# Some Generalized Formulas of Hadamard-Type Fractional Integral Inequalities

Authors: Xiujun Zhang; Ghulam Farid; Hafsa Yasmeen; Kamsing Nonlaopon
Journal: Journal of Function Spaces
 (2022)

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## Abstract

This paper is aimed at establishing the generalized forms of Riemann-Liouville fractional inequalities of the Hadamard type for a class of functions known as strongly exponentiallyα,h−m-p-convex functions. These inequalities provide some general formulas from which one can get associated inequalities for various types of exponentially convex and strongly convex functions. Refinements of well-known inequalities are also deducible from the established theorems.


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## Body

## 1. Introduction

The notion of convexity is utilized significantly for finding solutions of essential mathematical problems in subjects of science and engineering. Leading with major developments in several branches of mathematics, convexity made its way in statistics, geometric function theory, graph theory, and economics. In recent decades, classes of functions related to convex functions are frequently used in proving new fractional integral inequalities in the form of numerous refinements and generalizations of classical inequalities.LetI be an interval of real numbers. A function f:I⟶ℝ satisfying fxt+1−ty≤tfx+1−tfy, for all x,y∈I and t∈0,1, is called convex function.A convex function satisfies the well-known Hadamard inequality:(1)fx+y2≤1y−x∫xyfξdξ≤fx+fy2.Iff is concave function, then, (1) holds in a reverse order. The inequality (1) had/has been studied by many researchers and consequently obtained a lot of its variants by introducing new classes of functions. For example, in [1], it is studied for s-convex functions; in [2], it is studied for p−h-convex functions; in [3, 4], it is studied for harmonically convex functions; in [5], it is studied for strongly harmonically convex and strongly p-convex functions. Our goal in this paper is to study the inequality (1) for strongly exponentially α,h−m-p-convex functions.Definition 1 (see [6]).
A functionf:0,b⟶ℝ is called strongly exponentially (α, h-m)-p-convex with modulus c≥0, if f is nonnegative and
(2)ftxp+m1−typ1/p≤htαfxeηx+mh1−tαfyeηy−cmhtαh1−tαyp−xp2eηxp+yp,holds, while J⊆ℝ is an interval containing 0,1 and h:J⟶ℝ is a nonnegative function along with x,y,m−1y,txp+m1−typ1/p∈0,b, t∈0,1, p∈ℝ\0, and α,m∈0,12.By using (2), one can find various classes of functions closely related with the convex function and strongly convex functions already defined by different authors. Strongly convex functions provide the refinements of convex functions.In [2], Theorem 5, if we take I⊂0,∞, p∈ℝ\0, and ht=t, then, we have the following theorem.Theorem 2.
Letf:a,b⊂0,∞⟶ℝ be a positive function such that f∈L1a,b. If f is a p-convex function on a,b, p∈ℝ\0. Then, the following integral inequality holds:
(3)fap+bp21/p≤pbp−ap∫abftt1−pdt≤fa+fb2.Our aim in this paper is the derivation of compact forms of Hadamard-type inequalities for strongly exponentiallyα,h−m-p-convex functions via Riemann-Liouville fractional integrals involving monotone functions. The established formulas will generate Hadamard-type inequalities for fractional Riemann-Liouville integrals which have been published by various authors in the recent past (see Remarks 11 & 23). Also, Hadamard-type inequalities are deducible for some new classes of functions (see Corollaries 12–32). In the following, we give the definition of Riemann-Liouville fractional integrals:Definition 3.
Letf∈L1a,b. Then, Riemann-Liouville fractional integral operators of order μ for a function f, where Rμ>0, are given by
(4)Ia+μfx=1Γμ∫axx−tμ−1ftdt,x>a,Ib−μfx=1Γμ∫xbt−xμ−1ftdt,x<b.
Next, we give Hadamard-type inequalities via Riemann-Liouville fractional integrals of convex functions as follows:Theorem 4 (see [7]).
Letf:a,b⟶ℝ be a positive function with 0≤a<b and f∈L1a,b. If f is a convex function on a,b, then, the following fractional integral inequality holds:
(5)fa+b2≤Γμ+12b−aμIa+μfb+Ib−μfa≤fa+fb2,with μ>0.Theorem 5 (see [8]).
With the assumptions given in Theorem4, one can have the fractional integral inequality as follows:
(6)fa+b2≤2μ−1Γμ+1b−aμIa+b/2+μfb+Ia+b/2−μfa≤fa+fb2,with μ>0.Theorem 6 (see [7]).
Letf:a,b⟶ℝ be a differentiable mapping on a,b with a<b. If f′ is convex on a,b, then, the following fractional integral inequality holds:
(7)fa+fb2−Γμ+12b−aμIa+μfb+Ib−μfa≤b−a2μ+11−12μf′a+f′b.The definition ofk-fractional Riemann-Liouville integral operators is given as follows:Definition 7 (see [9]).
Letf∈L1a,b, k>0. Then, k-fractional Riemann-Liouville integrals for a function f of order μ where Rμ>0 are given by
(8)kIa+μfx=1kΓkμ∫axx−tμ/k−1ftdt,x>a,kIb−μfx=1kΓkμ∫xbt−xμ/k−1ftdt,x<b,where Γk. is defined as follows:
(9)Γkμ=∫0∞tμ−1e−tk/kdt,Rμ>0.The generalized form of Riemann-Liouville fractional integrals is given in the following definition:Definition 8 (see [10]).
