# Hadamard-Typek-Fractional Integral Inequalities for Exponentially α,h−m-Convex Functions

Authors: Chahn Yong Jung; Ghulam Farid; Sidra Bibi; Kottakkaran Sooppy Nisar; Shin Min Kang
Journal: Journal of Mathematics
 (2020)

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## Abstract

The class of exponentiallyα,h−m-convex functions has been discovered to unify different kinds of convexities. This paper finds new Riemann–Liouville fractional Hadamard-type inequalities for this generalized class of convex functions. These results further have their consequences which are presented in the form of corollaries. Moreover, some known results are identified in the form of remarks.


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## Body

## 1. Introduction

Fractional calculus has opened a new era in the theoretic and application point of view. Many subjects of science and engineering get wonderful advancements in modeling and solving complex systems with the help of fractional integral and derivative operators. Besides making advancements in other fields, fractional integral and derivative operators have been proved very useful in generalizing and extending mathematical inequalities. In this paper, we will extend the Hadamard-type inequalities for Riemann–Liouville (RL) fractional integral operators.Definition 1.
Letf∈L1a,b. The left and right RL fractional integrals Ia+αf and Ib−αf of order α∈ℂ, ℜα>0, of f are defined by(1)Ia+αfx=1Γα∫axx−t1−αftdt,x>a,Ib−αfx=1Γα∫xbt−x1−αftdt,x<b,respectively, where Γα is the Euler’s gamma function and Ia+0fx=Ib−0fx=fx.
Thek-analogue RL fractional integral operators are studied by Mubeen and Habibullah in [1].Definition 2.
Letf∈L1a,b. The left and right RL k-fractional integrals Ia+α,kf and Ib−α,kf of order α∈ℂ,ℜα>0,k>0, of f are defined by(2)Ia+α,kfx=1kΓkα∫axx−tα/k−1ftdt,x>a,Ib−α,kfx=1kΓkα∫xbt−xα/k−1ftdt,x<b,where Γkα is the k-gamma function and Ia+0,1fx=Ib−0,1fx=fx.
Fractional integral inequalities are generalizations of classical inequalities which play a key role in the solutions and their uniqueness in fractional boundary value problems. The Hadamard inequality is one of the classical inequalities for convex functions, and it has been studied by a lot of researchers for different kinds of fractional integral and derivative operators, see [2–11] and references therein. It is stated in the undermentioned theorem.Theorem 1 (see [11]).
Iff:I⟶ℝ is a convex function on the interval I of real numbers and a,b∈I with a<b then(3)fa+b2≤1b−a∫abfxdx≤fa+fb2.Definition 3.
A functionf:I⟶ℝ, where I is an interval in ℝ, is said to be convex function if(4)frx+1−ry≤rfx+1−rfy,holds for all x,y∈I and r∈0,1.
The Hadamard inequality is the geometric interpretation of convex functions which has been analyzed by many researchers for fractional integral and differentiation operators. The objective of this paper is to obtaink-fractional integral inequalities for a generalized class of convex functions, namely, exponentially α,h−m-convex functions. Convex functions proved very useful for the establishment of very known and vital inequalities. An important and significant generalization of convex functions is exponentially α,h−m-convex functions. It is defined by He et al. in [12] as follows.Definition 4.
LetJ⊆ℝ be an interval containing 0,1 and let h:J⟶ℝ be a nonnegative function. A function f:0,b⟶ℝ is called an exponentially α,h−m-convex function, if f is nonnegative, and for all x,y∈0,b, t∈0,1, α,m∈0,12, and η∈ℝ, one has(5)ftx+m1−ty≤htαfxeηx+mh1−tαfyeηy.Remark 1.
