# Study of Fractional Integral Operators Containing Mittag-Leffler Functions via Stronglyα,m-Convex Functions

Authors: Yanliang Dong; Maryam Saddiqa; Saleem Ullah; Ghulam Farid
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

The main aim of this paper is to give refinement of bounds of fractional integral operators involving extended generalized Mittag-Leffler functions. A new definition, namely, stronglyα,m-convex function is introduced to obtain improvements of bounds of fractional integral operators for convex, m-convex, and α,m-convex functions. The results of this paper will provide simultaneous generalizations as well as refinements of various published results.


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## Body

## 1. Introduction

Convexity is one of the most important and key concept in mathematics, and many researchers have extended, generalized, and refined it in different ways. Numerous generalizations and extensions have been produced in recent past, for example, in generalization and extension point of views,m-convexity, α,m-convexity, s-convexity, s,m-convexity, h-convexity, and h,m-convexity are remarkable, and in refinement point of view, the strongly convexity is the tremendous notion. In this paper, we have introduced the notion of strongly α,m-convex function. By utilizing this refined form of convex function, we obtain refinements of the bounds of fractional integral operators involving Mittag-Leffler functions in their kernels. Therefore, the results of this paper are refinements of all the results proved in [1]. First, we give definitions of convex, strongly convex, and α,m-convex functions.Definition 1.
(see [2]). Let I be an interval on real line. A function f:I⟶ℝ is said to be convex function if the following inequality holds:(1)ftu1+1−tu2≤tfu1+1−tfu2,for all u1,u2∈I and t∈0,1.Definition 2.
(see [3]). Let I be an interval on real line. A real-valued function f is said to be strongly convex with modulus λ≥0 on I if, for each u1,u2∈I and t∈0,1, we have(2)ftu1+1−tu2≤tfu1+1−tfu2−λt1−tu2−u12.Definition 3.
(see [4]). A function f:0,b⊆ℝ⟶ℝ is said to be α,m-convex function, where α,m∈0,12 and b>0 if, for every u1,u2∈0,b and t∈0,1, we have(3)ftu1+m1−tu2≤tαfu1+m1−tαfu2.
The well-known Mittag-Leffler function is denoted byEξ. and defined as follows (see [5]):(4)Eξt=∑n=0∞tnΓξn+1,where t,ξ∈ℂ,ℜξ>0, and Γ. is the gamma function. It is a natural extension of exponential, hyperbolic, and trigonometric functions. This function and its extensions appear as solution of fractional integral equations and fractional differential equations. For a detailed study of Mittag-Leffler function and its extensions, see [6–10]. The following extended generalized Mittag-Leffler function is introduced by Andrić et al.Definition 4.
(see [11]). Let μ,ξ,l,γ,c∈ℂ, ℜμ,ℜξ,ℜl>0, and ℜc>ℜγ>0 with p≥0, δ>0, and 0<k≤δ+ℜμ. Then, the Mittag-Leffler function Eμ,ξ,lγ,δ,k,ct;p is defined by(5)Eμ,ξ,lγ,δ,k,ct;p=∑n=0∞βpγ+nk,c−γβγ,c−γcnkΓμn+αtnlnδ,where βp is defined by(6)βpx,y=∫01tx−11−ty−1e−p/t1−tdt,and.cnk=Γc+nk/Γc.
A derivative formula of the extended generalized Mittag-Leffler function is given in the following lemma.Lemma 1.
(see [11]). If m∈ℕ,ω,μ,ξ,l,γ,c∈ℂ,ℜμ,ℜξ,ℜl>0,andℜc>ℜγ>0 with p≥0,δ>0, and 0<k<δ+ℜμ, then(7)ddtmtξ−1Eμ,ξ,lγ,δ,k,cωtμ;p=tξ−m−1Eμ,ξ−m,lγ,δ,k,cωtμ;p,ℜξ>m.Remark 1.
The extended generalized Mittag-Leffler function (5) produces the related functions defined in [8–10, 12, 13], see Remark 1.3 in [14].
Next, we have the definition of the generalized fractional integral operator containing the extended generalized Mittag-Leffler function (5).Definition 5.
