# Hermite-Hadamard-Fejér Inequality Related to Generalized Convex Functions via Fractional Integrals

Authors: M. Rostamian Delavar; S. Mohammadi Aslani; M. De La Sen
Journal: Journal of Mathematics
 (2018)

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## Abstract

This paper deals with Hermite-Hadamard-Fejér inequality for(η1,η2)-convex functions via fractional integrals. Some mid-point and trapezoid type inequalities related to Hermite-Hadamard inequality when the absolute value of derivative of considered function is (η1,η2)-convex functions are obtained. Furthermore, a refinement for classic Hermite-Hadamard inequality via fractional integrals is given when a positive (η1,η2)-convex function is increasing.


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## Body

## 1. Introduction and Preliminaries

In 1906, L. Fejér [1] proved the following integral inequalities which are known in the literature as Fejér inequality:(1)fa+b2∫abgxdx≤∫abfxgxdx≤fa+fb2∫abgxdx,where f:[a;b]→R is convex and g:[a,b]→[0,∞)=R+ is integrable and symmetric to x=((a+b)/2)(g(x)=g(a+b-x),∀x∈[a,b]).If we consider in (1) that g≡1, then we recapture the known Hermite-Hadamard inequality. (2)fa+b2≤1b-a∫abfxdx≤fa+fb2.On the other hand, the concept of(η1,η2)-convex has been introduced in [2, 3] as a generalization of preinvex functions [4–6] and η-convex functions [7–9]. We remind these introductory concepts in the following with some geometric interpretation.(Preinvex Functions)Definition 1.
A setI⊆R is invex with respect to a real bifunction η:I×I→R, if (3)x,y∈I,λ∈0,1⟹y+ληx,y∈I. If I is an invex set with respect to η, then a function f:I→R is said to be preinvex if x,y∈I and λ∈[0,1] implies (4)fy+ληx,y≤λfx+1-λfy.In fact, in an invex set for anyx,y∈I, there is a path starting from y to y+η(x,y) which is contained in I. The point x is not necessarily the end point of the path. If for every x,y∈I we need that x should be an end point of the path, then I reduces to a convex set.( η -Convex Functions)Definition 2.
Consider a convex setI⊆R and a bifunction η:f(I)×f(I)→R. A function f:I→R is called convex with respect to η (briefly η-convex), if (5)fλx+1-λy≤fy+ληfx,fy,for all x,y∈I and λ∈[0,1].Geometrically, it says that if a function isη-convex on I, then, for any x,y∈I, its graph is on or under the path starting from (y,f(y)) and ending at (x,f(y)+η(f(x),f(y))). If f(x) should be the end point of the path for every x,y∈I, then we have η(x,y)=x-y and the function reduces to a convex one.(( η 1 , η 2 )-Convex Functions)Definition 3.
LetI⊆R be an invex set with respect to η1:I×I→R. Consider f:I→R and η2:f(I)×f(I)→R. The function f is said to be (η1,η2)-convex if (6)fx+λη1y,x≤fx+λη2fy,fx,for all x,y∈I and λ∈[0,1].Remark 4.
An(η1,η2)-convex function reduces to(i)
aη-convex function if we consider η1(x,y)=x-y for all x,y∈I,(ii)
a preinvex function if we considerη2(x,y)=x-y for all x,y∈f(I),(iii)
a convex function if satisfying (i) and (ii).Example 5 (see [2]).
Consider the functionf:R+→R+ by (7)fx=x,0≤x≤1;1,x>1.Define two bifunctions η1:R+×R+→R and η2:R+×R+→R+ by (8)η1x,y=-y,0≤y≤1;x+y,y>1,and (9)η2x,y=x+y,x≤y;2x+y,x>y.Then f is an (η1,η2)-convex function. But f is not preinvex with respect to η1 and it is not convex (consider x=0, y=2, and λ>0).Also we need the following short preliminaries about the fractional calculus theory which are used throughout the paper.Definition 6 (see [10]).
Considerf∈L1[a,b]. The Riemann-Liouville integrals Ja+αf and Jb-αf of order α>0 with a≥0 are defined by (10)Ja+αfx=1Γα∫axx-tα-1ftdt,x>a, and (11)Jb-αfx=1Γα∫xbt-xα-1ftdt,x<b, respectively, where (12)Γα=∫0∞e-ttα-1dtis Gamma function and Ja+0f(x)=Jb-0f(x)=f(x).Fejér inequality for convex functions related to fractional integrals has been obtained in [11] as the following theorem.Theorem 7.
Letf:[a,b]→R be convex function with a<b and f∈L1[a,b]. If g:[a,b]→R is nonnegative, integrable, and symmetric to (a+b)/2, then the following inequalities hold for fractional integrals: (13)fa+b2Ja+αgb+Jb-αga≤Ja+αfgb+Jb-αfga≤fa+fb2Ja+αgb+Jb-αga, with α>0.Motivated by above works and results, in this paper, we obtain Fejér inequality for(η1,η2)-convex functions via fractional integrals. Also we give some mid-point and trapezoid-type inequalities related to Hermite-Hadamard inequality when the absolute value of the derivative of the considered function is (η1,η2)-convex functions. Furthermore, we prove that when a positive (η1,η2)-convex function is increasing, there exists a refinement for classic Hermite-Hadamard inequality via fractional integrals.


