# Some New Refinements of Hermite–Hadamard-Type Inequalities Involvingψk-Riemann–Liouville Fractional Integrals and Applications

Authors: Muhammad Uzair Awan; Sadia Talib; Yu-Ming Chu; Muhammad Aslam Noor; Khalida Inayat Noor
Journal: Mathematical Problems in Engineering
 (2020)

---

## Abstract

The main objective of this article is to establish some new fractional refinements of Hermite–Hadamard-type inequalities essentially using newψk-Riemann–Liouville fractional integrals, where k>0. Using this new fractional integral, we also derive two new fractional integral identities. Applications of the obtained results are also discussed.


---

## Body

## 1. Introduction and Preliminaries

Letf:I=a,b⊂ℝ⟶ℝ be a convex function; then,(1)fa+b2≤1b−a∫abfxdx≤fa+fb2.The above inequality is known as Hermite–Hadamard’s inequality [1–5]. This inequality provides us a necessary and sufficient condition for a function to be convex. It can be considered as one of the most extensively studied results pertaining to convexity. Since the appearance of this result in the literature, it gained popularity, and many new generalizations for this classical result have been obtained. This can be attributed to its applications in various other fields such as in numerical analysis and in mathematical statistics. For more details on generalizations of convexity, Hermite–Hadamard-like inequalities, and its applications, see [6–14].Fractional calculus is a calculus in which we study about the integrals and derivatives of any arbitrary real or complex order. The history of fractional calculus is not very much old, but in the short span of time, it experienced a rapid development. Recently, the generalizations [15–25], extensions [26–32], and applications [33–46] for fractional calculus have been made by many researchers. The Riemann–Liouville fractional integrals are defined as follows.Definition 1.
(see [47]). Let f∈L1a,b. Then, Riemann–Liouville integrals Ja+αf and Jb−αf of order α>0 with a≥0 are defined by(2)Ja+αfx=1Γα∫axx−tα−1ftdt,x>a,(3)Jb−αfx=1Γα∫xbt−xα−1ftdt,x<b,where(4)Γα=∫0∞e−xxα−1dx,is the well-known gamma function.
Sarikaya et al. [10] elegantly utilized this concept in establishing fractional analogue of Hermite–Hadamard’s inequality. This idea motivated other researchers, and consequently, many new generalizations of Hermite–Hadamard’s inequality have been obtained using the concept of Riemann–Liouville fractional integrals.
Sarikaya and Karaca [12] introduced k-analogue of Riemann–Liouville fractional integrals and discussed some of its basic properties. They defined this concept in the following way: to be more precise, let f be piecewise continuous on I∗=0,∞ and integrable on any finite subinterval of I=0,∞. Then, for t>0, we consider k-Riemann–Liouville fractional integral of f of order α as(5)Jkaαfx=1kΓkα∫axx−tα/k−1ftdt,x>a,k>0.
Ifk⟶1, then k-Riemann–Liouville fractional integrals reduce to classical the Riemann–Liouville fractional integral. It is worth to mention here that the concept of the k-Riemann–Liouville fractional integral is a significant generalization of Riemann–Liouville fractional integrals; as for k≠1, the properties of k-Riemann–Liouville fractional integrals are quite different from the classical Riemann–Liouville fractional integrals.
Another important generalization of Riemann–Liouville fractional integrals isψk-Riemann–Liouville fractional integrals.Definition 2.
(see [6]). Let a,b be a finite interval of the real line ℝ and α>0. Also, let ψx be an increasing and positive monotone function on a,b, having a continuous derivative ψ′x on a,b. Then, the left- and right-sided ψ-Riemann–Liouville fractional integrals of a function f with respect to another function ψ on a,b are defined as(6)Ia+α;ψfx=1Γα∫axψ′tψx−ψtα−1ftdt,Ib−α;ψfx=1Γα∫xbψ′tψt−ψxα−1ftdt,respectively; Γ⋅ is the gamma function.
For some recent research works, see [48].
Recently, Liu et al. [14] obtained some interesting results pertaining to Hermite–Hadamard’s inequality involving ψk-Riemann–Liouville fractional integrals. Motivated by the research work of Liu et al. [14], we obtain some new refinements of fractional Hermite–Hadamard’s inequality essentially using ψk-Riemann–Liouville fractional integrals. We also discuss applications of the obtained results to means. We show that our results represent significant generalization of some previous results.


