# Derivation of Bounds of an Integral Operator via Exponentially Convex Functions

Authors: Hong Ye; Ghulam Farid; Babar Khan Bangash; Lulu Cai
Journal: Journal of Mathematics
 (2020)

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## Abstract

In this paper, bounds of fractional and conformable integral operators are established in a compact form. By using exponentially convex functions, certain bounds of these operators are derived and further used to prove their boundedness and continuity. A modulus inequality is established for a differentiable function whose derivative in absolute value is exponentially convex. Upper and lower bounds of these operators are obtained in the form of a Hadamard inequality. Some particular cases of main results are also studied.


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## Body

## 1. Introduction

We start with the definition of convex function.Definition 1.
(see [1]). A function f:a,b⟶ℝ is said to be convex if(1)ftx+1−ty≤tfx+1−tfy,holds for all x,y∈a,b and t∈0,1. If inequality (1) is reversed, then the function f will be concave on a,b.
Convex functions are very useful in many areas of mathematics and other subjects due to their fascinating properties and characterizations. Their geometric and analytic interpretations provide straightforward proofs of many mathematical inequalities including Hadamard, Jensen, Hölder, and Minkowski [1–3]. Theoretically, convex functions have been generalized and extended as h-convex, m-convex, s-convex, α,m-convex, h−m-convex, s,m-convex, etc. Awan et al. [4] defined the function named exponentially convex function as follows:Definition 2.
A functionf:K⊆ℝ⟶ℝ, where K is an interval, is said to be an exponentially convex function if(2)fta+1−tb≤tfaeαa+1−tfbeαb,holds for all a,b∈K , t∈0,1 and α∈ℝ. If the inequality in (2) is reversed, then f is called exponentially concave.
Ifα=0, then (2) gives inequality (1). For some recent citations and utilization of exponentially convex functions, one can see [5–14] and references therein. Our goal in this paper is to prove generalized integral inequalities for exponentially convex functions by using integral operators given in Definition 7. In the following, we give definitions of Riemann–Liouville fractional integrals:Definition 3.
Letf∈L1a,b. Then, the left-sided and right-sided Riemann–Liouville fractional integral operators of order μ∈ℂℛμ>0 are defined by(3)Iμa+fx=1Γμ∫axx−tμ−1ftdt,x>a,(4)Iμb−fx=1Γμ∫xbt−xμ−1ftdt,x<b.
Ak-fractional analogue is given as follows:Definition 4.
(see [15]). Let f∈L1a,b. Then, for k>0, the k-fractional integral operators of f of order μ∈ℂ,ℛμ>0 are defined by(5)Iμa+kfx=1kΓκμ∫axx−tμ/k−1ftdt,x>a,(6)Iμb−kfx=1kΓkμ∫xbt−xμ/k−1ftdt,x<b.
A more general definition of the Riemann–Liouville fractional integral operators is given as follows:Definition 5.
(see [16]). Let f: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∈ℂℛμ>0 are defined by(7)Igμa+fx=1Γμ∫axgx−gtμ−1g′tftdt,x>a,(8)Igμb−fx=1Γμ∫xbgt−gxμ−1g′tftdt,x<b,where Γ. is the gamma function.Definition 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 k-fractional integral operators, k>0, of a function f with respect to another function g on a,b of order μ,k∈ℂ,ℛμ>0 are defined by(9)Igμa+kfx=1kΓkμ∫axgx−gtμ/k−1g′tftdt,x>a,(10)Igμb−kfx=1kΓkμ∫xbgt−gxμ/k−1g′tftdt,x<b,where Γk. is the k-gamma function.
A compact form of integral operators defined above is given as follows:Definition 7.