Letf∈L1a,b. Also, let ψ be an increasing and positive monotone function on a,b, having a continuous derivative ψ′ on a,b. The left-sided and right-sided fractional integrals of a function f with respect to another function ψ on a,b of order μ where Rμ>0 are given by
(10)Ia+μ,ψfx=1Γμ∫axψ′tψx−ψtμ−1ftdt,x>a,Ib−μ,ψfx=1Γμ∫xbψ′tψt−ψxμ−1ftdt,x<b.The definition of thek-analogue of the abovementioned definition is given as follows:Definition 9 (see [11]).
Letf,ψ,μ be the same as in the abovementioned definition. Then, for k>0, the k-analogue of (10) and is given by
(11)kIa+μ,ψfx=1kΓkμ∫axψ′tψx−ψtμ/k−1ftdt,x>a,kIb−μ,ψfx=1kΓkμ∫xbψ′tψt−ψxμ/k−1ftdt,x<b.
Using the fact thatΓkμ=kμ/k−1Γμ/k in (10) after replacing μ by μ/k, we get
(12)k−μ/kIa+μ,ψfx=kIa+μ,ψfx,k−μ/kIb−μ,ψfx=kIb−μ,ψfx.
For further detailed study on fractional integrals, we refer the readers to [12, 13]. In the next section, we formulate the Hadamard-type inequalities for strongly exponentially α,h−m-p-convex function via integrals (10) which are compact forms of a plenty of well-known Hadamard-type inequalities holding for classes of convex, strongly convex, and exponentially convex functions. Specifically, one can have refinements of the inequalities proved in recent decades. Several special case inequalities in the form of corollaries are also given.


## 2. Main Results

We will use the following notations for terms which will appear frequently in the results of this section(13)cμ,mψpa,ψpb=cmμμ+1ψpb−ψpa2+2ψpam−mψpb2+2μψpb−ψpaψpam−mψpb,Rα,μ,m,ηh,Hψpa,ψpb=cmμg2ηh12αψpb−ψpa2eηψpa+ψpb+mg3ηH12ψpb−ψpa/m22eηψpa/m2+ψpb∫01htαHttμ−1dt,Fμ,mψpa,ψpb=cmμμ+1ψpb−ψpa2+μ2+5μ+8ψpam−mψpb2+2μμ+3×ψpb−ψpaψpam−mψpb,Aα,μ,m,ηh,Hfψa,fψb=μg2ηh12αfψaeηψa+mg3ηH12fψbeηψb∫01htαtμ−1dt,Bα,μ,m,ηh,Hfψa,fψb=mμg2ηh12αfψbeηψb+mg3ηH12fψpa/m2eηψa/m2∫01Httμ−1dt.Theorem 10.
Letf, ψ:ap,mbp⊂0,∞⟶ℝ, range ψ⊂ap,mbp be the positive functions such that f∈L1ap,mbp, and ψ be a differentiable and strictly increasing. If f is strongly exponentially α,h−m-p-convex function on ap,mbp such that p,η∈ℝ and p≠0, then, for α,m∈0,12, the following fractional integral inequalities hold:
(i)
Ifp>0,(14)fψpa+mψpb21/p+cμ,mψpa,ψpbg1ηh1/2αH1/2μ+1μ+2≤Γμ+1mψpb−ψpaμg2ηh12α×Iψ−1ψpa+μ,ψf∘ϕψ−1mψpb+mμ+1g3ηH12Iψ−1ψpb−μ,ψf∘ϕψ−1ψpam≤Aα,μ,m,ηh,Hfψa,fψb+Bα,μ,m,ηh,Hfψa,fψb−Rα,μ,m,ηh,Hψpa,ψpb,with μ>0, Ht=h1−tα, ϕt=ψ1/pt for all t∈ap,mbp and
(15)g1η=e−ηψpa+ψpb,ifη>0,e−ηmψpb+ψpa/m,ifη<0,g2η=e−ηmψpb1/p,ifη<0,e−ηψa,ifη>0,g3η=e−ηψb,ifη<0,e−ηψpa/m1/p,ifη>0(ii)
Forp<0, one can have(16)fψpa+mψpb21/p+cμ,mψpa,ψpbg1ηh1/2αH1/2μ+1μ+2≤Γμ+1mψpb−ψpaμg2ηh12α×Iψ−1ψpa−μ,ψf∘ϕψ−1mψpb+mμ+1g3ηH12Iψ−1ψpb+μ,ψf∘ϕψ−1ψpam≤Aα,μ,m,ηh,Hfψa,fψb+Bα,μ,m,ηh,Hfψa,fψb−Rα,μ,m,ηh,Hψpa,ψpb,with μ>0, Ht=h1−tα, ϕt=ψ1/pt for all t∈mbp,ap and
(17)g1η=e−ηψpa+ψpb,ifη<0,e−ηmψpb+ψpa/m,ifη>0,g2η=e−ηmψpb1/p,ifη>0,e−ηψa,ifη<0,g3η=e−ηψb,ifη>0,e−ηψpa/m1/p,ifη<0Proof.
i The following inequality holds for a strongly exponentially α,h−m-p-convex function
(18)fψpx+mψpy21/p≤h12αfψxeηψx+mH12fψyeηψy−cmh1/2αH1/2ψpy−ψpx2eηψpx+ψpy.
