀠

q

󸀠

󸀠

Here we use the following fact due to Minkowski’s inequality for p ≥ 1: 1

1 p

p

α

1

αp

1 p

1

p

(∫(t + α(1 − λ)) dt) ≤ (∫ t dt) + (∫(α(1 − λ)) dt) 0

0

1 p

0

and 1

󵄨q

1 q

1

1 q

1

1 q

󵄨 (∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨 dt) ≤ (∫ t qs |f 󸀠 (b)|q dt) + (∫(1 − t)qs |f 󸀠 (a)|q dt) . 0

0

0

The proof is completed. Theorem 128. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with a < b. If |f 󸀠 | ∈ L[a, b] and |f 󸀠 | is an s-convex function, then, for any 0 < α ≤ 1 and a < x < b, the following inequality for fractional integrals holds: 󵄨󵄨 (x − a)[(a + x)f (x) − xf (a)] + (b − x)[(x + b)f (x) − xf (b)] 󵄨󵄨 󵄨󵄨 󵄨󵄨 (b − a)2 󵄨󵄨 Γ(α + 1) a b 󵄨󵄨 α α ( − RL Jx− f (a) + RL Jx+ f (b))󵄨󵄨󵄨 α−1 α−1 2 (b − x) (b − a) (x − a) 󵄨 a x aΓ(α + 1)Γ(s + 1) x (x − a)2 󸀠 [( + )|f (x)| + ( + )|f 󸀠 (a)|] ≤ Γ(α + s + 2) s+1 (b − a)2 α + s + 1 s + 1 +

(b − x)2 bΓ(α + 1)Γ(s + 1) x b x [( + )|f 󸀠 (b)| + ( + )|f 󸀠 (x)|]. Γ(α + s + 2) s+1 α+s+1 s+1 (b − a)2

Proof. Using Lemma 64 via f 󸀠 ∈ L[a, b] and |f 󸀠 | is an s-convex function, we have 󵄨󵄨 (x − a)[(a + x)f (x) − xf (a)] + (b − x)[(x + b)f (x) − xf (b)] 󵄨󵄨 󵄨󵄨 󵄨󵄨 (b − a)2 󵄨󵄨 Γ(α + 1) a b 󵄨󵄨 α α − ( RL Jx− f (a) + RL Jx+ f (b))󵄨󵄨󵄨 α−1 α−1 2 (b − x) (b − a) (x − a) 󵄨

130 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

(x − a)2 󵄨 󵄨 ≤ ∫(at α + x)󵄨󵄨󵄨f 󸀠 (tx + (1 − t)a)󵄨󵄨󵄨dt (b − a)2 0

+

1

2

≤

1

(b − x)2 󵄨 󵄨 ∫(b(1 − t)α + x)󵄨󵄨󵄨f 󸀠 (tb + (1 − t)x)󵄨󵄨󵄨dt (b − a)2 0

(x − a) ∫(at α + x)(t s |f 󸀠 (x)| + (1 − t)s |f 󸀠 (a)|)dt (b − a)2 0

+

1

(b − x)2 ∫(b(1 − t)α + x)(t s |f 󸀠 (b)| + (1 − t)s |f 󸀠 (x)|)dt (b − a)2 0

(x − a)2 a x aΓ(α + 1)Γ(s + 1) x = [( + )|f 󸀠 (x)| + ( + )|f 󸀠 (a)|] 2 α+s+1 s+1 Γ(α + s + 2) s+1 (b − a) +

x b x (b − x)2 bΓ(α + 1)Γ(s + 1) [( + )|f 󸀠 (b)| + ( + )|f 󸀠 (x)|]. 2 Γ(α + s + 2) s + 1 α + s + 1 s + 1 (b − a)

The proof is completed. Theorem 129. Let f : [a, b] → ℝ be a differentiable mapping. If |f 󸀠 |q ∈ L[a, b] and |f 󸀠 |q (q > 1) is an s-convex function. Then, for any 0 < α ≤ 1, a < x < b, the following inequality for fractional integrals holds: 󵄨󵄨 (x − a)[(a + x)f (x) − xf (a)] + (b − x)[(x + b)f (x) − xf (b)] 󵄨󵄨 󵄨󵄨 󵄨󵄨 (b − a)2 󵄨󵄨 Γ(α + 1) a b 󵄨󵄨 α α − ( J J − f (a) + + f (b))󵄨 RL RL x x 󵄨󵄨 (b − x)α−1 (b − a)2 (x − a)α−1 󵄨 1

≤

1

p |f 󸀠 (a)|q + |f 󸀠 (x)|q q (x − a)2 2p−1 ap + 2p−1 xp ) ( ) ( 2 s+1 (b − a) pα + 1 1

1

p (b − x)2 2p−1 bp |f 󸀠 (b)|q + |f 󸀠 (x)|q q + ( + 2p−1 x p ) ( ) , 2 s+1 (b − a) pα + 1

where

1 p

+

1 q

= 1.

Proof. Using Lemmas 64 and 37, Hölder’s inequality via |f 󸀠 |q ∈ L[a, b] and |f 󸀠 |q is an s-convex function, we have 󵄨󵄨 (x − a)[(a + x)f (x) − xf (a)] + (b − x)[(x + b)f (x) − xf (b)] 󵄨󵄨 󵄨󵄨 󵄨󵄨 (b − a)2 󵄨󵄨 Γ(α + 1) a b 󵄨󵄨 α α ( J J − − f (a) + + f (b))󵄨 RL RL x x 󵄨󵄨 (b − x)α−1 (b − a)2 (x − a)α−1 󵄨 1

(x − a)2 󵄨 󵄨 ≤ ∫(at α + x)󵄨󵄨󵄨f 󸀠 (tx + (1 − t)a)󵄨󵄨󵄨dt (b − a)2 0

4.3 Inequalities via s-convex functions | 131 1

(b − x)2 󵄨 󵄨 + ∫(b(1 − t)α + x)󵄨󵄨󵄨f 󸀠 (tb + (1 − t)x)󵄨󵄨󵄨dt (b − a)2 0

1 p

1

1

(x − a)2 p 󵄨q 󵄨 ≤ (∫(at α + x) dt) (∫ 󵄨󵄨󵄨f 󸀠 (tx + (1 − t)a)󵄨󵄨󵄨 dt) (b − a)2 0

1 q

0

1 p

1

1

(b − x)2 p 󵄨q 󵄨 (∫(b(1 − t)α + x) dt) (∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)x)󵄨󵄨󵄨 dt) + (b − a)2 0

1 q

0

1 p

1

1

(x − a)2 󵄨 󵄨q (∫ 2p−1 (ap t αp + xp )dt) (∫ 󵄨󵄨󵄨f 󸀠 (tx + (1 − t)a)󵄨󵄨󵄨 dt) ≤ (b − a)2 0

1 q

0

1 p

1

1

(b − x)2 󵄨 󵄨q + (∫ 2p−1 (bp (1 − t)αp + xp )dt) (∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)x)󵄨󵄨󵄨 dt) 2 (b − a) 0

0

1 p

1

(x − a)2 2p−1 ap |f 󸀠 (a)|q + |f 󸀠 (x)|q q p−1 p = ( ( + 2 x ) ) s+1 (b − a)2 pα + 1 1

1

p (b − x)2 2p−1 bp |f 󸀠 (b)|q + |f 󸀠 (x)|q q p−1 p + ( + 2 x ) ( ) , s+1 (b − a)2 pα + 1

where we use the following fact due to [178, Theorem 2], 1

|f 󸀠 (a)|q + |f 󸀠 (x)|q 󵄨 󵄨q , ∫ 󵄨󵄨󵄨f 󸀠 (tx + (1 − t)a)󵄨󵄨󵄨 dt ≤ s+1 0

1

|f 󸀠 (b)|q + |f 󸀠 (x)|q 󵄨 󵄨q . ∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)x)󵄨󵄨󵄨 dt ≤ s+1 0

The proof is completed. Corollary 130. In Theorem 129, if we choose x =

a+b , 2

we have

󵄨󵄨 (a + b)(4f ( a+b ) − f (a) − f (b)) 󵄨󵄨 2 󵄨󵄨 4(b − a) 󵄨󵄨 −

󵄨󵄨 2α−1 Γ(α + 1) 󵄨󵄨 α α (a J a+b − f (a) + b RL J a+b + f (b))󵄨󵄨 RL ( 2 ) ( 2 ) 󵄨󵄨 (b − a)α+1 1

1

󸀠 q 󸀠 a+b q 1 2p−1 ap (a + b)p p |f (a)| + |f ( 2 )| q + ) ( ) ≤ ( 4 pα + 1 2 s+1 1

1

󸀠 q 󸀠 a+b q 1 2p−1 bp (a + b)p p |f (b)| + |f ( 2 )| q + ( + ) ( ) . 4 pα + 1 2 s+1

1 q

132 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals Theorem 131. Let f : [a, b] → ℝ be a differentiable on (a, b) with a < b, |f 󸀠 |q ∈ L[a, b] and |f 󸀠 |q (q > 1) is s-convex function, then for any 0 < α ≤ 1, a < x < b we have 󵄨󵄨 (x − a)[(a + x)f (x) − xf (a)] + (b − x)[(x + b)f (x) − xf (b)] 󵄨󵄨 󵄨󵄨 󵄨󵄨 (b − a)2 󵄨󵄨 Γ(α + 1) a b 󵄨󵄨 α α − ( RL Jx− f (a) + RL Jx+ f (b))󵄨󵄨󵄨 α−1 α−1 2 (b − x) (b − a) (x − a) 󵄨 1− q1

(x − a)2 a ≤ ( + x) (b − a)2 α + 1

1− q1

b (b − x)2 ( ψ(α, s, x) + + x) (b − a)2 α + 1

ω(α, s, x),

where 1

q x aΓ(α + 1)Γ(s + 1) x a + )|f 󸀠 (x)|q + ( + )|f 󸀠 (a)|q ) , ψ(α, s, x) = (( α+s+1 s+1 Γ(α + s + 2) s+1 1

q bΓ(α + 1)Γ(s + 1) x b x ω(α, s, x) = (( + )|f 󸀠 (b)|q + ( + )|f 󸀠 (x)|q ) . Γ(α + s + 2) s+1 α+s+1 s+1

Proof. Using Lemma 64 and the well-known power mean inequality and |f 󸀠 |q is s-convex function, we have 󵄨󵄨 (x − a)[(a + x)f (x) − xf (a)] + (b − x)[(x + b)f (x) − xf (b)] 󵄨󵄨 󵄨󵄨 󵄨󵄨 (b − a)2 󵄨󵄨 Γ(α + 1) a b 󵄨󵄨 α α − ( J J − f (a) + + f (b))󵄨 RL RL x x 󵄨󵄨 (b − x)α−1 (b − a)2 (x − a)α−1 󵄨 1

(x − a)2 󵄨 󵄨 ≤ ∫(at α + x)󵄨󵄨󵄨f 󸀠 (tx + (1 − t)a)󵄨󵄨󵄨dt (b − a)2 0

+

1

(b − x)2 󵄨 󵄨 ∫(b(1 − t)α + x)󵄨󵄨󵄨f 󸀠 (tb + (1 − t)x)󵄨󵄨󵄨dt (b − a)2 0

1

(x − a)2 (∫(at α + x)dt) ≤ (b − a)2

1− q1

0

1

󵄨 󵄨q (∫(at + x)󵄨󵄨󵄨f 󸀠 (tx + (1 − t)a)󵄨󵄨󵄨 dt) 0

1

α

1− q1

(b − x)2 + (∫(b(1 − t)α + x)dt) (b − a)2 0

1

(x − a)2 ≤ (∫(at α + x)dt) (b − a)2 0

1

1− q1

1

󵄨 󵄨q (∫(b(1 − t) + x)󵄨󵄨󵄨f 󸀠 (tb + (1 − t)x)󵄨󵄨󵄨 dt) α

0

α

s

q

s

q

(∫(at + x)(t |f (x)| + (1 − t) |f (a)| )dt) 0

1− q1

(b − x)2 + (∫(b(1 − t)α + x)dt) (b − a)2 0

1

1 q

󸀠

󸀠

1 q

1 q

4.3 Inequalities via s-convex functions | 133

1

× (∫(b(1 − t)α + x)(t s |f 󸀠 (b)|q + (1 − t)s |f 󸀠 (x)|q )dt) 0

≤

1− q1

(x − a)2 a ( + x) (b − a)2 α + 1

ψ(α, s, x) +

1 q

1− q1

b (b − x)2 ( + x) (b − a)2 α + 1

ω(α, s, x).

The proof is completed. Theorem 132. Let f : [a, b] → ℝ be twice differentiable. If f 󸀠󸀠 ∈ L[a, b] and |f 󸀠󸀠 | is an s-convex function, then for any 0 < α ≤ 1 and a < x < b we have 󵄨󵄨 (x(α + 1) + b)f (x) − bf (b) (b(α + 1) + x)f (x) − xf (a) 1 󵄨󵄨 [ + ] 󵄨󵄨 󵄨󵄨 (b − a)2 b−x x−a 󵄨󵄨 (α + 1)Γ(α + 1) b x 󵄨󵄨 α α − J J ( + f (b) + − f (a))󵄨 RL RL x x 󵄨󵄨 (b − x)α+1 (x − a)α+1 (b − a)2 󵄨 x b xΓ(α + 2)Γ(s + 1) b−x 󸀠󸀠 [( + )|f (x)| + ( ≤ Γ(α + s + 3) (b − a)2 α + s + 2 s + 2 b x−a bΓ(α + 2)Γ(s + 1) + )|f 󸀠󸀠 (b)|] + [( (s + 1)(s + 2) Γ(α + s + 3) (b − a)2 b x x )|f 󸀠󸀠 (a)| + ( + )|f 󸀠󸀠 (x)|]. + (s + 1)(s + 2) α+s+2 s+2 Proof. Using Lemma 65 via |f 󸀠󸀠 | is an s-convex function, we have 󵄨󵄨 (x(α + 1) + b)f (x) − bf (b) (b(α + 1) + x)f (x) − xf (a) 1 󵄨󵄨 [ + ] 󵄨󵄨 󵄨󵄨 (b − a)2 b−x x−a 󵄨󵄨 (α + 1)Γ(α + 1) x b 󵄨󵄨 α α − ( RL Jx+ f (b) + RL Jx− f (a))󵄨󵄨󵄨 α+1 α+1 2 (b − x) (x − a) (b − a) 󵄨 1

≤

(b − x) 󵄨 󵄨 ∫(t α+1 x + tb)󵄨󵄨󵄨f 󸀠󸀠 (tx + (1 − t)b)󵄨󵄨󵄨dt (b − a)2 0

1

(x − a) 󵄨 󵄨 + ∫((1 − t)α+1 b + (1 − t)x)󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)x)󵄨󵄨󵄨dt (b − a)2 1

≤

0

(b − x) ∫(t α+1 x + tb)(t s |f 󸀠󸀠 (x)| + (1 − t)s |f 󸀠󸀠 (b)|)dt (b − a)2 0

+

1

(x − a) ∫((1 − t)α+1 b + (1 − t)x)(t s |f 󸀠󸀠 (a)| + (1 − t)s |f 󸀠󸀠 (x)|)dt (b − a)2 0

x b−x b xΓ(α + 2)Γ(s + 1) = [( + )|f 󸀠󸀠 (x)| + ( Γ(α + s + 3) (b − a)2 α + s + 2 s + 2 b x−a bΓ(α + 2)Γ(s + 1) + )|f 󸀠󸀠 (b)|] + [( (s + 1)(s + 2) Γ(α + s + 3) (b − a)2

134 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals

+

b x x )|f 󸀠󸀠 (a)| + ( + )|f 󸀠󸀠 (x)|]. (s + 1)(s + 2) α+s+2 s+2

The proof is completed.

Theorem 133. Let f : [a, b] → ℝ be twice differentiable. If |f 󸀠󸀠 |q ∈ L[a, b] and |f 󸀠󸀠 |q (q > 1) is an s-convex function, then for any 0 < α ≤ 1 and a < x < b, the following inequality for fractional integrals holds: 󵄨󵄨 1 (x(α + 1) + b)f (x) − bf (b) (b(α + 1) + x)f (x) − xf (a) 󵄨󵄨 [ + ] 󵄨󵄨 b−x x−a 󵄨󵄨 (b − a)2 󵄨󵄨󵄨 x b (α + 1)Γ(α + 1) α α 󵄨 ( − RL Jx+ f (b) + RL Jx− f (a))󵄨󵄨󵄨 α+1 α+1 2 (b − x) (x − a) (b − a) 󵄨 1

1

2p−1 xp 2p−1 bp p |f 󸀠󸀠 (x)|q |f 󸀠󸀠 (b)|q q b−x ( + ) ( + ) ≤ p+1 s+1 s+1 (b − a)2 p(α + 1) + 1 1

1

x−a 2p−1 bp 2p−1 x p p |f 󸀠󸀠 (a)|q |f 󸀠󸀠 (x)|q q + ( + ) ( + ) , p+1 s+1 s+1 (b − a)2 p(α + 1) + 1 where

1 p

+

1 q

= 1.

Proof. Using Lemmas 37 and 65, Holder’s inequality via and f 󸀠󸀠 |q is an s-convex function, we have 󵄨󵄨 (x(α + 1) + b)f (x) − bf (b) (b(α + 1) + x)f (x) − xf (a) 1 󵄨󵄨 [ + ] 󵄨󵄨 b−x x−a 󵄨󵄨 (b − a)2 󵄨󵄨 x (α + 1)Γ(α + 1) b 󵄨󵄨 α α ( − RL Jx+ f (b) + RL Jx− f (a))󵄨󵄨󵄨 α+1 α+1 2 (b − x) (x − a) (b − a) 󵄨 1

≤

(b − x) 󵄨 󵄨 ∫(t α+1 x + tb)󵄨󵄨󵄨f 󸀠󸀠 (tx + (1 − t)b)󵄨󵄨󵄨dt (b − a)2 0

1

(x − a) 󵄨 󵄨 + ∫((1 − t)α+1 b + (1 − t)x)󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)x)󵄨󵄨󵄨dt (b − a)2 0

1 p

1

1

b−x p 󵄨 󵄨q (∫(t α+1 x + tb) dt) (∫ 󵄨󵄨󵄨f 󸀠󸀠 (tx + (1 − t)b)󵄨󵄨󵄨 dt) ≤ 2 (b − a) 0

0

1 p

1

1

1 q

x−a p 󵄨 󵄨q + (∫((1 − t)α+1 b + (1 − t)x) dt) (∫ 󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)x)󵄨󵄨󵄨 dt) (b − a)2 0

1

1 p

1 q

0

1

b−x ≤ (∫ 2p−1 (x p t p(α+1) + t p bp )dt) (∫(t s |f 󸀠󸀠 (x)|q + (1 − t)s |f 󸀠󸀠 (b)|q )dt) (b − a)2 0

+

1

0

1 x−a (∫ 2p−1 ((1 − t)p(α+1 )bp + (1 − t)p x p )dt) p (b − a)2

0

1 q

4.3 Inequalities via s-convex functions | 135

1

× (∫(t s |f 󸀠󸀠 (a)|q + (1 − t)s |f 󸀠󸀠 (x)|q )dt) 0

=

1 q

1

1

b−x 2p−1 xp 2p−1 bp p |f 󸀠󸀠 (x)|q |f 󸀠󸀠 (b)|q q ( + ) ( + ) 2 p+1 s+1 s+1 (b − a) p(α + 1) + 1 1

1

2p−1 xp p |f 󸀠󸀠 (a)|q |f 󸀠󸀠 (x)|q q x−a 2p−1 bp + ) ( + ) . + ( 2 p+1 s+1 s+1 (b − a) p(α + 1) + 1 The proof is completed. Corollary 134. In Theorem 133, if we choose x =

a+b , 2

then we have

󵄨󵄨 ((a + 3b)(α + 1) + (a + 3b))f ( a+b ) − 2bf (b) − (a + b)f (a) (α + 1)Γ(α + 1) 󵄨󵄨 2 − 󵄨󵄨 󵄨󵄨 (b − a)3 (b − a)2 󵄨󵄨 2α (a + b) b2α+1 󵄨󵄨 α α ( RL J( a+b )+ f (b) + RL J( a+b )− f (a))󵄨󵄨󵄨 α+1 α+1 (b − a) (b − a) 2 2 󵄨 1

1

󸀠󸀠 a+b q 󸀠󸀠 q (a + b)p 2p−1 bp p |f ( 2 )| + |f (b)| q 1 ( + ) ( ) ≤ 2(b − a) 2(p(α + 1) + 1) p+1 s+1 1

1

󸀠󸀠 a+b q 󸀠󸀠 q 1 2p−1 bp (a + b)p p |f ( 2 )| + |f (a)| q + ( + ) ( ) . 2(b − a) p(α + 1) + 1 2(p + 1) s+1

4.3.2 Applications to some special means Now, we give some applications to special means of real numbers (see Proposition 4.1– 4.3, [232]).

Proposition 135. Let a, b ∈ ℝ+ , a < b, s ∈ (0, 1] and q > 1. Then 1 2(b−a) (1 − 21+s )A(a−2 , b−2 ), { s+2 { 1 1 { 1 { 1 p 1 p 1 q { 󵄨󵄨 󵄨󵄨 {(b − a)( p+1 ) (1 − 2p ) ( (s+1)2s+1 ) −1 −1 −1 󵄨󵄨A(a , b ) − L (a, b)󵄨󵄨 ≤ { 1 { { × (1 + (2s+1 − 1) q )A(a−2 , b−2 ), { { 1 1 1 { 1 1− q 1 q 1 q −2 −2 {2(b − a)( 4 ) ( s+2 ) (1 − 21+s ) A(a , b ),

where

1 p

(4.74)

= 1 − q1 .

Proof. Applying Theorem 121, Corollary 123, and Corollary 125 respectively, for f (x) = and α = 1, one can obtain the result immediately. The proof is completed.

1 x

136 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals Proposition 136. Let a, b ∈ ℝ+ , a < b, n ∈ Z, |n| ≥ 2, s ∈ (0, 1] and q > 1. Then 2|n|(b−a) 1 (1 − 21+s )A(an−1 , bn−1 ), { s+2 { 1 1 1 { { 1 p 1 p 1 q { 󵄨󵄨 󵄨󵄨 {|n|(b − a)( p+1 ) (1 − 2p ) ( (s+1)2s+1 ) n n n 󵄨󵄨A(a , b ) − Ln (a, b)󵄨󵄨 ≤ { 1 { { × (1 + (2s+1 − 1) q )A(an−1 , bn−1 ), { { 1 1 1 { 1 1− q 1 q 1 q n−1 n−1 {2|n|(b − a)( 4 ) ( s+2 ) (1 − 21+s ) A(a , b ),

where

1 p

(4.75)

= 1 − q1 .

Proof. Applying Theorem 121, Corollary 123, and Corollary 125, respectively, for f (x) = xn and α = 1, one can obtain the result immediately. The proof is completed. Proposition 137. Let a, b ∈ ℝ+ , (a < b), a−1 > b−1 . For n ∈ Z, |n| ≥ 2, s ∈ (0, 1] and q > 1, we have 󵄨󵄨 −1 −1 −1 −1 −1 −1 󵄨 󵄨󵄨H (b , a ) − L (b , a )󵄨󵄨󵄨 1 2(b−a) (1 − 21+s )H −1 (a−2 , b−2 ), { ab(s+2) { 1 1 { 1 { 1 p 1 1 p q { { b−a (1 − ( ( ) ) ) p s+1 2 (s+1)2 ≤ { ab p+1 1 { { × (1 + (2s+1 − 1) q )H −1 (a−2 , b−2 ), { { 1 { 2(b−a) 1 1− 1 1 1 1 q −1 −2 −2 q q { ab ( 4 ) ( s+2 ) (1 − 21+s ) H (a , b ),

(4.76)

󵄨󵄨 −1 n n n −1 −1 󵄨 󵄨󵄨H (b , a ) − Ln (b , a )󵄨󵄨󵄨

2|n|(b−a) 1 (1 − 21+s )H −1 (an−1 , bn−1 ), { ab(s+2) { 1 1 1 { { 1 1 p q { { |n|(b−a) ( p+1 ) (1 − 21p ) p ( (s+1)2 s+1 ) ab ≤{ 1 { { × (1 + (2s+1 − 1) q )H −1 (an−1 , bn−1 ), { { 1 { 2|n|(b−a) 1 1− 1 1 1 1 q −1 n−1 n−1 q q { ab ( 4 ) ( s+2 ) (1 − 21+s ) H (a , b ),

where

1 p

(4.77)

= 1 − q1 .

Proof. Making the substitutions a → b−1 , b → a−1 in the inequalities (4.74) and (4.75), one can obtain desired inequalities (4.76) and (4.77), respectively. The proof is completed. Theorem 138. For some s ∈ (0, 1], n ∈ ℤ \ {−1, 0}, 0 ≤ a < b, the following inequality for fractional integrals hold: 1

1

p aq + bq q 1 󵄨󵄨 2 2 n 2 2 󵄨 ) ( ) , 󵄨󵄨A(a , b ) − Ln (a , b )󵄨󵄨󵄨 ≤ 2(b − a)( p+1 1 + sq

where

1 p

+

1 q

= 1, 1 < q < ∞.

Proof. Applying Theorem 126 for f (x) = x 2 , α = 1, λ = 1, one can obtain the result immediately. The proof is completed.

4.3 Inequalities via s-convex functions | 137

Theorem 139. For some s ∈ (0, 1], 0 ≤ a < b, the following inequality for fractional integrals holds: 1

1

󵄨󵄨 1 p p 1 1 󵄨󵄨󵄨󵄨 b2q + a2q 󵄨󵄨 − ) ( 2q 2q ) , 󵄨󵄨 ≤ (b − a)( 󵄨󵄨 󵄨󵄨 H(a, b) L(a, b) 󵄨󵄨 p+1 a b (1 + sq) where

1 p

+

1 q

= 1, 1 < q < ∞. 1 , x

Proof. Applying Theorem 126 for f (x) = immediately. The proof is completed.

α = 1, λ = 1, one can obtain the result

Theorem 140 (see Theorem 4.1, [51]). For some s ∈ (0, 1], n ∈ ℤ \ {−1, 0}, 0 ≤ a < b, the following inequality for fractional integrals holds: 󵄨󵄨 (4( a+b )n − an − bn ) a a+b 󵄨󵄨 2 A(a, b) − ( L̄ n (a, ) 󵄨󵄨 󵄨󵄨 2(b − a) 2(b − a) n 2 󵄨󵄨 b a+b 󵄨 + L̄ nn ( , b))󵄨󵄨󵄨 󵄨󵄨 2(b − a) 2 n−1

≤

a a+b a+b n [( + )( ) 4 2 + s 2(s + 1) 2 +

+(

aΓ(2)Γ(s + 1) a+b + )an−1 ] Γ(3 + s) 2(s + 1)

n−1

a+b b a+b a+b n bΓ(2)Γ(s + 1) [( + )bn−1 + ( + )( ) 4 Γ(3 + s) 2(s + 1) 2 + s 2(s + 1) 2

Proof. Applying Theorem 128 for f (t) = t n , α = 1 and x = immediately. The proof is completed.

].

a+b , one can obtain the result 2

Theorem 141 (see Theorem 4.3, [51]). For some s ∈ (0, 1], n ∈ Z \ {−1, 0}, 0 ≤ a < b, the following inequality for fractional integrals holds: 󵄨󵄨 (4( a+b )n − an − bn ) a a+b 󵄨󵄨 2 A(a, b) − ( L̄ n (a, ) 󵄨󵄨 󵄨󵄨 2(b − a) 2(b − a) n 2 󵄨󵄨 b a+b 󵄨 + L̄ nn ( , b))󵄨󵄨󵄨 󵄨󵄨 2(b − a) 2 1

n 2p−1 ap (a + b)p p a + ) ( ≤ ( 4 p+1 2 1

+ where

1 p

+

1 q

q(n−1)

+ ( a+b )q(n−1) 2 s+1

q(n−1)

n 2p−1 bp (a + b)p p b ( + ) ( 4 p+1 2

1 q

)

+ ( a+b )q(n−1) 2 s+1

1 q

) ,

= 1, 1 < q < ∞.

Proof. Applying Theorem 129 for f (t) = t n , α = 1 and x = immediately. The proof is completed.

a+b , one can obtain the result 2

138 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals Theorem 142 (see Theorem 4.5, [51]). For some s ∈ (0, 1], n ∈ ℤ \ {−1, 0}, 0 ≤ a < b, the following inequality for fractional integrals holds: 󵄨󵄨 3( a+b )n (a + 3b) − 2bn+1 − (a + b)an 󵄨󵄨 2 󵄨󵄨 󵄨󵄨 (b − a)3 󵄨󵄨 2(a + b) ̄ n a + b 4b ̄ n (a, a + b ))󵄨󵄨󵄨 −( L L ( , b) + n n 󵄨󵄨 2 2 (b − a)3 (b − a)3 󵄨 1

≤

1

n(n − 1) (a + b)p 2p−1 bp p (a + b)(n−2)q b(n−2)q q + ( + ) ( (n−2)q ) 2(b − a) 2(2p + 1) p+1 s+1 2 (s + 1) 1

1

n(n − 1) 2p−1 bp (a + b)p p (a + b)(n−2)q a(n−2)q q + ( + ) ( (n−2)q + ) , 2(b − a) 2p + 1 2(p + 1) s+1 2 (s + 1) where

1 p

+

1 q

= 1, 1 < q < ∞.

Proof. Applying Theorem 133 for f (t) = t n , α = 1 and x = immediately. The proof is completed.

a+b , one can obtain the result 2

Theorem 143 (see Theorem 4.6, [51]). For some s ∈ (0, 1], 0 ≤ a < b, the following inequality for fractional integrals holds: 󵄨󵄨 6(a + 3b) a+b 2 󵄨󵄨 − − 󵄨󵄨 󵄨󵄨 (a + b)(b − a)3 a(b − a)3 (b − a)3 󵄨󵄨 2(a + b) 4b 1 1 󵄨󵄨 −( + ) 󵄨 ̄ a+b ) 󵄨󵄨󵄨 (b − a)3 L(̄ a+b , b) (b − a)3 L(a, 2

p

p−1 p

2

1 p

1

q 1 (a + b) 2 b 8q 1 + 3q ) ( + ) ( ≤ 3q 2(b − a) 2(2p + 1) p+1 (a + b) (s + 1) b s + 1 1

1

q 2p−1 bp (a + b)p p 8q 1 1 ( + ) ( + 3q ) , + 3q 2(b − a) 2p + 1 2(p + 1) (a + b) (s + 1) a s + 1

where

1 p

+

1 q

= 1, 1 < q < ∞.

Proof. Applying Theorem 133 for f (t) = 1t , α = 1 and x = immediately. The proof is completed.

a+b , one can obtain the result 2

4.4 Inequalities via m-convex functions The results in this section are taken from [216]. Theorem 144. Let f : [0, b∗ ] → ℝ be a twice-differentiable mapping with b∗ > 0. If |f 󸀠󸀠 |q is measurable and m-convex on [a, mb ] for some fixed q ≥ 1, 0 ≤ a < b and m ∈ (0, 1]

4.4 Inequalities via m-convex functions | 139

with

b m

≤ b∗ , then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J + f (b) + RL Jb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 2 2(b − a)α 󵄨 1

≤

|f 󸀠󸀠 (a)|q + m|f 󸀠󸀠 ( mb )|q q α(b − a)2 [ ] . 2(α + 1)(α + 2) 2

Proof. Firstly, we suppose that q = 1. From Lemma 47, we have 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α [ J − + f (b) + RL Jb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 2 2(b − a)α 󵄨 1

(b − a)2 1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 ≤ ∫ 󵄨󵄨f (ta + (1 − t)b)󵄨󵄨󵄨dt, 2 α+1

(4.78)

0

because (1 − t)α+1 + t α+1 ≤ 1 for any t ∈ [a, b]. Since |f 󸀠󸀠 | is m-convex on [a, mb ], we know that for any t ∈ [0, 1] 󵄨 󵄨󵄨 󵄨󵄨 󸀠󸀠 󵄨 󵄨 󸀠󸀠 b 󵄨󵄨 󸀠󸀠 󵄨󵄨f (ta + (1 − t)b)󵄨󵄨󵄨 ≤ t|f (a)| + m(1 − t)󵄨󵄨󵄨f ( )󵄨󵄨󵄨. m 󵄨󵄨 󵄨󵄨 Therefore, 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) 󵄨 󵄨󵄨 [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 − 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) ≤

1 󵄨󵄨 (b − a)2 1 − (1 − t)α+1 − t α+1 b 󵄨󵄨󵄨 󵄨 (t|f 󸀠󸀠 (a)| + m(1 − t)󵄨󵄨󵄨f 󸀠󸀠 ( )󵄨󵄨󵄨)dt ∫ 󵄨󵄨 2 α+1 m 󵄨󵄨 0

=

|f 󸀠󸀠 (a)| + m|f 󸀠󸀠 ( mb )| α(b − a)2 ( ), 2(α + 1)(α + 2) 2

which completes the proof for this case. Secondly, we suppose that q > 1. Using Lemma 47 and the power mean inequality for q, we obtain 1

󵄨 󵄨 ∫(1 − (1 − t)α+1 − t α+1 )󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 0

1

α+1

≤ (∫(1 − (1 − t) 0 1

−t

α+1

1− q1

)dt) 1 q

󵄨 󵄨q × (∫(1 − (1 − t)α+1 − t α+1 )󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) . 0

(4.79)

140 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals Since |f 󸀠󸀠 |q is m-convex on [a, mb ], we know that for any t ∈ [0, 1] 󵄨󵄨 󵄨q 󵄨 󸀠󸀠 b 󵄨󵄨 󵄨q 󵄨󵄨 󸀠󸀠 󸀠󸀠 q 󵄨󵄨f (ta + (1 − t)b)󵄨󵄨󵄨 ≤ t|f (a)| + m(1 − t)󵄨󵄨󵄨f ( )󵄨󵄨󵄨 . 󵄨󵄨 m 󵄨󵄨

(4.80)

Hence, from (4.78), (4.79) and (4.80), we obtain 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 1

(b − a)2 (∫(1 − (1 − t)α+1 − t α+1 )dt) ≤ 2(α + 1)

1− q1

0

1

󵄨 󵄨q × (∫(1 − (1 − t)α+1 − t α+1 )󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) 0

1

(b − a)2 (∫(1 − (1 − t)α+1 − t α+1 )dt) ≤ 2(α + 1)

1 q

1− q1

0

1

1

q q 󵄨󵄨 b 󵄨󵄨󵄨 󵄨 × (∫(1 − (1 − t)α+1 − t α+1 )[t|f 󸀠󸀠 (a)|q + m(1 − t)󵄨󵄨󵄨f 󸀠󸀠 ( )󵄨󵄨󵄨 ]dt) 󵄨󵄨 m 󵄨󵄨

0

1− q1

α (b − a)2 ( ) = 2(α + 1) α + 2

1

1

q |f 󸀠󸀠 (a)|q + m|f 󸀠󸀠 ( mb )|q q α ) [ ] ( α+2 2 1

|f 󸀠󸀠 (a)|q + m|f 󸀠󸀠 ( mb )|q q α(b − a)2 [ ] . = 2(α + 1)(α + 2) 2

(4.81)

The proof is completed. Remark 145 (with the same assumptions in Theorem 144). If |f 󸀠󸀠 (x)| ≤ M on [a, mb ], we obtain 1

󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) Mα(b − a)2 1+m q 󵄨󵄨 󵄨󵄨 α α − [ J f (b) + J f (a)] ≤ ( ) , − + 󵄨󵄨 󵄨 RL RL a b 󵄨 α 󵄨󵄨 󵄨 2 2(b − a) 2 󵄨 2(α + 1)(α + 2) for q ≥ 1. Now, we are ready to state the second theorem in this section. Theorem 146. Let f : [0, b∗ ] → ℝ be a twice-differentiable mapping with b∗ > 0. If |f 󸀠󸀠 |q is measurable and m-convex on [a, mb ] for some fixed q > 1, 0 ≤ a < b and m ∈ (0, 1] with mb ≤ b∗ , then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨

4.4 Inequalities via m-convex functions | 141 1

1

p |f 󸀠󸀠 (a)|q + m|f 󸀠󸀠 ( mb )|q q (b − a)2 2 ≤ (1 − ) ( ) , 2(α + 1) p(α + 1) + 1 2

where

1 p

+

1 q

= 1.

Proof. From Lemma 47 and using the well-known Hölder’s inequality, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) 1

≤

(b − a)2 1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 ∫ 󵄨󵄨f (ta + (1 − t)b)󵄨󵄨󵄨dt 2 α+1 0

1 p

1

1

(b − a)2 p 󵄨 󵄨q ≤ (∫(1 − (1 − t)α+1 − t α+1 ) dt) (∫ 󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) 2(α + 1) 0

1 q

0

1

(b − a)2 ≤ (∫(1 − (1 − t)p(α+1) − t p(α+1) )dt) 2(α + 1)

1 p

0

1

1

q 󵄨󵄨 󵄨q 1 󵄨󵄨 󸀠󸀠 b 󵄨󵄨󵄨 󸀠󸀠 q × (|f (a)| ∫ tdt + m󵄨󵄨f ( )󵄨󵄨 ∫(1 − t)dt) 󵄨󵄨 m 󵄨󵄨

0

0

1 p

1

|f (a)|q + m|f 󸀠󸀠 ( mb )|q q 2 (b − a) (1 − ) ( ) . = 2(α + 1) p(α + 1) + 1 2 2

󸀠󸀠

Here we use q

(1 − (1 − t)α+1 − t α+1 ) ≤ 1 − (1 − t)q(α+1) − t q(α+1) , for any t ∈ [0, 1], which follows from (A − B)q ≤ Aq − Bq , for any A > B ≥ 0 and q ≥ 1. The proof is completed. Corollary 147. With the above assumptions, given that |f 󸀠󸀠 (x)| ≤ M on [a, mb ], we obtain 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J f (b) + J f (a)] + − 󵄨󵄨 󵄨󵄨 RL RL a b 󵄨󵄨 󵄨󵄨 2 2(b − a)α 1

1

p M(b − a)2 2 1+m q ≤ (1 − ) ( ) . 2(α + 1) p(α + 1) + 1 2

Another Hermit-Hadamard type inequality for powers in terms of the second derivatives is obtained as follows.

142 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals Theorem 148. Let f : [0, b∗ ] → ℝ be a twice-differentiable mapping with b∗ > 0. If |f 󸀠󸀠 |q is measurable and m-convex on [a, mb ] for some fixed q ≥ 1, 0 ≤ a < b and m ∈ (0, 1] with mb ≤ b∗ , then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) 1

1

󸀠󸀠 q 󸀠󸀠 b q (b − a)2 q(α + 1) − 1 q |f (a)| + m|f ( m )| q ≤ ( ) ( ) . 2(α + 1) q(α + 1) + 1 2

Proof. From Lemma 47 and using the well-known Hölder’s inequality, we have 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J f (b) + J f (a)] + − 󵄨󵄨 󵄨󵄨 RL RL a b 󵄨󵄨 󵄨󵄨 2 2(b − a)α 1

(b − a)2 1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 ≤ ∫ 󵄨󵄨f (ta + (1 − t)b)󵄨󵄨󵄨dt 2 α+1 0

1 p

1

1

(b − a)2 q󵄨 󵄨q (∫ 1dt) (∫(1 − (1 − t)α+1 − t α+1 ) 󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) ≤ 2(α + 1) 0

≤

1

0

(b − a)2 (|f 󸀠󸀠 (a)|q ∫[t − (1 − t)q(α+1) t − t q(α+1)+1 ]dt 2(α + 1) 0

1

󵄨q 1

q 󵄨󵄨 b 󵄨󵄨 󵄨 + m󵄨󵄨󵄨f 󸀠󸀠 ( )󵄨󵄨󵄨 ∫[(1 − t) − (1 − t)q(α+1)+1 − t q(α+1) (1 − t)]dt) 󵄨󵄨 m 󵄨󵄨

0

1

1

󸀠󸀠 q 󸀠󸀠 b q (b − a) q(α + 1) − 1 q |f (a)| + m|f ( m )| q ( ) ( ) . = 2(α + 1) q(α + 1) + 1 2 2

Here we use q

(1 − (1 − t)α+1 − t α+1 ) ≤ 1 − (1 − t)q(α+1) − t q(α+1) , for any t ∈ [0, 1], which follows from (A − B)q ≤ Aq − Bq , for any A > B ≥ 0 and q ≥ 1. The proof is completed. Remark 149. From Theorems 144, 146 and 148, we have

where

󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 ≤ min{K1 , K2 , K3 }, 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 1

|f 󸀠󸀠 (a)|q + m|f 󸀠󸀠 ( mb )|q q α(b − a)2 [ ] , K1 = 2(α + 1)(α + 2) 2

1 q

4.4 Inequalities via m-convex functions | 143 1

1

p |f 󸀠󸀠 (a)|q + m|f 󸀠󸀠 ( mb )|q q (b − a)2 2 K2 = (1 − ) ( ) , 2(α + 1) p(α + 1) + 1 2 1

1

󸀠󸀠 q 󸀠󸀠 b q (b − a)2 q(α + 1) − 1 q |f (a)| + m|f ( m )| q K3 = ( ) ( ) . 2(α + 1) q(α + 1) + 1 2

We start to list some theorems containing Hermite-Hadmard type inequality via m-convex functions. Theorem 150. Let f : [0, b∗ ] → ℝ be a twice-differentiable mapping with b∗ > 0. If |f 󸀠󸀠 |q is measurable and m-convex on [a, mb ] for some fixed q ≥ 1, 0 ≤ a < b and m ∈ (0, 1] with mb ≤ b∗ , r > 0 then 󵄨󵄨 󵄨󵄨 f (a) + f (b) 2 a+b Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α + f( )− [ J + f (b) + RL Jb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 r(r + 1) r+1 2 r(b − a)α 󵄨 1

|f 󸀠󸀠 (a)|q + m|f 󸀠󸀠 ( mb )|q q 1 α + )( ) . r(α + 1)(α + 2) 4(r + 1) 2

≤ (b − a)2 (

Proof. Case 1: we suppose that q = 1. From Lemma 57, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 2 a+b Γ(α + 1) 󵄨󵄨 󵄨 + f( )− [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 r(r + 1) 󵄨󵄨 r+1 2 r(b − a) 2

1

󵄨 󵄨 ≤ (b − a) ∫ |k(t)|󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt, 0

due to (1 − t)α+1 + t α+1 ≤ 1 for any t ∈ [0, 1]. Since |f 󸀠󸀠 | is m-convex on [a, mb ], we know that for any t ∈ [0, 1] 󵄨 󵄨󵄨 󵄨󵄨 󸀠󸀠 󵄨 󵄨 󸀠󸀠 b 󵄨󵄨 󸀠󸀠 󵄨󵄨f (ta + (1 − t)b)󵄨󵄨󵄨 ≤ t|f (a)| + m(1 − t)󵄨󵄨󵄨f ( )󵄨󵄨󵄨. m 󵄨󵄨 󵄨󵄨 Therefore, 󵄨󵄨 󵄨󵄨 f (a) + f (b) 2 a+b Γ(α + 1) 󵄨 󵄨󵄨 + f( )− [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 r+1 2 r(b − a) 󵄨󵄨 r(r + 1) 1 2

󵄨󵄨 b 󵄨󵄨󵄨 󵄨 ≤ (b − a)2 ∫ |k1 (t)|(t|f 󸀠󸀠 (a)| + m(1 − t)󵄨󵄨󵄨f 󸀠󸀠 ( )󵄨󵄨󵄨)dt 󵄨󵄨 m 󵄨󵄨 0

1

󵄨󵄨 b 󵄨󵄨󵄨 󵄨 + (b − a) ∫ |k2 (t)|(t|f 󸀠󸀠 (a)| + m(1 − t)󵄨󵄨󵄨f 󸀠󸀠 ( )󵄨󵄨󵄨)dt, 󵄨󵄨 m 󵄨󵄨 1 2

2

where k1 (t) =

1 − (1 − t)α+1 − t α+1 t − , r(α + 1) r+1

144 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals

k2 (t) =

1 − (1 − t)α+1 − t α+1 1 − t − . r(α + 1) r+1

Denote 1 2

󵄨󵄨 b 󵄨󵄨󵄨 󵄨 I1 = ∫ |k1 (t)|(t|f 󸀠󸀠 (a)| + m(1 − t)󵄨󵄨󵄨f 󸀠󸀠 ( )󵄨󵄨󵄨)dt, 󵄨󵄨 m 󵄨󵄨 0

1

󵄨󵄨 b 󵄨󵄨󵄨 󵄨 I2 = ∫ |k2 (t)|(t|f 󸀠󸀠 (a)| + m(1 − t)󵄨󵄨󵄨f 󸀠󸀠 ( )󵄨󵄨󵄨)dt. 󵄨󵄨 m 󵄨󵄨 1 2

Integrating the above equalities by parts respectively, we have 1 2

I1 ≤ ∫[ 0

󵄨󵄨 1 − (1 − t)α+1 − t α+1 t b 󵄨󵄨󵄨 󵄨 + ](t|f 󸀠󸀠 (a)| + m(1 − t)󵄨󵄨󵄨f 󸀠󸀠 ( )󵄨󵄨󵄨)dt 󵄨󵄨 r(α + 1) r+1 m 󵄨󵄨 1 2

1 2

|f 󸀠󸀠 (a)| 2 |f 󸀠󸀠 (a)| = ∫ [1 − (1 − t)α+1 − t α+1 ]tdt + ∫ t dt r(α + 1) r+1 0

+

0

m|f 󸀠󸀠 ( mb )| r(α + 1)

1 2

∫[1 − (1 − t)α+1 − t α+1 ](1 − t)dt + 0

m|f 󸀠󸀠 ( mb )| r+1

1 2

∫ t(1 − t)dt 0

1 |f 󸀠󸀠 (a)| 1 1 |f (a)| 1 ]+ [ − + α+2 r(α + 1) 8 (α + 2)(α + 3) (α + 2)(α + 3)2 24 r + 1 󸀠󸀠

=

+

󸀠󸀠 b m|f 󸀠󸀠 ( mb )| 3 1 1 1 m|f ( m )| [ − − ] + , r(α + 1) 8 α + 3 (α + 2)(α + 3)2α+2 12 r + 1

and 1

I2 ≤ ∫[ 1 2

󵄨󵄨 1 − (1 − t)α+1 − t α+1 1 − t b 󵄨󵄨󵄨 󵄨 + ](t|f 󸀠󸀠 (a)| + m(1 − t)󵄨󵄨󵄨f 󸀠󸀠 ( )󵄨󵄨󵄨)dt 󵄨󵄨 r(α + 1) r+1 m 󵄨󵄨 1

1

|f 󸀠󸀠 (a)| |f 󸀠󸀠 (a)| = ∫[1 − (1 − t)α+1 − t α+1 ]tdt + ∫(1 − t)tdt r(α + 1) r+1 1 2

+

=

m|f 󸀠󸀠 ( mb )| r(α + 1)

1 2

1

∫[1 − (1 − t)α+1 − t α+1 ](1 − t)dt + 1 2

m|f 󸀠󸀠 ( mb )| r+1

|f 󸀠󸀠 (a)| 3 1 1 1 |f 󸀠󸀠 (a)| [ − − ] + r(α + 1) 8 α + 3 (α + 2)(α + 3)2α+2 12 r + 1 +

1

∫(1 − t)2 dt 1 2

󸀠󸀠 b m|f 󸀠󸀠 ( mb )| 1 1 1 1 m|f ( m )| [ − + ] + . r(α + 1) 8 (α + 2)(α + 3) (α + 2)(α + 3)2α+2 24 r + 1

4.4 Inequalities via m-convex functions | 145

Therefore, 󵄨󵄨 f (a) + f (b) 󵄨󵄨 2 a+b Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α + f( )− [ J f (b) + J f (a)] + − 󵄨󵄨 󵄨󵄨 RL b 󵄨󵄨 r(r + 1) 󵄨󵄨 r+1 2 r(b − a)α RL a 2 2 (b − a) 󸀠󸀠 1 1 1 (b − a) 󸀠󸀠 ≤ |f (a)|( − )+ |f (a)| r(α + 1) 2 α+2 8 r+1 m(b − a)2 󵄨󵄨󵄨󵄨 󸀠󸀠 b 󵄨󵄨󵄨󵄨 1 1 1 m(b − a)2 󵄨󵄨󵄨󵄨 󸀠󸀠 b 󵄨󵄨󵄨󵄨 + )+ 󵄨󵄨f ( )󵄨󵄨( − 󵄨f ( )󵄨󵄨 r(α + 1) 󵄨󵄨 m 󵄨󵄨 2 α + 2 8 r + 1 󵄨󵄨󵄨 m 󵄨󵄨 α 1 = (b − a)2 |f 󸀠󸀠 (a)|( + ) 2r(α + 1)(α + 2) 8(r + 1) 󵄨󵄨 1 b 󵄨󵄨󵄨 α 󵄨 + m(b − a)2 󵄨󵄨󵄨f 󸀠󸀠 ( )󵄨󵄨󵄨( + ) 󵄨󵄨 m 󵄨󵄨 2r(α + 1)(α + 2) 8(r + 1) = (b − a)2

|f 󸀠󸀠 (a)| + m|f 󸀠󸀠 ( mb )| 2

(

α 1 + ), r(α + 1)(α + 2) 4(r + 1)

which completes the proof for this case. Case 2: Suppose that q > 1. By (3.39) via the power mean inequality for q, it is easy to see 󵄨󵄨 󵄨󵄨 f (a) + f (b) 2 a+b Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α [ J + f( )− + f (b) + RL Jb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 r(r + 1) r+1 2 r(b − a)α 󵄨 2

1

≤ (b − a) (∫ |k(t)|dt)

1− q1

0

1

󵄨 󵄨q × (∫ |k(t)|󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt)

1 q

0

1 2

1

1− q1

= (b − a)2 (∫ |k1 (t)|dt + ∫ |k2 (t)|dt) 0

1 2

1 2

1

1 q

󵄨 󵄨 󵄨q 󵄨q × (∫ |k1 (t)|󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt + ∫ |k2 (t)|󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) . 0

1 2

Clearly, 1 2

∫ |k1 (t)|dt ≤ 0

1

∫ |k2 (t)|dt ≤ 1 2

α 1 + , 2r(α + 1)(α + 2) 8(r + 1) α 1 + . 2r(α + 1)(α + 2) 8(r + 1)

Since |f 󸀠󸀠 |q is m-convex on [a, mb ], we know that for any t ∈ [0, 1], 󵄨󵄨 󵄨q 󵄨󵄨 󸀠󸀠 󵄨q 󵄨 󸀠󸀠 b 󵄨󵄨 󸀠󸀠 q 󵄨󵄨f (ta + (1 − t)b)󵄨󵄨󵄨 ≤ t|f (a)| + m(1 − t)󵄨󵄨󵄨f ( )󵄨󵄨󵄨 . m 󵄨󵄨 󵄨󵄨

146 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals Thus, 1 2

󵄨q 󵄨 ∫ |k1 (t)|󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt 0

1 2

≤ ∫( 0

=

q 󵄨󵄨 b 󵄨󵄨󵄨 1 − (1 − t)α+1 − t α+1 t 󵄨 + )(t|f 󸀠󸀠 (a)|q + m(1 − t)󵄨󵄨󵄨f 󸀠󸀠 ( )󵄨󵄨󵄨 )dt 󵄨󵄨 r(α + 1) r+1 m 󵄨󵄨

|f 󸀠󸀠 (a)|q 1 1 1 |f 󸀠󸀠 (a)|q [ − + ] + r(α + 1) 8 (α + 2)(α + 3) (α + 2)(α + 3)2α+2 24(r + 1) +

m|f 󸀠󸀠 ( mb )|q 3 m|f 󸀠󸀠 ( mb )|q 1 1 ] + [ − − , r(α + 1) 8 α + 3 (α + 2)(α + 3)2α+2 12(r + 1)

and 1

󵄨 󵄨q ∫ |k2 (t)|󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt 1 2

≤

|f 󸀠󸀠 (a)|q |f 󸀠󸀠 (a)|q 3 1 1 ] + [ − − r(α + 1) 8 α + 3 (α + 2)(α + 3)2α+2 12(r + 1) +

m|f 󸀠󸀠 ( mb )|q m|f 󸀠󸀠 ( mb )|q 1 1 1 ] + [ − + . r(α + 1) 8 (α + 2)(α + 3) (α + 2)(α + 3)2α+2 24(r + 1)

Therefore, 󵄨󵄨 󵄨󵄨 f (a) + f (b) 2 a+b Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α + f( )− [ J + f (b) + RL Jb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 r(r + 1) r+1 2 r(b − a)α 󵄨 1

|f 󸀠󸀠 (a)|q + m|f 󸀠󸀠 ( mb )|q q 1 α + )( ) . ≤ (b − a) ( r(α + 1)(α + 2) 4(r + 1) 2 2

The proof is completed. Remark 151. With the same assumptions in Theorem 150. If |f 󸀠󸀠 (x)| ≤ M on [a, mb ], we obtain 󵄨󵄨 f (a) + f (b) 󵄨󵄨 2 a+b Γ(α + 1) 󵄨󵄨 󵄨 + f( )− [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 r(r + 1) 󵄨󵄨 r+1 2 r(b − a) 1

1 1+m q α + )( ) , ≤ M(b − a) ( r(α + 1)(α + 2) 4(r + 1) 2 2

q ≥ 1.

Now, we begin to state the second theorem in this section. Theorem 152. Let f : [0, b∗ ] → ℝ be a twice-differentiable mapping with b∗ > 0. If |f 󸀠󸀠 |q is measurable and m-convex on [a, mb ] for some fixed q > 1, 0 ≤ a < b and m ∈ (0, 1]

4.4 Inequalities via m-convex functions | 147

with

b m

≤ b∗ , r > 0, then 󵄨󵄨 f (a) + f (b) 󵄨󵄨 2 a+b Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α + f( )− [ J + f (b) + RL Jb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 r(r + 1) r+1 2 r(b − a)α 󵄨 1

1

p |f 󸀠󸀠 (a)|q + m|f 󸀠󸀠 ( mb )|q q (b − a)2 2 ≤ (1 − ) ( ) , r(α + 1) p(α + 1) + 1 2

where

1 p

+

1 q

= 1.

Proof. From Lemma 57 and using the well-known Hölder’s inequality, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 2 a+b Γ(α + 1) 󵄨󵄨 󵄨 + f( )− [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 r(r + 1) 󵄨󵄨 r+1 2 r(b − a) 1 p

1

1 q

1

󵄨 󵄨q ≤ (b − a)2 (∫ |k(t)|p dt) (∫ 󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) .

(4.82)

0

0

On the one hand, 1 2

p

1 2

∫ |k1 (t)| dt ≤ ∫( 0

0

p

1 − (1 − t)α+1 − t α+1 ) dt r(α + 1) 1 2

1 p = p ∫(1 − (1 − t)α+1 − t α+1 ) dt p r (α + 1) 0

1 2

1 ≤ p ∫(1 − (1 − t)p(α+1) − t p(α+1) )dt r (α + 1)p 0

1 1 1 ( − ), = p r (α + 1)p 2 p(α + 1) + 1

(4.83)

and 1

∫ |k2 (t)|p dt ≤ 1 2

1 1 1 ( − ), r p (α + 1)p 2 p(α + 1) + 1

(4.84)

where we use the fact q

(1 − (1 − t)α+1 − t α+1 ) ≤ 1 − (1 − t)q(α+1) − t q(α+1) ,

(4.85)

for any t ∈ [0, 1], which follows from (A − B)q ≤ Aq − Bq , for any A > B ≥ 0 and q ≥ 1. On the other hand, 1

1 q 󵄨󵄨 b 󵄨󵄨󵄨 󵄨󵄨 󸀠󸀠 󵄨󵄨q 󵄨 ∫ 󵄨󵄨f (ta + (1 − t)b)󵄨󵄨 dt ≤ ∫[|f 󸀠󸀠 (a)|q t + m(1 − t)󵄨󵄨󵄨f 󸀠󸀠 ( )󵄨󵄨󵄨 ]dt 󵄨󵄨 m 󵄨󵄨 0

0

148 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 󵄨q 1 󵄨󵄨 󵄨󵄨 󸀠󸀠 b 󵄨󵄨󵄨 = |f (a)| ∫ tdt + m󵄨󵄨f ( )󵄨󵄨 ∫(1 − t)dt 󵄨󵄨 m 󵄨󵄨 =

󸀠󸀠

q

󸀠󸀠

q

1

0

|f (a)| +

m|f 󸀠󸀠 ( mb )|q

2

0

(4.86)

.

Finally, substituting (4.83), (4.84) and (4.86) into (4.82). The proof is completed. Remark 153. With the same assumptions as in Theorem 152. If |f 󸀠󸀠 (x)| ≤ M on [a, mb ], we obtain 󵄨󵄨 󵄨󵄨 f (a) + f (b) 2 a+b Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α + f( )− [ J + f (b) + RL Jb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 r(r + 1) r+1 2 r(b − a)α 󵄨 1

≤ where

1 p

+

1 q

1

p 2 1+m q M(b − a)2 (1 − ) ( ) r(α + 1) p(α + 1) + 1 2

= 1.

Another Hermit-Hadamard type inequality for powers in terms of the second derivatives is obtained as follows. Theorem 154. Let f : [0, b∗ ] → ℝ be a twice-differentiable mapping with b∗ > 0. If |f 󸀠󸀠 |q is measurable and m-convex on [a, mb ] for some fixed q > 1, 0 ≤ a < b and m ∈ (0, 1] with mb ≤ b∗ , r > 0, then 󵄨󵄨 f (a) + f (b) 󵄨󵄨 2 a+b Γ(α + 1) 󵄨󵄨 󵄨 + f( )− [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 r(r + 1) 󵄨󵄨 r+1 2 r(b − a) 1

1

󸀠󸀠 q 󸀠󸀠 b q (b − a)2 q(α + 1) − 1 q |f (a)| + m|f ( m )| q ( ) ( ) . ≤ r(α + 1) q(α + 1) + 1 2

Proof. From Lemma 57 and using the well-known Hölder’s inequality, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 2 a+b Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α + f( )− [ J + f (b) + RL Jb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 r(r + 1) r+1 2 r(b − a)α 󵄨 1 p

1

1

󵄨 󵄨q ≤ (b − a) (∫ 1dt) (∫ 󵄨󵄨󵄨k(t)f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) 2

1 q

0

0

1

1

q q 1 󵄨󵄨 b 󵄨󵄨󵄨 󵄨 ≤ (b − a) (|f (a)| ∫ t|k(t)| dt + m󵄨󵄨󵄨f 󸀠󸀠 ( )󵄨󵄨󵄨 ∫(1 − t)|k(t)|q dt) . 󵄨󵄨 m 󵄨󵄨

2

󸀠󸀠

q

q

0

0

Calculating by parts, we have 1

1 2

0

q

1

∫ t|k(t)| dt = ∫ t|k1 (t)| dt + ∫ t|k2 (t)|q dt 0

q

1 2

(4.87)

4.4 Inequalities via m-convex functions | 149

≤

1 1 1 [ − ], r q (α + 1)q 2 q(α + 1) + 1

1

1 2

0

0

(4.88)

1

∫(1 − t)|k(t)|q dt = ∫(1 − t)|k1 (t)|q dt + ∫(1 − t)|k2 (t)|q dt

≤

1 2

1 1 1 [ − ]. r q (α + 1)q 2 q(α + 1) + 1

(4.89)

Substituting (4.88) and (4.89) to (4.87) via (4.85), one can obtain the result. The proof is completed. Remark 155. With the same assumptions in for Theorem 154. If |f 󸀠󸀠 (x)| ≤ M on [a, mb ], we obtain 󵄨󵄨 f (a) + f (b) 󵄨󵄨 2 a+b Γ(α + 1) 󵄨󵄨 󵄨 + f( )− [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 r+1 2 r(b − a) 󵄨󵄨 r(r + 1) 1

1

M(b − a)2 q(α + 1) − 1 q 1 + m q ( ) ( ) ≤ r(α + 1) q(α + 1) + 1 2 where

1 p

+

1 q

= 1.

Remark 156. From Theorems 150, 152 and 154, we have 󵄨󵄨 󵄨󵄨 f (a) + f (b) 2 a+b Γ(α + 1) 󵄨 󵄨󵄨 + f( )− [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 ≤ min{K1 , K2 , K3 } 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 r(r + 1) r+1 2 r(b − a) where 1

|f 󸀠󸀠 (a)|q + m|f 󸀠󸀠 ( mb )|q q α 1 K1 = (b − a) ( + )( ) , r(α + 1)(α + 2) 4(r + 1) 2 2

1

1

p |f 󸀠󸀠 (a)|q + m|f 󸀠󸀠 ( mb )|q q (b − a)2 2 K2 = (1 − ) ( ) , r(α + 1) p(α + 1) + 1 2 1

K3 =

1

󸀠󸀠 q 󸀠󸀠 b q (b − a)2 q(α + 1) − 1 q |f (a)| + m|f ( m )| q ( ) ( ) . r(α + 1) q(α + 1) + 1 2

From Theorem 152 and Theorem 154, we use a particular technique for shrinking an inequality and now we use another shrinking technique. Theorem 157. Let f : [0, b∗ ] → ℝ be a twice-differentiable mapping with b∗ > 0. If |f 󸀠󸀠 |q is measurable and m-convex on [a, mb ] for some fixed q > 1, 0 ≤ a < b and m ∈ (0, 1] with mb ≤ b∗ , r > 0, then 󵄨󵄨 f (a) + f (b) 󵄨󵄨 2 a+b Γ(α + 1) 󵄨 󵄨󵄨 + f( )− [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 r+1 2 r(b − a) 󵄨󵄨 r(r + 1)

150 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

p+1

p (b − a)2 1 2 r+1 ≤( ) [( + ) p+1 r+1 2 r(α + 1)

×( where

1 p

+

1 q

|f 󸀠󸀠 (a)|q + m|f 󸀠󸀠 ( mb )|q 2

1 q

1 p

p+1

r+1 −( ) r(α + 1)

]

)

= 1.

Proof. From Lemma 57 and using the well-known Hölder’s inequality, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 2 a+b Γ(α + 1) 󵄨󵄨 󵄨 + f( )− [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 r+1 2 r(b − a) 󵄨󵄨 r(r + 1) 1 p

1

1

1 q

󵄨q

󵄨 ≤ (b − a)2 (∫ |k(t)|p dt) (∫ 󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) .

(4.90)

0

0

Note that (1 − t)α+1 + t α+1 ≤ 1 for any t ∈ [0, 1]. Calculating by parts, we find 1 2

1 2

0

0

∫ |k1 (t)|p dt ≤ ∫( 1 2

p

1 − (1 − t)α+1 − t α+1 t + ) dt r(α + 1) r+1 p

≤ ∫( 0

1 t + ) dt r(α + 1) r + 1 1 2

=

p

1 r+1 + t) dt, ∫( (r + 1)p r(α + 1) 0

1 2

p

∫( 0

p+1

1 1 r+1 r+1 + t) dt = ( + ) r(α + 1) p + 1 2 r(α + 1)

p+1

−

1 r+1 ( ) p + 1 r(α + 1)

,

and 1

1

∫ |k2 (t)|p dt ≤ ∫( 1 2

1 2

1

≤ ∫( 1 2

p

1 − (1 − t)α+1 − t α+1 1 − t + ) dt r(α + 1) r+1 p

1 1−t + ) dt r(α + 1) r + 1 1

p

1 r+1 = + 1 − t) dt, ∫( (r + 1)p r(α + 1) 1 2

1

∫( 1 2

p

p+1

r+1 1 1 r+1 + 1 − t) dt = ( + ) r(α + 1) p + 1 2 r(α + 1)

p+1

−

1 r+1 ( ) p + 1 r(α + 1)

.

4.4 Inequalities via m-convex functions | 151

Thus, 1

∫ |k(t)|p dt ≤ 0

p+1

1 r+1 2 [( + ) (r + 1)p (p + 1) 2 r(α + 1)

p+1

−(

r+1 ) r(α + 1)

(4.91)

].

Moreover, 1 1 q 󵄨󵄨 b 󵄨󵄨󵄨 󵄨 󵄨q 󵄨 ∫ 󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt ≤ ∫(|f 󸀠󸀠 (a)|q t + m(1 − t)󵄨󵄨󵄨f 󸀠󸀠 ( )󵄨󵄨󵄨 )dt 󵄨󵄨 m 󵄨󵄨 0

0

1

󵄨󵄨 󵄨q 1 󵄨󵄨 󸀠󸀠 b 󵄨󵄨󵄨 = |f (a)| ∫ tdt + m󵄨󵄨f ( )󵄨󵄨 ∫(1 − t)dt 󵄨󵄨 m 󵄨󵄨 =

󸀠󸀠

q

󸀠󸀠

q

0

|f (a)| +

m|f 󸀠󸀠 ( mb )|q

2

0

(4.92)

.

Now substituting (4.91) and (4.92) into (4.90). The proof is completed. Remark 158. With the same assumptions in Theorem 157. If |f 󸀠󸀠 (x)| ≤ M on [a, mb ], we obtain 󵄨󵄨 f (a) + f (b) 󵄨󵄨 2 a+b Γ(α + 1) 󵄨󵄨 󵄨 + f( )− [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 r+1 2 r(b − a) 󵄨󵄨 r(r + 1) 1

p+1

p 2 1 r+1 M(b − a)2 ( ) [( + ) ≤ r+1 p+1 2 r(α + 1)

where

1 p

+

1 q

p+1

r+1 −( ) r(α + 1)

1 p

1

1+m q ] ( ) 2

= 1.

Another Hermit-Hadamard type inequality for powers in terms of the second derivatives is obtained as follows. Theorem 159. Let f : [0, b∗ ] → ℝ be a twice-differentiable mapping with b∗ > 0. If |f 󸀠󸀠 |q is measurable and m-convex on [a, mb ] for some fixed q > 1, 0 ≤ a < b and m ∈ (0, 1] with mb ≤ b∗ , r > 0, then 󵄨󵄨 f (a) + f (b) 󵄨󵄨 2 a+b Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α + f( )− [ J f (b) + J f (a)] + − 󵄨󵄨 󵄨󵄨 RL RL a b 󵄨󵄨 r(r + 1) 󵄨󵄨 r+1 2 r(b − a)α 1

q+1

q (b − a)2 1 r+1 2 ) [( + ) ≤( q+1 r+1 2 r(α + 1)

×(

|f 󸀠󸀠 (a)|q + m|f 󸀠󸀠 ( mb )|q 2

1 q

1 q

q+1

r+1 −( ) r(α + 1)

]

) .

Proof. From Lemma 57 and using the well-known Hölder’s inequality, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 2 a+b Γ(α + 1) 󵄨󵄨 󵄨 + f( )− [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 r+1 2 r(b − a) 󵄨󵄨 r(r + 1)

152 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1 p

1

1

󵄨q 󵄨 ≤ (b − a)2 (∫ 1dt) (∫ 󵄨󵄨󵄨k(t)f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) 0

1 q

0

1

q q 1 󵄨󵄨 b 󵄨󵄨󵄨 󵄨 ≤ (b − a) (|f (a)| ∫ t|k(t)| dt + m󵄨󵄨󵄨f 󸀠󸀠 ( )󵄨󵄨󵄨 ∫(1 − t)|k(t)|q dt) . 󵄨󵄨 m 󵄨󵄨

2

󸀠󸀠

q

1

q

0

0

Calculating by parts, we have 1

1 2

0

0

1

∫ t|k(t)|q dt = ∫ t|k1 (t)|q dt + ∫ t|k2 (t)|q dt 1 2

with 1 2

1 2

0

0

∫ t|k1 (t)|q dt ≤ ∫( 1 2

≤ ∫( 0

q

1 − (1 − t)α+1 − t α+1 t + ) tdt r(α + 1) r+1 q

1 t + ) tdt r(α + 1) r + 1 1 2

=

q

1 r+1 + t) tdt, ∫( (r + 1)p r(α + 1) 0

and 1

1

∫ t|k2 (t)|q dt ≤ ∫( 1 2

1 2

1

≤ ∫( 1 2

q

1 − (1 − t)α+1 − t α+1 1 − t + ) tdt r(α + 1) r+1 q

1−t 1 + ) tdt r(α + 1) r + 1 1

q

1 r+1 = + 1 − t) tdt, ∫( (r + 1)q r(α + 1) 1 2

where 1 2

∫( 0

q

q+1

r+1 1 1 r+1 + t) tdt = ( + ) r(α + 1) 2(q + 1) 2 r(α + 1)

q+2

−

1 1 r+1 ( + ) (q + 1)(q + 2) 2 r(α + 1)

(4.93)

4.4 Inequalities via m-convex functions | 153 q+2

+

r+1 1 ( ) (q + 1)(q + 2) r(α + 1)

,

and 1

q

∫( 1 2

=−

r+1 + 1 − t) tdt r(α + 1) q+1

1 r+1 ( ) q + 1 r(α + 1)

+

q+2

−

1 r+1 ( ) (q + 1)(q + 2) r(α + 1) q+1

1 1 r+1 ( + ) 2(q + 1) 2 r(α + 1)

+

q+2

1 1 r+1 ( + ) (q + 1)(q + 2) 2 r(α + 1)

.

Thus, 1

∫ t|k(t)|q dt = 0

q+1

1 1 r+1 [( + ) (q + 1)(r + 1)q 2 r(α + 1)

q+1

−(

r+1 ) r(α + 1)

(4.94)

].

Clearly, 1

∫(1 − t)|k(t)|q dt 0 1 2

1

= ∫(1 − t)|k1 (t)|q dt + ∫(1 − t)|k2 (t)|q dt 0

≤

1 2

q+1

1 1 r+1 [( + ) (q + 1)(r + 1)q 2 r(α + 1)

q+1

−(

r+1 ) r(α + 1)

(4.95)

].

Now, substituting (4.94) and (4.95) into (4.93), one can obtain the result. The proof is completed. Remark 160 (with the same assumptions as in Theorem 159). If |f 󸀠󸀠 (x)| ≤ M on [a, mb ], we obtain 󵄨󵄨 f (a) + f (b) 󵄨󵄨 2 a+b Γ(α + 1) 󵄨󵄨 󵄨 + f( )− [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 r+1 2 r(b − a) 󵄨󵄨 r(r + 1) 1

q+1

q M(b − a)2 1 r+1 2 ) [( + ) ≤( q+1 r+1 2 r(α + 1)

where

1 p

+

1 q

q+1

r+1 −( ) r(α + 1)

1 q

1

1+m q ] ( ) , 2

= 1.

Remark 161. From Theorems 157 and 159, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 2 a+b Γ(α + 1) 󵄨 󵄨󵄨 + f( )− [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 ≤ min{N1 , N2 } 󵄨󵄨 α 󵄨󵄨 r+1 2 r(b − a) 󵄨󵄨 r(r + 1)

154 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals where 1

N1 = (

p+1

p (b − a)2 1 r+1 2 ) [( + ) p+1 r+1 2 r(α + 1)

×(

|f 󸀠󸀠 (a)|q + m|f 󸀠󸀠 ( mb )|q 2

1

1 q

×(

q+1

2

r+1 ) r(α + 1)

]

) ,

q (b − a)2 1 r+1 2 ) [( + ) N2 = ( q+1 r+1 2 r(α + 1)

|f 󸀠󸀠 (a)|q + m|f 󸀠󸀠 ( mb )|q

1 p

p+1

−(

1 q

1 q

q+1

r+1 −( ) r(α + 1)

]

) .

4.5 Inequalities via (s, m)-convex functions The results in this section are adopted from [216, 243, 95].

4.5.1 Main results Theorem 162. Let f : [a, b] → ℝ be a twice-differentiable mapping with a < b. If |f 󸀠󸀠 |q is measurable and (s, m)-convex on [a, b] for some fixed q ≥ 1 and (s, m) ∈ (0, 1]2 , then the following inequality for fractional integrals holds: 󵄨󵄨 󵄨󵄨 f (a) + f (mb) Γ(α + 1) 󵄨󵄨 󵄨󵄨 α − [RL Jaα+ f (mb) + RL Jmb − f (a)]󵄨 󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(mb − a) 󵄨 1− q1

(mb − a)2 α ≤ ( ) 2(α + 1) α + 2

[|f 󸀠󸀠 (a)|q (

1 1 − B(s + 1, α + 2) − ) s+1 α+s+2 1

q 2 1 1 − + B(s + 1, α + 2) + )] , + m|f (b)| (1 − s+1 α+2 α+s+2

󸀠󸀠

q

where B(x, y), x, y > 0 is the beta function. Proof. First, we suppose that q = 1. From Lemma 48, we have 󵄨󵄨 f (a) + f (mb) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α − [RL Jaα+ f (mb) + RL Jmb − f (a)]󵄨 󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(mb − a) 󵄨 1

(mb − a)2 1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 ≤ ∫ 󵄨󵄨f (ta + m(1 − t)b)󵄨󵄨󵄨dt. 2 α+1 0

Since |f 󸀠󸀠 | is (s, m)-convex on [a, b], we know that for any t ∈ [0, 1] 󵄨󵄨 󸀠󸀠 󵄨 s 󸀠󸀠 s 󸀠󸀠 󵄨󵄨f (ta + m(1 − t)b)󵄨󵄨󵄨 ≤ t |f (a)| + m(1 − t )|f (b)|.

(4.96)

4.5 Inequalities via (s, m)-convex functions | 155

Therefore 󵄨󵄨 󵄨󵄨 f (a) + f (mb) Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J + J f (a)] f (mb) − + 󵄨󵄨 󵄨󵄨 RL RL mb a 󵄨󵄨 󵄨󵄨 2 2(mb − a)α 1

(mb − a)2 1 − (1 − t)α+1 − t α+1 s 󸀠󸀠 ≤ (t |f (a)| + m(1 − t s )|f 󸀠󸀠 (b)|)dt ∫ 2 α+1 0

2

=

1 1 (mb − a) [|f 󸀠󸀠 (a)|( − B(s + 1, α + 2) − ) 2(α + 1) s+1 α+s+2 2 1 1 − + B(s + 1, α + 2) + )], + m|f 󸀠󸀠 (b)|(1 − s+1 α+2 α+s+2

which completes the proof for this case. Second, we suppose that q > 1. Using Lemma 48 and the power mean inequality for q, we obtain 1

󵄨 󵄨 ∫(1 − (1 − t)α+1 − t α+1 )󵄨󵄨󵄨f 󸀠󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨dt 0

1− q1

1

≤ (∫(1 − (1 − t)α+1 − t α+1 )dt) 0 1

󵄨q

1 q

󵄨 × (∫(1 − (1 − t)α+1 − t α+1 )󵄨󵄨󵄨f 󸀠󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨 dt) . 0

Since |f 󸀠󸀠 | is (s, m)-convex on [a, b], we know that for any t ∈ [0, 1] 󵄨󵄨 󸀠󸀠 󵄨q s 󸀠󸀠 q s 󸀠󸀠 q 󵄨󵄨f (ta + (1 − t)b)󵄨󵄨󵄨 ≤ t |f (a)| + m(1 − t )|f (b)| . Hence, from (4.96) and (4.97), we obtain 󵄨󵄨 󵄨󵄨 f (a) + f (mb) Γ(α + 1) 󵄨󵄨 󵄨󵄨 α − [RL Jaα+ f (mb) + RL Jmb − f (a)]󵄨 󵄨󵄨 󵄨󵄨 α 2 2(mb − a) 󵄨 󵄨󵄨 1

(mb − a)2 ≤ (∫(1 − (1 − t)α+1 − t α+1 )dt) 2(α + 1)

1− q1

0

1

α+1

× (∫(1 − (1 − t) 0 1

−t

α+1

󵄨 󵄨q )󵄨󵄨󵄨f 󸀠󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨 dt)

(mb − a)2 ≤ (∫(1 − (1 − t)α+1 − t α+1 )dt) 2(α + 1) 0

1− q1

1 q

(4.97)

156 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

× (∫(1 − (1 − t)α+1 − t α+1 )[t s |f 󸀠󸀠 (a)|q + m(1 − t s )|f 󸀠󸀠 (b)|q ]dt) 0

=

1− q1

(mb − a)2 α ( ) 2(α + 1) α + 2

[|f 󸀠󸀠 (a)|q (

1 q

1 1 − B(s + 1, α + 2) − ) s+1 α+s+2 1

q 1 2 1 + m|f (b)| (1 − − + B(s + 1, α + 2) + )] . s+1 α+2 α+s+2

q

󸀠󸀠

The proof is completed. Corollary 163. In Theorem 162, if we choose s = m = 1, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 1− q1

α (b − a)2 ( ) ≤ 2(α + 1) α + 2

1 1 [|f 󸀠󸀠 (a)|q ( − B(2, α + 2) − ) 2 α+3 1

q 2 1 1 + B(2, α + 2) + )] . + |f (b)| ( − 2 α+2 α+3

󸀠󸀠

q

Theorem 164. Let f : [a, b] → ℝ be a twice-differentiable mapping with a < b. If |f 󸀠󸀠 |q is measurable and (s, m)-convex on [a, b] for some fixed q > 1 and (s, m) ∈ (0, 1]2 , then the following inequality for fractional integrals holds: 󵄨󵄨 󵄨󵄨 f (a) + f (mb) Γ(α + 1) 󵄨󵄨 󵄨󵄨 α [RL Jaα+ f (mb) + RL Jmb − − f (a)]󵄨 󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(mb − a) 󵄨 1

1

p q 1 s (mb − a)2 2 ≤ (1 − ) (|f 󸀠󸀠 (a)|q + m|f 󸀠󸀠 (b)|q ) , 2(α + 1) p(α + 1) + 1 s+1 s+1

where

1 p

+

1 q

= 1.