By selecting a suitable functionh and a particular value of parameters m and η, the above definition produces the functions comprised in the following remarks:(i)
By settingη=0, α,h−m-convex function [4] can be obtained(ii)
By settingη=0 and α=1, h−m-convex function [13] can be obtained(iii)
By settingη=0 and htα=tα, α,m-convex function [14] can be obtained(iv)
By settingη=0,α=1, and m=1, h-convex function [15] can be obtained(v)
By settingη=0, α=1, and ht=t, m-convex function [16] can be obtained(vi)
By settingη=0, α=1, m=1, and ht=t, convex function [17] can be obtained(vii)
By settingη=0, m=1, α=1, and ht=1, p-function [18] can be obtained(viii)
By settingα=1 and ht=ts, exponentially s,m-convex function [19] can be obtained(ix)
By settingα=1, m=1, and ht=ts, exponentially s-convex function [20] can be obtained(x)
By settingα=1 and ht=t, exponentially m-convex function [21] can be obtained(xi)
By settingα=1, m=1, and ht=t, exponentially convex function [22] can be obtained(xii)
By settingη=0, α=1, and ht=ts, s,m-convex function [23] can be obtained(xiii)
By settingα=1, η=0, m=1, and ht=ts, s-convex function [20] can be obtained(xiv)
By settingα=1, η=0, m=1, and ht=1/t, Godunova–Levin function [24] can be obtained(xv)
By settingα=1, η=0, m=1, and ht=1/ts, s-Godunova–Levin function of the second kind [25] can be obtained
The article is organized in the following manner: in Section2, we prove the k-fractional integral inequality of Hadamard type for exponentially α,h−m-convex functions and deduce some related results. In section 3, we prove a version of the k-fractional integral inequality of Hadamard type for differentiable function f so that f′ is exponentially α,h−m-convex. In section 4, we prove the k-fractional integral inequality of Hadamard type for the product of two exponentially α,h−m-convex functions.


## 2. Main Results

First, we give ak-fractional integral inequality for exponentially α,h−m-convex functions as follows.Theorem 2.
Letf:0,∞⟶ℝ be an exponentially α,h−m-convex function with α,m∈0,1×0,1, also let f∈L1a,b, 0≤a<mb. Then, we will have(6)kΓkαb−aα/kIa+α,kfb+Ib−α,kfa≤faeηa+fbeηb∫01tα/k−1htαdt+mfb/meηb/m+fa/meηa/m∫01tα/k−1h1−tαdt≤1α/kp−p+11/pfaeηa+fbeηb∫01htαqdt1/q+mfb/meηb/m+fa/meηa/m∫01htαqdt1/q,where p−1+q−1=1,p>1, and η∈ℝ.Proof.
Sincef is exponentially α,h−m-convex on a,b, for t∈0,1, we have(7)fta+1−tb+f1−ta+tb≤htαfaeηa+fbeηb+mh1−tαfb/meηb/m+fa/meηa/m,from which multiplying both sides with tα/k=−1 and integrating over 0,1, we will have(8)∫01tα/k−1fta+1−tb+f1−ta+tbdt≤faeηa+fbeηb∫01tα/k−1htαdt+mfb/meηb/m+fa/meηa/m∫01tα/k−1h1−tαdt.
By changing variables, we will have(9)kΓkαb−aα/kIa+α,kfb+Ib−α,kfa≤faeηa+fbeηb∫01tα/k−1htαdt+mfb/meηb/m+fa/meηa/m∫01tα/k−1h1−tαdt.
This completes the proof of first inequality in (6). The second inequality in (6) follows by using Holder’s inequality:(10)∫01tα/k−1htαdt≤1α/kp−p+11/p∫01htαqdt1/q,∫01tα/k−1h1−tαdt≤1α/kp−p+11/p∫01h1−tαqdt1/q.
Thus from (9), we will get (6).Some special cases of the above theorem are discussed in the following corollaries.Corollary 1.
The following inequality holds for an exponentiallyα,h−m-convex function via an RL fractional integral by setting k=1 in inequality (6):(11)Γαb−aαIa+αfb+Ib−αfa≤faeηa+fbeηb∫01tα−1htαdt+mfb/meηb/m+fa/meηa/m∫01tα−1h1−tαdt≤1αp−p+11/pfaeηa+fbeηb∫01htαqdt1/q+mfb/meηb/m+fa/meηa/m∫01h1−tαqdt1/q.Corollary 2.
The following inequality holds for theα,h−m-convex function via an RL k-fractional integral by setting η=0 in inequality (6):(12)kΓkαb−aα/kIa+α,kfb+Ib−α,kfa≤fa+fb∫01tα/k−1htαdt+mfbm+fam∫01tα/k−1h1−tαdt≤1α/kp−p+11/pfa+fb∫01htαqdt1/q+mfbm+fam∫01h1−tαqdt1/q.Corollary 3.
