# Bounds of a Unified Integral Operator via Exponentiallys,m-Convexity and Their Consequences

Authors: Yi Hu; Ghulam Farid; Zijiang; Kahkashan Mahreen
Journal: Journal of Function Spaces
 (2020)

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## Abstract

Various known fractional and conformable integral operators can be obtained from a unified integral operator. The aim of this paper is to find bounds of this unified integral operator via exponentiallys,m-convex functions. The resulting bounds provide compact formulas for the bounds of associated fractional and conformable integral operators. Several Hadamard-type inequalities have been produced from a compact version for unified integral operators for exponentially s,m-convex functions.


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## Body

## 1. Introduction

Fractional integral inequalities are useful in the field of fractional calculus. Many mathematicians have introduced fractional differential, fractional integral, and fractional conformable integral operators in this field (see [1–14]). Recently, several mathematical inequalities have been introduced via s,m-convexity (see [15, 16]). The goal of this paper is to obtain the bounds of all integral operators explained in Remarks 5 and 6 in a unified form for exponentially s,m-convex functions.Definition 1 (see [6]).
Letf:a,b⟶ℝ be an integrable function. Also let g be an increasing and positive function on a,b, having a continuous derivative g′ on a,b. The left-sided and right-sided fractional integrals of a function f with respect to another function g on a,b of order μ where Rμ>0 are defined by
(1)gμIa+fx=1Γμ∫axgx−gtμ−1g′tftdt,x>a,(2)gμIb−fx=1Γμ∫xbgt−gxμ−1g′tftdt,x<b,where Γ. is the gamma function.Definition 2 (see [17]).
Letf:a,b⟶ℝ be an integrable function. Also let g be an increasing and positive function on a,b, having a continuous derivative g′ on a,b. The left-sided and right-sided fractional integrals of a function f with respect to another function g on a,b of order μwhereRμ>0, k>0, are defined by
(3)gμIa+kfx=1kΓkμ∫axgx−gtμ/k−1g′tftdt,x>a,(4)gμIb−kfx=1kΓkμ∫xbgt−gxμ/k−1g′tftdt,x<b,where Γk. is the gamma function given as follows [18]:
(5)Γkx=∫0∞tx−1e−tk/kdt,Rx>0.A fractional integral operator containing an extended generalized Mittag-Leffler function in its kernel is defined as follows:Definition 3 (see [19]).
Letω,μ,α,l,γ,c∈ℂ, Rμ,Rα,Rl>0, Rc>Rγ>0withp≥0, δ>0, and0<k≤δ+Rμ. Letf∈L1a,bandx∈a,b. Then the generalized fractional integral operators εμ,α,l,ω,a+γ,δ,k,cfandεμ,α,l,ω,b−γ,δ,k,cf are defined by
(6)εμ,α,l,ω,a+γ,δ,k,cfx;p=∫axx−tα−1Eμ,α,lγ,δ,k,cωx−tμ;pftdt,(7)εμ,α,l,ω,b−γ,δ,k,cfx;p=∫xbt−xα−1Eμ,α,lγ,δ,k,cωt−xμ;pftdt,where
(8)Eμ,α,lγ,δ,k,ct;p=∑n=0∞βpγ+nk,c−γβγ,c−γcnkΓμn+αtnlnδis the extended generalized Mittag-Leffler function.Recently, a unified integral operator is defined as follows:Definition 4 (see [20]).
Letf,g:a,b⟶ℝ, 0<a<b, be the functions such that f be positive and f∈L1a,b, andg be differentiable and strictly increasing. Also let ϕ/x be an increasing function on a,∞andα,l,γ,c∈ℂ, p,μ,δ≥0, and0<k≤δ+μ. Then, forx∈a,b, the left and right integral operators are defined by
(9)gFμ,α,l,a+ϕ,γ,δ,k,cfx,ω;p=∫axKxyEμ,α,lγ,δ,k,c,g;ϕfydgy,(10)gFμ,α,l,b−ϕ,γ,δ,k,cfx,ω;p=∫xbKyxEμ,α,lγ,δ,k,c,g;ϕfydgy,where KxyEμ,α,lγ,δ,k,c,g;ϕ=ϕgx−gy/gx−gyEμ,α,lγ,δ,k,cωgx−gyμ;p.For suitable settings of functionsϕ, g, and certain values of parameters included in Mittag-Leffler function (8), some interesting consequences can be obtained which are comprised in the upcoming remarks.Remark 5.