## 2. Fejér Inequality

In this section, we obtain(η1,η2)-convex version of the Fejér inequality related to fractional integrals. For convenience, we separate this inequality to the left and right.Theorem 8 (Fejér’s left inequality).
LetI⊆R be an invex set with respect to η1 such that(14)η1x2+t2η1x1,x2,x2+t1η1x1,x2=t2-t1η1x1,x2,for all x1,x2∈I and t1,t2∈[0,1]. Also let f:I→R be an (η1,η2)-convex function, where η2 is an integrable bifunction on f(I)×f(I). For any a,b∈I with η1(b,a)>0, suppose that f∈L1[a,a+η1(b,a)] and the function g:[a,a+η1(b,a)]→R+ is integrable and symmetric to a+(1/2)η1(b,a). Then, for α>0, the following inequality holds: (15)f2a+η1b,a2Ja+αga+η1b,a+Ja+η1b,a-αga-12Γα∫aa+η1b,ax-aα-1+a+η1b,a-xα-1η2fx,f2a+η1b,a-xgxdx≤Ja+αfga+η1b,a+Ja+η1b,a-αfga.Proof.
Using condition (14) and the (η1,η2)-convexity of f, we have(16)f2a+η1b,a2=f2a+1+tη1b,a2-t2η1b,a=f2a+1+tη1b,a2+12η1a+1-t2η1b,a,a+1+t2η1b,a=f2a+1+tη1b,a2+12η12a+η1b,a-tη1b,a2,2a+η1b,a+tη1b,a2≤f2a+1+tη1b,a2+12η2f2a+1-tη1b,a2,f2a+1+tη1b,a2,and with the same argument as above we have(17)f2a+η1b,a2≤f2a+1-tη1b,a2+12η2f2a+1+tη1b,a2,f2a+1-tη1b,a2.
By the use of Definition6 and two changes of variable (18)x=2a+1-tη1b,a2, and (19)x=2a+1+tη1b,a2, in (16) and (17), respectively, we obtain the following inequalities:(20)K1=ΓαJa+η1b,a-αfaga=∫aa+η1b,ax-aα-1fxgxdx=∫aa+1/2η1b,ax-aα-1fxgxdx+∫a+1/2η1b,aa+η1b,ax-aα-1fxgxdx=η1b,a2α∫011-tα-1f2a+1-tη1b,a2g2a+1-tη1b,a2dt+∫011+tα-1f2a+1+tη1b,a2g2a+1+tη1b,a2dt≥η1b,a2α∫011-tα-1f2a+η1b,a2g2a+1-tη1b,a2dt-12∫011-tα-1η2f2a+1+tη1b,a2,f2a+1-tη1b,a2×g2a+1-tη1b,a2dt+∫011+tα-1f2a+η1b,a2g2a+1+tη1b,a2dt-12∫011+tα-1η2f2a+1-tη1b,a2,f2a+1+tη1b,a2×g2a+1+tη1b,a2dt.The simple form of (20) along with the fact that g is symmetric to a+(1/2)η1(b,a) leads to the following relations:(21)K1≥f2a+η1b,a2∫aa+1/2η1b,ax-aα-1gxdx-12∫aa+1/2η1b,aa+η1b,a-xα-1η2fx,f2a+η1b,a-xgxdx+f2a+η1b,a2∫a+1/2η1b,aa+η1b,ax-aα-1gxdx-12∫a+1/2η1b,aa+η1b,aa+η1b,a-xα-1η2fx,f2a+η1b,a-xgxdx=f2a+η1b,a2∫aa+η1b,ax-aα-1gxdx-12∫aa+η1b,aa+η1b,a-xα-1η2fx,f2a+η1b,a-xgxdx=Γαf2a+η1b,a2Ja+η1b,a-αga-12∫aa+η1b,aa+η1b,a-xα-1η2fx,f2a+η1b,a-xgxdx.Also with the same argument as above we have(22)K2=ΓαJa+αfa+η1b,aga+η1b,a=∫aa+η1b,aa+η1b,a-xα-1fxgxdx≥Γαf2a+η1b,a2Ja+αga+η1b,a-12∫aa+η1b,ax-aα-1η2fx,f2a+η1b,a-xgxdx. Now adding K1 to K2 implies the result.To obtain the right part of the Fejér inequality related to fractional integrals, we need a primary lemma.Lemma 9.
LetI⊆R be an invex set with respect to η1:I×I→R and a,b∈I with η1(b,a)>0. If g:[a,a+η1(b,a)]→R is integrable and symmetric to a+(1/2)η1(b,a), then (23)Ja+αga+η1b,a=Ja+η1b,a-αga=12Ja+αga+η1b,a+Ja+η1b,a-αga.Proof.