(see [11]). Let ω,μ,ξ,l,γ,c∈ℂ, ℜμ,ℜξ,ℜl>0, and ℜc>ℜγ>0 with p≥0, δ>0, and 0<k≤δ+ℜμ. Let f∈L1a,b and x∈a,b. Then, the generalized fractional integral operators containing Mittag-Leffler function are defined by(8)ϵμ,ξ,l,ω,a+γ,δ,k,cfx;p=∫axx−tξ−1Eμ,ξ,lγ,δ,k,cωx−tμ;pftdtand(9)ϵμ,ξ,l,ω,b−γ,δ,k,cfx;p=∫xbt−xξ−1Eμ,ξ,lγ,δ,k,cωt−xμ;pftdt.Remark 2.
The operators defined in (8) and (9) produce several kinds of known fractional integral operators, see Remark 1.4 in [14].
The classical Riemann–Liouville fractional integral operator is defined as follows.Definition 6.
(see [13]). Let f∈L1a,b. Then, Riemann–Liouville fractional integral operators of order ξ>0 are defined by(10)Ia+ξfx=1Γξ∫xbx−tξ−1ftdt,x>a,(11)Ib−ξfx=1Γξ∫axt−xξ−1ftdt,x<b.
It can be noted thatϵμ,ξ,l,0,a+γ,δ,k,cfx;0=Ia+ξfx and ϵμ,ξ,l,0,b−γ,δ,k,cfx;0=Ib−ξfx. From fractional integral operators (8) and (9), we can have(12)Jξ,a+x;p≔ϵμ,ξ,l,ω,a+γ,δ,k,c1x;p=x−aξEμ,ξ+1,lγ,δ,k,cwx−aμ;p,(13)Jη,b−x;p≔ϵμ,η,l,ω,b−γ,δ,k,c1x;p=b−xηEμ,η+1,lγ,δ,k,cwb−xμ;p.
In view of wide applications of Riemann–Liouville fractional integrals and derivatives, the problems which involve this integral operator are studied extensively by many authors. The aim of this paper is to provide fractional integral inequalities which are generalizations of Riemann–Liouville fractional integral inequalities. These inequalities also give associated inequalities for fractional integral operators containing Mittag-Leffler functions with different parameters.
The bounds of fractional integrals (8) and (9) for α,m-convex functions are given in the following theorems.Theorem 1.
(see [1]). Let f:a,b⟶ℝ be a real-valued function. If f is positive α,m-convex, then for ξ,η≥1, the following fractional integral inequality for generalized integral operators (8) and (9) holds:(14)ϵμ,ξ,l,ω,a+γ,δ,k,cfx0;p+ϵμ,η,l,ω,b−γ,δ,k,cfx0;p≤fa+mαfx0/mα+1x0−aJξ−1,a+x0;p+fb+mαfx0/mα+1b−x0Jη−1,b−x0;p,x0∈a,b.Theorem 2.
(see [1]). Let f:a,b⟶ℝ be a real-valued function. If f is differentiable and f′ is α,m-convex, then, for ξ,η≥1, the following fractional integral inequality for generalized integral operators (8) and (9) holds:(15)ϵμ,ξ+1,l,ω,a+γ,δ,k,cfx0;p+ϵμ,η+1,l,ω,b−γ,δ,k,cfx0;p−Jξ−1,a+x0;pfa+Jη−1,b−x0;pfb≤f′a+mαf′x0/mα+1x0−aJξ−1,a+x0;p+f′b+mαf′x0/mα+1b−x0Jη−1,b−x0;p,x0∈a,b.Theorem 3.
(see [1]). Let f:a,b⟶ℝ, a>b, be a real-valued function. If f is positive, α,m-convex, and fa+mb−x0/m=fx0, then, for ξ,η>0, the following fractional integral inequality for generalized integral operators (8) and (9) holds:(16)11/2α+m1−1/2αfa+mb2Jη+1,b−a;p+Jξ+1,a+b;p≤ϵμ,η+1,l,ω,b−γ,δ,k,cfa;p+ϵμ,ξ+1,l,ω,a+γ,δ,k,cfb;p≤Jη−1,b−a;p+Jξ−1,a+b;pb−a2fb+mαfa/mα+1.
In Section2, by using definition of strongly α,m-convex function, we establish new refinements of the bounds of generalized fractional integral operators. Also, the refinements of bounds of these operators are presented in the form of Hadamard-like inequality by using strongly α,m-convex functions. The results of this paper are connected with several well-known inequalities.