(see [17]). Let f,g:a,b⟶ℝ,0<a<b be the functions such that f be positive and f∈L1a,b  and g be differentiable and strictly increasing. Also, let ϕ/x be an increasing function on a,∞. Then, for x∈a,b, the left- and right-sided integral operators are defined by(11)Fa+ϕ,gfx=∫axKgx,t;ϕftdgt,x>a,(12)Fb−ϕ,gfx=∫xbKgt,x;ϕftdgt,x<b,where Kgx,y;ϕ=ϕgx−gy/gx−gy.
Integral operators defined in (11) and (12) produce several fractional and conformable integral operators defined in [16, 18–25].Remark 1.
Integral operators given in (11) and (12) produce fractional and conformable integral operators as follows:(i)
If we considerϕt=tμ/k/kΓkμ, then (11) and (12) integral operators coincide with (9) and (10) fractional integral operators.(ii)
If we considerϕt=tμ/Γμ, μ>0, then (11) and (12) integral operators coincide with (7) and (8) fractional integral operators.(iii)
If we considerϕt=tμ/k/kΓkμ and g as identity function (11) and (12), integral operators coincide with (5 and 6) fractional integral operators.(iv)
If we considerϕt=tμ/Γμ, μ>0, and along with g as identity function (11) and (12), integral operators coincide with (3 and 4) fractional integral operators.(v)
If we considerϕt=tμ/k/kΓkμ, k=1, and gx=xρ/ρ,ρ>0, then (11) and (12) produce Katugampola fractional integral operators defined by Chen and Katugampola in [18].(vi)
If we considerϕt=tμ/k/kΓkμ, k=1, and gx=xτ+s/τ+s,s>0, then (11) and (12) produce generalized conformable integral operators defined by Khan and Khan in [22].(vii)
If we considerϕt=tμ/k/kΓkμ and gx=x−as/s,s>0, in (11) and ϕt=tμ/k/kΓkμ and gx = −b−xs/s,s>0, in (12), respectively, then conformable k,s-fractional integrals are achieved as defined by Habib et al. in [20].(viii)
If we considerϕt=tμ/k/kΓkμ and gx=x1+s/1+s, then (11) and (12) produce conformable fractional integrals defined by Sarikaya et al. in [24].(ix)
If we considerϕt=tμ/k/kΓkμ and gx=x−as/s,s>0, in (11) and ϕt=tμ/k/kΓkμ and gx=−b−xs/s,s>0, in (12) with k=1, respectively, then conformable fractional integrals are achieved as defined by Jarad et al. in [21].(x)
If we considerϕt=tμ/kℱρ,λσ,kwtρ, then (11) and (12) produce generalized k-fractional integral operators defined by Tunc et al. in [25].(xi)
If we considerϕt=exp−At/μandA=1−μ/μ,μ>0, then following generalized fractional integral operators with exponential kernel are obtained [19]:(13)Egμa+fx=1μ∫axexp−1−μμgx−gtftdt,x>a,Egμb−fx=1μ∫xbexp−1−μμgx−gtftdt,x<b.(xii)
If we considerϕt=tμ/Γμ and gt=lnt, then Hadamard fractional integral operators will be obtained [16, 23].(xiii)
If we considerϕt=tμ/Γμ and gt=−t−1, then Harmonic fractional integral operators given in [16] will be obtained and given as follows:(14)Rμa+fx=tμΓμ∫axx−tμ−1fttμ+1dt,x>a,Rμb−fx=tμΓμ∫axt−xμ−1fttμ+1dt,x<b.(xiv)
If we considerϕt=tμlnt, then left- and right-sided logarithmic fractional integrals are obtained in [19] and given as follows:(15)ℒgμa+fx=∫axgx−gtμ−1lngx−gtg′tdt,x>a,ℒgμb−fx=∫axgt−gxμ−1lngx−gtg′tdt,x<b.
In the upcoming section, we will derive bounds of sum of the left- and right-sided integral operators defined in (11) and (12) for exponentially convex functions. These bounds lead to produce results associated to several kinds of well-known operators for exponentially convex functions, some of the results are presented in particular cases. Further in Section 3, bounds are presented in the form of a Hadamard inequality; several Hadamard type inequalities are obtained.