By settingψx=ψpat+m1−tψpb1/p, ψy=ψpa/m1−t+ψpbt1/p in (18) and then integrating on 0,1 after multiplying with tμ−1, one can get
(19)1μfψpa+mψpb21/p≤h12α∫01fψpat+m1−tψpb1/peηψpat+m1−tψpb1/ptμ−1dt+mH12×∫01fψpa/m1−t+ψpbt1/peηψpa/m1−t+ψpbt1/ptμ−1dt−cmh12αH12∫011−tψpa/m−mψpb+tψpb−ψpa2eη1−tψpa/m+mψpb+tψpb+ψpa1/ptμ−1dt.
Settingψu=ψpat+m1−tψpb and ψv=ψpa/m1−t+ψpbt in (19) and multiplying by μ, after applying Definition 3, the following inequality can be obtained:
(20)fψpa+mψpb21/p≤Γμ+1mψpb−ψpaμg2ηh12αIψpa+μf∘ϕmψpb+mμ+1g3ηH12×Iψpb−μf∘ϕψpam−cμ,mψpa,ψpbg1ηh1/2αH1/2μ+1μ+2.
Now, by using definition of strongly exponentiallyα,h−m-p-convex function for f and then integrating the resulting inequality on 0,1 after multiplying with tμ−1, one can get
(21)g2ηh12α∫01fψpat+m1−tψpb1/ptμ−1dt+mg3ηH12∫01fψpat+m1−tψpb1/ptμ−1dt≤Aα,μ,m,ηh,Hfψa,fψbμ+Bα,μ,m,ηh,Hfψa,fψbμ−Rα,μ,m,ηh,Hψpa,ψpbμ.
Again, using substitution as considered in (20) leads to the second inequality of (14)
(ii) The proof is followed on same lines as the proof of (i)Remark 11.
The aforementioned version of the Hadamard inequalities gives (i) [4], Theorem 4 for p=−1, m=α=1, h=ψ=I, and c=η=0; (ii) [3], Theorem 2.4 for p=−1, m=α=μ=1, h=ψ=I, and c=η=0; (iii) [14], Theorem 3.10 for h=ψ=I and c=η=0; (iv) [15], Corollary 2.2 for α=p=1, ψ=I, and c=η=0; (v) Theorem 2 for α=m=p=1, h=ψ=I, and c=η=0; (vi) [16], Theorem 2.1 for α=p=1, h=ψ=I, and c=η=0; (vii) [14], Theorem 2.2 for ψ=I and c=η=0; (viii) Theorem 1 for α=μ=m=1, h=ψ=I, and c=η=0; (ix) [17], Theorem 2.1 for α=μ=m=1, p=−1, ht=ts, ψ=I, and c=η=0; (x) [1], Theorem 2.1 for α=μ=m=p=1, ht=ts, ψ=I, and c=η=0; and (xi) [6], Theorem 3 for ψ=I. Moreover, the refinements of all the deduced results will occur for c>0.Corollary 12.
i For p>0, one can have for the strongly α,h−m-p-convex function the following fractional integral inequality:
(22)fψpa+mψpb21/p+cμ,mψpa,ψpbh1/2αH1/2μ+1μ+2≤Γμ+1mψpb−ψpaμh12α×Iψ−1ψpa+μ,ψf∘ϕψ−1mψpb+mμ+1H12Iψ−1ψpb−μ,ψf∘ϕψ−1ψpam≤Aα,μ,m,0h,Hfψa,fψb+Bα,μ,m,0h,Hfψa,fψb−Rα,μ,m,0h,Hψpa,ψpbProof.
The abovementioned inequality can be deduced by settingη=0 in (14).
(ii) Forp<0, one can have for the strongly α,h−m-p-convex function the following fractional integral inequality:
(23)fψpa+mψpb21/p+cμ,mψpa,ψpbh1/2αH1/2μ+1μ+2≤Γμ+1mψpb−ψpaμh12α×Iψ−1ψpa−μ,ψf∘ϕψ−1mψpb+mμ+1H12Iψ−1ψpb+μ,ψf∘ϕψ−1ψpam≤Aα,μ,m,0h,Hfψa,fψb+Bα,μ,m,0h,Hfψa,fψb−Rα,μ,m,0h,Hψpa,ψpbProof.
The abovementioned inequality can be deduced by settingη=0 in (16).Corollary 13.
i For p>0, one can have for the strongly exponentially h−m-p-convex function the following fractional integral inequality:
(24)fψpa+mψpb21/p+cμ,mψpa,ψpbg1ηh21/2μ+1μ+2≤h1/2Γμ+1mψpb−ψpaμg2η×Iψ−1ψpa+μ,ψf∘ϕψ−1mψpb+mμ+1g3ηH12Iψ−1ψpb−μ,ψf∘ϕψ−1ψpam≤A1,μ,m,ηh,Hfψa,fψb+B1,μ,m,ηh,Hfψa,fψb−R1,μ,m,ηh,Hψpa,ψpbProof.
The abovementioned inequality can be deduced by settingα=1 in (14).
(ii) Forp<0, one can have for the strongly exponentially h−m-p-convex function the following fractional integral inequality:
(25)fψpa+mψpb21/p+cμ,mψpa,ψpbg1ηh21/2μ+1μ+2≤h1/2Γμ+1mψpb−ψpaμg2η×Iψ−1ψpa−μ,ψf∘ϕψ−1mψpb+mμ+1g3ηH12Iψ−1ψpb+μ,ψf∘ϕψ−1ψpam≤A1,μ,m,ηh,Hfψa,fψb+B1,μ,m,ηh,Hfψa,fψb−R1,μ,m,ηh,Hψpa,ψpbProof.