Proof. From Lemma 48 and using the well-known Hölder’s inequality, we have 󵄨󵄨 f (a) + f (mb) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α − [RL Jaα+ f (mb) + RL Jmb − f (a)]󵄨 󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(mb − a) 󵄨 1

(mb − a)2 1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 ≤ ∫ 󵄨󵄨f (ta + m(1 − t)b)󵄨󵄨󵄨dt 2 α+1 0

1

1 p

1

(mb − a)2 p 󵄨 󵄨q ≤ (∫(1 − (1 − t)α+1 − t α+1 ) dt) (∫ 󵄨󵄨󵄨f 󸀠󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨 dt) 2(α + 1) 0

1

0

(mb − a)2 ≤ (∫(1 − (1 − t)p(α+1) − t p(α+1) )dt) 2(α + 1) 0

1 p

1 q

4.5 Inequalities via (s, m)-convex functions | 157

1

1

× (|f 󸀠󸀠 (a)|q ∫ t s dt + m|f 󸀠󸀠 (b)|q ∫(1 − t s )dt) 0

=

1 p

1 q

0

1

q (mb − a)2 2 s 1 (1 − ) (|f 󸀠󸀠 (a)|q + m|f 󸀠󸀠 (b)|q ) . 2(α + 1) p(α + 1) + 1 s+1 s+1

The proof is completed. Corollary 165. In Theorem 164, if we choose s = m = 1, we obtain 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α [ J f (b) − + J f (a)] + − 󵄨󵄨 󵄨󵄨 RL RL a b 󵄨󵄨 󵄨󵄨 2 2(b − a)α 1

1

p (b − a)2 2 |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q ≤ (1 − ) ( ) . 2(α + 1) p(α + 1) + 1 2

Theorem 166. Let f : [a, b] → ℝ be a twice-differentiable mapping a < b. If |f 󸀠󸀠 |q is measurable and (s, m)-convex on [a, b] for some fixed q > 1 and (s, m) ∈ (0, 1]2 , then the following inequality for fractional integrals holds: 󵄨󵄨 󵄨󵄨 f (a) + f (mb) Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α [ J f (mb) − + J f (a)] + − 󵄨󵄨 󵄨󵄨 RL mb 󵄨󵄨 󵄨󵄨 2 2(mb − a)α RL a 2 (mb − a) 1 1 ≤ [|f 󸀠󸀠 (a)|q ( − B(s + 1, q(α + 1) + 1) − ) 2(α + 1) s+1 q(α + 1) + s + 1 2 1 − + m|f 󸀠󸀠 (b)|q (1 − s + 1 q(α + 1) + 1 1

+ B(s + 1, q(α + 1) + 1) +

q 1 )] . q(α + 1) + s + 1

Proof. From Lemma 48 and using the well-known Hölder’s inequality, we have 󵄨󵄨 󵄨󵄨 f (a) + f (mb) Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J f (mb) + J f (a)] − + 󵄨󵄨 󵄨󵄨 RL RL a mb 󵄨󵄨 󵄨󵄨 2 2(mb − a)α 1

(mb − a)2 1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 ≤ ∫ 󵄨󵄨f (ta + m(1 − t)b)󵄨󵄨󵄨dt 2 α+1 0

1

1 p

1

(mb − a)2 q󵄨 󵄨q ≤ (∫ 1dt) (∫(1 − (1 − t)α+1 − t α+1 ) 󵄨󵄨󵄨f 󸀠󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨 dt) 2(α + 1) 0

≤

1

0

(mb − a)2 (|f 󸀠󸀠 (a)|q ∫[t s − (1 − t)q(α+1) t s − t q(α+1)+s ]dt 2(α + 1) 0

1

+ m|f 󸀠󸀠 (b)|q ∫[(1 − t s ) − (1 − t)q(α+1) (1 − t s ) − t q(α+1) (1 − t s )]dt) 0

1 q

1 q

158 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals

=

(mb − a)2 󸀠󸀠 1 1 [|f (a)|q ( − B(s + 1, q(α + 1) + 1) − ) 2(α + 1) s+1 q(α + 1) + s + 1 1 2 + m|f 󸀠󸀠 (b)|q (1 − − s + 1 q(α + 1) + 1 1

q 1 )] . + B(s + 1, q(α + 1) + 1) + q(α + 1) + s + 1

The proof is completed. Corollary 167. In Theorem 166, if we choose s = m = 1, we obtain 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) 󵄨 󵄨󵄨 − [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 2 (b − a) 1 1 ≤ [|f 󸀠󸀠 (a)|q ( − B(2, q(α + 1) + 1) − ) 2(α + 1) 2 q(α + 1) + 2

1

q 1 2 1 + |f (b)| ( − + B(2, q(α + 1) + 1) + )] . 2 q(α + 1) + 1 q(α + 1) + 2

q

󸀠󸀠

Remark 168. From Theorems 162, 164 and 166, we have 󵄨󵄨 f (a) + f (mb) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α [RL Jaα+ f (mb) + RL Jmb − − f (a)]󵄨 ≤ min{N1 , N2 , N3 }, 󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(mb − a) 󵄨 where N1 =

1− q1

α (mb − a)2 ( ) 2(α + 1) α + 2

[|f 󸀠󸀠 (a)|q (

1 1 − B(s + 1, α + 2) − ) s+1 α+s+2 1

q 1 2 1 + m|f (b)| (1 − − + B(s + 1, α + 2) + )] , s+1 α+2 α+s+2

󸀠󸀠

q

1

1

p q (mb − a)2 2 1 s N2 = (1 − ) (|f 󸀠󸀠 (a)|q + m|f 󸀠󸀠 (b)|q ) , 2(α + 1) p(α + 1) + 1 s+1 s+1

N3 =

(mb − a)2 󸀠󸀠 1 1 [|f (a)|q ( − B(s + 1, q(α + 1) + 1) − ) 2(α + 1) s+1 q(α + 1) + s + 1 1 2 + m|f 󸀠󸀠 (b)|q (1 − − s + 1 q(α + 1) + 1 1

+ B(s + 1, q(α + 1) + 1) +

q 1 )] . q(α + 1) + s + 1

Theorem 169. Let f : [a, b] → ℝ be a twice-differentiable mapping with a < b. If |f 󸀠󸀠 |q is measurable and (s, m)-convex on [a, b] for some fixed q ≥ 1 and (s, m) ∈ (0, 1]2 , r > 0, then 󵄨󵄨 f (a) + f (mb) 󵄨󵄨 2 a + mb Γ(α + 1) 󵄨󵄨 󵄨󵄨 α + f( )− [RL Jaα+ f (mb) + RL Jmb − f (a)]󵄨 󵄨󵄨 󵄨󵄨 α r+1 2 r(mb − a) 󵄨 󵄨󵄨 r(r + 1)

4.5 Inequalities via (s, m)-convex functions | 159 1− q1

α 1 ≤ (mb − a) ( + ) r(α + 1)(α + 2) 4(r + 1) 2

1

q α 1 × [|f (a)| I + m|f (b)| ( + − I)] , r(α + 1)(α + 2) 4(r + 1)

󸀠󸀠

q

q

󸀠󸀠

(4.98)

where I=

1 1 − B(s + 1, α + 2) r(s + 1)(s + α + 2) r(α + 1) s+1

1 1 (1 − ( ) (r + 1)(s + 1)(s + 2) 2

+

).

(4.99)

Proof. Case 1: we suppose that q = 1. From Lemma 58, we have 󵄨󵄨 󵄨󵄨 f (a) + f (b) 2 a+b Γ(α + 1) 󵄨 󵄨󵄨 [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 + f( )− 󵄨󵄨 α 󵄨󵄨 r+1 2 r(b − a) 󵄨󵄨 r(r + 1) 1

󵄨 󵄨 ≤ (mb − a)2 ∫ 󵄨󵄨󵄨k(t)f 󸀠󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨dt.

(4.100)

0

Since |f 󸀠󸀠 | is (s, m)-convex on [a, b], we know that for any t ∈ [0, 1] 󵄨󵄨 󸀠󸀠 󵄨 s 󸀠󸀠 s 󸀠󸀠 󵄨󵄨f (ta + m(1 − t)b)󵄨󵄨󵄨 ≤ t |f (a)| + m(1 − t )|f (b)|. Therefore (4.100) turns into 󵄨󵄨 󵄨󵄨 f (a) + f (b) 2 a+b Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α + f( )− [ J + f (b) + RL Jb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 r(r + 1) r+1 2 r(b − a)α 󵄨 2 ≤ (mb − a) (I1 + I2 ), where 1 2

I1 = ∫ |k1 (t)|(t s |f 󸀠󸀠 (a)| + m(1 − t s )|f 󸀠󸀠 (b)|)dt, 0

1

I2 = ∫ |k2 (t)|(t s |f 󸀠󸀠 (a)| + m(1 − t s )|f 󸀠󸀠 (b)|)dt. 1 2

Calculating by parts, we have 1 2

I1 ≤ ∫[ 0

1 − (1 − t)α+1 − t α+1 t + ](t s |f 󸀠󸀠 (a)| + m(1 − t s )|f 󸀠󸀠 (b)|)dt r(α + 1) r+1 s+1

|f 󸀠󸀠 (a)| 1 1 ≤ [ ( ) r(α + 1) s + 1 2

s+α+2

1 1 − ( ) s+α+2 2

1 2

− ∫ t s (1 − t)α+1 dt] 0

160 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals s+2

+

1 |f 󸀠󸀠 (a)| 1 ( ) r+1 s+2 2

α+s+2

1 1 + ( ) α+s+2 2

and 1

I2 ≤ ∫[ 1 2

m|f 󸀠󸀠 (b)| 1 1 1 [ − ( ) r(α + 1) 2 α + 2 2 α+2

1 1 + ( ) α+2 2

m|f (b)| 1 1 1 [ − ( ) r+1 8 s+2 2

s+1

1 1 ( ) s+1 2

−

1 2

1 − + ∫(1 − t)α+1 t s dt] α+2 0

s+2

󸀠󸀠

+

α+2

+

],

1 − (1 − t)α+1 − t α+1 1 − t s 󸀠󸀠 + ](t |f (a)| + m(1 − t s )|f 󸀠󸀠 (b)|)dt r(α + 1) r+1 s+1

≤

|f 󸀠󸀠 (a)| 1 1 1 1 [ − − ( ) r(α + 1) s + 1 s + α + 2 s + 1 2 s+α+2

+

1 1 ( ) s+α+2 2

1

− ∫ t s (1 − t)α+1 dt] 1 2

s+1

s+2

+

|f 󸀠󸀠 (a)| 1 1 1 1 [ − − ( ) r+1 s+1 s+2 s+1 2

+

m|f 󸀠󸀠 (b)| 1 1 1 1 1 1 [ − − + + ( ) r(α + 1) 2 α + 2 s + 1 α + s + 2 α + 2 2

−

1 1 ( ) α+s+2 2

α+s+2

+

α+2

−

1 1 ( ) α+2 2

+

1 1 ( ) s+2 2

] α+2

s+1

+

1 1 ( ) s+1 2

1

+ ∫(1 − t)α+1 t s dt] 1 2

s+1

1 1 1 1 m|f 󸀠󸀠 (b)| 1 [ − + + ( ) r+1 8 s+1 s+2 s+1 2

s+2

−

1 1 ( ) s+2 2

].

Therefore 󵄨󵄨 f (a) + f (mb) 󵄨󵄨 2 a + mb Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α + f( )− [ J + f (mb) + RL Jmb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 r(r + 1) r+1 2 r(mb − a)α 󵄨 1

α+1 |f 󸀠󸀠 (a)| ( − ∫ t s (1 − t)α+1 dt)] ≤ (mb − a) [ r(α + 1) (s + 1)(s + α + 2) 2

0

s+1

+ (mb − a)2 [

|f (a)| 1 1 1 ( − ( ) r + 1 (s + 1)(s + 2) (s + 1)(s + 2) 2

+ (mb − a)2 [

m|f 󸀠󸀠 (b)| 2 1 1 (1 − − + + ∫(1 − t)α+1 t s dt)] r(α + 1) α+2 s+1 α+s+2

+ (mb − a)2 [

m|f (b)| 1 1 1 1 ( − + ( ) r+1 4 (s + 1)(s + 2) (s + 1)(s + 2) 2

󸀠󸀠

)] 1

0

󸀠󸀠

s+1

)]

4.5 Inequalities via (s, m)-convex functions | 161 1

1 1 − = (mb − a) [|f (a)|( ∫ t s (1 − t)α+1 dt r(s + 1)(s + α + 2) r(α + 1) 2

󸀠󸀠

0

s+1

+

1 1 (1 − ( ) (r + 1)(s + 1)(s + 2) 2

+ (mb − a)2 [m|f 󸀠󸀠 (b)|(

))]

1 2 1 − − r(α + 1) r(α + 1)(α + 2) r(s + 1)(α + s + 2)

1

1 1 1 + − ∫ t s (1 − t)α+1 dt + r(α + 1) 4(r + 1) (r + 1)(s + 1)(s + 2) 0

s+1

1 × (1 − ( ) 2

))],

1

because ∫0 t s (1 − t)α+1 dt = B(s + 1, α + 2). Note that (4.99), one can derive 󵄨󵄨 󵄨󵄨 f (a) + f (mb) 2 a + mb Γ(α + 1) 󵄨󵄨 󵄨󵄨 α [RL Jaα+ f (mb) + RL Jmb + f( )− − f (a)]󵄨 󵄨󵄨 󵄨󵄨 α 󵄨󵄨 r(r + 1) r+1 2 r(mb − a) 󵄨 1 α 2 󸀠󸀠 󸀠󸀠 + − I)], ≤ (mb − a) [|f (a)|I + m|f (b)|( r(α + 1)(α + 2) 4(r + 1) which completes the proof for this case. Case 2: Suppose that q > 1. Using Lemma 58 and the power mean inequality for q, we obtain 1

󵄨 󵄨 ∫ 󵄨󵄨󵄨k(t)f 󸀠󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨dt 0

1

1− q1

≤ (∫ |k(t)|dt)

1

1 q

󵄨q

󵄨 × (∫ 󵄨󵄨󵄨k(t)f 󸀠󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨 dt) .

(4.101)

0

0

Since |f 󸀠󸀠 | is (s, m)-convex on [a, b], we know that for any t ∈ [0, 1] 󵄨󵄨 󸀠󸀠 󵄨q s 󸀠󸀠 q s 󸀠󸀠 q 󵄨󵄨f (ta + (1 − t)b)󵄨󵄨󵄨 ≤ t |f (a)| + m(1 − t )|f (b)| .

(4.102)

Hence, from (4.101) and (4.102), we obtain 󵄨󵄨 󵄨󵄨 f (a) + f (mb) 2 a + mb Γ(α + 1) 󵄨󵄨 󵄨󵄨 α + f( )− [RL Jaα+ f (mb) + RL Jmb − f (a)]󵄨 󵄨󵄨 󵄨󵄨 α r+1 2 r(mb − a) 󵄨 󵄨󵄨 r(r + 1) 1

≤ (mb − a)2 (∫ |k(t)|dt) 0

1− q1

1

󵄨 󵄨q × (∫ 󵄨󵄨󵄨k(t)f 󸀠󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨 dt) 0

1 q

162 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1− q1

1

= (mb − a)2 (∫ |k(t)|dt) 0 1 2

1 q

1

󵄨q 󵄨q 󵄨 󵄨 × (∫ |k1 (t)|󵄨󵄨󵄨f 󸀠󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨 dt + ∫ |k2 (t)|󵄨󵄨󵄨f 󸀠󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨 dt) . 0

1 2

Calculating by parts, we have 1

∫ |k(t)|dt ≤ 0

α 1 + , r(α + 1)(α + 2) 4(r + 1)

and 1 2

󵄨 󵄨q ∫ |k1 (t)|󵄨󵄨󵄨f 󸀠󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨 dt 0

1 2

≤ ∫[ 0

1 − (1 − t)α+1 − t α+1 t + ](t s |f 󸀠󸀠 (a)|q + m(1 − t s )|f 󸀠󸀠 (b)|q )dt r(α + 1) r+1 s+1

s+α+2

|f 󸀠󸀠 (a)|q 1 1 = [ ( ) r(α + 1) s + 1 2

s+2

q

1 |f (a)| 1 ( ) r+1 s+2 2 󸀠󸀠

+

1 1 − ( ) s+α+2 2

α+s+2

+

1 1 ( ) α+s+2 2

+

m|f (b)| 1 1 1 [ − ( ) r+1 8 s+2 2 󸀠󸀠

0

1 2

1 1 ( ) α+2 2

−

0

],

and 1

󵄨 󵄨q ∫ |k2 (t)|󵄨󵄨󵄨f 󸀠󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨 dtdt 1 2

s+1

≤

1 1 1 1 |f 󸀠󸀠 (a)|q [ − − ( ) r(α + 1) s + 1 s + α + 2 s + 1 2 s+α+2

1 1 + ( ) s+α+2 2

1

− ∫ t s (1 − t)α+1 dt] 1 2

s+1

−

1 1 ( ) s+1 2

1 + ∫(1 − t)α+1 t s dt] α+2

s+2

q

α+2

m|f (b)| 1 1 1 [ − ( ) r(α + 1) 2 α + 2 2 α+2

+

− ∫ t s (1 − t)α+1 dt]

q

󸀠󸀠

+

1 2

4.5 Inequalities via (s, m)-convex functions | 163 s+1

+ +

+

1 1 ( ) s+2 2

] α+2

s+1

m|f 󸀠󸀠 (b)|q 1 1 1 1 1 1 [ − − + + ( ) r(α + 1) 2 α + 2 s + 1 α + s + 2 α + 2 2 α+s+2

1 1 − ( ) α+s+2 2 +

s+2

1 1 1 1 |f 󸀠󸀠 (a)|q [ − − ( ) r+1 s+1 s+2 s+1 2

α+2

1 1 − ( ) α+2 2

+

1 1 ( ) s+1 2

1

+ ∫(1 − t)α+1 t s dt] 1 2

s+1

1 1 1 1 m|f 󸀠󸀠 (b)|q 1 [ − + + ( ) r+1 8 s+1 s+2 s+1 2

s+2

−

1 1 ( ) s+2 2

].

Therefore, using the above facts, one can obtain (4.98). The proof is completed. Remark 170. In Theorem 169, if we choose s = m = 1, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 2 a+b Γ(α + 1) 󵄨󵄨 󵄨 + f( )− [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 r(r + 1) 󵄨󵄨 r+1 2 r(b − a) 1− q1

1 α + ) ≤ (b − a) ( r(α + 1)(α + 2) 4(r + 1) 2

× [|f 󸀠󸀠 (a)|q I 󸀠 + |f 󸀠󸀠 (b)|q (

1

q α 1 + − I 󸀠 )] r(α + 1)(α + 2) 4(r + 1)

where I󸀠 =

1 1 1 − B(2, α + 2) + . 2r(α + 3) r(α + 1) 8(r + 1)

Theorem 171. Let f : [a, b] → ℝ be a twice-differentiable mapping with a < b. If |f 󸀠󸀠 |q is measurable and (s, m)-convex on [a, b] for some fixed q > 1 and (s, m) ∈ (0, 1]2 , r > 0, then 󵄨󵄨 f (a) + f (mb) 󵄨󵄨 2 a + mb Γ(α + 1) 󵄨󵄨 󵄨󵄨 α + f( )− [RL Jaα+ f (mb) + RL Jmb − f (a)]󵄨 󵄨󵄨 󵄨󵄨 α r+1 2 r(mb − a) 󵄨󵄨 r(r + 1) 󵄨 1

1

p q 1 2 ms 󸀠󸀠 (mb − a)2 (1 − ) ( |f 󸀠󸀠 (a)|q + |f (b)|q ) , ≤ r(α + 1) p(α + 1) + 1 s+1 s+1

where

1 p

+

1 q

= 1.

Proof. From Lemma 58 and using the well-known Hölder’s inequality, we have 󵄨󵄨 f (a) + f (mb) 󵄨󵄨 2 a + mb Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α + f( )− [ J + f (mb) + RL Jmb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 r(r + 1) r+1 2 r(mb − a)α 󵄨 2

1

󵄨 󵄨 ≤ (mb − a) ∫ 󵄨󵄨󵄨k(t)f 󸀠󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨dt 0

164 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1 p

1

1 q

1

󵄨q 󵄨 ≤ (mb − a)2 (∫ |k(t)|p dt) (∫ 󵄨󵄨󵄨f 󸀠󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨 dt) . 0

0

Note that 1

∫ |k(t)|p dt ≤ 0

2 1 (1 − ), r p (α + 1)p p(α + 1) + 1

where we use (1 − (1 − t)α+1 − t α+1 )q ≤ 1 − (1 − t)q(α+1) − t q(α+1) , for any t ∈ [0, 1]. Moreover, 1

1

1

0

0

󵄨 󵄨q ∫󵄨󵄨󵄨f 󸀠󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨 dt ≤ |f 󸀠󸀠 (a)|q ∫ t s dt + m|f 󸀠󸀠 (b)|q ∫(1 − t s )dt 0

1 ms 󸀠󸀠 = |f 󸀠󸀠 (a)|q + |f (b)|q . s+1 s+1

Therefore, 󵄨󵄨 󵄨󵄨 f (a) + f (mb) 2 a + mb Γ(α + 1) 󵄨󵄨 󵄨󵄨 α [RL Jaα+ f (mb) + RL Jmb + f( )− − f (a)]󵄨 󵄨󵄨 󵄨󵄨 α r+1 2 r(mb − a) 󵄨 󵄨󵄨 r(r + 1) 1

1

p q (mb − a)2 2 1 ms 󸀠󸀠 ≤ (1 − ) ( |f 󸀠󸀠 (a)|q + |f (b)|q ) . r(α + 1) p(α + 1) + 1 s+1 s+1

The proof is completed. Remark 172. In Theorem 171, if we choose s = m = 1, we obtain 󵄨󵄨 󵄨󵄨 f (a) + f (b) 2 a+b Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α + f( )− [ J + f (b) + RL Jb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 r(r + 1) r+1 2 r(b − a)α 󵄨 1

≤

1

p 2 (b − a)2 |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q (1 − ) ( ) . r(α + 1) p(α + 1) + 1 2

Theorem 173. Let f : [a, b] → ℝ be a twice-differentiable mapping a < b. If |f 󸀠󸀠 |q is measurable and (s, m)-convex on [a, b] for some fixed q > 1 and (s, m) ∈ (0, 1]2 , r > 0, then 󵄨󵄨 f (a) + f (mb) 󵄨󵄨 2 a + mb Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α + f( )− [ J f (mb) + J f (a)] + − 󵄨󵄨 󵄨󵄨 RL mb 󵄨󵄨 r(r + 1) 󵄨󵄨 r+1 2 r(mb − a)α RL a 2 1 1 (mb − a) [|f 󸀠󸀠 (a)|q ( − − B(s + 1, q(α + 1) + 1)) ≤ r(α + 1) s + 1 q(α + 1) + s + 1 s 2 1 + m|f 󸀠󸀠 (b)|q ( − + s + 1 q(α + 1) + 1 q(α + 1) + s + 1 + B(s + 1, q(α + 1) + 1))].

(4.103)

4.5 Inequalities via (s, m)-convex functions | 165

Proof. From Lemma 58 and using the well-known Hölder’s inequality, we have 󵄨󵄨 f (a) + f (mb) 󵄨󵄨 2 a + mb Γ(α + 1) 󵄨󵄨 󵄨󵄨 α + f( )− [RL Jaα+ f (mb) + RL Jmb − f (a)]󵄨 󵄨󵄨 󵄨󵄨 α 󵄨󵄨 r(r + 1) r+1 2 r(mb − a) 󵄨 1 p

1

1

󵄨q

󵄨 ≤ (mb − a)2 (∫ 1dt) (∫ 󵄨󵄨󵄨k(t)f 󸀠󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨 dt) 0

1 q

0

1

1

0

0

1 q

≤ (mb − a)2 (|f 󸀠󸀠 (a)|q ∫ |k(t)|q t s + m|f 󸀠󸀠 (b)|q ∫ |k(t)|q (1 − t s )dt) . Calculating by parts, we have 1

1

1 1 1 [ − − ∫ t s (1 − t)q(α+1) dt], ∫ t |k(t)| dt ≤ q r (α + 1)q s + 1 q(α + 1) + s + 1 s

q

0

0

and 1

∫(1 − t s )|k(t)|q dt 0

1

1 1 2 1 [1 − ≤ q − + + ∫ t s (1 − t)q(α+1) dt]. r (α + 1)q s + 1 q(α + 1) + 1 q(α + 1) + s + 1 0

Therefore, using the above facts via (1 − (1 − t)α+1 − t α+1 )q ≤ 1 − (1 − t)q(α+1) − t q(α+1) for any t ∈ [0, 1], one can derive (4.103). The proof is completed. Remark 174. In Theorem 173, if we choose s = m = 1, we obtain 󵄨󵄨 f (a) + f (b) 󵄨󵄨 2 a+b Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α + f( )− [ J + f (b) + RL Jb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 r(r + 1) r+1 2 r(b − a)α 󵄨 1 (b − a)2 󸀠󸀠 q 1 [|f (a)| ( − − B(2, q(α + 1) + 1)) ≤ r(α + 1) 2 q(α + 1) + 2 2 1 1 + |f 󸀠󸀠 (b)|q ( − + + B(2, q(α + 1) + 1))]. 2 q(α + 1) + 1 q(α + 1) + 2 Remark 175. From Theorems 169, 171 and 173, we have 󵄨󵄨 f (a) + f (mb) 󵄨󵄨 2 a + mb Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α + f( )− [ J f (mb) + J f (a)] + − 󵄨󵄨 󵄨󵄨 RL mb 󵄨󵄨 r(r + 1) 󵄨󵄨 r+1 2 r(mb − a)α RL a ≤ min{K1 , K2 , K3 } where 1− q1

1 α + ) K1 = (mb − a) ( r(α + 1)(α + 2) 4(r + 1) 2

166 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

q α 1 × [|f (a)| I + m|f (b)| ( + − I)] , r(α + 1)(α + 2) 4(r + 1)

q

󸀠󸀠

󸀠󸀠

q

1

1

p q 2 ms 󸀠󸀠 (mb − a)2 1 (1 − ) ( |f 󸀠󸀠 (a)|q + |f (b)|q ) , K2 = r(α + 1) p(α + 1) + 1 s+1 s+1

K3 =

1 1 (mb − a)2 󸀠󸀠 [|f (a)|q ( − − B(s + 1, q(α + 1) + 1)) r(α + 1) s + 1 q(α + 1) + s + 1 s 2 1 + m|f 󸀠󸀠 (b)|q ( − + s + 1 q(α + 1) + 1 q(α + 1) + s + 1 + B(s + 1, q(α + 1) + 1))],

where I is defined in (4.99). From Theorem 171 and Theorem 173, we use one particular technique for shrinking an inequality, and we now use another shrinking technique. Theorem 176. Let f : [a, b] → ℝ be a twice-differentiable mapping with a < b. If |f 󸀠󸀠 |q is measurable and (s, m)-convex on [a, b] for some fixed q > 1 and (s, m) ∈ (0, 1]2 , r > 0, then 󵄨󵄨 f (a) + f (mb) 󵄨󵄨 2 a + mb Γ(α + 1) α α 󵄨󵄨󵄨 󵄨󵄨󵄨 + f ( ) − [ J f (mb) + J f (a)] + − RL RL a mb 󵄨󵄨 r(r + 1) 󵄨󵄨 r+1 2 r(mb − a)α 󵄨 󵄨 1

p+1

p 2 1 r+1 (mb − a)2 ( ) [( + ) ≤ r+1 p+1 2 r(α + 1) 1

×( where

1 p

+

1 q

1 p

p+1

r+1 −( ) r(α + 1)

]

q 1 ms 󸀠󸀠 |f 󸀠󸀠 (a)|q + |f (b)|q ) s+1 s+1

= 1.

Proof. From Lemma 58 and using the well-known Hölder’s inequality, we have 󵄨󵄨 f (a) + f (mb) 󵄨󵄨 2 a + mb Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α + f( )− [ J f (mb) + J f (a)] + − 󵄨󵄨 󵄨󵄨 RL mb 󵄨󵄨 r(r + 1) 󵄨󵄨 r+1 2 r(mb − a)α RL a 1

󵄨 󵄨 ≤ (mb − a)2 ∫ 󵄨󵄨󵄨k(t)f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 0

1

1 p

1

󵄨 󵄨q ≤ (mb − a) (∫ |k(t)| dt) (∫ 󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) 2

0

2

≤

p

1 p

1 q

0

p+1

(mb − a) 2 1 r+1 ( ) [( + ) r+1 p+1 2 r(α + 1) 1

q 1 ms 󸀠󸀠 ×( |f 󸀠󸀠 (a)|q + |f (b)|q ) , s+1 s+1

1 p

p+1

−(

r+1 ) r(α + 1)

]

because (1 − t)α+1 + t α+1 ≤ 1, for any t ∈ [0, 1]. The proof is completed.

4.5 Inequalities via (s, m)-convex functions | 167

Remark 177. With the same assumptions in Theorem 176, if we choose s = m = 1, we obtain 󵄨󵄨 f (a) + f (b) 󵄨󵄨 2 a+b Γ(α + 1) 󵄨󵄨 󵄨 + f( )− [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 r+1 2 r(b − a) 󵄨󵄨 r(r + 1) 1

p+1

p (b − a)2 2 1 r+1 ≤ ( ) [( + ) r+1 p+1 2 r(α + 1) 1

1 p

p+1

r+1 −( ) r(α + 1)

]

|f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q ×( ) 2 where

1 p

+

1 q

= 1.

Another Hermit-Hadamard type inequality for powers in terms of the second derivatives is obtained as follows. Theorem 178. Let f : [a, b] → ℝ be a twice-differentiable mapping a < b. If |f 󸀠󸀠 |q is measurable and (s, m)-convex on [a, b] for some fixed q > 1 and (s, m) ∈ (0, 1]2 , r > 0, then the following inequality for fractional integrals holds: 󵄨󵄨 󵄨󵄨 f (a) + f (mb) 2 a + mb Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α [ J + f( )− + f (mb) + RL Jmb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 r(r + 1) r+1 2 r(mb − a)α 󵄨 q+1

≤

2 1 r+1 (mb − a)2 󸀠󸀠 [|f (a)|q H + m|f 󸀠󸀠 (b)|q ( ( + ) r+1 q + 1 2 r(α + 1) q+1

−

2 r+1 ( ) q + 1 r(α + 1)

− H)],

(4.104)

where 1 2

q

1

q

r+1 r+1 H = ∫( + t) t s dt + ∫( + 1 − t) t s dt. r(α + 1) r(α + 1) 0

1 2

Proof. From Lemma 58 and using the well-known Hölder’s inequality, we have 󵄨󵄨 f (a) + f (mb) 󵄨󵄨 2 a + mb Γ(α + 1) 󵄨󵄨 󵄨󵄨 α + f( )− [RL Jaα+ f (mb) + RL Jmb − f (a)]󵄨 󵄨󵄨 󵄨󵄨 α 󵄨󵄨 r(r + 1) r+1 2 r(mb − a) 󵄨 2

1

󵄨 󵄨 ≤ (mb − a) ∫ 󵄨󵄨󵄨k(t)f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 0

1

1 p

1

󵄨 󵄨q ≤ (mb − a) (∫ 1dt) (∫ 󵄨󵄨󵄨k(t)f 󸀠󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨 dt) 2

0

0

1

1

0

0

1 q

1 q

≤ (mb − a)2 (|f 󸀠󸀠 (a)|q ∫ t s |k(t)|q dt + m|f 󸀠󸀠 (b)|q ∫(1 − t s )|k(t)|q dt) .

168 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals Calculating by parts, we have 1

1 2

1

∫ t |k(t)| dt = ∫ t s |k1 (t)|q dt + ∫ t s |k2 (t)|q dt, s

q

0

0

1 2

where 1 2

1 2

0

0

∫ t s |k1 (t)|q dt ≤ ∫(

q

1 t + ) t s dt r(α + 1) r + 1 1 2

q

1 r+1 = + t) t s dt, ∫( (r + 1)q r(α + 1) 1

1

∫ |k2 (t)|q t s dt ≤ ∫( 1 2

1 2

0

q

1 1−t + ) t s dt r(α + 1) r + 1 1

q

r+1 1 ≤ + 1 − t) t s dt. ∫( q (r + 1) r(α + 1) 1 2

Thus, 1

1 2

0

0

1

q

q

r+1 r+1 1 [∫( + t) t s dt + ∫( + 1 − t) t s dt]. ∫ t s |k(t)|q dt = q (r + 1) r(α + 1) r(α + 1) 1 2

Similarly, 1

s

1 2

q

1

∫(1 − t )|k(t)| dt = ∫(1 − t )|k1 (t)| dt + ∫(1 − t s )|k2 (t)|q dt 0

s

q

0

1 2

q+1

≤

1 1 r+1 2 ( + ) [ (r + 1)q q + 1 2 r(α + 1) 1 2

q

1

q+1

−

2 r+1 ( ) q + 1 r(α + 1) q

r+1 r+1 − ∫( + t) t s dt − ∫( + 1 − t) t s dt]. r(α + 1) r(α + 1) 0

1 2

Now using the just-stated facts, one can obtain (4.104). The proof is completed. Remark 179. With the same assumptions in Theorem 178, if we choose s = m = 1, we obtain 󵄨󵄨 󵄨󵄨󵄨 f (a) + f (b) 2 a+b Γ(α + 1) 󵄨󵄨 α α 󵄨󵄨 󵄨󵄨 r(r + 1) + r + 1 f ( 2 ) − r(b − a)α [RL Ja+ f (b) + RL Jb− f (a)]󵄨󵄨󵄨 󵄨 󵄨

4.5 Inequalities via (s, m)-convex functions | 169 q+1

≤

(b − a)2 󸀠󸀠 2 1 r+1 [|f (a)|q H 󸀠 + |f 󸀠󸀠 (b)|q ( ( + ) r+1 q + 1 2 r(α + 1) q+1

− where

1 p

+

1 q

r+1 2 ( ) q + 1 r(α + 1)

− H 󸀠 )],

= 1, and H󸀠 =

q+1

1 1 r+1 [( + ) q + 1 2 r(α + 1)

q+1

−(

r+1 ) r(α + 1)

].

Remark 180. From Theorems 176 and 178, we have 󵄨󵄨 󵄨󵄨 f (a) + f (mb) 2 a + mb Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α [ J f (mb) + J f (a)] + f( )− + − 󵄨󵄨 󵄨󵄨 RL RL a mb 󵄨󵄨 󵄨󵄨 r(r + 1) r+1 2 r(mb − a)α ≤ min{N1 , N2 }, where 1

p+1

p 1 2 r+1 (mb − a)2 ( ) [( + ) N1 = r+1 p+1 2 r(α + 1) 1

q 1 ms 󸀠󸀠 ×( |f 󸀠󸀠 (a)|q + |f (b)|q ) , s+1 s+1

N2 =

1 p

p+1

r+1 −( ) r(α + 1)

]

q+1

(mb − a)2 󸀠󸀠 2 1 r+1 [|f (a)|q H + m|f 󸀠󸀠 (b)|q ( ( + ) r+1 q + 1 2 r(α + 1) q+1

−

2 r+1 ( ) q + 1 r(α + 1)

− H)],

where H is defined in Theorem 178. Theorem 181. Let f : [a, b] → R be twice-differentiable on (a, b) with a ≥ 0, f 󸀠󸀠 ∈ L[a, b] and a < mb < b. If |f 󸀠󸀠 | is (s, m)-convex function, then for any 0 < α ≤ 1, we have 󵄨󵄨 f (mb) + f (a) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − ( J f (a) + J f (mb)) − + 󵄨󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 2 2(mb − a)α RL mb 2 mα 󸀠󸀠 (mb − a) [ω(α, s)(|f 󸀠󸀠 (a)| − m|f 󸀠󸀠 (b)|) + |f (b)|], ≤ 2(α + 1) α+2 where ω(α, s) =

1 1 Γ(α + 2)Γ(s + 1) − − . s+1 α+s+2 Γ(α + s + 3)

Proof. Using Lemma 63 via |f 󸀠󸀠 | is (s, m)-convex function, we have 󵄨󵄨 f (mb) + f (a) 󵄨󵄨 Γ(α + 1) 󵄨 󵄨󵄨 α α − (RL Jmb − f (a) + RL J + f (mb))󵄨󵄨 󵄨󵄨 a 󵄨󵄨 α 2 2(mb − a) 󵄨󵄨 󵄨

170 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

(mb − a)2 (1 − (1 − t)α+1 − t α+1 ) 󵄨󵄨 󸀠󸀠 󵄨 ≤ ∫ 󵄨󵄨f (ta + m(1 − t)b)󵄨󵄨󵄨dt 2 α+1 0

1

≤

(mb − a)2 ∫(1 − (1 − t)α+1 − t α+1 )(t s |f 󸀠󸀠 (a)| + m(1 − t s )|f 󸀠󸀠 (b)|)dt 2(α + 1) 0

1

2

(mb − a) ≤ [(|f 󸀠󸀠 (a)| − m|f 󸀠󸀠 (b)|) ∫(t s − t s (1 − t)α+1 − t α+s+1 )dt 2(α + 1) 0

1

+ m ∫(1 − (1 − t)α+1 − t α+1 )dt] 0

(mb − a)2 mα 󸀠󸀠 = [ω(α, s)(|f 󸀠󸀠 (a)| − m|f 󸀠󸀠 (b)|) + |f (b)|]. 2(α + 1) α+2 The proof is completed. Theorem 182. Let f : [a, b] → R be twice-differentiable on (a, b) with a ≥ 0, |f 󸀠󸀠 |q (q > 1) ∈ L[a, b] and a < mb < b. If |f 󸀠󸀠 | is (s, m)-convex function, then 󵄨󵄨 󵄨󵄨 f (mb) + f (a) Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α ( J f (a) − + J f (mb)) − + 󵄨󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 2 2(mb − a)α RL mb 1

(mb − a)2 p(α + 1) − 1 p ≤ ( ) ψ(s, q, m), 2(α + 1) p(α + 1) + 1 where

1 p

+

1 q

= 1 and 1

1

|f 󸀠󸀠 (a)|q q qsmq |f 󸀠󸀠 (b)|q q ψ(s, q, m) = ( ) +( ) . qs + 1 qs + 1 Proof. Using Lemma 63, Holder’s inequality and the fact that |f 󸀠󸀠 | is (s.m)-convex function, one has 󵄨󵄨 󵄨󵄨 f (mb) + f (a) Γ(α + 1) 󵄨 󵄨󵄨 α α − (RL Jmb − f (a) + RL J + f (mb))󵄨󵄨 󵄨󵄨 a 󵄨󵄨 α 󵄨󵄨 2 2(mb − a) 󵄨 1

(mb − a)2 (1 − (1 − t)α+1 − t α+1 ) 󵄨󵄨 󸀠󸀠 󵄨 ≤ ∫ 󵄨󵄨f (ta + m(1 − t)b)󵄨󵄨󵄨dt 2 α+1 0

1

(mb − a)2 p ≤ (∫(1 − (1 − t)α+1 − t α+1 ) dt) 2(α + 1) 0

1

q

× (∫(t s |f 󸀠󸀠 (a)| + m(1 − t s )|f 󸀠󸀠 (b)|) dt) 0

1 p

1 q

4.6 Inequalities via preinvex convex functions |

1

(mb − a)2 ≤ (∫(1 − (1 − t)p(α+1) − t p(α+1) )dt) 2(α + 1) 0

1

qs

q

1 q

1

q

171

1 p

s q

q

1 q

× [(∫ t |f (a)| dt) + (∫ m |f (b)| (1 − t ) dt) ] 󸀠󸀠

0

1 p

1

1

|f 󸀠󸀠 (a)|q q qsmq |f 󸀠󸀠 (b)|q q (mb − a) p(α + 1) − 1 ( ) [( ) +( ) ]. 2(α + 1) p(α + 1) + 1 qs + 1 qs + 1 2

=

0

󸀠󸀠

Here we use the following fact based on Minkowski’s inequality for q > 1: 1

s

s

(∫(t |f (a)| + m(1 − t )|f (b)|)dt) 󸀠󸀠

0 1

1 q

1 q

󸀠󸀠

1

s q

1 q

≤ (∫ t qs |f 󸀠󸀠 (a)|q dt) + (∫ mq |f 󸀠󸀠 (b)|q (1 − t ) dt) . 0

0

The proof is completed.

4.5.2 Applications to some special means (please disconnect) Theorem 183. For some s, m ∈ (0, 1], n ∈ ℤ \ {−1, 0}, 0 ≤ a < b, the following inequality for fractional integrals holds: 1

1

1

q (mb − a)2 2p − 1 p 1 qsmq q 󵄨󵄨 2 2 2 n 2 2 2 󵄨 ( ) (( ) + ) , 󵄨󵄨A(a , m b ) − Ln (a , m b )󵄨󵄨󵄨 ≤ 2 2p + 1 1 + sq 1 + sq

where

1 p

+

1 q

= 1, 1 < q < ∞.

Proof. Applying Theorem 182 for f (x) = x2 , α = 1, λ = 1 one can obtain the result immediately.

4.6 Inequalities via preinvex convex functions 4.6.1 Right-hand side inequalities via preinvex convex functions 4.6.1.1 Main results Theorem 184 (see Theorem 7, [61]). Suppose that f : [a, b] → ℝ is a preinvex function and f ∈ L1 [a, a + η(b, a)]. For any x, y ∈ [a, b] and any t1 , t2 ∈ [0, 1], η satisfies η(y + t2 η(x, y), y + t1 η(x, y)) = (t2 − t1 )η(x, y). Then the following inequality for fractional

172 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals integrals holds: f(

2a + η(b, a) Γ(α + 1) α )≤ α [ J α+ f (a + η(b, a)) + RL J(a+η(b,a)) − f (a)] 2 2η (b, a) RL a f (a) + f (a + η(b, a)) . ≤ 2

(4.105)

Proof. In Theorem 7 [61], let b = a + η(b, a), we can obtain the results. The proof is completed. Theorem 185. Suppose that f : [a, b] → ℝ be a twice-differentiable mapping and f 󸀠󸀠 ∈ L[a, a + η(b, a)]. If |f 󸀠󸀠 |q is preinvex function on [a, b] for some fixed q ≥ 1, then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (a + η(b, a)) Γ(α + 1) 󵄨󵄨 󵄨󵄨 󵄨󵄨 α − α [RL Jaα+ f (a + η(b, a)) + RL J(a+η(b,a)) − f (a)]󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 2η (b, a) 󵄨 1

αη2 (b, a) |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q [ ] . ≤ 2(α + 1)(α + 2) 2

(4.106)

Proof. Case 1: we suppose that q = 1. From Lemma 49, we have 󵄨󵄨 󵄨󵄨 f (a) + f (a + η(b, a)) Γ(α + 1) 󵄨󵄨 󵄨󵄨 α − α [RL Jaα+ f (a + η(b, a)) + RL J(a+η(b,a)) − f (a)]󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 2η (b, a) 󵄨 1

≤

η2 (b, a) 1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 ∫ 󵄨󵄨f (a + tη(b, a))󵄨󵄨󵄨dt, 2 α+1

(4.107)

0

where we use the fact (1 − t)α+1 + t α+1 ≤ 1 for any t ∈ [a, b]. Since |f 󸀠󸀠 | is preinvex on [a, b], we know that for any t ∈ [0, 1], 󵄨󵄨 󸀠󸀠 󵄨 󸀠󸀠 󸀠󸀠 󵄨󵄨f (ta + (1 − t)b)󵄨󵄨󵄨 ≤ (1 − t)|f (a)| + t|f (b)|. Therefore 󵄨󵄨 󵄨󵄨 f (a) + f (a + η(b, a)) Γ(α + 1) 󵄨󵄨 󵄨󵄨 α − α [RL Jaα+ f (a + η(b, a)) + RL J(a+η(b,a)) − f (a)]󵄨 󵄨󵄨 󵄨󵄨 2 2η (b, a) 󵄨 󵄨󵄨 1

≤

η2 (b, a) 1 − (1 − t)α+1 − t α+1 ((1 − t)|f 󸀠󸀠 (a)| + t|f 󸀠󸀠 (b)|)dt ∫ 2 α+1 0

αη2 (b, a) |f 󸀠󸀠 (a)| + |f 󸀠󸀠 (b)| = ( ), 2(α + 1)(α + 2) 2 which completes the proof for this case. Case 2: we suppose that q > 1. Using Lemma 49 and the power mean inequality for q, we obtain 1

󵄨 󵄨 ∫(1 − (1 − t)α+1 − t α+1 )󵄨󵄨󵄨f 󸀠󸀠 (a + tη(b, a))󵄨󵄨󵄨dt 0

4.6 Inequalities via preinvex convex functions |

173

1− q1

1

≤ (∫(1 − (1 − t)α+1 − t α+1 )dt) 0 1

α+1

× (∫(1 − (1 − t)

−t

α+1

0

1 q

󵄨q 󵄨 )󵄨󵄨󵄨f 󸀠󸀠 (a + tη(b, a))󵄨󵄨󵄨 dt) .

(4.108)

󸀠󸀠 q

Since |f | is preinvex on [a, b], we know that for any t ∈ [0, 1]

󵄨󵄨 󸀠󸀠 󵄨q 󸀠󸀠 q 󸀠󸀠 q 󵄨󵄨f (a + tη(b, a))󵄨󵄨󵄨 ≤ (1 − t)|f (a)| + t|f (b)| .

(4.109)

Hence, from (4.107), (4.108) and (4.109), we obtain 󵄨󵄨 󵄨󵄨 f (a) + f (a + η(b, a)) Γ(α + 1) 󵄨󵄨 󵄨󵄨 α − α [RL Jaα+ f (a + η(b, a)) + RL J(a+η(b,a)) − f (a)]󵄨 󵄨󵄨 󵄨󵄨 2 2η (b, a) 󵄨 󵄨󵄨 1− q1

1

η2 (b, a) (∫(1 − (1 − t)α+1 − t α+1 )dt) ≤ 2(α + 1) 0

1

󵄨 󵄨q × (∫(1 − (1 − t)α+1 − t α+1 )󵄨󵄨󵄨f 󸀠󸀠 (a + tη(b, a))󵄨󵄨󵄨 dt) 0

1 q

1− q1

1

η2 (b, a) ≤ (∫(1 − (1 − t)α+1 − t α+1 )dt) 2(α + 1) 0

1

α+1

× (∫(1 − (1 − t) 0 2

1− q1

η (b, a) α ( ) = 2(α + 1) α + 2

−t

α+1

q

q

)[(1 − t)|f (a)| + t|f (b)| ]dt) 󸀠󸀠

󸀠󸀠

1

1 q

1

q α |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q ( ) [ ] α+2 2 1

αη2 (b, a) |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q [ ] . = 2(α + 1)(α + 2) 2

The proof is completed.

Next, we present the second theorem in this section. Theorem 186. Suppose that f : [a, b] → ℝ be a twice-differentiable mapping and f 󸀠󸀠 ∈ L1 [a, a + η(b, a)]. If |f 󸀠󸀠 |q is preinvex function on [a, b] for some fixed q > 1, then the following inequality for fractional integrals holds: 󵄨󵄨 󵄨󵄨󵄨 f (a) + f (a + η(b, a)) Γ(α + 1) 󵄨󵄨 α 󵄨󵄨 − α [RL Jaα+ f (a + η(b, a)) + RL J(a+η(b,a)) − f (a)]󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨 2 2η (b, a) 1

1

p η2 (b, a) 2 |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q ≤ (1 − ) ( ) , 2(α + 1) p(α + 1) + 1 2

where

1 p

+

1 q

= 1.

(4.110)

174 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals Proof. It follows from Lemma 49 and Hölder’s inequality that 󵄨󵄨 f (a) + f (a + η(b, a)) Γ(α + 1) 󵄨󵄨 󵄨󵄨 󵄨󵄨 α − α [RL Jaα+ f (a + η(b, a)) + RL J(a+η(b,a)) − f (a)]󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 2η (b, a) 󵄨 1

η2 (b, a) 1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 ≤ ∫ 󵄨󵄨f (a + tη(b, a))󵄨󵄨󵄨dt 2 α+1 0

1 p

1

1

η2 (b, a) p 󵄨 󵄨q ≤ (∫(1 − (1 − t)α+1 − t α+1 ) dt) (∫ 󵄨󵄨󵄨f 󸀠󸀠 (a + tη(b, a))󵄨󵄨󵄨 dt) 2(α + 1)

1 q

0

0

1

η2 (b, a) ≤ (∫(1 − (1 − t)p(α+1) − t p(α+1) )dt) 2(α + 1) 0

1

1

× (|f 󸀠󸀠 (a)|q ∫(1 − t)dt + |f 󸀠󸀠 (b)|q ∫ tdt) 0

1 p

1 q

0

1 p

1

η2 (b, a) 2 |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q = (1 − ) ( ) . 2(α + 1) p(α + 1) + 1 2 Here we use q

(1 − (1 − t)α+1 − t α+1 ) ≤ 1 − (1 − t)q(α+1) − t q(α+1) , for any t ∈ [0, 1], which follows from (A − B)q ≤ Aq − Bq , for any A > B ≥ 0 and q ≥ 1. The proof is completed. To complete this section, we give another fractional version Hermit-Hadamard inequality. Theorem 187. Suppose that f : [a, b] → ℝ be a twice-differentiable mapping and f 󸀠󸀠 ∈ L1 [a, a + η(b, a)]. If |f 󸀠󸀠 |q is preinvex function on [a, b] for some fixed q > 1, then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (a + η(b, a)) Γ(α + 1) 󵄨󵄨 󵄨󵄨 󵄨󵄨 α − α [RL Jaα+ f (a + η(b, a)) + RL J(a+η(b,a)) − f (a)]󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 2η (b, a) 󵄨 1

≤ where

1 p

1

η2 (b, a) q(α + 1) − 1 q |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q ( ) ( ) , 2(α + 1) q(α + 1) + 1 2

+

1 q

= 1.

(4.111)

4.6 Inequalities via preinvex convex functions |

175

Proof. It follows from Lemma 49 and Hölder’s inequality that 󵄨󵄨 f (a) + f (a + η(b, a)) Γ(α + 1) 󵄨󵄨 󵄨󵄨 󵄨󵄨 α − α [RL Jaα+ f (a + η(b, a)) + RL J(a+η(b,a)) − f (a)]󵄨 󵄨󵄨 󵄨󵄨 2 2η (b, a) 󵄨󵄨 󵄨 1

≤

η2 (b, a) 1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 ∫ 󵄨󵄨f (a + tη(b, a))󵄨󵄨󵄨dt 2 α+1 0

1 p

1

1

η2 (b, a) q󵄨 󵄨q ≤ (∫ 1dt) (∫(1 − (1 − t)α+1 − t α+1 ) 󵄨󵄨󵄨f 󸀠󸀠 (a + tη(b, a))󵄨󵄨󵄨 dt) 2(α + 1) 0

≤

1

1 q

0

η2 (b, a) (|f 󸀠󸀠 (a)|q ∫[(1 − t) − (1 − t)q(α+1)+1 − t q(α+1) (1 − t)]dt 2(α + 1) 0

q

1

q(α+1)

+ |f (b)| ∫[t − (1 − t) 󸀠󸀠

0 2

t−t

q(α+1)+1

1 q

]dt)

1

1

η (b, a) q(α + 1) − 1 q |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q = ( ) ( ) . 2(α + 1) q(α + 1) + 1 2 Here we use q

(1 − (1 − t)α+1 − t α+1 ) ≤ 1 − (1 − t)q(α+1) − t q(α+1) , for any t ∈ [0, 1], which follows from (A − B)q ≤ Aq − Bq , for any A > B ≥ 0 and q ≥ 1. The proof is completed. Corollary 188. If we set α = 1 and η(b, a) = b − a in Theorems 185–187, then we have b 󵄨󵄨 󵄨󵄨 󵄨󵄨 f (a) + f (b) 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 ≤ min{v1 , v2 , v3 }, − f (x)dx ∫ 󵄨󵄨 󵄨󵄨 2 b − a 󵄨󵄨 󵄨󵄨 a

where 1

(b − a)2 |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q [ ] , v1 = 12 2 1

v2 =

1

p (b − a)2 2 |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q (1 − ) [ ] , 4 2p + 1 2 1

1

(b − a)2 2q − 1 q |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q v3 = ( ) [ ] . 4 2q + 1 2

176 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 4.6.1.2 Applications to some special means Now, using the above theory results, we present some applications to special means of real numbers. Proposition 189. Let a, b ∈ ℝ, a < b and 0 does not belong to [a, b] and n ∈ ℤ, |n| ≥ 2. Then 󵄨󵄨 󵄨 n n n 󵄨󵄨A(a , b ) − Ln (a, b)󵄨󵄨󵄨 ≤ min{v1 , v2 , v3 }, where 1

v1 =

n(n − 1)(b − a)2 |a|(n−2)q + |b|(n−2)q q ( ) , 12 2 1

1

p n(n − 1)(b − a)2 2 |a|(n−2)q + |b|(n−2)q q v2 = (1 − ) ( ) , 4 2p + 1 2 1

1

n(n − 1)(b − a)2 2q − 1 q |a|(n−2)q + |b|(n−2)q q v2 = ( ) ( ) . 4 2q + 1 2 Proof. Applying Corollary 188 for f (x) = x n , one can obtain the result immediately. The proof is completed. Proposition 190. Let a, b ∈ ℝ, a < b and 0 does not belong to [a, b]. Then |H −1 (a, b) − L−1 (a, b)| ≤ min{v1 , v2 , v3 }, where 1

(b − a)2 |a|−3q + |b|−3q q v1 = ( ) , 6 2 1

v2 =

1

p (b − a)2 2 |a|−3q + |b|−3q q (1 − ) ( ) , 2 2p + 1 2 1

1

(b − a)2 2q − 1 q |a|−3q + |b|−3q q v2 = ( ) ( ) . 2 2q + 1 2 Proof. Applying Corollary 188 for f (x) = x1 , one can obtain the result immediately. The proof is completed. Proposition 191. Let a, b ∈ ℝ, a < b and 0 does not belong to [a, b]. Then |ln G(a, b) − ln I(a, b)| ≤ min{v1 , v2 , v3 }, where 1

(b − a)2 |a|−2q + |b|−2q q v1 = [ ] , 12 2

4.6 Inequalities via preinvex convex functions | 1

177

1

p (b − a)2 2 |a|−2q + |b|−2q q v2 = (1 − ) [ ] , 4 2p + 1 2 1

1

(b − a)2 2q − 1 q |a|−2q + |b|−2q q v3 = ( ) [ ] . 4 2q + 1 2 Proof. Applying Corollary 188 for f (x) = ln x, one can obtain the result immediately. The proof is completed. 4.6.2 Left-hand side inequalities via Preinvex convex functions 4.6.2.1 Main results In this section, we will introduce another new left-hand side identity via RiemannLiouville fractional integrals and twice-differentiable mappings. Theorem 192. Suppose that f : [a, b] → ℝ be a twice-differentiable mapping and f 󸀠󸀠 ∈ L1 [a, a + η(b, a)]. If |f 󸀠󸀠 | is preinvex function on [a, b], then the following inequality for fractional integrals holds: 󵄨󵄨 Γ(α + 1) 2a + η(b, a) 󵄨󵄨󵄨󵄨 󵄨󵄨 α [RL Jaα+ f (a + η(b, a)) + RL J(a+η(b,a)) )󵄨󵄨 − f (a)] − f ( 󵄨󵄨 α 󵄨󵄨 2 󵄨󵄨 2η (b, a) 2 η (b, a) 1 α ( − )(|f 󸀠󸀠 (a)| + |f 󸀠󸀠 (b)|). (4.112) ≤ 4 4 (α + 1)(α + 2) Proof. From Lemma 55, we have 󵄨󵄨 Γ(α + 1) 2a + η(b, a) 󵄨󵄨󵄨󵄨 󵄨󵄨 α [RL Jaα+ f (a + η(b, a)) + RL J(a+η(b,a)) )󵄨󵄨 − f (a)] − f ( 󵄨󵄨 α 󵄨󵄨 2 󵄨󵄨 2η (b, a) 1 2

η2 (b, a) 1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 [∫(t − )󵄨󵄨f (a + tη(b, a))󵄨󵄨󵄨dt ≤ 2 α+1 0

1

+ ∫(1 − t − 1 2

:=

1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 )󵄨󵄨f (a + tη(b, a))󵄨󵄨󵄨dt] α+1

η2 (b, a) (K1 + K2 ). 2

Using the preinvex of |f 󸀠󸀠 |, we obtain 1 2

K1 = ∫(t − 0

1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 )󵄨󵄨f (a + tη(b, a))󵄨󵄨󵄨dt α+1

1 2

≤ ∫[(t − 0

1 − (1 − t)α+1 − t α+1 )(1 − t)|f 󸀠󸀠 (a)| α+1

(4.113)

178 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1 − (1 − t)α+1 − t α+1 )t|f 󸀠󸀠 (b)|]dt α+1 1 3 1 1 1 ( − − =[ − )]|f 󸀠󸀠 (a)| 12 α + 1 8 α + 3 (α + 2)(α + 3)2α+2 1 1 1 1 ( − +[ − 24 α + 1 8 (α + 2)(α + 3) 1 )]|f 󸀠󸀠 (b)|, + (α + 2)(α + 3)2α+2 + (t −

and 1

K2 = ∫(1 − t − 1 2

(4.114)

1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 )󵄨󵄨f (a + tη(b, a))󵄨󵄨󵄨dt α+1

1

≤ ∫[(1 − t − 1 2

1 − (1 − t)α+1 − t α+1 )(1 − t)|f 󸀠󸀠 (a)| α+1

1 − (1 − t)α+1 − t α+1 )t|f 󸀠󸀠 (b)|]dt α+1 1 1 1 1 1 =[ − )]|f 󸀠󸀠 (a)| ( − + 24 α + 1 8 (α + 2)(α + 3) (α + 2)(α + 3)2α+2 1 3 1 1 1 ( − − +[ − )]|f 󸀠󸀠 (b)|. 12 α + 1 8 α + 3 (α + 2)(α + 3)2α+2 + (1 − t −

(4.115)

Substituting (4.114) and (4.115) into (4.113), we obtain the inequality (4.112). The proof is completed. Theorem 193. Suppose that f : [a, b] → ℝ be a twice-differentiable mapping and f 󸀠󸀠 ∈ L1 [a, a + η(b, a)]. If |f 󸀠󸀠 |q is preinvex function on [a, b] for some fixed q > 1, then the following inequality for fractional integrals holds: 󵄨󵄨 Γ(α + 1) 2a + η(b, a) 󵄨󵄨󵄨󵄨 󵄨󵄨 α [RL Jaα+ f (a + η(b, a)) + RL J(a+η(b,a)) )󵄨󵄨 − f (a)] − f ( 󵄨󵄨 α 󵄨󵄨 2η (b, a) 󵄨󵄨 2 1− q1

η2 (b, a) 1 α ( − ) ≤ 2 8 2(α + 1)(α + 2) × {[

1

q 1 1 3 1 1 − ( − − )] α+2 12 α + 1 8 α + 3 (α + 2)(α + 3)2

1

q 1 1 1 1 1 +[ − ( − + )] } α+2 24 α + 1 8 (α + 2)(α + 3) (α + 2)(α + 3)2

× (|f 󸀠󸀠 (a)| + |f 󸀠󸀠 (b)|).

Proof. From Lemma 55, we have 󵄨󵄨󵄨 Γ(α + 1) 2a + η(b, a) 󵄨󵄨󵄨󵄨 α α 󵄨󵄨 )󵄨󵄨 󵄨󵄨 2ηα (b, a) [RL Ja+ f (a + η(b, a)) + RL J(a+η(b,a))− f (a)] − f ( 󵄨󵄨 2 󵄨

(4.116)

4.6 Inequalities via preinvex convex functions |

179

1 2

η2 (b, a) 1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 ≤ [∫(t − )󵄨󵄨f (a + tη(b, a))󵄨󵄨󵄨dt 2 α+1 0

1

+ ∫(1 − t − 1 2

:=

1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 )󵄨󵄨f (a + tη(b, a))󵄨󵄨󵄨dt] α+1

η2 (b, a) (K1 + K2 ). 2

(4.117)

Using the preinvex of |f 󸀠󸀠 |q and the power mean inequality for q, we obtain 1 2

K1 = ∫(t − 0

1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 )󵄨󵄨f (a + tη(b, a))󵄨󵄨󵄨dt α+1

1 2

1− q1

1 − (1 − t)α+1 − t α+1 ≤ [∫(t − )dt] α+1 0

1 2

1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨q × [∫(t − )󵄨󵄨f (a + tη(b, a))󵄨󵄨󵄨 dt] α+1 0

1 q

1− 1

q 1 α − ) 8 2(α + 1)(α + 2) 1 3 1 1 1 ( − − )]|f 󸀠󸀠 (a)|q × {[ − 12 α + 1 8 α + 3 (α + 2)(α + 3)2α+2

=(

1

q 1 1 1 1 1 ( − + +[ − )]|f 󸀠󸀠 (b)|q } α+2 24 α + 1 8 (α + 2)(α + 3) (α + 2)(α + 3)2

1− q1

1 α ≤( − ) 8 2(α + 1)(α + 2)

1

q 1 1 3 1 1 − ( − − )] |f 󸀠󸀠 (a)| α+2 12 α + 1 8 α + 3 (α + 2)(α + 3)2 1 1 1 1 ( − +[ − 24 α + 1 8 (α + 2)(α + 3)

× {[

1

q 1 + )] f 󸀠󸀠 (b)}, α+2 (α + 2)(α + 3)2

and 1

K2 = ∫(1 − t − 1 2

1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 )󵄨󵄨f (a + tη(b, a))󵄨󵄨󵄨dt α+1

(4.118)

180 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1− q1

1

1 − (1 − t)α+1 − t α+1 ≤ [∫(1 − t − )dt] α+1 1 2

1

1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨q × [∫(1 − t − )󵄨󵄨f (a + tη(b, a))󵄨󵄨󵄨 dt] α+1

1 q

1 2

1− q1

α 1 ) =( − 8 2(α + 1)(α + 2) × {[

1 1 1 1 1 − ( − + )]|f 󸀠󸀠 (a)|q 24 α + 1 8 (α + 2)(α + 3) (α + 2)(α + 3)2α+2 1

q 1 1 3 1 1 )]|f 󸀠󸀠 (b)|q } +[ − ( − − α+2 12 α + 1 8 α + 3 (α + 2)(α + 3)2

1− q1

1 α ≤( − ) 8 2(α + 1)(α + 2)

1

q 1 1 1 1 1 × {[ − ( − + )] |f 󸀠󸀠 (a)| α+2 24 α + 1 8 (α + 2)(α + 3) (α + 2)(α + 3)2 1

q 1 1 3 1 1 +[ − ( − − )] f 󸀠󸀠 (b)}. 12 α + 1 8 α + 3 (α + 2)(α + 3)2α+2

(4.119)

Submitting (4.118) and (4.119) to (4.117), we obtain the inequality (4.116). The proof is completed. Corollary 194. If we choose α = 1 and η(b, a) = b − a in Theorems 192–193, then we have b 󵄨󵄨 󵄨 󵄨󵄨 1 a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 f (x)dx − f ( ) ∫ 󵄨󵄨 ≤ min{k1 , k2 }, 󵄨󵄨 b − a 󵄨󵄨 2 󵄨󵄨 󵄨 a

where k1 =

(b − a)2 󸀠󸀠 (|f (a)| + |f 󸀠󸀠 (b)|), 48 1

1

(b − a)2 5 q 3 q k2 = [( ) + ( ) ](|f 󸀠󸀠 (a)| + |f 󸀠󸀠 (b)|). 48 8 8 4.6.2.2 Applications to some special means Proposition 195. Let a, b ∈ ℝ, a < b and 0 does not belong to [a, b] and n ∈ ℤ, |n| ≥ 2. Then 󵄨󵄨 n n n 󵄨 󵄨󵄨Ln (a, b) − A(a , b )󵄨󵄨󵄨 ≤ min{k1 , k2 },

4.7 Inequalities via (β, m)-geometrically convex functions | 181

where k1 =

n(n − 1)(b − a)2 (|a|n−2 + |b|n−2 ), 48 1

1

n(n − 1)(b − a)2 5 q 3 q k2 = [( ) + ( ) ](|a|n−2 + |b|n−2 ). 48 8 8 Proof. Applying Corollary 194 for f (x) = xn , one can obtain the result immediately. The proof is completed. Proposition 196. Let a, b ∈ ℝ, a < b and 0 does not belong to [a, b]. Then |L−1 (a, b) − A−1 (a, b)| ≤ min{k1 , k2 }, where k1 =

(b − a)2 (|a|−3 + |b|−3 ), 24 1

1

3 q (b − a)2 5 q k2 = [( ) + ( ) ](|a|−3 + |b|−3 ). 24 8 8 Proof. Applying Corollary 194 for f (x) = x1 , one can obtain the result immediately. The proof is completed. Proposition 197. Let a, b ∈ ℝ, a < b, 0 does not belong to [a, b]. Then |ln I(a, b) − ln A(a, b)| ≤ min{k1 , k2 }, where k1 =

(b − a)2 (|a|−2 + |b|−2 ), 48 1

1

(b − a)2 5 q 3 q [( ) + ( ) ](|a|−2 + |b|−2 ). k2 = 48 8 8 Proof. Applying Corollary 194 for f (x) = ln x, one can obtain the result immediately. The proof is completed.

4.7 Inequalities via (β, m)-geometrically convex functions Now we are ready to establish some Hermite-Hadamard type inequalities for fractional integrals via (β, m)-geometrically convex functions. Theorem 198. Let (β, m) ∈ (0, 1] × (0, 1], α ∈ (0, ∞) and 1 ≤ p, q < ∞ with

1 p

+

1 q

= 1. For

f : [a, b] ⊆ [1, ∞) → ℝ+ , if |f 󸀠 |q is decreasing and (β, m)-geometrically convex function

182 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals on [a, b], then 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 α α − [RL Ja+ f (b) + RL Jb− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 1 p b−a 󸀠 2 1 ≤ ] |f (b)|m [ − 2 α + 1 2α−1 (α + 1) ∞

× [∑

1−β

(− ln k)i−1 max{k 2

i=1

−1

− k(−1)i−1 , k(−1)i−1 − 1}

β( αβ + β1 )i

1 q

] ,

where k=(

q

|f 󸀠 (a)| ) . |f 󸀠 (b)|m

Proof. Using Lemma 42, Lemma 36, power mean inequality and Definition 23, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 α α − [RL Ja+ f (b) + RL Jb− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 1

≤

b−a 󵄨 󵄨 ∫ |(1 − t)α − t α |󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 0

1 p

1

1

b−a 󵄨 󵄨q ≤ (∫ |(1 − t)α − t α |dt) (∫ |(1 − t)α − t α |󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) 2 0

1 p

0

1

2 1 b−a 󵄨 󵄨q ( − ) (∫ |(1 − t)α − t α |󵄨󵄨󵄨f 󸀠 (at bm(1−t) )󵄨󵄨󵄨 dt) ≤ 2 α + 1 2α−1 (α + 1) 1 p

1 q

1 q

0 1

β β b−a 2 1 ) (∫ |(1 − t)α − t α ||f 󸀠 (a)|qt |f 󸀠 (b)|qm(1−t ) dt) ≤ ( − 2 α + 1 2α−1 (α + 1)

0

1

1

qt β

p 2 1 |f 󸀠 (a)| b−a 󸀠 |f (b)|m [ − α−1 ≤ ] [∫ |(1 − t)α − t α |( 󸀠 ) 2 α + 1 2 (α + 1) |f (b)|m 1

≤

0

p b−a 󸀠 2 1 |f (b)|m [ − α−1 ] 2 α + 1 2 (α + 1)

∞

× [∑

1−β

(− ln k)i−1 max{k 2

i=1

−1

− k(−1)i−1 , k(−1)i−1 − 1}

β( αβ + β1 )i

where we have used the following inequalities, 1

β

∫ t α k t dt = 0

1

α 1 1 + −1 ∫ t β β k t dt β

0

1 ∞ (− ln k)i−1 = k∑ α 1 , β i=1 ( + )i β

β

1 q

] ,

dt]

1 q

1 q

4.7 Inequalities via (β, m)-geometrically convex functions | 183

and 1

1

α tβ

β

∫(1 − t) k dt = ∫ t α k (1−t) dt 0

0

1

1−β

≤ ∫ tα k2 0

=k

21−β

−t β

dt

1

tβ

∫ t α (k −1 ) dt 0

= k2

1−β

∞ (− ln k −1 )i−1 1 ⋅ k −1 ∑ α 1 β i=1 ( + )i β

21−β ∞

=

β

i−1

k (ln k) ∑ α 1 , kβ i=1 ( + )i β β

and 1

1

β

β

∫(1 − t)α k t dt = ∫ t α k (1−t) dt 0

0

1

β

≥ ∫ t α k 1−t dt 0

1

tβ

= k ∫ t α (k −1 ) dt 0 ∞

=

1 (ln k)i−1 ∑ α 1 . β i=1 ( + )i β β

The proof is completed. Theorem 199. Let (β, m) ∈ (0, 1] × (0, 1], α ∈ (0, ∞) and 1 ≤ p, q < ∞ with

1 p

+

1 q

= 1. For

f : [a, b] ⊆ [1, ∞) → ℝ , if |f | is decreasing and (β, m)-geometrically convex function on [a, b], then 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 α α − f (a)]󵄨󵄨󵄨 [RL Ja+ f (b) + RL Jb− 󵄨󵄨 α 2 2(b − a) 󵄨󵄨 󵄨󵄨 +

󸀠

1 q

(b − a)|f 󸀠 (b)|m 1 ∞ (− ln k)i−1 [ k∑ ] . ≤ 1 β i=1 ( 1 )i (αp + 1) p β

Proof. Using Lemma 42, Lemma 36, the well-known Hölder inequality and Definition 23, it follows that 󵄨󵄨 󵄨󵄨󵄨 f (a) + f (b) Γ(α + 1) 󵄨 α α 󵄨󵄨 − [RL Ja+ f (b) + RL Jb− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨

184 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

1

b − a α 󵄨󵄨 󸀠 b−a 󵄨 󵄨 󵄨 ≤ ∫(1 − t)α 󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt + ∫ t 󵄨󵄨f (ta + (1 − t)b)󵄨󵄨󵄨dt 2 2 0

≤

1

1

0

0

0

b−a b − a α 󵄨󵄨 󸀠 t m(1−t) 󵄨󵄨 󵄨 󵄨 )󵄨󵄨dt ∫(1 − t)α 󵄨󵄨󵄨f 󸀠 (at )bm(1−t) 󵄨󵄨󵄨dt + ∫ t 󵄨󵄨f (a b 2 2 1

1

β β β β b−a b−a α 󸀠 ≤ ∫(1 − t)α |f 󸀠 (a)|t |f 󸀠 (b)|m(1−t ) dt + ∫ t |f (a)|t |f 󸀠 (b)|m(1−t ) dt 2 2

0

0

1 p

1

1

tβ q

(b − a)|f 󸀠 (b)|m |f 󸀠 (a)| ) ≤ [∫(1 − t)αp dt] [∫( 󸀠 2 |f (b)|m 0

1

1 p

1

tβ q

|f 󸀠 (a)| (b − a)|f 󸀠 (b)|m [∫ t αp dt] [∫( 󸀠 ) + 2 |f (b)|m 0

=

(b − a)|f 󸀠 (b)|m 1

(αp + 1) p

0

1

dt]

0

tβ q

|f 󸀠 (a)| [∫ ( 󸀠 ) |f (b)|m

dt]

0

dt]

1 q

1 q

1 q

1 q

(b − a)|f 󸀠 (b)|m 1 ∞ (− ln k)i−1 = [ k∑ ] . 1 β i=1 ( 1 )i (αp + 1) p β

The proof is completed. Corollary 200. Under the assumptions of Theorem 199 with β = 1, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 α α − [RL Ja+ f (b) + RL Jb− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨

|f (a)| q (b − a)|f 󸀠 (b)|m ( |f 󸀠 (b)|m ) − 1 q [ ] . ≤ 󸀠 (a)| 1 q ln( |f|f󸀠 (b)| (αp + 1) p m) 1

󸀠

Theorem 201. Let α ∈ (0, ∞), (β, m) ∈ (0, 1] × (0, 1], and 1 ≤ p, q < ∞ with 󸀠 q

1 p

+

1 q

= 1,

f : [a, b] ⊆ [1, ∞) → ℝ , |f | be decreasing and (β, m)-geometrically convex function, then we have: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J f (b) + J f (a)] 󵄨󵄨 󵄨󵄨 RL RL a+ b− 󵄨󵄨 󵄨󵄨 2 2(b − a)α +

1 q

(b − a)|f 󸀠 (b)|m 1 ∞ (− ln k)i−1 ≤ [ k∑ ] . 1 β i=1 ( 1 )i (αp + 1) p β

Proof. Using Lemma 42, Lemma 36, the well-known Hölder inequality and Definition 23, it follows that 󵄨󵄨 󵄨󵄨󵄨 f (a) + f (b) Γ(α + 1) 󵄨 α α 󵄨󵄨 − [RL Ja+ f (b) + RL Jb− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨

4.8 Inequalities via geometrical-arithmetically s-convex functions | 185 1

1

b − a α 󵄨󵄨 󸀠 b−a 󵄨 󵄨 󵄨 ≤ ∫(1 − t)α 󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt + ∫ t 󵄨󵄨f (ta + (1 − t)b)󵄨󵄨󵄨dt 2 2 0

0

1 p

1

1

b−a 󵄨q 󵄨 ≤ (∫(1 − t)αp dt) (∫ 󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) 2 0

0

1 p

1

1

b−a 󵄨q 󵄨 (∫ t αp dt) (∫ 󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) + 2 0

≤

≤

=

b−a

1

(αp + 1) p b−a (αp + 1)

0

1

t m(1−t) 󵄨󵄨q

󵄨 (∫ 󵄨󵄨󵄨f 󸀠 (a b 1

1 p

)󵄨󵄨 dt)

0

β

β

0

1

(αp + 1) p

1

tβ q

|f 󸀠 (a)| ) [∫( 󸀠 |f (b)|m 0

1 q

1 q

(∫ |f 󸀠 (a)|qt |f 󸀠 (b)|qm(1−t ) dt)

(b − a)|f 󸀠 (b)|m

1 q

dt]

1 q

1 q

1 q

(b − a)|f 󸀠 (b)|m 1 ∞ (− ln k)i−1 = ] . [ k∑ 1 β i=1 ( 1 )i (αp + 1) p β

The proof is completed. Corollary 202. Under the assumptions of Theorem 201 with β = 1, we have 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) 󵄨 󵄨󵄨 α α [RL Ja+ f (b) + RL Jb− − f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a)

|f (a)| q (b − a)|f 󸀠 (b)|m ( |f 󸀠 (b)|m ) − 1 q ≤ [ ] . 󸀠 (a)| 1 q ln( |f|f󸀠 (b)| (αp + 1) p m) 󸀠

1

4.8 Inequalities via geometrical-arithmetically s-convex functions The results in this section are due to [96, 97, 99].

4.8.1 Main results Theorem 203. Let f : [0, b] → ℝ be a differentiable mapping. If |f 󸀠 | is measurable and |f 󸀠 | is decreasing and geometric-arithmetically s-convex on [0, b] for some fixed α ∈ (0, ∞), s ∈ (0, 1], 0 ≤ a < b, then the following inequality for fractional inte-

186 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals grals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨󵄨 α − [J f (b) + J f (a)] + − 󵄨󵄨 󵄨󵄨 b 󵄨󵄨 󵄨󵄨 2 2(b − a)α a α+s 󸀠 󸀠 󸀠 (b − a)(2 |f (b)| − |f (a)| − |f (b)|) ≤ (α + s + 1)2α+s+1

+ (b − a)|f 󸀠 (a)|[0.5B(s + 1, α + 1) − B0.5 (α + 1, s + 1)]

+ (b − a)|f 󸀠 (b)|[B0.5 (s + 1, α + 1) − 0.5B(s + 1, α + 1)], where 1

B(s + 1, α + 1) = ∫ t s (1 − t)α dt, 0

and 0.5

B0.5 (s + 1, α + 1) = ∫ t s (1 − t)α dt. 0

Proof. By using Definition 21, Definition 3, Lemma 36 and 42, we have 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) α 󵄨 󵄨󵄨 − [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 1

≤

b−a 󵄨 󵄨 ∫ |(1 − t)α − t α |󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 0

1

b−a 󵄨 󵄨 ≤ ∫(t α − (1 − t)α )󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 1 2

1 2

b−a 󵄨 󵄨 + ∫((1 − t)α − t α )󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 0

1 2

1

≤

b−a b−a 󵄨 󵄨 󵄨 󵄨 ∫(t α − (1 − t)α )󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨dt + ∫((1 − t)α − t α )󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨dt 2 2 0

1 2

1

≤

b−a ∫(t α − (1 − t)α )[t s |f 󸀠 (a)| + (1 − t)s |f 󸀠 (b)|]dt 2 1 2

1 2

b−a + ∫((1 − t)α − t α )[t s |f 󸀠 (a)| + (1 − t)s |f 󸀠 (b)|]dt 2 0

4.8 Inequalities via geometrical-arithmetically s-convex functions | 187 1

1

(b − a)|f 󸀠 (a)| s (b − a)|f 󸀠 (a)| α+s ≤ ∫ t dt − ∫ t (1 − t)α dt 2 2 1 2

1 2

1

−

1

(b − a)|f 󸀠 (b)| (b − a)|f 󸀠 (b)| α ∫(1 − t)α+s dt + ∫ t (1 − t)s dt 2 2 1 2

1 2

1 2

(b − a)|f (a)| α+s (b − a)|f (a)| s ∫ t dt + ∫ t (1 − t)α dt 2 2 󸀠

−

1 2

󸀠

0

0

1 2

1 2

(b − a)|f 󸀠 (b)| (b − a)|f 󸀠 (b)| α + ∫(1 − t)α+s dt − ∫ t (1 − t)s dt 2 2 1

0

0

1

≤ (b − a)|f 󸀠 (a)| ∫ t α+s dt − (b − a)|f 󸀠 (a)| ∫ t s (1 − t)α dt 1 2

1 2

1

1

− (b − a)|f 󸀠 (b)| ∫(1 − t)α+s dt + (b − a)|f 󸀠 (b)| ∫ t α (1 − t)s dt 1 2

1 2

1

− (b − a)|f (a)| ∫ t 󸀠

0

+

1

α+s

1

(b − a)|f 󸀠 (a)| s dt + ∫ t (1 − t)α dt 2 0

1

(b − a)|f 󸀠 (b)| (b − a)|f 󸀠 (b)| α ∫(1 − t)α+s dt − ∫ t (1 − t)s dt 2 2 0

0

1 ≤ −(b − a)|f (a)| (α + s + 1)2α+s+1 1 + (b − a)|f 󸀠 (a)| − (b − a)|f 󸀠 (a)|B0.5 (α + 1, s + 1) α+s+1 1 − (b − a)|f 󸀠 (b)| + (b − a)|f 󸀠 (b)|B0.5 (s + 1, α + 1) (α + s + 1)2α+s+1 󸀠

(b − a)|f 󸀠 (a)| 1 + B(s + 1, α + 1) α+s+1 2 (b − a)|f 󸀠 (b)| 1 (b − a)|f 󸀠 (b)| + − B(s + 1, α + 1) 2 α+s+1 2 (b − a)(2α+s |f 󸀠 (b)| − |f 󸀠 (a)| − |f 󸀠 (b)|) ≤ (α + s + 1)2α+s+1 − (b − a)|f 󸀠 (a)|

+ (b − a)|f 󸀠 (a)|[0.5B(s + 1, α + 1) − B0.5 (α + 1, s + 1)]

+ (b − a)|f 󸀠 (b)|[B0.5 (s + 1, α + 1) − 0.5B(s + 1, α + 1)]. The proof is completed.

