Lian Tie-yan,Tang Weiand Zhou Rui

(1.Faculty of Light Industry&Energy,Shaanxi University of Science&Technology,Xi’an,710021)

(2.College of Electrical&Information Engineering,Shaanxi University of Science&Technology,Xi’an,710021)

(3.College of Arts and Sciences,Shaanxi University of Science&Technology,Xi’an,710021)

Communicated by Ji You-qing

Abstract:In this article,we extend some estimates of the right-hand side of the Hermite-Hadamard type inequality for preinvex functions with fractional integral.The notion of logarithmically s-Godunova-Levin-preinvex function in second sense is introduced and then a new Hermite-Hadamard inequality is derived for the class of logarithmically s-Godunova-Levin-preinvex function.

Key words:Hermite-Hadamard’s integral inequality,Riemann-Liouville fractional integral,H¨older’s integral inequality

1 Introduction

Let R be the set of real numbers,I?R,Iois the interior of I.It is common knowledge in mathematical analysis that a function f:I?R→R is said to be convex on an interval I if the inequality

is valid for all x,y∈I and λ∈[0,1].

Many inequalities have been established for convex functions but the most famous is the Hermite-Hadamard’s integral inequality,due to its rich geometrical significance and applications,which is stated as follow(see[1]):

If f:I?R→R is a convex function on I and a,b∈I with a

hold.

In[2],Dragomir and Agarwal proved the following results connected with the right part of(1.2).

Lemma 1.1[2]Let f:Io?R→R be a differentiable mapping on Io,a,b∈Iowith a

Theorem 1.1[2]Let f:Io?R→R be a differentiable mapping on Io,a,b∈Iowith a

In[3],Sarikaya et al.established Hermite-Hadamard’s inequalities for Riemann-Liouville fractional integral.And furthermore,(1.3)and(1.4)for fractional integral type were all obtained.

Fractional calculus is a theory of integral and differential operators of non-integral order.Many mathematicians,like Liouville,Riemann and Weyl,made major contributions to the theory of fractional calculus.The study on the fractional calculus continued with the contributions from Fourier,Abel,Lacroix,Leibniz,Grunwald and Letnikov.For details,see[4]–[6].A first formulation of an integral operator of fractional order in reliable form is named the Riemann-Liouville fractional integral operator.

In recent years,several extensions and generalizations have been considered for classical convexity.A significant generalization of convex functions is that of invex functions introduced by Hanson[7].Weir and Mond[8]introduced the concept of preinvex functions and applied it to the establishment of the sufficient optimality conditions and duality in nonlinear programming.Noor[9],[10]introduced the Hermite-Hadamard inequality for preinvex and log-preinvex functions.

In this paper,we promote all the results of literature[3]for preinvex functions.Then we need the following definitions.For more details,one can consult[4]–[17].

Definition 1.1[3]Let f∈L1[a,b].The Riemann-Liouville integralsorder α >0 with a ≥ 0 are defined by respectively.Here,Γ(α)is the Gamma function and

Definition 1.2[7]A set S?R is said to be invex with respect to the map η:S×S→R,if for every x,y∈S and t∈[0,1],one has

It is obvious that every convex set is invex with respect to the map η(x,y)=x ? y,but there exist invex sets which are not convex(see[11]).

The mapping η is said to be satisfies the condition(C)if for every x,y ∈ S and t∈ [0,1],

Note that for every x,y∈S and t1,t2∈[0,1]from condition(C)we have

We can see[8]and[16]for details.

Definition 1.3[4]Let S?R be an invex set with respect to η:S×S→R.A continuous function f:S → R is said to be preinvex with respect to η if

for every x,y∈S and t∈[0,1].

Every convex function is an preinvex with respect to the map η(x,y)=x ? y,but the converse does not holds.For properties and applications of preinvex functions,see[8],[16]and[17]and references therein.

Definition 1.4[12]Let S?R be an invex set with respect to η:S×S →R.A continuous function f:S → R is said to be m-preinvex with respect to η if

for every x,y∈S,t∈[0,1]and m∈(0,1].

2 Hermite-Hadamard Fractional Integral Inequalities for Preinvex Functions

Hermite-Hadamard’s inequalities for preinvex functions can be represented in fractional integral forms as follows.

Theorem 2.1S?R be invex with respect to the map η:S×S→R,and a,b∈S with a

Proof. Since η satisfied condition(C),by(1.6)for t∈ [0,1],we have

Since f is preinvex with respect to η,we obtain

Multiplying both sides of(2.3)by tα?1,then integrating the resulting inequality with respect to t over[0,1],it follows that

Letting u=a+(1?t)η(b,a)),we have

Letting u=a+tη(b,a)),we have

Substituting(2.5)and(2.6)to(2.4),we get

In addition,using the preinvexity of f,we have

Therefore,by using(2.5)–(2.7)and(2.9),we get(2.1),which completes the proof of the theorem.