The abovementioned inequality can bed deduced by settingα=1 in (16).Corollary 14.
i For p>0, one can have for the strongly exponentially s,m-p-Godunova-Levin function the following fractional integral inequality:
(26)fψpa+mψpb21/p+cμ,mψpa,ψpbg1η22sμ+1μ+2≤Γμ+12smψpb−ψpaμg2ηIψpa+μ,ψf∘ϕψ−1mψpb+mμ+1g3ηIψpb−μ,ψf∘ϕψ−1ψpam≤Aα,μ,m,ηt−s,1−t−sfψa,fψb+Bα,μ,m,ηt−s,1−t−sfψa,fψb−Rα,μ,m,ηt−s,1−t−sψpa,ψpbProof.
The abovementioned inequality can be deduced by settingα=1 and ht=t−s in (2.1).
(ii) Forp<0, one can have for the strongly exponentially s,m-p-Godunova-Levin function the following fractional integral inequality:
(27)fψpa+mψpb21/p+cμ,mψpa,ψpbg1η22sμ+1μ+2≤Γμ+12smψpb−ψpaμg2ηIψpa−μ,ψf∘ϕψ−1mψpb+mμ+1g3ηIψpb+μ,ψf∘ϕψ−1ψpam≤Aα,μ,m,ηt−s,1−t−sfψa,fψb+Bα,μ,m,ηt−s,1−t−sfψa,fψb−Rα,μ,m,ηt−s,1−t−sψpa,ψpbProof.
The abovementioned inequality can be deduced by settingα=1 and ht=t−s in (16).Corollary 15.
i For p>0, one can have for the strongly exponentially s,m-p-convex function in the third sense the following fractional integral inequality:
(28)fψpa+mψpb21/p+cμ,mψpa,ψpbg1η22sμ+1μ+2≤Γμ+1mψpb−ψpaμg2ηIψpa+μ,ψf∘ϕψ−1mψpb+mμ+1g3ηIψpb−μf∘ϕψ−1ψpam≤Aα,μ,m,ηts,1−tsfψa,fψb+Bα,μ,m,ηts,1−tsfψa,fψb−Rα,μ,m,ηts,1−tsψpa,ψpbProof.
The abovementioned inequality can be deduced by settingα=1 and ht=ts in (14).
(ii) Forp<0, one can have for the strongly exponentially s,m-p-convex function in third sense the following fractional integral inequality:
(29)fψpa+mψpb21/p+cμ,mψpa,ψpbg1η22sμ+1μ+2≤Γμ+1mψpb−ψpaμg2ηIψpa−μ,ψf∘ϕψ−1mψpb+mμ+1g3ηIψpb+μf∘ϕψ−1ψpam≤Aα,μ,m,ηts,1−tsfψa,fψb+Bα,μ,m,ηts,1−tsfψa,fψb−Rα,μ,m,ηts,1−tsψpa,ψpbProof.
The abovementioned inequality can be deduced by settingα=1 and ht=ts in (16).Corollary 16.
i For p>0, one can have for the strongly exponentially α,m-p-convex function the following fractional integral inequality:
(30)fψpa+mψpb21/p+cμ,mψpa,ψpbg1η2α−122αμ+1μ+2≤Γμ+1mψpb−ψpaμg2ηIψ−1ψpa+μ,ψf∘ϕψ−1mψpb+mμ+1g3η2α−1Iψ−1ψpb−μ,ψf∘ϕψpam≤Aα,μ,m,ηtα,1−tαfψa,fψb+Bα,μ,m,ηtα,1−tαfψa,fψb−Rα,μ,m,ηtα,1−tαψpa,ψpbProof.
The abovementioned inequality can be deduced by settinght=t in (14).
(ii) Forp<0, one can have for the strongly exponentially α,m-p-convex function the following fractional integral inequality:
(31)fψpa+mψpb21/p+cμ,mψpa,ψpbg1η2α−122αμ+1μ+2≤Γμ+1mψpb−ψpaμg2ηIψ−1ψpa−μ,ψf∘ϕψ−1mψpb+mμ+1g3η2α−1Iψ−1ψpb+μ,ψf∘ϕψpam≤Aα,μ,m,ηtα,1−tαfψa,fψb+Bα,μ,m,ηtα,1−tαfψa,fψb−Rα,μ,m,ηtα,1−tαψpa,ψpbProof.
The abovementioned inequality can be deduced by settinght=t in (16).Corollary 17.
i For p>0, one can have for the strongly exponentially p-convex function the following fractional integral inequality:
(32)fψpa+ψpb21/p+cg1ημ2−μ+2ψpb−ψpa24μ+1μ+2≤Γμ+1ψpb−ψpaμg2ηIψ−1ψpa+μ,ψf∘ϕψ−1ψpb+g3ηIψ−1ψpb−μ,ψf∘ϕψ−1ψpa≤A1,μ,m,ηt,1−tfψa,fψb+B1,μ,m,ηt,1−tfψa,fψb−R1,μ,m,ηt,1−tψpa,ψpbProof.
The abovementioned inequality can be deduced by settingα=m=1 and ht=t in (14).
(ii) Forp<0, one can have for strongly exponentially p-convex function the following fractional integral inequality
(33)fψpa+ψpb21/p+cg1ημ2−μ+2ψpb−ψpa24μ+1μ+2≤Γμ+1ψpb−ψpaμg2ηIψ−1ψpa−μ,ψf∘ϕψ−1ψpb+g3ηIψ−1ψpb+μ,ψf∘ϕψ−1ψpa≤A1,μ,m,ηt,1−tfψa,fψb+B1,μ,m,ηt,1−tfψa,fψb−R1,μ,m,ηt,1−tψpa,ψpbProof.