The following inequality holds for theh−m-convex functions via RL k-fractional integrals by setting η=0 and α=1 in inequality (6):(13)kΓkαb−aα/kIa+α,kfb+Ib−α,kfa≤fa+fb∫01tα/k−1htdt+mfbm+fam∫01tα/k−1h1−tdt≤1α/kp−p+11/p∫01htqdt1/qfa+fb+mfbm+fam.Corollary 4.
The following inequality holds for theh-convex function via RL fractional integrals by setting η=0, k=1, m=1, and α=1 in inequality (6):(14)Γαb−aαIa+αfb+Ib−αfa≤fa+fb∫01tα−1ht+h1−tdt≤2fa+fbαp−p+11/p∫01htqdt1/q.
The next two special cases are already proved in [26].Corollary 5.
The following inequality holds for the exponentiallys,m-convex function via RL k-fractional integrals:(15)kΓkαb−aα/kIa+α,kfb+Ib−α,kfa≤faeηa+fbeηbkα+ks+mfb/meηb/m+fa/meηa/mβαk,s+1≤faeηa+fbeηbkα+ks+fb/meηb/m+fa/meηa/mmα/kp−p+11/pqs+11/q.Proof.
By settingα=1 and h1/2=1/2s in inequality (6) of Theorem 2, we get the above inequality (15) which is given in [26] (Theorem 1).Corollary 6.
The following inequality holds for thes,m-convex function via RL k-fractional integrals:(16)kΓkαb−aα/kIa+α,kfb+Ib−α,kfa≤fa+fbkα+ks+mfbm+famβαk,s+1≤fa+fbkα+ks+fbm+fammα/kp−p+11/pqs+11/q.Proof.
By settingη=0, α=1, and h1/2=1/2s in inequality (6) of Theorem 2, we get the above inequality (16) which is given in [26] (Theorem 4). □
The next result holds for exponentiallyα,h−m-convex functions, whereas the function h is superadditive.Theorem 3.
Letf:0,∞⟶ℝ be exponentially α,h−m-convex functions with α,m∈0,1×0,1, also let h be superadditive and f∈L1a,b, 0≤a<mb. Then, for RL k-fractional integrals, we have(17)Γkα+kb−aα/kIa+α,kfb+Ib−α,kfa≤h12faeηa+fbeηb+mfb/meηb/m+fa/meηa/m,η∈ℝ.Proof.
Sincef is an exponentially α,h−m-convex function on a,b, for t∈0,1, we get(18)fta+1−tb+f1−ta+tb≤htα+h1−tα2faeηa+fbeηb+mfb/meηb/m+fa/meηa/m,since h is superadditive; therefore,(19)fta+1−tb+f1−ta+tb≤h12faeηa+fbeηb+mfb/meηb/m+fa/meηa/m.
By multiplying both sides of the above inequality withtα/k−1 and integrating over 0,1, we will have(20)∫01tα/k−1fta+1−tb+f1−ta+tbdt≤h12faeηa+fbeηb+mfb/meηb/m+fa/meηa/m∫01tα/k−1dt.
By substitutingw=ta+1−tb in the left side of the above inequality, it leads to (17).Some special cases are given in the form of following corollaries.Corollary 7.
The following inequality holds for the exponentiallyα,h−m-convex function via RL fractional integrals by setting k=1 in inequality (17):(21)Γkα+kb−aα/kIa+α,kfb+Ib−α,kfa≤h12faeηa+fbeηb+mfb/meηb/m+fa/meηa/m.Corollary 8.
The following inequality holds for theα,h−m-convex function via RL k-fractional integrals by setting η=0 in inequality (17):(22)Γkα+kb−aα/kIa+α,kfb+Ib−α,kfa≤h12fa+fb+mfbm+fam.Corollary 9.
The following inequality holds forh−m-convex functions via RL k-fractional integrals:(23)Γkα+kb−aα/kIa+α,kfb+Ib−α,kfa≤h1fa+fb2+mfb/m+fa/m2.Proof.
By settingη=0 and α=1 in inequality (17) of Theorem 3, we get the above inequality (22) which is given in [4] (Theorem 2.6).


