Appl. Math. Inf. Sci.9, No. 1, 233-243 (2015) 233

Applied Mathematics & Information SciencesAn International Journal

http://dx.doi.org/10.12785/amis/090129

Generalized Convexity and Integral InequalitiesMuhammad Aslam Noor∗, Khalida Inayat Noor and Muhammad Uzair Awan

Mathematics Department, COMSATS Institute of InformationTechnology, Park Road, Islamabad, Pakistan

Received: 29 Mar. 2014, Revised: 29 Jun. 2014, Accepted: 30 Jun. 2014Published online: 1 Jan. 2015

Abstract: In this paper, we consider a very useful and significant classof convex sets and convex functions that is relative convex setsand relative convex functions which was introduced and studied by Noor [20]. Several new inequalities of Hermite-Hadamard type forrelative convex functions are established using differentapproaches. We also introduce relativeh-convex functions and is shown thatrelativeh-convex functions include Noor relative convex functions as special cases. Results obtained in this paper may inspire futureresearch in convex analysis and related optimization fields.

Keywords: convex sets, relative convex sets, convex functions, relative convex functions, relativeh-convex functions, fractionalintegrals, Hermite-Hadamard inequality2010 AMS Subject Classification:26D15, 26A51, 26A33, 49J40, 90C33

1 Introduction

It is well known [29] that modern analysis directly orindirectly involve the applications of convexity. Due to itsapplications and significant importance, the concept ofconvexity has been extended and generalized in severaldirections, see [1,2,5,6,7,8,9,17,18,19,20,21,22,24,25,26,33,35]. Inspired and motivated by research going inthis dynamic and fascinating field, Noor [20] introducedand studied a new class of convex set and convex functionwith respect to an arbitrary function; which is called asrelative convex set and relative convex function. Theserelative convex sets and relative convex functions arenonconvex. It is worth mentioning that these relativeconvex sets and relative convex functions are quitedifferent than those of relative convex sets and relativeconvex functions considered by Youness [35]. Cristescuet al. [4] has explored the applications of relative convexsets of Noor in the fields of transportation, colored imageprocessing and computer aided design; see example 1 andexample 2. Noor [20] has shown that optimalityconditions of the differentiable relative convex functionscan be characterized by a class of variational inequality.This aspect has motivated Noor [20] to introduce a newclass of variational inequality, which is called generalvariational inequality. For the numerical methods andother aspects of general variational inequalities, see [20]

and the references theirin.

The concept of convexity and its variant forms haveplayed a fundamental role in the development of variousfields. Hermite (1883) and Hadamard (1896)independently have shown that the convex functions arerelated to an integral inequality, this inequality is knownas Hermite-Hadamard inequality.

Let f : I ⊆ R → R be a convex function witha < b anda,b∈ I . Then

f

(

a+b2

)

≤1

b−a

b∫

a

f (x)dx≤f (a)+ f (b)

2.

This double inequality is known as classicalHermite-Hadamard inequality. For differentgeneralizations and extensions of Hermite-Hadamardinequalities interested readers are referred to [1,6,7,8,10,11,12,14,17,18,19,22,24,25,26,28,29,30,31,32,34].

In this paper, we consider the Noor’s relative convexfunctions. We derive several new Hermite-Hadamardinequality for relative convex functions and for its variantforms. Finally we also introduce the concept of relativeh-convex function. This class also generalizes the class ofrelative convex functions. The new ideas and techniquesused in this paper may motivate the interested readers to

∗ Corresponding author e-mail:noormaslam@hotmail.com

c© 2015 NSPNatural Sciences Publishing Cor.

234 M. A. Noor et. al. : Generalized Convexity and Integral Inequalities

explore the applications of relative convex functions invarious branches of pure and applied sciences.

2 Basic concepts and results

In this section, we recall the definition of relative convexsets and relative convex functions respectively. Wediscuss some previously known results for relative convexfunctions.

Definition 1([20]). Let Kg be any set in H. The set Kgis said to be relative convex with respect to an arbitraryfunction g: H → H such that

(1− t)u+ tg(v)∈ Kg, ∀u,v∈ H : u,g(v) ∈ Kg, t ∈ [0,1].

Note that every convex set is relative convex, but theconverse is not true, see [20]. If g = I , the identityfunction, then the definition of relative convex setrecaptures the definition of classical convex set.We remark that this type of relative convex sets are calledNoor convex sets [3,4] and are distinctly different thanthat of Youness’s relative convex set [35].

Now we give some examples of relative convex sets,which shows the significance of relative convex sets. Theseexamples are mainly due to Cristescu et al [4].

Example 1([4]). One of the most important goals of theInternational Union of Railways (U.I.C.) is to enable therailway companies to measure the impact of their activityon the environment (see the U.I.C. guide). Theenvironment indicators in the domain of railway transportdefined under U.I.C. and presented into the abovementioned guide include the level of noise, which shouldbe, in normal conditions, in the interval[0,50]db(A). Theactual noise level produced by wagons is[125,130]db(A).The noise level around the railway stations located intowns is represented by the set[0,50] ∪ [125,130]. Byrelocating the railway transport system outside the towns,the resulted level of noise becomes[0,50]. Let us denoteby g : R→R the function defined by

g(x) =

{

x i f x∈ [0,50]0 otherwise (1)

which is the function describing the efforts of kipping thenormal level of sound, which works under this project.Then the set[0,50]∪ [125,130] is g-convex.