The abovementioned inequality can be deduced by settingα=m=1 and ht=t in (16).Corollary 18.
For the strongly exponentiallyα,h−m-HA-convex function, the following inequality holds:
(34)f2ψaψbψb+mψa+cmh1/2αH1/2μ+1μ+2μμ+11ψb−1ψa2+21mψa−mψb2+2μ1ψb−1ψa×1mψa−mψb≤Γμ+1ψaψbμψb−mψaμh12αIψ−11/ψa−μ,ψf∘ϕψ−1mψb+mμ+1H12×Iψ−11/ψb+μ,ψf∘ϕψ−11mψa≤Aα,μ,m,ηh,Hfψa,fψb+Bα,μ,m,ηh,Hfψa,fψb−Rα,μ,m,ηh,H1ψa,1ψb.Proof.
The abovementioned inequality can be deduced by settingp=−1 in (16).Corollary 19.
For the strongly exponentiallyα,m-HA-convex function, the following inequality holds:
(35)f2ψaψbψb+ψam+cmg1η2α−122αμ+1μ+2μμ+1ψb−ψaψaψb2+2ψb−ψam2ψaψbm2+2μψa−ψbψb−ψam2mψaψb2≤Γμ+1ψaψbμ2αψb−mψaμg2ηIψ−11/ψa−μ,ψf∘ϕψ−1mψb+mμ+1g3η2α−1Iψ−11/b+μ,ψf∘ϕψ−11mψa≤Aα,μ,m,ηtα,1−tαfψa,fψb+Bα,μ,m,ηtα,1−tαfψa,fψb−Rα,μ,m,ηtα,1−tαψpa,ψpb.Proof.
The abovementioned inequality can be deduced by settingp=−1 and ht=t in (16).Corollary 20.
For the strongly exponentiallys,m-HA-convex function, the following inequality holds:
(36)f2ψaψbψam+ψb+cmg1η22sμ+1μ+2μμ+1ψb−ψaψaψb2+2ψb−ψam2ψaψbm2+2μψa−ψbψbψam2mψaψb2≤Γμ+1ψaψbμψbψamμg2ηI1/ψa−μ,ψf∘ϕψ−1mb+mμ+1g3ηIψ−11/ψb+μ,ψf∘ϕψ−11mψa≤A1,μ,m,ηts,1−tsfψa,fψb+B1,μ,m,ηts,1−tsfψa,fψb−R1,μ,m,ηts,1−tsψpa,ψpb.Proof.
The abovementioned inequality can be deduced by settingp=−1, α=1, and ht=ts in (16).Corollary 21.
For the Godunova-Levin type of strongly exponentiallys,m-HA-convex function, the following inequality holds:
(37)f2ψaψbψb+mψa+22sg1ηcmμ+1μ+2μμ+11ψb−1ψa2+21mψa−mψb2+2μ1ψb−1ψa×1mψa−mψb≤2sΓμ+1ψaψbμψb−mψaμg2ηIψ−11/ψa−μ,ψf∘ϕψ−1mψb+mμ+1g3η×Iψ−11/ψb+μ,ψf∘ϕψ−11mψa≤A1,μ,m,ηt−s,1−t−sfψa,fψb+B1,μ,m,ηt−s,1−t−sfψa,fψb−R1,μ,m,ηt−s,1−t−sψpa,ψpb.Proof.
The abovementioned inequality can be deduced by settingp=−1, α=1, and ht=t−s in (2.2).The second variant of Hadamard inequality for strongly exponentiallyα,h−m-p-convex function is proved as follows.Theorem 22.
Under the assumption of Theorem10, the following fractional integral inequalities hold:
(i)
Forp>0, one can have(38)fψpa+mψpb21/p+Fμ,mψpa,ψpbg1ηh1/2αH1/24μ+1μ+2≤2μΓμ+1mψpb−ψpaμg2ηh12α×Iψ−1ψpa+mψpb/2+μ,ψf∘ϕψ−1mψpb+mμ+1H12g3ηIψ−1ψpa+mψpb/2m−μ,ψf∘ϕψ−1ψpam≤Aα,μ,m,ηh,Hfψa,fψb+Bα,μ,m,ηh,Hfψa,fψb−Rα,μ,m,ηh,Hψpa,ψpb,with μ>0, Ht=h1−tα, ϕz=z1/p for all z∈ap,mbp and
(39)g1η=e−ηψpa+ψpb,ifη>0,e−ηmψpb+ψpa/m,ifη<0,g2η=e−ηmψpb1/p,ifη<0,e−ηψa,ifη>0,g3η=e−ηψb,ifη<0,e−ηψpa/m1/p,ifη>0(ii)
Forp<0, one can have(40)fψpa+mψpb21/p+Fμ,mψpa,ψpbg1ηh1/2αH1/24μ+1μ+2≤2μΓμ+1mψpb−ψpaμg2ηh12α×Iψ−1ψpa+mψpb/2−μ,ψf∘ϕψ−1mψpb+mμ+1H12g3ηIψ−1ψpa+mψpb/2m+μ,ψf∘ϕψ−1ψpam≤Aα,μ,m,ηh,Hfψa,fψb+Bα,μ,m,ηh,Hfψa,fψb−Rα,μ,m,ηh,Hψpa,ψpb,with μ>0, ϕz=z1/p for all z∈mbp,ap and
(41)g1η=e−ηψpa+ψpb,ifη<0,e−ηmψpb+ψpa/m,ifη>0,g2η=e−ηmψpb1/p,ifη>0,e−ηψa,ifη<0,g3η=e−ηψb,ifη>0,e−ηψpa/m1/p,ifη<0Proof.