188 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals Theorem 204. Let f : [0, b] → ℝ be a differentiable mapping and 1 < q < ∞. If |f 󸀠 |q is measurable and |f 󸀠 |q is decreasing and geometric-arithmetically s-convex on [0, b] for some fixed α ∈ (0, ∞), s ∈ (0, 1], 0 ≤ a < b, then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨 − [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 1

1

b − a |f 󸀠 (a)|q + |f 󸀠 (b)|q q 2 − 21−pα p ( ) ( ) , ≤ 2 s+1 pα + 1 where

1 p

+

1 q

= 1.

Proof. By using Definition 21, Hölder inequality, Lemma 36 and 42, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨󵄨 α [J f (b) − + J f (a)] + − 󵄨󵄨 󵄨󵄨 b 󵄨󵄨 󵄨󵄨 2 2(b − a)α a 1

b−a 󵄨 󵄨 ≤ ∫ |(1 − t)α − t α |󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 0

1 p

1

1

b−a 󵄨 󵄨q (∫ |(1 − t)α − t α |p dt) (∫ 󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) ≤ 2 0

0

1 p

1

1

b−a 󵄨 󵄨q (∫ |(1 − t)α − t α |p dt) (∫ 󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨 dt) ≤ 2 0

1 q

0

1 p

1

1 q

1

b−a ≤ (∫ |(1 − t)α − t α |p dt) (∫[t s |f 󸀠 (a)|q + (1 − t)s |f 󸀠 (b)|q ]dt) 2 0

0

1 q

1

b − a |f 󸀠 (a)|q + |f 󸀠 (b)|q ≤ ( ) (∫ |(1 − t)α − t α |p dt) 2 s+1

1 q

1 p

0

1 q

1 2

1

b − a |f 󸀠 (a)|q + |f 󸀠 (b)|q p p ≤ ( ) (∫(t α − (1 − t)α ) dt + ∫((1 − t)α − t α ) dt) 2 s+1 0

1 2

1

1 p

1 2

1

b − a |f 󸀠 (a)|q + |f 󸀠 (b)|q q ≤ ( ) (∫(t pα − (1 − t)pα )dt + ∫((1 − t)pα − t pα )dt) 2 s+1 0

1 2

1 q

≤

1 p

b − a |f 󸀠 (a)|q + |f 󸀠 (b)|q 2 − 21−pα ( ) ( ) . 2 s+1 pα + 1

The proof is completed.

1 p

4.8 Inequalities via geometrical-arithmetically s-convex functions | 189

Theorem 205. Let f : [0, b] → ℝ be a differentiable mapping. If |f 󸀠 | is measurable and |f 󸀠 | is decreasing and geometric-arithmetically s-convex on [0, b] for some fixed α ∈ (0, ∞), s ∈ (0, 1], 0 ≤ a < b, then the following inequality for fractional integrals holds: a+b Γ(α + 1) α α [J f (b) + Jb− f (a)] − f ( ) 2(b − a)α a+ 2

≤ 0.5(b − a)(|f 󸀠 (a)| + |f 󸀠 (b)|)

× (B0.5 (α + 1, s + 1) − B0.5 (s + 1, α + 1) +

2−α−1 1 + ). α+s+1 s+1

Proof. By using Definition 21, Lemma 36 and Lemma 50, we have 󵄨󵄨󵄨 Γ(α + 1) α a + b 󵄨󵄨󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2(b − a)α [Ja+ f (b) + Jb− f (a)] − f ( 2 )󵄨󵄨󵄨 󵄨 󵄨 1

≤

b−a 󵄨 󵄨 ∫ |h(t) − (1 − t)α + t α |󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 0

1

b−a 󵄨 󵄨 ≤ ∫ |−1 − (1 − t)α + t α |󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 1 2

1 2

b−a 󵄨 󵄨 + ∫ |1 − (1 − t)α + t α |󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 1

≤

0

b−a 󵄨 󵄨 ∫(1 + (1 − t)α − t α )󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 1 2

1 2

b−a 󵄨 󵄨 + ∫(1 − (1 − t)α + t α )󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 1

≤

0

b−a 󵄨 󵄨 ∫(1 + (1 − t)α − t α )󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨dt 2 1 2

1 2

b−a 󵄨 󵄨 + ∫(1 − (1 − t)α + t α )󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨dt 2 1

≤

0

b−a ∫(1 + (1 − t)α − t α )[t s |f 󸀠 (a)| + (1 − t)s |f 󸀠 (b)|]dt 2 1 2

1 2

b−a + ∫(1 − (1 − t)α + t α )[t s |f 󸀠 (a)| + (1 − t)s |f 󸀠 (b)|]dt 2 0

190 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

1

1

(b − a)|f 󸀠 (a)| s (b − a)|f 󸀠 (a)| s (b − a)|f 󸀠 (a)| α+s ≤− ∫ t dt + ∫ t dt + ∫ t (1 − t)α dt 2 2 2 1 2

+

1

1 2

1

(b − a)|f 󸀠 (b)| (b − a)|f 󸀠 (b)| α ∫(1 − t)α+s dt − ∫ t (1 − t)s dt 2 2 1 2

1 2

1

+

1 2

(b − a)|f 󸀠 (b)| (b − a)|f 󸀠 (a)| α+s ∫ t dt ∫(1 − t)s dt + 2 2 1 2 1 2

0

1 2

(b − a)|f (a)| s (b − a)|f (a)| s ∫ t (1 − t)α dt + ∫ t dt 2 2 󸀠

−

1 2

0

1 2

󸀠

0 1 2

(b − a)|f 󸀠 (b)| (b − a)|f 󸀠 (b)| α − ∫(1 − t)α+s dt + ∫ t (1 − t)s dt 2 2 0

1 2

+

0

(b − a)|f 󸀠 (b)| ∫(1 − t)s dt 2 0

(b − a)|f 󸀠 (a)| 1 − 2−α−s−1 (b − a)|f 󸀠 (a)| 1 − 2−s−1 ≤− + 2 α+s+1 2 s+1 (b − a)|f 󸀠 (b)| 2−α−s−1 (b − a)|f 󸀠 (a)| B0.5 (α + 1, s + 1) + + 2 2 α+s+1 (b − a)|f 󸀠 (b)| (b − a)|f 󸀠 (b)| 2−s−1 − B0.5 (s + 1, α + 1) + 2 2 s+1 (b − a)|f 󸀠 (a)| 2−α−s−1 + 2 α+s+1 (b − a)|f 󸀠 (a)| 2−s−1 (b − a)|f 󸀠 (a)| B0.5 (s + 1, α + 1) + − 2 2 s+1 (b − a)|f 󸀠 (b)| 1 − 2−α−s−1 − 2 α+s+1 (b − a)|f 󸀠 (b)| 1 − 2−s−1 (b − a)|f 󸀠 (b)| + + B0.5 (α + 1, s + 1) 2 s+1 2 (b − a)|f 󸀠 (a)| 2−α−s − 1 (b − a)|f 󸀠 (a)| ≤ + [B0.5 (α + 1, s + 1) − B0.5 (s + 1, α + 1)] 2 α+s+1 2 󸀠 −α−s 󸀠 (b − a)|f (a)| (b − a)|f (b)| 2 −1 + + 2(s + 1) 2 α+s+1 (b − a)|f 󸀠 (b)| (b − a)|f 󸀠 (b)| + [B0.5 (α + 1, s + 1) − B0.5 (s + 1, α + 1)] + 2 2(s + 1) ≤ 0.5(b − a)(|f 󸀠 (a)| + |f 󸀠 (b)|) 2−α−1 1 × (B0.5 (α + 1, s + 1) − B0.5 (s + 1, α + 1) + + ). α+s+1 s+1 The proof is completed.

4.8 Inequalities via geometrical-arithmetically s-convex functions | 191

Theorem 206. Let f : [0, b] → ℝ be a differentiable mapping and 1 < q < ∞. If |f 󸀠 |q is measurable and |f 󸀠 |q is decreasing and geometric-arithmetically s-convex on [0, b] for some fixed α ∈ (0, ∞), s ∈ (0, 1], 0 ≤ a < b, then the following inequality for fractional integrals holds: 󵄨󵄨 Γ(α + 1) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 α α [J )󵄨󵄨 f (b) + J f (a)] − f ( 󵄨󵄨 a+ b− 󵄨󵄨 󵄨󵄨 2(b − a)α 2 1

1

b − a |f 󸀠 (a)|q + |f 󸀠 (b)|q q (1 + 21−α )p 2p (1 − 2−pα ) p ≤ max{ ( ) ( − ) , 2 s+1 2 (pα + 1) 1

1

|f 󸀠 (a)|q + |f 󸀠 (b)|q q p−1 2p (1 − 2−pα−1 ) p (b − a)( ) (2 − ) }, s+1 pα + 1 where

1 p

+

1 q

= 1.

Proof. To achieve our aim, we divide our proof into two cases. Case 1: α ∈ (0, 1). By using Definition 21, Lemma 39, Lemma 50 and Hölder inequality, we have 󵄨󵄨 Γ(α + 1) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 α α [Ja+ f (b) + Jb− f (a)] − f ( )󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 󵄨󵄨 2(b − a) 1

≤

b−a 󵄨 󵄨 ∫ |h(t) − (1 − t)α + t α |󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 0

1 p

1

1

b−a 󵄨 󵄨q (∫ |h(t) − (1 − t)α + t α |p dt) (∫ 󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) ≤ 2 0

1 p

1

0

1

b−a 󵄨 󵄨q ≤ (∫ |h(t) − (1 − t)α + t α |p dt) (∫ 󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨 dt) 2 0

1 p

1

1 q

1 q

0

1

b−a ≤ (∫ |h(t) − (1 − t)α + t α |p dt) (∫[t s |f 󸀠 (a)|q + (1 − t)s |f 󸀠 (b)|q ]dt) 2 0

0

1

1 q

b − a |f 󸀠 (a)|q + |f 󸀠 (b)|q ≤ ( ) (∫ |h(t) − (1 − t)α + t α |p dt) 2 s+1 ≤

q

q

b − a |f (a)| + |f (b)| ( ) 2 s+1 󸀠

1

󸀠

1 q

0

1 2

α p

α p

× (∫(1 − t α + (1 − t) ) dt + ∫(1 − (1 − t)α + t ) dt) 0

1 2

≤

b − a |f 󸀠 (a)|q + |f 󸀠 (b)|q ( ) 2 s+1

1 q

1 p

1 p

1 q

192 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

α

× (∫[1 − t + 2

1−α

1 2

α p

α

α

p

− t ] dt + ∫[(1 + t ) − (1 − t )] dt)

1 p

0

1 2

q

q

1 q

1 2

1

b − a |f (a)| + |f (b)| p ( ) (∫[(1 + 21−α ) − 2p t pα ]dt + ∫ 2p t pα dt) ≤ 2 s+1 󸀠

󸀠

1 p

0

1 2

1 q

1

b − a |f 󸀠 (a)|q + |f 󸀠 (b)|q (1 + 21−α )p 2p (1 − 2−pα ) p ≤ ( ) ( − ) . 2 s+1 2 (pα + 1) Case 2: α ∈ [1, ∞). By using Definition 21, Lemma 36, Lemma 39 and Lemma 50, we have 󵄨󵄨 Γ(α + 1) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 α [J α f (b) + Jb− f (a)] − f ( )󵄨󵄨 󵄨󵄨 󵄨󵄨 2(b − a)α a+ 󵄨󵄨 2 1

b − a |f 󸀠 (a)|q + |f 󸀠 (b)|q q ( ) ≤ 2 s+1 1

1 2

α p

α p

× (∫(1 − t α + (1 − t) ) dt + ∫(1 − (1 − t)α + t ) dt)

1 p

0

1 2 1

1

|f 󸀠 (a)|q + |f 󸀠 (b)|q q p ) (∫(1 − t α + 1 − t α ) dt) ≤ (b − a)( s+1

1 p

1 2

1

1

|f 󸀠 (a)|q + |f 󸀠 (b)|q q ≤ (b − a)( ) (∫ 2p (1 − t pα )dt) s+1

1 p

1 2

1

1

|f 󸀠 (a)|q + |f 󸀠 (b)|q q p−1 2p (1 − 2−pα−1 ) p ≤ (b − a)( ) (2 − ) . s+1 pα + 1 The proof is completed. Theorem 207. Let f : [0, b] → ℝ be a differentiable mapping. If |f 󸀠 | is measurable and |f 󸀠 | is decreasing and geometric-arithmetically s-convex on [0, b] for some fixed α ∈ (0, ∞), s ∈ (0, 1], 0 ≤ a < x < b, then the following inequality for fractional integrals holds: 󵄨󵄨 (x − a)α + (b − x)α 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨 f (x) − [Jx− f (a) + Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 b−a b−a α+1 α+1 󸀠 |(x − a) − (b − x) | |f (a)| ≤ [ + |f 󸀠 (b)|B(α + 1, s + 1)], b−a α+s+1

4.8 Inequalities via geometrical-arithmetically s-convex functions | 193

where 1

B(s + 1, α + 1) = ∫ t s (1 − t)α dt. 0

Proof. By using Definition 21, Lemma 36 and Lemma 56, we have 󵄨󵄨 (x − a)α + (b − x)α 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨 f (x) − [Jx− f (a) + Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 b−a b−a 1

|(x − a)α+1 − (b − x)α+1 | α 󵄨󵄨 󸀠 󵄨 ≤ ∫ t 󵄨󵄨f (tx + (1 − t)a)󵄨󵄨󵄨dt b−a 0

1

≤

|(x − a)α+1 − (b − x)α+1 | α 󵄨󵄨 󸀠 t 1−t 󵄨󵄨 ∫ t 󵄨󵄨f (x a )󵄨󵄨dt b−a 0

≤

|(x − a)

α+1

α+1

− (b − x) b−a

|

1

∫ t α [t s |f 󸀠 (a)| + (1 − t)s |f 󸀠 (b)|]dt 0

1

|(x − a)α+1 − (b − x)α+1 | ≤ ∫[t α+s |f 󸀠 (a)| + t α (1 − t)s |f 󸀠 (b)|]dt b−a 0

|(x − a)α+1 − (b − x)α+1 | |f 󸀠 (a)| ≤ [ + |f 󸀠 (b)|B(α + 1, s + 1)]. b−a α+s+1 The proof is completed. Theorem 208. Let f : [0, b] → ℝ be a differentiable mapping and 1 < q < ∞. If |f 󸀠 |q is measurable and |f 󸀠 |q is decreasing and geometric-arithmetically s-convex on [0, b] for some fixed α ∈ (0, ∞), s ∈ (0, 1], 0 ≤ a < b, then the following inequality for fractional integrals holds: 󵄨󵄨 (x − a)α + (b − x)α 󵄨󵄨 Γ(α + 1) α 󵄨 󵄨󵄨 f (x) − [Jx− f (a) + Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 b−a b−a 1

≤

b−a |f 󸀠 (a)|q + |f 󸀠 (b)|q q ( ) . 2(pα + 1) s+1

Proof. By using Definition 21, Lemma 36, Hölder inequality and Lemma 56, we have 󵄨󵄨 󵄨󵄨 (x − a)α + (b − x)α Γ(α + 1) α 󵄨 󵄨󵄨 f (x) − [Jx− f (a) + Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 b−a b−a 1

|(x − a)α+1 − (b − x)α+1 | α 󵄨󵄨 󸀠 󵄨 ≤ ∫ t 󵄨󵄨f (tx + (1 − t)a)󵄨󵄨󵄨dt b−a 1

1 p

0

1

b−a 󵄨 󵄨q ≤ (∫ t pα dt) (∫ 󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) 2 0

0

1 q

194 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

b−a 󵄨 󵄨q ≤ (∫ 󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) 2(pα + 1) 0

1

b−a 󵄨 󵄨q ≤ (∫ 󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨 dt) 2(pα + 1)

1 q

1 q

0

1

b−a ≤ (∫[t s |f 󸀠 (a)|q + (1 − t)s |f 󸀠 (b)|q ]dt) 2(pα + 1) 0

≤ where

1 p

+

1 q

1 q

1

b−a |f 󸀠 (a)|q + |f 󸀠 (b)|q q ( ) , 2(pα + 1) s+1

= 1. The proof is completed.

Theorem 209. Let f : [0, b] → ℝ be a differentiable mapping. If |f 󸀠󸀠 | is measurable and |f 󸀠󸀠 | is decreasing and geometric-arithmetically s-convex on [0, b] for some fixed α ∈ (0, ∞), s ∈ (0, 1], 0 ≤ a < b, then the following inequality for fractional integrals holds: 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) α 󵄨 󵄨󵄨 [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 − 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 (b − a)2 (|f 󸀠󸀠 (a)| + |f 󸀠󸀠 (b)|) 1 1 ≤ ( − ). 2(α + 1) s+1 α+s+2 Proof. By using Definition 21, Lemma 36 and Lemma 47, we have 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) α 󵄨 󵄨󵄨 [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 − 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 1 (b − a)2 󵄨󵄨󵄨󵄨 1 − (1 − t)α+1 − t α+1 󵄨󵄨󵄨󵄨󵄨󵄨 󸀠󸀠 󵄨 ≤ ∫ 󵄨󵄨 󵄨󵄨󵄨󵄨f (ta + (1 − t)b)󵄨󵄨󵄨dt 󵄨󵄨 2 α+1 󵄨󵄨 0

2

1

(b − a) 󵄨 󵄨 ≤ ∫(1 − (1 − t)α+1 − t α+1 )󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2(α + 1) 0

1

≤

(b − a)2 󵄨 󵄨 ∫(1 − (1 − t)α+1 − t α+1 )󵄨󵄨󵄨f 󸀠󸀠 (at b1−t )󵄨󵄨󵄨dt 2(α + 1) 0

2

≤

1

(b − a) ∫(1 − (1 − t)α+1 − t α+1 )[t s |f 󸀠󸀠 (a)| + (1 − t)s |f 󸀠󸀠 (b)|]dt 2(α + 1) 0

1

1

0

0

(b − a)2 |f 󸀠󸀠 (a)| α+s+1 (b − a)2 |f 󸀠󸀠 (a)| s ≤− dt + ∫t ∫ t dt 2(α + 1) 2(α + 1)

4.8 Inequalities via geometrical-arithmetically s-convex functions | 195 1

(b − a)2 |f 󸀠󸀠 (a)| s − ∫ t (1 − t)α+1 dt 2(α + 1) 0

−

2

+

󸀠󸀠

0

0

1

0

(b − a) |f (a)| 1 (b − a)2 |f 󸀠󸀠 (a)| 1 + 2(α + 1) α+s+2 2(α + 1) s+1

− − − ≤

1

(b − a) |f (b)| ∫(1 − t)s dt 2(α + 1) 2

≤−

1

(b − a)2 |f 󸀠󸀠 (b)| (b − a)2 |f 󸀠󸀠 (b)| α+1 ∫(1 − t)α+s+1 dt − ∫ t (1 − t)s dt 2(α + 1) 2(α + 1)

󸀠󸀠

(b − a)2 |f 󸀠󸀠 (a)| B(s + 1, α + 2) 2(α + 1)

(b − a)2 |f 󸀠󸀠 (b)| 1 (b − a)2 |f 󸀠󸀠 (b)| 1 + 2(α + 1) α+s+2 2(α + 1) s+1 (b − a)2 |f 󸀠󸀠 (b)| B(s + 1, α + 2) 2(α + 1)

(b − a)2 (|f 󸀠󸀠 (a)| + |f 󸀠󸀠 (b)|) 1 1 ( − ). 2(α + 1) s+1 α+s+2

The proof is completed. Theorem 210. Let f : [0, b] → ℝ be a differentiable mapping and 1 < q < ∞. If |f 󸀠󸀠 |q is measurable and |f 󸀠󸀠 |q is decreasing and geometric-arithmetically s-convex on [0, b] for some fixed α ∈ (0, ∞), s ∈ (0, 1], 0 ≤ a < b, then the following inequality for fractional integrals holds: 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) α 󵄨󵄨 󵄨󵄨 α [J f (b) + J f (a)] − + − 󵄨󵄨 󵄨󵄨 b 󵄨󵄨 󵄨󵄨 2 2(b − a)α a

1

(b − a)2 max{1 − 21−α , 21−α − 1} |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q ( ) . ≤ 2(α + 1) s+1 Proof. To achieve our aim, we divide our proof into two cases. Case 1: α ∈ (0, 1). By using Definition 21, Lemma 39, Lemma 36, Hölder inequality and Lemma 47, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨󵄨 α − [J + f (b) + Jb− f (a)]󵄨 󵄨󵄨 a 󵄨󵄨 󵄨󵄨 2 2(b − a)α 󵄨 1 α+1 α+1 2 󵄨 󵄨 (b − a) 󵄨󵄨 1 − (1 − t) − t 󵄨󵄨󵄨󵄨󵄨 󸀠󸀠 󵄨 ≤ ∫ 󵄨󵄨󵄨 󵄨󵄨󵄨󵄨f (ta + (1 − t)b)󵄨󵄨󵄨dt 󵄨󵄨 󵄨󵄨 2 α+1 0

1

1 p

1

(b − a)2 󵄨 󵄨q ≤ (∫ |1 − (1 − t)α − t α |p dt) (∫ 󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) 2(α + 1) 0

0

1 q

196 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1 p

1

1

(b − a)2 󵄨q 󵄨 ≤ (∫ |1 − (1 − t)α − t α |p dt) (∫ 󵄨󵄨󵄨f 󸀠󸀠 (at b1−t )󵄨󵄨󵄨 dt) 2(α + 1) 0

1 p

1

1 q

0

1

(b − a)2 ≤ (∫ |1 − (1 − t)α − t α |p dt) (∫[t s |f 󸀠󸀠 (a)|q + (1 − t)s |f 󸀠󸀠 (b)|q ]dt) 2(α + 1) 0

0

1

1 q

(b − a)2 |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q p ≤ ( ) (∫[(1 − t)α + t α − 1] dt) 2(α + 1) s+1 0

1

1 q

(b − a)2 |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q p ≤ ( ) (∫[21−α − 1] dt) 2(α + 1) s+1 0

2

1−α

1 p

1 p

1

(b − a) (2 − 1) |f (a)| + |f (b)|q q ≤ ( ) , 2(α + 1) s+1 󸀠󸀠

q

1 q

󸀠󸀠

where p1 + q1 = 1. Case 2: α ∈ [1, ∞). By using Definition 21, Lemma 39, Lemma 36, Hölder inequality and Lemma 47, we have 󵄨󵄨󵄨 󵄨󵄨󵄨 f (a) + f (b) Γ(α + 1) α 󵄨󵄨 − [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 2 2(b − a) 󵄨󵄨 󵄨 1

1

(b − a)2 |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q p ( ) (∫[1 − (1 − t)α − t α ] dt) ≤ 2(α + 1) s+1 0

2

1−α

(b − a) (1 − 2 ≤ 2(α + 1)

The proof is completed.

1

) |f (a)| + |f (b)|q q ( ) . s+1 󸀠󸀠

q

1 p

󸀠󸀠

Theorem 211. Let f : [0, b] → ℝ be a differentiable mapping. If |f 󸀠󸀠 | is measurable and |f 󸀠󸀠 | is decreasing and geometric-arithmetically s-convex functions on [0, b] for some fixed α ∈ (0, ∞), s ∈ (0, 1], 0 ≤ a < b, then the following inequality for fractional integrals holds: 󵄨󵄨󵄨 Γ(α + 1) α a + b 󵄨󵄨󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2(b − a)α [Ja+ f (b) + Jb− f (a)] − f ( 2 )󵄨󵄨󵄨 󵄨 󵄨 −s−1 −s−1 2 󸀠󸀠 −2 α+1 1 (b − a) |f (a)| α − α2 [ − + 2B(s + 1, α + 2) + ] ≤ 2(α + 1) 1+s 2+s α+s+2 + where

(b − a)2 |f 󸀠󸀠 (b)| α2−s−1 + 2−s−1 − 1 1 [ + + 2B(α + 2, s + 1)], 2(α + 1) 1+s α+s+2 1

∫ t s (1 − t)α+1 dt = B(s + 1, α + 2). 0

4.8 Inequalities via geometrical-arithmetically s-convex functions | 197

Proof. By using Definition 21, Lemma 36 and Lemma 53, we have 󵄨󵄨 Γ(α + 1) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 α α [J f (b) + J f (a)] − f ( )󵄨󵄨 󵄨󵄨 b− a+ 󵄨󵄨 󵄨󵄨 2(b − a)α 2 1

≤

(b − a)2 󵄨 󵄨 ∫ |m(t)|󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 2

≤

0 1

(b − a) 󵄨 󵄨 ∫ |m(t)|󵄨󵄨󵄨f 󸀠󸀠 (at b1−t )󵄨󵄨󵄨dt 2 0 1

≤

(b − a)2 ∫ |m(t)|[t s |f 󸀠󸀠 (a)| + (1 − t)s |f 󸀠󸀠 (b)|]dt 2 2

≤

0 1

(b − a) 1 − (1 − t)α+1 − t α+1 s 󸀠󸀠 |[t |f (a)| + (1 − t)s |f 󸀠󸀠 (b)|]dt ∫ |1 − t − 2 α+1 1 2

1 2

(b − a)2 1 − (1 − t)α+1 − t α+1 s 󸀠󸀠 |[t |f (a)| + (1 − t)s |f 󸀠󸀠 (b)|]dt + ∫ |t − 2 α+1 1

≤

0

(b − a)2 ∫ |α − tα − t + (1 − t)α+1 + t α+1 |[t s |f 󸀠󸀠 (a)| + (1 − t)s |f 󸀠󸀠 (b)|]dt 2(α + 1) 1 2

1 2

(b − a)2 + ∫ |tα + t − 1 + (1 − t)α+1 + t α+1 |[t s |f 󸀠󸀠 (a)| + (1 − t)s |f 󸀠󸀠 (b)|]dt 2(α + 1) 0

≤

1

(b − a)2 󸀠󸀠 |f (a)| ∫[αt s − (α + 1)t s+1 + t s (1 − t)α+1 + t α+s+1 ]dt 2(α + 1) 1 2

1

(b − a)2 󸀠󸀠 + |f (b)| ∫[α(1 − t)s − (α + 1)t(1 − t)s + (1 − t)α+s+1 + t α+1 (1 − t)s ]dt 2(α + 1) 1 2

2

+

1 2

(b − a) 󸀠󸀠 |f (a)| ∫[−t s + (α + 1)t s+1 + t s (1 − t)α+1 + t α+s+1 ]dt 2(α + 1) 0

1 2

(b − a)2 󸀠󸀠 + |f (b)| ∫[−(1 − t)s + (α + 1)t(1 − t)s + (1 − t)α+s+1 + t α+1 (1 − t)s ]dt 2(α + 1) 0

(b − a)2 󸀠󸀠 1 − 2−s−1 1 − 2−s−2 |f (a)|[α − (α + 1) + B(s + 1, α + 2) ≤ 2(α + 1) 1+s 2+s 1 − 2−α−s−2 + ] α+s+2

198 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals

+ + +

(b − a)2 󸀠󸀠 2−s−1 2−α−s−2 |f (b)|[α − (α + 1)B(2, s + 1) + + B(α + 2, s + 1)] 2(α + 1) s+1 α+s+2

(b − a)2 󸀠󸀠 2−s−1 2−s−2 2−α−s−2 |f (a)|[− + (α + 1) + B(α + 2, s + 1) + ] 2(α + 1) s+1 s+2 α+s+2 (b − a)2 󸀠󸀠 1 − 2−s−1 1 − 2−α−s−2 |f (b)|[− + (α + 1)B(2, s + 1) + 2(α + 1) 1+s α+s+2

+ B(α + 2, s + 1)] ≤

(b − a)2 |f 󸀠󸀠 (a)| α − α2−s−1 − 2−s−1 α + 1 1 [ − + 2B(s + 1, α + 2) + ] 2(α + 1) 1+s 2+s α+s+2 +

1 (b − a)2 |f 󸀠󸀠 (b)| α2−s−1 + 2−s−1 − 1 [ + + 2B(α + 2, s + 1)]. 2(α + 1) 1+s α+s+2

The proof is completed. Theorem 212. Let f : [0, b] → ℝ be a differentiable mapping. |f 󸀠󸀠 | is measurable and 1 < q < ∞. If |f 󸀠󸀠 |q is decreasing and geometric-arithmetically s-convex on [0, b] for some fixed α ∈ (0, ∞), s ∈ (0, 1], 0 ≤ a < b, then the following inequality for fractional integrals holds: Γ(α + 1) α a+b α [J f (b) + Jb− f (a)] − f ( ) 2(b − a)α a+ 2 1

1

(b − a)2 |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q (α + 1)2−p−1 + (α + 0.5)p+1 − αp+1 p ≤ ( ) ( ) , 2(α + 1) s+1 p+1 where

1 p

+

1 q

= 1.

Proof. By using Definition 21, Lemma 36, Hölder inequality and Lemma 53, we have 󵄨󵄨 Γ(α + 1) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 α α )󵄨󵄨 [Ja+ f (b) + Jb− f (a)] − f ( 󵄨󵄨 α 󵄨󵄨 2(b − a) 󵄨󵄨 2 1

(b − a)2 󵄨 󵄨 ≤ ∫ |m(t)|󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 0

1

1 p

1

(b − a)2 󵄨 󵄨q (∫ |m(t)|p dt) (∫ 󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) ≤ 2 0

1

1 p

0

1

(b − a)2 󵄨 󵄨q ≤ (∫ |m(t)|p dt) (∫ 󵄨󵄨󵄨f 󸀠󸀠 (at b1−t )󵄨󵄨󵄨 dt) 2 0

1

1 p

1 q

1 q

0

1

(b − a)2 ≤ (∫ |m(t)|p dt) (∫[t s |f 󸀠󸀠 (a)|q + (1 − t)s |f 󸀠󸀠 (b)|q ]dt) 2 0

0

1 q

4.8 Inequalities via geometrical-arithmetically s-convex functions | 199

1

1

(b − a)2 |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q ≤ ( ) (∫ |m(t)|p dt) 2 s+1

1 p

0

1

1

2 p 󵄨󵄨 1 − (1 − t)α+1 − t α+1 󵄨󵄨󵄨󵄨 (b − a)2 |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q 󵄨 ≤ ( ) (∫ 󵄨󵄨󵄨t − 󵄨󵄨 dt 󵄨󵄨 󵄨󵄨 2 s+1 α+1

0

1

1

p p 󵄨󵄨 1 − (1 − t)α+1 − t α+1 󵄨󵄨󵄨󵄨 󵄨 + ∫ 󵄨󵄨󵄨1 − t − 󵄨󵄨 dt) 󵄨󵄨 󵄨󵄨 α+1 1 2

1

1 2

(b − a)2 |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q ( ) (∫ |αt + t − 1 + (1 − t)α+1 + t α+1 |p dt ≤ 2(α + 1) s+1 0

1

α+1

+ ∫ |α − t + (1 − t)

+t

α+1 p

| dt)

1 p

1 2

2

q

q

1 q

1 2

1

(b − a) |f (a)| + |f (b)| ≤ ( ) ((α + 1) ∫ t p dt + ∫(α − t + 1)p dt) 2(α + 1) s+1 󸀠󸀠

󸀠󸀠

0

1

1 p

1 2

1

(b − a)2 |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q (α + 1)2−p−1 + (α + 0.5)p+1 − αp+1 p ( ) ( ) . ≤ 2(α + 1) s+1 p+1 The proof is completed. Theorem 213. Let f : [0, b] → ℝ be a differentiable mapping. If |f 󸀠󸀠 | is measurable and |f 󸀠󸀠 | is decreasing and geometric-arithmetically s-convex on [0, b] for some fixed α ∈ (0, ∞), s ∈ (0, 1], 0 ≤ a < b, then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 2 a+b Γ(α + 1) α 󵄨󵄨 󵄨 + f( )− [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 r+1 2 r(b − a) 󵄨󵄨 r(r + 1) −s−1 󸀠󸀠 −s−1 2 |f (a)| (1 − 2 )|f 󸀠󸀠 (b)| (b − a)2 max{[r + 1 − (r + 1)2−α ][ + ] ≤ r(r + 1)(α + 1) s+1 s+1 − r(α + 1)[ r(α + 1)[ +

2−s−2 |f 󸀠󸀠 (a)| + B0.5 (2, s + 1)|f 󸀠󸀠 (b)|], s+2

2−s−2 |f 󸀠󸀠 (a)| + B0.5 (2, s + 1)|f 󸀠󸀠 (b)|]} s+2

(b − a)2 max{[r + 1 − (r + 1)2−α − r(α + 1)] r(r + 1)(α + 1)

×[

(1 − 2−s−1 )|f 󸀠󸀠 (a)| 2−s−1 |f 󸀠󸀠 (b)| − ] s+1 s+1

200 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals (1 − 2−s−2 )|f 󸀠󸀠 (a)| + B0.5 (s + 1, 2)|f 󸀠󸀠 (b)|], s+2

+ r(α + 1)[ r(α + 1)[ −

(1 − 2−s−1 )|f 󸀠󸀠 (a)| − 2−s−1 |f 󸀠󸀠 (b)| s+1

(1 − 2−s−2 )|f 󸀠󸀠 (a)| − B0.5 (s + 1, 2)|f 󸀠󸀠 (b)|]}. s+2

Proof. By using Definition 21, Definition 3, Lemma 36 and Lemma 57, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 2 a+b Γ(α + 1) α 󵄨󵄨 󵄨 + f( )− [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 r(r + 1) 󵄨󵄨 r+1 2 r(b − a) 1

2

󵄨 󵄨 ≤ (b − a) ∫ |k(t)|󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 0

1

󵄨 󵄨 ≤ (b − a)2 ∫ |k(t)|󵄨󵄨󵄨f 󸀠󸀠 (at b1−t )󵄨󵄨󵄨dt 0

1

≤ (b − a) ∫ |k(t)|[t s |f 󸀠󸀠 (a)| + (1 − t)s |f 󸀠󸀠 (b)|]dt 2

0 1 2

󵄨󵄨 1 − (1 − t)α+1 − t α+1 t 󵄨󵄨󵄨󵄨 s 󸀠󸀠 󵄨 s 󸀠󸀠 − ≤ (b − a) ∫ 󵄨󵄨󵄨 󵄨[t |f (a)| + (1 − t) |f (b)|]dt 󵄨󵄨 r(α + 1) r + 1 󵄨󵄨󵄨 2

0

1

󵄨󵄨 1 − (1 − t)α+1 − t α+1 1 − t 󵄨󵄨 󵄨 󵄨󵄨 s 󸀠󸀠 s 󸀠󸀠 − + (b − a) ∫ 󵄨󵄨󵄨 󵄨[t |f (a)| + (1 − t) |f (b)|]dt 󵄨󵄨 r(α + 1) r + 1 󵄨󵄨󵄨 1 2

2

1 2

(b − a)2 ≤ ∫ |r + 1 − (r + 1)(t α+1 + (1 − t)α+1 ) − tr(α + 1)| r(r + 1)(α + 1) s

0

× [t |f (a)| + (1 − t)s |f 󸀠󸀠 (b)|]dt 󸀠󸀠

1

+

(b − a)2 ∫ |r + 1 + tr(α + 1) − (r + 1)(t α+1 + (1 − t)α+1 ) r(r + 1)(α + 1) 1 2

− r(α + 1)|[t s |f 󸀠󸀠 (a)| + (1 − t)s |f 󸀠󸀠 (b)|]dt

≤

(b − a)2 2−s−1 |f 󸀠󸀠 (a)| (1 − 2−s−1 )|f 󸀠󸀠 (b)| max{[r + 1 − (r + 1)2−α ][ + ] r(r + 1)(α + 1) s+1 s+1 − r(α + 1)[ r(α + 1)[ +

2−s−2 |f 󸀠󸀠 (a)| + B0.5 (2, s + 1)|f 󸀠󸀠 (b)|], s+2

2−s−2 |f 󸀠󸀠 (a)| + B0.5 (2, s + 1)|f 󸀠󸀠 (b)|]} s+2

(b − a)2 max{[r + 1 − (r + 1)2−α − r(α + 1)] r(r + 1)(α + 1)

4.8 Inequalities via geometrical-arithmetically s-convex functions | 201

×[

(1 − 2−s−1 )|f 󸀠󸀠 (a)| 2−s−1 |f 󸀠󸀠 (b)| − ] s+1 s+1

(1 − 2−s−2 )|f 󸀠󸀠 (a)| + B0.5 (s + 1, 2)|f 󸀠󸀠 (b)|], s+2

+ r(α + 1)[ r(α + 1)[

(1 − 2−s−1 )|f 󸀠󸀠 (a)| − 2−s−1 |f 󸀠󸀠 (b)| (1 − 2−s−2 )|f 󸀠󸀠 (a)| − s+1 s+2

− B0.5 (s + 1, 2)|f 󸀠󸀠 (b)|]}, where we have used the following inequalities: 1 2

∫[r + 1 − (r + 1)(t α+1 + (1 − t)α+1 ) − tr(α + 1)][t s |f 󸀠󸀠 (a)| + (1 − t)s |f 󸀠󸀠 (b)|]dt 0 1 2

≤ [r + 1 − (r + 1)2−α ] ∫[t s |f 󸀠󸀠 (a)| + (1 − t)s |f 󸀠󸀠 (b)|]dt 0 1 2

− r(α + 1) ∫[t s+1 |f 󸀠󸀠 (a)| + t(1 − t)s |f 󸀠󸀠 (b)|]dt 0

≤ [r + 1 − (r + 1)2−α ][ − r(α + 1)[

2−s−1 |f 󸀠󸀠 (a)| (1 − 2−s−1 )|f 󸀠󸀠 (b)| + ] s+1 s+1

2−s−2 |f 󸀠󸀠 (a)| + B0.5 (2, s + 1)|f 󸀠󸀠 (b)|], s+2

and 1 2

∫[−r − 1 + (r + 1)(t α+1 + (1 − t)α+1 ) + tr(α + 1)][t s |f 󸀠󸀠 (a)| + (1 − t)s |f 󸀠󸀠 (b)|]dt 0 1 2

≤ [−r − 1 + (r + 1)] ∫[t s |f 󸀠󸀠 (a)| + (1 − t)s |f 󸀠󸀠 (b)|]dt 0 1 2

+ r(α + 1) ∫[t s+1 |f 󸀠󸀠 (a)| + t(1 − t)s |f 󸀠󸀠 (b)|]dt ≤ r(α + 1)[

2

0 −s−2

|f 󸀠󸀠 (a)| + B0.5 (2, s + 1)|f 󸀠󸀠 (b)|], s+2

and 1

∫[r + 1 + tr(α + 1) − (r + 1)(t α+1 + (1 − t)α+1 ) 1 2

202 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals − r(α + 1)][t s |f 󸀠󸀠 (a)| + (1 − t)s |f 󸀠󸀠 (b)|]dt 1

≤ [r + 1 − (r + 1)2

−α

− r(α + 1)] ∫[t s |f 󸀠󸀠 (a)| + (1 − t)s |f 󸀠󸀠 (b)|]dt 1 2

1

+ r(α + 1) ∫[t s+1 |f 󸀠󸀠 (a)| + t(1 − t)s |f 󸀠󸀠 (b)|]dt 1 2

≤ [r + 1 − (r + 1)2−α − r(α + 1)][ + r(α + 1)[

(1 − 2−s−1 )|f 󸀠󸀠 (a)| 2−s−1 |f 󸀠󸀠 (b)| − ] s+1 s+1

(1 − 2−s−2 )|f 󸀠󸀠 (a)| + B0.5 (s + 1, 2)|f 󸀠󸀠 (b)|], s+2

and 1

∫[−r − 1 − tr(α + 1) + (r + 1)(t α+1 + (1 − t)α+1 ) 1 2

+ r(α + 1)][t s |f 󸀠󸀠 (a)| + (1 − t)s |f 󸀠󸀠 (b)|]dt 1

≤ [−r − 1 + (r + 1) + r(α + 1)] ∫[t s |f 󸀠󸀠 (a)| + (1 − t)s |f 󸀠󸀠 (b)|]dt 1 2

1

− r(α + 1) ∫[t s+1 |f 󸀠󸀠 (a)| + t(1 − t)s |f 󸀠󸀠 (b)|]dt 1 2

≤ r(α + 1)[

(1 − 2−s−1 )|f 󸀠󸀠 (a)| 2−s−1 |f 󸀠󸀠 (b)| − ] s+1 s+1

− r(α + 1)[ ≤ r(α + 1)[

(1 − 2−s−2 )|f 󸀠󸀠 (a)| + B0.5 (s + 1, 2)|f 󸀠󸀠 (b)|] s+2

(1 − 2−s−1 )|f 󸀠󸀠 (a)| − 2−s−1 |f 󸀠󸀠 (b)| (1 − 2−s−2 )|f 󸀠󸀠 (a)| − s+1 s+2

− B0.5 (s + 1, 2)|f 󸀠󸀠 (b)|]. The proof is completed. Theorem 214. Let f : [0, b] → ℝ be a differentiable mapping. |f 󸀠󸀠 | is measurable and 1 < q < ∞. If |f 󸀠󸀠 |q is decreasing and geometric-arithmetically s-convex on [0, b] for some fixed α ∈ (0, ∞), s ∈ (0, 1], 0 ≤ a < b, then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 2 a+b Γ(α + 1) α 󵄨󵄨 󵄨 + f( )− [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 r+1 2 r(b − a) 󵄨󵄨 r(r + 1)

4.8 Inequalities via geometrical-arithmetically s-convex functions | 203 1

(b − a)2 |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q ≤ ) ( −1 s+1 [r(r + 1)(α + 1)]1+p p+1

× (max{[r + 1 − (r + 1)2−α ]

[r(α + 1)]

p+1 −p−1

2

}

p+1

+ max{[r + 1 − (r + 1)2−α ] [0.5r(α + 1)] where

1 p

+

1 q

p+1

1 p

p+1

− [1 + 0.5r(1 − α) − (1 + r)2−α ]

p+1

− [1 + 0.5r(1 − α) + (1 + r)2−α ]

,

,

}) ,

= 1.

Proof. By using Definition 21, Lemma 36, Hölder inequality and Lemma 57, we have 󵄨󵄨 󵄨󵄨 f (a) + f (b) 2 a+b Γ(α + 1) α 󵄨 󵄨󵄨 [Ja+ f (b) + Jbα− f (a)]󵄨󵄨󵄨 + f( )− 󵄨󵄨 α 󵄨󵄨 r+1 2 r(b − a) 󵄨󵄨 r(r + 1) 1

󵄨 󵄨 ≤ (b − a)2 ∫ |k(t)|󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 0

1

1 p

1

󵄨 󵄨q ≤ (b − a) (∫ |k(t)| dt) (∫ 󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)b)󵄨󵄨󵄨 dt) 2

p

0

1

1 p

0

1

󵄨 󵄨q ≤ (b − a) (∫ |k(t)| dt) (∫ 󵄨󵄨󵄨f 󸀠󸀠 (at b1−t )󵄨󵄨󵄨 dt) 2

p

0

2

1

p

1 p

1 q

1 q

0

1

s

q

s

q

≤ (b − a) (∫ |k(t)| dt) (∫[t |f (a)| + (1 − t) |f (b)| ]dt) 0

󸀠󸀠

0 1

1

|f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q ≤ (b − a) ( ) (∫ |k(t)|p dt) s+1 2

󸀠󸀠

1 q

1 p

0

≤ (b − a)2 (

1 q

1 2

p 󵄨󵄨 1 − (1 − t)α+1 − t α+1 |f (a)| + |f (b)| t 󵄨󵄨󵄨󵄨 󵄨 ) (∫ 󵄨󵄨󵄨 − 󵄨󵄨 dt 󵄨󵄨 s+1 r(α + 1) r + 1 󵄨󵄨 󸀠󸀠

q

󸀠󸀠

q

0

1

1

󵄨p

p 󵄨󵄨 1 − (1 − t)α+1 − t α+1 1 − t 󵄨 󵄨 󵄨󵄨 − + ∫ 󵄨󵄨󵄨 󵄨󵄨 dt) r(α + 1) r + 1 󵄨󵄨 󵄨󵄨 1 2

1

(b − a)2 |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q ≤ ( ) r(r + 1)(α + 1) s+1 1 2

× (∫ |r + 1 − (r + 1)(t α+1 + (1 − t)α+1 ) − tr(α + 1)|p dt 0

204 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

+ ∫ |r + 1 + tr(α + 1) − (r + 1)(t α+1 + (1 − t)α+1 ) − r(α + 1)|p dt)

1 p

1 2 1

≤

(b − a)2 |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q ( ) r(r + 1)(α + 1) s+1 × (max{

[r + 1 − (r + 1)2−α ]p+1 − [1 + 0.5r(1 − α) − (1 + r)2−α ]p+1 , r(α + 1)(p + 1)

[r(α + 1)]p+1 2−p−1 } r(α + 1)(p + 1) + max{

[r + 1 − (r + 1)2−α ]p+1 − [1 + 0.5r(1 − α) + (1 + r)2−α ]p+1 , r(α + 1)(p + 1) 1

p [0.5r(α + 1)]p+1 }) r(α + 1)(p + 1)

1

(b − a)2 |f 󸀠󸀠 (a)|q + |f 󸀠󸀠 (b)|q q ≤ ( ) −1 s+1 [r(r + 1)(α + 1)]1+p p+1

× (max{[r + 1 − (r + 1)2−α ] p+1 −p−1

[r(α + 1)]

2

}

+ max{[r + 1 − (r + 1)2−α ] p+1

[0.5r(α + 1)]

1 p

p+1

− [1 + 0.5r(1 − α) − (1 + r)2−α ]

p+1

p+1

− [1 + 0.5r(1 − α) + (1 + r)2−α ]

,

,

}) ,

where we have used the following inequalities: 1 2

p

∫[r + 1 − (r + 1)(t α+1 + (1 − t)α+1 ) − tr(α + 1)] dt 0

≤

[r + 1 − (r + 1)2−α ]p+1 − [1 + 0.5r(1 − α) − (1 + r)2−α ]p+1 , r(α + 1)(p + 1)

and 1 2

p

∫[−r − 1 + (r + 1)(t α+1 + (1 − t)α+1 ) + tr(α + 1)] dt ≤ 0

[r(α + 1)]p+1 2−p−1 , r(α + 1)(p + 1)

and 1

p

(∫[r + 1 + tr(α + 1) − (r + 1)(t α+1 + (1 − t)α+1 ) − r(α + 1)] dt) 1 2

≤

[r + 1 − (r + 1)2−α ]p+1 − [1 + 0.5r(1 − α) + (1 + r)2−α ]p+1 , r(α + 1)(p + 1)

1 p

4.8 Inequalities via geometrical-arithmetically s-convex functions | 205

and 1

(∫[−r − 1 − tr(α + 1) + (r + 1)(t

α+1

α+1

+ (1 − t)

1 p

p

) + r(α + 1)] dt) ≤

1 2

[0.5r(α + 1)]p+1 . r(α + 1)(p + 1)

The proof is completed. By Definition 21, Lemma 37, Lemma 38 and Lemma 36 we have the following results. Theorem 215. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with 0 ≤ a < b. If |f 󸀠 | is measurable, decreasing and geometric-arithmetically s-convex on [a, b] for some fixed α ∈ (0, ∞), s ∈ (0, 1], then the following inequality for fractional integrals holds: 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) α 󵄨󵄨 󵄨󵄨 α − [J f (b) + J f (a)] 󵄨󵄨 ≤ Ks , 󵄨󵄨 b− 󵄨󵄨 󵄨󵄨 2 2(b − a)α a+

(4.120)

where α

1−s

b − a (1 − s)(2 1−s − 1) [ ] (|f 󸀠 (a)| + |f 󸀠 (b)|), α+1−s 22s+α (2α − 1)(b − a) 󸀠 (|f (a)| + |f 󸀠 (b)|), for s = 1. Ks := 2α+1 (α + 1)

Ks :=

for 0 < s < 1,

Proof. (i) Case 1: 0 < s < 1. From Definition 21, Lemma 37, Lemma 38 and Lemma 36, we have 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) α 󵄨 󵄨󵄨 α [Ja+ f (b) + Jb− f (a)]󵄨󵄨󵄨 − 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 󵄨󵄨 1 b − a 󵄨󵄨 1−t 1+t ≤ α+2 󵄨󵄨󵄨 ∫[(1 − t)α − (1 + t)α ]f 󸀠 ( a+ b)dt 󵄨󵄨 2 2 2 󵄨0 1

+ ∫[(1 + t)α − (1 − t)α ]f 󸀠 ( 0

≤

󵄨󵄨 󵄨󵄨 1+t 1−t b+ a)dt 󵄨󵄨󵄨 󵄨󵄨 2 2 󵄨

1 󵄨󵄨 1 − t 󵄨󵄨󵄨󵄨 b−a 1+t 󵄨 [ |(1 − t)α − (1 + t)α |󵄨󵄨󵄨f 󸀠 ( a+ b)󵄨󵄨dt ∫ α+2 󵄨󵄨 󵄨󵄨 2 2 2 1

0

󵄨󵄨 1 − t 󵄨󵄨󵄨󵄨 1+t 󵄨 b+ a)󵄨󵄨dt] + ∫ |(1 + t)α − (1 − t)α |󵄨󵄨󵄨f 󸀠 ( 󵄨󵄨 󵄨󵄨 2 2 0

1

1−t 1+t b−a 󵄨 󵄨 ≤ α+2 [∫ |(1 − t)α − (1 + t)α |󵄨󵄨󵄨f 󸀠 (a 2 b 2 )󵄨󵄨󵄨dt 2

0

206 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

1+t 1−t 󵄨 󵄨 + ∫ |(1 + t)α − (1 − t)α |󵄨󵄨󵄨f 󸀠 (b 2 a 2 )󵄨󵄨󵄨dt]

0

≤

1

s

0

1

+ ∫ |(1 + t)α − (1 − t)α |[( 0

=

s

b−a 1+t 1−t {∫ |(1 − t)α − (1 + t)α |[( ) |f 󸀠 (a)| + ( ) |f 󸀠 (b)|]dt 2 2 2α+2 s

s

1+t 1−t ) |f 󸀠 (b)| + ( ) |f 󸀠 (a)|]dt} 2 2

1

b−a {∫[(1 + t)α − (1 − t)α ][(1 + t)s |f 󸀠 (a)| + (1 − t)s |f 󸀠 (b)|]dt 2s+α+2 0

1

+ ∫[(1 + t)α − (1 − t)α ][(1 + t)s |f 󸀠 (b)| + (1 − t)s |f 󸀠 (a)|]dt} 0

1

b−a = s+α+2 (|f 󸀠 (a)| + |f 󸀠 (b)|) ∫[(1 + t)α − (1 − t)α ][(1 + t)s + (1 − t)s ]dt 2 0

≤

1

1 b−a 󸀠 (|f (a)| + |f 󸀠 (b)|)(∫[(1 + t)α − (1 − t)α ] 1−s dt) s+α+2 2

0

1

s

1 s

s

1−s

s

× (∫[(1 + t) + (1 − t) ] dt) 0

1−s

1

α α b−a ≤ s+α+2 (|f 󸀠 (a)| + |f 󸀠 (b)|)(∫[(1 + t) 1−s − (1 − t) 1−s ]dt) 2

× (2

1 −1 s

s

∫[(1 + t) + (1 − t)]dt) 0

=

0

1

α

1−s

b − a (1 − s)(2 1−s − 1) [ ] α+1−s 22s+α

(|f 󸀠 (a)| + |f 󸀠 (b)|).

(ii) Case 2: s = 1. Like in Case 1, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨 α − [Ja+ f (b) + Jb− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) 1

≤

1+t 1−t b−a 󵄨 󵄨 [∫ |(1 − t)α − (1 + t)α |󵄨󵄨󵄨f 󸀠 (a 2 b 2 )󵄨󵄨󵄨dt α+2 2

1

0

1+t 1−t 󵄨 󵄨 + ∫ |(1 + t)α − (1 − t)α |󵄨󵄨󵄨f 󸀠 (b 2 a 2 )󵄨󵄨󵄨dt]

0

4.8 Inequalities via geometrical-arithmetically s-convex functions | 207 1

1−t b−a 1+t )|f 󸀠 (a)| + ( )|f 󸀠 (b)|]dt ≤ α+2 {∫ |(1 − t)α − (1 + t)α |[( 2 2 2 1

0

+ ∫ |(1 + t)α − (1 − t)α |[( 0

1+t 1−t )|f 󸀠 (b)| + ( )|f 󸀠 (a)|]dt} 2 2

1

b−a = α+3 [∫[(1 + t)α − (1 − t)α ][(1 + t)|f 󸀠 (a)| + (1 − t)|f 󸀠 (b)|]dt 2 1

0

+ ∫[(1 + t)α − (1 − t)α ][(1 + t)|f 󸀠 (b)| + (1 − t)|f 󸀠 (a)|]dt] 0

=

1

b−a 󸀠 (|f (a)| + |f 󸀠 (b)|) ∫[(1 + t)α − (1 − t)α ]dt 2α+2 0

α

=

(2 − 1)(b − a) 󸀠 (|f (a)| + |f 󸀠 (b)|). 2α+1 (α + 1)

The proof is completed. Remark 216. If we take α = 1 in Theorem 215, then the equality (4.120) becomes the following inequality: b 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 f (a) + f (b) 1 󵄨󵄨 ≤ Ks , 󵄨󵄨 f (x)dx − ∫ 󵄨󵄨 󵄨󵄨 2 b−a 󵄨󵄨 󵄨󵄨 a

where 1

1−s

b − a (1 − s)(2 1−s − 1) [ ] (|f 󸀠 (a)| + |f 󸀠 (b)|), 2−s 22s+1 b−a 󸀠 Ks := (|f (a)| + |f 󸀠 (b)|), for s = 1. 8 Ks :=

for 0 < s < 1,

Theorem 217. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with 0 ≤ a < b. If |f 󸀠 | is measurable, decreasing and geometric-arithmetically s-convex on [a, b] for some fixed α ∈ (0, ∞), s ∈ (0, 1], then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨 α − [Ja+ f (b) + Jb− f (a)]󵄨󵄨󵄨 ≤ Kα , 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) where Kα :=

2(2α+s − 1) 1 b−a 󸀠 󸀠 (|f (a)| + |f (b)|)[ + + B(α + 1, s + 1) α+s+1 s+1 2s+α+2 1 − 2s−1 ( + B(s + 1, α + 1))], for 0 < α ≤ 1, α+1

(4.121)

208 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals

Kα :=

2(2α+s − 1) 1 b−a 󸀠 (|f (a)| + |f 󸀠 (b)|)[ + 2α−1 ( + B(α + 1, s + 1)) s+α+2 α+s+1 s+1 2 1 − 2s−1 ( + B(s + 1, α + 1))], for α > 1, α+1

where B(⋅, ⋅) denotes the Beta function.

Proof. (i) Case 1: 0 < α ≤ 1. From Definition 21, Lemma 37, Lemma 38 and Lemma 36, we have 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) α 󵄨 󵄨󵄨 α − [Ja+ f (b) + Jb− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) 󵄨󵄨 1 b − a 󵄨󵄨 1−t 1+t ≤ α+2 󵄨󵄨󵄨 ∫[(1 − t)α − (1 + t)α ]f 󸀠 ( a+ b)dt 󵄨󵄨 2 2 2 󵄨0 1

+ ∫[(1 + t)α − (1 − t)α ]f 󸀠 ( 0

󵄨󵄨 󵄨󵄨 1+t 1−t b+ a)dt 󵄨󵄨󵄨 󵄨󵄨 2 2 󵄨

1 󵄨󵄨 1+t 1 − t 󵄨󵄨󵄨󵄨 b−a 󵄨 a+ b)󵄨󵄨dt ≤ α+2 [∫ |(1 − t)α − (1 + t)α |󵄨󵄨󵄨f 󸀠 ( 󵄨󵄨 󵄨󵄨 2 2 2 1

0

󵄨󵄨 1 − t 󵄨󵄨󵄨󵄨 1+t 󵄨 b+ a)󵄨󵄨dt] + ∫ |(1 + t)α − (1 − t)α |󵄨󵄨󵄨f 󸀠 ( 󵄨󵄨 󵄨󵄨 2 2 0

1

1−t 1+t b−a 󵄨 󵄨 ≤ α+2 [∫ |(1 − t)α − (1 + t)α |󵄨󵄨󵄨f 󸀠 (a 2 b 2 )󵄨󵄨󵄨dt 2

1

0

1+t 1−t 󵄨 󵄨 + ∫ |(1 + t)α − (1 − t)α |󵄨󵄨󵄨f 󸀠 (b 2 a 2 )󵄨󵄨󵄨dt]

0

≤

1

s

1

0

+ ∫ |(1 + t)α − (1 − t)α |[( 0

=

s

1−t b−a 1+t {∫ |(1 − t)α − (1 + t)α |[( ) |f 󸀠 (a)| + ( ) |f 󸀠 (b)|]dt 2 2 2α+2 s

s

1+t 1−t ) |f 󸀠 (b)| + ( ) |f 󸀠 (a)|]dt} 2 2

1

b−a {∫[(1 + t)α − (1 − t)α ][(1 + t)s |f 󸀠 (a)| + (1 − t)s |f 󸀠 (b)|]dt 2s+α+2 1

0

+ ∫[(1 + t)α − (1 − t)α ][(1 + t)s |f 󸀠 (b)| + (1 − t)s |f 󸀠 (a)|]dt} 0

1

1

0

0

b−a = s+α+2 (|f 󸀠 (a)| + |f 󸀠 (b)|)[∫[(1 + t)α+s − (1 − t)α+s ]dt + ∫(1 + t)α (1 − t)s dt 2

4.8 Inequalities via geometrical-arithmetically s-convex functions | 209 1

− ∫(1 − t)α (1 + t)s dt] 0

b−a ≤ s+α+2 (|f 󸀠 (a)| + |f 󸀠 (b)|) 2 ×[

1

1

0

0

2(2α+s − 1) + ∫(1 + t α )(1 − t)s dt − 2s−1 ∫(1 + t s )(1 − t)α dt] α+s+1

b−a = s+α+2 (|f 󸀠 (a)| + |f 󸀠 (b)|) 2 1

1

1

0

0

0

2(2α+s − 1) ×[ + ∫(1 − t)s dt + ∫ t α (1 − t)s dt − 2s−1 ∫(1 − t)α dt α+s+1 1

− 2s−1 ∫ t s (1 − t)α dt] 0

b−a = s+α+2 (|f 󸀠 (a)| + |f 󸀠 (b)|) 2 2(2α+s − 1) 1 1 ×[ + + B(α + 1, s + 1) − 2s−1 ( + B(s + 1, α + 1))]. α+s+1 s+1 α+1 (ii) Case 2: α > 1. Like in Case 1, we have 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) α 󵄨 󵄨󵄨 α − [Ja+ f (b) + Jb− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 1

≤

b−a 󸀠 (|f (a)| + |f 󸀠 (b)|)[∫[(1 + t)α+s − (1 − t)α+s ]dt 2s+α+2 1

1

0

0

0

+ ∫(1 + t)α (1 − t)s dt − ∫(1 + t)s (1 − t)α dt] b−a ≤ s+α+2 (|f 󸀠 (a)| + |f 󸀠 (b)|) 2 ×[

1

1

0

0

2(2α+s − 1) + 2α−1 ∫(1 + t α )(1 − t)s dt − 2s−1 ∫(1 + t s )(1 − t)α dt] α+s+1

b−a = s+α+2 (|f 󸀠 (a)| + |f 󸀠 (b)|) 2 1

×[

1

2(2α+s − 1) + 2α−1 ∫(1 − t)s dt + 2α−1 ∫ t α (1 − t)s dt α+s+1 1

0

1

− 2s−1 ∫(1 − t)α dt − 2s−1 ∫ t s (1 − t)α dt] 0

0

0

210 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals

=

b−a 󸀠 (|f (a)| + |f 󸀠 (b)|) 2s+α+2 2(2α+s − 1) 1 ×[ + 2α−1 ( + B(α + 1, s + 1)) α+s+1 s+1 1 + B(s + 1, α + 1))]. − 2s−1 ( α+1

The proof is completed. Remark 218. If we take α = 1 in Theorem 217, then the equality (4.121) becomes the following inequality: b 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 f (a) + f (b) 1 󵄨󵄨 󵄨󵄨 f (x)dx − ∫ 󵄨󵄨 󵄨󵄨 2 b−a 󵄨󵄨 󵄨󵄨 a

≤

2(2s+1 − 1) 1 b−a 󸀠 (|f (a)| + |f 󸀠 (b)|)[ + s+3 s+2 s+1 2 1 + B(2, s + 1) − 2s−1 ( + B(s + 1, 2))]. 2

Theorem 219. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with 0 ≤ a < b. If |f 󸀠 |q , q > 1 is measurable, decreasing and geometric-arithmetically s-convex on [a, b] for some fixed α ∈ (0, ∞), s ∈ (0, 1], then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨󵄨 α − [J f (b) + J f (a)] 󵄨󵄨 󵄨󵄨 b− 󵄨󵄨 󵄨󵄨 2 2(b − a)α a+ qα

(q − 1)(2 q−1 − 1) [ ] ≤ α+s+ q1 +1 qα + q − 1 2 b−a

1

1− q1

q 1 +( ) ](|f 󸀠 (a)| + |f 󸀠 (b)|). s+1

Proof. From Definition 21, Lemma 37 and Lemma 38, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) α 󵄨󵄨 󵄨 α − [Ja+ f (b) + Jb− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(b − a) 󵄨󵄨 1 1−t 1+t b − a 󵄨󵄨 a+ b)dt ≤ α+2 󵄨󵄨󵄨 ∫[(1 − t)α − (1 + t)α ]f 󸀠 ( 󵄨󵄨 2 2 2 󵄨0 1

+ ∫[(1 + t)α − (1 − t)α ]f 󸀠 ( 0

󵄨󵄨 󵄨󵄨 1+t 1−t b+ a)dt 󵄨󵄨󵄨 󵄨󵄨 2 2 󵄨

1 󵄨󵄨 b−a 1+t 1 − t 󵄨󵄨󵄨󵄨 󵄨 ≤ α+2 [∫ |(1 − t)α − (1 + t)α |󵄨󵄨󵄨f 󸀠 ( a+ b)󵄨󵄨dt 󵄨󵄨 󵄨󵄨 2 2 2 0

1

2s+1 − 1 q [( ) s+1 (4.122)

4.8 Inequalities via geometrical-arithmetically s-convex functions | 211 1

󵄨󵄨 1+t 1 − t 󵄨󵄨󵄨󵄨 󵄨 b+ a)󵄨󵄨dt] + ∫ |(1 + t)α − (1 − t)α |󵄨󵄨󵄨f 󸀠 ( 󵄨󵄨 󵄨󵄨 2 2 0

1

q b−a ≤ α+2 [(∫ |(1 − t)α − (1 + t)α | q−1 dt) 2

1− q1

0

1

+ (∫ |(1 + t)α − (1 − t)α |

q q−1

1− q1

dt)

0

1

1

q q 󵄨󵄨 1+t 1 − t 󵄨󵄨󵄨󵄨 󵄨 a+ b)󵄨󵄨 dt) (∫ 󵄨󵄨󵄨f 󸀠 ( 󵄨󵄨 󵄨󵄨 2 2

0

1

1

q q 󵄨󵄨 1+t 1 − t 󵄨󵄨󵄨󵄨 󵄨 b+ a)󵄨󵄨 dt) ] (∫ 󵄨󵄨󵄨f 󸀠 ( 󵄨󵄨 󵄨󵄨 2 2

0

1

q b−a ≤ α+2 [(∫ |(1 − t)α − (1 + t)α | q−1 dt) 2

1− q1

0

1

α

α

+ (∫ |(1 + t) − (1 − t) |

q q−1

1− q1

dt)

1

1−t 1+t 󵄨 󵄨q (∫ 󵄨󵄨󵄨f 󸀠 (a 2 b 2 )󵄨󵄨󵄨 dt)

0

1 q

1

1+t 1−t 󵄨 󵄨q (∫ 󵄨󵄨󵄨f 󸀠 (b 2 a 2 )󵄨󵄨󵄨 dt) ]

0

0

1− q1

1

q b−a ≤ α+2 {(∫[(1 + t)α − (1 − t)α ] q−1 dt) 2

0

1

s

s

1−t 1+t ) |f 󸀠 (a)|q + ( ) |f 󸀠 (b)|q ]dt) × (∫[( 2 2

1 q

0

1

α

α

+ (∫[(1 + t) − (1 − t) ]

q q−1

1− q1

dt)

0 1

s

1 q

s

1+t 1−t × (∫[( ) |f 󸀠 (b)|q + ( ) |f 󸀠 (a)|q ]dt) } 2 2 0

1− q1

1

q b−a = α+s+2 [(∫[(1 + t)α − (1 − t)α ] q−1 dt) 2

0

1

s

q

s

q

× (∫[(1 + t) |f (a)| + (1 − t) |f (b)| ]dt) 󸀠

󸀠

1 q

0 1

α

α

+ (∫[(1 + t) − (1 − t) ]

q q−1

1− q1

dt)

0 1

1 q

× (∫[(1 + t)s |f 󸀠 (b)|q + (1 − t)s |f 󸀠 (a)|q ]dt) ] 0

1 q

212 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1− q1

1

q b−a = α+s+2 (∫[(1 + t)α − (1 − t)α ] q−1 dt) 2

0

1

s

q

s

q

× [(∫[(1 + t) |f (a)| + (1 − t) |f (b)| ]dt) 󸀠

󸀠

0 1

s

q

s

1 q

1 q

q

+ (∫[(1 + t) |f (b)| + (1 − t) |f (a)| ]dt) ] 󸀠

󸀠

0

1− q1

1

qα qα b−a ≤ α+s+2 (∫[(1 + t) q−1 − (1 − t) q−1 ]dt) 2

0

1

s

q

s

q

× [(∫[(1 + t) |f (a)| + (1 − t) |f (b)| ]dt) 󸀠

󸀠

0 1

s

q

1 q

1 q

s

󸀠

q

1− q1

|f 󸀠 (a)|q s+1 |f 󸀠 (b)|q q [( (2 − 1) + ) s+1 s+1

+ (∫[(1 + t) |f (b)| + (1 − t) |f (a)| ]dt) ] 󸀠

0

qα b−a 2(q − 1) = α+s+2 ( (2 q−1 − 1)) qα + q − 1 2

+(

1

|f 󸀠 (b)|q s+1 |f 󸀠 (a)|q q (2 − 1) + ) ] s+1 s+1 1− q1

qα

(q − 1)(2 q−1 − 1) [ ] ≤ 1 α+s+ q +1 qα + q − 1 2 b−a

1

+(

1

1

1

q 2s+1 − 1 q 󸀠 1 [( ) |f (a)| + ( ) |f 󸀠 (b)| s+1 s+1 1

q 2s+1 − 1 q 󸀠 1 ) |f (b)| + ( ) |f 󸀠 (a)|] s+1 s+1

1− q1

qα

(q − 1)(2 q−1 − 1) = [ ] 1 α+s+ q +1 qα + q − 1 2 b−a

1

× [(

1

q 2s+1 − 1 q 1 ) +( ) ](|f 󸀠 (a)| + |f 󸀠 (b)|). s+1 s+1

The proof is completed.

Remark 220. If we take α = 1 in Theorem 219, then the equality (4.122) becomes the following inequality: b 󵄨󵄨 󵄨󵄨 󵄨󵄨 f (a) + f (b) 󵄨󵄨 1 󵄨󵄨 − ∫ f (x)dx 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 2 b − a 󵄨󵄨 󵄨 a

b−a

q

(q − 1)(2 q−1 − 1) ≤ [ ] s+ q1 +2 2q − 1 2

1− q1

1

1

q 2s+1 − 1 q 1 [( ) +( ) ](|f 󸀠 (a)| + |f 󸀠 (b)|). s+1 s+1

4.8 Inequalities via geometrical-arithmetically s-convex functions | 213

4.8.2 Applications to some special means Next we apply the obtained results to provide some applications to special means of real numbers. Theorem 221. For some s ∈ (0, 1], n ∈ ℤ \ {−1, 0}, 0 ≤ a < b, the following inequality for fractional integrals holds: 1

1

n(b − a) a(n−1)q + b(n−1)q q 2 − 21−p p 󵄨󵄨 n n n n n 󵄨 ( ) ( ) , 󵄨󵄨A(a , b ) − Ln (a , b )󵄨󵄨󵄨 ≤ 2 s+1 p+1 where 1 1 + = 1, p q

1 < q < ∞.

Proof. Applying Theorem 204 for f (x) = xn , α = 1, one can obtain the result immediately. Theorem 222. For some s ∈ (0, 1], n ∈ ℤ \ {−1, 0}, 0 ≤ a < b, the following inequality for fractional integrals holds: 󵄨󵄨 n n n 󵄨 n 󵄨󵄨Ln (a , b ) − A (a, b)󵄨󵄨󵄨

1

a(n−1)q + b(n−1)q q ) ≤ (b − a)( s+1 p−2

× max{(2

1

1

2p−1 − 0.5 p 2p − 0.5 p − ) , (2p−1 − ) }, p+1 p+1

where 1 1 + = 1, p q

1 < q < ∞.

Proof. Applying Theorem 206 for f (x) = xn , α = 1, one can obtain the result immediately. Theorem 223. For some s ∈ (0, 1], n ∈ ℤ \ {−1, 0}, 1 < q < ∞, 0 ≤ a < x < b, the following inequality for fractional integrals holds: 1

b − a |f 󸀠 (a)|q + |f 󸀠 (b)|q q 󵄨󵄨 n n n n󵄨 ( ) . 󵄨󵄨Ln (a , b ) − x 󵄨󵄨󵄨 ≤ 2(p + 1) s+1 Proof. Applying Theorem 208 for f (x) = xn , α = 1, one can obtain the result immediately.

214 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals Theorem 224. For some s ∈ (0, 1], 0 ≤ a < b, the following inequality for fractional integrals holds: 1

1

󵄨󵄨 1 q 1 󵄨󵄨󵄨󵄨 b − a a2q + b2q 2 − 21−p p 󵄨󵄨 ) − ( ( ) , 󵄨󵄨 ≤ 󵄨󵄨 󵄨󵄨 H(a, b) L(a, b) 󵄨󵄨 2 p+1 (s + 1)a2q b2q where 1 1 + = 1, p q

1 < q < ∞.