## 2. Hermite–Hadamard’s Inequality

In this section, we derive a new refinement of Hermite–Hadamard’s inequality via theψk-Riemann–Liouville fractional integral.Definition 3.
Letk>0, a,b be a finite interval of the real line ℝ, and α>0. Also, let ψx be an increasing and positive monotone function on a,b, having a continuous derivative ψ′x on a,b. Then, the left- and right-sided ψk-Riemann–Liouville fractional integrals of a function f with respect to another function ψ on a,b are defined as(7)Ika+α;ψfx=1kΓkα∫axψ′tψx−ψtα/k−1ftdt,Ikb−α;ψfx=1kΓkα∫xbψ′tψt−ψxα/k−1ftdt,respectively;(8)Γkx=∫0∞tx−1e−tk/kdt,ℜx>0,is the k-analogue of gamma function.
Thek-analogues of beta function and incomplete beta function are, respectively, defined as(9)Bkx,y=1k∫01tx/k−11−tx/k−1dt,(10)Bkz;x,y=1k∫0ztx/k−11−tb/y−1dt.
We now derive the main result of this section.Theorem 1.
Let0≤e<f and g:e,f⟶ℝ be a positive function and g∈L1e,f. Also, suppose that g is a convex function on e,f, ψx is an increasing and positive monotone function on e,f, having a continuous derivative ψ′x on e,f, and α∈0,1. Then, for k>0, the following k-fractional integral inequalities hold:(11)ge+f2≤Γkα+k2f−eα/kIkψ−1e+α;ψg∘ψψ−1f+Ikψ−1f−α;ψg∘ψψ−1e≤ge+gf2.Proof.
Using the convexity ofg, we have(12)2ge+f2≤gtc+1−tf+g1−te+td.
Multiplying both sides bytα/k−1 and then integrating with respect to t on 0,1, we have(13)2kαge+f2≤∫01tα/k−1gtc+1−tfdt+∫01tα/k−1g1−te+tddt.
Now, making the substitutiont=ψv−f/e−f,s=ψv−e/f−e, we have(14)Γkα+k2f−eα/kIkψ−1e+α;ψg∘ψψ−1f+Ikψ−1f−α;ψg∘ψψ−1e=Γkα+k2f−eα/k1kΓkα∫ψ−1eψ−1ff−ψvα/kg∘ψvψ′vdv+∫ψ−1eψ−1fψv−eα/kg∘ψvψ′vdv=α2k∫01tα/k−1gtc+1−tfdt+∫01tα/k−1g1−te+tddt≥ge+f2.
Also, using the convexity property ofg, we have(15)gtc+1−tf+g1−te+td≤ge+gf.
Multiplying both sides bytα/k−1 and then integrating it with respect to t on 0,1, we obtain(16)∫01tα/k−1gtc+1−tfdt+∫01tα/k−1g1−te+tddt≤kαge+gf.
This implies(17)Γkα+k2f−eα/kIkψ−1e+α;ψg∘ψψ−1f+Ikψ−1f−α;ψg∘ψψ−1e≤ge+gf2.
The proof is completed.