Sinceg is symmetric to a+(1/2)η1(b,a), we have g(2a+η1(b,a)-x)=g(x) for all x∈[a,a+η1(b,a)]. Then (24)Ja+αga+η1b,a=1Γα∫aa+η1b,aa+η1b,a-xα-1gxdx=1Γα∫aa+η1b,at-aα-1g2a+η1b,a-tdt=1Γα∫aa+η1b,at-aα-1gtdt=Ja+η1b,a-αga.Theorem 10 (Fejér’s right inequality).
LetI⊆R be an invex set with respect to η1 and let f:I→R be an (η1,η2)-convex function, where η2 is an integrable bifunction on f(I)×f(I). For any a,b∈I with η1(b,a)>0, suppose that the function g:[a,a+η1(b,a)]→R+ is integrable and symmetric to a+(1/2)η1(b,a) and f∈L1[a,a+η1(b,a)]. Then, for α>0, the following inequality holds:(25)Ja+αfga+η1b,a+Ja+η1b,a-αfga≤2fa+η2fb,fa2Ja+αga+η1b,a+Ja+η1b,a-αga.Proof.
From(η1,η2)-convexity of f, using the changes of variables x=a+tη1(b,a) and x=a+(1-t)η1(b,a), respectively, we obtain the two following inequalities:(26)Γαη1αb,aJa+η1b,a-αfaga=1η1αb,a∫aa+η1b,ax-aα-1fxgxdx≤∫01tα-1fa+tη2fb,faga+tη1b,adt=fa∫01tα-1ga+tη1b,adt+η2fb,fa∫01tαga+tη1b,adt,and(27)Γαη1αb,aJa+αfa+η1b,aga+η1b,a=1η1αb,a∫aa+η1b,aa+η1b,a-xα-1fxgxdx≤∫01tα-1fa+1-tη2fb,faga+1-tη1b,adt=fa∫01tα-1ga+1-tη1b,adt+η2fb,fa∫01tα-11-tga+1-tη1b,adt.
Now adding (26) to (27) with the fact that g is symmetric to a+(1/2)η1(b,a) implies that (28)Γαη1αb,aJa+αfa+η1b,aga+η1b,a+Ja+η1b,a-αfaga≤2fa∫01tα-1ga+tη1b,adt+η2fb,fa∫01tα-1ga+tη1b,adt=2fa+η2fb,fa∫01tα-1ga+tη1b,adt=Γαη1αb,a2fa+η2fb,faJa+η1b,a-αga.Now by the use of Lemma 9 we have (29)Γαη1αb,aJa+αfa+η1b,aga+η1b,a+Ja+η1b,a-αfaga≤Γαη1αb,a2fa+η2fb,fa2Ja+αga+η1b,a+Ja+η1b,a-αga, which implies the respected inequality.Corollary 11.
If in Theorems8 and 10 we consider
(i)α=1, then the following inequality holds, which is the classical form of Fejér inequality related to (η1,η2)-convex functions:(30)f2a+η1b,a2∫aa+η1b,agxdx-12∫aa+η1b,aη2fx,f2a+η1b,a-xgxdx≤∫aa+η1b,afxgxdx≤fa+η2fb,fa2∫aa+η1b,agxdx.
(ii)g≡1, then we get Hermite-Hadamard inequality for (η1,η2)-convex functions as follows:(31)f2a+η1b,a2≤Γα+12b-aαJa+αfb+Jb-αfa≤2fa+η2fb,fa2, which is a generalization of inequality (2.1) in [12].Corollary 12.
If in Theorems8 and 10 we set η1(x,y)=x-y for all x,y∈I, then we obtain Fejér inequality for fractional integrals related to η-convex functions.(32)fa+b2Ja+αgb+Jb-αga≤Ja+αfgb+Jb-αfga≤2fa+η2fb,fa2Ja+αgb+Jb-αga.Corollary 13.
If in Theorems8 and 10 we set η2(x,y)=x-y for all x,y∈f(I), then we obtain Fejér inequality for fractional integrals related to preinvex functions.(33)f2a+η1b,a2Ja+αga+η1b,a+Ja+η1b,a-αga≤Ja+αfga+η1b,a+Ja+η1b,a-αfga≤fa+fb2Ja+αga+η1b,a+Ja+η1b,a-αga.Corollary 14 (see [11]).
With all conditions of Corollaries12 and 13, we have the classic Fejér inequality for fractional integrals.(34)fa+b2Ja+αgb+Jb-αga≤Ja+αfgb+Jb-αfga≤fa+fb2Ja+αgb+Jb-αga.Corollary 15 (see [12]).
If in (34) we consider g≡1, then we recapture Hermite-Hadamard inequality for fractional integrals in convex case.(35)fa+b2≤Γα+12b-aαJa+αfb+Jb-αfa≤fa+fb2.