Proof. Applying Theorem 204 for f (x) = x−1 , α = 1, one can obtain the result immediately. Theorem 225. For some s ∈ (0, 1], 0 ≤ a < b, the following inequality for fractional integrals holds: 󵄨󵄨 1 1 󵄨󵄨󵄨󵄨 󵄨󵄨 − 󵄨 󵄨󵄨 󵄨󵄨 H(a, b) A(a, b) 󵄨󵄨󵄨

1

1

q (1 + 21−α )p 2p−1 (1 − 2−pα ) p a2q + b2q ) − ) , max{( ≤ (b − a)( 4 (pα + 1) (s + 1)a2q b2q 1

(2p−1 −

2p (1 − 2−pα−1 ) p ) }, pα + 1

where 1 1 + = 1, p q

1 < q < ∞.

Proof. Applying Theorem 206 for f (x) = x−1 , α = 1, one can obtain the result immediately. Theorem 226. For some s ∈ (0, 1], 0 ≤ a < x < b, the following inequality for fractional integrals holds: 1

󵄨󵄨 1 q 1 󵄨󵄨󵄨 a2q + b2q b−a 󵄨󵄨 − 󵄨󵄨󵄨 ≤ ( ) , 󵄨󵄨 󵄨󵄨 H(a, b) x 󵄨󵄨 2(p + 1) (s + 1)a2q b2q where 1 1 + = 1, p q

1 < q < ∞.

Proof. Applying Theorem 208 for f (x) = x−1 , α = 1, one can obtain the result immediately.

4.9 Inequalities via (α, m)-logarithmically convex functions | 215

4.9 Inequalities via (α, m)-logarithmically convex functions The results in this section are based on [34, 98]. Now we are ready to present the main results in this section. Theorem 227. Let f : [0, b] → ℝ be a differentiable mapping. If |f 󸀠 | is measurable and (α, m)-logarithmically convex on [a, b] for some fixed α ∈ (0, 1], 0 ≤ a < mb ≤ b, then the following inequality for fractional integrals holds: 󵄨󵄨 󵄨󵄨 f (a) + f (mb) Γ(α + 1) 󵄨󵄨 󵄨󵄨 α − [RL Jaα+ f (mb) + RL Jmb − f (a)]󵄨 ≤ Ik , 󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(mb − a) 󵄨 where k2 (mb − a)|f 󸀠 (b)|m ∞ (− ln k)i−1 αi (k − iα ) Ik = ∑[ α [α; i] 2 i=1 −α

αi (− ln k)i−1 k 2 k2 k + ( iα+1 + iα−α+2 − )], [α; i − 1] 2 2 2 󸀠 m α |f (b)| (mb − a)(2 − 1) Ik = , for k = 1, (α + 1)2α |f 󸀠 (a)| k= 󸀠 , |f (b)|m −α

−α

for k ≠ 1,

and [α; 0] := 1,

[α; i] := (α + 1)(2α + 1) ⋅ ⋅ ⋅ (iα + 1),

i ∈ N.

Proof. (i) Case 1: k ≠ 1. By Definition 24, Lemma 33 and Lemma 39, we have 󵄨󵄨 f (a) + f (mb) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J + f (mb) + RL Jmb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 2 2(mb − a)α 󵄨 1

mb − a 󵄨 󵄨 ≤ ∫ |(1 − t)α − t α |󵄨󵄨󵄨f 󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨dt 2 0

1

≤

α α mb − a ∫ |(1 − t)α − t α ||f 󸀠 (a)|t |f 󸀠 (b)|m−mt dt 2

0

1

α (mb − a)|f 󸀠 (b)|m ≤ ∫ |(1 − t)α − t α |k t dt 2

0

1 2

1

α α (mb − a)|f 󸀠 (b)|m (mb − a)|f 󸀠 (b)|m ≤ ∫ |(1 − t)α − t α |k t dt + ∫ |(1 − t)α − t α |k t dt 2 2

0

1 2

216 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1 2

α α (mb − a)|f 󸀠 (b)|m ≤ ∫((1 − t)α − t α )(k t + k (1−t) )dt 2

0

≤

1 2

m

α α (mb − a)|f (b)| ∫((1 − t)α k t + (1 − t)α k (1−t) )dt 2

󸀠

0

1 2

α α (mb − a)|f 󸀠 (b)|m + ∫(−t α k t − t α k (1−t) )dt 2

0

1 2

1 2

α α (mb − a)|f 󸀠 (b)|m (mb − a)|f 󸀠 (b)|m ≤ ∫(1 − t)α k t dt + ∫(1 − t)α k (1−t) dt 2 2

0

−

0

1 2

1 2

0

0

α α (mb − a)|f 󸀠 (b)|m (mb − a)|f 󸀠 (b)|m ∫ t α k t dt − ∫ t α k (1−t) dt 2 2 1 2

≤

1

α α (mb − a)|f 󸀠 (b)|m (mb − a)|f 󸀠 (b)|m ∫(21−α − t α )k t dt + ∫ t α k t dt 2 2

0

1 2

1 2

1

α α (mb − a)|f 󸀠 (b)|m (mb − a)|f 󸀠 (b)|m − ∫(1 − t)α k t dt ∫ t α k t dt − 2 2

0

1 2

1 2

1

α α (mb − a)|f 󸀠 (b)|m (mb − a)|f 󸀠 (b)|m ≤ ∫ t α k t dt ∫(21−α − t α )k t dt + 2 2

0

1 2

1 2

1

α α (mb − a)|f 󸀠 (b)|m (mb − a)|f 󸀠 (b)|m − ∫ t α k t dt − ∫(1 − t α )k t dt 2 2

0

1 2

1 2

1

α α 2(mb − a)|f 󸀠 (b)|m 2(mb − a)|f 󸀠 (b)|m ≤ ∫ t α k t dt − ∫ t α k t dt 2 2

0

1 2

+ 21−α

1 2

1

α α (mb − a)|f 󸀠 (b)|m (mb − a)|f 󸀠 (b)|m ∫ k t dt − ∫ k t dt 2 2

0

1

α

1 2

1 2

α

≤ (mb − a)|f 󸀠 (b)|m ∫ t α k t dt − (mb − a)|f 󸀠 (b)|m ∫ t α k t dt 1 2

0

4.9 Inequalities via (α, m)-logarithmically convex functions | 217 1 2

1

α (mb − a)|f 󸀠 (b)|m + 2 (mb − a)|f (b)| ∫ k dt − ∫ k t dt 2

−α

m

󸀠

tα

0

1 2

1 2α

1

1 1 1 1 ≤ (mb − a)|f (b)| ∫ t ( α +1)−1 k t dt − (mb − a)|f 󸀠 (b)|m ⋅ ∫ t ( α +1)−1 k t dt α α

󸀠

m

0

1 2α

1 2α

1

1 1 1 (mb − a)|f 󸀠 (b)|m 1 + 2−α (mb − a)|f 󸀠 (b)|m ⋅ ∫ t α −1 k t dt − ⋅ ∫ t α −1 k t dt α 2 α

0

1 2α

1 2α

1

1 1 (mb − a)|f 󸀠 (b)|m ≤ ( ∫ t ( α +1)−1 k t dt − ∫ t ( α +1)−1 k t dt) α

0

1 2α

+

m

2 (mb − a)|f (b)| α −α

󸀠

1 2α

1

∫ t α −1 k t dt − 0

1

1 (mb − a)|f 󸀠 (b)|m ∫ t α −1 k t dt 2α 1 2α

1 2α

1

1 1 (mb − a)|f 󸀠 (b)|m (∫ t ( α +1)−1 k t dt − 2 ∫ t ( α +1)−1 k t dt) ≤ α

0

0

1 2α

+

1 2−α (mb − a)|f 󸀠 (b)|m ∫ t α −1 k t dt α

0

1 2α

1

1 1 (mb − a)|f 󸀠 (b)|m − (∫ t α −1 k t dt − ∫ t α −1 k t dt) 2α

0

0

1

1

i−1

∞ ∞ (− ln k) 1 α (− ln k)i−1 1 α (mb − a)|f 󸀠 (b)|m [k ∑ 1 − 2( α ) k 2α ∑ 2 1 ≤ α 2 ( + 1) ( + 1)i i i=1 i=1 +1

α

α

1

1

]

i−1

2−α (mb − a)|f 󸀠 (b)|m 1 α 21α ∞ (− 2α ln k) + ( α) k ∑ α 2 ( 1 )i i=1 α

1

1

i−1

∞ (mb − a)|f 󸀠 (b)|m (− ln k)i−1 1 α 1 ∞ (− α ln k) − (k ∑ − ( α ) k 2α ∑ 2 1 1 2α 2 ( )i ( )i i=1 i=1 α

≤

α

∞ ∞ −α (mb − a)|f 󸀠 (b)|m (− ln k)i−1 αi (− ln k)i−1 αi [k ∑ − k2 ∑ ] α [α; i] [α; i]2iα i=1 i=1

k 2 (mb − a)|f 󸀠 (b)|m ∞ αi (− ln k)i−1 ∑ iα+1 α i=1 [α; i − 1]2 −α

+

)

218 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals

−

∞ i ∞ −α α (− ln k)i−1 αi (− ln k)i−1 (mb − a)|f 󸀠 (b)|m (k ∑ − k2 ∑ ) iα−α+1 2α [α; i − 1] i=1 i=1 [α; i − 1]2

(mb − a)|f 󸀠 (b)|m ∞ (− ln k)i−1 αi k2 (k − iα ) ∑[ α [α; i] 2 i=1 −α

≤

αi (− ln k)i−1 k 2 k2 k ( iα+1 + iα−α+2 − )]. [α; i − 1] 2 2 2 −α

+

−α

(ii) Case 2: k = 1. By Definition 24, we have 󵄨󵄨 f (a) + f (mb) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J + f (mb) + RL Jmb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 2 2(mb − a)α 󵄨 1

mb − a 󵄨 󵄨 ≤ ∫ |(1 − t)α − t α |󵄨󵄨󵄨f 󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨dt 2 0 1

≤

α α mb − a ∫ |(1 − t)α − t α ||f 󸀠 (a)|t |f 󸀠 (b)|m−mt dt 2

0

≤

1

(mb − a)|f 󸀠 (b)|m ∫ |(1 − t)α − t α |dt 2 0

m

1 2

≤ (mb − a)|f (b)| ∫ |(1 − t)α − t α |dt 󸀠

0

|f 󸀠 (b)|m (mb − a)(2α − 1) . ≤ (α + 1)2α

The proof is completed.

Theorem 228. Let f : [0, b] → ℝ be a differentiable mapping and 1 < q < ∞. If |f 󸀠 |q is measurable and (α, m)-logarithmically convex on [a, b] for some fixed α ∈ (0, 1], 0 ≤ a < mb ≤ b, then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J + f (b) + RL Jb− f (a)]󵄨 ≤ Ik , 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 2 2(b − a)α 󵄨 where

1

(mb − a)|f 󸀠 (a)| 1 − 2−pα p ( ) Ik = 1 1 pα + 1 2q αq

1 q

(mq ln |f 󸀠 (b)| − q ln |f 󸀠 (a)|)i−1 αi × [∑ ] , [α; i − 1] i=1 ∞

1

(b − a)|f 󸀠 (b)|m 1 − 2−pα p Ik = ( ) , 1 pα + 1 2q and k =

|f 󸀠 (a)|q 1 , |f 󸀠 (b)|mq p

+

1 q

= 1.

for k = 1,

for k ≠ 1,

4.9 Inequalities via (α, m)-logarithmically convex functions | 219

Proof. (i) Case 1: k ≠ 1. By Definition 24, Lemma 33, and using Hölder inequality, we have 󵄨󵄨 󵄨󵄨 f (a) + f (mb) Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J + f (b) + RL Jmb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 2 2(mb − a)α 󵄨 1

mb − a 󵄨 󵄨 ≤ ∫ |(1 − t)α − t α |󵄨󵄨󵄨f 󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨dt 2 0

1 p

1

1

mb − a 󵄨 󵄨q ≤ [∫ |(1 − t)α − t α |p dt] [∫ 󵄨󵄨󵄨f 󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨 dt] 2 0

1 q

0

1 2

1 p

1

mb − a p 󵄨 󵄨q = [2 ∫((1 − t)α − t α ) dt] [∫ 󵄨󵄨󵄨f 󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨 dt] 2 0

0

≤

mb − a 2

1− p1

1 q

1 2

1 p

1

󵄨q

󵄨 [∫((1 − t)pα − t pα )dt] [∫ 󵄨󵄨󵄨f 󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨 dt] 0

0

1 p

1

mb − a 1 − 2−pα 󵄨 󵄨q ( ) [∫ 󵄨󵄨󵄨f 󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨 dt] = 1− p1 pα + 1 2

1 q

0

1 p

1

α α mb − a 1 − 2−pα ≤ ( ) [∫ |f 󸀠 (a)|qt |f 󸀠 (b)|mq−mqt dt] 1 pα + 1 2q

0

1

1

α (mb − a)|f 󸀠 (b)|m 1 − 2−pα p ) [∫ k t dt] ≤ ( 1 pα + 1 2q

1 q

1 q

0

1 p

∞ (mb − a)|f 󸀠 (b)|m 1 − 2−pα (− ln k)i−1 ≤ ( ) ] [k ∑ 1 1 pα + 1 ( 1 )i i=1 2q αq α

1 p

∞ (− ln k)i−1 (mb − a)|f 󸀠 (a)| 1 − 2−pα ≤ ( ) [∑ ] 1 1 pα + 1 ( 1 )i i=1 2q αq

1 q

1 q

α

1 p

1 q

1 q

(mb − a)|f (a)| 1 − 2 (mq ln |f 󸀠 (b)| − q ln |f 󸀠 (a)|)i−1 αi ≤ [ ( ) ] . ∑ 1 1 pα + 1 [α; i − 1] i=1 2q αq 󸀠

−pα

∞

(ii) Case 2: k = 1. By Definition 24, Lemma 33, and using Hölder inequality, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 [RL Jaα+ f (b) + RL Jbα− f (a)]󵄨󵄨󵄨 − 󵄨󵄨 α 󵄨󵄨 2 2(b − a) 󵄨󵄨 1

1

α α b − a 1 − 2−pα p ≤ ( ) [∫ |f 󸀠 (a)|qt |f 󸀠 (b)|mq−mqt dt] 1 pα + 1 2q

0

1 q

220 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

(b − a)|f 󸀠 (b)|m 1 − 2−pα p ≤ ( ) . 1 pα + 1 2q The proof is completed. Theorem 229. Let f : [0, b] → ℝ be a differentiable mapping. If |f 󸀠󸀠 | is measurable and (α, m)-logarithmically convex on [a, b] for some fixed α ∈ (0, 1], 0 ≤ a < mb ≤ b, then the following inequality for fractional integrals holds: 󵄨󵄨 f (a) + f (mb) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J f (mb) + J f (a)] + − 󵄨󵄨 󵄨󵄨 ≤ Ik , RL mb 󵄨󵄨 󵄨󵄨 2 2(mb − a)α RL a where Ik =

|f 󸀠󸀠 (a)|(mb − a)2 (2α − 1) ∞ (mα ln |f 󸀠󸀠 (b)| − α ln |f 󸀠󸀠 (a)|)i−1 , ∑ [α; i − 1] 2α+1 (α + 1) i=1

for k ≠ 1,

1

p 2 (b − a)2 |f 󸀠󸀠 (b)|m (1 − ) , Ik = 2(α + 1) pα + p + 1

and k =

for k = 1,

|f 󸀠󸀠 (a)| . |f 󸀠󸀠 (b)|m

Proof. (i) Case 1: k ≠ 1. By Definition 24, Lemma 32 and Lemma 39, we have 󵄨󵄨 f (a) + f (mb) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J + f (mb) + RL Jmb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 2 2(mb − a)α 󵄨 1

(mb − a)2 1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 ≤ ∫ 󵄨󵄨f (ta + m(1 − t)b)󵄨󵄨󵄨dt 2 α+1 0

1

≤

α α (mb − a)2 1 − (1 − t)α+1 − t α+1 󸀠󸀠 |f (a)|t |f 󸀠󸀠 (b)|m−mt dt ∫ 2 α+1

0

2

1

α (mb − a) |f 󸀠󸀠 (b)|m ≤ ∫[1 − (2−α − t α+1 ) − t α+1 ]k t dt 2(α + 1)

0

1

α (mb − a)2 |f 󸀠󸀠 (b)|m (2α − 1) ≤ ∫ k t dt 2α+1 (α + 1)

0

1

≤

1 (mb − a)2 |f 󸀠󸀠 (b)|m (2α − 1) ∫ t α −1 k t dt α+1 2 α(α + 1)

0

(mb − a)2 |f 󸀠󸀠 (b)|m (2α − 1) ∞ (− ln k)i−1 ≤ k∑ 2α+1 α(α + 1) ( 1 )i i=1 α

|f 󸀠󸀠 (a)|(mb − a)2 (2α − 1) ∞ (m ln |f 󸀠󸀠 (b)| − ln |f 󸀠󸀠 (a)|)i−1 αi ≤ ∑ [α; i − 1] 2α+1 α(α + 1) i=1

4.9 Inequalities via (α, m)-logarithmically convex functions | 221

≤

|f 󸀠󸀠 (a)|(mb − a)2 (2α − 1) ∞ (mα ln |f 󸀠󸀠 (b)| − α ln |f 󸀠󸀠 (a)|)i−1 . ∑ [α; i − 1] 2α+1 (α + 1) i=1

(ii) Case 2: k = 1. By Definition 24, Lemma 32, we have 󵄨󵄨 f (a) + f (mb) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J + f (mb) + RL Jmb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 2 2(mb − a)α 󵄨 1

α α (mb − a)2 1 − (1 − t)α+1 − t α+1 󸀠󸀠 ≤ |f (a)|t |f 󸀠󸀠 (b)|m−mt dt ∫ 2 α+1

0

≤

1

(mb − a)2 |f 󸀠󸀠 (b)|m ∫[1 − (2−α − t α+1 ) − t α+1 ]dt 2(α + 1) 0

(mb − a)2 |f 󸀠󸀠 (b)|m (2α − 1) . ≤ 2α+1 (α + 1) The proof is completed. Theorem 230. Let f : [0, b] → ℝ be a differentiable mapping and 1 < q < ∞. If |f 󸀠󸀠 |q is measurable and (α, m)-logarithmically convex on [a, b] for some fixed α ∈ (0, 1], 0 ≤ a < mb ≤ b, then the following inequality for fractional integrals holds: 󵄨󵄨 󵄨󵄨 f (a) + f (mb) Γ(α + 1) 󵄨󵄨 󵄨󵄨 α [RL Jaα+ f (mb) + RL Jmb − − f (a)]󵄨 ≤ Ik , 󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(mb − a) 󵄨 where 1

(mb − a)2 |f 󸀠󸀠 (a)| 1 p Ik = (1 − pα ) 2(α + 1) 2

1 q

(mqα ln |f 󸀠 (b)| − qα ln |f 󸀠 (a)|)i−1 ] , × [∑ [α; i − 1] i=1 ∞

for k ≠ 1,

1

(mb − a)2 |f 󸀠󸀠 (b)| 1 p Ik = (1 − pα ) , 2(α + 1) 2 and k =

|f 󸀠󸀠 (a)|q 1 , |f 󸀠󸀠 (b)|mq p

+

1 q

for k = 1,

= 1.

Proof. (i) Case 1: k ≠ 1. By Definition 24, Lemma 32, Lemma 39 and using Hölder inequality, we have 󵄨󵄨 f (a) + f (mb) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J + f (mb) + RL Jmb− f (a)]󵄨 󵄨󵄨 RL a 󵄨󵄨 󵄨󵄨 2 2(mb − a)α 󵄨 1

(mb − a)2 1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 ≤ ∫ 󵄨󵄨f (ta + m(1 − t)b)󵄨󵄨󵄨dt 2 α+1 0

222 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1 p

1

1

(mb − a)2 p 󵄨q 󵄨 ≤ (∫(1 − (1 − t)α+1 − t α+1 ) dt) (∫ 󵄨󵄨󵄨f 󸀠󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨 dt) 2(α + 1) 0

0

1

1 p

α α (mb − a)2 1 ≤ (1 − pα ) (∫ |f 󸀠󸀠 (a)|qt |f 󸀠󸀠 (b)|mq−mqt dt) 2(α + 1) 2

0

1

1

α (mb − a)2 |f 󸀠󸀠 (b)|m 1 p ≤ (1 − pα ) (∫ k t dt) 2(α + 1) 2

≤

≤

(mb − a)2 |f 󸀠󸀠 (a)| 1

2(α + 1)α q (mb − a)2 |f 󸀠󸀠 (a)| 1

2(α + 1)α q

(1 −

1

2pα

∞

) (∑ i=1

1 p

1 q

1 q

0

1 p

1 q

(− ln k)i−1

)

( α1 )i

(mq ln |f 󸀠 (b)| − q ln |f 󸀠 (a)|)i−1 αi (1 − pα ) [∑ ] 2 [α; i − 1] i=1 1

1 q

∞

1 q

1 q

1

1 p ∞ (mqα ln |f 󸀠 (b)| − qα ln |f 󸀠 (a)|)i−1 (mb − a)2 |f 󸀠󸀠 (a)| (1 − pα ) [∑ ] , ≤ 2(α + 1) 2 [α; i − 1] i=1 where we use the following inequality p

p

(1 − (1 − t)α+1 − t α+1 ) ≤ 1 − [(1 − t)α+1 + t α+1 ] p

≤ 1 − (2−α ) 1 = 1 − pα . 2

(ii) Case 2: k = 1. By Definition 24, Lemma 32 and using Hölder inequality, we have 󵄨󵄨 f (a) + f (mb) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α − [RL Jaα+ f (mb) + RL Jmb − f (a)]󵄨 󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(mb − a) 󵄨 1

(mb − a)2 1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 ≤ ∫ 󵄨󵄨f (ta + m(1 − t)b)󵄨󵄨󵄨dt 2 α+1 0

1 p

1

1

(mb − a)2 p 󵄨 󵄨q ≤ (∫(1 − (1 − t)α+1 − t α+1 ) dt) (∫ 󵄨󵄨󵄨f 󸀠󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨 dt) 2(α + 1) 0

0

1 p

1

α α (mb − a)2 1 ≤ (1 − pα ) (∫ |f 󸀠󸀠 (a)|qt |f 󸀠󸀠 (b)|mq−mqt dt) 2(α + 1) 2

0

2

m

1

(mb − a) |f (b)| 1 p ≤ (1 − pα ) . 2(α + 1) 2 󸀠󸀠

The proof is completed.

1 q

1 q

4.9 Inequalities via (α, m)-logarithmically convex functions | 223

All these main results are presented above by using Lemma 43. Now, we apply Lemma 52 to derive some left Hermite-Hadamard type inequalities for differentiable (α, m)-logarithmically convex functions. Theorem 231. Let f : [0, b] → ℝ be a differentiable mapping. If |f 󸀠 | is measurable and (α, m)-logarithmically convex on [a, b] for some fixed α ∈ (0, 1], 0 ≤ a < mb ≤ b, then the following inequality for fractional integrals holds: 󵄨󵄨 Γ(α + 1) a + mb 󵄨󵄨󵄨󵄨 󵄨󵄨 α α [Ja+ )󵄨󵄨 ≤ Ik , f (mb) + Jmb− f (a)] − f ( 󵄨󵄨 α 󵄨󵄨 2 󵄨󵄨 2(mb − a)

where

m

1−α

Ik = (mb − a)|f (b)| (1 + 2 󸀠

(− ln k)i−1 αi (k2iα+1−α − k 2 ) )∑ [α; i − 1]2iα+2−α i=1 −α

∞

(− ln k)i−1 αi (k 2 − k2iα ) , + (mb − a)|f (b)| ∑ [α; i]2iα i=1 󸀠

m

∞

−α

(mb − a)|f 󸀠 (b)|m (α2α−1 + 2α−1 + 1 − 2α ) , (α + 1)2α |f 󸀠 (a)| , k= 󸀠 |f (b)|m

Ik =

for k ≠ 1,

for k = 1,

and [α; 0] := 1,

[α; i] := (α + 1)(2α + 1) ⋅ ⋅ ⋅ (iα + 1),

i ∈ N.

Proof. (i) Case 1: k ≠ 1. By Definition 24, Lemma 33, Lemma 39 and Lemma 52, we have 󵄨󵄨 Γ(α + 1) a + mb 󵄨󵄨󵄨󵄨 󵄨󵄨 α α [Ja+ f (mb) + Jmb− f (a)] − f ( )󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 󵄨󵄨 2(mb − a) 1

≤

mb − a 󵄨 󵄨 ∫ |h(t) − (1 − t)α + t α |󵄨󵄨󵄨f 󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨dt 2 0

1

α α mb − a ≤ ∫ |h(t) − (1 − t)α + t α ||f 󸀠 (a)|t |f 󸀠 (b)|m−mt dt 2

0

≤

1

α (mb − a)|f 󸀠 (b)|m ∫ |h(t) − (1 − t)α + t α |k t dt 2

0

1 2

α (mb − a)|f 󸀠 (b)|m ≤ ∫ |1 − (1 − t)α + t α |k t dt 2

0

1

α (mb − a)|f 󸀠 (b)|m + ∫ |−1 − (1 − t)α + t α |k t dt 2 1 2

224 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1 2

α (mb − a)|f 󸀠 (b)|m ≤ ∫ |1 − (1 − t)α + t α |k t dt 2

0

1 2

α (mb − a)|f 󸀠 (b)|m + ∫ |−1 − t α + (1 − t)α |k (1−t) dt 2

0

1 2

≤

α α (mb − a)|f 󸀠 (b)|m ∫ |1 − (1 − t)α + t α |(k t + k (1−t) )dt 2

0

≤

m

1 2

m

1 2

α (mb − a)|f (b)| (mb − a)|f (b)| ∫ k t dt + ∫ k (1−t) dt 2 2

󸀠

α

0

󸀠

0

1 2

1 2

α α (mb − a)|f 󸀠 (b)|m (mb − a)|f 󸀠 (b)|m − ∫(1 − t)α k t dt − ∫(1 − t)α k (1−t) dt 2 2

0

+

0

1 2

1 2

0

0

α α (mb − a)|f 󸀠 (b)|m (mb − a)|f 󸀠 (b)|m ∫ t α k t dt + ∫ t α k (1−t) dt 2 2 1 2

1

α α (mb − a)|f 󸀠 (b)|m (mb − a)|f 󸀠 (b)|m ≤ ∫ k t dt + ∫ k t dt 2 2

0

1 2

1 2

1

α α (mb − a)|f 󸀠 (b)|m (mb − a)|f 󸀠 (b)|m − ∫(1 − t α )k t dt − ∫ t α k t dt 2 2

0

1 2

1 2

1

α α (mb − a)|f 󸀠 (b)|m (mb − a)|f 󸀠 (b)|m + ∫ t α k t dt + ∫(1 − t)α k t dt 2 2

0

1 2

1 2

1

α α (mb − a)|f 󸀠 (b)|m (mb − a)|f 󸀠 (b)|m ≤ ∫ k t dt + ∫ k t dt 2 2

0

1 2

1 2

1

α α (mb − a)|f 󸀠 (b)|m (mb − a)|f 󸀠 (b)|m − ∫(1 − t α )k t dt − ∫ t α k t dt 2 2

0

1 2

1 2

1

α α (mb − a)|f 󸀠 (b)|m (mb − a)|f 󸀠 (b)|m + ∫ t α k t dt + ∫(21−α − t α )k t dt 2 2

0

1 2

4.9 Inequalities via (α, m)-logarithmically convex functions | 225 1

α (mb − a)|f 󸀠 (b)|m (1 + 21−α ) ∫ k t dt ≤ 2 1 2

1

1 2

α

α

− (mb − a)|f 󸀠 (b)|m ∫ t α k t dt + (mb − a)|f 󸀠 (b)|m ∫ t α k t dt 0

1 2

1

≤

1 (mb − a)|f 󸀠 (b)|m (1 + 21−α ) ∫ t α −1 k t dt 2 1 2α

1

− (mb − a)|f 󸀠 (b)|m ∫ t

1 2α

( α1 +1)−1

1

k t dt + (mb − a)|f 󸀠 (b)|m ∫ t ( α +1)−1 k t dt 0

1 2α 1 2α

1

1 1 (mb − a)|f 󸀠 (b)|m ≤ (1 + 21−α )(∫ t α −1 k t dt − ∫ t α −1 k t dt) 2

0

1

m

− (mb − a)|f (b)| ∫ t 󸀠

0

( α1 +1)−1 t

m

× (k ∑

(− ln k)i−1 ( α1 )i

i=1

1

1

i−1

1 α 1 ∞ (− α ln k) − ( α ) k 2α ∑ 2 1 2 ( )i i=1 ∞

− (mb − a)|f 󸀠 (b)|m k ∑

1

0

0

(mb − a)|f 󸀠 (b)|m (1 + 21−α ) ≤ 2 ∞

1 2α

k dt + 2(mb − a)|f (b)| ∫ t ( α +1)−1 k t dt 󸀠

(− ln k)i−1 ( α1 + 1)i

i=1

1

)

α

1

i−1

∞ (− ln k) 1 α 1 α + 2(mb − a)|f (b)| ( α ) k 2α ∑ 2 1 2 ( + 1)i i=1 m

󸀠

+1

α

∞ ∞ −α (mb − a)|f 󸀠 (b)|m (− ln k)i−1 αi (− ln k)i−1 αi ≤ (1 + 21−α )(k ∑ − k2 ∑ ) iα+1−α 2 [α; i − 1] i=1 i=1 [α; i − 1]2 ∞

− (mb − a)|f 󸀠 (b)|m k ∑ i=1

m

∞ −α (− ln k)i−1 αi (− ln k)i−1 αi + 2(mb − a)|f 󸀠 (b)|m k 2 ∑ iα+1 [α; i] i=1 [α; i]2

∞ (mb − a)|f (b)| (− ln k)i−1 αi (k2iα+1−α − k 2 ) ≤ (1 + 21−α ) ∑ 2 [α; i − 1]2iα+1−α i=1 󸀠

−α

(− ln k)i−1 αi (k 2 − k2iα ) + (mb − a)|f (b)| ∑ [α; i]2iα i=1 m

󸀠

m

−α

∞

1−α

≤ (mb − a)|f (b)| (1 + 2 󸀠

(− ln k)i−1 αi (k2iα+1−α − k 2 ) )∑ [α; i − 1]2iα+2−α i=1 ∞

−α

226 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals (− ln k)i−1 αi (k 2 − k2iα ) + (mb − a)|f (b)| ∑ . [α; i]2iα i=1 m

󸀠

−α

∞

(ii) Case 2: k = 1. By Definition 24 and Lemma 52, we have 󵄨󵄨 Γ(α + 1) a + mb 󵄨󵄨󵄨󵄨 󵄨󵄨 α α [Ja+ )󵄨󵄨 f (mb) + Jmb− f (a)] − f ( 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2(mb − a) 2 1

≤

mb − a 󵄨 󵄨 ∫ |h(t) − (1 − t)α + t α |󵄨󵄨󵄨f 󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨dt 2 0

1

≤

α α mb − a ∫ |h(t) − (1 − t)α + t α ||f 󸀠 (a)|t |f 󸀠 (b)|m−mt dt 2

0

1

(mb − a)|f 󸀠 (b)|m ≤ ∫ |h(t) − (1 − t)α + t α |dt 2 0

≤

1 2

m

(mb − a)|f (b)| ∫ |1 − (1 − t)α + t α |dt 2 󸀠

0

1

(mb − a)|f (b)|m + ∫ |−1 − (1 − t)α + t α |dt 2 󸀠

1 2

1 2

(mb − a)|f 󸀠 (b)|m ≤ ∫ |1 − (1 − t)α + t α |dt 2 0

m

+

1 2

(mb − a)|f (b)| ∫ |−1 − t α + (1 − t)α |dt 2 󸀠

0

1 2

≤ (mb − a)|f 󸀠 (b)|m ∫ |1 − (1 − t)α + t α |dt 0 m

1 2

≤ (mb − a)|f (b)| ∫[1 − (1 − t)α + t α ]dt 󸀠

0

(mb − a)|f 󸀠 (b)|m (α2α−1 + 2α−1 + 1 − 2α ) ≤ . (α + 1)2α The proof is completed. Theorem 232. Let f : [0, b] → ℝ be a differentiable mapping and 1 < q < ∞. If |f 󸀠 |q is measurable and (α, m)-logarithmically convex on [a, b] for some fixed α ∈ (0, 1], 0 ≤ a
1) is an s-Godunova-Levin function, then for any 0 < α ≤ 1, a < x < b we have 󵄨󵄨 (x − a)[(a + x)f (x) − xf (a)] + (b − x)[(x + b)f (x) − xf (b)] 󵄨󵄨 󵄨󵄨 󵄨󵄨 (b − a)2 󵄨󵄨 Γ(α + 1) a b 󵄨󵄨 α α − ( RL Jx− f (a) + RL Jx+ f (b))󵄨󵄨󵄨 α−1 α−1 2 (b − x) (b − a) (x − a) 󵄨 1

1

p (x − a)2 |f 󸀠 (x)|q |f 󸀠 (a)|q q ap ≤ (( ) + x)( + ) 2 pα + 1 1−s 1−s (b − a)

244 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

1

p bp (b − x)2 |f 󸀠 (b)|q |f 󸀠 (x)|q q (( + ) + x)( + ) , 2 pα + 1 1−s 1−s (b − a)

where

1 p

+

1 q

= 1.

Proof. Using Lemma 64, Holder’s inequality and the fact that |f 󸀠 |q is an s-GodunovaLevin function, one has 󵄨󵄨 (x − a)[(a + x)f (x) − xf (a)] + (b − x)[(x + b)f (x) − xf (b)] 󵄨󵄨 󵄨󵄨 󵄨󵄨 (b − a)2 󵄨󵄨 Γ(α + 1) a b 󵄨󵄨 α α − ( RL Jx− f (a) + RL Jx+ f (b))󵄨󵄨󵄨 α−1 α−1 2 (b − x) (b − a) (x − a) 󵄨 1

(x − a)2 󵄨 󵄨 ≤ ∫(at α + x)󵄨󵄨󵄨f 󸀠 (tx + (1 − t)a)󵄨󵄨󵄨dt (b − a)2 0

1

(b − x)2 󵄨 󵄨 + ∫(b(1 − t)α + x)󵄨󵄨󵄨f 󸀠 (tb + (1 − t)x)󵄨󵄨󵄨dt (b − a)2 0

1 p

1

1

(x − a)2 p 󵄨 󵄨q (∫(at α + x) dt) (∫ 󵄨󵄨󵄨f 󸀠 (tx + (1 − t)a)󵄨󵄨󵄨 dt) ≤ (b − a)2 0

1 q

0

1 p

1

1

(b − x)2 p 󵄨 󵄨q + (∫(b(1 − t)α + x) dt) (∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)x)󵄨󵄨󵄨 dt) 2 (b − a) 0

0

1 p

1

1 p

1

(x − a)2 [(∫ ap t pα dt) + (∫ x p dt) ] ≤ (b − a)2 0

0

1

× (∫(t −s |f 󸀠 (x)|q + (1 − t)−s |f 󸀠 (a)|q )dt)

1 q

0 1 p

1

1 p

1

(b − x)2 + [(∫ bp (1 − t)pα dt) + (∫ x p dt) ] (b − a)2 0

1

0

q

q

× (∫(t |f (b)| + (1 − t) |f (x)| )dt) −s

󸀠

−s

󸀠

1 q

0 1

1

p (x − a)2 ap |f 󸀠 (x)|q |f 󸀠 (a)|q q = (( ) + x)( + ) pα + 1 1−s 1−s (b − a)2 1

1

p (b − x)2 |f 󸀠 (b)|q |f 󸀠 (x)|q q bp + (( ) + x)( + ) , 2 pα + 1 1−s 1−s (b − a)

1 q

4.10 Inequalities via s-Godunova-Levin functions | 245

where we use the following fact due to Minkowski’s inequality for p > 1: 1

α

1 p

p

1

p pα

1 p

1

1 p

p

(∫(at + x) dt) ≤ (∫ a t dt) + (∫ x dt) , 0 1

α

1 p

p

0 1

0

p

pα

1 p

1

1 p

p

(∫(b(1 − t) + x) dt) ≤ (∫ b (1 − t) dt) + (∫ x dt) . 0

0

0

The proof is completed. Corollary 243. In Theorem 242, if we choose x =

a+b , 2

we have

󵄨󵄨 (a + b)(4f ( a+b ) − f (a) − f (b)) 󵄨󵄨 2 󵄨󵄨 4(b − a) 󵄨󵄨 󵄨󵄨 2α−1 Γ(α + 1) 󵄨󵄨 α α (a J − f (a) + b J f (b)) 󵄨󵄨 a+b a+b − + RL ( RL ( ) ) 󵄨󵄨 (b − a)α+1 2 2 1

1

󸀠 a+b q 󸀠 q p ap a + b |f ( 2 )| + |f (a)| q 1 ) + )( ) ≤ (( 4 pα + 1 2 1−s 1

+

1

󸀠 a+b q 󸀠 q p 1 bp a + b |f ( 2 )| + |f (b)| q (( ) + )( ) . 4 pα + 1 2 1−s

Theorem 244. Let f : [a, b] → ℝ be twice differentiable. If f 󸀠󸀠 ∈ L[a, b] and |f 󸀠󸀠 | is an s-Godunova-Levin function, then for any 0 < α ≤ 1 and a < x < b, we have 󵄨󵄨 (x(α + 1) + b)f (x) − bf (b) (b(α + 1) + x)f (x) − xf (a) 1 󵄨󵄨 [ + ] 󵄨󵄨 b−x x−a 󵄨󵄨 (b − a)2 󵄨󵄨 x b (α + 1)Γ(α + 1) 󵄨󵄨 α α ( − RL Jx+ f (b) + RL Jx− f (a))󵄨󵄨󵄨 α+1 α+1 2 (b − x) (x − a) (b − a) 󵄨 b x b−x + )|f 󸀠󸀠 (x)| ≤ [( (b − a)2 α − s + 2 2 − s xΓ(α + 2)Γ(1 − s) b +( + )|f 󸀠󸀠 (b)|] Γ(α − s + 3) (1 − s)(2 − s) x−a x bΓ(α + 2)Γ(1 − s) + [( + )|f 󸀠󸀠 (a)| Γ(α − s + 3) (1 − s)(2 − s) (b − a)2 b x +( + )|f 󸀠󸀠 (x)|]. α−s+2 2−s Proof. Using Lemma 65 via |f 󸀠󸀠 | is an s-Godunova-Levin function, we have 󵄨󵄨 (x(α + 1) + b)f (x) − bf (b) (b(α + 1) + x)f (x) − xf (a) 1 󵄨󵄨 [ + ] 󵄨󵄨 󵄨󵄨 (b − a)2 b−x x−a 󵄨󵄨 (α + 1)Γ(α + 1) x b 󵄨󵄨 α α − ( RL Jx+ f (b) + RL Jx− f (a))󵄨󵄨󵄨 α+1 α+1 2 (b − x) (x − a) (b − a) 󵄨

246 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

(b − x) 󵄨 󵄨 ≤ ∫(t α+1 x + tb)󵄨󵄨󵄨f 󸀠󸀠 (tx + (1 − t)b)󵄨󵄨󵄨dt (b − a)2 0

1

(x − a) 󵄨 󵄨 + ∫((1 − t)α+1 b + (1 − t)x)󵄨󵄨󵄨f 󸀠󸀠 (ta + (1 − t)x)󵄨󵄨󵄨dt (b − a)2 0

1

≤

(b − x) ∫(t α+1 x + tb)(t −s |f 󸀠󸀠 (x)| + (1 − t)−s |f 󸀠󸀠 (b)|)dt (b − a)2 0

1

(x − a) + ∫((1 − t)α+1 b + (1 − t)x)(t −s |f 󸀠󸀠 (a)| + (1 − t)−s |f 󸀠󸀠 (x)|) (b − a)2 0

=

x b b−x [( + )|f 󸀠󸀠 (x)| 2 α − s + 2 2 − s (b − a) +( +

xΓ(α + 2)Γ(1 − s) b + )|f 󸀠󸀠 (b)|] Γ(α − s + 3) (1 − s)(2 − s)

bΓ(α + 2)Γ(1 − s) x x−a [( + )|f 󸀠󸀠 (a)| Γ(α − s + 3) (1 − s)(2 − s) (b − a)2

+(

x b + )|f 󸀠󸀠 (x)|]. α−s+2 2−s

The proof is completed.

4.10.2 Applications to some special means Theorem 245. For some s ∈ (0, 1], 0 ≤ a < b, the following inequality for fractional integrals holds: 󵄨󵄨 (4( a+b )n − an − bn ) 󵄨󵄨 2 A(a, b) 󵄨󵄨 󵄨󵄨 2(b − a) −(

󵄨󵄨 a a+b b a+b 󵄨 Lnn (a, )+ Lnn ( , b))󵄨󵄨󵄨 󵄨󵄨 2(b − a) 2 2(b − a) 2 n−1

≤

a a+b a+b n [( + )( ) 4 2 − s 2(1 − s) 2 +

+(

aΓ(2)Γ(1 − s) a+b + )an−1 ] Γ(3 − s) 2(1 − s)

n−1

n bΓ(2)Γ(1 − s) a+b b a+b a+b [( + )bn−1 + ( + )( ) 4 Γ(3 − s) 2(1 − s) 2 − s 2(1 − s) 2

Proof. Applying Theorem 239 for f (t) = t n , α = 1 and x = immediately.

].

a+b , one can obtain the result 2

4.11 Inequalities via (s, m)-Godunova-Levin functions | 247

Theorem 246. For some s ∈ (0, 1], 0 ≤ a < b, the following inequality for fractional integrals holds: 󵄨󵄨 ( 1 − 1 − 1 ) 󵄨󵄨 b a 1 1 󵄨󵄨 2(a+b) a b 󵄨󵄨 + ) A(a, b) − ( 󵄨󵄨 󵄨 󵄨󵄨 2(b − a) 2(b − a) L(a, a+b ) 2(b − a) L( a+b , b) 󵄨󵄨󵄨 2 2 1

≤

1

q 1 ap p a + b 4q 1 (( ) + )( + 2q ) 2q 4 p+1 2 (1 − s)(a + b) a (1 − s) 1

1

q 1 1 bp p a + b 4q + 2q + (( ) + )( ) , 2q 4 p+1 2 (1 − s)(a + b) b (1 − s)

where

1 p

+

1 q

= 1, 1 < q < ∞.

Proof. Applying Theorem 242 for f (t) = 1t , α = 1 and x = immediately.

a+b , one can obtain the result 2

4.11 Inequalities via (s, m)-Godunova-Levin functions The results in this section are taken from [95]. 4.11.1 Main results Theorem 247. Let f : [a, b] → ℝ be twice differentiable on (a, b) with a ≥ 0, f 󸀠󸀠 ∈ L[a, b] and a < mb ≤ b. If |f 󸀠󸀠 | is an (s, m)-Godunova-Levin function, then for any 0 < α ≤ 1 we have 󵄨󵄨 f (mb) + f (a) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − (RL Jmb − f (a) + RL Ja+ f (mb))󵄨 󵄨󵄨 󵄨󵄨 α 2 2(mb − a) 󵄨 󵄨󵄨 󸀠󸀠 2 (mb − a) |f (b)| (Φ1 (α, s)|f 󸀠󸀠 (a)| + Φ2 (α, s) ), ≤ 2 m where Φ1 (α, s) =

1 1 Γ(α + 2)Γ(1 + s) − − , 1−s α−s+2 Γ(α − s + 3)

and Φ2 (α, s) =

s+2 1 Γ(1 + s)Γ(α + 2) 2(α + 2) + s − − − . s+1 a+2 Γ(α + s + 3) (α + 2)(α + s + 2)

Proof. Using Lemma 63 via |f 󸀠󸀠 | is an (s, m)-Godunova-Levin function, we have 󵄨󵄨 f (mb) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 α α − (RL Jmb − f (a) + RL J + f (mb))󵄨󵄨 󵄨󵄨 a 󵄨󵄨 α 2 2(mb − a) 󵄨󵄨 󵄨

248 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1

(mb − a)2 1 − (1 − t)α+1 − t α+1 󵄨󵄨 󸀠󸀠 󵄨 ≤ ∫ 󵄨󵄨f (ta + m(1 − t)b)󵄨󵄨󵄨dt 2 α+1 0

1

≤

(mb − a)2 |f 󸀠󸀠 (a)| |f 󸀠󸀠 (b)| + )dt ∫(1 − (1 − t)α+1 − t α+1 )( s 2(α + 1) t m(1 − t s ) 0

(mb − a)2 |f 󸀠󸀠 (b)| = (Φ1 (α, s)|f 󸀠󸀠 (a)| + Φ2 (α, s) ). 2 m The proof is completed. Theorem 248. Let f : [a, b] → ℝ be twice differentiable on (a, b) with a ≥ 0, f 󸀠󸀠 ∈ L[a, b] and a < mb < b. If |f 󸀠󸀠 |q (q ≥ 1) is an (s, m)-Godunova-Levin function, then for any 0 < α ≤ 1 we have 󵄨󵄨 f (mb) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 α α − (RL Jmb − f (a) + RL J + f (mb))󵄨󵄨 󵄨󵄨 a 󵄨󵄨 α 2 2(mb − a) 󵄨󵄨 󵄨 1

1

(mb − a)2 p(α + 1) − 1 p |f 󸀠󸀠 (a)|q (s + 2)|f 󸀠󸀠 (b)|q q ( ) ( + ) , ≤ 2(α + 1) p(α + 1) + 1 1−s mq (1 + s) where

1 p

+

1 q

= 1.

Proof. Using Lemma 63, Holder’s inequality and the fact that |f 󸀠󸀠 |q is an (s.m)-GodunovaLevin function, one has 󵄨󵄨 f (mb) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 α α (RL Jmb − − f (a) + RL J + f (mb))󵄨󵄨 󵄨󵄨 a 󵄨󵄨 α 2 2(mb − a) 󵄨 󵄨󵄨 1

≤

(mb − a)2 (1 − (1 − t)α+1 − t α+1 ) 󵄨󵄨 󸀠󸀠 󵄨 ∫ 󵄨󵄨f (ta + m(1 − t)b)󵄨󵄨󵄨dt 2 α+1 0

1 p

1

1

(mb − a)2 p 󵄨 󵄨q ≤ (∫(1 − (1 − t)α+1 − t α+1 ) dt) (∫ 󵄨󵄨󵄨f 󸀠󸀠 (ta + m(1 − t)b)󵄨󵄨󵄨 dt) 2(α + 1) 0

0

1

(mb − a)2 (∫(1 − (1 − t)p(α+1) − t p(α+1) )dt) ≤ 2(α + 1) 0

1

|f 󸀠󸀠 (a)|q |f 󸀠󸀠 (b)|q × (∫( + q )dt) s t m (1 − t s ) 0

1

1 p

1 q

1

(mb − a)2 p(α + 1) − 1 p |f 󸀠󸀠 (a)|q (s + 2)|f 󸀠󸀠 (b)|q q = ( ) ( + ) . 2(α + 1) p(α + 1) + 1 1−s mq (1 + s) The proof is completed.

1 q

4.12 Inequalities via AG(log)-convex functions | 249

4.11.2 Applications to some special means Theorem 249. For some s, m ∈ (0, 1], n ∈ ℤ \ {−1, 0}, 0 ≤ a < b, the following inequality for fractional integrals holds: 1

1

q (mb − a)2 2p − 1 p 1 s+2 󵄨󵄨 2 2 2 n 2 2 2 󵄨 ( ) ( + q ) , 󵄨󵄨A(a , m b ) − Ln (a , m b )󵄨󵄨󵄨 ≤ 2 2p + 1 1 − s m (1 + s)

where

1 p

+

1 q

= 1, 1 < q < ∞.

Proof. Applying Theorem 248 for f (x) = x2 , α = 1, λ = 1, one can obtain the result immediately. Theorem 250. For some s, m ∈ (0, 1], 0 ≤ a < b, the following inequality for fractional integrals holds: 1

1

󵄨󵄨 󵄨󵄨 (mb − a)2 2p − 1 p q 1 1 s+2 1 󵄨󵄨 󵄨󵄨 − ( ) ( 3q + 3q q ) , 󵄨󵄨 󵄨󵄨 ≤ 2 2p + 1 a (1 − s) b m (1 + s) 󵄨󵄨 H(a, mb) L(a, mb) 󵄨󵄨 where

1 p

+

1 q

= 1, 1 < q < ∞.

Proof. Applying Theorem 248 for f (x) = immediately.

1 , x

α = 1, λ = 1, one can obtain the result

4.12 Inequalities via AG(log)-convex functions The results in this section are adopted from [105]. Now, we give the inequalities for AG(log)-convex functions. Theorem 251. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with a < b. If f 󸀠 ∈ L[a, b] and |f 󸀠 | is AG(log)-convex functions, |f 󸀠 (a)| > 0, |f 󸀠 (b)| > 0, |f 󸀠 (a)| ≠ |f 󸀠 (b)|, then for any 0 < α ≤ 1 and 0 < λ ≤ 1, p > 1, q > 1, p1 + q1 = 1, the following inequality for fractional integrals holds: 󵄨󵄨 󵄨󵄨 󵄨󵄨(1 + λ(1 − α))f (b) + (1 − λα)f (a) + λ(2α − 1)f ((1 − α)b + αa) 󵄨󵄨 󵄨󵄨 Γ(α + 1) 󵄨󵄨 α α ( J − + f (b) + RL Jb− f (a))󵄨 RL a 󵄨󵄨 (b − a)α 󵄨 ≤ (b − a){Q󸀠1 + Q󸀠2 + Q󸀠3 }, where Q󸀠1 =

1−α

|f 󸀠 (b)| λα‖f 󸀠 (a)| {( 󸀠 ) 󸀠 󸀠 ln |f (b)| − ln |f (a)| |f (a)|

− 1},

250 | 4 Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals 1−α

|f 󸀠 (b)| λ(1 − α)|f 󸀠 (a)| ( ) ln |f 󸀠 (b)| − ln |f 󸀠 (a)| |f 󸀠 (a)|

Q󸀠2 = Q󸀠3

1 − ( 21 )αp

1 p

α

{(

1

|f 󸀠 (b)| ) − 1}. |f 󸀠 (a)|

|f 󸀠 (b)| 2 =( ) {1 + ( 󸀠 ) } αp + 1 |f (a)|

q

1

q |f 󸀠 (a)|q |f 󸀠 (b)| 2 ×[ (( 󸀠 ) − 1)] . 󸀠 󸀠 q(ln |f (b)| − ln |f (a)|) |f (a)|

Proof. Using Lemma 60 via f 󸀠 ∈ L[a, b] and |f 󸀠 | is a AG(log)-convex function, we have 󵄨󵄨 󵄨󵄨 󵄨󵄨(1 + λ(1 − α))f (b) + (1 − λα)f (a) + λ(2α − 1)f ((1 − α)b + αa) 󵄨󵄨 󵄨󵄨 Γ(α + 1) 󵄨 (RL Jaα+ f (b) + RL Jbα− f (a))󵄨󵄨󵄨 − α 󵄨󵄨 (b − a) 1

1−α

󵄨 󵄨 󵄨 󵄨 ≤ (b − a){λα ∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt + λ(1 − α) ∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt 1−α

0

1

󵄨 󵄨 + ∫ 󵄨󵄨󵄨(t α − (1 − t)α )f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt} 0

≤ (b − a){Q󸀠1 + Q󸀠2 + Q󸀠3 }.

(4.123)

Integrating by parts and using Lemma 61, 1−α

󵄨 󵄨 Q󸀠1 := λα ∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt 0

1−α

≤ λα ∫ |f 󸀠 (b)|t |f 󸀠 (a)|1−t dt 0

≤

1−α

|f 󸀠 (b)| λα|f 󸀠 (a)| {( 󸀠 ) 󸀠 − ln |f (a)| |f (a)|

ln |f 󸀠 (b)|

− 1},

(4.124)

and similarly we get 1−α

λ(1 − α)|f 󸀠 (a)| |f 󸀠 (b)| ( ) ln |f 󸀠 (b)| − ln |f 󸀠 (a)| |f 󸀠 (a)|

Q󸀠2 ≤

α

{(

Next, using Hölder inequality, we have 1

󵄨 󵄨 Q󸀠3 := ∫ 󵄨󵄨󵄨(t α − (1 − t)α )f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt 0

1 2

󵄨 󵄨 ≤ ∫((1 − t)α − t α )󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt 0

|f 󸀠 (b)| ) − 1}. |f 󸀠 (a)|

(4.125)

4.12 Inequalities via AG(log)-convex functions |

251

1

󵄨 󵄨 + ∫(t α − (1 − t)α )󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨dt 1 2

1 2

1 2

1 p

󵄨q 󵄨 ≤ (∫((1 − t) − t ) dt) (∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨 dt) α p

α

0

1 q

0

1

1 p

1

󵄨q 󵄨 + (∫((1 − t) − t ) dt) (∫ 󵄨󵄨󵄨f 󸀠 (tb + (1 − t)a)󵄨󵄨󵄨 dt) α p

α

1 2

≤(

1 q

1 2

1 − ( 21 )αp αp + 1 1 󸀠

1 2

1 p

󸀠

qt

󸀠

q(1−t)

) {(∫ |f (b)| |f (a)| 0 qt

󸀠

+ (∫ |f (b)| |f (a)|

q(1−t)

dt)

1 q

1 q

dt) }

1 2

1 − ( 21 )αp

1 p

1

|f 󸀠 (b)| 2 ) {1 + ( 󸀠 ) } ≤( αp + 1 |f (a)| ×[

q

1

q |f 󸀠 (a)|q |f 󸀠 (b)| 2 (( 󸀠 ) − 1)] . 󸀠 󸀠 q(ln |f (b)| − ln |f (a)|) |f (a)|

Substituting (4.124), (4.125) and (4.126) into (4.123), we have the result.

(4.126)

5 Hermite-Hadamard inequalities involving Hadamard fractional integrals 5.1 Inequalities via convex functions The results in this section are due to [217, 220, 143, 174].

5.1.1 Hermite-Hadamard’s inequalities Hermite-Hadamard’s inequalities for Hadamard fractional integrals can be represented as follows. Theorem 252. Let f : [a, b] → ℝ be a positive function with 0 < a < b and f ∈ L1 [a, b]. If f is a nondecreasing and convex function on [a, b], then the following inequality for fractional integrals holds: f (√ab) ≤

Γ(α + 1) [ J α+ f (b) + H Jbα− f (a)] ≤ f (b). 2(ln b − ln a)α H a

(5.1)

Proof. Since f is a nondecreasing and convex function on [a, b], we have for x, y ∈ [a, b] with λ = 21 , f (√xy) ≤ f (

x+y f (x) + f (y) )≤ . 2 2

(5.2)

Set x = eln b−t(ln b−ln a) and y = eln a+t(ln b−ln a) for 0 < t < 1, then 2f (√ab) = 2f (√eln b+ln a ) ≤ f (eln b−t(ln b−ln a) ) + f (eln a+t(ln b−ln a) ).

(5.3)

Multiplying both sides of (5.3) by t α−1 , then integrating the resulting inequality with respect to t over [0, 1], we obtain 2 √ 2 f ( ab) = f (√eln b+ln a ) α α 1

≤ ∫t

α−1

f (e

ln b−t(ln b−ln a)

0

=

1

)dt + ∫ t α−1 f (eln a+t(ln b−ln a) )dt 0

a

α−1

1 ln b − ln u ) ∫( ln a − ln b ln b − ln a

f (u)

b

b

α−1

1 ln v − ln a + ) ∫( ln b − ln a ln b − ln a a

https://doi.org/10.1515/9783110523621-005

du u

f (v)

dv v

254 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals

=

Γ(α) [ J α+ f (b) + H Jbα− f (a)], (ln b − ln a)α H a

which implies that Γ(α + 1) [ J α+ f (b) + H Jbα− f (a)]. 2(ln b − ln a)α H a

f (√ab) ≤

On the other hand, note that f is nondecreasing, we have f (eln b−t(ln b−ln a) ) + f (eln a+t(ln b−ln a) ) ≤ 2f (eln b ) = 2f (b).

(5.4)

Then multiplying both sides of (5.4) by t α−1 and integrating the resulting inequality with respect to t over [0, 1], we obtain 1

∫t

α−1

f (e

ln b−t(ln b−ln a)

1

)dt + ∫ t α−1 f (eln a+t(ln b−ln a) )dt

0

0

Γ(α) = [ J α+ f (b) + H Jbα− f (a)] (ln b − ln a)α H a ≤ 2f (e

ln b

1

) ∫ t α−1 dt

2 = f (b), α

0

which yields Γ(α + 1) [ J α+ f (b) + H Jbα− f (a)] ≤ 2f (b). (ln b − ln a)α H a The proof is completed. Example 253. Let a = 1, b = e, α = 2 and f (x) = x 2 . Then all the assumptions in Theorem 252 are satisfied. Clearly, 2 H J1+ f (e)

2 H Je− f (1)

e

1

= ∫(1 − ln t)tdt = ∫ se2(1−s) ds = 1

e

1

1

0

0

= ∫ t ln tdt = ∫ se2s ds =

1 2 3 e − , 4 4

1 2 1 e + . 4 4

Thus, (5.1)

⇐⇒

e
b−1 . For n ∈ ℤ, |n| ≥ 2, we have (i) ln b − ln a (ln b − ln a) 󵄨󵄨 −1 −1 −1 󵄨 [6( ) 󵄨󵄨H (b, a) − L(b , a )󵄨󵄨󵄨 ≤ 8a 2

−1

a + √ ], b

(ii) (ln b − ln a) 󵄨󵄨 −1 −1 −1 󵄨 , 󵄨󵄨H (b, a) − L(b , a )󵄨󵄨󵄨 ≤ 4a (iii) 󵄨󵄨 a−1 − b−1 n −1 −1 󵄨󵄨󵄨󵄨 󵄨󵄨 −1 n+1 n+1 L (b , a )󵄨󵄨 󵄨󵄨H (a , b ) − 󵄨󵄨 󵄨󵄨 ln b − ln a n −1 a (n + 1)(ln b − ln a) ln b − ln a ) + √ ], [6( ≤ n+1 2 b 8a (iv) 󵄨󵄨 a−1 − b−1 n −1 −1 󵄨󵄨󵄨󵄨 (n + 1)(ln b − ln a) 󵄨󵄨 −1 n+1 n+1 L (b , a )󵄨󵄨 ≤ . 󵄨󵄨H (a , b ) − 󵄨󵄨 󵄨󵄨 ln b − ln a n 4an+1 Proof. Making the substitutions a → b−1 , b → a−1 into the inequalities (5.16)–(5.19), one can obtain the desired inequalities respectively. Now, we give some applications to special means of real numbers (see Proposition 4.1–4.4, [220]).

274 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals Proposition 268. Let a, b ∈ ℝ+ \ {0}, a < b. Then √ab(ln b − ln a) 1 5b ], |L(a, b) − A(a, b)| ≤ [ − a + 2 2 4 1 b(ln b − ln a) |L(a, b) − A(a, b)| ≤ [b − a + ], 2 2 1 a 6 󵄨󵄨 󵄨 b(ln b − ln a) (4 + + √ ), 󵄨󵄨L(a, b) − [A(a, b)H(a, b)] 2 󵄨󵄨󵄨 ≤ 8 ln b − ln a b

(5.20) (5.21) (5.22)

and 1 󵄨 3b(ln b − ln a) 󵄨󵄨 . 󵄨󵄨L(a, b) − [A(a, b)H(a, b)] 2 󵄨󵄨󵄨 ≤ 4

(5.23)

Proof. Applying Theorems 257, 258, 259 and 260 for f (x) = x and α = 1, one can obtain the results immediately. Proposition 269. Let a, b ∈ ℝ+ \ {0}, a < b. Then n−1 √ab(ln b − ln a) 5b 󵄨󵄨 n n 󵄨 nb n [ −a+ ], 󵄨󵄨L(a , b ) − A (a, b)󵄨󵄨󵄨 ≤ 2 2 4

(5.24)

n−1 b(ln b − ln a) 󵄨󵄨 n n 󵄨 nb n [b − a + ], (5.25) 󵄨󵄨L(a , b ) − A (a, b)󵄨󵄨󵄨 ≤ 2 2 n n 6 a 󵄨󵄨 n n 󵄨 nb (ln b − ln a) (4 + + √ ), (5.26) 󵄨󵄨L(a , b ) − [A(a, b)H(a, b)] 2 󵄨󵄨󵄨 ≤ 8 ln b − ln a b

and n n 󵄨󵄨 n n 󵄨 3nb (ln b − ln a) . 󵄨󵄨L(a , b ) − [A(a, b)H(a, b)] 2 󵄨󵄨󵄨 ≤ 4

(5.27)

Proof. Applying Theorems 257, 258, 259 and 260 for f (x) = x n and α = 1, one can obtain the results immediately. Proposition 270. Let a, b ∈ ℝ+ \ {0}, a < b and n ∈ ℤ, |n| ≥ 2. Then n √ab(ln b − ln a) 󵄨󵄨 󵄨 (n + 1)b 5b n n+1 [ −a+ ], (5.28) 󵄨󵄨L(a, b)Ln (a, b) − A (a, b)󵄨󵄨󵄨 ≤ 2 2 4 n b(ln b − ln a) 󵄨󵄨 󵄨 (n + 1)b n n+1 [b − a + ], (5.29) 󵄨󵄨L(a, b)Ln (a, b) − A (a, b)󵄨󵄨󵄨 ≤ 2 2 n+1 󵄨󵄨 󵄨 n 󵄨󵄨L(a, b)Ln (a, b) − [A(a, b)H(a, b)] 2 󵄨󵄨󵄨 (n + 1)bn+1 (ln b − ln a) 6 a (4 + + √ ), (5.30) ≤ 8 ln b − ln a b

and n+1 n+1 󵄨󵄨 󵄨 3(n + 1)b (ln b − ln a) n . 󵄨󵄨L(a, b)Ln (a, b) − [A(a, b)H(a, b)] 2 󵄨󵄨󵄨 ≤ 4

(5.31)

5.1 Inequalities via convex functions | 275

Proof. Applying Theorems 257, 258, 259 and 260 for f (x) = x n+1 and α = 1, x ∈ ℝ, z ∈ ℤ, |n| ≥ 2, one can obtain the result immediately. Proposition 271. Let a, b ∈ ℝ+ \ {0}, (a < b), a−1 > b−1 . Then we have, for n ∈ ℤ, |n| ≥ 2 (c1) 1 ln b − ln a 󵄨 1 5 󵄨󵄨 −1 −1 −1 ), 󵄨󵄨L(b , a ) − H (b, a)󵄨󵄨󵄨 ≤ ( − + 2 2a b 4√ab (c2) 1 b − a ln b − ln a 󵄨󵄨 −1 −1 󵄨 −1 ( + ), 󵄨󵄨L(b , a ) − H (b, a)󵄨󵄨󵄨 ≤ 2a b 2 (c3) a 6 ln b − ln a −1 󵄨 󵄨󵄨 −1 −1 (4 + + √ ), 󵄨󵄨L(b , a ) − [A(a, b)H(a, b)] 2 󵄨󵄨󵄨 ≤ 8a ln b − ln a b (c4) 3(ln b − ln a) −1 󵄨 󵄨󵄨 −1 −1 , 󵄨󵄨L(b , a ) − [A(a, b)H(a, b)] 2 󵄨󵄨󵄨 ≤ 4a (c5) 5 1 ln b − ln a n 󵄨󵄨 −n −n 󵄨 −n ), 󵄨󵄨L(b , a ) − H (b, a)󵄨󵄨󵄨 ≤ n−1 ( − + 2a b 2a 4√ab (c6) n b − a ln b − ln a 󵄨󵄨 −n −n 󵄨 −n + ), 󵄨󵄨L(b , a ) − H (b, a)󵄨󵄨󵄨 ≤ n ( 2a b 2 (c7) a 6 n(ln b − ln a) −n 󵄨 󵄨󵄨 −n −n + √ ), (4 + 󵄨󵄨L(b , a ) − [A(a, b)H(a, b)] 2 󵄨󵄨󵄨 ≤ 8an ln b − ln a b (c8) 3n(ln b − ln a) −n 󵄨 󵄨󵄨 −n −n , 󵄨󵄨L(b , a ) − [A(a, b)H(a, b)] 2 󵄨󵄨󵄨 ≤ 4an (c9) 1 ln b − ln a 󵄨󵄨 −1 −1 n −1 −1 󵄨 n+1 5 −n−1 (b, a)󵄨󵄨󵄨 ≤ ( − + ), 󵄨󵄨L(b , a )Ln (b , a ) − H 2an 2a b 4√ab (c10) 󵄨󵄨 −1 −1 n −1 −1 󵄨 n + 1 b − a ln b − ln a −n−1 (b, a)󵄨󵄨󵄨 ≤ n+1 ( + ), 󵄨󵄨L(b , a )Ln (b , a ) − H b 2 2a

276 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals (c11) − n+1 󵄨 󵄨󵄨 −1 −1 n −1 −1 󵄨󵄨L(b , a )Ln (b , a ) − [A(a, b)H(a, b)] 2 󵄨󵄨󵄨 (n + 1)(ln b − ln a) 6 a ≤ (4 + + √ ), ln b − ln a b 8an+1

(c12) 3(n + 1)(ln b − ln a) − n+1 󵄨 󵄨󵄨 −1 −1 n −1 −1 . 󵄨󵄨L(b , a )Ln (b , a ) − [A(a, b)H(a, b)] 2 󵄨󵄨󵄨 ≤ 4an+1 Proof. Making the substitutions a → b−1 , b → a−1 into the (5.20)–(5.31), one can obtain desired inequalities respectively, where A−1 (a−1 , b−1 ) = H(a, b) = 1 2 1 , b−1 < a−1 . a

+b

Now, we give some applications to special means of real numbers (see Proposition 4.1–4.4, [143]). Proposition 272. Let a, b ∈ ℝ+ \ {0}, 0 ≤ a < b, x ∈ [0, b]. Then |A(a, b) − L(a, b)| ≤

a(ln a − ln b) − (√a − √b)2 a + b + , ln b − ln a 2

1

1

|A(a, b) − L(a, b)| ≤

1 p+1 p aq (ln a − ln b)q − aq + bq q ln b − ln a 2 − 2( 2 ) ) ( ) [( 2 p+1 q2 (ln a − ln b)2 1

aq − bq aq (ln a − ln b)q − aq + bq q +( − ) ], q(ln a − ln b) q2 (ln a − ln b)2 (2 ln x − ln a − ln b − 2)x + a + b |A(x, x) − L(a, b)| ≤ , ln b − ln a 1

1

p (ln x − ln a)2 1 x q (ln x − ln a)q − x q + aq q ( ) [( ) |A(x, x) − L(a, b)| ≤ ln b − ln a p + 1 q2 (ln x − ln a)2 1

1

p aq (ln a − ln x)q − aq + x q q (ln b − ln x)2 1 ( ) +( ) ] + 2 2 ln b − ln a p + 1 q (ln x − ln a) 1

1

bq (ln bx )q − bq + x q q xq (ln x − ln b)q − x q + bq q × [( + ( ) ) ]. q2 (ln x − ln b)2 q2 (ln x − ln b)2 Proof. Applying Theorems 261, 262, 263 and 264, for f (x) = x and α = 1, one can obtain the results immediately. Proposition 273. Let a, b ∈ ℝ+ \ {0}, 0 ≤ a < b, x ∈ [0, b], n ≥ 2. Then 󵄨󵄨 󵄨󵄨 1 󵄨󵄨 󵄨 n n n−1 󵄨󵄨A(a , b ) − L(a, b)Ln−1 (a, b)󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 2 1 (√a − √b)2 1 ≤ (nan−1 − 3nbn−1 )(a − √ab + ) + nbn−1 (√a − √b)2 2 ln a − ln b 2

5.1 Inequalities via convex functions | 277

+ (nan−1 − nbn−1 )(−a +

√ab 2√ab − 2a 2(√a − √b)2 ), − − 2 ln a − ln b (ln a − ln b)2

󵄨󵄨 󵄨󵄨 1 󵄨󵄨 󵄨 n n n−1 󵄨󵄨A(a , b ) − L(a, b)Ln−1 (a, b)󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 2 1

1

1 p+1 p ln b − ln a 2 − 2( 2 ) aq (ln a − ln b)q − aq + bq q ≤ ( ) [nan−1 ( ) 2 p+1 q2 (ln a − ln b)2 1

n−1

+ nb

aq − bq aq (ln a − ln b)q − aq + bq q ( − ) ], q(ln a − ln b) q2 (ln a − ln b)2

󵄨󵄨 󵄨 n n n−1 󵄨󵄨A(x , x ) − L(a, b)Ln−1 (a, b)󵄨󵄨󵄨 nxn (ln x − ln a)2 − n(ln x − ln a)(an−1 x + an − 2xn ) + 2n(x − a)(−an−1 + x n−1 ) ≤ (ln b − ln a)(ln x − ln a) +

nxn (ln x − ln b)2 − n(ln x − ln b)(bn−1 x + bn − 2xn ) + 2n(x − b)(−bn−1 + x n−1 ) , (ln b − ln a)(ln x − ln b)

󵄨󵄨 󵄨 n n n−1 󵄨󵄨A(x , x ) − L(a, b)Ln−1 (a, b)󵄨󵄨󵄨 1

1

p 1 (ln x − ln a)2 xq (ln x − ln a)q − x q + aq q ( ) [nxn−1 ( ) ≤ ln b − ln a p + 1 q2 (ln x − ln a)2 1

n−1

+ na

1

p (ln b − ln x)2 aq (ln a − ln x)q − aq + xq q 1 ) ] + ( ( ) 2 2 ln b − ln a p + 1 q (ln x − ln a) 1

× [nx

n−1

1

q q q q xq (ln x − ln b)q − xq + bq q n−1 b (ln b − ln x)q − b + x ( ) + nb ( ) ]. q2 (ln x − ln b)2 q2 (ln x − ln b)2

Proof. Applying Theorems 261, 262, 263 and 264, for f (x) = x n and α = 1, one can obtain the results immediately. Proposition 274. Let a, b ∈ ℝ+ \ {0}, 0 ≤ a < b, x ∈ [0, b], n ≥ 2. Then 󵄨󵄨 1 1 1 1 1 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨A( , ) − L( , )󵄨󵄨󵄨 󵄨󵄨 a b 2 a b 󵄨󵄨 (√a − √b)2 1 1 3 1 ) + 2 (√a − √b)2 ≤ ( 2 − 2 )(a − √ab − 2 a ln a − ln b b 2b 1 1 1√ 2√ab − 2a 2(√a − √b)2 + ( 2 − 2 )(−a + ab − − ), 2 ln a − ln b (ln a − ln b)2 a b 󵄨󵄨 1 1 1 1 1 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨A( , ) − L( , )󵄨󵄨󵄨 󵄨󵄨 a b 2 a b 󵄨󵄨

1

1

1 p+1 p ln b − ln a 2 − 2( 2 ) 1 aq (ln a − ln b)q − aq + bq q ≤ ( ) [ 2( ) 2 p+1 a q2 (ln a − ln b)2

278 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals 1

aq − bq 1 aq (ln a − ln b)q − aq + bq q ) ], + 2( − b q(ln a − ln b) q2 (ln a − ln b)2 󵄨󵄨 1 1 1 1 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨A( , ) − L( , )󵄨󵄨󵄨 a b 󵄨󵄨 󵄨󵄨 x x ≤

1 1 x(ln x − ln a)2 − 2x(ln x − ln a) + 2x − 2a 1 [( 2 − 2 ) ln b − ln a x ln x − ln a a + ×

x(ln x − ln a) − x + a 1 1 1 ]+ [( − ) ln b − ln a x 2 b2 a2

x(ln x − ln b)2 − 2x(ln x − ln b) + 2x − 2b x(ln x − ln b) − x + b + ], ln x − ln b b2

󵄨󵄨 1 1 1 1 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨A( , ) − L( , )󵄨󵄨󵄨 a b 󵄨󵄨 󵄨󵄨 x x

1

1

p 1 (ln x − ln a)2 1 xq (ln x − ln a)q − x q + aq q ( ) [ 2( ) ≤ ln b − ln a p + 1 x q2 (ln x − ln a)2 1

+

1

p (ln b − ln x)2 1 aq (ln a − ln x)q − aq + xq q 1 ( ) ]+ ( ) 2 2 2 ln b − ln a p + 1 a q (ln x − ln a) 1

1

1 xq (ln x − ln b)q − xq + bq q 1 bq (ln b − ln x)q − bq + x q q ×[ 2( ) + ( ) ]. x q2 (ln x − ln b)2 b2 q2 (ln x − ln b)2 Proof. Applying Theorems 261, 262, 263 and 264, for f (x) = the results immediately.