Corollary 2.1If the mapping η satisfies η(b,a)=b ? a in Theorem 2.1,then(2.1)reduces to the following inequality in[3]:

Remark 2.1In Theorem 2.1,we let η(b,a)=b ? a and α =1,then inequality(2.1)becomes inequality(1.2).

3 Hermite-Hadamard Type Fractional Integral Inequalities for m-preinvex Functions

Theorem 3.1Let S? R be invex with respect to the map η:S×S→ R and a,b∈S with a

Proof. By integration by parts and making use of the substitution u=a+tη(b,a),we have

Adding(3.2)and(3.3),we get the desired equality.This completes the proof of the theorem.

Corollary 3.1If the mapping η satisfies η(b,a)=b?a in Theorem 3.1,then(3.1)reduces to the following inequality which was used to obtain some fractional integral inequalities in[3]:

Remark 3.1In Theorem 3.1,we let η(b,a)=b ? a and α =1.Then equality(3.1)becomes equality(1.3)of Lemma 1.1.

Theorem 3.2Let S?R be invex with respect to the map η:S×S→R and a,b∈S with a

Proof. By using(3.1)and the m-preinvexity of|f′|,we have

Calculating I1and I2,we have

By using(3.6)and(3.7)in(3.5),we get the inequality(3.4),which completes the proof of the theorem.

Corollary 3.2If the mapping η satisfies η(b,a)=b?a and m=1 in Theorem 3.2,then(3.4)reduces to the following inequality which was used to obtain some fractional integral inequalities in[3]:

Remark 3.2In Theorem 3.2,let η(b,a)=b ? a,m=1 and α =1.Then inequality(3.4)becomes inequality(1.4)of Theorem 1.1.

Theorem 3.3Let S?R be invex with respect to the map η:S×S→R and a,b∈S with a1,then we have the following inequality:

Proof.By using(3.1)and the well known H¨older’s integral inequality,we obtain

Since|f′|qis m?preinvex on S,for every a,b∈ [a,b]with a

In addition,

By using the fact(a+b)p≥ap+bp,a,b≥0,p>1 and(3.12),(3.13),we have

Substituting(3.11)and(3.14)into(3.10),we get the desired result(3.9).Hence the proof of the Theorem is completed.

Corollary 3.3If the mapping η satisfies η(b,a)=b?a and m=1 in Theorem 3.3,then the following fractional integral inequality holds:

Corollary 3.4If α=1 in Theorem 3.3,then the following integral inequality holds for m-preinvex function:

Corollary 3.5Let α =1.If the mapping η satisfies η(b,a)=b?a and m=1 in Theorem 3.3,then the following inequality holds:

Theorem 3.4Let S?R be invex with respect to the map η:S×S→R and a,b∈S with a

Proof.By using(3.1)and the well known power-mean integral inequality,we obtain

Since|f′|qis m-preinvex on S for every a,b∈ [a,b]with a

In addition,

By making use of the last four formulas in(3.19),we get(3.18).Hence the proof of the Theorem is completed.

Corollary 3.6Let α=1 in Theorem 3.4.Then the following inequality holds:

Remark 3.3This is the Theorem 6 of[12].We can discuss Theorem 3.4 with other conditions,such as α =1,η(b,a)=b? a or η(b,a)=b? a etc.,some of which have been found in literature[12].

4 Inequality for Logarithmically s-Godunova-Levinpreinvex Function of Second Kind

In[18],Noor et al.have introduced a new class of convex functions which is called logarithmic s-Godunova-Levin function of second kind.

Definition 4.1[18]A function f→ (0,∞)is said to be logarithmic s-Godunova-Levin function of second kind,if s∈[0,1],we have

From above inequality,it follows that

We will define a new generalization of logarithmic s-Godunova-Levin function of second kind,which is called the logarithmically s-Godunova-Levin-preinvex function of second kind with respect to η.A new Hermite-Hadamard inequality is obtained.

Definition 4.2Let S?R be an invex set with respect to η:S×S→R.A continuous function f:S→R is said to be logarithmically s-Godunova-Levin-preinvex function of second kind with respect to η if

for every x,y∈S,s∈[0,1]and t∈(0,1).

Theorem 4.1If f:I→ (0,∞)is logarithmically s-Godunova-Levin-preinvex function of second kind with respect to η,then for s ∈ (0,1),we have

Proof.Since f is logarithmically s-Godunova-Levin-preinvex function of second kind,we have the following inequality by making use of(2.2):

Taking the logarithms on both sides of the above inequality yields

Integrating inequality(4.3)with respect to t∈(0,1)procures

In addition,since f is logarithmically s-Godunova-Levin-preinvex function of second kind,we have

Integrating on both sides of(4.5)with respect to t∈(0,1)generates

Combining(4.4)and(4.6)gives

Taking the power of e on all sides of(4.7)produces the inequalities(4.1).This completes the proof.

Remark 4.1In Theorem 4.1,let η(b,a)=b? a.We get the result proved in[18].