## 3. Some More Fractional Inequalities of Hermite–Hadamard Type

We now derive two new fractional integral identities involvingψk-Riemann–Liouville fractional integrals. These results will serve as auxiliary results for obtaining our next results.Lemma 1.
Lete<f and g:e,f⟶ℝ be a differentiable mapping on e,f. Also, suppose that g′∈Le,f, ψx is an increasing and positive monotone function on e,f, having a continuous derivative ψ′x on e,f, and α∈0,1. Then, for k>0, the following identity holds:(18)ge+gf2−Γkα+k2f−eα/kIkψ−1e+α;ψg∘ψψ−1f+Ikψ−1f−α;ψg∘ψψ−1e=12f−eα/k∫ψ−1eψ−1fψv−eα/k−f−ψvα/kg′∘ψvψ′vdv.Proof.
ConsiderJ1=Γkα+k/2f−eα/kIkψ−1e+α;ψg∘ψψ−1f and J2=Γkα+k/2f−eα/kIkψ−1f−α;ψg∘ψψ−1e.
Now,(19)J1=α2kf−eα/k∫ψ−1eψ−1ff−ψvα/k−1g∘ψvψ′vdv=−12kf−eα/k∫ψ−1eψ−1fg∘ψvdf−ψvα/k=ge2+12f−eα/k∫ψ−1eψ−1ff−ψvα/kg′∘ψvψ′vdv.
Similarly,(20)J2=α2kf−eα/k∫ψ−1eψ−1fψv−eα/k−1g∘ψvψ′vdv=12kf−eα/k∫ψ−1eψ−1fg∘ψvdψv−eα/k=gf2−12f−eα/k∫ψ−1eψ−1fψv−eα/kg′∘ψvψ′vdv.
It follows that(21)ge+gf2−J1+J2=12f−eα/k∫ψ−1eψ−1fψv−eα/k−f−ψvα/kg′∘ψvψ′vdv.Example 1.
Letc=2,d=3,α=1/2,k=2,gx=x2,ψx=x. Then, all the assumptions in Lemma 1 are satisfied. Observe that gc+gd/2=13/2.(22)Γkα+k2d−cα/kIkψ−1c+α;ψg∘ψψ−1d+Ikψ−1d−α;ψg∘ψψ−1c=Γ21/221Γ21/2∫23v23−v−3/4dv+1Γ21/2∫23v2v−2−3/4dv=57790.
This implies(23)gc+gd2−Γkα+k2d−cα/kIkψ−1c+α;ψg∘ψψ−1d+Ikψ−1d−α;ψg∘ψψ−1c=445.
Also,(24)12d−cα/k∫ψ−1cψ−1dψv−cα/k−d−ψvα/kg′∘ψvψ′vdv=∫23vv−21/4dv−∫23v3−v1/4dv=445.Example 2.
Letc=2,d=3,α=1/2,k=1/2,gx=x2,ψx=x. Then, all the assumptions in Lemma 1 are satisfied. Observe that gc+gd/2=13/2.(25)Γkα+k2d−cα/kIkψ−1c+α;ψg∘ψψ−1d+Ikψ−1d−α;ψg∘ψψ−1c=Γ1/21/221Γ1/21/2∫23v2dv+1Γ1/21/2∫23v2dv=193.
This implies(26)gc+gd2−Γkα+k2d−cα/kIkψ−1c+α;ψg∘ψψ−1d+Ikψ−1d−α;ψg∘ψψ−1c=16.
Also,(27)12d−cα/k∫ψ−1cψ−1dψv−cα/k−d−ψvα/kg′∘ψvψ′vdv=∫23vv−2dv−∫23v3−vdv=16.Lemma 2.
Lete<f and g:e,f⟶ℝ be a differentiable mapping on e,f. Also, suppose that g′∈Le,f, ψx is an increasing and positive monotone function on e,f, having a continuous derivative ψ′x on e,f, and α∈0,1. Then, for k>0, the following identity holds:(28)Γkα+k2f−eα/kIkψ−1e+α;ψg∘ψψ−1f+Ikψ−1f−α;ψg∘ψψ−1e−ge+f2=∫ψ−1eψ−1fhg′∘ψvψ′vdv+12f−eα/k∫ψ−1eψ−1fψv−eα/k−f−ψvα/kg′∘ψvψ′vdv,where(29)h=12,forψ−1e+f2≤v≤ψ−1f,−12,forψ−1e≤v≤ψ−1e+f2.Proof.
Suppose(30)I1=−12∫ψ−1eψ−1e+f/2g′∘ψvψ′vdv=−12ge+f2+ge2,I2=12∫ψ−1e+f/2ψ−1fg′∘ψvψ′vdv=−12ge+f2+gf2,I3=12f−eα/k∫ψ−1eψ−1ff−ψvα/kg′∘ψvψ′vdv=−ge2+α2kf−eα/k∫ψ−1eψ−1ff−ψvα/k−1g∘ψvψ′vdv=−ge2+Γkα+k2f−eα/kIkψ−1e+α;ψg∘ψψ−1f,(31)I4=−12f−eα/k∫ψ−1eψ−1fψv−eα/kg′∘ψvψ′vdv=−gf2+α2kf−eα/k∫ψ−1eψ−1fψv−eα/k−1g∘ψvψ′vdv=−gf2+Γkα+k2f−eα/kIkψ−1f−α;ψg∘ψψ−1e.
SummingI1,I2,I3, and I4, we get the required result.Example 3.
Letc=2,d=3,α=1/2,k=2,gx=x2,ψx=x. Then, all the assumptions in Lemma 2 are satisfied. Note that gc+d/2=25/4.(32)Γkα+k2d−cα/kIkψ−1c+α;ψg∘ψψ−1d+Ikψ−1d−α;ψg∘ψψ−1c=Γ1/21/281Γ1/21/2∫23v23−v−3/4dv+1Γ1/21/2∫23v2v−2−3/4dv=57790.
This implies(33)Γkα+k2d−cα/kIkψ−1c+α;ψg∘ψψ−1d+Ikψ−1d−α;ψg∘ψψ−1c−gc+d2=29180.