## 3. Mid-Point and Trapezoid-Type Inequalities

In this section, we obtain, respectively, the mid-point and trapezoid-type inequalities related to (31) when the absolute value of the derivative of the considered function is (η1,η2)-convex. In fact, by mid-point-type inequality we mean estimating the difference between left and middle parts of (31) and by trapezoid-type inequality we mean estimating the difference between right and middle parts of (31).The following lemma is generalization of Lemma 1 obtained in [13] to the preinvex case.Lemma 16.
LetI⊆R be an open invex set with respect to η1:I×I→R and let f:I→R be a differentiable function. For any a,b∈I with η1(b,a)>0, if f∈L1[a,a+η1(b,a)], then the following equality for fractional integrals holds: (36)f2a+η1b,a2-Γα+12η1αb,aJa+αfa+η1b,a+Ja+η1b,a-αfa=η1b,a2∑k=14Ik,where (37)I1=∫01/2tαf′a+tη1b,adt,I2=∫01/2-tαf′a+1-tη1b,adt,I3=∫1/21tα-1f′a+tη1b,adt,I4=∫1/211-tαf′a+1-tη1b,adt.Proof.
Integrating by parts inI1 implies that (38)I1=∫01/2tαf′a+tη1b,adt=tαfa+tη1b,aη1b,a01/2-αη1b,a∫01/2tα-1fa+tη1b,adt=2-αη1b,af2a+η1b,a2-αη1b,a∫01/2tα-1fa+tη1b,adt. Similarly, we have (39)I2=2-αη1b,af2a+η1b,a2-αη1b,a∫01/2tα-1fa+1-tη1b,adt,I3=∫1/21tα-1f′a+tη1b,adt=tα-1fa+tη1b,aη1b,a1/21-αη1b,a∫1/21tα-1fa+tη1b,adt=1-2-αη1b,af2a+η1b,a2-αη1b,a∫1/21tα-1fa+tη1b,adt,I4=1-2-αη1b,af2a+η1b,a2-αη1b,a∫1/21tα-1fa+1-tη1b,adt. Now, by adding all of above equalities, we get to the desired result. (40)I1+I2+I3+I4=2η1b,af2a+η1b,a2-αη1b,a∫01tα-1fa+tη1b,adt+∫01tα-1fa+1-tη1b,adt=2η1b,af2a+η1b,a2-Γα+1η1α+1b,aJa+η1b,a-αfa+Ja+αfa+η1b,a.The mid-point-type inequality related to (31) is obtained in the following.Theorem 17.
LetI⊆R be an open invex set with respect to η1:I×I→R and let f:I→R be a differentiable function. Suppose that f′ is an (η1,η2)-convex function on I and, for any a,b∈I with η1(b,a)>0, f′∈L1[a,a+η1(b,a)]. Then (41)f2a+η1b,a2-Γα+12η1αb,aJa+αfa+η1b,a+Ja+η1b,a-αfa≤η1b,a2α+1α+1f′a+f′b+12η2f′a,f′b+12η2f′b,f′a,for 0<α≤1.Proof.
Using Lemma16, we get (42)f2a+η1b,a2-Γα+12η1αb,aJa+αfa+η1b,a+Ja+η1b,a-αfa≤η1b,a2∑k=14Ik. Now, using (η1,η2)-convexity of f′, we obtain (43)I1≤∫01/2tαf′a+tη1b,adt≤∫01/2tαf′adt+∫01/2tα+1η2f′b,f′adt=12α+1α+1f′a+12α+2α+2η2f′b,f′a.Analogously, (44)I2≤12α+1α+1f′b+12α+2α+2η2f′a,f′b. Also, using the fact that (45)t1α-t2α≤t1-t2αfor all α∈(0,1] and t1,t2∈[0,1], we have (46)I3≤12α+1α+1f′a+α+32α+2α+1α+2η2f′b,f′a,and (47)I4≤12α+1α+1f′b+α+32α+2α+1α+2η2f′a,f′b. Now adding all of above inequalities implies the required result.Corollary 18.