Other examples are easy to find in the domain ofimage processing, in which a transformation of the realplaneR

2 into a set of grid-points,Z2 for example, isnecessary. In order to present this type of examples weneed to choose a transformation of the space, whichperforms the space digitization. The general definition ofthis kind of transformations is

Definition 2. A function f:Rn →Zn is said to be a method

of digitization ofRn intoZn if f (x) = x whenever x∈ Z

n.

In what follows we assume thatn = 2 andg = E = f isthe digitization method used in black and white pictureprocessing by Rosenfeld (1969) and in colored imageprocessing by Chassery (1978). It is defined byf : R2 → Z

2, f (x,y) = (i, j), i ∈ Z, j ∈ Zwhenever(x,y) ∈[i −1/2, i +1/2)× [ j −1/2, j +1/2).

Example 2([4]). The setA := B∪ < (i, j),(i + m, j) >,whenever i ∈ Z, j ∈ Z,m ∈ Z andB⊆ [i −1/2, i +m+1/2)× [ j−1/2, j +1/2) is an unionof triangles having one side< (i, j),(i + m, j) > isrelative convex. Indeed, considering two pointsx,y ∈ A,there are two numbersk ∈ Z and l ∈ Z such thatx ∈ [i + k − 1/2, i + k + 1/2)× [ j − 1/2, j + 1/2) andy ∈ [i + l − 1/2, i + l + 1/2) × [ j − 1/2, j + 1/2).Thereforeg(y) = (i + l , j). Then for anyt ∈ [0,1], there isthe integers betweenk and l such thattx+(1− t)g(y) ∈[i + s−1/2, i + s+1/2)× [ j −1/2, j +1/2) ⊆ A sinceBis an union of triangles having one side< (i, j),(i +m, j) >. It means thatA is relative convex. Inthe same manner one can take vertical columns of pixelsand obtain relative convex sets.

Definition 3([20]). A function f : Kg → H is said to berelative convex, if there exists an arbitrary functiong : H → H such that

f ((1− t)u+ tg(v))≤ (1− t) f (u)+ t f (g(v)),

∀u,v∈ H : u,g(v) ∈ Kg, t ∈ [0,1]. (2)

Clearly every convex function is relative convex, but theconverse is not true. For the properties of relative convexfunctions, see [18,19,20].Note that fort = 1

2 in (2), we have the definition of Jensen’srelative convex function. That is

f

(

a+g(b)2

)

≤f (a)+ f (g(b))

2.

Definition 4([19]). The function f: Kg → (0,∞) is said tobe relative logarithmic convex, if there exists an arbitraryfunction g: H → H such that

f ((1− t)u+ tg(v))≤ ( f (u))1−t( f (g(v)))t ,

∀u,v∈ H : u,g(v) ∈ Kg, t ∈ [0,1]. (3)

This implies that

log f ((1− t)u+ tg(v))≤ (1− t) log f (u)+ t log f (g(v)),

∀u,v∈ H : u,g(v) ∈ Kg, t ∈ [0,1].

Definition 5([19]). The function f: Kg → H is said to berelative quasi convex, if there exists an arbitrary functiong : H → H such that

f ((1− t)u+ tg(v))≤ max{ f (u), f (g(v))},

∀u,v∈ H : u,g(v) ∈ Kg, t ∈ [0,1]. (4)

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Appl. Math. Inf. Sci.9, No. 1, 233-243 (2015) /www.naturalspublishing.com/Journals.asp 235

Remark.From inequalities (2), (3) and (4), it follows that

f ((1− t)u+ tg(v)) ≤ ( f (u))1−t( f (g(v)))t

≤ (1− t) f (u)+ t f (g(v))

≤ max{ f (u), f (g(v))}.

This shows that

relativelogconvex function⇒ relative convex functions

⇓

relative quasi convex functions,

but the converse is not true.

Definition 6([18]). Let Kg = [a,g(b)]⊆ R be the intervaland g: H →H be any arbitrary function. Then f is relativeconvex function, if and only if,∣

∣

∣

∣

∣

∣

1 1 1a x g(b)

f (a) f (x) f (g(b))

∣

∣

∣

∣

∣

∣

≥ 0; a≤ x≤ g(b).

From the above definition one can obtain the followingequivalent forms:

1.f is relative convex function.2.f (x) ≤ f (a)+ f (g(b))− f (a)

g(b)−a (x−a).

3. f (x)− f (a)x−a ≤ f (g(b))− f (a)

g(b)−a ≤ f (g(b))− f (x)g(b)−x .

4. f (a)(x−a)(g(b)−a) +

f (x)(g(b)−x)(x−a) +

f (g(b))(g(b)−a)(g(b)−x) ≥ 0.

5.(g(b)−x) f (a)+(g(b)−a) f (x)+(x−a) f (g(b)) ≥ 0.

Now we recall the definition of similarly orderedfunctions.

Definition 7([29]). Two functions f and w are said to besimilarly ordered ( f is w-monotone), if

〈 f (x)− f (y),g(x)−g(y)〉 ≥ 0, ∀x,y∈ R.

Theorem 1.The product of two relative convex functionsf and w is again relative convex provided if f and w aresimilarly ordered.

Proof.The proof is obvious. ⊓⊔

We now discuss the optimality condition of thedifferentiable relative convex functions on the relativeconvex sets. This result is due to Noor [20]. We include itfor the sake of completeness.

Theorem 2([20]). Let Kg be a relative convex set in Hand let f be a differentiable relative convex function. Then,u∈ Kg is the minimum of f on Kg, if and only if, u∈ Kg,satisfies

〈 f ′(u),g(v)−u〉 ≥ 0, ∀v∈ H : g(v) ∈ Kg,

where f′(u) is the differential of f at u∈ Kg. Inequality ofthe above type is called the general variational inequality.