Also,(34)∫ψ−1cψ−1dhg′∘ψvψ′vdv=14,where h is defined in Lemma 2.(35)12d−cα/k∫ψ−1cψ−1dd−ψvα/k−ψv−cα/kg′∘ψvψ′vdv=∫23vv−21/4dv−∫23v3−v1/4dv=−445.
This implies(36)∫ψ−1cψ−1dhg′∘ψvψ′vdv+12d−cα/k∫ψ−1cψ−1dd−ψvα/k−ψv−cα/kg′∘ψvψ′vdv=29180.Example 4.
Letc=2,d=3,α=1/2,k=1/2,gx=x2,ψx=x. Then, all the assumptions in Lemma 2 are satisfied. Note that gc+d/2=25/4.(37)Γkα+k2d−cα/kIkψ−1c+α;ψg∘ψψ−1d+Ikψ−1d−α;ψg∘ψψ−1c=Γ1/21/221Γ1/21/2∫23v2dv+1Γ21/2∫23v2dv=193.
This implies(38)Γkα+k2d−cα/kIkψ−1c+α;ψg∘ψψ−1d+Ikψ−1d−α;ψg∘ψψ−1c−gc+d2=112.
Also,(39)∫ψ−1cψ−1dhg′∘ψvψ′vdv=14,where h is defined in Lemma 2.(40)12d−cα/k∫ψ−1cψ−1dd−ψvα/k−ψv−cα/kg′∘ψvψ′vdv=∫23vv−2dv−∫23v3−vdv=−16.
This implies(41)∫ψ−1cψ−1dhg′∘ψvψ′vdv+12d−cα/k∫ψ−1cψ−1dd−ψvα/k−ψv−cα/kg′∘ψvψ′vdv=112.
Before proceeding to next results, let us recall the definition ofs-convex function of Breckner type.Definition 4.
(see [49]). A function g:0,∞⟶0,∞ is said to be s-convex function of Breckner type if(42)g1−tx+ty≤1−tsgx+tsgy,∀x,y∈0,∞,t∈0,1,s∈0,1.Theorem 2.
Lete<f and g:e,f⟶ℝ be a differentiable mapping on e,f. Also, suppose that g′ is Breckner type of s-convex on e,f, ψx is an increasing and positive monotone function on e,f, having a continuous derivative ψ′x on e,f, and α∈0,1. Then, for k>0, the following inequality holds:(43)ge+gf2−Γkα+k2f−eα/kIkψ−1e+α;ψg∘ψψ−1f+Ikψ−1f−α;ψg∘ψψ−1e≤f−e2L1g′e+L2g′f,where(44)L1≔2kBk12;1+s,k+αk+k1−2−ks+α/kk+ks+α−Bk1+s,k+αk,(45)L2≔k1−2−ks+α/kk+ks+α−2kBk12;k+αk,1+s−Bkk+αk,1+s,respectively.Proof.
Using Lemma1 and the fact that g′ is Breckner type of s-convex function, we have(46)ge+gf2−Γkα+k2f−eα/kIkψ−1e+α;ψg°ψψ−1f+Ikψ−1f−α;ψg°ψψ−1e≤12f−eα/k∫ψ−1eψ−1fψv−eα/k−f−ψvα/kg′∘ψvψ′vdv=f−e2∫011−tα/k−tα/kg′tc+1−tfdt≤f−e2∫011−tα/k−tα/ktsg′e+1−tsg′fdt=f−e2g′e∫01/2ts1−tα/k−tα/kdt+g′f∫01/21−ts1−tα/k−tα/kdt+g′e∫1/21tstα/k−1−tα/kdt+g′f∫1/211−tstα/k−1−tα/kdt=f−e2L1g′e+L2g′f,where(47)L1≔H1+H3=∫01/2ts1−tα/k−tα/kdt+∫1/21tstα/k−1−tα/kdt=2kBk12;1+s,k+αk+k1−2−ks+α/kk+ks+α−Bk1+s,k+αk,(48)L2≔H2+H4=∫01/21−ts1−tα/k−tα/kdt+∫1/211−tstα/k−1−tα/kdt=k1−2−ks+α/kk+ks+α−2kBk12;k+αk,1+s−Bkk+αk,1+s.
This completes the proof.Theorem 3.
Letg:e,f⟶ℝ be a differentiable function on e,f with e<f. Also, suppose that g′ is Breckner type of s-convex function. If ψx is an increasing and positive monotone function on e,f, having a continuous derivative ψ′x on e,f and α∈0,1, then for k>0, the following inequality holds:(49)Γkα+k2f−eα/kIkψ−1e+α;ψg∘ψψ−1f+Ikψ−1f−α;ψg∘ψψ−1e−ge+f2≤gf−ge2+f−e2L1g′e+L2g′f,where L1 and L2 are given by (44) and (45), respectively.Proof.
Using Lemma2, the property of modulus, and the given hypothesis of the theorem, we have(50)Γkα+k2f−eα/kIkψ−1e+α;ψg°ψψ−1f+Ikψ−1f−α;ψg°ψπ−ge+f2≤∫ψ−1eψ−1fhg′∘ψvψ′vdv+12f−eα/k∫ψ−1eψ−1ff−ψvα/k−ψv−eα/kg′∘ψvψ′vdv=I1+I2.
Using substitutiont=ψv−e/f−e and the fact that g′ is Breckner type of s-convex function, we have(51)I1≤f−e2L1g′e+L2g′f,where L1 and L2 are given by (44) and (45), respectively. And(52)I2=gf−ge2.
This completes the proof.