If in Theorem17 we consider η1(x,y)=x-y for all x,y∈I, then (48)fa+b2-Γα+12b-aαJa+αfb+Jb-αfa≤b-a2α+1α+1f′a+f′b+12η2f′a,f′b+12η2f′b,f′a. Furthermore, if we set η2(x,y)=x-y for all x,y∈f(I), then we recapture Theorem 2 in [13].The following result has been obtained in [14].Lemma 19.
LetI⊆R be an open invex set with respect to η1:I×I→R. Also, suppose that f:I→R is a differentiable function. For any a,b∈I with η1(b,a)>0, if f∈L1[a,a+η1(b,a)], then the following equality for fractional integral holds: (49)fa+fa+η1b,a2-Γα+12η1αb,aJa+αfa+η1b,a+Ja+η1b,a-αfa=η1b,a2∫01tα-1-tαf′a+tη1b,adt.Now we give the trapezoid-type inequality related to (31).Theorem 20.
LetI⊆R be an open invex set with respect to η1:I×I→R and let f:I→R be a differentiable function. Suppose that f′ is an (η1,η2)-convex function on I and, for any a,b∈I with η1(b,a)>0, f′∈L1[a,a+η1(b,a)]. Then the following inequality for fractional integrals holds: (50)fa+fa+η1b,a2-Γα+12η1αb,aJa+αfa+η1b,a+Ja+η1b,a-αfa≤η1b,a2α+11-12α2f′a+η2f′b,f′a.Proof.
Using Lemma19 and (η1,η2)-convexity of f′, we get (51)fa+fa+η1b,a2-Γα+12η1αb,aJa+αfa+η1b,a+Ja+η1b,a-αfa≤η1b,a2∫01tα-1-tαf′a+tη1b,adt≤η1b,a2∫01tα-1-tαf′a+tη2f′b,f′adt=η1b,a2∫01/21-tα-tαf′a+tη2f′b,f′adt+∫1/21tα-1-tαf′a+tη2f′b,f′adt=η1b,a2α+11-12α2f′a+η2f′b,f′a, where the last equality can be obtained after some calculations in corresponding integrals and utilizing them.Corollary 21.
If in Theorem20 we consider
(i)η1(x,y)=x-y for all x,y∈I, then we obtain (52)fa+fb2-Γα+12b-aαJa+αfb+Jb-αfa≤b-a2α+11-12α2f′a+η2f′b,f′a.
(ii)η2(x,y)=x-y for all x,y∈f(I), then we have Theorem 2.5 in [14].
(iii) conditions of (i) and (ii) together, then we recapture Theorem 3 in [12].As the last result by using Theorem 2 in [15], we obtain a refinement of Hermite-Hadamard inequality in connection with fractional integrals related to the increasing (η1,η2)-convex functions.Theorem 22 (see [15]).