## 2. Main Results

First, we define stronglyα,m-convex function.Definition 7.
A functionf:0,+∞⟶ℝ is said to be strongly α,m-convex function, with modulus λ≥0, for α,m∈0,12, if(17)ftu1+m1−tu2≤tαfu1+m1−tαfu2−λmtα1−tαu2−u12holds, for all u1,u2∈0,+∞ and t∈0,1.Remark 3.
(i)
By settingα=1 in (17), strongly m-convex function can be obtained [15](ii)
By settingλ=0 in (17), α,m-convex function can be obtained(iii)
By settingα=m=1 and λ=0 in (17), convex function can be obtained(iv)
By settingα=1 and λ=0 in (17), m-convex function can be obtained(v)
By settingα=m=1 in (17), strongly convex function can be obtained
In the following, by using stronglyα,m-convex functions the refinement of already proved results are given.Theorem 4.
Letf:a,b⟶ℝ be a real-valued function. If f is positive and strongly α,m-convex, then, for ξ,η≥1, the following fractional integral inequality for generalized integral operators (8) and (9) holds:(18)ϵμ,ξ,l,ω,a+γ,δ,k,cfx0;p+ϵμ,η,l,ω,b−γ,δ,k,cfx0;p≤fa+mαfx0/mα+1−λαx0−ma2mα+12α+1x0−aJξ−1,a+x0;p+fb+mαfx0/mα+1−λαmb−x02mα+12α+1b−x0Jη−1,b−x0;p,x0∈a,b.Proof.
Letx0∈a,b. Then, for t∈a,x0 and ξ≥1, one can have the following inequality:(19)x0−tξ−1Eμ,ξ,lγ,δ,k,cωx0−tμ;p≤x0−aξ−1Eμ,ξ,lγ,δ,k,cωx0−aμ;p.
The functionf is strongly α,m-convex; therefore, one can obtain(20)ft≤x0−tx0−aαfa+m1−x0−tx0−aαfx0m−λmx0−tx0−aα1−x0−tx0−aαx0−mam2.
By multiplying (19) and (20) and then integrating over a,x0, we obtain(21)∫ax0x0−tξ−1Eμ,ξ,lγ,δ,k,cωx0−tμ;pftdt≤x0−aξ−1Eμ,ξ,lγ,δ,k,cωx0−aμ;pfa∫ax0x0−tx0−aαdt+mfx0m∫ax01−x0−tx0−aαdt−λmx0−ma2∫ax0x0−tx0−aα1−x0−tx0−aαdt.
From which we have that the left integral operator satisfies the following inequality:(22)ϵμ,ξ,l,ω,a+γ,δ,k,cfx0;p≤x0−aJξ−1,a+x0;pfa+mαfx0/mα+1−λαx0−ma2mα+12α+1.
Now, on the contrary, fort∈x0,b and η≥1, one can have the following inequality:(23)t−x0η−1Eμ,η,lγ,δ,k,cωt−x0μ;p≤b−x0η−1Eμ,η,lγ,δ,k,cωb−x0μ;p.
Again from stronglyα,m-convexity of f, we have(24)ft≤t−x0b−x0αfb+m1−t−x0b−x0αfx0m−λmt−x0b−x0α1−t−x0b−x0αmb−x0m2.
By multiplying (23) and (24) and then integrating over x0,b, we have(25)∫x0bt−x0η−1Eμ,η,lγ,δ,k,cωt−x0μ;pftdt≤b−x0η−1Eμ,η,lγ,δ,k,cωb−x0μ;p×fa∫x0bt−x0b−x0αdt+mfx0m∫x0b1−t−x0b−x0αdt−λmmb−x02∫x0bt−x0b−x0α1−t−x0b−x0αdt.
The right integral operator satisfies the following inequality:(26)ϵμ,η,l,ω,b−γ,δ,k,cfx0;p≤b−x0Jη−1,b−x0;pfb+mαfx0/mα+1−λαmb−x02mα+12α+1.
By adding (22) and (26), the required inequality (18) is established.Remark 4.