## 4. Applications

In this section, we discuss some applications of Theorem2 to means by considering a particular example of s-convexity. First of all, we recall some previously known concepts related to means [50].For arbitrary real numbersα,β, α≠β, we define the following:(1)
Arithmetic mean:(53)Aα,β=α+β2,α,β∈ℝ;(2)
Logarithmic mean:(54)L¯α,β=β−αlnβ−lnα,α,β∈ℝ\0;(3)
Generalized log-mean:(55)Lnα,β=βn+1−αn+1n+1β−α1/n,n∈ℕ,n≥1,α,β∈ℝ,α<β.We now give the main results of this section.Proposition 1.
Lete,f∈ℝ+ with e<f; then,(56)Aes,fs−Lsse,f≤sf−e2W1es−1+W2fs−1,where(57)W1≔2B12;1+s,2+1−2−1−s2+s−B1+s,2,(58)W2≔1−2−1−s2+s−2B12;2,1+s−B2,1+s,respectively.Proof.
Applying Theorem2 for gx=xs, ψx=x, and α=1=k, we obtain the required result.Proposition 2.
Lete,f∈ℝ+ with e<f; then,(59)Aes,fs−Lsse,f≤fs−gs2+sf−e2W1es−1+W2fs−1,where W1 and W2 are given by (57) and (58), respectively.Proof.
Applying Theorem3 for gx=xs, ψx=x, and α=1=k, we obtain the required result.


## 5. Conclusion

In this article, we obtain some new fractional estimates of Hermite–Hadamard’s inequality essentially using a newk-analogue of ψk-fractional integrals. We derive two new fractional integral identities in the setting of k-fractional calculus. In order to check the validity of these identities, we discuss some particular examples. In the final section, we have discussed applications of Theorems 2 and 3 to means.



---

*Source: 3051920-2020-04-25.xml*