Iff1 and f2 are positive increasing functions on [0,1], then (53)∫01f1xdx∫01f2xdx≤∫01f1xf2xdx.Also iff1 and f2 are positive decreasing functions on [0,1] and K is an upper bound for f1 and f2, then K-f1 and K-f2 are positive increasing functions and we have (54)∫01K-f1xdx∫01K-f2xdx≤∫01K-f1xK-f2xdx, which gives again (55)∫01f1xdx∫01f2xdx≤∫01f1xf2xdx.Theorem 23.
LetI⊆R be an invex set with respect to η1:I×I→R. Also let f:I→R be an increasing positive (η1,η2)-convex function. If, for each a,b∈I with η1(b,a)>0, f∈L1[a,a+η1(b,a)] and η2 is integrable on f(I)×f(I), then, for α>1, the following inequalities hold: (56)f2a+η1b,a2≤1η1b,a∫aa+η1b,afxdx≤Γα+12η1αb,aJa+αfa+η1b,a+Ja+η1b,a-αfa≤fa+η2fb,fa2.Proof.
From(η1,η2)-convexity of f, we have, (57)fa+tη1b,a≤fa+tη2fb,fa, and (58)fa+1-tη1b,a≤fa+1-tη2fb,fa. By adding these inequalities, we get(59)fa+tη1b,a+fa+1-tη1b,a≤2fa+η2fb,fa.Multiplying both sides of (59) by tα-1 and integrating the resulting inequality with respect to t over [0,1] and using Theorem 22, we obtain (60)∫01tα-1dt·∫01fa+tη1b,adt+∫01tα-1dt·∫01fa+1-tη1b,adt≤∫01tα-1fa+tη1b,adt+∫01tα-1fa+1-tη1b,adt≤2fa+η2fb,fa∫01tα-1dt. Using the changes of variables x=a+tη1(b,a) and x=a+(1-t)η1(b,a), respectively, in above integrals, we have (61)1η1b,a∫01tα-1dt∫aa+η1b,afxdx+∫aa+η1b,af2a+η1b,a-xdx≤Γαη1αb,aJa+αfa+η1b,a+Ja+η1b,a-αfa≤2fa+η2fb,fa∫01tα-1dt.Since (62)∫01tα-1dt=1α, and (63)∫aa+η1b,afxdx=∫aa+η1b,af2a+η1b,a-xdx,we get (64)1η1b,a∫aa+η1b,afxdx≤Γα+12η1αb,aJa+αfa+η1b,a+Ja+η1b,a-αfa≤fa+η2fb,fa2. Furthermore, since f is (η1,η2)-convex, then (65)f2a+η1b,a2≤1η1b,a∫aa+η1b,afxdx, which completes the proof.Corollary 24.
Letf:[a,b]→R be an increasing positive convex function. Then, for α>1, the following inequalities hold: (66)fa+b2≤1b-a∫abfxdx≤Γα+12b-aαJa+αfb+Jb-αfa≤fa+fb2.


## 4. Conclusions

The convexity of a function is the basis for many inequalities in mathematics. Note that, in new problems related to the convexity, generalized notions about convex functions are required to obtain applicable results. One of these generalizations is the notion of(η1,η2)-convex functions which can generalize many inequalities related to convex functions such as Hermite-Hadamard inequality, Fejér inequality, and trapezoid-type and mid-point-type inequalities.



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