(i) Letψx=ψpat/2+m2−t/2ψpb1/p, ψy=ψpa/m2−t/2+ψpbt/21/p in (18) and integrating the resulting inequality over the interval 0,1 after multiplying with tμ−1, we get
(42)1μfψpa+mψpb21/p≤h12α∫01fψpbt/2+m2−t/2ψpb1/peηψpbt/2+m2−t/2ψpb1/ptμ−1dt+mH12∫01fψpa/m2−t/2+ψpbt/21/peηψpa/m2−t/2+ψpbt/21/ptμ−1dt−cmh12αH12∫01t/2ψpb−ψpa+2−t/2ψpa/m−mψpb2eη2−t/2ψpa/m+mψpb+t/2ψpb+ψpa1/ptμ−1dt.
Settingψu=ψpbt/2+m2−t/2ψpb and ψv=ψpa/m2−t/2+ψpbt/2 in (42), then, by applying Definition 3, we get the following inequality:
(43)fψpa+mψpb21/p≤2μΓμ+1mψpb−ψpaμh12αg2ηIψ−1ψpa+mψpb/2+μ,ψf∘ϕψ−1mψpb+mμ+1g3ηH12×Iψ−1ψpa+mψpb/2m−μf∘ϕψ−1ψpam−Fμ,mψpa,ψpbg1ηh1/2αH1/24μ+1μ+2.
Now, again, using strongly exponentiallyα,h−m-p-convexity of f and integrating the resulting inequality over 0,1 after multiplying with tμ−1, we get
(44)g2ηh12α∫01fψpat2+m2−t2ψpb1/ptμ−1dt+mg3ηH12∫01fψpat2+m2−t2ψpb1/ptμ−1dt≤Aα,μ,m,ηh,Hfψa,fψbμ+Bα,μ,m,ηh,Hfψa,fψbμ−Rα,μ,m,ηh,Hψpa,ψpbμ.
Again, using substitution as considered in (42) leads to the second inequality of (38)
(ii) The proof is followed on the same lines as the proof of (i)Remark 23.
The aforementioned version of the Hadamard inequalities give (i) [18], Theorem 2.4 for c=0 and ψ=I; (ii) [14], Theorem 2.4 for c=η=0 and ψ=I; (iii) [19], Theorem 7 for α=m=1, c=η=0, and ht=t; (iv) [19], Theorem 7 for α=m=1, c=η=0, and h=ψ=I; (v) Theorem 3 for α=m=p=1, c=η=0, and h=ψ=I; (vi) [20], Theorem 2.1 for α=1=p, c=η=0, and h=ψ=I; (vii) [21], Theorem 4 for α=m=1, p=−1, c=η=0, and h=ψ=I; (viii) [3], Theorem 2.4 for p=−1, α=μ=m=1, c=η=0, and h=ψ=I; (ix) [22], Theorem 7 for α=μ=m=p=1, c=η=0, ht=t−s, and ψ=I; (x) [23], Theorem 3.1 for α=μ=m=1, p=−1, c=η=0, ht=t−s, and ψ=I; (xi) Theorem 1 for α=μ=m=1, c=η=0, and h=ψ=I; (xii) [24], Theorem 2.3 for α=μ=m=1, η=0, and h=ψ=I; (xiii) [25], Theorem 6 for α=μ=m=p=1, η=0, and h=ψ=I; (xiv) [17], Theorem 2.1 for α=μ=m=1, p=−1, c=η=0, ht=ts, and ψ=I; (xv) [1], Theorem 2.1 for α=μ=p=m=1, c=η=0, ht=ts, and ψ=I; and (xvi) [5], Theorem 2.1 for α=μ=m=1, p=−1, η=0, and h=ψ=I. Moreover, the refinements of all the deduced results will occur for c>0.Corollary 24.
i For p>0, one can have for the strongly α,h−m-p-convex function the following fractional integral inequality:
(45)fψpa+mψpb21/p+cμ,mψpa,ψpbh1/2αH1/2μ+1μ+2≤Γμ+1mψpb−ψpaμh12α×Iψ−1ψpa+μ,ψf∘ϕψ−1mψpb+mμ+1H12Iψ−1ψpb−μ,ψf∘ϕψ−1ψpam≤Aα,μ,m,0h,Hfψa,fψb+Bα,μ,m,0h,Hfψa,fψb−Rα,μ,m,0h,Hψpa,ψpbProof.
The abovementioned inequality can be deduced by settingη=0 in (14).
(ii) Forp<0, one can have for the strongly α,h−m-p-convex function the following fractional integral inequality:
(46)fψpa+mψpb21/p+cμ,mψpa,ψpbh1/2αH1/2μ+1μ+2≤Γμ+1mψpb−ψpaμh12α×Iψ−1ψpa−μ,ψf∘ϕψ−1mψpb+mμ+1H12Iψ−1ψpb+μ,ψf∘ϕψ−1ψpam≤Aα,μ,m,0h,Hfψa,fψb+Bα,μ,m,0h,Hfψa,fψb−Rα,μ,m,0h,Hψpa,ψpbProof.