## 3. Fractional Hadamard-Type Inequalities for Functions Whose Derivatives in Absolute Values are Exponentiallyα,h−m-Convex

In the following,k-fractional integral inequalities of Hadamard type for exponentially α,h−m-convex functions in terms of the first derivatives have been obtained. For the next result, we use the following lemma.Lemma 1 (see [4]).
Letf:a,mb⟶ℝ be a differentiable mapping on the interval a,mb with a<mb. If f′∈L1a,mb, then for k-fractional integrals, we will have(24)fmb+fa2−kΓkα/k+k2mb−aα/kIa+α,kfmb+Imb−α,kfa=mb−a2∫011−tα/k−tα/kf′m1−tb+tadt.Theorem 4.
Letf:a,mb⟶ℝ be a function such that a,mb⊆0,∞ and f∈L1a,mb. If f′ is exponentially α,h−m-convex functions with α,m∈0,12 and hq∈L10,1, q>1. Then, for RL k-fractional integrals, we have(25)fmb+fa2−kΓkα/k+k2mb−aα/kIa+α,kfmb+Imb−α,kfa≤mb−a22α/kp+1−12α/kp+1α/kp+11/p−12α/kp+1α/kp+11/p×f′aeηa∫01/2htαqdt1/q+∫1/21htαqdt1/q+mf′beηb∫01/2h1−tαqdt1/q+∫1/21h1−tαqdt1/q,where 1/p+1/q=1,η∈ℝ.Proof.
By using the property of modulus from Lemma1, we will get(26)fmb+fa2−kΓkα/k+k2mb−aα/kIa+α,kfmb+Imb−α,kfa≤mb−a2∫011−tα/k−tα/kf′m1−tb+tadt.
By exponentiallyα,h−m-convexity of f′, we have(27)fmb+fa2−kΓkα/k+k2b−aα/kIa+α,kfmb+Imb−α,kfa≤mb−a2∫01/21−tα/k−tα/kmh1−tαf′beηb+htαf′aeηadt+∫1/201−tα/k−tα/kmh1−tαf′beηb+htαf′aeηadt=mb−a2f′aeηa∫01/21−tα/khtαdt−∫01/2tα/khtαdt+mf′beηb∫01/21−tα/kh1−tαdt−∫01/2tα/kh1−tαdt+f′aeηa∫1/21tα/khtαdt−∫1/211−tα/khtαdt+mf′beηb∫1/21tα/kh1−tα/kdt−∫1/211−tα/kh1−tαdt.
Now, by using Holder inequality in the right-hand side of (27), we will get(28)fmb+fa2−kΓkα/k+k2mb−aα/kIa+α,kfmb+Imb−α,kfa≤mb−a2f′aeηa2α/kp+1−12α/kp+1α/kp+11/p−12α/kp+1α/kp+11/p∫01/2htαqdt1/q+2α/kp+1−12α/kp+1α/kp+11/p−12α/kp+1α/kp+11/p∫1/21htαqdt1/q+mf′beηb2α/kp+1−12α/kp+1α/kp+11/p−12α/kp+1α/kp+11/p∫01/2h1−tαqdt1/q+2α/kp+1−12α/kp+1α/kp+11p−12α/kp+1α/kp+11p∫121h1−tαqdt1q.
By some manipulation, one can get inequality (25).Corollary 10.
The following inequality holds for the exponentiallyα,h−m-convex function via RL fractional integrals by setting k=1 in inequality (25):(29)fmb+fa2−Γα+12mb−aαIa+αfmb+Imb−αfa≤mb−a22αp+1−12αp+1αp+11/p−12αp+1αp+11/p×f′aeηa∫01/2htαqdt1/q+∫1/21h1−tαqdt1/q+mf′beηb∫01/2h1−tαqdt1/q+∫1/21h1−tαqdt1/q.Corollary 11.
The following inequality holds for theα,h−m-convex function via RL k-fractional integrals by setting η=0 in inequality (25):(30)fmb+fa2−kΓkα/k+k2mb−aα/kIa+α,kfmb+Imb−α,kfa≤mb−a22α/kp+1−12α/kp+1α/kp+11/p−12α/kp+1α/kp+11/p×f′a∫01/2htαqdt1/q+∫1/21htαqdt1/q+mf′a∫012h1−tαqdt1/q+∫121h1−tαqdt1/q.The following special cases are proved in [4, 26].Corollary 12.