1 x

and α = 1, one can obtain

Proposition 275. Let a, b ∈ ℝ+ \ {0}, 0 ≤ a < b, x ∈ [0, b], n ≥ 2. Then (2 ln x + ln a + ln b − 2)x + a−1 + b−1 󵄨󵄨 −1 −1 󵄨 , 󵄨󵄨A(x, x) − L(b , a )󵄨󵄨󵄨 ≤ ln b − ln a 󵄨󵄨 −1 −1 󵄨 󵄨󵄨A(x, x) − L(b , a )󵄨󵄨󵄨 ≤

1

1

p 1 xq (ln x + ln b)q − x q + b−q q (ln x + ln b)2 ( ) [( ) ln b − ln a p + 1 q2 (ln x + ln b)2 1

1

p b−q (− ln b − ln x)q − b−q + xq q (ln a + ln x)2 1 +( ) ] + ( ) 2 2 ln b − ln a p + 1 q (ln x + ln b) 1

1

xq (ln x + ln a)q − xq + a−q q a−q (− ln a − ln x)q − a−q + x q q × [( ) + ( ) ], q2 (ln x + ln a)2 q2 (ln x + ln a)2 b−1 (ln a − ln b) − (√b−1 − √a−1 )2 b−1 + a−1 󵄨󵄨 −1 −1 −1 −1 󵄨 + , 󵄨󵄨A(b , a ) − L(b , a )󵄨󵄨󵄨 ≤ ln b − ln a 2

5.1 Inequalities via convex functions | 279

󵄨󵄨 −1 −1 −1 −1 󵄨 󵄨󵄨A(b , a ) − L(b , a )󵄨󵄨󵄨

1

1

1 p+1 p ln b − ln a 2 − 2( 2 ) b−q (ln a − ln b)q − b−q + a−q q ≤ ) ( ) [( 2 p+1 q2 (ln a − ln b)2 1

b−q − a−q b−q (ln a − ln b)q − b−q + a−q q +( ) ], − q(ln a − ln b) q2 (ln a − ln b)2 󵄨󵄨 n n −1 −1 n−1 −1 −1 󵄨 󵄨󵄨A(x , x ) − L(b , a )Ln−1 (b , a )󵄨󵄨󵄨 nxn (ln x + ln b)2 − n(ln x + ln b)(b1−n x + b−n − 2x n ) + 2n(x − b−1 )(−b1−n + x n−1 ) ≤ (ln b − ln a)(ln x + ln b) +

nxn (ln x + ln a)2 − n(ln x + ln a)(a1−n x + a−n − 2x n ) + 2n(x − a−1 )(−a1−n + x n−1 ) , (ln b − ln a)(ln x + ln a)

󵄨󵄨 n n −1 −1 n−1 −1 −1 󵄨 󵄨󵄨A(x , x ) − L(b , a )Ln−1 (b , a )󵄨󵄨󵄨 1

1

p 1 xq (ln x + ln b)q − x q + b−q q (ln x + ln b)2 ( ) [nxn−1 ( ) ≤ ln b − ln a p + 1 q2 (ln x + ln b)2 1

1−n

+ nb

1

p (ln x + ln a)2 1 b−q (− ln b − ln x)q − b−q + xq q ) ]+ ( ) ( 2 2 ln b − ln a p + 1 q (ln x + ln b) 1

× [nxn−1 ( + na1−n (

xq (ln x + ln a)q − xq + a−q q ) q2 (ln x + ln a)2

1

a−q (− ln a − ln x)q − a−q + xq q ) ], q2 (ln x + ln a)2

󵄨󵄨 󵄨 1 󵄨󵄨 −1 −1 n−1 −1 −1 󵄨󵄨 −n −n 󵄨󵄨A(b , a ) − L(b , a )Ln−1 (b , a )󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 2 √ ( b−1 − √a−1 )2 1 ) ≤ (nb1−n − 3na1−n )(b−1 − √b−1 a−1 + 2 ln a − ln b 1 2 + na1−n (√b−1 − √a−1 ) + (nb1−n − na1−n ) 2 √b−1 a−1 2√b−1 a−1 − 2a−1 2(√b−1 − √a−1 )2 × (−b−1 + − − ), 2 ln a − ln b (ln a − ln b)2 󵄨󵄨 󵄨 1 󵄨󵄨 −1 −1 n−1 −1 −1 󵄨󵄨 −n −n 󵄨󵄨A(b , a ) − L(b , a )Ln−1 (b , a )󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 2 1 p+1

ln b − ln a 2 − 2( 2 ) ( ≤ 2 p+1 1−n

+ na

1 p

1−n

) [nb

1

b−q (ln a − ln b)q − b−q + a−q q ( ) q2 (ln a − ln b)2 1

b−q − a−q b−q (ln a − ln b)q − b−q + a−q q ( − ) ], q(ln a − ln b) q2 (ln a − ln b)2

280 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals 󵄨󵄨 1 1 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨A( , ) − L(b, a)󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 x x 1 1 x(ln x + ln b)2 − 2x(ln x + ln b) + 2x − 2b−1 ≤ [( 2 − b2 ) ln b − ln a x ln x + ln b 1 [a2 (x(ln x + ln a) − x + a−1 ) + b2 (x(ln x + ln b) − x + b−1 )] + ln b − ln a 1 x(ln x + ln a)2 − 2x(ln x + ln a) + 2x − 2a−1 + ( 2 − a2 ) ], ln x + ln a x 󵄨󵄨 1 1 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨A( , ) − L(b, a)󵄨󵄨󵄨 󵄨󵄨 x x 󵄨󵄨

1

1

p 1 1 xq (ln x + ln b)q − x q + b−q q (ln x + ln b)2 ) ( ) [ 2( ≤ ln b − ln a p + 1 x q2 (ln x + ln b)2 1

+ b2 (

1

p b−q (− ln b − ln x)q − b−q + xq q (ln a + ln x)2 1 ) ]+ ( ) 2 2 ln b − ln a p + 1 q (ln x + ln b) 1

1

−q −q 1 xq (ln x + ln a)q − xq + a−q q + xq q 2 a (− ln a − ln x)q − a ×[ 2( ) + a ( ) ], x q2 (ln x + ln a)2 q2 (ln x + ln a)2

󵄨󵄨 󵄨󵄨 1 󵄨󵄨 󵄨 󵄨󵄨A(b, a) − L(b, a)󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 2 (√b−1 − √a−1 )2 1 2 )+ ≤ (b − 3a2 )(b−1 − √b−1 a−1 − 2 ln a − ln b 2√b−1 a−1 − 2b−1 1 + (b2 − a2 )(−b−1 + √b−1 a−1 − 2 ln a − ln b

1 2 √ −1 √ −1 2 a( b − a ) 2 2(√b−1 − √a−1 )2 − ), (ln a − ln b)2

󵄨󵄨 󵄨󵄨 1 󵄨󵄨 󵄨 󵄨󵄨A(b, a) − L(b, a)󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 2 1 1 p+1 p1 −q −q 2 − 2( ) + a−q q ln b − ln a 2 b (ln a − ln b)q − b 2 ( ) [b ( ) ≤ 2 p+1 q2 (ln a − ln b)2 1

b−q − a−q b−q (ln a − ln b)q − b−q + a−q q +a ( − ) ]. q(ln a − ln b) q2 (ln a − ln b)2 2

Proof. Making the substitutions a → b−1 , b → a−1 into Proposition 272, Proposition 273 and Proposition 274, one can obtain desired inequalities respectively.

5.1.3 Hadamard fractional Fejér inequalities We first derive two Hadamard fractional Fejér inequalities. Theorem 276. Let f : [a, b] → ℝ be a positive, convex and nondecreasing function on [a, b] with 0 < a < b, f ∈ L[a, b]. If g : [a, b] → ℝ is nonnegative, integrable and

5.1 Inequalities via convex functions | 281

symmetric to √ab, the order 0 < α ≤ 1, then f (√ab)((H Jaα+ g)(b) + (H Jbα− g)(a)) ≤ [(H Jaα+ fg)(b) + (H Jbα− fg)(a)]

≤ f (b)((H Jaα+ g)(b) + (H Jbα− g)(a)).

Proof. Note that 1 2

1 2

a b = (a

1−t 2

t 2

t 2

b )(a b

1−t 2

t

1−t 2

t

1−t

b 2 )2 + (a 2 b 2 )2 )≤ 2 (a1−t bt ) + (at b1−t ) ≤ , 0 ≤ t ≤ 1. 2 (a

Since f is a nondecreasing and convex function on [a, b], we have f (√ab) ≤ f (

(a1−t bt ) + (at b1−t ) f (at b1−t ) + f (a1−t bt ) )≤ , 2 2

0 ≤ t ≤ 1.

(5.32)

Multiplying both sides of (5.32) by t α−1 g(a1−t bt ), then integrating the resulting inequality with respective to t over [0, 1], we obtain 1

f (√ab) ∫ t α−1 g(a1−t bt )dt 0

1

1

1 ≤ [∫ t α−1 f (at b1−t )g(a1−t bt )dt + ∫ t α−1 f (a1−t bt )g(a1−t bt )dt], 2 0

0

let x = at b1−t , y = a1−t bt , so we have b

α−1

ln y − ln a ) f (√ab) ∫( ln b − ln a a

a

α−1

1 ln b − ln x ) ≤ [∫( 2 ln b − ln a

f (x)g(

b

b

+ ∫( a

g(y)d(

α−1

ln y − ln a ) ln b − ln a

ln y − ln a ) ln b − ln a

ab ln b − ln x )d( ) x ln b − ln a

f (y)g(y)d(

ln y − ln a )]. ln b − ln a

And then, we have b

dy f (√ab) ∫(ln y − ln a)α−1 g(y) y a

b

b

a

a

1 ab dx dy ≤ [∫(ln b − ln x)α−1 f (x)g( ) + ∫(ln y − ln a)α−1 f (y)g(y) ], 2 x x y

282 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals and using the Lemma 74 an symmetric and definition of Hadamard fractional integrals, we can obtain the left-hand side of the result of the inequality. On the other side, we have f (at b1−t ) + f (a1−t bt ) ≤ 2f (b), multiplying both sides by t α−1 g(a1−t bt ), then integrating the resulting inequality with respective to t over [0, 1], we have 1

∫t

α−1

t 1−t

f (a b

1

1−t t

)g(a

b )dt + ∫ t

0

α−1

1−t t

f (a

1−t t

b )dt ≤ 2f (b) ∫ t α−1 g(a1−t bt )dt.

b )g(a

0 t 1−t

1

0

1−t t

Let x = a b , y = a b , and using the Lemma 74, symmetric and definition of Hadamard fractional integrals, so we have the result of the right-side inequality [(H Jaα+ fg)(b) + (H Jbα− fg)(a)] ≤ f (b)((H Jaα+ g)(b) + (H Jbα− g)(a)). The proof is completed. Remark 277. In Theorem 276, if g(x) = 1, then (see [220, Theorem 1.1]) f (√ab) ≤

Γ(α + 1) [( J α+ f )(b) + (H Jbα− f )(a)] ≤ f (b). 2(ln b − ln a)α H a

Theorem 278. Let f : [a, b] → ℝ be a positive, convex and nondecreasing function on [a, b], f ∈ L[a, b], 0 < a < b, 1 ≤ b. If g : [a, b] → ℝ is nonnegative, integrable and symmetric to √ab, the order 0 < α ≤ 1, then we have f (√ab)((H Jaα+ g)(b) + (H Jbα− g)(a)) ≤ (H Jaα+ fg)(b) + (H Jbα− fg)(a) f (a) + f (b) ≤ ((H Jaα+ g)(b) + (H Jbα− g)(a)). 2 Proof. The left result of this theorem is the same as Theorem 276. So we only need to test the right-side result. Using the Lemma 36, we have at b1−t ≤ ta + (1 − t)b, 1−t t

a

b ≤ (1 − t)a + tb,

0 ≤ t ≤ 1,

0 ≤ t ≤ 1.

Since f is a nondecreasing and convex function on [a, b], we have f (at b1−t ) + f (a1−t bt ) ≤ f (ta + (1 − t)b) + f ((1 − t)a + tb) ≤ f (a) + f (b),

0 ≤ t ≤ 1.

(5.33)

Multiplying both sides of (5.33) by t α−1 g(a1−t bt ), then integrating the resulting inequality with respect to t over [0, 1], we obtain 1

∫t 0

α−1

t 1−t

f (a b

)g(a

1−t t

1

b )dt + ∫ t α−1 f (a1−t bt )g(a1−t bt )dt 0

5.1 Inequalities via convex functions | 283 1

≤ (f (a) + f (b)) ∫ t α−1 g(a1−t bt )dt. 0

Let x = at b1−t , y = a1−t bt , and using the Lemma 74, symmetric and definition of Hadamard fractional integrals, we can get right result of the inequality. Remark 279. In Theorem 278, if g(x) = 1, then we have f (a) + f (b) Γ(α + 1) [( J α+ f )(b) + (H Jbα− f )(a)] ≤ . 2(ln b − ln a)α H a 2

f (√ab) ≤

Next, we give two important fractional integral equalities. Lemma 280. Let f : [a, b] → ℝ be a differentiable function on [a, b], f 󸀠 ∈ L[a, b], 0 < a < b. If g : [a, b] → ℝ is integrable and Symmetric to √ab, 0 < a < b, the order 0 < α ≤ 1, then we have f (a) + f (b) ((H Jaα+ g)(b) + (H Jbα− g)(a)) − [(H Jaα+ fg)(b) + (H Jbα− fg)(a)] 2 b

t

b

a

a

t

1 ds ds = ∫[∫(ln b − ln s)α−1 g(s) − ∫(ln s − ln a)α−1 g(s) ]f 󸀠 (t)dt. Γ(α) s s Proof. Denote b

t

I = ∫[∫(ln b − ln s) a

a

b

t

a

a

α−1

b

ds ds g(s) − ∫(ln s − ln a)α−1 g(s) ]f 󸀠 (t)dt s s t

= ∫[∫(ln b − ln s)α−1 g(s)

b

b

a

t

ds 󸀠 ds ]f (t)dt + ∫[− ∫(ln s − ln a)α−1 g(s) ]f 󸀠 (t)dt s s

= I1 + I2 . Integrating by parts, b

t

I1 := ∫[∫(ln b − ln s)α−1 g(s) a

a

t

α−1

= [∫(ln b − ln s) a

ds 󸀠 ]f (t)dt s b

ds dt g(s) ]f (t)|ba − ∫(ln b − ln t)α−1 g(t)f (t) s t a

= Γ(α)[f (b)(H Jaα+ g)(b) − (H Jaα+ fg)(b)], and similarly, we have b

b

a

t

I2 := ∫[− ∫(ln s − ln a)α−1 g(s)

ds 󸀠 ]f (t)dt s

284 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals b

= [− ∫(ln s − ln a)

α−1

t

b

ds dt g(s) ]f (t)|ba − ∫(ln t − ln a)α−1 g(t)f (t) s t a

= Γ(α)[f (a)(H Jbα− g)(a) − (H Jbα− fg)(a)]. Using Lemma 74, we have I = Γ(α){

f (a) + f (b) ((H Jaα+ g)(b) + (H Jbα− g)(a)) 2

− [(H Jaα+ fg)(b) + (H Jbα− fg)(a)]}, multiplying both sides of (5.34) by pleted.

1 , Γ(α)

(5.34)

then we obtain the result. The proof is com-

Lemma 281. Let f : [a, b] → ℝ be a differentiable function on [a, b], f 󸀠 ∈ L[a, b], 0 < a < b. If g : [a, b] → ℝ is integrable and Symmetric to √ab, the order 0 < α ≤ 1, then we have the equality f (b)((H Jaα+ g)(b) + (H Jbα− g)(a)) − [(H Jaα+ fg)(b) + (H Jbα− fg)(a)] b

t

α−1

1 b ∫[∫(ln ) Γ(α) s

=

a

t

g(s)

a

α−1

ds s + ∫(ln ) s a

g(s)

a

ds 󸀠 ]f (t)dt. s

Proof. Denote b

t

α−1

b L = ∫[∫(ln ) s a

a

b

t

α−1

a

α−1

ds s + ∫(ln ) s a

g(s)

a

b = ∫[∫(ln ) s a

t

g(s)

b

ds 󸀠 ]f (t)dt s

t

α−1

ds s g(s) ]f 󸀠 (t)dt + ∫[∫(ln ) s a a

= L1 + L2 .

a

g(s)

ds 󸀠 ]f (t)dt s

Integrating by parts, b

t

α−1

b L1 := ∫[∫(ln ) s a

g(s)

a

b

α−1

b = [∫(ln ) s a

ds 󸀠 ]f (t)dt s b

a

= Γ(α)[f (b)(H Jaα+ g)(b) − (H Jaα+ fg)(b)], and similarly, we get b

t

α−1

s L2 := ∫[∫(ln ) a a

a

α−1

ds b g(s) ]f (b) − ∫(ln ) s s

g(s)

ds 󸀠 ]f (t)dt s

g(s)f (s)

ds s

5.1 Inequalities via convex functions | 285 b

α−1

s = [∫(ln ) a a

b

α−1

ds s g(s) ]f (b) − ∫(ln ) s a a

g(s)f (s)

ds s

= Γ(α)[f (b)(H Jbα− g)(a) − (H Jbα− fg)(a)]. Then, we obtain L = Γ(α){f (b)((H Jaα+ g)(b) + (H Jbα− g)(a)) − [(H Jaα+ fg)(b) + (H Jbα− fg)(a)]}. Multiplying both sides of (5.35) by pleted.

1 , Γ(α)

(5.35)

then we obtain the result. The proof is com-

Now we are ready to present new Hermite-Hadamard-Fejér inequalities. Theorem 282. Let f : [a, b] → ℝ be a differentiable function on [a, b], f 󸀠 ∈ L[a, b], 0 < a < b. If |f 󸀠 | is a nondecreasing and convex function on [a, b], g : [a, b] → ℝ is continuous and symmetric to √ab, the order 0 < α ≤ 1, then we have the inequality 󵄨󵄨 f (a) + f (b) 󵄨󵄨 󵄨󵄨 󵄨 ((H Jaα+ g)(b) + (H Jbα− g)(a)) − [(H Jaα+ fg)(b) + (H Jbα− fg)(a)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 α ‖g‖∞ (ln b − ln a) ≤ {(|f 󸀠 (b)| − |f 󸀠 (a)|)K1 (t; a, b, α) (b − a)2α Γ(α + 1) + (b|f 󸀠 (a)| − a|f 󸀠 (b)|)K2 (t; a, b, α)},

where ‖g‖∞ = sup |g(x)|, t∈[a,b]

K1 (t; a, b, α) =

i=∞ i=∞ b−a (ln b − ln a)i−1 (ln a − ln b)i−1 [2(b + a) − αa ∑ − αb ∑ ], 2 (α)i (α)i i=1 i=1 i=∞

K2 (t; a, b, α) = (√b − √a)[2(√b + √a) − α√a ∑ i=1

i=∞

− α√b ∑ i=1

(ln b − ln a)i−1 2i−1 (α)i

(ln a − ln b) 2i−1 (α)i

i−1

],

(α)i = α(α + 1)(α + 2) ⋅ ⋅ ⋅ (α + i − 1). Proof. Using Lemma 280, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 󵄨󵄨 󵄨 ((H Jaα+ g)(b) + (H Jbα− g)(a)) − [(H Jaα+ fg)(b) + (H Jbα− fg)(a)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 b󵄨 t b 󵄨 󵄨󵄨 1 ds ds 󵄨󵄨󵄨 󵄨 ≤ ∫ 󵄨󵄨󵄨 ∫(ln b − ln s)α−1 g(s) − ∫(ln s − ln a)α−1 g(s) 󵄨󵄨󵄨|f 󸀠 (t)|dt, Γ(α) 󵄨󵄨󵄨 s s 󵄨󵄨󵄨 a a t

286 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals |f 󸀠 | is convex on [a, b], so we have 󵄨󵄨 b−t t − a 󵄨󵄨󵄨󵄨 b − t 󸀠 t−a 󸀠 󵄨 |f 󸀠 (t)| = 󵄨󵄨󵄨f 󸀠 ( a+ b)󵄨 ≤ |f (a)| + |f (b)|. 󵄨󵄨 b−a b − a 󵄨󵄨󵄨 b − a b−a Since g : [a, b] → ℝ is symmetric to √ab, we write b

∫(ln s − ln a)α−1 t

a

g(s) ab ds = ∫ (ln − ln a) s v ab t

a

= ∫ (ln b − ln v)α−1 g( ab t

α−1

g( ab ) v ab v

d(

ab ) v

ab t

ab −1 g(v) )v 2 dv = ∫ (ln b − ln v)α−1 dv. v v v a

And then, we have b 󵄨󵄨 󵄨󵄨 t 󵄨 󵄨󵄨 󵄨󵄨 ∫(ln b − ln s)α−1 g(s) ds − ∫(ln s − ln a)α−1 g(s) ds󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 s s 󵄨󵄨 󵄨󵄨 a t ab

t 󵄨󵄨 t 󵄨 󵄨󵄨 g(s) 󵄨󵄨󵄨󵄨 g(s) = 󵄨󵄨󵄨 ∫(ln b − ln s)α−1 ds − ∫ (ln b − ln s)α−1 ds󵄨󵄨 󵄨󵄨 󵄨󵄨 s s 󵄨a 󵄨 a ab

󵄨 󵄨󵄨 t 󵄨󵄨 g(s) 󵄨󵄨󵄨󵄨 ds󵄨󵄨 = 󵄨󵄨󵄨 ∫ (ln b − ln s)α−1 󵄨󵄨 󵄨󵄨 s 󵄨 󵄨t t 󵄨 α−1 g(s) 󵄨󵄨 󵄨 ds, t ∈ [√ab, b] { {∫abt 󵄨󵄨(ln b − ln s) s 󵄨󵄨 ≤ { ab . { t 󵄨󵄨 α−1 g(s) 󵄨󵄨 √ ab] ds, t ∈ [a, (ln b − ln s) ∫ 󵄨 󵄨 s 󵄨 { t 󵄨 So we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 󵄨󵄨 󵄨 ((H Jaα+ g)(b) + (H Jbα− g)(a)) − [(H Jaα+ fg)(b) + (H Jbα− fg)(a)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 ab

ab t α−1 󵄨󵄨 g(s) 󵄨󵄨󵄨󵄨 t−a 󸀠 1 b b−t 󸀠 󵄨 ≤ |f (a)| + |f (b)|)dt ∫ ( ∫ 󵄨󵄨󵄨(ln ) 󵄨󵄨ds)( 󵄨 󵄨 Γ(α) s s 󵄨 b−a b−a 󵄨 a √

t

b

t α−1 󵄨󵄨 1 g(s) 󵄨󵄨󵄨󵄨 b b−t 󸀠 t−a 󸀠 󵄨 + |f (a)| + |f (b)|)dt ∫ ( ∫ 󵄨󵄨󵄨(ln ) 󵄨ds)( 󵄨󵄨 Γ(α) s s 󵄨󵄨󵄨 b−a b−a ab √ab

t

√ab

α

α

‖g‖∞ b t ≤ { ∫ ((ln ) − (ln ) )((b − t)|f 󸀠 (a)| + (t − a)|f 󸀠 (b)|)dt (b − a)Γ(α + 1) t a a

5.1 Inequalities via convex functions | 287 b

α

α

t b + ∫ ((ln ) − (ln ) )((b − t)|f 󸀠 (a)| + (t − a)|f 󸀠 (b)|)dt} a t √ab

≤

‖g‖∞ {(|f 󸀠 (b)| − |f 󸀠 (a)|)K1 (t; a, b, α) (b − a)Γ(α + 1) + (b|f 󸀠 (a)| − a|f 󸀠 (b)|)K2 (t; a, b, α)},

where √ab

√ab

α

b

α

b

α

α

b t t b K1 (t; a, b, α) = ∫ (ln ) tdt − ∫ (ln ) tdt + ∫ (ln ) tdt − ∫ (ln ) tdt, t a a t a

a

√ab

√ab

α

√ab

√ab

b

α

b

α

α

b t t b K2 (t; a, b, α) = ∫ (ln ) dt − ∫ (ln ) dt + ∫ (ln ) dt − ∫ (ln ) dt. t a a t a

a

√ab

√ab

As for the left two terms of K1 (t; a, b, α), √ab

√ab

α

α

b t H1 = ∫ (ln ) tdt − ∫ (ln ) tdt, t a a

a

letting u = ln bt , v = ln at , we have 2

1 2

H1 = −b

ln

b a

α −2u

∫ u e

du − a

1 2

ln

b a

∫ vα e2v dv 0

b a

ln

2

ln

b a

ln

b a

1 2

ln

b a

b2 a2 = α+1 ∫ (2u)α e−2u d2u − α+1 ∫ (2v)α e2v d2v, 2 2 1 2

0

letting x = 2u, y = 2v, we have 2

2 ln

b a

2

ln

b a

a b ∫ xα e−x dx − α+1 ∫ yα ey dy, 2α+1 2

H1 =

ln

0

b a

letting x = z + ln ba , we have ln

b a

α

ln

b a

b b2 b a2 H1 = α+1 ∫ (z + ln ) e−z−ln a dz − α+1 ∫ yα ey dy. a 2 2

0

0

288 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals Using Lemma 37, we obtain ln

b a

ln

α

b a

b a2 b2 a H1 ≤ α+1 ∫ (z α + (ln ) )e−z dz − α+1 ∫ yα ey dy a 2 b 2 0

0

b a

ln

ln

b a

b a

ln

α

a2 b2 a b ≤ α+1 [ ∫ z α e−z dz + ∫ (ln ) e−z dz] − α+1 ∫ yα ey dy a 2 b 2 0

≤

0

ln

0

b a

α

ln

b a

b ab ln [−(z α e−z |0 − α ∫ z α−1 e−z dz) + (ln ) ∫ e−z dz] a 2α+1 b a

0

0

b a

ln

a2 ln b − α+1 [yα ey |0 a − α ∫ yα−1 ey dy], 2 0

using Lemma 61 and integrating, we can get H1 ≤

i=∞ a2 (ln b − ln a)i−1 b α + − 1] (ln b − ln a) [−1 + α ∑ (α)i a 2α+1 i=1 i=∞ ab (ln a − ln b)i−1 α (ln b − ln a) [−1 + α ], ∑ (α)i 2α+1 i=1

+

as the same method, we obtain b

b

α

α

t b H2 = ∫ (ln ) tdt − ∫ (ln ) tdt a t √ab

≤

√ab

i=∞ a (ln a − ln b)i−1 b2 (ln b − ln a)α [1 − α ∑ +1− ] α+1 (α) b 2 i i=1

+

i=∞ ab (ln b − ln a)i−1 α (ln b − ln a) [1 − α ]. ∑ (α)i 2α+1 i=1

As for the left two terms of K2 (t; a, b, α), √ab

α

√ab

α

b t H3 = ∫ (ln ) dt − ∫ (ln ) dt, t a a

a

letting u = ln bt , v = ln at , we can get ln

b a

1 2

ln

b a

H3 = b ∫ u e du − a ∫ vα ev dv, 1 2

ln

b a

α −u

0

5.1 Inequalities via convex functions | 289

letting u = z + 21 ln ba , we have 1 2

ln

b a

H3 = b ∫ (z + 0

α

1 b ln ) e 2 a

−z− 21

ln

b a

1 2

ln

b a

dz − a ∫ vα ev dv. 0

Using Lemma 37, we have the inequality 1 2

H3 ≤ b√

ln

b a

1 2

α

ln

b a

a 1 b ∫ (z α + ( ln ) )e−z dz − a ∫ yα ey dy b 2 a 0 1 2

0

b a

ln

1 2

b a

ln

α

1 2

ln

b a

1 b a ≤ b√ [ ∫ z α e−z dz + ∫ ( ln ) e−z dz] − a ∫ yα ey dy b 2 a 0

0

0

1 2

ln

b a

α

1 a 1 b ln b ≤ b√ [−(z α e−z |02 a − α ∫ z α−1 e−z dz) + ( ln ) b 2 a

1 2

ln

∫ e−z dz]

0

1

− a[yα ey |02

ln

b a

1 2

ln

b a

0

b a

− α ∫ yα−1 ey dy], 0

using Lemma 61 and integrating, we can obtain H3 ≤ b(

α

a a i=∞ (ln b − ln a)i−1 a a ln b − ln a − +√ ] ) [− + α ∑ 2 b b i=1 b b 2i−1 (α)i

− a(

α

ln b − ln a b b i=∞ (ln a − ln b)i−1 ) [√ − α√ ∑ ]. 2 a a i=1 2i−1 (α)i

As the same method, we have b

b

α

α

t b H4 = ∫ (ln ) dt − ∫ (ln ) dt a t √ab

≤ b(

α

√ab

i=∞ ln b − ln a (ln a − ln b)i−1 a ) [1 − α ∑ +1−√ ] i−1 (α) 2 b 2 i i=1

+ √ab(

α

i=∞ ln b − ln a (ln b − ln a)i−1 ) [1 − α ∑ ]. 2 2i−1 (α)i i=1

Substituting H1 –H4 into K1 (t; a, b, α) and K2 (t; a, b, α), we get the result. The proof is completed.

290 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals Theorem 283. Let f : [a, b] → ℝ be a differentiable function on [a, b], f 󸀠 ∈ L[a, b], 0 < a < b. If |f 󸀠 | is a nondecreasing and convex function on [a, b], g : [a, b] → ℝ is continuous and symmetric to √ab, the order 0 < α ≤ 1, then we have the inequality f (b)((H Jaα+ g)(b) + (H Jbα− g)(a)) − [(H Jaα+ fg)(b) + (H Jbα− fg)(a)] ≤

‖g‖∞ (ln b − ln a)α 󸀠 {|f (a)|S1 (t; a, b, α) + |f 󸀠 (b)|S2 (t; a, b, α)}, (b − a)Γ(α + 1)

where ‖g‖∞ = sup |g(x)|, t∈[a,b]

S1 (t; a, b, α) = b2 [1 − α

i=∞ (ln a − ln b)i−1 a i=∞ (ln b − ln a)i−1 +1−α ∑ ] ∑ b i=1 (α)i (α)i i=1

− [b − a + +

i=∞ b2 (ln a − ln b)i−1 )], (1 − α ∑ 2 2i−1 (α)i i=1

S2 (t; a, b, α) = −ab[1 − α

i=∞ a i=∞ (ln b − ln a)i−1 (ln a − ln b)i−1 +1−α ∑ ] ∑ b i=1 (α)i (α)i i=1

+ [b − a + +

i=∞ (ln b − ln a)i−1 a2 (1 − α ∑ ) 2 2i−1 (α)i i=1

i=∞ (ln b − ln a)i−1 a2 ) (1 − α ∑ 2 2i−1 (α)i i=1

i=∞ b2 (ln a − ln b)i−1 (1 − α ∑ )], 2 2i−1 (α)i i=1

(α)i = α(α + 1)(α + 2) ⋅ ⋅ ⋅ (α + i − 1).

Proof. Using Lemma 281, the process of proof Theorem 283 is similar to Theorem 282, so we do not repeat again.

5.1.4 Further results We first present a new Hadamard type fractional integral inequalities for convex function. Theorem 284. Let f : [a, b] → ℝ be a convex function with 0 ≤ a < b and α > 0. Then the following inequality for fractional integrals holds: f(

a+b Γ(α + 1) f (a) + f (b) α a )≤ [ J α a + (f ∘ ln)(eb ) + H J(e . b )− (f ∘ ln)(e )] ≤ 2 2(b − a)α H (e ) 2

(5.36)

5.1 Inequalities via convex functions | 291

Proof. Since f is a convex function on [a, b], we have f(

f (x) + f (y) a+b )≤ . 2 2

Set x = ta + (1 − t)b and y = tb + (1 − t)a for 0 ≤ t ≤ 1, then 2f (

a+b ) ≤ f (ta + (1 − t)b) + f (tb + (1 − t)a). 2

(5.37)

Multiplying both sides of (5.37) by t α−1 , then integrating the resulting inequality with respect to t over [0, 1], we obtain 1

1

2 a+b f( ) ≤ ∫ t α−1 f (ta + (1 − t)b)dt + ∫ t α−1 f (tb + (1 − t)a)dt α 2 0

0

eb

=

α−1

1 b − ln u ) ∫( b−a b−a

f (ln u)

ea

eb

α−1

1 ln u − a + ) ∫( b−a b−a

du u

f (ln u)

ea

du u

Γ(α) α a [ J α a + (f ∘ ln)(eb ) + H J(e = b )− (f ∘ ln)(e )], (b − a)α H (e ) which implies that 2 a+b Γ(α) α a [ J α a + (f ∘ ln)(eb ) + H J(e f( )≤ b )− (f ∘ ln)(e )]. α 2 (b − a)α H (e ) Since f is a convex function, then for t ∈ [0, 1], it yields f (ta + (1 − t)b) ≤ tf (a) + (1 − t)f (b),

f (tb + (1 − t)a) ≤ tf (b) + (1 − t)f (a). By adding the above two inequalities, we get f (ta + (1 − t)b) + f (tb + (1 − t)a) ≤ f (a) + f (b).

(5.38)

Then multiplying both sides of (5.38) by t α−1 and integrating the inequality, we obtain 1

∫t

α−1

1

f (ta + (1 − t)b)dt + ∫ t

0

0

α−1

1

f (tb + (1 − t)a)dt ≤ ∫[f (a) + f (b)]t α−1 dt. 0

So we can get the following result: [f (a) + f (b)] Γ(α) α a [ J α a + (f ∘ ln)(eb ) + H J(e . b )− (f ∘ ln)(e )] ≤ (b − a)α H (e ) α The proof is completed.

292 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals Using the right-sided inequality result of (5.36), we have the following conclusion. Remark 285. Let f : [a, b] → ℝ be a differentiable function with 0 ≤ a < b and f 󸀠 ∈ L[a, b]. Then for all x ∈ [a, b], λ ∈ [0, 1] and α > 0, we have 1 [I (b, λ, α; a, b) + If (a, λ, α; a, b)] 2 f Γ(α + 1) f (a) + f (b) α b − [ J α b − (f ∘ ln)(ea ) + H J(e = a )+ (f ∘ ln)(e )] 2 2(b − a)α H (e ) 1

=

(b − a)α+1 ∫(t α − λ)[f 󸀠 (tb + (1 − t)a) − f 󸀠 (ta + (1 − t)b)]dt. 2

(5.39)

0

Theorem 286. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with 0 < a < b such that f 󸀠 ∈ L[a, b]. If |f 󸀠 |q is a convex function for some fixed q ≥ 1, then for all x ∈ [a, b], λ ∈ [0, 1] and α > 0, we have 1− q1

|If (x, λ, α; a, b)| ≤ A1

1

(α, λ){(x − a)α+1 [A2 (α, λ)|f 󸀠 (x)|q + A3 (α, λ)|f 󸀠 (a)|q ] q 1

+ (b − x)α+1 [A2 (α, λ)|f 󸀠 (x)|q + A3 (α, λ)|f 󸀠 (b)|q ] q },

(5.40)

where 1

2αλ1+ α + 1 − λ, A1 (α, λ) = α+1 α 1+ α2 1 λ A2 (α, λ) = λ + − , α+1 α+2 2 2 1+ α2 1 λ 2α 1+ α1 λ − λ + − . A3 (α, λ) = α+1 α+2 (α + 2)(α + 1) 2 Proof. By Lemma 75, we have |If (x, λ, α; a, b)| ≤ If1 (x, λ, α; a, b) + If2 (x, λ, α; a, b),

(5.41)

where α+1

If1 (x, λ, α; a, b) := (x − a)

1

󵄨 󵄨 ∫ |t α − λ|󵄨󵄨󵄨f 󸀠 (tx + (1 − t)a)󵄨󵄨󵄨dt, 0

α+1

If2 (x, λ, α; a, b) := (b − x)

1

󵄨 󵄨 ∫ |t α − λ|󵄨󵄨󵄨f 󸀠 (tx + (1 − t)a)󵄨󵄨󵄨dt. 0

Using Hölder inequality, we have 1

If1 (x, λ, α; a, b) ≤ (x − a)α+1 (∫ |t α − λ|dt) 0

1− q1

1

λα

{ ∫ (λ − t α )[t|f 󸀠 (x)|q + (1 − t)|f 󸀠 (a)|q ]dt 0

5.1 Inequalities via convex functions | 293

1

+ ∫ (t α − λ)[t|f 󸀠 (x)|q + (1 − t)|f 󸀠 (a)|q dt]}

1 q

1

λα α+1

= (x − a)

1− q1

1

2αλ1+ α + 1 ( − λ) α+1

[(

1 λ α 1+ α2 λ + − )|f 󸀠 (x)|q α+1 α+2 2 1

q 2α 1+ α1 λ 2 1+ α2 1 +( λ λ − )|f 󸀠 (a)|q ] . − + α+1 α+2 (α + 2)(α + 1) 2

(5.42)

Similarly, we obtain α+1

If2 (x, λ, α; a, b) ≤ (b − x)

1− q1

1

2αλ1+ α + 1 − λ) ( α+1

[(

α 1+ α2 1 λ λ + − )|f 󸀠 (x)|q α+1 α+2 2 1

q 2α 1+ α1 2 1+ α2 1 λ +( λ − λ + − )|f 󸀠 (a)|q ] . α+1 α+2 (α + 2)(α + 1) 2

(5.43)

From above, one can substitute (5.42) and (5.43) into (5.41) to derive the result. The proof is completed. Remark 287. Under the assumptions of Theorem 286 with x = inequality (5.40), we can get 󵄨󵄨 󵄨󵄨 a + b 󵄨 󵄨󵄨 , 0, α; a, b)󵄨󵄨󵄨 󵄨󵄨If ( 󵄨󵄨 󵄨󵄨 2 1− 1

α+1

a+b , 2

λ = 0, from the

1

q q b−a 1 1 ) [ ( ) ] ≤( α+1 α+2 2 1 1 q 󵄨 󵄨󵄨 󵄨q 󵄨󵄨 q q 1 1 󵄨󵄨 󸀠 a + b 󵄨󵄨󵄨 󵄨󵄨 󸀠 a + b 󵄨󵄨󵄨 󸀠 q 󸀠 q )󵄨󵄨 + |f (a)| ] + [󵄨󵄨f ( )󵄨󵄨 + |f (b)| ] }. × {[󵄨󵄨f ( 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 α+1 2 α+1

Remark 288. Let f : [a, b] → ℝ be a differentiable function with 0 ≤ a < b. If q = 1, then for all x ∈ [a, b], λ ∈ [0, 1] and α > 0, we have 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 [If (a, λ, α; a, b) + If (b, λ, α; a, b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 2 1 α+1 󸀠 󸀠 ≤ (b − a) [|f (a)| + |f (b)|][A2 (α, λ) + A3 (α, λ)]. 2 Further, we have the following result. Remark 289. Let f : [a, b] → ℝ be a twice-differentiable mapping on (a, b) with a < b. If f 󸀠󸀠 ∈ L[a, b], then the following equality for fractional integrals holds 1 [I (b, λ, α, a, b) + If (a, λ, α, a, b)] 2 f =

(b − a)α+1 1 [( − λ)[f 󸀠 (b) − f 󸀠 (a)] 2 α+1

294 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals 1

(b − a)α+2 − ∫(t α+1 − λ(α + 1)t)[f 󸀠󸀠 (tb + (1 − t)a) + f 󸀠󸀠 (ta + (1 − t)b)]dt]. α+1 0

Theorem 290. Let f : [a, b] → ℝ be a twice-differentiable mapping on (a, b) with a < b and f 󸀠󸀠 ∈ L[a, b]. If |f 󸀠󸀠 |q is a convex function for some fixed q ≥ 1 on [a, b], then the following equality for fractional integrals holds: |If (x, λ, α; a, b)| 󵄨󵄨 󵄨󵄨 1 󵄨 󵄨 ≤ 󵄨󵄨󵄨( − λ)f 󸀠 (x)[(x − a)α+1 − (b − x)α+1 ]󵄨󵄨󵄨 󵄨󵄨 α + 1 󵄨󵄨 α+2 1 1 1− (x − a) [A5 (α, λ)|f 󸀠󸀠 (x)|q + A6 (α, λ)|f 󸀠󸀠 (a)|q ] q + A4 q (α, λ){ α+1 +

1 (b − x)α+2 [A5 (α, λ)|f 󸀠󸀠 (x)|q + A6 (α, λ)|f 󸀠󸀠 (b)|q ] q }, α+1

(5.44)

(5.45)

where α λ(α + 1) 1 1+ 2 [λ(α + 1)] α − + , α+2 2 α+2 3 2α 1 λ(α + 1) 1+ A5 (α, λ) = [λ(α + 1)] α + − , 3(α + 3) α+3 3 α 2α 1 λ(α + 1) 1+ 2 1+ 3 A6 (α, λ) = [λ(1 + α)] α − [λ(1 + α)] α + − . α+2 3(α + 3) (α + 2)(α + 3) 6

A4 (α, λ) =

Proof. Follows from (3.94), we have 󵄨󵄨 󵄨󵄨 1 󵄨 󵄨 |If (x, λ, α; a, b)| ≤ 󵄨󵄨󵄨( − λ)f 󸀠 (x)[(x − a)α+1 − (b − x)α+1 ]󵄨󵄨󵄨 󵄨󵄨 α + 1 󵄨󵄨 α+2 α+2 (b − x) (x − a) |Jf1 (x, λ, α; a, b)| + |Jf2 (x, λ, α; a, b)|, + α+1 α+1 where 1

Jf1 (x, λ, α; a, b) = ∫[t α+1 − λ(α + 1)t]f 󸀠󸀠 (tx + (1 − t)a)dt, 0

1

Jf2 (x, λ, α; a, b) = ∫[t α+1 − λ(α + 1)t]f 󸀠󸀠 (tx + (1 − t)b)dt. 0

By using Hölder inequality, we can get |Jf1 (x, λ, α; a, b)| 1

󵄨 󵄨 ≤ ∫ |t α+1 − λ(α + 1)t|󵄨󵄨󵄨f 󸀠󸀠 (tx + (1 − t)a)󵄨󵄨󵄨dt 0

(5.46)

5.1 Inequalities via convex functions | 295 1− q1

1

≤ (∫ |t α+1 − λ(α + 1)t|dt) 0 1

[λ(α+1)] α

×{

∫

[λ(α + 1)t − t α+1 ](t|f 󸀠󸀠 (x)|q + (1 − t)|f 󸀠󸀠 (a)|q )dt

0 1

+

[t α+1 − λ(α + 1)t](t|f 󸀠󸀠 (x)|q + (1 − t)|f 󸀠󸀠 (a)|q )dt}

∫

1 q

1

[λ(α+1)] α

1− q1

1

≤ (∫ |t α+1 − λ(α + 1)t|dt) 0 1

[λ(α+1)] α

q

󸀠󸀠

× {|f (x)| [

∫

[λ(α + 1)t − t

α+2

1

]dt +

0

[t α+2 − λ(α + 1)t 2 ]dt]

∫ 1

[λ(α+1)] α 1

+ |f 󸀠󸀠 (a)|q [

[λ(α+1)] α

∫

[λ(α + 1)t − λ(α + 1)t 2 − t α+1 + t α+2 ]dt

0 1

+

[t α+1 − t α+2 − λ(α + 1)t + λ(α + 1)t 2 ]dt]}

∫

1 q

1

[λ(α+1)] α

1− q1

1

= A4 (α, λ)[A5 (α, λ)|f 󸀠󸀠 (x)|q + A6 (α, λ)|f 󸀠󸀠 (a)|q ] q .

(5.47)

Using the same method, we get 1− 1

1

|Jf2 (x, λ, α; a, b)| ≤ A4 q (α, λ)[A5 (α, λ)|f 󸀠󸀠 (x)|q + A6 (α, λ)|f 󸀠󸀠 (b)|q ] q .

(5.48)

On the other hand, 1

∫ |t α+1 − λ(α + 1)t|dt 0 1

[λ(α+1)] α

=

∫ 0

=

[λ(α + 1)t − t

α+1

1

]dt +

[t α+1 − λ(α + 1)t]dt

∫ 1

[λ(α+1)] α

α λ(α + 1) 1 1+ 2 [λ(α + 1)] α − + . α+2 2 α+2

By substituting (5.47), (5.48) and (5.49) into (5.46), the proof is completed.

(5.49)

296 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals Remark 291. Let f : [a, b] → ℝ be a twice-differentiable function with 0 ≤ a < b. If q = 1, then for all x ∈ [a, b], λ ∈ [0, 1] and α > 0, we have 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 [If (a, λ, α; a, b) + (b, λ, α; a, b)]󵄨󵄨󵄨 󵄨󵄨 2 󵄨󵄨 α+1 󵄨󵄨 󵄨󵄨 1 (b − a) 󵄨󵄨 󵄨 ≤ − λ)[f 󸀠 (b) − f 󸀠 (a)]󵄨󵄨󵄨 󵄨󵄨( 󵄨󵄨 α + 1 󵄨󵄨 2 α+2 (b − a) + [|f 󸀠󸀠 (a)| + |f 󸀠󸀠 (b)|][A5 (α, λ) + A6 (α, λ)]. 2(α + 1) Remark 292. Under the assumptions of Theorem 290 with x = inequality (5.44), we can get 󵄨󵄨 a + b 󵄨󵄨 󵄨󵄨 󵄨 , 0, α; a, b)󵄨󵄨󵄨 󵄨󵄨If ( 󵄨󵄨 󵄨󵄨 2

a+b , 2

λ = 0, from the

1

1

1

1− q α+2 q 󵄨󵄨 q q 1 1 b−a a + b 󵄨󵄨󵄨󵄨 1 1 󵄨 ( ) ( ) ( ) {[󵄨󵄨󵄨f 󸀠󸀠 ( )󵄨󵄨 + |f 󸀠󸀠 (a)|q ] ≤ 󵄨󵄨 󵄨󵄨 α+1 α+2 α+3 2 2 α+2 1 q 󵄨󵄨 q 1 a + b 󵄨󵄨󵄨󵄨 󵄨 + [󵄨󵄨󵄨f 󸀠󸀠 ( )󵄨󵄨 + |f 󸀠󸀠 (b)|q ] }. 󵄨󵄨 󵄨󵄨 2 α+2

Remark 293. Let f : [a, b] → ℝ be a twice differentiable mapping on (a, b) with 0 ≤ a < b and f 󸀠󸀠 ∈ L[a, b]. If |f 󸀠󸀠 (x)| ≤ M for all x ∈ [a, b] and M > 0. Then the following inequality for fractional integrals holds 󵄨󵄨 αM(b − a)α+2 󵄨󵄨 1 󵄨 󵄨󵄨 . 󵄨󵄨 [If (b, λ, α, a, b) + If (a, λ, α, a, b)]󵄨󵄨󵄨 ≤ 󵄨󵄨 2(α + 1)(α + 2) 󵄨󵄨 2 Proof. By using the mean value theorem for f 󸀠 , we have 1 󵄨󵄨 󵄨󵄨 󵄨󵄨 (b − a)α+1 󵄨󵄨 α 󸀠 󸀠 󵄨󵄨 ∫(t − λ)[f (tb + (1 − t)a) − f (ta + (1 − t)b)]dt 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 2 󵄨󵄨 󵄨 0

=

󵄨 1 󵄨󵄨 󵄨󵄨 (b − a)α+1 󵄨󵄨󵄨󵄨 α 󸀠 󸀠 󵄨󵄨 ∫(t − λ)[f (tb + (1 − t)a) − f (ta + (1 − t)b)]dt 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 2 󵄨 󵄨 0

1 󵄨󵄨 󵄨󵄨 󵄨󵄨 α 󵄨󵄨 ∫(t − λ)M[(tb + (1 − t)a) − (ta + (1 − t)b)]dt 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨

α+1 󵄨󵄨󵄨

(b − a) ≤ 2 =

0

󵄨 1 󵄨󵄨 󵄨󵄨 M(b − a)α+2 󵄨󵄨󵄨󵄨 α 󵄨󵄨 (t − λ)(2t − 1)dt ∫ 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 󵄨󵄨 󵄨 0

αM(b − a)α+2 . = 2(α + 1)(α + 2) The proof is completed.

5.2 Inequalities via s-e-condition functions | 297

Now, using the theory results, we provide some applications to special means of real numbers. Proposition 294. Let a, b ∈ ℝ+ , a < b. Then 1 󵄨󵄨 a b a b 󵄨 2 a b 󵄨󵄨A(e , e ) − L(e , e )󵄨󵄨󵄨 ≤ (b − a) (e + e ). 6 Proof. Applying Remark 288, for f (ln x) = x, λ = 1 and α = 1, one can obtain the results immediately. Proposition 295. Let a, b ∈ ℝ+ , a < b. Then 1 󵄨󵄨 −1 a b 2 −a −b −a −b 󵄨 󵄨󵄨H (e , e ) − L(e , e )󵄨󵄨󵄨 ≤ (b − a) (e + e ). 6 Proof. Applying Remark 288, for a−1 > b−1 , f (ln x) = x1 , λ = 1 and α = 1, one can obtain the results immediately. Proposition 296. Let a, b ∈ ℝ+ , a < b. Then 󵄨󵄨 eb − ea n a b 󵄨󵄨󵄨󵄨 1 󵄨󵄨 a(n+1) b(n+1) L (e , e )󵄨󵄨 ≤ (b − a)2 (n + 1)[ea(n+1) + eb(n+1) ]. ,e )− 󵄨󵄨A(e 󵄨󵄨 󵄨󵄨 6 b−a n Proof. Applying Remark 288, for f (ln x) = xn+1 , λ = 1 and α = 1, one can obtain the results immediately.

5.2 Inequalities via s-e-condition functions The results in this section are based on [220, 221]. Theorem 297. Let f : [a, b] ⊂ (0, ∞) → ℝ be a differentiable mapping on (a, b) with 0 < a < b such that f 󸀠 ∈ L1 (a, b). If |f 󸀠 | satisfies the s-e-condition on [a, b] for some fixed s ∈ (0, 1] and |f 󸀠 (x)| ≤ M, x ∈ [a, b], then the following inequality for fractional integrals with α > 0 holds: 󵄨󵄨 α α α α 󵄨 󵄨󵄨Γ(α + 1)[H Jx− f (a) + H Jx+ f (b)] − [f (a)(ln x − ln a) + f (b)(ln b − ln x) ]󵄨󵄨󵄨 Γ(α + 1)Γ(s + 1) (ln x − ln a)α+1 + (ln b − ln x)α+1 2α − ] . ≤ Mb[1 + s+1 Γ(α + s + 1) α+s+1 Proof. From Lemma 71 and since |f 󸀠 | satisfies the s-e-condition on [a, b], we have 󵄨󵄨 α α α α 󵄨 󵄨󵄨Γ(α + 1)[H Jx− f (a) + H Jx+ f (b)] − [f (a)(ln x − ln a) + f (b)(ln b − ln x) ]󵄨󵄨󵄨 ≤ (ln b − ln x)

α+1

1

󵄨 󵄨 ∫(1 − t α )et ln x+(1−t) ln b 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln b )󵄨󵄨󵄨dt 0

298 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals

α+1

+ (ln x − ln a)

1

1

󵄨 󵄨 ∫(1 − t α )et ln x+(1−t) ln a 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln a )󵄨󵄨󵄨dt 0

≤ (ln b − ln x)α+1 ∫(1 − t α )x t b1−t (t s |f 󸀠 (x)| + (1 − t)s |f 󸀠 (b)|)dt 0 α+1

+ (ln x − ln a)

1

∫(1 − t α )xt a1−t (t s |f 󸀠 (x)| + (1 − t)s |f 󸀠 (a)|)dt 0

1

≤ (ln b − ln x)α+1 Mb ∫[t s (1 − t α ) + (1 − t α )(1 − t)s ]dt 0

1

+ (ln x − ln a)α+1 Mb ∫[t s (1 − t α ) + (1 − t α )(1 − t)s ]dt 0

= Mb[(ln x − ln a)

α+1

α+1

+ (ln b − ln x)

1

] ∫[t s (1 − t α ) + (1 − t)s − t α (1 − t)s ]dt 0

2 1 Γ(α + 1)Γ(s + 1) = Mb[ − − ][(ln x − ln a)α+1 + (ln b − ln x)α+1 ] s+1 s+α+1 Γ(α + s + 2) = Mb[1 +

2α Γ(α + 1)Γ(s + 1) (ln x − ln a)α+1 + (ln b − ln x)α+1 − ] , s+1 Γ(α + s + 1) α+s+1

where we use the fact that xt a1−t ≤ x ≤ b and x t b1−t ≤ b via 1

∫ t s (1 − t α )dt = 0

1 1 − , s+1 s+α+1

and 1

∫[(1 − t)s − t α (1 − t)s ]dt = 0

1 1 Γ(α + 1)Γ(s + 1) − 𝔹(α + 1, s + 1) = − . s+1 s+1 Γ(α + s + 2)

So using the reduction formula Γ(n + 1) = nΓ(n) (n > 0) for Euler gamma function, the proof is completed. Theorem 298. Let f : [a, b] ⊂ (0, ∞) → ℝ be a differentiable mapping on (a, b) with 0 < a < b such that f 󸀠 ∈ L1 (a, b). If |f 󸀠 |q satisfies the s-e-condition on [a, b] for some fixed s ∈ (0, 1] and |f 󸀠 (x)| ≤ M, x ∈ [a, b], then the following inequality for fractional integrals with α > 0 holds: 󵄨󵄨 α α α α 󵄨 󵄨󵄨Γ(α + 1)[H Jx− f (a) + H Jx+ f (b)] − [f (a)(ln x − ln a) + f (b)(ln b − ln x) ]󵄨󵄨󵄨 1

1

p 1 2α Γ(α + 1)Γ(s + 1) q ≤ Mb(1 − ) [1 + − ] α+1 s+1 Γ(α + s + 1)

5.2 Inequalities via s-e-condition functions | 299

× where

1 p

+

1 q

(ln x − ln a)α+1 + (ln b − ln x)α+1 1

(α + s + 1) q

,

= 1.

Proof. Using Lemma 71 and the well-known Hölder inequality and since |f 󸀠 |q satisfies the s-e-condition on [a, b], we have

󵄨󵄨 α α α α 󵄨 󵄨󵄨Γ(α + 1)[H Jx− f (a) + H Jx+ f (b)] − [f (a)(ln x − ln a) + f (b)(ln b − ln x) ]󵄨󵄨󵄨 α+1

≤ (ln b − ln x)

1

󵄨 󵄨 ∫(1 − t α )et ln x+(1−t) ln b 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln b )󵄨󵄨󵄨dt 0

1

󵄨 󵄨 + (ln x − ln a)α+1 ∫(1 − t α )et ln x+(1−t) ln a 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln a )󵄨󵄨󵄨dt 0

1

󵄨 󵄨 ≤ (ln b − ln x)α+1 b ∫(1 − t α )󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln b )󵄨󵄨󵄨dt 0

α+1

+ (ln x − ln a)

1

󵄨 󵄨 b ∫(1 − t α )󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln a )󵄨󵄨󵄨dt 0

1 p

1

1

󵄨 󵄨q ≤ (ln b − ln x)α+1 b(∫(1 − t α )dt) (∫(1 − t α )󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln b )󵄨󵄨󵄨 dt)

1 q

0

0

1 p

1

1

󵄨 + (ln x − ln a)α+1 b(∫(1 − t α )dt) (∫(1 − t α )󵄨󵄨󵄨f 󸀠 (e 0

t ln x+(1−t) ln a 󵄨󵄨q

)󵄨󵄨 dt)

0

1

1

1 q

p 1 ≤ (ln b − ln x)α+1 b(1 − ) (∫(1 − t α )[t s |f 󸀠 (x)|q + (1 − t)s |f 󸀠 (b)|q ]dt) α+1

1 q

0

1

1 p

1 + (ln x − ln a)α+1 b(1 − ) (∫(1 − t α )[t s |f 󸀠 (x)|q + (1 − t)s |f 󸀠 (a)|q ]dt) α+1 0 1

1 p

1 ≤ (ln b − ln x)α+1 Mb(1 − ) (∫(1 − t α )[t s + (1 − t)s ]dt) α+1

1 q

0

1 p

1

1 + (ln x − ln a)α+1 Mb(1 − ) (∫(1 − t α )[t s + (1 − t)s ]dt) α+1 1 p

0

1

1 2α Γ(α + 1)Γ(s + 1) q = Mb(1 − ) [1 + − ] α+1 s+1 Γ(α + s + 1)

1 q

1 q

300 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals (ln x − ln a)α+1 + (ln b − ln x)α+1

×

1

(α + s + 1) q

.

The proof is completed. Theorem 299. Let f : [a, b] ⊂ (0, ∞) → ℝ be a differentiable mapping on (a, b) with 0 < a < b such that f 󸀠 ∈ L1 (a, b). If |f 󸀠 | is convex and nondecreasing on [a, b] and |f 󸀠 (x)| ≤ M, x ∈ [a, b], then the following inequality for fractional integrals with α > 0 holds: 󵄨󵄨 α α α α 󵄨 󵄨󵄨Γ(α + 1)[H Jx− f (a) + H Jx+ f (b)] − [f (a)(ln x − ln a) + f (b)(ln b − ln x) ]󵄨󵄨󵄨 (ln x − ln a)α+1 + (ln b − ln x)α+1 . ≤ αMb α+1 Proof. Using Remark 26 and repeating the same procedures in Theorem 297, we can derive the results immediately. Theorem 300. Let f : [a, b] ⊂ (0, ∞) → ℝ be a differentiable mapping on (a, b) with 0 < a < b such that f 󸀠 ∈ L1 (a, b). If |f 󸀠 |q is convex and nondecreasing on [a, b] and |f 󸀠 (x)| ≤ M, x ∈ [a, b], then the following inequality for fractional integrals with α > 0 holds: 󵄨󵄨 α α α α 󵄨 󵄨󵄨Γ(α + 1)[H Jx− f (a) + H Jx+ f (b)] − [f (a)(ln x − ln a) + f (b)(ln b − ln x) ]󵄨󵄨󵄨 1

p 1 (ln x − ln a)α+1 + (ln b − ln x)α+1 ) , ≤ α Mb(1 − 1 α+1 (α + 1) q 1 q

where

1 p

+

1 q

= 1.

Proof. Using Remark 26 and repeating the same procedures in Theorem 298, we can derive the results immediately. Theorem 301. Let f : [a, b] ⊂ (0, ∞) → ℝ, be a differentiable mapping on (a, b) with 0 < a < b such that f 󸀠 ∈ L[a, b]. If |f 󸀠 | satisfies the s-e-condition on [a, b] for some fixed s ∈ (0, 1] and |f 󸀠 (x)| ≤ M, x ∈ [a, b], then the following inequality for fractional integrals with α > 0 holds: 󵄨󵄨 󵄨󵄨 (ln x − ln a)α + (ln b − ln x)α Γ(α + 1) 󵄨 󵄨󵄨 f (x) − [H Jxα− f (a) + H Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a 󵄨󵄨 α+1 α+1 Γ(α + 1)Γ(s + 1) (ln x − ln a) + (ln b − ln x) Mb (1 + ) , ≤ ln b − ln a Γ(α + s + 1) α+s+1 for any x ∈ (a, b). Proof. From Lemma 68 and since |f 󸀠 | satisfies the s-e-condition on [a, b], we have 󵄨󵄨 (ln x − ln a)α + (ln b − ln x)α 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 f (x) − [H Jxα− f (a) + H Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a 󵄨󵄨

5.2 Inequalities via s-e-condition functions | 301 1

(ln x − ln a)α+1 󵄨 󵄨 ≤ ∫ t α et ln x+(1−t) ln a 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln a )󵄨󵄨󵄨dt ln b − ln a 0

+

α+1

≤

(ln x − ln a) ln b − ln a +

1

(ln b − ln x)α+1 󵄨 󵄨 ∫ t α et ln x+(1−t) ln b 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln b )󵄨󵄨󵄨dt ln b − ln a 0

1

∫ t α xt a(1−t) (t s |f 󸀠 (x)| + (1 − t)s |f 󸀠 (a)|)dt 0

1

(ln b − ln x)α+1 ∫ t α xt b(1−t) (t s |f 󸀠 (x)| + (1 − t)s |f 󸀠 (b)|)dt ln b − ln a 0

α+1

≤

1

(ln x − ln a) Mb ∫(t α+s + t α (1 − t)s )dt ln b − ln a 0

1

(ln b − ln x)α+1 Mb + ∫(t α+s + t α (1 − t)s )dt ln b − ln a 0

Mb 1 Γ(α + 1)Γ(s + 1) ≤ ( + )[(ln x − ln a)α+1 + (ln b − ln x)α+1 ], ln b − ln a α + s + 1 Γ(α + s + 2) where we use the fact that xt a(1−t) ≤ x ≤ b and xt b(1−t) ≤ b via 1

∫t

α+s

0

1 dt = α+s+1

1

and

∫ t α (1 − t)s dt = 0

Γ(α + 1)Γ(s + 1) . Γ(α + s + 2)

So using the reduction formula Γ(n + 1) = nΓ(n) (n > 0) for Euler gamma function, the proof is completed. Theorem 302. Let f : [a, b] ⊂ (0, ∞) → ℝ, be a differentiable mapping on (a, b) with 0 < a < b such that f 󸀠 ∈ L[a, b]. If |f 󸀠 |q satisfies the s-e-condition on [a, b] for some fixed s ∈ (0, 1] and |f 󸀠 (x)| ≤ M, x ∈ [a, b], then the following inequality for fractional integrals with α > 0 holds: 󵄨󵄨 (ln x − ln a)α + (ln b − ln x)α 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 f (x) − [H Jxα− f (a) + H Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a 󵄨󵄨 1

q 2 (ln x − ln a)α+1 + (ln b − ln x)α+1 ( ) , ≤ 1 ln b − ln a (1 + pα) p s + 1

Mb

for any x ∈ (a, b), where

1 p

+

1 q

= 1, α > 0.

Proof. From Lemma 68 and using the well-known Hölder inequality, we have 󵄨󵄨 (ln x − ln a)α + (ln b − ln x)α 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 f (x) − [H Jxα− f (a) + H Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a 󵄨󵄨

(5.50)

302 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals 1

(ln x − ln a)α+1 󵄨 󵄨 ≤ ∫ t α et ln x+(1−t) ln a 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln a )󵄨󵄨󵄨dt ln b − ln a 0

+

1

(ln b − ln x)α+1 󵄨 󵄨 ∫ t α et ln x+(1−t) ln b 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln b )󵄨󵄨󵄨dt. ln b − ln a 0

α+1

≤

1

(ln x − ln a) b α 󵄨󵄨 󸀠 t ln x+(1−t) ln a 󵄨󵄨 )󵄨󵄨dt ∫ t 󵄨󵄨f (e ln b − ln a 0

+

1

(ln b − ln x)α+1 b α 󵄨󵄨 󸀠 t ln x+(1−t) ln b 󵄨󵄨 )󵄨󵄨dt. ∫ t 󵄨󵄨f (e ln b − ln a 0

1 p

1

1

(ln x − ln a)α+1 b 󵄨 󵄨q (∫ t pα dt) (∫ 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln a )󵄨󵄨󵄨 dt) ≤ ln b − ln a 0

0

1 p

1

1 q

1 q

1

(ln b − ln x)α+1 b 󵄨 󵄨q (∫ t pα dt) (∫ 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln b )󵄨󵄨󵄨 dt) . + ln b − ln a 0

0

Since |f 󸀠 |q satisfies the s-e-condition on [a, b] and |f 󸀠 (x)| ≤ M, we get 1

2M q 󵄨 󵄨q ∫ 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln a )󵄨󵄨󵄨 dt ≤ s+1

1

and

0

2M q 󵄨 󵄨q . ∫ 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln b )󵄨󵄨󵄨 dt ≤ s+1 0

On the other hand, 1

∫ t pα dt = 0

1 . pα + 1

Hence, we have 󵄨󵄨 (ln x − ln a)α + (ln b − ln x)α 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 f (x) − [H Jxα− f (a) + H Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a 1

q 2 (ln x − ln a)α+1 + (ln b − ln x)α+1 ( ) . ≤ 1 ln b − ln a (1 + pα) p s + 1

Mb

The proof is completed. Theorem 303. Let f : [a, b] ⊂ (0, ∞) → ℝ, be a differentiable mapping on (a, b) with 0 < a < b such that f 󸀠 ∈ L[a, b]. If |f 󸀠 |q satisfies the s-e-condition on [a, b] for some fixed s ∈ (0, 1] and |f 󸀠 (x)| ≤ M, x ∈ [a, b], then the following inequality for fractional integrals with α > 0 holds: 󵄨󵄨 (ln x − ln a)α + (ln b − ln x)α 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 f (x) − [H Jxα− f (a) + H Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a 󵄨󵄨

5.2 Inequalities via s-e-condition functions | 303 1

1

q 1 Γ(αq + 1)Γ(s + 1) q (ln x − ln a)α+1 + (ln b − ln x)α+1 ≤ Mb( ) (1 + ) , 1 + αq + s Γ(αq + s + 1) ln b − ln a

for any x ∈ (a, b), where

1 p

+

1 q

= 1, α > 0.

Proof. From Lemma 68 and using the well-known power mean inequality, we have 󵄨󵄨 (ln x − ln a)α + (ln b − ln x)α 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 f (x) − [H Jxα− f (a) + H Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a 1

(ln x − ln a)α+1 󵄨 󵄨 ≤ ∫ t α et ln x+(1−t) ln a 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln a )󵄨󵄨󵄨dt ln b − ln a 0

+

1

(ln b − ln x)α+1 󵄨 󵄨 ∫ t α et ln x+(1−t) ln b 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln b )󵄨󵄨󵄨dt ln b − ln a 0

α+1

≤

1

(ln x − ln a) b α 󵄨󵄨 󸀠 t ln x+(1−t) ln a 󵄨󵄨 )󵄨󵄨dt ∫ t 󵄨󵄨f (e ln b − ln a 0

1

(ln b − ln x)α+1 b α 󵄨󵄨 󸀠 t ln x+(1−t) ln b 󵄨󵄨 )󵄨󵄨dt + ∫ t 󵄨󵄨f (e ln b − ln a 0

1

(ln x − ln a)α+1 b 󵄨 󵄨q ≤ (∫ t αq 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln a )󵄨󵄨󵄨 dt) ln b − ln a

1 q

0

1 q

1

(ln b − ln x)α+1 b 󵄨 󵄨q (∫ t αq 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln b )󵄨󵄨󵄨 dt) . + ln b − ln a 0

Since |f 󸀠 |q satisfies the s-e-condition on [a, b] and |f 󸀠 (x)| ≤ M, we get 1

αq 󵄨󵄨 󸀠

∫ t 󵄨󵄨f (e 0

1

t ln x+(1−t) ln a 󵄨󵄨q

)󵄨󵄨 dt ≤ ∫ t 0

=

αq+s

q

1

|f (x)| dt + ∫ t αq (1 − t)s |f 󸀠 (a)|q dt 󸀠

0

1

|f 󸀠 (x)|q + |f 󸀠 (a)|q ∫ t αq (1 − t)s dt αq + s + 1 0

|f 󸀠 (x)|q = + |f 󸀠 (a)|q β(αq + 1, s + 1) αq + s + 1

|f 󸀠 (x)|q Γ(αq + 1)Γ(s + 1) + |f 󸀠 (a)|q αq + s + 1 Γ(αq + s + 2) q M Γ(αq + 1)Γ(s + 1) ≤ (1 + ), αq + s + 1 Γ(αq + s + 1)

=

304 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals and 1

αq 󵄨󵄨 󸀠

∫ t 󵄨󵄨f (e

t ln x+(1−t) ln b

0

1

1

0

0

󵄨 )󵄨󵄨󵄨dt ≤ ∫ t αq+s |f 󸀠 (x)|q dt + ∫ t αq (1 − t)s |f 󸀠 (b)|q dt Mq Γ(αq + 1)Γ(s + 1) ≤ (1 + ), αq + s + 1 Γ(αq + s + 1)

where β is an Euler beta function, and we used the fact that β(x, y) =

Γ(x)Γ(y) Γ(x + y)

and

Γ(n + 1) = nΓ(n)

(n > 0).

Hence, we have 󵄨󵄨 (ln x − ln a)α + (ln b − ln x)α 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 f (x) − [H Jxα− f (a) + H Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a 1

1

q Γ(αq + 1)Γ(s + 1) q (ln x − ln a)α+1 + (ln b − ln x)α+1 1 ) (1 + ) . ≤ Mb( 1 + αq + s Γ(αq + s + 1) ln b − ln a

The proof is completed. Corollary 304. Let f : [a, b] ⊂ (0, ∞) → ℝ, be a differentiable mapping on (a, b) with 0 < a < b such that f 󸀠 ∈ L[a, b]. If |f 󸀠 | is convex and nondecreasing on [a, b] and |f 󸀠 (x)| ≤ M, x ∈ [a, b], then the following inequality for fractional integrals with α > 0 holds: 󵄨󵄨 (ln x − ln a)α + (ln b − ln x)α 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 f (x) − [H Jxα− f (a) + H Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a α+1 α+1 Mb (ln x − ln a) + (ln b − ln x) ≤ , ln b − ln a α+1 for any x ∈ (a, b). Proof. Noting Remark 27 and Theorem 301, one can obtain the results immediately. Corollary 305. Let f : [a, b] ⊂ (0, ∞) → ℝ, be a differentiable mapping on (a, b) with 0 < a < b such that f 󸀠 ∈ L[a, b]. If |f 󸀠 |q is convex and nondecreasing on [a, b], p, q > 1 and |f 󸀠 (x)| ≤ M, x ∈ [a, b], then the following inequality for fractional integrals with α > 0 holds: 󵄨󵄨 󵄨󵄨 (ln x − ln a)α + (ln b − ln x)α Γ(α + 1) 󵄨󵄨 󵄨 f (x) − [H Jxα− f (a) + H Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a 󵄨󵄨 Mb (ln x − ln a)α+1 + (ln b − ln x)α+1 ≤ , 1 ln b − ln a (1 + pα) p for any x ∈ (a, b), where

1 p

+

1 q

= 1.

5.2 Inequalities via s-e-condition functions | 305

Proof. Noting Remark 27 and Theorem 302, one can obtain the results immediately. Theorem 306. Under the assumptions of Theorem 301, the following inequality for fractional integrals with α > 0 holds: 󵄨󵄨 (ln x − ln a)α + (ln b − ln x)α 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 f (x) − [H Jxα− f (a) + H Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a i−1

≤

∞ (ln x − ln a)α+1 x xM ∑(− ln ) ln b − ln a a i=1

+

[

1 1 + ] (α + s + 1)i (α + 1)i

i−1

∞ (ln b − ln x)α+1 x xM ∑(− ln ) ln b − ln a b i=1

[

1 1 + ], (α + s + 1)i (α + 1)i

for any x ∈ (a, b). Proof. Noting Lemma 32 and using s-e-conditions, we have 󵄨󵄨 (ln x − ln a)α + (ln b − ln x)α 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 f (x) − [H Jxα− f (a) + H Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a 1

(ln x − ln a)α+1 󵄨 󵄨 ≤ ∫ t α et ln x+(1−t) ln a 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln a )󵄨󵄨󵄨dt ln b − ln a 0

+

1

α+1

≤

1

(ln b − ln x)α+1 󵄨 󵄨 ∫ t α et ln x+(1−t) ln b 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln b )󵄨󵄨󵄨dt ln b − ln a

(ln x − ln a) ln b − ln a

0

∫ t α xt a(1−t) (t s |f 󸀠 (x)| + (1 − t)s |f 󸀠 (a)|)dt 0

1

(ln b − ln x)α+1 + ∫ t α xt b(1−t) (t s |f 󸀠 (x)| + (1 − t)s |f 󸀠 (b)|)dt ln b − ln a 0

α+1

≤

aM(ln x − ln a) ln b − ln a +

x i−1 x i−1 x ∞ (− ln a ) x ∞ (− ln a ) [ ∑ + ∑ ] a i=1 (α + s + 1)i a i=1 (α + 1)i

x i−1 x i−1 x ∞ (− ln b ) bM(ln b − ln x)α+1 x ∞ (− ln b ) [ ∑ + ∑ ] ln b − ln a b i=1 (α + s + 1)i b i=1 (α + 1)i i−1

≤

∞ (ln x − ln a)α+1 x xM ∑(− ln ) ln b − ln a a i=1

+

[

i−1

∞ (ln b − ln x)α+1 x xM ∑(− ln ) ln b − ln a b i=1

The proof is completed.

1 1 + ] (α + s + 1)i (α + 1)i [

1 1 + ]. (α + s + 1)i (α + 1)i

306 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals Theorem 307. Under the assumptions of Theorem 302, the following inequality for fractional integrals with α > 0 holds: 󵄨󵄨 (ln x − ln a)α + (ln b − ln x)α 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 f (x) − [H Jxα− f (a) + H Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a ≤

xM(ln x − ln a)α+1 1

(ln b − ln a)p p +

xM(ln b − ln x)α+1 1

(ln b − ln a)p p 1 p

for any x ∈ (a, b), where

1 q

+

1

1

∞ (− ln x )i−1 pi p q 2 a ( ) [∑ ] s+1 (αp + 1)i i=1

1

1

∞ (− ln x )i−1 pi p q 2 b ( ) [∑ ] , s+1 (αp + 1)i i=1

= 1.

Proof. Noting Lemma 32 and using Hölder inequality and s-e-conditions, we have 󵄨󵄨 󵄨󵄨 (ln x − ln a)α + (ln b − ln x)α Γ(α + 1) 󵄨 󵄨󵄨 f (x) − [H Jxα− f (a) + H Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a 󵄨󵄨 󵄨󵄨 1

≤

t

a(ln x − ln a)α+1 x 󵄨 󵄨 ∫ t α ( ) 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln a )󵄨󵄨󵄨dt ln b − ln a a 0

1

α+1

α+1

≤

a(ln x − ln a) ln b − ln a

t

x 󵄨 󵄨 ∫ t α ( ) 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln b )󵄨󵄨󵄨dt b

b(ln b − ln x) + ln b − ln a

0

1

1 p

pt

1

x 󵄨 󵄨q (∫ t αp ( ) dt) (∫ 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln a )󵄨󵄨󵄨 dt) a 0

1 q

0

1

1 p

pt

1

x b(ln b − ln x)α+1 󵄨 󵄨q (∫ t αp ( ) dt) (∫ 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln b )󵄨󵄨󵄨 dt) + ln b − ln a b 0

1

0

1 p

pt

1 q

1

a(ln x − ln a)α+1 x ≤ (∫ t αp ( ) dt) (∫ t s |f 󸀠 (x)|q + (1 − t)s |f 󸀠 (a)|q dt) ln b − ln a a 0

1 q

0

1

pt

1 p

1

b(ln b − ln x)α+1 x + (∫ t αp ( ) dt) (∫ t s |f 󸀠 (x)|q + (1 − t)s |f 󸀠 (b)|q dt) ln b − ln a b 0

0

p ∞

x i−1 αp+i ) p a

(− ln a(ln x − ln a)α+1 1 x ≤ [( ) ∑ 1 ln b − ln a pα+ p a i=1 (αp + 1)i

1 p

1

] (

2M q q ) s+1

1

1

p ∞ (− ln bx )i−1 pαp+i p 2M q q b(ln b − ln x)α+1 1 x + [( ) ∑ ] ( ) 1 ln b − ln a pα+ p b i=1 (αp + 1)i s+1

≤

xM(ln x − ln a)α+1 1

(ln b − ln a)p p

1

1

∞ (− ln x )i−1 pi p q 2 a ( ) [∑ ] s+1 (αp + 1)i i=1

1 q

5.2 Inequalities via s-e-condition functions | 307

+

xM(ln b − ln x)α+1 1

(ln b − ln a)p p

1

1

∞ (− ln x )i−1 pi p q 2 b ( ) [∑ ] . s+1 (αp + 1)i i=1

The proof is completed.

Theorem 308. Under the assumptions of Theorem 303, the following inequality for fractional integrals with α > 0 holds: 󵄨󵄨 (ln x − ln a)α + (ln b − ln x)α 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 f (x) − [H Jxα− f (a) + H Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a ≤

i−1

xM(ln x − ln a)α+1 (ln b − ln a)q +

∞ x ∑(− ln ) a i=1

1 q

xM(ln b − ln x)α+1 (ln b − ln a)q

1 q

qi [ i−1

∞ x ∑(− ln ) b i=1

1 1 + ] (αq + s + 1)i (αq + 1)i

qi [

1 1 + ], (αq + s + 1)i (αq + 1)i

for any x ∈ (a, b) and for q > 0. Proof. Noting Lemma 32 and using Hölder inequality and s-e-conditions, we have 󵄨󵄨 󵄨󵄨󵄨 (ln x − ln a)α + (ln b − ln x)α Γ(α + 1) 󵄨 󵄨󵄨 f (x) − [H Jxα− f (a) + H Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a 󵄨 󵄨󵄨 1

≤

(ln x − ln a)α+1 󵄨 󵄨 ∫ t α et ln x+(1−t) ln a 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln a )󵄨󵄨󵄨dt ln b − ln a 0

α+1

(ln b − ln x) + ln b − ln a

1

󵄨 󵄨 ∫ t α et ln x+(1−t) ln b 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln b )󵄨󵄨󵄨dt 0

1

tq

x 󵄨 a(ln x − ln a)α+1 󵄨q [∫ t αq ( ) 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln a )󵄨󵄨󵄨 dt] ≤ ln b − ln a a

1 q

0

1

tq

b(ln b − ln x)α+1 x 󵄨 󵄨q + [∫ t αq ( ) 󵄨󵄨󵄨f 󸀠 (et ln x+(1−t) ln b )󵄨󵄨󵄨 dt] ln b − ln a b

1 q

0

1

tq

a(ln x − ln a)α+1 x ≤ [∫ t αq ( ) (t s |f 󸀠 (x)|q + (1 − t)s |f 󸀠 (a)|q )dt] ln b − ln a a

1 q

0

1

tq

b(ln b − ln x)α+1 x + [∫ t αq ( ) (t s |f 󸀠 (x)|q + (1 − t)s |f 󸀠 (b)|q )dt] ln b − ln a b 0

α+1

≤

aM(ln x − ln a) ln b − ln a +

( ax )q ∞ ∑ qαq+1 i=1

(− ln

[

( ax )q

∞

qαq+s+1

x i−1 ) αq+i a

(αq + 1)i

q

∑ i=1 1 q

]

(− ln ax )i−1

(αq + s + 1)i

qαq+s+i

1 q

308 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals

+

+ ≤

x q x i−1 bM(ln b − ln x)α+1 ( b ) ∞ (− ln b ) [ αq+s+1 ∑ qαq+s+i ln b − ln a (αq + s + 1) q i i=1

( bx )q

qαq+1

∞

∑ i=1

(− ln bx )i−1 (αq + 1)i

xM(ln x − ln a)α+1 (ln b − ln a)q +

1 q

1 q

] i−1

∞ x ∑(− ln ) a i=1

xM(ln b − ln x)α+1 (ln b − ln a)q

q

αq+i

1 q

qi [ i−1

∞ x ∑(− ln ) b i=1

1 1 + ] (αq + s + 1)i (αq + 1)i

qi [

1 1 + ]. (αq + s + 1)i (αq + 1)i

The proof is completed.