Now we define generalized Riemann-Liouville fractionalintegrals, with respect to an arbitrary functiong.

Definition 8. Left-sided and right-sided generalizedRiemann-Liouville fractional integrals of orderα ∈ R

+

are respectively defined as

Jαa+ f (x) =

1Γ (α)

x∫

a

(x− t)α−1 f (t)dt,

0≤ a< x≤ g(b), (5)

and

Jαg(b)− f (x) =

1Γ (α)

g(b)∫

x

(t − x)α−1 f (t)dt,

0≤ a< x≤ g(b), (6)

whereΓ (x) =∞∫

0e−ttx−1dt, x> 0 is the gamma function.

It is clear from (5) and (6) thatJαa+ f (a) = 0, and

Jαg(b)− f (g(b)) = 0. For g = I , where I is the identity

function then above definition of generalizedRiemann-Liouville fractional integrals reduces to thedefinition of classical Riemann-Liouville fractionalintegrals, see [13,15,16].

3 Hermite-Hadamard inequalities

In this section we derive some Hermite-Hadamard type ofintegral inequalities for relative convex functions. Fromnow onward Throughout this sectionKg = [a,g(b)] be theinterval unless otherwise specified.

Theorem 3([19]). Let f : Kg → R be a relative convexfunction. Then, we have

f

(

a+g(b)2

)

≤1

g(b)−a

g(b)∫

a

f (x)dx≤f (a)+ f (g(b))

2. (7)

The inequality (7) is the extension of classicalHermite-Hadamard inequality for relative convexfunctions. For g = I where I is identity function,inequality (7) reduces to classical Hermite-Hadamardinequality.

Theorem 4. Let f,w : Kg → R be relative convexfunctions. Then for all t∈ [0,1], we have

2 f

(

a+g(b)2

)

w

(

a+g(b)2

)

−

[

16

M( f ,w;a,g(b))+12

N( f ,w;a,g(b))

]

≤1

g(b)−a

g(b)∫

a

f (x)w(x)dx

≤13

M( f ,w;a,g(b))+16

N( f ,w;a,g(b)),

where

M( f ,w;a,g(b)) = f (a)w(a)+ f (g(b))w(g(b)), (8)

c© 2015 NSPNatural Sciences Publishing Cor.

236 M. A. Noor et. al. : Generalized Convexity and Integral Inequalities

and

N( f ,w;a,g(b)) = f (a)w(g(b))+ f (g(b))w(a). (9)

Proof.Let

f

(

a+g(b)2

)

w

(

a+g(b)2

)

= f

(

ta+(1− t)g(b)+(1− t)a+ tg(b)2

)

×w

(

ta+(1− t)g(b)+(1− t)a+ tg(b)2

)

≤12[ f (ta+(1− t)g(b))+ f ((1− t)a+ tg(b))]

×12[w(ta+(1− t)g(b))+w((1− t)a+ tg(b))]

=14[ f (ta+(1− t)g(b))w(ta+(1− t)g(b))

+ f ((1− t)a+ tg(b))w((1− t)a+ tg(b))]

+14[ f (ta+(1− t)g(b))w((1− t)a+ tg(b))

+ f ((1− t)a+ tg(b))w(ta+(1− t)g(b))]

≤14[ f (ta+(1− t)g(b))w(ta+(1− t)g(b))

+ f ((1− t)a+ tg(b))w((1− t)a+ tg(b))]

+14[t f (a)+(1− t) f (g(b)))((1− t)w(a)+ tw(g(b)))

+((1− t) f (a)+ t f (g(b)))(tw(a)+(1− t)w(g(b)))]

=14[ f (ta+(1− t)g(b))w(ta+(1− t)g(b))

+ f ((1− t)a+ tg(b))w((1− t)a+ tg(b))]

+14[2t(1− t)( f (a)w(a)+ f (g(b))w(g(b)))

+(t2+(1− t)2)( f (g(b))w(a)+ f (a)w(g(b)))]

.

Integrating with respect tot on [0,1], we have

f

(

a+g(b)2

)

w

(

a+g(b)2

)

≤14

2g(b)−a

g(b)∫

a

f (x)w(x)dx

+12

[

16

M( f ,w;a,g(b))+13

N( f ,w;a,g(b))

]

.

This implies that

2 f

(

a+g(b)2

)

w

(

a+g(b)2

)

−

[

16

M( f ,w;a,g(b))+13

N( f ,w;a,g(b))

]

≤1

g(b)−a

g(b)∫

a

f (x)w(x)dx.

Also

f (ta+(1− t)g(b))w(ta+(1− t)g(b))

≤ [t f (a)+ (1− t) f (g(b))][tw(a)+ (1− t)w(g(b))].

Integrating above inequality with respect tot on [0,1], wehave

1g(b)−a

g(b)∫

a

f (x)w(x)dx

≤13

M( f ,w;a,g(b))+16

N( f ,w;a,g(b)).

This completes the proof.⊓⊔

Theorem 5. Let f,w : Kg → R be similarly ordered andrelative convex functions on Kg. Then for all t∈ [0,1], wehave

2 f

(

a+g(b)2

)

w

(

a+g(b)2

)

−14

M( f ,w;a,g(b))

≤1

g(b)−a

g(b)∫

a

f (x)w(x)dx≤12

M( f ,w;a,g(b)),

where M( f ,w;a,g(b)) is given by(8).