(i)
Inequality (18) provides the refinement of Theorem 2.1 in [1](ii)
Ifα=1 in (18), then result for strongly m-convex will be obtained(iii)
Ifα=m=1 and λ=0 in (18), then Corollary 1 [16] is obtained(iv)
Ifα=m=1, λ=0, and ω=p=0 in (18), then Theorem 1 in [17] is obtained(v)
Ifα=1 and t=ts in (18), then Theorem 3 in [18] is obtainedCorollary 1.
If we setξ=η in (18), then the following inequality is obtained:(27)ϵμ,ξ,l,ω,a+γ,δ,k,cfx0;p+ϵμ,ξ,l,ω,b−γ,δ,k,cfx0;p≤fa+mαfx0/mα+1−λαx0−ma2mα+12α+1x0−aJξ−1,a+x0;p+fb+mαfx0/mα+1−λαmb−x02mα+12α+1b−x0Jξ−1,b−x0;p,x0∈a,b.Remark 5.
Inequality (27) provides the refinement of Corollary 2.1 in [1].Corollary 2.
Along with assumptions of Theorem1, if f∈L∞a,b, then the following inequality is obtained:(28)ϵμ,ξ,l,ω,a+γ,δ,k,cfx0;p+ϵμ,η,l,ω,b−γ,δ,k,cfx0;p≤f∞1+mαα+1x0−aJξ−1,a+x0;p+b−x0Jη−1,b−x0;p−λαx0−ma2x0−amα+12α+1Jξ−1,a+x0;p+λαmb−x02b−x0mα+12α+1Jη−1,b−x0;p.Corollary 3.
Forξ=η in (28), we get the following result:(29)ϵμ,ξ,l,ω,a+γ,δ,k,cfx0;p+ϵμ,ξ,l,ω,b−γ,δ,k,cfx0;p≤f∞1+mαα+1x0−aJξ−1,a+x0;p+b−x0Jξ−1,b−x0;p−λαx0−ma2x0−amα+12α+1Jξ−1,a+x0;p+λαmb−x02b−x0mα+12α+1Jξ−1,b−x0;p.Theorem 5.
Letf:a,b⟶ℝ be a real-valued function. If f is differentiable and f′ is strongly α,m-convex, then for ξ,η≥1, the following fractional integral inequality for generalized integral operators (8) and (9) holds:(30)ϵμ,ξ+1,l,ω,a+γ,δ,k,cfx0;p+ϵμ,η+1,l,ω,b−γ,δ,k,cfx0;p−Jξ−1,a+x0;pfa+Jη−1,b−x0;pfb≤f′a+mαf′x0/mα+1−λαx0−ma2mα+12α+1x0−aJξ−1,a+x0;p+f′b+mαf′x0/mα+1−λαmb−x02mα+12α+1b−x0Jη−1,b−x0;p,x0∈a,b.Proof.
Asx0∈a,b and t∈a,x0, by using strongly α,m-convexity of f′, we have(31)f′t≤x0−tx0−aαf′a+m1−x0−tx0−aαf′x0m−λmx0−tx0−aα1−x0−tx0−aαx0−mam2.
From (31), one can have(32)f′t≤x0−tx0−aαf′a+m1−x0−tx0−maαf′x0m−λmx0−tx0−aα1−x0−tx0−aαx0−am2.
The product of (19) and (32) gives the following inequality:(33)x0−tξ−1Eμ,ξ,lγ,δ,k,cωx0−tμ;pf′tdt≤x0−aξ−1Eμ,ξ,lγ,δ,k,cωx0−aμ;px0−tx0−aαf′a+m1−x0−tx0−aαf′x0m−λmx0−tx0−aα1−x0−tx0−aαx0−mam2.
After integrating above inequality overa,x0, we obtain(34)∫ax0x0−tξ−1Eμ,ξ,lγ,δ,k,cωx0−tμ;pf′tdt≤x0−aξ−1Eμ,ξ,lγ,δ,k,cωx0−aμ;pf′a∫ax0x0−tx0−aαdt+mf′x0m∫ax01−x0−tx0−aαdt−λmx0−ma2∫ax0x0−tx0−aα1−x0−tx0−aαdt=x0−aξEμ,ξ,lγ,δ,k,cωx0−aμ;pα+1f′a+mαf′x0m−λαx0−ma2m2α+1.
The left-hand side of (34) is calculated as follows:(35)∫ax0x0−tξ−1Eμ,ξ,lγ,δ,k,cωx0−tμ;pf′tdt.