(i)
Letϕx=xβ/kΓβ/kΓkβ, k>0, β>k, and p=ω=0, in unified integral operators (9) and (10). Then, generalized Riemann-Liouville fractional integral operators (3) and (4) are obtained.(ii)
Fork=1, (3) and (4) fractional integrals coincide with (1) and (2) fractional integrals, which further produce the following fractional and conformable integrals:
(1)
By takingg as identity function, (3) and (4) fractional integrals coincide with k-fractional Riemann-Liouville integrals defined by Mubeen et al. in [7](2)
Fork=1, along with g as the identity function, (3) and (4) fractional integrals coincide with Riemann-Liouville fractional integrals [6](3)
Fork=1 and gx=xρ/ρ, ρ>0, (3) and (4) produce fractional integrals defined by Chen and Katugampola in [1](4)
Fork=1 and gx=xτ+s/τ+s, (3) and (4) produce generalized conformable fractional integrals defined by Khan and Khan in [5](5)
If we takegx=x−as/s, s>0, in (3) and gx=−b−xs/s, s>0, in (4), then conformable k,s-fractional integrals are achieved as defined by Habib et al. in [3](6)
If we takegx=x1+s/1+s, then conformable fractional integrals are achieved as defined by Sarikaya et al. in [11](7)
If we takegx=x−as/s, s>0, in (3) and gx=−b−xs/s, s>0, in (4) with k=1, then conformable fractional integrals are achieved as defined by Jarad et al. in [4](8)
If we takeω=p=0 and ϕt=ΓμFρ,λσ,kωtρ in (9) and (10), then generalized k-fractional integral operators are achieved as defined by Tunc et al. in [13](9)
If we takek=1 and ϕt=Γμexp−At/μ, A=1−μ/μ, μ>0, in (3) and (4), then generalized fractional integral operators with an exponential kernel are obtained [2]Remark 6.
Letϕx=xβandgx=x, β>0, in unified integral operators (9) and (10). Then, fractional integral operators (6) and (7) are obtained, which along with different settings of k,δ,l,c,γ in generalized Mittag-Leffler function give the following integral operators:
(1)
By settingp=0, fractional integral operators (6) and (7) reduce to the fractional integral operators defined by Salim and Faraj in [10](2)
By settingl=δ=1, fractional integral operators (6) and (7) reduce to the fractional integral operators defined by Rahman et al. in [9](3)
By settingp=0 and l=δ=1, fractional integral operators (6) and (7) reduce to the fractional integral operators defined by Srivastava and Tomovski in [12](4)
By settingp=0 and l=δ=k=1, fractional integral operators (6) and (7) reduce to the fractional integral operators defined by Prabhakar in [8](5)
By settingp=ω=0, fractional integral operators (6) and (7) reduce to the left-sided and right-sided Riemann-Liouville fractional integrals(6)
By settingp=ω=0 in fractional integral operators (9) and (10) we get 1/ΓμFa+ϕ,gfx=gFμ,α,l,a+ϕ,γ,δ,k,cfx,0;0 and 1/ΓμFb−ϕ,gfx=gFμ,α,l,b−ϕ,γ,δ,k,cfx,0;0 where Fa+ϕ,gfx and Fb−ϕ,gfx are defined in [21]
Convex functions play a very vital role in the theory of inequalities. Motivated by its analytical interpretation, many other extended and generalized notions have been defined in literature. These notions are used to extend and generalize a lot of classical results (see [15, 16, 22, 23] and references therein). A generalized notion called exponentially s,m-convexity is defined in [24].Definition 7.
Lets∈0,1 and I⊆0,∞ be an interval. A function f:I⟶ℝ is said to be exponentially s,m-convex in the second sense, if
(11)ftx+m1−ty≤tsfxeαx+m1−tsfyeαy,holds for all x,y∈I,m∈0,1 and α∈ℝ.One can note the deducible definitions in the following remark:Remark 8.