## 2. Bounds of Integral Operators and Their Consequences

Theorem 1.
Letf:a,b⟶ℝ be a positive and exponentially convex function and g:a,b⟶ℝ be a differentiable and strictly increasing function. Also, let ϕ/x be an increasing function on a,b. Then, for x∈a,b the following inequality for integral operators (11) and (12) holds(16)Fa+ϕ,gfx+Fb−ϕ,gfx≤ϕgx−gafxeαx+faeαa+ϕgb−gxfxeαx+fbeαb.Proof.
For the kernel of integral operator (11), we have(17)Kgx,t;ϕg′t≤Kgx,a;ϕg′t,x∈a,bandt∈a,x.
An exponentially convex function satisfies the following inequality:(18)ft≤x−tx−afaeαa+t−ax−afxeαx.
Inequalities (17) and (18) lead to the following integral inequality:(19)∫axKgx,t;ϕg′tftdt≤Kgx,a;ϕfaeαa∫axx−tx−ag′tdt+fxeαx∫axt−ax−ag′tdt,while (19) gives(20)Fa+ϕ,gfx≤ϕgx−gafxeαx+faeαa.
Again, for the kernel of integral operator (12), we have(21)Kgt,x;ϕg′t≤Kgb,x;ϕg′t,t∈x,bandx∈a,b.
An exponentially convex function satisfies the following inequality:(22)ft≤t−xb−xfbeαb+b−tb−xfxeαx.
Inequalities (21) and (22) lead to the following integral inequality:(23)∫xbKgt,x;ϕg′tftdt≤Kgb,x;ϕfbeαb∫xbt−xb−xg′tdt+fxeαx∫xbb−tb−xg′tdt,while (23) gives(24)Fb−ϕ,gfx≤ϕgb−gxfxeαx+fbeαb.
By adding (20) and (24), (16) can be achieved.
The following remark connected the abovementioned theorem with already known results.Remark 2.
(1)
Forϕt=tμ/Γμ,μ>0, and α=0 in (16), Corollary 1 in [26] can be achieved.(2)
Forϕt=tμ/Γμ,μ>0, gx=x, and α=0 in (16), Corollary 1 in [27] can be achieved.(3)
Forα=0, in (16), Theorem 1 in [28] can be achieved.
Next results indicate upper bounds of several known fractional and conformable integral operators.Proposition 1.
Letϕt=tμ/Γμ,μ>0. Then, (11) and (12) produce the fractional integral operators (7) and (8) as follows:(25)Fa+tμ/Γμ,gfx≔Igμa+fx,Fb−tμ/Γμ,gfx≔Igμb−fx.
Further they satisfy the following bound forμ≥1:(26)Igμa+fx+Igμb−fx≤gx−gaμΓμfxeαx+faeαa+gb−gxμΓμfxeαx+fbeαb.Proposition 2.
Letgx=Ix=x. Then, (11) and (12) produce integral operators defined in [29] as follows:(27)Fa+ϕ,Ifx≔Ia+ϕfx=∫axϕx−tx−tftdt,Fb−ϕ,Ifx≔Ib−ϕfx=∫xbϕt−xt−xftdt.
Further they satisfy the following bound:(28)Ia+ϕfx+Ib−ϕfx≤ϕx−afxeαx+faeαa+ϕb−xfxeαx+fbeαb.Corollary 1.
If we takeϕt=tμ/k/kΓkμ, then (11) and (12) produce the fractional integral operators (9) and (10) as follows:(29)Fa+tμ/k/kΓkμ,gfx≔Igμa+kfx,Fb−tμ/k/kΓkμ,gfx≔Igμb−kfx.From (16), the following bound holds for μ≥k:(30)Igμa+kfx+Igμb−kfx≤gx−gaμ/kkΓkμfxeαx+faeαa+gb−gxμ/kkΓkμfxeαx+fbeαb.Corollary 2.
If we takeϕt=tμ/Γμ, μ>0, and gx=Ix=x, then (11) and (12) produce left- and right-sided Riemann–Liouville fractional integral operators (3) and (4) as follows:(31)Fa+tμ/Γμ,Ifx≔Iμa+fx,Fb−tμ/Γμ,Ifx≔Iμb−fx.
From (16), the following bound holds for μ≥1:(32)Iμa+fx+Iμb−fx≤x−aμΓμfxeαx+faeαa+b−xμΓμfxeαx+fbeαb.Corollary 3.
If we takeϕt=tμ/k/kΓkμ and gx=Ix=x, then (11) and (12) produce the fractional integral operators (5) and (6) as follows:(33)Fa+tμ/k/kΓkμ,Ifx≔Iμa+kfx,Fb−tμ/k/kΓkμ,Ifx≔Iμb−kfx.
From (16), the following bound holds for μ≥k:(34)Iμa+kfx+Iμb−kfx≤x−aμ/kkΓkμfxeαx+faeαa+b−xμ/kkΓkμfxeαx+fbeαb.Corollary 4.