5.3 Inequalities via geometric-geometric coordinated convex function Hermite-Hadamard’s inequalities for double Hadamard fractional integrals can be represented as follows. Theorem 309. If f : Δ → ℝ+ with a, c > 0 be geometric-geometric coordinated convex on Δ and α, β > 0, then the following inequalities for double Hadamard fractional integrals holds Γ(α)Γ(β) α,β α,β [(H Ja+ ,c+ f )(b, d) + (H Ja+ ,d− f )(b, c) (ln b − ln a)α (ln d − ln c)β α,β

α,β

+ (H Jb− ,c+ f )(a, d) + (H Jb− ,d− f )(a, c)] ∞ i−1 ∞

i=1 j=0 l=1

i−1−j

× (ln

f (b, c) ) f (b, d)

∞ i−1 ∞

+ f (a, d) ∑ ∑ ∑ i=1 j=0 l=1

f (b, d) ) f (b, c)

∞ i−1 ∞

+ f (b, c) ∑ ∑ ∑ i=1 j=0 l=1

f (a, c) ) f (a, d)

j

(ln[f (b, d)] − ln[f (a, d)])l−1 (α + j)l

j

(−1)i−1 j f (b, c)f (a, d) C (ln ) (β)i i−1 f (a, c)f (b, d)

i−1−j

× (ln

(ln[f (b, c)] − ln[f (a, c)])l−1 (α + j)l

(−1)i−1 j f (a, d)f (b, c) C (ln ) (β)i i−1 f (b, c)f (a, c)

i−1−j

× (ln

j

f (a, c)f (b, d) (−1)i−1 j C (ln ) (β)i i−1 f (b, c)f (a, d)

≤ f (a, c) ∑ ∑ ∑

(ln[f (a, c)] − ln[f (b, c)])l−1 (α + j)l

5.3 Inequalities via geometric-geometric coordinated convex function j

∞ i−1 ∞

(−1)i−1 j f (b, d)f (a, c) Ci−1 (ln ) (β)i f (a, d)f (b, c)

+ f (b, d) ∑ ∑ ∑ i=1 j=0 l=1

i−1−j

× (ln

| 309

f (a, d) ) f (a, c)

(ln[f (a, d)] − ln[f (b, d)])l−1 , (α + j)l

j

where Ci−1 is a binomial coefficient. Proof. Firstly, we set x = at1 b1−t1 , y = cs1 d1−s1 . By using Definition 8, we have 1 1

β−1

∫ ∫ t1α−1 s1 f (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 0 0

=

a c

α−1

ln b − ln x 1 ) ∫ ∫( (ln b − ln a)(ln d − ln c) ln b − ln a b d

β−1

(

ln d − ln y ) ln d − ln c

f (x, y)

b d

=

dydx yx

dydx 1 ∫ ∫(ln b − ln x)α−1 (ln d − ln y)β−1 f (x, y) yx (ln b − ln a)α (ln d − ln c)β a c

Γ(α)Γ(β) α,β (H Ja+ ,c+ f )(b, d). = (ln b − ln a)α (ln d − ln c)β

Similarly, we have

1 1

β−1

∫ ∫ t1α−1 s1 f (at1 b1−t1 , ds1 c1−s1 )ds1 dt1 0 0

= and

Γ(α)Γ(β) α,β (H Ja+ ,d− f )(b, c), (ln b − ln a)α (ln d − ln c)β 1 1

β−1

∫ ∫ t1α−1 s1 f (bt1 a1−t1 , cs1 d1−s1 )ds1 dt1 0 0

= and

Γ(α)Γ(β) α,β (H Jb− ,c+ f )(a, d), (ln b − ln a)α (ln d − ln c)β 1 1

β−1

∫ ∫ t1α−1 s1 f (bt1 a1−t1 , ds1 c1−s1 )ds1 dt1 0 0

=

Γ(α)Γ(β) α,β (H Jb− ,d− f )(a, c). (ln b − ln a)α (ln d − ln c)β

From these equations, we obtain 1 1

β−1

∫ ∫ t1α−1 s1 [f (at1 b1−t1 , cs1 d1−s1 ) + f (at1 b1−t1 , ds1 c1−s1 )]ds1 dt1 0 0

310 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals 1 1

β−1

+ ∫ ∫ t1α−1 s1 [f (bt1 a1−t1 , cs1 d1−s1 ) + f (bt1 a1−t1 , ds1 c1−s1 )]ds1 dt1 0 0

=

Γ(α)Γ(β) α,β α,β [(H Ja+ ,c+ f )(b, d) + (H Ja+ ,d− f )(b, c) (ln b − ln a)α (ln d − ln c)β α,β

α,β

+ (H Jb− ,c+ f )(a, d) + (H Jb− ,d− f )(a, c)].

Denote A󸀠 = ln[f (a, c)] + ln[f (b, d)] − ln[f (b, c)] − ln[f (a, d)], B󸀠 = ln [f (b, c)] − ln[f (b, d)], k=[

t

f (a, c)f (b, d) 1 f (b, c) ] . f (a, d)f (b, c) f (b, d)

By using the Binomial Theorem, Definition 31 and Lemma 32, we have 1 1

β−1

∫ ∫ t1α−1 s1 f (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 0 0

1 1

ts

β−1

t (1−s1 )

≤ ∫ ∫ t1α−1 s1 (f (a, c)) 1 1 (f (a, d)) 1 0 0 1

t

1−t1

∫ s1 k s1 ds1 dt1

1−t1

∞

= ∫ t1α−1 [f (a, d)] 1 [f (b, d)] 0

1

t

= ∫ t1α−1 [f (a, c)] 1 [f (b, c)] 0

∞

= f (b, c) ∑ i=1

1

(f (b, c))

(1−t1 )s1

(f (b, d))

(1−t1 )(1−s1 )

ds1 dt1

β−1

0

∑ i=1

(− ln k)i−1 dt1 (β)i

1

t

(−1)i−1 f (a, c) 1 i−1 ] (t1 A󸀠 + B󸀠 ) dt1 ∫ t1α−1 [ (β)i f (b, c) 0

1

t

f (a, c) 1 i−1 j (−1) j = f (b, c) ∑ ] ∑ Ci−1 (t1 A󸀠 ) B󸀠 i−1−j dt1 ∫ t1α−1 [ (β) f (b, c) i i=1 j=0 ∞

i−1

0

1

t

(−1)i−1 i−1 j 󸀠 j 󸀠 i−1−j α+j−1 f (a, c) 1 = f (b, c) ∑ [ ] dt1 ∑C A B ∫ t1 (β)i j=0 i−1 f (b, c) i=1 ∞

∞

= f (b, c) ∑ i=1

i−1 i−1

f (a, c) ∞ (ln[f (b, c)] − ln[f (a, c)])l−1 (−1) j ∑ Ci−1 A󸀠 j B󸀠 i−1−j ∑ (β)i j=0 f (b, c) l=1 (α + j)l

∞ i−1 ∞

= f (a, c) ∑ ∑ ∑ i=1 j=0 l=1

×

0

j

i−1−j

(−1)i−1 j f (a, c)f (b, d) f (b, c) Ci−1 (ln ) (ln ) (β)i f (b, c)f (a, d) f (b, d)

(ln[f (b, c)] − ln[f (a, c)])l−1 . (α + j)l

5.3 Inequalities via geometric-geometric coordinated convex function | 311

Similarly, we have 1 1

β−1

∫ ∫ t1α−1 s1 f (at1 b1−t1 , ds1 c1−s1 )ds1 dt1 0 0

1 1

ts

β−1

t (1−s1 )

≤ ∫ ∫ t1α−1 s1 (f (a, d)) 1 1 (f (a, c)) 1

(f (b, d))

(1−t1 )s1

0 0

∞ i−1 ∞

= f (a, d) ∑ ∑ ∑ i=1 j=0 l=1

(1−t1 )(1−s1 )

ds1 dt1

(f (a, d))

(1−t1 )(1−s1 )

ds1 dt1

(f (a, c))

(1−t1 )(1−s1 )

ds1 dt1

j

f (a, d)f (b, c) (−1)i−1 j C (ln ) (β)i i−1 f (b, d)f (a, c)

i−1−j

× (ln

(f (b, c))

f (b, d) ) f (b, c)

(ln[f (b, d)] − ln[f (a, d)])l−1 , (α + j)l

and 1 1

β−1

∫ ∫ t1α−1 s1 f (bt1 a1−t1 , cs1 d1−s1 )ds1 dt1 0 0

1 1

ts

β−1

t (1−s1 )

≤ ∫ ∫ t1α−1 s1 (f (b, c)) 1 1 (f (b, d)) 1

(f (a, c))

(1−t1 )s1

0 0

j

∞ i−1 ∞

f (b, c)f (a, d) (−1)i−1 j C (ln ) (β)i i−1 f (a, c)f (b, d)

= f (b, c) ∑ ∑ ∑ i=1 j=0 l=1

i−1−j

× (ln

f (a, c) ) f (a, d)

(ln[f (a, c)] − ln[f (b, c)])l−1 , (α + j)l

and 1 1

β−1

∫ ∫ t1α−1 s1 f (bt1 a1−t1 , ds1 c1−s1 )ds1 dt1 0 0

1 1

ts

β−1

t (1−s1 )

≤ ∫ ∫ t1α−1 s1 (f (b, d)) 1 1 (f (b, c)) 1 0 0

∞ i−1 ∞

= f (b, d) ∑ ∑ ∑ i=1 j=0 l=1

f (a, d) ) f (a, c)

(1−t1 )s1

j

(−1)i−1 j f (b, d)f (a, c) C (ln ) (β)i i−1 f (a, d)f (b, c)

i−1−j

× (ln

(f (a, d))

(ln[f (a, d)] − ln[f (b, d)])l−1 . (α + j)l

So we obtain Γ(α)Γ(β) α,β α,β [(H Ja+ ,c+ f )(b, d) + (H Ja+ ,d− f )(b, c) (ln b − ln a)α (ln d − ln c)β

312 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals α,β

α,β

+ (H Jb− ,c+ f )(a, d) + (H Jb− ,d− f )(a, c)] ∞ i−1 ∞

i=1 j=0 l=1

i−1−j

× (ln

f (b, c) ) f (b, d)

∞ i−1 ∞

+ f (a, d) ∑ ∑ ∑ i=1 j=0 l=1

f (b, d) ) f (b, c)

∞ i−1 ∞

+ f (b, c) ∑ ∑ ∑ i=1 j=0 l=1

f (a, c) ) f (a, d)

∞ i−1 ∞

+ f (b, d) ∑ ∑ ∑ i=1 j=0 l=1

(ln[f (b, d)] − ln[f (a, d)])l−1 (α + j)l

j

(ln[f (a, c)] − ln[f (b, c)])l−1 (α + j)l

j

(−1)i−1 j f (b, d)f (a, c) C (ln ) (β)i i−1 f (a, d)f (b, c)

i−1−j

× (ln

j

(−1)i−1 j f (b, c)f (a, d) C (ln ) (β)i i−1 f (a, c)f (b, d)

i−1−j

× (ln

(ln[f (b, c)] − ln[f (a, c)])l−1 (α + j)l

(−1)i−1 j f (a, d)f (b, c) C (ln ) (β)i i−1 f (b, d)f (a, c)

i−1−j

× (ln

j

f (a, c)f (b, d) (−1)i−1 j C (ln ) (β)i i−1 f (b, c)f (a, d)

≤ f (a, c) ∑ ∑ ∑

f (a, d) ) f (a, c)

(ln[f (a, d)] − ln[f (b, d)])l−1 . (α + j)l

The proof is completed. Theorem 310. Let f : Δ → (1, ∞) and ln f with a, c > 0 be geometric-geometric coordinated convex on Δ and α, β > 0, then the following inequalities for double Hadamard fractional integrals holds: 1 ln f (√ab, √cd) αβ Γ(α)Γ(β) α,β (H Ja+ ,c+ ln f )(b, d) ≤ (ln b − ln a)α (ln d − ln c)β Γ(α)Γ(β) α,β + (H Ja+ ,d− ln f )(b, c) (ln b − ln a)α (ln d − ln c)β Γ(α)Γ(β) α,β + (H Jb− ,c+ ln f )(a, d) (ln b − ln a)α (ln d − ln c)β Γ(α)Γ(β) α,β + (H Jb− ,d− ln f )(a, c) (ln b − ln a)α (ln d − ln c)β ≤

j

∞ i−1 ∞ 1 (−1)i−1 j ln f (a, c) ln f (b, d) [ln f (a, c) ∑ ∑ ∑ Ci−1 (ln ) 4 (β) ln f (b, c) ln f (a, d) i i=1 j=0 l=1 i−1−j

× (ln

ln f (b, c) ) ln f (b, d)

(ln[ln f (b, c)] − ln[ln f (a, c)])l−1 (α + j)l

5.3 Inequalities via geometric-geometric coordinated convex function | 313 ∞ i−1 ∞

+ ln f (a, d) ∑ ∑ ∑ i=1 j=0 l=1

j

(−1)i−1 j ln f (a, d) ln f (b, c) Ci−1 (ln ) (β)i ln f (b, c) ln f (a, c)

i−1−j

ln f (b, d) ) ln f (b, c)

× (ln

∞ i−1 ∞

+ ln f (b, c) ∑ ∑ ∑ i=1 j=0 l=1

∞ i−1 ∞

+ ln f (b, d) ∑ ∑ ∑ i=1 j=0 l=1

(ln[ln f (a, c)] − ln[ln f (b, c)])l−1 (α + j)l

j

(−1)i−1 j ln f (b, d) ln f (a, c) Ci−1 (ln ) (β)i ln f (a, d) ln f (b, c)

i−1−j

ln f (a, d) ) ln f (a, c)

× (ln

j

(−1)i−1 j ln f (b, c) ln f (a, d) Ci−1 (ln ) (β)i ln f (a, c) ln f (b, d)

i−1−j

ln f (a, c) ) ln f (a, d)

× (ln

(ln[ln f (b, d)] − ln[ln f (a, d)])l−1 (α + j)l

(ln[ln f (a, d)] − ln[ln f (b, d)])l−1 ]. (α + j)l

Proof. We set 1 x = at1 b1−t1 , y = bt1 a1−t1 , u = cs1 d1−s1 , w = ds1 c1−s1 , s = t = . 2 By using Definition 31 we have f (x t y1−t , us w1−s ) ≤ √(f (at1 b1−t1 , cs1 d1−s1 ))√(f (at1 b1−t1 , ds1 c1−s1 )) 4

4

× √(f (bt1 a1−t1 , cs1 d1−s1 ))√(f (bt1 a1−t1 , ds1 c1−s1 )). 4

4

Thus, ln f (√ab, √cd) 1 ≤ [ln f (at1 b1−t1 , cs1 d1−s1 ) + ln f (at1 b1−t1 , ds1 c1−s1 ) 4 + ln f (bt1 a1−t1 , cs1 d1−s1 ) + ln f (bt1 a1−t1 , ds1 c1−s1 )]. From 1 1

β−1

∫ ∫ t1α−1 s1 f (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 0 0

=

Γ(α)Γ(β) α,β (H Ja+ ,c+ f )(b, d), (ln b − ln a)α (ln d − ln c)β

we know 1 1

β−1

∫ ∫ t1α−1 s1 0 0

ln f (at1 b1−t1 , cs1 d1−s1 )ds1 dt1

314 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals

=

Γ(α)Γ(β) α,β (H Ja+ ,c+ ln f )(b, d). (ln b − ln a)α (ln d − ln c)β

Similarly, we have 1 1

β−1

∫ ∫ t1α−1 s1

ln f (at1 b1−t1 , ds1 c1−s1 )ds1 dt1

0 0

=

Γ(α)Γ(β) α,β (H Ja+ ,d− ln f )(b, c), (ln b − ln a)α (ln d − ln c)β

and 1 1

β−1

∫ ∫ t1α−1 s1

ln f (bt1 a1−t1 , cs1 d1−s1 )ds1 dt1

0 0

=

Γ(α)Γ(β) α,β (H Jb− ,c+ ln f )(a, d), (ln b − ln a)α (ln d − ln c)β

and 1 1

β−1

∫ ∫ t1α−1 s1

ln f (bt1 a1−t1 , ds1 c1−s1 )ds1 dt1

0 0

=

Γ(α)Γ(β) α,β (H Jb− ,d− ln f )(a, c). (ln b − ln a)α (ln d − ln c)β

Then we have 1 1

β−1

∫ ∫ t1α−1 s1 0 0

≤

ln f (√ab, √cd)ds1 dt1

1 1

1 β−1 [∫ ∫ t1α−1 s1 ln f (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 4 0 0

1 1

β−1

ln f (at1 b1−t1 , ds1 c1−s1 )ds1 dt1

β−1

ln f (bt1 a1−t1 , cs1 d1−s1 )ds1 dt1

β−1

ln f (bt1 a1−t1 , ds1 c1−s1 )ds1 dt1 ]

+ ∫ ∫ t1α−1 s1 0 0

1 1

+ ∫ ∫ t1α−1 s1 0 0

1 1

+ ∫ ∫ t1α−1 s1 0 0

=

Γ(α)Γ(β) α,β (H Ja+ ,c+ ln f )(b, d) (ln b − ln a)α (ln d − ln c)β

5.3 Inequalities via geometric-geometric coordinated convex function | 315

Γ(α)Γ(β) α,β (H Ja+ ,d− ln f )(b, c) (ln b − ln a)α (ln d − ln c)β Γ(α)Γ(β) α,β + (H Jb− ,c+ ln f )(a, d) (ln b − ln a)α (ln d − ln c)β Γ(α)Γ(β) α,β + (H Jb− ,d− ln f )(a, c). (ln b − ln a)α (ln d − ln c)β

+

From these, we obtain 1 ln f (√ab, √cd) αβ Γ(α)Γ(β) α,β ≤ (H Ja+ ,c+ ln f )(b, d) (ln b − ln a)α (ln d − ln c)β Γ(α)Γ(β) α,β (H Ja+ ,d− ln f )(b, c) + (ln b − ln a)α (ln d − ln c)β Γ(α)Γ(β) α,β + (H Jb− ,c+ ln f )(a, d) (ln b − ln a)α (ln d − ln c)β Γ(α)Γ(β) α,β + (H Jb− ,d− ln f )(a, c) (ln b − ln a)α (ln d − ln c)β 1 1

=

1 β−1 [∫ ∫ t1α−1 s1 ln f (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 4 0 0

1 1

β−1

ln f (at1 b1−t1 , ds1 c1−s1 )ds1 dt1

β−1

ln f (bt1 a1−t1 , cs1 d1−s1 )ds1 dt1

β−1

ln f (bt1 a1−t1 , ds1 c1−s1 )ds1 dt1 ].

+ ∫ ∫ t1α−1 s1 0 0

1 1

+ ∫ ∫ t1α−1 s1 0 0

1 1

+ ∫ ∫ t1α−1 s1 0 0

Now we need to verify that the right-hand side of the inequality holds. In fact, by observing the proof of Theorem 309, we have 1 1

β−1

∫ ∫ t1α−1 s1

ln f (at1 b1−t1 , cs1 d1−s1 )ds1 dt1

0 0

j

∞ i−1 ∞

(−1)i−1 j ln f (a, c) ln f (b, d) C (ln ) (β)i i−1 ln f (b, c) ln f (a, d)

≤ ln f (a, c) ∑ ∑ ∑ i=1 j=0 l=1

i−1−j

× (ln

ln f (b, c) ) ln f (b, d)

(ln[ln f (b, c)] − ln[ln f (a, c)])l−1 , (α + j)l

316 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals and 1 1

β−1

∫ ∫ t1α−1 s1

ln f (at1 b1−t1 , ds1 c1−s1 )ds1 dt1

0 0

j

∞ i−1 ∞

ln f (a, d) ln f (b, c) (−1)i−1 j C (ln ) (β)i i−1 ln f (b, c) ln f (a, c)

≤ ln f (a, d) ∑ ∑ ∑ i=1 j=0 l=1

i−1−j

× (ln

ln f (b, d) ) ln f (b, c)

(ln[ln f (b, d)] − ln[ln f (a, d)])l−1 , (α + j)l

and 1 1

β−1

∫ ∫ t1α−1 s1 f (bt1 a1−t1 , cs1 d1−s1 )ds1 dt1 0 0

j

∞ i−1 ∞

ln f (b, c) ln f (a, d) (−1)i−1 j C (ln ) (β)i i−1 ln f (a, c) ln f (b, d)

≤ ln f (b, c) ∑ ∑ ∑ i=1 j=0 l=1

i−1−j

× (ln

ln f (a, c) ) ln f (a, d)

(ln[ln f (a, c)] − ln[ln f (b, c)])l−1 , (α + j)l

and 1 1

β−1

∫ ∫ t1α−1 s1

ln f (bt1 a1−t1 , ds1 c1−s1 )ds1 dt1

0 0

∞ i−1 ∞

≤ ln f (b, d) ∑ ∑ ∑ i=1 j=0 l=1

i−1−j

× (ln

ln f (a, d) ) ln f (a, c)

j

ln f (b, d) ln f (a, c) (−1)i−1 j C (ln ) (β)i i−1 ln f (a, d) ln f (b, c)

(ln[ln f (a, d)] − ln[ln f (b, d)])l−1 . (α + j)l

So we obtain 1 1

1 β−1 [∫ ∫ t1α−1 s1 ln f (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 4 0 0

1 1

β−1

ln f (at1 b1−t1 , ds1 c1−s1 )ds1 dt1

β−1

ln f (bt1 a1−t1 , cs1 d1−s1 )ds1 dt1

β−1

ln f (bt1 a1−t1 , ds1 c1−s1 )ds1 dt1 ]

+ ∫ ∫ t1α−1 s1 0 0

1 1

+ ∫ ∫ t1α−1 s1 0 0

1 1

+ ∫ ∫ t1α−1 s1 0 0

5.3 Inequalities via geometric-geometric coordinated convex function | 317 j

≤

∞ i−1 ∞ (−1)i−1 j 1 ln f (a, c) ln f (b, d) [ln f (a, c) ∑ ∑ ∑ Ci−1 (ln ) 4 (β)i ln f (b, c) ln f (a, d) i=1 j=0 l=1 i−1−j

× (ln

ln f (b, c) ) ln f (b, d)

∞ i−1 ∞

+ ln f (a, d) ∑ ∑ ∑ i=1 j=0 l=1

ln f (b, d) ) ln f (b, c)

∞ i−1 ∞

+ ln f (b, c) ∑ ∑ ∑ i=1 j=0 l=1

ln f (a, c) ) ln f (a, d)

∞ i−1 ∞

+ ln f (b, d) ∑ ∑ ∑ i=1 j=0 l=1

ln f (a, d) ) ln f (a, c)

j

(ln[ln f (a, c)] − ln[ln f (b, c)])l−1 (α + j)l

j

(−1)i−1 j ln f (b, d) ln f (a, c) Ci−1 (ln ) (β)i ln f (a, d) ln f (b, c)

i−1−j

× (ln

(ln[ln f (b, d)] − ln[ln f (a, d)])l−1 (α + j)l

(−1)i−1 j ln f (b, c) ln f (a, d) Ci−1 (ln ) (β)i ln f (a, c) ln f (b, d)

i−1−j

× (ln

j

(−1)i−1 j ln f (a, d) ln f (b, c) Ci−1 (ln ) (β)i ln f (b, c) ln f (a, c)

i−1−j

× (ln

(ln[ln f (b, c)] − ln[ln f (a, c)])l−1 (α + j)l

(ln[ln f (a, d)] − ln[ln f (b, d)])l−1 ]. (α + j)l

The proof is completed. Using Lemma 73, we can obtain the following explicit estimate. Theorem 311. Let f : Δ ⊂ ℝ2 → ℝ be a partially differentiable mapping on Δ := [a, b] × 𝜕2 f [c, d] in ℝ2 with a, c > 0. If | 𝜕t𝜕s | is a geometric-geometric coordinated convex function on Δ and G > max{1, F1 , F, F 2 }, then the following inequality holds: 󵄨󵄨 Γ(α + 1)Γ(β + 1) α,β α,β 󵄨󵄨 [( J f )(a, c) + (H Ja+ ,d− f )(b, c) 󵄨󵄨 󵄨󵄨 bd(ln b − ln a)α+1 (ln d − ln c)β+1 H b− ,d− 󵄨󵄨 α,β α,β 󵄨 + (H Ja+ ,c+ f )(b, d) + (H Jb− ,c+ f )(a, d)] − A − B󵄨󵄨󵄨 󵄨󵄨 1

p ∞ 󵄨󵄨 󵄨󵄨 𝜕2 f (−p ln(EF))i−1 󵄨 󵄨 (b, d)󵄨󵄨󵄨G max{1, 21−α } max{EF[∑ ] ≤ 󵄨󵄨󵄨 󵄨󵄨 (pβ + 1)i 󵄨󵄨 𝜕t1 𝜕s1 i=1 1

[ln(FG)]1−q − (ln G)1−q q ×[ ] (1 − q) ln G 1 p

1

(p ln(EF))i−1 (ln G − 2 ln F)1−q − (ln G − ln F)1−q q − [∑ ] [ ] , (pβ + 1)i (q − 1) ln F i=1 ∞

1 p

1

(p ln(EF))i−1 (ln G − 2 ln F)1−q − (ln G − ln F)1−q q [∑ ] [ ] (pβ + 1)i (q − 1) ln F i=1 ∞

318 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals 1 p

1

(−p ln(EF))i−1 [ln(FG)]1−q − (ln G)1−q q − EF[∑ ] [ ] }, (pβ + 1)i (1 − q) ln G i=1 ∞

where

1 p

+

1 q

= 1, 1 < p < ∞ and 󵄨󵄨󵄨󵄨 𝜕2 f 󵄨󵄨󵄨−1 c 󵄨󵄨󵄨󵄨 𝜕2 f 󵄨󵄨 (b, c)󵄨󵄨󵄨󵄨󵄨󵄨 (b, d)󵄨󵄨󵄨 , 󵄨󵄨 󵄨󵄨󵄨󵄨 𝜕t1 𝜕s1 󵄨󵄨 d 󵄨󵄨 𝜕t1 𝜕s1 󵄨󵄨󵄨󵄨 𝜕2 f 󵄨󵄨−1 󵄨󵄨 𝜕2 f 󵄨󵄨−1 󵄨󵄨 𝜕2 f 󵄨󵄨 a 󵄨󵄨󵄨 𝜕2 f 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 F = 󵄨󵄨󵄨 (a, c)󵄨󵄨󵄨󵄨󵄨󵄨 (a, d)󵄨󵄨󵄨 󵄨󵄨󵄨 (b, c)󵄨󵄨󵄨 󵄨󵄨󵄨 (b, d)󵄨󵄨󵄨, 󵄨󵄨󵄨󵄨 𝜕t1 𝜕s1 󵄨󵄨 󵄨󵄨 𝜕t1 𝜕s1 󵄨󵄨 󵄨󵄨 𝜕t1 𝜕s1 󵄨󵄨 b 󵄨󵄨 𝜕t1 𝜕s1 −1 2 2 󵄨 󵄨 󵄨 󵄨 󵄨󵄨󵄨󵄨 𝜕 f 󵄨󵄨 a 󵄨󵄨 𝜕 f G = 󵄨󵄨󵄨 (a, d)󵄨󵄨󵄨󵄨󵄨󵄨 (b, d)󵄨󵄨󵄨 . 󵄨 󵄨󵄨 󵄨 󵄨 b 󵄨 𝜕t1 𝜕s1 󵄨󵄨 𝜕t1 𝜕s1 E=

Proof. Case 1: α ≥ 1. By using Lemma 73 and Definition 31, we have 󵄨󵄨 Γ(α + 1)Γ(β + 1) α,β α,β 󵄨󵄨 [( J f )(a, c) + (H Ja+ ,d− f )(b, c) 󵄨󵄨 󵄨󵄨 bd(ln b − ln a)α+1 (ln d − ln c)β+1 H b− ,d− 󵄨󵄨 α,β α,β 󵄨 + (H Ja+ ,c+ f )(b, d) + (H Jb− ,c+ f )(a, d)] − A − B󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 1 1 t1 s1 󵄨󵄨 c 𝜕2 f a (at1 b1−t1 , cs1 d1−s1 )dt1 ds1 = 󵄨󵄨󵄨 ∫ ∫(1 − t1 )α (1 − s1 )β ( ) ( ) 󵄨󵄨 b d 𝜕t 𝜕s 1 1 󵄨0 0 1 1

t

s

a 1 c 1 𝜕2 f − ∫ ∫ t1α (1 − s1 )β ( ) ( ) (at1 b1−t1 , cs1 d1−s1 )ds1 dt1 b d 𝜕s1 𝜕t1 0 0

1 1

t

s

1 c 1 𝜕2 f β a + ∫ ∫ t1α s1 ( ) ( ) (at1 b1−t1 , cs1 d1−s1 )dt1 ds1 b d 𝜕t1 𝜕s1

0 0

1 1 󵄨󵄨 t1 s 󵄨󵄨 c 1 𝜕2 f β a − ∫ ∫(1 − t1 )α s1 ( ) ( ) (at1 b1−t1 , cs1 d1−s1 )dt1 ds1 󵄨󵄨󵄨 󵄨󵄨 b d 𝜕t1 𝜕s1 󵄨 0 0

1 1

t

s1 󵄨

1 c β a ≤ ∫ ∫(1 − t1 )α |(1 − s1 )β − s1 |( ) ( ) b d

0 0

1 1

t

s1 󵄨

1 c β a + ∫ ∫ t1α |(1 − s1 )β − s1 |( ) ( ) b d

0 0

1 1

t

s1 󵄨

t

s1 󵄨

0 0

1 c β a ≤ ∫ ∫ |(1 − s1 )β − s1 |( ) ( ) b d

0 0

󵄨 󵄨󵄨 𝜕2 f t1 1−t1 s1 1−s1 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 𝜕t 𝜕s (a b , c d )󵄨󵄨󵄨dt1 ds1 󵄨 1 1 󵄨

󵄨 󵄨󵄨 𝜕2 f t1 1−t1 s1 1−s1 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 𝜕t 𝜕s (a b , c d )󵄨󵄨󵄨dt1 ds1 󵄨 1 1 󵄨

1 c β a ≤ ∫ ∫ |(1 − s1 )β − s1 |( ) ( ) b d

1 1

󵄨 󵄨󵄨 𝜕2 f t1 1−t1 s1 1−s1 󵄨󵄨󵄨 󵄨󵄨 (a b , c d ) 󵄨󵄨dt1 ds1 󵄨󵄨 𝜕t 𝜕s 󵄨󵄨 󵄨 1 1

󵄨󵄨t1 s1 󵄨󵄨 𝜕2 f 󵄨󵄨t1 (1−s1 ) 󵄨󵄨 𝜕2 f 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 𝜕t 𝜕s (a, c)󵄨󵄨󵄨 󵄨󵄨󵄨 𝜕t 𝜕s (a, d)󵄨󵄨󵄨 󵄨 1 1 󵄨 󵄨 1 1 󵄨

5.3 Inequalities via geometric-geometric coordinated convex function | 319

󵄨󵄨(1−t1 )s1 󵄨󵄨 𝜕2 f 󵄨󵄨(1−t1 )(1−s1 ) 󵄨󵄨 𝜕2 f 󵄨 󵄨 󵄨 󵄨󵄨 (b, c)󵄨󵄨󵄨 (b, d)󵄨󵄨󵄨 dt1 ds1 × 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 𝜕t1 𝜕s1 󵄨󵄨 𝜕t1 𝜕s1

󵄨󵄨 1 󵄨󵄨󵄨󵄨 𝜕2 f 󵄨󵄨−1 s1 󵄨󵄨 𝜕2 f c 󵄨󵄨󵄨 𝜕2 f β 󵄨 󵄨󵄨 󵄨 󵄨 = 󵄨󵄨󵄨 (b, d)󵄨󵄨󵄨 ∫ |(1 − s1 )β − s1 |( 󵄨󵄨󵄨 (b, c)󵄨󵄨󵄨󵄨󵄨󵄨 (b, d)󵄨󵄨󵄨 ) ds1 󵄨󵄨 󵄨󵄨󵄨󵄨 𝜕t1 𝜕s1 󵄨󵄨 󵄨󵄨 𝜕t1 𝜕s1 d 󵄨󵄨 𝜕t1 𝜕s1 0

1

󵄨󵄨s1 󵄨󵄨 𝜕2 f 󵄨󵄨1−s1 󵄨󵄨 𝜕2 f 󵄨󵄨−s1 󵄨󵄨 𝜕2 f 󵄨󵄨s1 −1 t1 a 󵄨󵄨󵄨 𝜕2 f 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 × ∫( 󵄨󵄨󵄨 (a, c)󵄨󵄨󵄨 󵄨󵄨󵄨 (a, d)󵄨󵄨󵄨 󵄨󵄨󵄨 (b, c)󵄨󵄨󵄨 󵄨󵄨󵄨 (b, d)󵄨󵄨󵄨 ) dt1 󵄨󵄨 󵄨󵄨 𝜕t1 𝜕s1 󵄨󵄨 󵄨󵄨 𝜕t1 𝜕s1 󵄨󵄨 󵄨󵄨 𝜕t1 𝜕s1 󵄨󵄨 b 󵄨󵄨 𝜕t1 𝜕s1 0

1 1 󵄨󵄨 𝜕2 f 󵄨󵄨󵄨 t β 󵄨 = 󵄨󵄨󵄨 (b, d)󵄨󵄨󵄨 ∫ |(1 − s1 )β − s1 |E s1 ds1 ∫(F s1 G) 1 dt1 󵄨󵄨 𝜕t1 𝜕s1 󵄨󵄨 0

0

󵄨󵄨 𝜕2 f 󵄨󵄨 1 F s1 G − 1 β 󵄨󵄨 󵄨 = 󵄨󵄨 (b, d)󵄨󵄨󵄨 ∫ |(1 − s1 )β − s1 |E s1 ds 󵄨󵄨 𝜕t1 𝜕s1 󵄨󵄨 s1 ln F + ln G 1 0

1

1

p ∞ 󵄨󵄨 𝜕2 f 󵄨󵄨 [ln(FG)]1−q − (ln G)1−q q (−p ln(EF))i−1 󵄨 󵄨 ≤ 󵄨󵄨󵄨 (b, d)󵄨󵄨󵄨G max{EF[∑ ] [ ] 󵄨󵄨 𝜕t1 𝜕s1 󵄨󵄨 (pβ + 1)i (1 − q) ln G i=1 1 p

1

(p ln(EF))i−1 (ln G − 2 ln F)1−q − (ln G − ln F)1−q q − [∑ ] [ ] , (pβ + 1)i (q − 1) ln F i=1 ∞

1 p

1

(p ln(EF))i−1 (ln G − 2 ln F)1−q − (ln G − ln F)1−q q [∑ ] [ ] (pβ + 1)i (q − 1) ln F i=1 ∞

1 p

1

(−p ln(EF))i−1 [ln(FG)]1−q − (ln G)1−q q − EF[∑ ] [ ] }, (pβ + 1)i (1 − q) ln G i=1 ∞

where we use the following necessary inequalities: 1

β

∫ s1 E s1 0

1

F s1 G ds s1 ln F + ln G 1

β

= G ∫ s1 (EF)s1 0

1 ds s1 ln F + ln G 1

1

≤

p β G[∫(s1 (EF)s1 ) ds1 ] 0 1

s

pβ

1 p

q 󵄨󵄨 󵄨󵄨q 1 󵄨 󵄨󵄨 [∫ 󵄨󵄨󵄨 󵄨󵄨 ds1 ] 󵄨󵄨 s1 ln F + ln G 󵄨󵄨 1 p

0

= G[∫ s1 [(EF)p ] 1 ds1 ] [ 0

1

1

1

[ln(FG)]1−q − (ln G)1−q q ] (1 − q) ln G 1 p

1

(− ln(EF)p )i−1 [ln(FG)]1−q − (ln G)1−q q = G[(EF) ∑ ] [ ] (pβ + 1)i (1 − q) ln G i=1 p

∞

320 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals 1 p

1

(−p ln(EF))i−1 [ln(FG)]1−q − (ln G)1−q q = EFG[∑ ] [ ] , (pβ + 1)i (1 − q) ln G i=1 ∞

and 1

∫(1 − s1 )β E s1 0

1

= ∫ sβ 0

F s1 G ds s1 ln F + ln G 1

(EF)1−s G ds (1 − s) ln F + ln G 1

s

= EFG ∫ sβ ((EF)−1 ) 0 1

1 ds s(− ln F) + (ln G − ln F) 1 p

1

1

q 󵄨󵄨 󵄨󵄨q 1 󵄨 󵄨󵄨 ≤ EFG[∫(sβ ((EF) ) ) ds] [∫ 󵄨󵄨󵄨 󵄨󵄨 ds] 󵄨󵄨 s(− ln F) + (ln G − ln F) 󵄨󵄨

−1 s p

0

1

1 p

0

1

1

q 󵄨󵄨 󵄨󵄨q 1 󵄨 󵄨󵄨 = EFG[∫ spβ ((EF) ) ds] [∫ 󵄨󵄨󵄨 󵄨󵄨 ds] 󵄨󵄨 s(− ln F) + (ln G − ln F) 󵄨󵄨

−p s

0

0

(− ln(EF)−p )i−1 = EFG[(EF)−p ∑ ] (pβ + 1)i i=1 ∞

1 p

1

(− ln F + ln G − ln F)1−q − (ln G − ln F)1−q q ×[ ] (1 − q)(− ln F) 1 p

1

(p ln(EF))i−1 (ln G − 2 ln F)1−q − (ln G − ln F)1−q q = G[∑ ] [ ] . (pβ + 1)i (q − 1) ln F i=1 ∞

Case 2: 0 < α < 1. By using Lemma 39, one can complete the rest of the proof with the same process as in the Case 1. The proof is completed.

5.4 Inequalities via geometric-geometric-convex functions The results in this section are taken from [101, 246]. 5.4.1 Main results In this section, we will use the results via GG-convex functions in Section 5.2 to derive our main results in this paper, where T is defined by Tf (x, y) =

x|f 󸀠 (x)| , y|f 󸀠 (y)|

5.4 Inequalities via geometric-geometric-convex functions | 321

If (a, x, y) := I(a, Tf (x, y))

If (a, x, y, z) := H(a, Tf (x, y))

(see Lemma 32),

(see Lemma 33).

Theorem 312. Let f : [0, b] → ℝ be a differentiable mapping. If |f 󸀠 | is measurable and |f 󸀠 | is GG-convex on [0, b] for some fixed α ∈ (0, ∞), 0 ≤ a < b, then the following inequality for fractional integrals holds: 󵄨󵄨 (ln x − ln a)α + (ln b − ln x)α 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 f (x) − [H Jxα− f (a) + H Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a i−1 ∞ α+1 (ln Tf (x, a)) (ln x − ln a) a|f 󸀠 (a)|Tf (x, a) ∑(−1)i−1 ≤ ln b − ln a (α + 1)i i=1 +

∞ (ln Tf (x, b))i−1 (ln b − ln x)α+1 󸀠 b|f (b)|Tf (x, b) ∑(−1)i−1 , ln b − ln a (α + 1)i i=1

for any x ∈ (a, b). Proof. By using Definition 19 and Lemma 32, we have 1

󵄨 󵄨 ∫ t α xt a1−t 󵄨󵄨󵄨f 󸀠 (xt a1−t )󵄨󵄨󵄨dt 0

1

≤ ∫ t α xt a1−t |f 󸀠 (x)|t |f 󸀠 (a)|1−t dt = a|f 󸀠 (a)|If (α + 1, x, a). 0

For the same reason, 1

󵄨 󵄨 ∫ t α xt b1−t 󵄨󵄨󵄨f 󸀠 (xt b1−t )󵄨󵄨󵄨dt = b|f 󸀠 (b)|If (α + 1, x, b). 0

By using Lemma 68, we obtain 󵄨󵄨 󵄨󵄨 (ln x − ln a)α + (ln b − ln x)α Γ(α + 1) 󵄨󵄨 󵄨 f (x) − [H Jxα− f (a) + H Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a 1 1 󵄨󵄨 󵄨󵄨 α+1 󵄨󵄨 (ln b − ln x)α+1 󵄨󵄨󵄨 (ln x − ln a) α t 1−t 󸀠 t 1−t α t 1−t 󸀠 t 1−t = 󵄨󵄨 ∫ t x a f (x a )dt − ∫ t x b f (x a )dt 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a 󵄨 󵄨 0 0 ≤

1

1

0

0

(ln b − ln x)α+1 (ln x − ln a)α+1 󵄨 󵄨 󵄨 󵄨 ∫ t α xt a1−t 󵄨󵄨󵄨f 󸀠 (x t a1−t )󵄨󵄨󵄨dt + ∫ t α x t b1−t 󵄨󵄨󵄨f 󸀠 (x t b1−t )󵄨󵄨󵄨dt ln b − ln a ln b − ln a

(ln x − ln a)α+1 󸀠 (ln b − ln x)α+1 󸀠 ≤ a|f (a)|If (α + 1, x, a) + b|f (b)|If (α + 1, x, b) ln b − ln a ln b − ln a ∞ (ln Tf (x, a))i−1 (ln x − ln a)α+1 󸀠 = a|f (a)|Tf (x, a) ∑(−1)i−1 ln b − ln a (α + 1)i i=1

322 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals

+

∞ (ln Tf (x, b))i−1 (ln b − ln x)α+1 󸀠 b|f (b)|Tf (x, b) ∑(−1)i−1 . ln b − ln a (α + 1)i i=1

The proof is completed. Theorem 313. Let f : [0, b] → ℝ be a differentiable mapping and 1 < q < ∞. If |f 󸀠 |q is measurable and |f 󸀠 |q is GG-convex on [0, b] for some fixed α ∈ (0, ∞), 0 ≤ a < b, then the following inequality for fractional integrals holds: 󵄨󵄨 (ln x − ln a)α + (ln b − ln x)α 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 f (x) − [H Jxα− f (a) + H Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a 󵄨󵄨 1 1 q α+1 p Tf (x, a) − 1 q 1 (ln x − ln a) a|f 󸀠 (a)|( ) ( ) ≤ ln b − ln a αp + 1 q ln(Tf (x, a)) p Tfq (x, b) − 1 q 1 (ln b − ln x)α+1 󸀠 b|f (b)|( ) ( ) , + ln b − ln a αp + 1 q ln(Tf (x, b)) 1

where

1 p

+

1 q

1

= 1.

Proof. By using Definition 19 and Hölder inequality, we obtain 1

󵄨 󵄨 ∫ t α xt a1−t 󵄨󵄨󵄨f 󸀠 (xt a1−t )󵄨󵄨󵄨dt 0

1 p

1

1

q

󵄨 󵄨 ≤ (∫ t αp dt) (∫(xt a1−t 󵄨󵄨󵄨f 󸀠 (xt a1−t )󵄨󵄨󵄨) dt) 0

1 q

0

1 p

1

1 ≤( ) (∫ xqt aq(1−t) |f 󸀠 (x)|qt |f 󸀠 (a)|q(1−t) dt) αp + 1 0

1 p

1

1 ≤( ) a|f 󸀠 (a)|(∫ Tfq (x, a)dt) αp + 1 0

1 q

1 q

Tfq (x, a) − 1 q 1 ) a|f 󸀠 (a)|( ) . ≤( αp + 1 q ln(Tf (x, a)) 1 p

1

In the same way, 1

󵄨 󵄨 ∫ t α xt b1−t 󵄨󵄨󵄨f 󸀠 (xt b1−t )󵄨󵄨󵄨dt ≤ ( 0

p Tfq (x, b) − 1 q 1 ) b|f 󸀠 (b)|( ) . αp + 1 q ln(Tf (x, b)) 1

By using Lemma 68, we have 󵄨󵄨 (ln x − ln a)α + (ln b − ln x)α 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 f (x) − [H Jxα− f (a) + H Jxα+ f (b)]󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a 󵄨󵄨

1

5.4 Inequalities via geometric-geometric-convex functions | 323 1 1 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 (ln x − ln a)α+1 (ln b − ln x)α+1 α t 1−t 󸀠 t 1−t α t 1−t 󸀠 t 1−t 󵄨 = 󵄨󵄨 ∫ t x a f (x a )dt − ∫ t x b f (x b )dt 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a ln b − ln a 󵄨 󵄨 0 0 1

≤

1

(ln x − ln a)α+1 (ln b − ln x)α+1 󵄨 󵄨 󵄨 󵄨 ∫ t α xt a1−t 󵄨󵄨󵄨f 󸀠 (x t a1−t )󵄨󵄨󵄨dt + ∫ t α x t b1−t 󵄨󵄨󵄨f 󸀠 (x t b1−t )󵄨󵄨󵄨dt ln b − ln a ln b − ln a 0

α+1

≤

(ln x − ln a) ln b − ln a

a|f 󸀠 (a)|(

Tfq (x, a) − 1

1 p

1 ) ( ) αp + 1 q ln(Tf (x, a))

0

1 q

p Tfq (x, b) − 1 q 1 (ln b − ln x)α+1 󸀠 b|f (b)|( ) ( ) . + ln b − ln a αp + 1 q ln(Tf (x, b)) 1

1

The proof is completed. Theorem 314. Let f : [0, b] → ℝ be a differentiable mapping. If |f 󸀠 | is measurable and |f 󸀠 | is GG-convex on [0, b] for some fixed α ∈ (0, ∞), 0 ≤ a < b, then the following inequality for fractional integrals holds: 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 󵄨 [H Jbα− f (a) + H Jaα+ f (b)] − f (√ab)󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2(ln b − ln a) 󵄨󵄨

1

2Tf (a, b) − 1 + (21−α + 1)Tf2 (a, b) ln b − ln a 󸀠 b|f (b)|[ ≤ max{ 2 ln(Tf (a, b)) 1 − 4If (α + 1, a, b, ) + 2If (α + 1, a, b, a)], 2 1

2Tf2 (a, b) − 1 1 ln b − ln a 󸀠 b|f (b)|[ − 4If (α + 1, a, b, ) + 2If (α + 1, a, b, a)]}. 2 ln(Tf (a, b)) 2 Proof. To achieve our aim, we divide our proof into two cases. Case 1: α ∈ (0, 1). By using Definition 19, Lemma 70, Lemma 32, Lemma 38, Lemma 39, Hölder inequality and Lemma 33, we have 󵄨󵄨 󵄨󵄨 Γ(α + 1) 󵄨 󵄨󵄨 [H Jbα− f (a) + H Jaα+ f (b)] − f (√ab)󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2(ln b − ln a) 1 1 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a [∫ kat b1−t f 󸀠 (at b1−t )dt − ∫[(1 − t)α − t α ]at b1−t f 󸀠 (at b1−t )dt]󵄨󵄨󵄨 = 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 2 󵄨 󵄨 0 0 ≤

1

1

0

0

ln b − ln a 󵄨 󵄨 󵄨 󵄨 [∫ at b1−t 󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨dt + ∫ |(1 − t)α − t α | ⋅ at b1−t ⋅ 󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨dt] 2 1

≤

ln b − ln a [∫ at b1−t |f 󸀠 (a)|t |f 󸀠 (b)|1−t dt 2 0

324 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals 1

+ ∫ |(1 − t)α − t α |at b1−t |f 󸀠 (a)|t |f 󸀠 (b)|1−t dt] 0

=

1

1

ln b − ln a [b|f 󸀠 (b)| ∫ Tft (a, b)dt + b|f 󸀠 (b)| ∫ |(1 − t)α − t α |Tft (a, b)dt] 2 0

=

1

1

0

0

1

1 2

0

0

0

ln b − ln a 󸀠 b|f (b)|[∫ Tft (a, b)dt + ∫ |(1 − t)α − t α |Tft (a, b)dt] 2

ln b − ln a 󸀠 b|f (b)|[∫ Tft (a, b)dt + ∫((1 − t)α − t α )Tft (a, b)dt = 2 1

+ ∫(−(1 − t)α + t α )Tft (a, b)dt] 1 2

1

1 2

0

0

ln b − ln a 󸀠 ≤ b|f (b)|[∫ Tft (a, b)dt + ∫(1 − t)α Tft (a, b)dt 2 1 2

1

1

− ∫ t α Tft (a, b)dt − ∫(1 − t)α Tft (a, b)dt + ∫ t α Tft (a, b)dt] 0

≤

1 2

1 2

1

1 2

0

0

ln b − ln a 󸀠 b|f (b)|[∫ Tft (a, b)dt + ∫(21−α − t α )Tft (a, b)dt 2 1 2

−

∫ t α Tft (a, b)dt 0

1

− ∫(1 − t

α

)Tft (a, b)dt

1

+ ∫ t α Tft (a, b)dt] 1 2

1 2

1

1 2

0

0

ln b − ln a 󸀠 = b|f (b)|[∫ Tft (a, b)dt + ∫(21−α − 2t α )Tft (a, b)dt 2 1

+ ∫(2t α − 1)Tft (a, b)dt] 1 2

1

1 2

0

0

ln b − ln a 󸀠 = b|f (b)|[∫ Tft (a, b)dt + ∫ 21−α Tft (a, b)dt 2 1 2

1

1

− 2 ∫ t α Tft (a, b)dt − ∫ Tft (a, b)dt + 2 ∫ t α Tft (a, b)dt] 0

1 2

1 2

5.4 Inequalities via geometric-geometric-convex functions | 325

1

1 2

0

0

ln b − ln a 󸀠 = b|f (b)|[∫ Tf (a, b)dt + 21−α ∫ Tf (a, b)dt 2 1 2

1

0

0

1

− ∫ Tf (a, b)dt − 4 ∫ t α Tf (a, b)dt + 2 ∫ t α Tf (a, b)dt] 1 2

1

1

1−α Tf (a, b) − 1 2 (Tf2 (a, b) − 1) Tf (a, b) − Tf2 (a, b) ln b − ln a 󸀠 b|f (b)|[ + − = 2 ln(Tf (a, b)) ln(Tf (a, b)) ln(Tf (a, b)) α+1

1 − 4( ) 2

∞

1

Tf2 (a, b) ∑

i=0

∞

+ 2Tf (a, b) ∑(−1)i−1 i=1

(− 21 ln(Tf (a, b)))i−1 (α + 1)i

(ln(Tft (a, b)))i−1 (α + 1)i

] 1

2Tf (a, b) − 1 + (21−α + 1)Tf2 (a, b) ln b − ln a 󸀠 = b|f (b)|[ 2 ln(Tf (a, b)) 1 − 4If (α + 1, a, b, ) + 2If (α + 1, a, b, a)]. 2

Case 2: α ∈ [1, ∞). By using Definition 19, Lemma 70, Lemma 32, Lemma 38, Lemma 39, Hölder inequality and Lemma 33 again, we have 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 󵄨 [H Jbα− f (a) + H Jaα+ f (b)] − f (√ab)󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2(ln b − ln a) 󵄨󵄨 1 1 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨󵄨 ln b − ln a α α t 1−t 󸀠 t 1−t t 1−t 󸀠 t 1−t [∫ ka b f (a b )dt − ∫[(1 − t) − t ]a b f (a b )dt]󵄨󵄨󵄨 = 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 󵄨 󵄨 0 0 ≤

1

1

0

0

ln b − ln a 󵄨 󵄨 󵄨 󵄨 [∫ at b1−t 󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨dt + ∫ |(1 − t)α − t α |at b1−t 󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨dt] 2 1

≤

ln b − ln a [∫ at b1−t |f 󸀠 (a)|t |f 󸀠 (b)|1−t dt 2 1

0

+ ∫ |(1 − t)α − t α |at b1−t |f 󸀠 (a)|t |f 󸀠 (b)|1−t dt] 0

=

1

0

=

1

ln b − ln a [b|f 󸀠 (b)| ∫ Tft (a, b)dt + b|f 󸀠 (b)| ∫ |(1 − t)α − t α |Tft (a, b)dt] 2 1

1

0

0

0

ln b − ln a 󸀠 b|f (b)|[∫ Tft (a, b)dt + ∫ |(1 − t)α − t α |Tft (a, b)dt] 2

326 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals 1

1 2

0

0

ln b − ln a 󸀠 = b|f (b)|[∫ Tft (a, b)dt + ∫((1 − t)α − t α )Tft (a, b)dt 2 1

+ ∫(−(1 − t)α + t α )Tft (a, b)dt] 1 2

1

1 2

1 2

0

0

ln b − ln a 󸀠 b|f (b)|[∫ Tft (a, b)dt + ∫(1 − t)α Tft (a, b)dt − ∫ t α Tft (a, b)dt = 2 0

1

1

− ∫(1 − t)α Tft (a, b)dt + ∫ t α Tft (a, b)dt] 1 2

1 2

1

≤

1 2

1 2

0

0

ln b − ln a 󸀠 b|f (b)|[∫ Tft (a, b)dt + ∫(1 − t α )Tft (a, b)dt − ∫ t α Tft (a, b)dt 2 0

1

1

− ∫(21−α − t α )Tft (a, b)dt + ∫ t α Tft (a, b)dt] 1 2

1 2

=

1

1 2

0

0

1

ln b − ln a b[∫ Tft (a, b)dt + ∫(1 − 2t α )Tft (a, b)dt + ∫(−1 + 2t α )Tft (a, b)dt] 2 1 2

1

1 2

0

0

1

1 2

ln b − ln a 󸀠 = b|f (b)|[∫ Tft (a, b)dt + ∫ Tft (a, b)dt − ∫ Tft (a, b)dt − 2 ∫ t α Tft (a, b)dt 2 1

1 2

0

0

1 2

0

+ 2 ∫ t α Tft (a, b)dt − 2 ∫ t α Tft (a, b)dt] 1

1 2

0

0

1

ln b − ln a 󸀠 = b|f (b)|[∫ Tft (a, b)dt + ∫ Tft (a, b)dt − ∫ Tft (a, b)dt 2 1 2

−

4 ∫ t α Tft (a, b)dt 0

1 2

1

+ 2 ∫ t α Tft (a, b)dt] 0

1

2Tf2 (a, b) − 1 ln b − ln a 󸀠 1 = b|f (b)|[ − 4If (α + 1, a, b, ) + 2If (α + 1, a, b, a)]. 2 ln(Tf (a, b)) 2 The proof is completed.

5.4 Inequalities via geometric-geometric-convex functions | 327

Theorem 315. Let f : [0, b] → ℝ be a differentiable mapping and 1 < q < ∞. If |f 󸀠 |q is measurable and |f 󸀠 |q is GG-convex on [0, b] for some fixed α ∈ (0, ∞), 0 ≤ a < b, then the following inequality for fractional integrals holds: 󵄨󵄨 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α [ J − f (a) + H Ja+ f (b)] − f (√ab)󵄨 󵄨󵄨 H b 󵄨󵄨 󵄨󵄨 2(ln b − ln a)α 󵄨 ≤ max{

q Tfq (a, b) ln b − ln a 󸀠 (1 + 21−α )p 2p (1 − 2−pα ) p b|f (b)|( − ) ( ) , 2 2 pα + 1 q ln Tf (a, b) 1

1

q Tfq (a, b) ln b − ln a 󸀠 2p+1 − 2p(1−α) p p b|f (b)|(2 − ) ( ) }, 2 pα + 1 q ln Tf (a, b) 1

where

1 p

+

1 q

1

= 1.

Proof. To achieve our aim, we divide our first step into two cases. Case 1: α ∈ (0, 1). By using Lemma 38 and Lemma 39, we obtain 1

∫ |k − (1 − t)α + t α |p dt 0 1 2

α

1

α p

p

= ∫(1 − (1 − t) + t ) dt + ∫(1 − t α + (1 − t)α ) dt 0

1 2

1 2

1

p

p

≤ ∫[(1 + t α ) − (1 − t α )] dt + ∫[1 − t α + 21−α − t α ] dt 0

1 2

1 2

1

p

= ∫ 2p t pα dt + ∫[(1 + 21−α ) − 2p t pα ]dt 0

=

1 2

(1 + 21−α )p 2p (1 − 2−pα ) − . 2 pα + 1

Case 2: α ∈ [1, ∞). By using Lemma 38 and Lemma 39, we obtain 1

∫ |k − (1 − t)α + t α |p dt 0 1 2

α

α p

1

p

= ∫(1 − (1 − t) + t ) dt + ∫(1 − t α + (1 − t)α ) dt 0

1 2

1

α p

1

p

= 2 ∫(1 − t + (1 − t) ) dt ≤ 2 ∫(1 − t α + 1 − t α ) dt 1 2

α

1 2

328 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals

p+1

≤2

1

α p

∫(1 − t ) dt ≤ 2

p+1

1

∫(1 − t αp )dt 1 2

1 2

1

1 2p+1 − 2p(1−α) = 2p+1 ( − ∫ t pα dt) ≤ 2p − . 2 pα + 1 1 2

Since |f 󸀠 (x)|q is GG-convex on [0, b], from Definition 19, we drive 1

1

0

0

q 󵄨 󵄨q ∫(at b1−t 󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨) dt ≤ ∫(at b1−t ) |f 󸀠 (a)|tq |f 󸀠 (b)|q(1−t) dt q

1

= (b|f (b)|) ∫ Tfq (a, b)dt 󸀠

0

q

󸀠

= (b|f (b)|)

Tfq (a, b)

q ln Tf (a, b)

.

By using Lemma 39 and Hölder inequality, we have 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 󵄨󵄨 α α [ J − f (a) + H Ja+ f (b)] − f (√ab)󵄨 󵄨󵄨 H b 󵄨󵄨 󵄨󵄨 2(ln b − ln a)α 󵄨 1 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨󵄨 ln b − ln a α α t 1−t 󸀠 t 1−t [∫(k − (1 − t) + t )a b f (a b )dt]󵄨󵄨󵄨 = 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 󵄨 󵄨 0 1

≤

ln b − ln a 󵄨 󵄨 ∫ |k − (1 − t)α + t α |at b1−t 󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨dt 2 0

1 p

1

1

ln b − ln a 󵄨 󵄨q ≤ (∫ |k − (1 − t)α + t α |p dt) (∫(at b1−t 󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨) dt) 2 0

0

1−α p

(1 + 2 ln b − ln a 󸀠 ≤ max{ b|f (b)|( 2 2

)

1 q

q Tfq (a, b) 2p (1 − 2−pα ) p − ) ( ) , pα + 1 q ln Tf (a, b) 1

1

q Tfq (a, b) ln b − ln a 󸀠 2p+1 − 2p(1−α) p p b|f (b)|(2 − ) ( ) }. 2 pα + 1 q ln Tf (a, b) 1

1

The proof is completed. Theorem 316. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with 0 ≤ a < b. If |f 󸀠 | is integrable and GG-convex on [a, b], then for 0 ≤ λ ≤ 1, x ∈ (a, b), the following inequality holds: x|f 󸀠 (x)| i−1 󵄨󵄨 ∞ (ln b|f 󸀠 (b)| ) (ln b − ln x)α 󵄨󵄨󵄨 󸀠 i−1 |If (x, λ, α, a, b)| ≤ |x|f (x)󵄨󵄨 ∑(−1) 󵄨󵄨 ln b − ln a (α + 1)i 󵄨 i=1

5.4 Inequalities via geometric-geometric-convex functions | 329

+λ

󵄨󵄨 󵄨󵄨 󸀠 󸀠 󵄨󵄨 (x|f (x)| − b|f (b)|) x|f 󸀠 (x)| 󵄨󵄨󵄨 ln b|f 󸀠 (b)| 󵄨 1

x|f (x)| i−1 󵄨󵄨 ∞ (ln a|f 󸀠 (a)| ) 󵄨󵄨 (ln x − ln a)α i−1 󸀠 󵄨 + |x|f (x)󵄨󵄨 ∑(−1) 󵄨 ln b − ln a (α + 1)i 󵄨󵄨 i=1 󸀠

+λ

󵄨󵄨 󵄨󵄨 󸀠 󸀠 󵄨󵄨. (x|f (x)| − a|f (a)|) 󵄨󵄨 x|f 󸀠 (x)| 󵄨󵄨 ln a|f 󸀠 (a)| 1

Proof. By using Definition 19 and Lemma 32, we have 1 󵄨󵄨 󵄨󵄨 (ln b − ln x)α 󵄨󵄨 (sα + λ)f 󸀠 (e(1−s) ln b+s ln x )e(1−s) ln b+s ln x ds ∫ 󵄨󵄨 ln b − ln a 󵄨󵄨 0

−

1 󵄨󵄨 󵄨󵄨 (ln x − ln a)α ∫(sα + λ)f 󸀠 (e(1−s) ln a+s ln x )e(1−s) ln a+s ln x ds󵄨󵄨󵄨 󵄨󵄨 ln b − ln a 󵄨 0

≤ |Uf (x, λ, α, a, b)| + |Vf (x, λ, α, a, b)|

1 󵄨󵄨 󵄨󵄨 󵄨󵄨 (ln b − ln x)α 󵄨󵄨 󵄨 󵄨 󵄨 ≤ 󵄨󵄨 ∫(sα + λ)󵄨󵄨󵄨f 󸀠 (e(1−s) ln b+s ln x )󵄨󵄨󵄨e(1−s) ln b+s ln x ds󵄨󵄨󵄨 󵄨󵄨 ln b − ln a 󵄨󵄨 󵄨 󵄨 0 1 󵄨󵄨 󵄨󵄨 󵄨󵄨 (ln x − ln a)α 󵄨󵄨 󵄨 󵄨 󵄨 + 󵄨󵄨 ∫(sα + λ)󵄨󵄨󵄨f 󸀠 (e(1−s) ln a+s ln x )󵄨󵄨󵄨e(1−s) ln a+s ln x ds󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a 󵄨 󵄨 0 1 󵄨 1 󵄨󵄨 (ln b − ln x)α 󵄨󵄨󵄨󵄨 α 󵄨󵄨 󸀠 1−s s 󵄨󵄨 1−s s 󵄨󵄨 󸀠 1−s s 󵄨󵄨 1−s s 󵄨󵄨󵄨 ≤ 󵄨 ∫ s 󵄨󵄨f (b x )󵄨󵄨b x ds + ∫ λ󵄨󵄨f (b x )󵄨󵄨b x ds󵄨󵄨 󵄨󵄨 ln b − ln a 󵄨󵄨󵄨󵄨 󵄨 0

0

1 󵄨 1 󵄨󵄨 (ln x − ln a)α 󵄨󵄨󵄨󵄨 α 󵄨󵄨 󸀠 1−s s 󵄨󵄨 1−s s 󵄨󵄨 󸀠 1−s s 󵄨󵄨 1−s s 󵄨󵄨󵄨 s f (a x ) a x ds + λ + f (a x ) a x ds ∫ ∫ 󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a 󵄨󵄨󵄨󵄨 󵄨 0 0

≤

1 󵄨 1 󵄨󵄨 (ln b − ln x)α 󵄨󵄨󵄨󵄨 α 󸀠 1−s 󸀠 s 1−s s 󸀠 1−s 󸀠 s 1−s s 󵄨󵄨 󵄨󵄨 ∫ s |f (b)| |f (x)| b x ds + ∫ λ|f (b)| |f (x)| b x ds󵄨󵄨󵄨 󵄨󵄨 ln b − ln a 󵄨󵄨󵄨 󵄨 0

0

1 󵄨 1 󵄨󵄨 (ln x − ln a)α 󵄨󵄨󵄨󵄨 α 󸀠 1−s 󸀠 s 1−s s 󸀠 1−s 󸀠 s 1−s s 󵄨󵄨 + 󵄨󵄨 ∫ s |f (a)| |f (x)| a x ds + ∫ λ|f (a)| |f (x)| a x ds󵄨󵄨󵄨 󵄨󵄨 ln b − ln a 󵄨󵄨󵄨 󵄨 0 0

≤

1 1 󵄨 s s 󵄨󵄨 󸀠 󵄨󵄨 x|f 󸀠 (x)| (ln b − ln x)α 󵄨󵄨󵄨󵄨 󸀠 α x|f (x)| ) ds + λb|f 󸀠 (b)| ∫( 󸀠 ) ds󵄨󵄨󵄨 󵄨󵄨b|f (b)| ∫ s ( 󸀠 󵄨 󵄨󵄨 ln b − ln a 󵄨󵄨 b|f (b)| b|f (b)| 󵄨 0

1

0

1 󵄨 s s 󵄨󵄨 󸀠 󵄨󵄨 x|f 󸀠 (x)| (ln x − ln a)α 󵄨󵄨󵄨󵄨 󸀠 α x|f (x)| 󸀠 ) ds + λa|f (a)| ∫( 󸀠 ) ds󵄨󵄨󵄨 + 󵄨󵄨a|f (a)| ∫ s ( 󸀠 󵄨󵄨 ln b − ln a 󵄨󵄨󵄨 a|f (a)| a|f (a)| 󵄨 0

0

330 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals x|f 󸀠 (x)| i−1 󵄨󵄨 ∞ (ln 󸀠 (b)| ) 󵄨 (ln b − ln x)α b|f 󵄨 ≤ |x|f 󸀠 (x)󵄨󵄨󵄨 ∑(−1)i−1 󵄨󵄨 ln b − ln a (α + 1)i 󵄨 i=1 󵄨󵄨󵄨 1 󵄨󵄨 󸀠 󸀠 (x|f (x)| − b|f (b)|) +λ 󵄨󵄨 x|f 󸀠 (x)| 󵄨󵄨 ln b|f 󸀠 (b)| 󵄨

x|f 󸀠 (x)| i−1 󵄨󵄨 ∞ (ln a|f 󸀠 (a)| ) (ln x − ln a)α 󵄨󵄨󵄨 i−1 󸀠 + |x|f (x)󵄨󵄨 ∑(−1) 󵄨󵄨 ln b − ln a (α + 1)i 󵄨 i=1 󵄨󵄨󵄨 1 󵄨 (x|f 󸀠 (x)| − a|f 󸀠 (a)|)󵄨󵄨󵄨. +λ 󸀠 x|f (x)| 󵄨󵄨 ln a|f 󸀠 (a)| 󵄨

The proof is completed. Theorem 317. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with 0 ≤ a < b. If |f 󸀠 |p , p > 1 is integrable and GG-convex on [a, b], then for 0 ≤ λ ≤ 1, x ∈ (a, b), the following inequality holds: |If (x, λ, α, a, b)|

1

1 󵄨 x q i−1 q ∞ p (ln b − ln x)α 󵄨󵄨󵄨󵄨 1 |f 󸀠 (x)|p − |f 󸀠 (b)|p q i−1 [ln( b ) ] ≤ ) ) [(x (−1) ( ∑ 󵄨 ln b − ln a 󵄨󵄨󵄨󵄨 p ln |f 󸀠 (x)| − ln |f 󸀠 (b)| (qα + 1)i i=1 1

󵄨󵄨 q 󵄨󵄨 λq xq − bq +( ) ]󵄨󵄨󵄨 󵄨󵄨 q ln x − ln b 󵄨

1

1 󵄨 x q i−1 q ∞ p (ln x − ln a)α 󵄨󵄨󵄨󵄨 1 |f 󸀠 (x)|p − |f 󸀠 (a)|p q i−1 [ln( b ) ] ) [(x (−1) ) + ∑ 󵄨󵄨( ln b − ln a 󵄨󵄨󵄨 p ln |f 󸀠 (x)| − ln |f 󸀠 (a)| (qα + 1)i i=1 1 󵄨󵄨 q λq xq − aq 󵄨󵄨 +( ) ]󵄨󵄨󵄨. 󵄨󵄨 q ln x − ln a 󵄨

Proof. By using Definition 19, Lemma 32 and Hölder inequality, we have 1 󵄨󵄨 󵄨󵄨 (ln b − ln x)α 󵄨󵄨 (sα + λ)f 󸀠 (e(1−s) ln b+s ln x )e(1−s) ln b+s ln x ds ∫ 󵄨󵄨 ln b − ln a 󵄨󵄨 0

1 󵄨󵄨 󵄨󵄨 (ln x − ln a)α − ∫(sα + λ)f 󸀠 (e(1−s) ln a+s ln x )e(1−s) ln a+s ln x ds󵄨󵄨󵄨 󵄨󵄨 ln b − ln a 󵄨 0

≤ |Uf (x, λ, α, a, b)| + |Vf (x, λ, α, a, b)|

1 󵄨󵄨 󵄨󵄨 󵄨󵄨 (ln b − ln x)α 󵄨󵄨 󵄨 󵄨 ≤ 󵄨󵄨󵄨 ∫(sα + λ)󵄨󵄨󵄨f 󸀠 (e(1−s) ln b+s ln x )󵄨󵄨󵄨e(1−s) ln b+s ln x ds󵄨󵄨󵄨 󵄨󵄨 ln b − ln a 󵄨󵄨 󵄨 󵄨 0

1 󵄨󵄨 󵄨󵄨 󵄨󵄨 (ln x − ln a)α 󵄨󵄨 󵄨 󵄨 󵄨 + 󵄨󵄨 ∫(sα + λ)󵄨󵄨󵄨f 󸀠 (e(1−s) ln a+s ln x )󵄨󵄨󵄨e(1−s) ln a+s ln x ds󵄨󵄨󵄨 󵄨󵄨 ln b − ln a 󵄨󵄨 󵄨 󵄨 0

5.4 Inequalities via geometric-geometric-convex functions | 331 1 󵄨󵄨 󵄨 1 (ln b − ln x)α 󵄨󵄨󵄨󵄨 α 󵄨󵄨 󸀠 1−s s 󵄨󵄨 1−s s 󵄨󵄨 󸀠 1−s s 󵄨󵄨 1−s s 󵄨󵄨󵄨 ≤ 󵄨 ∫ s 󵄨󵄨f (b x )󵄨󵄨b x ds + ∫ λ󵄨󵄨f (b x )󵄨󵄨b x ds󵄨󵄨 󵄨󵄨 ln b − ln a 󵄨󵄨󵄨󵄨 󵄨 0

0

1 󵄨󵄨 󵄨 1 (ln x − ln a)α 󵄨󵄨󵄨󵄨 α 󵄨󵄨 󸀠 1−s s 󵄨󵄨 1−s s 󵄨 󸀠 1−s s 󵄨 1−s s 󵄨󵄨 + 󵄨󵄨 ∫ s 󵄨󵄨f (a x )󵄨󵄨a x ds + ∫ λ󵄨󵄨󵄨f (a x )󵄨󵄨󵄨a x ds󵄨󵄨󵄨 󵄨󵄨 ln b − ln a 󵄨󵄨󵄨 󵄨 0 0 1

1

1 1 󵄨 q p qs ps (ln b − ln x)α 󵄨󵄨󵄨󵄨 q qα x |f 󸀠 (x)| 󸀠 p ≤ ) ds) 󵄨󵄨(b ∫ s ( ) ds) (|f (b)| ∫( 󸀠 ln b − ln a 󵄨󵄨󵄨 b |f (b)| 0

1

0

1 q

qs

1

1

ps

p 󵄨󵄨 󵄨󵄨 |f 󸀠 (x)| x + ((λb) ∫( ) ds) (|f 󸀠 (b)|p ∫( 󸀠 ) ds) 󵄨󵄨󵄨 󵄨󵄨 b |f (b)| 󵄨 0 0

q

1

1

1 1 󵄨 q p qs ps |f 󸀠 (x)| (ln x − ln a)α 󵄨󵄨󵄨󵄨 q qα x 󸀠 p ( (|f (a)| ) ds) ) ds) + s ( (a ∫ ∫ 󵄨 ln b − ln a 󵄨󵄨󵄨󵄨 a |f 󸀠 (a)| 0

1

qs

0

1 q

1

1

ps

p 󵄨󵄨 x |f 󸀠 (x)| 󵄨󵄨 + ((λa) ∫( ) ds) (|f 󸀠 (a)|p ∫( 󸀠 ) ds) 󵄨󵄨󵄨 󵄨󵄨 a |f (a)| 󵄨 0 0

q

1

󵄨 x q i−1 q q ∞ (ln b − ln x)α 󵄨󵄨󵄨󵄨 q x i−1 [ln( b ) ] ) ≤ 󵄨(b ( ) ∑(−1) ln b − ln a 󵄨󵄨󵄨󵄨 b i=1 (qα + 1)i × (|f 󸀠 (b)|p

1 (x)| p ln |f|f 󸀠 (b)| 󸀠

(

ps 󵄨󵄨1

|f 󸀠 (x)| ) |f 󸀠 (b)|

qs 󵄨󵄨1

x 1 + ((λb) x( ) q ln b b q

1

󵄨󵄨 p 󵄨󵄨 ) 󵄨󵄨 󵄨󵄨0

1 ps 󵄨󵄨1 p1 󵄨󵄨 󵄨󵄨 q 󵄨󵄨 󵄨󵄨 1 |f 󸀠 (x)| 󵄨󵄨 ) (|f 󸀠 (b)|p ) 󵄨󵄨󵄨 ) 󵄨󵄨󵄨 󸀠 (x)| ( 󵄨󵄨 󸀠 |f 󵄨 󵄨󵄨0 󵄨󵄨0 󵄨󵄨󵄨 p ln |f 󸀠 (b)| |f (b)| 1

󵄨 x q i−1 q q ∞ (ln x − ln a)α 󵄨󵄨󵄨󵄨 q x i−1 [ln( a ) ] ) + 󵄨󵄨(a ( ) ∑(−1) ln b − ln a 󵄨󵄨󵄨 a i=1 (qα + 1)i ps 󵄨󵄨1

1

|f 󸀠 (x)| × (|f (a)| ( 󸀠 ) 󸀠 (x)| |f (a)| p ln |f|f 󸀠 (a)| p

󸀠

1

󵄨󵄨 p 󵄨󵄨 ) 󵄨󵄨 󵄨󵄨0

1

1 󵄨 x q i−1 q ∞ p (ln b − ln x)α 󵄨󵄨󵄨󵄨 1 |f 󸀠 (x)|p − |f 󸀠 (b)|p q i−1 [ln( b ) ] ) ) ≤ [(x (−1) ∑ 󵄨󵄨( ln b − ln a 󵄨󵄨󵄨 p ln |f 󸀠 (x)| − ln |f 󸀠 (b)| (qα + 1)i i=1 1 1 󵄨󵄨 󵄨 q p 󵄨󵄨 (ln x − ln a)α 󵄨󵄨󵄨 1 |f 󸀠 (x)|p − |f 󸀠 (a)|p λq xq − bq 󵄨󵄨( +( ) ]󵄨󵄨󵄨 + ) 󵄨󵄨 q ln x − ln b ln b − ln a 󵄨󵄨󵄨󵄨 p ln |f 󸀠 (x)| − ln |f 󸀠 (a)| 󵄨

q

∞

x q i−1 i−1 [ln( b ) ]

× [(x ∑(−1) i=1

The proof is completed.