(i)
Form=1, (11) produces the definition of exponentially s-convex function(ii)
Forα=0, (11) produces the definition of s,m-convex function(iii)
Forα=0 and m=1, (11) produces the definition of s-convex function in the second sense(iv)
Forα=0 and m=1, (11) produces the definition of the convex function(v)
Forα=0 and s=1, (11) produces the definition of the m-convex functionIn the upcoming section, bounds of unified integral operators for exponentiallys,m-convex functions are given in different forms. Bounds of associated fractional and conformable integral operators which are known in literature are also deduced. The Hadamard inequality is derived for exponentially s,m-convex functions. Its diverse conformable and fractional versions are presented. A modulus inequality is established by using exponentially s,m-convexity of ∣f′∣. In Section 3, boundedness and continuity of these operators are given.


## 2. Main Results

Bounds of unified integral operators (9) and (10), by using exponentially s,m-convexity, are established in the following theorem:Theorem 9.
Letf:a,b⟶ℝ, 0≤a<b, be a positive exponentially s,m-convex function with m∈0,1 and g:a,b⟶ℝ be a differentiable and strictly increasing function. Also let ϕ/x be an increasing function on a,b. Ifα,l,γ,c∈ℂ, p,μ≥0, δ≥0, and 0<k≤δ+μ, then for x∈a,b, we have
(12)gFμ,α,l,a+ϕ,γ,δ,k,cfx,ω;p+gFμ,α,l,b−ϕ,γ,δ,k,cfx,ω;p≤KxaEμ,α,lγ,δ,k,c,g;ϕmfx/meαx/mgx−faeαaga−Γs+1x−asmfx/meαx/msIx−ga−faeαasIa+ga+KbxEμ,α,lγ,δ,k,c,g;ϕfbeαbgb−mfx/meαx/mgx−Γs+1b−xsfbeαbsIb−gx−mfx/meαx/msIx+gb.Proof.
Under the suppositions ofϕ and g, we can obtain
(13)ϕgx−gtgx−gt≤ϕgx−gagx−ga;t∈a,x,x∈a,b.
Multiplying withEμ,α,lγ,δ,k,cωgx−gtμ;pg′t, we can obtain
(14)KxtEμ,α,lγ,δ,k,c,g;ϕg′t≤ϕgx−gagx−gaEμ,α,lγ,δ,k,cωgx−gtμ;pg′t.
By usingEμ,α,lγ,δ,k,cωgx−gtμ;p≤Eμ,α,lγ,δ,k,cωgx−gaμ;p, the following inequality is obtained:
(15)KxtEμ,α,lγ,δ,k,c,g;ϕg′t≤KxaEμ,α,lγ,δ,k,c,g;ϕg′t.
Using exponentiallys,m-convexity of f, we have
(16)ft≤x−tx−asfaeαa+mt−ax−asfx/meαx/m.
Multiplying (15) and (16) and integrating over a,x, we can obtain
(17)∫axKxtEμ,α,lγ,δ,k,c,g;ϕftdgt≤faeαaKxaEμ,α,lγ,δ,k,c,g;ϕ∫axx−tx−asdgt+mfx/meαx/mKxaEμ,α,lγ,δ,k,c,g;ϕ∫axt−ax−asdgt.
By using Definition4 and integrating by parts, the following inequality is obtained:
(18)gFμ,α,l,a+ϕ,γ,δ,k,cfx,ω;p≤KxaEμ,α,lγ,δ,k,c,g;ϕmfx/meαx/mgx−faeαaga−Γs+1x−asmfx/meαx/msIx−ga−faeαasIa+ga.
Now on the other hand fort∈x,b and x∈a,b, the following inequality holds true:
(19)KxtEμ,α,lγ,δ,k,c,g;ϕg′t≤KbxEμ,α,lγ,δ,k,c,g;ϕg′t.
Using exponentiallys,m-convexity of f, we have
(20)ft≤t−xb−xsfbeαb+mb−tb−xsfx/meαx/m.
Adopting the same pattern as we did for (15) and (16), we obtained the following inequality from (19) and (20):
(21)gFμ,α,l,b−ϕ,γ,δ,k,cfx,ω;p≤KbxEμ,α,lγ,δ,k,c,g;ϕfbeαbgb−mfx/meαx/mgx−Γs+1b−xsfbeαbsIb−gx−mfx/meαx/msIx+gb.
By adding (18) and (21), (12) can be obtained.Remark 10.