The following inequality holds for theh−m-convex function via RL k-fractional integrals:(31)fmb+fa2−kΓkα/k+k2mb−aα/kIa+α,kfmb+Imb−α,kfa≤mb−af′a+mf′b22α/kp+1−12α/kp+1α/kp+11/p−12α/kp+1α/kp+11/p×∫01/2htαqdt1/q+∫1/21htαqdt1/q.Proof.
By settingη=0 and α=1 in inequality (25) of Theorem 4, we get the above inequality (31) which is given in [4] (Theorem 3.6).Corollary 13.
The following inequality holds for the exponentiallys,m-convex function via RL k-fractional integrals:(32)fmb+fa2−kΓkα/k+k2mb−aα/kIa+α,kfmb+Imb−α,kfa≤mb−amf′b/eηb+f′a/eηa22α/k+s+1−22α/k+s+1α/k+s+1−12α/kp+1α/kp+11/p2qs+1−12qs+1qs+11/q+2α/kp+1−12α/kp+1α/kp+11/p12qs+1qs+11/q.Proof.
By settingα=1 and ht=ts in inequality (25) of Theorem 4, we get the above inequality (32) which is given in [26] (Theorem 2).Corollary 14.
The following inequality holds for the exponentiallys,m-convex function via RL k-fractional integrals:(33)fmb+fa2−kΓkα/k+k2mb−aα/kIa+α,kfmb+Imb−α,kfa≤mb−amf′b−f′a22α/k+s+1−22α/k+s+1α/k+s+1−12α/kp+1α/kp+11/p2qs+1−12qs+1qs+11/q+2α/kp+1−12α/kp+1α/kp+11/p12qs+1qs+11/q,.Proof.
By settingη=0, α=1, and ht=ts in inequality (25) of Theorem 4, we get the above inequality (33) which is given in [26] (Theorem 5).


## 4. Fractional Hadamard-Type Inequalities for Product of Two Exponentiallyα,h−m-Convex Functions

Now, we obtain some Hadamard-type inequalities for the product of two exponentiallyα,h−m-convex functions via RL k-fractional integrals.Theorem 5.
Letf,g:0,∞⟶ℝ be functions such that fg∈L1a,b, 0≤a<mb. If function f is exponentially α,h−m1-convex and function g is exponentially α,h−m2-convex on 0,∞ with α,m1,α,m2∈0,12, then the following inequalities hold for RL k-fractional integrals:(34)kΓkαb−aα/kIa+α,kfbgb+Ib−α,kfaga≤fagae2ηa+fbgbe2ηb∫01tα/k−1h2tαdt+m2fagb/m2eηa+b/m2+fbga/m2eηb+a/m2+m1gafb/m1eηa+b/m1+gbfa/m1eηb+a/m1×∫01tα/k−1htαh1−tαdt+m1m2fb/m1gb/m2eηbm1+m2/m1m2+fa/m1ga/m2eηam1+m2/m1m2∫01tα/k−1h21−tαdt≤1α/kp−p+11/p∫01h2qtαdt1/qfagae2ηa+fbgbe2ηb+m2fagb/m2eηa+b/m2+m1gafb/m1eηa+b/m1+m2fbga/m2eηb+a/m2+m1gbfa/m1eηb+a/m1×∫01htαh1−tαqdt1/q+∫01h2q1−tαdt1/qm1m2fb/m1gb/m2eηbm1+m2/m1m2+fa/m1ga/m2eηam1+m2/m1m2,where p>1 and 1/p+1/q=1.Proof.
Since functionf is exponentially α,h−m1-convex and function g is exponentially α,h−m2-convex, for t∈0,1, we have(35)fta+1−tbgta+1−tb≤h2tαfagae2ηa+m2htαh1−tαfagb/m2eηa+b/m2+m1htαh1−tαgafb/m1eηa+b/m1+m1m2h21−tαfb/m1gb/m2eηbm1+m2/m1m2,by multiplying both sides of the above inequality with tα/k−1 and integrating over 0,1, we have(36)∫01tα/k−1fta+1−tbgta+1−tbdt≤fagae2ηa∫01tα/k−1h2tαdt+m2fagb/m2eηa+b/m2∫01tα/k−1htαh1−tαdtn+m1gafb/m1eηa+b/m1∫01tα/k−1htαh1−tαdt+m1m2fb/m2gb/m2eηbm1+m2/m1m2∫01tα/k−1h21−tαdt.