Putx0−t=z, that is, t=x0−z; also using the derivative property (7) of Mittag-Leffler function, we have(36)∫0x0−azξ−1Eμ,ξ,lγ,δ,k,cωzμ;pf′x0−zdz=x0−aξ−1Eμ,ξ,lγ,δ,k,cωx0−aμ;pfa−∫0x0−azξ−2Eμ,ξ,lγ,δ,k,cωzμ;pfx0−zdz.
Now, putx0−z=t in the second term of the right-hand side of the above equation, and then using (8), we obtain(37)∫0x0−azξ−1Eμ,ξ,lγ,δ,k,cωzμ;pf′x0−zdz=x0−aξ−1Eμ,ξ,lγ,δ,k,cωx0−aμ;pfa−ϵμ,ξ+1,l,ω,a+γ,δ,k,cfx0;p.
Therefore, (34) takes the following form:(38)Jξ−1,a+x0;pfa−ϵμ,ξ+1,l,ω,a+γ,δ,k,cfx0;p≤x0−aJξ−1,a+x0;pf′a+mαf′x0/mα+1−λαx0−ma2mα+12α+1.
Also, from (31), one can have(39)f′t≥−x0−tx0−aαf′a+m1−x0−tx0−aαf′x0m−λmx0−tx0−aα1−x0−tx0−aαx0−mam2.
Following the same procedure as we did for (32), one can obtain(40)ϵμ,ξ+1,l,ω,a+γ,δ,k,cfx0;p−Jξ−1,a+x0;pfa≤x0−aJξ−1,a+x0;pf′a+mαf′x0/mα+1−λαx0−ma2mα+12α+1.
From (38) and (40), we obtain(41)ϵμ,ξ+1,l,ω,a+γ,δ,k,cfx0;p−Jξ−1,a+x0;pfa≤x0−aJξ−1,a+x0;pf′a+mαf′x0/mα+1−λαx0−ma2mα+12α+1.
Now, we letx0∈a,b and t∈x0,b. Then, by using strongly α,m-convexity of f′, we have(42)f′t≤t−x0b−x0αf′b+m1−t−x0b−x0αf′x0m−λmt−x0b−x0α1−t−x0b−x0αmb−x0m2.
On the same lines as we have done for (19), (32), and (39), one can get, from (23) and (12), the following inequality:(43)ϵμ,η+1,l,ω,b−γ,δ,k,cfx0;p−Jη−1,b−x0;pfb≤b−x0Jη−1,b−x0;pf′b+mαf′x0/mα+1−λαmb−x02mα+12α+1.
From inequalities (41) and (43), (30) is obtained.Remark 6.
(i)
Inequality (30) provides the refinement of Theorem 2.2 [1](ii)
Ifα=1 in (30), then result for strongly m-convex will be obtained(iii)
Ifα=m=1 and λ=0 in (30), then Corollary 2 [16] is obtained(iv)
Ifα=m=1, λ=0, and ω=p=0 in (30), then Theorem 2 [17] is obtained(v)
Ifα=1 and t=ts in (30), then Theorem 5 in [18] is obtainedCorollary 4.
If we putξ=η in (30), then the following inequality is obtained:(44)ϵμ,ξ+1,l,ω,a+γ,δ,k,cfx0;p+ϵμ,ξ+1,l,ω,b−γ,δ,k,cfx0;p−Jξ−1,a+x0;pfa+Jξ−1,b−x0;pfb≤f′a+mαf′x0/mα+1−λαx0−ma2mα+12α+1x0−aJξ−1,a+x0;p+f′b+mαf′x0/mα+1−λαmb−x02mα+12α+1b−x0Jξ−1,b−x0;p,x0∈a,b.Remark 7.
Inequality (44) provides the refinement of Corollary 2.2 in [8].
To prove our next result, we consider following lemma.Lemma 2.
Letf:a,b⟶ℝ,a<mb, be strongly α,m-convex function. If fa+mb−x0/m=fx0,m≠0 and α,m∈0,12; then, the following inequality holds:(45)fa+mb2≤fx012α+m1−12α−λm2α−122α×a+mb−x0−mx02.Proof.