(1)
If we putp=ω=0 in (12), then result for exponentially s,m-convex functions can be obtained for the integral operators defined in [23].(2)
If we putp=ω=α=0 in (12), then Theorem 1 in [23] can be obtained(3)
If we puts,m=1,1 and α=0 in (12), Theorem 8 in [20] can be obtained(4)
If we puts,m=1,1 and p=ω=0 in (12), then the result for exponentially convex functions can be obtained for the integral operators defined in [23].(5)
If we puts,m=1,1, p=ω=0, and α=0 in (12), then the result for convex functions can be obtained for the integral operators defined in [23].(6)
If we putϕt=tμ,μ>0, α=0, p=ω=0, and s,m=1,1 in (12), then Proposition 10 in [20] can be obtained(7)
If we putϕt=tμ,μ>0, gx=x, α=0, p=ω=0, and s,m=1,1 in (12), then Corollary 1 in [26] can be obtainedProposition 11.
Letϕt=tμ, and p=ω=0. Then, (12) gives the following bound for fractional integral operators defined in [6] for μ≥1:
(22)gμIa+fx+gμIb−fx≤gx−gaμ−1Γμmfx/meαx/mgx−faeαaga−Γs+1x−asmfx/meαx/msIx−ga−faeαasIa+gx+gb−gxμ−1Γμfbeαbgb−mfx/meαx/mgx−Γs+1b−xsfbeαbsIb−gx−mfx/meαx/msIx+gb.Proposition 12.
Letgx=Ix=x and p=ω=0. Then, (12) gives the following bound for integral operators defined in [27]:
(23)a+Iϕfx+b−Iϕfx≤ϕx−ax−amfx/meαx/mgx−faeαaga−Γs+1x−asmfx/meαx/msIx−ga−faeαasIa+gx+ϕb−xb−xfbeαbgb−mfx/meαx/mgx−Γs+1b−xsfbeαbsIb−gx−mfx/meαx/msIx+gb.Corollary 13.
If we takeϕt=Γμtμ/k/kΓkμ and p=ω=0, then (12) gives the following bound for fractional integral operators defined in [17] for μ≥k:
(24)gμIa+kfx+gμIb−kfx≤gx−gaμ/k−1kΓkμmfx/meαx/mgx−faeαaga−Γs+1x−asmfx/meαx/msIx−ga−faeαasIa+gx+gb−gxμ/k−1kΓkμfbeαbgb−mfx/meαx/mgx−Γs+1b−xsfbeαbsIb−gx−mfx/meαx/msIx+gb.Corollary 14.
If we takeϕt=tμ, p=ω=0, and gx=Ix=x, then (12) gives the following bound for Riemann-Liouvelle fractional integrals defined in [6] for μ≥1:
(25)μIa+fx+μIb−fx≤x−aμ−1Γμmfx/meαx/mgx−faeαaga−Γs+1x−asmfx/meαx/msIx−ga−faeαasIa+gx+b−xμ−1Γμfbeαbgb−mfx/meαx/mgx−Γs+1b−xsfbeαbsIb−gx−mfx/meαx/msIx+gb.Corollary 15.
If we takeϕt=Γμtμ/k/kΓkμ, p=ω=0, and gx=Ix=x, then (12) gives the following bound for fractional integral operators defined in [7] for μ≥k:
(26)μIa+kfx+μIb−kfx≤x−aμ/k−1kΓkμmfx/meαx/mgx−faeαaga−Γs+1x−asmfx/meαx/msIx−ga−faeαasIa+gx+b−xs/μ−1kΓkμfbeαbgb−mfx/meαx/mgx−Γs+1b−xsfbeαbsIb−gx−mfx/meαx/msIx+gb.Corollary 16.
If we takeϕt=tμ, p=ω=0, and gx=xρ/ρ,ρ>0, then (12) gives the following bound for fractional integral operators defined in [1]:
(27)ρIa+μfx+ρIb−μfx≤xρ−aρμ−1Γμρμ−1mfx/meαx/mgx−faeαaga−Γs+1x−asmfx/meαx/msIx−ga−faeαasIa+gx+bρ−xρμ−1Γμρμ−1fbeαbgb−mfx/meαx/mgx−Γs+1x−asfbeαbsIb−gx−mfx/meαx/msIx+gb.The following lemma is essential to prove the next result.Lemma 17.