(qα + 1)i

1 q

) +(

1 󵄨󵄨 q λq xq − aq 󵄨󵄨 ) ]󵄨󵄨󵄨. 󵄨󵄨 q ln x − ln a 󵄨

332 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals Theorem 318. Let f : [0, b] → ℝ be a differentiable mapping. If |f 󸀠 | is measurable and |f 󸀠 | is GG-convex on [a, b] for some fixed α ∈ (0, ∞) and t ∈ [0, 1], 0 ≤ a < b, then the following integrals holds: 󵄨󵄨󵄨 f (a) + f (b) 󵄨󵄨󵄨 Γ(α + 1) 󵄨󵄨 − [H Jaα+ f (b) + H Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 2 2(ln b − ln a) 󵄨 󵄨󵄨 ln b − ln a 1 b|f 󸀠 (b)| b|f 󸀠 (a)| + a|f 󸀠 (b)| − 2b|f 󸀠 (b)| ≤ {(1 − α+1 )[ + ] 2 α+1 α+2 2 a|f 󸀠 (a)| − a|f 󸀠 (b)| − b|f 󸀠 (a)| a|f 󸀠 (a)| − b|f 󸀠 (a)| + − a|f 󸀠 (a)|B1 (3, α + 3) − α+3 (α + 3)2α+3 − (b|f 󸀠 (a)| + a|f 󸀠 (b)|)B1 (2, α + 2) + (2b|f 󸀠 (a)| + 2a|f 󸀠 (b)|)B0.5 (2, α + 2) + 2a|f 󸀠 (a)|B0.5 (3, α + 1)}. Proof. Noting that Definition 42, Definition 3, Lemma 19 and Lemma 36, we have 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J f (b) + J f (a)] + − 󵄨󵄨 󵄨󵄨 H H a b 󵄨󵄨 󵄨󵄨 2 2(ln b − ln a)α 1

ln b − ln a 󵄨 󵄨 ≤ ∫ |(1 − t)α − t α |et ln a+(1−t) ln b 󵄨󵄨󵄨f 󸀠 (et ln a+(1−t) ln b )󵄨󵄨󵄨dt 2 0

1

ln b − ln a 󵄨 󵄨 ≤ ∫ at b1−t [t α − (1 − t)α ]󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨dt 2 1 2

1 2

+

ln b − ln a 󵄨 󵄨 ∫ at b1−t [(1 − t)α − t α ]󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨dt 2 1

≤

0

ln b − ln a ∫[at + b(1 − t)][t α − (1 − t)α ]|f 󸀠 (a)|t |f 󸀠 (b)|1−t dt 2 1 2

1 2

ln b − ln a + ∫[at + b(1 − t)][(1 − t)α − t α ]|f 󸀠 (a)|t |f 󸀠 (b)|1−t dt 2 1

≤

0

ln b − ln a ∫[at + b(1 − t)][t α − (1 − t)α ][|f 󸀠 (a)|t + |f 󸀠 (b)|(1 − t)]dt 2 1 2

1 2

ln b − ln a + ∫[at + b(1 − t)][(1 − t)α − t α ][|f 󸀠 (a)|t + |f 󸀠 (b)|(1 − t)]dt 2 0

1

1

ln b − ln a a|f 󸀠 (a)| b|f 󸀠 (a)| ≤ { ∫(α + 3)t α+2 dt + ∫(α + 2)t α+1dt 2 α+3 α+2 1 2

1 2

5.4 Inequalities via geometric-geometric-convex functions | 333 1

1

a|f 󸀠 (b)| b|f 󸀠 (a)| − ∫(α + 3)t α+2 dt + ∫(α + 2)t α+1 dt α+3 α+2 1 2

1 2

1

1

b|f 󸀠 (b)| a|f 󸀠 (b)| − ∫(α + 3)t α+2 dt + ∫(α + 1)t α dt α+3 α+1 1 2

−

1 2

1

1

2b|f 󸀠 (b)| b|f 󸀠 (b)| ∫(α + 2)t α+1 dt + ∫(α + 3)t α+2 dt α+2 α+3 1 2

1

1 2

1

α 2

− a|f (a)| ∫(1 − t) t dt − b|f (a)| ∫ t(1 − t)α+1 dt 󸀠

󸀠

1 2

1 2

1

− a|f 󸀠 (b)| ∫ t(1 − t)α+1 dt − 1 2

1

b|f 󸀠 (b)| ∫(α + 3)(1 − t)α+2 dt} α+3 1 2

1 2

1 2

ln b − ln a + {a|f 󸀠 (a)| ∫ t 2 (1 − t)α dt + b|f 󸀠 (a)| ∫ t(1 − t)α+1 dt 2 0

1 2

+ a|f 󸀠 (b)| ∫ t(1 − t)α+1 dt + 0

0

1 2

b|f 󸀠 (b)| ∫(α + 3)(1 − t)α+2 dt α+3 0

1 2

1 2

a|f 󸀠 (a)| b|f 󸀠 (a)| − ∫(α + 3)t α+2 dt − ∫(α + 2)t α+1 dt α+3 α+2 0

󸀠

0

0

1 2

1 2

a|f (b)| b|f (b)| ∫(α + 3)t α+2 dt − ∫(α + 1)t α dt α+3 α+1 󸀠

+

1 2

b|f (a)| a|f (b)| ∫(α + 3)t α+2 dt − ∫(α + 2)t α+1 dt α+3 α+2 󸀠

+

0

1 2

0

󸀠

0

1 2

1 2

2b|f 󸀠 (b)| b|f 󸀠 (b)| + ∫(α + 2)t α+1 dt − ∫(α + 3)t α+2 dt} α+2 α+3 0

0

ln b − ln a a|f (a)| 1 b|f 󸀠 (a)| 1 b|f 󸀠 (a)| 1 { (1 − α+3 ) + (1 − α+2 ) − (1 − α+3 ) 2 α+3 α+2 α+3 2 2 2 󸀠

≤

+

a|f 󸀠 (b)| 1 a|f 󸀠 (b)| 1 b|f 󸀠 (b)| 1 (1 − α+2 ) − (1 − α+3 ) + (1 − α+1 ) α+2 α+3 α+1 2 2 2 1

2b|f 󸀠 (b)| 1 b|f 󸀠 (b)| 1 − (1 − α+2 ) + (1 − α+3 ) − a|f 󸀠 (a)| ∫(1 − t)α t 2 dt α+2 α+3 2 2 1 2

334 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals 1

α+1

− b|f (a)| ∫ t(1 − t) 󸀠

1

dt − a|f (b)| ∫ t(1 − t)α+1 dt 󸀠

1 2

1 2

−

1 2

1

b|f (b)| ln b − ln a {a|f 󸀠 (a)| ∫ t 2 (1 − t)α dt ∫(α + 3)(1 − t)α+2 dt} + α+3 2 󸀠

0

1 2

1 2

1 2

+ b|f 󸀠 (a)| ∫ t(1 − t)α+1 dt + a|f 󸀠 (b)| ∫ t(1 − t)α+1 dt + 0

b|f (a)| 1 b|f 󸀠 (a)| 1 a|f 󸀠 (b)| 1 a|f (a)| 1 − + − α+3 α+3 α+2 α+3 2 α+2 2 α+3 2 α + 2 2α+2 󸀠

󸀠

− + ≤

0

b|f 󸀠 (b)| 1 ( α+3 − 1) α+3 2

a|f 󸀠 (b)| 1 b|f 󸀠 (b)| 1 2b|f 󸀠 (b)| 1 b|f 󸀠 (a)| 1 − + − } α+1 α+3 α+2 α+3 2 α+1 2 α+2 2 α + 3 2α+3

ln b − ln a (a − b)[|f 󸀠 (a)| − |f 󸀠 (b)|] 1 b|f 󸀠 (b)| 1 { (1 − α+3 ) + (1 − α+1 ) 2 α+3 α+1 2 2 1

+

b[|f 󸀠 (a)| − |f 󸀠 (b)|] + (a − b)|f 󸀠 (b)| 1 (1 − α+2 ) − a|f 󸀠 (a)| ∫(1 − t)α t 2 dt α+2 2 0

1 2

1

0

0

1 2

+ a|f 󸀠 (a)| ∫(1 − t)α t 2 dt − b|f 󸀠 (a)| ∫ t(1 − t)α+1 dt + b|f 󸀠 (a)| ∫ t(1 − t)α+1 dt 0

1

1 2

0

0

− a|f (b)| ∫ t(1 − t)α+1 dt + a|f 󸀠 (b)| ∫ t(1 − t)α+1 dt} 󸀠

ln b − ln a {a|f 󸀠 (a)|B0.5 (3, α + 1) + b|f 󸀠 (a)|B0.5 (2, α + 2) + 2 b|f 󸀠 (b)| 1 a|f 󸀠 (b)| + a|f 󸀠 (b)|B0.5 (2, α + 2) + ( α+3 − 1) − α+3 2 (α + 3)2α+3 󸀠 󸀠 󸀠 󸀠 b|f (a)| + a|f (b)| − 2b|f (b)| b|f (b)| − − } (α + 1)2α+1 (α + 2)2α+2 ln b − ln a (a − b)[|f 󸀠 (a)| − |f 󸀠 (b)|] 1 b|f 󸀠 (b)| 1 ≤ { (1 − α+3 ) + (1 − α+1 ) 2 α+3 α+1 2 2 b[|f 󸀠 (a)| − |f 󸀠 (b)|] + (a − b)|f 󸀠 (b)| 1 + (1 − α+2 ) − a|f 󸀠 (a)|B1 (3, α + 1) α+2 2 + a|f 󸀠 (a)|B0.5 (3, α + 1) − b|f 󸀠 (a)|B1 (2, α + 2) + b|f 󸀠 (a)|B0.5 (2, α + 2)

− a|f 󸀠 (b)|B1 (2, α + 2) + a|f 󸀠 (b)|B0.5 (2, α + 2)} +

ln b − ln a {a|f 󸀠 (a)|B0.5 (3, α + 1) + b|f 󸀠 (a)|B0.5 (2, α + 2) 2

5.4 Inequalities via geometric-geometric-convex functions | 335

1 b|f 󸀠 (b)| a|f 󸀠 (b)| ( α+3 − 1) − α+3 2 (α + 3)2α+3 b|f 󸀠 (b)| b|f 󸀠 (a)| + a|f 󸀠 (b)| − 2b|f 󸀠 (b)| − } − α+2 (α + 1)2α+1 (α + 2)2 ln b − ln a 1 b|f 󸀠 (b)| b|f 󸀠 (a)| + a|f 󸀠 (b)| − 2b|f 󸀠 (b)| ≤ {(1 − α+1 )[ + ] 2 α+1 α+2 2 a|f 󸀠 (a)| − a|f 󸀠 (b)| − b|f 󸀠 (a)| a|f 󸀠 (a)| − b|f 󸀠 (a)| + − − a|f 󸀠 (a)|B1 (3, α + 3) α+3 (α + 3)2α+3 + a|f 󸀠 (b)|B0.5 (2, α + 2) +

− (b|f 󸀠 (a)| + a|f 󸀠 (b)|)B1 (2, α + 2) + (2b|f 󸀠 (a)| + 2a|f 󸀠 (b)|)B0.5 (2, α + 2) + 2a|f 󸀠 (a)|B0.5 (3, α + 1)}.

The proof is completed. Theorem 319. Let f : [0, b] → ℝ be a differentiable mapping. If |f 󸀠 |q is measurable and |f 󸀠 |q , (q > 1) is GG-convex on [a, b] for some fired α ∈ (0, ∞) and t ∈ [0, 1], 0 ≤ a < b, then the following integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [H Jaα+ f (b) + H Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(ln b − ln a) 1

bp 1 bp ln b − ln a |f 󸀠 (a)|q + |f 󸀠 (b)|q q ap − 2bp ( ) ( + − pα+1 ≤ 2 2 pα + 2 pα + 1 2 pα + 1 + where

1 p

+

1

2pα+2 1 q

1

p 2bp − ap − ap B1 (2, pα) + 2ap B0.5 (2, pα + 1) − bp B0.5 (pα + 1, 2)) , pα + 2

= 1.

Proof. Noting Definition 42, Definition 3, Lemma 19 and Lemma 36, we have 󵄨󵄨 󵄨󵄨 f (a) + f (b) Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J f (b) + J f (a)] − + 󵄨󵄨 󵄨󵄨 H H a b 󵄨󵄨 󵄨󵄨 2 2(ln b − ln a)α 1

ln b − ln a 󵄨 󵄨 ≤ ∫ |(1 − t)α − t α |et ln a+(1−t) ln b 󵄨󵄨󵄨f 󸀠 (et ln a+(1−t) ln b )󵄨󵄨󵄨dt 2 0

1

ln b − ln a 󵄨 󵄨 ≤ ∫ at b1−t |(1 − t)α − t α |󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨dt 2 0

1

1 p

1

ln b − ln a p 󵄨 󵄨q (∫(at b1−t |(1 − t)α − t α |) dt) (∫ 󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨 dt) ≤ 2 0

1

1 p

1 q

0

1

ln b − ln a ≤ (∫ apt bp(1−t) |(1 − t)α − t α |p dt) (∫ |f 󸀠 (a)|qt |f 󸀠 (b)|q(1−t) dt) 2 0

0

1 q

336 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals 1

ln b − ln a ≤ (∫[tap + (1 − t)bp ]|(1 − t)α − t α |p dt) 2 0

1

× (∫[t|f 󸀠 (a)|q + (1 − t)|f 󸀠 (b)|q ]dt) 0

≤

1 p

1 q

1

ln b − ln a (∫[tap + (1 − t)bp ](t pa − (1 − t)pa )dt 2 1 2

1 2

+ ∫[tap + (1 − t)bp ][(1 − t)pα − t pα ]dt) 0

1 p

1

|f 󸀠 (a)|q |f 󸀠 (b)|q q ×( + |f 󸀠 (b)|q − ) 2 2

1

1

1

ln b − ln a |f 󸀠 (a)|q + 2|f 󸀠 (b)|q − |f 󸀠 (b)|q q p pα+1 ≤ ( ) (a ∫ t dt − ap ∫ t(1 − t)pα dt 2 2 1 2

1

p

pα

p

1 2

1 2

1 2

0

0

1

+ b ∫ t (1 − t)dt − b ∫(1 − t)pα+1 dt + ap ∫ t(1 − t)pα dt − ap ∫ t pα+1 dt 1 2

1 2

1 2

p

pα+1

+ b ∫(1 − t)

p

1 2

pα

dt − b ∫ t (1 − t)dt)

0

0

1 p

1

ln b − ln a |f (a)| + 2|f 󸀠 (b)|q − |f 󸀠 (b)|q q ap ≤ ( ) ( − ap B1 (2, pα) 2 2 pα + 2 bp 1 bp bp + ap B0.5 (2, Pα + 1) + − pα+1 − pα + 1 2 pα + 1 pα + 2 1 bp 1 ap p + pα+2 + a B0.5 (2, pα + 1) − pα+2 pα + 2 pα + 2 2 2 󸀠

+ ≤

1

2pα+2

q

1

p bp bp − − bp B0.5 (pα + 1, 2)) pα + 2 pα + 2 1

ln b − ln a |f 󸀠 (a)|q + |f 󸀠 (b)|q q ap − 2bp bp 1 bp ( ) ( + − pα+1 2 2 pα + 2 pα + 1 2 pα + 1 +

1

2pα+2

1

p 2bp − ap − ap B1 (2, pα) + 2ap B0.5 (2, pα + 1) − bp B0.5 (pα + 1, 2)) . pα + 2

The proof is completed. Theorem 320. Let f : [0, b] → ℝ be a differentiable mapping. If |f 󸀠 | is measurable and |f 󸀠 | is GG-convex on [a, b] for some fired α ∈ (0, ∞), t ∈ [0, 1] and k > 0, 0 ≤ a < b, then

5.4 Inequalities via geometric-geometric-convex functions | 337

the following integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨󵄨 α α − [ J + f (b) + H Jb− f (a)]󵄨 󵄨󵄨 H a 󵄨󵄨 󵄨󵄨 2 2(ln b − ln a)α 󵄨

a|f (a)| i−1 ∞ [ln a|f 󸀠 (a)| ]i−1 [ln b|f 󸀠 (b)| ] ln b − ln a a|f 󸀠 (a)| ∞ b|f (b)| i−1 ≤ b|f (b)| { 󸀠 +∑ ∑(−1) 2 b|f (b)| i=1 (α + 1)i (α + 1)i i=1 󸀠

󸀠

󸀠

1

a|f (a)| 2 ( 21 )α−1 ( b|f 󸀠 (b)| ) 󸀠

−

α+1

󸀠

+2

1

a|f (a)| 2 1 − ( 21 )α ( b|f 󸀠 (b)| ) a|f (a)| ln b|f 󸀠 (b)| 󸀠

}.

Proof. By using Definition 19, Definition 3, Lemma 42, Lemma 32, Lemma 33 and Lemma 34, we have

󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [H Jaα+ f (b) + H Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(ln b − ln a) 1

≤

ln b − ln a 󵄨 󵄨 ∫ |(1 − t)α − t α |et ln a+(1−t) ln b 󵄨󵄨󵄨f 󸀠 (et ln a+(1−t) ln b )󵄨󵄨󵄨dt 2 0

1

ln b − ln a 󵄨 󵄨 ≤ ∫ at b1−t [t α − (1 − t)α ]󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨dt 2 1 2

1 2

ln b − ln a 󵄨 󵄨 + ∫ at b1−t [(1 − t)α − t α ]󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨dt 2 1

≤

0

ln b − ln a ∫ at b1−t [t α − (1 − t)α ]|f 󸀠 (a)|t |f 󸀠 (b)|1−t dt 2 1 2

1 2

ln b − ln a + ∫ at b1−t [(1 − t)α − t α ]|f 󸀠 (a)|t |f 󸀠 (b)|1−t dt 2 0

≤ b|f 󸀠 (b)|

1

t

ln b − ln a a|f 󸀠 (a)| ) [t α − (1 − t)α ]dt ∫( 󸀠 2 b|f (b)| 1 2

1 2

+ b|f 󸀠 (b)|

t

ln b − ln a a|f 󸀠 (a)| ) [(1 − t)α − t α ]dt ∫( 󸀠 2 b|f (b)| 0

1

t

1 2

t

ln b − ln a a|f 󸀠 (a)| α a|f 󸀠 (a)| α {∫( 󸀠 ) t dt − 2 ∫( 󸀠 ) t dt ≤ b|f 󸀠 (b)| 2 b|f (b)| b|f (b)| 0

0

338 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals 1

1 2

t

t

a|f 󸀠 (a)| a|f 󸀠 (a)| − ∫( 󸀠 ) (1 − t)α dt + 2 ∫( 󸀠 ) (1 − t)α dt} b|f (b)| b|f (b)| 0

0

a|f (a)| i−1 ∞ [ln a|f 󸀠 (a)| ]i−1 [ln b|f 󸀠 (b)| ] ln b − ln a a|f 󸀠 (a)| ∞ b|f (b)| ≤ b|f (b)| { 󸀠 −∑ ∑(−1)i−1 2 b|f (b)| i=1 (α + 1)i (α + 1)i i=1 󸀠

󸀠

󸀠

a|f 󸀠 (a)| i−1

1

1

1 a|f 󸀠 (a)| 2 ∞ [− 2 ln b|f 󸀠 (b)| ] −( ) ( 󸀠 ) ∑ 2 b|f (b)| i=1 (α + 1)i α

a|f (a)| i−1 [ln b|f 󸀠 (b)| ] 󸀠

∞

+2∑

α+i

(α + 1)i

1 [1 − ( ) 2

i=0

1

a|f 󸀠 (a)| 2 ( 󸀠 ) ]} b|f (b)|

a|f (a)| i−1 ∞ [ln( a|f 󸀠 (a)| )]i−1 [ln b|f 󸀠 (b)| ] ln b − ln a a|f 󸀠 (a)| ∞ b|f (b)| { 󸀠 −∑ ≤ b|f (b)| ∑(−1)i−1 2 b|f (b)| i=1 (α + 1)i (α + 1)i i=1 󸀠

󸀠

󸀠

a|f 󸀠 (a)|

1

1

i−1

1 a|f 󸀠 (a)| 2 ∞ [− 2 ln( b|f 󸀠 (b)| )] −( ) ( 󸀠 ) ∑ 2 b|f (b)| i=1 (α + 1)i α

a|f (a)| i−1 [ln b|f 󸀠 (b)| ] 󸀠

∞

+ 2∑ i=1

󸀠

+2

󸀠

1

1

a|f (a)| ln b|f 󸀠 (b)|

a|f 󸀠 (a)| i−1

1 a|f 󸀠 (a)| 2 ∞ [ 2 ln b|f 󸀠 (b)| ] −( ) ( 󸀠 ) ∑ 2 b|f (b)| i=1 (α + 1)i

(α + 1)i

α

1

a|f (a)| 2 1 − ( 21 )α ( b|f 󸀠 (b)| )

}

a|f (a)| i−1 ∞ [ln a|f 󸀠 (a)| ]i−1 [ln b|f 󸀠 (b)| ] ln b − ln a a|f 󸀠 (a)| ∞ b|f (b)| { 󸀠 +∑ ≤ b|f (b)| ∑(−1)i−1 2 b|f (b)| i=1 (α + 1)i (α + 1)i i=1 󸀠

󸀠

󸀠

󸀠

−

1

a|f (a)| 2 ( 21 )α−1 ( b|f 󸀠 (b)| )

α+1

1

a|f (a)| 2 1 − ( 21 )α ( b|f 󸀠 (b)| ) 󸀠

+2

The proof is completed.

a|f (a)| ln b|f 󸀠 (b)| 󸀠

}.

Theorem 321. Let f : [0, b] → ℝ be a differentiable mapping. If |f 󸀠 |q is measurable and |f 󸀠 |q , (q > 1) is GG-convex on [a, b] for some fired α ∈ (0, ∞), t ∈ [0, 1], and k > 0, 0 ≤ a < b, then the following integrals holds: 󵄨󵄨 f (a) + f (b) 󵄨󵄨 Γ(α + 1) 󵄨󵄨 󵄨 − [H Jaα+ f (b) + H Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 󵄨󵄨 2 2(ln b − ln a) p 1 a p i−1 a p i−1 ∞ ∞ [ln( b ) ] ( 21 )pα−1 (ab 2 ) ln b − ln a p i−1 [ 2 ln( b ) ] p (a ∑(−1) ≤ +b ∑ − 2 (pα + 1)i (pα + 1)i pα + 1 i=1 i=1 p −1

pα−1

1 a − [ln( ) ] [( ) b 2

p 2

p

1 p

(ab) + 2b ]) (

1

󵄨󵄨 f 󸀠 (a) 󵄨󵄨q q 󵄨󵄨 󵄨󵄨 [ − 1]) . 󵄨 󵄨 󸀠 󵄨 󵄨 󸀠 ln( |f 󸀠 (a)| )q 󵄨󵄨 f (b) 󵄨󵄨 |f 󸀠 (b)|q |f (b)|

Proof. By using Definition 19, Definition 3, Lemma 42, Lemma 32, Lemma 33 and Lemma 34, we have 󵄨󵄨 󵄨󵄨󵄨 f (a) + f (b) Γ(α + 1) 󵄨 󵄨󵄨 − [H Jaα+ f (b) + H Jbα− f (a)]󵄨󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 2(ln b − ln a) 󵄨

5.4 Inequalities via geometric-geometric-convex functions | 339 1

ln b − ln a 󵄨 󵄨 ≤ ∫ |(1 − t)α − t α |et ln a+(1−t) ln b 󵄨󵄨󵄨f 󸀠 (et ln a+(1−t) ln b )󵄨󵄨󵄨dt 2 0

1

≤

ln b − ln a 󵄨 󵄨 ∫ |(1 − t)α − t α |et ln a+(1−t) ln b 󵄨󵄨󵄨f 󸀠 (et ln a+(1−t) ln b )󵄨󵄨󵄨dt 2 0

1

≤

ln b − ln a 󵄨 󵄨 ∫ at b1−t |(1 − t)α − t α |󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨dt 2 0

1 p

1

1

ln b − ln a p 󵄨 󵄨q ≤ (∫(at b1−t |(1 − t)α − t α |) dt) (∫ 󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨 dt) 2 0

0

1

≤

1 q

1

pt

1 2

pt

pt

a a a ln b − ln a p (b ∫( ) t pα dt − bp ∫( ) (1 − t)pα dt + bp ∫( ) (1 − t)pα dt 2 b b b 1 2

1 2

0

1 2

1 p

pt

1

q

qt

a |f 󸀠 (b)|q |f 󸀠 (a)| |f 󸀠 (a)| − b ∫( ) t pα dt) ( ln( ) [ ] dt) ∫ 󸀠 (a)| b |f 󸀠 (b)| |f 󸀠 (b)| ln( |f|f 󸀠 (b)| )q 0 0 p

1 q

1 2

p pt ∞ [ln( a )p ]i−1 [ln( ba )p ]i−1 ln b − ln a p a ∞ a b ≤ (b ( ) ∑(−1)i−1 − bp ∑ − 2bp ∫( ) t pα dt 2 b i=1 (pα + 1)i (pα + 1)i b i=0 0

1 2

1 p

1

pt q q |f 󸀠 (b)|q 󵄨󵄨󵄨󵄨 f 󸀠 (a) 󵄨󵄨󵄨󵄨 a + 2b ∫( ) (1 − t)pα dt) ( [󵄨󵄨 󸀠 󵄨󵄨 − 1]) 󸀠 (a)| q 󵄨󵄨 f (b) 󵄨󵄨 b ln( |f|f 󸀠 (b)| ) 0 p

≤

p ∞ [ln( a )p ]i−1 [ln( ba )p ]i−1 ln b − ln a p a ∞ b (b ( ) ∑(−1)i−1 − bp ∑ 2 b i=1 (pα + 1)i (pα + 1)i i=1 p

1

p

p 1 a p i−1 pα pα+i ∞ [ln( a )p ]i−1 1 a 2 ∞ [− ln( b ) ] a 2 1 b − ( ) bp ( ) ∑ 2 + 2bp ∑ [1 − ( ) (1 − ) ]) 2 b (pα + 1)i (pα + 1)i b 2 i=1 i=0 1

󵄨󵄨 f 󸀠 (a) 󵄨󵄨q q 󵄨󵄨 󵄨󵄨 ×( [ 󵄨 󵄨 − 1]) |f 󸀠 (a)| q 󵄨󵄨󵄨 f 󸀠 (b) 󵄨󵄨󵄨 ln( 󸀠 ) |f 󸀠 (b)|q |f (b)|

≤

∞ [ln( a )p ]i−1 [ln( ba )p ]i−1 ln b − ln a p ∞ b (a ∑(−1)i−1 − bp ∑ 2 (pα + 1) (pα + 1)i i i=1 i=1 pα ∞ [− 1 ln( a )p ]i−1 ∞ [ln( a )p ]i−1 p 1 p 2 b b 2 − ( ) (ab) ∑ + 2b ∑ 2 (pα + 1) (pα + 1)i i i=1 i=0 p

pα+i

a 2 ∞ 1 − 2bp ( ) ∑ (1 − ) b 2 i=0

[ln( ba )p ]i−1

1 p

1

󵄨󵄨 f 󸀠 (a) 󵄨󵄨q q 󵄨󵄨 󵄨 ) ( [ 󵄨󵄨 󸀠 󵄨󵄨󵄨 − 1]) 󸀠 (a)| |f 󵄨 󵄨 (pα + 1)i ln( 󸀠 )q 󵄨 f (b) 󵄨 |f 󸀠 (b)|q |f (b)|

340 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals

≤

∞ [ln( a )p ]i−1 [ln( ba )p ]i−1 ln b − ln a p ∞ b (a ∑(−1)i−1 − bp ∑ 2 (pα + 1) (pα + 1)i i i=1 i=1

pα p −1 ∞ [− 1 ln( a )p ]i−1 ∞ [ln( a )p ]i−1 p 1 a b b − ( ) (ab) 2 ∑ 2 + 2bp ∑ + 2bp [ln( ) ] 2 (pα + 1)i (pα + 1)i b i=1 i=1 p

pα+i

a 2 ∞ 1 − 2b ( ) ∑ (1 − ) b 2 i=0 p

[ln( ba )p ]i−1 (pα + 1)i

1 p

) (

1

󵄨󵄨 f 󸀠 (a) 󵄨󵄨q q 󵄨󵄨 󵄨󵄨 − 1]) [ 󵄨 󵄨 󸀠 󵄨 󵄨 󸀠 ln( |f 󸀠 (a)| )q 󵄨󵄨 f (b) 󵄨󵄨 |f 󸀠 (b)|q |f (b)|

∞ [ln( a )p ]i−1 [ln( ba )p ]i−1 ln b − ln a p b (a ∑(−1)i−1 + bp ∑ ≤ 2 (pα + 1) (pα + 1)i i i=1 i=1 ∞

pα pα i−1 ∞ [− 1 ln( a )p ]i−1 ∞ p p [ln( ba )p ]i−1 1 1 1 b − ( ) (ab) 2 ∑ ( ) − ( ) (ab) 2 ∑ 2 2 (pα + 1)i 2 2 (pα + 1)i i=1 i=0 1

1

p −1 p q q a |f 󸀠 (b)|q 󵄨󵄨󵄨󵄨 f 󸀠 (a) 󵄨󵄨󵄨󵄨 + 2bp [ln( ) ] ) ( [ − 1]) 󵄨 󵄨 󸀠 (a)| 󵄨 󵄨 󸀠 |f 󵄨 󵄨 q b ln( |f 󸀠 (b)| ) 󵄨 f (b) 󵄨

≤

∞ [ln( a )p ]i−1 [ln( ba )p ]i−1 ln b − ln a p ∞ b + bp ∑ (a ∑(−1)i−1 2 (pα + 1) (pα + 1)i i i=1 i=1

pα pα ∞ [− 1 ln( a )p ]i−1 ∞ [ 1 ln( a )p ]i−1 p p 1 1 b b − ( ) (ab) 2 ∑ 2 − ( ) (ab) 2 ∑ 2 2 (pα + 1) 2 (pα + 1)i i i=1 i=1 pα−1

1 −( ) 2 ≤

p −1

a (ab) [ln( ) ] b p 2

1

1

p −1 p q q a |f 󸀠 (b)|q 󵄨󵄨󵄨󵄨 f 󸀠 (a) 󵄨󵄨󵄨󵄨 + 2b [ln( ) ] ) ( [ − 1]) 󵄨 󵄨 󸀠 󵄨 󵄨 󸀠 (a)| q 󵄨󵄨 f (b) 󵄨󵄨 b ln( |f|f 󸀠 (b)| ) p

pα ∞ [ln( a )p ]i−1 p [ln( ba )p ]i−1 1 ln b − ln a p ∞ 1 b + bp ∑ − 2( ) (ab) 2 (a ∑(−1)i−1 2 (pα + 1) (pα + 1) 2 pα +1 i i i=1 i=1 p −1

pα−1

a 1 − [ln( ) ] [( ) b 2

1 i−1 [ 2

ln b − ln a p ≤ (a ∑(−1) 2 i=1 ∞

p −1

pα−1

1 a − [ln( ) ] [( ) b 2

1 p

1

󵄨󵄨 f 󸀠 (a) 󵄨󵄨q q 󵄨 󵄨 (ab) + 2b ]) ( [󵄨󵄨󵄨 󸀠 󵄨󵄨󵄨 − 1]) 󸀠 ln( |f 󸀠 (a)| )q 󵄨󵄨 f (b) 󵄨󵄨 p 2

p

ln( ba )p ]i−1

(pα + 1)i

|f 󸀠 (b)|q |f (b)|

p

∞

+b ∑ 1 p

i=1

[ln( ba )p ]i−1 (pα + 1)i

p

−

( 21 )pα−1 (ab 2 ) pα + 1

1

󵄨󵄨 f 󸀠 (a) 󵄨󵄨q q 󵄨󵄨 󵄨󵄨 (ab) + 2b ]) ( [ 󵄨 󵄨 − 1]) . |f 󸀠 (a)| q 󵄨󵄨󵄨 f 󸀠 (b) 󵄨󵄨󵄨 ln( 󸀠 ) p 2

p

|f 󸀠 (b)|q |f (b)|

The proof is completed. Theorem 322. Let f : [0, b] → ℝ be a differentiable mapping. If |f 󸀠 | is measurable and |f 󸀠 | is GG-convex on [a, b] for some fired α ∈ (0, ∞) and t ∈ [0, 1], 0 ≤ a < b, then the following integrals holds: 󵄨󵄨󵄨 Γ(α + 1) a + b 󵄨󵄨󵄨󵄨 α α 󵄨󵄨 󵄨󵄨 2(ln b − ln a)α [H Ja+ f (b) + H Jb− f (a)] − f ( 2 )󵄨󵄨󵄨 󵄨 󵄨

5.4 Inequalities via geometric-geometric-convex functions |

≤

|f 󸀠 (b)| − |f 󸀠 (a)| 2 ln b − ln a 1 b|f 󸀠 (b)| b|f 󸀠 (a)| + a|f 󸀠 (b)| − 2b|f 󸀠 (b)| + {(1 − α+1 )[ + ] 2 α+1 α+2 2 a|f 󸀠 (a)| − a|f 󸀠 (b)| − b|f 󸀠 (a)| a|f 󸀠 (a)| − b|f 󸀠 (a)| + − α+3 (α + 3)2α+3 − a|f 󸀠 (a)|B1 (3, α + 3) − (b|f 󸀠 (a)| + a|f 󸀠 (b)|)B1 (2, α + 2)

+ (2b|f 󸀠 (a)| + 2a|f 󸀠 (b)|)B0.5 (2, α + 2) + 2a|f 󸀠 (a)|B0.5 (3, α + 1)}. Proof. By using Definition 19, Definition 3, Lemma 50 and Lemma 36, we have 󵄨󵄨 Γ(α + 1) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 [H Jaα+ f (b) + H Jbα− f (a)] − f ( )󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 󵄨󵄨 2(ln b − ln a) 1

≤

b − a 󵄨󵄨 󸀠 󵄨 ∫ 󵄨󵄨kf (ta + (1 − t)b)󵄨󵄨󵄨dt 2 0

1

ln b − ln a 󵄨 󵄨 + ∫ |(1 − t)α − t α |et ln a+(1−t) ln b 󵄨󵄨󵄨f 󸀠 (et ln a+(1−t)ln b )󵄨󵄨󵄨dt 2 0

1

−1 ln b − ln a 󵄨 󵄨 ≤ (|f (a)| − |f (b)|) + {∫ at b1−t [t α − (1 − t)α ]󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨dt 2 2 1 2

1 2

󵄨 󵄨 + ∫ at b1−t [(1 − t)α − t α ]󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨dt} 0

1

−1 ln b − ln a ≤ (|f (a)| − |f (b)|) + {∫ t α [at + b(1 − t)][t|f 󸀠 (a)| + (1 − t)|f 󸀠 (b)|]dt 2 2 1 2

1

− ∫(1 − t)α [at + b(1 − t)][t|f 󸀠 (a)| + (1 − t)|f 󸀠 (b)|]dt} 1 2 1 2

ln b − ln a + {∫(1 − t)α [at + b(1 − t)][t|f 󸀠 (a)| + (1 − t)|f 󸀠 (b)|]dt 2 0

1 2

− ∫ t α [at + b(1 − t)][t|f 󸀠 (a)| + (1 − t)|f 󸀠 (b)|]dt} 0

1

−1 ln b − ln a ≤ (|f (a)| − |f (b)|) + {|f 󸀠 (a)| ∫ t α+1 [at + b(1 − t)]dt 2 2 1 2

341

342 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals 1

1

α

+ |f (b)| ∫(1 − t)t [at + b(1 − t)]dt − |f (a)| ∫(1 − t)α t[at + b(1 − t)]dt 󸀠

󸀠

1 2

1 2

1

− |f 󸀠 (b)| ∫(1 − t)α+1 [at + b(1 − t)]dt} 1 2 1 2

1 2

ln b − ln a {|f 󸀠 (a)| ∫ t(1 − t)α [at + b(1 − t)]dt + |f 󸀠 (b)| ∫(1 − t)α+1 [at + b(1 − t)]dt 2

+

0

0

1 2

1 2

− |f 󸀠 (a)| ∫ t α+1 [at + b(1 − t)]dt − |f 󸀠 (b)| ∫(1 − t)t α [at + b(1 − t)]dt} 0

≤

0

|f (b)| − |f (a)| ln b − ln a a|f (a)| 1 b|f 󸀠 (a)| 1 + { (1 − α+3 ) + (1 − α+2 ) 2 2 α+3 α+2 2 2 󸀠

󸀠

󸀠

b|f 󸀠 (a)| (1 − α+3 b|f 󸀠 (b)| + (1 − α+1 −

1

a|f 󸀠 (b)| a|f 󸀠 (b)| 1 1 (1 − α+2 ) − (1 − α+3 ) α+2 α+3 2 2 1 1 b|f 󸀠 (b)| 1 2b|f 󸀠 (b)| (1 − ) + (1 − α+3 ) ) − α+1 α+2 α+2 α+3 2 2 2 1

2α+3

)+

α 2

1

α+1

− a|f (a)| ∫(1 − t) t dt − b|f (a)| ∫ t(1 − t) 󸀠

󸀠

1

dt − a|f (b)| ∫ t(1 − t)α+1 dt 󸀠

1 2

1 2

1 2 1 2

1

b|f 󸀠 (b)| ln b − ln a − {a|f 󸀠 (a)| ∫ t 2 (1 − t)α dt ∫(α + 3)(1 − t)α+2 dt} + α+3 2 0

1 2

1 2

α+1

+ b|f (a)| ∫ t(1 − t) 󸀠

1 2

dt + a|f (b)| ∫ t(1 − t)α+1 dt + 󸀠

0

0

b|f 󸀠 (b)| 1 ( − 1) α + 3 2α+3

b|f 󸀠 (a)| 1 b|f 󸀠 (a)| 1 a|f 󸀠 (b)| 1 a|f 󸀠 (a)| 1 − − + − α + 3 2α+3 α + 2 2α+2 α + 3 2α+3 α + 2 2α+2 + ≤

a|f 󸀠 (b)| 1 b|f 󸀠 (b)| 1 2b|f 󸀠 (b)| 1 b|f 󸀠 (a)| 1 − + − } α + 3 2α+3 α + 1 2α+1 α + 2 2α+2 α + 3 2α+3

1 |f 󸀠 (b)| − |f 󸀠 (a)| ln b − ln a (a − b)[|f 󸀠 (a)| − |f 󸀠 (b)|] + { (1 − α+3 ) 2 2 α+3 2 +

b|f 󸀠 (b)| 1 b[|f 󸀠 (a)| − |f 󸀠 (b)|] + (a − b)|f 󸀠 (b)| 1 (1 − α+1 ) + (1 − α+2 ) α+1 α+2 2 2 1

0

α 2

1 2

1

0

0

− a|f (a)| ∫(1 − t) t dt + a|f (a)| ∫(1 − t)α t 2 dt − b|f 󸀠 (a)| ∫ t(1 − t)α+1 dt 󸀠

󸀠

5.4 Inequalities via geometric-geometric-convex functions |

1 2

α+1

+ b|f (a)| ∫ t(1 − t) 󸀠

0

1

α+1

dt − a|f (b)| ∫ t(1 − t) 󸀠

0

343

1 2

dt + a|f (b)| ∫ t(1 − t)α+1 } 󸀠

0

ln b − ln a + {a|f 󸀠 (a)|B0.5 (3, α + 1) + b|f 󸀠 (a)|B0.5 (2, α + 2) 2 b|f 󸀠 (b)| 1 a|f 󸀠 (b)| + a|f 󸀠 (b)|B0.5 (2, α + 2) + ( α+3 − 1) − α+3 2 (α + 3)2α+3 󸀠 󸀠 󸀠 󸀠 b|f (b)| b|f (a)| + a|f (b)| − 2b|f (b)| − } − (α + 1)2α+1 (α + 2)2α+2 |f 󸀠 (b)| − |f 󸀠 (a)| ln b − ln a (a − b)[|f 󸀠 (a)| − |f 󸀠 (b)|] 1 ≤ + { (1 − α+3 ) 2 2 α+3 2 b|f 󸀠 (b)| 1 b[|f 󸀠 (a)| − |f 󸀠 (b)|] + (a − b)|f 󸀠 (b)| 1 + (1 − α+1 ) + (1 − α+2 ) α+1 α+2 2 2 − a|f 󸀠 (a)|B1 (3, α + 1) + a|f 󸀠 (a)|B0.5 (3, α + 1) − b|f 󸀠 (a)|B1 (2, α + 2)

+ b|f 󸀠 (a)|B0.5 (2, α + 2) − a|f 󸀠 (b)|B1 (2, α + 2) + a|f 󸀠 (b)|B0.5 (2, α + 2)} ln b − ln a {a|f 󸀠 (a)|B0.5 (3, α + 1) + b|f 󸀠 (a)|B0.5 (2, α + 2) 2 b|f 󸀠 (b)| 1 a|f 󸀠 (b)| + a|f 󸀠 (b)|B0.5 (2, α + 2) + ( α+3 − 1) − α+3 2 (α + 3)2α+3 󸀠 󸀠 󸀠 󸀠 b|f (a)| + a|f (b)| − 2b|f (b)| b|f (b)| − − } α+2 (α + 1)2α+1 (α + 2)2 1 b|f 󸀠 (b)| |f 󸀠 (b)| − |f 󸀠 (a)| ln b − ln a + {(1 − α+1 )[ ≤ 2 2 α+1 2 󸀠 󸀠 󸀠 󸀠 b|f (a)| + a|f (b)| − 2b|f (b)| a|f (a)| − a|f 󸀠 (b)| − b|f 󸀠 (a)| + ]+ α+2 α+3 󸀠 󸀠 a|f (a)| − b|f (a)| − − a|f 󸀠 (a)|B1 (3, α + 3) − (b|f 󸀠 (a)| + a|f 󸀠 (b)|)B1 (2, α + 2) (α + 3)2α+3 +

+ (2b|f 󸀠 (a)| + 2a|f 󸀠 (b)|)B0.5 (2, α + 2) + 2a|f 󸀠 (a)|B0.5 (3, α + 1)}. The proof is completed. Theorem 323. Let f : [0, b] → ℝ be a differentiable mapping. If |f 󸀠 |q is measurable and |f 󸀠 |q , (q > 1) is GG-convex on [a, b] for some fired α ∈ (0, ∞) and t ∈ [0, 1], 0 ≤ a < b, then the following integrals holds: 󵄨󵄨 Γ(α + 1) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 α α [ J )󵄨󵄨 + f (b) + H Jb− f (a)] − f ( 󵄨󵄨 H a 󵄨󵄨 󵄨󵄨 2(ln b − ln a)α 2

1

|f 󸀠 (b)| − |f 󸀠 (a)| ln b − ln a |f 󸀠 (a)|q + |f 󸀠 (b)|q q ≤ + ( ) 2 2 2

344 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals

×(

bp 1 1 2bp − ap ap − 2bp bp + − pα+1 + pα+2 pα + 2 pα + 1 2 pα + 1 2 pα + 2

p

p

1 p

p

− a B1 (2, pα) + 2a B0.5 (2, pα + 1) − b B0.5 (pα + 1, 2)) , where

1 p

+

1 q

= 1.

Proof. By using Definition 19, Definition 3, Lemma 50 and Lemma 36, we have 󵄨󵄨 Γ(α + 1) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 [H Jaα+ f (b) + H Jbα− f (a)] − f ( )󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2(ln b − ln a) 󵄨󵄨 2 1

≤

b − a 󵄨󵄨 󸀠 󵄨 ∫ 󵄨󵄨kf (ta + (1 − t)b)󵄨󵄨󵄨dt 2 0

1

ln b − ln a 󵄨 󵄨 + ∫ |(1 − t)α − t α |et ln a+(1−t) ln b 󵄨󵄨󵄨f 󸀠 (et ln a+(1−t)ln b )󵄨󵄨󵄨dt 2 0

1

≤

b−a 󵄨 󵄨 ∫ |k|󵄨󵄨󵄨f 󸀠 (at + b(1 − t))󵄨󵄨󵄨dt 2 0

+

1

ln b − ln a 󵄨 󵄨 ∫ |(1 − t)α − t α |et ln a+(1−t) ln b 󵄨󵄨󵄨f 󸀠 (et ln a+(1−t) ln b )󵄨󵄨󵄨dt 2 1

0

1

ln b − ln a (b − a) 󵄨󵄨 󸀠 󵄨 󵄨 󵄨 ≤ ∫ 󵄨󵄨f (at + b(1 − t))󵄨󵄨󵄨dt + ∫ at b1−t |(1 − t)α − t α |󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨dt 2 2 0

0

1

ln b − ln a −1 p (∫(at b1−t |(1 − t)α − t α |) dt) ≤ (|f 󸀠 (a)| − |f 󸀠 (b)|) + 2 2 1

󵄨 󵄨q × (∫ 󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨 dt)

1 p

0

1 q

0

1

−1 ln b − ln a ≤ (|f 󸀠 (a)| − |f 󸀠 (b)|) + (∫ apt bp(1−t) |(1 − t)α − t α |p dt) 2 2 1

× (∫ |f 󸀠 (a)|qt |f 󸀠 (b)|q(1−t) dt)

1 q

1 p

0

0 1

ln b − ln a −1 ≤ (|f 󸀠 (a)| − |f 󸀠 (b)|) + (∫[tap + (1 − t)bp ]|(1 − t)α − t α |p dt) 2 2 0

1

× (∫[t|f 󸀠 (a)|q + (1 − t)|f 󸀠 (b)|q ]dt) 0

1 q

1 p

5.4 Inequalities via geometric-geometric-convex functions | 345 1

ln b − ln a −1 ≤ (|f 󸀠 (a)| − |f 󸀠 (b)|) + (∫[tap + (1 − t)bp ](t pa − (1 − t)pa )dt 2 2 1 2

1 2

p

p

pα

+ ∫[ta + (1 − t)b ][(1 − t) 0

1 p

1

|f 󸀠 (a)|q |f 󸀠 (b)|q q − t ]dt) ( + |f 󸀠 (b)|q − ) 2 2 pα

1

−1 ln b − ln a |f 󸀠 (a)|q + 2|f 󸀠 (b)|q − |f 󸀠 (b)|q q ≤ (|f 󸀠 (a)| − |f 󸀠 (b)|) + ( ) 2 2 2 1

p

× (a ∫ t

pα+1

p

1 2

pα

p

1

1 2

pα

p

1 2 1 2

p

+ a ∫ t(1 − t) dt − a ∫ t 0

pα

1

dt − a ∫ t(1 − t) dt + b ∫ t (1 − t)dt − b ∫(1 − t)pα+1 dt

1 2

p

1

pα+1

p

1 2

1 2

pα+1

dt + b ∫(1 − t)

0

p

1 2

− b ∫ t (1 − t)dt)

0

0 q

pα

1

−1 ln b − ln a |f (a)| + 2|f (b)| − |f 󸀠 (b)|q q ≤ (|f 󸀠 (a)| − |f 󸀠 (b)|) + ( ) 2 2 2 bp 1 bp ap − ap B1 (2, pα) + ap B0.5 (2, Pα + 1) + − pα+1 ×( pα + 2 pα + 1 2 pα + 1 p bp bp a 1 1 − + + ap B0.5 (2, pα + 1) − pα+2 pα + 2 2pα+2 pα + 2 pα + 2 2 󸀠

+

1

2pα+2

q

1 p

󸀠

1

p bp bp − − bp B0.5 (pα + 1, 2)) pα + 2 pα + 2

1

|f 󸀠 (b)| − |f 󸀠 (a)| ln b − ln a |f 󸀠 (a)|q + |f 󸀠 (b)|q q ap − 2bp bp ≤ + ( ) ( + 2 2 2 pα + 2 pα + 1 p p p 1 2b − a 1 b + − ap B1 (2, pα) − pα+1 pα + 1 2pα+2 pα + 2 2 p

p

1 p

+ 2a B0.5 (2, pα + 1) − b B0.5 (pα + 1, 2)) . The proof is completed. Theorem 324. Let f : [0, b] → ℝ be a differentiable mapping. If |f 󸀠 | is measurable and |f 󸀠 | is GG-convex on [a, b] for some fired α ∈ (0, ∞) and t ∈ [0, 1], 0 ≤ a < b, then the following integrals holds: 󵄨󵄨 Γ(α + 1) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 [H Jaα+ f (b) + H Jbα− f (a)] − f ( )󵄨󵄨 󵄨󵄨 α 󵄨󵄨 2 󵄨󵄨 2(ln b − ln a)

a|f (a)| i−1 [ln b|f 󸀠 (b)| ] 1 ln b − ln a a|f 󸀠 (a)| ∞ ≤ (|f 󸀠 (b)| − |f 󸀠 (a)|) + b|f 󸀠 (b)| { 󸀠 ∑(−1)i−1 2 2 b|f (b)| i=1 (α + 1)i 󸀠

346 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals a|f (a)| i−1 [ln b|f 󸀠 (b)| ]

+∑ i=1

(α + 1)i

1

a|f (a)| 2 ( 21 )α−1 ( b|f 󸀠 (b)| )

󸀠

∞

󸀠

−

α+1

1

a|f (a)| 2 1 − ( 21 )α ( b|f 󸀠 (b)| ) 󸀠

+2

a|f (a)| ln b|f 󸀠 (b)| 󸀠

}.

Proof. By using Definition 19, Definition 3, Lemma 50, Lemma 32 and Lemma 33, we have 󵄨󵄨 Γ(α + 1) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 α α [ J + J f (a)] − f ( )󵄨󵄨 f (b) − + 󵄨󵄨 H H b a 󵄨󵄨 2(ln b − ln a)α 󵄨󵄨 2 1 󵄨󵄨 󵄨󵄨 󵄨󵄨 b − a 󵄨󵄨 ≤ 󵄨󵄨󵄨 ∫ kf 󸀠 (ta + (1 − t)b)dt 󵄨󵄨󵄨 󵄨󵄨 2 󵄨󵄨 󵄨 󵄨 0

1 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a α α t ln a+(1−t) ln b 󸀠 t ln a+(1−t)ln b 󵄨 f (e )dt 󵄨󵄨󵄨 + 󵄨󵄨 ∫[(1 − t) − t ]e 󵄨󵄨 󵄨󵄨 2 󵄨 󵄨 0 1

≤

ln b − ln a −1 󵄨 󵄨 (|f (a)| − |f (b)|) + {∫ at b1−t [t α − (1 − t)α ]󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨dt 2 2 1 2

1 2

󵄨 󵄨 + ∫ at b1−t [(1 − t)α − t α ]󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨dt} 0

1

ln b − ln a −1 ≤ (|f (a)| − |f (b)|) + ∫ at b1−t [t α − (1 − t)α ]|f 󸀠 (a)|t |f 󸀠 (b)|1−t dt 2 2 1 2

1 2

+

ln b − ln a ∫ at b1−t [(1 − t)α − t α ]|f 󸀠 (a)|t |f 󸀠 (b)|1−t dt 2 0

1

t

ln b − ln a a|f 󸀠 (a)| −1 ≤ (|f (a)| − |f (b)|) + b|f 󸀠 (b)| ) [t α − (1 − t)α ]dt ∫( 󸀠 2 2 b|f (b)| 1 2

1 2

t

ln b − ln a a|f 󸀠 (a)| + b|f (b)| ) [(1 − t)α − t α ]dt ∫( 󸀠 2 b|f (b)| 󸀠

0

1

t

1

t

1 ln b − ln a a|f 󸀠 (a)| α a|f 󸀠 (a)| ≤ (|f (b)| − |f (a)|) + b|f 󸀠 (b)| {∫( 󸀠 ) t dt − ∫( 󸀠 ) (1 − t)α dt 2 2 b|f (b)| b|f (b)| 0

1 2

− 2 ∫( 0

t

1 2

0

t

a|f 󸀠 (a)| α a|f 󸀠 (a)| ) t dt + 2 ( ) (1 − t)α dt} ∫ b|f 󸀠 (b)| b|f 󸀠 (b)| 0

a|f (a)| i−1 [ln b|f 󸀠 (b)| ] 1 ln b − ln a a|f 󸀠 (a)| ∞ ≤ (|f (b)| − |f (a)|) + b|f 󸀠 (b)| { 󸀠 ∑(−1)i−1 2 2 b|f (b)| i=1 (α + 1)i 󸀠

347

5.4 Inequalities via geometric-geometric-convex functions |

a|f (a)| i−1 [ln b|f 󸀠 (b)| ] 󸀠

∞

−∑

(α + 1)i

i=1

α−1

1 −( ) 2

a|f (a)| i−1 [ln b|f 󸀠 (b)| ] 󸀠

∞

+2∑

(α + 1)i

i=0

a|f 󸀠 (a)| i−1

1

1

a|f 󸀠 (a)| 2 ∞ [− 2 ln b|f 󸀠 (b)| ] ( 󸀠 ) ∑ b|f (b)| i=1 (α + 1)i α+i

1 [1 − ( ) 2

1

a|f 󸀠 (a)| 2 ( 󸀠 ) ]} b|f (b)|

a|f (a)| i−1 [ln b|f 󸀠 (b)| ] ln b − ln a a|f 󸀠 (a)| ∞ 1 { 󸀠 ≤ (|f 󸀠 (b)| − |f 󸀠 (a)|) + b|f 󸀠 (b)| ∑(−1)i−1 2 2 b|f (b)| i=1 (α + 1)i 󸀠

a|f (a)| i−1 [ln b|f 󸀠 (b)| ] 󸀠

∞

+∑

(α + 1)i

i=1

1

a|f (a)| 2 ( 21 )α−1 ( b|f 󸀠 (b)| ) 󸀠

−

α+1

󸀠

+2

1

a|f (a)| 2 1 − ( 21 )α ( b|f 󸀠 (b)| ) a|f (a)| ln b|f 󸀠 (b)| 󸀠

}.

The proof is completed. Theorem 325. Let f : [0, b] → ℝ be a differentiable mapping. If |f 󸀠 |q is measurable and |f 󸀠 |q , (q > 1) is GG-convex on [a, b] for some fired α ∈ (0, ∞) and t ∈ [0, 1], 0 ≤ a < b, then the following integrals holds: 󵄨󵄨 Γ(α + 1) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 α α [ J )󵄨󵄨 + f (b) + H Jb− f (a)] − f ( 󵄨󵄨 H a 󵄨󵄨 2(ln b − ln a)α 󵄨󵄨 2 1 ∞ [ ln( ba )p ]i−1 ln b − ln a p 1 (a ∑(−1)i−1 2 ≤ (|f (b)| − |f (a)|) + 2 2 (pα + 1)i i=1 ∞

p

+b ∑

[ln( ba )p ]i−1

i=1

(pα + 1)i

p

−

( 21 )pα−1 (ab 2 ) pα + 1

p −1

pα−1

a 1 − [ln( ) ] [( ) b 2

p 2

p

(ab) + 2b ])

1 p

1

󵄨󵄨 f 󸀠 (a) 󵄨󵄨q q 󵄨󵄨 󵄨󵄨 ×( [ 󵄨 󵄨 − 1]) , |f 󸀠 (a)| q 󵄨󵄨󵄨 f 󸀠 (b) 󵄨󵄨󵄨 ln( 󸀠 ) |f 󸀠 (b)|q |f (b)|

where

1 p

+

1 q

= 1.

Proof. By using Definition 19, Definition 3, Lemma 50, Lemma 32, Lemma 33 and Lemma 34, we have 󵄨󵄨 Γ(α + 1) a + b 󵄨󵄨󵄨󵄨 󵄨󵄨 )󵄨󵄨 [H Jaα+ f (b) + H Jbα− f (a)] − f ( 󵄨󵄨 α 󵄨󵄨 2(ln b − ln a) 󵄨󵄨 2 1

≤

b − a 󵄨󵄨 󸀠 󵄨 ∫ 󵄨󵄨kf (ta + (1 − t)b)󵄨󵄨󵄨dt 2 0

+

1

ln b − ln a 󵄨 󵄨 ∫ |(1 − t)α − t α |et ln a+(1−t) ln b 󵄨󵄨󵄨f 󸀠 (et ln a+(1−t)ln b )󵄨󵄨󵄨dt 2 1

0

1

b−a ln b − ln a 󵄨 󵄨 󵄨 󵄨 ≤ ∫ |k|󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt + ∫ at b1−t |(1 − t)α − t α |󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨dt 2 2 0

0

348 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals 1

b−a 󵄨 󵄨 ≤ ∫ |k|󵄨󵄨󵄨f 󸀠 (ta + (1 − t)b)󵄨󵄨󵄨dt 2 0

1 p

1

1

ln b − ln a p 󵄨q 󵄨 + (∫(at b1−t |(1 − t)α − t α |) dt) (∫ 󵄨󵄨󵄨f 󸀠 (at b1−t )󵄨󵄨󵄨 dt) 2 0

1 q

0

1

1

pt

pt

−1 a a ln b − ln a p ≤ (|f (a)| − |f (b)|) + (b ∫( ) t pα dt − bp ∫( ) (1 − t)pα dt 2 2 b b 1 2

1 2

1 2

pt

1 2

1 p

pt

1

qt

|f 󸀠 (a)| a a ] dt) + b ∫( ) (1 − t)pα dt − bp ∫( ) t pα dt) (|f 󸀠 (b)|q ∫[ 󸀠 b b |f (b)| p

0

1 q

0

0

1

1 2

pt

pt

a −1 ln b − ln a p a ≤ (|f (a)| − |f (b)|) + (b ∫( ) t pα dt − bp ∫( ) t pα dt 2 2 b b 0

1

1 2

pt

0

1 2

pt

pt

a a a − bp ∫( ) (1 − t)pα dt + ∫( ) (1 − t)pα dt + bp ∫( ) (1 − t)pα dt b b b 0

0

1 2

0

1 p

pt q a |f 󸀠 (b)|q 󵄨󵄨󵄨󵄨 f 󸀠 (a) 󵄨󵄨󵄨󵄨 [ − bp ∫( ) t pα dt) ( 󵄨 󵄨 󸀠 (a)| 󵄨 󸀠 󵄨 − 1]) b )q 󵄨󵄨 f (b) 󵄨󵄨 ln( |f|f 󸀠 (b)| 0

≤

1 q

p [ln( ba )p ]i−1 −1 ln b − ln a p a ∞ (|f (a)| − |f (b)|) + (b ( ) ∑(−1)i−1 2 2 b i=1 (pα + 1)i p

[ln( ba )p ]i−1

∞

−b ∑

(pα + 1)i

i=0

1 2

1 2

pt

pt

a a − 2b ∫( ) t pα dt + 2bp ∫( ) (1 − t)pα dt) b b p

0

1 p

0

1

󵄨󵄨 f 󸀠 (a) 󵄨󵄨q q |f (b)| 󵄨 󵄨 ×( [󵄨󵄨󵄨 󸀠 󵄨󵄨󵄨 − 1]) 󸀠 ln( |f 󸀠 (a)| )q 󵄨󵄨 f (b) 󵄨󵄨 q

󸀠

|f (b)|

≤

p [ln( ba )p ]i−1 −1 ln b − ln a p a ∞ (|f (a)| − |f (b)|) + (b ( ) ∑(−1)i−1 2 2 b i=1 (pα + 1)i p

∞

−b ∑

[ln( ba )p ]i−1

i=1

(pα + 1)i

p

a p i−1 1 pα 2 ∞ [− ln( ) ] 1 p a 2 b −( ) b ( ) ∑ 2 b (pα + 1)i i=1

[ln( ba )p ]i−1

p

pα+i

1 a 2 + 2bp ∑ [1 − ( ) (1 − ) (pα + 1)i b 2 i=0 ∞

≤

1

󵄨󵄨 f 󸀠 (a) 󵄨󵄨q q 󵄨󵄨 󵄨󵄨 ]) ( [ 󵄨 󵄨 − 1]) |f 󸀠 (a)| q 󵄨󵄨󵄨 f 󸀠 (b) 󵄨󵄨󵄨 ln( 󸀠 ) |f 󸀠 (b)|q |f (b)|

a p i−1 i−1 [ln( b ) ]

−1 ln b − ln a p (|f (a)| − |f (b)|) + (a ∑(−1) 2 2 i=1 ∞

1 p

(pα + 1)i

5.4 Inequalities via geometric-geometric-convex functions |

∞

− bp ∑ i=1

[ln( ba )p ]i−1 (pα + 1)i

pα ∞ [− 1 ln( a )p ]i−1 ∞ [ln( a )p ]i−1 p 1 b b − ( ) (ab) 2 ∑ 2 + 2bp ∑ 2 (pα + 1) (pα + 1)i i i=1 i=0

p

pα+i

a 2 ∞ 1 − 2b ( ) ∑ (1 − ) b 2 i=0 p

≤

349

[ln( ba )p ]i−1

1 p

1

󵄨󵄨 f 󸀠 (a) 󵄨󵄨q q 󵄨 󵄨󵄨 ) ( [ 󵄨󵄨 󸀠 󵄨󵄨󵄨 − 1]) 󸀠 (a)| |f (pα + 1)i ln( 󸀠 )q 󵄨󵄨 f (b) 󵄨󵄨 |f 󸀠 (b)|q |f (b)|

∞ [ln( a )p ]i−1 [ln( ba )p ]i−1 −1 ln b − ln a p ∞ b (|f (a)| − |f (b)|) + (a ∑(−1)i−1 − bp ∑ 2 2 (pα + 1) (pα + 1)i i i=1 i=1 pα p −1 ∞ [− 1 ln( a )p ]i−1 ∞ [ln( a )p ]i−1 p a 1 b b + 2bp ∑ + 2bp [ln( ) ] − ( ) (ab) 2 ∑ 2 2 (pα + 1)i (pα + 1)i b i=1 i=1 p

pα+i

a 2 ∞ 1 − 2bp ( ) ∑ (1 − ) b 2 i=0

[ln( ba )p ]i−1

1 p

1

󵄨󵄨 f 󸀠 (a) 󵄨󵄨q q 󵄨 󵄨 ) ( [󵄨󵄨󵄨 󸀠 󵄨󵄨󵄨 − 1]) 󸀠 (pα + 1)i ln( |f 󸀠 (a)| )q 󵄨󵄨 f (b) 󵄨󵄨 |f 󸀠 (b)|q |f (b)|

∞ [ln( a )p ]i−1 [ln( ba )p ]i−1 ln b − ln a p 1 b (a ∑(−1)i−1 + bp ∑ ≤ (|f (b)| − |f (a)|) + 2 2 (pα + 1)i (pα + 1)i i=1 i=1 ∞

pα pα ∞ [− 1 ln( a )p ]i−1 ∞ [ 1 ln( a )p ]i−1 p p 1 1 b b − ( ) (ab) 2 ∑ 2 − ( ) (ab) 2 ∑ 2 2 (pα + 1)i 2 (pα + 1)i i=1 i=1 pα−1

1 −( ) 2

p −1

a (ab) [ln( ) ] b p 2

1 p

1

p −1 q q a |f 󸀠 (b)|q 󵄨󵄨󵄨󵄨 f 󸀠 (a) 󵄨󵄨󵄨󵄨 + 2b [ln( ) ] ) ( [ − 1]) 󵄨 󵄨 󸀠 (a)| 󵄨 󵄨 󸀠 |f 󵄨 󵄨 b ln( |f 󸀠 (b)| )q 󵄨 f (b) 󵄨 p

[ 1 ln( ba )p ]i−1 1 ln b − ln a p ∞ ≤ (|f (b)| − |f (a)|) + (a ∑(−1)i−1 2 2 2 (pα + 1)i i=1 p

∞

+b ∑ i=1

[ln( ba )p ]i−1 (pα + 1)i

p

−

( 21 )pα−1 (ab 2 ) pα + 1

p −1

pα−1

a 1 − [ln( ) ] [( ) b 2

p 2

p

(ab) + 2b ])

1 p

1

󵄨󵄨 f 󸀠 (a) 󵄨󵄨q q 󵄨󵄨 󵄨󵄨 [ ×( 󵄨 − 1]) . 󵄨 |f 󸀠 (a)| q 󵄨󵄨󵄨 f 󸀠 (b) 󵄨󵄨󵄨 ln( 󸀠 ) |f 󸀠 (b)|q |f (b)|

The proof is completed.

5.4.2 Applications to special means Based on the previous conclusions, we give some applications to special means of real numbers (see Proposition 4.1–4.4, [101]). Proposition 326. Let a, b ∈ ℝ+ \ {0}, 0 ≤ a < b, x ∈ [0, b]. Then |A(x, x) − L(a, b)| ≤

(2 ln x − ln a − ln b − 2)x + a + b , ln b − ln a

350 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals 1

1

p q ( ax )q − 1 (ln x − ln a)2 1 |A(x, x) − L(a, b)| ≤ a( ) ( ) ln b − ln a p+1 q(ln x − ln a) 1

1

p q ( bx )q − 1 (ln b − ln x)2 1 + b( ) ( ) , ln b − ln a p+1 q(ln x − ln b)

1 (√a + √b)2 󵄨 󵄨󵄨 , 󵄨󵄨L(a, b) − [A(a, b)H(a, b)] 2 󵄨󵄨󵄨 ≤ a − b − ln a − ln b 1

1

1 q 1 2p+1 − 1 p 󵄨󵄨 󵄨 ln b − ln a a(2p − ) ( ) . 󵄨󵄨L(a, b) − [A(a, b)H(a, b)] 2 󵄨󵄨󵄨 ≤ 2 p+1 q(ln a − ln b)

Proof. Applying Theorems 312, 313, 314, and 315 for f (x) = x and α = 1, one can obtain the results immediately. Proposition 327. Let a, b ∈ ℝ+ \ {0}, 0 ≤ a < b, x ∈ [0, b], n ≥ 2. Then an + bn 2 󵄨󵄨 󵄨 n n n−1 )x n − , 󵄨󵄨A(x , x ) − L(a, b)Ln−1 (a, b)󵄨󵄨󵄨 ≤ (1 − n(ln b − ln a) n(ln b − ln a) 1

1 p ) n( p+1 1 q 󵄨󵄨 󵄨󵄨 n n n−1 [((ln x − ln a)2q−1 (xqn − aqn ) ) A(x , x ) − L(a, b)L (a, b) ≤ 󵄨󵄨 󵄨󵄨 n−1 ln b − ln a qn 1

1

2q−1

+ ((ln x − ln b) n

(x

qn

1 q − b ) ) ], qn qn

n

n 2( a ) 2 − ( ba )n − 1 a 2 󵄨󵄨 n n 󵄨 n ), 󵄨󵄨L(a , b ) − (H(a, b)A(a, b)) 2 󵄨󵄨󵄨 ≤ −nb (( ) − 1 + b2 b n (ln a − ln b) 1

1

p+1 n q 1 󵄨󵄨 n n 󵄨 ln b − ln a n p 2 − 1 p na (2 − ) ( ) . 󵄨󵄨L(a , b ) − (H(a, b)A(a, b)) 2 󵄨󵄨󵄨 ≤ 2 p+1 qn(ln a − ln b)

Proof. Applying Theorems 312, 313, 314 and 315 for f (x) = x n and α = 1, one can obtain the results immediately. Proposition 328. Let a, b ∈ ℝ+ \ {0}, 0 ≤ a < b, x ∈ [0, b], n ≥ 2. Then 󵄨󵄨 1 1 1 1 1 󵄨󵄨󵄨 1 2(ln a + ln b) − 4 ln x − 4 2 󵄨󵄨 )− , 󵄨󵄨A( , ) − L( , )󵄨󵄨󵄨 ≤ ( 󵄨󵄨 x x 2 a b 󵄨󵄨 x ln b − ln a a(ln b − ln a) 1

1

󵄨󵄨 1 1 p q ( ax )q − 1 1 1 1 󵄨󵄨󵄨󵄨 (ln x − ln a)2 2 1 󵄨󵄨 A( , ) − L( , ) ≤ ⋅ ⋅ ( ) ( ) 󵄨󵄨 󵄨󵄨 󵄨󵄨 x x 2 a b 󵄨󵄨 ln b − ln a a p+1 q(ln a − ln x) 1

1

p q ( bx )q − 1 (ln b − ln x)2 2 1 + ⋅ ⋅( ) ( ) , ln b − ln a b p+1 q(ln b − ln x)

5.4 Inequalities via geometric-geometric-convex functions | 351 1

b b 󵄨󵄨 1 1 1 1 1 󵄨󵄨󵄨 b 2( ) 2 − a − 1 󵄨󵄨 ), 󵄨󵄨L( , ) − G( , )󵄨󵄨󵄨 ≤ − (−1 + + a a b 󵄨󵄨 2b a ln b − ln a 󵄨󵄨 b a 1

1

󵄨󵄨 1 1 q 1 1 󵄨󵄨󵄨 ln b − ln a p 2p+1 − 1 p 1 󵄨󵄨 (2 − ) ( ) . 󵄨󵄨L( , ) − G( , )󵄨󵄨󵄨 ≤ a b 󵄨󵄨 4a p+1 q(ln b − ln a) 󵄨󵄨 b a

Proof. Applying Theorems 312, 313, 314 and 315 for f (x) = the results immediately.

1 x

and α = 1, one can obtain

Proposition 329. Let a, b ∈ ℝ+ \ {0}, 0 ≤ a < b, x ∈ [0, b], n ≥ 2. Then (2 ln x + ln a + ln b − 2)x + a−1 + b−1 󵄨󵄨 −1 −1 󵄨 , 󵄨󵄨A(x, x) − L(b , a )󵄨󵄨󵄨 ≤ ln b − ln a 1

1

p q (ln x + ln b)2 −1 1 (xb)q − 1 󵄨󵄨 −1 −1 󵄨 b ( ) ( ) 󵄨󵄨A(x, x) − L(b , a )󵄨󵄨󵄨 ≤ ln b − ln a p+1 q(ln x + ln b) 1

+

1

p q (ln a + ln x)2 −1 (xa)q − 1 1 a ( ) ( ) , ln b − ln a p+1 q(ln x + ln a)

1 (√a−1 + √b−1 )2 󵄨󵄨 −1 −1 󵄨 −1 −1 −1 −1 −1 −1 , 󵄨󵄨L(b , a ) − [A(b , a )H(b , a )] 2 󵄨󵄨󵄨 ≤ −a + b − ln a − ln b 1 󵄨 󵄨󵄨 −1 −1 −1 −1 −1 −1 󵄨󵄨L(b , a ) − [A(b , a )H(b , a )] 2 󵄨󵄨󵄨 1

1

q ln b − ln a −1 p 2p+1 − 1 p 1 ≤ b (2 − ) ( ) , 2 p+1 q(ln a − ln b)

a−n + b−n 2 󵄨󵄨 n n −1 −1 n−1 −1 −1 󵄨 )x n − , 󵄨󵄨A(x , x ) − L(b , a )Ln−1 (b , a )󵄨󵄨󵄨 ≤ (1 − n(ln b − ln a) n(ln b − ln a) 󵄨󵄨 n n −1 −1 n−1 −1 −1 󵄨 󵄨󵄨A(x , x ) − L(b , a )Ln−1 (b , a )󵄨󵄨󵄨 1 1 1 p ) n( p+1 1 q 2q−1 qn −qn [((ln x + ln b) (x − b ) ⋅ ) ≤ ln b − ln a qn 1

2q−1

+ ((ln x + ln a)

(x

qn

−a

−qn

1 q ) ], )⋅ qn n

n

n 2( a ) 2 − ( ba )n − 1 a 2 󵄨󵄨 −n −n 󵄨 −1 −1 −1 −1 −n ), 󵄨󵄨L(b , a ) − (H(b , a )A(b , a )) 2 󵄨󵄨󵄨 ≤ −na (( ) − 1 + b2 b n (ln a − ln b) n 󵄨󵄨 −n −n 󵄨 −1 −1 −1 −1 󵄨󵄨L(b , a ) − (H(b , a )A(b , a )) 2 󵄨󵄨󵄨 1

1

q ln b − ln a −n p 2p+1 − 1 p 1 ≤ nb (2 − ) ( ) , 2 p+1 qn(ln a − ln b)

352 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals 󵄨󵄨 1 −2(ln a + ln b) − 4 ln x − 4 󵄨󵄨 1 1 1 2b 󵄨 󵄨󵄨 )− , 󵄨󵄨A( , ) − L(b, a)󵄨󵄨󵄨 ≤ ( 󵄨󵄨 x 2 ln b − ln a (ln b − ln a) 󵄨󵄨 x x 1

1

1 q 󵄨󵄨 1 1 󵄨󵄨 (ln x + ln b)2 p q ( xb ) −1 1 1 󵄨󵄨 󵄨 ⋅ 2b ⋅ ( ) ( ) 󵄨󵄨A( , ) − L(b, a)󵄨󵄨󵄨 ≤ 󵄨󵄨 x x 󵄨 2 ln b − ln a p+1 q(− ln b − ln x) 󵄨 1

1

1 q p q ( xa ) −1 (ln a + ln x)2 1 + ⋅ 2a ⋅ ( ) ( ) , ln b − ln a p+1 q(− ln a − ln x) 1

b b b 2( a ) 2 − a − 1 ), |L(a, b) − G(b, a)| ≤ −2a(−1 + + a ln b − ln a 1

1

q 2p+1 − 1 p 1 1 ) ( ) . |L(a, b) − G(b, a)| ≤ b(ln b − ln a)(2p − 4 p+1 q(ln b − ln a)

Proof. Making the substitutions a → b−1 , b → a−1 in the Proposition 326, Proposi-

tion 327 and Proposition 328, one can obtain desired inequalities respectively. We give some applications to special means of real numbers. Example 330. Let a, b ∈ ℝ+ \ {0}, 0 ≤ a < b, x ∈ [0, b]. Then |A(a, b) − L(a, b)| ≤

ln b − ln a 63a − b , 2 192

|A(a, b) − L(a, b)| ≤

bp 2bp ln b − ln a 1 q ap − 2bp ( ) ( + − p+1 2 2 p+2 p + 1 2 (p + 1)

1

1

p bp − ap ap − p+2 − + 2ap B0.5 (2, p + 1)) , 2 (p + 2) (p + 1)(p + 2)

|L(a, b) − A(a, b)| ≤

ln b − ln a 63a − b , 2 192 1

ln b − ln a 1 q ap − 2bp bp 2bp |L(a, b) − A(a, b)| ≤ ( ) ( + − p+1 2 2 p+2 p + 1 2 (p + 1)

1

p bp − ap ap − p+2 − + 2ap B0.5 (2, p + 1)) , 2 (p + 2) (p + 1)(p + 2) 1

a ( a ) 2 − ba + ( ba )2 − 1 ln b − ln a b − 1 |A(a, b) − L(a, b)| ≤ b [ a +2 b ]. 2 ln b (ln ba )2

5.4 Inequalities via geometric-geometric-convex functions | 353

By using Lemma 32, Lemma 33 and Lemma 34, we have |A(a, b) − L(a, b)| ≤ b

p

a 1 ln b − ln a (I[p + 1] − H(p + 1, ( ) , ) 2 b 2 p

p

1

p a a 1 − 2J(p + 1, ( ) ) + 2bR(p + 1, ( ) , )) , b b 2 1

a ( a ) 2 − ba + ( ba )2 − 1 ln b − ln a b − 1 [ a +2 b |L(a, b) − A(a, b)| ≤ b ], 2 ln b (ln ba )2

|L(a, b) − A(a, b)| ≤ b

p

a 1 ln b − ln a (I[p + 1] − H(p + 1, ( ) , ) 2 b 2 p

p

1

p a a 1 − 2J(p + 1, ( ) ) + 2bR(p + 1, ( ) , )) . b b 2

Proof. Applying Theorems 318, 319, 320, 321, 322, 323, 324 and 325 for f (x) = x and α = 1, one can obtain the results immediately. Example 331. Let a, b ∈ ℝ+ \ {0}, 0 ≤ a < b, x ∈ [0, b], n ≥ 2. Then n n−1 n−1 n 󵄨󵄨 󵄨 ln b − ln a 22na + 12nba + 9nab − 12nb n n n−1 , 󵄨󵄨A(a , b ) − Ln−1 (a, b)L(a, b)󵄨󵄨󵄨 ≤ 2 96

󵄨󵄨 󵄨 n n n−1 󵄨󵄨A(a , b ) − Ln−1 (a, b)L(a, b)󵄨󵄨󵄨

1

bp 1 bp ln b − ln a nq aq(n−1) + nq bq(n−1) q ap − 2bp ( ) ( + − p+1 ≤ 2 2 p+2 p+1 2 p+1 1

+

p 1 bp − ap 1 1 − p+3 − ap B1 (2, p) + 2ap B0.5 (2, p + 1)) , p+2 p+2 p+3 2 2

󵄨󵄨 n−1 󵄨 n 󵄨󵄨Ln−1 (a, b)L(a, b) − A (x, y)󵄨󵄨󵄨 nbn−1 − nan−1 ln b − ln a 22nan + 12nban−1 + 9nabn−1 − 12nbn + , ≤ 2 2 96 󵄨󵄨 n−1 󵄨 n 󵄨󵄨Ln−1 (a, b)L(a, b) − A (x, y)󵄨󵄨󵄨

1

nbn−1 − nan−1 ln b − ln a nq aq(n−1) + nq bq(n−1) q + ( ) ≤ 2 2 2 ap − 2bp bp 1 bp 1 bp − ap ×( + − p+1 + p+2 p+2 p+1 2 p+1 2 p+2 −

1

2p+3

1

p 1 − ap B1 (2, p) + 2ap B0.5 (2, p + 1)) , p+3

354 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals n

n

n

a 4 an − 2 ban − 2 󵄨󵄨 󵄨 n n n−1 n ln b − ln a bn − 1 ( an + b ). 󵄨󵄨A(a , b ) − Ln−1 (a, b)L(a, b)󵄨󵄨󵄨 ≤ nb n 2 ln bn (ln ban )2

By using Lemma 32, Lemma 33 and Lemma 34, we have 󵄨󵄨 󵄨 n n n−1 󵄨󵄨A(a , b ) − Ln−1 (a, b)L(a, b)󵄨󵄨󵄨 p p ln b − ln a a 1 a ≤b (I[p + 1] − H(p + 1, ( ) , ) − 2J(p + 1, ( ) ) 2 b 2 b p

1

1

q

p q a 1 |nbn−1 |q an−1 + 2bR(p + 1, ( ) , )) ( [( n−1 ) − 1]) , n−1 b 2 ln( an−1 )q b

b

󵄨󵄨 n−1 󵄨 n 󵄨󵄨Ln−1 (a, b)L(a, b) − A (x, y)󵄨󵄨󵄨 ≤

n

n

n

a a a ln b − ln a bn − 1 4 bn − 2 bn − 2 nbn−1 − nan−1 + nbn ( an + ), n 2 2 ln bn (ln ban )2

󵄨󵄨 n−1 󵄨 n 󵄨󵄨Ln−1 (a, b)L(a, b) − A (x, y)󵄨󵄨󵄨 ≤

p

ln b − ln a a 1 nbn−1 − nan−1 +b (I[p + 1] − H(p + 1, ( ) , ) 2 2 b 2 p

p

1

q

1

p q a 1 |nbn−1 |q a an−1 ) − 1]) . − 2J(p + 1, ( ) ) + 2bR(p + 1, ( ) , )) ( [( n−1 n−1 a b b 2 ln( n−1 )q b

b

Proof. Applying Theorems 318, 319, 320, 321, 322, 323, 324 and 325 for f (x) = x n and α = 1, one can obtain the results immediately. Example 332. Let a, b ∈ ℝ+ \ {0}, 0 ≤ a < b, x ∈ [0, b], n ≥ 2. Then 1 a b 1 1 󵄨󵄨 −1 󵄨 ln b − ln a 450 a + 60 b2 − 30 a2 − 32 a + 42 a −2 , 󵄨󵄨H (a, b) − G (a, b)L(a, b)󵄨󵄨󵄨 ≤ 2 1920

󵄨󵄨 −1 󵄨 −2 󵄨󵄨H (a, b) − G (a, b)L(a, b)󵄨󵄨󵄨 1

bp 1 bp ln b − ln a a−2q + b−2q q ap − 2bp ( ) ( + − p+1 ≤ 2 2 p+2 p+1 2 p+1

1

p 1 2bp − ap + p+2 − ap B1 (2, p) + 2ap B0.5 (2, p + 1) − bp B0.5 (p + 1, 2)) , p+2 2

󵄨󵄨 −2 󵄨 −1 󵄨󵄨G (a, b)L(a, b) − A (a, b)󵄨󵄨󵄨 1 a b 1 1 b2 − a2 ln b − ln a 450 a + 60 b2 − 30 a2 − 32 a + 42 a ≤ + , 2 1920 2a2 b2

5.5 Inequalities via geometric-arithmetic-convex functions | 355

󵄨 󵄨󵄨 −2 −1 󵄨󵄨G (a, b)L(a, b) − A (a, b)󵄨󵄨󵄨

1

b2 − a2 ln b − ln a a−2q + b−2q q ≤ + ( ) 2 2 2a2 b2 ap − 2bp bp 1 bp 1 2bp − ap ×( + − p+1 + p+2 p+2 p+1 2 p+1 2 p+2 p

p

p

1 p

− a B1 (2, p) + 2a B0.5 (2, p + 1) − b B0.5 (p + 1, 2)) , 1

b b b 󵄨󵄨 −1 󵄨 ln b − ln a a − 1 4( a ) 2 − 2 a − 2 −2 ( b + ). 󵄨󵄨H (a, b) − G (a, b)L(a, b)󵄨󵄨󵄨 ≤ 2b ln a (ln ba )2

By using Lemma 32, Lemma 33 and Lemma 34, we have 󵄨󵄨 −1 󵄨 −2 󵄨󵄨H (a, b) − G (a, b)L(a, b)󵄨󵄨󵄨 p p a 1 a ln b − ln a (I[p + 1] − H(p + 1, ( ) , ) − 2J(p + 1, ( ) ) ≤b 2 b 2 b 1 1 1 q p q 2 󵄨 󵄨 p q | 2| 󵄨󵄨 b 󵄨󵄨 a 1 + 2bR(p + 1, ( ) , )) ( b 2 [󵄨󵄨󵄨 2 󵄨󵄨󵄨 − 1]) , b q 󵄨󵄨 a 󵄨󵄨 b 2 ln | 2 | a

1

b b b 2 2 ln b − ln a a − 1 4( a ) 2 − 2 a − 2 󵄨󵄨 −2 󵄨 b −a −1 + ( + ), 󵄨󵄨G (a, b)L(a, b) − A (a, b)󵄨󵄨󵄨 ≤ 2b 2a2 b2 ln ba (ln ba )2

󵄨󵄨 −2 󵄨 −1 󵄨󵄨G (a, b)L(a, b) − A (a, b)󵄨󵄨󵄨 p ln b − ln a a 1 b2 − a2 + b (I[p + 1] − H(p + 1, ( ) , ) ≤ 2 b 2 2a2 b2 1

1

p p p q | 12 |q 󵄨󵄨󵄨 b2 󵄨󵄨󵄨q a 1 a − 2J(p + 1, ( ) ) + 2bR(p + 1, ( ) , )) ( b 2 [󵄨󵄨󵄨 2 󵄨󵄨󵄨 − 1]) . b 󵄨 󵄨 b b 2 ln | 2 |q 󵄨 a 󵄨 a

Proof. Applying Theorems 318, 319, 320, 321, 322, 323, 324 and 325 for f (x) = α = 1, one can obtain the results immediately.