Proof. Using the fact thatf and w are similarly orderedfunctions, proof follows from Theorem 4.⊓⊔

Theorem 6.Let f be relative convex function, then for allλ ∈ (0,1), we have

f

(

a+g(b)2

)

≤ ∆1(λ ) ≤1

g(b)−a

g(b)∫

a

f (x)dx

≤ ∆2(λ )≤f (a)+ f (g(b))

2, (10)

where

∆1(λ ) = λ f

(

(2−λ )a+λg(b)2

)

+(1−λ ) f

(

(1−λ )a+(1+λ )g(b)2

)

, (11)

and

∆2(λ ) =f ((1−λ )a+λg(b))+λ f (a)+(1−λ ) f (g(b))

2. (12)

Now essentially using the technique of [12] one can provefollowing result. This result plays a key role in proving ournext result.

Lemma 1. If f (n)(x) for n∈ N exists and is integrable on[a,g(b)], then

Ξ (a,g(b);k;n; f )

=(g(b)−a)n

2n!

1∫

0

tn−1(n−2t) f (n)(ta+(1− t)g(b))dt,

where

Ξ (a,g(b);k;n; f ) =f (a)+ f (g(b))

2−

1g(b)−a

g(b)∫

a

f (x)dx

−n−1

∑k=2

(k−1)(g(b)−a)k

2(k+1)!f (k)(a)

c© 2015 NSPNatural Sciences Publishing Cor.

Appl. Math. Inf. Sci.9, No. 1, 233-243 (2015) /www.naturalspublishing.com/Journals.asp 237

Theorem 7.Let f : Kg → R be n-times differentiable andintegrable on Kg. If | f (n)|q where q> 1 is relative convexfunction, then

|Ξ (a,g(b);k;n; f )|

≤(g(b)−a)n

2n!

(

n−1n+1

)1− 1q (

G1| f(n)(a)|q+G2| f

(n)(g(b))|q)

1q,

where

G1 =n2−2

(n+1)(n+2),

and

G2 =n

(n+1)(n+2),

respectively.

Proof. Using Lemma 1, power-mean inequality and thefact that| f (n)|q is relative convex function, we have

|Ξ (a,g(b);k;n; f )|

=

∣

∣

∣

∣

∣

∣

(g(b)−a)n

2n!

1∫

0

tn−1(n−2t) f (n)(ta+(1− t)g(b))dt

∣

∣

∣

∣

∣

∣

≤(g(b)−a)n

2n!

1∫

0

tn−1(n−2t)dt

1− 1q

×

1∫

0

tn−1(n−2t)| f (n)(ta+(1− t)g(b))|qdt

1q

≤(g(b)−a)n

2n!

(

n−1n+1

)1− 1q

×

1∫

0

tn−1(n−2t)[t| f (n)(a)|q+(1− t)| f (n)(g(b))|q]dt

1q

=(g(b)−a)n

2n!

(

n−1n+1

)1− 1q

×

(

n2−2(n+1)(n+2)

| f (n)(a)|q+n

(n+1)(n+2)| f (n)(g(b))|q

)

1q

.

This completes the proof.⊓⊔

Corollary 1. Under the assumptions of Theorem 7, if q=1, then

|Ξ (a,g(b);k;n; f )|

≤(g(b)−a)n

2n!

×

(

n2−2(n+1)(n+2)

| f (n)(a)|+n

(n+1)(n+2)| f (n)(g(b))|

)

.

Now we derive Hermite-Hadamard type of inequalities forrelative logarithmic convex functions.

Theorem 8. Let f : Kg → (0,∞) be relative logarithmicconvex function, then for all t∈ [0,1], we have

f

(

a+g(b)2

)

≤ exp

1g(b)−a

g(b)∫

a

log f (x)dx

≤√

f (a) f (g(b)).

Theorem 9. Let f : Kg → (0,∞) be relative logarithmicconvex function, then for all t∈ [0,1], we have

f

(

a+g(b)2

)

≤ exp

1g(b)−a

g(b)∫

a

log f (x)dx

≤1

g(b)−a

g(b)∫

a

G( f (x), f (a+g(b)−x))dx

≤1

g(b)−a

g(b)∫

a

f (x)dx

≤ L[ f (g(b)), f (a)]

≤ A[ f (a), f (g(b))],

where

L[ f (g(b)), f (a)] =f (g(b))− f (a)

log f (g(b))− log f (a),

is the logarithmic mean,

A[ f (a), f (g(b))] =f (a)+ f (g(b))

2,

is the arithmetic mean, and

G[ f (a), f (g(b))] =√

f (a) f (g(b)),

is the Geometric mean respectively.

Proof. The proof of first inequality follows directly fromTheorem 8. In order to prove second inequality, weproceed as

G( f (x), f (a+g(b)−x)) = exp[logG( f (x), f (a+g(b)−x))] .

Integrating above inequality with respect tox on [a,g(b)],we have

1g(b)−a

g(b)∫

a

G( f (x), f (a+g(b)−x))dx

=1

g(b)−a

g(b)∫

a

exp[logG( f (x), f (a+g(b)−x))]dx

≥ exp

1g(b)−a

g(b)∫

a

logG( f (x), f (a+g(b)−x)dx

= exp

1g(b)−a

g(b)∫

a

log f (x)+ log f (a+g(b)−x)2

dx

= exp

1g(b)−a

g(b)∫

a

log f (x)dx

.

c© 2015 NSPNatural Sciences Publishing Cor.

238 M. A. Noor et. al. : Generalized Convexity and Integral Inequalities

UsingAM−GM inequality, we have

G( f (x), f (a+g(b)−x)) ≤f (x)+ f (a+g(b)−x)

2.