Letf:0,∞⟶ℝ be an exponentially s,m-convex function with m∈0,1. If 0≤a<b and fx/eαx=fa+b−x/m/eαa+b−x/m, then the following inequality holds:
(28)fa+b2≤12s1+mfxeαx,x∈a,b.Proof.
Sincef is an exponentially s,m-convex, therefore, the following inequality is valid:
(29)fa+b2≤12sfxeαx+m2sfa+b−x/meαa+b−x/m.
By usingfx/eαx=fa+b−x/m/eαa+b−x/m in the above inequality, we get (28).
The following result provides the Hadamard inequality for unified integral operators defined in (9) and (10).Theorem 18.
Under the assumptions of Theorem9, in addition if fx/eαx=fa+b−x/m/eαa+b−x/m, then we have
(30)hα2sfa+b/21+mgFμ,α,l,b−ϕ,γ,δ,k,c1a,ω;p+gFμ,α,l,a+ϕ,γ,δ,k,c1b,ω;p≤gFμ,α,l,b−ϕ,γ,δ,k,cfa,ω;p+gFμ,α,l,a+ϕ,γ,δ,k,cfb,ω;p≤2KbaEμ,α,lγ,δ,k,c,g;ϕfbeαbgb−mfa/meαa/mga−Γs+1b−asfbeαbsIb−ga−mfa/meαa/msIa+gb.Proof.
Under the assumptions ofϕ and g, we can obtain:
(31)ϕgx−gagx−ga≤ϕgb−gagb−ga.
Multiplying withEμ,α,lγ,δ,k,cωgx−gaμ;pg′x, we can obtain:
(32)KxaEμ,α,lγ,δ,k,c,g;ϕg′x≤ϕgb−gagb−gaEμ,α,lγ,δ,k,cωgx−gaμ;pg′x.
By usingEμ,α,lγ,δ,k,cωgx−gaμ;p≤Eμ,α,lγ,δ,k,cωgb−gaμ;p, the following inequality is obtained:
(33)KxaEμ,α,lγ,δ,k,c,g;ϕg′x≤KbaEμ,α,lγ,δ,k,c,g;ϕg′x.
Using exponentiallys,m-convexity of f, we have
(34)fx≤x−ab−asfbeαb+mb−xb−asfa/meαa/m.
Multiplying (33) and (34) and integrating the resulting inequality over a,b, we can obtain
(35)∫abKxaEμ,α,lγ,δ,k,c,g;ϕdgx≤mfa/meαa/mKbaEμ,α,lγ,δ,k,c,g;ϕ∫abb−xb−asdgx+fbeαbKbaEμ,α,lγ,δ,k,c,g;ϕ∫abx−ab−asdgx.
By using Definition4 and integrating by parts, the following inequality is obtained:
(36)gFμ,α,l,a+ϕ,γ,δ,k,cfb,ω;p≤KbaEμ,α,lγ,δ,k,c,g;ϕfbeαbgb−mfa/meαa/mga−Γs+1b−asfbeαbsIb−ga−mfa/meαa/msIa+gb.
On the other hand, forx∈a,b, the following inequality holds true:
(37)KbxEμ,α,lγ,δ,k,c,g;ϕg′x≤KbaEμ,α,lγ,δ,k,c,g;ϕg′x.
Adopting the same pattern of simplification as we did for (33) and (34), the following inequality can be observed for (34) and (37):
(38)gFμ,α,l,b−ϕ,γ,δ,k,cfa,ω;p≤KbaEμ,α,lγ,δ,k,c,g;ϕfbeαbgb−mfa/meαa/mga−Γs+1b−asfbeαbsIb−ga−mfa/meαa/msIa+gb.
By adding (36) and (38), following inequality can be obtained:
(39)gFμ,α,l,a+ϕ,γ,δ,k,cfb,ω;p+gFμ,α,l,b−ϕ,γ,δ,k,cfa,ω;p≤KbaEμ,α,lγ,δ,k,c,g;ϕfbeαbgb−mfa/meαa/mga−Γα+1b−asfbeαbsIb−gb−mfa/meαa/msIa+gb.
Multiplying both sides of (28) by KxaEμ,α,lγ,δ,k,c,g;ϕg′x, and integrating over a,b we have
(40)fa+b2∫abKxaEμ,α,lγ,δ,k,c,g;ϕdgx≤12s1+m∫abKxaEμ,α,lγ,δ,k,c,g;ϕfxeαxdgx.