If we takeϕt=tμ/Γμ,μ>0 and gx=xρ/ρ,ρ>0, then (11) and (12) produce the fractional integral operators defined in [18] as follows:(35)Fa+tμ/Γμ,gfx=Iρa+μfx=ρ1−μΓμ∫axxρ−tρμ−1tρ−1ftdt,Fb−tμ/Γμ,gfx=Iρb−μfx=ρ1−μΓμ∫xbtρ−xρμ−1tρ−1ftdt.From (16), they satisfy the following bound:(36)Iρa+μfx+Iρb−μfx≤xρ−aρμΓμρμfxeαx+faeαa+bρ−xρμΓμρμfxeαx+fbeαb.Corollary 5.
If we takeϕt=tμ/Γμ,μ>0, and gx=xn+1/n+1,n>0, then (11) and (12) produce the fractional integral operators defined as follows:(37)Fa+tμ/Γμ,gfx=Ina+μfx=n+11−μΓμ∫axxn+1−tn+1μ−1tnftdt,Fb−tμ/Γμ,gfx=Inb−μfx=n+11−μΓμ∫xbtn+1−xn+1μ−1tnftdt.
From (16), they satisfy the following bound:(38)Ina+μfx+Inb−μfx≤xn+1−an+1μΓμn+1μfxeαx+faeαa+xn+1−an+1μΓμn+1μfxeαx+fbeαb.Next, we prove boundedness and continuity of integral operators.Theorem 2.
Let the assumptions of Theorem1 are satisfied. If f∈L∞a,b, then integral operators defined in (11) and (12) are continuous.Proof.
From (20), we have(39)Fa+ϕ,gfx≤ϕgx−gaf∞1eαx+1eαa.
It is given thatϕ/x is increasing on a,b:(40)ϕgx−gagx−ga≤ϕgb−gagb−ga,ϕgx−ga≤ϕgb−gagx−gagb−ga.
Furtherg is increasing; therefore, we have(41)ϕgx−ga≤ϕgb−ga.
Therefore, (39) gives(42)Fa+ϕ,gfx≤ϕgb−gaf∞e−αx+e−αa.
Ifα>0, then e−αx is decreasing on a,b and we get(43)Fa+ϕ,gfx≤2e−αaϕgb−gaf∞.
Ifα<0, then e−αx is increasing on a,b and we get from (42)(44)Fa+ϕ,gfx≤2e−αbϕgb−gaf∞.
Hence,Fa+ϕ,gfx is bounded and it is linear, and therefore, Fa+ϕ,gfx is continuous.
Similarly, continuity ofFb−ϕ,gfx can be proved.
For a differentiable functionf, as f′ is exponentially convex, the following result holds:Theorem 3.
Letf:I⟶ℝ be a differentiable function. If f′ is exponentially convex and g:I⟶ℝ is a differentiable and strictly increasing function. Also, let ϕ/x be an increasing function on I, then for a,b∈I, a<b, the following inequalities for integral operators holds:(45)Fa+ϕ,gf∗gx≤ϕgx−gaf′aeαa+f′xeαx,(46)Fb−ϕ,gf∗gx≤ϕgb−gxf′beαb+f′xeαx,where(47)Fa+ϕ,gf∗gx=∫axKgx,t;ϕg′tf′tdt,Fb−ϕ,gf∗gx=∫xbKgt,x;ϕg′tf′tdt.Proof.
An exponentially convex functionf′ satisfies the following inequality:(48)f′t≤x−tx−af′aeαa+t−ax−af′xeαx.
From which, we can write(49)f′t≤x−tx−af′aeαa+t−ax−af′xeαx.
Inequalities (17) and (49) lead to the following integral inequality:(50)∫axKgx,t;ϕg′tf′tdt≤Kgx,a;ϕf′aeαa∫axx−tx−ag′tdt+f′xeαx∫axt−ax−ag′tdt,while (50) gives(51)Fa+ϕ,gf∗gx≤ϕgx−gaf′aeαa+f′xeαx.
From (48), we can write(52)f′t≥−x−tx−af′aeαa+t−ax−af′xeαx.
Adopting the same method as we did for (49), the following integral inequality holds:(53)Fa+ϕ,gf∗gx≥−ϕgx−gaf′aeαa+f′xeαx.
From (51) and (53), (45) can be achieved.
An exponentially convex functionf′ satisfies the following inequality:(54)f′t≤t−xb−xf′beαb+b−tb−xf′xeαx.
From which, we can write(55)f′t≤t−xb−xf′beαb+b−tb−xf′xeαx.
Inequalities (21) and (55) lead the following integral inequality:(56)∫xbKgt,x;ϕg′tf′tdt≤Kgb,x;ϕf′beαb∫xbx−tb−xg′tdt+f′xeαx∫xbb−tb−xg′tdt,while (56) gives(57)Fb−ϕ,gf∗gx≤ϕgb−gxf′beαb+f′xeαx.
From (54), we can write(58)f′t≥−t−xb−xf′beαb+b−tb−xf′xeαx.
Adopting the same method as we did for (55), the following inequality holds:(59)Fb−ϕ,gf∗gx≥−ϕgb−gxf′beαb+f′xeαx.
From (57) and (59), (46) can be achieved.