Integrating the above inequality with respect tox on[a,g(b)], we have

1g(b)−a

g(b)∫

a

G( f (x), f (a+g(b)−x))dx≤1

g(b)−a

g(b)∫

a

f (x)dx.

Since f is relative logarithmic convex function, so for allt ∈ [0,1], we have

f ((1− t)a+ tg(b)) ≤ [ f (a)]1−t [ f (g(b))]t .

Integrating above inequality with respect tot on [0,1], wehave

1g(b)−a

g(b)∫

a

f (x)dx ≤ f (a)

1∫

0

[

f (g(b))f (a)

]t

dt

=f (g(b))− f (a)

log f (g(b))− log f (a)

= L[ f (g(b)), f (a)]

≤f (a)+ f (g(b))

2= A[ f (a), f (g(b))].

This completes the proof.⊓⊔

Theorem 10.Let f,w : Kg → (0,∞) be relative logarithmicconvex functions, then

logw

(

a+g(b)2

)

−1

g(b)−a

g(b)∫

a

logw(x)dx

≤1

g(b)−a

g(b)∫

a

log f (x)dx− log f

(

a+g(b)2

)

.

Proof. Let f and g be relative logarithmic convexfunctions. Then

f

(

a+g(b)2

)

w

(

a+g(b)2

)

= f

(

(1− t)a+ tg(b)+ ta+(1− t)g(b)2

)

w

(

(1− t)a+ tg(b)+ ta+(1− t)g(b)2

)

≤ [ f ((1− t)a+ tg(b)) f (ta+(1− t)g(b))]12

[w((1− t)a+ tg(b))w(ta+(1− t)g(b))]12 .

Taking log on both sides, we have

log

[

f

(

a+g(b)2

)

w

(

a+g(b)2

)]

≤12[log f ((1− t)a+ tg(b))+ log f (ta+(1− t)g(b))

+ logw((1− t)a+ tg(b))+ logw(ta+(1− t)g(b))] .

Integrating both sides of above inequality with respect toton [0,1], we have the required result.⊓⊔

Theorem 11.Let f,w : Kg → R be relative logarithmicconvex functions, then

1g(b)−a

g(b)∫

a

f (x)w(a+g(b)−x)dx

≤[A( f (a), f (g(b)))]2+[A(w(a),w(g(b)))]2

2.

Proof.Since f ,w be relative logarithmic convex functions,then, we have

1g(b)−a

g(b)∫

a

f (x)w(a+g(b)−x)dx

=

1∫

0

f (ta+(1− t)g(b))w((1− t)a+ tg(b))dt

≤

1∫

0

[ f (a)]t [ f (g(b))]1−t [w(a)]1−t [w(g(b))]tdt

=

1∫

0

[ f (g(b))]

[

f (a)f (g(b))

]t

w(a)

[

w(g(b))w(a)

]t

dt

= f (g(b))w(a)

1∫

0

[

f (a)w(g(b))f (g(b))w(a)

]t

dt

= f (g(b))w(a)

f (a)w(g(b))− f (g(b))w(a)f (g(b))w(a)

log f (a)w(g(b))− log f (g(b))w(a)

=f (a)w(g(b))− f (g(b))w(a)

log f (a)w(g(b))− log f (g(b))w(a)

= L[ f (a)w(g(b)), f (g(b))w(a)]

≤f (a)w(g(b))+ f (g(b))w(a)

2= A[ f (a)w(g(b)), f (g(b))w(a)]

≤12

1∫

0

{

[ f (ta+(1− t)g(b))]2 +[w((1− t)a+ tg(b))]2}

dt

≤12

1∫

0

{

[ f (a)]t [ f (g(b))]1−t}2

dt

+12

1∫

0

{

[w(a)]1−t [w(g(b))]t}2

dt

=[ f (g(b))]2

4

2∫

0

[

f (a)f (g(b))

]u

du

+[w(a)]2

4

2∫

0

[

w(g(b))w(a)

]u

du

=14

[ f (a)]2− [ f (g(b))]2

log f (a)− log f (g(b))+

14

[w(a)]2− [w(g(b))]2

logw(a)− logw(g(b))

=12

[

f (a)+ f (g(b))2

f (a)− f (g(b))log f (a)− log f (g(b))

]

c© 2015 NSPNatural Sciences Publishing Cor.

Appl. Math. Inf. Sci.9, No. 1, 233-243 (2015) /www.naturalspublishing.com/Journals.asp 239

+12

[

w(a)+w(g(b))2

w(a)−w(g(b))logw(a)− logw(g(b))

]

=12[A[ f (a), f (g(b))]L[ f (a), f (g(b))]]

+12[A[w(a),w(g(b))]L[w(a),w(g(b))]]

≤12

[

f (a)+ f (g(b))2

f (a)+ f (g(b))2

]

+12

[

w(a)+w(g(b))2

w(a)+w(g(b))2

]

=[A( f (a), f (g(b)))]2+[A(w(a),w(g(b)))]2

2,

which is the required result.⊓⊔

Theorem 12. Let f,w : Kg → (0,∞) be increasing andrelative logarithmic convex functions on Kg. Then, wehave

f(

a+g(b)2

)

L[w(a),w(g(b))]+w(

a+g(b)2

)

L[ f (a), f (g(b))]

≤ 1g(b)−a

g(b)∫

af (x)w(x)dx+L[ f (a)w(a), f (g(b))w(g(b))].