By substitutingz=ta+1−tb in the left-hand side of the above inequality, we get(37)kΓkαb−aα/kIa+α,kfbgb≤fagae2ηa∫01tα/k−1h2tαdt+m2fagb/m2eηa+b/m2∫01tα/k−1htαh1−tαdt+m1gafb/m1eηa+b/m1∫01tα/k−1htαh1−tαdt+m1m2fb/m1gb/m2eηbm1+m2/m1m2∫01tα/k−1h21−tαdt.
By using Holder inequality, we will have(38)∫01tα/k−1h2tαdt≤1α/kp−p+11/p∫01h2qtαdt1/q,∫01tα/k−1htαh1−tαdt≤1α/kp−p+11/p∫01htαh1−tαqdt1/q,∫01tα/k−1h21−tαdt≤1α/kp−p+11/p∫01h2q1−tαdt1/q.
Thus, we will get(39)kΓkαb−aα/kIa+α,kfbgb≤fagae2ηa∫01tα/k−1h2tαdt+m2fagb/m2eηa+b/m2∫01tα/k−1htαh1−tαdt+m1gafb/m1eηa+b/m2∫01tα/k−1htαh1−tαdt+m1m2fb/m1gb/m2eηbm1+m2/m1m2∫01tα/k−1h21−tαdt≤1α/kp−p+11/p∫01h2qtαdt1/pfagae2ηa+∫01htαh1−tαqdt1/qm2fagb/m2eηa+b/m2+m1gafb/m1eηa+b/m1+∫01h2q1−tαdt1/qm1m2fb/m1gb/m2eηbm1+m2/m1m2.
Similarly, by changing the roles ofa and b, after a little computation, one can get(40)kΓkαb−aα/kIb−α,kfaga≤fbgbe2ηb∫01tα/k−1h2tαdt+m2fbga/m2eηb+a/m2∫01tα/k−1htαh1−tαdt+m1gbfa/m1eηb+a/m1∫01tα/k−1htαh1−tαdt+m1m2fa/m1ga/m2eηam1+m2/m1m2∫01tα/k−1h21−tαdt≤1α/kp−p+11/p∫01h2qtαdt1/qfbgbe2ηb+∫01htαh1−tαqdt1/qm2fbga/m2eηb+a/m2+m1gbfa/m1eηb+a/m1+∫01h2q1−tαdt1/qm1m2fa/m1ga/m2eηam1+m2/m1m2.
Adding (39) and (40), we get the required result.Corollary 15.
The following inequality holds for the exponentiallyα,h−m-convex function via RL fractional integrals:(41)Γαb−aαIa+αfbgb+Ib−αfaga≤fagae2ηa+fbgbe2ηb∫01tα−1h2tαdt+m2fagb/m2eηa+b/m2+fbga/m2eηb+a/m2+m1gafb/m1eηa+b/m1+gbfa/m1eηb+a/m1×∫01tα−1htαh1−tαdt+m1m2fb/m1gb/m2eηbm1+m2/m1m2+fa/m1ga/m2eηam1+m2/m1m2∫01tα−1h21−tαdt≤1α/kp−p+11/p∫01h2qtαdt1/qfagae2ηa+fbgbe2ηb+m2fagb/m2eηa+b/m2+m1gafb/m1eηa+b/m1+m2fbga/m2eηb+a/m2+m1gbfa/m1eηb+a/m1×∫01htαh1−tαqdt1/q+∫01h2q1−tαdt1/qm1m2fb/m1gb/m2eηbm1+m2/m1m2+fa/m1ga/m2eηam1+m2/m1m2.Proof.
By settingk=1 in inequality (34) of Theorem 5, we get the above inequality (41).Corollary 16.