The abovementioned inequality can be deduced by settingη=0 in (14); then, one can obtain the required inequality.Corollary 25.
i For p>0, one can have for the strongly exponentially h−m-p-convex function the following fractional integral inequality:
(47)fψpa+mψpb21/p+Fμ,mψpa,ψpbg1ηh21/24μ+1μ+2≤2μh1/2Γμ+1mψpb−ψpaμg2η×Iψ−1ψpa+mψpb/2+μ,ψf∘ϕψ−1mψpb+mμ+1g3ηIψ−1ψpa+mψpb/2m−μ,ψf∘ϕψ−1ψpam≤A1,μ,m,ηt,1−tfψa,fψb+B1,μ,m,ηt,1−tfψa,fψb−R1,μ,m,ηt,1−tψpa,ψpbProof.
Ifα=1 in (38), then, the abovementioned inequality is obtained.
(ii) Forp<0, one can have for the strongly exponentially h−m-p-convex function the following fractional integral inequality:
(48)fψpa+mψpb21/p+Fμ,mψpa,ψpbg1ηh21/24μ+1μ+2≤2μh1/2Γμ+1mψpb−ψpaμg2η×Iψ−1ψpa+mψpb/2−μ,ψf∘ϕψ−1mψpb+mμ+1g3ηIψ−1ψpa+mψpb/2m+μ,ψf∘ϕψ−1ψpam≤A1,μ,m,ηt,1−tfψa,fψb+B1,μ,m,ηt,1−tfψa,fψb−R1,μ,m,ηt,1−tψpa,ψpbProof.
The abovementioned inequality can be deduced by settingα=1 in (40).Corollary 26.
i For p>0, one can have for the strongly exponentially s,m-p-convex function the following fractional integral inequality:
(49)fψpa+mψpb21/p+Fμ,mψpa,ψpbg1η22s+2μ+1μ+2≤2μ−sΓμ+1mψpb−ψpaμg1ηIψpa+mψpb/2+μ,ψf∘ϕψ−1mb+mμ+1Iψ−1ψpa+mψpb/2m−μf∘ϕψ−1ψpam≤A1,μ,m,ηts,1−tsfψa,fψb+B1,μ,m,ηts,1−tsfψa,fψb−R1,μ,m,ηts,1−tsψpa,ψpbProof.
The abovementioned inequality can be deduced by settingα=1 and ht=ts in (38).
(ii) Forp<0, one can have for the strongly exponentially s,m-p-convex function the following fractional integral inequality
(50)fψpa+mψpb21/p+Fμ,mψpa,ψpbg1η22s+2μ+1μ+2≤2μ−sΓμ+1mψpb−ψpaμg1ηIψpa+mψpb/2−μ,ψf∘ϕψ−1mb+mμ+1Iψ−1ψpa+mψpb/2m+μf∘ϕψ−1ψpam≤A1,μ,m,ηts,1−tsfψa,fψb+B1,μ,m,ηts,1−tsfψa,fψb−R1,μ,m,ηts,1−tsψpa,ψpbProof.
The abovementioned inequality can be deduced by settingα=1 and ht=ts in (40).Corollary 27.
i For p>0, one can have for the strongly exponentially Godunova-Levin type of s,m-p-convex function the following fractional integral inequality:
(51)fψpa+mψpb21/p+22sFμ,mψpa,ψpbg1η4μ+1μ+2≤2μ+sΓμ+1mψpb−ψpaμg2ηIψ−1ψpa+mψpb/2+μ,ψf∘ϕψ−1mψpb+mμ+1g3ηIψ−1ψpa+mψpb/2m−μ,ψf∘ϕψ−1ψpam≤A1,μ,m,ηt−s,1−t−sfψa,fψb+B1,μ,m,ηt−s,1−t−sfψa,fψb−R1,μ,m,ηt−s,1−t−sψpa,ψpbProof.
The abovementioned inequality can be deduced by settingα=1 and ht=t−s in (38).
(ii) Forp<0, one can have for the strongly exponentially Godunova-Levin type of s,m-p-convex function the following fractional integral inequality:
(52)fψpa+mψpb21/p+22sFμ,mψpa,ψpbg1η4μ+1μ+2≤2μ+sΓμ+1mψpb−ψpaμg2ηIψ−1ψpa+mψpb/2−μ,ψf∘ϕψ−1mψpb+mμ+1g3ηIψ−1ψpa+mψpb/2m+μ,ψf∘ϕψ−1ψpam≤A1,μ,m,ηt−s,1−t−sfψa,fψb+B1,μ,m,ηt−s,1−t−sfψa,fψb−R1,μ,m,ηt−s,1−t−sψpa,ψpbProof.
The abovementioned inequality can be deduced by settingα=1 and ht=t−s in (40).Corollary 28.
i For p>0, one can have for the strongly exponentially α,m-p-convex function the following fractional integral inequality:
(53)fψpa+mψpb21/p+2α−1Fμ,mψpa,ψpbg1η22α+2μ+1μ+2≤2μ−αΓμ+1mψpb−ψpaμg2ηIψ−1ψpa+mψpb/2+μ,ψf∘ϕψ−1mψpb+mμ+1g3η2α−1Iψ−1ψpa+mψpb/2m−μ,ψf∘ϕψ−1ψpam≤A1,μ,m,ηtα,1−tαfψa,fψb+B1,μ,m,ηtα,1−tαfψa,fψb−R1,μ,m,ηtα,1−tαψpa,ψpbProof.
The abovementioned inequality can be deduced by settinght=t in (38).