Asf is strongly α,m-convex function, we have(46)fa+mb2≤f1−ta+mtb2α+m1−12αfta+m1−tbm−λm2α−122αta+m1−tbm−1−ta+mtb2.
Letx0=a1−t+mtb, and we have a+mb−x0=ta+m1−tb.(47)fa+mb2≤fx02α+m1−12αfa+mb−x0m−λm2α−122αa+mb−x0−mx02.
Hence, by usingfa+mb−x0/m=fx0,m≠0, inequality (45) can be obtained.Theorem 6.
Letf:a,b⟶ℝ, a>b, be a real-valued function. If f is positive, strongly α,m-convex, and fa+mb−x0/m=fx0,m≠0, then, for ξ,η>0, the following fractional integral inequality for generalized integral operators (8) and (9) holds: (48)11/2α+m1−1/2αfa+mb2Jη+1,b−a;p+Jξ+1,a+b;p+λm2α−122αM1+M2≤ϵμ,η+1,l,ω,b−γ,δ,k,cfa;p+ϵμ,ξ+1,l,ω,a+γ,δ,k,cfb;p≤Jη−1,b−a;p+Jξ−1,a+b;pb−a2fb+mαfa/mα+1−λαa−mb2mα+12α+1,where(49)M1=b−aη+2Jη+1,b−a;p+21+mb−aη+1Jη+2,b−a;p+21+m2Jη+3,b−a;p,M2=b−aξ+2Jξ+1,a+b;p+21+mb−aξ+1Jξ+2,a+b;p+21+m2Jξ+3,a+b;p.Proof.
Forx0∈a,b, we have(50)x0−aηEμ,η,lγ,δ,k,cωx0−aμ;p≤b−aηEμ,η,lγ,δ,k,cωb−aμ;p,η>0.
Asf is strongly α,m-convex, so, for x0∈a,b, we have(51)fx0≤x0−ab−aαfb+m1−x0−ab−aαfam−λmx0−ab−aα1−x0−ab−aαa−mbm2.
By multiplying (50) and (51) and then integrating over a,b, we obtain(52)∫abx0−aηEμ,η,lγ,δ,k,cωx0−aμ;pfx0dx0≤b−aηEμ,η,lγ,δ,k,cωb−aμ;pfb∫abx0−ab−aαdx0+mfam∫ab1−x0−ab−aαdx0−λma−mb2×∫abx0−ab−aα1−x0−ab−aαdx0.From which we have(53)ϵμ,η+1,l,ω,b−γ,δ,k,cfa;p≤b−aη+1Eμ,η,lγ,δ,k,cωb−aμ;p×fb+mαfa/mα+1−λαa−mb2mα+12α+1,that is,(54)ϵμ,η+1,l,ω,b−γ,δ,k,cfa;p≤b−a2Jη−1,b−a;p×fb+mαfa/mα+1−λαa−mb2mα+12α+1.
Now, on the contrary, forx0∈a,b, we have(55)b−x0ξEμ,ξ,lγ,δ,k,cωb−x0μ;p≤b−aξEμ,ξ,lγ,δ,k,cωb−aμ;p,ξ>0.
By multiplying (51) and (55) and then integrating over a,b, we obtain(56)∫abb−x0ξEμ,ξ,lγ,δ,k,cωb−x0μ;pfx0dx0≤b−aξEμ,ξ,lγ,δ,k,cωb−aμ;pfb∫abx0−ab−aαdx0+mfam∫ab1−x0−ab−aαdx0−λma−mb2∫abx0−ab−aα1−x0−ab−aαdx0.
From which we have(57)ϵμ,ξ+1,l,ω,a+γ,δ,k,cfb;p≤b−aξ+1Eμ,ξ,lγ,δ,k,cωb−aμ;p×fb+mαfa/mα+1−λαa−mb2mα+12α+1,that is,(58)ϵμ,ξ+1,l,ω,a+γ,δ,k,cfb;p≤b−a2Jξ−1,a+b;p×fb+mαfa/mα+1−λαa−mb2mα+12α+1.
Adding (54) and (58), we get the second inequality of (48). Multiplying (45) with x0−aηEμ,η,lγ,δ,k,cωx0−aμ;p and integrating over a,b, we obtain(59)fa+mb2∫abx0−aηEμ,η,lγ,δ,k,cωx0−aμ;pdx0≤12α+m1−12α∫abx0−aηEμ,η,lγ,δ,k,cωx0−aμ;pfx0dx0−λm2α−122α×∫abx0−aηEμ,η,lγ,δ,k,cωx0−aμ;pa+mb−x0−mx02dx0.