## 3. Hadamard Type Inequalities for Exponentially Convex Function

In this section, we prove the Hadamard type inequality for an exponentially convex function. In order to prove this inequality result, we need the following lemma.Lemma 1.
(see [30]). Let f:a,b⟶ℝ be an exponentially convex function. If f is exponentially symmetric, then the following inequality holds:(60)fa+b2≤fxeαx,x∈a,b.Theorem 4.
Letf:a,b⟶ℝ be positive, exponentially convex, and symmetric about a+b/2 and g:a,b⟶ℝ be a differentiable and strictly increasing function. Also, let ϕ/x be an increasing function on a,b. Then, for a,b∈I, a<b, the following estimations of Hadamard type are valid.(61)hαfa+b2Fb−ϕ,g1a+Fa+ϕ,g1b≤Fb−ϕ,gfa+Fa+ϕ,gfb≤2ϕgb−gafbeαb+faeαa,where hα=eαb for α<0 and hα=eαa for α≥0.Proof.
For the kernel of integral operator (11), we have(62)Kgx,a;ϕg′x≤Kgb,a;ϕg′x,x∈a,b.
An exponentially convex function satisfies the following inequality:(63)fx≤x−ab−afbeαb+b−xb−afaeαa.
Inequalities (62) and (63) lead the following integral inequality:(64)∫abKgx,a;ϕg′xfxdx≤Kgb,a;ϕfbeαb∫abx−ab−ag′xdx+faeαa∫abb−xb−ag′xdx,while (64) gives(65)Fb−ϕ,gfa≤ϕgb−gafbeαb+faeαa.
On the contrary, for the kernel of integral operator (12), we have(66)Kgb,x;ϕg′x≤Kgb,a;ϕg′x.
Inequalities (63) and (66) lead the following integral inequality:(67)∫abKgb,x;ϕg′xfxdx≤Kgb,a;ϕfbeαb∫abx−ab−ag′xdx+faeαa∫abb−xb−ag′xdx,while the abovementioned inequality gives(68)Fa+ϕ,gfb≤ϕgb−gafbeαb+faeαa.
From (65) and (68), the following inequality can be obtained:(69)Fa+ϕ,gfb+Fb−ϕ,gfa≤2ϕgb−gafbeαb+faeαa.
Now, using Lemma1 and multiplying (60) with Kgx,a;ϕg′x, then integrating over a,b, we have(70)∫abKgx,a;ϕfa+b2g′xdx≤∫ab1eαxKgx,a;ϕg′xfxdx.
From which, we have(71)fa+b2Fb−ϕ,g1a≤1hαFb−ϕ,gfa.
Again using Lemma1 and multiplying (60) with Kgb,x;ϕg′x, then integrating over a,b, we have(72)∫abKgb,x;ϕfa+b2g′xdx≤∫ab1eαxKgb,x;ϕg′xfxdx.
From which, we have(73)fa+b2Fa+ϕ,g1b≤1hαFa+ϕ,gfb.
From (71) and (73), the following inequality can be achieved:(74)fa+b2Fb−ϕ,g1a+Fa+ϕ,g1b≤1hαFb−ϕ,gfa+Fa+ϕ,gfb.
From (69) and (74), (61) can be achieved.Remark 3.
Forα=0, in (61), Theorem 3 in [28] can be achieved.Corollary 6.
If we putϕt=tμ/k/kΓkμ, then the inequality (61) produces the following Hadamard type inequality:(75)hαfa+b2Igμb−k1a+Igμa+k1b≤Igμb−kfa+Igμa+kfb≤2gb−gaμ/kkΓkμfbeαb+faeαa.Corollary 7.
If we putϕt=tμ/Γμ, then the inequality (61) produces the following Hadamard type inequality:(76)hαfa+b2Igμb−1a+Igμa+1b≤Igμb−fa+Igμa+fb≤2gb−gaμΓμfbeαb+faeαa.Corollary 8.
If we putϕt=tμ/k/kΓkμ and g as identity function, then the inequality (61) produces the following Hadamard type inequality:(77)hαfa+b2Iμb−k1a+Iμa+k1b≤Iμb−kfa+Iμa+kfb≤2b−aμ/kkΓkμfbeαb+faeαa.Corollary 9.
If we putϕt=tμ/Γμ and g as identity function, then the inequality (61) produces the following Hadamard type inequality:(78)hαfa+b2Iμb−1a+Iμa+1b≤Iμb−fa+Iμa+fb≤2b−aΓμfbeαb+faeαa.


## 4. Concluding Remarks

We have studied an integral operator for exponentially convex functions; this operator has direct consequences to several fractional and conformable integral operators. We have obtained bounds of the integral operator in different forms. In Theorem1, upper bounds of this operator are studied for an exponentially convex function and several special cases have been presented in the form of propositions and corollaries. The boundedness is studied in Theorem 2. In Theorem 3, we have obtained results for differentiable function f such that f′ is exponentially convex. A version of the Hadamard inequality is proved in Theorem 4 which leads to its several variants for fractional and conformable integral operators.



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