From Definition4, the following inequality is obtained:
(41)hαfa+b22s1+mgFμ,α,l,b−ϕ,γ,δ,k,c1a,ω;p≤gFμ,α,l,b−ϕ,γ,δ,k,cfa,ω;p.
Similarly multiplying both sides of (28) by KbxEμ,α,lγ,δ,k,c,g;ϕg′x and integrating over a,b, we have
(42)hαfa+b22s1+mgFμ,α,l,a+ϕ,γ,δ,k,c1b,ω;p≤gFμ,α,l,a+ϕ,γ,δ,k,cfb,ω;p.
By adding (41) and (42) following inequality is obtained:
(43)hαfa+b22s1+mgFμ,α,l,b−ϕ,γ,δ,k,c1a,ω;p+gFμ,α,l,a+ϕ,γ,δ,k,c1b,ω;p≤gFμ,α,l,b−ϕ,γ,δ,k,cfa,ω;p+gFμ,α,l,a+ϕ,γ,δ,k,cfb,ω;p.
Using (39) and (43), inequality (30) can be achieved.Remark 19.
(1)
If we putp=ω=0 in (30), then result for exponentially s,m-convex functions can be obtained for the integral operators defined in [23](2)
If we puts,m=1,1 and α=0 in (30), then Theorem 22 in [20] can be obtained(3)
If we puts,m=1,1 and p=ω=0 in (30), then the result for exponentially convex functions can be obtained for the integral operators defined in [23](4)
If we putp=ω=α=0 in (30), then Theorem 3 in [23] can be obtained(5)
If we puts,m=1,1 and p=ω=α=0 in (30), then the result for convex functions can be obtained for the integral operators defined in [23]Corollary 20.
If we putϕt=Γμtμ/k/kΓkμ and p=ω=0, then the inequality (30) produces the following Hadamard inequality for fractional integral operators defined in [17]:
(44)hα2s+1fa+b/2μΓkμm+1gb−gaμ/k≤gμIb−kfa+gμIa+kfb≤2gb−gaμ/k−1kΓkμfbeαbgb−mfa/meαa/mga−Γs+1b−asfbeαbsIb−ga−mfa/meαa/msIa+gb.Corollary 21.
If we putϕt=tμ and p=ω=0, then the inequality (30) produces the following Hadamard inequality for fractional integral operators defined in [6]:
(45)hα2s+1fa+b/2μΓμm+1gb−gbμ≤gμIb−fa+gμIa+fb≤2gb−gaμ−1Γμfbeαbgb−mfa/meαa/mga−Γs+1b−asfbeαbsIb−ga−mfa/meαa/msIa+gb.Corollary 22.
If we putϕt=Γμtμ/k/kΓkμ, p=ω=0 and take g as identity function, then the inequality (30) produces the following Hadamard inequality for fractional integral operators defined in [7]:
(46)hα2s+1fa+b/2μΓkμm+1b−aμ/k≤μIb−kfa+μIa+kfb≤2b−aμ/k−1kΓkμfbeαbgb−mfa/meαa/mga−Γs+1b−asfbeαbsIb−ga−mfa/meαa/msIa+gb.Corollary 23.
If we putϕt=tμ, p=ω=0 and take g as the identity function, then the inequality (30) produces the following Hadamard inequality for fractional integral operators defined in [6]:
(47)hα2s+1fa+b/2μΓμm+1gb−gaμ≤μIb−fa+μIa+fb≤2b−aμ−1Γμfbeαbgb−mfa/meαa/mga−Γs+1b−asfbeαbsIb−ga−mfa/meαa/msIa+gb.Theorem 24.