The following inequality holds for theα,h−m-convex function via RL k-fractional integrals:(42)kΓkαb−aα/kIa+α,kfbgb+Ib−α,kfaga≤faga+fbgb∫01tα/k−1h2tαdt+m2fagbm2+fbgam2+m1gafbm1+gbfam1×∫01tα/k−1htαh1−tαdt+m1m2fbm1gbm2+fam1gam2∫01tα/k−1h21−tαdt≤1α/kp−p+11/p∫01h2qtαdt1/qfaga+m2fagbm2+m1gafbm1+m2fbgam2+m1gbfam1×∫01htαh1−tαqdt1/q+∫01h2q1−tαdt1/qm1m2fbm1gbm2+fam1gam2.Proof.
By settingη=0 in inequality (34) of Theorem 5, we get the above inequality (42).Corollary 17.
The following inequality holds for theh−m-convex function via RL k-fractional integrals:(43)kΓkαb−aα/kIa+α,kfbgb+Ib−α,kfaga≤faga+fbgb∫01tα/k−1h2tdt+m2fagbm2+fbgam2+m1gafbm1+gbfam1×∫01tα/k−1hth1−tdt+m1m2fbm1gbm2+fam1gam2∫01tα/k−1h21−tdt≤1α/kp−p+11/p∫01h2qtdt1/qfaga+m1m2fbm1gbm2+fbgb+m1m2fbm1gbm2+∫01hth1−tqdt1/qm2fagbm2+m1gafbm1+m2fbgam2+m1gbfam1.Proof.
By settingη=0 and α=1 in inequality (34) of Theorem 5, we get the above inequality (43) which is given in [4] (Theorem 4.3).Corollary 18.
The following inequality holds for the exponentiallys,m-convex function via RL k-fractional integrals:(44)kΓkαb−aα/kIa+α,kfbgb+Ib−α,kfaga≤kα+2ksfagae2ηa+fbgbe2ηb+m1gbfa/m1eηb+a/m1+gafb/m1eηa+b/m1+m2fbga/m2eηb+a/m2+fagb/m2eηa+b/m2βα/k+s,s+1+m1m2fb/m1gb/m2eηbm1+m2/m1m2+fa/m1ga/m2eηam1+m2/m1m2βα/k,2s+1≤kα+2ksfagae2ηa+fbgbe2ηb+m1gbfa/m1eηb+a/m1+gafb/m1eηa+b/m1+m2fbga/m2eηb+a/m2+fagb/m2eηa+b/m2×kpα−k+ks+k1/p1qs+11/q+m1m2fb/m1gb/m2eηbm1+m2/m1m2+fa/m1ga/m2eηam1+m2/m1m2kαp−kp+k1/p12qs+11/q.Proof.
By settingα=1 and ht=ts in inequality (34) of Theorem 5, we get the above inequality (44) which is given in [26] (Theorem 3).Corollary 19.
The following inequality holds for thes,m-convex function via RL k-fractional integrals:(45)kΓkαb−aα/kIa+α,kfbgb+Ib−α,kfaga≤kα+2ksfaga+fbgb+m1gbfam1+gafbm1+m2fbgam2+fagbm2βαk+s,s+1+m1m2fbm1gbm2+fam1gam2βαk,2s+1≤kα+2ksfaga+fbgb+m1gbfam1+gafbm1+m2fbgam2+fagbm2×kpα−k+ks+k1/p1qs+11/q+m1m2fbm1gbm2+fam1gam2kαp−kp+k1/p12qs+11/q.Proof.
By settingη=0, α=1, and ht=ts in inequality (34) of Theorem 5, we get the above inequality (45) which is given in [26] (Theorem 6).Theorem 6.
Letf,g:0,∞⟶ℝ be functions such that fg∈L1a,b, 0≤a<mb. If function f is exponentially α1,h−m1-convex and function g is exponentially α2,h−m2-convex on 0,∞ with α,m1,α,m2∈0,12, then the following inequalities hold for RL k-fractional integrals:(46)kΓkαb−aα/kIa+α,kfbgb+Ib−α,kfaga≤fagae2ηa+fbgbe2ηb∫01tα/k−1htα1htα2dt+m1m2fb/m1gb/m2eηbm1+m2/m1m2+fa/m1ga/m2eηam1+m2/m1m2×∫01tα/k−1h1−tα1h1−tα2dt+m2fagb/m2eηa+b/m2+fbga/m2eηb+a/m2+m1gafb/m1eηa+b/m1+gbfa/m1eηb+a/m1∫01tα/k−1htα2h1−tα1dt.Proof.