Proof. Let f and w are relative logarithmic convexfunctions. Then, we have

f (ta+(1− t)g(b)) ≤ [ f (a)]t [ f (g(b))]1−t

w(ta+(1− t)g(b)) ≤ [w(a)]t [w(g(b))]1−t .

Now, using〈x1 − x2,x3 − x4〉 ≥ 0, (x1,x2,x3,x4 ∈ R) andx1 < x2 < x3 < x4, we have

f (ta+(1− t)g(b))[w(a)]t [w(g(b))]1−t

+w(ta+(1− t)g(b))[ f (a)]t [ f (g(b))]1−t

≤ f (ta+(1− t)g(b))w(ta+(1− t)g(b))

+[ f (a)]t [ f (g(b))]1−t [w(a)]t [w(g(b))]1−t .

Integrating above inequalities with respect tot on[0,1], wehave

1∫

0

f (ta+(1− t)g(b))[w(a)]t [w(g(b))]1−tdt

+

1∫

0

w(ta+(1− t)g(b))[ f (a)]t [ f (g(b))]1−tdt

≤

1∫

0

f (ta+(1− t)g(b))w(ta+(1− t)g(b))dt

+

1∫

0

[ f (a)]t [ f (g(b))]1−t [w(a)]t [w(g(b))]1−tdt.

Now, sincef andw are increasing, we have

1∫

0

f (ta+(1− t)g(b))dt

1∫

0

[w(a)]t [w(g(b))]1−tdt

+

1∫

0

w(ta+(1− t)g(b))dt

1∫

0

[ f (a)]t [ f (g(b))]1−tdt

≤

1∫

0

f (ta+(1− t)g(b))w(ta+(1− t)g(b))dt

+

1∫

0

[ f (a)]t [ f (g(b))]1−t [w(a)]t [w(g(b))]1−tdt.

Now computing the simple integration, we have

1g(b)−a

g(b)∫

a

f (x)dxL[w(a),w(g(b))]

+1

g(b)−a

g(b)∫

a

w(x)dxL[ f (a), f (g(b))]

≤1

g(b)−a

g(b)∫

a

f (x)w(x)dx+L[ f (a)w(a), f (g(b))w(g(b))].

Now, using the left hand side of Hermite-Hadamard’sinequality for relative logarithmic convex functions, wehave

f

(

a+g(b)2

)

L[w(a),w(g(b))]

+w

(

a+g(b)2

)

L[ f (a), f (g(b))]

≤1

g(b)−a

g(b)∫

a

f (x)w(x)dx+L[ f (a)w(a), f (g(b))w(g(b))].

The desired result.⊓⊔

4 Fractional Hermite-Hadamard inequalities

In this section, we give some Hermite-Hadamard typeinequalities for relative convex functions viaRiemann-Liouville fractional integrals.

Theorem 13. Let f be positive and relative convexfunction also f∈ L[a,g(b)], then we have the followinginequality

f

(

a+g(b)2

)

≤Γ (α +1)

2(g(b)−a)α [Jαa+ f (g(b))+Jα

g(b)− f (a)]

≤f (a)+ f (g(b))

2.

Proof.Since f is relative convex function, then, we have

2 f

(

a+g(b)2

)

≤ f (ta+(1− t)g(b))+ f ((1− t)a+ tg(b)).

Multiplying both sides of above inequality bytα−1 andthen integrating it with respect tot on [0,1], we have

2α

f

(

a+g(b)2

)

≤

1∫

0

tα−1 f (ta+(1− t)g(b))dt+

1∫

0

tα−1 f ((1− t)a+ tg(b))dt.

c© 2015 NSPNatural Sciences Publishing Cor.

240 M. A. Noor et. al. : Generalized Convexity and Integral Inequalities

Let u= g(b)− t(g(b)−a) andv= a+ t(g(b)−a). Then,from above inequality, we have

2α

f

(

a+g(b)2

)

≤

g(b)∫

a

(

g(b)−ug(b)−a

)α−1

f (u)du

g(b)−a

+

g(b)∫

a

(

v−ag(b)−a

)α−1

f (v)dv

g(b)−a

=1

(g(b)−a)α

g(b)∫

a

(g(b)−u)α−1 f (u)du

+

g(b)∫

a

(v−a)α−1 f (v)dv

=Γ (α)

(g(b)−a)α

1Γ (α)

g(b)∫

a

(g(b)−u)α−1 f (u)du

+1

Γ (α)

g(b)∫

a

(v−a)α−1 f (v)dv

=Γ (α)

(g(b)−a)α [Jαa+ f (g(b))+Jα

g(b)− f (a)]. (13)

Also

f (ta+(1− t)g(b))+ f ((1− t)a+ tg(b))

≤ t f (a)+(1− t) f (g(b))+(1− t) f (a)+ t f (g(b))

= f (a)+ f (g(b)),

Multiplying both sides of above inequality bytα−1 andintegrating with respect tot on [0,1], we have

1∫

0

tα−1 f (ta+(1− t)g(b))dt+

1∫

0

tα−1 f ((1− t)a+ tg(b))dt

≤ { f (a)+ f (g(b))}

1∫

0

tα−1dt,

Now by simple computation and using the definitions ofRiemann-Liouville integrals, we have

Γ (α)

(g(b)−a)α [Jαa+ f (g(b))+Jα

g(b)− f (a)]≤f (a)+ f (g(b))

α. (14)

After combining (13) and (14) and simplerearrangements, we have the required result.⊓⊔

Now using the techniques of [31,34], one can prove thefollowing results respectively.