(ii) Forp<0, one can have for the strongly exponentially α,m-p-convex function the following fractional integral inequality:
(54)fψpa+mψpb21/p+2α−1Fμ,mψpa,ψpbg1η22α+2μ+1μ+2≤2μ−αΓμ+1mψpb−ψpaμg2ηIψ−1ψpa+mψpb/2−μ,ψf∘ϕψ−1mψpb+mμ+1g3η2α−1Iψ−1ψpa+mψpb/2m+μ,ψf∘ϕψ−1ψpam≤A1,μ,m,ηtα,1−tαfψa,fψb+B1,μ,m,ηtα,1−tαfψa,fψb−R1,μ,m,ηtα,1−tαψpa,ψpbProof.
The abovementioned inequality can be deduced by settinght=t in (40).Corollary 29.
i For p>0, one can have for strongly exponentially p-convex function the following fractional integral inequality:
(55)fψpa+ψpb21/p+cg1ηψpb−ψpa22μ+1μ+2≤2μ−1Γμ+1ψpb−ψpaμg2ηIψ−1ψpa+mψpb/2+μ,ψf∘ϕψ−1ψpb+g3ηIψ−1ψpa+ψpb/2−μ,ψf∘ϕψ−1ψpa≤A1,μ,1,ηtα,1−tαfψa,fψb+B1,μ,1,ηtα,1−tαfψa,fψb−R1,μ,1,ηtα,1−tαψpa,ψpbProof.
The abovementioned inequality can be deduced by settingα=m=1 and ht=t in (38).
(ii) Forp<0, one can have for the strongly exponentially p-convex function the following fractional integral inequality:
(56)fψpa+ψpb21/p+cg1ηψpb−ψpa22μ+1μ+2≤2μ−1Γμ+1ψpb−ψpaμg2ηIψ−1ψpa+mψpb/2−μ,ψf∘ϕψ−1ψpb+g3ηIψ−1ψpa+mψpb/2+μ,ψf∘ϕψ−1ψpa≤A1,μ,1,ηtα,1−tαfψa,fψb+B1,μ,1,ηtα,1−tαfψa,fψb−R1,μ,1,ηtα,1−tαψpa,ψpbProof.
The abovementioned inequality can be deduced by settingα=m=1 and ht=t in (40).Corollary 30.
For the strongly exponentiallyα,m-HA-convex function, the following inequality holds:
(57)f2ψaψbψa+mψb+2α−1cmg1η22α+2μ+1μ+2μμ+1ψb−ψaψaψb2+μ2+5μ+8ψb−ψam2ψaψbm2+2μμ+3ψa−ψbmψaψb2≤2μ−αΓμ+1ψaψbμψb−ψamμg2ηIψb+mψa/2−μ,ψf∘ϕψ−1mψb+mμ+1g3η×2α−1Iψa+mψb/2ψaψbm+μf∘ϕψ−11ψam≤A1,μ,m,ηtα,1−tαfψa,fψb+B1,μ,m,ηtα,1−tαfψa,fψb−R1,μ,m,ηtα,1−tα1ψa,1ψb.Proof.
The abovementioned inequality can be deduced by settingp=−1 and ht=t in (40).Corollary 31.
For the strongly exponentiallys,m-HA-convex function, the following inequality holds:
(58)f2ψaψbψa+mψb+cmg1η22s+2μ+1μ+2μμ+1ψb−ψaψaψb2+μ2+5μ+8ψb−ψam2ψaψbm2+2μμ+3ψa−ψbmψaψb2≤2μ−sΓμ+1ψaψbμψb−ψamμg2ηIψb+mψa/2−μ,ψf∘ϕψ−1mψb+mμ+1g3η×Iψa+mψb/2ψaψbm+μ,ψf∘ϕψ−11am≤A1,μ,m,ηts,1−tsfψa,fψb+B1,μ,m,ηts,1−tsfψa,fψb−R1,μ,m,ηts,1−ts1ψa,1ψb.Proof.
The abovementioned inequality can be deduced by settingp=−1, α=1, and ht=ts in (40).Corollary 32.
For the Godunova-Levin type of the strongly exponentiallys,m-HA-convex function, the following inequality holds:
(59)f2ψaψbψa+mψb+cmg1η22s+2μ+1μ+2μμ+1ψb−ψaψaψb2+μ2+5μ+8ψb−ψam2ψaψbm2+2μμ+3ψa−ψbmψaψb2≤2μ+sΓμ+1ψaψbμψb−ψamμg2ηIψb+mψa/2−μ,ψf∘ϕψ−1mψb+mμ+1g3ηIψa+mψb/2ψaψbm+μf∘ϕψ−11ψam≤A1,μ,m,ηt−s,1−t−sfψa,fψb+B1,μ,m,ηt−s,1−t−sfψa,fψb−R1,μ,m,ηt−s,1−t−s1ψa,1ψb.Proof.
The abovementioned inequality can be deduced by settingp=−1, α=1, and ht=t−s in (40).Remark 33.
Using (12) with replacement of μ by μ/k in all the abovementioned inequalities, the k-fractional versions of all the abovementioned results can be obtained.


## 3. Conclusion

Some inequalities of the Hadamard type for the strongly exponentiallyα,h−m-p-convex functions using generalized Riemann-Liouville fractional integrals have been proved. These inequalities give refinements of different Hadamard-type inequalities related to various types of convexities. The outcomes of this paper can also provide the k-fractional versions of established inequalities via parametric substitution along with constant multiplier.



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