Since functionf is exponentially α1,h−m1-convex and function g is exponentially α2,h−m2-convex, then for t∈0,1, we have(47)fta+1−tbgta+1−tb≤htα1htα2fagae2ηa+m2htα1h1−tα2fagb/m2eηa+b/m2+m1htα2h1−tα1gafb/m1eηa+b/m1+m1m2h1−tα1h1−tα2fb/m1gb/m2eηbm1+m2/m1m2.
By multiplying both sides of the above inequality withtα/k−1 and integrating over 0,1, we have(48)∫01tα/k−1fta+1−tbgta+1−tbdt≤fagae2ηa∫01tα/k−1htα1htα2dt+m2fagb/m2eηa+b/m2∫01tα/k−1htα1h1−tα2dt+m1gafb/m1eηa+b/m1∫01tα/k−1htα2h1−tα1dt+m1m2fb/m1gb/m2eηbm1+m2/m1m2∫01tα/k−1h1−tα1h1−tα2dt.
By substitutingw=ta+1−tb in the left-hand side of the above inequality, we get(49)kΓkαb−aα/kIa+α,kfbgb≤fagae2ηa∫01tα/k−1htα1htα2dt+m2fagb/m2eηa+b/m2∫01tα/k−1htα1h1−tα2dt+m1gafb/m1eηa+b/m1∫01tα/k−1htα2h1−tα1dt+m1m2fb/m1gb/m2eηbm1+m2/m1m2∫01tα/k−1h1−tα1h1−tα2dt.
Similarly, by changing the rules ofa and b, after a little computation, one can get(50)kΓkαb−aα/kIb−α,kfaga≤fbgbe2ηb∫01tα/k−1htα1htα2dt+m2fbga/m2eηb+a/m2∫01tα/k−1htα1h1−tα2dt+m1gbfa/m1eηb+a/m1∫01tα/k−1htα2h1−tα1dt+m1m2fa/m1ga/m2eηam1+m2/m1m2∫01tα/k−1h1−tα1h1−tα2dt.
Adding (49) and (50), we get the required result.Corollary 20.
The following inequality holds for the exponentiallyα,h−m-convex function via RL fractional integrals:(51)Γαb−aαIa+αfbgb+Ib−αfaga≤fagae2ηa+fbgbe2ηb∫01tα−1htα1htα2dt+m1m2fb/m1gb/m2eηbm1+m2/m1m2+fa/m1ga/m2eηam1+m2/m1m2×∫01tα−1h1−tα1h1−tα2dt+m2fagb/m2eηa+b/m2+fbga/m2eηb+a/m2+m1gafb/m2eηa+b/m2+gbfa/m1eηb+a/m1∫01tα−1htα2h1−tα1dt.Proof.
By settingk=1 in inequality (49) of Theorem 6, we get the above inequality (51).Corollary 21.
The following inequality holds for theα,h−m-convex function via RL k-fractional integrals:(52)kΓkαb−aα/kIa+α,kfbgb+Ib−α,kfaga≤faga+fbgb∫01tα/k−1htα1htα2dt+m1m2fbm1gbm2+fam1gam2×∫01tα/k−1h1−tα1h1−tα2dt+m2fagbm2+fbgam2+m1gafbm1+gbfam1∫01tα/k−1htα2h1−tα1dt.Proof.
By settingη=0 in inequality (49) of Theorem 6, we get the above inequality (52) which is given in [4] (Theorem 4.6).Corollary 22.
The following inequality holds for theα,m-convex function via RL fractional integrals:(53)Γαb−aαIa+αfbgb≤1α1+α2+αfaga+α1α+α1α1+α2+αm2fagbm2+α1α+α1α1+α2+αm1gafbm1+1α−1α+α1−1α+α2+1α1+α2+αm1m2fbm1gbm2,and(54)Γαb−aαIb−αfaga≤1α1+α2+αfbgb+α1α+α1α1+α2+αm2fbgbm2+α1α+α1α1+α2+αm1gbfbm1+1α−1α+α1−1α+α2+1α1+α2+αm1m2fam1gam2.Proof.
From (49) by setting η=0, k=1, and ht=t, we get (53). Similarly, using η=0, k=1, and ht=t in (50), we get (54) which is given in [27] (Theorem 12).



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*Source: 6815056-2020-10-24.xml*