Lemma 2.Let f : Kg → R be a differentiable function onK◦

g . If f ′ ∈ L[a,g(b)], then

f (a)+ f (g(b))2

−Γ (α +1)

2(g(b)−a)α

[

Jαa+ f (g(b))+Jα

(g(b))−f (a)

]

=g(b)−a

2

1∫

0

((1− t)α − tα ) f ′(ta+(1− t)g(b))dt.

Lemma 3.Let f : Kg →R be twice differentiable functionon K◦

g. If f ′′ ∈ L[a,g(b)], then

f (a)+ f (g(b))2

−Γ (α +1)

2(g(b)−a)α

[

Jαa+ f (g(b))+Jα

(g(b))−f (a)

]

=(g(b)−a)2

2

1∫

0

1− (1− t)α+1− tα+1

α +1f ′′(ta+(1− t)g(b))dt.

Theorem 14.Let f : Kg → R be a differentiable functionon K◦

g and f′ ∈ L[a,g(b)]. If | f ′| is relative convex on Kg,then∣

∣

∣

∣

f (a)+ f (g(b))2

−Γ (α +1)

2(g(b)−a)α

[

Jαa+ f (g(b))+Jα

(g(b))−f (a)

]

∣

∣

∣

∣

≤g(b)−a2(α +1)

(

1−1

2α

)

[

| f ′(a)|+ | f ′(g(b))|]

.

Proof. Using Lemma 2 and the fact that| f ′| is relativeconvex, we have∣

∣

∣

∣

f (a)+ f (g(b))2

−Γ (α +1)

2(g(b)−a)α

[

Jαa+ f (g(b))+Jα

(g(b))−f (a)

]

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

g(b)−a2

1∫

0

((1− t)α − tα ) f ′(ta+(1− t)g(b))dt

∣

∣

∣

∣

∣

∣

≤g(b)−a

2

1∫

0

|(1− t)α − tα || f ′(ta+(1− t)g(b))|dt

≤g(b)−a

2

1∫

0

|(1− t)α − tα |[t| f ′(a)|+(1− t)| f ′(g(b))|]dt

≤g(b)−a

2

12

∫

0

[(1− t)α − tα ][t| f ′(a)|+(1− t)| f ′(g(b))|]dt

+

1∫

12

[(tα − (1− t)α )][t| f ′(a)|+(1− t)| f ′(g(b))|]dt

=g(b)−a

2[I1+ I2]. (15)

Now

I1 =

12

∫

0

[(1− t)α − tα ][t| f ′(a)|+(1− t)| f ′(g(b))|]dt

= | f ′(a)|

[

1(α +1)(α +2)

−

( 12

)α+1

α +1

]

+| f ′(g(b))|

[

1α +2

−

( 12

)α+1

α +1

]

, (16)

and

I2 =

1∫

12

[tα(1− t)α ][t| f ′(a)|+(1− t)| f ′(g(b))|]dt

c© 2015 NSPNatural Sciences Publishing Cor.

Appl. Math. Inf. Sci.9, No. 1, 233-243 (2015) /www.naturalspublishing.com/Journals.asp 241

= | f ′(a)|

[

1α +2

−

( 12

)α+1

α +1

]

+| f ′(g(b))|

[

1(α +1)(α +2)

−

( 12

)α+1

α +1

]

. (17)

Combining (15), (16) and (17) completes the proof.⊓⊔

Theorem 15. Let f : Kg → R be twice differentiablefunction on K◦g and f′′ ∈ L[a,g(b)]. If | f ′′| is relativeconvex on Kg, then∣

∣

∣

∣

f (a)+ f (g(b))2

−Γ (α +1)

2(g(b)−a)α

[

Jαa+ f (g(b))+Jα

(g(b))−f (a)

]

∣

∣

∣

∣

=α(g(b)−a)2

2(α +1)(α +2)

(

| f ′′(a)|+ | f ′′(g(b))|2

)

.

Proof. Using Lemma 3, the proof directly follows fromTheorem 14. ⊓⊔

5 Relativeh-convex functions

In this section, we investigate the class of relativeh-convexfunctions and also discuss some special cases.

Definition 9. A function f : Kg → H is said to be relativeh-convex function with respect to two functions h: [0,1]→(0,∞) and g: H → H such that Kg is a relative convex set,we have

f ((1− t)x+ tg(y))≤ h(1− t) f (x)+h(t) f (g(y)),

∀x,y∈ H : x,g(y) ∈ Kg, t ∈ (0,1). (18)

Special cases:

I. If we take h(t) = t then we have the definition ofrelative convex function.II. If we take h(t) = t−1 in (18), then the definition ofrelative h-convex function reduces to the definition ofrelative Godunova-Levin function.

Definition 10. A function f : Kg → H is said to be relativeGodunova-Levin function with respect to g: H → H, if

f ((1− t)x+ tg(y))≤ (1− t)−1 f (x)+ t−1 f (g(y)),

∀x,y∈ H : x,g(y) ∈ Kg, t ∈ (0,1). (19)

III. If we take h(t) = ts in (18), then the definition ofrelative h-convex function reduces to the definition ofrelatives-convex function.

Definition 11. A function f : Kg → [0,∞) is said to berelative s-convex function where s∈ (0,1] with respect tofunction g: H → H, such that

f ((1− t)x+g(y))≤ (1− t)s f (x)+ ts f (g(y))

∀x,y∈ [0,∞) : x,g(y) ∈ Kg, t ∈ [0,1]. (20)

IV. If we take h(t) = 1 in (18), then the definition ofrelative h-convex function reduces to the definition ofrelativeP-function.