Letf:I⟶ℝ be a differentiable function. If ∣f′∣ is exponentially s,m-convex with m∈0,1 and g:I⟶ℝ be a differentiable and strictly increasing function. Also let ϕ/x be an increasing function on I then fora,b∈I, a<b. Ifα,l,γ,c∈ℂ, p,μ≥0, δ≥0 and 0<k≤δ+μ, then for x∈a,b, we have
(48)gFμ,α,l,a+ϕ,γ,δ,k,cf∗gx,ω;p+gFμ,α,l,b−ϕ,γ,δ,k,cf∗gx,ω;p≤KxaEμ,α,lγ,δ,k,c,g;ϕm∣f′x/m∣eαx/mgx−∣f′a∣eαaga−Γs+1x−asm∣f′x/m∣eαx/msIx−ga−∣f′a∣eαasIa+gx+KbxEμ,α,lγ,δ,k,c,g;ϕ∣f′b∣eαbgb−m∣f′x/m∣eαx/mgx−Γs+1x−as∣f′b∣eαbsIb−gx−m∣f′x/m∣eαx/msIx+gb,where
(49)gFμ,α,l,a+ϕ,γ,δ,k,cf∗gx,ω;p≔∫axKxtEμ,α,lγ,δ,k,c,g;ϕf′tdgt,gFμ,α,l,b−ϕ,γ,δ,k,cf∗gx,ω;p≔∫xbKtxEμ,α,lγ,δ,k,c,g;ϕf′tdgt.Proof.
Using exponentiallys,m-convexity of ∣f′∣, we have
(50)∣f′t∣≤x−tx−as∣f′a∣eαa+mt−ax−as∣f′x/m∣eαx/m.
The inequality (50) can be written as follows:
(51)−x−tx−as∣f′a∣eαa+mt−ax−as∣f′x/m∣eαx/m≤f′t≤x−tx−as∣f′a∣eαa+mt−ax−as∣f′x/m∣eαx/m.
Let us consider the second inequality of (51)
(52)f′t≤x−tx−as∣f′a∣eαa+mt−ax−as∣f′x/m∣eαx/m.
Multiplying (15) and (52) and integrating over a,x, we can obtain:
(53)∫axKxtEμ,α,lγ,δ,k,c,g;ϕftdgt≤∣f′a∣eαaKxaEμ,α,lγ,δ,k,c,g;ϕ∫axx−tx−asdgt+m∣f′x/m∣eαx/m×KxaEμ,α,lγ,δ,k,c,g;ϕ∫axt−ax−asdgt.
By using (9) of Definition 4 and integrating by parts, the following inequality is obtained:
(54)gFμ,α,l,a+ϕ,γ,δ,k,cfx,ω;p≤KxaEμ,α,lγ,δ,k,c,g;ϕ×m∣f′x/m∣eαx/mgx−∣f′a∣eαaga−Γs+1x−asm∣f′x/m∣eαx/msIx−ga−∣f′a∣eαasIa+gx.
Now we consider the left-hand side from the inequality (51), and adopting the same pattern as we did for the right-hand side inequality, we have
(55)gFμ,α,l,a+ϕ,γ,δ,k,cfx,ω;p≥−KxaEμ,α,lγ,δ,k,c,g;ϕ×m∣f′x/m∣eαx/mgx−∣f′a∣eαaga−Γs+1x−asm∣f′x/m∣eαx/msIx−ga−∣f′a∣eαasIa+gx.
From (54) and (55), the following inequality is observed:
(56)gFμ,α,l,a+ϕ,γ,δ,k,cf∗gx,ω;p≤KxaEμ,α,lγ,δ,k,c,g;ϕ×m∣f′x/m∣eαx/mgx−∣f′a∣eαaga−Γs+1x−asm∣f′x/m∣eαx/msIx−ga−∣f′a∣eαasIa+gx.
Now using exponentiallys,m-convexity of ∣f′∣, we have
(57)∣f′t∣≤t−xb−xs∣f′b∣eαb+mb−tb−xs∣f′x/m∣eαx/m.
On the same pattern as we did for (15) and (50), one can obtain the following inequality from (19) and (57):
(58)gFμ,α,l,b−ϕ,γ,δ,k,cf∗gx,ω;p≤KbxEμ,α,lγ,δ,k,c,g;ϕ×∣f′b∣eαbgb−m∣f′x/m∣eαx/mgx−Γs+1x−as∣f′b∣eαbsIb−gx−m∣f′x/m∣eαx/msIx+gb.
By adding (56) and (58), inequality (48) can be achieved.Remark 25.
(1)
If we putα=0 in (48), then Theorem 2 in [23] can be obtained(2)
If we puts,m=1,1 in (48), then the result for exponentially convex functions can be obtained for the integral operators defined in [23](3)
If we putp=ω=0 in (48), then the result for exponentially s,m-convex functions can be obtained for the integral operators defined in [23](4)
If we puts,m=1,1 and α=0 in (48), then Theorem 25 in [20] can be obtained


## 3. Boundedness and Continuity

Theorem 26.