1 x

and

5.5 Inequalities via geometric-arithmetic-convex functions The results in this section are adopted from [246]. 5.5.1 Main results Theorem 333. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with 0 ≤ a < b. If |f 󸀠 | is integrable and GA-convex on [a, b], then for 0 ≤ λ ≤ 1, x ∈ (a, b) and α > 0, the

356 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals following inequality holds: 󵄨 x i−1 ∞ (ln b − ln x)α 󵄨󵄨󵄨󵄨 󸀠 i−1 (ln b ) − x[|f 󸀠 (b)| − |f 󸀠 (x)|] |If (x, λ, α, a, b)| ≤ 󵄨󵄨x|f (b)| ∑(−1) ln b − ln a 󵄨󵄨󵄨 (α + 1)i i=1 ∞

× ∑(−1)i−1

(ln bx )i−1

+

󵄨󵄨 󵄨󵄨 λ [x|f 󸀠 (x)| − 2λ|f 󸀠 (b)| + b|f 󸀠 (x)|]󵄨󵄨󵄨 󵄨󵄨 ln x − ln b 󵄨

+

󵄨󵄨 λ 󵄨󵄨 [x|f 󸀠 (x)| − 2λ|f 󸀠 (a)| + a|f 󸀠 (x)|]󵄨󵄨󵄨. 󵄨󵄨 ln x − ln a 󵄨

(α + 2)i 󵄨 x i−1 ∞ (ln x − ln a)α 󵄨󵄨󵄨󵄨 󸀠 i−1 (ln a ) + − x[|f 󸀠 (a)| − |f 󸀠 (x)|] 󵄨󵄨x|f (a)| ∑(−1) ln b − ln a 󵄨󵄨󵄨 (α + 1) i i=1 i=1

∞

× ∑(−1)i−1

(ln ax )i−1

i=1

(α + 2)i

Proof. By using Definition 20 and Lemma 32, we have |If (x, λ, α, a, b)|

≤ |Uf (x, λ, α, a, b)| + |Vf (x, λ, α, a, b)|

1 󵄨 1 󵄨󵄨 (ln b − ln x)α 󵄨󵄨󵄨󵄨 α 󵄨󵄨 󸀠 1−s s 󵄨󵄨 1−s s 󵄨 󸀠 1−s s 󵄨 1−s s 󵄨󵄨 ≤ 󵄨󵄨 ∫ s 󵄨󵄨f (b x )󵄨󵄨b x ds + ∫ λ󵄨󵄨󵄨f (b x )󵄨󵄨󵄨b x ds󵄨󵄨󵄨 󵄨󵄨 ln b − ln a 󵄨󵄨󵄨 󵄨 0

0

1 󵄨 1 󵄨󵄨 (ln x − ln a)α 󵄨󵄨󵄨󵄨 α 󵄨󵄨 󸀠 1−s s 󵄨󵄨 1−s s 󵄨 󸀠 1−s s 󵄨 1−s s 󵄨󵄨 + 󵄨󵄨 ∫ s 󵄨󵄨f (a x )󵄨󵄨a x ds + ∫ λ󵄨󵄨󵄨f (a x )󵄨󵄨󵄨a x ds󵄨󵄨󵄨 󵄨󵄨 ln b − ln a 󵄨󵄨󵄨 󵄨 0 0

󵄨 1 (ln b − ln x)α 󵄨󵄨󵄨󵄨 α 󸀠 󸀠 1−s s ≤ 󵄨 ∫ s [(1 − s)|f (b)| + s|f (x)|]b x ds ln b − ln a 󵄨󵄨󵄨󵄨 0

1

󸀠

0

+

󵄨󵄨 󵄨󵄨 x ds󵄨󵄨󵄨 󵄨󵄨 󵄨

1−s s

+ ∫ λ[(1 − s)|f (b)| + s|f (x)|]b 󸀠

󵄨 1 (ln x − ln a)α 󵄨󵄨󵄨󵄨 α 󸀠 󸀠 1−s s 󵄨 ∫ s [(1 − s)|f (a)| + s|f (x)|]a x ds ln b − ln a 󵄨󵄨󵄨󵄨 0

1

󵄨󵄨 󵄨󵄨 + ∫ λ[(1 − s)|f 󸀠 (a)| + s|f 󸀠 (x)|]a1−s x s ds󵄨󵄨󵄨 󵄨󵄨 󵄨 0

≤

1 1 󵄨 s s (ln b − ln x)α 󵄨󵄨󵄨󵄨 󸀠 󸀠 α+1 x α x ) ds − b|f (b)| s ( ) ds b|f (b)| s ( ∫ ∫ 󵄨 ln b − ln a 󵄨󵄨󵄨󵄨 b b 0

1

s

1

0

s

1

s

x x x + b|f 󸀠 (x)| ∫ sα+1 ( ) ds + λb|f 󸀠 (b)| ∫( ) ds − λb|f 󸀠 (b)| ∫ s( ) ds b b b 0

1

s 󵄨󵄨 x 󵄨󵄨 + λb|f 󸀠 (x)| ∫ s( ) ds󵄨󵄨󵄨 󵄨󵄨 b 󵄨 0

0

0

5.5 Inequalities via geometric-arithmetic-convex functions | 357 1 1 󵄨 s s (ln x − ln a)α 󵄨󵄨󵄨󵄨 󸀠 α x 󸀠 α+1 x + 󵄨󵄨a|f (a)| ∫ s ( ) ds − a|f (a)| ∫ s ( ) ds ln b − ln a 󵄨󵄨󵄨 a a 0

1

s

1

0

1

s

s

x x x + a|f 󸀠 (x)| ∫ sα+1 ( ) ds + λa|f 󸀠 (a)| ∫( ) ds − λa|f 󸀠 (a)| ∫ s( ) ds a a a 0

0

1

0

s

󵄨󵄨 󵄨󵄨 x + λa|f 󸀠 (x)| ∫ s( ) ds󵄨󵄨󵄨 󵄨󵄨 a 󵄨 0 x i−1 α 󵄨󵄨󵄨 (ln b − ln x) 󵄨󵄨 󸀠 x x ∞ i−1 (ln b ) ≤ − [b|f 󸀠 (b)| − b|f 󸀠 (x)|] 󵄨󵄨b|f (b)| ∑(−1) 󵄨 ln b − ln a 󵄨󵄨 b i=1 (α + 1)i b x i−1 ∞ (ln b ) 1 x 1 1 + λb|f 󸀠 (b)| x − λb|f 󸀠 (b)| x − λb|f 󸀠 (b)| x × ∑(−1)i−1 (α + 2) b ln ln ln i i=1 b b b 󵄨󵄨 x i−1 ∞ α 󵄨󵄨󵄨 ) (ln 󵄨 1 󵄨 (ln x − ln a) 󵄨󵄨 󸀠 x i−1 a + λb|f 󸀠 (x)| x 󵄨󵄨󵄨 + 󵄨a|f (a)| ∑(−1) ln b − ln a 󵄨󵄨󵄨󵄨 a i=1 (α + 1)i ln b 󵄨󵄨󵄨 x i−1 x ∞ 1 x 󸀠 󸀠 i−1 (ln a ) − [a|f (a)| − a|f (x)|] ∑(−1) + λa|f 󸀠 (a)| x a i=1 (α + 2)i ln a a 󵄨 1 1 1 󵄨󵄨󵄨 − λa|f 󸀠 (a)| x − λa|f 󸀠 (a)| x + λa|f 󸀠 (x)| x 󵄨󵄨󵄨 ln a ln a ln a 󵄨󵄨󵄨 󵄨 x i−1 ∞ (ln b − ln x)α 󵄨󵄨󵄨󵄨 󸀠 i−1 (ln b ) − x[|f 󸀠 (b)| − |f 󸀠 (x)|] ≤ 󵄨󵄨x|f (b)| ∑(−1) ln b − ln a 󵄨󵄨󵄨 (α + 1)i i=1 󵄨󵄨 ∞ (ln bx )i−1 󵄨󵄨 1 × ∑(−1)i−1 + λ x [x|f 󸀠 (x)| − 2λ|f 󸀠 (b)| + b|f 󸀠 (x)|]󵄨󵄨󵄨 󵄨󵄨 (α + 2)i ln b i=1 󵄨 󵄨 x i−1 ∞ α 󵄨󵄨 ) (ln (ln x − ln a) 󵄨󵄨 󸀠 i−1 a − x[|f 󸀠 (a)| − |f 󸀠 (x)|] + 󵄨x|f (a)| ∑(−1) ln b − ln a 󵄨󵄨󵄨󵄨 (α + 1) i i=1 󵄨󵄨 x i−1 ∞ (ln a ) 󵄨󵄨 1 × ∑(−1)i−1 + λ x [x|f 󸀠 (x)| − 2λ|f 󸀠 (a)| + a|f 󸀠 (x)|]󵄨󵄨󵄨. 󵄨󵄨 (α + 2) ln i i=1 a 󵄨 The proof is completed. Theorem 334. Let f : [a, b] → ℝ be a differentiable mapping on (a, b) with 0 ≤ a < b. If |f 󸀠 |q , q > 1 is integrable and GA-convex on [a, b], then for 0 ≤ λ ≤ 1, x ∈ (a, b), the following inequality holds: 1

1 󵄨 x q i−1 q (ln b − ln x)α 󵄨󵄨󵄨󵄨 q ∞ |f 󸀠 (b)| + |f 󸀠 (x)|p p i−1 (ln( b ) ) |If (x, λ, α, a, b)| ≤ ) ( ) 󵄨(x ∑(−1) ln b − ln a 󵄨󵄨󵄨󵄨 (qα + 1)i 2 i=1 1 1 󵄨 q q |f 󸀠 (b)| + |f 󸀠 (x)|p p 󵄨󵄨󵄨󵄨 1 x 1 q + ((λb) ) 󵄨󵄨 x ( ) − (λb) x) ( 󵄨󵄨 2 q ln b b q ln b 󵄨

q

358 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals 1

1 󵄨 x q i−1 q (ln x − ln a)α 󵄨󵄨󵄨󵄨 q ∞ |f 󸀠 (a)| + |f 󸀠 (x)|p p i−1 (ln( a ) ) + ) ( ) 󵄨(x ∑(−1) ln b − ln a 󵄨󵄨󵄨󵄨 (qα + 1)i 2 i=1 1 1 󵄨 q q x 1 1 |f 󸀠 (a)| + |f 󸀠 (x)|p p 󵄨󵄨󵄨󵄨 q ( ) − (λa) ) ( + ((λa) ) 󵄨󵄨. 󵄨󵄨 2 q ln ax a q ln ax 󵄨

q

Proof. By using Definition 20 and Lemma 32 and Hölder inequality, we have |If (x, λ, α, a, b)|

≤ |Uf (x, λ, α, a, b)| + |Vf (x, λ, α, a, b)|

1 󵄨󵄨 󵄨󵄨 󵄨󵄨 (ln b − ln x)α 󵄨󵄨 󸀠 (1−s) ln b+s ln x 󵄨󵄨 (1−s) ln b+s ln x 󵄨󵄨󵄨 α 󵄨 ≤ 󵄨󵄨 )󵄨󵄨e ds󵄨󵄨 ∫(s + λ)󵄨󵄨f (e 󵄨󵄨 ln b − ln a 󵄨󵄨 󵄨 󵄨 0

1 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 (ln x − ln a)α 󵄨 󵄨 + 󵄨󵄨󵄨 ∫(sα + λ)󵄨󵄨󵄨f 󸀠 (e(1−s) ln a+s ln x )󵄨󵄨󵄨e(1−s) ln a+s ln x ds󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 ln b − ln a 󵄨 󵄨 0

1 󵄨 1 󵄨󵄨 (ln b − ln x)α 󵄨󵄨󵄨󵄨 α 󵄨󵄨 󸀠 1−s s 󵄨󵄨 1−s s 󵄨 󸀠 1−s s 󵄨 1−s s 󵄨󵄨 ≤ 󵄨󵄨 ∫ s 󵄨󵄨f (b x )󵄨󵄨b x ds + ∫ λ󵄨󵄨󵄨f (b x )󵄨󵄨󵄨b x ds󵄨󵄨󵄨 󵄨󵄨 ln b − ln a 󵄨󵄨󵄨 󵄨 0

0

1 󵄨 1 󵄨󵄨 (ln x − ln a)α 󵄨󵄨󵄨󵄨 α 󵄨󵄨 󸀠 1−s s 󵄨󵄨 1−s s 󵄨󵄨 󸀠 1−s s 󵄨󵄨 1−s s 󵄨󵄨󵄨 + 󵄨 ∫ s 󵄨󵄨f (a x )󵄨󵄨a x ds + ∫ λ󵄨󵄨f (a x )󵄨󵄨a x ds󵄨󵄨 󵄨󵄨 ln b − ln a 󵄨󵄨󵄨󵄨 󵄨 0 0 1

1

1 1 󵄨 1 q p qs (ln b − ln x)α 󵄨󵄨󵄨󵄨 󸀠 p 󸀠 p q qα x ≤ 󵄨󵄨(∫ b s ( ) ds) (∫(1 − s)|f (b)| ds + ∫ s|f (x)| ds) ln b − ln a 󵄨󵄨󵄨 b 0

1

1 q

qs

0

0

1

1

x + (∫(λb)q ( ) ds) (∫(1 − s)|f 󸀠 (b)|p ds + ∫ s|f 󸀠 (x)|p ds) b 0

0

1 q

1 p

0

󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨

0

1

qs

1 q

1

1

1 󵄨 p qs (ln x − ln a)α 󵄨󵄨󵄨󵄨 󸀠 p 󸀠 p q qα x + 󵄨󵄨(∫ a s ( ) ds) (∫(1 − s)|f (a)| ds + ∫ s|f (x)| ds) ln b − ln a 󵄨󵄨󵄨 a 1

0

0

1

1

x + (∫(λa)q ( ) ds) (∫(1 − s)|f 󸀠 (a)|p ds + ∫ s|f 󸀠 (x)|p ds) a 0

0

0

1 q

1 p

󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨

1 󵄨 x q i−1 (ln b − ln x)α 󵄨󵄨󵄨󵄨 q ∞ |f 󸀠 (b)| |f 󸀠 (x)|p p 󸀠 i−1 (ln( b ) ) ) (|f (b)| − + ) ≤ 󵄨(x ∑(−1) ln b − ln a 󵄨󵄨󵄨󵄨 (qα + 1)i 2 2 i=1

+ ((λb)q

1 1 󵄨 q q 1 x 1 |f 󸀠 (b)| |f 󸀠 (x)|p p 󵄨󵄨󵄨󵄨 q 󸀠 ( ) − (λb) ) (|f (b)| − + ) 󵄨󵄨 󵄨󵄨 2 2 q ln( bx ) b q ln bx 󵄨 1

1 󵄨 x q i−1 q (ln x − ln a)α 󵄨󵄨󵄨󵄨 q ∞ |f 󸀠 (a)| |f 󸀠 (x)|p p i−1 (ln( a ) ) 󸀠 + (x ) (|f (a)| − + ) (−1) ∑ 󵄨 ln b − ln a 󵄨󵄨󵄨󵄨 (qα + 1)i 2 2 i=1

5.5 Inequalities via geometric-arithmetic-convex functions | 359 1 1 󵄨 q q 1 x 1 |f 󸀠 (a)| |f 󸀠 (x)|p p 󵄨󵄨󵄨󵄨 q 󸀠 + ((λa) ( ) − (λa) ) (|f (a)| − + ) 󵄨󵄨 󵄨󵄨 2 2 q ln( ax ) a q ln ax 󵄨

q

1

1 󵄨 x q i−1 q (ln b − ln x)α 󵄨󵄨󵄨󵄨 q ∞ |f 󸀠 (b)| + |f 󸀠 (x)|p p i−1 (ln( b ) ) ≤ ) ( ) 󵄨󵄨(x ∑(−1) ln b − ln a 󵄨󵄨󵄨 (qα + 1)i 2 i=1 1

1

󵄨 q q 1 x 1 |f 󸀠 (b)| + |f 󸀠 (x)|p p 󵄨󵄨󵄨󵄨 q + ((λb) ( ) − (λb) ) ) ( 󵄨󵄨 󵄨󵄨 2 q ln( bx ) b q ln bx 󵄨 q

1

1 󵄨 x q i−1 q (ln x − ln a)α 󵄨󵄨󵄨󵄨 q ∞ |f 󸀠 (a)| + |f 󸀠 (x)|p p i−1 (ln( a ) ) + ) ( ) 󵄨(x ∑(−1) ln b − ln a 󵄨󵄨󵄨󵄨 (qα + 1)i 2 i=1

1 1 󵄨 q q 1 x |f 󸀠 (a)| + |f 󸀠 (x)|p p 󵄨󵄨󵄨󵄨 1 q + ((λa) ( ) − (λa) ) ( ) 󵄨󵄨. 󵄨󵄨 2 q ln( ax ) a q ln ax 󵄨

q

The proof is completed.

5.5.2 Applications to special means We give some applications to special means of real numbers (see Proposition 4.1– 4.2, [246]). Proposition 335. Let a, b ∈ ℝ+ \ {0}, 0 ≤ a < b, x ∈ [0, b]. Then (i)

(ii)

|λA(b, a)H(b, a) − (1 + λ)H(b, a)G(x, x) + 2L(b, a)| 1 − b 󵄨󵄨󵄨󵄨 ln b − ln x 󵄨󵄨󵄨󵄨 󸀠 x(ln x − ln b) − x + b +λ ≤ 󵄨󵄨|f (b)| 󵄨 2 󵄨 ln b − ln a 󵄨 ln x − ln b 󵄨󵄨󵄨 (ln x − ln b) 1 − a 󵄨󵄨󵄨󵄨 ln x − ln a 󵄨󵄨󵄨󵄨 󸀠 x(ln x − ln a) − x + a + λ + 󵄨󵄨|f (a)| 󵄨, ln b − ln a 󵄨󵄨 ln x − ln a 󵄨󵄨󵄨 (ln x − ln a)2 |λA(b, a)H(b, a) − (1 + λ)H(b, a)G(x, x) + 2L(b, a)| 1 1 (λx)q − (λb)q q 󵄨󵄨󵄨󵄨 ln b − ln x 󵄨󵄨󵄨󵄨 q q ] 󵄨 ≤ 󵄨[b I(q + 1)] + [ ln b − ln a 󵄨󵄨󵄨 q(ln x − ln b) 󵄨󵄨󵄨 1 1 (λx)q − (λa)q q 󵄨󵄨󵄨󵄨 ln x − ln a 󵄨󵄨󵄨󵄨 q q + [ ] + [a I(q + 1)] 󵄨 󵄨. ln b − ln a 󵄨󵄨󵄨 q(ln x − ln a) 󵄨󵄨󵄨

Proof. Applying Theorems 333 and 334, for f (x) = x, λ ∈ [0, 1] and α = 1, one can obtain the results immediately. Proposition 336 (see Proposition 4.2, [246]). Let a, b ∈ ℝ+ \ {0}, 0 ≤ a < b, x ∈ [0, b]. Then (i)

󵄨󵄨 󵄨 −1 −1 −1 󵄨󵄨λA(a , b )H(b, a) − (1 + λ)H(b, a)G(x, x) + 2H(b, a)K (a, b)󵄨󵄨󵄨

360 | 5 Hermite-Hadamard inequalities involving Hadamard fractional integrals

≤

(ii)

1 1 x(ln x − ln b − x) + b ln b − ln x 󵄨󵄨󵄨󵄨 x(ln x − ln b − x) + b + (λ − 2)( 2 − 2 ) 󵄨 ln b − ln a 󵄨󵄨󵄨 b2 (ln x − ln b)2 x b (ln x − ln b)2 1 λ (x − b) 󵄨󵄨󵄨󵄨 ln x − ln a 󵄨󵄨󵄨󵄨 x(ln x − ln a − x) + a 1 + x( 2 − 2 ) + 2 󵄨+ 󵄨 x b b ln x − ln b 󵄨󵄨󵄨 ln b − ln a 󵄨󵄨󵄨 a2 (ln x − ln a)2 1 1 x(ln x − ln a − x) + a 1 1 λ (x − a) 󵄨󵄨󵄨󵄨 + (λ − 2)( 2 − 2 ) + x( − ) + 󵄨, x a (ln x − ln a)2 x 2 a2 a2 ln x − ln a 󵄨󵄨󵄨

󵄨 󵄨󵄨 −1 −1 −1 󵄨󵄨λA(a , b )H(b, a) − (1 + λ)H(b, a)G(x, x) + 2H(b, a)K (a, b)󵄨󵄨󵄨 1 1 1 1 ln b − ln x 󵄨󵄨󵄨󵄨 I q (q − 1) x2p + b2p p λ(x q − bq ) q x 2p + b2p p 󵄨󵄨󵄨󵄨 ≤ ( ) + ( ) 󵄨󵄨 󵄨 󵄨󵄨 ln b − ln a 󵄨󵄨󵄨 bx2 2 2 (ln x − ln b)b2 x 2 1 1 1 1 ln x − ln a 󵄨󵄨󵄨󵄨 I q (q − 1) x2p + a2p p λ(x q − aq ) q x 2p + a2p p 󵄨󵄨󵄨󵄨 + ( ) + ( ) 󵄨󵄨 󵄨󵄨. ln b − ln a 󵄨󵄨 ax2 2 2 (ln x − ln a)a2 x 2 󵄨󵄨

Proof. Applying Theorems 333 and 334, for f (x) = x −1 , λ ∈ [0, 1] and α = 1, one can obtain the results immediately.

Bibliography [1] [2] [3]

[4]

[5]

[6] [7] [8] [9]

[10]

[11] [12] [13]

[14] [15] [16] [17] [18] [19] [20]

G. Abbas, K. A. Khan, G. Farid, A. U. Farid, Generalizations of some fractional integral inequalities via generalized Mittag-Leffler function, J. Inequal. Appl., 2017(2017), 1–10. S. Abramovich, J. Barić, J. Pec̆arić, Fejer and Hermite-Hadamard type inequalities for superquadratic functions, J. Math. Anal. Appl., 344(2008), 1048–1056. B. Ahmad, S. K. Ntouyas, A. Alsaedi, A study of nonlinear fractional differential equations of arbitrary order with Riemann-Liouville type multistrip boundary conditions, Math. Probl. Eng., 2013(2013), Art. ID 320415, 9 pages. A. O. Akdemir, M. Gürbüz, E. Set, Integral inequalities for different kinds of convex functions involving Riemann-Liouville fractional integrals, Karaelmas Fen ve Müh. Derg., 7(2017), 140–144. A. Ali, G. Gulshan, R. Hussain, A. Latif, Generalized inequalities of the type of Hermite-Hadamard-Fejer with quasi-convex functions by way of k-fractional derivatives, J. Comput. Appl. Math., 22(2017), 1208–1219. T. Ali, M. A. Khan, Y. Khurshidi, Hermite-Hadamard inequality for fractional integrals via η-convex functions, Acta Math. Univ. Comen., 86(2017), 153–164. M. Alomari, M. Darus, Co-ordinated s-convex function in the first sense with some Hadamard-type inequalities. Int. J. Contemp. Math. Sci., 3(2008), 1557–1567. M. Alomari, M. Darus, On the Hadamard’s inequality for log-convex functions on the coordinates, J. Inequal. Appl., 2009 (2009), 1–13. M. Alomari, M. Darus, S. S. Dragomir, New inequalities of Hermite-Hadamard type for functions whose second derivatives absolute values are quasi-convex, Tamkang J. Math., 41(2010), 353–359. N. Alp, M. Z. Sarikaya, M. Kunt, et al., q-Hermite Hadamard inequalities and quantum estimates for midpoint type inequalities via convex and quasi-convex functions, J. King Saud Univ., Sci., (2016), doi:10.1016/j.jksus.2016.09.007. G. A. Anastassiou, Fractional Differentiation Inequalities, Springer, Berlin, 2009. A. Arslan, F. M. Atici, Discrete Hermite-Hadamard inequality and its applications, Appl. Anal., 10(2016), 366–377. L. Avazpoura, T. Allahviranloo, S. Islam, Uncertain Hermite-Hadamard inequality for functions with (s, m)-Godunova-Levin derivatives via fractional integral, J. Nonlinear Sci. Appl., 9(2016), 3333–3347. M. U. Awan, M. A. Noor, M. V. Mihai, K. I. Noor, Fractional Hermite-Hadamard inequalities for differentiable s-Godunova-Levin functions, Filomat, 30(2016), 3235–3241. R. Bai, F. Qi, B. Xi, Hermite-Hadamard type inequalities for the m- and (α, m)-logarithmically convex functions, Filomat, 27(2013), 1–7. Y. Baia, F. Qi, Some integral inequalities of the Hermite-Hadamard type for log-convex functions on co-ordinates, J. Nonlinear Sci. Appl., 9(2016), 5900–5908. T. Bakkyaraj, R. Sahadevan, Invariant analysis of nonlinear fractional ordinary differential equations with Riemann-Liouville fractional derivative, Nonlinear Dyn., 80(2015), 447–455. M. K. Bakula, M. E. Özdemir, J. Pec̆rić, Hadamard type inequalities for m-convex and (α, m)-convex functions, J. Inequal. Pure Appl. Math., 9(2008), Art. 96. D. Baleanu, J. A. T. Machado, A. C. J. Luo, eds., Fractional Dynamics and Control, Springer, New York, 2012. A. Barani, F. Malmir, Hermite-Hadamard inequality for geometrically quasiconvex functions on co-ordinates, Int. J. Nonlinear Anal. Appl., 8(2017), 47–60.

https://doi.org/10.1515/9783110523621-006

362 | Bibliography

[21] [22] [23]

[24]

[25] [26] [27] [28] [29] [30] [31] [32] [33] [34] [35] [36] [37] [38] [39] [40] [41] [42] [43]

M. Bessenyei, The Hermite-Hadamard inequality in Beckenbach’s setting, J. Math. Anal. Appl., 364(2010), 366–383. G. W. S. Blair, M. Reiner, The rheological law underlying the Nutting equation, Appl. Sci. Res., 2(1951), 225–234. K. Boukerrioua, T. Chiheb, B. Meftah, Fractional Hermite-Hadamard type inequalities for functions whose second derivative are (s, r)-convex in the second sense, Kragujev. J. Math., 40(2016), 172–191. H. Budak, M. Z. Sarikaya, On Hermite-Hadamard type inequalities for s-convex mappings via fractional integrals of a function with respect to another function, Fasc. Math., 57(2016), 25–36. J. Cal, J. Carcamob, L. Escauriaza, A general multidimensional Hermite-Hadamard type inequality, J. Math. Anal. Appl., 356(2009), 659–663. F. Chen, On Hermite-Hadamard type inequalities for s-convex functions on the coordinates via Riemann-Liouville fractional integrals, J. Appl. Math., 2014(2014), Art. ID 248710, 8 pages. F. Chen, Extensions of the Hermite-Hadamard inequality for harmonically convex functions via fractional integrals, Appl. Math. Comput., 268(2015), 121–128. F. Chen, Extensions of the Hermite-Hadamard inequality for convex functions via fractional integrals, J. Math. Inequal., 10(2016), 75–81. H. Chen, U. N. Katugampola, Hermite-Hadamard and Hermite-Hadamard-Fejér type inequalities for generalized fractional integrals, J. Math. Anal. Appl., 446(2017), 1274–1291. F. Chen, S. Wu, Several complementary inequalities to inequalities of Hermite-Hadamard type for s-convex functions, J. Nonlinear Sci. Appl., 9 (2016), 705–716. Q. Chen, J. Wang, F. Chen, Y. Zhang, Periodic solutions of quadratic Weyl fractional integral equations, Commun. Nonlinear Sci. Numer. Simul., 19(2014), 1945–1955. Y. Chu, M. A. Khan, T. U. Khan, T. Alib, Generalizations of Hermite-Hadamard type inequalities for MT-convex functions, J. Nonlinear Sci. Appl., 9 (2016), 4305–4316. L. Ciurdariu, Hermite-Hadamard type inequalities for fractional integrals operators, Appl. Math. Sci., 35(2017), 1745–1754. J. Deng, J. Wang, Fractional Hermite-Hadamard inequalities for (α, m)-logarithmically convex functions, J. Inequal. Appl., 364(2013), 1–11. K. Diethelm, The Analysis of Fractional Differential Equations, Lecture Notes in Mathematics, Springer, Berlin, 2010. X. Dong, J. Wang, Y. Zhou, On nonlocal problems for fractional differential equations in Banach spaces, Opusc. Math., 31(2011), 341–357. S. S. Dragomir, On Hadamard inequality for convex functions on the co-ordinates in a rectangle from the plane Taiwan. J. Math., 4(2001), 775–788. S. S. Dragomir, Hermite-Hadamard’s type inequalities for operator convex functions, Appl. Math. Comput., 218(2011), 766–772. S. S. Dragomir, Hermite-Hadamard’s type inequalities for convex functions of selfadjoint operators in Hilbert spaces, Linear Algebra Appl., 436(2012), 1503–1515. S. S. Dragomir, Inequalities of Hermite-Hadamard type for h-convex functions on linear spaces, Proyecciones, 34(2015), 323–341. S. S. Dragomir, n-Points inequalities of Hermite-Hadamard type for h-convex functions on linear spaces, Armen. J. Math., 8(2016), 38–57. S. S. Dragomir, Double integral inequalities of Hermite-Hadamard type for h-convex functions on linear spaces, Analysis, 37(2017), 13–22. S. Dragomir, Inequalities of Hermite-Hadamard type for functions of selfadjoint operators and matrices, J. Math. Inequal., 11(2017), 241–259.

Bibliography | 363

[44] [45] [46] [47] [48]

[49] [50] [51]

[52] [53] [54] [55] [56] [57] [58] [59]

[60] [61] [62] [63] [64] [65] [66]

G. Farid, M. Marwan, A. U. Rehman, New mean value theorems and generalization of Hadamard inequality via coordinated m-convex functions, J. Inequal. Appl., 2015(2015), 283. G. Farid, A. U. Rehman, M. Zahra, On Hadamard inequalities for relative convex functions via fractional integrals, Nonlinear Anal. Forum, 21(2016), 77–86. M. Fečkan, Y. Zhou, J. Wang, On the concept and existence of solution for impulsive fractional differential equations, Commun. Nonlinear Sci. Numer. Simul., 17(2012), 3050–3060. M. Fečkan, J. Wang, Y. Zhou, Controllability of fractional functional evolution equations of Sobolev type via characteristic solution operators, J. Optim. Theory Appl., 156(2013), 79–95. M. Fečkan, J. Wang, Y. Zhou, Response to “Comments on the concept of existence of solution for impulsive fractional differential equations [Commun Nonlinear Sci Numer Simul 2014;19:401-3.]”, Commun. Nonlinear Sci. Numer. Simul., 19(2014), 4213–4215. M. Fečkan, J. Wang, M. Pospišil, Fractional Order Equations and Inclusions, Walter de Gruyter GmbH, Berlin/Boston, 2017. Z. Gao, X. Yu, J. Wang, Nonlocal problems for Langevin-type differential equations with two fractional-order derivatives, Bound. Value Probl., 2016(2016), 52. Z. Gao, M. Li, J. Wang, On some fractional Hermite-Hadamard inequalities via s-convex and s-Godunova-Levin functions and their applications, Bol. Soc. Mat. Mexicana, 23(2017), 691–711. B. Gavrea, A Hermite-Hadamard type inequality with applications to the estimation of moments of continuous random variables, Appl. Math. Comput., 254(2015), 92–98. A. G. Ghazanfari, A. Barani, Some Hermite-Hadamard type inequalities for the product of two operator preinvex functions, Banach J. Math. Anal., 9(2015), 9–20. Z. Q. Gong, D. L. Qian, P. Guo, C. P. Li, On the Hadamard type fractional differential system, Fractional Dynamics and Control, Springer, Berlin, 2011, Chapter 13, 159–171. M. A. Hanson, On sufficiency of the Kuhn-Tucker conditions, J. Math. Anal. Appl., 80(1981), 545–550. O. Hölder, Ueber einen Mittelwerthsatz Nachr. Ges. Wiss. Göttingen, Göttingen 1889, 38–47. K. C. Hsu, Refinements of Hermite-Hadamard type inequalities for differentiable co-ordinated convex functions and applications, Taiwan. J. Math., 19(2015), 133–157. H. Hudzik, L. Maligranda, Some remarks on s-convex functions, Aequ. Math., 48(1994), 100–111. M. Iqbal, M. I. Bhatti, K. Nazeer, Generalization of inequalities analogous to Hermite-Hadamard inequality via fractional integrals, Bull. Korean Math. Soc., 52(2015), 707–716. I. Işcan, New general integral inequalities for quasi-geometrically convex functions via fractional integrals, J. Inequal. Appl., 2013(2013), 491. İ. İşcan, Hermite-Hadamard’s inequalities for preinvex functions via fractional integrals and related fractional inequalities, Am. J. Math. Anal., 3(2013), 33–38. İ. İşcan, Generalization of different type integral inequalities for s-convex functions via fractional integrals, Appl. Anal., 93(2014), 1846–1862. İ. İşcan, Generalization of different type integral inequalities for (α, m)-convex functions via fractional integrals, Appl. Math. Sci., 9(2015), 2925–2939. İ. İşcan, Hermite-Hadamard type inequalities for p-convex functions, Int. J. Anal. Appl., 11(2016), 137–145. İ. İşcan, S. Wu, Hermite-Hadamard type inequalities for harmonically convex functions via fractional integrals, Appl. Math. Comput., 238(2014), 237–244. J. L. W. V. Jensen, On konvexe funktioner og uligheder mellem middlvaerdier, Nyt Tidsskr. Math. B, 16(1905), 49–69.

364 | Bibliography

[67] [68] [69] [70]

[71] [72] [73] [74]

[75] [76] [77] [78] [79] [80] [81] [82]

[83]

[84] [85] [86] [87] [88]

M. Jleli, B. Samet, On Hermite-Hadamard type inequalities via fractional integrals of a function with respect to another function, J. Nonlinear Sci. Appl., 9(2016), 1252–1260. M. Jleli, D.Óregan, B. Samet, On Hermite-Hadamard type inequalities via generalized fractional integrals, Turk. J. Math., 40(2016), 1221–1230. M. Jleli, D. O’Regan, B. Samet, Some fractional integral inequalities involving m-convex functions, Aequ. Math., 91(2017), 479–490. M. A. Khan, T. Ali, S. S. Dragomir, et al., Hermite-Hadamard type inequalities for conformable fractional integrals, Rev. R. Acad. Cienc. Exactas Fís. Nat., Ser. A Mat., (2017), https://doi.org/10.1007/s13398-017-0408-5. A. A. Kilbas, H. M. Srivastava, J. J. Trujillo, Theory and Applications of Fractional Differential Equations, Elsevier Science B.V., Amsterdam, 2006. V. Kiryakova, Generalized Fractional Calculus and Applications, Pitman Res. Notes Math. Ser., vol. 301, Longman, New York, 1994. M. Kunt, İ. İşcan. Hermite-Hadamard type inequalities for harmonically (α, m)-convex functions via fractional integrals, Konuralp J. Math., 5(2017), 201–213. M. Kunt, İ. İşcan, N. Alp, et al., (p, q)-Hermite-Hadamard inequalities and (p, q)-estimates for midpoint type inequalities via convex and quasi-convex functions, Rev. R. Acad. Cienc. Exactas Fís. Nat., Ser. A Mat., (2017), https://doi.org/10.1007/s13398-017-0402-y. M. Kunta, İ. İşcan, Hermite-Hadamard type inequalities for p-convex functions via fractional integrals, Moroccan J. Pure Appl. Anal., 3(2017), 22–35. V. Lakshmikantham, S. Leela, J. V. Devi, Theory of Fractional Dynamic Systems, Cambridge Scientific Publishers, Cambridge, 2009. M. Latif, Inequalities of Hermite-Hadamard type for functions whose derivatives in absolute value are convex with applications, Arab J. Math. Sci., 21(2015), 84–97. M. A. Latif, M. Alomari, Hadamard-type inequalities for product two convex functions on the co-ordinates, Int. Math. Forum, 4(2009), 2327–2338. M. A. Latif, M. Alomari, On the Hadamard-type inequalities for h-convex functions on the co-ordinates. Int. J. Math. Anal., 3(2009), 1645–1656. M. A. Latif, S. S. Dragomir, On Hermite-Hadamard type integral inequalities for n-times differentiable log-preinvex functions, Filomat, 29(2015), 1651–1661. M. A. Latif, S. Hussain, New inequalities of Ostrowski type for co-ordinated convex functions via fractional integrals, J. Fract. Calc. Appl., 2(2012), 1–15. M. A. Latif, S. S. Dragomir, E. Momoniat, Some φ-analogues of Hermite-Hadamard inequality for s-convex functions in the second sense and related estimates, J. Math., 48(2016), 147–166. M. A. Latif, S. S. Dragomir, E. Momoniat, Some q-analogues of Hermite-Hadamard inequality of functions of two variables on finite rectangles in the plane, J. King Saud Univ., Sci., 29(2017), 263–273. C. P. Li, W. H. Deng, Remarks on fractional derivatives, Appl. Math. Comput., 187(2007), 777–784. C. K. Li, C. P. Li, On defining the distributions δk and (δ󸀠 )k by fractional derivatives, Appl. Math. Comput., 246(2014), 502–513. C. K. Li, C. P. Li, Remarks on fractional derivatives of distributions, Tbilisi Math. J., 10(2017), 1–18. C. P. Li, Y. Ma, Fractional dynamical system and its linearization theorem, Nonlinear Dyn., 71(2013), 621–633. C. P. Li, S. Sarwar, Existence and continuation of solutions for Caputo type fractional differential equations, Electron. J. Differ. Equ., 2016(2016), No.207, 1–14.

Bibliography | 365

[89] [90] [91] [92] [93] [94] [95] [96]

[97]

[98] [99] [100] [101] [102] [103] [104] [105] [106] [107] [108] [109] [110] [111] [112]

M. Li, J. Wang, Analysis of nonlinear Hadamard fractional differential equations via properties of Mittag-Leffler functions, J. Appl. Math. Comput., 51(2016), 487–508. M. Li, J. Wang, Finite time stability of fractional delay differential equations, Appl. Math. Lett., 64(2017), 170–176. C. P. Li, F. H. Zeng, Numerical Methods for Fractional Calculus, CRC Press, Boca Raton, 2015. C. P. Li, F. R. Zhang, J. Kurths, F. H. Zeng, Equivalent system for a multiple-rational-order fractional differential system, Phil. Trans. R. Soc. A., 371(1990), 20120156 (30 pages), 2013. C. P. Li, X. H. Dao, P. Guo, Fractional derivatives in complex plane, Nonlinear Anal., 71(2009), 1857–69. C. P. Li, Z. Q. Gong, D. L. Qian, Y. Q. Chen, On the bound of the Lyapunov exponents for the fractional differential systems, Chaos, 20(2010), 013127. M. Li, J. Wang, W. Wei, Some fractional Hermite-Hadamard inequalities for convex and Godunova-Levin functions, Facta Univ., Ser. Math. Inform., 30(2015), 195–208. Y. Liao, J. Deng, J. Wang, Riemann-Liouville fractional Hermite-Hadamard inequalities. Part I: for once differentiable geometric-arithmetically s-convex functions, J. Inequal. Appl., 443(2013), 1–13. Y. Liao, J. Deng, J. Wang, Riemann-Liouville fractional Hermite-Hadamard inequalities. Part II: for twice differentiable geometric-arithmetically s-convex functions, J. Inequal. Appl., 517(2013), 1–13. Y. Liao, J. Deng, J. Wang, On the fractional Hermite-Hadamard type inequalities for (α, m)-logarithmically convex functions, Filomat, 29(2015), 1565–1580. Z. Lin, J. Wang, New Riemma-Liouville fractional Hermite-Hadamard inequalities via two kinds of convex functions, J. Interdiscip. Math., 20(2017), 357–382. Z. Lin, J. Wang, W. Wei, Fractional Hermite-Hadamard inequalities through r-convex functions via power means, Facta Univ., Ser. Math. Inform., 30(2015), 129–145. S. Liu, J. Wang, Hermite-Hadamard type fractional integral inequalities for geometric–geometric convex functions, Matematiche, LXX(2015), 3–20. S. Liu, J. Wang, Fractional order iterative learning control with randomly varying trial lengths, J. Franklin Inst., 354(2017), 967–992. F. Liu, P. Zhuang, Q. Liu, Numerical Methods for Fractional Partial Differential Equations and Their Applications, Science Press, Beijing, 2015. W. Liu, W. Wen, J. Park, Hermite-Hadamard type inequalities for MT-convex functions via classical integrals and fractional integrals, J. Nonlinear Sci. Appl., 9(2016), 766–777. Z. Luo, J. Wang, Fractional type Hermite-Hadamard inequalities for convex and AG(Log)-convex functions, Facta Univ., Ser. Math. Inform., 30(2015), 649–662. Z. Luo, W. Wei, J. Wang, Robust convergence analysis of iterative learning control for impulsive Riemann-Liouville fractional-order systems, Adv. Differ. Equ., 2016(2016), 256. L. Lv, J. Wang, W. Wei, Existence and uniqueness results for fractional differential equations with boundary value conditions, Opusc. Math., 31(2011), 629–643. L. Ma, C. P. Li, On Hadamard fractional calculus, Fractals, 25(2017), 1–16. L. Ma, C. P. Li, On finite part integrals and Hadamard-type fractional derivatives, J. Comput. Nonlinear Dyn., (2017), doi: 10.1115/1.4037930. J. Makó, A. Házy, On approximate Hermite-Hadamard type inequalities, J. Convex Anal., 24(2017), 349–363. M. W. Michalski, Derivatives of Noninteger Order and Their Applications, Dissertationes Mathematicae, vol. CCCXXVIII, Inst. Math., Polish Acad. Sci., Warsaw, 1993. M. V. Mihai, F. C. Mitroi, Hermite-Hadamard type inequalities obtained via Riemann-Liouville fractional calculus, Acta Math. Univ. Comen., 83(2014), 209–215.

366 | Bibliography

[113] K. S. Miller, B. Ross, An Introduction to the Fractional Calculus and Differential Equations, John Wiley, New York, 1993. [114] S. N. Mishra, P. K. Das, G. C. Nayak, Some results associated with Hermite-Hadamard type inequalities in invariant harmonic convex Set, Glob. J. Pure Appl. Math., 13(2017), 1241–1248. [115] D. S. Mitrinović, I. B. Lacković, Hermite and convexity, Aequ. Math., 28(1985), 229–232. [116] S. Mubeen, S. Iqbal, Grüss type integral inequalities for generalized Riemann-Liouville k-fractional integrals, J. Inequal. Appl., 2016(2016), 109. [117] C. P. Niculescu, Convexity according to the geometric mean, Math. Inequal. Appl., 3(2000), 155–167. [118] C. P. Niculescu, The Hermite-Hadamard inequality for log-convex functions, Nonlinear Anal., 75(2012), 662–669. [119] M. A. Noor, K. I. Noor, M. U. Awan, S. Khan, Fractional Hermite-Hadamard inequalities for some new classes of Godunova-Levin functions, Appl. Math. Inf. Sci., 8(2014), 2865–2872. [120] M. A. Noor, K. I. Noor, M. U. Awan, Generalized convexity and integral inequalities, Appl. Math. Inf. Sci., 9(2015), 233–243. [121] M. A. Noor, K. I. Noor, M. U. Awan, Some quantum estimates for Hermite-Hadamard inequalities, Appl. Math. Comput., 251(2015), 675–679. [122] M. A. Noor, K. I. Noor, M. U. Awan, et al., Hermite-Hadamard type inequalities for differentiable h-preinvex functions, Arab. J. Math., 4(2015), 63–76. [123] M. A. Noor, K. I. Noor, M. U. Awan, New fractional estimates of Hermite-Hadamard inequalities and applications to means, Stud. Univ. Babeş–Bolyai, Math., 61(2016), 3–15. [124] M. A. Noor, K. I. Noor, M. U. Awan, Fractional Hermite-Hadamard inequalities for two kinds of s-preinvex functions, Nonlinear Sci. Lett. A, 8(2017), 11–24. [125] M. A. Noor, K. I. Noor, S. Iftikhar, et al., Hermite-hadamard inequalities for co-ordinated harmonic convex functions, U.P.B. Sci. Bull. Series A, 79(2017), 25–34. [126] M. Nowicka, A. Witkowski, A refinement of the right-hand side of the Hermite-Hadamard inequality for simplices, Aequ. Math., 91(2017), 121–128. [127] P. G. Nutting, A new general law of deformation, J. Franklin Inst., 191(1921), 679–685. [128] P. G. Nutting, A general stress-strain-time formula, J. Franklin Inst., 235(1943), 513–524. [129] M. E. Ödemir, M. Avci, E. Set, On some inequalities of Hermite-Hadamard type via m-convexity, Appl. Math. Lett., 23(2010), 1065–1070. [130] M. E. Ödemir, M. Avci, H. Kavurmaci, Hermite-Hadamard-type inequalities via (α, m)-convexity, Comput. Math. Appl., 61(2011), 2614–2620. [131] M. E. Özdemir, Some inequalities for the s-Godunova-Levin type functions, Math. Sci., 9(2015), 27–32. [132] M. E. Özdemir, E. Set, M. Z. Sarikaya, New some Hadamard’s type inequalities for co-ordinated m-convex and (α, m)-convex functions, Hacet. J. Math. Stat., 40(2011), 219–229. [133] J. Park, Fractional Hermite-Hadamard-like type inequalities for convex functions, Int. J. Math. Anal., 9(2015) 1415–1429. [134] J. Park, On some integral inequalities for twice differentiable quasi-convex and convex functions via fractional integrals, Appl. Math. Sci., 9(2015), 3057–3069. [135] J. Park, Hermite-Hadamard-like type inequalities for s-convex functions and s-Godunova-Levin functions of two Kinds, Appl. Math. Sci., 9(2015), 3431–3447. [136] J. Park, Generalized simpson-like type integral inequalities for twice differentiable convex functions via Riemann-Liouville integrals, Int. J. Math. Anal., 9(2015), 767–777. [137] Z. Pavić, Improvements of the Hermite-Hadamard inequality, J. Inequal. Appl., 2015(2015), 222. [138] C. E. M. Pearce, J. E. Pec̆arić, Inequalities for differentiable mappings with application to special means and quadrature formula, Appl. Math. Lett., 13(2000), 51–55.

Bibliography | 367

[139] C. E. M. Pearce, J. Pec̆arić, V. S̆imić, Stolarsky means and Hadamard’s inequality, J. Math. Anal. Appl., 220(1998), 99–109. [140] K. Pearson, E. S. Pearson, N. L. Johnson, Tables of the Incomplete Beta Function, Cambridge University Press, Cambridge, 2nd edn, 1968. [141] J. E. Pec̆arić, F. Proschan, Y. L. Tong, Convex Functions, Partial Orderings, and Statistical Applications, Academic Press Inc., San Diego, 1992. [142] S. Peng, J. Wang, Cauchy problem for nonlinear fractional differential equations with positive constant coefficient, J. Appl. Math. Comput., 51(2016), 341–351. [143] S. Peng, W. Wei, J. Wang, On the Hermite-Hadamard inequalities for convex functions via Hadamard fractional integrals, Facta Univ., Ser. Math. Inform., 29(2014), 55–75. [144] I. Podlubny, Fractional Differential Equations, Academic Press, San Diego, 1999. [145] A. P. Prudnikov, Y. A. Brychkov, O. I. Marichev, Integral and Series, in: Elementary Functions, vol. 1, Nauka, Moscow, 1981. [146] S. Qaisar, M. Iqbal, M. Muddassar, New Hermite-Hadamard inequalities for preinvex functions via fractional integrals, J. Comput. Anal. Appl., 20(2016), 1318–1328. [147] K. Qiu, J. Wang, A fractional integral identity and its application to fractional Hermite-Hadamard type inequalities, J. Interdiscip. Math., 21(2018), 1–16. [148] R. A. Santnoianu, Improved GA-convexity inequalities, J. Inequal. Pure Appl. Math., 3(2002), Article 82. [149] M. Z. Sarikaya, On the Hermite-Hadamard-type inequalities for co-ordinated convex function via fractional integrals, Integral Transforms Spec. Funct., 25(2014), 134–147. [150] M. Z. Sarikaya, N. Aktan, On the generalization of some integral inequalities and their applications, Math. Comput. Model., 54(2011), 2175–2182. [151] M. Z. Sarikaya, H. Budak, Some generalization of integral inequalities for twice differentiable mappings involving fractional integrals, Acta Univ. Sapientiae Math., 7(2015), 251–264. [152] M. Z. Sarikaya, H. Budak, On generalized Hermite-Hadamard inequality for generalized convex function, RGMIA Res. Rep. Collect., 18(2015), 1–15. [153] M. Z. Sarikaya, H. Budak, Generalized Hermite-Hadamard type integral inequalities for fractional integrals, Filomat, 30(2016), 1315–1326. [154] M. Z. Sarikaya, S. Erden, New weighted integral inequalities for twice differentiable convex functions, Kragujev. J. Math., 40(2016), 15–33. [155] M. Z. Sarikaya, M. E. Kiris, Some new inequalities of Hermite-Hadamard type for s-convex functions, Miskolc Math. Notes, 16(2015), 491–501. [156] M. Z. Sarikaya, H. Yaldiz, On the Hadamard’s type inequalities for L-Lipschitzian mapping, Konuralp J. Math., 2(2013) 33–40, 2010, arXiv:1005.4138. [157] M. Z. Sarikaya, H. Yildirim. On Hermite-Hadamard type inequalities for Riemann-Liouville fractional integrals, Miskolc Math. Notes, 17(2016), 1049–1059. [158] M. Z. Sarikaya, E. Set, M. E. Ozdemir, S. S. Dragomir, New some Hadamard type inequalities for co-ordinated convex functions, Tamsui Oxf. J. Inf. Math. Sci., 28(2012), 137–152. [159] M. Z. Sarikaya, E. Set, H. Yaldiz, N. Başak, Hermite-Hadamard’s inequalities for fractional integrals and related fractional inequalities, Math. Comput. Model., 57(2013), 2403–2407. [160] M. Z. Sarikaya, Z. Dahmani, M. E. Kiris, F. Ahmad, On the k-Riemann-Liouville fractional integral and applications, Int. J. Stat. Math., 1(2014), 33–43. [161] M. Z. Sarikaya, H. Filiz, M. E. Kiris, On some generalized integral inequalities for Riemann-Liouville fractional integrals, Filomat, 29(2015), 1307–1314. [162] E. Set, New inequalities of Ostrowski type for mappings whose derivatives are s-convex in the second sense via fractional integrals, Comput. Math. Appl., 63(2012), 1147–1154. [163] E. Set, B. Çelik, Fractional Hermite-Hadamard type inequalities for quasi-convex functions, Ordu Univ. J. Sci. Tech., 6(2016), 137–149.

368 | Bibliography

[164] E. Set, İ. İşcan, M. Z. Sarikaya, et al., On new inequalities of Hermite-Hadamard-Fejér type for convex functions via fractional integrals, Appl. Math. Comput., 259(2015), 875–881. [165] E. Set, İ. İşcan, H. H. Kara, Hermite-Hadamard-Fejer type inequalities for s-convex function in the second sense via fractional integrals, Filomat, 30(2016), 3131–3138. [166] E. Set, A. Gözpinar, J. Choi, Hermite-Hadamard type inequalities for twice differentiable m-convex functions via conformable fractional integrals, Far East J. Math. Sci., 101(2017), 873. [167] E. Set, M. A. Noor, M. U. Awan, et al., Generalized Hermite-Hadamard type inequalities involving fractional integral operators, J. Inequal. Appl., 2017(2017), 169. [168] E. Set, M. E. Özdemir, E. A. Alan, On new fractional Hermite-Hadamard type inequalities for n-time differentiable quasi-convex functions and P-functions, AIP Conf. Proc., 1833(2017), 020031. [169] E. Set, M. E. Özdemir, M. Z. Sarikaya, F. Karakoç, Hermite-Hadamard type inequalities for (α, m)-convex functions via fractional integrals, Moroccan J. Pure Appl. Anal., 3(2017), 15–21. [170] E. Set, M. Z. Sarikaya, N. Uygun, On new inequalities of Simpson’s type for generalized quasi convex functions, Adv. Inequal. Appl., 2017 (2017), Article ID 3. [171] Y. Shuang, H. P. Yin, F. Qi, Hermite-Hadamard type integral inequalities for geometric-arithmetically s-convex functions, Analysis, 33(2013), 197–208. [172] Z. Z. Sun, G. H. Gao, The Finite Difference Methods for Fractional Differential Equations, Chinese Scientific Press, Beijing, 2015. [173] V. E. Tarasov, Fractional Dynamics: Application of Fractional Calculus to Dynamics of Particles, Fields and Media, Springer, HEP, Beijing, 2011. [174] Y. Tian, J. Wang, On some Hermite-Hadamard type inequalities for convex functions via Hadamard fractional integrals, Prog. Fract. Differ. Appl., 2(2015), 103–110. [175] M. Tomar, E. Set, S. Maden, Hermite-Hadamard type inequalities for log-convex stochastic processes, J. New Theory, 2(2015), 23–32. [176] K. Tseng, S. Hwang, K. Hsu, Hadamard-type and Bullen-type inequalities for Lipschitzian functions and their applications, Comput. Math. Appl., 64(2012), 651–660. [177] M. Tunç, On new inequalities for h-convex functions via Riemmann-Liouville fractional integration, Filomat, 4(2013), 559–565. [178] M. Tunç, Some Hadamard-like inequalities via convex and s-convex functions and their applications for special means, Mediterr. J. Math., 11(2014), 1047–1059. [179] S. H. Wang, Hermite-Hadamard type inequalities for operator convex functions on the co-ordinates, J. Nonlinear Sci. Appl., 10(2017), 1116–1125. [180] J. Wang, M. Fečkan, Ulam-Hyers-Rassias stability for semilinear equations, Discontinuity, Nonlinearity, and Complexity, 3(2014), 379–388. [181] J. Wang, M. Fečkan, Practical Ulam-Hyers-Rassias stability for nonlinear equations, Math. Bohem., 142(2017), 47–56. [182] J. Wang, A. G. Ibrahim, Existence and controllability results for nonlocal fractional impulsive differential inclusions in Banach spaces, J. Funct. Spaces Appl., 2013(2013), 1–16. [183] J. Wang, X. Li, Eα -Ulam type stability of fractional order ordinary differential equations, Appl. Math. Comput., 45(2014), 449–459. [184] S. Wang, F. Qi, Hermite-Hadamard type inequalities for s-convex functions via Riemann-Liouville fractional integrals, J. Comput. Anal. Appl., 22(2017), 1124–1134. [185] J. Wang, W. Wei, A class of nonlocal impulsive problems for integrodifferential equations in Banach spaces, Results Math., 58(2010), 379–397. [186] J. Wang, Y. Zhang, Ulam-Hyers-Mittag-Leffler stability of fractional order delay differential equations, Optimization, 63(2014), 1181–1190.

Bibliography | 369

[187] J. Wang, Y. Zhang, On the concept and existence of solutions for fractional impulsive systems with Hadamard derivatives, Appl. Math. Lett., 39(2015), 85–90. [188] J. Wang, Y. Zhou, A class of fractional evolution equations and optimal controls, Nonlinear Anal., Real World Appl., 12(2011), 262–272. [189] J. Wang, Y. Zhou, Existence and controllability results for fractional semilinear differential inclusions, Nonlinear Anal., Real World Appl., 12(2011), 3642–3653. [190] J. Wang, Y. Zhou, Analysis of nonlinear fractional control systems in Banach spaces, Nonlinear Anal., 74(2011), 5929–5942. [191] J. Wang, Y. Zhou, Mittag-Leffer-Ulam stabilities of fractional evolution equations, Appl. Math. Lett., 25(2012), 723–728. [192] J. Wang, Y. Zhou, Complete controllability of fractional evolution systems, Commun. Nonlinear Sci. Numer. Simul., 17(2012), 4346–4355. [193] J. Wang, X. Xiang, W. Wei, A class of nonlinear integrodifferential impulsive periodic systems of mixed type and optimal controls on Banach space, J. Appl. Math. Comput., 34(2010), 465–484. [194] J. Wang, M. Fečkan, Y. Zhou, On the new concept of solutions and existence results for impulsive fractional evolution equations, Dyn. Partial Differ. Equ., 8(2011), 345–361. [195] J. Wang, L. Lv, Y. Zhou, Ulam stability and data dependence for fractional differential equations with Caputo derivative, Electron. J. Qual. Theory Differ. Equ., 63(2011), 1–10. [196] J. Wang, Y. Zhou, M. Fečkan, Alternative results and robustness for fractional evolution equations with periodic boundary conditions, Electron. J. Qual. Theory Differ. Equ., 97(2011), 1–15. [197] J. Wang, Y. Zhou, W. Wei, A class of fractional delay nonlinear integrodifferential controlled systems in Banach spaces, Commun. Nonlinear Sci. Numer. Simul., 16(2011), 4049–4059. [198] J. Wang, Y. Zhou, W. Wei, Impulsive fractional evolution equations and optimal controls in infinite dimensional spaces, Topol. Methods Nonlinear Anal., 38(2011), 17–43. [199] J. Wang, X. Dong, Y. Zhou, Existence, attractiveness and stability of solutions for quadratic Urysohn fractional integral equations, Commun. Nonlinear Sci. Numer. Simul., 17(2012), 545–554. [200] J. Wang, X. Dong, Y. Zhou, Analysis of nonlinear integral equations with Erdelyi-Kober fractional operator, Commun. Nonlinear Sci. Numer. Simul., 17(2012), 3129–3139. [201] J. Wang, Z. Fan, Y. Zhou, Nonlocal controllability of semilinear dynamic systems with fractional derivative in Banach spaces, J. Optim. Theory Appl., 154(2012), 292–302. [202] J. Wang, M. Fečkan, Y. Zhou, Ulam’s type stability of impulsive ordinary differential equations, J. Math. Anal. Appl., 395(2012), 258–264. [203] J. Wang, L. Lv, Y. Zhou, New concepts and results in stability of fractional differential equations, Commun. Nonlinear Sci. Numer. Simul., 17(2012), 2530–2538. [204] J. Wang, W. Wei, M. Fečkan, Nonlocal Cauchy problems for fractional evolution equations involving Volterra-Fredholm type integral operators, Miskolc Math. Notes, 13(2012), 127–147. [205] J. Wang, Y. Zhou, M. Fečkan, Nonlinear impulsive problems for fractional differential equations, Comput. Math. Appl., 64(2012), 3389–3405. [206] J. Wang, Y. Zhou, M. Fečkan, On recent developments in the theory of boundary value problems for impulsive fractional differential equations, Comput. Math. Appl., 64(2012), 3008–3020. [207] J. Wang, Y. Zhou, M. Medved’, On the solvability and optimal controls of fractional integrodifferential evolution systems with infinite delay, J. Optim. Theory Appl., 152(2012), 31–50. [208] J. Wang, Y. Zhou, M. Medved’, Picard and weakly Picard operators technique for nonlinear differential equations in Banach spaces, J. Math. Anal. Appl., 389(2012), 261–274.

370 | Bibliography

[209] J. Wang, Y. Zhou, W. Wei, Optimal feedback control for semilinear fractional evolution equations in Banach spaces, Syst. Control Lett., 61(2012), 472–476. [210] J. Wang, Y. Zhou, W. Wei, Fractional Schrödinger equations with potential and optimal controls, Nonlinear Anal., Real World Appl., 13(2012), 2755–2766. [211] J. Wang, J. Deng, M. Fečkan, Hermite-Hadamard type inequalities for r-convex functions via Riemann-Liouville fractional integrals, Ukr. Math. J., 65(2013), 193–211. [212] J. Wang, M. Fečkan, Y. Zhou, Nonexistence of periodic solutions and asymptotically periodic solutions for fractional differential equations, Commun. Nonlinear Sci. Numer. Simul., 18(2013), 246–256. [213] J. Wang, M. Fečkan, Y. Zhou, Fractional order iterative functional differential equations with parameter, Appl. Math. Model., 37(2013), 6055–6067. [214] J. Wang, M. Fečkan, Y. Zhou, Presentation of solutions of impulsive fractional Langevin equations and existence results, Eur. Phys. J. Spec. Top., 222(2013), 1855–1872. [215] J. Wang, M. Fečkan, Y. Zhou, Relaxed controls for nonlinear fractional impulsive evolution equations, J. Optim. Theory Appl., 156(2013), 13–32. [216] J. Wang, X. Li, M. Fečkan, Y. Zhou, Hermite-Hadamard-type inequalities for Riemann-Liouville fractional integrals via two kinds of convexity, Appl. Anal., 92(2013), 2241–2253. [217] J. Wang, X. Li, C. Zhu, Refinements of Hermite-Hadamard type inequalities involving fractional integrals, Bull. Belg. Math. Soc. Simon Stevin, 20(2013), 655–666. [218] J. Wang, Y. Zhou, M. Fečkan, Abstract Cauchy problem for fractional differential equations, Nonlinear Dyn., 71(2013), 685–700. [219] J. Wang, Ch. Zhu, M. Fečkan, Existence, uniqueness and limit property of solutions to quadratic Erdelyi-Kober type integral equations of fractional order, Cent. Eur. J. Phys., 11(2013), 779–791. [220] J. Wang, C. Zhu, Y. Zhou, New generalized Hermite-Hadamard type inequalities and applications to special means, J. Inequal. Appl., 325(2013), 1–15. [221] J. Wang, J. Deng, M. Fečkan, Exploring s-e-condition and applications to some Ostrowski type inequalities via Hadamard fractional integrals, Math. Slovaca, 64(2014), 1381–1396. [222] J. Wang, M. Fečkan, Y. Zhou, Weakly Picard operators method for modified fractional iterative functional differential equations, Fixed Point Theory, 15(2014), 297–310. [223] J. Wang, Y. Zhou, M. Fečkan, On the nonlocal Cauchy problem for semilinear fractional order evolution equations, Cent. Eur. J. Math., 12(2014), 911–922. [224] J. Wang, Y. Zhou, Z. Lin, On a new class of impulsive fractional differential equations, Appl. Math. Comput., 242(2014), 649–657. [225] J. Wang, Ch. Zhu, M. Fečkan, Analysis of Abel-type nonlinear integral equations with weakly singular kernels, Bound. Value Probl., 2014(2014), 1–20. [226] J. Wang, Ch. Zhu, M. Fečkan, Solvability of fully nonlinear functional equations involving Erdelyi-Kober fractional integrals on the unbounded interval, Optimization, 63(2014), 1235–1248. [227] J. Wang, A. G. Ibrahim, M. Fečkan, Nonlocal impulsive fractional differential inclusions with fractional sectorial operators on Banach spaces, Appl. Math. Comput., 257(2015), 103–118. [228] J. Wang, A. G. Ibrahim, M. Fečkan, Nonlocal Cauchy problems for semilinear differential inclusions with fractional order in Banach spaces, Commun. Nonlinear Sci. Numer. Simul., 27(2015), 281–293. [229] J. Wang, M. Fečkan, Y. Zhou, A survey on impulsive fractional differential equations, Fract. Calc. Appl. Anal., 19(2016), 806–831. [230] J. Wang, A. G. Ibrahim, M. Fečkan, Differential inclusions of arbitrary fractional order with anti-periodic conditions in Banach spaces, Electron. J. Qual. Theory Differ. Equ., 2016(2016), 1–22.

Bibliography | 371

[231] W. Wang, İ İscan, H. Zhou, Fractional integral inequalities of hermite-hadamard type for m−HH convex functions with applications, Adv. Stud. Contemp. Math., 26(2016), 501–512. [232] J. Wang, X. Li, Y. Zhou, Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals via s-convex functions and Applications to special means, Filomat, 30(2016), 1143–1150. [233] Y. Wang, B. Y. Xi, F. Qi, Integral inequalities of Hermite-Hadamard type for functions whose derivatives are strongly α-preinvex, Acta Math. Acad. Paedagog. Nyházi., 32(2016), 79–87. [234] J. Wang, Q. Chen, C. Zhu, Periodic solutions of Weyl fractional order integral systems, Math. Methods Appl. Sci., 40(2017), 137–145. [235] J. Wang, M. Fečkan, Y. Zhou, Center stable manifold for planar fractional damped equations, Appl. Math. Comput., 296(2017), 257–269. [236] T. Weir, B. Mond, Preinvex functions in multiple objective optimization, J. Math. Anal. Appl., 136(1998), 29–38. [237] B. Xi, R. Bai, F. Qi, Hermite-Hadamard type inequalities for the m-and (α, m)-geometrically convex functions, Aequ. Math., 84(2012), 261–269. [238] B. Y. Xi, T. Y. Zhang, F. Qi, Some inequalities of Hermite-Hadamard type for m-harmonic-arithmetically convex functions, Science Asia, 41(2015), 357–361. [239] R. Xiang, Refinements of Hermite-Hadamard type inequalities for convex functions via fractional integrals, J. Appl. Math. Inform., 33(2015), 119–125. [240] Z. Xiao, Z. Zhang, Y. Wu, On weighted Hermite-Hadamard inequalities, Appl. Math. Comput., 218(2011), 1147–1152. [241] H. Xiao, Y. T. Ma, C. P. Li, Chaotic vibration in fractional maps, J. Vib. Control, 20(2014), 964–972. [242] C. Yildiz, M. E. Ozdemir, On generalized inequalities of Hermite-Hadamard type for convex functions, Int. J. Anal. Appl., 14(2017), 52–63. [243] Y. Zhang, J. Wang, On some new Hermite-Hadamard inequalities involving Riemann-Liouville fractional integrals, J. Inequal. Appl., 220(2013), 1–27. [244] Y. Zhang, J. Wang, Nonlocal Cauchy problems for a class of implicit impulsive fractional relaxation differential systems, J. Appl. Math. Comput., 52(2016), 323–343. [245] Y. Zhang, J. Wang, Existence and finite-time stability results for impulsive fractional differential equations with maxima, J. Appl. Math. Comput., 51(2016), 67–79. [246] Z. Zhang, W. Wei, J. Wang, Generalization of Hermite-Hadamard inequalities involving Hadamard fractional integrals, Filomat, 29(2015), 1515–1524. [247] S. Zheng, T. S. Du, S. S. Zhao, et al., New Hermite-Hadamard inequalities for twice differentiable φ-MT-preinvex functions, J. Nonlinear Sci. Appl., 9(2016), 5648–5660. [248] Y. Zhou, Basic Theory of Fractional Differential Equations, World Scientific, Singapore, 2014. [249] Y. Zhou, Fractional Evolution Equations and Inclusions: Analysis and Control, Academic Press, San Diego, 2016. [250] Y. Zhou, J. Wang, L. Zhang, Basic Theory of Fractional Differential Equations, 2nd edn, World Scientific, Singapore, 2016. [251] C. Zhu, M. Fečkan, J. Wang, Fractional integral inequalities for differentiable convex mappings and applications to special means and a midpoint formula, J. Appl. Math. Stat. Inform., 8(2012), 21–28.

About the authors JinRong Wang is a Professor of Mathematics at the Department of Mathematics in the School of Mathematics and Statistics at the Guizhou University in Guiyang, P. R. China. He obtained his Ph.D. degree from the same university. His research field is applied mathematics, including fractional Hermite-Hadamard inequality, fractional differential equations and inclusions, impulsive differential equations and inclusions, impulsive delay differential equations, iterative learning controls and multi-agent systems. He is included in Highly Cited Researcher in Mathematics from Clarivate Analytics. Michal Fečkan is a Professor of Mathematics at the Department of Mathematical Analysis and Numerical Mathematics on the Faculty of Mathematics, Physics and Informatics at the Comenius University in Bratislava, Slovak Republic. He obtained his Ph.D. (mathematics) from the Mathematical Institute of Slovak Academy of Sciences in Bratislava, Slovak Republic. He is interested in nonlinear functional analysis, bifurcation theory, dynamical systems and fractional calculus with applications to mechanics, vibrations and economics.

https://doi.org/10.1515/9783110523621-007

Index s-e-condition 15 Co-ordinated convex – co-ordinated convex 15 – geometric-geometric co-ordinated convex 16 Convex – (α, m)-logarithmically convex 15 – (β, m)-geometrically convex 14 – (s, m)-convex 14 – h-convex 14 – m-convex 13 – m-geometrically convex 14 – r-convex 14 – s-convex in the first sense 13 – s-convex in the second sense 13 – arithmetic-geometric convex 14 – convex 13 – geometric-arithmetically convex 14 – geometric-arithmetically s-convex 14 – geometric-geometric convex 14 – log-convex 14 – quasi-convex 13 – quasi-geometrically convex 13

Fractional integral – double Hadamard fractional integrals 12 – Hadamard fractional integrals 12 – Riemann-Liouville fractional integrals 11

invex – invex 15 – preinvex 15

Special function – Beta function 11 – Gamma function 11 – incomplete Beta function 11 Special mean – arithmetic mean 19 – geometric mean 19 – harmonic mean 19 – identric mean 19 – logarithmic mean 19 – p-logarithmic mean 19

Fractional Calculus in Applied Sciences and Engineering Volume 4 Kecai Cao, YangQuan Chen Fractional Order Crowd Dynamics. Cyber-Human System Modeling and Control, 2018 ISBN 978-3-11-047281-3, e-ISBN (PDF) 978-3-11-047398-8, e-ISBN (EPUB) 978-3-11-047283-7 Volume 3 Michal Fečkan, JinRong Wang, Michal Pospíšil Fractional-Order Equations and Inclusions, 2017 ISBN 978-3-11-052138-2, e-ISBN (PDF) 978-3-11-052207-5, e-ISBN (EPUB) 978-3-11-052155-9 Volume 2 Bruce J. West Nature’s Patterns and the Fractional Calculus, 2017 ISBN 978-3-11-053411-5, e-ISBN (PDF) 978-3-11-053513-6, e-ISBN (EPUB) 978-3-11-053427-6 Volume 1 Dingyü Xue Fractional-Order Control Systems. Fundamentals and Numerical Implementations, 2017 ISBN 978-3-11-049999-5, e-ISBN (PDF) 978-3-11-049797-7, e-ISBN (EPUB) 978-3-11-049719-9