Definition 12. A function f : Kg → H is said to be relativeP-function with respect to function g: H → H, such that

f ((1− t)x+ tg(y))≤ f (x)+ f (g(y))

∀x,y∈ Kg : x,g(y) ∈ Kg, t ∈ [0,1]. (21)

V. If we take h(t) = t−s in (18), then the definition ofrelative h-convex function reduces to the definition ofrelatives-Godunova-Levin function.

Definition 13. A function f : Kg → H is said to be relatives-Godunova-Levin function where s∈ [0,1] with respect tofunction g: H → H, such that

f ((1− t)x+ tg(y))≤ (1− t)−s f (x)+ t−s f (g(y))

∀x,y∈ Kg : x,g(y) ∈ Kg, t ∈ [0,1]. (22)

VI. If we takeg= I in (18), then the definition of relativeh-convex function reduces to the definition ofh-convexfunction [33].

We now discuss some Hermite-Hadamard inequalitiesrelated to relativeh-convex functions.

Theorem 16. Let f : Kg → R be a relative h-convexfunction, such that h(1

2) 6= 0, then, we have

1

2h(12)

f

(

a+g(b)2

)

≤1

g(b)−a

g(b)∫

a

f (x)dx

≤ [ f (a)+ f (g(b))]

1∫

0

h(t)dt.

Remark.For h(t) = t−1,h(t) = ts, h(t) = 1 andh(t) = t−s

in Theorem 16 we have results for relativeGodunova-Levin functions, relatives-convex functions,relative P-functions and relative s-Godunova-Levinfunctions, respectively.

Theorem 17.Let f be relative h1-convex function and wbe relative h2-convex function such that h1(

12) 6= 0 and

h2(12) 6= 0, then

[

1

2h1(12)h2(

12)

f

(

a+g(b)2

)

w

(

a+g(b)2

)

]

−

M( f ,w;a,g(b))

1∫

0

h1(t)h2(1− t)dt

+N( f ,w;a,g(b))

1∫

0

h1(t)h2(t)dt

≤1

g(b)−a

g(b)∫

a

f (x)w(x)dx

≤

M( f ,w;a,g(b))

1∫

0

h1(t)h2(t)dt

+N( f ,w;a,g(b))

1∫

0

h1(t)h2(1− t)dt

,

c© 2015 NSPNatural Sciences Publishing Cor.

242 M. A. Noor et. al. : Generalized Convexity and Integral Inequalities

where, M( f ,w;a,g(b)) = f (a)w(a)+ f (g(b))w(g(b)) andN( f ,w;a,g(b)) = f (a)w(g(b))+ f (g(b))w(a).

Proof.The proof follows directly from Theorem 4.

Acknowledgement

The authors are grateful to Dr. S. M. Junaid Zaidi, Rector,COMSATS Institute of Information Technology, Pakistanfor providing excellent research and academicenvironment.This research is supported by HEC NRPUproject No: 20-1966/R&D/11-2553.

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c© 2015 NSPNatural Sciences Publishing Cor.

Appl. Math. Inf. Sci.9, No. 1, 233-243 (2015) /www.naturalspublishing.com/Journals.asp 243

M. Aslam Noor earnedhis PhD degree from BrunelUniversity, London, UK(1975) in the field of AppliedMathematics(NumericalAnalysis and Optimization).He has vast experienceof teaching and researchat university levels in variouscountries including Pakistan,

Iran, Canada, Saudi Arabia and United Arab Emirates.His field of interest and specialization is versatile innature. It covers many areas of Mathematical andEngineering sciences such as Variational Inequalities,Operations Research and Numerical Analysis. He hasbeen awarded by the President of Pakistan: President’sAward for pride of performance on August 14, 2008,in recognition of his contributions in the field ofMathematical Sciences. He was awarded HEC BestResearch paper award in 2009. He has supervisedsuccessfully several Ph.D and MS/M.Phil students. He iscurrently member of the Editorial Board of severalreputed international journals of Mathematics andEngineering sciences. He has more than 750 researchpapers to his credit which were published in leadingworld class journals.

Khalida Inayat Nooris a leading world-knownfigure in mathematics andis presently employed as HECForeign Professor at CIIT,Islamabad. She obtained herPhD from Wales University(UK). She has a vastexperience of teaching andresearch at university levelsin various countries including

Iran, Pakistan, Saudi Arabia, Canada and United ArabEmirates. She was awarded HEC best research paperaward in 2009 and CIIT Medal for innovation in 2009.She has been awarded by the President of Pakistan:Presidents Award for pride of performance on August 14,2010 for her outstanding contributions in mathematicalsciences and other fields. Her field of interest andspecialization is Complex analysis, Geometric functiontheory, Functional and Convex analysis. She introduced anew technique, now called as Noor Integral Operatorwhich proved to be an innovation in the field of geometricfunction theory and has brought new dimensions in therealm of research in this area. She has been personallyinstrumental in establishing PhD/MS programs at CIIT.Dr. Khalida Inayat Noor has supervised successfullyseveral Ph.D and MS/M.Phil students. She has been aninvited speaker of number of conferences and haspublished more than 400 (Four hundred ) research articlesin reputed international journals of mathematical andengineering sciences. She is member of editorial boardsof several international journals of mathematical andengineering sciences.

Muhammad Uzair Awanhas earned his MS degreefrom COMSATS Instituteof Information Technology,Islamabad, Pakistan underthe supervision of Prof.Dr. Muhammad Aslam Noor.His field of interest is ConvexAnalysis and NumericalOptimization. He haspublished several papers in

outstanding International journals of Mathematics andEngineering sciences.

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