Under the assumptions of Theorem9, the following inequality holds for m-convex functions:
(59)gFμ,α,l,a+ϕ,γ,δ,k,cfx,ω;p+gFμ,α,l,b−ϕ,γ,δ,k,cfx,ω;p≤KxaEμ,α,lγ,δ,k,c,g;ϕgx−gamfxm+fa+KbxEμ,α,lγ,δ,k,c,g;ϕgb−gxmfxm+fb.Proof.
If we puts=1 and α=0 in (17), we have
(60)∫axKxtEμ,α,lγ,δ,k,c,g;ϕftdgt≤faKxaEμ,α,lγ,δ,k,c,g;ϕ×∫axx−tx−adgt+mfxmKxaEμ,α,lγ,δ,k,c,g;ϕ∫axt−ax−adgt.
By using Definition4 and integrating by part, the following inequality is obtained:
(61)gFμ,α,l,a+ϕ,γ,δ,k,cfx,ω;p≤KxaEμ,α,lγ,δ,k,c,g;ϕmgxfxm+gafa−mgxfxm−gafa,which can be written as
(62)gFμ,α,l,a+ϕ,γ,δ,k,cfx,ω;p≤KxaEμ,α,lγ,δ,k,c,g;ϕgx−gamfxm+fa.
Similarly from (21), the following inequality holds:
(63)∫xbKtxEμ,α,lγ,δ,k,c,g;ϕftdgt≤fbKbxEμ,α,lγ,δ,k,c,g;ϕ×∫xbt−xb−xdgt+mfxmKbxEμ,α,lγ,δ,k,c,g;ϕ∫xbb−tb−xdgt,which further simplifies as follows:
(64)gFμ,α,l,b−ϕ,γ,δ,k,cfx,ω;p≤KbxEμ,α,lγ,δ,k,c,g;ϕgb−gxmfxm+fb.
From (62) and (64), (59) can be obtained.Corollary 27.
If we takem=1 in Theorem 26, then the following inequality holds for convex functions:
(65)gFμ,α,l,a+ϕ,γ,δ,k,cfx,ω;p+gFμ,α,l,b−ϕ,γ,δ,k,cfx,ω;p≤KxaEμ,α,lγ,δ,k,c,g;ϕgx−gafx+fa+KbxEμ,α,lγ,δ,k,c,g;ϕgb−gxfx+fb.Theorem 28.
With assumptions of Theorem26, if f∈L∞a,b, then unified integral operators for m-convex functions are bounded and continuous.Proof.
From (62), we have
(66)gFμ,α,l,a+ϕ,γ,δ,k,cfx,ω;p≤KbaEμ,α,lγ,δ,k,c,g;ϕgb−gam+1∥f∥∞,which further gives
(67)gFμ,α,l,a+ϕ,γ,δ,k,cfx,ω;p≤K∥f∥∞,where K=gb−gam+1KbaEμ,α,lγ,δ,k,c,g;ϕ.
Similarly, from (64), the following inequality holds:
(68)gFμ,α,l,b−ϕ,γ,δ,k,cfx,ω;p≤K∥f∥∞.
The boundedness is established; also, they are linear therefore the continuity is obtained.Corollary 29.
If we takem=1in Theorem 28, then unified integral operators for convex functions are bounded and continuous and following inequalities hold:
(69)gFμ,α,l,a+ϕ,γ,δ,k,cfx,ω;p≤K∥f∥∞,gFμ,α,l,b−ϕ,γ,δ,k,cfx,ω;p≤K∥f∥∞,where K=2gb−gaKbaEμ,α,lγ,δ,k,c,g;ϕ.


## 4. Concluding Remarks

The class of exponentiallys,m-convex functions contains s,m-convex, exponentially convex, s-convex, m-convex, and convex functions. The unified integral operators defined in (9) and (10) produce almost all fractional and conformable integral operators which have been defined independently in recent decades. We have established the bounds of sum of unified integral operators of exponentially s,m-convex functions. Also a Hadamard inequality for these operators is established. In conclusion, the presented results reproduce plenty of results for operators composed in Remarks 5 and 6 for functions deduced in Remark 8.



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