By using (9) and (12), we obtain(60)fa+mb2Jη+1,b−a;p≤12α+m1−12α×ϵμ,η+1,l,ω,b−γ,δ,k,cfa;p−λm2α−122α×ϵμ,η+1,l,ω,b−γ,δ,k,ca+mb−x0−mx02a;p.
The integral operator appearing in the last term is computed as follows:(61)ϵμ,η+1,l,ω,b−γ,δ,k,ca+mb−x0−mx02a;p=∑n=0∞βpγ+nk,c−γβγ,c−γ×cnkΓμn+α1lnδ∫bax0−aη+μna+mb−x0−mx02dx0=∑n=0∞βpγ+nk,c−γβγ,c−γcnkΓμn+α1lnδb−aη+μn+3η+μn+1+21+mb−aη+μn+3η+μn+1η+μn+2+21+m2b−aη+μn+3η+μn+1η+μn+2η+μn+3=b−aη+2Jη+1,b−a;p+21+mb−aη+1Jη+2,b−a;p+21+m2Jη+3,b−a;p.
Now, inequality (43) becomes(62)fa+mb2Jη+1,b−a;p≤12α+m1−12α×ϵμ,η+1,l,ω,b−γ,δ,k,cfa;p−λm2α−122αb−aη+2Jη+1,b−a;p+21+mb−aη+1Jη+2,b−a;p+21+m2Jη+3,b−a;p.
By multiplying (45) with b−x0ξEμ,ξ,lγ,δ,k,cωb−x0μ;p and integrating over a,b, also using (8) and (12), we obtain(63)fa+mb2Jξ+1,a+b;p≤12α+m1−12αϵμ,ξ+1,l,ω,a+γ,δ,k,cfb;p−λm2α−122αϵμ,ξ+1,l,ω,a+γ,δ,k,ca+mb−x0−mx02b;p.
The integral operator appearing in the last term is computed as follows:(64)ϵμ,ξ+1,l,ω,a+γ,δ,k,ca+mb−x0−mx02b;p=b−aξ+2Jξ+1,a+b;p+21+mb−aξ+1Jξ+2,a+b;p+21+m2Jξ+3,a+b;p.
Now, inequality (63) becomes(65)fa+mb2Jξ+1,a+b;p≤12α+m1−12αϵμ,ξ+1,l,ω,a+γ,δ,k,cfb;p−λm2α22α−1b−aξ+2Jξ+1,a+b;p+21+mb−aξ+1Jξ+2,a+b;p+21+m2Jξ+3,a+b;p.
By adding (62) and (65), first inequality of (48) can be obtained.Remark 8.
(i)
Inequality (48) provides the refinement of Theorem 2.3 in [1](ii)
Ifα=1 in (48), then result for strongly m-convex will be obtained(iii)
Ifα=m=1 and λ=0 in (48), then Corollary 2 in [16] is obtained(iv)
Ifα=m=1, λ=0, and ω=p=0 in (48), then Theorem 2 in [17] is obtained(v)
Ifα=1 and t=ts in (48), then Theorem 7 in [18] is obtainedCorollary 5.
If we putξ=η in (48), then the following inequality is obtained:(66)11/2α+m1−1/2αfa+mb2Jξ+1,b−a;p+Jξ+1,a+b;p+λm2α−122αM1+M2≤ϵμ,ξ+1,l,ω,b−γ,δ,k,cfa;p+ϵμ,ξ+1,l,ω,a+γ,δ,k,cfb;p≤Jξ−1,b−a;p+Jξ−1,a+b;pb−a2fb+mαfa/mα+1−λαa−mb2mα+12α+1.Remark 9.
Inequality (66) provides the refinement of Corollary 2.3 in [1].


## 3. Concluding Remarks

The presented results are the refinements of the bounds of generalized fractional integral operators given in (8) and (9) for strongly α,m-convex functions. From the presented results, one can obtain already proved results for convex, m-convex, and α,m-convex functions. Moreover, the refinements of some known fractional versions of the Hadamard inequality are also given.



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